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Sturm-Liouville operators, mixed spectral types, spectral function
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\begin{document}
\title{Sturm--Liouville operators in the half axis with shifted potentials }
\author{Rafael del Rio \thanks{
IIMAS--UNAM,
Circuito Escolar, Ciudad Universitaria,
04510 M\'exico, D.F. Telephone: (52-55) 56 22 35 95 Fax: (52-55) 56 22 35 96      %\hfill\break \indent
{\em E-mail:} delrio@servidor.unam.mx }   
\and Carmen A. Mart\'{\i}nez  \thanks{ Facultad de Ciencias--UNAM,
Circuito Exterior, Ciudad Universitaria,
04510 M\'exico, D.F. {\em E-mail: camt@servidor.unam.mx}  }
 }
\date{}
\maketitle

\begin{abstract}    \noindent
We consider Sturm--Liouville operators in the half axis generated by shifts of the potential and prove that Lebesgue measure is equivalent to a measure defined as an average of spectral measures which correspond to these operators. This is then used to obtain results on stability of spectral behavior under change of parameters such as boundary conditions, local perturbations and shifts.

\end{abstract}

\section {Introduction} \label{intro}
In several proofs of localization phenomena a key step has been to establish absolute continuity of measures
$\mu  (\cdot ) = \int \rho_{\lambda } (\cdot ) d \lambda $ generated as averages of spectral measures 
$\rho_{\lambda }$ which correspond to selfadjoint operators, particularly to Schr\"odinger operators.
Using various versions of an argument known as Kotani's trick it can be shown that the continuous spectrum is absent
in some models.  Depending on the case, the averaging parameter $\lambda $ could be a boundary condition, a 
coupling constant in a family $H_0 + \lambda  W$ or number related to a shift of the potential. See \cite{Kir_1}, \cite{Kir_2},\cite{Ko1}, \cite{Ko2}, \cite{Simon} and \cite{Sto1}.

Consider for example a family of self-adjoint operators $L_{\theta } (q)$ in $L_2 (0, \infty )$ generated by the differential expression

$$l u = - u^{\prime \prime } + q (x) u$$

\noindent and the boundary condition 

$$ u (0) \cos \theta - u^{\prime} (0) \sin \theta = 0, \qquad \theta \in [0, \pi ), $$

\noindent and let $\rho_{\theta } $ be the spectral measure corresponding to $L_{\theta }$. The average $\mu  (\cdot ) := \int^{\pi }_0 \rho_{\theta } (\cdot ) d \theta $ is not only absolutely continuous with respect to Lebesgue measure, but is equal to it! (See e.g. \cite[Theorem 1]{Pearson}, \cite{Jav}).

Using the absolute continuity of $\mu $ the following can be proven (see e.g. \cite[Corollary 3.2]{dRSiSto}).

\begin{theorem} \label{thm_0}
 Let $I$ be an open set in $\mathbb{R}$. If for almost every $E \in I $, there exists a nontrivial $L^2$ solution of $l u = E u$, then

$$ \sigma_c (L_{\theta }) \cap I = \phi $$

\noindent for almost every $\theta \in [0, \pi ) $, where $ \sigma_c $ denotes the continuous spectrum.
\end{theorem} 

After realizing the usefulness of the absolute continuity of $\mu $ it is natural to ask  if Lebesgue measure is absolutely continuous with respect to $\mu $ and if this could be of any use to understand spectral properties of the operators involved.

In this paper we give conditions which imply that Lebesgue measure and a properly chosen average measure $\mu $ have the same zero sets and apply this to prove stability results concerning spectra, when potentials of half line Schr\"odinger operators are shifted. In particular, if some spectral property holds for a set of positive Lebesgue measure of shifts then the same property will hold for a set of positive Lebesgue measure  of other parameters, such as boundary conditions and local perturbations.

This will be a consequence of our main result Theorem  \cite{dRSiSto}. See for example Corollary \ref{cor_1} below.

In previous work \cite{dRO} a similar approach was used to study local perturbations with compact support.  Here we shall consider situations where both shift and local perturbations are present.  In Section \ref{mainresults} our main result is proven.  The techniques rely heavily on the Pr\"ufer transform which is an important tool in this paper.  In Section \ref{generalizations} we show how more general results can be obtained and some applications to $\alpha $-singular and $\alpha $-continuous spectrum. Some of the methods used have their origin in \cite{Ko2} and \cite{Sto1}.

We shall use the notation $\mathbb{R}^+$ for the non negative reals, i.e. $\mathbb{R}^ += \{ a \in \mathbb{R} / a \geq 0 \}$.

\section {Main Results} \label{mainresults}

For each $(a, \theta ) \in \mathbb{R}^+  \times  [0, \pi )$
let us consider the selfadjoint operator $L_{a \theta} $ in $L_2 (0, \infty )$ generated by the differential expression

$$l_a u = \frac{- d^2}{d x^2} + V_a (x),$$

\noindent where $V_a (x) := \left\{ \begin{array}{cl} q (x - a) & \textrm{if } x \geq  a \\ 0 & 0 \leq  x < a \end{array} \right.$  
as follows: 

\medskip

$L_{a \theta}  u = l_a u$ for $u$ in the domain of $L_{a \theta }$ given by

$$ \begin{array}{rr}
D (L_{a \theta})  = & \{ f \in L^2 (0, \infty ) : f, f^{\prime} $$ $$ \textrm{ locally absolutely continuous in } (0, \infty ),  \\
& l_a f \in L_2 (0, \infty ),  \quad f (0) \cos \theta - f^{\prime} (0) \sin \theta = 0,  \\
& \theta \in [0, \pi )\}. \end{array} $$ 

\noindent We assume $l u = - u^{\prime \prime } + q (x) u$ is in the limit point case at $\infty $.
The function $q (x)$ is locally in $L_1$.

\medskip

If $\mu$ and $\nu$ are two measures we use the notation $\mu \prec\prec \nu$ when $\mu$ is absolutely continuous with respect to $\nu$, that is if $\nu (A) = 0$ implies $\mu (A) = 0$.  If $\mu \prec\prec \nu $ and $\nu \prec\prec \mu$  we say that the two measures are equivalent. We use $\vert \cdot \vert$, to denote the Lebesgue measure.

Let $a_2 > a_1 > 0 $ and $E_0 > 0 $.  For a Borel set $A \subset [E_0, \infty )$ define

$$ \mu_{\theta }  (A) := \int\limits^{a_2}_{a_1} \rho_{a \theta } (A) d a, $$

\noindent where $\rho_{a \theta }$ is the spectral function associated to the operator $L_{a \theta}$.

\bigskip
\begin{theorem} \label{thm_1}
\hfill\break
\begin{enumerate} 
\item[I)] $\mu_{\theta } (\cdot ) \prec \prec \mid \cdot \mid $

\item[II)] If $a_2 - a_1 \geq  \frac{\pi }{\sqrt{E_0}}$ then $\mu_{\theta } (\cdot ) \succ \succ  \mid \cdot \mid $.

\end{enumerate}

\end{theorem} 

%%%\cite[]{} 
\medskip                                                                                                          
Before we prove the theorem let us introduce notation and recall some results.

\medskip

Consider real solutions $u \not\equiv  0$ of $l_a u = E u$ such that for fixed $c \in \mathbb{R}^+$

$$\begin{array}{l} u (c) = \sin \theta \\ u^{\prime } (c) = \cos \theta. \end{array}$$

If we write the vector $\big( u^{\prime} (x), u (x) \big) \;$  for $x > 0$ in polar coordinates we obtain 

$$\begin{array}{ll} u (x) = r_c (x) \sin  \phi_c (x) \\ u^{\prime } (x) = r_c (x) \cos \phi_c (x)  \end{array}$$

\noindent where $\phi_c (x, \theta, E, V_a)$ and $r_c (x, \theta, E, V_a)$ are called the Pr\"{u}fer phase and the Pr\"{u}fer amplitud of  $u$, respectively.

We fix a unique value of $\phi_c$ by requiring $\phi_c (c, \theta, E) = \theta$ and continuity in $x$. These functions $r_c$ and $\phi_c$ are jointly continuous in $x, E$  (use arguments similar to 
\cite [Thm 2.1]{Weidmann}, \cite [Thm Ch.2.4]{CoLe}, \cite [Thm Ch.V.3]{Hartman} ). 
This will be important in what follows.

\medskip

We shall need the next results

\begin{enumerate}
\item[a)]  

$$\rho_{a \theta }  \big( (E_1, E_2) \big) = \lim\limits_{b \rightarrow \infty } \frac{1}{\pi }
\int\limits^{E_2}_{E_1} r_0 (b, \theta, E, V_a)^{-2} d E $$

\noindent if $E_1$ and $E_2$ are not discrete points of $\rho_{a \theta }$. See \cite[Thm 2]{Pearson}, $E_1 < E_2$.

\item[b)] 

For any $a, x, \theta, E \in \mathbb{R} $

$$\frac{1}{\pi } \int\limits^{\theta + \pi }_{\theta} r_a (x, \beta, E)^{-2} d \beta = 1. $$

See \cite[Corollary 12]{Sto1} and \cite[Appendix B]{Sto2}.

\item[c)] 
$$\frac{\partial \phi _0}{\partial x} = \cos^2 \phi _0 (x) + (E -V_a) \sin^2 \phi_0 (x). $$ 

%(see \cite[formula (2.5) in proof of Thm 2.1, p.~212]{CL}).

\end{enumerate}

\noindent This follows from a straightforward calculation.

\subsection*{Proof (of Theorem \ref{thm_1})}          %\hfill\break

We use an argument similar to the one used in \cite{Sto1}  
where I) was proven for the case of the whole line. Given any $E_1 < E_2, 0 < E_0 \leq E_1$ the measure $\rho_{a \theta}$ is continuous in $E_1$ and $E_2$ for almost any $a $.

\medskip

From a) above we know that

\begin{equation}	\label{ec_1}
\mu \big( (E_1, E_2) \big) = \int\limits^{a_2}_{a_1} \rho_{a \theta }  \big( (E_1, E_2) \big) d a = \int\limits^{a_2}_{a_1} \left[ \lim\limits_{b\rightarrow \infty } \frac{1}{\pi } \int\limits_{E_1}^{E_2}  r_0 (b + a, \theta, E, V_a)^{-2} d E \right] d a
\end{equation}

We shall forget for a while the limit which appears in the expression (\ref{ec_1}). Our aim is to get first the estimate  (\ref{ec_5}) below.

\medskip

Since

\begin{equation} \label{ec_2} 
\begin{array}{lcl}
r_0 \big( b + a, \theta, E, V_a \big) & =  &
r_0 \big( a, \theta , E, 0 \big) r_a \big( b + a, \phi_0 (a, \theta , E, 0), E, V_a \big) \\ [1ex]
& = & r_0 \big( a, \theta , E, 0 \big) r_0 \big( b, \phi_0 (a, \theta , E, 0), E, q (x) \big)
\end{array}
\end{equation}

\noindent we obtain

$$
\int\limits^{a_2}_{a_1} \frac{1}{\pi } \left[ \int\limits_{E_1}^{E_2} \big( r_0 (b + a, \theta, E, V_a) \big)^{-2} d E \right] d a = $$
$$
\int\limits^{a_2}_{a_1} d a \frac{1}{\pi } \left[ \int\limits_{E_1}^{E_2} d E \Big( r_0 \big( a, \theta, E, 0 \big)^{-2} r_0 \big( b, \phi_0 (a, \theta, E, 0) , E, q(x) \big)^{-2} \Big) d E \right] $$

Since $r_0 (a, \theta, E, 0)$ is uniformly bounded for $(a, E) \in [a_1, a_2] \times [E_1, E_2]$ (use joint continuity) and interchanging the order of the integrals, we get that

\begin{eqnarray} \label{ec_3}
& & 
C_1 \int\limits_{E_1}^{E_2} d E \int\limits_{a_1}^{a_2}  d a \, r_0 \big( b, \phi_0 (a, \theta, E, 0), E, q (x) \big)^{-2}   \nonumber \\
& \geq  & \int\limits^{a_2}_{a_1} \frac{1}{\pi } \left[ \int\limits_{E_1}^{E_2} \big( r_0 (b + a, \theta, E, V_a) \big)^{-2} d E \right] d a \\
& \geq & C_2 \int\limits_{E_1}^{E_2} d E \int\limits^{a_2}_{a_1} d a \, r_0 \big( b, \phi_0 (a, \theta, E, 0), E, q (x) \big)^{-2}  \nonumber
\end{eqnarray}

\noindent where we can choose, for example, $C_1$ and $C_2$ to be the $\sup$, respectively the $\inf$, of 
$r_0 (a, \theta, E, 0)$ when $(a, E) \in [a_1, a_2] \times [E_1, E_2]$.

\medskip

Now if we denote

$$ \beta (a) := \phi_0 (a, \theta, E, 0) $$

\noindent and change variables we obtain

$$
\int\limits^{a_2}_{a_1} d a \, r_0 \big( b, \beta  (a), E, q (x) \big)^{-2}  = 
\int\limits_{\beta (a_1)}^{\beta (a_2)} \frac{d \beta }{\beta^{\prime } (a)} r_0 \big( b, \beta  (a), E, q (x) \big)^{-2}
$$

Since

$$ \min \{ 1, E \} \leq  \beta^{\prime } (a) \leq \max \{ 1, E \}, $$

\noindent (see c) above) and recalling that $E \in [E_1, E_2]$ then

\begin{eqnarray} \label{ec_4}
& & 
C_3 \int\limits_{\beta (a_1)}^{\beta (a_2)} r_0 \big( b, \beta , E, q (x) \big)^{-2} d \beta \nonumber \\
& \geq & \int\limits^{a_2}_{a_1} d a \, r_0 \big( b, \beta  (a), E, q (x) \big)^{-2} \\
& \geq & C_4 \int\limits_{\beta (a_1)}^{\beta (a_2)} r_0 \big( b, \beta , E, q (x) \big)^{-2} d \beta \nonumber
\end{eqnarray}

\noindent for suitable positive constants $C_3, C_4$.

\medskip

From inequalities (\ref{ec_3}) and (\ref{ec_4}) we obtain

\begin{eqnarray} \label{ec_5}
& &
C_5 \int\limits_{E_1}^{E_2} d E \left[ \int\limits_{\beta (a_1)}^{\beta (a_2)} r_0 \big( b, \beta , E, q (x) \big)^{-2} d \beta \right]  \nonumber \\
& \geq &  \int\limits^{a_2}_{a_1} \frac{1}{\pi } \left[ \int\limits_{E_1}^{E_2} \big( r_0 (b + a, \theta, E, V_a) \big)^{-2} d E \right] d a\\
& \geq &  C_6 \int\limits_{E_1}^{E_2} d E \left[  \int\limits_{\beta (a_1)}^{\beta (a_2)} r_0 \big( b, \beta , E, q (x) \big)^{-2} d \beta \right]  \nonumber
\end{eqnarray}

\noindent for some positive constants $C_5, C_6$.

\begin{proof}[\large{Proof of I)}] \hfill\break

Splitting the interval $[\beta  (a_1), \beta  (a_2)]$ into intervals of length at most $\pi $ yields by b) above

$$\int\limits_{\beta (a_1)}^{\beta (a_2)} r_0 \big( b, \beta , E, q (x) \big)^{-2} d \beta \leq \pi 
\left( \frac{\beta (a_2) - \beta (a_1)}{\pi } + 1\right), $$

\noindent therefore from the first inequality in (\ref{ec_5}) we get

$$C_7 (E_2 - E_1) \geq \int\limits^{a_2}_{a_1} d a \left[ \frac{1}{\pi } \int\limits_{E_1}^{E_2} r_0 (b +  a, \theta, E, V_a)^{-2} d E \right]
$$

Applying Fatou's lemma we obtain

$$C_7 (E_2 - E_1) \geq \mu_{\theta } (E_1,E_2) $$

Using countable additivity part I) of the theorem follows for general Borel sets.

\bigskip

\noindent{\large{\it Proof of II)} }\hfill\break

Let us look now for conditions on $a_1, a_2$ which imply $\beta (a_2) -  \beta (a_1) \geq \pi $. Since in the definition of $\beta$ the potential zero was used, it is possible to calculate $\beta$ explicitly. 

\medskip

Observe that $ u (x) := \sin (\alpha  + \sqrt{E} x ) $, where $\alpha  := \arctan ( \sqrt{E} \tan \theta ) \in [0, \pi )$, satisfies the equation

$$ - u^{\prime \prime } (x) = E u (x) $$

\noindent and the conditions

$$ \begin{array}{rcl}
u (0) & = & \sin \theta \\
u^{\prime } (0)  & = & \cos \theta. 
\end{array}
$$

We can then write

\begin{eqnarray*}
\beta  (x) + n \pi  & = & \phi_0 (x, \theta, E, 0) + n \pi  \\
& = &    \arg   \big( u^{\prime } (x) + i u (x) \big) + n \pi \\
& = &    \arg  \big( \sqrt{E} \cos (\alpha + \sqrt{E} x ) + i \sin (\alpha  + \sqrt{E} x ) \big) + n \pi \\
& = &    \arg   \big( \sqrt{E} \cos (\alpha + \sqrt{E} x  + n \pi ) + i \sin (\alpha  + \sqrt{E} x  + n \pi ) \big) \\
& = &    \beta \Big(   x + \frac{n \pi }{\sqrt{E} }  \Big)
\end{eqnarray*}

Since $\beta  (x)$ is increasing (see c) above) then $a_2 - a_1 \geq \frac{\pi}{\sqrt{E}}$ implies $\beta (a_2) -  \beta (a_1) \geq \pi $.

\medskip

Therefore using b) above we get

$$ \int\limits_{\beta (a_1)}^{\beta (a_2)} r_0 \big( b, \beta , E, q (x) \big)^{-2} d \beta \geq \pi \quad
\textrm{ if } \quad a_2 - a_1 \geq \frac{\pi }{\sqrt{E}} $$

We can conclude that

$$ \int\limits^{a_2}_{a_1} \frac{1}{\pi } \left[  \int\limits_{E_1}^{E_2} r_0 (b +  a, \theta, E, V_a)^{-2} d E
\right] d a \geq  C_8 (E_2 - E_1) \quad \textrm{ if } \quad a_2 - a_1 \geq \frac{\pi }{\sqrt{E_0}} $$

To be able to consider the limit which appears in  (\ref{ec_1}) we shall bound

$$ \tilde{F}_b (a) := \int\limits_{E_1}^{E_2} r_0 (b +  a, \theta, E, V_a)^{-2} d E $$

\noindent for every $b > 0$ and apply the Lebesgue dominated convergence theorem.  Observe that using the decomposition (\ref{ec_2}) it is enough to bound

$$F_b (a) = \int\limits_{E_1}^{E_2} r_0 \big( b, \beta (a), E, q (x) \big)^{-2} d E. $$

From Lemma 1 of \cite{Pearson} we know that

$$F_b (a) = \int\limits^{\pi }_0 \mu^b_{\beta  (a) \gamma } (E_1, E_2) d \gamma $$

\noindent where $\mu^b_{\beta \gamma }$ is the spectral measure associated to the regular problem in $[0, b]$ with boundary conditions

$$ \begin{array}{rcl}
u (0) \cos \beta & - & u^{\prime } (0) \sin \beta = 0 \\ 
u (b) \cos \gamma  & - & u^{\prime } (b) \sin \gamma = 0
\end{array}
$$

In the same paper \cite{Pearson} it is observed (see proof of Corollary 3) that these measures are uniformly bounded in $b, \beta, \gamma $. Hence the boundedness of $F$ follows. Therefore

$$\mu_{\theta } (E_1, E_2) \geq C_8 (E_2 - E_1) $$

\noindent and using countable additivity we get II) for general Borel sets. 

\end{proof}


\begin{corollary} \label{cor_1}
Let $I:= (E_1, E_2) \subset \mathbb{R}^+$ open and define $L_{a \theta }$ as above. 
For any $a\in \mathbb{R}^+$, the operator $L_{a \theta }$ has singular continuous spectrum in $I$ for a set of positive Lebesgue measure of $\theta$'s, if and only if for any $\theta \in [0, \pi), L_{a\theta}$ has singular continuous spectrum in $I$ for a set $B$ of $a$'s of positive Lebesgue measure. 

\medskip

Moreover $ \big\vert B \cap [a_1, a_2 ] \big\vert > 0 \,$ if  $\, a_2 - a_1 \geq \frac{\pi}{\sqrt{E_0}} $ where $0 < E_0 < E_1$.
 
\end{corollary}

\begin{proof}
 $\Rightarrow$) Let $S$ be the set of points $E$ for which there are subordinate solutions of $l_a u = E u$ which are not in $L_2$. It is known that this set is a support of the singular continuous part of $L_{a \theta }$ and it does not depend on $a$  and $\theta$.  Since $\rho_{a \theta } (S\cap I) > 0$ for a set of positive measure in $\theta $ by hypothesis, using equality

$$ \vert S\cap I \vert = \int\limits^{\pi }_0 \rho_{a \theta } (S\cap I) d \theta $$

\noindent we deduce $\vert S\cap I \vert > 0 $. This implies using Theorem \ref{thm_1} II) that 
$ \mu_{\theta}(S\cap I) > 0 $ for any $\theta$ if $a_2 - a_1 \geq \frac{\pi}{\sqrt{E_0}}$. 
>From here we know $\rho _{a \theta } (S\cap I) > 0$ for $a \in B$ where $B$ satisfies

$$\big\vert B \cap [a_1, a_2 ] \big\vert > 0 $$

\noindent $\Leftarrow$) Assume $L_{a{\theta}}$ has singular continuous spectrum in $I$ for a set $B$ of  $a$'s of positive Lebesgue measure. Then $\rho_{a \theta } (S \cap I ) > 0 \textrm{ for } a \in B$ and


$$\int\limits_B \rho_{a \theta } (S \cap I) d a > 0 $$

\noindent Therefore there exists an interval $J = [a_1, a_2]$ such that

$$
\int\limits_J \rho_{a \theta } (S\cap I) d a \geq \int\limits_{B \cap J} \rho_{a \theta } (S\cap I) d a > 0
$$

\noindent Using Theorem \ref{thm_1} I) we obtain $\vert S\cap I\vert > 0$ and therefore

$$\int\limits_0^{\pi} \rho_{a \theta } (S \cap I) d\theta = \vert S \cap I \vert > 0 $$

\noindent for every fixed $a$. Therefore $L_{a\theta}$ has singular continuous spectrum  in $I$ for a set of positive Lebesgue measure in $\theta$.

\end{proof}


\begin{remark}
If instead of taking the support $S$ as above we take the set $P$ which corresponds to subordinate solutions which are in $L_2$ we get the same result for the pure point part and taking $P\cup S$, we obtain the result for the singular part of $L_{a \theta }$.
\end{remark}

\section{Generalizations.} \label{generalizations}

Recall the construction of $L_{a \theta }$ given in Section 2 and choose $q (x) = V (x) + \lambda  W (x)$ where we assume $W (x) > 0 $ for a.e. $ x \in [0, c]$ and $W (x)  = 0$ for every $x \not\in [0, c]$.
The corresponding operator and spectral function will be denoted by $L_{a \theta \lambda }$ and $\rho_{a \lambda  \theta} $ respectively.  Fixing two of the three parameters $a, \lambda, \theta $ we define lines in $\mathbb{R}^+  \times  \mathbb{R} \times [0, \pi )$ as follows

$$ \begin{array}{rl}
l = & \{ (a,\lambda, \theta ) \in \mathbb{R}^+  \times  \mathbb{R} \times [0, \pi ) /  \\
&  \textrm{ two parameters are fixed. In case $(a, \theta )$ } \\
&  \textrm{ are fixed set $a = 0$} \}
\end{array} $$

The set  of all lines defined in this way will be denoted by ${\cal M}$.  Now consider the set ${\cal P}, {\cal P} \subset {\cal M}$ defined in the following way

$$ \begin{array}{rl}
{\cal P} = & \{ l \in {\cal M} / \exists \; B \subset l, \vert B \vert > 0  \\
&  \textrm{ such that $\sigma (L_{a, \lambda, \theta} ) \cap  I \not= \phi $ } \\
&  \textrm{ for $(a,\lambda, \theta ) \in B$} \}
\end{array} $$

Here $\vert \cdot \vert$ denotes the one dimensional Lebesgue measure and $\sigma  = \sigma_{sc},
\sigma_{pp} \textrm{ or } \sigma_s$ where, as usual, $\sigma_{sc},\sigma_{pp}, \sigma_s$ denote the singular continuous, pure point and singular spectrum respectively.

\vskip 0.5cm

\begin{theorem} \label{thm_2}

$${\cal P} = {\cal M} \textrm{ or } {\cal P} = \phi  $$

\end{theorem}

\begin{proof}
 
 
If $I$ is open we know that

\begin{equation}   \label{ec_6} 
\sigma (L_{a, \lambda, \theta} ) \cap  I \not= \phi \Longleftrightarrow 
\rho_{a, \lambda, \theta} (A \cap I) >0
\end{equation} 

\noindent where $A = S, P \textrm{ or } S \cup  P$, as defined in Corollary above and the Remark following it, depending on whether $\sigma $ is $\sigma_{sc}, \sigma_{pp} \textrm{ or } \sigma_s$. See for example
\cite[Corollary 2.8]{dRSiSto}.
On the other hand we know


\begin{equation} \label{ec_7}
\vert \cdot \vert\sim \int\limits^{\lambda_2}_{\lambda_1} \rho_{0 \lambda  \theta } ( \cdot ) d \lambda \sim \int\limits^{a_2}_{a_1} \rho_{a \lambda  \theta } ( \cdot ) d a \sim \int\limits^{\pi}_0 
\rho_{a \lambda  \theta } ( \cdot ) d \theta 
\end{equation} 

\noindent where $\sim $ denotes equivalence of measures (two measures are equivalent if they have the same sets of measure zero),  if we take $\lambda_2 - \lambda_1, a_2 - a_1 $ large enough.  If not, then the measures defined as averages are just absolutely continuous with respect to Lebesgue measure.

The first equivalence is the main result in \cite{dRO},
the second is Theorem \ref{thm_1} above, and the third goes back to \cite{Jav}.

The theorem then follows from (\ref{ec_6}) and (\ref{ec_7}).

Let us for example fix $a= 0, \theta  = \theta_0$ and let $\lambda $ vary in $\mathbb{R} $.  Assume there is a set $B \subset \mathbb{R}$ of positive Lebesgue measure such that for $\lambda \in B $

$$\sigma  (L_{0, \lambda, \theta_0}) \cap  I \not= \phi. $$

In other words, assume that the line $\{ (0, \lambda, \theta_0 ) / \lambda  \in \mathbb{R}\}$ is in ${\cal P}$.  Then we know by (\ref{ec_6}), that $\rho_{0, \lambda , \theta_0} (A \cap  I) > 0$ for $\lambda  \in B$.
Using that $\int\limits^{\lambda _2}_{\lambda _1} \rho_{0 \lambda \theta_0} (\cdot ) d \lambda  \prec \prec \vert
\cdot \vert $ (with no restrictions on the length $\lambda_2 - \lambda_1 $) it follows that $\vert A \cap I \vert > 0$ and from (\ref{ec_7}) we obtain 

$$\sigma  (L_{a, \lambda, \theta}) \cap  I \not= \phi $$

\noindent for a set of $a$'s of positive measure. The other cases are proven analogously.
 
\end{proof}

In the case $\sigma  = \sigma_{pp}$, we can choose

$$q (x) = V (x) + \lambda  W (x) + U (x) $$

\noindent where $U$ is locally $L_1$ and

$$\int \vert U (x) \vert e^{A\vert x \vert} d x < \infty \qquad \textrm{for all } A > 0. $$

Using \cite[Colollary 1.8]{KiLaSi} %%%\cite{} 
then Theorem \ref{thm_2} can be proven in this case too.

\begin{proposition}
Let $I$ be an open interval $I \subset [E_0, \infty ) \, E_0 > 0$ and define

$$B:= \{ a \in \mathbb{R}^+ / \sigma  (L_{a \lambda_0  \theta_0 })\cap  I \not= \phi \} $$

If $\vert B \vert > 0$ then

$$\vert B \cap [a_1, a_2] \vert > 0$$

\noindent wherever $a_2 - a_1 \geq \frac{\pi }{\sqrt{E_0}}$. In particular $B$ is unbounded.

\end{proposition}

\begin{proof}From (\ref{ec_6}), if $a \in B$ then $\rho_{a \lambda_0  \theta_0 }  (A\cap I) > 0 $ and $0 < \int\limits^{a_2}_{a_1} \rho_{a \lambda_0  \theta_0 } (A\cap I) d a $ for some $a_1 < a_2$ if $\vert B \vert > 0 $.

From Theorem \ref{thm_1} I) it follows $\vert A \cap I \vert > 0 $, and taking $a_1 < a_2 $ such that $a_2 - a_1 \geq 
\frac{\pi }{\sqrt{E_0}}$ then from Theorem \ref{thm_1} II) we get

$$\int\limits^{a_2}_{a_1} \rho_{a \lambda_0  \theta_0 } (A\cap I) d a > 0 $$

\noindent and therefore $\vert B \cap  [a_1, a_2] \vert > 0$ using (\ref{ec_6}) again.

\end{proof}


Similar results can be obtained for the $\alpha$-continuous and $\alpha$-singular spectrum. Recall that for $\alpha \in [0, 1]$ the $\alpha$-dimensional Hausdorff measure is defined for Borel sets $A$ by 

$$h^\alpha (A) \equiv \lim\limits_{\delta\to 0} \, \inf\limits_{\delta-{\textrm{covers} }}
\sum\limits^\infty_{v=1} \vert b_v\vert^\alpha,$$

\noindent where a $\delta$-cover is a countable collection of intervals each of length at most $\delta $ so $A \subset \cup^\alpha_{v=1} b_v$.

Given $\alpha  \in [0, 1]$ we define a measure  $\mu$ to be $\alpha$-continuous $(\alpha c)$ if 
$\mu (S) = 0$ for any set $S$ with $h^\alpha (S)=0$ and $\alpha$-singular  $(\alpha s)$ if it is supported on a set of $S$ with $h^\alpha (S)= 0$.  For every such $\alpha$ and any measure $\mu$, one can uniquely decompose $\mu  = \mu^{\alpha c} + \mu^{\alpha s}$ with $\mu^{\alpha c} 
\alpha$-continuous and $\mu^{\alpha s}, \alpha$-singular.

Denote $\rho : = \rho_{a \lambda \theta }$.  It is possible to find sets $A_\alpha$ and $B_\alpha$ such that 

\begin{eqnarray*}
d\rho^{\alpha c}  & = & d \rho (A_\alpha \cap . ) \\
d\rho^{s c} & = & d \rho (B_\alpha \cap . )
\end{eqnarray*}

\noindent and it happens that $A_\alpha$ and $B_\alpha$ are independent of $\theta, a $ and $\lambda$. See \cite{KiLaSi} 
and references therein.

Using the same reasoning as above one can prove Theorem \ref{thm_2} 
for the $\alpha$-singular and $\alpha$-continuous part of the spectral measure.
We get for example

\begin{theorem} \label{thm_4}
If $\rho^{\alpha c}_{a \lambda_0  \theta_0 }  (I) > 0 $for $a \in B$ where $I$ is an open interval, $\vert B \vert > 0$ and $\lambda_0, \theta_0$ are fixed then $\rho^{\alpha c}_{a_0 \lambda_0  \theta }  (I) > 0 $ for $\theta \in \tilde B, \vert \tilde B \vert > 0$ for any $a_0, \lambda_0$ fixed.


\end{theorem}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%available electronically in %\hfill\break 
%{\tt http://www.ma.utexas.edu/mp\_arc-bin/mpa?yn=02-394}.

%\baselineskip=28pt

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