\input amstex \documentstyle{amsppt} \magnification=1200 \baselineskip=1.5 pt \TagsOnRight \NoBlackBoxes \topmatter \title The Xi Function \endtitle \author F.~Gesztesy$^{1}$ and B.~Simon$^{2}$ \endauthor \leftheadtext{F.~Gesztesy and B.~Simon} \thanks $^{1}$ Department of Mathematics, University of Missouri, Columbia, MO 65211. E-mail: mathfg\@mizzou1.\linebreak missouri.edu \endthanks \thanks $^{2}$ Division of Physics, Mathematics, and Astronomy, California Institute of Technology, 253-37,\linebreak Pasadena, CA 91125. This material is based upon work supported by the National Science Foundation under Grant No.~DMS-9101715. The Government has certain rights in this material. \endthanks \thanks To be submitted to {\it Ann.~Math.} \endthanks \endtopmatter \vskip 0.5in \document \flushpar {\bf \S1. Introduction} Despite the fact that spectral and inverse spectral properties of one-dimensional Schr\"o-dinger operators $H=-\frac{d^2}{dx^{2}}+V$ have been extensively studied for seventy-five years, there remain large areas where our knowledge is limited. For example, while the inverse theory for operators on $L^{2}(-\infty, \infty)$ is well understood in case $V$ is periodic [12,24,25,35,39--42,49], it is not understood in case $\lim\limits_{|x|\to\infty}\,V(x)=\infty$ and $H$ has discrete spectrum. Our goal here is to introduce a special function $\xi(x, \lambda)$ on $\Bbb R\times\Bbb R$ associated to $H$ which we believe will be a valuable tool in the spectral and inverse spectral theory. In a sense we'll make precise, it complements the Weyl $m$-functions, $m_{\pm}(x, \lambda)$. A main application of $\xi$ which we will make here concerns a generalization of the trace formula for Schr\"odinger operators to general $V$'s. Recall the well-known trace formula for periodic potentials: Let $V(x)=V(x+1)$. Then, by Floquet theory (see, e.g., [10,37,44]) $$\text{spec}(H)=[E_{0}, E_{1}]\cup [E_{2}, E_{3}]\cup\dots$$ a set of bands. If $V$ is $C^1$, one can show that the sum of the gap sizes is finite, that is, $$\sum^{\infty}_{n=1}|E_{2n}-E_{2n-1}|<\infty. \tag 1.1$$ For fixed $y$, let $H_y$ be the operator on $-\frac{d^2}{dx^2}+V$ on $L^{2}([y, y+1])$ with $u(y)=u(y+1)=0$ boundary conditions. Its spectrum is discrete, that is, there are eigenvalues $\{\mu_{n}(y)\}^{\infty}_{n=1}$ with $$E_{2n-1}\leq\mu_{n}(y)\leq E_{2n}. \tag 1.2$$ The trace formula says $$V(y)=E_{0}+\sum^{\infty}_{n=1} [E_{2n}+E_{2n-1}-2\mu_{n}(y)]. \tag 1.3$$ By (1.2), $$|E_{2n}+E_{2n-1}-2\mu_{n}(y)|\leq |E_{2n}-E_{2n-1}|$$ so (1.1) implies the convergence of the sum in (1.3). The earliest trace formula for Schr\"odinger operators was found on a finite interval in 1953 by Gel'fand and Levitan [15] with later contributions by Dikii [8], Gel'fand [13], Halberg-Kramer [23], and Gilbert-Kramer [22]. The first trace formula for periodic $V$ was obtained in 1965 by Hochstadt [24], who showed that for finite-gap potentials $$V(x)-V(0)=2\sum^{g}_{n=1}[\mu_{n}(0)-\mu_{n}(x)].$$ Dubrovin [9] then proved (1.3) for finite-gap potentials. The general formula (1.3) under the hypothesis that $V$ is periodic and $C^{\infty}$ was proven in 1975 by McKean-van Moerbeke [41], and Flaschka [12], and later for general $C^3$ potentials by Trubowitz [49]. Formula (1.3) is a key element of the solution of inverse spectral problems for periodic potentials [9,12,24,35,39,41,42,49]. There have been two classes of potentials for which (1.3) has been extended. Certain almost periodic potentials are studied in Levitan [34,35], Kotani-Krishna [31], and Craig [5]. In 1979, Deift-Trubowitz [7] proved that if $V(x)$ decays sufficiently rapidly at infinity and $-\frac{d^2}{dx^2}+V$ has no negative eigenvalues, then $$V(x)=\frac{2i}{\pi}\int\limits^{\infty}_{-\infty}\,dk\,k\, \ln\biggl[1+R(k)\frac{f_{+}(x, k)}{f_{-}(x, k)}\biggr] \tag 1.4$$ (where $f_{\pm}(x, k)$ are the Jost functions at energy $E=k^{2}$ and $R(k)$ is a reflection coefficient) which, as we will see, is an analog of (1.3). Recently, Venakides [49] studied a trace formula for $V$, a positive smooth potential of compact support, by writing (1.3) for the periodic potential $$V_{L}(x)=\sum^{\infty}_{n=-\infty}V(x+nL)$$ and then taking $L$ to $\infty$. He found an integral formula which, although he didn't realize it, is precisely (1.4)! The basic definition of $\xi$ depends on the theory of the Krein spectral shift [32]. If $A$ and $B$ are self-adjoint operators with $A\geq\eta$, $B\geq\eta$ for some real $\eta$ and so that $[(A+i)^{-1}- (B+i)^{-1}]$ is trace class, then there exists a measurable function $\xi(\lambda)$ associated with the pair $(B, A)$ so that $$\text{Tr}[f(A)-f(B)]=-\int\limits_{\Bbb R}f'(\lambda)\xi(\lambda)\, d\lambda \tag 1.5$$ for $f$'s which are sufficiently smooth and which decay sufficiently rapidly at infinity, and, in particular for $f(\lambda)=e^{-t\lambda}$ for any $t>0$; and so that $$\xi(\lambda)=0\qquad \text{if } \lambda<\eta. \tag 1.6$$ Moreover, (1.5), (1.6) uniquely determine $\xi(\lambda)$ for a.e.~$\lambda$, and if $[(A+i)^{-1}-(B+i)^{-1}]$ is rank $n$, then $$|\xi(\lambda)|\leq n$$ and if $B\geq A$, then $\xi(\lambda)\geq 0$. For the rank one case of importance in this paper, an extensive study of $\xi$ can be found in [48] and a brief discussion in the appendix to this paper. Let $V$ be a continuous function on $\Bbb R$ which is bounded from below. Let $H=-\frac{d^2}{dx^2}+V$ which is self-adjoint on $C^{\infty}_{0} (\Bbb R)$ (see, e.g., [43]) and let $H_{D; x}$ be the operator on $L^{2}(-\infty, x)\oplus L^{2}(x, \infty)$ with $u(x)=0$ Dirichlet boundary conditions. Then $[(H_{D;x}+i)^{-1}-(H+i)^{-1}]$ is rank one, so there results a Krein spectral shift $\xi(x, \lambda)$ for the pair $(H_{D; x}, H)$ which in particular obeys: $$\text{Tr}(e^{-tH}-e^{-tH_{D; x}})=t\int\limits^{\infty}_{0} e^{-t\lambda}\xi(x, \lambda)\, d\lambda. \tag 1.7$$ While $\xi$ is defined in terms of $H$ and $H_{D; x}$, there is a formula that only involves $H$, or more precisely, the Green's function $G(x,y; z)$ defined by $$((H-z)^{-1}f)(x)=\int\limits_{\Bbb R} G(x, y, z)f(y)\, dy \tag 1.8$$ for $\text{Im }z\neq 0$. Then by general principles, $\lim\limits_ {\epsilon\downarrow 0}\, G(x, y; \lambda+i\epsilon)$ exists for a.e.~$\lambda\in\Bbb R$, and \proclaim{Theorem 1.1} $$\xi(x, \lambda)=\frac{1}{\pi}\,\text{\rom{Arg}}\bigl(\lim\limits_ {\epsilon\downarrow 0}\, G(x, x, \lambda+i\epsilon)\bigr).$$ \endproclaim This is {\it formally} equivalent to formulae that Krein [32] has for $\xi$ but in a singular setting (i.e., corresponding to an infinite coupling constant). It follows from equations (A.8)--(A.10) in the appendix. With this definition out of the way, we can state the general trace formula: $$V(x)=\lim\limits_{\alpha\downarrow 0}\, \biggl[E_{0}+\int\limits ^{\infty}_{E_0} e^{-\alpha\lambda}[1-2\xi(x, \lambda)]\, d\lambda \biggr], \tag 1.9$$ where $E_{0}\leq\inf\,\text{spec}(H)$. In particular, if $\int\limits^{\infty}_{E_0}|1-2\xi(x, \lambda)|\,d\lambda<\infty$, then $$V(x)=E_{0}+\int\limits^{\infty}_{E_0}[1-2\xi(x, \lambda)]\, d\lambda. \tag 1.10$$ For certain almost periodic potentials, Craig [5] used a regularization similar to the $\alpha$-regularization in (1.9). We will prove (1.9) in \S3 if $V$ is continuous, bounded below, and obeys a bound $$|V(x)|\leq C_{1}e^{C_{2}x^{2}}. \tag 1.11$$ In a subsequent paper [17], we'll allow any $V$ which is bounded below and even drop the continuity property ((1.9) will then hold at points, $x$, of Lebesgue continuity for $V$). That paper will also discuss higher order trace formulae,'' familiar from the context of the Korteweg-de Vries hierarchy, that is, formulae where the left side has suitable polynomials in derivatives of $V$ at $x$. Basically, (1.9) will follow from (1.7) and an asymptotic formula, $$\text{Tr}(e^{-tH}-e^{-tH_{D;x}})=\frac{1}{2}[1-tV(x)+o(t)]. \tag 1.12$$ Examples of the trace formula can be found in \S3 including the case $V(x)\to\infty$ as $|x|\to\infty$. In [16], we'll prove that (1.4) is a special case of (1.9). The proof in \S3 depends on technical preliminaries in \S2. We discuss the case of Jacobi matrices (discrete Schr\"odinger operators) in \S4 including a result for $\Bbb Z^n$. A general $\Bbb R^n$ result that is a kind of analog of (1.12) can be found in [18]. In \S5 we turn to some continuity properties of $\xi(x, \lambda)$ in the potential $V$ and use them to find a new proof (and generalization) of a recent striking result of Last [33]. In particular, we establish $\xi(x, \lambda)$ as a new tool in spectral theory and derive a novel criterion for the essential support of the absolutely continuous spectrum of one-dimensional Schr\"odinger operators and (multi-dimensional) Jacobi matrices. In \S6, we discuss an overview of the connection of the function $\xi$ to inverse problems, including a generalized trace formula that shows how to recover the diagonal Green's function $g(x, z):=G(x, x, z)$ (a Herglotz function w.r.t.~$z$) from $\xi(x, \lambda)$. \bigpagebreak \flushpar {\bf \S2. First Order Asymptotics of the Heat Kernel Trace} As we have seen, a basic role is played by the asymptotics of $$\text{Tr}(e^{-tH}-e^{-tH_{D; x}})\qquad\text{as } t\downarrow 0.$$ In this section we'll prove: \proclaim{Theorem 2.1} Let $V$ be a continuous function which is bounded from below and which obeys $$|V(x)|\leq C_{1}e^{C_{2}x^{2}}. \tag 2.1$$ Then $$\text{\rom{Tr}}(e^{-tH}-e^{-tH_{D; x}})=\frac12 [1-tV(x)+o(t)]. \tag 2.2$$ \endproclaim Under hypothesis (2.1), one can prove this result using the method of images and a DuHamel expansion of $e^{-tH}$ in terms of $e^{-tH_0}$. We will instead use a path integral expansion. The advantage of this approach is that by a more detailed analysis of the path space measure, one can automate higher order expansions in $t$ as $t\downarrow 0$ and can drop the growth condition (2.1). This will be described in a subsequent paper [17]. By translation invariance, we henceforth set the point $x$ in (2.2) to $x=0$. So, our first step in proving Theorem 2.1 is to define a process we will call the xi process, that is, a probablity measure on paths $\omega:[0, 1]\to\Bbb R$. Recall [46] that the Brownian bridge is the Gaussian process $\{\alpha(s)\mid 0\leq s\leq 1\}$ with $E(\alpha(s))=0$, $E(\alpha(s)\alpha(t))=s(1-t)$ if $0\leq s\leq t\leq 1$. The Brownian bridge is of interest because of the following Feynman-Kac formula [46]: $$e^{-tH}(x, x) =(4\pi t)^{-1/2} E\biggl(\exp\biggl(- t\int\limits^{1}_{0} V(x+\sqrt{2t}\, \alpha(s))\, ds\biggr)\biggr), \tag 2.3$$ $$e^{-tH_{D; 0}}(x, x) =(4\pi t)^{-1/2} E\biggl(\exp\biggl(- t\int\limits^{1}_{0} V(x+\sqrt{2t}\, \alpha(s))\, ds\biggr) \chi(\alpha\mid x+\sqrt{2t}\,\alpha(s)\neq 0 \text{ all } s)\biggr), \tag 2.4$$ where $\chi(\alpha\mid x+\sqrt{2t}\,\alpha(s)\neq 0 \text{ all } s)$ is the characteristic function of those $\alpha$ for which $x+\sqrt{2t}\,\alpha(s)$ is non-vanishing for all $s$ in $[0, 1]$, so since paths are continuous, those $\alpha$ with $x+\sqrt{2t}\,\alpha(s)>0$ for all $s$ if $x>0$. There is a $\sqrt{2t}$ in (2.3/2.4) rather than the $\sqrt{t}$ in [46] because [46] considers $-\frac12\, \frac{d^2}{dx^2}$ where we consider $-\frac{d^2}{dx^2}$. Consider the measure $d\kappa$ on $\Omega=\Bbb R\times C([0, 1])$ given by $(4\pi)^{-1/2}\, dx\otimes \Cal D\alpha$ where $dx$ is Lebesgue measure, and define $\omega$ on $\Omega$ by $\omega(s)=x+\alpha(s)$. Since $\alpha(0)=\alpha(1)\equiv 0$, we have $\omega(0)=\omega(1)=x$. $\omega$ defines a natural map of $\Omega$ to $C([0, 1])$ and we henceforth view everything on that space. Let $\Omega_{0}= \{\omega\mid\omega(s)=0 \text{ for some$s$in } [0, 1]\}$. \proclaim{Proposition 2.2} The $\kappa$ measure of $\Omega_0$ is $\frac12$. \endproclaim \demo{Proof} Let $\Delta=\frac{d^2}{dx^2}$ and $\Delta_D$ be the same operator with a Dirichlet boundary condition at $x=0$. By (2.3--4) with $V=0$, \align \kappa(\Omega_0) &=\int\limits_{\Bbb R} dx\, [e^{\Delta/2}(x, x)- e^{\Delta_{D}/2}(x, x)] \\ &=\int\limits_{\Bbb R} dx\, e^{\Delta/2}(x, -x) \\ &=\int\limits_{\Bbb R} dx\, e^{\Delta/2}(2x, 0) \\ &=\frac12\int\limits_{\Bbb R} dx\, e^{\Delta/2}(0, x)=\frac12 , \endalign where the second equality is by the method of images. \qed \enddemo We will call the probability measure $2\chi_{\Omega_{0}}\, d\kappa$ on $C([0, 1])$ the xi process and denote its expectations as $E_\omega$. (2.3--4) and the regularity of the integral kernel immediately imply that \proclaim{Proposition 2.3} For any $V$ which is bounded from below: $$\text{\rom{Tr}}(e^{-tH}-e^{-tH_{D; 0}})=\frac12 E_{\omega} \biggl(\exp \biggl(-t\int\limits^{1}_{0} V(\sqrt{2t}\, \omega(s))\, ds\biggr)\biggr).$$ \endproclaim We note the $t^{-1/2}$ in front of (2.3--4) is absorbed in the change of variables from $x+\sqrt{2t}\, \alpha(s)$ to $\sqrt{2t}\, (x+\alpha(s))$. \proclaim{Lemma 2.4} If $C<\frac12$, then $E_{\omega}(e^{C\omega(s)^{2}}) <\infty$ for each $s$ with a bound uniform in $s$. \endproclaim \demo{Proof} Let $f$ be a bounded even function on $\Bbb R$. Then $$\split E_{\omega}(f(\omega(s))&=2\int\limits \Sb x>0 \\ y>0 \endSb \left[e^{(1- s)\Delta/2}(x, y) f(y)\, e^{s\Delta/2}(y, x)\right. \\ &\qquad \left. - e^{(1-s)\Delta_{D}/2} (x, y)f(y)\, e^{s\Delta_{D}/2}(y, x)\right]\, dx\, dy \endsplit$$ and it easy to see using the method of images that for $f(y)=\min(R, e^{Cy^2})$ the integral remains finite as $R\to\infty$. \qed \enddemo \demo{Proof of Theorem {\rom{2.1}}} It is easy to see if we prove the formula for $V(x)$, it follows for $V$ replaced by $V(x)+C$. Thus, without loss we suppose $V\geq 0$. Let $a\leq 0$. Then by Taylor's theorem with remainder: $$|e^{a}-1-a|\leq\frac12\, a^2.$$ Thus, with $a=-t\int\limits^{1}_{0}V(\sqrt{2t}\, \omega(s))\, ds$ $$\left|\frac12 E_{\omega}(e^{a})-\frac12-\frac12\, E_{\omega}(a)\right| \leq\frac14\, E_{\omega}(a^{2}).$$ Using (2.1) and Lemma 2.4, it is easy to see that $$E_{\omega}(a^{2})=O(t^{2})$$ so it suffices to show that $$E_{\omega}\biggl(\int\limits^{1}_{0} V(\sqrt{2t}\, \omega(s))\,\biggr) =V(0)+o(1).$$ This follows from Lemma 2.4, (2.1), continuity of $V$ at $x=0$, and dominated convergence. \qed \enddemo \remark{Remark} This is crude analysis compared with the detailed path space argument in [17] but it is elementary and beyond the argument of previous authors who supposed that $V$ is bounded. \endremark \bigpagebreak \flushpar {\bf \S3. The Trace Formula: Schr\"odinger Case} Our main goal in this section is to prove: \proclaim{Theorem 3.1} Suppose $V$ is a continuous function bounded from below on $\Bbb R$. Let $\xi(x, \lambda)$ be the Krein spectral shift for the pair $(H_{D;x}, H)$ with $H_{D;x}$ the operator on $L^{2}(-\infty, x)\oplus L^{2}(x, \infty)$ obtained from $H=-\frac{d^2} {dx^2}+V$ with a Dirichlet boundary condition at $x$. Let $E_{0}\leq \inf\,\text{\rom{spec}}(H)$. Then $$V(x)=\lim_{\alpha\downarrow 0} \biggl[E_{0}+\int\limits^{\infty} _{E_0} e^{-\alpha\lambda} [1-2\xi(x, \lambda)]\, d\lambda\biggr]. \tag 3.1$$ \endproclaim \demo{Proof} Let $E_{1}=\inf\,\text{spec}(H)$. Then for $E_{0}\leq E_{1}$ $$E_{0}+\int\limits^{\infty}_{E_0} e^{-\alpha\lambda}(1-2\xi)\, d\lambda=E_{0}+\int\limits^{E_1}_{E_0} e^{-\alpha\lambda}\,d\lambda + \int\limits^{\infty}_{E_1} e^{-\alpha\lambda}(1-2\xi)$$ and $\lim\limits_{\alpha\downarrow 0}\,\int\limits^{E_1}_{E_0} e^{- \alpha\lambda}\,d\lambda=E_{1}-E_{0}$ so the formula for $E_1$ implies it for $E_0$; that is, without loss we suppose $E_{0}=E_{1}$. By Theorems 2.1 and eq.~(1.7), $$\alpha\int\limits^{\infty}_{E_0} e^{-\alpha\lambda}\xi(x, \lambda)\, d\lambda=\frac12\, [1-\alpha V(x)+o(\alpha)].$$ Moreover, $$\frac12\,\alpha\int\limits^{\infty}_{0} e^{-\alpha\lambda}\, d\lambda=\frac12 \tag 3.2$$ so $$\frac12\, \alpha\int\limits^{\infty}_{E_0}e^{-\alpha\lambda}\, d\lambda=\frac12\, [1-\alpha E_{0}+o(\alpha)]$$ and hence $$\alpha\int\limits^{\infty}_{E_0} e^{-\alpha\lambda}\biggl(\xi-\frac12 \biggr)\, d\lambda=-\frac12\, \alpha[V(x)-E_{0}]+o(\alpha)$$ which is (3.1). \qed \enddemo \example{Example 3.2} $V=0$. Then $g(x, \lambda)=\frac12 \,(- \lambda)^{-1/2}$ and so $\arg\,g(x, \lambda)=0$ (resp.~$\pi/2$) if $\lambda<0$ (resp.~$\lambda>0$). Thus, by Theorem 1.1, $\xi(x, \lambda)\equiv\frac12$ on $[0, \infty)$ and (3.2) is just Theorem 2.1 for $V=0$. When $\xi=\frac12$ on a subset of $\text{spec}(H)$, that set drops out of (3.1). \endexample \example{Example 3.3} Suppose that $V(x)\to\infty$ as $|x|\to\infty$. Then $H$ has eigenvalues $E_{0}-\infty$ ($\Bbb R$ case) or $\sup\limits_{(n, j)\in\Bbb N\times\Bbb Z} |V_{n}(j)|<\infty$ ($\Bbb Z$ case). \item"{(ii)}" For each $R<\infty$, $\sup\limits_{|x|\leq R} |V_{n}(x) -V(x)|\to 0$ as $n\to\infty$. \endroster \enddefinition \proclaim{Lemma 5.3} If $V_{n}\to V$ locally as $n\to\infty$ and $H_n, H$ are the corresponding Schr\"odinger operators \rom(resp.~Jacobi matrices\rom), then $(H_{n}-z)^{-1}\to (H-z)^{-1}$ strongly for $\text{Im}\,z\neq 0$ as $n\to\infty$. \endproclaim \demo{Proof} Let $\varphi\in C^{\infty}_{0}(\Bbb R)$ or a finite sequence in $\ell^{2}(\Bbb Z)$. Then $[(H_{n}-z)^{-1}-(H-z)^{-1}] (H-z)\varphi=(H_{n}-z)^{-1}(V-V_{n})\varphi\to 0$ as $n\to\infty$. But $\{(H- z)\varphi\}$ is a dense set (since $H$ is essentially self-adjoint on $C^{\infty}_{0}(\Bbb R)$ resp.~on finite sequences). \qed \enddemo \proclaim{Theorem 5.4} If $V_{n}\to V$ locally as $n\to\infty$ and $\xi_{n}(x,\lambda)$, $\xi(x, \lambda)$ are the corresponding xi functions for fixed $x$, then $\xi_{n}(x, \lambda)\, d\lambda$ converges to $\xi(x, \lambda)\, d\lambda$ weakly in the sense that $$\int\limits_{\Bbb R} f(\lambda)\xi_{n}(x, \lambda)\, d\lambda \to \int\limits_{\Bbb R} f(\lambda)\xi(x, \lambda)\, d\lambda \qquad \text{as } n\to\infty \tag 5.2$$ for any $f\in L^{1}(\Bbb R; d\lambda)$. \endproclaim \demo{Proof} By a simple density argument (using $|\xi(x, \lambda)|\leq 1$), it suffices to prove this for $f(\lambda)=(\lambda- z)^{-2}$ and all $z\in\Bbb C\backslash\Bbb R$. But by (A.$7'$): $$\int\limits_{\Bbb R} (\lambda-z)^{-2}\xi_{n}(x, \lambda)\,d\lambda =\frac{d}{dz}\,F_{n}(x, z),$$ where $F_{n}(x, z)=\ln\,g_{n}(x, z)$. Since the $F$'s are analytic and uniformly bounded, pointwise convergence of the $F$'s implies convergence of the derivatives $dF_{n}/dz$. Thus we need only show $$g_{n}(x, z)\operatornamewithlimits{\to}\limits_{n\to\infty} g(x, z).$$ This follows from Lemma 5.3 (and, in the Schr\"odinger case, some elliptic estimates to turn convergence of the operators to pointwise convergence of the integral kernels). \qed \enddemo \definition{Definition} For any $H$, let $|S_{\text{\rom{ac}}}(H)|$ denote the Lebesgue measure of the essential support of the absolutely continuous spectrum of $H$. \enddefinition \proclaim{Theorem 5.5} \rom{(For one-dimensional Schr\"odinger operators or Jacobi matrices)} Suppose $V_{n}\to V$ locally as $n\to\infty$ and each $V_n$ is periodic. Then for any interval $(a, b)\subset\Bbb R$: $$|(a, b)\cap S_{\text{\rom{ac}}}|\geq \varlimsup\limits_{n\to\infty} |(a, b)\cap S_{\text{\rom{ac}}} (H_{n})|.$$ \endproclaim \remark{Remark} The periods of $V_{n}$ need {\it not} be fixed; indeed, almost periodic $V$'s are allowed. \endremark \demo{Proof} By periodicity, $\xi_{n}(x, \lambda)$ is $0$, $\frac12$, $1$ for a.e.~$\lambda\in\Bbb R$. Let $A_{n}=\{\lambda\in (a, b)\mid\xi_{n}(x, \lambda)=0\}$ and $A=\{\lambda\in (a, b)\mid\xi(x, \lambda)=0\}$. Then, $\xi_{n}(x, \lambda)\geq\frac12$ on $A\backslash A_{n}$, so for any $a, b$: $$\int\limits_{A} \xi_{n}(x, \lambda)\, d\lambda \geq\frac12 |(A\backslash A_{n})|.$$ But by Theorem 5.4, $\int\limits_{A}\xi_{n}(x, \lambda)\, d\lambda \operatornamewithlimits{\to}\limits_{n\to\infty} \int\limits_{A}\xi(x, \lambda)\, d\lambda=0$. Thus, $\frac12 |A\backslash A_{n}|\to 0$, so $|A|\leq\varliminf\limits_{n\to\infty} |A_{n}|$. Similarly, using $1-\xi$, we get an inequality on $$|\{\lambda\in (a, b)|\xi (x,\lambda)=1\}|\leq\varliminf\limits_{n\to\infty} |\{\lambda\in (a, b)|\xi (x, \lambda)=1\}|.$$ This implies the result. \qed \enddemo \example{Example 5.6} Let $\alpha_n$ be a sequence of rationals and $\alpha=\lim\limits_{n\to\infty}\,\alpha_n$. Let $H_n$ be the Jacobi matrix with potential $\lambda\cos (2\pi\alpha_{n}+\theta)$ for $\lambda, \theta$ fixed. Then [2] have shown for $|\lambda|\leq 2$, $|S_{n}| \geq 4-2|\lambda|$. It follows from the last theorem that $|S|\geq 4-2|\lambda|$. This provides a new proof (and a strengthening) of an important result of Last [33]. \endexample \example{Example 5.7} Let $\{a_{m}\}_{m\in\Bbb N}$ be a sequence with $s=\sum\limits^{\infty}_{m=1}2^{m}|a_{m}|<2$. Let $V(n)=\sum\limits^{\infty}_{m=1}a_{m}\cos(2\pi n/2^{m})$, a limit periodic potential on $\Bbb Z$. Let $h$ be the corresponding Jacobi matrix. We claim that $$|\sigma_{\text{ac}}(h)|\geq 2(2-s). \tag 5.31$$ For let $V_{M}(n)=\sum\limits^{M}_{m=1} a_{m}\cos(2\pi n/2^{m})$ with $h_{M}$ the associated Jacobi matrix. Then the external edges of the spectrum move in at most by $\|V_{M}\|_{\infty}\leq\sum\limits^{M}_{m=1}|a_{m}|$. $V_{M-1}$ has at most $2^{M-1}-1$ gaps. They increase in size in going from $V_{M-1}$ to $V_{M}$ by $2|a_{M}|$. In addition, $V_{M}$ has $2^{M-1}$ new gaps. Thus, $\sigma(h_{M})\geq 4-2\|V\|_{\infty}-\sum\limits^{M}_{m=1} (2^{m}-1) |a_{m}|\geq 4-2s$, which yields (5.31) on account of Theorem 5.5. Knill-Last [29] have shown how to use our Theorem 5.5 to treat more general limit periodic potentials, including Schr\"odinger operators of the form studied by Chulaevsky [4], and have also treated quasi-periodic potentials of the form $V(n)=\sum\limits^{\infty}_{m=1}\lambda_{m}\cos([2\pi\alpha n +\theta]m)$ where they show $|\sigma_{\text{ac}}|\geq 4-6\sum \limits^{\infty}_{m=1}m|\lambda_{m}|$. \endexample \bigpagebreak \flushpar {\bf \S6. Inverse Problems} We want to give an overview of how we believe $\xi(x, \lambda)$ can be an important tool in the study of inverse problems and apply the philosophy in a few cases. Roles are played by $\xi(x_{0}, \lambda)$, the diagonal Green's function $g(x_{0}, \lambda)$, and the Weyl $m$-functions $m_{\pm}(x_{0}, z)$ (corresponding to the Dirichlet boundary condition at $x=x_{0}$). The relationship is that $\xi$ is closest to spectral and scattering information and it, under proper circumstances, determines $g(x_{0}, \lambda)$ and the derivative $g'(x_{0}, \lambda)$. They determine $m_{\pm}(x_{0}, \lambda)$, which in turn determine $V(x)$ for a.e.~$x\in\Bbb R$ by the Gel'fand-Levitan method [14,36]. The scheme underlying our philosophy is illustrated in Fig.~1. That $m_{\pm}(y, \lambda)$ for all $\lambda$ and a single $y$ determine $V(x)$ on $(-\infty, y)$ and $(y, \infty)$ is well known [38]. That $g(x, \lambda)$ and $\frac{d}{dx}\,g(x, \lambda)$ at a single point $x$ determine $m_{\pm}(x, \lambda)$ follows from the pair of formulae: \align g(x, \lambda) &= -[m_{+}(x, \lambda)+m_{-}(x, \lambda)]^{-1}, \tag 6.1 \\ g'(x, \lambda) &= -[m_{+}(x, \lambda)-m_{-}(x, \lambda)]\big/ [m_{+}(x, \lambda)+m_{-}(x, \lambda)]. \tag 6.2 \endalign (6.2) follows from (6.1) and the Riccati equations $$m'_{\pm}(x, \lambda)=\mp[m^{2}_{\pm}(x, \lambda)-V(x)+\lambda].$$ (6.2) is not new; it can be found, for example, in Johnson-Moser [26]. Thus, to recover $V(x)$ for all $x\in\Bbb R$ from $\xi(x_{0}, \lambda)$ for a fixed $x_{0}$ and all $\lambda$, we only need a method to compute $g(x_{0}, \lambda)$ and $g'(x_{0}, \lambda)$ from $\xi(x_{0}, \lambda)$. One can get $g$ in general from the following formula which follows from Theorem A.2 in the appendix and the proposition below: $$g(x, z)=(-z)^{-1/2}\lim\limits_{\gamma\to\infty}\, \exp\biggl( \int\limits^{\infty}_{E_{0}}\biggl[\frac{\xi(x, \lambda)-\frac{1}{2}} {z-\lambda}\biggr]\,\biggl[\frac{\gamma}{\gamma+\lambda}\biggr]\, d\lambda\biggr), \qquad E_{0}=\inf\,\text{spec}(H). \tag 6.3$$ The proposition we need is \proclaim{Proposition 6.1} Let $V$ be continuous and bounded from below and let $g(x, z)=\mathbreak G(x, x, z)$ be the diagonal Green's function for $H=-\frac{d^2}{dx^2}+V$. Then $$\lim\limits_{\lambda\to\infty}\, g(x, -\lambda)\big/(\lambda)^{-1/2} =1. \tag 6.4$$ \endproclaim \demo{Proof} Let $p(x, t)$ be the diagonal heat kernel for $H$. By the Feynman-Kac formula [46] $$p(x, t)=(4\pi t)^{-1/2} E\,\biggl(\exp\biggl(-t\int\limits^{1}_{0} V(x+\sqrt{2t}\,\alpha(s))\, ds\biggr)\biggr),$$ where $\alpha$ is the Brownian bridge. It follows by the dominated convergence theorem that $$p(x, t)\big/(4\pi t)^{-1/2}\to 1 \tag 6.5$$ as $t\downarrow 0$. Since $$g(x, -\lambda)=\int\limits^{\infty}_{0} e^{-\lambda t}p(x, t)\, dt$$ we obtain (6.4). \qed \enddemo \remark{Remarks} (i) (6.4) can also be read off of asymptotics of $m_{\pm}$ found in [1,11]. (ii) (6.5) can be used to prove the following stronger version of (6.3): $$g(x, z)=(-z)^{-1/2}\lim\limits_{\alpha\downarrow 0}\,\exp\biggl( \int\limits^{\infty}_{E_{0}}\biggl[\frac{\xi(x, \lambda)-\frac{1}{2}} {z-\lambda}\biggr]\,e^{-\alpha\lambda}\, d\lambda\biggr).$$ Thus, the solution of the inverse problem for going from $\xi(x_{0}, \cdot)$ at a single $x_{0}$ to $V(x)$ for all $x\in\Bbb R$ is connected to finding $g'(x_{0}, z)$ from $\xi(x_{0}, \lambda)$. In absolute generality, we are unsure how to proceed with this because we have no general theory for a differential equation that $\xi(x, \lambda)$ obeys for $\lambda$ in the essential spectrum of $H$. Indeed, for random $V$'s where typically $\text{spec}(H)=[\alpha, \infty)$ for some $\alpha$, $\xi(x, \lambda)=1$ or $0$ on $\Bbb R$ and $\overline{\{\lambda\in\Bbb R\mid \xi(x, \lambda)=1\}}=[\alpha, \infty)$, $\overline{\{\lambda\in\Bbb R\mid\xi (x, \lambda)=0\}}=\Bbb R$ and the $x$ dependence must be very complex. However, one class of potentials does allow some progress: \endremark \definition{Definition} We say that $V$ is discretely dominated if for all $x\in\Bbb R$, $\xi(x, \lambda)=\frac12$ for a.e.~$\lambda\in \sigma_{\text{ess}}(H)$. \enddefinition Examples include reflectionless (soliton) potentials in the short-range case, the periodic case, algebro-geometric finite-gap potentials and limiting cases thereof (such as solitons relative to finite-gap backgrounds), certain almost periodic potentials, and potentials with $V(x)\to\infty$ as $|x|\to\infty$. In this case if $E_{0}=\inf\,\text{spec} (H)$, $[E_{0}, \alpha)\backslash\text{spec}(H)=\operatornamewithlimits{\cup} \limits^{N}_{n=1} (\alpha_{n}, \beta_{n})$ where $N$ is finite or infinite. For each $x$, there is at most one eigenvalue for $H_{D; x}$ in each $(\alpha_{n}, \beta_{n})$; call it $\mu_{n}(x)$. If there is no eigenvalue in $(\alpha_{n}, \beta_{n})$, then $\xi(x, \lambda)$ is either $1$ on $(\alpha_{n}, \beta_{n})$ or $0$, and then we set $\mu_{n}(x)$ equal to $\beta_{n}$ or to $\alpha_{n}$. Thus $$\xi(x, \lambda) = \cases \frac12, &\quad \lambda\in\text{spec}(H) \\ 0, &\quad \mu_{n}(x)<\lambda<\beta_{n} \\ 1, &\quad \alpha_{n}<\lambda<\mu_{n}(x) \endcases$$ and the inverse formulae at a fixed $x$ say that $$V(x)=\lim\limits_{t\downarrow 0}\sum^{N}_{n=1} [2e^{-t\mu_{n}(x)} -e^{-t\alpha_{n}}-e^{-t\beta_{n}}]\big/t, \tag 6.6$$ $$g(x, z)=(E_{0}-z)^{-1/2}\lim\limits_{\gamma\to\infty}\, \biggl[\prod^{N}_{n=1}\biggl\{\frac{[z-\mu_{n}(x)]^{2}}{(z-\alpha_{n}) (z-\beta_{n})}\, \frac{(\gamma+\alpha_{n})(\gamma+\beta_{n})} {[\gamma-\mu_{n}(x)]^{2}}\biggr\}\biggr]^{1/2}. \tag 6.7$$ If $\sum\limits_{n}|\beta_{n}-\alpha_{n}|<\infty$, then $$g(x, z)=(E_{0}-z)^{-1/2}\biggl[\prod^{N}_{n=1}\,\frac{[z- \mu_{n}(x)]^{2}}{(z-\alpha_{n})(z-\beta_{n})}\biggr]^{1/2} \tag 6.8$$ (with an absolutely convergent product if $N=\infty$). The $\mu$'s obey a differential equation essentially that was found by Dubrovin [9] in 1975 for the finite-gap periodic case and extended later by McKean-Trubowitz [42], Trubowitz [49], Levitan [34,35], Kotani-Krishna [31], and Craig [5]. The form we give is the one in Kotani-Krishna [31]. Previous authors only considered the periodic or almost periodic case, so, in particular, our result is new in the case $|V(x)|\to\infty$ where the regulariziation (6.6) is needed since $\sum\limits_{n\in\Bbb N}|\beta_{n}-\alpha_{n}|=\infty$: \proclaim{Theorem 6.2} Let $\alpha_{n}<\mu_{n}(x_{0})<\beta_{n}$. Then $\mu_n$ is $C^1$ near $x_0$ and $$\left.\frac{d}{dx}\,\mu_{n}(x)\right|_{x=x_{0}} =\pm 1\bigg/\left.\frac{\partial g}{\partial\lambda}\,(x_{0}, \lambda)\right|_ {\lambda=\mu_{n}(x_{0})}, \tag 6.9$$ where $g$ is given by \rom{(6.7)} or \rom{(6.8)}. In \rom{(6.9)}, the $\pm 1$ is $+1$ \rom(resp.~$-1$\rom) if $\mu_{n}(x_{0})$ is an eigenvalue of $H_{x_{0}; D}$ on $(x_{0}, \infty)$ \rom(resp.~$(-\infty, x_{0})$\rom). \endproclaim \demo{Proof} The number $\mu_{n}(x)$ obeys $$g(x, \mu_{n}(x))=0.$$ It is easy to see that $g$ is strictly monotone; indeed, $\frac{\partial g}{\partial\lambda}>0$ on each $(\alpha_{n}, \beta_{n})$ and so by the implicit function theorem, $\mu_{n}(x)$ is $C^1$ and $$\frac{d\mu_{n}}{dx}=-\frac{\partial g/\partial x}{\partial g/\partial\lambda}$$ so (6.9) is equivalent to $\frac{\partial g}{\partial x}=\mp 1$ if the eigenvalue corresponds to the half-line $(x, \infty)$ (resp.~$(-\infty, x)$). But the associated eigenvector lies in $L^{2}(x_{0}, \infty)$, (resp.~$L^{2}(-\infty, x_{0})$) if and only if $m_{+}(x, \lambda)$ (resp.~$m_{-}(x, \lambda)$) is $\infty$ at $x=x_{0}$, $\lambda=\mu_{n}(x_{0})$. By (6.2), $\frac{\partial g}{\partial x}=\mp 1$ if $m_{\pm}=\infty$. \qed \enddemo The simple example of the unique discretely dominated potential with $\sigma(H)=\{-1\}\cup [0, \infty)$ (the one-soliton potential) is discussed in [48]. (6.8)/(6.9) become an elementary differential equation and $V$ is then given by (6.6). We will further explore this approach in future papers [20,21]. Analogs of $\xi$ in the related inverse cases are also useful. For example, we have shown that the $\xi$ function relating to half-line problems on $[0, \infty)$ with different boundary conditions at $0$ determines the potential uniquely a.e. This result was previously obtained independently by Borg [3] and Marchenko [38] in 1952 under the strong additional hypothesis that the corresponding spectra were purely discrete. Our approach allows us to dispense with the discrete spectrum hypothesis and applies to arbitrary spectra. \bigpagebreak \flushpar {\bf Appendix: Rank One Perturbations and the Krein Spectral Shift} In this appendix, we will give a self-contained approach to the Krein spectral shift in a slightly more general setting than usual and using more streamlined calculations. The lecture notes [48] contain more about this approach. Let $A\geq 0$ be a positive self-adjoint operator in some complex separable Hilbert space $\Cal H$ and let $\Cal H_{k}(A)(-\infty 0$ we have $0\leq\text{Arg}(\cdot)\leq\pi$ (and $\text{Im }F(z)>0$ if $\text{Im }z>0$) and thus $$0\leq\xi_{\alpha}(\lambda)\leq 1$$ in this case. Since $\text{Arg}(F(\lambda+i0))=\text{Im }\ln(F(\lambda+i0))$, an elementary contour integral argument ([48]) shows that (A.7) becomes $$\text{Tr}[(A-z)^{-1}-(A_{\alpha}-z)^{-1}]=\int\limits^{\infty}_ {E_{\alpha}}\frac{\xi_{\alpha}(\lambda)\, d\lambda}{(\lambda-z)^{2}}, \qquad E_{\alpha}=\inf\,\text{spec}(A_{\alpha}). \tag A.9$$ (A.9) is a special case of $$\text{Tr}[f(A)-f(A_{\alpha})]=-\int\limits^{\infty}_{E_{\alpha}} f'(\lambda)\xi_{\alpha}(\lambda)\, d\lambda \tag A.10$$ for the functions $f_{z}(\lambda)=(\lambda-z)^{-1}$. By analyticity in $z$, one sees immediately that $[(A-z)^{-n}-(A_{\alpha}-z)^{-n}]$ is trace class and (A.10) holds for $f_{z, n}(\lambda)=(\lambda-z)^{-n}$. A straightforward limiting argument lets one prove ([48]) that if $f$ is $C^2$ on $\Bbb R$ with $(1+|x|)^{2}\,\frac{d^{j}f}{dx^{j}}\in L^{2}(0, \infty)$ for $j=1,2$, then $[f(A)-f(A_{\alpha})]$ is trace class and (A.10) holds. In particular, $$\text{Tr}(e^{-At}-e^{-tA_{\alpha}})=t\int\limits^{\infty}_{E_{\alpha}} e^{-t\lambda}\xi_{\alpha}(\lambda)\, d\lambda. \tag A.11$$ For the case where $\alpha=\infty$ and $\varphi\in\Cal H$, $f(A_{\alpha})$ is interpreted as the operator on $\Cal H(A_{\infty})$ extended to $\Cal H$ by setting it equal to zero on $\Cal H(A_{\infty})^{\perp}$. This follows from the approximation argument since that is the meaning of $(A_{\infty}-z)^{-1}$. In particular, \proclaim{Theorem A.1} Let $A$ be a bounded operator and $\varphi$ a unit vector in $\Cal H$. Let $Q=\Bbb I-(\varphi, \cdot)\varphi$. Then $A-QAQ$ is finite rank and $$\text{Tr}(A-QAQ)=-\int\limits^{\infty}_{-\infty} f'(\lambda)\xi_{\infty} (\lambda)\, d\lambda,$$ where $\xi_{\infty}(\lambda)=\frac{1}{\pi}\text{Arg}(\varphi, (A- \lambda-i0)^{-1}\varphi)$ and $f$ is any function in $C^{\infty}_{0}$ with $f(x)=x$ for $x\in [-\|A\|_{\infty}, \|A\|_{\infty}]$. \endproclaim One cannot recover $F(z)$ from $\xi_{\infty}(A)$ without some additional information. For by (A.$7'$), $\xi_{\infty}$ determines $\frac{d}{dz}\, \ln\,F(z)$. There is then a constant needed to get $F$ by integration. However, asymptotics of $F$ at $-\infty$ are often enough to recover $F$ from $\xi_{\infty}$. This is what is needed in \S6. For generalizations, see [48]. \proclaim{Theorem A.2} Let $A\geq 0$. Suppose $(-z)^{1/2}F(z)\to 1$ as $z\to -\infty$ along the real axis. Then $$F(z)=(-z)^{-1/2}\lim\limits_{\gamma\to\infty}\,\exp\biggl[ \int\limits^{\infty}_{0}\biggl[\frac{\xi_{\infty}(\lambda)-\frac{1}{2}} {z-\lambda}\biggr]\,\biggl[\frac{\gamma}{\lambda+\gamma}\biggr]\, d\lambda\biggr].$$ \endproclaim \demo{Proof} Let $F^{(0)}(z)=(-z)^{-1/2}$. Then $$\frac{d}{dz}\,\ln\, F^{(0)}(z)=\frac{1}{2}\int\limits^{\infty}_{0} \frac{d\lambda}{(z-\lambda)^{2}}$$ so by (A.$7'$): $$\frac{d}{dz}\,\biggl(\biggl[\frac{F(z)}{F^{(0)}(z)}\biggr]\biggr)= \int\limits^{\infty}_{0}\frac{[\xi_{\infty}(\lambda)-\frac{1}{2}]} {(z-\lambda)^{2}}\, d\lambda$$ hence integrating, $$\ln\,\biggl[\frac{F(z)}{F^{(0)}(z)}\biggr] - \ln\,\biggl[\frac{F(- \gamma)}{F^{(0)}(-\gamma)}\biggr] =\int\limits^{\infty}_{0}\, \frac{[\xi_{\infty}(\lambda)-\frac{1}{2}]}{(\lambda-z)(\lambda+\gamma)}\, (\gamma+z)\,d\lambda.$$ By hypothesis, $\lim\limits_{\gamma\to\infty}\,\ln\,\bigl[\frac{F(-\gamma)} {F^{(0)}(-\gamma)}\bigr]=0$ and by dominated convergence for any fixed $z$, \break $\lim\limits_{\gamma\to\infty}\bigl[\int\limits^{\infty}_{0} \frac{[\xi_{\infty}(\lambda)-\frac{1}{2}]}{(\lambda-z)(\lambda+\gamma)}\, d\lambda\bigr]=0$, proving the theorem. \qed \enddemo As an example of the abstract theory, fix $V$, a continuous function on $\Bbb R$ which is bounded from below, and $x_{0}\in\Bbb R$. Let $A=-\frac{d^2}{dx^2}+V$. Let $F:Q(A)\to\Bbb C$ by $F(f)=f(x_{0})$. By a Sobolev estimate and using $\Cal H_{1}(A)\subset\Cal H_{1}(- \frac{d^2}{dx^2})$, $F$ is a functional in $\Cal H_{-1}$, so we write $F(f)=\langle\varphi, f\rangle$ with $\varphi(x)=\delta(x-x_{0})$. The form domain of $A_{\infty}$ is thus $f\in\Cal H_{1}(A)$ with $f(x_{0})=0$; thus $A_{\infty}$ is exactly the operator $H_{x_{0}; D}$ with a Dirichlet boundary condition at $x_{0}$ that we discuss in the body of the paper. \vskip 0.3in \example{Acknowledgments} We would like to thank H.~Holden, Y.~Last, and Z.~Zhao for discussions on this subject. 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