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Brown-Ravenhall operator, stability of matter, dimension two, relativistic sytems, Kato Inequality
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%% 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 j<l\leq N}\frac{e^2}{\vert x_j-x_l \vert}
+\sum_{1\leq j<k\leq K}\frac{Z_jZ_ke^2}{\vert R_j-R_k \vert}\label{eq:pot}
\e
where $x_j$, $R_k$ denote the coordinates of the $j$th electron and $k$th nucleus respectively, 
and $Z_k$ the charge of the $k$th nucleus. 

The case of $K$ nuclei and 1 electron was treated by Evans and Balinsky in \cite{bae1}. They show that the Brown-Ravenhall operator ${\bf B}_{1,K}$ satisties ${\bf B}_{1,K}\geq const.K$ if for all $1\leq k\leq K$,
$\alpha Z_k\leq \gamma_c
=\frac{2}{(\frac{2}{\pi}+\frac{\pi}{2})}$ and $\alpha\leq \gamma_c/(4+2\sqrt{1+\frac{\pi}{2})})$. Here
$\alpha=e^2/c\hbar$ denotes Sommerfeld's fine structure constant. On taking $\alpha^{-1}\approx 137$, the critical charge $Z_c$ in \cite{bae1} is 124 and hence all known elements are include in the range. The case of $K$ nuclei and $N$ electrons,
but zero mass, was subsequently treated by Hoever and Siedentop in \cite{hs1} where it is shown that there exists an $\alpha_0$ such that ${\bf B}_{N,K}\geq 0$ if for all $1\leq k\leq K$, $\alpha Z_k\leq \gamma_c$ and  $\alpha\leq \alpha_0$. Therefore, they
get stability up to  $Z_k\leq 88$ ($k=1,\cdots,K$) on taking $\alpha^{-1}\approx 137$. They do not include all known elements but this exceeds the maximal 
atomic number one would get by considering the zero mass operator ${\bf H}_{N,K}=\sum_{n=1}^N\sqrt{-\Delta_n}+V_c$ (see \cite{ly}). 

In the case of one electron and one nucleus, the results are more complete. Indeed it 
is proved by Evans, Perry and Siedentop \cite{eps} that the operator ${\bf B}_{1,1}$ is bounded from below 
if and only if $\alpha Z\leq \gamma_c$. Burenkov and Evans \cite{bue} and Tix \cite{t1,t2} 
proved that ${\bf B}_{1,1}$ is in fact strictly positive in this range $\alpha Z$.

 
In this paper, we are interested in the mathematical properties of the two-dimensional Brown-Ravenhall 
operator (\ref{eq:br}). We particularly focus on the problem of the boundedness from below which is related 
to the stability of relativistic systems in 2 dimension. More precisely, we are interested in 
finding the largest value of $\delta$ such that the sesquilinear form $(\psi,{\bf B}\psi)$, for $\psi\in{\cal H}_+$ 
a smooth rapidly decaying Dirac spinor, is bounded from below. The result is the following :

\begin{thm}\label{th}
The operator ${\bf B}$ is bounded from below if and only if
\b
\delta\leq\delta_c:=\frac{\hbar c}
{2\left(\frac{\Gamma(1/4)^4}{8\pi^2}+\frac{8\pi^2}{\Gamma(1/4)^4}\right)}.
\e
where $\Gamma(\cdot)$ is the usual Gamma function.
\end{thm}

\begin{remark}
An approximate value of $\delta_c/\hbar c$ is $0.189$. If we describe the one-electron atom 
by writing $\delta=Ze^2$, the Brown-Ravenhall operator ${\bf B}$ 
is then bounded from below for a critical charge $Z_c=25$ on taking $\alpha=\frac{e^2}{\hbar c}\approx 1/137$.
\end{remark}
To prove Theorem \ref{th}, we use a similar method to that in \cite{eps,ly}. As we will see the analysis 
is a bit different in 2 dimension. Essentially, the boundedness below of ${\bf B}$ will be deduced from an inequality of Hilbert \cite{hlp} (see also Remark \ref{legrmq}). 
Our analysis also yields a sharp Kato 
inequality. 
\begin{thm}\label{katoth}
The following sharp inequality holds :
\b
\sqrt{-\Delta}\geq \frac{4\pi^2}{\Gamma(1/4)^4}\frac{1}{\vert x\vert}.
\label{eq:kato}
\e
\end{thm}
\begin{remark} An approximate value of the constant involved in (\ref{eq:kato})
is  $0.124$.
\end{remark}
Two proofs of Theorem \ref{katoth} are presented in section 3. One is 
based on the techniques used to prove Theorem \ref{th} and one (more direct) is inspired by the work of \cite{ly}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%% PROOF %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\section{Proof of Theorem~\ref{th}}
We will follow the same strategy as Lieb and Yau \cite{ly} and Evans, Perry and Siedentop \cite{eps}. 
Let us recall the different steps.
The idea is to work in the {\it momentum representation}. The Fourier transform ${\cal F}$ is 
defined as follows :
$$
({\cal F}f)(\p):=\frac{1}{\sqrt{2\pi\hbar}}
\int_{\R^2}e^{-\frac{i}{\hbar}\p\cdot{\bf z}}
f({\bf z})d{\bf z},\;\;\;f\in{\cal S}(\R^2)
$$
where $\p=(p_1,p_2)$. The projection $\Lambda_+$ takes, in the momentum representation, 
the following simple form
$$
\Lambda_+(\p):={\cal F}\Lambda_+{\cal F}^{-1}=
\frac{1}{2}I_{\R^{2}}+\frac{1}{2E(\p)}D(\p)
$$
where $I_{\R^{2}}$ is the unit $2\times 2$ matrix and $E(\p)=(c^2|\p|^2+m^2c^4)^{1/2}$ is the positive
eigenvalue of the hermitian $2\times 2$ matrix
$$
D(\p):={\cal F}D{\cal F}^{-1}=c\bs{\sigma}\cdot \p+m\sigma_3=
\pmatrix{mc^2 & c(p_1-ip_2)\cr c(p_1+ip_2) & -mc^2}.
$$
The normalized eigenvector of $D(\p)$ corresponding to $E({\bf p})$ is 
$$
\xi({\bf p})= \frac{1}{N(\p)}\pmatrix{E(\p)+mc^2 \cr c(p_1+ip_2)}\in\C^2,
$$ 
where $N(\p)=(2E(\p)(E(\p)+mc^2))^{1/2}$.
The Fourier transform ${\cal F}\psi$ of any spinor $\psi$ 
in the positive spectral subspace of $D$ can then be written in the following form 
\b
({\cal F}\psi)(\p)=u(\p)\xi(\p),\label{eq:prop}
\e
where $u\in L^2(\R^2)$.
Conversely any Dirac spinor of the form (\ref{eq:prop}) is in the image of ${\cal H}_+$ under the 
Fourier transform.
To complete the description, note that the Fourier transform of the Coulomb potential remains 
Coulombian~:
$$
{\cal F}(\frac{1}{|\cdot|})(\p)=\frac{1}{\sqrt{2\pi\hbar}}\frac{1}{|\p|}.
$$

The first step is a {\it reduction of spinors}. 
We will write $\langle\cdot,\cdot\rangle$ for the inner product in $L^2(\R^2,\C^2)$ or 
in $L^2(\R^2,\C)$ as appropriate. Using the above description, for any $\psi$ in the positive 
spectral subspace, we then obtain that
\ba
\langle\psi,{\bf B}\psi\rangle&=&\int E(\p)| u(\p)|^2d\p-\frac{\delta}{2\pi\hbar}
\int\int u(\p)\bar{ u}(\p')K(\p,\p')d\p d\p'\nonumber\\
&:=&\langle u,{\bf b} u\rangle\label{eq:reduction}
\ea
where the kernel $K$ is the $2\times 2$ matrix-valued function given by
\b
K(\p,\p')=\frac{(E(\p)+mc^2)(E(\p')+mc^2))+c^2\p\bar{\p}'}{N(\p)N(\p')|\p-\p'|}.\label{eq:reduction1}
\e
The upper bar appearing in (\ref{eq:reduction1}) denotes the complex conjugate ($\bar{\p}'=p_1'-ip_2'$). 
The transition from $\psi\in L^2(\R^2,\C^2)$ to the reduced spinor $u\in L^2(\R^2,\C)$ through the introduction 
of the operator ${\bf b}$ is possible because we are working in ${\cal H}_+$. The equality (\ref{eq:reduction})
defines a self-adjoint operator ${\bf b}$ if we can show that the form $\langle u,{\bf b} u\rangle$ is bounded from 
below. Of course $\langle\psi,{\bf B}\psi\rangle$ is bounded from below if and only if $\langle u,{\bf b} u\rangle$ 
is bounded from below. In that case, (\ref{eq:reduction}) will define a self-adjoint operator ${\bf b}$. We will refer
to this operator as the Brown-Ravenhall operator. It is associated to the original operator ${\bf B}$ via
(\ref{eq:reduction}).

The second step is  to study the operator ${\bf b}$ by a {\it partial wave decomposition}. For any $u\in L^2(\R^2,\C)$,
we have that
\b 
u(\p)=\frac{1}{\sqrt{2\pi}}\sum_{k\in\Z}\frac{1}{\sqrt{r}}
a_k(r)e^{ik\theta}\label{eq:fourserie}
\e
where $r=\sqrt{p_1^2+p_2^2}$, $\theta=Arg(p_1+ip_2)$ $[2\pi]$ and where $a_k(\cdot)$ are complex functions 
defined on $\R^+$. So that 
$$
\sum_{k\in\Z}\int_0^{+\infty}\vert a_k(r)\vert^2 dr=\int_{\R^2}\vert  u(\p)\vert^2 d\p(:=\Vert u\Vert_2).
\label{eq:isometrie}
$$
We will write $e(r):=(c^2r^2+m^2c^4)^{1/2}$ and 
$n(r):=(2e(r)(e(r)+mc^2))^{1/2}$. 
Using the expansion (\ref{eq:fourserie}) of $u$, the operator ${\bf b}$ is splitted as follows.
\begin{lemma} For any spinor $u$, we have that
\b
\langle u,{\bf b} u\rangle=\langle a_0,b_0 a_0\rangle+2\sum_{k\geq 1}\langle a_k,b_k a_k\rangle,
\label{eq:channels1}
\e
such that for any $k\in\N$,
\b
\langle a_k,b_k a_k\rangle=\int_0^{+\infty}e(r)\vert a_k(r)\vert^2 dr
-\frac{\delta}{\pi\hbar}\int_0^{+\infty}\int_0^{+\infty}
 a_k(r) \bar{a_k}(r')K_{k}(r,r')drdr'.\label{eq:channels2}
\e
Moreover, the kernels are given by
$$
K_{k}(r,r')=\beta_1(r,r')\sqrt{rr'}I_k(r,r')
+ \beta_2(r,r')\sqrt{rr'}I_{k+1}(r,r')
\label{eq:3parties}
$$
where
$$
\beta_1(r,r'):=\frac{(e(r)+mc^2)(e(r')+mc^2)}{n(r)n(r')},\;\;\; \beta_2(r,r'):=\frac{c^2rr'}{n(r)n(r')}
$$
and 
\b
I_k(r,r'):=\int_0^{2\pi}\frac{\cos{(k\theta)}}{\sqrt{r^2+r'^2-2rr'\cos{(\theta)}}}d\theta.\label{eq:ik}
\e
\end{lemma}
\pf
Inserting the expansion (\ref{eq:fourserie}) of $u$, in the expression (\ref{eq:reduction}), 
yields\\
\vskip5pt
$\bd
\langle u,{\bf b} u\rangle=\sum_{k\in\Z}\int_0^{+\infty}e(r)\vert a_k(r)\vert^2 dr
\ed$
$$
-\frac{\delta}{4\pi^2\hbar}\sum_{k,k'\in\Z}\int_0^{+\infty}\int_0^{+\infty}
\int_0^{2\pi}\int_0^{2\pi}
 a_k(r) \bar{a_k}(r')K_{k,k'}(r,r',\theta,\theta')drdr'd\theta d\theta'
\label{eq:9}
$$
with 
$$
K_{k,k'}(r,r',\theta,\theta')=\frac{e^{ik\theta}e^{-ik'\theta'}}
{\sqrt{r^2+r'^2-2rr'\cos{(\theta-\theta')}}}
\left[
\frac{\sqrt{rr'}(e(r)+mc^2)(e(r')+mc^2)}{n(r)n(r')}+\frac{c^2(rr')^{3/2}e^{i\theta}e^{-i\theta'}}{n(r)n(r')}
\right].\label{eq:10}
$$
Using now the fact that
for any $q\in ]0,1[$ and any $l,l'\in\Z$, 
\b
\int_0^{2\pi}\int_0^{2\pi}\frac{e^{il\theta}e^{-il'\theta'}}
{\sqrt{1-q\cos{(\theta-\theta')}}}d\theta d\theta'
=4\pi\delta_{l,l'}\int_0^{2\pi}\frac{\cos{(l\theta)}}
{\sqrt{1-q\cos{(\theta)}}}d\theta\label{eq:simplif}
\e
and the parity of the cosine appearing in the right hand side of (\ref{eq:simplif}),
we get the lemma. 
\QED
Let us give now a result which permits to describe more precisely the properties of the kernels $K_k$ involved in 
each channel $k$.
\begin{lemma}\label{decroissance}
For any $k\in\N$, we have that
\b
I_k(r,r')=\frac{2}{\sqrt{rr'}}Q_{k-1/2}\left(\frac{1}{2}(\frac{r}{r'}+\frac{r'}{r})\right)\label{eq:legendre}
\e
where $Q_s$ is the Legendre function of the second kind. 
Consequently, $(I_k)_{k\in\N}$ is a positive decreasing sequence. 
\end{lemma}
\begin{remark}\label{legrmq}
As in the three dimensional case \cite{eps}, the kernels involved in each channel $k$ are described 
in terms of Legendre functions of the second kind. Nevertheless, in our situation,
they are not indexed by integers (but by half integers). Then, since we do not have the Neumann 
decomposition for Legendre polynomials (see \cite{ww}), 
the calculation of the critical constant $\delta_c$ in Theorem~\ref{th} has to be done in a different way.
We shall use the Mellin transform (see below). 
\end{remark}
\pf
Let $q=\frac{2rr'}{r^2+r'^2}$ and write 
\b
\sqrt{r^2+r'^2}I_k(r,r')=\int_0^{2\pi}\frac{e^{ik\theta}}{(1-q\cos(\theta))^{1/2}}d\theta.
\label{eq:relphi}
\e
Introduce now the following analytic function 
on $U=\C\setminus\R^+$ :
$$
f(z)=\sqrt{\frac{2}{q}}\frac{z^{k-\frac{1}{2}}}{\sqrt{(z-z_-)(z-z_+)}}
$$
where $z_\pm=\frac{1\pm\sqrt{1-q^2}}{q}$ are both positive numbers satisfying $0< z_-< 1< z_+$. 
On integrating $f$ on the usual "lock" path in $U$, we get that 
$$
\int_0^{2\pi}\frac{e^{ik\theta}}{(1-q\cos(\theta))^{1/2}}d\theta=J_k^1(q)+J_k^2(q)
$$
where
\b
J_k^1(q):=\sqrt{\frac{2}{q}}\int_0^{z_{-}}\frac{z^{  k-\frac{1}{2}}}{\sqrt{(z-z_-)(z-z_+)}}dz
+\sqrt{\frac{2}{q}}\int_{z_{-}}^1\frac{z^{  k-\frac{1}{2}}}{i\sqrt{(z-z_-)(z_+-z)}}dz\label{eq:j1}
\e
and
\b
J_k^2(q):=\sqrt{\frac{2}{q}}\int_0^{z_{-}}\frac{z^{  k-\frac{1}{2}}}{\sqrt{(z-z_-)(z-z_+)}}dz
+\sqrt{\frac{2}{q}}\int_{z_{-}}^1\frac{z^{  k-\frac{1}{2}}}{-i\sqrt{(z-z_-)(z_+-z)}}dz.\label{eq:j2}
\e
By  addition of (\ref{eq:j1}) and (\ref{eq:j2}) and setting $t=1/q$, we get 
\b
\frac{\sqrt{rr'}}{2}I_k(r,r')=\int_0^{t-\sqrt{t^2-1}}\frac{z^{k-\frac{1}{2}}}{\sqrt{z^2-2tz+1}}dz.
\label{eq:leg}
\e
The lemma is therefore proved since the right hand side of (\ref{eq:leg}) 
is the Legendre function of the second kind $Q_{k-1/2}$ (see \cite{ww} p. 334). Moreover, 
since $t-\sqrt{t^2-1}$ is strictly smaller than $1$ for $t>1$, 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 %%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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