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\begin{document}
\title{Bounds on embedded singular spectrum for one-dimensional
Schr\"odinger operators}
\author{Christian Remling}
\maketitle
\noindent
Universit\"at Osnabr\"uck,
Fachbereich Mathematik/Informatik,
49069 Osnabr\"uck, GERMANY
\\[0.2cm]
E-mail: cremling@mathematik.uni-osnabrueck.de\\[0.3cm]
1991 AMS Subject Classification: 34L40, 81Q10\\[0.3cm]
\begin{abstract}
We show that the solutions of the
one-dimensional Schr\"odinger equation
$-y''+Vy=Ey$ with potential $V(x)=O(x^{-\alpha})$
satisfy the WKB asymptotic formulae off a set
of energies $E$ of Hausdorff dimension $\le 2(1-\alpha)$.
This result gives restrictions on the structure of possible
embedded singular spectrum.
The proof relies on new norm estimates for the integral
transform associated with the WKB method.
\end{abstract}
\section{Introduction}
This paper is, in a sense, a sequel to (the first
part of) \cite{Remac}. I'm interested in one-dimensional
Schr\"odinger equations,
\begin{equation}
\label{1.1}
-y''(x)+V(x)y(x)=Ey(x),
\end{equation}
with potentials $V$ satisfying
\begin{equation}
\label{1.2}
|V(x)| \le \frac{C}{(1+x)^{\alpha}}.
\end{equation}
Eq.\ \eqref{1.1} describes the motion of a quantum mechanical
particle, subject to an external field $V$. The physics of
this system depends directly on the spectral properties of
the corresponding operators (consult \cite{RS3} for further
background information); here, we'll treat explicitly
only the half-line problem. So, we'll study the spectra
of the self-adjoint operators $H_{\beta}=
-\frac{d^2}{dx^2}+V(x),
\beta\in [0,\pi)$, on $L_2(0,\infty)$ with boundary conditions
$y(0)\cos\beta + y'(0)\sin\beta =0$ (see, e.g.,
\cite{WMLN} for the general theory).
If $V$ is asymptotically small in a suitable sense, then, from physical
considerations, one expects that $H_{\beta}$ has purely absolutely
continuous spectrum on $(0,\infty)$. In fact, if $V$ tends to zero
and doesn't oscillate too wildly, then one can use the
so-called WKB methods to solve \eqref{1.1}
asymptotically (cf.\ \cite{East}). One gets solutions of the
form
\begin{equation}
\label{1.3}
\left(
\begin{array}{c}
y(x,E) \\ y'(x,E)
\end{array} \right) =
\left( \begin{array}{c} 1 \\ i\sqrt{E} \end{array} \right)
e^{i\omega(x,E)/2} + o(1) \quad\quad (x\to\infty).
\end{equation}
The WKB phase $\omega$ will be defined below; it satisfies
$\omega(x,E) \sim 2\sqrt{E} x$.
Since these solutions are, in particular, bounded, we can
indeed deduce absence of singular spectrum \cite{Stolz}.
Clearly, assumption \eqref{1.2} alone does not prevent
$V$ from oscillating and hence does not ensure the
applicability of the WKB methods. Indeed, classical \cite{NW}
and more recent \cite{Nab,Simpp} counterexamples show that
there can be singular spectrum on $(0,\infty)$ if $\alpha\le 1$.
Moreover, if $\alpha\le 1/2$, there are (random) examples
with {\it purely} singular spectrum \cite{DSS,Simppp}.
However, as was recently proved by Christ-Kiselev \cite{CK}
and by myself
\cite{Remac} (see also the joint announcement \cite{CKR}),
this behavior is still exceptional if $\alpha>1/2$.
More precisely, let
\begin{equation}
\label{1.4}
S=\{E>0: \text{\eqref{1.1} does {\it not} have a solution $y$
satisfying \eqref{1.3}} \}.
\end{equation}
Then the results of \cite{CK,Remac}, specialized to the
situation where $V$ satisfies \eqref{1.2}, say:
\begin{Theorem}[\cite{CK,Remac}]
\label{T1.1}
Assume that \eqref{1.2} holds with $\alpha> 1/2$.
Then the exceptional set is of Lebesgue measure
zero: $|S|=0$.
\end{Theorem}
Again by \cite{Stolz}, this result in particular implies
that $\sigma_{ac}=[0,\infty)$.
The main goal of this paper is to deepen our understanding
of the behavior of the exceptional set $S$. We will
prove the following strengthening of Theorem \ref{T1.1}.
\begin{Theorem}
\label{T1.2}
Assume that \eqref{1.2} holds. Then the Hausdorff dimension
of the exceptional set satisfies
$\dim S\le 2(1-\alpha)$.
\end{Theorem}
I sincerely believe that this bound is optimal: It gives the
correct values in the borderline cases $\alpha=1/2$ and
$\alpha=1$, and it ``interpolates'' in a natural way.
However, it's unclear to me how to
construct
examples substantiating this opinion.
There's an interesting difference between Theorems \ref{T1.1},
\ref{T1.2}: Theorem \ref{T1.1} remains true if one modifies
$V$ by inserting arbitrary intervals with $V=0$ (this follows
from \cite{CK,Remac}). This is completely wrong for Theorem
\ref{T1.2}: If these intervals occur sufficiently frequently
and are long enough,
one always has $\dim S=1$ (see
\cite{Remsc}, where a stronger result is proved).
Of course, Theorem \ref{T1.2} is more than merely a statement
on the solutions of \eqref{1.1}. Since the singular
part of the spectral measure (restricted to $(0,\infty)$)
is supported by $S$ (note that this holds for all boundary
conditions $\beta$ simultaneously), Theorem \ref{T1.2}
immediately gives restrictions on the structure of possible
embedded singular spectrum.
In fact, the picture is perhaps somewhat different from
what could be expected naively. If $\alpha>1$, then
the spectrum is purely absolutely continuous
on $(0,\infty)$. For $\alpha\le 1$,
there may be singular spectrum as well, but
the singular spectral measures that occur first
(namely, for $\alpha=1$) are, so to speak, at
the opposite end of the scale of measures: They
are zero-dimensional! As $\alpha$ decreases further,
the absolutely continuous spectrum persists as long as
$\alpha>1/2$, while the ``gap'' of forbidden dimensions
$(2(1-\alpha),1)$ becomes smaller. It disappears at
$\alpha=1/2$.
It should also be kept in mind that it's not known
if there really are potentials satisfying
\eqref{1.2} with $\alpha >1/2$ with
non-empty singular {\it continuous} spectrum.
Theorem \ref{T1.2} also has the following consequence:
\begin{Theorem}
\label{T1.3}
Assume that \eqref{1.2} holds. Then $H_{\beta}$ is
purely absolutely continuous on $(0,\infty)$ for all
$\beta\in [0,\pi)$ with the possible exception of
a set of $\beta$'s of Hausdorff dimension $\le 2(1-\alpha)$.
\end{Theorem}
The plan of this paper is as follows.
The proof of Theorem \ref{T1.3}, given Theorem \ref{T1.2},
is based on the methods developed
in \cite{dRJLS}; it will be given in the last section.
The proof of Theorem \ref{T1.2} rests on two pillars:
First of all, we will establish a new norm estimate (= Theorem
\ref{T3.1}) on what I'd like
to call the ``WKB transform''
\[
(Tf)(E) \equiv \int f(x)e^{i\omega(x,E)}\, dx.
\]
The importance of this operator was first recognized in
\cite{Kis2/3}.
Here, it will be vitally important to work in $L_2$-spaces
with {\it singular} measures,
in order to obtain the finer resolution needed
to pass from Theorem \ref{T1.1} to Theorem \ref{T1.2}.
We will prove this norm estimate in Sections 2, 3, which
are very much at the heart of this paper.
Second, we will need the machinery of \cite{Remac}.
In fact, Theorem \ref{T1.2} will follow from combining
the methods of \cite{Remac} with Theorem \ref{T3.1},
plus some measure theory. This will be
done in Section 4. To keep the
length of this paper within reasonable bounds, we will
have to be rather sketchy here; some further details
will be provided in the Appendix.
\section{WKB transforms of $D$-dimensional measures}
Throughout this paper, $\mu$ will denote a finite Borel
measure whose support is contained in some compact subset
of $(0,\infty)$. Such a $\mu$ is said to be $D$-dimensional,
with $D\in [0,1]$, if $\mu(I)\le C|I|^D$ for all intervals
$I\subset\mathbb R$.
It will be convenient to work with wavenumbers
$k=\sqrt{E}$. Of course, the transformed set $\widetilde{S}
=\{ k>0: k^2 \in S\}$ has the same Hausdorff dimension as $S$.
Given a potential $V$, we define the corresponding
WKB phase $\omega=\omega_V$ by
\[
\omega(x,k)=2kx -\frac{1}{k} \int_0^x V(t)\, dt.
\]
As explained in the preceding section,
this quantity plays a central role in the perturbation theory
of the Schr\"odinger equation \eqref{1.1}.
Our goal in this section is to prove the following result on
WKB transforms of $D$-dimensional measures.
\begin{Theorem}
\label{T2.1}
Suppose that $\mu$ is $D$-dimensional with $D\in [0,1)$ and
$|V(x)|\le C_0(1+x)^{-\alpha}$ for some $\alpha>1/2$. Then there
is a constant $C=C(\mu,C_0,\alpha)$ so that
\[
\int_0^Ldx\, \left|\int d\mu(k)\, f(k) e^{i\omega_V(x,k)}
\right|^2 \le C L^{1-D} \int d\mu(k) |f(k)|^2
\]
for all $f\in L_2(\mathbb R,d\mu), L\ge 0$.
\end{Theorem}
{\it Remarks.} 1. In the special case $V\equiv 0$, the
Theorem gives estimates on the Fourier transform of $D$-dimensional
measures. These results are
due to Strichartz \cite{Strich}; the proof is
easier in this case. We will use some ideas from the
proof of Strichartz's Theorem, as given in \cite{Last}.
2. Since there will be so many different constants, we will,
as usual, use the same letter $C$ for all of them.
{\it Proof.} The assertion is obvious for $L<1$, so
we may assume $L\ge 1$ throughout.
Clearly,
\begin{align}
\int_0^L & dx\, \left|\int d\mu(k)\, f(k) e^{i\omega(x,k)}
\right|^2 \nonumber\\
& \le e \int_{-\infty}^{\infty} dx\, e^{-x^2/L^2}
\left|\int d\mu(k)\, f(k) e^{i\omega(x,k)}
\right|^2 \nonumber\\
& = e \int\!\!\! \int d\mu(k)\, d\mu(k')\, f(k)\overline{f(k')}
\int_{-\infty}^{\infty} dx\, e^{-x^2/L^2} e^{2i(k-k')x}
e^{i\left( \frac{1}{k'}-\frac{1}{k}\right) \int_0^x V(s)\, ds}.
\label{2.1}
\end{align}
We integrate by parts in the last integral; in order to avoid
convergence issues, we set $V=0$ on $(-\infty,0)$.
\begin{subequations}
\begin{align}
\label{2.2a}
\int_{-\infty}^{\infty} & dx\, e^{-x^2/L^2} e^{2i(k-k')x}
e^{i\left( \frac{1}{k'}-\frac{1}{k}\right) \int_0^x V(s)\, ds}
= \int_{-\infty}^{\infty} dx\, e^{-x^2/L^2} e^{2i(k-k')x}
+ \\
\label{2.2b}
& i\left( \frac{1}{k'}-\frac{1}{k} \right)
\int_{-\infty}^{\infty} dx\,
V(x)e^{i\left( \frac{1}{k'}-\frac{1}{k}\right) \int_0^x V(s)\, ds}
\int_x^{\infty} dt\, e^{-t^2/L^2} e^{2i(k-k')t}
\end{align}
\end{subequations}
In the first integral \eqref{2.2a},
we've gotten
rid of the $V$ in the exponent, and thus the corresponding
term can be estimated as in the case of the Fourier transform.
We repeat this argument here for completeness: First of all,
the $x$-integral can be evaluated, and after inserting it into
\eqref{2.1} we get
\begin{multline*}
C L \int\!\!\!\int d\mu(k)\, d\mu(k')\, f(k)\overline{f(k')}
e^{-(k-k')^2L^2} \\ \le
CL \int d\mu(k)\, |f(k)|^2 \int d\mu(k')\,
e^{-(k-k')^2L^2}
\end{multline*}
Now,
\begin{align*}
\int d\mu(k')\,
e^{-(k-k')^2L^2} & = \sum_{n=0}^{\infty} \int_{n/L\le |k-k'|
< (n+1)/L} d\mu(k')\, e^{-(k-k')^2L^2}
\\
& \le CL^{-D} \sum_{n=0}^{\infty} e^{-n^2} = CL^{-D},
\end{align*}
so we get indeed the desired bound on \eqref{2.1}.
It remains to study the contribution
coming from \eqref{2.2b}. The basic
strategy is as follows. We'll break up the double integral over
$k,k'$ into two parts, according to the size of $|k-k'|$.
Since the $t$-integral from \eqref{2.2b} is oscillatory, we
expect that it decays as $|k-k'|$ grows. Note that a similar effect
made the above argument work. For small
$|k-k'|$, there are no such cancellations, but then the
$k,k'$ integration is only over a small region.
We start by integrating by parts in the $t$-integral:
\begin{subequations}
\begin{align}
\label{2.3a}
\int_x^{\infty} dt\, e^{-t^2/L^2}e^{2i(k-k')t} & =
\frac{-1}{2i(k-k')} e^{2i(k-k')x} e^{-x^2/L^2} + \\
\label{2.3b}
&\quad \frac{1}{i(k-k')L^2} \int_x^{\infty} dt\, t e^{-t^2/L^2}
e^{2i(k-k')t}.
\end{align}
\end{subequations}
Let's first look at the contribution coming from
\eqref{2.3a}. So insert \eqref{2.3a} into \eqref{2.2b},
and then plug the resulting
term into \eqref{2.1}. We get
\begin{multline*}
C\int\!\!\!\int d\mu(k)\, d\mu(k')\, \frac{f(k)\overline{f(k')}}
{kk'} \int_{-\infty}^{\infty} dx\, V(x)
e^{-x^2/L^2} e^{2i(k-k')x}
e^{i\left( \frac{1}{k'}-\frac{1}{k}\right) \int_0^x V(s)\, ds}\\
= C \int_{-\infty}^{\infty} dx\, V(x) e^{-x^2/L^2} \left|
\int d\mu(k)\, \frac{f(k)}{k} e^{i\omega(x,k)}\right|^2,
\end{multline*}
and the absolute value of this expression can obviously be
estimated by
\begin{multline}
\label{2.4}
C\int_{-\infty}^{\infty} dx\, (1+x^2)^{-\alpha/2} e^{-x^2/L^2} \left|
\int d\mu(k)\, \frac{f(k)}{k} e^{i\omega(x,k)}\right|^2 =\\
C\int\!\!\!\int d\mu(k)\, d\mu(k')\, \frac{f(k)\overline{f(k')}}
{kk'} \int_{-\infty}^{\infty} dx\, (1+x^2)^{-\alpha/2}
e^{-x^2/L^2} e^{2i(k-k')x}
e^{i\left( \frac{1}{k'}-\frac{1}{k}\right) \int_0^x V(s)\, ds}.
\end{multline}
As announced above, we now break up this triple integral into
two parts, according as $|k-k'|\le L^{\alpha/D-1}$ or
$|k-k'|> L^{\alpha/D -1}$. (Of course, we assume $D>0$ here;
if $D=0$, the assertion is trivial.)
Consider first the region $|k-k'|\le L^{\alpha/D-1}$:
We don't expect cancellations in the $x$-integral here,
so we simply use
\begin{multline*}
\left| \int_{-\infty}^{\infty} dx\, (1+x^2)^{-\alpha/2}
e^{-x^2/L^2} e^{2i(k-k')x}
e^{i\left( \frac{1}{k'}-\frac{1}{k}\right) \int_0^x V(s)\, ds}
\right| \\
\le \int_{-\infty}^{\infty} dx\, (1+x^2)^{-\alpha/2}
e^{-x^2/L^2} \le CL^{1-\alpha}.
\end{multline*}
This yields the required bound on the corresponding part of
\eqref{2.4}:
\begin{align*}
CL^{1-\alpha}\int\!\!\!\int_{|k-k'|\le L^{\alpha/D-1}}
& d\mu(k)\, d\mu(k')\, | f(k) f(k')|\\
& \le CL^{1-\alpha} \int d\mu(k) |f(k)|^2 \int_{|k-k'|\le L^{\alpha/D-1}}
d\mu(k') \\
& \le CL^{1-\alpha+(\alpha/D-1)D} \int d\mu(k) |f(k)|^2 \\
& = CL^{1-D} \int d\mu(k) |f(k)|^2.
\end{align*}
To analyze the part of \eqref{2.4} corresponding to
$|k-k'|>L^{\alpha/D-1}$, we use once again integration by
parts in the $x$-integral:
\begin{multline}
\label{2.5}
\int_{-\infty}^{\infty} dx\, (1+x^2)^{-\alpha/2}
e^{-x^2/L^2} e^{2i(k-k')x}
e^{i\left( \frac{1}{k'}-\frac{1}{k}\right) \int_0^x V(s)\, ds} =\\
\int_{-\infty}^{\infty} dx\, (1+x^2)^{-\alpha/2}
e^{i\left( \frac{1}{k'}-\frac{1}{k}\right) \int_0^x V(s)\, ds}
\left[ -\alpha x (1+x^2)^{-1} +
i\left( \frac{1}{k'}-\frac{1}{k}\right)
V(x) \right]\\
\times \int_x^{\infty} dt\, e^{-t^2/L^2} e^{2i(k-k')t}.
\end{multline}
Now integrate by parts in the $t$-integral (this is the usual method
to exploit oscillations). We did this calculation already
in \eqref{2.3a}, \eqref{2.3b}. It follows from these
formulae that
\[
\left| \int_x^{\infty} dt\, e^{-t^2/L^2} e^{2i(k-k')t}
\right| \le \frac{C}{|k-k'|},
\]
and thus \eqref{2.5} implies that also
\[
\left| \int_{-\infty}^{\infty} dx\, (1+x^2)^{-\alpha/2}
e^{-x^2/L^2} e^{2i(k-k')x}
e^{i\left( \frac{1}{k'}-\frac{1}{k}\right) \int_0^x V(s)\, ds}
\right| \le \frac{C}{|k-k'|}.
\]
Therefore, going back to \eqref{2.4}, we find the bound
\begin{align}
C\int\!\!\!&\int_{|k-k'|>L^{\alpha/D-1}} d\mu(k)\, d\mu(k')\,
\frac{|f(k)f(k')|}{|k-k'|} \nonumber\\
& \le C \int d\mu(k) |f(k)|^2 \int_{|k-k'|>L^{\alpha/D-1}}
\frac{d\mu(k')}{|k-k'|}\nonumber\\
&\le C L^{(1-D+\epsilon)(1-\alpha/D)}
\int d\mu(k) |f(k)|^2 \int_{|k-k'|>L^{\alpha/D-1}}
\frac{d\mu(k')}{|k-k'|^{D-\epsilon}}.
\label{2.6}
\end{align}
Now we need the following elementary fact:
\begin{Lemma}
\label{L2.1}
If $\mu$ is $D$-dimensional, then, for any $\epsilon\in (0,D)$,
\[
\sup_{x\in \mathbb R} \int\frac{d\mu(y)}{|x-y|^{D-\epsilon}}
<\infty.
\]
\end{Lemma}
{\it Proof.}
\begin{align*}
\int\frac{d\mu(y)}{|x-y|^{D-\epsilon}}
& = \sum_{n=1}^{\infty}\int_{(n+1)^{-1/\epsilon}<
|x-y|\le n^{-1/\epsilon}}
\frac{d\mu(y)}{|x-y|^{D-\epsilon}} + \int_{|x-y|>1}
\frac{d\mu(y)}{|x-y|^{D-\epsilon}} \\
& \le C\sum_{n=1}^{\infty} (n+1)^{(D-\epsilon)/\epsilon}
n^{(-1-1/\epsilon)D} + \mu(\mathbb R) \equiv C<\infty
\end{align*}
$\square$
Hence, if $\epsilon>0$ is small enough, then \eqref{2.6}
can indeed be estimated by $CL^{1-D}\|f\|^2_{L_2(\mathbb R,d\mu)}$.
This concludes the analysis of the contributions coming from
\eqref{2.3a}.
We now move on to studying \eqref{2.3b}. The same break-up as
above will be used. If $|k-k'|\le L^{\alpha/D-1}$, note that
\[
|\eqref{2.3b}|\le \frac{C}{|k-k'|L^2} \int_x^{\infty} dt\,
t e^{-t^2/L^2} \le \frac{C}{|k-k'|} e^{-x^2/(2L^2)},
\]
hence again
\[
\left| \left(\frac{1}{k'}-\frac{1}{k}\right)
\int_{-\infty}^{\infty} dx V(x) e^
{i\left( \frac{1}{k'}-\frac{1}{k}\right) \int_0^x V(s)\, ds}
\times \eqref{2.3b}\right| \le
C L^{1-\alpha},
\]
and proceed as above.
If $|k-k'|>L^{\alpha/D-1}$, we integrate by parts one
more time:
\[
\eqref{2.3b} = \frac{1}{2(k-k')^2L^2} \left( x e^{-x^2/L^2} e^{2i(k-k')x}
+ \int_x^{\infty} dt\, \left( 1-\frac{2t^2}{L^2}\right)
e^{-t^2/L^2}e^{2i(k-k')t} \right).
\]
It's easy to see that this gives an estimate on the corresponding
contribution to \eqref{2.2b} of the form $C/(|k-k'|L^{\alpha})$.
So we can again conclude the argument as above (in fact, we even
have an additional small factor $L^{-\alpha}$). $\square$
\section{Norm estimates}
The result of the previous section enables us to show
\begin{Theorem}
\label{T3.1}
Suppose that $\mu$ is $D$-dimensional ($D\in [0,1]$) and
$|V(x)|\le C_0(1+x)^{-\alpha}$ with $\alpha>1/2$. Then there is
a constant $C$ so that
\[
\int d\mu(k)\, \left| \int_c^d dx\, f(x) e^{i\omega_V(x,k)}\right|^2
\le C (d-c)^{1-D} \int_c^d dx\, |f(x)|^2
\]
for all $f\in L_2(c,d), 0\le c2(1-\alpha)$ ($\alpha$ is from \eqref{1.2}, of course).
Then
$\mu(S)=0$, where $S$ is the exceptional set
defined in \eqref{1.4}.
Accepting this, we can now prove Theorem
\ref{T1.2} as follows. Assume that, contrary to the claim,
$\dim S > 2(1-\alpha)$. Then, by definition of the
Hausdorff dimension, there exists
$D>2(1-\alpha)$ so that $h^D(S)=\infty$
($h^D$ denotes the $D$-dimensional
Hausdorff measure).
It's easy to see that $S$ is
a Borel set. Hence, by \cite[Theorem 5.6]{Fal}, we
can find a compact subset $S'\subset S$, such that
$00\}
\]
(where $\rho_{\beta}$ is the spectral measure of $H_{\beta}$).
Then $\dim A \le \dim T$.
\end{Theorem}
{\it Proof.} Let
\[
B=\{ -\cot\beta: \beta\in A, \beta\not= 0\}.
\]
Clearly, $\dim A=\dim B$, so we may as well work with
$B$. Now, assume that
$h^D(B)=\infty$ for some $D<1$.
Then, invoking again the argument based on
\cite[Theorem 5.6]{Fal}, we see that there's a $D$-dimensional
measure $\mu$, supported by $B$ and with $0<\mu(B)<\infty$.
Define another measure $\nu$ by
\[
\nu(F) = \int \rho_{-\cot^{-1} x}(F)\, \frac{d\mu(x)}{1+x^2}.
\]
Denote the Borel transform of a measure $\eta$ by
$M_{\eta}$:
\[
M_{\eta}(z) \equiv \int \frac{d\eta(t)}{t-z}.
\]
Then \cite[Lemma 5.3]{dRJLS} (slightly modified to apply
to the case at hand) says that
\begin{equation}
\label{5.1}
M_{\nu}(z) = M_{\mu}(m_0(z)),
\end{equation}
where $m_0$ is the usual Titchmarsh-Weyl $m$-function of
$H_0$.
Since $\mu$ is $D$-dimensional, we have that
\begin{equation}
\label{5.2}
| M_{\mu}(E+i\epsilon)| \le C \epsilon^{D-1}.
\end{equation}
This is well-known
and can be shown by an argument similar to the proof of
Lemma \ref{L2.1}. Combining \eqref{5.1} and \eqref{5.2}
yields
\[
\epsilon^{1-D} |M_{\nu}(E+i\epsilon)| \le
C \left( \frac{\epsilon}{\text{Im }m_0(E+i\epsilon)}
\right)^{1-D}.
\]
Since
\[
\frac{\text{Im }m_0(E+i\epsilon)}{\epsilon} =
\int \frac{d\rho_0(t)}{(t-E)^2+\epsilon^2},
\]
we see that in particular
\[
\limsup_{\epsilon\to 0+}
\epsilon^{1-D} |M_{\nu}(E+i\epsilon)| < \infty
\]
for all $E\in \mathbb R$.
This implies (cf.\ \cite[Theorem 3.1]{dRJLS})
\[
(D_D\nu)(E) \equiv \limsup_{\epsilon\to 0+}
\frac{\nu(E-\epsilon,E+\epsilon)}{(2\epsilon)^D} < \infty
\]
for all $E$. Hence $\nu(F)=0$ for every set $F$ with $\dim F0$, thus $\dim T\ge D$. The only assumption
on $D$ was $h^D(B)=\infty$, and this holds for any
$D<\dim B$, so we conclude that indeed
$\dim T\ge \dim B$.
$\square$
\begin{appendix}
\section{The method of \cite{Remac}}
First of all, we rewrite \eqref{1.1} using (modified)
Pr\"ufer variables. So, if $y(x,k)$ is a solution of
\eqref{1.1} with $E=k^2$ (fixed via initial values at
$x=0$), write $y=R \sin \psi/2, y'=kR \cos\psi/2$.
The equations for $R$ and $\theta\equiv\psi-\omega$ are, in integrated
form,
\begin{align}
\label{A.1}
2k (\ln R(y,k)-\ln R(x,k)) & = \int_x^y V(t)\sin\psi(t,k)\, dt,\\
\label{A.2}
k(\theta(y,k)-\theta(x,k)) & = \int_x^y V(t)\cos\psi(t,k)\, dt.
\end{align}
We want
to show that $R$ and $\theta$
tend to limits as $x\to\infty$,
for $\mu$-a.e. $k$.
Let $x_n\ge 0$ be an increasing sequence that tends to
infinity, and write $V_n$ for the restriction of $V$ to the
interval $(x_{n-1},x_n)$. Eqs.\ \eqref{A.1}, \eqref{A.2} suggest
that we study
\[
\sum_{n=1}^{\infty} \int V_n(x) e^{i\psi(x,k)}\, dx
= \sum_{n=1}^{\infty} \int V_n(x) e^{i\omega(x,k)}
e^{i\theta(x,k)}\, dx.
\]
In fact, it turns out that for appropriately chosen
$x_n$'s, these sums are (absolutely)
convergent for almost every $k$ with respect to $\mu$ if
$\mu$ is as in
Sect.\ 4.
To prove this,
one uses integration by parts, integrating $V_n e^{i\omega}$
and differentiating $e^{i\theta}$. This gives
\begin{multline*}
\sum_{n=1}^{\infty} \left(
e^{i\theta(x_n,k)} \int V_n(x) e^{i\omega(x,k)}\, dx \quad - \right. \\
\left. \frac{i}{k} \int dx\, V_n(x) \cos(\omega(x,k)+\theta(x,k))
e^{i\theta(x,k)} \int_{x_{n-1}}^x dt\, V_n(t) e^{i\omega(t,k)}
\right) .
\end{multline*}
Now Theorem \ref{T3.1} together with the
Cauchy-Schwarz inequality (in $L_2(d\mu)$)
show that the $\mu$-integral of
the first term in the sum can be estimated by
\[
\int d\mu(k)\,\left| \int V_n(x) e^{i\omega(x,k)}\, dx
\right| \le C (x_n-x_{n-1})^{(1-D)/2} \|V_n\|_2.
\]
In particular, if the $x_n$'s are chosen in such a way that
this bound is summable (and we will do so shortly), then also
\[
\sum_{n=1}^{\infty} \left|
\int V_n(x) e^{i\omega(x,k)}\, dx \right| <\infty
\]
for $\mu$-almost every $k$.
As for the remaining terms, it's possible to set up an
iteration scheme also based on these two devices:
integration by parts and the control over integrals
with respect to $d\mu$ provided by Theorem \ref{T3.1}.
More specifically, one starts with the intervals
$(x_{n-1},x_n)$, and at each step, every interval
is subdivided into smaller subintervals. The
point is that the double
integrals are
over smaller and smaller regions (modulo contributions
that are summable $\mu$-almost everywhere). Finally, an elementary
estimate shows that these terms are also summable
almost everywhere (after a sufficiently large number
of steps).
In the case at hand, this procedure works if we take
(sticking to the notation of \cite{Remac})
$x_n=n^a$ with $a>\frac{D}{D-2(1-\alpha)}$
(here, we need the basic assumption $D>2(1-\alpha)$
to ensure that
the denominator is positive) and $N_n=[n^b]$ with
$0**2, 1/p+1/q=1$,
\[
\|M_n\|_{L_q(\mathbb R,d\mu)} \le C_q (x_n-x_{n-1})^{(1-D)/q}
\|V_n\|_p.
\]
This is proved by interpolation: Since, trivially, the WKB
transform is bounded as a map from
$L_1$ to $L_{\infty}$ with norm $1$, the Riesz-Thorin Theorem
\cite[Chapter XII, 1.11]{Zyg} shows that the claimed
estimate holds for the WKB transform itself. Then the
results of
\cite{Kisint} and
\cite[Sect.\ 3]{CK} guarantee that it holds for the
maximal function as well.
Now, if we take $q$ sufficiently close
to $2$, we can argue as in \cite[Sect.\ 3]{Remac} to
prove that also
\[
\lim_{n\to\infty} \max_{x_{n-1}\le \xi \le x_n}
\left| \int_{x_{n-1}}^{\xi} V(x)e^{i\psi(x,k)}\, dx \right|
=0
\]
for $\mu$-almost every $k$.
\end{appendix}
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\end{document}
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