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\def\ref#1#2#3#4#5#6{#1, {\it #2,} #3 {\bf #4} (#5), #6.}
%References
\def \jdam {1} % J. d'Analyse Math.
\def\mrl {2} % Math. Rev. Lett.
\def\jfa {3} % Infinite coupling
\def\how {4} % Howland
\def\lev {5} % Levitan book
\def\van {6} % Vancouver lecture
\def\sw {7} % Simon-Wolff
\topmatter
\title Point Spectrum and Mixed Spectral Types for Rank One Perturbations
\endtitle
\author Rafael del Rio$^1$ and Barry Simon$^2$
\endauthor
\leftheadtext{R.~del Rio and B.~Simon }
\thanks$^1$ IIMAS-UNAM, Apdo.~Postal 20-726, Admon.~No.~20, 01000 Mexico
D.F., Mexico
\endthanks
\thanks$^2$ Division of Physics, Mathematics, and Astronomy, California Institute of
Technology, Pasadena, CA 91125. This material is based upon work supported by the
National Science Foundation under Grant No.~DMS-9401491. The Government has
certain rights in this material.
\endthanks
\thanks To be submitted to {\it{Proc.~Amer.~Math.~Soc.}}
\endthanks
\date June 27, 1996
\enddate
\abstract We consider examples $A_\lambda = A+\lambda (\varphi, \,\cdot\,)\varphi$ of
rank one perturbations with $\varphi$ a cyclic vector for $A$. We prove that for any
bounded measurable set $B\subset I$, an interval, there exists $A, \varphi$ so that $\{E\in
I \mid\text{some $A_\lambda$ has $E$ as an eigenvalue}\}$ agrees with $B$ up to sets of Lebesgue measure zero. We also show that there exist examples where $A_\lambda$ has
a.c.~spectrum $[0,1]$ for all $\lambda$, and for sets of $\lambda$'s of positive Lebesgue
measure, $A_\lambda$ also has point spectrum in $[0,1]$, and for a set of $\lambda$'s
of positive Lebesgue measure, $A_\lambda$ also has singular continuous spectrum in
$[0,1]$.
\endabstract
\endtopmatter
\document
\flushpar{\bf \S 1. Introduction}
\vskip 0.1in
In this note we will consider families of operators
$$
A_\lambda = A+ \lambda (\varphi, \,\cdot\,)\varphi
$$
where $A$ is a self-adjoint operator on a separable Hilbert space $\Cal H$ and
$\varphi\in\Cal H$ is a cyclic vector for $A$. It will be convenient to consider also the
value $\lambda=\infty$, which is the operator $QAQ$ on $Q\Cal H$ where $Q$ is the
projection onto the operators orthogonal to $\varphi$. Let $d\mu_\lambda$ be the
spectral measure for $A_\lambda$ with vector $\varphi$ and $d\rho_\lambda = (1 +
\lambda^2) d\mu_\lambda$. It it known [\jfa] that $d\rho_\lambda$ has a weak limit as
$\lambda \to\infty$, $d\rho_\infty$, which is a spectral measure for $A_\infty$.
Define for $x\in\Bbb R$,
$$
G_\lambda(x) = \int\frac{d\rho_\lambda (y)}{(x-y)^2}
$$
where $G$ may be infinite.
Also define for $z\in\Bbb C$ with $\text{Im}\,z >0$,
$$
F_\lambda (z) =\int\frac{d\rho_\lambda (E)}{E-z} = (1+\lambda)^2
(\varphi, (A_\lambda -z)^{-1}\varphi).
$$
(This differs from the standard $F$ [\van] by a factor of $(1+\lambda^2)$.) It is
known [\mrl, \van] that
\proclaim{Theorem 0} The sets,
$$\align
P &= \{E\mid G_\lambda (E) <\infty\}\cup \{E\mid E\text{ is an eigenvalue of $A_\lambda$}\} \\
L &= \{E\mid\lim_{\epsilon\downarrow 0} F_\lambda (E+i\epsilon)\equiv F_\lambda
(E+i0) \text{ exists and }\text{\rom{Im}}\, F_\lambda (E+i0) > 0\} \\
S &= \Bbb R \backslash P\cup L
\endalign
$$
are $\lambda$ independent for $\lambda\in \Bbb R$, and for every $\lambda\in\Bbb R\cup
\{\infty\}$:
$$\align
\rho^{\text{\rom{pp}}}_\lambda (\,\cdot\,) &= \rho_\lambda (\,\cdot\,\cap P) \tag 1a \\
\rho^{\text{\rom{ac}}}_\lambda (\,\cdot\,) &= \rho_\lambda (\,\cdot\, \cap L) \tag 1b \\
\rho^{\text{\rom{sc}}}_\lambda (\,\cdot\,) &= \rho_\lambda (\,\cdot\, \cap S) \tag 1c
\endalign
$$
where $\rho^{\text{\rom{pp}}}_\lambda$, $\rho^{\text{\rom{ac}}}_\lambda$,
$\rho^{\text{\rom{sc}}}_\lambda$ are the pure point, absolutely continuous, and
singular continuous parts of the measure $\rho_\lambda$. Moreover,
$$
P = \bigcup\limits_{\lambda\in\Bbb R\cup\{\infty\}} \{E\mid E \text{ is an eigenvalue
of $A_\lambda$}\}
$$
and for any set $C$,
$$
\int \frac{\rho_\lambda (C)}{(1+\lambda^2)} \, d\lambda = |C| \tag 2
$$
the Lebesgue measure of $C$. In particular, by {\rom{(1a)}}
$$
\int\frac{\rho^{\text{\rom{pp}}}_\lambda (C)}{(1+\lambda^2)}\, d\lambda =
|C\cap P| \tag 3
$$
and similarly for $L$ and $S$.
\endproclaim
One can ask what kind of sets can occur as a $P$. We have a partial answer given
in Section 2:
\proclaim{Theorem 1} For any bounded measurable set $B$ and any interval $I\supset B$,
there exists a measure $d\mu$ on $I$ so that \rom(where a.e.~means with respect to Lebesgue
measure\rom)
$$
G_0 (x) = \cases <\infty & \text{\rom{a.e. }} x\in B \\
=\infty & \text{\rom{a.e. }} x\in I\backslash B. \endcases
$$
The measure $d\mu$ may be chosen purely a.c., or purely s.c., or purely p.p.
\endproclaim
\remark{Remarks} 1. By Theorem 0, this says something about allowed sets of eigenvalues.
2. We will also show that if $B$ is open, we can drop the a.e. We believe that this can be
done for an arbitrary $F_\delta$, but have not proven it.
\endremark
Using a technical result in Section 3, we will prove our second main result in Section 4:
\proclaim{Theorem 2} There exists an example $A$ so that
\roster
\item"\rom{(i)}" $\sigma_{\text{\rom{ac}}}(A_\lambda) = [0,1]$ for all $\lambda$.
\item"\rom{(ii)}" $\{\lambda\mid\sigma_{\text{\rom{pp}}}(A_\lambda) \cap [0,1]
\neq \emptyset\}$
has positive Lebesgue measure; indeed, for any interval $I\subset [0,1]$, $\{\lambda\mid
\sigma_{\text{\rom{pp}}}(A_\lambda)\cap I \neq\emptyset\}$ has positive measure.
\item"{(iii)}" $\{\lambda\mid\sigma_{\text{\rom{sc}}}(A_\lambda) \neq\emptyset\}$
has positive Lebesgue measure; indeed, for any interval $I\subset [0,1]$, $\{\lambda\mid
\sigma_{\text{\rom{sc}}}(A_\lambda)\cap I\neq\emptyset\}$ has positive measure.
\endroster
There also exist examples where \rom{(i)} is replaced by $\sigma_{\text{\rom{ac}}}
(A_\lambda)=\emptyset$.
\endproclaim
One can translate these results into ones for variations on boundary conditions for
Schr\"odinger operators $-u'' + Vu$ on $[0,\infty)$ in two steps:
\roster
\item"\rom{(a)}" Extend the theory to $\varphi\in\Cal H_{-1}(A)$ and rewrite the
Sturm-Liouville/Schr\"odinger operator in this language [\van].
\item"\rom{(b)}" Appeal to the Gel'fand-Levitan construction [\lev], which implies that for
any measure $\mu$ on a bounded interval $I$, we can find a continuous $V$ on $[0,\infty)$
with $-u''+Vu$ limit point at infinity and boundary condition $\theta$ at $x=0$ so that
the spectral measure $d\rho_\theta$ restricted to $I$ is $d\mu$. Typical of the result is:
\endroster
\proclaim{Theorem 1$^\prime$} For any bounded measurable set $B$ and interval
$I\supset B$, there is a continuous function $V$ on $[0,\infty)$ so that up to sets of
Lebesgue measure zero,
$$
\{E\mid -u''+Vu =Eu \text{ has a solution $L^2$ at infinity}\}
$$
is precisely $B$.
\endproclaim
Because the Gel'fand-Levitan construction gives no information on $V$ at infinity (for example,
it could be unbounded below), we regard these translations as being of limited interest.
\vskip 0.3in
\flushpar{\bf \S 2. The Set Where $\boldkey G$ Is Finite}
\vskip 0.1in
Recall that a perfect set is a closed set with no isolated points. We will also need the
following notion.
\definition{Definition} A closed subset $C\subset\Bbb R$ will be called {\it{minimal}} if
and only if for all $x\in C$ and $\epsilon >0$, $|(x-\epsilon, x+\epsilon)\cap C| >0$.
\enddefinition
The name comes from the fact that among all closed sets $D$ with $|D\triangle C|$=0,
$C$ is the minimal such set. We will see below that any closed set $D$ has a minimal
closed set $C$ contained in it so that $|D\backslash C|=0$.
We also define $G_\mu$ by
$$
G_\mu (x) =\int\frac{d\mu (y)}{(x-y)^2}.
$$
With these notions out of the way, we can state the two main theorems of this section:
\proclaim{Theorem 2.1}
\roster
\runinitem"\rom{(a)}" Let $C$ be any closed set in $\Bbb R$. Then there
exists a pure point measure $\mu$ supported on $C$ so that $\{x\mid G_\mu (x) =\infty\}
=C$.
\item"\rom{(b)}" Let $C$ be any perfect set. Then there exists a singular continuous
measure $\mu$ supported on $C$ so that $\{x\mid G_\mu (x) =\infty\}=C$.
\item"\rom{(c)}" Let $C$ any minimal closed set. Then there exists an absolutely continuous
measure $\mu$ supported on $C$ so that $\{x\mid G_\mu (x) =\infty\} =C$.
\endroster
\endproclaim
\remark{Remarks} 1. The assumptions on the closed sets are optimal in that if $x$ is an
isolated point of $C$, then $G_\mu (x) <\infty$ for any singular continuous measure $\mu$
supported on $C$; and if $x\in C$ is a point with $|(x-\epsilon, x+\epsilon)\cap C| =0$ for
some $\epsilon >0$, then $G_\mu (x)<\infty$ for any a.c.~measure supported on $C$.
2. In general, $\{x\mid G_\mu (x)=\infty\}$ is only a $G_\delta$, not a closed set. It is
open if ``closed" in this theorem can be replaced by $G_\delta$.
3. If $B$ is any measurable set, we can apply the methods of proof below and get a $\mu$
supported on $B$ with $\{x\mid G_\mu (x)=\infty\} \supset B$. If $B$ is arbitrary, we
can take $\mu$ pure point. If $B$ has no isolated points, we can take $\mu$ singular
continuous, and if $B$ has no essentially isolated points (i.e., no points $x$ with
$|(x-\epsilon, x+\epsilon)\cap B| =0$ for some $\epsilon >0$), we can take $\mu$
absolutely continuous.
\endremark
If we are willing to throw out sets of measure zero, we can go beyond Theorem 2.1. We
write $A\equiv B$ to mean $|A\triangle B|=0$. Then we will prove that:
\proclaim{Theorem 2.2 ($\equiv$ Theorem 1)} For $B$ an arbitrary measurable subset
of an interval $I$, we can find $\mu$ supported on $I$ so that
$$
\{x\in I\mid G_\mu (x) <\infty\} \equiv B.
$$
$\mu$ can be chosen to be purely absolutely continuous or purely singular continuous
or pure point. In the a.c.~case, $\mu$ can be chosen so that the essential support of
$\mu$ is $I\backslash B$.
\endproclaim
In understanding perfect and minimal closed sets, it is useful to have the following pair
of results, which we will also need in proving Theorem 2.2.
\proclaim{Proposition 2.3} Any closed set $S$ in $\Bbb R$ can be written as $S=C\cup D$
where $C$ is perfect and $D$ is countable.
\endproclaim
\demo{Proof} Let $C=\{x\in S\mid\forall\epsilon >0,\, (x-\epsilon, x+\epsilon)\cap S
\text{ is uncountable}\}$ and $D=S\backslash C$. It is easy to see that $C$ is closed. If we
show $D$ is countable, then each $(x-\epsilon, x+\epsilon)\cap C$ is uncountable, so not
empty and $C$ is perfect.
If $x\notin C$, we can find $a$ and $b$ rational so $x\in (a,b)$ and $(a,b)\cap S$ is
countable. Since there are only countably many $(a,b)$ with $a,b$ rational, we can find a
countable family of $\{O_n\}_{n=1}$ with each $O_n \cap S$ countable, so $D\subset
\operatornamewithlimits{\cup}\limits_n (O_n \cap S)$ is countable. \qed
\enddemo
\proclaim{Proposition 2.4} Any closed set $S$ in $\Bbb R$ can be written as $S= C\cup D$
where $C$ is minimal closed and $|D|=0$.
\endproclaim
\demo{Proof} Let $C=\{x\in S\mid \forall \epsilon >0, \, |(x-\epsilon, x+\epsilon) \cap S|
>0\}$ and $D=S\backslash C$. Now just mimic the proof of Proposition 2.3. \qed
\enddemo
We need one more preliminary:
\proclaim{Proposition 2.5}
\roster
\runinitem"\rom{(a)}" For any non-empty closed set $C$, there exists a point measure
supported by $C$.
\item"\rom{(b)}" For any non-empty perfect set $C$, there exists a singular continuous
measure supported by $C$.
\item"\rom{(c)}" For any non-empty minimal closed set $C$, there is an absolutely
continuous measure supported by $C$.
\endroster
\endproclaim
\demo{Proof} (a) is trivial and stated for parallelism. (c) is also trivial (take $d\mu =
\chi_C dx$). That leaves (b); so let $C$ be perfect. If $C$ contains an entire interval
$[a,b]$, place a scaled Cantor measure on $(a,b)$ and use that for $d\mu$. So we need
only consider a nowhere dense perfect set. By intersecting it with a suitable bounded
interval and scaling, we will suppose it is a subset of $[0,1]$.
We claim such a $C$ is homeomorphic to $\{0,1\}^\Bbb N$, the infinite sequences of $0$'s
and $1$'s. Use that homeomorphism to transfer the two mutually singular measures
$d\alpha_1 = \operatornamewithlimits{\otimes}\limits^\infty_{n=1} [\frac{1}{2} (\delta_0
+ \delta_1)]$ and $d\alpha_2 = \operatornamewithlimits{\otimes}\limits^\infty_{n=1} (\frac{1}
{3} \delta_0 + \frac{2}{3} \delta_1)$. $d\alpha_1$ may be purely absolutely continuous
(as it is if $C$ is a symmetric positive measure Cantor set), but then $d\alpha_2$ is purely
singular continuous. Either way, either $d\alpha_1$ or $d\alpha_2$ has a non-zero singular
continuous component.
To prove the claim (known, but the proof is so short that we give it) that a nowhere closed
perfect subset $C$ of $[0,1)$ is homeomorphic to $\{0, 1\}^{\Bbb N}$, let $a_- =\min (C)$,
$a_+ =\max(C)$, and $\ell_1 = a_+ - a_-$, the length of $C$. Since $C$ is perfect, $\ell_1 >0$. Let $J = (\frac{a_- + a_+}{2} - \frac{\ell_1}{6}, \frac{a_- + a_+}{2} + \frac{\ell_1}{6})$,
the middle third of $(a_-, a_+)$. Since $C$ is nowhere dense, we can find $x_1\in J\backslash
C$. Let $C_0 = C\cap (-\infty, x_1)$, $C_1 = C\cap (x_1, \infty)$. Then $C_0, C_1$ are
perfect and $\text{diam}(C_1) \leq \frac{2}{3}$. Now repeat this process, and so find
$C_{m_1 \dots m_\ell} (m_i \in \{0,1\})$ inductively so that $\text{diam}(C_{m_1\dots
m_\ell}) \leq (\frac{2}{3})^\ell$, $C_{m_1 \dots m_\ell} = C_{m_1 \dots m_\ell 0} \cup
C_{m_1\dots m_\ell 1}$, each $C_{m_1\dots m_\ell}$ is perfect. Define $a_\ell : C\to
\{0, 1\}$ by $a_\ell =0$ on each $C_{m_1 \dots m_{\ell-1}0}$ and $a_\ell =1$ on each
$C_{m_1 \dots m_{\ell -1}1}$. Each $a_\ell$ is continuous since each $C_{m_1 \dots
m_\ell}$ is closed. Map $C\to \{0, 1\}^\ell$ by $x\to (a_1 (x), a_2 (x), \dots )$. This map
is onto since for any fixed $m_1, \dots, \operatornamewithlimits{\cap} \limits^{\infty}_{\ell=1}
C_{m_1 \dots m_\ell} \neq \emptyset$ by compactness. This map is one-one since $\text{diam} (C_{m_1 \dots m_\ell}) \to 0$ to $\ell\to\infty$ uniformly in the choice of $m_\ell$. A continuous
bijection is a homeomorphism. \qed
\enddemo
\demo{Proof of Theorem {\rom{2.1}}} This is motivated by a construction in [\sw]. For
$n=1, 2, \dots$ and $j=0, \dots, 2^n -1$, let $C^{(n)}_j =\overline{(\frac{j}{2^n}, \frac
{j+1}{2^n})\cap C}$ which is $C\cap [\frac{j}{2^n}, \frac{j+1}{2^n}]$ with the
endpoints dropped if they would be isolated. Then if $C$ is perfect (minimal), so is
each non-empty $C^{(n)}_j$. For such non-empty $C^{(n)}_j$, let $\mu^{(n)}_j$ be a
measure of the requisite type (i.e., pure point, singular continuous, or absolutely continuous)
of unit measure and supported on $C^{(n)}_j$. Such measures exist by Proposition 2.5. Let
$$
\mu =\sum^\infty_{n=1} n^{-2} 2^{-n} \sum^{2^n} \Sb j=1 \\ j \text{ so that} \\
C^{(n)}_j \neq \emptyset\endSb \mu^{(n)}_j.
$$
Then $\mu$ is a finite measure of the requisite type supported on $C$. If $y\notin C$,
then $G_\mu (y) \leq \text{dist}(y, C)^{-2} \int d\mu <\infty$ since $C$ is closed. On the
other hand, if $y\in C$ and $y\in (\frac{j}{2^n}, \frac{j+1}{2^n})$, then $C^{(n)}_j
\neq\emptyset$ and $\int\frac{d\mu^{(n)}_j (x)}{(x-y)^2} \geq (2^{-n})^2$, and if
$y\in \{\frac{j}{2^n}\}^{2^n}_{j=0}\cap C$, either $C^{(n)}_j$ or $C^{(n)}_{j-1}$
is non-empty. It follows that
$$
\int \frac{d\mu(x)}{(x-y)^2} \geq \sum^\infty_{n=1} 2^{2n} n^{-2} 2^{-n} =\infty,
$$
so $\{y\mid G_\mu (y) =\infty\}= C$. \qed
\enddemo
\demo{Proof of Theorem {\rom{2.2}}} This uses an explicit version of an argument of
Howland [\how] as in [\jdam]. Since Lebesgue measure is inner regular, we can find $C_1,
\dots C_n, \dots$ and $K_1, \dots, \mathbreak K_n, \dots$ closed with $C_1 \subset C_2
\subset \cdots \subset I \backslash B$ and $K_1 \subset K_2 \subset \cdots \subset B$ and
with $|B\backslash \operatornamewithlimits{\cup}\limits_n K_n|=0$, $|(I\backslash B)
\backslash \operatornamewithlimits{\cup}\limits_n C_n|=0$.
By Proposition 2.3, we can suppose that $C_n$'s are minmal closed (and so, perfect)
without loss of generality. We can also suppose each $C_n$ is non-empty (if $|I\backslash
B|=0$, we just take $\mu =0$).
Let $\mu_n$ be a unit measure of the requisite type supported on $C_n$ with
$$
C_n = \{x\mid G_{\mu_n} (x) =\infty\}.
$$
Let
$$
\mu =\sum^\infty_{n=1} 2^{-n} \text{dist}(K_n, C_n)^2 \mu_n.
$$
Since $K_n$ and $C_n$ are compact and disjoint, $\text{dist}(K_n, C_n) >0$ and
thus, $G_\mu (x) \geq \mathbreak 2^{-n}\text{dist}(K_n, C_n)^2 G_{\mu_n}(x)=
\infty$ on $C_n$ and so on $\cup C_n$ and so a.e.~on $I\backslash B$.
On the other hand, since $K_n\subset K_{n+1}, \dots$, $\text{dist}(K_n, C_m)\geq
\text{dist}(K_m, C_m)$ if $m\geq n$ and so if $x\in K_n$,
$$
G_\mu (x) =\sum^{n-1}_{\ell=1} 2^{-\ell} \text{dist}(K_\ell, C_\ell)^2
G_{\mu_\ell} (x) +\sum^\infty_{\ell=n} 2^{-n} < \infty,
$$
and so $G_\mu <\infty$ on $\cup K_n$ and thus a.e.~on $B$.
In the a.c.~case, we can take $\mu_n =\frac{1}{|C_n|} \chi_{C_n} dx$, in which case
it is evident that the essential support of $\mu$ is $\cup C_n = I\backslash B$ as claimed.
\qed
\enddemo
\vskip 0.3in
\flushpar {\bf \S 3. Essentially Dense Sets}
\vskip 0.1in
\definition{Definition} A measurable set $S\subset I$, an interval, is called essentially dense
if for every subinterval $J\subset I$, we have $|J\cap S|>0$.
\enddefinition
\proclaim{Theorem 3.1} There exist disjoint measurable subsets $A,B,C\subset [0,1]$ whose
union is $[0,1]$ so that each is essentially dense.
\endproclaim
\remark{Remarks} 1. Our proof shows that one can assert the same for sets $A_1, \dots,
A_n$ rather than three sets or even construct a countable disjoint decomposition, each
of which is essentially dense.
2. Our construction is related to a construction in [\mrl].
\endremark
\demo{Proof} Let $n_j = (2j+1)^2$, the square of the $j^{\text{\rom{th}}}$ odd number.
Given $x\in [0,1]$, we define $a_j (x)$ by requiring
$$
x=\sum^\infty_{j=1} \frac{a_j (x)}{n_1 \dots n_j}
$$
with $a_j(x) \in \{0,1,\dots, n_j -1\}$ and the requirement that if $x$'s expansion can end in
all $0$'s, we do that (to settle the ambiguity between $\dots a(n_j -1)(n_{j+1}-1)\dots$
and $\dots (a+1) 0\,0 \dots$). This is a standard positive measure Cantor set construction.
Define $m_j =\frac{1}{2} (n_j -1)$. Let
$$\align
A &= \{x\mid\text{ the number of $j$'s with $a_j (x) =m_j$ is $1,4,\dots$ or infinite}\} \\
B &= \{x\mid\text{ the number of $j$'s with $a_j (x) =m_j$ is $2,5,8,\dots$}\} \\
C &= \{x\mid\text{ the number of $j$'s with $a_j (x)=m_j$ is $3,6,9,\dots$}\}.
\endalign
$$
This is obviously a decomposition. We need only to show that each set is essentially dense.
It suffices to show that $|B\cap J|>0$ for any interval of the form $J=\{x\mid a_1 (x) =
\alpha_1, \dots, a_k (x) = \alpha_k\}$ since every interval contains such a $J$. By increasing
$k$ by $1$ or $2$ and shrinking $J$ by taking $\alpha_{k+1}=m_{k+1}$ (and perhaps $\alpha_{k+2} = m_{k+2}$), we can suppose that $\#\{j\in \{1, \dots, k\}\mid \alpha_j = m_j\}\equiv 2 \mod 3$. In that case, by looking at $x$'s with no further $a_\ell (x)=m_\ell$,
we have
$$
|B\cap J| \geq \prod\limits^{\infty}_{\ell=k+1} \biggl( 1-\frac{1}{n_\ell}\biggr) > 0
$$
since $\sum\frac{1}{n_\ell}<\infty$. \qed
\enddemo
\vskip 0.3in
\flushpar {\bf \S 4. Mixed Spectra}
\vskip 0.1in
\demo{Proof of Theorem {\rom{2}}} Decompose $[0,1]=A\cup B\cup C$ into three
disjoint essentially dense sets. Pick a measure $d\mu_1$ which is absolutely continuous
with essential support $A$ so that $G_{\mu_1}(x) <\infty$ a.e.~on $B\cup C$ and a
s.c.~measure $\mu_2$ supported on $B$ so that $G_{\mu_2}(x) <\infty$ on $A\cup
C$ and $G_{\mu_2}(x)=\infty$ a.e.~on $B$. Let $d\mu =d\mu_1 + d\mu_2$.
By Theorem 0 (recall $X\equiv Y$ means $|X\triangle Y|=0$),
$$\align
P &\equiv C \cup (\Bbb R\backslash [0,1]) \\
L &\equiv A \\
S &\equiv B.
\endalign
$$
By equation (3) and its analogs for a.c.~ and s.c., we have the claimed assertions (i)--(iii).
For the examples with $\sigma_{\text{\rom{ac}}}(A_\lambda)=\emptyset$, just use
$d\mu =d\mu_2$. \qed
\enddemo
\vskip 0.3in
\example{Acknowledgment} R.d.R. would like to thank Professor Alejandro Bravo for
very useful discussions.
\endexample
\vskip 0.3in
\Refs
\endRefs
\item{\jdam.} R. del Rio, S.~Jitomirskaya, Y.~Last, and B.~Simon, {\it{Operators
with singular continuous spectrum, IV. Hausdorff dimensions, rank one perturbations,
and localization}}, to appear in J.~d'Analyse Math.
\gap
\item{\mrl.} \ref{R.~del Rio, B.~Simon, and G.~Stolz}{Stability of spectral types for
Sturm-Liouville operators}{Math.~Research Lett.}{1}{1994}{437--450}
\gap
\item{\jfa.}\ref{F.~Gesztesy and B.~Simon}{Rank one perturbations at infinite coupling}
{J.~Funct.~Anal.}{128}{1995}{245--252}
\gap
\item{\how.}\ref{J.~Howland}{On a theorem of Carey and Pincus}{J.~Math.~Anal.}
{145}{1990}{562--565}
\gap
\item{\lev.} B.M.~Levitan, {\it{Inverse Sturm-Liouville Problems}}, VNU Science Press,
Utrecht, 1987.
\gap
\item{\van.} B.~Simon, {\it{Spectral analysis of rank one perturbations and applications}},
in CRM Proc. and Lecture Notes, (J.~Feldman, R.~Froese, and L.~Rosen, eds.), Vol.~8,
pp.~109--149, Amer.~Math.~Society, Providence, RI, 1995.
\gap
\item{\sw.} \ref{B.~Simon and T.~Wolff}{Singular continuous spectrum under rank one
perturbations and localization for random Hamiltonians}{Commun.~Pure Appl.~Math.}
{39}{1986}{911--918}
\enddocument