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Spectral Measure, Sturm-Liouville, mixed types, rank one perturbations
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\documentclass[12pt]{koval}
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\newcommand{\ERE}{\hbox{\BBFONT R}}
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\newcommand{\pequere}{\hbox{\bbfont R}}
\usepackage{eucal}
\newtheorem{thm}{Theorem}
\newtheorem{lem}{Lemma}
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\newtheorem{cor}{Corollary}
\theoremstyle{definition}
\newtheorem{exmp}{Example}
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\begin{document}
\title[Families of spectral measures with mixed types]{Families of spectral measures with mixed types}
\author{R. del Rio}
\address{IIMAS-UNAM,
Circuito Escolar, Ciudad Universitaria,
04510, M\'exico, D.F., M\'exico.}
\email{delrio@servidor.unam.mx }
\thanks{R. dR. was partially supported by projects IN-102998 PAPIIT-UNAM and 27487E CONACyT}
\author{S. Fuentes}
\curraddr{}
\author{A. Poltoratski}
\address{Department of Mathematics,
College Station, Texas 77843--3368,
U.S.A.}
\email{alexei.poltoratski@math.tamu.edu}
\thanks{A. P. was supported in part by
the NSF grant DMS-9970151}
\date{November 2, 2000}
\maketitle{ }
\begin{abstract} Consider a family of Sturm-Liouville operators $H_\theta$ on
the half-axis defined as $$H_\theta u = -u^{\prime\prime} + q (x) u \qquad 0
\leq x < \infty$$
with the boundary
condition
$$u (0) \cos \theta + u^\prime (0) \sin \theta = 0 \qquad 0 \leq \theta <
\pi$$
and the limit point case at infinity. We show that it is possible for all
$H_\theta$ to have dense absolutely continuous and dense singular spectrum.
The construction is based on integral representations of Pick functions in the
upper half-plane. We also discuss applications to the Krein spectral
shift.
\end{abstract}
\section {Introduction}
This note studies the interplay between various types of
spectra under small perturbations or a change of the boundary condition.
Let $A_0$ be a cyclic self-adjoint operator,
$\varphi$ its cyclic vector and $\mu$ the corresponding spectral measure.
Denote by $A_\lambda$ the rank one perturbations of $A_0$:
$$A_\lambda=A+\lambda(\cdot,\varphi)\varphi, \ \ \lambda\in\ERE.$$
Let $\mu_\lambda$ be the spectral measures of $A_\lambda$ corresponding to
$\varphi$. We study the following general problem: What can happen to
$\mu_\lambda$ when the parameter (the coupling constant) $\lambda$ is
changing?
It is well known that the same problem can be formulated in terms of
Sturm--Liouville operators on the half-axis. Let the operator $H_\theta$
be defined as
$$H_\theta u = -u^{\prime\prime} + q (x) u \qquad 0 \leq x < \infty$$
where $q$ is real valued, locally integrable function defined in $[0,
\infty) )$. The domain is restricted by the boundary condition
at zero
$$u (0) \cos \theta + u^\prime (0) \sin \theta = 0 \qquad 0 \leq \theta <
\pi.$$
We assume that the limit point case occurs at infinity.
For each $\theta$ one can define the so-called Weyl--m function $m_\theta (z)$
which is analytic and has positive imaginary part in the upper half-plane.
The imaginary part of $m_\theta (z)$ has the following integral
representation
$$Im \ m_\theta (z) = \int_{\ERE} \frac{y}{(t-x)^2 + y^2} d \rho_\theta
(t) \qquad z= x + iy$$
The measures $\rho_\theta$ are called the Weyl spectral
measures of the boundary value problem.
The following relation is satisfied:
\begin{equation}
m_\theta (z) = \frac{ -\sin (\theta-\beta) + m_\beta (z) \cos (\theta-\beta) }{ \ \cos (\theta-\beta)
+ m_\beta (z) \sin (\theta-\beta) }, \label{uno}
\end{equation}
see \cite{Rio_Pol} and \cite{Eas_Kal}.
In these settings our general question transforms
into: How does the spectrum of $H_\theta$ (the measure $\rho_\theta$) depend
on the parameter $\theta$ in the boundary condition? The same question can
also be reformulated in terms of self-adjoint extensions of a symmetric
operator with deficiency indices (1,1) or discrete Schr\"odinger operators.
For more on these connections see \cite{Don_2} and \cite{Sim}.
In this paper we focus on the correlations between the absolutely continuous
and singular parts of the spectrum under the change of the coupling constant
$\lambda$ (the boundary parameter $\theta$). By the Kato-Rosenblum
theorem the absolutely continuous spectrum is invariant
under such small perturbations. At the same time, examples obtained by many
researchers in recent years show that the singular spectrum can be extremely
unstable under the change of the coupling constant $\lambda$ in particular in
"mixed" situations when both singular and absolutely continuous spectrum are
present.
In most cases, the singular spectrum of $A_\lambda$ is located
in the gaps of the absolutely continuous spectrum. With the change of
$\lambda$ the isolated eigenvalues of $A_\lambda$ move inside their respective
gaps and often dissappear when they hit the edge of the gap. But what
happens if there are no gaps in the absolutely continuous spectrum? It seems
that in this case there is no space for the singular spectrum. Even if some of
$A_\lambda$'s have nontrivial singular components, squeezed somewhere in the
midst of the absolutely continuous spectrum, they must be easily destroyed by
the change of $\lambda$.
In particular, it seems unlikely that all operators can have dense absolutely
continuous and dense singular spectrum. All the examples we know seem to
support this intuitive argument: the
examples of Naboko \cite{Nab} and Simon \cite{Sim_2}
exhibit singular spectra only for a set of boundary conditions
of Lebesgue measure zero (in $\theta$), and the examples of Remling \cite{Rem}
do not have dense singular spectra. In \cite{Rio_Sim}
such a coexistence of spectra is shown for a set of positive measure in the
coupling constant but not for all coupling constants. Also, in \cite{Rio_Mak_Sim} it is
shown that two other types of spectrum, the pure point and
continuous, cannot coexist for all coupling constants if the spectrum of $A_0$
is dense. Other papers where coexistence is studied are \cite{Alb_Bra_Nei} and \cite{Fig_Pas}.
In this paper we show that, despite all the evidence mentioned above, it is
possible for $all$ $A_\lambda$'s to have everywhere dense absolutely
continuous $and$ everywhere dense singular spectrum. The corresponding example
is constructed in Section 2.
In Section 1 we develop our machinery. It is based on the
integral representations of analytic Pick functions in the upper half-plane.
Our main tool is Lemma 2, which reveals the relations between the families of
Pick functions appearing in Perturbation Theory. In addition to the main
example, in Section 2 we explain how one can use Lemma 2 to obtain the singular
components of the spectral measures $\mu_\lambda$ directly from the
corresponding Krein function.
\section{Preliminaries}
Every Pick function $F (z)$ (an analytic function which takes the upper
half-plane into itself) has an integral representation of the form
\begin{equation}
F (z) = a +bz + \int\limits_{\pequere}
\left[ \frac{1}{t - z}
- \frac{t }{1 + t ^2}
\right] d \mu (t )
\label{two}
\end{equation}
\noindent where $a, b \in \ERE, b \geq 0$ and $\mu $ is a non negative Borel measure which
satisfies
\begin{equation}
\mid \mu \mid := \int\limits_{\pequere} \frac{d \mu (t )}{1 + t ^2} < \infty. \label{odin}
\end{equation}
\noindent Conversely, any function of this form is analytic and takes the
upper half-plane into itself, see \cite{Don} and \cite{Ros_Rov}. The integral
on the righthand side of (\ref{two}) is the Cauchy integral of $\mu$ in the
upper half-plane. We will denote it by $K\mu$. We will also denote by $P\mu$
the Poisson integral of $\mu$: $$P\mu(x+iy)= I m
K\mu(x+iy)=\int\limits_{\pequere} \frac{y}{(x-t)^2 + y^2}d \mu (t ).$$
The Poisson integral is a so called approximative identity: its kernel
$\frac{y}{(x-\epsilon)^2 + y^2}$
is positive, tends to zero uniformally outside of any neighborhood of $x$ as
$y\to 0$ and its $L^1$-norm is constant. This implies the following version
of the Lebesgue Theorem (about the Lebesgue points of a summable
function), see \cite{Pol}:
\begin{lem}
If $\mu$ is a complex Borel measure on $\ERE$ such that $|\mu|$ satisfies
(\ref{odin}) and $f\in L^1(\mu)$ then
$$\lim\limits_{y \downarrow 0} \frac{Pf\mu(x+iy)}{P\mu(x+iy)}=f(x)$$
for $\mu$-a. e. $x$.
\end{lem}
\begin{cor}
If $\mu$ and $\nu$ are complex Borel measures on $\ERE$ such that $|\mu|$ and
$|\nu|$ satisfy (\ref{odin}) and $\mu=f\nu+\eta$, where $f\in L^1(\nu)$ and
$\eta\perp\nu$ ($\eta$ and $\nu$ are mutually singular),
then
$$\lim\limits_{y
\downarrow 0} \frac{P\mu(x+iy)}{P\nu(x+iy)}=f(x)$$ for $\nu$-a. e. $x$.
\end{cor}
\noindent{\em Proof:}
Since
$$\lim\limits_{y
\downarrow 0} \frac{P\mu(x+iy)}{P\nu(x+iy)}=
\lim\limits_{y
\downarrow 0} \frac{Pf\nu(x+iy)}{P\nu(x+iy)}+
\lim\limits_{y
\downarrow 0} \frac{P\eta(x+iy)}{P\nu(x+iy)}$$
by Lemma 1 it is enough to show that the last summand tends to 0 $\nu^s$-a.
e. Consider $f\in L^1(\nu+\eta)$ defined as 1 $\eta$-a.e. and as 0 $\nu$-a.e.
Then by Lemma 1
$$
\lim\limits_{y
\downarrow 0} \frac{P\eta(x+iy)}{P(\nu+\eta)(x+iy)}=
\lim\limits_{y
\downarrow 0} \frac{Pf(\nu+\eta)(x+iy)}{P(\nu+\eta)(x+iy)}=0
$$
$\nu$-a.e. Therefore
$$\left[\frac{P\eta(x+iy)}{P(\nu+\eta)(x+iy)}\right]^{-1}=1+\frac{P\nu(x+iy)}{P(\eta)(x+iy)}\to \infty$$
$\nu$-a.e. and we obtain our statement.
\rightline{\sc q.e.d.}
To construct our main example we will also need the following lemma.
\noindent Consider the family of Pick functions
\begin{equation}
f_{\theta} (z):=\frac{\cos\theta + z\sin\theta}{\sin\theta - z\cos\theta},\ \ \theta\in\ERE.
\label{dos}
\end{equation}
Let $L (z)$ be another Pick function such that $0 \leq I m L(z) \leq \pi$.
For any $\alpha\in\ERE$ denote $$M_{\alpha} (z) := f_{\alpha} (L (z) ) \;
\mbox{ and } \; N_\alpha (z) := f_{\alpha} (\exp L (z) ).$$
Both $M_{\alpha} (z)$ and $N_\alpha (z)$ are Pick functions admitting
representations similar to (\ref{two}). We denote by
$\mu_{\alpha}$ and $\nu_{\alpha }$ the measures appearing in the representations for
$M_\alpha$ and $N_\alpha $ respectively.
Then the singular parts of these measures, $\nu^s_{\alpha }$ and $\mu^s_{\beta
}$, enjoy the following relation:
\bigskip
\begin{lem}
Let $\nu_{\alpha }$ and $\mu_{\beta }$ be as above. Define the function
$\alpha(\beta)$ as
$$ \alpha (\beta ) = tg^{-1} (\exp tg \beta ), \quad \beta \in (- \pi /2, \pi /2)$$
\noindent then
$$ \mu^s_{\beta} = \alpha^{\prime } (\beta ) \; \nu ^s_{\alpha(\beta)}.$$
\end{lem}
\bigskip
\noindent{\em Proof:}
\medskip
Let us first show that
$$ \lim\limits_{y \downarrow 0} \frac{I m M_{\beta}(x + i y)}{I m
N_{\alpha }(x + i y)} = \alpha^{\prime } (\beta ) $$
\noindent for $\mu^s_{\beta} \: a.e.\ \ x$.
From the definition of $M_{\beta}$ and $N_{\alpha}$ we have $(z = x +
i y)$
$$\lim\limits_{y \downarrow 0} \frac{I m M_{\beta}(z)}{I m
N_{\alpha}(z)}= \lim\limits_{y \downarrow 0} \frac{I m L (z)}{I m \exp
L(z)} \left\vert \frac{\sin \alpha - \exp L(z) \cos \alpha}{\sin \beta - L
(z) \cos \beta} \right\vert^2 .$$
From the definition of $\alpha $
\begin{equation}
\frac{\sin \alpha - \exp L(z) \cos \alpha}{\sin \beta - L
(z) \cos \beta} =
\frac{\cos \alpha }{\cos \beta } \cdot \frac{\exp (tg \beta) -\exp L (z)}{tg
\beta - L (z)}. \label{tres}
\end{equation}
It is well known that for $ \mu^s_{\beta } $-a.e. $x$ the Cauchy integral of
$\mu_\beta$ at $x+i\varepsilon$ tends to infinity as $\varepsilon\to 0$.
Therefore (by (\ref{two})) $M_\beta(x + i \varepsilon ) \stackrel{\varepsilon
\downarrow 0}{\longrightarrow } \infty$. The formula for $f_\beta$ and the
definition of $M_\beta$ now imply that for $ \mu^s_{\beta }$- a.e. $x$ $L(x +
i \varepsilon ) \stackrel{\varepsilon \downarrow 0}{\longrightarrow } tg
\beta $.
Hence the expression in the righthand side of (\ref{tres}) tends to
$$ \frac{\cos \alpha }{\cos \beta } \quad \exp tg \beta \quad \mbox{ when } \varepsilon \downarrow 0$$
\noindent and we get
$$ \left\vert \frac{\sin \alpha - \exp L(z) \cos \alpha}{\sin \beta - L (z)
\cos \beta} \right\vert^2 \stackrel{\varepsilon \downarrow 0}{\longrightarrow
} \left( \frac{\cos \alpha }{\cos \beta }\right)^2 ( \exp tg \beta )^2
$$
\noindent for $\mu^s_{\beta} \: a.e.\ \ x$.
\medskip
Now if $L = a(z) + i b (z)$ then
$$ \frac{I m L (z)}{I m \exp L (z)} = \frac{b (z)}{e^{a(z)} \sin b (z)}.$$
\noindent For $\mu^s_{\beta} \: a.e.\ x \quad b \stackrel{\varepsilon
\downarrow 0}{\longrightarrow } 0$ and therefore $\displaystyle{ \frac{b}{\sin
b} } \stackrel{\varepsilon \downarrow 0}{\longrightarrow } 1$. Since $ e^{a +
i b} \stackrel{\varepsilon \downarrow 0}{\longrightarrow } tg \alpha $ then
$ e^a \stackrel{\varepsilon \downarrow 0}{\longrightarrow } tg \alpha $.
Hence
$$ \frac{I m L}{I m \exp L} \stackrel{\varepsilon \downarrow
0}{\longrightarrow } \frac{1}{tg \alpha } \quad \mbox{ for } \mu^s_{\beta}
\: a.e.\ \ x .$$
Therefore we obtain
$$
\lim\limits_{y \downarrow 0} \frac{I m M_{\beta}(z)}{I m
N_{\alpha}(z)} = \left( \frac{\cos \alpha }{\cos \beta }\right)^2
(tg \alpha )
=\alpha^{\prime } (\beta )
$$
\noindent for $\mu^s_{\beta} \: a.e.\ \ x$. Since the Poisson integral
$P\mu_\beta(x+i\varepsilon)$ tends to infinity at $\mu^s_{\beta} \: a.e.\ \
x$, the last equation implies
\begin{equation}
\lim\limits_{y \downarrow 0} \frac{Im M _{\beta}(x+iy)}{Im
N_{\alpha}(x+iy)} = \lim\limits_{y \downarrow 0} \frac{Im K
\mu_\beta(x+iy)+C_1y}{Im K\nu_\alpha(x+iy)+C_2y} =\lim\limits_{y \downarrow 0}
\frac{P_{\mu _{\beta}}(z)}{P{\nu_{\alpha}}(z)} =\alpha^{\prime } (\beta
)\label{dva}
\end{equation}
\noindent for $\mu^s_{\beta} \: a.e.\ \ x$. If $\mu_\beta=f\nu_\alpha+\eta$,
where $f\in L^1(\nu_\alpha)$ and $\eta\perp\nu_\alpha$, then Corollary 1 and
(\ref{dva}) imply $f= \alpha^{\prime } (\beta )>0$ $\mu^s_\beta$-a.e.
Thus $\mu^s_\beta<<\nu^s_\alpha$. In the same way one can show that
$$
\lim\limits_{y \downarrow 0}
\frac{P_{\nu _{\alpha}}(z)}{P{\mu_{\beta}}(z)} =\frac 1{\alpha^{\prime }
(\beta )} $$
\noindent for $\nu^s_{\alpha} \: a.e.\ \ x$ and therefore
$\nu^s_\alpha<<\mu^s_\beta$. Hence $f\nu^s_\alpha=\mu^s_\beta$. Again by
(\ref{dva}) and Lemma 1, $f\equiv \alpha^{\prime } (\beta )$.
\rightline{\sc q.e.d.}
\bigskip
\begin{rem}
In the definition of the function $N_\alpha$ in the above lemma instead of
$\exp (z)$ one can use any other function analytic in the neighborhood of
the strip $S=\{0 0$, for every subinterval $J \subset I \quad \beta \in (- \pi/2, \pi/2)$
\item[b)] $\mu^{a c}_{\beta } (J) > 0$, for every subinterval $J \subset I \quad \beta \in (- \pi/2, \pi/2]$
\end{enumerate}
\medskip
\noindent {\em Proof:}
a) Let
\begin{equation} K{\nu_{\pi/2}} (z) : = \exp ( K{\mu_{\pi/2}} (z) ).
\label{siete}\end{equation}
Then
$$ u (x) = \frac{1}{\pi } \arg K{\nu_{\pi/2}} ( x + i 0) \quad \mbox{ for } a.e.x \in I $$
\noindent and using the definition of $u$ it follows that
$$ I m K{\nu_{\pi/2}} ( x + i 0) = 0 \quad \mbox{ for } \: a.e.x \in I $$
Since the support of the absolutely continuous part of $\nu_{\alpha }$ is the set
$$ \{ x / I m K{\nu_{\pi/2}} (x + i 0) > 0 \} $$
\noindent (see \cite{Gil_Pea}),
$\nu_{\alpha }$ is purely singular in $I$ for every $ \alpha \in (- \pi/2, \pi/2)$.
Given an interval $J \subset I$ assume that $\nu_{\pi /2} (J) = 0$. Then $K{\nu_{\pi/2}} ( z)$
can be extended analytically across $J$ and from (\ref{siete})
the same follows for $K{\mu_{\pi/2}} ( z)$. Since $\mu_{\pi/2}>0$, this
implies $\mu_{\pi /2} (J) = 0$, which contradicts the construction of $\mu_{\pi
/2} $. Hence $\nu_{\pi /2}(J) > 0$ for every $J\subset I$. From this we obtain
$\nu^s_{\alpha } (J) > 0$ for every $ \alpha \in (- \pi/2, \pi/2)$ (see, for
instance, \cite[p. 38, theorem 2.52 ]{Eas_Kal}).
Now to obtain a) we just recall that from Theorem 1 we have
$$ \mu^s_{\beta} (F) = \alpha ^{\prime } (\beta) \; \nu^s_{\alpha} (F) $$
when $\alpha (\beta ) = tg^{-1} (\exp tg \beta )$ for every Borel set F. $\beta \in (- \pi/2, \pi/2),
\alpha \in (0, \pi/2)$
\medskip
b) Follows from the well known stability of the absolutely continuous part of the spectrum (see \cite[theorem 2.1 ]{Sim}) since $\mu_{\pi /2} $ is a.c. by construction.
\rightline{\sc q.e.d.}
\end{exmp}
\bigskip
\begin{rem}
Note that in our construction the absolutely continuous spectra is recurrent
(see \cite{Avr_Sim}).
\end{rem}
\bigskip
\begin{rem}
In \cite{Rio_Sim_Sto}
five examples are given of families of measures $\{\mu _{\beta } \}$ where
$d \mu = \chi_{\scriptscriptstyle B} (x)d x$, $B$ is Lebesgue measurable set
and
\end{rem}
\[ \chi_{\scriptscriptstyle B} (x) =
\left\{ \begin{array}{cl}
1 & x \in B \\
0 & x \not\in B
\end{array} \right.
\]
The occurrence of the singular spectrum embedded in the a.c. spectrum is only
shown for a set of $\beta ^{\prime } s$ of positive Lebesgue measure. The
construction above proves in \cite[examples 4 and 5]{Rio_Sim_Sto},
coexistence for all $\beta ^{\prime } s $ with the exception of one ($\pi/2$).
\medskip
To construct a family of measures such that a) holds for all $\beta \in (- \pi/2, \pi/2]$
observe that
$$
f_{\theta} (z) = i \frac{-e^{2 i \theta} + \varphi (z)}{-e^{2 i \theta} - \varphi (z)}
\mbox{ where } \varphi (z) = \frac{z - i}{z + i}.
$$
Also,
$$f_{\theta} (z) = f_{\theta + \pi} (z)$$
and
$$ i \frac{-e^{2 i \theta} + \varphi^2 (z)}{-e^{2 i \theta} - \varphi^2 (z)} =
\frac{1}{2} \; f_{\frac{\theta}{2} + \frac{\pi}{4}} (z) + \frac{1}{2} \; f_{\frac{\theta}{2} -
\frac{\pi}{4}} (z) $$
Denote by $\tilde \mu_{\beta }$ the measure corresponding to the Pick function
$$ i \frac{-e^{2 i \beta } + \varphi^2 (K{\mu_{\pi/2}}(z))}
{-e^{2 i \beta } + \varphi^2 (K{\mu_{\pi/2}}(z))}. $$
Then
\begin{equation}
\tilde \mu_{\beta } = \frac{1}{2} \: \mu_{ \frac{\beta }{2} +
\frac{\pi }{4}}+ \frac{1}{2} \: \mu_{ \frac{\beta }{2} - \frac{\pi }{4}}. \label{tri}
\end{equation}
The measures $\tilde \mu_{\beta }$ satisfy the decay condition
and can be realized as spectral measures for a family of Sturm-Liouville
operators. Also, from (\ref{tri}) it follows that $\tilde \mu_{\beta } $ have
dense singular and absolutely continuous parts on $I$ for every $ \beta \in
(- \pi/2, \pi/2]$.
\bigskip
\begin{rem}
If we multiply the measure $\mu _{\pi /2}$ used in
Example 1 by a constant less than 1, then we get singular components only for
some of the coupling constants $\beta $. More precisely,
let $H_{\pi/2}$ be the Sturm-Liouville operator whose spectral measure is
defined as $\gamma_{\pi/2} = \alpha \mu_{\pi/2}=\alpha u (x) dx $ where $u$
is as in Example 1 and $0 < \alpha< 1$.
Let $\gamma_{\alpha }$ be the spectral measures of $H_\alpha$'s.
Then using the same methods as in Example 1 one can prove the following claim:
\medskip
For the family of measures $\gamma_{\alpha }$ we have $\gamma^s_{\alpha } (J) > 0$
for every subinterval $J \subset I$ if $\alpha \in (0, \pi/2)$. If $\alpha \in (- \pi/2,
0]$ then $\gamma_{\alpha }$ is absolutely continuous.
\end{rem}
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\end{document}
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