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%References
\def\atk {1} % Atkinson
\def\ber {2} % Berezinskii
\def\borg {3} % Borg
\def\dl {4} % Danielyan-Levitan
\def\eve {5} % Everitt
\def\gel {6} % Gel'fand
\def\gl {7} % Gel'fand-Levitan
\def\gsmf {8} % Gesztesy-Simon, m-function paper
\def\gsds{9} % Gesztesy-Simon, discrete spectrum
\def\gsip{10} % Gesztesy-Simon, in preparation
\def\lev {11} % Levitan
\def\mara {12} % Marchenko article
\def\marb {13} % Marchenko book
\def\pov {14} % Povzner
\def\rs {15} % Reed-Simon II
\def\simon {16} % Simon book
\def\simvan {17} % Simon Vancouver lecture
\def\sti {18} % Stieltjes
\topmatter
\title A New Approach to Inverse Spectral Theory, \\ I.
Fundamental Formalism
\endtitle
\rightheadtext{Inverse Spectral Theory: Fundamental Formalism}
\author Barry Simon
\endauthor
\leftheadtext{B.~Simon}
\affil Division of Physics, Mathematics, and Astronomy \\
California Institute of Technology \\
Pasadena, CA 91125
\endaffil
\date December 18, 1997
\enddate
\abstract We present a new approach (distinct from
Gel'fand-Levitan) to the theorem of Borg-Marchenko that the
$m$-function (equivalently, spectral measure) for a finite
interval or half-line Schr\"odinger operator determines the
potential. Our approach is an analog of the continued fraction
approach for the moment problem. We prove there is a
representation for the $m$-function $m(-\kappa^2) = -\kappa -
\int_0^b A(\alpha) e^{-2\alpha\kappa}\, d\alpha +
O(e^{-(2b-\varepsilon)\kappa})$. $A$ on $[0,a]$ is a function
of $q$ on $[0,a]$ and vice-versa. A key role is played by a
differential equation that $A$ obeys after allowing $x$-dependence:
$$
\frac{\partial A}{\partial x} = \frac{\partial A}{\partial \alpha}
+\int_0^\alpha A(\beta, x) A(\alpha -\beta, x)\, d\beta.
$$
Among our new results are necessary and sufficient conditions
on the $m$-functions for potentials $q_1$ and $q_2$ for $q_1$
to equal $q_2$ on $[0,a]$.
\endabstract
\endtopmatter
\document
\vskip 0.1in
\flushpar{\bf \S 1. Introduction}
\vskip 0.1in
Inverse spectral methods have been actively studied in the past
years both via their relevance in a variety of applications and
their connection to the KdV equation. A major role is played by
the Gel'fand-Levitan equations. Our goal in this paper is to
present a new approach to their basic results that we expect will
lead to resolution of some of the remaining open questions in
one-dimensional inverse spectral theory. We will introduce a
new basic object (see (1.24) below), the remarkable equation,
(1.28), it obeys and illustrate with several new results.
To present these new results, we will first describe the
problems we discuss. We will consider differential operators on
either $L^2 (0,b)$ with $b<\infty$ or $L^2 (0,\infty)$ of the
form
$$
-\frac{d^2}{dx^2} + q(x). \tag 1.1
$$
If $b$ is finite, we suppose
$$
\beta_1 \equiv \int_0^b |q(x)|\, dx < \infty \tag 1.2
$$
and place a boundary condition
$$
u'(b) + hu(b) =0, \tag 1.3
$$
where $h\in\Bbb R \cup \{\infty\}$ with $h=\infty$ shorthand
for the Dirichlet condition $u(b)=0$. If $b=\infty$, we suppose
$$
\int_y^{y+1} |q(x)|\, dx < \infty \qquad \text{for all } y
\tag 1.4
$$
and
$$
\beta_2 \equiv \sup_{y>0} \int_y^{y+1} \max(q(x), 0)\, dx <
\infty . \tag 1.5
$$
Under condition (1.5), it is known that (1.1) is limit point at
infinity [\rs].
In either case, for each $z\in\Bbb C \backslash [\beta, \infty)$
with $-\beta$ sufficiently large, there is a unique solution (up
to an overall constant), $u(x,z)$, of $-u''+qu =zu$ which obeys
(1.3) at $b$ if $b<\infty$ or which is $L^2$ at $\infty$ if
$b=\infty$. The principal $m$-function $m(z)$ is defined by
$$
m(z) =\frac{u'(0,z)}{u(0,z)}\, . \tag 1.6
$$
We will sometimes need to indicate the $q$-dependence explicitly
and write $m(z;q)$. If $b<\infty$, ``$q$" is intended to
include all of $q$ on $(0,b)$, $b$, and the value of $h$.
If we replace $b$ by $b_1 = b-x_0$ with $x_0\in (0,b)$ and let
$q(s) = q(x_0+s)$ for $s\in (0,b_1)$, we get a new $m$-function
we will denote by $m(z,x_0)$. It is given by
$$
m(z,x) = \frac{u'(x,z)}{u(x,z)}\, . \tag 1.7
$$
$m(z,x)$ obeys the Ricatti equation
$$
\frac{dm}{dx} = q(x) - z - m^2 (z,x). \tag 1.8
$$
Obviously, $m(z,x)$ only depends on $q$ on $(x,b)$ (and on $h$ if
$b<\infty)$. A basic result of the inverse theory says that the
converse is true:
\proclaim{Theorem 1.1 (Borg [\borg], Marchenko [\mara])} $m$
determines $q$. Explicitly, if $q_1, q_2$ are two potentials and
$m_1 (z)=m_2(z)$, then $q_1 \equiv q_2$ \rom(including $h_1 =
h_2$\rom).
\endproclaim
We will improve this as follows:
\proclaim{Theorem 1.2} If $(q_1, b_1, h_1)$, $(q_2, b_1, h_2)$
are two potentials and $a <\min(b_1, b_2)$ and if
$$
q_1 (x) = q_2 (x) \qquad \text{on } (0,a), \tag 1.9
$$
then as $\kappa\to\infty$,
$$
m_1 (-\kappa^2) - m_2 (-\kappa^2) = \tilde O(e^{-2\kappa a}).
\tag 1.10
$$
Conversely, if {\rom{(1.10)}} holds, then {\rom{(1.9)}} holds.
\endproclaim
In (1.10), we use the symbol $\tilde O$ defined by $f=\tilde O(g)$
as $x\to x_0$ (where $\lim_{x\to x_0} g(x)=0$) if and only if
$\lim_{x\to x_0} \frac{|f(x)|}{|g(x)|^{1-\varepsilon}} =0$ for
all $\varepsilon >0$.
>From a results point of view, this local version of the
Borg-Marchenko uniqueness theorem is our most significant new
result, but a major thrust of this paper are the new methods.
Theorem~1.2 says that $q$ is determined by the asymptotics of
$m(-\kappa^2)$ as $\kappa \to \infty$. We can also read off
differences of the boundary condition from these asymptotics.
We will also prove that
\proclaim{Theorem 1.3} Let $(q_1, b_1, h_1)$, $(q_2, b_2, h_2)$
be two potentials and suppose that
$$
b_1 = b_2 \equiv b <\infty, \quad |h_1| + |h_2| < \infty,
\quad q_1(x) = q_2 (x) \qquad\text{on } (0,b). \tag 1.11
$$
Then
$$
\lim_{\kappa\to\infty} e^{2b\kappa}
|m_1 (-\kappa^2) - m_2 (-\kappa^2)| = 4(h_1 - h_2). \tag 1.12
$$
Conversely, if {\rom{(1.12)}} holds for some $b <\infty$ with a
limit in $(0,\infty)$, then {\rom{(1.11)}} holds.
\endproclaim
\remark{Remark} That (1.11) implies (1.12) is not so hard to see.
It is the converse that is interesting.
\endremark
To understand our new approach, it is useful to recall briefly
the two approaches to the inverse problem for Jacobi matrices on
$\ell^2 (\{0,1,2,\dots, \})$ [\ber,\gsmf,\sti]:
$$
A = \pmatrix b_0 & a_0 & 0 & 0 & \dots \\
a_0 & b_1 & a_1 & 0 & \dots \\
0 & a_1 & b_2 & a_2 & \dots \\
\dots & \dots & \dots & \dots & \dots \endpmatrix
$$
with $a_i >0$. Here the $m$-function is just $(\delta_0,
(A-z)^{-1}\delta_0) = m(z)$ and, more generally, $m_n(z)=
(\delta_n, (A^{(n)}-z)^{-1}\delta_n)$ with $A^{(n)}$ on
$\ell^2 (\{n, n+1, \dots, \})$ obtained by truncating the
first $n$ rows and $n$ columns of $A$. Here $\delta_n$ is the
Kronecker vector, that is, the vector with $1$ in slot $n$ and
$0$ in other slots. The fundamental theorem in this case is that
$m(z) \equiv m_0(z)$ determines the $b_n$'s and $a_n$'s.
$m_n(z)$ obeys an analog of the Ricatti equation (1.8):
$$
a^2_n m_{n+1}(z) = b_n - z - \frac{1}{m_n(z)}\, . \tag 1.13
$$
One solution of the inverse problem is to turn (1.13) around to
see that
$$
m_n (z)^{-1} = -z + b_n - a^2_n m_{n+1} (z) \tag 1.14
$$
which, first of all, implies that as $z\to\infty$, $m_n(z) =
-z^{-1} + O(z^{-2})$, so (1.14) implies
$$
m_n (z)^{-1} = -z + b_n + a^2_n z^{-1} + O(z^{-2}). \tag 1.15
$$
Thus, (1.15) for $n=0$ yields $b_0$ and $a^2_0$ and so $m_1(z)$
by (1.13), and then an obvious induction yields successive
$b_k$, $a^2_k$, and $m_{k+1}(z)$.
A second solution involves orthogonal polynomials. Let $P_n(z)$
be the eigensolutions of the formal $(A-z)P_n =0$ with boundary
conditions $P_{-1}(z)=0$, $P_0(z)=1$. Explicitly,
$$
P_{n+1}(z) = a^{-1}_n [(z-b_n)P_n (z)] -a_{n-1} P_{n-1}.
\tag 1.16
$$
Let $d\rho (x)$ be the spectral measure for $A$ and vector
$\delta_0$ so that
$$
m(z) = \int \frac{d\rho (x)}{x-z}. \tag 1.17
$$
Then one can show that
$$
\int P_n(x) P_m(x)\, d\mu(x) =\delta_{nm}, \qquad
n,m=0,1,\dots . \tag 1.18
$$
Thus, $P_n(z)$ is a polynomial of degree $n$ with positive
leading coefficients determined by (1.18). These orthonormal
polynomials are determined via Gram-Schmidt from $\rho$ and by
(1.17) from $m$. Once one has the $P_n$, one can determine the
$a$'s and $b$'s from the equation (1.16).
Of course, these approaches via Ricatti equation and orthogonal
polynomials are not completely disjoint. The Ricatti solution
gives the $a_n$'s and $b_n$'s as continued fractions. The
connection between continued fractions and orthogonal polynomials
goes back a hundred years to Stieltjes' work on the moment problem
[\sti].
The Gel'fand-Levitan-Marchenko [\gl,\lev,\mara,\marb] approach
to the continuum case is a direct analog of this orthogonal
polynomial case. One looks at solutions $U(x,k)$ of
$$
-U'' + q(x)U = k^2 U(x) \tag 1.19
$$
obeying $U(0)=1$, $U'(0)=ik$, and proves that they obey a
representation
$$
U(x,k)=e^{ikx} + \int_{-x}^x K(x,y) e^{iky}\, dy, \tag 1.20
$$
the analog of $P_n(z)=cz^n +$ lower order. One defines $s(x,k)
=(2ik)^{-1} [U(x,k) - U(x,-k)]$ which obeys (1.19) with $s(0)
=0$, $s'(0)=1$.
The spectral measure $d\rho$ associated to $m(z)$ by $d\rho
(\lambda) = \lim_{\varepsilon\downarrow 0} [(2\pi)^{-1}
\text{ Im } m(\lambda + i\varepsilon)\, d\lambda]$ obeys
$$
\int s(x,k) s(y,k)\, d\rho(k^2) = \delta (x-y), \tag 1.21
$$
at least formally. (1.20) and (1.21) yield an integral equation
for $K$ depending only on $d\rho$ and then once one has $K$,
one can find $U$ and so $q$ via (1.19) (or via another relation
between $K$ and $q$).
Our goal in this paper is to present a new approach to the
continuum case, that is, an analog of the Ricatti equation
approach to the discrete inverse problem. The simple idea for
this is attractive but has a difficulty to overcome. $m(z,x)$
determines $q(x)$ at least if $q$ is continuous by the known
asymptotics ([\dl]):
$$
m(-\kappa^2, x) = -\kappa - \frac{q(x)}{2\kappa} + o(\kappa^{-1}).
\tag 1.22
$$
We can therefore think of (1.8) with $q$ defined by (1.22) as an
evolution equation for $m$. The idea is that using a suitable
underlying space and uniqueness theorem for solutions of
differential equations, (1.8) should uniquely determine $m$ for
all positive $x$, and so $q(x)$ by (1.22).
To understand the difficulty, consider a potential $q(x)$ on
the whole real line. There are then functions $u_\pm (x,z)$
defined for $z\in\Bbb C\backslash [\beta, \infty)$ which are
$L^2$ at $\pm\infty$ and two $m$-functions $m_\pm (z,x) =
\frac{u'_\pm (x,z)}{u_\pm (x,z)}$. Both obey (1.8), yet $m_+
(0,z)$ determines and is determined by $q$ on $(0,\infty)$
while $m_- (0,z)$ has the same relation to $q$ on $(-\infty, 0)$.
Put differently, $m_+ (0,z)$ determines $m_+ (x,z)$ for $x>0$
but not at all for $x<0$. $m_-$ is the reverse. So uniqueness
for (1.8) is one-sided and either side is possible! That this
does not make the scheme hopeless is connected with the fact
that $m_-$ does not obey (1.22), rather
$$
m_- (-\kappa^2, x)=\kappa + \frac{q(x)}{2\kappa} +
o(\kappa^{-1}). \tag 1.23
$$
We will see the one-sidedness of the solubility is intimately
connected with the sign of the leading $\pm\kappa$ term in
(1.22/1.23).
The key object in this new approach is a function $A(\alpha)$
defined for $\alpha \in (0,b)$ related to $m$ by
$$
m(-\kappa^2) = -\kappa - \int_0^a A(\alpha) e^{-2\alpha\kappa}
\, d\alpha +\tilde O(e^{-2a\kappa}) \tag 1.24
$$
as $\kappa\to\infty$. We have written $A(\alpha)$ as a function
of a single variable but we will allow similar dependence on
other variables. Since $m(-\kappa^2, x)$ is also an $m$-function,
(1.24) has an analog with a function $A(\alpha, x)$. We will
also sometimes consider the $q$-dependence explicitly, using
$A(\alpha, x;q)$ or for $\lambda$ real and $q$ fixed $A(\alpha,
x;\lambda)\equiv A(\alpha,x; \lambda q)$. If we are interested
in $q$-dependence but not $x$, we will sometimes use $A(\alpha;
\lambda)$. The semicolon and context distinguish between
$A(\alpha, x)$ and $A(\alpha;\lambda)$.
By uniqueness of inverse Laplace transforms (see Theorem~A.2.2
in Appendix 2), (1.24) and $m$ near $-\infty$ uniquely determine
$A(\alpha)$.
Not only will (1.24) hold but, in a sense, $A(\alpha)$ is close
to $q(\alpha)$. Explicitly, in Section~3 we will prove that
\proclaim{Theorem 1.4} Let $m$ be the $m$-function of the
potential $q$. Then there is a function $A(\alpha)\in L^1
(0,b)$ if $b< \infty$ and $A(\alpha)\in L^1 (0,a)$ for all
$a<\infty$ if $b=\infty$ so that {\rom{(1.24)}} holds for any
$a\leq b$ with $a<\infty$. $A(\alpha)$ only depends on $q(y)$
for $y\in [0,\alpha]$. Moreover, $A(\alpha)=q(\alpha) +
E(\alpha)$ where $E(\alpha)$ is continuous and obeys
$$
|E(\alpha)|\leq \biggl( \int_0^\alpha |q(y)|\,
dy\biggr)^2 \exp\biggl(\alpha \int_0^\alpha |q(y)|\, dy\biggr).
\tag 1.25
$$
\endproclaim
Restoring the $x$-dependence, we see that $A(\alpha, x) =
q(\alpha + x) + E(\alpha, x)$ where
$$\lim_{\alpha\downarrow 0}\, \sup_{0\leq x\leq a}
|E(\alpha, x)|=0
$$
for any $a>0$, so
$$
\lim_{\alpha\downarrow 0} A(\alpha, x) = q(x), \tag 1.26
$$
where this holds in general in $L^1$ sense. If $q$ is
continuous, (1.26) holds pointwise. In general, (1.26) will hold
at any point of right Lebesgue continuity of $q$.
Because $E$ is continuous, $A$ determines any discontinuities
or singularities of $q$. More is true. If $q$ is $C^k$, then
$E$ is $C^{k+2}$ in $\alpha$, and so $A$ determines
$k^{\text{th}}$ order kinks in $q$. Much more is true. In
Section~7, we will prove
\proclaim{Theorem 1.5} $q$ on $[0,a]$ is only a function of
$A$ on $[0,a]$. Explicitly, if $q_1, q_2$ are two potentials,
let $A_1, A_2$ be their $A$-functions. If $a0$ and $0\frac12 \|q\|_1$, then
$$
m(-\kappa^2)=-\kappa -\int_0^\infty A(\alpha) e^{-2\alpha\kappa}
\, d\alpha. \tag 1.29
$$
In Section~3, we use this and localization estimates from
Appendix~1 to prove Theorem~1.4 in general. Section~4 is an aside
to study implications of (1.24) for asymptotic expansions. In
particular, we will see that
$$
m(-\kappa^2)=-\kappa -\int_0^a q(\alpha) e^{-2\alpha\kappa}
\, d\alpha + o(\kappa^{-1}), \tag 1.30
$$
which is essentially a result of Atkinson [\atk]. In Section~5,
we turn to proofs of Theorem~1.6 and Theorem~1.3. Indeed, we will
prove an analog of (1.27) for any $a<\infty$. If $a\frac12 \|q\|_1$, then
$$
m(-\kappa^2)=-\kappa -\int_0^\infty A(\alpha) e^{-2\alpha\kappa}
\, d\alpha. \tag 2.3
$$
Moreover, if $q,\tilde q$ are both in $L^1$, then
$$
|A(\alpha;q) - A(\alpha; \tilde q)| \leq \| q-\tilde q\|_1
[Q(\alpha)+\tilde Q(\alpha)] \exp(\alpha [Q(\alpha) +
\tilde Q(\alpha)]). \tag 2.4
$$
\endproclaim
We begin the proof with several remarks. First, since
$m(-\kappa^2)$ is analytic in $\Bbb C\backslash [\beta,\infty)$,
we need only prove (2.3) for all sufficiently large $\kappa$.
Second, since $m(-\kappa^2; q_n)\to m(-\kappa^2; q)$ as $n\to
\infty$ if $\|q_n-q\|_1 \to 0$, we can use (2.4) to see that it
suffices to prove the theorem if $q$ is a continuous function of
compact support, which we do henceforth. So suppose $q$ is
continuous and supported in $[0,B]$.
We will prove the following:
\proclaim{Lemma 2.2} Let $q$ be a continuous function supported
on $[0,B]$. For $\lambda\in\Bbb R$, let $m(z;\lambda)$ be the
$m$-function for $\lambda q$. Then for any $z\in\Bbb C$
with $\text{{\rom{dist}}}(z, [0,\infty))>\lambda \|q\|_\infty$,
$$
m(z;\lambda) =-\kappa - \sum_{n=1}^\infty
M_n (z; q)\lambda^n, \tag 2.5a
$$
where for $\kappa >0$,
$$
M_n (-\kappa^2 ;q) =\int_0^{nB} A_n (\alpha)
e^{-2\kappa \alpha}\, d\alpha, \tag 2.5b
$$
where
$$
A_1 (\alpha) =q(\alpha) \tag 2.6
$$
and for $n\geq 2$, $A_n(\alpha)$ is a continuous function obeying
$$
|A_n(\alpha)| \leq Q(\alpha)^n \frac{\alpha^{n-2}}
{(n-2)!}\, . \tag 2.7
$$
Moreover, if $\tilde q$ is a second such potential and $n\geq 2$,
$$
|A_n (\alpha;q) - A_n (\alpha;\tilde q)| \leq
(Q(\alpha) +\tilde Q(\alpha))^{n-1} \biggl[\, \int_0^\alpha
|q(y)-\tilde q(y)|\, dy \biggr] \frac{\alpha^{n-2}}
{(n-2)!}\, . \tag 2.8
$$
\endproclaim
\demo{Proof of Theorem 2.1 given Lemma 2.2} By (2.7),
$$
\int_0^\infty \sum_{n=2}^\infty |A_n(\alpha)| e^{-2\kappa\alpha}
\, d\alpha <\infty
$$
if $\kappa >\frac12 \|q\|_1$. Thus in (2.5a) for $\lambda=1$,
we can interchange the sum and integral to get the representation
(2.3). (2.7) then implies (2.1) and (2.8) implies (2.4). \qed
\enddemo
\demo{Proof of Lemma 2.2} Let $H_\lambda$ be $-\frac{d^2}{dx^2}
+ \lambda q(x)$ on $L^2(0,\infty)$ with $u(0)=0$ boundary
conditions at $0$. Then $\|(H_0 -z)^{-1}\|=\text{dist}(z,
[0,\infty))^{-1}$. So, in the sense of $L^2$ operators, if
$\text{dist}(z, [0,\infty))>\lambda \|q\|_\infty$, the expansion
$$
(H_\lambda -z)^{-1} = \sum_{n=0}^\infty (-1)^n
(H_0 -z)^{-1} [\lambda q (H_0 -z)^{-1}]^n \tag 2.9
$$
is absolutely convergent.
As is well known, $G_\lambda (x,y;z)$, the integral kernel of
$(H_\lambda -z)^{-1}$, can be written down in terms of the
solution $u$ which is $L^2$ at infinity, and the solution
$w$ of
$$
-w'' +qw =zw \tag 2.10
$$
obeying $w(0)=0$, $w'(0)=1$
$$
G_\lambda (x,y;z) =w(\min(x,y))\, \frac{u(\max(x,y))}{u(0)}\, .
\tag 2.11
$$
In particular,
$$
m(z) = \lim\Sb xFrom this and (2.9), we see that (using $\left.
\frac{\partial G_0}{\partial x}(x,y)\right|_{x=0} =
e^{-\kappa y}$)
$$
m(-\kappa^2;\lambda)=-\kappa -\lambda \int e^{-2\kappa y}
q(y)\, dy + \lambda^2 \langle \varphi_\kappa ,
(H_\lambda +\kappa^2)^{-1} \varphi_\kappa\rangle ,
$$
where $\varphi_\kappa (y)=q(y) e^{-\kappa y}$. Since
$\varphi_\kappa \in L^2$, we can use the convergent expansion
(2.9) and so conclude that (2.5a) holds with (for $n\geq 2$)
$$\split
M_n &(-\kappa^2; q) = \\
&\quad (-1)^{n-1} \int e^{-\kappa x_1}
q(x_1) G_0 (x_1, x_2) q(x_2)
\dots G_0 (x_{n-1}, x_n)
q(x_n) e^{-\kappa x_n}\, dx_1 \dots dx_n.
\endsplit
\tag 2.13
$$
Now use the following representation for $G_0$:
$$\align
G_0 (x,y; -\kappa^2) &= \frac{\text{sinh}(\kappa \min(x,y))}
{\kappa}\, e^{-\kappa\max(x,y)} \\
&= \frac12 \int_{|x-y|}^{x+y} e^{-\ell \kappa}\, d\ell \tag 2.14
\endalign
$$
to write
$$\multline
M_n (-\kappa^2; q) = \\
\frac{(-1)^{n-1}}{2^{n-1}} \int_{R_n}
q(x_1) \dots q(x_n) e^{-2\alpha (x_1, x_n, \ell_1,
\dots, \ell_{n-1})\kappa} \, dx_1 \dots dx_n d\ell_1 \dots
d\ell_{n-1}, \endmultline \tag 2.15
$$
where $\alpha$ is shorthand for the linear function
$$
\alpha = \frac12 \biggl( x_1 + x_n + \sum_{j=1}^{n-1}
\ell_j \biggr) \tag 2.16
$$
and $R_n$ is the region
$$\split
R_n = \{(x_1, \dots, x_n, &\ell_1, \dots, \ell_{n-1}) \in
\Bbb R^{2n-1}\mid 0\leq x_i \leq B \text{ for } i=1,\dots, n; \\
&\qquad |x_i - x_{i+1}| \leq \ell_i \leq x_i + x_{i-1}
\text{ for } i-1, \dots, n-1 \}.
\endsplit
$$
In the region $R_n$, notice that
$$
\alpha\leq \frac12 \biggl( x_1 + x_n + \sum_{j=1}^{n-1}
(x_j + x_{j+1})\biggr) = \sum_{j=1}^n x_j \leq nB.
$$
Change variables by replacing $\ell_{n-1}$ by $\alpha$ using
the linear transformation (2.16) and use $\ell_{n-1}$ for the
linear function
$$
\ell_{n-1} (x_1, x_n, \ell_1, \dots, \ell_{n-2},
\alpha) = 2\alpha - x_1 - x_n - \sum_{j=1}^{n-2} \ell_j.
\tag 2.17
$$
Thus, (2.5b) holds where
$$
A_n(\alpha) = \frac{(-1)^{n-1}}{2^{n-2}} \int_{R_n(\alpha)}
q(x_1) \dots q(x_n) \, dx_1 \dots dx_n d\ell_1 \dots
d\ell_{n-2}. \tag 2.18
$$
$2^{n-1}$ has become $2^{n-2}$ because of the Jacobian of the
transition from $\ell_{n-1}$ to $\alpha$. $R_n(\alpha)$ is the
region
$$\multline
R_n(\alpha) = \{(x_1, \dots, x_n, \ell_1, \dots, \ell_{n-2}) \in
\Bbb R^{2n-2} \mid 0\leq x_i \leq B \text{ for } i=1, \dots, n; \\
|x_i - x_{i+1}| \leq \ell_i \leq x_i + x_{i+1} \text{ for }
i=1,\dots, n-2; \\
|x_{n-1}+x_n| \leq \ell_{n-1}
(x_1, \dots, x_n, \ell_1, \dots, \ell_{n-2},\alpha)
\leq x_{n-1}+x_n\}
\endmultline
\tag 2.19
$$
with $\ell_{n-1}$ the functional given by (2.17).
We claim that
$$
R_n(\alpha) \subset \tilde R_n (\alpha) = \biggl\{(x_1,
\dots, x_n, \ell_1, \dots, \ell_{n-2})\in\Bbb R^{2n-2}
\bigg| 0\leq x_i \leq \alpha; \,
\ell_i \geq 0; \sum_{i=1}^{n-2} \ell_i \leq 2\alpha \biggr\}.
\tag 2.20
$$
Accepting (2.20) for a moment, we note by (2.18) that
$$\align
|A_n(\alpha)| &\leq \frac{1}{2^{n-2}} \int_{\tilde R_n(\alpha)}
|q(x_1)| \dots |q(x_n)|\, dx_1 \dots d\ell_{n-2} \\
&= \biggl( \int_0^\alpha |q(x)|\, dx\biggr)^n
\frac{\alpha^{n-2}}{(n-2)!}
\endalign
$$
since $\int_{\sum y_i =b; y_i \geq 0} dy_1 \dots dy_n =
\frac{b^n}{n!}$ by a simple induction. This is just (2.7).
To prove (2.8), we note that
$$\align
|A_n (\alpha; q) - A_n (\alpha, \tilde q)| &\leq 2^{-n-2}
\int_{\tilde R_n(\alpha)} |q(x_1)\dots q(x_n) - \tilde q(x_1)
\dots \tilde q(x_n)| \, dx_1 \dots d\ell_{n-2} \\
& \leq \frac{\alpha^{n-2}}{(n-2)!} \sum_{j=0}^{n-1}
Q(\alpha)^j \biggl[ \int_0^\alpha |q(y) - \tilde q(y)|\, dy
\biggr] \tilde Q(\alpha)^{n-j-1}.
\endalign
$$
Since $\sum_{j=0}^m a^j b^{m-j} \leq \sum_{j=0}^m \binom{m}{j}
a^j b^{m-j} = (a+b)^m$, (2.8) holds.
Thus, we need only prove (2.20). Suppose $(x_1, \dots, x_n,
\ell_1, \dots, \ell_{n-2})\in R_n (\alpha)$. Then
$$\align
2x_m &\leq |x_1 - x_m| + |x_n - x_m| + x_1 + x_n \\
& \leq x_1 + x_n + \sum_{j=1}^{n-1} |x_{j+1} - x_j| \\
&\leq x_1 + x_n + \sum_{j=1}^{n-2} \ell_j + \ell_{n-1}
(x_1, \dots, x_n, \ell_1, \dots, \ell_{n-2}; \alpha) =2\alpha
\endalign
$$
so $0\leq x_j \leq \alpha$, proving that part of the condition
$(x_1,\ell_{n-2}) \subset \tilde R_n(\alpha)$. For the second
part, note that
$$
\sum_{j=1}^{n-2} \ell_j = 2\alpha - x_1 - x_n - \ell_{n-1}
(x_1, \dots, x_n, \ell_1, \dots, \ell_{n-2}) \leq 2\alpha
$$
since $x_1$, $x_n$, and $\ell_{n-2}$ are non-negative on
$R_n (\alpha)$. \qed
\enddemo
We want to say more about the smoothness of the functions
$A_n(\alpha)$ and $A_n(\alpha,x)$ defined for $x\geq 0$ and
$n\geq 2$ by
$$
A_n (\alpha, x) = \frac{(-1)^{n-1}}{2^{n-2}} \int_{R_n(\alpha)}
q(x+x_1) \dots q(x+x_n)\, dx_1 \dots dx_n d\ell_1 \dots
d\ell_{n-2} \tag 2.21
$$
so that $A(\alpha, x) =\sum_{n=0}^\infty A_n (\alpha,x)$ is the
$A$-function associated to $m(-\kappa^2,x)$. We begin with
$\alpha$ smoothness for fixed $x$.
\proclaim{Proposition 2.3} $A_n(\alpha,x)$ is a $C^{n-2}$-function
in $\alpha$ and obeys for $n\geq 3$
$$
\left| \frac{d^j A_n(\alpha)}{d\alpha^j}\right| \leq
\frac{1}{(n-2-j)!}\, \alpha^{n-2-j} Q(\alpha)^n;
\qquad j=1,\dots, n-2. \tag 2.22
$$
\endproclaim
\demo{Proof} Write
$$
A_n (\alpha) = \frac{(-1)^{n-1}}{2^{n-1}} \int_{R_n}
q(x_1) \dots q(x_n) \, \delta\biggl(2\alpha - x_1 -
x_n -\sum_{m=1}^{n-1} \ell_i \biggr) \,
dx_1 \dots dx_n d\ell_j \dots d\ell_{n-1}.
$$
Thus, formally,
$$
\frac{d^j A_n(\alpha)}{d\alpha^j} = \frac{(-1)^{n-1}2^j}{2^{n-2}}
\int_{R_n} q(x_1) \dots q(x_n) \, \delta^{(j)} \biggl(
2\alpha - x_1 - x_n - \sum_{m=1}^{n-1} \ell_i \biggr) \,
dx_1 \dots d\ell_{n-1}. \tag 2.23
$$
Since $j+1 \leq n-1$, we can successively integrate out
$\ell_{n-1}, \ell_{n-2}, \dots, \ell_{n-j-1}$ using
$$
\int_a^b \delta^j (c-\ell)\, d\ell = \delta^{j-1} (c-a)
-\delta^{j-1} (c-b) \tag 2.24
$$
and
$$
\int_a^b \delta(c-\ell)\, d\ell = \chi_{(a,b)}(c). \tag 2.25
$$
Then we estimate each of the resulting $2^j$ terms as in
the previous lemma, getting
$$
\left| \frac{d^j A_n(\alpha)}{d\alpha^j}\right| \leq
\frac{2^{j}}{2^{n-2}}\, Q(\alpha)^n
\frac{(2\alpha)^{n-j-2}}{(n-j-2)!}
$$
which is (2.22).
(2.24), (2.25), while formal, are a way of bookkeeping for
legitimate movement of hyperplanes. In (2.25), there is a
singularity at $c=a$ and $c=b$, but since we are integrating
in further variables, these are irrelevant. \qed
\enddemo
\proclaim{Proposition 2.4} If $q$ is $C^m$, then $A_n (\alpha)$
is $C^{m+(2n-2)}$.
\endproclaim
\demo{Proof} Write $R_n$ as $n!$ terms with orderings $x_{\pi(1)}
<\cdotsa$. By Theorem~A.1.1,
$m-\tilde m =\tilde O(e^{-2a\kappa})$, and by Theorem~2.1,
$\tilde m$ has a representation of the form (3.3). \qed
\enddemo
\vskip 0.3in
\flushpar{\bf \S 4. Asymptotic Formula}
\vskip 0.1in
While our interest in the representation (1.24) is primarily for
inverse theory and, in a sense, it provides an extremely complete
form of asymptotics, the formula is also useful to recover and
extend results of others on more conventional asymptotics.
In this section, we will explain this theme. We begin with a result
related to Atkinson [\atk] (who extended Everitt [\eve]).
\proclaim{Theorem 4.1} For any $q$ \rom(obeying
{\rom{(1.2)--(1.5)}}\rom), we have that
$$
m(-\kappa^2)=-\kappa-\int_0^b q(x) e^{-2x\kappa}\, dx
+ o(\kappa^{-1}). \tag 4.1
$$
\endproclaim
\remark{Remarks} 1. Atkinson's ``$m$" is the negative inverse of
our $m$ and he uses $k=i\kappa$, and so his formula reads ((4.3)
in [\atk])
$$
m_{\text{Atk}}(k^2) = ik^{-1}+k^{-2} \int_0^b e^{2ikx} q(x)\,
dx + o(|k|^{-3}).
$$
2. Atkinson's result is stronger in that he allows cases where
$q$ is not bounded below (and so he takes $|z|\to\infty$
staying away from the negative real axis also). [\gsip] will extend
(4.1) to some such situations.
3. Atkinson's method breaks down on the real $x$ axis where
our estimates hold, but one could use Phragm\'en-Lindel\"of
methods and Atkinson's results to prove Theorem~4.1.
\endremark
\demo{Proof} By Theorem~3.1, $(A-q)\to 0$ as $\alpha\downarrow
0$ so $\int_0^a e^{-2a\kappa}(A(\alpha)-q(\alpha))\, d\alpha =
o(\kappa^{-1})$. Thus, (3.3) implies (4.1). \qed
\enddemo
\proclaim{Corollary 4.2}
$$
m(-\kappa^2) = -\kappa + o(1)
$$
\endproclaim
\demo{Proof} Since $q\in L^1$, dominated convergence implies
that $\int_0^b q(x) e^{-2\kappa x}\, dx = o(1)$. \qed
\enddemo
\proclaim{Corollary 4.3} If $\lim_{x\downarrow 0} q(x) =a$
\rom(indeed, if $\frac1s \int_0^s q(x)\, dx\to a$ as $s
\downarrow 0$\rom), then
$$
m(-\kappa^2) = -\kappa - \frac{a}{2}\,
\kappa^{-1} + o(\kappa^{-1}).
$$
\endproclaim
\proclaim{Corollary 4.4} If $q(x) = cx^{-\alpha} + o(x^{-\alpha})$
for $0<\alpha<1$, then
$$
m(-\kappa^2) = -\kappa - c[2^{a-1} \Gamma (1-\alpha)]
\kappa^{\alpha-1} + o(\kappa^{\alpha-1}).
$$
\endproclaim
We can also recover the result of Danielyan and Levitan [\dl]:
\proclaim{Theorem 4.5} Let $q(x)\in C^n [0,\delta)$ for some
$\delta>0$. Then as $\kappa\to\infty$, for suitable $\beta_0,
\dots, \beta_n$, we have that
$$
m(-\kappa^2) = -\kappa - \sum_{m=0}^n \beta_j \kappa^{-j-1}
+ O(\kappa^{-n-1}). \tag 4.2
$$
\endproclaim
\remark{Remarks} 1. Our $m$ is the negative inverse of their
$m$.
2. Our proof does not require that $q$ is $C^n$. It suffices
that $q(x)$ has an asymptotic series $\sum_{m=0}^n a_m x^m
+o(x^n)$ as $x\downarrow 0$.
\endremark
\demo{Proof} By Theorems~3.1 and 2.5, $A(\alpha)$ is $C^n$ on
$[0,\delta)$. It follows that $A(\alpha) = \sum_{m=0}^n b_j
\alpha^j \mathbreak + o(\alpha^j)$. Since $\int_0^\delta \alpha^j
e^{-2\alpha\kappa}\, d\alpha = \kappa^{-j-1} 2^{-j-1} j!
+ \tilde O(e^{-2\delta\kappa})$, we have (4.2) $\beta_j =
2^{j-1} j! b_j = 2^{j-1} \frac{\partial^j A}{\partial \alpha^j}
(\alpha=0)$. \qed
\enddemo
Later we will prove that $A$ obeys (1.28). This immediately
yields a recursion formula for $\beta_j(x)$, viz:
$$\align
\beta_{j+1}(x)&= \frac12\,\frac{\partial\beta_j}{\partial x}
+\frac12 \sum_{\ell=0}^j \beta_\ell (x) \beta_{j-\ell}(x),
\qquad j\geq 0 \\
\beta_0 (x) &= \frac12\, q(x),
\endalign
$$
see also [\gsds, Sect.~2].
\vskip 0.3in
\flushpar{\bf \S 5. Reading Boundary Conditions}
\vskip 0.1in
Our goal in this section is to prove Theorem~1.6 and then
Theorem~1.3. Indeed, we will prove the following stronger result:
\proclaim{Theorem 5.1} Let $m$ be the $m$-function for a potential
$q$ with $b<\infty$. Then there exists a measurable function
$A(\alpha)$ on $[0,\infty)$ which is $L^1$ on any finite interval
$[0,R]$, so that for each $N=1,2,\dots$ and any $a<2Nb$,
$$
m(-\kappa^2) = -\kappa -\int_0^a A(\alpha) e^{-2\alpha\kappa}\,
d\alpha - \sum_{j=1}^N A_j \kappa e^{-2\kappa bj} -
\sum_{j=1}^N B_j e^{-2\kappa bj} + \tilde O(e^{-2a\kappa}),
\tag 5.1
$$
where
\roster
\item"\rom{(a)}" If $h=\infty$, then $A_j =2$ and $B_j =
-2j \int_0^b q(y)\, dy$.
\item"\rom{(b)}" If $|h|<\infty$, then $A_j =2(-1)^j$ and $B_j
= 2(-1)^{j+1} j[2h+\int_0^b q(y)\, dy]$.
\endroster
\endproclaim
\remark{Remarks} 1. The combination $2h+\int_0^b q(y)\, dy$ is
natural when $|h|<\infty$. It also enters into the formula for
eigenvalue asymptotics [\lev,\marb].
2. One can think of (5.1) as saying that
$$
m(-\kappa^2)=-\kappa-\int_0^a \tilde A(\alpha) e^{-2\alpha\kappa}
\, d\alpha + \tilde O(e^{-2a\kappa})
$$
for any $a$ where now $\tilde A$ is only a distribution of the
form $\tilde A(\alpha) = A(\alpha) +\frac12 \sum_{j=1}^\infty
A_j \delta' (\alpha-jb)+\sum_{j=1}^\infty B_j \delta (\alpha -jb)$
where $\delta'$ is the derivative of a delta function.
3. As a consistency check on our arithmetic, we note that if
$q(y)\to q(y)+c$ and $\kappa^2\to\kappa^2 -c$ for some $c$, then
$m(-\kappa^2)$ should not change. $\kappa^2\to\kappa^2 -c$ means
$\kappa \to\kappa - \frac{c}{2\kappa}$ and so $\kappa
e^{-2\kappa bj}\to \kappa e^{-2\kappa bj} + cbj e^{-2\kappa bj}
+ O(\kappa^{-1})$ terms. That means that under $q\to q+c$, we
must have that $B_j\to B_j -cbjA_j$, which is the case.
\endremark
\demo{Proof} Consider first the free Green's function for
$-\frac{d^2}{dx^2}$ with Dirichlet boundary conditions at $0$
and $h$-boundary condition at $b$. It has the form
$$
G_0 (x,y) = \frac{\text{sinh}(\kappa x)\, u_+(y)}
{\kappa \, u_+(0)} \, , \qquad xb$,
then the limit (1.12) is zero, so $h_1 \neq h_2$ implies either
$b_1$ or $b_2$ is $b$. If only one is $b$, then the difference
has a $\delta'$ term and the limit in (1.12) is infinite.
Therefore, $b_1 = b_2 = b$.
Since $A_1 = A_2$ on $[0,b]$, Theorem~1.2 implies that $q_1(x)
=q_2(x)$ on $[0,b]$. If both $h_1$ and $h_2$ are infinite, then
the limit is zero. If only one is infinite, then there is a
$\delta'$ term and the limit is infinite. Thus, a limit on
$(0,\infty)$ implies $h_1$ and $h_2$ are both finite and so,
by Theorem~5.1, the limit is $4(h_1 - h_2)$ as claimed. \qed
\enddemo
\vskip 0.3in
\flushpar{\bf \S 6. The $\boldkey A$-Equation}
\vskip 0.1in
In this section, we will prove equation (1.28). We begin with
the case where $q$ is $C^1$. In general, given $q$ (i.e.,
$q$, $b$, and $h$ if $b<\infty)$, we can define $m(z,x) =
\frac{u'_+ (x,z)}{u_+ (x,z)}$ for $x\in [0,b)$ and $z\in
\Bbb C\backslash [\beta,\infty)$ for suitable $\beta\in\Bbb R$.
By Theorem~3.1, there is a function $A(\alpha, x)$ defined for
$(\alpha,x)\in \{(\alpha,x)\in\Bbb R^2 \mid 0\leq x < b;\,
0<\alpha < b-x\}\equiv S$ so that for any $a0$, there is a
$\Lambda$ so that for almost every $\alpha$ with $0<\alpha<
\Lambda$, we have $\int_0^a |g_\alpha (x) - q(x)|\, dx
<\varepsilon$). Alternatively, one can pick a concrete
realization of $q$ and then use the fact that $A-q$ is continuous
to define $A(x,\alpha) - q(x+\alpha)$ for all $x,\alpha$ and
then (6.2) holds in traditional sense. Indeed, if $q$ is
continuous, it holds pointwise.
\proclaim{Theorem 6.1} If $q$ is $C^1$, then $A$ is jointly
$C^1$ on $S$ and obeys
$$
\frac{\partial A}{\partial x} = \frac{\partial A}{\partial\alpha}
+\int_0^\alpha A(\beta,x) A(\alpha -\beta, x)\, d\beta. \tag 6.3
$$
\endproclaim
\demo{Proof} That $A$ is jointly $C^1$ when $q$ is $C^1$
of compact support follows from the arguments in Section~2
(and then the fact that $A$ on $[0,a)$ is only a function of $q$
on $[0,a)$ lets us extend this to all $C^1$ $q$'s). Moreover,
by Theorem~2.7,
$$
\frac{\partial m}{\partial x}\, (-\kappa^2, x) = -\int_0^a
\frac{\partial A}{\partial x}\, (\alpha, x) e^{-2\alpha\kappa}
\, d\alpha + \tilde O(e^{-2a\kappa}) \tag 6.4
$$
for all $a0$. Then
$$
A(\alpha) = \frac{\sqrt\gamma}{\alpha}\, I_1 \left(2\alpha \sqrt\gamma
\right),
$$
where $I_1$ is the standard Bessel function denoted by $I_1 (\, \cdot \, )$.
Since
$$
I_1 (z) = \tfrac12\, z \sum_{k=0}^\infty \frac{(\frac14 z^2)^k}
{k! (k+1)!}\, ,
$$
the $\frac{1}{n!}$ bounds in (2.7) are not good as $n\to\infty$ if $q$
is bounded. This is discussed further in [\gsip].
\vskip 0.3in
\flushpar{\bf Appendix 1: Localization of Asymptotics}
\vskip 0.1in
Our goal in this appendix is to prove one direction of
Theorem~1.2, viz:
\proclaim{Theorem A.1.1} If $(q_1, b_1, h_1), (q_2, b_2, h_2)$
are two potentials and $a<\min (b_1, b_2)$ and if
$$
q_1(x) = q_2 (x) \qquad \text{on } (0,a), \tag A.1.1
$$
then as $\kappa\to\infty$,
$$
m_1 (-\kappa^2) - m_2 (-\kappa^2) = \tilde O(e^{-2\kappa a}).
\tag A.1.2
$$
\endproclaim
While we know of no explicit reference for this form of the
result, the closely related Green's function bounds have long
been in the air, going back at least to ideas of Donoghue,
Kac, and McKean over thirty years ago. A basic role in our
proof will be played by the Neumann analog of the Dirichlet
relation (2.2). Explicitly, if $G^D (x,y;z,q)$ and $G^N
(x,y;z,q)$ are the integral kernels of $(H-z)^{-1}$ with
$H=-\frac{d^2}{dx^2}+q(x)$ on $L^2 (0,\infty)$ with $u(0) =0$
(Dirichlet) and $u'(0)=0$ (Neumann) boundary conditions,
respectively, then
$$
m(z) = \lim\Sb x0$, there
exists $C_\varepsilon >0$ (depending only on the $\beta_2$ for
$q_1, q_2$), so that
$$
|P(0,0; t, q_1) - P(0,0; t,q_2)| \leq C_\varepsilon
\exp(-(1-\varepsilon) a^2/t). \tag A.1.10
$$
(A.1.9) implies (A.1.6) since $\int_0^\infty 2(4\pi t)^{-1/2}
e^{-\kappa^2 t}\, dt = \kappa^{-1} \int_0^\infty (\pi t)^{-1/2}
e^\kappa\, dt = \kappa^{-1}$.
To obtain (A.1.7), we use (A.1.8), (A.1.10), and
$$
|P(0,0; t,q_j)| \leq C_1 e^{Dt}
$$
since
$$\align
\int_0^1 e^{a^2/t} e^{-\kappa^2 t}\, dt &= e^{-2\kappa a}
\int_0^1 e^{-(a^{-1/2} -\kappa t^{1/2})^2}\, dt \\
&= O(e^{-2\kappa a}).
\endalign
$$
\qed
\enddemo
Next, we consider a situation where $b<\infty$, $q$ is given in
$L^1 (0,b)$, and $h$ is $0$ or $\infty$. Define $\tilde q$ on
$\Bbb R$ by requiring that
$$\alignat2
\tilde q(x+2mb) &= \tilde q(x) \qquad && m=0, \pm1, \pm2, \dots,
\text{all }x\in\Bbb R \\
\tilde q(-x) &=\tilde q(x) \qquad && \text{all }x\in\Bbb R \\
\tilde q(x) &= q(x) \qquad && x\in [0,b]
\endalignat
$$
which uniquely defines $\tilde q$ (since each orbit $\{\pm
x+2mb\}$ contains one point in $[0,b]$). Let $G^{(N,N)}$ and
$G^{(N,D)}$ be the Green's functions of $-\frac{d^2}{dx^2}
+ q(x)$ on $L^2 (0,b)$ with $u'(0)=0$ boundary conditions
at zero and $u'(b)=0$ ($(N,N)$ case) or $u(b)=0$ ($(N,D)$
case) boundary conditions at $b$. Let $\tilde G$ be the
Green's function for $-\frac{d^2}{dx^2} + \tilde q$ on $L^2
(\Bbb R)$. Let $P$ be the corresponding integral kernels for
$e^{-tH}$.
By the method of images for $x,y \in [0,b]$:
$$\align
G^{(N,N)} (x,y; -\kappa^2) &= \sum_{m=-\infty}^\infty
\tilde G(x, i_m (y); -\kappa^2) \tag A.1.11 \\
G^{(N,D)} (x,y; -\kappa^2) &= \sum_{m=-\infty}^\infty
\sigma_m \tilde G(x, i_m (y); -\kappa^2) \tag A.1.12
\endalign
$$
where
$$\alignat2
i_m (y) &= y+mb \qquad && m=0,\pm 2, \pm \cdots \\
&= -y + mb + b \qquad && m=\pm 1, \pm 3, \pm \cdots \\
\sigma_m &= -1 \qquad && m=1,2,5,6,9,10, \dots, -2, -3, -6, -7,
\dots \\
&= 1 \qquad && \text{otherwise}
\endalignat
$$
(i.e., $\sigma_m = -1$, if and only if $m=1,2$ mod 4).
By a simple path integral (or other) estimate on $P$ and Laplace
transform, we have
$$
|\tilde G (x,y; -\kappa^2)| \leq C_\varepsilon
e^{-\kappa |x-y|(1-\varepsilon)} \tag A.1.13
$$
for any $\varepsilon >0$ and $\kappa$ sufficiently large. Since
the images of $0$ are $\pm 2b, \pm 4b, \dots$, (A.1.11) and
(A.1.2) imply
\proclaim{Proposition A.1.3}
$$
|G^{(N,N)} (0,0; -\kappa^2) -\tilde G(0,0; -\kappa^2)| =
\tilde O (e^{-2b\kappa}) \tag A.1.14
$$
and similarly for $|G^{(N,D)} (0,0; -\kappa^2) - \tilde G
(0,0; -\kappa^2)|$.
\endproclaim
\remark{Remark} (A.1.14) and (A.1.6) imply (A.1.2) for the pairs
$q_1 = \tilde q$, $b_1 = \infty$ and $q_2=q$, $b_2 =b$, and
$h_2 =0$ or $\infty$.
\endremark
Finally, we compare $b<\infty$ fixed for any two finite values of
$h$:
\proclaim{Proposition A.1.4} Let $q\in L^1 (0,b)$. For $h <
\infty$, let $G^h$ be the integral kernel for $(-\frac{d^2}{dx^2}
+ q-z)^{-1}$ with boundary conditions $u(0)=0$ and $u'(b)+hu(b)
=0$. Then
$$
|G^h (0,0; -\kappa^2) - G^{h=0}(0,0; -\kappa^2)| =
\tilde O(e^{-2b\kappa}). \tag A.1.15
$$
\endproclaim
\demo{Proof} Let $H$ be the $h=0$ operator and $H_h$ the operator
for $h<\infty$. By the analysis of rank one perturbations (see,
e.g., [\simvan]),
$$
H_h = H + h(\delta_b, \, \cdot\, ) \delta_b,
$$
where $\delta_b \in \Cal H_{-1} (H)$ is the function
$(\delta_b, g)=g(b)$.
Again, by the theory of rank one perturbations [\simvan],
let $F(z,h)=G^h (b,b;z)$. Then
$$
F(z,h) = \frac{F(z,0)}{1+hF(z,0)}
$$
and
$$\aligned
G^h (0,0;z) - G^{h=0} (0,0;z) &= -h G^{h=0} (0,b;z)
G^{h=0} (b,0;z) [1-hF(z,h)] \\
&= -h G^{h=0} (0,b;z) G^{h=0} (b,0;z) [1 + hF(z,0)]^{-1}.
\endaligned \tag A.1.16
$$
Now $F(-\kappa^2,0) = \kappa^{-1} +o(\kappa^{-1})$ (this is
essentially (A.1.6)) while (A.1.11) and (A.1.13) imply that
$$
G^{h=0} (0,b;z) = \tilde O(e^{-\kappa b}). \tag A.1.17
$$
(A.1.16) and (A.1.17) imply (A.1.15). \qed
\enddemo
Transitivity and Propositions A.1.2--A.2.4 imply Theorem~A.1.1.
We close the appendix with two remarks:
1. Do not confuse the Laplace transform in (1.24) (which is in
$2\kappa$) with that in (A.1.8) (which is $\kappa^2$).
2. We used path integrals above. As long as $q(x) =O(e^{b|x|})$
for some $b<\infty$, one can instead use more elementary
Green's function estimates.
\vskip 0.3in
\flushpar{\bf Appendix 2: Some Results on Laplace Transforms}
\vskip 0.1in
In this paper, I need some elementary facts about Laplace
transforms. While I'm sure that these facts must be in the
literature, I was unable to locate them in the precise form
needed, so I will give the simple proofs below.
\proclaim{Lemma A.2.1} Let $f\in L^1(0,a)$. Suppose that $g(z)
\equiv \int_0^a f(y) e^{-zy}\, dy$ obeys
$$
g(x) = \tilde O(e^{-ax}) \tag A.2.1
$$
as $x\to\infty$. Then $f\equiv 0$.
\endproclaim
\demo{Proof} Suppose first that $f$ is real-valued. $g(z)$ is
an entire function which obeys
$$
|g(z)| \leq \|f\|_1 e^{a\text{\rom{Re}}_- (z)},
$$
where $\text{Re}_-\,(z)$ is the negative part of $\text{Re}\, z$.
Moreover, along the real axis, $g$ obeys (A.2.1). Because of this,
$$
h(w) = \int_0^\infty g(x) e^{iwx} \, dx
$$
is an analytic function of $w$ in the region $\text{Im}\,
w>-a$. Now for $r>0$:
$$\align
h(ir) &= \int_0^\infty g(x) e^{-rx}\, dx \\
&= \int_0^\infty \biggl(\int_0^a f(y)e^{-x(y+r)}\, dy \biggr)\,
dx \\
&= \int_0^a \frac{f(y)}{y+r}\, dy, \tag A.2.2
\endalign
$$
where the interchange of integration variables is easy to
justify. (A.2.2) implies that
$$
h(w) = \int_0^a \frac{f(y)}{y-iw}\, dy \tag A.2.3
$$
holds for $w$ with $\text{Im}\, w>0$ and then allows analytic
continuation into the region $\Bbb C\backslash\{is \mid
s<0\}$. (A.2.3) and the reality of $f$ implies that for a.e.~$r
\in (0,a)$, $f(r) = \lim_{\varepsilon\downarrow 0} \frac{1}
{2\pi i} [h(\varepsilon - ir) - h(-\varepsilon-ir)]$, so the
analyticity of $h$ in $\text{Im}\,w >-a$ implies that $f\equiv 0$.
For general complex valued $f$, consider the real and imaginary
parts separately. \qed
\enddemo
An immediate consequence of this is the uniqueness of inverse
Laplace transforms.
\proclaim{Theorem A.2.2} Suppose that $f,g\in L^1 (0,a)$ and for
some $b\leq a$, $\int_0^a f(y) e^{-xy}\, dy - \int_0^a g(y)
e^{-xy}\, dy = \tilde O(e^{-bx})$. Then $f\equiv g$ on $[0,b)$.
\endproclaim
The other fact we need is that the set of Laplace transforms has
a number of closure properties. Let $\Cal L_a$ be the set of
functions, $f$, analytic in some region $\{ z\mid \,
|\text{Arg}(z)| <\varepsilon\}\equiv R_\varepsilon$ obeying
$$
f(z) = 1+ \int_0^a g(\alpha) e^{-\alpha z}\, d\alpha
+ \tilde O(e^{-a\text{\rom{Re}}\,z})
$$
in that region for some $g\in L^1 (0,a)$. Denote $g$ by
$\Cal I(f)$.
\proclaim{Theorem A.2.3} If $f,h\in\Cal L_a$ so are $fh$, $f +
h-1$, and $f^{-1}$.
\endproclaim
\demo{Proof} $f+h-1$ is trivial. $fh$ is elementary; indeed,
$$
\Cal I (fh)(\alpha) = \Cal I (f)(\alpha) + \Cal I(h)(\alpha)
+ \int_0^\alpha \Cal I(f)(\beta)\Cal I(h)(\alpha-\beta)\,
d\beta.
$$
For the inverse, we start by seeking $k$ obeying (where $g=
\Cal I(f)$)
$$
g(\alpha) + k(\alpha) + \int_0^\alpha d\beta \ k(\beta)
g(\alpha-\beta)=0.
$$
This Volterra equation always has a solution (by iteration).
Let $h(z)=1 + \int_0^a k(\alpha) e^{-\alpha z}\,
d\alpha$. Then
$$
fh = 1+\tilde O(e^{-a\text{\rom{Re}}(z)})
$$
and so
$$\align
f^{-1} &= h(1+\tilde O(e^{-a\text{\rom{Re}}(z)}))^{-1} \\
&= h+\tilde O(e^{-a\text{\rom{Re}}(z)})
\endalign
$$
as required. \qed
\enddemo
\vskip 0.3in
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\enddocument