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%References
\def\borg{1} % Borg
\def\dgs {2} % del-Rio-Gesztesy-Simon
\def\gsun {3} % Gesztesy-Simon, TAMS
\def\gsac {4} % Gesztesy-Simon, ac paper
\def\gsmf {5} % Gesztesy-Simon, m-function paper
\def\gsds {6} % Gesztesy-Simon, ds paper
\def\levin {7} % Levin
\def\lev {8} % Levitan 68
\def\levbook {9} % Levitan book
\def\lg {10} % Levitan-Gasymov
\def\ls {11} % Levitan-Sargsjan
\def\mar {12} % Marchenko
\def\piv {13} % Pivovarchik
\def\simon {14} % Simon
\def\tit {15} % Titchmarsh
\topmatter
\title On the Determination of a Potential from Three Spectra
\endtitle
\author Fritz Gesztesy$^1$ and Barry Simon$^2$
\endauthor
\leftheadtext{F.~Gesztesy and B.~Simon}
\thanks$^1$ Department of Mathematics, University of Missouri,
Columbia, MO~65211, USA. E-mail: fritz\@\linebreak
math.missouri.edu
\endthanks
\thanks
Partially supported by the National Science Foundation under
Grant No.~DMS-9623121.
\endthanks
\thanks$^2$ Division of Physics, Mathematics, and Astronomy,
California Institute of Technology, Pasadena, CA~91125, USA.
E-mail: bsimon\@caltech.edu
\endthanks
\thanks To appear in the Birman Birthday Volume in
{\it{Advances in Mathematical Sciences}}, V.~Buslaev and
M.~Solomyak (eds.), Amer.~Math.~Soc., Providence, RI.
\endthanks
\date August 19, 1997
\enddate
\dedicatory
Dedicated to M.S.~Birman on the occasion of his
seventieth birthday
\enddedicatory
\keywords Inverse spectral theory, Schr\"odinger operators,
Weyl-Titchmarsh $m$-functions
\endkeywords
\subjclass Primary 34A55, 34B20; Secondary 34L05, 34L40
\endsubjclass
\abstract We prove that under suitable circumstances, the spectra
of a Schr\"odinger operator on the three intervals $[0,1]$, $[0,a]$,
and $[a,1]$ for some $a\in (0,1)$ uniquely determine the potential
$q$ on $[0,1]$.
\endabstract
\endtopmatter
\document
\vskip 0.1in
\flushpar{\bf \S 1. Introduction}
\vskip 0.1in
This is a paper in our series [\dgs,\gsac,\gsmf,\gsds]
on the use of Weyl-Titchmarsh $m$-function methods
to obtain information on what spectral information
uniquely determines the potential $q$
in a one-dimensional Schr\"odinger operator
$-\frac{d^2}{dx^2} + q$. Typical of our results is:
\proclaim{Theorem 1} Fix $c,d\in \Bbb R$ with $c0$, $k=1,\dots,N-1$. Let $A^{[i,j]}$ be the submatrix
of $A$ obtained by keeping rows and columns $i, i+1, \dots, j-1,
j$. In [\gsmf] we considered to what extent $A$ is determined by
$g(z,k)$, the $kk$ matrix element of $(A-z)^{-1}$ (for all $z\in
\Bbb C\backslash\text{spec}(A)$). We found that generically
there were $\binom{N-1}{k-1}$ possible $A$'s consistent with
a given $g(z,k)$.
The proof of this fact depends on the argument that looked at
the eigenvalues of $A^{[1, k-1]}$ and $A^{[k+1, N]}$. The function
$g(z,k)$ determined the union of these sets. Then $\binom{N-1}{k-1}$
possible values depended on the choice of which were actually
eigenvalues of $A^{[1, k-1]}$ and which of $A^{[k+1, N]}$. If one
a priori knows which are which (the hypothesis of Theorem~1), one
has uniqueness.
The non-generic case in [\gsmf] occurs precisely when $A^{[1,k]}$
and $A^{[k+1,N]}$ share an eigenvalue, in which case there is a
manifold of possible $A$'s consistent with $g(z,k)$.
In a sense, Theorem~1 can be thought of as a continuum analog
of a part of the result in [\gsmf].
We actually prove a more general result than Theorem~1. Let $h_c,
h_d\in\Bbb R \cup \{\infty\}$. We let $H(c,d; h_c, h_d; q)$ be
the operator $-\frac{d^2}{dx^2}+q$ on $L^2((c,d))$ with boundary
conditions
$$
u'(c) + h_c u(c)= 0, \quad u'(d) + h_d u(d)=0,
$$
where $h_{x_0} =\infty$ is a shorthand notation for the Dirichlet
boundary condition at $x=x_0$ (i.e., $u(x_0)=0$). Let
$S(c,d; h_c, h_d; q)$ be the set of eigenvalues (i.e., the
spectrum) of $H(c,d; h_c, h_d; q)$.
We will prove
\proclaim{Theorem 2} Fix $a\in (0,1)$ and $h_0, h_1, h_a \in
\Bbb R\cup \{\infty\}$. Suppose $q_1, q_2 \in L^1 ((0,1))$
are real-valued and
\roster
\item"\rom{(i)}" $S(0,1; h_0, h_1; q_1) =S(0,1; h_0, h_1; q_2)$,
$S(0,a; h_0, h_a; q_1)=S(0,a; h_0, h_a; q_2)$, and
$S(a,1; h_a, h_1; q_1)=S(a,1; h_a, h_1; q_2)$.
\item"\rom{(ii)}" The sets $S(0,1; h_0, h_1; q_1)$,
$S(0,a; h_0, h_a; q_k)$, and $S(a,1; h_a, h_1; q_k)$ are
pairwise disjoint.
\endroster
Then $q_1 = q_2$ a.e.~on $[0,1]$.
\endproclaim
\remark{Remark} The proof actually shows that not only is $q$
determined by $S(0,1)$, $S(0,a;h_a)$, and $S(a,1;h_a)$, but so
are $h_0$ and $h_1$.
\endremark
\vskip 0.1in
The structure of this paper is as follows: In Section~2, we prove
several results which illustrate when Green's functions are
determined by zeros, poles, and residues. In Section~3, we prove
Theorem~2 when $h_a =\infty$ (including Theorem~1); and in
Section~4, we prove Theorem~2 when $|h_a|< \infty$. In Section~5,
we discuss the case where condition (ii) fails. In Section~6, we
consider some cases where $q$ is defined on all of $\Bbb R$.
\vskip 0.1in
It is a great pleasure to dedicate this paper as a seventieth
birthday present to M.S.~Birman, whose work has long
inspired us. In our use of Green's functions and analytic
function theory, the reader will see echoes of his influence.
\vskip 0.1in
We thank V.~Pivovarchik for sending us his manuscript [\piv] prior
to publication. F.G.~is indebted to A.~S.~Kechris and C.~W.~Peck
for a kind invitation to Caltech for a month during the summer of
1997. The extraordinary hospitality and support by the Department
of Mathematics at Caltech are gratefully acknowledged. B.S.~would
like to thank M.~Ben-Artzi for the hospitality of Hebrew
University where some of this work was done.
\vskip 0.3in
\flushpar{\bf \S 2. Some Uniqueness Theorems of Meromorphic
Herglotz Functions}
\vskip 0.1in
One could prove the basic result of this paper using the theorems
in [\dgs,\gsds] on the determination of an entire function by its
values on a set of suitable density. Instead we will use some
alternative theorems that allow ready extension to $q$'s on all
of $\Bbb R$, a typical one being
\proclaim{Theorem 2.1} Let $00
\quad \text{\rom{for }} z\in\Bbb C \setminus \Bbb R \tag 2.1
$$
and hence a Herglotz function. Moreover, any meromorphic function
$f(z)$ satisfying {\rom{(2.1)}}
with zeros precisely at $\{z_j\}_{j=1}^\infty$ and poles precisely
at $\{w_j\}_{j=1}^\infty$ is a positive multiple of $g(z)$.
\endproclaim
\remark{Remarks} 1. Theorems of this genre can be found in
Levin [\levin].
2. This is a variant of the standard theorem on the convergence
of alternating series.
3. One can easily accommodate situations where there are also
zeros and poles alternating towards $-\infty$.
4. Any meromorphic Herglotz function (i.e., any meromorphic
function satisfying (2.1)) can be seen to satisfy $f'(z)>0$ away
from its polar singularities, so its zeros and poles are simple,
its zeros and poles alternate, and residues at poles are negative.
Thus Theorem~2.1 describes all meromorphic Herglotz functions
which are positive on $(-\infty, w_1)$ for some $w_1 >0$.
\endremark
\demo{Proof} Let $g_N(z)=\prod_{j=1}^N (1-z/z_j)/
\prod_{j=1}^{N+1} (1-z/w_j)$. Then $g_N$ has simple poles at
$w_1, w_2, \dots, w_{N+1}$ and because of the alternating nature
of the $z_j$'s and $w_j$'s, each residue is negative. Since
$g_N(z)\to 0$ as $|z|\to\infty$, it follows that $g_N(z) =
\sum_{j=1}^{N+1} \frac{\alpha_j^{(N)}}{w_j-z}$ with
$\alpha_j^{(N)}>0$, $j=1,\dots,N+1$.
Thus, each $g_N$ is a Herglotz function and so $g_N$ maps
$\Bbb C\backslash [0,\infty)$ to $\Bbb C \backslash
(-\infty, 0]$. Let $H$ be a biholomorphic map of $\Bbb C
\backslash (-\infty, 0]$ to the open unit disk (e.g., $H(w)=
\frac{\sqrt w -1}{\sqrt w +1}$). By applying the Vitali
convergence theorem (see, e.g., [\tit], Ch.~5) to $H \circ
g_N$, we see it suffices to show $g_N (x)$ converges for each
$x\in(-\infty,0)$ to conclude that $g_N(z)$ converges as
$N\to\infty$ for $z\in \Bbb C\backslash (0,\infty)$.
Since $w_j < z_j$, we have $(1-x/z_j) / (1-x/w_j) <1$, and since
$w_{j+1} >z_j$, we have $(1-x/z_j) / (1-x/w_{j+1}) > 1$ assuming
$x<0$. Thus $g_1 (x) < g_2 (x) < \cdots < g_N (x) < g_{N+1}(x)<1$,
so $\lim_{N\to\infty} g_N (x)$ exists for $x<0$.
Once we have convergence on $\Bbb C \backslash (0,\infty)$, it
is easy to extend the argument to $\Bbb C \backslash
\{w_j\}_{j=1}^\infty$.
Finally, let $f(z)$ be a Herglotz function with the stated zeros
and poles. Then $f(z)/g(z)$ is an entire non-vanishing function,
and on $\Bbb C \backslash [0,\infty)$, $|\text{Im}\,
(\ln (f(z)/g(z)))| \leq 2\pi$ since \linebreak $|\text{Im}\,
(\ln (f(z)))| \leq \pi$ and $|\text{Im}\, (\ln (g(z)))|\leq\pi$
on $\Bbb C\setminus [0,\infty)$. It follows that $f(z)/g(z)$ is
constant. \qed
\enddemo
In exactly the same way one infers
\proclaim{Theorem 2.2} Let $0 < z_1 < w_1 < z_2 < w_2 < \cdots$
with $\lim_{n\to\infty} w_n =\infty$. Then
$$
g(z) = \lim_{n\to\infty} \prod_{j=1}^n (1-z/z_j) \bigg/
\prod_{j=1}^n (1-z/w_j)
$$
exists for any $z$ in $\Bbb C \backslash \{w_j\}_{j=1}^\infty$
with convergence uniform on compact subsets of $\Bbb C
\backslash \{w_j\}_{j=1}^\infty$. $g(z)$ is a meromorphic
function with $\frac{\text{Im}\, (g(z))}{\text{Im}\, (z)} < 0$
for $z\in\Bbb C \setminus \Bbb R$.
Moreover, any meromorphic function $f(z)$ satisfying {\rom{(2.1)}}
with zeros precisely at $\{z_j\}_{j=1}^\infty$ and poles precisely
at $\{w_j\}_{j=1}^\infty$ is a negative multiple of $g(z)$.
\endproclaim
We also have theorems on asymptotics, poles, and residues
determining a meromorphic Herglotz function.
\proclaim{Theorem 2.3} Let $f_1(z), f_2(z)$ be two meromorphic
Herglotz functions with identical sets of poles and residues,
respectively. If
$$
f_1 (ix) - f_2 (ix) \to 0 \text{ as } x\to\infty, \tag 2.2
$$
then $f_1 = f_2$.
\endproclaim
\demo{Proof} By the Herglotz representation theorem, if $f(z)$
is a meromorphic Herglotz function with poles at
$\{ w_j\}_{j=1}^\infty$ in $\Bbb R$ and residues $-\alpha_k <0$
at $z=w_k$, then for some constants $A\geq 0$ and $B\in\Bbb R$,
$$
f(z) = Az + B + \sum_{j=1}^\infty \alpha_j
\biggl[ \frac{1}{w_j -z} - \frac{w_j}{1+w^2_j}\biggr],
$$
where the sum is absolutely convergent since $\sum_{j=1}^\infty
\frac{\alpha_j}{1+w^2_j} < \infty$.
Thus $f_1(z) - f_2(z) = {\tilde A}z - {\tilde B}$ for some
${\tilde A},{\tilde B} \in \Bbb R$, and
therefore, {\rom{(2.2)}} implies ${\tilde A}={\tilde B}=0$. \qed
\enddemo
In applications, either $f_1 (ix)$ and $f_2 (ix)$ are both
$o(1)$ as $x\to\infty$ or else, $f_1 (ix)$ and $f_2 (ix)$ are
both $\sqrt{ix} + o(1)$ as $x\to\infty$.
\vskip 0.3in
\flushpar{\bf \S 3. The Case of a Dirichlet Boundary Condition
$h_a=\infty$}
\vskip 0.1in
We want to prove Theorem~2 when $h_a = \infty$. If $h_0 < \infty$,
let $u_- (z,x; q)$ solve $-u'' + qu = zu$ with boundary conditions
$u_- (z,0; q) =1$, $u'_- (z,0; q)=-h_0$. If $h_0 = \infty$, let
it satisfy $u_- (z,0; q)=0$, $u'_- (z,0; q)=1$. As is well known
(see, e.g., [\ls], Ch.~1), $u_-$ is an entire function of $z$.
Similarly, $u_+$ satisfies the $h_1$ boundary condition at $1$.
Let
$$
W(z;q) = u'_-(z,x;q) u_+(z,x;q) - u_-(z,x;q) u'_+(z,x;q),
$$
which is independent of $x$. The zeros of $W$ are precisely the
points $w_i$ of $S(0,1; h_0, h_1;q)$, that is, the eigenvalues of
$H := H(0,1; h_0, h_1; q)$.
Fix $a\in (0,1)$ and $q$. Let $g(z)=G(z,a,a)$ be the Green's
function of $H$ in $L^2 ((0,1))$ at $(a,a)$, that is, the
integral kernel of $(H-z)^{-1}$ at $(a,a)$. (We also use the
notation $g(z;q)$ for $g(z)$ whenever the dependence of $g(z)$
on $q$ needs to be underscored.) Then, by a standard formula for
the Green's function of $H$,
$$
g(z;q) = \frac{u_- (z,a; q) u_+ (z,a; q)}{W(z;q)}\, . \tag 3.1
$$
The zeros of $u_+ (z,a;q)$ are precisely the points of
$S(a,1; h_a =\infty, h_1;q)$ and the zeros of $u_- (z,a;q)$
are precisely the points of $S(0,a; h_0, h_a =\infty; q)$. The
hypothesis (ii) on disjointness of the $S$ sets in Theorem~2 says
that the poles of $g(z)$ are precisely the points of $S(0,1)$,
and the zeros, the points of $S(0,a) \cup S(a,1)$. (If the sets
are not disjoint, there are cancellations between zeros and
poles.)
By Theorem~2.1 (adding a constant to $q$ if need be, we can
assume all poles and zeros are positive), the zeros and poles
of $g(z)$ and the known asymptotics $g(-\kappa^2;q)=
(2\kappa)^{-1}$ ($1+o(1)$) as $\kappa\to\infty$ determine $g$,
that is, $g(z;q_1) = g(z; q_2)$.
Next we use the $m$-functions $m_\pm$ defined by $m_\pm (z;q)
= \pm u'_\pm (z,a;q)/u_\pm (z,a;q)$. By (3.1),
$$
g(z;q) = -\frac{1}{[m_+ (z;q) + m_- (z;q)]}\, . \tag 3.2
$$
Moreover, the poles of $m_+$ (resp.~$m_-$) are precisely the
points $\lambda$ of $S(a,1; h_a =\infty, h_1; q)$ (resp.~$S
(0,a; h_0, h_a =\infty; q)$). And the residues of the poles
are determined by $g$. Explicitly, if $\lambda_0$ is a pole of
$m_+$, by hypothesis (ii) in Theorem~2, it is not a pole of
$m_-$, and so its residue is $\left. -1/\frac{\partial g}
{\partial z}\right|_{z=\lambda_0}$.
By Theorem~2.2 and the asymptotics $m_\pm (-\kappa^2; q) =
-\kappa + o(1)$ as $\kappa\to\infty$, the poles and residues
determine $m_\pm$; that is, $m_\pm (z;q_1) = m_\pm (z;q_2)$.
Finally, the uniqueness result of Borg [\borg] and Marchenko
[\mar] guarantees that $m_\pm (z;q)$ uniquely determine $g$ on
$[0,a]$ and $[a,1]$, so $q_1 = q_2$ a.e.~on $[0,1]$.
\vskip 0.3in
\flushpar{\bf \S 4. The Case $h_a\in\Bbb R$}
\vskip 0.1in
The changes in the proof when $|h_a| <\infty$ are minimal. Define
$u_\pm$ as in the last section, but instead of (3.1), define
$$
g(z;q) = \frac{[u'_- (z,a;q) + h_a u_- (z,a;q)]
[u'_+ (z,a; q) + h_a u_+ (z,a;q)]}{W(z;q)}\, . \tag 4.1
$$
Since $W= (u'_- + h_a u_-) u_+ - u_- (u'_+ +h_a u_+)$, (3.2)
becomes
$$
g(z;q) = \frac{1}{\frac{1}{m_+(z;q) + h_a}
+ \frac{1}{m_-(z;q) - h_a}}\, . \tag 4.2
$$
The spectra determine the zeros and poles of $g$ which, together
with the asymptotics $g(-\kappa^2; q) = -\frac12 \kappa (1 +
o(1))$ as $\kappa\to\infty$, determine $g$ by Theorem~2.1 or 2.2.
By hypothesis (ii) of Theorem~2, the poles of
$(m_\pm \pm h_a)^{-1}$ are distinct and so their residues
are determined by (4.2) and the knowledge of $g$. The poles and
residues of $-(m_\pm \pm h_a)^{-1}$ and the fact that
$|m_\pm (ix)| \to \infty$ as $x\to\infty$ determine
$(m_\pm \pm h_a)^{-1}$ by Theorem~2.3. The Borg-Marchenko
uniqueness theorem then completes the proof.
\vskip 0.3in
\flushpar{\bf \S 5. Examples of Non-Uniqueness}
\vskip 0.1in
Our goal here is to show that if condition (ii) fails, then the
uniqueness result in Theorem~2 can also fail. We will take an
extreme case where $S(0,\frac12)=S(\frac12, 1)$ for simplicity;
but we have no doubt that a single point in common suffices to
construct counterexamples to the extension of Theorem~2 with
(ii) absent. We note that $S(0,1)\cap S(0,\frac12) = S(0,1) \cap
S(\frac12, 1)=S(0,\frac12) \cap S(\frac12, 1)$ so that if two
$S$'s fail to be disjoint, each pair has non-zero intersection.
To begin we note
\proclaim{Lemma 5.1} Let $f$ be a continuous map of $Q:=[0,1]
\times [0,1]$ to the unit circle. Then, there exists a pair
of points $p_0, p_1 \in Q$ with $p_0\neq p_1$ and $f(p_0) =
f(p_1)$.
\endproclaim
\demo{Proof} If $f(0,0) = f(1,1)$, we have the required points.
If not, reparametrize the circle so that $f(0,0)=1$, $f(1,1) =
-1$. Consider the images $f(\gamma_j(t))$, $t\in[0,1]$, $j=
0,1,2$ of the three curves $\gamma_0, \gamma_1, \gamma_2$
given by $\gamma_j (t) = (t, t+ (j-1)\pi^{-1} \sin(\pi t))$,
$t\in [0,1]$, $j=0,1,2$. If two of these images contain the
point $(0,-1)$ on the unit circle, then that value is taken
twice. If at most one of these images contains $(0, -1)$, then
by the intermediate value theorem, two images must contain
$(0,1)$. \qed
\enddemo
As explained in [\gsds], by results of Levitan [\lev], [\levbook],
Ch.~3 and Levitan-Gasymov [\lg], one can prove
\proclaim{Proposition 5.2} Suppose that $x_0 < y_0 < x_1 < y_1
<\cdots$ so that for $n$ sufficiently large, $x_n = [(2n)\pi]^2$,
$y_n = [(2n+1)\pi]^2$. Then there exists a unique $h_1$ and a
$C^\infty$-function $q$ on $[\frac12, 1]$ so that
$$
-\frac{d^2}{dx^2} + q \text{ in } L^2 ((\tfrac12, 1));
\quad u'(\tfrac12)=0, \quad u'(1) + h_1 u(1) =0
$$
has eigenvalues $\{x_n\}_{n=0}^\infty$ and
$$
-\frac{d^2}{dx^2} + q \text{ in } L^2 ((\tfrac12, 1)];
\quad u(\tfrac12)=0, \quad u'(1) + h_1 u(1) =0
$$
has eigenvalues $\{y_n\}_{n=0}^\infty$. Moreover, if a finite
subset of $x$'s and $y$'s is varied, $h_1$ varies continuously
as a function of these numbers.
\endproclaim
Consider now fixing $y_n = [(2n+1)\pi]^2$ for all $n\in\Bbb N_0$
($=\Bbb N\cup\{0\}$) and $x_n = [(2n)\pi]^2$ for $n\geq 2$ and
varying $(x_0, x_1)$ in $[0,1] \times [20,21]$. By Lemma~5.1
and Proposition~5.2, we can find $(x^{(0)}_0, x^{(0)}_1) \neq
(x^{(1)}_0, x^{(1)}_1)$ so that the corresponding values of
$h_1$ are equal. Set $\tilde q_0, \tilde q_1$ as the
corresponding $q$'s and $h$ as the common value of $h_1$.
Let $q_1, q_2$ be defined on $[0,1]$ by
$$\alignat2
q_1 (x) &= \tilde q_0 (1-x), \qquad && 0\leq x \leq \tfrac12, \\
&= \tilde q_1 (x), \qquad && \tfrac12 \leq x \leq 1, \\
q_2 (x) &= q_0 (1-x).
\endalignat
$$
Then $q_1 \neq q_2$ but $S(0,\frac12; h_0 =-h, h_{\frac12} =
\infty; q_1) = S(0,\frac12; h_0 = -h, h_{\frac12} = \infty; q_2)
= S(\frac12, 1; h_{\frac12} = \infty, h_1 = h; q_1) =
S(\frac12, 1; h_{\frac12}=\infty, h_1 =h; q_2) =
\{((2n+1)\pi)^2 \}_{n\in\Bbb N}$, and by reflection
symmetry:
$$
S(0,1; h_0 =-h, h_1 =h; q_1) = S(0,1; h_0 =-h, h_1 =h; q_2).
$$
Since $q_1 \neq q_2$, this provides the required counterexample.
(There is no particular significance in our choice of $x_1 \in
[20,21]$. Any interval of length one contained in $(y_0, y_1)
= (\pi^2, 9\pi^2)$ would be admissible.)
As in the finite-difference case [\gsmf], we believe an analysis
of the situation where $S(0,\frac12)\cap S(\frac12, 1)$ has
$k$-points will yield $k$-parameter sets of $q$'s (as long as
we are allowed to vary $h_0, h_1$ as well as $q$) consistent
with the given sets of eigenvalues.
\vskip 0.3in
\flushpar{\bf \S 6. The Whole Line Case}
\vskip 0.1in
In this section, we will extend Theorem~2 to the situation where
$[0,1]$ is replaced by $\Bbb R$, but the spectrum of the
corresponding Schr\"odinger operator $H$ in $L^2(\Bbb R)$
is purely discrete and
bounded from below. Typical situations are, for instance, $q\in
L^1_{\text{\rom{loc}}}(\Bbb R)$ real-valued with $q(x)\to\infty$
as $|x|\to\infty$ or, $q\in L^1_{\text{\rom{loc}}}(\Bbb R)$
real-valued, $q$ bounded from below, and $\lim_{x\to\pm\infty}
\int_x^{x+a} dy \, q(y) =\infty$ for any $a>0$ (cf.~[\ls],
Sect.~4.1). In this case, the maximal operator $H$ in $L^2
(\Bbb R)$ associated with the differential expression
$-\frac{d^2}{dx^2} + q$ on $\Bbb R$ (with domain $\Cal D(H)
=\{f\in L^2(\Bbb R) \mid f,f' \text{ locally absolutely
continuous on }\Bbb R; (-f'' + qf)\in L^2(\Bbb R)\}$) is
self-adjoint. In [\gsds] our extensions required a hypothesis
on $q$ that $q(x) \geq C |x|^{2+\varepsilon}+1$ for some
$C,\varepsilon >0$. This was because we used results on densities
of zeros. Here, because we rely on Theorems~2.1, 2.2, we note
that the following result holds by the identical proof to
Theorem~2:
\proclaim{Theorem 3} Suppose $q\in L^1_{\text{loc}}(\Bbb R)$ is
real-valued and $H$ in $L^2(\Bbb R)$ is bounded from below with
purely discrete spectrum $S(-\infty, \infty; q)$. Let
$S(-\infty, 0; h_0;q)$ denote the spectrum of the corresponding
\rom(maximally defined\rom) operator in $L^2((-\infty, 0))$
with $u'(0) + h_0u(0)=0$ boundary conditions, and similarly
for $S(0,\infty; h_0;q)$. Suppose that $q_1, q_2$ are given
and we have a fixed $h_0\in\Bbb R\cup \{0\}$ so that
\roster
\item"\rom{(i)}" $S(-\infty,\infty; q_1) = S(-\infty, \infty; q_2)$,
$S(-\infty,0;h_0; q_1) = S(-\infty, 0; h_0; q_2)$, and \linebreak
$S(0,\infty; h_0; q_1) = S(0, \infty; h_0; q_2)$
\item"\rom{(ii)}" The sets $S(-\infty, \infty; q_1)$,
$S(-\infty, 0; h_0; q_1)$, and $S(0, \infty;h_0; q_1)$ are pairwise
disjoint.
\endroster
Then $q_1 = q_2$ a.e.~on $\Bbb R$.
\endproclaim
As noted in Remark~2 following Theorem~2.1, this result extends to
Schr\"odinger operators $H$ with purely discrete spectra
accumulating at $+\infty$ and $-\infty$. In particular, it extends
to cases where $H$ is in the limit circle case at $+\infty$ and/or
$-\infty$ as long as the corresponding (separated) boundary
condition at $+\infty$ and/or $-\infty$ is kept fixed for all
three operators on $\Bbb R$, $(-\infty, 0)$, and $(0,\infty)$.
The reader might want to contrast Theorem~3 with Corollary~3.4
in [\gsun], where we obtained uniqueness of $q$ from three
(discrete) spectra of operator realizations of $-\frac{d^2}
{dx^2}+q$ on $\Bbb R$. There one of the three spectra is
$S(-\infty,\infty;q)$ as above in Theorem~3; the other two,
$S(-\infty,\infty;\beta_j, q)$, $j=1,2$, are associated with
$-\frac{d^2}{dx^2}+q$ on $\Bbb R$ and the boundary conditions
$\lim_{\varepsilon\downarrow 0} [u' (\pm\varepsilon) + \beta_j
u(\pm\varepsilon)]=0$, where $\beta_j\in\Bbb R\cup \{\infty\}$,
$j=1,2$, $\beta_1\neq\beta_2$, $(\beta_1,\beta_2)\neq (0,\infty)$,
$(\infty, 0)$.
\vskip 0.3in
\Refs
\endRefs
\vskip 0.1in
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\gap
\item{\dgs.} R.~del Rio, F.~Gesztesy and B.~Simon, {\it{Inverse
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\gap
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\gap
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\gap
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\gap
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\gap
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\gap
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\gap
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\gap
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\gap
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\gap
\enddocument