% The following TeXfile comprises part of

%     MEASURE THEORY, by D.H.Fremlin
%          (University of Essex, Colchester, England;
%                   e-mail  fremdh@essex.ac.uk)

% Please contact me with any difficulties not covered by the remarks
% below.

% The file is set up for processing by plain TeX in the presence of
% files  amstex.tex ,  amssym.tex  and  amsppt.sty  suitably supported by the
% appropriate fonts.   If these are unavailable, a readable version
% can be obtained by removing the comment symbol  %  from the line

%        %   \amstexsupportedfalse

% below.   Some systems do not allow access to files  amstex.tex  and
% amsppt.sty , but instead employ a command "amstex <filename.tex>";
% for such systems it is sometimes right to delete or disable the
% commands

%        \input amstex
%        \documentstyle{amsppt}

% (I have little experience with this and should welcome comments.)

% Various formats of output are obtainable.   The principal variations
% are

%    1. "Standard" output uses the command
%               \smallprint
% and gives output adapted to A4 paper (21 cm x 29.7 cm).

%    2.  Using the command
%               \USsmallprint
% gives output adapted to 8.5" x 11" .

%    3.  Using the command
%               \noproofs
% produces an abridged version, in which proofs are omitted.

%    4.   Using the command
%               \resultsonly
% produces a super-abridged version, omitting comments and exercises
% as well as proofs.

% These commands are grouped together at the end of the introductory
% definitions;  they may be activated by deleting the initial comment
% symbols  % .

% Any diagrams included will be called for by inputting postscript files.
% (You can look for these by hunting for   \picture  in the code below.)
% To get these you will need the relevant  .ps  files with the expected
% names  ( \picture{mt133g}{...}  demands the file  mt133g.ps , for instance).


%(If you can read TeX fluently, you may wish to fast-forward to a line
%beginning   %file starts here   or   \frfilename
%if one of these is present.)

%TeXfiles with this heading are designed for processing with AmSTeX,
%accessed through TeXfiles  amstex.tex (with amssym.tex  and
% amsppt.sty  and possibly  amsppt.sti)  during compilation.   
%If these are unavailable, a readable
%version can (I hope) be produced by deleting the  %  before
% \amstexsupportedfalse  below.

%Some special difficulties arise with the importation of  .ps  files
%in diagrams.   Look for  \picture  below for notes on these.

%The following \count registers are used for special purposes:
%   \count1=\stylenumber  to specify style
%   \count2=\footnotenumber  to count footnotes

%AmSTeX declaration
\newif\ifamstexsupported
     \amstexsupportedtrue
%     \amstexsupportedfalse               %use when AMSTeX unavailable
\newif\ifDHFamstexloaded        %for detection of a non-standard AMSTeX
     \DHFamstexloadedfalse    
\newif\ifDHFamspptloaded        
     \DHFamspptloadedfalse    
\newif\iffourteensavailable
     \fourteensavailabletrue
    \fourteensavailablefalse
\newif\ifathome
     \athometrue     %to cope with a quirk     of my son's Ghostscript
     \athomefalse
\newif\iffirstpage
     \firstpagetrue

%Style assignments
\countdef\stylenumber=1
\countdef\footnotenumber=2
\def\fullsize{\magnification=\magstep1
     \def\pagewidth{390pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=1}
\def\smallprint{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{468pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=2}
\def\printersize{\magnification=\magstep1
     \baselineskip24pt
     \overfullrule0pt
     \def\pagewidth{330pt}
     \def\pageheight{8.5truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=3}
\def\refsize{\magnification=\magstep1
     \overfullrule0pt
     \def\pagewidth{390pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=4}
\def\rrversion{\magnification=\magstep1
     \overfullrule0pt
     \def\pagewidth{330pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=5}
\def\bigsize{\magnification=\magstep2
     \overfullrule0pt
     \def\pagewidth{325pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=6}
\def\bbigsize{\magnification=\magstep3
     \overfullrule0pt
     \def\pagewidth{271pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=7}
\def\USsmallprint{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{482pt}
     \def\pageheight{8.7truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=8}
\def\halfmagsize{\magnification=\magstephalf
     \def\pagewidth{427pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=9}
           % register  stylenumber  is used to control the output format

%addresses, letterheads
\def\deptaddress{\hskip 3truein Department of Mathematical Sciences\par
   \noindent e-mail: {\tt fremdh}\hskip 2.9truein\today\par
   \medskip}
\long\def\univaddress#1{\hskip 3truein Department of Mathematical Sciences\par
   \hskip 3truein University of Essex\par
   \hskip 3truein Colchester CO4 3SQ\par
   \hskip 3truein England\par
   %\hskip 3truein van Vleck Hall\par
   %\hskip 3truein 480 Lincoln Drive\par
   %\hskip 3truein Madison\par
   %\hskip 3truein WI 53706-1388, U.S.A.\par
   \hskip 3.2truein#1\par
   \noindent e-mail: {\tt david\@fremlin.org}
       %{\tt fremlin\@math.wisc.edu}
   \bigskip}
\long\def\homeaddress#1{\hskip 3truein 25 Ireton Road\par
   \hskip 3truein Colchester CO3 3AT\par
   \hskip 3truein England\par
   \hskip 3.2truein#1\par
   \noindent e-mail: {\tt david\@fremlin.org}
   \bigskip}
\long\def\letterhead#1#2#3{\univaddress{#1}
   {\parindent=0pt
    #2
   }
   \medskip

   \noindent Dear #3,
     }
\def\shortaddress{\noindent Departmentof Mathematical Sciences, University of Essex,
   Colchester CO4 3SQ, England\par
   \noindent e-mail: david\@fremlin.org}
\def\valediction#1#2{\bigskip
   \hskip 3truein #1,
   \vskip 1truein
   \hskip 3.3truein #2
   \bigskip}

%picture importation
   %It is clear that there is no general standard.   The syntax I
   %use when calling for a  .ps  file is
   %    \def\Caption{<...>}
   %    \picture{<filename>}{<dim>}
   %which is supposed to import  <filename>.ps  at height  <dim> ,
   %centered, and adding a one-line centered caption  <...> .
   %The basic macro listed in  \def\picture  is that which
   %used to work at Essex.   I now generally use the macro in
   % \def\picturewithpsfig , which requires a previous input
   % psfig.tex  or  psfig.sty ;
   %if you interpolate  \atUEssex  early in your file it will
   %give you the latter.
   %   I find that  psfig  is erratic in its centering, especially
   %under magnification, so I have written a routine
   % \centerpicturewithpsfig  (which may be called from
   % \centerpicture ) with the syntax
   %    \centerpicture{<filename>}{<dim>}{<dim>}
   %the first dimension being width and the second height.

   % A second package is called for by  \usegraphicx .   This requires
   % miniltx.tex  and  graphicx.sty ;  miniltx.tex  is currently (April 2016)
   % available from
   % http://mirror.ox.ac.uk/sites/ctan.org/macros/plain/graphics/miniltx.tex
   % The advantage of this is that it seems to allow manipulation of the
   % dvi>pdf stage using  dvipdfmx .

\def\picture#1#2{\bigskip
  \bigskip
  \immediate\write0{importing diagram #1.ps at height #2
         with dvitops}
  \vbox{\vskip #2
    \special{dvitops: import #1.ps \pagewidth #2}
    \Centerline{\Caption}
    }
   \medskip
   \def\Caption{}}
%an alternative version, without caption
\def\centerpicture#1#2#3{\bigskip
  \bigskip
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with dvitops}
  \vbox{\vskip #3
    \special{dvitops: import #1.ps \pagewidth #3}
  }}
%to put pictures on same line
\def\sideshiftedpicture#1#2#3#4{\bigskip
   \bigskip
   \immediate\write0{importing diagram #1.ps at height #4
        and width #3 with dvitops}
   \vbox{\special{dvitops:  import #1.ps #3 #4}}
   \hskip #2 plus 0pt minus 0pt
   }
\def\startsideshiftedpicture{}
\newcount\scaleup
\newcount\scaledown
\scaleup=1
\scaledown=1
\def\scalesideshiftedpicture#1#2#3#4{
  \dimen4=#4
  \dimen3=#3
  \multiply\dimen4 by\scaleup
  \divide\dimen4 by \scaledown
  \multiply\dimen3 by\scaleup
  \divide\dimen3 by \scaledown
  \sideshiftedpicture{#1}{#2}{\dimen3}{\dimen4}}

\def\picturewithpsfig#1#2{\bigskip
   \immediate\write0{importing diagram #1.ps at height #2 with psfig}
   \vbox{\hsize\pagewidth
   \centerline{\psfig{figure=#1.ps,height=#2}}
   \medskip
   \centerline{\Caption}}
   \medskip
   \def\Caption{}}
\def\centerpicturewithpsfig#1#2#3{
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with psfig}
  \dimen 3 = #2
  \dimen 4 = #3
  \dimen 2 = \pagewidth
  \advance\dimen2 by -\dimen3
  \divide\dimen2 by 2
  \discrversionB{\shrinkonestep}{}
  \discrversionH{\shrinkonestep}{}
  \discrversionF{\enlargeonestep}{}
  \discrversionG{\enlargeonestep\enlargeonestep}{}
  \dimen 6 = \dimen4
  \divide\dimen6 by 3
  \vskip\dimen6
  \ifdim\dimen2>0\hskip\dimen2\fi
     \vbox{\psfig{file=#1.ps,width=\dimen3,height=\dimen4}}
  \discrversionG{\vskip-\dimen6}{}
  }
\def\enlargeonestep{\dimen5 = 5truein
   \dimen6 = 7truein
   \ifdim\dimen3<\dimen5
      \ifdim\dimen4<\dimen6
         \divide\dimen3 by 5
         \multiply\dimen3 by 6
         \divide\dimen4 by 5
         \multiply\dimen4 by 6
      \fi\fi}
\def\shrinkonestep{\divide\dimen3 by 6
         \multiply\dimen3 by 5
         \divide\dimen4 by 6
         \multiply\dimen4 by 5}
\def\sideshiftedpicturewithpsfig#1#2#3#4{
   \immediate\write0{importing diagram #1.ps at height #4
      and width #3 with psfig}
   \dimen 4 = #4
   \dimen 3 = #3
   \discrversionB{\shrinkonestep}{}
   \discrversionH{\shrinkonestep}{}
   \discrversionF{\enlargeonestep}{}
   \discrversionG{\enlargeonestep\enlargeonestep}{}
%   \vbox{\psfig{figure=#1.ps,height=\dimen4,width=\dimen3}}
   \psfig{figure=#1.ps,height=\dimen4,width=\dimen3}
   \hskip #2 plus 0pt minus 0pt}

\def\picturewithgraphicx#1#2{\bigskip
   \immediate\write0{importing diagram #1.ps at height #2 with graphicx}
   \vbox{\hsize\pagewidth
   \centerline{\includegraphics[height=#2]{#1.ps}}
   \medskip
   \centerline{\Caption}}
   \medskip
   \def\Caption{}}
\def\centerpicturewithgraphicx#1#2#3{
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with graphicx}
  \dimen 3 = #2
  \dimen 4 = #3
  \dimen 2 = \pagewidth
  \advance\dimen2 by -\dimen3
  \divide\dimen2 by 2
  \discrversionB{\shrinkonestep}{}
  \discrversionH{\shrinkonestep}{}
  \discrversionF{\enlargeonestep}{}
  \discrversionG{\enlargeonestep\enlargeonestep}{}
  \dimen 6 = \dimen4
  \divide\dimen6 by 3
  \vskip\dimen6
  \ifdim\dimen2>0\hskip\dimen2\fi
     \vbox{\includegraphics[width=\dimen3,height=\dimen4]{#1.ps}}
  \discrversionG{\vskip-\dimen6}{}
  }
\def\enlargeonestep{\dimen5 = 5truein
   \dimen6 = 7truein
   \ifdim\dimen3<\dimen5
      \ifdim\dimen4<\dimen6
         \divide\dimen3 by 5
         \multiply\dimen3 by 6
         \divide\dimen4 by 5
         \multiply\dimen4 by 6
      \fi\fi}
\def\shrinkonestep{\divide\dimen3 by 6
         \multiply\dimen3 by 5
         \divide\dimen4 by 6
         \multiply\dimen4 by 5}
\def\sideshiftedpicturewithgraphicx#1#2#3#4{
   \immediate\write0{importing diagram #1.ps at height #4
      and width #3 with graphicx}
   \dimen 4 = #4
   \dimen 3 = #3
   \discrversionB{\shrinkonestep}{}
   \discrversionH{\shrinkonestep}{}
   \discrversionF{\enlargeonestep}{}
   \discrversionG{\enlargeonestep\enlargeonestep}{}
%   \vbox{\includegraphics[height=\dimen4,width=\dimen3]{#1.ps}}
   \includegraphics[height=\dimen4,width=\dimen3]{#1.ps}
   \hskip #2 plus 0pt minus 0pt}

%alternative picture importation code;  I've been told about this
  %but have not used it successfully myself
%\def\Caption{}
%\input epsf
%\epsfverbosetrue
%\def\picture#1#2{\bigskip
%    \bigskip
%    \immediate\write0{importing diagram #1.ps at height #2 with epsf}
%    \epsfysize=#2
%    \centerline{\epsffile{#1.ps}}
%    \medskip
%    \Centerline{\Caption}
%    \medskip
%    }

%in addition it may be necessary to amend the imported  .ps  file
%by replacing the   %%BoundingBox   header by  %%BoundingBox:

\def\atUEssex{
   \input psfig.sty
   \def\picture##1##2{\picturewithpsfig{##1}{##2}}
   \def\sideshiftedpicture##1##2##3##4{
     \sideshiftedpicturewithpsfig{##1}{##2}{##3}{##4}}
   \def\centerpicture##1##2##3{
       \centerpicturewithpsfig{##1}{##2}{##3}}
   \def\startsideshiftedpicture{
       \sideshiftedpicture{empty}{1pt}{1pt}{1pt}}
   }
\def\usegraphicx{
   \input miniltx
   \def\Gin@driver{dvips.def}
   \input graphicx.sty
   \def\picture##1##2{\picturewithgraphicx{##1}{##2}}
   \def\sideshiftedpicture##1##2##3##4{
     \sideshiftedpicturewithgraphicx{##1}{##2}{##3}{##4}}
   \def\centerpicture##1##2##3{
       \centerpicturewithgraphicx{##1}{##2}{##3}}
   \def\startsideshiftedpicture{
       \sideshiftedpicture{empty}{1pt}{1pt}{1pt}}
}


%special fonts
\def\biggerscriptfonts{\scriptfont0=\eightrm
     \scriptfont1=\eighti
     \scriptfont2=\eightsy
     \scriptfont3=\eightex
     \scriptfont\bffam=\eightbf
     \scriptfont\msbfam=\eightmsb
     \scriptfont\eufmfam=\eighteufm
     }
\def\Loadcyrillic{\ifamstexsupported
      \input cyracc.def
      \font\tencyr=wncyr10
      \def\cyr{\tencyr\cyracc} \def\itcyr{\tenitcyr\cyracc}
    \else\def\cyr{\smc}\def\cdprime{u\i}\def\Cdprime{UI}\fi}
%http://wkweb3.cableinet.co.uk/linux/LDP/HOWTO/Cyrillic-HOWTO-7.html
%cyrillic.tex

\def\Loadfourteens{\iffourteensavailable
       \font\fourteenrm=cmr14
       \font\fourteenbf=cmbx14
       \font\fourteenit=cmsl14
   \else\font\fourteenrm=cmr10 scaled \magstep2
      \font\fourteenbf=cmbx10 scaled \magstep2
      \font\fourteenit=cmsl10 scaled \magstep2
   \fi}
\def\smallerfonts{\eightpoint
   \textfont\msbfam=\eightmsb
   \textfont\eufmfam=\eighteufm
   \textfont\bffam=\eightbf
   \textfont\eurmfam=\eighteurm
   \textfont\eusmfam=\eighteusm
   }
\font\twelveex=cmex10 scaled \magstep1


%Copyright declarations
\newcount\oldcopyrightdate
\newcount\newcopyrightdate
\oldcopyrightdate=0
\def\copyrightdate#1{\newcopyrightdate=#1
       \ifnum\newcopyrightdate=\oldcopyrightdate\else
       \oldcopyrightdate=#1
       \oldfootnote{}{\eightrm \copyright\ #1 D.~H.~Fremlin}
       \immediate\write0{Copyright #1}
       \fi}

%hyphenation
\hyphenation{equi-ver-id-ical}

%miscellaneous definitions
\def\Aut{\mathop{\text{Aut}}}
\def\backstep#1{\hskip -0.#1em plus 0 pt minus 0 pt}
\def\Bang{\vthsp\hbox{\bf X\hskip0.07em\llap X\hskip0.07em\llap X\rm}}
\def\BanG{\Bang{\hskip 1.111pt plus 3.333pt minus 0pt}}
%\def\BbbN{\mathchoice{\hbox{$\Bbb N$\hskip0.02em}}
%  {\hbox{$\Bbb N$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb N$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb N$\hskip0.04em}}}
%\def\BbbQ{\mathchoice{\hbox{$\Bbb Q$\hskip0.02em}}
%  {\hbox{$\Bbb Q$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb Q$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb Q$\hskip0.04em}}}
%\def\BbbR{\mathchoice{\hbox{$\Bbb R$\hskip0.02em}}
%  {\hbox{$\Bbb R$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb R$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb R$\hskip0.04em}}}
%  %do not use with $\Bbb R^X$
%\def\BbbZ{\mathchoice{\hbox{$\Bbb Z$\hskip0.02em}}
%  {\hbox{$\Bbb Z$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb Z$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb Z$\hskip0.04em}}}
\def\BbbN{\openBbb{N}}
\def\BbbQ{\openBbb{Q}}
\def\BbbR{\openBbb{R}}
  %do not use with $\Bbb R^X$  do not use in $_{\Bbb R^r}$
\def\BbbZ{\openBbb{Z}}
\def\Bbbone{\hbox{1\hskip0.13em\llap{1}}}
\def\biggerint{\textfont3=\twelveex\intop\nolimits}
  % you can get this size into a displayed formula with
  %  $$\hbox{$\biggerint_{\Cal S}\pmb{u}\,d\pmb{v}$}$$
  % but it may then seem cramped
  % or try \hbox{$\biggerint_{\Cal S}\,$}
\def\biggeriint{\textfont3=\twelveex\iint}
  % remarkable fact:  this seems to infect all \int within a given
  % $..$ ; but  $\biggerint$$\int$ seems to leave the $\int$ normal
  % really  \textfont3=\twelveex  is more reliable, since  \biggerint
  % as written gets turned off rather easily (e.g. by  \comment,
  % or by an extra  {...})
\def\bover#1#2{{{#1}\over{#2}}}
\def\Bover#1#2{\backstep4\hbox{
     \biggerscriptfonts${\shortstrut#1}\over{\shortstrut#2}$}}
  % when there are superscripts in the denominator, as in \Bover1{2^m}
  % or  \Bover1{\sqrt{2\pi}^r} , these can sometimes come out with better
  % horizontal alignment if you use  \Bover1{{\sshortstrut 2}^m}  or
  % \Bover1{{\sqrt{2\pi}}^r}
\def\bsover#1#2{\textstyle\bover{#1}{#2}}
\def\bsp{/\penalty-100}
  % \penalty seems to behave oddly in a \tt string in \fullsize mode
\def\Caption{}
\def\Center#1{

   \Centerline{#1}

   \noindent}
\def\Centerline#1{\vskip 5pt
     {\def\int{\text{}$$\textfont3=\twelveex\intop\nolimits}
      \def\dashint{\text{}$$\textfont3=\twelveex
         \mskip2mu\vrule height2.65pt width4.5pt depth-2.35pt\mskip-11mu
	 \intop\nolimits\mskip-2mu}
      \def\idashint{\text{}$$\int\mskip-7mu\dashint}	 
      \def\dashiint{\text{}$$\dashint\mskip-6mu\int}	 
      \def\dashidashint{\text{}$$\dashint\mskip-6mu\dashint}	 
      \def\underlineint{
           \text{}$$\textfont3=\twelveex\underline{\intop\nolimits}}
      \def\overlineint{
           \text{}$$\textfont3=\twelveex\overline{\intop\nolimits}}
      \def\iint{\text{}$$\textfont3=\twelveex
           \intop\nolimits\hskip-0.5em\intop\nolimits}
      \def\sum{\text{}$$\sumop}
      \def\prod{\text{}$$\prodop}
      \def\bigcap{\text{}$$\bigcapop}
      \def\bigcup{\text{}$$\bigcupop}
   \centerline{#1}}\vskip 5pt}
   \mathchardef\bigcapop="1354
   \mathchardef\bigcupop="1353
   \mathchardef\prodop="1351
   \mathchardef\sumop="1350
   %in  \Centerline{$\overline{}$, $\cmmnt{}$} must use \sumop etc
\def\cf{\mathop{\text{cf}}}
\def\circumflex#1{{\accent"5E #1}}
\def\closeplus{\hbox{+}}
\long\def\cmmnt#1{\ifresultsonly\else{#1}\fi}
\def\coint#1{\left[#1\right[}
\def\curl{\mathop{\text{curl}}}
\def\dashint{\mathchoice{-\mskip-20mu\intop\nolimits}
   {{\vrule height2.75pt width4.5pt depth-2.45pt\mskip-10mu\intop\nolimits}}
      {-\mskip-16.5mu\intop\nolimits} %not yet used
      {-\mskip-16.5mu\intop\nolimits}} %not yet used
\def\idashint{\mathchoice{\int\mskip-6mu\dashint}
      {\int\mskip-5mu\dashint}
      {\int\mskip-5mu\dashint} %not yet used
      {\int\mskip-5mu\dashint}} %not yet used
\def\dashiint{\mathchoice{\dashint\mskip-10mu\int}
      {\dashint\mskip-6mu\int}
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\dashidashint{\mathchoice{\dashint\mskip-10mu\dashint}
      {\dashint\mskip-6mu\dashint}
      {\dashint\mskip-6mu\dashint} %not yet used
      {\dashint\mskip-6mu\dashint}} %not yet used
\def\dashidashiint{\mathchoice{\dashint\mskip-8mu\dashint\mskip-10mu\int}
      {\dashint\mskip-4mu\dashint\mskip-8mu\int} 
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\dashiidashint{\mathchoice{\dashint\mskip-10mu\int\mskip-8mu\dashint}
      {\dashint\mskip-4mu\dashint\mskip-8mu\int} 
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\idashidashint{\mathchoice{\int\mskip-8mu\dashint\mskip-8mu\dashint}
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\diam{\mathop{\text{diam}}}
\def\discrpage{\ifnum\stylenumber=1{\frnewpage}  %redefined in mt.tex
           \else\ifnum\stylenumber=3{\frnewpage}
           \else\ifnum\stylenumber=5{\frnewpage}
           \else\bigskip\fi\fi\fi}
\long\def\discrversionA#1#2{\ifnum\stylenumber=1#1\else#2\fi}
\long\def\discrversionB#1#2{\ifnum\stylenumber=2#1\else#2\fi}
\long\def\discrversionC#1#2{\ifnum\stylenumber=3#1\else#2\fi}
\long\def\discrversionD#1#2{\ifnum\stylenumber=4#1\else#2\fi}
\long\def\discrversionE#1#2{\ifnum\stylenumber=5#1\else#2\fi}
\long\def\discrversionF#1#2{\ifnum\stylenumber=6#1\else#2\fi}
\long\def\discrversionG#1#2{\ifnum\stylenumber=7#1\else#2\fi}
\long\def\discrversionH#1#2{\ifnum\stylenumber=8#1\else#2\fi}
           % facility for special interpolations in particular output
           % versions
\def\displaycause#1{\noalign{\noindent (#1)}}
\def\diverg{\mathop{\text{div}}}
\def\dom{\mathop{\text{dom}}}
\def\eae{=_{\text{a.e.}}}

\def\esssup{\mathop{\text{ess sup}}}
\mathchardef\Exists="3239

\def\family#1#2#3{\langle#3\rangle_{#1\in#2}}
\def\familyi#1#2{\family{i}{#1}{#2}}
\def\familyiI#1{\familyi{I}{#1}}

\def\frfilename#1{\ifnum\stylenumber=1{\noindent filename #1}
               \else\ifnum\stylenumber=6{\noindent{\fiverm filename #1}}
               \else\ifnum\stylenumber=7{\noindent{\fiverm filename #1}}
               \else\fi\fi\fi}
\def\Footnote#1#2{\footnote"#1"{#2}}
\mathchardef\Forall="3238
\def\frakc{\frak c\phantom{.}}
\def\frnewpage{\vskip \pageheight plus 0pt minus\pageheight
           \eject
           \gdef\topparagraph{}
           \gdef\bottomparagraph{}
           }
\def\frsmallcirc{{\hskip 1pt plus 0pt minus 0pt}
     {\raise 1pt\hbox{$\smallcirc$}}{\hskip 1pt plus 0pt minus 0pt}}

\def\geae{\ge_{\text{a.e.}}}
\def\grad{\mathop{\text{grad}}}
\def\grheada{{\bf ($\pmb{\alpha}$)}}
\def\grheadb{{\bf ($\pmb{\beta}$)}}
\def\grheadc{{\bf ($\pmb{\gamma}$)}}
\def\grheadd{{\bf ($\pmb{\delta}$)}}
\def\grheade{{\bf ($\pmb{\epsilon}$)}}
\def\grheadz{{\bf ($\pmb{\zeta}$)}}
\def\grv#1{{\accent"12 #1}}

\def\hfillall{\hskip\pagewidth plus0pt minus\pagewidth}

\def\Imag{\mathop{\Cal I\text{m}}}
\long\def\inset#1{{\narrower\narrower{\vskip 1pt plus 1pt minus 0pt}
     #1
     {\vskip 1pt plus 1pt minus 0pt}}}
\def\Int#1#2{\displaystyle{\int_{#1}^{#2}}}
\def\interior{\mathop{\text{int}}}

\def\leae{\le_{\text{a.e.}}}
\long\def\leaveitout#1{}
\def\LLcorner{\mathchoice{\,\hbox{\vrule height7pt width0.5pt depth0pt
       \vrule height0.5pt width5pt depth0pt}\,}
   {\,\hbox{\vrule height7pt width0.5pt depth0pt
       \vrule height0.5pt width5pt depth0pt}\,}
   {\,\hbox{\vrule height5pt width0.3pt depth0pt
       \vrule height0.3pt width4pt depth0pt}\,}
   {\,\hbox{\vrule height4pt width0.25pt depth0pt
       \vrule height0.25pt width3pt depth0pt}\,}
  }
\def\Ln{\mathop{\text{Ln}}}
\def\Loadfont#1{\font#1}
\def\loibr{\hbox{\hskip0.1em]\hskip0.05em}}
\def\Matrix#1{\pmatrix#1\endpmatrix}
\def\Mu{\text{M}}
\def\noamslogo{\def\amslogo{}}
\def\normalsubgroup{\vartriangleleft}
\def\Nu{\text{N}}
\def\ocint#1{\left]#1\right]}
\def\ofamily#1#2#3{\langle#3\rangle_{#1<#2}}
\let\oldcases=\cases
\let\oldfootnote=\footnote
\def\On{\mathop{\text{On}}}
\def\ooint#1{\left]#1\right[}
\def\openBbb#1{\Bbb#1\mskip1mu}
\def\pd#1#2{\bover{\partial#1}{\partial#2}}
\def\Pd#1#2{\Bover{\partial#1}{\partial#2}}
\def\pdd#1#2{{{\partial^2#1}\over{\partial#2^2}}}
\def\Pdd#1#2{\Bover{\partial^2#1}{\partial#2^2}}
\def\pde#1#2#3{{{\partial^2#1}\over{\partial#2\partial#3}}}
\def\Pde#1#2#3{\Bover{\partial^2#1}{\partial#2\partial#3}}
\def\Prf{\vthsp\hbox{\bf P\hskip0.07em\llap P\hskip0.07em\llap P\rm}}
\def\pushbottom#1{\hbox{\vrule height0pt depth#1 width0pt}}
\def\Qed{\vthsp\hbox{\bf Q\hskip0.07em\llap Q\hskip0.07em\llap Q\rm}}
\def\QeD{\Qed{\hskip 1.111pt plus 3.333pt minus 0pt}}
\def\Quer{\vthsp\hbox{\bf ?\hskip0.05em\llap ?\hskip0.05em\llap
   ?\thinspace\rm}}
\def\query{\discrversionA{\immediate\write0{query}
    \global\advance\footnotenumber by 1
    \oldfootnote{$^{\the\footnotenumber}$}{query}}{}}
\def\Real{\mathop{\Cal R\text{e}}}
\def\restr{\mathchoice
  {\hbox{$\restriction$}\mskip 1mu}
  {\hbox{$\restriction$}\mskip 1mu}
  {\hbox{$\scriptstyle\restriction$}\mskip 1mu}
  {\hbox{$\scriptscriptstyle\restriction$}\mskip 1mu}}
\def\restrp{\mathchoice
  {\hbox{$\restriction$}\mskip 2mu}
  {\hbox{$\restriction$}\mskip 2mu}
  {\mskip-1mu\hbox{$\scriptstyle\restriction$}\mskip 2mu}
  {\mskip-1mu\hbox{$\scriptscriptstyle\restriction$}\mskip 1mu}}
    %use in \restrp V \restrp W \restrp Y
    % \restrp\Cal P \restrp\Bbb R \restrp\frak
    % \restrp\Omega \restrp\Tau \restrp\Upsilon
    % \restrp[ \restrp\widehat{}
\def\Rho{\text{P}}
\def\roibr{\hbox{\hskip0.05em[\hskip0.1em}}
\def\RoverC{\hbox{\biggerscriptfonts${{\Bbb R}\atop{\Bbb C}}$}}
\def\sequence#1#2{\family{#1}{\Bbb N}{#2}}
\def\sequencen#1{\sequence{n}{#1}}
\def\setover#1#2{\overset{#1}\to{#2}}
\def\shortstrut{\hbox{\vrule height7.5pt depth2.5pt width0pt}}
\def\sshortstrut{\hbox{\vrule height5.8pt depth2.5pt width0pt}}
\def\smallcirc{{\scriptstyle\circ}}
\def\ssbullet{{\scriptscriptstyle\bullet}}
\def\symmdiff{\triangle}
\def\thicken#1{#1\hskip0.03em\llap #1\hskip0.03em\llap
   #1\hskip0.03em\llap #1}
\def\Tau{{\text{T}}}
\def\tbf#1{\text{\bf #1}}
\def\terror#1#2#3{#1:  `#2' for `#3'}
\def\today{\number\day.\number\month.\number\year}
\def\UEMDRR{University of Essex Mathematics Department Research Report}
\def\UEMDWP{University of Essex Mathematics Department Working Paper}
\def\versiondate#1{
    \immediate\write0{Version of #1}
    \def\vsdate{#1}
               \ifnum\stylenumber=1\hfill Version of #1
                              \medskip
               \else\ifnum\stylenumber=2\hfill Version of #1
                              \medskip
               \else\ifnum\stylenumber=6\hfill{\fiverm Version of #1}
                              \medskip
               \else\ifnum\stylenumber=7\hfill{\fiverm Version of #1}
                              \medskip
               \else\ifnum\stylenumber=8\hfill Version of #1
                              \medskip
               \else\fi\fi\fi\fi\fi}
\def\versionhead{\ifnum\stylenumber=3 \hfill PRINTER'S VERSION

                        \bigskip
                   \else\ifnum\stylenumber=4 \hfill REFEREE'S VERSION

                        \bigskip
                   \else \fi\fi}
\def\versionref{\ifnum\stylenumber=1\workingref
           \else\ifnum\stylenumber=2\workingref
           \else\ifnum\stylenumber=8\workingref
           \else\printerref\fi\fi\fi}
\def\vthsp{\hskip0.001em}

%Missing AmSTeX can be coped with by using the following lines:

\ifamstexsupported\else
   \immediate\write0{AmSTeX-unsupported version of  fremtex.tex }
   \fourteensavailablefalse
   \def\@{@}
   \def\\{\cr}
   \def\AmSTeX{AmS\TeX}
   \def\Bbb#1{\underline{\hbox{\bf #1}}}
   \def\biggerscriptfonts{}
   \def\bumpeq{=_{\text{approx}}}
   \def\Cal#1{{\cal#1}}
   \def\curlyeqprec{\preceq}
   \def\eightit{\it}
   \def\eightrm{\rm}
   \def\eightsmc{\smc}
   \def\eurm#1{\underline{\cal#1}}
   \def\Footnote#1#2{\footnote#1#2}
   \def\frac#1#2{{#1}\over{#2}}
   \def\frak#1{{\hbox{\bf #1\hskip0.1em\llap #1}}}
   \def\frnewpage{\vfill\eject}
   \def\iint{\int\!\!\!\int}
   \def\iiint{\int\!\!\!\int\!\!\!\int}
   \def\iiiint{\int\!\!\!\int\!\!\!\int\!\!\!\int}
   \def\llcorner{\lfloor}
   \def\loadeurm{}
   \def\loadeusm{}
%   \def\loadeusmx#1{\def\eusm#1{\underline{\cal#1}}}
   \def\ltimes{|\hskip-0.5em\times}
   \def\eusm#1{\underline{\cal#1}}
   \def\ltimes{|\times}
   \def\Loadfont#1{}
   \def\Matrix#1{\pmatrix{#1}}
   \def\noamslogo{}
   \def\nrightarrow{\not\rightarrow}
   \def\preccurlyeq{\preceq}
   \def\query{\discrversionA{\immediate\write0{query}
                             \footnote{$^*$}{query}}{}}
   \def\restriction{|}
   \def\rtimes{\times|}
   \def\setover#1#2{#2^{#1}}
   \def\shortstrut{\strut}
   \def\smallerfonts{\eightrm}
   \def\smallfrown{\wedge}
   \def\smc{}
   \def\spcheck{\check{\phantom{x}}}
   \def\sphat{\hat{\phantom{x}}}
   \def\square{\mathchoice\sqr64\sqr64\sqr43\sqr33}
   \def\sqr#1#2{{\vcenter{\vbox{\hrule height.#2pt
          \hbox{\vrule width.#2pt height#1pt \kern#1pt
                  \vrule width.#2pt}
          \hrule height.#2pt}}}}
   \def\pmb#1{\underline{#1}}
   \def\text#1{\hbox{#1}}
   \def\therefore{\ssbullet\raise.9ex\hbox{$\ssbullet$}\ssbullet}
   \def\twelvebf{\bf}
   \def\vartriangleleft{\triangleleft}
   \def\Vdash{|\vdash}
   \def\vDash{|=}
   \fi

%End of lines for missing AmSTeX

\ifamstexsupported\input amstex
%              \documentstyle{amsppt}
               \input amsppt.sty
               \PSAMSFonts
	       \tenpoint
               %\font\eightmsb=msym8
               \font\eightmsb=msbm10
               \font\eighteufm=eufm10 at 8 pt
               \font\eighteurm=eurm8
               \font\eighteusm=eusm8
               \font\varseveneufm=eufm10 at 7 pt
               \scriptfont\eufmfam=\varseveneufm
   \def\.#1{{\accent"5F #1}}
   \def\=#1{{\accent"16 #1}}
   \gdef\nopagenumbers{\NoPageNumbers}
   \else\fi

\newcount\footmarkcount
\ifDHFamstexloaded
    \loadmsBm
    \def\eightBbb#1{{\fam\msBfam#1}}
    \loadmsFm
    \def\eightfrak#1{{\fam\msFfam#1}}
  \else
    \def\eightBbb#1{{\scriptstyle\Bbb#1}}
    \def\eightfrak#1{{\scriptstyle\frak#1}}
  \fi
\ifathome\athome\fi

%interword glue (\tenrm standard):
%  (space) 3.333 + 1.667 - 1.111
%  (,)     3.333 + 2.083 - 0.889
%  (.)     4.444 + 5.    - 0.370

%very odd fact:  \discrversionA{2.}{}  doesn't work,
%   \discrversionA{ 2.}{} does

%The false line feed introduced by my Fortran programs, leading to
%a surplus  ae  at the top of the next page, seems to be associated
%with  Fk(\032)3081  or something similar in the PostScript file.


\immediate\write0{Extract from MEASURE THEORY,
   by D.H.Fremlin, University of Essex, Colchester.
   This material is copyright.   It is issued under the terms of the
   Design Science License as published in
   http://dsl.org/copyleft/dsl.txt.   This is a development version and the
   source files are not permanently archived.
   For further information contact david\@fremlin.org.}

%hyphenations
\hyphenation{Archi-me-dean}

%new style
\def\smallprintersize{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{468pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \nologo
     \def\copyrightdate##1{}
     \def\folio{\ifnum\pageno<0{}\else\number\pageno\fi}
     \def\mtfootline{}
     \stylenumber=10
     \pageno=-1000}

%With-proofs, results-only versions
\newif\ifwithproofs
\withproofstrue
\newif\ifresultsonly
\resultsonlyfalse
\def\noproofs{\withproofsfalse}
\def\resultsonly{\resultsonlytrue\withproofsfalse}
\long\def\proof#1{\ifwithproofs{\medskip

     \noindent{\bf proof} #1}\else\fi}
\long\def\prooflet#1{\ifwithproofs{#1}\else\fi}
%\long\def\Prooflet#1{\prooflet{\Prf\ #1 \Qed\ }}
\long\def\endnotes#1{\ifresultsonly\else{#1}\fi}
\long\def\exercises#1{\ifresultsonly\else{#1}\fi}
\long\def\references#1{\ifresultsonly\else{#1}\fi}
\long\def\dvro#1#2{\ifresultsonly{#1}\else{#2}\fi}

%Headline assignments

\def\topparagraph{}
\def\bottomparagraph{}
\def\newparagraph{}
\newif\ifpageturned

\newdimen\rhoffset
\newdimen\lhoffset
\lhoffset=0cm
\rhoffset=0cm

\newcount\volumeno
\volumeno=5
\newcount\luluvolumeno
\luluvolumeno=0

\let\plainpagebody=\pagebody
\def\pagebody{\ifodd\pageno\global\hoffset=\rhoffset
         \else\global\hoffset=\lhoffset\fi
     \plainpagebody
     \turnpage}
\def\turnpage{
     \iffirstpage\else{\baselineskip=24pt\line{\mtfootline}}\fi
%     \immediate\write0{turning page \folio,
%           top paragraph \topparagraph,   bottom \bottomparagraph,
%           new \newparagraph, pagetotal \the\pagetotal}
     \global\pageturnedtrue
     \global\firstpagefalse
     \global\let\topparagraph=\bottomparagraph
     \global\let\bottomparagraph=\newparagraph
     }

\newdimen\pagetotalplus
\newdimen\pagetotalminus

%wheader parameters:  #1 new paragraph name
 % #2#3#4  \vskip parameters if page break not forced (no. of points)
 % #5  spare height demanded if page break not to be forced
\long\def\wheader#1#2#3#4#5{\gdef\newparagraph{#1}
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by #5
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  %\showthe\pagetotal \showthe\pageshrink \showthe\pagegoal
  \global\pageturnedfalse
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal
%     \immediate\write0{forcing page break}
     \eject
     \else\vskip #2pt plus #3pt minus #4pt\fi
     \else\vskip #2pt plus #3pt minus #4pt\fi
  \ifpageturned\global\let\topparagraph=\newparagraph
%     \immediate\write0{paragraph change on turning page}
     \fi
%  \immediate\write0{top paragraph \topparagraph, bottom
%       \bottomparagraph, pagetotal \the\pagetotal}
  \ifdim\pagetotal>\pagegoal\else\gdef\bottomparagraph{#1}\fi
%   \immediate\write0{top paragraph \topparagraph, bottom
%       \bottomparagraph, pagetotal \the\pagetotal}
  }
\def\header#1{
     \wheader{#1}{10}{4}{4}{24pt}}
\def\leader#1#2{\immediate\write0{#1 #2}
     \wheader{#1}{14}{7}{5}{24pt}
     {\bf #1 #2}}
\def\vleader#1#2#3{\immediate\write0{#2 #3}
     \wheader{#2}{14}{7}{5}{#1}
     {\bf #2 #3}}
\def\Notesheader#1{\immediate\write0{#1 Notes and comments}
     \wheader{#1 Notes}{16}{8}{5}{48pt}
     \noindent{\bf #1 Notes and comments}}
\def\vNotesheader#1#2{\immediate\write0{#2 Notes and comments}
     \wheader{#2 Notes}{16}{8}{5}{#1}
     \noindent{\bf #2 Notes and comments}}
\def\spheader#1#2#3#4#5{\header{#1#2#3#4#5}{\bf (#5)}}
\def\spstheader#1#2#3#4#5{\header{*#1#2#3#4#5}{\bf (#5)}}
\def\vspheader#1#2#3#4#5#6{
     \wheader{#2#3#4#5#6}{10}{4}{4}{#1}{\bf (#6)}}
\def\sqheader#1#2#3#4#5{\header{#1#2#3#4#5}{$\pmb{>}${\bf (#5)}}}
\def\errheader#1#2#3{
     \wheader{#3}{6}{4}{2}{24pt}
     {\bf p #1 l #2}}
\long\def\woddheader#1#2#3#4#5{
   \ifodd\pageno\wheader{#1}{#2}{#3}{#4}{#5}
   \else\vskip #2pt plus #3pt minus #4pt\fi}

\def\indexmedskip{
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by 18 pt
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal
      \else\medskip\fi\else\medskip\fi}

\def\indexheader#1{\immediate\write0{#1}\vindexheader{#1}{24}}
\def\vindexheader#1#2{
  \gdef\newparagraph{#1}
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by #2 pt
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  %\showthe\pagetotal \showthe\pageshrink \showthe\pagegoal
  \global\pageturnedfalse
%  \ifdim\pagetotal>\pagegoal
%      \immediate\write0{warning: pagetotal greater than pagegoal}
%      \immediate\write0{...   last paragraph reached is #1}
%      \immediate\write0{...   check for underfull vbox here}
%      \else\fi
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal{\eject}
      \fi\fi
  \ifdim\pagetotal>\pagegoal\else\gdef\bottomparagraph{#1}\fi
  \ifpageturned{\global\let\topparagraph=\newparagraph
%     \immediate\write0{paragraph change on turning page}
      }\else\fi
%  \immediate\write0{top paragraph \topparagraph, bottom
%    \bottomparagraph}
  }

\def\leftheadline{\rlap{\folio}\hfill{\eightit\chaptername}\hfill
       \llap{\eightbf\topparagraph}}
\def\rightheadline{\rlap{\eightbf\bottomparagraph}\hfill
       {\eightit\headlinesectionname}\hfill\llap{\folio}}

%Footline assignments
\def\mtfootline{\ifodd\pageno\rightfootline\else\leftfootline\fi}
\def\rightfootline{\hfill\eightsmc D.H.Fremlin}
\def\leftfootline{\ifwithproofs{\eightsmc Measure Theory\hfill}
    \else{\eightsmc Measure Theory \eightit (abridged version)\hfill}
    \fi}

%Style assignments
\def\newchapter#1{\gdef\topparagraph{}
   \gdef\bottomparagraph{Chap.\ #1 {\it intro.}}
   \gdef\newparagraph{Chap.\ #1 {\it intro.}}
   \footmarkcount=0
   \gdef\sectionname{Introduction}
   \gdef\headlinesectionname{Introduction}
   \centerline{\bf Chapter #1}
   \medskip
   \centerline{\bf \chaptername}
   \medskip
   }
\def\newsection#1{\gdef\bottomparagraph{\S#1 {\it intro.}}
   \noindent{\bf #1 \sectionname}
   \smallskip
   \let\headlinesectionname=\sectionname}
\def\newvolume#1{\gdef\topparagraph{}
   \gdef\bottomparagraph{Vol.\ #1 {\it intro.}}
   \Loadfont{\twelvebf=cmbx12}
   \def\chaptername{Volume #1}
   \def\sectionname{\volumename}
   \def\headlinesectionname{\volumename}
   \centerline{\twelvebf Volume #1:  \volumename}
   \bigskip
   }

%font variations
\newif\ifvolwp
\volwpfalse
\def\tenrmstretch#1{\fontdimen3\tenrm=#1}
\font\elevenex=cmex10 at 11pt

%contents and introductions
\newif\iflargelogo
\def\vtmpb#1{}
\def\newvolume#1{}
\def\pagereference#1#2{\hfill\hbox{\ifresultsonly#2\qquad \hskip 1truein
                                                 \else#1\qquad \fi}}
\long\def\section#1#2#3#4#5#6{\ifresultsonly
    \qquad #1 #2\pagereference{#4}{#5}
  \else\vskip 2pt plus 1pt minus 0pt
    \vbox{\qquad #1 #2\pagereference{#4}{#5}\par
    \abstr{#6}
    }\vskip -1pt plus 1pt minus 0pt
  \fi}
\long\def\chapintrosection#1#2#3{\ifresultsonly
    \qquad Introduction\pagereference{#2}{#3}
  \else\vskip 1pt plus 1pt minus 0pt
    \qquad Introduction\pagereference{#2}{#3}
    \vskip 1pt plus 1pt minus 0pt
  \fi}
\def\abstr#1{\ifresultsonly{}
   \else{\advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \vskip1pt plus 0pt minus0pt
     \parindent=0pt
     \smallerfonts #1%
     \hbox{\vrule height0pt depth2pt width0pt}
     \vskip1pt plus 0pt minus0pt}\fi}
     
%language variations
\newif\ifUSEnglish
\USEnglishfalse
%\USEnglishtrue

%"Later editions only"
\newcount\latereditionpage
\newcount\latereditionmark
\latereditionpage=-1
\latereditionmark=0
\newcount\newpageno

%removing the \write0 instruction altogether disorganizes it!
\def\Latereditions{\newpageno=\pageno%
  \ifdim\pagetotal>\pagegoal\advance\newpageno by 1\fi%
     \divide\newpageno by 2%
  \immediate\write0{%
%     "Later editions only";
%      pagetotal \the\pagetotal, pagegoal \the\pagegoal,
%      pageno \number\pageno,\par newpageno \number\newpageno,
%      latereditionpage \number\latereditionpage%
  }\ifnum\latereditionpage=\newpageno%
     \ifDHFamspptloaded$^{\number\latereditionmark}$%
     \else\footnote{Later editions only.}\fi%
  \else\footnote{Later editions only.}%
  \global\latereditionpage=\newpageno%
  \global\latereditionmark=\footmarkcount\fi}%
%  \immediate\write0{newpageno \number\newpageno,
%      latereditionpage \number\latereditionpage}

   
%general definitions
\def\action{\mathchoice{\lower-.3ex\hbox{$\ssbullet$}}
                       {\lower-.3ex\hbox{$\ssbullet$}}
                       {\ssbullet}
                       {\ssbullet}}  %see also \varaction
\def\add{\mathop{\text{add}}\nolimits}
\long\def\allowmorestretch#1#2{
  \ifdim\pagewidth<#1pt\else
    \fontdimen3\tenrm=2pt\fontdimen4\tenrm=1.67pt\fi
  #2
    \fontdimen3\tenrm=1.67pt\fontdimen4\tenrm=1.11pt}
\def\am{atom\discretionary{-}{}{}lessly-measurable}
\def\AmuA{\Aut_{\bar\mu}\mskip-2mu\frak A}
\def\ASint{\mathchoice{\Bbb A\mskip-5mu}
                  {\Bbb A\mskip-3mu}
                  {\Bbb A\mskip-2mu}
                  {\Bbb A\mskip-2mu}\Sint}
\def\Asisfour{$(\frak A,\bar\mu,T,\family{t}{T}{\frak A_t})$}
\def\Asissix{$(\frak A,\bar\mu,T,
   \penalty-100\family{t}{T}{\frak A_t},
   \penalty-100\Cal T,\family{\tau}{\Cal T}{\frak A_{\tau}})$}
\def\Asisfourt{$(\frak A,\bar\mu,\coint{0,\infty},
   \langle\frak A_t\rangle_{t\ge 0})$}
\def\Asissixt{$(\frak A,\bar\mu,\coint{0,\infty},
   \penalty-100\langle\frak A_t\rangle_{t\ge 0},
   \penalty-100\Cal T,\family{\tau}{\Cal T}{\frak A_{\tau}})$}
\def\Atop#1#2{{\scriptstyle #1}\atop{\scriptstyle #2}}
\def\Aut{\mathop{\text{Aut}}}

\def\Bcap{\Booleanop{\cap}}
\def\Bcup{\Booleanop{\cup}}
\def\Booleanop#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\hskip.2em\scriptfont2=\eightsy$\scriptstyle#1$\hskip.2em}}
    {\hbox{\hskip.2em\scriptfont2=\eightsy$\scriptstyle#1$\hskip.2em}}
 %   {{\mskip2mu\scriptstyle#1\mskip2mu}}  %until 15.10.17
    {{\scriptscriptstyle#1}}  %from 15.10.17
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\Bsetminus{\Booleanop{\setminus}}
\def\Bsubseteq{\hskip.1em\Booleanop{\subseteq}\hskip.1em}
\def\Bsubset{\Booleanop{\subset}}
\def\Bsupseteq{\Booleanop{\supseteq}}
\def\Bsupset{\Booleanop{\supset}}
\def\Bsymmdiff{\Booleanop{\symmdiff}}
\def\Bcapshort{\Booleanopshort{\cap}}
\def\Bcupshort{\Booleanopshort{\cup}}
\def\Booleanopshort#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {\hskip-1pt\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {{\scriptstyle#1}}
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\Bsetminusshort{\Booleanopshort{\setminus}}
\def\BSint{\mathchoice{\Bbb B\mskip-5mu}
                  {\Bbb B\mskip-3mu}
                  {\Bbb B\mskip-2mu}
                  {\Bbb B\mskip-2mu}\Sint}
%\def\Bsubseteqshort{\hskip.1em\Booleanopshort{\subseteq}\hskip.1em}
\def\Bsubseteqshort{\hskip.1em\Booleanopshort{\subseteq}}
\def\Bsubsetshort{\Booleanopshort{\subset}}
\def\Bsupseteqshort{\Booleanopshort{\supseteq}}
\def\Bsymmdiffshort{\Booleanopshort{\symmdiff}}

\def\closeBooleanop#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {{\scriptstyle#1}}
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\closeBcup{\closeBooleanop{\cup}}
\def\closeBcap{\closeBooleanop{\cap}}
\def\closeBsetminus{\closeBooleanop{\setminus}}
\def\closeBsymmdiff{\closeBooleanop{\symmdiff}}

\def\biggerHeint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 2.6pt{\sevenrm H}\hss}
   \intop\nolimits}
\def\biggerPfint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 1.3pt{\sevenrm Pf}\hss}
   \intop\nolimits\hskip 1.2pt}
\def\biggerSint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 3.6pt{\sevenrm S}\hss}
   \intop\nolimits}
\def\bu{\mathop{\text{bu}}}
\def\Bvalue#1{\mathchoice
   {\hbox{$[\![#1]\!]$}}
   {\hbox{$[\![#1]\!]$}}
   {\hbox{$\scriptstyle[\![#1]\!]$}}
   {\hbox{$\scriptscriptstyle[\![#1]\!]$}}}
\def\cadlag{c\`adl\`ag}
\def\CalBa{\Cal{B}\eurm{a}}
   %note: neither  \cal  nor  \eusm  has lower case
\def\CalFr{\Cal F_{\text{Fr}}}
\def\callal{c\`all\`al}
\def\CalNwd{\Cal{N}\hskip-1.2pt\eurm{w}\hskip-.8pt\eurm{d}}

\def\CalRbg{\Cal{R}\mskip-1mu\eurm{b}\mskip-1mu\eurm{g}}
\def\CalSmz{\Cal{S}\mskip-2mu\eurm{mz}}
\def\CalUn{\Cal{U}\eurm{n}}
\def\capacity{\mathop{\text{cap}}\nolimits}
\def\Caratheodory{Cara\-th\'eo\-dory}
\def\Cdlg{C_{\text{dlg}}}
\def\cff{\cf\mskip1mu}
\def\checkpmb#1{\check{\pmb #1}}
\def\ci{\mathop{\text{ci}}}
\def\cl{\mathop{\text{cl}}}
\def\Clll{\hbox{$C$
  \hskip-4.5pt\raise4pt\hbox{\char'23
  \hskip-1.6pt\vrule height6.5pt width.5pt depth0pt
  \hskip.7pt\vrule height6.5pt width.5pt depth0pt
  \hskip-2.5pt\lower5.4pt\hbox{\char'23}}}}
\def\clMsimp{\overline{M}_{\text{simp}}}
\def\closeuparrow{\mathchar"0222}
\def\closedownarrow{\mathchar"0223}
\def\clstar{\text{cl*}}
\def\cov{\mathop{\text{cov}}}
\def\covar#1#2{\mathchoice
  {[#1{\lower4pt\hbox{${\scriptstyle *}\atop{\vrule height 18pt depth -10pt}$}}#2]}
  {[#1{*\atop{\vrule height 8pt depth 0pt}}#2]}
  {[#1{*\atop{\vrule height 6pt depth 0pt}}#2]}
  {[#1{*\atop{\vrule height 6pt depth 0pt}}#2]}}
\def\covSh{\mathop{\text{cov}}\nolimits_{\text{Sh}}}
\def\cdownarrow{c\mskip2mu^{\downarrow}}
\def\cycle#1{(\overleftarrow{#1})}
\def\ddaggerloc{$(\ddagger)_{\text{loc}}$}
\def\ddownarrow{d\mskip2mu^{\downarrow}}
%\def\DiniD{\setover{
%  \hbox{\hskip.1em\vrule height-1pt width.5em depth1.4pt\hfil}}D}
%\def\Dinid{\hskip.1em
%  \hbox{\hskip.05em\vrule height-1.3pt width.5em depth1.7pt}\hskip-.65em D}
\def\Dinid{\hbox{$D$\hskip-8.3pt\vrule height-1pt width7pt depth1.85pt}
   \hskip1pt}
\def\DiniD{\hbox{$D$\hskip-8.3pt\vrule height9.2pt width7ptdepth-8.35pt}
   \hskip1pt}

\def\discrcenter#1#2{\ifdim\pagewidth>#1{#2}\else{

  \Centerline{#2}

  \noindent}\fi}
    %note:  in \discrcenter, make sure that #2 ends in a space
    % (if you want one to appear) and that the continuation
    % follows without
    % a space after the final } of #2, so that it will be
    % truly unindented
\def\discrpage{\ifnum\stylenumber=1{\frnewpage}  %redefined in mt.tex
        \else\ifnum\stylenumber=3{\frnewpage}
        \else\ifnum\stylenumber=5{\frnewpage}
        \else\wheader{}{18}{14}{2}{72pt}\fi\fi\fi}
\def\displaycause#1{\noalign{\noindent (#1)}}
\def\Displaycause#1{\cr\displaycause{#1}&}
\def\dotproduct{\hskip.1em\tbf{.}\hskip.1em}
\def\doubleheadrightarrow{\rightarrow\mskip-12mu\rightarrow}
\long\def\doubleinset#1{\inset{\inset{\parindent=-20pt #1}}}
\def\duparrow{d\mskip2mu^{\uparrow}}
\def\dvAformerly#1{\discrversionA{\footnote{Formerly #1.}}{}}
\def\dvAnew#1{\discrversionA{\footnote{New #1.}}{}}
\def\dvArevised#1{\discrversionA{\footnote{Revised #1.}}{}}
\def\dvrocolon#1{\dvro{#1: }{#1}}
\def\eightVVdash{\mskip5mu\vrule height 6pt depth 2pt width 0.4pt
  \mskip2mu\vrule height 6pt depth 2pt width 0.4pt
  \vrule height 2.25pt depth -1.8pt width 3.2pt\mskip2mu}
\def\embedsinto{\mskip5mu\raise1.5pt\hbox{$\subset$}\mskip-15mu
   \lower3.5pt\hbox{$\rightarrow$}\mskip5mu}
\def\energy{\mathop{\text{energy}}\nolimits}
%\eqalign:  see \S478 for breaks during \eqalign
\def\equivGT{\equiv_{\text{GT}}}
\def\equivT{\equiv_{\text{T}}}
\def\Eta{\text{H}}
\def\Expn{\Bbb E}
\long\def\Extract{\oldfootnote{}{
   \noindent{\eightit
       \ifresultsonly{Extract from {\eightsmc Measure Theory,}
          results-only version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else\ifwithproofs{Extract from {\eightsmc Measure Theory,}
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else{Extract from {\eightsmc Measure Theory,} abridged version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \fi\fi
   This material is copyright.   It is issued under the terms of the
   Design Science License as published in
   {\eighttt http://dsl.org/copyleft/dsl.txt}.
   This is a development version and the
   source files are not permanently archived, but current versions are
normally
   accessible through
   {\eighttt https://www1.essex.ac.uk/maths/people/fremlin/mt.htm}.
   For further information contact {\eighttt david\@fremlin.org}.}
   }}
\def\Fn{\mathop{\text{Fn}}\nolimits}
\def\FN{\mathop{\text{FN}}\nolimits}
\def\formerly#1{\footnote{Formerly #1.}}
  %use only once in a chapter, never in same volume as reference
\def\fraction#1{\mathchoice{\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$\scriptstyle<$}#1\hbox{$\scriptstyle>$}}
     {\hbox{$\scriptscriptstyle<$}#1\hbox{$\scriptscriptstyle>$}}}
\def\frakmctbl{\frak m_{\text{countable}}}
\def\frmedcirc{{\raise 2.5pt\hbox{$_{\bigcirc}$}}}
\def\grad{\mathop{\text{grad}}\nolimits}
\def\hatpmb#1{\hat{\pmb{#1}}}
\def\hattilde#1{\skew2\hat{\tilde#1}}
\def\hattildepmb#1{\skew1\hat{\tildepmb#1}}
\def\hcov{\mathop{\text{hcov}}\nolimits}
\def\Heint{\mathchoice{\hbox to 0pt{\hskip 3.1pt H\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     \intop\nolimits}
\def\high#1{\lceil#1\rceil}
\def\highoverline#1{\overline{\vrule height6pt width0pt depth 0pt#1}}
\def\Hint#1{({\it Hint\/}: #1)}
\def\hL{\mathop{\text{hL}}}
\def\Hmp{\text{Hom}_{\bar\mu,\pi}}
\def\hp{\mathop{\text{hp}}\nolimits}
\def\hwtr{hereditarily weakly $\theta$-{\vthsp}refinable}
\def\iiv{ii_{\pmb{v}}}
\def\iivu{\iiv(\pmb{u})}
\def\imp{inverse-{\vthsp}measure-{\vthsp}preserving}
\def\innerprod#1#2{(#1|#2)}
\def\intstar{\text{int*}}
\def\CalIS{\Cal I(\Cal S)}
\def\jump{\mathop{\text{jump}}\nolimits}
\def\Jump{\mathop{\text{Jump}}\nolimits}
\def\Krein{Kre\v{\i}n}
\def\Lambdapv{\Lambda_{\text{pv}}}
\def\leRB{\le_{\text{RB}}}
\def\leRK{\le_{\text{RK}}}
\def\link{\mathord{\text{link}}}
\def\locmafp{\widehat{\otimes}_{\text{loc}}}
\def\Locmafp{\widehat{\bigotimes}^{\text{loc}}}
\def\low#1{\lfloor#1\rfloor}
\def\lti{left-\vthsp translation-\vthsp invariant}
\def\magnitude{\mathop{\text{mag}}}
\def\Mah{\mathop{\text{Mah}}}
\def\Mahcr{\Mah_{\text{cr}}}
\def\MahcrR{\Mah_{\text{crR}}}
\def\MahqR{\Mah_{\text{qR}}}
\def\MahR{\Mah_{\text{R}}}
\def\MahspcrR{\Mah_{\text{spcrR}}}
\def\Malm{M_{\text{alm}}}
\def\Marik{Ma\v r\'\i k} 
\def\Mbv{M_{\text{bv}}}
\def\Mcdlg{M_{\text{cdlg}}}
\def\Mdm{M_{\text{D-M}}}
\def\med{\mathop{\text{med}}\nolimits}
\def\Mfa{M_{\text{fa}}}
\def\Mhsr{\mathop{\text{Mhsr}}\nolimits}
\def\Mjf{\mathop{M_{\text{j-f}}}\nolimits}
\def\Mlbv{M_{\text{lbv}}}
\def\Mlob{M_{\text{lo-b}}}
\def\Mlns{M_{\text{ln-s}}}
\def\Mmo{M_{\text{mo}}}
\def\Mlmo{M_{\text{lmo}}}
\def\Mns{M_{\text{n-s}}}
\def\Mnsu{M_{\text{n-s}}^{\uparrow}}
%\def\Mnsu{M_{\text{n-s}\uparrow}}
\def\MMah{M_{\text{Mah}}}
\def\Mob{M_{\text{o-b}}}
\def\Mpob{M_{\text{po-b}}}
\def\Mpv{M_{\text{pv}}}
\def\MSi{M_{\text{S-i}}}
\def\Msimp{M_{\text{simp}}}
\def\Mth{Ma\-ha\-ram-{\vthsp}type-{\vthsp}homogeneous}
\def\Mvlm{M_{\text{vlm}}}
\def\negquad{\hskip-1em\relax}
\def\NN{\BbbN^{\Bbb N}}
\def\non{\mathop{\text{non}}}
\def\notBooleanop#1{
   \ifamstexsupported{\mathchoice
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
    {{\not\scriptstyle#1}}
    {{\not\scriptscriptstyle#1}}}
   \else{\mathchoice{{\not\scriptstyle#1}}{{\not\scriptstyle#1}}
    {{\not\scriptstyle#1}}{{\not\scriptscriptstyle#1}}}
    \fi}
\def\notBsubseteq{\notBooleanop{\subseteq}}
\def\notBsupseteq{\notBooleanop{\supseteq}}
\def\notVVdash{\not\mskip4mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \mskip2.5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \vrule height 2.75pt depth -2.25pt width 4pt\mskip-2mu}
\def\nuprime{\nu^{\mskip0.7mu\prime}} %better  \nu'  in superscripts
\def\nw{\mathop{\text{nw}}}
\def\Osclln{\mathop{\text{Osclln}}\nolimits}
\def\otp{\mathop{\text{otp}}}
\def\partstar{\mathchoice{\partial\hbox{*}}
     {\partial\hbox{*}}
     {\partial\hbox{\sevenrm *}}
     {\partial\hbox{\fiverm *}}}
\def\per{\mathop{\text{per}}}
\def\Pfint{\mathchoice{\hbox to 0pt{\hskip 3.1pt Pf\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.3pt{\sevenrm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     \intop\nolimits
     \mathchoice{\hskip 2pt}{\hskip 1.2pt}{\hskip 1pt}{\hskip 1pt}}
\def\Pou{\mathop{\text{Pou}}\nolimits}
\def\prforc{\preccurlyeq_{\text{forcing}}}
\def\prGT{\preccurlyeq_{\text{GT}}}
\def\prT{\preccurlyeq_{\text{T}}}
\def\pssqa{power set $\sigma$-quotient algebra}
\def\qm{quasi-{\vthsp}measurable}
\def\Reverse#1{\setover
   {\lower0.4ex\hbox{$\scriptscriptstyle\leftrightarrow$}}{#1}}
\def\rhoKR{\rho_{\text{KR}}} 
\def\rhotv{\rho_{\text{tv}}} 
\def\Rint{\mathchoice{\hbox to 0pt{\hskip 3.1pt R\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     \intop\nolimits}
\def\RO{\text{RO}}
\def\rti{right-{\vthsp}translation-{\vthsp}invariant}
\def\rvm{real-{\vthsp}valued-{\vthsp}measurable}
\def\sat{\mathop{\text{sat}}}
\def\sfamilysS#1{\family{\sigma}{\Cal S}{#1_{\sigma}}}
\def\sfcc{$\sigma$-finite-cc}
\def\sgn{\mathop{\text{sgn}}}
\def\shr{\mathop{\text{shr}}\nolimits}
\def\Sii{\text{S}ii}
\def\Siiv{\Sii_{\pmb v}}
\def\Sint{\mathchoice{\hbox to 0pt{\hskip 4.1pt\ninerm S\hss}} 
     {\raise 0.6pt\hbox to 0pt{\hskip 3.0pt{\sevenrm S}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     \intop\nolimits}
\def\Sinteightpoint{\raise 0.6pt\hbox to 0pt{\hskip 2.3pt{\fiverm S}\hss}
     \intop\nolimits}
\def\SL{\text{SL}}
\def\sm{semi-martingale} %hyphen to remind me that it's the "virtual" kind
\def\spread{\mathop{\text{spr}}}
\def\ssplrarrow{^{\scriptscriptstyle\leftrightarrow}}
\def\ssptilde{{\raise1.2ex\hbox{$\scriptscriptstyle\sim$}}}
%\def\sspvec{{\raise1.2ex\hbox{$\scriptscriptstyle\rightarrow$}}}
\def\sspvec{\mskip3mu\vec{\null}\mskip5mu}
\def\Sti{\mathop{\text{Sti}}\nolimits}
\def\strbar{^{\scriptscriptstyle\pmb{-}}}
\def\strplus{^{\scriptscriptstyle\pmb{+}}}
\def\strprime{^{\mathchoice{\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1mu\pmb{\prime}}}}
%\def\strprime{\,\pmb{\prime}}
\def\supp{\mathop{\text{supp}}}
\def\tausm{\tau_{\sigma\text{-m}}}
\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_<$}
   \phantom{\tbf 1^{(#1)}}}
%\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_-$}
%   \phantom{\tbf 1^{(#1)}}}
\def\tensorhat{\widehat{\otimes}}
\def\Tensorhat{\widehat{\bigotimes}}
\def\tensorhatsigma{\widehat{\otimes}_{\sigma}}
\def\tildeClll{\hbox{$\tilde C$
  \hskip-4.5pt\raise4pt\hbox{\char'23
  \hskip-1.6pt\vrule height6.5pt width.5pt depth0pt
  \hskip.7pt\vrule height6.5pt width.5pt depth0pt
  \hskip-2.5pt\lower5.4pt\hbox{\char'23}}}}
\def\tildehat#1{\skew2\tilde{\hat#1}}
\def\tildehatpmb#1{\tilde{\hatpmb#1}}
\def\tildepmb#1{\tilde{\pmb{#1}}}
\def\Tr{\mathop{\text{Tr}}}
\def\trs{^{\top}\hskip-1pt}
\def\triplepc#1#2#3{(#1,#2,\hbox{$<$}#3)}
\def\TSi{\frak T_{\text{S-i}}}
\def\undphi{\underline{\phi}{}}
\def\undpsi{\underline{\psi}{}}
\def\undtheta{\underline{\theta}{}}
\def\updown{\mathchoice
  {\hbox{\hskip-1.5pt\char'23
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-3.3pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-2.5pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23  %OK for subscript, fails for superscript
    \hskip-1.7pt\vrule height6.5pt width.7pt depth0pt
    \hskip.9pt\vrule height6.5pt width.7pt depth0pt
    \hskip-2.5pt\lower5.4pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23  %not tested
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-2.5pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}}
\def\updownarrows{\uparrow\mskip-4mu\downarrow}
\def\updowncgt{$\updown$-convergent}
\def\upr{\mathop{\text{upr}}\nolimits}
\def\ureg{\mathop{\text{ureg}}}

\def\Var{\mathop{\text{Var}}\nolimits}

\def\varaction{\mathchoice{\lower-.2ex\hbox{$\smallcirc$}}
                       {\lower-.2ex\hbox{$\smallcirc$}}
                       {\ssbullet}
                       {\ssbullet}}
\def\varcheck#1{{\setover
       {\lower 0.4ex\hbox{$\scriptscriptstyle\vee$}}{#1}}}
\def\varcheckf{{\setover{\,\lower 0.4ex
              \hbox{$\scriptscriptstyle\vee$}}{f}}}
\def\varchecki{\mskip-3mu\text{\it{\accent"14{\hbox{\hskip-3.5pt\i}}}}\mskip2mu}
\def\varspcheck{^{\scriptscriptstyle\vee}}
%\def\varspcheck{{\setover{\scriptscriptstyle\vee}{\phantom{g}}}}
%\def\varspcheckf{{\setover{\lower 0.4ex
%              \hbox{$\scriptscriptstyle\vee$}}{\phantom{f}}}}
\def\varhat#1{{\setover{\lower 0.4ex
              \hbox{$\scriptscriptstyle\wedge$}}{#1}}}
\def\varhatf{{\setover{\,\lower 0.4ex
              \hbox{$\scriptscriptstyle\wedge$}}{f}}}
\def\varinnerprod#1#2{#1\dotproduct#2}
\def\varsphat{^{\scriptscriptstyle\wedge}}
% or use  \hbox{\^{\null}}
%\def\varsphat{{\setover{\scriptscriptstyle\wedge}{\phantom{g}}}}
%\def\varsphatf{{\setover{\lower 0.4ex
%              \hbox{$\scriptscriptstyle\wedge$}}{\phantom{f}}}}
\def\varBspcheck{\mathchoice
  {\hskip-2pt
  {\fourteenrm\hbox{\lower2pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2pt
  {\fourteenrm\hbox{\lower2pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2.5pt
  {\fourteenrm\hbox{\lower4pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2.5pt
  {\fourteenrm\hbox{\lower4pt\hbox{\v{\null}}}}\hskip-1pt}}
\def\varCspcheck{\mathchoice
  {\hskip-1pt\check{\phantom x}}   %not tested
  {\hskip-1pt\check{\phantom x}}
  {\hskip-1.2pt\check{\phantom x}}
  {\hskip-1.5pt\check{\phantom x}}}  %not tested
\def\varCsphat{\mathchoice
  {\hskip-1pt\hat{\phantom x}}   %not tested
  {\hskip-1pt\hat{\phantom x}}
  {\hskip-1.2pt\hat{\phantom x}}  %not tested
  {\hskip-1.5pt\hat{\phantom x}}}  %not tested
\def\VeqL{$V\hbox{$=$}L$}
\def\VVdash{\mskip5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \mskip2.5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \vrule height 2.75pt depth -2.25pt width 4pt\mskip2mu}
\def\VVdP{\VVdash_{\Bbb P}}
\def\VVdPk{\VVdash_{\Bbb P_{\kappa}}}
\def\wdistr{\mathop{\text{wdistr}}}
\def\weight{\mathop{\text{wt}}}
\def\w1lim{\mathop{\text{w1lim}}}
\def\wsid{weakly $(\sigma,\infty)$-{\vthsp}distributive}
\def\1lim{\mathop{1\mskip-1.5mu\text{lim}}}
\def\2vm{two-{\vthsp}valued-{\vthsp}measurable}

%Cara\-th\'e\-o\-dory

\def\chaptername{}
\def\sectionname{}
\def\headlinesectionname{}

\loadeusm\loadeurm
\Loadfourteens
\Loadfont{\ninerm=cmr9}

%note:  $..\cmmnt{=..}..$  is better as  \cmmnt{\mskip5mu
 
  
%----------------------------------------------------------%  
  
\smallprint  
%\USsmallprint  
%\noproofs  
%\resultsonly  
%\atUEssex  
\usegraphicx
  
\Extract  
  
%----------------------------------------------------------%  

\frfilename{mt28.tex} 
\versiondate{17.1.15} 
\copyrightdate{1994} 
 
\def\chaptername{Fourier analysis} 
\def\sectionname{Introduction} 
 
\newchapter{28} 
 
For the last chapter of this volume, I attempt a brief account of one of 
the most important topics in analysis.   This is a bold enterprise, and 
I cannot hope to satisfy the reasonable demands of anyone who knows and 
loves the subject as it deserves.   But I also cannot pass it by without 
being false to my own subject, since problems contributed by the 
study of Fourier series and transforms have led measure theory 
throughout its history.   What I will try to do, therefore, is to give 
versions of those results which everyone ought to know in language 
unifying them with the rest of this treatise, aiming to open up a 
channel for the transfer of intuitions and techniques between the 
abstract general study of measure spaces, which is the centre of our 
work, and this particular family of applications of the theory of 
integration. 
 
I have divided the material of this chapter, conventionally enough, into 
three parts:  Fourier series, Fourier transforms and the characteristic 
functions of probability theory.   While it will be obvious that many 
ideas are common to all three, I do not think it useful, at this stage, 
to try to formulate an explicit generalization to unify them;  that 
belongs to a more general theory of harmonic analysis on groups, which 
must wait until Volume 4.   I begin however with a section on the 
Stone-Weierstrass theorem (\S281), which is one of the basic tools of 
functional analysis, as well as being useful for this chapter.   The 
final section (\S286), a proof of Carleson's theorem, is at a rather 
different level from the rest. 
 
\discrpage 
 
\frfilename{mt281.tex} 
\versiondate{4.12.12} 
\copyrightdate{1994} 
      
\def\chaptername{Fourier analysis} 
\def\sectionname{The Stone-Weierstrass theorem} 
      
\newsection{281} 
      
Before we begin work on the real subject of this chapter, it will be 
helpful to have a reasonably general statement of a fundamental theorem 
on the approximation of continuous functions.   In fact I give a variety 
of forms (281A, 281E, 281F and 281G, together with 281Ya, 281Yd and 
281Yg), all of which are sometimes useful.   I end the section with a 
version of Weyl's Equidistribution Theorem (281M-281N). 
      
\leader{281A}{Stone-Weierstrass theorem:  first form} 
Let $X$ be a topological space and $K$ a compact subset of $X$.   Write 
$C_b(X)$ for the space of all bounded continuous real-valued functions 
on $X$, so that $C_b(X)$ is a linear space over $\Bbb R$. 
Let $A\subseteq C_b(X)$ be such that 
      
\inset{$A$ is a linear subspace of $C_b(X)$;} 
      
\inset{$|f|\in A$ for every $f\in A$;} 
      
\inset{$\chi X\in A$;} 
      
\inset{whenever $x$, $y$ are distinct points of $K$ there is an $f\in A$ 
such that $f(x)\ne f(y)$.} 
      
\noindent Then for every continuous $h:K\to\Bbb R$ and $\epsilon>0$ 
there is an $f\in A$ such that 
      
\inset{$|f(x)-h(x)|\le\epsilon$ for every $x\in K$,} 
      
\inset{if $K\ne\emptyset$, $\inf_{x\in X}f(x)\ge\inf_{x\in K}h(x)$ and 
$\sup_{x\in X}f(x)\le\sup_{x\in K}h(x)$.} 
      
\cmmnt{\medskip 
      
\noindent{\bf Remark} I have stated this theorem in its natural context, 
that of general topological spaces.   But if these are unfamiliar to 
you, you do not in fact need to know what they are.   If you read 
`let $X$ be a topological space' as `let $X$ be a subset of $\Bbb R^r$'  
and `$K$ is a compact subset of $X$' as `$K$ is a subset of 
$X$ which is closed and bounded in $\BbbR^r$', you will have enough for 
all the applications in this chapter.   In order to follow the proof, of 
course, you will need to know a little about compactness in $\BbbR^r$; 
I have written out the necessary facts in \S2A2. 
} 
      
\proof{{\bf (a)} If $K$ is empty, then we can take $f=\tbf{0}$ to be the 
constant function with value $0$.   So henceforth let us suppose that 
$K\ne\emptyset$. 
      
\medskip 
      
{\bf (b)} The first point to note is that if $f$, $g\in A$ then  
$f\wedge g$ and $f\vee g$ belong to $A$, where 
      
\Centerline{$(f\wedge g)(x)=\min(f(x),g(x))$,\quad 
$(f\vee g)(x)=\max(f(x),g(x))$} 
      
\noindent for every $x\in X$;  this is because 
      
\Centerline{$f\wedge g=\Bover12(f+g-|f-g|)$,\quad 
$f\vee g=\Bover12(f+g+|f-g|)$.} 
      
\noindent It follows by induction on $n$ that $f_0\wedge\ldots\wedge f_n$  
and $f_0\vee\ldots\vee f_n$ belong to $A$ for all $f_0,\ldots,f_n\in A$. 
      
\medskip 
      
{\bf (c)} If $x$, $y$ are distinct points of $K$, and $a$, $b\in\Bbb R$, 
there is an $f\in A$ such that $f(x)=a$ and $f(y)=b$.   \Prf\ Start from 
$g\in A$ such that $g(x)\ne g(y)$;  this is the point at which we use 
the last of the list of four hypotheses on $A$.   Set 
      
\Centerline{$\alpha=\Bover{a-b}{g(x)-g(y)}$,\quad 
$\beta=\Bover{bg(x)-ag(y)}{g(x)-g(y)}$,\quad 
$f=\alpha g+\beta\chi X\in A$.   \Qed} 
      
\medskip 
      
{\bf (d)} (The heart of the proof lies in the next two paragraphs.)  Let 
$h:K\to\coint{0,\infty}$ be a continuous function and $x$ 
any point of $K$.   For any $\epsilon>0$, there is an $f\in A$ such that 
$f(x)=h(x)$ and $f(y)\le h(y)+\epsilon$ for every $y\in K$.   \Prf\ 
Let $\Cal G_x$ be the family of those open sets $G\subseteq X$ for which 
there is some $f\in A$ such that $f(x)=h(x)$ and $f(w)\le h(w)+\epsilon$ 
for every $w\in K\cap G$.   I claim that $K\subseteq\bigcup\Cal G_x$. 
To see this, take any $y\in K$.   By (c), there is an $f\in A$ such that 
$f(x)=h(x)$ and $f(y)=h(y)$.   Now $h-f\restr K:K\to\Bbb R$ is a 
continuous function, taking the value $0$ at $y$, so there is an open 
subset $G$ of $X$, containing $y$, such that 
$(h-f\restr K)(w)\ge-\epsilon$ for every $w\in G\cap K$, that is, 
$f(w)\le h(w)+\epsilon$ for every $w\in G\cap K$.   Thus $G\in\Cal G_x$ 
and $y\in\bigcup\Cal G_x$, as required. 
      
Because $K$ is compact, $\Cal G_x$ has a finite subcover 
$G_0,\ldots,G_n$ 
say.   For each $i\le n$, take $f_i\in A$ such that $f_i(x)=h(x)$ and 
$f_i(w)\le h(w)+\epsilon$ for every $w\in G_i\cap K$.   Then 
      
\Centerline{$f=f_0\wedge f_1\wedge\ldots\wedge f_n\in A$,} 
      
\noindent by (b), and evidently $f(x)=h(x)$, while if $y\in K$ there is 
some $i\le n$ such that $y\in G_i$, so that 
      
\Centerline{$f(y)\le f_i(y)\le h(y)+\epsilon$. \Qed} 
      
      
\medskip 
      
{\bf (e)} If $h:K\to\Bbb R$ is any continuous function and $\epsilon>0$, 
there is an $f\in A$ such that $|f(y)-h(y)|\le\epsilon$ for every $y\in 
K$.   \Prf\ This time, let $\Cal G$ be the set of those open subsets 
$G$ of $X$ for which there is some $f\in A$ such that $f(y)\le 
h(y)+\epsilon$ for every $y\in K$ and $f(x)\ge h(x)-\epsilon$ for every 
$x\in G\cap K$.   Once again, $\Cal G$ is an open cover of $K$.   To 
see this, take any $x\in K$.   By (d), there is an $f\in A$ such that 
$f(x)=h(x)$ and $f(y)\le h(y)+\epsilon$ for every $y\in K$.   Now 
$h-f\restr K:K\to\Bbb R$ is a continuous function which is zero at $x$, 
so there is an open subset $G$ of $X$, containing $x$, such that 
$(h-f\restr K)(w)\le\epsilon$ for every $w\in G\cap K$, that is, 
$f(w)\ge 
h(w)-\epsilon$ for every $w\in G\cap K$.   Thus $G\in\Cal G$ and 
$x\in\bigcup\Cal G$, as required. 
      
Because $K$ is compact, $\Cal G$ has a finite subcover $G_0,\ldots,G_m$ 
say.   For each $j\le m$, take $f_j\in A$ such that $f_j(y)\le 
h(y)+\epsilon$ for every $y\in K$ and $f_j(w)\ge h(w)-\epsilon$ for 
every $w\in G_j\cap K$.   Then 
      
\Centerline{$f=f_0\vee f_1\vee\ldots\vee f_m\in A$,} 
      
\noindent by (b), and evidently $f(y)\le h(y)+\epsilon$ for every $y\in 
K$,  while if $x\in K$ there is some $j\le m$ such that $x\in G_j$, so 
that 
      
\Centerline{$f(x)\ge f_j(x)\ge h(x)-\epsilon$.} 
      
\noindent Thus $|f(x)-h(x)|\le\epsilon$ for every $x\in K$, as required. 
\Qed 
      
\medskip 
      
{\bf (f)} Thus we have an $f$ satisfying the first of the two 
requirements of the theorem.   But for the second, set 
$M_0=\inf_{x\in K}h(x)$ and $M_1=\sup_{x\in K}h(x)$, and 
      
\Centerline{$f_1=\med(M_0\chi X,f,M_1\chi X)
=(M_0\chi X)\vee(f\wedge M_1\chi X)$;} 
      
\noindent $f_1$ satisfies the second condition as well as the first. 
(I am tacitly assuming here what is 
in fact the case, that $M_0$ and $M_1$ are finite;  this is because $K$ 
is compact -- see 2A2G or 2A3N.) 
}%end of proof of 281A 
      
\leader{281B}{}\cmmnt{ We need some simple 
tools, belonging to the basic theory of normed spaces;  but I hope they 
will be accessible even if you have not encountered `normed spaces' 
before, if you keep a finger at the beginning of \S2A4 as you 
read the next lemma. 
      
\medskip 
      
\noindent}{\bf Lemma} Let $X$ be any set.   Write $\ell^{\infty}(X)$ 
for the set of bounded functions from $X$ to $\Bbb R$.   For 
$f\in\ell^{\infty}(X)$, set 
      
\Centerline{$\|f\|_{\infty}=\sup_{x\in X}|f(x)|$,} 
      
\noindent counting the supremum as $0$ if $X$ is empty.   Then 
      
(a) $\ell^{\infty}(X)$ is a normed space. 
      
(b) Let $A\subseteq\ell^{\infty}(X)$ be a subset and 
$\overline{A}$ its closure\cmmnt{ (2A3D)}. 
      
\quad (i) If $A$ is a linear subspace of $\ell^{\infty}(X)$, so is 
$\overline{A}$. 
      
\quad (ii) If $f\times g\in A$ whenever $f$, $g\in A$, then  
$f\times g\in\overline{A}$ whenever $f$, $g\in\overline{A}$. 
      
\quad (iii) If $|f|\in A$ whenever $f\in A$, then $|f|\in\overline{A}$ 
whenever $f\in\overline{A}$. 
      
\proof{{\bf (a)} This is a routine verification.   To confirm 
that $\ell^{\infty}(X)$ is a linear space over $\Bbb R$, we have to 
check that $f+g$, $cf$ belong to $\ell^{\infty}(X)$ whenever $f$, 
$g\in\ell^{\infty}(X)$ and $c\in\Bbb R$;   simultaneously we can 
confirm that $\|\,\|_{\infty}$ is a norm on $\ell^{\infty}(X)$ by 
observing that 
      
\Centerline{$|(f+g)(x)|\le|f(x)|+|g(x)| 
\le\|f\|_{\infty}+\|g\|_{\infty}$,} 
      
\Centerline{$|cf(x)|=|c||f(x)|\le|c|\|f\|_{\infty}$} 
      
\noindent whenever $f$, $g\in\ell^{\infty}(X)$ and $c\in\Bbb R$. 
It is worth noting at the same time that if $f$, $g\in\ell^{\infty}(X)$, 
then 
      
\Centerline{$|(f\times g)(x)| 
=|f(x)||g(x)|\le\|f\|_{\infty}\|g\|_{\infty}$} 
      
\noindent for every $x\in X$, so that 
$\|f\times g\|_{\infty}\le\|f\|_{\infty}\|g\|_{\infty}$. 
      
(Of course all these remarks are very elementary special cases of parts 
of \S243;  see 243Xl.) 
      
\medskip 
      
{\bf (b)} Recall that 
      
\Centerline{$\overline{A}=\{f:f\in\ell^{\infty}(X),\Forall 
\epsilon>0\Exists f_1\in A, 
\,\|f-f_1\|_{\infty}\le\epsilon\}$} 
      
\noindent (2A3Kb).   Take $f$, $g\in\overline{A}$ and $c\in\Bbb R$, 
and let $\epsilon>0$.   Set 
      
\Centerline{$\eta 
=\min(1,\Bover{\epsilon}{2+|c|+\|f\|_{\infty}+\|g\|_{\infty}})>0$.} 
      
\noindent Then there are $f_1$, $g_1\in A$ such that 
$\|f-f_1\|_{\infty}\le\eta$ and $\|g-g_1\|_{\infty}\le\eta$. 
      
Now 
      
$$\eqalignno{\|(f+g)-(f_1+g_1)\|_{\infty} 
&\le\|f-f_1\|_{\infty}+\|g-g_1\|_{\infty} 
\le 2\eta 
\le\epsilon,\cr 
\cr 
\|cf-cf_1\|_{\infty} 
&=|c|\|f-f_1\|_{\infty} 
\le|c|\eta 
\le\epsilon,\cr 
\cr 
\|(f\times g)-(f_1\times g_1)\|_{\infty} 
&=\|(f-f_1)\times g+f\times(g-g_1)-(f-f_1)\times(g-g_1)\|_{\infty}\cr 
&\le\|(f-f_1)\times g\|_{\infty}+\|f\times(g-g_1)\|_{\infty} 
  +\|(f-f_1)\times(g-g_1)\|_{\infty}\cr 
&\le\|f-f_1\|_{\infty}\|g\|_{\infty}+\|f\|_{\infty}\|g-g_1)\|_{\infty} 
  +\|f-f_1\|_{\infty}\|g-g_1\|_{\infty}\cr 
&\le\eta(\|g\|_{\infty}+\|f\|_{\infty}+\eta) 
\le\eta(\|g\|_{\infty}+\|f\|_{\infty}+1) 
\le\epsilon,\cr 
\cr 
\||f|-|f_1|\|_{\infty} 
&\le\|f-f_1\|_{\infty} 
\le\eta 
\le\epsilon.\cr}$$ 
      
\quad{\bf (i)} If $A$ is a linear subspace, then $f_1+g_1$ and $cf_1$ 
belong to $A$.   As $\epsilon$ is arbitrary, $f+g$ and $cf$ belong to 
$\overline{A}$.   As $f$, $g$ and $c$ are arbitrary, $\overline{A}$ is a 
linear subspace of $\ell^{\infty}(X)$. 
      
\quad{\bf (ii)} If $A$ is closed under multiplication, then $f_1\times 
g_1\in A$.   As $\epsilon$ is arbitrary, $f\times g\in \overline{A}$. 
      
\quad{\bf (iii)} If the absolute values of functions in $A$ belong to 
$A$, then $|f_1|\in A$.   As $\epsilon$ is arbitrary, 
$|f|\in\overline{A}$. 
}%end of proof of 281B 
      
\leader{281C}{Lemma} There is a sequence $\sequencen{p_n}$ of real 
polynomials such that $\lim_{n\to\infty}p_n(x) 
\ifdim\pagewidth=390pt\break\fi 
=|x|$ uniformly for 
$x\in[-1,1]$. 
      
\proof{{\bf (a)} By the Binomial Theorem we have 
      
\Centerline{$(1-x)^{1/2} 
=1-\Bover12x-\Bover1{4{\cdot}2!}x^2 
  -\Bover{1{\cdot}3}{2^3{\cdot}3!}x^3-\ldots 
=-\sum_{n=0}^{\infty}\Bover{(2n)!}{(2n-1)(2^nn!)^2}x^n$} 
      
\noindent whenever $|x|<1$, with the convergence being uniform on any 
interval $[-a,a]$ with $0\le a<1$.   (For a proof of this, see almost 
any book on real or complex analysis.   If you have no favourite text to 
hand, you can try to construct a proof from the following facts:  (i) 
the radius of convergence of the series is $1$, so on any interval 
$[-a,a]$, with $0\le a<1$, it is uniformly absolutely summable (ii) 
writing $f(x)$ for the sum of the series for $|x|<1$, use Lebesgue's 
Dominated Convergence Theorem to find expressions for the indefinite 
integrals $\int_0^xf$, $-\int_{-x}^0f$ and show that these are 
$\bover23(1-(1-x)f(x))$, $\bover23(1-(1+x)f(-x))$ for $0\le x<1$ (iii) 
use the Fundamental Theorem of Calculus to show that 
$f(x)+2(1-x)f'(x)=0$ 
(iv) show that $\bover{d}{dx}\bigl(\bover{f(x)^2}{1-x}\bigr)=0$ and 
hence (v) that $f(x)^2=1-x$ whenever $|x|<1$.   Finally, show that 
because $f$ is continuous and non-zero in $\ooint{-1,1}$, $f(x)$ must be 
the positive square root of $1-x$ throughout.) 
      
We have a further fragment of information.   If we set 
      
\Centerline{$q_0(x)=1$, \quad$q_1(x)=1-\Bover12x$, 
\quad$q_n(x)=-\sum_{k=0}^n\Bover{(2k)!}{(2k-1)(2^kk!)^2}x^k$} 
      
\noindent for $n\ge 2$ and $x\in[0,1]$, so that $q_n$ is the $n$th partial sum of the binomial series for $(1-x)^{1/2}$, then we have 
$\lim_{n\to\infty}q_n(x)=(1-x)^{1/2}$ for every $x\in\coint{0,1}$.   But 
also every $q_n$ is non-increasing on $[0,1]$, and 
$\sequencen{q_n(x)}$ is a non-increasing sequence for 
each $x\in[0,1]$.   So we must have 
      
\Centerline{$\sqrt{1-x}\le q_n(x) 
\Forall n\in\Bbb N,\,x\in\coint{0,1}$,} 
      
\noindent and therefore, because all the $q_n$ are continuous, 
      
\Centerline{$\sqrt{1-x}\le q_n(x) 
\Forall n\in\Bbb N,\,x\in[0,1]$.} 
      
\noindent Moreover, given $\epsilon>0$, set $a=1-{1\over 4}\epsilon^2$, 
so that $\sqrt{1-a}=\bover{\epsilon}2$.   Then there is an 
$n_0\in\Bbb N$ such that $q_n(x)-\sqrt{1-x}\le\bover{\epsilon}2$ for 
every $x\in[0,a]$ and $n\ge n_0$.   In particular, $q_n(a)\le\epsilon$, 
so $q_n(x)\le\epsilon$ and $q_n(x)-\sqrt{1-x}\le\epsilon$ whenever
$x\in[a,1]$ and $n\ge n_0$.   This means that 
      
\Centerline{$0\le q_n(x)-\sqrt{1-x}\le\epsilon 
\Forall n\ge n_0,\,x\in[0,1]$;} 
      
\noindent as $\epsilon$ is arbitrary, 
$\sequencen{q_n(x)}\to\sqrt{1-x}$ uniformly on $[0,1]$. 
      
\medskip 
      
{\bf (b)} Now set $p_n(x)=q_n(1-x^2)$ for $x\in\Bbb R$.   Because each 
$q_n$ is a real polynomial of degree $n$, each $p_n$ is a real 
polynomial of degree $2n$.   Next, 
      
$$\eqalign{\sup_{|x|\le 1}|p_n(x)-|x|| 
&=\sup_{|x|\le 1}|q_n(1-x^2)-\sqrt{1-(1-x^2)}|\cr 
&=\sup_{y\in[0,1]}|q_n(y)-\sqrt{1-y}| 
\to 0\cr}$$ 
      
\noindent as $n\to\infty$, so $\lim_{n\to\infty}p_n(x)=|x|$ uniformly 
for $|x|\le 1$, as required. 
}%end of proof of 281C 
      
\leader{281D}{Corollary} Let $X$ be a set, and $A$ a norm-closed linear 
subspace of $\ell^{\infty}(X)$ containing $\chi X$ and such that 
$f\times g\in A$ whenever $f$, $g\in A$.   Then $|f|\in A$ for every 
$f\in A$. 
      
\proof{ Set 
      
\Centerline{$f_1=\Bover1{1+\|f\|_{\infty}}f$,} 
      
\noindent so that $f_1\in A$ and $\|f_1\|_{\infty}\le 1$.   Because $A$ 
contains $\chi X$ and is closed under multiplication, 
$p\frsmallcirc f_1\in A$ 
for every polynomial $p$ with real coefficients.   In particular, 
$g_n=p_n\frsmallcirc f_1\in A$ for every $n$, where $\sequencen{p_n}$ is 
the 
sequence of 281C.   Now, because $|f_1(x)|\le 1$ for every $x\in X$, 
      
\Centerline{$\|g_n-|f_1|\|_{\infty} 
=\sup_{x\in X}|p_n(f_1(x))-|f_1(x)|| 
\le\sup_{|y|\le 1}|p_n(y)-|y|| 
\to 0$} 
      
\noindent as $n\to\infty$.   Because $A$ is $\|\,\|_{\infty}$-closed, 
$|f_1|\in A$;  consequently $|f|\in A$, as claimed. 
}%end of proof of 281D 
      
\leader{281E}{Stone-Weierstrass theorem:  second form} Let $X$ be a 
topological space and $K$ a compact subset of $X$.   Write $C_b(X)$ for 
the space of all bounded continuous real-valued functions on $X$.   Let 
$A\subseteq C_b(X)$ be such that 
      
\inset{$A$ is a linear subspace of $C_b(X)$;} 
      
\inset{$f\times g\in A$ for every $f$, $g\in A$;} 
      
\inset{$\chi X\in A$;} 
      
\inset{whenever $x$, $y$ are distinct points of $K$ there is an $f\in A$ 
such that $f(x)\ne f(y)$.} 
      
\noindent Then for every continuous $h:K\to\Bbb R$ and $\epsilon>0$ 
there is an $f\in A$ such that 
      
\inset{$|f(x)-h(x)|\le\epsilon$ for every $x\in K$,} 
      
\inset{if $K\ne\emptyset$, $\inf_{x\in X}f(x)\ge\inf_{x\in K}h(x)$ and 
$\sup_{x\in X}f(x)\le\sup_{x\in K}h(x)$.} 
      
\proof{ Let $\overline{A}$ be the 
$\|\,\|_{\infty}$-closure of $A$ in $\ell^{\infty}(X)$.   It is helpful 
to know that $\overline{A}\subseteq C_b(X)$;  this is because the 
uniform limit of continuous functions is continuous.   (But if this is 
new to you, or your memory has faded, don't take time to look it up 
now;  just read `$\overline{A}\cap C_b(X)$' in place of 
`$\overline{A}$' in the rest of this argument.)   By 281B-281D, 
$\overline{A}$ is a linear subspace of $C_b(X)$ and $|f|\in\overline{A}$ 
for every $f\in\overline{A}$, so the conditions of 281A apply to 
$\overline{A}$. 
      
Take a continuous $h:K\to\Bbb R$ and an $\epsilon>0$.   The cases in 
which $K=\emptyset$ or $h$ is constant are trivial, because all constant 
functions belong to $A$;  so I suppose that $M_0=\inf_{x\in K}h(x)$ and 
$M_1=\sup_{x\in K}h(x)$ are defined and distinct.   As observed at the 
end of the proof of 281A, $M_0$ and $M_1$ are finite.   Set 
      
\Centerline{$\eta=\min(\bover13\epsilon,\bover12(M_1-M_2))>0$,\quad 
$\tilde h(x)=\med(M_0+\eta,h(x),M_1-\eta)$ for $x\in K$} 
      
\noindent (definition:  2A1Ac), 
so that $\tilde h:K\to\Bbb R$ is continuous and  
$M_0+\eta\le\tilde h(x)\le M_1-\eta$ for every $x\in K$.      By 281A,  there is an 
$f_0\in\overline{A}$ such that 
$|f_0(x)-\tilde h(x)|\le\eta$ for every $x\in K$ and $M_0+\eta\le 
f_0(x)\le M_1-\eta$ for every $x\in X$.   Now there is an $f\in A$ such 
that 
$\|f-f_0\|_{\infty}\le \eta$, so that 
      
\Centerline{$|f(x)-h(x)|\le|f(x)-f_0(x)|+|f_0(x)-\tilde h(x)| 
+|\tilde h(x)-h(x)|\le 3\eta\le\epsilon$} 
      
\noindent for every $x\in K$, while 
      
\Centerline{$M_0\le f_0(x)-\eta\le f(x)\le f_0(x)+\eta\le M_1$} 
      
\noindent for every $x\in X$. 
}%end of proof of 281E 
      
\leader{281F}{Corollary:  Weierstrass' theorem} Let $K$ be any closed 
bounded subset of 
$\Bbb R$.   Then every continuous $h:K\to\Bbb R$ can be uniformly 
approximated on $K$ by polynomials. 
      
\proof{ Apply 281E with $X=K$ (noting that $K$, being closed and 
bounded, is compact), and $A$ the set of polynomials with real 
coefficients, regarded as functions from $K$ to $\Bbb R$. 
}%end of proof of 281F 
      
\leader{281G}{Stone-Weierstrass theorem:  third form} Let $X$ be a 
topological space and $K$ a compact subset of $X$.   Write 
$C_b(X;\Bbb C)$ 
for the space of all bounded continuous complex-valued functions on 
$X$, so that $C_b(X;\Bbb C)$ is a linear space over $\Bbb C$.   Let 
$A\subseteq C_b(X;\Bbb C)$ be such that 
      
\inset{$A$ is a linear subspace of $C_b(X;\Bbb C)$;} 
      
\inset{$f\times g\in A$ for every $f$, $g\in A$;} 
      
\inset{$\chi X\in A$;} 
      
\inset{the complex conjugate $\bar f$ of $f$ belongs to $A$ for every 
$f\in A$;} 
      
\inset{whenever $x$, $y$ are distinct points of $K$ there is an $f\in A$ 
such that $f(x)\ne f(y)$.} 
      
\noindent Then for every continuous $h:K\to\Bbb C$ and $\epsilon>0$ 
there is an $f\in A$ such that 
      
\inset{$|f(x)-h(x)|\le\epsilon$ for every $x\in K$,} 
      
\inset{if $K\ne\emptyset$, $\sup_{x\in X}|f(x)|\le\sup_{x\in K}|h(x)|$.} 
      
\proof{ If $K=\emptyset$, or $h$ is identically zero, we 
can take $f=\tbf{0}$.   So let us suppose that 
$M=\sup_{x\in K}|h(x)|>0$. 
      
\medskip 
      
{\bf (a)} Set 
      
\Centerline{$A_{\Bbb R}=\{f:f\in A,\,f(x)$ is real for every $x\in 
X\}$.} 
      
\noindent Then $A_{\Bbb R}$ satisfies the conditions of 281E.   \Prf\ 
(i) Evidently $A_{\Bbb R}$ is a subset of $C_b(X)=C_b(X;\Bbb R)$, is 
closed under addition, multiplication by real scalars and pointwise 
multiplication of functions, and contains $\chi X$.  If $x$, $y$ are 
distinct points of $K$, there is an $f\in A$ such that $f(x)\ne f(y)$. 
Now 
      
\Centerline{$\Real f=\Bover12(f+\bar f)$, 
\quad$\Imag f=\Bover1{2i}(f-\bar 
f)$} 
      
\noindent both belong to $A$ and are real-valued, so belong to $A_{\Bbb 
R}$, and at least one of them takes different values at $x$ and $y$. 
\Qed 
      
\medskip 
      
{\bf (b)} Consequently, given a continuous function $h:K\to\Bbb C$ and 
$\epsilon>0$, we may apply 281E twice to find $f_1$, $f_2\in A_{\Bbb R}$ 
such that 
      
\Centerline{$|f_1(x)-\Real(h(x))|\le\eta$,\quad 
$|f_2(x)-\Imag(h(x))|\le\eta$} 
      
\noindent for every $x\in K$, where 
$\eta=\min(\bover12,M,\bover16\epsilon)>0$.   
Setting $g=f_1+if_2$, we have $g\in A$ 
and 
$|g(x)-h(x)|\le2\eta$ for every $x\in K$. 
      
\medskip 
      
{\bf (c)} Set $M_1=\|g\|_{\infty}$.   If $M_1\le M$ we can take $f=g$ and 
stop.   Otherwise, consider the function 
      
\Centerline{$\phi(t)=\Bover{M-\eta}{\max(M,\sqrt{t})}$} 
%\eta\le M so \phi\ge 0
      
\noindent for $t\in [0,M_1^2]$.   By Weierstrass' theorem 
(281F), there is a real polynomial $p$ such that 
$|\phi(t)-p(t)|\le\Bover{\eta}{M_1}$ whenever $0\le t\le M_1^2$.   Note 
that $|g|^2=g\times\bar g\in A$, so that 
      
\Centerline{$f=g\times p(|g|^2)\in A$.} 
      
\noindent Now 
      
\Centerline{$|p(t)|\le\phi(t)+\Bover{\eta}{M_1} 
\le\phi(t)+\Bover{\eta}{\max(M,\sqrt{t})} 
=\Bover{M}{\max(M,\sqrt{t})}$} 
      
\noindent whenever $0\le t\le M_1^2$, so 
      
      
\Centerline{$|f(x)|\le|g(x)|\Bover{M}{\max(M,|g(x)|)}\le M$} 
      
\noindent for every $x\in X$.    Next, if $0\le t\le\min(M_1,M+2\eta)^2$, 
      
\Centerline{$|1-p(t)|\le\Bover{\eta}{M_1}+1-\phi(t) 
\le\Bover{\eta}M+1-\Bover{M-\eta}{M+2\eta} 
\le\Bover{4\eta}M$.} 
      
\noindent    Consequently, if $x\in K$, so that 
      
\Centerline{$|g(x)|\le\min(M_1,|h(x)|+2\eta)\le 
\min(M_1,M+2\eta)$,} 
      
\noindent we shall have 
      
\Centerline{$|1-p(|g(x)|^2)|\le\Bover{4\eta}M$,} 
      
\noindent and 
      
$$\eqalign{|f(x)-h(x)| 
&\le|g(x)-h(x)|+|g(x)||1-p(|g(x)|^2)|\cr 
&\le 2\eta+\Bover{4\eta}M(M+2\eta) 
\le 2\eta+\Bover{4\eta}M(M+1) 
%\eta\le\bover12
\le\epsilon,\cr}$$ 
%\eta\le\bover16\epsilon
      
\noindent as required. 
}% end of proof of 281G 
      
\cmmnt{ 
\medskip 
      
\noindent{\bf Remark} Of course we could have saved ourselves effort by 
settling for 
      
\Centerline{$\sup_{x\in X}|f(x)|\le 2\sup_{x\in K}|h(x)|$,} 
      
\noindent which would be quite good enough for the applications below. 
} 
      
\leader{281H}{Corollary} Let $[a,b]\subseteq\Bbb R$ be a 
non-empty bounded closed interval and $h:[a,b]\to\Bbb C$ a continuous 
function.   Then for any $\epsilon>0$ there are $y_0,\ldots,y_n\in\Bbb 
R$ and $c_0,\ldots,c_n\in\Bbb C$ such that 
      
\Centerline{$|h(x)-\sum_{k=0}^nc_ke^{iy_kx}|\le\epsilon$ for every 
$x\in[a,b]$,} 
      
\Centerline{$\sup_{x\in\Bbb 
R}|\sum_{k=0}^nc_ke^{iy_kx}|\le\sup_{x\in[a,b]}|h(x)|$.} 
      
\proof{ Apply 281G with $X=\Bbb R$, $K=[a,b]$ and $A$ the 
linear span of the functions $x\mapsto e^{iyx}$ as $y$ runs over $\Bbb 
R$.} 
      
\leader{281I}{Corollary} Let $S^1$ be the unit circle 
$\{z:|z|=1\}\subseteq\Bbb C$.   Then for any continuous function 
$h:S^1\to\Bbb C$ and $\epsilon>0$, there are $n\in\Bbb N$ and 
$c_{-n},c_{-n+1},\ldots,c_0,\ldots,c_n\in\Bbb C$ such that 
$|h(z)-\sum_{k=-n}^nc_kz^k|\le\epsilon$ for every $z\in S^1$. 
      
\proof{ Apply 281G with $X=K=S^1$ and $A$ the linear span 
of the functions $z\mapsto z^k$ for $k\in\Bbb Z$. 
} 
      
\leader{281J}{Corollary} Let $h:[-\pi,\pi]\to\Bbb C$ be a continuous 
function such that $h(\pi)=h(-\pi)$.   Then for any $\epsilon>0$ there 
are $n\in\Bbb N$, $c_{-n},\ldots,c_n\in\Bbb C$ such that 
$|h(x)-\sum_{k=-n}^nc_ke^{ikx}|\le\epsilon$ for every $x\in[-\pi,\pi]$. 
      
\proof{ The point is that $\tilde h:S^1\to\Bbb C$ is 
continuous on $S^1$, where $\tilde h(z)=h(\arg z)$;  this is because 
$\arg$ is 
continuous everywhere except at $-1$, and 
      
\Centerline{$\lim_{x\downarrow-\pi}h(x) 
=h(-\pi)=h(\pi)=\lim_{x\uparrow\pi}h(x)$,} 
      
\noindent so 
      
\Centerline{$\lim_{z\in S^1,z\to-1}\tilde h(z)=h(\pi)=\tilde h(-1)$.} 
      
\noindent Now by 281I there are $c_{-n},\ldots,c_n\in\Bbb C$ such that 
$|\tilde h(z)-\sum_{k=-n}^nc_kz^k|\le\epsilon$ for every $z\in S^1$, and 
these coefficients 
serve equally for $h$. 
} 
      
\vleader{60pt}{281K}{Corollary} Suppose that $r\ge 1$ and that 
$K\subseteq\BbbR^r$ is a non-empty closed bounded set.   Let 
$h:K\to\Bbb C$ be a continuous 
function, and $\epsilon>0$.   Then there are $y_0,\ldots,y_n\in\Bbb Q^r$ 
and $c_0,\ldots,c_n\in\Bbb C$ such that 
      
\Centerline{$|h(x)-\sum_{k=0}^nc_ke^{iy_k\dotproduct x}|\le\epsilon$ for 
every $x\in K$,} 
      
\Centerline{$\sup_{x\in\Bbb R^r}|\sum_{k=0}^nc_ke^{iy_k\dotproduct x}| 
\le\sup_{x\in K}|h(x)|$,} 
      
\noindent writing $y\dotproduct x=\sum_{j=1}^r\eta_j\xi_j$ when 
$y=(\eta_1,\ldots,\eta_r)$ and $x=(\xi_1,\ldots,\xi_r)$ belong to 
$\BbbR^r$. 
      
      
\proof{ Apply 281G with $X=\BbbR^r$ and $A$ the linear 
span of the functions $x\mapsto e^{iy\dotproduct x}$ as $y$ runs over 
$\BbbQ^r$.} 
      
\leader{281L}{Corollary}  Suppose that $r\ge 1$ and that 
$K\subseteq\BbbR^r$ is a non-empty closed bounded set.   Let 
$h:K\to\Bbb R$ be a continuous 
function, and $\epsilon>0$.   Then there are $y_0,\ldots,y_n\in\BbbR^r$ 
and $c_0,\ldots,c_n\in\Bbb C$ such that, writing 
$g(x)=\sum_{k=0}^nc_ke^{iy_k\dotproduct x}$, $g$ is real-valued and 
      
\Centerline{$|h(x)-g(x)|\le\epsilon$ for every $x\in K$,} 
      
\Centerline{$\inf_{y\in K}h(y)\le g(x)\le\sup_{y\in K}h(y)$ for every 
$x\in\BbbR^r$.} 
      
\proof{ Apply 281E with $X=\BbbR^r$ and $A$ the set of 
{\it real-}valued functions on $\BbbR^r$ which are {\it complex} linear 
combinations of the functions $x\mapsto e^{iy\dotproduct x}$;  as 
remarked 
in 
part (a) of the proof of 281G, $A$ satisfies the conditions of 281E. 
} 
      
\leader{281M}{Weyl's Equidistribution Theorem}\cmmnt{ We are now ready 
for one of the basic results of number theory.   I shall actually apply 
it to provide an example in \S285 below, but (at least in the 
one-variable case) it is 
surely on the (rather long) list of things which every pure 
mathematician should know.    For the sake of the application I have in 
mind, I give the full $r$-dimensional version, but you may wish to take 
it in the first place with $r=1$. 
      
It will be helpful to have a notation for `fractional part'.}    For 
any real number $x$, write $\fraction{x}$ for that number in 
$\coint{0,1}$ such that $x-\fraction{x}$ is an integer.   \cmmnt{Now 
for the 
theorem.} 
      
\leader{281N}{Theorem} Let $\eta_1,\ldots,\eta_r$ be real numbers such 
that $1,\eta_1,\ldots,\eta_r$ are 
linearly independent over $\Bbb Q$.   Then whenever 
$0\le \alpha_j\le\beta_j\le 1$ for each $j\le r$, 
      
\Centerline{$\lim_{n\to\infty}\Bover1{n+1}\#(\{m:m\le 
n,\,\fraction{m\eta_j}\in[\alpha_j,\beta_j]$ for every $j\le r\}) 
=\prod_{j=1}^r(\beta_j-\alpha_j)$.} 
      
\cmmnt{\medskip 
      
\noindent{\bf Remark} Thus the theorem says that the long-term 
proportion of the $r$-tuples $(\fraction{m\eta_1}, 
\ifdim\pagewidth=390pt\break\fi 
\ldots,\fraction{m\eta_r})$ 
which belong to the interval $[a,b]\subseteq[\tbf{0},\tbf{1}]$ is just 
the Lebesgue measure $\mu[a,b]$ of the interval.   Of course the 
condition 
`$1,\eta_1,\ldots,\eta_r$ are linearly independent over $\Bbb Q$' is 
necessary as well as sufficient (281Xg).} 
      
\proof{{\bf (a)} Write $y=(\eta_1,\ldots,\eta_r)\in\BbbR^r$, 
      
\Centerline{$\fraction{my} 
=(\fraction{m\eta_1},\ldots,\fraction{m\eta_r}) 
\in\coint{\tbf{0},\tbf{1}}=\coint{0,1}^r$} 
      
\noindent for each $m\in\Bbb N$.   Set 
$I=[\tbf{0},\tbf{1}]=[0,1]^r$, and for any function $f:I\to\Bbb R$ write 
      
\Centerline{$\overline{L}(f) 
=\limsup_{n\to\infty}\Bover1{n+1}\sum_{m=0}^{n}f(\fraction{my})$,} 
      
\Centerline{$\underline{L}(f) 
=\liminf_{n\to\infty}\Bover1{n+1}\sum_{m=0}^{n}f(\fraction{my})$;} 
      
\noindent and for $f:I\to\Bbb C$ write 
      
\Centerline{$L(f) 
=\lim_{n\to\infty}\Bover1{n+1}\sum_{m=0}^nf(\fraction{my})$} 
      
\noindent if the limit exists.    It will be worth noting that for 
non-negative functions $f$, $g$, $h:I\to\Bbb R$ such that $h\le f+g$, 
      
\Centerline{$\overline{L}(h)\le\overline{L}(f)+\overline{L}(g)$,} 
      
\noindent and that $L(cf+g)=cL(f)+L(g)$ for any two functions $f$, 
$g:I\to\Bbb C$ such that $L(f)$ and $L(g)$ exist, and any $c\in\Bbb C$. 
      
\medskip 
      
{\bf (b)} I mean to show that $L(f)$ exists and is equal to $\int_{I}f$ 
for (many) continuous functions $f$.   The key step is to consider 
functions of the form 
      
\Centerline{$f(x)=e^{2\pi ik\dotproduct x}$,} 
      
\noindent where $k=(\kappa_1,\ldots,\kappa_r)\in\Bbb Z^r$.   In this 
case, if $k\ne\tbf{0}$, 
      
\Centerline{$k\dotproduct y=\sum_{j=1}^r\kappa_j\eta_j\notin\Bbb Z$} 
      
\noindent because $1,\eta_1,\ldots,\eta_r$ are linearly independent over 
$\Bbb Q$.   So 
      
$$\eqalignno{L(f) 
&=\lim_{n\to\infty}\Bover1{n+1} 
  \sum_{m=0}^ne^{2\pi ik\dotproduct \fraction{my}} 
=\lim_{n\to\infty}\Bover1{n+1}\sum_{m=0}^ne^{2\pi imk\dotproduct y}\cr 
\noalign{\noindent (because 
$mk\dotproduct y-k\dotproduct \fraction{my} 
=\sum_{j=1}^r\kappa_j(m\eta_j-\fraction{m\eta_j})$ is an integer)} 
&=\lim_{n\to\infty} 
\bover{1-e^{2\pi i(n+1)k\dotproduct y}}{(n+1)(1-e^{2\pi ik\dotproduct 
y})}\cr 
\noalign{\noindent (because $e^{2\pi ik\dotproduct y}\ne 1$)} 
&=0,\cr}$$ 
      
\noindent because $|1-e^{2\pi i(n+1)k\dotproduct y}|\le 2$ for every 
$n$.   Of course we can also calculate the integral of $f$ over $I$, 
which is 
      
$$\eqalignno{\int_If(x)dx 
&=\int_Ie^{2\pi ik\dotproduct x}dx 
=\int_I\prod_{j=1}^re^{2\pi i\kappa_j\xi_j}dx\cr 
\noalign{\noindent (writing $x=(\xi_1,\ldots,\xi_r)$)} 
&=\int_{0}^1\ldots\int_0^1\prod_{j=1}^r 
  e^{2\pi i\kappa_j\xi_j}d\xi_r\ldots d\xi_1\cr 
&=\int_{0}^1e^{2\pi i\kappa_r\xi_r}d\xi_r 
  \ldots\int_0^1e^{2\pi i\kappa_1\xi_1}d\xi_1 
=0\cr}$$ 
      
\noindent because at least one $\kappa_j$ is non-zero, and for this $j$ 
we must have 
      
\Centerline{$\int_0^1e^{2\pi i\kappa_j\xi_j}d\xi_j 
=\Bover{1}{2\pi i\kappa_j}(e^{2\pi i\kappa_j}-1)=0$.} 
      
\noindent So we have $L(f)=\int_If=0$ when $k\ne\tbf{0}$.   On the other 
hand, if $k=\tbf{0}$, then $f$ is constant with value $1$, so 
      
\Centerline{$L(f) 
=\lim_{n\to\infty}\Bover1{n+1}\sum_{m=0}^nf(\fraction{my}) 
=\lim_{n\to\infty}1=1=\int_If(x)dx$.} 
      
\medskip 
      
{\bf (c)} Now write 
$\partial I=[\tbf{0},\tbf{1}]\setminus\ooint{\tbf{0},\tbf{1}}$, the 
boundary of 
$I$.   If $f:I\to\Bbb C$ is continuous and $f(x)=0$ for 
$x\in\partial I$, then $L(f)=\int_If$.   \Prf\ As in 281I, let $S^1$ be 
the unit circle $\{z:z\in\Bbb C,\,|z|=1\}$, and set 
$K=(S^1)^r\subseteq\Bbb C^r$. 
If we think of $K$ as a subset of $\BbbR^{2r}$, it is closed and 
bounded.   Let $\phi:K\to I$ be given by 
      
\Centerline{$\phi(\zeta_1,\ldots,\zeta_r) 
=(\Bover12+\Bover{\arg\zeta_1}{2\pi},\ldots, 
\Bover12+\Bover{\arg\zeta_r}{2\pi})$} 
      
\noindent for $\zeta_1,\ldots,\zeta_r\in S^1$.   Then $h=f\phi:K\to\Bbb 
C$ is continuous, because $\phi$ is continuous on 
$(S^1\setminus\{-1\})^r$ and 
      
\Centerline{$\lim_{w\to z}f\phi(w)=f\phi(z)=0$} 
      
\noindent for any $z\in K\setminus(S^1\setminus\{-1\})^r$.   (Compare 
281J.)   Now apply 281G with $X=K$ and $A$ the set of polynomials in 
$\zeta_1,\ldots,\zeta_r,\zeta_1^{-1},\ldots,\zeta_r^{-1}$ to see that, 
given $\epsilon>0$, there is a function of the form 
      
\Centerline{$g(z)=\sum_{k\in 
J}c_k\zeta_1^{\kappa_1}\ldots\zeta_r^{\kappa_r}$,} 
      
\noindent for some finite set $J\subseteq\Bbb Z^r$ and constants 
$c_k\in\Bbb C$ for $k\in J$, such that 
      
\Centerline{$|g(z)-h(z)|\le\epsilon$ for every $z\in K$.} 
      
\noindent Set 
      
\Centerline{$\tilde g(x) 
=g(e^{\pi i(2\xi_1-1)},\ldots,e^{\pi i(2\xi_r-1)}) 
=\sum_{k\in J}c_ke^{\pi ik\dotproduct (2x-\tbf{1})} 
=\sum_{k\in J}(-1)^{k\tbf{.1}}c_ke^{2\pi ik\dotproduct x}$,} 
      
\noindent so that $\tilde g\phi = g$, and see that 
      
\Centerline{$\sup_{x\in I}|\tilde g(x)-f(x)| 
=\sup_{z\in K}|g(z)-h(z)|\le\epsilon$.} 
      
Now $\tilde g$ is of the form dealt with in (a), so we must have 
$L(\tilde g)=\int_I\tilde g$.    Let $n_0$ be such that 
      
\Centerline{$\bigl|\int_I\tilde g 
  -\Bover1{n+1}\sum_{m=0}^n\tilde g(\fraction{my})\bigr| 
\le\epsilon$} 
      
\noindent for every $n\ge n_0$.   Then 
      
\Centerline{$|\int_If-\int_I\tilde g|\le\int_I|f-\tilde g|\le\epsilon$} 
      
\noindent and 
      
$$\eqalign{|\Bover1{n+1}\sum_{m=0}^{n}\tilde g(\fraction{my}) 
-\Bover1{n+1}\sum_{m=0}^{n}f(\fraction{my})| 
&\le\Bover1{n+1}\sum_{m=0}^{n} 
 |\tilde g(\fraction{my})-f(\fraction{my})|\cr 
&\le\Bover1{n+1}(n+1)\epsilon 
=\epsilon\cr}$$ 
      
\noindent for every $n\in\Bbb N$.   So for $n\ge n_0$ we must have 
      
\Centerline{$|\Bover1{n+1}\sum_{m=0}^{n}f(\fraction{my}) 
-\int_If|\le3\epsilon$.} 
      
\noindent As $\epsilon$ is arbitrary, $L(f)=\int_If$, as required. 
\Qed 
      
\medskip 
      
{\bf (d)} Observe next that if $a$, 
$b\in\ooint{\tbf{0},\tbf{1}}=\ooint{0,1}^r$, and $\epsilon>0$, there are 
continuous functions $f_1$, $f_2$ such that 
      
\Centerline{$f_1\le\chi[a,b]\le f_2\le\chi\ooint{\tbf{0},\tbf{1}}$, 
\quad $\int_If_2-\int_If_1\le\epsilon$.} 
      
\noindent\Prf\ This is elementary.   For $n\in\Bbb N$, define 
$h_n:\Bbb R\to[0,1]$ by setting $h_n(\xi)=0$ if $\xi\le 0$, $2^n\xi$ if 
$0\le\xi\le 2^{-n}$ and $1$ if $\xi\ge 2^{-n}$.   Set 
      
\Centerline{$f_{1n}(x) 
=\prod_{j=1}^rh_n(\xi_j-\alpha_j)h_n(\beta_j-\xi_j)$,} 
      
\Centerline{$f_{2n}(x) 
=\prod_{j=1}^r(1-h_n(\alpha_j-\xi_j))(1-h_n(\xi_j-\beta_j))$} 
      
\noindent for $x=(\xi_1,\ldots,\xi_r)\in\BbbR^r$.   (Compare the proof 
of 242Oa.)   Then 
$f_{1n}\le\chi[a,b]\le f_{2n}$ for each $n$, 
$f_{2n}\le\chi\ooint{\tbf{0},\tbf{1}}$ for all $n$ so large that 
      
\Centerline{$2^{-n} 
\le\min(\min_{j\le r}\alpha_j,\min_{j\le r}(1-\beta_j))$,} 
      
\noindent and $\lim_{n\to\infty}f_{2n}(x)-f_{1n}(x)=0$ for every $x$, so 
      
\Centerline{$\lim_{n\to\infty}\int_If_{2n}-\int_If_{1n}=0$.} 
      
\noindent Thus we can take $f_1=f_{1n}$, $f_2=f_{2n}$ for any $n$ large 
enough.   \Qed 
      
\medskip 
      
{\bf (e)} It follows that if $a$, $b\in\ooint{\tbf{0},\tbf{1}}$ and 
$a\le b$, $L(\chi[a,b])=\mu[a,b]$.   \Prf\ Let $\epsilon>0$.   Take 
$f_1$, $f_2$ as in (d).   Then, using (c), 
      
\Centerline{$\overline{L}(\chi[a,b])\le\overline{L}(f_2) 
=L(f_2)=\int_If_2\le\int_If_1+\epsilon\le\mu[a,b]+\epsilon$,} 
      
\Centerline{$\underline{L}(\chi[a,b])\ge\underline{L}(f_1) 
=L(f_1)=\int_If_1\ge\int_If_2-\epsilon\ge\mu[a,b]-\epsilon$,} 
      
\noindent so 
      
\Centerline{$\mu[a,b]-\epsilon\le\underline{L}(\chi[a,b]) 
\le\overline{L}(\chi[a,b])\le\mu[a,b]+\epsilon$.} 
      
\noindent   As $\epsilon$ is arbitrary, 
      
\Centerline{$\mu[a,b]=\overline{L}(\chi[a,b])=\underline{L}(\chi[a,b]) 
=L(\chi[a,b])$,} 
      
\noindent as required.\ \Qed 
      
\medskip 
      
{\bf (f)} To complete the proof, take any $a$, $b\in I$ with $a\le b$. 
For $0<\epsilon\le\bover12$, set 
$I_{\epsilon}=[\epsilon\tbf{1},(1-\epsilon)\tbf{1}]$, so that 
$I_{\epsilon}$ is a closed interval included in 
$\ooint{\tbf{0},\tbf{1}}$ and $\mu I_{\epsilon}=(1-2\epsilon)^r$.   Of 
course $L(\chi I)=\mu I=1$, so 
      
\Centerline{$L(\chi(I\setminus I_{\epsilon})) 
=L(\chi I)-L(\chi I_{\epsilon}) 
=1-\mu I_{\epsilon}$,} 
      
\noindent and 
      
$$\eqalign{\mu[a,b]-1+\mu I_{\epsilon} 
&\le\mu[a,b]+\mu I_{\epsilon}-\mu([a,b]\cup I_{\epsilon}) 
=\mu([a,b]\cap I_{\epsilon})\cr 
&=L(\chi([a,b]\cap I_{\epsilon})) 
\le\underline{L}(\chi([a,b]))\cr 
&\le\overline{L}(\chi([a,b])) 
\le\overline{L}(\chi([a,b]\cap I_{\epsilon})) 
  +\overline{L}(\chi(I\setminus I_{\epsilon}))\cr 
&=L(\chi([a,b]\cap I_{\epsilon}))+1-\mu I_{\epsilon}\cr 
&=\mu([a,b]\cap I_{\epsilon})+1-\mu I_{\epsilon} 
\le\mu[a,b]+1-\mu I_{\epsilon}.\cr}$$ 
      
\noindent As $\epsilon$ is arbitrary, 
      
\Centerline{$\mu[a,b]=\overline{L}(\chi[a,b])=\underline{L}(\chi[a,b]) 
=L(\chi[a,b])$,} 
      
\noindent as stated. 
}%end of proof of 281N 
      
\exercises{ 
\leader{281X}{Basic exercises (a)} 
%\spheader 281Xa 
Let $A$ be the set of those bounded continuous 
functions  $f:\BbbR^r\times\BbbR^r\to\Bbb R$ which are expressible in 
the form $f(x,y)=\sum_{k=0}^ng_k(x)g'_k(y)$, where all the $g_k$, $g'_k$ 
are continuous functions from $\BbbR^r$ to $\Bbb R$.   Show that for 
any 
bounded continuous function $h:\BbbR^r\times\BbbR^r\to\Bbb R$ and any 
bounded set $K\subseteq\BbbR^r\times\BbbR^r$ and any $\epsilon>0$, 
there is an $f\in A$ such that 
$|f(x,y)-h(x,y)|\le\epsilon$ for every $(x,y)\in K$ and 
$\sup_{x,y\in\BbbR^r}|f(x,y)|\le\sup_{x,y\in \BbbR^r}|h(x,y)|$. 
%281E 
      
\spheader 281Xb Let $K$ be a closed bounded set in $\BbbR^r$, 
where $r\ge 1$, and $h:K\to\Bbb R$ a continuous function.   Show that 
for any $\epsilon>0$ there is a polynomial $p$ in $r$ variables such 
that $|h(x)-p(x)|\le\epsilon$ for every $x\in K$. 
%281F 
      
\sqheader 281Xc Let $[a,b]$ be a non-empty closed 
interval of $\Bbb R$ and $h:[a,b]\to\Bbb R$ a continuous function. 
Show that for any $\epsilon>0$ there are 
$y_0,\ldots,y_n,a_0,\ldots,a_n,b_0,\ldots,b_n\in\Bbb R$ such that 
      
\Centerline{$|h(x)-\sum_{k=0}^n(a_k\cos y_kx+b_k\sin y_kx)|\le\epsilon$ 
for every $x\in [a,b]$,} 
      
\Centerline{$\sup_{x\in\Bbb R}|\sum_{k=0}^n(a_k\cos y_kx+b_k\sin 
y_kx)|\le\sup_{x\in [a,b]}|h(x)|$.} 
%281H 
      
\spheader 281Xd Let $h$ be a complex-valued function on 
$\ocint{-\pi,\pi}$ such that $|h|^p$ is integrable, where 
$1\le p<\infty$.   Show that for every $\epsilon>0$ there is a function 
of the form $x\mapsto f(x)=\sum_{k=-n}^nc_ke^{ikx}$, where 
$c_{-k},\ldots,c_k\in\Bbb C$, such that 
$\int_{-\pi}^{\pi}|h-f|^p\le\epsilon$.   (Compare 244H.) 
%281H 
      
\sqheader 281Xe Let $h:[-\pi,\pi]\to\Bbb R$ be a continuous 
function such that $h(\pi)=h(-\pi)$, and $\epsilon>0$.   Show that there 
are $a_0,\ldots,a_n,b_1,\ldots,b_n\in\Bbb R$ such that 
      
\Centerline{$|h(x)-\Bover12a_0-\sum_{k=1}^n(a_k\cos kx+b_k\sin kx)| 
\le\epsilon$} 
      
\noindent for every $x\in[-\pi,\pi]$. 
%281J 
      
\spheader 281Xf Let $K$ be a non-empty closed bounded set in $\BbbR^r$, 
where $r\ge 1$, and $h:K\to\Bbb R$ a continuous function.   Show that 
for any $\epsilon>0$ there are $y_0,\ldots,y_n\in\BbbR^r$, 
$a_0,\ldots,a_n,b_0,\ldots,b_n\in\Bbb R$ such that 
      
\Centerline{$|h(x) 
  -\sum_{k=0}^n(a_k\cos(y_k\dotproduct x)+b_k\sin(y_k\dotproduct x))| 
\le\epsilon$ for every $x\in K$,} 
      
\Centerline{$\sup_{x\in\Bbb R} 
 |\sum_{k=0}^n(a_k\cos(y_k\dotproduct x)+b_k\sin(y_k\dotproduct x))| 
\le\sup_{x\in K}|h(x)|$,} 
      
\noindent interpreting $y\dotproduct x$ as in 281K. 
%281K 
      
\spheader 281Xg Let $y_1,\ldots,y_r$ be real numbers such that
$1,y_1,\ldots,y_r$ are 
not linearly independent over $\Bbb Q$.   Show that there is a 
non-trivial interval $[a,b]\subseteq[\tbf{0},\tbf{1}]\subseteq\BbbR^r$ 
such that $(\fraction{my_1},\ldots,\fraction{my_r})\notin[a,b]$ for 
every $m\in\Bbb Z$. 
%281M 
      
\spheader 281Xh Let $\eta_1,\ldots,\eta_r$ be real numbers such 
that $1,\eta_1,\ldots,\eta_r$ are 
linearly independent over $\Bbb Q$.   Suppose that 
$0\le\alpha_j\le\beta_j\le 1$ for each $j\le r$.   
Show that for every $\epsilon>0$ there is an $n_0\in\Bbb N$ such that 
      
\Centerline{$|\prod_{j=1}^r(\beta_j-\alpha_j)
 -\Bover1{n+1}\#(\{m:k\le m\le k+n,\,\fraction{m\eta_j}
 \in[\alpha_j,\beta_j]$ for every $j\le r\})|\le\epsilon$} 
 
\noindent whenever $n\ge n_0$ and $k\in\Bbb N$.   ({\it Hint\/}: 
in the proof of 281N, set  
 
\Centerline{$\overline{L}(f) 
=\limsup_{n\to\infty}\sup_{k\in\Bbb N}\Bover1{n+1}\sum_{m=k}^{k+n}f(\fraction{my})$.)} 
%281N 
      
\leader{281Y}{Further exercises (a)} Show that under the hypotheses of 
281A, there is 
an $f\in\overline{A}$, the $\|\,\|_{\infty}$-closure of $A$ in $C_b(X)$, 
such that $f\restr K=h$.   ({\it Hint\/}:  take $f=\lim_{n\to\infty}f_n$ 
where 
      
\Centerline{$\|f_{n+1}-f_n\|_{\infty}\le\sup_{x\in K}|f_n(x)-h(x)| 
\le 2^{-n}$} 
      
\noindent for every $n\in\Bbb N$.) 
%281A 
      
      
\spheader 281Yb Let $X$ be a topological space and 
$K\subseteq X$ a compact subset.   Suppose that for any distinct points 
$x$, $y$ of $K$ there is a continuous function $f:X\to\Bbb R$ such that 
$f(x)\ne f(y)$.   Show that for any $r\in\Bbb N$ and any continuous 
$h:K\to\BbbR^r$ there is a continuous $f:X\to\BbbR^r$ extending $h$. 
\Hint{consider $r=1$ first.} 
%281Ya, 281A 
      
\spheader 281Yc Let $\langle X_i\rangle_{i\in I}$ be any family 
of compact Hausdorff spaces, and $X$ their product as topological 
spaces.   For each $i$, write $C(X_i)$ for the set of continuous 
functions from $X_i$ to $\Bbb R$, and $\pi_i:X\to X_i$ for the 
coordinate map.   Show that the subalgebra of $C(X)$ generated by  
$\{f\pi_i:i\in I,\,f\in C(X_i)\}$ is 
$\|\,\|_{\infty}$-dense in $C(X)$.   ({\it Note\/}:  you will need to 
know that $X$ is compact, and that if $Z$ is any compact Hausdorff space 
then for any distinct $z$, $w\in Z$ there is an $f\in C(Z)$ such that 
$f(z)\ne f(w)$.   For references see 3A3J and 3A3Bf in the next volume.) 
%281A 
      
\spheader 281Yd Let $X$ be a topological space and $K$ a compact 
subset of $X$.   Let $A$ be a linear subspace of the space $C_b(X)$ of 
bounded
real-valued continuous functions on $X$ such that $|f|\in A$ for every 
$f\in A$.   Let $h:K\to\Bbb R$ be a continuous function such that 
whenever $x$, $y\in K$ there is an $f\in A$ such that $f(x)=h(x)$ and 
$f(y)=h(y)$.   Show that for every $\epsilon>0$ there is an $f\in A$ 
such that $|f(x)-h(x)|\le\epsilon$ for every $x\in K$. 
%281A 
      
\spheader 281Ye Let $X$ be a compact topological space and write 
$C(X)$ for the set of continuous functions from $X$ to $\Bbb R$. Suppose 
that $h\in C(X)$, and let $A\subseteq C(X)$ be such that 
      
\inset{$A$ is a linear subspace of $C(X)$;} 
      
\inset{{\it either} $|f|\in A$ for every $f\in A$ {\it or} 
$f\times g\in A$ for every $f$, $g\in A$ 
{\it or} $f\times f\in A$ for every $f\in A$;} 
      
\inset{whenever $x$, $y\in X$ and $\delta>0$ there is an $f\in A$ such 
that $|f(x)-h(x)|\le\delta$ and $|f(y)-h(y)|\le\delta$.} 
      
\noindent Show that for every $\epsilon>0$ there is an $f\in A$ such 
that $|h(x)-f(x)|\le\epsilon$ for every $x\in X$. 
%281E 
      
\spheader 281Yf Let $X$ be a compact topological space and $A$ a 
$\|\,\|_{\infty}$-closed linear subspace of the space $C(X)$ of 
continuous functions from $X$ to $\Bbb R$.   Show that the following are 
equiveridical: 
      
\quad (i) $|f|\in A$ for every $f\in A$; 
      
\quad (ii) $f\times f\in A$ for every $f\in A$; 
      
\quad (iii) $f\times g\in A$ for all $f$, $g\in A$, 
      
\noindent and that in this case $A$ is closed in $C(X)$ for the topology 
defined by the pseudometrics 
      
\Centerline{$(f,g)\mapsto|f(x)-g(x)|: 
  C(X)\times C(X)\to\coint{0,\infty}$} 
      
\noindent as $x$ runs over $X$ (the `topology of pointwise convergence' 
on $C(X)$). 
%281E 
      
\spheader 281Yg Show that under the hypotheses of 281G there is 
an $f\in\overline{A}$, the $\|\,\|_{\infty}$-closure of $A$ in 
$C_b(X;\Bbb C)$, such that $f\restr K=h$ and (if $K\ne\emptyset$) 
$\|f\|_{\infty}=\sup_{x\in K}|h(x)|$. 
%281G 
      
\spheader 281Yh Let $y\in\Bbb R$ be irrational.   Show that for 
any Riemann integrable function $f:[0,1]\to\Bbb R$, 
      
\Centerline{$\int_0^1f(x)dx 
=\lim_{n\to\infty}\Bover1{n+1}\sum_{m=0}^nf(\fraction{my})$,} 
      
\noindent writing $\fraction{my}$ for the fractional part of $my$. 
({\it Hint\/}:  recall {\it Riemann's criterion:} for any $\epsilon>0$, 
there are $a_0,\ldots,a_n$ with $0=a_0\le a_1\le \ldots\le a_n=1$ and 
      
\Centerline{$\sum\{a_{j}-a_{j-1}:j\le n,\,\sup_{x\in[a_{j-1},a_j]}f(x) 
-\inf_{x\in[a_{j-1},a_j]}f(x)\ge\epsilon\}\le\epsilon$.)} 
%281N 
      
\ifdim\pagewidth>467pt\fontdimen3\tenrm=1.84pt 
  \fontdimen4\tenrm=1.22pt\fi 
\spheader 281Yi Let $\sequencen{t_n}$ be a sequence in $[0,1]$.   Show 
that the following are equiveridical: 
(i) $\lim_{n\to\infty}\bover1{n+1}\sum_{k=0}^nf(t_k) 
\ifdim\pagewidth>467pt\penalty-100\fi 
=\int_0^1f$ for every continuous function $f:[0,1]\to\Bbb R$; 
(ii) 
\ifdim\pagewidth=390pt\break\fi 
$\lim_{n\to\infty}\bover1{n+1}\sum_{k=0}^nf(t_k)=\int_0^1f$ for 
every Riemann integrable function $f:[0,1]\to\Bbb R$; 
(iii) $\liminf_{n\to\infty}\bover1{n+1}\#(\{k:k\le n,\,t_k\in G\}) 
\ge\mu G$ for every open set $G\subseteq[0,1]$, where $\mu$ is Lebesgue
measure on $\Bbb R$; 
(iv) $\lim_{n\to\infty}\bover1{n+1}\#(\{k:k\le n,\,t_k\le\alpha\}) 
=\alpha$ for every $\alpha\in[0,1]$; 
(v) $\lim_{n\to\infty}\bover1{n+1}\#(\{k:k\le n,\,t_k\in E\})=\mu E$ for 
every $E\subseteq[0,1]$ such that $\mu(\interior E)=\mu\overline{E}$ 
(vi) $\lim_{n\to\infty}\bover1{n+1}\sum_{k=0}^ne^{2\pi imt_k}=0$ for 
every $m\ge 1$. 
(Cf. 273J.   Such sequences $\sequencen{t_n}$ are called {\bf 
equidistributed} or {\bf uniformly distributed}.) 
%281N 
\fontdimen3\tenrm=1.67pt\fontdimen4\tenrm=1.11pt 
      
\spheader 281Yj Show that the sequence $\sequencen{\fraction{\ln(n+1)}}$ 
is not equidistributed. 
%281Yi, 281N 
      
\spheader 281Yk Give $[0,1]^{\Bbb N}$ its product measure $\lambda$. 
Show that $\lambda$-almost every sequence 
$\sequencen{t_n}\in[0,1]^{\Bbb N}$ is equidistributed in the sense of 
281Yi.   \Hint{273J.} 
%281Yi, 281N 
      
\spheader 281Yl Let $f:[0,1]^2\to\Bbb C$ be a continuous function. 
Show that if $\gamma\in\Bbb R$ is irrational then 
$\lim_{a\to\infty}\bover1a\int_0^af(\fraction{t},\fraction{\gamma t})dt 
\penalty-100=\int_{[0,1]^2}f$. 
\Hint{first consider functions of the form 
$x\mapsto e^{2\pi ik\dotproduct x}$.} 
%281N 
 
\spheader 281Ym A sequence $\sequencen{t_n}$ in $[0,1]$ is  
{\bf well-distributed} (with respect to Lebesgue measure $\mu$) if  

\Centerline{$\liminf_{n\to\infty}\inf_{l\in\Bbb N}
  \Bover1{n+1}\#(\{k:l\le k\le l+n$,  
$t_k\in G\})\ge\mu G$}

\noindent for every open set $G\subseteq[0,1]$    
(i) Show that $\sequencen{t_n}$ is well-distributed iff  
$\lim_{n\to\infty}\sup_{l\in\Bbb N} 
|\int_0^1f-\bover1{n+1}\sum_{k=l}^{l+n}f(t_k)|=0$ for every continuous  
$f:[0,1]\to\Bbb R$.    
(ii) Show that $\sequencen{\fraction{n\alpha}}$ is well-distributed for  
every irrational $\alpha$. 
%281Xh 281Yi 281N 
}%end of exercises 
      
\endnotes{\Notesheader{281} I have given three statements (281A, 
281E and 281G) of the Stone-Weierstrass theorem, with an acknowledgment 
(281F) of Weierstrass' own version, and three further forms (281Ya, 
281Yd, 281Yg) in the exercises.   Yet another will appear in \S4A6 in 
Volume 4.   Faced with such a multiplicity, you may wish to try 
your own hand at writing out theorems which will cover some or all of 
these versions.   I myself see no way of doing it without setting up a 
confusing list of alternative hypotheses and conclusions.   At which 
point, I ask `what is a theorem, anyway?', and answer, it is a 
stopping-place on our journey;  it is a place where we can rest, and 
congratulate ourselves on our achievement;  it is a place which we can 
learn to recognise, and use as a starting point for new adventures;  it 
is a place we can describe, and share with others.   For some theorems, 
like Fermat's last theorem, there is a canonical statement, an exactly 
locatable 
point.   For others, like the Stone-Weierstrass theorem here, we reach a 
mass of closely related results, all depending on some arrangement of 
the arguments laid out in 281A-281G and 281Ya (which introduces a new 
idea), and all useful in different ways.   I suppose, indeed, that most 
authors would prefer the versions 281Ya and 281Yg, which eliminate the 
variable 
$\epsilon$ which appears in 281A, 281E and 281G, at the expense of 
taking a closed subspace $A$.   But I find that the corollaries which 
will be useful later (281H-281L) are more naturally expressed in terms 
of linear subspaces which are not closed. 
      
The applications of the theorem, or the theorems, 
or the method -- choose your own expression -- are legion; 
only a few of them are here. 
An apparently innocent one is in 281Xa and, in a different variant, in 
281Yc;  these are enormously important in their own domains.   In this 
volume the principal application will be to 285L below, depending on 
281K, and it is perhaps right to note that there is an alternative 
approach to this particular result, based on ideas in 282G.   But I 
offer Weyl's equidistribution theorem 
(281M-281N) as evidence that we can expect to find good use for these 
ideas in almost any branch of mathematics. 
} 
      
\discrpage 
      
\frfilename{mt282.tex}
\versiondate{24.9.09}
\copyrightdate{1996}

\def\chaptername{Fourier analysis}
\def\sectionname{Fourier series}
\def\Var{\mathop{\text{Var}}}

\newsection{282}

Out of the enormous theory of Fourier series, I extract a few results
which may at least provide a basis for further study.   I give the
definitions of Fourier and Fej\'er sums (282A), with five of the most
important results concerning their convergence (282G, 282H, 282J, 282L,
282O).   On the way I include the Riemann-Lebesgue lemma (282E).   I end
by mentioning convolutions (282Q).

\vleader{48pt}{282A}{Definition} Let $f$ be an integrable complex-valued
function defined almost everywhere in $\ocint{-\pi,\pi}$.

\header{282Aa}{\bf (a)} The {\bf
Fourier coefficients} of $f$ are the complex numbers

$$c_k=\Bover1{2\pi}\int_{-\pi}^{\pi}f(x)e^{-ikx}dx$$

\noindent for $k\in\Bbb Z$.

\header{282Ab}{\bf (b)} The {\bf Fourier sums} of $f$ are the functions

$$s_n(x)=\sum_{k=-n}^nc_ke^{ikx}$$

\noindent for $x\in\ocint{-\pi,\pi}$, $n\in\Bbb N$.

\header{282Ac}{\bf (c)} The {\bf Fourier
series} of $f$ is the series $\sum_{k=-\infty}^{\infty}c_ke^{ikx}$,
or\cmmnt{ (because
we ordinarily consider the symmetric partial sums $s_n$)} the series
$c_0+\sum_{k=1}^{\infty}(c_ke^{ikx}+c_{-k}e^{-ikx})$.

\header{282Ad}{\bf (d)} The {\bf Fej\'er sums} of $f$ are the functions

$$\sigma_m=\Bover1{m+1}\sum_{n=0}^ms_n$$

\noindent for $m\in\Bbb N$.

\header{282Ae}{\bf (e)}\cmmnt{ It will be convenient to have a further
phrase available.}   If $f$ is
any function with $\dom f\subseteq\ocint{-\pi,\pi}$, its {\bf periodic
extension} is the function $\tilde f$, with domain
$\bigcup_{k\in\Bbb Z}(\dom f+2k\pi)$, such that $\tilde f(x)=f(x-2k\pi)$
whenever $k\in\Bbb Z$ and $x\in\dom f+2k\pi$.

\cmmnt{
\leader{282B}{Remarks} I have made two more or less arbitrary choices
here.

\header{282Ba}{\bf (a)} I have chosen to express Fourier series in their
`complex' form rather than
their `real' form.   From the point of view of pure measure theory
(and, indeed, from the point of view of the nineteenth-century origins
of the subject) there are gains in elegance from directing attention to
real functions $f$ and looking at the real coefficients

$$a_k=\Bover1{\pi}\int_{-\pi}^{\pi}f(x)\cos kx\,dx\text{ for }k\in\Bbb
N,$$

$$b_k=\Bover1{\pi}\int_{-\pi}^{\pi}f(x)\sin kx\,dx\text{ for }k\ge
1.$$

\noindent If we do this we have

\Centerline{$c_0=\Bover12a_0$,}

\noindent and for $k\ge 1$ we have

\Centerline{$c_k=\Bover12(a_k-ib_k)$,\quad $c_{-k}=\Bover12(a_k+ib_k)$,
\quad
$a_k=c_k+c_{-k}$,\quad $b_k=i(c_k-c_{-k})$,}

\noindent so that the Fourier sums become

$$s_n(x)=\Bover12a_0+\sum_{k=1}^na_k\cos kx+b_k\sin kx.$$

\noindent The advantage of this is that real functions $f$ correspond to
real coefficients $a_k$, $b_k$, so that it is obvious that if $f$ is
real-valued so are its Fourier and Fej\'er sums.   The disadvantages are
that we have to
use a variety of trigonometric equalities which are rather more
complicated than the properties of the complex exponential function
which they reflect, and that we are farther away from the natural
generalizations to locally compact abelian groups.   So both electrical
engineers and harmonic analysts tend to prefer the coefficients $c_k$.

\header{282Bb}{\bf (b)} I have taken the functions $f$ to be defined on
the interval $\ocint{-\pi,\pi}$ rather than on the circle
$S^1=\{z:z\in\Bbb C,\,|z|=1\}$.   There would be advantages in elegance
of language in using $S^1$, though I do not recall often seeing the
formula

\Centerline{$c_k=\int z^kf(z)dz$}

\noindent which is the natural translation of
$c_k=\bover1{2\pi}\int e^{ikx}f(x)dx$ under the substitution $x=\arg z$,
$dx=2\pi\nu(dz)$.   However, applications of the theory tend to deal
with periodic functions on the real line, so I work with
$\ocint{-\pi,\pi}$, and accept the fact that its group operation
$+_{2\pi}$, writing
$x+_{2\pi}y$ for whichever of $x+y$, $x+y+2\pi$, $x+y-2\pi$ belongs to
$\ocint{-\pi,\pi}$, is less familiar than multiplication on
$S^1$.

\header{282Bc}{\bf (c)} The remarks in (b) are supposed to remind you of
\S255.

\header{282Bd}{\bf (d)} Observe that if $f\eae g$ then $f$ and $g$ have
the same Fourier coefficients, Fourier sums and Fej\'er sums.   This
means that we could, if we wished, regard the $c_k$, $s_n$ and
$\sigma_m$ as associated with a member of $L_{\Bbb C}^1$, the space of
equivalence classes of integrable functions (\S242), rather than as
associated with a particular function $f$.   Since however the $s_n$ and
$\sigma_m$ appear as actual functions, and since many of the questions
we are interested in refer to their values at particular points, it is
more natural to express the theory in terms of integrable functions $f$
rather than in terms of members of $L_{\Bbb C}^1$.
}%end of comment

\leader{282C}{The problems (a)} Under what conditions, and in what
senses, do the Fourier and Fej\'er sums $s_n$ and $\sigma_m$ of a
function $f$ converge to $f$?

\header{282Cb}{\bf (b)} How do the properties of the double-ended
sequence $\langle c_k\rangle_{k\in\Bbb Z}$ reflect the properties of
$f$, and vice versa?

\cmmnt{\medskip

\noindent{\bf Remark} The theory of Fourier series has been one of the
leading topics of analysis for nearly two hundred years, and innumerable
further problems have contributed greatly to our understanding.   (For
instance:  can one characterize those sequences $\langle
c_k\rangle_{k\in\Bbb Z}$ which are the Fourier coefficients of some
integrable
function?)   But in this outline I will concentrate on the question (a)
above, with one and a half results (282K, 282Rb) addressing (b), which
will give us more than enough material to work on.

While most people would feel that the Fourier sums are somehow closer to
what we really want to know, it turns out that the Fej\'er sums are
easier to analyse, and there are advantages in dealing with them first.
So while you may wish to look ahead to the statements of 282J, 282L and
282O for an idea of where we are going, the first half of this section
will be largely about Fej\'er sums.   Note that in any case in which we
know that the Fourier sums converge (which is quite common;  see, for
instance, the examples in 282Xh and 282Xo), then if we know that the
Fej\'er sums converge to $f$, we can deduce that the Fourier sums also
do, by 273Ca.

The first step is a basic lemma showing that both the Fourier and
Fej\'er sums of a function $f$ can be thought of as convolutions of $f$
with kernels describable in terms of familiar functions.
}%end of comment

\leader{282D}{Lemma} Let $f$ be a complex-valued function which is
integrable over $\ocint{-\pi,\pi}$, and

\Centerline{$c_k=\bover1{2\pi}\int_{-\pi}^{\pi}f(x)e^{-ikx}dx$,
\quad$s_n(x)=\sum_{k=-n}^nc_ke^{ikx}$,
\quad$\sigma_m(x)=\bover1{m+1}\sum_{n=0}^ms_n(x)$}

\noindent its Fourier coefficients, Fourier sums and Fej\'er sums.
Write $\tilde f$ for the periodic extension of $f$\cmmnt{ (282Ae)}.   For
$m\in\Bbb N$, write

$$\psi_m(t)=\Bover{1-\cos(m+1)t}{2\pi(m+1)(1-\cos t)}$$

\noindent for $0<|t|\le\pi$.   \cmmnt{(If you like, you can set
$\psi_m(0)=\bover{m+1}{2\pi}$ to make $\psi_m$ continuous on
$[-\pi,\pi]$.)}

(a) For each $n\in\Bbb N$, $x\in\ocint{-\pi,\pi}$,

$$\eqalign{s_n(x)
&=\Bover1{2\pi}\int_{-\pi}^{\pi}f(t)
\Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})(x-t)}
  {\sin{1\over 2}(x-t)}dt\cr
&=\Bover1{2\pi}\int_{-\pi}^{\pi}\tilde f(x+t)
\Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}dt\cr
&=\Bover1{2\pi}\int_{-\pi}^{\pi}f(x-_{2\pi}t)
\Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}dt,\cr}$$

\noindent writing $x-_{2\pi}t$ for whichever of $x-t$, $x-t-2\pi$,
$x-t+2\pi$ belongs to $\ocint{-\pi,\pi}$.

(b) For each $m\in\Bbb N$, $x\in\ocint{-\pi,\pi}$,

$$\eqalign{\sigma_m(x)
&=\int_{-\pi}^{\pi}\tilde f(x+t)\psi_m(t)dt\cr
&=\int_{0}^{\pi}(\tilde f(x+t)+\tilde f(x-t))\psi_m(t)dt\cr
&=\int_{-\pi}^{\pi}f(x-_{2\pi}t)\psi_m(t)dt.\cr}$$

(c) For any $n\in\Bbb N$,

$$\Bover1{2\pi}\int_{-\pi}^{0}
  \Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}dt
=\Bover1{2\pi}\int_{0}^{\pi}
\Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}dt
=\Bover12,
\quad\Bover1{2\pi}\int_{-\pi}^{\pi}
\Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}dt
=1.$$

(d) For any $m\in\Bbb N$,

\quad (i) $0\le\psi_m(t)\le\Bover{m+1}{2\pi}$ for every $t$;

\quad (ii) for any $\delta>0$, $\lim_{m\to\infty}\psi_m(t)=0$ uniformly
on $\{t:\delta\le|t|\le\pi\}$;

\quad (iii) $\int_{-\pi}^0\psi_m=\int_0^{\pi}\psi_m=\Bover12$,
\quad $\int_{-\pi}^{\pi}\psi_m=1$.


\proof{ Really all that these amount to is summing geometric series.

\medskip

{\bf (a)} For (a), we have

$$\eqalign{\sum_{k=-n}^ne^{-ikt}
&=\bover{e^{int}-e^{-i(n+1)t}}{1-e^{-it}}\cr
&=\bover{e^{i(n+{1\over 2})t}-e^{-i(n+{1\over 2})t}}
{e^{{1\over 2}it}-e^{-{1\over 2}it}}
=\bover{\sin(n+{1\over 2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}.\cr}$$

\noindent So

$$\eqalign{s_n(x)
&=\sum_{k=-n}^nc_ke^{ikx}
=\Bover1{2\pi}\int_{-\pi}^{\pi}f(t)
  \bigl(\sum_{k=-n}^ne^{ik(x-t)}\bigr)dt\cr
&=\Bover1{2\pi}\int_{-\pi}^{\pi}f(t)
  \Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})(x-t)}
     {\sin{1\over 2}(x-t)}dt
=\Bover1{2\pi}\int_{-\pi}^{\pi}\tilde f(t)
  \Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})(x-t)}
     {\sin{1\over 2}(x-t)}dt\cr
&=\Bover1{2\pi}\int_{-\pi-x}^{\pi-x}\tilde f(x+t)
  \Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}dt
=\Bover1{2\pi}\int_{-\pi}^{\pi}\tilde f(x+t)
  \Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}dt\cr}$$

\noindent because $\tilde f$ and $t\mapsto
\bover{\sin(n+{1\over 2})t}{\sin{1\over 2}t}$ are periodic with period
$2\pi$, so that the integral from $-\pi-x$ to $-\pi$ must be the same as
the integral from $\pi-x$ to $\pi$.

For the expression in terms of $f(x-_{2\pi}t)$, we have

$$\eqalignno{s_n(x)
&=\Bover1{2\pi}\int_{-\pi}^{\pi}\tilde f(x+t)
\Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}dt
=\Bover1{2\pi}\int_{-\pi}^{\pi}\tilde f(x-t)
\Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})(-t)}
   {\sin{1\over 2}(-t)}dt\cr
\noalign{\noindent (substituting $-t$ for $t$)}
&=\Bover1{2\pi}\int_{-\pi}^{\pi}f(x-_{2\pi}t)
\Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}dt\cr}$$

\noindent because (for $x$, $t\in\ocint{-\pi,\pi}$)
$f(x-_{2\pi}t)=\tilde f(x-t)$ whenever
either is defined, and $\sin$ is an odd function.

\medskip

{\bf (b)} In the same way, we have

$$\eqalign{\sum_{n=0}^m\sin(n+\Bover12)t
&=\Imag\bigl(\sum_{n=0}^me^{i(n+{1\over 2})t}\bigr)
=\Imag\bigl(e^{{1\over 2}it}\sum_{n=0}^me^{int}\bigr)\cr
&=\Imag\bigl(e^{{1\over 2}it}\bover{1-e^{i(m+1)t}}{1-e^{it}}\bigr)
=\Imag\bigl(\bover{1-e^{i(m+1)t}}{e^{-{1\over 2}it}
   -e^{{1\over2}it}}\bigr)\cr
&=\Imag\bigl(\bover{1-e^{i(m+1)t}}{-2i\sin{1\over 2}t}\bigr)
=\Imag\bigl(\bover{i(1-e^{i(m+1)t})}{2\sin{1\over 2}t}\bigr)\cr
&=\bover{1-\cos(m+1)t}{2\sin{1\over 2}t}.\cr}$$

\noindent So

\Centerline{$\sum_{n=0}^m
  \Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}
=\Bover{1-\cos(m+1)t}{2\sin^2{1\over 2}t}
=\Bover{1-\cos(m+1)t}{1-\cos t}
=2\pi(m+1)\psi_m(t)$.}

\noindent Accordingly,

$$\eqalignno{\sigma_m(x)
&={\Bover1{m+1}}\sum_{n=0}^ms_n(x)\cr
&={\Bover1{m+1}}\sum_{n=0}^m{\Bover1{2\pi}}\int_{-\pi}^{\pi}
   \tilde f(x+t){\Bover{\sin(n+{1\over
2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}}dt\cr
&={\Bover1{2\pi}}\int_{-\pi}^{\pi}
   \tilde f(x+t)\bigl({\Bover1{m+1}}\sum_{n=0}^m
   {\Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}}
     \bigr)dt\cr
&=\int_{-\pi}^{\pi}\tilde f(x+t)\psi_m(t)dt
=\int_{-\pi}^{\pi}f(x-_{2\pi}t)\psi_m(t)dt\cr}$$

\noindent as in (a), because $\cos$ and $\psi_m$ are even functions.
For the same reason,

$$\int_{0}^{\pi}\tilde f(x-t)\psi_m(t)dt
=\int_{-\pi}^{0}\tilde f(x+t)\psi_m(t)dt,$$

\noindent so

$$\sigma_m(x)=\int_0^{\pi}(\tilde f(x+t)+\tilde f(x-t))\psi_m(t)dt.$$


\medskip

{\bf (c)} We need only look at where the formula
$\bover{\sin(n+{1\over 2})t}{\sin{1\over 2}t}$ came from to see that

$$\eqalign{\Bover1{2\pi}\int_I
  \Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}dt
&=\Bover1{2\pi}\int_I\sum_{k=-n}^ne^{ikt}dt\cr
&=\Bover1{2\pi}\int_I(1+2\sum_{k=1}^n\cos kt)dt
=\Bover12\cr}$$

\noindent for both $I=[-\pi,0]$ and $I=[0,\pi]$, because
$\int_I\cos kt\,dt=0$ for every $k\ne 0$.

\medskip

{\bf (d)(i)} $\psi_m(t)\ge 0$ for every $t$ because
$1-\cos(m+1)t$, $1-\cos t$ are always greater than or equal to $0$.
For the upper bound, we have, using the constructions in (a) and (b),

$$\bigl|\Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})t}
   {\sin{1\over 2}t}\bigr|
=\bigl|\sum_{k=-n}^ne^{ikt}\bigr|
\le 2n+1$$

\noindent for every $n$, so

$$\eqalign{\psi_m(t)
&=\Bover1{2\pi(m+1)}\sum_{n=0}^m\Bover{\sin(n+{1\over
2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}\cr
&\le\Bover1{2\pi(m+1)}\sum_{n=0}^m\,2n+1
=\Bover{m+1}{2\pi}.\cr}$$

\medskip

\quad{\bf (ii)} If $\delta\le|t|\le\pi$,

\Centerline{$\psi_m(t)\le\Bover1{\pi(m+1)(1-\cos t)}
\le\Bover1{\pi(m+1)(1-\cos\delta)}\to 0$}

\noindent as $m\to\infty$.

\medskip

\quad{\bf (iii)} also follows from the construction in (b), because

$$\int_I\psi_m
={\Bover1{2\pi(m+1)}}\sum_{n=0}^m
  \int_I\Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}dt
={\Bover1{m+1}}\sum_{n=0}^m\Bover12
=\Bover12$$

\noindent for both $I=[-\pi,0]$ and $I=[0,\pi]$, using (c).
}%end of proof of 282D

\cmmnt{\medskip

\noindent{\bf Remarks} For a discussion of substitution in integrals, if
you feel any need to justify the manipulations in part (a) of the proof,
see 263J.

The functions

\Centerline{$t\mapsto\Bover{\sin(n+{1\over
2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}$,
\quad $t\mapsto \Bover{1-\cos(m+1)t}{(m+1)(1-\cos t)}$}

\noindent are called respectively the {\bf Dirichlet kernel} and the
{\bf Fej\'er kernel}.

I give the formulae in terms of $f(x-_{2\pi}t)$ in (a) and (b) in order
to provide a link with the work of 255O.
}

\leader{282E}{}\cmmnt{ The next step is a vital lemma, with a suitably
distinguished name which (you will be glad to know) reflects its
importance rather than its difficulty.

\medskip

\noindent}{\bf The Riemann-Lebesgue lemma} Let $f$ be a
complex-valued function which is integrable over $\Bbb R$.   Then

\Centerline{$\lim_{y\to\infty}\int f(x)e^{-iyx}dx
=\lim_{y\to-\infty}\int f(x)e^{-iyx}dx
=0$.}

\proof{{\bf (a)} Consider first the case in which $f=\chi\ooint{a,b}$,
where $a<b$.   Then

\Centerline{$|\int f(x)e^{-iyx}dx|
=|\int_a^{b}e^{-iyx}dx|
=|\Bover1{-iy}(e^{-iyb}-e^{-iya})|
\le\Bover2{|y|}$}

\noindent if $y\ne 0$.   So in this case the result is obvious.

\medskip

{\bf (b)} It follows at once that the result is true if $f$ is a
step-function with bounded support, that is, if there are $a_0\le
a_1\ldots\le a_n$ such that
$f$ is constant on every interval
$\ooint{a_{j-1},a_j}$ and zero outside $[a_0,a_n]$.

\medskip

{\bf (c)} Now, for a given integrable $f$ and $\epsilon>0$, there is
a step-function $g$ such that $\int|f-g|\le\epsilon$
(242Oa).   So

\Centerline{$|\int f(x)e^{-iyx}dx-\int g(x)e^{-iyx}dx|
\le\int|f(x)-g(x)|dx\le\epsilon$}

\noindent for every $y$, and

\Centerline{$\limsup_{y\to\infty}|\int f(x)e^{-iyx}dx|\le\epsilon$,}

\Centerline{$\limsup_{y\to-\infty}|\int f(x)e^{-iyx}dx|\le\epsilon$.}

\noindent As $\epsilon$ is arbitrary, we have the result.
}%end of proof of 282E

\leader{282F}{Corollary} (a) Let $f$ be a complex-valued
function which is integrable over $\ocint{-\pi,\pi}$, and
$\langle c_k\rangle_{k\in\Bbb Z}$
its sequence of Fourier coefficients.   Then
$\lim_{k\to\infty}c_k=\lim_{k\to-\infty}c_k=0$.

(b) Let $f$ be a complex-valued function which is integrable over
$\Bbb R$.   Then
\penalty -100
$\lim_{y\to\infty}\int f(x)\sin yx\,dx=0$.

\wheader{282F}{6}{2}{2}{36pt}

\proof{{\bf (a)} We need only identify

$$c_k=\Bover1{2\pi}\int_{-\pi}^{\pi}f(x)e^{-ikx}dx$$

\noindent with $\int g(x)e^{-ikx}dx$, where $g(x)=f(x)/2\pi$ for
$x\in\dom f$ and $0$ for $|x|>\pi$.

\medskip

{\bf (b)} This is just because

\Centerline{$\int f(x)\sin yx\,dx
=\Bover1{2i}(\int f(x)e^{iyx}dx-\int f(x)e^{-iyx}dx)$.}
}%end of proof of 282F

\leader{282G}{}\cmmnt{ We are now ready for theorems on the
convergence of Fej\'er sums.
I start with an easy one, almost a warming-up exercise.

\medskip

\noindent}{\bf Theorem} Let $f:\ocint{-\pi,\pi}\to\Bbb C$ be a
continuous function such that $\lim_{t\downarrow-\pi}f(t)=f(\pi)$.
Then its sequence $\sequence{m}{\sigma_m}$ of
Fej\'er sums converges uniformly to $f$ on $\ocint{-\pi,\pi}$.

\proof{   The conditions on $f$ amount just to saying that
its periodic extension $\tilde f$ is defined and continuous everywhere
on $\Bbb R$.   Consequently it is bounded and uniformly continuous on
any bounded interval, in particular, on the interval $[-2\pi,2\pi]$.
Set
$K=\sup_{|t|\le 2\pi}|\tilde f(t)|=\sup_{t\in\ocint{-\pi,\pi}}|f(t)|$.
Write

\Centerline{$\psi_m(t)=\Bover{1-\cos(m+1)t}{2\pi(m+1)(1-\cos t)}$}

\noindent for $m\in\Bbb N$,  $0<|t|\le\pi$, as in 282D.

Given $\epsilon>0$ we can find a $\delta\in\ocint{0,\pi}$ such that
$|\tilde f(x+t)-\tilde f(x)|\le\epsilon$ whenever $x\in[-\pi,\pi]$ and
$|t|\le\delta$.   Next, we can find an $m_0\in\Bbb N$ such that
$M_m\le\bover{\epsilon}{4\pi K}$ for every $m\ge m_0$, where
$M_m=\sup_{\delta\le|t|\le\pi}\psi_m(t)$ (282D(d-ii)).   Now suppose
that $m\ge m_0$ and $x\in\ocint{-\pi,\pi}$.   Set
$g(t)=\tilde f(x+t)-f(x)$ for $|t|\le\pi$.   Then $|g(t)|\le 2K$ for all
$t\in[-\pi,\pi]$ and $|g(t)|\le\epsilon$ if $|t|\le\delta$, so

$$\eqalign{\bigl|\int_{-\pi}^{\pi}g\times\psi_m\bigr|
&\le\int_{-\pi}^{-\delta}|g|\times\psi_m
  +\int_{-\delta}^{\delta}|g|\times\psi_m
  +\int_{\delta}^{\pi}|g|\times\psi_m\cr
&\le 2M_mK(\pi-\delta)
  +\epsilon\int_{-\delta}^{\delta}\psi_m
  +2M_mK(\pi-\delta)\cr
&\le 4\pi M_mK+\epsilon
\le 2\epsilon.\cr}$$

\noindent Consequently, using 282Db and 282D(d-iii),

$$|\sigma_m(x)-f(x)|
=|\int_{-\pi}^{\pi}(\tilde f(x+t)-f(x))\psi_m(t)dt|
\le 2\epsilon$$

\noindent for every $m\ge m_0$;  and this is true for every
$x\in\ocint{-\pi,\pi}$.   As $\epsilon$ is arbitrary,
$\sequence{m}{\sigma_m}$ converges to $f$ uniformly on
$\ocint{-\pi,\pi}$.
}%end of proof of 282G

\leader{282H}{}\cmmnt{ I come now to a theorem describing the
behaviour of the
Fej\'er sums of general functions $f$.   The hypothesis of the theorem
may take a little bit of digesting;  you can get an idea of its intended
scope by glancing at Corollary 282I.

\medskip

\noindent}{\bf Theorem} Let $f$ be a complex-valued function which is
integrable over $\ocint{-\pi,\pi}$, and $\sequence{m}{\sigma_m}$ its
sequence of Fej\'er sums.
Suppose that $x\in\ocint{-\pi,\pi}$ and $c\in\Bbb C$ are such that

$$\lim_{\delta\downarrow
0}\Bover1{\delta}\int_0^{\delta}|\tilde f(x+t)+\tilde f(x-t)-2c|dt=0,$$

\noindent writing $\tilde f$ for the periodic extension of $f$\cmmnt{, as
usual};  then $\lim_{m\to\infty}\sigma_m(x)=c$.

\proof{ Set $\phi(t)=|\tilde f(x+t)+\tilde f(x-t)-2c|$ when
this is defined, which is almost everywhere, and
$\Phi(t)=\int_0^t\phi$, which is defined for every $t\ge 0$,
because $\tilde f$ is integrable over $\ocint{-\pi,\pi}$ and therefore
over every bounded interval.

As in 282D, set

\Centerline{$\psi_m(t)=\Bover{1-\cos(m+1)t}{2\pi(m+1)(1-\cos t)}$}

\noindent for $m\in\Bbb N$, $0<|t|\le\pi$.    We have

$$|\sigma_m(x)-c|
=|\int_0^{\pi}(\tilde f(x+t)+\tilde f(x-t)-2c)\psi_m(t)dt|
\le\int_0^{\pi}\phi(t)\psi_m$$

\noindent by (b) and (d) of 282D.

Let $\epsilon>0$.   By hypothesis, $\lim_{t\downarrow 0}\Phi(t)/t=0$;
let $\delta\in\ocint{0,\pi}$ be such that $\Phi(t)\le\epsilon t$ for
every $t\in[0,\delta]$.   Take any $m\ge \pi/\delta$.   I break the
integral $\int_0^{\pi}\phi\times\psi_m$ up into three parts.

\medskip

{\bf (i)} For the integral from $0$ to $1/m$, we have

$$\int_0^{1/m}\phi\times\psi_m
\le\int_0^{1/m}\Bover{m+1}{2\pi}\phi
=\Bover{m+1}{2\pi}\Phi(\Bover1m)
\le\Bover{\epsilon(m+1)}{2\pi m}
\le\epsilon,$$

\noindent because $\psi_m(t)\le\bover{m+1}{2\pi}$ for every $t$
(282D(d-i)).

\medskip

{\bf (ii)} For the integral from $1/m$ to $\delta$, we have

$$\eqalignno{\int_{1/m}^{\delta}\phi\times\psi_m
&\le\bover1{2\pi(m+1)}\int_{1/m}^{\delta}\phi(t)\bover{1}{1-\cos t}dt
\le\bover{\pi}{4(m+1)}\int_{1/m}^{\delta}\bover{\phi(t)}{t^2}dt\cr
\noalign{\noindent (because $1-\cos t\ge\Bover{2t^2}{\pi^2}$ for
$|t|\le\pi$)}
&=\bover{\pi}{4(m+1)}\bigl(\bover{\Phi(\delta)}{\delta^2}
-\bover{\Phi({1\over m})}{({1\over m})^2}
+\int_{1/m}^{\delta}\bover{2\Phi(t)}{t^3}dt\bigr)\cr
\noalign{\noindent (integrating by parts -- see 225F)}
&\le\bover{\pi}{4(m+1)}\bigl(\bover{\epsilon}{\delta}
+\int_{1/m}^{\delta}\bover{2\epsilon}{t^2}dt\bigr)\cr
\noalign{\noindent (because $\Phi(t)\le\epsilon t$ for $0\le
t\le\delta$)}
&\le\bover{\pi}{4(m+1)}\bigl(\bover{\epsilon}{\delta}
   +2\epsilon m\bigr)
\le\bover{\pi\epsilon}{4(m+1)\delta}+\bover{\pi\epsilon}2
\le\bover{\epsilon}4+\bover{\pi\epsilon}2
\le 2\epsilon.\cr}$$

\medskip

{\bf (iii)} For the integral from $\delta$ to $\pi$, we have

$$\int_{\delta}^{\pi}\phi\times\psi_m
\le\int_{\delta}^{\pi}\Bover1{\pi(m+1)(1-\cos\delta)}\phi
\to 0\text{ as }m\to\infty$$

\noindent because $\phi$ is integrable over $[-\pi,\pi]$.   There must
therefore be an $m_0\in\Bbb N$ such that

$$\int_{\delta}^{\pi}\phi\times\psi_m\le\epsilon$$

\noindent for every $m\ge m_0$.

\medskip

Putting these together, we see that

$$\int_0^{\pi}\phi\times\psi_m
\le\epsilon+2\epsilon
+\epsilon= 4\epsilon$$

\noindent for every $m\ge \max(m_0,\bover{\pi}{\delta})$.   As
$\epsilon$ is arbitrary,
$\lim_{m\to\infty}\sigma_m(x)=c$, as claimed.
}%end of proof of 282H

\leader{282I}{Corollary} Let $f$ be a complex-valued
function which is integrable over $\ocint{-\pi,\pi}$, and
$\sequence{m}{\sigma_m}$ its
sequence of Fej\'er sums.

(a) $f(x)=\lim_{m\to\infty}\sigma_m(x)$ for almost every
$x\in\ocint{-\pi,\pi}$.

(b) $\lim_{m\to\infty}\int_{-\pi}^{\pi}|f-\sigma_m|=0$.

(c) If $g$ is another integrable function with the same Fourier
coefficients, then $f\eae g$.

(d) If $x\in\ooint{-\pi,\pi}$ is such that
$a=\lim_{t\in\dom f,t\uparrow
x}f(t)$ and $b=\lim_{t\in\dom f,t\downarrow x}f(t)$ are both defined in
$\Bbb C$, then

\Centerline{$\lim_{m\to\infty}\sigma_m(x)=\Bover12(a+b)$.}

(e) If $a=\lim_{t\in\dom f,t\uparrow \pi}f(t)$ and $b=\lim_{t\in\dom
f,t\downarrow -\pi}f(t)$ are both defined in $\Bbb C$, then

\Centerline{$\lim_{m\to\infty}\sigma_m(\pi)=\Bover12(a+b)$.}

(f) If $f$ is defined and continuous at $x\in\ooint{-\pi,\pi}$, then

\Centerline{$\lim_{m\to\infty}\sigma_m(x)=f(x)$.}

(g) If $\tilde f$, the periodic extension of $f$, is defined and
continuous at $\pi$, then

\Centerline{$\lim_{m\to\infty}\sigma_m(\pi)=f(\pi)$.}

\proof{{\bf (a)} We have only to recall that by 223D

$$\eqalign{\limsup_{\delta\downarrow 0}\Bover1{\delta}\int_0^{\delta}
&|f(x+t)+f(x-t)-2f(x)|dt\cr
&\le\limsup_{\delta\downarrow 0}
\Bover1{\delta}\bigl(\int_0^{\delta}|f(x+t)-f(x)|dt
+\int_0^{\delta}|f(x-t)-f(x)|dt\bigr)\cr
&=\limsup_{\delta\downarrow 0}
\Bover1{\delta}\int_{-\delta}^{\delta}|f(x+t)-f(x)|dt
=0\cr}$$

\noindent for almost every $x\in\ooint{-\pi,\pi}$.

\medskip

{\bf (b)} Next observe that, in the language of 255O,

\Centerline{$\sigma_m=f*\psi_m$,}

\noindent by the last formula in 282Db.   Consequently, by 255Od,

\Centerline{$\|\sigma_m\|_1\le\|f\|_1\|\psi_m\|_1$,}

\noindent writing $\|\sigma_m\|_1=\int_{-\pi}^{\pi}|\sigma_m|$.
But this means that we have

\Centerline{$f(x)
=\lim_{m\to\infty}\sigma_m(x)$ for almost every $x$,
\quad$\limsup_{m\to\infty}\|\sigma_m\|_1\le\|f\|_1$;}

\noindent and it follows from 245H that
$\lim_{m\to\infty}\|f-\sigma_m\|_1=0$.

\medskip

{\bf (c)} If $g$ has the same Fourier coefficients as $f$, then it has
the same Fourier and Fej\'er sums, so we have

\Centerline{$g(x)=\lim_{m\to\infty}\sigma_m(x)=f(x)$}

\noindent almost everywhere.

\medskip

{\bf (d)-(e)} Both of these amount to considering $x\in\ocint{-\pi,\pi}$
such that

\Centerline{$\lim_{t\in\dom\tilde f,t\uparrow x}\tilde f(t)=a$,\quad
$\lim_{t\in\dom\tilde f,t\downarrow x}\tilde f(t)=b$.}

\noindent Setting $c={1\over 2}(a+b)$,
$\phi(t)=|\tilde f(x+t)+\tilde f(x-t)-2c|$ whenever this is defined, we
have $\lim_{t\in\dom\phi,t\downarrow 0}\phi(t)=0$, so surely
$\lim_{\delta\downarrow 0}\bover1{\delta}\int_0^{\delta}\phi=0$, and the
theorem applies.

\medskip

{\bf (f)-(g)} are special cases of (d) and (e).
}%end of proof of 282I

\leader{282J}{}\cmmnt{ I now turn to conditions for the convergence of
Fourier sums.    Probably the easiest result -- one which
is both striking and satisfying -- is the following.

\medskip

\noindent}{\bf Theorem} Let $f$ be a complex-valued function which is
square-integrable over $\ocint{-\pi,\pi}$.   Let
$\langle c_k\rangle_{k\in\Bbb Z}$ be its Fourier coefficients and
$\sequencen{s_n}$ its Fourier sums\cmmnt{ (282A)}.   Then

(i) $\sum_{k=-\infty}^{\infty}|c_k|^2
=\Bover1{2\pi}\int_{-\pi}^{\pi}|f|^2$,

(ii) $\lim_{n\to\infty}\int_{-\pi}^{\pi}|f-s_n|^2=0$.

\proof{{\bf (a)} I recall some notation from 244N/244P.   Let
$\eusm L_{\Bbb C}^2$
be the space of square-integrable complex-valued functions on
$\ocint{-\pi,\pi}$.   For $g$, $h\in\eusm L_{\Bbb C}^2$, write

$$(g|h)=\int_{-\pi}^{\pi}g\times\bar h,\quad
\|g\|_2=\sqrt{(g|g)}.$$

\noindent Recall that $\|g+h\|_2\le\|g\|_2+\|h\|_2$ for all $g$,
$h\in\eusm L_{\Bbb C}^2$ (244Fb/244Pb).   For $k\in\Bbb Z$,
$x\in \ocint{-\pi,\pi}$ set $e_k(x)=e^{ikx}$, so that

$$(f|e_k)=\int_{-\pi}^{\pi}f(x)e^{-ikx}dx=2\pi c_k.$$

\noindent Moreover, if $|k|\le n$,

$$(s_n|e_k)=\sum_{j=-n}^nc_j\int_{-\pi}^{\pi}e^{ijx}e^{-ikx}dx
=2\pi c_k,$$

\noindent because

$$\eqalign{\int_{-\pi}^{\pi}e^{ijx}e^{-ikx}dx&=2\pi\text{ if }j=k,\cr
&=0\text{ if }j\ne k.\cr}$$

\noindent  So

\Centerline{$(f-s_n|e_k)=0$ whenever $|k|\le n$;}

\noindent in particular,

$$(f-s_n|s_n)=\sum_{k=-n}^n\bar c_k(f-s_n|e_k)=0$$

\noindent for every $n\in\Bbb N$.

\medskip

{\bf (b)} Fix $\epsilon>0$.   The next element of the proof is the fact
that there are $m\in\Bbb N$, $a_{-m},\ldots,a_m\in\Bbb C$ such that
$\|f-h\|_2\le\epsilon$, where $h=\sum_{k=-m}^ma_ke_k$.   \Prf\ By
244Hb/244Pb we know that there is a continuous function
$g:[-\pi,\pi]\to\Bbb C$ such that $\|f-g\|_2\le{{\epsilon}\over 3}$.
Next, modifying $g$ on a suitably short interval
$\ocint{\pi-\delta,\pi}$, we can find a continuous function
$g_1:[-\pi,\pi]\to\Bbb C$ such that $\|g-g_1\|_2\le{{\epsilon}\over 3}$
and $g_1(-\pi)=g_1(\pi)$.
(Set $M=\sup_{x\in[-\pi,\pi]}|g(x)|$, take $\delta\in\ocint{0,2\pi}$
such that $(2M)^2\delta\le(\epsilon/3)^2$, and set
$g_1(\pi-t\delta)=tg(\pi-\delta)+(1-t)g(-\pi)$ for $t\in[0,1]$.)
Either by
the Stone-Weierstrass theorem (281J), or by 282G above, there are
$a_{-m},\ldots,a_m$ such
that $|g_1(x)-\sum_{k=-m}^ma_ke^{ikx}|\le\Bover{\epsilon}{3\sqrt{2\pi}}$
for every $x\in[-\pi,\pi]$;  setting $h=\sum_{k=-m}^ma_ke_k$, we have
$\|g_1-h\|_2\le\bover13\epsilon$, so that

\Centerline{$\|f-h\|_2\le\|f-g\|_2+\|g-g_1\|_2+\|g_1-h\|_2
\le\epsilon$.   \Qed}

\medskip

{\bf (c)} Now take any $n\ge m$.   Then $s_n-h$ is a linear combination
of  $e_{-n},\ldots,e_n$, so $(f-s_n|s_n-h)=0$.   Consequently

$$\eqalign{\epsilon^2
&\ge(f-h|f-h)\cr
&=(f-s_n|f-s_n)+(f-s_n|s_n-h)+(s_n-h|f-s_n)+(s_n-h|s_n-h)\cr
&=\|f-s_n\|_2^2+\|s_n-h\|_2^2
\ge\|f-s_n\|_2^2.\cr}$$

\noindent Thus $\|f-s_n\|_2\le\epsilon$ for every $n\ge m$.   As
$\epsilon$ is arbitrary, $\lim_{n\to\infty}\|f-s_n\|_2^2=0$, which
proves (ii).

\medskip

{\bf (d)} As for (i), we have

$$\sum_{k=-n}^n|c_k|^2=\Bover1{2\pi}\sum_{k=-n}^n\bar c_k(s_n|e_k)
=\Bover1{2\pi}(s_n|s_n)=\Bover1{2\pi}\|s_n\|_2^2.$$

\noindent But of course

\Centerline{$\bigl|\|s_n\|_2-\|f\|_2\bigr|\le\|s_n-f\|_2\to 0$}

\noindent as $n\to\infty$, so

$$\sum_{k=-\infty}^{\infty}|c_k|^2
=\Bover1{2\pi}\lim_{n\to\infty}\|s_n\|_2^2=\Bover1{2\pi}\|f\|_2^2
=\Bover1{2\pi}\int_{-\pi}^{\pi}|f|^2,$$

\noindent as required.
}%end of proof of 282J

\leader{282K}{Corollary} Let $L_{\Bbb C}^2$ be the Hilbert space of
equivalence classes of square-integrable complex-valued functions on
$\ocint{-\pi,\pi}$, with the inner product

$$(f^{\ssbullet}|g^{\ssbullet})
=\int_{-\pi}^{\pi}f\times\bar g$$

\noindent and norm

$$\|f^{\ssbullet}\|_2
=\bigl(\int_{-\pi}^{\pi}|f|^2\bigr)^{1/2},$$

\noindent writing $f^{\ssbullet}\in L_{\Bbb C}^2$ for the equivalence
class of a
square-integrable function $f$.   Let $\ell_{\Bbb C}^2(\Bbb Z)$ be the
Hilbert space of
square-summable double-ended complex sequences, with the inner product

$$(\pmb{c}|\pmb{d})=\sum_{k=-\infty}^{\infty}c_k\bar d_k$$

\noindent and norm

$$\|\pmb{c}\|_2
=\bigl(\sum_{k=-\infty}^{\infty}|c_k|^2\bigr)^{1/2}$$

\noindent for $\pmb{c}=\langle c_k\rangle_{k\in\Bbb Z}$,
$\pmb{d}=\langle d_k\rangle_{k\in\Bbb Z}$ in $\ell_{\Bbb C}^2(\Bbb Z)$.
Then we have an inner-product-space isomorphism
$S:L_{\Bbb C}^2\to\ell_{\Bbb C}^2(\Bbb Z)$
defined by saying that

\Centerline{$S(f^{\ssbullet})(k)
=\Bover1{\sqrt{2\pi}}\int_{-\pi}^{\pi}f(x)e^{-ikx}dx$}

\noindent for every square-integrable function $f$ and every
$k\in\Bbb Z$.

\proof{{\bf (a)} As in 282J, write $\eusm L_{\Bbb C}^2$ for the space of
square-integrable functions.   If $f$, $g\in\eusm L_{\Bbb C}^2$ and
$f^{\ssbullet}=g^{\ssbullet}$, then $f\eae g$, so

\Centerline{$\Bover1{\sqrt{2\pi}}\int_{-\pi}^{\pi}f(x)e^{-ikx}dx
=\Bover1{\sqrt{2\pi}}\int_{-\pi}^{\pi}g(x)e^{-ikx}dx$}

\noindent for every $k\in\Bbb N$.    Thus $S$ is well-defined.

\medskip

{\bf (b)}  $S$ is linear.   \Prf\ This is elementary.   If $f$,
$g\in\eusm L_{\Bbb C}^2$ and $c\in\Bbb C$,

$$\eqalign{S(f^{\ssbullet}+g^{\ssbullet})(k)
&=\Bover1{\sqrt{2\pi}}\int_{-\pi}^{\pi}(f(x)+g(x))e^{-ikx}dx\cr
&=\Bover1{\sqrt{2\pi}}\int_{-\pi}^{\pi}f(x)e^{-ikx}dx
  +\Bover1{\sqrt{2\pi}}\int_{-\pi}^{\pi}g(x)e^{-ikx}dx\cr
&=S(f^{\ssbullet})(k)+S(g^{\ssbullet})(k)\cr}$$

\noindent for every $k\in\Bbb Z$, so that
$S(f^{\ssbullet}+g^{\ssbullet})=S(f^{\ssbullet})+S(g^{\ssbullet})$.
Similarly,

\Centerline{$S(cf^{\ssbullet})(k)
=\Bover1{\sqrt{2\pi}}\int_{-\pi}^{\pi}cf(x)e^{-ikx}dx
=\Bover{c}{\sqrt{2\pi}}\int_{-\pi}^{\pi}f(x)e^{-ikx}dx
=cS(f^{\ssbullet})(k)$}

\noindent for every $k\in\Bbb Z$, so that
$S(cf^{\ssbullet})=cS(f^{\ssbullet})$.   \Qed

\medskip

{\bf (c)} If $f\in\eusm L_{\Bbb C}^2$ has Fourier coefficients $c_k$,
then
$S(f^{\ssbullet})=\langle c_k\sqrt{ 2\pi}\rangle_{k\in\Bbb Z}$, so by 282J(i)

$$\|S(f^{\ssbullet})\|^2_2
=2\pi\sum_{k=-\infty}^{\infty}|c_k|^2
=\int_{-\pi}^{\pi}|f|^2
=\|f^{\ssbullet}\|_2^2.$$

\noindent Thus $Su\in\ell_{\Bbb C}^2(\Bbb Z)$ and $\|Su\|_2=\|u\|_2$ for
every $u\in L_{\Bbb C}^2$.   Because $S$ is linear and norm-preserving,
it is surely injective.

\medskip

{\bf (d)}  It now follows that $(Sv|Su)=(v|u)$ for every $u$, $v\in
L_{\Bbb C}^2$.   \Prf\ (This is of course a standard fact about Hilbert
spaces.)   We know that for any $t\in\Bbb R$

$$\eqalign{\|u\|_2^2+2\Real(e^{it}(v|u))+\|v\|_2^2
&=(u|u)+e^{it}(v|u)+e^{-it}(u|v)+(v|v)\cr
&=(u+e^{it}v|u+e^{it}v)\cr
&=\|u+e^{it}v\|_2^2
=\|S(u+e^{it}v)\|_2^2\cr
&=\|Su\|_2^2+2\Real(e^{it}(Sv|Su))+\|Sv\|_2^2\cr
&=\|u\|_2^2+2\Real(e^{it}(Sv|Su))+\|v\|_2^2,\cr}$$

\noindent so that $\Real(e^{it}(Sv|Su))=\Real(e^{it}(v|u))$.   As $t$ is
arbitrary, $(Sv|Su)=(v|u)$.   \Qed

\medskip

{\bf (e)}    Finally, $S$ is surjective.   \Prf\ Let
$\pmb{c}=\langle c_k\rangle_{k\in\Bbb Z}$ be any member of
$\ell_{\Bbb C}^2(\Bbb Z)$.   Set
$c^{(n)}_k=c_k$ if $|k|\le n$, $0$ otherwise, and
$\pmb{c}^{(n)}=\langle c^{(n)}_k\rangle_{k\in\Bbb N}$.   Consider

$$s_n=\sum_{k=-n}^nc_ke_k,\quad u_n=s_n^{\ssbullet}$$

\noindent where I write $e_k(x)=\bover1{\sqrt{2\pi}}e^{ikx}$ for
$x\in\ocint{-\pi,\pi}$.   Then $Su_n=\pmb{c}^{(n)}$, by the same
calculations as in part (a) of the proof of 282J.   Now

\Centerline{$\|\pmb{c}^{(n)}-\pmb{c}\|_2=\sqrt{\sumop_{|k|>n}|c_k|^2}
\to 0$}

\noindent as $n\to\infty$, so

\Centerline{$\|u_m-u_n\|_2=\|\pmb{c}^{(m)}-\pmb{c}^{(n)}\|_2
\to 0$}

\noindent as $m$, $n\to\infty$, and $\sequencen{u_n}$ is a Cauchy
sequence in $L_{\Bbb C}^2$.   Because $L_{\Bbb C}^2$ is complete
(244G/244Pb), $\sequencen{u_n}$ has a limit $u\in L_{\Bbb C}^2$, and now

\Centerline{$Su=\lim_{n\to\infty}Su_n=\lim_{n\to\infty}\pmb{c}^{(n)}
=\pmb{c}$.   \Qed}

Thus $S:L_{\Bbb C}^2\to\ell_{\Bbb C}^2(\Bbb Z)$ is an
inner-product-space isomorphism.
}%end of proof of 282K

\cmmnt{\medskip

\noindent{\bf Remark} In the language of Hilbert spaces, all that is
happening here is that $\langle e_k^{\ssbullet}\rangle_{k\in\Bbb Z}$ is
a `Hilbert space basis' or `complete orthonormal sequence' in
$L_{\Bbb C}^2$, which is matched by $S$ with the standard basis of
$\ell^2_{\Bbb C}(\Bbb Z)$.   The only step which calls on non-trivial
real analysis, as
opposed to the general theory of Hilbert spaces, is the check that the
linear subspace generated by $\{e_k^{\ssbullet}:k\in\Bbb Z\}$ is dense;
this is part (b) of the proof of 282J.

Observe that while $S:L^2\to\ell^2$ is readily described, its inverse is
more of a problem.   If $\pmb{c}\in\ell^2$, we should like to say that
$S^{-1}\pmb{c}$ is the equivalence class of $f$, where
$f(x)=\bover1{\sqrt{2\pi}}\sum_{k=-\infty}^{\infty}c_ke^{ikx}$ for every
$x$.   This works very well if $\{k:c_k\ne 0\}$ is finite, but for the
general case it is less clear how to interpret the sum.   It is in fact
the case that if $\pmb{c}\in\ell^2$ then

\Centerline{$g(x)
=\Bover1{\sqrt{2\pi}}\lim_{n\to\infty}\sum_{k=-n}^nc_ke^{ikx}$}

\noindent is defined for almost every $x\in\ocint{-\pi,\pi}$, and that
$S^{-1}\pmb{c}=g^{\ssbullet}$ in $L^2$;  this is, in effect,  Carleson's
theorem (286V).   A proof of Carleson's theorem is out of our reach for
the moment.   What is covered by the results of this section is that

\Centerline{$h(x)
=\Bover1{\sqrt{2\pi}}\lim_{m\to\infty}\Bover1{m+1}\sum_{n=0}^m
\sum_{k=-n}^nc_ke^{ikx}$}

\noindent is defined for almost every $x\in\ocint{-\pi,\pi}$, and that
$h^{\ssbullet}=S^{-1}\pmb{c}$.   (The point is that we know from the
result just proved that there is {\it some} square-integrable $f$ such
that $\pmb{c}$ is the sequence of Fourier coefficients of $f$;  now
282Ia declares that the Fej\'er sums of $f$ converge to $f$ almost
everywhere, that is, that $h\eae\bover1{\sqrt{2\pi}}f$.)
}%end of comment

\leader{282L}{}\cmmnt{ The next result is the easiest, and one of the
most useful, theorems concerning pointwise convergence of Fourier sums.

\medskip

\noindent}{\bf Theorem} Let $f$ be a complex-valued function which is
integrable over $\ocint{-\pi,\pi}$, and $\sequencen{s_n}$ its
sequence of Fourier sums.

(i) If $f$ is differentiable at $x\in\ooint{-\pi,\pi}$, then
$f(x)=\lim_{n\to\infty}s_n(x)$.

(ii) If the periodic extension $\tilde f$ of $f$ is differentiable at
$\pi$, then $f(\pi)=\lim_{n\to\infty}s_n(\pi)$.

\wheader{282L}{0}{0}{0}{60pt}

\proof{{\bf (a)} Take $x\in\ocint{-\pi,\pi}$ such that $\tilde
f$ is differentiable at $x$;  of course this covers both parts.    We
have

$$s_n(x)=\Bover1{2\pi}\int_{-\pi}^{\pi}
\Bover{\tilde f(x+t)}{\sin{1\over 2}t}\sin(n+\Bover12)t\,dt$$

\noindent for each $n$, by 282Da.

\medskip

{\bf (b)} Next,

$$\int_{-\pi}^{\pi}\Bover{\tilde f(x+t)-\tilde f(x)}{t}dt$$

\noindent exists in $\Bbb C$, because there is surely some
$\delta\in\ocint{0,\pi}$
such that $(\tilde f(x+t)-\tilde f(x))/t$ is bounded on
$\{t:0<|t|\le\delta\}$, while

$$\int_{-\pi}^{-\delta}
\Bover{\tilde f(x+t)-\tilde f(x)}{t}dt,
\quad\int_{\delta}^{\pi}\Bover{\tilde f(x+t)-\tilde f(x)}{t}dt$$

\noindent exist because $1/t$ is bounded on those intervals.   It
follows that

$$\int_{-\pi}^{\pi}
\Bover{\tilde f(x+t)-\tilde f(x)}{\sin{1\over 2}t}dt$$

\noindent exists, because $|t|\le\pi|\sin{1\over 2}t|$ if
$|t|\le\pi$.   So by the Riemann-Lebesgue lemma (282Fb),

$$\lim_{n\to\infty}\int_{-\pi}^{\pi}
\Bover{\tilde f(x+t)-\tilde f(x)}{\sin{1\over 2}t}\sin(n+\Bover12)t\,dt
=0.$$

\medskip

{\bf (c)} Because

$$\Bover1{2\pi}\int_{-\pi}^{\pi}\tilde f(x)
  \Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})t}{\sin{1\over 2}t}dt
=\tilde f(x)$$

\noindent for every $n$ (282Dc),

$$s_n(x)
=\tilde f(x)+\Bover{1}{2\pi}\int_{-\pi}^{\pi}
  \Bover{\tilde f(x+t)-\tilde f(x)}{\sin{1\over 2}t}
  \sin(n+\Bover12)t\,dt
\to\tilde f(x)$$

\noindent as $n\to\infty$, as required.
}%end of proof of 282L

\leader{282M}{Lemma} Suppose that $f$ is a complex-valued
function, defined almost everywhere and of bounded variation on
$\ocint{-\pi,\pi}$.   Then $\sup_{k\in\Bbb Z}|kc_k|<\infty$, where $c_k$
is the $k$th Fourier coefficient of $f$, as in 282A.

\proof{ Set

\Centerline{$M
=\lim_{x\in\dom f,x\uparrow \pi}|f(x)|+\Var_{\ooint{-\pi,\pi}}(f)$.}

\noindent By 224J,

$$\eqalign{|kc_k|
&=\Bover{1}{2\pi}\bigl|\int_{-\pi}^{\pi}kf(t)e^{-ikt}dt\bigr|
\le\Bover1{2\pi}M
   \sup_{c\in[-\pi,\pi]}\bigl|\int_{-\pi}^cke^{-ikt}dt\bigr|\cr
&=\Bover{M}{2\pi}\sup_{c\in[-\pi,\pi]}|e^{-ikc}-e^{ik\pi}|
\le\Bover{M}{\pi}\cr}$$

\noindent for every $k$.
}%end of proof of 282M

\leader{282N}{}\cmmnt{ I give another lemma, extracting the technical
part of
the proof of the next theorem.   (Its most natural application is in
282Xn.)

\medskip

\noindent}{\bf Lemma} Let $\sequence{k}{d_k}$ be a complex sequence, and
set $t_n=\sum_{k=0}^nd_k$, $\tau_m=\bover1{m+1}\sum_{n=0}^mt_n$ for $n$,
$m\in\Bbb N$.   Suppose that $\sup_{k\in\Bbb N}|kd_k|=M<\infty$.   Then
for any $j\ge 1$ and any $c\in\Bbb C$,

\Centerline{$|t_n-c|\le\Bover{M}{j}+(2j+3)\sup_{m\ge n-n/j}|\tau_m-c|$}

\noindent for every $n\ge j^2$.

\proof{{\bf (a)} The first point to note is that for any $n$,
$n'\in\Bbb N$,

\Centerline{$|t_n-t_{n'}|\le\Bover{M|n-n'|}{1+\min(n,n')}$.}

\noindent\Prf\ If $n=n'$ this is trivial.   Suppose that $n'<n$.   Then

$$|t_n-t_{n'}|
=|\sum_{k=n'+1}^{n}d_k|
\le\sum_{k=n'+1}^n\Bover{M}{k}
\le\Bover{M(n-n')}{n'+1}
=\Bover{M|n-n'|}{1+\min(n',n)}.$$

\noindent Of course the case $n<n'$ is identical.   \Qed


\medskip

{\bf (b)} Now take any $n\ge j^2$.
Set $\eta=\sup_{m\ge n-n/j}|\tau_m-c|$.   Let $m\ge j$ be such that
$jm\le n<j(m+1)$;  then $n<jm+m$;  also

\Centerline{$n(1-\bover1{j})\le m(j+1)(1-{1\over j})\le mj$.}

\noindent    Set

$$\tau^*=\Bover1m\sum_{n'=jm+1}^{jm+m}t_{n'}
=\Bover{jm+m+1}{m}\tau_{jm+m}-\Bover{jm+1}{m}\tau_{jm}.$$

\noindent Then

$$\eqalign{|\tau^*-c|
&=|\Bover{jm+m+1}{m}\tau_{jm+m}-\Bover{jm+1}{m}\tau_{jm}-c|\cr
&=|\Bover{jm+m+1}{m}(\tau_{jm+m}-c)
   -\Bover{jm+1}{m}(\tau_{jm}-c)|\cr
&\le\Bover{jm+m+1}{m}\eta+\Bover{jm+1}{m}\eta
\le(2j+3)\eta.\cr}$$

\noindent On the other hand,

$$\eqalign{|\tau^*-t_n|
&=\bigl|\Bover1m\sum_{n'=jm+1}^{jm+m}(t_{n'}-t_n)\bigr|
\le\Bover1m\sum_{n'=jm+1}^{jm+m}\Bover{M|n-n'|}{1+\min(n,n')}\cr
&\le\Bover1m\sum_{n'=jm+1}^{jm+m}\Bover{Mm}{1+jm}
=\Bover{Mm}{1+jm}
\le\Bover{M}{j}.\cr}$$

\noindent Putting these together, we have

\Centerline{$|t_n-c|\le|t_n-\tau^*|+|\tau^*-c|
\le\Bover{M}{j}+(2j+3)\eta
=\Bover{M}{j}+(2j+3)\sup_{m\ge n-n/j}|\tau_m-c|$,}

\noindent as required.
}%end of proof of 282N

\leader{282O}{Theorem} Let $f$ be a complex-valued function of bounded
variation, defined almost everywhere in $\ocint{-\pi,\pi}$, and let
$\sequencen{s_n}$ be its sequence of Fourier sums.

(i) If $x\in\ooint{-\pi,\pi}$, then

\Centerline{$\lim_{n\to\infty}s_n(x)
=\Bover12(\lim_{t\in\dom f,t\uparrow x}f(t)
  +\lim_{t\in\dom f,t\downarrow x}f(t))$.}

(ii) $\lim_{n\to\infty}s_n(\pi)
=\Bover12(\lim_{t\in\dom f,t\uparrow\pi}f(t)
  +\lim_{t\in\dom f,t\downarrow-\pi}f(t))$.

(iii) If $f$ is defined throughout $\ocint{-\pi,\pi}$, is continuous,
and $\lim_{t\downarrow -\pi}f(t)=f(\pi)$, then $s_n(x)\to f(x)$
uniformly on $\ocint{-\pi,\pi}$.

\proof{{\bf (a)} Note first that 224F shows that the limits
$\lim_{t\in\dom f,t\downarrow x}f(t)$,
$\lim_{t\in\dom f,t\uparrow x}f(t)$ required in the formulae above
always exist.   We know also from 282M that $M=
\penalty-100
\sup_{k\in\Bbb Z}|kc_k|<\infty$, where $c_k$ is the $k$th
Fourier coefficient of $f$.

Take any $x\in\ocint{-\pi,\pi}$, and set

\Centerline{$c=\bover12(\lim_{t\in\dom f,t\uparrow x}\tilde
f(t)+\lim_{t\in\dom\tilde f,t\downarrow x}\tilde f(t))$,}

\noindent writing $\tilde f$ for the periodic extension of $f$, as
usual.   We know from 282Id-282Ie that
$c=\lim_{m\to\infty}\sigma_m(x)$, writing $\sigma_m$ for the Fej\'er
sums of $f$.   Let $\epsilon>0$.   Take any $j\ge\max(2,2M/\epsilon)$, and
$m_0\ge 1$ such that
$|\sigma_m(x)-c|\le\epsilon/(2j+3)$ for every $m\ge m_0$.

Now if $n\ge\max(j^2,2m_0)$, apply Lemma 282N with

\Centerline{$d_0=c_0$,
\quad$d_k=c_ke^{ikx}+c_{-k}e^{-ikx}$ for $k\ge 1$,}

\noindent so that $t_n=s_n(x)$, $\tau_m=\sigma_m(x)$ and $|kd_k|\le 2M$
for every $k$, $n$, $m\in\Bbb N$.   We have $n-n/j\ge{1\over 2}n\ge
m_0$, so

\Centerline{$\eta=\sup_{m\ge n-n/j}|\tau_m-c|
\le\sup_{m\ge m_0}|\tau_m-c|\le\Bover{\epsilon}{2j+3}$.}

\noindent So 282N tells us that

\Centerline{$|s_n(x)-c|=|t_n-c|\le
\Bover{2M}{j}+(2j+3)\sup_{m\ge n-n/j}|\tau_m-c|\le\epsilon+(2j+3)\eta
\le 2\epsilon$.}

\noindent As $\epsilon$ is arbitrary, $\lim_{n\to\infty}s_n(x)=c$, as
required.

\medskip

{\bf (b)} This proves (i) and (ii) of this theorem.
Finally, for (iii), observe that under these conditions $\sigma_m(x)\to
f(x)$ uniformly as $m\to\infty$, by 282G.   So given $\epsilon>0$
we choose $j\ge \max(2,2M/\epsilon)$ and $m_0\in\Bbb N$ such
that $|\sigma_m(x)-f(x)|\le\epsilon/(2j+3)$ whenever $m\ge m_0$ and
$x\in\ocint{-\pi,\pi}$.   By the same calculation as before,

\Centerline{$|s_n(x)-f(x)|\le 2\epsilon$}

\noindent for every $n\ge\max(j^2,2m_0)$ and every
$x\in\ocint{-\pi,\pi}$.   As $\epsilon$ is arbitrary,
$\lim_{n\to\infty}s_n(x)=f(x)$ uniformly for $x\in\ocint{-\pi,\pi}$.
}%end of proof of 282O

\leader{282P}{Corollary} Let $f$ be a complex-valued
function which is integrable over $\ocint{-\pi,\pi}$, and
$\sequencen{s_n}$ its sequence of
Fourier sums.

(i) Suppose that $x\in\ooint{-\pi,\pi}$ is such that $f$ is of bounded
variation on some neighbourhood of $x$.   Then

\Centerline{$\lim_{n\to\infty}s_n(x)=\Bover12(\lim_{t\in\dom f,t\uparrow
x}f(t)+\lim_{t\in\dom f,t\downarrow x}f(t))$.}

(ii) If there is a $\delta>0$ such that $f$ is of bounded variation on
both $\ocint{-\pi,-\pi+\delta}$ and $[\pi-\delta,\pi]$, then

\Centerline{$\lim_{n\to\infty}s_n(\pi)
=\Bover12(\lim_{t\in\dom f,t\uparrow \pi}f(t)
  +\lim_{t\in\dom f,t\downarrow-\pi}f(t))$.}

\proof{ In case (i), take $\delta>0$ such that $f$ is of
bounded variation on $[x-\delta,x+\delta]$ and set $f_1(t)=f(t)$ if
$x\in\dom f\cap[x-\delta,x+\delta]$, $0$ for other
$t\in\ocint{-\pi,\pi}$;   in case (ii), set $f_1(t)=f(t)$ if $t\in\dom
f$ and $|t|\ge\pi-\delta$, $0$ for other $t\in\ocint{-\pi,\pi}$, and say
that $x=\pi$.   In either case, $f_1$ is of bounded variation, so by
282O the Fourier sums $\sequencen{s'_n}$ of $f_1$ converge at
$x$ to the value given by the formulae above.   But now observe that,
writing $\tilde f$ and $\tilde f_1$ for the periodic extensions of $f$
and $f_1$, $\tilde f-\tilde f_1=0$ on a neighbourhood of $x$, so

$$\int_{-\pi}^{\pi}
  \Bover{\tilde f(x+t)-\tilde f_1(x+t)}{\sin{1\over 2}t}dt$$

\noindent exists in $\Bbb C$, and by 282Fb

$$\lim_{n\to\infty}\int_{-\pi}^{\pi}
 \bover{\tilde f(x+t)-\tilde f_1(x+t)}{\sin{1\over 2}t}
\sin(n+\Bover12)t\,dt=0,$$

\noindent that is, $\lim_{n\to\infty}s_n(x)-s'_n(x)=0$.   So
$\sequencen{s_n}$ also converges to the right limit.
}%end of proof of 282P

\leader{282Q}{}\cmmnt{ I cannot leave this section without mentioning
one of the most important facts about Fourier series, even though I have
no space here to discuss its consequences.

\medskip

\noindent}{\bf Theorem} Let $f$ and $g$ be complex-valued
functions which are integrable over $\ocint{-\pi,\pi}$, and
$\langle c_k\rangle_{k\in\Bbb N}$,
$\langle d_k\rangle_{k\in\Bbb N}$ their Fourier coefficients.   Let
$f*g$ be their convolution, defined by the formula

$$(f*g)(x)=\int_{-\pi}^{\pi}f(x-_{2\pi}t)g(t)dt
=\int_{-\pi}^{\pi}\tilde f(x-t)g(t)dt,$$

\noindent\cmmnt{as in 255O,} writing $\tilde f$ for the periodic
extension of $f$.   Then the Fourier coefficients of $f*g$
are $\langle 2\pi c_kd_k\rangle_{k\in\Bbb Z}$.

\proof{ By 255O(c-i),

$$\eqalign{\Bover1{2\pi}\int_{-\pi}^{\pi}(f*g)(x)e^{-ikx}dx
&=\Bover1{2\pi}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}
e^{-ik(t+u)}f(t)g(u)dtdu\cr
&=\Bover1{2\pi}\int_{-\pi}^{\pi}e^{-ikt}f(t)dt\int_{-\pi}^{\pi}
e^{-iku}g(u)du
=2\pi c_kd_k.\cr}$$
}%end of proof of 282Q

\leader{*282R}{}\cmmnt{ In my hurry to get to the theorems on
convergence of Fej\'er and Fourier sums, I have rather neglected the
elementary manipulations which are essential when applying the theory.
One basic result is the following.

\medskip

\noindent}{\bf Proposition} (a) Let $f:[-\pi,\pi]\to\Bbb C$ be an
absolutely continuous function such that $f(-\pi)=f(\pi)$, and
$\family{k}{\Bbb Z}{c_k}$ its sequence of Fourier coefficients.   Then
the Fourier coefficients of $f'$ are $\family{k}{\Bbb Z}{ikc_k}$.

(b)\discrversionA{\footnote{Corrected and refined 2009.}}{}
Let $f:\Bbb R\to\Bbb C$ be a differentiable function such that $f'$
is absolutely continuous on $[-\pi,\pi]$, and $f(\pi)=f(-\pi)$.
If $\family{k}{\Bbb Z}{c_k}$ are the Fourier coefficients of
$f\restr\ocint{-\pi,\pi}$, then
$\sum_{k=-\infty}^{\infty}|c_k|$ is finite.

\proof{{\bf (a)} By 225Cb, $f'$ is integrable over $[-\pi,\pi]$;  by
225E, $f$ is an indefinite integral of $f'$.   So 225F tells us that

\Centerline{$\int_{-\pi}^{\pi}f'(x)e^{-ikx}dx
=f(\pi)e^{-ik\pi}-f(-\pi)e^{ik\pi}+ik\int_{-\pi}^{\pi}f(x)e^{-ikx}dx
=ikc_k$}

\noindent for every $k\in\Bbb Z$.

\medskip

{\bf (b)(i)} Suppose first that $f'(\pi)=f'(-\pi)$.
By (a), applied twice, the Fourier coefficients of $f''$ are
$\family{k}{\Bbb Z}{-k^2c_k}$, so $\sup_{k\in\Bbb Z}k^2|c_k|$ is finite;
because $\sum_{k=1}^{\infty}\Bover1{k^2}<\infty$,
$\sum_{k=-\infty}^{\infty}|c_k|<\infty$.

\medskip

\quad{\bf (ii)} Next, suppose that $f(x)=x^2$ for every $x$.   Then, for
$k\ne 0$,

$$\eqalign{c_k
&=\Bover1{2\pi}\int x^2e^{-ikx}dx
=\Bover1{2\pi}\bigl(-\Bover1{ik}(\pi^2e^{-ik\pi}-\pi^2e^{ik\pi})
   +\int_{-\pi}^{\pi}\Bover{2x}{ik}e^{-ikx}dx\bigr)\cr
&=\Bover1{ik\pi}\bigl(-\Bover1{ik}(\pi e^{-ik\pi}+\pi e^{ik\pi})
   +\Bover1{ik}\int_{-\pi}^{\pi}e^{ikx}dx\bigr)
=\Bover{2}{k^2}(-1)^k,\cr}$$

\noindent so
$\sum_{k\in\Bbb Z}|c_k|\le|c_0|+4\sum_{k=1}^{\infty}\Bover1{k^2}$
is finite.

\medskip

\quad{\bf (iii)} In general, we can express $f$ as $f_1+cf_2$ where
$f_2(x)=x^2$ for every $x$, $c=\Bover1{4\pi}(f'(\pi)-f'(-\pi))$, and
$f_1$ satisfies the conditions of (i);  so that
$\family{k}{\Bbb Z}{c_k}$ is the sum of two summable sequences and is
itself summable.
}%end of proof of 282R

\exercises{
\leader{282X}{Basic exercises $\pmb{>}$(a)}
%\spheader 282Xa
Suppose that $\langle c_k\rangle_{k\in\Bbb N}$
is an absolutely summable double-ended sequence of complex numbers.
Show that $f(x)=\sum_{k=-\infty}^{\infty}c_ke^{ikx}$ exists for every
$x\in\Bbb R$, that $f$ is continuous and periodic, and that its Fourier
coefficients are the $c_k$.
%282A

\spheader 282Xc Set $\phi_n(t)=\bover2t\sin(n+\bover12t)$ for $t\ne 0$.
(This is sometimes called the {\bf modified Dirichlet kernel}.)
Show that for any integrable function $f$ on $\ocint{-\pi,\pi}$, with
Fourier sums $\sequencen{s_n}$ and periodic extension $\tilde f$,

\Centerline{$\lim_{n\to\infty}|s_n(x)-\bover1{2\pi}
  \int_{-\pi}^{\pi}\phi_n(t)\tilde f(x+t)dt|
=0$}

\noindent for every $x\in\ocint{-\pi,\pi}$.   \Hint{show that
$\bover2t-\bover{1}{\sin{1\over 2}t}$ is bounded, and use 282E.}
%282E

\spheader 282Xd Give a proof of 282Ib from 242O, 255O and 282G.
%282I

\spheader 282Xe Give another proof of 282Ic, based on 222D, 281J and an
idea in the proof of 242O instead of on 282H.
%282I

\spheader 282Xf Use the idea of 255Ya to shorten one of the steps in the
proof of 282H, taking
\discrcenter{468pt}{$g_m(t)
=\min(\bover{m+1}{2\pi},\bover{\pi}{4(m+1)t^2})$ }for $|t|\le\delta$, so
that $g_m\ge\psi_m$ on $[-\delta,\delta]$.
%282H

\sqheader 282Xg(i) Let $f$ be a real square-integrable function
on $\ocint{-\pi,\pi}$, and $\sequence{k}{a_k}$,
$\langle b_k\rangle_{k\ge 1}$ its real Fourier coefficients (282Ba).
Show that
$\bover12a_0^2+\sum_{k=1}^{\infty}(a_k^2+b_k^2)
=\bover1{\pi}\int_{-\pi}^{\pi}|f|^2$.  (ii) Show that
$f\mapsto(\sqrt{\bover{\pi}2}a_0,\sqrt{\pi}a_1,
\ifdim\pagewidth>467pt\penalty-100\fi
\sqrt{\pi}b_1,\ldots)$
defines an inner-product-space isomorphism between the real Hilbert
space $L^2_{\Bbb R}$ of equivalence classes of real square-integrable
functions on $\ocint{-\pi,\pi}$ and the real Hilbert space
$\ell^2_{\Bbb R}$ of square-summable sequences.
%282K

\spheader 282Xh Show that
$\bover{\pi}4=1-\bover13+\bover15-\bover17+\ldots$.   \Hint{find the
Fourier series of $f$ where $f(x)=x/|x|$, and compute the sum of the
series at $\bover{\pi}2$.   Of course there are other methods, e.g.,
examining the Taylor series for $\arctan\bover{\pi}4$.}
%282L

\spheader 282Xi Let $f$ be an integrable complex-valued function
on $\ocint{-\pi,\pi}$, and $\sequencen{s_n}$ its sequence of Fourier
sums.   Suppose that $x\in\ooint{-\pi,\pi}$, $a\in\Bbb C$ are such that
$\biggerint_{-\pi}^{\pi}\Bover{f(t)-a}{t-x}dt$ exists and is finite.
Show that $\lim_{n\to\infty}s_n(x)=a$.
Explain how this generalizes 282L.   What modification is appropriate to
obtain a limit $\lim_{n\to\infty}s_n(\pi)$?
%282L

\spheader 282Xj Suppose that $\alpha>0$, $K\ge 0$ and
$f:\ooint{-\pi,\pi}\to\Bbb C$ are such that
$|f(x)-f(y)|\le K|x-y|^{\alpha}$ for all $x$, $y\in\ooint{-\pi,\pi}$.
(Such functions are called {\bf H\"older continuous}.)
Show that the Fourier sums of $f$ converge to $f$ everywhere in
$\ooint{-\pi,\pi}$.   \Hint{use 282Xi.}  (Compare 282Yb.)
%282Xi 282L

\spheader 282Xk In 282L, show that it is enough if $\tilde f$ is
differentiable with respect to its domain at $x$ or $\pi$ (see 262Fb),
rather than differentiable in the strict sense.
%282L

\spheader 282Xl Show that $\lim_{a\to\infty}\int_0^a\bover{\sin t}tdt$
exists and is finite.   \Hint{use 224J to estimate
$\int_a^b\bover{\sin t}tdt$ for $0<a\le b$.}
%

\spheader 282Xm Show that
$\int_0^{\infty}\bover{|\sin t|}tdt=\infty$.   \Hint{show that
$\sup_{a\ge 0}|\int_1^a\bover{\cos 2t}tdt|<\infty$, and therefore that
$\sup_{a\ge 0}\int_1^a\bover{\sin^2t}{t}dt\penalty-100=\infty$.}
%282Xl

\sqheader 282Xn Let $\sequence{k}{d_k}$ be a sequence in $\Bbb C$ such
that $\sup_{k\in\Bbb N}|kd_k|<\infty$ and

\Centerline{$\lim_{m\to\infty}\Bover1{m+1}\sum_{n=0}^m\sum_{k=0}^nd_k
=c\in\Bbb C$.}

\noindent Show that $c=\sum_{k=0}^{\infty}d_k$.   \Hint{282N.}
%282N

\sqheader 282Xo Show that
$\sum_{n=1}^{\infty}\bover1{n^2}=\bover{\pi^2}6$.   \Hint{(b-ii) of the
proof of 282R.}
%282Q

\spheader 282Xp Let $f$ be an integrable
complex-valued function on $\ocint{-\pi,\pi}$, and $\sequencen{s_n}$ its
sequence of Fourier sums.    Suppose that $x\in\ooint{-\pi,\pi}$ is such
that

(i) there is an $a\in\Bbb C$ such that

\inset{{\it either}
$\int_{-\pi}^x\Bover{a-f(t)}{x-t}dt$ exists in $\Bbb C$}

\inset{{\it or} there is some $\delta>0$ such that $f$ is of bounded
variation on $[x-\delta,x]$, and $a=\lim_{t\in\dom f,t\uparrow x}f(t)$}

(ii) there is a $b\in\Bbb C$ such that

\inset{{\it either}
$\int_x^{\pi}\Bover{f(t)-b}{t-x}dt$ exists in $\Bbb C$}

\inset{{\it or} there is some $\delta>0$ such that $f$ is of bounded
variation on $[x,x+\delta]$, and $b=\lim_{t\in\dom f,t\downarrow
x}f(t)$.}

\noindent Show that $\lim_{n\to\infty}s_n(x)={1\over 2}(a+b)$.
What modification is appropriate to obtain a limit
$\lim_{n\to\infty}s_n(\pi)$?
%282Xi, 282O

\sqheader 282Xq Let $f$, $g$ be integrable complex-valued
functions on $\ocint{-\pi,\pi}$, and
$\pmb{c}=\langle c_k\rangle_{k\in\Bbb Z}$,
$\pmb{d}=\langle d_k\rangle_{k\in\Bbb Z}$
their sequences of Fourier coefficients.   Suppose that {\it either}
$\sum_{k=-\infty}^{\infty}|c_k|<\infty$ {\it or}
$\sum_{k=-\infty}^{\infty}|c_k|^2+|d_k|^2<\infty$.   Show that the
sequence of
Fourier coefficients of $f\times g$ is just the convolution
$\pmb{c}*\pmb{d}$ of $\pmb{c}$ and $\pmb{d}$ (255Xk).
%282Q

\spheader 282Xr In 282Ra, what happens if $f(\pi)\ne f(-\pi)$?
%282R

\spheader 282Xs  Suppose that $\langle c_k\rangle_{k\in\Bbb N}$
is a double-ended sequence of complex numbers such that
$\sum_{k=-\infty}^{\infty}|kc_k|<\infty$.   Show that
$f(x)=\sum_{k=-\infty}^{\infty}c_ke^{ikx}$ exists for every
$x\in\Bbb R$ and that $f$ is differentiable everywhere.
%282R

\spheader 282Xt Let $\langle c_k\rangle_{k\in\Bbb Z}$ be a
double-ended sequence of complex numbers such that
$\sup_{k\in\Bbb Z}|kc_k|<\infty$.   Show that there is a
square-integrable function $f$
on $\ocint{-\pi,\pi}$ such that the $c_k$ are the
Fourier coefficients of $f$, that $f$ is the limit almost everywhere of
its Fourier sums, and that $f*f*f$ is differentiable.   \Hint{use
282K to show that there is an $f$, and 282Xn to show that its Fourier
sums converge wherever its Fej\'er sums do;  use 282Q and 282Xs to
show that $f*f*f$ is differentiable.}
%282Xs, 282Q

\leader{282Y}{Further exercises (a)}
%\spheader 282Ya
Let $f$ be a non-negative integrable function on
$\ocint{-\pi,\pi}$, with Fourier coefficients
$\langle c_k\rangle_{k\in\Bbb Z}$.   Show that

\Centerline{$\sum_{j=0}^n\sum_{k=0}^na_j\bar a_kc_{j-k}\ge 0$}

\noindent for all complex numbers $a_0,\ldots,a_n$.   (See also 285Xu
below.)
%282A

\spheader 282Yb Let $f:\ocint{-\pi,\pi}\to\Bbb C$, $K\ge 0$, $\alpha>0$
be such that $|f(x)-f(y)|\le K|x-y|^{\alpha}$ for all $x$,
$y\in\ocint{-\pi,\pi}$.   Let $c_k$, $s_n$ be the Fourier coefficients
and sums of $f$.   (i) Show that
$\sup_{k\in\Bbb Z}|k|^{\alpha}|c_k|<\infty$.   \Hint{show that
$c_k=\bover1{4\pi}\int_{-\pi}^{\pi}(f(x)-\tilde f(x+\bover{\pi}k))
e^{-ikx}dx$.}   (ii) Show that if $f(\pi)=\lim_{x\downarrow-\pi}f(x)$
then $s_n\to f$ uniformly.   (Compare 282Xj.)
%282E

\spheader 282Yc Let $f$ be a measurable
complex-valued function on $\ocint{-\pi,\pi}$, and suppose that
$p\in\coint{1,\infty}$ is such that
$\int_{-\pi}^{\pi}|f|^p<\infty$.   Let
$\sequence{m}{\sigma_m}$ be the sequence of Fej\'er sums of $f$.   Show
that $\lim_{m\to\infty}\int_{-\pi}^{\pi}|f-\sigma_m|^p=0$.
\Hint{use 245Xl, 255Yk and the ideas in 282Ib.}
%282I

\spheader 282Yd Construct a continuous function
$h:[-\pi,\pi]\to\Bbb R$ such that $h(\pi)=h(-\pi)$ but the Fourier sums
of $h$ are unbounded at $0$, as follows.   Set
$\alpha(m,n)=\int_0^{\pi}
\bover{\sin(m+{1\over2})t\sin(n+{1\over 2})t}{\sin{1\over 2}t}dt$.
Show that $\lim_{n\to\infty}\alpha(m,n)=0$
for every $m$, but $\lim_{n\to\infty}\alpha(n,n)=\infty$.   Set
$h_0(x)=\sum_{k=0}^{\infty}\delta_k\sin(m_k+{1\over 2})x$ for
$0\le x\le\pi$, $0$ for $-\pi\le x\le 0$, where
$\delta_k>0$, $m_k\in\Bbb N$ are such that
($\alpha$) $\delta_k\le 2^{-k}$, $\delta_k|\alpha(m_k,m_n)|\le 2^{-k}$
for every $n<k$ (choosing $\delta_k$) ($\beta$)
$\delta_k\alpha(m_k,m_k)\ge k$, $\delta_n|\alpha(m_k,m_n)|\le 2^{-n}$
for every $n<k$ (choosing
$m_k$).   Now modify $h_0$ on $\coint{-\pi,0}$ by adding a function of
bounded variation.
%282*

\spheader 282Ye(i) Show that $\lim_{n\to\infty}\int_{-\pi}^{\pi}
|\bover{\sin(n+{1\over2})t}{\sin{1\over 2}t}|dt=\infty$.  \Hint{282Xm.}
(ii) Show that for any $\delta>0$ there are $n\in\Bbb N$, $f\ge 0$ such
that $\int_{-\pi}^{\pi}f\le\delta$, $\int_{-\pi}^{\pi}|s_n|\ge 1$, where
$s_n$ is the $n$th Fourier sum of $f$.   \Hint{take $n$ such that
$\Bover1{2\pi}\int_{-\pi}^{\pi}
|\bover{\sin(n+{1\over2})t}{\sin{1\over 2}t}|dt>\Bover1{\delta}$ and set
$f(x)=\Bover{\delta}{\eta}$ for $0\le x\le\eta$, $0$ otherwise, where
$\eta$ is small.}   (iii) Show that there is an integrable function
$f:\ocint{-\pi,\pi}\to\Bbb R$ such that $\sup_{n\in\Bbb N}\|s_n\|_1$ is
infinite, where $\sequencen{s_n}$ is the sequence of Fourier sums of
$f$.   \Hint{it helps to know the `Uniform Boundedness Theorem' of
functional analysis, but $f$ can also be constructed bare-handed by the
method of 282Yd.}
%282*

\spheader 282Yf Let $u:[-\pi,\pi]\to\Bbb R$ be an absolutely continuous
function such that $u(\pi)=u(-\pi)$ and $\int_{-\pi}^{\pi}u=0$.   Show
that $\|u\|_2\le\|u'\|_2$.   (This is {\bf Wirtinger's inequality}.)
%282J

\spheader 282Yg For $0\le r<1$, $t\in\Bbb R$ set
$A_r(t)=\Bover{1-r^2}{1-2r\cos t+r^2}$.
($A_r$ is the {\bf Poisson kernel};  see 478Xl\Latereditions\ in
Volume 4.)
(i) Show that $\Bover1{2\pi}\int_{-\pi}^{\pi}A_r=1$.
(ii) For a real function $f$ which is
integrable over $\ocint{-\pi,\pi}$, with real Fourier coefficients
$a_k$, $b_k$ (282Ba), set
$S_r(x)=\Bover12a_0+\sum_{k=1}^{\infty}r^k(a_k\cos kx+b_k\sin kx)$ for
$x\in\ocint{-\pi,\pi}$, $r\in\coint{0,1}$.   Show that
$S_r(x)=\Bover1{2\pi}\int_{-\pi}^{\pi}A_r(x-t)f(t)dt$ for every
$x\in\ocint{-\pi,\pi}$.   \Hint{$A_r(t)=1+2\sum_{n=1}^{\infty}r^n\cos nt$.}
(iii) Show that $\lim_{r\uparrow 1}S_r(x)=f(x)$ for every
$x\in\ooint{-\pi,\pi}$ which is in the Lebesgue set of $f$.
\Hint{223Yg.}  (iv) Show that
$\lim_{r\uparrow 1}\int_{-\pi}^{\pi}|S_r-f|=0$.   (v) Show that if $f$ is
defined everywhere on $\ocint{-\pi,\pi}$, is continuous, and
$f(\pi)=\lim_{x\downarrow-\pi}f(x)$, then
$\lim_{r\uparrow 1}\sup_{x\in\ocint{-\pi,\pi}}|S_r(x)-f(x)|=0$.
%282+
}%end of exercises

\endnotes{
\Notesheader{282} This has been a long section with a
potentially confusing collection of results, so perhaps I should
recapitulate.   Associated with any integrable function on
$\ocint{-\pi,\pi}$ we have the corresponding Fourier sums, being the
symmetric partial sums $\sum_{k=-n}^nc_ke^{ikx}$ of the complex series
$\sum_{k=-\infty}^{\infty}c_ke^{ikx}$, or, equally, the partial sums
${1\over 2}a_0+\sum_{k=1}^na_k\cos kx+b_k\sin kx$ of the real series
${1\over 2}a_0+\sum_{k=1}^{\infty}a_k\cos kx+b_k\sin kx$.   The Fourier
coefficients $c_k$, $a_k$, $b_k$ are the only natural ones, because
if the series is to converge with any regularity at all then

\Centerline{$\Bover1{2\pi}\int_{-\pi}^{\pi}
\bigl(\sum_{k=-\infty}^{\infty}c_ke^{ikx}\bigr)e^{-ilx}dx$}

\noindent ought to be simultaneously

\Centerline{$\sum_{k=-\infty}^{\infty}\Bover1{2\pi}\int_{-\pi}^{\pi}
c_ke^{ikx}e^{-ilx}dx=c_l$}

\noindent and

\Centerline{$\Bover1{2\pi}\int_{-\pi}^{\pi}f(x)e^{-ilx}dx$.}

\noindent (Compare the calculations in 282J.)   The effect of taking
Fej\'er sums $\sigma_m(x)$ rather than the Fourier sums $s_n(x)$ is to
smooth the sequence out;   recall that if $\lim_{n\to\infty}s_n(x)=c$
then $\lim_{m\to\infty}\sigma_m(x)=c$, by 273Ca in the last chapter.

Most of the work above is concerned with the question of when Fourier or
Fej\'er sums converge, in some sense, to the original function $f$.   As
has happened before, in \S245 and elsewhere, we have more than
one kind of convergence to consider.   {\it Norm} convergence, for
$\|\,\|_1$ or $\|\,\|_2$ or $\|\,\|_{\infty}$, is the simplest;  the
three theorems 282G, 282Ib and 282J at least are
relatively straightforward.   (I have given 282Ib as a corollary of
282Ia;  but there is an easier proof from 282G.   See 282Xd.)
Respectively, we have

\inset{if $f$ is continuous (and matches at $\pm\pi$, that is,
$f(\pi)=\lim_{t\downarrow-\pi}f(t)$) then $\sigma_m\to
f$ uniformly, that is, for $\|\,\|_{\infty}$ (282G);}

\inset{if $f$ is any integrable function, then $\sigma_m\to f$ for
$\|\,\|_1$ (282Ib);}

\inset{if $f$ is a square-integrable function, then $s_n\to f$ for
$\|\,\|_2$ (282J);}

\inset{if $f$ is continuous and of bounded
variation (and matches at $\pm\pi$), then $s_n\to f$ uniformly (282O).}

\noindent There are some similar results for other $\|\,\|_p$ (282Yc);
but note that the Fourier sums need not converge for $\|\,\|_1$ (282Ye).

{\it Pointwise} convergence is harder.   The results I give are

\inset{if $f$ is any integrable function, then $\sigma_m\to f$ almost
everywhere (282Ia);}

\noindent this relies on some careful calculations in 282H, and also on
the deep result 223D.   Next we have the results which look at the
average of the limits of $f$ from the two sides.   Suppose I write

\Centerline{$f^{\pm}(x)=\Bover12(\lim_{t\uparrow
x}f(t)+\lim_{t\downarrow x}f(t))$}

\noindent whenever this is defined, taking $f^{\pm}(\pi)
=\bover12(\lim_{t\uparrow\pi}f(t)+\lim_{t\downarrow-\pi}f(t))$.   Then
we have

\inset{if $f$ is any integrable function, $\sigma_m\to f^{\pm}$ wherever
$f^{\pm}$ is defined (282I);}

\inset{if $f$ is of bounded variation, $s_n\to f^{\pm}$ everywhere
(282O).}

\noindent Of course these apply at any point at which $f$ is continuous,
in which case $f(x)=f^{\pm}(x)$.
Yet another result of this type is

\inset{if $f$ is any integrable function, $s_n\to f$ at any point at
which $f$ is differentiable (282L);}

\noindent in fact, this can be usefully extended for very little extra
labour (282Xi, 282Xp).

I cannot leave this list without mentioning the theorem I have {\it not}
given.   This is {\bf Carleson's theorem}:

\inset{if $f$ is square-integrable, $s_n\to f$ almost everywhere}

\noindent ({\smc Carleson 66}).   I will come to this in \S286.   There
is an elementary special case in 282Xt.   The result is in fact valid
for many other $f$ (see the notes to \S286).

The next glaring lacuna in the exposition here is the absence of any
examples to show how far these results are best possible.   There is no
suggestion, indeed, that there are any natural necessary and sufficient
conditions for

\inset{$s_n\to f$ at every point.}

\noindent Nevertheless, we have to make an effort to find a continuous
function
for which this is not so, and the construction of an example by du
Bois-Reymond ({\smc Bois-Reymond 1876}) was an important moment in
the history of analysis, not least because it forced mathematicians to
realise that some comfortable assumptions about  the classification of
functions -- essentially, that functions are either `good' or so bad
that one needn't trouble with them -- were false.  The example is
instructive but I have had to omit it for lack of space;  I give an
outline of a possible method in 282Yd.   (You can find a detailed
construction in {\smc K\"orner 88}, chapter 18, and a proof that such a
function exists in {\smc Dudley 89}, 7.4.3.)    If you allow general
integrable functions, then you can do much better, or perhaps I should
say much worse;  there is an integrable $f$ such that $\sup_{n\in\Bbb
N}|s_n(x)|=\infty$ for every $x\in\ocint{-\pi,\pi}$
({\smc Kolmogorov 1926};  see {\smc Zygmund 59}, $\S\S$VIII.3-4).

In 282C I mentioned two types of problem.   The first -- when is a
Fourier
series summable? -- has at least been treated at length, even though I
cannot pretend to have given more than a sample of what is known.   The
second -- how do properties of the $c_k$ reflect properties of $f$? -- I
have hardly touched on.   I do give what seem to me to be the three most
important results in this area.   The first is

\inset{if $f$ and $g$ have the same Fourier coefficients, they are equal
almost everywhere (282Ic).}

\noindent This at least tells us that we ought in principle to be able
to learn almost anything about $f$ by looking at its Fourier series.
(For instance, 282Ya describes a necessary and sufficient condition for
$f$ to be non-negative almost everywhere.)   The second is

\inset{$f$ is square-integrable iff
$\sum_{k=-\infty}^{\infty}|c_k|^2<\infty$;}

\noindent in fact,

\inset{$\sum_{k=-\infty}^{\infty}|c_k|^2
=\Bover1{2\pi}\int_{\pi}^{\pi}|f|^2$
(282J).}

\noindent Of course this is fundamental, since it shows that Fourier
coefficients provide a natural Hilbert space isomorphism between $L^2$
and $\ell^2$ (282K).   I should perhaps remark that while the real
Hilbert spaces $L^2_{\Bbb R}$, $\ell^2_{\Bbb R}$ are isomorphic as inner
product spaces (282Xg), they are certianly not isomorphic as Banach
lattices;  for instance, $\ell^2_{\Bbb R}$ has `atomic' elements
$\pmb{c}$ such that if $0\le\pmb{d}\le\pmb{c}$ then $\pmb{d}$ is a
multiple of $\pmb{c}$, while $L^2_{\Bbb R}$ does not.   Perhaps even
more important is

\inset{the Fourier coefficients of a convolution $f*g$ are just a scalar
multiple of the
products of the Fourier coefficients of $f$ and $g$ (282Q);}

\noindent  but to use this effectively we need to study the Banach
algebra structure of $L^1$, and I have no choice but to abandon this
path immediately.   (It will form a conspicuous part of Chapter 44 in
Volume 4.)   282Xt gives an elementary
consequence, and 282Xq a very partial description of the
relationship between a product $f\times g$ of two functions and the
convolution product of their sequences of Fourier coefficients.

The Fej\'er sums considered in this section are one way of working
around the convergence difficulties associated with Fourier sums.   When we
come to look at Fourier transforms in the next two sections we shall need
some further manoeuvres.   A different type of smoothing is
obtained by using the Poisson kernel in place of the Dirichlet or Fej\'er
kernel (282Yg).

I end these notes with a remark on the number $2\pi$.   This enters
nearly every formula involving Fourier series, but could I think be
removed totally from the present section, at least, by re-normalizing
the measure of $\ocint{-\pi,\pi}$.   If instead of Lebesgue measure
$\mu$ we took the measure $\nu=\bover1{2\pi}\mu$ throughout, then every
$2\pi$ would disappear.   (Compare the remark in 282Bb concerning the
possibility of doing integrals over $S^1$.)   But I think most of us
would prefer to remember the location of a $2\pi$ in every formula than
to deal with an unfamiliar measure.
}%end of notes

\discrpage

\frfilename{mt283.tex}
\versiondate{31.3.13}

\def\chaptername{Fourier analysis}
\def\sectionname{Fourier transforms I}
\copyrightdate{1994}

\newsection{283}

I turn now to the theory of Fourier transforms on $\Bbb R$.   In the
first of two sections on the subject, I present those parts of the
elementary theory which can be dealt with using the methods of the
previous section on Fourier series.   I find no way of making sense of
the theory, however, without introducing a fragment of L.Schwartz'
theory of distributions, which I present in \S284.
As in \S282, of course, this treatment also is nothing but a start in
the topic.

The whole theory can also be done in $\BbbR^r$.   I leave this
extension to the exercises, however, since there are few new ideas, the
formulae are significantly more complicated, and I shall not, in this
volume at least, have any use for the multidimensional versions of these
particular theorems, though some of the same ideas will appear, in
multidimensional form, in \S285.

\leader{283A}{Definitions} Let $f$ be a complex-valued function which is
integrable over $\Bbb R$.


\spheader 283Aa The {\bf Fourier transform} of $f$ is the
function $\varhatf:\Bbb R\to\Bbb C$ defined by setting

\Centerline{$\varhatf(y)=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-iyx}f(x)dx$}

\noindent for every $y\in\Bbb R$.   \cmmnt{(Of course the integral is
always defined because $x\mapsto e^{-iyx}$ is bounded and continuous,
therefore measurable.)}

\spheader 283Ab The {\bf inverse Fourier transform} of $f$ is
the function $\varcheckf:\Bbb R\to\Bbb C$ defined by setting

\Centerline{$\varcheckf(y)=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{iyx}f(x)dx$}

\noindent for every $y\in\Bbb R$.

\cmmnt{
\leader{283B}{Remarks (a)} It is a mildly vexing feature of the theory
of Fourier transforms -- vexing, that is,
for outsiders like myself -- that there is in fact
no standard definition of `Fourier transform'.
The commonest definitions are, I think,

\Centerline{$\varhatf(y)=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{\mp iyx}f(x)dx$,}

\Centerline{$\varhatf(y)=\int_{-\infty}^{\infty}
e^{\mp iyx}f(x)dx$,}

\Centerline{$\varhatf(y)=\int_{-\infty}^{\infty}
e^{\mp 2\pi iyx}f(x)dx$,}

\noindent corresponding to inverse transforms

\Centerline{$\varcheckf(y)=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{\pm iyx}f(x)dx$,}

\Centerline{$\varcheckf(y)=\Bover1{{2\pi}}\int_{-\infty}^{\infty}
e^{\pm iyx}f(x)dx$,}

\Centerline{$\varcheckf(y)=\int_{-\infty}^{\infty}
e^{\pm 2\pi iyx}f(x)dx$.}

\noindent I leave it to you to check that the whole theory can be
carried through with any of these six pairs, and to investigate other
possibilities (see 283Xa-283Xb below).

\spheader 283Bb The phrases `Fourier transform', `inverse
Fourier transform' make it plain that $(\varhatf)\varspcheck$ is
supposed to be $f$, at least some of the time.   This is indeed the
case, but the class of $f$ for which this is true in the literal sense
is somewhat constrained, and we shall have to wait a little while before
investigating it.

\spheader 283Bc No amount of juggling with constants, in the
manner of (a) above, can make $\varhatf$ and $\varcheckf$ quite the
same.   However, on the definitions I have chosen, we do have
$\varcheckf(y)=\varhatf(-y)$ for every $y$, so that $\varcheckf$ and
$\varhatf$ will share essentially all the properties of interest to us
here;  in particular, everything in the next proposition will be valid
with $\varspcheck$ in place of $\varsphat$, if you change signs at the
right points in parts (c), (h) and (i).
}%end of comment

\leader{283C}{Proposition} Let $f$ and $g$ be complex-valued functions
which are integrable over $\Bbb R$.

(a) $(f+g)\varsphat=\varhatf+\varhat{g}$.

(b) $(cf)\varsphat=c\varhatf$ for every $c\in\Bbb C$.

(c) If $c\in\Bbb R$ and $h(x)=f(x+c)$ whenever this is
defined, then $\varhat{h}(y)=e^{icy}\varhatf(y)$ for every $y\in\Bbb R$.

(d) If $c\in\Bbb R$ and $h(x)=e^{icx}f(x)$ for every
$x\in\dom f$, then $\varhat{h}(y)=\varhatf(y-c)$ for every
$y\in\Bbb R$.

(e) If $c>0$ and $h(x)=f(cx)$ whenever this is defined,
then $\varhat{h}(y)=\Bover1c\varhatf(\Bover{y}{c})$
for every $y\in\Bbb R$.

(f) $\varhatf:\Bbb R\to\Bbb C$ is continuous.

(g) $\lim_{y\to\infty}\varhatf(y)=\lim_{y\to-\infty}\varhatf(y)=0$.

(h) If $\int_{-\infty}^{\infty}|xf(x)|dx<\infty$, then $\varhatf$ is
differentiable, and its derivative is

\Centerline{$\varhatf\vthsp'(y)=-\Bover{i}{\sqrt{2\pi}}
\int_{-\infty}^{\infty}e^{-iyx}xf(x)dx$}

\noindent for every $y\in\Bbb R$.

(i) If $f$ is absolutely continuous on every bounded interval and $f'$
is integrable, then $(f')\varsphat(y)=iy\varhatf(y)$ for every $y\in\Bbb
R$.

\proof{{\bf (a)} and {\bf (b)} are trivial, and {\bf (c)}, {\bf (d)} and
{\bf (e)} are elementary substitutions.

\medskip

{\bf (f)} If $\sequencen{y_n}$ is any convergent sequence in $\Bbb R$
with limit $y$, then

$$\eqalign{\varhatf(y)
&=\Bover1{\sqrt{2\pi}}
\int_{-\infty}^{\infty}\lim_{n\to\infty}e^{-iy_nx}f(x)dx\cr
&=\lim_{n\to\infty}\Bover1{\sqrt{2\pi}}
\int_{-\infty}^{\infty}e^{-iy_nx}f(x)dx
=\lim_{n\to\infty}\varhatf(y_n)\cr}$$

\noindent by Lebesgue's Dominated Convergence Theorem, because
$|e^{-iy_nx}f(x)|\le|f(x)|$ for every $n\in\Bbb N$ and $x\in\dom f$.   As
$\sequencen{y_n}$ is arbitrary, $\varhatf$ is continuous.

\medskip

{\bf (g)} This is just the Riemann-Lebesgue lemma (282E).

\medskip

{\bf (h)} The point is that $|\pd{}{y}e^{-iyx}f(x)|=|xf(x)|$ whenever
$x\in\dom f$ and $y\in\Bbb R$.   So by 123D

$$\eqalign{\varhatf\vthsp'(y)
&=\Bover1{\sqrt{2\pi}}\Bover{d}{dy}
     \int_{-\infty}^{\infty}e^{-iyx}f(x)dx
=\Bover1{\sqrt{2\pi}}\Bover{d}{dy}\int_{\dom f}
     e^{-iyx}f(x)dx\cr
&=\Bover1{\sqrt{2\pi}}\int_{\dom f}
     \Bover{\partial}{\partial y}e^{-iyx}f(x)dx
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
     -ixe^{-iyx}f(x)dx\cr
&=-\Bover{i}{\sqrt{2\pi}}\int_{-\infty}^{\infty}
     xe^{-iyx}f(x)dx.\cr}$$

\medskip

{\bf (i)} Because $f$ is absolutely continuous on every bounded
interval,

\Centerline{$f(x)=f(0)+\int_0^xf'$ for $x\ge 0$,\quad
$f(x)=f(0)-\int_x^0f'$ for $x\le 0$.}

\noindent Because $f'$ is integrable,

\Centerline{$\lim_{x\to\infty}f(x)=f(0)+\int_0^{\infty}f'$,\quad
$\lim_{x\to-\infty}f(x)=f(0)-\int_{-\infty}^0f'$}

\noindent both exist.   Because $f$ also is integrable, both limits must
be zero.   Now we can integrate by parts (225F) to see that

$$\eqalign{(f')\varsphat(y)
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-iyx}f'(x)dx
=\Bover1{\sqrt{2\pi}}\lim_{a\to\infty}\int_{-a}^{a}e^{-iyx}f'(x)dx\cr
&=\Bover1{\sqrt{2\pi}}\bigl(\lim_{a\to\infty}e^{-iya}f(a)
-\lim_{a\to-\infty}e^{-iya}f(a)\bigr)
+\Bover{iy}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-iyx}f(x)dx\cr
&=iy\varhatf(y).\cr}$$
}%end of proof of 283C

\leader{283D}{Lemma} (a) $\lim_{a\to\infty}\int_{0}^a\bover{\sin
x}{x}dx=\bover{\pi}2$, $\lim_{a\to\infty}\int_{-a}^a\bover{\sin
x}{x}dx=\pi$.

(b) There is a $K<\infty$ such that $|\int_a^b\bover{\sin cx}{x}dx|\le
K$ whenever $a\le b$ and $c\in\Bbb R$.

\proof{{\bf (a)(i)} Set

\Centerline{$F(a)=\int_0^a\Bover{\sin x}{x}dx$ if $a\ge 0$,\quad
$F(a)=-\int_{-a}^0\Bover{\sin x}{x}dx$ if $a\le 0$,}

\noindent so that $F(a)=-F(-a)$ and
$\int_a^b\bover{\sin x}{x}dx=F(b)-F(a)$ for all $a\le b$.

If $0<a\le b$, then by 224J

\Centerline{$|\int_a^b\Bover{\sin x}{x}dx|
\le(\Bover1b+\Bover1a-\Bover1b)\sup_{c\in[a,b]}
  |\int_a^c\sin x\,dx|
\le\Bover1a\sup_{c\in[a,b]}|\cos a-\cos c|
\le\Bover2a$.}

\noindent In particular, $|F(n)-F(m)|\le\bover2{m}$ if $0<m\le n$ in
$\Bbb N$, and $\sequencen{F(n)}$ is a Cauchy sequence with limit
$\gamma$ say;  now

\Centerline{$|\gamma - F(a)|
=\lim_{n\to\infty}|F(n)-F(a)|\le\Bover2a$}

\noindent for every $a>0$, so $\lim_{a\to\infty}F(a)=\gamma$.
Of course we also have

\Centerline{$\lim_{a\to\infty}\int_{-a}^a\Bover{\sin x}{x}dx
=\lim_{a\to\infty}(F(a)-F(-a))
=\lim_{a\to\infty}2F(a)=2\gamma$.}

\medskip

\quad{\bf (ii)} So now I have to calculate $\gamma$.    For this,
observe first that

\Centerline{$2\gamma
=\lim_{a\to\infty}\int_{-\pi a}^{\pi a}\Bover{\sin x}{x}dx
=\lim_{a\to\infty}\int_{-\pi}^{\pi}\Bover{\sin at}{t}dt$}

\noindent (substituting $x=t/a$).   Next,

\Centerline{$\lim_{t\to 0}\Bover1t-\Bover1{2\sin{1\over 2}t}=\lim_{u\to
0}\Bover{\sin u-u}{2u\sin u}=0$,}

\noindent so

\Centerline{$\int_{-\pi}^{\pi}
  \bigl|\Bover1t-\Bover1{2\sin{1\over 2}t}\bigr|dt<\infty$,}

\noindent and by the Riemann-Lebesgue lemma (282Fb)

\Centerline{$\lim_{a\to\infty}\int_{-\pi}^{\pi}
   \bigl(\Bover1t-\Bover1{2\sin{1\over 2}t}\bigr)\sin at\,dt=0$.}

\noindent But we know that

\Centerline{$\int_{-\pi}^{\pi}
\Bover{\sin(n+{1\over 2})t\pushbottom{3.5pt}}{2\sin{1\over 2}t}dt
=\pi$}

\noindent for every $n$ (using 282Dc), so we must have

$$\eqalign{\lim_{a\to\infty}\int_{-a}^a\Bover{\sin t}{t}dt
&=\lim_{a\to\infty}\int_{-\pi}^{\pi}\Bover{\sin at}{t}dt
=\lim_{a\to\infty}\int_{-\pi}^{\pi}
      \Bover{\sin at}{2\sin{1\over 2}t}dt\cr
&=\lim_{n\to\infty}\int_{-\pi}^{\pi}
    \Bover{\sin(n+{1\over 2})t\pushbottom{3.5pt}}{2\sin{1\over 2}t}dt
=\pi,\cr}$$

\noindent and $\gamma=\pi/2$, as claimed.

\medskip

{\bf (b)} Because $F$ is continuous and

\Centerline{$\lim_{a\to\infty}F(a)=\gamma=\Bover{\pi}2$, \quad
$\lim_{a\to-\infty}F(a)=-\gamma=-\Bover{\pi}2$,}

\noindent $F$ is bounded;  say $|F(a)|\le K_1$ for all $a\in\Bbb R$.
Try $K=2K_1$.   Now suppose that $a<b$ and $c\in\Bbb R$.   If $c>0$,
then

\Centerline{$|\int_a^b\Bover{\sin cx}{x}dx|
=|\int_{ac}^{bc}\Bover{\sin t}{t}dt|
=|F(bc)-F(ac)|\le 2K_1=K$,}

\noindent substituting $x=t/c$.   If $c<0$, then

\Centerline{$|\int_a^b\Bover{\sin cx}{x}dx|
=|-\int_a^b\Bover{\sin(-c)x}{x}dx|
\le K$;}

\noindent while if $c=0$ then

\Centerline{$|\int_a^b\Bover{\sin cx}{x}dx|
=0\le K$.}
}%end of proof of 283D

\leader{283E}{}\cmmnt{ The hardest work of this section will lie in
the `pointwise inversion theorems' 283I and 283K below.   I begin
however with a relatively easy, and at least equally important, result,
showing (among other things) that an integrable function $f$ can
(essentially) be recovered from its Fourier transform.

\medskip

\noindent}{\bf Lemma} Whenever $c<d$ in $\Bbb R$,

$$\eqalign{\lim_{a\to\infty}\int_{-a}^{a}
     e^{-iyx}\,\Bover{e^{idy}-e^{icy}}{y}dy
&=2\pi i\text{ if }c<x<d,\cr
&=\pi i\text{ if }x=c\text{ or }x=d,\cr
&=0\text{ if }x<c\text{ or } x>d.\cr}$$

\proof{ We know that for any $b>0$

\Centerline{$\lim_{a\to\infty}\int_{-a}^{a}\Bover{\sin bx}{x}dx
=\lim_{a\to\infty}\int_{-ab}^{ab}\Bover{\sin t}{t}dt
=\pi$}

\noindent (subsituting $x=t/b$), and therefore that for any $b<0$

\Centerline{$\lim_{a\to\infty}\int_{-a}^{a}\Bover{\sin bx}{x}dx
=-\lim_{a\to\infty}\int_{-a}^{a}\Bover{\sin(-b)x}{x}dx
=-\pi$.}

\noindent Now consider, for $x\in\Bbb R$,

\Centerline{$\lim_{a\to\infty}\int_{-a}^{a}e^{-iyx}
\,\Bover{e^{idy}-e^{icy}}{y}dy$.}

\noindent First note that all the integrals $\int_{-a}^a$ exist, because

\Centerline{$\lim_{y\to 0}\Bover{e^{idy}-e^{icy}}{y}=i(d-c)$}

\noindent is finite, and the integrand is certainly continuous except at
$0$.   Now we have

$$\eqalignno{\int_{-a}^ae^{-iyx}&\Bover{e^{idy}-e^{icy}}{y}dy\cr
&=\int_{-a}^a\Bover{e^{i(d-x)y}-e^{i(c-x)y}}{y}dy\cr
&=\int_{-a}^a\Bover{\cos(d-x)y-\cos(c-x)y}{y}dy
 +i\int_{-a}^a\Bover{\sin(d-x)y-\sin(c-x)y}{y}dy\cr
&=i\int_{-a}^a\Bover{\sin(d-x)y-\sin(c-x)y}{y}dy\cr}$$

\noindent because $\cos$ is an even function, so

\Centerline{$\int_{-a}^{a}\Bover{\cos(d-x)y-\cos(c-x)y}{y}dy=0$}

\noindent for every $a\ge 0$.   (Once again, this integral exists
because

\Centerline{$\lim_{y\to 0}\Bover{\cos(d-x)y-\cos(c-x)y}{y}=0$.)}

\noindent   Accordingly

$$\eqalign{\lim_{a\to\infty}\int_{-a}^a
e^{-iyx}\Bover{e^{idy}-e^{icy}}{y}dy
&=i\lim_{a\to\infty}\int_{-a}^a\Bover{\sin(d-x)y}{y}dy
   -i\lim_{a\to\infty}\int_{-a}^a\Bover{\sin(c-x)y}{y}dy\cr
&=i\pi-i\pi=0\text{ if }x<c,\cr
&=i\pi-0=\pi i\text{ if }x=c,\cr
&=i\pi+i\pi=2\pi i\text{ if }c<x<d,\cr
&=0+i\pi=\pi i\text{ if }x=d,\cr
&=-i\pi+i\pi=0\text{ if }x>d.\cr}$$
}%end of proof of 283E

\leader{283F}{Theorem} Let $f$ be a complex-valued function which is
integrable over $\Bbb R$, and $\varhatf$ its Fourier transform.   Then
whenever $c\le d$ in $\Bbb R$,

$$\int_c^df=\Bover{i}{\sqrt{2\pi}}\lim_{a\to\infty}
\int_{-a}^{a}\Bover{e^{icy}-e^{idy}}{y}\varhatf(y)dy.$$

\proof{ If $c=d$ this is trivial;  let us suppose that $c<d$.

\medskip

{\bf (a)} Writing

$$\theta_a(x)=\biggerint_{-a}^{a}e^{-iyx}
\,\Bover{e^{idy}-e^{icy}}{y}dy$$

\noindent for $x\in\Bbb R$ and $a\ge 0$, 283E tells us that

\Centerline{$\lim_{a\to\infty}\theta_a(x)=2\pi i\theta(x)$}

\noindent where $\theta=\bover12(\chi[c,d]+\chi\ooint{c,d})$ takes the
value $1$ inside the interval $[c,d]$, $0$ outside and the value
$\bover12$ at the endpoints.   At the same time,

$$\eqalignno{|\theta_a(x)|
&=|\int_{-a}^a\Bover{\sin(d-x)y-\sin(c-x)y}{y}dy|\cr
\displaycause{see the proof of 283E}
&\le|\int_{-a}^a\Bover{\sin(d-x)y}{y}dy|
  +|\int_{-a}^a\Bover{\sin(c-x)y}{y}dy|
\le 2K\cr}$$

\noindent for all $a\ge 0$ and $x\in\Bbb R$, where $K$ is the constant of
283Db.   Consequently $|f\times\theta_a|\le 2K|f|$ everywhere on
$\dom f$, for every $a\ge 0$, and
(applying Lebesgue's Dominated Convergence
Theorem to sequences $\sequencen{f\times\theta_{a_n}}$, where
$a_n\to\infty$)

\Centerline{$\lim_{a\to\infty}
\int f\times\theta_a=2\pi i\int f\times\theta=2\pi i\int_c^df$.}

\medskip

{\bf (b)} Now consider the limit in the statement of the theorem.   We
have

$$\eqalignno{\int_{-a}^a
\bover{e^{icy}-e^{idy}}{y}\varhatf(y)dy
&=\bover1{\sqrt{2\pi}}\int_{-a}^a
\int_{-\infty}^{\infty}
\bover{e^{icy}-e^{idy}}{y}e^{-iyx}f(x)dxdy\cr
&=\bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}\int_{-a}^a
\bover{e^{icy}-e^{idy}}{y}e^{-iyx}f(x)dydx\cr
&=-\bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}f\times\theta_a\cr}$$

\noindent by Fubini's and Tonelli's theorems (252H), using the fact that
$(e^{icy}-e^{idy})/y$ is bounded to see that

\Centerline{$\int_{-\infty}^{\infty}\int_{-a}^{a}
  \bigl|\bover{e^{icy}-e^{idy}}{y}e^{-iyx}f(x)\bigr|dydx$}

\noindent is finite.   Accordingly

$$\eqalign{\bover{i}{\sqrt{2\pi}}\lim_{a\to\infty}\int_{-a}^a
\bover{e^{icy}-e^{idy}}{y}\varhatf(y)dy
&=-\bover{i}{2\pi}\lim_{a\to\infty}
\int_{-\infty}^{\infty}f\times\theta_a\cr
&=-\bover{i}{2\pi}2\pi i\int_c^df
=\int_c^df,\cr}$$

\noindent as required.
}%end of proof of 283F

\leader{283G}{Corollary} If $f$ and $g$ are
complex-valued functions which are integrable over $\Bbb R$, then
$\varhatf=\varhat{g}$ iff $f\eae g$.

\proof{ If $f\eae g$ then of course

\Centerline{$\varhatf(y)
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-iyx}f(x)dx
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-iyx}g(x)dx
=\varhat{g}(y)$}

\noindent for every $y\in\Bbb R$.   Conversely, if
$\varhatf=\varhat{g}$, then by the last theorem

\Centerline{$\int_c^df=\int_c^dg$}

\noindent for all $c\le d$, so $f=g$ almost everywhere, by 222D.
}%end of proof of 283G

\leader{283H}{Lemma} Let $f$ be a complex-valued function which is
integrable over $\Bbb R$, and $\varhatf$ its Fourier transform.   Then

\Centerline{$\Bover1{\sqrt{2\pi}}\int_{-a}^{a}e^{ixy}\varhatf(y)dy
=\Bover1{\pi}\int_{-\infty}^{\infty}\Bover{\sin a(x-t)}{x-t}f(t)dt
=\Bover1{\pi}\int_{-\infty}^{\infty}\Bover{\sin at}{t}f(x-t)dt$}

\noindent whenever $a>0$ and $x\in\Bbb R$.

\proof{ We have

\Centerline{$\int_{-a}^a\int_{-\infty}^{\infty}|e^{ixy}e^{-iyt}f(t)|dtdy
\le 2a\int_{-\infty}^{\infty}|f(t)|dt<\infty$,}

\noindent so (because the function $(t,y)\mapsto e^{ixy}e^{-iyt}f(t)$ is
surely jointly measurable) we may reverse the order of integration, and
get

$$\eqalign{\Bover1{\sqrt{2\pi}}\int_{-a}^ae^{ixy}\varhatf(y)dy
&=\Bover1{2\pi}\int_{-a}^a\int_{-\infty}^{\infty}
   e^{ixy}e^{-iyt}f(t)dt\,dy\cr
&=\Bover1{2\pi}\int_{-\infty}^{\infty}f(t)\int_{-a}^a
   e^{i(x-t)y}dy\,dt\cr
&=\Bover1{2\pi}\int_{-\infty}^{\infty}
   \Bover{2\sin(x-t)a}{x-t}f(t)dt
=\Bover1{\pi}\int_{-\infty}^{\infty}
   \Bover{\sin au}{u}f(x-u)du,\cr}$$

\noindent substituting $t=x-u$.
}%end of proof of 283H


\leader{283I}{Theorem} Let $f$ be a complex-valued function which is
integrable over $\Bbb R$, and suppose that $f$ is differentiable at
$x\in\Bbb R$.   Then

\Centerline{$f(x)
=\Bover{1}{\sqrt{2\pi}}
\lim_{a\to\infty}\int_{-a}^ae^{ixy}\varhatf(y)dy
=\Bover{1}{\sqrt{2\pi}}
\lim_{a\to\infty}\int_{-a}^ae^{-ixy}\varcheckf(y)dy$.}

\proof{ Set $g(u)=f(x)$ if $|u|\le 1$, $0$ otherwise, and observe that
$\lim_{u\to 0}\bover1u(f(x-u)-g(u))=-f'(x)$ is finite, so that there is
a $\delta\in\ocint{0,1}$ such that

\Centerline{$K
=\sup_{0<|u|\le\delta}\bigl|\Bover{f(x-u)-g(u)}{u}\bigr|<\infty$.}

\noindent Consequently

$$\eqalign{\int_{-\infty}^{\infty}\bigl|\Bover{f(x-u)-g(u)}{u}\bigr|du
&\le\Bover1{\delta}\int_{-\infty}^{-\delta}|f(x-u)|du
   +\Bover1{\delta}\int_{-1}^1|g|\cr
&{\hskip 8em plus 0pt minus 0pt}
   +\int_{-\delta}^{\delta}K
   +\Bover1{\delta}\int_{\delta}^{\infty}|f(x-u)|du\cr
&\le\Bover{1}{\delta}\int_{-\infty}^{\infty}|f|
   + \Bover2{\delta}|f(x)|+ 2\delta K
<\infty.\cr}$$

\noindent By the Riemann-Lebesgue lemma (282Fb again),

\Centerline{$\lim_{a\to\infty}\int_{-\infty}^{\infty}
\Bover{\sin au}{u}(f(x-u)-g(u))du=0$.}

\noindent If we now examine $\int\bover{\sin au}{u}g(u)du$, we get

$$\eqalign{\int_{-\infty}^{\infty}\Bover{\sin au}{u}g(u)du
&=\int_{-1}^1\Bover{\sin au}{u}f(x)du
=f(x)\int_{-a}^a\Bover{\sin v}{v}dv,\cr}$$

\noindent substituting $u=v/a$.   So we get

$$\eqalign{\lim_{a\to\infty}\int_{-\infty}^{\infty}\Bover{\sin
au}{u}f(x-u)du
&=\lim_{a\to\infty}\int_{-\infty}^{\infty}\Bover{\sin au}{u}g(u)du\cr
&=\lim_{a\to\infty}f(x)\int_{-a}^a\Bover{\sin v}{v}dv
=\pi f(x),\cr}$$

\noindent by 283Da.   Accordingly

$$\Bover{1}{\sqrt{2\pi}}
\lim_{a\to\infty}\int_{-a}^ae^{ixy}\varhatf(y)dy
=\Bover{1}{\pi}\lim_{a\to\infty}\int_{-\infty}^{\infty}\Bover{\sin
au}{u}f(x-u)du
= f(x),$$

\noindent using 283H.    As for the second equality,

$$\eqalignno{\Bover1{\sqrt{2\pi}}\lim_{a\to\infty}\int_{-a}^a
e^{-ixy}\varcheckf(y)dy
&=\Bover1{\sqrt{2\pi}}\lim_{a\to\infty}\int_{-a}^a
e^{-ixy}\varhatf(-y)dy\cr
&=\Bover1{\sqrt{2\pi}}\lim_{a\to\infty}\int_{-a}^a
e^{ixu}\varhatf(u)du
=f(x)\cr}$$

\noindent (substituting $y=-u$).
}%end of proof of 283I

\cmmnt{\medskip

\noindent{\bf Remark} Compare 282L.}%end of comment

\leader{283J}{Corollary} Let $f:\Bbb R\to\Bbb C$ be an integrable
function such that $f$ is differentiable and $\varhatf$ is integrable.
Then $f=(\varhatf)\varspcheck=(\varcheckf)\varsphat$.

\proof{ Because $\varhatf$ is integrable,

\Centerline{$\varhatf\varcheck{\phantom{f}}(x)=\lim_{a\to\infty}
\Bover1{\sqrt{2\pi}}\int_{-a}^ae^{ixy}\varhatf(y)dy = f(x)$}

\noindent for every $x\in\Bbb R$.   Similarly,

\Centerline{$\varcheckf\varhat{\phantom{f}}(x)=\lim_{a\to\infty}
\Bover1{\sqrt{2\pi}}\int_{-a}^ae^{-ixy}\varcheckf(y)dy = f(x).$}
}%end of proof of 283J

\cmmnt{\medskip

\noindent{\bf Remark} See also 283Wk below.
}

\leader{283K}{}\cmmnt{ The next proposition gives a class of functions
to which the last corollary can be applied.

\medskip

\noindent}{\bf Proposition} Suppose that $f$ is a twice-differentiable
function from $\Bbb R$ to $\Bbb C$ such that $f$, $f'$ and $f''$ are all
integrable.   Then $\varhatf$ is integrable.

\proof{ Because $f'$ and $f''$ are integrable, $f$ and $f'$
are absolutely continuous on any bounded interval (225L).   So by 283Ci
we have

\Centerline{$(f'')\varsphat(y)=iy(f')\varsphat(y)=-y^2\varhatf(y)$}

\noindent for every $y\in\Bbb R$.   At the same time, by 283Cf-283Cg,
$(f'')\varsphat$ and $\varhatf$ must be bounded;  say
$|\varhatf(y)|+|(f'')\varsphat(y)|\le K$ for every $y\in\Bbb R$.   Now

\Centerline{$|\varhatf(y)|\le\Bover{K}{1+y^2}$}

\noindent for every $y$, so that

\Centerline{$\int_{-\infty}^{\infty}|\varhatf|
\le K\int_{-\infty}^{-1}\Bover{1}{y^2}dy
  +2K+K\int_{1}^{\infty}\Bover{1}{y^2}dy
=4K<\infty$.}
}%end of proof of 283K

\cmmnt{\medskip

\noindent{\bf Remark} Compare 282Rb.}

\leader{283L}{}\cmmnt{ I turn now to the result corresponding to 282O,
using a slightly different approach.

\medskip

\noindent}{\bf Theorem} Let $f$ be a complex-valued function which is
integrable over $\Bbb R$, with Fourier transform $\varhatf$ and inverse
Fourier transform $\varcheckf$, and suppose that $f$ is of bounded
variation on some neighbourhood of $x\in\Bbb R$.   Set
$a=\lim_{t\in\dom f,t\uparrow x}f(t)$,
$b=\lim_{t\in\dom f,t\downarrow x}f(t)$.   Then

\Centerline{$\Bover1{\sqrt{2\pi}}\lim_{\gamma\to\infty}
     \int_{-\gamma}^{\gamma}e^{ixy}\varhatf(y)dy
=\Bover1{\sqrt{2\pi}}\lim_{\gamma\to\infty}
     \int_{-\gamma}^{\gamma}e^{-ixy}\varcheckf(y)dy
=\Bover12(a+b)$.}

\proof{{\bf (a)} The limits $\lim_{t\in\dom f,t\uparrow x}f(t)$ and
$\lim_{t\in\dom f,t\downarrow x}f(t)$ exist because $f$ is of bounded
variation near $x$ (224F).   Recall from 283Db that there is a constant
$K<\infty$ such that

\Centerline{$|\int_{\gamma}^{\delta}\Bover{\sin cx}{x}dx|\le K$}

\noindent whenever $\gamma\le \delta$ and $c\in\Bbb R$.

\medskip

{\bf (b)} Let $\epsilon>0$.   The hypothesis is that there is some
$\delta>0$ such that $\Var_{[x-\delta,x+\delta]}(f)<\infty$.
Consequently

\Centerline{$\lim_{\eta\downarrow 0}\Var_{\ocint{x,x+\eta}}(f)
=\lim_{\eta\downarrow 0}\Var_{\coint{x-\eta,x}}(f)=0$}

\noindent (224E).   There is therefore an $\eta>0$ such that

\Centerline{$\max(\Var_{\coint{x-\eta,x}}(f),
               \Var_{\ocint{x,x+\eta}}(f))\le\epsilon$.}

\noindent Of course

\Centerline{$|f(t)-f(u)|\le\Var_{\coint{x-\eta,x}}(f)\le\epsilon$}

\noindent whenever $t$, $u\in\dom f$ and $x-\eta\le t\le u<x$, so we
shall have

\Centerline{$|f(t)-a|\le\epsilon$ for every
$t\in\dom f\cap\coint{x-\eta,x}$,}

\noindent and similarly

\Centerline{$|f(t)-b|\le\epsilon$ whenever $t\in\dom f
\cap\ocint{x,x+\eta}$.}

\medskip

{\bf (c)} Now set

\Centerline{$g_1(t)=f(t)$ when $t\in\dom f$ and $|x-t|>\eta$, $0$
otherwise,}

\Centerline{$g_2(t)=a$ when $x-\eta\le t<x$, $b$ when $x<t\le x+\eta$,
$0$ otherwise,}

\Centerline{$g_3=f-g_1-g_2$.}

\noindent Then $f=g_1+g_2+g_3$;  each $g_j$ is integrable;  $g_1$ is
zero on a neighbourhood of $x$;

\Centerline{$\sup_{t\in\dom g_3,t\ne x}|g_3(t)|\le\epsilon$,}

\Centerline{$\Var_{\coint{x-\eta,x}}(g_3)\le\epsilon$,
\quad $\Var_{\ocint{x,x+\eta}}(g_3)\le\epsilon$.}

\medskip

{\bf (d)} Consider the three parts $g_1$, $g_2$, $g_3$ separately.

\medskip

\quad{\bf (i)} For the first, we have

\Centerline{$\lim_{\gamma\to\infty}\Bover1{\sqrt{2\pi}}
   \int_{-\gamma}^{\gamma}e^{ixy}\varhat{g}_1(y)dy=0$}

\noindent by 283I.

\medskip

\quad{\bf (ii)} Next,

$$\eqalignno{\Bover{1}{\sqrt{2\pi}}
        \int_{-\gamma}^{\gamma}e^{ixy}\varhat{g}_2(y)dy
&=\Bover1{\pi}\int_{-\infty}^{\infty}
        \Bover{\sin(x-t)\gamma}{x-t}g_2(t)dt\cr
\noalign{\noindent (by 283H)}
&=\Bover{a}{\pi}\int_{x-\eta}^{x}\Bover{\sin(x-t)\gamma}{x-t}dt
   +\Bover{b}{\pi}\int_{x}^{x+\eta}\Bover{\sin(x-t)\gamma}{x-t}dt\cr
&=\Bover{a}{\pi}\int_0^{\gamma\eta}\Bover{\sin u}{u}du
   +\Bover{b}{\pi}\int_{0}^{\gamma\eta}\Bover{\sin u}{u}du\cr
\noalign{\noindent (substituting $t=x-\bover{1}{\gamma}u$ in the first
integral, $t=-x+\bover1{\gamma}u$ in the second)}
&\to \Bover{a+b}{2}\text{ as }\gamma\to\infty\cr}$$

\noindent by 283Da.

\medskip

\quad{\bf (iii)} As for the third, we have, for any $\gamma>0$,

$$\eqalignno{\bigl|\Bover1{\sqrt{2\pi}}\int_{-\gamma}^{\gamma}e^{ixy}
     \varhat{g}_3(y)dy\bigr|
&=\Bover1{\pi}\bigl|\int_{-\infty}^{\infty}
     \Bover{\sin(x-t)\gamma}{x-t}g_3(t)dt\bigr|
=\Bover1{\pi}\bigl|\int_{-\infty}^{\infty}
     \Bover{\sin t\gamma}{t}g_3(x-t)dt\bigr|\cr
&\le\Bover1{\pi}\bigl|\int_{-\eta}^0
     \Bover{\sin t\gamma}{t}g_3(x-t)dt\bigr|
   +\Bover1{\pi}\bigl|\int_0^{\eta}
     \Bover{\sin t\gamma}{t}g_3(x-t)dt\bigr|\cr
&\le \Bover{K}{\pi}\Bigl(\sup_{t\in\dom g_3\cap\ooint{x-\eta,x}}|g_3(t)|
   +\Var_{\ooint{x-\eta,x}}(g_3)\cr
  &{\hskip 8em plus 0pt minus 0pt}
   +\sup_{t\in\dom g_3\cap\ooint{x,x+\eta}}|g_3(t)|
   +\Var_{\ooint{x,x+\eta}}(g_3)\Bigr)\cr
&\le 4\epsilon\Bover{K}{\pi},\cr}$$

\noindent using 224J again
to bound the integrals in terms of the variation
and supremum of $g_3$ and integrals of $\bover{\sin\gamma t}{t}$ over
subintervals.

\medskip

{\bf (e)} We therefore have

$$\eqalign{\limsup_{\gamma\to\infty}\bigl|\Bover1{\sqrt{2\pi}}
        \int_{-\gamma}^{\gamma}
   &e^{ixy}\varhatf(y)dy-\Bover{a+b}{2}\bigr|\cr
&\le\limsup_{\gamma\to\infty}\Bover1{\sqrt{2\pi}}
     \bigl|\int_{-\gamma}^{\gamma}e^{ixy}\varhat{g}_1(y)dy\bigr|\cr
&{\hskip 6em plus 0pt minus 0pt}
     +\limsup_{\gamma\to\infty}\bigl|\Bover1{\sqrt{2\pi}}
     \int_{-\gamma}^{\gamma}e^{ixy}\varhat{g}_2(y)dy
     -\Bover{a+b}{2}\bigr|\cr
&{\hskip 6em plus 0pt minus 0pt}
     +\limsup_{\gamma\to\infty}\Bover1{\sqrt{2\pi}}
     \bigl|\int_{-\gamma}^{\gamma}e^{ixy}\varhat{g}_3ydy\bigr|\cr
&\le 0+0+\Bover{4K}{\pi}\epsilon\cr}$$

\noindent by the calculations in (d).   As $\epsilon$ is arbitrary,

\Centerline{$\lim_{\gamma\to\infty}\Bover1{\sqrt{2\pi}}
     \int_{-\gamma}^{\gamma}e^{ixy}\varhatf(y)dy-\Bover{a+b}{2}=0$.}

\medskip

{\bf (f)} This is the first half of the theorem.   But of course the
second half follows at once, because

$$\eqalign{\Bover1{\sqrt{2\pi}}\lim_{\gamma\to\infty}
    \int_{-\gamma}^{\gamma}e^{-ixy}\varcheckf(y)dy
&=\Bover1{\sqrt{2\pi}}\lim_{\gamma\to\infty}
    \int_{-\gamma}^{\gamma}e^{-ixy}\varhatf(-y)dy\cr
&=\Bover1{\sqrt{2\pi}}\lim_{\gamma\to\infty}
    \int_{-\gamma}^{\gamma}e^{ixy}\varhatf(y)dy
=\Bover{a+b}2.\cr}$$
}%end of proof of 283L

\cmmnt{\medskip

\noindent{\bf Remark} You will see that this argument uses some of the
same ideas as those in 282O-282P.   It is more direct because (i) I am
not using any concept corresponding to Fej\'er sums (though a very
suitable one is available;  see 283Xf)
(ii) I do not trouble to give the result concerning uniform convergence
of the Fej\'er integrals when $f$ is continuous and of bounded variation
(283Xj)
(iii) I do not give any pointer to the significance of the fact that if
$f$ is of bounded variation then $\sup_{y\in\Bbb
R}|y\varhatf(y)|<\infty$ (283Xk).
}%end of comment

\leader{283M}{}\cmmnt{ Corresponding to 282Q, we have the following.

\medskip

\noindent}{\bf Theorem} Let $f$ and $g$ be complex-valued functions
which are integrable over $\Bbb R$, and $f*g$ their convolution product,
defined by setting

\Centerline{$(f*g)(x)=\int_{-\infty}^{\infty}f(t)g(x-t)dt$}

\noindent whenever this is defined\cmmnt{ (255E)}.   Then

\Centerline{$(f*g)\varsphat(y)
=\sqrt{2\pi}\varhatf(y)\varhat{g}(y)$,\quad
$(f*g)\varspcheck(y)
=\sqrt{2\pi}\varcheckf(y)\varcheck{g}(y)$}

\noindent for every $y\in\Bbb R$.

\proof{ For any $y$,

$$\eqalignno{(f*g)\varsphat(y)
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-iyx}(f*g)(x)dx\cr
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
e^{-iy(t+u)}f(t)g(u)dtdu\cr
\noalign{\noindent (using 255G)}
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-iyt}f(t)dt
\int_{-\infty}^{\infty}e^{-iyu}g(u)du
=\sqrt{2\pi}\varhatf(y)\varhat{g}(y).\cr}$$

\noindent Now, of course,

\Centerline{$(f*g)\varspcheck(y)
=(f*g)\varsphat(-y)
=\sqrt{2\pi}\varhatf(-y)\varhat{g}(-y)
=\sqrt{2\pi}\varcheckf(y)\varcheck{g}(y)$.}
}%end of proof of 283M

\leader{283N}{}\cmmnt{ I show how to compute a special Fourier
transform, which
will be used repeatedly in the next section.

\medskip

\noindent}{\bf Lemma} For $\sigma>0$, set
$\psi_{\sigma}(x)=\bover1{\sigma\sqrt{2\pi}}e^{-x^2/2\sigma^2}$ for
$x\in\Bbb R$.   Then its Fourier transform and inverse Fourier transform
are

\Centerline{$\varhat{\psi}_{\sigma}=\varcheck{\psi}_{\sigma}
=\Bover1{\sigma}\psi_{1/\sigma}$.}

\noindent In particular, $\varhat{\psi}_1=\psi_1$.

\proof{{\bf (a)} I begin with the special case $\sigma=1$, using the
Maclaurin series

\Centerline{$e^{-iyx}=\sum_{k=0}^{\infty}\Bover{(-iyx)^k}{k!}$}

\noindent and the expressions for
$\int_{-\infty}^{\infty}x^ke^{-x^2/2}dx$ from \S263.

Fix $y\in\Bbb R$.    Writing

\Centerline{$g_k(x)=\Bover{(-iyx)^k}{k!}e^{-x^2/2}$,\quad
$h_n(x)=\sum_{k=0}^ng_k(x)$, \quad$h(x)=e^{|yx|-x^2/2}$,}

\noindent we see that

\Centerline{$|g_k(x)|\le \Bover{|yx|^k}{k!}e^{-x^2/2}$,}

\noindent so that

\Centerline{$|h_n(x)|\le\sum_{k=0}^{\infty}|g_k(x)|
\le e^{|yx|}e^{-x^2/2}=h(x)$}

\noindent for every $n$;  moreover, $h$ is integrable, because
$|h(x)|\le e^{-|x|}$ whenever $|x|\ge 2(1+|y|)$.   Consequently, using
Lebesgue's Dominated Convergence Theorem,

$$\eqalignno{\varhat{\psi}_1(y)
&=\Bover1{{2\pi}}\int_{-\infty}^{\infty}
  \lim_{n\to\infty}h_n\
=\Bover1{{2\pi}}\lim_{n\to\infty}\int_{-\infty}^{\infty}
  h_n\cr
&=\Bover1{{2\pi}}\sum_{k=0}^{\infty}
  \int_{-\infty}^{\infty}g_k
=\Bover1{{2\pi}}\sum_{k=0}^{\infty}\Bover{(-iy)^k}{k!}
  \int_{-\infty}^{\infty}x^ke^{-x^2/2}dx\cr
&=\Bover1{{2\pi}}\sum_{j=0}^{\infty}\Bover{(-iy)^{2j}}{(2j)!}
  \Bover{(2j)!}{2^jj!}\sqrt{2\pi}\cr
\noalign{\noindent (by 263H)}
&=\Bover1{\sqrt{2\pi}}\sum_{j=0}^{\infty}\Bover{(-y^2)^j}{2^jj!}
=\Bover1{\sqrt{2\pi}}e^{-y^2/2}
=\psi_1(y),\cr}$$

\noindent as claimed.

\medskip

{\bf (b)} For the general case,
$\psi_{\sigma}(x)=\Bover1{\sigma}\psi_1(\Bover{x}{\sigma})$, so that

\Centerline{$\varhat{\psi}_{\sigma}(y)
=\Bover1{\sigma}\cdot\sigma\varhat{\psi}_1(\sigma y)
=\Bover1\sigma\psi_{1/\sigma}(y)$}

\noindent by 283Ce.   Of course we now have

\Centerline{$\varcheck{\psi}_{\sigma}(y)=\varhat{\psi}_{\sigma}(-y)
=\Bover1{\sigma}\psi_{1/\sigma}(y)$}

\noindent because $\psi_{1/\sigma}$ is an even function.
}%end of proof of 283N


\leader{283O}{}\cmmnt{ To lead into the ideas of the next section, I
give the following very simple fact.

\medskip

\noindent}{\bf Proposition} Let $f$ and $g$ be two
complex-valued functions which are integrable over $\Bbb R$.   Then
$\int_{-\infty}^{\infty}f\times\varhat{g}
=\int_{-\infty}^{\infty}\varhatf\times g$ and
$\int_{-\infty}^{\infty}f\times\varcheck{g}
  =\int_{-\infty}^{\infty}\varcheckf\times g$.

\proof{  Of course

\Centerline{$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
|e^{-ixy}f(x)g(y)|dxdy
=\int_{-\infty}^{\infty}|f|\int_{-\infty}^{\infty}|g|<\infty$,}

\noindent so

$$\eqalign{\int_{\infty}^{\infty}f\times\varhat{g}
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
f(y)e^{-iyx}g(x)dxdy\cr
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
f(y)e^{-ixy}g(x)dydx
=\int_{-\infty}^{\infty}\varhatf\times g.\cr}$$

\noindent For the other half of the proposition, replace every
$e^{-ixy}$ in the argument by $e^{ixy}$.
}%end of proof of 283O

\exercises{\leader{283W}{Higher dimensions} I offer a series of
exercises designed to provide hints on how the work of this section may
be done in the
$r$-dimensional case, where $r\ge 1$.

\spheader 283Wa Let $f$ be an integrable complex-valued function
defined almost everywhere in $\BbbR^r$.   Its {\bf Fourier transform}
is the function $\varhatf:\BbbR^r\to\Bbb C$ defined by the formula

\Centerline{$\varhatf(y)
=\Bover1{(\sqrt{2\pi})^r}\int e^{-iy\dotproduct x}f(x)dx$,}

\noindent writing $y\dotproduct x=\eta_1\xi_1+\ldots+\eta_r\xi_r$ for
$x=(\xi_1,\ldots,\xi_r)$ and $y=(\eta_1,\ldots,\eta_r)\in\BbbR^r$, and
$\int\ldots dx$ for integration with respect to Lebesgue measure on
$\BbbR^r$.   Similarly, the {\bf inverse Fourier transform} of $f$ is
the function $\varcheckf$ given by

\Centerline{$\varcheckf(y)
=\Bover1{(\sqrt{2\pi})^r}\int e^{iy\dotproduct x}f(x)dx=\varhatf(-y)$.}

\noindent Show that, for any integrable complex-valued function $f$ on
$\BbbR^r$,

\quad (i) $\varhatf:\BbbR^r\to\Bbb C$ is continuous;

\quad (ii) $\lim_{\|y\|\to\infty}\varhatf(y)=0$, writing
$\|y\|=\sqrt{y\dotproduct y}$ as usual;

\quad (iii) if $\int\|x\||f(x)|dx<\infty$, then $\varhatf$ is
differentiable, and

\Centerline{$\Pd{}{\eta_j}\varhatf(y)
=-\Bover{i}{(\sqrt{2\pi})^r}\int e^{-iy\dotproduct x}\xi_jf(x)dx$}

\noindent for $j\le r$, $y\in\BbbR^r$, always taking $\xi_j$ to be the
$j$th coordinate of $x\in\BbbR^r$;

\quad (iv) if $j\le r$ and $\pd{f}{\xi_j}$ is defined everywhere and is
integrable, then
$(\pd{f}{\xi_j})\varsphat(y)=i\eta_j\varhatf(y)$
for every $y\in\Bbb R^r$.   \prooflet{(Use 225L to show that if
$e\in\BbbR^r$ is a unit vector, then
$\gamma\mapsto f(x+\gamma e)$ is absolutely continuous on every bounded
interval for almost every $x$.)}
%283C

\spheader 283Wb Show that if $f_1,\ldots,f_r$ are integrable
complex-valued functions on $\Bbb R$ with Fourier transforms
$g_1,\ldots,g_r$, and we write $f(x)=f_1(\xi_1)\ldots f_r(\xi_r)$ for
$x=(\xi_1,\ldots,\xi_r)\in\BbbR^r$, then the Fourier transform of $f$
is $y\mapsto g_1(\eta_1)\ldots g_r(\eta_r)$.

\spheader 283Wc Let $f$ be an integrable complex-valued function
on $\BbbR^r$, and $\varhatf$ its Fourier transform.   If $c\le d$ in
$\BbbR^r$, show that

$$\int_{[c,d]}f
=(\bover{i}{\sqrt{2\pi}})^r
\lim_{\alpha_1,\ldots,\alpha_r\to\infty}
\int_{[-a,a]}\prod_{j=1}^r
\bover{e^{i\gamma_j\eta_j}-e^{i\delta_j\eta_j}}{\eta_j}\varhatf(y)dy,$$

\noindent setting $a=(\alpha_1,\ldots)$, $c=(\gamma_1,\ldots)$,
$d=(\delta_1,\ldots)$.
%283F

\spheader 283Wd
Let $f$ be an integrable complex-valued function on $\BbbR^r$, and
$\varhatf$ its Fourier transform.   Show that if we write

\Centerline{$B_{\infty}(\tbf{0},a)=\{y:|\eta_j|\le a
  \text{ for every }j\le r\}$,}

\noindent then

\Centerline{$\Bover1{(\sqrt{2\pi})^r}\int_{B_{\infty}(\tbf{0},a)}
e^{ix\dotproduct y}\varhatf(y)dy
=\int\phi_a(t)f(x-t)dt$}

\noindent for every $a\ge 0$, where

\Centerline{$\phi_a(t)
=\Bover1{\pi^r}\prod_{j=1}^r\Bover{\sin a\tau_j}{\tau_j}$}

\noindent for $t=(\tau_1,\ldots,\tau_r)\in\BbbR^r$.
%283H

\spheader 283We Show that
$\biggerint_{\BbbR^r}\Bover{1}{1+\|x\|^{r+1}}dx<\infty$.

\spheader 283Wf Let $f:\BbbR^r\to\Bbb C$ be an integrable function.
Show that if all the partial derivatives
$\Bover{\partial^kf}{\partial\xi_j^k}$, for $k\le r+1$ and
$j\le r$, are defined almost everywhere and integrable,
then $\varhat{f}$ is integrable.
%283Wa(iii) 283We 283K

\spheader 283Wg\dvAformerly{2{}83Wj}
Show that if $f$ and $g$ are integrable
complex-valued functions on $\BbbR^r$, then (defining convolution as in
255L) $(f*g)\varsphat=(\sqrt{2\pi})^r\varhatf\times\varhat{g}$.
%283M

\spheader 283Wh Let $f$ and $g$ be integrable complex-valued
functions on $\BbbR^r$.   Show that
$f*\varcheck{g}=(\sqrt{2\pi})^r(\varhatf\times g)\varspcheck$.
%283M 283Wg

\spheader 283Wi\dvAformerly{2{}83We} For $\sigma>0$, define
$\psi_{\sigma}:\Bbb R^r\to\Bbb C$ by setting

\Centerline{$\psi_{\sigma}(x)=\Bover1{(\sigma\sqrt{2\pi})^r}
e^{-x\dotproduct x/2\sigma^2}$,
\quad$(\varhat{\psi}_{\sigma})\varspcheck=\psi_{\sigma}$.}

\noindent for every $x\in\BbbR^r$.   Show that

\Centerline{$\varhat{\psi}_{\sigma}
=\varcheck{\psi}_{\sigma}=\Bover1{\sigma^r}\psi_{1/\sigma}$.}
%283N

\spheader 283Wj Defining $\psi_{\sigma}$ as in (e), show that
$\lim_{\sigma\to 0}(f*\psi_{\sigma})(x)=f(x)$ whenever $x\in\BbbR^r$ and
$f:\BbbR^r\to\Bbb C$ is continuous and either integrable or bounded.
(Cf.\ 261Ye, 262Yi.)
%283Wi

\spheader 283Wk Show that if $f:\BbbR^r\to\Bbb C$ is continuous
and integrable, and $\varhatf$ also is integrable, then
$f=\varhatf\varcheck{\phantom{f}}$.
({\it Hint\/}:  Show that both are equal at every point to
$\lim_{\sigma\to 0}(\sqrt{2\pi})^r
  (\varhatf\times\varhat{\psi}_{\sigma})\varspcheck$.)
%283Wj 283Wh 283J

\spheader 283Wl Show that if $f$ and $g$ are integrable
complex-valued functions on $\BbbR^r$, then $\int
f\times\varhat{g}=\int\varhatf\times g$.

\spheader 283Wm(i) Show that
$\int_{2k\pi}^{2(k+1)\pi}\bover{\sin t}{t\sqrt{t}}dt>0$ for every
$k\in\Bbb N$, and hence that
$\int_0^{\infty}\bover{\sin t}{t\sqrt{t}}dt>0$.

\quad{(ii)} Set $f_1(\xi)=1/\sqrt{|\xi|}$ for $0<|\xi|\le 1$, $0$ for
other $\xi$.   Show that
$\lim_{a\to\infty}\bover1{\sqrt{a}}\int_{-a}^a\varhatf_1(\eta)d\eta$
exists in $\Bbb R$ and is greater than $0$.

\quad{(iii)} Construct an integrable function $f_2$, zero on some
neighbourhood of $0$, such that there are infinitely many $m\in\Bbb N$
for which $|\int_{-m}^m\varhatf_2(\eta)d\eta|\ge\bover1{\sqrt{m}}$.
\Hint{take $f_2(\xi)=2^{-k}\sin m_k\xi$ for $k+1\le\xi<k+2$, for a
sufficiently rapidly increasing sequence $\sequence{k}{m_k}$.}

\quad{(iv)} Set $f(x)=f_1(\xi_1)f_2(\xi_2)$ for $x\in\BbbR^2$.   Show
that $f$ is integrable, that $f$ is zero in a neighbourhood of
$\tbf{0}$, but that

\Centerline{$\limsup_{a\to\infty}\Bover1{2\pi}|\int_{B_{\infty}(\tbf{0},
a)}\varhatf(y)dy|>0$,}

\noindent defining $B_{\infty}$ as in 283Wd.
}%end of 283W

\exercises{
\leader{283X}{Basic exercises (a)}
%\spheader 283Xa
Confirm that the six alternative
definitions of the transforms $\varhatf$, $\varcheckf$ offered in 283B
all lead to the same theory;  find the constants involved in the new
versions of 283Ch, 283Ci, 283L, 283M and 283N.
%283B

\spheader 283Xb If we redefined $\varhatf(y)$ to be
$\alpha\int_{-\infty}^{\infty}e^{i\beta xy}f(x)dx$, what would
$\varcheckf(y)$ be?
%283B

\spheader 283Xc Show that nearly every $2\pi$ would disappear
from the theorems of this section if we defined a measure $\nu$ on
$\Bbb R$ by saying that $\nu E=\bover1{\sqrt{2\pi}}\mu E$ for every Lebesgue measurable set $E$, where $\mu$ is Lebesgue measure, and wrote

\Centerline{$\varhatf(y)=\int_{-\infty}^{\infty}e^{-iyx}f(x)\nu(dx)$,
\quad$\varcheckf(y)=\int_{-\infty}^{\infty}e^{iyx}f(x)\nu(dx)$,}

\Centerline{$(f*g)(x)=\int_{-\infty}^{\infty}f(t)g(x-t)\nu(dt)$.}

\noindent What is
$\lim_{a\to\infty}\int_{-a}^a\bover{\sin t}{t}\nu(dt)$?
%283B+

\sqheader 283Xd Let $f$ be an integrable complex-valued function
on $\Bbb R$, with Fourier transform $\varhatf$.   Show that
(i) if $g(x)=f(-x)$ whenever this
is defined, then $\varhat{g}(y)=\varhatf(-y)$ for
every $y\in\Bbb R$;
(ii) if $g(x)=\overline{f(x)}$ whenever this is defined, then
$\varhat{g}(y)=\overline{\varhatf(-y)}$ for every $y$.
%283C

\spheader 283Xe Let $f$ be an integrable complex-valued function
on $\Bbb R$, with Fourier transform $\varhatf$.   Show that

\Centerline{$\int_c^d\varhatf(y)dy=\Bover{i}{\sqrt{2\pi}}
\int_{-\infty}^{\infty}\Bover{e^{-idx}-e^{-icx}}{x}f(x)dx$}

\noindent whenever $c\le d$ in $\Bbb R$.
%283F

\sqheader 283Xf For an integrable complex-valued function $f$ on
$\Bbb R$, let its {\bf Fej\'er integrals} be

\Centerline{$\sigma_c(x)
=\Bover1{c\sqrt{2\pi}}\int_0^c
  \bigl(\int_{-a}^{a}e^{ixy}\varhatf(y)dy\bigr)da$}

\noindent for $c>0$.   Show that

\Centerline{$\sigma_c(x)=
\Bover1{\pi}\int_{-\infty}^{\infty}
  \Bover{1-\cos ct}{ct^2}f(x-t)dt$.}
%283F

\spheader 283Xg Show that
$\biggerint_{-\infty}^{\infty}\Bover{1-\cos at}{at^2}dt=\pi$ for every $a>0$.   \Hint{integrate by parts and use 283Da.}   Show that

\Centerline{$\lim_{a\to\infty}\int_{\delta}^{\infty}
         \Bover{1-\cos at}{at^2}dt
=\lim_{a\to\infty}\sup_{t\ge\delta}\Bover{1-\cos at}{at^2}
=0$}

\noindent for every $\delta>0$.
%283Xf, 283D, 283F

\wheader{283Xh}{0}{0}{0}{72pt}

\spheader 283Xh Let $f$ be an integrable complex-valued function
on $\Bbb R$, and define its Fej\'er integrals $\sigma_a$ as in 283Xf
above.   Show that if $x\in\Bbb R$, $c\in\Bbb C$ are such that

\Centerline{$\lim_{\delta\downarrow 0}\Bover1{\delta}\int_0^{\delta}
    |f(x+t)+f(x-t)-2c|dt=0$,}

\noindent then $\lim_{a\to\infty}\sigma_a(x)=c$.   \Hint{adapt
the argument of 282H.}
%283Xf, 283Xg

\sqheader 283Xi Let $f$ be an integrable complex-valued function
on $\Bbb R$, and define its Fej\'er integrals $\sigma_a$ as in 283Xf
above.   Show that $f(x)=\lim_{a\to\infty}\sigma_a(x)$ for almost every
$x\in\Bbb R$.
%283Xf, 283Xh

\spheader 283Xj Let $f:\Bbb R\to\Bbb C$ be a continuous integrable
complex-valued function of bounded variation, and define its Fej\'er
integrals $\sigma_a$ as in 283Xf above.   Show that
$f(x)=\lim_{a\to\infty}\sigma_a(x)$ uniformly for $x\in\Bbb R$.
%283Xf, 283Xi

\sqheader 283Xk Let $f$ be an integrable complex-valued function
of bounded variation on $\Bbb R$, and $\varhatf$ its Fourier transform.
Show that $\sup_{y\in\Bbb R}|y\varhatf(y)|<\infty$.
%283F

\spheader 283Xl Let $f$ and $g$ be integrable complex-valued
functions on $\Bbb R$.   Show that
$f*\varcheck{g}=\sqrt{2\pi}(\varhatf\times g)\varspcheck$.
%283M

\spheader 283Xm Let $f$ be an integrable complex-valued function
on $\Bbb R$, and fix $x\in\Bbb R$.    Set

\Centerline{$\hat f_x(y)=\int_{-\infty}^{\infty}f(t)\cos y(x-t)dt$}

\noindent for $y\in\Bbb R$.   Show that

\quad (i) if $f$ is differentiable at $x$,

\Centerline{$f(x)
=\Bover1{\pi}\lim_{a\to\infty}\int_{0}^a\hat f_x(y)dy$;}

\quad (ii) if there is a neighbourhood of $x$ in which $f$ has bounded
variation, then

\Centerline{$\Bover1{\pi}\lim_{a\to\infty}\int_{0}^a\hat f_x(y)dy
=\Bover12(\lim_{t\in\dom f,t\uparrow 0}f(t)
 +\lim_{t\in\dom f,t\downarrow 0}f(t))$;}

\quad (iii) if $f$ is twice differentiable and $f'$, $f''$ are
integrable then $\hat f_x$ is integrable and
$f(x)=\bover1{\pi}\int_0^{\infty}\hat f_x$.
(The formula

\Centerline{$f(x)=\Bover1{\pi}\int_{0}^{\infty}\bigl(
\int_{-\infty}^{\infty}f(t)\cos y(x-t)dt\bigr)dy$,}

\noindent valid for such functions $f$, is called {\bf Fourier's
integral formula}.)

\spheader 283Xn Show that if $f$ is a complex-valued function of
bounded variation, defined almost everywhere in $\Bbb R$, and converging
to $0$ (along its domain) at $\pm\infty$, then

\Centerline{$g(y)
=\Bover1{\sqrt{2\pi}}\lim_{a\to\infty}\int_{-a}^ae^{-iyx}f(x)dx$}

\noindent is defined in $\Bbb C$ for every $y\ne 0$, and that the limit
is uniform in any region bounded away from $0$.

\spheader 283Xo Let $f$ be an integrable complex-valued function
on $\Bbb R$.   Set

\Centerline{$\varhatf_c(y)=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
\cos yx\,f(x)dx$,\quad
$\varhatf_s(y)=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
\sin yx\,f(x)dx$}

\noindent for $y\in\Bbb R$.   Show that

\Centerline{$\Bover1{\sqrt{2\pi}}\int_{-a}^ae^{ixy}\varhatf(y)dy
=\sqrt{\Bover2{\pi}}\int_0^a\cos xy\,\varhatf_c(y)dy
+\sqrt{\Bover2{\pi}}\int_0^a\sin xy\,\varhatf_s(y)dy$}

\noindent for every $x\in\Bbb R$ and $a\ge 0$.

\spheader 283Xp Use the fact that
$\int_{0}^a\int_{0}^{\infty}e^{-xy}\sin
y\,dxdy=\int_{0}^{\infty}\int_{0}^ae^{-xy}\sin y\,dydx$ whenever
$a\ge 0$ to show that 
$\int_0^{\infty}\bover1{1+x^2}dx=\lim_{a\to\infty}\int_0^a\bover{\sin y}{y}dy$.

\sqheader 283Xq Show that if $f(x)=e^{-\sigma|x|}$, where
$\sigma>0$, then
$\varhatf(y)=\bover{2\sigma}{\sqrt{2\pi}(\sigma^2+y^2)}$.
Hence, or otherwise, find the Fourier transform of
$y\mapsto\Bover{1}{1+y^2}$.

\spheader 283Xr Find the inverse Fourier transform of the
indicator function of a bounded interval in $\Bbb R$.   Show that
in a formal sense 283F can be regarded as a special case of 283O.

\spheader 283Xs Let $f$ be a non-negative integrable function on
$\Bbb R$, with Fourier transform $\varhatf$.   Show that
\discrcenter{468pt}{$\sum_{j=0}^n\sum_{k=0}^n
  a_j\bar a_k\varhatf(y_j-y_k)\ge 0$ }whenever $y_0,\ldots,y_n$ in
$\Bbb R$ and $a_0,\ldots,a_n\in\Bbb C$.

\spheader 283Xt Let $f$ be an integrable complex-valued function on
$\Bbb R$.   Show that $\tilde f(x)=\sum_{n=-\infty}^{\infty}f(x+2\pi n)$
is defined in $\Bbb C$ for almost every $x$.
\Hint{$\sum_{n=-\infty}^{\infty}\int_{-\pi}^{\pi}|f(x+2\pi n)|dx
<\infty$.}   Show that $\tilde f$ is periodic.   Show that the
Fourier coefficients of $\tilde f\restr\ocint{-\pi,\pi}$ are
$\langle\bover{1}{\sqrt{2\pi}}\varhatf(k)\rangle_{k\in\Bbb Z}$.

\leader{283Y}{Further exercises (a)}
%\spheader 283Ya
Show that if $f:\Bbb R\to\Bbb C$ is
absolutely continuous in every bounded interval, $f'$ is of bounded
variation on $\Bbb R$, and
$\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=0$, then

\Centerline{$g(y)=\Bover1{\sqrt{2\pi}}\lim_{a\to\infty}
\int_{-a}^ae^{-iyx}f(x)dx
=-\Bover{i}{y\sqrt{2\pi}}\lim_{a\to\infty}
\int_{-a}^ae^{-iyx}f'(x)dx$}

\noindent is defined, with

\Centerline{$y^2|g(y)|\le\Bover4{\sqrt{2\pi}}\Var_{\Bbb R}(f')$,}

\noindent for every $y\ne 0$.
%283C

\spheader 283Yb Let $f:\Bbb R\to\Bbb C$ be an integrable
function which is absolutely continuous on every bounded interval, and
suppose that its derivative $f'$ is of bounded variation on $\Bbb R$.
Show that $\varhatf$ is integrable and that
$f=\varhatf\varcheck{\phantom{f}}$.   \Hint{225Yd, 283Ci, 283Xk.}
%283Xk 283K

\spheader 283Yc Let $f:\Bbb R\to\coint{0,\infty}$ be an even
function such that $f$ is convex on $\coint{0,\infty}$
and $\lim_{x\to\infty}f(x)=0$.

\quad{(i)} Show that, for any $y>0$ and $k\in\Bbb N$,
$\int_{-2k\pi/y}^{2k\pi/y}e^{-iyx}f(x)dx\ge 0$.

\quad{(ii)} Show that
$g(y)=\bover1{\sqrt{2\pi}}\lim_{a\to\infty}\int_{-a}^ae^{-iyx}f(x)dx$
exists in $\coint{0,\infty}$ for every $y\ne 0$.

\quad{(iii)} For $n\in\Bbb N$, set $f_n(x)
=e^{-|x|/(n+1)}f(x)$ for every $x$.   Show that $f_n$ is integrable and
convex on $\coint{0,\infty}$.

\quad{(iv)} Show that $g(y)=\lim_{n\to\infty}\varhatf_{n}(y)$ for
every $y\ne 0$.

\quad{(vi)} Show that if $f$ is integrable then

\Centerline{$\int_{-a}^a\varhatf
=\Bover4{\sqrt{2\pi}}\int_{0}^{\infty}\Bover{\sin at}{t}f(t)dt
\le\Bover{4a}{\sqrt{2\pi}}\int_0^{\pi/a}f
\le 2\sqrt{2\pi}f(0)
$}

\noindent for every $a\ge 0$.   Hence show that whether $f$ is
integrable or not, $g$ is integrable and $f_n=(\varhatf_n)\varspcheck$
for every $n$.

\quad{(vii)} Show that
$\lim_{a\downarrow 0}\sup_{n\in\Bbb N}\int_{-a}^a\varhatf_n=0$.

\quad(viii) Show that if $f'$ is bounded (on its domain) then
$\{\varhatf_n:n\in\Bbb N\}$ is uniformly integrable ({\it hint\/}: use
(vii) and 283Ya), so that $\lim_{n\to\infty}\|\varhat f_n-g\|_1=0$ and
$f=\varcheck{g}$.

\quad(ix) Show that if $f'$ is unbounded then for every $\epsilon>0$ we
can find $h_1$, $h_2:\Bbb R\to\coint{0,\infty}$, both even, convex and
converging to $0$ at $\infty$, such that $f=h_1+h_2$, $h'_1$ is bounded,
$\int h_2\le\epsilon$ and $h_2(0)\le\epsilon$.   Hence show that in this
case also $f=\varcheck{g}$.

\spheader 283Yd Suppose that $f:\Bbb R\to\Bbb R$ is even,
twice differentiable and convergent to $0$ at $\infty$, that $f''$ is
continuous and that $\{x:f''(x)=0\}$
is bounded in $\Bbb R$.   Show that $f$ is the Fourier transform of an
integrable function.   \Hint{use 283Yc and 283Yb.}
%283 notes

\spheader 283Ye Let $g:\Bbb R\to\Bbb R$ be an odd function of
bounded variation such that $\int_1^{\infty}\bover1xg(x)dx=\infty$.
Show that $g$ cannot be the Fourier transform of any integrable function
$f$.   ({\it Hint\/}:  show that if $g=\varhatf$ then
%283Yd

\Centerline{$-i\int_0^1f
=\Bover{2}{\sqrt{2\pi}}\lim_{a\to\infty}
  \int_0^a\Bover{1-\cos x}{x}g(x)dx=\infty$.)}
}%end of exercises

\endnotes{
\Notesheader{283}
I have tried in this section to give the elementary theory of Fourier
transforms of integrable functions on $\Bbb R$, with an eye to the
extension of the concept which will be attempted in the next section.
Following \S282, I have given prominence to two theorems (283I and
283L) describing conditions for the inversion of the Fourier transform
to return to the original function;   we find ourselves looking at
improper integrals
$\lim_{a\to\infty}\int_{-a}^a$, just as earlier we needed to look at
symmetric sums $\lim_{n\to\infty}\sum_{k=-n}^n$.   I do not go quite so
far as in \S282, and in particular I leave the study of
square-integrable functions for the moment, since their Fourier
transforms may not be describable by the simple formulae used here.

One of the most fundamental obstacles in the subject is the lack of any
effective criteria for determining which functions are the Fourier
transforms of integrable functions.   (Happily, things are better for
square-integrable functions;  see 284O-284P.)   In 283Yc-283Yd I sketch
an argument showing that `ordinary' non-oscillating {\it even} functions
which converge to $0$ at $\pm\infty$ are Fourier transforms of
integrable functions.   Strikingly, this is not true of {\it odd}
functions;  thus $y\mapsto\Bover{1}{\ln(e+y^2)}$ is the Fourier
transform of an integrable function, but
$y\mapsto\Bover{\arctan y}{\ln(e+y^2)}$ is not (283Ye).

In 283W I sketch the corresponding theory of Fourier transforms in
$\Bbb R^r$.   There are few surprises.   One point to note is that where
in the one-dimensional case we ask for a well-behaved second derivative,
in the $r$-dimensional case we may need to differentiate $r+1$ times
(283Wf).   Another is that we lose the `localization principle'.   In
the one-dimensional case, if $f$ is integrable and zero on an interval
$\ooint{c,d}$, then
$\lim_{a\to\infty}\int_{-a}^ae^{ixy}\varhatf(y)dy=0$ for every
$x\in\ooint{c,d}$;  this is immediate from either 283I or 283L.   But in
higher dimensions the most natural formulation of a corresponding result
is false (283Wm).




}%end of notes

\discrpage

\frfilename{mt284.tex}
\versiondate{30.8.13}
\copyrightdate{1994}

\def\rdtf{rapidly decreasing test function}
\def\Ft{Fourier transform}

\def\chaptername{Fourier analysis}
\def\sectionname{Fourier transforms II}

\newsection{284}

The basic paradox
of Fourier transforms is the fact that while for certain functions (see
283J-283K) we have $(\varhatf)\varspcheck=f$, `ordinary' integrable
functions $f$ (for instance, the indicator functions of non-trivial
intervals) give rise to non-integrable Fourier transforms $\varhatf$
for which there is no direct definition available for
$\varhatf\varcheck{\phantom{f}}$, making it a puzzle to decide in what
sense the formula
$f=\varhatf\varcheck{\phantom{f}}$ might be true.   What now seems by
far the most
natural resolution of the problem lies in declaring the Fourier
transform to be an operation on {\it distributions} rather than on {\it
functions}.   I shall not attempt to describe this theory properly
(almost any book on `Distributions' will cover the ground better than
I can possibly do here), but will try to convey the fundamental ideas,
so far as they are relevant to the questions dealt with here, in
language which will make the transition to a fuller treatment
straightforward.   At the same time, these methods make it easy to prove
strong versions of the `classical' theorems concerning Fourier
transforms.

\leader{284A}{Test functions:  Definition} Throughout this section, a
{\bf rapidly decreasing test function} or {\bf Schwartz function}
will be a function  $h:\Bbb R\to\Bbb C$ such
that $h$ is {\bf smooth}, that is, differentiable everywhere any finite
number of times, and moreover

\Centerline{$\sup_{x\in\Bbb R}|x|^k|h^{(m)}(x)|<\infty$}

\noindent for all $k$, $m\in\Bbb N$, writing $h^{(m)}$ for the $m$th
derivative of $h$.

\leader{284B}{}\cmmnt{ The following elementary facts will be useful.

\medskip

\noindent}{\bf Lemma} (a) If $g$ and $h$ are rapidly decreasing test
functions, so are $g+h$ and $ch$, for any $c\in\Bbb C$.

(b) If $h$ is a rapidly decreasing test function and $y\in\Bbb R$, then
$x\mapsto h(y-x)$ is a rapidly decreasing test function.

(c) If $h$ is any rapidly decreasing test function, then $h$ and $h^2$
are integrable.

(d) If $h$ is a \rdtf, so is its derivative $h'$.


(e) If $h$ is a \rdtf, so is the function $x\mapsto xh(x)$.

(f) For any $\epsilon>0$, the function
$x\mapsto e^{-\epsilon x^2}$ is a rapidly decreasing test function.

\proof{{\bf (a)} is trivial.

\medskip

{\bf (b)} Write $g(x)=h(y-x)$ for $x\in\Bbb R$.   Then
$g^{(m)}(x)=(-1)^mh^{(m)}(y-x)$ for every $m$, so $g$ is smooth.   For
any $k\in\Bbb N$,

\Centerline{$|x|^k\le 2^k(|y|^k+|y-x|^k)$}

\noindent for every $x$, so

$$\eqalign{\sup_{x\in\Bbb R}|x|^k|g^{(m)}(x)|
&=\sup_{x\in\Bbb R}|x|^k|h^{(m)}(y-x)|\cr
&\le 2^k|y|^k\sup_{x\in\Bbb R}|h^{(m)}(y-x)|
   +2^k\sup_{x\in\Bbb R}|y-x|^k|h^{(m)}(y-x)|\cr
&=2^k|y|^k\sup_{x\in\Bbb R}|h^{(m)}(x)|
   +2^k\sup_{x\in\Bbb R}|x|^k|h^{(m)}(x)|
<\infty.\cr}$$

\medskip

{\bf (c)} Because

\Centerline{$M=\sup_{x\in\Bbb R}|h(x)|+x^2|h(x)|$}

\noindent is finite, we have

\Centerline{$\int|h|\le\int\Bover{M}{1+x^2}dx<\infty$.}

\noindent Of course we now have $|h^2|\le M|h|$, so $h^2$ also is
integrable.

\medskip

{\bf (d)} This is immediate from the definition, as every derivative of
$h'$ is a derivative of $h$.

\medskip

{\bf (e)} Setting $g(x)=xh(x)$, $g^{(m)}(x)=xh^{(m)}(x)+mh^{(m-1)}(x)$
for $m\ge 1$, so

\Centerline{$\sup_{x\in\Bbb R}|x^kg^{(m)}(x)|
\le\sup_{x\in\Bbb R}|x^{k+1}h^{(m)}(x)|
   +m\sup_{x\in\Bbb R}|x^kh^{(m-1)}(x)|$}

\noindent is finite, for all $k\in\Bbb N$, $m\ge 1$.

\medskip

{\bf (f)} If $h(x) = e^{-\epsilon x^2}$, then for each $m\in\Bbb N$ we
have $h^{(m)}(x)=p_m(x)h(x)$, where $p_0(x)=1$ and
$p_{m+1}(x)=p'_m(x)-2\epsilon xp_m(x)$, so that $p_m$ is a polynomial.
Because $e^{\epsilon x^2}\ge\epsilon^{k+1}x^{2k+2}/(k+1)!$ for all $x$,
$k\ge 0$,

\Centerline{$\lim_{|x|\to\infty}|x|^kh(x)
=\lim_{x\to\infty}x^k/e^{\epsilon x^2}=0$}

\noindent for every $k$, and $\lim_{|x|\to\infty}p(x)h(x)=0$ for every
polynomial $p$;  consequently

\Centerline{$\lim_{|x|\to\infty}x^kh^{(m)}(x)
=\lim_{|x|\to\infty}x^kp_m(x)h(x)=0$}

\noindent for all $k$, $m$, and $h$ is a rapidly decreasing test
function.
}%end of proof of 284B


\leader{284C}{Proposition} Let $h:\Bbb R\to\Bbb C$ be a rapidly
decreasing test function.   Then $\varhat{h}:\Bbb R\to\Bbb C$ and
$\varcheck{h}:\Bbb R\to\Bbb C$ are rapidly
decreasing test functions, and $\varhat{h}\varcheck{\phantom{h}} =
\varcheck{h}\varhat{\phantom{h}} = h$.

\proof{{\bf (a)} Let $k$, $m\in\Bbb N$.   Then
$\sup_{x\in\Bbb R}(|x|^m+|x|^{m+2})|h^{(k)}(x)|<\infty$ and
$\int_{-\infty}^{\infty}|x^mh^{(k)}(x)|dx
\ifdim\pagewidth=390pt\penalty-100\fi
<\infty$.   We may therefore
use 283Ch-283Ci to see that
$y\mapsto i^{k+m}y^k\varhat{h}\vthsp^{(m)}(y)$ is the Fourier
transform of $x\mapsto x^mh^{(k)}(x)$, and therefore that
$\lim_{|y|\to\infty}y^k\varhat{h}\vthsp^{(m)}(y)=0$, by 283Cg, so that
(because $\varhat{h}\vthsp^{(m)}$ is continuous)
$\sup_{y\in\Bbb R}|y^k\varhat{h}\vthsp^{(m)}(y)|$ is finite.   As $k$
and $m$ are arbitrary, $\varhat{h}$ is a rapidly decreasing test
function.

\medskip

{\bf (b)} Since $\varcheck{h}(y)=\varhat{h}(-y)$ for every $y$, it
follows at once that $\varcheck{h}$ is a rapidly decreasing test
function.

\medskip

{\bf (c)} By 283J, it follows from (a) and (b) that
$\varhat{h}\varcheck{\phantom{h}}=\varcheck{h}\varhat{\phantom{h}}=h$.
}%end of proof of 284C

\leader{284D}{Definition} I will use the phrase {\bf tempered function}
on $\Bbb R$ to mean
a measurable complex-valued function $f$, defined almost everywhere in
$\Bbb R$, such that

\Centerline{$\int_{-\infty}^{\infty}
  \Bover1{1+|x|^k}|f(x)|dx<\infty$}

\noindent for some $k\in\Bbb N$.

\leader{284E}{}\cmmnt{ As in 284B I spell out some elementary facts.

\medskip

\noindent}{\bf Lemma} (a) If $f$ and $g$ are tempered functions, so are
$|f|$, $f+g$ and $cf$, for any $c\in\Bbb C$.

(b) If $f$ is a tempered function then it is integrable over any bounded
interval.

(c) If $f$ is a tempered function and $x\in\Bbb R$, then $t\mapsto
f(x+t)$ and $t\mapsto f(x-t)$ are both tempered functions.

\proof{{\bf (a)} is elementary;  if

\Centerline{$\int_{-\infty}^{\infty}
  \Bover1{1+|x|^j}f(x)dx<\infty$,
\quad$\int_{-\infty}^{\infty}\Bover1{1+|x|^k}g(x)dx<\infty$,}

\noindent then

\Centerline{$\int_{-\infty}^{\infty}\Bover1{1+|x|^{j+k}}
|(f+g)(x)|dx<\infty$}

\noindent because

\Centerline{$1+|x|^{j+k}\ge\max(1,|x|^{j+k})\ge\max(1,|x|^j,|x|^k)
\ge\Bover12\max(1+|x|^j,1+|x|^k)$}

\noindent for all $x$.

\medskip

{\bf (b)} If

\Centerline{$\int_{-\infty}^{\infty}\Bover1{1+|x|^k}|f(x)|dx
=M<\infty$,}

\noindent then for any $a\le b$

\Centerline{$\int_a^b|f|\le M(1+|a|^k+|b|^k)(b-a)<\infty$.}

\medskip

{\bf (c)} The idea is the same as in 284Bb.   If $k\in\Bbb N$ is such
that

\Centerline{$\int_{-\infty}^{\infty}\Bover1{1+|t|^k}|f(t)|dt
=M<\infty$,}

\noindent then we have

\Centerline{$1+|x+t|^k
\le 2^k(1+|x|^k)(1+|t|^k)$}

\noindent so that

\Centerline{$\Bover1{1+|t|^k}
\le 2^k(1+|x|^k)\Bover1{1+|x+t|^k}$}

\noindent for every $t$, and

\Centerline{$\int_{-\infty}^{\infty}\Bover{|f(x+t)|}{1+|t|^k}dt
\le 2^k(1+|x|^k)\int_{-\infty}^{\infty}
   \Bover{|f(x+t)|}{1+|x+t|^k}dt
\le 2^k(1+|x|^k)M<\infty$.}

\noindent Similarly,

\Centerline{$\int_{-\infty}^{\infty}\Bover{|f(x-t)|}{1+|t|^k}dt
\le 2^k(1+|x|^k)M<\infty$.}
}%end of proof of 284E

\leader{284F}{}\cmmnt{ Linking the two concepts, we have the
following.

\medskip

\noindent}{\bf Lemma} Let $f$ be a tempered function on $\Bbb R$ and $h$
a rapidly decreasing test function.   Then $f\times h$ is integrable.

\proof{ Of course $f\times h$ is measurable.   Let
$k\in\Bbb N$ be such that
$\int_{-\infty}^{\infty}
\bover1{1+|x|^k}|f(x)|dx<\infty$.
There is an $M$ such that $(1+|x|^k)|h(x)|\le M$ for every
$x\in\Bbb R$, so that

\Centerline{$\int_{-\infty}^{\infty}|f\times h|
\le M\int_{-\infty}^{\infty}\Bover1{1+|x|^k}|f(x)|dx<\infty$.}
}%end of proof of 284F

\leader{284G}{Lemma} Suppose that $f_1$ and $f_2$ are tempered functions
and that $\int f_1\times h=\int f_2\times h$ for every rapidly
decreasing test function $h$.   Then $f_1\eae f_2$.

\proof{{\bf (a)} Set $g=f_1-f_2$;  then $\int g\times h=0$ for every
rapidly decreasing test function $h$.   Of course $g$ is a tempered
function, so is integrable over any bounded interval.   By 222D, it will
be enough if I can show that $\int_a^bg=0$ whenever $a<b$, since then we
shall have $g=0$ a.e.\ on every bounded interval and $f_1\eae f_2$.

\medskip

{\bf (b)} Consider the function $\tilde\phi(x)=e^{-1/x}$ for $x>0$.   Then
$\tilde\phi$ is differentiable arbitrarily often everywhere in
$\ooint{0,\infty}$, $0<\tilde\phi(x)<1$ for every $x>0$, and
$\lim_{x\to\infty}\tilde\phi(x)=1$.   Moreover, writing $\tilde\phi^{(m)}$ for
the $m$th derivative of $\tilde\phi$,

\Centerline{$\lim_{x\downarrow 0}\tilde\phi^{(m)}(x)
=\lim_{x\downarrow 0}\Bover1x\tilde\phi^{(m)}(x)=0$}

\noindent for every $m\in\Bbb N$.   \Prf\ (Compare 284Bf.)   We have
$\tilde\phi^{(m)}(x)=p_m(\bover1x)\tilde\phi(x)$, where $p_0(t)=1$ and
$p_{m+1}(t)=t^2(p_m(t)-p_m'(t))$, so that $p_m$ is a polynomial for each
$m\in\Bbb N$.   Now for any $k\in\Bbb N$,

\Centerline{$0\le\limsup_{t\to\infty}t^ke^{-t}
\le\lim_{t\to\infty}\Bover{(k+1)!t^k}{t^{k+1}}=0$,}

\noindent so

\Centerline{$\lim_{x\downarrow 0}\tilde\phi^{(m)}(x)
=\lim_{t\to\infty}p_m(t)e^{-t}=0$,}

\Centerline{$\lim_{x\downarrow 0}\Bover1x\tilde\phi^{(m)}(x)
=\lim_{t\to\infty}tp_m(t)e^{-t}=0$. \Qed}

\medskip

{\bf (c)} Consequently, setting $\phi(x)=0$ for $x\le 0$, $e^{-1/x}$ for
$x>0$, $\phi$ is smooth, with $m$th derivative

\Centerline{$\phi^{(m)}(x)=0$ for $x\le 0$,
\quad$\phi^{(m)}(x)=\tilde\phi^{(m)}(x)$ for $x>0$.}

\noindent (The proof is an easy induction on $m$.)   Also
$0\le\phi(x)\le 1$ for every $x\in\Bbb R$, and
$\lim_{x\to\infty}\phi(x)=1$.

\medskip

{\bf (d)} Now take any $a<b$, and for $n\in\Bbb N$ set

\Centerline{$\phi_n(x)=\phi(n(x-a))\phi(n(b-x))$.}

\noindent Then $\phi_n$ will be smooth and  $\phi_n(x)=0$ if
$x\notin\ooint{a,b}$, so surely $\phi_n$ is a rapidly decreasing test
function, and

\Centerline{$\int_{-\infty}^{\infty}g\times\phi_n=0$.}

\noindent Next,  $0\le\phi_n(x)\le 1$ for every $x$, $n$, and if $a<x<b$
then $\lim_{n\to\infty}\phi_n(x)=1$.   So

\Centerline{$\int_a^bg=\int g\times\chi(\ooint{a,b})
=\int g\times(\lim_{n\to\infty}\phi_n)
=\lim_{n\to\infty}\int g\times\phi_n=0$,}

\noindent using Lebesgue's Dominated Convergence Theorem.   As $a$ and
$b$ are arbitrary, $g=0$ a.e., as required.
}%end of proof of 284G

\leader{284H}{Definition} Let $f$ and $g$ be tempered
functions\cmmnt{ in
the sense of 284D}.   Then I will say that $g$ {\bf represents the
Fourier transform of} $f$ if

\Centerline{$\int_{-\infty}^{\infty}g\times h
=\int_{-\infty}^{\infty}f\times\varhat{h}$}

\noindent for every rapidly decreasing test function $h$.

\leader{284I}{Remarks (a)}\cmmnt{ As usual, when shifting definitions
in this way, we have some checking to do.}
If $f$ is an integrable complex-valued function on $\Bbb R$ and $\varhatf$
is its Fourier transform, then\cmmnt{ surely $\varhatf$ is a tempered
function,
being a bounded continuous function;  and if $h$ is any rapidly
decreasing test function, then
$\int\varhatf\times h=\int f\times\varhat{h}$
by 283O.   Thus} $\varhatf$ `represents the Fourier transform of
$f$'\cmmnt{ in
the sense of 284H above}.

\header{284Ib}{\bf (b)} Note also that\cmmnt{ 284G assures us that} if
$g_1$, $g_2$ are two tempered functions both representing the Fourier
transform of $f$, then $g_1\eae g_2$\cmmnt{, since we must have

\Centerline{$\int g_1\times h=\int f\times\varhat{h}=\int g_2\times h$}

\noindent for every rapidly decreasing test function $h$.}

\spheader 284Ic It is I suppose obvious that if $f_1$, $f_2$, $g_1$ and
$g_2$ are tempered functions and $g_i$ represents the Fourier transform of
$f_i$ for both $i$, then $cg_1+g_2$ represents the Fourier transform of
$cf_1+f_2$ for every $c\in\Bbb C$.

\cmmnt{\spheader 284Id
Of course the value of this indirect approach is that we can assign
Fourier transforms, in a sense, to many more functions.   But we must
note at once that if $g$ `represents the Fourier transform of $f$'
then so will any function equal almost everywhere to $g$;  we can no
longer expect to be able to speak of `the' Fourier transform of $f$
as a function.   We could say that `the' Fourier transform of $f$ is
a functional
$\phi$ on the space of rapidly decreasing test functions, defined by
setting $\phi(h)=\int f\times\varhat{h}$;   alternatively, we could say
that `the' Fourier transform of $f$ is a member of $L_{\Bbb C}^0$, the
space of equivalence classes of almost-everywhere-defined measurable
functions (241J).
}%end of comment

\spheader 284Ie\cmmnt{ It is now natural to say that} {\bf $g$
represents the inverse Fourier transform of $f$}\cmmnt{ just when $f$
represents the Fourier transform of $g$;  that is,}  when $\int f\times
h=\int g\times\varhat{h}$ for every rapidly decreasing test function
$h$.   \cmmnt{Because
$\varhat{h}\varcheck{\phantom{h}}=\varcheck{h}\varhat{\phantom{h}}=h$
for every such $h$ (284C), this is the same thing as saying that
$\int g\times h=\int f\times\varcheck{h}$ for every rapidly decreasing test
function $h$, which is the other natural expression of what it might
mean to say that `$g$ represents the inverse Fourier transform of
$f$'.}

\spheader 284If If $f$, $g$ are tempered functions and we write
$\Reverse{g}(x)=g(-x)$ whenever this is defined, then\cmmnt{ $\Reverse{g}$ will also
be a tempered function, and we shall always have

\Centerline{$\int\Reverse{g}\times\varhat{h}
=\int g(-x)\varhat{h}(x)dx
=\int g(x)\varhat{h}(-x)dx
=\int g\times\varcheck{h}$,}

\noindent so that}

\qquad $g$ represents the Fourier transform of $f$

\cmmnt{\qquad\qquad$\iff\,\int g\times h=\int f\times\varhat{h}$ for
every test function $h$

\qquad\qquad$\iff\,\int g\times\varcheck{h}=\int
f\times\varcheck{h}\varhat{\phantom{h}}$ for every $h$

\qquad\qquad$\iff\,\int\Reverse{g}\times\varhat{h}=\int f\times h$ for every
$h$
}%end of comment

\qquad\qquad$\iff\,\Reverse{g}$ represents the inverse Fourier transform of
$f$.

\cmmnt{\noindent Combining this with (d), we get

\qquad $g$ represents the \Ft\ of $f$

%\qquad\qquad$\iff\,(\Reverse{f})\ssplrarrow=f$
%represents the inverse \Ft\ of $g$

\qquad\qquad$\iff\,\Reverse{\Reverse{f}}=f$
represents the inverse \Ft\ of $g$

\qquad\qquad$\iff\,\Reverse{f}$ represents the \Ft\ of $g$.
}%end of comment

\cmmnt{\spheader 284Ig Yet again, we ought to be conscious that a check
is called for:  if
$f$ is integrable and $\varcheckf$ is its inverse Fourier transform as
defined in 283Ab, then

\Centerline{$\int\varcheckf\times\varhat{h}
=\int f\times\varhat{h}\varcheck{\phantom{h}}
=\int f\times h$}

\noindent for every rapidly decreasing test function $h$, so
$\varcheckf$ `represents the inverse Fourier transform of $f$' in the
sense given here.
}%end of comment

\leader{284J}{Lemma} Let $f$ be any tempered function and $h$ a rapidly
decreasing test function.   Then $f*h$, defined by the formula

\Centerline{$(f*h)(y)=\int_{-\infty}^{\infty}f(t)h(y-t)dt$,}

\noindent is defined everywhere.

\proof{ Take any $y\in\Bbb R$.   By 284Bb,
$t\mapsto h(y-t)$ is a rapidly decreasing test function, so the integral
is always defined in $\Bbb C$, by 284F.
}%end of proof of 284J

\leader{284K}{Proposition} Let $f$ and $g$ be tempered functions such
that $g$ represents the Fourier transform of $f$, and $h$ a rapidly
decreasing test function.

(a) The Fourier transform of the integrable function $f\times h$ is
$\bover1{\sqrt{2\pi}}g*\varhat{h}$.

(b) The Fourier transform of the continuous function $f*h$ is
represented by the
product $\sqrt{2\pi}g\times\varhat{h}$.

\proof{{\bf (a)} Of course $f\times h$ is integrable, by 284F,
while $g*\varhat{h}$ is defined everywhere, by 284C and 284J.

Fix $y\in\Bbb R$.   Set $h_1(x)=\varhat{h}(y-x)$ for $x\in\Bbb R$;
then $h_1$ is a rapidly decreasing test function because $\varhat{h}$
is (284Bb).   Now

$$\eqalign{\varhat{h}_1(t)
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
  e^{-itx}\varhat{h}(y-x)dx
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
  e^{-it(y-x)}\varhat{h}(x)dx\cr
&=\Bover{1}{\sqrt{2\pi}}e^{-ity}
   \int_{-\infty}^{\infty}e^{itx}\varhat{h}(x)dx
=e^{-ity}\varhat{h}\varcheck{\phantom{h}}(t)
=e^{-ity}h(t),\cr}$$

\noindent using 284C.   Accordingly

$$\eqalignno{(f\times h)\varsphat(y)
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ity}f(t)h(t)dt\cr
&=\Bover1{\sqrt{2\pi}}
  \int_{-\infty}^{\infty}f(t)\varhat{h}_1(t)dt
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(t)h_1(t)dt\cr
\noalign{\noindent (because $g$ represents the Fourier transform of
$f$)}
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(t)\varhat{h}(y-t)dt
=\Bover1{\sqrt{2\pi}}(g*\varhat{h})(y).\cr}$$

\noindent As $y$ is arbitrary, $\Bover1{\sqrt{2\pi}}g*\varhat{h}$ is the
Fourier transform of $f\times h$.


\medskip

{\bf (b)} Write $f_1$ for the Fourier transform of $g\times\varhat{h}$,
$\Reverse{f}(x)=f(-x)$ when this is defined, and $\Reverse{h}(x)=h(-x)$ for every $x$,
so that $\Reverse{f}$ represents the Fourier transform of $g$, by 284If, and
$\Reverse{h}$ is the Fourier transform of $\varhat{h}$.   By (a), we have
$f_1=\Bover1{\sqrt{2\pi}}\Reverse{f}*\Reverse{h}$.
This means that the inverse Fourier transform of
$\sqrt{2\pi}g\times\varhat{h}$ must be
$\sqrt{2\pi}\Reverse{f}_1=(\Reverse{f}*\Reverse{h})\ssplrarrow$;
and as

$$\eqalign{(\Reverse{f}*\Reverse{h})\ssplrarrow(y)
&=(\Reverse{f}*\Reverse{h})(-y)
=\int_{-\infty}^{\infty}\Reverse{f}(t)\Reverse{h}(-y-t)dt\cr
&=\int_{-\infty}^{\infty}f(-t)h(y+t)dt
=\int_{-\infty}^{\infty}f(t)h(y-t)dt
=(f*h)(y),\cr}$$

\noindent the inverse Fourier transform of
$\sqrt{2\pi}g\times\varhat{h}$ is $f*h$ (which is therefore continuous),
and $\sqrt{2\pi}g\times\varhat{h}$ must represent the Fourier transform
of $f*h$.
}%end of proof of 284K

\cmmnt{\medskip

\noindent{\bf Remark} Compare 283M.   It is typical of the theory of
Fourier transforms that we have formulae valid in a wide variety of
contexts, each requiring a different interpretation and a different
proof.
}%end of comment

\leader{284L}{}\cmmnt{ We are now ready for a result corresponding to
282H.   I use a different method, or at least a different arrangement of
the ideas, through the following fact, which is important in other ways.

\medskip

\noindent}{\bf Proposition}  Let $f$ be a tempered function.
Writing $\psi_{\sigma}(x) =\bover1{\sigma\sqrt{2\pi}}e^{-x^2/2\sigma^2}$
for $x\in\Bbb R$ and $\sigma>0$, then

\Centerline{$\lim_{\sigma\downarrow 0}(f*\psi_{\sigma})(x)=c$}

\noindent whenever $x\in\Bbb R$ and $c\in\Bbb C$ are such that

\Centerline{$\lim_{\delta\downarrow 0}
\Bover1{\delta}\int_{0}^{\delta}|f(x+t)+f(x-t)-2c|dt=0$.}

\proof{{\bf (a)} By 284Bf, every $\psi_{\sigma}$ is a rapidly
decreasing test function, so that $f*\psi_{\sigma}$ is defined
everywhere, by 284J.   We need to know that
$\int_{-\infty}^{\infty}\psi_{\sigma}=1$;  this is because (substituting
$u=x/\sigma$)

\Centerline{$\int_{-\infty}^{\infty}\psi_{\sigma}
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-u^2/2}du=1$,}

\noindent by 263G.   The argument now follows the lines of 282H.   Set

\Centerline{$\phi(t)=|f(x+t)+f(x-t)-2c|$}

\noindent when this is defined, which is almost everywhere, and
$\Phi(t)=\int_0^t\phi$, defined for all $t\ge 0$ because $f$ is
integrable over every bounded interval (284Eb).   We have

$$\eqalignno{|(f*\psi_{\sigma})(x)-c|
&=|\int_{-\infty}^{\infty}f(x-t)\psi_{\sigma}(t)dt
-c\int_{-\infty}^{\infty}\psi_{\sigma}(t)dt|\cr
&=|\int_{-\infty}^0(f(x-t)-c)\psi_{\sigma}(t)dt
+\int_{0}^{\infty}(f(x-t)-c)\psi_{\sigma}(t)dt|\cr
&=|\int_{0}^{\infty}(f(x+t)-c)\psi_{\sigma}(t)dt
+\int_{0}^{\infty}(f(x-t)-c)\psi_{\sigma}(t)dt|\cr
\noalign{\noindent (because $\psi_{\sigma}$ is an even function)}
&=|\int_{0}^{\infty}(f(x+t)+f(x-t)-2c)\psi_{\sigma}(t)dt|\cr
&\le\int_{0}^{\infty}|f(x+t)+f(x-t)-2c|\psi_{\sigma}(t)dt
=\int_0^{\infty}\phi\times\psi_{\sigma}.\cr}$$

\medskip

{\bf (b)} I should explain why this last integral is finite.   Because
$f$ is a tempered function, so are the functions $t\mapsto f(x+t)$,
$t\mapsto f(x-t)$ (284Ec);  of course constant functions are tempered,
so $t\mapsto\phi(t)=|f(x+t)+f(x-t)-2c|$ is tempered, and because
$\psi_{\sigma}$ is a rapidly decreasing test function we may
apply 284F
to see that the product is integrable.

\medskip

{\bf (c)} Let $\epsilon>0$.   By hypothesis, $\lim_{t\downarrow
0}\Phi(t)/t=0$;
let $\delta>0$ be such that $\Phi(t)\le\epsilon t$ for
every $t\in[0,\delta]$.   Take any $\sigma\in\ocint{0,\delta}$.   I
break the
integral $\int_0^{\infty}\phi\times\psi_{\sigma}$ up into three parts.

\medskip

\quad{\bf (i)} For the integral from $0$ to $\sigma$, we have

$$\int_0^{\sigma}\phi\times\psi_{\sigma}
\le\int_0^{\sigma}\Bover{1}{\sigma\sqrt{2\pi}}\phi
=\Bover1{\sigma\sqrt{2\pi}}\Phi(\sigma)
\le\Bover{\epsilon\sigma}{\sigma\sqrt{2\pi}}\le\epsilon,$$

\noindent because $\psi_{\sigma}(t)\le\bover1{\sigma\sqrt{2\pi}}$ for
every $t$.

\medskip

\quad{\bf (ii)} For the integral from $\sigma$ to $\delta$, we have

$$\eqalignno{\int_{\sigma}^{\delta}\phi\times\psi_{\sigma}
&\le\Bover1{\sigma\sqrt{2\pi}}\int_{\sigma}^{\delta}
  \phi(t)\Bover{2\sigma^2}{t^2}dt\cr
\noalign{\noindent (because $e^{-t^2/2\sigma^2}
=1/e^{t^2/2\sigma^2}\le
1/(t^2/2\sigma^2)=2\sigma^2/t^2$ for every $t\ne 0$)}
&=\sigma\sqrt{\Bover2{\pi}}\int_{\sigma}^{\delta}
  \Bover{\phi(t)}{t^2}dt
=\sigma\sqrt{\Bover2{\pi}}\bigl(\Bover{\Phi(\delta)}{\delta^2}
   -\Bover{\Phi(\sigma)}{\sigma^2}
   +\int_{\sigma}^{\delta}\Bover{2\Phi(t)}{t^3}dt\bigr)\cr
\noalign{\noindent (integrating by parts -- see 225F)}
&\le\sigma\bigl(\Bover{\epsilon}{\delta}
   +\int_{\sigma}^{\delta}\Bover{2\epsilon}{t^2}dt\bigr)\cr
\noalign{\noindent (because $\Phi(t)\le\epsilon t$ for $0\le
t\le\delta$ and $\sqrt{2/\pi}\le 1$)}
&\le\sigma\bigl(\Bover{\epsilon}{\delta}
   +\Bover{2\epsilon}{\sigma}\bigr)
\le 3\epsilon.\cr}$$

\medskip

\quad{\bf (iii)} For the integral from $\delta$ to $\infty$, we have

$$\eqalign{\int_{\delta}^{\infty}\phi\times\psi_{\sigma}
&=\bover1{\sqrt{2\pi}}\int_{\delta}^{\infty}
  \phi(t)\bover{e^{-t^2/2\sigma^2}}{\sigma}dt.\cr}$$

\noindent Now for any $t\ge\delta$,

\Centerline{$\sigma\mapsto\Bover1{\sigma}
e^{-t^2/2\sigma^2}:\ocint{0,\delta}\to\Bbb R$}

\noindent is monotonically increasing, because its derivative

\Centerline{$\Bover{d}{d\sigma}\Bover1{\sigma}e^{-t^2/2\sigma^2}
=\Bover1{\sigma^2}\bigl(\Bover{t^2}{\sigma^2}-1\bigr)
e^{-t^2/2\sigma^2}$}

\noindent is positive, and

\Centerline{$\lim_{\sigma\downarrow 0}\Bover1{\sigma}
e^{-t^2/2\sigma^2}=
\lim_{a\to\infty}ae^{-a^2t^2/2}=0$.}

\noindent So we may apply Lebesgue's Dominated Convergence Theorem to
see that

$$\lim_{n\to\infty}\int_{\delta}^{\infty}
   \phi(t)\bover{e^{-t^2/2\sigma_n^2}}{\sigma_n}dt=0$$

\noindent whenever $\sequencen{\sigma_n}$ is a sequence in
$\ocint{0,\delta}$ converging to $0$, so that

$$\lim_{\sigma\downarrow 0}\int_{\delta}^{\infty}
   \phi(t)\bover{e^{-t^2/2\sigma^2}}{\sigma}dt=0.$$

\noindent There must therefore
be a $\sigma_0\in\ocint{0,\delta}$ such that

\Centerline{$\int_{\delta}^{\infty}\phi\times\psi_{\sigma}
\le\epsilon$}

\noindent for every $\sigma\le \sigma_0$.

\medskip

\quad{\bf (iv)} Putting these together, we see that

\Centerline{$|(f*\psi_{\sigma})(x)-c|
\le\int_0^{\infty}\phi\times\psi_{\sigma}
\le\epsilon+3\epsilon+\epsilon=5\epsilon$}

\noindent whenever $0<\sigma\le\sigma_0$.   As $\epsilon$ is arbitrary,
$\lim_{\sigma\downarrow 0}(f*\psi_{\sigma})(x)=c$, as claimed.
}%end of proof of 284L

\vleader{60pt}{284M}{Theorem} Let $f$ and $g$ be tempered functions such that
$g$ represents the Fourier transform of $f$.   Then

(a)(i) $g(y)
=\lim_{\epsilon\downarrow 0}\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-iyx}e^{-\epsilon x^2}f(x)dx$ for almost every $y\in\Bbb R$.

\quad(ii) If $y\in\Bbb R$ is such that $a=\lim_{t\in\dom g,t\uparrow
y}g(t)$ and $b=\lim_{t\in\dom g,t\downarrow y}g(t)$ are both defined in
$\Bbb C$, then

\Centerline{$\lim_{\epsilon\downarrow 0}\Bover1{\sqrt{2\pi}}
\int_{-\infty}^{\infty}
e^{-iyx}e^{-\epsilon x^2}f(x)dx=\Bover12(a+b)$.}

(b)(i) $f(x)
=\lim_{\epsilon\downarrow 0}\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{ixy}e^{-\epsilon y^2}g(y)dy$ for almost every $x\in\Bbb R$.

\quad(ii) If $x\in\Bbb R$ is such that $a=\lim_{t\in\dom f,t\uparrow
x}f(t)$ and $b=\lim_{t\in\dom f,t\downarrow x}f(t)$ are both defined in
$\Bbb C$, then

\Centerline{$\lim_{\epsilon\downarrow 0}\Bover1{\sqrt{2\pi}}
\int_{-\infty}^{\infty}
e^{ixy}e^{-\epsilon y^2}g(y)dy=\Bover12(a+b)$.}

\proof{{\bf (a)(i)}  By 223D,

\Centerline{$\lim_{\delta\downarrow 0}\Bover1{2\delta}
\int_{-\delta}^{\delta}|g(y+t)-g(y)|dt=0$}

\noindent for almost every $y\in\Bbb R$, because $g$ is integrable over
any bounded interval.   Fix any such $y$.   Set
$\phi(t)=|g(y+t)+g(y-t)-2g(y)|$ whenever this is defined.   Then, as in
the proof of 282Ia,

\Centerline{$\int_0^{\delta}\phi
\le\int_{-\delta}^{\delta}|g(y+t)-g(y)|dt$,}

\noindent so $\lim_{\delta\downarrow
0}\bover1{\delta}\int_0^{\delta}\phi=0$.   Consequently, by 284L,

\Centerline{$g(y)=\lim_{\sigma\to\infty}(g*\psi_{1/\sigma})(y)$.}

\noindent We know from 283N that the Fourier transform of
$\psi_{\sigma}$ is $\bover1{\sigma}\psi_{1/\sigma}$ for any $\sigma>0$.
Accordingly, by 284K, $g*\psi_{1/\sigma}$ is the Fourier transform of
$\sigma\sqrt{2\pi} f\times\psi_{\sigma}$, that is,

\Centerline{$(g*\psi_{1/\sigma})(y)
=\int_{-\infty}^{\infty}e^{-iyx}\sigma\psi_{\sigma}(x)f(x)dx$.}

\noindent So

$$\eqalign{g(y)
&=\lim_{\sigma\to\infty}
   \int_{-\infty}^{\infty}e^{-iyx}\sigma\psi_{\sigma}(x)f(x)dx\cr
&=\lim_{\sigma\to\infty}\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
   e^{-iyx}e^{-x^2/2\sigma^2}f(x)dx\cr
&=\lim_{\epsilon\downarrow 0}\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
   e^{-iyx}e^{-\epsilon x^2}f(x)dx.\cr}$$

\noindent And this is true for almost every $y$.

\medskip

\quad{\bf (ii)} Again, setting $c=\bover12(a+b)$,
$\phi(t)=|g(y+t)+g(y-t)-2c|$ whenever this is defined, we have
$\lim_{t\in\dom\phi,t\downarrow 0}\phi(t)\penalty-100=0$, so of course
$\lim_{\delta\downarrow 0}\bover{1}{\delta}\int_0^{\delta}\phi=0$, and

$$c
=\lim_{\sigma\to\infty}(g*\psi_{1/\sigma})(y)
=\lim_{\epsilon\downarrow 0}\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
   e^{-iyx}e^{-\epsilon x^2}f(x)dx$$

\noindent as before.


\medskip

{\bf (b)} This can be shown by similar arguments;  or it may be actually
deduced from (a), by observing that $x\mapsto\Reverse{f}(x)=f(-x)$ represents
the Fourier transform of $g$ (see 284Ie), and applying (a) to $g$ and
$\Reverse{f}$.
}%end of proof of 284M

\leader{284N}{$L^2$ \dvrocolon{spaces}}\cmmnt{ We are now ready for
results corresponding to 282J-282K.

\medskip

\noindent}{\bf Lemma} Let $\eusm L_{\Bbb C}^2$ be the space of
square-integrable
complex-valued functions on $\Bbb R$, and $\eusm S$ the space of rapidly
decreasing test functions.   Then for every $f\in\eusm L_{\Bbb C}^2$ and
$\epsilon>0$ there is an $h\in\eusm S$ such that $\|f-h\|_2\le\epsilon$.

\proof{ Set $\phi(x)=e^{-1/x}$ for $x>0$, zero for $x\le
0$;  recall from the proof of 284G that $\phi$ is smooth.   For any
$a<b$, the functions

\Centerline{$x\mapsto\phi_n(x)=\phi(n(x-a))\phi(n(b-x))$}

\noindent provide a sequence of test functions converging to
$\chi\ooint{a,b}$ from below, so (as in 284G)

\Centerline{$\inf_{h\in\eusm S}\,\|\chi\ooint{a,b}-h\|^2_2
\le\lim_{n\to\infty}\int_a^b|1-\phi_n|^2
=0$.}

\noindent Because $\eusm S$ is a linear space (284Ba), it follows that
for every step-function $g$ with bounded
support and every $\epsilon>0$ there is an $h\in\eusm S$ such that
$\|g-h\|_2\le\bover12\epsilon$.   But we know from 244H/244Pb
that for every
$f\in\eusm L_{\Bbb C}^2$ and $\epsilon>0$ there is  a step-function $g$
with bounded
support such that $\|f-g\|_2\le\bover12\epsilon$;  so there must be an
$h\in\eusm S$ such that

\Centerline{$\|f-h\|_2\le\|f-g\|_2+\|g-h\|_2\le \epsilon$.}

\noindent As $f$ and $\epsilon$ are arbitrary, we have the result.
}%end of proof of 284N

\leader{284O}{Theorem} (a) Let $f$ be any complex-valued function which
is square-integrable over $\Bbb R$.   Then $f$ is a tempered function
and its Fourier transform is represented by another square-integrable
function $g$, and $\|g\|_2=\|f\|_2$.

(b) If $f_1$ and $f_2$ are complex-valued functions, square-integrable
over $\Bbb R$, with Fourier transforms represented by functions $g_1$, $g_2$, then

\Centerline{$\int_{-\infty}^{\infty}f_1\times\bar f_2
=\int_{-\infty}^{\infty}g_1\times\bar g_2$.}

(c) If $f_1$ and $f_2$ are complex-valued functions, square-integrable
over $\Bbb R$, with Fourier transforms represented by functions $g_1$, $g_2$, then the integrable function $f_1\times f_2$ has Fourier transform $\bover1{\sqrt{2\pi}}g_1*g_2$.

(d) If $f_1$ and $f_2$ are complex-valued functions, square-integrable
over $\Bbb R$, with Fourier transforms represented by functions $g_1$, $g_2$, then $\sqrt{2\pi}g_1\times g_2$ represents the Fourier transform of the continuous function $f_1*f_2$.

\proof{{\bf (a)(i)} Consider first the case in which $f$ is a
rapidly decreasing test function and $g$ is its Fourier transform;  we
know that $g$ is also a \rdtf, and that $f$ is the inverse \Ft\ of $g$
(284C).   Now the complex conjugate $\overline{g}$ of $g$ is given by
the formula

\Centerline{$\overline{g}(y)
=\overline{\Bover1{\sqrt{2\pi}}
  \intop\nolimits_{-\infty}^{\infty}e^{-iyx}f(x)dx}
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{iyx}\overline{f}(x)dx$,}

\noindent so that $\overline{g}$ is the inverse Fourier transform of
$\overline{f}$.   Accordingly

\Centerline{$\int f\times\overline{f}
=\int\varcheck{g}\times\overline{f}
=\int g\times\varcheck{\overline{f}}
=\int g\times\overline{g}$,}

\noindent using 283O for the middle equality.

\medskip

\quad{\bf (ii)}  Now suppose that $f\in\eusm L_{\Bbb C}^2$.   I said
that $f$ is a tempered function;  this is simply because

\Centerline{$\int_{-\infty}^{\infty}
\bigl(\Bover{1}{1+|x|}\bigr)^2dx<\infty$,}

\noindent so

\Centerline{$\int_{-\infty}^{\infty}
\Bover{|f(x)|}{1+|x|}dx<\infty$}

\noindent (244Eb).   By 284N, there is a sequence $\sequencen{f_n}$
of rapidly decreasing test functions such that
$\lim_{n\to\infty}\|f-f_n\|_2=0$.   By (i),

\Centerline{$\lim_{m,n\to\infty}\|\varhatf_m-\varhatf_n\|_2
=\lim_{m,n\to\infty}\|f_m-f_n\|_2=0$,}

\noindent and the sequence $\sequencen{{\varhatf}\vthsp_n^{\ssbullet}}$
of equivalence classes is a Cauchy sequence in $L_{\Bbb C}^2$.
Because $L_{\Bbb C}^2$ is complete (244G/244Pb),
$\sequencen{{\varhatf}\vthsp_n^{\ssbullet}}$ has a limit in
$L_{\Bbb C}^2$, which is
representable as $g^{\ssbullet}$ for some $g\in\eusm L_{\Bbb C}^2$.
Like $f$, $g$ must be a tempered function.   Of course

\Centerline{$\|g\|_2=\lim_{n\to\infty}\|\varhatf_n\|_2
=\lim_{n\to\infty}\|f_n\|_2=\|f\|_2$.}

\noindent Now if $h$ is any rapidly decreasing test function,
$h$ and $\varhat{h}$ are square-integrable (284Bc, 284C), so we shall have

\Centerline{$\int g\times h
=\lim_{n\to\infty}\int\varhatf_n\times h
=\lim_{n\to\infty}\int f_n\times\varhat{h}
=\int f\times\varhat{h}$.}

\noindent So $g$ represents the Fourier transform of $f$.

\medskip

{\bf (b)} By 284Ib, any functions representing the Fourier transforms of
$f_1$ and $f_2$ must be equal almost everywhere to square-integrable
functions, and therefore square-integrable, with the right norms.   It
follows as in 282K (part (d) of the proof) that if $g_1$, $g_2$
represent the Fourier
transforms of $f_1$, $f_2$, so that $ag_1+bg_2$ represents the Fourier
transform of $af_1+bf_2$ and $\|ag_1+bg_2\|_2=\|af_1+bf_2\|_2$ for all
$a$, $b\in\Bbb C$, we must have

\Centerline{$\int f_1\times\overline{f}_2=(f_1|f_2)
=(g_1|g_2)=\int g_1\times\overline{g}_2$.}

\medskip

{\bf (c)} Of course $f_1\times f_2$ is integrable because it is the
product of two square-integrable functions (244E/244Pb).

\medskip

\quad{\bf (i)} Let $y\in\Bbb R$ and set $f(x)=\overline{f_2(x)}e^{iyx}$
for $x\in\Bbb R$.   Then $f\in\eusm L_{\Bbb C}^2$.   We need to know
that the Fourier transform of $f$ is represented by $g$, where
$g(u)=\overline{g_2(y-u)}$.   \Prf\ Let $h$ be a rapidly decreasing test
function.   Then

$$\eqalign{\int g\times h
&=\int\overline{g_2(y-u)}h(u)du
=\int \overline{g_2(u)}h(y-u)du\cr
&=\overline{\int g_2\times h_1}
=\overline{\int f_2\times \varhat{h}_1},\cr}$$

\noindent where $h_1(u)=\overline{h(y-u)}$.   To compute $\varhat{h}_1$,
we have

$$\eqalign{\varhat{h}_1(v)
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ivu}h_1(u)du
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
  e^{-ivu}\overline{h(y-u)}du\cr
&=\overline{\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
  e^{ivu}h(y-u)du}
=\overline{\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
  e^{iv(y-u)}h(u)du}
=\overline{e^{ivy}\varhat{h}(v)}.\cr}$$

\noindent So

\Centerline{$\int g\times h
=\overline{\intop f_2\times\varhat{h}_1}
=\int\overline{f_2(v)\varhat{h}_1(v)}dv
=\int\overline{f_2(v)}e^{ivy}\varhat{h}(v)dv
=\int f\times\varhat{h}$;}

\noindent as $h$ is arbitrary, $g$ represents the Fourier transform of
$f$. \Qed

\medskip

\quad{\bf (ii)} We now have

$$\eqalignno{(f_1\times f_2)\varsphat(y)
&=\Bover1{\sqrt{2\pi}}
  \int_{-\infty}^{\infty}e^{-iyx}f_1(x)f_2(x)dx\cr
&=\Bover1{\sqrt{2\pi}}
  \int_{-\infty}^{\infty}f_1\times\bar f
=\Bover1{\sqrt{2\pi}}
  \int_{-\infty}^{\infty}g_1\times\bar g\cr
\noalign{\noindent (using part (b))}
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}g_1(u)g_2(y-u)du
=\Bover1{\sqrt{2\pi}}(g_1*g_2)(y).\cr}$$

\noindent As $y$ is arbitrary,
$(f_1\times f_2)\varsphat=\bover1{\sqrt{2\pi}}g_1*g_2$, as claimed.

\medskip

{\bf (d)} By (c), the Fourier transform of $\sqrt{2\pi}g_1\times g_2$ is
$\Reverse{f}_1*\Reverse{f}_2$, writing $\Reverse{f}_1(x)=f_1(-x)$,
so that $\Reverse{f}_1$ represents
the Fourier transform of $g_1$.   So the inverse Fourier transform of
$\sqrt{2\pi}g_1\times g_2$ is $(\Reverse{f}_1*\Reverse{f}_2)\ssplrarrow$.
But, just as in the
proof of 284Kb, $(\Reverse{f}_1*\Reverse{f}_2)\ssplrarrow=f_1*f_2$,
so $f_1*f_2$ is the
inverse Fourier transform of $\sqrt{2\pi}g_1\times g_2$, and
$\sqrt{2\pi}g_1\times g_2$ represents the Fourier transform of
$f_1*f_2$, as claimed.   Also $f_1*f_2$, being the Fourier transform of
an integrable function, is continuous (283Cf;  see also 255K).
}%end of proof of 284O

\leader{284P}{Corollary} Writing $L_{\Bbb C}^2$ for the Hilbert space of
equivalence classes of square-integrable complex-valued functions on
$\Bbb R$, we have a linear isometry $T:L_{\Bbb C}^2\to L_{\Bbb C}^2$
given by saying that
$T(f^{\ssbullet})=g^{\ssbullet}$ whenever $f$, $g\in\eusm L_{\Bbb C}^2$
and $g$ represents the Fourier transform of $f$.

\cmmnt{
\leader{284Q}{Remarks (a)} 284P corresponds, of course, to 282K, where
the similar isometry between $\ell_{\Bbb C}^2(\Bbb Z)$ and
$L_{\Bbb C}^2(\ocint{-\pi,\pi})$
%terms reversed for sake of line break
is described.   In that case there was a marked asymmetry which is
absent from the present situation;  because the relevant measure on
$\Bbb Z$, counting measure, gives non-zero mass to every point, members
of $\ell_{\Bbb C}^2$ are true functions, and it is not surprising that
we have a
straightforward formula for $S(f^{\ssbullet})\in\ell_{\Bbb C}^2$ for
every $f\in\eusm L_{\Bbb C}^2(\ocint{-\pi,\pi})$.   The difficulty of
describing
$S^{-1}:\ell_{\Bbb C}^2(\Bbb Z)\to L_{\Bbb C}^2(\ocint{-\pi,\pi})$ is
very similar to the difficulty of describing
$T:L_{\Bbb C}^2(\Bbb R)\to L_{\Bbb C}^2(\Bbb R)$ and its inverse.
284Yg and 286U-286V show just how close this similarity is.

\header{284Qb}{\bf (b)} I have spelt out parts (c) and (d) of 284O in
detail, perhaps in unnecessary detail, because they give me an
opportunity to insist on the difference between
`$\sqrt{2\pi}g_1\times g_2$ {\it represents} the Fourier transform of
$f_1*f_2$' and
`$\bover1{\sqrt{2\pi}}g_1*g_2$ {\it is} the Fourier transform of
$f_1\times f_2$'.   The actual functions $g_1$ and $g_2$ are not
well-defined by the hypothesis that they represent the Fourier
transforms of $f_1$ and $f_2$, though their equivalence classes
$g_1^{\ssbullet}$, $g_2^{\ssbullet}\in L_{\Bbb C}^2$ are.   So the
product
$g_1\times g_2$ is also not uniquely defined as a function, though its
equivalence class
$(g_1\times g_2)^{\ssbullet}=g_1^{\ssbullet}\times g_2^{\ssbullet}$ is
well-defined as a member of $L_{\Bbb C}^1$.   However the
continuous function $g_1*g_2$ is unaffected by changes to $g_1$ and
$g_2$ on negligible sets, so is well defined as a function;  and since
$f_1\times f_2$ is integrable, and has a true Fourier transform, it is
to be expected that $(f_1\times f_2)\varsphat$ should be exactly equal
to $\bover1{\sqrt{2\pi}}g_1*g_2$.

This distinction between `being' a Fourier transform and `representing' a
Fourier transform echoes a question which arose in 233D concerning
conditional expectations.   I spoke there of `a' conditional expectation
on $\Tau$ of a function $f$
as being `a $\mu\restr\Tau$-integrable function $g$ such that
$\int_Fg\,d\mu=\int_Ffd\mu$ for every $F\in\Tau$';  the point being that
any $\mu\restr\Tau$-integrable function equal almost everywhere to $g$
would equally be a conditional expectation of $f$.   Here we see that if
$g$ represents the Fourier transform of $f$ then any function almost
everywhere equal to $g$ will also represent the Fourier transform of $f$.
In 242J I suggested resolving this complication by regarding conditional
expectation as a map between $L^1$ spaces rather than between $\eusm L^1$
spaces.   Here, similarly, we could think of the Fourier transform
considered in 284H as being a linear operator
defined on a certain subspace of $L^0(\mu)$.

In the case of conditional expectations, I think that there are solid
reasons for taking the operators on $L^1$ spaces as the real embodiment of
the idea;  I will expand on these in Chapter 36 of the next volume.   In
the case of Fourier transforms, I do not think the arguments have the same
force.   In 284R below, and in \S285, we
shall see that there are important cases in which we want to talk about
Fourier transforms which cannot be represented by members of $L^0$, so
that this would still be only a half-way house.

\spheader 284Qc Of course 284Oc-284Od also exhibit a
characteristic
feature of arguments involving Fourier transforms, the extension by
continuity of relations valid for test functions.

\spheader 284Qd 284Oa is a version of {\bf Plancherel's
theorem}.   The formula $\|f\|_2=\|\varhatf\|_2$ is {\bf Parseval's
identity}.
}%end of comment

\cmmnt{
\leader{284R}{Dirac's delta function} Consider the tempered
function $\chi\Bbb R$ with constant value $1$.   In what sense, if any, can
we assign a Fourier transform to $\chi\Bbb R$?

If we examine $\int\chi\Bbb R\times\varhat{h}$, as suggested in 284H, we
get

\Centerline{$\int_{-\infty}^{\infty}\chi\Bbb R\times\varhat{h}
=\int_{-\infty}^{\infty}\varhat{h}
=\sqrt{2\pi}\varhat{h}\varcheck{\phantom{h}}(0)
=\sqrt{2\pi}h(0)$}

\noindent for every rapidly decreasing test function $h$.   Of course
there is no {\it function} $g$ such that
$\int g\times h=\sqrt{2\pi}h(0)$ for every rapidly decreasing test
function $h$, since
(using the arguments of 284G) we should have to have
$\int_{a}^bg=\sqrt{2\pi}$ whenever $a<0<b$, so that the indefinite
integral of $g$ could not be continuous at $0$.   However there is a
{\it measure} on $\Bbb R$ with exactly the right property, the Dirac
measure $\delta_0$ concentrated at $0$;  this is a Radon probability
measure (257Xa), and $\int h\,d\delta_0=h(0)$ for
every function $h$ defined at $0$.   So we shall have

\Centerline{$\int_{-\infty}^{\infty}\chi\Bbb R\times\varhat{h}
=\sqrt{2\pi}\int h\,d\delta_0$}

\noindent for every rapidly decreasing test function $h$, and we can
reasonably say that the measure $\nu=\sqrt{2\pi}\delta_0$ `represents
the Fourier transform of $\chi\Bbb R$'.

We note with pleasure at this point that

\Centerline{$\Bover1{\sqrt{2\pi}}\int e^{ixy}\nu(dy)=1$}

\noindent for every $x\in\Bbb R$, so that $\chi\Bbb R$ can be called the
inverse Fourier transform of $\nu$.

If we look at the formulae of Theorem 284M, we get ideas consistent with
this pairing of $\chi\Bbb R$ with $\nu$.   We have

\Centerline{
$\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
  e^{-iyx}e^{-\epsilon x^2}\chi\Bbb R(x)dx
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-iyx}e^{-\epsilon x^2}dx
=\Bover1{\sqrt{2\epsilon}}e^{-y^2/4\epsilon}$}

\noindent for every $y\in\Bbb R$, using 283N with
$\sigma=1/\sqrt{2\epsilon}$.   So

\Centerline{$\lim_{\epsilon\downarrow 0}\Bover1{\sqrt{2\pi}}
\int_{-\infty}^{\infty}e^{-iyx}e^{-\epsilon x^2}\chi\Bbb R(x)dx=0$}

\noindent for every $y\ne 0$, and the Fourier transform of $\chi\Bbb R$
should be
zero everywhere except at $0$.   On the other hand, the functions
$y\mapsto \Bover1{\sqrt{2\epsilon}}e^{-y^2/4\epsilon}$ all have integral
$\sqrt{2\pi}$, concentrated more and more closely about $0$ as
$\epsilon$ decreases to $0$, so also point us directly to $\nu$, the
measure which gives mass $\sqrt{2\pi}$ to $0$.

Thus allowing measures, as well as functions, enables us to extend the
notion of Fourier transform.   Of course we can go very much farther
than this.   If $h$ is any rapidly decreasing test function, then (because
$\varcheck{h}\varhat{\phantom{h}}=h$)

\Centerline{$\int_{-\infty}^{\infty}x\varhat{h}(x)dx
=-i\sqrt{2\pi}h'(0)$,}

\noindent so that the identity function $x\mapsto x$ can be assigned, as
a Fourier transform, the operator $h\mapsto -i\sqrt{2\pi}h'(0)$.

At this point we are entering the true theory of (Schwartzian)
distributions or `generalized functions', and I had better stop.
The `Dirac delta function' is most naturally regarded as the measure
$\delta_0$ above;  alternatively, as
$\Bover1{\sqrt{2\pi}}\varhat{\chi\Bbb R}$.
}%end of comment

\exercises{\leader{284W}{The multidimensional case} As in \S283, I give
exercises
designed to point the way to the $r$-dimensional generalization.

\header{284Wa}{\bf (a)} A {\bf rapidly decreasing test function} on
$\BbbR^r$ is a function $h:\BbbR^r\to\Bbb C$ such that (i) $h$ is {\bf
smooth}, that is, all repeated partial derivatives

\Centerline{$\Bover{\partial^m h}{\partial
\xi_{j_1}\ldots\partial\xi_{j_m}}$}

\noindent are defined and continuous everywhere in $\BbbR^r$ (ii)

\Centerline{$\sup_{x\in\BbbR^r}\|x\|^k|h(x)|<\infty$,
\quad $\sup_{x\in\Bbb
R^r}\|x\|^k|\Bover{\partial^m h}{\partial
\xi_{j_1}\ldots\partial\xi_{j_m}}(x)|<\infty$}

\noindent for every $k\in\Bbb N$, $j_1,\ldots,j_m\le r$.   A {\bf
tempered function} on $\BbbR^r$ is a measurable complex-valued function
$f$, defined almost everywhere in $\BbbR^r$, such that, for some
$k\in\Bbb N$,

\Centerline{$\int_{\Bbb R^r}\Bover1{1+\|x\|^k}|f(x)|dx<\infty$.}

\noindent Show that if $f$ is a tempered function on $\BbbR^r$ and $h$
is a rapidly decreasing test function on $\BbbR^r$ then $f\times h$ is
integrable.

\header{284Wb}{\bf (b)} Show that if $h$ is a rapidly decreasing test
function on $\BbbR^r$ so is $\varhat{h}$, and that in this case
$\varhat{h}\varcheck{\phantom{h}}=h$.

\header{284Wc}{\bf (c)} Show that if $f$ is a tempered function on
$\Bbb R^r$ and $\int f\times h=0$ for every rapidly decreasing test
function $h$ on $\BbbR^r$, then $f=0$ a.e.

\header{284Wd}{\bf (d)} If $f$ and $g$ are tempered functions on
$\Bbb R^r$, I say that {\bf  $g$ represents the Fourier transform of $f$}
if $\int g\times h=\int f\times\varhat{h}$ for every rapidly decreasing
test function $h$ on $\BbbR^r$.   Show that if $f$ is integrable then
$\varhatf$ represents the Fourier transform of $f$ in this sense.

\header{284We}{\bf (e)} Let $f$ be any tempered function on $\BbbR^r$.
Writing $\psi_{\sigma}(x)
=\bover1{(\sigma\sqrt{2\pi})^r}e^{-x\dotproduct x/2\sigma^2}$
for $x\in\BbbR^r$, show that
$\lim_{\sigma\downarrow 0}(f*\psi_{\sigma})(x)=c$ whenever
$x\in\BbbR^r$, $c\in\Bbb C$ are such that
$\lim_{\delta\downarrow 0}\bover1{\delta^r}
\int_{B(x,\delta)}|f(t)-c|dt=0$, writing
$B(x,\delta)=\{t:\|t-x\|\le\delta\}$.

\header{284Wf}{\bf (f)} Let $f$ and $g$ be tempered functions on
$\BbbR^r$ such that $g$ represents the Fourier transform of $f$, and $h$
a rapidly decreasing test function.   Show that (i) the Fourier
transform of $f\times h$ is $\bover1{(\sqrt{2\pi})^r}g*\varhat{h}$ (ii)
$(\sqrt{2\pi})^rg\times\varhat{h}$ represents the Fourier transform of
$f*h$.

\header{284Wg}{\bf (g)} Let $f$ and $g$ be tempered functions on
$\BbbR^r$ such that $g$ represents the Fourier transform of $f$.   Show
that

\Centerline{$g(y)
=\lim_{\epsilon\downarrow 0}\Bover1{(\sqrt{2\pi})^r}
  \int_{\Bbb R^r}
  e^{-iy\dotproduct x}e^{-\epsilon x\dotproduct x}f(x)dx$}

\noindent for almost every $y\in\BbbR^r$.

\header{284Wh}{\bf (h)} Show that for any square-integrable
complex-valued function $f$ on $\BbbR^r$ and any $\epsilon>0$ there is
a \rdtf\ $h$ such that $\|f-h\|_2\le\epsilon$.

\header{284Wi}{\bf (i)} Let $\eusm L_{\Bbb C}^2$ be the space of
square-integrable complex-valued functions on $\BbbR^r$.   Show that

\quad (i) for every $f\in\eusm L_{\Bbb C}^2$ there is a
$g\in\eusm L_{\Bbb C}^2$ which
represents the Fourier transform of $f$, and in this case
$\|g\|_2=\|f\|_2$;

\quad (ii) if $g_1$, $g_2\in\eusm L_{\Bbb C}^2$ represent the Fourier
transforms of $f_1$, $f_2\in\eusm L_{\Bbb C}^2$, then
$\bover1{(\sqrt{2\pi})^r}g_1*g_2$ is
the Fourier transform of $f_1\times f_2$, and
$(\sqrt{2\pi})^rg_1\times g_2$ represents the Fourier transform of
$f_1*f_2$.

\spheader 284Wj Let $T$ be an invertible real $r\times r$ matrix, regarded
as a linear operator from $\BbbR^r$ to itself.   (i) Show that
$\varhatf=|\det T|(fT)\varsphat T\trs$ for every integrable
complex-valued function $f$ on $\BbbR^r$.   (ii) Show that $hT$
is a \rdtf\ for
every \rdtf\ $h$.   (iii) Show that if $f$, $g$ are a tempered functions
and $g$ represents the Fourier transform of $f$, then
$\Bover1{|\det T|}g(T\trs)^{-1}$ represents the Fourier transform of $fT$;
so that if $T$ is orthogonal, then $gT$ represents the
Fourier transform of $fT$.
%mt28bits
}%end of exercises 284W

\exercises{
\leader{284X}{Basic exercises (a)} Show that if $g$ and $h$ are rapidly
decreasing test functions, so is $g\times h$.
%284A

\spheader 284Xb Show that there are non-zero continuous integrable
functions $f$, $g:\Bbb R\to\Bbb C$ such that $f*g=0$ everywhere.
\Hint{take them to be \Ft s of suitable test functions.}
%284C

\spheader 284Xc Suppose that $f:\Bbb R\to\Bbb C$ is a
differentiable function such that its derivative $f'$ is a tempered
function and, for some $k\in\Bbb N$,

\Centerline{$\lim_{x\to\infty}x^{-k}f(x)
=\lim_{x\to-\infty}x^{-k}f(x)=0$.}

\noindent   (i) Show that $\int f\times h'=-\int f'\times h$ for every
rapidly decreasing test function $h$.   (ii) Show that if $g$ is a
tempered function representing the Fourier transform of $f$, then
$y\mapsto iyg(y)$ represents the Fourier transform of $f'$.
%284H

\spheader 284Xd\dvAnew{2013}
For a tempered function $f$ and $\alpha\in\Bbb R$, set

\Centerline{$(S_{\alpha}f)(x)=f(x+\alpha)$,
\quad$(M_{\alpha}f)(x)=e^{i\alpha x}f(x)$,
\quad$(D_{\alpha}f)(x)=f(\alpha x)$}

\noindent whenever these are defined.   (i) Show that $S_{\alpha}f$,
$M_{\alpha}f$ and (if $\alpha\ne 0$) $D_{\alpha}f$ are tempered functions.
(ii) Show that if $g$ is a tempered function which represents the Fourier
transform of $f$, then $M_{-\alpha}g$ represents the Fourier transform of
$S_{\alpha}f$, $S_{-\alpha}g$ represents the Fourier transform of
$M_{\alpha}f$, $\bar{\Reverse{g}}=\Reverse{\bar g}$ represents the 
Fourier transform of
$\bar f$, and if $\alpha\ne 0$ then $\Bover1{|\alpha|}D_{1/\alpha}g$
represents the Fourier transform of $D_{\alpha}f$.
%284I

\spheader 284Xe Show that if $h$ is a rapidly decreasing test
function and $f$ is any measurable complex-valued function, defined
almost everywhere in $\Bbb R$, such that
$\int_{-\infty}^{\infty}|x|^k|f(x)|dx<\infty$ for every $k\in\Bbb N$,
then the convolution $f*h$ is a rapidly decreasing test function.
\Hint{show that the Fourier transform of $f*h$ is a test
function.}
%284J

\sqheader 284Xf Let $f$ be a tempered function such that
$\lim_{a\to\infty}\int_{-a}^af$ exists in $\Bbb C$.   Show that
this limit is also equal to
$\lim_{\epsilon\downarrow 0}\int_{-\infty}^{\infty}
e^{-\epsilon x^2}f(x)dx$.
\Hint{set $g(x)=f(x)+f(-x)$.   Use 224J to show that if $0\le a\le b$
then $|\int_a^bg(x)e^{-\epsilon x^2}dx|\le\sup_{c\in[a,b]}|\int_a^cg|$,
so that $\lim_{a\to\infty}\int_0^ag(x)e^{-\epsilon x^2}dx$ exists
uniformly in $\epsilon$, while
$\lim_{\epsilon\downarrow 0}\int_0^ag(x)e^{-\epsilon x^2}dx
=\int_0^ag$ for every $a\ge 0$.}
%284L

\sqheader 284Xg Let $f$ and $g$ be tempered functions on
$\Bbb R$ such that $g$ represents the
Fourier transform of $f$.   Show that

\Centerline{$g(y)
=\lim_{a\to\infty}\Bover1{\sqrt{2\pi}}\int_{-a}^ae^{-iyx}f(x)dx$}

\noindent at almost all points $y$ for which the limit exists.
\Hint{284Xf, 284M.}
%284M, 284Xf, 284L

\sqheader 284Xh Let $f$ be an integrable complex-valued function
on $\Bbb R$ such that $\varhatf$ also is integrable.   Show that
$\varhatf\varcheck{\phantom{f}}=f$ at any point at which $f$ is
continuous.
%284M

\spheader 284Xi Show that for every $p\in\coint{1,\infty}$,
$f\in\eusm L_{\Bbb C}^p$ and $\epsilon>0$ there is a rapidly decreasing
test function $h$ such that $\|f-h\|_p\le\epsilon$.
%284N

\sqheader 284Xj Let $f$ and $g$ be square-integrable complex-valued
functions on $\Bbb R$ such that $g$ represents the Fourier transform of
$f$.   Show that

\Centerline{$\int_c^df=\Bover{i}{\sqrt{2\pi}}
\int_{-\infty}^{\infty}\Bover{e^{icy}-e^{idy}}{y}g(y)dy$}

\noindent whenever $c<d$ in $\Bbb R$.
%284O

\spheader 284Xk Let $f$ be a measurable complex-valued function,
defined almost everywhere in $\Bbb R$, such that $\int|f|^p<\infty$,
where $1<p\le 2$.   Show that $f$ is a tempered function and that there
is a tempered function $g$ representing the Fourier transform of $f$.
\Hint{express $f$ as $f_1+f_2$, where $f_1$ is integrable and
$f_2$ is square-integrable.}  ({\bf Remark} Defining $\|f\|_p$,
$\|g\|_q$ as in 244D, where $q=p/(p-1)$, we have
$\|g\|_q\le(2\pi)^{(p-2)/2p}\|f\|_p$;  see {\smc Zygmund 59}, XVI.3.2.)
%284O

\spheader 284Xl Let $f$, $g$ be square-integrable
complex-valued functions on $\Bbb R$ such that $g$ represents the
Fourier transform of $f$.

\quad{(i)} Show that

\Centerline{$\Bover1{\sqrt{2\pi}}\int_{-a}^ae^{ixy}g(y)dy
=\Bover1{\pi}\int_{-\infty}^{\infty}\Bover{\sin at}{t}f(x-t)dt$}

\noindent whenever $x\in\Bbb R$ and $a>0$.   \Hint{find the
inverse Fourier transform of $y\mapsto e^{-ixy}\chi[-a,a](y)$, and use
284Ob.}

\quad{(ii)} Show that if $f(x)=0$ for $x\in\ooint{c,d}$ then

\Centerline{$\Bover1{\sqrt{2\pi}}\lim_{a\to\infty}\int_{-a}^a
e^{ixy}g(y)dy=0$}

\noindent for $x\in\ooint{c,d}$.

\quad{(iii)} Show that if $f$ is differentiable at $x\in\Bbb R$,
then

\Centerline{$\Bover1{\sqrt{2\pi}}\lim_{a\to\infty}\int_{-a}^a
e^{ixy}g(y)dy=f(x)$.}

\quad{(iv)} Show that if $f$ has bounded variation over some
interval properly containing $x$, then

\Centerline{$\Bover1{\sqrt{2\pi}}\lim_{a\to\infty}\int_{-a}^a
e^{ixy}g(y)dy=\Bover12(\lim_{t\in\dom f,t\uparrow x}f(t)+\lim_{t\in\dom
f,t\downarrow x}f(t))$.}
%284O

\spheader 284Xm Let $f$ be an integrable complex function on $\Bbb R$.
Show that if $\varhatf$ is square-integrable, so is $f$.
%284O

\spheader 284Xn Let $f_1$, $f_2$ be square-integrable complex-valued
functions on $\Bbb R$ with \Ft s represented by $g_1$, $g_2$.   Show
that
$\int_{-\infty}^{\infty}f_1(t)f_2(-t)dt
=\int_{-\infty}^{\infty}g_1(t)g_2(t)dt$.
%284O

\spheader 284Xo Suppose $x\in\Bbb R$.
Write $\delta_x$ for Dirac measure on $\Bbb R$ concentrated
at $x$.   Describe a sense in which $\sqrt{2\pi}\delta_x$ can
be regarded as
the Fourier transform of the function $t\mapsto e^{ixt}$.
%284R

\spheader 284Xp For any tempered function $f$ and $x\in\Bbb R$,
let $\delta_x$ be the Dirac measure on $\Bbb R$ concentrated at $x$,
and set

\Centerline{$(\delta_x*f)(u)=\int f(u-t)\delta_x(dt)=f(u-x)$}

\noindent for every $u$ for which $u-x\in\dom f$ (cf.\ 257Xe).   If $g$
represents
the Fourier transform of $f$, find a corresponding representation of the
Fourier transform of $\delta_x*f$, and relate it to the product of $g$
with the Fourier transform of $\delta_x$.
%284R, 284Xo

\spheader 284Xq(i) Show that

\Centerline{$\lim_{\delta\downarrow 0,a\to\infty}
\bigl(\int_{-a}^{-\delta}\Bover1x e^{-iyx}dx
+\int_{\delta}^a\Bover1x e^{-iyx}dx\bigr)
=-\pi i\sgn y$}

\noindent for every $y\in\Bbb R$, writing $\sgn y=y/|y|$ if $y\ne 0$ and
$\sgn 0=0$.   \Hint{283Da.}

\quad{(ii)} Show that

\Centerline{$\lim_{c\to\infty}\Bover1c\int_0^c
  \int_{-a}^ae^{ixy}\sgn y\,dy\,da
=\Bover{2i}x$}

\noindent for every $x\ne 0$.

\quad{(iii)} Show that for any rapidly decreasing test function $h$,

$$\eqalign{\int_0^{\infty}\Bover1x(\varhat{h}(x)-\varhat{h}(-x))dx
&=\lim_{\delta\downarrow 0,a\to\infty}\bigl(\int_{-a}^{-\delta}
\Bover1x\varhat{h}(x)dx
+\int_{\delta}^{a}\Bover1x\varhat{h}(x)dx\bigr)\cr
&=-\Bover{i\pi}{\sqrt{2\pi}}\int_{-\infty}^{\infty}h(y)\sgn
y\,dy.\cr}$$

\quad{(iv)} Show that for any rapidly decreasing test function $h$,

\Centerline{$\Bover{i\pi}{\sqrt{2\pi}}\int_{-\infty}^{\infty}
  \varhat{h}(x)\sgn x\,dx
=\int_0^{\infty}\Bover1y(h(y)-h(-y))dy$.}
%284notes

\spheader 284Xr Let $\sequencen{h_n}$ be a sequence of rapidly
decreasing test functions such that
$\phi(f)=\lim_{n\to\infty}\int_{-\infty}^{\infty}h_n\times f$ is defined
for every rapidly decreasing test function $f$.   Show that
$\lim_{n\to\infty}\int_{-\infty}^{\infty}h'_n\times f$,
$\lim_{n\to\infty}\int_{-\infty}^{\infty}\varhat{h}_n\times f$ and
$\lim_{n\to\infty}\int_{-\infty}^{\infty}(h_n*g)\times f$ are defined
for all rapidly decreasing test functions $f$ and $g$, and are zero if
$\phi$ is identically zero.   \Hint{255G will help with the last.}
%284+

\leader{284Y}{Further exercises (a)}
%\spheader 284Ya
Let $f$ be an integrable complex-valued function
on $\ocint{-\pi,\pi}$, and $\tilde f$ its periodic extension, as in
282Ae.   Show that $\tilde f$ is a tempered function.   Show that for
any rapidly decreasing test function $h$,
$\int\tilde f\times\varhat{h}
=\sqrt{2\pi}\sum_{k=-\infty}^{\infty}c_kh(k)$, where
$\sequence{k}{c_k}$ is the sequence of Fourier coefficients of $f$.
({\it Hint\/}:  begin with the case $f(x)=e^{inx}$.   Next show that

\Centerline{$M=\sum_{k=-\infty}^{\infty}|h(k)|
+\sum_{k=-\infty}^{\infty}
\sup_{x\in[(2k-1)\pi,(2k+1)\pi]}|\varhat{h}(x)|<\infty$,}

\noindent and that

\Centerline{$|\int\tilde f\times\varhat{h}
-\sqrt{2\pi}\sum_{k=-\infty}^{\infty}c_kh(k)|
\le M\|f\|_1$.}

\noindent Finally apply 282Ib.)
%284C

\header{284Yb}{\bf (b)} Let $f$ be a complex-valued function, defined
almost everywhere in $\Bbb R$, such that $f\times h$ is integrable for
every rapidly decreasing test function $h$.   Show that $f$ is tempered.
%284F

\spheader 284Yc Let $f$ and $g$ be tempered functions on $\Bbb
R$ such that $g$ represents the Fourier transform of $f$.   Show that

$$\int_c^df
=\Bover{i}{\sqrt{2\pi}}\lim_{\sigma\to\infty}\int_{-\infty}^{\infty}
\Bover{e^{icy}-e^{idy}}{y}e^{-y^2/2\sigma^2}g(y)dy$$

\noindent whenever $c\le d$ in $\Bbb R$.   \Hint{set
$\theta=\chi[c,d]$.   Show that both sides are
$\lim_{\sigma\to\infty}\int f\times(\theta*\psi_{1/\sigma})$, defining
$\psi_{\sigma}$ as in 283N and 284L.}
%284H

\spheader 284Yd Show that if $g:\Bbb R\to\Bbb R$ is an odd
function of bounded variation such that
$\int_1^{\infty}\bover1x g(x)dx=\infty$, then $g$ does not represent the
Fourier transform of any tempered function.   \Hint{283Ye, 284Yc.}
%284Yc, 284H

\ifdim\pagewidth>467pt\fontdimen3\tenrm=1.84pt
  \fontdimen4\tenrm=1.22pt\fi
\spheader 284Ye
Let $\eusm S$ be the space of
rapidly decreasing test functions.   For $k$, $m\in\Bbb N$ set
$\tau_{km}(h)=\sup_{x\in\Bbb R}|x|^k|h^{(m)}(x)|$ for every
$h\in\eusm S$, writing $h^{(m)}$ for the $m$th derivative of $h$ as
usual.   (i) Show
that each $\tau_{km}$ is a seminorm and that $\eusm S$ is
complete and separable for
the metrizable linear space topology $\frak T$ they define.
(ii) Show that $h\mapsto\varhat{h}:\eusm S\to\eusm S$
is continuous for $\frak T$.   (iii) Show that if $f$ is any
tempered function, then $h\mapsto\int f\times h$ is
$\frak T$-continuous.   (iv) Show that if $f$ is an integrable
function such that $\int|x^kf(x)|dx<\infty$ for every $k\in\Bbb N$, then
$h\mapsto f*h:\eusm S\to\eusm S$ is $\frak T$-continuous.
%284J
\fontdimen3\tenrm=1.67pt\fontdimen4\tenrm=1.22pt

\spheader 284Yf Show that if $f$ is a tempered function on $\Bbb R$ and

\Centerline{$\gamma
=\lim_{c\to\infty}\Bover1c\int_{0}^c\int_{-a}^af(x)dxda$}

\noindent is defined in $\Bbb C$, then $\gamma$ is also

\Centerline{$\lim_{\epsilon\downarrow 0}\int_{-\infty}^{\infty}
f(x)e^{-\epsilon|x|}dx$.}
%284Xf, 284L

\spheader 284Yg Let $f$, $g$ be square-integrable
complex-valued functions on $\Bbb R$ such that $g$ represents the
Fourier transform of $f$.   Suppose that $m\in\Bbb Z$ and that
$(2m-1)\pi<x<(2m+1)\pi$.   Set $\tilde f(t)=f(t+2m\pi)$ for those
$t\in\ocint{-\pi,\pi}$ such that $t+2m\pi\in\dom f$.   Let
$\langle c_k\rangle_{k\in\Bbb Z}$ be the sequence of Fourier
coefficients of $\tilde f$.   Show that

\Centerline{$\Bover1{\sqrt{2\pi}}\lim_{a\to\infty}\int_{-a}^a
e^{ixy}g(y)dy
=\lim_{n\to\infty}\sum_{k=-n}^nc_ke^{ikx}$}

\noindent in the sense that if one limit exists in $\Bbb C$ so does the
other, and they are then equal.   \Hint{284Xl(i), 282Da.}
%284O, 284Xl

\spheader 284Yh Show that if $f$ is integrable over $\Bbb R$ and there
is some $M\ge 0$ such that $f(x)=\varhatf(x)=0$ for $|x|\ge M$, then
$f=0$ a.e.   \Hint{reduce to the case $M=\pi$.   Looking at the Fourier
series of $f\restr\ocint{-\pi,\pi}$, show that $f$ is expressible in the
form $f(x)=\sum_{k=-m}^mc_ke^{ikx}$ for almost every
$x\in\ocint{-\pi,\pi}$.   Now compute $\varhatf(2n+\bover12)$ for large
$n$.}

\spheader 284Yi Let $\nu$ be a Radon measure on $\Bbb R$ 
which is `tempered' in the sense that
$\int_{-\infty}^{\infty}\Bover1{1+|x|^k}\nu(dx)$ is finite for some
$k\in\Bbb N$.   (i) Show that every rapidly decreasing test function is
$\nu$-integrable.   (ii) Show that if $\nu$ has bounded support
(definition:  256Xf), and $h$ is a rapidly decreasing test function,
then $\nu*h$ is a rapidly decreasing test function, where
$(\nu*h)(x)=\int_{-\infty}^{\infty}h(x-y)\nu(dy)$ for $x\in\Bbb R$.
(iii) Show that there is a sequence $\sequencen{h_n}$ of rapidly
decreasing test functions such that
$\lim_{n\to\infty}\int_{-\infty}^{\infty}h_n\times f
=\int_{-\infty}^{\infty}fd\nu$ for every rapidly decreasing test
function $f$.
%284Xr

\spheader 284Yj Let $\phi:\eusm S\to\Bbb R$ be a functional defined by
the formula of 284Xr.   Show that $\phi$ is continuous for the topology
of 284Ye.
({\it Note\/}:  it helps to know a little more about metrizable linear
topological spaces than is covered in \S2A5.)
%284Xr

}%end of exercises

\cmmnt{
\Notesheader{284} Yet again I must warn you that the
material above gives a very restricted view of the subject.   I have
tried to indicate how the theory of Fourier transforms of `good'
functions -- here taken to be the rapidly decreasing test
functions -- may be extended, through a kind of duality,
to a very much wider class
of functions, the `tempered functions'.   Evidently, writing $\eusm S$
for the linear space of rapidly decreasing test functions, we can seek
to investigate a Fourier transform of any linear functional
$\phi:\eusm S\to\Bbb C$, writing $\varhat\phi(h)=\phi(\varhat{h})$ for
any $h\in\eusm S$.   (It is actually commoner at this point to restrict
attention to functionals $\phi$ which are continuous for the standard
topology on $\eusm S$, described in 284Ye;  these are called 
{\bf tempered (Schwartzian) distributions}.)   
By 284F-284G, we can identify some of these
functionals with equivalence classes of tempered functions, and then set
out to investigate those tempered functions whose Fourier transforms can
again be represented by tempered functions.

I suppose the structure of the theory of Fourier transforms is best laid
out through the formulae involved.   Our aim is to set up pairs
$(f,g)=(f,\varhatf)=(\varcheck{g},g)$ in such a way that we have

\quad {\it Inversion\/}:  $\varhat{h}\varcheck{\phantom{h}}
=\varcheck{h}\varhat{\phantom{h}}=h$;   %284If

\quad {\it Reversal\/}:  $\varcheck{h}(y)=\varhat{h}(-y)$;  %284If

\quad {\it Linearity\/}:
$(h_1+h_2)\varsphat=\varhat{h}_1+\varhat{h}_2$,
\quad$(ch)\varsphat=c\varhat{h}$;  %284Ic

\quad {\it Differentiation\/}: $(h')\varsphat(y)=iy\varhat{h}(y)$;

\quad {\it Shift\/}:  if $h_1(x)=h(x+c)$ then
$\varhat{h}_1(y)=e^{iyc}\varhat{h}(y)$;  %284Xd

\quad {\it Modulation\/}:  if $h_1(x)=e^{icx}h(x)$ then
$\varhat{h}_1(y)=\varhat{h}(y-c)$;  %284Xd

\quad {\it Symmetry\/}:  if $h_1(x)=h(-x)$ then $\varhat{h}_1(y)
=\varhat{h}(-y)$;  %284Xd

\quad {\it Complex Conjugate\/}:
$(\overline{h})\varsphat(y)=\overline{\varhat{h}(-y)}$;  %284Xd

\quad {\it Dilation\/}:  if $h_1(x)=h(cx)$, where $c>0$, then
$\varhat{h}_1(y)=\Bover1c\varhat{h}(\Bover{y}{c})$;  %284Xd

\quad {\it Convolution\/}:
$(h_1*h_2)\varsphat=\sqrt{2\pi}\varhat{h}_1\times\varhat{h}_2$, %284Kb
\quad $(h_1\times h_2)\varsphat
=\Bover1{\sqrt{2\pi}}\varhat{h}_1*\varhat{h}_2$; %284Ka

\quad {\it Duality\/}: $\int_{-\infty}^{\infty}h_1\times\varhat{h}_2
=\int_{-\infty}^{\infty}\varhat{h}_1\times h_2$;

\quad {\it Parseval\/}:  $\int_{-\infty}^{\infty}h_1\times\overline{h_2}
=\int_{-\infty}^{\infty}\varhat{h}_1\times\overline{\varhat{h}_2}$; %284Ob

\noindent and, of course,

\Centerline{$\varhat{h}(y)
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-iyx}h(x)dx$,} %284Xg

\Centerline{$\int_c^d\varhat{h}(y)dy
=\Bover{i}{\sqrt{2\pi}}\int_{-\infty}^{\infty}
\Bover{e^{-icy}-e^{-idy}}{y}h(y)dy$.}  %284Xj

\noindent (I have used the letter $h$ in the list above to suggest what
is in fact the case, that all the formulae here are valid for rapidly
decreasing test functions.)   On top of all this, it is often important
that the operation $h\mapsto\varhat{h}$ should be continuous in some
sense.

The challenge of the `pure' theory of Fourier transforms is to find
the widest possible variety of objects $h$ for which the formulae above
will be valid, subject to appropriate interpretations of $\varsphat$,
$*$ and $\int_{-\infty}^{\infty}$.
I must of course remark here that from the very beginnings, the subject
has been enriched by its applications in other parts of mathematics, the
physical sciences and the social sciences, and that again and again
these have suggested further possible pairs $(f,\varhatf)$, making new
demands on our power to interpret the rules we seek to follow.   Even
the theory of distributions does not seem to give a full canonical
account of what can be done.   First, there are great difficulties in
interpreting the `product' of two arbitrary distributions, making
several of the formulae above problematic;  and second, it is not
obvious that only one kind of distribution need be considered.   In this
section I have looked at just one space of `test functions', the
space $\eusm S$ of rapidly decreasing test functions;  but at least two
others are significant, the space $\eusm D$ of smooth functions with
bounded support and the space $\eusm Z$ of Fourier transforms of
functions in
$\eusm D$.   The advantage of starting with $\eusm S$ is that it gives a
symmetric theory, since $\varhat{h}\in\eusm S$ for every $h\in\eusm S$;
but it is easy to find objects (e.g., the function $x\mapsto e^{x^2}$,
or the function $x\mapsto 1/|x|$) which cannot be interpreted as
functionals on $\eusm S$, so that their Fourier transforms must be
investigated by other methods, if at all.   In 284Xq I sketch some of
the arguments which can be used to justify the assertion that the
Fourier transform of the function $x\mapsto 1/x$ is, or can be
represented by, the function
$y\mapsto -i\sqrt{\bover{\pi}2}\sgn y$;  the general principle in this
case being that we approach both $0$ and $\infty$ symmetrically.
For a variety of such matching pairs, established by arguments based on
the idea in 284Xr, see {\smc Lighthill 59}, chap.\ 3.

Accordingly it seems that, after two centuries,
% Fourier "Analytic Theory of Heat" 1822
%    1807 Mémoire sur la propagation de la chaleur dans les corps solides
% Wikipedia:  "When Fourier submitted a later competition essay in 1811,
% the committee (which included Lagrange, Laplace, Malus and Legendre,
% among others) concluded: ...the manner in which the author arrives at
% these equations is not exempt of difficulties and...his analysis to
% integrate them still leaves something to be desired on the score of
% generality and even rigour [citation needed]"
we must still
proceed by carefully examining particular classes of function, and
checking appropriate interpretations of the formulae.   In the work
above I have repeatedly used the concepts

\Centerline{$\lim_{a\to\infty}\int_{-a}^af$,
\quad$\lim_{\epsilon\downarrow 0}\int_{-\infty}^{\infty}
  e^{-\epsilon x^2}f(x)dx$}

\noindent as alternative interpretations of $\int_{-\infty}^{\infty}f$.
(Of course they are closely related;  see 284Xf.)   The reasons for
using the particular kernel $e^{-\epsilon x^2}$ are that it belongs to
$\eusm S$, it is an even function,
its Fourier transform is calculable and easy to manipulate, and it is
associated with the normal probability density function
$\bover1{\sigma\sqrt{2\pi}}e^{-x^2/2\sigma^2}$, so that any
miscellaneous facts we gather have a chance of being valuable elsewhere.
But there are applications in which alternative kernels are more
manageable -- e.g., $e^{-\epsilon|x|}$ (283Xq, 283Yc, 284Yf).

One of the guiding principles here is that purely formal manipulations,
along the lines of those in the list above, and (especially) changes in
the order of integration, with other exchanges of limit, again and again
give rise to formulae which, suitably interpreted, are valid.   First
courses in analysis are often inhibitory;  students are taught to
distrust any manipulation which they cannot justify.   To my own eye,
the delight of this topic lies chiefly in the variety of the arguments
demanded by a rigorous approach, the ground constantly shifting with the
context;  but there is no doubt that cheerful sanguinity is often the
best guide to the manipulations which it will be right to try to
justify.

This being a book on measure theory, I am of course particularly
interested in the possibility of a measure appearing as a Fourier
transform.   This is what happens if we seek the Fourier transform of
the constant function $\chi\Bbb R$ (284R).   More generally, any periodic
tempered function $f$ with period $2\pi$ can be assigned a Fourier
transform which is a `signed measure' (for our present purposes, a
complex linear combination of measures) concentrated on $\Bbb Z$, the
mass at each $k\in\Bbb Z$ being determined by the corresponding Fourier
coefficient of $f\restr\ocint{-\pi,\pi}$ (284Xo, 284Ya).   In the next
section I will go farther in this direction, with particular reference
to probability distributions on $\BbbR^r$.   But the reason why {\it
positive} measures have not forced themselves on our attention so far is
that we do not expect to get a positive function as a Fourier transform
unless some very special conditions are satisfied, as in 283Yc.

As in \S282, I have used the Hilbert space structure of $L_{\Bbb C}^2$
as the basis of the discussion of Fourier transforms of functions in
$\eusm L_{\Bbb C}^2$ (284O-284P).   But as with Fourier series,
Carleson's theorem (286U) provides a more direct description.

In 284Wj, I offer a calculation based on the change-of-variable formula in
263A to present a multidimensional
version of Reversal and Dilation.   But what I am really trying to do is
to show that Fourier transforms on
$\BbbR^r$ are based on the geometry of the
Euclidean inner product, not on the Cartesian coordinate system.
}%end of notes

\discrpage

\frfilename{mt285.tex}
\versiondate{18.9.14}
\copyrightdate{1994}

\def\chaptername{Fourier analysis}
\def\sectionname{Characteristic functions}

\newsection{285}

I come now to one of the most effective applications of Fourier
transforms, the use of `characteristic functions' to analyse
probability distributions.   It turns out not only that the Fourier
transform of a probability distribution determines the distribution
(285M) but that many of the things we want to know about a
distribution are easily calculated from its
transform\cmmnt{ (285G, 285Xi)}.    Even more strikingly, pointwise
convergence of
Fourier transforms corresponds (for sequences) to convergence for the
vague topology in
the space of distributions, so they provide a new and extremely powerful
method for proving such results as the Central Limit Theorem and
Poisson's theorem (285Q).

As the applications of the ideas here mostly belong to probability
theory, I return to probabilists' terminology, as in Chapter 27.
There will nevertheless be many points at which it is appropriate to
speak of integrals, and there will often be more than one measure in play;
so I should say directly that an integral $\int f(x)dx$ will be with
respect to Lebesgue measure (usually, but not always, one-dimensional),
as in the rest of this chapter,
while integrals with respect to other measures will be expressed
in the forms $\int fd\nu$ or $\int f(x)\nu(dx)$.

\leader{285A}{Definition (a)} Let $\nu$ be a Radon probability measure
on $\BbbR^r$\cmmnt{ (256A)}.   Then the {\bf characteristic function}
of $\nu$ is the function $\varphi_{\nu}:\BbbR^r\to\Bbb C$ given by the
formula

\Centerline{$\varphi_{\nu}(y)=\int e^{iy\dotproduct x}\nu(dx)$}

\noindent for every $y\in\BbbR^r$, writing
$y\dotproduct x=\eta_1\xi_1+\ldots+\eta_r\xi_r$ if
$y=(\eta_1,\ldots,\eta_r)$ and $x=(\xi_1,\ldots,\xi_r)$.

\header{285Ab}{\bf (b)} Let $X_1,\ldots,X_r$ be real-valued random
variables on the same probability space.   The {\bf characteristic
function} of $\pmb{X}=(X_1,\ldots,X_r)$ is the characteristic function
$\varphi_{\pmb{X}}=\varphi_{\nu_{\pmb{X}}}$ of their joint probability
distribution $\nu_{\pmb{X}}$ as defined in 271C.

\cmmnt{
\leader{285B}{Remarks (a)} By one of the ordinary accidents of history,
the definitions of `characteristic function' and `Fourier
transform' have evolved independently.   In 283Ba I remarked that the
definition of the Fourier transform remains unfixed, and that the
formulae

\Centerline{$\varhatf(y)=\int_{-\infty}^{\infty}e^{iyx}f(x)dx$,}

\Centerline{$\varcheckf(y)
=\Bover1{2\pi}\int_{-\infty}^{\infty}e^{-iyx}f(x)dx$}

\noindent are sometimes used.   On the other hand, I think that nearly
all authors agree on the definition of the characteristic function as
given above.   You may feel therefore that I should have followed their
lead, and chosen the definition of Fourier transform which best matches
the definition of characteristic function.   I did not do so largely
because I wished to emphasise the symmetry between the Fourier transform
and the inverse Fourier transform, and the correspondence between
Fourier transforms and Fourier series.   The principal advantage of
matching the definitions up would be to make the constants in such
theorems as 283F, 285Xk the same, and would be balanced by the need
to remember different constants for $\varhatf$ and $\varcheckf$ in such
results as 283M.

\header{285Bb}{\bf (b)} A secondary reason for not trying too hard to
make the formulae of this section match directly those of \S\S283-284
is that the $r$-dimensional case is at the heart of some of the most
important
applications of characteristic functions, so that it seems right to
introduce it from the beginning;  and
consequently the formulae of this section will necessarily have new
features compared with those in the body of the work so far.
}

\leader{285C}{}\cmmnt{ Of course there is a direct way to describe the
characteristic function of a family $(X_1,\ldots,X_r)$ of random
variables, as follows.

\medskip

\noindent}{\bf Proposition} Let $X_1,\ldots,X_r$ be real-valued random
variables on the same probability space, and
$\nu_{\pmb{X}}$ their joint distribution.   Then their characteristic
function $\varphi_{\nu_{\pmb{X}}}$ is given by

\Centerline{$\varphi_{\nu_{\pmb{X}}}(y)=\Expn(e^{iy\dotproduct \pmb{X}})
=\Expn(e^{i\eta_1X_1}e^{i\eta_2X_2}\ldots e^{i\eta_rX_r})$}

\noindent for every $y=(\eta_1,\ldots,\eta_r)\in\BbbR^r$.

\proof{ Apply 271E to the functions $h_1$, $h_2:\Bbb R^r\to\Bbb R$
defined by

\Centerline{$h_1(x)=\cos(y\dotproduct x)$,\quad
$h_2(y)=\sin(y\dotproduct x)$,}

\noindent to see that

$$\eqalign{\varphi_{\nu_{\pmb{X}}}(y)
&=\int h_1(x)\nu_{\pmb{X}}(dx) + i\int h_2(x)\nu_{\pmb{X}}(dx)\cr
&=\Expn(h_1(\pmb{X})) + i\Expn(h_2(\pmb{X}))
=\Expn(e^{iy\dotproduct \pmb{X}}).\cr}$$
}%end of proof of 285C

\vleader{72pt}{285D}{}\cmmnt{ I ought to spell out the correspondence
between Fourier transforms, as defined in 283A, and characteristic
functions.

\medskip

\noindent}{\bf Proposition} Let $\nu$ be a Radon probability measure on
$\Bbb R$.   Write

\Centerline{$\varhat{\nu}(y)
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
e^{-iyx}\nu(dx)$}

\noindent for every $y\in\Bbb R$, and $\varphi_{\nu}$ for the
characteristic function of $\nu$.


(a) $\varhat{\nu}(y)=\Bover1{\sqrt{2\pi}}\varphi_{\nu}(-y)$ for every
$y\in\Bbb R$.

(b) For any Lebesgue integrable complex-valued function $h$ defined
almost everywhere in $\Bbb R$,

\Centerline{$\int_{-\infty}^{\infty}\varhat{\nu}(y)h(y)dy
=\int_{-\infty}^{\infty}\varhat{h}(x)\nu(dx)$.}

(c) For any rapidly decreasing test function $h$ on $\Bbb R$\cmmnt{ (see
\S284)},

\Centerline{$\int_{-\infty}^{\infty}h(x)\nu(dx)
=\int_{-\infty}^{\infty}\varcheck{h}(y)\varhat{\nu}(y)dy$.}

(d) If $\nu$ is an indefinite-integral measure over Lebesgue measure,
with Radon-Nikod\'ym derivative $f$, then $\varhat{\nu}$ is the Fourier
transform of $f$.

\proof{{\bf (a)} This is immediate from the definitions of
$\varphi_{\nu}$ and $\varhat{\nu}$.

\medskip

{\bf (b)} Because

\Centerline{$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
|h(y)|\nu(dx)dy=\int_{-\infty}^{\infty}|h(y)|dy<\infty$,}

\noindent we may change the order of integration to see that

$$\eqalign{\int_{-\infty}^{\infty}\varhat{\nu}(y)h(y)dy
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
e^{-iyx}h(y)\nu(dx)dy\cr
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
e^{-iyx}h(y)dy\,\nu(dx)
=\int_{-\infty}^{\infty}\varhat{h}(x)\nu(dx).\cr}$$

\medskip

{\bf (c)} This follows immediately from (b), because $\varcheck{h}$ is
integrable and $\varcheck{h}\varhat{\phantom{h}}=h$ (284C).

\medskip

{\bf (d)} The point is just that

\Centerline{$\int h\,d\nu=\int h(x)f(x)dx$}

\noindent for every bounded Borel measurable $h:\Bbb R\to\Bbb R$ (235K),
and therefore for the functions $x\mapsto e^{-iyx}:\Bbb R\to\Bbb C$.
Now

\Centerline{$\varhat{\nu}(y)
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-iyx}\nu(dx)
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-iyx}f(x)dx
=\varhatf(y)$}

\noindent for every $y$.
}% end of proof of 285D

\leader{285E}{Lemma} Let $X$ be a normal random variable with expectation
$a$ and variance $\sigma^2$, where $\sigma>0$.   Then the characteristic
function of $X$ is given by the formula

\Centerline{$\varphi(y)=e^{iya}e^{-\sigma^2y^2/2}$.}

\proof{ This is just 283N with the constants changed.   We have

$$\eqalignno{\varphi(y)
&=\Expn(e^{iyX})
=\Bover1{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty}
  e^{iyx}e^{-(x-a)^2/2\sigma^2}dx\cr
\noalign{\noindent (taking the density function for $X$ given in 274Ad,
and applying 271Ic)}
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
  e^{iy(\sigma t+a)}e^{-t^2/2}dt\cr
\noalign{\noindent (substituting $x=\sigma t+a$)}
&=e^{iya}\sqrt{2\pi}\varhat{\psi}_1(-y\sigma)\cr
\noalign{\noindent (setting $\psi_1(x)=\bover1{\sqrt{2\pi}}e^{-x^2/2}$,
as in 283N)}
&=e^{iya}e^{-\sigma^2y^2/2}.\cr}$$
}%end of proof of 285E

\leader{285F}{}\cmmnt{ I now give results corresponding to parts of
283C, with an
extra refinement concerning independent random variables (285I).

\medskip

\noindent}{\bf Proposition} Let $\nu$ be a Radon probability measure on
$\BbbR^r$, and $\varphi$ its characteristic function.

(a) $\varphi(0)=1$.

(b) $\varphi:\BbbR^r\to\Bbb C$ is uniformly continuous.

(c) $\varphi(-y)=\overline{\varphi(y)}$, $|\varphi(y)|\le 1$ for every $y\in\Bbb
R^r$.

(d) If $r=1$ and $\int|x|\nu(dx)<\infty$, then $\varphi'(y)$ exists and is
equal to $i\int xe^{ixy}\nu(dx)$ for every $y\in\Bbb R$.

(e) If $r=1$ and $\int x^2\nu(dx)<\infty$, then $\varphi''(y)$ exists and
is equal to $-\int x^2e^{ixy}\nu(dx)$ for every $y\in\Bbb R$.

\proof{{\bf (a)} $\varphi(0)=\int\chi\BbbR^r\,d\nu=\nu(\BbbR^r)=1$.

\medskip

{\bf (b)} Let $\epsilon>0$.   Let $M>0$ be such that

\Centerline{$\nu\{x:\|x\|\ge M\}\le\epsilon$,}

\noindent writing $\|x\|=\sqrt{x\dotproduct x}$ as usual.   Let
$\delta>0$
be such that $|e^{ia}-1|\le\epsilon$ whenever $|a|\le\delta$.   Now
suppose that $y$, $y'\in\BbbR^r$ are such that $\|y-y'\|\le\delta/M$.
Then whenever $\|x\|\le M$,

\Centerline{$|e^{iy\dotproduct x}-e^{iy'\dotproduct x}|
=|e^{iy'\dotproduct x}||e^{i(y-y')\dotproduct x}-1|
=|e^{i(y-y')\dotproduct x}-1|
\le\epsilon$}

\noindent because

\Centerline{$|(y-y')\dotproduct x|\le\|y-y'\|\|x\|\le\delta$.}

\noindent Consequently, writing $B$ for $\{x:\|x\|\le M\}$,

$$\eqalign{|\varphi(y)-\varphi(y')|
&\le\int_B|e^{iy\dotproduct x}-e^{iy'\dotproduct x}|\nu(dx)
\cr&\qquad\qquad
  +\int_{\BbbR^r\setminus B}|e^{iy\dotproduct x}|\nu(dx)
  +\int_{\BbbR^r\setminus B}|e^{iy'\dotproduct x}|\nu(dx)\cr
&\le\epsilon+\epsilon+\epsilon
=3\epsilon.\cr}$$

\noindent As $\epsilon$ is arbitrary, $\varphi$ is uniformly continuous.


\medskip

{\bf (c)} This is elementary;

\Centerline{$\varphi(-y)=\int e^{-iy\dotproduct x}\nu(dx)
=\overline{\intop e^{iy\dotproduct x}\nu(dx)}=\overline{\varphi(y)}$,}

\Centerline{$|\varphi(y)|=|\int e^{iy\dotproduct x}\nu(dx)|
\le\int|e^{iy\dotproduct x}|\nu(dx)
=1$.}

\medskip

{\bf (d)} The point is that $|\pd{}{y}e^{iyx}|=|x|$ for every $x$,
$y\in\Bbb R$.   So by 123D (applied, strictly speaking, to the real and
imaginary parts of the function)

\Centerline{$\varphi'(y)
=\Bover{d}{dy}\int e^{iyx}\nu(dx)
=\int\Pd{}{y}e^{iyx}\nu(dx)
=\int ixe^{iyx}\nu(dx)$.}

\medskip

{\bf (e)} Since we now have $|\pd{}{y}xe^{iyx}|=x^2$ for every $x$, $y$,
we can repeat the argument to get

\Centerline{$\varphi''(y)
=i\Bover{d}{dy}
     \int xe^{iyx}\nu(dx)
=i\int
     \Pd{}{y}xe^{iyx}\nu(dx)
=-\int
     x^2e^{iyx}\nu(dx)$.}
}%end of proof of 285F

\vleader{48pt}{285G}{Corollary} (a) Let $X$ be a real-valued random
variable
with finite expectation, and $\varphi$ its characteristic function.   Then
$\varphi'(0)=i\Expn(X)$.

(b) Let $X$ be a real-valued random variable with
finite variance, and $\varphi$ its characteristic function.   Then
$\varphi''(0)=-\Expn(X^2)$.

\proof{ We have only to match $X$ to its distribution
$\nu$, and say that

\Centerline{`$X$ has finite expectation'}

\noindent corresponds to

\Centerline{`$\int |x|\nu(dx)=\Expn(|X|)<\infty$',}

\noindent so that

\Centerline{$\varphi'(0)=i\int x\,\nu(dx)=i\Expn(X)$,}

\noindent and that

\Centerline{`$X$ has finite variance'}

\noindent corresponds to

\Centerline{`$\int x^2\nu(dx)=\Expn(X^2)<\infty$',}

\noindent so that

\Centerline{$\varphi''(0)=-\int x^2\,\nu(dx)=-\Expn(X^2)$,}

\noindent as in 271E.
}%end of proof of 285G

\cmmnt{
\leader{285H}{Remark} Observe that there is no result
corresponding to 283Cg (`$\lim_{|y|\to\infty}\varhatf(y)
\ifdim\pagewidth=390pt\penalty-100\fi
=0$').   If $\nu$ is the Dirac measure on $\Bbb R$ concentrated at $0$,
that is, the distribution of a random variable
which is zero almost everywhere, then $\varphi(y)=1$ for every $y$.
}

\leader{285I}{Proposition} Let $X_1,\ldots,X_n$ be independent
real-valued random variables, with characteristic functions
$\varphi_{1},\ldots,\varphi_{n}$.   Let $\varphi$ be the characteristic function
of their sum $X=X_1+\ldots+X_n$.   Then

\Centerline{$\varphi(y)=\prod_{j=1}^n\varphi_j(y)$}

\noindent for every $y\in\Bbb R$.

\proof{ Let $y\in\Bbb R$.   By 272E, the variables

\Centerline{$Y_j=e^{iyX_j}$}

\noindent are independent, so by 272R

\Centerline{$\varphi(y)=\Expn(e^{iyX})=\Expn(e^{iy(X_1+\ldots+X_n)})
=\Expn(\prod_{j=1}^nY_j)=\prod_{j=1}^n\Expn(Y_j)
=\prod_{j=1}^n\varphi_j(y)$,}

\noindent as required.
}%end of proof of 285I

\cmmnt{
\medskip

\noindent{\bf Remark} See also 285R below.
}

\leader{285J}{}\cmmnt{ There is an inversion theorem for
characteristic
functions, corresponding to 283F;  I give it in 285Xk, with an
$r$-dimensional version in 285Yb.   However, this does not seem to be as
useful as the following group of results.

\medskip

\noindent}{\bf Lemma} Let $\nu$ be a Radon probability measure on
$\Bbb R^r$, and $\varphi$ its characteristic function.
Then for $1\le j\le r$ and $a>0$,

\Centerline{$\nu\{x:|\xi_j|\ge a\}
\le 7a\int_0^{1/a}(1-\Real\varphi(te_j))dt$,}

\noindent where $e_j\in\BbbR^r$ is the $j$th unit vector.

\wheader{285J}{0}{0}{0}{84pt}

\proof{ We have

$$\eqalignno{7a\int_0^{1/a}(1-\Real\varphi(te_j))dt
&=7a\int_0^{1/a}
   \bigl(1-\Real\int_{\Bbb R^r}e^{it\xi_j}\nu(dx)\bigr)dt\cr
&=7a\int_0^{1/a}\int_{\Bbb R^r}1-\cos(t\xi_j)\nu(dx)dt\cr
&=7a\int_{\Bbb R^r}\int_0^{1/a}1-\cos(t\xi_j)dt\,\nu(dx)\cr
\displaycause{because $(x,t)\mapsto 1-\cos(t\xi_j)$ is bounded and
$\nu\BbbR^r\cdot\Bover1a$ is finite}
&=7a\int_{\Bbb R^r}
  \bigl(\bover1a-\bover1{\xi_j}\sin\bover{\xi_j}a\bigr)\nu(dx)\cr
&\ge7a\int_{|\xi_j|\ge a}
\bigl(\bover1a-\bover1{\xi_j}\sin\bover{\xi_j}a\bigr)\nu(dx)\cr
\noalign{\noindent (because $\Bover1{\xi}\sin\Bover{\xi}{a}\le\Bover1a$
for every $\xi\ne 0$)}
&\ge\nu\{x:|\xi_j|\ge a\},\cr}$$

\noindent because

\Centerline{$\Bover{\sin\eta}{\eta}\le\Bover{\sin 1}{1}\le\Bover67$ if
$\eta\ge 1$,}

\noindent so

\Centerline{$a(\Bover1a-\Bover1{\xi_j}\sin\Bover{\xi_j}a)\ge\Bover17$}

\noindent if $|\xi_j|\ge a$.
}%end of proof of 285J

\cmmnt{
\leader{285K}{Characteristic functions and the vague topology}
The time has come to return to ideas mentioned briefly in 274L.   Fix
$r\ge 1$ and let $P$ be the set of all Radon probability measures on
$\BbbR^r$.   For any bounded continuous function $h:\BbbR^r\to\Bbb R$,
define $\rho_h:P\times P\to\Bbb R$ by setting

\Centerline{$\rho_h(\nu,\nuprime)=|\int h\,d\nu-\int h\,d\nuprime|$}

\noindent for $\nu$, $\nuprime\in P$.   Then the vague topology on $P$ is
the topology generated by the pseudometrics $\rho_h$ (274Ld).
}

\leader{285L}{Theorem} Let $\nu$, $\sequencen{\nu_n}$ be Radon
probability measures on $\BbbR^r$, with characteristic functions
$\varphi$, $\sequencen{\varphi_n}$.   Then the following are equiveridical:

(i) $\nu=\lim_{n\to\infty}\nu_n$ for the vague topology;

(ii) $\int h\,d\nu=\lim_{n\to\infty}\int h\,d\nu_n$ for every bounded
continuous $h:\BbbR^r\to\Bbb R$;

(iii) $\lim_{n\to\infty}\varphi_n(y)=\varphi(y)$ for every $y\in\BbbR^r$.

\proof{{\bf (a)} The equivalence of (i) and (ii) is virtually
the definition of the vague topology;  we have

$$\eqalignno{\lim_{n\to\infty}
&\nu_n=\nu\text{ for the vague topology}\cr
&\iff\,\lim_{n\to\infty}\rho_h(\nu_n,\nu)=0
\text{ for every bounded continuous }h\cr
\noalign{\noindent (2A3Mc)}
&\iff\,\lim_{n\to\infty}|\int h\,d\nu_n-\int h\,d\nu|=0
   \text{ for every bounded continuous }h.\cr}$$

\medskip

{\bf (b)} Next, (ii) obviously implies (iii), because

\Centerline{$\Real\varphi(y)
=\int h_y\,d\nu
=\lim_{n\to\infty}h_y\,d\nu_n
=\lim_{n\to\infty}\Real\varphi_n(y)$,}

\noindent setting $h_y(x)=\cos(x\dotproduct y)$ for each $x$, and
similarly

\Centerline{$\Imag\varphi(y)=\lim_{n\to\infty}\Imag\varphi_n(y)$}

\noindent for every $y\in\BbbR^r$.

\medskip

{\bf (c)} So we are left to prove that (iii)$\Rightarrow$(ii).   I start
by showing that, given $\epsilon>0$, there is a closed bounded set $K$
such that

\Centerline{$\nu_n(\BbbR^r\setminus K)\le\epsilon$ for every
$n\in\Bbb N$.}

\noindent\Prf\ We know that $\varphi(0)=1$ and that $\varphi$ is continuous at
$0$ (285Fb).   Let $a>0$ be so large that whenever $j\le r$ and
$|t|\le 1/a$ we have

\Centerline{$1-\Real\varphi(te_j)\le\Bover{\epsilon}{14r}$,}

\noindent writing $e_j$ for the $j$th unit vector, as in 285J.   Then

\Centerline{$7a\int_0^{1/a}
(1-\Real\varphi(te_j))dt\le\Bover{\epsilon}{2r}$}

\noindent for each $j\le r$.   By Lebesgue's Dominated Convergence
Theorem (since of course the functions $t\mapsto 1-\Real\varphi_n(te_j)$
are uniformly bounded on $[0,\bover1a]$), there is an $n_0\in\Bbb N$
such that

\Centerline{$7a\int_0^{1/a}
(1-\Real\varphi_n(te_j))dt\le\Bover{\epsilon}{r}$}

\noindent for every $j\le r$ and $n\ge n_0$.   But 285J tells us that now

\Centerline{$\nu_n\{x:|\xi_j|\ge a\}\le\Bover{\epsilon}{r}$}

\noindent for every $j\le r$, $n\ge n_0$.   On the other hand, there is
surely a $b\ge a$ such that

\Centerline{$\nu_n\{x:|\xi_j|\ge b\}\le\Bover{\epsilon}{r}$}

\noindent for every $j\le r$, $n<n_0$.   So, setting
$K=\{x:|\xi_j|\le b\text{ for every }j\le r\}$,

\Centerline{$\nu_n(\BbbR^r\setminus K)\le\epsilon$}

\noindent for every $n\in\Bbb N$, as required.  \Qed

\medskip

{\bf (d)} Now take any bounded continuous $h:\BbbR^r\to\Bbb R$ and
$\epsilon>0$.   Set $M=1+\sup_{x\in\BbbR^r}|h(x)|$, and let $K$ be a
bounded closed set such that

\Centerline{$\nu_n(\BbbR^r\setminus K)\le\Bover{\epsilon}M$ for every
$n\in\Bbb N$, \quad$\nu(\BbbR^r\setminus K)\le\Bover{\epsilon}M$,}

\noindent using (b) just above.   By the
Stone-Weierstrass theorem (281K) there are $y_0,\ldots,y_m\in\Bbb Q^r$
and $c_0,\ldots,c_m\in\Bbb C$ such that

\Centerline{$|h(x)-g(x)|\le\epsilon$ for every $x\in K$,}

\Centerline{$|g(x)|\le M$ for every $x\in\BbbR^r$,}

\noindent writing $g(x)=\sum_{k=0}^mc_ke^{iy_k\dotproduct x}$ for
$x\in\BbbR^r$.   Now

\Centerline{$\lim_{n\to\infty}\int g\,d\nu_n
=\lim_{n\to\infty}\sum_{k=0}^mc_k\varphi_n(y_k)
=\sum_{k=0}^mc_k\varphi(y_k)
=\int g\,d\nu$.}

\noindent On the other hand, for every $n\in\Bbb N$,

\Centerline{$|\int g\,d\nu_n-\int h\,d\nu_n|
\le\int_K|g-h|d\nu_n+2M\nu_n(\Bbb R\setminus K)
\le 3\epsilon$,}

\noindent and similarly $|\int g\,d\nu-\int h\,d\nu|\le 3\epsilon$.
Consequently

\Centerline{$\limsup_{n\to\infty}|\int h\,d\nu_n-\int h\,d\nu|\le
6\epsilon$.}

\noindent As $\epsilon$ is arbitrary,

\Centerline{$\lim_{n\to\infty}\int h\,d\nu_n=\int h\,d\nu$,}

\noindent and (ii) is true.
}%end of proof of 285L


\vleader{72pt}{285M}{Corollary} (a) Let $\nu$, $\nuprime$ be two Radon
probability
measures on $\BbbR^r$ with the same characteristic functions.   Then
they are equal.

(b) Let $(X_1,\ldots,X_r)$ and $(Y_1,\ldots,Y_r)$ be two families of
real-valued random variables.   If

\Centerline{$\Expn(e^{i\eta_1X_1+\ldots+i\eta_rX_r})
=\Expn(e^{i\eta_1Y_1+\ldots+i\eta_rY_r})$}

\noindent for all $\eta_1,\ldots,\eta_r\in\Bbb R$, then
$(X_1,\ldots,X_r)$ has the same joint distribution as
$(Y_1,\ldots,Y_r)$.

\proof{{\bf (a)} Applying 285L with $\nu_n=\nuprime$ for every $n$,
we see that $\int h\,d\nuprime=\int h\,d\nu$ for every bounded continuous
$h:\BbbR^r\to\Bbb R$.   By 256D(iv), $\nu=\nuprime$.

\medskip

{\bf (b)} Apply (a) with $\nu$, $\nuprime$ the two joint distributions.
}%end of proof of 285M

\cmmnt{
\leader{285N}{Remarks} Probably the most important application of this
theorem is to the standard proof of the Central Limit Theorem.   I
sketch the ideas in 285Xq and 285Yl-285Yo;  details may be found in most
serious probability texts;  two on my shelf are {\smc Shiryayev 84},
\S III.4, and {\smc Feller 66}, \S XV.6.   However, to get the full
strength of Lindeberg's version of the Central Limit Theorem we have to
work quite hard, and I therefore propose to illustrate the method with a
version of Poisson's theorem (285Q) instead.   I begin with two lemmas
which are very frequently used in results of this kind.
}

\leader{285O}{Lemma} Let
$c_0,\ldots,c_n,d_0,\ldots,d_n$ be complex numbers of
modulus at most $1$.   Then

\Centerline{$|\prod_{k=0}^nc_k-\prod_{k=0}^nd_k|
\le\sum_{k=0}^n|c_k-d_k|$.}

\proof{ Induce on $n$.   The case $n=0$ is trivial.   For
the case $n=1$ we have

$$\eqalign{|c_0c_1-d_0d_1|
&=|c_0(c_1-d_1)+(c_0-d_0)d_1|\cr
&\le|c_0||c_1-d_1|+|c_0-d_0||d_1|
\le|c_1-d_1|+|c_0-d_0|,\cr}$$

\noindent which is what we need.   For the inductive step to $n+1$, we
have

$$\eqalignno{|\prod_{k=0}^{n+1}c_k-\prod_{k=0}^{n+1}d_k|
&\le|\prod_{k=0}^nc_k-\prod_{k=0}^nd_k|
+|c_{n+1}-d_{n+1}|\cr
\noalign{\noindent (by the case just done, because
$c_{n+1}$, $d_{n+1}$, $\prod_{k=0}^nc_k$ and $\prod_{k=0}^nd_k$ all have
modulus at most $1$)}
&\le\sum_{k=0}^n|c_k-d_k|+|c_{n+1}-d_{n+1}|\cr
\noalign{\noindent (by the inductive hypothesis)}
&=\sum_{k=0}^{n+1}|c_k-d_k|,\cr}$$

\noindent so the induction continues.
}%end of proof of 285O

\leader{285P}{Lemma} Suppose that $M\ge 0$ and $\epsilon>0$.
Then there are
$\eta>0$ and $y_0,\ldots,y_n\in\Bbb R$ such that whenever $X$, $Z$ are
two real-valued random variables with $\Expn(|X|)\le M$,
$\Expn(|Z|)\le M$ and
$|\varphi_X(y_j)-\varphi_Z(y_j)|\le\eta$ for every $j\le n$, then
$F_X(a)\le F_Z(a+\epsilon)+\epsilon$ for every $a\in\Bbb R$, where I
write
$\varphi_X$ for the characteristic function of $X$ and $F_X$ for the
distribution function of $X$.

\proof{ The case $M=0$ is trivial, as then both $X$ and $Z$ are zero a.e.,
so I will suppose henceforth that $M>0$.
Set $\delta=\Bover{\epsilon}7>0$, $b=\Bover{M}{\delta}$.

\medskip

{\bf (a)} Define $h_0:\Bbb R\to[0,1]$ by setting
$h_0(x)=\med(0,1-\Bover{x}{\delta},1)$ for $x\in\Bbb R$.
Then $h_0$ is continuous.   Let $m=\lfloor\bover{b}{\delta}\rfloor$ 
be the integer part of $\bover{b}{\delta}$, and for $-m\le k\le m+1$ set
$h_k(x)=h_0(x-k\delta)$.

By the Stone-Weierstrass theorem (281K again), there are
$y_0,\ldots,y_n\in\Bbb R$ and $c_0,\ldots,c_n\in\Bbb C$ such that,
writing $g_0(x)=\sum_{j=0}^nc_je^{iy_jx}$,

\Centerline{$|h_0(x)-g_0(x)|\le\delta$
  for every $x\in[-b-(m+1)\delta,b+m\delta]$,}

\Centerline{$|g_0(x)|\le 1$ for every $x\in\Bbb R$.}

\noindent For $-m\le k\le m+1$, set

\Centerline{$g_k(x)=g_0(x-k\delta)
=\sum_{j=0}^nc_je^{-iy_jk\delta}e^{iy_jx}$.}

\noindent Set $\eta=\delta/(1+\sum_{j=0}^n|c_j|)>0$.

\medskip

{\bf (b)} Now suppose that $X$, $Z$ are random variables such that
$\Expn(|X|)\le M$, $\Expn(|Z|)\le M$ and
$|\varphi_X(y_j)-\varphi_Z(y_j)|\le\eta$ for every $j\le n$.   Then for any
$k$ we have

\Centerline{$\Expn(g_k(X))
=\Expn(\sum_{j=0}^nc_je^{-iy_jk\delta}e^{iy_jX})
=\sum_{j=0}^nc_je^{-iy_jk\delta}\varphi_X(y_j)$,}

\noindent and similarly

\Centerline{$\Expn(g_k(Z))
=\sum_{j=0}^nc_je^{-iy_jk\delta}\varphi_Z(y_j)$,}

\noindent so

\Centerline{$|\Expn(g_k(X))-\Expn(g_k(Z))|
\le\sum_{j=0}^n|c_j||\varphi_X(y_j)-\varphi_Z(y_j)|
\le\sum_{j=0}^n|c_j|\eta\le\delta$.}

\noindent Next,

\Centerline{$|h_k(x)-g_k(x)|\le\delta$ for every
$x\in[-b-(m+1)\delta+k\delta,b+m\delta+k\delta]\supseteq[-b,b]$,}

\Centerline{$|h_k(x)-g_k(x)|\le 2$ for every $x$,}

\Centerline{$\Pr(|X|\ge b)\le\Bover{M}{b}=\delta$,}

\noindent so $\Expn(|h_k(X)-g_k(X)|)\le 3\delta$;  and similarly
$\Expn(|h_k(Z)-g_k(Z)|)\le 3\delta$.   Putting these together,

\Centerline{$|\Expn(h_k(X))-\Expn(h_k(Z))|\le 7\delta=\epsilon$}

\noindent whenever $-m\le k\le m+1$.

\medskip

{\bf (c)} Now suppose that $-b\le a\le b$.   Then there is a $k$ such
that $-m\le k\le m+1$ and $a\le k\delta\le a+\delta$.   Since

\Centerline{$\chi\ocint{-\infty,a}
\le\chi\ocint{-\infty,k\delta}
\le h_k
\le\chi\ocint{-\infty,(k+1)\delta}
\le\chi\ocint{-\infty,a+2\delta}$,}

\noindent we must have

\Centerline{$\Pr(X\le a)\le\Expn(h_k(X))$,}

\Centerline{$\Expn(h_{k}(Z))\le\Pr(Z\le a+2\delta)\le\Pr(Z\le
a+\epsilon)$.}

\noindent But this means that

\Centerline{$\Pr(X\le a)\le\Expn(h_k(X))\le\Expn(h_k(Z))+\epsilon
\le\Pr(Z\le a+\epsilon)+\epsilon$}

\noindent whenever $a\in[-b,b]$.

\medskip

{\bf (d)} As for the cases $a\ge b$, $a\le -b$,
we surely have

\Centerline{$b(1-F_Z(b))=b\Pr(Z>b)\le\Expn(|Z|)\le M$,}

\noindent so if $a\ge b$ then

\Centerline{$F_X(a)\le 1\le F_Z(a)+1-F_Z(b)\le F_Z(a)+\Bover{M}{b}
=F_Z(a)+\delta\le F_Z(a+\epsilon)+\epsilon$.}

\noindent Similarly,

\Centerline{$bF_X(-b)\le \Expn(|X|)\le M$,}

\noindent so

\Centerline{$F_X(a)\le\delta\le F_Z(a+\epsilon)+\epsilon$}

\noindent  for every $a\le -b$.
This completes the proof.
}%end of proof of 285P

\vleader{72pt}{285Q}{Law of Rare Events:  Theorem} For any $M\ge 0$ and
$\epsilon>0$ there is a $\delta>0$
such that whenever $X_0,\ldots,X_n$ are independent $\{0,1\}$-valued
random variables with $\Pr(X_k=1)=p_k\le\delta$ for every $k\le n$ and
$\sum_{k=0}^np_k=\lambda\le M$, and $X=X_0+\ldots+X_n$, then

\Centerline{$|\Pr(X=m)-\Bover{\lambda^m}{m!}e^{-\lambda}|\le\epsilon$}

\noindent for every $m\in\Bbb N$.

\proof{{\bf (a)} We should begin by calculating some
characteristic functions.   First, the characteristic function $\varphi_k$
of $X_k$ will be given by

\Centerline{$\varphi_k(y)=(1-p_k)e^{iy0}+p_ke^{iy1}=1+p_k(e^{iy}-1)$.}

\noindent Next, if $Z$ is a Poisson random variable with parameter
$\lambda$ (that is, if $\Pr(Z=m)=\lambda^me^{-\lambda}/m!$ for every
$m\in\Bbb N$;  all you need to know at this point about the Poisson
distribution is
that $\sum_{m=0}^{\infty}\lambda^me^{-\lambda}/m!=1$), then its
characteristic function $\varphi_Z$ is given by

\Centerline{$\varphi_Z(y)
=\sum_{m=0}^{\infty}\Bover{\lambda^m}{m!}e^{-\lambda}e^{iym}
=e^{-\lambda}\sum_{m=0}^{\infty}\Bover{(\lambda e^{iy})^m}{m!}
=e^{-\lambda}e^{\lambda e^{iy}}
=e^{\lambda(e^{iy}-1)}$.}

\medskip

{\bf (b)} Before getting down to $\delta$'s and $\eta$'s, I show how to
estimate $\varphi_X(y)-\varphi_Z(y)$.   We know that

\Centerline{$\varphi_X(y)=\prod_{k=0}^n\varphi_k(y)$}

\noindent (using 285I), while

\Centerline{$\varphi_Z(y)=\prod_{k=0}^ne^{p_k(e^{iy}-1)}$.}

\noindent Because $\varphi_k(y)$, $e^{p_k(e^{iy}-1)}$ all have modulus at
most $1$ (we have

\Centerline{$|e^{p_k(e^{iy}-1)}|=e^{-p_k(1-\cos y)}\le 1$,)}

\noindent 285O tells us that

\Centerline{$|\varphi_X(y)-\varphi_Z(y)|
\le\sum_{k=0}^n|\varphi_k(y)-e^{p_k(e^{iy}-1)}|
=\sum_{k=0}^n|e^{p_k(e^{iy}-1)}-1-p_k(e^{iy}-1)|$.}

\medskip

{\bf (c)} So we have a little bit of analysis to do.   To estimate
$|e^z-1-z|$ where $\Real z\le 0$, consider the function

\Centerline{$g(t)=\Real(c(e^{tz}-1-tz))$}

\noindent where $|c|=1$.   We have $g(0)=g'(0)=0$ and

\Centerline{$|g''(t)|=|\Real(c(z^2e^{tz}))|\le|c||z^2||e^{tz}|\le|z|^2$}

\noindent for every $t\ge 0$, so that

\Centerline{$|g(1)|\le\Bover12|z|^2$}

\noindent by the (real-valued) Taylor theorem with remainder, or
otherwise.   As $c$ is arbitrary,

\Centerline{$|e^{z}-1-z|\le\Bover12|z|^2$}

\noindent whenever $\Real z\le 0$.   In particular,

\Centerline{$|e^{p_k(e^{iy}-1)}-1-p_k(e^{iy}-1)|
\le\Bover12p_k^2|e^{iy}-1|^2\le 2p_k^2$}

\noindent for each $k$, and

\Centerline{$|\varphi_X(y)-\varphi_Z(y)|
\le\sum_{k=0}^n|e^{p_k(e^{iy}-1)}-1-p_k(e^{iy}-1)|
\le 2\sum_{k=0}^np_k^2$}

\noindent for each $y\in\Bbb R$.

\medskip

{\bf (d)} Now for the detailed estimates.   Given $M\ge 0$ and
$\epsilon>0$, let $\eta>0$ and $y_0,\ldots,y_l\in\Bbb R$ be such that

\Centerline{$\Pr(X\le a)\le\Pr(Z\le a+\Bover12)+\Bover{\epsilon}2$}

\noindent whenever $X$, $Z$ are real-valued random variables,
$\Expn(|X|)\le M$, $\Expn(|Z|)\le M$ and
$|\varphi_X(y_j)-\varphi_X(y_j)|\le\eta$ for every $j\le l$ (285P).   Take
$\delta=\bover{\eta}{2M+1}$
and suppose that $X_0,\ldots,X_n$ are independent
$\{0,1\}$-valued random variables with $\Pr(X_k=1)=p_k\le\delta$ for
every $k\le n$, where $\lambda=\sum_{k=0}^np_k$ is less than or equal to
$M$.   Set $X=X_0+\ldots+X_n$
and let $Z$ be a Poisson random variable with parameter $\lambda$;  then
by the arguments of (a)-(c),

\Centerline{$|\varphi_X(y)-\varphi_Z(y)|\le 2\sum_{k=0}^np_k^2
\le 2\delta\sum_{k=0}^np_k=2\delta\lambda\le\eta$}

\noindent for every $y\in\Bbb R$.   Also

\Centerline{$\Expn(|X|)=\Expn(X)=\sum_{k=0}^np_k=\lambda\le M$,}

\Centerline{$\Expn(|Z|)=\Expn(Z)
=\sum_{m=0}^{\infty}m\Bover{\lambda^m}{m!}e^{-\lambda}
=e^{-\lambda}\sum_{m=1}^{\infty}\Bover{\lambda^m}{(m-1)!}
=e^{-\lambda}\sum_{m=0}^{\infty}\Bover{\lambda^{m+1}}{m!}
=\lambda\le M$.}

\noindent So

\Centerline{$\Pr(X\le a)\le\Pr(Z\le a+\Bover12)+\Bover{\epsilon}2$,}

\Centerline{$\Pr(Z\le a)\le\Pr(X\le a+\Bover12)+\Bover{\epsilon}2$}

\noindent for every $a$.   But as both $X$ and $Z$ take all their values
in $\Bbb N$,

\Centerline{$|\Pr(X\le m)-\Pr(Z\le m)|\le\Bover{\epsilon}2$}

\noindent for every $m\in\Bbb N$, and

\Centerline{$|\Pr(X=m)-\Bover{\lambda^m}{m!}e^{-\lambda}|
=|\Pr(X=m)-\Pr(Z=m)|
\le\epsilon$}

\noindent for every $m\in\Bbb N$, as required.
}%end of proof of 285Q

\leader{285R}{Convolutions}\dvro{ If}{ Recall from 257A that if}
$\nu$, $\tilde\nu$ are Radon probability measures on $\BbbR^r$ then
%\tilde\nu  not  \nuprime
%because we have  \tilde f  for its R-N deriviative
\dvro{$\varphi_{\nu*\tilde\nu}(y)
=\varphi_{\nu}(y)\varphi_{\tilde\nu}(y)$
for every $y\in\BbbR^r$.}{they have a convolution
$\nu*\tilde\nu$ defined by writing

\Centerline{$(\nu*\tilde\nu)(E)=(\nu\times\tilde\nu)\{(x,y):x+y\in E\}$}

\noindent for every Borel set $E\subseteq\BbbR^r$, which is also a
Radon probability measure.   We can readily compute the characteristic
function $\varphi_{\nu*\tilde\nu}$ from 257B:  we have

$$\eqalign{\varphi_{\nu*\tilde\nu}(y)
&=\int e^{iy\dotproduct x}(\nu*\tilde\nu)(dx)
=\int e^{iy\dotproduct (x+x')}\nu(dx)\tilde\nu(dx')\cr
&=\int e^{iy\dotproduct x}e^{iy\dotproduct x'}\nu(dx)\tilde\nu(dx')
=\int e^{iy\dotproduct x}\nu(dx)
  \int e^{iy\dotproduct x'}\tilde\nu(dx')
=\varphi_{\nu}(y)\varphi_{\tilde\nu}(y)}$$

\noindent for every $y\in\BbbR^r$.
(Thus convolution of measures corresponds to pointwise
multiplication of characteristic functions, just as convolution of
functions corresponds to pointwise multiplication of Fourier
transforms.)   Recalling that the sum of independent random variables
corresponds to convolution of their distributions (272T), this
gives another way of looking at 285I.   Remember also that if $\nu$,
$\tilde\nu$ have Radon-Nikod\'ym derivatives $f$, $\tilde f$ with
respect to Lebesgue
measure then $f*\tilde f$ is a Radon-Nikod\'ym derivative of
$\nu*\tilde\nu$ (257F).
}%end of dvro

\leader{285V}{Proposition}\dvAnew{2018} Let $\nu$ be a Radon probability measure
on $\BbbR^r$ such that $\nu*\nu=\nu$.   Then $\nu$ is the Dirac measure
$\delta_0$ concentrated at $0$.
%out of order query

\proof{ By 285R, $\varphi_{\nu}^2=\varphi_{\nu}$, so $\varphi_{\nu}$ is $\{0,1\}$-valued;
as $\varphi_{\nu}(0)=1$ (285Fa) and $\varphi_{\nu}$ is continuous (285Fb),  
$\varphi_{\nu}(y)=1$ for every $y\in\Bbb R$, that is, $\varphi_{\nu}=\varphi_{\delta_0}$
(285H).   By 285Ma, $\nu=\delta_0$.
}%end of proof of 285V

\leader{285S}{The vague topology and pointwise
convergence of characteristic functions}\dvro{ Write}{ In 285L we saw
that a sequence
$\sequencen{\nu_n}$ of Radon probability measures on $\BbbR^r$
converges in the vague topology to a Radon probability measure $\nu$ if
and only if

\Centerline{$\lim_{n\to\infty}\int e^{iy\dotproduct
x}\nu_n(dx)=\int
e^{iy\dotproduct x}\nu(dx)$}

\noindent for every $y\in\BbbR^r$;  that is, iff

\Centerline{$\lim_{n\to\infty}\rho'_y(\nu_n,\nu)=0$ for every $y\in\Bbb
R^r$,}

\noindent writing}

\Centerline{$\rho'_y(\nu,\nuprime)=|\int e^{iy\dotproduct x}\nu(dx)
-\int e^{iy\dotproduct x}\nuprime(dx)|$}

\noindent for Radon probability measures $\nu$, $\nuprime$ on $\BbbR^r$ and
$y\in\BbbR^r$.   \dvro{Write}{It is natural to ask whether the
pseudometrics
$\rho'_y$ actually define the vague topology.   Writing} $\frak T$ for
the vague topology and $\frak S$ for the topology defined by
$\{\rho'_y:y\in\BbbR^r\}$\cmmnt{, we surely have
$\frak S\subseteq\frak T$,
just because every $\rho'_y$ is one of the pseudometrics used in the
definition of $\frak T$.   Also we know that $\frak S$ and $\frak T$
give the same convergent sequences, and incidentally that $\frak T$ is
metrizable (see 285Xt).   But all this does not quite amount to saying
that the two topologies are the same, and indeed they are not, as the
next result shows.}%end of comment

\leader{285T}{Proposition} Suppose that $y_0,\ldots,y_n\in\Bbb R$
and $\eta>0$.   Then there are infinitely many $m\in\Bbb N$ such that
$|1-e^{iy_km}|\le\eta$ for every $k\le n$.

\proof{ Let $\eta_1,\ldots,\eta_r\in\Bbb R$ be such that
$1=\eta_0,\eta_1,\ldots,\eta_r$ are linearly independent over $\Bbb Q$
and every $y_k/2\pi$ is a linear combination of the $\eta_j$ over
$\Bbb Q$;  say $y_k=2\pi\sum_{j=0}^rq_{kj}\eta_j$ where every
$q_{kj}\in\Bbb Q$.   Express the $q_{kj}$ as $p_{kj}/p$ where each
$p_{kj}\in\Bbb Z$
and $p\in\Bbb N\setminus\{0\}$.   Set
$M=\max_{k\le n}\sum_{j=0}^r|p_{kj}|$.

Take any $m_0\in\Bbb N$ and let $\delta>0$ be such that
$|1-e^{2\pi ix}|\le\eta$ whenever $|x|\le2\pi M\delta$.   By Weyl's
Equidistribution
Theorem (281N), there are infinitely many $m$ such that
$\fraction{m\eta_j}\le\delta$ whenever $1\le j\le r$;  in particular,
there is such an $m\ge m_0$.    Set $m_j=\lfloor m\eta_j\rfloor$, 
so that $|m\eta_j-m_j|\le\delta$ for $0\le j\le r$.   Then

\Centerline{$|mpy_k-2\pi\sum_{j=0}^rp_{kj}m_j|
\le2\pi\sum_{j=0}^r|p_{kj}||m\eta_j-m_j|
\le 2\pi M\delta$,}

\noindent so that

\Centerline{$|1-e^{iy_kmp}|
=|1-\exp(i(mpy_k-2\pi\sum_{j=0}^rp_{kj}m_j))|
\le\eta$}

\noindent for every $k\le n$.
As $mp\ge m_0$ and $m_0$ is arbitrary, this proves the result.
}%end of proof of 285T

\leader{285U}{Corollary} The topologies $\frak S$ and $\frak T$ on the
space of Radon probability measures on $\Bbb R$, as described in 285S,
are different.

\proof{ Let $\delta_x$ be the Dirac measure on $\Bbb R$ concentrated at
$x$.   By 285T, every
member of $\frak S$ which contains $\delta_0$ also contains $\delta_m$
for infinitely many $m\in\Bbb N$.   On the other hand, the set

\Centerline{$G=\{\nu:\int e^{-x^2}\nu(dx)>\Bover12\}$}

\noindent is a member of $\frak T$, containing $\delta_0$, which does
not contain $\delta_m$ for any integer $m\ne 0$.
So $G\in\frak T\setminus\frak S$ and $\frak T\ne\frak S$.
}%end of proof of 285V


\exercises{
\leader{285X}{Basic exercises (a)}
%\spheader 285Xa
Let $\nu$ be a Radon probability
measure on $\BbbR^r$, where $r\ge 1$, and suppose that
$\int\|x\|\nu(dx)<\infty$.   Show that the characteristic function
$\varphi$ of $\nu$ is differentiable (in the full sense of 262Fa) and
that $\pd{\varphi}{\eta_j}(y)=i\int\xi_je^{iy\dotproduct x}\nu(dx)$ for
every
$j\le r$ and $y\in\BbbR^r$, using $\xi_j$, $\eta_j$ to represent the
coordinates of $x$ and $y$ as usual.
%285A

\sqheader 285Xb Let $\pmb{X}=(X_1,\ldots,X_r)$ be a family of
real-valued random variables, with characteristic function
$\varphi_{\pmb{X}}$.   Show that the characteristic function $\varphi_{X_j}$
of $X_j$ is given by

\Centerline{$\varphi_{X_j}(y)=\varphi_{\pmb{X}}(ye_j)$ for every $y\in\Bbb
R$,}

\noindent where $e_j$ is the $j$th unit vector of $\BbbR^r$.
%285A

\sqheader 285Xc Let $X$ be a real-valued random variable and
$\varphi_X$ its characteristic function.   Show that

\Centerline{$\varphi_{aX+b}(y)=e^{iyb}\varphi_X(ay)$}

\noindent for any $a$, $b$, $y\in\Bbb R$.
%285A

\spheader 285Xd\dvAnew{2014}
Let $X$ be a real-valued random variable which is not
essentially constant, and $\varphi$ its characteristic function.   Show that
$|\varphi(y)|<1$ for all but countably many $y\in\Bbb R$.
\Hint{the support (256Xf) of the distribution of $X$ has distinct
points $x$, $x'$ and if $e^{iyx}\ne e^{iyx'}$ then $|\varphi(y)|<1$.}
%285D

\spheader 285Xe Let $X$ be a real-valued random variable and
$\varphi$ its characteristic function.

\quad{(i)} Show that for any integrable complex-valued function $h$
on $\Bbb R$,

\Centerline{$\Expn(\varhat{h}(X))=\Bover1{\sqrt{2\pi}}
\int_{-\infty}^{\infty}\varphi(-y)h(y)dy$,}

\noindent writing $\varhat{h}$ for the Fourier transform of $h$.

\quad{(ii)} Show that for any rapidly decreasing test function $h$,

\Centerline{$\Expn(h(X))
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
  \varphi(y)\varhat{h}(y)dy$.}
%285D

\spheader 285Xf Let $\nu$ be a Radon probability measure on
$\Bbb R$, and suppose that its characteristic function $\varphi$ is
square-integrable.   Show that $\nu$ is an indefinite-integral measure
over Lebesgue measure and that its Radon-Nikod\'ym derivatives are also
square-integrable.   \Hint{use
284O to find a square-integrable $f$ such that
$\int f\times h=\bover1{\sqrt{2\pi}}\int\varphi\times\varhat{h}$ for every
rapidly
decreasing test function $h$, and ideas from the proof of 284G to show
that $\int_a^bf=\nu\ooint{a,b}$ whenever $a<b$ in $\Bbb R$.}
%285D 285Xe

\spheader 285Xg\dvAformerly{2{}85Yp} Let $\nu$ be a Radon probability
measure on $\BbbR^r$ with bounded support (definition: 256Xf).   Show that
its characteristic function is smooth.
%285D

\spheader 285Xh Let $X$ be a normal random variable with expectation $a$
and variance $\sigma^2$.   Show that $\Expn(e^X)=\exp(a+\bover12\sigma^2)$.
%285E

\sqheader 285Xi Let $\pmb{X}=(X_1,\ldots,X_r)$ be a family of
real-valued random variables with characteristic function
$\varphi_{\pmb{X}}$.   Suppose that $\varphi_{\pmb{X}}$ is expressible in the
form

\Centerline{$\varphi_{\pmb{X}}(y)=\prod_{j=1}^r\varphi_j(\eta_j)$}

\noindent for some functions $\varphi_1,\ldots,\varphi_r$, writing
$y=(\eta_1,\ldots,\eta_r)$ as usual.   Show that $X_1,\ldots,X_r$ are
independent.   \Hint{show that the $\varphi_j$ must be multiples of the
characteristic functions of the $X_j$;  now show that the distribution
of $\pmb{X}$ has the same characteristic function as the product of the
distributions of the $X_j$.}
%285I

\spheader 285Xj Let $X_1$, $X_2$ be independent real-valued random
variables with the same distribution, and $\varphi$ the characteristic
function of $X_1-X_2$.   Show that $\varphi(t)=\varphi(-t)\ge 0$ for every
$t\in\Bbb R$.
%285I

\spheader 285Xk Let $\nu$ be a Radon probability measure on
$\Bbb R$, with characteristic function $\varphi$.   Show that

\Centerline{$\Bover12(\nu[c,d]+\nu\ooint{c,d})
=\Bover{i}{2\pi}\lim_{a\to\infty}
  \int_{-a}^a\Bover{e^{-idy}-e^{-icy}}{y}\varphi(y)dy$}

\noindent whenever $c<d$ in $\Bbb R$.
\Hint{use part (a) of the proof of 283F.}
%285J

\spheader 285Xl Let $X$ be a real-valued random variable and
$\varphi_X$ its characteristic function.   Show that

\Centerline{$\Pr(|X|\ge a)
\le 7a\int_0^{1/a}(1-\Real(\varphi_X(y))dy$}

\noindent for every $a>0$.
%285J

\spheader 285Xm We say that a set $Q$ of Radon probability
measures on $\Bbb R$ is {\bf uniformly tight} if for every $\epsilon>0$
there is
an $M\ge 0$ such that $\nu(\Bbb R\setminus[-M,M])\le\epsilon$ for every
$\nu\in Q$.    Show that if $Q$ is any uniformly tight family of Radon
probability
measures on $\Bbb R$, and $\epsilon>0$, then there are $\eta>0$ and
$y_0,\ldots,y_n\in\Bbb R$ such that

\Centerline{$\nu\ocint{-\infty,a}
\le\nuprime\ocint{-\infty,a+\epsilon}+\epsilon$}

\noindent whenever $\nu$, $\nuprime\in Q$ and
$|\varphi_{\nu}(y_j)-\varphi_{\nuprime}(y_j)|\le\eta$ for every $j\le n$, writing
$\varphi_{\nu}$ for the characteristic function of $\nu$.
%285M

\spheader 285Xn\dvArevised{2018}
Let $\sequencen{\nu_n}$ be a sequence of Radon
probability measures on $\Bbb R$.   Show that it converges for the
vague topology to a Radon probability measure $\nu$ iff 
$\{\nu\}\cup\{\nu_n:n\in\Bbb N\}$ is uniformly tight in the sense of
285Xm and $\limsup_{n\to\infty}\nu_n\ocint{-\infty,a}
\le\liminf_{n\to\infty}\nu_n\ocint{-\infty,b}$ whenever $a<b$ in $\Bbb R$.
\Hint{Setting $g(x)=\inf\liminf_{n\to\infty}\nu_n\ocint{-\infty,b}:x<b\}$
for $x\in\Bbb R$, show that the Lebesgue-Stieltjes measure $\nu$ associated 
with $g$ is a probability measure and $\sequencen{\nu_n}$ converges to $\nu$ 
for the vague topology.}
%285M

\sqheader 285Xo Let $\nu$, $\nuprime$ be two totally finite Radon measures
on $\BbbR^r$ which agree on all closed half-spaces, that is, sets of
the form $\{x:x\dotproduct y\ge c\}$ where $y\in\BbbR^r$ is non-zero and
$c\in\Bbb R$.   Show that $\nu=\nuprime$.   
\Hint{reduce to the case $\nu\BbbR^r=\nuprime\BbbR^r=1$ and use 285M.}
%285M

\sqheader 285Xp\dvAformerly{2{}85Xm}
For $\gamma>0$, the {\bf Cauchy distribution} with
centre $0$ and scale parameter $\gamma$ is the Radon probability measure
$\nu_{\gamma}$ defined by the formula

\Centerline{$\nu_{\gamma}(E)
=\Bover{\gamma}{\pi}\int_E\Bover1{\gamma^2+t^2}dt$.}

\noindent (i) Show that if $X$ is a random variable with distribution
$\nu_{\gamma}$ then $\Pr(X\ge 0)=\Pr(|X|\ge\gamma)=\bover12$.
(ii) Show that the characteristic function of
$\nu_{\gamma}$ is $y\mapsto e^{-\gamma|y|}$.   \Hint{283Xq.}   (iii)
Show that if $X$ and $Y$ are independent random variables with Cauchy
distributions, both centered at $0$ and with scale parameters $\gamma$,
$\delta$ respectively, and $\alpha$, $\beta$ are not both $0$, then
$\alpha X+\beta Y$ has a Cauchy distribution centered at $0$ and with
scale parameter $|\alpha|\gamma+|\beta|\delta$.   (iv) Show that if $X$
and $Y$ are independent normally distributed random variables with expectation $0$ then $X/Y$ has a Cauchy distribution.
%285M

\sqheader 285Xq Let $X_1,X_2,\ldots$ be an independent identically
distributed sequence of random variables, all with zero expectation and
variance $1$;  let $\varphi$ be their common characteristic function.   For
each $n\ge 1$, set $S_n=\bover1{\sqrt{n}}(X_1+\ldots+X_n)$.

\quad{(i)} Show that the characteristic function $\varphi_n$ of $S_n$
is given by the formula $\varphi_n(y)=(\varphi(\Bover{y}{\sqrt{n}}))^{n}$
for each $n$.

\quad{(ii)} Show that
$|\varphi_n(y)-e^{-y^2/2}|\le n|\varphi(\Bover{y}{\sqrt{n}})-e^{-y^2/2n}|$
for $n\ge 1$ and $y\in\Bbb R$.

\ifdim\pagewidth>467pt\fontdimen3\tenrm=3.33pt\fi
\quad(iii) Setting $h(y)=\varphi(y)-e^{-y^2/2}$, show that
$h(0)=h'(0)=h''(0)=0$ and therefore that
$\lim_{n\to\infty} nh(y/\sqrt{n})=0$, so that
$\lim_{n\to\infty}\varphi_n(y)=e^{-y^2/2}$ for every $y\in\Bbb R$.
\fontdimen3\tenrm=1.67pt

\quad{(iv)} Show that $\lim_{n\to\infty}\Pr(S_n\le a)
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^ae^{-x^2/2}dx$ for every
$a\in\Bbb R$.
%285N

\sqheader 285Xr\dvAformerly{2{}85Xm}
A random variable $X$ has a {\bf Poisson distribution}
with parameter $\lambda>0$ if $\Pr(X=n)=e^{-\lambda}\lambda^n/n!$ for
every $n\in\Bbb N$.    (i) Show that in this case
$\Expn(X)=\Var(X)=\lambda$.   (ii) Show that if $X$ and $Y$ are
independent random variables with Poisson distributions then $X+Y$ has a
Poisson distribution.   (iii) Find a proof of (ii) based on 285Q.
%285Q

\sqheader 285Xs For $x\in\BbbR^r$, let $\delta_x$ be the Dirac measure
on $\BbbR^r$ concentrated at $x$.
Show that $\delta_x*\delta_y=\delta_{x+y}$ for all $x$, $y\in\BbbR^r$.
%285R

\spheader 285Xt Let $P$ be the set of Radon probability measures
on $\BbbR^r$.   For $y\in\BbbR^r$, set
$\rho'_y(\nu,\nuprime)=|\varphi_{\nu}(y)-\varphi_{\nuprime}(y)|$ for all $\nu$,
$\nuprime\in P$, writing $\varphi_{\nu}$ for the characteristic function of
$\nu$.   Set $\psi(x)=\bover1{(\sqrt{2\pi})^r}e^{-x\dotproduct x/2}$ for
$x\in\BbbR^r$.    Show that the vague topology on $P$ is
defined by the family $\{\rho_{\psi}\}\cup\{\rho'_y:y\in\Bbb Q^r\}$,
defining $\rho_{\psi}$ as in 285K, and is therefore metrizable.
\Hint{281K;  cf.\ 285Xm.}
%285S

\sqheader 285Xu\dvAformerly{2{}85Xr}
Let $\varphi:\BbbR^r\to\Bbb C$ be the
characteristic function of a Radon probability measure on $\BbbR^r$.
Show that $\varphi(0)=1$ and that
$\sum_{j=0}^n\sum_{k=0}^nc_j\bar c_k\varphi(a_j-a_k)\ge 0$ whenever
$a_0,\ldots,a_n\in\BbbR^r$ and $c_0,\ldots,c_n\in\Bbb C$.
(`Bochner's theorem' states that these conditions are
sufficient, as well as necessary, for $\varphi$ to be a characteristic
function;  see 445N in Volume 4.)
%285*

\spheader 285Xv\dvAnew{2018} Let $\sequencen{\nu_n}$ be a sequence of 
Radon probability measures on $\Bbb R$ such that 
$\psi(y)=\lim_{n\to\infty}\varphi_{\nu_n}(y)$ is defined for every $y\in\Bbb R$ and
$\psi:\Bbb R\to\Bbb C$ is continuous at $0$.   Show that $\psi$ is the
characteristic function of a Radon probability measure $\nu$ and that
$\sequencen{\nu_n}$ converges to $\nu$ for the vague topology.   \Hint{as in
part (c) of the proof of 285L, show that $\{\nu_n:n\in\Bbb N\}$ is uniformly 
tight.   Show that there is a
subsequence $\sequence{k}{\nu_{n_k}}$ such that 
$f(q)=\lim_{k\to\infty}\nu_{n_k}\ocint{-\infty,q}$ is defined for every
$q\in\Bbb Q$.   Use 285Xn to show that $\sequence{k}{\nu_{n_k}}$
converges for the vague topology.}
%285Xm 285Xn 285M

\leader{285Y}{Further exercises (a)}
%\spheader 285Ya
Let $\nu$ be a Radon probability measure on
$\BbbR^r$.   Write

\Centerline{$\varhat{\nu}(y)=\Bover1{(\sqrt{2\pi})^r}
\int e^{-iy\dotproduct x}\nu(dx)$}

\noindent for every $y\in\BbbR^r$.

\quad{(i)} Writing $\varphi_{\nu}$ for the characteristic function of
$\nu$, show that $\varhat{\nu}(y)
=\Bover1{(\sqrt{2\pi})^r}\varphi_{\nu}(-y)$ for every $y\in\BbbR^r$.

\quad{(ii)} Show that
$\int h(y)\varhat{\nu}(y)dy=\int\varhat h(x)\nu(dx)$
for any Lebesgue integrable complex-valued function $h$ on
$\BbbR^r$, defining the Fourier transform $\varhat{h}$ as in 283Wa.

\quad{(iii)} Show that $\int h(x)\nu(dx)
=\int\varcheck{h}(y)\varhat{\nu}(y)dy$ for any rapidly decreasing test
function $h$ on $\BbbR^r$.

\quad{(iv)} Show that if $\nu$ is an indefinite-integral measure over
Lebesgue measure, with Radon-Nikod\'ym derivative $f$, then
$\varhat{\nu}$ is the Fourier transform of $f$.
%285D

\spheader 285Yb Let $\nu$ be a Radon probability
measure on $\BbbR^r$, with characteristic function $\varphi$.   Show that
whenever $c\le d$ in $\BbbR^r$ then

$$\bigl(\bover{i}{2\pi}\bigr)^r\lim_{\alpha_1,\ldots,\alpha_r\to\infty}
\int_{[-a,a]}\bigl(\prod_{j=1}^r
\Bover{e^{-i\delta_j\eta_j}-e^{-i\gamma_j\eta_j}}{\eta_j}\bigr)
\varphi(y)dy$$

\noindent exists and lies between $\nu\ooint{c,d}$ and $\nu[c,d]$,
writing $a=(\alpha_1,\ldots,\alpha_r)$ and
$\ooint{c,d}=\prod_{j\le r}\ooint{\gamma_j,\delta_j}$ if
$c=(\gamma_1,\ldots,\gamma_r)$ and $d=(\delta_1,\ldots,\delta_r)$.
%285Xk, 285J, 283Wc

\spheader 285Yc Let $\sequencen{X_n}$ be an independent identically
distributed sequence of
(not-essentially-constant) random variables.   Show that
$\lim_{n\to\infty}\Pr(|\sum_{k=0}^nX_k|\le\alpha)=0$ for every
$\alpha\in\Bbb R$.
%285Xd, 285Xk, 285J

\spheader 285Yd For Radon probability measures $\nu$, $\nuprime$ on
$\BbbR^r$ set

$$\eqalign{\rho(\nu,\nuprime)=\inf\{\epsilon:\epsilon\ge 0,\,
\nu\ocint{-\infty,a}\le\nuprime\ocint{-\infty,a+\epsilon\tbf{1}}+\epsilon
&\le\nu\ocint{-\infty,a+2\epsilon\tbf{1}}+2\epsilon\cr
&\text{for every }a\in\BbbR^r\},\cr}$$

\noindent writing $\ocint{-\infty,a}
=\{(\xi_1,\ldots,\xi_r):\xi_j\le\alpha_j$ for every $j\le r\}$
when $a=(\alpha_1,\ldots,\alpha_r)$, and
$\tbf{1}=(1,\ldots,1)\in\BbbR^r$.   Show that $\rho$ is a metric on the
set of Radon probability measures on $\BbbR^r$, and that the topology it
defines is the vague topology.   (Cf.\ 274Yc.)
%285K

\spheader 285Ye Let $r\ge 1$ and let $P$ be the set of Radon
probability measures on $\BbbR^r$.   For $m\in\Bbb N$ let $\rho^*_m$ be
the pseudometric on $P$ defined by setting
$\rho^*_m(\nu,\nuprime)=\sup_{\|y\|\le m}|\varphi_{\nu}(y)-\varphi_{\nuprime}(y)|$
for $\nu$, $\nuprime\in P$, writing $\varphi_{\nu}$ for the
characteristic function of $\nu$.   Show that $\{\rho^*_m:m\in\Bbb N\}$
defines the vague topology on $P$.
%285K

\spheader 285Yf Let $r\ge 1$.   We say that a set $Q$ of Radon
probability measures on $\BbbR^r$ is {\bf uniformly tight} if for every
$\epsilon>0$ there is a compact set $K\subseteq\BbbR^r$ such that
$\nu(\BbbR^r\setminus K)\le\epsilon$ for every $\nu\in Q$.
Show that if $Q$ is any uniformly tight family of Radon probability
measures on $\BbbR^r$, and $\epsilon>0$, then there are $\eta>0$,
$y_0,\ldots,y_n\in\BbbR^r$ such that
$\nu\ocint{-\infty,a}\le\nuprime\ocint{-\infty,a+\epsilon\tbf{1}}+\epsilon$
whenever $\nu$, $\nuprime\in Q$ and $a\in\BbbR^r$ and
$|\varphi_{\nu}(y_j)-\varphi_{\nuprime}(y_j)|\le\eta$ for every $j\le n$, writing
$\varphi_{\nu}$ for the characteristic function of $\nu$.
%285M, 285Xm

\spheader 285Yg Show that for any $M\ge 0$ the set of Radon
probability measures $\nu$ on $\BbbR^r$ such that
$\int\|x\|\nu(dx)\le M$ is uniformly tight in the sense of 285Yf.
%285Yf, 285Xm, 285M

\spheader 285Yh Let $C_b(\BbbR^r)$ be the Banach space of
bounded continuous real-valued functions on $\BbbR^r$.

\quad{(i)} Show that any Radon probability measure $\nu$ on
$\Bbb R^r$ corresponds to a continuous linear functional
$h_{\nu}:C_b(\Bbb R^r)\to\Bbb R$, writing $h_{\nu}(f)=\int fd\nu$ for
$f\in C_b(\Bbb R^r)$.

\quad{(ii)} Show that if $h_{\nu}=h_{\nuprime}$ then $\nu=\nuprime$.

\quad{(iii)} Show that the vague topology on the set of Radon
probability measures corresponds to the weak* topology on the dual
$(C_b(\BbbR^r))^*$ of $C_b(\BbbR^r)$.
%285M mt28bits

\spheader 285Yi Let $r\ge 1$ and let $P$ be the set of Radon
probability measures on $\BbbR^r$.   For $m\in\Bbb N$ let
$\tilde\rho^*_m$ be the pseudometric on $P$ defined by setting

\Centerline{$\tilde\rho^*_m(\nu,\nuprime)
=\int_{\{y:\|y\|\le m\}}|\varphi_{\nu}(y)-\varphi_{\nuprime}(y)|dy$}

\noindent for $\nu$, $\nuprime\in P$, writing $\varphi_{\nu}$ for the
characteristic function of $\nu$.   Show that
$\{\tilde\rho^*_m:m\in\Bbb N\}$ defines the vague topology on $P$.
%285M mt28bits

\spheader 285Yj Let $(\Omega,\Sigma,\mu)$ be a probability space.
Suppose that $\sequencen{X_n}$ is a sequence of real-valued random
variables on $\Omega$,
and $X$ another real-valued random variable on $\Omega$;  let $\varphi_{X_n}$,
$\varphi_X$ be the corresponding characteristic functions.   Show that the
following are equiveridical:
(i) $\lim_{n\to\infty}\Expn(h(X_n))=\Expn(h(X))$ for every bounded
continuous function $h:\Bbb R\to\Bbb R$;  (ii)
$\lim_{n\to\infty}\varphi_{X_n}(y)=\varphi_X(y)$ for every $y\in\Bbb R$.
%285Xm 285M

\spheader 285Yk Let $(\Omega,\Sigma,\mu)$ be a probability space, and $P$
the set of Radon probability measures on $\Bbb R$.   (i) Show that we have
a function $\psi:L^0(\mu)\to P$ defined by saying that
$\psi(X^{\ssbullet})$ is the distribution of $X$ whenever $X$ is a
real-valued random variable on $\Omega$.   (ii) Show that $\psi$ is
continuous for the topology of convergence in measure on $L^0(\mu)$ and the
vague topology on $P$.   (Compare 271Yd.)
%285Yj 285M

\spheader 285Yl Let $X$ be a real-valued random variable with
finite variance.   Show that for any $\eta\ge 0$,

\Centerline{$|\varphi(y)-1-iy\Expn(X)+\Bover12y^2\Expn(X^2)|
\le\Bover16\eta|y|^3\Expn(X^2)
+y^2\Expn(\psi_{\eta}(X))$,}

\noindent writing $\varphi$ for the characteristic function of $X$ and
$\psi_{\eta}(x)=0$ for $|x|\le\eta$, $x^2$ for $|x|>\eta$.
%285N

\spheader 285Ym Suppose that $\epsilon\ge\delta>0$ and that
$X_0,\ldots,X_n$ are independent
real-valued random variables such that

\Centerline{$\Expn(X_k)=0$ for every $k\le n$,
\quad$\sum_{k=0}^n\Var(X_k)=1$,
\quad$\sum_{k=0}^n\Expn(\psi_{\delta}(X_k))\le\delta$}

\noindent (writing $\psi_{\delta}(x)=0$ if $|x|\le\delta$, $x^2$ if
$|x|>\delta$).   Set $\gamma=\epsilon/\sqrt{\delta^2+\delta}$, and let
$Z$ be a standard normal random variable.   Show that

\Centerline{$|\varphi(y)-e^{-y^2/2}|
\le\Bover13\epsilon|y|^3+y^2(\delta+\Expn(\psi_{\gamma}(Z)))$}

\noindent for every $y\in\Bbb R$, writing $\varphi$ for the characteristic
function of $X=\sum_{k=0}^nX_k$.
% mt28bits 285Yl, 285N

\spheader 285Yn Show that for every $\epsilon>0$ there is a
$\delta>0$ such that whenever $X_0,\ldots,X_n$ are independent
real-valued random variables such that

\Centerline{$\Expn(X_k)=0$ for every $k\le n$,
\quad$\sum_{k=0}^n\Var(X_k)=1$,
\quad$\sum_{k=0}^n\Expn(\psi_{\delta}(X_k))\le\delta$}

\noindent (writing $\psi_{\delta}(x)=0$ if $|x|\le\delta$, $x^2$ if
$|x|>\delta$), then $|\varphi(y)-e^{-y^2/2}|\le\epsilon(y^2+|y^3|)$
for every $y\in\Bbb R$, writing $\varphi$ for the characteristic
function of $X=X_0+\ldots+X_n$.
%285Ym, 285N

\spheader 285Yo Use 285Yn to prove Lindeberg's theorem (274F).
%285Yn, 285N

\spheader 285Yp Let $r\ge 1$ and let $P$ be the set of Radon
probability measures on $\BbbR^r$.   Show that convolution, regarded as
a map from $P\times P$ to $P$, is continuous when $P$ is given the vague
topology.
%285R \Hint{281Xa and 257B will help.}

\spheader 285Yq Let $\frak S$ be the topology on $\Bbb R$
defined by $\{\rho'_y:y\in\Bbb R\}$, where
$\rho'_y(x,x')=|e^{iyx}-e^{iyx'}|$ (compare 285S).   Show that addition
and subtraction are continuous for $\frak S$ in the sense of 2A5A.
%285S

\spheader 285Yr\dvAnew{2014} Let $\nu$ be a probability measure on
$\Bbb R$.   Show that
$|\varphi_{\nu}(y)-\varphi_{\nu}(y')|^2\le 2(1-\Real\varphi_{\nu}(y-y'))$ for any
$y$, $y'\in\Bbb R$.
%|\varphi_{\nu}(y)-\varphi_{\nu}(y')|^2\le\int|e^{iyx)-e^{iy'x}|^2\nu(dx)
%285J

\spheader 285Ys\dvAnew{2014} Let $\sequencen{\nu_n}$ be a sequence of
probability measures on $\Bbb R$.   Set $E=\{y:y\in\Bbb R$,
$\lim_{n\to\infty}\varphi_{\nu_n}(y)=1\}$.   (i) Show that $E-E$ and
$E+E$ are included in $E$.   (ii) Show that if $E$ is not Lebesgue
negligible it is the whole of $\Bbb R$.
%285Yr 285J

\spheader 285Yt\dvAformerly{2{}85Xs}
Let $\sequencen{X_n}$ be an independent sequence of
real-valued random variables and set $S_n=\sum_{j=0}^nX_j$ for each
$n\in\Bbb N$.
Suppose that the sequence $\sequencen{\nu_{S_n}}$ of distributions is
convergent for the vague topology to a distribution.
Show that $\sequencen{S_n}$ converges in measure, therefore a.e.
%\Hint{285J, 273B, 285Ys.}
%285I  char fns of  X_n  converge
%  to  1  whenever char fns of  S_n  don't converge to  0
}

\endnotes{
\Notesheader{285} Just as with Fourier transforms, the
power of methods which use the characteristic functions of distributions
is based on three points:  (i) the characteristic function of a
distribution determines the distribution (285M); (ii) the properties of
interest in a distribution are reflected in accessible properties of its
characteristic function (285G, 285I, 285J) (iii) these properties of the
characteristic function are actually {\it different} from the
corresponding properties of the distribution, and are amenable to
different kinds of investigation.   Above all, the fact that (for
sequences!) convergence in the vague topology of distributions
corresponds to pointwise convergence for characteristic functions (285L)
provides us with a path to the classic limit theorems, as in 285Q and
285Xq.   In 285S-285U I show that this result for sequences does not
correspond immediately to any alternative characterization of the vague
topology, though it can be adapted in more than one way to give such a
characterization (see 285Ye-285Yi).

Concerning the Central Limit Theorem there is one conspicuous difference
between the method suggested here and that of \S274.   The previous
approach offered at least a theoretical possibility of giving an
explicit formula for $\delta$ in 274F as a function of $\epsilon$, and
hence an estimate of the rate of convergence to be expected in the
Central Limit Theorem.   The arguments in the present chapter, involving
as they do an entirely
non-constructive compactness argument in 281A, leave us with no way of
achieving such an estimate.   But in fact the method of characteristic
functions, suitably refined, is the basis of the best estimates known,
such as the Berry-Ess\'een theorem (274Hc).

In 285D I try to show how the characteristic function $\varphi_{\nu}$ of a
Radon probability measure can be related to a `Fourier transform'
$\varhat{\nu}$ of $\nu$ which corresponds directly to the Fourier
transforms of functions discussed in \S\S283-284.   If $f$ is a
non-negative Lebesgue integrable function and we take $\nu$ to be the
corresponding indefinite-integral measure, then $\varhat{\nu}=\varhatf$.
Thus the concept
of `Fourier transform of a measure' is a natural extension of the
Fourier transform of an integrable function.   Looking at it from the
other side, the formula of 285Dc shows that $\nu$ can be thought of as
representing the inverse Fourier transform of $\varhat{\nu}$ in the
sense of 284H-284I.   Taking $\nu$ to be the measure which assigns a
mass $1$ to the point $0$, we get the Dirac delta function, with Fourier
transform the constant function $\chi\Bbb R$.   These ideas can be extended
without difficulty to handle convolutions of measures (285R).

It is a striking fact that while there is no satisfactory
characterization of the functions which are Fourier transforms of
integrable functions, there is a characterization of the characteristic
functions of probability distributions.   This is `Bochner's
theorem'.   I give the condition in 285Xu, asking you to prove its
necessity as an exercise;  we already have three-quarters of the
machinery to prove its sufficiency, but the last step will have to wait
for Volume 4.
}

\discrpage

\frfilename{mt286.tex}
\versiondate{30.3.16}
\copyrightdate{2000}
\newdimen\sparewidth

\def\chaptername{Fourier analysis}
\def\sectionname{Carleson's theorem}

\def\energy{\mathop{\text{energy}}\nolimits}
\def\Innerprod#1#2{\bigl(#1\bigr|#2\bigr)}
\def\mass{\mathop{\text{mass}}\nolimits}
\def\recheck{\discrversionA{\immediate\write0{query}
    \global\advance\footnotenumber by 1
    \oldfootnote{$^{\the\footnotenumber}$}{recheck}}{}}
\def\vartildef{\tilde{\hbox{$f$}}}

%try:  f\in\eusm L^2
%      h rapidly decr test fn
%      g other

\newsection{286}

Carleson's theorem ({\smc Carleson 66}) was the (unexpected)
%Dorothy Maharam once said
solution to
a long-standing problem.   Remarkably, it can be proved by `elementary'
methods.   The hardest part of the work below, in 286J-286L, demands
only the laborious verification of inequalities.   How the inequalities
were chosen is a different matter;  for once, some of the ideas of
the proof are embodied in the statements of the lemmas.   The argument here is a
greatly expanded version of {\smc Lacey \& Thiele 00}.

The Hardy-Littlewood Maximal Theorem (286A) is important, and worth
learning even if you
leave the rest of the section as an unexamined monument.   I bring
286B-286D %286B 286C 286D
forward to the beginning of the section, even though they are little
more than worked exercises, because they also have potential uses in
other contexts.

%mention subsupersubscripts?

\cmmnt{The complexity of the argument is such that it is useful to introduce a
substantial number of special notations.   Rather than include
these in
the general index, I give a list in 286W.   Among them are ten
constants $C_1,\ldots,C_{10}$.
The values of these numbers are of no significance.   The method of
proof here is quite inappropriate if we want to estimate rates of
convergence.   I give recipes for the calculation of the $C_n$ only
for the sake of the linear logic in which this treatise is written, and
because they occasionally offer clues concerning the tactics being used.
}

In this section all integrals are with respect to Lebesgue measure $\mu$
on $\Bbb R$ unless otherwise stated.

\leader{286A}{The Maximal Theorem}
% Hardy \& Littlewood 30, but not best constant
Suppose that $1<p<\infty$ and that
$f\in\eusm L_{\Bbb C}^p(\mu)$\cmmnt{ (definition:  244P)}.   Set

\Centerline{$f^*(x)=\sup\{\Bover1{b-a}\int_a^b|f|:a\le x\le b$, $a<b\}$}

\noindent for $x\in\Bbb R$.   Then
$\|f^*\|_p\le\Bover{2^{1/p}p}{p-1}\|f\|_p$.

\proof{{\bf (a)} It is enough to consider the case $f=|f|$.   Note that
if $E\subseteq\Bbb R$ has finite measure, then

\Centerline{$\int_Ef
=\int(f\times\chi E)\times\chi E
\le\|f\times\chi E\|_p(\mu E)^{1/q}\le\|f\|_p(\mu E)^{1/q}$}

\noindent is finite, where $q=\Bover{p}{p-1}$, by H\"older's inequality
(244Eb).   Consequently, if $t>0$ and
$\int_Ef\ge t\mu E$, we must have
$t\mu E\le\|f\times\chi E\|_p(\mu E)^{1/q}$,
$t(\mu E)^{1/p}\le\|f\times\chi E\|_p$ and

\Centerline{$\mu E\le\Bover1{t^p}\|f\times\chi E\|^p_p
=\Bover1{t^p}\int_Ef^p$.}

\medskip

{\bf (b)} For $t>0$, set

\Centerline{$G_t=\{x:t(y-x)<\int_x^yf$ for some $y>x\}$.}

\medskip

\quad{\bf (i)} $G_t$ is an open set.  \Prf\ For any $y\in\Bbb R$,

\Centerline{$G_{ty}=\{x:x<y$, $t(y-x)<\int_x^yf\}$}

\noindent is open, because $x\mapsto t(y-x)$ and $x\mapsto\int_x^yf$ are
continuous (225A);  so $G_t=\bigcup_{y\in\Bbb R}G_{ty}$ is open.\ \Qed

\medskip

\quad{\bf (ii)} By 2A2I, there is a partition $\Cal C$ of $G_t$ into
open intervals.
Now $C$ is bounded and $t\mu C\le\int_Cf$ for every $C\in\Cal C$.

\medskip

\quad\Prf\grheada\ For $x\in C$, consider 
$F_x=\{y:y\ge x$, $t(y-x)\le\int_x^yf\}$.
$x\in F_x$
and $y-x\le\Bover1{t^p}\int_{-\infty}^{\infty}f^p$ for every $y\in F_x$, 
by (a), so
$F_x$ is bounded above.   Set $z_x=\sup F_x$.   Because
$y\mapsto t(y-x)-\int_x^yf$ is continuous, $z_x\in F_x$.   
\Quer\ If $z_x\in G_t$, there is a
$y>z_x$ such that $t(y-z_x)<\int_{z_x}^yf$;  but now 

\Centerline{$t(y-x)
\le\int_x^{z_x}f+\int_{z_x}^yf=\int_x^yf$}

\noindent and $y\in F_x$, which is impossible.\ \BanG\  Thus $z_x\notin G_t$
and $z_x\notin C$, so that $z_x$ is an upper bound of $C$.

\medskip

\qquad\grheadb\ This shows that 

\Centerline{$\sup C\le z_x\le x+\Bover1{t^p}\int_{-\infty}^{\infty}f^p$}

\noindent for every $x\in C$.   So in fact $C$ is
bounded and is of the form $\ooint{a,b}$ where $a<b$ in $\Bbb R$.   
\Quer\ If $t(b-a)>\int_a^bf$, there is an $x\in\ooint{a,b}$ such that
$t(b-x)>\int_x^bf$.   Now we know that $b\le z_x$ and $b\notin G_t$, 
so we have $t(z_x-b)\ge\int_b^{z_x}f$.   Adding,
$t(z_x-x)>\int_x^{z_x}f$ and $z_x\notin F_x$.\ \Bang

\medskip

\qquad\grheadc\ Thus $t\mu C\le\int_Cf$, as claimed.\ \Qed

\medskip

\quad{\bf (iii)} Accordingly, because $\Cal C$ is countable and
$f$ is non-negative, we can apply (a) in its full strength to see that

\Centerline{$\mu G_t=\sum_{C\in\Cal C}\mu C
\le\sum_{C\in\Cal C}\Bover1{t^p}\int_Cf^p
\le\Bover1{t^p}\int_{-\infty}^{\infty}f^p$}

\noindent is finite, and

\Centerline{$\int_{G_t}f=\sum_{C\in\Cal C}\int_Cf
\ge\sum_{C\in\Cal C}t\mu C=t\mu G_t$.}

\medskip

{\bf (c)} All this is true for every $t>0$.   Now if we set

\Centerline{$f_1^*(x)=\sup_{b>x}\Bover1{b-x}\int_b^af$}

\noindent for $x\in\Bbb R$, we have $\{x:f_1^*(x)>t\}=G_t$ for
every $t>0$.

For any $t>0$,

\Centerline{$\Bover1pt\mu G_t
=(1-\Bover1q)t\mu G_t
\le\int_{G_t}f-\Bover1q t\chi\Bbb R
\le\int_{-\infty}^{\infty}(f-\Bover1q t\chi\Bbb R)^+$.}

\noindent So

$$\eqalignno{\int_{-\infty}^{\infty}(f_1^*)^p
&=\int_0^{\infty}\mu\{x:f_1^*(x)^p>t\}dt\cr
\displaycause{see 252O}
&=p\int_0^{\infty}u^{p-1}\mu\{x:f_1^*(x)>u\}du\cr
\displaycause{substituting $t=u^p$}
&=p\int_0^{\infty}u^{p-1}\mu G_udu
\le p^2\int_0^{\infty}u^{p-2}
  \bigl(\int_{-\infty}^{\infty}(f-\Bover1q u\chi\Bbb R)^+\bigr)du\cr
&=p^2\int_{-\infty}^{\infty}\int_0^{\infty}
  \max(0,f(x)-\Bover1q u)u^{p-2}dudx\cr
\displaycause{by Fubini's theorem, 252B, because
$(x,u)\mapsto u^{p-2}\max(0,f(x)-\bover1qu)$ is measurable and
non-negative}
&=p^2\int_{-\infty}^{\infty}\int_0^{qf(x)}
  u^{p-2}(f(x)-\Bover1q u)dudx\cr
&=\Bover{p^2q^{p-1}}{p(p-1)}\int_{-\infty}^{\infty}f^p
=(\Bover{p}{p-1})^p\|f\|_p^p.\cr}$$

\medskip

{\bf (d)} Similarly, setting $f_2^*(x)=\sup_{a<x}\Bover1{x-a}\int_a^xf$
for $x\in\Bbb R$,
$\int_{-\infty}^{\infty}(f_2^*)^p\le(\Bover{p}{p-1})^p\|f\|_p^p$.   But
$f^*=\max(f_1^*,f_2^*)$.   \Prf\ Of course $f_1^*\le f^*$ and
$f_2^*\le f^*$.   But also, if $f^*(x)>t$, there must be a non-trivial
interval $I$ containing $x$ such that $\int_If>t\mu I$;  if $a=\inf I$
and $b=\sup I$, then either $\int_a^xf>(x-a)t$ and $f_2^*(x)>t$, or
$\int_x^bf>(b-x)t$ and $f_1^*(x)>t$.   As $x$ and $t$ are arbitrary,
$f^*=\max(f_1^*,f_2^*)$.\ \Qed

Accordingly

$$\eqalign{\|f^*\|_p^p
&=\int_{-\infty}^{\infty}(f^*)^p
=\int_{-\infty}^{\infty}\max((f_1^*)^p,(f_2^*)^p)\cr
&\le\int_{-\infty}^{\infty}(f_1^*)^p+(f_2^*)^p
\le 2(\Bover{p}{p-1})^p\|f\|^p_p.\cr}$$

\noindent Taking $p$th roots, we have the inequality we seek.
}%end of proof of 286A

\leader{286B}{Lemma} Let $g:\Bbb R\to\coint{0,\infty}$ be a function
which is non-decreasing on $\ocint{-\infty,\alpha}$, non-increasing on
$\coint{\beta,\infty}$ and constant on $[\alpha,\beta]$, where
$\alpha\le\beta$.   Then for any measurable function
$f:\Bbb R\to[0,\infty]$,
$\int_{-\infty}^{\infty}f\times g\le\int_{-\infty}^{\infty}g
  \cdot\sup_{a\le\alpha,b\ge\beta,a<b}\Bover1{b-a}\int_a^bf$.

\proof{ Set
$\gamma=\sup_{a\le\alpha,b\ge\beta,a<b}\Bover1{b-a}\int_a^bf$.   For
$n$, $k\in\Bbb N$ set
$E_{nk}=\{x:\alpha-2^n\le x\le\beta+2^n$, $g(x)\ge 2^{-n}(k+1)\}$, so
that $E_{nk}$
is either empty or a bounded interval including $[\alpha,\beta]$,
and $\int_{E_{nk}}f\le\gamma\mu E_{nk}$.   For
$n\in\Bbb N$, set $g_n=2^{-n}\sum_{k=0}^{4^n-1}\chi E_{nk}$;  then
$\sequencen{g_n}$ is a non-decreasing sequence of functions with
supremum $g$, and

$$\eqalign{\int_{-\infty}^{\infty} f\times g
&=\sup_{n\in\Bbb N}\int_{-\infty}^{\infty} f\times g_n
=\sup_{n\in\Bbb N}2^{-n}\sum_{k=0}^{4^n-1}\int_{E_{nk}}f\cr
&\le\sup_{n\in\Bbb N}2^{-n}\sum_{k=0}^{4^n-1}\gamma\mu E_{nk}
=\sup_{n\in\Bbb N}\gamma\int_{-\infty}^{\infty} g_n
=\gamma\int_{-\infty}^{\infty} g,\cr}$$

\noindent as claimed.
}%end of proof of 286B

\cmmnt{\medskip

\noindent{\bf Remark} Compare 224J.}

\vleader{48pt}{286C}{Shift, modulation and dilation}\cmmnt{ Some of
the
calculations below will be easier if we use the following formalism.}
For any function $f$ with domain included in $\Bbb R$, and
$\alpha\in\Bbb R$, we can define

\Centerline{$(S_{\alpha}f)(x)=f(x+\alpha)$,
\quad$(M_{\alpha}f)(x)=e^{i\alpha x}f(x)$,
\quad$(D_{\alpha}f)(x)=f(\alpha x)$}

\noindent whenever the right-hand sides are defined.   \cmmnt{In the
case of $S_{\alpha}f$ and $D_{\alpha}f$ it is sometimes convenient to
allow $\pm\infty$ as a value of the function.   We have the following
elementary facts.}

\spheader 286Ca $S_{-\alpha}S_{\alpha}f=f$, $D_{1/\alpha}D_{\alpha}f=f$
if $\alpha\ne 0$.

\spheader 286Cb
$S_{\alpha}(f\times g)=S_{\alpha}f\times S_{\alpha}g$,
$D_{\alpha}(f\times g)=D_{\alpha}f\times D_{\alpha}g$.

\spheader 286Cc $D_{\alpha}|f|=|D_{\alpha}f|$.

\spheader 286Cd If $f$ is integrable, then

\Centerline{$(M_{\alpha}f)\varsphat=S_{-\alpha}\varhatf$,
\quad$(S_{\alpha}f)\varsphat=M_{\alpha}\varhatf$,
\quad$(S_{\alpha}f)\varspcheck=M_{-\alpha}\varcheckf$;}

\noindent if moreover $\alpha>0$, then

\Centerline{$\alpha(D_{\alpha}f)\varsphat
=D_{1/\alpha}\varhatf$,
\quad$\alpha(D_{\alpha}f)\varspcheck
=D_{1/\alpha}\varcheckf$\dvro{.}{}}

\cmmnt{\noindent (283Cc-283Ce).}

\spheader 286Ce If $f$ belongs to
$\eusm L^1_{\Bbb C}=\eusm L^1_{\Bbb C}(\mu)$, so do $S_{\alpha}f$,
$M_{\alpha}f$ and (if $\alpha\ne 0$) $D_{\alpha}f$, and in this case

\Centerline{$\|S_{\alpha}f\|_1=\|M_{\alpha}f\|_1=\|f\|_1$,
\quad$\|D_{\alpha}f\|_1=\Bover1{|\alpha|}\|f\|_1$.}

\spheader 286Cf If $f$ belongs to
$\eusm L^2_{\Bbb C}$ so do $S_{\alpha}f$, $M_{\alpha}f$ and (if
$\alpha\ne 0$) $D_{\alpha}f$, and in this case

\Centerline{$\|S_{\alpha}f\|_2=\|M_{\alpha}f\|_2=\|f\|_2$,
\quad$\|D_{\alpha}f\|_2=\Bover1{\sqrt{|\alpha|}}\|f\|_2$.}

\spheader 286Cg If $h$ is a rapidly decreasing test
function\cmmnt{ (284A)}, so are
$M_{\alpha}h$ and $S_{\alpha}h$ and (if $\alpha\ne 0$) $D_{\alpha}h$.

\leader{286D}{Lemma} Suppose that $g:\Bbb R\to[0,\infty]$ is a
measurable function such that, for some constant $C\ge 0$,
$\int_Eg\le C\sqrt{\mu E}$ whenever $\mu E<\infty$.   Then $g$ is finite
almost everywhere and
$\biggerint_{-\infty}^{\infty}\Bover1{1+|x|}g(x)dx$ is finite.

\proof{ For any $n\ge 1$, set $E_n=\{x:|x|\le n,\,g(x)\ge n\}$;  then

\Centerline{$n\mu E_n\le\int_{E_n}g\le C\sqrt{\mu E_n}$,}

\noindent so $\mu E_n\le\Bover{C^2}{n^2}$ and

\Centerline{$\{x:g(x)=\infty\}=\bigcap_{n\ge 1}\bigcup_{m\ge n}E_m$}

\noindent has measure at most
$\inf_{n\ge 1}\sum_{m=n}^{\infty}\mu E_m=0$.

As for the integral,
set $G(x)=\int_0^xg$ for $x\ge 0$.   Then, for any $a\ge 0$,

$$\eqalignno{\int_0^a\Bover{g(x)}{1+x}dx
&=\Bover{G(a)}{1+a}+\int_0^a\Bover{G(x)}{(1+x)^2}dx\cr
\displaycause{225F}
&\le C\bigl(\Bover{\sqrt{a}}{1+a}
  +\int_0^a\Bover{\sqrt{x}}{(1+x)^2}dx\bigr)
\le C\bigl(1+\int_0^{\infty}\Bover{\sqrt{x}}{(1+x)^2}dx\bigr),\cr}$$

\noindent so

\Centerline{$\int_0^{\infty}\Bover{g(x)}{1+x}dx
\le C\bigl(1+\int_0^{\infty}\Bover{\sqrt{x}}{(1+x)^2}dx\bigr)$}


\noindent is finite.   Similarly,
$\biggerint_{-\infty}^0\Bover{g(x)}{1-x}dx$
is finite, so we have the result.
}%end of proof of 286D

\allowmorestretch{500}{
\leader{286E}{The Lacey-Thiele construction (a)}
Let $\Cal I$ be the family of all {\bf dyadic intervals} of the form
\penalty-100$\coint{2^kn,2^k(n+1)}$ where $k$, $n\in\Bbb Z$.   The essential
geometric property of $\Cal I$ is that if $I$, $J\in\Cal I$ then either
$I\subseteq J$ or $J\subseteq I$ or $I\cap J=\emptyset$.   Let $Q$ be
the set of all pairs $\sigma=(I_{\sigma},J_{\sigma})\in\Cal I^2$ such
that $\mu I_{\sigma}\cdot\mu J_{\sigma}=1$.   For $\sigma\in Q$, let
$k_{\sigma}\in\Bbb Z$
be such that $\mu J_{\sigma}=2^{k_{\sigma}}$ and
$\mu I_{\sigma}=2^{-k_{\sigma}}$;  let $x_{\sigma}$
be the midpoint of $I_{\sigma}$, $y_{\sigma}$ the midpoint of
$J_{\sigma}$, $J^l_{\sigma}\in\Cal I$ the left-hand half-interval of
$J_{\sigma}$, $J^r_{\sigma}\in\Cal I$ the
right-hand half-interval of $J_{\sigma}$, and $y^l_{\sigma}$ the lower
quartile of $J_{\sigma}$\cmmnt{, that is, the midpoint of
$J^l_{\sigma}$}.
}%end of allowmorestretch

\spheader 286Eb There is a rapidly decreasing test function $\phi$ such
that $\varhat{\phi}$ is real-valued and
$\chi[-\bover16,\bover16]\le\varhat{\phi}\le\chi[-\bover15,\bover15]$.
\prooflet{\Prf\ Look at parts (b)-(d) of the proof
of 284G.   The process there can be used to provide us with a smooth
function $\psi_1$ which is zero outside the interval
$[\bover16,\bover15]$ and strictly positive on
$\ooint{\bover16,\bover15}$;  multiplying by a suitable factor, we
can arrange that $\int_{-\infty}^{\infty}\psi_1=1$.   So if we set
$\psi_2(x)=1-\int_{-\infty}^x\psi_1$ for $x\in\Bbb R$, $\psi_2$ will be
smooth, and
$\chi\ocint{-\infty,\bover16}\le\psi_2\le\chi\ocint{-\infty,\bover15}$.
Now set $\psi_0(x)=\psi_2(x)\psi_2(-x)$ for $x\in\Bbb R$, and
$\phi=\varcheck{\psi}_0$;  $\varhat{\phi}=\psi_0$ (284C) will have
the required property.\ \Qed}
%wouldn't it be nicer if $\|\phi\|_2=1$?  but it is helpful at one
%point to have $\varhat\phi(0)=1$

For $\sigma\in Q$, set\cmmnt{ $\phi_{\sigma}
=2^{k_{\sigma}/2}M_{y^l_{\sigma}}S_{-x_{\sigma}}D_{2^{k_{\sigma}}}\phi$,
so that}

\Centerline{$\phi_{\sigma}(x)
=\sqrt{\mu J_{\sigma}}e^{iy^l_{\sigma}x}
  \phi((x-x_{\sigma})\mu J_{\sigma})$.}

\noindent\cmmnt{Observe that }$\phi_{\sigma}$ is a rapidly decreasing
test function.   \cmmnt{Now
$\varhat{\phi}_{\sigma}=2^{-k_{\sigma}/2}S_{-y^l_{\sigma}}
M_{-x_{\sigma}}D_{2^{-k_{\sigma}}}\varhat{\phi}$, that is,}

\Centerline{$\varhat{\phi}_{\sigma}(y)
=\sqrt{\mu I_{\sigma}}e^{-ix_{\sigma}(y-y^l_{\sigma})}
  \varhat{\phi}((y-y^l_{\sigma})\mu I_{\sigma})$,}

\noindent which is zero
unless\cmmnt{ $|y-y^l_{\sigma}|\le\bover15\mu J_{\sigma}$;  since the
length of $J^l_{\sigma}$ is $\bover12\mu J_{\sigma}$, this can be so
only when} $y\in J^l_{\sigma}$.   \cmmnt{We have the following simple
facts.}

\quad(i) $\|\phi_{\sigma}\|_2
=\cmmnt{\sqrt{\mu J_{\sigma}}\cdot\sqrt{\mu I_{\sigma}}\|\phi\|_2
=}\|\phi\|_2$
for every $\sigma\in Q$.

\quad(ii) $\|\varhat{\phi}_{\sigma}\|_1
=\cmmnt{\sqrt{\mu I_{\sigma}}\cdot\mu J_{\sigma}\|\varhat{\phi}\|_1
=}\sqrt{\mu J_{\sigma}}\|\varhat{\phi}\|_1$ for every $\sigma\in Q$.

\quad(iii) If $\sigma$, $\tau\in Q$ and
$J^l_{\sigma}\cap J^l_{\tau}=\emptyset$ then

\Centerline{$\innerprod{\phi_{\sigma}}{\phi_{\tau}}
=\innerprod{\varhat{\phi}_{\sigma}}{\varhat{\phi}_{\tau}}=0$\dvro{.}{,}}

\noindent\cmmnt{by 284Ob.   }(For $f$,
$g\in\eusm L_{\Bbb C}^2$, I
write $\innerprod{f}{g}$ for $\int_{-\infty}^{\infty}f\times\bar g$.)

\quad(iv) If $\sigma$, $\tau\in Q$ and
$J_{\sigma}\ne J_{\tau}$ and $J^r_{\sigma}\cap J^r_{\tau}$ is
non-empty, then\cmmnt{ $J^l_{\sigma}\cap J^l_{\tau}=\emptyset$ so}
$\innerprod{\phi_{\sigma}}{\phi_{\tau}}=0$.

\spheader 286Ec Set $w(x)=\Bover1{(1+|x|)^3}$ for $x\in\Bbb R$.
For $\sigma\in Q$, set\cmmnt{ $w_{\sigma}
=2^{k_{\sigma}}S_{-x_{\sigma}}D_{2^{k_{\sigma}}}w$, so that}

\Centerline{$w_{\sigma}(x)=w((x-x_{\sigma})\mu J_{\sigma})\mu J_{\sigma}
\le\mu J_{\sigma}=2^{k_{\sigma}}$}

\noindent for every
$x$.   Note that $w_{\sigma}=w_{\tau}$ whenever $I_{\sigma}=I_{\tau}$.

\leader{286F}{A partial order (a)} For $\sigma$, $\tau\in Q$ say that
$\tau\le\sigma$ if $J_{\tau}\subseteq J_{\sigma}$ and
$I_{\sigma}\subseteq I_{\tau}$.   Then $\le$ is a partial order on $Q$.
\cmmnt{We have the following elementary facts. }

\quad(i) If $\tau\le\sigma$, then $k_{\tau}\le k_{\sigma}$.

\quad(ii) If $\sigma$ and $\tau$ are incomparable\cmmnt{ (that is,
$\sigma\not\le\tau$ and $\tau\not\le\sigma$)}, then
$(I_{\sigma}\times J_{\sigma})\cap(I_{\tau}\times J_{\tau})$ is empty.
\prooflet{\Prf\ We may suppose that $k_{\sigma}\le k_{\tau}$.   If
$J_{\sigma}\cap J_{\tau}\ne\emptyset$, then
$J_{\sigma}\subseteq J_{\tau}$, because both are dyadic intervals, and
$J_{\sigma}$ is the shorter;  but as $\sigma\not\le\tau$, this means
that $I_{\tau}\not\subseteq I_{\sigma}$ and
$I_{\sigma}\cap I_{\tau}=\emptyset$.\ \Qed}

\quad(iii) If $\sigma$, $\sigma'$ are incomparable and both greater
than or equal to $\tau$, then
$I_{\sigma}\cap I_{\sigma'}=\emptyset$\cmmnt{, because
$J_{\tau}\subseteq J_{\sigma}\cap J_{\sigma'}$}.

\quad(iv) If $\tau\le\sigma$ and $k_{\tau}\le k\le k_{\sigma}$, then
there is
a (unique) $\upsilon$ such that $\tau\le\upsilon\le\sigma$ and
$k_{\upsilon}=k$.   \prooflet{(The point is that there is a unique
$I\in\Cal I$ such that $I_{\sigma}\subseteq I\subseteq I_{\tau}$ and
$\mu I=2^{-k}$;  and similarly there is just one candidate for
$J_{\upsilon}$.)}

\spheader 286Fb\dvro{ If}{ It will be convenient to have a shorthand for
the following:  if} $R\subseteq Q$, say that 

\Centerline{$R^+=\bigcup_{\tau\in R}\{\sigma:\tau\le\sigma\in Q\}$.}

 
\spheader 286Fc For $\tau\in Q$ set

\Centerline{$T_{\tau}=\{\sigma:\sigma\in Q$, $\tau\le\sigma$,
$J^r_{\tau}\subseteq J^r_{\sigma}\}$.}

\noindent Note that if $\sigma$, $\sigma'\in T_{\tau}$
and $k_{\sigma}\ne k_{\sigma'}$ then\cmmnt{ $J_{\sigma}\ne J_{\sigma'}$ and
$J^r_{\sigma}\cap J^r_{\sigma'}\ne\emptyset$, so}
$\innerprod{\phi_{\sigma}}{\phi_{\sigma'}}=0$\cmmnt{ (286E(b-iv))}.

\vleader{72pt}{286G}{}\cmmnt{ We shall need the results of some elementary
calculations.   The first four are nearly trivial.

\medskip

\noindent}{\bf Lemma} (a)
$\int_{-\infty}^{\infty}w_{\sigma}=\int_{-\infty}^{\infty}w=1$ for every
$\sigma\in Q$.

(b) For any $m\in\Bbb N$,
$\sum_{n=m}^{\infty}w(n+\bover12)\le\Bover1{2(1+m)^2}$.

(c) Suppose that $\sigma\in Q$ and that $I$ is an interval not
containing $x_{\sigma}$ in its interior.   Then
$\int_Iw_{\sigma}\ge w_{\sigma}(x)\mu I$, where $x$ is the midpoint of
$I$.

(d) For any $x\in\Bbb R$,
$\sum_{n=-\infty}^{\infty}w(x-n)\le 2$.

(e) There is a constant $C_1\ge 0$ such that
$|\phi(x)|\le C_1\min(w(3),w(x)^2)$ for every $x\in\Bbb R$ and

\Centerline{$|\phi_{\sigma}(x)|
\le C_1\sqrt{\mu I_{\sigma}}w_{\sigma}(x)
  \min(1,w_{\sigma}(x)\mu I_{\sigma})$}

\noindent for every $x\in\Bbb R$ and $\sigma\in Q$.

(f) There is a constant $C_2\ge 0$ such that
$\int_{-\infty}^{\infty}w(x)w(\alpha x+\beta)dx\le C_2w(\beta)$ whenever
$0\le\alpha\le 1$ and $\beta\in\Bbb R$.

(g) There is a constant $C_3\ge 0$ such that
$|\innerprod{\phi_{\sigma}}{\phi_{\tau}}|
\le C_3\sqrt{\mu I_{\sigma}}\sqrt{\mu J_{\tau}}\int_{I_{\tau}}w_{\sigma}$
whenever $\sigma$, $\tau\in Q$ and $k_{\sigma}\le k_{\tau}$.

(h) There is a constant $C_4\ge 0$ such that

\Centerline{$\sum_{\sigma\in Q,\sigma\ge\tau,k_{\sigma}=k}
\int_{\Bbb R\setminus I_{\tau}}w_{\sigma}\le C_4$}

\noindent whenever $\tau\in Q$ and $k\in\Bbb Z$.

\proof{{\bf (a)} Immediate from the definition in 286Ec,
the formulae in 286Ce and the fact that
$\biggerint_0^{\infty}\Bover1{(1+x)^3}dx=\Bover12$.

\medskip

{\bf (b)} The point is just that $w$ is convex on
$\ocint{-\infty,0}$ and
$\coint{0,\infty}$.   So we can apply 233Ib with $f(x)=x$, or
argue directly from the fact that
$w(n+\bover12)\le\bover12(w(n+\bover12+x)+w(n+\bover12-x))$ for
$|x|\le\bover12$, to see that $w(n+\bover12)\le\int_n^{n+1}w$
for every $n\ge 0$.   Accordingly

\Centerline{$\sum_{n=m}^{\infty}w(n+\bover12)
\le\int_m^{\infty}w=\Bover1{2(1+m)^2}$.}

\medskip

{\bf (c)} Similarly, because $I$ lies all on the same side of
$x_{\sigma}$, $w_{\sigma}$ is convex on $I$, so the same inequality
yields $w_{\sigma}(x)\mu I\le\int_Iw_{\sigma}$.

\medskip

{\bf (d)} Let $m$ be such that $|x-m|\le\bover12$.   Then,
using the same inequalities as before to estimate $w(x-n)$ for
$n\ne m$, we have

$$\eqalign{\sum_{n=-\infty}^{\infty}w(x-n)
&\le w(x-m)+\int_{-\infty}^{x-m-\bover12}w
  +\int_{x-m+\bover12}^{\infty}w\cr
&\le 1+\int_{-\infty}^{\infty}w
=2.\cr}$$

\medskip

{\bf (e)} Because
$\lim_{x\to\infty}x^6\phi(x)=\lim_{x\to-\infty}x^6\phi(x)=0$,
there is a $C_1>0$ such that
$|\phi(x)|\le C_1\min(w(3),w(x)^2)$ for every $x\in\Bbb R$.  Now
$|\phi(x)|\le C_1w(x)^2=C_1w(x)\min(1,w(x))$ for every $x$, so

$$\eqalign{|\phi_{\sigma}(x)|
&=\sqrt{\mu J_{\sigma}}|\phi((x-x_{\sigma})\mu J_{\sigma}|
\le C_1\sqrt{\mu J_{\sigma}}w((x-x_{\sigma})\mu J_{\sigma})
   \min(1,w((x-x_{\sigma})\mu J_{\sigma}))\cr
&=C_1\sqrt{\mu J_{\sigma}}w_{\sigma}(x)\mu I_{\sigma}
   \min(1,w_{\sigma}(x)\mu I_{\sigma})
=C_1\sqrt{\mu I_{\sigma}}w_{\sigma}(x)
   \min(1,w_{\sigma}(x)\mu I_{\sigma})\cr}$$

\noindent whenever $\sigma\in Q$ and $x\in\Bbb R$.

\medskip

{\bf (f)(i)} The first step is to note that

$$\bover{w(\bover12(1+\beta))}{w(\beta)}
=\bover{8(1+\beta)^3}{(3+\beta)^3}\le 8$$

\noindent for every $\beta\ge 0$.   Now
$\alpha w(\alpha+\alpha\beta)\le 4w(\beta)$ whenever $\beta\ge 0$ and
$\alpha\ge\bover12$.   \Prf\ For $t\ge\bover12$,

\Centerline{$\Bover{d}{dt}tw(t+t\beta)
=\Bover{1-2t(1+\beta)}{(1+t+t\beta)^4}\le 0$,}

\noindent so

\Centerline{$\alpha w(\alpha+\alpha\beta)
\le\bover12w(\bover12+\bover12\beta)\le 4w(\beta)$. \Qed}

Of course this means that

\Centerline{$\Bover1{\alpha}w(\Bover{1+\beta}{2\alpha})\le 8w(\beta)$}

\noindent whenever $\beta\ge 0$ and $0<\alpha\le 1$.

\medskip

\quad{\bf (ii)} Try $C_2=16$.   If $0<\alpha\le 1$ and $\beta\ge 0$, set
$\gamma=\Bover{1+\beta}{2\alpha}$.   Then, for any $x\ge-\gamma$,

\Centerline{$1+\alpha x+\beta
=(1+\beta)(1+\Bover{\alpha x}{1+\beta})\ge\Bover12(1+\beta)$,}

\noindent so $w(\alpha x+\beta)\le 8w(\beta)$ and

\Centerline{$\int_{-\gamma}^{\infty}w(x)w(\alpha x+\beta)dx
\le 8w(\beta)\int_{-\gamma}^{\infty}w\le 8w(\beta)$.}

\noindent On the other hand,

$$\eqalign{\int_{-\infty}^{-\gamma}w(x)w(\alpha x+\beta)dx
&\le w(\gamma)\int_{-\infty}^{\infty}w(\alpha x+\beta)dx\cr
&=\Bover1{\alpha}w(\Bover{1+\beta}{2\alpha})
  \int_{-\infty}^{\infty}w
\le 8w(\beta).\cr}$$

\noindent Putting these together,
$\int_{-\infty}^{\infty}w(x)w(\alpha x+\beta)dx
\le 16w(\beta)$;  and this is true whenever $0<\alpha\le 1$ and
$\beta\ge 0$.

\medskip

\quad{\bf (iii)} If $\alpha=0$, then

\Centerline{$\int_{-\infty}^{\infty}w(x)w(\alpha x+\beta)dx
=w(\beta)\int_{-\infty}^{\infty}w=w(\beta)\le C_2w(\beta)$}

\noindent for any $\beta$.   If $0<\alpha\le 1$ and $\beta<0$, then

$$\eqalignno{\int_{-\infty}^{\infty}w(x)w(\alpha x+\beta)dx
&=\int_{-\infty}^{\infty}w(-x)w(-\alpha x-\beta)dx\cr
\displaycause{because $w$ is an even function}
&=\int_{-\infty}^{\infty}w(x)w(\alpha x-\beta)dx
\le C_2w(-\beta)\cr
\displaycause{by (ii) above}
&=C_2w(\beta).\cr}$$

\noindent So we have the required inequality in all cases.

\medskip

{\bf (g)} Set $C_3=\max(C_1^2C_2,\|\phi\|_2^2/\int_{-1/2}^{1/2}w)$.

\medskip

\quad{\bf (i)} It is worth disposing immediately of the case
$\sigma=\tau$.   In this case,

\Centerline{$|\innerprod{\phi_{\sigma}}{\phi_{\tau}}|
=\|\phi_{\sigma}\|_2^2=\|\phi\|_2^2$,}

\noindent while

$$\eqalign{\int_{I_{\tau}}w_{\sigma}
&=\mu J_{\sigma}\int_{x_{\sigma}-\bover12\mu I_{\sigma}}
  ^{x_{\sigma}+\bover12\mu I_{\sigma}}w((x-x_{\sigma})\mu J_{\sigma})dx
=\int_{-1/2}^{1/2}w,\cr}$$

\noindent so certainly
$|\innerprod{\phi_{\sigma}}{\phi_{\tau}}|
\le C_3\int_{I_{\tau}}w_{\sigma}$.

\medskip

\quad{\bf (ii)} If $\sigma\ne\tau$ and $I_{\sigma}=I_{\tau}$ then
$J_{\sigma}\cap J_{\tau}=\emptyset$ so
$\innerprod{\phi_{\sigma}}{\phi_{\tau}}=0$, by 286E(b-iii).

\medskip

\quad{\bf (iii)} Now suppose that $I_{\sigma}\ne I_{\tau}$.   In this
case, because $\mu I_{\tau}\le\mu I_{\sigma}$,
$I_{\tau}$ must lie all on the same side of $x_{\sigma}$, so
$\int_{I_{\tau}}w_{\sigma}\ge w_{\sigma}(x_{\tau})\mu I_{\tau}$, by (c).
Accordingly

$$\eqalignno{|\innerprod{\phi_{\sigma}}{\phi_{\tau}}|
&\le\int_{-\infty}^{\infty}|\phi_{\sigma}|\times|\phi_{\tau}|
\le C_1^2\sqrt{\mu I_{\sigma}}\sqrt{\mu I_{\tau}}
  \int_{-\infty}^{\infty}w_{\sigma}\times w_{\tau}
\Displaycause{using (e) twice}
=C_1^2\sqrt{\mu J_{\sigma}}\sqrt{\mu J_{\tau}}\int_{-\infty}^{\infty}
  w((x-x_{\sigma})\mu J_{\sigma})w((x-x_{\tau})\mu J_{\tau})dx\cr&
=C_1^2\sqrt{\mu J_{\sigma}}\sqrt{\mu I_{\tau}}\int_{-\infty}^{\infty}
  w(x\mu J_{\sigma}\mu I_{\tau}
    +(x_{\tau}-x_{\sigma})\mu J_{\sigma})w(x)dx\cr&
\le C_1^2C_2\sqrt{\mu J_{\sigma}}\sqrt{\mu I_{\tau}}
  w((x_{\tau}-x_{\sigma})\mu J_{\sigma})
\Displaycause{by (f), since $\mu J_{\sigma}\mu I_{\tau}\le 1$}
\le C_3\sqrt{\mu I_{\sigma}}\sqrt{\mu I_{\tau}}w_{\sigma}(x_{\tau})
\le C_3\sqrt{\mu I_{\sigma}}\sqrt{\mu J_{\tau}}
   \int_{I_{\tau}}w_{\sigma},\cr}$$

\noindent as required.

\medskip

{\bf (h)} Set
$C_4=2\sum_{j=0}^{\infty}\int_{j+\bover12}^{\infty}w$;  this is
finite because $\int_{\alpha}^{\infty}w=\Bover1{2(1+\alpha)^2}$
for every $\alpha\ge 0$.

If $k<k_{\tau}$ then $k_{\sigma}\ne k$ for any $\sigma\ge\tau$, so the
result is trivial.   If $k\ge k_{\tau}$, then for each dyadic
subinterval $I$ of $I_{\tau}$ of length $2^{-k}$ there is exactly one
$\sigma\ge\tau$ such that $I_{\sigma}=I$, since $J_{\sigma}$ must be the
unique dyadic interval of length $2^k$ including $J_{\tau}$.
List these as
$\sigma_0,\ldots$ in ascending order of the centres $x_{\sigma_j}$, so
that if $I_{\tau}=\coint{m\mu I_{\tau},(m+1)\mu I_{\tau}}$ then
$x_{\sigma_j}=m\mu I_{\tau}+2^{-k}(j+\bover12)$, for
$j<2^{k-k_{\tau}}$.   Now

$$\eqalign{\sum_{j=0}^{2^{k-k_{\tau}}-1}
  \int_{-\infty}^{m\mu I_{\tau}}w_{\sigma_j}
&=2^k\sum_{j=0}^{2^{k-k_{\tau}}-1}\int_{-\infty}^{m\mu I_{\tau}}
  w(2^k(x-m\mu I_{\tau})-j-\Bover12)dx\cr
&=\sum_{j=0}^{2^{k-k_{\tau}}-1}
  \int_{-\infty}^0w(x-j-\Bover12)dx\cr
&\le\sum_{j=0}^{\infty}
  \int_{j+\bover12}^{\infty}w
=\Bover12C_4.\cr}$$

\noindent Similarly (since $w$ is an even function, so the whole picture
is symmetric about $x_{\tau}$)

\Centerline{$\sum_{j=0}^{2^{k-k_{\tau}}-1}
  \int_{(m+1)\mu I_{\tau}}^{\infty}w_{\sigma_j}
\le\Bover12C_4$,}

\noindent and

\Centerline{$\sum_{\sigma\ge\tau,k_{\sigma}=k}
  \int_{\Bbb R\setminus I_{\tau}}w_{\sigma}\le C_4$,}

\noindent as required.
}%end of proof of 286G

\leader{286H}{`Mass' and `energy'}\cmmnt{ ({\smc Lacey \& Thiele 00})}
If $P$ is a subset of $Q$, $E\subseteq\Bbb R$ is measurable,
$g:\Bbb R\to\Bbb R$ is measurable, and $f\in\eusm L^2_{\Bbb C}$, set

\Centerline{$\mass_{Eg}(P)
=\sup_{\sigma\in P,\tau\in Q,\tau\le\sigma}
  \int_{E\cap g^{-1}[J_{\tau}]}w_{\tau}
\le\sup_{\tau\in Q}\int_{-\infty}^{\infty}w_{\tau}=1$,}

\Centerline{$\Delta_f(P)
=\sum_{\sigma\in P}|\innerprod{f}{\phi_{\sigma}}|^2$,}

\Centerline{$\energy_f(P)
=\sup_{\tau\in Q}\sqrt{\mu J_{\tau}}\sqrt{\Delta_f(P\cap T_{\tau})}$.}

\noindent If $P'\subseteq P$ then $\mass_{Eg}(P')\le\mass_{Eg}(P)$ and
$\energy_f(P')\le\energy_f(P)$.   \cmmnt{Note that}
$\energy_f(\{\sigma\})=\sqrt{\mu J_{\sigma}}|\innerprod{f}{\phi_{\sigma}}|$
for any $\sigma\in Q$\cmmnt{, since if $\sigma\in T_{\tau}$ then
$\mu J_{\tau}\le\mu J_{\sigma}$}.
%end of 286H

\leader{286I}{Lemma} If $P\subseteq Q$ is finite and
$f\in\eusm L^2_{\Bbb C}$, then

(a) $\Delta_f(P)\le\|\sum_{\sigma\in P}
  \innerprod{f}{\phi_{\sigma}}\phi_{\sigma}\|_2\|f\|_2$,

(b) $\sum_{\sigma,\tau\in P,J_{\sigma}=J_{\tau}}
  \bigl|\innerprod{f}{\phi_{\sigma}}
    \innerprod{\phi_{\sigma}}{\phi_{\tau}}
    \innerprod{\phi_{\tau}}{f}\bigr|\le C_3\Delta_f(P)$.

\proof{{\bf (a)}

\Centerline{$\Delta_f(P)
=\sum_{\sigma\in P}
  \innerprod{f}{\phi_{\sigma}}\innerprod{\phi_{\sigma}}{f}
=\Innerprod{\sum_{\sigma\in P}
  \innerprod{f}{\phi_{\sigma}}\phi_{\sigma}}{f}
\le\|\sum_{\sigma\in P}
  \innerprod{f}{\phi_{\sigma}}\phi_{\sigma}\|_2\|f\|_2$}

\noindent by Cauchy's inequality (244Eb).

\medskip

{\bf (b)}

$$\eqalignno{\sum_{\Atop{\sigma,\tau\in P}{J_{\sigma}=J_{\tau}}}
  \bigl|\innerprod{f}{\phi_{\sigma}}
    \innerprod{\phi_{\sigma}}{\phi_{\tau}}
    \innerprod{\phi_{\tau}}{f}\bigr|
&\le\sum_{\Atop{\sigma,\tau\in P}{J_{\sigma}=J_{\tau}}}
  \Bover12\bigl(|\innerprod{f}{\phi_{\sigma}}|^2
             +|\innerprod{f}{\phi_{\tau}}|^2\bigr)
  |\innerprod{\phi_{\sigma}}{\phi_{\tau}}|\cr
\displaycause{because $|\xi\zeta|\le\bover12(|\xi|^2+|\zeta|^2)$ for
all complex numbers $\xi$, $\zeta$}
&=\sum_{\sigma\in P}\sum_{\Atop{\tau\in P}{J_{\sigma}=J_{\tau}}}
  |\innerprod{f}{\phi_{\sigma}}|^2
  |\innerprod{\phi_{\sigma}}{\phi_{\tau}}|\cr
&\le\sum_{\sigma\in P}|\innerprod{f}{\phi_{\sigma}}|^2
  \sum_{\Atop{\tau\in P}{J_{\sigma}=J_{\tau}}}
     C_3\int_{I_{\tau}}w_{\sigma}\cr
\displaycause{by 286Gg, since $k_{\sigma}=k_{\tau}$ if
$J_{\sigma}=J_{\tau}$}
&\le\sum_{\sigma\in P}|\innerprod{f}{\phi_{\sigma}}|^2
  C_3\int_{-\infty}^{\infty}w_{\sigma}\cr
\displaycause{because if $\tau$, $\tau'$ are distinct members of $P$ and
$J_{\tau}=J_{\tau'}$, then $I_{\tau}$ and $I_{\tau'}$ are disjoint}
&=C_3\sum_{\sigma\in P}|\innerprod{f}{\phi_{\sigma}}|^2
=C_3\Delta_f(P).\cr}$$
}%end of proof of 286I

\leader{286J}{Lemma} Set $C_5=2^{12}$.   If $P\subseteq Q$ is finite,
$E\subseteq\Bbb R$ is
measurable, $g:\Bbb R\to\Bbb R$ is measurable, and
$\gamma\ge\mass_{Eg}(P)$, then we can find 
$R\subseteq Q$ such that
$\gamma\sum_{\tau\in R}\mu I_{\tau}\le C_5\mu E$
and\cmmnt{ (in the notation of 286Fb)} 
$\mass_{Eg}(P\setminus R^+)\le\bover14\gamma$.

\proof{{\bf (a)} If $\gamma=0$ we can take $R=\emptyset$.
Otherwise, set 
$P_1=\{\sigma:\sigma\in P$, $\mass_{Eg}(\{\sigma\})>\bover14\gamma\}$.   
For each $\sigma\in P_1$ let
$\sigma'\in Q$ be such that $\sigma'\le\sigma$ and
$\int_{E\cap g^{-1}[J_{\sigma'}]}w_{\sigma'}>\bover14\gamma$.   Let
$R$ be the set of elements of $\{\sigma':\sigma\in P_1\}$
which are minimal for $\le$.   Then
$P\setminus R^+\subseteq\{\sigma:\mass_{Eg}(\{\sigma\})\le\bover14\gamma\}$
so $\mass_{Eg}(P\setminus R^+)\le\bover14\gamma$.

\medskip

{\bf (b)} For $k\in\Bbb N$ set

\Centerline{$R_k
=\{\tau:\tau\in R$,
  $\mu J_{\tau}\cdot\mu(E\cap g^{-1}[J_{\tau}]\cap I^{(k)}_{\tau})
  \ge 2^{2k-9}\gamma\}$,}

\noindent where $I^{(k)}_{\tau}$ is the half-open
interval with the same centre as
$I_{\tau}$ and $2^k$ times its length.   Now
$R=\bigcup_{k\in\Bbb N}R_k$.   \Prf\ Take $\tau\in R$.   If
$k\in\Bbb N$ and $x\in\Bbb R\setminus I^{(k)}_{\tau}$, then
$|x-x_{\tau}|\ge\bover12\mu I^{(k)}_{\tau}=2^{k-1}\mu I_{\tau}$, so

\Centerline{$w_{\tau}(x)
=w((x-x_{\tau})\mu J_{\tau})\mu J_{\tau}
\le w(2^{k-1})\mu J_{\tau}
=(1+2^{k-1})^{-3}\mu J_{\tau}$.}

\noindent Accordingly

$$\eqalign{\Bover14\gamma
&<\int_{E\cap g^{-1}[J_{\tau}]}w_{\tau}
=\int_{E\cap g^{-1}[J_{\tau}]\cap I_{\tau}}w_{\tau}
  +\sum_{k=0}^{\infty}\int_{E\cap g^{-1}[J_{\tau}]
    \cap I^{(k+1)}_{\tau}\setminus I^{(k)}_{\tau}}w_{\tau}\cr
&\le \mu J_{\tau}\cdot\mu(E\cap g^{-1}[J_{\tau}]\cap I_{\tau})
  +\sum_{k=0}^{\infty}(1+2^{k-1})^{-3}\mu J_{\tau}\cdot
   \mu(E\cap g^{-1}[J_{\tau}]\cap I^{(k+1)}_{\tau}).\cr}$$

\noindent It follows that either

\Centerline{$\mu J_{\tau}\cdot\mu(E\cap g^{-1}[J_{\tau}]\cap I_{\tau})
\ge\Bover18\gamma$}

\noindent and $\tau\in R_0$, or there is some $k\in\Bbb N$ such that

\Centerline{$(1+2^{k-1})^{-3}\mu J_{\tau}
  \cdot\mu(E\cap g^{-1}[J_{\tau}]\cap I^{(k+1)}_{\tau})
  \ge 2^{-k-4}\gamma$}

\noindent and

\Centerline{$\mu J_{\tau}
  \cdot\mu(E\cap g^{-1}[J_{\tau}]\cap I^{(k+1)}_{\tau})
\ge(1+2^{k-1})^32^{-k-4}\gamma
\ge 2^{2k-7}\gamma$,}

\noindent so that $\tau\in R_{k+1}$.\ \Qed

\medskip

{\bf (c)} For every $k\in\Bbb N$,
$\gamma\sum_{\tau\in R_k}\mu I_{\tau}\le 2^{11-k}\mu E$.   \Prf\ If
$R_k=\emptyset$, this is trivial.   Otherwise, enumerate $R_k$ as
$\langle\tau_j\rangle_{j\le n}$ in such a way that
$k_{\tau_j}\le k_{\tau_l}$ if $j\le l\le n$.   Define
$q:\{0,\ldots,n\}\to\{0,\ldots,n\}$ inductively by the rule

\Centerline{$q(l)
=\min(\{l\}\cup\{j:j<l$, $q(j)=j$,
$(I^{(k)}_{\tau_j}\times J_{\tau_j})
  \cap(I^{(k)}_{\tau_l}\times J_{\tau_l})\ne\emptyset\})$}

\noindent for each $l\le n$.   Note that, for $l\le n$, 
$q(q(l))=q(l)\le l$ and
$I^{(k)}_{\tau_{q(l)}}\cap I^{(k)}_{\tau_l}\ne\emptyset$,
so that

\Centerline{$I_{\tau_l}\subseteq I^{(k)}_{\tau_l}
\subseteq I^{(k+2)}_{\tau_{q(l)}}$, }

\noindent because $\mu I_{\tau_l}^{(k)}\le\mu I_{\tau_{q(l)}}^{(k)}$.
Moreover, if $j<l\le n$ and $q(j)=q(l)$, then both $J_{\tau_j}$ and
$J_{\tau_l}$ meet $J_{\tau_{q(j)}}$, therefore include it, and
$J_{\tau_j}\subseteq J_{\tau_l}$.   But as $\tau_j$ and $\tau_l$ are
distinct members of $R$, $\tau_j\not\le\tau_l$ and
$I_{\tau_j}\cap I_{\tau_l}$ must be empty.

Set $M=\{q(j):j\le n\}$.   We have

$$\eqalign{\gamma\sum_{\tau\in R_k}\mu I_{\tau}
&=\gamma\sum_{m\in M}\sum_{\Atop{j\le n}{q(j)=m}}\mu I_{\tau_j}
\le\gamma\sum_{m\in M}\mu I^{(k+2)}_{\tau_m}
=2^{k+2}\gamma\sum_{m\in M}\mu I_{\tau_m}\cr
&\le 2^{k+2}\sum_{m\in M}
  2^{9-2k}\mu(E\cap g^{-1}[J_{\tau_m}]\cap I^{(k)}_{\tau_m})\cr
&\le 2^{k+2}\cdot 2^{9-2k}\mu E
=2^{11-k}\mu E\cr}$$

\noindent because if $l$, $m\in M$ and $l<m$ then
$I^{(k)}_{\tau_l}\times J_{\tau_l}$ and
$I^{(k)}_{\tau_m}\times J_{\tau_m}$ are disjoint (since $q(m)=m$)),
so that $g^{-1}[J_{\tau_l}]\cap I^{(k)}_{\tau_l}$ and
$g^{-1}[J_{\tau_m}]\cap I^{(k)}_{\tau_m}$ are disjoint.\ \Qed

\medskip

{\bf (d)} Accordingly

\Centerline{$\gamma\sum_{\tau\in R}\mu I_{\tau}
\le\gamma\sum_{k=0}^{\infty}\sum_{\tau\in R_k}\mu I_{\tau}
\le 2^{12}\mu E$,}

\noindent as required.
}%end of proof of 286J

\leader{286K}{Lemma}
Set $C_6=4(C_3+4C_3\sqrt{2C_4})$.   Suppose that $P\subseteq Q$ is a finite set,
$f\in\eusm L^2_{\Bbb C}$, $\|f\|_2=1$ and $\gamma\ge\energy_f(P)$.
Then we can find $R\subseteq Q$ such that 
$\gamma^2\sum_{\tau\in R}\mu I_{\tau}\le C_6$ and 
$\energy_f(P\setminus R^+)\le\bover12\gamma$.

\proof{{\bf (a)} We may suppose that $\gamma>0$ and that
$P\ne\emptyset$, since otherwise we can take $R=\emptyset$.

\medskip

\quad{\bf (i)}
There are only finitely many sets of the form $P\cap T_{\tau}$
for $\tau\in Q$;  let $\tilde R\subseteq Q$ be a non-empty finite set
such that whenever $\tau\in Q$
and $P\cap T_{\tau}$ is not empty, there is a $\tau'\in \tilde R$ such that
$P\cap T_{\tau}=P\cap T_{\tau'}$ and $k_{\tau'}\ge k_{\tau}$;
this is possible because if $A\subseteq P$ is not empty then
$k_{\tau}\le\min_{\sigma\in A}k_{\sigma}$ whenever $T_{\tau}\supseteq A$.

\medskip

\quad{\bf (ii)} Choose $\tau_0,\tau_1,\ldots$, $P_0,P_1,\ldots$
inductively, as follows.   $P_0=P$.   Given that $P_j\subseteq P$ is
not empty, consider

\Centerline{$R_j
=\{\tau:\tau\in\tilde R$, $\Delta_f(P_j\cap T_{\tau})
  \ge\Bover14\gamma^2\mu I_{\tau}\}$.}

\noindent If $R_j=\emptyset$, stop the induction and set $n=j$ and
$R=\{\tau_l:l<j\}$.   Otherwise, among the members of
$R_j$ take one with $y_{\tau}$ as far to the left as possible, and call
it $\tau_j$;  set $P_{j+1}=P_j\setminus\{\tau_j\}^+$, and
continue.   Note that as $R_{j+1}\subseteq R_j$ for every $j$,
$y_{\tau_{j+1}}\ge y_{\tau_j}$ for every $j$.

The induction must stop at a finite stage because if it does not stop
with $n=j$ then
$\Delta_f(P_j\cap T_{\tau_j})>0$, so $P_j\cap T_{\tau_j}$ is not empty
and $P_{j+1}\subseteq P_j\setminus T_{\tau_j}$ is a proper subset of
$P_j$, while $P_0=P$ is finite.   Since $R_n=\emptyset$,

$$\eqalign{\energy_f(P\setminus R^+)
&=\energy_f(P_n)
=\sup_{\tau\in Q}\sqrt{\mu J_{\tau}}\sqrt{\Delta_f(P_n\cap T_{\tau})}\cr&
=\max_{\tau\in\tilde R}\sqrt{\mu J_{\tau}}\sqrt{\Delta_f(P_n\cap T_{\tau})}
\le\Bover12\gamma.\cr}$$

\medskip

\quad{\bf (iii)} Set
$P'_j=P_j\cap T_{\tau_j}\subseteq P_j\setminus P_{j+1}$ for $j<n$,
so that $\ofamily{j}{n}{P'_j}$ is disjoint, and
$P'=\bigcup_{j<n}P'_j\subseteq P$.
Then if $\sigma\in P'$, $j<n$ and $J_{\tau_j}\subseteq J^l_{\sigma}$,
$I_{\sigma}\cap I_{\tau_j}=\emptyset$.   \Prf\
Let $l<n$ be such that $\sigma\in P'_l$.   Then
$y_{\tau_j}\in J_{\tau_j}\subseteq J^l_{\sigma}$ and
$y_{\tau_l}\in J^r_{\tau_l}\subseteq J^r_{\sigma}$, so
$y_{\tau_j}<y_{\tau_l}$ and $j<l$.   Accordingly $P_{j+1}\supseteq P_l$
contains $\sigma$, so $\sigma\not\ge\tau_j$;  as
$J_{\tau_j}\subseteq J_{\sigma}$, $I_{\sigma}\not\subseteq I_{\tau_j}$,
while $\mu I_{\sigma}\le\mu I_{\tau_j}$, so $I_{\sigma}$ is disjoint from
$I_{\tau_j}$.\ \Qed

It follows that if $\sigma$, $\tau\in P'$ are distinct and
$J^l_{\sigma}\cap J^l_{\tau}$ is not empty, then
$I_{\sigma}\cap I_{\tau}=\emptyset$.   \Prf\ If $J_{\sigma}=J_{\tau}$
this is true just because $\sigma\ne\tau$.   Otherwise, since
$J_{\sigma}$ and $J_{\tau}$ intersect, one is included in the other;
suppose that $J_{\sigma}\subset J_{\tau}$.   Since $J_{\sigma}$ meets
$J^l_{\tau}$, $J_{\sigma}\subseteq J^l_{\tau}$.   Now let $j<n$ be such
that $\sigma\in P'_j$;  then $\sigma\ge\tau_j$, so
$J_{\tau_j}\subseteq J_{\sigma}\subseteq J^l_{\tau}$, and
$I_{\sigma}\cap I_{\tau}\subseteq I_{\tau_j}\cap I_{\tau}=\emptyset$ by
the last remark.\ \Qed

\medskip

{\bf (b)} Now let us estimate

\Centerline{$\gamma^2\sum_{j<n}\mu I_{\tau_j}
\le 4\sum_{j<n}\Delta_f(P'_j)=4\Delta_f(P')=4\alpha$}

\noindent say.   Because $\|f\|_2=1$, we have
$\alpha\le\|\sum_{\sigma\in P'}
  \innerprod{f}{\phi_{\sigma}}\phi_{\sigma}\|_2$ (286Ia).    So

$$\eqalignno{\alpha^2
&\le\|\sum_{\sigma\in P'}
  \innerprod{f}{\phi_{\sigma}}\phi_{\sigma}\|^2_2
=\sum_{\sigma,\tau\in P'}
  \innerprod{f}{\phi_{\sigma}}\innerprod{\phi_{\sigma}}{\phi_{\tau}}
  \innerprod{\phi_{\tau}}{f}\cr
&=\sum_{\Atop{\sigma,\tau\in P'}{J_{\sigma}=J_{\tau}}}
  \innerprod{f}{\phi_{\sigma}}\innerprod{\phi_{\sigma}}{\phi_{\tau}}
  \innerprod{\phi_{\tau}}{f}\cr
&\mskip100mu
 +\sum_{\Atop{\sigma,\tau\in P'}{J_{\sigma}\subseteq J^l_{\tau}}}
  \innerprod{f}{\phi_{\sigma}}\innerprod{\phi_{\sigma}}{\phi_{\tau}}
  \innerprod{\phi_{\tau}}{f}
 +\sum_{\Atop{\sigma,\tau\in P'}{J_{\tau}\subseteq J^l_{\sigma}}}
  \innerprod{f}{\phi_{\sigma}}\innerprod{\phi_{\sigma}}{\phi_{\tau}}
  \innerprod{\phi_{\tau}}{f}\cr}$$

\noindent because $\innerprod{\phi_{\sigma}}{\phi_{\tau}}=0$ unless
$J^l_{\sigma}\cap J^l_{\tau}\ne\emptyset$, as noted in 286E(b-iii).

Take
these three terms separately.   For the first, we have

\Centerline{$\sum_{\sigma,\tau\in P',J_{\sigma}=J_{\tau}}
  \bigl|\innerprod{f}{\phi_{\sigma}}
    \innerprod{\phi_{\sigma}}{\phi_{\tau}}
    \innerprod{\phi_{\tau}}{f}\bigr|
\le C_3\alpha$}

\noindent by 286Ib.   For the second term, we have

$$\eqalign{\sum_{\Atop{\sigma,\tau\in P'}{J_{\sigma}\subseteq J^l_{\tau}}}
  \bigl|\innerprod{f}{\phi_{\sigma}}
   \innerprod{\phi_{\sigma}}{\phi_{\tau}}
   \innerprod{\phi_{\tau}}{f}\bigr|
&\le\sum_{\sigma\in P'}|\innerprod{f}{\phi_{\sigma}}|
  \sum_{\Atop{\tau\in P'}{J_{\sigma}\subseteq J^l_{\tau}}}
   |\innerprod{\phi_{\sigma}}{\phi_{\tau}}
   \innerprod{\phi_{\tau}}{f}|\cr
&\le\sqrt{\sum_{\sigma\in P'}|\innerprod{f}{\phi_{\sigma}}|^2}
  \sqrt{\sum_{\sigma\in P'}
   \bigl(\sum_{\Atop{\tau\in P'}{J_{\sigma}\subseteq J^l_{\tau}}}
     |\innerprod{\phi_{\sigma}}{\phi_{\tau}}
     \innerprod{\phi_{\tau}}{f}|\bigr)^2}\cr
&=\sqrt{\alpha}\sqrt{\sum_{j<n}H_j},\cr}$$

\noindent where for $j<n$ I set

$$H_j=\sum_{\sigma\in P'_j}
   \bigl(\sum_{\Atop{\tau\in P'}{J_{\sigma}\subseteq J^l_{\tau}}}
     |\innerprod{\phi_{\sigma}}{\phi_{\tau}}
     \innerprod{\phi_{\tau}}{f}|\bigr)^2.$$

\noindent Now we can estimate $H_j$ by observing that, for any
$\tau\in P'$,

\Centerline{$|\innerprod{\phi_{\tau}}{f}|
=\sqrt{\mu I_{\tau}}\energy_f(\{\tau\})\le\gamma\sqrt{\mu I_{\tau}}$,}

\noindent while if $\sigma$, $\tau\in P'$ and
$J_{\tau}^l\supseteq J_{\sigma}$ then

\Centerline{$|\innerprod{\phi_{\sigma}}{\phi_{\tau}}|
\le C_3\sqrt{\mu I_{\sigma}}\sqrt{\mu J_{\tau}}\int_{I_{\tau}}w_{\sigma}$}

\noindent by 286Gg.   We also need to know that if
$\sigma\in P'_j$ and $\tau$, $\tau'$ are distinct elements of
$P'$ such that $J_{\sigma}\subseteq J^l_{\tau}\cap J^l_{\tau'}$, then
$I_{\tau}$, $I_{\tau'}$ and $I_{\tau_j}$ are all disjoint, by (a-iii)
above, because $J_{\tau_j}\subseteq J_{\sigma}$.   So we have

$$\eqalignno{H_j
&\le\sum_{\sigma\in P'_j}
   \bigl(\sum_{\Atop{\tau\in P'}{J_{\sigma}\subseteq J^l_{\tau}}}
    \gamma\sqrt{\mu I_{\tau}}
      \cdot C_3\sqrt{\mu I_{\sigma}}\sqrt{\mu J_{\tau}}
      \int_{I_{\tau}}w_{\sigma}\bigr)^2\cr
&=C_3^2\gamma^2\sum_{\sigma\in P'_j}\mu I_{\sigma}
   \bigl(\sum_{\Atop{\tau\in P'}{J_{\sigma}\subseteq J^l_{\tau}}}
     \int_{I_{\tau}}w_{\sigma}\bigr)^2
\le C_3^2\gamma^2\sum_{\sigma\in P'_j}\mu I_{\sigma}
   (\int_{\Bbb R\setminus I_{\tau_j}}w_{\sigma})^2\cr
&\le C_3^2\gamma^2\sum_{k=k_{\tau_j}}^{\infty}2^{-k}
   \sum_{\Atop{\sigma\in P'_j}{k_{\sigma}=k}}
   \int_{\Bbb R\setminus I_{\tau_j}}w_{\sigma}
   \cdot\int_{-\infty}^{\infty}w_{\sigma}
\le C_3^2\gamma^2\sum_{k=k_{\tau_j}}^{\infty}2^{-k}C_4\cr
\displaycause{by 286Ga and 286Gh, since $\sigma\ge\tau_j$ for every
$\sigma\in P'_j$}
&=C_3^2\gamma^22^{-k_{\tau_j}+1}C_4
=2C_3^2C_4\gamma^2\mu I_{\tau_j}.\cr}$$

\noindent Accordingly

\Centerline{$\sum_{j<n}H_j
\le 2C_3^2\gamma^2C_4\sum_{j<n}\mu I_{\tau_j}
\le 2C_3^2C_4\cdot 4\alpha$,}

\noindent and

\Centerline{$\sum_{\sigma,\tau\in P',J_{\sigma}\subseteq J^l_{\tau}}
  |\innerprod{f}{\phi_{\sigma}}
   \innerprod{\phi_{\sigma}}{\phi_{\tau}}
   \innerprod{\phi_{\tau}}{f}|
\le\sqrt{\alpha\sumop_{j<n}H_j}
\le 2C_3\alpha\sqrt{2C_4}$.}

Similarly,

\Centerline{$\sum_{\sigma,\tau\in P',J_{\tau}\subseteq J^l_{\sigma}}
  |\innerprod{f}{\phi_{\sigma}}
   \innerprod{\phi_{\sigma}}{\phi_{\tau}}
   \innerprod{\phi_{\tau}}{f}|
\le 2C_3\alpha\sqrt{2C_4}$;}

\noindent putting these together,

\Centerline{$\alpha^2
\le\alpha(C_3+4C_3\sqrt{2C_4})
=\Bover14\alpha C_6$}

\noindent and $\alpha\le\bover14C_6$.   But this means that

\Centerline{$\gamma^2\sum_{j<n}\mu I_{\tau_j}
\le 4\alpha\le C_6$,}

\noindent and $R=\{\tau_j:j<n\}$ has both the properties required.
}%end of proof of 286K

\leader{286L}{Lemma} Set

\Centerline{$C_7
=C_1\bigl(\Bover{7}2+\Bover87+\Bover{28}{w(3/2)}
  +\Bover{4\sqrt{14 C_3}}{w(3/2)}\bigr)$.}

\noindent Suppose that $P$ is a finite subset of $Q$ with a lower bound
$\tau$ in $Q$ for the ordering $\le$,
$E\subseteq\Bbb R$ is measurable,
$g:\Bbb R\to\Bbb R$ is measurable and $f\in\eusm L^2_{\Bbb C}$.   Then

\Centerline{$\sum_{\sigma\in P}|\innerprod{f}{\phi_{\sigma}}
  \int_{E\cap g^{-1}[J^r_{\sigma}]}\phi_{\sigma}|
\le C_7\energy_f(P)\mass_{Eg}(P)\mu I_{\tau}$.}

\proof{ Set $\gamma=\energy_f(P)$, $\gamma'=\mass_{Eg}(P)$.   If
$P=\emptyset$ the result is trivial, so suppose that $P\ne\emptyset$.

\medskip

{\bf (a)(i)} Note that $\bigcup_{\sigma\in P}I_{\sigma}\subseteq I_{\tau}$,
$J_{\tau}\subseteq\bigcap_{\sigma\in P}J_{\sigma}$ and
$k_{\tau}\le\min_{\sigma\in P}k_{\sigma}$.
So if $\sigma$, $\sigma'\in P$ are distinct and
$\mu I_{\sigma}=\mu I_{\sigma'}$, then $J_{\sigma}=J_{\sigma'}$ and
$I_{\sigma}\cap I_{\sigma'}=\emptyset$.

\medskip

\quad{\bf (ii)}
For a dyadic interval $I$ let
$I^*$ be the half-open interval with the same
centre as $I$ and three times its length.   Let $\Cal J$ be the family
of those $I\in\Cal I$ such that $I_{\sigma}\not\subseteq I^*$ for any
$\sigma\in P$ such that $\mu I_{\sigma}\le\mu I$.   Because $P$ is
finite, all sufficiently small intervals belong to $\Cal J$, and
$\bigcup\Cal J=\Bbb R$;  let $\Cal K$ be the set of maximal members of
$\Cal J$, so that $\Cal K$ is disjoint.  Then $\bigcup\Cal K=\Bbb R$.
\Prf\ The point is that $P\ne\emptyset$;  fix $\sigma\in P$ for the
moment.   If $I\in\Cal J$, consider for each $n\in\Bbb N$ the interval
$\tilde I^{(n)}\in\Cal I$ including $I$ with length $2^n\mu I$.   Then
there is some $n\in\Bbb N$ such that
$\mu\tilde I^{(n)}\ge\mu I_{\sigma}$ and
$I_{\sigma}\subseteq(\tilde I^{(n)})^*$, so that
$\tilde I^{(k)}\notin\Cal J$ for any
$k\ge n$, and there must be some $k<n$ such that
$\tilde I^{(k)}\in\Cal K$.   Thus
$I\subseteq\tilde I^{(k)}\subseteq\bigcup\Cal K$;  as $I$ is arbitrary,
$\bigcup\Cal K=\bigcup\Cal J=\Bbb R$.\ \Qed

\medskip

\quad{\bf (iii)} For $K\in\Cal K$, let $l_K\in\Bbb Z$ be such that
$\mu K=2^{-l_K}$.   If $l_K\ge k_{\tau}$, that is,
$\mu K\le\mu I_{\tau}$, then $K$ must lie within the half-open
interval $\hat I$
with centre $x_{\tau}$ and length $7\mu I_{\tau}$, since otherwise we
should have $I_{\tau}\cap\tilde K^*=\emptyset$, where $\tilde K$ is
the dyadic interval of length $2\mu K$ including $K$, and $\tilde K$
would belong to $\Cal J$.   But this means that

\Centerline{$\sum_{K\in\Cal K,\mu K\le\mu I_{\tau}}\mu K
  \le\mu\hat I=7\mu I_{\tau}$,}

\noindent because $\Cal K$ is disjoint.

\medskip

\quad{\bf (iv)} For any $l<k_{\tau}$, there are just three members $K$
of $\Cal K$ such that $l_K=l$.   \Prf\ If $I\in\Cal I$ and
$\mu I>\mu I_{\tau}$, then either $I_{\tau}\subseteq I^*$ or
$I_{\tau}\cap I^*=\emptyset$, and $I\in\Cal J$ iff $I_{\tau}\cap I^*$ is
empty.   This means that if $K\in\Cal I$ and $\mu K=2^{-l}$,
$K\in\Cal K$ iff $I_{\tau}\cap K^*$ is empty and
$I_{\tau}\subseteq\tilde K^*$.   So if
$I_{\tau}\subseteq\coint{2^{-l}n,2^{-l}(n+1)}$ and
$K=\coint{2^{-l}m,2^{-l}(m+1)}$, we shall have $K\in\Cal K$ iff

\Centerline{{\it either} $m=n-2$
{\it or} $m=n+2$
{\it or} $m=n-3$ is even
{\it or} $m=n+3$ is odd;}

\noindent which for any given $n$ happens for just three values of
$m$.\ \Qed

\medskip

{\bf (b)} For $\sigma\in P$, let $\zeta_{\sigma}$ be a complex number of
modulus $1$ such that
$\zeta_{\sigma}\innerprod{f}{\phi_{\sigma}}
   \int_{E\cap g^{-1}[J^r_{\sigma}]}\phi_{\sigma}$ is real and
non-negaive.   Set $W=P\times\Cal K$.   For $(\sigma,K)\in W$, set

\Centerline{$\alpha_{\sigma K}
=\innerprod{f}{\phi_{\sigma}}
  \int_{E\cap g^{-1}[J^r_{\sigma}]\cap K}\phi_{\sigma}$.}

\noindent The aim of the proof is to estimate

\Centerline{$\sum_{\sigma\in P}\bigl|\innerprod{f}{\phi_{\sigma}}
  \int_{E\cap g^{-1}[J^r_{\sigma}]}\phi_{\sigma}\bigr|
=\sum_{(\sigma,K)\in W}\zeta_{\sigma}\alpha_{\sigma K}$.}

\noindent It will be helpful to note straight away that

\Centerline{$\sum_{(\sigma,K)\in W}|\alpha_{\sigma K}|
\le\sum_{\sigma\in P}|\innerprod{f}{\phi_{\sigma}}|
  \int_{-\infty}^{\infty}|\phi_{\sigma}|$}

\noindent is finite.

Set

\Centerline{$W_0
=\{(\sigma,K):\sigma\in P$, $K\in\Cal K$,
$\mu I_{\sigma}\le\mu K\le\mu I_{\tau}\}$,}

\Centerline{$W_1
=\{(\sigma,K):\sigma\in P$, $K\in\Cal K$, $\mu I_{\tau}<\mu K\}$,}

\Centerline{$W_2
=\{(\sigma,K):\sigma\in P\setminus T_{\tau}$, $K\in\Cal K$,
$\mu K<\mu I_{\sigma}\}$,}

\Centerline{$W_3
=\{(\sigma,K):\sigma\in P\cap T_{\tau}$, $K\in\Cal K$,
$\mu K<\mu I_{\sigma}\}$.}

\noindent Because $\mu I_{\sigma}\le\mu I_{\tau}$
for every $\sigma\in P$,
$W=W_0\cup W_1\cup W_2\cup W_3$.   I will give estimates for

\Centerline{$\alpha_j
=\sum_{(\sigma,K)\in W_j}\zeta_{\sigma}\alpha_{\sigma K}$}

\noindent for each $j$;  the four components in the expression for
$C_7$ given above are bounds for $|\alpha_0|, |\alpha_1|$, $|\alpha_2|$
and $|\alpha_3|$ respectively.

\wheader{286L}{6}{2}{2}{36pt}

{\bf (c)(i)} If $K\in\Cal K$ and $l_K=l$, then for any $k\ge l$

\Centerline{$\sum_{\sigma\in P,k_{\sigma}=k}|\alpha_{\sigma K}|
\le 2^{-k}C_1\gamma\gamma'(1+2^{k-l})^{-2}
\le 2^{-k-2}C_1\gamma\gamma'$.}

\noindent\Prf\ For any $\sigma\in P$,

\Centerline{$|\innerprod{f}{\phi_{\sigma}}|
=\sqrt{\mu I_{\sigma}}\energy_f(\{\sigma\})
\le\gamma\sqrt{\mu I_{\sigma}}$}

\noindent as noted in 286H, and

$$\eqalignno{\int_{E\cap g^{-1}[J^r_{\sigma}]\cap K}|\phi_{\sigma}|
&\le C_1\mu I_{\sigma}\sqrt{\mu I_{\sigma}}
  \int_{E\cap g^{-1}[J^r_{\sigma}]\cap K}w_{\sigma}^2\cr
\displaycause{286Ge}
&\le C_1\mu I_{\sigma}\sqrt{\mu I_{\sigma}}
  \int_{E\cap g^{-1}[J_{\sigma}]}w_{\sigma}
  \cdot\sup_{x\in K}w_{\sigma}(x)\cr
&\le C_1\mu I_{\sigma}\sqrt{\mu I_{\sigma}}\gamma'
  \sup_{x\in K}w_{\sigma}(x)
=C_1\sqrt{\mu I_{\sigma}}\gamma'w(\mu J_{\sigma}\rho(x_{\sigma},K)),\cr}$$

\noindent where I write $\rho(x_{\sigma},K)$ for
$\inf_{x\in K}|x-x_{\sigma}|$.   So, for $k\ge l$,

$$\eqalign{\sum_{\Atop{\sigma\in P}{k_{\sigma}=k}}
  |\alpha_{\sigma K}|
&\le\sum_{\Atop{\sigma\in P}{k_{\sigma}=k}}
  C_1\gamma\gamma'\mu I_{\sigma}w(\mu J_{\sigma}\rho(x_{\sigma},K))\cr
&=2^{-k}C_1\gamma\gamma'
  \sum_{\Atop{\sigma\in P}{k_{\sigma}=k}}
    w(2^k\rho(x_{\sigma},K))
\le 2^{-k}C_1\gamma\gamma'
  \cdot 2\sum_{n=2^{k-l}}^{\infty}w(n+\hbox{$\bover12$})\cr}$$

\noindent because the $x_{\sigma}$, for $\sigma\in P$ and
$k_{\sigma}=k$, are all distinct (see (a-i) above)
and all a distance at least $\mu K=2^{k-l}2^{-k}$
from $K$ (because $I_{\sigma}\not\subseteq K^*$);  so there are at most
two such
$\sigma$ with $\rho(x_{\sigma},K)=2^{-k}(n+\bover12)$ for each
$n\ge 2^{k-l}$.   So we have

\Centerline{$\sum_{\sigma\in P,k_{\sigma}=k}
  |\alpha_{\sigma K}|
\le 2^{-k}C_1\gamma\gamma'(1+2^{k-l})^{-2}
\le 2^{-k-2}C_1\gamma\gamma'$}

\noindent by 286Gb.\ \Qed

\medskip

\quad{\bf (ii)} Now

$$\eqalign{|\alpha_0|
&\le\sum_{(\sigma,K)\in W_0}|\alpha_{\sigma K}|
=\sum_{\Atop{K\in\Cal K}{\mu K\le\mu I_{\tau}}}
  \sum_{\Atop{\sigma\in P}{\mu I_{\sigma}\le\mu K}}
   |\alpha_{\sigma K}|\cr
&=\sum_{\Atop{K\in\Cal K}{\mu K\le\mu I_{\tau}}}\sum_{k=l_K}^{\infty}
   \sum_{\Atop{\sigma\in P}{k_{\sigma}=k}}
   |\alpha_{\sigma K}|
\le\sum_{\Atop{K\in\Cal K}{\mu K\le\mu I_{\tau}}}\sum_{k=l_K}^{\infty}
   2^{-k-2}C_1\gamma\gamma'\cr
&=C_1\gamma\gamma'\sum_{\Atop{K\in\Cal K}{\mu K\le\mu I_{\tau}}}
  2^{-l_K-1}
=\Bover12C_1\gamma\gamma'
   \sum_{\Atop{K\in\Cal K}{\mu K\le\mu I_{\tau}}}\mu K
\le\Bover{7}2C_1\gamma\gamma'\mu I_{\tau}\cr}$$

\noindent by the formula in (a-iii).   This deals with $\alpha_0$.

\medskip

{\bf (d)} Next consider $W_1$.   We have

$$\eqalignno{|\alpha_1|
&\le\sum_{(\sigma,K)\in W_1}|\alpha_{\sigma K}|
=\sum_{l=-\infty}^{k_{\tau}-1}\sum_{k=k_{\tau}}^{\infty}
   \sum_{\Atop{K\in\Cal K}{l_K=l}}\sum_{\Atop{\sigma\in P}{k_{\sigma}=k}}
   |\alpha_{\sigma K}|\cr
&\le\sum_{l=-\infty}^{k_{\tau}-1}\sum_{k=k_{\tau}}^{\infty}
   \sum_{\Atop{K\in\Cal K}{l_K=l}}
   2^{-k}C_1\gamma\gamma'(1+2^{k-l})^{-2}\cr
\displaycause{by (c-i) above}
&=3C_1\gamma\gamma'\sum_{l=-\infty}^{k_{\tau}-1}\sum_{k=k_{\tau}}^{\infty}
   2^{-k}(1+2^{k-l})^{-2}\cr
\displaycause{by (a-iv)}
&\le 3C_1\gamma\gamma'\sum_{l=-\infty}^{k_{\tau}-1}
  \sum_{k=k_{\tau}}^{\infty}2^{-3k}2^{2l}
=3C_1\gamma\gamma'2^{2(k_{\tau}-1)}2^{-3k_{\tau}}
  \sum_{l=0}^{\infty}2^{-2l}\sum_{k=0}^{\infty}2^{-3k}\cr
&=\Bover34C_1\gamma\gamma'2^{-k_{\tau}}\cdot\Bover43\cdot\Bover87
=\Bover87C_1\gamma\gamma'\mu I_{\tau}.\cr}$$

\noindent This deals with $\alpha_1$.

\medskip

{\bf (e)} For $K\in\Cal K$, set $G_K
=K\cap E\cap\bigcup_{\sigma\in P,\mu I_{\sigma}>\mu K}g^{-1}[J_{\sigma}]$.
Then $\mu G_K\le 2\gamma'\mu K/w(\bover32)$.   \Prf\
If $\mu I_{\tau}\le\mu K$, then $G_K=\emptyset$, so we may suppose that
$\mu K<\mu I_{\tau}$.   Let $\tilde K\in\Cal I$ be the dyadic interval
containing $K$ and with twice the length.   Then
$\tilde K\notin\Cal J$, so there is a $\sigma\in P$ such that
$\tilde K^*\supseteq I_{\sigma}$ and

\Centerline{$\mu I_{\sigma}\le\mu\tilde K=2\mu K\le\mu I_{\tau}$.}

\noindent Let $\upsilon\in Q$ be such
that $\tau\le\upsilon\le\sigma$ and
$\mu I_{\upsilon}=2\mu K$ (286F(a-iv)).   Then
$I_{\upsilon}$ meets $\tilde K^*$, so $\tilde K$ is either equal to
$I_{\upsilon}$ or adjacent to it, and
$|x-x_{\upsilon}|\le\bover32\cdot\mu I_{\upsilon}$ for every
$x\in\tilde K$, therefore for every $x\in K$.   Accordingly

\Centerline{$w_{\upsilon}(x)\ge w(\bover32)\mu J_{\upsilon}
=w(\bover32)/2\mu K$}

\noindent for every
$x\in K$.   On the other hand, because $\sigma\in P$ and
$\upsilon\le\sigma$,
$\int_{E\cap g^{-1}[J_{\upsilon}]}w_{\upsilon}\le\gamma'$.   So

\Centerline{$\mu(E\cap g^{-1}[J_{\upsilon}]\cap K)
\le 2\gamma'\mu K/w(\bover32)$.}

Now suppose that $\sigma'\in P$ and $\mu I_{\sigma'}>\mu K$.   Then
$k_{\sigma'}\le k_{\upsilon}$ and
$J_{\sigma'}$ is the dyadic interval of length $2^{k_{\sigma'}}$
including $J_{\tau}$.   But $J_{\upsilon}$ is the dyadic interval of
length $2^{k_{\upsilon}}$ including $J_{\tau}$, so includes
$J_{\sigma'}$, and
$g^{-1}[J_{\sigma'}]\subseteq g^{-1}[J_{\upsilon}]$.   As $\sigma'$ is
arbitrary, $G_K\subseteq E\cap g^{-1}[J_{\upsilon}]\cap K$ and
$\mu G_K\le 2\gamma'\mu K/w(\bover32)$, as claimed.\ \Qed

\medskip

{\bf (f)(i)} If $\sigma$, $\upsilon\in P\setminus T_{\tau}$ and
$k_{\sigma}\ne k_{\upsilon}$, then
$J^r_{\sigma}\cap J^r_{\upsilon}=\emptyset$.   \Prf\
It is enough to consider the case $\mu J_{\sigma}<\mu J_{\upsilon}$, so
that $\mu J_{\sigma}\le\mu J^r_{\upsilon}$.   As $J_{\sigma}$ includes
$J_{\tau}$, but $J^r_{\upsilon}$ does not, $J_{\sigma}$ is disjoint from
$J^r_{\upsilon}$ and we have the result.\ \Qed

\medskip

\quad{\bf (ii)} For $x\in\Bbb R$, set

\Centerline{$v_2(x)=\bigl|\sum_{(\sigma,K)\in W_2}
   \zeta_{\sigma}\innerprod{f}{\phi_{\sigma}}\phi_{\sigma}(x)
   \chi(E\cap g^{-1}[J^r_{\sigma}]\cap K)(x)\bigr|$.}

\noindent (The sum is finite because there is at most one $K\in\Cal K$
containing $x$.)   Then for any $x\in\Bbb R$ there is a
$k\ge k_{\tau}$ such that

\Centerline{$v_2(x)
=\bigl|\sum_{\sigma\in P,k_{\sigma}=k}
   \zeta_{\sigma}\innerprod{f}{\phi_{\sigma}}\phi_{\sigma}(x)\bigl|$.}

\noindent\Prf\ If $v_2(x)=0$, any sufficiently large $k$ will serve.
Otherwise, $x\in E$ and we have a pair
$(\upsilon,L)\in W_2$ such that
$x\in g^{-1}[J^r_{\upsilon}]\cap L$.   Try $k=k_{\upsilon}$.
$L$ is the only member of $\Cal K$ containing $x$, so

\Centerline{$v_2(x)
=\bigl|\sum_{\sigma\in P_x}
  \innerprod{f}{\phi_{\sigma}}\phi_{\sigma}(x)\bigr|$,}

\noindent where $P_x=\{\sigma:\sigma\in P\setminus T_{\tau}$,
$\mu I_{\sigma}>\mu L$, $g(x)\in J^r_{\sigma}\}$.   Now if
$\sigma\in P$ and $k_{\sigma}=k$, then
$\mu I_{\sigma}=\mu I_{\upsilon}>\mu L$, $J_{\sigma}=J_{\upsilon}$ and
$J^r_{\sigma}=J^r_{\upsilon}$ does not include $J^r_{\tau}$, so that
$\sigma\in P\setminus T_{\tau}$, $g(x)\in J^r_{\sigma}$ and
$\sigma\in P_x$.   On the other hand, (i) above tells us that
$k_{\sigma}=k$ whenever $\sigma\in P\setminus T_{\tau}$ and
$g(x)\in J^r_{\sigma}$.   So
$P_x=\{\sigma:\sigma\in P$, $k_{\sigma}=k\}$ and

\Centerline{$v_2(x)
=\bigl|\sum_{\sigma\in P,k_{\sigma}=k}
   \zeta_{\sigma}\innerprod{f}{\phi_{\sigma}}\phi_{\sigma}(x)\bigl|$.
\Qed}

\medskip

\quad{\bf (iii)} It follows that $v_2(x)\le 2C_1\gamma$ for every
$x\in\Bbb R$.   \Prf\ If $v_2(x)=0$ this is trivial.   Otherwise, take
$k$ from (ii).   Then

$$\eqalignno{v_2(x)
&\le\sum_{\Atop{\sigma\in P}{k_{\sigma}=k}}
  |\innerprod{f}{\phi_{\sigma}}\phi_{\sigma}(x)|
\le\sum_{\Atop{\sigma\in P}{k_{\sigma}=k}}\sqrt{\mu I_{\sigma}}\gamma
  \cdot\sqrt{\mu I_{\sigma}}C_1w_{\sigma}(x)\cr
\displaycause{by 286H and 286Ge}
&=C_1\gamma\sum_{\Atop{\sigma\in P}{k_{\sigma}=k}}w(2^k(x-x_{\sigma}))
\le C_1\gamma\sum_{n=-\infty}^{\infty}w(2^kx-n-\hbox{$\bover12$})\cr
\displaycause{because the $x_{\sigma}$, for $\sigma\in P$ and
$k_{\sigma}=k$, are all distinct and of the form $2^{-k}(n+\bover12)$}
&\le 2C_1\gamma\cr}$$

\noindent by 286Gd.\ \Qed

\medskip

\quad{\bf (iv)} Note also that, if $v_2(x)>0$, there is a pair
$(\sigma,K)\in W_2$ such that $x\in g^{-1}[J_{\sigma}]\cap K$, so that
$\mu K<\mu I_{\sigma}\le\mu I_{\tau}$ and $x\in G_K$.   But now we have

$$\eqalignno{|\alpha_2|
&=\bigl|\sum_{(\sigma,K)\in W_2}
  \zeta_{\sigma}\innerprod{f}{\phi_{\sigma}}\int_{-\infty}^{\infty}
  \phi_{\sigma}\times\chi(E\cap g^{-1}[J^r_{\sigma}]\cap K)\bigr|\cr
&\le\int_{-\infty}^{\infty}v_2
\le\sum_{\Atop{K\in\Cal K}{\mu K<\mu I_{\tau}}}\int_{G_K}v_2
\le\sum_{\Atop{K\in\Cal K}{\mu K<\mu I_{\tau}}}
  \bover{4C_1\gamma\gamma'\mu K}{w(\bover32)}\cr
\displaycause{putting the estimates in (e) and (iii) just above together}
&\le\bover{28\cdot C_1\gamma\gamma'\mu I_{\tau}}{w(\bover32)}\cr}$$

\noindent by (a-iii).   This deals with $\alpha_2$.

\medskip

{\bf (g)} Set $P'=P\cap T_{\tau}$ and
$\tilde f=\sum_{\sigma\in P'}
\zeta_{\sigma}\innerprod{f}{\phi_{\sigma}}\phi_{\sigma}$.   Then

\Centerline{$\|\tilde f\|^2_2
\le C_3\gamma^2\mu I_{\tau}$.}

\noindent\Prf\ If $\sigma$, $\sigma'\in P'$ and
$k_{\sigma}\ne k_{\sigma'}$, then
$\innerprod{\phi_{\sigma}}{\phi_{\sigma'}}=0$ (286Fc).   While if
$k_{\sigma}=k_{\sigma'}$, then $J_{\sigma}=J_{\sigma'}$, by (a-i).   So

$$\eqalignno{\|\tilde f\|_2^2
&=\sum_{\sigma,\sigma'\in P'}
  \zeta_{\sigma}\innerprod{f}{\phi_{\sigma}}
  \innerprod{\phi_{\sigma}}{\phi_{\sigma'}}
  \innerprod{\phi_{\sigma'}}{f}\bar\zeta_{\sigma'}\cr
&\le\sum_{\Atop{\sigma,\sigma'\in P'}{J_{\sigma}=J_{\sigma'}}}
  \bigl|\innerprod{f}{\phi_{\sigma}}
  \innerprod{\phi_{\sigma}}{\phi_{\sigma'}}
  \innerprod{\phi_{\sigma'}}{f}\bigr|
\le C_3\Delta_f(P')\cr
\displaycause{286Ib}
&\le C_3\gamma^2\mu I_{\tau}\cr}$$

\noindent by the definition of `energy', because $P'\subseteq T_{\tau}$.\
\Qed

\medskip

{\bf (h)} For $m\in\Bbb N$, set

\Centerline{$\tilde f_m=\sum_{\sigma\in P',k_{\sigma}\le m}
  \zeta_{\sigma}\innerprod{f}{\phi_{\sigma}}\phi_{\sigma}$.}

\noindent Then whenever $x$, $x'\in\Bbb R$ and $|x-x'|\le 2^{-m}$,
$|\tilde f_m(x)|\le\bover12C_1\tilde f^*(x')$, where

\Centerline{$\tilde f^*(x')
=\sup_{a\le x'\le b,a<b}\Bover1{b-a}\int_a^b|\tilde f|$}

\noindent as in 286A.   \Prf\ {\bf (i)} Since $k_{\sigma}\ge k_{\tau}$
for every $\sigma\in P'$, we may take it that $m\ge k_{\tau}$.   Let
$\hat J$ be the dyadic interval
of length $2^m$ including $J_{\tau}$, and $\hat y$ its midpoint.  Set
$\psi=S_{-\hat y}D_{2^{-m}/3}\varhat{\phi}$, that is,
$\psi(y)=\varhat{\phi}(\bover132^{-m}(y-\hat y))$ for $y\in\Bbb R$.

\medskip

\quad{\bf (ii)} If $\sigma\in P'$ and $k_{\sigma}\le m$ and
$\varhat{\phi}_{\sigma}(y)\ne 0$,
then $y\in J^l_{\sigma}$.   But $J_{\sigma}\cap\hat J\supseteq J_{\tau}$
is not empty, so $J_{\sigma}\subseteq\hat J$,
$|y-\hat y|\le\bover122^m$, $|\bover132^{-m}(y-\hat y)|\le\bover16$ and
$\psi(y)=1$.

\medskip

\quad{\bf (iii)} If $\sigma\in P'$ and $k_{\sigma}>m$ and
$\varhat{\phi}_{\sigma}(y)\ne 0$, then
$J^r_{\sigma}\cap\hat J\supseteq J^r_{\tau}$ is non-empty, so
$\hat J\subseteq J^r_{\sigma}$ and $y\le y_{\sigma}\le\hat y$;  now

\Centerline{$\hat y-y
=(\hat y-y_{\sigma})+(y_{\sigma}-y)
\ge\Bover12\cdot 2^m+\Bover1{20}\mu J_{\sigma}
\ge\Bover35\cdot 2^m$,
\quad$|\Bover13\cdot 2^{-m}(y-\hat y)|\ge\Bover15$}

\noindent and $\psi(y)=0$.
\medskip

\quad{\bf (iv)}  What this means is that if $\sigma\in P'$ then

$$\eqalign{\varhat{\phi}_{\sigma}\times\psi
&=\varhat{\phi}_{\sigma}\text{ if }k_{\sigma}\le m,\cr
&=0\text{ if }k_{\sigma}>m,\cr}$$

\noindent so that $\varhat{\vartildef}_m=\psi\times\varhat{\vartildef}$.

\medskip

\quad{\bf (v)} By 283M,
$\tilde f_m=\Bover1{\sqrt{2\pi}}\tilde f*\varcheck{\psi}$,
where $\tilde f*\varcheck{\psi}$ is the convolution of $\tilde f$ and
the inverse Fourier transform $\varcheck{\psi}$ of $\psi$.   (Strictly
speaking, 283M, with the help of 284C, tells us that $\tilde f_m$ and
$\Bover1{\sqrt{2\pi}}\tilde f*\varcheck{\psi}$ have the same Fourier
transforms.   By 283G, they are equal almost everywhere;  by 255K,
the convolution is defined everywhere and is continuous;  so in fact
they are the same function.)   Now

\Centerline{$\varcheck{\psi}
=3\cdot 2^mM_{\hat y}D_{3\cdot 2^m}
\varhat{\phi}\varcheck{\phantom{\phi}}
=3\cdot 2^mM_{\hat y}D_{3\cdot 2^m}\phi$,}

\noindent that is,

\Centerline{$\varcheck{\psi}(x)
=3\cdot 2^me^{ix\hat y}\phi(3\cdot 2^mx)$}

\noindent for $x\in\Bbb R$.

\medskip

\quad{\bf (vi)} Set $w_1(x)=\min(w(3),w(x))$ for $x\in\Bbb R$, so
that $w_1$ is
non-decreasing on $\ocint{-\infty,-3}$, non-increasing on
$\coint{3,\infty}$, and constant on $[-3,3]$, and
$|\phi(x)|\le C_1w_1(x)$ for every $x$, by the choice of $C_1$
(286Ge).   Take $x$, $x'\in\Bbb R$ such that $|x-x'|\le 2^{-m}$.
Then

$$\eqalignno{|\tilde f_m(x)|
&\le\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
  |\tilde f(x-t)||\varcheck{\psi}(t)|dt
=\Bover{3\cdot 2^m}{\sqrt{2\pi}}\int_{-\infty}^{\infty}
  |\tilde f(x-t)||\phi(3\cdot 2^mt)|dt\cr
&\le\Bover{3\cdot 2^m}{\sqrt{2\pi}}C_1\int_{-\infty}^{\infty}
  |\tilde f(x-t)|w_1(3\cdot 2^mt)dt
=\Bover{3\cdot 2^m}{\sqrt{2\pi}}C_1\int_{-\infty}^{\infty}
  |\tilde f(x+t)|w_1(3\cdot 2^mt)dt\cr
\displaycause{because $w_1$ is an even function}
&\le\Bover{3\cdot 2^m}{\sqrt{2\pi}}C_1
  \int_{-\infty}^{\infty}w_1(3\cdot 2^mt)dt
  \cdot\sup_{\Atop{a\le-2^{-m}}{b\ge 2^{-m}}}
  \Bover1{b-a}\int_a^b|\tilde f(x+t)|dt\cr
\displaycause{by 286B, because $t\mapsto w_1(3\cdot 2^mt)$ is
non-decreasing on $\ocint{-\infty,-2^{-m}}$, non-increasing on
$\coint{2^{-m},\infty}$ and constant on $[-2^{-m},2^{-m}]$}
&=\Bover1{\sqrt{2\pi}}C_1\int_{-\infty}^{\infty}w_1
  \cdot\sup_{\Atop{a\le x-2^{-m}}{b\ge x+2^{-m}}}
  \Bover1{b-a}\int_a^b|\tilde f|
\le\Bover12C_1\int_{-\infty}^{\infty}w
  \cdot\tilde f^*(x')\cr
\displaycause{because if $a\le x-2^{-m}$ and $b\ge x+2^{-m}$ then
$a\le x'\le b$}
&=\Bover12C_1\tilde f^*(x')\cr}$$

\noindent (286Ga), as required.\ \Qed

\medskip

{\bf (i)} For $x\in\Bbb R$, set

\Centerline{$v_3(x)=\bigl|\sum_{(\sigma,K)\in W_3}
   \zeta_{\sigma}\innerprod{f}{\phi_{\sigma}}\phi_{\sigma}(x)
   \chi(E\cap g^{-1}[J^r_{\sigma}]\cap K)(x)\bigr|$.}

\noindent Then whenever $L\in\Cal K$ and $x$, $x'\in L$,
$|v_3(x)|\le C_1\tilde f^*(x')$.   \Prf\ We may suppose that
$v_3(x)\ne 0$, so that, in particular, $x\in E$.   The only pairs
$(\sigma,K)$ contributing to the sum forming $v_3(x)$ are those in which
$x\in K$, so that $K=L$, and $g(x)\in J^r_{\sigma}$.   Moreover, since
we are looking only at $\sigma\in T_{\tau}$, so that
$J^r_{\tau}\subseteq J^r_{\sigma}$, $J^r_{\sigma}$ will always be
the dyadic interval of length $2^{k_{\sigma}-1}$ including $J^r_{\tau}$.
So these intervals are nested, and there will be some $m$ such that (for
$\sigma\in T_{\tau}$) $g(x)\in J^r_{\sigma}$ iff $k_{\sigma}\ge m$.
Accordingly

\Centerline{$v_3(x)
=\bigl|\sum_{\sigma\in P',m\le k_{\sigma}<l_L}
   \zeta_{\sigma}\innerprod{f}{\phi_{\sigma}}\phi_{\sigma}(x)\bigr|
=|\tilde f_{l_L-1}(x)-\tilde f_{m-1}(x)|$}

\noindent (we must have $m<l_L$ because $v_3(x)\ne 0$).   Now
$|x-x'|\le 2^{-l_L}\le 2^{-m}$, so (h) tells us that
both $|\tilde f_{l_L-1}(x)|$ and $|\tilde f_{m-1}(x)|$ are at most
$\bover12C_1\tilde f^*(x')$, and $v_3(x)\le C_1\tilde f^*(x')$, as
claimed.\ \Qed

It follows that $v_3(x)\le\Bover{C_1}{\mu L}\biggerint_L\tilde f^*$
for every $x\in L$.

\wheader{286L}{6}{2}{2}{108pt}

{\bf (j)} Now we are in a position to estimate

$$\eqalignno{|\alpha_3|
&=|\sum_{(\sigma,K)\in W_3}\zeta_{\sigma}\alpha_{\sigma K}|
\le\int_{-\infty}^{\infty}v_3
\le\sum_{\Atop{K\in\Cal K}{\mu K<\mu I_{\tau}}}\int_{G_K}v_3\cr
\displaycause{because if $v_3(x)\ne 0$ there are $(\sigma,K)\in W_3$
such that $x\in K$, and in this case $x\in G_K$ and
$\mu K<\mu I_{\sigma}\le\mu I_{\tau}$}
&\le\sum_{\Atop{K\in\Cal K}{\mu K<\mu I_{\tau}}}
  \mu G_K\cdot\Bover{C_1}{\mu K}\int_K\tilde f^*\cr
\displaycause{by (i) above, because $G_K\subseteq K$}
&\le C_1\sum_{\Atop{K\in\Cal K}{\mu K<\mu I_{\tau}}}
  \bover{2\gamma'}{w(\bover32)}\int_K\tilde f^*\cr
\displaycause{by (e)}
&\le\bover{2C_1\gamma'}{w(\bover32)}\int_{\hat I}\tilde f^*\cr
\displaycause{because if $\mu K<\mu I_{\tau}$ then $K\subseteq\hat I$, as
noted in (a-iii)}
&\le\bover{2C_1\gamma'}{w(\bover32)}\sqrt{\mu\hat I}
  \cdot\|\tilde f^*\|_2\cr
\displaycause{by Cauchy's inequality}
&\le\bover{2C_1\gamma'}{w(\bover32)}\sqrt{7\mu I_{\tau}}
  \cdot\sqrt{8}\|\tilde f\|_2\cr
\displaycause{by the Maximal Theorem, 286A}
&\le\bover{4C_1\gamma'\sqrt{14}}{w(\bover32)}\sqrt{\mu I_{\tau}}
  \cdot\gamma\sqrt{C_3\mu I_{\tau}}\cr
\displaycause{by (g)}
&=\bover{4C_1\sqrt{14 C_3}}{w(\bover32)}
  \gamma\gamma'\mu I_{\tau}.\cr}$$

\medskip

{\bf (k)} Assembling these,

$$\eqalign{\sum_{\sigma\in P}\bigl|\innerprod{f}{\phi_{\sigma}}
  \int_{E\cap g^{-1}[J^r_{\sigma}]}\phi_{\sigma}\bigr|
&=\sum_{\Atop{\sigma\in P}{K\in\Cal K}}\zeta_{\sigma}\alpha_{\sigma K}
=\sum_{j=0}^3\sum_{(\sigma,K)\in W_j}\zeta_{\sigma}\alpha_{\sigma K}
\le\sum_{j=0}^3|\alpha_j|\cr
&\le\Bover72\cdot C_1\gamma\gamma'\mu I_{\tau}
  +\Bover87\cdot C_1\gamma\gamma'\mu I_{\tau}
  +28\cdot C_1\gamma\gamma'\mu I_{\tau}/w(\hbox{$\bover32$})\cr
&\qquad\qquad\qquad\qquad
  +4\sqrt{14C_3}\cdot
      C_1\gamma\gamma'\mu I_{\tau}/w(\hbox{$\bover32$})\cr
&=C_7\gamma\gamma'\mu I_{\tau},\cr}$$

\noindent as claimed.
}%end of proof of 286L

\leader{286M}{The Lacey-Thiele lemma} Set $C_8=3C_7(C_5+C_6)$.   Then

\Centerline{$\sum_{\sigma\in Q}|\innerprod{f}{\phi_{\sigma}}
  \int_{E\cap g^{-1}[J^r_{\sigma}]}\phi_{\sigma}|
\le C_8$}

\noindent whenever $f\in\eusm L^2_{\Bbb C}$, $\|f\|_2=1$,
$\mu E\le 1$ and $g:\Bbb R\to\Bbb R$ is measurable.

\proof{%
{\bf (a)} The first step is to combine 286J and 286K, as follows:
if $P\subseteq Q$ is finite and
$\max(\sqrt{\mass_{Eg}(P)},\energy_f(P))\penalty-100\le\gamma$, there is an
$R\subseteq Q$ such that
$\gamma^2\sum_{\tau\in R}\mu I_{\tau}\le C_5+C_6$ and
$\max(\sqrt{\mass_{Eg}(P\setminus R^+)},
\energy_f(P\setminus R^+))
\le\bover12\gamma$.
\Prf\ Since $\mass_{Eg}(P)\le\gamma^2$, 286J tells us that there is an
$R_0\subseteq Q$ such that
$\gamma^2\sum_{\tau\in R_0}\mu I_{\tau}\le C_5$ and
$\mass_{Eg}(P\setminus R_0^+)\le\bover14\gamma^2$.
Turn to 286K:  since
$\energy_f(P\setminus R_0^+)\le\energy_f(P)\le\gamma$, we can find
$R_1\subseteq Q$ such that
$\gamma^2\sum_{\tau\in R_1}\mu I_{\tau}\le C_6$ and
$\energy_f((P\setminus R_0^+)\setminus R_1^+)\le\bover12\gamma$.
Set $R=R_0\cup R_1$.   Then

\Centerline{$\gamma^2\sum_{\tau\in R}\mu I_{\tau}\le C_5+C_6$, 
\quad$\mass_{Eg}(P\setminus R^+)\le\mass_{Eg}(P\setminus R_0^+)
\le\bover14\gamma^2$}

\noindent so
$\max(\sqrt{\mass_{Eg}(P\setminus R^+)},\energy_f(P\setminus R^+))
\le\bover12\gamma$.\ \Qed
\fontdimen3\tenrm=1.67pt

\medskip

{\bf (b)} Now take any finite $P\subseteq Q$.
Let $k\in\Bbb N$ be such that
$\max(\sqrt{\mass_{Eg}(P)},\energy_f(P))\le 2^k$.   By (a), we can
choose $\sequencen{P_n}$, $\sequencen{R_n}$ inductively such that
$P_0=P$ and, for each $n\in\Bbb N$,

\Centerline{$P_{n+1}=P_n\setminus R_n^+$,}

\Centerline{$2^{2k-2n}\sum_{\tau\in R_n}\mu I_{\tau}\le C_5+C_6$,
\quad$\max(\sqrt{\mass_{Eg}(P_n)},\energy_f(P_n))\le 2^{k-n}$.}

\noindent Since
$\energy_f(\{\sigma\})=\sqrt{\mu J_{\sigma}}|\innerprod{f}{\phi_{\sigma}}|>0$
whenever $\innerprod{f}{\phi_{\sigma}}\ne 0$ (286H),
$\innerprod{f}{\phi_{\sigma}}=0$ whenever
$\sigma\in\bigcap_{n\in\Bbb N}P_n$, and

$$\eqalignno{\sum_{\sigma\in P}\bigl|\innerprod{f}{\phi_{\sigma}}
  \int_{E\cap g^{-1}[J^r_{\sigma}]}\phi_{\sigma}\bigr|
&=\sum_{\sigma\in\bigcup_{n\in\Bbb N}P_n\setminus P_{n+1}}
  \bigl|\innerprod{f}{\phi_{\sigma}}
  \int_{E\cap g^{-1}[J^r_{\sigma}]}\phi_{\sigma}\bigr|\cr
&=\sum_{n=0}^{\infty}\sum_{\sigma\in P_n\setminus P_{n+1}}
  \bigl|\innerprod{f}{\phi_{\sigma}}
  \int_{E\cap g^{-1}[J^r_{\sigma}]}\phi_{\sigma}\bigr|\cr
&\le\sum_{n=0}^{\infty}\sum_{\tau\in R_n}
  \sum_{\Atop{\sigma\in P_n}{\sigma\ge\tau}}
     \bigl|\innerprod{f}{\phi_{\sigma}}
       \int_{E\cap g^{-1}[J^r_{\sigma}]}\phi_{\sigma}\bigr|\cr
&\le\sum_{n=0}^{\infty}\sum_{\tau\in R_n}C_7
  \energy_f(P_n)\mass_{Eg}(P_n)\mu I_{\tau}\cr
\displaycause{by 286L}
&\le C_7\sum_{n=0}^{\infty}2^{k-n}\min(1,2^{2k-2n})
  \sum_{\tau\in R_n}\mu I_{\tau}\cr
\displaycause{because $\mass_{Eg}(P_n)\le 1$ for every $n$, as noted in
286H}
&\le C_7\sum_{n=0}^{\infty}2^{k-n}\min(1,2^{2k-2n})2^{2n-2k}(C_5+C_6)
  \cr
&=C_7(C_5+C_6)\sum_{n=0}^{\infty}\min(2^{n-k},2^{k-n})\cr
&\le C_7(C_5+C_6)\sum_{n=-\infty}^{\infty}\min(2^n,2^{-n})
=3C_7(C_5+C_6).\cr}$$

\medskip

{\bf (c)} Since this true for every finite $P\subseteq Q$,

\Centerline{$\sum_{\sigma\in Q}|\innerprod{f}{\phi_{\sigma}}
  \int_{E\cap g^{-1}[J^r_{\sigma}]}\phi_{\sigma}|
\le 3C_7(C_5+C_6)=C_8$,}

\noindent as claimed.
}%end of proof of 286M

\leader{286N}{Lemma} Set $C_9=C_8\sqrt{2}$.   Suppose that
$f\in\eusm L^2_{\Bbb C}$, $g:\Bbb R\to\Bbb R$ is
measurable and $\mu F<\infty$.   Then

\Centerline{$\sum_{\sigma\in Q}|\innerprod{f}{\phi_{\sigma}}
  \int_{F\cap g^{-1}[J^r_{\sigma}]}\phi_{\sigma}|
\le C_9\|f\|_2\sqrt{\mu F}$.}

\proof{ This is trivial if $\|f\|_2=0$, that is, $f=0$ a.e.   So we may
take it that $\|f\|_2>0$.   Dividing both sides by $\|f\|_2$, we may
suppose that $\|f\|_2=1$.

Let $k\in\Bbb Z$ be such that $2^{k-1}<\mu F\le 2^k$.   We have a
permutation $\sigma\mapsto\sigma^*:Q\to Q$ defined by saying that
$\sigma^*=(2^{-k}I_{\sigma},2^kJ_{\sigma})$;  so that
$k_{\sigma^*}=k_{\sigma}+k$, $x_{\sigma^*}=2^{-k}x_{\sigma}$,
$y^l_{\sigma^*}=2^ky^l_{\sigma}$, $J^r_{\sigma^*}=2^kJ^r_{\sigma}$, and
for every $x\in\Bbb R$

$$\eqalign{\phi_{\sigma}(2^kx)
&=\sqrt{\mu J_{\sigma}}e^{2^kiy^l_{\sigma}x}
  \phi((2^kx-x_{\sigma})\mu J_{\sigma})\cr
&=\sqrt{\mu J_{\sigma}}e^{iy^l_{\sigma^*}x}
  \phi(2^{k_{\sigma}+k}(x-2^{-k}x_{\sigma}))\cr
&=2^{-k/2}\sqrt{\mu J_{\sigma^*}}e^{iy^l_{\sigma^*}x}
  \phi((x-x_{\sigma^*})\mu J_{\sigma^*})
=2^{-k/2}\phi_{\sigma^*}(x).\cr}$$

\noindent Write $\tilde F=2^{-k}F$, so that $\mu\tilde F\le 1$, and
$\tilde g(x)=2^kg(2^kx)$ for every $x$.   Then, for $\sigma\in Q$,

$$\eqalign{F\cap g^{-1}[J^r_{\sigma}]
&=\{x:x\in F,\,g(x)\in J^r_{\sigma}\}
=\{x:2^{-k}x\in\tilde F,\,2^{-k}\tilde g(2^{-k}x)\in J^r_{\sigma}\}\cr
&=\{x:2^{-k}x\in\tilde F,\,\tilde g(2^{-k}x)\in J^r_{\sigma^*}\}
=2^k\{x:x\in\tilde F,\,\tilde g(x)\in J^r_{\sigma^*}\}.\cr}$$

\noindent Write $\tilde f(x)=2^{k/2}f(2^kx)$, so that

\Centerline{$\|\tilde f\|_2=2^{k/2}\|D_{2^k}f\|_2=\|f\|_2=1$,}

\noindent while

\Centerline{$\innerprod{f}{\phi_{\sigma}}
=\int_{-\infty}^{\infty}f\times\bar\phi_{\sigma}
=2^k\int_{-\infty}^{\infty}
f(2^kx)\overline{\phi_{\sigma}(2^kx)}dx
=\innerprod{\tilde f}{\phi_{\sigma^*}}$}

\noindent for every $\sigma\in Q$.   Putting all these together,

$$\eqalignno{\sum_{\sigma\in Q}\bigl|\innerprod{f}{\phi_{\sigma}}
  \int_{F\cap g^{-1}[J^r_{\sigma}]}\phi_{\sigma}\bigr|
&=2^k\sum_{\sigma\in Q}\bigl|\innerprod{f}{\phi_{\sigma}}
  \int_{2^{-k}(F\cap g^{-1}[J^r_{\sigma}])}
     \phi_{\sigma}(2^kx)dx\bigr|\cr
&=2^{k/2}\sum_{\sigma\in Q}\bigl|\innerprod{\tilde f}{\phi_{\sigma^*}}
  \int_{\tilde F\cap\tilde g^{-1}[J^r_{\sigma^*}]}
    \phi_{\sigma^*}\bigr|\cr
&=2^{k/2}\sum_{\tau\in Q}\bigl|\innerprod{\tilde f}{\phi_{\tau}}
  \int_{\tilde F\cap\tilde g^{-1}[J^r_{\tau}]}\phi_{\tau}\bigr|\cr
&\le 2^{k/2}C_8\cr
\displaycause{by the Lacey-Thiele lemma, applied to $\tilde g$,
$\tilde F$ and $\tilde f$}
&\le C_9\sqrt{\mu F}
=C_9\|f\|_2\sqrt{\mu F}.\cr}$$
}%end of proof of 286N

\leader{286O}{Lemma}\dvArevised{2013} (a) For $z\in\Bbb R$, define
$\theta_z:\Bbb R\to[0,1]$ by setting

\Centerline{$\theta_z(y)=\varhat{\phi}(2^{-k}(y-\hat y))^2$}

\noindent whenever there is a dyadic interval $J\in\Cal I$ of length
$2^k$ such that $z$ belongs to the right-hand half of $J$ and $y$
belongs to the left-hand half of $J$ and $\hat y$ is the lower quartile
of $J$, and zero if there is no such $J$.   
Then $(y,z)\mapsto\theta_z(y)$ is Borel measurable,
$0\le\theta_z(y)\le 1$
for all $y$, $z\in\Bbb R$, and $\theta_z(y)=0$ if $y\ge z$.

(b) For $k\in\Bbb Z$, set $Q_k=\{\sigma:\sigma\in Q$, $k_{\sigma}=k\}$.
Let $[Q]^{<\omega}$ be the set of finite subsets of
$Q$, $[\Bbb Z]^{<\omega}$ the set of finite subsets of $\Bbb Z$
and $\Cal L$ the set of subsets $L$ of $Q$ such that $L\cap Q_k$ is
finite for every $k$.
If $K\in[\Bbb Z]^{<\omega}$ and $L\in\Cal L$, set

$$\eqalign{\Cal P_{KL}
&=\{P:P\in[Q]^{<\omega},\,P\cap Q_k\supseteq L\cap Q_k
  \text{ whenever }k\in\Bbb Z
\cr&\mskip200mu  
   \text{and either }k\in K\text{ or }P\cap Q_k\ne\emptyset\};\cr}$$
   
\noindent set

\Centerline{$\Cal F=\{\Cal P:\Cal P\subseteq[Q]^{<\omega}$ and there are 
  $K\in[\Bbb Z]^{<\omega}$, $L\in\Cal L$ such that 
  $\Cal P\supseteq\Cal P_{KL}\}$.}
  
\noindent Then $\Cal F$ is a filter on $[Q]^{<\omega}$ and

\Centerline{$2\pi\int_F(\varhat h\times\theta_z)\varspcheck
=\lim_{P\to\Cal F}\sum_{\sigma\in P,z\in J^r_{\sigma}}
   \innerprod{h}{\phi_{\sigma}}\int_F\phi_{\sigma}$}
   
\noindent for every $z\in\Bbb R$ and rapidly decreasing test function $h$ and measurable set
$F\subseteq\Bbb R$ of finite measure.

\proof{{\bf (a)(i)} I had better start by explaining why the recipe above
defines a function $\theta_z$.   Let $M$ be the set of those
$k\in\Bbb Z$ such that $z$ belongs to the right-hand half of the dyadic
interval $\hat J_k$ of length $2^k$ containing $z$.
For $k\in M$, let $\hat y_k$ be the midpoint of the left-hand half
$\hat J^l_k$ of $\hat J_k$, and set
$\psi_k(y)=\varhat{\phi}(2^{-k}(y-\hat y_k))^2$ for $y\in\Bbb R$;  then
$\psi_k$ is smooth and zero outside $\hat J^l_k$.   But now observe that
if $k$, $k'$ are distinct members of $M$, then $\hat J^l_k$ and
$\hat J^l_{k'}$ are disjoint, as remarked in
286E(b-iv).   So $\theta_z$ is just the sum $\sum_{k\in M}\psi_k$.
Because $\varhat{\phi}$ takes values in $[0,1]$, so does $\theta_z$.
If $y\ge z$, then of course $y\notin\hat J^l_k$ for any $k\in M$, so
$\theta_z(y)=0$.

\medskip

\quad{\bf (ii)} To see that $(y,z)\mapsto\theta_z(y)$ is Borel
measurable, observe that

\Centerline{$\{(y,z):\theta_z(y)\ge\gamma\}
=\bigcup_{\sigma\in Q}\{(y,z):
  \hat\phi((y-y^l_{\sigma})\mu I_{\sigma})^2\ge\gamma$,
  $z\in J^r_{\sigma}\}$}

\noindent for every $\gamma\in\Bbb R$.

\medskip

{\bf (b)(i)} $\emptyset$ belongs to both $[\Bbb Z]^{<\omega}$ and $\Cal L$ and
$[Q]^{<\omega}=\Cal P_{\emptyset\emptyset}$ belongs to $\Cal F$.
If $K\in[\Bbb Z]^{<\omega}$ and $L\in\Cal L$ then
$\bigcup_{k\in K}L\cap Q_k$ belongs to $\Cal P_{KL}$.   So no
$\Cal P_{KL}$ is empty and $\emptyset\notin\Cal F$.

If $\Cal P$, $\Cal P'\in\Cal F$, there are
$K$, $K'\in[\Bbb Z]^{<\omega}$ and $L$, $L'\in\Cal L$ 
such that $\Cal P_{KL}\subseteq\Cal P$ and $\Cal P_{K'L'}\subseteq\Cal P'$.
Now $K\cup K'\in[\Bbb Z]^{<\omega}$, $L\cup L'\in\Cal L$ and

\Centerline{$\Cal P_{K\cup K',L\cup L'}
\subseteq\Cal P_{KL}\cap\Cal P_{K'L'}\subseteq\Cal P\cap\Cal P'$,}

\noindent so $\Cal P\cap\Cal P'\in\Cal F$.

If $\Cal P\in\Cal F$ and $\Cal P\subseteq\Cal P'\subseteq[Q]^{<\omega}$,
then of course $\Cal P'\in\Cal F$.   So $\Cal F$ is a filter on
$[Q]^{<\omega}$.

\medskip

\quad{\bf (ii)} Now fix on $z\in\Bbb R$, a rapidly decreasing test function $h$ and a set $F$ of
finite measure.   Take $M$ and
$\psi_k$, $\hat J_k$, $\hat y_k$ for $k\in M$ from (a-i) above;
it will be convenient to set $\psi_k=0$ for $k\in\Bbb Z\setminus M$, so
that $\theta_z=\sum_{k\in\Bbb Z}\psi_k$.

For $k\in\Bbb Z$,

\Centerline{$2\pi\int_F(\varhat h\times\psi_k)\varspcheck
=\sum_{\sigma\in Q_k,z\in J^r_{\sigma}}
  \innerprod{h}{\phi_{\sigma}}\int_F\phi_{\sigma}$.}
  
\noindent\Prf\ If $k\notin M$, then $z\notin J^r_{\sigma}$ for any
$\sigma\in Q_k$, while $\psi_k=0$, so the result is trivial.   So I will
suppose that $k\in M$ and that $\hat y_k$ is defined.
If $\sigma\in Q_k$ and $z\in J^r_{\sigma}$, 
$y^l_{\sigma}=\hat y_k$ and $x_{\sigma}$ is
of the form $2^{-k}(n+\bover12)$ for some $n\in\Bbb Z$.   So

$$\eqalignno{\innerprod{h}{\phi_{\sigma}}
&=\int_{-\infty}^{\infty}
  \varhat h\times\bar\varhat{\phi}_{\sigma}\cr
\displaycause{284O}
&=\int_{-\infty}^{\infty}\varhat h(t)
  \cdot 2^{-k/2}e^{2^{-k}i(n+\bover12)(t-\hat y_k)}
  \varhat\phi(2^{-k}(t-\hat y_k))dt\cr
\displaycause{by the formula in 286Eb, because $\varhat{\phi}$ is
real-valued}
&=2^{k/2}\int_{-\infty}^{\infty}\varhat h(2^kt+\hat y_k)
  e^{i(n+\bover12)t}\varhat\phi(t)dt
=2^{k/2}\int_{-\pi}^{\pi}\varhat h(2^kt+\hat y_k)
  e^{i(n+\bover12)t}\varhat\phi(t)dt\cr
\displaycause{because $\varhat{\phi}(t)=0$ if $|t|\ge\bover15$}
&=2^{k/2}\int_{-\pi}^{\pi}g(t)e^{int}dt,\cr}$$

\noindent where $g(t)=\varhat h(2^kt+\hat y_k)e^{it/2}\varhat{\phi}(t)$
for $-\pi<t\le\pi$.   So if we set
$c_n=\Bover1{2\pi}\int_{-\pi}^{\pi}g(t)e^{-int}dt$, as in 282A, we have

\Centerline{$\innerprod{h}{\phi_{\sigma}}=2^{k/2}\cdot 2\pi c_{-n}$}

\noindent when $\sigma\in Q_k$, $z\in J^r_{\sigma}$
and $x_{\sigma}=2^{-k}(n+\bover12)$.
Note that as $g$ is smooth and zero outside
$[-\bover15,\bover15]$, $\sum_{n=-\infty}^{\infty}|c_n|<\infty$
(282Rb).

Now, for any $y\in\hat J^l_k$, writing
$R_k$ for 

\Centerline{$\{\sigma:\sigma\in Q_k$, $z\in J^r_{\sigma}\}
=\{\sigma:\sigma\in Q_k$, $J_{\sigma}=\hat J_k\}
=\{(I,\hat J_k):I\in\Cal I$, $\mu I=2^{-k}\}$,}

\noindent we have

$$\eqalignno{\sum_{\sigma\in R_k}
  \innerprod{h}{\phi_{\sigma}}\varhat{\phi}_{\sigma}(y)
&=\sum_{n=-\infty}^{\infty}2^{k/2}\cdot 2\pi c_{-n}
   \cdot 2^{-k/2}e^{-2^{-k}i(n+\bover12)(y-\hat y_k)}
   \varhat{\phi}(2^{-k}(y-\hat y_k))\cr
&=2\pi\varhat{\phi}(2^{-k}(y-\hat y_k))
   e^{-2^{-k-1}i(y-\hat y_k)}
   \sum_{n=-\infty}^{\infty}c_{-n}e^{-2^{-k}in(y-\hat y_k)}\cr
&=2\pi\varhat{\phi}(2^{-k}(y-\hat y_k))
   e^{-2^{-k-1}i(y-\hat y_k)}
   \sum_{n=-\infty}^{\infty}c_ne^{in 2^{-k}(y-\hat y_k)}\cr
&=2\pi\varhat{\phi}(2^{-k}(y-\hat y_k))
   e^{-2^{-k-1}i(y-\hat y_k)}g(2^{-k}(y-\hat y_k))\cr
\displaycause{by 282L(i), because $2^{-k}|y-\hat y_k|\le\bover14<\pi$
and $g$ is smooth}
&=2\pi e^{-2^{-k-1}i(y-\hat y_k)}\varhat{\phi}(2^{-k}(y-\hat y_k))
  \varhat h(y)e^{2^{-k-1}i(y-\hat y_k)}
  \varhat{\phi}(2^{-k}(y-\hat y_k))\cr
&=2\pi\varhat h(y)\psi_k(y).\cr}$$

\noindent On the other hand, if $y\in\Bbb R\setminus\hat J^l_k$,
$\psi_k(y)=\varhat{\phi}_{\sigma}(y)=0$ for every $\sigma\in R_k$, so
again
$\sum_{\sigma\in R_k}
  \innerprod{h}{\phi_{\sigma}}\varhat{\phi}_{\sigma}(y)
=2\pi\varhat h(y)\psi_k(y)$.

Next,

\Centerline{$\sum_{\sigma\in R_k}|\innerprod{h}{\phi_{\sigma}}|
=2\pi\cdot 2^{k/2}\sum_{n=-\infty}^{\infty}|c_n|$}

\noindent and

\Centerline{$\sup_{\sigma\in R_k}\int_{-\infty}^{\infty}
  |\varhat{\phi}_{\sigma}|
  =2^{k/2}\int_{-\infty}^{\infty}|\varhat{\phi}|$}

\noindent are finite, while of course $\varhat{\chi F}$ is bounded.
So

$$\eqalignno{2\pi\int_F(\varhat h\times\psi_k)\varspcheck
&=2\pi\innerprod{\varhat h\times\psi_k)\varspcheck}{\chi F}
=2\pi\innerprod{(\varhat h\times\psi_k)\varspcheck)\varsphat}
  {\varhat{\chi F}}\cr
\displaycause{284Ob again}
&=2\pi\innerprod{\varhat h\times\psi_k}{\varhat{\chi F}}
=\int_{-\infty}^{\infty}
    2\pi\varhat h\times\psi_k\times\overline{\varhat{\chi F}}
\cr&=\int_{-\infty}^{\infty}\sum_{\sigma\in R_k}
   \innerprod{h}{\phi_{\sigma}}\varhat{\phi}_{\sigma}
   \times\overline{\varhat{\chi F}}
=\sum_{\sigma\in R_k}\innerprod{h}{\phi_{\sigma}}\int_{-\infty}^{\infty}
   \varhat{\phi}_{\sigma}\times\overline{\varhat{\chi F}}\cr
\displaycause{226E}   
&=\sum_{\sigma\in R_k}\innerprod{h}{\phi_{\sigma}}
   \int_F{\phi}_{\sigma}
=\sum_{\sigma\in Q_k,z\in J^r_{\sigma}}\innerprod{h}{\phi_{\sigma}}
   \int_F{\phi}_{\sigma}.  \text{ \Qed}\cr}$$
   
\medskip

\quad{\bf (iii)} In the last sentence of the argument just above, 
I quoted B.Levi's
theorem in the form 226E, even though $R_k$ has a natural
enumeration, because I shall specifically want to say later that

\inset{\noindent
for every $\epsilon>0$ there is a finite $L_0\subseteq R_k$ such that

\centerline{$|2\pi\int_F(\varhat h\times\psi_k)\varspcheck
-\sum_{\sigma\in L}\innerprod{h}{\phi_{\sigma}}
   \int_F{\phi}_{\sigma}|\le\epsilon$}
   
\noindent whenever $L\subseteq R_k$ is finite and $L\supseteq L_0$;}

\noindent it follows at once that
   
\inset{\noindent
for every $\epsilon>0$ there is a finite $L_0\subseteq Q_k$ such that

\centerline{$|2\pi\int_F(\varhat h\times\psi_k)\varspcheck
-\sum_{\sigma\in L,z\in J^r_{\sigma}}\innerprod{h}{\phi_{\sigma}}
   \int_F{\phi}_{\sigma}|\le\epsilon$}
   
\noindent whenever $L\subseteq Q_k$ is finite and $L\supseteq L_0$.}

\medskip

\quad{\bf (iv)} Now let us consider $(\varhat h\times\theta_z)\varspcheck$.
Because every $\psi_k$ is non-negative, $\theta_z=\sum_{k\in\Bbb Z}\psi_k$
is bounded above by $1$, and $\varhat h$ is integrable,

$$\eqalign{\int_F(\varhat h\times\theta_z)\varspcheck
&=\int_{-\infty}^{\infty}\varhat h\times\theta_z\times\varhat{\chi F}
\cr&=\sum_{k\in\Bbb Z}\int_{-\infty}^{\infty}\varhat h\times\psi_k\times\varhat{\chi F}
=\sum_{k\in\Bbb Z}\int_F(\varhat h\times\psi_k)\varspcheck.\cr}$$

\noindent So here we can say

\inset{\noindent for every $\epsilon>0$ there is a 
$K_0\in[\Bbb Z]^{<\omega}$ such that 

\centerline{$|\int_F(\varhat h\times\theta_z)\varspcheck
  -\sum_{k\in K}\int_F(\varhat h\times\psi_k)\varspcheck|\le\epsilon$}

\noindent whenever $K\in[\Bbb Z]^{<\omega}$ and $K\supseteq K_0$.}  

\medskip

\quad{\bf (v)} To express the facts above in terms of a limit along the
filter $\Cal F$, we can argue as follows.   Take any $\epsilon>0$.
For each $k\in\Bbb Z$, (iii) tells us that there is a finite set
$L_k\subseteq Q_k$ such that 

\Centerline{$|2\pi\int_F(\varhat h\times\psi_k)\varspcheck
-\sum_{\sigma\in L',z\in J^r_{\sigma}}\innerprod{h}{\phi_{\sigma}}
   \int_F{\phi}_{\sigma}|\le 2^{-|k|}\epsilon$}
   
\noindent whenever $L'\subseteq Q_k$ is finite and $L'\supseteq L_k$;  of
course we can suppose that every $L_k$ is non-empty.
Set $L=\bigcup_{k\in\Bbb Z}L_k$, so that $L\cap Q_k=L_k$ is finite for each
$k$, and $L\in\Cal L$.   Next, there is a $K\in[\Bbb Z]^{<\omega}$ such
that

\Centerline{$|\int_F(\varhat h\times\theta_z)\varspcheck
  -\sum_{k\in K'}\int_F(\varhat h\times\psi_k)\varspcheck|
  \le\epsilon$}

\noindent whenever $K'\in[\Bbb Z]^{<\omega}$ and $K'\supseteq K$.
Take any $P\in\Cal P_{KL}$.   Setting $K'=\{k:P\cap Q_k\ne\emptyset\}$,
we have $K'\supseteq K$ and $P\cap Q_k\supseteq L\cap Q_k$ for
every $k\in K'$.   Accordingly

$$\eqalign{&|2\pi\int_F(\varhat h\times\theta_z)\varspcheck
   -\sum_{\sigma\in P,z\in J^r_{\sigma}}
      \innerprod{h}{\phi_{\sigma}}\int_F\phi_{\sigma}|
\cr&\mskip100mu
=|2\pi\int_F(\varhat h\times\theta_z)\varspcheck
   -\sum_{k\in K'}\sum_{\Atop{\sigma\in P\cap Q_k}{z\in J^r_{\sigma}}}
      \innerprod{h}{\phi_{\sigma}}\int_F\phi_{\sigma}|
\cr&\mskip100mu
\le 2\pi|\int_F(\varhat h\times\theta_z)\varspcheck
   -\sum_{k\in K'}\int_F(\varhat h\times\psi_k)\varspcheck|      
\cr&\mskip200mu
+\sum_{k\in K'}|2\pi\int_F(\varhat h\times\psi_k)\varspcheck
   -\sum_{\Atop{\sigma\in P\cap Q_k}{z\in J^r_{\sigma}}}
      \innerprod{h}{\phi_{\sigma}}\int_F\phi_{\sigma}|
\cr&\mskip100mu
\le 2\pi\epsilon+\sum_{k\in K'}2^{-|k|}\epsilon
\le (2\pi+3)\epsilon.\cr}$$

\noindent As $\Cal P_{KL}\in\Cal F$, and $\epsilon$ was arbitrary,

\Centerline{$2\pi\int_F(\varhat h\times\theta_z)\varspcheck
=\lim_{P\to\Cal F}\sum_{\sigma\in P,z\in J^r_{\sigma}}
   \innerprod{h}{\phi_{\sigma}}\int_F\phi_{\sigma}$}

\noindent as claimed.
}%end of proof of 286O
   
\leader{286P}{Lemma}\dvArevised{2013} 
Suppose that $h$ is a rapidly decreasing test function.   
For $x\in\Bbb R$, set

\Centerline{$Ah(x)
=\sup_{z\in\Bbb R}|2\pi(\varhat h\times\theta_z)\varspcheck(x)|$.}

\noindent Then $Ah:\Bbb R\to[0,\infty]$ is Borel measurable, and
$\int_FAh\le 4C_9\|h\|_2\sqrt{\mu F}$ whenever $\mu F<\infty$.

\proof{{\bf (a)} As $(\varhat h\times\theta_z)\varspcheck$ is continuous
for every $z$, $Ah$ is lower semi-continuous, therefore Borel measurable,
by 256Ma.   By 256Mb,

\Centerline{$\int_FAh
=\sup\{\int_F\sup_{i\le n}
  |2\pi(\varhat h\times\theta_{z_i})\varspcheck|:
  z_0,\ldots,z_n\in\Bbb R\}$.}

\medskip

{\bf (b)} Fix $z_0,\ldots,z_n\in\Bbb R$ for the moment.

\medskip

\quad{\bf (i)} Set  
$v_i=2\pi(\varhat h\times\theta_{z_i})\varspcheck$ for $i\le n$, and
$v=\sup_{i\le n}|v_i|$.   Set 
$E_i=\{x:v(x)=|v_i(x)|\}\setminus\bigcup_{j<i}\{x:v(x)=|v_j(x)|\}$ for
$i\le n$, so that $(E_0,\ldots,E_n)$ is a partition of $\Bbb R$ into
Borel sets, and 

\Centerline{$\int_Fv
=\int_F\sum_{i=0}^n|v_i|\times\chi E_i
=\int_F|\sum_{i=0}^nv_i\times\chi E_i|
\le 4|\int_{F'}\sum_{i=0}^nv_i\times\chi E_i|$}

\noindent for a suitable measurable $F'\subseteq F$ (246K).
Setting $g(x)=z_i$ for $x\in E_i$, $g:\Bbb R\to\Bbb R$ is Borel measurable.

\medskip

\quad{\bf (ii)} For each $i\le n$,

$$\eqalignno{\int_{F'}v_i\times\chi E_i
&=\int_{F'\cap E_i}v_i
=\lim_{P\to\Cal F}\sum_{\sigma\in P,z_i\in J^r_{\sigma}}
  \innerprod{h}{\phi_{\sigma}}\int_{F'\cap E_i}\phi_{\sigma}\cr
\displaycause{where $\Cal F$ is the filter on $[Q]^{<\omega}$ described in
286O}  
&=\lim_{P\to\Cal F}\int_{F'\cap E_i}
  \sum_{\sigma\in P,z_i\in J^r_{\sigma}}
  \innerprod{h}{\phi_{\sigma}}\phi_{\sigma}
\cr&=\lim_{P\to\Cal F}\int_{F'\cap E_i}
  \sum_{\sigma\in P,g(x)\in J^r_{\sigma}}
  \innerprod{h}{\phi_{\sigma}}\phi_{\sigma}(x)dx.\cr}$$
  
\noindent So

$$\eqalign{\int_{F'}\sum_{i=0}^nv_i\times\chi E_i  
&=\lim_{P\to\Cal F}\sum_{i=0}^n\int_{F'\cap E_i}
  \sum_{\sigma\in P,g(x)\in J^r_{\sigma}}
  \innerprod{h}{\phi_{\sigma}}\phi_{\sigma}(x)dx
\cr&=\lim_{P\to\Cal F}\int_{F'}
  \sum_{\sigma\in P,g(x)\in J^r_{\sigma}}
  \innerprod{h}{\phi_{\sigma}}\phi_{\sigma}(x)dx.\cr}$$
  
\noindent Now for any finite set $P\subseteq Q$,

\Centerline{$\int_{F'}\sum_{\sigma\in P,g(x)\in J^r_{\sigma}}
  \innerprod{h}{\phi_{\sigma}}\phi_{\sigma}(x)dx
=\sum_{\sigma\in P}\int_{F'\cap g^{-1}[J^r_{\sigma}]}
  \innerprod{h}{\phi_{\sigma}}\phi_{\sigma}$;}
  
\noindent if you like, you can think of this as an application of
Fubini's theorem, if you give counting measure to $Q$ and look at the 
function

$$\eqalign{(x,\sigma)
&\mapsto\innerprod{h}{\phi_{\sigma}}\phi_{\sigma}(x)
  \text{ if }x\in F',\,\sigma\in P\text{ and }g(x)\in J^r_{\sigma},
\cr&\mapsto 0\text{ otherwise}.\cr}$$  

\noindent But this means that

\Centerline{$|\int_{F'}\sum_{\sigma\in P,g(x)\in J^r_{\sigma}}
  \innerprod{h}{\phi_{\sigma}}\phi_{\sigma}(x)dx|
\le\sum_{\sigma\in P}|\innerprod{h}{\phi_{\sigma}}
   \int_{F'\cap g^{-1}[J^r_{\sigma}]}\phi_{\sigma}|
\le C_9\|h\|_2\sqrt{\mu F'}$}

\noindent by 286N.   Taking the limit as $P\to\Cal F$,

\Centerline{$|\int_{F'}\sum_{i=0}^nv_i\times\chi E_i|
\le C_9\|h\|_2\sqrt{\mu F'}$.}

\medskip

\quad{\bf (iii)} Thus we have

$$\eqalign{\int_F\sup_{i\le n}
   |2\pi(\varhat h\times\theta_{z_i})\varspcheck|
=\int_Fv
&\le 4|\int_{F'}\sum_{i=0}^nv_i\times\chi E_i|
\cr&\le 4C_9\|h\|_2\sqrt{\mu F'}
\le 4C_9\|h\|_2\sqrt{\mu F}.\cr}$$

\medskip

{\bf (c)} As $z_0,\ldots,z_n$ were arbitrary,
$\int_FAh\le 4C_9\|h\|_2\sqrt{\mu F}$, as claimed.
}%end of proof of 286P

\leader{286Q}{Lemma} For $\alpha>0$ and $y$, $z$, $\beta\in\Bbb R$, set
$\theta'_{z\alpha\beta}(y)=\theta_{\alpha z+\beta}(\alpha y+\beta)$.
Then

(a) the function $(\alpha,\beta,y,z)\mapsto\theta'_{z\alpha\beta}(y):
\ooint{0,\infty}\times\BbbR^3\to[0,1]$ is Borel measurable;

(b) $\theta'_{z\alpha\beta}(y)=0$ whenever $y\ge z$;

(c) for any rapidly decreasing test function $h$, and any $z\in\Bbb R$,

\Centerline{$2\pi|(\varhat h\times\theta'_{z\alpha\beta})\varspcheck|
\le D_{1/\alpha}AM_{\beta}D_{\alpha}h$}

\noindent\cmmnt{(in the notation of 286C)} at every point.

\proof{{\bf (a)} We need only recall that
$(y,z)\mapsto\theta_z(y):\BbbR^2\to\Bbb R$
is Borel measurable (286Oa), and that
$(\alpha,\beta,y,z)\mapsto\theta'_{z\alpha\beta}(y)$ is built up from
this, $+$ and $\times$.

\medskip

{\bf (b)} Again, this is immediate from 286Oa, because $\alpha>0$.

\medskip

{\bf (c)} Set $v=\alpha z+\beta$, so that
$\theta'_{z\alpha\beta}=D_{\alpha}S_{\beta}\theta_v$.   Then

$$\eqalign{\varhat h\times\theta'_{z\alpha\beta}
&=\varhat h\times D_{\alpha}S_{\beta}\theta_v
=D_{\alpha}S_{\beta}(S_{-\beta}D_{1/\alpha}\varhat h\times\theta_v)\cr
&=\alpha D_{\alpha}S_{\beta}
  (S_{-\beta}(D_{\alpha}h)\varsphat\times\theta_v)
=\alpha D_{\alpha}S_{\beta}
  ((M_{\beta}D_{\alpha}h)\varsphat\times\theta_v),\cr}$$

\noindent so

$$\eqalign{(\varhat h\times\theta'_{z\alpha\beta})\varspcheck
&=\alpha\bigl(D_{\alpha}S_{\beta}
  ((M_{\beta}D_{\alpha}h)\varsphat\times\theta_v)\bigr)\varspcheck\cr
&=D_{1/\alpha}\bigl(S_{\beta}
  ((M_{\beta}D_{\alpha}h)\varsphat\times\theta_v)\bigr)\varspcheck
=D_{1/\alpha}M_{-\beta}\bigl(
  (M_{\beta}D_{\alpha}h)\varsphat\times\theta_v\bigr)\varspcheck\cr}$$

\noindent and

\Centerline{$2\pi|(\varhat h\times\theta'_{z\alpha\beta})\varspcheck|
=2\pi D_{1/\alpha}\bigl|\bigl(
  (M_{\beta}D_{\alpha}h)\varsphat
  \times\theta_v\bigr)\varspcheck\bigr|
\le D_{1/\alpha}A(M_{\beta}D_{\alpha}h)$.}
}%end of proof of 286Q

\leader{286R}{Lemma} For any $y$, $z\in\Bbb R$,

\Centerline{$\tilde\theta_z(y)
=\int_1^2\Bover1{\alpha}\bigl(\lim_{n\to\infty}
  \Bover1n\int_0^n\theta'_{z\alpha\beta}(y)d\beta\bigr)d\alpha$}

\noindent is defined, and

$$\eqalign{\tilde\theta_z(y)&=\tilde\theta_1(0)>0\text{ if }y<z,\cr
&=0\text{ if }y\ge z.\cr}$$

\proof{{\bf (a)} The case $y\ge z$ is trivial, because if $y\ge z$ then
$\theta'_{z\alpha\beta}(y)=0$ for all $\alpha>0$ and
$\beta\in\Bbb R$ (286Qb) and $\tilde\theta_z(y)=0$.   For the rest of the proof,
therefore, I look at the case $y<z$.

\medskip

{\bf (b)(i)} Given $y<z\in\Bbb R$ and $\alpha>0$, set
$l=\lfloor\log_2(20\alpha(z-y))\rfloor$.   Then
$\theta'_{z,\alpha,\beta+2^l}(y)=\theta'_{z\alpha\beta}(y)$ for every
$\beta\in\Bbb R$.   \Prf\ If
$\theta'_{z\alpha\beta}(y)=\theta_{\alpha z+\beta}(\alpha y+\beta)$ is
non-zero, there must be $k$, $m\in\Bbb Z$ such that

\Centerline{$2^k(m+\Bover12)\le\alpha z+\beta<2^k(m+1)$}

\noindent and

\Centerline{$\varhat{\phi}(2^{-k}(\alpha y+\beta)-(m+\Bover14))^2
=\theta'_{z\alpha\beta}(y)\ne 0$,}

\noindent so

\Centerline{$2^km\le\alpha y+\beta\le 2^k(m+\Bover9{20})$}

\noindent because $\varhat\phi$ is zero outside $[-\bover15,\bover15]$.
In this case,
$\Bover1{20}\cdot 2^k<\alpha(z-y)$, so that $k\le l$.   We therefore
have

\Centerline{$2^k(m+2^{l-k}+\Bover12)\le\alpha z+\beta+2^l
<2^k(m+2^{l-k}+1)$,}

\Centerline{$2^k(m+2^{l-k})\le\alpha y+\beta+2^l
<2^k(m+2^{l-k}+\Bover12)$,}

\noindent so

\Centerline{$\theta'_{z,\alpha,\beta+2^l}(y)
=\varhat{\phi}(2^{-k}(\alpha y+\beta+2^l)-(m+2^{l-k}+\Bover14))^2
=\theta'_{z\alpha\beta}(y)$.}

\noindent Similarly,

\Centerline{$2^k(m-2^{l-k}+\Bover12)\le\alpha z+\beta-2^l
<2^k(m-2^{l-k}+1)$,}

\Centerline{$2^k(m-2^{l-k})\le\alpha y+\beta-2^l
<2^k(m-2^{l-k}+\Bover12)$,}

\noindent so

\Centerline{$\theta'_{z,\alpha,\beta-2^l}(y)
=\varhat{\phi}(2^{-k}(\alpha y+\beta-2^l)-(m-2^{l-k}+\Bover14))^2
=\theta'_{z\alpha\beta}(y)$.}

\noindent What this shows is that
$\theta'_{z,\alpha,\beta+2^l}(y)=\theta'_{z\alpha\beta}(y)$ if either is
non-zero, so we have the equality in any case.\ \Qed

\medskip

{\bf (ii)} It follows that $g(\alpha,y,z)
=\lim_{b\to\infty}\Bover1b\biggerint_0^b\theta'_{z\alpha\beta}(y)d\beta$ is
defined.   \Prf\ Set

\Centerline{$\gamma
=2^{-l}\int_0^{2^l}\theta'_{z\alpha\beta}(y)d\beta$.}

\noindent From (i) we see that

\Centerline{$\gamma
=2^{-l}\int_{2^lm}^{2^l(m+1)}\theta'_{z\alpha\beta}(y)d\beta$}

\noindent for every $m\in\Bbb Z$, and therefore that

\Centerline{$\gamma
=\Bover1{2^lm}\int_0^{2^lm}\theta'_{z\alpha\beta}(y)d\beta$}

\noindent for every $m\ge 1$.   Now $\theta'_{z\alpha\beta}(y)$ is always
greater than or equal to $0$, so if $2^lm\le b\le 2^l(m+1)$ then

\Centerline{$\Bover{m}{m+1}\gamma
=\Bover1{2^l(m+1)}\int_0^{2^lm}\theta'_{z\alpha\beta}
\le\Bover1b\int_0^b\theta'_{z\alpha\beta}
\le\Bover1{2^lm}\int_0^{2^l(m+1)}\theta'_{z\alpha\beta}
=\Bover{m+1}{m}\gamma$,}

\noindent which approach $\gamma$ as $b\to\infty$.\ \Qed

\medskip

{\bf (c)} Because $(\alpha,y)\mapsto\theta'_{z\alpha\beta}(y)$ is
always Borel measurable, each of the functions
$\alpha\mapsto\Bover1n\int_0^n\theta'_{z\alpha\beta}$, for
$n\ge 1$, is Borel measurable (putting 251M and 252P together), and
$\alpha\mapsto g(\alpha,y,z):\ooint{0,\infty}\to\Bbb R$ is Borel
measurable;  at the same time, since
$0\le\theta'_{z\alpha\beta}(y)\le 1$ for all $\alpha$ and $\beta$,
$0\le g(\alpha,y,z)\le 1$ for every $\alpha$, and
$\tilde\theta_z(y)=\int_1^2\Bover1{\alpha}g(\alpha,y,z)d\alpha$ is
defined in $[0,1]$.

\medskip

{\bf (d)} For any $y<z$, $\gamma\in\Bbb R$ and $\alpha>0$,
$g(\alpha,y+\gamma,z+\gamma)=g(\alpha,y,z)$.   \Prf\ It is enough to
consider the case $\gamma\ge 0$.   In this case

$$\eqalign{g(\alpha,y+\gamma,z+\gamma)
&=\lim_{b\to\infty}\Bover1b\int_0^b
  \theta'_{z+\gamma,\alpha,\beta}(y+\gamma)d\beta\cr
&=\lim_{b\to\infty}\Bover1b\int_0^b
  \theta_{\alpha z+\alpha\gamma+\beta}
    (\alpha y+\alpha\gamma+\beta)d\beta\cr
&=\lim_{b\to\infty}\Bover1b\int_{\alpha\gamma}^{b+\alpha\gamma}
  \theta_{\alpha z+\beta}(\alpha y+\beta)d\beta
=\lim_{b\to\infty}\Bover1b\int_{\alpha\gamma}^{b+\alpha\gamma}
  \theta'_{z\alpha\beta}(y)d\beta,\cr}$$

\noindent so

$$\eqalign{|g(\alpha,y+\gamma,z+\gamma)-g(\alpha,y,z)|
&=\lim_{b\to\infty}\Bover1b\bigl
  |\int_b^{b+\alpha\gamma}\theta'_{z\alpha\beta}(y)d\beta
  -\int_0^{\alpha\gamma}\theta'_{z\alpha\beta}(y)d\beta\bigr|\cr
&\le\lim_{b\to\infty}\Bover{2\alpha\gamma}{b}=0. \text{ \Qed}\cr}$$

It follows that whenever $y<z$ and $\gamma\in\Bbb R$,

\Centerline{$\tilde\theta_{z+\gamma}(y+\gamma)
=\int_{0}^{1}\Bover1{\alpha}g(\alpha,y+\gamma,z+\gamma)d\alpha
=\int_{0}^{1}\Bover1{\alpha}g(\alpha,y,z)d\alpha
=\tilde\theta_z(y)$.}

\medskip

{\bf (e)} The next essential fact to note is that $\theta_{2z}(2y)$ is
always equal to $\theta_z(y)$.   \Prf\ If $\theta_z(y)\ne 0$, then (as
in (b) above) there are $k$, $m\in\Bbb Z$ such that

\Centerline{$2^k(m+\Bover12)\le z<2^k(m+1)$,
\quad$2^km\le y<2^k(m+\Bover12)$,
\quad$\theta_z(y)=\varhat{\phi}(2^{-k}y-(m+\Bover14))^2$.}

\noindent In this case,

\Centerline{$2^{k+1}(m+\Bover12)\le 2z<2^{k+1}(m+1)$,
\quad$2^{k+1}m\le 2y<2^{k+1}(m+\Bover12)$,}

\noindent so

\Centerline{$\theta_{2z}(2y)
=\varhat{\phi}(2^{-k-1}\cdot 2y-(m+\Bover14))^2
=\theta_z(y)$.}

\noindent Similarly,

\Centerline{$2^{k-1}(m+\Bover12)\le \Bover12z<2^{k-1}(m+1)$,
\quad$2^{k-1}m\le \Bover12y<2^{k-1}(m+\Bover12)$,}

\noindent so

\Centerline{$\theta_{\bover12z}(\Bover12y)
=\varhat{\phi}(2^{-k+1}\cdot\Bover12y-(m+\Bover14))^2
=\theta_z(y)$.}

\noindent This shows that $\theta_{2z}(2y)=\theta_z(y)$ if either is
non-zero, and therefore in all cases.\ \Qed

Accordingly

\Centerline{$\theta'_{z,2\alpha,2\beta}(y)
=\theta_{2\alpha z+2\beta}(2\alpha y+2\beta)
=\theta_{\alpha z+\beta}(\alpha y+\beta)
=\theta'_{z\alpha\beta}(y)$}

\noindent for all $y$, $z$, $\beta\in\Bbb R$ and all $\alpha>0$.

\medskip

{\bf (f)} Consequently

$$\eqalign{g(2\alpha,y,z)
&=\lim_{b\to\infty}\Bover1b\int_0^b\theta'_{z,2\alpha,\beta}(y)d\beta
=\lim_{b\to\infty}\Bover2b\int_0^{b/2}
  \theta'_{z,2\alpha,2\beta}(y)d\beta\cr
&=\lim_{b\to\infty}\Bover2b\int_0^{b/2}
  \theta'_{z\alpha\beta}(y)d\beta
=\lim_{b\to\infty}\Bover1b\int_0^b\theta'_{z\alpha\beta}(y)d\beta
=g(\alpha,y,z)\cr}$$

\noindent whenever $\alpha>0$ and $y$, $z\in\Bbb R$.   It follows that

\Centerline{$\int_{\gamma}^{\delta}
   \Bover1{\alpha}g(\alpha,y,z)d\alpha
=\int_{\gamma}^{\delta}\Bover1{\alpha}g(2\alpha,y,z)d\alpha
=\int_{2\gamma}^{2\delta}\Bover1{\alpha}g(\alpha,y,z)d\alpha$}

\noindent whenever $0<\gamma\le\delta$, and therefore that

\Centerline{$\int_{\gamma}^{2\gamma}
  \Bover1{\alpha}g(\alpha,y,z)d\alpha
=\int_1^2\Bover1{\alpha}g(\alpha,y,z)d\alpha$}

\noindent for every $\gamma>0$.   \Prf\ Take $k\in\Bbb Z$ such that
$2^k\le\gamma<2^{k+1}$.   Then

$$\eqalign{\int_{\gamma}^{2\gamma}\Bover1{\alpha}g(\alpha,y,z)d\alpha
&=\int_{2^k}^{2^{k+1}}\Bover1{\alpha}g(\alpha,y,z)d\alpha
  -\int_{2^k}^{\gamma}\Bover1{\alpha}g(\alpha,y,z)d\alpha
  +\int_{2^{k+1}}^{2\gamma}\Bover1{\alpha}g(\alpha,y,z)d\alpha\cr
&=\int_{2^k}^{2^{k+1}}\Bover1{\alpha}g(\alpha,y,z)d\alpha
=\int_1^2\Bover1{\alpha}g(\alpha,y,z)d\alpha.\text{  \Qed}\cr}$$

\medskip

{\bf (g)} Now if $\alpha$, $\gamma>0$ and $y<z$,

\Centerline{$g(\alpha,\gamma y,\gamma z)
=\lim_{b\to\infty}\Bover1b\int_0^b
  \theta_{\alpha\gamma z+\beta}(\alpha\gamma y+\beta)d\beta
=g(\alpha\gamma,y,z)$.}

\noindent So if $\gamma>0$ and $y<z$,

$$\eqalign{\tilde\theta_{\gamma z}(\gamma y)
&=\int_1^2\Bover1{\alpha}g(\alpha,\gamma y,\gamma z)d\alpha
=\int_1^2\Bover1{\alpha}g(\alpha\gamma,y,z)d\alpha\cr
&=\int_{\gamma}^{2\gamma}\Bover1{\alpha}g(\alpha,y,z)d\alpha
=\int_1^2\Bover1{\alpha}g(\alpha,y,z)d\alpha
=\tilde\theta_z(y).\cr}$$

\noindent Putting this together with (d), we see that if $y<z$ then

\Centerline{$\tilde\theta_z(y)
=\tilde\theta_{z-y}(0)
=\tilde\theta_1(0)$.}

\medskip

{\bf (h)} I have still to check that $\tilde\theta_1(0)$ is not zero.
But suppose that $1\le\alpha<\bover76$ and that there is some
$m\in\Bbb Z$ such that $2(m+\bover1{12})\le\beta\le 2(m+\bover5{12})$.
Then $2(m+\bover12)\le\alpha+\beta<2(m+1)$, while
$|\bover12\beta-(m+\bover14)|\le\bover16$, so

\Centerline{$\theta_{\alpha+\beta}(\beta)
=\varhat{\phi}(\Bover12\beta-(m+\Bover14))^2
=1$.}

\noindent What this means is that, for $1\le\alpha<\bover76$,

$$\eqalign{g(\alpha,0,1)
&=\lim_{m\to\infty}\Bover1{2m}\int_0^{2m}
  \theta_{\alpha+\beta}(\beta)d\beta\cr
&\ge\lim_{m\to\infty}\Bover{1}{2m}
  \sum_{j=0}^{m-1}\mu[2(j+\Bover1{12}),2(j+\Bover5{12})]
=\Bover13.\cr}$$

\noindent So

\Centerline{$\tilde\theta_1(0)
=\int_1^2\Bover1{\alpha}g(\alpha,0,1)d\alpha
\ge\Bover13\int_1^{7/6}\Bover1{\alpha}d\alpha
>0$.}

\noindent This completes the proof.
}%end of proof of 286R

\leader{286S}{Lemma} Suppose that $h$ is a rapidly decreasing test function.

(a) For every $x\in\Bbb R$,

\Centerline{$(\tilde Ah)(x)
=\liminf_{n\to\infty}\Bover1n\int_1^2\Bover1{\alpha}\int_0^n
  (D_{1/\alpha}AM_{\beta}D_{\alpha}h)(x)d\beta d\alpha$}

\noindent is defined in $[0,\infty]$, and
$\tilde Ah:\Bbb R\to[0,\infty]$ is Borel measurable.

(b) $\int_F\tilde Ah\le 3C_9\|h\|_2\sqrt{\mu F}$ whenever $\mu F<\infty$.

(c) If $z\in\Bbb R$,
$2\pi|(\varhat h\times\tilde\theta_z)\varspcheck|\le\tilde Ah$ at every
point.

\proof{{\bf (a)} The point here is that the function

\Centerline{$(\alpha,\beta,x)
\mapsto(D_{1/\alpha}AM_{\beta}D_{\alpha}h)(x):
  \ooint{0,\infty}\times\BbbR^2\to[0,\infty]$}

\noindent is Borel measurable.   \Prf\

$$\eqalign{(D_{1/\alpha}AM_{\beta}D_{\alpha}h)(x)
&=(AM_{\beta}D_{\alpha}h)(\Bover{x}{\alpha})
\cr&=\sup_{z\in\Bbb R}
  |2\pi((M_{\beta}D_{\alpha}h)\varsphat\times\theta_z)\varspcheck
     (\Bover{x}{\alpha})|
\cr&=\Bover{2\pi}{\alpha}\sup_{z\in\Bbb R}
  |(S_{-\beta}D_{1/\alpha}\varhat h\times\theta_z)\varspcheck
     (\Bover{x}{\alpha})|.
\cr}$$
  
\noindent Now, for any $z\in\Bbb R$, 

\Centerline{$(S_{-\beta}D_{1/\alpha}\varhat h\times\theta_z)\varspcheck
     (\bover{x}{\alpha})
=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{ixy/\alpha}
   \varhat h(\bover{y-\beta}{\alpha})\theta_z(y)dy$.}

\noindent We know that $\varhat h$ is a rapidly decreasing test function, 
so there is some 
$\gamma\ge 0$ such that $|\varhat h(t)|\le\Bover{\gamma}{1+t^2}$ for every
$t\in\Bbb R$.   This means that if $\alpha>0$ and $\beta\in\Bbb R$
and $\sequencen{\alpha_n}$,
$\sequencen{\beta_n}$ are sequences in $\ocint{0,2\alpha}$ and
$[\beta-1,\beta+1]$, converging to $\alpha$, $\beta$ respectively, 
and we set $g(t)
=\sup_{n\in\Bbb N}|\varhat h(\Bover{t-\beta_n}{\alpha_n})\theta_z(t)|$,
then 

$$\eqalign{g(t)
&\le\Bover{4\gamma\alpha^2}{(|t|-|\beta|-1)^2}\text{ if }|t|\ge|\beta|+2,
  \cr
&\le\gamma\text{ otherwise},\cr}$$

\noindent and $g$ is integrable.   (Remember that 
$0\le\theta_z(y)\le 1$ for every $y$, as noted in 286Oa.)   So Lebesgue's
Dominated Convergence Theorem tells us that if
$\sequencen{\alpha_n}\to\alpha$ and $\sequencen{\beta_n}\to\beta$ and
$\sequencen{x_n}\to x$, then

\Centerline{$\int_{-\infty}^{\infty}e^{ix_ny/\alpha_n}
   \varhat h(\bover{y-\beta_n}{\alpha_n})\theta_z(y)dy
\to\int_{-\infty}^{\infty}e^{ixy/\alpha}
   \varhat h(\bover{y-\beta}{\alpha})\theta_z(y)dy$.}
   
\noindent Thus 
$(\alpha,\beta,x)
\mapsto(S_{-\beta}D_{1/\alpha}\varhat h\times\theta_z)\varspcheck
(\bover{x}{\alpha})$ is continuous;  and this is true for every
$z\in\Bbb R$.   Consequently

\Centerline{$(\alpha,\beta,x)\mapsto\sup_{z\in\Bbb R}
  |(S_{-\beta}D_{1/\alpha}\varhat h\times\theta_z)\varspcheck
     (\Bover{x}{\alpha})|$}   

\noindent and $(\alpha,\beta,x)
\mapsto(D_{1/\alpha}AM_{\beta}D_{\alpha}h)(x)$ are lower semi-continuous,
therefore Borel measurable, by 256Ma again.\ \Qed

It follows that the repeated integrals

\Centerline{$\int_1^2\Bover1{\alpha}\int_0^n
(D_{1/\alpha}AM_{\beta}D_{\alpha}h)(x)d\beta d\alpha$}

\noindent are defined in $[0,\infty]$ and are Borel measurable functions
of $x$ (252P again), so that $\tilde Af$ is Borel measurable.

\medskip

{\bf (b)} For any $n\in\Bbb N$,

$$\eqalignno{\int_F\Bover1n\int_1^2\Bover1\alpha\int_0^n
&(D_{1/\alpha}AM_{\beta}D_{\alpha}h)(x)d\beta d\alpha dx\cr
&=\Bover1n\int_1^2\Bover1\alpha\int_0^n\!\int_F
  (D_{1/\alpha}AM_{\beta}D_{\alpha}h)(x)dx d\beta d\alpha\cr
\displaycause{by Fubini's theorem, 252H}
&=\Bover1n\int_1^2\!\int_0^n\!\int_F\Bover1\alpha
  (AM_{\beta}D_{\alpha}h)(\Bover{x}{\alpha})dx d\beta d\alpha\cr
&=\Bover1n\int_1^2\!\int_0^n\int_{\alpha^{-1}F}
  (AM_{\beta}D_{\alpha}h)(x)dx d\beta d\alpha\cr
&\le\Bover1n\int_1^2\!\int_0^n 4C_9
  \|M_{\beta}D_{\alpha}h\|_2\sqrt{\mu(\alpha^{-1}F)} d\beta d\alpha\cr
\displaycause{286P}
&=4C_9\cdot\Bover1n\int_1^2\!\int_0^n
  \Bover1{\sqrt{\alpha}}\|h\|_2\cdot\Bover1{\sqrt{\alpha}}\sqrt{\mu F}
    d\beta d\alpha\cr
&=4C_9\|h\|_2\sqrt{\mu F}\cdot\Bover1n\int_1^2\Bover1{\alpha}\int_0^n
 d\beta d\alpha\cr
&=4C_9\|h\|_2\sqrt{\mu F}\ln 2
\le 3C_9\|h\|_2\sqrt{\mu F}.\cr}$$

\noindent So

$$\eqalignno{\int_F\tilde Ah
&=\int_F\liminf_{n\to\infty}\Bover1n\int_1^2\Bover1\alpha\int_0^n
  (D_{1/\alpha}AM_{\beta}D_{\alpha}h)(x)d\beta d\alpha dx\cr
&\le\liminf_{n\to\infty}\int_F\Bover1n\int_1^2\Bover1\alpha\int_0^n
  (D_{1/\alpha}AM_{\beta}D_{\alpha}h)(x)d\beta d\alpha dx\cr
\displaycause{by Fatou's lemma}
&\le 3C_9\|h\|_2\sqrt{\mu F}.\cr}$$

\medskip

{\bf (c)} For any $x\in\Bbb R$,

\Centerline{$\int_{-\infty}^{\infty}
  |\varhat h(y)|\int_1^2\Bover1{\alpha}
  \bigl(\sup_{n\in\Bbb N}\Bover1n\int_0^n
  \theta'_{z\alpha\beta}(y)d\beta\bigr)d\alpha dy
\le\ln 2\cdot\int_{-\infty}^{\infty}|\varhat h|$}

\noindent is finite.   So

$$\eqalignno{(\varhat h\times\tilde\theta_z)\varspcheck(x)
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
  e^{ixy}\varhat h(y)\tilde\theta_z(y)dy\cr
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
  e^{ixy}\varhat h(y)\int_1^2\Bover1{\alpha}\lim_{n\to\infty}
  \Bover1n\int_0^n\theta'_{z\alpha\beta}(y)d\beta d\alpha dy\cr
&=\Bover1{\sqrt{2\pi}}\lim_{n\to\infty}\int_{-\infty}^{\infty}
  e^{ixy}\varhat h(y)\int_1^2\Bover1{\alpha n}
  \int_0^n\theta'_{z\alpha\beta}(y)d\beta d\alpha dy\cr
\displaycause{by Lebesgue's Dominated Convergence Theorem}
&=\Bover1{\sqrt{2\pi}}\lim_{n\to\infty}\int_1^2\Bover1{\alpha n}
  \int_0^n\int_{-\infty}^{\infty}
  e^{ixy}\varhat h(y)\theta'_{z\alpha\beta}(y)dy d\beta d\alpha\cr
\displaycause{by Fubini's theorem}
&=\lim_{n\to\infty}\int_1^2\Bover1{\alpha n}
  \int_0^n(\varhat h\times\theta'_{z\alpha\beta})\varspcheck(x)
  d\beta d\alpha,\cr}$$

\noindent and

$$\eqalignno{2\pi|(\varhat h\times\tilde\theta_z)\varspcheck(x)|
&=2\pi\bigl|\lim_{n\to\infty}\int_1^2\Bover1{\alpha n}
  \int_0^n(\varhat h\times\theta'_{z\alpha\beta})\varspcheck(x)
  d\beta d\alpha\bigr|\cr
&\le 2\pi\liminf_{n\to\infty}\int_1^2\Bover1{\alpha n}
  \int_0^n|(\varhat h\times\theta'_{z\alpha\beta})\varspcheck(x)|
  d\beta d\alpha\cr
&\le\liminf_{n\to\infty}\int_1^2\Bover1{\alpha n}
  \int_0^n(D_{1/\alpha}AM_{\beta}D_{\alpha}h)(x)
  d\beta d\alpha\cr
\displaycause{286Qb}
&=(\tilde Ah)(x).\cr}$$
}%end of proof of 286S

\leader{286T}{Lemma} Set $C_{10}=3C_9/\pi\tilde\theta_1(0)$.   For
$f\in\eusm L^2_{\Bbb C}$, define $\hat Af:\Bbb R\to[0,\infty]$ by
setting

\Centerline{$(\hat Af)(y)
=\sup_{a\le b}\Bover1{\sqrt{2\pi}}|\int_a^be^{-ixy}f(x)dx|$}

\noindent for each $y\in\Bbb R$.   Then
$\int_F\hat Af\le C_{10}\|f\|_2\sqrt{\mu F}$ whenever $\mu F<\infty$.

\proof{{\bf (a)} As usual, the first step is to confirm that $\hat Af$
is measurable.   \Prf\ For $a\le b$,
$y\mapsto|\Bover1{\sqrt{2\pi}}\int_a^be^{-ixy}f(x)dx|$ is continuous (by
283Cf, since $f\times\chi[a,b]$ is integrable), so $\hat Af$ is lower
semi-continuous, therefore Borel measurable (256Ma once more).\ \Qed

\medskip

{\bf (b)} Suppose that $h$ is a rapidly decreasing test function.   Then

\Centerline{$(\hat Ah)(y)
\le\Bover1{\pi\tilde\theta_1(0)}(\tilde A\varcheck h)(-y)$}

\noindent for every $y\in\Bbb R$.   \Prf\ If $a\in\Bbb R$ then

$$\eqalignno{\Bover1{\sqrt{2\pi}}|\int_{-\infty}^ae^{-ixy}h(x)dx|
&=\Bover1{\tilde\theta_1(0)\sqrt{2\pi}}|\int_{-\infty}^{\infty}
  e^{-ixy}\tilde\theta_a(x)h(x)dx|\cr
\displaycause{286R}
&=\Bover1{\tilde\theta_1(0)}|(h\times\tilde\theta_a)\varspcheck(-y)|
=\Bover1{\tilde\theta_1(0)}
  |(\varcheck h\varhat{\phantom{h}}
    \times\tilde\theta_a)\varspcheck(-y)|\cr
\displaycause{284C once more}
&\le\Bover1{2\pi\tilde\theta_1(0)}(\tilde A\varcheck{h})(-y)\cr}$$

\noindent (286Sc).   So if $a\le b$ in $\Bbb R$,

\Centerline{$\Bover1{\sqrt{2\pi}}|\int_{a}^{b}e^{-ixy}h(x)dx|
\le\Bover1{\pi\tilde\theta_1(0)}(\tilde A\varcheck{h})(-y)$;}

\noindent taking the supremum over $a$ and $b$, we have the result.\
\Qed

It follows that

$$\eqalignno{\int_F\hat Ah
&\le\Bover1{\pi\tilde\theta_1(0)}\int_{-F}\tilde A\varcheck h
\le\Bover3{\pi\tilde\theta_1(0)}C_9\|h\|_2\sqrt{\mu(-F)}\cr
\displaycause{286Sb, 284Oa}
&=C_{10}\|h\|_2\sqrt{\mu F}.\cr}$$

\medskip

{\bf (c)} For general square-integrable $f$, take any $\epsilon>0$ and
any $n\in\Bbb N$.   Set

\Centerline{$(\hat A_nf)(y)
=\sup_{-n\le a\le b\le n}\Bover1{\sqrt{2\pi}}|\int_a^be^{-ixy}f(x)dx|$}

\noindent for each $y\in\Bbb R$.   Let $h$ be a rapidly decreasing test
function such that $\|f-h\|_2\le\epsilon$ (284N).   Then

\Centerline{$\hat Ah
\ge\hat A_nh
\ge\hat A_nf-\Bover{\sqrt{2n}}{\sqrt{2\pi}}\epsilon$}

\noindent (using Cauchy's inequality), so

\Centerline{$\int_F\hat A_nf
\le\int_F\hat Ah+\sqrt{\bover{n}{\pi}}\epsilon\mu F
\le C_{10}(\|f\|_2+\epsilon)\sqrt{\mu F}
  +\sqrt{\bover{n}{\pi}}\epsilon\mu F$.}

\noindent As $\epsilon$ is arbitrary,
$\int_F\hat A_nf\le C_{10}\|f\|_2\sqrt{\mu F}$;  letting $n\to\infty$,
we get $\int_F\hat Af\le C_{10}\|f\|_2\sqrt{\mu F}$.
}%end of proof of 286T

\vleader{72pt}{286U}{Theorem} If $f\in\eusm L^2_{\Bbb C}$ then

\Centerline{$g(y)
=\lim_{a\to-\infty,b\to\infty}
  \Bover1{\sqrt{2\pi}}\int_a^be^{-ixy}f(x)dx$}

\noindent is defined in $\Bbb C$ for almost every $y\in\Bbb R$, and $g$
represents the Fourier transform of $f$.

\proof{{\bf (a)} For $n\in\Bbb N$, $y\in\Bbb R$ set

\Centerline{$\gamma_n(y)=\sup_{a\le -n,b\ge n}\Bover1{\sqrt{2\pi}}
  \bigl|\int_a^be^{-ixy}f(x)dx-\int_{-n}^ne^{-ixy}f(x)dx\bigr|$.}

\noindent Then $g(y)$ is defined whenever
$\inf_{n\in\Bbb N}\gamma_n(y)=0$.   \Prf\ If
$\inf_{n\in\Bbb N}\gamma_n(y)=0$ and $\epsilon>0$, take $m\in\Bbb N$
such that $\gamma_m(y)\le\bover12\epsilon$;  then
$\Bover1{\sqrt{2\pi}}
|\int_a^be^{-ixy}f(x)dx-\int_{-n}^ne^{-ixy}f(x)dx|\le\epsilon$ whenever
$n\ge m$, $a\le-n$ and $b\ge n$.   But this means, first, that
$\sequencen{\int_{-n}^ne^{-ixy}f(x)dx}$ is a Cauchy sequence, so has a
limit $\zeta$ say, and, second, that
$\zeta=\lim_{a\to-\infty,b\to\infty}\int_a^be^{-ixy}f(x)dx$, so that
$g(y)=\Bover{\zeta}{\sqrt{2\pi}}$ is defined.\ \Qed

Also each $\gamma_n$ is lower-semicontinuous (cf.\ part (a) of the
proof of 286T).

\medskip

{\bf (b)} \Quer\ Suppose, if possible, that
$\{y:\inf_{n\in\Bbb N}\gamma_n(y)>0\}$ is not negligible.   Then

\Centerline{$\lim_{m\to\infty}\mu\{y:|y|\le m,\,
  \inf_{n\in\Bbb N}\gamma_n(y)\ge\Bover1m\}
>0$,}

\noindent so there is an $\epsilon>0$ such that

\Centerline{$F=\{y:|y|\le\Bover1{\epsilon},\,
  \inf_{n\in\Bbb N}\gamma_n(y)\ge\epsilon\}$}

\noindent has measure greater than $\epsilon$.   Let $n\in\Bbb N$ be
such that

\Centerline{$4C^2_{10}
  (\int_{-\infty}^{\infty}|f(x)|^2dx-\int_{-n}^n|f(x)|^2dx)
\le\epsilon^3$,}

\noindent and set $f_1=f-f\times\chi[-n,n]$;  then
$2C_{10}\|f_1\|_2\le\epsilon^{3/2}$.

We have

$$\eqalign{\gamma_n(y)
&=\sup_{a\le -n,b\ge n}\Bover1{\sqrt{2\pi}}
  \bigl|\int_a^be^{-ixy}f_1(x)dx-\int_{-n}^ne^{-ixy}f_1(x)dx\bigr|\cr
&\le 2\sup_{a\le b}\Bover1{\sqrt{2\pi}}
  |\int_a^be^{-ixy}f_1(x)dx|
\le 2(\hat Af_1)(y),\cr}$$

\noindent so that

$$\eqalignno{\epsilon\mu F
&\le\int_F\gamma_n
\le 2\int_F\hat Af_1
\le 2C_{10}\|f_1\|_2\sqrt{\mu F}\cr
\displaycause{286T}
&\le\epsilon^{3/2}\sqrt{\mu F}\cr}$$

\noindent and $\mu F\le\epsilon$;  but we chose $\epsilon$ so that
$\mu F$ would be greater than $\epsilon$.\ \Bang

\medskip

{\bf (c)} Thus $g(y)$ is defined for almost every $y\in\Bbb R$.  Now $g$
represents the Fourier transform of $f$.   \Prf\ Let $h$ be a rapidly
decreasing test function.   The restriction of $\hat Af$ to the set
on which it is finite is a tempered function, by 286D, so
$\int_{-\infty}^{\infty}(\hat Af)\times|h|$ is finite, by 284F.   Now

$$\eqalignno{\int_{-\infty}^{\infty}g\times h
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}
  \bigl(\lim_{n\to\infty}\int_{-n}^ne^{-ixy}f(x)dx\bigr)h(y)dy\cr
&=\Bover1{\sqrt{2\pi}}\lim_{n\to\infty}\int_{-\infty}^{\infty}
  \int_{-n}^ne^{-ixy}f(x)h(y)dxdy\cr
\displaycause{because
$\Bover1{\sqrt{2\pi}}|\int_{-n}^ne^{-ixy}f(x)dx|\le\hat Af(y)$ for every
$n$ and $y$, so we can use Lebesgue's Dominated Convergence Theorem}
&=\Bover1{\sqrt{2\pi}}\lim_{n\to\infty}
  \int_{-n}^n\int_{-\infty}^{\infty}e^{-ixy}f(x)h(y)dydx\cr
\displaycause{because $\int_{-\infty}^{\infty}\int_{-n}^n|f(x)h(y)|dxdy$
is finite for each $n$}
&=\lim_{n\to\infty}
  \int_{-n}^nf\times\varhat{h}
=\int_{-\infty}^{\infty}f\times\varhat{h}\cr}$$

\noindent because $f\times\varhat{h}$ is certainly integrable.   As $h$
is arbitrary, $g$ represents the Fourier transform of $f$.\ \Qed
}%end of proof of 286U

\leader{286V}{Theorem} For any square-integrable complex-valued function
on $\ocint{-\pi,\pi}$, its sequence of
Fourier sums converges to it almost everywhere.

\proof{ Suppose that
$f\in\eusm L^2_{\Bbb C}(\mu_{\ocint{-\pi,\pi}})$.
Set $f_1(x)=f(x)$ for $x\in\dom f$, $0$ for
$x\in\Bbb R\setminus\ocint{-\pi,\pi}$;  then
$f_1\in\eusm L^2_{\Bbb C}(\mu)$.   Let $g\in\eusm L^2_{\Bbb C}(\mu)$
represent the inverse Fourier transform of $f_1$ (284O).   Then 286U
tells us that
$f_2(x)=\lim_{a\to\infty}\Bover1{\sqrt{2\pi}}\int_{-a}^ae^{-ixy}g(y)dy$
is defined for almost every $x$, and that $f_2$ represents the Fourier
transform of $g$, so is equal almost everywhere to $f_1$ (284Ib).

Now, for any $a\ge 0$, $x\in\Bbb R$,

$$\eqalignno{\int_{-a}^ae^{-ixy}g(y)dy
&=\innerprod{g}{h_{ax}}\cr
\displaycause{where $h_{ax}(y)=e^{ixy}$ if $|y|\le a$, $0$ otherwise}
&=\innerprod{f_2}{\varhat{h}_{ax}}\cr
\displaycause{284Ob}
&=\Bover1{\sqrt{2\pi}}\int_{-\infty}^{\infty}f_2(t)
  \overline{\int_{-\infty}^{\infty}e^{-ity}h_{ax}(y)dy}\,dt\cr
&=\Bover2{\sqrt{2\pi}}\int_{-\infty}^{\infty}
  \Bover{\sin(x-t)a}{x-t}f_2(t)dt
=\Bover2{\sqrt{2\pi}}\int_{-\pi}^{\pi}
  \Bover{\sin(x-t)a}{x-t}f(t)dt.\cr}$$

\noindent So

\Centerline{$f(x)=f_2(x)
=\lim_{a\to\infty}\Bover1{\pi}\int_{-\pi}^{\pi}
  \Bover{\sin(x-t)a}{x-t}f(t)dt$}

\noindent for almost every $x\in\ocint{-\pi,\pi}$.

On the other hand, writing $\sequencen{s_n}$ for the sequence of Fourier
sums of $f$, we have, for any $x\in\ooint{-\pi,\pi}$,

\Centerline{$s_n(x)
=\Bover1{2\pi}\int_{-\pi}^{\pi}f(t)
\Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})(x-t)}
  {\sin{1\over 2}(x-t)}dt$}

\noindent for each $n$, by 282Da.   Now

$$\eqalign{\Bover1{2\pi}\int_{-\pi}^{\pi}f(t)
  &\Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})(x-t)}
  {\sin{1\over 2}(x-t)}dt
-\Bover1{\pi}\int_{-\pi}^{\pi}f(t)
  \Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})(x-t)}
  {x-t}dt\cr
&=\Bover1{\pi}\int_{-\pi}^{\pi}f(t)
  \bigl(\Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})(x-t)}
    {2\sin{1\over 2}(x-t)}
  -\Bover{\sin(n+\bover12\pushbottom{3.5pt})(x-t)}{x-t}\bigr)dt\cr
&=\Bover1{\pi}\int_{x-\pi}^{x+\pi}\bigl
  (\Bover1{2\sin{1\over 2}t}-\Bover1{t}\bigr)
  f(x-t)\sin(n+\hbox{$\bover12$})t\,dt.\cr}$$

\noindent But if we look at the function

$$\eqalign{p_x(t)
&=\bigl(\Bover1{2\sin\bover12t}-\Bover1t\bigr)f(x-t)
  \text{ if }x-\pi<t<x+\pi\text{ and }t\ne 0,\cr
&=0\text{ otherwise},\cr}$$

\noindent $p_x$ is integrable, because $f$ is integrable over
$\ocint{-\pi,\pi}$ and
$\lim_{t\to 0}\Bover1{2\sin\bover12t}-\Bover1t=0$, so
$\sup_{t\ne 0,x-\pi\le t\le x+\pi}|\Bover1{2\sin\bover12t}-\Bover1t|$ is
finite.   (This is where we need to know that $|x|<\pi$.)   So

$$\eqalign{\lim_{n\to\infty}s_n(x)-\Bover1{\pi}\int_{-\pi}^{\pi}f(t)
  \Bover{\sin(n+{1\over 2}\pushbottom{3.5pt})(x-t)}
  {x-t}dt
=\lim_{n\to\infty}\int_{-\infty}^{\infty}
       p_x(t)\sin(n+\hbox{$\bover12$})t\,dt
=0\cr}$$

\noindent by the Riemann-Lebesgue lemma (282Fb).   But this means that
$\lim_{n\to\infty}s_n(x)=f(x)$ for any $x\in\ooint{-\pi,\pi}$ such that
$f(x)=\lim_{a\to\infty}\Bover1{\pi}\int_{-\pi}^{\pi}
  \Bover{\sin(x-t)a}{x-t}f(t)dt$, which is almost every
$x\in\ocint{-\pi,\pi}$.
}%end of proof of 286V

\cmmnt{
\vleader{108pt}{286W}{Glossary} The following special notations are
used in more than one paragraph of this section:

\def\vta{$\mu$ for Lebesgue measure on $\Bbb R$.}
\def\vtb{286A:  $f^*$.}
\def\vtc{286C:  $S_{\alpha}f$, $M_{\alpha}f$, $D_{\alpha}f$.}
\def\vtd{286Ea:  $\Cal I$, $Q$, $I_{\sigma}$, $J_{\sigma}$,
  $k_{\sigma}$, $x_{\sigma}$, $y_{\sigma}$, $J^l_{\sigma}$,
  $J^r_{\sigma}$, $y^l_{\sigma}$.}
\def\vte{286Eb:  $\phi$, $\phi_{\sigma}$, $\innerprod{f}{g}$.}
\def\vtf{286Ec:  $w$, $w_{\sigma}$.}
\def\vtg{286F:  $\le$, $R^+$, $T_{\tau}$.}
\def\vth{286G:  $C_1$, $C_2$, $C_3$, $C_4$.}
\def\vti{286H:  $\mass$, $\Delta_f$, $\energy$.}
\def\vtj{286J:  $C_5$.}
\def\vtk{286K:  $C_6$.}
\def\vtl{286L:  $C_7$.}
\def\vtm{286M:  $C_8$.}
\def\vtn{286N:  $C_9$.}
\def\vto{286O:  $\theta_z$, $\Cal F$.}
\def\vtp{286P:  $Ah$.}
\def\vtq{286Q:  $\theta'_{z\alpha\beta}$.}
\def\vtr{286R:  $\tilde\theta_z$.}
\def\vts{286S:  $\tilde Ah$.}
\def\vtt{286T:  $C_{10}$, $\hat Af$.}
\sparewidth=\pagewidth
\advance\sparewidth by -370pt
\divide\sparewidth by 4

\medskip

\halign{\hskip\sparewidth#\hfil&\hskip\sparewidth#\hfil
&\hskip\sparewidth#\hfil\cr
\vta&\vth&\vto\cr
\vtb&\vti&\vtp\cr
\vtc&\vtj&\vtq\cr
\vtd&\vtk&\vtr\cr
\vte&\vtl&\vts\cr
\vtf&\vtm&\vtt\cr
\vtg&\vtn\cr}
}%end of comment

\exercises{\leader{286X}{Basic exercises (a)}
%\spheader 286Xa
Use 284Oa and 284Xg to shorten part (c) of the proof of 286U.
%286U

\spheader 286Xb Show that if $\sequence{k}{c_k}$ is a sequence of
complex numbers such that $\sum_{k=0}^{\infty}|c_k|^2$ is finite, then
$\sum_{k=0}^{\infty}c_ke^{ikx}$ is defined in $\Bbb C$ for almost all
$x\in\Bbb R$.
%286V

\leader{286Y}{Further exercises (a)}
%\spheader 286Ya
Show that if $f$ is a square-integrable function on
$\BbbR^r$, where $r\ge 2$, then

\Centerline{$g(y)
=\Bover1{(\sqrt{2\pi})^r}
 \lim_{\alpha_1,\ldots,\alpha_r\to-\infty,
  \beta_1,\ldots,\beta_r\to\infty}
 \int_{a}^{b}e^{-iy\dotproduct x}f(x)dx$}

\noindent is defined in $\Bbb C$ for almost every $y\in\BbbR^r$, and
that $g$ represents the Fourier transform of $f$.
}%end of exercises

\endnotes{
\Notesheader{286} This is not the longest single section in this
treatise as a whole, but it is by a substantial margin the longest in
the present volume, and thirty pages of sub-superscripts must tax the
endurance of the most enthusiastic.   You will easily understand why
Carleson's theorem is not usually presented at this level.   But I am
trying in this book to present complete proofs of the principal
theorems, there is no natural place for Carleson's theorem in later
volumes as at present conceived, and it is (just) accessible at this
point;  so I take the space to do it here.

The proof here divides naturally into two halves:  the `combinatorial'
part in 286E-286M, up to the Lacey-Thiele lemma, followed by the
`analytic' part in 286N-286V, in which the averaging process

\Centerline{$\int_1^2\Bover1{\alpha}\lim_{b\to\infty}
\Bover1b\int_0^b\ldots d\beta d\alpha$}

\noindent is used to transform the geometrically coherent, but
analytically irregular, functions $\theta_z$ into the indicator
functions $\Bover1{\tilde\theta_1(0)}\tilde\theta_z$.   From the
standpoint of ordinary Fourier analysis, this second part is essentially
routine;  there are many paths we could follow, and we have only to take
the ordinary precautions against illegitimate operations.   
\footnote{I ought at this
point to confess that I blundered badly in the 2001 edition of this
volume, and failed to notice my error until it was brought to my attention
by A.Derighetti at the end of 2013.   I hope that the version presented
here is essentially correct.}

Carleson ({\smc Carleson 66}) stated his theorem in the Fourier-series
form of 286V;  but it had long been understood that this was
equiveridical
with the Fourier-transform version in 286U.   There are of course many
ways of extending the theorem.   In particular, there are
corresponding
results for functions in $\eusm L^p$ for any $p>1$, and even for
functions $f$ such that $f\times\ln(1+|f|)\times\ln\ln\ln(16+|f|)$ is
integrable ({\smc Antonov 96}).   The methods here do not seem to
reach so far.   I ought also to remark that if we define $\hat Af$ as in
286T, then there is for every $p>1$ a constant $C$ such that
$\|\hat Af\|_p\le C\|f\|_p$ for every $f\in\eusm L^p_{\Bbb C}$
({\smc Hunt 67}, {\smc Mozzochi 71}, {\smc J{\o}rsboe \& Mejlbro 82},
{\smc Arias de Reyna 02}, {\smc Lacey 05}).

Note that the point of Carleson's theorem, in either form, is that we
take special limits.   In the formulae

\Centerline{$\varhatf(y)
=\Bover1{\sqrt{2\pi}}\lim_{a\to-\infty,b\to\infty}
\int_a^be^{-ixy}f(x)dx$,}

\Centerline{$f(x)=\lim_{n\to\infty}\sum_{-n}^nc_ke^{ikx}$,}

\ifdim\pagewidth>467pt\tenrmstretch{3pt}\fi
\noindent valid almost everywhere for square-integrable functions $f$,
we are not taking the ordinary integral
$\int_{-\infty}^{\infty}e^{-ixy}f(x)dx$ or the unconditional sum
$\sum_{k\in\Bbb Z}c_ke^{ikx}$.   If $f$ is not integrable, or
$\sum_{k=-\infty}^{\infty}|c_k|$ is infinite, these will not be defined
at even one point.
% Weyl's equidistribution theorem
Carleson's theorem makes sense only because we have a natural
preference for particular kinds of improper integral and conditional
sum.   So when we return, in Chapter 44 of Volume 4, to Fourier analysis
on general
topological groups, there will simply be no language in which to express
the theorem, and while versions have been proved for other groups (e.g.,
{\smc Schipp 78}),
they necessarily depend on some structure beyond the simple notion of
`locally compact Hausdorff abelian topological group'.   Even in
$\BbbR^2$, I understand that it is still unknown whether
\tenrmstretch{1.67pt}

\Centerline{$\lim_{a\to\infty}\Bover1{2\pi}\int_{B(\tbf{0},a)}
e^{-iy\dotproduct x}f(x)dx$}

\noindent will be defined a.e.\ for any square-integrable function $f$,
if we use ordinary Euclidean balls $B(\tbf{0},a)$ in place of the
rectangles in 286Ya.
}%end of notes

\frnewpage


%\ifresultsonly\else
\frfilename{mt28conc.tex}
\versiondate{6.1.10}
\copyrightdate{2010}

\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}

\def\chaptername{Fourier analysis}
\def\sectionname{Concordance}

\newsection{}

I list here the section and paragraph numbers which have (to my
knowledge) appeared in print in references to this chapter, and which
have since been changed.

\medskip

\leader{2{}85Xm}{Cauchy distribution} The exercise introducing the
Cauchy distribution, referred to in the 2002, 2004 and 2012 printings of
Volume 3, is now 285Xp.

\leader{2{}85Xo}{Poisson distribution} The exercise naming the Poisson
distribution, referred to in the 2003, 2006 and 2013 printings of 
Volume 4, is now 285Xr.

\leader{2{}85Xr}{Bochner's theorem} The exercise on a special case of
Bochner's theorem, referred to in the 2003, 2006 and 2013 printings 
of Volume 4, is now 285Xu.

\leader{2{}86U}{Carleson's theorem} The sequential form, 
referred to in {\smc Bogachev 07}, is now in 286V.

\bigskip

\noindent{\bf References}

\medskip

Bogachev V.I.\ [07] {\it Measure theory.}  Springer, 2007.

\frnewpage
%\fi

\references{
\frfilename{mt2r.tex}
\versiondate{17.12.12}
\copyrightdate{2000}

% except for references in ...X., ...Y., ...Z. give locations in form
%  342L  rather than  342Lb
%  \S511 {\it notes\/}, \S512 {\it intro.}]

\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}
\gdef\chaptername{References}
\gdef\sectionname{References}
\gdef\headlinesectionname{References}

\centerline{\bf References for Volume 2}

\medskip

%A

Alexits G.\ [78] (ed.) {\it Fourier Analysis and Approximation Theory.}
North-Holland, 1978 (Colloq.\ Math.\ Soc.\ Janos Bolyai 19).

Antonov N.Yu.\ [96] `Convergence of Fourier series', East J.\ Approx.\ 7 (1996) 187-196.
\cmmnt{[\S286 {\it notes\/}.]}

Arias de Reyna J.\ [02] {\it Pointwise Convergence of Fourier Series.} Springer, 2002 (Lecture Notes in Mathematics 1785).
\cmmnt{[\S286 {\it notes\/}.]}

\medskip%B

Baker R.\ [04] `{}``Lebesgue measure'' on $\Bbb R^{\infty}$, II',
Proc.\ Amer.\ Math.\ Soc.\ 132 (2004) 2577-2591.
\cmmnt{[254Yb.]}

Bergelson V., March P.\ \& Rosenblatt J.\ [96] (eds.) {\it Convergence
in Ergodic Theory and Probability.}  de Gruyter, 1996.

\indexheader{Bogachev}
Bogachev V.I.\ [07] {\it Measure theory.}  Springer, 2007.

du Bois-Reymond P.\ [1876] `Untersuchungen \"uber die Convergenz und
Divergenz der Fouriersche Darstellungformeln', Abh.\ Akad.\ M\"unchen 12
(1876) 1-103.   \cmmnt{[\S282 {\it notes}.]}

Bourbaki N.\ [66] {\it General Topology.}  Hermann/Addison-Wesley, 1968.
\cmmnt{[2A5F.]}

Bourbaki N.\ [87] {\it Topological Vector Spaces.}  Springer, 1987.
\cmmnt{[2A5E.]}

\medskip%C

Carleson L.\ [66] `On convergence and growth of partial sums of
Fourier series', Acta Math.\ 116 (1966) 135-157.  \cmmnt{[\S282 {\it
notes}, \S286 {\it intro.}, \S286 {\it notes}.]}

Clarkson J.A.\ [1936] `Uniformly convex spaces', Trans.\ Amer.\ Math.\ Soc.\
40 (1936) 396-414.
\cmmnt{[244O.]}

\medskip%D

Defant A.\ \& Floret K.\ [93] {\it Tensor Norms and Operator Ideals},
North-Holland, 1993.   \cmmnt{[\S253 {\it notes}.]}

Doob J.L.\ [53] {\it Stochastic Processes.}  Wiley, 1953.
%used for 275Ye

Dudley R.M.\ [89] {\it Real Analysis and Probability.} Wadsworth \&
Brooks/Cole, 1989.   \cmmnt{[\S282 {\it notes}.]}

Dunford N.\ \& Schwartz J.T.\ [57] {\it Linear Operators I.}  Wiley,
1957 (reprinted 1988).  \cmmnt{[\S244 {\it notes}, 2A5J.]}

\medskip%E

Enderton H.B.\ [77] {\it Elements of Set Theory.}   Academic, 1977.
\cmmnt{[\S2A1.]}

Engelking R.\ [89] {\it General Topology.} Heldermann, 1989 (Sigma
Series in Pure Mathematics 6).   \cmmnt{[2A5F.]}

Etemadi N.\ [96] `On convergence of partial sums of independent random
variables', pp.\ 137-144 in {\smc Bergelson March \& Rosenblatt 96}.
\cmmnt{[272V.]}

Evans L.C.\ \& Gariepy R.F.\ [92] {\it Measure Theory and Fine
Properties of Functions.}   CRC Press, 1992.
\cmmnt{[263Ec, \S265 {\it notes}.]}
%Kirszbraun's theorem mentioned but not proved

\medskip%F

Federer H.\ [69] {\it Geometric Measure Theory.}  Springer, 1969
(reprinted 1996).
\cmmnt{[262C, 263Ec, \S264 {\it notes}, \S265 {\it notes},
\S266 {\it notes}.]}

Feller W.\ [66] {\it An Introduction to Probability Theory and its
Applications,} vol.\ II.   Wiley, 1966.  \cmmnt{[Chap.\ 27 {\it
intro.}, 274H, 275Xc, 285N.]}

Fremlin D.H.\ [74] {\it Topological Riesz Spaces and
Measure Theory.}   Cambridge U.P., 1974.
\cmmnt{[\S232 {\it notes}, 241F, \S244 {\it notes}, \S245 {\it notes},
\S247 {\it notes}.]}

Fremlin D.H.\ [93] `Real-valued-measurable cardinals', pp.\ 151-304 in
{\smc Judah 93}.  \cmmnt{[232H.]}

\medskip%G\medskip%H

Haimo D.T.\ [67] (ed.) {\it Orthogonal Expansions and their Continuous
Analogues.}   Southern Illinois University Press, 1967.

Hall P.\ [82] {\it Rates of Convergence in the Central Limit Theorem.}
Pitman, 1982.  \cmmnt{[274H.]}

Halmos P.R.\ [50] {\it Measure Theory.}  Van Nostrand, 1950.
\cmmnt{[\S251 {\it notes}, \S252 {\it notes}, 255Yn.]}

Halmos P.R.\ [60] {\it Naive Set Theory.} Van Nostrand, 1960.
\cmmnt{[\S2A1.]}

Hanner O.\ [56] `On the uniform convexity of $L^p$ and $l^p$',
Arkiv f\"or Matematik 3 (1956) 239-244.
\cmmnt{[244O.]}

Henle J.M.\ [86] {\it An Outline of Set Theory.}  Springer, 1986.
\cmmnt{[\S2A1.]}

Hoeffding W.\ [63]
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\cmmnt{[272W.]}

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\medskip%I\medskip%J

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\medskip%K

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\medskip%L

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%expanded from 
% Publ.\ Mat.\ 48 (2004) 251-307
\cmmnt{[\S286 {\it notes}.]}

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\medskip%M

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\medskip%N

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%anon04.PDF

\medskip%O\medskip%P\medskip%Q\medskip%R

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\medskip%S

%Saks S.\ [1924] 'Sur les nombres d\'eriv\'es des fonctions', 
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% D-Y-S theorem
% books/saks.24.pdf   http://matwbn.icm.edu.pl/ksiazki/fm/fm5/fm5112.pdf

Saks S.\ [1924] 'Sur les nombres d\'eriv\'es des fonctions', Fund

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\medskip%T\medskip%U\medskip%V\medskip%W\medskip%Y\medskip%Z

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Zygmund A. [59] {\it Trigonometric Series.}   Cambridge U.P, 1959.
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\frnewpage

}%end of references

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