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Schr&ouml;dinger operator, spectral theory, inverse spectral theory
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\begin{document}
\title[Schr\"odinger operators with measures]
{Direct and inverse spectral theory of one-dimensional
Schr\"odinger operators with measures}
\author{Ali Ben Amor and Christian Remling}
\address{Universit\"at Osnabr\"uck\\
Fachbereich Mathematik/Informatik\\
49069 Osnabr\"uck\\
Germany}
\email{abamor@mathematik.uni-osnabrueck.de}
\email{cremling@mathematik.uni-osnabrueck.de}
\urladdr{www.mathematik.uni-osnabrueck.de/staff/phpages/abamor.rdf.html}
\urladdr{www.mathematik.uni-osnabrueck.de/staff/phpages/remlingc.rdf.html}
\date{September 4, 2003}
\thanks{2000 {\it Mathematics Subject Classification.} Primary 34L15 34L40;
secondary 34L16}
\keywords{Schr\"odinger operator,
spectral measure}
\thanks{Ben Amor's work was supported by the Deutsche
Forschungs\-gemein\-schaft}
\thanks{Remling's work was supported by the Heisenberg program
of the Deutsche Forschungs\-gemein\-schaft}
%\thanks{to appear in {\it Commun.\ Math.\ Phys.} }
\begin{abstract}
We present a direct and rather elementary
method for defining and analyzing
one-dimensional Schr\"odinger operators $H=-d^2/dx^2+\mu$
with measures as potentials. The basic idea is to let the
(suitably interpreted) equation $-f''+\mu f=zf$ take center
stage. We show that the basic results from direct and
inverse spectral theory then carry over to Schr\"odinger
operators with measures.
\end{abstract}
\maketitle
\section{Introduction}
In this paper we will discuss the direct and inverse
spectral theory of generalized
Schr\"odinger operators
\begin{equation}
\label{so}
H=-\frac{d^2}{dx^2}+\mu,
\end{equation}
with a signed Borel measure $\mu$ as the potential.
In the case where $\mu$ is absolutely continuous,
$d\mu(x)=V(x)\, dx$, the theory of course reduces to the theory
of the usual Schr\"odinger operator $-d^2/dx^2+V(x)$.
Operators of the form \eqref{so} have been used to model
singular interactions, which are located on small sets.
See, for example, \cite{AGHH,Brasche-Exner,BFT,GK,KL}.

We will make essential use of the fact that the problem
is one-dimensional. Namely, we will use a direct approach based
on the solutions of the (suitably interpreted) equation
$-f''+\mu f = zf$ instead of quadratic forms, which is the customary
tool in the theory of Schr\"odinger operators with measures.
As one advantage of this approach, we do not
need semi-boundedness assumptions on $\mu$; the
drawback is that our method only works in
the one-dimensional case.

We will discuss the general direct and inverse spectral
theory of the operators $-d^2/dx^2+\mu$, using this approach.
The setup and basic definitions may be found in the following
section. Given this material, it will then be rather
straightforward to extend the classical theory of the
spectral representation of one-dimensional Schr\"odinger
operators (based on the Titchmarsh-Weyl $m$ function).
Therefore, our discussion of the direct spectral theory
in Sect.~3 will be rather sketchy. However, things get
more challenging in inverse spectral theory. Here, the
smoothness of the potential plays a decisive role, and
clearly, a measure $\mu$ can be even more singular than
a potential. We will
take the recent treatment of \cite{Rem03} as a guideline.
More specifically, we will discuss and use the tool of
de~Branges spaces, and we will introduce a so-called
$\phi$ function as the spectral data. Then, the inverse problem
is to recover $\mu$ from $\phi$. It turns out, as expected,
that $\phi$ has less smoothness than in the case where
$\mu$ is absolutely continuous. Namely, $\phi$ is no longer
absolutely continuous but (in general) only of bounded
variation. This larger set of $\phi$ functions is in
one-to-one correspondence to the class of (signed) measures
$\mu$. These results will be discussed in Sect.~4--6.
\section{Schr\"odinger operators with measures}
Let $\mu$ be a signed measure on the Borel subsets
of $[0,\infty)$ with $|\mu|([0,N])<\infty$ for all $N>0$.
We will define the operator $-d^2/dx^2+\mu$ as
a self-adjoint operator on the Hilbert space $L_2(0,\infty)$
using a direct method rather than quadratic forms.

For $0\le a<b<\infty$, denote by $AC(a,b)$
the space of complex valued absolutely continuous
functions on $(a,b)$. So $f\in AC(a,b)$ means there
exists $g\in L_1(a,b)$ so that $f(x)=f(c)+\int_c^x g(t)\, dt$
for all $x\in (a,b)$ and with $c\in (a,b)$.
$AC(0,\infty)$ is defined locally, that is,
$f\in AC(0,\infty)$ if $f\in AC(0,b)$
for all $b>0$.

Fix $a>0$ and define, for $f\in AC(0,\infty)$,
\[
(Af)(x)= f'(x) - \int_a^x f(t)\, d\mu(t) ;
\]
here, $\int_a^x \cdots$ is short-hand for
\[
\int_a^x f(t)\, d\mu(t) = \begin{cases}
\int_{[a,x]} f(t)\, d\mu(t) & x\ge a\\
-\int_{(x,a)} f(t)\, d\mu(t) & x<a \end{cases} .
\]
Clearly, $Af$ is only defined as an element of $L_{1,loc}(0,\infty)$,
or, in other words, almost everywhere with respect to Lebesgue measure.
$Af$ will play the role of a quasi-derivative of $f$. We can now
define the maximal operator associated with the differential
expression $-d^2/dx^2+\mu$. Let
\begin{align*} 
D(T)& =\left\{ f\in L_2(0,\infty): f, Af
\in AC(0,\infty),
(Af)'\in L_2(0,\infty)\right\} \\
Tf & = -(Af)' .
\end{align*}
It is clear that neither the domain of $T$, $D(T)$, nor the
definition of $T$ depend on the choice of $a>0$. We want to
establish regularity at $x=0$; in particular, we can then take
$a=0$ in the above definitions.
\begin{Proposition}
\label{P2.1}
If $f\in D(T)$,
then $f,Af \in AC[0,\infty)$.
\end{Proposition}
Here, we say that $f\in AC[0,\infty)$ if there
is a function $g\in L_{1,loc}([0,\infty))$, so that
$f(x)=f(0)+\int_0^x g(t)\, dt$. Put more precisely, Proposition
\ref{P2.1} then claims that the functions $f,Af$ have extensions
to $[0,\infty)$ which are in $AC[0,\infty)$.
\begin{proof}
To prove the statement on $Af$, we only need to observe that
$(Af)'\in L_2(0,\infty)\subset L_{1,loc}([0,\infty))$ and let
$c\to 0+$ in
\[
(Af)(x)=(Af)(c)+\int_c^x (Af)'(t)\, dt .
\]
As for $f$, we have that
\[
f(x) = f(c)+ \int_c^x f'(t)\, dt
= f(c)+\int_c^x (Af)(t)\, dt + \int_c^x dt \int_c^t d\mu(s)\, f(s) .
\]
By using Fubini's Theorem in the double integral, we thus see that
\[
\|f\|_{L_{\infty}(\epsilon,c)} \le C + \|f\|_{L_{\infty}(\epsilon,c)}
\int_{(\epsilon,c)} (s-\epsilon)\, d|\mu|(s) .
\]
The constant $C$ is
independent of $\epsilon$,
so by letting $\epsilon\to 0+$ and then taking $c>0$ sufficiently
small, we see that $f$ is bounded on
$(0,c)$. Hence $f'(x)=(Af)(x)+\int_c^x f(t)\, d\mu(t)$ is
in $L_{1,loc}([0,\infty))$.
\end{proof}
This argument was local, so it is also true that $f,Af\in
AC[0,N)$ if $\varphi f\in D(T)$ for some function $\varphi$
which is equal to $1$ on $(0,N)$.

Proposition \ref{P2.1} allows us to take $a=0$ in the definition
of $Af$. So from now on, we will define $Af$ for $f\in AC[0,N)$ by
\begin{equation}
\label{2.2}
(Af)(x)=f'(x) - \int_{[0,x]} f(t)\, d\mu(t) .
\end{equation}

To develop the theory of the self-adjoint realizations of
$-d^2/dx^2+\mu$ along the usual lines
(see for example \cite{WMLN}),
we need Green's identity. Fix $N>0$ and let
\begin{align*}
D(T_N) &= \{ f\in L_2(0,N): f, Af \in AC(0,N), (Af)'\in L_2(0,N)\} , \\
T_N f & = -(Af)' .
\end{align*}
By Proposition \ref{P2.1} and the remark following its proof
(and also an analogous version for $x=N$ instead of $x=0$),
we automatically have that $f,Af \in AC[0,N]$ for
$f\in D(T_N)$. For $f\in D(T_N)$,
we also fix a particular representative of
$f'\in L_1(0,N)$ by
\begin{equation}
\label{deff'}
f'(x) := (Af)(x) + \int_{[0,x]} f(t)\, d\mu(t) \quad (0\le x\le N);
\end{equation}
here, we take the unique continuous representative of $Af$.
\begin{Theorem}[Green's identity]
\label{T2.2}
Suppose that $f,g \in D(T_N)$. Then
\[
\langle f, T_N g \rangle - \langle T_N f,g \rangle =
\left. \left( \overline{f'(x)}g(x) - \overline{f(x)}g'(x)
\right) \right|_{x=0}^{x=N} .
\]
\end{Theorem}
\begin{proof}
Integration by parts shows that
\begin{multline*}
\langle f, T_N g \rangle - \langle T_N f,g \rangle =
\left. \left( \overline{(Af)(x)}g(x) - \overline{f(x)}(Ag)(x)
\right) \right|_{x=0}^{x=N} \\ + \int_0^N \left( \overline{f'(x)}
(Ag)(x) - \overline{(Af)(x)}g'(x)\right) \, dx.
\end{multline*}
In the integral, we plug in $Af$, $Ag$ from \eqref{2.2} and
use Fubini's theorem:
\begin{align*}
\int_0^N \left( \overline{f'}
(Ag) - \overline{(Af)}g'\right) \, dx & =
\int_0^N dx\, \int_{[0,x]} d\mu(t)\, \left( \overline{f(t)}g'(x)
- \overline{f'(x)}g(t) \right) \\
& = \int_{[0,N]} d\mu(t)\, \int_t^N dx\,
\left( \overline{f(t)}g'(x)
- \overline{f'(x)}g(t) \right) \\
& = g(N) \int_{[0,N]} \overline{f(t)}\, d\mu(t) -
\overline{f(N)} \int_{[0,N]} g(t)\, d\mu(t)
\end{align*}
Since $\overline{(Af)(0)}g(0) - \overline{f(0)}(Ag)(0)=
\overline{f'(0)}g(0) - \overline{f(0)}g'(0)$,
the asserted identity now follows.
\end{proof}
The theory of self-adjoint extensions depends on Green's identity,
but not on the particular form of
the differential expression itself, so we have the same theory here
as in the classical case (where $d\mu(x)=V(x)\, dx$). For a detailed
exposition of this theory, see \cite{WMLN}.
$T_N$ is self-adjoint on a domain $D\subset D(T_N)$
if $D(T_N)/D$ is two-dimensional and the \textit{Lagrange form}
$L(f,g)=\left. \left( \overline{f'(x)}g(x) - \overline{f(x)}g'(x)
\right) \right|_{x=0}^{x=N}$ vanishes on $D$. These domains $D$
can be described in terms of boundary conditions. Here, we are
interested in \textit{separated} boundary conditions, that is,
we require that the contributions from $x=0$ and $x=N$ vanish
separately on $D$. An elementary argument shows that then the
admissable boundary conditions are precisely given by
\begin{equation}
\label{bc}
f(0)\cos\alpha + f'(0)\sin\alpha=0, \quad
f(N)\cos\beta + f'(N)\sin\beta=0,
\end{equation}
with $\alpha,\beta\in [0,\pi)$. Here, $f(0)$ and $f(N)$ are well
defined because $f\in AC[0,N]$, and $f'(0)$ and $f'(N)$ are determined
from \eqref{deff'}.

Now observe that the operators just defined are independent of
$\mu(\{ 0 \} )$. Indeed, if $f\in D(T_N)$, the values of
$f(0)$ and $f'(0)$ do not change if we vary $\mu(\{ 0 \} )$.
However, this does not necessarily mean that one should dismiss
$\mu(\{ 0 \} )$ and view $\mu$ as a measure on $(0,N)$. Rather,
we can make good use of $\mu(\{ 0 \} )$ in the following way:
By the observation made at the end of the
proof of Theorem \ref{T2.2}, the boundary conditions
at $x=0$ may also be written in the form $f(0)\cos\gamma+
(Af)(0)\sin\gamma=0$, where $\gamma$ varies over $[0,\pi)$.
Let us fix the boundary condition $(Af)(0)=0$. Since
$(Af)(0)=f'(0)-\mu(\{ 0 \} ) f(0)$, we now get all boundary
conditions except Dirichlet boundary conditions ($f(0)=0$)
by adjusting $\mu(\{ 0 \} )$. Dirichlet boundary conditions
play a special role for other reasons as well. So from now on,
we work with the boundary conditions
\begin{equation}
\label{bc1}
(Af)(0)=0, \quad f(N)\cos\beta + f'(N)\sin\beta=0 .
\end{equation}
Of course, this is only a minor detail, but we will find it
very convenient to use \eqref{bc1} and vary $\mu(\{ 0 \} )$
rather than use \eqref{bc} and vary $\alpha$ ($\alpha\not= 0$).

Next, we study the solutions of the equation $-f''+\mu f =zf+g$.
Clearly, this needs to be interpreted properly. The following
definition suggests itself.
\begin{Definition}
\label{D2.1}
Let $g\in L_1(0,N)$, $z\in\C$.
We say that a function $f:(0,N)\to\C$ solves the equation 
\begin{equation}
-f''+\mu f=zf+g
\label{Eqq1}
\end{equation}
on $(0,N)$ if $f,Af\in AC(0,N)$ and $-(Af)'=zf+g$ almost everywhere
(with respect to Lebesgue measure) on $(0,N)$.
\end{Definition} 
We have the usual existence and uniqueness results for the
initial value problems associated with \eqref{Eqq1}.
\begin{Theorem}
\label{T2.3}
Let $g\in L_1(0,N)$, $z\in\C$ and $c,c'\in\C$. Then the equation
\eqref{Eqq1} has a unique continuous solution on $[0,N]$
such that $f(0)=c$, $(Af)(0)=c'$. Moreover,
$f(x,z)$ is entire in $z$ for every fixed $x\in [0,N]$.
\end{Theorem}
Proposition \ref{P2.1} and the remark following its proof show
that if $f$ solves \eqref{Eqq1} on $(0,N)$, then in fact
$f,Af\in AC[0,N]$.
\begin{proof}[Sketch of proof]
The proof is essentially the same as the one for differential
equations (see, e.g., \cite[Chapter 1]{CodLev}), so we only
sketch the proof.

By integrating twice and using Fubini's theorem, we see that
\eqref{Eqq1} together with the initial conditions $f(0)=c$,
$(Af)(0)=c'$ is equivalent to the integral equation
\begin{equation}
\label{Eqq2}
f(x)=c+c'x+\int_{[0,x]}(x-t)f(t)(d\mu(t)-z\,dt)
-\int_0^x(x-t)g(t)\,dt.
\end{equation}
Here, we seek continuous solutions $f$. If such an $f$ solves
\eqref{Eqq2}, then $f,Af$ are automatically absolutely continuous.

The right-hand side of \eqref{Eqq2} defines a contractive mapping
on $C[0,\eta]$, provided $\eta>0$ is chosen sufficiently small.
This yields existence and uniqueness on the interval $[0,\eta]$,
and this argument may be repeated (with $\eta$ independent of the
left endpoint) to obtain existence and uniqueness on all of
$[0,N]$. Holomorphic dependence on $z$ also follows from this method.
\end{proof}
We believe that the above interpretation of $-f''+\mu f=zf+g$ is
natural; this is also confirmed by the following fact.
\begin{Theorem}
\label{T2.4}
If the function $f$ solves \eqref{Eqq1} in the sense of
Definition \ref{D2.1} on $(0,N)$, then
\[
-f''+f\mu=zf+g
\]
in the sense of distributions on $(0,N)$.
\end{Theorem}
\begin{proof}
We must show that the distributional derivative of the
function $F(x):=\int_{[0,x]} f(t)\, d\mu(t)$ is equal to
the measure $F'=f\mu$. Indeed, it follows from this fact
that $(Af)'=f''-f\mu$ (distributional derivatives), so the
claim holds.

Let $\varphi\in C_0^{\infty}(0,N)$ be a test function.
By the definition of the distributional derivative,
\begin{align*}
(F',\varphi) & = -(F,\varphi') = -\int_0^N dx\, \varphi'(x)
\int_{[0,x]} f(t)\, d\mu(t) \\
& = -\int_{[0,N]} f(t)\, d\mu(t) \int_t^N \varphi'(x)
\, dx = \int_{(0,N)} \varphi(t)f(t)\, d\mu(t)
= (f\mu,\varphi) ,
\end{align*}
as desired.
\end{proof}
We also see from the preceding result and its proof that $f''=(Af)'
+f\mu$ is a measure if $f$ solves an equation of the form \eqref{Eqq1}.
In particular, this is the case for $f\in D(T)$ or $f\in D(T_N)$.
So, $f'$ is then of bounded variation, with jumps precisely at
the discrete points of the measure $f\mu$. The one-sided limits
$f'(x\pm):=\lim_{t\to x\pm} f'(t)$ exist everywhere. The choice
\eqref{deff'} (which we also use if $f$ solves \eqref{Eqq1})
implies that $f'(x+)=f'(x)$, and thus
$f'(x)-f'(x-)=f(x)\mu(\{ x \} )$.

With this notation, we also have that $(Af)(0)=f'(0-)$. This
remark throws additional light on our discussion above of boundary
conditions at $x=0$.

If $f,g$ are solutions of the homogenous equation
$-y''+\mu y=zy$, then their \textit{Wronskian} is defined as
$W(f,g)=f(x)g'(x)-f'(x)g(x)$.
\begin{Proposition}
\label{P2.2}
The Wronskian of two solutions of the same equation is
independent of $x$.
\end{Proposition}
\begin{proof}
This follows from the following computation:
\begin{align*}
0 & = \int_0^N \left( f(x)(Ag)'(x) - (Af)'(x)g(x)\right) \, dx \\
& = \left( f(x)(Ag)(x)-(Af)(x)g(x)\right) \bigr|_{x=0}^{x=N}
+ \int_0^N \left( (Af)(x)g'(x) - f'(x)(Ag)(x) \right) \, dx \\
&  = W(N)-W(0) -f(N)\int_{[0,N]} g(t)\, d\mu(t) + g(N)
\int_{[0,N]} f(t)\, d\mu(t) \\
& \quad\quad\quad\quad\quad\quad\quad\quad
\quad\quad\quad\quad\quad
+ \int_0^N dx \int_{[0,x]} d\mu(t)\,
\left( f'(x)g(t)-f(t)g'(x) \right) \\
& = W(N)-W(0) -f(N)\int_{[0,N]} g(t)\, d\mu(t) + g(N)
\int_{[0,N]} f(t)\, d\mu(t) \\
& \quad\quad\quad\quad\quad\quad\quad\quad
\quad\quad\quad\quad\quad
+ \int_{[0,N]} d\mu(t) \int_t^N dx\,
\left( f'(x)g(t)-f(t)g'(x) \right) \\
& = W(N) - W(0)
\end{align*}
\end{proof}
Finally, we introduce the notation $u(x,z)$ for the solution
$u$ of $-u''+\mu u = zu$ with the initial values $u(0)=1$,
$(Au)(0)=0$. Note that $u$ satisfies the boundary condition
at $x=0$. This solution $u$ will become important in the
following sections.
\section{Spectral representation}
Given the material from the preceding section, the theory of
the spectral representation of the operators associated with
$-d^2/dx^2+\mu$ can be developed as in the classical
case where $\mu$
is absolutely continuous. We will therefore just compile
some basic results and say relatively little about the proofs,
which are completely analogous to the classical ones (see
\cite{CodLev,WMLN}).

The operators on $L_2(0,N)$ have purely discrete spectrum. This
follows from the fact that the resolvent is a compact operator,
and this, in turn, follows from an explicit formula for the
integral kernel of the resolvent in terms of the solutions
of $-u''+\mu u=zu$.

A spectral representation is thus obtained by expanding in
eigenfunctions. Since these eigenfunctions must be multiples
of the solutions $u(x,z)$ introduced at the end of Sect.\ 2,
we can proceed as follows. Define
\begin{equation}
\label{defU}
(Uf)(E) = \int u(x,E)f(x)\, dx .
\end{equation}
Then, by the above remarks, $U$ maps the original Hilbert space
$L_2(0,N)$ unitarily onto $L_2(\mathbb R, d\rho_N^{(\beta)})$, where
\[
\rho_N^{(\beta)} = \sum \frac{\delta_E}{\|u(\cdot,E)\|^2} .
\]
The sum is over the eigenvalues $E$ (which depend on $N$ and the
boundary condition $\beta$ at $x=N$) and $\delta_E$ denotes the
Dirac measure at $E$, so $\delta_E(\{ E \} )=1$,
$\delta_E(\mathbb R\setminus \{ E \} )=0$. The denominator is
necessary because the $u$'s are not normalized.

For operators on the half-axis $(0,\infty)$, the situation is
considerably more complicated, because the spectrum need not be
discrete. However, as already explained, one may proceed exactly
as in the case of a potential ($d\mu(x)=V(x)\, dx$). There is the
distinction between the limit point and limit circle case at
infinity. In the latter case, one needs a boundary condition at
infinity to obtain self-adjoint operators. In either case, one
can define a Titchmarsh-Weyl $m$ function in the usual way. The
measure $\rho$ from the Herglotz representation
of this $m$ function is a spectral measure
in the sense that $U$ from \eqref{defU} still maps $L_2(0,\infty)$
unitarily onto $L_2(\mathbb R, d\rho)$. Moreover, the transformed
operator $UHU^*$ is multiplication by the variable in
$L_2(\mathbb R, d\rho)$, but this will be less important for us.

It is, of course, no coincidence that these well known methods
carry over to Schr\"odinger operators with measures. Rather, the
treatment of \cite[Chapter 9]{CodLev} does not depend on the precise
form of the underlying differential equation, but only on Green's
identity (see Theorem \ref{T2.2}).
\section{de~Branges spaces}
In this section, we want to follow \cite{Rem03} and use de~Branges
spaces as a tool in the spectral analysis of Schr\"odinger operators.
For the definition and the properties of de~Branges functions
and spaces we refer the reader to \cite{deBranges,Rem03}.

We fix $N>0$ and, as in \cite{Rem03},
set $E_N(z):=u(N,z)+iu'(N,z)$. Here, $u$ still is the solution defined
at the end of Sect.~2.
\begin{Proposition}
\label{P4.1}
The function $E_N$ is a de~Branges function.
\end{Proposition}
\begin{proof}
It is clear that $E_N$ is entire, so it remains to show
that $|E_N(z)|>|E_N(\overline{z})|$ for all $z\in \C^+=
\{ z\in\C : \text{Im }z>0\}$.
Let $\zeta,z\in\C$.
Green's identity (Theorem \ref{T2.2}), applied with $f=u(\cdot,z)$,
$g=u(\cdot,\zeta)$ gives
\begin{equation}
\label{4.1}
\frac{\overline{u(N,z)}{u'(N,\zeta)}-\overline{u'(N,z)}u(N,\zeta)}
{\overline z-\zeta}=\int_0^N\overline{u(N,z)}u(x,\zeta)\,dx.
\end{equation}
From the integral equation satisfied by $u(x,z)$
and the uniqueness of the solution we get
$\overline{u(x,z)}=u(x,\overline z)$. Therefore the left-hand side
of \eqref{4.1} is equal to
\[
\frac{\overline{E_N(z)}E_N(\zeta)-E_N(\overline z)
\overline{E_N(\overline\zeta)}}{2i(\overline z-\zeta)} .
\]
Hence for $z=\zeta\in\C^+$ we have that $|E_N(z)|>|E_N(\overline z)|$,
as required.
\end{proof}

Given a de~Branges function $E$, one can form the
\textit{de~Branges space} $B(E)$ based on $E$. One possible
definition is given by
\[
B(E)= \{ F:\C\to\C : F \text{ entire, } F/E, F^{\#}/E\in H_2 \} .
\]
Here, $F^{\#}(z)=\overline{F(\overline z)}$, and $H_2$ is
the Hardy space on the upper half plane, that is, $f\in H_2$
precisely if $f$ is (defined and) holomorphic on the upper half plane
and $\sup_{y>0} \int_{-\infty}^{\infty} |f(x+iy)|^2\, dx < \infty$.
$B(E)$ is a Hilbert space with the scalar product
$[F,G]=(1/\pi ) \int_{\R} \overline{F}G/|E|^2\, d\lambda$.
These spaces $B(E)$ may be used in the spectral representation
of Schr\"odinger operators. We will write $B(E_N)=:S_N$ (S for
Schr\"odinger).
Set 
\[
V:S_N\longrightarrow L_2(\R,d\rho_N^{(\beta)}),\ VF=F\bigr|_{\R}.
\]
By using the arguments from \cite[Sect.~3]{Rem03},
we conclude that $V$ is unitary. So we may replace the
space $L_2(\R,d\rho_N^{(\beta)})$ in the spectral
representation by the de~Branges space $S_N$. We get an induced
unitary map $U:L_2(0,N)\to S_N$
(which we still denote by $U$).
It is given by
\[
(Uf)(z)= \int u(x,z)f(x)\, dx .
\]
In particular,
\[
S_N=\left\{ F(z)=\int_0^N u(x,z)f(x)\,dx: f\in L_2(0,N)\right\} .
\]
Finally, de~Branges spaces have reproducing kernels $J_z$ (so
$[J_z,F]=F(z)$ for all $F\in B(E)$). In the case of $S_N$,
they are given by
\[
J_z(\zeta)=\int_0^N\overline{u(x,z)}u(x,\zeta)\,dx .
\]

The function $u'(x,z)$ has jumps at the discrete points of
$u(x,z)\, d\mu(x)$. More precisely, as observed in Sect.~2,
we have that $u'(x,z)-u'(x-,z)=u(x,z)\mu(\{ x \} )$.
So it would have been equally natural
to work with $\widetilde E_N(z)=u(N,z)+iu'(N-,z)$. However,
this new de~Branges function leads to the same de~Branges space:
$B(E_N)=B(\widetilde E_N)$. This follows at once from
\cite[Theorem 7.2]{Rem03} because (using the notation from
that reference) $\widetilde A(z)=A(z)$, $\widetilde B(z)=
B(z)-\mu(\{ N \} ) A(z)$. It is also quite clear form another
point of view because replacing $u'(N,z)$ by $u'(N-,z)$ amounts
to a change of boundary conditions at $x=N$, but the de~Branges
space $S_N$ is independent of this boundary condition.

As in \cite{Rem03}, we will now describe the spaces $S_N$
in more detail. We prepare for this with two lemmas which
describe the large $z$ asymptotics of $u(N,z)$ and $u'(N,z)$.
\begin{Lemma}
\label{L4.2}
Write $z=k^2$. Then $u$ satisfies the integral equation
\begin{equation}
\label{4.2}
u(x,z)=\cos kx+\frac{1}{k}\int_{[0,x]}
u(t,z)\sin k(x-t)\, d\mu(t).
\end{equation}
\end{Lemma}
Note that only even functions of $k$ occur, so there is
no problem with the fact that there are two solutions to
$z=k^2$ if $z\not= 0$. Moreover, these functions are also
entire, so everything makes sense for $z=k=0$ as well.
\begin{proof}
Fix $z=k^2\not=0$. For $k=0$, the following calculation
also works but it must be slightly modified.
From the equation statisfied by $u$ we get
\[
-\int_0^x (Au)'(t) \sin k(x-t)\, dt= k^2 \int_0^x u(t,z)
\sin k(x-t)\, dt .
\]
Integrating the first term by parts and using Fubini's theorem,
we get
\begin{multline*}
-ku(x)+k\cos kx+k^2\int_0^x u(t)\sin k(x-t)\,dt
+\int_{[0,x]}u(t)\sin k(x-t)\, d\mu(t) \\ = k^2\int_0^x u(t)
\sin k(x-t)\, dt,
\end{multline*}
which is the desired equation
\end{proof}
\begin{Lemma}
\label{compare}
Write $z=k^2$. Then, for large $|k|$, we have that
\begin{align*}
\left|u(x,z)-\cos kx\right|& \le \frac{C_N}{|k|}\,
\exp (|\text{\rm Im } k|\, x), \\
\left|u'(x,z)+(1/k)\sin kx\right| & \le C_N
\exp (|\text{\rm Im } k|\, x) .
\end{align*}
These estimates hold uniformly in $x\in [0,N]$.
\end{Lemma}
\begin{proof}
Write $k=r+it$ and
let $f(x):=e^{-|t|x}u(x,z)$. Then from Lemma \ref{L4.2}, we get
\[
M\equiv \sup_{x\in [0,N]}|f(x)|\le 1 + \frac{1}{|k|}|\mu|([0,N]) M .
\]
Hence $M\le 2$, say, for large enough $k$. By using this
estimate in \eqref{4.2} again, we get the desired estimate,
with $C_N=2|\mu|([0,N])$.

To prove the estimate on $u'$, note that Lemma \ref{L4.2} implies that
\[
u'(x,z)= -k\sin kx - \int_{[0,x]} u(t,z)\cos k(x-t)\, d\mu(t) .
\]
Indeed, the right-hand side is the distributional derivative
of the right-hand side of \eqref{4.2} and it is also the
correct representative because it is right continuous.
The asserted estimate on $u'$ now follows immediately.
\end{proof}
\begin{Theorem}
As a set, the space $S_N$ is given by
\[
S_N=\left\{ F(z)=\int_0^N f(t)\cos\sqrt{z}t\,dt:
f\in L_2(0,N)\right\} .
\]
\label{decrire}
\end{Theorem}
This is the generalization of \cite[Theorem 4.1]{Rem03},
and, given Lemma \ref{compare}, it has the same proof.
\section{The direct spectral theorem}
It remains to analyze the possible scalar products
on the de~Branges spaces coming
from generalized Schr\"odinger equations.
To this end, we introduce a function $\phi$ that describes
these scalar products and
may thus be thought of as representing
the spectral data.

We need some notation. We write $BV(-2N,2N)$ for the
set of real valued functions on $(-2N,2N)$
that are of bounded variation.
Given an even function $\phi\in BV(-2N,2N)$, we set
\[
K(s,t):=\frac{1}{2}\left( \phi(s-t)+\phi(s+t)\right) .
\]
Define the integral operator ${\mathcal K}_{\phi}$ on $L_2(0,N)$ by 
\[
\left( {\mathcal K}_{\phi}f\right)(t):=\int_0^N K(t,s)f(s)\,ds.
\]
Then ${\mathcal K}_{\phi}$ is a self-adjoint
Hilbert-Schmidt operator.
For the proof of the direct spectral theorem we will need
an asymptotic formula for the Titchmarsh-Weyl $m$ function. We
consider the problem on the half line $(0,\infty)$ with the
modified measure $\mu_N=\chi_{[0,N]}\mu$. The $m$ function of
this problem (with boundary condition $(Au)(0)=0$) will be denoted
by $m_N$, and $m_0$ will be the $m$ function for $\mu=0$, that
is, $m_0(z)=(-z)^{1/2}$, where the square root is determined
by the requirement that $\text{Im }m(z)>0$ for $z\in\C^+$.
If $z=k^2$ with $\text{Im }k>0$ and $f$ is the solution
of $-f''+\mu_N f=k^2 f$ with $f(x,k)=e^{ikx}$ for $x>N$, then
$m_N(k^2)=-f(0,k)/(Af)(0,k)$.

Finally, we introduce $M_N(k)=m_N(k^2)$. The functions $m_N$,
$m_0$ (originally defined on $\C^+$) may be holomorphically
continued to $\C\setminus (-\infty, 0]$, and $M_N$ extends to
a meromorphic function on $\C$. In the following lemma, we work
with these extensions. We refer the reader to \cite{Rem03} for
a more careful discussion of these issues.
\begin{Lemma}  
a) The limit $\lim_{k\to 0}kM_N(k)$ exists.\\
b) For $\text{Im }k\ge 0,\ k\notin(-\infty,0]$, we have
\[
m_N(k^2)-m_0(k^2)=\frac{1}{k^2}\int_{[0,N]}
e^{2ikx}\,d\mu(x)+O(|k|^{-3}) .
\]
\label{asymptotic}
\end{Lemma}
\begin{proof}
Part a) is proved as in \cite{Rem03}. We use the constancy
of the Wronskian established in Proposition \ref{P2.2}.

To prove part b), we also proceed as in the proof of
\cite[Lemma 4.3]{Rem03}. We claim that $g(x,k)\equiv
f(x,k)e^{-ikx}$ is the unique solution of the integral equation
\begin{equation}
\label{5.1}
g(x,k)=1+\frac{1}{2ik} \int_{(x,N]} \left( e^{2ik(t-x)} - 1
\right) g(t,k)\, d\mu(t) .
\end{equation}
To show this, let $I$ be the integral on the right-hand side.
We may write this integral in the form
\[
I=\int_{(x,N]} d\mu(t)\, g(t,k) \int_x^t ds\, e^{2ik(t-s)}
= \int_x^N ds \int_{(s,N]} d\mu(t)\, g(t,k) e^{2ik(t-s)} .
\]
This last expression shows that $I$ and hence also the
solution $g$ of \eqref{5.1}
is absolutely continuous
and $I'=g'=-\int_{(x,N]} g(t,k)e^{2ik(t-x)}\, d\mu(t)$.
Note that it is the right choice to take the interval
open at $x$ because this makes $g'$ right continuous, as
it should be.
As for $f=ge^{ikx}$, we thus have that
\begin{gather}
\label{5.2}
(Af)(x)= ike^{ikx} - \int_{[0,N]} \omega(x,t) g(t,k)\, d\mu(t),\\
\omega(x,t):=
\begin{cases}
0 & 0\le t\le x \\
(1/2) \left( e^{ik(2t-x)} + e^{ikx} \right) & x<t\le N
\end{cases} . \nonumber
\end{gather}
Since $\partial\omega/\partial x$ is uniformly bounded,
we obtain
\[
(Af)'(x)=-k^2e^{ikx} - \int_{(x,N]} \frac{\partial\omega}
{\partial x}\, g(t,k)\, d\mu(t) = -k^2 f(x,k) ,
\]
so $f$ solves $-f''+\mu f = k^2 f$, and clearly $f(x)=e^{ikx}$
for $x>N$.

Now given \eqref{5.1}, we can complete the proof as in
\cite{Rem03}. First of all, \eqref{5.1} implies that
$\|g\|_{\infty}\le 2$ if $\text{Im }k\ge 0$ and
$|k|\ge 2|\mu|([0,N])$.
Hence $g(x,k)=1+O(|k|^{-1})$ for these $k$.
Eq.~\eqref{5.2} now shows that
\begin{align*}
(Af)(0) & = ik - \mu(\{ 0 \} ) g(0) -\frac{1}{2}
\int_{(0,N]} \left( e^{2ikt} + 1 \right) g(t) \, d\mu(t) \\
& = ik - \mu(\{ 0 \} ) -\frac{1}{2}
\int_{(0,N]} \left( e^{2ikt} + 1 \right) \, d\mu(t)
+ O(|k|^{-1}) .
\end{align*}
Similarly,
\[
f(0)=g(0)= 1 + \frac{1}{2ik} \int_{(0,N]}
\left( e^{2ikt} - 1 \right) \, d\mu(t)
+ O(|k|^{-2}) .
\]
The asserted asymptotic formula for $m_N=-f(0,k)/(Af)(0,k)$
follows by combining these equations.
\end{proof}
The next result gives restrictions
on the scalars products on de Branges spaces coming
from a generalized Schr\"odinger equation.  
\begin{Theorem}
There is an even function $\phi\in BV(-2N,2N)$ with
$\phi(0)=-\mu(\{0\})$, such that for every $F\in S_N$,
\[
\|F\|_{S_N}^2=\langle f,(1+\mathcal K_{\phi})f\rangle_{L_2(0,N)}.
\]
Here $f$ is related to $F$ as in Theorem \ref{decrire}.
\label{direct-spectral}
\end{Theorem}
Note that since $\phi$ is even, it is continuous at $x=0$.

We will not discuss the details of the proof here.
One uses the same method as in the proof of
\cite[Theorem 4.2]{Rem03}, with Lemma \ref{asymptotic}
as an important input. So the sought function $\phi$
is defined formally as
\[
\phi(x)=\int \cos\sqrt{\lambda} x\, d(\rho_N-\rho_0)(\lambda) ,
\]
where $\rho_N$ and $\rho_0$ are the spectral measures of
the half line problems with $\mu_N\equiv \chi_{[0,N]}\mu$
and zero potential, respectively. More precisely, the
integral over $(-\infty,0]$ may be evaluated directly
(it is a finite sum), and the integral over $(0,\infty)$
is defined as distribution. One then shows by analyzing the
Fourier transform with the help of Lemma \ref{asymptotic}b)
that this distribution is in fact a function of bounded variation.

As in \cite{Rem03}, we have the formula
\[
m_N(-y^2)-m_0(-y^2) = \frac{1}{y} \int_0^{\infty}
\phi(t)e^{-yt}\, dt,
\]
which is valid for sufficiently large $y>0$. (The integral
makes sense because the method of \cite{Rem03} gives a
function $\phi$ defined on all of $\mathbb R$.)
Fubini's theorem thus shows that
\[
m_N(-y^2)-m_0(-y^2) = \frac{\phi(0)}{y^2} + \frac{1}{y^2}
\int_{(0,\infty)} e^{-ys}\, d\phi(s) =
\frac{\phi(0)}{y^2} + o(y^{-2})
\]
as $y\to\infty$. Comparison with Lemma \ref{asymptotic}
now shows that $\phi(0)=-\mu(\{ 0 \} )$, as claimed.
\section{The inverse spectral theorem}
In Theorem \ref{direct-spectral}, we associated
to each generalized Schr\"odinger equation a
$\phi$ function that determines the scalar product
on the corresponding de Branges spaces. Our next goal
is to prove the converse: Every function $\phi$ having the
properties stated in Theorem \ref{direct-spectral} comes
from a generalized Schr\"odinger equation.

Let
\[
\Phi_N:=\left\{ \phi\in BV(-2N,2N) :
\phi \text{ even},
1+\mathcal K_{\phi}>0 \right\}
\]
be the set of functions that could in principle be
a $\phi$ function in the sense of Theorem \ref{direct-spectral}.
In the last condition, we require that $1+\mathcal K_{\phi}$
be positive definite as an operator on $L_2(0,N)$. This clearly
is a necessary condition on $\phi$ if $\langle f,
(1+\mathcal K_{\phi}) f\rangle$ is to define a norm.

Our principal result in this section states that indeed
every $\phi\in\Phi_N$ occurs as the $\phi$ function (in
the sense of Theorem \ref{direct-spectral}) of some signed
Borel measure $\mu$ on $[0,N)$. We exclude the point
$x=N$ here because we have already seen that $\mu(\{ N \} )$
has no influence on the de~Branges space $S_N$.
\begin{Theorem}
For every $\phi\in\Phi_N$, there is a signed Borel measure
$\mu$ on $[0,N)$ such that the norm on the
de~Branges space $S_N$
associated with $-d^2/dx^2+\mu$ on $(0,N)$ is given by
\[
\|F\|_{S_N}^2=\langle f,(1+\mathcal K_{\phi}) f
\rangle_{L_2(0,N)} ,
\]
where $F(z)=\int f(t)\cos\sqrt{z}t\,dt$.
\label{inverse-spectral}
\end{Theorem}
The first impression is that the method of \cite{Rem03},
suitably extended, should also suffice to prove this result.
However, we have not been able to make this approach work.
More specifically, we have not succeeded in proving the
needed analogs of the smoothness results from
\cite[Sect.~14]{Rem03}. Rather, we will use an approximation
argument that allows us to circumvent these difficulties.

Before proving Theorem \ref{inverse-spectral}, we note
that we have uniqueness in both directions of the correspondence
$\mu\leftrightarrow\phi$. This will be used in the proof
of Theorem \ref{inverse-spectral}; of course, it is also of
independent interest.
\begin{Theorem}
\label{T6.2}
a) If a (finite, signed) Borel measure $\mu$ on $[0,N)$ is given,
the $\phi\in\Phi_N$ from Theorem \ref{direct-spectral} is unique.

b) If $\phi\in\Phi_N$ is given, the $\mu$ from
Theorem \ref{inverse-spectral} is unique.
\end{Theorem}
This is proved in the same way as the corresponding result
(Theorem 5.2) from \cite{Rem03}. Then, in the proof of part b),
one now needs the fact that $u(x,z)$ for $x\in [0,N]$, $z\in
\mathbb C$ determines $\mu$ on $[0,N)$. This indeed holds because
$u\mu =  u''+zu$ and $u(\cdot,z)$ has no zeros for $z\in
\mathbb C\setminus\mathbb R$.

\begin{proof}[Proof of Theorem \ref{inverse-spectral}]
For absolutely continuous $\phi$, this is proved in
\cite{Rem03}. In fact, this is not literally true, because
in \cite{Rem03}, it is assumed that $\phi(0)=0$, but the
extension to general values of $\phi(0)$ (and thus to
general boundary conditions) is rather straightforward.
We will now approximate the given
$\phi\in\Phi_N$ by \textit{smooth} functions $\phi_n\in\Phi_N$.
More precisely, we choose the $\phi_n\in C_0^{\infty}(-2N,2N)$
so that $\sup_n\|\phi'_n\|_{L_1(-2N,2N)} <\infty$ and
$\phi_n'\to\phi'$ in the $w^*$-topology of the space of
Borel measures on $(-2N,2N)$. Recall that
$\phi'$, the
distributional derivative of $\phi$, is a measure because $\phi$
itself is of bounded variation. In the sequel, we will also denote
this measure by $\nu=\phi'$.

To construct a sequence $\phi'_n$ with the above
properties, we can proceed as
follows. Fix $\psi\in C_0^{\infty}(0,1)$ with $\int \psi = 1$.
For $n\in\mathbb N$ (typically large), subdivide $(-2N,2N)$
into $2n$ subintervals of length $\ell=2N/2n$. If $I$ is such
a subinterval, with endpoints $a$ and $b$, say,
define $\phi'_n$ on $I$ by
\[
\phi'_n(x)= \frac{\nu(I)}{\ell} \, \psi\left(
\frac{x-a}{\ell}\right) .
\]
Then $\phi'_n\in C_0^{\infty}(-2N,2N)$, these functions
are automatically odd, and
\[
\int f(x)\phi'_n(x)\, dx \to \int f(x)\, d\nu(x)
\]
as $n\to\infty$ for all $f\in C_0(-2N,2N)$; in other words,
$\phi'_n\to\nu$ in the weak $*$-topology. Moreover,
$\int_I |\phi'_n| \le |\nu(I)| \int |\psi|$, hence
$\|\phi'_n\|_{L_1(-2N,2N)} \lesssim \sum_I |\nu(I)|$. Since
$\sum_I |\nu(I)| \le |\nu|((-2N,2N))$, the sequence
$\phi'_n$ is bounded in $L_1(-2N,2N)$.

The functions $\phi_n$ are still only determined up to
a constant. We make the obvious choice, namely, we require
that $\phi_n(0)=\phi(0)$. Because of the weak $*$-convergence
and the norm boundedness in $L_1$ of the derivatives, we have
that the functions $\phi_n(x)$ converge \textit{boundedly} almost
everywhere to $\phi(x)$. More precisely, we will certainly have
convergence whenever $\phi(x)=\phi(x-)$, or, equivalently,
$\nu(\{ x \} )=0$.

By dominated convergence this implies that
the operators $\mathcal K_{\phi_n}$ converge to
$\mathcal K_{\phi}$ in Hilbert-Schmidt norm. In particular,
they converge in operator norm and consequently,
$1+\mathcal K_{\phi_n}>0$ for sufficiently large $n$.
Hence $\phi_n\in\Phi_N$ for these $n$.

By \cite[Theorem 5.1]{Rem03} (more precisely, by the extension of
this result to arbitrary non-Dirichlet boundary conditions),
there are potentials $V_n$ such that the
de~Branges spaces associated to $-d^2/dx^2+V_n(x)$ are described
by the functions $\phi_n$. If we again describe the boundary
condition at $x=0$ by $\mu(\{ 0 \} )$, then we actually obtain
the measures $\mu_n = -\phi(0)\delta_0 + V_n$.

More explicitly, the potential $V_n$ is given by 
\begin{equation}
\label{6.1}
V_n(x)=\frac{p_n''(x,x)}{p_n(x,x)},
\end{equation}
where $p_n$ is the solution of the integral equation
\begin{equation}
\label{ie}
p_n(x,t)+\int_0^xK_{\phi_n}(t,s)p_n(x,s)\,ds=g_n(t)
\end{equation}
(see \cite[Sect.\ 18]{Rem03}.
In fact, there are two such equations, corresponding to
$g_n\equiv 1$ and $g''_n=\phi_n$, $g_n(0)=0$, $g'_n(0)=1$.
With either choice of $g_n$, \eqref{6.1} holds. Moreover,
the zeros of the solutions $p_n(x,x)$ form a discrete set,
and it does not happen that both solutions have a zero at the
same point. So for at least one choice of $p_n$, \eqref{6.1}
is applicable.

We now want to show that $(V_n)$ is bounded in $L_1(0,N)$.
By the Banach-Alaoglu theorem, this will allow us to extract
a subsequence $(\mu_{n_k})$ that converges to some measure
$\mu$ in the $w^*$-topology. We will then conclude the proof
by showing that $\phi$ is the $\phi$-function of
$-d^2/dx^2+\mu$.

To prove that $\sup_n\|V_n\|_{L_1(0,N)}<\infty$ it suffices
to show that $\sup_n\|p''_n\|_{L_1(0,N)}<\infty$ because of
\eqref{6.1} and the remarks following \eqref{ie}.
Denote by
$\mathcal K_n^{(x)}:=\mathcal K_{\phi_n}^{(x)}$
the integral operator
on $L_2(0,x)$ generated by the kernel
\[
K_{\phi_n}(t,s)= \frac{1}{2}
\left( \phi_n(t-s)+\phi_n(t+s)\right) .
\]
We will consider
$\mathcal K_n^{(x)}$
as an operator on $C[0,x]$ and on $L_1(0,x)$.
On both of these spaces,
we have convergence in operator norm to the corresponding limit
operator as $n\to\infty$. For instance, to compute the $B(C)$ norm
of $\mathcal K_n^{(x)}-\mathcal K^{(x)}$,
we must analyze expressions of the form
\begin{multline*}
\sup_{\|f\|_{\infty}=1} \sup_{0\le t\le x}
\left| \int_0^x \left( \phi_n(t-s)-\phi(t-s)\right) f(s)\, ds \right|
\\ \le \sup_{0\le t\le x} \int_0^x \left| \phi_n(t-s)-\phi(t-s)
\right|\, ds
\le \int_{-x}^x \left| \phi_n(u)-\phi(u)\right| \, du .
\end{multline*}
So we can indeed estimate uniformly in $x\in [0,N]$:
\[
\|\mathcal K_n^{(x)}-\mathcal K^{(x)}\|\le
\|\phi_n-\phi\|_{L_1(-2N,2N)} .
\]
This estimate also holds for the operator norm in $L_1(0,x)$.

Since inversion is a continuous operation, $1+\mathcal K_n^{(x)}$
is boundedly invertible for sufficiently large $n$ in $C[0,x]$
and in $L_1(0,x)$, and the operator norms of the inverses are
uniformly bounded in $n$.

Finally, note
that $\|d^kg_n/dx^k\|_{\infty}$ also remains bounded in $n$
for $k=0,1,2$ and either choice of $g_n$.

Since $(1+\mathcal K_n^{(x)}) p_n(x,\cdot)=g_n$, we obtain as
a first consequence of the above observations that
\[
|p_n(x,t)| \le C \quad (0\le t\le x \le N, n\ge n_0) .
\]
Moreover, since $p_{x,n}:=\partial p_n/\partial x$ satisfies
\[
p_{x,n}=-p_n(x,x)(1+\mathcal K_n^{(x)})^{-1} K_n(\cdot, x) ,
\]
this function is also uniformly bounded. The other first order
partial derivative satisfies
\[
p_{t,n}(x,t)=-\int_0^x K_{t,n}(t,s)p_n(x,s)\,ds+g'_n(t).
\]
Hence 
\[
\sup_{0\le t\le x}\left| p_{t,n}(x,t)\right|\lesssim
\|\phi'_n\|_{L_1(-2N,2N)}+\|g'_n\|_{\infty},
\]
and since $\|\phi'_n\|_{L_1(-2N,2N)}$ also remains bounded, we
get a uniform bound on $p_{t,n}$ as well.

The second order derivatives are not necessarily bounded pointwise,
so here we work with $L_1$ bounds. Apart from this, the reasoning
is similar to the arguments used above. By differentiating
\eqref{ie} twice with respect to $x$ and solving for $p_{xx,n}$,
we obtain
\begin{multline*}
p_{xx,n}(x,t) = \\
\left( 1+\mathcal K_n^{(x)}\right)^{-1}
\bigl( -K_{x,n}(t,x)p_n(x,x) - K_n(t,x)\left( 2p_{x,n}(x,s) +
p_{s,n}(x,s)\right) \bigr|_{s=x} \bigr) .
\end{multline*}
The expression in parantheses on the right-hand side remains
bounded in $L_1(0,x)$, uniformly in $x\in [0,N]$, as $n$ varies.
Hence $\sup_{n,x} \|p_{xx,n}(x,\cdot)\|_{L_1(0,x)} < \infty$.

Now we are ready to bound $p''_n(x,x)$. To simplify the notation,
we will drop the index $n$ in this final step. By \eqref{ie},
\begin{align*}
p'(x,x) & =\left( p_x(x,t)+p_t(x,t)\right) \bigr|_{t=x}\\
& = -K(x,x)p(x,x) - \int_0^x K(x,s)p_x(x,s)\, ds + g'(x) -
\int_0^x K_x(x,s)p(x,s)\, ds .
\end{align*}
We differentiate one more time. This gives
\begin{multline*}
p''(x,x)=-\phi'(2x)p(x,x) - K(x,x)p'(x,x) + g''(x) - K(x,x)
p_x(x,s) \bigr|_{s=x} \\
- 2 \int_0^x K_x(x,s)p_x(x,s)\, ds
- \int_0^x K(x,s)p_{xx}(x,s)\, ds \\
- K_x(x,s)\bigr|_{s=x}
\cdot p(x,x) - \int_0^x K_{xx}(x,s)p(x,s)\, ds .
\end{multline*}
Our previous results show that everything on the right-hand
side except possibly the last term is bounded in $L_1$.
In the last term, the problem is the occurence of $\phi''_n$;
we do not have control on these functions. However, since
$K_{xx}=K_{ss}$, we can integrate by parts:
\begin{align*}
\int_0^x K_{xx}(x,s)p(x,s)\, ds & = \int_0^x K_{ss}(x,s)
p(x,s)\, ds \\
& = K_s(x,s)\bigr|_{s=x} \cdot p(x,x) - \int_0^x K_s(x,s)
p_s(x,s)\, ds
\end{align*}
So this term is bounded, too. In conclusion, we have that
$\sup_n \|p''_n(x,x)\|_{L_1(0,N)} < \infty$, as we wanted
to show.

As explained above, we may now assume, by passing to a
subsequence if necessary (we will later see that this is
actually \textit{not} necessary), that $V_n(x)\, dx -
\phi(0)\delta_0$ converges to a (finite, signed) Borel measure
$\mu$ on $[0,N]$ in the weak $*$-topology. Since
$\sup_n \|V_n\|_{L_1(0,x)}\to 0$ as $x\to 0+$, the limit
measure must also satisfy $\mu(\{ 0 \} )=-\phi(0)$.

The final step in the proof consists of showing that $\phi$ is
the $\phi$ function of $-d^2/dx^2+\mu$. We will use the
following result.
\begin{Lemma}
\label{L6.1}
Let $\mu_n$ be a bounded sequence of measures
(so $\sup |\mu_n|([0,N])<\infty$) that converges to the
measure $\mu$ in the weak $*$-topology. Let $\rho_n$ and
$\rho$ be the spectral measures of $-d^2/dx^2+\mu_n$ and
$-d^2/dx^2+\mu$, respectively, on $[0,N]$,
with boundary conditions
$(Af)(0)=0$, $f'(N)=0$. Then $(1+\lambda^2)^{-1}\,
d\rho_n(\lambda)\to (1+\lambda^2)^{-1}\,d\rho(\lambda)$ in
the weak $*$-topology.
\end{Lemma}
Here, the weight $(1+\lambda^2)^{-1}$ is introduced to
get finite measures. Assuming Lemma \ref{L6.1}, we can
complete the proof of Theorem \ref{inverse-spectral}
as follows. We apply the Lemma with $\mu_n=V_n(x)\, dx -
\phi(0)\delta_0$. For $f\in C_0^{\infty}(0,N)$, the
transform $F(\lambda)=\int f(x)\cos \sqrt{\lambda} x\, dx$
goes to zero rapidly as $\lambda\to\infty$. Moreover,
since $\mu_n$ is bounded, there is an $E_0\in\mathbb R$
so that the supports
of the $\rho_n$'s and $\rho$ are all contained in $(E_0,\infty)$.
Therefore, we can cut off $F(\lambda)$ to the left of $E_0$ without
changing the integrals $\int |F|^2\, d\rho_n$,
$\int |F|^2\, d\rho$. So, $(1+\lambda^2)
|F|^2$ is an admissable test function
and thus
\[
\int_{\mathbb R} |F(\lambda)|^2\, d\rho_n(\lambda) \to
\int_{\mathbb R} |F(\lambda)|^2\, d\rho(\lambda).
\]
On the other hand, $\int |F|^2\, d\rho_n=\langle f, (1+
\mathcal K_{\phi_n}) f\rangle$ by the identification of $L_2(
\mathbb R, d\rho_n)$ with $S_N^{(n)}$ (see Sect.~4). Since
\[
\langle f, (1+\mathcal K_{\phi_n}) f\rangle\to
\langle f, (1+\mathcal K_{\phi}) f\rangle
\]
and $\int |F|^2\, d\rho= \| F \|_{S_N}^2$, the uniqueness of
$\phi$ (Theorem \ref{T6.2}b)) together with the fact that $f$
varies over a dense subset of $L_2(0,N)$ now show that $\phi$
is the $\phi$ function of $-d^2/dx^2+\mu$.

It remains to prove Lemma \ref{L6.1}. We will prove that
the corresponding $m$ functions converge, $m_n(z)\to m(z)$,
uniformly on compact subsets of $\mathbb C^+$. As is
well known, this implies
weak $*$-convergence of the (renormalized) spectral measures
(compare \cite{Don}). The $m$ functions
can be computed as follows: $m(z)=-f(0,z)/(Af)(0,z)$, where
$f$ solves $-f''+\mu f=zf$ and $f'(N,z)=0$, and similarly for
$m_n(z)$. We normalize by further requiring that $f(N,z)=1$
(a multiplicative constant of course drops out anyway when
$m$ is computed). Then, in analogy to Lemma \ref{L4.2} (compare
also \eqref{5.2}), $f$ is the unique solution of the integral equation
\begin{equation}
\label{6.2}
f(x,k)=\cos k(x-N) + \frac{1}{k} \int_{(x,N]} \sin k(t-x)
f(t,k)\, d\mu(t) .
\end{equation}
Here, $z=k^2$ (since only even functions of $k$ occur, there
is no problem with the choice of the square root), and, of course,
the $f_n$'s obey the same equations with $\mu$ replaced
by $\mu_n$. Now we can argue as in the proof of Theorem \ref{T2.3}.
Namely, given a compact subset $K\subset\mathbb C$, we pick
$\eta>0$ so that
\[
\sup_{0<s\le\eta} \sup_{z\in K} \left| \frac{\sin ks}{k}
\right| \, |\mu|([0,N]) < 1 .
\]
Then \eqref{6.2} shows that $f(x,k)$ remains bounded for
$x\in [N-\eta, N]$ and $z\in K$. But now a similar argument
may be applied to the interval $[N-2\eta,N-\eta]$ etc.\ so that
$f$ is eventually
seen to be uniformly bounded on $[0,N]$. Clearly, this
argument has in fact shown that $f_n$ is also bounded, uniformly
in $n$. Finally, a slight modification of the whole analysis
shows that $f_n\to f$, uniformly in $x\in [0,N]$ and $z\in K$.

Eq.\ \eqref{6.2} implies that
\[
(Af)(0,k)= k\sin kN - \int_{[0,N]} \cos kt f(t,k)\, d\mu(t) .
\]
Hence we also have that $(Af_n)(0,k)\to (Af)(0,k)$ uniformly
on compact sets. Since it is not possible that $(Af)(0,z)=0$
for non-real $z$, it now follows that $m_n(z)\to m(z)$ locally
uniformly, as desired.
\end{proof}
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\end{document}
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