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absolutely continuous spectrum,
Hamiltonian system, dissipative operators
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\title{Absence of the absolutely continuous spectrum of a first-order
non-selfadjoint Dirac-like system for slowly decaying
perturbations}
\author{
Marco Marletta
\\ School of Mathematics,
\\ Cardiff University,
\\ PO Box 926, Cardiff CF24 4YH
\\ e-mail: Marco.Marletta@cs.cardiff.ac.uk
\and Roman Romanov\thanks{
On leave from Laboratory of Quantum
Networks, Institute for Physics, Saint Petersburg State
University. The work of R. R. has been partially supported by
EPSRC Grant GR/ R20885 and by RFBR Grant 03-01-00090}
\\ Department of Computer Science,
\\ Cardiff University,
\\ PO Box 916, Cardiff CF24 3XF
\\ e-mail: r.v.romanov@cs.cf.ac.uk}
\begin{document}
\maketitle
\begin{abstract}
We prove that the absolutely continuous subspace of the completely
non-selfadjoint part of a first-order dissipative Dirac-like
system is trivial when the imaginary part of the potential is
non-integrable.
\end{abstract}
%\input{amssym}
\noindent {\sc AMS(MOS) Subject classifications:} 34L40, 47B44.
\noindent {\bf Keywords:} absolutely continuous spectrum,
Hamiltonian system, dissipative operators.
\section*{Introduction}\label{section:1}
In this article we analyze the structure of the essential spectrum
of dissipative operator realizations of the ordinary differential
expression \be \ell_Q := J\frac{d}{dx} + Q(x), \;\;\; x \in
[0,\infty), \label{eq:1.1} \ee where \be J =
\left(\begin{array}{cc} 0 & i \\ i & 0
\end{array}\right), \;\;\; Q(x) = \left(\begin{array}{cc} q_{1}(x)
& 0 \\ 0 &
q_{2}(x) \end{array}\right), \label{eq:1.2}
\ee in which $ Q ( x ) $ is a bounded function such that $\Im
q_1(x)$ and $\Im q_{2}(x)$ are non-negative a. e. We shall prove
the following result.
\bigskip
\noindent {\bf Theorem 1.} {\it Let $L$ be the operator in
$L^2(\Rr^+,\Cc^2)$ given by the expression (\ref{eq:1.1}) and with
domain determined by a self-adjoint boundary condition at $0$. If
the absolutely continuous subspace of the completely
non-selfadjoint part of $L$ is nontrivial, then $\Im q_{1}$ and
$\Im q_{2}$ both lie in $L^1(\Rr^+)$.}
\bigskip
This assertion is an analogue of the following theorem in the
paper of Romanov \cite{Rom} for Schr\"{o}dinger operators on the
half-line: if a bounded potential $ q $ on the half-line $ x \ge 0
$ such that $ \Im q ( x ) \ge 0 $, $ \Im q ( x ) \to 0 $ as $ x
\to \infty $, satisfies $ \Im q \notin L^1 $, then the dissipative
operator $ l $, $ l u = - u'' + q u $, corresponding to a
self-adjoint boundary condition at zero has trivial absolutely
continuous subspace.
Together with the latter theorem, the result of the present paper
suggests that the existence of an absolutely continuous spectrum
for dissipative ordinary differential operators is equivalent to
having a scattering theory. Let us mention here that the abstract
local nuclear scattering theory developed in \cite{N,Nab} shows
that for any pair of operators $ B, B_0 $ such that $ B_0 $ is
selfadjoint and $ B - B_0 $ is of (relative) trace class the
absolutely continuous parts of $ B $ and $ B_0 $ are quasi-similar
by means of wave operators and their spectra coincide. In the
situation under consideration this implies that the absolutely
continuous spectrum of $ L $ coincides with that of the
selfadjoint Dirac operator $ \Re L $ when $ \Im Q \in L^1 $. This
is to be compared with the situation of the self-adjoint theory,
where the existence of modified wave operators for the
Schr\"odinger operator in dimension $1$ has only been established
for specific classes of slowly decaying potentials $ q $ (see, e.
g. \cite{Kiselev}), while the absolutely continuous spectrum is
known to coincide with the positive real line for any real $ q \in
L^2 $ \cite{DK}.
The strategy of the proof of the main result follows that in the
paper \cite{Rom} with one notable exception. We start with an
abstract re-formulation of triviality of the absolutely continuous
subspace of a dissipative operator, given in Proposition
\ref{criterion}, and then obtain in Corollary \ref{corollary:1} a
convenient sufficient condition in the situation under
consideration which is expressed in terms of the asymptotics of
solutions of the equation $ \ell_{ Q^* } y = k y $ with real $ k
$. Specifically, it says that the absolutely continuous subspace
of the completely non-selfadjoint part of $L$ is trivial if $
\int^\infty \( \Im q_1 | y_1 |^2 + \Im q_2 | y_2 |^2 \) dx $ is
infinite for a.e. real $ k $. Here $ y = \( y_1 , y_2 \)^T $ is
the solution of the Cauchy problem for $ \ell_{ Q^* } y = k y $
with the initial data $ y(0) = \left( 1 , 0 \right)^T $ (or any
other $ k $-independent non-zero initial data satisfying a
selfadjoint boundary condition). In the case of a Schr\"odinger
operator \cite{Rom}, the next step in the argument used spectral
averaging \cite{SimonLast} for the selfadjoint problem with the
potential $ \Re q $. The spectral averaging method, as developed
by Last-Simon, seems to be only applicable to operators
semi-bounded below, while a selfadjoint operator corresponding to
a differential expression of the form (\ref{eq:1.1}) is never
semi-bounded. To circumvent this difficulty, we first establish
the result for potentials with "spread" $ \Im Q $ (see Proposition
\ref{theorem:weak1} and a remark after it), and then show that an
arbitrary potential can be represented as a sum of a potential
with spread $ \Im Q $ and an $ L^1 $-potential. An appropriate
variant of the trace class method, which we develop in the
abstract part of the paper in Lemma \ref{trace}, allows us to
handle the $ L^1 $ - perturbation. This argument is similar to
that used in \cite{Rom} for proving a discrete variant of the main
result.
Throughout the paper $ {\bf S}^1 $ stands for the trace class of
operators. Given an operator $ B $ we write $ B^\prime $ for the
completely non-selfadjoint part of $ B $. For any vector $ v $ let
$ v^T $ stand for its transpose.
\refstepcounter{section}
\subsection*{\S 1. Preliminaries: Absolutely Continuous Subspace}
Let $ H $ be a Hilbert space and $ L $ be a maximal dissipative
operator in $ H $ of the form $ L = A + i V $, $ A = A^* $, $ V
\ge 0 $ is bounded. We shall assume that\footnote{By $\sigma_{ess}(L)$
we mean the union of the set of all non-isolated points of $\sigma(L)$
and the set of all isolated points of $\sigma(L)$ of infinite
multiplicity.} $ \sigma_{ ess } ( L )
\subset \Rr $. Define $ H_0 $ to be the maximal
reducing subspace of $ L $ on which it induces a selfadjoint
operator, which will be denoted by $ L_0 $. Let $ L^\prime $ be
the completely non-selfadjoint part of $ L $, $ L^\prime = \left.
L \right|_{ H^\prime } $, $ H^\prime = H \ominus H_0 $ \cite{N}.
Define the Hardy classes of vector functions $ \H^2_\pm $ to be
the collections of $H$-valued analytic functions $f$ on $ \Cc_\pm
= \{ z \in \Cc \, | \, \pm \Im z > 0 \} $, which satisfy $\sup_{
\von
> 0 } \int_\R \len f ( k \pm i\von ) \rin^2 dk < \infty $, respectively.
\begin{definition}
{\it The absolutely continuous (a. c.) subspace}, $ H_{ac} (
L^\prime ) $, of the completely non-selfadjoint operator $
L^\prime $ is defined as follows \cite{N,Sachn} \be\label{Hac}
H_{ac} ( L^\prime ) = {\rm{clos}} \left\{ u \in H:
\begin{array}{l} 1) \( L - z \)^{ -1 } u \mbox{ is analytic in } \C_+
\\ 2) \, \, V^{1/2} \( L - z \)^{ -1 } u \in \H^2_+
\end{array} \right\} . \ee
By the a. c. subspace of the operator $ L $ we call the subspace
$$ H_{ ac } ( L ) = H_{ac} ( L^\prime ) \oplus H_{ac} ( L_0 ) $$
where $ H_{ac} ( L_0 ) $ is the a. c. subspace of the self-adjoint
operator $ L_0 $ defined in the standard way. \end{definition}
The a. c. subspace of $ L $ is known \cite{N} to be regular
invariant, that is, $ \overline{( L - z )^{-1} H_{ac} } = H_{ac} $
for all $ z \in \rho (L) $.
Various motivations of this definition and its relation to
scattering theory are given in \cite{N,Nab,Sachn,Pavli,Pav1}. We
only mention here\footnote{Theorem \ref{equivl} is not used in the
paper.} the following "weak" characterization of the subspace $
H_{ac} ( L ) $.
\begin{theorem} \label{equivl}
\cite{equival} $$ H_{ac} ( L ) = {\rm{clos }} \left\{ u \in H: \;
\begin{array}{l} 1) \( L - z \)^{ -1 } u \mbox{ is analytic in } \C_+ \\
2) \left. \left\langle \( L - z \)^{ -1 } u , v \right\rangle
\right|_{ \C_\pm } \in \H^2_\pm \mbox{ for all } v \in H \cr
\end{array} \right\}. $$
\end{theorem}
This theorem shows, in particular, that (\ref{Hac}) can be
considered as a generalization of the definition of the a. c.
subspace in the self-adjoint theory.
The triviality of the subspace $ H_{ac} ( L^\prime ) $ in the
example studied in this paper will be established on the basis of
the following criterion which is implicit in \cite{Pav1}.
\begin{proposition} \label{criterion}
$ H_{ac} ( L^\prime ) = \{ 0 \} $ iff for a. e. $ k \in \R $ we
have ($ z = k + i \von $) \be\label{crit} D ( z ) \equiv \sqrt
\von \( L^* - z \)^{ -1 } \sqrt V \stackrel{s} \longrightarrow 0
\ee as $ \von \searrow 0 $.
\end{proposition}
Different elementary proofs of the "if" part of the criterion can
be found in \cite{Rom,equival}. The "only if" part is not used in
this paper.
\medskip
\noindent {\bf Remark} {\it \cite{Pav1} The function $ D ( z ) $
satisfies $ \| D ( z ) \| \le 1/2 $ for all $ z \in \C_+ $.}
\medskip
The proof of this fact is by direct calculation. A similar
calculation is contained in (\ref{contr}) in Lemma \ref{trace}
below.
As is well known, if two self-adjoint operators $ {\cal L}_{ 1,2 }
$ satisfy $ {\cal L}_1 - {\cal L}_2 \in \bS^1 $, then their a. c.
spectra coincide. In the non-selfadjoint case, the question
whether $ L_1 - L_2 \in \bS^1 $ for a pair of dissipative
operators $ L_1 $, $ L_2 $ implies $ \sigma \( L_1 |_{ H_{ac} (
L_1 ) } \) = \sigma \( L_2 |_{ H_{ac} ( L_2 ) } \) $ appears to be
open if $ \Im L_{ 1,2 } $ are not of $ \bS^1 $ separately. We
shall need a result of this type in a special situation.
\begin{lemma} \label{trace} Let $ L $ and $ \tL $ be two dissipative
operators such that \be\label{trres} \( L - z \)^{ -1 } - \( \tL -
z \)^{ -1 } \in \bS^1 , \; \; z \in\rho ( L ) \cap \rho ( \tL ) .
\ee Write $\tL - L = i \Gamma $, and assume that $ \Gamma \ge 0 $
and there exists a bounded operator $ \Pi \ge 0 $ such that $
\Gamma = \tV \Pi $, $ \tV = \Im \tL $. If $ H_{ac} ( \tL^\prime )
= \{ 0 \} $, then $ H_{ac} ( L^\prime ) = \{ 0 \} $.
\end{lemma}
This lemma is a variant of Lemma 2.3 from \cite{Rom}, where $
\Gamma $ was assumed to be trace class.
\noindent {\bf Proof}\, We shall actually prove that if $
\widetilde D ( k + i \von ) \equiv \sqrt \von \( \tL^* - z \)^{ -1
} \sqrt \tV \stackrel{s} \longrightarrow 0 $ as $ \von \to 0 $ for
a.e. $ k \in \R $, then $ D ( k + i \von ) \stackrel{s}
\longrightarrow 0 $ for a. e. $ k \in \R $ as well. Taking into
account the assumed factorization of $ \Gamma $ we find that $$
\sqrt \von \( \tL^* - z \)^{ -1 } \sqrt \Gamma \stackrel{s}
\longrightarrow 0 $$ for a. e. $ k \in \R $. Let us show that $
\sqrt \von \( L^* - z \)^{ -1 } \sqrt \Gamma \stackrel{s}
\longrightarrow 0 $ for a. e. $ k $. We have, $$ \sqrt \von \( L^*
- z \)^{ -1 } \sqrt \Gamma \cdot G ( z ) = \sqrt \von \( \tL^* - z
\)^{ -1 } \sqrt \Gamma , $$ where $$ G ( z ) = I + i \sqrt \Gamma
\( \tL^* - z \)^{ -1 } \sqrt \Gamma . $$ Let $ V = \Im L $. Then
the following calculation shows that $ G ( z ) $ is a contraction
for all $ z \in \C_+ $,
\begin{eqnarray} \label{contr} I - G^* ( z ) G ( z ) = I - \( I - i
\sqrt \Gamma \( \tL - \overline z \)^{ -1 } \sqrt \Gamma \) \( I +
i \sqrt \Gamma \( \tL^* - z \)^{ -1 } \sqrt \Gamma \) = \nonumber
\\ i \sqrt \Gamma \( \tL - \overline z \)^{ -1 } \( \( \tL^* - z \) -
\( \tL - \overline z \) + i\Gamma \) \( \tL^* - z \)^{ -1 } \sqrt
\Gamma = \nonumber
\\ \sqrt \Gamma \( \tL - \overline z \)^{ -1 } ( 2 \von + 2 V + \Gamma ) \(
\tL^* - z \)^{ -1 } \sqrt \Gamma .
\end{eqnarray}
Since the right hand side is obviously non-negative, we conclude
that $ G ( z ) $ is a contractive analytic operator-function in $
\C_+ $. We shall now show that this function admits a scalar
multiple, that is, there exists a scalar contractive analytic
function, $ g \not\equiv 0 $, such that $ G ( z ) \Omega ( z ) = g
( z ) \cdot I $ for a certain bounded analytic function $ \Omega
$. Let $ \left\{ X_N \right\}_{ N=1 }^\infty $ be an arbitrary
family of finite-dimensional subspaces in $ H $ such that $
X_{N-1} \subset X_N $ and $ \bigvee_N X_N = H $. We shall use the
subscript $ N $ with operators to denote their truncation to $ X_N
$, $ B_N \colon = \left. P_N B \right|_{ X_N } $ where $ P_N $ is
the orthogonal projection on $ X_N $. Define $ g_N ( z ) = \det
G_N ( z ) $. Our aim is to show that there exists a subsequence of
$ N $ such that $ g_N ( z ) $ converges when $ N \to \infty $
along the subsequence for all $ z \in \C_+ $ and $ g( z ) = \lim
g_N ( z ) $ is the scalar multiple. Notice that the operator $ G
(z) $ is boundedly invertible for $ \Im z $ large enough, for $
G(z) \to I $ in the operator norm when $ \Im z \to + \infty $
since $ \tL $ has a bounded imaginary part. Fix an arbitrary $ z_0
\in \C_+ $ such that $ \len I - G ( z_0 ) \rin \le 1/2 $, so that
$ T = G ( z_0 ) $ is boundedly invertible. Obviously, $ | \det T_N
| \le 1 $ since $ T $ is a contraction, and $ T_N $ is boundedly
invertible for all $ N $. Fix arbitrarily a subsequence of $ N $
such that $ \det T_N $ converges. To keep notation to a minimum,
we do not introduce the corresponding subscript and will always
assume that $ N \to \infty $ along this subsequence. Then, $ g ( z
) = \lim g_N ( z ) $ is the scalar multiple. Indeed, the limit
exists for all $ z \in \C_+ $: we have $$ g_N = \det G_N = \det
T_N \cdot \det \( I_N+ T_N^{ -1 } \( G_N - T_N \) \) $$; then,
\begin{eqnarray*} G ( z ) - T = i \sqrt \Gamma \( \( \tL^* - z
\)^{ -1 } - \( \tL^* - z_0 \)^{ -1 } \) \sqrt \Gamma = & & \\ &
\hspace{-3cm} i ( z - z_0 ) \sqrt \Gamma \( \tL^* - z \)^{ -1 }
\cdot \( \tL^* - z_0 \)^{ -1 } \sqrt \Gamma \in \bS^1 &
\end{eqnarray*}
since $ \sqrt \Gamma \( \tL^* - z \)^{ -1 } $ is Hilbert-Schmidt,
which is easily seen to be equivalent to the assumption
(\ref{trres}); hence, $ G_N - T_N \rightarrow G - T $ in the $
\bS^1 $-norm, and therefore $$ \det \( I_N+ T_N^{ -1 } \( G_N -
T_N \) \) \longrightarrow \det \( I+ T^{ -1 } \( G - T \) \) . $$
Since $ \det T_N $ converges by construction, we obtain that the
limit $ g( z ) $ exists. Obviously, the limit is a bounded
function of $ z $ in $ \C_+ $ because $ | g_N ( z ) | \le 1 $ for
all $ N $. An application of the Montel theorem shows that $ g ( z
) $ is an analytic function.
Let us check that $ g \not\equiv 0 $. If $ \det \( I+ T^{ -1 } \(
G ( z ) - T \) \) = 0 $, then $ \ker G ( z ) $ is nontrivial which
is only possible for a discrete set of values of $ z \in \C_+ $
since $ G ( z ) $ is a compact operator. It remains to show that $
\lim \det T_N \ne 0 $. Indeed, consider $ \left| \det T_N
\right|^2 = \det \( T_N^* T_N \) $. It follows from (\ref{contr})
that $ I - T^* T \in \bS^1 $, and a calculation similar to
(\ref{contr}) gives
$$ I_N - T_N^* T_N = \left. P_N \sqrt \Gamma
\( \tL - \overline z_0 \)^{ -1 } \( 2 \von + 2 \tV - \sqrt \Gamma
P_N \sqrt \Gamma \) \( \tL^* - z_0 \)^{ -1 } \sqrt \Gamma
\right|_{ X_N }. $$
This shows that the right hand side (extended to
$ X_N^\perp $ by zero) converges in the trace norm to $ I - T^* T
$. Therefore, $ \lim \det \( T_N^* T_N \) = \det T^* T \ne 0 $
since $ \ker T $ is trivial by the choice of $ z_0 $.
Now, $ \Omega ( z ) = g ( z ) G^{ -1 } ( z ) $ satisfies $ \len
\Omega ( z ) \rin \le 1 $ since $ | g_N ( z ) | \len G^{ -1 }_N (
z ) \rin \le 1 $ for each $ N $ and all $ z $ such that $ g_N ( z
) \ne 0 $, and thus $ g (z) $ is a scalar multiple.
It follows from the existence of a scalar multiple and the Fatou
theorem that the strong boundary values, $ G^{ -1 } ( k ) =
\mathop{\mbox{s-lim}}_{ \von \downarrow 0 } G^{ -1 } ( k + i \von
) $, of the function $ G^{ -1 } $ exist for a. e. $ k \in \R $. We
infer that for a. e. $ k \in \R $ $$ \sqrt \von \( L^* - z \)^{ -1
} \sqrt \Gamma = \sqrt \von \( \tL^* - z \)^{ -1 } \sqrt \Gamma
\cdot G^{ -1 }( z ) \mathop{\stackrel{s} \longrightarrow}_{ \von
\downarrow 0 } 0 . $$
One can now verify the condition (\ref{crit}) for the operator $ L
$. We have, \be\label{lv} \sqrt \von \( L^* - z \)^{ -1 } \sqrt V
= \sqrt \von \( \tL^* - z \)^{ -1 } \sqrt V - i \sqrt \von \( L^*
- z \)^{ -1 } \sqrt \Gamma \cdot Q ( z ) , \ee where $ Q ( z ) =
\sqrt \Gamma \( \tL^* - z \)^{ -1 } \sqrt V $ is a bounded
analytic function in $ \C_+ $. The latter follows from a
calculation similar to (\ref{contr}) giving that $ I - G ( z ) G^*
( z ) $ equals the right hand side of (\ref{contr}) with $ \(
\tL - \overline z \)^{ -1 } $ and $ \( \tL^* - z \)^{ -1 } $
swapped, and therefore $ Q ( z ) Q^* ( z ) \le I - G ( z ) G^* ( z
) \le I $. The first term on the right hand side of (\ref{lv})
converges strongly to zero by the assumption, and an application
of the Fatou theorem to the function $ Q (z) $ shows that the
second term converges strongly to zero as well. \hfill $\Box$
In the next section we establish triviality of the absolutely
continuous subspace for an operator $ L $ by verifying the
condition (\ref{crit}) for an operator $ \tL $ obeying the
assumptions of this lemma. Since the {\it proof} of the lemma
consists of demonstrating that the condition (\ref{crit}) for the
operator $ L $ is satisfied if it is satisfied for the operator $
\tL $, the final result then does not depend on the "only if" part
of the criterion of Proposition \ref{criterion}.
\refstepcounter{section}
\subsection*{\S 2. The operator corresponding to $ \ell_Q $}
Let $\TTh(\cdot,z)$ and $\TPh(\cdot,z)$ be the solutions of the
initial value problems
\[ \ell_{Q^*}\TTh = z \TTh, \;\;\; \TTh(0,z) = \left(\begin{array}{c}
1 \\ 0 \end{array}\right), \]
\[ \ell_{Q^*}\TPh = z\TPh, \;\;\; \TPh(0,z) = \left(\begin{array}{c}
0 \\ 1 \end{array}\right). \]
It is standard that the functions $ \TTh ( x , z ) $ and $ \TPh (
x , z ) $ are entire functions of $ z $ for all $ x $ and are
continuous in $ z $ uniformly in $ x \in I $ for any compact
interval $ I $.
\begin{lemma} \label{lemma1}For each $z$ in the upper half plane
$\Cc_{+} $, there exists a unique (up to scalar multiples)
solution of the differential equation $\ell_{Q^*}\TPs = z \TPs$
which lies in $L^2(\Rr^+,\Cc^2)$, and which can be chosen to be of
the form \be \TPs(x,z) = \TTh(x,z) + m (z)\TPh(x,z) .
\label{eq:2.1} \ee
\end{lemma}
\noindent {\bf Proof}\, Multiplying the equation $\ell_{Q^*} y =
zy$ by $J$ we obtain an equation of the form $y' = Ay$ in which
$A$ has zero diagonal. The fundamental matrix of this equation
therefore has a constant determinant; it follows that any two
linearly independent solutions, say $u=\left(\begin{array}{c} u_1
\\ u_2
\end{array}\right)$ and $v=\left(\begin{array}{c} v_1 \\ v_2 \end{array}\right)
$ satisfy $W[u,v] \equiv u_1(x)v_2(x)-u_2(x)v_1(x) = const. \neq 0$.
Integrating this identity in $x$ and applying the Schwarz inequality
shows that at most one of the solutions lies in $L^2(\Rr^+,\Cc^2)$.
This is a version of a standard argument: see, e.g., Levitan and
Sargsjan \cite{kn:LevitanSargsjan}.
Let $L_{min}$ be the closure of the operator defined by the
differential expression $\ell_{Q}$ on $ C_0^\infty( \Rr^+,\Cc^2 )
$ in the space $ L^2(\Rr^+ , \Cc^2 ) $. An integration by parts
shows that $ L_{min} $ is a dissipative operator. Since $ L_{min}
$ is obviously not maximal dissipative, $\dim\ker(L_{min}^*-zI)$
is either 1 or 2 for all $z\in \Cc_+$. By the argument of the
previous paragraph, dimension 2 is impossible. Thus there exists
exactly one solution $\TPs$ of the equation $\ell_{Q^*}y = zy$ in
$L^2$.
To show that $\TPs$ can be chosen of the form (\ref{eq:2.1}) it
remains to note that $\TPh(\cdot,z)\not\in L^2(\Rr^+,\Cc^2)$ for
all $z\in \Cc_{+}$: if it were, for some $z$, then $ -z $ would be
an eigenvalue in the lower half plane of a dissipative operator
corresponding to the differential expression $ -\ell_{ Q^* } $.
\hfill $\Box$
\begin{definition}
Let $h\in i\Rr$ be fixed. By $L$ we denote the operator in the
Hilbert space $ H = L^2(\Rr^+ , \Cc^2 ) $ given by the
differential expression $ \ell_Q $ with domain
\[ D(L) = \left\{ y = \left(\begin{array}{c} y_1 \\ y_2 \end{array}\right)
\in H \, | \, y\in AC_{loc}(\Rr^+), \;\;
\ell_Qy \in H, \;\; y_2(0) = h y_1(0) \right\} . \]
\end{definition}
Using Lemma \ref{lemma1} it is easy to see that $ L $ is a maximal
dissipative operator, and so $ \sigma ( L ) \subset \overline
{\C_+ } $. As the following example shows, this operator is not
completely non-selfadjoint in general.
\medskip
\noindent{\bf Example.} We construct a potential $Q$ with $\Im Q
\neq 0$, such that $L$ has a real eigenvalue. The boundary
condition will be $y_1(0)=0$, corresponding to $h= \infty $.
Let $X>0$ be fixed and let $ y $ be given on $[0,X]$ by
\[ y(x) = \left(\begin{array}{c} 0 \\ 1 \end{array}\right),
\;\;\; 0 \leq x \leq X. \]
Fix a real non-zero $k$ and choose $q_2 = k$ on $[0,X]$; let
$q_1$ satisfy $\Im(q_1)\not\equiv 0$ and be otherwise arbitrary on
$[0,X]$. Clearly $y$ satisfies the differential equation
$\ell_{Q}y = k y$ on $[0,X]$ together with the boundary condition.
For $x>X$, let $y_2(x) = 1/\cosh(x-X)$ and $q_1(x)\equiv 0$.
Define $y_1(x)$ for $x>X$ by
\[ y_1(x) = \frac{iy_2'(x)}{k} = -\frac{i\sinh(x-X)}{k\cosh^2(x-X)}. \]
Obviously $y_1$ and $y_2$ are continuous at $x=X$ and decay exponentially
for large $x$; therefore $y\in D(L)$. The equation $iy_2'=(k-q_1)y_1$ is
satisfied for all $x$ by construction. If we now choose $q_2$ for
$x>X$ so that the equation $iy_1'= (k-q_2)y_2$ holds, then $ y $ will be
an eigenfunction of $ L $. We thus let
\[ q_2 = k + k^{-1}\frac{y_2''}{y_2}. \]
The function $q_2$ is asymptotically a real constant at infinity and so
$Q$ satisfies all our hypotheses. \hfill $\Box$
\medskip
We shall require an expression for the action of the resolvent
$(L^*-z)^{-1} $ on compactly supported vectors.
\begin{definition} Let $\TPs_h(\cdot,z)$ denote the solution of the differential
equation $\ell_{Q^*}\TPs_h = z \TPs_h$ with the initial condition
given by $\TPs_h(0,z) = \left(\begin{array}{c} 1
\\ h \end{array}\right) $, so that
\[ \TPs_h(\cdot,z) = \TTh(\cdot,z) + h\TPh(\cdot,z) . \] \end{definition}
Notice that $ m (z) \ne h $ for any $ z \in \Cc_+ $, for otherwise
$ \TPs ( x , z ) $ would be an eigenfunction of $ L^* $, and $ z $
would be an eigenvalue. Let $ E = \mbox{diag} ( 1 , -1 ) $. Then a
straightforward calculation taking into account that $ \det (
\TPs_h , \TPs ) = m ( z ) - h $ shows that for any compactly
supported $ u \in H $ \be\label{resol} \( L^* - z \)^{ -1 } u =
\frac i{m(z)-h}\left\{ \TPs_h(x,z)\int_x^\infty\TPs^T(s,z)E u(s)ds
+ \TPs(x,z)\int_0^x\TPs_h^T(s,z)E u(s)ds\right\}. \ee
\begin{lemma}\label{lemma:3}
The function $ m(z)$ defined in lemma \ref{lemma1} is analytic in
$\Cc_{+}$. For all $ z \in \C_+ $ it satisfies $\Re m (z) < 0$.
Moreover, for all $ z $ with $ \epsilon = \Im z > 0 $,
\begin{eqnarray} - \Re m (z) & = &\epsilon \int_0^\infty
\left( |\TPs_1(x,z)|^2 + |\TPs_2(x,z)|^2 \right) dx + \nonumber \\
& & \int_0^\infty \left( \Im q_{1}(x) |\TPs_1(x, z )|^2 + \Im
q_{2}(x) |\TPs_2(x, z)|^2 \right) dx , \label{eq:2.4}
\end{eqnarray}
and therefore
\[ \limsup_{\epsilon \searrow
0 }\( \epsilon \len \TPs ( \cdot , k + i \epsilon ) \rin_H^2 \)
\le - \Re m ( k ) \] whenever the limit $ m ( k ) = \lim_{
\epsilon \searrow 0 } m ( k + i \epsilon )$ exists.
\end{lemma}
\noindent {\bf Proof}\, It follows from (\ref{resol}) that for any
compactly supported $ u,v \in H $ $$ F ( z ) \equiv \llangle \(
L^* - z \)^{ -1 } u , v \rrangle = \frac {A m ( z ) + B }{m(z)-h}
$$ where $ A, B $ are complex constants depending on $ u $ and $ v
$. Then, $ B \ne - h A $ for a suitable choice of $ u $ and $ v $,
since otherwise the fact that $ F ( z ) \to 0 $ when $ \Im z \to
\+ \infty $ would imply that $ F ( z ) \equiv 0 $ for any
compactly supported $ u, v \in H $, and therefore $ \( L^* - z
\)^{ -1 } = 0 $. We now infer that $ m ( z ) $ is analytic in $
\C_+ $ from the analyticity of $ F ( z ) $. Alternatively, the
analyticity of $m$ can be proved through the nesting circles
analysis in the same way as in the case of a dissipative second
order Schr\"{o}dinger operator \cite{Sims}.
We shall now establish (\ref{eq:2.4}). The differential equations
satisfied by $\TPs_1(x,k+i\epsilon)$ and $\TPs_2(x,k+i\epsilon)$
are
\[ i \TPs_2' + (\Re q_1(x) - i \Im q_1(x) - k - i\epsilon)\TPs_1 = 0, \]
\[ i \TPs_1' + (\Re q_2(x) - i \Im q_2(x) -k - i\epsilon)\TPs_2 = 0. \]
Multiply the first equation by $\overline{\TPs_1}$ and the complex
conjugate of the second equation by $\TPs_2$, subtract and obtain
\[ - i (\Im q_1+\epsilon)|\TPs_1|^2 - i(\Im
q_2+\epsilon)|\TPs_2^2|^2
+ (\Re q_1 -k)|\TPs_1|^2 - (\Re q_2 -k)|\TPs_2|^2
+ i (\overline{\TPs_1}\TPs_2)' = 0. \]
Now on taking imaginary parts and integrating over $[0, X ] $, \[
\Re (\overline{\TPs_1 ( X ) }\TPs_2 ( X )) = \int_0^X \left\{ (\Im
q_1+\epsilon)|\TPs_1|^2 + (\Im q_2+\epsilon)|\TPs_2|^2 \right\} dx
+ \Re m ( k + i \epsilon ) . \]
This shows that the left hand side has a limit, finite or
infinite, when $ X \to \infty $. This limit must be zero, because
$ \TPs \in L^2 $, and therefore, $ \overline{\TPs_1 }\TPs_2 \in
L^1 $. The result follows. \hfill $\Box$
In order to use Proposition \ref{criterion} we shall need an
asymptotic of $ \( L^* - z \)^{ -1 } u $ when $ \epsilon \searrow
0 $ for compactly supported $ u $, given by the following lemma.
\begin{lemma}\label{lemma:6}
For any $ u \in L^2 ( \Rr_+ , \C^2 ) $ having compact support and
$z=k+i\epsilon$, $ \epsilon > 0 $, \be \begin{array}{c} \( L^* - z
\)^{ -1 } u = \beta_z [u] \TPs(x,z) + r ( x , z ) , \\ \\
{\displaystyle
\beta_z [u] = \frac{i}{m (z)-h} \int_0^\infty \TPs_h^T(s,z)Eu(s)ds }
, \end{array} \label{eq:2.5}
\ee where $ r ( x , z ) $ satisfies $ \limsup_{ \epsilon \searrow
0 } \len r ( \cdot , z ) \rin_H < \infty $ for all $ k $ such that
the finite boundary value $ m ( k ) = \lim_{ \epsilon \searrow 0 }
m ( k + i \epsilon )$ exists and $ m ( k ) \neq h $.
\end{lemma}
\noindent {\bf Proof}\, Let $ X < \infty $ be such that $ u $ is
supported on $ [ 0 , X ] $, and consider the expression
(\ref{resol}) for $ \( L^* - z \)^{ -1 } u $. Notice that the
first term in the curly brackets vanishes for $ x > X $, and
therefore $ r = \( L^* - z \)^{ -1 } u -\beta_z \TPs $ is
supported on $ [0,X] $ for any $ z \in \Cc_+ $. Then, the standard
perturbation theory for initial value problems implies that $
\TPs_h ( x, z ) $ and $ \TPs ( x, z ) $ converge to $ \TPs_h ( x,
k ) $ and $ \TPs ( x, k ) $ when $ \epsilon \searrow 0 $ uniformly
in $ x < X $ for all $ k $ such that the finite boundary value $ m
( k ) $ exists. Combining these facts, we obtain that $ r ( x , z
) $ converges when $ \von \searrow 0 $ uniformly in $ x < X $,
provided that the common factor in the right hand side of
(\ref{resol}) converges to a finite limit. The result follows.
\hfill $ \Box $
\refstepcounter{section}
\subsection*{\S 3. Proof of the main result}
Define the non-negative functions $r_1$ and $r_2$ by
\[ r_1(x)^2 := \Im q_1(x), \;\; r_2(x)^2 := \Im q_2(x) . \]
Then the operator $ \( \Im L \)^{ 1/2 } $ is given by
mulptiplication by the matrix \be R(x) = \left(\begin{array}{cc}
r_1(x) & 0
\\ 0 & r_2(x)
\end{array}\right).
\label{eq:Rdef} \ee
\begin{lemma} \label{theorem:2}
Let $k\in\Rr$ be such that $ m (k)$ exists finitely and is not
equal to $h$. If $R \TPs_h(\cdot,k) \not\in L^2(\Rr^+,\Cc^2)$,
then $
D(k+i\epsilon) \stackrel{s}{\rightarrow} 0 $ as $\epsilon \searrow 0 $.
\end{lemma}
\noindent {\bf Proof}\, Observe that the linear set \[ {\cal D} =
\left\{ u \in L^2(\Rr^+,\Cc^2) \, | \, u \mbox{ is compactly
supported and } \int_0^\infty \TPs_h^T (s,k) E R(s)u(s)ds = 0
\right\}
\] is dense in $ L^2(\Rr^+,\Cc^2)$ if $
R(\cdot)\TPs_h(\cdot,k)\not\in L^2(\Rr^+,\Cc^2)$. Since the
function $ D ( z ) $ is bounded in $ \C_+ $ by a remark after
Proposition \ref{criterion}, it suffices to prove \be\label{Dlim}
\lim_{\epsilon\rightarrow 0}D(k+i\epsilon)u = 0 \ee for $u\in
{\cal D}$. From Lemma \ref{lemma:6}, we have for any $u \in {\cal
D}$, with $ z = k + i\epsilon $, $$ D(k+i\epsilon)u = \epsilon^{
1/2 } \beta_z [ R u ] \TPs( \cdot ,z) + o ( 1 ) , \; \; \epsilon
\searrow 0 , $$ where the $ o $-symbol refers to the $ L^2 $-norm.
Now, Lemma \ref{lemma:3} shows that $ \epsilon^{ 1/2 } \len \TPs (
k + i \epsilon ) \rin_H $ is bounded above when $ \epsilon
\searrow 0 $. Since $ m ( k ) \neq h $, one can pass to the limit
$ \epsilon \searrow 0 $ in the expression (\ref{eq:2.5}) for $
\beta_z $ to find that $ \beta_{ k + i \epsilon } $ converges to a
multiple of $ \int_0^\infty \TPs_h^T(s,k)E R ( s ) u(s)ds $ which
is zero because $ u \in {\cal D} $. Combining these, we obtain
(\ref{Dlim}). \hfill $ \Box $
Note that the complement of the set of all $k\in \Rr$ for which $
m (k)$ exists and is not equal to $h$ is a set of (Lebesgue)
measure 0, by the uniqueness theorem for Nevanlinna functions.
Taking into account the criterion of Proposition \ref{criterion},
we arrive at the following.
\begin{corollary}\label{corollary:1}
If $ R\TPs_h(\cdot,k)\not\in L^2(\Rr^+,\Cc^2)$ for Lebesgue almost
all $k\in\Rr$, then\footnote{Recall that $ L^\prime $ denotes the
completely non-selfadjoint part of the operator $ L $.} $ H_{ac} (
L^\prime ) = \{ 0 \}$.
\end{corollary}
Thus, our main result will be established if we prove that
$R(\cdot)\TPs_h(\cdot,k)\not\in L^2(\Rr^+,\Cc^2)$ for a. a.
$k\in\Rr$ if $ \Im Q\notin L^1 . $ Before doing this, we would
like to indicate a simple argument showing that $ H_{ ac } $ is
trivial when $ \Im Q \notin L^1 $ for a class of potentials $ Q $.
Notice first that if $ m ( k ) $ exists finitely, then the
solution $ \TPs ( x , k ) = \lim_{ \epsilon \searrow 0 } \TPs (x,k
+ i \epsilon ) $ to $ \ell_{ Q^* } \Psi = k \Psi $ exists. As is
shown in the asymptotic theory of linear differential systems
\cite{kn:Eastham}, if the potential $ Q $ decays at infinity in a
sufficiently regular way the asymptotics of solutions of the
equation $ \ell_{ Q^* } \Psi = k \Psi $ can be calculated
explicitly. In particular, an application of the general theory in
the situation under consideration gives the following.
\begin{proposition} \label{wkbcond} \cite[Theorem
1.8.3]{kn:Eastham} Let $ \Re Q = 0 $, and suppose that $ Q \to 0
$, $ Q^\prime \in L^1 $ and $ Q \in L^2 $. Then for any $ k \in \R
$ there exist two solutions, $ \Psi_\pm $, of the equation $
\ell_{ Q^* } \Psi = k \Psi $ with the asymptotics of the form (the
WKB asymptotic) \be\label{wkb} \Psi_\pm \sim \exp \left[ \pm i \(
k x + \frac 12 \int^x \trace Q ( \xi ) d \xi \) \right]
\left(\begin{array}{c} 1
\\ \mp 1 \end{array}\right), \; \; x
\to \infty . \ee \end{proposition}
Assume that the asymptotics (\ref{wkb}) holds for a. e. $ k \in \R
$, and that $ \Im Q \notin L^1 $. Then, obviously, the solution $
\Psi_- $ is growing, $ | \Psi_- | \sim \exp \( \frac 12 \int^x
\trace ( \Im Q ) \) $, while $ \Psi_+ $ decays. Notice that if the
asymptotics holds for a given $ k $, and a solution $ v $ to $
\ell_{ Q^* } \Psi = k \Psi $ satisfies $ R v \in L^2 $, then $ v $
must be a multiple of $ \Psi_+ $, for
\begin{eqnarray*} \int^\infty \left\| R \Psi_- \right\|^2 dx \sim
\int^\infty \trace \( \Im Q ( x )\) \exp \( C \int^x \trace \( \Im
Q ( \xi ) \) \, d \xi \) dx = \\ \exp \( C \int^\infty \trace \(
\Im Q ( x ) \) \, dx \) = \infty . \end{eqnarray*} It follows that
if (\ref{wkb}) holds for a $ k \in \R $ such that $ m ( k ) $
exists and $ R \TPs_h ( \cdot , k ) \in L^2 $, then the solutions
$ \TPs_h ( \cdot , k ) $ and $ \TPs ( \cdot , k ) $ must coincide.
This is, however, only possible on the set of $ k $ such that $ m
( k ) = h $, which has zero measure. Therefore $ R \TPs_h ( \cdot
, k ) \notin L^2 $ for a. e. $ k $. Applying corollary
\ref{corollary:1}, we thus obtain the main result for potentials $
Q $ such that the asymptotics (\ref{wkb}) holds for a. e. $ k \in
\R $. Notice that although the conditions of Proposition
\ref{wkbcond} can be sharpened, the WKB asymptotics cannot be
justified for generic potentials, and the argument above has
nothing to do with the actual proof of Theorem 1. Moreover, even
in the case of Schr\"odinger operators it is well-known
\cite{Pearson} that the WKB asymptotic may fail for a. e. positive
value of the spectral parameter.
We shall establish Theorem 1 by successively reducing it to a
partial case which we now proceed to prove.
\begin{proposition}\label{theorem:weak1}
If $r_1r_2\notin L^1(\Rr^+)$ then $ R \TPs_h ( \cdot , k ) \notin
L^2 $ for a. e. $ k \in \R $.
\end{proposition}
\noindent{\bf Proof} \, We prove the result in the contrapositive
form. Suppose that $R\TPs_h (\cdot,k)\in L^2 $ for a certain real
$ k $ such that $ m ( k ) \neq h $. Multiply the Wronskian
identity $\det(\TPs_h(x,k),\TPs(x,k)) = m (k)-h \ne 0 $ by
$r_1r_2$ and integrate in $ x $ to obtain $$ \(\mbox{const.}\)
\int^N \hspace{-2mm} r_1(x)r_2(x) dx \leq
\int^N |r_1 \TPs_{h,1}|\cdot |r_2 \TPs_2| dx
+ \int^N |r_2\TPs_{h,2}|\cdot |r_1\TPs_1| dx . $$
Now $r_1
\TPs_{h,1} \in L^2(\Rr^+)$ and $r_2\TPs_{h,2}\in L^2(\Rr^+)$ by
choice of $k$, while $r_1\TPs_{1}\in L^2(\Rr^+)$ and
$r_2\TPs_{2}\in L^2(\Rr^+)$ by Lemma \ref{lemma:3}. By the Schwarz
inequality it follows that $r_1r_2\in L^1(\Rr^+)$. \hfill $\Box$
\vspace{2mm}
In particular, this proposition shows that if $r_1r_2\notin
L^1(\Rr^+)$, then $H_{ac}(L^\prime) = \{ 0 \}$ in view of
Corollary \ref{corollary:1}.
We would now like to reduce the problem under consideration to a
problem where only one of the entries of $ \Im Q $ is non-zero. We
start with a Gronwall-type lemma comparing the solutions $ \TPs $
and $ \TPs_h $ corresponding to different $ \Im Q $ at large $ x
$.
\begin{lemma} \label{lemma:5.2}
Let $k\in\Rr$ be such that $ m ( k ) $ exists finitely, and $ m (
k ) \neq h $. Let the function $$ \hQ (x) =
\left(\begin{array}{cc} \hat{ q_1 } (x) & 0
\\ 0 & \hat{q_2} (x)
\end{array}\right) $$ satisfy $ 0
\le \Im \hQ ( x ) \le \Im Q ( x ) $ and $ \Re \hQ ( x ) = \Re Q (
x ) $ for a.e. $ x $. If $R(\cdot)\TPs_h(\cdot,k)\in
L^2(\Rr^+,\Cc^2)$, then the equation $\ell_{\hQ^* }y = k y$ has
solutions $\hat{\Psi}^{(1)}$ and $\hat{\Psi}^{(2)}$ such that
\be
\left.
\begin{array}{rcl}
\hat{\Psi}^{(1)}(x,k) & = & \TPs_h(x,k) + o \( \left| \TPs_h(x,k)
\right| + \left| \TPs(x,k) \right|\) \\ \hat{\Psi}^{(2)}(x,k) & =
& \TPs(x,k) + o \( \left| \TPs_h(x,k) \right| + \left| \TPs(x,k)
\right|\)
\end{array}\right\} \;\; (x \rightarrow \infty).
\label{eq:5.5} \ee
\end{lemma}
\noindent{\bf Proof}\, Let the function $ B ( x ) \ge 0 $ be
defined by $ B^2 = \Im ( Q - \hQ ) $, and $ Y $ denote the $ 2
\times 2 $-matrix $(\TPs_h,\TPs)$. Write
$(\hat{\Psi}^{(1)},\hat{\Psi}^{(2)}) = Y Z $. Elementary
manipulations show that the matrix $Z$ must satisfy the
differential equation \be JYZ' = -i B^2 YZ. \label{eq:5.6} \ee
As the argument from the beginning of the proof of Lemma
\ref{lemma1} shows\footnote{This is the first time that we use in
an essential way the fact that we deal with diagonal $ \Im Q $.},
the determinant of the matrix $ Y $ is constant in $ x $, $ \det Y
= m ( k ) - h \ne 0 $. Now, let $ J_0 = \left(\begin{array}{cc} 0
& -1
\\ 1 & 0 \end{array}\right) $. A straightforward calculation
shows that
\[ Y^T J_0 Y = ( \det Y ) J_0 . \]
Thus, $(JY)^{-1} = J_0^{-1}Y^TJ_0J^{-1}/\det Y$ and hence from
(\ref{eq:5.6}), together with the fact that
$J_0J^{-1}=iE=i\,\mbox{diag}(1,-1)$,\be Z' = - \frac{i}{\det Y}
J_0^{-1}Y^TJ_0J^{-1}B^2 YZ = - \frac{1}{\det Y}J_0(BY)^TE(BY)Z.
\label{eq:5.7} \ee
It follows from Lemma \ref{lemma:3} that \[ \limsup_{ \epsilon
\searrow 0 } \int_0^\infty \len R ( x ) \TPs (x, z ) \rin^2 dx \]
is finite when $ m ( k ) $ exists finitely. Hence, by the Fatou
lemma $R\TPs(\cdot,k) \in L^2(\Rr^+,\Cc^2)$. Together with the
hypothesis of the current lemma this implies that $ R Y\in L^2 $,
and, moreover, $ B Y \in L^2 $, since $ B ( x ) \le R ( x ) $ a.e.
Thus, $(BY)^T E (BY) $ is summable, and by the Levinson theorem
(see, e.g., \cite[Theorem 1.3]{kn:Eastham}) it follows that
(\ref{eq:5.7}) possesses a solution $Z$ satisfying \be Z(x) = I +
o(1), \;\;\; x\rightarrow\infty \label{eq:5.8} , \ee and
(\ref{eq:5.5}) is proved. \hfill $\Box$
Let $ \hat{R} = \( \Im \hat{Q} \)^{ 1/2 } $. The asymptotic
(\ref{eq:5.5}) shows, in particular, that in the situation of the
lemma, $ \hat{R} \hat{\Psi} \in L^2 $ for any solution $ \hat \Psi
$ to $\ell_{\hQ^* }y = k y$. Applying Corollary \ref{corollary:1},
we now obtain the following assertion.
\begin{corollary}\label{lemma:5.7} Let $L_j$ denote the operators defined by the same boundary
conditions as $L$ but given by
\[ L_j y = \ell_{Q_j}y, \]
where
\[ Q_1(x) = \left(\begin{array}{cc} q_{1}(x) & 0 \\
0 & \Re q_{2}(x) \end{array}\right), \;\; Q_2(x) =
\left(\begin{array}{cc} \Re q_{1}(x) & 0 \\ 0 & q_{2}(x)
\end{array}\right) , \] and let $ R_j = \( \Im Q_j \)^{ 1/2 } $,
and $ \TPs_{h,j} ( x , k ) $ be the solutions to $\ell_{Q_j^*}y =
k y $ with the Cauchy data $ y ( 0 ) = \( 1 , h \)^T $. If at
least one of the operators $ L_j $ satisfies the condition $ R_j
\TPs_{h,j} ( \cdot , k ) \notin L^2 $ for a. e. $ k \in \R $ (and
so the corresponding subspace $ H_{ac} (L_j^\prime) $ is trivial),
then $ H_{ ac } ( L^\prime ) $ is trivial.
\end{corollary}
\noindent {\bf Proof of Theorem 1}\, In view of Corollary
\ref{lemma:5.7} it is sufficient to show that $ R \TPs_h ( \cdot ,
k ) \notin L^2 $ for a. e. $ k \in \R $ for the operator $ L $
corresponding to a potential $ Q $ such that $\Im(q_1)\equiv 0$
and $\Im(q_2)\notin L^1$. Since $\Im(q_2)\notin L^1$ we have $r_2
\notin L^2$ and so by the Uniform Boundedness Principle there
exists a function $d\in L^2$ such that $ r_2 d \notin L^1 $.
Obviously we can choose $d$ to be non-negative. Let $L_d $ denote
the operator with the potential
\[ Q_d = \left(\begin{array}{cc} \Re q_{1} + id^2 & 0 \\
0 & q_{2}(x) \end{array}\right) , \] and let $ R_d = \( \Im Q_d
\)^{ 1/2 } $, and $ \TPs_h^d ( x , k ) $ be the solution of the
differential equation $\ell_{Q^*_d}\TPs_h = z \TPs_h$ with the
initial condition $\TPs_h^d (0,z) = \left(\begin{array}{c} 1
\\ h \end{array}\right)$. Since
$r_2 d \notin L^1$, we have $ R_d \TPs_h^d ( \cdot , k ) \notin
L^2 $ for a. e. $ k \in \R $ by Proposition \ref{theorem:weak1},
and therefore $ H_{ ac } ( L_d^\prime ) = \{ 0 \} $. We now wish
to apply Lemma \ref{trace} with $\tL = L_d $ so that $ \Gamma $ is
given by multiplication by $ \mbox{diag} \( d^2 , 0 \) $. We only
have to check that $ \( L - z \)^{ -1 } - \( L_d - z \)^{ -1 } \in
\bS^1 $, since all the other assumptions of the Lemma are
satisfied trivially. Let $ A_0 $ be the operator corresponding to
the potential $ Q \equiv 0 $. The fact that $ d\in L^2$ and the
assumption that $ \Re Q $ is bounded imply that, in the notation
of Lemma \ref{trace}, the operator $ \sqrt \Gamma \( A_0 - z \)^{
-1 } $ is Hilbert-Schmidt for any non-real $ z $ since the
integral matrix kernel $K(x,s)$ of this operator satisfies $$ \len
K(x,s) \rin_{ \mbox{Mat}_2 ( \C ) } \leq C
d(x)\exp(-\epsilon|x-s|), \; \; \; \epsilon = | \Im z |
> 0,
$$ for all $x$ and $s$, where $C$ is a constant depending on $ z $
only. Applying the resolvent identity repeatedly, we now have (for
any non-real $ z \in \rho ( L_d ) \cap \rho ( L ) $) $$ \( L - z
\)^{ -1 } - \( L_d - z \)^{ -1 } = \Xi_1 ( z ) \( A_0 - z \)^{ -1
} \sqrt \Gamma \cdot \sqrt \Gamma \( A_0 - z \)^{ -1 } \Xi_2 ( z )
$$ where $ \Xi_{ 1,2 } ( z ) $ are bounded operators. This formula
shows that its left hand side is of the trace class. The result
follows. \hfill $\Box$
Apparently, Theorem 1 cannot be obtained in this way without a
condition on the behavior of $ \Re Q $ at infinity, which
guarantees that $ \( \Re L - z \)^{ -1 } - \( A_0 - z \)^{ -1 } $
is a bounded operator and makes it possible to apply the resolvent
identity to show that the difference of the resolvents of $ L $
and $ L_d $ is trace class.
\section*{Concluding Remarks}
Notice that the general matrix differential operator $ L $
corresponding to the differential expression $ \ell_V :=
J\frac{d}{dx} + V(x) $ with the same boundary conditions as above
and arbitrary bounded matrix potential $ V = \left(
\begin{array}{cc} v_{11} & v_{12} \\ v_{21} &
v_{22} \end{array} \right) $ is unitarily
equivalent to an operator with a diagonal potential $ Q $ if $ v_{
12 } = v_{ 21 } $ and is real. The equivalence is given by a
unitary gauge transformation $ U $ (see \cite{kn:LevitanSargsjan}) defined by
\[ U y = \exp\( i\int_0^x v_{12}(t)dt \) y . \]
Then $ U^* L U = L_Q $ where $ L_Q $ is operator corresponding to
the differential expression $ \ell_Q $ with
\[ Q(x) = \left(\begin{array}{cc} v_{11}(x) & 0 \\ 0 &
v_{22}(x) \end{array}\right).
\]
Since the operator $ U $ transforms the absolutely continuous
subspaces of $ L_Q $ into that of $ L $,
$H_{ ac } ( L ) = U H_{ac} ( L_Q ) $.
\bibliographystyle{plain}
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\end{document}
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