\input amstex
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\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 $E_{0}<E_{1}<E_{2}<\cdots$ and $H^{D}_{x}$ has 
eigenvalues $\{\mu_{j}(x)\}^{\infty}_{j=1}$ with $E_{j-1}\leq\mu_{j}(x)\leq 
E_{j}$.  We have
$$
\xi(x, \lambda) = \cases 1, &\qquad E_{j-1}<\lambda <\mu_{j}(x) \\
0, &\qquad \lambda<E_{0} \text{ or } \mu_{j}(x)<\lambda<E_{j}.
\endcases
$$
Thus (3.1) becomes:
$$
V(x)=E_{0}+\lim_{\alpha\downarrow 0}\, \biggl[\sum^{\infty}_{j=1}
\bigl(\left.2e^{-\alpha\mu_{j}(x)}-e^{-\alpha E_{j}}-e^{-\alpha E_{j-
1}}\bigr)\right/\alpha\biggr]. \tag 3.3
$$
If we could take $\alpha$ to zero inside the sum, we would get
$$
V(x)=E_{0}+\sum^{\infty}_{j=1}\,[E_{j}+E_{j-1}-2\mu_{j}(x)]
\qquad \text{(formal)} \tag 3.4
$$
which is just a limit of the periodic formula (1.3) in the limit of 
vanishing band widths.  (3.3) is just a kind of abelianized summation 
procedure applied to (3.3).

As a special case of this example, consider $V(x)=x^{2}-1$.  Then 
$E_{j}=2j$ and $\{\mu_{j}(0)\}$ is the set $\{2, 2, 6, 6, 10, 10, 14, 
14, \dots \}$ of $j$ odd eigenvalues, each doubled.  Thus (3.4) is the 
formal sum
$$
-1=-2+2-2-2\dots \qquad \text{(formal)}
$$
with (3.3)
$$\align
-1 &=\lim_{\alpha\downarrow 0}\,\{1-e^{-2\alpha})/\alpha\}\,\{1-e^{-2\alpha}
+e^{-4\alpha}\dots\} \\
&=\lim_{\alpha\downarrow 0}\,\{(1-e^{-2\alpha})/\alpha\}\, 
1/(1+e^{-2\alpha})
\endalign
$$
with a true abelian summation.
\endexample

\example{Example 3.4}  Suppose $V(x)=V(x+1)$.  Let $E_{j}, \mu_{j}(x)$ 
be the band edges and Dirichlet eigenvalues as in (1.2), (1.3).  Then it 
follows from results in Kotani [30] (see also Deift and Simon [6]) and 
the fact that $g(x, \lambda):=G(x, x, \lambda+i0)=-[m_{+}(x, \lambda)+m_{-}
(x, \lambda)]^{-1}$ in terms of the Weyl $m$-functions, that $g(x, \lambda)$ 
is purely imaginary on $\text{spec}(H)$; that is, $\xi(x, \lambda)=\frac12$ 
there, so
$$
\xi(x, \lambda)= \cases \frac12, &\quad E_{2n}<\lambda<E_{2n+1} \\
1, &\quad E_{2n-1}<\lambda<\mu_{n}(x) \\
0, &\quad \mu_{n}(x)<\lambda<E_{2n}.
\endcases
$$
It follows that
$$
\int\limits^{\infty}_{E_0} |1-2\xi(x, \lambda)|\, dx=\sum^{\infty}_{n=1}
|E_{2n}-E_{2n-1}|
$$
is finite if (1.1) holds.  In that case one can take the limit inside 
the integral in (3.1) and so recover (1.3).
\endexample

\example{Example 3.5}  In [16] we will show that if $V$ is short-range, 
that is, $V\in H^{2, 1}(\Bbb R)$, then, $\int\limits^{\infty}_{E_0}|
1-2\xi(x, \lambda)|\, d\lambda<\infty$ and we can take the limit in (3.1) 
inside the integral.  This recovers Venakides' result [50] with an explicit 
form for $\xi$ in terms of the Green's function (see Theorem 1.1).  
Similarly, one can treat short-range perturbations $W$ of periodic 
background potentials $V$ (modeling scattering off defects or impurities, 
described by $W$, in one-dimensional solids) and ``cascading'' potentials, 
that is, potentials approaching different 
spatial asymptotes sufficiently fast [16].
\endexample

\bigpagebreak
\flushpar {\bf \S4. The Trace Formula: Jacobi Case}

Our goal here is the proof of an analog for Theorem 3.1 for Jacobi 
matrices.  It will be a special case of the following:

\proclaim{Theorem 4.1}  Let $A$ be a bounded self-adjoint operator 
in some complex separable Hilbert space $\Cal H$ with $\alpha=\inf\,
\text{\rom{spec}}(A)$, $\beta=\sup\,\text{\rom{spec}}(A)$.  Let 
$\varphi\in\Cal H$ be an arbitrary unit vector and let $\xi(\lambda)$ be the 
Krein spectral shift for the pair $(A_{\infty},A)$, $A_{\infty}:=A+\infty 
(\varphi, \cdot)\varphi$ \rom(where the infinite coupling perturbation 
is discussed in the appendix\rom).  Then for any $E_{-}\leq\alpha$ and 
$E_{+}\geq\beta$:
$$\align
(\varphi, A\varphi) &=E_{-}+\int\limits^{E_{+}}_{E_{-}} [1-\xi(\lambda)]\, 
d\lambda \tag 4.1 \\
&= E_{+}-\int\limits^{E_{+}}_{E_{-}} \xi(\lambda)\, d\lambda \tag 4.2 \\
&=\frac12\, (E_{+}+E_{-})+\frac12 \int\limits^{E_{+}}_{E_{-}} 
[1-2\xi(\lambda)]\, d\lambda. \tag 4.3
\endalign
$$
\endproclaim

\demo{Proof}  (4.1) follows from (4.2) by integrating 1 from $E_{-}$ 
to $E_{+}$ and (4.3) is the average of (4.1) and (4.2).  Moreover, since 
$\xi(\lambda)=1$ for $\lambda\geq\beta$ and $\xi(\lambda)=0$ for 
$\lambda\leq\alpha$, it is easy to see that it suffices to prove the 
result for $E_{-}=\alpha$ and $E_{+}=\beta$.  Thus, we are reduced to 
proving (4.2) for $E_{+}=\beta$, $E_{-}=\alpha$.

Let $f\in C^{\infty}_{0}(\Bbb R)$ with $f=x$ on $[\alpha, \beta]$.  
Then $f(A)=A$, $f(A_{\infty})=QAQ$ and with $Q=1-(\varphi, 
\cdot)\varphi$, $\text{Tr}[f(A)-f(A_{\infty})]=(\varphi, A\varphi)$ (see the 
appendix).  Thus
$$
(\varphi, A\varphi)=\int\limits^{\alpha}_{-\infty} (-f'(\lambda))
\xi(\lambda)\, d\lambda +\int\limits^{\beta}_{\alpha} (-f'(\lambda))
\xi(\lambda)\, d\lambda +\int\limits^{\infty}_{\beta}(-f'(\lambda))
\xi(\lambda)\, d\lambda.
$$
Since $\xi(\lambda)=0$ on $(-\infty, \alpha)$, the first integral is 
zero.  Since $f'(\lambda)=1$ on $[\alpha, \beta]$, the second integral 
is $-\int\limits^{\beta}_{\alpha}\xi(\lambda)\, d\lambda$.  Since 
$f\equiv 0$ near infinity and $\xi(\lambda)\equiv 1$ on $(\beta, 
\infty)$, the third integral is $f(\beta)=\beta$. \qed
\enddemo

\proclaim{Corollary 4.2}  Let $H$ be a Jacobi matrix on $\ell^{2}(\Bbb 
Z^{\nu})$, that is, for a bounded function $V$ on $\Bbb Z^{\nu}$:
$$
(Hu)(n)=\sum_{|n-m|=1} u(m)+V(n)u(n), \qquad n\in\Bbb Z^{\nu}. \tag 4.4
$$
For $r\in\Bbb Z^{\nu}$, let $H^{D}_{r}$ be the operator on $L^{2}(\Bbb 
Z^{\nu}\backslash\{r\})$ given by \rom{(4.4)} with $u(r)=0$ boundary 
conditions.  Let $\xi(r, \lambda)$ be the spectral shift for the pair
$(H^{D}_{r}, H)$.  Then
$$\align
V(r) &= E_{-}+\int\limits^{E_{+}}_{E_{-}} [1-\xi(r, \lambda)]\, 
d\lambda \tag 4.5 \\
&= E_{+}-\int\limits^{E_{+}}_{E_{-}} \xi(r, \lambda)\, d\lambda \tag 4.6 \\
&= \frac12\, (E_{+}+E_{-})+\frac12 \int\limits^{E_{+}}_{E_{-}}
[1-2\xi(r, \lambda)]\, d\lambda \tag 4.7
\endalign
$$
for any $E_{-}\leq\inf\,\text{\rom{spec}}(H)$, 
$E_{+}\geq\sup\,\text{\rom{spec}}(H)$.
\endproclaim

\remark{Remark}  Only when $\nu=1$ does this have an interpretation as a 
formula using Dirichlet problems on the half-line.
\endremark

\bigpagebreak
\flushpar {\bf \S 5. Absolutely Continuous Spectrum}

We will also show that the $\xi(x, \lambda)$ function for a single fixed 
$x\in\Bbb R$ determines the absolutely continuous spectrum of a 
one-dimensional Schr\"odinger operator or Jacobi matrix.  We begin with a 
result that holds for a higher dimensional Jacobi matrix as well:

\proclaim{Proposition 5.1}  \rom{(i)} For an arbitrary Jacobi matrix, 
$H$, on $\Bbb Z^{\nu}$, $\operatornamewithlimits{\cup}\limits_
{j\in\Bbb Z^{\nu}}\{\lambda\in\Bbb R\mid 0<\xi(j, \lambda)<1\}$ is an 
essential support for the absolutely continuous spectrum of $H$.

\rom{(ii)} For a one-dimensional Schr\"odinger operator, 
$H=-\frac{d^2}{dx^2}+V$ \rom(with $V$ continuous and bounded from
below\rom), $\operatornamewithlimits{\cup}\limits_{x\in\Bbb Q}\{\lambda
\in\Bbb R\mid 0<\xi(x, \lambda)<1\}$ is an essential support for the 
absolutely continuous spectrum of $H$.
\endproclaim

\remark{Remark}  Recall that every absolutely continuous measure, 
$d\mu$, has the form $f(E)\,dE$.  $S\equiv\{E\mid f(E)\neq 0\}$ is 
called an essential support.  Any Borel set which differs from $S$ by 
sets of zero Lebesgue measure is also called an essential support. If 
$A$ is a self-adjoint operator on $\Cal H$ and $\varphi_{n}$, an 
orthonormal basis for $\Cal H$, and $d\mu_{n}$, the spectral measure for 
the pair, $A, \varphi_{n}$ (i.e., $(\varphi_{n}, e^{isA}\varphi_{n})=
\int\limits_{\Bbb R} e^{isE}\,d\mu_{n}(E)$) and if $d\mu^{\text{\rom{ac}}}_{n}$ 
is the absolutely continuous component of $d\mu_{n}$ with $S_n$ its 
essential support, then $\operatornamewithlimits{\cup}\limits_{n}S_n$ is the 
essential support of the absolutely continuous spectrum for $A$.
\endremark

\demo{Proof}  (i)  Let $g(j, z)$ be the diagonal Green's function 
$(\delta_{j}, (A-z)^{-1}\delta_{j})$ for $\text{Im}\, z\neq 0$.  Thus $g(j, 
z)=\int\limits_{\Bbb R} d\mu_{j}(E)(E-z)^{-1}$.  By general properties of 
Borel transforms of measures (see, e.g., [28,48]), for a.e.~$\lambda\in\Bbb R$, 
$\lim\limits_{\epsilon\downarrow 0} g(j, \lambda+i\epsilon)$ exists and 
is non-zero; and $S_j$, the essential support of $d\mu_{j, \text{\rom{ac}}}$, 
is given by
$$
S_{j}=\{\lambda\in\Bbb R\mid\text{Im}\,g(j, \lambda+i0)\neq 0\}.
$$
But if $g(j, \lambda+i0)\neq 0$, then $\text{Im}\,g(j, \lambda +i0)\neq 0$ 
is equivalent to $0<\text{Arg}(g(j, \lambda+i0))<\pi$, so, up to 
sets of measure $0$,
$$
S_{j}=\{\lambda\in\Bbb R\mid 0<\xi(j, \lambda)<1\}.
$$
Since $\{\delta_{j}\}_{j\in\Bbb Z^{\nu}}$ are an orthonormal basis for 
$\Bbb Z^{\nu}$, the result is proven.

(ii)  Let $\Cal H_{-1}$ be the minus one space in the scale of spaces 
associated to $H$ (see, e.g., [43]).  Then, $\delta_{x}$, the delta 
function supported at $x$ is in $\Cal H_{-1}$ and the diagonal Green's function 
$g(x, z)$ is just $(\delta_{x}, (H-z)^{-1}\delta_{x})$.  Since 
$\{\delta_{x}\}_{x\in\Bbb Q}$ are total in $\Cal H_{-1}$, the argument is 
essentially the same as in (i). \qed
\enddemo

In one dimension though, a single $x$ suffices:

\proclaim{Theorem 5.2}  For one-dimensional Schr\"odinger operators 
resp.~Jacobi matrices, $\{\lambda\in\Bbb R\mid 0<\xi(x, \lambda)<1\}$ 
is an essential support for the absolutely continuous measure for \rom{any } 
fixed $x\in\Bbb R$ resp.~$\Bbb Z$.
\endproclaim

\demo{Proof}  Consider the Schr\"odinger case first.  Let $m_{\pm}(x, 
z)$ be the Weyl $m$-functions (see, e.g., [48]) for $-\frac{d^2}{dx^2}+V$ 
and let $H_{\pm, x}$ be the Dirichlet operators on $L^{2}(x, \pm\infty)$.  
Then
$$
g(x, z)=-1\big/[m_{+}(x, z)+m_{-}(x, z)] \tag 5.1
$$
and $H_{\pm, x}$ is unitarily equivalent to multiplication by 
$\lambda$ on $L^{2}(\Bbb R; d\mu_{x, \pm})$, where $d\mu_{x, \pm}$ is a 
limit of the measures $m_{\pm}(x, \lambda+i\epsilon)\, d\lambda$ as 
$\epsilon\downarrow 0$. Thus, up to sets of measure zero:
$$\align
\{\lambda\in\Bbb R\mid 0<\xi(x, \lambda)<1\} &= \{\lambda\in\Bbb R
\mid\text{Im}\, g(x, \lambda+i0)\neq 0\} \\
&= \{\lambda\in\Bbb R\mid\text{Im}\, m_{+}(x, \lambda+i0)\neq 0\} \\
& \qquad \cup\{\lambda\in\Bbb R\mid\text{Im}\, m_{-}(x, \lambda+i0)\neq 0\} \\
&= S_{x, +}\cup S_{x, -}
\endalign
$$
with $S_{x, \pm}$ the essential support of the a.c.~part of 
$d\mu_{\pm, x}$.  Thus, $\{\lambda\in\Bbb R\mid 0<\xi(x, \lambda)<1\}$ is an 
essential support for the a.c.~spectrum of $H_{+, x}\oplus H_{-, x}$.  
But $H$ and $H_{+, x}\oplus H_{-, x}$ have resolvents differing by a 
rank one perturbation and so equivalent absolutely continuous spectrum 
by the theory of trace class perturbations [27,47].

The Jacobi case is similar but requires (5.1) to be replaced by
$$
g(j, z)=-1\big/[m_{+}(j, z)+m_{-}(j, z)+z-V(j)]. \qed
$$
\enddemo

These results are of particular interest because of their implications 
for a special kind of semi-continuity of the spectrum.  We begin by noting 
a lemma (that requires a preliminary definition).

\definition{Definition} Let $\{V_{n}\},V$ be continuous potentials on 
$\Bbb R$ (resp.~on $\Bbb Z$).  We say that $V_n$ converges to $V$ locally 
as $n\to\infty$ if and only if
\roster
\item"{(i)}"  $\inf\limits_{(n, x)\in\Bbb N\times\Bbb R}V_{n}(x)>-\infty$ 
($\Bbb R$ 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 
$\Cal H_{k}(A)(-\infty <k<\infty)$ be the usual scale of spaces associated 
to $A$ [43].  Let $\varphi\in\Cal H_{-1}(A)$.

Then $(\varphi, \cdot)\varphi$ defines a form bounded perturbation of 
$A$ with relative bound zero, so for any $\alpha\in\Bbb R$, we can define
$$
A_{\alpha}=A+\alpha(\varphi, \cdot)\varphi \tag A.1
$$
as a closed form on $\Cal H_{+1}(A)$ with an associated self-adjoint 
operator also denoted by $A_\alpha$.  For $\text{Im }z\neq 0$ define
$$
F(z)=(\varphi, (A-z)^{-1}\varphi), \qquad
F_{\alpha}(z)=(\varphi, (A_{\alpha}-z)^{-1}\varphi). \tag A.2
$$
By the second resolvent formula
$$
(A_{\alpha}-z)^{-1}=(A-z)^{-1}-\alpha((A_{\alpha}-\bar z)^{-1}\varphi,
\cdot)(A-z)^{-1}\varphi \tag A.3
$$
so taking expectations in $\varphi$ and solving for $F_\alpha$, we 
find
$$
F_{\alpha}(z)=F(z)\big/[1+\alpha F(z)] \tag A.4
$$
and then applying (A.3) to $\varphi$:
$$
(A_{\alpha}-z)^{-1}\varphi = [1+\alpha F(z)]^{-1}(A-z)^{-1}\varphi
\tag A.5
$$
so by (A.3) again,
$$
(A_{\alpha}-z)^{-1}=(A-z)^{-1}-\frac{\alpha}{1+\alpha F(z)}\,
((A-\bar z)^{-1}\varphi, \cdot)(A-z)^{-1}\varphi. \tag A.6
$$
In particular,
$$\align
\text{Tr}\bigl[(A-z)^{-1}-(A_{\alpha}-z)^{-1}\bigr] &= \frac{\alpha}
{[1+\alpha F(z)]}\, (\varphi, (A-z)^{-2}\varphi) \\
&= \frac{d}{dz} \ln(1+\alpha F(z)). \tag A.7
\endalign
$$

By (A.6), $\lim\limits_{\alpha\to\infty}\, (A_{\alpha}-z)^{-1}$ exists 
in norm.  Since $A_\alpha$ is a monotone increasing sequence of forms we 
can identify the limit, which we will call $A_{\infty}$, explicitly
[27,45].  If $\varphi\notin\Cal H$, then $A_{\infty}$ is the 
self-adjoint operator associated to the densely defined closed form 
$A$ restricted to $\{\eta\in\Cal H_{+1}(A)\mid (\varphi, \eta)=0\}$ and 
$\lim(A_{\alpha}-z)^{-1}=(A_{\infty}-z)^{-1}$.  If $\varphi\in\Cal H$, 
then one looks at the self-adjoint operator $A_\infty$ on $\Cal 
H(A_{\infty})\equiv \{\eta\in\Cal H\mid(\varphi, \eta)=0\}$ whose quadratic 
form is $A$ restricted to $\Cal H(A_{\infty})$.  Extend $(A_{\infty}-
z)^{-1}$ to all of $\Cal H$ by setting it to $0$ on $\Cal H(A_{\infty})^
{\perp}=\{c\varphi\mid c\in\Bbb C\}$.  Then $\lim\limits_{\alpha\to\infty}
(A_{\alpha}-z)^{-1}=(A_{\infty}-z)^{-1}$ still holds.  Convergence 
properties of $d\mu_{\alpha}$, the spectral measure for $\varphi$ 
associated to $A_{\alpha}$ (i.e., $F_{\alpha}(z)=\int\limits_{\Bbb R}
(x-z)^{-1}\,d\mu_{\alpha}(x)$) are studied in detail in [19].

Taking $\alpha$ to infinity in (A.6) and repeating the proof of (A.7), 
we get
$$
\text{Tr}[(A-z)^{-1}-(A_{\infty}-z)^{-1}]=\frac{d}{dz}\,\ln\,F(z).
\tag A.$7'$
$$

$F(z)$ is the Borel transform of a measure, so by general principles
([28,48]), there exist boundary values $F(\lambda+i0)$ for 
a.e.~$\lambda\in\Bbb R$ and $F(\lambda +i0)$ takes any given value $-
1/\beta$ on a set of measure zero.  Thus we can define
$$\hskip 1.5 truein
\xi_{\alpha}(\lambda) = \left\{\alignedat2 &\frac{1}{\pi}\, \text{Arg}
(1+\alpha F(\lambda+i0)), &&\qquad \alpha \text{ finite} \hskip 1.27truein
\text{(A.8)} \\  
&\frac{1}{\pi}\, \text{Arg}(F(\lambda+i0)), &&\qquad \alpha=\infty
\hskip 1.27truein\text{(A.$8'$)}\endalignedat
\right.
$$
for each $\alpha\in\Bbb R$ and a.e.~$\lambda\in\Bbb R$.  For $\alpha >0$
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.  F.G.~is indebted to the 
Department of Mathematics at Caltech for its hospitality and support 
during the summers of 1992 and 1993 where some of this work was done.
\endexample

\vskip 0.3in
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\enddocument