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\begin{document}
\author{M Krishna \\ Institute of Mathematical Sciences \\ Taramani Chennai 600 113 \\ India}
\title{Regularity of the Interband Light Absorption Coefficient}
\maketitle
\date
\begin{abstract} 
In this paper we consider the Interband Light Absorption Coefficient (ILAC), in a symmetric form,  in the case of random operators on the 
$d$-dimensional lattice.  
We show that the symmetrized version of ILAC is either continuous or has a component which has the same modulus of continuity as the density of states.
\end{abstract}

\section{Introduction}
In the theory of disordered systems, one of the quantities that is widely studied is the integrated
density of states, whose continuity properties and its behaviour near band edges (Lifshitz tails) were
of great interest.  Another quantity that is of interest is the interband light absorption coefficient
ILAC, which is an important quantitative characteristic of semiconductors.

When photons with sufficient energy are incident on a pure semiconductor crystal
absorption of photons takes place with simultaneous creation of electron-hole
pairs, which means excitation of electrons from valance band to the conduction
band. This process is intrinsic interband absorption. The threshold electron
energy required is related to the basic band gap.  The absorption coefficient
increases above the threshold.

To get the absorption coefficient one considers 
the transition of an electron between states in the same or a different band. 
The rate of absorption is then calculated using perturbation theory and
the Fermi golden rule. 

The theory of interband light absorption can be found in a book such as
\cite{balwal}. 

On the other hand the presence of impurities cause electronic states to be 
produced in the forbidden band and this reduction of the band gap and 
the associated effect on the inter band light absorption coefficient 
is discussed in \cite{Eff}.

In experimental studies the absorption coefficient is a means to study the band gaps at different temperatures for a given material.

In mathematical terms this means that when there is a periodic potential
(=pure crystal) there are bands and gaps and when one adds random potential
to such a periodic back ground, spectrum extends to into the original gaps.

The literature on the density of states is vast so we refer the reader to 
the books \cite{CarLac}, \cite{CFKS}, \cite{IvaVes} and the review \cite{KirMet}.  
The continuity properties of the density of states and its Lifshitz tails
behaviour in various models is widely understood.  The physics literature is 
abound with works on the ILAC starting from \cite{Eff} and for example 
\cite{AtKarSar}.  On the other hand rigorous work in this area seem to be minimal,  see for example \cite{KirPas1}, \cite{KirPas2}, \cite{KhoKirPas}, 
\cite{KirPasStor}.  

We consider a borel probability space $(\Omega, \bb, \PP)$ with $\ZZ^d$ 
acting on $\Omega$ such that $\PP$ is invariant and ergodic with respect 
to this action.  Let $V : \Omega \rightarrow \RR^{\ZZ^d}$ 
(so that each $V(n)$ is measurable).     
We consider a self-adjoint operator $H_0 = \Delta$ (given in equation (\ref{eqn01}) ) on $\ell^2(\ZZ^d)$ and  
consider the family of operators, 
\begin{equation}\label{themodel}
H_\omega^\pm = H_0 \pm V^\omega, ~ (V^\omega u)(n) = V^\omega(n) u(n), ~ u \in 
\ell^2(\ZZ^d),
\end{equation}
such that $V^\omega$ are covariant in the sense made precise in Hypothesis (\ref{hyp1}) below.

We denote by $\delta_n$ the elements of the standard basis of $\ell^2(\ZZ^d)$
in the rest of the paper.

We define the density of states measures $n_{\pm}$ associated with 
$H_\omega^\pm$ by
\begin{equation}\label{eqn000}
n_\pm = \EE \left( \langle \delta_0,  E_{H_\omega^\pm}(\cdot) \delta_0\rangle \right).
\end{equation}

Suppose $H_{\omega, \Lambda}^\pm$ are the restrictions of 
$H_\omega^\pm$ to $\ell^2(\Lambda)$, where $\Lambda \subset \ZZ^d$ is
a finite set (usually taken to be a lattice cube centred at the origin) 
and $\lambda_\pm, u_{\lambda_\pm}$ are eigenvalues and eigen functions of $H_{\Lambda, \omega}^\pm$.  

Then the Interband Light Absorption Coefficient $A_{\Lambda}$ for such finite volume models
can be defined by taking the measure $\rho_{\Lambda, \omega}$ as given below.
 
\begin{equation}\label{eqn0}
\rho_{_{\Lambda, \omega}} = \frac{1}{|\Lambda|} \sum_{\lambda_\pm \in \sigma(H_{\omega,\Lambda}^\pm)} |\langle u_{\lambda_+}, v_{\lambda_-}\rangle|^2 \delta_{\lambda_+, \lambda_-},
\end{equation}  

\begin{equation}\label{eqn00}
A_{\Lambda}(\lambda) = \rho_{_{\Lambda, \omega} }(\{(\lambda_+, \lambda_-): \lambda_+ + \lambda_- \leq \lambda \} ),
\end{equation}  

The operators $H_\omega^\pm$ may be unbounded however,
the finite volume operators $H_{\Lambda, \omega}$ are symmetric finite 
dimensional matrices when $\Lambda$ is a finite set, so their eigenvalues are 
finite in number and the eigen functions $\{ u_{\lambda_\pm} : \lambda_\pm \in \sigma(H_{\omega, \Lambda}^\pm) \}$ are orthonormal respectively 
(for each sign $\pm$).  These properties show that indeed
the measure $\rho_{\Lambda, \omega}$ is a probability measure on $\RR^2$, since
$$
\sum_{\lambda_- \in \sigma(H_{\omega, \Lambda}^-)} |\langle u_{\lambda_+}, v_{\lambda_-}\rangle|^2 \delta_{\lambda_+, \lambda_-}(\RR^2) = \|u_{\lambda_+}\|^2 = 1.
$$ 
and the second sum (over $\lambda_+$ ) is normalized by the size of the set
$|\Lambda|$ which is precisely the number of eigenvalues  of  $H_{\Lambda, \omega}^+$. 

There are several earlier works for example Bellissard-van Elst - Schulze-Baldes \cite{besb}, Bouchlet-Germinet-Klein-Schenker \cite{bgks} defining
the density of states given in equation (\ref{eqn000}) as average trace 
per unit volume, namely
$$
n_\pm(\cdot) = \lim_{\Lambda \uparrow \ZZ^d} n_{\pm, \Lambda}(\cdot), ~ n_{\pm, \Lambda}(\cdot) = \frac{1}{|\Lambda|} Tr (\chi_{\Lambda}E_{H_\omega^\pm}(\cdot)).
$$

The above limit exits a.e. $\omega$, in the weak sense for measures, by using Birkoff's ergodic theorem and the expression  in equation (\ref{eqn1}) is arrived at using covariance of the spectral measures.

However more classical definitions of density of states involves taking the
operators $H_{\omega, \Lambda}^\pm = \chi_\lambda H_{\omega}^\pm \chi_\Lambda$
on $\ell^2(\Lambda)$, considering the average spectral measure, counting
multiplicity, 
$$
n^\pm(\cdot) = \lim_{\Lambda \uparrow \ZZ^d} n_{\omega,\Lambda}^\pm(\cdot), ~ n_{\omega,\Lambda}^\pm(\cdot) = \frac{1}{|\Lambda|} \sum_{\lambda \in \sigma(H_{\omega, \Lambda}^\pm)} \delta_\lambda
$$
and taking their limits.  In the limit both these definitions agree with that
given in equation (\ref{eqn000}).  

It is well known that for the models such as the one considered in equation
(\ref{themodel}), $n_\pm = n^\pm$. 

The trace per unit volume definition also allows one to define the limits a.e.
$$
\lim_{\Lambda \uparrow \ZZ^d} \frac{1}{|\Lambda|}Tr(\chi_\Lambda A_\omega B_\omega \chi_\Lambda)
$$ 
for a pair of covariant operators, satisfying an integrability condition, 
which results in the limit being equal to $\EE (\langle \delta_0, A_\omega B_\omega \delta_0\rangle)$ and one also gets, 
$$
\EE (\langle \delta_0, A_\omega B_\omega \delta_0\rangle) 
= \EE (\langle \delta_0, B_\omega A_\omega \delta_0\rangle).
$$

In the case of continuous models (i.e. models on $L^2(\RR^d)$) 
Kirsch-Pastur obtained in Theorem 2.1(i) \cite{KirPas1},   
limits of the finite dimensional ILACS, using sub additivity
properties of such finite dimensional quantities and they did not
have to use the "trace per unit volume" definition.


However in the present case the existence of such a limit is unclear for
the quantities defined in equation (\ref{eqn0}), when $\Lambda \uparrow \ZZ^d$.
It would be nice to show such a result, to do which one might need to
show that for nice functions $f$,
$$
\lim_{\Lambda \uparrow \ZZ^d} \frac{1}{|\Lambda|} Tr(f(\chi_\Lambda H_\omega \chi_\Lambda)) = \lim_{\Lambda \uparrow \ZZ^d} \frac{1}{|\Lambda|} Tr(\chi_\Lambda f( H_\omega )\chi_\Lambda ).
$$
in other words obtain  Szego type asymptotics.  We do not attempt this here
since this is not the main aim of the paper. 

Therefore we take directly the definition given in equation (\ref{eqn1})
 of the correlation measure in the infinite lattice case  and define 
the ILAC as a marginal in direct analogy with
the finite volume case, these definitions are consistent with the ones
obtained in the continuous case.

The main theorems of this paper are Theorems \ref{thm2} and \ref{thm4}. 
Theorem \ref{thm2} obtains estimates on the correlation measure of balls of 
radius $a$ given in terms of the uniform modulus of continuity of the 
density of states.  

So taking the correlation measure $\rho$ as in 
equation (\ref{eqn1}) and denoting the density of states as $n$,  if $\rho((a, b)\times \RR) \leq |b-a|^\alpha$,
uniformly for all $(a,b)$,then our theorem is that 
$\rho(\{x \in \RR^2: |x-b| \leq r \}) \leq 
c r^{\alpha/2}$.  So the regularity along a line through the origin
 says something for the
whole measure, but not enough to conclude the regularity of marginals
along other lines.  (This is a general fact valid for any finite
borel measure on $\RR^n$, regularity along one line through the origin implies 
some regularity for the whole measure. )  

As an example even if the density of states are absolutely continuous,
it will only imply that the measure $\rho$ is $\half$-H\"older continuous
on $\RR^2$ and this still leaves room for the measure to be supported
on some lines, which means some marginals of the measure could have
atomic components.

This is the main obstruction to obtaining regularity
of the ILAC, which happens to be the distribution function of 
a marginal of the measure $\rho$ along
a diagonal direction.  However, given that $\rho$ has some modulus of 
continuity, it follows that, if along some direction it has a component that 
is not continuous at all, such a component should have continuity along an 
orthogonal direction.

This is the feature we exploit for Theorem \ref{thm4}.

In Theorem \ref{thm4} we address the above question, observing that in
the case when the operators $H_\omega^\pm$ have some further symmetry
the marginals defined along the two orthogonal directions $\{\lambda \in \RR^2 : \lambda_1 = \lambda_2\}$
and $\{x: x_1 = - x_2\}$ actually agree.  Therefore we take a symmetric
definition of the ILAC and work with it.  In view of the comments made above,
such a symmetric definition enables us to conclude some regularity of some
components, though at first such a theorem seems surprising. 

\section{The Symmetric and Asymmetric ILAC}

In this section we define the Interband Light Absorption Coefficient in 
analogy with the case of continuum models using a correlation
measure.  We argue that in some cases
when the spectra of the operators in question have some symmetry properties,
the ILAC can be taken to be the distribution function of the average of the
marginals of the correlation measure along two diagonal lines. 
the 

We denote by $U_i, i\in\ZZ^d$ the unitary operators $(U_iu)(n) = u(n-i), ~ u \in \ell^2(\ZZ^d).$ 

\begin{hyp}\label{hyp1} 
\begin{enumerate}
\item (Covariance) The potential $V^\omega$ satisfies
$U_i^* V^\omega U_i = V^{T_i\omega}$, where $T_i\omega(n) = \omega(n+i)$.
\item There is a bijection $R$ of $\Omega$ to itself such that 
$V^{R\omega} = - V^\omega$ and $\PP$ is invariant under $R$. 
\item The operators $H_\omega^\pm$ are self-adjoint with
a common dense domain for a set of full measure in $\omega$.
\item The density of states measures $n_\pm$ are continuous. 
\end{enumerate}
\end{hyp}


\begin{exam}\label{exam0}
Here are two extreme examples of operators satisfying the
above conditions. Of course there are many more of various varaities.
\begin{enumerate}
\item The Anderson Model:
\begin{equation}\label{eqn01}
V^\omega(n) = \omega(n),  ~ (\Delta u)(n) = \sum_{|i|=1} u(n+i), ~ ~ u \in \ell^2(\ZZ^d)
\end{equation}
and $R\omega = - \omega$ and $\PP =\times \mu$ with a probability measure
$\mu$ on $\RR$. If $\mu$ is continuous, then the density of states
is continuous. We take $\mu$ to satisfy $\mu(B) = \mu(-B)$ for all borel
subsets of $\RR$ and take $R \omega = -\omega.$  Then $\PP$ is invariant 
under $R$.  
\item The Almost Mathieu model:
Take $d=1$ and take $\Omega = \TT$, $V^\omega(n) = \lambda \cos(\alpha n + \omega)$, $R\omega = \omega + \pi$ and $\PP$ the rotation invariant measure on 
$\TT$.  The density of states of this model is absolutely continuous, when
$\alpha$ is not rational and for $|\lambda| \neq 2$, see \cite{AviDam}.
\end{enumerate}
\end{exam}

\begin{rem}\label{rem1}
We note that, using the definition of $H_\omega^\pm$ and $V^\omega$ and the bijection $R$ mentioned in the Hypothesis \ref{hyp1},  that 
\begin{equation}
\label{eqn5}
H_\omega^- = H_{R\omega}^+, ~ ~ H_\omega^+ = H_{R\omega}^-.  
\end{equation}
Therefore if $\PP$ satisfies Hypothesis \ref{hyp1} (2), then for any integrable
function $f$ of $\omega$, we have
$$
\EE \left( f(\omega) \right)  = \EE \left( f(R\omega) \right).
$$
\end{rem}

The immediate consequence of our hypothesis is the equality of spectra
of $H_\omega^\pm.$

\begin{thm}\label{thm00}
Let $H_\omega^\pm$ be as in Hypothesis \ref{hyp1}.  Then we have
$$
\sigma(H_\omega^+) = \sigma(H_\omega^-), ~ a.e. \omega.
$$
\end{thm}
\pf Under the assumptions of Hypothesis \ref{hyp1}, it is well known that the
spectrum of the associated operators $\sigma(H_\omega^\pm)$ are constant
sets almost everywhere (Proposition V.2.4, Carmona-Lacroix \cite{CarLac}).
The Hypothesis \ref{hyp1}(2)
implies that $H_\omega^+ = H_{R\omega}^-$ and also that the support of 
$\PP$ is invariant under $R$.  Therefore we have
$$
\sigma(H_\omega^+) = \sigma(H_{R\omega}^+) = \sigma(H_\omega^-), ~ a.e. \omega,
$$
proving the result. \qed

We consider $H_\omega^\pm$ as in equation (\ref{themodel}), their  
spectral measures $E_{H_\omega^\pm}$ and define the measure $\rho$ as follows.
\begin{equation}\label{eqn1}
\rho = \EE \left( \langle \delta_0, E_{H_\omega^+}(\cdot) E_{H_\omega^-}(\cdot) \delta_0\rangle \right)
\end{equation}
on $\RR^2$. Let 
$$
\ii = \RR \cup \{(a, b]: a, b \in \RR\} \cup \{(a, \infty) : a \in \RR\} \cup  \{(-\infty, a] : a \in \RR\}.$$  This collection of sets forms a boolean semi-algebra on $\RR$.  We then consider the boolean semi algebra $\ii \times \ii$ 
and there define the set function $\rho$ by
$$
\rho(\cup_{i=1}^k I_i\times J_i) = \sum_{i=1}^k \EE \left( \langle \delta_0, E_{H_\omega^+}(I_i) E_{H_\omega^-}(J_i) \delta_0\rangle \right), ~ I_i, J_i \in \ii,
$$
where the $\{I_i\times J_i\:i=1,\dots,k\}$ are mutually disjoint rectangles.  Then this
$\rho$, takes values in $[0,1]$ and satisfies $\rho(\RR \times \RR) =1$.
The positivity of $\rho$ follows from Proposition \ref{prop1} (2), below
and since intersection of rectangles of the form considered are again rectangles of the same form, $\rho$ is also seen to be well defined.
Hence it extends to a unique probability measure on the boolean algebra
generated by $\ii\times \ii$, see Exercises 1.4.4 - 1.4.6 and Proposition 1.4.7
of \cite{Krp}.
The unique extension of this to a probability measure on the borel
$\sigma$-algebra on $\RR^2$ is again standard, see Proposition 2.5.1 \cite{Krp}.)

\begin{prop}\label{prop1}
Consider the operators $H_\omega^\pm$, with $\omega \in supp (\PP)$ and let
$\rho$ be as in equation (\ref{eqn1}). Then for any borel subsets $B, C$ of $\RR$,
\begin{enumerate}
\item 
$\rho(B\times C) = \EE \left( \langle \delta_0, E_{H_\omega^-}(C) E_{H_\omega^+}(B) \delta_0\rangle \right)$
\item 
$\rho(B\times C) = \EE \left( \langle \delta_0, E_{H_\omega^-}(C) E_{H_\omega^+}(B) E_{H_\omega^-}(C)\delta_0\rangle \right)$
\item 
$\rho(B\times C) = \EE \left( \langle \delta_0, E_{H_\omega^+}(B) E_{H_\omega^-}(C) E_{H_\omega^+}(B)\delta_0\rangle \right)$
\item The following inequalities are valid
$$
\rho(B\times C) \leq n_+(B), ~ \rho(B \times C) \leq n_-(C).
$$
\end{enumerate} 
\end{prop}
\pf When we consider operators $H_\omega^\pm$ satisfying the Hypothesis \ref{hyp1}, they form a covariant family of operators in the sense of Hypothesis 1 of \cite{KirKri} (taking $G=L=\ZZ^d$ there). Then the proof of this proposition is 
the same as that given in Proposition 1 of \cite{KirKri}, so we omit it. \qed

To define the ILAC we need to look at the marginal of the measure $\rho$ along
the diagonal directions $\{(\lambda_1, \lambda_2): \lambda_1 =\pm \lambda_2\}$.
We rotate the coordinate axes of $\RR^2$ so that these directions form 
the coordinate axes and to enable this we define the rotation $T$ and look
at the measure $\rho$ from this new perspective.  

The marginals of $\rho \circ T$ along the coordinate axes are precisely the
marginals of $\rho$ along the diagonal directions.

Let $T$ be a transformation from $\RR^2$ to itself given by the matrix 
$$ \frac{1}{\sqrt{2}}\left(
\begin{matrix}
1 & 1 \\ 1 & -1
\end{matrix} \right).
$$
Then $T$ is an orthogonal matrix with $T = T$ and we have 
$$T\left(\begin{matrix}\lambda_1\\ \lambda_2\end{matrix}\right) = \left(\begin{matrix}\frac{\lambda_1 + \lambda_2}{\sqrt{2}}\\ \frac{\lambda_1 - \lambda_2}{\sqrt{2}}\end{matrix}\right).$$

Using these we define:
\begin{defin}\label{defi0} We consider the measure $\rho$ defined in equation
(\ref{eqn1}) and set the  asymmetric ILAC as: 
\begin{equation}
\label{eqn2}
A_{as}(\lambda) = \nu((-\infty, \lambda]), ~ \mathrm{where} ~  
\nu(B) = \rho \circ T (B\times \RR) 
\end{equation}
and the symmetric ILAC as :
\begin{equation}
\label{eqn3}
A_{s}(\lambda) = \sigma((-\infty, \lambda]), ~ \mathrm{where} ~  
\sigma(B) = \half \left(\rho \circ T (B\times \RR) + \rho \circ T (\RR \times B)\right). 
\end{equation}
\end{defin}

In the above definitions and in our model we have dropped all the physical 
constants and also 
have dropped the band gap $E_g$ that customarily appears in the definition
since they play no role in the regularity properties as seen in the proofs
of our theorems.  

The reason we consider a symmetrized version of ILAC is that, in the case of
disordered models where the spectrum is symmetric about $0$, 
if $\lambda$ is in the spectrum then $-\lambda$ is also in the spectrum.
Therefore given a $E$ we can have $\lambda^+ + \lambda^- = E$ 
and also $\lambda_+ - (\tilde{\lambda}_-) = E$ (of course $\tilde{\lambda_-}$
would be $- \lambda_-$).  Therefore in the definition of the finite
distribution functions in equation (\ref{eqn00}) we could also have taken
the sum over $\lambda_+ - \lambda_- \leq E$.  The distribution functions, 
however differ for these two different definitions.  Therefore it might be
more meaningful to take a symmetric definition.  

%It is not hard to see that the $A_{as}(\lambda)$ and $A_s(\lambda)$ are the limits of the
%distribution functions 
%$$A_{as}(\lambda) = \lim_{\Lambda \rightarrow \ZZ^d} \rho_{_{\Lambda, \omega}}(\lambda_+ + \lambda_- \leq \sqrt{2} \lambda)$$ 
%and
% $$A_{s}(\lambda) = \lim_{\Lambda \rightarrow \ZZ^d} \half \left(\rho_{_{\Lambda, \omega}}(\lambda_+ + \lambda_- \leq \sqrt{2} \lambda) +
%\rho_{_{\Lambda, \omega}}(\lambda_+ - \lambda_- \leq \sqrt{2} \lambda) \right).$$
 
\section{Regularity Properties}

In this section we show a regularity of a symmetrized ILAC.  The idea behind the
proofs is the following.  The ILAC is the distribution function of the  marginal
of a two dimensional measure taken along the principal diagonal, 
while the marginals along the coordinate axes are the density of states.
This measure $\rho$ itself acquires a part of the regularity of the density of 
states. However even if $\rho$ is smooth, it is possible for marginals
along some directions have atomic components.  

This is possible only if the measure itself has its
support (not the topological support) $\Sigma_1 \cup \Sigma_2$, with $\Sigma_1$ being a subset of a straight line which is disjoint from $\Sigma_2$.  
If this happens
then restricted to the straight line containing $\Sigma_1$, the measure must be as 
regular as the density of states.  This is precisely our conclusion.

\begin{lemma}\label{lem1}
Consider $H_\omega^\pm$ satisfying Hypothesis \ref{hyp1}.  Then, $n_+ = n_-$ and in this case
$\rho$ is symmetric, i.e. $\rho(A\times B) = \rho(B \times A)$.
\end{lemma}
\pf The hypothesis \ref{hyp1}(2) says that for any integrable function $f$, $\EE(f(\omega)) = \EE(f(R\omega))$.
Therefore taking $f(\omega) = \langle \delta_0, E_{H_\omega^+}(B) \delta_0\rangle$, 
for a fixed borel set $B$, we see that it is integrable and satisfies $f(R\omega) = 
\langle \delta_0, E_{H_\omega^-}(B) \delta_0\rangle$.
Therefore $n_+(B) = \EE(f(\omega)) = \EE(f(R\omega)) = n_-(B)$.  This being valid for any borel set
$B$ the measures $n_+$ and $n_-$ agree.  

The symmetry of $\rho$ follows from the following equalities, using the invariance of $\PP$ under $R$. 
\begin{equation}\label{eqn7}
\begin{split}
\rho(B\times C) &= \EE \left( \langle \delta_0, E_{H_\omega^+}(B) E_{H_\omega^-}(C) \delta_0\rangle \right) \\
 &= \EE \left( \langle \delta_0, E_{H_{R\omega}^-}(B) E_{H_{R\omega}^+}(C) \delta_0\rangle \right) \\
 &= \EE \left( \langle \delta_0, E_{H_{\omega}^-}(B) E_{H_{\omega}^+}(C) \delta_0\rangle \right) = \rho(C\times B). 
\end{split}
\end{equation}


In the following we shall denote the marginals of $\rho \circ T$ by,  
\begin{equation}\label{eqn9}
\nu_1 = \frac{1}{2} \rho \circ T (B \times \RR), ~ \nu_2 = \frac{1}{2} \rho \circ T (\RR \times B). 
\end{equation}
Then, clearly
\begin{equation}\label{eqn91}
A_{as}(\lambda) = 2 \nu_{1}((-\infty, \lambda]) ~ \mathrm{and} ~ 
A_{s}(\lambda) = (\nu_{1} + \nu_2) ((-\infty, \lambda]). 
\end{equation}

\begin{rem}\label{rem2}
The measure $\rho$ is quite nice and we can say more about it.  We shall denote by $B_a(x)$ 
a ball of radius $a$ with centre $x \in \RR^2$. We denote by $\kappa$ the marginal $\rho(\cdot \times \RR)$ and note that $\kappa = n_+$.  In the case when $\rho(B\times \RR) = \rho(\RR \times B)$ for all borel $B$, then we have $\kappa(B) = n_+(B) = n_-(B)$, from the definitions of $n_\pm, \rho$ and $\kappa$.
\end{rem}

\begin{defin}\label{def1}
Given a probability measure $\mu$ and a bounded continuous function $h$ on $[r, \infty)$, positive
on $(r, \infty)$ and  vanishing at $r$, we say
that $\mu$ has modulus of continuity $h$ at a point $x$ if
$$
\limsup_{a >0} \frac{ \mu(x-a, x+a) }{h(a+r)} < \infty.
$$
We say that $\mu$ is uniformly $h$-continuous if the above condition is valid independent of $x$.
\end{defin}

\begin{exam}\label{exam1}
\begin{enumerate}
\item Let $r=0$, $h(x) =x^\alpha , 0 \leq x \leq 1, h(x) = 1, x >1$ for some $0 < \alpha \leq 1$. 
Then $h$-continuity of $\mu$ for this $h$ is called $\alpha$-H\"older 
continuity.
\item If $r=1$ and $h(x) = |(\ln(x))^{-\alpha}|, 0 \leq x \leq 1/2$ 
and some positive bounded continuous function on $(1/2,\infty)$ then 
$h$-continuity for this $h$ is called $\alpha$-log H\"older continuity.  
\item Let $r=0$. Let $h(a) = \tau\left( (y-a, y+a) \right), y \in \RR$, for
a probability $\tau$, then $h$-continuity with this $h$ means
the modulus of continuity of $\mu$ at $x$ is the same
as that of $\tau$ at $y$.
\end{enumerate}
\end{exam}

\begin{rem}\label{rem3}
In the theorems below we will only present the case when $r=0$, the theorems
easily follow even when we take $r \neq 0$ by taking $h(\cdot +r)$ to replace
$h$.  
\end{rem}

\begin{thm}\label{thm2}
Consider $H_\omega^\pm$ satisfying Hypothesis \ref{hyp1}.  Suppose the density of states $n=n_+=n_-$
is uniformly $h$-continuous for some $h$ as in Definition \ref{def1} with 
$r=0.$ Then, if $B_a(x)$ is a ball of radius $a$ centred at $x$, 
$$\limsup_{a\rightarrow 0} \frac{\rho(B_a(x))}{h(a)} < C, ~ ~ \mathrm{for ~ all }~ ~ x .$$
\end{thm}
\pf We consider the function $\psi(x) = \frac{1}{1 + \|x\|^2}, ~ x \in \RR^2$,
where $\|x\|^2 = x_1^2 + x_2^2, ~ x = (x_1, x_2)$.  Then $\psi$ is integrable with respect to the
probability measure $\rho$ on $\RR^2$. This $\psi$ satisfies
$\psi(x) \geq \half$, whenever $\|x\| \leq 1$. So taking $\delta =1$ in
Theorem \ref{thm3}, it is 
enough to show that 
$$
\limsup_{a \rightarrow 0} \frac{1}{h(a)} \int \psi_{a}(y-x) d\rho(y) < \infty. 
$$
To see this we note that 
$$
\psi_a(y-x) \leq \frac{1}{(1 + \frac{(y_1 - x_1)^2}{a^2}) }, 
$$
so that 
\begin{eqnarray}\nonumber
\frac{1}{h(a)} \int \psi_a(y-x) ~ d\rho(y)  &\leq \frac{1}{h(a)} \int \frac{1}{(1 + \frac{(y_1 - x_1)^2}{a^2}) } d\kappa(y_1) \\ &=
\frac{1}{h(a)} \int \phi_a(y_1-x_1) ~ dn(y_1),
\end{eqnarray}
where we have integrated over the variable $y_2$ on the right hand side and used the definition 
of the measure $\kappa$, remark \ref{rem2}  and have taken $\phi(y) = 1/(1 + y^2), \phi_a(y) = \phi(y/a), y\in \RR$.   

Then using Theorem \ref{thm5}, we see that the $limsup$ of the right had side is 
finite for all $x_1$ once $n$ is uniformly $h$-continuous.  Therefore the $limsup$ of
the left hand side is finite for all $x$. \qed


This theorem shows that $\rho$ has no atoms, that is for any point
$x \in \RR^2, \rho(\{x\}) = 0$. The marginals of $\rho$ along
the axes, namely
$\rho(A\times \RR)$ and $\rho(\RR\times A)$ both equal the density of states,
as seen by using equation (\ref{eqn7}) and the fact that 
$E_{H_\omega^\pm}(\RR) = I$, 
and hence are continuous if the density of states has no atoms. 

However it is possible that some marginal taken along other directions 
in $\RR^2$ may have atoms.  Consider for example a measure on $\RR^2$
supported on the $y$-axis $\{ (x_1,x_2)\in \RR^2 : x_1 = 0 \}$, then
the marginal of this measure along the $x$-axis is atomic with an atom
at the point $0$.


\begin{thm}\label{thm4} Consider $H_\omega^\pm$ satisfying 
Hypothesis \ref{hyp1} and 
suppose the density of states $n$ is uniformly $h$-continuous for some $h$ as in
Definition \ref{def1}.  

If  $\nu_1$ or $\nu_2$ defined in equation (\ref{eqn9}) has an
atom, then the function $A_{s}(\lambda)$ defined in equation (\ref{eqn91})
has a uniformly $h$-continuous component.
\end{thm}
\pf Suppose $\rho \circ T(A\times B) \neq 0$ for a pair of borel subsets
$A, B$ of $\RR$. Then $\rho \circ T(A\times \cdot)$ and $\rho \circ T(\cdot \times B)$
are both non-trivial finite positive measures on borel subsets of $\RR$. 
We also note that if $\rho \circ T(A\times \RR) \neq 0$,
for a given borel set $A$, then there must be a borel set 
$B \subsetneq \RR$, such
that $\rho \circ T (A\times B) \neq 0$. (Otherwise if $\rho\circ T(A\times C) = 0$ for all borel $C \subsetneq \RR$, then taking any $C \neq \emptyset$,
we see that $\rho\circ T(A\times \RR) = \rho\circ T(A\times C ) 
+ \rho \circ T(A\times C^c) = 0$). 
A similar statement is valid when
$\rho(\RR \times B) \neq 0$.)  

Therefore if the marginal $\rho \circ T(A \times \RR)$  
has an atom at a point $x$, then we can decompose the other marginal measure
$\rho \circ T(\RR \times B)$ as 
\begin{eqnarray*}
\rho \circ T(\RR \times B) &= \rho \circ T((\RR\setminus \{x\}) \times B) + 
\rho \circ T(\{x\} \times B). 
\end{eqnarray*}
Let $S_{\rho}$ denotes a finite subset of the set of atoms of $\rho \circ T(A \times \RR)$, then we can write 
\begin{eqnarray}\label{eqn10}\nonumber
\rho \circ T(\RR \times B) &= \rho \circ T(\RR\setminus S_\rho \times B) +
\rho \circ T(S_\rho \times B) \\ &= \rho_1(B) + \rho_2(B). \end{eqnarray}

Similarly if $S_\sigma$ is some finite subset of the set of atoms of $\rho \circ T(\RR \times B)$,
then we can write
\begin{eqnarray}\label{eqn11}\nonumber
\rho \circ T(A \times \RR) &= \rho \circ T(A \times \RR\setminus S_\sigma) +
\rho \circ T(A \times S_\sigma) \\ &= \sigma_1(A) + \sigma_2(A). \end{eqnarray}

We have for each $A, B$, the following relations, which is easy to see
from the above argument.
\begin{eqnarray}\label{eqn12}\nonumber
\rho_2(B) &= \sum_{x \in S_\rho} \rho \circ T(\{x\} \times B), \\
\sigma_2(A) &= \sum_{x \in S_\sigma} \rho \circ T(A \times \{x\}).
\end{eqnarray}
Using the definition of $A_s$ from equation (\ref{eqn91}), decomposition in equations (\ref{eqn11}, \ref{eqn12}), we can write
\begin{eqnarray}\label{eqn13}\nonumber
A_{s}(\lambda) &= \half ( \rho_1 + \sigma_1)\left( (-\infty, \lambda] \right) +
\half ( \rho_2 + \sigma_2)\left( (-\infty, \lambda] \right) \\
&= A_{s, 1}(\lambda) + A_{s, 2}(\lambda), 
\end{eqnarray}
where $A_{s,1}, A_{s,2}$ are non-zero functions, as seen by the preceding arguments. 

Now the result follows from Lemma \ref{lem2} below. \qed

\begin{lemma}\label{lem2}
Assume the conditions of theorem \ref{thm4}.  Consider the function $A_{s,2}$
defined in equation (\ref{eqn13}).  If the density of states $n$ is 
uniformly $h$-continuous for some $h$ (as in Definition \ref{def1}), 
then $A_{s,2}$ is uniformly $h$-continuous for the same $h$. 
\end{lemma}
\pf We will prove that $\rho_{2}$ is uniformly $h$-continuous, the proof
for $\sigma_2$ is similar.  From these two statements the uniform $h$-continuity of $A_{s,2}(\lambda)$ is clear.  
Let the cardinality of $S_\rho$ be $N$ and let $E_1, \dots, E_N$ be the elements
of $S_\rho$.  Then
\begin{equation}\label{eqn14}
\begin{split}
& \frac{1}{h(a)} \int \phi_a(y-x) ~ d\rho_2(y)  = \sum_{j=1}^N \frac{1}{h(a)} \int
 \phi_a(y-x) d\rho\circ T(E_j, y) \\ 
 & ~ ~ = \sum_{j=1}^N \frac{1}{h(a)} \int_{T(\{E_j\} \times \RR)} \phi_a( w -(x+E_j) ) d\rho(\frac{w}{\sqrt{2}}, \frac{w-2E_j}{\sqrt{2}}) \\ 
\end{split}
\end{equation}
Since ${T(\{E_j\} \times \RR)} \subset \RR \times \RR$, the right hand side is
bounded by 
\begin{eqnarray*}
&\sum_{j=1}^N \frac{1}{h(a)} \int_{\RR \times \RR} \phi_a( w -(x+E_j) ) ~  d\rho(\frac{w}{\sqrt{2}}, \frac{z}{\sqrt{2}}) \\
& ~ ~ = \sum_{j=1}^N \frac{1}{h(a)} \int_{\RR} \phi_a( \sqrt{2} w -(x+E_j) ) ~  dn(w),
\end{eqnarray*}
where we used the fact that $\rho(\cdot \times \RR) = n(\cdot).$ The uniform 
$h$-continuity of the density of states $n$ shows that the right hand side is bounded uniformly in $x$, proving
the lemma. \qed

\vspace{.5cm}

{\noindent \bf Acknowledgement:} I thank Prof Werner Kirsch for going through
this paper and his comments.  I thank the referee of an earlier version of the
paper for extensive and very helpful comments and point out the references
\cite{besb} and \cite{bgks}.

\section{Appendix}

We present here some results that we use in the main part, whose proofs
are essentially available elsewhere.



We have an abstract theorem that extends the theorem of Jensen-Krishna in \cite{DemKri}.
In the following let $(X, \|\cdot \|)$ be a normed vector space over complex numbers and $\rho$ a 
probability measure on $X$ with respect to
the borel $\sigma$-algebra.  Denote by $B_a(x)$ the ball with centre $x$ of radius $a$. Let $\psi$ 
be a positive bounded continuous function on $X$ taking value $1$ at $0$.  Denote by 
$\psi_a(x) = \psi(x/a), ~ a >0$.  

\begin{thm}\label{thm3}
Let $h$ be a function as in Definition \ref{def1}. Suppose 
$$
\limsup_{a > 0} \frac{1}{h(a)} \int \psi_a(y-x) d\rho(y)  <\infty,
$$ 
Then, there are constants $C$, $\delta >0$, depending upon $\psi$, such that 
$$
\rho(B_{a\delta}(x)) \leq C h(a), ~  a >0, ~ x \in X.
$$
\end{thm}
\pf Since $\psi$ is continuous and is $1$ at $0$, there is a $\delta >0$ such that 
$\psi(y) \geq \half$, whenever $\|x\| < \delta$.  So we have
$$
 \frac{1}{h(a)} \int \psi_a(y-x) d\rho(y)  \geq \frac{1}{2h(a)} \rho(B_{a\delta}(x)).
$$   
We have used the fact that $\|x\|/ a \leq \delta \iff \|x\| \leq a\delta$.  
Taking $sup$ first on the left hand side, which is finite since the $limsup$ of the left hand
side is finite by assumption,  and then taking $sup$ over $ a$, for a fixed
$\delta$ on the right hand side shows that the right hand side is finite for all $x$. 
\qed

Given a function $\psi$ satisfying:

\begin{hyp}
\label{hyp1.1}
Let $\psi$ be a continuous function on $\RR$ with
$\psi(0) = 1$ and $A_\psi = \int \psi(x) dx \neq 0$.
Further assume that
\begin{enumerate}
\item $\psi$ is bounded and positive.
\item $\psi$ is differentiable, even and satisfies
$$
|\psi(x)| + |x\psi^\prime(x)| \leq \jap{x}^{-\delta}, ~ \text{for some} ~  \delta > 1,
$$
where $\jap{x} = (1 + x^2)^{1/2}$.
\item Let $h$ be as in the definition \ref{def1}.  Let
$K(y) = \sup_{0 < a < 1} |\frac{h(ay)}{h(a)}|, y >0$.  Then
$\int K(y) |\psi^\prime(y)| dy <\infty.$
\end{enumerate}
\end{hyp}

In most cases the assumption (3) on $\psi$ above follows from (2), 
but we include it for generality.


We set, given a $\psi$  
$$
C^h_{\mu, \psi}(x) = \limsup_{a > 0} \frac{\psi_a *\mu}{h(\delta a)}(x), 
D^h_{\mu, \psi}(x) = \limsup_{a > 0} \frac{\mu\left( (x-a, x+a) \right)}{h(a)}(x), 
$$

\begin{thm}\label{thm5}
Let $\mu$ be a probability measure and let $\psi$ satisfy the
Hypothesis \ref{hyp1.1}.  Then $C_{\mu, \psi}^h$ is finite for
any $x$, iff $D_\mu^h(x)$ is finite for the same $x$.
\end{thm}

\pf We note that as in equation (1.3.4) \cite{DemKri}, we have by integration 
by parts
$$
\frac{1}{h(a)} \psi_a * \mu(x) = - \frac{1}{h(a)} \int_0^\infty \psi^{\prime}(y)~  
\frac{h(ay)}{h(a)} \frac{\Phi_\mu(x+ay) - \Phi_\mu(x-ay)}{h(ay)} ~ dy,
$$ 
where $\Phi_\mu$ is the distribution function of $\mu$. Then taking
limsup as $a \rightarrow 0$, using the condition that $K(y)\psi^\prime(y)$
is integrable by assumption we see that the finiteness of $D^h_{\mu, \psi}(x)$
implies that of $C^h_{\mu, \psi}(x)$.  

To see the other direction, note that since $\psi$ is a positive continuous
function on $\RR$, it attains a positive minimum on $[-1,1]$, say $\beta$.
Then we have the estimate
$$ 
\frac{1}{h(a)} \psi_a * \mu(x) \geq \frac{1}{h(a)} \int_{-a}^a \psi_a(y) d\mu(x+y) \geq \beta \frac{\mu( (x-a, x+a) )}{h(a)},
$$
from which we conclude that finiteness of $C^h_{\mu, \psi}(x)$ implies that
of  $D^h_{\mu, \psi}(x)$. \qed

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\end{document}
