\input amstex
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\baselineskip=15 pt
%\NoBlackBoxes
\TagsOnRight
\def\gap{\vskip 0.1in\noindent}
\def\ref#1#2#3#4#5#6{#1, {\it #2,} #3 {\bf #4} (#5), #6.}
%References
\def \atk {1} % Atkinson
\def\borg {2} % Borg
\def\borgun {3} % Borg unique
\def\cl {4} % Coddington-Levinson
\def\dl {5} % Danielyan-Levitan
\def\eve {6} % Everitt
\def\ghsz {7} % Gesztesy-Holden-Simon-Zhao
\def\gsun {8} % Gesztesy-Simon, uniqueness paper
\def\gsac {9} % Gesztesy-Simon, ac paper
\def\gsjac {10} % Gesztesy-Simon, Jacobi paper
\def\haldma {11} % Hald Mantle
\def\hsturm {12} % Hochstadt 73
\def\hmixed {13} % Hochstadt - mixed given data
\def\hl {14} % Hochstadt-Lieberman
\def\iwa {15} % Iwasaki 87
\def\jay {16} % Jayawardena
\def\kha {17} % Khaled
\def\kr {18} % Krein
\def\levindist {19} % Levin 72
\def\levs {20} % Levinson
\def\lev {21} % Levitan - AMS Transl.
\def\levbook {22} % Levitan book
\def\lg {23} % Levitan-Gasymov
\def\ls {24} % Levitan-Sargsjan
\def\mal {25} % Malamud
\def\maun {26} % Marchenko
\def\mar {27} % Marchenko
\def\marku {28} % Markushevich
\def\pt {29} % Poschl-Trubowitz
\def\rs {30} % Reed-Simon IV
\def\delrio {31} % Del Rio
\def\rs {32} % Rundell-Sacks
\def\suz {33} % Suzuki 82
\def\suzgel {34} % Suzuki 85
\def\suzinv {35} % Suzuki 86
\def\tit {36} % Titchmarsh.
\topmatter
\title Inverse Spectral Analysis With Partial Information on the Potential,
II. The Case
of Discrete Spectrum
\endtitle
\rightheadtext{Inverse Spectral Analysis: The Case of Discrete Spectrum}
\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$^2$ Division of Physics, Mathematics, and Astronomy, California
Institute of
Technology, Pasadena, CA~91125, USA. E-mail: bsimon\@caltech.edu
\endthanks
\thanks
This material is based upon work supported by the National Science
Foundation under
Grant Nos.~DMS-9623121 and DMS-9401491.
\endthanks
\date March 3, 1997
\enddate
\abstract We discuss results where the discrete spectrum (or partial
information on the discrete
spectrum) and partial information on the potential $q$ of a one-dimensional
Schr\"odinger operator $H=-\frac{d^2}{dx^2}+q$ determine the potential
completely. Included are theorems for finite intervals
and for the whole line. In particular, we pose and solve a new type of inverse
spectral problem involving fractions of the eigenvalues of $H$ on a finite
interval and knowledge of $q$ over a corresponding fraction of the interval.
The methods employed rest on Weyl $m$-function techniques and densities of
zeros of a class of entire functions.
\endabstract
\endtopmatter
\document
\vskip 0.1in
\flushpar{\bf \S 1. Introduction}
\vskip 0.1in
In 1978, Hochstadt-Lieberman [\hl] proved the following remarkable theorem:
\proclaim{Theorem 1.1} Let $h_0 \in \Bbb R$, $h_1 \in \Bbb R \cup \{\infty\}$
and assume $q_1, q_2 \in L^1 ((0,1))$ to be real-valued. Consider
the Schr\"odinger operators $H_1, H_2$ in $L^2 ((0,1))$ given by
$$
H_j = -\frac{d^2}{dx^2}+ q_j, \quad j=1,2,
$$
with the boundary conditions
$$\align
u'(0) + h_0 u(0) & = 0, \tag 1.1a \\
u'(1) + h_1 u(1) & = 0. \tag 1.1b
\endalign
$$
Let $\sigma(H_j)=\{\lambda_{j,n}\}$ be the \rom(necessarily simple\rom)
spectra of
$H_j, j=1,2$. Suppose that $q_1 = q_2$ \rom(a.e.\rom) on $[0,\frac12]$ and
that
$\lambda_{1,n} = \lambda_{2,n}$ for all $n$. Then $q_1 = q_2$ \rom(a.e.\rom)
on $[0,1]$.
\endproclaim
Here, in obvious notation, $h_1 =\infty$ in (1.1b) singles out the
Dirichlet boundary
condition $u(1)=0$.
For each $\varepsilon >0$, there are simple examples where $q_1 = q_2$ on
$[0,
\frac12 - \varepsilon]$ and $\sigma (H_1)$ = $\sigma (H_2)$ but $q_1 \neq
q_2$.
(Choose $h_0=-h_1$, $q_1 (x)=0$ for $x \in (0, \frac12 - \varepsilon] \cup
[\frac12, 1]$ and nonzero on $(\frac12 - \varepsilon, \frac12)$, and $q_2
(x) =
q_1 (1-x)$. See also Theorem I$^\prime$ in the appendix of [\suzinv].)
Later refinements of Theorem~1.1 in [\haldma, \suzinv] (see also the
summary in
[\suz]) showed that the boundary condition for $H_1$ and $H_2$ at $x=1$ need
not be assumed a priori to be the same, and that if $q$ is continuous, then
one only
needs $\lambda_{1,n} = \lambda_{2,m(n)}$ for all values of $n$ but one.
([\suzinv]
claims the result does not require continuity of $q$, but we will see in
Section~3
that this assertion is false.) The same boundary condition for $H_1$ and $H_2$
at $x=0$, however, is crucial for Theorem~1.1 to hold (see [\haldma, \delrio]).
Moreover, analogs of Theorem~1.1 for certain Schr\"odinger operators are
considered in [\kha] and the interval $[0,\frac12]$ replaced by different
subsets of
$[0,1]$ was studied in [\jay] (see also [\pt], Ch.~4). Reconstruction
techniques for
$q(x)$ in this context are discussed in [\rs].
Our purpose in this paper is to provide a new approach to Theorem~1.1 that we
feel is more transparent and, moreover, capable of vast generalizations. To
state our
generalizations, we will introduce a shorthand notation to paraphrase
Theorem~1.1
by saying ``$q$ on $[0, \frac12]$ and the eigenvalues of $H$ uniquely
determine
$q$." This is just a shorthand notation for saying $q_1 =q_2$ if the obvious
conditions hold.
Unless explicitly stated otherwise, all potentials $q, q_1$, and $q_2$ will be
real-valued and in $L^1 ((0,1))$ for the remainder of this paper. Moreover,
to avoid
too many case distinctions in the proofs below, we shall assume $h_0 , h_1 \in
\Bbb R$ in (1.1) throughout the main body of this paper. In particular, for
$h_0,
h_1 \in \Bbb R$ we index the corresponding eigenvalues $\lambda_n$ of $H$ by
$n \in \Bbb N_0 = \Bbb N \cup \{0\}$. The case of Dirichlet boundary
conditions,
where $h_0 = \infty$ and/or $h_1 = \infty$ will be dealt with in Appendix~A.
Here are some of the generalizations we will prove for Schr\"odinger operators
on $[0,1]$:
\proclaim{Theorem 1.2} Let $H=-\frac{d^2}{dx^2}+ q$ in $L^2 ((0,1))$ with
boundary conditions {\rom{(1.1)}} and $h_0,h_1 \in \Bbb R$. Suppose $q$ is
$C^{2k}((\frac12 - \varepsilon, \frac12 + \varepsilon))$ for some
$k=0,1,\dots$\,
and for some $\varepsilon > 0$. Then $q$ on $[0,\frac12]$, $h_0$, and all the
eigenvalues of $H$ except for $(k+1)$ uniquely determine $h_1$ and $q$ on all
of $[0,1]$.
\endproclaim
\remark{Remarks} 1. The case $k=0$ in Theorem~1.2 is due to Hald [\haldma].
2. In the non-shorthand form of this theorem, we mean that both $q_1$ and
$q_2$ are $C^{2k}$ near $x=\frac12$.
3. One need not know which eigenvalues are missing. Since the eigenvalues
asymptotically satisfy
$$
\lambda_n = (\pi n)^2 + 2(h_1 - h_0) + \int_0^1 dx\, q(x) +o(1) \qquad
\text{\rom{as }} n \to \infty, \tag 1.2
$$
given a set of candidates for the spectrum, one can tell how many are missing.
4. For the sake of completeness we mention the precise definition of $H$ in
$L^2 ((0,1))$ for real-valued $q \in L^1((0,1))$ and boundary condition
parameters
$h_0, h_1 \in \Bbb R \cup \{\infty \}$ in (1.1):
$$\align
H=&-\frac{d^2}{dx^2} + q, \\
D(H)=&\{g \in L^2((0,1)) \, | \, g,g' \in AC([0,1]); \, (-g'' + qg) \in
L^2((0,1)); \\
& \quad \quad \quad \quad g'(0) + h_0g(0)=0, \, g'(1) + h_1g(1) = 0 \},
\tag 1.3
\endalign
$$
where $AC([0,1])$ denotes the set of absolutely continuous functions on
$[0,1]$
and $h_{x_0}=\infty$ represents the Dirichlet boundary condition $g(x_0)=0$
for $x_0 \in \{0,1\}$ in (1.3).
%4. It is only required that $q$ be $C^k$ at $x=\frac12$, that is, that $q$ is
% $C^{k-1}$ near $x=\frac12$,
%and that
%$$
%\lim_{x\to\frac12}\, \frac{q^{(k-1)}(x) - q^{(k-1)}(\frac12)}{x-\frac12}
%$$
%exists. Indeed, it suffices that a Taylor expansion
%$q(s+\frac12) = \sum^k_{j=0}
% a_j s^j +
%o(s^k)$ holds at $x=\frac12$.
\endremark
In Section~3, we discuss examples which show that Theorem~1.2 is optimal in
the
sense that if $q$ is only assumed to be $C^{2k-1}$ near $x=\frac12$ for
some $k
\geq 1$, then it is not uniquely determined by $q\restriction [0,\frac12]$
and all the
eigenvalues but $(k+1)$.
Theorem~1.2 works because the condition that $q$ is $C^{2k}$ near $x=\frac12$
gives us partial information about $q$ on $[\frac12, 1]$; namely, we know
$q(\frac12), q'(\frac12), \dots, q^{(2k)}(\frac12)$ computed on $[\frac12,
1]$ since
we can compute them on $[0,\frac12]$. This suggests that knowing $q$ on more
than $[0,\frac12]$ should let one dispense with a finite density of
eigenvalues.
That this is indeed the case is the content of the following theorem:
\proclaim{Theorem 1.3} Let $H = -\frac{d^2}{dx^2}+q$ in $L^2 ((0,1))$ with
boundary conditions {\rom{(1.1)}} and $h_0,h_1 \in \Bbb R$. Then $q$ on $[0,
\frac12 +\frac\alpha2]$ for some $\alpha \in (0,1)$, $h_0$, and a subset $S
\subseteq \sigma(H)$ of all the eigenvalues $\sigma(H)$ of $H$ satisfying
$$
\#\{\lambda\in S \mid \lambda\leq \lambda_0 \} \geq (1-\alpha)
\# \{\lambda\in \sigma(H) \mid \lambda\leq \lambda_0 \} + \tfrac\alpha2
\tag 1.4
$$
for all sufficiently large $\lambda_0 \in \Bbb R$, uniquely determine $h_1$
and
$q$ on all of $[0,1]$.
\endproclaim
\remark{Remarks} 1. As a typical example, knowing slightly more than half the
eigenvalues and knowing $q$ on $[0,\frac34]$ determines $q$ uniquely on all of
$[0,1]$. To the best of our knowledge, Theorem~1.3 solves a new type of
inverse
spectral problem. In particular, we are not aware of any inverse spectral
result involving fractions of the set of eigenvalues as in (1.4).
2. As in the case $\alpha =0$, we have an extension of the same type as
Theorem~1.2. Explicitly, if $q$ is assumed to be $C^{2k}$ near $x=\frac12 +
\frac\alpha2$, we only need
$$
\# \{\lambda\in S \mid \lambda\leq \lambda_0 \} \geq (1-\alpha)
\# \{\lambda\in \sigma(H) \mid \lambda\leq \lambda_0 \} + \tfrac\alpha2 -
(k+1) \tag 1.5
$$
instead of (1.4).
\endremark
We can also derive results about problems on all of $\Bbb R$. In Section~5,
we will prove
\proclaim{Theorem 1.4} Suppose that $q\in L^1_{\text{\rom{loc}}} (\Bbb R)$
obeys
\roster
\item"\rom{(i)}" $q(x) \geq C|x|^{2+\varepsilon} -D$ for some
$C,\varepsilon, D>0$, and that
\item"\rom{(ii)}" $q(-x) \geq q(x) \quad x\geq 0$.
\endroster
Then $q$ on $[0,\infty)$ and the spectrum of $H = -\frac{d^2}{dx^2} + q$ in
$L^2 (\Bbb R)$ uniquely determine $q$ on all of $\Bbb R$.
\endproclaim
In Section~5, we will also present further conjectures and explain how
condition (i) is related to the class of entire functions of type less than
one.
All these results are related to two other papers we have written. In
[\gsjac], we
consider, among other topics, analogs of Theorems~1.1 and 1.3 for finite
tridiagonal
(Jacobi) matrices extending a result in [\hmixed]. The approach there is
very similar
to the current one except that the somewhat subtle theorems on zeros of entire
functions in this paper are replaced by the elementary fact that a
polynomial of
degree at most $N$ with $N+1$ zeros must be identically zero. In [\gsac], we
consider results related to Theorem~1.4 in that for Schr\"odinger operators on
$(-\infty, \infty)$, ``spectral" information plus the potential on one of
the half-lines
determine the potential on all of $(-\infty,\infty)$. In that paper, we
consider
situations where there are scattering states for some set of energies and the
``spectral" data are given by a reflection coefficient on a set of positive
Lebesgue
measure in the a.c.~spectrum of $H$. The approach is not as close to this
paper as
is [\gsjac], but $m$-function techniques (see also [\gsun]) are critical in
all
three papers.
Hochstadt-Lieberman [\hl] use the details of the inverse spectral theory in
their proof.
In a sense, we only use the main uniqueness theorem of that theory due to
Marchenko
[\maun], which we now describe. For $q \in L^1((a,b))$ real-valued,
$-\infty 0 \Longrightarrow \text{Im}\, (m_- (z,b)) > 0,
\quad \text{Im}\, (m_+ (z,a)) > 0. \tag 1.11
$$
Marchenko's [\maun] fundamental uniqueness theorem of inverse spectral theory
then reads as follows:
\proclaim{Theorem 1.5} $m_+ (z,a)$ uniquely determines $h_b$ as well as
$q$ \rom(a.e.\rom) on $[a,b]$.
\endproclaim
If $q \in L^1_{\text{\rom{loc}}}([a,\infty))$ is real-valued (with
$|a|<\infty$) and
$-\frac{d^2}{dx^2} + q$ is in the limit point case at infinity, one can
still define
a unique
$m_+ (z,a)$ function but now for $\text{Im}\, (z) \neq 0$ rather than all
$z \in
\Bbb C$. For such $z$, there is a unique function $u_+(z, \,\cdot\,)$ which is
$L^2$ at infinity (unique up to an overall scale factor which drops out of
$m_+ (z,a)$ defined by (1.8)). Again, one has the following uniqueness result
independently proved by Borg [\borgun] and Marchenko [\maun]
\proclaim{Theorem 1.6} $m_+ (z,a)$ uniquely determines $q$ \rom(a.e.\rom)
on $[a,\infty)$.
\endproclaim
It is useful to have $m_-(z,b)$ because of the following basic fact:
\proclaim{Theorem 1.7} Let $H = -\frac{d^2}{dx^2} + q$ be a Schr\"odinger
operator in $L^2 ((a,b))$ with boundary conditions {\rom{(1.7), (1.9)}} and
let
$G(z,x,y)$ be the integral kernel of $(H-z)^{-1}$. Suppose $c \in (a,b)$
and let
$m_+ (z,c)$ be the corresponding $m_+$-function for $[c,b]$ and $m_-(z,c)$
the $m_-$-function for $[a,c]$. Then
$$
G(z,c,c) = -\frac{1}{m_+ (z,c) + m_- (z,c)}\, . \tag 1.12
$$
\endproclaim
Theorems~1.5 and 1.6 are deep facts; Theorem~1.7 is an elementary calculation
from the explicit formula for the integral kernel of $(H-z)^{-1}$,
$$
G(z,x,y) = \frac{u_- (z,\min (x,y)) u_+ (z, \max (x,y))}{W (u_- (z), u_+
(z))(x)}\, ,
$$
where $W(\,\cdot\, ,\,\cdot\,)$ is the Wronskian defined by
$$
W(f,g)(x) = f' (x) g(x) - f(x)g'(x).
$$
An analog of Theorem~1.7 holds in case $[a,b]$ is replaced by
$(-\infty,\infty)$.
We can now describe the strategy of our proofs of Theorems~1.1--1.4.
$G(z,c,c)$
has poles at the eigenvalues of $H$ (this is not quite true; see below), so
by (1.12),
at eigenvalues $\lambda_n$ of $H$:
$$
m_+ (\lambda_n,c) = -m_- (\lambda_n,c). \tag 1.13
$$
If we know $q$ on a left partial interval $[a,c]$ and we know some eigenvalue
$\lambda_n$, then we know $m_- (z,c)$ exactly; so by (1.13), we know the value
of $m_+ (\lambda_n,c)$ at the point $\lambda_n$. In Appendix~B we discuss when
knowing the values of $f(\lambda_n)$ of an analytic function of the type of
the
$m$-functions uniquely determines $f(z)$. If $m_+(z,c)$ is determined, then by
Theorem~1.5, $q$ is determined on $[a,b]$ and so is $h_b$.
So the logic of the argument for a theorem like Theorem~1.1 is the following:
\roster
\item"\rom{(i)}" $q$ on $[0,\frac12]$ and $h_0$ determine $m_- (z,\tfrac12)$
by direct spectral theory.
\item"\rom{(ii)}" The $\lambda_n$ and (1.13) determine $m_+ (\lambda_n,
\tfrac12)$, and then by suitable theorems in complex analysis, $m_+ (z,
\tfrac12)$ is uniquely determined for all $z$.
\item"\rom{(iii)}" $m_+ (z,\tfrac12)$ uniquely determines $q$ (a.e.) on
$[\frac12,
1]$ and $h_1$ by inverse spectral theory.
\endroster
It is clear from this approach why $h_0$ is required and $h_1$ is free in the
context of Theorem~1.1 (see [\delrio] for examples where $h_1$ and $q
\restriction [0,\frac12]$ do not determine $q$); without $h_0$ we cannot
compute
$m_-(z, \tfrac12)$ and so start the process.
As indicated before (1.13), $G(z,c,c)$ may not have a pole at an eigenvalue
$\lambda_n$ of $H$. It will if $u_n (c) \neq 0$, but if $u_n (c) =0$, then
$G(z,c,c) = 0$ rather than $\infty$. Here $u_n$ denotes the eigenfunction
of $H$
associated with the (necessarily simple) eigenvalue $\lambda_n$. Nevertheless,
(1.13) holds at points where $u_n (c) =0$ since then $u_- (c)=u_+ (c) = 0$,
and so
both sides of (1.13) are infinite. (In spite of (1.13), $m_+ + m_-$ is also
infinite at $z
= \lambda_n$ and so $G(\lambda_n, c, c) = 0$.) We summarize this discussion in
the following
\proclaim{Theorem 1.8} For any $c\in (a,b)$, {\rom{(1.13)}} holds at any
eigenvalue $\lambda_n$ of $H_{[a,b]}$ \rom(with the possibility of both
sides of
{\rom{(1.13)}} being infinite\rom).
\endproclaim
An alternative way of proving (1.13) is that $\lambda_n$ is an eigenvalue
if and
only if the Wronskian of $u_+$ and $u_-$ is zero, which is precisely (1.13).
Here is a sketch of the contents of this paper. In Section~2 we present our
proofs
of Theorems~1.1 and 1.2. In Section~3 we discuss an example that delimits
Theorem~1.2 and shows that Theorem~1.2 is optimal with respect to smoothness
conditions on $q$. In Section~4 we prove Theorem~1.3, and in Section~5 we
prove Theorem~1.4. Appendix~A is devoted to the case
of Dirichlet boundary conditions, and Appendix~B presents some facts on entire
functions that are necessary to prove our principal results.
\vskip 0.3in
\flushpar{\bf \S 2. Theorems for a Half Interval}
\vskip 0.1in
In this section, we will prove the original Hochstadt-Lieberman theorem
(Theorem~1.1) and our extension of it (Theorem~1.2) for $h_0,h_1 \in \Bbb R$.
Consider a problem on $[0,1]$ with boundary condition (1.2) at $x=1$. Let
$u_+ (z,x)$ be defined by $-u''_+ + qu_+ = zu_+$ and
$$
u'_+ (z,1) = -h_1, \quad u_+ (z,1) =1. \tag 2.1
$$
Then $u_+$ is known to have the following properties:
\roster
\item"\rom{(1)}" For each $x \in [0,1]$, $u_+ (z,x), u'_+ (z,x)$ are entire
functions
of $z$. (This follows from the fact that $u_+ (z,1)=1$ and $u'_+
(z,1)=-h_1$ are
independent of $z$, see, e.g., [\cl], Theorem~I.8.4, Problem~I.7, and p.~226.)
\item"\rom{(2)}"
$$\align
u_+ (z,x) &= \cos \big( \sqrt z \, (1-x)\big) + O\biggl(
\frac{e^{\text{\rom{Im}}\,
(\sqrt z) \, (1-x)}}{\sqrt z} \biggr), \tag 2.2 \\
u'_+ (z,x) &= \sqrt z \, \sin \big( \sqrt z \, (1-x)\big) + O \bigl(
e^{\text{\rom{Im}}
\, (\sqrt z) \, (1-x)}\big) \tag 2.3
\endalign
$$
as $|z| \to \infty$ for all $x\in [0,1]$, where $\sqrt z$ is the square
root branch with
$\text{Im}(\sqrt z) \, \geq 0$ (see, e.g., [\mar], Sect.~1.4).
\item"\rom{(3)}" The zeros of $u_+ (\,\cdot\, , x)$ and $u'_+ (\, \cdot\,
, x)$ are
all real for any $x\in [0,1]$ and they all lie in some $\lambda$-interval
$[c, \infty)$
(this is because these zeros are eigenvalues of self-adjoint boundary value
problems
for Schr\"odinger operators in $L^2 ((0,1))$ bounded from below).
\endroster
The final pair of preliminary results we need concerns the high-energy
asymptotics
of the $m_+$-function,
$$
m_+ (z,x) = \frac{u'_+ (z,x)}{u_+ (z,x)}\, .
$$
\roster
\item"\rom{(4)}" It is known [\atk, \eve] that under the general
hypothesis $q \in
L^1((0,1))$,
$$
m_+ (z,x)^{-1} = -i (\sqrt{z}\, )^{-1} (1 + o(z^{-1/2})) \tag 2.4
$$
uniformly in $x \in [0,1-\delta]$, $\delta>0$ as $|z| \to \infty$ in any
sector
$\varepsilon < \text{Arg}(z) < 2\pi - \varepsilon$, $\varepsilon > 0$.
\item"\rom{(5)}" If $q$ is $C^{2k}$ near $x_0\in (0,1)$, $k=0,1,2,\dots$,
then $m_+ (z,x_0)$ and $m_+ (z,x_0)^{-1}$ are known to have asymptotic
expansions of the form [\dl],
$$\alignat2
m_+ (z,x_0) &= i(\sqrt{z}\, )\biggl( \, \sum^{2k+2}_{\ell=0} C_{\ell}(x_0)
z^{-\ell /2} + o(z^{-k-1}) \biggr), &&\quad C_0 (x_0) = 1, \tag 2.5 \\
m_+ (z,x_0)^{-1} &= -i(\sqrt{z}\, )^{-1} \biggl(\, \sum^{2k+2}_{\ell=0}
D_{\ell}(x_0) z^{-\ell /2} + o(z^{-k-1}) \biggr), &&\quad D_0 (x_0) = 1,
\tag 2.6
\endalignat
$$
as $|z| \to \infty$ in any sector $\varepsilon < \text{Arg}(z) < 2\pi -
\varepsilon$,
$\varepsilon > 0$, where $C_{\ell}(x_0)$ and $D_{\ell}(x_0)$ are universal
functions of $q(x_0), \dots, q^{(\ell-2)}(x_0)$. In fact, $C_{\ell}(x)$ and
$D_{\ell}(x_0)$ have a well-known connection to the conserved densities of the
KdV hierarchy [\ghsz] and they can be computed recursively as follows.
Consider
the Riccati-type equations for $m_+ (z,x)$ and $m_+ (z,x)^{-1}$,
$$\gather
m'_+ (z, x)+m_+ (z, x)^{2}=q(x)-z, \tag 2.7 \\
[m_+ (z,x)^{-1}]' + m_+ (z,x)^{-2}[q(x)-z] = 1. \tag 2.8
\endgather
$$
Inserting the asymptotic expansions (2.5) and (2.6) into (2.7) and (2.8)
then yields the
recursion relations
$$
\align
C_0 (x) &=1, \quad C_1 (x)=0, \quad C_2 (x)=-\tfrac12 \, q(x), \\
C_j (x) &=\tfrac{i}2 \, C'_{j-1} (x) - \tfrac12 \, \sum^{j-1}_{\ell=1}
C_{\ell} (x)C_{j-\ell} (x), \quad j \geq 3, \tag 2.9 \\
D_0 (x) &=1, \quad D_1 (x)=0, \quad D_2 (x)=\tfrac12 \, q(x), \\
D_j (x) &=\tfrac{i}2\, D'_{j-1} (x) + \tfrac12 \, q(x) \sum^{j-2}_{\ell=0}
D_{\ell} (x)D_{j-\ell-2} (x) - \tfrac12 \sum^{j-1}_{\ell=1} D_{\ell}
(x)D_{j-\ell}
(x), \quad j \geq 3. \tag 2.10
\endalign
$$
\endroster
With these preliminaries out of the way, let $q$ be given (a.e.) on
$[0,\frac12]$
and let $q_1, q_2$ be two candidates for $q$ extended to all of $[0,1]$.
Let $\sigma
(H_1) = \{\lambda_{1,n}\}^\infty_{n=0}$ be the set of all the eigenvalues
of $H_1
= -\frac{d^2}{dx^2} + q_1$. Define for $j=1,2$,
$$\align
P_j (z) &= u_{j,+} (z, \tfrac12), \tag 2.11 \\
Q_j (z) &= u'_{j,+} (z, \tfrac12), \tag 2.12 \\
f_j (z) &= \frac{P_j (z)}{Q_j (z)} = m_{j,+}(z,\tfrac12)^{-1}, \tag 2.13 \\
g(z) &= u'_{1,+} (z,0) + h_0 u_{1,+} (z,0), \tag 2.14
\endalign
$$
so that $\{\lambda_{1,n}\}^\infty_{n=0}$ are precisely the zeros of $g(z)$.
(Note in
this context that $u_{1,+} (z,x)$ satisfies (1.1b) at $x=1$ for all $z \in
\Bbb C$.
Thus, if and only if $g(\hat \lambda) = 0$, $u_{1,+} (\hat \lambda, x)$
also satisfies
(1.1a) at $x=0$ and hence $\hat \lambda \in \sigma (H_1)$.) Here $u_{j,
\pm}$ are
the corresponding solutions of $-u'' +q_j u = zu$ used in (1.8) and (1.10).
By adding
a sufficiently large constant to $q_1$ and $q_2$, we can suppose all the
zeros of
$P_j, Q_j$, and $g$ are in $[1,\infty)$.
By (1)--(5) above, we infer:
\roster
\item"\rom{(a)}" $P_j, Q_j$, and $g$ are all of the form (see, e.g.,
[\levindist],
Ch.~I; [\marku], Sect.~II.48)
$$
c \prod^\infty_{n=0} \bigg( 1-\frac{z}{x_n}\biggr) \tag 2.15
$$
for suitable $\{x_n\}^\infty_{n=0} \subseteq [1, \infty)$ (which a priori
could
differ for the five functions).
\item"\rom{(b)}" $P_j, Q_j$, and $g$ are all bounded by $C_1 \exp (C_2
|z|^{1/2})$ for some $C_1,C_2>0$.
\item"\rom{(c)}" $|f_1 (iy) - f_2 (iy)| = o (|y|^{-1})$ as $y \text{ (real)}\to
\pm\infty$.
\item"\rom{(d)}" If $q_j \in C^{2k}$ near $x = \frac12$, then as $y \text{
(real)}
\to\pm\infty$,
$$
|f_1 (iy) - f_2 (iy)| = o(|y|^{-(2k+3)/2}).
$$
\item"\rom{(e)}" $|Q_j (iy)| = \frac12 |y|^{1/2} \,|\exp (\frac12
\text{\rom{Im}} \,
(\sqrt{i})\, |y|^{1/2})| (1+o(1))$ as $y\text{ (real)} \to\infty$.
\item"\rom{(f)}" $|g(iy)| = \frac12 |y|^{1/2} \, |\exp (\text{\rom{Im}} \,
(\sqrt {i}) \, |y|^{1/2})| \, (1+o(1))$ as $y\text{ (real)}\to\infty$.
\item"\rom{(g)}" For $n$ sufficiently large, $\inf_{\theta \in [0,2\pi]}
|g((\pi (n+
\frac12 ))^2 e^{i\theta})| \geq \pi n + O(1)$.
\endroster
Part (d) holds by (2.6) and (2.10) because $q^{(\ell)}_1 (\frac12) =
q^{(\ell)}_2
(\frac12)$ for all $0 \leq \ell \leq 2k$ by the regularity of $q$ near
$x=\frac12$ and
hence the terms $D_0(x_0),\dots,D_{2k+2}(x_0)$ in (2.6) in connection with
$q_1$ and $q_2$ will cancel when inserted into $[f_1 (iy)-f_2 (iy)]$. (g)
follows
from (2.3) since by (2.15), the infimum is taken at $\theta =0$.
\demo{Proof of Theorem {\rom{1.1 (}}for $h_0,h_1 \in \Bbb R$\rom)} Define
$$
F(z) = \frac{[P_1 (z) Q_2(z) - P_2 (z) Q_1 (z)]}{g(z)}\, . \tag 2.16
$$
By Theorem~1.8, (1.13) holds at the points $\lambda_{1,n}$. Hence
$\frac{Q_1(z)}{P_1(z)} = \frac{Q_2(z)}{P_2(z)}$ at $z = \lambda_{1,n}$.
Moreover, at points where both sides are infinite, one infers $P_1 = P_2 =
0$. Thus,
the cross ratio $P_1 Q_2 - P_2 Q_1$ vanishes at each point where $g$ vanishes,
and since $g$ necessarily has simple zeros ($H_1$ has simple spectrum), $F$
is an
entire function.
In addition, by (b) and (g), $F(z)$ satisfies
$$
|F(z)| \leq C_1 \exp (C_2 |z|^{1/2}) \tag 2.17
$$
since (2.17) first holds when $|z| = (\pi (n + \frac12 ))^2$ for $n$
sufficiently large
(by (f)) and then by the maximum modulus principle for all $z$. By
Proposition~B.6
(a Phragm\'en-Lindel\"of argument) and (2.17), if we show that $|F(iy)| \to
0$ as
$y \to \infty$ ($y$ real), then $F \equiv 0$.
But
$$
F(z) = \frac{Q_1 (z) Q_2 (z)}{g(z)} \, [f_1 (z) - f_2 (z)], \tag 2.18
$$
so by (c), (e), and (f),
$$
|F(iy)| = \tfrac12 \, |y|^{1/2} \, [o(|y|^{-1})] (1+o(1)) = o(|y|^{-1/2})
\quad \text{\rom{as }} y \text{\rom{ (real)}} \to \infty, \tag 2.19
$$
goes to zero as required.
Once $F\equiv 0$, we can multiply by $\frac{g(z)}{Q_1 (z) Q_2 (z)}$ (which has
isolated zeros and poles) to conclude that $f_1 = f_2$, and so by Theorem~1.5,
$q_1 = q_2$ (a.e.). \qed
\enddemo
\remark{Remark} There is a (patchable) gap in the paper of
Hochstadt-Lieberman [\hl].
They consider an entire function $\psi (z) = \frac{H(z)}{\omega (z)}$ where
they show
$|H(z)| \leq M \exp (\text{\rom{Im}}\, (\sqrt{|z|)}\, )$ and $\omega (z) =
C\sqrt z \,
\sin (\sqrt z\, ) + O(e^{\text{\rom{Im}} \, (\sqrt z)})$ and then claim
$|\psi(z)| = O\bigl(\frac{1}{\sqrt{|z|}}\bigr)$ without comment. Because of
the zeros of $\sin
(\,\cdot \,)$, this is not evident and one needs a Phragm\'en-Lindel\"of-type
argument to complete their proof.
\endremark
\demo{Proof of Theorem {\rom{1.2}}} Let $\{
\lambda_{\ell}\}^{k+1}_{\ell=1}$ be
the $k+1$ eigenvalues of $-\frac{d^2}{dx^2}+ q_1$ in $L^2 ((0,1))$ which a
priori
are not assumed to be the same for the two potentials. Now define $\tilde
F(z)$ by
$$
\tilde F(z) = \frac{[P_1(z) Q_2(z) - P_2 (z) Q_1 (z)]}{g(z)}\,
\prod^{k+1}_{\ell=1} (z- \lambda_{\ell}) \tag 2.20
$$
instead of (2.16).
(2.17) still holds, and as in (2.18) and (2.19), one now infers from (d),
(e), and (f),
$$
|\tilde F(iy)| = O(|y|^{k+1}) |y|^{1/2} \, [o(|y|^{-(2k+3)/2})] (1+o(1)) =
o(1)
\quad \text{\rom{as }} y \text{\rom{ (real)}} \to \infty.
$$
Thus $m_+(z,\tfrac12)$ determines $h_1$ (cf.~Theorem~1.5) and $q_1 = q_2$
(a.e.)
follows as in the previous proof. \qed
\enddemo
\vskip 0.3in
\flushpar{\bf \S 3. An Example}
\vskip 0.1in
Our goal in this section is to construct, for each $k$, a function $q$ on
$[0,1]$ with
the following properties:
\roster
\item"\rom{(1)}" $q$ is $C^\infty$ on $[0,\frac12]$ and $[\frac12, 1]$; $q$ is
$C^{2k-1}$ on $[0,1]$.
\item"\rom{(2)}" $\frac{d^{2k}q}{dx^{2k}}$ is discontinuous at $x = \frac12$.
\item"\rom{(3)}" $q = 0$ on $[0,\frac12]$.
\item"\rom{(4)}" For a suitable boundary condition parameter $h_1 \in \Bbb R$,
the eigenvalues of $-\frac{d^2}{dx^2} + q$ in $L^2 ((0,1))$ with $u'(0) = 0$,
$u'(1) + h_1 u(1)=0$ boundary conditions agree with those for
$-\frac{d^2}{dx^2}$
in $L^2 ((0,1))$ with $u'(0) = u'(1) = 0$ boundary conditions with
precisely $(k+1)$
exceptions.
\endroster
For $k=0$, (2) means that $q$ is discontinuous at $x=\frac12$.
This example shows that in Theorem~1.2, one cannot weaken the continuity
requirement on $q$. In particular, it provides a counterexample to the
claim in
Suzuki [\suzinv] that his Theorem~I in his Appendix only requires $q\in L^1
((0,1))$.
Continuity of $q$ at $x=\frac12$ is critical for his result to hold.
Our results depend on the following well-known fact:
\proclaim{Proposition~3.1} Suppose that $x_0 < y_0 < x_1 < y_1 < \cdots$ are
given so that for $n$ sufficiently large,
$$
x_n = [(2n)\pi ]^2, \quad y_n = [(2n+1)\pi]^2. \tag 3.1
$$
Then, there exists \rom(a unique\rom) $h_1$ and a $C^\infty$ function $q$ on
$[\frac12, 1]$ so that
$$
-\frac{d^2}{dx^2} + q \quad \text{\rom{in }} L^2 ((\tfrac12, 1));
\qquad u' (\tfrac12) =0, \ u'(1) + h_1 u(1) = 0 \tag 3.2
$$
has eigenvalues $\{ x_n\}^\infty_{n=0}$ and
$$
-\frac{d^2}{dx^2} + q \quad \text{\rom{in }} L^2 ((\tfrac12, 1));
\qquad u (\tfrac12) =0, \ u'(1) + h_1 u(1)=0 \tag 3.3
$$
has eigenvalues $\{y_n\}^\infty_{n=0}$. \rom(By {\rom{(1.2)}} and
{\rom{(A.5c)}}, $h_1=-\tfrac12 \int_{1/2}^1 dxq(x)$.\rom)
\endproclaim
This is just a special case of the construction of Levitan and Gasymov [\lg].
Historically, this classical two-spectra inverse problem goes back to Borg's
seminal paper [\borg]. Subsequently, Levinson [\levs] found considerable
simplifications of Borg's uniqueness arguments, and Krein [\kr] developed his
own solution of these inverse spectral problems. This circle of ideas was
further
developed in [\hsturm], [\lev], [\levbook], Ch.~3, [\lg], [\ls], Sect.~6.11
and continues to
generate interest (see, e.g., [\iwa], [\mal], [\suzgel]).
We also need the elementary:
\proclaim{Lemma 3.2} Suppose that $h_0$, $h_1$, and $q$ are given, and for
some
$\hat \lambda$, there exists an $h_{1/2}$ \rom(with the Dirichlet boundary
condition
$h_{1/2} = \infty$ at $x = \tfrac12$ allowed; $u'(\frac12) + h_{1/2}
u(\frac12) = 0$
is then interpreted as $u(\frac12) = 0$\rom) so that $\hat \lambda$ is an
eigenvalue of
both
$$
-\frac{d^2}{dx^2} + q \quad \text{in } L^2 ((0,\tfrac12)); \qquad
u'(0) + h_0 u(0) = 0, \ u' (\tfrac12) + h_{1/2} u (\tfrac12) = 0
$$
and
$$
-\frac{d^2}{dx^2} + q \quad \text{in } L^2 ((\tfrac12,1)); \qquad
u' (\tfrac12) + h_{1/2} u (\tfrac12) = 0, \ u'(1) + h_1 u(1) = 0.
$$
Then $\hat \lambda$ is also an eigenvalue of
$$
-\frac{d^2}{dx^2} + q \quad \text{in } L^2 ((0,1)); \qquad
u'(0) + h_0 u(0) = 0, \ u' (1) + h_{1} u (1) = 0.
$$
\endproclaim
\demo{Proof} One can match the solutions in the two halves so that they and
their
first derivatives become absolutely continuous near $x = \frac12$. \qed
\enddemo
Let $x^{(0)}_n = [(2n)\pi ]^2$ and $y^{(0)}_n = [(2n+1)\pi ]^2$, $n =
0,1,\dots$
be the eigenvalues that lead to $q = 0$ on $[\frac12, 1]$ (and $h^{(0)}_1 =
0$)
in Proposition~3.1. To construct our example, we will take $x_n =
x^{(0)}_n$ for all
$n = 0,1,\dots$ and $y_n = y^{(0)}_n$ for $n = 0,1 \dots$ with $k+1$
exceptions,
say, $y_{n_0} \neq y^{(0)}_{n_0}, y_{n_1} \neq y^{(0)}_{n_1}, \dots, y_{n_k}
\neq y^{(0)}_{n_k}$. Make the choice so that for $\ell = 1,2, \dots, k$,
$$
\sum^k_{j=0} y^\ell_{n_j} = \sum^k_{j=0} [y^{(0)}_{n_j}]^\ell . \tag 3.4
$$
Choices satisfying (3.4) can certainly be made. For example, we can take
$y_{n_0}
= y^{(0)}_{n_0} + \varepsilon$ with $\varepsilon$ small, and solve the $k$
equations, (3.4), for $y_{n_1}, \dots, y_{n_k}$ using the fact that the
Jacobian
determinant that needs to be non-zero to apply the inverse function theorem is
essentially just a $k \times k$ Vandermonde determinant, $\det(a)$, with
$a_{j\ell} = \ell (y^{(0)}_{n_j})^{\ell-1}$.
Let $m^{(0)}_{N,+} (z,\frac12) = -\frac{1}{\sqrt z} \cot (\frac12 \sqrt
z)$, the
Neumann $m_+$-function on $[\frac12, 1]$ for $q=0$ (and $h^{(0)}_1 =0$).
Let $m_{N,+} (z)$ be the corresponding Neumann $m_+$-function on
$[\frac12,1]$ for the $q$ constructed in Proposition~3.1 (whose poles and
zeros
are given by the eigenvalues $x_n$ and $y_n$, $n=0,1 \dots$ of (3.2) and
(3.3)).
We claim that
$$
m_{N,+} (z,\tfrac12) = \prod^k_{j=0}
\biggl( \frac{z-y_{n_j}}{z-y^{(0)}_{n_j}} \biggr) m^{(0)}_{N,+} (z,\tfrac12).
\tag 3.5
$$
Indeed, the two sides have the same zeros and poles and both are ratios of
functions of order $\frac12$; thus they are constant multiples of each other.
Since both sides behave asymptotically like $\sim -|z|^{-1/2}$ as $z\to
-\infty$,
the constant multiple must be $1$.
Because of (3.4),
$$\align
\ln \biggl( \prod^k_{j=0} \biggl( \frac{z-y_{n_j}}{z-y^{(0)}_{n_j}}
\biggr) \biggr) =
\sum^k_{j=1} \biggl[ \ln \biggl( 1-\frac{y_{n_j}}{z}\biggr) -
\ln \biggl( 1-\frac{y^{(0)}_{n_j}}{z} \biggr) \biggr]
= O(z^{-k-1}). \tag 3.6
\endalign
$$
Since $m^{(0)}_{N,+}(z, \frac12) = -\frac{i}{\sqrt z} (1+ o(z^{-K}))$ for all
$K$ consistent with $q=0$ in (2.10), (3.5) and (3.6) imply that
$$
m_{N,+} (z, \tfrac12) = -\frac{i}{\sqrt z}\, (1+ O(z^{-k-1})).
$$
Thus in (2.6), $D_\ell (\frac12) = 0$ for $\ell = 1,2, \dots, 2k+1$. But by
(2.10)
and induction, this implies that
$$
q^{(m)} (\tfrac12) = 0 \qquad \text{for } m=0,1,2,\dots, 2k-1. \tag 3.7
$$
Next, let $q(x)$ be defined a.e.~on $[0,1]$ by
$$\alignat2
q(x) & = 0, &&\qquad 0\leq x < \tfrac12 \\
&= \text{constructed } q \restriction (\tfrac12,1),
&& \qquad \tfrac12 < x \leq 1.
\endalignat
$$
By (3.7), $q$ is $C^{2k-1}$ at $x = \frac12$. By Lemma~3.2, $-\frac{d^2}
{dx^2} + q$ in $L^2 ((0,1))$, with $u'(0)=0$, $u'(1) + h_1u(1)=0$ boundary
conditions, has $\{x^{(0)}_n\}^\infty_{n=0} \cup \{y^{(0)}_n\}^\infty_{n=0,
n \neq n_0, n_1, \dots, n_k}$ as eigenvalues, so at most $(k+1)$ eigenvalues
differ from the free Neumann case (i.e., $q=0$ on $[0,1], \, h_0 = h_1 =
0$). If fewer
than $k+1$ eigenvalues differed, then by Theorem~1.2, $q\equiv 0$ and $h_1
=0$.
Since the $y_{n_i}$ are not $y^{(0)}_{n_i}$, this cannot be true. Thus,
exactly
$k+1$ eigenvalues differ. If $q^{(2k)}(\frac12)$ were zero, then by
Theorem~1.2,
again $q\equiv 0$ and $h_1 =0$, so $q$ is not $C^{2k}$.
There remains an interesting open question: Can one replace information on the
missing eigenvalue by knowledge of the boundary condition $h_1$?
\vskip 0.3in
\flushpar{\bf \S 4. The Case of Partially Known Spectra}
\vskip 0.1in
Our goal in this section is to prove Theorem~1.3. Define
$$
g_S (z) = \prod_{\lambda_n\in S} \biggl( 1-\frac{z}{\lambda_n}\biggr), \quad
g_{\sigma(H)} (z) = \prod_{n=0}^{\infty} \biggl( 1-\frac{z}{\lambda_n}\biggr),
\quad S \subseteq \sigma(H)=\{\lambda_n\}_{n=0}^{\infty}.
$$
By the hypothesis (1.4) on $S$ and $\sigma(H)$ in Theorem~1.3 and the
method of
proof of Theorem~B.4 (see the critical equality (B.16)), we infer
$$
\ln (| g_S (iy)|) \geq (1-\alpha) \ln (|g_{\sigma(H)} (iy)|)
+ \tfrac\alpha4 \ln(1+y^2)+C_0. \tag 4.1
$$
Since $\sigma(H)$ is a complete set of eigenvalues for a self-adjoint
problem on
$[0,1]$, we know that asymptotically
$$
|g_{\sigma(H)} (iy)| \sim \tfrac12 \, |y|^{1/2}\, \bigl|e^{\text{\rom{Im}}
(\sqrt i) \, |y|^{1/2}}\bigr| \quad \text{as } y \text{ (real)} \to \infty.
$$
Thus by (4.1), there exists a constant $C>0$ such that
$$
|g_S (iy)| \geq C|y|^{1/2} \, \bigl| e^{\text{\rom{Im}}
(\sqrt i) (1-\alpha) |y|^{1/2}} \bigr| \tag 4.2
$$
for $|y|$ sufficiently large.
Let $P_j(z)=u_{j,+} (z,\frac12 + \frac\alpha2)$, $Q_j (z) = u'_{j,+}
(z,\frac12
+ \frac\alpha2)$, $j=1,2$ for the two candidate potentials. Then, since $1-
(\frac12 + \frac\alpha2) = \frac12 (1-\alpha)$, we use (2.3) to infer
asymptotically
$$
|Q_j (iy)| \sim \tfrac12 \, |y|^{1/2} \bigl| e^{\text{\rom{Im}} (\sqrt i)
\, (1/2)
(1-\alpha) |y|^{1/2}} \bigr| \quad \text{as } y \text{ (real)} \to
\infty. \tag 4.3
$$
With (4.2), (4.3), the arguments in Section~2 extend to prove Theorem~1.3. \qed
\vskip 0.3in
\flushpar{\bf \S 5. Theorems for the Whole Real Line}
\vskip 0.1in
Our main goal in this section is to prove Theorem~1.4. So we suppose that
$q_1 (x),
q_2(x)$ are two potentials on $\Bbb R$ satisfying
$$
q_j (x) \geq C|x|^{2+\varepsilon} +1, \quad j=1,2 \tag 5.1
$$
for some $C,\varepsilon >0$. (The condition in Theorem~1.4 has $-D$ in
place of
$1$. Just add $D+1$ to initial $q_j$'s to get (5.1) if need be.) Thus for
any $z\in
\Bbb C$, there exist solutions $u_{j,\pm} (x,z)$ of $-u''(x) + q_j (x) u(x) =
zu(x)$ which are $L^2$ near $\pm\infty$ and
$$
m_{j,\pm} (z) := \pm \frac{u'_{j,\pm}(0,z)}{u_{j,\pm} (0,z)}, \quad j=1,2
$$
are meromorphic functions of $z$. Let $\{\lambda_m\}^\infty_{m=1}$ be the
eigenvalues of $-\frac{d^2}{dx^2} + q_1 (x)$ on $(-\infty,\infty)$ and
denote by
$\{\mu_{j,m} \}^\infty_{m=1}$ the eigenvalues of $-\frac{d^2}{dx^2} +
q_j (x)$ on $(-\infty, 0]$ with Dirichlet (i.e., $u(0)=0$) boundary
conditions at $x=0$. We claim:
\proclaim{Proposition 5.1} {\rom{(i)}} Let $\rho_0 = 1 -\frac{\varepsilon}
{(4+2\varepsilon)}$. Then for all $\rho > \rho_0, \, j=1,2$,
$$\align
\sum_{m=1}^{\infty} \lambda^{-\rho}_m &< \infty, \tag 5.2 \\
\sum_{m=1}^{\infty} (\mu_{j,m})^{-\rho} &< \infty. \tag 5.3
\endalign
$$
{\rom{(ii)}
$$
\mu_{j,m} \geq \lambda_{2m} \, \, , \quad m\in \Bbb N. \tag 5.4
$$
\endproclaim
\demo{Proof} Let $\alpha_m$ be the $m^{\text{\rom{th}}}$ eigenvalue of
$-\frac{d^2}{dx^2} + Cx^{2+\varepsilon}$ on $(-\infty,\infty)$. The large
$m$ asymptotics of
$\alpha_m$ is given by a classical phase space argument (see, e.g.,
Theorem~XIII.81 in [\rs], or Section~7.1 in [\tit]), that is, for an explicit
constant $K\in (0,\infty)$, $\lim_{m\to \infty}
\alpha_m /m^{1/\rho_0} = K$. Thus, $\sum_{m\in\Bbb N} \alpha^{-\rho}_m <
\infty$ if
$\rho > \rho_0$, and so (5.2) holds since $\lambda_m \geq \alpha_m$.
Let $\beta_{j,m}$ be the $m^{\text{\rom{th}}}$ eigenvalue
of $-\frac{d^2}{dx^2}+ q_j (-|x|)$ on $(-\infty,\infty)$. By the hypothesis
$q_j (-|x|) \geq q_j (x)$, we
infer that $\beta_{j,m} \geq \lambda_m$. But by Dirichlet-Neumann alternation,
$\mu_{j,m}= \beta_{j,2m}$ for $m=1,2,\dots$ proving (5.4). (5.3) then
follows from (5.4). \qed
\enddemo
Define
$$\align
Q_j (z) &= \prod^\infty_{m=1} \biggl( 1-\frac{z}{\mu_{j,m}}\biggr), \\
g(z) &= \prod^\infty_{m=1} \biggl( 1 -\frac{z}{\lambda_m}\biggr).
\endalign
$$
\proclaim{Proposition 5.2} For all $y\in\Bbb R$,
$$
\frac{| Q_1 (iy) Q_2 (iy)|}{|g(iy)|} \leq 1. \tag 5.5
$$
\endproclaim
\demo{Proof} $| 1-\frac{iy}{w}| = (1+\frac{y^2}{w^2})^{1/2}$ for $y,w$
real, is monotone decreasing in $w$, so by (5.4),
$$
\biggl| 1- \frac{iy}{\mu_{j,m}} \biggr| \leq \biggl| 1 -\frac{iy}{\lambda_{2m}}
\biggr| \leq \biggl| 1 -\frac{iy}{\lambda_{2m-1}}\biggr|.
$$
It follows that
$$
|Q_j (iy)| \leq \prod^\infty_{m=1} |1-\frac{iy}{\lambda_{2m}}|
\leq \prod^\infty_{m=1} |1-\frac{iy}{\lambda_{2m-1}}|,
$$
so (5.5) holds. \qed
\enddemo
Now let
$$
P_j (z) = m_{j,-} (z) Q_j (z), \quad j=1,2,
$$
which are entire functions, and
$$
f_j (z) = m_{j,-} (z), \quad j=1,2.
$$
Define
$$
F(z) = \frac{P_1 (z) Q_2 (z) - Q_1 (z) P_1 (z)}{g(z)} = \frac{Q_1 (z) Q_2(z)}
{g(z)} \, (f_1 (z) - f_2 (z)). \tag 5.6
$$
\demo{Proof of Theorem {\rom{1.4}}} At $z=\lambda_k$, the eigenfunctions
on the left half-line for both $q_1$ and $q_2$ must match to the common
eigenfunctions on the right, so $P_1 (\lambda_k) Q_2 (\lambda_k) - P_2
(\lambda_k) Q_1 (\lambda_k) = 0$, that is, $F(z)$ is an entire function.
By (5.2), $g(z)$ is a function of $m$-type as defined in Appendix~B. Thus by
Proposition~B.5, there exists a sequence $R_k \to\infty$ so that $\sup
\{\frac{1}
{|g(z)|} \mid \, |z| = R_k \} \leq C_1 \exp(+C_2 R^\rho_k)$ for some $\rho
<1$.
By (5.3) and a similar estimate for the Dirichlet eigenvalues, $P_j, Q_j$ are
functions of $m$-type. It follows by Proposition~B.6 that if
$$
\lim_{|y|\to\infty} |F(iy)| =0, \tag 5.7
$$
then $F\equiv 0$. If we prove that, then $m_{1,-} (z) = m_{2,-} (z)$, and thus
$q_1 = q_2$ a.e. on $(-\infty, 0]$ and hence on $\Bbb R$. Thus, we need
only prove (5.7).
By (2.4), which holds for half-line $m$-functions [\atk, \eve], $|f_1 (iy) -
f_2 (iy)| = o(1)$. Thus, Proposition~5.2 and (5.6) show that (5.7) holds. \qed
\enddemo
Several questions remain open. We do not believe that hypothesis (i) is
needed in Theorem~1.4:
\proclaim{Conjecture 5.3} Theorem~{\rom{1.4}} remains true if {\rom{(i)}} is
replaced by $\lim_{|x|\to\infty} q(x) = \infty$.
\endproclaim
This will require dealing with entire functions of type larger than $1$.
We also believe:
\proclaim{Conjecture 5.4} Suppose $\lim_{x\to\infty} \frac{q(-x)}{q(x)} =
\infty$ and
$\lim_{|x|\to\infty} q(x) = \infty$. Then $q$ near $+\infty$ and a finite
density subset
of eigenvalues for $-\frac{d^2}{dx^2} + q(x)$ uniquely determine $q$ on
$\Bbb R$.
\endproclaim
\vskip 0.3in
\flushpar {\bf Acknowledgments.} We would like to thank R.~del Rio for
discussions
and pertinent hints to the literature. F.G.~is indebted to A.S.~Kechris and
C.W.~Peck
for a kind invitation to Caltech during the summer of 1996 where some of
this work
was done. The extraordinary hospitality and support by the Department of
Mathematics
at Caltech are gratefully acknowledged. B.S.~would like to thank
M.~Ben-Artzi and
the Hebrew University during a visit where some of this work was done.
\vskip 0.3in
\flushpar{\bf Appendix A: Dirichlet Boundary Conditions}
\vskip 0.1in
In this appendix we provide some details in the remaining cases, which involve
Dirichlet boundary conditions at $x=0$ and/or $x=1$. We need to distinguish
three cases (cf.~(1.1)):
\roster
\item"\rom{(I)}" $H$ has a Dirichlet boundary condition at $x=0$ and $x=1$,
that is,
$$
\align
u(0) & = 0, \tag A.1a \\
u(1) & = 0. \tag A.1b
\endalign
$$
\item"\rom{(II)}" $H$ has a non-Dirichlet boundary condition at $x=0$ and a
Dirichlet boundary condition at $x=1$, that is,
$$
\align
u'(0) + h_0 u(0) & = 0, \quad h_0 \in \Bbb R,\tag A.2a \\
u(1) & = 0. \tag A.2b
\endalign
$$
\item"\rom{(III)}" $H$ has a Dirichlet boundary condition at $x=0$ and a
non-Dirichlet boundary condition at $x=1$, that is,
$$
\align
u(0) & = 0, \tag A.3a \\
u'(1) + h_1 u(1) & = 0, \quad h_1 \in \Bbb R. \tag A.3b
\endalign
$$
\endroster
Since later on, $q$ is supposed to be known on $[0,\frac12 + \varepsilon]$ for
some $0\leq\varepsilon < \frac12$, cases II and III represent inequivalent
situations and need to be treated separately in connection with
Theorems~A.1--A.3. Depending on the case at hand, we index the corresponding
eigenvalues $\lambda_n$ of $H$ by
$$
\{\lambda_n\}_{n=1}^\infty \quad \text{\rom{in case I \quad and}} \quad
\{\lambda_n\}_{n=0}^\infty \quad \text{\rom{in cases II and III}}. \tag A.4
$$
The asymptotic expansion (1.2) then becomes as $n \to \infty$,
$$\alignat2
\lambda_n &= (\pi n)^2 + \int_0^1 dx\, q(x) +o(1) &&\quad \text{in case I},
\tag A.5a \\
\lambda_n &= (\pi (n+\tfrac12))^2 - 2h_0 + \int_0^1 dx\, q(x) +o(1)
&& \quad \text{in case II}, \tag A.5b \\
\lambda_n &= (\pi (n+\tfrac12))^2 + 2h_1 + \int_0^1 dx\, q(x) +o(1)
&& \quad \text{in case III}. \tag A.5c
\endalignat
$$
Let $u_+ (z,x)$ be defined by $-u''_+ + qu_+ = zu_+$ subject to the boundary
conditions and normalizations
$$\alignat3
u'_+ (z,1) &= 1, &\quad u_+ (z,1) &=0 &&\qquad \text{in cases I and II},
\tag A.6a \\
u'_+ (z,1) &= -h_1, &\quad u_+ (z,1) &=1 &&\qquad \text{in case III}. \tag A.6b
\endalignat
$$
Next we note that items (1) and (3)--(5) of Section~2 remain valid in the
present
cases I--III, whereas item (2) becomes
$$\align
u_+ (z,x) &= - \frac {\sin \big( \sqrt z \,(1-x)\big)}{\sqrt z}
+ O \biggl( \frac{e^{\text{\rom{Im}} \, (\sqrt z) \, (1-x)}}{z} \biggr),
\tag A.7 \\
u'_+ (z,x) &= \cos \big( \sqrt z \, (1-x)\big) +
O\biggl( \frac{e^{\text{\rom{Im}}\, (\sqrt z) \, (1-x)}}{\sqrt z} \biggr)
\tag A.8
\endalign
$$
in cases I and II, and
$$\align
u_+ (z,x) &= \cos \big( \sqrt z \, (1-x)\big) +
O\biggl( \frac{e^{\text{\rom{Im}}\, (\sqrt z) \, (1-x)}}{\sqrt z} \biggr),
\tag A.9 \\
u'_+ (z,x) &= \sqrt z \, \sin \big( \sqrt z \, (1-x)\big)+
O \bigl( e^{\text{\rom{Im}} \, (\sqrt z) \, (1-x)}\big) \tag A.10
\endalign
$$
in case III, as $|z| \to \infty$ for all $x\in [0,1]$.
Introducing $P_j, Q_j, f_j$, $j=1,2$ as in (2.11)--(2.13), and $g$ by
$$\alignat2
g(z)&=u_{1,+}(z,0) &&\quad \text{in cases I and III}, \tag A.11a \\
g(z)&=u'_{1,+} (z,0) + h_0 u_{1,+} (z,0) &&\quad\text{in case II},
\tag A.11b
\endalignat
$$
one infers again that $\{\lambda_{1,n}\}^\infty_{n=1}$ in case I and
$\{\lambda_{1,n}\}^\infty_{n=0}$ in cases II and III are precisely the
zeros of
$g(z)$. The corresponding properties (a)--(g) listed in Section~2, suitably
adapted to the present cases I--III, then read as follows:
\roster
\item"\rom{(a$^\prime$)}" Item (a) remains valid except that $n$ in (2.15)
runs
through $\Bbb N$ in case I and through $\Bbb N_0$ in cases II and III.
\item"\rom{(b$^\prime$--d$^\prime$)}" Items (b), (c), and (d) remain valid.
\item"\rom{(e$^\prime$)}" As $y \text{ (real)} \to\infty$,
$$\alignat2
|Q_j (iy)| &= \tfrac12 \, |\exp (\tfrac12 \, \text{Im } (\sqrt i)\,
|y|^{1/2})|
(1+o(1)) &&\quad \text{in cases I and II}, \\
|Q_j (iy)| &= \tfrac12 \, |y|^{1/2} \, |\exp (\tfrac12 \, \text{Im } (\sqrt
i)\,
|y|^{1/2}) |(1+o(1)) &&\quad \text{in case III}.
\endalignat
$$
\item"\rom{(f$^\prime$)}" As $y \text{ (real)} \to\infty$,
$$\alignat2
|g(iy)| &= \tfrac12 \, |y|^{-1/2} \,|\exp (\text{Im } (\sqrt i) \, |y|^{1/2})|
\, (1+o(1)) &&\quad \text{in case I}, \\
|g(iy)| &= \tfrac12 \, |\exp (\text{Im } (\sqrt i) \, |y|^{1/2})| \, (1+o(1))
&&\quad \text{in cases II and III}.
\endalignat
$$
\item"\rom{(g$^\prime$)}" For $n$ sufficiently large, one obtains,
$$\alignat2
\inf_{\theta \in [0,2\pi]} |g((\pi (n+ \tfrac12 ))^2 e^{i\theta})| &\geq
\tfrac{1}{\pi n} + O(\tfrac{1}{n^2}) &&\quad\text{in case I}, \\
\inf_{\theta \in [0,2\pi]} |g((\pi (n+1))^2 e^{i\theta})| &\geq
1 + O(\tfrac{1}{n}) &&\quad \text{in cases II and III}.
\endalignat
$$
\endroster
Introducing $F(z)$ as in (2.16) and (2.18), one verifies, using
(b$^\prime$) and
(g$^\prime$), that (2.17) remains valid. Items (c$^\prime$), (e$^\prime$), and
(f$^\prime$), however, yield the following modification of (2.19) as $y
\text{ (real)} \to\infty$,
$$\alignat2
|F(iy)| &= \tfrac12 \, |y|^{1/2} \, [o(|y|^{-1})] (1+o(1)) = o(|y|^{-1/2})
&&\quad \text{in case I}, \tag A.12a \\
|F(iy)| &= \tfrac12 [o(|y|^{-1})] (1+o(1)) = o(|y|^{-1})
&&\quad \text{in case II}, \tag A.12b \\
|F(iy)| &= \tfrac12 \, |y| [o(|y|^{-1})] (1+o(1)) = o(1)
&&\quad \text{in case III}. \tag A.12c
\endalignat
$$
Following the arguments in (2.16)--(2.19) step-by-step and taking into account
the eigenvalue asymptotics in (A.5), the remaining Dirichlet cases in
Theorem~1.1
then read as follows.
\proclaim{Theorem A.1} Let $H=-\frac{d^2}{dx^2}+ q$ in $L^2 ((0,1))$ with
boundary conditions {\rom{(A.1)}}, {\rom{(A.2)}}, or {\rom{(A.3)}} according to
cases \rom{I}, {\rom{II}}, or {\rom{III}}. Then $q$ on $[0,\frac12]$,
together with
the knowledge of $h_0=\infty$ or $h_0\in\Bbb R$, and all the eigenvalues of
$H$,
uniquely determine $h_1$ \rom(i.e., $h_1=\infty$ in cases \rom{I} and
{\rom{II}}
and $h_1\in\Bbb R$ in case {\rom{III)}} and $q$ \rom(a.e.\rom) on all of
$[0,1]$.
\endproclaim
\remark{Remark} Case I for $q \in L^2((0,1))$ appears to be due to
P\"oschel and
Trubowitz [\pt], Ch.~4. Much to our surprise, the extension of case I to $q
\in
L^1((0,1))$ in Theorem~A.1 seems to be new. Case II is originally due to
Hochstadt
and Lieberman [\hl] as recorded in Theorem~1.1. To the best of our knowledge,
case III is a new result.
\endremark
The analog of Theorem~1.2 is now obtained as follows. Replace the
definition of
$\tilde F$ in (2.20) by
$$\alignat2
\tilde F(z) &= \frac{[P_1(z) Q_2(z) - P_2 (z) Q_1 (z)]}{g(z)} \,
\prod^{k+1}_{\ell=1} (z- \lambda_{\ell}) &&\quad \text{in cases I and II},
\tag A.13a \\
\tilde F(z) &= \frac{[P_1(z) Q_2(z) - P_2 (z) Q_1 (z)]}{g(z)} \,
\prod^{k}_{\ell=1} (z- \lambda_{\ell}) &&\quad \text{in case III}. \tag A.13b
\endalignat
$$
Here $\{\lambda_{\ell}\}_{\ell =1}^{k+1}$ in cases I and II, and
$\{\lambda_{\ell}\}_{\ell =1}^k$ in case III represent eigenvalues of $H_1$
and
$H_2$ which are not a priori assumed to be equal.
The asymptotic behavior of $|\tilde F(iy)|$ as $y \text{ (real) } \to\infty$,
$$\alignat2
|\tilde F(iy)| &= o(1) &&\quad \text{in case I}, \tag A.14a \\
|\tilde F(iy)| &= o(|y|^{-1/2}) &&\quad \text{in cases II and III}, \tag A.14b
\endalignat
$$
then yields the following new result.
\proclaim{Theorem A.2} Let $H=-\frac{d^2}{dx^2}+ q$ in $L^2 ((0,1))$
with boundary conditions {\rom{(A.1)}}, {\rom{(A.2)}}, or {\rom{(A.3)}}
according to cases \rom{I}, {\rom{II}}, or {\rom{III}}. Suppose $q$ is $C^{2k}
((\frac12 -\varepsilon, \frac12 +\varepsilon))$, $k=0,1,\dots$\, for some
$\varepsilon
> 0$. Then $q$ on $[0,\frac12]$, together with the knowledge of
$h_0=\infty$ or
$h_0\in\Bbb R$, and all the eigenvalues of $H$ except for $(k+1)$ in cases
\rom{I}
and {\rom{II}} and $k$ in case {\rom{III}}, uniquely determine $h_1$
\rom(i.e.,
$h_1 =\infty$ in cases \rom{I} and {\rom{II}} and $h_1\in\Bbb R$ in case
{\rom{III)}} and $q$ on all of $[0,1]$.
\endproclaim
Finally, we consider the analog of Theorem~1.3 in the Dirichlet context.
\proclaim{Theorem A.3} Let $H=-\frac{d^2}{dx^2}+q$ in $L^2 ((0,1))$ with
boundary conditions {\rom{(A.1)}}, {\rom{(A.2)}}, or {\rom{(A.3)}}. Then $q$
on $[0, \frac12 +\frac\alpha2]$ for some $\alpha \in (0,1)$, $h_0=\infty$
or $h_0
\in\Bbb R$, and a subset $S \subseteq \sigma(H)$ of all the eigenvalues
$\sigma
(H)$ of $H$ satisfying
$$
\#\{\lambda\in S \mid \lambda\leq \lambda_0 \} \geq (1-\alpha)
\# \{\lambda\in\sigma(H) \mid \lambda\leq \lambda_0 \} + \tfrac\alpha2
\tag A.15
$$
for all sufficiently large $\lambda_0 \in \Bbb R$, uniquely determine $h_1$
\rom(i.e., $h_1=\infty$ in cases \rom{I} and {\rom{II}} and $h_1\in\Bbb R$ in
case {\rom{III)}} and $q$ on all of $[0,1]$.
\endproclaim
\demo{Proof} Following the arguments employed in Section~4, we introduce again
$$\alignat2
g_S (z) &= \prod_{\lambda_n\in S} \biggl( 1-\frac{z}{\lambda_n}\biggr),
&&\quad S \subseteq \sigma(H), \\
g_{\sigma(H)} (z) &= \prod_{n=1}^{\infty} \biggl( 1-\frac{z}{\lambda_n}\biggr)
&&\quad \text{in case I}, \\
g_{\sigma(H)} (z) &= \prod_{n=0}^{\infty} \biggl( 1-\frac{z}{\lambda_n}\biggr)
&&\quad \text{in cases II and III},
\endalignat
$$
where
$$
\sigma(H)=\{\lambda_n\}_{n=1}^{\infty} \quad \text{in case I \quad and}
\quad \sigma(H)=\{\lambda_n\}_{n=0}^{\infty} \quad \text{in cases II and III}.
$$
Then (A.15) and the method of proof of Theorem~B.4 yield
$$
\ln (| g_S (iy)|) \geq (1-\alpha) \ln (|g_{\sigma(H)} (iy)|)
+ \tfrac\alpha4 \ln(1+y^2)+C_0. \tag A.16
$$
Since asymptotically (cf.~(f$^\prime$)) for $|y|$ large enough,
$$\alignat2
|g_{\sigma(H)} (iy)| &\sim \tfrac12 \, |y|^{-1/2} \bigl| e^{\text{\rom{Im}}
(\sqrt i) \, |y|^{1/2}}\bigr| &&\quad \text{in case I}, \\
|g_{\sigma(H)} (iy)| &\sim \tfrac12 \, \bigl| e^{\text{\rom{Im}} (\sqrt i) \,
|y|^{1/2}}\bigr| &&\quad \text{in cases II and III},
\endalignat
$$
one infers from (A.16) that for some $C>0$
$$\alignat2
|g_S (iy)| &\geq C|y|^{-1/2} \bigl| e^{\text{\rom{Im}} (\sqrt i)
(1-\alpha) |y|^{1/2}} \bigr| &&\quad \text{in case I}, \tag A.17a \\
|g_S (iy)| &\geq C \bigl| e^{\text{\rom{Im}} (\sqrt i)
(1-\alpha) |y|^{1/2}} \bigr| &&\quad \text{in cases II and III} \tag A.17b
\endalignat
$$
for $|y|$ sufficiently large.
Introducing again $P_j(z)=u_{j,+} (z,\frac12 + \frac\alpha2)$, $Q_j (z) =
u'_{j,+} (z,\frac12 + \frac\alpha2)$, $j=1,2$ for the two candidate
potentials,
noticing $1-(\frac12 + \frac\alpha2) = \frac12 (1-\alpha)$, we infer
asymptotically
as $y \text{ (real) } \to\infty$,
$$\alignat2
|Q_j (iy)| &\sim \tfrac12 \, \bigl| e^{\text{\rom{Im}} (\sqrt i) \, (1/2)
(1-\alpha) |y|^{1/2}} \bigr| &&\quad \text{in cases I and II}, \tag A.18a \\
|Q_j (iy)| &\sim \tfrac12 |y|^{1/2} \bigl| e^{\text{\rom{Im}} (\sqrt i) \,
(1/2)
(1-\alpha) |y|^{1/2}} \bigr| &&\quad \text{in case III}. \tag A.18b
\endalignat
$$
Given (A.17) and (A.18), one can now finish the proof of Theorem~A.3 in the
same
way as that of Theorem~1.3 in Section~4. \qed
\enddemo
\remark{Remark} As in the case $\alpha =0$, we have an extension of the
same type
as Theorem~A.2. Explicitly, if $q$ is assumed to be $C^{2k}$ near $x=\frac12 +
\frac\alpha2$, we only need
$$
\#\{\lambda\in S\mid \lambda\leq \lambda_0 \} \geq (1-\alpha)
\#\{\lambda\in \sigma(H) \mid \lambda\leq \lambda_0 \} + \tfrac\alpha2 - N(k)
\tag A.19
$$
instead of (A.15), where $N(k)=k+1$ in cases I and II and $N(k)=k$ in case III.
\endremark
\vskip 0.3in
\flushpar {\bf Appendix B: Zeros of Entire Functions}
\vskip 0.1in
In discussing extensions of Hochstadt's discrete (finite matrix) version
[\hmixed]
of the\linebreak Hochstadt-Lieberman theorem in [\gsjac], we made use of the
following simple lemma which is an elementary consequence of the fact that
any polynomial of degree $d$ with $d+1$ zeros must be the zero polynomial:
\proclaim{Lemma B.1} Suppose $f_1 = \frac{P_1}{Q_1}$ and $f_2 =
\frac{P_2}{Q_2}$ are two rational fractions where the polynomials satisfy
$\deg (P_1) = \deg (P_2)$ and $\deg (Q_1)=\deg (Q_2)$. Suppose that $d =
\deg (P_1) + \deg (Q_1)$ and that $f_1 (z_n) = f_2 (z_n)$ for $d+1$ distinct
points $\{z_n\}^{d+1}_{n=1}\in \Bbb C$. Then $f_1 = f_2$.
\endproclaim
Our main goal in this appendix is to prove an analogous theorem for a class of
entire functions. The theorem is sharp in the sense that it includes Lemma~B.1
(at least the case of Lemma~B.1 where the zeros of the entire functions
involved
and the $z_n$ are all positive).
We will be interested here in entire functions of the form
$$
f(z) = C \prod^\infty_{n=0} \biggl( 1- \frac{z}{x_n}\biggr), \tag B.1
$$
where $0 \rho_0$}
\tag B.3
$$
then the product in {\rom{(B.1)}} defines an entire function $f$ with
$$
|f(z)| \leq C_1 \exp (C_2 |z|^\rho ) \qquad \text{for all $\rho > \rho_0$}.
\tag B.4
$$
\item"\rom{(ii)}" Conversely, if $f$ is an entire function satisfying
{\rom{(B.4)}}
with all its \rom(complex\rom) zeros on $(0, \infty)$, then its zeros
$\{x_n \}^\infty_{n=0}$ satisfy {\rom{(B.3)}}, and $f$ has the canonical
product
expansion {\rom{(B.1)}}.
\endroster
Moreover, {\rom{(B.3)}} holds if and only if
$$
N(t) \leq C |t|^\rho \qquad \text{for all $\rho > \rho_0$}. \tag B.5
$$
\endproclaim
Given this theorem, we single out:
\definition{Definition B.3} A function $f$ is called of $m$-type if and
only if $f$
is an entire function satisfying (B.4) (of type $0<\rho<1$ in the usual
definition)
with all the zeros of $f$ on $(0, \infty)$.
\enddefinition
Our choice of ``$m$-type" in Definition~B.3 comes from the fact that in many
cases we discuss in this paper, the $m$-function is a ratio of functions of
$m$-type.
By Theorem~B.2, $f$ in Definition~B.3 has the form (B.1) and $N(t)$, which we
will denote as $N_f (t)$, satisfies (B.5). We are heading toward a proof of
\proclaim{Theorem B.4} Let $f_1 ,f_2 ,g$ be three functions of $m$-type so
that:
\roster
\item"\rom{(i)}" $f_1 (z) = f_2 (z)$ at any point $z$ with $g(z) =0$.
\item"\rom{(ii)}" For all sufficiently large $t$,
$$
\max (N_{f_1} (t), N_{f_2} (t)) \leq N_g (t) -1.
$$
\endroster
Then, $f_1 =f_2$.
\endproclaim
\proclaim{Proposition B.5} Let $f$ be a function of $m$-type. Then there
exists
a $0 < \rho < 1$ and a sequence $\{R_k\}_{k=1}^{\infty}$, $R_k \to \infty$ as
$k \to \infty$, so that
$$
\inf \{ |f(z)| \mid |z| = R_k\} \geq C_1 \exp (-C_2 R^{\rho}_k).
$$
\endproclaim
\demo{Proof} By hypothesis, for some $0 <\rho' < 1$,
$$
N_f (t) \leq Ct^{\rho'}. \tag B.6
$$
This implies
$$
x_n \geq \biggl(\frac{n}{C}\biggr)^{1/{\rho'}}. \tag B.7
$$
If for all $n \geq n_0$,
$$
|x_n - x_{n-1}| \leq 2,
$$
then
$$
x_n \leq x_{n_0} + 2(n-n_0),
$$
which contradicts (B.7). Thus for an infinite sequence
$\{n(k)\}_{k=1}^{\infty}$,
$n(k) \to \infty$ as $k \to \infty$, we necessarily must have
$$
x_{n(k)} - x_{n(k)-1} \geq 2. \tag B.8
$$
We will pick
$$
R_k = \tfrac12\, (x_{n(k)} + x_{n(k)-1}). \tag B.9
$$
For any $\alpha >0$, $| 1-\alpha e^{i\theta}|^2 = 1 +\alpha^2 - 2\alpha
\cos(\theta)$ takes its minimum value at $\theta =0$, so
$$
\inf \{ |f(z)| \mid z=R_k \} = |f(R_k)|. \tag B.10
$$
By Theorem~B.2, $f$ has the form (B.1). We will write
$$
\ln (|f(R_k)|) = A_1 + A_2, \tag B.11
$$
where
$$
A_1 = \sum_{n\mid x_n \geq 2x_{n(k)}} \ln\biggl(\, \biggl| 1- \frac{R_k}{x_n}
\biggr| \, \biggr), \quad A_2 = \sum_{n\mid x_n < 2x_{n(k)}} \ln \biggl( \,
\biggl| 1-\frac{R_k}{x_n}\biggr| \, \biggr).
$$
We estimate $A_1$ by writing the sum as a Stieltjes integral, integrating
by parts,
and using (B.6):
$$\align
A_1 &= \int^\infty_{2x_{n(k)}} \ln \biggl( 1- \frac{R_k}{t}\biggl)\, dN_f
(t) \\
&= - \int^\infty_{2x_{n(k)}} \biggl( \frac{1}{1-\frac{R_k}{t}}\biggr) \,
\frac{R_k}{t^2} \, [N_f (t) - N_f (2x_{n(k)})]\,dt \\
&\geq -\int^\infty_{2R_k} \frac{2R_k}{t^2} \, Ct^{\rho'} \, dt
= -CR^{\rho'}_k, \tag B.12
\endalign
$$
where we have used $C$ to represent a positive constant that varies from
formula to formula.
For $A_2$, we write,
$$\align
\ln \biggl( \, \biggl|1-\frac{R_k}{x_n}\biggr| \, \biggr) &= \ln (|x_n -
R_k|) -
\ln (|x_n|)
\geq - \ln (|x_n|) \tag B.13 \\
&\geq - \ln (4R_k), \tag B.14
\endalign
$$
where (B.13) follows from (B.8) and (B.9), and (B.14) follows from
$$
|x_n| < 2x_{n(k)} \leq 2(2R_k).
$$
Thus by (B.6),
$$
A_2 \geq -N_f (2x_{n(k)}) \ln (4R_k)
\geq -C R^{\rho'}_k \ln (4R_k) \geq -CR^{\rho''}_k \tag B.15
$$
for some $1> \rho'' > \rho'$ and suitable positive constants $C$. (B.10),
(B.11),
(B.12), and (B.15) prove the proposition. \qed
\enddemo
\proclaim{Proposition B.6} Let $F$ be an entire function that satisfies
\roster
\item"\rom{(i)}" $\sup_{|z|=R_k} |F(z)| \leq C_1 \exp (C_2 R^\rho_k)$ for
some $0 \leq \rho <1$, $C_1, C_2 >0$, and some sequence $R_k \to \infty$ as
$k \to \infty$.
\item"\rom{(ii)}" $\lim_{|x|\to\infty; x\in\Bbb R} |F(ix)| =0$.
\endroster
Then $F \equiv 0$.
\endproclaim
\demo{Proof} A standard Phragm\'en-Lindel\"of estimate separately applied to
$\text{Re}(z)>0$ and $\text{Re}(z)<0$ (i.e., to an angle of opening $\pi$;
see,
e.g., [\marku], Sect.~II.34) shows that $F$ is bounded. Liouville's theorem
implies that $F$ is constant, and then the fact that $|F(ix)| \to 0$ as $x \to
\infty$ ($x$ real), shows that $F=0$. \qed
\enddemo
\demo{Proof of Theorem {\rom{B.4}}} By hypothesis (i), $\frac{f_1 (z)-f_2
(z)}{g(z)}:=
Q(z)$ is an entire function. By (B.4) (applied to $f_1$ and $f_2$) and
Proposition~B.5 (applied to $g$), there is a sequence $R_k \to \infty$ as
$k \to \infty$ so that $Q(z)$ satisfies condition (i) of Proposition~B.6.
Thus, it
suffices to prove $\lim_{x\to\pm\infty}\frac{f_1 (ix)}{g(ix)} =
\lim_{x\to\pm\infty} \frac{f_2 (ix)}{g(ix)} =0$. We will prove the $f_1$
result
for definiteness. In fact, our proof will show that $\frac{f_1 (ix)}{g (ix)} =
O(x^{-1})$ as $|x| \to \infty$. Without loss of generality, we will assume
that
$f_1,g$ satisfy (B.1) with $C=1$, that is, that $f_1 (0)=g(0)=1$. We will also
suppose that $N_{f_1} (t)=N_g (t)=0$ if $t\leq 1$, which can be arranged by
appropriate scaling.
Notice first that
$$\align
\ln (|f_1 (ix)|) &= \sum_{n=0}^{\infty} \tfrac12 \,
\ln \biggl( 1 + \frac{x^2}{x^2_n} \biggr) \\
&= \tfrac12 \int^\infty_0 \ln \biggl( 1+ \frac{x^2}{t^2}\biggr) \, dN_{f_1}
(t) \\
&= \tfrac12 \int^\infty_0 \frac{1}{1+\frac{x^2}{t^2}}
\, \frac{2x^2}{t^3}\, N_{f_1} (t) \, dt \\
&= \int^\infty_0 \frac{x^2}{t^3 + tx^2} \, N_{f_1} (t) \, dt. \tag B.16
\endalign
$$
The boundary term at $t=0$ in the integration by parts step vanishes since
$N_{f_1} (0)=0$ and the one at $t=\infty$ vanishes by the estimate (B.5) and
the fact that for $x$ fixed, $\ln (1+ \frac{x^2}{t^2})=O(t^{-2})$ as $t \to
\infty$.
By hypothesis (ii) of the theorem, there are $t_0 \geq 1$ and $C \geq 0$
such that
$$\alignat2
N_{f_1} (t) &\leq N_g (t) -1, && \qquad t\geq t_0 \tag B.17a\\
&\leq N_g (t) + C, && \qquad t \leq t_0. \tag B.17b
\endalignat
$$
Hence, by (B.16),
$$\align
\ln \biggl(\, \biggl| \frac{f_1 (ix)}{g(ix)}\biggr| \, \biggr) &\leq (C+1)
\int^{t_0}_1 \frac{x^2}{t^3 + tx^2} \, dt - \int^\infty_1
\frac{x^2}{t^3 + tx^2}\, dt \\
&\leq (C+1) \ln (t_0) - \tfrac12 \ln (1+x^2),
\endalign
$$
since $\frac{x^2}{t^3 + tx^2} = -\frac{d}{dt} [\frac12 \ln (1 +
\frac{x^2}{t^2})]$. Thus, as claimed,
$$
\frac{f_1 (ix)}{g(ix)} = O(x^{-1})
$$
as $|x| \to \infty$. \qed.
\enddemo
One can replace (B.17a) by the following pair of conditions for $t\geq t_0
\geq 1$, $t_0$ sufficiently large:
$$\gather
N_{f_1}(t) \leq N_g (t) + D \quad \text{for some } D>-1, \tag B.18 \\
\varlimsup_{t\to\infty} t^{-(D+1)^{-1}} \, | \{ s \in [t_0, t] \mid
N_{f_1} (s) > N_g (s) -1 \}| = 0, \tag B.19
\endgather
$$
where $| \cdot |$ abbreviates Lebesgue measure. Indeed, denoting by
$$
\gamma (t) = | \{ s \in [t_0, t] \mid N_{f_1} (s) > N_g (s) -1 \}|
$$
the Lebesgue measure of the set in (B.19), the method of proof in Theorem~B.4, together with (B.18) and (B.19), imply$$\align\ln \biggl( \, \biggr|\frac{f_
1 (ix)}{g(ix)}\biggr| \, \biggr) &= \int^\infty_0
\frac{x^2}{t^3 + tx^2} \, [N_{f_1}(t) - N_g(t)] \, dt \\
&\leq C \int^{t_0}_1 \frac{x^2}{t^3 + tx^2} \, dt + \int^\infty_{t_0}
\frac{x^2}{t^3 + tx^2} \, [N_{f_1}(t) - N_g(t)] \, dt \\
&= C \int^{t_0}_1 \frac{x^2}{t^3 + tx^2} \, dt + \int_{\{t \in [t_0,\infty)
\mid
N_{f_1} (t) \leq N_g (t) -1 \}} \frac{x^2}{t^3 + tx^2} \, [N_{f_1}(t) -
N_g(t)] \, dt \\
&\quad + \int_{\{t \in [t_0,\infty) \mid N_{f_1} (t) > N_g (t) -1 \}}
\frac{x^2}{t^3 + tx^2} \, [N_{f_1}(t) - N_g(t)] \, dt \\
&\leq C \int^{t_0}_1 \frac{x^2}{t^3 + tx^2} \, dt - \int^{\infty}_{t_0}
\frac{x^2}{t^3 + tx^2} \, dt \\
&\quad + (D+1) \int_{\{t \in [t_0,\infty) \mid N_{f_1} (t) > N_g (t) -1 \}}
\frac{x^2}{t^3 + tx^2} \, dt \\
&\leq (C+1) \ln (t_0) - \tfrac12 \ln (1+x^2) \\
&\quad + (D+1) \int^{\infty}_x \frac{x^2}{t^3 + tx^2} \, dt + (D+1)
\int_{\{t \in [1,x] \mid N_{f_1} (t) > N_g (t) -1 \}}
\frac{x^2}{t^3 + tx^2} \, dt \\
&\leq (C+1) \ln (t_0) - \tfrac12 \ln (1+x^2) + \tfrac12 (D+1) \ln (2) \\
&\quad + (D+1) \int^{\gamma(x)+1}_1 \frac{x^2}{t^3 + tx^2} \, dt \\
&\leq (C+1) \ln (t_0) - \tfrac12 \ln (1+x^2) + \tfrac12 (D+1) \ln (2)
+ (D+1) \int^{\gamma(x)+1}_1 \frac{dt}{t} \\
&\leq - \tfrac12 \ln (1+x^2) + \tfrac12 (D +1) [\ln (2) +
\ln ((1+\gamma(x))^2)] + (C+1) \ln (t_0). \tag B.20
\endalign
$$
In particular, (B.19) is precisely the result needed in (B.20) to ensure
that the
limit $|x| \to \infty$ of $|\frac{f_1(ix)}{g(ix)}|$ is zero. In (B.20) we
used the
obvious inequality $\frac{x^2}{[t^3 + tx^2]} < (\frac1t)$ for $t>0$ and the
fact that
$$
\int_{\Omega} f(t) \,dt \leq \int^{|\Omega|+1}_1 f(t) \, dt
$$
whenever $\Omega \subseteq [1,\infty)$ has finite Lebesgue measure, $|\Omega|
< \infty$, and $f$ is monotone decreasing on $[1,\infty)$.
An interesting case is $D = 0$ in (B.18)--(B.20).
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\enddocument