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\begin{document}

\title{\sch operators with fairly arbitrary spectral features}
\author{M. Krishna and V.S. Sunder\\
{}\\
The Institute of Mathematical Sciences\\
Madras 600113\\
INDIA \\ }

\maketitle

\begin{abstract} 
It is shown, using methods of inverse-spectral theory, that there
exist Schr\"{o}dinger operators on the line with fairly general
spectral features. 
Thus, for instance, it follows from the main theorem, that if
~$\Sigma$~ is any perfect subset of ~$(-\infty,0]$, then there exist
potentials ~$q_j, 
j=1,2$~ such that the associated Schr\"{o}dinger operators ~$H_j$~ are
self-adjoint and satisfy : ~$\sigma (H_j) = \Sigma \cup [0,\infty),
\sigma_{ac}(H_j) = [0,\infty), \sigma_{pp}(H_1) = \sigma_{sc}(H_2) =
\Sigma.$  The main result also implies the existence of states with
interesting transport properties. 
\end{abstract}

\newpage
\section{Introduction}

Inverse spectral theory for Jacobi/Schr\"odinger
operators has a long history.  However almost all the effort has gone
in understanding the operators with absolutely continuous spectra
and recovering, to some extent, detailed information about the potential.
We refer to the monographs of Newton \cite{new}, Chadan-Sabatier 
\cite{chsa}, Marchenko \cite{mar},
Levitan \cite{lev} and Carmona-Lacroix \cite{cl}. In
addition, recent advances in the theory can be found in the
works of Kotani \cite{sk} (who developed an inverse spectral
theory for random Schr\"odinger operators), 
Kotani-Krishna \cite{kk}, Craig \cite{cr}
and Gesztesy-Simon \cite{gs, gs2}, Gesztesy-Teschl \cite{gt} for
Schr\"odinger operators and Antony-Krishna \cite{ank} and
Sodin-Yuditskii \cite{sy1, sy2, sy} for Jacobi operators.
In particular, the recent lectures of Simon \cite{sim1} and the 
article of Gesztesy \cite{ges} give an overview of the
recent advances.

In contrast very little is known regarding inverse 
spectral theory for singular spectra; but, more recently, after the
first examples of 
Pearson \cite{pe} of \sch ~operators with singular continuous spectra,
the direct spectral problem for operators with singular continuous
spectra has been gaining attention 
in a substantial way - for example, in the works of Gordon
\cite{go}, Del Rio-Makaraov-Simon\cite{dms}, Del
Rio-Jitomirskaya-Last-Simon\cite{djls} and Simon\cite{sim2}.  We refer to 
\cite{sim1, sim2, djls} for a collection, as well as history, of
results in this direction. (The preceding list of references is, no
doubt, influenced by the authors' limitations, and lays no claim to
any sort of completeness.) 

This paper is motivated partly by a desire to understand the structure of
inverse theories for singular spectra and partly by the work of
Last \cite{la} and
Del Rio-Jitomirskaya-Last-Simon \cite{djls} - where operators with
singular continuous spectra, in the context of rank one perturbations,
are analyzed, for the geometry of spectrum, and the transport properties of
the physical systems corresponding to such operators. 

In the case of Jacobi operators,  Gesztesy-Krishna-Teschl
\cite{gkt} worked out the iso-spectral set for reflection-less bounded
Jacobi operators with the singular spectrum given 
by a countable set.  Recently we came to
know of the preprint of Last-Jitomirskaya \cite{laji} constructing
explicit Jacobi matrices with $\alpha$ dimensional singular spectra.  

The aim of this paper is to prove the following result.

\begin{Theorem}\label{mt}
Let $\mu_0$~ be a compactly supported probability measure with ~$supp
~\mu_0 \\
=~\Sigma ~\subset ~[-M,M],$ say, and such that ~$\mu_0$~ is
singular with respect to 
Lebesgue measure. Then there exists a function ~$q : \R \rightarrow
\R$~ such that (i) $q$~ is continuous away from 0, and (ii) $q$~ has
at most a jump discontinuity at 0, and such that the symmetric
operator ~$(\frac{-d^2}{dx^2} + q(x))$~ (defined on the space \coinf
~of compactly supported smooth functions) is essentially self-adjoint,
and its unique self-adjoint extension
~$H$~ in ~\lr~ has the following spectral properties:

(a) ~$\sigma (H) ~=~ \Sigma ~\cup~ [0,\infty);$ and

(b) there exists a vector ~$v$~ in the domain of $H$ such that the
associated `spectral measure' (defined by ~$\nu(E) ~=~ \langle
1_E(H)v,v \rangle)$ is mutually absolutely continuous with the measure
~$\mu_0 + m,$ where ~$m$~ denotes - here and throughout this paper - the
measure defined by ~$m(E) ~=~ \int_{E \cap [0, \infty)} \frac{1}{2 \pi
\sqrt{\lambda}} ~d\lambda ~).$

Further, if it is the case that ~$\Sigma ~\subset ~(-\infty,0],$ then,
~$ \sigma_s(H) ~=~ \Sigma ,$ where ~$\sigma_s(H)$~ denotes the
singular spectrum of ~$H.$ 

\hfill$\Box$
\end{Theorem}


The proof of the theorem relies on inverse-spectral theory {\em \`{a}
la} Gelfand-Levitan. 

The paper is broken up into two sections, as follows. 
The first section is devoted to gathering together (for the
convenience of the reader) various facts - mostly known, and an
occasional fact which might, perhaps, not occur in existing literature
in quite the form we have stated - concerning (i) Herglotz
functions, (ii) symmetric operators, (iii) \sch operators, and (iv) the
Gelfand-Levitan theory. The second section contains the proof of the
theorem, and concludes with a discussion of some
examples (which, among other things, justify the assertions in Remark
\ref{rmk2} below concerning perfect sets of arbitrary Hausdorff
measure). 

\vskip 1em

We make a few remarks to justify the assertions made in the
abstract. 

\begin{Remark}\label{rmk1}
If ~$\Sigma$~ is any compact set, then there exists an atomic measure
~$\mu$ - i.e., ~$\mu ~=~ \mu_{pp}$ - such that ~$supp ~\mu ~=~ \Sigma.$
This shows that if we had chosen
~$\Sigma$~ to be a subset of ~$(-\infty, 0],$ then, by Theorem
\ref{mt}, there does indeed exist a potential ~$q$~ on ~$\R$~ 
such that the associated \sch~ operator ~$H$~ satisfies
~$\sigma(H) ~=~ \Sigma ~\cup ~[0,\infty),$ and
~$\sigma_{pp}(H) ~=~ \Sigma.$
\end{Remark}

\vskip 1em

\begin{Remark}\label{rmk2}
If ~$\Sigma$~ is a perfect set - i.e., ~$\Sigma$~ is compact and has no
isolated points - then there exists a singular continuous measure
~$\mu$ - i.e., ~$\mu ~=~ \mu_{sc}$ - such that ~$\Sigma ~=~ supp ~\mu.$
(Reason : if ~$\Sigma$~ has empty interior, this can be done by
adapting the construction of the Cantor function, to come up with a
continuous monotonic function which `rises' only on ~$\Sigma$~ and is
consequently the distribution function of the desired measure ~$\mu.$
If ~$\Sigma$~ has interior, we may construct several probability
measures supported on perfect nowhere dense sets whose union is dense
in ~$\Sigma$ and take an appropriate weighted average of the
associated countable family of probability measures.)

It follows now from Theorem \ref{mt} that if ~$\Sigma$~ is a perfect
set in ~$(-\infty, 0],$ then we can (start with a singular
continuous probability measure ~$\mu_0$ with support given by
~$\Sigma$ and) find a potential ~$q$~ on ~$\R$~ such that the associated
\sch ~operator ~$H$~ has ~$\sigma(H) ~=~ \Sigma \cup [0, \infty)$~ and
~$\sigma_{sc}(H) ~=~ \Sigma.$

Finally, the perfect set $\Sigma$ can be chosen to have any
preassigned Hausdorff dimension ~$0 < \alpha < 1,$ since there do
exist - see \cite{fa} - perfect sets 
~$\Sigma$~ (contained in ~$(-\infty,0])$ with Hausdorff dimension
exactly equal to ~$\alpha.$ 
\end{Remark}

\begin{Remark}\label{rmk3}
One final remark
concerns the transport properties of the
states associated with the singular spectra of the type we
constructed.  The results of Last \cite{la}
on the lower bounds on transport will be valid for 
the Shcr\"odinger operators we constructed with the $\alpha$
dimensional singular spectra,  
since in this case the operator $\chi_E(H)\chi_{[0,R]}(|x|)$ can be 
shown to be Hilbert-Schmidt, for any compact subset E (where 
$\chi_B(.)$ is the characteristic function of the
Borel set B).
\end{Remark}

\section{Preliminaries}

Throughout this paper, the word `measure' will mean a ~$\sigma$-finite
positive measure defined on the ~$\sigma$-algebra ~$\CB_\R$~ of Borel
sets in the real line ~$\R.$  For any measure, we shall write
\be \label{lebdec}
\mu ~=~ \mu_{ac} + \mu_{pp} + \mu_{sc}
\ee

\noindent for the usual Lebesgue decomposition (where the subscripts
are, of course, acronyms for `absolutely continuous', `pure point' and
`singular continuous', respectively). We shall also use the notation
~$\mu_s ~=~ \mu_{pp} + \mu_{sc}.$

Given a self-adjoint operator ~$H$~ in a Hilbert space ~$\H,$ we shall
use the 
symbols ~$B_b(H)$~ (resp., ~$C_c(H)$) to denote the algebra
given by ~$\{\phi(H): \phi $ is a bounded Borel measurable function on
~$\R\}$ (resp., ~$\{\phi(H): \phi $ is a continuous function on
~$\R$~ which vanishes at ~$\infty\}.$ For ~$x \in \H,$ the unique
measure ~$\mu_x$~ satisfying ~$\langle \phi(H)x, x \rangle ~=~ \int
\phi d \mu_x ~\forall \phi(H) \in C_c(H)$~ will be referred to as the
`spectral measure of ~$H$~ associated with the vector ~$x'.$

A subset ~$S \subset \H$~ is said to be `cyclic' for the self-adjoint
operator ~$H$~ if the set
~$\{Tx : T \in C_b(H), x \in S\}$~ is total in ~$\H$ - meaning that
the linear subspace generated by the latter set is dense in ~$\H.$ By
the multiplicity of ~$H$~ is meant the number ~$min ~\{$ card $S~:~ S$
is cyclic for ~$H \}.$ If ~$H$~ is of finite multiplicity, and if
~$S $~ is any finite set which is cyclic for ~$H,$ then any measure
which is mutually equivalent to ~$\sum_{x \in S} \mu_x$~ will be
called a total spectral measure for ~$H.$ In more conventional
terminology, we have: ~$\mu$~ is a total spectral measure for ~$H$~ if
and only if the projection-valued measure ~$E \mapsto 1_E(H)$~ and
~$\mu$~ have the same class of null sets.

Finally, if ~$H$~ is as above, and if ~$\mu$~ is a total spectral
measure for ~$H,$ then we write ~$\sigma(H)$~ for the spectrum of
~$H,$ and we write
\[ \sigma_{ac}(H) = supp~\mu_{ac}, ~
\sigma_{pp}(H) = supp~\mu_{pp}, ~
\sigma_{sc}(H) = supp~\mu_{sc} \]

\noindent where ~$supp~\mu$~ denotes the closed support of
the measure ~$\mu.$

\subsection{Herglotz functions}

Recall that a {\em Herglotz} function is, by definition, an analytic
map ~$f$~ of the upper half-plane ~$\Pi^+$~ into itself. For the sake
of convenient reference, we state below the standard representation
of such functions as well as some other simple properties, as a
proposition. 

\begin{Proposition}\label{Herg}
(a) Let ~$\mu$~ be a positive measure (not identically equal to zero)
such that  ~$\int_\R \frac{1}{1 + \lambda^2} d\mu (\lambda) ~<~
\infty;$ then the equation 
\[ F(z) ~=~ \int_\R \left[ \frac{1}{\lambda - z} ~-~ \frac{\lambda}{1 +
\lambda^2 } \right] ~d\mu (\lambda) ~~,\]

\noindent defines a Herglotz function ~$F;$

(b) Conversely if ~$F$~ is a Herglotz function, there exist unique
real constants ~$a,b$~ with ~$b \geq 0,$ and a unique positive measure
~$\mu$~ which satisfies the condition ~$\int_\R \frac{1}{1 +
\lambda^2} d\mu (\lambda) ~<~ \infty,$  such that
\be\label{heq}
F(z) ~=~ a + bz + \int_\R \left[ \frac{1}{\lambda - z} ~-~
\frac{\lambda}{1 + \lambda^2 } \right] ~d\mu (\lambda)~~;
\ee

\noindent in particular (thanks to the monotone convergence theorem), 
~$Im ~F(\lambda + i0) ~=~ \lim_{\epsilon
\downarrow 0} Im ~F(\lambda + i \epsilon)$~ exists (and is a non-negative
extended real number) for every ~$\lambda \in \R;$ 

(c) If ~$\mu$~ and $F$ are as in (b) above, then, we have:

(i) ~$supp~\mu_{ac} ~=~ \{\lambda \in \R : 0 < Im ~F(\lambda + i 0) <
\infty\}^-,$ where the superscript ${^-}$ denotes closure; and
\[ d\mu_{ac} ~=~ \frac{1}{\pi} ~Im ~F(\lambda + i 0) ~d\lambda~; \]

(ii) ~$\mu(\{\lambda\}) ~=~ \frac{1}{\pi} ~\lim_{\epsilon \downarrow
0} \epsilon F(\lambda + i \epsilon),$ and consequently
~$supp~\mu_{pp} ~=~ \{\lambda \in \R : ~\lim_{\epsilon \downarrow
0} \epsilon F(\lambda + i \epsilon) \neq 0 \}^-;$ 

(iii) ~$supp~\mu_{s} ~=~ \{\lambda \in \R : Im ~F(\lambda + i 0) ~=~
\infty\}^-;$ and consequently,

(iv) ~$\lambda_0 \notin ~supp~\mu~\Leftrightarrow ~Im ~F(\lambda + i
0) = 0$~ for all ~$\lambda$~ in some neighbourhood of ~$\lambda_0.$

\hfill$\Box$
\end{Proposition}

\begin{Remark}\label{bt}
If ~$F$~ and ~$\mu$~ are related as in Proposition \ref{Herg} (a),
then ~$F$~ is called the {\em Borel transform} of ~$\mu.$ 
If ~$F$~ and ~$\mu$~ are related as in Proposition \ref{Herg} (b),
then we shall write ~$\mu ~=~ \mu_F.$
\end{Remark}

\subsection{Symmetric operators}

We start with some preliminary facts about symmetric operators and
their symmetric, resp., self-adjoint, extensions. For all these facts,
we refer the reader to the book \cite{ds} by Dunford and Schwartz.

Suppose ~$T$~ is a closed symmetric operator with (dense) domain contained in
the Hilbert space ~$\H;$ thus, ~$T \subset T^*.$ Let us write  ~$\CD
~=~ dom ~T^*.$ Then, any symmetric extension
~$S$~ of ~$T$~ satisfies ~$S \subset T^*.$ Define
the subspaces 
\be \label{dpmi}
\CD_\pm ~=~ \{ x \in \CD ~:~ T^*x = \pm i x \}~~,
\ee
\noindent where, of course, ~$i ~=~ \sqrt{-1}.$

The numbers ~$n_\pm ~=~ dim ~\CD_\pm$~ are called the deficiency
indices of the operator ~$T.$ 

Since ~$T^*$~ is a closed operator, the space ~$\CD$~ becomes a
Hilbert space with respect to the inner-product defined by
\[ (x , y)_* ~=~ \langle x , y \rangle ~+~ \langle T^*x, T^*y \rangle
~~.\]

It is then a fact that, with respect to the inner-product ~$( \cdot ,
\cdot )_*,$ the Hilbert space ~$\CD$~ admits the orthogonal
decomposition ~$\CD ~=~ dom ~T ~\oplus ~\CD_+ ~ \oplus ~\CD_-.$

A crucial fact is that ~$T$~ admits self-adjoint extensions if and
only if ~$n_+ = n_-,$ and that self-adjoint extensions of ~$T$~ are in
bijective correspondence with unitary operators from ~$\CD_+$~ onto
~$\CD_-.$ More precisely, if ~$V : \CD_+ \rightarrow \CD_-$~ is
unitary, then the operator given by 
\be \label{gensa}
\S \subset T^*~,~dom ~S ~=~ \{ x \in \CD ~:~ (x, y - Vy)_* = 0
~\forall ~y \in \CD_+ \}
\ee
\noindent is self-adjoint, and conversely, every self-adjoint
extension of ~$T$~ arises in this fashion. 

By a {\em boundary value for} ~$T$~ is meant a bounded linear
functional on the Hilbert space ~$\CD$~ which vanishes on the closed
subspace ~$dom ~T.$ Thus, in view of the last paragraph, a boundary
value for ~$T$~ is just a map of the form ~$\CD ~\ni ~x ~\mapsto ~(x ,
y)_*,$ where ~$y \in ~\CD_+ ~ \oplus ~\CD_-.$

Thus, every self-adjoint extension of ~$T$~ arises by restricting
~$T^*$~ to a sub-domain which is characterised by ~$n = n_+ = n_-$~
`boundary conditions'. The reason for the use of the expressions
`boundary values' and `boundary conditions' stems from the case of
differential operators.

\subsection{\sch ~operators}

Let ~$\tau$~ denote the differential expression ~$\lap~ + q(x),$ where
~$q$~ is a real-valued function on some interval ~$I = (a,b), ~- \infty \leq
a < b \leq \infty.$  (Later, we will impose more stringent conditions
on ~$q,$ when needed.) We shall write ~$T_0$~ and ~$T_1$~ for the
operators defined by $T_jf = \tau f$ on ~$L^2(I)$~ with domains given,
respectively, by 
\begin{eqnarray}\label{di}
\CD_0 & = & C^\infty_c(I) \\
\CD & = & \{u \in L^2(I) ~:~ u, u^\prime \mbox{ absolutely
continuous}, ~\tau u \in L^2(I)\}
\end{eqnarray}

Then, it is well-known - see \cite{rs} Proposition 2, Appendix to
Chapter X.1, for instance - that ~$T_0$~ is a symmetric operator and that
~$T_0^* = T_1.$ We shall need to apply the preceding general analysis
to the closed symmetric operator ~$T = T_0^-,$ i.e., the closure of
~$T_0$~ or equivalently the operator defined by ~$T \subset T_1$~ where
~$dom ~T$~ consists of those ~$f \in \CD$~ for which there exists a
sequence ~$ \{f_n\} \subset \CD_0$~ such that both ~$\{f_n\}$~ and
~$\{\tau f_n\}$~ are Cauchy sequences in ~$L^2(I).$

The deficiency subspaces ~$\CD_\pm$~ are at most ~$2$-dimensional,
since a linear differential equation of second order can have at most
two linearly independent solutions; further, since ~$q$~ is real. it
follows that ~$f \in \CD_+ ~\Leftrightarrow ~\overline{f} \in \CD_-,$
and consequently ~$n_+ = n_-,$ whence ~$T$~ has self-adjoint
extensions. Since ~$0 \leq n_\pm \leq 2,$ there are three
possibilities.

\vskip 1em

\noindent {\em Case (i):} ~$n_\pm = 0.$ 

In this case, it follows from ~$\CD_\pm = \{0\}$~ that ~$T = T_1$~ (and
that consequently $T_0$~ is essentially self-adjoint).

\vskip 1em

\noindent {\em Case (ii):} ~$n_\pm = 1.$ 

In this case, there exists a unique solution ~$u_\pm$~ of the equation
~$\tau u = \pm i u$~ which satisfies ~$u \in ~L^2(I).$

\vskip 1em

\noindent {\em Case (iii):} ~$n_\pm = 2.$ 

In this case, every solution ~$u_\pm$~ of the equation
~$\tau u = \pm i u$~ satisfies ~$u \in ~L^2(I).$

\vskip 1em

We pause to mention here - for ease of reference - the very useful
Green's formula, which states, in the above notation, (and follows from
twice integrating by parts) that if ~$f, g \in \CD,$ then
\be \label{grfor}
\langle T^*f, g \rangle - \langle f, T^*g \rangle ~=~ \lim_{x
\uparrow b,  ~y \downarrow a}\{[f,\overline{g}](x) -
[f,\overline{g}](y)\}~~, 
\ee

\noindent (where ~$[f,g](x) ~=~ f(x) g^\prime(x) -
f^\prime(x) g(x)$~ denotes the Wronskian). For typographical
convenience, we shall simply write ~$[f,g]^
b_a$~ to denote the
expression ~$lim_{x \uparrow b, ~y \downarrow a}\{[f,g](x) -
[f,g](y)\}.$ 

The following lemmas will be needed in the sequel.

\begin{Lemma}\label{*wr}
If ~$T_0, ~\CD, ~\CD_\pm$~ are as above, then, for arbitrary ~$f \in
\CD$~ and ~$u \in \CD_+ \oplus \CD_-,$ we have
\be \label{*wro}
(f,u)_* ~=~ [f,T^* \overline{u}]^b_a ~~.
\ee
\end{Lemma}

\vskip 1em

{\bf Proof:} We may, without loss of generality, assume that ~$u \in
\CD_\pm$. Then,
\begin{eqnarray*}
(f,u)_* & = & \langle f, u \rangle ~+~ \langle T^*f ,T^*u \rangle\\
& = & \langle f, u \rangle ~+~ \langle T^*f ,\pm i u \rangle\\
& = & \langle f, u \rangle ~\mp~i [f,\overline{u}]^b_a ~\mp~i \langle f, T^* u
\rangle\\
& = & \langle f, u \rangle ~\mp~ i [f,\overline{u}]^b_a ~\mp~ i \langle f,
\pm i u \rangle\\ 
& = & \langle f, u \rangle ~\mp~i [f,\overline{u}]^b_a ~-~ \langle f, u
\rangle\\
& = & \mp~i [f,\overline{u}]^b_a \\
& = & [f, T^* \overline{u}]^b_a ~~,
\end{eqnarray*}

\noindent as desired.

\hfill$\Box$

\begin{Lemma}\label{wrab}
Let ~$- \infty < a < b \leq \infty.$ Suppose ~$q$~ is a real-valued
continuous function on ~$(a,b)$ such that ~$q \in L^2(a, a+ \epsilon).$
Let ~$T_0$~ be the symmetric operator on ~$L^2(a,b)$~ defined by
~$dom ~T_0 ~=~ C^\infty_c(a,b), ~T_0 u ~=~ - u^{\prime \prime} + qu,$
and let ~$\CD ~=~ dom ~T^*$~ and ~$\CD_\pm$~ be as above. 

(i) ~$u \in \CD ~\Rightarrow ~\CD \ni f \mapsto ~[f,u]^b_a$~ 
defines a continuous linear functional on ~$\CD$~ (with respect to the
norm coming from ~$( \cdot , \cdot )_*~)$ which vanishes on ~$dom
~\overline{T_0},$ and is consequently a `boundary value' in the sense
discussed earlier (in the context of general symmetric operators).


(ii) If ~$f \in \CD$~ is arbitrary, then both ~$f$~ and ~$f^\prime$~
have limits (from the right) at ~$a$~ and further ~$\CD \ni f ~\mapsto
~f(a)$~ and ~$\CD \ni f ~\mapsto ~f^\prime(a)$~ (which make sense by
the preceding assertion) are `boundary values' in the sense of (i) above.

(iii) If ~$u \in \CD,$ then ~$\CD \ni f ~\mapsto ~lim_{x
\uparrow b} [f,u](x)$~ defines a boundary value. (In the sequel, we
shall simply write ~$[f,u](b)$~ to denote the preceding limit.)
\end{Lemma}

\vskip 1em

{\bf Proof:} (i) This is an immediate consequence of Green's formula
\ref{grfor}. 

(ii) Suppose ~$\phi \in C^\infty_c(a,b)$~ be such that ~$\phi$~ is
identically equal to 1 in a neighbourhood of ~$a$~ and identically
equal to 0 in a neighbourhood of ~$b.$ Then, clearly ~$\phi \in \CD,$ by the
assumed hypothesis on ~$q.$ If ~$f \in \CD,$ note that ~$[f,\phi]^b_a
~=~ lim_{x \downarrow a}\{f(x) \phi^\prime(x) - f^\prime(x)
\phi(x)\};$ since ~$(\phi(x), \phi^\prime(x)) = (1,0)$~ in a
neighbourhood of ~$a,$ it follows from (i) ~$lim_{x \downarrow a}
f^\prime(x)$~ does indeed exist and the limit defines a boundary value in the
sense of (i).

Now let ~$\psi \in ~C^\infty_c[a,b)$~ be any function such that
~$\psi(x) = x - a$~ in a neighbourhood of ~$a.$ Again, ~$\psi \in
\CD,$ and we find that if ~$f \in \CD,$ then, ~$[f,\psi]^b_a ~=~
lim_{x \downarrow a} f(x).$

(iii) This is an immediate consequence of (i) and (ii).

\hfill$\Box$

\begin{Lemma}\label{u0=0}
Suppose ~$q$~ is a continuous function on ~$\R^+ ~=~ (0, \infty)$
which is square-integrable in a neighbourhood of 0.
Let ~$T_0$~ be the symmetric operator on ~$L^2(\R^+)$~ defined by
~$dom ~T_0 ~=~ C^\infty_c(\R^+), ~T_0 u ~=~- u^{\prime \prime} + qu,$
and let ~$\CD ~=~ dom ~T^*$~ as above. Also, as before, let ~$\CD_\pm
~=~ ker(T^* \mp i),$ and let ~$n ~= ~ dim ~ker ~(T^* \pm i).$ Suppose
~$H$~ is a self-adjoint extension of ~$T,$ and suppose it is true that
~$dom ~H ~\subset ~\{u \in \CD ~:~ u(0) = 0\},$ where we have used the
short-hand notation $u(0)$ for the more accurate $\lim_{x \downarrow
0} u(x)$.  Then,

(a) ~$n \neq 0;$

(b) if ~$n = 1,$ then ~$dom ~H ~= ~\{u \in \CD ~:~ u(0) = 0\};$ and

(c) if ~$n = 2,$ then there exists ~$u_1 \in \CD_+ \oplus \CD_-$~ such
that ~$dom ~H ~=~ \{u \in \CD ~:~ u(0) = 0, ~[u,u_1](\infty) ~(=~
\lim_{R \rightarrow \infty} [u, u_1](R)~) ~=~ 0\}.$  
\end{Lemma}

\vskip 1em

{\bf Proof} (a) If ~$n = 0,$ then ~$\overline{T_0} = T^*$~ - see Case
(i) just before Green's formula \ref{grfor}. This, together with the
assumed self-adjointness of ~$H$~ and the fact that ~$H \subset T^*$
would imply that ~$H = T^*,$ which contradicts our assumption about
~$dom ~H.$

(b) If ~$n=1,$ then it follows from equation \ref{gensa} (and the
remark following it) that there exists a non-zero vector ~$v_0 \in \CD$~ such
that ~$dom ~H ~=~ \{f \in \CD ~:~ (f, v_0)_* = 0\}.$ On the other hand,
we know from Lemma \ref{wrab} (ii) (and the Riesz representation
theorem applied to the Hilbert space ~$\CD)$~ that there exists a
vector ~$u_0 \in \CD$~ such that ~$(f, u_0)_* = f(0).$  Our assumption
on ~$dom ~H$~ then translates into the statement that ~$f \in \CD,
(f,v_0)_* = 0 ~\Rightarrow (f, u_0)_* = 0.$ Since ~$u_0$~ is clearly
non-zero, this implies that ~$u_0$~ and ~$v_0$~ are non-zero multiples
of one another, which yields the desired conclusion.

(c) If ~$n =2,$ appeal to equation \ref{gensa} to conclude that there
exists a two-dimensional subspace ~$\CM$~ of ~$\CD_+ \oplus \CD_-$~
such that ~$dom ~H ~=~ \{f \in \CD ~:~ (f,u)_* = 0 ~\forall ~u \in
\CM\}.$ The assumption on ~$dom ~H$~ implies that ~$u_0 \in \CM,$
where ~$u_0$~ is as in the proof of (b) above. Pick ~$v_1 \in \CM$~
such that ~$(v_1,u_0)_* = 0,~ \|v_1\|_* = 1.$ Then ~$v_1(0) ~=~
(v_1,v_0)_* = 0,$ and ~$dom ~H ~=~ \{f \in \CD ~:~ f(0) ~=~ (f,v_1)_* =
0\}.$ Now, appeal to equation \ref{*wro}, set ~$u_1 ~=~T^*
\overline{v_1},$ and conclude that ~$f \in ~dom ~H \Rightarrow
(f,v_1)_* ~=~ [f,u_1]^\infty_0 ~=~ [f,u_1](\infty),$ where we have
used the fact that ~$u_1(0) ~=~0$~ as well as the fact - see Lemma
\ref{wrab}(ii) - that ~$f$~ and ~$f^\prime$~ have finite right-limits
at 0, to conclude that ~$[f,u_1](0) = 0.$

\hfill$\Box$

\vskip 1em

In this paper, we will have to deal with the \sch ~operator ~$H = \lap
+ q,$ where the potential ~$q$~ has the form
\be \label{qdefn}
q(x) ~=~ \left\{ \begin{array}{ll} q^+(x) & \mbox{if }~x > 0\\
0 & \mbox{if } ~x = 0\\
q^-(-x) & \mbox{if }~x < 0
\end{array} \right.
\ee

\noindent where ~$q^\pm$~ are continuous
functions on ~$(0,\infty)$~ with finite limits at zero - i.e., the limits
~$q^\pm(0 +) ~=~ \lim_{x \downarrow 0} q^\pm(x)$ exist and are finite.
Throughout this section, this is the potential we shall be working with.

We shall encounter the following situation; if we write ~$T_0,$
(resp., ~$T_0^\pm)$ for the operator on ~$L^2(\R)$~ (resp.,
~$L^2(\R^\pm)~)$ with domain given by ~$C^\infty_c(\R)$ (resp.,
~$C^\infty_c(\R^\pm)$ and defined by the differential operator given
by ~$\tau ~=~ \lap $\\$~+~ q(x),$ it will be the case that both operators
~$T_0^\pm$~ will have deficiency indices (1,1), and will give rise to
self-adjoint operators ~$H_\pm$~ on ~$L^2(\R^\pm)$~ given
by ~$H_\pm \subset T_\pm^*$~ and
\be \label{domhpm}
\CD(H_\pm) ~=~ \{ u \in L^2(\R^\pm) : -u^{\prime \prime} + q u \in
L^2(\R^\pm), u(0) = 0 \} ~~.
\ee

It will, further be the case that the \sch ~operator we shall be
interested in will be given by ~$H = T_0^*;$~ i.e.,
\be \label{domh}
\CD (H) ~=~ \{u \in L^2(\R) : -u^{\prime \prime} + qu \in L^2(\R)\} ~~.
\ee

Temporarily fix ~$z \in \C$~ with ~$Im ~z \neq 0$ and consider the
differential equation 
\be \label{eveq}
-u^{\prime \prime }(x) + q(x) u(x) ~=~ z u(x) ~~.
\ee

\noindent This second-order differential equation has a
two-dimensional linear space of solutions; let ~$\phi( \cdot , z),
\psi( \cdot , z) $~ denote the unique solutions satisfying ~$\phi(0,z)
~=~$\\ $ \psi^\prime(0,z) ~=~ 0,$
~$\phi^\prime(0,z) ~=~ \psi(0,z) = 1,$ where we write
~$\phi^\prime(0,z)$~ to denote ~$\frac{\partial \phi}{\partial x}(0,z).$

The fact that ~$T_0^\pm$ have deficiency indices (1,1) implies 
that there exist unique scalars ~$m_\pm(z)$~ such that if
\be \label{mpm}
u^\pm(x,z) = \psi(x,z) \pm m_\pm(z) \phi(x,z)~~,
\ee

\noindent then ~$u^\pm( \cdot ,z) \in L^2(\R^\pm).$

Recall that the Green's function for ~$H$~ is given by 
\be \label{grn}
g(z;x,y) ~=~ \frac{1}{[u^+(x,z), u^-(x,z)]}
\left\{ \begin{array}{ll}
u^-(x,z) u^+(y,z) & if ~x \leq y\\
u^-(y,z) u^+(x,z) & if ~x \geq y 
\end{array} \right. 
\ee

It is a fact that the Weyl $m$-functions ~$m_\pm(z),$ when restricted
to ~$\Pi^+,$ are Herglotz functions, and that they are related to the
Green's function by
\be \label{gmm}
g(z;0,0) ~=~ \frac{-1}{m_+(z) + m_-(z)} 
\ee

\noindent and

\be \label{g2mm}
\lim_{0 < x < y \downarrow 0} \frac{\partial^2}{\partial x \partial y}
g(z;x,y) ~=~ \frac{m^+(z) m_-(z)}{m_+(z) + m_-(z)} ~~.
\ee

Further, it is true that half-space Green's functions
- which are given by 
\[ g_\pm(z;x,y) ~=~ \frac{\pm 1}{[u^\pm(x,z), \phi(x,z)]}
\left\{ \begin{array}{ll}
\phi(x,z) u^\pm (y,z) & if ~\pm x \leq \pm y\\
\phi(y,z) u^\pm (x,z) & if ~\pm x \geq \pm y 
\end{array} \right. \]
\noindent - are related to the Weyl $m$-functions by 
\be \label{gpmmm}
\lim_{0 < x < y \downarrow 0} \frac{\partial^2}{\partial x \partial y}
g^\pm (z;x,y) ~=~ m^\pm (z) ~~.
\ee

The following known fact - which we single out as a
proposition, for convenience of reference - is a consequence of
equation \ref{gpmmm}. 

\begin{Proposition}\label{spmhmm}
With the foregoing notation, define 
\be \label{hG}
F ~=~ \frac{-1}{m^+ + m^-} ~~,~~ G ~=~ \frac{m^+ m^-}{m^+ + m^-}~~;
\ee
\noindent then

(i) ~$F$~ and ~$G$~ are Herglotz functions;

(ii) there exists vectors ~$x_\pm \in L^2(\R^+)$~ such that the
singleton set 
~$\{x_\pm\}$~ is cyclic for ~$H^\pm,$ and the associated spectral
measure for ~$H^\pm$ (which, in view of the cyclicity of ~$x_\pm,$ is
a total spectral measure for ~$H^\pm)$ is equivalent to the measure
given by ~$\mu^\pm ~=~ \mu_{x_\pm};$ and

(iii) there exists vectors ~$x_1, x_2 \in L^2(\R)$~ such that the
two-element set ~$\{x_1,
x_2\}$~ is cyclic for ~$H,$ and the associated spectral measures are
equivalent to the measures
given by ~$\mu_1 ~=~ \mu_{x_1} ~=~ \mu_F$~ and ~$\mu_2 ~=~ \mu_{x_2}
~=~ \mu_G;$ and consequently ~$\mu_1 + \mu_2$~ is a total spectral
measure for ~$H.$
\end{Proposition}

\vskip 1em

{\bf Sketch of Proof:} (i) This is because ~$m^\pm$~ are Herglotz
functions. 

(ii) In view of our stated conditions on ~$H_\pm,$ it follows that
what we have called ~$T_0^\pm$~ is `in the limit-point case at ~$\pm
\infty ';$ the desideratum follows from \cite{cole} Chapter 9, Theorem 3.1
and Chapter 9, Problems 12 and 14. (The roles of ~$\phi$~ and ~$\psi$~
in our discussion and in the discussion in \cite{cl} are reversed
because of the boundary condition at 0 that we work with.)

(iii) See \cite{cl} (top of page 108).

\hfill$\Box$

\subsection{Gelfand-Levitan theory}

The Gelfand-Levitan theory starts with appropriate measures ~$\mu$~
and proves the existence of a \sch ~operator on ~$L^2(\R^+)$ with
continuous potential, and with total
spectral measure given by ~$\mu.$ We briefly outline the facts of this
theory that we shall need. The conditions on the measure vary
according to the boundary conditions imposed in the domain of the
\sch ~operator. 

We present only the case of the Dirichlet boundary condition at 0.

The result we shall use - see \cite{gl} and \cite{lg} - is the
following - which we state much more elaborately than in the
original, in order to facilitate later reference.

\begin{Theorem}\label{gell}({\bf Gelfand-Levitan})
Let ~$\nu$~ be the measure with distribution function given by
\[ \nu(- \infty, x]) ~=~ \left\{ \begin{array}{ll}
0 & \mbox{if }~x \leq 0\\
\frac{2}{3 \pi} x^{\frac{3}{2}} & \mbox{if }~x > 0
\end{array} \right. \]

\noindent
Suppose ~$\mu$~ is a measure satisfying the following conditions:

(i) ~$\int_{-\infty}^0 e^{a\sqrt{- \lambda}} d \mu (\lambda) ~<
\infty~~\forall~a > 0~~;$

(ii) put ~$\sigma ~=~ \mu - \nu;$ suppose the function defined by
\be \label{afun}
a(x) ~=~ \int_1^\infty ~\frac{cos(\sqrt{\lambda}x)}{\lambda^2} ~d
\sigma (\lambda)
\ee

\noindent (where the integral is interpreted as an improper integral)
satisfies ~$a \in C^4(0,\infty).$ 

\vskip 1em

\noindent Then, 

(a) Define
\be \label{ffun}
f(x,y) ~=~ \int_\R ~ \frac{sin \sqrt{\lambda}x}{\sqrt{\lambda}}
\frac{sin \sqrt{\lambda}y}{\sqrt{\lambda}} ~d \sigma (\lambda)~~,
\ee

\noindent where the integral is again interpreted as an improper integral;
then ~$f \in C^1(\R^+ \times \R^+),$ and the integral equation

\be \label{inteq}
K(x,y) ~+~ f(x,y) ~+~ \int_0^x ~K(x,s) f(y,s) ds ~=~ 0 ~, 0 < y \leq x
\ee

\noindent has a unique solution ~$K(x,y);$ further, ~$K \in
C^1(\{(x,y) ~:~ 0 \leq y \leq x\}$.

\vskip 1em

(b) Define ~$q(x) ~=~ 2 ~\frac{d}{dx} K(x,x),$ and
\be \label{phifun}
\phi(x,\lambda) ~=~ \frac{sin \sqrt{\lambda}x}{\sqrt{\lambda}} ~+~
\int_0^x ~K(x,y) \frac{sin \sqrt{\lambda}y}{\sqrt{\lambda}} ~dy~~, x >
0, \lambda \in \R~;
\ee

\noindent then ~$\phi$~ satisfies the eigenvalue equation \ref{eveq}
for ~$x > 0,$ with the boundary conditions ~$\phi(0,\lambda) = 0,
~\phi^\prime(0,\lambda) = 1.$

\vskip 1em

(c) Finally, the equation
\be \label{ueq}
(Ug)(\lambda) ~=~ \int_0^\infty ~g(x) \phi(x,\lambda) ~dx
\ee

\noindent (with the right-hand side interpreted as a limit, as ~$R
\rightarrow \infty,$ in ~$L^2(\mu)$ of the functions defined by
~$(U_Rg)(\lambda) ~=~ \int_0^R ~g(x) \phi(x,\lambda) ~dx)$
defines a unitary operator ~$U : L^2(\R^+, dx) ~\rightarrow ~L^2(\R,\mu).$

The inverse (or adjoint) of the operator ~$U$~ is given by
\be \label{u*}
(U^*g)(x) ~=~ \int_\R g(\lambda) \phi(x,\lambda) d\mu(\lambda)
\ee
\noindent where this equation is also interpreted in the improper
~$L^2$-sense as was equation \ref{ueq}
\end{Theorem}

Before proceeding to discuss consequences of this theorem, we wish to
point out that the version of their theorem which we have stated here
corresponds to what they call the case ~$h = \infty,$ and that
although they do not quite state their theorem in this fashion, the
proof of their theorem establishes this version.

We state the consequences of this theorem that we shall need as a
proposition. 

\begin{Proposition}\label{ourgl}
Suppose ~$\mu$~ is a measure satisfying conditions (i) and (ii) of
Theorem \ref{gell}. Suppose, also, that the following conditions are
satisfied: 

(i) ~$\mu$~ is not purely atomic;

(ii) ~$\int_\R \frac{1}{(1 + |\lambda|)^2} ~d\mu ~< \infty;$

(iii) ~$supp ~\mu ~\subset ~[c,\infty),$ for some ~$c > - \infty;$ and

(iv) the function ~$K$~ given by equation \ref{inteq} is bounded in
~$\{(x,y) : 0 \leq y \leq x \leq 1\};$

(v) the potential ~$q$~ constructed as in Theorem
\ref{gell}(b) has a finite right-limit at 0.

\vskip 1em

Then, the operator ~$T$~ on ~$L^2(\R^+)$~ defined by ~$dom~T
~=~ \{ u \in L^2(\R^+) :~ u,u^\prime ~absolutely~ continuous,~\tau u
\in L^2(\R^+),~ u(0) ~=~0\},~ (T u) ~=~ \tau u$~ \\
(where ~$\tau u
=(\lap ~+ q)~u)$ is self-adjoint and ~$\mu$~ is a total spectral
measure for ~$T.$ (In fact, ~$\tau$~ has deficiency indices (1,1) and
is in the limit-point case at ~$\infty.)$
\end{Proposition}

\vskip 1em

{\bf Proof:} We break the proof into a sequence of steps.

\vskip 1em

{\em Step 1:} Let ~$T_0$~ be the operator given by ~$T_0f ~=~ \tau f,
~dom~T_0 ~=~ C^\infty_c(\R^+).$ If ~$U$~ is as
in Theorem \ref{gell}(c), then
\be \label{toueq}
(U T_0 U^* f)(\lambda) ~=~ \lambda f(\lambda) ~ for ~\mu ~a.a.
\lambda~~.
\ee

This follows from the definition of ~$U$~ and the properties of the
function ~$\phi$~ stated in Theorem \ref{gell}(b).

\vskip 1em

{\em Step 2:} Let ~$T ~=~ U^* M_\lambda U,$ where ~$M_\lambda$~
denotes the (self-adjoint) operator of multiplication by the
independent variable. Then,
\[ dom~T ~\subset ~\{f \in ~dom~ T_0^* ~:~\lim_{x \downarrow 0} ~f(x) ~=~0\}. \]

{\em Reason:}  With ~$c$~ as in assumption (iii) of the proposition,
it is a consequence of assumption (iv) and equation \ref{phifun} that
\be \label{phibd}
\sup_{0 \leq x \leq 1, c \leq \lambda < \infty} ~|\phi(x,\lambda)| ~<~
\infty~.
\ee

Also, note that assumption (iv) shows that ~$\phi(x,\lambda)
~\rightarrow 0$~ as ~$x \rightarrow 0,$ for all ~$\lambda \in \R.$
An application of equation \ref{u*} (and the Cauchy-Schwarz inequality
and bounded convergence theorem) shows that
\begin{eqnarray*}
f \in ~dom~T & \Rightarrow & g ~=~ Uf ~\in ~dom ~M_\lambda\\
& \Rightarrow & \int_c^\infty (1 + |\lambda|)^2 |g(\lambda)|^2 ~d\mu(\lambda)
~< ~\infty\\
& \Rightarrow & |f(x)| \leq (\int (1 + |\lambda|)^2 |g(\lambda)|^2
~d\mu(\lambda)~)^\frac{1}{2}(\int (1 + |\lambda|)^{-2} |\phi(x,\lambda)|^2
~d\mu(\lambda)~)^\frac{1}{2} 
\end{eqnarray*}
\noindent which establishes the desired limit assertion, since the
first integral is finite, and the second integral converges to 0.

\vskip 1em

{\em Step 3:} ~$\tau$~ has deficiency indices (1,1) and is in
the limit-point case at ~$\infty.$

\vskip 1em

{\em Reason:} In view of Lemma \ref{u0=0} - which is applicable because
continuity of ~$q$~ is guaranteed by Theorem \ref{gell}, and because
of assumption (v) and the already established Step 2 above - we only
need to show that ~$\tau$~ does not have deficiency indices (2,2).
However, \cite{mhs} Theorem 10.19 states that if ~$\tau$~ had
deficiency indices (2,2), then the operator ~$T$~ should have only
point spectrum, and this is ruled out by assumption (i). The
assumption (v) guarantees that ~$\tau$~ is in the limit-circle case at
0, and hence the fact that the deficiency indices of ~$\tau$~ are
(1,1) imply that ~$\tau$~ is necessarily in the limit-point case at
~$\infty.$ 

\vskip 1em

{\em Step 4:} ~$dom ~T ~=~ ~\{f \in ~dom~ T_0^* ~:~\lim_{x \downarrow
0} ~f(x) ~=~0\}. $

\vskip 1em

This follows from Lemma \ref{u0=0} and the proof of the proposition is
complete. 

\hfill$\Box$

Before concluding this section, we would like to point out the
following interesting criterion for essential self-adjointness of the
one-dimensional \sch ~operator, which can be proved as in the proof of
Step 3 above.

\begin{Proposition}\label{esa}
Let ~$q \in C(\R)$~ be a continuous potential. Suppose the
differential operators ~$T_0, T_0^+$~ and ~$T_-$~ defined by ~$u
~\mapsto~ -u^{\prime \prime} + q u$~, with 
domains ~$C^\infty_c(\R), C^\infty_c((- \infty, a))$ and
~$C^\infty_c((a,\infty)) $~ have self-adjoint extensions ~$H, ~H_+,$~
and ~$H_-$~ respectively. Assume that neither ~$H_+$~ nor ~$H_-$~ has
pure point spectrum. Then ~$T_0$~ is essentially self-adjoint.
\end{Proposition}

It should be remarked here that stronger versions of this statement
are known in the literature - see, for example, Gesztesy \cite{gz2}.

\section{Proof of Theorem 1}

Before going into details, we quickly outline the structure of the
proof. 

With ~$\mu_0, m$~ as in the statement of Theorem \ref{mt}, consider
the function
\[ F(z) ~=~ \frac{-1}{2 \sqrt{-z}} ~+~ \int_\R \frac{1}{\lambda - z}
~d(\mu_0) (\lambda) ~,\] 
\noindent (where we have chosen that branch of the square root which
is   positive on the negative axis), and note that ~$\mu_F ~=~m + \mu_0.$
Define (the Herglotz function) 
~$f(z) ~=~ \frac{-1}{F(z)}.$ Then we have, by Theorem \ref{Herg}(b),
\be \label{hfun}
f(z) ~=~ a + bz + \int_\R \left[ \frac{1}{\lambda - z} ~-~
\frac{\lambda}{1 + \lambda^2 } \right] ~d\mu_f (\lambda)~~,
\ee
\noindent for some uniquely determined constants ~$a,b.$ (We have used
the notational convention described in Remark \ref{bt}, and we shall
continue, in the sequel, to use this convention.)

It is easy to see that ~$(\mu_f)_{ac} \cong m;$ define ~$\mu_+ ~=~
\frac{1}{2} (\mu_f)_{ac},~ \mu_- ~=~ \mu_f ~-~ \mu_+$~ and note that
~$\mu_\pm$~ are both non-zero positive measures and satisfy
~$(\mu_+)_{ac} ~=~ (\mu_-)_{ac}$ \\
$\cong~ m.$ Now define
\begin{eqnarray}\label{mpmdef}
m_+ & = & \int_\R \left[ \frac{1}{\lambda - z} ~-~
\frac{\lambda}{1 + \lambda^2 } \right] ~d\mu_+ (\lambda)\\
m_- & = & a + bz + \int_\R \left[ \frac{1}{\lambda - z} ~-~
\frac{\lambda}{1 + \lambda^2 } \right] ~d\mu_- (\lambda)
\end{eqnarray}


\noindent and note that b = 0 and  ~$m_\pm$~ are Herglotz functions such that ~$f
~=~ m_+ + m_-.$ 

We then verify that the measures ~$\mu_\pm$~ satisfy the assumptions
of Proposition \ref{ourgl}, and deduce (from that proposition) the
existence of potentials ~$q_\pm$~ on ~$\R^+$~ such that the associated
\sch ~operators with Dirichlet boundary condition at 0 are
self-adjoint (and that ~$q_\pm$~ are in the
limit-point case at ~$\infty.)$

We finally put these two potentials together to construct a potential
~$q$~ on ~$\R$~ as in equation \ref{qdefn} and then verify that the
associated \sch ~operator ~$H$~ satisfies the conclusions of the
theorem.

\vskip 1em

{\em Step 0:} We first verify that the measures ~$\mu_\pm$~ satisfy
conditions (i) and (ii) of Theorem \ref{gell}.

\vskip 1em

We only verify condition (ii) since condition (i) of Theorem
\ref{gell} is easily seen to follow from Condition (iii) of
Proposition \ref{ourgl}, which we will verify later in Step 3.

We first assert that there exist (real) constants ~$A, B$~ and a
function ~$e(\lambda),$ and a constant ~$R > 0$~ such that
\be \label{fasy}
Im ~f(\lambda + i 0) ~=~ 2 \lambda^\frac{1}{2} ~+A
\lambda^\frac{-1}{2} ~+ e(\lambda) ~,~\forall ~\lambda > R
\ee

\noindent and

\be \label{eest}
| e (\lambda )| ~\leq ~ B \lambda^\frac{-3}{2} ~,~\forall ~\lambda >
R~.
\ee

For this, we first define

\be \label{c0}
C_0(\lambda) ~=~ ~\int_\R \frac{1}{x-\lambda} ~d \mu_0(x)~~\forall
~\lambda > K
\ee

\noindent and note that since ~$supp ~\mu_0 ~\subset ~[-M,M],$ it
follows that ~$C_0(\lambda)$~ is real, where defined. Note further
that

\be \label{c01}
C_0(\lambda) ~=~ - \lambda^{-1} ~-~ (\int_\R x d\mu_0(x))~
\lambda^{-2}~-~ (\int_\R x d\mu_0(x))^2~ \lambda^{-3} ~-~ e_1(\lambda)
~,
\ee

\noindent which implies that we can find a sufficiently large ~$R >
M$~ such that the following inequalities are valid for any ~$\lambda >
R~:$ 
\begin{eqnarray} \label{ineqs}
2 \lambda^\frac{1}{2} |C_0(\lambda)| ~< \frac{1}{2}~,&
|C_0(\lambda)| ~< \frac{2}{\lambda} ~,\\
(C_0(\lambda))^2 ~=~ \lambda^{-2} ~+~ A_1 \lambda^{-3} ~+~
O(\lambda^{-4}) & |e_1(\lambda)| ~\leq ~B_1 \lambda^{-4}
~,
\end{eqnarray}

\noindent for some constants ~$A_1, B_1.$

Note next that since ~$(\mu_f)_{ac} ~\cong ~(\mu_F)_{ac} ~=~ m,$ it
follows that ~$Im ~f(\lambda + i 0)~$ is finite and strictly positive
on ~$(M,\infty).$ 
Now , for ~$\lambda > R,$ compute thus, using the geometric series:
\begin{eqnarray}\label{imfl}
f(\lambda + i0) & = & 2\sqrt{-\lambda} ~\left(1 ~-~ 2 \sqrt{-\lambda}
~\int_\R \frac{1}{x-\lambda} ~d \mu_0(x) ~\right)^{-1} \nonumber \\
& = & \sum_{n=0}^4 ~(2 \sqrt{-\lambda})^{n+1} C_0(\lambda)^n
~+~k(\lambda)~, 
\end{eqnarray}

\noindent where ~$k(\lambda) ~=~ (2 \sqrt{-\lambda})^{6} C_0(\lambda)^5
~\left( 1 - 2 \sqrt{-\lambda} C_0(\lambda) \right)^{-1}.$

The desired asymptotic assertion \ref{fasy} follows upon taking
imaginary parts in equation \ref{imfl} and appealing to equation
\ref{c01} and the estimates \ref{eest}.

\vskip 1em
Since ~$(\mu_\pm)_{ac} ~=~ \frac{1}{2}(\mu_f)_{ac},$ it follows that
~$(\mu_\pm)_{ac}$~ has density ~$\frac{1}{2 \pi} ~Im ~f(\lambda +
i0);$ we now define 
\be \label{sigpm}
d\sigma_{\pm}(\lambda) = d\mu_{\pm}(\lambda) -
\frac{1}{\pi}\chi_{(0,\infty)}(\lambda)\sqrt{\lambda}d\lambda.
\ee
Then, we write 
\be
 a(x) = \int_1^{R} \frac{\cos\sqrt{\lambda}x}{\lambda^2} d\sigma_{\pm}(\lambda) ~and~ 
 b(x) = \int^{\infty}_{R} \frac{\cos\sqrt{\lambda}x}{\lambda^2}
d\sigma_{\pm}(\lambda), 
\ee
\noindent with $R$ chosen as in equation \ref{fasy}, so that
\[ \int_1^{\infty} \frac{\cos\sqrt{\lambda}x}{\lambda^2} d\sigma_{\pm}(\lambda)  ~=~ a(x) + b(x). \]
It is clear that $a(x) \in C^{\infty}(0,\infty)$ so that to verify condition (ii) of Theorem
\ref{gell} we need to only verify that $b(x)$ has the required smoothness properties.
We note next that in the region $(R, \infty)$, the (signed) measures $\sigma_{\pm}$ are absolutely
continuous and have densities given by 
\be \label{sigpmac}
2 \pi d\sigma_{\pm}(\lambda) ~=~  ~\frac{A}{ \sqrt{\lambda}}d\lambda ~+ e(\lambda)d\lambda 
\ee 
which follows from the definition of $\sigma_{\pm}$ and the equation \ref{fasy}. 
We now write b(x) as a sum of two parts $\frac{1}{2 \pi}(b_1(x) + b_2(x))$, where
$$
  b_1(x) ~=~ A \int_{R}^{\infty} \frac{\cos\sqrt{\lambda}x}{\lambda^2} \frac{1}{\sqrt{\lambda}}
d\lambda
 ~~and~~
  b_2(x) ~=~ \int_{R}^{\infty} \frac{\cos\sqrt{\lambda}x}{\lambda^2} e(\lambda)d\lambda. 
$$ 
Then it is clear from the estimate \ref{eest} on $e(\lambda)$ that $b_2(x)$ satisfies condition (ii) of
theorem \ref{gell}.  Finally the fact that $b_1(x)$ verifies condition (ii) of theorem
\ref{gell} follows from an explicit computation where we first change variables to $k =
\sqrt{\lambda}$ and then do  integration by parts twice to get, for $x > 0$: 
$$
b_1(x) ~=~2 A\left\{ \frac{-\sin\sqrt{R}x}{xR^2} ~+~ \frac{5\cos\sqrt{R}x}{x^2\sqrt{R^5}} ~-~
\frac{30}{x^2}\int_{\sqrt{R}}^{\infty} \frac{\cos kx}{k^6}dk\right\}.
$$ 
\noindent It is now clear that $b_2(x)$ satisfies the conditions (ii) of theorem \ref{gell} for
$x > 0.$  We now verify the other conditions of the proposition \ref{ourgl} in the following
steps. 

\vskip 1em

{\em Step 1:} Since ~$(\mu_\pm)_{ac} \cong m,$ it is clear that
~$\mu_\pm$~ satisfy condition (i) of Proposition \ref{ourgl}.

\vskip 1em

{\em Step 2:} Since ~$m_\pm$~ are Herglotz functions, it follows from
Proposition \ref{Herg}(b) that
the measures ~$\mu_\pm$~ satisfy condition (ii) of
Proposition \ref{ourgl}.

\vskip 1em

{\em Step 3:} Since the measure ~$\mu_0$~ is supported in ~$[-M,M],$ note
from the definition of ~$F$~ that ~$F$~ is analytic on ~$\C ~-~
[-M,\infty),$ real on ~$(-\infty , -M]$~ and ~$lim_{\lambda \rightarrow
-\infty} F(\lambda) ~=~ 0.$ Also, it is clear by differentiating the
expression defining ~$F$~ that the real function ~$F(x)$~ is a
strictly increasing function for ~$x < -M.$ Hence ~$F(x) \neq 0
~\forall ~x < -M.$ Then, a simple computation shows that for ~$x < -M,
\lim_{\epsilon \downarrow 0} Im ~ f(x + i \epsilon) ~=~ 0,$ and
consequently, ~$supp ~\mu_f ~\subset ~[-M,\infty),$ by Proposition
\ref{Herg}(iv). Since ~$\mu_\pm$~ are absolutely continuous with
respect to ~$\mu_f,$ it follows that ~$supp ~\mu_\pm ~\subset
~[-M,\infty).$ 

\vskip 1em

{\em Step 4:} Condition (iv) of Proposition \ref{ourgl} is a direct
consequence of the continuity of the kernel ~$K(x,y)$~ - see Theorem
\ref{gell}(a). 

\vskip 1em

{\em Step 5:} The smoothness properties of the functions ~$f, K,$~ -
stated in Theorem \ref{gell} - the definition of ~$q_\pm$~ in terms
of the kernel - see Theorem \ref{gell}(b)
- and the integral equation \ref{inteq}
show (after a routine differentiation with respect to ~$x$~ of the
equation obtained by setting ~$y = x$~ in \ref{inteq}) that indeed
the kernels ~$q_\pm$~ have finite right-limits at 0.

\vskip 1em

To complete the proof, we note that Steps 0-5 permit us to conclude
from Proposition \ref{ourgl} that there exist potentials ~$q_\pm$~ on
~$\R^+$~ such that the associated \sch ~operators with Dirichlet
boundary condition at 0 are self-adjoint, and such that ~$q_\pm$~ are
in the limit-point case at ~$\infty.$ This means that the potential
~$q$~ defined by equation \ref{qdefn} is in the limit-point case at
both ~$+ \infty$~ and ~$- \infty$. We may now appeal to \cite{rs}
Theorem X.7 (in Appendix to X.1) to deduce that the differential
operator ~$\lap ~+~ q$~ is essentially self-adjoint on
~$C^\infty_c(\R).$ 

\vskip 1em

In view of Proposition \ref{spmhmm} (iii) and the discussion leading
up to that proposition, the main theorem will be finally proved once
we establish that ~$supp ~\mu_G ~\subset supp~\mu_F ~=~ supp~(\mu_0 +
m),$ where of course ~$F, G$~ are given as in Proposition
\ref{spmhmm}, with ~$m_\pm$~ having the meanings attributed to them in
this section. To do this, (since ~$supp ~\mu_F ~\supset ~[0,\infty),)$
it suffices to establish the following:

(i) if ~$(a,b) ~\subset ~([-M,0] - ~supp~\mu_F),$ then ~$\lim_{\epsilon
\downarrow 0} Im ~G(x + i \epsilon) ~=~0, ~\forall ~x \in (a,b);$ 

(ii) if ~$(a,b) ~\subset ~([0,M] - ~supp~\mu_0),$ then ~$\infty
> \lim_{ \downarrow 0} Im ~G(x + i \epsilon) ~>~0, ~\forall ~x \in (a,b);$ 
and
 
(iii) therefore  ~$supp ~(\mu_G)_s ~\subset ~\Sigma.$ 


\vskip 1em

For (i), note that ~$F$~ is real and strictly increasing in ~$(a,b)$~
and consequently has at most one 0 - say ~$x_0$ - in ~$(a,b).$ 

Suppose ~$a < x \neq x_0 < b;$ then ~$f ~=~ \frac{-1}{F}$~ is analytic
in a neighbourhood of ~$x$~ and ~$f(x)$~ is real; on the other hand,
since ~$supp ~\mu_+ ~\subset ~[0,\infty),$ also ~$m_+$~ is analytic in
a neighbourhood of ~$x$~ and ~$m_+(x)$~ is real. This implies that
~$m_-$~ inherits these properties from ~$f~$ and ~$m_+.$ Hence, so
also does ~$G ~=~ - m_+ m_- f.$

Now, consider ~$x_0$~ - assuming it exists. Note that ~$f$~ has a
simple pole at ~$x_0,$ while ~$m_+$~ is still analytic in a neighbourhood
of ~$x_0$~ and ~$m_+(x_0) \in \R.$ Since ~$G ~=~
\frac{-m_+}{(\frac{m_+}{m_-}) + 1},$ it follows that ~$\lim_{\epsilon
\downarrow 0} ~G(x_0 + i \epsilon) ~=~ -m_+(x_0) ~\in \R,$ thus
completing the proof of (i) above.

As for (ii), we again see that ~$F$~ has finite positive  imaginary
part  and finite real part in $(a,b)$, since $~supp ~\mu_0$ does not
intersect $(a,b)$ by assumption.  This implies that the imaginary parts
of  $~f~$ and $m_{\pm}$ are positive and
finite in $(a.b),$ while the real parts of $~f~$, $m_{\pm}$
are finite in $(a,b).$   This immediately implies that the
imaginary part of $~G~$ is finite in $(a,b)$ since $G =
-m_+m_-f.$  This shows that the singular part of $~\mu_G$ is not
supported in $(a,b).$        
\hfill$\Box$

{\bf Acknowledgement :} M K wishes to thank Walter Craig for
discussions on the Gelfand-Levitan theory and the department of
mathematics, Brown University, Providence and the Erwin
Schr\"odinger Institute, Vienna for invitations where 
parts of this work was done.  We wish to thank Werner Kirsch and
Peter Hislop for comments which resulted in the remarks in the
introduction. 

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\end{document}

