Content-Type: multipart/mixed; boundary="-------------0702081748818" This is a multi-part message in MIME format. ---------------0702081748818 Content-Type: text/plain; name="07-33.comments" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="07-33.comments" Dipartimento di Matematica, Universita' di Bologna, Italy (graffi@dm.unibo.it) Instituto de Matematica, UNAM Cuernavaca, Mexico (villegas@matcuer.unam.mx) ---------------0702081748818 Content-Type: text/plain; name="07-33.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="07-33.keywords" Normal form, quantization, convergence, Cherry's theorem ---------------0702081748818 Content-Type: application/x-tex; name="VBG2d.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="VBG2d.tex" \documentclass[11pt]{article} %\documentclass[a4paper,11pt,reqno]{amsart} %\documentclass[a4paper,draft,reqno]{amsart} \input{amssym.def} \input{amssym} %\usepackage[notref]{showkeys} \setlength{\textwidth}{15.0cm} %%DB margin change%% \setlength{\textheight}{23.0cm} \hoffset=-1.0cm \voffset=-2.0cm \newtheorem{theorem}{Theorem} \newtheorem{proposition}{Proposition} \newtheorem{lemma}{Lemma} \newtheorem{corollary}{Corollary} \newtheorem{definition}{Definition} \renewcommand{\thesection} {\arabic{section}} \renewcommand{\thetheorem} {\thesection.\arabic{theorem}} \renewcommand{\theproposition} {\thesection.\arabic{proposition}} \renewcommand{\thelemma} {\thesection.\arabic{lemma}} \renewcommand{\thedefinition} {\thesection.\arabic{definition}} \renewcommand{\thecorollary} {\thesection.\arabic{corollary}} \renewcommand{\theequation} {\thesection.\arabic{equation}} \newcommand{\begsection}[1]{\setcounter{equation}{0}\section{#1}} \newcommand{\finsection}{\vskip20pt} \newcommand{\hindsp}{\hspace{2em}} %\def\C{{\mathcal C}} \def\L{{\mathcal L}} \def\B{{\mathcal B}} \def\D{{\mathcal D}} \def\H{{\mathcal H}} \def\P{{\mathcal P}} \def\R{{\mathcal R}} \def\G{{\mathcal G}} %\def\S{{\mathcal S}} \def\vf{\varphi} \def\ve{\varepsilon} \def\quadratino{\hfill \vbox{\hrule\hbox{\vrule\vbox to 7 pt {\vfill\hbox to 7 pt {\hfill\hfill} \vfill}\vrule}\hrule}\par} \def\R{\Bbb R} \def\Z{\Bbb Z} \def\N{\Bbb N} \def\T{\Bbb T} \def\C{\Bbb C} \def\A{{\mathcal A}} \def\res{{\mathcal R}} \def\etab{\overline{\eta}} \def\ha{Ha\-mil\-to\-nian} \def\Sc{Schr\"o\-din\-ger} \def\hp{{\hbar}} \def\hpp{{h^{\prime}}} \def\zbar{\overline{z}} \def\wbar{\overline{w}} \def\la{\langle} \def\be{\begin{equation}} \def\ee{\end{equation}} \def\ra{\rangle} \def\ds{\displaystyle} \def\chit{\tilde{\chi}} \def\om{\omega} \def\Om{\Omega} \def\ep{\epsilon} \def\gk{\tilde{g}_k} \def\op#1{{ Op}^{W}\left(#1\right)} \def\norma#1{\left\Vert#1\right\Vert} \def\imma{{\rm Im}} \def\F{{\mathcal F}_{\rho,\sigma}} \def\As{{\mathcal A}_{\om,\sigma}} \def\Ars{{\mathcal A}_{\om,\rho,\sigma}} \def\Aris{{\mathcal A}_{i\om,\rho,\sigma}} \newcommand{\Nb}{\overline{\N}} \newcommand{\Rb}{\overline{\R}} \newcommand{\re}{{\rm Re}\,} \newcommand{\im}{{\rm Im}\,} \newcommand{\power}[1]{{\frakb P}\left(#1\right)} \newcommand{\id}{{\rm Id}} \newcommand{\nsp}[1]{{\scr N}\left(#1\right)} \newcommand{\rge}[1]{{\scr R}\left(#1\right)} \newcommand{\Hom}[2]{{\rm Hom}\left(#1;#2\right)} \newcommand{\matr}[1]{{\scr M}_{#1}} \newcommand{\trasp}[1]{#1^{t}\,} \newcommand{\tr}{{\rm tr}\,} \newcommand{\hau}{{\scr H}} \newcommand{\leb}{{\scr L}} \newcommand{\esssup}{\mathop{\rm ess\,sup}} \newcommand{\essinf}{\mathop{\rm ess\,inf}} \newcommand{\supt}{{\rm supt}} \newcommand{\reg}[1]{{\scr R}_{#1}} %\binoppenalty=10000 % per non spezzare le cose tra dollari %\relpenalty=10000 % per non spezzare le cose tra dollari \overfullrule=5pt \def\perogni{\forall\hskip1pt} \def\esiste{\exists\hskip1.3pt} \def\nota#1{\hskip15pt {\tt #1}\hskip15pt} \def\norma#1{\left\Vert#1\right\Vert} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%% %%%%%%%% begin %%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{document} \baselineskip=19pt \begin{center} {\large\bf A UNIFORM QUANTUM VERSION OF THE \\ CHERRY THEOREM} \end{center} \vskip 13pt \begin{center} Sandro Graffi\footnote{Dipartimento di Matematica, Universit\`a di Bologna (graffi@dm.unibo.it)}, Carlos Villegas Blas\footnote{ Instituto de Matematica, Universit{a}d Nacional de Mexico, Cuernavaca (Mexico).} \end{center} \begin{abstract} \noindent Consider in $L^2(\R^2)$ the operator family $H(\epsilon):=P_0(\hbar,\omega)+\epsilon F_0$. $P_0$ is the quantum harmonic oscillator with diophantine frequency vector $\om$, $F_0$ a bounded pseudodifferential operator with symbol decreasing to zero at infinity in phase space, and $\ep\in\C$. Then there exist $\ep^\ast >0$ independent of $\hbar$ and an open set $\Omega\subset\C^2\setminus\R^2$ such that if $|\ep|<\ep^\ast$ and $\om\in\Om$ the quantum normal form near $P_0$ converges uniformly with respect to $\hbar$. This yields an exact quantization formula for the eigenvalues, and for $\hbar=0$ the classical Cherry theorem on convergence of Birkhoff's normal form for complex frequencies is recovered. \end{abstract} \vskip 1cm % %\date{\today} %\subjclass{???} \keywords{????} % %\maketitle %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% Approximate solutions %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begsection{Introduction and statement of the results} \setcounter{equation}{0}% \setcounter{theorem}{0}% \setcounter{proposition}{0}% \setcounter{lemma}{0}% \setcounter{corollary}{0}% \setcounter{definition}{0}% Consider in the phase space $\R^{2l}$ with canonical coordinates denoted $(x,\xi)$ the Hamiltonian system defined by the principal function \begin{eqnarray} \label{Ham1} p_\ep(x,\xi;\om)&:=&p_0(x,\xi)+\ep f_0(x,\xi) \\ \label{azioni} p_0(x,\xi;\om)&:=&\frac12(|\xi|^2+|\om x|^2)\,=\, \sum_{k=1}^l\om_kI_k(x,\xi), \\ I_k(x,\xi)&:=&\frac{1}{2\om_k}[\xi_k^2+\om^2_kx_k^2], \quad k=1,\ldots,l. \end{eqnarray} Here $f_0:\R^{2l}\to\R$ is analytic; $f_0=O([|\xi|^2+|\om x|^2]^{s/2})$, $s\geq 3$, as $|x|+|\xi|\to 0$, and $\ep\in\R$. Any analytic Hamiltonian near a non-degenerate elliptic equilibrium point can be written in the form (\ref{Ham1}). Let the {\it frequencies} $\om:=(\om_1,\ldots,\om_l)$ fulfill a diophantine condition, i.e %.: $\exists$ %$ %\tau>l-1,\gamma >0$ such that \begin{equation} \label{Diofanto} \la{ \om}, { k}\ra \geq {\gamma}{|{k} |^{-\tau}}, \quad \forall { k}\in \Z^l\setminus\{0\}, \; |k|:=|k_1|+\ldots+|k_l|,\; ,\;\gamma>0,\;\tau >l-1. \end{equation} Under these circumstances the Birkhoff theorem holds, namely (see e.g.\cite{SM}, \S 30): \par\noindent {\it $\forall\,N\in\N$, $\forall\,p\in\N$, $\forall\,\ep\in\R$ one can construct an analytic, canonical bijection $(y,\eta)=\chi_{\ep,N}(x,\xi): \R^{2l}\leftrightarrow \R^{2l}$ and a sequence of analytic functions $Y_p(I;\om): \R_+^l\to\R$ such that: \begin{eqnarray} p_\ep\circ \chi_{\ep,N}^{-1}(y,\eta)=\sum_{k=1}^l\om_kI_k(y,\eta)+\sum_{p=1}^{N-1} Y_p(I(y,\eta);\om)\ep^p+\ep^{N}R_N(y,\eta;\ep). \end{eqnarray} } The $l$ functions $I:=(I_k(y,\eta): k=1,\ldots,l)$, the mechanical actions, are thus first integrals of the transformed Hamiltonian up to an error of order $\ep^N$. Hence the system is integrable if the remainder in (1.4) vanishes as $N\to\infty$, namely if the {\it Birkhoff normal form} \be \label{Bi} B(I;\om,\ep):=\la\om,I\ra+\sum_{p=1}^{\infty}Y_p(I;\om)\ep^p, \quad \la\om,I\ra:=\sum_{k=1}^l\om_kI_k \ee converges when the actions belong to some ball $|I|0$. Here: \be \label{normas} \|f\|_{\sigma}:=\int_{R^{2}\times\R^2}|\widehat{f}(s)|e^{\sigma |s|}\,ds<+\infty \ee \par\noindent 4. ${\cal A}_{\Gamma,\rho,\sigma}:=\{f\in L^1(\R^2\times \R^2)\cap C(\C^2\times \C^2)\,|\,\|f\|_{\Gamma,\rho,\sigma}<+\infty\}$, $\rho>0$, $\sigma >0$. Here: \begin{eqnarray} \label{norma4} \|f\|_{\Gamma,\rho,\sigma}:=\sup_{\om\in\Gamma}\sum_{\nu\in\Z^2}\,e^{\rho|\nu|}\|f_{\nu,\om}\|_{\sigma}; \end{eqnarray} We can now state our assumption on the perturbation. \begin{itemize} \item[(A3)] $F_0$ is a semiclassical pseudodifferential operator of order $\leq 0$ with (Weyl) symbol $f_0\in{\cal A}_{\Gamma,\rho,\sigma}$ for some $\rho>0$, $ \sigma>0$. Explicitly: (notation as in \cite{Ro}) $F_0= Op^W_h(f_0)$, \begin{eqnarray} \label{Weyl} (F_0\psi)(x)=\frac{1}{h^2}\int\!\!\!\int_{\R^2\times\R^2}e^{i\la (x-y),\xi\ra/\hbar} f_0((x+y)/2,\xi)\psi(y)\,dyd\xi,\quad \psi\in{\cal S}(\R^2). \end{eqnarray} \end{itemize} {\bf Remarks} \begin{enumerate} \item Since (\cite{Ro}, \S II.4) $ \|F\|_{L^2\to L^2}\leq \|\widehat{f}\|_{L^1}$, $F_0$ extends to a conti\-nuous operator in $L^2(\R^2)$ because: \be \label{cont} \|F_0\|_{L^2\to L^2} \leq \|\widehat{f}_0\|_{L^1} \leq \|f_0\|_{\sigma}\leq \|f_0\|_{\Gamma,\rho,\sigma}. \ee \item Any $f\in {\cal A}_{\Gamma,\rho,\sigma}$ admits a holomorphic continuation from $u=(x,\xi)\in\R^2\times \R^2$ to the strip $\{z=(z_1,z_2)\in\C^2\times\C^2\,|\, |{\rm Im}\,z|<\sigma\} $. Obviously this holomorphic continuation can be different from the function $f\circ\Psi_{\phi,\om}(z_1,z_2):\C^2\times \C^2\to\R$, as in the example $f=e^{-|z|^2}P(z):\C^2\times \C^2\to\C$, $P$ any polynomial, discussed in Appendix. \end{enumerate} Since $F_0$ is bounded, $H(\ep)$ defined on $D(P_0)$ is closed with pure-point spectrum $\forall\,\ep\in\C$, and is self-adjoint for $\ep\in\R$ if $\om\in\R_+^2$. Moreover, $P_0$ can be considered a semiclassical pseudodifferential operator of order $2$ with symbol $p_0(x,\xi;\om)$. \vskip 0.3cm\noindent \begin{theorem} \label{mainth} Let (A1-A3) be verified and let $h^\ast>0$. Then there exists $\ep^\ast >0$ independent of $\hbar\in[0,\hbar^\ast]$ such that if $|\ep|<\ep^\ast$ the spectrum of $H(\ep)$ is given by the quantization formula \begin{eqnarray} \label{quantiz} E_n(\hbar,\ep)&=&\la \om,n\ra\hbar+\frac12(\om_1+\om_2)\hbar+ {\cal N}(n\hbar,\hbar;\ep). \\ \label{nserie} {\cal N}(n\hbar,\hbar;\ep)&=&\sum_{p=1}^\infty \,{\cal N}_p(n\hbar,\hbar)\ep^p \end{eqnarray} Here $n=(n_1,n_2) $, $n_i=0,1,\ldots$, and: \par\noindent 1. ${\cal N}_p(I,\hbar):\R^2_+\times[0,h^\ast]\to\C$ is analytic in $I$ and continuous in $\hbar$; \par\noindent 2. The series (\ref{quantiz}) has convergence radius $\ep^\ast$ uniformy with respect to $(I,\hbar)\in\Om\times [0,h^\ast]$. Here $\Om$ is any compact of $\R^2_+$; \par\noindent 3. ${\cal N}_p(I,\hbar): p=1,2,\ldots$ admits an asymptotic expansion to all orders in $\hbar$; the order $0$ term is the coefficient $Y_p(I)$ of the Birkhoff normal form. \end{theorem} \noindent {\bf Remarks} \begin{enumerate} \item The conditions of the Cherry theorem are much less restrictive than the present ones. In particular, the standard Schr\"odinger operator in which $f_0$ depends only on $x$ is excluded. On the other hand, in the classical case $\hbar=0$ we obtain an improved version of the theorem: indeed, in our conditions the Birkhoff normal form converges, for $\epsilon$ small enough, in {\it any} compact of $\R^2$. To our knowledge this result is new. \item Taking $\hbar=0$ in ${\cal N}_p(I,\hbar)$ (\ref{quantiz}) becomes $\ds E_\nu^{BS}(\hbar,\ep):=\la\om,n\ra\hbar+\frac12(\om_1+\om_2)\hbar+\sum_{p=1}^\infty \,Y_p(n\hbar)\ep^p$, namely the Bohr-Sommerfeld quantization of the Birkhoff normal form. Formula (\ref{quantiz}) yields all corrections needed to recover the eigenvalues $E_n(\hbar,\ep)$. \item For any fixed $n$ and $\hbar$ the series (\ref{quantiz}) coincides with the Rayleigh-Schr\"odinger perturbation expansion near the simple eigenvalue $\ds \la\om,n\ra\hbar+\frac12(\om_1+\om_2)\hbar$ of $P_0$ \cite{GP}. \item The eigenvalues $E_n(\hbar,\ep)$ admit the interpretation of quantum resonances of a self-adjoint Schr\"odinger operator. For this matter the reader is referred to \cite{MS}, where under much more general conditions on $f_0$ the eigenvalues are obtained by an exact quantization of the KAM iteration scheme. \end{enumerate} We thank Dario Bambusi for a critical reading of the manuscript and Andr\'e Martinez for providing us a first proof of Lemma 2.3. % \vskip 1.0cm\noindent %%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%% \begsection{Proof of the results} \setcounter{equation}{0}% \setcounter{theorem}{0}% \setcounter{proposition}{0}% \setcounter{lemma}{0}% \setcounter{corollary}{0}% \setcounter{definition}{0}% The proof is to be obtained in four steps. \par\noindent {\it 1. Perturbation theory: the formal construction} \par\noindent Look for a unitary transformation $\ds U(\om,\ep,\hbar)=e^{i W(\ep)/\hbar}: L^2\leftrightarrow L^2$, $W(\ep)=W^\ast(\ep)$, $\ep\in\R$, such that: \begin{eqnarray} \label{passo1bis} S(\ep):=UH(\ep)U^{-1}=P_0(\hbar,\om)+\ep Z_1+\ep^2 Z_2+\ldots+ \ep^k R_k(\ep) \end{eqnarray} where $[Z_p,P_0]=0$, $p=1,\ldots,k-1$. Recall the formal commutator expansion: \be \label{commu} e^{it W(\ep)/\hbar}He^{-it W(\ep)/\hbar}=\sum_{l=0}^\infty t^lH_l,\quad H_0:=H,\quad H_l:=\frac{[W,H_{l-1}]}{i\hbar l}, \;l\geq 1 \ee Looking for $W(\ep)$ under the form of a power series, $W(\ep)=\ep W_1+\ep^2W_2+\ldots$, (\ref{commu}) becomes: \be \label{Explicitq1} S=\sum_{s=0}^{k}\ep^s P_s +\ep^{k+1}{R}^{(k+1)} \ee where \be \label{Explicitq2} {P}_s=\frac{[W_s,P_0]}{i\hbar}+F_s,\quad s\geq 1, \;F_1\equiv F_0 \ee \begin{eqnarray*} F_s &=&\sum_{r=2}^s\frac{1}{r!}\sum_{{j_1+\ldots+j_r=s}\atop {j_l\geq 1}}\frac{[W_{j_1},[W_{j_2},\ldots,[W_{j_r},P_0]\ldots]}{(i\hbar)^r} \\ &+&\sum_{r=2}^{s-1}\frac{1}{r!}\sum_{{j_1+\ldots+j_r=s-1}\atop {j_l\geq 1}}\frac{[W_{j_1},[W_{j_2},\ldots,[W_{j_r},F_0]\ldots]}{(i\hbar)^r} \end{eqnarray*} %\begin{eqnarray*} %{R}^{(k+1)}&=&\sum_{l=0}^{\infty}\ep^l \Big[ %\sum_{r=2}^{k+1+l}\frac{1}{r!}\sum_{{j_1+\ldots+j_r=k+1+l} %\atop %{1\leq j_l\leq k}}\frac{[W_{j_1},[W_{j_2},\ldots,[W_{j_r},P_0]\ldots]} %{(i\hbar)^l} %\\ %&+&\sum_{r=2}^{k+l}\frac{1}{r!}\sum_{{j_1+\ldots+j_r=k+l}\atop {1\leq j_l\leq %k}}\frac{[W_{j_1},[W_{j_2},\ldots,[W_{j_r},Q_0]\ldots] }{(i\hbar)^l} \Big] %\end{eqnarray*} Since $F_s$ depends on $W_1,\ldots,W_{s-1}$, (\ref{passo1bis}) yields the recursive homological equations: \be \label{qhomeq} \frac{[W_s,P_0]}{i\hbar} +F_s=Z_s, \qquad [P_0,Z_s]=0 \ee To solve for $S$, $W_s$, $Z_s$, we can equivalently look for their symbols; from now on, we denote by the same letter, but in small case, the symbol $\sigma(A)$ of an operator $A$, except for the symbol of $S$, denoted $\Sigma$. Let us now recall the following relevant results (see e.g. \cite{Fo},\S $3.4$): \begin{enumerate} \item $\ds \sigma([A,B]/i\hbar)=\{a,b\}_M$, where $ \{a,b\}_M$ is the Moyal bracket of $a$ and $b$. \item Given $(g,g^{\prime})\in\As$, their Moyal bracket $\{g,g^{\prime}\}_M$ is defined as $$ \{g,g^{\prime}\}_M=g\# g^{\prime}-g^{\prime}\#g, $$ where $\#$ is the composition of $g, g^{\prime}$ considered as Weyl symbols. \item In the Fourier transform representation, used throughout the paper, the Moyal bracket has the expression \be \label{twisted} (\{g,g^{\prime}\}_M)^{\wedge}(s)= \frac{2}{\hbar'}\int_{\R^{2n}}\widehat{g}(s^1) \widehat{g^{\prime}}(s-s^1) \sin{\left[{\hp}(s-s^1)\wedge s^1/{2}\right]}\,ds^1, \ee where, given two vectors $s=(v,w)$ and $s^1=(v^1,w^1)$, $s\wedge s^1:=\la w,v_1\ra-\la v,w_1\ra$.\par\noindent \item $\{g,g^{\prime}\}_M=\{g,g^{\prime}\}$ if either $g$ or $g^{\prime}$ is quadratic in $(x,\xi)$. \end{enumerate} The equations (\ref{commu},\ref{Explicitq1},\ref{Explicitq2}) then become, once written for the symbols: \begin{eqnarray} \label{Explicits1} \sigma(e^{i W(\ep)/\hbar}He^{-i W(\ep)/\hbar})&=&\sum_{l=0}^\infty {\cal H}_l,\; {\cal H}_0:=p_0+\ep f_0,\; {\cal H}_l:=\frac{\{w,{\cal H}_{l-1}\}_M}{ l}, \;l\geq 1 \\ \Sigma(\ep)&=&\sum_{s=0}^{k}\ep^s p_s +\ep^{k+1}{r}^{(k+1)} \end{eqnarray} where \be \label{Explicits2} {p}_s:=\{w_s,p_0\}_M+f_s,\quad s=1, f_1\equiv f_0 \ee \begin{eqnarray} \label{q_s} f_s &:=&\sum_{r=2}^s\frac{1}{r!}\sum_{{j_1+\ldots+j_r=s}\atop {j_l\geq 1}}\{w_{j_1},\{w_{j_2},\ldots,\{w_{j_r},p_0\}_M\ldots\}_M \\ \nonumber &+& \sum_{r=2}^{s-1}\frac{1}{r!}\sum_{{j_1+\ldots+j_r=s-1}\atop {j_l\geq 1}}\{w_{j_1},\{w_{j_2},\ldots,\{w_{j_r},f_0\}_M\ldots\}_M, \quad s>1 \end{eqnarray} In turn, the recursive homological equations become: \be \label{qhomsq} \{w_s,p_0\}_M +f_s=\zeta_s, \qquad \{p_0,\zeta_s\}_M =0 \ee \par\noindent {\it 2. Solution of the homological equation and estimates of the solution} \vskip 3pt\noindent $f\in {\cal A}_{\om,\rho,\sigma}$ clearly entails the existence of the Fourier expansion of $f_{\phi,\om}(u)$, and its uniform convergence with respect to $\phi\in\T^2$, $u$ on compacts of $\R^{2}\times\R^2$, and $\om\in\Gamma$, namely: \be \label{Fourier} f_{\phi,\om}(u)=\sum_{\nu\in\Z^l}f_{\nu,\om}(u)e^{i\la \nu,\phi\ra}\Longrightarrow f(u)=\sum_{\nu\in\Z^l}f_{\nu,\om}(u). \ee We further denote, for $\om\in\Gamma$, and $\rho>0$: \begin{eqnarray} \|f\|_{\om,\sigma}&:=&\sum_{\nu\in\Z^2}\,\|f_{\nu,\om}\|_{\sigma};\quad {\cal A}_{\om,\sigma}:=\{f(u)\in{\cal F}_\sigma\,|\,\|f(u)\|_{\om,\sigma} <+\infty \} \\ \label{norma2} \|f\|_{\om,\rho,\sigma}&:=&\sum_{\nu\in\Z^2}\,e^{\rho|\nu|}\|f_{\nu,\om}\|_{\sigma};\quad {\cal A}_{\om,\rho,\sigma}:=\{f(u)\in{\cal A}_{\om,\sigma}\,|\,\|f(u)\|_{\om,\rho,\sigma} <+\infty \} \\\label{norma3} \|f\|_{\Gamma,\sigma}&:=&\sup_{\om\in\Gamma}\|f\|_{\om,\sigma};\quad {\cal A}_{\Gamma,\sigma}:=\{f(u)\in{\cal F}_\sigma\,|\,\|f(u)\|_{\Gamma,\sigma} <+\infty \} \\ \label{norma5} \|f\|_{\Gamma,\rho,\sigma}&:=&\sup_{\om\in\Gamma}\|f\|_{\om,\rho,\sigma}; \end{eqnarray} Hence $ {\cal A}_{\Gamma,\rho,\sigma}=\{f(u)\in{\cal F}_\sigma\,|\,\|f(u)\|_{\Gamma,\rho,\sigma} <+\infty \} $ and clearly ${\cal A}_{\Gamma,\rho,\sigma}\subset {\cal A}_{\Gamma,\sigma}\subset{\cal F}_\sigma$. Moreover the following inequalities obviously hold: \begin{eqnarray} \label{stima1} \sup_{u\in\R^2\times \R^2}|f_{\nu,\om}(u)|&\leq& \|\hat{f}_{\nu,\om}(s)\|_{L^1}\leq \|{f}_{\nu,\om}\|_\sigma\leq \|{f}\|_{\Gamma,\sigma}\leq \|{f}\|_{\Gamma,\rho,\sigma} \\ \label{stima1bis} \|\hat{f}\|_{L^1}\leq \|f\|_\sigma &\leq& \|f\|_\sigma \leq \|{f}\|_{\Gamma,\sigma}\leq \|{f}\|_{\Gamma,\rho,\sigma} \end{eqnarray} Now the key remark is that $\{a,p_0\}_M=\{a,p_0\}$ for any symbol $a$ because $p_0$ is quadratic in $(x,\xi)$. The homological equation (\ref{qhomsq}) becomes therefore \be \label{chomsq} \{w_s,p_0\} +f_s=\zeta_s, \qquad \{p_0,\zeta_s\} =0 \ee We then have: \par\noindent \begin{proposition} \label{lemmaom} Let $f\in{\mathcal A}_{\Gamma,\rho,\sigma}$. Then the equation \be \label{cchomsq} \{w,p_0\} +f=\zeta, \qquad \{p_0,\zeta\} =0\ee admits the solutions $\zeta\in {\mathcal A}_{\Gamma,\sigma}$, $w\in{\mathcal A}_{\Gamma,\rho,\sigma}$ \be \label{soluzioni} \label{Zg} \zeta:=f_{0,\om}; \qquad w:=\sum_{\nu\neq 0}\frac{f_{\nu,\om}}{i\la\om,\nu\ra}, \ee with the property $\zeta\circ\Psi_{\phi}=\zeta$; i.e., $\zeta$ depends only on $I_1,I_2$. Moreover: \begin{eqnarray} \label{stimaom} \|\zeta\|_{\Gamma,\sigma}\leq \|f\|_{\Gamma,\sigma}; \quad \|w\|_{\Gamma,\rho,\sigma} \leq \|f\|_{\Gamma,\rho,\sigma}, \quad \|\nabla w\|_{\Gamma,\rho,\sigma} \leq \frac{4C}{\sigma}\|f\|_{\Gamma,\rho,\sigma} \end{eqnarray} for some $C(\Gamma,\delta)>0$. \end{proposition} To prove the Proposition we need a preliminary result. \begin{lemma} \label{sserie} Let $w$ be defined by (\ref{soluzioni}), and $\Psi_{\phi,\om}(x,\xi)$ by (\ref{aazione}). Set: \be \label{flusso2} \Xi_{\phi,\om}(x,\xi):=\Phi_{i\phi,i\om}(x,\xi), \ee that is: $\Xi_{\phi,\om}(x,\xi):=(x^\prime_k,\xi^\prime_k)$, where: \be \left\{\begin{array}{l}{\ds x^\prime_k=x_k\cosh\phi_k+\frac{\xi_k}{\om_k}\sinh\phi_k}\\ {\ds \xi^\prime_k= \xi_k\cosh\phi_k+\om_kx_k\sinh\phi_k}\end{array}\right.\qquad k=1,2 \ee Then one has, uniformly with respect to $(x,\xi)$ on compacts of $\R^4$: \begin{eqnarray} \label{serie1} w\circ \Psi_{\phi,\om}(x,\xi) &=& \sum_{\nu\neq 0}\frac{f_{\nu,\om}(x,\xi)}{i\la\om,\nu\ra}e^{i\la\nu,\phi\ra},\quad \phi\in\T^2 \\ \label{serie2} w\circ \Xi_{\phi,\om}(x,\xi) &=& \sum_{\nu\neq 0}\frac{f_{\nu,i\om}(x,\xi)}{\la\om,\nu\ra}e^{-\la\nu,\phi\ra}, \quad |\phi|\leq \rho-\eta, \;\forall\,0<\eta<\rho \end{eqnarray} Moreover there is $C(\delta)>0$ such that: \be \label{stimaw} \|w\|_{\om,\rho,\sigma}\leq C\|f\|_{\om,\rho,\sigma}; \quad \|w\|_{i\om,\rho,\sigma}\leq C\|f\|_{i\om,\rho,\sigma} \ee \end{lemma} {\bf Proof} \newline Let us first prove that (\ref{soluzioni}), whose convergence is proved below, solves (\ref{cchomsq}), and that $w\circ \Psi_{\phi,\om}(x,\xi)$ admits the representation (\ref{serie1}). Following the argument of (\cite{BGP}), Lemma 3.6, let us write: \begin{eqnarray*} \{p_0,pw\}(x,\xi)=\left.\frac{d}{dt}\right|_{t=0}w\circ\Psi_{\om t,\om}(x,\xi)=\left.\frac{d}{dt}\right|_{t=0} \sum_{0\neq\nu\in\Z^2}\,\frac{f_{\nu,\om}\circ \Psi_{\om t,\om}(u)}{i\la\om,\nu\ra} \\ =\left.\frac{d}{dt}\right|_{t=0} \sum_{0\neq\nu\in\Z^2}\,\frac{f_{\nu,\om}\circ \Psi_{\om t,\om}(u)}{i\la\om,\nu\ra}=\left.\frac{d}{dt}\right|_{t=0} \sum_{0\neq\nu\in\Z^2}\,\frac{f_{\nu,\om}(u)e^{i\la \nu,\om t\ra}}{i\la\om,\nu\ra}= \sum_{0\neq\nu\in\Z^2}\,f_{\nu,\om}(u)\end{eqnarray*} Clearly, this equality also entails $\zeta=f_{0,\om}$. Consider now the expansions (\ref{serie1}, \ref{serie2}). First, it is easy to check that $\om\in\Gamma$ if and only if $i\om\in\Gamma$. Now we have: $$ w_{\nu,\om}=\frac{f_{\nu,\om}(x,\xi)}{i\la\om,\nu\ra} $$ and therefore, by a straightforward application of Lemma \ref{larged}: $$ \|w_{\nu,\om}\|_\sigma \leq C\|f_{\nu,\om}\|_\sigma . $$ Hence: $$ \|w\|_{\omega,\rho,\sigma}=\sum_{\nu\in\Z^2}\,e^{\rho |\nu|}\|w_{\nu,\om}\|_\sigma\leq C \sum_{\nu\in\Z^2}\,e^{\rho |\nu|}\|f_{\nu,\om}\|_\sigma =\|f\|_{\omega,\rho,\sigma}\quad\forall\,\om\in\Gamma. $$ Therefore $q\in {\mathcal A}_{\Gamma,\rho,\sigma}$ entails $w\circ\Psi_{\om,\phi}\in {\mathcal A}_{\Gamma,\rho,\sigma}$, whence the uniform convergence of the series (\ref{serie1}). Now $i\om\in\Gamma$ if $\om\in\Gamma$; hence $w\circ\Psi_{i\om,\phi}\in {\mathcal A}_{\Gamma,\rho,\sigma}$. On the other hand, the replacement $\phi\to i\phi$ maps $\Psi_{\phi,i\om}(x,\xi)$ into $\Xi_{\phi,\om}(x,\xi)$, and the series (\ref{serie2}) is uniformly convergent if $|{\rm Im}\,\phi |<\rho-\eta$, $0<\eta<\rho$. Formula (\ref{serie2}) is therefore proved. This concludes the proof of the Lemma. \vskip 0.3cm\noindent {\bf Proof of Proposition 2.1} \newline Let us first prove that $\zeta$ depends only on $I_1,I_2$. Consider for the sake of simplicity $u=(x,\xi)\in\R^2$. Since $f\in{\mathcal A}_{\Gamma,\rho,\sigma}$, we can write: \begin{eqnarray*} f_{\phi,\om}(x,\xi)= \sum_{m.n=0}^\infty\,\frac{a_{mn}}{2^{m+n}}\left[(x+\frac{\xi}{i\om})e^{i\phi}+ (x-\frac{\xi}{i\om})e^{-i\phi}\right]^m\left[(-i\om x+\xi)e^{i\phi}+ (i\om x +\xi)e^{-i\phi}\right]^n \end{eqnarray*} The average over $\phi$ eliminates all terms but those proportional to $$ [(x+\frac{\xi}{i\om})(x-\frac{\xi}{i\om})]^k[(-i\om x+\xi)(i\om x +\xi)]^l $$ i.e. to $I^kI^l$. The estimate $\|\zeta\|_{\om,\sigma}\leq \|f\|_{\om,\sigma}$ is obvious, and entails $\|\zeta\|_{\Gamma,\sigma}\leq \|f\|_{\Gamma,\sigma}$. The second estimate in (\ref{stimaom}) has been proved in Lemma 2.1 above. To prove the third one, consider the function $f\circ\Psi_{\phi,\om}(z)$ and compute, for $j=1,2$: \begin{eqnarray*} \frac{d}{d\phi_j}w\circ\Psi _{\phi,\om}(z)|_{\phi=0}&=&\left.\frac{\partial w}{\partial x_j} \frac{\partial x^\prime_j}{\partial \phi_j}+\frac{\partial w}{\partial \xi_j} \frac{\partial \xi^\prime_j}{\partial \phi_j}\right|_{\phi=0} = \\ \frac{\partial w}{\partial x_j}\frac{\xi_j}{\om_j}- \frac{\partial w}{\partial \xi_j}\om_jx_j&=&\sum_{0\neq \nu\in\Z^2} \frac{\nu_jf_{\nu,\om}}{i\la\om,\nu\ra} \end{eqnarray*} Therefore, once more by Lemma \ref{larged}, \begin{eqnarray*} \left\|\frac{\partial w}{\partial x_j}\frac{\xi_j}{\om_j}- \frac{\partial w}{\partial \xi_j}\om_jx_j\right\|_{\om,\rho,\sigma} &\leq& \sum_{0\neq \nu\in\Z^2}\,e^{\rho|\nu|} \frac{|\nu_j|}{|\la\om,\nu\ra|} \|f_{\nu,\om}\|_{\om,\sigma} \\ &\leq& C \sum_{0\neq \nu\in\Z^2}\,e^{\rho|\nu|} \|f_{\nu,\om}\|_{\om,\sigma}= C\|f\|_{\om,\rho,\sigma}. \end{eqnarray*} This yields: \be \label{stima5} \left\|\frac{\partial w}{\partial x_j}\frac{\xi_j}{\om_j}- \frac{\partial w}{\partial \xi_j}\om_jx_j\right\|_{\Gamma,\rho,\sigma} \leq C\|f\|_{\Gamma,\rho,\sigma}. \ee In the same way: \begin{eqnarray*} \frac{d}{d\phi_j}w\circ\Xi _{\phi,\om}(z)|_{\phi=0}&=&\left.\frac{\partial w}{\partial x_j} \frac{\partial x^\prime_j}{\partial \phi_j}+\frac{\partial w}{\partial \xi_j} \frac{\partial \xi^\prime_j}{\partial \phi_j}\right|_{\phi=0} = \\ \frac{\partial w}{\partial x_j}\frac{\xi_j}{\om_j}+ \frac{\partial w}{\partial \xi_j}\om_jx_j&=&\sum_{0\neq \nu\in\Z^2} \frac{\nu_jf_{\nu,i\om}}{\la\om,\nu\ra} \end{eqnarray*} whence, by Lemma \ref{larged}, \begin{eqnarray*} \left\|\frac{\partial w}{\partial x_j}\frac{\xi_j}{\om_j}+ \frac{\partial w}{\partial \xi_j}\om_jx_j\right\|_{i\om,\rho,\sigma} &\leq& \sum_{0\neq \nu\in\Z^2}\,e^{\rho|\nu|} \frac{|\nu_j|}{|\la\om,\nu\ra|} \|f_{\nu,i\om}\|_{i\om,\sigma} \\ &\leq& C \sum_{0\neq \nu\in\Z^2}\,e^{\rho|\nu|} \|f_{\nu,i\om}\|_{i\om,\sigma}= C\|f\|_{i\om,\rho,\sigma}. \end{eqnarray*} Recalling that $\om\in\Gamma$ if and only if $i\om\in\Gamma$ we get: \be \label{stima6} \left\|\frac{\partial w}{\partial x_j}\frac{\xi_j}{\om_j}+ \frac{\partial w}{\partial \xi_j}\om_jx_j\right\|_{\Gamma,\rho,\sigma} \leq C\|f\|_{\Gamma,\rho,\sigma}. \ee Denote now $s_j$, $t_j$ the Fourier dual variables of $(x_j,\xi_j)$, $j=1,2$. Then, by definition (we drop for the sake of simplicity the dependence of $\om$): $$ \left\|\frac{\partial w}{\partial x_j}{\xi_j}\right\|_{\sigma}=\int_{\R^4}\left|s_j\frac{\partial \widehat{w}(s_j,t_j)}{\partial t_j}\right|e^{\sigma(|s|+|t|)}\,dsdt $$ Applying Lemma \ref{Poinc} to the integration over $t_j$ we get: \begin{eqnarray*} \left\|\frac{\partial w}{\partial x_j}\right\|_{\om,\sigma}&=&\sum_{\nu\in\Z^2}\,\int_{\R^4}\left|s_j \widehat{w}_{\nu,\om}(s_j,t_j)\right |e^{\sigma(|s|+|t|)}\,dsdt \leq \\ &\leq& \frac{2}{\sigma}\sum_{\nu\in\Z^2}\,\int_{\R^4}\left|s_j\frac{\partial \widehat{w}_{\nu,\om}(s_j,t_j)}{\partial t_j}\right|e^{\sigma(|s|+|t|)}\,dsdt \\ &=& \frac{2}{\sigma}\sum_{\nu\in\Z^2}\,\left\|\frac{\partial w_{\nu,\om}}{\partial x_j}{\xi_j}\right\|_\sigma = \frac{2}{\sigma}\left\|\frac{\partial w}{\partial x_j}\xi_j\right\|_{\om,\sigma} \end{eqnarray*} Therefore, by (\ref{stima5},\ref{stima6}) $$ \left\|\frac{\partial w}{\partial x_j}\right\|_{\Gamma,\rho,\sigma}\leq \frac{2C|\om_j|}{\sigma}\|f\|_{\Gamma,\omega,\sigma} $$ Analogously, applying this time Lemma \ref{Poinc} to the integration over $s_j$: $$ \left\|\frac{\partial w}{\partial \xi_j}\right\|_{\Gamma,\rho,\sigma}\leq \frac{2C}{\sigma{|\om_j|}}\|f\|_{\Gamma,\omega,\sigma}. $$ This is enough to prove the Proposition. \vskip 0.3cm\noindent {\it 3. Iterative Lemma} \begin{proposition} Set: $$ \mu:=\frac{4\ep\|{f}_0\|_{\Gamma,\rho,\sigma}}{\sigma}. $$ Let $\mu<1/4$ and consider for $k=1,2,\ldots$ the function \begin{eqnarray} \label{sigmak} \Sigma_k:=p_0+\ep{\cal Z}_k+v_k \end{eqnarray} with ${\cal Z}_k, v_k\in{\mathcal A}_{\Gamma,\rho,\sigma}$, and let ${\cal Z}_k$ depend on $(I_1,I_2)$ only. \newline Assume moreover: \begin{eqnarray} \label{iter1} \|{\cal Z}_k\|_{\Gamma,\sigma}&\leq& \left\{\begin{array}{cc} \ds 0 & {\rm if}\;k=0 \\ \ds \sum_{s=0}^{k-1}(2\mu)^s &{\rm if}\;k\geq 1 \end{array}\right. \\ \label{iter2} \|v_k\|_{\Gamma,\rho,\sigma} &\leq&\ep(2\mu)^k\|f_0\|_{\Gamma,\rho,\sigma} \end{eqnarray} Let $S_k$ be the Weyl quantization of $\Sigma_k$. Then there exists a unitary map $T_k:L^2\to L^2$, $\ds T_k:=e^{i\ep W/\hbar}$ such that the Weyl symbol of the transformed operator $T_kS_kT^\ast_k:=S_{k+1}$ is given by (\ref{sigmak}) with $k+1$ in place of $k$ and satisfies (\ref{iter1},\ref{iter2}) with $k+1$ in place of $k$. \end{proposition} {\bf Proof} As in \cite{BGP}, Proposition 3.2, the homological equation: \be \label{hom5} \{p_0,w\}+v_k=\ep {\cal V}_k \ee determines the symbol $w$ of $W$. Here the second unknown ${\cal V}_k$ has to depend on $(x,\xi)$ only through $I_1,I_2$. Applying Proposition 1 we find that $w$ and ${\cal V}_k$ exist and fulfill the estimates $$ \|{w}\|_{\Gamma,\rho,\sigma}\leq \ep \|{f}_0\|_{\Gamma,\rho,\sigma}(2\mu)^k ;\quad \|{\nabla w}\|_{\Gamma,\rho,\sigma}\leq (2\mu)^{k+1};\quad\|{\cal V}_k\|_{\Gamma,\rho,\sigma}\leq \|{f}_0\|_{\Gamma,\rho,\sigma}(2\mu)^k. $$ Define now: \begin{eqnarray*} {\cal Z}_{k+1}&:=&{\cal Z}_{k}+{\cal V}_{k};\quad v_{k+1}:=\ep\sum_{l\geq 1}{\cal Z}_{k}^l+ \sum_{l\geq 1}{v}_{k}^l+\sum_{l\geq 1}p_{l0} \\ {\cal Z}_{k}^0&:=&{\cal Z}_{k};\quad {\cal Z}_{k}^l:=\frac1{l}\{w,{\cal Z}_{k}^{l-1}\}_M \end{eqnarray*} and analogous definitions for $v_k^l$ and $p_{l0}$. Clearly $v_{k+1}\in {\cal A}_{\Gamma,\rho,\sigma}$ by Lemma \ref{lemmaiter} below. Then the symbol of the transformed operator has the form (\ref{sigmak}) with $k+1$ in place of $k$. To get the estimates, for $k\geq 1$ we can write, by Proposition 1 and Lemmas 2.2. 2.3. 2.4: \begin{eqnarray*} \sum_{l\geq 1}\|({v}_{k}^l)\|_{\Gamma,\rho,\sigma}&\leq& \ep(2\mu)^k\sum_{l\geq 1}(2\mu)^l=\frac{\ep(2\mu)^{k+1}}{1-2\mu}\leq \ep (2\mu)^{k+1} \\ \sum_{l\geq 1}\|{{\cal Z}_{k}^l}\|_{\Gamma,\sigma}&\leq &\|{{\cal Z}_{k}^l}\|_{\Gamma,\sigma}\cdot\frac{\mu}{1-\mu}\leq 2\mu, \quad \sum_{l\geq 2}\|{{p}_{l0}}\|_{\Gamma,\rho,\sigma}\leq \ep(2\mu)^{k+1} \end{eqnarray*} whence the assertion in a straightforward way. \par\noindent {\bf Proof of Theorem 1} \par\noindent By Proposition 2 there is $\ep^\ast>0$ such that $$ \lim_{k\to\infty} p_0+\ep {\cal Z}_k:=\Sigma(\ep) $$ exists in the $\\|\cdot\|_{\Gamma,\rho,\sigma}$ norm if $|\ep|<\ep^\ast$. Then $S(\ep):=Op_h^W(\Sigma(\ep))$ is unitarily equivalent to $H(\ep)$. Since ${\cal Z}_k$ is a polynomial of order $k-1$ in $\ep$, we can write $\ds \Sigma_k=p_0+\sum_{l=1}^k\zeta^{(l)}\ep^l+v_k$, where $\zeta^{(l)}(I_1,I_2)$ are solutions of the homological equations (\ref{qhomsq}); therefore $S(\ep)$ has the form (\ref{passo1bis}). Note that $\ds \lim_{k\to\infty}\|v_k\|_{\Gamma,\rho,\sigma}=0$ entails $\ds \lim_{k\to\infty}\|R_k\|_{L^2\to L^2}=0$. To sum up, the Weyl symbol $\Sigma(\ep,\hbar)$ has the convergent (uniform with respect to $\hbar$) normal form $$ \Sigma(\ep,\hbar)=p_0(I)+\sum_{n=1}^\infty {\cal Z}_n(I,\hbar)\ep^n $$ Then the assertions of Theorem 1 follow exactly as in \cite{Sj} (see also \cite{BGP}). This concludes the proof. \vskip 0.3cm\noindent {\it 4. Auxiliary results} \begin{lemma} \label{Mo1} Let $(g,g^{\prime}, \nabla g,\nabla g^{\prime})\in {\cal F}_{\sigma}$. Then: \be \|\{g,g^{\prime}\}_M\|_{\sigma} \leq \|\nabla g\|_{\sigma} \|\nabla g^{\prime}\|_{\sigma}. \ee If $(g,g^{\prime}, \nabla g,\nabla g^{\prime})\in {\cal A}_{\om,\rho,\sigma}$ then \be \|\{g,g^{\prime}\}_M\|_{\om,\rho,\sigma} \leq \|\nabla g\|_{\om,\rho,\sigma} \|\nabla g^{\prime}\|_{\om,\rho,\sigma}. \ee and if $(g,g^{\prime}, \nabla g,\nabla g^{\prime})\in {\cal A}_{\Gamma,\rho,\sigma}$: \be \|\{g,g^{\prime}\}_M\|_{\Gamma,\rho,\sigma} \leq \|\nabla g\|_{\Gamma,\rho,\sigma} \|\nabla g^{\prime}\|_{\Gamma,\rho,\sigma}. \ee \end{lemma} {\bf Proof} \newline We repeat the argument of \cite{BGP}, Lemma 3.1. We have $\ds |s\wedge s^1|\leq |s|\cdot |s^1|$. Hence by (\ref{twisted}) and the definition of the $\sigma-$ norm we get: \begin{eqnarray*} \|\{g,g^{\prime}\}_M\|_{\sigma} &=& \frac{2}{\hbar}\int_{\R^{2l}}e^{\sigma|s|}\,ds\int_{\R^{2l}}|\hat g (s)\hat{g^{\prime}}(s-s^1)|\cdot |{\rm sinh}(\hbar (s-s^1)\wedge s^1)/2|\,ds^1 \\ &\leq& \frac{2}{\hbar}\int_{\R^{2l}}\,ds\int_{\R^{2l}}e^{\sigma(|s|+|s^1|)}|\hat g (s)\hat{g^{\prime}}(s^1)|\cdot |{\rm sinh}(\hbar s\wedge s^1)/2|\,ds^1 \\ &\leq& \int_{\R^{2l}}e^{\sigma|s|}|\hat g (s)|\,ds \int_{\R^{2l}}e^{\sigma|s^1|}|\hat{g^{\prime}}(s^1)|\cdot | s\wedge s^1|\,ds^1= \\ &\leq& \int_{\R^{2l}}e^{\sigma|s|}|\hat g (s)||s|\,ds \int_{\R^{2l}}e^{\sigma|s^1|}|\hat{g^{\prime}}(s^1)|\cdot | s^1|\,ds^1 =\|\nabla g\|_{\sigma} \|\nabla g^{\prime}\|_{\sigma}. \end{eqnarray*} The remaining two inequalities follow from the first one by exactly the same argument of \cite{BGP}, Lemma 3.4. This concludes the proof of the Lemma. \vskip 0.2cm\noindent \begin{lemma} \label{Poinc} Let $g\in {\cal F}_{\sigma}$, $u=(x,\xi)\in\R^{2l}$. Then: \be \label{Po} \|g\|_\sigma\leq \frac1{\sigma}\|ug\|_\sigma \ee \end{lemma} {\bf Proof} \noindent Setting $f(s):=\hat{g}(s)$ (\ref{Po}) is clearly equivalent to \be \label{Po1} \int_{\R^{2l}}e^{\sigma|s|}| f(s)|\,ds\leq \frac1{\sigma}\int_{\R^{2l}}e^{\sigma|s|}| \nabla f(s)|\,ds \ee We may limit ourselves to prove this inequality in the one-dimensional case, namely to show that: \be \label{Po2} \int_{\R}e^{\sigma|s|}| f(s)|\,ds\leq \frac1{\sigma}\int_{\R}e^{\sigma|s|}|f^{\prime}(s)|\,ds \ee To see this, first write, for $s>0$: $$ e^{\sigma s} f(s)=-\int_s^\infty e^{\sigma t}f^\prime(t)e^{\sigma(s-t)}\,dt $$ whence, for $A>0$: \begin{eqnarray*} \int_A^\infty |e^{\sigma s} f(s)|\,ds &\leq& \int\!\!\!\int_{A\leq s\leq t\leq\infty}\,|f^\prime(t)|e^{\sigma s}\,dsdt=\int_A^\infty|f^\prime(t)|\int_A^t e^{\sigma s}\,dsdt= \\ &=&\sigma^{-1}\int_A^\infty|f^\prime(t)|(e^{\sigma t}-e^{\sigma A})\,dt\leq \sigma^{-1}\int_A^\infty|f^\prime(t)|e^{\sigma t}\,dt \end{eqnarray*} Likewise, for $s<0$, $A<0$: $$ e^{-\sigma s} f(s)=\int_{-\infty}^s e^{-\sigma t}f^\prime(t)e^{-\sigma(s-t)}\,dt $$ \begin{eqnarray*} \int_{-\infty}^A |e^{-\sigma s} f(s)|\,ds &=& \int\!\!\!\int_{-\infty\leq t\leq s\leq A}\,|f^\prime(t)|e^{-\sigma s}\,dsdt=\int_{-\infty}^A|f^\prime(t)|\int_t^A e^{-\sigma s}\,dsdt= \\ &=&\sigma^{-1}\int_{-\infty}^A|f^\prime(t)|(e^{-\sigma t}-e^{-\sigma A})\,dt \leq \sigma^{-1}\int_{-\infty}^A|f^\prime(t)|e^{-\sigma t}\,dt \end{eqnarray*} Performing the limit $A\to 0$ in both inequalities we get (\ref{Po2}). This concludes the proof of the Lemma. \vskip 0.2cm\noindent \begin{lemma} \label{lemmaiter} \par\noindent Let $g\in {\mathcal A}_{\Gamma,\rho,\sigma}$, $w\in {\mathcal A}_{\Gamma,\rho,\sigma}$. \par\noindent 1. Define $$ g_r:=\frac{1}{r}\{w,g_{r-1}\}_{M}, \qquad r\geq 1; \;\;g_0:=g. $$ Then $g_r\in {\mathcal A}_{\Gamma,\rho,\sigma}$ and the following estimate holds \be \label{stimaindiv} \|{g_r}\|_{\Gamma,\rho,\sigma}\leq \left(4\frac{{\nabla w}_{\Gamma,\rho,\sigma}}{\sigma}\right)^r \|{g}\|_{\Gamma,\rho,\sigma}. \ee 2. Let $w$ solve the homological equation (\ref{qhomsq}). Define the sequence $p_{r0}: r=0,1,\ldots$: $$ p_{00}:=p_0; \qquad p_{r0}:=\frac{1}{r}\{w,p_{r-10}\}_M, \;r\geq 1. $$ Then $p_{r0}\in\As$ and fulfills the following estimate \be \label{stimaindiv1} \|{p_{r0}}\|_{\Gamma,\rho,\sigma}\leq \left(4\sigma^{-1}\|{\nabla w}\|_{\Gamma,\rho,\sigma}\right)^{r-1}\|{f_0}\|_{\Gamma,\rho,\sigma}, \quad r\geq 1. \ee \end{lemma} {\bf Proof} \par\noindent Both estimates (\ref{stimaindiv},\ref{stimaindiv1}) are straightforward consequences of Lemmas 2.2 and 2.3: as far as (\ref{stimaindiv1}) is concerned, it is indeed enough to note that $\{w,p_0\}=\zeta-q$ whence $$ \|{p_{10}}\|_{\Gamma,\rho,\sigma}+\|{\nabla p_{10}}\|_{\Gamma,\rho,\sigma}\leq \frac{4\|{f_0}\|_{\Gamma,\rho,\sigma}}{\sigma}. $$ \vskip 0.2cm\noindent \begin{lemma} \label{larged} If (A3) holds there is $C_\delta>0$ independent of $\om\in\Gamma$ such that \be \label{grandid} |\om_1\nu_1+\om_2\nu_2|\geq C_\delta\sqrt{v_1^2+\nu_2^2} \ee \end{lemma} {\bf Proof} \newline We have to show the existence of $C_\delta>0$ such that \be \label{min} f(\nu_1,\nu_2):=\frac{|\om_1\nu_1+\om_2\nu_2|^2}{v_1^2+\nu_2^2}\geq C_\delta,\quad \forall\,(\nu_1,\nu_2)\in\Z^2, (\nu_1,\nu_2)\neq (0,0) \ee Notice that $f$ is homogeneous of degree $0$, namely $f(\mu\nu_1,\mu\nu_2)=f(\nu_1,\nu_2)$ $\forall\,(\nu_1,\nu_2)\in\Z^2, (\nu_1,\nu_2)\neq (0,0)$, $\forall\,\mu\in\R$, $\mu\neq 0$. Hence it is enough to show that \be \label{min1} F(x,y):={|\om_1x+\om_2 y|^2}\geq C_\delta,\quad \forall\,(x,y)\in S^1 \ee or, writing $x=\cos\theta, y=\sin\theta$: $$ F(\theta):=\frac12[|\om_1|^2+|\om_2|^2]+\frac12[|\om_1|^2-|\om_2|^2]\cos{2\theta} +\la\om_1,\om_2\ra\sin{2\theta}\geq C $$ Note that $F(0)=F(2\pi)=|\om_1|^2$. A simple study of the function $F(\theta): S^1\to\R$ under the assumption (A2) shows the existence of $C_\delta\downarrow 0$ as $\delta\uparrow 1$ such that $|F(\theta)|\geq C_\delta$ $\forall\,\theta\in S^1$. We omit the elementary details. \vskip 1.0cm\noindent %\vfill\eject\noindent {\bf \Large Appendix} \par\noindent Consider the function $f:\C^4\to \R$ $$ f(z):=e^{-|z|^2}P_n(z), \quad z\in\C^4,\;|z|=\sum |z_k|^2. $$ Here $P_n(z)$ is a polynomial of degree $n$. \par\noindent Let us verify that $f$ belongs to ${\cal A}_{\Gamma,\rho,\sigma}$; namely, there are $\rho>0$, $\sigma>0$ such that: $$ \sup_{\om\in\Gamma}\sum_{\nu\in\Z^2}\,e^{\rho |\nu|}\|f_{\nu,\om}(u)\|_\sigma <+\infty. $$ It is clearly enough to consider the case $u=(x,\xi)\in\R^2$, $n=0$. \newline Set: $\om:=\gamma e^{i\theta}, 0\leq \theta\leq 2\pi$, $\delta_1\leq \gamma\leq \delta_2$. Then: $$ |\Psi_{\phi,\om}(u)|^2=\left|x{\cos}\phi+\frac{\xi}{\om}{\sin}\phi\right|^2+|\xi{\cos}\phi-\om x{\sin}\phi|^2=Ax^2+Bx\xi+C\xi^2 $$ $$ A:=\cos^2\phi+\gamma^2\sin^2\phi; \quad B:=\cos\theta(\gamma^{-1}-\gamma)\sin2\phi, \quad C:=\cos^2\phi+\gamma^{-2}\sin^2\phi $$ Therefore we can write: \begin{eqnarray*} f_{\phi,\om}(u)&:=&f\circ \Psi_{\om,\phi}(u)= e^{-\la Q(\gamma,\theta,\phi)u,u\ra}, \qquad Q(\gamma,\theta,\phi):=\left(\begin{array}{cc} \ds A & \frac12 B \\ \frac12 B & C\end{array}\right) \\ \det{Q}&=&\cos^4\phi+\sin^4\phi+[(\gamma^{-2}+\gamma^2)-\cos^2\theta(\gamma^{-1}-\gamma)^2]\sin^2\phi\cos^2\phi \\ &=& 1+\kappa (1-{\cos}^2\theta){\sin}^2\phi{\cos^2}\phi \\ {\rm Tr}\,{Q}&=& 2+\kappa{\sin}^2\phi \\ \kappa&:=&\gamma^{-2}+\gamma^2-2\geq 0 \end{eqnarray*} whence, $\forall\,(\theta,\phi)\in [0,2\pi]\times [0,2\pi]$ \begin{eqnarray*} 1\leq \lambda_1\lambda_2 \leq 1+\kappa, \quad 2\leq \lambda_1+\lambda_2\leq 2+\kappa \end{eqnarray*} where $0<\lambda_{1}(\gamma,\theta,\phi)\leq \lambda_{2}(\gamma,\theta,\phi)$ denote the eigenvalues of $Q(\gamma,\theta,\phi)>0$. This easily yields the uniform estimate: $$ \frac1{D}\leq \lambda_{1}(\gamma,\theta,\phi)\leq \lambda_{2}(\gamma,\theta,\phi)\leq D, \quad D:=\frac12[2+\kappa+\sqrt{(2+\kappa)^2-4}]. $$ \newline Consider now the Fourier coefficients $f_{\nu,\om}(u)=f_{\nu,\gamma,\theta}(u)$: \begin{eqnarray*} f_{\nu,\gamma,\theta}(u):=\frac1{2\pi}\int_0^{2\pi}f\circ \Psi_{\om,\phi}(u)e^{-i\nu\phi}\,d\phi= \frac1{2\pi}\int_0^{2\pi}e^{-\la Q(\gamma,\theta,\phi)u,u\ra}e^{-i\nu\phi}\,d\phi \end{eqnarray*} and compute their Fourier transform: \begin{eqnarray*} \hat{f}_{\nu,\gamma,\theta}(s) &=&\frac1{2(\pi)^2}\int_{\R^2}\int_0^{2\pi}e^{-\la Q(\gamma,\theta,\phi)u,u\ra}e^{-i\nu\phi}e^{-i\la u,s\ra}\,d\phi\,du \\ &=&\frac{2}{(2\pi)^2 \sqrt{\det{Q}} }\int_0^{2\pi}e^{-\la Q^{-1}(\gamma,\theta,\phi)s,s\ra/2}e^{-i\nu\phi}\,d\phi, \quad s\in\R^2 \\ Q^{-1}(\gamma,\theta,\phi)&=&\frac{1}{\det{Q}}\left(\begin{array}{cc} \ds C & -\frac12 B \\ -\frac12 B & A\end{array}\right). \end{eqnarray*} Since $$ \la s,Q^{-1}(\gamma,\theta,\phi)s\ra\geq \lambda_2^{-1}s^2\geq\frac{s^2}{D} $$ $\forall\,(\theta,\phi)\in[0,2\pi]\times [0,2\pi]$ we get the $(\nu,\theta,\phi)$-independent estimate $$ |\hat{f}_{\nu,\gamma,\theta}(s)|\leq \frac{2}{(2\pi)^2}e^{-|s|^2/D}\int_0^{2\pi}\,d\phi=\frac{1}{\pi}e^{-|s|^2/D} $$ Therefore $\|f_{\nu,\om}\|_\sigma<+\infty$ $\forall\,\sigma >0$, $\forall\,\nu\in\Z^2$. \par\noindent Let now $\phi\in\C$. Writing: $$ \det{Q(\gamma,\theta,\phi)}=1+\frac{A(\gamma,\theta)}{4}\sin^2(2\phi),\quad A(\gamma,\theta):=\kappa(1-\cos^2\theta)\geq 0 $$ we get (omitting the elementary details): $$ \det{Q(\gamma,\theta,\phi)}\neq 0,\qquad |{\rm Im}\,\phi |<\frac{1}{4}{\rm arc cosh}(1+8/\kappa)\,. $$ Therefore the function $$ \phi\mapsto \frac{e^{-\la Q^{-1}(\gamma,\theta,\phi)s,s\ra}}{\sqrt{\det{Q(\gamma,\theta,\phi)}}}:=G_{\gamma,\theta,s}(\phi) $$ is analytic with respect to $\phi$ in the strip $\ds |{\rm Im}\,\phi |<\frac{1}{4}{\rm arc cosh}(1+8/\kappa):=m(\kappa)$ uniformly with respect to $(\gamma,\theta,s)\in [\delta_1,\delta_2]\times [0,2\pi]\times \R^2$. \newline In turn the analyticity entails, as is well known, that for any $\ds 0<\eta< m(\kappa)$ there exists $\ds \rho_1>m(\kappa)-\eta $ independent of $(\gamma,\theta,s)\in [\delta_1,\delta_2]\times [0,2\pi]\times \R^2$ such that $$ |\hat{f}_{\nu,\gamma,\theta}(s)| \leq \sup_{|{\rm Im}\,\phi |\leq \eta}|G_{\gamma,\theta,s}(\phi)|{e^{-\rho_1|\nu|}}. $$ Since $\ds \det{Q(\gamma,\theta,\phi)}\neq 0$ for $ |{\rm Im}\,\phi |\leq \eta$, there exist $K_1(\eta)>, K_2(\eta)>0$ independent of $(\gamma,\theta)$ such that: $$ | \la Q^{-1}(\gamma,\theta,\phi)s,s\ra |\geq K_1|s|^2, \quad \frac{1}{|\sqrt{\det{Q(\gamma,\theta,\phi)|}}}0$ independent of $\nu$ such that, $\forall\,\sigma >0$: $$ \|f_{\nu,\om}\|_\sigma =\int_{\R^2}e^{\sigma |s|} |\hat{f}_{\nu,\gamma,\theta}(s)|\,ds \leq K_3 e^{-\rho_1|\nu|}. $$ Hence, $\forall\,0<\rho<\rho_1$: $$ \|f\|_{\om,\rho,\sigma}=\sum_{\nu\in\Z^2}e^{\rho|\nu|}\|f_{\nu,\om}\|_\sigma0$ independent of $\omega\in\Gamma$. We can thus conclude that $$ \|f\|_{\Gamma,\rho,\sigma}=\sup_{\omega\in\Gamma}\sum_{\nu\in\Z^2}e^{\rho|\nu|}\|f_{\nu,\om}\|_\sigma