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%%     journal="Proc. Amer. Math. Soc. 126, 2873-2881 (1998)",
%%     doi="10.1090/S0002-9939-98-04362-7",
%%     copyright="G. Teschl".
%%     }


\documentclass{proc-l}
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\issueinfo{126}% volume number
  {10}%        % issue number
  {October}%   % month
  {1998}%      % year
\PII{S \ISSN(98)04362-7}
\copyrightinfo{1997}{G. Teschl}
\pagespan{2873}{2881}
\commby{Palle E. T. Jorgensen}
\date{February 21, 1997}

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\newtheorem{coro}[thm]{Corollary}
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\newtheorem{rem}[thm]{Remark}

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\begin{document}

\title[Deforming the Point Spectra of Dirac Operators]{Deforming the Point Spectra
of One-Dimensional Dirac Operators}

\author{Gerald Teschl}
\address{Institut f\"ur Reine und Angewandte Mathematik\\
RWTH Aachen\\ 52056 Aachen\\ Germany}
\curraddr{Institut f\"ur Mathematik\\
Strudlhofgasse 4\\ 1090 Wien\\ Austria}
\email{Gerald.Teschl@univie.ac.at}
\urladdr{http://www.mat.univie.ac.at/\string~gerald/}

%\thanks{{\it To appear in Proc. of the AMS}}

\keywords{Spectral theory, Dirac operators, eigenvalues}
\subjclass{Primary 34L40, 34L05; Secondary 34B05, 47B25}


\begin{abstract}
We provide a method of inserting and removing any finite number of prescribed eigenvalues
into spectral gaps of a given one-dimensional Dirac operator. This is done in such a way
that the original and deformed operator are unitarily equivalent when restricted to
the complement of the subspace spanned by the newly inserted eigenvalue.
Moreover, the unitary transformation operator which links the
original operator to its deformed version is explicitly determined.
\end{abstract}

\maketitle


\section{Introduction}


Methods of inserting (and removing) eigenvalues in spectral gaps of a given
one-dimensional Schr\"odinger operator $H$ have  been quite popular recently.
This is due to their important role in diverse fields such as the inverse scattering
approach introduced by Deift and Trubowitz \cite{dt},  level comparison theorems
(cf.\ \cite{ba} and the literature cited therein), and as a tool for constructing
soliton solutions of the Korteweg--de Vries hierarchy relative to known
background solutions (see, e.g., \cite{gs}, and the references therein). For more
information and a brief historic account we refer the reader to \cite{gcom}, \cite{gtdc}.

It is surprising that even though Dirac operators are as important in applications
as Sturm--Liouville operators, no analogous methods are available for these operators
(except for the case of supersymmetric Dirac operators where results from
Sturm--Liouville operators apply). This lack is clearly connected to the fact that Dirac
operators are not  bounded from below and hence cannot be factored into a product of
type $A^* A$ (which would be necessarily non-negative). However, this factorization lies
at the heart of methods for inserting eigenvalues into the spectra of Sturm--Liouville
operators (cf.\ \cite{de}). This shows that, for inserting eigenvalues into the spectra of
Dirac operators, an entirely new strategy is needed. Our new approach is modeled after
``Hilbert's hotel". That is, our idea is to use a transformation operator which
``compresses" the underlying Hilbert space a little such that the range has codimension
one. This way we create a one-dimensional subspace to accommodate the new eigenvalue.
On the remainder of the Hilbert space we require the transform to be unitary such that
all other spectral features of the original operator are preserved.

Clearly, not any transformation can be used since, in general, the transformed operator
will not be a Dirac operator. However, a generalized version of a transformation found in
\cite{gtdc} will do the trick.

Let $I=(a,b) \subseteq \R$ (with $-\infty \le
a < b \le \infty$) be an arbitrary interval, $m \in \R^+ =[0,\infty)$, and
$\phi_{\rm am}, \phi_{\rm el}, \phi_{\rm sc} \in L^1_{loc}(I,\R)$ real-valued.
Consider the Dirac differential expression
\begin{equation}
\tau = \sig_2 \frac{1}{\I} \frac{d}{dx} + \phi(x).
\end{equation}
Here
\begin{equation}
\phi(x) =  \phi_{\rm el}(x)\id  + \phi_{\rm am}(x)\sig_1 + (m+ \phi_{\rm sc}(x)) \sig_3,
\end{equation}
where $\sigma_1$, $\sigma_2$, $\sigma_3$ denote the Pauli matrices
\begin{equation}
\sig_1=\left(\ba{cc} 0 & 1 \\ 1 & 0\ea\right), \quad 
\sig_2=\left(\ba{cc} 0 & -\I \\ \I & 0\ea\right), \quad 
\sig_3=\left(\ba{cc} 1 & 0 \\ 0 & -1\ea\right)
\end{equation}
and $m$, $\phi_{\rm sc}$, $\phi_{\rm el}$, and $\phi_{\rm am}$ are interpreted as
mass, scalar potential, electrostatic potential, and anomalous
magnetic moment, respectively (see \cite{th}, Chapter~4). We don't include a
magnetic moment since it can be easily eliminated by a simple gauge transformation
(there is also a gauge transformation which gets rid of $\phi_{\rm am}$; see \cite{ls},
Section~7.1.1).

If $\tau$ is limit point ($l.p.$) at both $a$ and $b$ (cf., e.g., \cite{ls},
\cite{wdl}, \cite{wd2}), then $\tau$ gives rise to a unique self-adjoint operator
$H$ when defined maximally. Otherwise, we fix a boundary condition at each
endpoint where $\tau$ is limit circle ($l.c.$).

By $u_+(z,x)$ (resp. $u_-(z,x)$) we will denote (non identically vanishing)
solutions of the differential equation $\tau u = z u$, $z\in\C$, which are integrable
near $b$ (resp.\ $a$) and  fulfill the boundary condition of $H$ at $b$
(resp.\ $a$) if any (i.e$.$, if $\tau$ is limit circle at $b$ (resp.\ $a$)).
A sufficient criterion for $u_\pm(z,x)$ to exist is $z\in\C\bs\sig_{ess}(H_{c,\pm})$
or $z\in\sig_{p}(H_{c,\pm})$, where $\sig_{p}(.)$, $\sig_{ess}(.)$ denotes the point,
essential spectrum, respectively. Here $H_{c,-}$ (resp.\ $H_{c,+}$), $c\in I$ denotes the
self-adjoint operators associated with $\tau$ on $L^2((a,c),\C^2)$ (resp.\ 
$L^2((c,b),\C^2)$) obtained from $H$ by imposing the additional boundary condition
$f_1(c)=0$. Then $H_{c,-}\oplus H_{c,+}$ is a rank one resolvent perturbation of $H$ and
hence $\sig_{ess}(H) = \sig_{ess}(H_{c,-}) \cup
\sig_{ess}(H_{c,+})$ (cf.\ \cite{wd2}, Korollar~6.2).

Using this notation, the operator $H$ is explicitly given by
\begin{equation}
H: \ba[t]{lcl} \db(H) &\to& L^2(I,\C^2) \\ f &\mapsto& \tau f \ea ,
\end{equation}
where
\begin{equation} \label{domH}
\db(H) = \{ f \in L^2(I,\C^2) | \ba[t]{l} f \in AC_{loc}(I,\C^2), \, \tau f \in
L^2(I,\C^2),\\ W_a(u_-(z_0),f) = W_b(u_+(z_0),f) =0 \} \ea
\end{equation}
with
\begin{equation}
W_x(f,g) = f_1(x) g_2(x) - f_2(x) g_1(x)
\end{equation}
the usual Wronskian (we remark that the the limit $W_{a,b}(.,..) = \lim_{x \to
a,b} W_x(.,..)$ exists for functions as in (\ref{domH})).



\section{Construction of a transformation operator}



Fix $n\in\N$ and let $k$ be a positive definite $n$ by $n$ matrix with coefficients in
$L^1_{loc}(I)$. We pick $\hr = L^2(I,\C^n; k\,dx)$ to be the 
underlying Hilbert space. The scalar product and norm in $\hr$ are given by
\begin{equation}
\spr{f}{g} = \int_a^b \ol{f(t)} k(t) g(t) dt, \quad \|f\|^2 = \spr{f}{f}.
\end{equation}
Denote by $\hr_-$ (resp.\ $\hr_+$) functions in $L^2_{loc}(I,\C^n;k\,dx)$
which are in $\hr$ near $a$ (resp.\ $b$) and choose a function $u \in\hr_-$
plus a constant $\gam\in[-\|u\|^{-2},\infty) \cup \{\infty\}$. Define 
\begin{equation}
c_\gam(x) = \frac{1}{\gam} + \spr{u}{u}_a^x, \quad \gam\ne 0
\end{equation}
(setting $\infty^{-1}=0$), where
\begin{equation}
\spr{f}{g}_y^x = \int_y^x \ol{f(t)} k(t) g(t) dt.
\end{equation}

Consider the following linear transformation
\begin{equation} \label{unitary1}
\ba{llcl} U_\gam: &\hr & \to & L^2_{loc}(I,\C^n, k\,dx) \\
& f(x) &\mapsto&  f(x) - u_\gam(x) \spr{u}{f}_a^x\ea,
\end{equation}
($U_0=\id$), where
\begin{equation} \label{psigam}
u_\gam(x) =\frac{u(x)}{c_\gam(x)},
\end{equation}
($u_0=0$). We note that $U_\gam$ can be defined on $\hr_-$
and $U_\gam u = \gam^{-1} u_\gam$, $\gam \ne 0$. Furthermore,
\begin{equation}
\ol{u_\gam(x)} k(x) u_\gam(x) = -\frac{d}{dx} \frac{1}{c_\gam(x)},
\end{equation}
and hence
\begin{equation}
\|u_\gam \|^2 = \left\{ \ba{cl} \gam, & u\not\in\hr\\
\frac{\gam^2\|u\|^2}{1+\gam\|u\|^2}, & u\in\hr\ea\right.
\end{equation}
implying $u_\gam\in\hr$ if $-\|u\|^{-2} < \gam < \infty$. If
$\gam= -\|u\|^{-2}$, $\gam=\infty$ we only have that $u_\gam$ is
in $\hr_-$, $\hr_+$, respectively. In addition, we remark that for
$f_\gam = U_\gam f$ we have
\bea
\ol{u_\gam(x)} k(x) f_\gam(x) &=& \frac{d}{dx}
\frac{\spr{u}{f}_a^x}{c_\gam(x)},\\
\ol{f_\gam(x)} k(x) f_\gam(x) &=& \ol{f(x)} k(x) f(x) -\frac{d}{dx}
\frac{|\spr{u}{f}_a^x|^2}{c_\gam(x)}.
\eea
Integrating over $x$ and taking limits (if $\gam=\infty$ use Cauchy-Schwarz)
shows
\bea \label{proporth}
\spr{u_\gam}{f_\gam}_a^x &=& \left\{ \ba{cl} 
c_\gam(x)^{-1} \spr{u}{f}_a^x, & \gam\in\R\\
\frac{\spr{u}{f}}{\|u\|^2}- c_\infty(x)^{-1} \spr{u}{f}_x^b,
& \gam=\infty \ea\right., \\ \label{propun}
\spr{f_\gam}{f_\gam}_a^x &=& \spr{f}{f}_a^x -
\frac{|\spr{u}{f}_a^x|^2}{c_\gam(x)}.
\eea
Clearly, the last equation implies $U_\gam: \hr \to \hr$.
In addition, we remark that this also shows $U_\gam : \hr_- \to \hr_-$.

Denote by $P,P_\gam$ the orthogonal projections onto 
the one-dimensional subspaces of $\hr$ spanned by $u, 
u_\gam$ (set $P, P_\gam = 0$ if $u,u_\gam\not\in \hr$), respectively.
Define
\begin{equation} \label{transfUf}
\ba{llcl} U_\gam^{-1}: & \hr &\to& L^2_{loc}(I,\C^n, k\,dx) \\
& g(x) &\mapsto& \left\{ \ba{cl} g(x) 
+ u(x) \spr{u_\gam}{g}_a^x, & \gam\in\R\\
g(x) - u(x) \spr{u_\infty}{g}_x^b, & \gam=\infty\ea\right. \ea
\end{equation}
and note
\begin{equation}
c_\gam^{-1}(x) = \left\{ \ba{cl} \gam - \spr{u_\gam}{u_\gam}_a^x, &
\gam\in\R \\
\| u \|^{-2} + \spr{u_\infty}{u_\infty}_x^b, & \gam=\infty
\ea\right. .
\end{equation}
As before one can show $U_\gam^{-1}:(\id-P_\gam)\hr \to\hr$ and
one verifies
\begin{equation}
\ba{ll} U_\gam U_\gam^{-1} = \id,\: U_\gam^{-1} U_\gam = \id, & \gam\in\R,\\ \\
U_\infty U_\infty^{-1} = \id,\: U_\infty^{-1} U_\infty = \id-P,\quad &
\gam=\infty,\ea
\end{equation}
If $P=0$, $\gam\in(-\|u\|^{-2},\infty)$, then $U_\gam U_\gam^{-1} = \id$
should be replaced by $U_\gam U_\gam^{-1} = \id_{(\id-P_\gam)\hr}$ since
$U_\gam^{-1} u_\gam \not\in \hr$ by
\begin{equation}
U_\gam^{-1} u_\gam = \left\{\ba{cl} \gam u, & \gam\in\R\\ \|u\|^{-2}u, &
\gam=\infty \ea\right. .
\end{equation}
Summarizing,

\begin{lemma} \label{lemuni}
The operator $U_\gam$ is unitary from $(\id -P) \hr$ onto $(\id -P_\gam) \hr$ with
inverse $U_\gam^{-1}$. If $P, P_\gam \ne 0$, then $U_\gam$ can be extended
to a unitary transformation $\ti{U}_\gam$ on $\hr$ by
\begin{equation}
\ti{U}_\gam = U_\gam(\id-P_\gam) + \sqrt{1+ \gam\|u\|^2} \,U_\gam P_\gam.
\end{equation}
\end{lemma}

\begin{proof}
Equation (\ref{proporth}) shows that $U_\gam$ maps $(\id-P) \hr$ onto $(\id-P_\gam)
\hr$. Unitarity follows from (\ref{propun}) and
\begin{equation} \label{sprfpsiopsis}
\lim_{x \to b} \frac{|\spr{u}{f}_a^x|^2}{\spr{u}{u}_a^x} =0
\end{equation}
for any $f \in \hr$ if $u \not\in\hr$. In fact, suppose $\| f \|=1$, pick
$y$ and $x>y$ so large that $\spr{f}{f}_y^b \le \eps/2$ and
$\spr{u}{u}_a^y / \spr{u}{u}_a^x\le \eps/2$.
Splitting up the sum in the numerator and applying Cauchy's inequality then shows
that the limit of (\ref{sprfpsiopsis}) is smaller than $\eps$.
\end{proof}

We remark that (\ref{propun}) plus the polarization identity implies
\begin{equation} \label{sprgam}
\spr{f_\gam}{g_\gam}_{a}^x = \spr{f}{g}_a^x - \frac{
\spr{f}{u}_a^x \spr{u}{g}_a^x}{c_\gam(x)},
\end{equation}
where $f_\gam = U_\gam f$, $g_\gam = U_\gam g$.


\section{Inserting a single eigenvalue}


Now we turn to the Dirac operator $H$ defined
in the Introduction. We will choose $(\lam_1,\gam_1)$ satisfying

\begin{hypo} \label{hypodc}
Suppose $(\lam,\gam) \in\R^2$ satisfies the following 
conditions.\\
(i). $u_-(\lam,x)$ exists.\\
(ii). $\gam\in[-\|u_-(\lam)\|^{-2},\infty) \cup \{\infty\}$.\\
(iii). If $u_-(\lam) \in\hr$, then $\lam\in\sig_p(H)$.
\end{hypo} 

and use Lemma~\ref{lemuni} with $u=u_-(\lam_1)$, $\gam=\gam_1$ to prove

\begin{thm} \label{thm}
Suppose (H.\ref{hypodc}) and let $H_{\gam_1}$ be the operator associated with
\bea
&H_{\gam_1} f = \tau_{\gam_1} f, \quad \db(H_{\gam_1}) = \{ \ba[t]{l} f \in 
\hr |\, f\in AC_{loc}(I,\C^2) ; \tau_{\gam_1} f \in \hr  ;\\
W_a(u_{\gam_1,-}(\lam_1),f)=W_b(u_{\gam_1,-}(\lam_1),f)=0 \}, \ea&
\eea
where
\begin{equation} \label{phigam}
\phi_{\gam_1} = \phi + \frac{u_-(\lam_1) \otimes_\sig u_-(\lam_1)}{
c_{\gam_1}(\lam_1,x)},
\quad f \otimes_{\sig} g = \frac{f \otimes (\sig_2 g) + (\sig_2 f) \otimes g}{\I},
\end{equation}
and
\begin{equation}
u_{\gam_1,-}(\lam_1,x) = \frac{u_-(\lam_1,x)}{c_{\gam_1}(\lam_1,x)},
\quad c_{\gam_1}(\lam_1,x) = \frac{1}{\gam_1} + \spr{u_-(\lam_1)}{u_-(\lam_1)}_a^x.
\end{equation}
Then 
\begin{equation}
H_{\gam_1} (\id-P_{\gam_1}(\lam_1)) = U_{\gam_1} H U_{\gam_1}^{-1}
(\id-P_{\gam_1}(\lam_1))
\end{equation}
and $\tau_{\gam_1} u_{\gam_1,-}(\lam_1) = \lam_1 u_{\gam_1,-}(\lam_1)$.
\end{thm}

\begin{proof}
Only the case $\gam\ne 0$ is of interest. The claim $\tau_{\gam_1} u_{\gam_1,-}(\lam_1) =
\lam_1 u_{\gam_1,-}(\lam_1)$ is straightforward and implies that $\tau_{\gam_1}$ is
$l.p.$ at $a,b$ if $\gam=\infty, -\|u_-(\lam_1)\|^{-2}$, respectively. Moreover, let $f
\in\db(H)$ then another straightforward calculation shows
\begin{equation}
\tau_{\gam_1} (U_{\gam_1} f) = U_{\gam_1} (\tau f)
\end{equation}
and it remains to compute $U_{\gam_1} \db(H)$. It suffices to vindicate
\begin{equation}
(\id -P_{\gam_1}(\lam_1)) U_{\gam_1} \db(H) \subseteq (\id -P_{\gam_1}(\lam_1))
\db(H_{\gam_1})
\end{equation}
since $(\id -P_{\gam_1}(\lam_1)) U_{\gam_1} \db(H)$ cannot be properly contained in
$(\id -P_{\gam_1}(\lam_1)) \db(H_{\gam_1})$ by the property of self-adjoint
operators being maximal. Only the boundary conditions are not obvious.
If $\gam\in\R$ the formula
\begin{equation} \label{wrongamp}
W_x(u_{\gam_1,-}(\lam_1),U_{\gam_1} f) = \frac{W_x(u_-(\lam_1),f)}{
c_{\gam_1}(\lam_1,x)}
\end{equation}
reveals $W_a(u_{\gam_1,-}(\lam_1),U_{\gam_1} f) =0$ for $f \in \db(H)$ (if
$\gam=\infty$, then $\tau_{\gam_1}$ is $l.p.$ at $a$). For the boundary condition
at $b$ we can assume $\gam \ne -\|u_-(\lam_1)\|^{-2}$. If $u_-(\lam_1) \in\hr$,
then (\ref{wrongamp}) shows $W_b(u_{\gam_1,-}(\lam_1),U_{\gam_1} f) = 0$ for
$f \in\db(H)$. Otherwise, that is, if $u_-(\lam_1) \not\in\hr $ we use
\begin{equation}
|W_x(u_{\gam_1,-}(\lam_1),U_{\gam_1} f)|^2 = \frac{|\spr{u_-(\lam_1)}{
(\tau-\lam_1) f}_a^x|^2}{c_{\gam_1}(\lam_1,x)^2},
\end{equation}
which tends to zero as $x\to b$ for $f\in\db(H)$ by (\ref{sprfpsiopsis}). 
\end{proof}

We remark that explicitly (\ref{phigam}) reads
\bea
\phi_{\gam_1,\rm el}(x) &=& \phi_{\rm el}(x),\\
\phi_{\gam_1,\rm am}(x) &=& \phi_{\rm am}(x) + \frac{u_{-,1}(\lam_1,x)^2
- u_{-,2}(\lam_1,x)^2}{c_{\gam_1}(\lam_1,x)},\\
\phi_{\gam_1,\rm sc}(x) &=& \phi_{\rm sc}(x) -2 \frac{u_{-,1}(\lam_1,x)
u_{-,2}(\lam_1,x)}{c_{\gam_1}(\lam_1,x)}.
\eea

\begin{coro} \label{corspec}
Suppose $u_-(\lam_1) \not\in \hr$.

(i). If $\gam_1>0$ then $H$ and $(\id-P_{\gam_1}(\lam_1)) H_{\gam_1}$ are
unitarily equivalent. Moreover, $H_{\gam_1}$ has the additional eigenvalue $\lam_1$
with eigenfunction $u_{\gam_1,-}(\lam_1)$.

(ii). If $\gam_1=\infty$ then $H$ and $H_{\gam_1}$ are unitarily equivalent.\\[1mm]

Suppose $u_-(\lam_1) \in \hr$ (i.e., $\lam_1$ is an eigenvalue of $H$).

(i). If $\gam_1\in(-\|u_-(\lam_1)\|^{-2},\infty)$ than $H$ and
$H_{\gam_1}$ are unitarily equivalent (using $\ti{U}_{\gam_1}$).

(ii). If $\gam_1=-\|u_-(\lam_1)\|^{-2},\infty$ then $(\id-P(\lam_1)) H$,
$H_{\gam_1}$ are unitarily equivalent, that is, the eigenvalue $\lam_1$ is removed.
\end{coro}

The following can be verified directly.

\begin{lemma}
Let $u \in AC_{loc}(I,\C^2)$ fulfill $\tau u = z u$ (with $z \in \C \bs \{ \lam_1 \}$)
and let
\begin{equation}
v(z,x) = u(z,x) + \frac{u_{\gam_1,-}(\lam_1,x)}{z-\lam_1} W_x(u_-(\lam_1),u(z)).
\end{equation}
Then $v \in AC_{loc}(I,\C^2)$ and $v$ fulfills $\tau_{\gam_1} v = z v$.
If $u$ is square integrable near $a$ and fulfills the boundary condition at
$a$ (if any) we have $v=U_{\gam_1}u$. We also note
\begin{equation} \label{normv}
|v(z,x)|^2 = |u(z,x)|^2 - \frac{1}{|z-\lam_1|^2} \frac{d}{dx}
\left( \frac{|W_x(u_-(\lam_1),u(z))|^2}{c_{\gam_1}(\lam_1,x)} \right),
\end{equation}
and if $\hat{u}$, $\hat{v}$ are constructed analogously then
\bea \nn
W_x(v(z),\hat{v}(\hat{z})) &=& W_x(u(z),\hat{u}(\hat{z}))
- \frac{1}{c_{\gam_1}(\lam_1,x)} \times \\ &&
\frac{z-\hat{z}}{(z-\lam_1)(\hat{z}-\lam_1)} 
W_x(u_-(\lam_1),u(z)) W_x(u_-(\lam_1),\hat{u}(\hat{z})).
\eea
In addition, the solutions 
\begin{equation} \label{ugampm}
u_{\pm,\gam_1}(z,x) = u_\pm(z,x) + \frac{u_{\gam_1,-}(\lam_1,x)}{z-\lam_1} 
W_x(u_-(\lam_1),u_\pm(z)).
\end{equation}
are square integrable  near $a,b$ and satisfy the boundary condition of $H_{\gam_1}$
at $a,b$, respectively.
\end{lemma}

\begin{rem} \label{remwm}
{}From (\ref{ugampm}) one can easily compute the Weyl $m$-functions corresponding to
$H$. Proceeding as in \cite{gcom}, \cite{gstdef} one can then
obtain an alternate proof for Corollary~\ref{corspec}.
\end{rem}

Since we have already seen, that our method does not preserve $l.p.$/$l.c.$
properties we want to discuss conditions for $\tau_{\gam_1}$ to be $l.p.$ at
$a,b$. Let $c\in I$ and let $H_{+,c}$ denote a self-adjoint operator 
associated with $\tau$ on $(c,b)$ and the boundary condition induced by
$u_-(\lam_1)$ at $c$ (i.e., $W_c(f,u_-(\lam_1))=0$, $f \in \db(H_{c,+})$).

\begin{hypo} \label{hypolp}
Suppose one of the following spectral conditions (i)--(ii) holds.
\begin{list}{(\roman{me}).}{\usecounter{me}}
\item $\sig_{ess} (H_{+,c}) \ne \emptyset$.
\item $\sig(H_{+,c})=\sig_d(H_{+,c})=\{\lam_n\}_{n\in \Z}$ with $\sum_{n\in
\Z} (1+\lam^2_{n})^{-1}=\infty$.
\end{list}
\end{hypo}

All conditions (i)--(ii) imply that $\tau$ is $l.p.$ at $b$ (since $\tau$ $l.c.$ at $b$
implies that the resolvent of $H_{+,c}$ is Hilbert--Schmidt) and we easily obtain (using
(\ref{normv}))

\begin{thm} \label{thmlp}
\begin{list}{(\roman{me}).}{\usecounter{me}\leftmargin2mm}
\item If $\gam_1\ne\infty$, then $\tau_{\gam_1}$ is $l.p.$ at $a$ if and
only if $\tau$ is. Otherwise, that is, if $\gam=\infty$, then
$\tau_{\gam_1}$ is $l.p.$ at $a$.
\item If $\gam_1\ne-\|u_-(\lam_1)\|^{-2}$, then $\tau_{\gam_1}$ is $l.c.$ at $b$ if
$\tau$ is. If $\gam_1 = -\|u_-(\lam_1)\|^{-2}(\ne 0)$, then $\tau_{\gam_1}$ is
$l.p.$ at $b$.
\item Assume (H.\ref{hypolp}), then $\tau_{\gam_1}$ is $l.p.$ at $b$.
\end{list}
\end{thm}

\begin{rem}
(i). Removing an eigenvalue from an operator which
is $l.c.$ at $b$ yields an operator which is $l.p.$. This shows that one cannot insert
additional eigenvalues into an operator which is $l.c.$ at $b$ (remove this
eigenvalue again to obtain a contradiction).\\
(ii). Clearly we can interchange the role of $a$ and
$b$. One only has to interchange $a$, $b$ in the text and $\spr{.}{..}_a^x$,
$\spr{.}{..}_x^b$ in the formulas.\\
(iii). As long as $u_-(\lam_1)$ exists (e.g., $\tau$ is $l.c.$ at $a$) our method can be
used to insert additional eigenvalues into the spectrum of $H$ (cf.\ \cite{rc}).\\
(iv). It is well-known that methods of inserting eigenvalues into the spectra of
one-dimensional Schr\"odinger operators are connected with B\"acklund (Darboux)
transformations of the (modified) Korteweg--de Vries hierarchy. This raises the question
whether our method is connected with transformations of the AKNS hierarchy. We recall
that the AKNS hierarchy is associated with the differential expression
\begin{equation}
\hat{\tau} = -\sig_3 \frac{1}{\I} \frac{d}{dx} + \frac{1}{2} \sig_2 (p(x) + q(x))
+ \frac{\I}{2} \sig_1 (p(x) - q(x)).
\end{equation}
Transforming $\hat{\tau}$ to our representation yields
\begin{equation} \label{tauakns}
\tau = U \hat{\tau} U^{-1}= \sig_2 \frac{1}{\I} \frac{d}{dx} -
\frac{1}{2} \sig_3 (p(x) + q(x)) + \frac{\I}{2} \sig_1 (p(x) - q(x)),
\end{equation}
where
\begin{equation}
U = \frac{1}{\sqrt{2}} \left(\ba{cc} 1 & \I \\ \I & 1 \ea\right).
\end{equation}
Since we are interested in the self-adjoint case we require $p=\ol{q}$.
This corresponds to the nonlinear Schr\"odinger equation. If in addition $\re(p)=0$ we
have the case of supersymmetric operators which is connected to the modified Korteweg--de
Vries hierarchy. Since our transformation does not leave (\ref{tauakns}) invariant, it
cannot correspond to a transformation of the AKNS system. This becomes even more evident in
the supersymmetric case, since inserting one eigenvalue $\lam_1 \ne 0$ must necessarily
destroy supersymmetry.
\end{rem}






\section{Inserting finitely many eigenvalues}




Finally we demonstrate how to iterate this method. We choose a given
background operator $H$ and pick $(\lam_1,\gam_1)$ according to (H.\ref{hypodc}).
Now define the transformation $U_{\gam_1}$ and the operator $H_{\gam_1}$ as in the
previous section. Next we choose $(\lam_2,\gam_2)$ and define
$u_{-,\gam_1}(\lam_2) = U_{\gam_1} u_-(\lam_2)$ and corresponding
operators $U_{\gam_1,\gam_2}$ and $H_{\gam_1,\gam_2}$. Applying this
procedure $N$ times leads to

\begin{thm} \label{dcommN}
Let $H$ be the background operator and let $( \lam_\ell,\gam_\ell)$, $1 \le \ell \le N$
satisfy (H.\ref{hypodc}). Define the following matrices ($1 \le \ell \le N$)
\begin{equation}
C^\ell(x) =  \left( \frac{\delta_{r,s}}{\gam_r}
+ \spr{u_-(\lam_r)}{u_-(\lam_s)}_a^x \right)_{1 \le r,s \le \ell},
\end{equation}
\begin{equation} \label{cij}
C^\ell(f,g)(x) = \left( \ba{c@{\quad}l}
C^{\ell-1}(x)_{r,s}, & {\scriptstyle r,s \le \ell-1}\\
\spr{f}{u_-(\lam_s)}_a^x, & 
{\scriptstyle s\le \ell-1, r=\ell}\\
\spr{u_-(\lam_r)}{g}_a^x, & {\scriptstyle r \le \ell-1, s=\ell}\\
\spr{f}{g}_a^x, & {\scriptstyle r=s=\ell}
\ea \right)_{1 \le r,s \le \ell} ,
\end{equation}
\begin{equation}
U^\ell(f)(x) = \left( \ba{c@{\quad}l}
C^\ell(x)_{r,s}, & {\scriptstyle r,s \le \ell}\\
\spr{u_-(\lam_s)}{f}_a^x, & {\scriptstyle s \le \ell, r=\ell+1}\\
u_-(\lam_r,x), & {\scriptstyle r \le \ell, s=\ell+1}\\
f, & {\scriptstyle r=s=\ell+1}
\ea \right)_{1 \le r,s \le \ell+1} .
\end{equation}
Then we have (set $C^0(x)=1$, $U^0(f) = f$)
\begin{equation} \label{sprl}
\spr{U_{\gam_1,\dots,\gam_{\ell-1}} \cdots U_{\gam_1} f}{
U_{\gam_1,\dots,\gam_{\ell-1}} \cdots U_{\gam_1} g}_a^x =
\frac{\det C^\ell(f,g)(x)}{\det C^{\ell-1}(x)}
\end{equation}
and
\begin{equation} \label{phil}
U_{\gam_1,\dots,\gam_\ell} \cdots U_{\gam_1} f(x) =
\frac{\det U^\ell(f)(x)}{\det C^\ell(x)}.
\end{equation}
In particular, we obtain
\bea \nn
c_{\gam_\ell}(\lam_\ell,x) &=& \frac{1}{\gam_\ell} + 
\spr{U_{\gam_1,\dots,\gam_{\ell-1}} \cdots U_{\gam_1} u_-(\lam_\ell)}{
U_{\gam_1,\dots,\gam_{\ell-1}} \cdots U_{\gam_1} u_-(\lam_\ell)}_a^x\\
\label{cln} &=& \frac{\det C^\ell(x)}{\det C^{\ell-1}(x)}.
\eea
The corresponding operator $H_{\gam_1,\dots,\gam_N}$ is
associated with
\begin{equation}
\phi_{\gam_1,\dots,\gam_N} =  \phi + \sum_{\ell=0}^N
\frac{\det U^{\ell-1}(u_-(\lam_\ell)) \otimes_\sig
\det U^{\ell-1}(u_-(\lam_\ell))}{\det C^\ell(x)}.
\end{equation}
and we have
\bea \nn
&& H_{\gam_1,\dots,\gam_N} (\id-\sum\limits_{j=1}^N
P_{\gam_1,\dots,\gam_N}(\lam_j))
\\ && \quad = (U_{\gam_1,\dots,\gam_N} \cdots U_{\gam_1}) H  (U_{\gam_1}^{-1}
\cdots U_{\gam_1,\dots,\gam_N}^{-1}) (\id- \sum\limits_{j=1}^N
P_{\gam_1,\dots,\gam_N}(\lam_j)),
\eea
where $P_{\gam_1,\dots,\gam_N}(\lam_j)$ denotes the projection onto the 
one-dimensional subspace span\-ned by
\begin{equation}
u_{\gam_1,\dots,\gam_N,-}(\lam_\ell,x) = \gam_\ell
\frac{\det U^\ell(u_-(\lam_\ell))(x)}{\det C^\ell(x)}
\end{equation}
(the last equation has to be understood as a limit if $\gam_\ell=\infty$).
\end{thm}

\begin{proof}
It suffices to prove (\ref{sprl}), (\ref{phil}) which requires a
straightforward induction argument using Sylvester's determinant identity
(\cite{gant}, Sect.\ II.3). The resulting identity
\bea \nn
\det C^\ell \det C^\ell(f,g) - \det C^\ell(u_-(\lam_\ell),g)
\det C^\ell(f,u_-(\lam_\ell)) \\ = \det C^{\ell-1} \det C^{\ell+1}(f,g),
\eea
together with (\ref{sprgam}) then proves (\ref{sprl}). Similarly,
\bea \nn
\det C^\ell \det U^{\ell-1}(f) - \det
U^{\ell-1}(u_-(\lam_\ell)) \det C^\ell(u_-(\lam_\ell),f)\\ =
\det C^{\ell-1} \det U^\ell(f),
\eea
and (\ref{unitary1}) prove (\ref{phil}). The rest then
follows from these two equations and Theorem~\ref{thm}.
\end{proof}

\begin{rem}
(i). The ordering of the pairs $( \lam_j,\gam_j)$, $1 \le j \le N$ is
clearly irrelevant (interchanging row $i,j$ and column $i,j$ leaves all
determinants unchanged). Moreover, if $\lam_i = \lam_j$, then $( \lam_i,\gam_i)$,
$( \lam_j,\gam_j)$ can be replaced by $( \lam_i,\gam_i+\gam_j)$ (by the
first assertion it suffices to verify this for $N=2$).\\
(ii). Equation (\ref{phil}) can be rephrased as
\bea \nn
&&(u_{\gam_1,\dots,\gam_N,-}(\lam_1,x), \dots,
u_{\gam_1,\dots,\gam_N,-} (\lam_N,x)) \\ && \quad =
(C^N(x))^{-1} (u_-(\lam_1,x), \dots, u_-(\lam_N,x) ),
\eea
where $(C^N(x))^{-1}$ denotes the inverse matrix of $C^N(x)$.
\end{rem}



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\end{document}
