Content-Type: multipart/mixed; boundary="-------------0109100844865" This is a multi-part message in MIME format. ---------------0109100844865 Content-Type: text/plain; name="01-324.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="01-324.keywords" Brown-Ravenhall operator, stability of matter, dimension two, relativistic sytems, Kato Inequality ---------------0109100844865 Content-Type: application/x-tex; name="D2BR.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="D2BR.tex" \documentclass[11pt]{article} \usepackage{latexsym,amsbsy,amssymb} \renewcommand\baselinestretch{2} \newcommand{\nc}{\newcommand} \newtheorem{lemma}{Lemma}[section] \newtheorem{thm}[lemma]{Theorem} \newtheorem{proposition}[lemma]{Proposition} \newtheorem{corollary}[lemma]{Corollary} \newtheorem{remark}[lemma]{Remark} \newtheorem{example}[lemma]{Example} \newenvironment{Remark}{\begin{remark}\rm }{\end{remark}\medskip} \newtheorem{definition}[lemma]{Definition} \newenvironment{Definition}{\begin{definition}\rm }{\end{definition}\medskip} \nc{\QED}{\mbox{}\hfill \raisebox{-2pt}{\rule{5.6pt}{8pt}\rule{4pt}{0pt}} \medskip\par} \newenvironment{Proof}{\noindent \parindent=0pt\abovedisplayskip = 0.5\abovedisplayskip \belowdisplayskip=\abovedisplayskip{\bf Preuve. }}{\QED} \newenvironment{Example}{\begin{example}\rm }{\QED\end{example}\medskip} %%%%%%%%% \newcommand{\bs}[1] {{\boldsymbol{#1}}} %%%%%%%%%%%%%%%%%% %\let\Log\log %\def\log{\Log{}} \newfont{\mbf}{cmmib10} \nc{\N}{{\rm I\mkern-4.0mu N}} \nc{\R}{{\rm I\mkern-4.0mu R}} \nc{\Z}{{\sf Z\mkern-6.5mu Z}} \nc{\Id}{{\mathchoice {\rm 1\mskip-4mu l} {\rm 1\mskip-4mu l} {\rm 1\mskip-4.5mu l} {\rm 1\mskip-5mu l}}} \nc{\C}{{\mathchoice {\setbox0=\hbox{$\displaystyle\rm C$}\hbox{\hbox to0pt{\kern0.4\wd0\vrule height0.9\ht0\hss}\box0}} {\setbox0=\hbox{$\textstyle\rm C$}\hbox{\hbox to0pt{\kern0.4\wd0\vrule height0.9\ht0\hss}\box0}} {\setbox0=\hbox{$\scriptstyle\rm C$}\hbox{\hbox to0pt{\kern0.4\wd0\vrule height0.9\ht0\hss}\box0}} {\setbox0=\hbox{$\scriptscriptstyle\rm C$}\hbox{\hbox to0pt{\kern0.4\wd0\vrule height0.9\ht0\hss}\box0}}}} \def\b{\begin{equation}} \def\e{\end{equation}} \def\bd{\begin{displaystyle}} \def\ed{\end{displaystyle}} \def\ba{\begin{eqnarray}} \def\ea{\end{eqnarray}} \def\baa{\begin{eqnarray*}} \def\eaa{\end{eqnarray*}} \def\p{{\bf p}} \def\pf{\noindent{\bf Proof :\ }} \textwidth 6in \hoffset=-5mm \textheight 8.5in \topmargin -0.1truein \oddsidemargin 0.4truein \evensidemargin 0.8truein \raggedbottom \font\subsect=cmbx10 scaled \magstep1 \font\sect=cmbx10 scaled\magstep2 \font\titre=cmbx10 scaled\magstep3 \font\Titre=cmbx10 scaled\magstep4 %%%%%%%%%%%%%%% %\newcommand{\sect}[1]{\setcounter{equation}{0}\section{#1}} \renewcommand{\theequation}{\thesection.\arabic{equation}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%% DEBUT DU DOCUMENT %%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{document} \begin{center} {\bf \Large Stability of the 2D Brown-Ravenhall operator} \footnote {Work supported by the European Union under the TMR grant FMRX-CT 96-0001.} \end{center} \vskip1.2cm \begin{center} {\large A. Bouzouina}\\ \vskip5pt {\it D\'epartement de Math\'ematiques, UMR 6056\\ Facult\'e des Sciences de Reims,\\ Moulin de la Housse B.P 1039\\ F-51687 Reims cedex 2, France\\ {\tt Email:abdelkader.bouzouina@univ-reims.fr} } \end{center} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%% ABSTRACT %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \vskip1cm \begin{abstract} We prove that the two-dimensional Brown-Ravenhall operator is bounded from below when the coupling constant is below a specified critical value --- property also referred to as stability. As a consequence, the operator is then self-adjoint. The proof is based on the strategy followed in \cite{ly} and \cite{eps} with some relevant changes characteristic of the dimension. Our analysis also yields a sharp Kato inequality. \end{abstract} \noindent {\bf Mathematical Subject Classifications (1990) :} 47N50 (Primary). %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%% SECTION 1: INTRODUCTION %%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Introduction and Results} We consider here the two dimensional Brown-Ravenhall operator \cite{brr} : \b {\bf B}=\Lambda_+(D-\frac{\delta}{|x|})\Lambda_+\;\;\;(\delta>0).\label{eq:br} \e In (\ref{eq:br}), the notation is as follows : \begin{itemize} \item $D$ is the free Dirac operator : $$ D=c\bs{\sigma}\cdot\frac{\hbar}{i}\nabla+mc^2\sigma_3=\frac{c\hbar}{i}\sigma_1\frac{\partial}{\partial x_1} +\frac{c\hbar}{i}\sigma_2\frac{\partial}{\partial x_2}+mc^2\sigma_3 \label{eq:2Dirac} $$ where $\hbar$ is the Planck constant, $m$ the electron mass, $c$ is the velocity of light and $\bs{\sigma}=(\sigma_1,\sigma_2)$ and $\sigma_3$ are the Pauli matrices : $$ \sigma_1=\pmatrix{0&1\cr 1&0}, \sigma_2=\pmatrix{0&-i\cr i&0}, \sigma_3=\pmatrix{1&0\cr 0&-1}. $$ \item $\Lambda_+:=\chi_{(0,+\infty)}(D)$, where $\chi_{(0,+\infty)}$ is the characteristic function of the interval $(0,+\infty)$, is the projection of $L^2(\R^2,\C^2)$ onto the positive spectral subspace of $D$. \end{itemize} \noindent The underlying Hilbert space in which ${\bf B}$ acts is $$ {\cal H}_+=\Lambda_+(L^2(\R^2,\C^2)). $$ Due to the algebraic properties of the Pauli matrices, the Dirac operator $D$ in $\R^2$ is a symmetric operator and satisfies $$ D^2=-\hbar^2c^2\Delta+m^2c^4. $$ Moreover, it has similar properties to that in $\R^3$ (see \cite{th} for example) : it is a self-adjoint operator in $L^2(\R^2,\C^2)$ with domain ${\cal D}(D)=H^1(\R^2,\C^2)$ and its spectrum is $(-\infty,-m]\cup [m,+\infty)$. In particular, it is unbounded from below, so that it is difficult to define the ground state of atoms and molecules. In order to study the stability of relativistic systems from the mathematical point of view, various models have been introduced. In the simplest one, $D$ is replaced by the positive definite Hamiltonian $\sqrt{-\hbar^2c^2\Delta+m^2c^4}$. Another alternative is the Brown-Ravenhall operator ${\bf B}$ introduced in \cite{brr}. If (\ref{eq:br}) is to model the 1-electron atom, then we take $\delta=Ze^2$, where $Z$ and $-e$ are the nuclear and the electron charge respectively. In the three dimensional situation, several recent papers have been devoted to the study of the Brown-Ravenhall operator. If we consider the many particle case (N electrons in the field of K nuclei), it takes the following form : \b {\bf B}_{N,K}=\Lambda_{+,N}(\sum_{n=1}^N D^{(n)}-V_c)\Lambda_{+,N}\label{eq:3Ddirac} \e and acts in the $N$-fold antisymmetric tensor product ${\cal H}_N$ of $\Lambda_+(L^2(\R^3,\C^4))$. In (\ref{eq:3Ddirac}), $D^{(n)}$ denotes the free 3 dimension Dirac operator acting on the $n$-th particle, $\Lambda_{+,N}$ is the projection onto ${\cal H}_N$ and the potential $V_c$ is the usual Coulomb interaction of the particles \b V_c(x_1,\cdots,x_N)=-\sum_{j=1}^N\sum_{k=1}^K \frac{Z_ke^2}{\vert x_j-R_k \vert}+\sum_{1\leq j1$, it follows from (\ref{eq:leg}) that the sequence $(I_k)_{k\in\N}$ is decreasing. \QED \vskip5pt The following result implies that the channel $k=0$ in (see equations (\ref{eq:channels1})-(\ref{eq:channels2})) makes the biggest contribution to the energy : \begin{corollary}\label{zerochannel} \baa & &\inf \{\langle u,{\bf b} u\rangle/ u\in L^2(\R^2,(1+\vert \p\vert^2)^{1/2}d\p),\Vert u\Vert_2=1\}\\ &=&\inf \{\langle f,b_0 f\rangle/f\in L^2(\R^+,(1+r^2)^{1/2}dr),\Vert f\Vert_2=1\}. \eaa \end{corollary} \pf The idea is the same as in \cite{eps}. Let us first note that using the positivity of the $I_k$'s, for any $a\in L^2(\R^+,dr)$, $\langle a,b_k a\rangle\geq\langle\vert a\vert,b_k\vert a\vert\rangle$. Thus we can restrict ourselves to positive functions when minimizing $\langle a,b_k a\rangle$. Now, since $(I_k)_{k\in\N}$ is a decreasing sequence (Lemma~\ref{decroissance}), we may assume in (\ref{eq:channels1})-(\ref{eq:channels2}) that the coefficients $a_k$ are zero unless $k=0$ when minimizing. The channel corresponding to $k=0$ then yields the lowest energy. \QED \vskip10pt Before giving the proof of Theorem~\ref{th}, let us recall the definition of the Mellin transform and its behavior with regard to the properties of Legendre functions of the second kind that we shall need. The Mellin transform ${\cal M}$ is defined on $L^2((0,+\infty);\frac{dr}{r})$ by \b ({\cal M}\psi)(w):=\frac{1}{\sqrt{2\pi}}\int_0^\infty r^{-1-iw}\psi(r)dr,\;\;\;\;w\in\R.\label{eq:def} \e It is a unitary map from $L^2((0,+\infty);\frac{dr}{r})$ onto $L^2(\R)$. The Mellin transform of the Legendre functions of the second kind satisfies (see \cite{mos}) : \b ({\cal M}Q_s\left(\frac{1}{2}(\cdot+\frac{1}{\cdot})\right))(w):=\frac{1}{2}\sqrt{\frac{\pi}{2}} \frac{\Gamma([s+1-iw]/2)\Gamma([s+1+iw]/2)}{\Gamma([s+2-iw]/2)\Gamma([s+2+iw]/2)}\label{eq:magnus} \e where $\Gamma(\cdot)$ is the Gamma function. \vskip10pt \noindent{\bf Proof of Theorem~\ref{th}} :\\ In this part, we underline the $m$-dependance and write ${\bf b}_m$ for ${\bf b}$. We will also denote by $Q({\bf b}_m)$ the form domain of ${\bf b}_m$. In view of Corollary~\ref{zerochannel}, one can minimize $\langle u,{\bf b}_m u\rangle$ just by looking at the expression \\ \vskip2pt $\bd \langle u,{\bf b}_m u\rangle=\int_0^{+\infty} e(r)a(r)^2dr \ed$ \b -\frac{\delta}{\pi\hbar} \int_0^{+\infty}\int_0^{+\infty}a(r)a(r') \sqrt{rr'}\{\beta_1(r,r')I_0(r,r')+ \beta_2(r,r')I_1(r,r')\}drdr'\label{eq:hl},\;\;\;u\in Q({\bf b}_m), \e where $a(\cdot)$ is a non-negative function. We prove the theorem in two steps. We consider first the massless case which is homogeneous and then conclude with the case of non-zero mass by estimating the difference ${\bf b}_m-{\bf b}_0$ in a suitable way.\\ \vskip3pt \noindent\underline{Massless case} :\\ When $m=0$, $\beta_1(r,r')=\beta_2(r,r')=1/2$ so that, the expression (\ref{eq:hl}) becomes : \b \langle u,{\bf b}_0 u\rangle=\int_0^{+\infty} cra(r)^2dr-\frac{\delta}{2\pi\hbar} \int_0^{+\infty}\int_0^{+\infty}a(r)a(r') \sqrt{rr'}\{I_0(r,r')+I_1(r,r')\}drdr'.\label{eq:hl1} \e The massless case being homogeneous, it allows us to use the Hilbert inequality (see \cite{hlp}) to the second integral on the right hand side of (\ref{eq:hl1}). It follows that \b \langle u,{\bf b}_0 u\rangle\geq (1-\frac{\delta\Omega}{2\pi\hbar c}) \int_0^{+\infty} cra(r)^2dr\label{eq:hilbert} \e where $$ \Omega=\int_0^{\infty}\frac{1}{\sqrt{r}}\left(I_0(1,r)+I_1(1,r)\right)dr. $$ Using Lemma \ref{decroissance} and definition (\ref{eq:def}), $\Omega$ can be rewritten as follows : $$ \Omega=2\sqrt{2\pi}\left\{({\cal M}Q_{-1/2}\left(\frac{1}{2}(\cdot+\frac{1}{\cdot})\right))(0) +({\cal M}Q_{1/2}\left(\frac{1}{2}(\cdot+\frac{1}{\cdot})\right))(0) \right\}. $$ Then using the property (\ref{eq:magnus}) combined with the facts that $\bd\Gamma(5/4)=\frac{1}{4}\Gamma(1/4)\ed$ and $\bd\Gamma(3/4)=\frac{\pi\sqrt{2}}{\Gamma(1/4)}\ed$, we get that \b \Omega=\frac{\Gamma(1/4)^4}{2\pi} +\frac{32\pi^3}{\Gamma(1/4)^4}.\label{eq:omega} \e According to (\ref{eq:hilbert}) and (\ref{eq:omega}), we then have that $\langle u,{\bf b}_0 u\rangle\geq 0$ on $Q({\bf b}_0)=L^2(\R^2;\sqrt{1+\vert\p\vert^2}d\p)$ if and only if $\bd\delta\leq\delta_c:=\frac{\hbar c} {\frac{\Gamma(1/4)^4}{4\pi^2}+\frac{16\pi^2}{\Gamma(1/4)^4}} \ed$. \vskip5pt \noindent\underline{Mass case} :\\ The boundedness below of the operator ${\bf b}_m$ will follow directly from the masseless case by use of the following result (see \cite{bae2,t2}) : \begin{lemma} For any $m\in\R^+$, the operator ${\bf b}_m-{\bf b}_0$ admits a bounded extension in $L^2(\R^2)$. Moreover, we have that, for any $ u\in L^2(\R^2)$, $$ \Vert {\bf b}_m( u)-{\bf b}_0( u)\Vert_2 \leq mc^2(1+\frac{3\delta\Omega}{\pi\hbar c})\Vert u\Vert_2. $$ \end{lemma} \pf Write first $$ K_k^m-K_k^0=\sqrt{rr'}\left\{ (\beta_1(r,r')-\frac{1}{2})I_k(r,r') +(\beta_2(r,r')-\frac{1}{2})I_{k+1}(r,r')\right\}. $$ By a straightforward calculation, using essentially the fact that $$ \sqrt{2}(e(r)+mc^2)\geq n(r)\geq\sqrt{2}e(r)\geq\sqrt{2}cr, $$ we have for any $r,r'\in\R^+$, $$ 0\leq \beta_1(r,r')-\frac{1}{2} \leq \frac{3}{4}mc\left(\frac{1}{r}+\frac{1}{r'}\right) $$ and $$ 0\leq \frac{1}{2}- \beta_2(r,r') \leq \frac{3}{2}mc\left(\frac{1}{r}+\frac{1}{r'}\right). $$ So that, using (\ref{eq:channels1})-(\ref{eq:channels2}) and the fact that $0\leq e_m(r)-e_0(r)\leq mc^2$, we have that for any rapidly decaying $u$,\\ \vskip2pt $\bd\vert \langle u, {\bf b}_m u\rangle-\langle u, {\bf b}_0 u\rangle\vert\leq mc^2\Vert u\Vert^2_2+ \frac{3mc\delta}{\pi\hbar}\ed$ \b \times\sum_k\int_0^{+\infty}\int_0^{+\infty} \left(\sqrt{\frac{r'}{r}}+\sqrt{\frac{r}{r'}}\right) (\frac{I_k(r,r')}{4}+\frac{I_{k+1}(r,r')}{2}) \vert a_k(r)\vert\vert a_k(r')\vert drdr'. \label{eq:integral} \e Splitting the integral (\ref{eq:integral}) into two parts and applying the Cauchy-Schwarz inequality, we get for the first one\\ \vskip2pt $\bd \int_0^{+\infty}\int_0^{+\infty} \sqrt{\frac{r'}{r}} (\frac{I_k(r,r')}{4}+\frac{I_{k+1}(r,r')}{2}) \vert a_k(r)\vert\vert a_k(r')\vert drdr'\ed$ \b \leq\int_0^{+\infty}\int_0^{+\infty} \vert a_k(r)\vert^2 \sqrt{\frac{r'}{r}} (\frac{I_k(r,r')}{4}+\frac{I_{k+1}(r,r')}{2})drdr' \label{eq:integral1} \e and for the second one\\ \vskip2pt $\bd \int_0^{+\infty}\int_0^{+\infty} \sqrt{\frac{r}{r'}} (\frac{I_k(r,r')}{4}+\frac{I_{k+1}(r,r')}{2}) \vert a_k(r)\vert\vert a_k(r')\vert drdr'\ed$ \b \leq\int_0^{+\infty}\int_0^{+\infty} \vert a_k(r')\vert^2 \sqrt{\frac{r}{r'}} (\frac{I_k(r,r')}{4}+\frac{I_{k+1}(r,r')}{2})drdr'. \label{eq:integral2} \e Then, by a change of variables and Lemma~\ref{decroissance}, the right hand side in (\ref{eq:integral1}) becomes\\ \vskip2pt $\bd \int_0^{+\infty}\int_0^{+\infty} \vert a_k(r)\vert^2 \sqrt{\frac{r'}{r}} (\frac{I_k(r,r')}{4}+\frac{I_{k+1}(r,r')}{2})drdr' \ed$ \b \leq\frac{1}{2}\left(\int_0^{+\infty}\frac{1}{\sqrt{u}}(I_{0}(1,u)+I_{1}(1,u))du \right)\int_0^{+\infty} \vert a_k(r)\vert^2dr. \e Similarly for the right hand side in (\ref{eq:integral2}), and we have that $$ \vert \langle u, {\bf b}_m u\rangle-\langle u, {\bf b}_0 u\rangle\vert\leq (mc^2+\frac{3mc\delta\Omega}{\pi\hbar})\Vert u\Vert^2_2. $$ And then for $\delta\leq \delta_c$, the above sesquilinear form (with form domain $Q({\bf b}_m)=Q({\bf b}_0)$) can be extended to a bounded sesquilinear form on $L^2(\R^2)\times L^2(\R^2)$. It follows that ${\bf b}_m$ and ${\bf b}_0$ have the same domain and this in turn implies the result. \QED %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%% KATO %%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Proof of Theorem~\ref{katoth} } \noindent{\bf Proof 1 :} We suppose here that $\hbar=1$. Consider the operator $\bd H=\sqrt{-\Delta}-\frac{1}{\vert x\vert}\ed$. We use the same method as previously for the Brown-Ravenhall operator. For any $\psi\in L^2(\R^2,\C)$, set $$ ({\cal F}\psi)(\p)=\frac{1}{\sqrt{2\pi}}\sum_{k\in\Z} \frac{1}{\sqrt{r}}v_k(r)e^{ik\theta}, $$ where the functions $v_k(\cdot)$ satisfy $$ \sum_{k\in\Z}\int_0^{+\infty}\vert v_k(r)\vert^2 dr=\int_{\R^2}\vert ({\cal F}\psi)(\p)\vert^2 d\p. $$ Thus, \b \langle \psi,H\psi\rangle= \sum_{k\in\Z}\langle v_k,d_kv_k\rangle \e where, with $I_k$ given by (\ref{eq:ik}), \b \langle v_k,d_k v_k\rangle=\int_0^{+\infty}r\vert v_k(r)\vert^2 dr -\frac{1}{2\pi}\int_0^{+\infty}\int_0^{+\infty}\sqrt{rr'} v_k(r) \bar{v_k}(r')I_{k}(r,r')drdr'.\label{eq:kato1} \e As for the Brown-Ravenhall operator, we need only consider the channel corresponding to $k=0$ and positive functions $v_0$. And then, we apply the Hilbert inequality to get \b \langle v_0,d_0 v_0\rangle\geq\left\{1-\frac{1}{\pi}\int_0^{+\infty}\frac{1}{r}Q_{-1/2}\left(\frac{1}{2} (r+\frac{1}{r})\right)dr\right\}\int_0^{+\infty}r\vert v_0(r)\vert^2 dr.\label{eq:kato2} \e Since the kernel involved in (\ref{eq:kato1}) (particularly for $k=0$) is moreover {\it a.e} positive, the inequality~(\ref{eq:kato2}) is sharp. The conclusion follows from (\ref{eq:magnus}). \QED %%%%%%%%%%%%%%%%%%%%%%%%%2eme preuve%%%%%%%%%%%%%%%%%%%%%%% \vskip0.6cm \noindent{\bf Proof 2 :} We also suppose here that $\hbar=1$. On working in the momentum representation, we have, for all $\psi\in L^2(\R^2)$, $$ \langle\psi,\vert x\vert^{-1}\psi\rangle= \frac{1}{2\pi}\int_{\R^4}\frac{\bar{u}(\p)u(\p')}{\vert \p\vert^{1/2}\vert \p'\vert^{1/2}\vert \p-\p'\vert}d\p d\p' $$ where $u(\p)=\vert \p\vert^{1/2}\hat{\psi}(\p)$. To prove the Theorem, we will show that the integral operator $$ Tu(\p)=\frac{1}{2\pi}\int_{\R^2}\frac{u(\p')}{\vert \p\vert^{1/2}\vert \p'\vert^{1/2}\vert \p-\p'\vert}d\p' $$ is bounded in $L^2(\R^2)$. For any $\omega\in S^1$ and $r\in\R^+$, write $ u_\omega(r)=ru(rw) $ so that $$ \Vert u\Vert_2^2=\int_{S^1\times\R^+}\vert u_\omega(r)\vert^2\frac{dr}{r}d\omega. $$ Write also $T$ as follows \b (Tu)_\omega(r)=\int_{S^1\times\R^+}A_{\omega,\omega'}(\frac{r'}{r})u_{\omega'}(r')\frac{dr'}{r'}d\omega' \label{eq:kernel} \e with $$ A_{\omega,\omega'}(t)=\frac{1}{2\pi}\frac{t^{1/2}}{\vert \omega-t\omega'\vert}. $$ The kernel $A_{\omega,\cdot}(\cdot)$ is in $ L^1(S^1\times\R^+,\frac{dt}{t}\otimes d\omega')$ and its $L^1$ norm doesn't depend on $\omega$. Indeed, using Lemma \ref{decroissance} and (\ref{eq:def})-(\ref{eq:magnus}), we have \b \frac{1}{2\pi}\int_{S^1\times\R^+}A_{\omega,\omega'}(t)\frac{dt}{t}d\omega' =\frac{1}{\pi}\int_0^{+\infty}\frac{1}{t}Q_{-1/2}(\frac{1}{2}(t+\frac{1}{t}))dt=\frac{\Gamma(1/4)^4}{4\pi^2}. \label{eq:egalite} \e Then, by simple use of Schur's Lemma, we get $$ \int_{S^1\times\R^+}\vert(Tu)_\omega(r)\vert^2\frac{dr}{r}d\omega\leq \left(\frac{\Gamma(1/4)^4}{4\pi^2}\right)^2\int_{S^1\times\R^+}\vert u_\omega(r)\vert^2\frac{dr}{r}d\omega= \left(\frac{\Gamma(1/4)^4}{4\pi^2}\right)^2\Vert u\Vert_2^2. $$ Since the kernel in (\ref{eq:kernel}) is positive, this inequality is sharp. \QED \vskip1cm \noindent{\bf Acknowledgment :} {\tt The author is grateful to A. Balinsky and W. D. Evans for\\ introducing him to the subject and for helpful discussions. The author wishes also to thank the referee for suggesting the second proof of Theorem \ref{katoth}.} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%% REFERENCES %%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{thebibliography}{abcd} \baselineskip=15pt \bibitem[1]{brr} Brown, G. E., Ravenhall, D. G., {\em On the interaction of two electrons}, Proc. Roy. Soc. London, A {\bf 208}, (1952), 552-559. \bibitem[2]{bue} Burenkov, V. I., Evans, W. D., {\em On the evaluation of the norm of an integral operator associated with the stability of one-electron atoms}, Proc. Roy. Soc. Edin., A {\bf 128} (1998), 993-1005. \bibitem[3]{bae1} Balinsky, A. A., Evans, W. D., {\em Stability of one-electron molecules in the Brown-ravenhall model}, Commun. Math. Phys., {\bf 202} (1999), 481-500. \bibitem[4]{bae2} Balinsky, A. A., Evans, W. 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N., {\em A Course of Modern Analysis}, Cambridge University Press, 4th ed., 1927. \end{thebibliography} \end{document} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%% FIN DE DOCUMENT %%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% ---------------0109100844865--