%%%%%%%%%%Paper Marmi, Moussa, Yoccoz %%%%%%%%
\def\service{T}

% ******************** output macros ********************************
\catcode`\@=11
%%% saclay A4 paper:
\def\unredoffs{\voffset=11mm \hoffset=0.5mm}
\def\redoffs{\voffset=-12.5truemm\hoffset=-6truemm}
\def\speclscape{\special{landscape}}
%
%---------------------------------------------------------------------%
\newbox\leftpage \newdimen\fullhsize \newdimen\hstitle \newdimen\hsbody
\newdimen\hdim
\tolerance=400\pretolerance=800
%\tolerance=1000\hfuzz=2pt
%\def\fontflag{cm}
%
%
\newif\ifsmall \smallfalse
\newif\ifdraft \draftfalse
\newif\iffrench \frenchfalse
\newif\ifeqnumerosimple \eqnumerosimplefalse
%
%
\nopagenumbers
\headline={\ifnum\pageno=1\hfill\else\hfil{\headrm\folio}\hfil\fi}
\def\draftstart{
\ifsmall
\message{(Reduced size)}
\let\l@r=L
\magnification=1000 \vsize=190truemm
\redoffs%
\hstitle=115truemm\hsbody=115truemm\fullhsize=10truein\hsize=\hsbody
%
\output={\ifnum\pageno=0 %%% This is the HUTP version
\shipout\vbox{\speclscape{\hsize\fullhsize\makeheadline}
\hbox to \fullhsize{\hfill\pagebody\hfill}}\advancepageno
\else
\almostshipout{\leftline{\vbox{\pagebody\makefootline}}}\advancepageno
\fi}
\headline={\hfil\oddpage\hfil\hfil\headrm\folio\hfil}
\gdef\oddpage{}
\def\almostshipout##1{\if L\l@r \count1=1 \message{[\the\count0.\the\count1]}
\global\setbox\leftpage=##1 \global\let\l@r=R
\xdef\oddpage{\ifnum\count0=1\else\headrm\the\count0\fi}
\else \count1=2
\shipout\vbox{\speclscape{\hsize\fullhsize\makeheadline}
\hbox to\fullhsize{\box\leftpage\hfil##1}}  \global\let\l@r=L\fi}
\else
\message{(Normal size)}
\magnification=1200 \unredoffs\hsize=130mm\vsize=190mm
\hsbody=\hsize \hstitle=\hsize %take default values for unreduced format
\fi
\ifdraft
\special{! userdict begin /bop-hook {gsave 100 160 translate 50 rotate
0 0 moveto /Times-Roman findfont 50 scalefont setfont
0.95 setgray (PRELIMINARY VERSION) show grestore} def end}
\footline={{\bf\hfil Version \today}}
\writelabels
\else
\nolabels
\overfullrule=0pt
\fi
\iffrench
% \fhyph
\dicof
\else
\dicoa
\fi
}
%**************** MAC.TEX ***********************************
%************************************************************
% origine: harvmac + modifications J. Zinn-Justin
%  + modifications J.-M. Drouffe
% fonts, Dirac slash

\font\elevrm=cmr9
\font\elevit=cmti9
\font\subrm=cmr7
\newdimen\chapskip
\font\twbf=cmssbx10 scaled 1200
\font\ssbx=cmssbx10
\font\twbi=cmmib10 scaled 1200
\font\caprm=cmr9
\font\capit=cmti9
\font\capbf=cmbx9
\font\capsl=cmsl9
\font\capmi=cmmi9
\font\capex=cmex9
\font\capsy=cmsy9
\chapskip=17.5mm
\def\makeheadline{\vbox to 0pt{\vskip-22.5pt
\line{\vbox to8.5pt{}\the\headline}\vss}\nointerlineskip}
%***************************************************
\font\tbfi=cmmib10
\font\tenbi=cmmib7
\font\fivebi=cmmib5
\textfont4=\tbfi
\scriptfont4=\tenbi
\scriptscriptfont4=\fivebi
\font\headrm=cmr10
\font\headit=cmti10
\font\twmi=cmmi10 scaled 1200
%****************************
\font\eightrm=cmr6
\font\sixrm=cmr5
\font\eightmi=cmmi6
\font\sixmi=cmmi5
\font\eightsy=cmsy6
\font\sixsy=cmsy5
\font\eightbf=cmbx6
\font\sixbf=cmbx5
\skewchar\capmi='177 \skewchar\eightmi='177 \skewchar\sixmi='177
\skewchar\capsy='60 \skewchar\eightsy='60 \skewchar\sixsy='60

\def\elevenpoint{
\textfont0=\caprm \scriptfont0=\eightrm \scriptscriptfont0=\sixrm
\def\rm{\fam0\caprm}
\textfont1=\capmi \scriptfont1=\eightmi \scriptscriptfont1=\sixmi
\textfont2=\capsy \scriptfont2=\eightsy \scriptscriptfont2=\sixsy
\textfont3=\capex \scriptfont3=\capex \scriptscriptfont3=\capex
\textfont\itfam=\capit \def\it{\fam\itfam\capit} % \it is family 4
\textfont\slfam=\capsl  \def\sl{\fam\slfam\capsl} % \sl is family 5
\textfont\bffam=\capbf \scriptfont\bffam=\eightbf
\scriptscriptfont\bffam=\sixbf
\def\bf{\fam\bffam\capbf} % \bf is family 6
\textfont4=\tbfi \scriptfont4=\tenbi \scriptscriptfont4=\tenbi
\normalbaselineskip=13pt
\setbox\strutbox=\hbox{\vrule height9.5pt depth3.9pt width0pt}
\let\big=\elevenbig \normalbaselines \rm}

\catcode`\@=11

\font\tenmsa=msam10
\font\sevenmsa=msam7
\font\fivemsa=msam5
\font\tenmsb=msbm10
\font\sevenmsb=msbm7
\font\fivemsb=msbm5
\newfam\msafam
\newfam\msbfam
\textfont\msafam=\tenmsa  \scriptfont\msafam=\sevenmsa
  \scriptscriptfont\msafam=\fivemsa
\textfont\msbfam=\tenmsb  \scriptfont\msbfam=\sevenmsb
  \scriptscriptfont\msbfam=\fivemsb

\def\hexnumber@#1{\ifcase#1 0\or1\or2\or3\or4\or5\or6\or7\or8\or9\or
	A\or B\or C\or D\or E\or F\fi }

%  The following 13 lines establish the use of the Euler Fraktur font.
%  To use this font, remove % from beginning of these lines.
\font\teneuf=eufm10
\font\seveneuf=eufm7
\font\fiveeuf=eufm5
\newfam\euffam
\textfont\euffam=\teneuf
\scriptfont\euffam=\seveneuf
\scriptscriptfont\euffam=\fiveeuf
\def\frak{\ifmmode\let\next\frak@\else
 \def\next{\Err@{Use \string\frak\space only in math mode}}\fi\next}
\def\goth{\ifmmode\let\next\frak@\else
 \def\next{\Err@{Use \string\goth\space only in math mode}}\fi\next}
\def\frak@#1{{\frak@@{#1}}}
\def\frak@@#1{\fam\euffam#1}
%  End definition of Euler Fraktur font.

\edef\msa@{\hexnumber@\msafam}
\edef\msb@{\hexnumber@\msbfam}

\def\msb{\tenmsb\fam\msbfam}
\def\Bbb{\ifmmode\let\next\Bbb@\else
 \def\next{\errmessage{Use \string\Bbb\space only in math mode}}\fi\next}
\def\Bbb@#1{{\Bbb@@{#1}}}
\def\Bbb@@#1{\fam\msbfam#1}
\font\sacfont=eufm10 scaled 1440
\catcode`\@=12
%
\def\sla#1{\mkern-1.5mu\raise0.4pt\hbox{$\not$}\mkern1.2mu #1\mkern 0.7mu}
\def\Dbar{\mkern-1.5mu\raise0.4pt\hbox{$\not$}\mkern-.1mu {\rm D}\mkern.1mu}
\def\Abar{\mkern1.mu\raise0.4pt\hbox{$\not$}\mkern-1.3mu A\mkern.1mu}
% *******************************************************************
%       Dictionnaires francais et anglais
\def\dicof{
\gdef\Resume{RESUME}
\gdef\Toc{Table des mati\`eres}
\gdef\soumisa{Soumis \`a:}
}
\def\dicoa{
\gdef\Resume{ABSTRACT}
\gdef\Toc{Table of Contents}
\gdef\soumisa{Submitted to}
}
% ****** extrait de definit.tex (obsolete ?)
\def\fileth{\noalign{\hrule}}
\def\saut{\noalign{\smallskip}}
\def\alignement{\offinterlineskip\halign}
\def\filetv{\vrule}
\def\colgauche{\strut\ }
\def\coldroite{\ }
\def\filetdroit{\cr}
\def\filetvide{height2pt}
\def\colvide{\omit}
\def\fintableau{}
\def\uniset{\rlap{\elevrm 1}\kern.15em 1}
\def\bkR{{\rm I\kern-.17em R}}
\def\bkC{{\rm \kern.24em
            \vrule width.05em height1.4ex depth-.05ex
            \kern-.26em C}}
% ********* A few math symbols
\def\e{\mathop{\rm e}\nolimits}
\def\sgn{\mathop{\rm sgn}\nolimits}
\def\Im{\mathop{\rm Im}\nolimits}
\def\Re{\mathop{\rm Re}\nolimits}
\def\d{{\rm d}}
\def\ud{{\textstyle{1\over 2}}}
\def\tr{\mathop{\rm tr}\nolimits}
\def\frac#1#2{{\textstyle{#1\over#2}}}
\def\today{\number\day/\number\month/\number\year}
\def\leaderfill{\leaders\hbox to 1em{\hss.\hss}\hfill}
% ******************** LOGOS **********************************************
\def\saclay{\if S\service \spec \else \spht \fi}
\def\spht{
\centerline{CEA, Service de Physique Th\'eorique, CE-Saclay}
\centerline{F-91191 Gif-sur-Yvette Cedex, FRANCE}}
\def\spec{
\centerline{CEA/DSM/DRECAM/Service de Physique de l'Etat Condens\'e}
\centerline{CE Saclay, F-91191 Gif-sur-Yvette Cedex, FRANCE}}
%
\def\logo{
\if S\service % Logo SPEC
\font\sstw=cmss10 scaled 1200
\font\ssx=cmss8
\vtop{\hsize 9cm
{\sstw {\twbf P}hysique de l'{\twbf E}tat {\twbf C}ondens\'e \par}
\ssx SPEC -- DRECAM -- DSM\par
\vskip 0.5mm
\sstw CEA -- Saclay \par
}
\else % Logo SPHT
\vtop{\hsize 9cm
\special{" /Helvetica-Bold findfont 9 scalefont setfont
0 -80 translate
 2 73 moveto (PHYSIQUE\ \ THEORIQUE) show
 35 38 moveto (CEA-DSM) show
0.7 setgray
/Helvetica-Bold findfont 26.5 scalefont setfont
 0 50 moveto (SACLAY) show
0 setgray 1.5 setlinewidth
0  41 moveto 32 41 lineto stroke
80 41 moveto 110 41 lineto stroke}}
\fi }
% *************************************************************************
\catcode`\@=11
% ************** double alignment in eqalignno style **********************
\def\deqalignno#1{\displ@y\tabskip\centering \halign to
\displaywidth{\hfil$\displaystyle{##}$\tabskip0pt&$\displaystyle{{}##}$
\hfil\tabskip0pt &\quad
\hfil$\displaystyle{##}$\tabskip0pt&$\displaystyle{{}##}$
\hfil\tabskip\centering& \llap{$##$}\tabskip0pt \crcr #1 \crcr}}
% ************** double eqalign ******************************************
\def\deqalign#1{\null\,\vcenter{\openup\jot\m@th\ialign{
\strut\hfil$\displaystyle{##}$&$\displaystyle{{}##}$\hfil
&&\quad\strut\hfil$\displaystyle{##}$&$\displaystyle{{}##}$
\hfil\crcr#1\crcr}}\,}
%***************************************************************************
%********* titlepage, headline, section, subsection, sub, appendix *********
%***************************************************************************
%********* introduce equation number file: for non-causal quotation
\openin 1=\jobname.sym
\ifeof 1\closein1\message{<< (\jobname.sym DOES NOT EXIST) >>}\else%
\input\jobname.sym\closein 1\fi
%
\newcount\nosection
\newcount\nosubsection
\newcount\neqno
\newcount\notenumber
\newcount\figno
\newcount\tabno
\def\content{\jobname.toc}
\def\symbols{\jobname.sym}
%\def\Figures{\jobname.fig}
%\def\Tables{\jobname.tab}
\newwrite\toc
\newwrite\sym
%\newwrite\Fig
%\newwrite\Tab
% ******************* titlepage **********************************
%\def\authorname#1{\maketitle{\bf #1}\smallskip}
\def\authorname#1{\centerline{\bf #1}\smallskip}
\def\address#1{ #1\medskip}
%
\newdimen\hulp
\def\maketitle#1{
\edef\oneliner##1{\centerline{##1}}
\edef\twoliner##1{\vbox{\parindent=0pt\leftskip=0pt plus 1fill\rightskip=0pt
plus 1fill
                     \parfillskip=0pt\relax##1}}
\setbox0=\vbox{#1}\hulp=0.5\hsize
                 \ifdim\wd0<\hulp\oneliner{#1}\else
                 \twoliner{#1}\fi}
\def\pacs#1{{\bf PACS numbers:} #1\par}
\def\submitted#1{{\it {\soumisa} #1}\par}
% **************** beginning
\def\title#1{\gdef\titlename{#1}
\maketitle{
%\ssbx\uppercase\expandafter
\twbf
{\titlename}}
\vskip3truemm\vfill
\nosection=0
\neqno=0
\notenumber=0
\figno=1
\tabno=1
\def\prefix{}
\def\eqprefix{}
\mark{\the\nosection}
\message{#1}
\immediate\openout\sym=\symbols
}
\def\preprint#1{\vglue-10mm
\line{ \logo \hfill {#1} }\vglue 20mm\vfill}
\def\abstract{\vfill\centerline{\Resume} \smallskip \begingroup\narrower
\elevenpoint\baselineskip10pt}
\def\endabstract{\par\endgroup \bigskip}
% ***************** input table of contents
\def\mktoc{\centerline{\bf \Toc} \medskip\caprm
\parindent=2em
\openin 1=\jobname.toc
\ifeof 1\closein1\message{<< (\jobname.toc DOES NOT EXIST. TeX again)>>}%
\else\input\jobname.toc\closein 1\fi
 \bigskip}
%******************************* section ***********************************
\def\section#1\par{\vskip0pt plus.1\vsize\penalty-100\vskip0pt plus-.1
\vsize\bigskip\vskip\parskip
\message{ #1}
\ifnum\nosection=0\immediate\openout\toc=\content%
\edef\ecrire{\write\toc{\par\noindent{\ssbx\ \titlename}
\string\leaderfill{\noexpand\number\pageno}}}\ecrire\fi% ajout
\advance\nosection by 1\nosubsection=0
\ifeqnumerosimple
\else \xdef\eqprefix{\prefix\the\nosection.}\neqno=0\fi
\vbox{\noindent\bf\prefix\the\nosection\ #1}
\mark{\the\nosection}\bigskip\noindent
\xdef\ecrire{\write\toc{\string\par\string\item{\prefix\the\nosection}
#1
\string\leaderfill {\noexpand\number\pageno}}}\ecrire}

% appendix
\def\appendix#1#2\par{\bigbreak\nosection=0
\notenumber=0
\neqno=0
\def\prefix{A}
\mark{\the\nosection}
\message{\appendixname}
\leftline{\ssbx APPENDIX}
\leftline{\ssbx\uppercase\expandafter{#1}}
\leftline{\ssbx\uppercase\expandafter{#2}}
\bigskip\noindent\nonfrenchspacing
\edef\ecrire{\write\toc{\par\noindent{{\ssbx A}\
{\ssbx#1\ #2}}\string\leaderfill{\noexpand\number\pageno}}}\ecrire}%

% **************************** \subsection *************************
\def\subsection#1\par {\vskip0pt plus.05\vsize\penalty-100\vskip0pt
plus-.05\vsize\bigskip\vskip\parskip\advance\nosubsection by 1
\vbox{\noindent\it\prefix\the\nosection.\the\nosubsection\
\it #1}\smallskip\noindent
\edef\ecrire{\write\toc{\string\par\string\itemitem
{\prefix\the\nosection.\the\nosubsection} {#1}
\string\leaderfill{\noexpand\number\pageno}}}\ecrire
}
%
\def\note #1{\advance\notenumber by 1
\footnote{$^{\the\notenumber}$}{\sevenrm #1}}
% ?????
\def\sub#1{\medskip\vskip\parskip
{\indent{\it #1}.}}
%\parindent=1em
%\newinsert\margin
%\dimen\margin=\maxdimen
%\count\margin=0 \skip\margin=0pt
% ********************* references harvmac style
\def\nolabels{\def\wrlabel##1{}\def\eqlabel##1{}\def\reflabel##1{}}
\def\writelabels{\def\wrlabel##1{\leavevmode\vadjust{\rlap{\smash%
{\line{{\escapechar=` \hfill\rlap{\sevenrm\hskip.03in\string##1}}}}}}}%
\def\eqlabel##1{{\escapechar-1\rlap{\sevenrm\hskip.05in\string##1}}}%
\def\reflabel##1{\noexpand\llap{\noexpand\sevenrm\string\string\string##1}}}
%*********
%\catcode`\@=11
\global\newcount\refno \global\refno=1
\newwrite\rfile
%
\def\ref{[\the\refno]\nref}
\def\nref#1{\xdef#1{[\the\refno]}\writedef{#1\leftbracket#1}%
\ifnum\refno=1\immediate\openout\rfile=\jobname.ref\fi
\global\advance\refno by1\chardef\wfile=\rfile\immediate
\write\rfile{\noexpand\item{#1\ }\reflabel{#1\hskip.31in}\pctsign}\findarg}
%	horrible hack to sidestep tex \write limitation
\def\findarg#1#{\begingroup\obeylines\newlinechar=`\^^M\pass@rg}
{\obeylines\gdef\pass@rg#1{\writ@line\relax #1^^M\hbox{}^^M}%
\gdef\writ@line#1^^M{\expandafter\toks0\expandafter{\striprel@x #1}%
\edef\next{\the\toks0}\ifx\next\em@rk\let\next=\endgroup\else\ifx\next\empty%
\else\immediate\write\wfile{\the\toks0}\fi\let\next=\writ@line\fi\next\relax}}
\def\striprel@x#1{}
\def\em@rk{\hbox{}}
%
\def\semi{;\hfil\break}
\def\addref#1{\immediate\write\rfile{\noexpand\item{}#1}} %now unnecessary
%
\def\listrefs{
\ifnum\refno=1 \else
\immediate\closeout\rfile\writestoppt\baselineskip=14pt%
\vskip0pt plus.1\vsize\penalty-100\vskip0pt plus-.1
\vsize\bigskip\vskip\parskip\centerline{{\bf References}}\bigskip%
{\frenchspacing%
\parindent=20pt\escapechar=` \input \jobname.ref\vfill\eject}%
\nonfrenchspacing
\fi}
%
\def\startrefs#1{\immediate\openout\rfile=\jobname.ref\refno=#1}
%
\def\xref{\expandafter\xr@f}\def\xr@f[#1]{#1}
\def\refs#1{[\r@fs #1{\hbox{}}]}
\def\r@fs#1{\ifx\und@fined#1\message{reflabel \string#1 is undefined.}%
\xdef#1{(?.?)}\fi \edef\next{#1}\ifx\next\em@rk\def\next{}%
\else\ifx\next#1\xref#1\else#1\fi\let\next=\r@fs\fi\next}
%************************
%
\newwrite\lfile
{\escapechar-1\xdef\pctsign{\string\%}\xdef\leftbracket{\string\{}
\xdef\rightbracket{\string\}}\xdef\numbersign{\string\#}}
\def\writedefs{\immediate\openout\lfile=labeldef.tmp \def\writedef##1{%
\immediate\write\lfile{\string\def\string##1\rightbracket}}}
%
\def\writestop{\def\writestoppt{\immediate\write\lfile{\string\pageno%
\the\pageno\string\startrefs\leftbracket\the\refno\rightbracket%
\string\def\string\secsym\leftbracket\secsym\rightbracket%
\string\secno\the\secno\string\meqno\the\meqno}\immediate\closeout\lfile}}
%
\def\writestoppt{}\def\writedef#1{}
%*************************************************************************
%Macro de numerotation automatique
%*************************************************************************
% numbering without naming
\def\eqnn{\global\advance\neqno by 1 \ifinner\relax\else%
\eqno\fi(\eqprefix\the\neqno)}
%
% numbering and attaching a name: \eqnd{\ename}
\def\eqnd#1{\global\advance\neqno by 1 \ifinner\relax\else%
\eqno\fi(\eqprefix\the\neqno)\eqlabel#1
{\xdef#1{($\eqprefix\the\neqno$)}}
\edef\ewrite{\write\sym{\string\def\string#1{($\eqprefix%
\the\neqno$)}}%
}\ewrite%
}
%
% for eqalignno, allows (1a) (1b)...
\def\eqna#1{\wrlabel#1\global\advance\neqno by1
{\xdef #1##1{\hbox{$(\eqprefix\the\neqno##1)$}}}
\edef\ewrite{\write\sym{\string\def\string#1{($\eqprefix%
\the\neqno$)}}%
}\ewrite%
}
%
\def\em@rk{\hbox{}}
\def\xeqn{\expandafter\xe@n}\def\xe@n(#1){#1}
\def\xeqna#1{\expandafter\xe@na#1}\def\xe@na\hbox#1{\xe@nap #1}
\def\xe@nap$(#1)${\hbox{$#1$}}
% \eqns allows to quote several equations, suppressing unnecessary ()
\def\eqns#1{(\e@ns #1{\hbox{}})}
\def\e@ns#1{\ifx\und@fined#1\message{eqnlabel \string#1 is undefined.}%
\xdef#1{(?.?)}\fi \edef\next{#1}\ifx\next\em@rk\def\next{}%
\else\ifx\next#1\xeqn#1\else\def\n@xt{#1}\ifx\n@xt\next#1\else\xeqna#1\fi
\fi\let\next=\e@ns\fi\next}
%*************************** figure macros ****************************
\def\fig{fig.~\the\figno\nfig}
\def\nfig#1{\xdef#1{\the\figno}%
\immediate\write\sym{\string\def\string#1{\the\figno}}%
\global\advance\figno by1}%
\def\xfig{\expandafter\xf@g}\def\xf@g fig.\penalty\@M\ {}%
\def\figs#1{figs.~\f@gs #1{\hbox{}}}%
\def\f@gs#1{\edef\next{#1}\ifx\next\em@rk\def\next{}\else%
\ifx\next#1\xfig #1\else#1\fi\let\next=\f@gs\fi\next}%
%
\long\def\figure#1#2#3{\midinsert
#2\par
{\elevenpoint
\setbox1=\hbox{#3}
\ifdim\wd1=0pt\centerline{{\bf Figure\ #1}\hskip7.5mm}%
\else\setbox0=\hbox{{\bf Figure #1}\quad#3\hskip7mm}
\ifdim\wd0>\hsize{\narrower\noindent\unhbox0\par}\else\centerline{\box0}\fi
\fi}
\wrlabel#1\par
\endinsert}
%*************************** table macros ****************************
\def\tab{table~\uppercase\expandafter{\romannumeral\the\tabno}\ntab}
\def\ntab#1{\xdef#1{\the\tabno}
\immediate\write\sym{\string\def\string#1{\the\tabno}}
\global\advance\tabno by1}
\long\def\table#1#2#3{\topinsert
#2\par
{\elevenpoint
\setbox1=\hbox{#3}
\ifdim\wd1=0pt\centerline{{\bf Table
\uppercase\expandafter{\romannumeral#1}}\hskip7.5mm}%
\else\setbox0=\hbox{{\bf Table
\uppercase\expandafter{\romannumeral#1}}\quad#3\hskip7mm}
\ifdim\wd0>\hsize{\narrower\noindent\unhbox0\par}\else\centerline{\box0}\fi
\fi}
\wrlabel#1\par
\endinsert}
%***********************************************************************
\catcode`@=12
\def\draftend{\immediate\closeout\sym\immediate\closeout\toc
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\draftstart
\preprint{T95/016}
\title{Continued fraction transformations, Brjuno functions and BMO spaces}
\authorname{S. Marmi}
\address{\centerline{Dipartimento di Matematica \lq\lq U. Dini\rq\rq ,
Universit\`a di}
\centerline{Firenze, }
\centerline{Viale Morgagni 67/A, 50134 Firenze, ITALY}
}
\authorname{P. Moussa}
\address{\saclay}
\authorname{J.-C. Yoccoz}
\address{\centerline{Universit\'e de Paris-Sud, Math\'ematiques}
\centerline{B\^at.425, 91405 Orsay, FRANCE}
}
\abstract
The small divisors problem which is raised by  stability questions
in classical mechanics has an analog in holomorphic dynamical
systems, namely the existence of Siegel disks. The size of these
disks is well represented by the Brjuno functions. We analyse
the relation between these functions and the various continued
fraction transformations, and display the functional equation
which is fulfilled by these highly singular functions.
The analysis of this functional equation shows that the Brjuno
function belong to the BMO space, and that a regular
perturbation of this equation leads to a modification of the
singular function which is 1/2-H\"older continuous. This leads
us to believe that the most singular part of the size of
the stability domains as function of the rotation number, is
`universal' up to a 1/2-H\"older continuous function.
\endabstract
\vfill
\submitted{Note CEA}
\eject
%%%% Marmi, Moussa, Yoccoz %%%%%
%%%%%%%%  macros %%%%%
\input amssym.def
\input amssym.tex
\magnification 1200
\pageno=1

\catcode`\@=11

\hsize=125 mm   \vsize =187mm
\hoffset=4mm    \voffset=10mm
\pretolerance=500 \tolerance=1000 \brokenpenalty=5000

\catcode`\;=\active
\def;{\relax\ifhmode\ifdim\lastskip>\z@
\unskip\fi\kern.2em\fi\string;}

\catcode`\:=\active
\def:{\relax\ifhmode\ifdim\lastskip>\z@\unskip\fi
\penalty\@M\ \fi\string:}

\catcode`\!=\active
\def!{\relax\ifhmode\ifdim\lastskip>\z@
\unskip\fi\kern.2em\fi\string!}

\catcode`\?=\active
\def?{\relax\ifhmode\ifdim\lastskip>\z@
\unskip\fi\kern.2em\fi\string?}

\def\^#1{\if#1i{\accent"5E\i}\else{\accent"5E #1}\fi}
\def\"#1{\if#1i{\accent"7F\i}\else{\accent"7F #1}\fi}

%\frenchspacing

\catcode`\@=12

\newif\ifpagetitre      \pagetitretrue
\newtoks\hautpagetitre  \hautpagetitre={\hfil}
\newtoks\baspagetitre   \baspagetitre={\hfil}

\newtoks\auteurcourant  \auteurcourant={\hfil}
\newtoks\titrecourant   \titrecourant={\hfil}

\newtoks\hautpagegauche \newtoks\hautpagedroite
\hautpagegauche={\hfil\the\auteurcourant\hfil}
\hautpagedroite={\hfil\the\titrecourant\hfil}

\newtoks\baspagegauche  \baspagegauche={\hfil\tenrm\folio\hfil}
\newtoks\baspagedroite  \baspagedroite={\hfil\tenrm\folio\hfil}

\headline={\ifpagetitre\the\hautpagetitre
\else\ifodd\pageno\the\hautpagedroite
\else\the\hautpagegauche\fi\fi}

\footline={\ifpagetitre\the\baspagetitre
\global\pagetitrefalse
\else\ifodd\pageno\the\baspagedroite
\else\the\baspagegauche\fi\fi}

\def\nopagenumbers{\def\folio{\hfil}}
\hautpagetitre={\hfill\tenrm  \hfill}
\hautpagegauche={\tenrm\folio\hfill\tenrm\the\auteurcourant}
\hautpagedroite={\tenrm\the\titrecourant\hfill\tenrm\folio}
\baspagegauche={\hfil} \baspagedroite={\hfil}
\auteurcourant{Marmi, Moussa, Yoccoz}
\titrecourant{Continued fractions, Brjuno functions and BMO spaces}
\def\mois{\ifcase\month\or January\or February\or March\or April\or
May\or June\or July\or August\or September\or October\or November\or
December\fi}
\def\Date{\rightline{\mois\ /\ \the\day\ /\/ \the\year}}
\hfuzz=0.3pt
\font\tit=cmb10 scaled \magstep1
\def\dst{\displaystyle}
\def\sst{\scriptstyle}
\def\hfb{\hfill\break\indent}
\def\ie{{\it i.e.\ }}
\def\R{\Bbb R}
\def\T{\Bbb T}
\def\Z{\Bbb Z}
\def\Q{\Bbb Q}
\def\C{\Bbb C}
\def\al{\alpha}
\def\be{\beta}
\def\ga{\gamma}
\def\de{\delta}
\def\la{\lambda}
\def\Lloc{L^1_{\rm loc}(\R)}
\def\mean#1#2{{1\over |#1|}\int_{#1}#2\,dx}
\def\Mean#1#2{{1\over |#1|}\int_{#1}|#2|\,dx}
\def\BM#1{\hbox{${\rm BMO}(#1)$}}
\def\norm#1#2{||#1||_{*,#2}}
\def\Norm#1#2#3{||#1||_{*,#3,#2}}
\def\lnorm#1#2#3{||#1||_{#3,#2}}
\def\Dnorm#1#2{\sup_{#1}{1\over |I|}\int_{I}|#2-#2_I|\,dx}
\def\hnorm#1#2{\hbox{$\vert#1\vert_{#2}$}}
\def\Hnorm#1#2{\hbox{$||#1||_{#2}$}}
\def\remark#1{\noindent{\it Remark\ }#1\ }
\def\proof{\noindent{\it Proof.\ }}
\def\qed{\hfill$\square$\par\smallbreak}
\def\Proc#1#2\par{\medbreak \noindent {\bf #1\enspace }{\sl #2}%
\par\ifdim \lastskip <\medskipamount \removelastskip%
\penalty 55\medskip \fi}
%%% end macros %%%%%
%%%%% Section  1 %%%%
%\Date
\vglue 2.5 truecm
\centerline{\tit Continued fraction transformations, Brjuno functions and
BMO spaces}
\vskip 1truecm
\centerline{S. Marmi\footnote{$^1$}{ Dipartimento di Matematica
``U. Dini'', Universit\`a di Firenze, Viale Morgagni 67$/$A, 50134
Firenze, Italy}, P. Moussa \footnote{$^2$}{Service de Physique
Th\'eorique, C.E. Saclay, 91191 Gif-Sur-Yvette, France},
and J.-C. Yoccoz \footnote{$^3$}{Universit\'e de Paris-Sud,
Math\'ematiques. B\^at. 425, 91405-Orsay, France}}
\vskip 1.5 truecm
\beginsection{\bf Abstract}\par
The small divisors problem which is raised by  stability questions
in classical mechanics has an analog in holomorphic dynamical
systems, namely the existence of Siegel disks. The size of these
disks is well represented by the Brjuno functions. We analyse
the relation between these functions and the various continued
fraction transformations, and display the functional equation
which is fulfilled by these highly singular functions.
The analysis of this functional equation shows that the Brjuno
function belong to the BMO space, and that a regular
perturbation of this equation leads to a modification of the
singular function which is 1/2-H\"older continuous. This leads
us to believe that the most singular part of the size of
the stability domains as function of the rotation number, is
`universal' up to a 1/2-H\"older continuous function.\par
\rightline{\hphantom{November 23 1994}}
\vfill\eject
\vglue 2.5 truecm
\beginsection{\bf 0. Introduction}\par
The Brjuno condition tells which invariant disks (called Siegel disks)
persist when an irrational
rotation is analytically perturbed. The Brjuno function tells more:
it gives an estimate of minus the logarithm of the size of the Siegel disks
as a function of the rotation number [Yo].
In this work, we first analyse the relation between the Brjuno function
and the various kind of continued fractions. We establish the functional
equation fulfilled by the Brjuno function, and show that its solution
requires the inversion of an operator $T$. We show that $T$ is a
contracting operator for  all $L^p$ norms, and also for the BMO
(Bounded Mean Oscillation) space.
The Brjuno function is obtained as the action of $(1-T)^{-1}$ on
a logarithmic function which belongs to the BMO space. Therefore
the Brjuno function is also in this space.
Noticing that the adjoint of $T$ is nothing else than the
Ruelle-Frobenius-Perron operator associated to the dynamical system which
generate the continued fraction, the identification of the space
adapted to $T$, seems to us promising for the dynamical properties.
Finally, the action of $T$ on continuous
function is decribed according to H\"older's properties.
We show that regular perturbations (at least $C^{1/2}$) of the
logarithmic term do modify the solution only by a $C^{1/2}$
contribution, so that the most singular part remain unchanged.
We anticipate that  this result might be much  more general:
the geometric renormalisation for holomorphic dynamical systems
will likely produce only $C^1$ perturbations to the renormalisation
equation, and the most singular part of minus the logarithm of
the size of the stability domains as function of the rotation number
could be universally (that is modulo $C^{1/2}$) described by the Brjuno
function.\par
\vfill \eject
\beginsection \vbox{\bf\noindent 1. On a family of continued fraction
transformations and the related Brjuno functions}\par
Let $\alpha \in [1/2,1]$ and let $x \in \Bbb R$. We define
$$
[x]_\alpha = \min \{ p \in \Bbb Z \mid x < \alpha + p\} \eqno(1.1)
$$
that is
$$
[x]_\alpha = p \hbox{ iff } \;\;\;\alpha - 1 + p \le x < \alpha + p \; .
$$
Note that
$$
 [x]_\alpha = [x-\alpha +1]
$$
where $[\;]=[\;]_1$ denotes the usual integer part of a real number.
We will consider the iteration of
$$
A_\alpha : (0,\alpha) \mapsto [0,\alpha] \eqno(1.2)
$$
defined by
$$
A_\alpha (x) = \left|\ {\ 1\ \over x} -
\left[\ {\ 1\ \over x}\ \right]_\alpha\ \right| \;. \eqno(1.3)
$$
\Proc{Theorem 1.1.}{The dynamical system defined by the iteration of
(1.3) preserves an absolutely continuous (w.r.t. Lebesgue) probability
measure $m_\alpha$ with density $c_\alpha \rho_\alpha (x)$:
$m_\alpha (dx) = c_\alpha \rho_\alpha (x) dx$. The density is  given by :
\item{(i)} if ${\sqrt{5} -1 \over 2} \le \alpha \le 1$
$$
\eqalign{ c_\alpha &= {1\over \log (1 + \alpha)} \; , \cr
            \rho_\alpha (x) & = {1 \over 1+x}
            \chi_{({1-\alpha \over \alpha},\alpha)}(x) +
                                {1 \over 2+x}
            \chi_{(1-\alpha, {1-\alpha \over \alpha}]} (x) \cr
                            & + \left( {1 \over x+2} + {1 \over 2-x}\right)
            \chi_{(0,1 - \alpha]} (x) \; , \cr}
            \eqno(1.4)
$$
\item{(ii)} if $2 - \sqrt{2} \le \alpha < {\sqrt{5} -1 \over 2}$
$$
\eqalign{ c_\alpha &= {1\over \log G} \cr
            \rho_\alpha (x) & = {1 \over G+x}
            \chi_{({2 \alpha - 1 \over 1-\alpha},\alpha)}(x) +
                                {1 \over 2+x}
            \chi_{(1-\alpha , {2 \alpha -1 \over 1-\alpha}]} (x) \cr
                    & + \left( {1 \over x+2} + {1 \over G+1-x} \right)
            \chi_{({2 \alpha -1 \over \alpha},1 - \alpha]} (x) \cr
                    & + \left( {1 \over x+2} + {1 \over 2-x} \right)
            \chi_{(0,{2 \alpha -1 \over \alpha}]} (x) \; , \cr}
            \eqno(1.5)
$$
\item{(iii)} if ${1 \over 2} \le \alpha < 2 - \sqrt{2} $
$$
\eqalignno{ c_\alpha =& {1\over \log G} \; , \cr
            \rho_\alpha (x)  = &{1 \over G+x} \chi_{(1-\alpha,\alpha)}(x) +
            \left( {1 \over G+x} + {1 \over G+1-x}\right)
            \chi_{({2 \alpha
            -1 \over 1-\alpha}, 1-\alpha]} (x) \cr
                     + \left( {1 \over x+2}\right. &+ \left.{1 \over G+1-x}
            \right) \chi_{({2 \alpha -1 \over \alpha},{2 \alpha -1 \over
            1 - \alpha}]} (x) +
            \left( {1 \over x+2} + {1 \over 2-x}
            \right) \chi_{(0,{2 \alpha -1 \over \alpha}]} (x) \; ,\cr
            &&(1.6)\cr}
$$
where $G={\sqrt{5}+1 \over 2}$ and $\chi_{(a,b)} (x)$ denotes the
characteristic function of the interval $(a,b)$.}
\par
\medskip
\remark{1.2.} Note that if $\alpha =1$ one finds Gauss' result:
$c_1\rho_1(x) = {1\over (1+x) \log 2 }$.
\par
\medskip
\proof
By Theorem 4.1.1 of [LM] $m_\alpha$ will be an invariant probability
measure for $A_\alpha$ if and only if its density $\rho_\alpha$ is
a fixed point of the Perron-Frobenius operator $P_\alpha$ associated
to $A_\alpha$: if $B$ denotes any measurable subset of $(0,\alpha )$,
and $f$ any summable function on $(0,\alpha )$, $P_\alpha$ is defined by
$$
\int_B (P_\alpha f)(x) dx = \int_{A_\alpha^{-1}(B)} f(x) dx \; .
$$
The Perron-Frobenius operator for all these maps has the following
form:
$$
(P_\alpha f)(x) = \sum_{m\ge m_+}^{+\infty} {1\over (m+x)^2}
f\left({1\over m+x}\right) + \sum_{m\ge m_-}^{+\infty} {1\over (m-x)^2}
f\left({1\over m-x}\right) \; ,
$$
where $m_+$ and $m_-$ depend both on $\alpha$ and on $x$.
\par
If $\alpha =1$ one has $m_+=1$ and $m_- = +\infty$ and it is
immediate to check that ${1\over 1+x}$ is a fixed point of $P_1$.
\par
If ${\sqrt{5} -1 \over 2} < \alpha < 1$ one has three possible cases:
\item{(a)} $x\in (0,1-\alpha ]$ then $m_+ = 2$, $m_- =2$ and
$\rho_\alpha (x)= {1\over x+2}+{1\over 2-x}$;
\item{(b)} $x\in (1-\alpha , {1-\alpha\over\alpha}]$ then
$m_+=2$, $m_-=+\infty$ and $\rho_\alpha (x)={1\over x+2}$;
\item{(c)} $x \in ({1-\alpha\over\alpha}, \alpha )$ then $m_+=1$,
$m_-=+\infty$ and $\rho_\alpha (x) = {1\over 1+x}$.
\par
Let us consider case (c). By (1.4) it follows that the two sums
in the definition of ${\cal P}_\alpha$ must
be splitted into three parts according to which interval contains
$1/(m\pm x)$:
$$
\eqalign{
(P_\alpha \rho_\alpha )(x) &= \sum_{m=2}^{m_1} {1\over (m+x)^2}
\displaystyle{1\over 1+ \displaystyle{1\over m+x}} +
\sum_{m=2}^{m_2} {1\over (m-x)^2}
\displaystyle{1\over 1+ \displaystyle{1\over m-x}} \cr
& + {1\over m_1+x+1}-{1\over m_1+x+{3\over 2}} +
{1\over m_2-x+1}-{1\over m_2-x+{3\over 2}}\cr
& + \sum_{m=m_1+2}^\infty {1\over (m+x)^2}\left(
\displaystyle{1\over 2+ \displaystyle{1\over m+x}} +
\displaystyle{1\over 2- \displaystyle{1\over m+x}}\right) \cr
& + \sum_{m=m_2+2}^\infty {1\over (m-x)^2}\left(
\displaystyle{1\over 2+ \displaystyle{1\over m-x}} +
\displaystyle{1\over 2- \displaystyle{1\over m-x}}\right)\cr}\; .
$$
Using
$$
\eqalign{
{1\over (m\pm x)^2}
\displaystyle{1\over 1+ \displaystyle{1\over m\pm x}} &=
{1\over m\pm x} - {1\over m+1\pm x} \cr
{1\over (m+x)^2}  \left(
\displaystyle{1\over 2+ \displaystyle{1\over m+x}} +
\displaystyle{1\over 2- \displaystyle{1\over m+x}}\right)
& = {1\over m+x-{1\over 2}}-{1\over m+x+{1\over 2}} \cr
{1\over (m-x)^2}  \left(
\displaystyle{1\over 2+ \displaystyle{1\over m-x}} +
\displaystyle{1\over 2- \displaystyle{1\over m-x}}\right)
& = {1\over m-x-{1\over 2}}-{1\over m-x+{1\over 2}} \cr}
$$
it is easy to show that all terms cancel except for ${1\over x+2} +
{1\over 2-x}$.
\par
The remaining cases (a) and (b) are simpler, and the proof of (ii) and
(iii) follows the same kinds of ideas (note that $G^2=G+1$,
$G-1=1/G$ and $2-G=1/G^2$). \qed
\par
\medskip
\remark{1.3.} It is well known that the Kolmogorov-Sinai entropy
of these maps is given by
$$
h(\alpha ) = -2\int_0^\alpha c_\alpha \rho_\alpha (x)\log x\,dx \; .
$$
If $\alpha =1$ one obtains
$$
h(1) = {\pi^2\over 6\log 2}\; ,
$$
for $\alpha =1/2$ one obtains,
$$
h(1/2) = {\pi^2\over 6\log G}\; ,
$$
a result already given by Rieger [Ri].
For general $\alpha$, we find that
$c_\alpha h(\alpha )$ does not depend on $\alpha$,
thus
$$
h(\alpha ) = \cases{ {\pi^2\over 6\log (1+\alpha )} &if
${\sqrt{5}-1\over 2}\le \alpha \le 1\;$, \cr
{\pi^2\over 6\log G} & if ${1\over 2}\le \alpha\le {\sqrt{5}-1\over 2}\;$,
\cr}
$$
as already shown by  Nakada [Na]. These results show the existence of a
phase transition at $\alpha =
{\sqrt{5}-1\over 2}$.
\par
\medskip
To each $x \in \Bbb R \setminus \Bbb Q$ we associate a continued fraction
expansion by iterating $A_\alpha$ as follows. Let
$$
\eqalign{x_0 & = | x - [x]_\alpha| \cr
            a_0 & = [x]_\alpha \cr} \eqno(1.7)
$$
then one obviously has
$$
x_0 = a_0 + \varepsilon_0 x_0 \eqno(1.8)
$$
where
$$
\varepsilon_0 = \cases{ +1 \hbox{ iff } x \ge [x]_\alpha \cr
                          -1 \hbox{ otherwise } \cr} \eqno(1.9)
$$
We now define inductively for all $n \ge 0$
$$
\eqalign{x_{n+1} & = A_\alpha(x_n) \cr
            a_{n+1} & = \left[ {1 \over x_n} \right]_\alpha  \ge 1\cr}
            \eqno(1.10)
$$
thus
$$
x_{n}^{-1} = a_{n+1} + \varepsilon_{n+1} x_{n+1} \eqno(1.11)
$$
where
$$
\varepsilon_{n+1} = \cases{ +1 \hbox{ iff } {1 \over x_n} \ge a_{n+1}\cr
                          -1 \hbox{ otherwise } \cr} \eqno(1.12)
$$
Therefore we have
$$
x=a_0 + \varepsilon_0 x_0=a_0+{\varepsilon_0 \over a_1 + \varepsilon_1
     x_1}= \ldots =a_0 + \displaystyle{\varepsilon_0 \over a_1
     + \displaystyle{\varepsilon_1  \over a_2 + \ddots +
     \displaystyle{\varepsilon_{n-1} \over a_n + \varepsilon_n x_n}}}
     \eqno(1.13)
$$
and we will write
$$
x=[(a_0,\varepsilon_0),(a_1,\varepsilon_1),\ldots ,(a_n,\varepsilon_n),
     \ldots] \;. \eqno(1.14)
$$
Note that for $\alpha = 1$ we recover the standard continued fraction
expansion defined through the iteration of Gauss' map $x \mapsto x^{-1}
\hbox{ mod } 1$, and all $\varepsilon_n = +1$. When $\alpha = 1/2$ one
has the
so-called nearest integer continued fraction, and $a_n \ge 2$ for all
$n \ge 1$.
\par
The nth-convergent is defined by
$$
{p_n \over q_n} = [(a_0,\varepsilon_0),(a_1,\varepsilon_1),\ldots ,
                     (a_n,\varepsilon_n)] =
                     a_0 + \displaystyle{\varepsilon_0 \over a_1
     + \displaystyle{\varepsilon_1  \over a_2 + \ddots +
     \displaystyle{\varepsilon_{n-1} \over a_n }}}
     \;. \eqno(1.15)
$$
and it is immediate to check that the numerators $p_n$ and denominators
$q_n$ are recursively determined by
$$
p_{-1}=q_{-2}=1 \;\;,\;\;\;p_{-2}=q_{-1}=0 \;\;,\eqno(1.16)
$$
and for all $n \ge 0$
$$
\eqalign{ p_n &= a_n p_{n-1} + \varepsilon_{n-1} p_{n-2} \; , \cr
            q_n &= a_n q_{n-1} + \varepsilon_{n-1} q_{n-2} \; . \cr}
           \eqno(1.17)
$$
Moreover
$$
\eqalignno{x &= {p_n + p_{n-1} \varepsilon_n x_n \over q_n + q_{n-1}
       \varepsilon_n x_n } &(1.18) \cr
              x_n &= - \varepsilon_n {q_n x -p_n \over q_{n-1} x - p_{n-1}}
              &(1.19) \cr
              q_n p_{n-1} - p_n q_{n-1} &= (-1)^n \varepsilon_0 \ldots
              \varepsilon_{n-1} &(1.20) \cr}
$$
Let
$$
\beta_n = \Pi_{i=0}^n x_i = (-1)^n \varepsilon_0 \ldots
  \varepsilon_n (q_n x - p_n) \eqno(1.21)
$$
Then
$$
\eqalign{x_n &= {\beta_n \over \beta_{n-1}} \cr
            \beta_{n-2} &= a_n \beta_{n-1} + \varepsilon_n \beta_n \cr}
            \eqno(1.22)
$$
{}From the definitions given one easily proves by induction
the following proposition \par
\medskip
\Proc{Proposition 1.4.} {Given $\alpha \in [1/2,1]$, for all
$x \in \Bbb R \setminus \Bbb Q$ and for all $n \ge 1$ one has
\item{(i)}\qquad $q_{n+1} > q_n > 0$;
\item{(ii)}\qquad $p_n > 0$ when $x>0$ and $p_n< 0$ when $x<0$;
\item{(iii)}\qquad $ \left|q_n x - p_n\right|
={\dst 1\over\dst q_{n+1}+\varepsilon_{n+1}q_nx_{n+1}}$,
so that ${\dst 1\over\dst 1+\alpha}<\beta_nq_{n+1}<{\dst 1\over\dst\alpha}$\ ;
\item{(iv)}\qquad if $\alpha>{\dst\sqrt{5}-1\over\dst2}\ ,\ \beta_n\le\alpha
\left({\dst\sqrt{5}-1\over\dst2}\right)^n$;
\item{(v)}\qquad if $\alpha>{\dst\sqrt{5}-1\over\dst2}\ ,\ q_n\ge{\dst1
\over\dst\alpha(1+\alpha)}\left({\dst\sqrt{5}+1\over\dst2}\right)^{(n-1)}$;
\item{(vi)}\qquad
if $\alpha\le{\dst\sqrt{5}-1\over\dst2}\ ,\ \beta_n\le\alpha
\left(\sqrt{2}-1)\right)^n$;
\item{(vii)}\qquad if $\alpha\le{\dst\sqrt{5}-1\over\dst2}\ ,\ q_n
\ge{\dst1\over\dst\alpha(1+\alpha)}\left(\sqrt{2}+1)\right)^{(n-1)}$.}
\par
\medskip
\remark{1.5.} We only quote here some aspects of the proof
of the above theorem: one gets parts (i) and (ii)
by recursion using (1.17). Part (iii) is easily obtained from (1.18),
and is used to deduce part (v) from (iv) and part (vii) from (vi).
Part iv) and vi) are easy to prove when $\alpha$ is close to one or
one half, respectively. However the proof is much more intricate
around $\alpha=(\sqrt{5}-1)/2$ (see [MMY]).
\par
\medskip
\remark{1.6.} From (iii) one gets
$${1\over 2q_nq_{n+1}}< {1 \over q_n (q_n + q_{n+1})}\le
 {1 \over q_n (\alpha q_n + q_{n+1})}< \left| x - {p_n \over q_n} \right| <
   {1 \over q_n q_{n+1}} \eqno(1.23)$$
if $\varepsilon_{n+1} = +1$, whereas
$$
 {1 \over q_n q_{n+1}} < \left| x - {p_n \over q_n} \right| <
   {1 \over q_n ( q_{n+1} - (1-\alpha)q_n )}<{1\over\alpha q_n^2} \eqno(1.24)
$$
if $\varepsilon_{n+1} = -1$.
\par
\medskip
\remark{1.7.} By (v) and (vii), Proposition 1.4,
there exists two positive constants $c_1$ and $c_2$ such that
$$
\eqalign{
\sum_{k=0}^\infty {\log q_{k}\over q_{k}} & \le c_1 \; , \cr
\sum_{k=0}^\infty {\log 2\over q_{k}} & \le c_2\; , \cr}
$$
for all $\alpha\in [1/2,1]$ and for all $x\in (0,\alpha )$.
\par
\medskip
Following Yoccoz [Yo] we define a (generalized) Brjuno function:
\par
\medskip
\Proc{Definition 1.8.}{The {\it $\alpha$-Brjuno function}
$B_\alpha \,: \Bbb R \setminus \Bbb Q \to \bar\Bbb R$ is defined by
the formula
$$
B_\alpha (x) = - \sum_{i=0}^\infty \beta_{i-1} \log x_i
\eqno(1.25)
$$
where we have posed $\beta_{-1} = 1$.}
\par
\medskip
\remark{1.9.} The Brjuno function defined in [Yo] corresponds to
$B_{1/2}$, the one defined by the nearest integer continued fraction
map $A_{1/2}$.
\par
\medskip
\Proc{Proposition 1.10.} {Given $\alpha \in [1/2,1]$ one has
\item{(i)} $B_\alpha (x) = B_\alpha (x+1) $  for all $x \in \Bbb R
            \setminus \Bbb Q$;
\item{(ii)} For all $x \in (0,\alpha) \cap \Bbb R \setminus \Bbb Q$
$$
B_\alpha (x) = - \log x + x B_\alpha \left( {1 \over x} \right)
$$
\item{(iii)} if $x \in [\alpha - 1, 0) \cap \Bbb R \setminus \Bbb Q$
then $B_\alpha (-x) = B_\alpha (x)$.
\item{(iv)} There exists a constant $C_\alpha >0$ such that for all
$x \in \Bbb R \setminus \Bbb Q$ one has
$$
\left| B_\alpha (x) - \sum_{j=0}^\infty {\log q_{j+1} \over q_j}
\right| \le C_{\alpha}
$$
where $\{q_j\}_{j \ge 0}$ denotes the sequence of the denominators
of the convergents to $x$ of the $\alpha$-continued fraction expansion.}
\par
\medskip
\proof Given $x\in \Bbb R\setminus \Bbb Q$, the sequences
$(x_i)_{i\ge 0}$ and $(\beta_i)_{i\ge 0}$ associated to $x$ and
$x+1$ are the same, which proves (i). The same is true for $x$
and $-x$ if $x\in (\alpha -1,0)$, which proves (iii).
\par
If $x\in (0,\alpha )$, let $y=1/x$ and denote by $y_i$, $a_i(y)$,
$\beta_i(y)$, and $x_i$, $a_i(x)$, $\beta_i (x)$ the sequences
(1.10) and (1.21) associated to $y$ and to $x$ respectively.
{}From (1.7) and (1.8) it follows that $x_0=x$, $a_0(y)=a_1(x)$,
$y_0=x_1$ and by induction for all $n\ge 0$
$y_n=x_{n+1}$ and $\beta_n(y)={\beta_{n+1} (x)\over x}$. Thus
$$
\eqalign{
B_\alpha (y)
&= -\sum_{i=0}^\infty\beta_{i-1}(y)\log y_i
=-\log y_0-\sum_{i=1}^\infty {1\over x}\beta_i(x)\log x_{i+1} \cr
& = -{1\over x}\sum_{i=1}^\infty \beta_{i-1}(x)\log x_i
= {1\over x}[B_\alpha (x)+\log x]\; ,\cr}
$$
which proves (ii).
\par
To prove (iv) we first remark that
$$
q_i\beta_{i-1}+\varepsilon_iq_{i-1}\beta_i=1
$$
for all $i\ge 0$.  Then
$$
\eqalign{
-B_\alpha (x) &+\sum_{i=0}^\infty {\log q_{i+1}\over q_i}
= \sum_{i=0}^\infty \beta_{i-1}\log {\beta_i\over\beta_{i-1}}
+\sum_{i=0}^\infty \left(\beta_{i-1}+\varepsilon_i{q_{i-1}\over q_i}
\beta_i\right)\log q_{i+1} \cr
&= \sum_{i=0}^\infty \beta_{i-1}\log\beta_iq_{i+1} -
\sum_{i=0}^\infty \beta_{i-1}\log\beta_{i-1} +
\sum_{i=0}^\infty \varepsilon_i{q_{i-1}\over q_i}\beta_i\log
q_{i+1}\; , \cr}
$$
but by (1.21), (1.23) and (1.24) one has
$$
\eqalign{
\left|\sum_{i=0}^\infty \beta_{i-1}\log\beta_iq_{i+1}\right| &\le
2\sum_{i=0}^\infty {\log 2\over q_i} \le 2c_2 \; , \cr
\left|\sum_{i=0}^\infty \beta_{i-1}\log\beta_{i-1} \right| &\le
2\sum_{i=0}^\infty {\log 2+\log q_i\over q_i}\le 2(c_1+c_2)\; , \cr
\left|\sum_{i=0}^\infty \varepsilon_i{q_{i-1}\over q_i}\beta_i\log
q_{i+1}\right| &\le
2\sum_{i=0}^\infty {\log q_{i+1}\over q_{i+1}}\le 2 c_1\; , \cr}
$$
from which it follows that
$$
\left|B_\alpha (x)-\sum_{i=0}^\infty {\log q_{i+1}\over q_i} \right|
\le 4(c_1+c_2)\; . \;\;\;
$$\qed
\par
\medskip
Let $P_n/Q_n$ denote the n-th convergent to $x$ according to the
standard continued fraction expansions (i.e. obtained by the iteration of
the Gauss map $A_1$). Using the results of [Bo]  one relates
the n-th convergents of the $\alpha$--continued fractions to $P_n/Q_n$:
\par
\medskip
\Proc{Lemma 1.11.} {Let
$k^\alpha\, : \Bbb N \to \Bbb N$ be the
arithmetic function inductively defined by k(-1)=-1 and
$$
k(n+1) = \cases{ k(n)+1 &if $\varepsilon_{n+1}=+1$,\cr
                 k(n)+2 &if $\varepsilon_{n+1}=-1$,\cr}
$$
where $\varepsilon_{k}$ is defined as in (1.10) and (1.11).
Then $k$ is strictly increasing and for all $n\in \Bbb N$
$$
{p_n\over q_n}={P_{k(n)}\over Q_{k(n)}}\; .
$$
Moreover, when $k(n+1)=k(n)+2$, we have for the denominators of
the convergent of Gauss'continued fraction
$Q_{k(n+1)} = Q_{k(n)+2}=Q_{k(n)+1}+Q_{k(n)}$}
\par
\medskip
By means of Lemma 1.11 one can prove the following
\par
\Proc{Theorem 1.12.}{There exists a positive constant $C>0$ such that
for all $\alpha \in [1/2,1]$ and for all $x \in \Bbb R \setminus \Bbb Q$
one has
$$
\left| B_\alpha (x) - \sum_{j=0}^\infty {\log Q_{j+1} \over Q_j} \right|
   \le C \eqno(1.26)
$$}
\par
\medskip
\proof Thanks to (iv), Proposition 1.10, it suffices to compare
$\sum_{j=0}^\infty {\log q_{j+1} \over q_j}$ with
$\sum_{j=0}^\infty {\log Q_{j+1} \over Q_j}$. By Lemma 1.11, one has
$q_j=Q_{k(j)}$ for all $j$ thus
$$
\sum_{j=0}^\infty {\log q_{j+1} \over q_j} =
\sum_{k(j+1)=k(j)+1} {\log Q_{k(j+1)} \over Q_{k(j)}}
+ \sum_{k(j+1)=k(j)+2} {\log Q_{k(j+1)} \over Q_{k(j)}}\; .
$$
Using the fact that
$Q_{k(j+1)} = Q_{k(j)+2}=Q_{k(j)+1}+Q_{k(j)}$
we have
$$
{\log  Q_{k+2}\over Q_{k}} =  {\log  (Q_{k+1}+Q_k)\over Q_{k}}
={\log Q_{k+1}\over Q_{k}} +
{\log \left(1+{Q_k\over Q_{k+1}}\right) \over Q_k}
$$
but
$$
0 \le {\log \left(1+{Q_k\over Q_{k+1}}\right)\over Q_k}
\le {\log 2\over Q_k} \; ,
$$
By applying the estimates of Remark 1.7 one gets the result:
$$
\left| \sum_{j=0}^\infty {\log q_{j+1} \over q_j} -
\sum_{j=0}^\infty {\log Q_{j+1} \over Q_j} \right| \le 2c_2 + c_1 \; .
\; \;
$$\qed
\par
\medskip
\remark{1.13.} The {\it Brjuno numbers} [Br] are usually
defined by the condition
$$
\sum_{i=0}^\infty {\log Q_{i+1}\over Q_i} < +\infty\; .
$$
Theorem 1.12 shows that the $\alpha$-Brjuno functions $B_\alpha$
are finite at $x$ if and only if $x$ is a Brjuno number and that
all the generalized Brjuno functions differ one from the other
for a $L^\infty$ function.
\par
On the other hand, the advantage of the functions $B_\alpha$ with respect
to the Brjuno condition is that they verify a nice functional equation
under the action of the modular group $\hbox{SL}\,(2,\Bbb Z)$.
\par
\medskip
Another important characterization of the generalized Brjuno functions
comes from their ``uniqueness'', as it is stated by Corollary 1.15 below.
\par
\medskip
Let us consider the operator
$$
(Tf)(x)=xf\left({1\over x}\right)\; ,
\eqno(1.27)
$$
if $x\in (0,\alpha )$, defined for the moment on measurable functions
of $\Bbb R$ which verify
$$
f(x)=f(x+1) \;  \hbox{for almost every}\, x\in \Bbb R\; ,
\;\;\;f(-x)=f(x)\;\hbox{ for a.e.}\, x\in (0,1-\alpha)\; .
\eqno(1.28)
$$
It is understood that the function $Tf$ is completed outside $(0,\alpha)$
by imposing on $Tf$ the same parity and periodicity conditions which
are expressed for $f$ in (1.28).
\par
The functional equation for the $\alpha$-Brjuno function can be written
in the form
$$
[(1-T)B_\alpha ](x)=-\log x\; ,
\eqno(1.29)
$$
for all $x\in (0,\alpha )$, complemented with the periodicity and
symmetry conditions (1.28). This suggest to study the operator
$T$ on the Banach spaces
$$
X_{\alpha ,p}= \left\{ f : \Bbb R \to \Bbb R \mid f \,\hbox{verifies (1.28)}
\; , \;\; f\in L^p(0,\alpha),\right\}
\eqno(1.30)
$$
endowed with the norm of $L^p(0,\alpha )$, as $\alpha$ varies in
$(1/2,1)$ and $p\in [1,\infty ]$. Note that if $p<p'$
one has the obvious inclusion $X_{\alpha ,p'}\subset X_{\alpha ,p}$
and let
$$
X_\alpha = \cap_{p\ge 1}X_{\alpha ,p}\; .
\eqno(1.31)
$$
Clearly
$$
X_{\alpha ,\infty}\subsetneqq X_\alpha \; ,
$$
since, for example, $\log |x-[x]_\alpha |\in X_\alpha\setminus
X_{\alpha ,\infty}$.
\par
\medskip
\Proc{Theorem 1.14.}{$T$ is a linear bounded operator from
$X_{\alpha ,p}$ into itself for all $\alpha\in [{1\over 2},1]$ and
for all $p\in [1,\infty ]$. Indeed we have
$$
|| T||_{{\cal L}(X_{\alpha ,\infty})} = \alpha \; ,
\eqno(1.32)
$$
moreover, if\ \ ${1\over 2}\le \alpha\le {\sqrt{5}-1\over 2}\ $, then
$$|| T||_{{\cal L}(X_{\alpha ,p})}\le
[\zeta (2+p)-1+\zeta (2+p,\alpha )-\alpha^{-(p+2)}]^{1\over p}\ ,
\eqno(1.33)$$
and if\ \  ${\sqrt{5}-1\over 2}<\alpha \le 1\ $, then
$$|| T||_{{\cal L}(X_{\alpha ,p})}\le
[\zeta (2+p)-1+\zeta (2+p,\alpha )-\alpha^{-(p+2)} +\alpha^{p+2}]^{1\over p}
\ ,\eqno(1.34)$$
where $\zeta (s)=\sum_{k=1}^\infty k^{-s}$ and
$\zeta (s,\alpha )=\sum_{k=0}^\infty (k+\alpha )^{-s}$
are Riemann's and Hurwitz's zeta functions.}
\par
\medskip
\proof (1.32) is immediately established by remarking that
the constant function $1$ on $(0,\alpha )$, extended to $\Bbb R$
by means of (1.28), belongs to $X_{\alpha ,\infty}$, has norm equal to $1$
whereas $(T1)(x)=x-[x]_\alpha$,
the norm of which is evidently equal to $\alpha$.
\par
If $f\in X_{\alpha ,p}$ and ${1\over 2}\le \alpha\le {\sqrt{5}-1\over 2}\;$,
since ${1\over \alpha}\ge\alpha +1$
$$
\eqalignno{
\int_0^\alpha |(Tf)(x)|^pdx &= \int_{1\over\alpha}^\infty
{|f(y)|^p\over y^{p+2}}dy \cr
& \le \sum_{k=2}^\infty\left[\int_0^\alpha {|f(y)|^p\over (k+y)^{p+2}}dy
+ \int_0^{1-\alpha }{|f(y)|^p\over (k-y)^{p+2}}dy\right]\cr
& \le || f||_{L^p(0,\alpha )}^p\left(\sum_{k=2}^\infty k^{-(p+2)}
+\sum_{k=1}^\infty (k+\alpha )^{-(p+2)}\right) \; ,&(1.35)\cr}
$$
which proves (1.33).
\par
If $ {\sqrt{5}-1\over 2}<\alpha \le 1\;$, ${1\over \alpha}<\alpha +1$
and by the same argument of above one gets
$$
\eqalignno{
\int_0^\alpha |(Tf)(x)|^pdx &\le || f||_{L^p(0,\alpha )}^p
\left(\sum_{k=2}^\infty k^{-(p+2)}
+\sum_{k=1}^\infty (k+\alpha )^{-(p+2)}\right) \cr
& + \int_{1\over\alpha}^{\alpha +1} {|f(y)|^p\over y^{p+2}}dy \cr
&\le || f||_{L^p(0,\alpha )}^p
\left(\sum_{k=2}^\infty k^{-(p+2)}
+\sum_{k=1}^\infty (k+\alpha )^{-(p+2)}+\alpha^{p+2}\right)\; . \;
\quad\quad&(1.36)\cr}
$$\qed
\par
\medskip
\Proc{Corollary 1.15.} {For all $\alpha\in [{1\over 2},1)$,
the $\alpha$-Brjuno function $B_\alpha$
is the unique solution of (1.29) which belongs to $X_\alpha$.}
\par
\medskip
\proof   The idea of the proof is to show that $T$ is a contraction on
$X_{\alpha,p}$ for $p$ sufficiently large. Then $B_{\alpha}$ is uniquely
defined by $B_{\alpha}(x)=\sum_{n=0}^{\infty}T^n(\log x)$.
If $\alpha \le {\sqrt{5}-1\over 2}$ one can show that $T$ is a
contraction on $X_{\alpha ,p}$ for
all $p\ge 1$: it suffices to remark that
$$
\zeta (p+2)-1+\zeta (p+2,\alpha )-\alpha^{-(p+2)}\le
\zeta (3)-1+\zeta (3,1/2)-8\; ,
$$
but
$$
\zeta (s,1/2)=(2^s-1)\zeta (s) \; ,
$$
thus for the norm of the linear operator $T$ we have
$$
|| T||_{{\cal L}(X_{\alpha ,p})} \le
8\zeta (3) - 9 \le .6164553 \; ,
$$
where we have used the inequality (see [AS]) $\zeta (3)\le 1.2020569032$.
\par
If ${\sqrt{5}-1\over 2}<\alpha <1$ one needs a slighlty more
refined argument to prove that $T$ is a contraction on
$X_{\alpha ,p}$ if $p$ is  large enough. First of all we remark that
$$
\eqalign{
\zeta (s) &= \sum_{k=1}^n k^{-s} + {n^{-s+1}\over s-1}
-s \int_n^\infty {x-[x]\over x^{s+1}}dx \; , \cr
\zeta (s,v) &= \sum_{k=0}^n (k+v)^{-s} + {(n+v)^{-s+1}\over s-1}
-s \int_n^\infty {x-[x]\over (x+v)^{s+1}}dx \; , \cr}
\eqno(1.37)
$$
as it is immediate to show  integrating by parts. Since
$$
-s\int_n^\infty {x-[x]\over x^{s+1}}dx <0\; , \; \; \;
-s \int_n^\infty {x-[x]\over (x+v)^{s+1}}dx<0\; ,
$$
by applying (1.37) with $v=\alpha$ and $n=2$ one gets
$$
\eqalign{
\zeta (2+p)-1 &\le {2^{-p-1}\over p+1}+2^{-p-2}\; , \cr
\zeta (2+p,\alpha ) -\alpha^{-p-2} &\le
{(2+\alpha )^{-p-1}\over p+1} +(2+\alpha )^{-p-2}+(1+\alpha )^{-p-2}
\; , \cr}
$$
thus the condition for $T$ being a contraction reads
$$
{2^{-p-1}\over p+1}+2^{-p-2}+{(2+\alpha )^{-p-1}\over p+1}
+(2+\alpha )^{-p-2}+(1+\alpha )^{-p-2}+\alpha^{p+2}<1 \; ,
$$
which is evidently satisfied for any fixed $\alpha <1$ by choosing
$p$ large enough.\qed
\par
\medskip
\remark{1.16.} By slightly refining the proof one can very easily
show that $T$ is a contraction on $X_{\alpha ,p}$ if:
\item{1)} $p \ge 1$ and $\alpha <.726$;
\item{2)} $p\ge 2$ and $\alpha <.909$;
\item{3)} $p\ge 3$ and $\alpha < .960$;
\par
and so on, and our bounds are still far from being optimal. On the
other hand it is easy to convince oneself by considering some
examples that $T$ cannot be a contraction on $X_{1,p}$.
\par
\medskip
\remark{1.17.} One can easily compute the adjoint $T^*\in
{\cal L}(X_{\alpha ,q})$ of $T\in {\cal L}(X_{\alpha ,p})$, where
$p^{-1}+q^{-1}=1$: let $f\in X_{\alpha ,p}$, $g\in X_{\alpha ,q}$
$$
\eqalign{
\int_0^\alpha (Tf)(x)g(x)dx &= \int_0^\alpha xf\left({1\over x}\right)
g(x)dx = \int_{1\over\alpha}^\infty f(y)g\left({1\over y}\right)
{dy\over y^3} \cr
&= \int_0^\alpha f(y)(T^*g)(y)dy \; , \cr}
$$
where
$$
(T^*g)(y) = \sum_{m\ge m_+}{1\over (m+y)^3}g\left({1\over m+y}\right)
+\sum_{m\ge m_-}{1\over (m-y)^3}g\left({1\over m-y}\right)\; ,
$$
and $m_+$, $m_-$ depend both on $\alpha$ and on $y$, as explained above in
the proof of Theorem 1.1. Note that the
adjoint of $T$ belongs to the class of the so-called
Perron-Frobenius-Ruelle operators [Me1,Me2] associated to $A_\alpha$
$$
({\cal P}_{\alpha ,\phi }g)(y) = \sum_{m\ge m_+}
\phi\left({1\over (m+y)}\right) g\left({1\over m+y}\right)
+\sum_{m\ge m_-}\phi\left({1\over (m-y)}\right)
g\left({1\over m-y}\right)\; .
$$
If $\phi (y)=|A_\alpha '(y)|^{-1}$ then ${\cal P}_{\alpha ,\phi }=
{\cal P}_\alpha$; if $\phi (y)=|A_\alpha '(y)|^{-3/2}$ then
${\cal P}_{\alpha ,\phi }=T^*$. \par\medskip
\remark{1.18.} When $\alpha=1/2$, we have $1/\alpha=2$. Therefore
in (1.35), the second sum in the right hand side needs only to run
over $k=2$ to $\infty$. As a consequence, we have
$$\eqalignno{||T||_{{\cal L}(X_{1/2 ,p})}&\le
\left(\sum_{k=2}^{\infty}\left( k^{-(p+2)}+(k+1/2)^{-(p+2)}
\right)\right)^{1\over p}&(1.38)\cr
&=\left(\zeta (2+p)-1+\zeta (2+p,1/2)-2^{p+2} -\left({2\over3}
\right)^{p+2}\right)^{1\over p}\ .&\quad(1.39)\cr}$$
This can be rewritten as
$$||T||_{{\cal L}(X_{1/2 ,p})}\le \left(2^{p+2}(\zeta(p+2)-1)-1
-\left({2\over3}\right)^{p+2}\right)^{1\over p}\ .\eqno(1.40)$$
The bound in the above equation is sharper than the bound in (1.33).
The explicit values for $p=2$ are
$$||T||_{{\cal L}(X_{1/2 ,2})}\le\left({8\pi^4\over45}-{1393\over81}
\right)^{1\over2}\sim 0.34589\ .\eqno(1.41)$$
By comparison, the numerical value obtained from Equation (1.33) would
be $0.56318$.\par\medskip
\remark{1.19.} The results obtained for the norm of $T$ can easily be
extended to the operators $T_{(\nu)}$ defined as follows for $\nu>0$
$$(T_{(\nu)}f)(x)=x^{\nu}f\left({1\over x}\right)\; ,\eqno(1.42)$$
if $x\in (0,\alpha )$, and completed as in (1.28) outside $(0,\alpha)$.
The adjoint operators $T^*_{(\nu)}$ also
belong to the class of the Perron-Frobenius-Ruelle operators
associated to $A_\alpha$ (see remark 1.17 above).\par
Similarly as equations (1.33), (1.34), and (1.40),
we get the following results:\par
if\ \ ${1\over 2}<\alpha\le {\sqrt{5}-1\over 2}\ $, then
$$|| T_{(\nu)}||_{{\cal L}(X_{\alpha ,p})}\le
[\zeta (2+p\nu)-1+\zeta (2+p\nu,\alpha )-\alpha^{-(p\nu+2)}]^{1\over p}\ ,
\eqno(1.43)$$\par
if\ \  ${\sqrt{5}-1\over 2}<\alpha \le 1\ $, then
$$|| T_{(\nu)}||_{{\cal L}(X_{\alpha ,p})}\le
[\zeta (2+p\nu)-1+\zeta (2+p\nu,\alpha )-\alpha^{-(p\nu+2)}
+\alpha^{p\nu+2}]^{1\over p}\ ,\eqno(1.44)$$\par
and finally if $\alpha={1\over2}$
$$||T_{(\nu)}||_{{\cal L}(X_{1/2 ,p})}\le \left(2^{p\nu+2}(\zeta(p\nu+2)-1)-1
-\left({2\over3}\right)^{p\nu+2}\right)^{1\over p}\ .\eqno(1.45)$$
For $p=2$, this last estimate can be rewritten as
$$||T_{(\nu)}||_{{\cal L}(X_{1/2 ,2})}\le \left(2^{2\nu+2}(\zeta(2\nu+2)-1)-1
-\left({2\over3}\right)^{2\nu+2}\right)^{1\over 2}\ .\eqno(1.46)$$
Proceeding as in the proof of corollary 1.15, one easily checks
that $T_{(\nu)}$ is
a contraction on $X_{\alpha,p}$ for $p$ sufficiently large.\par
\vfill \eject
%%%%%%%% Section 2  %%%%%%
\beginsection 2. BMO and the Brjuno function\par
\vskip .5 truecm
In this section we want to prove that the Brjuno functions $B_\alpha$
belong to the space BMO (see [Gr], and also Appendix A for a short
account on some fundamental results concerning BMO). Since they all
differ by a $L^\infty$ function, and $L^\infty\subset\,$BMO, it will
be enough to prove that $B_\alpha$ is BMO for a fixed value of
$\alpha$.
\par
In this section we fix $\alpha =1/2$ and denote $B_{1/2}$ simply by $B$.
\par
\medskip
Let $I\subset [0,1/2]$ be any interval, and $f\in L^1([0,1/2])$.
We denote by
$$
f_I = {1\over |I|}\int_I f(x)dx
\eqno(2.1)
$$
the mean of $f$ over $I$. Using Lemma A.10, we also define the quadratic
oscillation ${\cal O}_I(f)$ as follows
$$
{\cal O}_I(f) = \left({1\over 2|I|^2}\int_I\int_I(f(s)-f(t))^2dsdt
\right)^{1\over2}\;.\eqno(2.2)$$
\par
As in the previous Section, we define for $1\le p\le\infty$
the space $X_p\equiv X_{1/2,p}$\ :
$$
X_p = \{f\in L^p([0,1/2]) \mid f(x+1)=f(x)\; \forall x\in \Bbb R\; ,
\; f(-x)=f(x) \;\forall x\in [0,1/2]\}\; , \eqno(2.3)
$$
with the usual $L^p$-norm on $[0,1/2]$. In the above definition, $f$
is extended from $[0,1/2]$ to $\R$, in such a way that it is even and
periodic with period 1.\par
We now consider the space
$$
X_* = \{f\in\hbox{BMO}(\Bbb R)\, \mid f(x+1)=f(x)\; \forall x\in \Bbb R\; ,
\; f(-x)=f(x) \;\forall x\in [0,1/2]\}\; , \eqno(2.4)
$$
with the norm
$$
|| f||_* =A |f|_* +B\lnorm{f}{[0,1/2]}{2}  \; ,
\eqno(2.5)
$$
where $A>0$ and $B>0$ are fixed, $\lnorm{f}{[0,1/2]}{2}=
|| f||_{L^2(0,1/2)}$, and
$$
|f|_* = \sup_{I\subset [0,1/2]} {\cal O}_I(f)\; .\eqno(2.6)
$$
Therefore we have
$$||f||_*=A\sup_{I\subset[0,1/2]}\left({1\over 2|I|^2}\int_I
\int_I(f(s)-f(t))^2\,dsdt\right)^{1\over 2}+
B\left(\int_0^{1/2}(f(x))^2\,dx\right)^{1\over 2}\ .\eqno(2.7)$$
\par
\medskip
\remark{2.1.}
Using
John-Nirenberg's theorem and its corollary, the norm we use is equivalent
to the usual BMO norm (see Appendix A, Propositions A.4, A.6, A.8, and A.10).
\par
\medskip
\remark {2.2.}
One has the obvious inclusions $X_{\infty}\subsetneqq X_*
\subsetneqq X_{1}$. By elementary (but tedious) calculations, one checks
that the even and periodic function wich coincides with $\log x$ on
$(0,1/2]$ is in $X_*$, which can be written as
$\log |x-[x]_{1/2}|\in X_*$. \par\medskip
We now recall the definition of the operator $T$. For
$x\in\R$, let $[x]_{1/2}$ be the distance from $x$ to the nearest
integer. For $f\in L^1([0,1/2])$, we define $Tf$ as in (1.27 and 28)
$$(Tf)(x)=xf\left(\left|{1\over x}-\left[{1\over x}\right]_{1/2}\right|
\right)\quad\hbox{for}\quad x\in [0,1/2]\ .\eqno(2.8)$$
We continue $f$ , and $Tf$, to the full real axis so that they are
even and periodic, and (2.8) can be rewritten as $(Tf)(x)=xf(1/x)$,
for $0\le x\le 1/2$.
As already mentioned in Theorem 1.14, $Tf$ is a linear operator from
$X_{p}$ to $X_{p}$ for all $p\ge1$ including $\infty$. More precisely,
the norm of the operator $T$ in the various $X_p$ can be estimated.
We have
$$||T||_{{\cal L}(X_{\infty})}={1\over 2}\ .\eqno(2.9)$$
For finite $p\ge 1$, we use the bound described in Remark 1.18 above
when $\al=1/2$, namely
$$||T||_{{\cal L}(X_{p})}\le C_p\ ,\eqno(2.10)$$
with
$$C_p=\left(\sum_{n=2}^{\infty}\left({1\over n^{p+2}}+
{1\over (n+1/2)^{p+2}}\right)\right)^{{1\over p}}\ .\eqno(2.11)$$
By comparing the sums in (2.11) to an integral over
the interval $(3/2,\infty)$, which we cut at every integer and half-integer
points, we get an easy estimate
$$C_p\le\left(2\int_{3/2}^{\infty}{dx\over x^{p+2}}\right)^{1\over p}
=\left({2\over p+1}\right)^{1\over p}\left({2\over 3}\right)^{1+{1\over p}}
\le{2\over3}\ \ \ \hbox{for $p\ge 1$}\ .\eqno(2.12)$$
Therefore, as already stated in Theorem 2.14, the operator $T$
is a contraction on all $X_p$, for $1\le p\le\infty$.
We shall now consider the action of $T$ on $X_*$.
We have the following theorem.
\Proc{Theorem 2.3.}{$T$ is a
bounded linear operator $T:X_*\mapsto X_*$. Moreover it is a contraction
for the norm (2.7) with $A=1$ and $B\ge 7.405$ (as in (2.41) below), that is
$$||T||_{{\cal L}(X_{*})}\le c_*<1\ \ ,\ \  \hbox{\sl with}\ \  c_*=
{40\sqrt{3}\over 81}\sim 0.855 .\eqno(2.13)$$
In fact, if we allow to increase $B$ to a sufficiently large value,
one can reduce $c_*$ to be as close as one wants to
$(2/5)\sqrt{3}\sim 0.6928$.}\par
\proof\par
  a) Let $f\in X_*$, we shall show that for the seminorm (2.6) we have
$|Tf|_*<\infty$, which will  prove that $T$ is a linear operator
$T:X_*\mapsto X_*$.
For $I=[a,b]\subset[0,1/2]$, with $b>a$, we have from (2.6)
$$\eqalignno{({\cal O}_I(Tf))^2&={1\over2|I|^2}\int_{I\times I}
\left((Tf)(s)-(Tf)(t)\right)^2\,dsdt\cr=
&{1\over2|I|^2}\left[2|I|\int_{I}((Tf)(s))^2\,ds-2\left(\int_I
(Tf)(s)\,ds\right)^2\right]\le{1\over|I|}\int_I((Tf)(s))^2\,ds\cr
&\le{1\over|I|}\int_{J}(f(s))^2{ds\over s^4}\ ,&(2.14)}$$
where $J=[1/b,1/a]\subset [2,\infty]$. Let now $[m/2,n/2]$ be the smallest
interval with half integer edges, which contains $J$.
We have
$$1/b=\be+m/2\quad,\quad 1/a=n/2-\al\quad,\quad 0\le\al<1/2\quad,
\quad0\le\be<1/2\ .\eqno(2.15)$$
Since  $b-a>0$, we have for the integers $m$ and $n$, $4\le m<m+1\le n$.
We get
$$\eqalignno{
({\cal O}_I(Tf))^2&\le{(m+2\be)(n-2\al)\over2(n-m-2\be-2\al)}
\int_{m/2}^{n/2}(f(s))^2{ds\over s^4}&(2.16)\cr
&\le\lnorm{f}{[0,1/2]}{2}^2\,{(m+2\be)(n-2\al)\over2(n-m-2\be-2\al)}
\sum_{k=0}^{n-m-1}{2^4\over(m+k)^4}\ .&(2.17)\cr}$$
Comparing the last sum with the corresponding integral gives
$$\eqalignno{
({\cal O}_I(Tf))^2&\le\lnorm{f}{[0,1/2]}{2}^2\,
{(m+2\be)(n-2\al)\over2(n-m-2\be-2\al)}
\int_{m}^{n}{2^4ds\over (s-1)^4}\cr
&\le\lnorm{f}{[0,1/2]}{2}^2\,
{8(m+2\be)(n-2\al)\over3(n-m-2\be-2\al)}
\left({1\over(m-1)^3}-{1\over(n-1)^3}\right)\quad\quad\quad&(2.18)\cr}$$
However, since $n>m$,
$$\eqalign{{1\over(m-1)^3}-{1\over(n-1)^3}&=
{(n-m)((n-1)^2+(n-1)(m-1)+(m-1)^2)\over(n-1)^3(m-1)^3}\cr
&\le{3(n-m)\over(n-1)(m-1)^3}\ ,\cr}$$
moreover, $m+2\be\le m+1$, and $n-2\al\le n$,
which gives
$$({\cal O}_I(Tf))^2\le8\lnorm{f}{[0,1/2]}{2}^2
\left({n-m\over n-m-2\al-2\be}\right)
\left({n\over n-1}\right)\left({m+1\over (m-1)^3}\right)\ .$$
We have $m\ge4$, and $n\ge5$,  therefore
$$\left({m+1\over (m-1)^3}\right)\le{5\over27}\quad,\quad\hbox{and},\quad
\left({n\over n-1}\right)\le {5\over4}\ .$$
In addition, we have $2|J|=n-m-2\al-2\be$, and $0\le\al+\be<1$,
therefore
$$({\cal O}_I(Tf))^2\le{50\over27}\left(1+{1\over|J|}\right)
\lnorm{f}{[0,1/2]}{2}^2\ .$$
For intervals $I$ such that the inverse interval $J$ is not too small,
that is $|J|\ge\delta$,
we have a uniform bound for ${\cal O}_I(Tf)$ in terms of the
$L^2$-norm of $f$. For $I=[a,b]$,
$$\hfill\hbox{if\ }\left({1\over a}-{1\over b}\right)\ge
\delta\ ,\ \hbox{we have}\ \ \ \ {\cal O}_I(Tf)\le{5\over3}
\sqrt{2\over3}\left(1+{1\over\delta}\right)^{1\over2}
\lnorm{f}{[0,1/2]}{2}\ .\eqno(2.19)$$
\par\smallskip
b) Now we must bound ${\cal O}_I(Tf)$ when $|J|<\delta$.
For convenience, we will assume $\delta\le1/2$. From (2.6), we have
$${\cal O}_I(Tf)=\left({1\over2|I|^2}\int_{I\times I}\left[(s-t)f(1/s)
+t(f(1/s)-f(1/t))\right]^2\,dsdt\right)^{1\over2}\ .$$
Now, using the triangular inequality for the norm $L^2(I\times I)$,
one gets
$${\cal O}_I(Tf)\le\Gamma_1+\Gamma_2\ ,\eqno(2.20)$$
where
$$\eqalignno{\Gamma_1=&\left({1\over2|I|^2}\int_{I\times I}(s-t)^2\left[
f(1/s)\right]^2\,dsdt\right)^{1\over2}\ ,&(2.21)\cr
\Gamma_2=&\left({1\over2|I|^2}\int_{I\times I}t^2\left[
f(1/s)-f(1/t)\right]^2\,dsdt\right)^{1\over2}\ .&(2.22)}$$
\par\smallskip
c) We first find an upper bound for $\Gamma_1$. In the integral in
formula (2.21), we have $|s-t|<|I|$, and $|I|\le 1/2$, therefore
one gets
$$\eqalign{
\Gamma_1^2\ \le& {1\over4}\int_I\left(f\left({1\over s}\right)\right)^2\,ds=
\int_J{(f(s))^2\over4s^2}\,ds\cr
\le&\int_2^{\infty}{(f(s))^2\over4s^2}\,ds\le\sum_{n=0}^{\infty}
{1\over4(2+n/2)^2}\int_{2+n/2}^{3/2+n/2}(f(s))^2\, ds\cr&\le
{\lnorm{f}{[0,1/2]}{2}^2\over4}\sum_{n=0}^{\infty}{1\over(2+n/2)^2}
\le{\lnorm{f}{[0,1/2]}{2}^2\over2}\int_{-1}^{\infty}{1\over(2+s/2)^2}
\,ds\ ,\cr}$$
where we have used the comparison of the sum with an integral.
Finally, one gets
$$\Gamma_1\le\sqrt{2\over3}\lnorm{f}{[0,1/2]}{2}\ .\eqno(2.23)$$
The contribution $\Gamma_1$ to ${\cal O}_I(Tf)$ is estimated in terms
of The $L^2$-norm of $f$.\par\smallskip
d) Now we will estimate $\Gamma_2$. From (2.22), one gets
$$\Gamma_2=\left({1\over2|I|^2}\int_{J\times J}\left[
(f(s)-f(t))\right]^2\,{dsdt\over s^2t^4}\right)^{1\over2}\ ,\eqno(2.24)$$
where, once more, $J=[1/b,1/a]$ is the image of $I$ under inversion.
We assume $|J|<\delta\le 1/2$, and we still define $m$, $n$, $\al$, $\be$,
as in (2.15). Then we have $n-m-2\al-2\be<2\de\le1$.
We then have either $n=m+1$, or $n=m+2$.\par
If $n=m+1$, there is no points with half-integer coordinates in
$J\times J$, and therefore $J\times J$ is contained in the interior
of one cell of the lattice of half integer points. We bound from below
$s^2t^4$ by $(m/2+\be)^6$ in (2.24), and using the parity and periodicity
of $f$, in order to shift the integration over $[0,1/2]\times [0,1/2]$,
we get from (2.6)
$$\Gamma_2\le{1\over(m/2+\be)^3}\left({|J|\over|I|}\right)
|f|_*={m/2+1/2-\al\over(m/2+\be)^2}|f|_* \ ,\eqno(2.25)$$
and
$$\Gamma_2\le\left({1\over m+2\be}+{2\de\over(m+2\be)^2}\right)2|f|_*
\le\left({1\over2}+{\de\over4}\right)|f|_*
\le{5\over8}|f|_*\ .\eqno(2.26)$$
\indent If $n=m+2$, there is one, and only one point with half-integer
coordinates $(s,t)$, which is interior to $J\times J$. This point has
cordinates $(q,q)$, with $q=p/2$ and $p=m+1=n-1$. We have $q\ge 5/2$,
and we set $\al'=1/2-\al$ and $\be'=1/2-\be$.\par
The square $J\times J$ has non empty
intersection with four cells of the lattice of half integer points.
The lower left edge of these four cells are $(q-1/2,q-1/2)$
and $(q,q)$ for the `diagonal' ones, and  $(q-1/2,q)$ and
$(q,q-1/2)$ for the `off diagonal' ones. The integral representation
for $\Gamma_2^2$ deduced by squaring (2.24) can be split into four
contributions corresponding to each cell.\par
The intersection of $J\times J$ with each diagonal cell is a square,
and in the corresponding integral, we first bound from below $s$ and $t$
by the coordinates of its lower left edge. The remaining integrals
are nothing else than $(|K|/|I|)^2({\cal O}_K(f))^2$, with
$K=[q-\be',q]$ or $K=[q,q+\al']$ respectively.
Then, due to the parity and periodicity property of $f$, these
integrals are bounded by  $(|K|/|I|)^2|f|_*^2$, with
$|K|=\be'\ \hbox{or}\ \al'$ respectively.\par
The intersection of $J\times J$ with the off diagonal cells are rectangular
but we still can bound from below $s$ and $t$ in the integrals by the
coordinates of the lower left edge. Then, we enlarge each rectangle
up to the smallest square in which it is contained, and having
with the rectangle the common edge $(q,q)$. Due to the parity and
periodicity property of $f$, the two remaining integrals are both
equal to $(|K|/|I|)^2({\cal O}_K(f))^2$, with $K=[q,q+\ga']$
and $\ga'=\sup(\al',\be')$. They can be bounded by $(|K|/|I|)^2|f|_*^2$,
with $|K|=\ga'$. Putting everything together, one gets
$$\Gamma_2\le C_2|f|_*\ ,\eqno(2.27)$$
with
$$|I|^2C_2^2\le {\be'^2\over(q-\be')^6}+{\al'^2\over q^6}+
{\ga'^2\over q^2(q-\be')^4}+{\ga'^2\over q^4(q-\be')^2}
\ .\eqno(2.28)$$
{}From (2.15), we have
$$|I|=b-a={1\over\be+m/2}-{1\over n/2-\al}={\al'+\be'\over(q-\be')(q+\al')}
\ .$$
Therefore
$$C_2^2={(q+\al')^2\over(q-\be')^4}\left({\be'^2+\al'^2(1-\be'/q)^6+
\ga'^2(1-\be'/q)^4+\ga'^2(1-\be'/q)^2 \over(\al'+\be')^2}\right)\ ,$$
which implies, since $q\ge 5/2$, and $\be'<1/2$,
$$C_2^2={(q+\al')^2\over(q-\be')^4}\left({\be'^2+\al'^2+
2\ga'^2\over(\al'+\be')^2}\right)\ .\eqno(2.29)$$
However, $\be'^2+\al'^2+2\ga'^2=3\be'^2+\al'^2\ \hbox{or}\ \be'^2+3\al'^2$,
depending on whether $\be'\ge\al'$, or $\be'\le\al'$. In both cases,
it is easy to check that $\be'^2+\al'^2+2\ga'^2\le3(\al'+\be')^2$,
since $\al'$, $\be'$ (and $\ga'$) are non negative. Then
$$C_2^2=3\left({(q+\al')^2\over(q-\be')^4}\right)\ ,\eqno(2.30)$$
so that,
$$\Gamma_2\le \sqrt{3}\left({(q+\al')\over(q-\be')^2}
\right)|f|_*\ ,\eqno(2.31)$$
which is valid for $q\ge 5/2$, $0\le\al' <1/2$,
$0\le\be'<1/2$, and, since $n-m=2$, for $\al'+\be'<\de$.
Now (2.31) implies
$$\Gamma_2\le \sqrt{3}\left({1\over(q-\be')}+{\al'+\be'\over(q-\be')^2}
\right)|f|_*\le\sqrt{3}\left({1\over(q-\de)}+{\de\over(q-\de)^2}
\right)|f|_*\ ,$$
that is
$$\Gamma_2\le \sqrt{3}\left({q\over(q-\de)^2}\right)|f|_*
\le\sqrt{3}\left({10\over(5-2\de)^2}\right)|f|_*\ .\eqno(2.32)$$
\par\smallskip
e) Putting together (2.20), (2.23), (2.26) and (2.32), we get
the following statement.\par\noindent
For $I=[a,b]$, if $\left({\dst1\over\dst a}-{\dst1\over\dst b}\right)
<\delta$ we have
$${\cal O}_I(Tf)\le\sqrt{2\over3}\lnorm{f}{[0,1/2]}{2}+\sup\left({1\over2}
+{\de\over4},{10\sqrt{3}\over(5-2\de)^2}\right)|f|_*\ .$$
A bit of analysis shows that the supremum is always reached by the
second argument when $0\le\de\le 1/2$. Therefore
$${\cal O}_I(Tf)\le\sqrt{2\over3}\lnorm{f}{[0,1/2]}{2}+\left(
{10\sqrt{3}\over(5-2\de)^2}\right)|f|_*\ .\eqno(2.33)$$
The Equations (2.19) and (2.33) show that $|Tf|_*$
is finite, when $|f|_*$ and $\lnorm{f}{[0,1/2]}{2}$ are finite,
which completes the proof that  $f\in X_*$ implies $Tf\in X_*$.
\par\smallskip
f) We will now show that $T$ is a bounded linear operator $X_*\to X_*$
for a suitable choice of the coefficients $A$ and $B$ in the norm
(2.7).\par
First we note that (2.10) and (2.12) imply
$$\lnorm{Tf}{[0,1/2]}{2}\le c_1\lnorm{f}{[0,1/2]}{2}\ ,\eqno(2.34)$$
with $c_1=4/9$. Then, collecting results of (2.19) and (2.33)
we get for some choice of the parameter $\de$,
$$|Tf|_*\le c_*|f|_*+c_2 \lnorm{Tf}{[0,1/2]}{2}\ .\eqno(2.35)$$
Therefore,
$$A|Tf|_*+B\lnorm{Tf}{[0,1/2]}{2}\le c_*\left(A|f|_*+\left({Ac_2\over c_*}+
{Bc_1\over c_*}\right)\lnorm{f}{[0,1/2]}{2}\right)\ .$$
However can choose $A>0$ and $B>0$ such that
$$Bc_*\ge Ac_2+Bc_1\ ,\eqno(2.36)$$
because $c_*>c_1=4/9$. In fact the latter inequality results from
(2.33) which shows that $c_*>1/2$.
With such a choice for $A$ and $B$, we would get
$$A|Tf|_*+B\lnorm{Tf}{[0,1/2]}{2}\le c_*\left(A|f|_*+B
\lnorm{f}{[0,1/2]}{2}\right)\ ,$$
that is, in view of (2.7),
$$||Tf||_*\le c_*||f||_*\ .\eqno(2.37)$$
The latter equation shows that $T$ is a bounded operator in $X_*$,
for the norms $||f||_*$ with $$A=1\quad,\quad \hbox{and}
\quad\quad B\ge {c_2\over c_*-c_1}\ .\eqno(2.38)$$
With this norm on $X_*$, the norm of the operator $T$
fulfils $||T||_{{\cal L}(X_*)}\le c_*$.\par\smallskip
g) We now show that by a suitable choice of $\de$, we will
get $c_*<1$, which will show that $T$ is a contraction on $X_*$.
we have from (2.33)
$$c_*=\left({10\sqrt{3}\over(5-2\de)^2}\right)\ .\eqno(2.39)$$
Comparing (2.19) and (2.33) shows that
$$c_2={5\over3}\sqrt{2\over3}\left(1+{1\over\delta}\right)^{1\over2}
\ .\eqno(2.40)$$
However, as seen from (2.39), $c_*$ is a monotone
function of $\de$, which increases from $0.6928$ to $1.0825$ when $\de$
increases from $0$ to $1/2$. It reaches one for $\de$ around $0.4$.
Taking $\de $ smaller than $0.4$, proves that $T$ is a contraction.
In order to give numbers, we take $\de=1/4$,
wich gives
$$c_2={5\over3}\sqrt{10\over3}\quad\quad,\quad\quad c_*={40\sqrt{3}
\over 81}\sim 0.855$$
The corresponding values of $B$ are
$$B\ge\left({5\over3}\sqrt{2\over3}\left(1+{1\over\de}\right)^{1\over2}\right)
\left({10\sqrt{3}\over(5-2\de)^2}-{4\over9}\right)^{-1}={45\over2}
\sqrt{5\over2}{1\over10-3\sqrt{3}}\sim 7.405 .\eqno(2.41)$$
If now one takes $\de$ very small, one can reduce $c_*$ as close as one wants
to its value
for $\de=0$, which is $(2/5)\sqrt{3}=0.6928$, but meanwhile, $B$ must
become very large, and increase proportionnaly with $\de^{-(1/2)}$.
This completes the proof of Theorem 2.3.\qed
\par An immediate consequence of Theorem 2.3 is the following theorem
\Proc{Theorem 2.4}{The Brjuno function $B(x)\equiv B_{1/2}(x)$ belongs
to $X_*$, and therefore to \BM{\R}. The Brjuno functions $B_{\al}(x)$
belong to \BM{\R} for $1/2\le\al\le 1$.}\par
\proof $B(x)$ is the unique solution of (1.29) for $\al=1/2$, that is
$$((1-T)B)(x)=-\log x\quad, \quad\hbox{when}\quad 0\le x\le 1/2\ .$$
However $T$ is a contraction in $X_*$, and  $\log x$ is in \BM{[0,1/2]}.
Therefore, $1-T$ is invertible in \BM{[0,1/2]}, and $B\in\BM{[0,1/2]}$.
{}From Corollary A.9, we deduce $B\in\BM{\R}$.
Now, from Theorem 1.12, for $1/2\le\al\le1$ we have
$B_{\al}-B_{1/2}\in L^{\infty}(\R)$,
which implies $B_{\al}\in\BM{\R}$, since
$L^{\infty}(\R)\subset\BM{\R}$.\qed
Let us now go back to Theorem 2.3. The contraction property of $T$ depends
on the norm we have taken on $X_*$. Indeed it is not valid in all possible
equivalent norms. However, the spectral radius of $T$ in  $X_*$ does not
depend on the choice of the norm in the equivalence class. This makes the
following Theorem 2.5 relevant. \par\medskip
\Proc{Theorem 2.5.}{ The spectral radius of $T$ in $X_*$ is at most equal
to $\sqrt{2}-1$\ .}\par
\proof\par
a) We already know from Theorem 2.3 that the spectral radius of
$T$, which is defined as $\rho(T)=\limsup ||T^n||^{1/n}$, fulfils
$$\rho(T)\le{2\sqrt{3}\over5}\sim 0.6928\ .\eqno(2.42)$$\par\smallskip
b) We first show that $||T||_{{\cal L}(X_2)}<\sqrt{2}-1$. But that is just
where we need the estimate of Equation (1.40). Indeed we have
for $f\in X_2$
$$\lnorm{Tf}{[0,1/2]}{2}\le c_1\lnorm{f}{[0,1/2]}{2}\ ,\ \ \ \hbox{so that}
\ \ \ ||T||_{{\cal L}(X_2)}\le c_1\ ,\eqno(2.43)$$
with
$$c_1=\left({8\pi^4\over45}-{1393\over81}
\right)^{1\over2}\sim 0.34589...\le\sqrt{2}-1\sim 0.4142\ .\eqno(2.44)$$
Obviously we have
$$\lnorm{T^nf}{[0,1/2]}{2}\le c_1^n\lnorm{f}{[0,1/2]}{2}\ .\eqno(2.45)$$
\par\smallskip c) Assume now $f\in X_*$, and let $\de$ be a positive
number such that $\de\le1/2$. We consider intervals
$I=[a,b]\subset [0,1/2]$.
Let $J=[1/b,1/a]$ be the image of $I$ under the change $x\to 1/x$.
There are two possibilities:
\item{c1)} $J$ has length greater or equal to $\de$.
\item{c2)} $J$ has length smaller than $\de$.\par
We will now estimate in case c1) the quadratic oscillation ${\cal O}_I(Tf)$
as defined in (2.6). For this purpose, the argument in part a) of the
proof of Theorem 2.3 which leads to (2.19) may be used without
changes.
Therefore we have for $I=[a,b]$ and $\de\le1/2$,
$$\hfill\hbox{if\ }\ \ \left({1\over a}-{1\over b}\right)\ge\delta\ ,
\ \hbox{we have}\ \ \ \ {\cal O}_I(Tf)\le{5\over3}
\sqrt{2\over3}\left(1+{1\over\delta}\right)^{1\over2}
\lnorm{f}{[0,1/2]}{2}\ .\eqno(2.46)$$
Obviously from (2.43), we have (still in case c1) above) for $n\ge1$,
$${\cal O}_I(T^nf)\le{5\over3}\sqrt{2\over3}\left(1+{1\over\delta}
\right)^{1\over2}c_1^{n-1}\lnorm{f}{[0,1/2]}{2}\ .\eqno(2.47)$$
We must now deal with case c2), that is we assume
$(1/a-1/b)<\de\le1/2$.
The argument of part b) of the proof of Theorem 2.3 applies without changes
and we get as in (2.20)
$${\cal O}_I(Tf)\le\Gamma_1+\Gamma_2\ ,\eqno(2.48)$$
where $\Gamma_1$ and $\Gamma_2$ are given by (2.21) and (2.22) respectively.
We get a bound for $\Gamma_1$ as in part c) of the proof of Theorem 2.3,
namely
$$\Gamma_1\le\sqrt{2\over3}\lnorm{f}{[0,1/2]}{2}\ .\eqno(2.49)$$
which is nothing else than Equation (2.23).\par
Now we use (2.24) to estimate $\Gamma_2$, and we will introduce
the transformation
$\tau: I\to\tau I$, defined on intervals $I$ which are subintervals of
$[0,1/2]$, and such that the inverse interval $J$ has
length less than $\de\le1/2$. When these conditions are not fulfilled
(that is in case c1) above), we shall say that $\tau I$ is
{\it undefined}.
Let now $[r/2,s/2]$ be the smallest interval with integer or half integer
edges, which
contains $J=[1/b,1/a]\subset[2,\infty]$.
We have
$$1/b=r/2+\be\quad,\quad 1/a=s/2-\al\quad,\quad 0\le\al<1/2\quad,
\quad0\le\be<1/2\ .\eqno(2.50)$$
Since  $b-a>0$, we have for the integers $r$ and $s$, $4\le r<r+1\le s$.
Since we assume $|J|<\delta\le 1/2$, we have $s-r-2\al-2\be<2\de\le1$.
We then have either $s=r+1$, or $s=r+2$.\par
If $s=r+1$, there is no integer or half-integer point in the interior of $J$,
and we can define $\tau I$
as the interval contained in $[0,1/2]$ deduced from $J$ by translation or
by translation followed by a symmetry around zero.
We get $\tau I=[\be,1/2-\al]$ when $r$ is even, and $\tau I=[\al,1/2-\be]$
when $r$ is odd. We have $|\tau I|=|J|=1/2-\al-\be<\de\le 1/2$. It
is easy to check that, in the present case, $\tau I$ is nothing
else than the image of $I$ under the map $A_{1/2}$ defined in Section 1,
Equation (1.3). Then, following the argument which leaded us to (2.25)
we get
$$\Gamma_2\le{1\over (r/2+\be)^3}\left({|\tau I|\over|I|}\right)
{\cal O}_{\tau I}(f)={r/2+1/2-\al\over(r/2+\be)^2}
{\cal O}_{\tau I}(f)={b^2\over a}{\cal O}_{\tau I}(f)\ .\eqno(2.51)$$\par
If $s=r+2$, we have $J=[(r+1)/2-(1/2-\be),(r+1)/2+(1/2-\al)]$, so that
the point $(r+1)/2$ is in $J$. When $r$ is odd, by translation, we can
shift $J$ to $J'=[\be-1/2,1/2-\al]$ which contains $0$. In this case,
we define $\tau I=[0,\sup(1/2-\al,1/2-\be)]$. When $r$ is even, by
translation, we can shift $J$ to $J'=[1/2-(1/2-\be),1/2+(1/2-\al)]$
which contains $1/2$. In this case we define
$\tau I=[1/2-\sup(1/2-\al,1/2-\be),1/2]$. In both cases, we have
$|\tau I|=\sup(1/2-\al,1/2-\be)$, and
$$|J|/2\le(1-\al-\be)/2\le|\tau I|\le 1-\al-\be=|J|<\de\ .\eqno(2.52)$$
In the present case $\tau I$ is contained in the image of $I$ through
the map $A_{1/2}$. Once more, following the argument which leaded us to
(2.25) we get, since $|J|=|J'|$
$$\Gamma_2\le{1\over (r/2+\be)^3}\left({|J|\over|I|}\right)
{\cal O}_{J'}(f)={r/2+1-\al\over(r/2+\be)^2}
{\cal O}_{J'}(f)\ .\eqno(2.53)$$
By Lemma A.11, we have ${\cal O}_{J'}(f)\le 2 {\cal O}_{\tau I}(f)$,
and
$$\Gamma_2\le{2\over (r/2+\be)^3}\left({|J|\over|I|}\right)
{\cal O}_{\tau I}(f)={2(r/2+1-\al)\over(r/2+\be)^2}
{\cal O}_{\tau I}(f)={2b^2\over a}{\cal O}_{\tau I}(f)\ .\eqno(2.54)$$
Note than when $s=r+2$, one of the edges of $\tau I$ is either $0$ or
$1/2$. We summarize these results:
$$\eqalignno{&\hbox{if $\tau I$ is undefined}\quad{\cal O}_I(Tf)\le{5\over3}
\sqrt{2\over3}\left(1+{1\over\delta}\right)^{1\over2}\lnorm{f}{[0,1/2]}{2}
&(2.55)\cr&\hbox{if $\tau I$ is defined}\quad{\cal O}_I(Tf)\le\sqrt{2\over3}
\lnorm{f}{[0,1/2]}{2}+{\la b^2\over a}{\cal O}_{\tau I}(f)\ ,
&(2.56)\cr}$$
where $\la =1\hbox{\ or\ }2$, depending on whether $s=r+1$, in which
case (2.51) holds, or $s=r+2$, in which case (2.54) holds.
When $\tau I$ is defined, $a$ is strictly positive, and conversely
when $a=0$, $\tau I$ is undefined, because $|J|$ is then infinite. When
$\la=2$, either $0$ or $1/2$ belongs to $\tau I$.
Setting
$$C={5\over3}\sqrt{2\over3}\left(1+{1\over\delta}\right)^{1\over2}>
\sqrt{2\over3}\ ,\eqno(2.57)$$
we rewrite (2.55) and (2.56) in one single equation
$${\cal O}_I(Tf)\le C\lnorm{f}{[0,1/2]}{2}+{\la b^2\over a}
{\cal O}_{\tau I}(f)\ ,\eqno(2.58)$$
with the convention that $\la=0$ if $\tau I$ is undefined,
so that the second term in the above equation is absent (this
happens in particular when $a=0$).\par
Note that $1/a-1/b<\de$ implies $b/a<1+b\de$, and since $b\le1/2$,
$${2b^2\over a}<\ga=1+{\de\over2}\ .\eqno(2.59)$$
We have $|\tau I|=|J|$ when $\la=1$, and from (2.52), $|J|/2\le
|\tau I|\le|J|$ when $\la=2$. Also $|I|/|J|=ab\le1/4$,
therefore we have
$$\left.\matrix{\hbox{when $\la=1$}\hfill\quad&{\dst|\tau I|\over\dst|I|}=
{\dst1\over\dst ab}\ge4\quad\hfill\cr\hbox{when $\la=2$}\quad\hfill&{\dst
|\tau I|\over\dst|I|}\ge{\dst1\over\dst 2ab}\ge2\quad\hfill\cr}\right\}
\ \ .\eqno(2.60)$$\par\smallskip
d) We are now ready to iterate the procedure. We start from
$$\eqalign{{\cal O}_I(T^nf)&\le C\lnorm{T^{n-1}f}{[0,1/2]}{2}+
{\la b^2\over a}{\cal O}_{\tau I}(T^{n-1}f)\cr&\le
Cc_1^{n-1}\lnorm{f}{[0,1/2]}{2}+{\la b^2\over a}
{\cal O}_{\tau I}(T^{n-1}f)\ ,\cr}$$
and use once more the equation when $\tau I$ is defined, shifting $n$ to
$n-1$, and so one. We get after $k$ steps (for $k\le n-1$)
$${\cal O}_I(T^{n}f)\le C_k\lnorm{f}{[0,1/2]}{2}+\left(\prod_{i=0}^k
{\la_i b_i^2\over a_i}\right){\cal O}_{\tau^{k+1} I}
(T^{n-k-1}f)\ .\eqno(2.61)$$
The definition of $\tau^k I$ in the above equation, is recursive.
Obviously, if $\tau^k I$ is undefined, $\tau^{k+1} I$ is also undefined.
We sart from
$\tau^0 I=I$, then we set $\tau^{k+1} I=\tau(\tau^k I)$ if it is
defined, and the value of $\la_k$ is $1$ or $2$ as explained above
after (2.56). If $\tau(\tau^k I)$ is undefined, we set $\la_k=0$.
We also have set $\tau^k I=[a_k,b_k]$, and $I=[a_0,b_0]$.
The recursion starts with
$$C_0=c_1^{n-1}C\ ,$$
and proceeds as
$$C_{k+1}=C_k+c_1^{n-k-2}C\left(\prod_{i=0}^k{\la_i b_i^2\over a_i}
\right)\ ,$$
so that
$$C_{k+1}=C_0+C\sum_{j=0}^kc_1^{n-j-2}\left(\prod_{i=0}^j{\la_i b_i^2
\over a_i}\right)\le C_0+C\sum_{j=0}^kc_1^{n-j-2}\ga^{j+1}\ ,$$
that is
$$C_{k+1}=Cc_1^{n-k-2}{\ga^{k+2}-c_1^{k+2}\over\ga-c_1}\ .$$
Setting $k=n-2$, we get
$$C_{n-1}=Cc_1^{-2}{\ga^{n}-c_1^{n}\over\ga-c_1}\ ,\eqno(2.62)$$
where $\ga$ is given by (2.59) and $c_1$ by (2.44). From (2.61) we get
$${\cal O}_I(T^{n}f)\le C_{n-1}\lnorm{f}{[0,1/2]}{2}+\left(\prod_{i=0}^{n-1}
{\la_i b_i^2\over a_i}\right){\cal O}_{\tau^n I}(f)\ .\eqno(2.63)$$
The important feature of the above equation is that the coefficient
of $\lnorm{f}{[0,1/2]}{2}$ does not depend on $I$.
We will now bound the coefficient of ${\cal O}_{\tau^n I}(f)$.
For this purpose, we observe that in the sequence $\la_0,\la_1,\ldots
\la_{n-1}$, there is at most one element equal to $2$. Indeed, assume
that $\la_r=2$ for some $r$. We have already noticed after (2.56)
that if $\la_r=2$, then $\tau^{r+1} I$ contains either $0$ or $1/2$.
If it contains zero, its inverse is infinite, then $\la_{r+1}=0$
since then $\tau^{r+1} I$ is undefined and all following $\la's$ vanish.
If $\tau^{r+1} I$ contains $1/2$, then its inverse contains $2$,
and either $\tau^{r+1} I$ is undefined in which case $\la_{r+1}=0$,
or it has $0$ as lower edge, and then
$\la_{r+1}=1$. By the same argument $\tau^{r+2} I$ is then undefined,
$\la_{r+2}=0$ and all following $\la's$ vanish. If all $\la_i\neq0$
for $0\le i\le n-1$, then all $\la_i=1$ except may be $\la_{n-2}$ or
$\la_{n-1}$. Therefore we have
$${\cal O}_I(T^{n}f)\le C_{n-1}\lnorm{f}{[0,1/2]}{2}+\left(2\prod_{i=0}^{n-1}
{b_i^2\over a_i}\right){\cal O}_{\tau^n I}(f)\ .\eqno(2.64)$$
Taking as in (2.6) the supremum over all possible intervals $I$, we get
$$|T^{n}f|_*\le C_{n-1}\lnorm{f}{[0,1/2]}{2}+C_{*,n}|f|_*\ ,\eqno(2.65)$$
where $C_{*,n}$ is given by
$$C_{*,n}=\sup_{I}\left(2\prod_{i=0}^{n-1}{b_i^2\over a_i}\right)
\ ,\eqno(2.66)$$
where the supremum is taken over intervals $I$ such that the intervals
$\tau^k I=[a_k,b_k]$ are defined for $1\le k \le n$.\par\smallskip
e) We shall now prove there exists a constant $K>1$ independant on $n$
such that
$$C_{*,n}\le 2K (\sqrt{2}-1)^n\ .\eqno(2.67)$$\par
{}From the definition of $\tau$ given in part c) above, it is easy to see
that one can find for $0\le i \le n-1$ a sequence of numbers $x_i$, such that
$x_i\in\tau^i I=[a_i,b_i]$, and $x_{i+1}=A_{1/2}(x_i)$,
for $0\le i \le n-2$, where
$$A_{1/2}(x)=\left|{1\over x}-\left[{1\over x}\right]_{1/2}\right|
\ ,\eqno(2.68)$$
as in (1.3). For that purpose, take $x_{n-1}$ arbitrarily in the interior
of $\tau^{n-1} I$, and compute backward $x_k$ from $x_{k+1}$, adding at
each step the suitable integer part.
We choose $x_{n-1}$ irrational to avoid the discontinuity points
of the integer part $[x]_{1/2}$ defined in (1.1).
We obviously have
$$|b_i-a_i|\le |\tau^i I|\quad,\quad
|x_i-a_i|\le |\tau^i I|\quad,\quad  |x_i-b_i|\le |\tau^i I|\ .\eqno(2.69)$$
{}From (2.60), we get for $0\le k\le n-1$,
$$ |\tau^k I|\le 2^{-(n-k)}|\tau^{n} I|\le 2^{-(n-k)}\de\le2^{-(n-k+1)}
\ ,\eqno(2.70)$$
since $\de\le 1/2$. In fact we also get a better bound
$$ |\tau^k I|\le 2a_kb_k|\tau^{k+1} I|\le a_kb_k2^{-(n-k-2)}\de\le
a_kb_k2^{-(n-k-1)}\ .\eqno(2.71)$$
Then we get
$${b_i^2\over a_i}=x_i\left(1+{b_i-a_i\over a_i}\right)\left(1+
{b_i-x_i\over x_i}\right)\le x_i\left(1+{b_i-a_i\over a_i}\right)^2\ ,$$
so that using (2.69) and (2.71), we have
$${b_i^2\over a_i}\le x_i\left(1+{b_i\over 2^{n-1-i}}\right)^2
\le x_i\left(1+{1\over 2^{n-i}}\right)^2\ .$$
Therefore
$$\eqalign{\prod_{i=0}^{n-1}{b_i^2\over a_i}&\le \exp\left(2\sum_{i=0}^{n-1}
\log(1+2^{-(n-i)})\right)\prod_{i=0}^{n-1}x_i \le\exp\left(2\sum_{i=0}^{n-1}
2^{-(n-i)}\right)\prod_{i=0}^{n-1}x_i\cr&\le \exp\left(\sum_{i=0}^{n-1}
2^{-i}\right)\prod_{i=0}^{n-1}x_i\le\exp\left(\sum_{i=0}^{\infty}2^{-i}\right)
\prod_{i=0}^{n-1}x_i\le e^2\prod_{i=0}^{n-1}x_i\ .\cr}$$
Finally,
$$C_{*,n}\le 2e^2\sup_{x_0}\left(\prod_{i=0}^{n-1}x_i\right)
=2e^2\sup_{x_0}\left(\prod_{i=0}^{n-1}A^k_{1/2}(x_0)\right)\ ,\eqno(2.72)$$
where now the supremum is taken over $x_0$, that is over all possible
trajectories of the map $x\to A_{1/2}(x)$. The estimate (2.72) deserves some
comments. Although the map $\tau$ which transforms the interval $[a_i,b_i]$
into $[a_{i+1},b_{i+1}]$ give the same result as the action of $A_{1/2}$
when $\la_i$ is equal to $1$, the sequences $a_0,a_1,\ldots,a_n$,
and $b_0,b_1, \ldots,b_n$ are not in general trajectories for $A_{1/2}$,
because at each step $a_{i+1}$ is defined as the lower edge of
$\tau [a_i,b_i]$. Moreover, when $\la=2$, there is a further restriction
of the interval. Nevertheless these sequences can be approximated in a
suitable way by a true trajectory of $A_{1/2}$, because this map is
expansive. In particular, this approximation prevents the $a_i$'s to become
too small. The determination of the trajectory of $A_{1/2}$ is in fact
an appliction of the so-called Shadowing Lemma, which is a classical
trick in expansive dynamical systems.\par
The  estimate (2.67) then resuts from Proposition 1.4. which  is an
almost classical result in the theory of continued fractions described
in Section 1.\par
f) From (2.45) we have
$$\lnorm{T^nf}{[0,1/2]}{2}\le c_1^n\lnorm{f}{[0,1/2]}{2}\ ,\eqno(2.73)$$
and from (2.64) and (2.67) we get
$$|T^{n}f|_*\le C_{n-1}\lnorm{f}{[0,1/2]}{2}+2K(\sqrt{2}-1)^n|f|_*\ ,
\eqno(2.74)$$
By an argument similar to part f) of the proof of Theorem 2.3, we
use a norm of the family (2.7) with $A=1$ and
$$B>{C_{n-1}\over2K(\sqrt{2}-1)^n-c_1^n}\ ,$$
which is positive in view of (2.44). With this norm,
$$||T^nf||_*\le 2K(\sqrt{2}-1)^n||f||_*\ .\eqno(2.75)$$
Since $\limsup\ (2K)^{1/n}=1$, the spectral radius of $T$ in $X_*$ fulfils
$$||T||_{{\cal L}(X_*)}\le \sqrt{2}-1\ .\eqno(2.76)$$
which completes the proof of Theorem 2.5.\qed
\medskip\noindent
\remark {2.6.} The same argument can be applied to the operator $T_{(\nu)}$
introduced in Remark 1.18. The main change in the argument is the replacement
of $(b^2/a)$ in the equations (2.51) and (2.56) by $(b^{1+\nu}/a)$, so that
(after recomputing all constants!) one finally
gets instead of (2.74) and estimate of the kind.
$$|T^{n}_{(\nu)}f|_*\le C_{n-1,\nu}\lnorm{f}{[0,1/2]}{2}+K_{\nu}
(\sqrt{2}-1)^{n\nu}|f|_*\ ,\eqno(2.77)$$
A careful examination of (1.35) for $\al=1/2$
shows that one can reduce the $L_2$-norm estimate up to
$$\eqalignno{||T_{(\nu)}||_{{\cal L}(X_{1/2 ,2})}&\le
\left(2\sum_{n=2}^{\infty}{1\over (m+1/2)^{2+2\nu}}\right)^{1\over 2}\cr
&\le \left(2(2^{2+2\nu}-1)\zeta(2+2\nu)-2^{3+2\nu}-2\left({2\over3}
\right)^{2+2\nu}\right))^{1\over 2}&(2.78)\cr}$$
Therefore, the spectral radius of $T_{(\nu)}$ is less or equal to the
largest of the two following numbers: the BMO-norm estimate
$(\sqrt{2}-1)^{\nu}$ and the above
$L_2$-norm estimate (2.78), which is better than (1.46).
We already know that the BMO-norm estimate is
smaller when $\nu=1$, we also see that the $L_2$-norm will be smaller for
large $\nu$ since it behaves as  $(2/5)^{\nu}$.\par\bigskip
A more concise way toward the spectral radius of $T$ can be
obtained using the invariant measure described in Section 1
above. This can be found in [MMY].\par
\vfill \eject
%%%%%% Section 3 %%%%%%%
\beginsection 3. Regularity properties for the functional equation of the
Brjuno function\par
\vskip .5 truecm
The functional equation (1.29) for the Brjuno function for $\alpha=1/2$ is
$$
[(1-T)B_{1/2}](x)=-\log x\; ,
\eqno(3.1)
$$
for all $x\in (0,1/2)$, complemented with the condition that $B=B_{1/2}$
is even
and periodic.  In this Section we will suppose that the right hand side of
this equation is pertubed, by an additional term $f$, which is
less singular than the
logarithmic function. We want to study the singular properties of the
perturbed solution. Since  the equation is linear, we only need to consider
the action on $f$ of $T$ and $(1-T)^{-1}$, which
will be conveniently called the
Brjuno operator ${\bf B}$. We will consider even and periodic functions $f$
which are {\it continuous}. It is sufficient to know the value of $f$ on
$[0,1/2]$, so we assume $f\in C^0_{[0,1/2]}$. One can check  that
$Tf$ is also continuous provided we set $Tf(0)=0$. We need now the usual
H\"older's type semi-norms for continuous functions.
\Proc{Definition 3.1.}{Let $f\in C^0_{[0,1/2]}$, Then we define the
H\"older's $\ga$-norm as
$$\hnorm{f}{\ga}=
\sup_{0\le x<y\le1/2} {\dst|f(x)-f(y)|\over\dst|x-y|^{\ga}}\ ,\eqno(3.2)$$
with $0<\ga\le 1$.
This is a seminorm since it vanishes on constant functions, so that we
introduce the norm:
$$\Hnorm{f}{\ga}=A\hnorm{f}{\ga}+B\hnorm{f}{\infty}\ ,\eqno(3.3)$$
with $A$ and $B$ positive constants. We say that $f\in C^{\ga}$,
if $f\in C^0_{[0,1/2]}$ and $\hnorm{f}{\ga}$ is finite.}\par
We now have:
\Proc{Proposition 3.2.}{\item{i)} $T$ is a bounded operator in $C^{\ga}$,
for the norm $\Hnorm{f}{\ga}$, provided $0<\ga\le 1/2$, for $B/A$
large enough: if $B/A>(2^{\ga}-2^{-\ga})^{-1}$, the norm of $T$
corresponding to the norm (3.3) satisfies
$\Hnorm{T}{\ga}\le 2^{(2\ga-1)}\le1$.
\item{ii)} For $0<\ga<1/2$, $T$ is a contraction.}\par
We need the following Lemma
\Proc{Lemma 3.3}{Let $0< y<x\le 1/2$, and define $x_1$ and $y_1$ by the
following conditions
$$y={1\over n+y_1}\quad,\quad x={1\over m+x_1}\ ,\eqno(3.4)$$
with $n\ge2$ and $ m\ge 2$, and $-1/2\le x_1<   1/2 $ and
$-1/2\le y_1<   1/2$, then we have
$$\left\vert\vert x_1\vert- \vert y_1\vert\right\vert\le{|x-y|\over|x||y|}
\ .\eqno(3.5)$$}\par
\noindent{\it Proof of Lemma 3.3.} Since $y<x$, we have
$n-m>x_1-y_1>   -1$. therefore $n\ge m$. Let $n-m=p\ge0$.
We have $x-y=xy(p+y_1-x_1)$. So that we need to prove
$||x_1|-|y_1||\le |p+y_1-x_1|$. This is obvious when $p=0$.
We always have $||x_1|-|y_1||\le 1/2$, so the required inequality also holds
when $p\ge 2$. In the remaining case $p=1$, we set
$\eta= \hbox{sign}(y_1)$ and
$\epsilon=\hbox{sign}(x_1)$, and we need to check that
$||x_1|-|y_1||\le |1+\eta|y_1|-\epsilon|x_1||$. Still because the left hand
side is smaller or equal to $1/2$, this last inequality is not obvious
only when $\eta=-1$ and $\epsilon=+1$. It therefore remains to show that
$||x_1|-|y_1||\le |1-|y_1|-|x_1||$. Setting $u=1/2-|x_1|$ and $v=1/2-|y_1|$,
the last inequality is equivalent to $|1-v/u|\le|1+v/u|$, which is readily
checked since $u/v$ is real and non-negative.  \qed
\noindent{\it Proof of Proposition 3.2} Let $0<   y<x\le1/2$, and $x_1$ and
$y_1$  as in the preceding lemma. We have
$$\eqalign{|Tf(x)-Tf(y)|&=|xf(1/x)-yf(1/y)|=|xf(|x_1|)-yf(|y_1|)|\cr
&\le|x-y||f(|x_1|)|+|y||f(|x_1|)-f(|y_1|)|\cr
&\le|x-y|\hnorm{f}{\infty}+|y|\hnorm{f}{\ga}||x_1|-|y_1||^{\ga}\cr
&\le|x-y|\hnorm{f}{\infty}+\hnorm{f}{\ga}{|x-y|^{\ga}\over
|x|^{\ga}|y|^{\ga-1}}\ ,\cr}$$
where we have used $f\in C^{\ga}$, and Lemma 3.3. Therefore
$$\eqalign{{|Tf(x)-Tf(y)|\over|x-y|^{\ga}}&\le
|x-y|^{1-\ga}\hnorm{f}{\infty}+\left({|y|\over|x|}\right)^{\ga}
|y|^{1-2\ga}\hnorm{f}{\ga}\cr &\le (1/2)^{1-\ga}\hnorm{f}{\infty}+
(1/2)^{1-2\ga}\hnorm{f}{\ga} \ ,\cr}$$
since $0<   y<x\le 1/2$, and $\ga\le 1/2$. For $y=0$,
the right hand side can be replaced
by its first term $(1/2)^{1-\ga}\hnorm{f}{\infty}$,  and
the above inequality extends to the case where $y$ vanishes, so that
$$\hnorm{Tf}{\ga}\le K_{\ga}(f)=2^{\ga-1}\hnorm{f}{\infty}+
2^{2\ga-1}\hnorm{f}{\ga}\ .\eqno(3.6)$$
For the norm, we get
$$\eqalign{\Hnorm{Tf}{\ga}=&A\hnorm{Tf}{\ga}+B\hnorm{Tf}{\infty}\cr
&\le 2^{2\ga-1}\left[A\hnorm{f}{\ga}+(2^{-2\ga} B+2^{-\ga}A)\hnorm{f}{\infty}
\right]\ ,\cr
&\le 2^{(2\ga-1)}\Hnorm{f}{\ga}\ ,\cr}$$
provided $2^{-2\ga} B+2^{-\ga}A\le B$, that is
$$A/B\le 2^{\ga}-2^{-\ga}$$
which completes the proof.\qed
\par\medskip\noindent
The above proposition has two obvious consequences
\item{i)} Since $C^{\ga}\subset C^{\ga'}$ whenever $\ga'\le\ga$, we have
$$\eqalign{&f\in C^{\ga}\quad\hbox{and}\quad\ga\ge 1/2\quad\quad
\Longrightarrow\quad Tf\in C^{1/2}\cr
&f\in C^{\ga} \quad\hbox{and}\quad\ga\ge \ga_0\quad,\quad \ga_0\le1/2\quad
 \quad \Longrightarrow\quad Tf\in C^{\ga_0}\ .\cr}$$
\item{ii)} When $\ga<1/2$, $T$ is a contraction on $C^{\ga}$. Therefore
$1-T$ is invertible and ${\bf B}=\sum_0^{\infty} T^n =(1-T)^{-1}$ preserves
$C^{\ga}$,
and we have.
$$\eqalign{&f\in C^{\ga}\quad\hbox{and}\quad 0<\ga< 1/2\quad\quad
\Longrightarrow\quad {\bf B}f\in C^{\ga}\cr
&f\in C^{1/2} \quad\quad \Longrightarrow\quad {\bf B}f\in C^{\ga}\quad,\quad
\forall \ga \quad\hbox{such that}\quad 0<\ga<1/2\ .\cr}$$
\par \medskip
The Brjuno function which we have studied in the previous section is
nothing else than ${\bf B}\ell$, where $\ell$ is equal to minus the
logarithmic fonction restricted to $[0,1/2]$. When made even and periodic,
this function is not continuous. Suppose that we perturb $\ell$ by a function
with enough regularity properties (for example $C^1$ or $C^{1/2}$,
the change in  ${\bf B}\ell$ will be continuous and even
$C^{1/2-\epsilon}$ for any small $\epsilon$.
In this sense, the `most singular ' part of the Brjuno function
is stable or `universal', roughly speaking modulo H\"older one-half
continuous contributions.
\par\medskip
In fact we have a slightly better result for ${\bf B}$ than for $T$, as
shown in the next proposition, which shows that the $C^{1/2}$
property is effectively reached.
\Proc{Proposition 3.4.}{If $f\in C^{\ga}$\ , and $\ga>1/2$, then
${\bf B}f\in C^{1/2}$.}\par
\proof We use ${\bf B}f(x)=\sum_{n=0}^{\infty}(T^nf)(x)=
\sum_{n=0}^{\infty}\be_{n-1}(x)f(x_n)$,
which is analogous to (1.25), using convention $\be_{-1}(x)=1$.
Therefore we have
$$|{\bf B}f(x)-{\bf B}f(y)|=\left\vert\sum_{n=0}^{\infty}\be_{n-1}(x)f(x_n)-
\be_{n-1}(y)f(y_n)\right\vert\ ,\eqno(3.7)$$
where $x_i$ and $y_i$  are the respective trajectories of $x$ and $y$
under $A_{1/2}$.\par
Now, for each pair of numbers $(x,y)$, such that $x\neq y$,
we define the number $N$ which tells how far the
continued fraction expansions of $x$ and $y$ do coincide, more precisely:
we have $x_k^{-1}=a_{k+1}(x)+\varepsilon_{k+1}(x)x_{k+1}$, and
$y_k^{-1}=a_{k+1}(y)+\varepsilon_{k+1}(y)y_{k+1}$. We assume that
$a_{k}(x)=a_{k}(y)$, $\varepsilon_{k}(x)=\varepsilon_{k}(y)$
as long as $k\le N$. At least one of these equalities fail for $k>N$.
In order to avoid ambiguities in  the rational case, where we have $x_n=0$
for some $n$, we take $\varepsilon_n(x)=+1$ and $x_k=0$
whenever $k>n$, and let for $k>n$, $a_k$ and $\varepsilon_k$ be  undefined
(one could let $a_k$ be infinite). Therefore, if $x_n$ or  $y_n$ vanish
for some $n$, we get $N\le n$. We then have
$${1\over x_k}-{1\over y_k}=\varepsilon_{k+1}(x_{k+1}-y_{k+1})\quad
\hbox{when}\quad k+1\le N\ .\eqno(3.8)$$
Therefore we get, still for $k+1\le N$,
$$|x_{k+1}-y_{k+1}|={|x_k-y_k|\over|x_k||y_k|}={|x-y|\over\be_k(x)\be_k(y)}
\ .\eqno(3.9)$$
We also see that the numerators $p_k$, and the denominators $q_k$ of
the truncated continued fraction expansion of $x$ and $y$ coincide
as long as $k\le N$, therefore, we first have
$$\hbox{for}\ \ k\le N\quad,\quad |\beta_k(x)-\beta_k(y)|=
q_k|x-y|\ ,\eqno(3.10)$$
and also, due to Proposition (1.4, iii),
$$\hbox{for}\ \ k\le N-1\quad,\quad{1\over3}\ \le
\ {\be_k(x)\over\be_k(y)} \ \le\  3\ .\eqno(3.11)$$
We separate (3.7) into four contributions and divide by $|x-y|^{1/2}$,
that is,
$${|{\bf B}f(x)-{\bf B}f(y)|\over|x-y|^{1/2}} \le E_1+E_2+E_3+E_4\eqno(3.12)$$
with
$$\eqalignno{E_1=&|x-y|^{-1/2}\sum_{n=0}^{N}|\be_{n-1}(x)-\be_{n-1}(y)|
|f(x_n)|\ ,&(3.13)\cr
E_2=&|x-y|^{-1/2}\sum_{n=0}^{N}|\be_{n-1}(y)||f(x_n)-f(y_n)|\ ,&(3.14)\cr
E_3=&|x-y|^{-1/2}\sum_{n=N+1}^{N+2}|\be_{n-1}(x)f(x_n)
-\be_{n-1}(y)f(y_n)|\ .&(3.15)\cr
E_4=&|x-y|^{-1/2}\sum_{n=N+3}^{\infty}|\be_{n-1}(x)f(x_n)
-\be_{n-1}(y)f(y_n)|\ .&(3.16)\cr}$$
Our proof will be completed if we can bound $E_1$ to $E_4$ in terms
of $\hnorm{f}{\infty}$ and $\hnorm{f}{\ga}$ for $f\in C^{\ga}$ and
$\ga>1/2$.\par
1) Bound for $E_1$. From (3.10) we get $E_1\le|x-y|^{1/2}\hnorm{f}{\infty}
\sum_{n=1}^{N}q_{n-1}$, which using (3.8), gives
$E_1\le|x_N-y_N|^{1/2}\hnorm{f}{\infty}(\be_{N-1}(x)\be_{N-1}(y))^{1/2}
\sum_{n=0}^{N-1}q_{n}$,
but from Proposition 1.4., we have for $n\le N$,
$q_{n}<2(\be_{n-1}(x))^{-1}$ and $q_{n}<2(\be_{n-1}(y))^{-1}$
from which we get $q_{n}<2 (\be_{n-1}(x)\be_{n-1}(y))^{-1/2}$.
Therefore,
$$E_1\le2|x_N-y_N|^{1/2}\hnorm{f}{\infty}
\sum_{n=0}^{N-1}\left({\be_{N-1}(x)\over \be_{n-1}(x)}\right)^{1/2}
\left({\be_{N-1}(y)\over \be_{n-1}(y)}\right)^{1/2}\ ,\eqno(3.17)$$
However, using Proposition 1.4.,
$$\left({\be_{N-1}(x)\over \be_{n-1}(x)}\right)=\be_{N-n-1}(x_{n})
\le (\sqrt{2}-1)^{N-n-1}\ ,$$
and the sum in (3.14) is bounded by the converging series
$\sum_{n=0}^{\infty}(\sqrt{2}-1)^n=(2-\sqrt{2})^{-1}$.
Furthermore, $|x_N-y_N|\le1/2$
and $E_1$ is bounded by a constant time $\hnorm{f}{\infty}$.\par
2) Bound for $E_2$. We have $E_2\le |x-y|^{-1/2}\hnorm{f}{\ga}
\sum_{n=0}^{N}|\be_{n-1}(y)||x_n-y_n|^{\ga}$, so that, using (3.9),
$$E_2\le \hnorm{f}{\ga}
\sum_{n=0}^{N}\left({\be_{n-1}(y)\over\be_{n-1}(x)}\right)^{1/2}
|x_n-y_n|^{\ga-1/2} \ .\eqno(3.18)$$
Using (3.11), we get
$E_2\le 3^{1/2}\hnorm{f}{\ga} \sum_{n=0}^{N}
|x_n-y_n|^{\ga-1/2} \ .$
However for $n\le N$, we see from (3.9) that
$$|x_n-y_n|=|x_N-y_N|\left({\be_{N-1}(x)\be_{N-1}(y)\over\be_{n-1}(x)
\be_{n-1}(y)}\right)\le (\sqrt{2}-1)^{2N-2n-2}|x_N-y_N|\ .$$
Then $$|x_n-y_n|\le {(\sqrt{2}-1)^{2N-2n-2}\over2}\ ,$$
since $x_N-y_N\le1/2$, and the sum from 0 to $N$ is
bounded since  the geometric series
$\sum_{n=0}^{\infty}(\sqrt{2}-1)^{2n(\ga-1/2)}$ converges.
As a consequence, $E_2$ is bounded by  a constant times $\hnorm{f}{\ga}$.
Note that we have explicitely used here the hypothesis $\ga>1/2$. \par
3) Bound for $E_3$. We observe that since $\ga>1/2$, $f\in C_{\ga}
\Rightarrow f\in C_{1/2}$, and
in fact, we have $\hnorm{f}{1/2}\le 2^{(1/2)-\ga}\hnorm{f}{\ga}$.
We first bound the term of order $N+1$.
$$\eqalign{|\be_{N}&(x)f(x_{N+1}) -\be_{N}(y)f(y_{N+1})|=
|\be_{N-1}(x)(Tf)(x_N)-\be_{N-1}(y)(Tf)(y_N)| \cr&\le
|(\be_{N-1}(x) -\be_{N-1}(y))(Tf)(x_{N})|+
|\be_{N-1}(y)((Tf)(x_{N})-(Tf)(y_{N}))| \cr&\le
(1/2)q_{N-1}\hnorm{f}{\infty}|x-y|+K_{1/2}(f)|x_N-y_N|^{1/2}\be_{N-1}(y)
\ ,\cr} $$
where we have used (3.10) and (3.6) for the exponent $1/2$.
Therefore, using (3.9), one gets
$$\eqalign{{|\be_{N}(x)f(x_{N+1})-\be_{N}(y)f(y_{N+1})|\over|x-y|^{1/2}}
&\cr\le (1/2)\hnorm{f}{\infty}q_{N-1}&|x-y|^{1/2}+{K_{1/2}(f)\be_{N-1}(y)
\over|\be_{N-1}(x)\be_{N-1}(y)|^{1/2}}\ .\cr} $$
However, from (3.9), one gets
$|x-y|=|x_N-y_N|x_{N-1}y_{N-1}\be_{N-2}(x)\be_{N-2}(y)$, and  from
Proposition (1.4), $q_{N-1}\be_{N-2}\le 2$. Then (3.11) shows that both
terms of the right hand side in the above inequality are bounded
(in terms of $\hnorm{f}{\infty}$ and $\hnorm{f}{1/2}$).\par
We now bound the term of order $N+2$ in $E_3$.
We have
$$\eqalign{|\be_{N+1}(x)f(x_{N+2}) -\be_{N+1}(y)f(y_{N+2})|=&
|\be_{N}(x)((Tf)(x_{N+1})-(Tf)(y_{N+1}))|\  \cr+
|(\be_{N-1}(x)-\be_{N-1}(y))y_N(Tf)(y_{N+1})|+&
|\be_{N-1}(x)(Tf)(y_{N+1})(x_N-y_N)|\ ,  \cr\le&
\be_N(x)K_{1/2}(f)|x_{N+1}-y_{N+1}|^{1/2}\cr+y_Ny_{N+1}
\hnorm{f}{\infty}q_{N-1}|x-y|+&y_Ny_{N+1}\hnorm{f}{\infty}
\be_{N-1}(x)|x_N-y_N|\ .\cr}$$
We now use Lemma 3.3 to get $x_Ny_N|x_{N+1}-y_{N+1}|\le |x_N-y_N|$,
and we get
$$\eqalign{& {|\be_{N+1}(x)f(x_{N+2})-\be_{N+1}(y)f(y_{N+2})|
\over|x-y|^{1/2}} \cr
\le&  K_{1/2}(f)
\left({y_N\over x_N}\right)^{1/2}\be_{N-1}(x)
\left|{x_N-y_N \over x-y}\right|^{1/2}
\cr +&(1/4)\hnorm{f}{\infty}
\left(q_{N-1}|x-y|^{1/2}+\be_{N-1}(x){|x_N-y_N|
\over|x-y|^{1/2}}\right)\ .\cr}$$
In the above expression, we have got three terms to bound. In the first
we have the factor $(y_N/x_N)^{1/2}$ which is bounded when $y_N\le x_N$.
In the other case it is sufficient to permute $x$ and $y$.
It then remains the factor $\be_{N-1}(x)(|x_N-y_N|/|x-y|)^{1/2}$, which
is easily  bounded using (3.8) and (3.10). The same combination occurs in
the third term, with an additional bounded factor $|x_N-y_N|^{1/2}$.
The middle term $q_{N-1}|x-y|^{1/2}$ has been already treated for the
previous term (of order $N+1$).
\par
4) Bound for $E_4$. We have $$E_4=|x_N-y_N|^{-1/2}(\be_{N-1}(x)
\be_{N-1}(y))^{-1/2}\sum_{n=N+3}^{\infty}|\be_{n-1}(x)f(x_n)
-\be_{n-1}(y)f(y_n)|\ .\eqno(3.19)$$
{}From the above definition of $N$, we know that $2x_N^{-1}$  and
$2y_N^{-1}$ do not  both belong to the same interval  in the family  of
intervals including  for all $M\ge 2$,  $[2M,2M+1)$ and $[2M+1,2M+2)$.
Therefore the interval $[2x_N^{-1},2y_N^{-1}]$ contains at least one
integer point (we assume  $x_N>y_N$, the other possibility
being treated by permuting $x$ and $y$). Let now $r$ and $R$, be
the rationals between $x$ and $y$ having, up to order $N$ included, the
same continued fraction expansion coefficient as $x$ and $y$, and such that
$2r_N^{-1}$ and $2R_N^{-1}$ are respectively the smallest and the largest
integer such that
$2x_N^{-1}\le 2r_N^{-1}\le 2R_N^{-1}\le y_N^{-1}$. We now have
$$|{\bf B}f(x)-{\bf B}f(y)|\le |{\bf B}f(x)-{\bf B}f(r)|+
|{\bf B}f(r)-{\bf B}f(R)|+ |{\bf B}f(R)-{\bf B}f(y)|$$
and since we have either $x\le r\le R\le y$ or $y\le R\le r \le x$,
we see that we need only to get a bound  for $E_4$ in the two following cases:
either $x=r$ and $y=R$ are both rational and  $2x_N^{-1}$ and $2y_N^{-1}$
are different integers, or one of these (say $y=r$) has these properties, and
$|x_N^{-1}-y_N^{-1}|< 1/2$. Indeed the bound already obtained for
$E_1$, $E_2$, and $E_3$, apply to the present situations.\par
Now we observe that when $2r_N^{-1}$ is even, we have $r_{N+1}=0$,
and when $2r_N^{-1}$ is odd, we have $r_{N+1}=1/2$ and $r_{N+2}=0$,
so that the sum in (3.19) contains no terms coming
from $r$ or $R$. More precisely, there is no $E_4$ type contribution
in $|r-R|^{-1/2}|{\bf B}f(r)-{\bf B}f(R)|$, which was the motivation
to split the sum  for $n>N$ in (3.7) between $E_3$ and $E_4$. For the same
reason, in the other relevant
case---where $y=r$, and $-1/2< x_N^{-1}-r_N^{-1}< 1/2$---Equation (3.19)
reduces to
$$\eqalignno{E_4=&|x_N-r_N|^{-1/2}(\be_{N-1}(x)
\be_{N-1}(r))^{-1/2}\sum_{n=N+3}^{\infty}\be_{n-1}(x)|f(x_n)|\cr
\le&{\sqrt{3}\hnorm{f}{\infty}x_Nx_{N+1}x_{N+2}\over |x_N-r_N|^{1/2}}
\sum_{n=N+3}^{\infty}\be_{n-N+3}(x_{N+3})\cr
\le&\left({\sqrt{3}\hnorm{f}{\infty}\over 2-\sqrt{2}}\right)
\left({x_Nx_{N+1}x_{N+2}\over |x_N-r_N|^{1/2}}\right)
\ ,&(3.20)\cr}$$
where we have used (3.11), proposition (1.4), and the sum of the
converging series $\sum_{n=0}^{\infty}(\sqrt{2}-1)^n$.
We now define $\theta$ such that
$${1\over x_N}={1\over r_N}+\theta\quad,\quad -1/2<\theta<1/2
\ ,\eqno(3.21)$$
so that $|x_N-r_N|=r_Nx_N|\theta|$, and we have now to get a bound
for the remaining factor
$${x_Nx_{N+1}x_{N+2}\over|x_N-r_N|^{1/2}}=
{x_Nx_{N+1}x_{N+2}\over x_N^{1/2}r_N^{1/2}|\theta|^{1/2}}=
{x_{N+1}x_{N+2}\over|\theta|^{1/2}}\left({x_N\over r_N}\right)^{1/2}\ .$$
Notice that since $2x_N^{-1}$ and $2r_N^{-1}$ belong to the same
interval $[M,M+1]$, with $M\ge2$, their ratio remain bounded,
that is $2/3\le x_N/r_N\le 3/2$, so we need therefore only to get a bound
for $x_{N+1}x_{N+2}|\theta|^{-1/2}$.\par
In the case where $2r_N^{-1}$ is even, we have $x_{N+1}=|\theta|$,
so that $x_{N+1}x_{N+2}\theta^{-1/2}<2^{-3/2}$. In the case
where $2r_N^{-1}=2P+1$ is odd, we have $x_{N+1}=1/2-|\theta|$.
Either  $|\theta|\ge 1/10$  which implies $x_{N+1}\le2/5$
and $x_{N+1}x_{N+2}|\theta|^{-1/2}\le(2/5)^{1/2}$
or $|\theta|< 1/10$  and $2/5<x_{N+1}\le1/2$, in which case
$x_{N+2}=2-x_{N+1}^{-1}=2|\theta|/(1/2-|\theta|)$,and
$x_{N+1}x_{N+2}|\theta|^{-1/2}=2\sqrt{\theta}\le (2/5)^{1/2}$.
This concludes the proof of the proposition. \qed
\medskip
The regularity results of the present section might provide an
explanation for the numerical results [Ma], which suggested
a relation between the Brjuno function and the size of the
stability domain as function of the rotation number in
some holomorphic area preserving maps. Note that some rigorous results
have been obtained in the case of the semistandard map [Ma,Da].
We anticipate that  this result might be much  more general:
the geometric renormalisation for holomorphic dynamical systems
will likely produce only $C^1$ perturbations to the renormalisation
equation, and the most singular part of minus the logarithm of
the size of the stability domains as function of the rotation number
could be universally (that is modulo $C^{1/2}$) described by the Brjuno
function.\par\bigskip
%%%%%%%%%% Appendix A %%%%
\noindent {\bf Appendix A. Norms on B.M.O. spaces}\par
\bigskip
\beginsection 1. B. M. O. Norms\par
Let $f\in\Lloc$. We define the mean value $f_I$ of $f$ on the interval
$I$ as:
$$ f_I=\mean{I}{f}\ ,\eqno({\rm A}.1)$$
where $|I|$ is the length of the interval $I$. Then we define
for any interval $U$
$$ \norm{f}{U}=\Dnorm{I\subset U}{f}\ .\eqno({\rm A}.2)$$
We then say $f$ belongs to the space \BM{U} if $\norm{f}{U}<\infty$,
\ie is finite. $BMO$ is an abbreviation for `bounded mean oscillation'.
$\norm{f}{U}$ is a seminorm on \BM{U}, since for any constant $c$,
we have $\norm{f+c}{U}=\norm{f}{U}$. In particular $\norm{f}{U}=0$
if $f$ is constant on $U$. This applies to $U=\R$ and leads
to the space \BM{\R}, abbreviated as $BMO$ and the seminorm $\norm{f}{\R}$
on \BM{\R} will simply be written  $\norm{f}{}$.
In fact this seminorm is a norm on the quotient space of function
in $\Lloc$ modulo the constant functions. With this norm, the quotient
space is complete.\par\smallskip
We list now some classical results and lemmas [Gr,GRCF].
\Proc{Proposition A.1.}{If there exists a positive number $M$ such
that for any interval
$I\subset U$, we can find a number $\al_I$ with
$\Mean{I}{f-\al_I}\le M\ ,$ then $\vert f_I-\al_I\vert\le M$
and $\norm{f}{U}\le 2M\ .$}\par
\proof We have $(f_I-\al_I)=\mean{I}{(f-\al_I)}$, thus
$|f_I-\al_I|\le\Mean{I}{f-\al_I}\le M$, and
$\Mean{I}{f-f_I}=\Mean{I}{f-\al_I+\al_I-f_I}\le\Mean{I}{f-\al_I}
+|\al_I-f_I|\le 2M$. Therefore $\norm{f}{U}\le 2M$.\qed
\Proc{Proposition A.2.}{The space $L^{\infty}(U)$ is a subspace of \BM{U}, and
$\norm{f}{U}\le\inf_{c}\lnorm{f-c}{U}{\infty}$}.\par
\proof From H\"older's inequality, we get for any interval $I\subset U$,
$\Mean{I}{f-f_I}\le\left(\mean{I}{|f-f_I|^2}\right)^{1/2}$.
But $\mean{I}{(f-f_I)^2}=\mean{I}{f^2}-(f_I)^2
\le\mean{I}{f^2}\le\left(\lnorm{f}{U}{\infty}\right)^2$. As a result,
$\norm{f}{U}\le\lnorm{f}{U}{\infty}$, the $L^{\infty}$ norm of $f$ on $U$.
Since $\norm{f}{U}$ is equal for any $c$ to $\norm{f+c}{U}$, we immediately
deduce that $\norm{f}{U}\le\lnorm{f-c}{U}{\infty}$ on $U$, for
any $c$.\qed
\Proc{Proposition A.3.}{Assume $f\in \BM{U}$, with seminorm
$\norm{f}{U}$. Let $I$
and $J$ be two intervals in $U$ . Let $m$ be an arbitrary
integer such that $m>1$, we have\hfb
\quad\quad a) If $I\subset J$ and $1\le{\dst |J|\over\dst |I|}\le m $,
\ \ \ then \ \ $|f_I-f_J|\le m\norm{f}{U}$.\hfb
\quad\quad b) If $I\subset J$ and  $m<{\dst|J|\over\dst|I|}$,\ \ \ then
$$|f_I-f_J|\le \left(1+{\dst 1\over\dst \log m}\log\left({\dst
|J|\over\dst|I|}\right)\right) m\norm{f}{U}
\le{\dst 2m\over\dst \log m}\log\left({\dst
|J|\over\dst|I|}\right)\norm{f}{U}\ .$$
\quad\quad c) If $|I|=|J|$, let $K$ be the smallest interval containing
both $I$ and $J$. Assume first $|K|\le 2|I|=2|J|$, which means that $I$
and $J$ overlap or are consecutive intervals, then
$|f_I-f_J|\le 4\norm{f}{U}$. On the contrary, if
$|K|> 2|I|=2|J|$, which means that $I$ and $J$ are disjoint intervals,
then $|f_I-f_J|\le {\dst 8\over\dst \log 2}\norm{f}{U}\log\left({\dst
|K|\over\dst|I|}\right)={\dst8\over\dst\log 2}\norm{f}{U}\log
\left(2+{\dst\delta(I,J)\over\dst|I|}\right)$
where $\delta(I,J)$ is the length of the interval between $I$ and $J$.}\par
\proof a) $|f_I-f_J|\le\Mean{I}{f-f_J}={|J|\over|I|}{1\over |J|}\int_I
|f-f_J|\,dx\le {|J|\over|I|}{1\over|J|}\int_J|f-f_J|\,dx\le {|J|\over|I|}
\norm{f}{U}\le m\norm{f}{U}\ .$\par
b) Let $t={|J|\over|I|}$, and for $n> 1$, let $\lambda=t^{1/n}$.
It then easy to find intervals $I_2, I_3,\ldots,I_n$
chosen in an arbitrary way, but subjected to the conditions
$I=I_1\subset I_2\subset I_3\ldots\subset I_n\subset I_{n+1}=J$,
$|I_{k+1}|/|I_k|=\lambda$ for $1\le k\le n$, and
$${|J|\over |I|}={|I_{n+1}|\over|I_{n}|}{|I_{n}|\over|I_{n-1}|}\ldots
{|I_{2}|\over|I_{1}|}=\lambda^n=t\ .$$
Using $|f_I-f_J|=|f_{I_{1}}-f_{I_{2}}|+|f_{I_{2}}-f_{I_{3}}|+\ldots
+|f_{I_{n}}-f_{I_{n+1}}|$, we get
$$|f_I-f_J|\le \left(\sum_{i=k}^{n}{|I_{k+1}|\over|I_{k}|}\right)
\norm{f}{U}=nt^{1\over n}\norm{f}{U}\quad\hbox{as in the proof of a).}\ $$
For $t>m$, let now $n>1$ be the integer such that
$m^{n-1}<t\le m^n$, so that $t^{1/n}\le m$, and
$$n<1+{1\over\log m}\log t\le{2\over\log m}\log t\ .$$
Therefore, from $|f_I-f_J|\le nt^{1/n}\norm{f}{U}$, we get
$$|f_I-f_J|\le\left(1+{1\over\log m}\log
\left({|J|\over|I|}\right)\right)m\norm{f}{U}\le{2m\over\log m}\log
\left({|J|\over|I|}\right)\norm{f}{U}\ .$$
In fact we have got much better than statements a) and b): for
$I\subset J\subset U$, we have $|f_I-f_J|\le S\left({|J|\over|I|}\right)
\norm{f}{U}$,
where the function $S(t)$ is defined for $t\ge 1$ as
$S(t)=\inf_{n\ge 1}nt^{1/n}$. After some calculations, we find
$S(t)=t$ for $1\le t\le 4$, and $S(t)\le(2/\log 2)\log t$ for $t\ge 4$.
For $t$ large, we get $S(t)\sim e \log t$.\par
c) We have $|f_I-f_J|\le|f_K-f_I|+|f_K-f_J|$ and the statement follows from
parts a) or b) with $m=2$.\qed
The preceding proposition shows that when $|I|$ goes to zero, the mean
$f_I$ cannot grow faster than $|\log(|I|)|$ times some constant.
\Proc{Proposition A.4.}{We have the following `magic
reverse H\"older's inequality':
let $f\in\Lloc$ and suppose that for some interval $U\subset\R$,
the seminorm $\norm{f}{U}$ is finite, then for any bounded real $p\ge 1$,
there exists a constant $A_p$ such that
$$\sup_{I\subset U}\left(\mean{I}{|f-f_I|^p}\right)^{\sst1\over\sst p}
\le A_p\norm{f}{U}\ .$$}\par
\proof  In fact it is a corollary of the John-Nirenberg
theorem, see Garnett [Gr]. The proof there is for $U=\R$,
but a careful reading show that it works for any $U$. The constant
$A_p$ {\it does not depend on }$U$, and may be shown to be smaller
than $pC$  with an explicit constant $C$ (our evaluation
gives $C<10$). Note that the inequality does not work in the limit
$p\to\infty$.\qed
The preceding proposition shows that replacing the $L^1$ norm in the
definition of the $BMO$ norm $\norm{f}{U}$, by the analogous $L^p$
norm (with $p$ finite), leads to the same $BMO$ space. More precisely
using the usual $L^p$ norm
$$\lnorm{f}{U}{p}=\left(\int_U |f|^p\,dx\right)^{\sst1\over\sst p}\ ,$$
we define
$$\Norm{f}{U}{p}=\sup_{I\subset U}
\left(\mean{I}{|f-f_I|^p}\right)^{\sst1\over\sst p}=\sup_{I\subset U}
|I|^{-{\sst1\over\sst p}}\lnorm{f-f_I}{I}{p}\ .\eqno({\rm A}.3)$$
We then have
\Proc{Proposition A.5.}{The space \BM{U}, is a subspace of $L^p(U)$
when $U$ is a bounded interval.}\par
\proof Let $f\in\BM{U}$, then Proposition A.4 shows that
$\lnorm{f-f_U}{U}{p}\le |U|^{(1/p)} A_p\norm{f}{U}$. But
$\lnorm{f}{U}{p}\le\lnorm{f-f_U}{U}{p}+\lnorm{f_U}{U}{p}\le
|U|^{(1/p)}(A_p\norm{f}{U}+|f_U|)$, and since $f\in L^1(U)$,
$|f_U|\le |U|^{-1}\times\lnorm{f}{U}{1}$. Therefore,
$\lnorm{f}{U}{p}\le|U|^{(1/p)}A_p\norm{f}{U}+|U|^{(1/p)-1}\lnorm{f}{U}{1}$,
and $\lnorm{f}{U}{p}$ is bounded.\qed
Therefore, \BM{U} is a subspace of $\cap_{p=1}^{\infty}L^{p}(U)$, but
{\it not} a subspace of $L^{\infty}(U)$.
In fact, on \BM{U}, we have a family of equivalent norms, as shown in the
following proposition.
\Proc{Proposition A.6.}{On \BM{U}, where $U$ a bounded interval,
define for any real
$\al>0$, and $\be>0$, and for any integer $p\ge 1$ (finite), the
following family of norms $$N(f,\al,\be,p)=\al\Norm{f}{U}{p}+\be
\lnorm{f}{U}{p}\ ,\eqno({\rm A}.4)$$ then\hfb
\quad\quad a) these norms are all equivalent for various $\al$ and $\be$
and $p$ fixed. \hfb\quad\quad
b) these norms are all equivalent for any $p$.}\par
\proof a) is a consequence of the identities
$$(\al x +\be y)={\al\over\al'}(\al'x+\be'y)+
\left({\be\over\be'}-{\al\over\al'}\right)\be' y
={\be\over\be'}(\al'x+\be'y)-
\left({\be\over\be'}-{\al\over\al'}\right)\al' x\ ,\eqno({\rm A}.5)$$
which show that when all numbers $\al,\al',\be,\be'$  are
positive, and $x,y$ nonnegative, then
$\inf(\al/\al',\be/\be')(\al'x+\be'y)\le\al x+\be y\le
\sup(\al/\al',\be/\be')(\al'x+\be'y)\ .$
\par b) Using Proposition A.4 and
the argument in the proof of Proposition A.5, we have $N(f,\al,\be,p)
\le \al A_p\norm{f}{U} +\be(|U|^{(1/p)}A_p\norm{f}{U}+|U|^{(1/p)-1}
\lnorm{f}{U}{1}) $. Therefore,
$N(f,\al,\be,p)\le (\al +\be |U|^{(1/p)})A_p\norm{f}{U}
+\be|U|^{(1/p)-1}\lnorm{f}{U}{1}$.
On the other hand H\"older's inequality shows that
$N(f,\al,\be,p)\ge \al\norm{f}{U}+\be |U|^{(1/p)-1}\lnorm{f}{U}{1}$.
So that we have:
$$ N(f,\al,\be |U|^{(1/p)-1},1)\le N(f,\al,\be,p)\le
N(f,\al +\be |U|^{(1/p)},\be|U|^{(1/p)-1},1)\ .$$
Using Eq. (A.5), one easily finds positive constants $C_-$ and $C_+$ such that
$$N(f,\al +\be |U|^{(1/p)},\be|U|^{(1/p)-1},1)\le
C_+N(f,\al,\be,1)$$ and $$ N(f,\al,\be
|U|^{(1/p)-1},1)\ge C_-N(f,\al,\be,1)\ .$$ Therefore
$$C_-N(f,\al,\be,1)\le N(f,\al,\be,p)\le C_+N(f,\al,\be,1)\ .$$
and the norms $N(f,\al,\be,p)$ and $N(f,\al,\be,1)$ are equivalent.\qed
Remark that these quantities $N(f,\al,\be,p)$ are norms and not only
seminorms when $\be\neq 0$.
With respect to all these norms $N(f,\al,\be,p)$ (with $\al>0$ and $\be>0$),
the space \BM{U} is complete, since any Cauchy sequence with respect
to $N(f,\al,\be,p)$ is also a Cauchy sequence in $L^p(U)$ and
therefore in $L^1(U)$ for $U$ bounded. Moreover the $L^1$-limit of a sequence
of functions in \BM{U} is obviously also in \BM{U}. \par
We now define the oscillation $O_I(f)$ of $f$ on the interval $I$ as
$$O_I(f)=\Mean{I}{f-f_I}\ ,\eqno({\rm A}.6)$$
such that $\norm{f}{U}=\sup_{I\subset U}O_I(f)$. We then have
\Proc{Proposition A.7.}{Let $I_1$ and $I_2$ be two consecutive
intervals, which mean
that the closure  $\overline{I_1}$ and $\overline{I_2}$ have one and
only one common point, and let $I=\overline{I_1}\cup\overline{I_2}$.
Let $a_1=|I_1|$, $a_2=|I_2|$ so that $|I|=a_1+a_2$. Then
$$f_I={a_1 f_{I_1}+a_2 f_{I_2}\over a_1+a_2}\ ,\eqno({\rm A}.7)$$
and
$$O_I(f)={a_1\over a_1+a_2}O_{I_1}(f)+
{a_2\over a_1+a_2}O_{I_2}(f)+
{2a_1a_2\over(a_1+a_2)^2}|f_{I_1}-f_{I_2}|\ .\eqno({\rm A}.8)$$}
\par
\proof An elementary calculation gives Eq. (A.7) as well as
$$(f_I-f_{I_1})={a_2\over a_1+a_2}(f_{I_2}-f_{I_1})\quad
\hbox{and},\quad(f_I-f_{I_2})={a_1\over
a_1+a_2}(f_{I_1}-f_{I_2})\ .\eqno({\rm A}.9)$$
Now $$O_I(f)=\Mean{I}{f-f_I}={|I_1|\over|I|}\Mean{I_1}{f-f_I}+
{|I_2|\over|I|}\Mean{I_2}{f-f_I}\ ,$$
so that
$$\eqalign{\hfill O_I(f)\quad\le&\quad\hphantom{+}{|I_1|\over|I|}
\left(|f_{I_1}-f_I|+\Mean{I_1}{f-f_{I_1}}\right)\cr
&\quad+{|I_2|\over|I|}\left(|f_{I_2}-f_I|+\Mean{I_2}{f-f_{I_2}}\right)\cr
= {a_1\over a_1+a_2}&O_{I_1}(f)+
{a_2\over a_1+a_2}O_{I_2}(f)+
{a_1\over a_1+a_2}|f_I-f_{I_1}|+
{a_2\over a_1+a_2}|f_I-f_{I_2}|\ ,\cr}$$
and the proposition follows from Eq. (A.9).\qed
The above Proposition shows that in order to bound the oscillation
on the union of two consecutive intervals, one needs to control the difference
between the mean values on the two intervals.
\beginsection 2. B. M. O. Norms for periodic functions\par
In \BM{\R} we can consider the various seminorms $\norm{f}{U}$
defined in Eq. (A.2) above, for any
bounded interval $U$. We will also define the seminorm $\norm{f}{\T}$
as follows
$$\norm{f}{\T}=\Dnorm{|I|\le 1}{f}\ ,\eqno({\rm A}.10)$$
where the supremum is taken over all intervals $I\subset \R$ with length
less or equal to 1. The seminorm $\norm{f}{\T}$ is convenient for periodic
functions with period 1.
For any $f\in\BM{\R}$ we obviously have:
$$\norm{f}{[-(1/2),+(1/2)]}\le\norm{f}{\T}\le\norm{f}{\R}\ .
\eqno({\rm A}.11)$$
This observation will be useful if we now consider
functions $f\in\BM{\R}$ {\it which are even and periodic
with period 1}. For such periodic functions, we have the
following result
\Proc{Proposition A.8.}{There exist constant $C_1>1$, $C_2>1$, $C_3>1$
such that for any
$f\in\BM{\R}$, which is even and periodic with period 1, we have
$$\matrix{\hbox{a)}\quad\hfill &\norm{f}{\R}\le C_1\norm{f}{\T}\hfill&\cr
\hbox{b)}\quad\hfill &\norm{f}{\T}\le C_2\norm{f}{[0,1]}\hfill&\cr
\hbox{b)}\quad\hfill &\norm{f}{[0,1]}\le C_3\norm{f}{[0,1/2]}\hfill&\ .\cr
}$$}\par
\proof a) We observe first that the mean value and the oscillation of $f$
on any interval $K$ with length 1 does not depend on $K$. Let us call
these quantities $\overline f$ and $\overline O(f)$ respectively.
Now consider an arbitrary interval $I$. It is always possible to cut
this interval into two consecutive subintervals $I_1$ and $I_2$,
such that $I=I_1\cup I_2$, $|I_1|=n\ge 0$ is the integer part
of $|I|$, and $|I_2|=a<1$.
If $I_1$ is not empty, it is made of $n$ complete periods of $f$,
so that $f_{I_1}=\overline f$ and $O_{I_1}(f)=\overline O(f)<\norm{f}{\T}$.
If $n=0$, we have $O_I(f)\le \norm{f}{\T}$.
{}From Proposition A.7 we get for $n>0$
$$O_I(f)={n\over n+a}\overline O(f)+{a\over n+a}O_{I_2}(f)
+{2na\over (n+a)^2}|f_{I_2}-\overline f|\ . $$
but $O_{I_2}(f)\le\norm{f}{\T}$ and also $\overline O(f)\le\norm{f}{\T}$.
On the other hand, $I_2$ is contained in some interval with length equal to
1, and one can bound $|f_{I_2}-\overline f|$ using Proposition A.3 a) or b).
We get $|f_{I_2}-\overline f|\le 2\norm{f}{\T}$ for $a\ge(1/2)$,
and for $a<(1/2)$,
$$|f_{I_2}-\overline f|\le\left(1+ {|\log a|\over
\log 2 }\right)2\norm{f}{\T}\ .$$
Therefore for $a\ge(1/2)$
$$O_I(f)\le \left(1+{4na\over(n+a)^2}\right)\norm{f}{\T}\ ,$$
and for  $a<(1/2)$
$$O_I(f)\le \left(1+{4na\over(n+a)^2}\left(1+{|\log a|\over\log 2 }\right)
\right)\norm{f}{\T}\ .$$
An elementary study of the function $-x\log(x)$ on $[0,1]$ shows
that it is bounded by $e^{-1}$,and for $n\ge 1$ we have:
$$\hbox{for $a\ge 1/2$}\ ,\ {n\over(n+a)^2}\le {4\over 9}\le 1\ ,
\ \hbox{and for $a<1/2$}\ ,\ {n\over(n+a)^2}\le{1\over n}\le 1\ .$$
Therefore in both cases,
$$O_I(f)\le \left(3+{4\over e\log 2 }\right)\norm{f}{\T}=C_1
\norm{f}{\T}\ ,$$
and part a) holds with $C_1\sim 5.13$. Using the improved bounds
mentioned at the end of the proof of part b) of Proposition A.3,
it is possible to get $C_1\le 3.13$.
Note that part a) of the Proposition
does not require the hypothesis that $f$ is even.\par
b) Consider an arbitrary interval $I=(a,b)$ with length $(b-a)\le 1$.
If $b-a=1$, then $O_I(f)=\overline O(f)\le\norm{f}{[0,1]}$. If $(b-a)<1$
and if the integer parts of $[a]$ of $a$ and $[b]$ of $b$ are equal,
then the interval $I$ corresponds by periodicity to $([a],[b])\subset
[0,1]$ and $O_I(f)=O_{([a],[b])}(f)\le\norm{f}{[0,1]}$.
If $(b-a)<1$, and if $[b]=[a]+1$, then $I=I_1\cup I_2$, where
$I_1=(a,[a]+1]$ and $I_2=[[a]+1,b)$. Let $a'=[a]+1-a=|I_1|$,
$b'=b-[b]=|I_2|$. Then Eq. (A.8), shows that
$$O_I(f)={a'\over a'+b'}O_{I_1}(f)+{b'\over a'+b'}O_{I_2}(f)+
{2a'b'\over(a'+b')^2}|f_{I_1}-f_{I_2}|\ .\eqno({\rm A}.12)$$
By periodicity both   $O_{I_1}(f)$ and $O_{I_2}(f)$ are bounded
by $\norm{f}{[0,1]}$, then we get
$$O_I(f)=\norm{f}{[0,1]}+{2a'b'\over(a'+b')^2}|f_{I_1}-f_{I_2}|
=\norm{f}{[0,1]}+{2a'b'\over(a'+b')^2}|f_{(1-a',1]}-f_{[0,b')}|\ .$$
However, since $f$ is even, $f_{(1-a',1]}=f_{[0,a')}$. By symmetry,
we can assume $b'\ge a'$. Let $x=a'/b'<1$, then
from Proposition A.3, if $x^{-1}\le 2$, we get
$$|f_{[0,a')}-f_{[0,b')}|\le 2\norm{f}{[0,1]}\ ,$$
and
$$O_I(f)=\left(1+{4x\over(1+x)^2}\right)\norm{f}{[0,1]}\le
2\norm{f}{[0,1]}\ .\eqno({\rm A}.13)$$
If $x^{-1}>2$
$$|f_{[0,a')}-f_{[0,b')}|\le \left(1+{|\log x|\over \log 2}\right)
2\norm{f}{[0,1]}\ .$$
{}From this we get
$$O_I(f)=\left(1+{4x\over(1+x)^2}+{4x|\log x|\over(1+x)^2\log 2}\right)
\norm{f}{[0,1]}\ .\eqno({\rm A}.14)$$
Now $4x/((1+x)^2)< 8/9$ and $1/((1+x)^2)< 1$.
Therefore we get the expected result with
$$C_2\le \left(1+{8\over 9}+ {4\over e\log 2}\right)\le 4.012\ ,
\eqno({\rm A}.15)$$
by an argument similar to the one given in part a).
Once more, improved bounds would reduce $C_2$ up to $3.13$.\par
c) Let $I\subset[0,1]$. If either $I\subset [0,1/2]$ or
$I\subset [1/2,1]$, we get $O_I(f)\le\norm{f}{[0,1/2]}$,
using the parity property in the second case. On the other hand,
if $I$ contains $1/2$ in its interior, one divides $I$ into two parts
$I_1$ and $I_2$ contained in $[0,1/2]$ or $[1/2,1]$ repectively.
Using the same argument and setting again $a'=|I_1|$ and $b'=|I_2|$, we
still have Eq. (A.12).
Now, we have
$$|f_{I_1}-f_{I_2}|=|f_{(1/2-a',1/2]}-f_{[1/2,1/2+b')}|=
|f_{(1/2-a',1/2]}-f_{[1/2,1/2-b')}|\ ,$$
since $f$ is even and periodic. Using Proposition A.3, we get here the same
estimates as Eqs. (A.13) and (A.14),
which completes the proof with $C_3=C_2$.\qed
Note that parts b) and c) are not true if the periodic function $f$
is not even. A non trivial, but immediate consequence of these results
is the following corollary.
\Proc{Corollary A.9.}{Let $f$ be a function defined in $[0,1/2]$,
which belongs to \BM{[0,1/2]}. The function $g$ which is even and
periodic with period 1, and which coincides with $f$ on $[0,1/2]$
is in \BM{\R}.}\par
Using the above Proposition A.8, we see  that a natural
choice for a norm on the space of functions $f\in \BM{\R}$ which are even
and periodic with period 1 is $N(f,\al,\be,p)$ defined in (A.4), with $p=2$
and $U=[0,1/2]$, for some fixed $\al>0$ and $\be>0$. Changing the values of
$p$, $\al$, and $\be$ will lead to equivalent norms. Changing $U$ into
$[0,1]$, or $\T$ as in Eq. (A.10), will again give an equivalent norm.
Also equivalent would be a norm like $\al\norm{g}{\R}+
\be\lnorm{g}{[0,1]}{1}$ for the function $g$ which is even and periodic with
period 1, and which coincides with $f$ on $[0,1/2]$. The $L^1$-norm
$\lnorm{g}{[0,1]}{1}$ of $g$ on $[0,1]$, can obviously replaced by
the similar norm on $[0,1/2]$ that is $\lnorm{g}{[0,1/2]}{1}$.\par
\medskip\noindent
We will now quote two lemmas which are useful when using the
$L^2$-norm.
Let $I\subset [0,1/2]$ be any interval, and $f\in L^1([0,1/2])$.
We denote by
$$
f_I = {1\over |I|}\int_I f(x)dx
\eqno(A.16))
$$
the mean of $f$ over $I$.
\par
\medskip
\Proc{Lemma A.10.}{If $f\in L^2([0,1/2])$ and $I\subset [0,1/2]$ is
any interval
$$
\int_I (f-f_I)^2 ds = {1\over 2|I|} \int_I\int_I(f(s)-f(t))^2 dsdt \; .
\eqno(A.17)
$$}
\par
\smallskip
\proof It is a simple calculation: first one observes that
$$
\int_I (f-f_I)^2 ds =\int_I f(s)^2ds -{1\over |I|}\left(\int_I f(s)ds
\right)^2 = {1\over |I|}\int_I\int_I(f(s)^2-f(s)f(t))dsdt \; ,
$$
and then since one obviously has
$$
{1\over |I|}\int_I\int_I(f(s)^2-f(s)f(t))dsdt = {1\over |I|}\int_I\int_I
(f(t)^2-f(s)f(t))dsdt
$$
one immediately gets (A.17). \qed
\par
\medskip
The following lemma deals with the control of the oscillation in
$L_2$-norm.
\Proc{Lemma A.11.}{Let $f$ be an even function in $\Lloc$. Let
$0<a<b$. For any finite interval $I\in\R$ define
$${\cal O}_I(f) = \left({1\over 2|I|^2}\int_{I\times I}(f(s)-f(t))^2dsdt
\right)^{1\over2}\ .$$
Then we have
$${\cal O}_{[-a,b]}(f)\le 2{\cal O}_{[0,b]}(f)\ .$$}\par
\proof Set $I=[-a,b]$, $I_+=[0,b]$, $I_-=[-a,0]$. Then we have
$$2(a+b)^2({\cal O}_{[-a,b]}(f))^2=\int_{I\times I} (f(s)-f(t))^2dsdt\ .$$
The integral over $I\times I$ can be splitted as a sum over four integrals
over $I_+\times I_+$, $I_-\times I_-$, $I_+\times I_-$, $I_-\times I_+$
respectively. However, due to the parity property of $f$, each of these
four integrals is less than (or equal to) the integral over
$I_+\times I_+$. Therefore,
$$2(a+b)^2({\cal O}_{[-a,b]}(f))^2\le 8b^2 ({\cal O}_{[0,b]}(f))^2\ ,$$
and
$${\cal O}_{[-a,b]}(f)\le {2b\over a+b} {\cal O}_{[0,b]}(f)\le
2{\cal O}_{[0,b]}(f)\ .$$
\qed
The preceding Lemma shows that the quadratic oscillation of an even function
$f$ on an interval $I$ containing $0$ in its interior, is at most twice
the quadratic oscillation over the largest among the positive or negative
subintervals of $I$.\par
\vfill \eject
%%%%% references %%%%
\beginsection References \par
\item{[AS]} M. Abramowitz, I. Stegun ``Handbook of mathematical
functions'' Dover, New York (1972).
\item{[Bo]} W. Bosma ``Optimal continued
fractions'', {\it Indag. Math.}, {\bf A90}, 1987, 353-379.
\item{[Br]} A. D. Brjuno ``Analytical form of differential equations''
{\it Trans. Moscow Math. Soc.} {\bf 25} (1971), 131-288; {\bf 26}
(1972), 199-239.
\item{[Da]} A. M. Davie ``The critical function for the semistandard map''
{\it Nonlinearity} {\bf 7} (1994) 219-229.
\item{[Ga]} E. F. Gauss ``Collected Works'' Teubner, Leipzig, (1917), Vol.
X$_1$, p. 372.
\item{[Gr]} J. B. Garnett ``Bounded Analytic Functions'' Academic Press,
New York, (1981).
\item{[GCRF]} J. Garcia--Cuerva and J.L. Rubio de Francia ``Weighted Norm
Inequalities and Related Topics'' North Holland Mathematical
Studies {\bf 116}, Amsterdam, (1985).
\item{[LM]} A. Lasota, M. C. Mackey ``Probabilistic properties
of deterministic systems'' Cambridge University Press, Cambridge,
(1985).
\item{[Ma]} S. Marmi ``Critical functions for complex analytic maps''
{\it J. Phys. A: Math. Gen.} {\bf 23} (1990) 3447-74
\item{[MMY]} S. Marmi, P. Moussa, J.-C. Yoccoz ``The Brjuno
functions and their regularity properties'' Preprint SPhT,
Saclay, 1995.
\item{[Me1]} D. H. Meyer ``On a $\zeta$ function related to the
continued fraction transformation'' {\it Bull. Soc. Math. France}
{\bf 104} (1976), 195-203
\item{[Me2]} D. H. Meyer ``On the Thermodynamic Formalism for
the Gauss Map'' {\it Commun. Math. Phys.} {\bf 130} (1990), 311-333
\item{[Na]} H. Nakada ``On the invariant measures and the entropies
for continued fraction transformations'' {\it Keio Math. Rep.} {\bf 5}
(1980), 37-44.
\item{[Ri]} G. J. Rieger ``Mischung und Ergodizit\"ata bei Kettenbruchen
nach n\"achsten genzen'' {\it J. Reine Angew. Math.} {\bf 310} (1979),
171-181.
\item{[Yo]} J.C. Yoccoz ``Th\'eor\`eme de Siegel, polyn\^omes quadratiques
et nombres de Brjuno'' {\it Ast\'erisque} to appear (1994).
\bye
%%%% end  paper %%%%%