\documentclass[reqno]{amsart} \usepackage{amssymb} \usepackage{latexsym} % Standard sets and numbers \newcommand{\N}{{\mathbb{N}}} % natural numbers \newcommand{\Z}{{\mathbb{Z}}} % integers \newcommand{\Q}{{\mathbb{Q}}} % rationals \newcommand{\R}{{\mathbb{R}}} % real numbers \newcommand{\e}{{\mathrm{e}}} % Euler \date{September 12, 2000} % Theorem enviroments \newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{prop}[theorem]{Proposition} \newtheorem{coro}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \newtheorem{remarks}[theorem]{Remarks} % QED Symbol \renewcommand\qedsymbol{\hbox {\hskip 1pt \vrule width 6pt height 6pt depth 1.5pt \hskip 1pt}} \newcommand{\ind}{{\mathrm{ind}}} \newcommand{\tr}{{\rm tr}} \newcommand{\abs}[1]{\lvert#1\rvert} \sloppy \begin{document} \title[An optimal $\boldsymbol{L^p}$-bound on the Krein spectral shift function]{% An optimal $\boldsymbol{L^p}$-bound on the \\ Krein spectral shift function} \author[D.~Hundertmark and B.~Simon]{Dirk Hundertmark$^1$ and Barry Simon$^{1,2}$} \dedicatory{Dedicated to the memory of Tom Wolff,\\ analyst extraordinaire and friend} \thanks{$^1$ Department of Mathematics 253--37, California Institute of Technology, Pasadena, CA 91125, U.S.A.; E-mail: dirkh@caltech.edu, bsimon@caltech.edu} \thanks{$^2$ Supported in part by NSF grant DMS-9707661. } \subjclass{47A30, 81Q10} \keywords{Krein spectral shift function, $L^p$-bounds} \thanks{\copyright 2000 by the authors. Reproduction of this article, in its entirety, by any means is permitted for non-commercial purposes.} \thanks{To appear in {\it{J. d'Analyse Math.}}} \begin{abstract} [{\it Note}: This paper supplants B. Simon's preprint, $L^p$ bounds on the Krein spectral shift," which has been withdrawn.] \medskip Let $\xi_{A,B}$ be the Krein spectral shift function for a pair of operators $A,B$, with $C=A-B$ trace class. We establish the bound \begin{displaymath} \int F(\abs{\xi_{A,B}(\lambda)})\, d\lambda \le \int F(\abs{\xi_{\abs{C},0}(\lambda)})\, d\lambda = \sum_{j=1}^\infty \big[F(j)-F(j-1)]\mu_j(C), \end{displaymath} where $F$ is any non-negative convex function on $[0,\infty)$ with $F(0)=0$ and $\mu_j(C)$ are the singular values of $C$. Specializing to $F(t)=t^p$, $p\ge 1$ this improves a recent bound of Combes, Hislop, and Nakamura. \end{abstract} \maketitle %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Introduction}\label{sec:intro} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Let $A,B$ be bounded self-adjoint operators such that their difference $A-B$ is trace class. The Krein spectral shift function $\xi_{A,B}$ for the pair $A$, $B$ is determined by \begin{equation*}%\label{eq:krein} \tr(f(A)-f(B)) = \int f'(\lambda)\xi_{A,B}(\lambda)\,d\lambda \end{equation*} for all functions $f\in C^\infty_0(\R)$ and $\xi(\lambda)=0$ if $\abs{\lambda}$ is large enough. The two bounds $$\label{eq:Lone} \int \abs{\xi_{A,B}(\lambda)}\, d\lambda \le \tr(\abs{A-B})$$ and $$\label{eq:Linfinity} \vert\xi_{A,B}(\lambda)\vert \leq n \qquad \text{if A-B is rank n}$$ are well known; see, for example, \cite{Simon1} or \cite{Yafaev}. The Krein spectral shift function can also be defined for unbounded self-adjoint operators $A, B$ and enjoys the same properties as long as their difference is trace class. The results of this paper extend to general unbounded operators $A$ and $B$ (as long as their difference is trace class) but for simplicity, we will suppose that $A$ and $B$ are bounded. For applications of the spectral shift function in scattering theory, see, for example, the survey \mbox{article \cite{BY}}. The spectral shift function also found applications in the theory of random Schr\"odinger operators. Kostrykin and Schrader \cite{KS1,KS2} constructed a spectral shift density for random Schr\"odinger operators with Anderson-type potentials. More recently, Combes, Hislop, and Nakamura \cite{CHN} realized that $L^p$-bounds on the Krein spectral shift function can serve as a basic tool for a proof of H\"older continuity of the integrated density of states for a large class of continuous random Schr\"odinger operators. In terms of the singular values of the difference $C=A-B$, their bound reads $$\label{eq:CHN} \Vert \xi_{A,B}\Vert_p := \Big( \int \abs{\xi_{A,B}(\lambda)}^p\,d\lambda \Big)^{1/p} \le \sum_{j=1}^\infty \mu_j(C)^{1/p}$$ for $1\le p<\infty$. Note that (\ref{eq:CHN}) includes the endpoint cases (\ref{eq:Lone}) and (\ref{eq:Linfinity}) for $p=1$ and and in the limit $p\to\infty$, respectively. To see what type of bound is the correct one for an $L^p$-bound on the Krein spectral shift function, we consider a special case: Let $C$ be a positive trace class operator with eigenvalues $\mu_j$. Calculating \begin{displaymath} \tr[f(C)-f(0)] = \sum_{j=1}^\infty \int_0^{\mu_j} f'(\lambda)\, d\lambda , \end{displaymath} we see that the spectral shift function for the pair $C,0$ is simply given by $$\label{eq:xiC0} \xi_{C,0}(\lambda)= n \text{ if } \mu_{n+1}\le \lambda < \mu_n \qquad \text{and} \qquad \xi_{C,0}(\lambda) =0 \text{ if } \lambda <0 \text{ or } \lambda \ge \mu_1.$$ For cases like this where $A$ and $B$ are finite rank, it is known that the spectral shift function is just the difference of the dimensions of the spectral subspaces --- which leads immediately to another way of seeing why (\ref{eq:xiC0}) is true. In particular, $\xi_{C,0}$ enjoys the following important properties: \begin{itemize} \item $\xi_{C,0}$ takes only values in $\N_0$ (or $\Z$ if $C$ is not non-negative). \item For any non-negative function $F$ on $[0,\infty)$ with $F(0)=0$, we have $$\label{eq:example} \int F(\abs{\xi_{C,0}(\lambda)})\, d\lambda = \sum_{j=1}^\infty F(j)\big(\mu_j - \mu_{j+1}\big) .$$ \item In addition, if $F$ is monotone increasing, then $$\label{eq:exampleconvex} \int F(\abs{\xi_{C,0}(\lambda)})\, d\lambda = \sum_{j=1}^\infty \big[ F(j)-F(j\!-\!1) \big] \mu_j .$$ \end{itemize} The first two claims follow immediately from (\ref{eq:xiC0}). Formally, the last claim follows from the second by summation by parts. However, since \begin{displaymath} \sum_{j=1}^N F(j)\big(\mu_j -\mu_{j+1}\big) = \sum_{j=1}^N \big(F(j)-F(j\!-\! 1)\big) \mu_j - F(N)\mu_{N+1} , \end{displaymath} this usually poses some growth restriction on $F(j)$ to do the limit $N\to\infty$. We prefer not to do this but use positivity instead. Notice \begin{align*} \sum_{j=1}^\infty F(j)\big( \mu_j -\mu_{j+1}\big) &= \sum_{j=1}^\infty \sum_{n=1}^j \big(F(n)-F(n\!-\! 1)\big)\big( \mu_j -\mu_{j+1}\big) \\ &= \sum_{1\le n\le j} \big(F(n)-F(n\!-\! 1)\big)\big( \mu_j -\mu_{j+1}\big) . \end{align*} By the assumptions on $F$ and $\mu_j$, all terms in this double sum are non-negative. Hence we can use the Fubini-Tonelli theorem to freely interchange the summation and conclude \begin{align*} \sum_{j=1}^\infty F(j)\big( \mu_j -\mu_{j+1}\big) &= \sum_{n=1}^\infty \big(F(n)-F(n\!-\! 1)\big) \sum_{j=n}^\infty \big( \mu_j -\mu_{j+1}\big) \\ &= \sum_{n+1}^\infty \big(F(n)-F(n\!-\! 1)\big) \mu_n , \end{align*} where we used the fact that the last sum telescopes and $\mu_n\to 0$ as $n\to\infty$. In particular, the right-hand side of (\ref{eq:example}) and (\ref{eq:exampleconvex}) is finite if and only if the other is and then they are equal. Below we will use this type of argument to freely do summation by parts without a priori bounds on the boundary terms. Alternatively, one could consider the case of finite rank operators $C=A-B$ first and then use some approximation arguments. \medskip Our main result is that the above example (\ref{eq:exampleconvex}) is an extreme case: \begin{theorem} \label{thm:main} Let $F$ be a non-negative convex function on $[0,\infty)$ vanishing at zero. Given a non-negative compact operator $C$ with singular values $\mu_j(C)$, we have $$\label{eq:sharpbound} \begin{split} \int F(\abs{\xi_{A,B}(\lambda)})\, d\lambda &\le \int F(\abs{\xi_{C,0}(\lambda)})\, d\lambda \\ &= \sum_{j=1}^\infty \big[F(j)-F(j\!-\!1)]\mu_j(C) \end{split}$$ for all pairs of bounded operators $A,B$ with $\sum_{j=n}^\infty \mu_j(\abs{A-B}) \le \sum_{j=n}^\infty \mu_j(C)$ for all $n\in\N$. In particular, this is the case if $\abs{A-B}\le C$. \end{theorem} \noindent \textbf{Remark.} Moreover, if $F$ is strictly convex, the above inequality is strict if either the modulus of $\xi_{A,B}$ takes non-integer values on a set of positive Lebesgue measure or one does not have equality in Lemma \ref{lem:basic} below. However, it seems to be difficult to find necessary and sufficient conditions on $A$ and $B$ alone for the case of equality in (\ref{eq:sharpbound}). \medskip Specializing to $F(t)= t^p$ for some $p\ge 1$ we get as a corollary, \begin{coro} \label{cor:coro} Let $\xi_{A,B}$ be the Krein spectral shift function for the pair $A,B$. In terms of the singular values $\mu_j$ of the difference $A-B$, we have the $L^p$-bound $$\label{eq:Lpbound} \Vert \xi_{A,B}\Vert_p \le \Vert \xi_{\abs{A-B},0}\Vert_p = \Big(\sum_{n=1}^\infty \big[n^p -(n\!-\! 1)^p \big] \mu_n \Big)^{1/p} .$$ \end{coro} \noindent{\textbf{Remarks.}} i) There are two different ways to see that (\ref{eq:Lpbound}) is, indeed, stronger than the bound (\ref{eq:CHN}) by Combes et al. First, the direct argument: Rewrite \begin{displaymath} \Big(\sum_{n=1}^\infty \big[n^p -(n\!-\!1)^p \big] \mu_n \bigg)^{1/p} = \Big(\sum_{n=1}^\infty n^p (\mu_n-\mu_{n+1}) \Big)^{1/p} \end{displaymath} and consider the right-hand side as the $l^p$-norm of the function $n\to n^p$ in the weighted $l^p$-space $l^p(\mu)$ with measure $\mu(n):= \mu_n - \mu_{n+1}$ for $n\in\N$. Write $n = 1\!+\! (n\!-\!1)$ and use Minkowski's inequality for the $l^p(\mu)$-norm to get \begin{align*} \Big(\sum_{n=1}^\infty n^p \mu(n) \Big)^{1/p} &\le \Big(\sum_{n=1}^\infty \mu(n) \Big)^{1/p} + \Big(\sum_{n=2}^\infty (n\!-\! 1)^p \mu(j) \Big)^{1/p} \\ &= \mu_1^{1/p} + \Big(\sum_{n=2}^\infty (n\!-\! 1)^p \mu(n) \Big)^{1/p} \\ &\le \sum_{n=1}^N \mu_n^{1/p} + \bigg(\sum_{n=N}^\infty (n\!-\! N)^p \mu(n) \bigg)^{1/p} , \end{align*} where, for the last inequality, we repeated the first step $N$ times. Using monotone convergence for the limit $N\to\infty$, we conclude $$\label{eq:worse} \Big(\sum_{n=1}^\infty \big[n^p -(n\!-\! 1)^p \big] \mu_n \Big)^{1/p} \le \sum_{n=1}^\infty \mu_n^{1/p} .$$ For the soft argument, note that, according to (\ref{eq:exampleconvex}), \begin{displaymath} \Big(\sum_{n=1}^\infty \big[n^p -(n\!-\! 1)^p \big] \mu_n \Big)^{1/p} = \Vert \xi_{C,0}\Vert_p , \end{displaymath} where $C$ is any non-negative compact operator with singular values $\mu_n$. The bound (\ref{eq:CHN}) for $\xi_{C,0}$ immediately implies the inequality (\ref{eq:worse}). Of course, the direct argument is, in some sense, a reformulation of the inductive proof of Combes et al. ii) Our result shows that if $\sum_{n=1}^\infty n^{p-1}\mu_n<\infty$, then $\xi_{A,B}\in L^p(\R)$. Note that this \emph{cannot} be improved as far as only conditions on $\mu_n$ are used. It is also strictly better than the result by Combes et al. For example, if $\mu_n = n^{-p}\log(n+2)^{-\alpha}$, then Combes et al.\ require $\alpha\!>\!p$ to conclude $\xi_{A,B}\in L^p(\R)$, while our result only needs $\alpha \!>\! 1$. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Two Proofs}\label{sec:twoproofs} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \noindent We want to give two proofs of our main result, Theorem \ref{thm:main}, both depending on different aspects of the problem. First, some notation. For a complex-valued function $f$, let $m_f$ be its distribution function, that is, $m_f(t):= \vert\{\lambda: \abs{f(\lambda)} \!>\! t\}\vert$, with $\vert \mathcal{A} \vert$ the Lebesgue measure of a Borel set $\mathcal{A}\subset\R$. We will write $m_{A,B}$ for the distribution function of $\xi_{A,B}$. The following lemma is the core of both proofs: \begin{lemma}[Basic Lemma] \label{lem:basic} With $C= A-B$, we have for all $n\in\N_0$ \begin{displaymath} \int_n^\infty m_{A,B}(t)\,dt \le \sum_{j=n+1}^\infty \mu_j(C) =\int_n^\infty m_{\abs{C},0}(t)\,dt . \end{displaymath} \end{lemma} \noindent{\textbf{Remark.}} Setting $(x-s)_+ := \sup\{0,x-s\}$, we have $$\label{eq:m-f-formula} \int_s^\infty m_f(t)\, dt = \int (\abs{f(\lambda)}-s)_+\, d\lambda$$ for all $s\ge 0$. Hence Lemma \ref{lem:basic} is equivalent to \begin{displaymath} \int (\abs{\xi_{A,B}(\lambda)}-n)_+\, d\lambda \le \sum_{j=n+1}^\infty \mu_j(C) = \int (\abs{\xi_{\abs{C},0}(\lambda)}-n)_+\, d\lambda . \end{displaymath} \begin{proof}[Proof of Lemma \ref{lem:basic}:] By the formula (\ref{eq:xiC0}) for $\xi_{\abs{C},0}$, \begin{align*} \int (\abs{\xi_{\abs{C},0}}(\lambda)-n)_+\, d\lambda &= \sum_{j=n+1}^\infty (j-n) (\mu_j-\mu_{j+1}) \\ &= \sum_{l= n+1}^\infty \sum_{j=l}^\infty (\mu_j-\mu_{j+1}) = \sum_{l=n+1}^\infty \mu_l , \end{align*} proving the second part of the assertion in the lemma. For the inequality, let $\psi_j$, $\phi_j$ be two sets of orthonormal vectors such that $A\!-\!B= \sum_{j=1}^\infty \mu_j \langle \phi_j,\cdot\rangle \psi_j$. Note that $\phi_j=\pm \psi_j$ since $A$ and $B$ are self-adjoint. Set $C_0:= 0$ and $C_n:= \sum_{j=1}^n\mu_j \langle \phi_j,\cdot\rangle \psi_j$ for $n\in \N$. The spectral shift function is transitive. In particular, \begin{displaymath} \xi_{A,B}= \xi_{A,A+C_n} + \xi_{A+C_n, B} . \end{displaymath} By (\ref{eq:Linfinity}) we know $\abs{\xi_{A,A+C_n}}\le n$. Thus \begin{displaymath} (\abs{\xi_{A,B}(\lambda)}-n)_+ \le \abs{\xi_{A+C_n,B}(\lambda)} \end{displaymath} and hence, by (\ref{eq:m-f-formula}) and (\ref{eq:Lone}): \begin{displaymath} \int_n^\infty m_{A,B}(t)\,dt = \int (\abs{\xi_{A,B}(\lambda)}-n)_+\, d\lambda \le \tr \big( C-C_n \big) = \sum_{j=n+1}^\infty \mu_j . \end{displaymath} \end{proof} In the following we will write $m$ and $\xi$ for $m_{A,B}$, $\xi_{A,B}$, respectively. \begin{proof}[First proof of Theorem \ref{thm:main}:] For $n\in\N_0$, put $a_n := F(n\!+\! 1)-F(n)$, so we have $F(n)= \sum_{0\le l< n} a_l$. The $a_n$ are monotone increasing due to the convexity of $F$, that is, $a_n -a_{n-1}\ge 0$ for all $n\in\N$. Furthermore, set \begin{displaymath} x_n := \int_n^\infty m(t)\, dt \end{displaymath} and observe that, due to (\ref{eq:m-f-formula}), $$\label{eq:decomp0} \int\limits_{n< \abs{\xi} \le n\!+\! 1} (\abs{\xi_{A,B}(\lambda)}-n)\, d\lambda = x_n - x_{n+1} - m(n\!+\! 1).$$ Look at $$\label{eq:decomp1} \int F(\abs{\xi(\lambda)})\, d\lambda = \sum_{n=0}^\infty \, \int\limits_{n< \abs{\xi} \le n\!+\! 1} \!F(\abs{\xi(\lambda)})\, d\lambda .$$ By convexity of $F$, we have $F(s)\le F(n) +(s\!-\!n)\big(F(n\!+\! 1)- F(n)\big)$ for $s\in [n,n\!+\!1]$. Plugging this into (\ref{eq:decomp1}), we infer \begin{align} \label{eq:decomp2} \int F &(\abs{\xi(\lambda)})\, d\lambda \notag \\ &\le \sum_{n=0}^\infty \Big[ F(n) \int\limits_{n< \abs{\xi} \le n\!+\! 1} d\lambda + a_n \int\limits_{n< \abs{\xi} \le n\!+\! 1} \!(\abs{\xi(\lambda)}-n)\, d\lambda \Big] \notag \\ &= \sum_{n=0}^\infty \Big[ \big( \sum_{l=0}^{n-1} a_l \big) \big( m(n)-m(n\!+\!1)\big) + a_n \big(x_n-x_{n+1} - m(n+1)\big) \Big] \ \text{(by (\ref{eq:decomp0}))} \notag \\ &= \sum_{ l=0}^\infty a_l \sum_{n=l\!+\!1}^\infty\big(m(n)-m(n\!+\!1)\big) + \sum_{n=0}^\infty a_n \big(x_n-x_{n+1} - m(n\!+\!1)\big) \notag \\ &= \sum_{l=0}^\infty a_l m(l\!+\!1) + \sum_{n=0}^\infty a_n \big(x_n-x_{n\!+\!1} - m(n\!+\!1)\big) \notag \\ &= \sum_{n=0}^\infty a_n \big(x_n-x_{n\!+\!1}\big) . \end{align} Note that if $F$ is strictly convex, we have equality in this inequality if and only if $\xi$ takes only integer values! Using that the terms in this sum are non-negative, we can again reorder the summations freely to conclude, setting $a_{-1}:=0$, $$\label{eq:decomp3} \begin{split} \int F(\abs{\xi(\lambda)})\, d\lambda &\le \sum_{n=0}^\infty a_n (x_n-x_{n+1}) = \sum_{0\le l\le n} (a_l-a_{l-1}) (x_n-x_{n+1}) \\ &= \sum_{l=0}^\infty \big(a_l-a_{l-1}\big) x_l \le \sum_{l=0}^\infty \big(a_l-a_{l-1}\big) \sum_{j=l+1} \mu_j \\ &= \sum_{j=1}^\infty \big(F(j)-F(j\!-\!1)\big)\mu_j , \end{split}$$ where in the last inequality we used positivity of the increments $a_l -a_{l-1}$ (aka convexity of $F$) and Lemma \ref{lem:basic}. This proves Theorem \ref{thm:main}, including the case of equality since, if $F$ is strictly convex and $\xi$ takes non-integer values on a set of positive measure, the inequality in (\ref{eq:decomp2}) is strict. We will come back to this question in the second proof. \end{proof} For the second proof we need some preparatory lemmas: \begin{lemma}\label{lem:doublelayercake} For any non-negative, convex function $F$ on $[0,\infty)$ which vanishes at zero, there exists a non-negative, locally finite measure $\nu_F$ on $[0,\infty)$ such that \begin{displaymath} F(t) = \int_0^\infty (t-u)_+ \nu_F(du) \quad \text{for all } t\ge 0 . \end{displaymath} $F$ is strictly convex if and only if $\nu_F$ is strictly positive, that is, $\nu_F([a,b])>0$ for all $0\le at\}\vert$ and define \begin{equation*} Q_f(s):= \int_s^\infty m_f(t)\, dt = \int (\abs{f(\lambda)}-s)_+\, d\lambda . \end{equation*} \begin{lemma}\label{lem:monotoninQ} Let $F$ be any non-negative, convex function $F$ on $[0,\infty)$ which vanishes at zero. Given two functions $f$ and $g$, we have that $Q_f\le Q_g$ implies \begin{displaymath} \int F(\abs{f(\lambda)})\, d\lambda \le \int F(\abs{g(\lambda)})\, d\lambda . \end{displaymath} Moreover, if $F$ is strictly convex and $Q_f < Q_g$ on a set of positive Lebesgue measure, then the inequality above is strict. \end{lemma} \begin{proof} From Lemma \ref{lem:doublelayercake} we infer \begin{displaymath} \int F(\abs{f(\lambda)})\, d\lambda = \int_0^\infty Q_f(u) \, \nu_F(du) , \end{displaymath} which concludes the proof. \end{proof} \begin{lemma}\label{lem:discrete} Suppose that $g$ takes only values in an unbounded, discrete set $\mathcal{S}\subset [0,\infty)$ with $0\in\mathcal{S}$. Then the inequality $Q_f(s)\le Q_g(s)$ for $s\in\mathcal{S}$ implies $Q_f(t)\le Q_g(t)$ for all $t\in[0,\infty)$. \end{lemma} \begin{proof} Let \$\mathcal{S}= \{0\!=s_0\!<\!s_1