BODY \input amstex \documentstyle{amsppt} \NoRunningHeads \pageheight{20cm} %hauteur de la page d'impression \pagewidth{15cm} %largeur de la page d'impression \hcorrection{0.8cm} %deplace le centre de la page horizontalement; negatif vers la gauche; positif vers la droite %\vcorrection{-1cm} %deplace le centre de la page verticalement; negatif vers le bas positif vers le haut \topmatter \title A low concentration asymptotic expansion for the density of states of a random Schr\"odinger operator with Poisson disorder \endtitle \author Fr\'ed\'eric Klopp %\footnotemark \endauthor %\footnotetext{U.R.A 760 C.N.R.S} \affil Department of Mathematics, The Johns Hopkins University, 3400 N. Charles St., Baltimore, 21218 MD, U.S.A \\ \\ D\'epartement de Math\'ematique, B\^at. 425, Universit\'e de Paris-Sud, Centre d'Orsay, 91405 Orsay C\'edex, France \endaffil \email kloppf\@playmate.mat.jhu.edu \endemail \date December 1994 \enddate \keywords random Schr\"odinger operators, density of states, Poisson random potential \endkeywords \subjclass 35 Q 40, 47 B 80, 81 Q 10, 81 Q 20, 82 B 44 \endsubjclass \abstract In this paper, we study the density of states of a random Schr\"odinger operator of the form $H_\omega=-\Delta+V_\omega$ where $V_\omega$ is a Poisson potential (i.e a Poisson random field) of concentration $\mu$. We show that $N_\mu(d\lambda)$, the density of states of $H_\omega$, admits an asymptotic expansion in $\mu$ when $\mu\to0$. Then, we use this expansion to deduce the behaviour of the integrated density of states of $H_\omega$ in the energy interval $(-\infty,0)$ when $\mu$ goes to 0. \bigskip \par\noindent {\smc R\'esum\'e.} Dans ce travail, nous \'etudions la densit\'e d'\'etats d'un op\'erateur de Schr\"odinger al\'eatoire de la forme $H_\omega=-\Delta+V_\omega$ o\`u $V_\omega$ est un potentiel de Poisson (i.e. un champs al\'eatoire de Poisson) de concentration $\mu$. Nous d\'emontrons l'existence d'un d\'eveloppement asymptotique en $\mu$ pour $N_\mu(d\lambda)$, la densit\'e d'\'etat de $H_\omega$, ceci quand $\mu\to0$. Puis nous utilisons ce d\'eveloppement pour estimer la taille de la densit\'e d'\'etats int\'egr\'ee de $H_\omega$ dans l'intervalle d'\'energie $(-\infty,0)$ quand $\mu$ tend vers 0. \endabstract \endtopmatter \document %\hsize=40pc %\vsize=50pc %\magnification=1200 \loadbold \def\sgn{\text{sgn}} \def\L2{L^2({\Bbb R}^d)} \def\Rd{{\Bbb R}^d} \def\Zd{{\Bbb Z}^d} \def\R{{\Bbb R}} \def\tr{\text{Tr}} \def\Sp{{\Cal S}'(\R)} \def\S{{\Cal S}(\R)} \def\Co{{\Cal C}_o^\infty(\R)} \def\Dp{{\Cal D}'(\R)} \def\C{{\Bbb C}} \define\equ{\operatornamewithlimits{\sim}} \def\o{\omega} \def\O{\Omega} \def\e{\frak e} \def\m{\frak m} \subhead 0) Introduction \endsubhead \medskip In the present paper, we study the density of states of a random Poisson Schr\"odinger operator defined on $L^2(\Rd)$. We prove that this density of state admits an asymptotic expansion in powers of $\mu$ when $\mu\to0$, $\mu$ being the concentration parameter of the Poisson potential. We then use this expansion to study the behaviour of the integrated density of states of the random Poisson Schr\"odinger operator outside of the spectrum of $-\Delta$. \par More precisely, let $V_\o$ be a random potential obtained by putting impurities in $\Rd$ according to a Poisson process of concentration $\mu$. This means that $\mu$ is the expectation of the number of impurities found in set of Lebesgue measure 1. \par Consider the random Poisson Schr\"odinger operator $H_\o=-\Delta+V_\o$. One can define a density of states for $H_\o$ (see \cite{Pa-Fi}); we will denote it by $N_\mu(d\lambda)$. The support of $N_\mu$ is the almost sure spectrum of $H_\o$ which does not depend on $\mu$ as is seen form the results of \cite{Ki-Ma 2}; but, in the gaps of the spectrum of $-\Delta$, the density of states tends to 0 (in a weak sense) as $\mu$ tends to 0. \par We prove that $N_\mu(d\lambda)$ admits an asymptotic expansion (in the distributional sense) in powers of $\mu$ when $\mu$ tends to 0, i.e, there exists a sequence of distributions $(n_k)_{k\geq0}$ such that $$N_\mu(d\lambda)\equ\Sb \mu\to0\\ \mu>0\endSb\sum_{k\geq0}\mu^kn_k, \tag 0.1$$ the precise meaning of this asymptotic expansion being given in Theorem 1.1. \par We compute the distributions $(n_k)_{k\geq0}$. $n_0$ is the density of states of the free Laplace operator. For $k\geq1$, let $\Lambda$ be a set of $k$ sites (i.e $k$ distinct points in $\Rd$), and $H_\Lambda$ be the free Laplace operator perturbed by putting exactly one impurity at each site of $\Lambda$. This is a relatively compact perturbation of $H_0=-\Delta$; moreover one can define $\zeta(\lambda;\Lambda)$, the spectral shift function for the pair $(H_\Lambda,H_0)$ (cf \cite{Ya}). Then, for $n_k$ for $k\geq1$, we get the following formulae $$n_1(\lambda)=-\zeta'(\lambda;\{0\}), \tag 0.2 a$$ and, for $k\geq2$, $$n_k(\lambda)=\frac{(-1)^{k-1}}{k!}\int_{\Rd}\cdots\int_{\Rd}\left(\sum_{A\subset\{0,x_1,\dots,x_{k-1}\}}(-1)^{\#A}\zeta'(\lambda;A)\right)dx_1\dots dx_{k-1}.\tag 0.2 b$$ Here $\zeta'$ is the distributional derivative of $\zeta$. \par One can notice the analogy with the expansion found in \cite{Kl} for the density of states of a random Bernoulli Schr\"odinger operator. Here we just have a continuous version of it. \par In the case when the almost sure spectra of $H_\o$ is the whole real axis, we use the asymptotic expansion (0.1) and (0.2) to show that one can divide the negative real axis in union of intervals $(I_k)_{k\geq 1}$ such that, loosely speaking, in each $I_k$, the integrated density is of size $\mu^k$ when $\mu\to0$. \par Formally the proof of the asymptotic expansion (0.1) is quite easy. In \cite{Wa}, a Taylor expansion in $\mu$ is proved at the point $\mu=0$ for the Poisson process of concentration $\mu$. The coefficients of this expansion are combinations of Dirac masses. By \cite{Pa-Fi}, we know that the density of states may be expressed as the average of the spectral measure with respect to the Poisson process. So to get the expansion, we just have to integrate the spectral measure against each of the measure valued coefficients obtained in the Taylor expansion of the Poisson process. Unfortunately, this cannot be made rigorous in such a direct way. Instead, we choose to approximate the Poisson potential with Bernoulli potential, use the asymptotic expansion proved in \cite{Kl}, and show that all the terms are well enough behaved in the limit when the sequence of Bernoulli processes tends to the Poisson process. \par One of the main new technical difficulties comes from the fact that, if the impurities take negative values, then with probability 1, the realizations of the Poisson potential are unbounded below. So the proof may be simplified greatly, if one assumes that the impurities are non negative. \par The methods used here may be applied to obtain information on the density of states of other models of random Poisson Schr\"odinger operator; for example, in dimension 2, one may replace the free Laplace operator by a Laplace operator with a constant magnetic field. \medskip \subhead I) The main results \endsubhead \medskip \subhead a) The random Poisson Schr\"odinger operator \endsubhead \par Let $m(\o,dx)$ be a Poisson random measure of concentration $\mu$ on $\Rd$. By this we mean a function $m:\ \Omega\times{\Cal B}(\Rd)\mapsto\R$ (here ${\Cal B}(\Rd)$ denotes the set of bounded Borel subsets of $\Rd$, and $(\Omega,{\Bbb P}_\o,{\Cal T})$ is some probability space) such that: \par\noindent a) for any $B\in{\Cal B}(\Rd)$, $\o\in\O\mapsto m(\o,B)$ is an integer valued random variable with the following distribution $$ {\Bbb P}_\o(\{\o;\ m(\o,B)=n\})=e^{-\mu\mid B\mid}\frac{(\mu\mid B\mid)^n}{n!}\text{ for }n\in{\Bbb N},$$ \par\noindent b) for $B_1,\dots,B_n$ pairwise disjoint Borel sets in $\Rd$, the random variables $m(\o,B_1),\dots,m(\o,B_n)$ are independent. \smallskip \par\noindent As ${\Bbb E}_\o\{m(\o,B)\}=\mu\mid B\mid$, $\mu$ is called the concentration (${\Bbb E}_\o$ denotes the average with respect to ${\Bbb P}_\o$). \par Let $V\not\equiv0$ be a continuous function satisfying \smallskip \par\noindent (H.1) for some $\eta>0$, and some $C>0$, $\dsize\vert V(x)\vert\leq Ce^{-\eta\vert x\vert}$ for $x\in\Rd$. \smallskip \par For $\o\in\O$, define the Poisson random field $\dsize V_\o(x)=\int_{\Rd}V(x-y)m(\o,dy)$ and the Poisson random Schr\"odinger operator $$H_\o=-\Delta+V_\o.$$ \par By our assumptions on $V$, $H_\o$ is essentially self-adjoint on ${\Cal C}_0^\infty(\Rd)$ with probability 1 (cf \cite{Pa-Fi} or \cite{Ki-Ma 1} Proposition 2). Moreover $H_\o$ is ergodic (cf \cite{Pa-Fi}). Hence the spectrum of $H_\o$ is almost surely indepedent of $\o$. We denote it by $\Sigma$. Using the criterion of \cite{Ki-Ma 2}, it is not difficult to see that: \par\noindent a) if $\{x\in\Rd;\ V(x)<0\}=\emptyset$ then $\Sigma=[0,+\infty)$, \par\noindent b) if $\{x\in\Rd;\ V(x)<0\}\not=\emptyset$ then $\Sigma=\R$. \smallskip \subhead b) The density of states \endsubhead \smallskip \par For $H_\o$, one can define the density of states $N_\mu(d\lambda)$ (cf \cite{Pa-Fi}) in the following way: let $\Lambda_l$ be a cube in $\Rd$ centered at 0 and of sidelentgh $l$. Define $(H_\o)_l^D=H_{\o|\Lambda_l}$ i.e. $H_\o$ restricted to $\Lambda_l$ with Dirichlet boundary conditions. Define, for $\lambda\in\R$, $${\Cal N}^\o_l(\lambda)=\frac1{\text{Vol}(\Lambda_l)}\#\{\lambda_n;\ \lambda_n\text{ is an eigenvalue of }(H_\o)_l^D\text{ and }\lambda_n\leq\lambda\}$$ and $N^\o_l(d\lambda)=\partial_\lambda{\Cal N}^\o_l$, the corresponding discrete measure on $\R$. Then, e.g. by Theorem 5.20 of \cite{Pa-Fi}, there exists a non-random measure $N_\mu(d\lambda)$ such that, with probability 1, $$\lim_{l\to+\infty}N^\o_l(d\lambda)=N_\mu(d\lambda),$$ and $$N_\mu(d\lambda)={\Bbb E}_\o\{\tr(\chi_0 E(\o;d\lambda)\chi_0)\}$$ where $E(\o;d\lambda)$ is the spectral resolution of $H_\o$, $\chi_0$ is the characteristic function of the cube centered in 0 with sidelength 1, and $\tr(Op)$ is the trace of $Op$. \par Before we state our main result we need one more definition. Let $A$ be a finite subset of $\Rd$. Define $\dsize H(A)=-\Delta+\sum_{y\in A}V_y$ where, for $y\in A$, $V_y(x)=V(x-y)$. $H(A)$ is relatively compact perturbation of $-\Delta$ \par Moreover, for $q>d/2$ and $z\not\in\sigma(H(A))$, the operator $(z-H(A))^{-q}-(z+\Delta)^{-q}$ is trace class. Hence, we define $\zeta(\lambda;A)$ to be the {\it spectral shift function} for the pair of operators $H(A)$ and $-\Delta$; $\zeta(\lambda;A)$ is the distribution in $\Sp$ defined by, for $\phi\in\S$, $$Tr(\phi(H(A))-\phi(-\Delta))=\int_0^{+\infty}\zeta(\lambda;A)\phi'(\lambda)d\lambda$$ (see \cite{Ya} chapter 8 section 9). \par Our main result is \proclaim{Theorem 1.1} The density of states $N_\mu(d\lambda)$ admits an asymptotic expansion in $\Sp$ when $\mu$ tends to 0. More precisely there exists a sequence of distributions $(n_k)_{k\geq0}$ such that \par\noindent (a) for any $k\geq0$, $n_k\in\Sp$, \par\noindent (b) for any $\mu_0>0$ and any $N>0$, there exists $\vert\cdot\Vert$, a semi-norm in $\S$ such that, for any $\varphi\in\S$, one has $$\vert\langle N_\mu(d\lambda),\varphi\rangle-\sum_{k=0}^N \mu^k\langle n_k,\varphi\rangle\vert\leq \mu^{N+1}\vert\varphi\Vert\quad\text{for}\quad \mu\in[0,\mu_0].$$ Moreover the distributions $(n_k)_{k\in{\Bbb N}}$ are given by the following formulae: \par\noindent (c) if $k=0$: $n_0$ is the density of states for the unperturbed Laplace operator $-\Delta$, \par\noindent (d) if $k=1$: $n_1(\lambda)=-\zeta'(\lambda,0)$, the derivative of the spectral shift function for the pair of operators $(-\Delta+V,-\Delta)$, \par\noindent (e) if $k\geq2$: $n_k$ is given by the following convergent (in $\Sp$) integral: $$n_k(\lambda)=\frac{(-1)^{k-1}}{k!}\int_{\Rd}\cdots\int_{\Rd}\left(\sum_{A\subset\{0,x_1,\dots,x_{k-1}\}}(-1)^{\#A}\zeta'(\lambda;A)\right)dx_1\dots dx_{k-1},$$ where $\zeta'(\lambda;A)$ is the derivative of $\zeta(\lambda;A)$ with respect to $\lambda$. \endproclaim \remark{Remark} Here $\S$ is the space of rapidly decreasing smooth functions in $\R$, and $\Sp$, the Schwartz space of temperate distributions. \par Notice that, as supp$N_\mu\subset\Sigma$, for any $k\geq 0$, one has supp$n_k\subset\Sigma$. In the proof of Theorem 1.1, we also obtain an upper bound on the order of $n_k$ which is ord$(n_k)\leq (k+1)(d+1)+2$. The same way one can give an estimate upon the order of the semi-norm used to control the remainder in the asymptotic expansion; if the expansion is made up to order $N$, then the remainder is at most of order $(N+1)d+N+5$ (as an element in $\Sp$). \endremark \smallskip \subhead c) The behaviour of the integrated density of states for negative energies \endsubhead \smallskip This question is interesting only if $\Sigma=\R$ i.e if $\{x\in\Rd;\ V(x)<0\}\not=\emptyset$, for otherwise the support of $N_\mu$ is $[0,+\infty)$. \par For $x_1,\dots,x_k$ be $k$ points in $\Rd$, we define $\sigma(x_1,\dots,x_k)$ to be the spectrum of $H(x_1,\dots,x_k)$, $E(d\lambda;x_1,\dots,x_k)$ to be its spectral resolution. For $B$, a Borel subset of $\R$, let $E(B;x_1,\dots,x_k)$ be the spectral projection associated to the set of energies $B$ and the operator $H(\{x_1,\dots,x_k\})$. \par $H(\{x_1,\dots,x_k\})$ is jointly continuous in $(x_1,\dots,x_k)$ (in norm resolvent sense). Let $B$ be an interval. So by \cite{Re-Si 1}, if the endpoints of $B$ are not in the point spectrum of $H(x^0_1,\dots,x^0_k)$, $E(B;x_1,\dots,x_k)$ is continuous near $(x^0_1,\dots,x^0_k)$. \par If $B$ is some Borel subset of $(-\infty,0)$ such that $\overline B\subset (-\infty,0)$, then $E(B;x_1,\dots,x_k)$ is of finite rank (as $H(\{x_1,\dots,x_k\})$ is a relatively compact perturbation of $-\Delta$). The condition $\overline B\subset (-\infty,0)$ may be removed if the dimension $d\geq 3$ using the Lieb-Cwikel-Rosenblum bound (cf \cite{Re-Si 2}). \par Moreover, if $B$ is open, then $(x_1,\dots,x_k)\mapsto \tr(E(B;x_1,\dots,x_k))=\text{rank}((E(B;x_1,\dots,x_k))$ is lower semi-continuous, hence measurable. So, for any $B$ interval in $(-\infty,0)$, $(x_1,\dots,x_k)\mapsto \tr(E(B;x_1,\dots,x_k))$ is measurable. \par For $k\geq1$, $B$ open such that $\overline B\subset(-\infty,0)$, we define $${\frak n}_{k}(B)=\lim_{R\to+\infty}\left(\frac1{\text{Vol}(\{\mid x\mid\leq R\})}\int_{\mid x_1\mid\leq R}\cdots\int_{\mid x_k\mid\leq R}\tr(E(B;x_1,\dots,x_k))dx_1\dots dx_k\right).$$ Using the translation invariance of $-\Delta$, one has $${\frak n}_1(B)=\sum_{\lambda\in B\cap\sigma(-\Delta+V)}\text{multiplicity of }\lambda=\tr(E(B;{0}),$$ and $${\frak n}_{k}(B)=\int_{\Rd}\cdots\int_{\Rd}\tr(E(B;0,x_1,\dots,x_{k-1}))dx_1\dots dx_{k-1}\text{ if }k\geq2.$$ For $k\geq2$, ${\frak n}_{k}(B)$ may be infinite. \par For $k\geq 1$, let ${\frak B}_{k+1}$ be the closure of the union of the spectra of $H(\{x_1,\dots,x_k\})$ when $x_1,\dots,x_k$ run all over all possible points in $\Rd$ i.e, ${\frak B}_0=\sigma(-\Delta)$, and for $k\geq1$, $${\frak B}_{k}=\overline{\bigcup_{x_1\in\Rd}\cdots\bigcup_{x_k\in\Rd} \sigma(\{x_1,\dots,x_k\})}.$$ \par Using the exponential decrease of eigenfunction in the negative spectrum of $H(x_1,\dots,x_k)$, it is not difficult to see that, for $j\leq k$, ${\frak B}_j\subset{\frak B}_k$. When $\{x\in\Rd;\ V(x)<0\}\not=\emptyset$, one easily proves that $\dsize\cup_{k\in{\Bbb N}}{\frak B}_k=\R$. Moreover, if one supposes that, for almost every $x\in\Rd$, $V(x)\leq0$, then one easily proves that, for $j\not=k$, ${\frak B}_j\not={\frak B}_k$. \par The behaviour of the integrated density of states for negative energies is given by the \proclaim{Theorem 1.2} Let $I$ be an open interval such $\overline I\subset(-\infty,0)$. Then, for $k\geq 1$, if $I\cap{\frak B}_k\not=\emptyset$ and for $1\leq j\leq k-1$, $\overline{I}\cap{\frak B}_j=\emptyset$ then, $0<{\frak n}_k(I)\leq {\frak n}_k(\overline I)<+\infty$ and $$\limsup_{\mu\to0}\left\lbrack\mu^{-k} N_\mu(I)\right\rbrack=\limsup_{\mu\to0}\left\lbrack\mu^{-k}\int_I N_\mu(d\lambda)\right\rbrack\leq\frac{{\frak n}_k(\overline I)}{k!},$$ and $$\liminf_{\mu\to0}\left\lbrack\mu^{-k} N_\mu(I)\right\rbrack=\liminf_{\mu\to0}\left\lbrack\mu^{-k}\int_I N_\mu(d\lambda)\right\rbrack\geq\frac{{\frak n}_k(I)}{k!}.$$ \endproclaim \remark{Remark} If ${\frak n}_k(I)={\frak n}_k(\overline I)$ then we get that $\dsize N_\mu(I)=\frac{{\frak n}_k(I)}{k!}\mu^k(1+o(1))$ when $\mu\to0$. The condition ${\frak n}_k(I)={\frak n}_k(\overline I)$ is a condition on the behaviour of the negative eigenvalues of $H(x_1,\dots,x_k)$. For this condition to be satisfied, one just needs that, for any such eigenvalue $\lambda(x_1,\dots,x_k)$, the level set $\lambda^{-1}(\lambda_0)$ is of measure 0 if $\lambda_0$ is one of the ends of the interval $I$. \endremark \demo{Proof of Theorem 1.2} Let us first show some regularity result for ${\frak n}_k(I)$ for $I$ chosen as in Theorem 1.2. Assume $I=(a,b)$ and denote $I_{-\epsilon}=[a+\epsilon,b-\epsilon]$ and $I_{+\epsilon}=[a-\epsilon,b+\epsilon]$. Then $\dsize\lim_{\epsilon\to0^+}{\frak n}_k(I_{+\epsilon})={\frak n}_k(\overline I)$ and $\dsize\lim_{\epsilon\to0^+}{\frak n}_k(I_{-\epsilon})={\frak n}_k(I)$. \par Indeed, as for $1\leq j\leq k-1$, $\overline I\cap{\frak B}_j=\emptyset$ and the ${\frak B}_j$ are closed, for $\epsilon$ small enough, the functions $(x_1,\dots,x_k)\mapsto \tr(E(I;x_1,\dots,x_k))$, $(x_1,\dots,x_k)\mapsto \tr(E(\overline I;x_1,\dots,x_k))$, $(x_1,\dots,x_k)\mapsto \tr(E(I_{+\epsilon};x_1,\dots,x_k))$ and $(x_1,\dots,x_k)\mapsto \tr(E(I_{-\epsilon};x_1,\dots,x_k))$ vanish outside a compact set. Moreover the regularity of the spectral measure tells us that, for any $(x_1,\dots,x_k)$, $$\tr(E([a-\epsilon,a);x_1,\dots,x_k))\to0\text{ and }\tr(E((a,a+\epsilon];x_1,\dots,x_k))\to0\text{ when }\epsilon\to0.$$ Hence the claim is a direct consequence of Lebesgue's Dominated Convergence Theorem. \par Let $\chi_I$ be the characteristic function of $I$, and $\psi\in\Co$ such that $0\leq\psi\leq1$, $\psi\equiv1$ on $[-1,1]$, $\psi\equiv0$ outside of $[-2,2]$ and $\int_\R \psi=1$. Pick $\delta>0$. For $\epsilon>0$, define $\chi_{-\epsilon}$ to be the characteristic function of $I_{-\epsilon}$ and $\chi_{+\epsilon}$ to be the characteristic function of $I_{+\epsilon}$. Now define $\dsize \psi_\mu(x)=\frac1{\mu^\delta}\psi(\frac1{\mu^\delta}x)$, $\chi_\mu=\chi_I*\psi_\mu$, $\chi'_\mu=\chi_{-2\mu^\delta}*\psi_\mu$ and $\chi''_\mu=\chi_{+2\mu^\delta}*\psi_\mu$. Then one has, $\chi'_\mu\leq\chi_I\leq\chi''_\mu$ hence, as $N_\mu$ is a positive measure, $$\langle N_\mu,\chi'_\mu\rangle \leq \langle N_\mu,\chi_I\rangle=N_\mu(I)\leq \langle N_\mu,\chi''_\mu\rangle. $$ We then estimate $\langle N_\mu,\chi'_\mu\rangle$ and $\langle N_\mu,\chi''_\mu\rangle$ using the asymptotic expansion of Theorem 1.1. For $\mu$ small enough, for $1\leq j\leq k-1$, $I_{+3\mu^\delta}\cap{\frak B}_j=\emptyset$, so, for $1\leq j\leq k-1$, $$\langle n_j,\chi'_\mu\rangle=\langle n_j,\chi''_\mu\rangle=0.$$ Moreover, one has, as for $\mu$ small enough, $$\langle n_k,\chi'_\mu\rangle=\frac1{k!}\int_{\Rd}\cdots\int_{\Rd}-\langle\zeta'(\lambda;\{0,x_1,\dots,x_{k-1}\}),\chi'_\mu\rangle dx_1\dots dx_{k-1}\geq {\frak n}_k(I_{-\mu^\delta}),$$ and $$\langle n_k,\chi''_\mu\rangle=\frac1{k!}\int_{\Rd}\cdots\int_{\Rd}-\langle\zeta'(\lambda;\{0,x_1,\dots,x_{k-1}\}),\chi''_\mu\rangle dx_1\dots dx_{k-1}\leq {\frak n}_k(I_{+3\mu^\delta})$$ as, for $\lambda<0$ (i.e $\lambda\not\in\sigma(-\Delta)$), $\zeta'(\lambda;\{0,x_1,\dots,x_{k-1}\})=-\tr(E(d\lambda;x_1,\dots,x_{k-1}))$ \par One easily checks that, for any $n\geq0$, there exists $C_n>0$ such that $$\sup_{x\in\R}\vert\partial^n\chi''_\mu(x)\vert+\sup_{x\in\R}\vert\partial^n\chi''_\mu(x)\vert\leq C_n\mu^{\delta\cdot n}.$$ As we know an upper bound on the order needed to estimate the rest in the asymptotic expansion of Theorem 1.1 (see the remark following that theorem), we get, choosing $\delta$ enough small, $$\left\vert\langle N_\mu,\chi'_\mu\rangle-\langle n_k,\chi'_\mu\rangle\right\vert+\left\vert\langle N_\mu,\chi''_\mu\rangle-\langle n_k,\chi''_\mu\rangle\right\vert\to 0\text{ when }\mu\to0.$$ This completes the proof of Theorem 1.2. \qed\enddemo \medskip \subhead II) The proof of Theorem 1.1 \endsubhead \medskip The main idea of the proof of Theorem 1.1 is to approximate the Poisson random Schr\"odinger operator by suitably chosen Bernoulli random Schr\"odinger operators, to use the techniques and results of \cite{Kl} to get an asymptotic expansion for these models and to show that these asymptotic expansions converge to some new asymptotic expansion when one lets the Bernoulli approximation tend to the Poisson model. \par As a last remark, we point out that due to the many estimates needed and for the sake of simpleness, each estimate involving new constants, we chose to give the same name, $C$, to all these constants. So the value of $C$ is changing from one equation to another. \smallskip \subhead a) The binomial approximation \endsubhead \smallskip The main technical difficulty comes from the fact that the realizations of the Poisson potential are almost surely unbounded from below (at least if we assume $\{x\in\Rd;\ V(x)<0\}\not=\emptyset$). To overcome this difficulty, we first approximate the Poisson potential by a potential that is semi-bounded below. Let $\psi:\ \R\to\R$ be a ${\Cal C}^\infty$ function such that $\psi\geq0$, $\psi\equiv1$ in $[-1,1]$ and supp$\psi\subset[-2,2]$. For $n\in{\Bbb N}^*$ and $\o\in\O$, define $$V_\o^n(x)=V_\o(x)\psi\left(\frac1nV_\o(x)\right)\text{ and }H_\o^n=-\Delta+V^n_\o.$$ Then one has the \proclaim{Proposition 2.1} For any $\varphi\in\Co$, $$\lim_{n\to+\infty}{\Bbb E}_\o\{\tr(\chi_0\varphi(H_\o^n)\chi_0)\}={\Bbb E}_\o\{\tr(\chi_0\varphi(H_\o)\chi_0)\}.$$ \endproclaim Now to compute an asymptotic expansion for ${\Bbb E}_\o\{\tr(\chi_0\varphi(H_\o^n)\chi_0)\}$, we approximate $V_\o^n$ by some truncated Bernoulli random potential constructed in the following way: \par let $(t_\gamma)_{\gamma\in\frac1m\Zd}$ be a family of independent identically distributed Bernoulli random variables with common probability distribution $P(t_0=1)=p_m$ and $P(t_0=0)=1-p_m$ with $p_m=\mu m^{-d}$. Define $$V_m(t;x)=V_m(x)=\sum_{\gamma\in\frac1m\Zd}t_\gamma V(x-\gamma)\text{ and }V_m^n(t;x)=V_m^n(x)=V_m(x)\psi\left(\frac1nV_m(x)\right).$$ At last define, $$H_m^n(t)=H_m^n=-\Delta+V_m^n.$$ Then one proves \proclaim{Proposition 2.2} For any $\varphi\in\Co$, $$\lim_{m\to+\infty}{\Bbb E}_t\{\tr(\chi_0\varphi(H_m^n)\chi_0)\}={\Bbb E}_\o\{\tr(\chi_0\varphi(H_\o^n)\chi_0)\}.$$ Here ${\Bbb E}_t\{\cdot\}$ denotes the average with respect to the probability measure enduced by the random variables $(t_\gamma)_{\gamma\in\frac1m\Zd}$. \endproclaim Then, to get Theorem 1.1, we compute an asymptotic expansion of ${\Bbb E}_t\{\tr(\chi_0\varphi(H_m^n)\chi_0)\}$ for which we control the principal terms and the remainder when $m$ and $n$ go to $\infty$. This is done in the \proclaim{Proposition 2.3} Let $\mu_0>0$ and $N>1$ be an integer. There exists $C_{\mu_0,N}$ and $\vert\cdot\Vert$ a semi-norm in $\S$ of order at most $(N+1)d+N+5$ (depending only on $\mu_0$ and $N$), such that, for any $n\geq 2C\cdot N$, for any $m\geq1$ and for any $\varphi\in\Co$, one has $${\Bbb E}_t\{\tr(\chi_0\varphi(H_m^n)\chi_0)\}=\tr((\chi_0\varphi(-\Delta)\chi_0)+\sum_{j=1}^N \frac{\mu^j}{j!}\langle n_j^m,\varphi\rangle+\mu^{N+1}R_{m,N}^n(\varphi),$$ where $$\langle n_j^m,\varphi\rangle=m^{-jd}\sum\Sb \Lambda\subset\frac1m\Zd \\ \#\Lambda=j\endSb \sum_{A\subset\Lambda}(-1)^{j-\#A}\tr(\chi_0\varphi(H_A)\chi_0)\text{ and } H_A=-\Delta+\sum_{\gamma\in A}V(x-\gamma),$$ and $$\vert R_{m,k}^n(\varphi) \vert\leq C_{\mu_0,N}\vert \varphi \Vert\quad\text{for}\quad\mu\in[0,\mu_0].$$ \endproclaim \smallskip \subhead b) The end of the proof of Theorem 1.1 \endsubhead \smallskip To get Theorem 1.1, we now just need to compute the limits of $\langle n_j^m,\varphi\rangle$ when $m\to+\infty$ for $\varphi\in\Co$. Therefore we prove \proclaim{Proposition 2.4} Let $(x_1,\dots,x_k)\in(\Rd)^k$. Define $$\aligned \langle Z(0,x_1,\dots,x_k),\varphi\rangle&=-\sum_{A\subset\{0,x_1,\dots,x_k\}}(-1)^{\#A}\langle\zeta'(A),\varphi\rangle \\ &=\sum_{A\subset\{0,x_1,\dots,x_k\}}(-1)^{\#A}\tr(\varphi(H_A)-\varphi(-\Delta)).\endaligned$$ Then $(x_1,\dots,x_k)\mapsto \langle Z(0,x_1,\dots,x_k),\varphi\rangle$ is continuous and is in $L^1((\Rd)^k)$. Moreover, for some $n>0$ and $C>0$, one has, for any $\varphi\in\Co$, $$\int_{\Rd}\cdots\int_{\Rd}\vert\langle Z(0,x_1,\dots,x_k),\varphi\rangle\vert dx_1\dots dx_k\leq C \sum_{j,l=0}^n\sup_{x\in\R}\vert (1+x^2)^{j/2}\partial_x^l\varphi(x)\vert.$$ \endproclaim \remark{Remark} The constant $n$ in Proposition 2.4 may be bounded by $n\leq (k+1)(d+1)+2$. \endremark Let $C(m)=\{\gamma=(\gamma_1,\dots,\gamma_d)\in\frac1m\Zd;\ -1/2\leq\gamma_j<1/2\text{ for }1\leq j\leq d\}$, and $\chi_0^m$ be the characteristic function of a cube of center 0 and sidelength $1/m$. Then, using the translation invariance of $-\Delta$ and the commutation property of the trace, we compute $$\aligned \langle n_k^m,\varphi\rangle&=m^{-kd}\sum\Sb \Lambda\subset\frac1m\Zd \\ \#\Lambda=k\endSb \sum_{A\subset\Lambda}(-1)^{k-\#A}\tr((\varphi(H_A)-\varphi(-\Delta))\chi_0) \\&=m^{-kd}\sum\Sb \Lambda\subset\frac1m\Zd \\ \#\Lambda=k\endSb \sum_{\gamma\in C(m)}\sum_{A\subset\Lambda}(-1)^{k-\#A}\tr((\varphi(H_A)-\varphi(-\Delta))\chi_0^m(\cdot-\gamma)) \\&=m^{-kd}\sum_{\gamma\in C(m)} \sum\Sb \Lambda\subset\frac1m\Zd \\ \#\Lambda=k\text{ and }0\in\Lambda\endSb \sum_{\beta\in\frac1m\Zd}\sum_{A\subset\beta+\Lambda}(-1)^{k-\#A}\tr((\varphi(H_A)-\varphi(-\Delta))\chi_0^m(\cdot-\gamma)) \\&=m^{-kd}\sum_{\gamma\in C(m)} \sum\Sb \Lambda\subset\frac1m\Zd \\ \#\Lambda=k\text{ and }0\in\Lambda\endSb\sum_{A\subset\Lambda}(-1)^{k-\#A}\sum_{\beta\in\frac1m\Zd}\tr((\varphi(H_A)-\varphi(-\Delta))\chi_0^m(\cdot-\gamma-\beta)) \\&=m^{-(k-1)d}\sum\Sb \Lambda\subset\frac1m\Zd\\\#\Lambda=k\text{ and }0\in\Lambda\endSb\sum_{A\subset\Lambda}(-1)^{k-\#A}\tr(\varphi(H_A)-\varphi(-\Delta)) \endaligned$$ as $\dsize\sum_{\beta\in\frac1m\Zd}\chi_0^m(\cdot-\gamma-\beta)=1$ for any $\gamma\in\frac1m\Zd$ and $\#C(m)=m^d$. Hence $$\langle n_k^m,\varphi\rangle=(-1)^k m^{-(k-1)d}\sum\Sb \Lambda\subset\frac1m\Zd\\\#\Lambda=k\text{ and }0\in\Lambda\endSb \langle Z(0,x_1,\dots,x_{k-1}),\varphi\rangle$$ so, by Proposition 2.4, $$\lim_{m\to+\infty}\langle n_k^m,\varphi\rangle=(-1)^k\int_{\Rd}\cdots\int_{\Rd}\langle Z(0,x_1,\dots,x_{k-1}),\varphi\rangle dx_1\dots dx_{k-1}.$$ This ends the proof of Theorem 1.1 by Propositions 2.1, 2.2 and 2.3. \medskip \subhead III) The proofs of Proposition 2.1, 2.2, 2.3 and 2.4 \endsubhead \medskip We will begin with the proof of Proposition 2.3. \demo{Proof of Proposition 2.3} To find an expansion for ${\Bbb E}_t\{\tr(\chi_0\varphi(H_m^n)\chi_0)\}$, we will proceed as in \cite{Kl}. Let $\Lambda_l$ be a cube in $\Rd$ centered at 0 and of sidelentgh $l$. For any realization of $t$, we define $$H_{m,l}^n(t)=H+V_{m,l}^n\text{ where }V_{m,l}^n(t;x)=V_{m,l}(x)\cdot\psi\left(\frac1nV_{m,l}(x)\right)\text{ and }V_{m,l}(x)=\sum_{\gamma\in\Lambda_l}t_\gamma V_\gamma(x)$$ where $\psi$ has been defined in section 2. \par It is not difficult to see (cf \cite{Kl}) that $\chi_0\varphi(H_{m,l}^n)\chi_0\to\chi_0\varphi(H_m^n)\chi_0$ (in trace class norm) as $l\to+\infty$. Hence by Lebesgue's Dominated Convergence Theorem, $${\Bbb E}_t\{\tr(\chi_0\varphi(H_m^n)\chi_0)\}=\lim_{l\to+\infty}{\Bbb E}_t\{\tr(\chi_0\varphi(H_{m,l}^n)\chi_0)\}.$$ As $H_{m,l}^n$ only depends on a finite number of the random variables $(t_\gamma)_{\gamma\in\Zd}$, it is a polynomial in $p_m$. Hence, we may write its Taylor expansion. Using the computations done in \cite{Kl}, we get $$ {\Bbb E}_t\{\tr(\chi_0\varphi(H_{m,l}^n)\chi_0)\}=\sum_{j=0}^k\frac{f^{(j)}_l(\varphi,0)}{j!}(p_m)^j+\frac{(p_m)^{k+1}}{(k+1)!}\int_0^1f^{(k+1)}_l(\varphi,p_mv)(1-v)^kdv \tag 3.1$$ with $$f^{(j)}_l(\varphi,0)=\sum\Sb \Lambda\subset\frac1m\Zd\cap\Lambda_l \\ \#\Lambda=j\endSb\sum\Sb A\subset\Lambda\endSb(-1)^{\#(\Lambda\setminus A)}\tr(\chi_0\varphi(H_A^n)\chi_0)\} \tag 3.2 $$ where $$H_A^n=H+V_A^n\text{ and }V_A^n(t;x)=V_A(x)\cdot\psi\left(\frac1nV_A(x)\right)\text{ and }V_A(x)=\sum_{\gamma\in A}t_\gamma V_\gamma(x), $$ and, $$f^{(k+1)}_l(\varphi,p)=\sum\Sb \Lambda\subset\frac1m\Zd\cap\Lambda_l \\ \#\Lambda=k+1\endSb\int_{[0,1]^{\Lambda}}{\Bbb E}_u\left\{\tr\left(\chi_0\partial_\Lambda\left[\varphi(H_{m,\Lambda}^n(t,u))\right]\chi_0\right)\right\}dt_\Lambda, \tag 3.3 $$ where: \par\noindent (a) $H_{m,\Lambda}^n(t,u)=H+V_{m,\Lambda}^n(t,u)$ \par\noindent (b) $V_{m,\Lambda}^n(t,u)=V_{m,\Lambda}(t,u)\psi(\frac1nV_{m,\Lambda}(t,u))$ and $V_{m,\Lambda}(t,u)=\sum_{\gamma\in \Lambda}t_\gamma V_\gamma+\sum_{\gamma\in(\Lambda_l\setminus\Lambda)}u_\gamma V_\gamma$, \par\noindent (c) $(t_\gamma)_{\gamma\in\Lambda}$ are variables taking value in $[0,1]$, $\dsize\partial_\Lambda=\bigotimes_{\gamma\in\Lambda}\partial_{t_\gamma}$ et $\dsize dt_\Lambda=\bigotimes_{\gamma\in\Lambda}dt_\gamma$, \par\noindent (d) $(u_\gamma)_{\gamma\in\frac1m\Zd}$ are i.i.d random variables with common probability distribution $p\delta_1+(1-p)\delta_0$, \par\noindent (e) ${\Bbb E}_u$ is the expectation taken with respect to these random variables. \smallskip \par If we choose $n>2C\cdot(k+1)$ (C given in (H.1)) then, for $0\leq j\leq k$ and $\#A=j$, one has $\sup_{x\in\R}\mid V_A(x)\mid\leq n$, hence, $V_A^n=\sum_{\gamma\in A}t_\gamma V_\gamma(x)$. So, by \cite{Kl}, we know that $$\lim_{l\to+\infty}f^{(j)}_l(\varphi,0)=\sum\Sb \Lambda\subset\frac1m\Zd \\ \#\Lambda=j\endSb\sum\Sb A\subset\Lambda\endSb(-1)^{\#(\Lambda\setminus A)}\tr(\chi_0\varphi(H_A)\chi_0)\}.$$ The right hand side of this equation is a nice distribution in $\Sp$ as can be seen from \cite{Kl}. It is of order at most $(j+1)(d+1)+2$. \par To end the proof of Proposition 2.3, we then just are left with estimating the rest of expansion (3.1). Therefore, we recall formula (2.2) of \cite{Kl} (see also \cite{He-Sj}), that is $$\varphi(H_{m,\Lambda}^n(t,u))=\frac i{2\pi}\int_\C\frac{\partial\tilde\varphi}{\partial\overline z}(z)(z-H_{m,\Lambda}^n(t,u))^{-1}d\overline z\wedge dz, \tag 3.4$$ where $\tilde\varphi:\ \C\to\C$ is an almost analytic extension of $\varphi$. Derivating (3.4) with respect to $t$, we get $$\partial_\Lambda\left[\varphi(H_{m,\Lambda}^n(t,u))\right]=\frac i{2\pi}\int_\C\frac{\partial\tilde\varphi}{\partial\overline z}(z)\partial_\Lambda\left[(z-H_{m,\Lambda}^n(t,u))^{-1}\right]d\overline z\wedge dz.\tag 3.5$$ The interversion of teh derivation and the integration is easily justified using Lebesgue's Dominated Convergence Theorem, the fact that $\Vert\partial_\Lambda(z-H_{m,\Lambda}^n(t,u))^{-1}\Vert\leq C\vert\text{Im}z\vert^{-k-1}$ and the vanishing of $\dsize\frac{\partial\tilde\varphi}{\partial\overline z}$ to any order on $\R$. \par To estimate $f^{(k+1)}_l(\varphi,p_mv)$, we just need to estimate, for $z\in\C\setminus\R$ and $p=p_mv$, $$f^{(k+1)}_l(z,p)=\sum\Sb \Lambda\subset\frac1m\Zd\cap\Lambda_l \\ \#\Lambda=k+1\endSb\int_{[0,1]^{\Lambda}}{\Bbb E}_p\left\{\tr\left(\chi_0\partial_\Lambda\left[(z-H_{m,\Lambda}^n(t,u))^{-1}\right]\chi_0\right)\right\}dt_\Lambda, $$ One computes $$\partial_\Lambda\left[(z-H_{m,\Lambda}^n(t,u))^{-1}\right]=\sum_{\sigma\in{\frak S}(\Lambda)}(z-H_{m,\Lambda}^n(t,u))^{-1}\prod_{\gamma\in\Lambda}\left(\partial_{t_{\sigma(\gamma)}}\left(V_{m,\Lambda}^n(t,u)\right)(z-H_{m,\Lambda}^n(t,u))^{-1}\right).$$ where ${\frak S}(\Lambda)$ is the group of permutations of $\Lambda$. \par\noindent By Lemma 4.1, $(z-H_{m,\Lambda}^n(t,u))^{-1}$ is in ${\Cal T}_q$, the $q$-th Schatten class for $q>d/2$, hence as $k+1>d/2$, $\partial_\Lambda\left[(z-H_{m,\Lambda}^n(t,u))^{-1}\right]$ is trace class; so by (3.3), for $v\in[0,1]$, $$\vert f^{(k+1)}_l(z,p_mv)\vert\leq {\Bbb E}_{p_m\cdot v}\left\{\Sigma_{m,l}^n\right\}, \tag 3.6$$ where $$\Sigma_{m,l}^n=\sum\Sb \Lambda\subset\frac1m\Zd \\ \Lambda\subset\Lambda_l \\ \#\Lambda=k+1\endSb \sum_{\sigma\in{\frak S}(\Lambda)} \sup_{t\in[0,1]^{\Lambda}}\left\Vert\chi_0(z-H_{m,\Lambda}^n(t,u))^{-1}\prod_{i=1}^{k+1}\left(\partial_{t_{\gamma_i}}V_{m,\Lambda}^n(t,u)(z-H_{m,\Lambda}^n(t,u))^{-1}\right)\chi_0\right\Vert_{{\Cal T}_1}.$$ \par Now we want to estimate the right hand side of (3.6) uniformly in $n$ and $l$ that is, we want to take into account the fact that $V_\o$ is almost surely unbounded. The crucial point therefore is to notice that nevertheless, loosely speaking, the probability that $V_\o$ get very negative is small. We can estimate this probability; this is done in Lemma 4.2. Hence, as in (3.6) we compute an average, either $V_\o$ is very negative which happens only with very small probability, or we have good estimates (given by Lemma 4.1) on $\Sigma_{m,l}^n$ uniformly in $n$ and $l$. This will permit us to get the needed estimates for (3.6). \par Let $(\gamma_i)_{1\leq i\leq k+1}$ are the points of $\Lambda$. We compute $$\partial_{t_{\gamma_i}}\left(V_{m,\Lambda}^n(t,u)\right)=V_{\gamma_i}\cdot\left(\psi\left(\frac1nV_{m,\Lambda}(t,u)\right)+\frac1nV_{m,\Lambda}(t,u)\cdot\psi\left(\frac1nV_{m,\Lambda}(t,u)\right)\right).$$ Notice that, for some $C>0$, for any $\Lambda$, $m$ and $n$, $$\sup_{x\in\Rd}\left\vert\psi\left(\frac1nV_{m,\Lambda}(t,u)(x)\right)+\frac1nV_{m,\Lambda}(t,u)(x)\cdot\psi'\left(\frac1nV_{m,\Lambda}(t,u)(x)\right)\right\vert\leq C.$$ By Lemma 4.1, for $\nu'\in(0,1)$, if $\vert V_{m,\Lambda}^n(t,u)(x)\vert\leq \vert x\vert^{\nu}+N$ for all $x\in\Rd$, one has, for some $C>0$ and some $\epsilon>0$, $$\aligned &\Vert\chi_0 (z-H_{m,\Lambda}^n(t,u))^{-1}\prod_{i=1}^{k+1}\left(\partial_{t_{\gamma_i}}\left(V_{m,\Lambda}^n(t,u)\right)(z-H_{m,\Lambda}^n(t,u))^{-1}\right)\chi_0\Vert_{{\Cal T}_1} \\ &\hskip 2cm\leq C\sum_{\beta_1\in\Zd}\cdots\sum_{\beta_{k+1}\in\Zd}\Vert\chi_0 (z-H_{m,\Lambda}^n(t,u))^{-1}\chi_{\beta_1}\Vert_{{\Cal T}_q}\cdot \\&\hskip4cm \cdot\left(\prod_{i=1}^{k}\Vert V_{\gamma_i}\chi_{\beta_i}\Vert\cdot\Vert\chi_{\beta_i}(z-H_{m,\Lambda}^n(t,u))^{-1}\chi_{\beta_{i+1}}\Vert_{{\Cal T}_q}\right)\cdot \\ &\hskip 6cm\cdot\Vert V_{\gamma_{k+1}}\chi_{\beta_{k+1}}\Vert\cdot\Vert\chi_{\beta_{k+1}}(z-H_{m,\Lambda}^n(t,u))^{-1}\chi_0\Vert_{{\Cal T}_q} \\&\hskip 2cm\leq \frac C{\eta(z,N)^{k+2}}\sum_{\beta_1\in\Zd}\cdots\sum_{\beta_{k+1}\in\Zd}\left(\prod_{i=1}^{k+1}e^{-\eta\mid\beta_i-\gamma_i\mid}\right)\cdot\left(\prod_{i=1}^{k+1}(1+\vert\beta_i\vert)^\nu\right)\cdot \\ &\hskip 4cm \cdot e^{-\epsilon\cdot\eta(z,N)\left(\vert1-\delta_\nu(\beta_1)\vert+\sum_{i=1}^{k}\ve! rt\delta_\nu(\beta_i)-\delta_\nu( where $\delta_\nu(\cdot)=(1+(\cdot)^2)^{(1-\nu)/2}$. So, as $\mid\delta_nu(x)-\delta_nu(y)\mid\leq 2\mid x-y\mid$, $$\aligned &\Vert\chi_0 (z-H_{m,\Lambda}^n(t,u))^{-1}\prod_{i=1}^{k+1}\left(\partial_{t_{\gamma_i}}V_{m,\Lambda}^n(t,u)(z-H_{m,\Lambda}^n(t,u))^{-1}\right)\chi_0\Vert_{{\Cal T}_1}\\ &\hskip 2cm\leq \frac C{\eta(z,N)^{k+2}}\left(\prod_{i=1}^{k+1} e^{-\inf(\frac\eta4,\frac{\epsilon}{k+1}\cdot\eta(z,N))\delta_\nu(\gamma_i)}\right)\cdot \\ &\hskip 6cm \cdot\left(\sum_{\beta_1\in\Zd}\cdots\sum_{\beta_{k+1}\in\Zd}\left(\prod_{i=1}^{k+1}e^{-\frac\eta4\mid\beta_i-\gamma_i\mid}(1+\vert\beta_i\vert)^\nu\right)\right)\\ &\hskip 2cm\leq \frac C{\eta(z,N)^{k+2}}\left(\prod_{i=1}^{k+1} e^{-\inf(\frac\eta4,\frac{\epsilon}{k+1}\cdot\eta(z,N))\delta_\nu(\gamma_i)}(1+\vert\gamma_i\vert)^\nu\right).\endaligned$$ Hence, if $u$ is such that $\dsize\vert\sum_{\gamma\in\frac1m\Zd}u_\gamma V_\gamma(x)\vert\leq\mid x\mid^\nu+N-k-1$ then $\sup_t\vert V_{m,\Lambda}^n(t,u)(x)\vert\leq \vert x\vert^{\nu}+N$; in this case we get $$\aligned \Sigma_{m,l}^n &\leq \frac C{\eta(z,N)^{k+2}}\left(\sum_{\gamma\in\frac1m\Zd}e^{-\inf(\frac\eta4,\frac{\epsilon}{k+1}\cdot\eta(z,N))\delta_\nu(\gamma)}(1+\vert\gamma\vert)^\nu\right)^{k+1}\\ &\leq\frac C{\eta(z,N)^{k+2}}\left(\sum_{\gamma\in\Zd}\sum_{\beta\in C(m)} e^{-\inf(\frac\eta4,\frac{\epsilon}{k+1}\cdot\eta(z,N))\delta_\nu(\gamma+\beta)}(1+\vert\gamma+\beta\vert)^\nu\right)^{k+1} \\ & \leq\frac{C\cdot m^{(k+1)d}}{\eta(z,N)^{k+1}}\left(\sum_{\gamma_1\in\Zd}e^{-\inf(\frac\eta4,\frac{\epsilon}{k+1}\cdot\eta(z,N))\delta_\nu(\gamma_i)}(1+\vert\gamma_i\vert)^\nu\right)^{k+1}\\ &\leq \frac{C\cdot m^{(k+1)d}}{\eta(z,N)^{k+1}}\cdot\frac1{\inf(\frac\eta4,\frac{\epsilon}{k+1}\cdot\eta(z,N))^{(k+1)\frac{d+\nu}{1-\nu}}}.\endaligned $$ \par We define the events $E_k$ and, for $N>k$, $E_N$ by $$E_k=\left\{u;\ \forall x\in\R^d\ \left\vert\sum_{\gamma\in\frac1m\Zd}u_\gamma V_\gamma(x)\right\vert\leq\mid x\mid^\nu+1\right\}.$$ and $$\multline E_N=\left\{u;\ \exists x_0\in\R^d\ \mid x_0\mid^\nu+N-k\leq\left\vert\sum_{\gamma\in\frac1m\Zd}u_\gamma V_\gamma(x_0)\right\vert \right. \\ \left.\text{ and }\forall x\in\R^d\ \left\vert\sum_{\gamma\in\frac1m\Zd}u_\gamma V_\gamma(x)\right\vert\leq\mid x\mid^\nu+N+1-k\right\}.\endmultline$$ Using the Borel-Cantelli lemma and the estimate (1) of Lemma 4.2, we know that, for $v\in[0,1]$ $${\Bbb P}_{p_m\cdot v}\left(\left\{u;\exists N>0\text{ such that }\forall x\in\R^d\ \left\vert\sum_{\gamma\in\frac1m\Zd}u_\gamma V_\gamma(x)\right\vert\leq\mid x\mid^\nu+N\right\}\right)=1.$$ This means that $\cup_{N\geq k}E_N$ is of total measure. So, to estimate (3.6), we compute, using Lemma 4.2 $$\aligned \vert f^{(k+1)}_l(z,p_mv)\vert &\leq {\Bbb E}_{p_m\cdot v}\left\{\Sigma_{m,l}^n\right\} \\ &\leq \sup_{E_k}\Sigma_{m,l}^n+\sum_{N\geq k+1}{\Bbb P}_{p_m\cdot v}(E_N)\sup_{E_N}\Sigma_{m,l}^n \\ &\leq \frac{C\cdot m^{(k+1)d}}{\eta(z,k+1)^{k+2}}\cdot\frac1{\inf(\frac\eta4,\frac{\epsilon}{k+1}\cdot\eta(z,k+1))^{(k+1)\frac{d+\nu}{1-\nu}}}+ \\ &\hskip 1cm +F_\nu(v\mu)\sum_{N\geq 1}\frac{(v\mu)^{\e(N/c)}}{(\e(N/c))!}\cdot\frac{C\cdot m^{(k+1)d}}{\eta(z,k+N)^{k+2}}\cdot\frac1{\inf(\frac\eta4,\frac{\epsilon}k\cdot\eta(z,k+N))^{(k+1)\frac{d+\nu}{1-\nu}}}\endaligned$$ where for $x\in\R$, $\e(x)=\inf\{n\in{\Bbb N};\ x\leq n\}$. \par If $\dsize \epsilon\leq\frac{(k+1)\eta}4$, then $\dsize\inf(\frac\eta4,\frac{\epsilon}{k+1}\cdot\eta(z,k+N))\geq\frac{\text{Im}(z)}{\mid z\mid+k+N}$; hence $$\aligned \vert f^{(k+1)}_l(z,p_mv)\vert &\leq C\cdot m^{(k+1)d}\left(\frac{\mid z\mid+1}{\mid\text{Im}z\mid}\right)^{k+2+(k+1)\frac{d+\nu}{1-\nu}}\cdot \\ &\hskip 2cm\cdot\left(1+F_\nu(v\mu)\sum_{N>1}\frac{(v\mu)^{\e(N/c)}}{(\e(N/c))!}(N+k)^{k+2+(k+1)\frac{d+\nu}{1-\nu}}\right) \\&\leq C\cdot m^{(k+1)d}\left(\frac{\mid z\mid+1}{\mid\text{Im}z\mid}\right)^{k+2+(k+1)\frac{d+\nu}{1-\nu}}.\endaligned$$ Plugging this into (3.4) and using the properties of almost analytic extensions (see \cite{Kl}), we get that, for some $C>0$ (depending on $\nu$, $k$) and some $\vert\cdot\Vert$, a semi-norm of $\Sp$ of order at most $k+4+(k+1)\frac{d+\nu}{1-\nu}$, such that for $m\geq1$, one has, for some $C_{k,\nu}>0$, $$\sup_{l>0}\sup_{v\in[0,1]}\sup_{n>0}\vert f^{(k+1)}_l(\varphi,p_mv)\vert\leq C_{k,\nu}\cdot m^{(k+1)d}\vert\varphi\Vert.$$ This ends the proof of Proposition 2.3 if one chooses $\nu$ close to 0. \qed\enddemo \demo{Proof of Proposition 2.1} To prove this, we will use Lebesgue's Dominated Convergence Theorem. Pick $\nu\in(0,1)$ and $q>d/2$. For $N>0$, let $$E_N=\{\o;\ \forall x\in\R^d\ \mid x\mid^\nu+N\leq\vert V_\o(x)\vert\leq\mid x\mid^\nu+N+1\}$$ and $$E_0=\{\o;\ \forall x\in\R^d\ \vert V_\o(x)\vert\leq\mid x\mid^\nu\}.$$ Using the Borel-Cantelli lemma and the estimate (2) of Lemma 4.2, we know that, $${\Bbb P}_\o\left(\{\o;\exists N>0\text{ such that }\forall x\in\R^d\ \vert V_\o(x)\vert\leq\mid x\mid^\nu+N\}\right)=1.$$ So, for almost every $\o$, there exists $N>0$ such that for all $n>0$ and $x\in\Rd$, $\vert V_\o^n(x)\vert\leq\mid x\mid^\nu+N$. For $n>N$, $z\in\C\setminus\R$ and $(\alpha,\beta)\in\Zd\times\Zd$, $$\aligned \Vert\chi_\alpha(z-H_\o^n)^{-1}\chi_\beta-\chi_\alpha(z-H_\o)^{-1}\chi_\beta\Vert_{{\Cal T}_q}&=\Vert\chi_\alpha(z-H_\o^n)^{-1}V_\o\left(\psi\left(\frac1nV_\o\right)-1\right)(z-H_\o)^{-1}\chi_\beta\Vert_{{\Cal T}_q}\\&\leq C n\Vert\chi_\alpha(z-H_\o^n)^{-1}\chi_{\{\vert x\vert\geq(n-N)^{1/\nu}\}}(z-H_\o)^{-1}\chi_\beta\Vert_{{\Cal T}_q} \endaligned $$ using the support properties of $\psi$. So Lemma 4.1 tells us that, for some $n_0>0$ and $\epsilon_0>0$, $$\multline \Vert\chi_\alpha(z-H_\o^n)^{-1}\chi_\beta-\chi_\alpha(z-H_\o)^{-1}\chi_\beta\Vert_{{\Cal T}_q}\leq \\ \leq Cn\left(\frac{\mid z\mid+N}{\mid\text{Im}z\mid}\right)^{n_0}e^{-\epsilon\frac{\mid\text{Im}z\mid}{\mid z\mid+N}(\vert \delta_\nu(\alpha)-n^{(1-\nu)/\nu}\vert+\vert \delta_\nu(\beta)-n^{(1-\nu)/\nu}\vert)}.\endmultline\tag 3.7 $$ Then using formula (3.4), if $\tilde\varphi$ is an almost analytic extension for the function $(i-x)^q\varphi(x)$, we get $$\multline \tr(\chi_0\varphi(H_\o^n)\chi_0)-\tr(\chi_0\varphi(H_\o)\chi_0) \\ =\frac i{2\pi}\int_\C\frac{\partial\tilde\varphi}{\partial\overline z}(z)\tr[\chi_0((z-H_\o^n)^{-1}(i-H_\o^n)^{-q}-(z-H_\o)^{-1}(i-H_\o)^{-q}\chi_0)]d\overline z\wedge dz.\endmultline$$ By (3.7), computing the same kind of sums as in the proof of Proposition 2.3, we estimate, for some $n_0>0$, $$\multline \Vert\chi_0\left((z-H_\o^n)^{-1}(i-H_\o^n)^{-q}-(z-H_\o)^{-1}(i-H_\o)^{-q}\right)\chi_0\Vert_{{\Cal T}_1}\leq \\ \leq C \left(\frac{\mid z\mid+N}{\mid\text{Im}z\mid}\right)^{n_0}\left( n\cdot e^{-\epsilon\frac{\mid\text{Im}z\mid}{\mid z\mid+N}(n^{(1-\nu)/\nu})}\right).\endmultline$$ Then, as $\dsize\frac{\partial\tilde\varphi}{\partial\overline z}(z)$ is vanishing at any order when Im$z$ vanishes, we get that, for almost every $\o$, $$\tr(\chi_0\varphi(H_\o^n)\chi_0)\}-\tr(\chi_0\varphi(H_\o)\chi_0)\to 0 \text{ when }n\to+\infty.$$ Using Lemma 4.1 and formula (3.4) in the same way as above, one shows that there exists $C_\nu>0$ and $n_\nu>0$ (depending on $\nu$ only) such that, for $V$ a potential satisfying $\forall x\in\Rd$, $\vert V_(x)\vert\leq\mid x\mid^\nu+N$ (for some $N>0$), one has $$\vert \tr(\chi_0\varphi(-\Delta+V)\chi_0)\vert\leq C_\nu(N+1)^{n_\nu}.$$ So, for almost every $\o$ and for any $n>0$ $$\vert\tr(\chi_0\varphi(H_\o^n)\chi_0)\}\vert\leq \sum_{N\geq0}\chi_{E_N}(\o)C_\nu(N+1)^{n_\nu}.$$ But, by Lemma 4.2 (2), for $N>0$, $${\Bbb E}_\o(\chi_{E_N}(\o))\leq{\Bbb P}(\{\o; \exists x\in\R^d\ \mid x\mid^\nu+N\leq\vert V_\o(x)\vert\}\leq \frac{\mu^{\e(N/C)}}{\e(N/C)!}G_\nu(\mu),$$ So the function $\o\mapsto \sum_{N\geq0}\chi_{E_N}(\o)C_\nu(N+1)^{n_\nu}$ is $L^1(\O,{\Bbb P}_\o)$. Having found an integrable majorant for $\vert\tr(\chi_0\varphi(H_\o^n)\chi_0)\}\vert$, we apply Lebesgue's Dominated Convergence Theorem to get Proposition 2.1. \qed\enddemo \demo{Proof of Proposition 2.2} To prove this proposition we will use general facts about the convergence of probability measures. We introduce as shortly as possible the needed notions. More details may be found in \cite{Ne} (essentially in chapter I) where our presentation comes from. Let ${\Cal M}_+(\Rd)$ be the set of positive Radon measures on $\Rd$ (endowed with the topology of weak convergence). Then the Poisson process and the family of random variables $(t_\gamma)_{\gamma\in\frac1m\Zd}$ define two probability measures on ${\Cal M}_+(\Rd)$. A precise definition of the relevant $\sigma$-algebra can be found in \cite{Ne}. These probability measures may characterised by their Laplace transforms denoted by $\Psi_\o$ for the Poisson process and $\Psi_t$ for the Bernoulli process. The Laplace transforms are defined in the following way, for $f$ a positive borelian function on $\Rd$, $$\Psi_\o(f)={\Bbb E}_\o\left(e^{-\int_{\Rd}f(x)dm(\o,dx)}\right)\text{ and }\Psi_t(f)={\Bbb E}_t\left(e^{-\sum_{\gamma\in\frac1m\Zd}f(\gamma)}\right).$$ $\Psi_\o$ may be computed in terms of the intensity $\mu$ (i.e the concentration) (see \cite{Ne} p. 260). It is easy to see that, for $f$ being the characteristic function of some cube in $\Rd$, one has $\Psi_t(f)\to\Psi_\o(f)$ as $m\to+\infty$. This is just the usual theorem about the convergence in law of the sum of $n$ independent identical Bernoulli trials (with a suitably chosen probability law) to a Poisson law. \par Hence, approximating any positive borelian function by a sequence of positive linear combinations of characteristic functions of cubes, one gets that, for any $f$, positive borelian function on $\Rd$, one has $\Psi_t(f)\to\Psi_\o(f)$ as $m\to+\infty$. By Proposition I.16 of \cite{Ne}, this implies that for any bounded weakly continuous function $F$ on ${\Cal M}_+(\Rd)$, one has ${\Bbb E}_t[F(dm(t,dx))]\to{\Bbb E}_\o[F(dm(\o,dx))]$ when $m\to+\infty$. \par To prove Proposition 2.2, let us first assume that $V$ is compactly supported. This assumption will be removed later on. Then, for any $\m$, a positive Radon measure on $\Rd$, we define $$V_\m(x)=\int_{\Rd}V(x-y)d\m\text{ and }V_\m^n(x)=V_\m(x)\psi\left(\frac1nV_\m(x)\right)\text{ and }H_\m=-\Delta+V_\m^n,$$ here $\psi$ is the function defined in section 2. \par If we show that, for $\varphi\in\Co$, $\m\mapsto\tr(\chi_0\varphi(H_\m)\chi_0)$ is weakly continuous and bounded on ${\Cal M}_+(\Rd)$, Proposition I.16 of \cite{Ne} will show that ${\Bbb E}_t\{\tr(\chi_0\varphi(H_m^n)\chi_0)\}\to{\Bbb E}_\o\{\tr(\chi_0\varphi(H_\o^n)\chi_0)\}$ when $m\to+\infty$. \par As $\sup_\m\sup_{x\in\Rd}\vert V^n_\m(x)\vert\leq n$, for any $\m$, the operator $H_\m$ is lower semi-bounded by $-n$. Hence, we only need to consider test function $\varphi$ supported in $[-n,+\infty)$. Then, for such a test function, we define $\Phi(x)=(n+1+x)^q\varphi(x)$; obviously, $\Phi\in\Co$. Using (3.4) for $\tilde\varphi$ is an analytic extension of $\Phi$ and the fact that $\varphi(H_\m)=\Phi(H_\m)(n+1+H_\m)^{-q}$, we get $$\tr(\chi_o\varphi(H_\m)\chi_o)=\frac i{2\pi}\int_\C\frac{\partial\tilde\varphi}{\partial\overline z}(z)\tr(\chi_0(z-H_\m)^{-1}(n+1+H_\m)^{-q}\chi_0)d\overline z\wedge dz \tag 3.8$$ as $\chi_0(z-H_\m)^{-1}(n+1+H_\m)^{-q}\chi_0$ is trace class (we take $q>d/2$). Then using the bound on $V_\m^n$, we can upper-bound $\tr(\chi_0(z-H_\m)^{-1}(n+1+H_\m)^{-q}\chi_0)$ uniformly in $\m$ by some polynomial in $\dsize \frac{\mid z\mid+1}{\mid\text{Im}z\mid}$ for any $z\in\C\setminus\R$ (using either Lemma 4.1 or the section 4 of \cite{Kl}). This gives us a uniform bound on $\tr(\chi_0\varphi(H_\m)\chi_0)$ for $\m\in{\Cal M}_+(\Rd)$ (using the infinite order of vanishing on the real line for almost analytic extensions). \par We then need to show that $\m\mapsto\tr(\chi_0\varphi(H_\m)\chi_0)$ is weakly continuous in $\m$. That is, if $(\m_l)_{l\geq0}$ is sequence of Radon measure converging weakly to $\m$ (i.e, for any $f$ continuous with compact support, $\dsize\int_{\Rd}f(x)d\m_l\to\int_{\Rd}f(x)d\m$ when $l\to+\infty$) then $\tr(\chi_0\varphi(H_{\m_l})\chi_0)\to\tr(\chi_0\varphi(H_\m)\chi_0)$ when $l\to+\infty$. \par Let $(\m_l)_{l\geq0}$ be a sequence of Radon measure converging weakly to $\m$. Then, for any $x\in\Rd$, $V^n_{\m_l}(x)\to V^n_\m(x)$ when $l\to+\infty$ and $\sup_{l\geq0}\sup_{x\in\Rd}\vert V^n_{\m_l}(x)\vert\leq n$. Using (3.8) and the $\m$-uniform estimates on the traces we consider , we see that by Lebesgue's Dominated Convergence Theorem, we only need to show that, for $z\in\C\setminus\R$, $$\tr(\chi_0(z-H_{\m_l})^{-1}(n+1+H_{\m_l})^{-q}\chi_0)\to\tr(\chi_0(z-H_\m)^{-1}(n+1+H_\m)^{-q}\chi_0)\text{ when }l\to+\infty. \tag 3.9$$ We evaluate $$\aligned &\vert\tr(\chi_0(z-H_{\m_l})^{-1}(n+1+H_{\m_l})^{-q}\chi_0)-\tr(\chi_0(z-H_\m)^{-1}(n+1+H_\m)^{-q}\chi_0)\vert\leq \\ &\hskip1cm \leq\vert\tr(\chi_0(z-H_{\m_l})^{-1}(V_{\m_l}^n-V_\m^n)(z-H_{\m})^{-1}(n+1+H_{\m})^{-q}\chi_0)\vert+ \\ &\hskip 2cm+\sum_{p=1}^{q}\vert\tr(\chi_0(z-H_{\m_l})^{-1}(n+1+H_{\m_l})^{-p}(V_{\m_l}^n-V_\m^n)(n+1+H_{\m})^{-q+p-1}\chi_0)\vert.\endaligned \tag 3.10$$ Pick $q>d$. Pick $1\leq p\leq q$. Let us prove that $$\vert\tr(\chi_0(z-H_{\m_l})^{-1}(n+1+H_{\m_l})^{-p}(V_{\m_l}^n-V_\m^n)(n+1+H_{\m})^{-q+p-1}\chi_0)\vert\to 0\text{ when }n\to+\infty.\tag 3.11$$ The proof of this fact is the same for the first term of the right hand side of (3.10). \par We write $$\aligned &\chi_0(z-H_{\m_l})^{-1}(n+1+H_{\m_l})^{-p}(V_{\m_l}^n-V_\m^n)(n+1+H_{\m})^{-q+p-1}\chi_0= \\&\hskip 2cm=\sum_{\alpha\in\Zd}\chi_0(z-H_{\m_l})^{-1}(n+1+H_{\m_l})^{-p}\chi_\alpha\cdot\\ &\hskip5cm\cdot\chi_\alpha(V_{\m_l}^n-V_\m^n)(n+1+H_{\m})^{-1}\cdot\chi_\beta\\ &\hskip8cm\cdot \chi_\beta(n+1+H_{\m})^{-q+p}\chi_0.\endaligned$$ Either $\chi_0(z-H_{\m_l})^{-1}(n+1+H_{\m_l})^{-p}\chi_\alpha$ or $\chi_\beta(n+1+H_{\m})^{-q+p}\chi_0$ are trace class. Assume $\chi_\beta(n+1+H_{\m})^{-q+p}\chi_0$ is trace class. Moreover, using (4.1) and classical estimates on their operator norm, we get, for $z\in\C\setminus\R$, $$\Vert\chi_\beta(n+1+H_{\m})^{-q+p}\chi_0\Vert_{{\Cal T}_1}\leq e^{-\epsilon\delta_{1/2}(\beta)},$$ and $$\sup_{l\geq1}\Vert\chi_0(z-H_{\m_l})^{-1}(n+1+H_{\m_l})^{-p}\chi_\alpha\Vert\leq P(z)e^{-\mid\alpha\mid/P(z)}$$ where $\epsilon>0$ and $P(z)$ is some polynomial in $\dsize \frac{\mid z\mid+1}{\mid\text{Im}z\mid}$. \par\noindent Then if we show that $\chi_\alpha(V_{\m_l}^n-V_\m^n)(n+1+H_{\m})^{-1}\chi_\beta\to 0$ as a bounded operator when $l\to+\infty$, as it is uniformly bounded, using Lebesgue's Dominated Convergence Theorem, we get (3.11). As $\chi_\alpha$ is compactly supported, $\chi_\alpha(V_{\m_l}^n-V_\m^n)\to0$ in any $L^p(\Rd)$ for $1\leq p<+\infty$. Hence also as a bounded operator from $L^q(\Rd)$ to $L^2(\Rd)$ for any $q>2$. But by \cite{Si} section B.2, $(n+1+H_{\m})^{-1}\chi_\beta$ is bounded from $L^2(\Rd)$ to $L^{2+\delta}(\Rd)$ for some $\delta>0$. This ends the proof of (3.9), hence the proof of Proposition 2.2 in the case when $V$ is compactly supported. \par To get the result when $V$ is not compactly supported but satisfies only assumption (H.1), we approximate $V$ by $V^r=V\phi_r$ where $\phi_r(x)=\phi(x^2/r)$ and $\phi\in\Co$, $0\leq\phi\leq1$ and $\phi\equiv1$ on $\mid x\mid\leq1$. Define $$V_\o^r(x)=\int_{\Rd}V^r(x-y)m(\o,dy),\ V_\o^{n,r}(x)=V_\o^r(x)\psi\left(\frac1nV_\o^r(x)\right)\text{ and }H_\o^r=-\Delta+V_\o^{n,r},$$ and $$V_m^r(x)=\sum_{\gamma\in\frac1m\Zd}t_\gamma V^r(x-\gamma),\ \ V_m^{n,r}(x)=V_m^r(x)\psi\left(\frac1nV_m^r(x)\right)\text{ and }H_m^r=-\Delta+V_m^{n,r}.$$ We then have to show that $${\Bbb E}_\o\{\tr(\chi_0\varphi(H_\o^r)\chi_0)\}\to{\Bbb E}_\o\{\tr(\chi_0\varphi(H_\o)\chi_0)\}\text{ when }r\to+\infty$$ and $${\Bbb E}_t\{\tr(\chi_0\varphi(H_m^r)\chi_0)\}\to{\Bbb E}_t\{\tr(\chi_0\varphi(H_m)\chi_0)\}\text{ uniformly in }m\text{ when }r\to+\infty.$$ Let us do it in the case of the Bernoulli random variables. The Poisson case is done in the same way. By (3.8), it enough to show that, for $z\in\C\setminus\R$, $$\multline \sup_{m\geq1}\vert{\Bbb E}_t\{\tr(\chi_0(z-H_m^r)^{-1}(n+1+H_m^r)^{-q}\chi_0)\}-{\Bbb E}_t\{\tr(\chi_0(z-H_m)^{-1}(n+1+H_m)^{-q}\chi_0)\}\vert\to0 \\ \text{ when }r\to+\infty.\endmultline$$ Using the decomposition (3.10) and the uniform (in $m$ and in the realization of $t$) bounds known for the trace class norm and operator norm of $\chi_\beta(z-H_m^r)^{-p}\chi_0$ and $\chi_\beta(z-H_m)^{-p}\chi_0$ (these bounds are sub-exponentially decreasing in $\beta$), we just have to show that, for $z\in\C\setminus\R$, $$\sup_{m\geq1}\vert{\Bbb E}_t\{\Vert\chi_\alpha\left\lbrack(z-H_m^r)^{-1}-(z-H_m)^{-1}\right\rbrack\chi_\beta\Vert\}\vert\to0\text{ when }r\to+\infty$$ here $\Vert\cdot\Vert\text{ denotes the operator norm}$. \par\noindent For $x\in\Rd$, by assumption (H.1) on $V$, one has $$\vert V_m^r(x)-V_m(x)\vert\leq\sum_{\gamma\in\frac1m\Zd}t_\gamma\vert V(x-\gamma)\vert\cdot(1-\phi_r(x-\gamma))\leq e^{-\eta r/2}\sum_{\gamma\in\frac1m\Zd}t_\gamma e^{\eta\vert x-\gamma\vert/2}.$$ Then, mimicking the proof of Lemma 4.2, one proves that, for some $C>0$ and $\epsilon>0$ such that $C\epsilon e^{r\eta/2}>2$ $${\Bbb P}(\{t; \exists x\in\R^d\text{ such that }\vert V_m^r(x)-V_m(x)\vert\geq\epsilon\vert x\vert\})\leq \frac{\mu^{\e(C\epsilon e^{r\eta/2})}}{\e(C\epsilon e^{r\eta/2})!} F(\mu) \tag 3.12$$ where $F$ is increasing and positive. \par\noindent Define the event ${\Cal E}_r=\{t;\ \forall \mid x\mid\leq e^{\eta r/8},\ \vert V_m^r(x)-V_m(x)\vert\leq e^{-\eta r/8}\}$. Choosing in (3.12), $\epsilon= e^{-\eta r/4}$, we get, for $r$ large enough and some $C>0$, $${\Bbb P}({\Cal E}_r)\geq 1-Ce^{-\eta r/4}.$$ Then, for some $C>0$, $$\aligned &\vert{\Bbb E}_t\{\Vert\left\lbrack\chi_\alpha(z-H_m^r)^{-1}-(z-H_m)^{-1}\right\rbrack\chi_\beta\Vert\}\vert=\vert{\Bbb E}_t\{\Vert\chi_\alpha(z-H_m^r)^{-1}(V_m^{n,r}-V_m^n)(z-H_m)^{-1}\chi_\beta)\Vert\}\vert \\ &\hskip 2cm\leq C{\Bbb P}(^c{\Cal E}_r)+\left\vert\int_{{\Cal E}_r}\Vert\chi_\alpha(z-H_m^r)^{-1}(V_m^{n,r}-V_m^n)(z-H_m)^{-1}\chi_\beta)\Vert d{\Bbb P}_t\right\vert \\& \hskip 2cm\leq Ce^{-\eta r/8}+\left\vert\int_{{\Cal E}_r}\Vert\chi_\alpha(z-H_m^r)^{-1}\chi_{\mid x\mid>e^{\eta r/8}}(V_m^{n,r}-V_m^n)(z-H_m)^{-1}\chi_\beta)\Vert d{\Bbb P}_t\right\vert+ \\ &\hskip 4.5cm+\left\vert\int_{{\Cal E}_r}\Vert\chi_\alpha(z-H_m^r)^{-1}\chi_{\mid x\mid\leq e^{\eta r/8}}(V_m^{n,r}-V_m^n)(z-H_m)^{-1}\chi_\beta)\Vert d{\Bbb P}_t\right\vert, \endaligned $$ where $\chi_E$ is the characteristic function of $E\subset\R$. \par\noindent We know that, for some $C>0$ and $\epsilon>0$, $\Vert\chi_\alpha(z-H_m^r)^{-1}\chi_{\mid x\mid>e^{\eta r/8}}\Vert\leq Ce^{-\epsilon\mid\mid\alpha\mid-e^{\eta r/8}\mid}$ and $\Vert\chi_{\mid x\mid>e^{\eta r/8}}(z-H_m)^{-1}\chi_\beta)\Vert\leq Ce^{-\epsilon\mid\mid\beta\mid-e^{\eta r/8}\mid}$. Moreover, for some $C>0$, $\mid V_m^{n,r}(x)-V_m^n(x)\mid\leq C\mid V_m^{r}(x)-V_m(x)\mid$. Hence, for some $C>0$ and $\epsilon>0$, we get $$\multline\sup_{m\geq1}\left(\vert{\Bbb E}_t\{\Vert\left\lbrack\chi_\alpha(z-H_m^r)^{-1}-(z-H_m)^{-1}\right\rbrack\chi_\beta\Vert\}\vert\right)\leq \\ \leq C\left(e^{-\eta r/8}+n\cdot e^{-\epsilon(\mid\mid\alpha\mid-e^{\eta r/16}\mid+\mid\mid\beta\mid-e^{\eta r/16}\mid)}+e^{-\eta r/4}\right).\endmultline$$ This ends the proof of Proposition 2.2. \qed\enddemo \demo{Proof of Proposition 2.4} As, for $\#A=k$, $H_A$ is semi-bounded below by $-k\cdot C$ ($C$ given in (H.1)), we rewrite (3.8) in the following form, for $n\geq k\cdot C$, $$\varphi(H_A)=\frac i{2\pi}\int_\C\frac{\partial\tilde\varphi}{\partial\overline z}(z)(z-H_A)^{-1}(n+1+H_A)^{-q}d\overline z\wedge dz.$$ where $\tilde \varphi$ is an almost analytic extension of $\Phi(x)=(n+1+x)^q\varphi(x)$. \par So, we get, for $q>d/2$, (see \cite{Kl}) $$\aligned &\langle Z(0,x_1,\dots,x_k),\varphi\rangle= \\ &\hskip2.5cm=\frac i{2\pi}\sum_{A\subset\{0,x_1,\dots,x_k\}}(-1)^{\#A}\int_\C\frac{\partial\tilde\varphi}{\partial\overline z}(z)\tr\left((z-H_A)^{-1}(n+1+H_A)^{-q}-\right. \\ &\hskip6cm \left.-(z+\Delta)^{-1}(n+1-\Delta)^{-q}\right)d\overline z\wedge dz\\ &\hskip2.5cm=\frac i{2\pi}\int_\C\frac{\partial\tilde\varphi}{\partial\overline z}(z)\tr\left(\sum_{A\subset\{0,x_1,\dots,x_k\}}(z-H_A)^{-1}(-1)^{\#A}(n+1+H_A)^{-q}\right)d\overline z\wedge dz \\&\hskip2.5cm=\frac i{2\pi}\int_{[0,1]^k}\int_\C\frac{\partial\tilde\varphi}{\partial\overline z}(z)\tr\left(\partial_A\left((z-H_t)^{-1}(n+1+H_t)^{-q}\right)\right)d\overline z\wedge dz dt\endaligned \tag 3.13$$ where $H_t=-\Delta+t_0V+\sum_{l=1}^kt_l V_l$ o\`u $(t_l)_{0\leq l\leq k}\in [0,1]^{k+1}$ et $\dsize\partial_A=\bigotimes_{0\leq l\leq k}\partial_{t_l}$. \par Then as $V$ is continuous, using the estimates of Lemma 4.1, for $z\in\C\setminus\R$, $(x_1,\dots x_k)\mapsto(z-H_t)^{-1}(n+1+H_t)^{-q}$ is continuous in ${\Cal T}_1$ uniformly in $t$. Hence, $(x_1,\dots x_k)\mapsto\langle Z(0,x_1,\dots,x_k),\varphi\rangle$ is continuous using (3.13) and the well controlled ${\Cal T}_1$-norm estimates for $(z-H_t)^{-1}(n+1+H_t)^{-q}$. \par Now estimating $\partial_A\left((z-H_t)^{-1}(n+1+H_t)^{-q}\right)$ as in \cite{Kl}, we get, for some $\epsilon>0$, $C>0$, $N>0$ and $z\in\C\setminus\R$, $$\Vert\partial_A\left((z-H_t)^{-1}(n+1+H_t)^{-q}\right)\Vert\leq C\left(\frac{\mid z\mid+1}{\mid\text{Im}z\mid}\right)^N e^{-\epsilon\frac{\mid\text{Im}z\mid}{\mid z\mid+1}\sum_{l=1}^k\mid x_l\mid}.$$ Hence using the properties of almost analytic extensions, we get that $\langle Z(0,x_1,\dots,x_k),\varphi\rangle$ is in $L^1((\Rd)^k)$. This ends the proof of Proposition 2.4. \qed\enddemo \medskip \subhead IV) Some technical lemmas \endsubhead \medskip \subhead a) Resolvent estimates \endsubhead \smallskip Here we estimate the norm of the truncated resolvent in appropriate Schatten classes for some operators with unbounded potentials. \proclaim{Lemma 4.1} Let $\nu\in(0,1)$ and $q>d/2$. Then, there exists $C_{\nu,q}>0$ and $\epsilon_{\nu,q}>0$ such that, for any $V$, a function that satisfies for some $N>0$, $$\mid V(x)\mid\leq \vert x\vert^{\nu}+N\text{ for all }x\in\Rd$$ and, for any $(\alpha,\beta)\in\Zd\times\Zd$ and $z\in\C\setminus\R$, one has $$\multline \Vert\chi_\alpha(z-(-\Delta+V))^{-1}\chi_\beta\Vert_{{\Cal T}_q}\leq C_{\nu,q}\frac{(1+\vert\beta\vert)^{\nu}}{\eta(z,N)}e^{-\epsilon_{\nu,q}\cdot\eta(z,N)\vert\delta_{\nu}(\alpha)-\delta_{\nu}(\beta)\vert} \\ \text{where }\delta_\nu(x)=(1+x^2)^{(1-\nu)/2}\text{ and }\eta(z,N)=\frac{\text{dist}(z,\sigma(-\Delta+V))}{\vert z\vert+N+1}\endmultline$$ here $\Vert\cdot\Vert_{{\Cal T}_q}$ denotes the norm in the $q$-th Schatten class and dist$(z,z')$ denotes the distance in $\C$. \endproclaim \demo{Proof} By \cite{Si} section B.9, we know that $\chi_\alpha\cdot(z-(-\Delta+V))^{-1}\cdot\chi_\beta\in{\Cal T}_q$. To prove the estimate on its norm we will use the idea used in \cite{Si}. For $\epsilon>0$, one has, $$e^{-\epsilon\delta_\nu}\Delta e^{\epsilon\delta_\nu}=\Delta-2\epsilon\nabla\delta_\nu\cdot\nabla+(\epsilon\nabla\delta_\nu)^2-\epsilon\Delta\delta_\nu,$$ hence, for $z\in\C\setminus\R$, $$\aligned & e^{\epsilon\delta_\nu}(z-(-\Delta+V))^{-1}e^{-\epsilon\delta_\nu}=\left(z+\Delta-V-2\epsilon\nabla\delta_\nu\cdot\nabla+(\epsilon\nabla\delta_\nu)^2-\epsilon\Delta\delta_\nu\right)^{-1}\\ &\qquad=\left(1-(z-(-\Delta+V))^{-1}\left(2\epsilon\nabla\delta_\nu\cdot\nabla-(\epsilon\nabla\delta_\nu)^2+\epsilon\Delta\delta_\nu\right)\right)^{-1}(z-(-\Delta+V))^{-1}.\endaligned$$ Let us estimate $$\aligned &\Vert (z-(-\Delta+V))^{-1}\left(2\epsilon\nabla\delta_\nu\cdot\nabla-(\epsilon\nabla\delta_\nu)^2+\epsilon\Delta\delta_\nu\right)\Vert\leq\epsilon^2\Vert(z-(-\Delta+V))^{-1}(\nabla\delta_\nu)^2\Vert+\\ &\hskip2cm +\epsilon\Vert(z-(-\Delta+V))^{-1}\Delta\delta_\nu\Vert+\epsilon\Vert (\Delta-1)^{-1}\nabla\delta_\nu\cdot\nabla\Vert+\\&\hskip 4cm+\epsilon\Vert(z-(-\Delta+V))^{-1}(z+1-V)(\Delta-1)^{-1}\nabla\delta_\nu\cdot\nabla\Vert.\endaligned$$ As $-\Delta+V$ is self-adjoint and $\Vert V\nabla\delta_\nu\Vert\leq N$, using \cite{Si} Theorem B.6.5, we get, for some $C>0$, $$ \Vert (z-(-\Delta+V))^{-1}\left(2\epsilon\nabla\delta_\nu\cdot\nabla-(\epsilon\nabla\delta_\nu)^2+\epsilon\Delta\delta_\nu\right)\Vert\leq C\epsilon\left(1+\frac1{\eta(z,N)}\right).$$ So if we choose $\epsilon$ of the form $\pm\epsilon_\nu\eta(z,N)$ with $\epsilon_\nu C<1/8$, we get $$\Vert e^{\epsilon\delta_\nu} \chi_\alpha(z-(-\Delta+V))^{-1}\chi_\beta e^{-\epsilon\delta_\nu}\Vert_{{\Cal T}_q}\leq 4\Vert(z-(-\Delta+V))^{-1}\chi_\beta\Vert_{{\Cal T}_q}.\tag 4.1$$ Using the growth assumption on $V$ and standard estimates for the free Laplace operator, we get $$\aligned \Vert(z-(-\Delta+V))^{-1}\chi_\beta\Vert_{{\Cal T}_q} &\leq\Vert (z-(-\Delta+V))^{-1}(z-V+1)(\Delta-1)^{-1}\chi_\beta+(\Delta-1)^{-1}\chi_\beta\Vert_{{\Cal T}_q}\\&\leq\sum_{\alpha\in\Zd}\Vert (z-(-\Delta+V))^{-1}(z-V+1)\chi_\alpha\Vert\cdot\Vert\chi_\alpha(\Delta-1)^{-1}\chi_\beta\Vert_{{\Cal T}_q}+\\ &\hskip 4cm+\Vert(\Delta-1)^{-1}\chi_\beta\Vert_{{\Cal T}_q}\\ &\leq C\frac{(1+\vert\beta\vert)^{\nu}}{\eta(z,N)},\endaligned$$ then using (4.1) for $\epsilon$ of the form $\pm\epsilon_\nu\eta(z,N)$, we get the announced result. \qed\enddemo \smallskip \subhead a) A probabilistic estimate \endsubhead \smallskip Let $V_\o$ be the Poisson potential defined in section 1, and ${\Cal P}_\o$ be the associated probability measure. Let $\dsize V_m(t;x)=\sum_{\gamma\in\frac1m\Zd}t_\gamma V_\gamma$ where $(t_\gamma)_{\gamma\in\frac1m\Zd}$ be a family of independent identically distributed Bernoulli random variables with common probability distribution $P(t_0=1)=p_m$ and $P(t_0=0)=1-p_m$ with $p_m=\mu m^{-d}$. Let ${\Bbb P}_m$ denote the probability measure induced on the space of potentials by these random variables. For $x>0$, we define $\e(x)=\inf\{n\in{\Bbb N};\ n\geq x\}$. Then one has \proclaim{Lemme 4.2} For $\nu\in(0,1)$ and $N\in{\Bbb N}^*$, for some $C>0$, one has $$\sup_{m\geq1}\left({\Bbb P}_m\left(\{t;\ \exists x\in\Rd\text{ such that }\vert V_m(t;x)\vert\geq\vert x\vert^\nu+N\}\right)\right)\leq\frac{\mu^{\e(N/C)}}{\e(N/C)!}F_\nu(\mu), \tag 1$$ and $${\Bbb P}_\o\left(\{\o;\ \exists x\in\Rd\text{ such that }\vert V_\o(x)\vert\geq\vert x\vert^\nu+N\}\right)\leq\frac{\mu^{\e(N/C)}}{\e(N/C)!}G_\nu(\mu), \tag 2$$ where $F_\nu$ and $G_\nu$ are positive increasing functions of $\mu$ that do not depend on $N$. \endproclaim \demo{Proof} Let us first prove (1). By assumption (H.1), one has, for some $C>0$ and $x\in\Rd$, $$\vert V_m(t;x)\vert\leq C\sum_{\beta\in\Zd}e^{-\eta\vert x-\beta\vert}T_\beta \text{ where }T_\beta=\sum_{\gamma\in C(m)}t_{\beta+\gamma}\text{ for }\beta\in\Zd.$$ Hence, $$\aligned &{\Bbb P}_m\left(\{t;\exists x\in\Rd\text{ such that }\vert V_m(t;x)\vert\geq\vert x\vert^\nu+N\}\right)\\&\hskip 3cm\leq{\Bbb P}_m\left(\{t;\exists x\in\Rd\ \exists\beta\in\Zd\text{ such that }T_\beta\geq\frac1C(\vert x\vert^\nu+N)e^{\eta\vert x-\beta\vert}\}\right)\\&\hskip 6cm\leq{\Bbb P}_m\left(\{t;\exists\beta\in\Zd\text{ such that }T_\beta\geq\frac1C(\vert \beta\vert^\nu+N)\}\right). \endaligned$$ By the definition of $C(m)$ and the assumptions on the $(t_\gamma)_{\gamma\in\frac1m\Zd}$, $(T_\beta)_{\beta\in\Zd}$ is a family of i.i.d random variables that satisfies the following estimates, for $M>0$, for any $\beta\in\Zd$, for any $m\geq1$, $$ {\Bbb P}_m\left(\{t; T_\beta\geq M\}\right)=\sum_{n=\e(M)}^{m^d}\binom{m^d}{n}(\mu\cdot m^{-d})^n(1-\mu\cdot m^{-d})^{m^d-n}\leq \frac{\mu^{\e(M)}}{(\e(M))!}e^\mu.$$ So, for any $m\geq1$, $$\aligned {\Bbb P}_m\left(\{t;\exists x\in\Rd\text{ such that }\vert V_m(t;x)\vert\geq\vert x\vert^\nu+N\}\right)&\leq\sum_{\beta\in\Zd}{\Bbb P}_m\left(\{t;\ T_\beta\geq \frac1C(\vert \beta\vert^\nu+N)\}\right) \\&\leq e^\mu\sum_{\beta\in\Zd}\frac{\mu^{\e((\vert \beta\vert^\nu+N)/C)}}{(\e((\vert \beta\vert^\nu+N)/C))!} \\& \leq\frac{\mu^{\e(N/C)}}{(\e(N/C))!}F_\nu(\mu).\endaligned$$ This ends the proof of (1). \par To get (2), using (H.1), we know that, for some $C>0$, $$\{\o;\ \exists x\in\Rd\text{ such that }\vert V_\o(x)\vert\geq\vert x\vert^\nu+N\}\subset\bigcup_{\gamma\in\Zd}\{\o;\ \vert V_\o(\gamma)\vert\geq\frac1C(\vert\gamma\vert^\nu+N)\}.$$ Moreover, if $C_\beta(1)$ denotes the cube centered in $\beta$ of sidelength 1, then $$\vert V_\o(\gamma)\vert\leq C\sum_{\beta\in\Zd}e^{-\eta\vert\beta-\gamma\vert}m(\o;C_\beta(1))=C\sum_{\beta\in\Zd}e^{-\eta\vert\beta\vert}m(\o;C_{\beta+\gamma}(1)).$$ Hence, for some $C>0$, one has $$ \{\o;\ \vert \ V_\o(\gamma)\vert\geq\frac1C(\vert\gamma\vert^\nu+N)\} \subset\bigcup_{\beta\in\Zd}\{\o;\ m(\o;C_{\beta+\gamma}(1))\geq\frac1C(\vert\gamma\vert^\nu+N)e^{\eta\vert\beta\vert}\}.$$ So, using the definition of the Poisson process, we get $$\aligned {\Bbb P}_\o\left(\{\o;\ \vert V_\o(\gamma)\vert\geq\frac1C(\vert\gamma\vert^\nu+N)\}\right) &\leq\sum_{\beta\in\Zd}{\Bbb P}_\o\left(\{\o;\ m(\o;C_{\beta+\gamma}(1))\geq\frac1C(\vert\gamma\vert^\nu+N)e^{\eta\vert\beta\vert}\}\right)\\ &\leq \sum_{\beta\in\Zd}\frac{\mu^{\e(\frac1C(\vert\gamma\vert^\nu+N)e^{\eta\vert\beta\vert})}}{(\e(\frac1C(\vert\gamma\vert^\nu+N)e^{\eta\vert\beta\vert}))!}\\ &\leq \sum_{\beta\in\Zd} \frac1{(\e(e^{\eta\vert\beta\vert})-1)!}\left(\frac{\mu^{\e(\frac1C(\vert\gamma\vert^\nu+N))}}{\e(\frac1C(\vert\gamma\vert^\nu+N))!)}\right)^{e^{\eta\vert\beta\vert}}\\ &\leq \frac{\mu^{\e((\vert\gamma\vert^\nu+N)/C)}}{\e((\vert\gamma\vert^\nu+N)/C)!}G(\mu), \endaligned$$ so $$\aligned {\Bbb P}_m\left(\{t;\exists x\in\Rd\text{ such that }\vert V_\o(x)\vert\geq\vert x\vert^\nu+N\}\right)&\leq\sum_{\gamma\in\Zd}\frac{\mu^{\e(\frac1C(\vert\gamma\vert^\nu+N))}}{\e(\frac1C(\vert\gamma\vert^\nu+N))!}G(\mu)\\ &\leq \frac{\mu^{\e(N/C)}}{\e(N/C)!}G_\nu(\mu).\endaligned$$ This ends the proof of Lemma 4.2. \qed\enddemo \Refs\nofrills{R\'ef\'erences} \widestnumber\key{F-M-S-S} \ref \key He-Sj \by B. 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