0$ and $f=\sum_i \alpha_i f_i$, where
$f_i$ are subharmonic functions which are potentials, then
$M_{f}(z) \leq \sum_i \alpha_i M_{f_i}(z)$ and
$C_{f}(z) = \sum_i \alpha_i C_{f_i}(z)$ so that
$B_{f}(z) \leq \sum_i \alpha_i B_{f_i}(z)$. More generally, if
$x \mapsto f_x(z)$ is a random variable over a probability
space $(X,\Acal,\mu)$ for all $z$ and $z \mapsto f_x(z)$ is subharmonic,
then $f(z) = \int_X f_x(z) \; d\mu(x)$ is subharmonic and
$B_f(z) \leq \int_{X} B_{f_x}(z) \; d\mu(x)$. \\
(iii) Especially, if $f(z) = \int_{\CC} f_w \; dk(w)$ where $dk$ is a finite,
positive Borel measure in $\CC$ and $f_w$ are subharmonic potentials for
which $B_{f_w}$ is bounded above, then $f$ is a subharmonic potential and
$B_{f}(z) \leq \int_{\CC} B_{f_w} \; dk(w)$. \\
(iv) If ${\rm supp}(dk) \subset \{ |z|=r \}$, then
$B_{f^{(k)}}(z) \leq \exp(-|\log(r/|z|)| 2^k)$. \\
Proof. From (i), the result follows for
measures on $\{ |z|=r \}$ which are finite point measures. The general
case follows from (iii) by integration over the angles. \\
(v) For a general finite Borel measure $\rho$ on $\RR^+$, we have
$$B_{f^{(k)}}(z) \leq \int_0^{\infty}
\exp(-|\log(\frac{t}{|z|})| 2^k)\; d\rho_f(t) \; . $$
Proof. By (ii) and (iv) by integration over the radius. \\
(vi) The claim of the theorem: from Lemma~(\ref{homogenization})
and (v)
$$A_f(z) \leq \sum_{k=0}^{\infty} B_{f^{(k)}}(z)
\leq \sum_{k=0}^{\infty}
\int_0^{\infty} \exp(-|\log(\frac{t}{|z|})| 2^k)\; d\rho(t)
= \Kcal(\rho)(z) \; $$
allowing the right hand side to be $+\infty$.
\section{The homogenization transformation}
The {\bf homogenization transformations}
$$ \Kcal_l: \rho \mapsto \frac{1}{2^l} \sum_{k=l}^{\infty}
\int_{\RR^+} \exp(-|\log(\frac{r}{t})| 2^k) \; d\rho(t) \; $$
are defined on the linear space of all finite Borel measures
$\rho$ on $\RR^+=[0,\infty)$.
The image of $\Kcal_l(\rho)$ is a $\RR^+ \cup \{+\infty\}$-valued function
on $\RR^+$.
\begin{propo}[$\Kcal_1$ for $\rho \in L^{\infty}(\RR^+)$.]
\label{Lipshitz}
If $\rho \in L^{\infty}(\RR^+)$ then
$\Kcal_1(\rho)(z) \leq 2 ||\rho||_{\infty} |z|$.
Especially, $\Kcal_1(\rho) \in L^{\infty}_{loc}(\RR^+ \setminus \{0\})$.
\end{propo}
\begin{proof}
Using $\int_0^r (t/r)^m \; dt=r/(1+m)$ and
$\int_r^{\infty} (r/t)^m \; dt = r/(m-1)$, we estimate
\begin{eqnarray*}
2 \Kcal_1(\rho)(r)
&=& \sum_{k=1}^{\infty} \int_0^r (\frac{t}{r})^{2^k} \rho(t) dt
+ \sum_{k=1}^{\infty} \int_r^{\infty} (\frac{r}{t})^{2^k} \rho(t) dt \\
&\leq& ||\rho||_{\infty}
(\sum_{k=1}^{\infty} \int_0^r (\frac{t}{r})^{2^k} dt
+ \sum_{k=1}^{\infty} \int_r^{\infty} (\frac{r}{t})^{2^k} dt ) \\
&=& 2 ||\rho||_{\infty}
r \sum_{k=1}^{\infty} \frac{2^k}{2^{2 k}-1}
\leq 2 ||\rho||_{\infty}
r \sum_{k=1}^{\infty} \frac{1}{(2^{k}-2^{-k})}
\leq 4 ||\rho||_{\infty} r \; .
\end{eqnarray*}
\end{proof}
The next proposition can be applied if $C_f$ is Lipschitz on some interval
$[r/a,ra]$ in which case $\rho_f$ is bounded on that interval.
\begin{propo}[$\Kcal$ for $\rho \in L^{\infty}_{loc}(\RR^+)$]
\label{Lipshitz2}
Assume $\rho(s) \leq M$ for almost all $s \in [r/a,a r]$, with $a>1$ and
let $c = \rho(\CC)$ be the total mass of $\rho$. Then
$\Kcal(\rho)(z) \leq 2 M r (1- 1/a) + (c/2)/(1-1/a)$. Especially,
$\Kcal(\rho) \in L^{\infty}_{loc}(\RR^+)$.
Taking $a=2$, we get $\Kcal_1(\rho)(r) \leq Mr+c$ for $r \in [r/2,2 r]$.
\end{propo}
\begin{proof}
We get from $\int_{r/a}^r (t/r)^m \; dt= (1 - 1/a^{m+1}) r/(1+m)
\leq (1-1/a) r/(1+m)$, $\int_r^{r+a} (r/t)^m \; dt = (1- 1/a^{m-1}) r/(m-1)
\leq (1- 1/a) r/(m-1)$ the first term.
% Integrate[(t/r)^m,{t,r/a,r}]
% Integrate[(r/t)^m,{t,r,r+a}]
The second term comes by assuming the worst case, where we
have a Dirac measure of maximal mass $c$ at $r/a$.
\end{proof}
\begin{propo}[$\Kcal_1$ for $\rho \in L^p(\RR^+)$]
\label{Lp}
Given $1p_n \to 1$.
% \int_0^1 x^m g_n dx = m! (1-p)^!/(1+m-p)!$.
Definition. Given a subharmonic function $f$, the potential
$f^*(r)=f_{\rho_f}(r)= \int_{0}^{\infty} \log|r-s| \; d\rho_f(s)$
is called the {\bf $*$-function} of $f$. Define
$C_{\rho}(r) = \frac{1}{2\pi} \int_{0}^{2\pi}
f^*(r \exp(i \theta)) \; d\theta$. The function $f_{\rho}$
is obtained from $f$ by sweeping all masses to a positive (or negative)
real axes. The $*$ function appears in results on growth estimates for
subharmonic functions.
\begin{propo}[Bounds by potential of $*$-function]
\label{LogHoelder}
$$ C_{\rho}(r) - f_{\rho}(r) \leq
\Kcal(\rho)(r) \leq 2 (C_{\rho}(r) - f_{\rho}(r)) \; . $$
If $f(z)=f(\xi_l z)$ for all $z \in \CC$, then
$\Kcal_l(\rho)(r) \leq 2^{-(l-1)}(C_{\rho}(r) - f_{\rho}(r))$.
\end{propo}
\begin{proof}
We can assume that $\rho$ has no atom at $r$ because in that case, all
three terms are $+\infty$. \\
(i)
\begin{eqnarray*}
\int_{0}^{r} \log|r-s| \; d\rho(s)
&=& \int_0^{r} \log|1-\frac{s}{r}| \; d\rho(s)
+ \int_0^r \log|r| \; d\rho(s) \\
\int_{r}^{\infty} \log|r-s| \; d\rho(s)
&=& \int_r^{\infty} \log|1-\frac{r}{s}| \; d\rho(s)
+ \int_r^{\infty} \log|s| \; d\rho(s) \; .
\end{eqnarray*}
(ii)
\begin{eqnarray*}
\int_0^{r} \log|1-\frac{s}{r}| \; d\rho(s)
&=& - \sum_{k=1}^{\infty} \int_0^r (\frac{s}{r})^k/k \; d\rho(s) \\
\int_r^{\infty} \log|1-\frac{r}{s}| \; d\rho(s)
&=& - \sum_{k=1}^{\infty} \int_r^{\infty} (\frac{r}{s})^k/k \; d\rho(s) \; .
\end{eqnarray*}
(iii) For $00$-dimensional
Hausdorff measure in an open interval containing $r$. \\
%3) The set
%$S_{\rho}=\{ t \in \RR^+, \Kcal(\rho)(t) = \infty \}$ is a $G_{\delta}$
%polar set. Especially, it has Lebesgue measure zero. \\
3) The potential $f(x)=\int_{\RR^+} \log|x-w| \; d\rho(w)$ is as a
function on the real line equal to the Hilbert transform of the function
$h(x) = \int_0^x d\rho(y)$. The Hilbert transform maps
$\alpha$-H\"older spaces or $L^p$ spaces into itself.
We see especially that if $\rho$ is bounded then $h$ is Lipschitz and
$f$ $\alpha$-H\"older for all $\alpha \in (0,1)$.
Because also $C_f$ is Lipschitz if $\rho$ is bounded,
Proposition~(\ref{LogHoelder}) implies that if $\rho$ is bounded, then
$\Kcal(\rho)$ is sandwiched between two H\"older continuous maps.
\begin{coro}
\label{Milloux-Schmidt}
There exists a real sequence $r_n \to 0$ such that
$\limsup_{n \to \infty} \Kcal(\rho)(r_n) = 0$.
\end{coro}
\begin{proof}
It is a consequence of the Milloux-Schmidt inequality \cite{Hayman}
that for any subharmonic function $f$ with $f(0)>-\infty$ one has
$\limsup_{r \to 0} {\inf_{\theta}} f(r \exp(i \theta)) = f(0)$.
Therefore $\limsup_{r \to 0} f_{\rho}(r) = f_{\rho}(0)$. Together with
$\lim_{r \to 0} C_{\rho}(r) = C_{\rho}(0)$ which follows from
the upper-semicontinuity and the submean inequality for
subharmonic functions, we get
$\limsup_{r \to 0} 2( f_{\rho}(r)-C_{\rho}(r) ) = 0$. The claim
follows from Proposition~(\ref{LogHoelder}).
\end{proof}
\section{Radially mollified subharmonic functions}
The homogenization theorem is especially useful if $d\rho_g$ is bounded.
It can be applied to cases, where $\rho$ is smeared out radially in such
a way that the mollified $\rho$ comes from a radially mollified subharmonic
function. \\
Definition. Let $\phi$ be smooth positive
function on $\RR^+$ of compact support $K_{\phi} \subset \RR^+$
satisfying $\phi(x)=\phi(1/x)$.
If $f$ is a subharmonic function, define a {\bf radially mollified function}
as $f_{\phi}(z) = \int_{\RR} f(\beta z) \; d\phi(\beta)$.
\begin{lemma}
\label{CoroLipshitz}
If $f$ is a subharmonic potential, then $f_{\phi}(z)$ is a subharmonic
potential, $\rho_{f_{\phi}}$ has the same smoothness as $\phi$ and
$|\rho_{f_{\phi}}(r)| \leq ||\phi_{\infty}|| \rho_f(K_{\phi} r)$.
\end{lemma}
\begin{proof}
A positive average of subharmonic functions $f_{\beta}(z) = f(\beta z)$
is subharmonic. \\
Taking the Laplacian in the distributional
sense on $f_{\phi}(z) = \int_{\RR} f(\beta z) \; d\phi(\beta)$ shows
$dk_{f_{\phi}} = \int_0^{\infty} dk(\beta z) \; d\phi(\beta)$ so that
$f_{\phi}$ is a potential. \\
The function
$\rho_{f_{\phi}}(r) = \int_0^{\infty} \rho(\beta r) \; d\phi(\beta)
= \int_0^{\infty} \rho(\beta^{-1} r) \; d\phi(\beta)
= \int_{-\infty}^{\infty} \tilde{\rho}(r'-\beta')
\tilde{\phi}(\beta') \; d\beta'
= (\tilde{\phi} \star \tilde{\rho})(r')$ is in logarithmic coordinates
$r'=\log(r)$ a convolution with $\tilde{\phi}(r')=\phi(\exp(r'))$ and has
therefore the same smoothness as $\phi$. \\
The estimate $|\rho_{f_{\phi}}(r)|
\leq ||\phi_{\infty}|| \rho(K_{\phi} r)$
is obtained by assuming $\rho$ to have an atom of mass $\rho(K_{\phi} r)$
at $r$ in which case we have equality.
\end{proof}
Remark. While we could also smooth $f$ using the standard smoothing theorem
in the theory of subharmonic functions, the radial smoothing will be more
convenient for the applications later on. \\
Notation. Denote by $|Y|$ the Lebesgue measure of a measurable subset $Y$
of $\RR$. For $Y \subset \RR$ and $x \in \RR$ call $\limsup_{\epsilon \to 0}
| Y \cap [x-\epsilon,x+\epsilon] |/(2 \epsilon)$ the {\bf Lebesgue density}
of $Y$ at $x$. (It is equal to $M 1_Y(x)$, where $M f$ denotes the
Hardy-Littlewood maximal function of $f$). \\
We have already used the consequence of Milloux-Schmidt \cite{Hayman} that for any
subharmonic function $f$
$\limsup_{\lambda \to 0} {\inf_{\theta}} f(\lambda \exp(i \theta)) = f(0)$.
The following Corollary is stronger and provides also a new proof to this
fact:
\begin{thm}
\label{Density}
Let $f$ be an arbitrary subharmonic function and let $z \in \CC$ given
for which $f(z)>-\infty$. For every $\delta>0$ and every line
$z+\lambda e^{i \theta}$ through $z$, the Lebesgue density
of $Y=\{ \lambda \in \RR \; | \;
f(z+\lambda e^{i \theta}) > f(z)-\delta \; \}$ at $\lambda=0$ is
equal to $1$.
\end{thm}
\begin{proof}
Because the claim is obviously true for harmonic functions, it is
enough to prove the statement for potentials
$f(z) = \int_{\CC} \log|z-w| \; dk(w)$. After a translation,
we can assume $z=0$. If $K$ is a compact subset of $\CC$,
then $dk(K)< \infty$, otherwise, $f$ would be constant $-\infty$.
We have $B_f(z) = M_f(z) - C_f(z) \to 0$ for $z \to 0$.
Let $I$ be an open ball around $0$ such that
$B_f(z) \leq \delta/3$ for $z \in I$ and therefore
$M_f(z)-f(0) \leq \delta/3$ for $z \in I$.
Define $g(z) = (f(z)+f(-z))/2$. We will show that the Lebesgue density of
$Y=\{ \lambda \; | \; g(\lambda e^{i \theta}) \geq g(0)-\delta/3 \}$
is $1$ near $0$. This implies the claim because in $Y \cap I$,
$g(\lambda e^{i \theta}) > g(0)-\delta/3 = f(0)-\delta/3$, implies
together with $M_f(-\lambda e^{i \theta}))-f(0) \leq \delta/3$ that
$f(\lambda e^{i \theta}) = 2 g(\lambda e^{i \theta})
- f(-\lambda e^{i \theta}) \geq 2 f(0)-2 \delta/3 - (f(0)- \delta/3)
= f(0)-\delta$. \\
Let $\phi$ be a radial mollifier function with support in $[1/2,2]$
and let $g_{\phi}$ be the corresponding radially mollified
subharmonic potential.
Applying Corollary~(\ref{CoroLipshitz}) and the Homogenization
theorem, we get
$$ A_{g_{\phi}}(z) \leq
\Kcal_1(\rho_{\phi})(r) \leq 2 ||\rho_g|| \rho([r/2,2r]) r $$
and the right hand side is $\leq \delta/3$ for small enough $r$. \\
Because $\sup_{\beta \in [1/2,2]} B_{g}(\beta r) \to 0$ for $r \to 0$
and $A_{g}(\beta r) \leq \delta/3$ for a set of $\beta \in [1/2,2]$
which reaches Lebesgue measure $3/2$ for $r \to 0$,
the normalized Lebesgue measure of parameters
$s \in [r/2,2r]$ with $A_{g}(s) \leq \delta/3$ approaches $1$ for
$r \to 0$.
\end{proof}
Remark. It was already known that if $f(z)>-\infty$,
then for almost all $\theta$
$\lim_{r \to 0} f(z+r \exp(i \theta)) \to f(z)$. \\
This result shows that for "most" points $w$ near $z$, $f(w)$ is near
$f(z)$. A subharmonic function $f$ behaves in
a probabilistic sense like a continuous function: if a sequence
$z_n \to z$ is chosen at 'random' in the product probability space of a nested
sequence of open intervals $I_n \subset \RR$ around $0$
(all equipped with the normalized Lebesgue measure and
$\bigcap_n I_n = \{0\})$ , then with probability $1$,
a point $x=\{x_n\}$ in the product probability space satisfies
$f(z + e^{i \theta} x_n) \to f(z)$. For example, if $f$ is the
at $0$ discontinuous subharmonic function
$f=\sum_{n=1}^{\infty} 2^{-n} \log|z-2^{-n}|$ and $x_n$ is an
independent, identically distributed sequence of random
variables with uniform distribution in $[-1,1]$ then
$f(x_n/n) \to f(0)$ with probability $1$.
\section{Lyapunov exponents of matrix cocycles}
Let $(X,\Acal,m)$ be a probability space and let $T: X \mapsto X$
be a measure preserving invertible transformation.
Given $g \in L^{\infty}(X)$ for which
$\int_{X} \log|g(x)| \; dm(x)$ is finite.
Given $E,\lambda \in \RR$, the matrix-valued map
$A_{\lambda}(x) = \left( \begin{array}{cc} E- \lambda g(x) & -1 \\
1 & 0 \end{array} \right)$
defines a cocycle
$(n,x) \mapsto A_{\lambda}^n(x)
= A_{\lambda}(T^{n-1}x) \cdots A_{\lambda}(x)$.
Denote by $\mu(A_{\lambda})
= \lim_{n \to \infty} n^{-1} \int_X \log||A_{\lambda}^n(x)|| \; dm(x)$ the
Lyapunov exponent of $A_{\lambda}$.
Define $f(\epsilon) = \mu(B_{\epsilon})$, where $\epsilon=1/\lambda$ and
$B_{\epsilon}(x) := \left( \begin{array}{cc} \epsilon E+ g(x) & -\epsilon \\
\epsilon & 0
\end{array} \right)$.
We have $\mu(A_{\lambda}) = f(\epsilon) - \log|\epsilon|
= f(\epsilon) + \log|\lambda|$. The aim is to estimate the subharmonic
function $\epsilon \mapsto f(\epsilon)$ from below,
using $f(0) = \int_{X} \log|g(x)| \; dm(x) > - \infty$. The later will be
a standing assumption in this section.
\begin{lemma}
\label{subharmoniccheck}
$f$ is a subharmonic function which is a potential: there exists a
constant $C$ such that
$f(\epsilon)= C+\int_{\CC} \log|\epsilon-w| \; dk(w)$.
The total mass of the Riesz measure $dk$ is $1$. If there exists
a measure preserving involution $S$ on $X$ such that $g(x)=-g(S(x))$
for almost all $x \in X$, then $f(\epsilon)=f(-\epsilon)$.
\end{lemma}
\begin{proof}
It is an observation of Herman that
the Lyapunov exponent of an analytically parameterized cocycle $A$
is subharmonic as the $\liminf$ of subharmonic functions
$f_n(\epsilon) = 2^{-n} \int_X \log||A_{\epsilon}^{2^n}(x)|| \; dm(x)$
satisfying $f_n\leq f_m$ for $n \leq m$. \\
By the Riesz decomposition theorem, the Lyapunov exponent can be written
as $f(\epsilon) = \int_{\CC} \log|\epsilon-w| \; dk(w) + h(\epsilon)$,
where $\epsilon \mapsto h(\epsilon)$ is harmonic. \\
With the constant $C=\mu(B_{0})$, we have for $\epsilon \to \infty$
$f(\epsilon) = C + \log|\epsilon| + O(1)$. This shows that the order
of $f$ (defined as $\limsup_{r \to \infty} \log|M_f(r)|/\log|r|$) is
equal to $1$ implying that the harmonic function $h$ in the Riesz
decomposition theorem must be constant (see \cite{Hayman}).
If $g(x)=-g(S(x))$, the pointwise Lyapunov exponent
$\mu(\lambda,x)=
\lim_{n \to \infty} n^{-1} \log||A_{\lambda}^n(x)||$ (which exists
according to Kingman's subadditive ergodic theorem $m$-almost everywhere),
satisfies \linebreak
$\mu(-\lambda,S(x)) = \mu(\lambda,x)$ so that the average over $X$
satisfies $\mu(-\lambda) = \mu(\lambda)$ implying $f(\epsilon)=f(-\epsilon)$.
\end{proof}
\begin{thm}
For a fixed dynamical system, there exists a sequence
$\lambda_m \to \infty$ such that
$$ \limsup_{m \to \infty}
\mu(A_{\lambda_m}) - \log|\lambda_m|
- \int_{X} \log|g(x)| \; dx \geq 0 \; . $$
\end{thm}
\begin{proof}
This is consequence of Corollary~(\ref{Milloux-Schmidt}) and
the homogenization theorem.
\end{proof}
Example. For $g(x)=2x+2 \cos(x)$ we obtain
$\limsup_{m \to \infty} \mu(A_{\lambda_m}) - \log(|\lambda_m|/2) \geq 0$.
\begin{thm}
Let $\phi$ be a radial mollifier function. Then
$$ \int_0^{\infty} \mu(A_{\lambda \beta}) \; d\phi(\beta)
\geq \log|\lambda| + \int_{X} \log|g(x)| \; dx
- 2 ||\phi||_{\infty}/\lambda \; . $$
Let $\phi$ be a radial mollifier function such that
$\rho_{\phi}(s) \leq M_{\phi}$ for almost all $s \in [r/2,2r]$. Then
$$ \int_0^{\infty} \mu(A_{\lambda \beta}) \; d\phi(\beta)
\geq \log(|\lambda|) + \int_{X} \log|g(x)| \; dx
- 2 M_{\phi}/\lambda - 2 c\; , $$
where $c=\rho(\CC)$.
\end{thm}
\begin{proof}
The statements follow from the Homogenization theorem,
Corollary~(\ref{CoroLipshitz}) and the estimates
in Proposition~(\ref{Lipshitz}) rsp. Proposition~(\ref{Lipshitz2})
which can be applied using Lemma~(\ref{subharmoniccheck}).
\end{proof}
Example. For $g(x)=2 \cos(x)$, where $S(x)=x+\pi$ produces
the symmetry $f(\epsilon)=f(-\epsilon)$, this gives
$\int_{1/2}^2 \mu(A_{\lambda \beta}) \; d\phi(\beta)
\geq \log(|\lambda|/2) - 2 |\phi|_{\infty}/|\lambda|$. \\
The set of parameters for which the cocycle has positive Lyapunov exponent
reaches full Lebesgue density at $\lambda \to \infty$. This follows from
Theorem~(\ref{Density}). We state this here in a more
quantitative way:
\begin{coro}
There exists a constant $C$ such that for all $\lambda_0$
the Lebesgue measure of parameters
$\lambda \in [\lambda_0/2,2 \lambda_0]$ for which $\mu(A_{\lambda})=0$ is
$\leq C (3\lambda_0/2)/\log|\lambda_0|$.
\end{coro}
\begin{proof}
For $\lambda \in
[\lambda_0/2,2 \lambda_0]$ we have
$\mu(A_{\lambda}) \leq c_1 + \log(|\lambda_0|) + C_1/|\lambda_0|$
and $\int_{1/2}^2 \mu(A_{\lambda \cos}) \; d\phi(\beta)
\geq c_2+\log(|\lambda_0|) - C_2/|\lambda_0|$. Therefore, for each
$\lambda_0$ the normalized Lebesgue measure $x$ of the subset
$\{\lambda \in [\lambda_0/2,2 \lambda_0] \; |
\; \mu(A_{\lambda}) = 0 \; \}$ satisfies
$(1-x) \geq
(c_1+\log(|\lambda_0|) - C_2/|\lambda_0|)/
(c_2+\log(|\lambda_0|) + C_1/|\lambda_0|)$ which implies that there
exists a constant $C$ such that
$x \leq C/\log(\lambda_0)$ for some $C$. To get the Lebesgue measure,
we have to multiply this by $|[\lambda_0/2,2 \lambda_0]| = 3\lambda/2$.
\end{proof}
Theorem~(\ref{Density}) has an application in spectral theory:
\begin{coro}
Assume the transfer cocycle $A_E$ of a discrete ergodic
selfadjoint Jacobi matrix has a positive Lyapunov exponent at some energy
$E \in \RR$ for which the spectrum $\sigma(L)$ has positive Lebesgue density
$\limsup_{\epsilon \to 0}
| \sigma(L) \cap \{ [E-\epsilon,E+\epsilon] \} |/(2 \epsilon)>0$.
Then $L$ has some spectrum which is not absolutely continuous.
\end{coro}
\begin{proof}
We apply Theorem~(\ref{Density}) to the complex parameterization
$E \mapsto A_E$. Let $S= \{ E \in \RR \; | \; \mu(A_E)>0 \}$.
If $\mu(A_E)>0$, then
$\limsup_{\epsilon \to 0}
| S \cap \{ [E-\epsilon,E+\epsilon] \}|/(2 \epsilon) = 1$. Therefore
$S \cap \sigma(L)$ must have positive Lebesgue measure.
Ishii-Pastur-Kotani theory tells that on the set $S \cap \sigma(L)$
the spectrum is not absolutely continuous.
\end{proof}
{\bf Acknowledgments.} This research was supported by the
the Swiss National Science Foundation. Thanks both to the department
of Mathematics and the Institute for Fusion Studies at the
University of Texas for hospitality.
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