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\begin{document}
\title{Some Schr\"odinger operators with power-decaying potentials
and pure point spectrum}
\author{Christian Remling\\
Universit\"at Osnabr\"uck\\
Fachbereich Mathematik/Informatik\\
D-49069 Osnabr\"uck\\
GERMANY\\[0.2cm]
E-mail: cremling@mathematik.uni-osnabrueck.de\\[0.3cm]
Current address (until May 31, 1997):\\
253-37 Department of Mathematics\\
California Institute of Technology\\
Pasadena, CA 91125\\
U.S.A.\\[0.2cm]
(to appear in {\it Comm.\ Math.\ Phys.})}
\maketitle
\begin{abstract}
We construct (deterministic) potentials $V(x)=O(x^{-c})$ such that the
Schr\"odinger equation $-y''+Vy=Ey$ on $x\in [0,\infty )$ has dense
pure point spectrum in $(0,\infty)$ for almost all boundary
conditions at $x=0$. As a by-product, we also obtain power-decaying
potentials for which the spectrum is purely singular continuous on
$(0,\infty)$ for all boundary conditions.
\end{abstract}
\section{Introduction}
In this paper, we will construct power-decaying potentials
$V(x)=O(x^{-c})\:(c>0)$ such that the Schr\"odinger equation
for positive energies $E=k^2$
\begin{equation}
\label{1}
-y''(x)+V(x)y(x)=k^2y(x)
\end{equation}
on the half-axis $x\in [0,\infty)$ has square integrable solutions
for almost all $k>0$ with respect to Lebesgue measure.
By well-known arguments (using spectral averaging,
see e.g.\ \cite{SW}), this implies that the corresponding
Hamiltonian $H_{\alpha}=-\frac{d^2}{dx^2}+V(x)$ with boundary
condition $y(0)\cos\alpha+y'(0)\sin\alpha=0$ has dense pure
point spectrum in $(0,\infty)$ (i.e. $\sigma_{ess}=[0,\infty),
\sigma_c(H_{\alpha})=\emptyset$) for almost all $\alpha\in [0,\pi)$.
As far as I am aware, this is even the first explicit
example of a Schr\"odinger operator with this spectral
type and $V\to 0$. The existence of such potentials,
however, is not a new result: The work on decaying
random potentials \cite{KU,S} has shown that, roughly speaking,
random potentials decaying as $V_{\omega}(x)\sim x^{-c}$
with $c<1/2$ almost surely lead to dense pure point spectrum
if the values of the potential at different sites are
independent with zero expectation. (In fact, $c=1/2$ is
generally believed to be the borderline between absolutely
continuous and singular spectrum in the sense that $V(x)=O(x^{-c})$
with $c>1/2$ should imply $\sigma_{ac}=[0,\infty)$. Kiselev has
recently proved this for $c>2/3$ \cite{Kis}, but in the region
$1/20,l_n\ge 0$, and let $W(x)$ be a bounded,
measurable and periodic function with period $p$. Define $a_1=0$,
\[
b_n=a_n+p\frac{L_n}{g_n},\quad a_{n+1}=b_n+l_n\quad\: (n\ge 1).
\]
The potential we will study is given by
\[
V(x)=\left\{ \begin{array}{lr} 0& \mbox{if }x\in (b_n,a_{n+1})\\
W_{g_n}(x-a_n)& \mbox{if }x\in (a_n,b_n) \end{array} \right.
\]
where $W_g$ is the rescaled function $W_g(x)=g^2 W(gx)$.
\begin{Theorem}
\label{T1}
$L_n,g_n,l_n$ and $W(x)$ can be chosen such that:\\
1) $V(x)=O(x^{-c})$ for some $c>0$,\\
2) For almost all $k>0$, (\ref{1}) has an $L_2$-solution.
\end{Theorem}
It follows from $V\to 0$ that $\sigma_{ess}=[0,\infty)$.
Moreover, as pointed out above, 2) implies that for almost
all boundary conditions $\alpha$, the corresponding half-line
operator $H_{\alpha}$ has no continuous spectrum, hence it must
have dense pure point spectrum in $(0,\infty)$. Note also that
by recently discovered general principles on the instability
of point spectrum \cite{dRMS}, there is a dense $G_{\delta}$
set of boundary conditions for which the spectrum is purely
singular continuous in $(0,\infty)$!
\section{Preliminary observations}
Write
\[
Y(x,k):=\left( \begin{array}{c} y(x,k) \\ y'(x,k)/k \end{array}\right)
\]
where $y$ is a solution of (\ref{1}) satisfying an initial condition
of the form $y(0,k)=-\sin\alpha, y'(0,k)=\cos\alpha$. We will use
the modified Pr\"ufer variables $R,\varphi$ defined by the equation
\[
Y(x,k)=R(x,k)\left( \begin{array}{c} \sin\varphi(x,k)\\
\cos\varphi(x,k) \end{array}\right)
\]
and by requiring that $\varphi$ be continuous and $R>0$. The
transfer matrix
$T(t,s;k)$ is defined by the property that it takes $Y(s,k)$
to $Y(t,k)$, i.e.\ $Y(t,k)=T(t,s;k)Y(s,k)$. Hence it is given by
\begin{equation}
\label{2}
T(t,s;k)=\left( \begin{array}{cc} u(t,k) & kv(t,k)\\
u'(t,k)/k & v'(t,k) \end{array}\right)
\end{equation}
where $u,v$ are the solutions of (\ref{1}) with $u(s)=v'(s)=1,
u'(s)=v(s)=0$.
If the potential is periodic, then obviously $T(t+p,s+p;k)=T(t,s;k)$,
(this observation is roughly one half of what is usually called
Floquet theory). Abbreviating $T(k):=T_W(p,0;k)$ (the transfer
matrix over one period for the potential $W$), we get for the
transfer matrix for the potential $V$ described in the preceding Section
\begin{Lemma}
\label{L1}
\[
T(a_n+m_2pg_n^{-1},a_n+m_1pg_n^{-1};k)=T(kg_n^{-1})^{m_2-m_1},\quad\quad
m_i\in \{0,1,\ldots,L_n\}
\]
\end{Lemma}
{\it Proof.} By ``Floquet theory'', it suffices to prove the
assertion for $m_1=0, m_2=1$. Moreover,
$T(a_n+pg_n^{-1},a_n;k)=\hat{T}(pg_n^{-1},0;k)$ where
$\hat{T}$ is the transfer matrix of $-y''+W_{g_n}(x)y=k^2y$.
For a function $y(x)$ and $g>0$, let $w(t):=y(t/g)$. Then
we have: $y$ solves $-\frac{d^2y}{dx^2}+W_g(x)y=k^2y$ and
satisfies $(y,dy/dx)(0)=(a,b)$ if and only if $w$ solves
$-\frac{d^2w}{dt^2}+W(t)w=(k/g)^2w$ and satisfies
$(w,dw/dt)(0)=(a,g^{-1}b)$. Hence, if $u_g,v_g$ denote
the solutions of $-y''+W_gy=k^2y$ with $u_g(0)=v'_g(0)=1,
u'_g(0)=v_g(0)=0$, then
\begin{eqnarray*}
u_g(x,k)=u_1(gx,k/g), &&\quad v_g(x,k)=g^{-1}v_1(gx,k/g)\\
u'_g(x,k)=gu'_1(gx,k/g), &&\quad v'_g(x,k)=v'_1(gx,k/g).
\end{eqnarray*}
Now the assertion follows from (\ref{2}). \hfill $\Box$
Next, we recall some basic facts about periodic potentials;
for further background information on this topic, please
consult \cite{E}. Since $\det T(k)=1$ by constancy of the
Wronskian, the eigenvalues of $T$ are determined by its trace
\[
D(k):=\mbox{tr }T(k)=u(p,k)+v'(p,k),
\]
($u,v$ solve $-y''+Wy=k^2y$ and $u(0)=v'(0)=1,u'(0)=v(0)=0$).
More explicitly, the eigenvalues are $\mu,\mu^{-1}$ with
\begin{equation}
\label{3}
\mu(k)=\frac{D(k)}{2}\pm \sqrt{\frac{D(k)^2}{4}-1}.
\end{equation}
We choose the sign so that $|\mu|\ge 1$. $\mu$ is real and
larger than $1$ in magnitude if and only if $|D|>2$; if $|D|<2$,
then $|\mu|=1, \mu\not= \pm 1$. In either case, the eigenvalues
are distinct and hence $T$ is diagonalizable. Let $U(k)$ be a
diagonalizing matrix, i.e.\
\[
U^{-1}(k)T(k)U(k)= \left( \begin{array}{cc} \mu(k) & 0 \\ 0
& \mu^{-1}(k) \end{array}\right).
\]
This representation leads to a simple estimate on the norm of
the transfer matrix. In the following Lemma, we use the $l_2$-norm
$\|(v_1,v_2)^t\|^2=|v_1|^2+|v_2|^2$.
\begin{Lemma}
\label{L2}
a) If $|D(k)|<2$, then
\[
\|T(k)^mv\|\ge\frac{\|v\|}{\|U(k)\|\, \|U^{-1}(k)\|}.
\]
b) If $|D(k)|>2$, then there is an angle $\beta(k)\in [0,\pi)$
such that $|\varphi-\beta(k)-n\pi|\ge \epsilon \quad\forall
n\in {\Bbb Z}$ implies
\[
\|T(k)^m e_{\varphi}\|\ge \frac{2\epsilon}{\pi}|\mu(k)|^m
\]
(where $e_{\varphi}=(\sin\varphi,\cos\varphi)^t$).
\end{Lemma}
{\it Proof.} Obviously,
\begin{equation}
\label{4}
T(k)^m=U(k)\left(\begin{array}{cc}\mu(k)^m & 0\\\ 0 &
\mu(k)^{-m}\end{array}\right)U^{-1}(k).
\end{equation}
Hence $\|T^{-m}w\|\le |\mu|^m\|U\|\;\|U^{-1}\|\;\|w\|$ which implies a)
since $|\mu|=1$ in this case. In order to prove b), notice that because
$T(k)$ is real and has real, distinct eigenvalues if $|D(k)|>2$, we may
take $U$ in the form
\[
U=\left( \begin{array}{cc} \sin\alpha & \sin\beta \\ \cos\alpha &
\cos\beta \end{array}\right)
\]
with $\alpha,\beta\in [0,\pi),\;\alpha\not= \beta$.
Now an elementary calculation using (\ref{4}) leads to (writing
$\gamma=\alpha-\beta$)
\[
\|T^me_{\varphi}\|^2=\sin^{-2}\gamma\left( \mu^{2m}\sin^2(\varphi-\beta)
+\mu^{-2m}\sin^2(\varphi-\alpha) -2\cos\gamma\sin(\varphi-\beta)
\sin(\varphi-\alpha)\right).
\]
Viewing the right-hand side as a (quadratic) function of
$\sin(\varphi-\alpha)$ and determining the minimum shows
$\|T^me_{\varphi}\|^2\ge \mu^{2m}\sin^2(\varphi-\beta)$.
This yields the desired estimate, since $|\sin x|\ge 2|x|/\pi$
if $|x|\le \pi/2$. \hfill$\Box$.
Now we are in a position to explain the general idea of our construction.
By Lemma \ref{L1}, the transfer matrix for $x\in (a_n,b_n)$ is given by
powers of $T(k/g_n)$. We will take a function $W$ with infinitely many
``gaps'' (i.e. intervals with $|D|>2$), and we will choose the $g_n$
such that for every $k$, $k/g_n$ is in some gap for sufficiently many
$n$. Then Lemma \ref{L2}b) shows that for those $n$ the solution grows
by a factor $|\mu|^{L_n}$ unless the phase $\varphi(a_n)$ is close to
$\beta$. If $k/g_n$ is not in a gap, then the solution can be controlled
with the aid of Lemma \ref{L2}a). The point of this argument is that
the estimate of Lemma \ref{L2}a) is independent of $m$. Therefore,
after everything else has been chosen, we can take the $L_n$
sufficiently large to obtain increasing solutions. Of course,
one then needs an additional argument to deduce the existence
of $L_2$-solutions. To this end, we will use a Ruelle type Theorem
(see \cite{Ru}) recently proved in \cite{LS}.
We will make no attempt to optimize the exponent $c$ from Theorem
\ref{T1} because it does not seem possible to reach the critical
value $1/2$. In particular, we will sometimes use rather crude
estimates when this does not destroy the power decay.
The following Lemma allows us to adjust the phases $\varphi(a_n,k)$
if necessary. For $k>0$ with $|D(k)|>2$, define
$S_{\epsilon}(k)=(\beta(k)-\epsilon,\beta(k)+\epsilon)$
with $\beta(k)$ from Lemma \ref{L2}b).
\begin{Lemma}
\label{L4}
Suppose $V(x)$ has been chosen for $x\le b_n$. Let $M$ be
a measurable subset of $\{ k\in [k_1,k_2]: |D(k/g_{n+1})|>2\}$.
Then there is an $l_n\in [0,\pi/k_2]$ such that
\[
\left| \{k\in M:\varphi(a_{n+1},k)\in S_{\epsilon}(k/g_{n+1})\}
\right|\le \frac{2k_2}{\pi k_1} |M|\epsilon.
\]
\end{Lemma}
Of course, the phase $\varphi$ has to be evaluated modulo $\pi$ here.\\
{\it Proof.} First note that since $V=0$ on $(b_n,a_{n+1})$,
we have $\varphi(a_{n+1},k)=\varphi(b_n,k)+(a_{n+1}-b_n)k=\varphi(k)
+l_nk$ (using the abbreviation $\varphi(k)=\varphi(b_n,k)$).
Denote the set defined in the Lemma by $J(l_n)$ and integrate
$|J(l_n)|$ with respect to $dl_n$. Fubini-Tonelli shows
\[
\int_0^{\pi/k_2}|J(l_n)|\,dl_n=\int_Mdk\int_0^{\pi/k_2}dl_n\,
\chi_{S_{\epsilon}(k/g_{n+1})}(\varphi(k)+kl_n).
\]
($\varphi,\beta$ are continuous in $k$, so there is certainly no
problem with measurability.) $\chi_S$ denotes the characteristic
function of the set $S$. The second integral on the right-hand side
can be estimated by $k_1^{-1}2\epsilon$, because
$\varphi(k)+lk\:(0\le l<\pi/k_2)$ takes on every value
in $[0,\pi)$ at most once. Hence
$\int_0^{\pi/k_2}|J(l_n)|\,dl_n\le k_1^{-1}|M|2\epsilon$,
and therefore the claimed inequality must hold for some $l_n$.
\hfill $\Box$
\section{Asymptotics}
According to the above remarks, we need to study the asymptotics of
$D(k)$ as $k\to\infty$. Since the higher order terms in this asymptotic
expansion depend in a complicated way on the Fourier coefficients and
also on higher moments of $W$ such as $\int_0^pW(x)xe^{in\pi x/p}\,dx$
for a general $W$ (cf.\ \cite{E}), we now specialize to
\[
W(x)=\left\{ \begin{array}{lr} 0 & (02$ can
only hold if $k=n\pi/2+(2n\pi)^{-1}+\delta$ with $n\in {\Bbb N}$ and
$\delta=O(n^{-2})$. For these $k$, a Taylor expansion leads to
(we omit this lengthy, but completely elementary computation)
\begin{equation}
\label{5}
D(k)=-2+4\delta^2-\frac{4}{n^4\pi^4}+O(n^{-6})
\end{equation}
if $n$ is odd and
\begin{equation}
\label{6}
D(k)=2-4\delta^2-\frac{8\delta^2}{n^2\pi^2}+\frac{1}{n^6\pi^6}+O(n^{-7})
\end{equation}
if $n$ is even. With these formulae, we can now show
\begin{Lemma}
\label{L5}
Let $k_n=n\pi/2+(2n\pi)^{-1}\: (n\in {\Bbb N})$ and
\[
\delta_n=\left\{ \begin{array}{lr}(n\pi)^{-2} \quad & (n \mbox{ odd})\\
2^{-1}(n\pi)^{-3} \quad & (n \mbox{ even})\end{array} \right. .
\]
Then, for sufficiently large $n$:\\
a) If $k_{n-1}+\delta_{n-1}+(n-1)^{-7/2}\le k\le k_n-\delta_n -
n^{-7/2}$, then $|D(k)|\le 2- \pi^{-2}n^{-13/2}$.\\
b) If $|k-k_n|\le \delta_n - n^{-7/2}$, then $|D(k)|\ge 2+\pi^{-2}
n^{-13/2}$.
\end{Lemma}
{\it Remark.} Subsequently, we will only need that $|D|-2$ can be
estimated by some power of $n$ if the distance of $k$ to the band edges
is $\ge Cn^{-3-\epsilon}\: (\epsilon>0)$. Thus, for simplicity, we did
not distinguish between odd and even $n$ in the statement; in fact,
for odd $n$, the exponent in b) is $11/2$ rather than $13/2$. Note
also that both estimates are very crude if $k$ is not close to one
of the band edges.
{\it Proof.} The Lemma is an easy consequence of (\ref{5}), (\ref{6}).
The details are left to the reader. In order to establish a), recall
that $D(k)$ is monotone in intervals where $|D|<2$
\cite[Theorem 2.3.1]{E}. \hfill $\Box$
In order to be able to use Lemma \ref{L2}, we also need to know the
asymptotics of the diagonalizing transformations $U(k)$:
\begin{Lemma}
\label{L8}
If $k$ satisfies either of the assumptions of Lemma \ref{L5}, then
$U(k)$ can be chosen such that $\|U(k)\|\,\|U^{-1}(k)\|\le Cn^{13/2}$.
\end{Lemma}
{\it Proof.} Write $T(k)=\left( \begin{array}{cc} t_1 & t_2 \\ t_3 & t_4
\end{array} \right)$. If $t_2\not= 0$, then
\[
U=\left( t_2(\mu^{-1}-\mu)\right)^{-1/2} \left( \begin{array}{cc} t_2 &
t_2 \\ \mu-t_1 & \mu^{-1}-t_1 \end{array} \right)
\]
is a diagonalizing transformation. Notice that $|\mu-\mu^{-1}|^2=|4-D^2|
\ge cn^{-13/2}$ because of Lemma \ref{L5}. Since $|t_i|,|\mu|$ are
bounded from above, we get
\[
\|U\|^2 \le C_1\frac{n^{13/4}}{|t_2|}.
\]
Moreover, $\det U=1$ implies $\|U^{-1}\|=\|U\|$. Similarly, if
$t_3\not= 0$, then
\[
U=\left( t_3(\mu-\mu^{-1})\right)^{-1/2} \left( \begin{array}{cc}
\mu - t_4 & \mu^{-1}-t_4 \\ t_3 & t_3 \end{array} \right)
\]
is also a diagonalizing transformation with $\det U=1$ and
$\|U\|^2\le C_2 n^{13/4} |t_3|^{-1}$. So it suffices to show that
$\max \{ |t_2|,|t_3|\}\ge C_3 n^{-13/4}$.
In the situation of Lemma \ref{L5}a), the conditions
$t_1t_4-t_2t_3=1,\: |t_1+t_4|\le 2-\pi^{-2}n^{-13/2}$ imply that
$|t_2t_3|\ge Cn^{-13/2}$, hence $\max \{ |t_2|,|t_3|\}\ge C' n^{-13/4}$,
as desired. If the hypothesis of Lemma \ref{L5}b) holds, consider
\[
t_2+t_3=\left( \frac{k}{\omega}-\frac{\omega}{k}\right) \sin\omega
\cos k .
\]
A Taylor expansion similar to that leading to (\ref{5}), (\ref{6})
shows $t_2+t_3=-2(n\pi)^{-3}+O(n^{-4})$, so in particular
$\max \{ |t_2|,|t_3|\}\ge Cn^{-3}$. \hfill $\Box$
\section{Choosing the parameters}
Up to now, we have fixed the basic periodic potential $W$. We
proceed by choosing the $g_n$. As discussed above, the choice has to be
made so that for every $k$, $k/g_n$ is in some gap for sufficiently
many $n$.
\begin{Lemma}
\label{L6}
Let $[k_1,k_2]$ be a subinterval of $(0,\infty)$. Then there are
constants $G_0,C>0$ such that for any $G\in (0,G_0]$ we can find
numbers $g_1,\ldots, g_N$ with the following properties:\\
1) $N\le CG^{-3}$\\
2) $G/2\le g_i \le G$ for all $i=1,\ldots, N$\\
3) For all $k\in [k_1,k_2]$, there is an $i\in \{1,\ldots,N\}$ so
that $|k/g_i - k_r|\le \delta_r-r^{-7/2}$ for some $r\in {\Bbb N}$.
\end{Lemma}
{\it Proof.} In this proof, we write
\begin{eqnarray*}
c_i&=&k_{2i+1}-\delta_{2i+1}+(2i+1)^{-7/2},\\
d_i&=&k_{2i+1}+\delta_{2i+1}-(2i+1)^{-7/2},
\end{eqnarray*}
(we work with the gaps corresponding to odd $n$ only, because
the gaps with even $n$ are by a factor $\sim n^{-1}$ smaller,
see Lemma \ref{L5}.)
For $G>0$, let
\begin{eqnarray*}
r_1& =&\max \{ i\in {\Bbb N}: Gc_i\le k_1\}, \\
r_2&= &\min \{ i\in {\Bbb N}: Gd_i\ge k_2\}.
\end{eqnarray*}
Obviously, $r_i\to \infty$ as $G$ tends to zero.
>From the definition of $r_i$ (and that of $k_i,\delta_i$) one
deduces easily that $r_i=k_i(\pi G)^{-1}+O(1)$ where $O(1)$
denotes an expression that remains bounded as $G\to 0$. We
take $g_1=G$ and $g_n=g_{n-1}c_{r_2}/d_{r_2}$ for $n=2,\ldots,m$
where $m$ is the smallest integer with $g_mc_{r_2}\le Gd_{r_2-1}$.
By construction, this covers $I:=[Gc_{r_2-1},Gd_{r_2}]$ in the sense
that for all $k\in I$ there is an $i\le m$ such that $k/g_i\in
[c_{r_2}, d_{r_2}]$. Moreover,
\begin{equation}
\label{7}
g_i=G\left( 1-\frac{1}{2\pi^3 r_2^3}+O(r_2^{-4})\right)^{i-1}
\end{equation}
provided that $r_2$ is large enough, i.e. $G$ has to be sufficiently
small. In this case, one infers easily from (\ref{7}) that $m$
can be estimated by $m\le 4\pi^3r_2^2$. Using this and again
(\ref{7}) shows $g_i\ge G(1-r_2^{-1})\ge G/2$ if, again, $G$ is
small enough.
The intervals $[Gc_i,Gd_{i+1}]$ with $r_1\le i\le r_2-2$ can now be
covered in the same way by appropriately scaled copies of
$[c_{i+1},d_{i+1}]$. Note that the total number of $g_i$'s
is
\[
N\le 4\pi^3\left( (r_1+1)^2+\ldots + r_2^2\right)\le 4\pi^3 r_2^3.
\]
Hence this set of $g_i$'s has the properties stated in the Lemma.
\hfill $\Box$
In order to treat the whole half-axis $k\in (0,\infty)$,
let $G_0(m),C(m)$ be the constants of Lemma \ref{L6} for
$[k_1,k_2]=[m^{-1},m]\: (m\in {\Bbb N})$. Given any functions
$F,G>0$ with $F(s)\to \infty, G(s)\to 0$, we can find integers
$m(s)\to\infty$ such that $m(s)\le F(s)$ and $C(m(s))\le F(s),
G_0(m(s))\ge G(s)$ for all $s\in {\Bbb N}$ with $m(s)>1$,
(take $m(1)=\ldots =m(s_0-1)=1$, then $m(s_0)=2$ where $s_0$
is sufficiently large so that the inequalities hold etc.).
We can now choose the whole sequence $\left(g_n\right)_{n=1}^{\infty}$:
Let $F(s),G(s)>0$ be prescribed sequences with $F\to\infty,G\to 0$.
Determine $m(s)$ as explained in the preceding paragraph, and for
$s=1,2,\ldots$, pick $g_{j(s)+1},\ldots,g_{j(s)+N(s)}$
(where $j(s)=\sum_{t=1}^{s-1}N(t)$) so that these $g$'s and $N=N(s)$
satisfy 1)-3) of Lemma \ref{L6} for $G=G(s)$ and
$[k_1,k_2]=[m(s)^{-1},m(s)]$. These definitions of $j(s),N(s),g_n$ as
well as the meaning of $F(s),G(s),m(s)$ should be kept in mind throughout
the rest of this paper. We will also write $s(n)$ for the unique
$s$ determined by $j(s)+1\le n\le j(s)+N(s)$.
Next, we investigate the set of $k$ for which $k/g_n$ is close to
one of the band edges infinitely often
(abbreviated as i.o.\ in the sequel), i.e. for infinitely many $n$.
More precisely, consider
\[
A_n=\{k>0: |k/g_n-k_i-\delta_i|*0: k\in A_n \mbox{ i.o.}\}|=0$.
\end{Lemma}
In the same way, we can control the measure of the set where
$|D|>2$ but at the same time the phase is close to $\beta$
(recall the definitions from Section 3). For a sequence $\epsilon(s)>0$,
let (again, with $s=s(n)$)
\[
B_n=\{k\in [m(s)^{-1},m(s)]: |k/g_n-k_i|\le\delta_i-i^{-7/2}
\mbox{ for some } i, \varphi(a_n,k)\in S_{\epsilon(s)}(k/g_n)\}.
\]
Lemma \ref{L4} says that $|B_n|\le \frac{2}{\pi} F(s)^2|M| \epsilon(s)$
for appropriate $l_{n-1}\in [0,\pi/m(s)]$. Here,
\[
M=\{ k\in [m(s)^{-1},m(s)]: |k/g_n-k_i|\le \delta_i - i^{-7/2}
\mbox{ for some } i \}.
\]
In order to estimate $|M|$, proceed as above. This leads to
$|B_n|\le CF(s)^3G(s)^2 \epsilon(s)$, and summing over $n$ proves
\begin{Lemma}
\label{L9}
Assume $\sum_{s=1}^{\infty} F(s)^4G(s)^{-1}\epsilon(s)<\infty$.
Then there are $l_n\to 0$ so that $|\{k>0: k\in B_n \mbox{ i.o.}\}|=0$.
\end{Lemma}
Note the somewhat complicated logic behind this statement: If the
summability assumption holds, then for any choice of the $L_n$
(see Section 2), we can find $l_n$ with the stated properties.
In particular, recall that $l_n$ depends on the values of the potential
$V(x)$ for $x\le b_n$. However, this does not cause any difficulties
because we may as well think of the parameters $L_n,g_n$ as being fixed
from the beginning (with the values given below, of course). The
representation given here tries to avoid unmotivated choices.
There will be no further conditions on $F,G,\epsilon$, therefore we now
fix these parameters. We take $F(s)=\ln s, G(s)=s^{-3},
\epsilon(s)=s^{-5}$ (we remark that powers are by no means the only
possible choice), hence $N(s)\le s^9 \ln s$ by Lemma \ref{L6} 1).
$L_n$ will also depend on $s$ only, thus we can write $L(s)$.
\section{Conclusion of the proof}
We are now in a position to estimate the solutions of (\ref{1}).
More specifically, we consider $R_{ni}:=R(x_{ni},k)$ at the points
$x_{ni}=a_n+2i/g_n\: (n\in {\Bbb N}, i=0,1,\ldots,L_n)$. Fix a $k$
which is not in the exceptional sets of Lemmas \ref{L7}, \ref{L9},
and let $n_0$ be sufficiently large so that $k\in [m(s)^{-1},m(s)]$
if $s\ge s(n_0)$. Furthermore, take $n_0$ also greater than the
finitely many $n$ for which $k\in A_n\cup B_n$, and let $i_0\in
\{0,1,\ldots,L_{n_0}\}$. For some of the deductions below, it may
be necessary to restrict $n_0$ to even larger values; we will not
mention this explicitly in the sequel. Finally, write $s_0=s(n_0),
R_0=R_{n_0i_0}$.
By combining our previous results, we will obtain the following
estimate: If $x_{ni}\ge x_{n_0i_0}$, then
\begin{equation}
\label{9}
\ln R_{ni}-\ln R_0\ge K_1\sum_{t=s_0+1}^{s(n)-1} L(t) t^{-39/4}
-K_2\sum_{t=s_0}^{s(n)} t^9 \ln^2 t .
\end{equation}
Here, $K_{1/2}$ are positive constants which are independent of
$n_0,i_0,n,i$. If $x_{ni}0$,
there is a solution $y$ satisfying
\begin{equation}
\label{f}
(y^2+(y'/k)^2)(x_{ni})=\rho_{ni}^2 \le C e^{-C's^{10}}.
\end{equation}
Because of Gronwall's Lemma (compare the argument above),
this estimate also holds for $x\in [x_{ni},x_{n,i+1}]$
(with a possibly larger $C$). The number of indices $n,i$
corresponding to the same $s$ and the length of the
intervals $[x_{ni},x_{n,i+1}]$ can both be bounded by
a power of $s$, hence (\ref{f}) guarantees that $y\in L_2$.
Finally, for $x\ge 0$, let $s$ be the index corresponding
to $x$, i.e. $a_{j(s)+1}\le x < a_{j(s+1)+1}$. Then by
construction of $V$ (see Section 2)
\[
x< a_{j(s+1)+1}\le \sum_{t=1}^s N(t)\left( \frac{4L(t)}{G(t)}+\pi \right)
\le Cs^{33}.
\]
The second term in the sum estimates the contribution from the
$l_n$. On the other hand, $0\le V(x)\le s^{-6}$, so indeed
$V(x)=O(x^{-2/11})$. This completes the proof of Theorem \ref{T1}.
\section{Singular continuous spectrum}
We modify $V$ as follows. For a growing sequence $s_n\to\infty$,
let
\[
V_{aux}(x)=\left\{ \begin{array}{cc} V(x) &\quad (a_{j(s_n)+1}\le x
\le b_{j(s_n)+N(s_n)})\\ 0 & \mbox{else} \end{array} \right. .
\]
Now the modified potential $\tilde{V}$ is obtained from $V_{aux}$
by readjusting the $l_n$ (more precisely, the subsequence of $l_n$
corresponding to the sequence $s_n$) according to Lemma \ref{L9}.
The remark following Lemma \ref{L9} shows that this may be necessary.
Since the transfer matrix for zero potential is unitary, the
above arguments also apply to $\tilde{V}$. In particular, an
estimate similar to (\ref{9}) still holds, the sums now being
over the subsequence $s_n$. Consequently, one can show along
the lines of the preceding Section that for almost all $k$
\begin{equation}
\label{10}
R(b_{j(s_n)+N(s_n)},k)\ge C_1(k)\exp \left( C_2(k)
\sum_{i=1}^n s_i^{10} \right) .
\end{equation}
A version of Schnol's Theorem (namely \cite[Proposition 9]{St2})
states that if $h\in L_2\cap L_{\infty}$,
then $h(\cdot)v(\cdot,k^2)\in L_2$ for spectrally
almost all $E=k^2$. Here $v$ is a solution of the Schr\"odinger
equation satisfying the boundary condition at $x=0$.
Let $D_n=a_{j(s_{n+1})+1}-b_{j(s_n)+N(s_n)}$ and
$h(x)=(n^2D_n)^{-1/2}$ if $b_{j(s_n)+N(s_n)}\le x\le a_{j(s_{n+1})+1}$,
and $h(x)=0$ otherwise. Then (\ref{10}) shows that $hv\notin L_2$ for
almost all $k$, hence the spectrum is supported by a set of Lebesgue
measure zero and therefore $\sigma_{ac}=\emptyset$.
On the other hand, if $D_n$ grows sufficiently fast (which can be
achieved by taking a sparse sequence $s_n$), then we can also
exclude the existence of $L_2$-solutions. More specifically,
choose the parameters by the following inductive procedure:
Take $s_1=1$ and pick the corresponding $l_n$'s. Next,
determine $D_1$ large enough so that for all $k\in [1^{-1},1]$
the following holds: If $y$ is any solution with $R(0,k)=1$,
then $D_1R^2(b_{j(s_1)+N(s_1)},k)\ge 1$. Pick any $s_2$ so that
$a_{j(s_2)+1}\ge b_{j(s_1)+N(s_1)}+D_1$ (for instance, take
the smallest number with this property) and adjust the $l_n$'s
corresponding to this $s_2$. Again, choose $D_2$ so that
$D_2R^2(b_{j(s_2)+N(s_2)},k)\ge 1$, but this time for all
$k\in [2^{-1},2]$ etc. We can also be more explicit; for
instance, it is not hard to show (using Gronwall's Lemma again)
that the sequence $s_n$ defined by $s_1=1$,
\[
s_{n+1}=2^{\left( 2^{s_n} \right) }
\]
has the required properties.
Finally, note that since the original $V$ satisfies $V(x)=O(x^{-c})$,
$\tilde{V}$ is also of order $O(x^{-c})$. We summarize our result as
\begin{Theorem}
The potential $\tilde{V}$ constructed above satisfies:\\
1) $\tilde{V}(x)=O(x^{-c})\quad (c>0)$\\
2) For all boundary conditions at $x=0$, the spectrum is purely
singular continuous in $(0,\infty)$.
\end{Theorem}
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\end{document}
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