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\topmatter
\title Operators with Singular Continuous Spectrum, V. \\
Sparse Potentials
\endtitle
\rightheadtext{Sparse potentials}
\author B.~Simon$^{1}$ and G.~Stolz$^{2}$
\endauthor
\leftheadtext{B.~Simon and G.~Stolz}
\thanks $^{1}$ Division of Physics, Mathematics and Astronomy,
California Institute of Technology, 253-37, \linebreak Pasadena, CA
91125. This material is based upon work supported by the National
Science Foundation under Grant No.~DMS-9101715. The Government has
certain rights in this material.
\endthanks
\thanks $^{2}$ Department of Mathematics, University of Alabama,
Birmingham, AL 35294.
\endthanks
\thanks To appear in {\it{Proc.~Amer.~Math.~Soc.}}
\endthanks
\abstract By presenting simple theorems for the absence of positive
eigenvalues for certain one-dimensional Schr\"odinger operators, we
are able to construct explicit potentials which yield purely singular
continuous spectrum.
\endabstract
\endtopmatter
\document
\flushpar {\bf \S 1. Introduction}
This is one of a series of papers (see [2,11,3,7,1,13,4,6,12]) that
discusses singular continuous spectrum in families of concrete
operators. Our goal here is to provide examples of half-line
Schr\"odinger operators which have singular continuous spectrum
$[0,\infty)$ for {\it{all}} boundary conditions at $x=0$. In fact,
singular continuity of the spectrum will be preserved under arbitrary
compactly supported perturbations for these examples. This is of
interest because it provides totally explicit operators without a need to
appeal to generic values of parameters and because it is a concrete
example of the fact that while dense point spectrum is {\it{always}}
unstable under rank one perturbations [3], singular continuous spectrum
{\it{may}} be stable.
To be totally explicit, it will follow from our discussion below that
if $x_{n}=e^{2n^{3/2}}$, $n=1,2,\dots$ and
$$
V(x)=\cases n & |x-x_{n}|<\tfrac{1}{2} \\
0 &\text{otherwise}, \endcases
$$
then for any boundary condition at $0$, $H\equiv -\frac{d^2}{dx^{2}}
+V(x)$ on $[0,\infty)$ has $\sigma(H)=\sigma_{\text{sc}}(H)=
[0,\infty)$ and $\sigma_{\text{ac}}(H)=\emptyset$. Depending on the
boundary condition, $\sigma_{\text{pp}}(H)$ is either $\emptyset$ or a single
negative eigenvalue. And the whole-line problem, defined by symmetric
extension of $V$, has $\sigma(H)=\sigma_{\text{sc}}(H)=[0,\infty)$,
$\sigma_{\text{ac}}(H)=\sigma_{\text{pp}}(H)=\emptyset$.
Our examples here will be sparse, that is, they will be ``mainly''
zero. Our main technical result, in Section 2, will show that sparse
potentials have no point spectrum in $[0,\infty)$ for any boundary
condition. The proof will be very elementary. The examples of singular
continuous spectrum in Section 3 will then come by combining the
results of Sections 2 and 3 with the theorems of Simon-Spencer [14]
and Stolz [15] on the absence of absolutely continuous spectrum for
certain potentials.
In Section 4 we give two other situations where our ideas apply: If
$V$ is mainly periodic instead of mainly zero, then one gets singular
continuity in the spectral bands of the underlying periodic potential.
It is also easy to extend our results to Jacobi matrices.
It has not escaped our notice that Pearson's examples [10] are sparse
and, indeed, our Section 2 implies the absence of point spectrum in his
examples (for any boundary condition!). However, we know of no way
yet to prove the absence of absolutely continuous spectrum in his
examples other than the one he uses. At least, combining our argument
with the one of Pearson shows that Pearson's examples also work on the
whole line, a fact not noted in [10].
We do note, however, that Last and Simon (work in progress) have
another way to construct potentials which decay at infinity and have
purely singular continuous spectrum and by [11], Baire generic $V$'s
vanishing at infinity have singular continuous spectrum on
$[0,\infty)$.
A result of the type discussed in this paper has been stated before in
[5], but with no proof. After we completed this work, we learned from
[9] of some apparently unpublished results of Gordon on Jacobi
matrices with sparse potentials with a similar flavor to what we do.
G.S.~would like to thank M.~Aschbacher and C.~Peck for the hospitality
of Caltech where most of this work was done.
\vskip 0.3in
\flushpar {\bf \S 2. Absence of Point Spectrum}
Let $V(x)$ be a measurable function on $[0,\infty)$ which is $L^1$ on
any interval $[0,R)$. For $y\geq 0$, one can look at solutions of the
differential equation
$$\gather
-u''(x) + V(x)u(x)=Eu(x) \\
u(y)=a \qquad u'(y)=b.
\endgather
$$
If $\Phi(x,y,E; V)$ is the solution with $\binom{a}{b}=\binom{1}{0}$ and
$\Psi (x,y,E; V)$ with $\binom{a}{b}=\binom{0}{1}$, then the $2\times 2$
matrix
$$
M(x,y,E;v)=\pmatrix
\Phi (x,y,E;V) & \Psi (x,y,E;V) \\
\Phi' (x,y,E; V) & \Psi' (x,y,E; V) \endpmatrix
$$
is called the transfer or fundamental matrix. Solutions of
$-u''+Vu=Eu$ obey
$$
\binom{u(x)}{u'(x)} = M(x,y,E;V)\binom{u(y)}{u'(y)}. \tag 1
$$
and
$$
M(x,y)\,M(y,z)=M(x,z) \tag 2
$$
Constancy of the Wronskian implies that
$$
\det(M)=1. \tag 3
$$
\proclaim{Theorem 2.1} If $V$ is bounded from below and $M$ obeys
$$
\int\limits^{\infty}_{0} \frac{dx}{\|M(x,0,E;V)\|^{2}} =\infty \tag 4
$$
for some $E$, then $-u''+Vu=Eu$ has no solution $u\in L^2(0,\infty)$.
\endproclaim
\demo{Proof} Since $M$ is unimodular ((1)) and $2\times 2$, we have
$$
\|M^{-1}\| =\|M\|. \tag 5
$$
By (1),
$$
\binom{u(0)}{u'(0)} = M(x,0)^{-1} \binom{u(x)}{u'(x)}.
$$
So, by (5)
$$
\biggl\|\binom{u(x)}{u'(x)}\biggr\| \geq \biggl\| \binom{u(0)}{u'(0)}
\biggr\| \bigg/ \| M(x,0)\|.
$$
Thus, (4) implies that $\|\binom{u}{u'}\|\notin L^2$. Then also $u\notin
L^2$.
Suppose, on the contrary, that $u\in L^2$. Differentiating $uu'$, we
get
$$\align
u(x)u'(x) &= u(0)u'(0) + \int\limits^{x}_{0} (u^{\prime 2} +u^{2}
(V-E))\, dt \\
&\geq u(0)u'(0) + \int\limits^{x}_{0} (u^{\prime 2} - Cu^{2})\, dt,
\endalign
$$
where semiboundedness of $V$ was used. If $u\in L^2$, but $u'\notin
L^2$, this yields $\frac{1}{2}(u^{2})'(x)=u(x)u'(x)\to\infty$ as
$x\to\infty$. This would imply the contradiction $u^{2}(x)\to\infty$.
\qed
\enddemo
We are going to apply this to what we'll call sparse potentials.
\definition{Definition} A sparse potential is a function $V(x)$ on
$[0,\infty)$ for which there exist $x_{n+1}>x_{n}\to\infty$,
$\alpha_{n}>0$ and $h_{n}<\infty$ so
\roster
\item"{(i)}" $\frac{x_{n+1}-x_{n}}{\alpha_{n}+\alpha_{n+1}+1}
\to\infty$,
\item"{(ii)}" $|V(x)|\leq h_{n}$ if $|x-x_{n}|\leq \alpha_{n}$,
$n=1,2,\dots$,
\item"{(iii)}" $|V(x)=0$ otherwise.
\endroster
\enddefinition
By (i), (iii), sparse potentials are zero ``most of the time.'' We
define $L_{n}=x_{n+1}-x_{n}-\alpha_{n}-\alpha_{n+1}$.
\proclaim{Theorem 2.2} Let $V$ be a sparse potential. Suppose that
$$
L_{n}\geq\exp(4Q_{n}) \tag 6
$$
for all large $n$ where
$$
Q_{n}\equiv n\,\ln\,n + \sum^{n}_{j=1} \alpha_{j} (h_{j} + \ln\, j).
$$
Then $-u''+Vu=Eu$ has no solutions with $u\in L^2$ for any $E>0$.
\endproclaim
\demo{Proof} $M'(x)=A(x)M(x)$ where
$$
A(x)=\pmatrix 0 & 1 \\
-1 & E-V(x) \endpmatrix ,
$$
so $\|A(x)\|= |V(x)|+E+1\leq h_{n}+E+1$ on $(x_{n}-\alpha_{n},
x_{n}+\alpha_{n})$ and
$$
\|M(x_{n}+\alpha_{n}, x_{n}-\alpha_{n})\|\leq e^{2(h_{n}+E+1)\alpha_{n}}.
$$
On the other hand, the transfer matrix when $V=0$ and $E=k^{2}>0$ is
$$
M_{0}(x,y)=
\pmatrix \cos k(x-y) & k^{-1}\sin k(x-y) \\
-k\sin k(x-y) & \cos k(x-y) \endpmatrix
$$
and has norm bounded by $\max(k, k^{-1})$.
Using (2), it follows that for $y\in (x_{n}+\alpha_{n}, x_{n+1}
-\alpha_{n+1})$, we have
$$
\|M(y,0)\|\leq [\max (k,k^{-1})]^{n+1} \exp\biggl(\sum^{n}_{j=1}
2\alpha_{j} (h_{j}+E+1)\biggr) \leq \exp(2Q_{n})
$$
for $n$ large (and any fixed $E$). Thus,
$$
\int\limits^{x_{n+1}-\alpha_{n+1}}_{x_{n}+\alpha_{n}} \frac{dy}
{\|M(y,0)\|^{2}} \geq \frac{L_{n}}{\exp(4Q_{n})} \geq 1
$$
by hypothesis. Thus, (4) holds and there are no $L^{2}$ solutions.
\qed
\enddemo
\example{Example 1} $h_{n}=n$, $\alpha_{n}=\frac{1}{2}$; (6) requires
$L_{n}\geq \exp((1+\epsilon)n^{2})$.
\endexample
\example{Example 2} $h_{n}=c$, $\alpha_{n}=n$; (6) requires $L_{n}\geq
\exp((1+\epsilon)n^{2})$ again.
\endexample
In the case of constant high barriers, like in Example 1, one actually
has the following result, which gives improved estimates
\proclaim{Theorem 2.3} Let $V$ be a sparse potential with
$$
V(x)=h_{n} \qquad \text{\rom{if }} |x-x_{n}|\leq a_{n}, n=1,2,\dots,
$$
$h_{n}\to\infty$ and
$$
L_{n}\geq n^{\delta n} \biggl(\prod\limits^{n}_{j=1} h_{j}\biggr)
\exp\biggl( 4\sum^{n}_{j=1} \alpha_{j} \sqrt{h_{j}}\biggr) \tag 7
$$
for some $\delta >0$. Then $-u''+Vu=Eu$ has no solutions with $u\in
L^{2}$ for any $E>0$.
\endproclaim
Looking at Example 1 again, we see that (7) only requires
$L_{n}\geq\exp((1+\epsilon)n^{3/2})$.
\demo{Proof} We follow the proof of Theorem 2.2, but for $n$ with
$h_{n}>E+1$ and $x,y\in [x_{n}-\alpha_{n}, x_{n}+\alpha_{n}]$, replace
$M$ by the modified transfer matrix
$$
\tilde M_{n}(x,y)=\pmatrix \sqrt{h_{n}-E}\,\Phi & \sqrt{h_{n}-E}\,
\Psi \\
\Phi' & \Psi'
\endpmatrix .
$$
We have
$$
\tilde{M}'_{n} =\sqrt{h_{n}-E}\,\pmatrix 0 & 1 \\
1 & 0 \endpmatrix\, \tilde M_n
$$
and therefore,
$$
\|\tilde{M}_{n}(x_{n}+\alpha_{n}, x_{n}-\alpha_{n})\| \leq
\exp\bigl(2\alpha_{n}\sqrt{h_{n}-E}\,\bigr).
$$
Since
$$
M(x_{n}+\alpha_{n}, x_{n}-\alpha_{n})=\pmatrix (h_{n}-E)^{-1/2}
& 0 \\
0 & 1 \endpmatrix \,\tilde M_{n}(x_{n}+\alpha_{n}, x_{n}-\alpha_{n})
\pmatrix (h_{n}-E)^{1/2} & 0 \\
0 & 1 \endpmatrix ,
$$
we get
$$\align
\|M(x_{n}+\alpha_{n}, x_{n}-\alpha_{n})\| &\leq \max((h_{n}-E)^{1/2},
(h_{n}-E)^{1/2})\exp\bigl(2\alpha_{n}\sqrt{h_{n}-E}\,\bigr) \\
&\leq \sqrt{h_{n}}\,\exp\bigl(2\alpha_{n}\sqrt{h_n}\,\bigr).
\endalign
$$
Now the proof is completed as before. \qed
\enddemo
\vskip 0.3in
\flushpar {\bf \S 3. Examples with Singular Continuous Spectrum}
As stated in the introduction, the idea behind these examples is to
use [14] and [15] to eliminate a.c.~spectrum and Theorems 2.2 and 2.3 to
eliminate point spectrum. Both results apply with arbitrary bounded
condition at $x=0$. Non-existence of a.c.~spectrum is also stable
under a compactly supported perturbation of $V$ (a trace class
perturbation of the resolvent, e.g., [4]). Obviously, the same is true
for non-existence of square-integrable solutions.
Therefore, we get examples with purely singular continuous spectrum
for all boundary conditions and under arbitrary local perturbations.
Whole-line problems differ from the direct sum of two half-line
problems only by adding boundary conditions at $0$, a finite rank
perturbation. Moreover, lack of $L^2$ solutions on either half-line
clearly implies no $L^2$ solutions on the whole line. This shows that
we also get whole-line examples of purely singular continuous spectrum.
The statements on $\sigma_{\text{\rom{pp}}}$ in $(-\infty, 0]$ made in
the introduction follow from elementary convexity considerations for
solutions.
\example{Example 3} Let $x_{n}=e^{2n^{2/3}}$ and
$$
V(x) =\cases n & \text{if }|x-x_{n}|<\tfrac{1}{2}, n=1,2,\dots \\
0 & \text{otherwise}.
\endcases
$$
Then $-\frac{d^2}{dx^2}+V(x)$ on $L^{2}(0,\infty)$ has purely singular
continuous spectrum in $(0,\infty)$ for any boundary condition at $x=0$.
For [14] says there is no a.c.~spectrum and Theorem 2.3 says no point
spectrum.
\endexample
\example{Example 4} Here we construct bounded potentials with purely
singular continuous spectrum in $(0,\infty)$. Let $W(x)$ be the random
potential given by a random constant $c_n$ in each interval $(n-1,n)$,
$n=1,2\dots$, where the $c_n$ are independent and each uniformly
distributed in $[0,1]$. Let
$$
V(x) =\cases W(x+\tfrac{n(n-1)}{2} -x_{n}), & x_{n}\leq x
\leq x_{n}+n \\
0 & \text{otherwise}
\endcases
$$
where $x_{n}=e^{2n^2}$. Essentially, we've broken $W(x)$ into pieces
of size $1,2,\dots$ and placed them at the points $x_{1},x_{2}\dots$.
Then for almost all choices of the random potential, $-\frac{d^2}{dx^2}
+V(x)$ on $L^{2}(0,\infty)$ has purely singular continuous spectrum for
all boundary conditions at $x=0$.
By Theorem 2.2 there is no point spectrum. To prove absence of
a.c.~spectrum, let $\chi$ be the characteristic function on $[0,1]$
and
$$
q_{L}(x)=\sum^{\infty}_{n=-\infty} \chi(x-nL).
$$
If $E_{0}=k^{2}_{0}>0$ is such that $\frac{k_{0}}{2\pi}$ is not
rational, then Theorem 3.4 of [8] guarantees the existence of an
integer $L=L(E_{0})$ and $\delta >0$ such that $(E_{0}-\delta,
E_{0}+\delta)$ is contained in a spectral gap of $-\frac{d^2}{dx^2}+
q_L$. Almost certainly, there are intervals $I_n$ with length tending to
$\infty$ such that
$$
\sup\limits_{x\in\cup I_{n}}\, |W(x)-q_{L}(x)|<\frac{\delta}{2}.
$$
By construction, the same holds for $V$. By Theorem 1 of [15], we have
$\sigma_{\text{\rom{ac}}}(-\frac{d^2}{dx^2}+V)\cap (E_{0}-
\frac{\delta}{2}, E_{0}+\frac{\delta}{2})=\emptyset$ with probability
one. A compactness argument and the fact that the countable set
$\{k^{2}_{0}:\frac{k_{0}}{2\pi}\text{ rational}\}$ cannot support
a.c.~spectrum finally show almost sure absence of a.c.~spectrum.
Of course, the above construction and argument apply to much more
general random potentials $W$.
\endexample
\newpage
\vskip 0.3in
\flushpar {\bf \S 4. ``Mainly'' Periodic Potentials and Jacobi
Matrices}
In Section 2 the important property of the regions with $V=0$ was
that, for a given $E>0$, the norm of the transfer matrix $M_{0}(x,y)$ is
uniformly bounded in $x,y$. We are in the same situation if we look at
the transfer matrix for $-u''+V_{0}u=Eu$, where $V_0$ is periodic and
$E$ is an interior point of one of the stability intervals for $-
\frac{d^2}{dx^2}+V_{0}$. Thus, all our results for sparse potentials
have suitable extensions to ``mainly'' periodic potentials. We
illustrate this with
\example{Example 5} Let $V_0$ be real and periodic,
$x_{n}=e^{2n^{3/2}}$ and
$$
V(x) =\cases n & |x-x_{n}|<\tfrac{1}{2}, n=1,2,\dots \\
V_{0}(x) &\text{otherwise}.
\endcases
$$
Then $-\frac{d^2}{dx^2}+V$ on $L^{2}(0,\infty)$ with any boundary
condition at $x=0$ has purely singular continuous spectrum in the
interior of the stability intervals of $-\frac{d^2}{dx^2}+V_0$.
\endexample
The above methods can easily be applied to Jacobi matrices $h$ on
$\ell^{2}(0,\infty)$ defined by
$$\align
(hu)(0) &= u(1)+v(0)u(0), \\
(hu)(x) &= u(x-1)+u(x+1)+v(x)u(x), \quad x=1,2,\dots
\endalign
$$
As an example, we give
\proclaim{Theorem 4.1} Let $h_{n}\to\infty$, $x_{n}$ be integers with
$x_{n+1}>x_{n}\to\infty$ and
$$
v(x)=\cases h_{n} & x=x_{n}, n=1,2,\dots \\
0 & \text{otherwise}.
\endcases
$$
Furthermore, let $\ell_{n}=x_{n+1}-x_{n}$ satisfy
$$
\ell_{n}\geq \prod\limits^{n}_{j=1} (h_{j}+\ln j), \quad n=1,2,\dots
\tag 8
$$
Then, the Jacobi matrix $h$ is purely singular continuous in $(-2,2)$.
\endproclaim
\demo{Proof} $\sigma_{\text{\rom{ac}}}(h)=\emptyset$ follows from
$h_{n}\to\infty$ and [14].
The transfer matrix to solutions of
$$
u(x-1)+u(x+1)+v(x)u(x)=eu(x), \quad x=1,2,\dots
$$
is given by
$$
M(x)=\prod\limits^{x}_{j=1} T(j)
$$
where
$$
T(j)=\pmatrix 0 & 1 \\
-1 & e-v(j) \endpmatrix .
$$
By an analog to Theorem 2.2, it suffices to show that for $e\in (-2,2)$
$$
\sum^{\infty}_{x=1} \frac{1}{\|M(x)\|^{2}}=\infty. \tag 9
$$
We have
$$
\|T(x_{n})\|\leq h_{n}+|e|+1
$$
and, diagonalizing $\bigl(\smallmatrix 0 & 1 \\ -1 & e \endsmallmatrix
\bigr)$ for $e\in (-2, 2)$,
$$
\biggl\| \prod\limits^{x_{n+1}-1}_{j=x_{n}+1} T(j) \biggr\|
\leq C(e)
$$
uniformly in $n$. Thus,
$$
\|M(x)\| \leq C(e)^{n+1} \prod\limits^{n}_{j=1} (h_{j}+|e|+1)
$$
for $x\in (x_{n}, x_{n+1})$. (9) follows from this and (8). \qed
\enddemo
\vskip 0.3in
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\enddocument