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\begin{document}
\begin{center}
\Large
\bf Absolutely continuous spectrum of one-dimensional Schr\"odinger
operators and Jacobi matrices with slowly decreasing potentials
\vspace{0.5cm}
\end{center}
\begin{center}
\large
Alexander Kiselev
\vspace{0.2cm}
\end{center}
\begin{center}
Division of Physics, Mathematics and Astronomy \\
California Institute of Technology, 253-37 \\
Pasadena, CA 91125
\end{center}
\begin{abstract}
We prove that for any one-dimensional Schr\"odinger operator with
potential $V(x)$ satisfying decay condition $|V(x)| \leq Cx^{-3/4-\epsilon},$ the
absolutely continuous spectrum fills the whole positive semi-axis.
The description of the set in $\R^{+}$ on which the singular part of the
spectral measure might be supported is also given. Analogous results
hold for Jacobi matrices.
\end{abstract}
\begin{center}
\bf Mathematics subject classification \\
\it Primary: \rm 34L40, 47B39; \it Secondary: \rm 42B10, 81Q10.
\end{center}
\begin{center}
\bf Keywords: \\
\it Schr\"odinger operators, absolutely continuous spectrum, Fourier
integral, Jacobi matrices.
\end{center}
\begin{center}
\bf Introduction
\end{center}
\rm Let $H_{V}=-\frac{d^{2}}{dx^{2}}+V(x)$ be the one-dimensional Schr\"odinger
operator acting on $L^{2}(0, \infty).$ We assume $V(x)$ is a real-valued
locally integrable function which goes to zero at infinity.
It is a well-known fact that if we fix some self-adjoint boundary
condition at zero, the expression $H_{V}$ has unique self-adjoint realization
in $L^{2}(0, \infty).$ The essential specrum of the
operator $H_{V},$ $\sigma_{\rm{ess}}(H_{V}),$ coincides with the positive semi-axis
since the potential vanishing at infinity constitutes a relatively compact
perturbation of the free Hamiltonian.
In this paper, we explore the problem of dependence of the spectral
properties of $H_{V}$ for positive energies on the rate of decay of the
potential $V.$ In particular, the interesting question is to determine
the critical rate of decay which can lead to the complete or partial
destruction of the absolutely continuous spectrum on the positive half-axis,
and, correspondingly, to find out which classes of potentials are not strong
enough to seriously affect the absolutely continuous spectrum inherent for the free
Hamiltonian. As is generally known,
if $V(x)$ belongs to $L^{1}(0, \infty )$ then the spectrum on the positive semi-axis
is purely absolutely continuous (see, e.g., \cite{Weid}). The situation
is not so clear for decreasing potentials which are not absolutely integrable.
There are many results on
the absolute continuity of the spectrum on the positive semi-axis
(except perhaps for a finite number of resonances in some cases)
for certain classes of decaying potentials, such as potentials of
bounded variation \cite{Weid} or specific oscillating potentials
(see, e.g., \cite{Ben}, \cite{Hin}, \cite{White} \cite{Mat} for further
references). But no general relations between the rate of decay and
spectral properties, apart from the absolutely integrable
class, seem to be known.
The results concerning the spectral properties of Schr\"odinger operators
with random potentials, however, suggest that there may be a general
relation between the rate of decay of the potential and the preservation
of the absolutely continuous spectrum on $\R^{+}=(0,\infty).$ Namely, Kotani-Ushiroya
\cite{KotUsh} show that when $q(x)=a(x)F(Y_{x}(\omega))$ where $a(x)$
is a smooth power decaying non-random factor, $Y_{x}(\omega)$ is
a Brownian motion on a compact Riemannian manifold $M$ with the
volume element $\mu$ and $F:M \to \R$ is a non-flattening $C^{\infty}$
function satisfying $\int_{M}Fd \mu = 0,$ then the question of whether the rate
of decay of $a(x)$ is faster or slower than $x^{-1/2}$ is crucial for
the spectral properties of the corresponding random Schr\"odinger
operator. When $a(x)=(1+|x|)^{-\alpha}$ with $0<\alpha< \frac{1}{2},$
the spectrum on $\R^{+}$ is pure point with probability one; when
$\alpha> \frac{1}{2},$ then the spectrum on the positive semi-axis
is a.e.~purely absolutely continuous.
The methods of \cite{KotUsh} are probabilistic in nature and cannot
provide information on what happens in general for potentials satisfying
$|V(x)| \leq C(1+|x|)^{-\alpha},$ $\alpha > \frac{1}{2}.$
Although the set
of potentials leading to purely absolutely continuous spectrum is ``big''
in a certain sense \cite{KotUsh}, examples with eigenvalues on
$\R^{+}$ show there may be exceptions. Moreover, if one could find
at least one potential satisfying $|V(x)| \leq C(1+|x|)^{-\alpha},$ for
certain $\alpha >
\frac{1}{2}$ and $C,$ which gives rise to purely singular spectrum
on the positive semi-axis, then by general principles of the genericity
of singular continuous spectrum \cite{Sim1}, there would exist another ``big''
(in a topological sense) set of potentials obeying same decay condition and yielding
purely singular continuous spectrum on $\R^{+}.$ Namely, this set would be
a dense $G_{\delta}$ in the space of all potentials
satisfying the power decay estimate $|V(x)| \leq C(1+|x|)^{\alpha},$ equipped
with the $L^{\infty}$ norm.
An analogous situation is exactly the case for $\alpha < \frac{1}{2},$
when the spectrum on the positive semi-axis is dense pure point with probability one by \cite{KotUsh},
but, at the same time, by the recent result of Simon \cite{Sim1}, there exists
a dense $G_{\delta}$ set of potentials leading to purely singular continuous
spectrum on the $R^{+}$.
To further illustrate the difficulty of the passage from random to deterministic
results, we note that \cite{KotUsh} implies that there exist ``many''
potentials with power decay (slower than $x^{-1/2}$) yielding
dense pure point spectrum on $\R^{+}.$ But, nevertheless, there are
no deterministic examples of potentials with power decay
even leading to just purely singular spectrum (to construct an explicit
example of a potential having dense pure point spectrum should be much
harder, since an arbitrarily small change in the boundary condition may
change the spectrum to purely singular continuous \cite{DRMS}, \cite{Gord}).
In fact, the only known explicit examples of decaying potentials
yielding purely singular spectrum on $\R^{+}$ are due to Pearson \cite{Pear}
and these potentials exhibit slower than power-rate decay.
The main result we prove in this paper says that all potentials decaying
faster than $Cx^{-3/4-\epsilon},$ with no additional conditions,
preserve absolutely continuous spectrum
on the positive semi-axis, although of course embedded singular spectrum
may appear. This result provides a new general class of decaying potentials
preserving absolutely continuous spectrum of the free Hamiltonian. It also
shows that there is indeed a determinstic analog of the random potentials
results, at least in the range of power decay $\alpha \in (\frac{3}{4}, 1].$
The main new idea we use in the proof is a combination of a certain ODE asymptotic
technique, which has been commonly used for the treatment of oscillating
potentials, with some results from harmonic analysis
related to the almost everywhere convergence of Fourier integrals.
Another interesting aspect of the spectral behaviour of Schr\"odinger operators
with decreasing potentials is a phenomena of positive eigenvalues.
Eastham-Kalf \cite{EaKa} show that if $V(x)=o(1/x)$ as $x \rightarrow \infty,$
then $H_{V}$ does not have eigenvalues above zero. If $V(x)=O(1/x)$,
there are no eigenvalues above a certain constant. On the other hand,
Eastham-Leod \cite{EaLe}, with further deveopments by
Thurlow \cite{Thur}, show how to construct potentials $V(x)$ of the
type $V(x)=\frac{C(x)}{x},$ with $C(x)$ converging to infinity as $x$
tends to infinity, such that a prescribed countable set of isolated points
represents embedded positive eigenvalues of $H_{V}.$ These authors use
the Gel'fand-Levitan approach.
Later, Naboko \cite{Nab} described a construction which allows for an arbitrary
countable set $T$ of rationally independent numbers in $(0, \infty)$ (and so
possibly a dense set) to find a potential $V(x)$ satisfying $|V(x)| \leq \frac
{C(x)}{x}$ with $C(x) \stackrel{x \rightarrow \infty}{\longrightarrow} \infty$
monotonously at an arbitrarily slow given rate, such that the corresponding Schr\"odinger
operator has the set $T$ among its eigenvalues. Recently, Simon \cite
{Sim2} has found
a different construction that does away with the rational independence assumption.
The constructions of Naboko and Simon do not give information about other kinds of
spectrum on $\R^{+}$ in such a situation. In particular, it was not
clear whether there is any
other spectrum but pure point in the case when the set $T$ of prescribed
eigenvalues is dense in $\R^{+}.$
The present paper settles the questions arising
from Naboko's and Simon's constructions. Moreover, together with these
works, it provides explicit examples of potentials yielding an arbitrary
dense (countable) set of eigenvalues embedded in the absolutely continuous
spectrum.
We should also mention that the results for random decaying potentials
for the discrete Schr\"odinger operators (Jacobi matrices) \cite{DeSiSo} raise parallel
questions in the discrete case. There is also a discrete analog to the
continuous case of Naboko's construction by
Naboko and Yakovlev \cite{NaYak} which allows one to find a potential
decaying arbitrarily slower than $\frac{1}{n}$ such that the corresponding
discrete Schr\"odinger operator has eigenvalues
dense in the essential spectrum $[-2,2].$
The paper is organized as follows. In the first section we prove our main
result for power decaying potentials.
In the second section we consider an application
of our method to certain more general classes of potentials,
inculding some potentials of the bump type.
In the third section we show similar results for Jacobi matrices.
\begin{center}
\bf 1. Main results for power decaying potentials
\end{center}
Let us first set up some notation we will need.
Suppose the function $f(x)$ belongs to $L^{2}(0,\infty).$ Then we
denote by
$\Phi (f)(k)$ the Fourier transform of the function $f,$
\[ \Phi(f)(k) = L^{2}-\lim_{N \rightarrow \infty}
\int\limits_{-N}^{N} \exp (ikt) f(t) \,dt. \]
We also use the notation $M^{+}(g)$ for the following function corresponding
to the function $g \in L^{p}(R)$, $1 \leq p \leq \infty :$
\[ M^{+}(g)(x) = \sup\limits_{h>0} \frac{1}{h} \int\limits_{0}^{h}|g(x+t)+g(x-t)|\,dt \]
and notation ${\cal M}^{+}(g)$ for the set
\[ {\cal M}^{+}(g)= \left\{ x \mid M^{+}(g)(x)< \infty \right\} \]
The function $M^{+}(g)$ is ``almost'' a maximal function of the function $g;$
in particular, $M^{+}(g)$ is finite whenever the maximal function of $g$
is finite. By well-known properties of the maximal function (see, e.g., \cite
{Rud})
we have then that $M^{+}(g)$ is finite a.e.~and therefore
the complement of the set ${\cal M}^{+}(g)$ has measure zero.
The main result of this section is the following theorem: \\
\noindent \bf Theorem 1.1. \it Suppose that potential $V(x)$ satisfies $|V(x)| \leq Cx^{-3/4-
\epsilon}$ for $x \in (a,\infty)$ with some positive constants $\epsilon,
a,C.$ Then the absolutely continuous spectrum of the operator $H_{V}$ fills
the whole positive semi-axis, in the sense that the absolutely
continuous component $\rho_{\rm{ac}}$ of the spectral measure $\rho$
satisfies $\rho_{\rm{ac}}(T)>0$
for any measurable set
$T \subset (0, \infty )$ with $|T|>0$ \rm(\it where $| \cdot |=$ Lebesgue measure\rm)\it.
The singular spectrum on $(0, \infty )$ may be
located only on the complement of the set \[ S= \frac{1}{4} ({\cal M}^{+}
(\Phi(V(x)x^{1/4})))^{2} \] \rm(\it i.e., quarters of squares of the points from
${\cal M}^{+}(\Phi(V(x)x^{1/4}))$\rm)\it , so that $\rho_{\rm{sing}}(S)=0.$
Moreover, for every energy $\lambda \neq 0$
from the set $S$ we have two linearly independent solutions $\phi_{\lambda},$
$\overline{\phi}_{\lambda}$ \rm(\it$=$complex conjugation of $\phi_{\lambda}$\rm) \it of
the equation $H_{V} \phi - \lambda \phi =0$ with the following asymptotics
as $x$ goes to infinity:
\begin{equation}
\phi_{\lambda}(x)= \exp \left( i\sqrt{\lambda}x-\frac{i}{2\sqrt{\lambda}}
\int\limits^{x}_{0}V(s)\,ds \right) \left( 1+O(x^{-\epsilon} \log x) \right)
\end{equation}
\rm(\it which is exactly the WKB formula\rm). \\
The main idea behind the proof is a combination of the following three ingredients:
(i) The recent studies on the connection between asymptotic
behavior of solutions of the Schr\"odinger equation and spectral
properties, which allow one to conclude the absolute continuity
of the spectrum on a certain set from the boundedness of all solutions
corresponding to the energies from this set;
(ii) The methods of studying the asymptotics of
solutions, namely the ``$I+Q$'' transformation technique introduced
by Harris and Lutz \cite{HaLu} and later used by many authors for treating
Schr\"odinger operators with oscillating potentials;
(iii) The results from the theory of Fourier integrals; in
particular, the question of a.e.~convergence of the partial integral
$\int^{N}_{-N} \exp (ikt)f(t)\, dt$ to the Fourier transform of
$f$ under certain conditions and an estimation of the rate of convergence.
As a preparation for the proof, we need several lemmas.
The first lemma allows us to reduce the proof of Theorem 1.1 to
the study of generalized eigenfunction asymptotics. \\
\bf Lemma 1.2. \it Suppose that for every $\lambda$ from the set $B,$
all solutions of the equation $H_{V} \phi - \lambda \phi =0$
are bounded. Then on the set $B,$
the spectral measure $\rho$ of the operator $H_{V}$ is purely absolutely
continuous in the following sense:
\rm{(i)} $\rho_{\rm{ac}}(A)>0$ for any $A \subseteq B$ with $|A|>0$
\rm{(ii)} $\rho_{\rm{sing}}(B)=0$ \rm \\
\noindent \bf Proof. \rm For a large class
of potentials, including those we consider here, this lemma follows from the
Gilbert and Pearson subordinacy theory \cite{GiPea}, as shown
by Stolz \cite{Sto}. Also, in a recent paper, Jitomirskaya
and Last \cite{JiLa} obtained a rather transparent proof of more
general results. For a direct simple proof of the lemma we
refer to a paper of Simon \cite{Sim3}. $\Box$
The complement of the set $S$ in the statement of the Theorem 1.1
has Lebesgue measure zero (which of course follows from the fact that
the complement of the set ${\cal M}^{+}(\Phi(x^{1/4}$ $V(x)))$ has measure zero).
Therefore, we see that (assuming Lemma 1.2) for the proof of Theorem 1.1,
it suffices to prove the stated asymptotics
of generalized eigenfunctions for the energies from the set $S.$
The second lemma we need deals with certain properties of the Fourier
integral. \\
\noindent \bf Lemma 1.3. \it Consider the function $f(x) \in L^{2}(\R).$ Then for every
$k_{0} \in {\cal M}^{+}(\Phi(f)),$ we have
\[ \int\limits^{N}_{-N} f(x) \exp (ik_{0}x) \,dx = O(\log N). \] \rm \\
Before giving the proof, let us point out the relation between the question we study and one
of the subtle problems of harmonic analysis. The Fourier transform
of the square integrable function $f(x)$ is usually defined as a
limit in $L^{2}$-norm as $N \rightarrow \infty$ of the functions
$\int^{N}_{-N}f(x) \exp (-ikx) \, dx.$ The question of whether these
integrals converge to the Fourier transform of $f$
in an ordinary sense for almost all values of $k$ is, roughly speaking,
equivalent to Lusin's hypothesis that the Fourier series of square
integrable function converge almost everywhere, resolved positively by Carleson
\cite{Car} in 1966. All that our simple lemma says is that we
have an estimate from above on the speed of divergence of partial
integrals, but for a rather explicitly described set of values of the parameter
$k$ of full measure. In the next section, which treats certain non-power
decaying potentials, we will need more refined results
on the a.e.~convergence of Fourier integral. \\
\noindent \bf Proof of Lemma 1.3. \rm The proof uses Parseval equality:
\begin{eqnarray*}
\int\limits^{N}_{-N}f(x) \exp (ik_{0}x) \, dx & = & \frac{1}{\pi}\int\limits_{R} \Phi(f)(k)
\frac{\sin N(k_{0}-k)}{k_{0}-k} \,dk \\
& = & \frac{1}{\pi}\int\limits^{\infty}_{0} \frac{
\sin Nk}{k}\,(\Phi(f)(k_{0}-k) + \Phi(f)(k_{0}+k))\,dk.
\end{eqnarray*}
We split the last integral into three parts and estimate them separately:
\[ \left| \int\limits^{\infty}_{1} \frac{\sin Nk}{k}\,(\Phi (f)(k_{0}-k) +
\Phi(f)(k_{0}+k))\,dk \right| \leq 4 \pi \left\| \frac{\sin Nk}{k} \right\|_{L^{2}(1, \infty)}
\parallel f \parallel _{L^{2}(-\infty, \infty)} \]
and so this part is bounded when $N \rightarrow \infty;$
\[ \left| \int\limits^{1}_{0} \frac{\sin Nk}{k} \,(\Phi (f)(k_{0}-k) + \Phi (f) (k_{0}+k))\,dk \right|
\leq \] \[ \leq N \int\limits^{1/N}_{0}|\Phi(f)(k_{0}+k)+\Phi(f)(k_{0}-k)|\,dk
+ \int\limits^{1}_{1/N}\frac{1}{k}|\Phi(f)(k_{0}+k)+\Phi(f)(k_{0}-k)|\,dk. \]
In the last expression the first summand is bounded by
$M^{+}(\Phi(f))(k_{0})$,
while in the second we perform integration by parts:
\[ \int\limits^{1}_{1/N}\frac{1}{k}\,|\Phi(f)(k_{0}+k)+\Phi(f)(k_{0}-k)|\,dk =
\int\limits^{1}_{1/N} |\Phi(f)(k_{0}+k)+\Phi(f)(k_{0}-k)|\,dk + \]
\[ + \int\limits^{1}_{1/N} \frac{1}{k^{2}} \int\limits^{k}_{1/N}|\Phi(f)(k_{0}+t)+
\Phi(f)(k_{0}-t)|\,dtdk \leq \]
\[ \leq M^{+}(\Phi(f))(k_{0}) + \int\limits^{1}_{1/N} \frac{1}{k}
\, M^{+}(\Phi(f))(k_{0})\,dk = O( \log N). \;\; \Box \] \\
To begin with the proof of the theorem, we rewrite equation $H_{V} \phi
-\lambda \phi=0$ as a system of first-order equations:
\begin{equation}
w'(x)
= \left(
\begin{array}{cc}
0 & 1 \\
V(x)- \lambda & 0 \end{array}
\right)
w(x),
\end{equation}
where $w$ and $\phi$ are clearly related by
\renewcommand{\arraystretch}{.7}
$w(x) = \left( \begin{array}{c}
\scriptstyle \phi (x) \\ \scriptstyle \phi '(x) \end{array} \right).$
\renewcommand{\arraystretch}{1}
We perform two transformations with the system (2), the first of which
is the variation of parameters formula
\begin{equation}
w(x) = \left( \begin{array}{cc} \psi_{1}(x) & \psi_{2}(x) \\
\psi '_{1}(x) & \psi '_{2}(x) \end{array} \right) y(x)
\end{equation}
where it is convenient for our purpose to choose $\psi_{1}(x)= \exp (i \sqrt{\lambda}x),$
$\psi_{2}(x)= \exp (-i \sqrt{\lambda}x).$ Substituting (3) into (2),
we get for $y(x)$:
\begin{equation}
y'(x)= \frac{i}{2\sqrt{\lambda}} \left( \begin{array}{cc} -V(x) &
-V(x) \exp (-2i \sqrt{\lambda} x) \\ V(x) \exp (2i \sqrt{\lambda} x) &
V(x) \end{array} \right) y(x).
\end{equation}
We can also write this system as
\begin{equation}
y' = ({\cal D} + {\cal W})y
\end{equation}
where ${\cal D}$ stays for the diagonal part of the system and ${\cal W}$
for the non-diagonal part which we would like to consider as a perturbation.
The matrices ${\cal D}$
and ${\cal W}$ have the form
\[ {\cal D}= \left( \begin{array}{cc} D(x) & 0 \\ 0 & \overline{D}(x) \end{array} \right),
\,\, {\cal W} = \left( \begin{array}{cc} 0 & W(x) \\ \overline{W}(x) & 0 \end{array}
\right) \]
with $D(x)= -\frac{i}{2 \sqrt{\lambda}} V(x)$ and $W(x)= -\frac{i}{2 \sqrt{\lambda}}
V(x) \exp (-2 i \sqrt{\lambda} x)$ in our case.
The main approach to the study of the asymptotics of solutions for
systems similar to (5) is to attempt to find some transformation which will
reduce the off-diagonal terms so that they will become absolutely integrable
and then try to apply Levinson's theorem \cite{Lev} on the $L^{1}$-perturbations
of the systems of linear differential equations.
It was discovered by Harris and Lutz \cite{HaLu} that when $W(x)$ is
a conditionally integrable function, the following simple transformation
of the system (5) works in some cases. We let
\begin{equation}
y(x)= (I + {\cal Q})z(x)
\end{equation}
where $I$ is an identity matrix, while ${\cal Q}$ satisfies ${\cal Q}'
={\cal W},$ that is,
\[ {\cal Q}(x)= \left( \begin{array}{cc} 0 & q(x) \\ \overline{q}(x) & 0
\end{array} \right) \]
with $q(x)= -\int^{\infty}_{x} W(x)\,dx.$
In this case $q(x) \stackrel{x \rightarrow \infty}{\longrightarrow} 0$,
so that for large enough $x$ the transformation (7) is non-singular
and preserves the asymptotics of the solutions. For the
new variable $z(x)$ we have:
\[ z'= (I + {\cal Q})^{-1}({\cal D} + {\cal DQ} + {\cal WQ})z \]
which after calculation leads to
\begin{equation}
z' = \left( \left( \begin{array}{cc} D & 0 \\ 0 & \overline{D}
\end{array} \right) + (1-|q|^{2})^{-1} \left(
\begin{array}{cc} \overline{W}q+ 2|q|^{2} \overline{D} &
2 \overline{q} \overline{D}- \overline{q}^{2}W \\
2 q D - q^{2} \overline{W} & 2 |q|^{2} D + \overline{q}
W \end{array} \right) \right) z.
\end{equation}
Since $q(x)$ decays at infinity, there is hope that
$q(x)D(x)$ and $q(x)^{2}W(x)$ may be both absolutely integrable,
even if initially $W(x)$ was not.
We now return to a particular case of the system (5) we consider.
Our $W(x)$ is equal to $-\frac{i}{2 \sqrt{\lambda}}V(x) \exp (-2i \sqrt
{\lambda} x)$, depending not only on $x$ but also on the energy $\lambda,$
and we are seeking to define $q(x, \lambda)= \frac{i}{2 \sqrt{\lambda}}
\int^{\infty}_{x} V(s) \exp (-2i \sqrt{\lambda} s).$ The next technical lemma
shows that under our assumption on the decay of potential we can do it
and, in fact, rather succesfully for every energy $\lambda$ which
belongs to the set $S$ in the statement of Theorem 1.1. \\
\noindent \bf Lemma 1.4. \it Suppose that $V(x) \in L^{1,\rm{loc}}$ satisfies
$V(x) \leq C|x|^{-3/4-\epsilon}$ for $|x|>a$ with some positive constants
$C, a, \epsilon.$ Then
for every $k \in {\cal M}^{+}(\Phi (V(x)x^{1/4})),$ the integral
$\int^{\infty}_{x} \exp (-iks)$ $V(s)\,ds$ converges and moreover
\[ \int\limits^{\infty}_{x} \exp (-iks) f(s)\,ds = O(x^{-1/4} \log x) \]
as $x \rightarrow \infty.$ \rm \\
\noindent \bf Proof. \rm Note that $V(x)x^{1/4}$ is square integrable and
therefore by Lemma 1.3, for every $k \in {\cal M}^{+}(\Phi (V(x)x^{1/4}))$
we have as $x \rightarrow \infty$
\[ \int\limits^{x}_{0} V(s)s^{1/4} \exp (-iks) \,ds = O( \log x) \]
(we change $ik$ in Lemma 1.3 to $-ik,$ but since $V$ is real, it
does not change the set $S$). Writing $V(s)=V(s)s^{-1/4}s^{1/4}$ and integrating by parts we get
\[ \int\limits^{x}_{0}V(s) \exp (-iks) \,ds = x^{-1/4} \int\limits^{x}_{0} V(s)s^{1/4}
\exp (-iks) \,ds + \]
\[ + \frac{1}{4} \int\limits^{x}_{0}t^{-5/4} \int\limits^{t}_{0}V(s)s^{1/4} \exp (-iks) \,ds dt. \]
The first summand clearly behaves at infinity like $O(x^{-1/4} \log x),$
while the second is absolutely convergent, since
\begin{equation}
\left| t^{-5/4} \int\limits^{t}_{0} V(s) s^{1/4} \exp(-iks) \,ds \right|
\leq C_{1} t^{-5/4} \log t
\end{equation}
by Lemma 1.3 for all $t>2$ with some constant $C_{1}.$ The integral over $(0,2)$
is finite since $V(x) \in L^{1, \rm{loc}}.$ Therefore, the
integral $\int^{x}_{0} V(s) \exp (-iks) \,ds$ is conditionally convergent and
its ``tail'' is equal to
\[ \int\limits^{\infty}_{x} V(s) \exp (-iks) \,ds = -x^{-1/4} \int\limits^{x}_{0} V(s)s^{1/4}
\exp (-iks) \,ds + \]
\[ + \frac{1}{4} \int\limits^{\infty}_{x}t^{-5/4} \int\limits^{t}_{0} V(s)s^{1/4} \exp (-iks) \,ds dt \]
which we can estimate for $x$ large enough using Lemma 1.3 and (8):
\[ \left| \int\limits^{\infty}_{x} V(s) \exp (-iks) \,ds \right| \leq \]
\[ \leq C_{1}x^{-1/4} \log x
+ C_{1} \int\limits^{\infty}_{x} t^{-5/4} \log t \,dt = O(x^{-1/4} \log x). \;\;\; \Box \]
Now we note that the condition $\lambda \in S$ is equivalent to
$2 \sqrt{\lambda} \in {\cal M}^{+}(\Phi (V(x)x^{1/4}))$ by the definition
of the set $S.$ Therefore, for every $\lambda \in S$ there is
a number $a_{\lambda}$ such that for any $x>a_{\lambda}$ the function
$q(x, \lambda)$ is less than $\frac{1}{2}.$ Applying the ``$I+{\cal Q}$'' transformation for
$x>a_{\lambda}$ for each $\lambda \in S,$ we get a system (7) for
$x>a_{\lambda}.$ The shift on the finite distance from the origin
certainly does not affect asymptotics since the evolution, corresponding to such
shift is just multiplication by some constant (for each
$\lambda$) matrix. Lemma 1.4 allows us to see that the non-diagonal
part and, in fact, the whole second summand of the matrix in the system (7)
is now absolutely integrable. Indeed, every element of this matrix
is equal to the product of some bounded function and the function
$V(x)q(x, \lambda)$, the latter being absolutely integrable and
moreover, by our assumption on $V$ and Lemma 1.4,
satisfying $|V(x)q(x, \lambda)|< C_{2}(\lambda)x^{-1-\epsilon}
\log x$ for every $\lambda \in S$ with the constant $C_{2}$ depending
on $\lambda.$ We could now apply Levinson's theorem, but in
our situation we do not need the whole power of this result.
Rewriting the system (7) for every $\lambda \in S$ as
\[ z'(x) = \left( \frac{i}{2 \sqrt{\lambda}} \left( \begin{array}{cc}
-V(x) & 0 \\ 0 & V(x) \end{array} \right) + R(x, \lambda) \right)z(x) \]
with $V(x)$ real and $||R(x, \lambda)|| \in L^{1}$ with
$\int^{\infty}_{x}||R(s, \lambda)||\,ds= O(x^{-\epsilon} \log x),$
we can in a standard way transfer this system into the system of
integral equations, apply the Gronwall lemma and prove (see \cite{ReSi} for the details)
that for each
$\lambda \in S$ there exist
solutions $z_{\lambda}(x),$ $\overline{z}_{\lambda}(x)$ with the asymptotics
\[ z_{\lambda}(x) = \exp \left( -\frac{i}{2 \sqrt{\lambda}} \int\limits^{x}_{0} V(x) dx \right)
(1+ O(x^{-\epsilon} \log x)). \]
Applying now transformations (6) and (3) to obtain the solution asymptotics of
the initial problem, we conclude the proof of Theorem 1.1.
\vspace{0.2cm}
\newline
\bf Remarks. \rm
1. One may apply the proven results to the study of
the absolutely continuous spectrum of Schr\"odinger operators with
spherically symmetric potentials
in $\R^{n},$ satisfying $|V(r)| \leq Cr^{-3/4-\epsilon}.$ In a standard way,
one decomposes the Schr\"odinger operator $H_{V}$ into a direct sum of
one-dimensional operators $H_{V,l}= -\frac{d^{2}}{dr^{2}}+(f_{n}(l)r^{-2}
+V(r))$ acting on different moment subspaces (see, e.g., \cite{ReSi2}).
It is easy to see that the set $S_{l}$ of energies
for which all solutions of the equation $H_{V,l} \phi - \lambda \phi$ are
bounded will be in fact independent of $l,$ since the term $f_{n}(l)r^{-2}$
decays fast at infinity.
Correspondingly, the singular spectrum of $H_{V}$ on
$\R^{+}$ may only be supported on the complement of the set $S.$
2. With very little effort, the introduced method yields
results for the whole-line problem for the Schr\"odinger operator $H_{V}$
with potential $V \in L^{1,\rm{loc}}$ satisfying $|V(x)| \leq C(1+ |x|)^{-3/4-\epsilon}$
for $|x|$ large enough. The substitution of Lemma 1.2 for the whole line
can be easily recovered from the remark in \cite{Sim3}
and says that on the set $S_{+} \cup S_{-},$
where $S_{+}$ and $S_{-}$ are the sets of energies for which all
solutions are bounded as $x$ approaches correspondingly plus or minus infinity,
the spectrum is purely absolutely continuous of multiplicity two (in the
sense of Lemma 1.2). Of course, Lemmas 1.3 and 1.4 can be used for studying
the asymptotics of solutions at $-\infty$ as well as at $+\infty.$
We get in this case that the whole positive half-axis is filled
by the absolutely continuous spectrum of multiplicity two
and the singular spectrum may only be supported on the complement
of $S_{+} \cup S_{-}.$ Moreover, it is a known fact \cite{Kac} that the
multiplicity of the singular spectrum may only be one for the whole-line
Sturm-Liouville operators.
3. In fact, Theorem 1.1 is more than a deterministic analog
of the Kotani-Ushiroya theorem in the power range $\alpha \in
(\frac{3}{4}, 1].$
Indeed, one can check that from the
assumption $\int_{M}F\,d \mu = 0$ in their random model follows that
a.e. potential is conditionally integrable and satisfies
\[ \int^{\infty}_{x} V(t, \omega)\, dt \leq C(\omega) (1+|x|)^{- \beta} \]
for every $\beta < \alpha - \frac{1}{2}$
with probability one. Assuming conditional integrability of $V$ and
certain power-decay estimate on the ``tail" of potential, we can
extend our result about the presence of the absolutely continuous spectrum
on potentials satisfying only $|V(x)| \leq Cx^{-2/3-\epsilon}.$
We treat this case in the Appendix. \\
As a byproduct of the computations we performed, let us formulate the
following proposition, which is in fact a slight variation of Theorem 2.1 from
Harris and Lutz \cite{HaLu}: \\
\bf Proposition 1.5. \it Suppose that for given energy $\lambda > 0,$ the
function $V(x) \int^{\infty}_{x} \exp (-2i\sqrt{\lambda}t)$ $V(t)\,dt$
is well-defined and belongs to $L^{1}(0,\infty).$ Then there exist
two linearly independent solutions $\phi_{\lambda},$ $\overline{\phi}
_{\lambda}$ of the equation $H_{V} \phi - \lambda \phi=0$ with
the following asymptotics as $x \rightarrow \infty:$
\[ \phi_{\lambda}(x)= \exp \left( i\sqrt{\lambda}x-\frac{i}{2\sqrt{\lambda}}
\int\limits^{x}_{0}V(s)\,ds \right) \left( 1+O \left( \int\limits^{\infty}_{x}
\left| V(s)\int\limits^{\infty}_{s} V(t)\exp
(-2i\sqrt{\lambda}t )\,dt \right| \, ds \right) \right). \]
In particular, all solutions are bounded. \rm \\
Based on the technique introduced in the proof of our main
theorem, we now prove the result showing that certain conditions
on the Fourier transform of potentials decaying faster than $x^{-3/4-\epsilon}$
are sufficient to ensure the absence of the singular component of the
spectrum on the positive semi-axis. \\
\noindent \bf Theorem 1.6. \it Suppose potential $V(x)$ satisfies $|V(x)|a$ and the Fourier transform $\Phi (x^{1/4}V(x))(k)$ belongs to $L^{p,\rm{loc}}$ for
some $p > 1/ \epsilon.$ Then the spectrum of the operator $H_{V}$ on the
positive semi-axis is purely absolutely continuous, and for every energy
$\lambda \in (0, \infty )$ there exist two solutions $\phi_{\lambda},$ $\overline{
\phi}_{\lambda}$ with the asymptotics as $x \rightarrow \infty$
\[ \phi_{\lambda}(x)= \exp \left( i\sqrt{\lambda}x-\frac{i}{2\sqrt{\lambda}}
\int\limits^{x}_{0}V(s)\,ds \right) \left( 1+O(x^{-\epsilon+1/p}) \right). \]
\rm It is clear that we can concentrate on proving the stated
asymptotics for every $\lambda$ in the positive half-axis. A slight modification
of Lemma 1.3 is needed: \\
\noindent \bf Lemma 1.7. \it Suppose that the Fourier transform $\Phi (f)(k)$ of the function
$f(x) \in L^{2}$ belongs to $L^{p,\rm{loc}},$ $p>2.$ Then for every value of $k$
\[ \int\limits^{N}_{-N}f(x) \exp (ikx) dx = O(N^{1/p}). \]
\bf Proof. \rm As in the proof of Lemma 1.3 making use of Parseval equality,
we get
\[ \int\limits^{N}_{-N}f(x) \exp (ikx) dx
= \int\limits^{\infty}_{0} \frac{
\sin Nt}{t}\,(\Phi(f)(k-t) + \Phi(f)(k+t))\,dt. \]
Again, the integral from $1$ to $\infty$ is bounded uniformly in $N$
by the product of $L^{2}$-norms of the functions under the integral.
The remaining part we split into two integrals and estimate them using
H\"older's inequality:
\[ \left| \: \int\limits^{1}_{1/N} \frac{\sin Nt}{t}(\Phi(f)(k-t) + \Phi(f)(k+t)) \,dt \right| \leq \]
\[ \leq \left( \: \int\limits^{1}_{1/N} (1/t)^{p'} \,dt \right) ^{1/p'} \left( \: \int\limits^{k+1}_{k-1}
|\Phi(f)(t)|^{p} \,dt \right) ^{1/p} = O(N^{1/p}) \]
where $p'$ is a conjugate exponent for $p:$ $p'= \frac{p}{p-1}.$
The second integral is estimated in a similar way:
\[ \left| \int\limits^{1/N}_{0} \frac{\sin Nt}{t}\,(\Phi(f)(k-t) + \Phi(f)(k+t)) \,dt \right| \leq \]
\[ \leq \left( \int\limits_{0}^{1/N} N^{p'} \,dt \right) ^{1/p'} \left( \: \int\limits^{k+1/N}_{k-1/N} |\Phi(f)(t)|
^{p} \,dt \right) ^{1/p} = O(N^{1/p}). \;\; \Box \] \\
\noindent \bf Proof of Theorem 1.6. \rm The same calculation which we performed proving Lemma 1.4 (integration by
parts) shows that under the conditions of Theorem 1.6 for every positive
$\lambda,$ we have
\[ q(x, \lambda)= \int\limits^{\infty}_{x} V(x) \exp (-i \sqrt{\lambda}x)\, dx
= O(x^{-1/4+1/p}) \]
as $x \rightarrow \infty.$ This implies that for all energies the
function $V(x)q(x, \lambda)$ is absolutely integrable and moreover satisfies
the estimate for large enough $x$
\[ |V(x)q(x, \lambda)| < C(\lambda) x^{-1-\epsilon +1/p}. \]
By Proposition 1.5, the proof is complete. $\Box$
\vspace{0.2cm}
\newline
\bf Remark. \rm It is easy to modify the proof of Lemma 1.7 and Theorem 1.6
to obtain a local criteria for the absence of singular spectrum.
That is, if $V$ satisfies the conditions of Theorem 1.6 and $\Phi(x^{1/4}V(x))
(k)$ belongs to $L^{p}(a,b),$ $b>a>0,$ then the spectrum of the operator
$H_{V}$ is purely absolutely continuous in the energy interval $(\frac{a^{2}}{4},
\frac{b^{2}}{4}).$ \\
We note that the conditions stated in the theorem are rather precise.
For example, in the celebrated Wigner-von Neumann example (historically
the first example of the decaying potential having positive eigenvalue
embedded in the absolutely continuous spectrum), the asymptotic behavior
of the potential at infinity is $V(x)=-8(\sin 2x)/x+O(x^{-2})$ (see, e.g., \cite{ReSi})
so that $\epsilon=\frac{1}{4},$ while the singularity of the Fourier transform of
$x^{1/4}V(x)$ is easily seen to be of the order $(k-2)^{-1/4}$ which belongs
to $L^{p,\rm{loc}}$ with $p<4.$ It is an open question whether one can replace
condition $\Phi(x^{1/4}V(x)) \in L^{p,\rm{loc}},$ $p>1/\epsilon$ with the
simpler one $\Phi(x^{1/4}V(x)) \in L^{1/\epsilon, \rm{loc}}$ so that the last theorem
still remains true.
\newpage
\begin{center}
\bf 2. Non-power decreasing potentials
\end{center}
In this section we apply the method described in the preceding
section of the paper to a wider class of potentials. This class
will include, in particular, certain potentials of the bump type,
which are ``mostly'' zero but have bumps decaying at infinity.
Let us introduce the class of potentials we will treat. \\
\noindent \bf Definition. \rm We say
that the potential $\tilde{V}(x) \in L^{\infty}(0, \infty)$ belongs to the class
${\cal P}_{-\alpha}(0, \infty)$
if there exists a potential $V(x) \in L^{\infty}(0, \infty)$
satisfying $|V(x)| \leq Cx^{-\alpha}$
for $x$ large enough and a countable collection of disjoint intervals in $(0, \infty)$
$\left\{ (a_{j}, b_{j}) \right\} _{j=1}^{\infty},$ $b_{j} \leq a_{j+1}$ $\forall j,$
such that
\[ \tilde{V}(x)= \left\{ \begin{array}{ll} 0, & x \in (a_{n}, b_{n}) \\
V(x- \sum\limits_{j=1}^{n} (b_{j}-a_{j})), & x \in (b_{n}, a_{n+1})
\end{array}. \right. \] \rm
Roughly, the potential $\tilde{V}(x)$ is obtained from $V(x)$ by inserting
a countable number of intervals on which $\tilde{V}(x)$ vanishes; while on the rest
of the axis, it is $V(x)$ shifted on the distance which is equal to
sum of the lengths of the intervals inserted so far. Of course, $\tilde{V}(x) \in {\cal P}_{-\alpha}
$ need not
decay faster than any power at infinity. However, if we ``compress''
$\tilde{V}(x)$ by collapsing all intervals on which it vanishes, we get a
potential which is bounded by $Cx^{-\alpha}$ for large $x.$
The following theorem holds for potentials from the class ${\cal P}_{-3/4-\epsilon}:$ \\
\bf Theorem 2.1. \it Suppose $\tilde{V}(x) \in {\cal P}_{-3/4-\epsilon},$ $\epsilon>0.$
Then the absolutely continuous part of the spectral measure fills the
whole positive semi-axis, in the sense that $\rho_{\rm{ac}}(T)>0$
for any measurable set $T \subset (0, \infty )$ with $|T|>0.$
For almost every energy $\lambda \in (0, \infty),$ there exist two solutions
$\phi_{\lambda}$, $\overline{\phi}_{\lambda}$ with the asymptotics as
$x \rightarrow \infty $
\[ \phi_{\lambda}(x)= \exp \left( i\sqrt{\lambda}x-\frac{i}{2\sqrt{\lambda}}
\int\limits^{x}_{0} \tilde{V} (s)\,ds \right) ( 1+o(1)). \]
\rm Proposition 1.5 implies that to prove the
stated result, we need only to show that the function $R(t)=\tilde{V}(x)
\int^{\infty}_{x}\tilde{V}(t) \exp (-2i\sqrt{\lambda}t) \,dt$ is well
defined and belongs to $L^{1}(0, \infty)$ for a.e.~$\lambda \in \R^{+}.$
To proceed with the proof, we need some further facts from the theory
of Fourier integral.
The following result is due to Zygmund \cite{Zyg}. \\
\bf Theorem (Zygmund). \it If $f \in L^{p}(-\infty, \infty )$ where $1 \leq p < 2,$
then the integral
\[ F(f)(k, N)= \frac{1}{\sqrt{2 \pi}} \int\limits^{N}_{-N} f(x) \exp (-ikx) dx \]
converges as $N \rightarrow \infty$, in an ordinary sense for almost
every value of $k.$ \rm \\
This will serve us as an analog of Lemma 1.3. However, it is a much more
sophisticated result by itself. One of the consequences is that we do
not have a description of the exceptional set on which convergence fails
(and correspondingly, where the singular spectrum may be supported).
For future reference, let us denote by $A(f)$ the set of full measure for
which the integral $F(f)(k,N)$ does converge.
The main idea now is the same as before: to perform in some ``clever'' way
integration by parts to get estimates on the tail
\[ \tilde{q}(x, \lambda)= \int\limits^{\infty}_{x} \tilde{V}(t)
\exp (-2i \sqrt{\lambda}t) \,dt \] for a.e.~$\lambda.$
Of course, there is no hope anymore that $q(x, \lambda)$ will, in general,
decay even as some power for potentials we now consider. However, the special
structure of the potentials allows us to overcome this problem. \\
\noindent \bf Proof of Theorem 2.1. \rm
Let us factorize $\tilde{V}(x)= \tilde{V}_{1}(x) \tilde{V}_{2}(x)$ in
the following way: if $\tilde{V}(x) = V(x, \{ (a_{j}, b_{j} \}_{j=1}^{j=\infty} )$
then
\[ \tilde{V}_{1}(x)= \left\{ \begin{array}{ll} 0, & x \in (a_{n}, b_{n}) \\
(x- \sum\limits_{j=1}^{n} (b_{j}-a_{j}))^{1/4}V(x- \sum\limits_{j=1}^{n} (b_{j}-a_{j})),
& x \in (b_{n}, a_{n+1})
\end{array} \right. \]
and
\[ \tilde{V}_{2}(x)= \left\{ \begin{array}{ll} (a_{n} -\sum\limits_{j=1}^{n-1}
(b_{j}-a_{j}))^{-1/4}, & x \in (a_{n}, b_{n}) \\
(x- \sum\limits_{j=1}^{n} (b_{j}-a_{j}))^{-1/4}, & x \in (b_{n}, a_{n+1}).
\end{array}. \right. \]
Therefore, $\tilde{V}_{1}(x)$ is obtained from the function $x^{1/4}V(x)$
in the same way as $\tilde{V}(x)$ is obtained from $V(x)$, while
the quotient $\frac{\tilde{V}(x)}{\tilde{V}_{1}(x)}=\tilde{V}_{2}(x)$
is a continuous piecewise differentiable non-increasing function.
Since $|V(x)|a. \]
For potentials of this type, Pearson \cite{Pear} has shown that if one chooses
the distances between bumps to be big enough, then if
$\sum^{\infty}_{n=1}g^{2}_{n}=\infty,$
the corresponding Schr\"odinger operator has purely singular continuous
spectrum on $\R^{+}$ and if $\sum^{\infty}_{n=1}g^{2}_{n}<\infty,$
the spectrum on the positive semi-axis is purely absolutely continuous.
Otherwise, there was essentially nothing known about
possible spectral behavior for Schr\"odinger operators with bump potentials
which are not absolutely integrable and not power decaying. From the
last theorem it follows that if $|g_{n}|0$ for any measurable set $T \subset (-2,2)$ with positive
Lebesgue measure. The singular component of the spectral measure
may be supported only on the complement of the set $S=2 \cos
(\frac{1}{2}{\cal M}^{+}(\Phi (n^{1/4} v(n))) \cap (-2,2)$ \rm(\it values of energy such that
$2 \arccos$ of half their value belongs to the set ${\cal M}^{+}(\Phi (n^{1/4} V(n));$
we fix the range of the $\arccos$ to be $[0, \pi].$ Moreover, for every $\lambda \in S$ there exist two linearly independent
solutions $\psi_{\lambda}(n)$, $\overline{\psi}_{\lambda}(n)$ with the
following asymptotics as $n \rightarrow \infty:$
\[ \psi_{\lambda}(n) = \exp \left( ikn+ \frac{i}{2 \sin k} \sum_{l=1}^{n}V_{l} \right)
(1+O(n^{-1/4} \log n)) \]
where $k= \arccos \frac{1}{2} \lambda.$ \rm \\
The strategy of the proof is the same as in the Schr\"odinger operators case.
The analogs of the three lemmas we used heavily are as follows: \\
\bf Lemma 3.2. \it Assume that for every $\lambda$ from the set $B,$
all solutions of the equation $h_{v} \phi - \lambda \phi$
are bounded. Then on the set $B,$
the spectral measure $\rho$ of the operator $h_{v}$ is purely absolutely
continuous in the following sense:
\rm{(i)} $\rho_{\rm{ac}}(A)>0$ for any $A \subseteq B$ with $|A|>0$
\rm{(ii)}$\rho_{\rm{sing}}(B)=0.$ \rm \\
\noindent \bf Proof. \rm This lemma follows from the subordinacy
theory for infinite matrices, developed by Khan and Pearson \cite{KhPe}.
Recently, Jitomirskaya and Last proved
more general results for Jacobi matrices \cite{JiLa}.
The reference for a simple direct proof of the lemma
is the paper of Simon \cite{Sim3}. $\Box$ \\
\noindent \bf Lemma 3.3. \it Consider the function $f(n) \in l^{2}(\Z).$ Then for every
$k_{0} \in {\cal M}^{+}(\Phi(f)),$ we have
\[ \sum^{N}_{l=-N} f(x) \exp (ik_{0}l) = O(\log N). \] \rm \\
\bf Proof. \rm Parseval equality in this case yields
\begin{eqnarray*}
\sum^{N}_{l=-N} f(l) \exp (ik_{0}l) & = & \frac{1}{2\pi} \int\limits^{\pi}_{-\pi}
\frac{ \sin (N+1/2)(k_{0}-k)}{\sin \frac{1}{2}(k_{0}-k)}\,
\Phi(v)(k) \,dk = \\
& = & \frac{1}{2\pi} \int\limits^{\pi}_{0} \frac{ \sin (N+1/2)(k)}{\sin \frac{1}{2}k}
\,(\Phi(v)(k_{0}+k)+\Phi(v)(k_{0}-k))\,dk.
\end{eqnarray*}
The final expression may be estimated exactly as in the proof
of Lemma 1.3. $\Box$ \\
\noindent \bf Lemma 3.4. \it Suppose that sequence $v(n)$ satisfies
$|v(n)|1/\epsilon.$ Then the spectrum of the operator $h_{v}$ on the segment
$(-2,2)$ is purely absolutely continuous.
Moreover, for every value of $\lambda \in (-2,2)$ there exist two
solutions $\psi_{\lambda}$ and $\overline{\psi}_{\lambda}$
of the equation $h_{v} \psi - \lambda \psi =0$
with the following asymptotics as $n \rightarrow \infty:$
\[ \psi_{\lambda}= \exp \left( ikn + \frac{i}{2 \sin k} \sum^{n}_{l=1}v(l) \right) (1+O
(n^{-\epsilon+1/p})) \]
where $k= \arccos \frac{1}{2} \lambda.$ \\
\noindent \bf Proof. \rm The proof is a complete analogy of the proof of Theorem 1.5.
One only needs to replace integration by parts with Abel's transformation
(summation by parts). $\Box$ \\
Finally, we discuss Jacobi matrices with non-power decaying potentials.
The class of potentials we treat is again potentials which are
``mostly'' zero and become power decaying after ``compression.''
Namely, we say that a potential $\tilde{v}(n)$ belongs to the class ${\cal D}_{\alpha}$
if there exists a potential $v(n)$ and two sequences of positive
integers $\{ a_{i} \}_
{i=1}^{\infty}$ and $\{ b_{i} \}_{i=1}^{\infty}$ satisfying $b_{i-1}<
a_{i}0.$ Moreover, for a.e.~$\lambda \in (-2,2)$ there exist
two linearly independent solutions $\psi_{\lambda},$ $\overline{\psi}_{\lambda}$
with the following asymptotics as $n \rightarrow \infty:$
\[ \psi_{\lambda}= \exp \left( ikn + \frac{i}{2 \sin k} \sum^{n}_{l=1}v(l) \right) (1+O
(n^{-\epsilon+1/p})) \]
where $k= \arccos \frac{1}{2} \lambda.$
\rm For the proof of this theorem we need an analog of Zygmund's
result for the case of Fourier series instead of Fourier integral.
We refer to the work of Menchoff \cite{Men} for the following result: \\
\noindent \bf Theorem (Menchoff). \it Suppose $\{ \phi_{n}(x) \}^{\infty}_{n=1}$
is an orthonormal system of functions on the interval $(a,b)$ and the
sequence $\{ c_{n} \}^{\infty}_{n=1} $ belongs to $l^{p}(\Z)$
$0 < p <2.$ Then the series
\[ \sum_{l=1}^{N} c_{n} \phi_{n}(x) \]
converges, in the ordinary sense, for almost every $x \in (a,b).$ \\ \rm
In particular, taking $\phi_{n}(x)=\exp (inx) $ and $(a,b)=(0,2 \pi ),$
we obtain an analog of Zygmund's theorem. \\
\noindent \bf Proof of Theorem 3.8. \rm Given Menchoff's theorem,
the proof essentially repeats the argument we gave to prove Theorem 2.1 in the second
section. $\Box$ \\
\newpage
\begin{center}
{\bf Appendix}
\end{center}
In this section we prove
\noindent \bf Theorem A.1 \it Suppose that potential $V$ satisfies
$|V(x)| \leq C_{1} x^{-\frac{2}{3}-\epsilon}$ and is conditionally
integrable with $| \int^{\infty}_{x} V(t)\,dt| \leq C_{2}x^{-\delta}$
for some positive $\delta.$ Then the absolutely continuous component
of the spectral measure of the operator $H_{V}$ fills the whole $\R^{+}.$ \\
\vspace{0.1cm}
\rm For the proof of the theorem, we need to introduce some
notation. Suppose that potential $V(x)$ satisfies $|V(x)| \leq C_{1}x^{-\alpha-\epsilon},$
with some $\alpha>\frac{1}{2}$ and $\epsilon >0.$ Then we denote by $S(V)$ the set of energies
\[ S(V)= \left\{ \lambda \left| |q(x, \lambda)|= \left |\int^{\infty}_{x}V(t) \exp (-2i
\sqrt{\lambda} t)\,dt \right| \leq C(\lambda) x^{-\alpha+\frac{1}{2}} \log x \right. \right\}. \]
Essentially repeating the proof of Lemma 1.4, one can readily see that $S(V)$
contains the set ${\cal M}^{+}(\Phi(x^{\alpha - \frac{1}{2}}V(x)))$ and hence
is a set of full measure. \\
\vspace{0.1cm}
\noindent \bf Proof. \rm To make the argument simpler, it is convenient
to modify slightly the ${\cal I}+{\cal Q}$ transformation we applied to the system
(4):
\[ y'(x)= \frac{i}{2\sqrt{\lambda}} \left( \begin{array}{cc} -V(x) &
-V(x) \exp (-2i \sqrt{\lambda} x) \\ V(x) \exp (2i \sqrt{\lambda} x) &
V(x) \end{array} \right) y(x). \]
Now we let
\[ y(x) = (1-|q|^{2})^{-1/2}({\cal I}+{\cal Q})z(x) \]
where ${\cal Q}$ and $q=q(x, \lambda)$ are the same as before.
As we did earlier
in Section 1, we will always assume that since we are interested in the asymptotics,
we perform the ${\cal I} + {\cal Q}$ transformation ``far enough" so that
$|q|<1$ for the $x$ we consider.
A calculation leads us to the following system for $z(x):$
\[ z' = \left( \left( \begin{array}{cc} D & 0 \\ 0 & \overline{D}
\end{array} \right) + (1-|q|^{2})^{-1} \left(
\begin{array}{cc} \frac{1}{2}(\overline{W}q-W\overline{q}) + 2|q|^{2} \overline{D} &
2 \overline{q} \overline{D}- \overline{q}^{2}W \\
2 q D - q^{2} \overline{W} & - \frac{1}{2}(\overline{W}q-W\overline{q}) + 2 |q|^{2} D
\end{array} \right) \right) z. \]
Here, as in the first section, $D$ stands for $-\frac{i}{2\sqrt{\lambda}}V(x)$
and $W$ for $-\frac{i}{2\sqrt{\lambda}}V(x) \exp (-2i \sqrt{\lambda}x).$
As we already mentioned above, on the set $S(V)$ of the full measure
we have $|q(x, \lambda)| \leq C(\lambda) x^{-1/6} \log x.$ Hence, for
all energies $\lambda \in S(V),$ the function $q^{2}(x, \lambda)V(x)$ is
absolutely integrable and $|\int^{\infty}_{x} q^{2}(x, \lambda)V(x)\,dx|
\leq C(\lambda) x^{-1-\epsilon} \log x$. Therefore, we can rewrite the
system in the following way:
\begin{equation}
z' = \left( \left( \begin{array}{cc} D + \frac{1}{2}(\overline{W}q-W\overline{q})
& 2 \overline{q} \overline{D} \\ 2qD & \overline{D} -\frac{1}{2}(\overline{W}q-W\overline{q})
\end{array} \right) + {\cal R}(x) \right) z,
\end{equation}
where all entries of the matrix ${\cal R}$ are from $L^{1}.$
The only dangerous terms are the off-diagonal terms in the matrix,
since the diagonal terms are purely imaginary and alone would not lead
to unbounded solutions. The main idea now is to iterate the ${\cal I} +
{\cal Q}$ transformation, improving the rate of decay of the off-diagonal
terms. To apply this procedure, we need first of all to ensure that
$qD=\frac{1}{4 \lambda}V(x)\int^{\infty}_{x}\exp (-2i\sqrt{\lambda}s)\, ds$
is an a.e. $\lambda$ integrable function. For any $\lambda \in S(V),$ we have:
\[ \int^{x}_{0} V(t) \int^{\infty}_{t} V(s)\exp (-2i\sqrt{\lambda}s) \, ds =
- \left( \int^{\infty}_{0}V(t)\,dt \right) \left( \int^{\infty}_{0}V(s) \exp (-2i\sqrt{\lambda}s)
\, ds \right) +\]
\[ + \left( \int^{\infty}_{x}V(t)\,dt \right) \left( \int^{\infty}_{x} V(s) \exp (-2i\sqrt{\lambda}s)
\, ds \right) - \int^{x}_{0} \left( V(t)\int^{\infty}_{t}V(s)\,ds \right) \exp (-2i\sqrt{\lambda}t)
\, dt. \]
Hence, it is easy to see that for the energies $\lambda$ which lie in both
$S(V(x))$ and $S(V(x)\int^{\infty}_{x}V(t)\,dt)$ we have (recall that by our assumption $|\int^{\infty}_{x}V(t)\,dt| \leq C_{1}x^{-\delta}$)
\[ \left| \int^{\infty}_{x} V(t) \int^{\infty}_{t} V(s)\exp (-2i\sqrt{\lambda}s)\,dsdt \right|
\leq C(\lambda)x^{-\frac{1}{6}-\delta} \log x. \]
Applying the modified ${\cal I} + {\cal Q}_{1}$-transformation
\[ z= ({\cal I} + {\cal Q}_{1})z_{1}\;\: \rm{with} \:\;
{\cal Q}_{1} = \left( \begin{array}{cc} 0 & q_{1} \\ \overline{q}_{1} & 0
\end{array} \right), \] where $q_{1}= \frac{1}{4\lambda} \int^{\infty}_{x} V(t)
\int^{\infty}_{t} V(s) \exp (2iks) \, dsdt,$ we get after a computation similar
to the one leading from the system (4) to (18)
\begin{equation}
z_{1}' = \left( \left( \begin{array}{cc} D + \frac{1}{2}(\overline{W}q-W\overline{q})
& 2 \overline{q}_{1} \overline{D} \\ 2q_{1}D & \overline{D} -\frac{1}{2}(\overline{W}q-W\overline{q})
\end{array} \right) + {\cal R}_{1}(x) \right) z_{1}.
\end{equation}
Here ${\cal R}_{1}$ is a matrix with entries from $L^{1}.$ The off-diagonal
terms in the system (19) have a rate of decay $|q_{1}(x,\lambda)V(x)| \leq
C(\lambda) x^{-\frac{5}{6}-\delta} \log x $ for a.e. $\lambda.$
To complete the proof, we need to apply the ${\cal I} + {\cal Q}$ transformation
several times. The following lemma shows that under the assumptions of the
theorem we can do this and it also
determines the number of necessary iterations and the set
of full measure for which we can derive the asymptotics of solutions.
\vspace{0.1cm}
\noindent \bf Lemma A.1. \it Under the assumptions of Theorem A.1, the function
\[ f_{n}(t_{1},\lambda)= V(t_{1}) \int^{\infty}_{t_{1}}V(t_{2})\int^{\infty}_
{t_{2}} V(t_{3})...\int^{\infty}_{t_{n}} V(t_{n+1}) \exp (-2i \sqrt{\lambda}
t_{n+1}) \, dt_{n+1} \]
is integrable for every $\lambda \in \tilde{S}_{n}=\bigcap^{n}_{j=0}S_{j}$, where
\[ S_{j}= S \left( V(t_{1}) \left( \int^{\infty}_{t_{1}}V(t_{2})\,dt_{2} \right) ^{j} \right) \]
and moreover,
\[ \left| \int^{\infty}_{x} f_{n}(t_{1}, \lambda)\,dt_{1} \right| \leq Cx^{-\frac
{1}{6}-n \delta} \log x. \]
\vspace{0.05cm}
\noindent \bf Proof. \rm The proof is by induction. We have already checked that for
$n=1$ the statement is true. For the sake of simplicity, we assume
integrability and give an apriori estimate for the tail integral.
Of course, one can easily prove integrability by essentially the
same (but longer) computation. Now, integrating by parts, we find that
\[ \int^{\infty}_{x} f_{n}(t_{1}, \lambda)\, dt_{1} =
\left( \int_{x}^{\infty} V(t_{1})\, dt_{1} \right) \left( \int^{\infty}_{x} f_{n-1}(t_{1},
\lambda)\,dt_{1} \right) - \int^{\infty}_{x} f_{n-1}(t_{1}, \lambda)
\left( \int^{\infty}_{t_{1}} V(t_{2})\, dt_{2} \right) \, dt_{1} \]
According to the induction hypothesis and our assumption on $V,$
the first summand on the right hand side is bounded by $C(\lambda)
x^{-\frac{1}{6}-n\delta} \log x$ for every $\lambda \in \tilde{S}_{n-1}.$
In the second summand we perform integration by parts, integrating
$V(t_{1})\int^{\infty}_{t_{1}}V(t_{2})\,dt_{2}.$ As a result we get:
\[
-\int^{\infty}_{x} f_{n-1}(t_{1}, \lambda) \int^{\infty}_{t_{1}}
V(t_{2})\, dt_{1}dt_{2} = -\frac{1}{2} \left( \int^{\infty}_{x}V(t_{1})\,dt_{1}
\right)^{2} \int^{\infty}_{x} f_{n-2}(t_{1}, \lambda) \, dt_{1} + \]
\[ + \frac{1}{2} \int^{\infty}_{x} \left( \int^{\infty}_{t_{1}}V(t_{2})\,dt_{2}
\right)^{2} f_{n-2}(t_{1}, \lambda)\, dt_{1} \]
As before, the first term decays as $Cx^{-\frac{1}{6}-n\delta}
\log x$ for every $\lambda \in \tilde{S}_{n-2}.$ We continue to integrate
by parts the second term, integrating $V(t_{1})(\int^{\infty}_{t_{1}}
V(t_{2}) dt_{2} )^{2};$ we again get a sum of two terms the first
of which (off-integral) is well-behaved while the second is again integrated
by parts. We perform such a procedure $n$ times and in the end, summarizing
the result of the whole calculation we find that
\[ \int^{\infty}_{x} f_{n}(t_{1}, \lambda)\, dt_{1} = g(x, \lambda)
+ \frac{(-1)^{n}}{n!} \int^{\infty}_{x} V(t_{1}) \left( \int^{\infty}_{t_{1}}
V(t_{2}) dt_{2} \right) ^{n} \exp (2i \sqrt{\lambda} t_{1})\, dt_{1}, \]
where $g(x, \lambda)$ satisfies the decay condition
\[ |g(x, \lambda)| \leq C(\lambda) x^{-\frac{1}{6}-n\delta} \log x \]
for any $\lambda \in \tilde{S}_{n-1}.$ The last term obviously satisfies
the same estimate for every $\lambda \in S_{n}.$ Hence, as claimed,
$\int^{\infty}_{x} f_{n}(t, \lambda) \leq C(\lambda)
x^{-\frac{1}{6}-n\delta} \log x$ for every $\lambda \in \tilde{S}_{n}.$ $\Box$
\vspace{0.2cm}
The proven lemma justifies the iteration of the ${\cal I} + {\cal Q}$ transformation,
since on the $n$-th iteration, to obtain $q_{n},$ we need to integrate $q_{n-1}D$
which, up to irrelevant energy dependent constants, is exactly $f_{n}$
from the statement of the lemma.
After the $n$th iteration we arrive at a system
\[ z_{n}' = \left( \left( \begin{array}{cc} D + \frac{1}{2}(\overline{W}q-W\overline{q})
& 2 \overline{q}_{n} \overline{D} \\ 2q_{n}D & \overline{D} -\frac{1}{2}(\overline{W}q-W\overline{q})
\end{array} \right) + {\cal R}_{n}(x) \right) z_{1}, \]
where the matrix ${\cal R}_{n}$ has absolutely integrable entries.
Also by the second statement of the lemma, $|q_{n}(x, \lambda)V(x)| \leq C(\lambda) x^{-\frac{5}{6}-n\delta} \log x$
for every $\lambda \in \tilde{S}_{n}$ and is therefore absolutely integrable
as soon as $n>\frac{1}{6 \delta}.$
Therefore, for the energies from the set
$\tilde{S}_{m}$ of full measure, $m= \left[ \frac{1}{6\delta} \right] +1$
iterations are enough to bring the system to the form where we can
apply Levinson's theorem (or, as was noticed in Section 1,
just use integral equation technique bearing in mind that our
unperturbed eigenfunctons are bounded). We also
note that for every $\lambda \in \tilde{S}_{m}$, transforming
back, we get solutions $\phi_{\lambda}(x)$ and $\overline{\phi_{\lambda}(x)}$
with the asymptotics
\begin{eqnarray*}
\phi_{\lambda}(x) & = & \left( \exp (i\sqrt{\lambda}x - \frac{i}{2\sqrt{\lambda}}
\int^{x}_{0}V(t)\,dt+ \right. \\ & & \left. + \frac{i}{4\lambda} \int^{x}_{0}V(t)\int^{\infty}_{t}
\sin (2 \sqrt{\lambda}(t-s))V(s)\, dsdt \right) \times \left( 1 + O(x^{-\rho} \log x) \right)
\end{eqnarray*}
where $\rho= \min (\epsilon, m\delta-\frac{1}{6}).$
The solutions $\phi_{\lambda}(x)$ and $\overline{\phi_{\lambda}(x)}$
are bounded and clearly linearly independent. This completes the proof
of the theorem. $\Box$
\vspace{1cm}
\begin{center}
{\bf Acknowledgments } \\
\end{center}
The author is very grateful to Professor B.~Simon for his support, stimulating
discussions and valuable comments.
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\end{document}