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Stark operators, absolutely continuous spectrum, decaying perturbations,
smooth perturbations
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% very minor changes by MC 1/26/2001
% modified by AK 1/24/2001
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\begin{document}
\newtheorem{theorem}{Theorem}[section]
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\author{Michael Christ}
\address{
Michael Christ\\
Department of Mathematics\\
University of California \\
Berkeley, CA 94720-3840, USA}
\email{mchrist@math.berkeley.edu}
\thanks{The first author was supported in part by NSF grant DMS-9970660
and completed this research while on appointment as a
Miller Research Professor in the Miller Institute for Basic Research
in Science.}
\author{Alexander Kiselev}
\address{ Alexander Kiselev\\
Department of Mathematics\\
University of Chicago\\
Chicago, Ill. 60637}
\email{kiselev@math.uchicago.edu}
\thanks{The second author was supported in part by NSF grant DMS-9801530}
\date\today{}
% 1/26/2001
\title[Absolutely continuous spectrum of Stark operators]
{Absolutely continuous spectrum \\ of Stark operators}
\begin{abstract}
We prove several new results on the absolutely continuous spectra of
perturbed one-dimensional
Stark operators. First, we find new classes of perturbations,
characterized mainly by smoothness
conditions, which preserve purely absolutely continuous spectrum. Then
we establish stability of
the absolutely continuous spectrum in more general situations, where
imbedded singular spectrum
may occur. We present two kinds of optimal conditions for the stability of
absolutely continuous
spectrum: decay and smoothness. In the decay direction, we show that a
sufficient (in the power scale) condition is
$|q(x)| \leq C(1+|x|)^{-\frac{1}{4}-\epsilon};$ in the smoothness
direction, a sufficient condition in H\"older classes is
$q \in C^{\frac{1}{2}+\epsilon}(\reals)$.
On the other hand, we show that there exist potentials which both satisfy
$|q(x)| \leq C(1+|x|)^{-\frac14}$
and belong to $C^{\frac12}(\reals)$ for
which the spectrum becomes purely singular
on the whole real axis, so that the above results are optimal within the
scales considered.
\end{abstract}
\maketitle
\section{Introduction}
In this paper we consider the Stark operator
\begin{equation}
\label{stark}
H_q = -\frac{d^2}{dx^2} -x +q(x)
\end{equation}
defined on the whole real line $\reals.$ This operator describes a charged quantum
particle in a constant electric field subject to an
additional electric potential $q(x).$ There exists an extensive physical and
mathematical literature on Stark operators; for a review, see e.g. \cite{CFKS}.
When $q(x)=0,$ the operator has purely absolutely continuous
spectrum. The question we wish to address is which classes of perturbations
$q$ preserve this property. We will consider two classes of conditions
that ensure preservation of the absolutely continuous spectrum:
smoothness and decay.
%The fact that the effects of decay and smoothness of potential on the
%spectral properties of Stark operators are somewhat similar, is well-known.
The first result on the smoothness condition was proven by Walter
\cite{Wal}, who showed that if the potential is bounded and
has two bounded derivatives,
the spectrum remains purely absolutely continuous. Similar results were
obtained by Bentosela, Carmona, Duclos, Simon, Souillard and Weder
in \cite{Ben} using Mourre's method. A corollary noted in \cite{Ben}
is a drastic change in the spectral properties of Schr\"odinger
operators of Anderson model type, say
\[ H_\omega = -\frac{d^2}{dx^2} + \sum\limits_n a_n(\omega) V(x-n) \]
where $a_n$ are independent identically distributed random variables
and $V \in C_0^2 ((0,1)),$ when a constant electric field is switched on.
The spectrum changes from almost surely pure point to purely absolutely
continuous.
Recently, Sahbani
\cite{Sa1,Sa2} relaxed smoothness
conditions of \cite{Wal} and \cite{Ben} (see the remark after Theorem~\ref{spac2}).
%Slightly weaker results were proven in \cite{Ki1}.
On the opposite side of the smoothness scale, Delyon, Simon and Souillard \cite{DSS}
showed that for a periodic array of $\delta$ function potentials with random couplings
in a constant electric field, the spectrum is purely singular.
Avron, Exner and Last
\cite{AEL} realized that the spectrum may be purely singular even
for a deterministic periodic array of very singular interactions, such
as $\delta'.$
Generalizations of these results, as well as other models with singular
potentials, were considered in \cite{Min,Ex,ADE,MS,Briet,ABD}.
There remained a gap, however, between the classes of
potentials for which localization was known to occur, and those
for which the
spectrum was known to remain absolutely continuous.
As far as decay conditions are concerned, it is well known that if
$q(x)$ satisfies $|q(x)| \leq C(1+|x|)^{-\alpha},$ $\alpha>1/2,$ then the
spectrum remains purely absolutely continuous \cite{Va}.
Moreover, there are examples where $|q(x)x^{1/2}|\leq C$ and
isolated imbedded eigenvalues appear. If
$|q(x)|x^{1/2} \rightarrow \infty,$ it was shown by Naboko and Pushnitski
\cite{NaPu} that
dense (imbedded) point spectrum
may appear on all of $\reals$. We remark that for the
operator without electric field, the decay threshold where imbedded
eigenvalues may appear is the power $-1;$ of course, it is physically natural
that it is more difficult to get an imbedded eigenvalue in the presence
of the constant electric field.
However, if we do not wish to rule out imbedded singular spectrum,
it has been shown in \cite{Ki1} that the absolutely continuous spectrum
of a perturbed Stark operator still fills the whole real axis
when $|q(x)| \leq C(1+|x|)^{-\alpha},$ $\alpha>1/3.$
The question what is the critical rate of decay for which
the spectrum may become purely singular remained open.
Our main goal in this paper is to prove two sharp results on the preservation
of the absolutely continuous spectrum of Stark operators.
Recall that $f(x)$ is
called H\"older continuous with exponent
$\alpha$ ($f \in C^\alpha(\reals)$)
if
\[ \|f\|_{C^\alpha} = {\rm sup}_x |f(x)| + {\rm sup}_{x,y}
\frac{|f(x)-f(y)|}{|x-y|^\alpha} < \infty. \]
We will analyze solutions of the generalized eigenfunction equation
\begin{equation} \label{starke}
-u'' -xu + q(x)u = Eu\ .
\end{equation}
Here $-x$ represents a background potential due to a constant
electrical field, while $q$ is some perturbation.
\begin{theorem}
\label{sac1}
Assume that the potential $q(x)$ is H\"older continuous with exponent $\alpha>1/2.$
Then an essential support of the absolutely continuous part of the spectral
measure coincides with the whole real axis. Moreover, for a.e.~$E,$ all
solutions $u(x,E)$ of equation
\eqref{starke} satisfy $u(x,E)=O(x^{-1/4}),$ $u'(x,E)=O(x^{1/4})$ as $x \rightarrow +\infty.$
%\begin{equation}
%\label{wkbas}
%u_+(x,E) = x^{-1/4} e^{i\int\limits_0^x \left[ \sqrt{1 +\frac{E}{s}}s^{1/2} -
%\frac{q(s)}{2s^{1/2} \sqrt{1+\frac{E}{s}}} \right]\,ds} \left(1+o(1)\right)
%\end{equation}
%as $x \rightarrow \infty.$
\end{theorem}
\it Remark. \rm 1. An essential support of $\mu$ is a set $S$
such that $\mu(\reals \setminus S)=0$ and
$\mu (S_1)>0$ for any $S_1 \subset S$ of positive Lebesgue measure. \\
2. In this and subsequent theorems, only behavior of $q(x)$
for $|x|$ large matters. We will always implicitly assume
$q$ to be locally integrable, and will state only additional
hypotheses which concern its behavior for large $x$.
On the negative part of the real axis, it is sufficient
for all our conclusions
to require that $q(x)-x \rightarrow +\infty$ as $x \rightarrow -\infty$.
We prefer to state the results in a slightly weaker form to avoid
making statements too cumbersome.
\begin{theorem}
\label{dac1}
Assume that the potential $q(x)$ is locally integrable,
and that $q(x^2) \in L^p$ for some $1 \leq p <2.$
Then an essential support of the absolutely continuous part of the
spectral measure coincides with the whole real axis.
Moreover, for a.e.~$E$, all solutions $u(x,E)$ of equation
\eqref{starke} satisfy $u(x,E)=O(x^{-1/4}),$ $u'(x,E)=O(x^{1/4})$
%there exists a solution $u_+(x,E)$ of equation
%\eqref{starke} with the asymptotic behavior
%\begin{equation}
%\label{wkbas}
%u_+(x,E) = x^{-1/4} e^{i\int\limits_0^x \left[ \sqrt{1 +\frac{E}{s}}s^{1/2} -
%\frac{q(s)}{2s^{1/2} \sqrt{1+\frac{E}{s}}} \right]\,ds} \left(1+o(1)\right)
%\end{equation}
as $x \rightarrow +\infty.$
\end{theorem}
\it Remark. \rm 1. In particular,
the assumption of Theorem~\ref{dac1} is satisfied if
$|q(x)| \leq C(1+|x|)^{-\alpha}$, for some $\alpha>1/4$.
\\
2. An explicit expression for the leading term in an asymptotic
expansion of $u(x,E)$, as $x\to+\infty$,
can also be derived under the hypotheses of
Theorems~\ref{sac1} and \ref{dac1};
see Theorem~\ref{genthm} and Section~\ref{sm}.
Both Theorems~\ref{sac1} and \ref{dac1} are direct corollaries of the
following more general result.
\begin{theorem}\label{genthm}
Consider a Stark operator $H_q$ on $\reals^1$.
Assume that the potential $q(x)$ admits a decomposition
$q=q_1+q_2,$ where both $q_1(x^2)$ and $x^{-1}q'_2(x^2)$ belong
to $(L^1+L^p)(\reals),$ and that there exists $\zeta<1$ such that
$|q_2(x)|\leq \zeta |x|$ for sufficiently large $|x|$.
Then for almost every
energy $E$ there exists a solution $u_+(x,E)$ of equation
\eqref{starke} with the asymptotic behavior
\begin{equation}
\label{wkbas}
u_+(x,E) = (x-q_2(x)+E)^{-1/4}
e^{i\phi(x,E)}
%e^{i\int\limits_0^x
%\left[\sqrt{t-q_2(t)+E}-\frac{q_1(t)}{2\sqrt{x-q_2(t)+E}}\right]}
\left(1+o(1)\right)
\end{equation}
as $x \rightarrow +\infty$,
where
\begin{equation*}
\phi(x,E)
= \int\limits_0^x
\Big[\sqrt{t-q_2(t)+E}-\frac{q_1(t)}{2\sqrt{x-q_2(t)+E}}\Big]
\,dt
\end{equation*}
\end{theorem}
\it Remark. \rm
A sufficient condition on the derivative is that
$|q'(x)| \leq C(1+|x|)^{\alpha}$, for some $\alpha < 1/4$.
A surprising aspect of this theorem is that the perturbation $q_2$
is allowed virtually as much growth as the constant electric field
potential, and more flexibility on the derivative.
Theorem~\ref{dac1} follows immediately; Theorem~\ref{sac1} requires a
simple argument showing that
any $C^\alpha$ potential with $\alpha>1/2$ can be represented as in
Theorem~\ref{genthm}.
We sketch this argument in Section~\ref{sm}.
We notice that in the case of a Schr\"odinger operator without constant
electric field, the absolutely continuous spectrum is preserved for potentials
with power decay rate $\alpha>1/2$ \cite{CK,Re,DK}.
There exist potentials $V$ satisfying $|V(x)x^{1/2}| \leq C$ for which
the absolutely continuous spectrum is destroyed \cite{KU,KLS},
so that $\alpha = 1/2$ is a sharp threshold.
It is natural that in the presence of a constant electrical field,
the absolutely continuous spectrum is preserved under more slowly decaying
perturbations of the potential.
The next result shows optimality of Theorems~\ref{sac1} and \ref{dac1}.
Fix $f \in C_0^\infty ((0,1)),$ not identically zero,
and let $a_n(\omega)$ be independent,
identically distributed random variables with uniform distribution in
$[0,2\pi].$ Denote
$c=\left({3}/{2}\right)^{2/3}.$ Let us define
\begin{equation}
\label{qx}
q(x) = c \sum\limits_{n=1}^\infty n^{-1/2} f(\sqrt{cx}-n)
\sin (\tfrac{4}{3} x^{\frac{3}{2}} + a_n(\omega)) \ .
\end{equation}
(There is nothing magic in the choice of $c;$ however, this choice
will simplify computations later.)
We have
\begin{theorem}
\label{count}
Let $q(x)$ be a random potential given by \eqref{qx}. Then for a.e.~$\omega,$
the spectrum of the corresponding perturbed Stark operator
is purely singular on the whole real line.
\end{theorem}
In particular, any realization of $q$ defined by \eqref{qx}
satisfies $|q(x)| \leq Cx^{-1/4}$ and
belongs to $C^{1/2}(\reals),$ so Theorem~\ref{count}
assures sharpness of Theorems~\ref{sac1} and \ref{dac1} in H\"older spaces
and in the power scale, respectively.
On a more detailed level, we may want to distinguish between
perturbations for which the absolutely continuous
spectrum is preserved, but imbedded
singular spectrum may appear, and perturbations
which preserve purely absolutely continuous spectrum.
Our final results provide two criteria ensuring pure absolute
continuity of the spectrum.
Recall that a function $g(x)$ defined
on a real line is called smooth in Zygmund's sense if
\[ \int\limits_0^1 {\rm sup}_{x \in \reals} |g(x+\epsilon)-2g(x)+
g(x-\epsilon)| \frac{d \epsilon}{\epsilon^2} < \infty. \]
\begin{theorem}
\label{spac1}
Assume that the potential $q(x)$ is bounded,
has bounded continuous first derivative and is smooth in Zygmund's sense.
Then the Stark operator \eqref{stark} has purely absolutely continuous
spectrum on the whole real axis.
\end{theorem}
We can also allow the potential to grow at a rate arbitrarily close
to that of the constant electric field, and still have purely absolutely
continuous spectrum, provided that we impose a slightly different condition
on smoothness.
\begin{theorem}
\label{spac2}
Assume that the potential $q(x)$ satisfies $q(x) = O(x^\alpha)$
as $|x|\to\infty$ for some
$\alpha<1,$ is differentiable, and that its derivative $q'(x)$ is Dini
continuous:
\[ \int\limits_0^1 {\rm sup}_x |q'(x+\epsilon)-q'(x-\epsilon)|
\frac{d \epsilon}{\epsilon} < \infty. \]
Then the spectrum of the perturbed Stark operator \eqref{stark} is purely
absolutely continuous on the whole real axis.
\end{theorem}
\it Remark. \rm The classes of potentials that
are smooth in Zygmund's sense or have Dini continuous derivative
were first considered in this context by Sahbani in \cite{Sa1, Sa2}.
Using the conjugate operator approach,
he proved that under conditions similar to Theorem~\ref{spac1} or
Theorem~\ref{spac2} (but with stronger growth restrictions) the spectrum
is absolutely continuous with perhaps some imbedded eigenvalues.
His results extend (in a slightly weaker form) to the higher dimensional
setting. \\
We will later discuss examples demonstrating that
the criteria given by Theorems~\ref{spac1} and \ref{spac2}
are fairly sharp.
We employ two different approaches to prove the stated results.
First,
to prove Theorems~\ref{spac1}, \ref{spac2}, and \ref{count},
we apply a Liouville transformation to reduce the Stark operator
to a form reminiscent of the Schr\"odinger operator without electric
field, but with the energy entering in a non-standard way.
We then use a Pr\"ufer transformation
to analyze the asymptotic behavior of solutions.
In the proofs of the other results, it is more convenient to
represent \eqref{starke} as a first-order system and to employ
estimates for the solution series. Some of the tools for these estimates
come from our recent work \cite{CK,CK1,CK2,CKN}.
As soon as we have control over the asymptotic behavior
of solutions, we can apply the whole axis version of subordinacy theory
due to Gilbert \cite{Gi}, or the approximate eigenvectors criterion of \cite{CKL},
to draw spectral conclusions.
\section{Preservation of purely absolutely continuous spectrum}
We begin by proving Theorems~\ref{spac1} and \ref{spac2} as a warm-up.
All the proofs of spectral properties in this paper rely on the
study of solutions of the equation \eqref{starke},
$-u'' -xu +q(x)u = Eu$.
The link between
the behavior of solutions and spectral results is provided by
Gilbert-Pearson subordinacy
theory, more particularly, by the whole-line version
of this theory due to Gilbert \cite{Gi}. Recall that a real solution
$u_1(x,E)$ of \eqref{starke} is called subordinate on the right
if for any other real linearly independent solution $u_2(x,E)$ we have
\[ \lim_{N \rightarrow \infty} \frac{\int\limits_0^N |u_1(x,E)|^2\,dx}
{\int\limits_0^N |u_2(x,E)|^2 \,dx} =0. \]
Subordinacy on the left is defined similarly.
Note that it is easy to see that for equation \eqref{starke}, under the
assumptions of any of our theorems, there is always a solution subordinate
(in fact, $L^2$) on the left since the potential goes to $+\infty$ there.
The main result of Gilbert implies that singular spectrum may only be
supported on the set of energies where there exists a solution
subordinate on both sides. Moreover, the set of the energies
where there exists a solution
subordinate on one side, but there is no subordinate solution on the other
side, is an essential support of the absolutely continuous spectrum, of
multiplicity one. Therefore, our goal is to prove that for all
energies (if we want to show pure absolute continuity), or for a.e. energy
(if we allow imbedded singular spectrum), there is no solution of
\eqref{starke} subordinate on the right.
In the equation \eqref{starke}, let us perform a Liouville transformation
given by (see, e.g. \cite{Ol})
\begin{equation}
\label{liou}
\xi(x) = \int\limits_0^x \sqrt{t} \, dt = \tfrac{2}{3}x^{3/2},
\,\,\,
\phi (\xi) = x(\xi)^{1/4} u(x(\xi)).
\end{equation}
This transformation introduces an irrelevant singularity at the origin;
henceforth we always work outside some neighborhood of $0$.
The resulting function $\phi$ satisfies the Schr\"odinger equation
\begin{equation}
\label{basic1}
-\phi'' + \left( \frac{5}{36 \xi^2} + \frac{q(c\xi^{2/3}) -E}
{c \xi^{2/3}} \right) \phi = \phi,
\end{equation}
where $c=({3}/{2})^{2/3}$. Let us introduce
a short-hand notation $V(\xi,E)$ for the expression in brackets in
\eqref{basic1}. Let us further apply a Pr\"ufer transformation
to the equation for $\phi,$ setting for each $E$
\begin{equation} \label{proofer}
\begin{split}
\phi(\xi,E) & = R(\xi,E) \sin (\theta(\xi,E)) \\
\phi'(\xi,E) & = R(\xi,E) \cos (\theta(\xi,E)).
\end{split}
\end{equation}
The equations for $R$ and $\theta$ are as follows:
\begin{align}
\label{pru1}
(\log R(\xi, E))'
& =
\tfrac{1}{2}V(\xi,E)\sin (2\theta(\xi,E)) \\
\label{pru2}
\theta'(\xi,E)
& =
1- \tfrac{1}{2}V(\xi,E)(1-\cos (2\theta(\xi,E))).
\end{align}
Our main goal in proof of Theorem~\ref{spac1} will
be to show the convergence of the integral
\begin{equation}
\label{keyint}
\int\limits_{1}^{N} \left( \frac{5}{36\xi^{2}} +
\frac{-E + q(c\xi^{\frac{2}{3}})}{c\xi^{\frac{2}{3}}}\right)
\sin (2\theta(\xi,E))\, d \xi
\end{equation}
as $N \rightarrow \infty$ for every $E.$ This goal is motivated by
\begin{proposition}
\label{GP}
Suppose that
\[ \limsup_{x \rightarrow \infty}|q(x)|/x = \zeta <1, \]
and for a given $E$ the integral
\eqref{keyint} converges for all initial values of $\theta(0,E).$
Then for this value of $E,$ there is no subordinate on the right solution
of the equation \eqref{starke}.
\end{proposition}
\begin{proof}
If for a given value of $E$ the integral \eqref{keyint}
converges, it follows from \eqref{pru1} and \eqref{pru2} that all solutions
of the equation \eqref{basic1} are bounded and, moreover,
any solution $\phi_\beta$ (where $\beta$ parametrizes
boundary condition) has the following asymptotic behavior
as $\xi \rightarrow +\infty:$
\begin{equation*}
\phi_{\beta}(\xi,E) = C_\beta \sin (\xi + g_{\beta}(\xi,E))(1+o(1)),
\end{equation*}
where $|g_{\beta}'(\xi,E)|<\zeta<\zeta_1<1$ for $\xi$ sufficiently large.
Going back to the original equation \eqref{starke},
we infer that
every solution $u_{\beta}(x,E)$
has the following asymptotic behavior as
$x \rightarrow \infty:$
\[ u_{\beta}( x, E) =
C_\beta x^{-\frac{1}{4}}
\sin (\tfrac{2}{3}x^{\frac{3}{2}}+ f_{\beta}(x,E))(1+o(1)), \]
where $|f_{\beta}'(x,E)|<\zeta_1 x^{1/2}$ for sufficiently large $x.$
For any $u_{\beta}(x,E)$ we find
\begin{multline} \label{normg}
\int\limits_{1}^{N} |u_{\beta}(x,E)|^{2}\,dx = C_\beta^2\int\limits_{1}^{N}
x^{-\frac{1}{2}}(\sin(\frac{2}{3}x^{\frac{3}{2}}+ f_{\beta}(x,E))^{2}\,
dx \left( 1+o(1)\right) \\
= C_\beta^2
\left( N^{\frac{1}{2}} - \frac{1}{2}\int\limits_{1}^{N}x^{-\frac{1}{2}}
\cos (\tfrac{4}{3}x^{\frac{3}{2}}+2 f_{\beta}(x,E))\, dx \right)
\left(1+o(1)\right).
\end{multline}
Consider the integral
\begin{equation}
\label{auxint}
I_\beta(N)= \int\limits_{1}^{N}x^{-\frac{1}{2}}
e^{\left(i\frac{4}{3}x^{\frac{3}{2}}+2 f_{\beta}(x,E)\right)}\, dx.
\end{equation}
Integrating by parts, with $e^{2i f_\beta(x,E)}$ being differentiated,
we obtain that \eqref{auxint} is equal to
\begin{equation*}
-\int\limits_N^\infty x^{-\frac{1}{2}}
e^{i\frac{4}{3}x^{\frac{3}{2}}} \, dx e^{2i f_\beta(N,E)}
+
2i \int\limits_1^N f'_\beta(x,E)
e^{2i f_\beta(x,E)} \int\limits_x^\infty t^{-\frac{1}{2}}
e^{i\frac{4}{3}t^{\frac{3}{2}}} \, dt dx.
\end{equation*}
Since
\[ \int\limits_x^\infty t^{-\frac{1}{2}}
e^{i\frac{4}{3}t^{\frac{3}{2}}} \, dt = \frac{1}{2}x^{-1}e^{i\frac{4}{3}
x^{\frac{3}{2}}}(1+o(1)), \]
we obtain that
\[ |I_\beta (N)| \leq 2\zeta_1 N^{1/2} \]
for $N$ sufficiently large. Returning to \eqref{normg}, it is straightforward
to conclude that any solution $u_\beta(x,E)$ satisfies for sufficiently
large $N$
\[ C_{\beta}^2(1-\zeta_1)
N^{1/2} \leq \int\limits_{1}^{N} |u_{\beta}(x,E)|^{2}\,dx
\leq C_\beta^2 (1+\zeta_1) N^{1/2}. \]
Therefore, all solutions have the same rate of $L^2$ norm growth
as $N \rightarrow \infty,$ and there is no subordinate solution.
\end{proof}
Now we establish convergence of \eqref{keyint} for every energy
under the assumptions of
Theorem~\ref{spac1}, thus completing the proof of this result.
\begin{proof}[Proof of Theorem~\ref{spac1}]
Let us write
\[ \theta_\beta (\xi,E) = \xi +g_\beta(\xi,E), \]
with $\beta$ parametrizing the initial condition at $0$ and
\begin{equation}
\label{control}
|g'_\beta(\xi,E)| \leq C\xi^{-2/3}
\end{equation}
uniformly in $\beta.$
Clearly we can ignore the short-range quadratic decay term and consider
only
\[ \int\limits_1^N \frac{E-q(c\xi^{2/3})}{c\xi^{2/3}} e^{2i(\xi +
g_\alpha(\xi,E))}\,d\xi. \]
Moreover, the integral
\begin{multline*}
\int\limits_1^N \xi^{-2/3} e^{2i(\xi +
g_\alpha(\xi,E))}\,d\xi
\\
= -\left( \int\limits_\xi^\infty
\frac{e^{2i\eta}}{\eta^{2/3}}\,d\eta \right) e^{2ig_\alpha(\xi,E)}\left|_0^N
\right.
+ \int\limits_1^N e^{2ig_\alpha(\xi,E)} g'_\alpha(\xi,E)
\int\limits_\xi^\infty \frac{e^{2i\eta}}{\eta^{2/3}}\,d\eta d\xi
\end{multline*}
is clearly convergent due to \eqref{control}. It remains to estimate
\[ \int\limits_{3}^{N} \frac{q(c\xi^{2/3})}{c\xi^{2/3}} e^{2i(\xi +
g_\alpha(\xi,E))}\,d\xi \]
uniformly as $N \rightarrow \infty$
(we shifted the region of integration for convenience).
Fix $h,$ $\pi/2 > h > 0,$ and consider the equality
\begin{multline*}
\left(e^{2ih}+e^{-2ih}-2 \right)
\int\limits_{3}^{N} \frac{q(c\xi^{2/3})}{c\xi^{2/3}} e^{2i(\xi +
g_\alpha(\xi,E))}\,d\xi
= O(1) +
\\
\int\limits_{3}^{N}
e^{2i\xi} \left[ \frac{q(c(\xi+h)^{2/3}) e^{2ig_\alpha
(\xi+h,E)}}{(\xi+h)^{2/3}}+
\frac{q(c(\xi-h)^{2/3}) e^{2ig_\alpha(\xi-h,E)}}{(\xi-h)^{2/3}}-
\frac{2q(c\xi^{2/3}) e^{2ig_\alpha(\xi,E)}}{\xi^{2/3}} \right]
\\
= O(1)
+ \int\limits_{3}^{N} \frac{e^{2i(\xi+g_\alpha(\xi,E))}}{\xi^{2/3}}
\left[ q(c(\xi+h)^{2/3})+q(c(\xi-h)^{2/3}) -2 q(c \xi^{2/3}) \right]\, d\xi.
\end{multline*}
Since we assumed that $q \in C^1,$ it suffices to control
\begin{equation}
\label{con2}
\int\limits_{3}^{N} \frac{1}{\xi^{2/3}}
\left| q(c(\xi^{2/3} + \tfrac{2}{3} \xi^{-1/3} h)) +
q(c(\xi^{2/3} + \frac{2}{3} \xi^{-1/3} h))- 2q(c\xi^{2/3})\right|\,d\xi.
\end{equation}
Set $\epsilon = \frac{2}{3} c \xi^{-1/3}h,$ then
uniformly in $N,$ the integral \eqref{con2} is bounded by
\[ C \int\limits_0^1 \left|q(c_1 \epsilon^{-2} +\epsilon) +
q(c_1 \epsilon^{-2} -\epsilon)- 2 q(c_1 \epsilon^{-2})\right|
\frac{d \epsilon}{\epsilon^2} \]
which is finite by assumption.
\end{proof}
\it Remark. \rm Without change, the proof goes through
even with the weaker growth assumption $q(x) = O(x^{\frac{1}{2}-\epsilon})$
for some $\epsilon>0.$
\begin{proof}[Proof of Theorem~\ref{spac2}]
The proof of this theorem is very similar to the preceding proof.
However, it is convenient to employ a slight variation of the Pr\"ufer
transformation. Namely, we let
\begin{eqnarray*}
\sqrt{1-V(\xi,E)}\phi (\xi,E) = \tilde{R} \sin(\tilde{\theta}(\xi,E)) \\
\phi'(\xi,E) = \tilde{R} \cos(\tilde{\theta}(\xi,E)).
\end{eqnarray*}
This transformation is well-defined for large $\xi$ where $V(\xi,E)<1,$
and this suffices for our purpose since we are interested in the asymptotic
behavior at $+\infty.$ The equations for $\tilde{R}$ and $\tilde{\theta}$
are
\begin{align}
\label{tilR}
(\log \tilde{R})'(\xi,E)
& = -\frac{V'(\xi, E)}{4(1-V(\xi,E))}(1-
\cos 2\tilde{\theta} (\xi,E)), \\
\label{tilthe}
\tilde{\theta}'(\xi,E)
& = \sqrt{1-V(\xi,E)} -
\frac{V'(\xi, E)}{4(1-V(\xi,E))}\cos 2\tilde{\theta} (\xi,E).
\end{align}
The role analogous to the integral \eqref{keyint} is played by
\begin{equation}
\label{keyint1}
\int\limits_1^N \frac{V'(\xi,E)}{1-V(\xi,E)} \cos 2\tilde{\theta}(x,E) \,d\xi;
\end{equation}
the other term on the right hand side of \eqref{tilR} can be integrated explicitly.
If the integral \eqref{keyint1} converges for a given energy $E$, then for this energy
there is no solution subordinate on the right. This can be shown
in a direct analogy to the proof of Proposition~\ref{GP}; the details are
left to the reader.
Expressing $V'/(1-V)$ in terms of $q,$
we see that it is enough to show the convergence of
$ \int_{a}^{N} q'(c \xi^{2/3}) \xi^{-1} e^{i \tilde{\theta}(\xi,E)}\,d\xi $
as $N \rightarrow \infty.$
From \eqref{tilthe} and the assumption of the theorem, it follows that
\[ \tilde{\theta}(\xi,E) = \xi + \tilde{g}_\alpha(\xi,E), \]
where $|\tilde{g}'_\alpha(\xi,E)| \leq C \xi^{-\delta}$ for some
$\delta>0.$
Now fix $h,$ $00$ such that $|x-y|\geq N,$ we have
uniformly in $E \in J$
\begin{equation}\label{phest}
\left| \int\limits_x^y \partial_E \sigma(s,E) \,ds \right|
\geq C_1 |x-y|, \,\,\,
\left| \int\limits_x^y \partial_E^\beta \sigma(s,E) \,ds \right|
\leq C_2|x-y|, \,\,\,\beta=2,3
\end{equation}
with positive constants $C_1,$ $C_2.$
Integrate by parts two times
in \eqref{kernel}, with
\[ e^{4i \int_{x}^{y} \sigma (s,E) \,ds }
\int_{x}^{y} \partial_E \sigma (s,E) \, ds\] being integrated.
Using \eqref{phest} and
\eqref{assa}
we obtain
\[ |K(x,y)| \leq C (1 + |x-y|^2)^{-1} \]
for all $x$ and $y,$ which implies the desired $L^2-L^2$ bound.
\end{proof}
The fact that $a(x,E)$ satisfies \eqref{assa} can be checked directly from definitions
\eqref{errdef} and \eqref{adef}.
Now we complete the proof of Theorem~\ref{dac1}.
\begin{proof}[Proof of Theorem~\ref{dac1}]
Proposition~\ref{l2b} ensures that
Proposition~\ref{series} applies under conditions of the theorem.
Notice that
\[ e^{i\psi(x,E)-i \int_0^x \scripte (t,E)\,dt}=
(x-q_2+E)^{-1/4} e^{i \int\limits_0^{x}
\left( \sqrt{t-q_2(t)+E}-\frac{q_1(t)}{2\sqrt{t-q_2(t)+E}} \right)
\,dt }. \]
Applying transformations \eqref{tran2}, \eqref{tran1} and
\eqref{liou} to the solution
$y_+(\xi,E)$ and using \eqref{keyasym}, we obtain for a.e.~$E \in J$ a
solution $u_+(x,E)$ of the equation \eqref{starke} with the
asymptotic behavior
\[ u_+(x,E) = (x-q_2+E)^{-1/4} e^{i \int\limits_0^{x}
\left( \sqrt{t-q_2(t)+E}-\frac{q_1(t)}{2\sqrt{t-q_2(t)+E}} \right)
\,dt }\left(1+o(1) \right), \]
coinciding with \eqref{wkbas}, and
\[ u'_+(x,E) = i(x-q_2+E)^{1/4} e^{i \int\limits_0^{x}
\left( \sqrt{t-q_2(t)+E}-\frac{q_1(t)}{2\sqrt{t-q_2(t)+E}} \right)
\,dt }\left(1+o(1) \right). \]
There also exists solution $u_-(x,E)$ which is just complex conjugate
of $u_+.$ In particular, for a.e.~$E,$ any solution $u(x,E)$ of \eqref{starke}
satisfies
\begin{equation}\label{decsolest}
|u(x,E)| \leq Cx^{-1/4}, \,\,\, |u'(x,E)| \leq Cx^{1/4}.
\end{equation}
By the results of \cite{CKL}, to show that the essential support of the absolutely continuous part
of the spectral measure fills all real
line, it suffices to show that for a.e.~$E$ there exists a sequence $\psi_n(x,E)$ such that
\begin{equation}\label{appcri}
\limsup_{n \rightarrow \infty} \frac{|\psi_n(0,E)|+
|\psi'_n(0,E)|}{\|\psi_n\| \| (H_q-E) \psi_n\|}>0.
\end{equation}
We can obtain an appropriate sequence $\psi_n(E)$ by taking a smooth cutoff function
$\eta(x),$ $1>\eta(x)>0,$ $\eta(x) =1$ if $x<1/2,$ $\eta(x)=0$ if $x>1,$ and letting
$\psi_n(x,E) = \tilde{u}(x,E) \eta(2^{-n}x),$ where $\tilde{u}$ is the solution of \eqref{starke}
which is $L^2$ at $-\infty.$ Then $|\psi_n(0,E)|+|\psi'(0,E)|>c(E)>0$ since $\tilde{u}$
is nonzero. Moreover, using \eqref{decsolest} gives
$\|\psi_n\| \sim 2^{n/4}$ and $\|(H_q-E)\psi_n\| \sim 2^{-n/4},$ so that
\[ \|\psi_n\| \| (H_q-E) \psi_n\| \leq C(E).\]
Therefore, \eqref{appcri} is satisfied at a.e.~$E.$
An alternative (but more technical) way to finish the proof is to apply subordinacy theory
using more detailed information on the asymptotic behavior, similarly
to Proposition~\ref{GP}.
\end{proof}
\section{Potentials in $C^{\frac{1}{2}+\epsilon}$}\label{sm}
In this section we are going to prove a result from which Theorem~\ref{sac1}
will follow. Define
\[ \overline{D}^{\alpha}f(x) = {\rm sup}_{y:\,|x-y|<1}
\frac{|f(x)-f(y)|}{|x-y|^\alpha}, \]
$0<\alpha \leq 1.$
\begin{theorem}\label{sacs}
Assume that potential $q(x)$ satisfies
$|q(x)| \leq \zeta |x|$ with $\zeta <1$ for all sufficiently
large $|x|,$ and
$x^{-\alpha} \overline{D}^{\alpha}q(x^2) \in L^p(1,\infty)$ for
some $p<2$ and $1 \geq \alpha>0.$
Then an essential support of the absolutely continuous part
of the spectral measure coincides with the whole real axis.
\end{theorem}
\noindent \it Remark. \rm In particular,
if $q \in C^\alpha(\reals)$ with $\alpha>1/2,$ then
$D^\alpha q \in L^\infty$
and the assumption of Theorem~\ref{sacs} is satisfied.
\begin{proof}
We will show that if $q$
satisfies the conditions of the theorem, then it can be represented
as a sum $q=q_1+q_2$ with $q_1(x^2) \in L^p(\reals)$ and $x^{-1}q'_2(x^2) \in L^p(\reals),$
$|q_2(x)| \leq \zeta_1 x$ with $\zeta_1 <1$ for large $x.$
Then the result follows from Theorem~\ref{genthm}.
Fix a function $\eta(x) \in C_0^\infty(0,1)$ such that $\int_\reals \eta(x) \,dx =1$.
Set
\begin{align*}
q_2(x) &= x^{1/2} \int_\reals \eta(x^{1/2}(x-y))q(y)\,dy,
\\
q_1(x) &= x^{1/2} \int_\reals \eta(x^{1/2}(x-y))(q(x)-q(y))\,dy.
\end{align*}
Then
\[ |q_1(x)| \leq
x^{(1-\alpha)/2} \int_\reals \eta(x^{1/2}(x-y))\overline{D}^{\alpha}q(x)\,dy
= x^{-\alpha/2}\overline{D}^{\alpha}q(x). \]
Therefore, by assumption, $q_1(x^2) \in L^p.$
The property $|q_2(x)| \leq \zeta_1 x$ with $\zeta_1<1$
for all large $x$ is clear from
the definition.
Also
\begin{eqnarray}
q'_2(x)& = & \frac{1}{2} x^{-1/2}
\int_\reals \eta(x^{1/2}(x-y))q(y)\,dy \nonumber \\
& & + x \int_\reals \eta'(x^{1/2}(x-y))
(q(y)-q(x))\,dy + \frac{1}{2} \int_\reals
\eta'(x^{1/2}(x-y))(x-y)q(y)\,dy. \label{q_2est}
\end{eqnarray}
Due to the bound $|q(x)| \leq \zeta x,$ the first term on the right hand side of \eqref{q_2est}
is bounded, and so harmless.
In the third term, $|x-y| \leq x^{-1/2}$ where the integrand is nonzero, hence it is also
bounded (using $x^{1/2}\int |\eta'(x^{1/2}(x-y)| \,dy \leq C$).
Finally, in the second term we estimate
\[ |q(x)-q(y)| \leq C x^{-\alpha/2} \overline{D}^\alpha q(x) \]
where $\eta$ is nonzero, and so this term is bounded by
$Cx^{(1-\alpha)/2}\overline{D}^\alpha q(x).$
Hence, in all, $x^{-1}q'_2(x^2) \in L^p.$
\end{proof}
\section{A Counterexample}
Our main goal in this section is to prove Theorem~\ref{count},
showing that
the results of Theorems~\ref{sac1} and \ref{dac1} are sharp.
Namely, we will prove that there exist potentials which both belong to
$C^{1/2}(\reals)$, and decay at the rate $x^{-1/4}$, yet which
lead to singular spectrum on the whole real axis. Hence one example
will show optimality of both theorems.
Apart from establishing optimality,
the example we are going to construct
is interesting in its own right, suggesting the mechanism
to get singular spectrum in a background constant electric field
``at the lowest cost'',
and giving potentials with interesting dynamical properties.
Our potentials will be random,
and the construction is inspired
by \cite{KLS}, although there are some notable differences.
In \cite{KLS}, random perturbations of the free operator of the form
\begin{equation}
\label{KLSR}
V(x) = \sum\limits_{n=1}^\infty a_n(\omega)n^{-\alpha} f(x-n)
\end{equation}
were considered (here $f$ and $a_n$ can be taken as in \eqref{qx}).
This model has a transition of spectral properties from absolutely
continuous to singular at $\alpha = 1/2$ (see also \cite{KU} for a similar
earlier model).
By subordinacy theory, it is enough to construct the potential on the
positive half-axis with the
above decay and smoothness properties such that for a.e.~$E \in \reals$
there exists a solution subordinate on the right. We are going to analyze
the equation \eqref{basic1}
\[-\phi'' + \left( \frac{5}{36 \xi^2} + \frac{q(c\xi^{2/3}) -E}
{c \xi^{2/3}} \right) \phi = \phi, \]
and its Pr\"ufer variables representation \eqref{pru1}, \eqref{pru2}
\begin{align*}
(\log R(\xi, E))'
& =
\frac{1}{2} \left( \frac{5}{36 \xi^2} + \frac{q(c\xi^{2/3}) -E}
{c \xi^{2/3}} \right) \sin (2\theta(\xi,E)) \\
\theta'(\xi,E) & =
1- \frac{1}{2} \left( \frac{5}{36 \xi^2} + \frac{q(c\xi^{2/3}) -E}
{c \xi^{2/3}} \right)(1-\cos (2\theta(\xi,E))).
\end{align*}
Let us denote by $\beta$ the boundary condition for $\theta$ at the
origin.
Let $a_n(\omega)$
be independent random variables with uniform distribution in $[0,2\pi].$
Fix a function $f \in C_0^\infty((0,1)),$ $f$ not identically zero.
Consider a family of random
potentials $q(x)$ chosen so that
\begin{equation}
\label{ranpot}
\frac{q(c\xi^{2/3})}{c\xi^{2/3}} = \xi^{-2/3} \sum\limits_{n=1}^{\infty}
n^{-1/2} f(\xi^{1/3}-n) \sin (2\xi +a_n(\omega)).
\end{equation}
Notice that in the coordinate $x$, according to the Liouville transformation,
our potential $q(x)$ looks exactly like \eqref{qx}:
\[ q(x) = c \sum\limits_{n=1}^\infty n^{-1/2} f(\sqrt{cx}-n)
\sin (\tfrac{4}{3} x^{\frac{3}{2}} + a_n(\omega)). \]
In particular, any such $q$ satisfies $|q(x)| \leq Cx^{-1/4},$ and
any such $q$ belongs to $C^{1/2}(\reals).$
We begin the proof of Theorem~\ref{count} with a series of auxiliary
statements.
The key, as before, will be the analysis of the asymptotic behavior
of solutions. First, we are going to show that in order to apply subordinacy
theory, it is sufficient to study the asymptotic behavior of $R(\xi,E).$
\begin{lemma}
\label{derivest}
For any $q(x)$ satisfying $|q(x)| \leq Cx,$
any energy $E$ and any boundary condition $\beta,$
we have
\[ \int\limits_1^L \phi^2(\xi,E,\beta)\,d\xi \leq
\int\limits_1^L R^2(\xi,E,\beta)\,d\xi \leq
C_q(E) \int\limits_1^L \phi^2(\xi,E,\beta)\,d\xi. \]
\end{lemma}
\begin{proof}
Notice that condition on $q$ ensures that $q(c\xi^{2/3})\xi^{-2/3}$
is bounded. By definition of $R,$ it is enough to show that the
$L^2$ norm of the derivative of $\phi$ on $[1,L]$
is controlled by some constant
$C_q(E)$ times the $L^2$ norm of $\phi.$
The proof of the latter fact is a simple elliptic regularity type argument,
and is well known. We refer, for example, to \cite{LS1}, proof of
Theorem 2.1C, for a proof of completely parallel result.
\end{proof}
\begin{theorem}
\label{rangrowth}
Consider the Pr\"ufer variables equations
\eqref{pru1}, \eqref{pru2}, with potential $q$ defined
by \eqref{ranpot}.
Then for any $E \in \reals$ and any boundary condition $\beta$ we have
for a.e.~$\omega$
\begin{equation}
\label{lyapbeh}
\lim\limits_{\xi \rightarrow \infty} \frac{\log R(\xi,E,\beta)}{\log \xi} =
\Lambda (E),
\end{equation}
where
\begin{equation}
\label{lyapexp}
\Lambda(E) = \frac{3\pi}{8} |\hat{f} (\frac{3E}{C})|^2
\end{equation}
(here $\hat{f}$ denotes the Fourier transform $\int \exp(i\xi x)f(x)\,dx$
of $f$).
\end{theorem}
\it Remark. \rm Notice that ``Lyapunov exponent'' $\Lambda(E)$ is
positive everywhere except for perhaps a discrete set of points
since the Fourier transform of $f$ is analytic. \\
We begin with a sequence of auxiliary lemmas.
Consider the unique solution of
\eqref{pru2} satisfying some fixed boundary condition $\beta$.
First, notice that the integral
\[ \int\limits_1^N
\left( \frac{5}{36 \xi^2} -\frac{E}{c \xi^{2/3} } \right) \sin
(2 \theta (\xi,E)) \, d \xi \]
stays bounded as $N \rightarrow \infty$ for any $\omega$ by the
argument in the proof of Theorem~\ref{spac1},
and so in the equation \eqref{pru1} for $R(\xi,E)$ it is enough to
consider the integral
\begin{equation}
\label{conint}
\frac{1}{2}\int\limits_1^N \frac{q(c \xi^{2/3})}{c \xi^{2/3}}
\sin (2\theta (\xi, E))\, d\xi.
\end{equation}
First we are going to analyze $R(\xi,E)$ along a sequence $\xi=n^3,$
as suggested by the form of the potential \eqref{ranpot}, since the parts
with independent phases are supported on $[n^3, (n+1)^3].$
Notice that
\begin{multline}
\label{devest}
\Big| \int\limits_{n^3}^{\xi}
\frac{q(c \xi^{2/3})}{c \xi^{2/3}} \sin (2\theta
(\xi, E))\, d\xi \Big|
\\
= \Big| n^{-1/2}
\int\limits_{n^3}^{\xi} \xi^{-2/3} f(\xi^{1/3}-n)
\sin (2\xi+ a_n(\omega)) \sin (2\theta(\xi,E)) \, d\xi \Big|
\leq Cn^{-1/2}
\end{multline}
for $\xi \in [n^3, (n+1)^3].$
Therefore, if we control $R(n^3,E),$ we will have sufficient control
for all $\xi,$ in particular
\begin{equation}
\label{intsame}
\lim\limits_{x \rightarrow \infty} \frac{R(x,E)}{\log x}
= \frac{1}{3} \lim\limits_{n \rightarrow \infty} \frac{R(n^3,E)}{\log n}.
\end{equation}
Denote by $\theta_n$ the value of $\theta$ at the beginning of the
$n$th interval, $\theta_n = \theta(n^3,E) = \theta(n^3,E,\beta,\omega)$.
Define
\begin{equation} \label{Fdefinition}
F(\xi,E) = \tfrac{1}{2}\int\limits_{n^3}^{\xi}\frac{E}{c\eta^{2/3}}\, d\eta
= \frac{3}{2c}E(\xi^{1/3}-n).
\end{equation}
\begin{lemma}
\label{thetarep}
We can represent $\theta (\xi,E)$ on $[n^3, (n+1)^3]$ in the following form:
\begin{multline}
\theta(\xi,E) = \theta_n + \xi + F(\xi,E) + \Theta(\xi,E)
\\
\label{theta}
+ \tfrac12 {n^{-1/2}} \int\limits_{n^3}^\xi \eta^{-2/3} f(\eta^{1/3}-n)
\sin (2\eta+ a_n(\omega)) \cos(2\eta + 2F(\eta, E)+2\theta_n)\, d\eta,
\end{multline}
where
\begin{equation*}
|\Theta(\xi,E)| \leq C(E) n^{-1}.
\end{equation*}
\end{lemma}
\begin{proof}
Let us analyze the terms arising in the equation \eqref{pru2} for the angle
$\theta.$ The proof reduces to showing that
\begin{multline*}
\Theta(\xi,E) = \frac{1}{2}\int\limits_{n^3}^\xi
\left[
\left(
\frac{5}{36 \eta^2}
+\frac{E}{c\eta^{2/3}}\right) \cos(2\theta(\eta,E)) -
\left( \frac{5}{36 \eta^2}+\frac{q(c\eta^{2/3})}{c\eta^{2/3}} \right)
\right.
\\
\left.
+\
\frac{q(c\eta^{2/3})}{c\eta^{2/3}}\left( \cos(2\theta(\eta,E))
- \cos(2\eta + 2F(\eta, E)+2\theta_n)\right)
\right]
\, d\eta
\end{multline*}
satisfies the required estimate.
The first term satisfies
\begin{equation}
\label{aaa1}
\Big| \int\limits_{n^3}^\xi \Big( \frac{5}{36 \eta^2}
+\frac{E}{c\eta^{2/3} }\Big) \cos(2\theta(\eta,E)) \, d\eta \Big|
\leq C n^{-2}
\end{equation}
by an integration by parts argument identical to that in the proof of
Theorem~\ref{spac1}. The second term satisfies
\[ \big| \int\limits_{n^3}^\xi \Big(
\frac{5}{36 \eta^2}+\frac{q(c\eta^{2/3})}{c\eta^{2/3}} \Big)\, d \eta
\Big| \leq Cn^{-3}, \]
also by a simple integration by parts estimate for the second summand.
For the last term, we have
\begin{multline} \label{longint}
\int\limits_{n^3}^\xi
\frac{q(c\eta^{2/3})}{c\eta^{2/3}}\left( \cos(2\theta(\eta,E))
- \cos(2\eta + 2F(\eta, E)+2\theta_n)\right)\, d\eta
\\
=
2\int\limits_{n^3}^\xi \frac{f(\eta^{1/3}-n)}{n^{1/2} \eta^{2/3}}
\sin(2\eta +a_n(\omega)) \sin(\theta(\eta,E)+\eta+F(\eta,E)+\theta_n)
\\
\times
\sin(\theta(\eta,E) - \eta -F(\eta,E)-\theta_n)\,d\eta.
\end{multline}
Now, we have by \eqref{pru2}
\begin{multline*}
|\theta (\eta, E) - \eta - F(\eta,E)-\theta_n|
\\ =
\left| \frac{1}{2}
\int\limits_{n^3}^\eta
\left[ \left(
\frac{q(ct^{2/3})}{ct^{2/3}}+
\frac{5}{36t^2}\right)(1-\cos 2\theta(t,E)) -
\right. \right.
\left. \left.
\frac{E}{ct^{2/3}} \cos(2\theta(t,E))
\right] \, dt \right|
\\
\leq C n^{-1/2}
\end{multline*}
for $\eta \in [n^3, (n+1)^3]$
by the decay of $q$ and the estimate \eqref{aaa1}. Putting this into
\eqref{longint}, we bound the last term in the expression
for $\Theta(\xi,E)$ by $Cn^{-1}$.
\end{proof}
Next we need Lemma 8.4 from \cite{KLS}, concerning series
of random variables. We will denote by $\rm{Exp}(X)$ the expectation
of the random variable $X.$
Assume $X_i(\omega)$ are independent random variables
with zero mean ${ \rm Exp}(X_i)=0,$ and let
\begin{equation}
\label{rantyp}
Z_n = X_n f_n(X_1, \dots, X_{n-1}),
\end{equation}
where $f_n$ are some measurable
functions. We have
\begin{lemma}[\cite{KLS}]
\label{ranser}
Suppose that ${\rm Exp}(Z_n^2)\leq Cn^{-2\alpha}.$ Then for a.e.~$\omega,$ \\
{\rm (1)} If $\alpha<\frac12$ and $\beta >\frac12 (1-2\alpha),$ then
\[ \lim\limits_{n \rightarrow \infty} \left| \sum\limits_{j=1}^n
Z_j \right|
n^{-\beta} =0. \]
{\rm (2)} If $\alpha = \frac12$ and $\beta>\frac12,$ then
\[ \lim\limits_{n \rightarrow \infty} \left| \sum\limits_{j=1}^n
Z_j \right|
(\log n)^{-\beta} =0. \]
{\rm (3)} If $\alpha>\frac12,$
\[ \lim\limits_{n \rightarrow \infty} \left| \sum\limits_{j=1}^n
Z_j \right|
=Y_\infty \]
exists, and for any $\beta < \alpha -\frac12,$
\[ \lim\limits_{n \rightarrow \infty} \left| \sum\limits_{j=n}^\infty
Z_j \right|
n^{\beta} =0. \]
\end{lemma}
Consider
\begin{multline}
\label{conint1}
\log \frac{R((n+1)^3,E)}{R(n^3,E)} = I_n (E, \omega)
\\
=
\frac{1}{2n^{1/2}} \int\limits_{n^3}^{(n+1)^3}
\frac{f(\xi^{1/3}-n)}{\xi^{2/3}}
\sin(2\xi +a_n(\omega)) \sin 2\theta(\xi,E)\,d\xi.
\end{multline}
We have
\begin{lemma}
\label{keyranle}
The expression $I_n(E,\omega)$ may be represented as
\begin{equation}
\label{ranrep1}
I_n(E, \omega) = \frac{9\pi}{8n} |\hat{f}(\frac{3E}{c})|^2 +
Z_n^{(1)}+Z_n^{(2)}
\end{equation}
where $|Z_n^{(1)}(\omega)| \leq Cn^{-3/2}$, $\{Z_n^{(2)}(\omega)\}$
is a sequence of random variables of type \eqref{rantyp},
and $(Z_n^{(2)})^2$ has expectation $\leq Cn^{-1}$.
\end{lemma}
\begin{proof}
The first observation is that we can replace
$2\theta(\xi,E)$ in \eqref{conint1} by
\[ 2\xi + 2F(\xi,E)+\theta_n+
\int\limits_{n^3}^\xi \frac{q(c\eta^{2/3})}{c \eta^{2/3}}
\cos (2\eta+ 2F(\eta,E) + 2\theta_n)\,d\eta. \]
By Lemma~\ref{thetarep}, the value in \eqref{conint1} will change by at
most $Cn^{-3/2},$ and we can put this difference into $Z_n^{(1)}.$
Also by an estimate similar to \eqref{devest},
\begin{eqnarray}
\nonumber \sin \left(
2\xi + 2F(\xi,E)+ 2\theta_n+
\int\limits_{n^3}^\xi \frac{q(c\eta^{2/3})}{c \eta^{2/3}}
\cos (2\eta+ 2F(\eta,E) + 2\theta_n)\,d\eta
\right)
\\
=
\label{terr} \cos (2\xi + 2F(\xi,E)+2\theta_n)
\int\limits_{n^3}^\xi \frac{q(c\eta^{2/3})}{c \eta^{2/3}}
\cos (2\eta+ 2F(\eta,E) + 2\theta_n)\,d\eta
\\ \nonumber
+
\sin \left( 2\xi + 2F(\xi,E)+2\theta_n \right) + O(n^{-1}),
\end{eqnarray}
and so it suffices to consider the contributions from the
first two terms on the right hand side,
putting the $O(n^{-1})$ terms (multiplied by $n^{-1/2}$) into $Z_n^{(1)}.$
We will write $Z_n^{(2)}$ as a sum of four parts,
each satisfying the conclusion of the lemma.
The first of these four is
\[
% Z_n^{(3)}(\omega) =
\frac{1}{2n^{1/2}} \int\limits_{n^3}^{(n+1)^3} \xi^{-2/3}
f(\xi^{1/3}-n) \sin(2\xi +a_n(\omega)) \sin \left( 2\xi + 2F(\xi,E)+
2\theta_n\right)\,d\xi. \]
Clearly its mean is zero, and the expectation of its
square is $O(n^{-2})$. It remains to discuss the
contribution arising from the first term on the right hand
side in \eqref{terr}:
\begin{multline}
\label{lastt}
\frac{1}{2n} \int\limits_{n^3}^{(n+1)^3} \xi^{-2/3}
f(\xi^{1/3}-n) \sin(2\xi +a_n(\omega)) \cos (2\xi + 2F(\xi,E)+2\theta_n)
\\
\qquad\qquad \times
\int\limits_{n^3}^\xi \eta^{-2/3}
f(\eta^{1/3}-n) \sin(2\eta +a_n(\omega))
\cos (2\eta+ 2F(\eta,E) + 2\theta_n)\,d\eta d\xi
\\
=
\frac{1}{4n} \left( \int\limits_{n^3}^{(n+1)^3} \xi^{-2/3}
f(\xi^{1/3}-n) \sin(2\xi +a_n(\omega)) \cos (2\xi + 2F(\xi,E)+2\theta_n)
\right)^2.
\end{multline}
Write the product of trigonometric functions as
\begin{multline*}
\sin(2\xi +a_n(\omega)) \cos (2\xi + 2F(\xi,E)+2\theta_n) \\
=
\frac{1}{2} \left(
\sin(4\xi +2F(\xi,E)+2\theta_n+
a_n(\omega))+ \sin ( 2F(\xi,E)+2\theta_n-a_n(\omega)) \right).
\end{multline*}
The first term, by integration by parts, gives a contribution of
the order $n^{-3}$.
Introducing notation
\begin{align*}
I_n^{\sin} &= \int\limits_{n^3}^{(n+1)^3} \xi^{-2/3} f(\xi^{1/3}-n)
\sin(2F(\xi,E) + 2\theta_n)\,d\xi, \\
I_n^{\cos} &= \int\limits_{n^3}^{(n+1)^3} \xi^{-2/3} f(\xi^{1/3}-n)
\cos(2F(\xi,E) + 2\theta_n)\,d\xi
\end{align*}
we obtain that, modulo small corrections that can be put into $Z_n^{(1)},$
the right-hand side of \eqref{lastt} is equal to
\begin{equation}
\label{termleft}
\frac{1}{8n}\left( (I_n^{\sin})^2 (\cos(a_n(\omega)))^2 +
(I_n^{\cos})^2 (\sin(a_n(\omega)))^2 + I_n^{\sin}I_n^{\cos}\sin(2a_n(\omega))
\right).
\end{equation}
The contribution of the last term in the brackets is one of the random
variables of type \eqref{rantyp} constituting $Z_n^{(2)}.$ Subtracting
from the first two terms their expectations, we get the other two constituents
of $Z_n^{(2)}$.
Finally, note that ${\rm Exp}((\cos(a_n(\omega)))^2)=
{\rm Exp}((\sin(a_n(\omega)))^2)=\pi,$ and so after subtracting $Z_n^{(2)}$
from \eqref{termleft} and taking expectations we deduce that the right-hand
side of \eqref{lastt}, modulo the sum of the various terms $Z_n$, equals
\[ \frac{\pi}{8n}\left( (I_n^{\sin})^2 +
(I_n^{\cos})^2 \right) = \frac{\pi}{8n} \Big|
\int\limits_{n^3}^{(n+1)^3} \xi^{-2/3} f(\xi^{1/3}-n) e^{2iF(\xi,E)}\,d\xi
\Big|^2. \]
Recalling that $F(\xi,E) = \frac{3}{2c}E(\xi^{1/3}-n),$ and making a change
of variable $t=\xi^{1/3},$ we obtain
\[ \frac{9\pi}{8n} \Big| \hat{f}({3E}/{c}) \Big|^2\ , \]
proving the lemma.
\end{proof}
%Now let us complete the proof of Theorem~\ref{rangrowth}.
\begin{proof}[Proof of Theorem~\ref{rangrowth}]
The proof follows directly from combining results of
Lemma~\ref{keyranle},
Lemma~\ref{ranser}, and the relation \eqref{intsame},
choosing $\beta<1$ in conclusion (2) of Lemma~\ref{ranser}.
\end{proof}
The final step in our construction is showing that for a.e.~$\omega,$
a.e.~$E$ there exists a subordinate solution.
\begin{proposition}
\label{decsolp}
Define the potential $q$ by \eqref{ranpot}. Consider any
energy $E$ such that $\hat{f}(\frac{3E}{c}) \ne 0.$ Then for a.e.~$\omega$
there exists a boundary condition $\beta(\omega)= \beta(\omega,E)$ such that
\begin{equation}
\label{decsol}
\lim\limits_{\xi \rightarrow \infty} \frac{\log R(\xi,E,\beta(\omega))}
{\log \xi} = -\Lambda (E),
\end{equation}
where $\Lambda(E)= (3\pi/8)\,|\hat f(3E/c)|^2$.
\end{proposition}
\begin{proof}
%The proof follows precisely the proof of Lemma 8.8 from \cite{KLS}.
By Lemma 8.7 of \cite{KLS} (which is formulated there in a
discrete setting but
extends trivially to our situation), it is sufficient to show that
for a.e.~$\omega,$
\[ \rho(\xi,E) = \log \frac{R(\xi,E,0)}{R(\xi,E,\pi/2)} \]
has a limit $\rho_\infty(E)$ as $\xi \rightarrow \infty,$ and that
\begin{equation}
\label{rhoest}
\limsup\limits_{\xi \rightarrow \infty} \frac{\log |\rho(\xi,E)-\rho_\infty(E)|}
{\log \xi} \leq -2\Lambda(E).
\end{equation}
Notice that by constancy of the Wronskian,
\[ R(\xi,E,0)R(\xi,E,\frac{\pi}{2}) \sin(\theta(\xi,E,0)-
\theta(\xi,E,\pi/2)) = \text{constant.} \]
Therefore, by Theorem~\ref{rangrowth} for a.e.~$\omega$
\begin{equation}
\label{shut}
\lim\limits_{\xi \rightarrow \infty} \frac{\log |\theta(\xi,E,0)-
\theta(\xi,E,\pi/2)|}{\log \xi} = -2\Lambda(E)
\end{equation}
On the other hand, by Lemma~\ref{thetarep},
\begin{multline*}
\rho((n+1)^3,E) = \sum\limits_{j=1}^{n+1} n^{-1/2}
\int\limits_{n^3}^{(n+1)^3} \xi^{-2/3} f(\xi^{1/3}-n) \sin (2\xi+a_n(\omega))
\\
\times
\left( \sin(2\xi+2F(\xi,E)+2\theta_{n}(\xi,E,0))-
\sin(2\xi+2F(\xi,E)+2\theta_{n}(\xi,E,\pi/2)) \right)
\end{multline*}
plus $ O(n^{-1})$.
Expanding $\sin (2\xi+a_n(\omega))$ into sum of products, and using
\eqref{shut}, we can apply
Lemma~\ref{ranser} to prove convergence to $\rho_\infty$
and estimate the rate of convergence obtaining \eqref{rhoest}.
The estimate \eqref{devest} as before allows us to pass from estimates
over the sequence $n^3$ to estimates over all $\xi$.
\end{proof}
Our main Theorem~\ref{count} follows from Proposition~\ref{decsolp},
Theorem~\ref{rangrowth} and Lemma~\ref{derivest}
immediately.
\begin{proof}[Proof of Theorem~\ref{count}]
Going back to the $x$ variable representation, we find that
for a.e.~$\omega,$ for a.e.~$E$ there exist both a decaying solution
$u_d(x,E)$ and a growing solution $u_g(x,E)$, and the asymptotic behavior
of their $L^2$ norms taken over $[0,L]$ is given by
\begin{eqnarray}
\label{decas}
\log \int\limits_0^L |u_d (x,E)|^2\,dx = {\rm max}(0,\frac{1}{2}-3
\Lambda(E))\log L (1+o(1)), \\
\label{groas}
\log \int\limits_0^L |u_g (x,E)|^2\,dx = (\frac{1}{2}+3
\Lambda(E)) \log L (1+o(1)).
\end{eqnarray}
Hence, given a generic $\omega,$ there is a solution subordinate on the right
for a.e.~$E.$
\end{proof}
We would like to conclude by making several remarks about some curious
properties of the random potential \eqref{qx}.
First of all, the structure of potential \eqref{qx}
indicates that to get singular or point spectrum
``at minimal cost'' in terms of decay or smoothness of random
potential, it is important to have correlations over large (increasing)
distances. In particular, in our example correlation distance grows
as a square root of the distance from the origin.
This contrasts with the free Schr\"odinger operator case \cite{KU,KLS}, where long
distance correlation would have been ineffectual. Another characteristic
feature is the presence of increasingly fast oscillations at infinity.
The half-line Stark operator with potential \eqref{qx} exhibits a number
of interesting spectral and dynamical properties which are similar
to the properties of potentials with the borderline decay rate in a model of \cite{KLS}
in the free case. Introduce notation $X$ for the coordinate operator acting
by $Xf(x) = |x|f(x).$
Denote also
$\langle \psi_1, \psi_2 \rangle$ the inner product of two vectors.
Given an initial state $\psi,$
$\psi (t)$ stands for its unitary evolution $e^{-iH_q t} \psi.$
%In the spectral representation we have
%\[ \psi(t) = \int e^{-iEt} d\mu^{\psi}(E), \]
%where $\mu^\psi$ is the spectral measure of $H_q$ corresponding to the vector
%$\psi$ (the unique Borel measure on $\reals$ satisfying obeying
%$\int f(E) d\mu^\psi(E)= \langle f(H_q)\psi,\psi \rangle$ for any
% Borel function $f$).
One reasonable
measure of propagation properties is the averaged moments of coordinate
operator
\[ \langle \langle X^m \psi(t), \psi(t) \rangle \rangle_T =
\frac{1}{T} \int\limits_0^T \langle \psi(t), X^m \psi(t) \rangle \, dt. \]
Scaling
\[ \langle \langle X^m \psi(t), \psi(t) \rangle \rangle_T \approx CT^m \]
corresponds to the constant velocity ballistic rate, while the behavior
\[ \langle \langle X^m \psi(t), \psi(t) \rangle \rangle_T \approx CT^{2m} \]
to the superballistic constant
acceleration rate, as for the free Stark operator.
We will denote by $P_c$ the spectral projector on the continuous part of the
spectrum of the operator $H_q.$ Also, given a Borel measure $\mu,$ we
will say that $\mu$ has
local Hausdorff dimension $d(E)$ at the point $E$
if for every $\epsilon >0$ there exists
$\delta>0$ such that the restriction
of $\mu$ to $[E-\delta, E+\delta]$ gives zero weight to any set of
Hausdorff dimension $d(E)-\epsilon,$ and is supported on a set of
Hausdorff dimension $d(E)+\delta.$ Let us denote by $H_{q,\beta}$
the Schr\"odinger operator \eqref{stark} defined on the positive half-line
with the boundary condition
$\beta u(0)-u'(0)=0.$ Finally, let $\mu_\beta$ be
the spectral measure associated with
$H_{q,\beta}$ in a canonical way, namely
\[ \langle (H_{q,\beta}-z)^{-1} \delta_0, \delta_0 \rangle =
\int \frac{d\mu_\beta(t)}{t-z}, \]
where $\delta_0$ is the delta-function distribution at $0$
(see, e.g. \cite{Simvan}, Sections I.1-I.6).
The following proposition holds.
\begin{proposition}
\label{exodyn}
Assume that $q_\omega(x)$ is given by \eqref{qx}.
%and let $H_q$ be a half-line Stark operator perturbed by $q.$
Then for a.e.~$\beta,$ $\omega$ the spectral
measure of $H_{q,\beta}$ is pure point on the set where $\Lambda (E) >1/6,$
and is singular continuous with local Hausdorff dimension
\[ d(E) = 1- 6\Lambda(E) \]
on the set where $\Lambda(E)<1/6.$
Moreover, for a.e.~$\beta,$ $\omega,$
for any $\psi$ such that $P_c(\beta,\omega) \psi \ne 0,$
we have
\[ \langle \langle X^m \psi(t), \psi(t) \rangle \rangle_T \geq C_{\epsilon,
m, \beta,\omega} T^{2m(1-\epsilon)} \]
for any $\epsilon >0.$
\end{proposition}
\begin{proof}
The proof of this proposition is analogous to the proof of Theorem 8.10 of
\cite{KLS} and Theorem 5.1
of \cite{KL}, where similar properties were established for a model
\eqref{KLSR} of \cite{KLS} with $\alpha = 1/2$ in a discrete setting.
Let us sketch the arguments for the sake of completeness.
First, let
\[ A = \left\{ E \left| \Lambda(E)> 1/6 \right. \right\}. \]
Notice that by \eqref{lyapexp} and properties of $f,$ $A$ is a finite union
of disjoint bounded intervals. By Proposition~\ref{decsolp}, for
a.e.~$\omega,$ for a.e.~$E \in A,$ there exists a solution
$u_{\beta(\omega)}(x,E,\omega) \in L^2.$ Also, the set $A$ belongs to the
spectrum of $H_{q,\beta}$ for any $\omega$ and $\beta$ since $q$ is decaying.
The general theory of rank one perturbations then implies that for
a.e.~$\omega,$ for a.e.~$\beta$ the spectrum on $A$ is dense pure point
(see, e.g. \cite{DRMS}, Theorem 5.1, or \cite{Siman,Simvan}).
Now let $E$ be such that $\Lambda(E)<1/6,$ and fix $\epsilon>0.$
Consider an interval $I_\delta = [E-\delta, E+\delta]$ such that
the values of the Hausdorff dimension function $d(E)=1-6\Lambda(E)$ on
$I_\delta$ belong to
the interval $(d(E)-\epsilon,d(E)+\epsilon)$.
We can choose $\delta>0$ since $\Lambda(E)$ is continuous.
Denote
\[ \|u\|^2_L = \int\limits_0^L |u(x)|^2\,dx, \]
and recall that
the upper $\alpha$ derivative of the measure $\mu$ is defined by
\[ D^\alpha \mu(E) = \limsup_{\delta \rightarrow 0}
\frac{\mu(E-\delta, E+\delta)}
{(2\delta)^\alpha}. \]
By the Jitomirskaya-Last
extension of the subordinacy theory \cite{JL},
the spectral measure $\mu_\beta$
satisfies
\begin{equation}
\label{jl}
D^\alpha \mu_\beta(E) =
\infty \,\,\,{\rm iff}\,\,\, \liminf_{L \rightarrow \infty}
\frac{\|u_\beta \|_L^{2-\alpha}}{\|u_{\beta^{\perp}} \|_L^\alpha}
=0,
\end{equation}
where $u_\beta$ is a nonzero solution satisfying the boundary
condition at zero,
and $u_{\beta^\perp}$ is the solution satifying an orthogonal boundary
condition $\beta^{-1}u(0)+u'(0)=0$ (in fact, any linearly independent
solution will do in place of $u_{\beta^\perp}$).
From the estimates \eqref{decas}, \eqref{groas}
it follows that for a.e.~$\omega$ and a.e.~$E \in I_\delta,$ there
exists a boundary condition $\beta(\omega)$ such that
\begin{equation}
\label{LHD}
{\rm inf} \left\{ \alpha \left| D^\alpha \mu_{\beta(\omega)}(E)=\infty
\right. \right\}=d(E).
\end{equation}
Recall two basic facts from the general rank one perturbation theory:
first,
\[ \int_\reals\mu_\beta (S) d\beta = m(S) \]
for any Borel measurable set $S,$
where $m$ is the Lebesgue measure. Second,
we can have $D^1 \mu_{\beta_1}(E)=D^1 \mu_{\beta_2}(E)=\infty$ for some
$\beta_1 \ne \beta_2$ only on a fixed (for a given $\omega$) exceptional
set of energies of Lebesgue measure zero.
The first fact,
in the context of Schr\"odinger operators,
is due to Simon and Wolff \cite{SW}. The second is a simple consequence
of an explicit formula relating Cauchy transforms of $\mu_{\beta_1}$ and
$\mu_{\beta_2}.$
See, for example, equation (I.13) of \cite{Simvan} for more details.
Using these properties, we see that \eqref{LHD}
implies that for a.e.~$\omega,$
and a.e.~$\beta$ we have
\[ {\rm inf} \left\{ \alpha \left| D^\alpha \mu_{\beta}(E)=\infty
\right. \right\}=d(E) \]
for every $E \in I_\delta$ in the support of $\mu_\beta.$
Similarly, we also obtain that for a.e.~$\omega,$ $\beta$ the solution
$u_\beta(x,E)$ satisfies the bound \eqref{decas} for the energies in the
support of $\mu_\beta.$
By the choice of $\delta$ and well-known relation between $D^\alpha$
derivatives and dimensional properties of Radon measures (see \cite{Ro})
it follows that for a.e.~$\omega,$ $\beta$ the restriction of the
spectral measure $\mu_\beta$ to $I_\delta$ is supported on a set of
Hausdorff dimension $\leq d(E)+\epsilon$ and gives zero weight to any set
of Hausdorff dimension $\leq d(E)-\epsilon.$
To show the dynamical bound, notice that Theorem 1.2 of \cite{KL} says
that if the
spectral measure $\mu$ is $\alpha-$continuous on a set $S,$ and for
every $E \in S,$
the generalized eigenfunction $u(x,E)$ satisfies
\[ \limsup_{L\rightarrow \infty} L^{-\gamma} \|u\|^2_L <\infty, \]
then for any vector $\psi$ with nonzero projection on $S,$
that is, with $P_S (\psi) \ne 0,$
\[ \langle \langle |X|^m \psi(t), \psi(t) \rangle \rangle_T
\geq CT^{\frac{m \alpha}{\gamma}}. \]
The proof is now completed as in Theorem 5.1 in \cite{KL},
noticing that in our context, $\alpha \sim 1-6\Lambda(E),$ while by
\eqref{decas}
$\gamma \sim \frac{1}{2} - 3 \Lambda(E).$
\end{proof}
\noindent \it Remarks \rm 1. The methods of \cite{KL} also give
a stronger dynamical estimate that for a.e.~$\beta,$ $\omega$ and for
every $\epsilon>0$ and $\rho>0,$ there exists a constant $C_{\beta,
\omega,\rho,\epsilon}$ such that if $R_T = C_{\beta,
\omega,\rho,\epsilon}T^{2-\epsilon},$ then
\begin{equation}\label{DE}
\langle \|\psi(t)\|_{R_T}^2 \rangle_T \leq \|\psi - P_c(\beta,\omega)\psi\|^2
+\rho \|\psi\|^2.
\end{equation}
From \eqref{DE} it follows that the whole part of the
wavepacket lying in the continuous spectral subspace travels at a rate
$\gtrsim T^{2(1-\epsilon)},$
even though we can choose $f$ in \eqref{qx} so that
the spectral dimension $d(E)$ is arbitrarily close to zero in some parts
of the support of the spectral measure. \\
%To the best of our knowledge, this is the first example of such behavior.
2. The Proposition~\ref{exodyn} is not true for a fixed boundary
condition. In fact, for a.e.~$\omega,$
for a dense $G_\delta$ set of boundary conditions $\beta \in \reals,$
the spectrum of $H_{q,\beta}$ is going to be purely singular continuous
\cite{DRMS}. \\
A final remark, which is more of academic interest, is that our method
can be modified in a straightforward way to treat perturbations of
background operators
\[ -\frac{d^2}{dx^2} - x^{\lambda}, \]
where
$0<\lambda<2$ (the operator becomes non-selfadjoint for larger $\lambda$).
In particular, one obtains that the absolutely continuous spectrum of such
operators is preserved for perturbations $q$ satisfying
$|q(x)| \leq C(1+|x|)^{-\frac{1}{4}(2-\lambda)-\epsilon}$ or
$ q \in C^{\frac{1}{2\lambda}(2-\lambda)+\epsilon}(\reals),$ where
$\epsilon$ is an arbitrary small positive number. These results are
optimal in the power scale and in H\"older spaces, respectively.
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