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absolutely continuous spectrum,  
 
dissipative Schrodinger operators, Jacobi matrices
---------------0407221539460
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\begin{document}

\title[On Instability of the Spectrum]{On Instability of the Absolutely Continuous Spectrum of
Dissipative Schr\"odinger Operators and Jacobi matrices}


\begin{abstract}
We prove the absence of the absolutely continuous spectrum for
dissipative Schrodinger operators and Jacobi matrices with slowly
decaying imaginary part of the potential.
\end{abstract}


\author{Roman Romanov}
\address{\hskip-\parindent
        School of Computer Science\\
        Cardiff University, Cardiff \\
        Queen's Buildings, PO Box 916\\
        Newport Road, Cardiff CF24 3XF\\
        UK}
\address{\hskip-\parindent
        Laboratory of Quantum Networks\\
        Institute for Physics \\
        Saint Petersburg State University \\
        198504 St. Petersburg \\
        RUSSIA}
\email{roma@rvr.stud.pu.ru , R.V.Romanov@cs.cf.ac.uk}


\thanks{This
work was supported in part by EPSRC Grant GR/ R20885 and RFBR 
Grant 00-01-00479} 

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\date{\today}

\maketitle

\section{Introduction}

This paper addresses the question of stability of the absolutely
continuous spectrum of one-dimensional Schr\"odinger operators
under slowly decaying perturbations. In the selfadjoint case the
absolutely continuous spectrum is known \cite{DK,Kiselev} to 
coincide with the positive real axis, the spectrum of the 
unperturbed operator, as long as the potential is $ L^2 $. In 
particular, this result means that the absolutely continuous 
spectrum preserves when the conventional scattering theory is not 
applicable anymore, which, in terms of the $ L^p $ scale, is the 
case when the potential is not $ L^1 $. Our aim here is to show 
that the picture is very different for non-selfadjoint 
Schr\"odinger operators. Namely, the following two theorems 
hold\footnote{See (\ref{Hac}) for the definition of the absolutely 
continuous subspace for the class of operators under 
consideration.}, 

\medskip
\smallskip

{\bf Theorem 1.} {\it Let $ \{ q_j \} $ be a sequence of complex 
numbers, $ \im q_j \ge 0 $, and $ \im q_j \to 0 $, and let $ l $ 
be the discrete Schr\"odinger operator in $ l^2 ( \N ) $ defined 
by the matrix $$ \begin{pmatrix} \ q_1 & 1 & 0 & \ldots \cr 1 & 
q_2 & 1 & \ddots \cr 0 & 1 & q_3 & \ddots \cr \vdots & \ddots & 
\ddots & \ddots \cr  
\end{pmatrix}. $$ If the absolutely continuous subspace of $ l $ 
is nontrivial, then $ \im q \in l^1 $.} 

\medskip
\smallskip

{\bf Theorem 2.} {\it Let $ q $ be a locally bounded function on
$ \R_+ $ with $ \im q \ge 0 $, $ \im q ( x ) \to 0 $ as $ x \to
\infty $. Assume that $ \re q $ is bounded below, $ \sup_n
\int_n^{ n+1 } \left| \re q \right|^2 dt $ is finite, and define
$ l $ to be the Schr\"odinger operator in $ L^2 ( \R_+ ) $ with
the potential $ q $, $ l u = - u'' + q u $, on the domain fixed
by
a self-adjoint boundary condition at zero. If the absolutely
continuous subspace of $ l $ is nontrivial, then $ \im q \in L^1
$.}

\medskip

The abstract local scattering theory for trace class perturbations 
of selfadjoint operators developed in \cite{N,Nab} shows that the 
absolutely continuous parts of arbitrary operators $ L, L_0 $ are 
quasi-similar by means of wave operators and their spectra 
coincide when $ L_0 $ is selfadjoint and $ L - L_0 $ is of trace 
class. In the situation under consideration this implies that the 
absolutely continuous spectrum of $ l $ coincides with that of the 
selfadjoint Schr\"odinger operator $ \re l $ when $ \im q \in L^1 
$ ($ \im q \in l^1 $ in the discrete case). Thus, the theorems 
above effectively say that the existence of the absolutely 
continuous spectrum for Schr\"odinger operators in the dissipative 
case is equivalent to having a scattering theory. To the best of 
author's knowledge, no non-trivial {\it negative} results on the 
absolutely continuous spectrum of non-selfadjoint differential 
operators with non-discrete spectrum have been obtained 
previously. 

Notice that although the absolutely continuous subspace of a 
dissipative operator have mostly been studied \cite{N,Sachn,Pavli} 
in terms of the Sz\"okefalvi-Nagy - Foias functional model, we  
avoid using the model explicitly. A model-oriented reader will, 
however, easily recognize the condition of isometricity a. e. on 
the real axis of the characteristic function of the operator under 
consideration in the basic criterion (\ref{crit}). 

The structure of the paper is as follows. In section 2 we define
the absolutely continuous subspace of a dissipative operator and 
provide necessary abstract preliminaries. Section 3 is devoted to 
the proof of theorem 1, section 4 to the proof of theorem 2. 

NOTATION:

$ \C_\pm = \{ z \colon \pm \im z > 0 \} $. 

$ H_{ac} ( L ) $ - the absolutely continuous subspace of an 
operator $ L $. The notation $ H_{ ac } $ without an argument is 
used when it is clear from the context which operator it refers 
to. 

$ \bS^1 $ - trace class.

$ T ( x , k ) $ - the transfer matrix for the differential 
equation $ -y'' + p y = k y $ on the semi-axis $ x \ge 0 $: $$ T ( 
x , k ) 
\begin{pmatrix} y ( 0 ) \cr y'( 0 ) 
\end{pmatrix} = \begin{pmatrix} y ( x ) \cr y'( x ) 
\end{pmatrix} $$ 
for any solution $ y $ to the differential equation. 

$ \len \cdot \rin_X = \len \cdot \rin_{ L^2 ( 0 , X ) } $. 

$ \sigma_{ ess } ( L ) $ - the essential spectrum of the operator 
$ L $. The expression $ \sigma_{ ess } ( L ) \subset \Omega $ 
means that the spectrum of $ L $ in the complement of the set $ 
\Omega $ consists of at most countably many eigenvalues of finite 
geometric multiplicity. 

Unless otherwise specified, the abbreviation a. e. refers to the 
Lebesgue measure. 
 
\section{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 $ \sigma_{ ess } ( L ) 
\subset \R $. It is also assumed throughout that the operator $ L 
$ is completely non-selfadjoint, that is, has no reducing 
subspaces on which it induces a self-adjoint operator. Define the 
Hardy classes of vector functions $ \H^2_\pm $ to be the 
collections of functions $f: \C_\pm \to H $ analytic in $ \C_\pm 
$, respectively, and satisfying $\sup_{ \von 
> 0 } \int_\R \len f ( k \pm i\von ) \rin^2 dk < \infty $.

\begin{definition}
{\it The absolutely continuous subspace}, $ H_{ac} ( L ) $, of the 
operator $ L $ is defined as follows \cite{N,Sachn} \be\la{Hac} 
H_{ac} ( L ) = \mbox{clos} \left. \begin{cases} u \in H: \; & 1) 
\( L - z \)^{ -1 } u \mbox{ is analytic in } \C_+ \cr & 2) \, \, 
V^{1/2} \( L - z \)^{ -1 } u \in \H^2_+ \cr 
\end{cases} \right\} . \ee 
\end{definition}

The absolutely continuous subspace is regular invariant \cite{N}, 
that is, $ \overline{( L - z )^{-1} H_{ac} } = H_{ac} $ for all $ 
z \in \rho (L) $. This implies that the restriction $ L|_{ H_{ac} 
} $ with the domain $ \cD (L) \cap H_{ac} $ is a closed operator 
in $ H_{ac} $. 

Various motivations of this definition and its relation to 
scattering theory are given in \cite{N,Nab,Sachn,Pavli,Pav1}. We 
only mention here the following "weak" characterization of the 
subspace $ H_{ac} ( L ) $ (it will not be used in the paper). 

\medskip
\smallskip

{\bf Theorem.}\cite{equival} $$ H_{ac} ( L ) = \mbox{clos } \left. 
\begin{cases} u \in H: \; & 1) \( L - z \)^{ -1 } u \mbox{ is 
analytic in } \C_+ \cr & 2) \left. \llangle \( L - z \)^{ -1 } u , 
v \rrangle \right|_{ \C_\pm } \in \H^2_\pm \mbox{ for all } v \in 
H \cr 
\end{cases} \right\}. $$

\medskip

This theorem shows, in particular, that (\ref{Hac}) can be 
considered as a generalization of the definition of the absolutely 
continuous subspace in the self-adjoint theory. 

The triviality of the subspace $ H_{ac} $ in the examples 
considered in this paper will be established on the basis of the 
following criterion which is implicit in \cite{Pav1}. 

\begin{proposition} \la{criterion}
$ H_{ac} ( L ) = \{ 0 \} $ iff for a. e. $ k \in \R $ we have ($ z 
= k + i \von $) \be\la{crit} D ( z ) \equiv \sqrt \von \( L^* - z 
\)^{ -1 } \sqrt V \stackrel{s} \longrightarrow 0 \ee as $ \von 
\downarrow 0 $. 
\end{proposition}

Notice that it suffices to verify the strong convergence on a 
dense set, since the function $ D ( z ) $ is bounded in $ \C_+ $, 
$ \| D ( z ) \| \le 1/2 $ (see \cite{Pav1} or the proof of lemma 
\ref{trace} below where a similar calculation is carried out). To 
make the exposition closed, we give here an elementary proof of 
the if part (the only if part is not used in this 
paper\footnote{The proof of the only if part \cite{Pav1} depends 
on the functional model for the operator $ L $.}). In the argument 
we shall use the following inequality from \cite{N}, 
\be\la{ineqL}\von \int_\R \len \( L^* - k - i \von \)^{ -1 } 
v\rin^2 dk \le \pi \len v \rin^2 , \ee which holds true of any 
maximal dissipative operator $ L $ and all $ v \in H $ and $ \von 
> 0 $. This inequality easily follows from the existence of a 
self-adjoint dilation \cite{Na} of the maximal dissipative 
operator $ -L^* $, see \cite{N} for details. 

\begin{proof}
Given a $ u \in H $, define the functions $ u_\pm ( z ) $, $ z \in 
\C_\pm $, to be the restrictions of the function $ \sqrt V \( L - 
z \)^{ -1 } u $, to $ \C_\pm $, respectively. According to a 
fundamental property of dissipative operators, $ u_- \in \H^2_- $ 
for all $ u \in H $ \cite[Theorem 1]{N}. Suppose that (\ref{crit}) 
holds, and $ u_+ \in \H^2_+ $ for a certain $ u \in H $. We shall 
show that $ u = 0 $. Define $ u_\pm ( k ) \in L^2 ( \R , H ) $ to 
be the boundary values of the functions $ u_\pm $ on the real 
axis. The required assertion will be established if we show that $ 
u_+ = u_- $ in $ L^2 ( \R , H ) $. For all $ z = k + i \von $, $ 
\von 
> 0 $, and $ e \in H $ we have
\bequnan \left| \llangle u_+ ( z ) - u_- ( \overline z ) , e
\rrangle \right| = 2 \von \left| \llangle \sqrt V \( L - \overline
z \)^{ -1 } \( L - z \)^{ -1 } u , e \rrangle \right| = \\ 2 \sqrt
\von \left| \llangle \( L - z \)^{ -1 } u , D ( z ) e \rrangle
\right| . \eequnan Then, for any compact interval $ \Delta \subset
\R $ \begin{eqnarray}\la{zeroone} \int_\Delta \left| \llangle u_+
( k + i \von ) - u_- ( k - i \von ) , e \rrangle \right| dk \le 2
\( \int_\Delta \len D ( k + i \von ) e \rin^2 dk \)^{1/2}
\nonumber
\\ \( \von \int_\Delta \len \( L - k - i \von \)^{ -1 } u \rin^2
dk \)^{ 1/2 }. \end{eqnarray} The first factor on the right hand 
side vanishes as $ \von \to 0 $ by the assumption and the Lebesgue 
theorem. Let us show that the second factor is bounded in $ \von 
$. An application of the resolvent identity gives $$ \sqrt \von 
\len \( L - z \)^{ -1 } u \rin \le \sqrt \von \len \( L^* - z \)^{ 
-1 } u \rin + \len D ( z ) \rin \cdot \len \sqrt V \( L - z \)^{ 
-1 } u \rin . $$ Integrating and taking into account the 
inequality (\ref{ineqL}) we find $$ \von \int_\R \len \( L - k - i 
\von \)^{ -1 } u \rin^2 dk \le 2\pi \len u \rin^2 + \int_\R \len 
\sqrt V \( L - k - i \von \)^{ -1 } u \rin^2 dk , $$ which is 
bounded by the assumption about $ u $. We obtain that the left 
hand side of (\ref{zeroone}) converges to $ 0 $ which implies that 
$ \langle u_+(k) - u_-(k) , e \rangle = 0 $ a. e. on $ \Delta $. 
From arbitrariness of $ \Delta $ and $ e $ we now conclude that $ 
u_+ = u_- $. \end{proof} 

In the case of trace class perturbations a local scattering theory 
exists for the pair $ ( L , A ) $. The following assertion is 
implicit in \cite{N,Nab}. 

\begin{theorem}
If $ V \in \bS^1 $ then $ \sigma \( L|_{ H_{ac} } \) = \sigma_{ac} 
( A ) $ where the absolutely continuous spectrum of the 
self-adjoint operator $ A $ is defined in the standard way. 
\end{theorem} 

{\it Proof (sketch).} According to the results in \cite{N,Pavli}, 
the assumption about $ V $ implies that to any Borel subset $ 
\omega \subset \R $ there corresponds an invariant subspace $ 
H_\omega $ of the operator $ L|_{ H_{ac} } $ such that $ \sigma \( 
L|_{ H_\omega } \) \subset \overline \omega $, and $ H_\omega $ is 
a maximal invariant subspace with this property. Moreover, if $ 
\omega = \cup_n \omega_n $, then $ \sigma \( \left. L\right|_{ 
H_\omega } \) = \overline{ \cup_n \sigma \( L|_{ H_{\omega_n} } \) 
} $; $ H_\R = H_{ac} $, and $ H_\omega $ is trivial if $ | \omega 
| = 0 $. It is shown in the course of the proof of theorem 5 in 
\cite{N} that there exists a scalar contractive analytic function, 
$ m $, in the upper half plane such that for any $ c > 0 $ the 
bounded local wave operators $ W_\pm^\omega ( A , L )\colon 
H_\omega \to H $ for the set $ \omega = \{ k \in \R : | m ( k ) | 
\ge c \} $ exist and are complete in the sense that $ \Ran 
W_\pm^\omega ( A , L ) = P_\omega H $. Here $ P_\omega $ is the 
spectral projection of the a.c. part of $ A $ corresponding to the 
set $ \omega $. This means that $ L|_{ H_{\omega} } $ and $ 
A_\omega = \left. A \right|_{ \Ran P_\omega } $ are similar, and 
thus $ \sigma \( L|_{ H_\omega } \) = \sigma_{ac } ( A_\omega ) $. 
Now, considering the exhausting sequence $ \{ \omega_n \} $ of the 
sets defined as above with $ c = 1/n $ and taking into account 
that the set $ \{ k : \, m ( k ) = 0 \} $ has Lebesgue measure 
zero, we conclude from the additivity property that $ \sigma \( 
L|_{ H_{ac} } \) = \sigma_{ac } ( A ) $. \hfill $ \Box $ 

Notice that the statement of the theorem holds if the weaker 
condition $ \( L - z \)^{ -1 } - \( A - z \)^{ -1 } \in \bS^1 $ is 
satisfied. The set $ \sigma \( L|_{ H_{ac} } \) $ is referred to 
as {\it the absolutely continuous spectrum of the operator} $ L $. 

In general, 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. The original work of this paper starts here. 

\begin{lemma} \la{trace} Let $ L $ and $ \tL $ be two dissipative
operators such that $ \tL - L = i \Gamma $ with a $ \Gamma \in 
\bS^1 $. Assume that $ \Gamma \ge 0 $ and there exists a bounded 
operator $ T \ge 0 $ such that $ \Gamma = \tV T $, $ \tV = \im \tL 
$. If $ H_{ac} ( \tL ) = \{ 0 \} $, then $ H_{ac} ( L ) = \{ 0 \} 
$. 
\end{lemma}

\begin{proof} 
Applying the criterion of proposition \ref{criterion} to $ \tL $ 
and 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 contractive function in $ \C_+ $, 
\begin{eqnarray} \la{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 \ge 0 .
\end{eqnarray}
Notice that the condition on $ \Gamma $ implies that $ \sqrt 
\Gamma \( \tL^* - z \)^{ -1 } \sqrt \Gamma \in \bS^1 $. Thus, $ G 
( z ) $ is a contractive analytic operator-function in $ \C_+ $ 
such that $ I - G ( z ) \in \bS^1 $. Moreover, 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. Hence, by the 
Szokefalvi-Nagy theorem \cite{Na}, the function $ G ( z ) $ admits 
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 $. By the Fatou theorem \cite{Na} it follows 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\la{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 
(\ref{contr}) as applied to the dissipative operator $ - \tL^* $, 
using the fact that the left hand side is not greater than unity. 
Finally, 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. 
\end{proof}

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 $ satisfying the 
assumptions of this lemma. As the {\it proof} of the lemma shows, 
the final result then does not depend on the "only if" part of the 
criterion of proposition \ref{criterion}. 

\section{Discrete Schr\"odinger Operator}

Let $ \{ q_n \} $ be an arbitrary sequence of complex numbers such 
that $ 0 \le \im q_n \le C $, $ C < \infty $. Define $ \Theta ( z 
) $ and $ \Phi ( z ) $, $ z \in \C $, to be the solutions of the 
difference equation \be\la{diffeq} u_{ n+1 } + u_{ n-1 } + q_n u_n 
= z u_n \ee for $ n 
> 1 $ with the initial conditions $ \Theta_1 = 0 $, $ \Theta_2 = 
-1 $, $ \Phi_1 = 1 $, $ \Phi_2 = z - q_1 $. 

The discrete Schr\"odinger operator, $ l $, acts in the Hilbert 
space $ H = l^2 ( \N ) $ and has the form \begin{eqnarray*} & l = 
S + S^* + Q , & \\ & \( S u \)_n = u_{ n+1 } , \; \( Q u \)_n = 
q_n u_n . & 
\end{eqnarray*} This formula defines $ l $ on the domain of the multiplication 
operator $ Q $ as a maximal dissipative operator which is easily 
seen to be completely nonself-adjoint if $ \im q \not\equiv 0 $. 
For $ z \in \C_- $ let $ u ( z ) = \( l - z \)^{ -1 } \delta_1 $ 
where $ H \ni \delta_1 = ( 1, 0, \dots ) $, so $ u ( z ) $ is an $ 
l^2 $ - solution to (\ref{diffeq}) for $ n 
> 1 $. The function $ u $ satisfies $$ u ( z ) = - \Theta ( z ) + 
m ( z ) \Phi ( z ) $$ where $ m ( z ) \stackrel{def} = \llangle \( 
l - z \)^{ -1 } \delta_1 , \delta_1 \rrangle $. Obviously, $ - m ( 
z ) $ is a Nevanlinna function in $ \C_- $ ($ \im m ( z ) \le 0 
$), and therefore the finite boundary values, $ m ( k ) = 
\mathop{\lim} \limits_{\von \downarrow 0} m ( k - i \von ) $, 
exist for almost all $ k \in \R $. At such $ k $'s, the solution $ 
u ( k ) $, $ u_j ( k ) = \mathop{\lim} \limits_{\von \downarrow 0} 
u_j ( k - i \von ) $, to (\ref{diffeq}) with $ z = k $ is defined. 

\begin{lemma}\la{dsobounds}
If the finite boundary value $ m ( k ) $ exists, then the sum $ 
\sum_j \im q_j \left| u_j ( k ) \right|^2 $ is finite, and 
\be\la{supsub} \mathop{\limsup} \limits_{ \von \downarrow 0 } \( 
\von \len u ( k - i \von ) \rin^2 \) \le -\im m ( k ) .\ee 
\end{lemma}

\begin{proof}
Both assertions result immediately from the identity ($ z = k - i 
\von $, $ \von > 0 $) $$ \sum_{ j = 1 }^\infty \im q_j \left| u_j 
( z ) \right|^2 + \von \sum_{ j = 1 }^\infty \left| u_j ( z ) 
\right|^2 = - \im m ( z ) $$ since each term on the left hand side 
is non-negative. To verify this identity we notice that the left 
hand side equals to $$ \im \langle ( l - z ) u ( z ) , u ( z ) 
\rangle = \im \langle \delta_1 , u ( z ) \rangle = \im \overline{ 
m ( z ) } = - \im m ( z ) . $$ 
\end{proof}

Our aim now is to apply the criterion from proposition 
\ref{criterion}. For a function $ \psi $ analytic in $ \C_- $ 
introduce the notation $ \wt \psi ( z ) = \overline { \psi \( 
\overline z \) } $. Let $ \alpha_j = \sqrt { \im q_j } $, $ \alpha 
= \mbox{diag} \{ \alpha_j \} $, and let $ D ( z ) $, $ z = k + i 
\von $, be the operator defined in (\ref{crit}) with $ L = l $. 

\begin{lemma} \la{notinl2}
Let $ k \in \R $ be such that $ m ( k ) $ exists finitely. If $ \{ 
\alpha_j \Phi_j ( k ) \} \notin l^2 $, then $ D ( k + i \von ) 
\mathop{\stackrel{s} \longrightarrow} \limits_{\von \downarrow 0 } 
0 $. 
\end{lemma}

\begin{proof}
We shall check the required strong limit on a certain dense set of 
compactly supported vectors. Let us consider the action of the 
resolvent of $ l^* $ on such vectors. A straightforward 
calculation shows that for any compactly supported $ f $ ($ z = k 
+ i \von $, $ \von > 0 $), $$ \left[ \( l^* - z \)^{ -1 } f 
\right]_j = {\wt \Phi}_j ( z ) \sum_{ n = j }^\infty { \wt u }_n ( 
z ) f_n + { \wt u }_j ( z ) \sum_{ n = 1 }^{ j-1 } {\wt \Phi}_n ( 
z ) f_n . $$ The first term on the right hand side vanishes for $ 
j $'s to the right of the support of $ f $. If $ k $ is such that 
$ m ( k ) $ exists finitely, then $ {\wt u }_n ( z ) $ converges 
when $ \von \to 0 $ for each $ n $, and one concludes that both 
terms on the right hand side are bounded when $ \von \to 0 $ 
uniformly in $ j $ ranging over the support of $ f $. Thus, for 
any such $ k $, we have 
\begin{eqnarray} \la{resD} \( l^* - z \)^{ -1 } f = c_\von [f] {\wt u 
} ( z ) + r ( \von ) , \\ c_\von [f] = \sum_{ n=1 }^\infty 
{\wt{\Phi}}_n ( k + i \von ) f_n , \nonumber 
\end{eqnarray} where $ r ( \von ) $ vanishes to the right of the support 
of $ f $ and is bounded uniformly and, moreover, in the $ l^2 
$-norm, when $ \von \to 0 $. On substituting (\ref{resD}) in the 
definition of $ D ( z ) $ and taking into account that $ \von^{ 
1/2 } \len \wt u ( z ) \rin $ is bounded when $ \von \to 0 $ by
(\ref{supsub}), we find that for any compactly supported $ v $ $$ 
\len D ( z ) v \rin \le C \left| c_\von [ \alpha v ] \right| + o ( 
1 ) , \; \; \von \to 0 . $$ 

Observe now that the linear set $ \cD = \left\{ v \in l^2 ( \N ) : 
v \text{ is compactly supported, and} \right. $ $ \left. \sum_j 
v_j \alpha_j {\wt \Phi}_j ( k ) = 0 \right\} $ is dense in $ l^2 $ 
when $ \{ \alpha_j \Phi_j ( k ) \} \notin l^2 $. Hence, it is 
enough to check that $ D ( k + i \von ) v \to 0 $ for $ v \in \cD 
$. Indeed, $ c_\von [ \alpha v ] \to \sum_n \alpha_n {\wt{\Phi}}_n 
( k ) v_n = 0 $ as $ \von \to 0 $ for $ v \in \cD $, and the 
required limit follows. 
\end{proof}

\begin{theorem}
\la{discrete} Let $ \{ q_j \} $ be a sequence of complex numbers 
such that $ 0 \le \im q_n \le C $, $ C < \infty $, and $ \sigma_{ 
ess } ( l ) \subset \R $. If $ \im q \notin l^1 $, then $ H_{ac} = 
\{ 0 \} $. 
\end{theorem}

\begin{proof}

From the previous lemma and proposition \ref{criterion}, it is 
enough to show that if $ \im q \notin l^1 $ then $ \sum_j \im q_j 
\left| \Phi_j ( k ) \right|^2 = \infty $ for a. e. $ k \in \R $. 
For $ k $ such that $ m ( k ) $ exists finitely consider the 
Wronskian of the solutions $ \Phi $ and $ u $, \be\la{Wron} 1 = W 
[ \Phi ( k ) , u ( k ) ] \equiv \Phi_{ j-1 } ( k ) u_j ( k ) - 
\Phi_j ( k ) u_{ j-1 } ( k ) , \; j 
> 1 . \ee Multiplying this by $ \alpha_j \alpha_{ j -1 } $ and 
summing over $ j $, we have, \bequnan \sum^n \alpha_j \alpha_{ j - 
1 } \le \sum^n \alpha_{j-1}\left| \Phi_{ j-1 } ( k ) \right| \cdot 
\alpha_j \left| u_j ( k ) \right| + \sum^n \alpha_j \left| \Phi_j 
( k ) \right| \cdot \alpha_{ j-1 } \left| u_{ j-1 }( k ) \right| 
\le 
\\ 2 \( \sum^n \im q_j \left| \Phi_j ( k ) \right|^2 \)^{1/2} \(
\sum^n \im q_j \left| u_j ( k ) \right|^2 \)^{1/2} . \eequnan By 
lemma \ref{dsobounds} the last factor on the right hand side is 
bounded in $ n $. We thus obtain that if $ \sum_j \alpha_j 
\alpha_{ j - 1 } = \infty $ then $ \sum_j \im q_j \left| \Phi_j ( 
k ) \right|^2 = \infty $ for a. e. $ k \in \R $, as required. 

The general case is reduced to the one just considered, since if $ 
\im q \notin l^1 $, then there exists a nonnegative $ d \in l^1 $ 
such that $ \wt q = q + id $ satisfies $ \sum_j {\wt \alpha }_j 
{\wt \alpha }_{ j-1 } = \infty $ with $ {\wt \alpha}_j = \( \im { 
\wt q }_j \)^{ 1/2 } $. Indeed, by the uniform boundedness 
principle, the assumption that $ \alpha \notin l^2 $ implies that 
there exists a $ \rho \in l^2 $ such that $ \sum_j \rho_j \alpha_j 
= \infty $, and, obviously, the $ \rho $ can be chosen to be 
nonnegative. Now one can set $ d_j = \rho_{ j-1 }^2 $, for $ {\wt 
\alpha }_j {\wt \alpha }_{ j-1 } \ge \rho_{ j-1 } \alpha_{ j-1 } 
$. An application of lemma \ref{trace} completes the proof. 
\end{proof}

Since $ \sigma_{ ess } ( l ) \subset \R $ when $ \im q_n \to 0 $, 
this theorem contains theorem 1 from the Introduction. Considering 
that in the case $ \im q \in l^1 $ the imaginary part of $ l $ is 
of the trace class, and therefore $ \sigma \( \left. l \right|_{ 
H_{ac} } \) = \sigma_{ ac } ( \re l ) $, it gives a complete 
description of the a. c. spectrum of dissipative discrete 
Schr\"odinger operator. 


The above proof extends straightforwardly to Jacobi matrices with 
selfadjoint non-diagonal part. Namely, given a bounded sequence, $ 
\{ \rho_n \} $, of positive numbers, define the Jacobi matrix $ J 
$ by $$ J = \Lambda S + S^* \Lambda + Q , $$ where $ \( \Lambda u 
\)_n = \rho_n u_n $. Then $ J $ is a maximal dissipative operator, 
and the corresponding generalization of theorem \ref{discrete} 
reads as follows. 

\begin{theorem}
\la{Jacobi} Let $ \{ q_j \} $ be a sequence of complex numbers 
such that $ 0 \le \im q_n \le C $, $ C < \infty $, and $ \sigma_{ 
ess } ( J ) \subset \R $. If $ \im q \notin l^1 $, then $ H_{ac} ( 
J ) = \{ 0 \} $. 
\end{theorem}

Indeed, assuming without loss of generality that $ \rho_1 = 1 $, 
one can define the functions $ \Theta $, $ \Phi $ to be the 
solutions of the equation $ \rho_{ n+ 1 } u_{ n+1 } + \rho_n u_{ 
n-1 } + q_n u_n = z u_n $ with the same boundary conditions as in 
the $ \rho \equiv 1 $ case, and let $ u ( z ) = \( J - z \)^{ -1 } 
\delta_1 $. The proof then proceeds verbatim as in the case $ \rho 
\equiv 1 $ until (\ref{Wron}), where we will have $$ W_j [ \Phi , 
u ] \equiv \Phi_{ j-1 } ( k ) u_j ( k ) - \Phi_j ( k ) u_{ j-1 } ( 
k ) = \rho_{ j-1 }^{ -1 } $$ instead. Obviously, the left hand 
side is bounded away from zero, since the $ \rho_j $'s are 
bounded, and the rest of the proof is the same as above. 

\section{Continuous case}

{\bf $ \S 1 $. Preliminaries and the beginning of the proof of the 
main result.} Let $ q $ be a locally bounded function on $ \R_+ $ 
satisfying $ \im q ( x ) \ge 0 $ a. e. and $ h $ be an arbitrary 
real number. We shall assume throughout that $ \im q \in L^\infty 
( \R_+ ) $, and $ \re q $ is bounded below. Define $ \Theta ( x , 
z ) $ and $ \Phi ( x , z ) $ to be the solutions to \be\la{eqn} - 
u'' + \overline q u = z u , \; z \in \C , \ee with the Cauchy data 
$ \Phi ( 0 , z ) = \Theta^\prime ( 0 , z ) = 0 $, $ \Phi^\prime ( 
0 , z ) = \Theta ( 0 , z ) = 1 $. It is standard that the 
solutions $ \Theta $ and $ \Phi $ are continuous in $ z $ 
uniformly in $ x \in I $ for any compact interval $ I $. 

The Schr\"odinger operator $ l $, $ l y = - y'' + q y $, is 
defined in the Hilbert space $ H = L^2 ( \R_+ ) $ on the domain of 
its real part, the Schr\"odinger operator with the potential $ \re 
q $, fixed by the self-adjoint boundary condition $ y'(0) = h y 
(0) $. With this definition, $ l $ is a completely nonself-adjoint 
maximal dissipative operator, provided that $ \im q \not\equiv 0 $ 
\cite{Sachn}. 

The following lemma is a continuous analog of lemma 
\ref{dsobounds}. 

\begin{lemma}\la{conbounds}
For all $ z \in \C_+ $ there exists exactly one linear independent 
$ L^2 $-solution, $ u ( x , z ) $, to (\ref{eqn}), which can be 
chosen in the form \be\la{ThmPh} u( x , z ) = \Theta ( x , z ) + m 
( z ) \Phi ( x , z ) \ee where $ m $ is an analytic function in $ 
\C_+ $. For almost all $ k \in \R $ the function $ m $ has finite 
boundary values, $ m ( k ) = \mathop{\lim} \limits_{\von 
\downarrow 0} m ( k + i \von ) $, and the solution $ u ( x , k ) = 
\mathop{\lim} \limits_{\von \downarrow 0} u ( x , k + i \von ) $ 
satisfies \be\la{imqi} \int \im q ( x ) \left| u ( x, k ) 
\right|^2 dx < \infty . \ee Moreover, for any $ \von > 0 $ 
\be\la{immid} \im m ( z ) = \int_0^\infty \( \im q ( t ) + \von \) 
\left| u ( t, z ) \right|^2 dt , \; z = k + i \von . \ee In 
particular, \be \la{supsubcon} \mathop{\limsup} \limits_{ \von 
\downarrow 0 } \( \von \len u ( \cdot, k + i \von ) \rin^2 \) \le 
\im m ( k ) \ee whenever $ m ( k ) $ exists. 
\end{lemma} 

\begin{proof}
As is well-known, see e.g. \cite{Pearson}, the assumption that $ 
\re q $ is bounded below implies that the differential expression 
corresponding to the potential $ \re q $ is in the limit-point 
case at infinity, that is, the minimal operator $ \wt l = \left. 
\re l \right|_{ C_0^\infty } $ has deficiency indices $ (1,1) $. 
It follows that the operator $ l_0 = \wt l + i \im q $ satisfies $ 
\dim \ker \( l_0^* - z \) = 1 $ for $ \im z $ large enough, and 
therefore for all $ z \in \C_+ $, because $ \C_- \subset \wh \rho 
( l_0 ) $. Thus, (\ref{eqn}) has exactly one linear independent $ 
L^2 $-solution for all $ z \in \C_+ $. 

Assume now that the representation (\ref{ThmPh}) for the solution 
holds\footnote{Once the uniqueness of an $ L^2 $ - solution is 
established, the representation (\ref{ThmPh}) and the identity 
(\ref{immid}) are results of the nesting circles analysis of the 
equation (\ref{eqn}) carried out in \cite{Sims}. For completeness, 
we provide here a simple derivation of these facts, except for the 
analyticity of $ m $.}. Consider the identity, \be\la{byparts} 0 = 
\int_0^x \( - u'' + ( \overline q - z ) u \) \overline u \, dt = - 
u' (x) \overline { u ( x ) } + m ( z ) + \int_0^x \( \overline q - 
z \) \left| u \right|^2 dt + \int_0^x \left| u' \right|^2 dt . \ee 
Since $ \re q $ is bounded below, it shows that the function $ 
\left| u' \overline u \right| $ has a limit, finite or infinite, 
when $ x \to + \infty $. On taking the real part, we find $$ \frac 
12 \frac d{dx} \left| u ( x ) \right|^2 = \mbox{const} + \int_0^x 
\re ( q - z ) \left| u \right|^2 dt + \int_0^x \left| u' \right|^2 
dt , $$ which implies that $ u^\prime \in L^2 $ (otherwise $ 
\left| u ( x ) \right| \to \infty $ as $ x \to + \infty $). This 
yields that $ u' (x) \overline { u ( x ) } \to 0 $, $ x \to + 
\infty $. On taking the imaginary part of (\ref{byparts}) and 
passing to the limit $ x \to + \infty $, we obtain (\ref{immid}). 
It shows that $ m ( z ) $ is a Nevanlinna function in $ \C_+ $ ($ 
\im m ( z ) \ge 0 $), and therefore has finite boundary values for 
a. e. $ k $ on the real axis, and for any such $ k $ the integral 
$ \int^x \im q ( t ) \left| u ( t, k ) \right|^2 dt $ is bounded 
in $ x $ (by $ \im m( k ) $). It follows that the integral in 
(\ref{imqi}) is finite. 

To establish the representation (\ref{ThmPh}) it is enough to 
notice that $ \Phi ( x , z ) $ cannot be the $ L^2 $ solution, for 
otherwise a calculation similar to the above would lead to the 
formula $ \int_0^\infty \( \im q + \von \) \left| \Phi \right|^2 
dt = 0 $ implying that $ \Phi \equiv 0 $. The analyticity of the 
function $ m $ was established in \cite{Sims}. 
\end{proof}

We shall often suppress the argument $ x $ in the notation and 
write $ u ( z ) $ for the element of the Hilbert space $ H $ 
defined by the function $ u ( x , z ) $ for each $ z \in \C_+ $. 

From now on, we shall assume that $ \im q ( x ) \to 0 $ as $ x \to 
\infty $, which guarantees that $ \sigma_{ ess } ( l ) \subset \R 
$ by a relative compactness argument. Then, $ l $ is an operator 
of the class considered in section 2. 

In the following lemma $ D ( z ) $ is defined by (\ref{crit}) with 
$ L = l $. Let $ u_h = \Theta + h \Phi $. 

\begin{lemma} \la{notinl2cont}
Let $ k \in \R $ be such that the finite boundary value $ m ( k ) 
$ exists, and $ m ( k ) \neq h $. Then 

(i) for any compactly supported $ v \in H $ ($ z = k + i \von $) 
\be \la{resDcont} D ( z ) v = \sqrt \von \, C_\von [ v ] u ( z ) + 
o ( 1 ) , \; \; \von \to 0 , \ee where \be \la{C0} C_\von [ v ] = 
\frac 1{ h - m ( k + i \von ) }\int_0^\infty u_h ( x, k + i \von ) 
\sqrt{ \im q ( x ) } v ( x ) dx , \ee and the $ o $-symbol refers 
to the $ L^2 $ - norm. 

(ii) If $ \sqrt { \im q } \, u_h ( \cdot , k ) \notin L^2 $, then 
$ D ( k + i \von ) \mathop{\stackrel{s} \longrightarrow} 
\limits_{\von \downarrow 0 } 0 $. 
\end{lemma}

{\it Proof} of this lemma is completely analogous to that of lemma 
\ref{notinl2}, so we just sketch it. Let $ f \in H $ be supported 
on a finite interval $ [ 0 , X ] $. It is straightforward to check 
that for all $ z \in \C_+ $ 
\begin{eqnarray}\la{resol} \( \( l^* - z \)^{ -1 } f \) (x) = & & 
\nonumber \\ & \frac 1{ h - m ( z ) } \left[ u ( x , z ) \int_0^x 
u_h ( s , z ) f ( s ) ds + u_h ( x , z ) \int_x^\infty u ( s , z ) 
f ( s ) ds \right] . & 
\end{eqnarray} Obviously, the function in the square brackets,
which will be denoted by $ g $, coincides with $$ \( \int_0^\infty 
u_h ( s, z ) f ( s ) ds \) u ( x , z ) $$ for $ x > X $. Moreover, 
$ g ( x , z ) $ converges when $ \von \to 0 $ uniformly in $ x \in 
[ 0 , X ] $. On taking into account that the condition $ m ( k ) 
\neq h $ secures the boundedness of the common factor in 
(\ref{resol}), this gives $$ \( l^* - z \)^{ -1 } f = \( \frac 1{ 
h - m ( z ) }\int_0^\infty u_h ( s, z ) f ( s ) ds \) u ( z ) + r 
( z ) $$ where the function $ r ( x , z ) $ vanishes for $ x 
> X $ and is uniformly bounded when $ \von \to 0 $. This implies $ 
(i) $. 

Notice that, as (\ref{supsubcon}) shows, $ \von^{ 1/2 } \len u ( k 
+ i \von ) \rin $ is bounded when $ \von \to 0 $, so that $ \len D 
( z ) v \rin \le C \left| C_\von [ v ] \right| + o ( 1 ) $ for any 
compactly supported $ v $. Now, define $ C_0 [ v ] $ by the 
formula (\ref{C0}) with $ \von = 0 $. If $ v $ satisfies $ C_0 [ v 
] = 0 $, then $ C_\von \to 0 $, and thus $ D ( z ) v \to 0 $. 
Since the set of compactly supported $ v $ such that $ C_0 [v] = 0 
$ is dense in $ L^2 $ when $ \sqrt{ \im q } u_h ( \cdot , k ) 
\notin L^2 $, $ (ii) $ follows. \hfill $ \Box $ 

Since the set of $ k $ such that $ m ( k ) = h $ has measure zero 
by the uniqueness theorem for Nevanlinna functions, we have the 
following

\begin{corollary} \la{absHac} If $ \sqrt { \im q } \, u_h ( 
\cdot , k ) \notin L^2 $ for a. e. $ k \in \R $, then $ H_{ac} ( L 
) = \{ 0 \} $. 
\end{corollary} 

Before proving the main result (theorem 2 in the Introduction) we 
would like to give a motivation of it in terms of the WKB 
asymptotics for the solutions. As is shown in the asymptotic 
theory of ordinary differential equations \cite{Eastham}, for a 
class of potentials $ q $ decaying at infinity in a sufficiently 
regular way the asymptotics of solutions of the equation 
(\ref{eqn}) at $ x \to \infty $ can be calculated explicitly. In 
particular, the theory asserts the existence of two solutions, $ 
u_\pm $, of the equation $ - u'' + \overline q u = k u $ for $ k 
> 0 $ with the asymptotics of the form (the WKB asymptotic) 
\be\la{wkb} u_\pm \sim \exp \left[ \pm i \( \sqrt k x - \frac 1{ 2 
\sqrt k } \int^x \overline{ q ( \xi )} d \xi \) \right] , \; \; x 
\to \infty , \ee if $ q $ is decaying and satisfies various 
additional assumptions. Obviously, for $ \im q \notin L^1 $ the 
solution $ u_- $ is growing, $ | u_- | \sim \exp \( \frac 1{ 2 
\sqrt k } \int^x \im q \, d \xi \) $, while $ u_+ $ decays. Notice 
that if the WKB asymptotics holds for a given $ k 
> 0 $, and $ v $, a solution to (\ref{eqn}) with $ z = k 
> 0 $, satisfies $ \sqrt { \im q } v \in L^2 $, then it must be a 
multiple of $ u_+ $, for \bequnan \int^\infty \im q \left| u_- 
\right|^2 dx \sim \int^\infty \im q ( x ) \exp \( C \int^x \im q ( 
\xi ) d \xi \) dx = \\ \exp \( C \int^\infty \im q ( x ) dx \) = 
\infty . \eequnan Let us call the function $ q $ {\it a WKB-class 
potential} if the asymptotic (\ref{wkb}) holds for a. e. $ k > 0 
$, and $ \sigma_{ ess } ( l ) = \R_+ $. It now follows that if $ q 
$ is a WKB-class potential, then the solutions $ u_h ( \cdot , k ) 
$ and $ u ( \cdot , k ) $ must coincide for a. e. $ k > 0 $ 
whenever $ m ( k ) $ exists and $ \sqrt { \im q } \, u_h ( \cdot , 
k ) \in L^2 $. This is, however, only possible on the set of $ k $ 
such that $ m ( k ) = h $, hence $ \sqrt { \im q } \, u_h ( \cdot 
, k ) \notin L^2 $ for a. e. $ k > 0 $. Applying corollary 
\ref{absHac}, we thus obtain theorem 2 for WKB-class potentials. A 
simple sufficient condition for a function $ q $ to be a WKB-class 
potential is $ q ( x ) \to 0 $, $ q \in L^2 $ and $ q' \in L^1 $ 
\cite{Eastham}, which in fact secures the asymptotics (\ref{wkb}) 
for all $ k > 0 $. In particular, this shows immediately that the 
operator with the potential $ q ( x ) = i x^{ -\beta } $, $ 0 < 
\beta \le 1 $, has trivial absolutely continuous subspace. Notice 
that the eigenvalues of this operator only accumulate at $ 0 $ by 
a complex scaling argument (a remark due to S. Naboko). 

We now proceed to prove theorem 2. The argument becomes less 
direct than in the discrete case since estimating the Wronskian of 
solutions $ u $ and $ u_h $ requires a bound on their derivatives 
which is only available "in the mean" and leads to some mild 
conditions on the real part of the potential. The proof involves 
spectral averaging for the self-adjoint problem corresponding to 
the potential $ \re q $ \cite{SimonLast} and the subordinacy 
theory \cite{Pearson}. The required part of the Simon-Last result 
reads as follows. 

\begin{theorem} \la{SimLast} \cite[Theorem 2.1C.]{SimonLast} Let $ V $
be a real potential locally bounded and bounded below, and let $ 
d\mu = \min \{ d\rho^D , d \rho^N \} $ where $ d\rho^{D/N} $ are 
the spectral measures for the self-adjoint Schr\"odinger operators 
with the potential $ V $ and the Dirichlet/Neumann boundary 
conditions at zero, respectively. Let $ \Theta $ and $ \Phi $ be 
the solutions to $ -y'' + V y = k y $ with the Cauchy data $ \Phi 
( 0 , k ) = \Theta^\prime ( 0 , k ) = 0 $, $ \Phi^\prime ( 0 , k ) 
= \Theta ( 0 , k ) = 1 $. Then for any compact interval $ I 
\subset \R $ there exist a constant $ C ( I ) $ such that for all 
$x$ \be\la{sl1} \int_I \( |\Theta (x,k)|^2 +| \Phi (x,k) |^2 \) 
d\mu (k) \leq C ( I ) , \ee \be\la{sl2} \int_I 
\left[\int^{x+1}_{x-1} \( |\Theta^\prime (y,k)|^2 +| \Phi^\prime 
(y,k) |^2 \) dy \right] d\mu (k) \leq C (I). \ee If additionally $ 
\sup_n \int_n^{ n+1 } \left| V ( t ) \right|^2 dt $ is finite, 
then \be\la{sl3} \int_I \( |\Theta^\prime (x,k)|^2 +| \Phi^\prime 
(x,k) |^2 \) d\mu (k) \leq C (I) . \ee 
\end{theorem}

In terms of the transfer matrix $ T ( x , k ) $ for the 
differential equation $ -y'' + V y = k y $, (\ref{sl1}) and 
(\ref{sl3}) together are equivalent to saying that \be\la{transf} 
\int_I \len T ( t, k ) \rin^2 d \mu ( k ) \le C(I) . \ee 

We shall need the following lemma of some interest of its own. 

\begin{lemma} \la{y0y}
Let $ y_0 $ be any solution to the equation $ - y'' + \overline q 
y = k y $ such that $ r = y_0^\prime ( 0 ) / y_0 ( 0 ) \in \R $ 
(or $ y_0 ( 0 ) = 0 $). Then for any solution $ y $ to this 
equation there exists a constant $ C $ such that $ | y ( x ) | \le 
C | y_0 ( x ) | $ for $ x $ large enough, provided that $ \im q 
\not \equiv 0 $. 
\end{lemma}

\begin{proof}
As we shall see later, the solution $ y_0 ( x ) $ does not vanish 
for $ x $ large enough\footnote{This can be also seen immediately 
from complete non-selfadjointness of $ l $ and the fact that a 
completely non-selfadjoint dissipative operator cannot have real 
eigenvalues.}. It is then enough to establish the estimate for the 
solution $$ y_1 ( x ) = \( \int^x \frac {d\xi}{ y_0^2 ( \xi ) } \) 
y_0 ( x ) $$ linear independent of $ y_0 $. On multiplying the 
equation for $ y_0 $ by $ \overline { y_0 } $ and integrating from 
$ 0 $ to $ x $, we obtain, $$ - \left. y_0^\prime ( x ) \overline 
{ y_0 ( x ) } \right|_0^x + \int_0^x \left| y_0^\prime \right|^2 
dt + \int_0^x ( \overline q - k ) \left| y_0 \right|^2 dt = 0 . $$ 
Taking the imaginary part and noticing that $ y_0^\prime ( 0 ) 
\overline { y_0 ( 0 ) } = r \left| y_0( 0 ) \right|^2 $ is real, 
we find $$ - \im \left[ y_0^\prime ( x ) \overline { y_0 ( x ) 
}\right] = \int_0^x \im q \, \left| y_0 \right|^2 dt . $$ Let us 
denote the left hand side by $ \rho( x ) $. This identity shows 
that for $ x $ large enough $ \rho ( x ) > 0 $, so that $ y_0 ( x 
) \ne 0 $, and one can introduce the variables $ R , \Psi $, $ R 
> 0 $, by $ y_0 = R e^{ - i \Psi } $. On calculating,
$ \rho = R^2 \Psi^\prime $, so that $ \Psi^\prime 
> 0 $, and the phase $ \Psi $ can be chosen to be a continuous 
monotonic increasing function of $ x $. We then have, $$ \int^x 
\frac {d\xi}{ y_0^2 ( \xi ) } = \int^x e^{ 2i \Psi ( \xi ) } \frac 
{ \Psi^\prime ( \xi ) d\xi}{ R^2 ( \xi ) \Psi^\prime ( \xi ) } = 
\int^{ \Psi ( x ) } \frac { e^{ 2i \Psi } d\Psi}{ \rho ( \xi ( 
\Psi ) ) } . $$ The nonnegative function $ \zeta ( s ) = 1 /\rho 
\( \xi( s ) \) $ is monotonic decreasing, which implies that the 
integral on the right hand side is uniformly bounded, and the 
result follows. 
\end{proof}

This lemma shows, in particular, that the solution $ u_h ( x , k ) 
$ estimates above the solution $ u ( x , k ) $ whenever the latter 
exists, $ u = O ( |u_h| ) $ at large $ x $. In the following lemma 
we use the standard asymptotic perturbation theory \cite{Eastham} 
to compare the asymptotics of solutions corresponding to the 
potentials $ q $ and $ \re q $. 

\begin{lemma} \la{pertur}
Let $ k \in \R $ be such that $ m ( k ) $ exists, and $ m ( k ) 
\neq h $. If $ \sqrt { \im q } \, u_h ( \cdot , k ) \in L^2 $, 
then the equation \be \la{eqn2} - y'' + \re q \, y = k y \ee has 
solutions $ y_{ 1,2 } $ such that $ y_1 = u_h ( 1 + o ( 1 )) $, $ 
y_2 = u + o ( | u_h |) $ as $ x \to \infty $, and the Wronskian 
\be\la{wrr} W [ y_1 , y_2 ] = W [ u , u_h ] = m ( k ) - h. \ee 
\end{lemma}

\begin{proof}
Rewriting the equation (\ref{eqn2}) in the matrix form, $$ Y' = 
\begin{pmatrix} 0 & 1 \cr \re q - k & 0 \end{pmatrix} Y , \; \; Y =
\begin{pmatrix} y_1 & y_2 \cr y'_1 & y'_2 
\end{pmatrix} , $$ and applying the variation of the parameter
formula, we obtain the equation \be\la{Levin} Z' = \frac i{h-m ( k 
)} \begin{pmatrix} \im q \, u_h u & \im q \, u^2 \cr -\im q \, 
u_h^2 & -\im q \, u_h u \end{pmatrix} Z , \ee where $$ Y = 
\begin{pmatrix} u_h & u \cr u_h^\prime & u^\prime \end{pmatrix} Z 
. $$ The fact that $ \sqrt { \im q } \, u ( \cdot , k ) \in L^2 $ 
(see (\ref{imqi}) and the assumption $ \sqrt { \im q } \, u_h ( 
\cdot , k ) \in L^2 $ imply that $$ 
\begin{pmatrix} \im q \, u_h u & \im q \, u^2 \cr -\im q \, 
u_h^2 & -\im q \, u_h u \end{pmatrix} \in L^1 . $$ Applying now 
the Levinson theorem to the equation (\ref{Levin}), we find that 
it has a solution $ Z $ satisfying $ Z ( x ) = I + o ( 1 ) $. On 
taking into account that $ u = O ( |u_h| ) $, we obtain the 
existence of the solutions $ y_{1,2 } $. Calculating the 
determinant of $ Y $ at infinity, we obtain (\ref{wrr}). 
\end{proof}

\medskip

{\it Proof of theorem 2 (beginning)}. Suppose that $ H_{ac} $ is 
nontrivial. Then, according to and proposition \ref{criterion}, 
there exists a set $ M \subset \sigma_{ ess } ( l ) $ of positive 
Lebesgue measure such that $ D ( k + i \von ) $ does not converge 
to $ 0 $ strongly when $ \von \downarrow 0 $ for $ k \in M $, and 
therefore the function $ \sqrt { \im q } \, u_h ( \cdot , k ) \in 
L^2 $ for a. e. $ k \in M $ by lemma \ref{notinl2cont}(ii). Let $ 
y_{ 1,2 } ( \cdot , k ) $ be the solutions to (\ref{eqn2}) defined 
in lemma \ref{pertur} for a. e. $ k \in M $. From the lemma it is 
immediate that the solutions $ y_{ 1,2 } ( \cdot , k ) $ are 
square-summable with the weight $ \im q $ for a.e. $ k \in M $. 
Let $ \mu $ be the a.c. measure given by $ d\mu = \min \{ d\rho^D 
, d \rho^N \} $ where $ d\rho^{D/N} $ are the spectral measures 
for the self-adjoint Schr\"odinger operators with the potential $ 
\re q $ and the Dirichlet/Neumann boundary conditions at zero, 
respectively. Fix arbitrary $ c_0 $, $ N $ and $ A $ such that the 
set $$ I = \left\{ k \in M \colon \; |m ( k )| \le c_0 , \; 
\int_0^\infty \im q \( \left| y_1 ( t,k ) \right|^2 + \left| y_2 ( 
t,k ) \right|^2 \) dt \le N \right\} \cap [ -A , A ] $$ has 
positive $ \mu $ - measure. This is possible since, as we shall 
show later, $ \mu ( M ) > 0 $. Let $ \rho ( k ) = 1 + \max \{ 
\left| y_{1,2} ( 0 , k ) \right| , \left| y_{1,2}^\prime ( 0 , k ) 
\right| \} $, and $ T ( x , k ) $ be the transfer matrix for the 
equation (\ref{eqn2}). Then, $ \left| y_{1,2}^\prime ( x , k ) 
\right| \le \rho ( k ) \| T ( x , k ) \| $. Taking into account 
(\ref{wrr}), we have, $$ | h - m ( k ) | = \left| W [ y_1 , y_2 ] 
\right| \le \rho ( k ) \| T ( x , k ) \| \( | y_1 | + y_2 | \) . 
$$ Multiplying this by $ \( \rho ( k ) \)^{ -1 } \im q $ and 
integrating, we obtain 
\begin{eqnarray} \int_I \frac { | h - m ( k ) |}{ \rho ( k ) } 
d \mu ( k ) \int^x \im q \, dt \le C \( \int_I d \mu ( k ) \int^x 
\im q \, \( \left| y_1 \right|^2 + \left| y_2 \right|^2 \) dt \)^{ 
1/2 } \nonumber \\ \( \int^x dt \, \im q ( t ) \int_I \len T ( t, 
k ) \rin^2 d \mu ( k ) \)^{ 1/2 } . \la{wr} \end{eqnarray} Now, $ 
\int_I \len T ( t, k ) \rin^2 d \mu ( k ) $ is bounded uniformly 
in $ t $ by theorem \ref{SimLast} (see (\ref{transf})). At this 
point we used the assumption that $ \re q $ is uniformly locally $ 
L^2 $-bounded. This gives \bequnan C \( \int^x \im q \, dt \)^{ 
1/2 } \le C \( \int_I d \mu ( k ) \int^x \im q \, \( \left| y_1 
\right|^2 + \left| y_2 \right|^2 \) dt \)^{ 1/2 } \le C \sqrt { N 
\mu ( I ) } ,\eequnan and so the integral on the left hand side is 
bounded. 

Thus, it remains to show that $ \mu ( M ) > 0 $. This will occupy 
the rest of the paper and requires developing an analog of the 
subordinacy theory \cite{Pearson} for the dissipative 
Schr\"odinger operator $ l $. 

{\bf $ \S 2 $. Subordinacy theory and the end of the proof of 
theorem 2.} Recall \cite{Pearson} that a solution $ u_1 $ to a 
second-order differential equation on the semi-axis $ [ 0 , \infty 
) $ is called {\it subordinate} if $$ \frac { \len u_1 \rin_N } { 
\len u_2 \rin_N } \to 0 $$ as $ N \to \infty $ for any solution $ 
u_2 $ linear independent of $ u_1 $. Here $ \len \cdot \rin_N 
\equiv \len \cdot \rin_{ L^2 ( 0 , N ) } $. 

\begin{theorem} \la{sub}
Let $ \im q \not\equiv 0 $. Then for a. e. $ k \in \R $ the 
following assertions are equivalent 

$(i)$ $ D ( k + i \von ) \mathop{\stackrel{s} \longrightarrow}
\limits_{\von \downarrow 0 } 0 $.

$(ii)$ There exists a subordinate solution to $ - y'' + q y = k y 
$. 

\end{theorem}

As the proof shows, the exceptional set in this theorem can be 
identified with the complement of the set of $ k $ such that $ m ( 
k ) $ exists finitely and $ m ( k ) \neq h $. We shall provide a 
complete proof for the implication $ (ii) \Rightarrow (i) $ only, 
for the reverse implication is not used in this paper. The proof 
is subdivided into several assertions. 

\begin{lemma} 
For any real $ k $ such that $ m ( k ) $ exists and $ m ( k ) \ne 
h $ the assertion $ ( i ) $ is equivalent to saying that \be 
\la{epssub} \von \len u ( \cdot , k + i \von ) \rin^2 \to 0 . \ee 
\end{lemma}

\begin{proof}
From lemma \ref{notinl2cont}(i) we obtain that for any compactly 
supported $ v \in H $ $$ \len D ( k + i \von ) v \rin \le C \von^{ 
1/2 } \len u ( k + i \von ) \rin + o ( 1 ) $$ since $ C_\von [v] $ 
given by (\ref{C0}) converges when $ \von \to 0 $. This shows that 
if (\ref{epssub}) is satisfied, then $ D ( k + i \von ) v \to 0 $ 
for any compactly supported $ v \in H $, and $ (i) $ follows. 

To establish the reverse implication, we choose $ v $ to be $ 
\overline{ u_h ( x , k )} \chi ( x ) $ where $ \chi $ is the 
indicator of an interval $ \Delta $. Then, $$ C_\von [ v ] 
\mathop{\longrightarrow}_{ \von \to 0 } \frac 1{ h - m ( k ) } 
\int_\Delta \left| u_h ( s , k ) \right|^2 \sqrt{ \im q } \, ds 
\ne 0 $$ if $ \Delta $ is large enough. Now, lemma 
\ref{notinl2cont}(i) shows that if $ D ( z ) v \to 0 $, then 
(\ref{epssub}) holds. 
\end{proof}


%\smallskip
The proof of the following assertion, which is central in the 
subordinacy theory, carries over from the self-adjoint theory 
\cite{Pearson} without any modification but a slight 
simplification. 

\medskip

\begin{proposition}
If the equation $ - y'' + \overline q y = k y $ has a subordinate 
solution for a certain $ k \in \R $ such that $ m ( k ) $ exists 
and $ m ( k ) \ne h $, then this solution is a multiple of $ u ( x 
, k ) $. 
\end{proposition}

{\it Proof}. Without loss of generality, one can assume that $ u ( 
\cdot , k ) \notin L^2 $. Let $ v $ be the subordinate solution 
with the normalization chosen so that $ v = \Theta + c_0 \Phi $, 
hence the Wronskian $ W [ \Phi , v ] = 1 $. This is possible 
since, as lemma \ref{y0y} shows, $ \Phi $ cannot be a subordinate 
solution. We have to prove that $ c_0 = m ( k ) $. A 
straightforward calculation shows that the solution $ u ( x , z ) 
$ satisfies the following integral equation, $$ g ( x ) = \Theta ( 
x , k ) + m ( z ) \Phi ( x , k ) - i \von \left[ \Phi ( x , k ) 
\int_0^x v ( s ) g ( s ) ds - v ( x ) \int_0^x \Phi ( s , k ) g ( 
s ) ds \right] . $$ Let us consider this equation in $ L^2 ( 0 , A 
( \von ) ) $ with a finite $ A ( \von ) $ to be chosen so that $ A 
( \von ) \to \infty $ as $ \von \to 0 $. If \be\la{itera} \von 
\len \Phi ( \cdot , k ) \rin_{ A ( \von ) } \len v \rin_{ A ( \von 
) } < \frac 12 , \ee then it can be iterated in $ L^2 ( 0 , A ( 
\von ) ) $ to give \bequnan \len u ( \cdot , z ) \rin_{ A ( \von ) 
} \asymp \len \Theta ( \cdot , k ) + m ( z ) \Phi ( \cdot , k ) 
\rin_{ A ( \von ) } = \\ \len v ( \cdot , k ) + ( m ( z ) - c_0 ) 
\Phi ( \cdot , k ) \rin_{ A ( \von ) } = 
\\  \( | m ( z ) - c_0 | + o ( 1 ) \) \len \Phi ( \cdot , k ) \rin_{ A ( \von ) }
\eequnan where we used the fact that $ v $ is subordinate, and 
therefore $ \len v \rin_{ A ( \von ) } / \len \Phi ( \cdot , k ) 
\rin_{ A ( \von ) } \to 0 $. Define $ A ( \von ) $ by \bequnan &
\len \Phi ( \cdot , k ) \rin_{ A ( \von ) } \len v \rin_{ A ( \von 
) } = \frac 1{ 3 C } \len u ( \cdot , k + i \von ) \rin^2 , &
\\ & C = \limsup_{ \von \to 0 } \im m ( k + i \von ) . & \eequnan Then 
(\ref{itera}) is satisfied in view of (\ref{supsubcon}), and $ A ( 
\von ) $ tends to infinity as $ \von \downarrow 0 $ since 
otherwise $ \len u ( \cdot , k ) \rin \le \liminf \len u ( \cdot , 
k + i \von ) \rin $ would be finite. We thus obtain, $$ | c_0 - m 
( z ) | \le \frac{\len u ( \cdot , k + i \von ) \rin }{\len \Phi ( 
\cdot , k ) \rin_{ A ( \von ) } } + o ( 1 ) = \( 3 C \frac { \len 
v \rin_{ A ( \von ) } } {\len \Phi ( \cdot , k ) \rin_{ A ( \von ) 
} } \)^{ 1/2 } + o ( 1 ) \to 0 . \; \; \hfill \Box $$ 

\smallskip
\medskip

{\it Proof of theorem \ref{sub} $ (ii) \Rightarrow (i) $}. To 
establish the implication it remains to show that if $ u ( \cdot , 
k ) $ is a subordinate solution, then (\ref{epssub}) is satisfied. 
It is clear from (\ref{immid}) that it suffices to show that $$ 
\int_0^\infty \im q ( x ) \left| u ( x , k ) \right|^2 dx = \im m 
( k ) . $$ The argument is similar to that used in the proof of 
lemma \ref{y0y}. Multiplying the identity $ - u'' + ( \overline q 
- k ) u = 0 $ by $ \overline u $, integrating and taking the 
imaginary part, we obtain, $$ \im \left[ u^\prime ( x , k ) 
\overline { u ( x , k ) }\right] = \im m ( k ) - \int_0^x \im q ( 
t ) \, \left| u ( t , k ) \right|^2 dt . $$ As (\ref{immid}) 
shows, the right hand side here is non-negative for all $ x $, and 
so the required assertion is immediate if the left hand side has a 
zero. Suppose now that the left hand side does not vanish for all 
$ x > 0 $, so, at least, $ u ( x ) \ne 0 $, and one can introduce 
the variables $ R , \Psi $, $ R 
> 0 $, by $ u = R e^{ i \Psi } $. Then the left hand side 
becomes $ \rho ( x )\colon \! = R^2 ( x ) \Psi^\prime ( x ) $. 
Since $ \rho ( x ) $ is positive, it follows that the phase $ \Psi 
$ can be chosen to be a continuous monotonic increasing function 
of $ x $. 

We then have for a solution $ v $ linear independent of $ u $,
\bequnan v ( x ) = \( \int_a^x \frac {d\xi}{ u^2 ( \xi ) } \) u (
x ) = \( \int^x e^{ -2i \Psi ( \xi ) } \frac { \Psi^\prime ( \xi )
d\xi}{ R^2 ( \xi ) \Psi^\prime ( \xi ) } \) u ( x ) = \\ \( \int^{
\Psi ( x ) } e^{ -2i \Psi } \zeta ( \Psi ) d\Psi \) u ( x ),
\eequnan where $ \zeta ( s ) = 1 /\rho \( \Psi^{ -1 } ( s ) \) $ 
is a positive monotone increasing function. It follows that $ 
\zeta ( s ) \to \infty $, otherwise the integral would be bounded 
and $ v $ would be bounded above by $ u $, which contradicts the 
assumption that $ u $ is subordinate. Thus, $ \int_0^x \im q 
\left| u \right|^2 dt \to \im m ( k ) $ as $ x \to \infty $, as 
required. \hfill $ \Box $ 

It should be mentioned that in the selfadjoint theory the 
assertion corresponding to the last part of the proof of theorem 
\ref{sub} (if $ u ( \cdot , k ) $ is a subordinate solution, then 
$ \im m ( k ) = 0 $) is {\it trivial}, since a subordinate 
solution normalized by the condition $ u ( 0 ) = 1 $ must be real 
when $ \im q \equiv 0 $. 

Notice that we have left unproven only one assertion in the 
implication $ (i) \Rightarrow (ii) $ of theorem \ref{sub}: if $ 
\von \len u ( k + i \von ) \rin^2 \to 0 $ for a certain $ k \in \R 
$ such that $ m ( k ) $ exists and $ m ( k ) \ne h $, then $ u ( x 
, k ) $ is a subordinate solution. The proof of this fact can be 
taken verbatim from the corresponding assertion in the 
self-adjoint theory \cite{Pearson}. 

\medskip

{\it Proof of theorem 2 (end)}. We have to show that $ \mu ( M ) 
> 0 $. Suppose, by contradiction, that $ \mu ( M ) = 0 $. Since $ 
\mu $ is equivalent to the a.c. part of the spectral measure of a 
self-adjoint Schr\"odinger operator with the potential $ \re q $, 
the following assertion of the selfadjoint Gilbert - Pearson 
theory applies. 
\medskip
\smallskip

{\bf Proposition.} \cite{Pearson} {\it Let $ V $ be a locally 
bounded real potential in the limit point case at infinity and let 
$ d\nu $ be the a.c. part of the spectral measure for the 
self-adjoint Schr\"odinger operator with the potential $ V $ and a 
selfadjoint boundary condition at zero. If $ \nu ( S ) = 0 $, then 
the equation $ -y'' + V y = k y $ has a subordinate solution for 
Lebesgue-almost-all $ k \in S $.} \medskip 

This gives that the equation (\ref{eqn2}) has a subordinate 
solution for a. e. $ k \in M $. Lemma \ref{pertur} now shows that 
the equation (\ref{eqn}) with $ z = k $ has a subordinate solution 
for a. e. $ k \in M $ as well. Indeed, let $ y = c_0 y_1 + y_2 $ 
be a subordinate solution to (\ref{eqn2}). Then, $ y \sim u + c_0 
u_h + o ( u_h ) $, hence $$ \frac{\len u + c_0 u_h \rin_N }{\len 
u_h \rin_N } \sim \frac{\len y \rin_N }{\len y_1 \rin_N } \to 0 , 
$$ so that $ u + c_0 u_h $ is a subordinate solution to 
(\ref{eqn}). Since $ y_1 $ cannot be a subordinate solution, we 
are done. Applying now theorem \ref{sub}, we obtain that $ D ( k + 
i \von ) \mathop{\stackrel{s} \longrightarrow} 0 $ for a. e. $ k 
\in M $, a contradiction. \hfill $ \Box $ 

Conditions on the real part of the potential in theorem 2 are 
required to apply spectral averaging. In particular, the condition 
that $ \re q $ is uniformly locally $ L^2 $-bounded allows to use 
the following standard estimate for solutions of a second-order 
differential equation (see, for instance, \cite[Lemma 
3.1]{Simon}): if $ \sup_n \int_n^{ n+1 } | V ( t ) |^2 dt $ is 
finite, then $$ \left| g^\prime ( x ) \right|^2 \le C ( 1 + k^2 ) 
\int_{ x - 1 }^{ x + 1 } \left| g ( s ) \right|^2 ds $$ for any 
solution $ g $ to $ -y'' + V y = k y $ with a constant $ C $ 
depending on the potential $ V $ only. As was noticed in 
\cite{SimonLast}, a combination of this bound with (\ref{sl1}) 
allows to drop the additional integration in $ y $ in (\ref{sl2}) 
leading to (\ref{sl3}). According to the following theorem, if we 
omit the condition on the positive part of $ \re q $, then the 
non-triviality of $ H_{ac} $ implies that $ \im q $ is $ L^1 $ "in 
a local mean". 

\begin{theorem} Let $ q $ be a locally bounded function on $ 
\R_+ $ with $ \im q \ge 0 $, $ \im q ( x ) \to 0 $ as $ x \to 
\infty $. Assume that $ \re q $ is bounded below. If the 
absolutely continuous subspace of the Schr\"odinger operator $ l $ 
is nontrivial, then $$ \sum_n \( \int_n^{ n+1 } \sqrt {\im q } \, 
dt \)^2 < \infty . $$ 
\end{theorem} 

\begin{proof}
The argument proceeds as before until (\ref{wr}) where we 
integrate in $ t $ over the interval $ [ n, n+1 ] $ instead of $ [ 
0 , x ] $ to obtain $$ \int_I \frac { | h - m ( k ) |}{ \rho ( k ) 
} d \mu ( k ) \int_n^{ n+1 } \sqrt{\im q } \, dt \le C \( \int_I d 
\mu ( k ) \int_n^{ n+1 } \im q \, \( \left| y_1 \right|^2 + \left| 
y_2 \right|^2 \) dt \)^{ 1/2 } . $$ Here we have taken (\ref{sl2}) 
into account. Taking square and summing over $ n $, we arrive at 
the result. 
\end{proof} 

We were not able to adapt the argument for an "individual" $ k $ 
from the proof of theorem \ref{discrete} to the continuous case, 
for we do not know a proper continuous analog of the spread 
potential $ \wt q $. If this were possible, then one could hope to 
relax conditions on $ \re q $ in theorem 2. It is also unclear 
whether it is possible to give a proof of the result in the 
discrete case analogous to the above proof of theorem 2. The 
obstacle here is that we do not know an appropriate discrete 
analog of the trigonometric substitution $ u = R e^{ i \Psi } $ in 
the proof of theorem \ref{sub}. 

Finally, it should be noticed that the condition of pointwise 
decay of $ \im q $ as $ x \to \infty $ in theorem 2 is only used 
to ensure that $ \sigma _{ ess } ( l ) \subset \R $ and can be 
replaced by any condition guaranteeing this inclusion. 

\section{Conclusions}

As is well-known \cite{Na}, triviality of the space $ H_{ac} $
implies that the characteristic function of the operator $ L $ is
inner. So far, the spectral analysis of operators with inner
characteristic functions has only been developed effectively for
the discrete part of the spectrum in the case of trace class
perturbations \cite{Nik}. The examples considered in this paper
suggest, however, that the appearance of a nontrivial "singular"
inner function is typical in applications where the trace class 
theory is not applicable. 

Finally, one should mention that the instability phenomenon
described in this paper is likely to be a specific feature of
dissipative Schr\"odinger operators. In the non-dissipative case, 
a potential of the WKB - class will have all solutions of 
(\ref{eqn}) bounded if its imaginary part is conditionally 
integrable. This suggests that in this case the absolutely 
continuous subspace can be nontrivial even if $ \im q \notin L^1 
$. 

\bigskip
{\bf Acknowledgements.} The author is indebted to S. Naboko for 
helpful discussions, encouragement and critical reading of the 
manuscript and to M. Marletta for noticing a mistake in the 
initial formulation of lemma \ref{pertur}. 

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