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\topmatter
\title A Trace Formula for Multidimensional Schr\"odinger Operators
\endtitle
\rightheadtext{A Trace Formula for Multidimensional Schr\"odinger
Operators}
\author F.~Gesztesy$^{1}$, H.~Holden$^{2}$, B.~Simon$^{3}$, and
Z.~Zhao$^{1}$
\endauthor
\leftheadtext{F.~Gesztesy, H.~Holden, B.~Simon, and Z.~Zhao}
\thanks $^{1}$ Department of Mathematics, University of Missouri,
Columbia, MO 65211. E-mail for F.G.: mathfg\linebreak\@mizzou1.missouri.edu
\,E-mail for Z.Z.: mathzz\@mizzou1.missouri.edu
\endthanks
\thanks $^{2}$ Department of Mathematical Sciences, The Norwegian
Institute of Technology, University of Trondheim, N-7034 Trondheim,
Norway. E-mail: holden\@imf.unit.no
\endthanks
\thanks $^{3}$ 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 J.~Amer.~Math.~Soc.}
\endthanks
\endtopmatter
\bigpagebreak
\document
\block
\tenrm{\smc{Abstract.}} We prove multidimensional analogs of the trace formula
obtained previously for one-dimensional Schr\"odinger operators. For
example, let $V$ be a continuous function on $[0, 1]^{\nu}\subset\Bbb
R^{\nu}$. For $A\subset\{1,\dots ,\nu\}$, let $-\Delta_{A}$ be the
Laplace operator on $[0, 1]^{\nu}$ with mixed Dirichlet-Neumann
boundary conditions
$$\alignat2
\varphi(x) &=0, &&\qquad x_{j}=0 \text{ or } x_{j}=1 \quad\text{for }
j\in A, \\
\frac{\partial\varphi}{\partial x_{j}}(x) &= 0, &&\qquad x_{j}=0
\text{ or } x_{j}=1 \quad\text{for } j\notin A.
\endalignat
$$
Let $|A|=$ number of points in $A$. Then we'll prove that
$$
\text{Tr}\biggl(\sum_{A\subset\{1,\dots ,\nu\}} (-1)^{|A|} e^{-t(-
\Delta_{A}+V)}\biggr)=1-t\langle V\rangle +o(t) \quad\text{as }t\downarrow 0
$$
with $\langle V\rangle$ the average of $V$ at the $2^{\nu}$ corners of
$[0, 1]^{\nu}$.
\endblock
\vskip 0.4in
\flushpar {\bf \S 1. Introduction}
This paper is devoted to extensions of the trace formula for the ODE
$-\frac{d^2}{dx^2}+V(x)$ to the corresponding PDE, $-\Delta+V(x)$.
The simplest of all the one-dimensional results is the trace formula
[1,6] for the periodic case. Suppose $V$ is a $C^1$ function on $\Bbb
R$ obeying $V(x+1)=V(x)$. Let $E_{1}0$ and the $-$ is used for $Q=D$ and $+$ for $Q=N$. (Here and in
the remainder of this paper $e^{-tH}(x, y)$, $t>0$ denotes the integral
kernel of the semigroup $e^{-tH}$.) From this it follows that
$\text{Tr}(e^{-tH}-e^{-tH^{D}_{x}})=\text{Tr}(e^{-tH^{N}_{x}}-e^{-tH})$
and so (5) becomes
$$
\text{Tr}\bigl(e^{-tH^{N}_{x}}-e^{-tH^{D}_{x}}\bigr)=1-tV(x)+o(t),
\quad x\in\Bbb R. \tag 6
$$
But it is easy to see that (6) for $V$ even about $x$ implies it for
arbitrary $V$ since the operators break up into direct sums (see Lemma
2.1 below).
It is (6) that we will generalize to $\nu$ dimensions. Explicitly,
given $A\subset\{1,\dots ,\nu\}$, let $H_{A; x}$ be defined as
follows: Let $B_{\alpha}$, $\alpha\subset\{1,\dots,\nu\}$ be the $2^\nu$
blocks obtained by removing the hyperplanes $P^{(x)}_{j}:=\{y\in
\Bbb R^{\nu}\mid y_{j}=x_{j}\}$ from $\Bbb R^{\nu}$, that is,
$B_{\alpha}=\{x\in\Bbb R^{\nu}\mid x_{i}>0\text{ if } i\in\alpha,\,
x_{i}<0\text{ if } i\notin\alpha\}$. $H_{A; x}$ is then
defined to be the operator on $\oplus L^{2}(B_{\alpha})$ with Dirichlet
boundary conditions on $\{P^{(x)}_{j}\}_{j\in A}$ and Neumann boundary
conditions on $\{P^{(x)}_{j}\}_{j\notin A}$.
Explicitly, for each $\alpha=1,\dots,2^{\nu}$, let $\Cal
D^{(A)}_{\alpha; x}$ be the set of functions, $\varphi$, on $B_{\alpha}$
which are $C^{\infty}$ on $B^{\text{int}}_{\alpha}$, with
derivatives continuous up to $\partial B_{\alpha}$ with bounded
support and which obey the boundary conditions:
$$\alignat2
\varphi(y) &=0, &&\qquad y\in P^{(x)}_{j} \quad\text{for } j\in A, \\
\frac{\partial\varphi}{\partial y_{j}}(y) &= 0, &&\qquad y\in P^{(x)}_{j}
\quad\text{for } j\notin A.
\endalignat
$$
Obviously, $\Cal D^{(A)}_{\alpha; x}$ is dense in $L^{2}(B_{\alpha})$.
Then $\varphi\mapsto -\triangle\varphi$ is essentially self-adjoint on
$\Cal D^{(A)}_{\alpha, x}$, $-\triangle_{A}$ is the direct sum
of these operators on $\oplus L^{2}(B_{\alpha})$, and $H_{A; x}=-
\triangle_{A}\dot{+}V$ as a form sum.
We will prove (see Theorem 4.1) that
$$
\text{Tr}\biggl(\sum_{A\subset\{1,\dots ,\nu\}}(-1)^{|A|}\exp (-tH_{A;
x})\biggr)=1-tV(x)+o(t), \quad x\in\Bbb R^{\nu}. \tag 7
$$
(Note that for $A=\{1,\dots,\nu\}$ resp.~$A=\emptyset$, $H_{A; x}$ has
exclusively Dirichlet resp.~Neumann boundary conditions on the
hyperplanes $P^{(x)}_{j}$, $1\leq j\leq\nu$.)
The paper is laid out as follows. In \S 2, we'll use the method of
images to reduce (7) to the study of integrals of the form
$$
2^{\nu}\int\limits_{\Bbb R^{\nu}} \exp (-tH)(y, -y)\, d^{\nu}y, \tag 8
$$
where $H=-\triangle+V$ is a Schr\"odinger operator in $L^{2}(\Bbb R^{\nu})$
without any boundary conditions. In \S 3, we introduce a Gaussian process
that provides a concise Feynman-Kac type formula for integrals of the form
(8), and we'll prove (7) in \S 4. We'll discuss the periodic version of (7)
in \S 5, then prove an abelianized version of a recent conjecture of Lax [5]
that motivated our work. Finally, in \S 6, we'll make a few remarks
on the issue of going beyond Abelian sums in the periodic case.
\bigpagebreak
\flushpar {\bf \S 2. The Method of Images}
We begin with some small arguments to simplify notation and some later
details. First, without loss, we'll suppose $x=0$, that is, all
boundary conditions are on planes through $0$ and the final formula is
for $V(0)$. Let then $P_j$ be the coordinate plane $\{x\in\Bbb R^{\nu}\mid
x_{j}=0\}$ and let $H_{A}\equiv H_{A; 0}$. Let $\pi_{j}$ be the
reflection in $P_j$, that is,
$$\alignat2
(\pi_{j}x)_{i} &=x_{i}, &&\qquad i\neq j \\
&=-x_{j}, &&\qquad i=j.
\endalignat
$$
Call $V$ symmetric if and only if $V\circ\pi_{j}=V$ for all $j$. For
$\alpha\subset\{1, \dots ,\nu\}$, let $B_{\alpha}$ be the blocks introduced
previously. Let $V_\alpha$ be that symmetric function with $V_{\alpha}=V$ on
$B_\alpha$. Because of symmetry of $V_{\alpha}$ for each $A$, $\exp(-
t(-\Delta_{A}+V_{\alpha}))$ is a direct sum of $2^{\nu}$ pieces (acting on
the different $L^{2}(B_{\beta})$) and each of the pieces is unitarily
equivalent to a single operator, $P_{A; \alpha, t}$. On the other
hand, $\exp(-tH_{A})$ is also a direct sum, clearly unitarily
equivalent to $\operatornamewithlimits{\oplus}
\limits_{\alpha\subset\{1,\dots , \nu\}} P_{A; \alpha, t}$. It
follows that:
\proclaim{Lemma 2.1} To prove \rom{(7)}, it suffices to suppose $x=0$
and that $V$ is symmetric.
\endproclaim
So, henceforth, we can suppose that $V$ is symmetric, which we do in
Lemma 2.3 and Theorems 2.4/2.5 below.
We are interested in writing the heat kernel for $H_A$ using the method of
images. Some group theoretic notation will be useful. $\Cal P_{\nu}$
is the set of subsets of $\{1, \dots, \nu\}$ which forms a group under
$(A, B)\mapsto A\bigtriangleup B$, the symmetric difference. The identity
is $\emptyset$, the empty set. As a finite abelian group, $\Cal
P_{\nu}$ is its own dual. The character $\chi_{A}$ associated to
$A\in\Cal P_{\nu}$ acts by
$$
\chi_{A}(B)=(-1)^{|A\cap B|}. \tag 9
$$
In particular, orthogonality of characters implies
\proclaim{Lemma 2.2} $$\sum_{B\in\Cal P_{\nu}}
\chi_{A}(B)\chi_{C}(B)=2^{\nu}\delta_{AC}.$$
\endproclaim
Given $B\in\Cal P_{\nu}$, define the reflection $R_{B}$ acting on
$\Bbb R^{\nu}$ by
$$
R_{B}=\prod\limits_{j\in B} \pi_{j}.
$$
The method of images formula then says
\proclaim{Lemma 2.3} If $x, y$ are in the same orthant, then
$$
\exp(-tH_{A})(x, y)=\sum_{B\in\Cal P_{\nu}}\chi_{B}(A)\exp(-tH)(x,
R_{B}y)
$$
and the integral kernel is zero if $x, y$ are in different orthants.
\endproclaim
This immediately yields:
\proclaim{Theorem 2.4} Let $C_{t}=\sum\limits_{A\in\Cal P_{\nu}}
(-1)^{|A|} \exp(-tH_{A})$. Then the integral kernel of $C_{t}$ is
$$
C_{t}(x, y)=\cases 2^{\nu}\exp(-tH)(x, -y), & \,x, y \text{\rom{ in the same
orthant}} \\
0, & \,x, y \text{\rom{ in different orthants}}.
\endcases
$$
\endproclaim
\demo{Proof} Let $B_{0}=\{1, \dots, \nu\}$ so $R_{B_0}y=-y$ and
$\chi_{B_0}(A)=(-1)^{|A|}$ so if $x, y$ lie in the same orthant:
$$\alignat2
C_{t}(x, y) &=\sum_{A\in\Cal P_{\nu}}\chi_{B_0}(A)\exp(-tH_{A})(x, y)\\
&=\sum\Sb A\in\Cal P_{\nu} \\ B\in\Cal P_{\nu} \endSb \chi_{B_0}(A)
\chi_{B}(A)\exp(-tH_{A})(x, R_{B}y)&&\qquad \text{(by Lemma 2.3)} \\
&=2^{\nu}\sum_{B\in\Cal P_{\nu}} \delta_{B_{0}B} \exp(-tH)(x, R_{B}y) &&\qquad
\text{(by Lemma 2.2)} \\
&= 2^{\nu} \exp(-tH)(x, -y). \qed
\endalignat
$$
\enddemo
\proclaim{Theorem 2.5} Let $V$ be bounded below and $t>0$. Then the
operator $C_t$ of Theorem \rom{2.4} is a trace class opertor in
$L^{2}(\Bbb R^{\nu})$ and
$$
\text{\rom{Tr}}(C_{t})=2^{\nu}\int\limits_{\Bbb R^{\nu}} \exp(-tH)(x, -x)\,
d^{\nu}x.
$$
\endproclaim
\demo{Proof} Let $S_{t}=\exp(-tH)$. For $\alpha\in\tilde{\Bbb
Z}^{\nu}\equiv\Bbb Z^{\nu}+(\frac12, \dots, \frac12)$ let
$\chi_{\alpha}$ be the characteristic function of $\{x\mid |x_{i}-
\alpha_{i}| <\frac12,\text{ all }i\}$, and let $P_\alpha$ be the projection
which is multiplication by $\chi_\alpha$. Let $\tilde{\Bbb
Z}^{\nu}_{+}=\{\alpha\in\tilde{\Bbb Z}^{\nu}\mid \alpha_{i}>0\}$.
$C_t$ is a direct sum of $2^\nu$ operators, each unitarily equivalent
to $\tilde C_{t}:= C_{t} \restriction\{x\mid x_{i}>0\}$. Let
$R_{B_0}$ be the reflection $x\to -x$. Then by Theorem 2.4:
$$
\tilde C_{t}=2^{\nu}\sum_{\alpha, \beta\in\tilde{\Bbb Z}^{\nu}_{+}}
P_{\alpha} R_{B_0} S_{t} P_{\beta}.
$$
By the lemma below, $P_{\alpha}R_{B_0}S_{t}P_{\beta}$ is trace class
with trace norm bound by \linebreak$C_{1}\exp(-C_{2}|\alpha+\beta|)$
and trace given by the integral of the diagonal integral kernel. Since
$\sum\limits_{\alpha, \beta\in\tilde{\Bbb Z}^{\nu}_{+}} \exp(-C_{2}
|\alpha +\beta|)<\infty$, the result follows. \qed
\enddemo
\proclaim{Lemma 2.6} In the notation of the last proof,
$P_{\alpha} R_{B_0} S_{t} P_{\beta}$ is trace class with trace norm
bounded by $C_{1}\exp(-C_{2}|\alpha+\beta|)$ and trace given by the
integral of the diagonal of the \rom(continuous\rom) integral kernel.
\endproclaim
\demo{Proof} $P_{\alpha} R_{B_0} S_{t} P_{\beta}=\sum
\limits_{\gamma\in\tilde{\Bbb Z}^{\nu}} R_{B_0} (P_{-\alpha} S_{t/2}
P_{\gamma})(P_{\gamma} S_{t/2} P_{\beta})$. By a standard estimate
(see, e.g., [9]):
$$
|S_{u}(x, y)|\leq C_{1, u}\exp(-C_{2, u}|x-y|)
$$
(of course one can even have $|x-y|^{2}$ but we don't need that), so
by integrating the square of the integral kernel:
$$
\|P_{\alpha} S_{t/2} P_{\gamma}\|_{2} \leq C_{3}\exp(-C_{2}|\alpha-
\gamma|),
$$
where $\|\,\cdot\,\|_{2}$ is the Hilbert-Schmidt norm. Summing over $\gamma$,
we obtain the trace class result and bound since
$$
\sum_{\gamma\in\tilde{\Bbb Z}^{\nu}} \exp(-C_{2}|\alpha+\gamma|)
\exp(-C_{2}|\beta-\gamma|) \leq C_{5} \exp(-C_{4}|\alpha+\beta|).
$$
Since the trace of a product of Hilbert-Schmidt operators is given by
the integral of the diagonal integral kernel, we obtain the trace
result. \qed
\enddemo
\bigpagebreak
\flushpar {\bf \S 3. A Gaussian Process}
In this section, we present a Feynman-Kac type formula for
$\text{Tr}(C_{t})$ where $C_t$ is the operator of Theorem 2.4. If $V$
is bounded, (7) is an immediate consequence of this formula. For $V$
unbounded (from above) at infinity, we will need an additional
estimate on the Gaussian process, $\bold L(t)$, used in this formula and
that estimate appears at the end of this section. (We shall employ
the notation used in [8], i.e., $E(f)=\int\limits_{\Omega}f\,d\mu$,
$E(A)=\int\limits_{A}d\mu=\mu(A)$, $E(f; A)=\int\limits_{A}f\,d\mu$,
etc., where $(\Omega, \Cal F, \mu)$ denotes a probability space,
$A\in\Cal F$, $f:\Omega\to\Bbb R$ is $\Cal F$-measurable.)
All Gaussian processes considered in this paper have mean zero and we
will suppose that without explicitly saying it each time. Recall [8]
that the Brownian bridge $\{\alpha(s)\}_{0\leq s\leq 1}$ is the
Gaussian process with covariance:
$$\align
E_{\alpha}(\alpha (s)\alpha (t)) &=\min(s, t)(1-\max(s, t)) \\
&=\frac12 (s+t-|s-t|)-st. \tag 10
\endalign
$$
If $b(s)$ is Brownian motion, then $\alpha (s)=b(s)-sb(1)$ is an
explicit realization of the Brownian bridge. The $\nu$-dimensional
Brownian bridge is $\nu$ independent copies of $\alpha (s)$ thought of
as a vector valued process and one still has for the $\nu$-dimensional
objects
$$
\boldsymbol\alpha (s)=\bold b (s)-s \bold b (1). \tag 11
$$
Let $\bold g_{\bold x, \bold y} (s)=s \bold x +(1-s) \bold y$ be the straight
line from $\bold x$ to $\bold y$. Then, [8] shows that for any $V$
bounded from below (and locally bounded from above, say), if $H=-
\Delta+V$ in $L^{2}(\Bbb R^{\nu})$,
$$
\exp(-tH)(\bold x, \bold y)=\exp(t\Delta)(\bold x, \bold y)E_{\alpha}\biggl(
\exp\biggl(-\int\limits^{t}_{0} V\biggl(\bold g_{\bold x, \bold y}
\biggl(\frac{s}{t}\biggr)+\sqrt{2t}\,\boldsymbol\alpha
\biggl(\frac{s}{t}\biggr) \biggr)\, ds \biggr)\biggr). \tag 12
$$
We have $\sqrt{2t}$ rather than the $\sqrt{t}$ on pg.~54 of [8]
because we use $-\Delta$ rather than $-\frac12\Delta$. Plugging (12)
into Theorem 2.5 we find that
$$
\text{Tr}(C_{t})=\int\limits_{\Bbb R^{\nu}} d^{\nu}x N_{t}(\bold x)
E_{\alpha}\biggl(\exp \biggl(-\int\limits^{t}_{0} V
\biggl(\bold g_{\bold x, - \bold x} \biggl(\frac{s}{t}\biggr) +\sqrt{2t}\,
\boldsymbol\alpha\biggl(\frac{s}{t}\biggr)\biggr)\, ds \biggr)\biggr),
$$
where $N_{t}(\bold x)=2^{\nu}\exp(t\Delta)(\bold x, -\bold x)=2^{\nu}
\exp(t\Delta)(2\bold x, 0)\equiv\prod\limits^{\nu}_{i=1} \tilde N_{t}(x_{i})$.
Notice that $\tilde{N_{t}}(x_{i})$ is a Gaussian probability distribution
with variance $\langle x^{2}_{t,i}\rangle=\frac{2t}{(2)^{2}}=
\frac{t}{2}=\frac14 (\sqrt{2t})^{2}$, so if we let $x_{0,i}$ be a Gaussian
variable of variance $\langle x^{2}_{0,i}\rangle =\frac14$, then
$x_{t,i}=\sqrt{2t}\,x_{0,i}$. Note that $g_{\bold x, -\bold x}(\frac{s}{t})=
\sqrt{2t}\,\bold x_{0}(2(\frac{s}{t})-1)$.
This suggests we define a new process
$$
\bold L(s)=\bold x_{0} (2s-1)+\boldsymbol\alpha(s), \qquad 0\leq s\leq 1,
\tag 13
$$
where the components of $\bold x_0$ are independent Gaussian variables with
$\langle x^{2}_{0, i}\rangle=\frac14$ and independent of $\boldsymbol\alpha$
so
$$\align
E(\bold L_{i}(s) \bold L_{j}(w))
&=E_{\alpha}(\boldsymbol\alpha_{i} (s)\boldsymbol\alpha_{j} (w))
+\frac{\delta_{ij}}{4} (2s-1)(2w-1) \\
&=\delta_{ij}\biggl[\frac14 - \frac12 |s-w|\biggr]. \tag 14
\endalign
$$
We have thus proven that:
\proclaim{Theorem 3.1} Let $\bold L$ be the Gaussian process with
covariance \rom{(14)}. Then \rom(with $C_t$ given by Theorem
\rom{2.4)}:
$$
\text{\rom{Tr}}(C_{t})=E\biggl(\exp\biggl(-\int\limits^{t}_{0}
V\biggl(\sqrt{2t}\, \bold L\biggl(\frac{s}{t}\biggr)\biggr)\, ds \biggr)
\biggr).
$$
\endproclaim
We need the following estimate on $\bold L$:
\proclaim{Theorem 3.2} $E\bigl(\bigl\{\bigl[\sup\limits_{0\leq s\leq 1}
|\bold L(s)|\bigr]\geq a\bigr\}\bigr)\leq C_{1}\exp(-C_{2}a^{2})$
for some $C_{1}, C_{2} >0$.
\endproclaim
\demo{Proof} By the realizations (13) and (11), $\bold L(s)=\bold b(s)-
s\bold b(1)+\bold x_{0}(2s-1)$ so
$$
\sup\limits_{0\leq s\leq 1} |\bold L(s)| \leq |\bold x_{0}| + |\bold b(1)| +
\sup\limits_{0\leq s\leq 1} |\bold b(s)| \tag 15
$$
and for the $\sup\limits_{0\leq s\leq 1} |\bold L(s)|$ to be larger than $a$,
one or more of the three terms on the right side of (15) must be larger than
$a/3$. Each has a Gaussian bound since $\bold x_0$ and $\bold b(1)$ are
Gaussian and $\sup\limits_{0\leq s \leq 1}|\bold b(s)|$ has a Levy inequality
estimate (see [8], pp.~64 ff). \qed
\enddemo
\remark{Remarks} 1. Each component of $\bold L(t)$ is an independent
copy of the one-dimensional $L(t)$. $L(t)$ is intimately related to
the xi process, $\omega$, we introduced in [2]; namely
$$\alignat2
L(t) &= \omega(t), &&\qquad t\leq T_{\omega} \\
&= -\omega(t), &&\qquad t\geq T_{\omega},
\endalignat
$$
where $T_\omega$ is the first time that $\omega(t)=0$, that is, $\omega$
and $L$ are related by reflection at a first hitting time. Theorem 3.2 is
thus another proof of the estimate we proved on the xi process in [2].
2. The covariance (14) associated with $L(t)$ is just the zero energy
Green's function for $-\frac{d^2}{dx^2}$ on $L^{2}([0, 1])$ with
antiperiodic boundary conditions, just as (10) is the Dirichlet Green's
function. Notice that $L(1)=-L(0)$ is related to the antiperiodicity.
\endremark
\bigpagebreak
\flushpar {\bf \S 4. The Main Result in Unbounded Space}
Given Theorems 2.5, 3.1, and 3.2, the proof of (7) is easy following
the methods of [2,8]:
\proclaim{Theorem 4.1} Let $V$ be bounded from below and continuous on
$\Bbb R^{\nu}$. Let $C_{t}=\sum\limits_{A\in\Cal P_{\nu}}
(-1)^{|A|}\mathbreak\exp(-tH_{A})$, $t>0$. Then $C_t$ is a trace
class operator in $L^{2}(\Bbb R^{\nu})$ and
$$
\text{\rom{Tr}}(C_{t})=1-tV(0)+o(t)\qquad\text{as } t\downarrow 0.
$$
\endproclaim
\demo{Proof} As noted in Lemma 2.1, we can suppose that $V$ is symmetric.
By Theorems 2.5 and 3.1:
$$\align
\text{Tr}(C_{t}) &= E\biggl(\exp\biggl(-\int\limits^{t}_{0}V\biggl(
\sqrt{2t}\, L \biggl(\frac{s}{t}\biggr)\biggr) \, ds \biggr)\biggr) \\
&=T_{1}(V)+T_{2}(V),
\endalign
$$
where
$$
T_{i}(V)=E\biggl(\chi_{i,t}(L)\exp\biggl(-\int\limits^{t}_{0} V
\biggl(\sqrt{2t}\, L\biggl(\frac{s}{t}\biggr)\biggr)\, ds \biggr)\biggr)
$$
and $\chi_{1,t}$ is the characteristic function of
$\{L\mid\sup\limits_{0\leq s\leq 1} |L(s)|0$ (i.e., $V\chi_{R}\in M_{2-\alpha, 1}$ for all $R>0$,
where $\chi_{R}$ denotes the characteristic function of the ball
$\{x\in\Bbb R^{\nu}\mid |x|\leq R\}$).
\item"{(c)}" 0 is a point of Lebesgue continuity for $V$ for averaging
over balls shrinking to zero.
\endroster
\endremark
\medpagebreak
Just as we could write the one-dimensional trace formula as either (5)
or (6), in low dimension one can use the method of images to write $C_t$ in
alternate ways that avoid mixed boundary conditions. For example, in
two dimensions, let $H^D$ be what we called $H_{A=\{1, 2\}}$ resp.
$H^{N}=H_{A=\emptyset}$ correspond to Dirichlet resp.~Neumann
boundary conditions on both axes. Then by the method of images (i.e.,
Lemma 2.3):
$$
e^{-tH^{D}}(x, x)+e^{-tH^{N}}(x, x)=2e^{-tH}(x, x)+2e^{-tH}(x, -x),
\qquad x\in\Bbb R^{2}
$$
so we have
\proclaim{Proposition 4.2} In two dimensions
$$
\text{\rom{Tr}}(C_{t})=2\int\limits_{\Bbb R^2}\bigl[e^{-tH^{D}}+
e^{-tH^{N}}-2e^{-tH}\bigr] (x, x)\, d^{2}x.
$$
\endproclaim
\remark{Remark} Note we have not stated in the proposition that $[\dots]$
in the last integral is trace class because it is not in general. For
example, if $V=0$, it is not even Hilbert-Schmidt because of the
contribution of the integral kernel $-2e^{-tH}(x, y)$ with
$x=(u_{1}$, $v_{1})$, $y=(-u_{2}$, $v_{2})$, $00$.
That $B=O(t^{3/2-\epsilon})$ follows by noting first that
$|f(y)|\leq Cy^{2}$ on $[-1, 1]$ and that $|g(\bold w_{t}, t)|\leq Ct$
uniformly in $\bold w$ so $|f(g)-f(\tilde g)|\leq Ct^{2}$ uniformly in
$\bold w$. On the other hand, if $\text{dist}(x, \partial [0, 1]^{2})
\geq t^{1/2-\epsilon}$, $\text{Prob}(x+\sqrt{2t}\, \alpha (\frac{s}{t})
\notin [0, 1]^{2}\text{ for some } s)\leq\exp(-C/t^{2\epsilon})$, so,
since $f(g)=f(\tilde g)$ if $\bold w_{t}(s)\in [0, 1]^{2}$ for all $s$,
we have
$$
E(|f(g)-f(\tilde g)|)\leq O(\exp(-t^{-2\epsilon}))+Ct^{2}t^{1/2-
\epsilon}
$$
and we conclude that $B=O(t^{3/2-\epsilon})$ as required.
$A=0$ because of a cancellation. Let $\rho_{t}(x)$ be the probability
density of $\bold w_{t}(s)$ where $s$ is uniformly distributed in $[0, 1]$.
Then
$$
A=-(4\pi)^{-1} \int\limits_{\Bbb R^{2}}\rho_{t}(x) [V(x)-\tilde V(x)]\,
d^{2}x. \tag 21
$$
For each $\alpha\in\Bbb Z^{2}$, let $\square_{\alpha} $ be the square
centered at $(\frac12, \frac12)+\alpha$. There is a symmetry
$S_{\alpha}:\square_{\alpha}\to\square_{-\alpha}$ so that $V\circ
S_{\alpha}=\tilde V$ and $\rho_{t}\circ S_{\alpha}=\rho_{t}$ so that the
contribution of $V$ over $\square_{\alpha}$ in (21) cancels the
contribution of $\tilde V$ over $\square_{-\alpha}$. \qed
\enddemo
A different kind of a two-dimensional trace formula for $V(x)$ by
comparing heat kernels for $H=-\triangle +V$ and $H_{o}=-\triangle$ with
Dirichlet boundary conditions on a rectangular box was recently studied in [7].
\bigpagebreak
\flushpar {\bf \S 6. Sums Without Abelian Summation}
An interesting issue on which we haven't much to report is the extent
to which a formula like (2) holds. We note:
\proclaim{Theorem 6.1} In the context of Theorem \rom{5.1}, let
$\{E_{n}\}^{\infty}_{n=0}$ be a listing of all the eigenvalues of
$\{H_{A}\mid |A| \text{ is even}\}$ and $\{\tilde
E_{n}\}^{\infty}_{n=1}$ for $\{H_{A}\mid |A| \text{ is odd}\}$ ordered so
$E_{n}\leq E_{n+1};\tilde E_{n}\leq\tilde E_{n+1}$. Suppose that
$$
\sum^{\infty}_{n=1} |E_{n}-\tilde E_{n}| < \infty. \tag 22
$$
Then
$$
\langle V\rangle =E_{0}+\sum^{\infty}_{n=1} (E_{n}-\tilde E_{n}).
$$
\endproclaim
\remark{Remark} Note that the counting of $E_n$ starts at 0, but for
$\tilde E_n$ at 1.
\endremark
\demo{Proof} Theorem 5.1 says that
$$
\langle V\rangle =\lim\limits_{t\downarrow 0}\, \biggl[t^{-1}(1-e^{-tE_{0}})+
\sum^{\infty}_{n=1} t^{-1} (e^{-t\tilde E_{n}}-e^{-tE_{n}})\biggr].
$$
If $0\leq t\leq 1$, then
$$
|e^{-ta}-e^{-tb}| \leq [e^{-t\min(a, b)}+1] |a-b|
$$
and
$$
\lim\limits_{t\downarrow 0}\, t^{-1} (e^{-ta}-e^{-tb}) = b-a
$$
so the result follows by dominated convergence. \qed
\enddemo
When $V=0$, it is easy to see that $E_{n}=\tilde E_{n}$, $n\in\Bbb N$ so
(22) holds. It remains to be seen if one can prove it for sufficiently
smooth $V$'s.
\vskip 0.5in
\example{Acknowledgments} We would like to thank Peter Lax for
telling us of his work prior to publication. F.G.~is indebted to
M.~Aschbacher and G.~Neugebauer for the hospitality at Caltech where
some of this work was done. F.G.~and H.H.~were also supported in part by
the Norwegian Research Council for Science and the Humanities (NAVF).
\endexample
\vskip 0.5in
\Refs
\endRefs
\bigpagebreak
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operator}, Arch.~Rat.~Mech.~Anal.~ {\bf 59} (1975), 293--309.
\item{[2]} F.~Gesztesy, H.~Holden, B.~Simon, and Z.~Zhao, {\it Higher
order trace relations for Schr\"o-dinger operators}, to be submitted
to Commun.~Pure Appl.~Math.
\item{[3]} F.~Gesztesy and B.~Simon, {\it The xi function}, to be submitted
to Ann.~Math.
\item{[4]} H.~Hochstadt, {\it On the determination of a Hill's
equation from its spectrum}, Arch.~Rat. Mech.~Anal. {\bf 19} (1965),
353--362.
\item{[5]} P.~Lax, {\it The trace formula for the Schr\"odinger
operator}, to appear in Commun.~Pure Appl.~Math.
\item{[6]} H.P.~McKean and P.~van Moerbeke, {\it The spectrum of Hill's
equation}, Invent.~Math. {\bf 30} (1975), 217--274.
\item{[7]} V.G.~Papanicolaou, {\it Trace formulas and the behavior of
large eigenvalues}, to appear in SIAM J.~Math.~Anal.
\item{[8]} B.~Simon, {\it Functional Integration and Quantum Physics},
Academic Press, New York, 1979.
\item{[9]} B.~Simon, {\it Schr\"odinger semigroups},
Bull.~Amer.~Math.~Soc. {\bf 7} (1982), 447--526.
\enddocument