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\topmatter
\title $\boldkey L^{\boldkey p}$ Norms of the Borel Transform and the 
Decomposition of Measures
\endtitle
\rightheadtext{$L^{p}$ Norms of the Borel Transform and the
Decomposition of Measures}
\author B.~Simon$^{*}$
\endauthor
\leftheadtext{B.~Simon}
\affil Division of Physics, Mathematics, and Astronomy \\
California Institute of Technology, 253-37 \\
Pasadena, CA 91125
\endaffil
\thanks $^{*}$ 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 To appear in {\it Proc.~Amer.~Math.~Soc.}
\endthanks
\endtopmatter
\document
\bigpagebreak
\block
{\smc{Abstract.}}  We relate the decomposition over $[a,b]$ of a measure 
$d\mu$ (on $\Bbb R$) into absolutely continuous, pure point, and 
singular continuous pieces to the behavior of integrals $\int\limits
^{b}_{a}(\text{Im}\,F(x+i\epsilon))^{p}\,dx$ as $\epsilon\downarrow 0$. 
Here $F$ is the Borel transform of $d\mu$, that is, $F(z)=\int (x-
z)^{-1}\,d\mu(x)$.
\endblock
\vskip 0.4in
\flushpar {\bf \S 1. Introduction}

Given any positive measure $\mu$ on $\Bbb R$ with
$$
\int\frac{d\mu(x)}{1+|x|} < \infty, \tag 1.1
$$
one can define its Borel transform by
$$
F(z)=\int\frac{d\mu(x)}{x-z}. \tag 1.2
$$
We have two goals in this note. One is to discuss the relation of the 
decomposition of $\mu$ into components ($d\mu = d\mu_{\text{\rom ac}}
+d\mu_{\text{\rom pp}}+d\mu_{\text{\rom sc}}$ with $d\mu_{\text{\rom ac}}
(x)=g(x)\,dx$, $d\mu_{\text{\rom pp}}$ a pure point measure, and 
$d\mu_{\text{\rom sc}}$ a singular continuous measure) to integrals of 
powers of $\text{Im}\,F(x+i\epsilon)$.  This is straightforward, and global 
results (e.g., involving $\int\limits^{\infty}_{-\infty}|\text{Im}\,
F(x+i\epsilon)|^{2}\,dx$) are well-known to harmonic analysts (see, 
e.g., Koosis [5, pg.~157])---but there seems to be a point in writing down 
elementary proofs of the local results (e.g., involving $\int\limits^{b}
_{a}|\text{Im}\,F(x+i\epsilon)|^{2}\,dx$).

Secondly, by proper use of these theorems, we can simplify the proofs 
in [7] that certain sets of operators are $G_\delta$'s in certain 
metric spaces.

In \S 2, we will see that $\int\limits^{b}_{a}|\text{Im}\,F(x+
i\epsilon)|^{p}\,dx$ with $p>1$ is sensitive to singular parts of 
$d\mu$ and can be used to prove they are absent.  In \S 3, we see the 
opposite results when $p<1$ and the singular parts are irrelevant, so 
that integrals can be used for a test of whether $\mu_{\text{\rom ac}}
=0$. Finally, in \S 4, we turn to the aforementioned results on $G_\delta$ 
sets of operators.

Since we only discuss $\text{Im}\,F(z)$ and
$$
\text{Im}\,F(x+i\epsilon)=\epsilon\int\frac{d\mu(y)}{(x-y)^{2}+
\epsilon^{2}}, \tag 1.3
$$
our results actually hold if (1.1) is replaced by
$$
\int\frac{d\mu(x)}{(1+|x|)^{2}}<\infty. \tag 1.4
$$

It is a pleasure to thank S.~Jitomirskaya, A.~Klein, and T.~Wolff for 
valuable discussions.

\bigpagebreak
\flushpar {\bf \S 2. $\boldkey p$-norms for $\boldkey p \boldkey> \boldkey 1$}
\medpagebreak

\proclaim{Theorem 2.1}  Fix $p>1$.  Suppose that
$$
\sup\limits_{0<\epsilon<1} \int\limits^{b}_{a}|\text{\rom{Im}}\,F(x+i
\epsilon)|^{p}\,dx <\infty. \tag 2.1
$$
Then $d\mu$ is purely absolutely continuous on $(a, b)$, $\frac{d\mu_
{\text{\rom{ac}}}}{dx}\in L^{p}(a, b)$; and for any $[c, d]\subset 
(a,b)$, $\frac{1}{\pi}\,\text{\rom{Im}}\,F(x+i\epsilon)$ converge to 
$\frac{d\mu_{\text{\rom{ac}}}}{dx}$ in $L^p$.  Conversely, if $[a,b]
\subset (e,f)$ with $d\mu$ purely absolutely continuous on $(e, f)$, 
and $\frac{d\mu_{\text{\rom{ac}}}}{dx}\in L^{p}(e, f)$, then \rom(2.1\rom) 
holds.
\endproclaim

\remark{Remarks} 1.  This criterion with $p=2$ is used by Klein [4], 
who has a different proof.

2. The $p=2$ results can be viewed as following from Kato's theory of 
smooth perturbations [2,6].

3.  It is easy to construct measures supported on $\Bbb R\backslash
(a, b)$ so that (2.1) fails or so that the $L^p$ norm oscillates, for 
example, suitable point measures $\sum\alpha_{n}\delta_{x_n}$ with 
$x_{n}\uparrow a$.  For this reason, we are forced to shrink/expand 
$(a,b)$ to $(c,d)/(e,f)$.
\endremark

\demo{Proof}  Let $d\mu_{\epsilon}(x)=\pi^{-1}\text{Im}\,F(x+i
\epsilon)\,dx$.  Then [8] $d\mu_{\epsilon}\to d\mu$ weakly, as 
$\epsilon\downarrow 0$, that is, $\lim\limits_{\epsilon\downarrow 0} \int
f(x)\,d\mu_{\epsilon}(x)=\int f(x)\,d\mu(x)$ for $f$ a continuous 
function of compact support.  Let $q$ be the dual index to $p$ and $f$ 
a continuous function supported in $(a, b)$.  Then
$$\align
\biggl|\int f\,d\mu\biggr| &=\lim\limits_{\epsilon\downarrow 0}\,
\biggl|\int f\,d\mu_{\epsilon}\biggr| \\
&\leq \varlimsup\limits_{\epsilon\downarrow 0}\, \biggl[\int\limits
^{b}_{a} |f(x)|^{q}\,dx\biggr]^{1/q}\, \biggl[\int\limits^{b}_{a}
\biggr(\frac{1}{\pi}\,\text{Im}\,F(x+i\epsilon)\biggr)^{p}\,dx
\biggr]^{1/p} \\
&\leq C \|f\|_{q}.
\endalign
$$
Thus, $f\mapsto\int f\,d\mu$ is a bounded functional on $L^q$, and thus 
$\chi_{(a, b)}\,d\mu = g\,dx$ for some $g\in L^{p}(a, b)$.

We claim that when $\chi_{(a, b)}\,d\mu=g\,dx$ with $g\in L^{p}(a, b)$, 
then for any $[c, d]\subset (a, b)$, $\frac{1}{\pi}\,\text{Im}\,F(x+
i\epsilon)\to g$ in $L^{p}(c, d)$---this implies the remaining parts 
of the theorem.

To prove the claim, write $F=F_{1}+F_{2}$ where $F_1$ comes from 
$d\mu_{1}\equiv\chi_{(a, b)}\,d\mu$ and $d\mu_{2}=(1-\chi_{(a,b)})\,
d\mu$.  $\frac{1}{\pi}\,\text{Im}\,F_{1}$ is a convolution of $g\,dx$ 
with an approximate delta function.  So, by a standard argument, $\frac
{1}{\pi}\,\text{Im}\,F_{1}\to g$ in $L^p$. On the other 
hand, since $\text{dist}([c,d], \Bbb R\backslash (a,b))>0$, one 
easily obtains a bound:
$$
|\text{Im}\,F_{2}(x+i\epsilon)|\leq C\epsilon \qquad \text{for }
x\in [c,d].
$$
So $\frac{1}{\pi}\,\text{Im}\,F_{2}\to 0$ in $L^p$. \qed
\enddemo

The following is a local version of Wiener's theorem:

\proclaim{Theorem 2.2}
$$
\lim\limits_{\epsilon\downarrow 0}\,\epsilon\int\limits^{b}_{a}|\text
{\rom{Im}}\,F(x+i\epsilon)|^{2}\,dx=\frac{\pi}{2}\biggl(\frac{1}{2}\,
\mu(\{a\})^{2}+\frac{1}{2}\,\mu(\{b\})^{2}+\sum_{x\in (a, b)}\mu 
(\{x\})^{2}\biggr). \tag 2.1
$$
\endproclaim

\demo{Proof} Using (1.3), we see that
$$
\epsilon\int\limits^{b}_{a}(\text{Im}\,F(x+i\epsilon))^{2}\,dx =
\int\int g_{\epsilon}(x, y)\,d\mu(x)\,d\mu(y),
$$
where
$$
g_{\epsilon}(x, y)=\int\limits^{b}_{a}\frac{\epsilon^{3}\,dw}
{((w-x)^{2}+\epsilon^{2})((w-y)^{2}+\epsilon^{2})}.
$$
It is easy to see that for $0< \epsilon <1$:
\roster
\item"{(i)}" $g_{\epsilon}(x, y)\leq \pi\,\frac{1}{\text
{dist}(x, [a,b])^{2}+1}$
\item"{(ii)}" $\lim\limits_{\epsilon\downarrow 0}\,g_{\epsilon}(x, y)
=0$ if $x\neq y$ or $x\notin [a,b]$
\item"{(iii)}" $\lim\limits_{\epsilon\downarrow 0}\, g_{\epsilon}(x, y)
=\frac{\pi}{2}$ if $x=y\in (a, b)$
\item"{(iv)}" $\lim\limits_{\epsilon\downarrow 0}\, g_{\epsilon}(x, y)
=\frac{\pi}{4}$ if $x=y$ is $a$ or $b$.
\endroster
Thus, the desired result follows from dominated convergence. \qed
\enddemo

\remark{Remarks} 1. It is not hard to extend this to $\epsilon^{p-1}
\int\limits^{b}_{a}|\text{Im}\,F(x+i\epsilon)|^{p}\,dx$ for any $p>1$.  
The limit has $\int\limits^{\infty}_{-\infty}(1+x^{2})^{-p/2}\,dx$ in 
place of $\pi$ (which can be evaluated exactly in terms of gamma functions) 
and $\mu (\{x\})^p$ in place of $\mu (\{x\})^2$. For the above proof 
extends to $p$ an even integer.  Interpolation then shows that the 
continuous part of $\mu$ makes no contribution to the limit, and a 
simple argument restricts the result to a finite sum of point measure 
where it is easy.  (Note:  For $1<p<2$, one interpolates between 
boundedness for $p=1$ and the zero limit if $p=2$ and $\mu$ is 
continuous.)

2. On the other hand, $\sup\limits_{0<\epsilon <1}\,\epsilon^{\alpha}
\int\limits^{b}_{a}\text{Im}\,F(x+i\epsilon)^{2}\,dx$ for $0<\alpha 
<1$ says something about how singular the singular part of $d\mu$ can 
be.  If the $\sup$ is finite, then $\mu(A)=0$ for any subset $A$ of 
$[a,b]$ with Hausdorff dimension $d<1-\alpha$.  This will be proven in 
[1].
\endremark

\proclaim{Corollary 2.3} $\mu$ has no pure points in $[a,b]$ if and 
only if $\varliminf\limits_{k\to\infty} \frac{1}{k}\int\limits^{b}_{a}(\text
{\rom{Im}}\,F(x+ik^{-1}))^{2}\,dx=0$.
\endproclaim

(Of course the limit exists but we'll need this form in \S 4.)

\bigpagebreak
\flushpar {\bf \S3. $\boldkey p$-norms for $\boldkey p \boldkey< \boldkey 1$}
\medpagebreak

\proclaim{Theorem 3.1} Fix $p<1$.  Then
$$
\lim\limits_{\epsilon\downarrow 0}\,\int\limits^{b}_{a}\,\biggl|\frac
{1}{\pi}\,\text{\rom{Im}}\,F(x+i\epsilon)\biggr|^{p}\,dx =
\int\limits^{b}_{a}\biggl(\frac{d\mu_{\text{\rom{ac}}}}{dx}\biggr)^{p}
\,dx.
$$
\endproclaim

\demo{First Proof}  Write $d\mu$ as three pieces: $d\mu_{1}=(1-
\chi_{[a-1, b+1]})\,d\mu$, $d\mu_{2}=g\,dx$ with $g\in\mathbreak L^{1}(a-1, 
b+1)$, and $d\mu_{3}$ singular and finite and concentrated on $[a-1, 
b+1]$ and correspondingly, $F=F_{1}+F_{2}+F_{3}$.  It is easy to 
see that $|\text{Im}\,F_{1}(x+i\epsilon)|\leq C\epsilon$ on $[a,b]$, 
so its contribution to the limit of the integral is $0$.  Since 
$\frac{1}{\pi}\,\text{Im}\,F_{2}(x+i\epsilon)$ is a convolution of $g$ 
with an approximate delta function, $\frac{1}{\pi}\,\text{Im}\,F_{2}\to g$ 
in $L^1$, and so by Holder's inequality, $\int\limits^{b}_{a}|\frac{1}
{\pi}\,\text{Im}\,F_{2}(x+i\epsilon)|^{p}\,dx\to\int\limits^{b}_{a}
g(x)^{p}\,dx$ for any $p<1$.  It thus suffices to prove that:
$$
\int\limits^{b}_{a}\,\biggl|\frac{1}{\pi}\,\text{Im}\,F_{3}(x+i\epsilon)
\biggr|^{p}\,dx\to 0. \tag 3.1
$$
Let $S$ be a set with $\mu_{3}(\Bbb R\backslash S)=0$ and $|S|=0$.  Given 
$\delta$, by regularity of measures, find $C\subset S\subset\Cal O$ with 
$C$ compact and $\Cal O\subset (a-2, b+1)$ open so $\mu(S\backslash C)<\delta$ 
and $|\Cal O\backslash S|<\delta$, so $\mu(\Bbb R\backslash C)<\delta$ and 
$|\Cal O|<\delta$.  Let $h$ be a continuous function which is $1$ on 
$\Bbb R\backslash\Cal O$ and $0$ on $C$.

By Holder's inequality (with index $\frac{1}{p}$):
$$
\int\limits_{A}\biggl(\frac{1}{\pi}\,\text{Im}\,F_{3}\biggr)^{p}\,dx 
\leq |A|^{1-p} \biggl[\int\limits_{A}\biggl(\frac{1}{\pi}\,\text{Im}\,
F_{3}\biggr)\biggr]^{p} \tag 3.2
$$
for any set $A$.  Noting that $\int\limits_{\Bbb R}(\frac{1}{\pi}\,\text{Im}\,
F_{3})\,dx = \mu_{3}(\Bbb R)<\infty$, we see that
$$
\int\limits_{\Cal O}\biggl(\frac{1}{\pi}\,\text{Im}\,F_{3}\biggr)^{p}\,
dx\leq\mu_{3}(\Bbb R)^{p}\delta^{1-p}. \tag 3.3
$$
On the other hand,
$$\align
\int\limits_{[a,b]\backslash\Cal O}\biggl(\frac{1}{\pi}\,\text{Im}\,
F_{3}\biggr)^{p}\,dx &\leq |b-a|^{1-p}\biggl[\ \int\limits_{[a,b]
\backslash\Cal O}\biggl(\frac{1}{\pi}\,\text{Im}\,F_{3}\biggr)\, 
dx\biggr]^{p} \\
&\leq |b-a|^{1-p}\biggl[\ \int\limits^{b}_{a}h(x)\biggl(\frac{1}{\pi}\,
\text{Im}\,F_{3}\biggr)(x+i\epsilon)\, dx \biggr]^{p}.
\endalign
$$
The last integral converges to $\int h(x)\,d\mu_{3}(x)\leq\int
\limits_{\Bbb R\backslash C}d\mu_{3}(x)=\mu_{3}(\Bbb R\backslash C)=
\delta$.  Thus
$$
\varlimsup\limits_{\epsilon\downarrow 0}\int\limits^{b}_{a}\frac{1}{\pi}\,
\text{Im}\,F_{3}(x+i\epsilon)^{p}\,dx\leq\mu_{3}(\Bbb R)^{p}\delta^{1-p}
+|b-a|^{1-p}\delta^{p}.
$$
Since $\delta$ is arbitrary, the $\varlimsup$ is a zero and so the 
limit is zero. \qed
\enddemo

\demo{Second Proof} (suggested to me by T.~Wolff) As in the first 
proof, by writing $\mu$ as a sum of a finite measure and a measure 
obeying (1.1) but supported away from $[a,b]$, we can reduce the 
result to the case where $\mu$ is finite.  Let $M_{\mu}(x)$ be the 
maximal function of $\mu$:
$$
M_{\mu}(x)=\sup\limits_{t>0}\,(2t)^{-1}\mu(x-t, x+t).
$$
By the standard Hardy-Littlewood argument (see, e.g., Katznelson 
[3]):
$$
|\{x\mid M_{\mu}(x)>t\}|\leq C\mu(\Bbb R)/t,
$$
which in particular implies
$$
\int\limits^{b}_{a}M_{\mu}(x)^{p}\,dx <\infty
$$
for all $p<1$.

Since $\frac{1}{\pi}\,\text{Im}\,F(x+i\epsilon)\leq M_{\mu}(x)$ for all 
$\epsilon$ and $\frac{1}{\pi}\,\text{Im}\,F(\cdot +i\epsilon)\to(\frac{d\mu_
{\text{\rom{ac}}}}{dx})(x)$ a.e.~in $x$, the desired result follows by 
the dominated convergence theorem. \qed
\enddemo

\remark{Remark}  The reader will note that the first proof is similar 
to the proof in [7] that the measures with no a.c.~part are a 
$G_\delta$.  In a sense, this part of our discussion in \S4 is a 
transform for the proof of [7] to this proof instead!
\endremark

\proclaim{Corollary 3.2}  A measure $\mu$ has no absolutely continuous 
part on $(a, b)$ if and only if
$$
\varliminf\limits_{k\to\infty} \int\limits^{b}_{a} \text{\rom{Im}}\,
F(x+ik^{-1})^{1/2}\,dx = 0.
$$
\endproclaim


\bigpagebreak
\flushpar {\bf \S4. $\boldkey G_{\boldsymbol\delta}$ properties of sets of 
measures and operators}
\medpagebreak

\proclaim{Lemma 4.1}  Let $X$ be a topological space and $f_{n}:X\to
\Bbb R$ a sequence of non-negative continuous functions.  Then 
$\bigl\{x\mid\varliminf\limits_{n\to\infty}\,F_{n}(x)=0\bigr\}$ is a 
$G_\delta$.
\endproclaim

\demo{Proof}
$$\align
\biggl\{x\mid\varliminf\limits_{n\to\infty}\, F_{n}(x)=0\biggr\} &=\biggl\{x\mid
\forall k\,\forall N\,\exists n\geq N \ F_{n}(x)<\frac{1}{k}\biggr\} \\
&=\bigcap\limits^{\infty}_{k=1}\,\bigcap\limits^{\infty}_{N=1}\,
\bigcup\limits^{\infty}_{n=N} \biggl\{x\mid F_{n}(x)<\frac{1}{k}
\biggr\}
\endalign
$$
is a $G_\delta$. \qed
\enddemo

As a corollary of this and Corollaries 2.3 and 3.2, we obtain a proof 
of the result of [9].

\proclaim{Theorem 4.1}  Let $M$ be the set of probability measures 
on $[a,b]$ in the topology of weak convergence \rom(this is a 
complete metric space\rom). Then $\{\mu\mid\mu \text{\rom{ is purely 
singular continuous}}\}$ is a dense $G_\delta$.
\endproclaim

\demo{Proof} By Corollary 3.2:
$$
\{\mu\mid\mu_{\text{\rom{ac}}}=0\}=\biggl\{\mu\mid\varliminf\limits_{k\to
\infty}\int\limits^{b}_{a}(\text{Im}\,F_{\mu}(x+ik^{-1})^{1/2}\,dx=0
\biggr\},
$$
and by Corollary 2.3:
$$
\{\mu\mid\mu_{\text{\rom{pp}}}=0\}=\biggl\{\mu\mid\varliminf\limits_{k\to
\infty}k^{-1}\int\limits^{b}_{a}\text{Im}\,F_{\mu}(x+ik^{-1})^{2}\,
dx=0\biggr\},
$$
so by Lemma 4.1, each is a $G_\delta$.  Here we use the fact that 
$\mu\mapsto F_{\mu}(x+i\epsilon)$ is weakly continuous for each $x$, 
$\epsilon >0$ and dominated above for each $\epsilon >0$ so the 
integrals are weakly continuous.  By the convergence of the 
Riemann-Stieltjes integrals, the point measures are dense in $M$, so 
$\{\mu\mid\mu_{\text{\rom{ac}}}=0\}$ is dense.  On the other hand, the 
fact that $\frac{1}{\pi}\,\text{Im}\,F_{\mu}(x+i\epsilon)\,dx$ converge 
in $M$ to $d\mu$ shows that the a.c.~measures are dense in $M$, so
$\{\mu\mid\mu_{\text{\rom{pp}}}=0\}$ is dense.  Thus, by the Baire 
category theorem, $\{\mu\mid\mu_{\text{\rom{pp}}}=0\}\cap\{\mu\mid
\mu_{\text{\rom{ac}}}=0\}$ is a dense $G_\delta$! \qed
\enddemo

Finally, we recover our results in [7].  We call a metric space $X$ 
of self-adjoint operators on a Hilbert space $\Cal H$ regular if and 
only if $A_{n}\to A$ in the metric topology implies that $A_{n}\to A$ in 
strong resolvent sense.  (Strong resolvent convergence of self-adjoint 
operators means $(A_{n}-z)^{-1}\varphi\overset \|\,\| \to\longrightarrow 
(A-z)^{-1}\varphi$ for all $\varphi$ and all $z$ with $\text{Im}\,z\neq 0$.
Notice this implies that for any $a,b,p$ and $\epsilon >0$ and any 
$\varphi\in\Cal H$, $A\mapsto \int\limits^{b}_{a}\text{Im}(\varphi, 
(A-x-i\epsilon)^{-1}\varphi)^{p}\,dx\equiv F_{a,b,p,\epsilon,\varphi}(A)$ 
is a continuous function in the metric topology.

\proclaim{Theorem 4.3} For any open set $\Cal O\subset\Bbb R$ and any 
regular metric space of operators, $\{A\mid A\text{\rom{ has no 
a.c.~spectrum in }}\Cal O\}$ is a $G_\delta$.
\endproclaim

\demo{Proof}  Any $\Cal O$ is a countable union of intervals so it suffices 
to consider the case $\Cal O=(a,b)$. Let $\varphi_{n}$ be an orthonormal 
basis for $\Cal H$.  Then,
$$
\{A\mid A\text{ has no a.c.~spectrum in }(a,b)\}=\bigcap\limits_{n}
\biggl\{A\mid\varliminf\limits_{k\to\infty}\, F_{a, b, 1/2, 1/k, 
\varphi_{n}} (A)\biggr\}
$$
is a $G_\delta$ by Lemma 4.1 and Corollary 3.2. \qed
\enddemo

Similarly, using Corollary 2.3, we obtain

\proclaim{Theorem 4.4}  For any interval $[a,b]$ and any regular 
metric space of operators, $\{A\mid A\text{\rom{ has no point spectrum 
in }}[a,b]\}$ is a $G_\delta$.
\endproclaim

\remark{Note} This is slightly weaker than the result in [7] but 
suffices for most applications.  One can recover the full result of 
[7], namely Theorem 4.4 with $[a,b]$ replaced by an arbitrary closed 
set $K$, by first noting that any closed set is a union of compacts, 
so it suffices to consider compact $K$.  For each $K$, let 
$K_{\epsilon}=\{x\mid\text{dist}(x, K)<\epsilon\}$. Then one can show 
that if $d\mu$ has no pure points in $K$, then
$$
\lim\limits_{\epsilon\downarrow 0}\,\epsilon\int\limits_{K_\epsilon}
(\text{Im}\,F_{\epsilon}(x+i\epsilon))^{2}\, dx = 0;
$$
and if it does have pure points in $K$, then
$$
\varliminf\limits_{k\to\infty}\,k^{-1}\int\limits_{K_\epsilon} |\text{Im}\,
F(x+ik^{-1})|^{2}\,dx>0
$$
and Theorem 4.4 extends.
\endremark

\vskip 0.4in
\Refs
\endRefs

\item{[1]} R.~del Rio, S.~Jitomirskaya, Y.~Last, and B.~Simon,
{\it Operators with singular continuous spectrum, IV.~Hausdorff 
dimensions, rank one perturbations, and localization}, preprint.
\item{[2]} T.~Kato, {\it Wave operators and similarity for some 
non-self-adjoint operators}, Math. Ann. {\bf 162} (1966), 258--279.
\item{[3]} Y.~Katznelson, {\it An Introduction to Harmonic 
Analysis}, Dover, New York, 1976.
\item{[4]} A.~Klein, {\it Extended states in the Anderson model on 
Bethe lattice}, preprint.
\item{[5]} P.~Koosis, {\it Introduction to $H_p$ Spaces}, London 
Mathematical Society Lecture Note Series {\bf 40}, Cambridge University 
Press, New York, 1980.
\item{[6]} M.~Reed and B.~Simon, {\it Methods of Modern Mathematical 
Physics, IV.~Analysis of Operators}, Academic Press, New York, 
1978.
\item{[7]} B.~Simon, {\it Operators with singular continuous 
spectrum, I.~General operators}, Ann.~of Math. {\bf 141} (1995),
131--145
\item{[8]} B.~Simon, {\it Spectral analysis of rank one 
perturbations and applications}, to appear in Proc.~1993 Vancouver 
Summer School in Mathematical Physics.
\item{[9]} T.~Zamfirescu, {\it Most monotone functions are 
singular}, Amer.~Math.~Monthly {\bf 88} (1981), 47--49.

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