\documentstyle[12pt]{article}
\input{mfont.tex}
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%%%%%%% MY MACROS %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% THE FOLLOWING ARE BLACK BOARD CHARACTERS%%%%%%%%%
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\def\CC{\BC}
\def\ZZ{\BZ}
\def\real{\BR}
\def\xz{X_{\ZZ}}
%%%%%%%%%%END BLACK BOARD CHARACTERS %%%%%%%%%%%%%%%%%%%%
\def\gkw{g^\omega_\lambda(n,n)}
\def\<{\langle}
\def\>{\rangle}
\def\gk{g_{\lambda}(n,n)}
\def\mpmw{m_{\pm ,n}^\omega(\lambda)}
\def\supp{{\rm supp} {\cal P}}
\def\mpm{m_{\pm ,n}(\lambda)}
\def\gke{\frac {1}{\sqrt
{(\lambda-\mu^0_0)~(\lambda-\lambda^m_0)}}~\prod_{i=1}^{\infty}\frac
{(\lambda-\xi_i(n))}{\sqrt {(\lambda-\tau_{2i-1})~(\lambda-\tau_{2i})}}}
\def\res{\sum _{i=1}^\infty c_{i,n}\frac{{\rm Res}(\xi_i(n))}{(\lambda-\xi_i(n))}}
\def\op{H^\omega}
\def\np{m_{+,n}(\lambda)}
\def\mm{m_{-,n}(\lambda)}
\def\me{\sqrt{(\lambda-\mu^0_0)(\lambda-\lambda^m_0)}\prod_{i=1}^\infty\frac{\sqrt
{(\lambda-\tau_{2i-1})(\lambda-\tau_{2i})}}{(\lambda-\xi_i(n))}}
\def\Mp{m_{+,n}(\lambda)}
\def\Mm{- m_{-,n}(\lambda) - (\lambda-\omega(n))}
\def\xmodz{{\rm X}/\xz}
\def\RR{{\cal R}}
\def\MM{\cal M}
\def\AA{{\cal A}}
\def\AB{{{\cal A}_1}}
\def\AC{{{\cal A}_2}}
\def\BB{{\cal B}}
\def\btheta{\vec{\theta}}
\def\proof{{\noindent \bf Proof: }}
\newtheorem{theorem}{\bf Theorem}[section]
\newtheorem{tf}{\bf Corollory}[section]
\newtheorem{prop}{\bf Proposition}[section]
\newtheorem{remark}{\bf Remark}[section]
\newtheorem{definition}{\bf Definition}[section]
%%%%%%% END MY MACROS %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\title{Nature of some random Jacobi matrices}
\author{ Anand J Antony \\
School of Mathematics,SPIC Science Foundation \\ 92, G N Chetty Road,
Madras 600017, INDIA \\{\small anand@ ssf.ernet.in}
\and M Krishna
\\ Institute of Mathematical Sciences \\ Taramani, Madras 600113, INDIA
\\{\small krishna@ imsc.ernet.in}}
\date{ 6 July 1993}
\begin{document}
\maketitle
\begin{abstract}
In this paper we consider a random Jacobi matrix having purely
absolutely continuous spectrum with a band structure and having
possibly finitely many accumulation points of the spectrum. We show that such
operators come from almost periodic potentials.
\end{abstract}
\section{Introduction}
In recent years, in the study of one-dimensional random Shrodinger
operators, much interest has been shown in studying the influence of the
nature of the spectrum on the randomness of the potential. Research in
this direction has been carried out in both discrete and continuous
settings. See for example the Kotani theory on the determinacy of
potentials having some absolutely continuous spectrum; Kotani \cite{kotani} for the
continuous case and Simon \cite{simon} for the discrete case. Further it was
shown by Kotani and Krishna \cite{kotani&krishna} and Craig \cite{craig} that some random
Shrodinger operators with purely absolutely continuous spectrum of a
certain type are almost periodic of necessity. Proofs for such results consisted
of setting up and solving the Dubrovin equation for some spectral
parameters and Jacobi inversion on a Riemann surface. This was done for
the continuous case by Levitan \cite{levitan}, Mc Kean and Moerbeke
\cite{mckean} and Mc Kean
and Trubowitz \cite{mckean&trubovitz}.
In the discrete case Moerbeke \cite{moerbecke} and Toda \cite{toda} have discussed the
nature of the matrices constructed via inverse spectral theory.
These turnout to be periodic. Carmona and Kotani \cite{carkot}
worked out the inverse spectral theory
of random potentials, they construct an invariant probability measure
associated with the relevant spectral functions of the random Jacobi
matrix, through approximation by periodic ones.
Kotani \cite{kot1} has shown that if
a random potential takes finitely many values and has some absolutely
continuous spectrum then it is periodic. In \cite{anand&krishna} it was shown
that for random Jacobi matrices , if the spectrum is purely
absolutely continuous and consists of finite number of bands then the
support of the random potential consists of only almost periodic
sequences. In this paper we extend this result to the case where there
are infinite number of bands with finite number of accumulation points.
In \cite{anand&krishna} the proof was via Jacobi inversion on a Riemann surface using the Riemann
theta function. But in the infinite bands case we have difficulty in
showing the convergence of the series in the appropriate definition of the
Riemann theta function unless we assume exponential decay of the gap
lengths. So we use a direct method similar to the one used
by Levitan in \cite{levitan}.
After this work was done, we came to know of the work of Oliver Knill
\cite{oliver} on the isospectral deformations of random Jacobi matrices. This
paper in particular identifies a relevent Abel-Jacobi map and answers a
question raised in \cite{oliver}, in a special case. We also received
the paper of Gesztesy, Holden,Simon and Zhao \cite{ges} who discuss the
relevent trace formulae and the inverse spectral theory in a more general
context than that concerns us here.
Subsequently this paper contains five sections.
In the first we define the random
Jacobi matrix, state the assumptions on the spectrum and state the
main theorem. In the second we describe the inverse spectral theory
needed in the proof of the theorem. In the third we define a class of
holomorphic functions on a subset of C and prove an interpolation
theorem. In the fourth we construct a Riemann surface associated
with the spectral functions and define a class of holomorphic differentials.
Then we define the Abel-Jacobi map on a Banach space X and prove
its properties. Finally we show the almost periodicity of the
potential.
{\noindent \bf Acknowledgments:} We would like to thank Prof Madhav
Nori and P N Srikanth for some useful discussions.
We especially thank Prof S Nag for reading through a previous
version of the paper and helping us with his comments and suggestions.
AJA would like to thank Prof C S Seshadri for encouragement.
\section{Preliminaries}
Consider $(\Omega,\cal B,\cal P)$, where
$\Omega$ = the space of real valued bounded sequences,
${\cal B} =$ the Borel $\sigma$ -algebra on $\Omega$
generated by the topology of pointwise convergence on it and ${\cal P} =$
a probability measure
on $(\Omega,{\cal B})$ which is
invariant and ergodic with respect to the right translation $T$ on
$\Omega$ given by $(Tu)(n) = u(n-1)$.
Corresponding to each $\omega\in\Omega$ we have a self-adjoint operator
acting on $\ell^2(\ZZ)$ given by
$$
\op =\Delta+\omega
$$
where
$$
(\Delta u)(n) = u(n-1) + u(n+1)$$
and
$$(\omega u)(n) = \omega(n) u(n);\quad u\in\ell^2(\ZZ)$$
It is a result of P astur \cite{pastur} that, under the assumptions on
$\cal P$, the spectrum
and spectral type of $\op$ are constant for almost all $\omega$, ( see
\cite{kotani} );
ie, there are subsets $\Sigma,\Sigma_{a.c},\Sigma_{p.p},\Sigma_{s.c}$
of R such that
\begin{eqnarray*}
\Sigma~=~\Sigma_{a.c}\cup\Sigma{p.p}\cup\Sigma{s.c}~~,~{\rm (disjoint~
union)}
\end{eqnarray*}
and there exists a subset ${\Omega}^\prime~{\rm of}~ \Omega$
with full measure such that for all $\omega\in{\Omega}^\prime$
$$\sigma(\op)~=~\Sigma,~{\sigma}_{a.c}(\op)~=~{\Sigma}_{a.c},
~{\sigma}_{p.p}(\op)~=~{\Sigma}_{p.p}\quad {\rm and} {\sigma}_{s.c}~=
~{\Sigma}_{s.c}$$
In this paper we we assume that $\Sigma$ = $\Sigma_{ac}$ and
\begin{eqnarray*}
\Sigma &=& \{~\cup_{i=0}^\infty[\lambda^m_{2i+1},\lambda^m_{2i}]~\}~~\cup~~
\{~\cup_{i=0}^\infty[\mu^0_{2i},\mu^0_{2i+1}]~\}\\
& & \cup_{k=1}^{m-1}\{\cup_{i=0}^\infty[\lambda^k_{2i+1},\lambda^k_{2i}]
~\cup_{i=0}^\infty[\mu^k_{2i},\mu^k_{2i+1}]\} \\
& & \cup_{k=1}^{m}\{\lambda^k_\infty\}
\end{eqnarray*}
\noindent where
$$
\lambda^k_i>\lambda^k_{i+1};~\mu^k_i<\mu^k_{i+1};~k=1,..,m-1;~i=0,..
$$
$$
\lambda^m_i>\lambda^m_{i+1};~i=1,..
$$
$$
\mu^0_i<\mu^0_{i+1};~i=1,..
$$
$$
\lambda^k_i<\mu^k_j;~i,j=1,...,~k=1,..,m-1.
$$
$$
\mu^{k-1}_i\rightarrow\lambda_\infty^k\leftarrow\lambda^k_i~{\rm as}~
i\rightarrow\infty;k=1,..,m.
$$
$$
\lambda^k_\infty<\lambda_\infty^{k+1};~k=1,..,m-1.
$$
\noindent The spectrum looks as follows:
\setlength{\unitlength}{1.8mm}
\begin{picture}(70,10)
\put(0,0){$\scriptstyle \mu_0^0$} \put(0,2) {\line(1,0){3.5}}
\put(3,0){$\scriptstyle \mu_1^0$}
\put(6,0){$\scriptstyle \mu_2^0$} \put(6.5,2){\line(1,0){2.5}}
\put(8.5,0){$\scriptstyle \mu_3^0$}
\put(11,2){\line(1,0){1.5}}
\put(13.0,2){\line(1,0){.5}}
\put(14,2){\line(1,0){.25}}
\put(14,0){$\scriptstyle \lambda^1_\infty$} \put(14.5,2){\tiny .}
\put(14.75,2){\line(1,0){.25}}
\put(15.5,2){\line(1,0){.5}}
\put(16.5,2){\line(1,0){1.5}}
\put(19,0){$\scriptstyle \lambda^1_3$} \put(19.5,2){\line(1,0){3}}
\put(22,0){$\scriptstyle \lambda^1_2$}
\put(25,0){$\scriptstyle \lambda^1_1$} \put(25.5,2){\line(1,0){3}}
\put(28,0){$\scriptstyle \lambda^1_0$}
\put(31,0){$\scriptstyle \mu^1_0$} \put(31.25,2){\line(1,0){3}}
\put(33,0){$\scriptstyle \mu^1_1$}
\put(36,0){$\scriptstyle \mu^1_2$} \put(36.5,2){\line(1,0){3}}
\put(39,0){$\scriptstyle \mu^1_3$}
\put(40,2){\line(1,0){1.5}}
\put(43,2){\line(1,0){.5}}
\put(44,2){\line(1,0){.25}}
\put(44,0){$\scriptstyle \lambda^2_\infty$} \put(44.5,2){\tiny .}
\put(44.75,2){\line(1,0){.25}}
\put(45.5,2){\line(1,0){.5}}
\put(46,2){\line(1,0){1.5}}
\put(48,2){\line(1,0){.5}}
\put(49,2){\line(1,0){.5}}
\put(49.75,2){\line(1,0){.25}}
\put(49.5,0){$\scriptstyle \lambda^{m}_\infty$} \put(50,2){\tiny .}
\put(50.25,2){\line(1,0){.25}}
\put(51,2){\line(1,0){.5}}
\put(52,2){\line(1,0){1.5}}
\put(56,0){$\scriptstyle \lambda^m_3$} \put(56.5,2){\line(1,0){3}}
\put(59,0){$\scriptstyle \lambda^m_2$}
\put(62,0){$\scriptstyle \lambda^m_1$} \put(62.5,2){\line(1,0){3}}
\put(65,0){$\scriptstyle \lambda^m_0$}
\end{picture}
\vspace{1cm}
\noindent The gaps between any two consecutive spectral bands is called
a spectral gap. We reindex the spectral gaps and write the i th gap as
$(\tau_{2i-1},\tau_{2i})$ and denote $\ell_i$ its length. It is clear
that $\sum_{i=1}^\infty\ell_i <\infty$.
\vspace{5mm}
\newtheorem{ass}{Assumptions}
\begin{ass}
There are points $\alpha_i$ and $\beta_i$ on the spectral bands that are
respectively to the left and right of the i'th spectral gap such that
$\alpha_i$ lies to the right of the mid-point of the left spectral
band and $\beta_i$ lies to the left of the mid- point of the right
spectral band ;
$$s_i \equiv \tau_{2i-1}-\alpha_i=\beta_i - \tau_{2i}$$
\noindent such that
$\sum_i \frac{q_i}{s_i} < \infty$, where
$q_i~=~\frac{\ell_i}{s_i}$.
\end{ass}
\vspace{2mm}
Note that the above $s_i$ will be used in the construction of a
Banach space later on.
Before we proceed further, let us say that the condition on the
summability of $\frac{q_i}{s_i}$ is used only in the proof of
Lemma (\ref{contours}), in controlling some derivatives of Weyl
functions around the accumulation points in the spectrum and
nowhere else.
\vspace{5mm}
\noindent We now state our main
\vspace{5mm}
\begin{theorem}
\label{main theorem}
Consider $\left(\Omega, {\cal B}, {\cal P} \right)$ be as above with
${\cal P}$ an ergodic probability measure. Suppose the spectrum of
$\op$ is purely absolutely continuous and satisfies assumptions 1.
Then any $\omega\in\supp$ is an almost periodic sequence.
\end{theorem}
\vspace{5mm}
\noindent Here we give a general outline of the proof of this theorem. Fix an
$\omega\in\supp$ and consider the Green's kernel
$$\gkw =\<{(\op-\lambda)}^{-1}\delta_n,\delta_n\>$$
\noindent It has exactly one zero $\xi_i (n)$ in the
i'th spectral gap. Then the trace formula is that the potential is
given by
$$\omega(n)={1/2}(\mu^0_0+\lambda^m_0)+{1/2}\sum(\tau_{2i}+\tau_{2i+1}-\xi
_i^\omega(n))$$
So it is enough to show that the quantity on the R.H.S. of the above
equation is an almost periodic sequence in n. Now each
$(\xi_i(n))$, apart from being a zero of the Green's kernel, is a simple
pole of exactly one of two Weyl functions which we will define in the
next section and given this information it can be identified as a point
on an infinite genus hyperelliptic Riemann surface $\cal R$ associated
with the spectrum.
We define the
Abel-Jacobi map on X and show that the image of
$(\xi_i^\omega(n))$ undergoes a linear motion. We then invert the
Abel-Jacobi map and use the trace formula to show that $(\omega(n))$ is
an almost periodic sequence.
\section{Inverse spectral theory}
Consider the operator $\op$ defined in section 2, where $\omega\in\supp$
the following facts are known from Simon \cite{simon}.
Any $\lambda\in{\CC^+}$ belongs to the resolvent of $\op$ and we
denote the greens kernel $\<{(H^\omega-\lambda)}^{-1}\delta_n,\delta_n\>$
by $\gk$. Here $\delta_n$ is the basis vector which is 1 at the lattice
point n and zero elsewhere. By Weyl theory, for the eigenvalue equation
$(\op-\lambda)~u=0$ there exists exactly one linearly independent solution
each, which are integrable to the left and to the right respectively. Denote
them by $u_{\pm,\lambda}^\omega$. Then we have
\begin{equation}\gkw = \frac{{u_{+,\lambda}^\omega(n)}
{u_{-,\lambda}^\omega(n)}}
{W}
\label{gk}
\end{equation}
where W is the Wronskian of the solutions $u_{\pm,\lambda}^\omega$ given
by $(u_{+,\lambda}^\omega(1)u_{-,\lambda}^\omega(0)-u_{+,\lambda}^\omega(
0)u_{-,\lambda}^\omega(1))$.
We also define the Weyl functions by
\begin{equation}\mpmw =-\frac{u_{\pm,\lambda}^\omega(n\pm
1)}{u_{\pm,\lambda}^\omega(n)}\label{wf1}\end{equation}
Then $\mpmw$ satisfies the equation
\begin{equation}\mpmw =\omega(n)-\lambda-{(m_{\pm,(n\mp 1)})}^{-1}
\label{wfr}\end{equation}
In terms of the operators $H_{\pm,n}^\omega$ of $\op$ restricted to the
subspaces $\{ u\in\ell^2(\ZZ^\pm)~;~u(n)=0\}$ we have
\begin{equation}\mpmw =\<{(H_{\pm,n}^\omega-\lambda)}^{-1}\delta_{n\pm
1},\delta_{n\pm 1}\>\label{wf2}\end{equation}
\vspace{5mm}
\begin{definition}
F(z) defined on $\CC^+$ is called Herglotz if it maps $\CC^+$ into itself
and is analytic.
\end{definition}
\vspace{5mm}
\begin{theorem}\label{herglotz}
Let F($\lambda$) be Herglotz. Then it has the representation
$$F(\lambda)=a\lambda+b+{1}/{\pi}\int_R(\frac{1}{\xi-\lambda}-\frac{\xi}
{1+\xi^2})~d\mu(\xi)$$
where $a, b \in R,~a\geq 0$ and $d\mu(\xi)$ is a non- negative measure such that
\begin{eqnarray*}
{\rm supp}~d\mu_{a.c} &=& \{~\xi~:~
\lim_{\epsilon\rightarrow 0}~{\rm Im}~F(\xi+i\epsilon)~{\rm exists~and~is~finite}~\}
\\{\rm and ~ supp}~d\mu_{sing} &=& \{~\xi~:~\lim_{\epsilon\rightarrow 0}~
{\rm Im~F}(\xi+i\epsilon)=\infty~\}\\
{\rm Moreover}~d\mu_{a.c}(\xi) &=& {\rm Im}~F (\xi+i~0).d\xi
\end{eqnarray*}
\end{theorem}
\vspace{5mm}
For a proof see \cite{kotani} or \cite{simon}.
\vspace{5mm}
\newtheorem{lemma}{Lemma}[section]
\begin{lemma}
The functions $\gkw$ and $\mpmw$ are Herglotz.\end{lemma}
\noindent {\bf Proof:}
Follows easily from the spectral representation for Green's kernels.
Using the continuity properties of the green functions, in the
potential, one can easily see the following theorem on the lines
of \cite{craig},\cite{kotani} or \cite{simon}.
\vspace{5mm}
\begin{theorem}
\label{vanishing Im part}
Suppose the spectrum of $\op$ is purely absolutely continuous and is the
union of closed intervals (finitely or infinitely many). Then
\begin{eqnarray*}
{\rm Re}g_{\lambda+i0}^\omega(n,n) &=& 0~~ \forall
~\omega\in\supp{\rm ~and}~\forall~n \in Z,\lambda \in
{\rm Int}~\sigma(\op)\\
{\rm Im}~g_{\lambda+i0}^\omega(n,n) &=& 0~\forall
\omega\in\supp~{\rm and}~\forall~n \in Z,\lambda \in R
-\sigma(\op)
\end{eqnarray*}
Here $g_{\lambda+i0}^\omega$ etc. denote the limits ${\rm lim}_{\epsilon
\rightarrow0}g_{\lambda+i\epsilon}^\omega$ etc. and {\rm Int} $\sigma(\op)$
denotes the interior of the spectral bands.
\end{theorem}
Henceforth we fix an $\omega\in{\rm supp}~P$ and drop the subscript
$\omega$ from the spectral functions. We also note that the green
function is positive in the interval ($-\infty, \mu_0^0$). We can use
this fact in fixing the sign of the square root in the following lemma.
We recall that $(\tau_{2i-1},\tau_{2i})$ is the i'th spectral gap.
\vspace{5mm}
\begin{lemma}
\label{infinite product}
$\gk$ has exactly one zero $\xi_i(n)$ in the i'th spectral gap and
has the expression
\begin{equation}\gke \quad ;\lambda\in \CC^+\label{gke}\end{equation}
\end{lemma}
\vspace{5mm}
\noindent {\bf Proof:}
First of all we show that the infinite product given above converges.
For that we note that
\begin{eqnarray*}\left|\frac{\lambda-\xi_i(n)}{\lambda-\tau_{2i-1}}\right| &\le&
1+\left|\frac{\tau_{2i-1}-\xi_i(n)}{\lambda-\tau_{2i-1}}\right|\\ {\rm and}~
\left|\frac{\tau_{2i-1}-\xi_i(n)}{\lambda-\tau_{2i-1}}\right| &\le& 1+q_i~ {\rm for~all
~i~ except~ finitely~ many}\\{\rm and~ similarly
}~\left|\frac{\lambda-\xi_i(n)}{\lambda-\tau_{2i}}\right| &\le& 1+q_i{\rm~for~ all~i~
except~finitely~ many.}\end{eqnarray*}
So the infinite product converges since the product
$\prod_{i=1}^\infty(1+q_i)$ converges.
Now in each spectral gap \hfil $\gk$ \hfil is an increasing function since
$\frac{d}{d\lambda}\gk =~\~>~0$ and hence we can have
at most one zero in each spectral gap. First let us assume that it has
,for each i, one zero $\xi_i(n)$in the i'th spectral gap;
i=1,2,.... Now $\gk$ and log $\gk$ are Herglotz functions. By
Theorem \ref{herglotz}, ${\rm lim}_{\epsilon\rightarrow
0}~g_{\lambda+i\epsilon}(n,n)$ exists and is finite for almost
all $\lambda\in R$. Now by the continuity of the function log,
${\rm lim}_{\epsilon\rightarrow 0}~{\rm log}~g_{\lambda+i\epsilon}(n,n)$ exists
wherever ${\rm lim}_{\epsilon\rightarrow 0}~g_{\lambda+i\epsilon}(n,n)$ exists
and is finite or infinite according as the latter limit is finite or infinite.
Now by Theorem \ref{vanishing Im part} the latter limit is finite at all points
on the real line except perhaps at the spectral end-points and at the
accumulation points. However these points won't contribute to the
measure appearing in the Herglotz representation for log $\gk$ since it
does not have exponential behaviour at infinity. So by Theorem
\ref{herglotz}
\begin{eqnarray}
\log \gk& =&a\lambda +b + \frac{1}{\pi} \int_R(\frac{1}{(\xi-\lambda)} -
\frac{\xi}{(1+ \xi^2)})~\\
&&{\rm Im}~\log~(g_{\xi+i0} (n,n))~d\xi \label{log gk}
\end{eqnarray}
Now by Theorem \ref{vanishing Im part}, Im $g_{\xi+i0}(n,n)=0$ for all $\xi\in
\real-\Sigma$
and Re $g_{xi+i0}$(n,n) = 0 for all $\xi\in\Sigma$. Hence noting
that $\gk$ is an increasing function of $\lambda$ in each spectral gap
we write down the density Im log $g_{\xi+i0}(n,n)$ as
\[{\rm Im}\log g_{\xi+i0}(n,n) =\left\{\begin{array}{lll}0 &{\rm for}
&\xi<\mu^0_0\\ \pi &{\rm for} &\tau_{2i-1}\le\xi\le\xi_i(n)\\ 0 &{\rm for}
&\xi_i(n)<\xi\le\tau_{2i}\\ \pi &{\rm for} &\xi\ge\lambda^m_0\\ \frac{\pi}{2} &{\rm
for} &\xi\in\Sigma\end{array}\right.\]
Substituting in (\ref{log gk}) we get
\begin{eqnarray*}
\log\gk &=& a\lambda+\tilde{b}+\frac{1}{2}\sum_{i=0}^\infty
\int_{\tau_{2i}}^{\tau_{2i+1}}
\frac{1}{(\xi-\lambda)}d\xi
+\sum_{i=1}^\infty\int_{\tau_{2i-1}}^{\xi_i(n)}\frac{1}{(\xi-\lambda)}d\xi\\
&& +\int_{\lambda^m_0}^\infty\left(\frac{1}{\xi-\lambda}-
\frac{\xi}{1+\xi^2}\right)d\xi\\
&=& a\lambda+\tilde{b}+\sum_{i=0}^\infty\log\left(\sqrt
{\frac{\tau_{2i+1}-\lambda}{\tau_{2i}-\lambda}}\right)
+\sum_{i=1}^\infty\log\left(\frac{\xi_i(n)-\lambda}{\tau_{2i-1}-\lambda}\right)\\
&& +{\left[\log\frac{(\xi-\lambda)}{\sqrt{1+\xi^2}}\right]}_{\lambda^m_0}^\infty\\
&=& a\lambda+d
+\log\gke
\end{eqnarray*}
Now as $\lambda\rightarrow\infty,~\gk \sim~-\frac{1}{\lambda}$.
Hence we get (\ref{gke}).
In the proof we have assumed that in each spectral
gap $(\tau_{2i-1},\tau_{2i})$ there is a zero $\xi_i(n)$. Suppose it is
not so. Then in such gaps $\gk$ is either non-negative or non-positive.
In either case we can write down the density Im $\log_{\xi+i0}(n,n)$,
perform the integration and will get the same expression as
(\ref{gke}) with $\xi_i(n)=\tau_{2i-1}~{\rm or}~\tau_{2i}$ according as
$\gk$ is non-negative or non-positive in the i'th spectral gap. Hence
the lemma.
As a corollary we have the following trace formula.
\vspace{5mm}
\begin{tf}
For any $\omega\in\supp$ we have
\begin{equation}\omega(n) = \frac{1}{2}(\mu^0_0+\lambda^m_0)+\frac{1}{2}\sum_{i=1}
^\infty\left((\tau_{2i}+\tau_{2i-1})-2\xi_i(n)\right)\label{trace}
\end{equation}
\end{tf}
\noindent {\bf Proof:}
$\omega(n)$ is the coeficient of $\frac{1}{\lambda^2}$ in the asymptotic
expansion of
\noindent $\<{(H^\omega-\lambda)}^{-1}\delta_n,\delta_n\>$ and the
R.H.S of (\ref{trace}) is the coefficient of $\frac{1}{\lambda^2}$ in the
asymptotic expansion of the expression for $\gk$
\vspace{5mm}
\begin{lemma} For $\lambda \in \CC\setminus\Sigma$,
$$-\gk^{-1}=\np +\mm +\lambda-\omega(n)$$\end{lemma}
\vspace{5mm}
\noindent {\bf Proof:}
Follows from the definition of the above spectral functions and the
Weyl formula (\ref{wf1}) for the green functions.
\vspace{5mm}
\begin{lemma}
Let the spectrum of $H^\omega$ be purely absolutely continuous and
consists of bands; finitely or infinitely many. Then
$\forall\omega\in\supp$ and for all $\lambda$ belonging to the
interior of any spectral band we have
$${\rm Im}~m_{+,n}^\omega(\lambda)~=~{\rm Im}~m_{-,n}^\omega(\lambda)$$
\end{lemma}
\noindent {\bf Proof:} See \cite{kotani} or \cite{simon}.
\vspace{5mm}
\begin{lemma}
\label{contours}
The Weyl-functions have the expressions
\begin{eqnarray*}\mpm = \frac{1}{2}\left\{\right.& & \me \\& & -(\lambda-\omega(n))
\pm\res\left.\right\} \end{eqnarray*}\end{lemma}
\vspace{5mm}
{\bf Proof:}
>From lemmas 3.3 and 3.2 we get an expression for
$\np + \mm ;\lambda\in \CC^+$. So it is enough to obtain $\np -\mm$
explicitly. From these two relations the lemma follows.
To do this we proceed as follows.
We note that since $\gk$ has a simple zero at $\xi_i(n)$ in each gap,
its inverse has simple poles at these points. Let Res$(\xi_i(n))$'s
denote the corresponding residues at these points. Define variables
$c_{i,n}:\ZZ^+\rightarrow\{-1,+1\}$ by $c_{i,n}=\pm 1$ according
as $\xi_i(n)$ is a pole of $\mpm$. Then by lemma (3.3) the residue of
$(\np -\mm)$ at $\xi_i(n)$ is given by $c_{i,n}~{\rm Res}\xi_i(n)$. It
will be proved
in lemma (3.6) that $\sum_i{\rm Res}\xi_i(n)$ is absolutely convergent.
Consider the function $f(\lambda)=\np-\mm-\res$. Apriori this is a
holomorphic function defined on $\CC\setminus\Sigma$. Now by lemma (3.4),
Im $f(\lambda)$=0 forall $\lambda$ lying in the interior of the spectral
bands. Hence $f(\lambda)$ has analytic continuation through the interior
of the spectral bands as well. We claim that it is analytic at the
end-points of the bands also. For that it is enough to show that it is
bounded in some neighbourhood of each of these points. If any of these
points is a zero of $\gk$ then it is a simple pole of $\np-\mm$ and
hence from the definition of $f(\lambda)$ we see that it is bounded
around that point. Now consider the end-points which are not zeroes of
$\gk$. We have the following relationship
\begin{equation}
{(\np-\mm-(\lambda-\omega(n)))}^2={\gk}^{-2}-4~
\frac{g_\lambda(n+1,n+1)}{\gk} \label{*}
\end{equation}
This follows easily from the definition of $\mpm$ and from the identity
${(a-b)}^2 ={(a+b)}^2 -4ab$. From the expressions for the quantities
appearing in the the R.H.S. of (\ref{*}) we see that they are bounded in some
neighbourhood of each of these end-points. Hence the L.H.S. of (\ref{*})
and therefore $f(\lambda)$ are bounded around these points.
Thus $f(\lambda)$ can be extended as a analytic
function on $\CC\setminus\cup_{k=1}^m\lambda^k_\infty$. We claim that it is analytic at
$\lambda^k_\infty$'s also. For that we show that the non-holomorphic terms
in the Laurent series expansion of $f(\lambda)$ vanish.
For each k;
k=1,...,m, define a
sequence of rectangles $R^k_i$ around $\lambda^k_\infty$ as follows:
$R^k_i$ has its sides parallel to the axes; the sides perpendicular to the
x-axis cuts through the mid-points of suitable spectral bands
to the left and right of $\lambda^k_\infty$ in such a way that the width of
$R^k_i$ shrinks to zero as $i\rightarrow\infty$, and the height
above and below the x-axis will be the maximum of the distance of
$\lambda^k_\infty$ from the sides parallel to the x-axis.
Let $\lambda\in {R}^k_i$, and consider
$$\left|\frac{\lambda-\tau_i}{\lambda-\xi_i}\right|=1+
\left|\frac{\xi_i-\tau_i}{\lambda-\xi_i}\right|\leq 1+q_i$$
Doing the same analysis for the other terms of ${\gk}^{-2}$ and using
the fact that the infinite product $\prod_i(1+q_i)$ converges we see
that the first term on the R.H.S. of (\ref{*}) is bounded uniformly on
the $R^k_i$. Using an exactly similar argument we see that the second
term also is bounded on the $R^k_i$. Now in the proof of the next lemma
it will be shown that ${\rm Res}(\xi_i(n))\sim\ell_i$. So $\res\leq {\rm
const.}~\sum_i q_i$. Thus $f(\lambda)$ is bounded on the $R^k_i$ and
hence $\int_{R^k_i}f(\lambda)\rightarrow
0~{\rm as}~i\rightarrow\infty$.
Thus the residue of $f(\lambda)$ at $\lambda^k_\infty$ is zero.
Now multiplying $f(\lambda)$ by various powers of
$(\lambda-\lambda^k_\infty)$ and doing the same analysis as above we see
that the negative terms in the Laurent series expansion of $f(\lambda)$
around $\lambda^k_\infty$ vanish for each k=1,..,m. Hence $f(\lambda)$
is an entire function. Also, it vanishes at
infinity since $\mpm$ vanish there. Hence by Liouville's theorem
$f(\lambda)\equiv 0$. From this and the expressions for the green's
kernels, the lemma follows.
\noindent We now prove lemma (3.6) which was required in the proof of the above
lemma.
\vspace{5mm}
\begin{lemma}
$\sum_i{\rm Res}(\xi_i(n))$ is absolutely convergent.\end{lemma}
{\bf Proof:}
\begin{eqnarray*}
{\rm Res}~\xi_i(n) &=&
\left(-\sqrt{(\xi_i(n)-\mu^0_0)(\xi_i(n)-\lambda^m_0)}\right)\\
&& \times \left(\prod_{j\neq i}\frac{\sqrt{(\xi_i(n)-\tau_{2j-1})(\xi_i(n)-\tau_{2j})}}
{(\xi_i(n)-\xi_j(n))}\right)\\
&& \times\left(\sqrt{(\xi_i(n)-\tau_{2i-1})(\xi_i(n)-\tau_{2i})}\right)
\end{eqnarray*}
The quantity in the first bracket is bounded by a constant independent
of i while that in the third bracket is bounded by $\ell_i$. Now for
$j\neq i$
$$\left|\frac{\xi_i(n)-\tau_{2j-1}}{\xi_i(n)-\xi_j(n)}\right|\leq
1+\frac{|\xi_i(n)-\tau_{2j-1}|}{|\xi_i(n)-\xi_j(n)|}\leq 1+q_j$$
Similarly $$\left|\frac{\xi_i(n)-\tau_{2j}}{\xi_i(n)-\xi_j(n)}\right|\leq
1+q_i$$
So the quantity in the second bracket is bounded by $\prod_j(1+q_j)$.
Hence $|{\rm Res}\xi_i(n)|\leq c \ell_i$, where c is a constant
independent of i. The lemma follows since $\sum_i\ell_i<\infty$.
\vspace{5mm}
We have as a corollory of lemma \ref{contours} the following ,
\begin{theorem}
\label{inversion}
Consider P an ergodic potential as above and $\Sigma$ its
absolutely continuous spectrum, then the quantities $\{\xi(0),
c_{i,0}\}$ determine the potential $\omega$ uniquely.
\end{theorem}
{\bf Proof} From the above proof it is clear that the data
$\Sigma, \{\xi_i (0), c_{i,0}\}$ determine the m functions uniquely
and also by the trace formula, $\omega(0)$ uniquely. Hence the
theorem.
As a consequence of the proof of lemma \ref{contours}, we have the
following interesting corollory.
\vspace{5mm}
\begin{tf}
\label{uniform bounds}
The m functions $m_{\pm,n}(\lambda)$ and $(m_{\pm,n}(\lambda))^{-1}$
are uniformly bounded on the rectangles $R_i^k$ defined in the lemma
\ref{contours}. Moreover the derivatives of $m_{\pm,n}(\lambda)$
are uniformly bounded on $R_i^k \cap (\BC\setminus\Sigma)$ for
each i and k.
\end{tf}
{\noindent \bf Proof:} Clearly the bounds obtained in the proof of
lemma \ref{contours} proves the corollory for $m_{\pm,n}(\lambda)$. On
the other hand for the inverses of the m functions, we use the
relations of equation \ref{wfr} on the portion of $R_i^k$ 's in the
resolvent of $H^{\pm}$ and extend the bounds by continuity of
$m_{\pm,n}$ on the points in the spectrum. We have the
expression for $m_{+,n}(\lambda)$ from Lemma 3.5 as
\begin{eqnarray*}
m_{+,n}(\lambda) = -g_{\lambda}(n,n)^{-1} + \sum{\frac{Res
\xi_i(n)}{\lambda - \xi_i(n)}}
\end{eqnarray*}
Therefore it is enough to show the boundedness of the derivatives
of each of the terms on the right hand side of the above
equation. As for the boundedness of the second term it follows
from the assumptions and the estimates as in Lemma \ref{contours}
and regarding the boundedness of the derivative of the inverse of
the green function, we note that it is a meromorphic funciton in $
\CC \setminus \Sigma$ and the derivative is given by
\begin{eqnarray*}
\frac{d}{d\lambda}g_{\lambda}(n,n)^{-1}&=&\frac{1}{2}
g_{\lambda}(n,n)^{-1}\times\left(
\frac{1}{(\lambda - \lambda_0^0)^2}+\frac{1}{(\lambda - \lambda_0^m)^{2}}
\right. \\
&& + \left. \sum_i \frac{\tau_{2i-1} -
\xi_i(n)}{(\lambda - \tau_{2i-1})(\lambda - \xi_i(n))} +
\frac{\tau_{2i} - \xi_i(n)}{(\lambda - \tau_{2i})(\lambda - \xi_i(n))}
\right)
\end{eqnarray*}
Now note that the above equation shows that using a computation similar
to the one used in the Lemma \ref{contours} , that apriori the
derivative is bounded on the rectangles lying in any open set of
$\CC \setminus \Sigma$, and with a uniform bound coming from the proof of Lemma
\ref{contours}. We note that hence, even the lim sup of the
derivative along any
sequence on the rectangle approaching the real axis, both from $\CC^+$
and from $\CC^-$ is uniformly bounded. Hence the result.
\vspace{5mm}
\begin{remark}
\label{single function}
For future use we note the following
\begin{eqnarray*}
\Mm &=&\frac{1}{2}
\left\{\right.\me\\
& &-\mbox{}(\lambda-\omega(n))+\res\left .\right\}
\end{eqnarray*}
That is, the expressions for $\Mp$ and $\Mm$ differ only in the sign of the radical appearing in them.
\noindent Also from (\ref{wf2}) we note that $\np\sim -\frac{1}{\lambda}$ as
$\lambda\rightarrow\infty$ and $-\mm -(\lambda-\omega(n)) \sim -\lambda$
as $\lambda\rightarrow\infty$.
\end{remark}
%%%%%%%%%%%%%%%%%%% Interpolation %%%%%%%%%%%%%%
\section{An interpolation theorem}
Consider the domain $\Pi$ = $\CC\setminus\Sigma$, $\Sigma$ the spectrum.
We define a class of
functions on $\Pi$ having zeroes ``not more than the number of spectral
gaps minus one''. For $i=1,2,..,\infty$, let $(\xi^i_j)$ be such that
$\xi^i_j$ belongs to
the i'th spectral gap. We consider functions of the form
\begin{eqnarray*}
\omega_{i,\xi} &=&\frac{1}{\sqrt{(\lambda-\mu^0_0)(\lambda-\lambda^m_0)}
}\left(\prod_{j\neq
i}\frac{(\lambda-\xi^i_j)}{\sqrt{(\lambda-\tau_{2j})(\lambda-\tau_{2j-1})}}
\right)\\
&& \times \frac{1}{\sqrt{(\lambda-\tau_{2i})(\lambda-\tau_{2i-1})}}
\end{eqnarray*}
\vspace{5mm}
\noindent Let $\kappa_i$ be complex numbers such that $|\kappa_i|0~\forall k=1,..,n$,
where $F_k(.)$ denotes the k'th coordinate of F.
Then F has a zero in the interior of I.
\end{lemma}
{\bf Proof:} Without loss of generality we can assume that $a_k<0$ and $b_k>0$
$\forall k$. The straight line homotopy $x\rightarrow tF(x)+(1-t)x;0\leq
t\leq 1$ gives a homotopy between F and the identity map \cite{klaus}.
The conditions $F(x_1,..,a_k,..,x_n)<0$ and $F(x_1,..,b_k,..,x_n)>0$
ensure that throughout the homotopic
deformation no point in the the boundary of I goes to zero. Hence the
topological degree of F is the same as that of the identity map which is
1. So F has at least one zero in the interior of I.
\vspace{5mm}
{\noindent \bf Proof of Lemma \ref{differentials}:}
Fix an i. Let $I_k$ denote the closure of the k'th spectral gap and
corresponding to each (n-1) tuple $(x_1,..,x_k,..,x_n);~x_k\in I_k;~k\neq i$
define a function of a real variable $\lambda$ by
\begin{eqnarray*}
\lefteqn{f^n(x_1,..,x_n)(\lambda)=}\\
& & \frac{1}{\sqrt{(\lambda-\mu^0_0)
(\lambda-\lambda^m_0)}}
\frac{1}{\sqrt{(\lambda-\tau_{2i-1}) (\lambda-\tau_{2i})}}
\left({\prod^n_{j\neq i}}_{j=1}\frac{(\lambda-x_j)}
{\sqrt {(\lambda-\tau_{2j-1})(\lambda-\tau_{2j})}}\right)
\end{eqnarray*}
Define $F^n:{\prod_{k=1}^n}_{k\neq i}I_k\rightarrow R^{n-1}$ by
$$F^n(x_1,...,x_n)=\left(\int_{I_1}f^n(x_1,..,x_n)(\lambda)d\lambda,...,
\int_{I_n}f^n(x_1,..,x_n)(\lambda)d\lambda\right)$$
We claim that $F^n$ has a zero.
It is clear that if we multiply the integrals occuring in the definition
of $F^n$ by scalars (in particular -1), the zeros of $F^n$ are unaffected.
So, after such multiplications if
necessary, we see that $F^n$ satisfies the conditions of the lemma
\ref{degree theory}. Hence it has a zero $(x^n_1,..,x^n_n)$. This way
corresponding to each n we get an n-tuple. Since
$x^n_1$'s move in a compact set as n varies, we can find a subsequence
$n_1$ such that the corresponding $x^{n_1}_1$'s converge to a point
$\eta^i_1$. From the above subsequence we can find another
subsequence $n_2$ such that the corresponding $x^{n_2}_2$'s converge to a point
$\eta^i_2$. This way inductively we can define a sequence of points
$\eta^i_k$'s. Using these $\eta_k^i$'s we define ${d\omega}_i$ as in
(\ref{diff.1}). We claim that
$$
\int_{I_k}{d\omega}_i=0,~\forall k\neq i
$$
Fix a k. Let $\epsilon>0$ be given. Using estimates similar to those in
lemma 3.2 we see that there exists an $n_0$ such that,
$\forall\lambda\in I_k$ all the tail
products of the R.H.S. of (\ref{diff.1}) after $n_o$ terms differ from 1 at
most by
$\epsilon$. So
$$
|\int_{I_k}{d\omega}_i-\int_{I_k}{d\omega}^n_i|<\epsilon.~
\ell_k;~\forall n>n_0
$$
where ${d\omega}^n_i$ denote the n'th partial product
of ${d\omega}_i$.
Again using estimates similar to those in lemma 3.2 we see that
uniformly in $I_k$ any two partial products of ${d\omega}^n_i$
or $f^n$ exceeding the $n_0$'th one differ only by $\epsilon$. So choosing n
large enough so that the $n_0$'th partial product of ${d\omega}^n_i$
and an $f^m(.,,.)(\lambda)$ differ uniformly in $I_k$ by $\epsilon$ we see
that
$$
|\int_{I_k}{d\omega}_i|<4\epsilon~.~\ell_k
$$
Since $\epsilon$ is arbitrary the lemma follows for j$\neq$i.
For the case j = i , it is clear that the integrand has a
definite sign as the beta cycle is traversed, since the change of
sign in the integrand as one changes sheets is compensated by the
reversal of the contour of integration.
\vspace{5mm}
\begin{lemma}
\label{normalize}
The differentials $d\omega_i$ satisfy the normalizations
\begin{enumerate}
\item $\int_{\beta_j} d\omega_i~= ~\pi ~\delta_{ij}$
\item $\int_{\alpha_j} d\omega_i~=~\pi_{ij}$
with $\pi_{ij}$ purely imaginary.
\end{enumerate}
\end{lemma}
\proof
Let $\int_{\alpha_j}d\omega_i~=~\pi_{ij}$.
Since $\int_{\beta_i}{d\omega_i}$ is real, the normalizing constant is
real. Therefore chosing $\sigma_i$ appropriately we will get the
stated normalization on the beta cycles in view of Lemma
(\ref{differentials}). Now we can deform $\alpha_i$
into the closed loop that starts from
$\tau_{2i}$, goes to $\lambda^m_0$ and comes back to $\tau_{2i}$ entirely
through the embedded real line in such
a way that while coming back the embedded spectral gaps it traverses are
the same as those of the onward journey whereas the embedded spectral
bands it traverses are the ones other than the ones already traversed.
Hence we see that $\pi_{ij}$ is two times the sum of the integrals of
$d\omega_i$ over the spectral bands lying to the right of $\tau_{2i}$.
So from (\ref{diff.1}) we see that $\pi_{ij}$ is purely
imaginary.
Now the following lemma shows that while taking estimates we can
omit this normalizing constant and in future, for $d\omega_i$ we use
the same expression as on the R.H.S. of (\ref{diff.1}).
\vspace{5mm}
\begin{lemma}
\label{normalization}
There exists constants $c_1,c_2>0$ such that
$$c_1<|(\int_{\beta_i}{d\omega_i})|0$ such that
$$c_1~<\int_{\tau_{2i-1}}^{\tau_{2i}}|{d\omega_i}|~k}s_i $ goes to zero as k goes to $\infty$.
Now given any sequence $x^n$, we can find coordinate
wise convergent subsequences. Therefore by a diagonal sequence argument
gives the result
\vspace{5mm}
\begin{lemma}
\label{cptness}
The linear operator $A_x$ on X given by
\begin{equation}
(A_x)_{ij} = f_{ij}(x_j) , i \neq j ~ and ~ = 0 ~ for ~ i = j
\end{equation}
is compact on X , where $f_{ij}$'s are as in equation (\ref{omegatheta}).
\end{lemma}
\proof We will show that the closure of the image of the unit ball under
A is compact. Clearly for any
$y \in X$,
$$
sup_i \vert\left(A_xy\right)_i\vert = \sup_i \vert \sum_j
\left(A_x\right)_{ij}y_j\vert = \sup_i \vert \sum_j
f_{ij}(x_j)y_j\vert
$$
$$
\leq sup_i sup_{j\neq i}\vert s_j^{-1}f_{ij}\vert \sum s_j \vert y_j \vert
\leq \left(sup_i sup_j \vert s_j^{-1}f_{ij}\vert\right)\|y\| < \infty
$$
in the above inequality we have used $s_i \geq q_i$ and the equation
(\ref{est}). From the above estimates the result follows.
In the following to distinguish vectors in X from points in $\real$ we
use the vector notaion. Let $\btheta \in $ X and consider the map
\begin{equation}
\AA (\btheta)(i) = \sum_{k = 1}^{\infty} \int_0^{\theta_k}f_{ik} (\theta) d\theta
, ~ i \in \ZZ^+
\end{equation}
where $f_{ik}$ 's are defined periodically on the whole of $\real$.
Then it is clear that in view of the normalizations, lemma
\ref{normalization}, we have, that $\AA $ maps X into itself and
for later convenience we write $\AA$ as $\AB + \AC$, where $\AB$
and $\AC$ are given by
\begin{equation}
\AB(\btheta)(i) = \int^{\theta_i} f_{ii}(\theta) d\theta ~ and ~
\AC(\btheta) = \AA(\btheta) - \AB(\btheta)
\end{equation}
These two maps satisfy the following property with respect to addition
by elements from $\xz$.
\begin{equation}
\label{linearity}
\AB(\btheta + \vec{m}) = \AB(\btheta) + \vec{m},
~ and ~ \AC(\btheta + \vec{m}) = \AC(\btheta), ~ for ~ each ~ \vec{m} \in
\xz
\end{equation}
Then we have the following theorem
\vspace{5mm}
\begin{theorem}
\label{abel-jacobi}
The map $\AA$,$\AB$ and $\AC$ have the following properties as maps
from X to X.
\begin{enumerate}
\item $\AB$ is one-one ,onto and $\AB$, ${\AB}^{-1}$ are
uniformly Lipschitz.
\item $\AC$, ${\AC}\cdot{\AB}^{-1}$ are strongly differentiable and
the derivatives at each x $\in$ X are compact operators.
\item $\BB$ = I + ${\AC}\cdot{\AB}^{-1}$ is strongly differentiable
with the derivative $(I + B_x)$ invertible for each x $\in X$ and
$sup_{x\in \xmodz} \|(I + B_x)^{-1} \| < \infty$.
\item $\AA$ and $\BB$ are one to one and onto.
\item $\AA^{-1}$ is uniformly Lipshitz on $\xmodz$.
\end{enumerate}
\end{theorem}
\proof
\begin{enumerate}
\item The map $\AB$ is diagonal and in view of the equation
(\ref{linearity}) above , it is enough to look at $\AB$ on $\xmodz$.
Here $\AB(x)(i) = g_i(x_i) = \int^{x_i} f_{ii}(\theta)d\theta$ is
an increasing function on $[0,\pi]$ with $g_i(0) = 0$ and $g_i(\pi) =
\pi$ in view of the normalization \ref{normalization}. Therefore it is
one one and onto coordinatewise.
By equation (\ref{est}), $f_{ii} (x_i)$ and their inverses are uniformly
bounded for each i and $x_i \in [0,\pi]$, hence the mean value theorem
applied to each $g_i$, we see that both $\AB$ and $\AB^{-1}$ are
uniformly Lipshitz.
coordinatewise.
\item We shall prove that $\AC$ is strongly differentiable , the proof
for $\AC\cdot{\AB}^{-1}$ is similar. Consider the matrix $A_{2,x}$ of
partial derivatives of $\AC$ at x. Explicitly we have
$$
(A_{2,x})_{ij} = f_{ij}(x_j) ~ for i \neq j ~ and ~ = 0 for i = j
$$
Then we have for any h $\in$ X
\begin{equation}
\|{\AC(x+h) - \AC(x) - A_{2,x}h}\|
\end{equation}
\begin{eqnarray*}
& = & \sum_{i = 1}^{\infty} s_i \mid
\AC(x+h)(i) - \AC(x)(i) - (A_{2,x}h)(i)\mid
\\ & = &
\sum_{i = 1}^{\infty} s_i \mid\sum_{k \neq i} [\int_{x_k}^{x_k+h_k}
f_{ik}(\theta_k)d\theta_k - f_{ik}(x_k)h_k]\mid
\\ & = &
\sum_{i = 1}^{\infty} s_i \mid\sum_{k \neq i}
[f_{ik}(\tilde\theta_k) - f_{ik}(x_k)]h_k\mid
\\ & \leq &
\sum_{i = 1}^{M(\epsilon)} s_i \mid\sum_{k \neq i}^{\ell(\epsilon)}
[f_{ik}(\tilde\theta_k) - f_{ik}(x_k)]h_k\mid \\ & + &
\sum_{i = M(\epsilon) + 1}^{\infty} s_i \mid\sum_{k \neq i}
[f_{ik}(\tilde\theta_k) - f_{ik}(x_k)]h_k\mid \\ & + &
\sum_{i = 1}^{M(\epsilon)} s_i \mid\sum_{k (\neq i)=\ell(\epsilon)+1}
[f_{ik}(\tilde\theta_k) - f_{ik}(x_k)]h_k\mid
\\ & \leq & \epsilon \|h\|
\end{eqnarray*}
where $\tilde\theta_k$ is some point in $[x_k , x_k + h_k]$ for each k.
The first sum in the last of the inequalities is small since
$h_k$ goes to zero for each k as $\|h\|$ goes to zero and that
each of the $f_{ik}$'s is uniformly continuous on compacts. The second
and the third of the inequalities are small from equations (\ref{est})
and the summability of the quantities $s_i$. The
compactness of the operator $A_{2,x}$ for each x follows from
lemma (\ref{cptness}).
\item The strong differentiablity of $\BB$ follows from (2)
above and its derivative is the operator $( I + A_2 \cdot {A_1}^{-1})$ for
each x in X. Let us call the compact operator $
A_2 \cdot {A_1}^{-1}$ as $B_x$ for each x for convenience. Then we
show the uniform boundedness of the inverse $(I + B_x)^{-1}$ as
follows. First note that since $B_x$ is a compact operator, by
the Fredholm alternative, it is enough to show that $I + B_x$ has
no zero eigen value to show the invertibility and the boundedness
of the inverse at x. Therefore consider
$$
(I + A_2\cdot(A_1)^{-1} ) y = 0 , ~for ~ y \in X
$$
Since $A_1$ is invertible and maps X to itself, it is enough
to show that
$$
(A_1 + A_2 ) y = 0 , ~for ~ y \in X
$$
Writing this explicitly in terms of the entries of the matrices
we have that
$$
\sum_{j = 1}^{\infty} f_{ij} (x_j) y_j = 0 ~ for ~ each ~ i
$$
Now from Lemma (\ref{unique}) we can choose numbers $C_i^k$ for
each k such that
\begin{equation}
\sum_{j} \left( \sum_i C_i^k f_{ij} (x_j)\right) y_j =
\left( \sum_i C_i^k f_{ik} (x_k)\right) y_k = 0 , ~ for ~ each ~ k
\end{equation}
Since the sum in the parantheses is non zero again by Lemma
(\ref{unique}), the result follows. Now
showing the continuity of $( I + B_x )^{-1}$ shows
the uniform boundedness of its norm on compacts, hence in particular
on $\xmodz$. The norm continuity
of $( I + B_x)^{-1}$ follows from the relation
$$
(I + B_y)^{-1} - (I + B_x)^{-1} = (I + B_x)^{-1} \left[ \left[ 1 + ( I +
B_x)^{-1} (B_y - B_x)\right] - I \right]
$$
As y goes to x in X, the right hand side of the above equation
is a bounded operator and goes to zero in norm , hence the
result.
\item We shall show that $\AA$ is one-one and that $\BB$ is
onto. Then from (1) it follows that $\BB$ is one one and since $\AA =
\BB\cdot\AB$ that $\AA$ is also onto. Now suppose there are x and y
in X such that
$$
\AA(x) = \AA(y)
$$
This would imply that
$$
\sum_j \int_{x_j}^{y_j} f_{ij} (\theta_j) d\theta_j = 0 ~ for each ~ i
$$
That is for $a_j = \tau_{2j-1} + \ell_j sin^2(x_j)$ and similarly
defining $b_j$ with $x_j$ replaced by $y_j$ , we have that
$$
\sum_j \int_{a_j}^{b_j} g_i(\lambda) d\lambda = 0 ~ for ~ each ~ i
$$
where $g_i$ are the functions appearing in equation (\ref{diff.1}).
By the interpolation theorem, then this would imply that for each
function in the class $\MM$ the above would be true. This gives a
contradiction as we can choose a specific function from that class with
prescribed zeros such that the integral is nonzero for distinct x and
y. To show that $\BB$ is onto , we note that (3) already implies, by
the inverse function theorem, that the range of $\BB$ is open. So it
is enough to that the range is closed to conclude $\BB$ is onto. Now
$\BB$ satisfies $\BB(\theta + \vec m) = \BB(\theta) + \vec m$ for
$\vec m \in \xz$. So by the uniqueness of the decomposition of
a number into it's fractional and integer parts, it is enough to show
that $Ran(\BB) \cap \xmodz$ is closed. To this end consider a sequence
$y^n$ in the above set. Let it converge to some point y in $\xmodz$.
Since, $y^n$ is in the range of $\BB$ there is a sequence $x^n$ such
that $\BB(x^n) = y^n$. It is easy to verify that such an $x^n$
satisfies $sup_n ~ sup_i ~ |x_i^n|$ is uniformly bounded. Hence as in
the proof of lemma (\ref{cptness}) , we can find a convergent
subsequence $x^{n_k}$ converging to some point x . Hence the result.
\item From (4) we see that $\AA^{-1}$ exists and also satisfies
$\AA^{-1}(\theta + \vec m) = \AA^{-1}(\theta) - \vec m$.
We shall show that $\BB^{-1}$ is uniformly Lipschitz
on $\xmodz$ and since $\AB$ is one one from $\xmodz$ onto itself, the
result for $\AA^{-1}$ follows. First note that $\xmodz$ is a closed
convex set. Consider any x, y in $\xmodz$. Then by the fundamental
theorem for integral calculus we have that
\begin{eqnarray*}
\BB^{-1}(x) - \BB^{-1}(y) & = & \int_0^1 \frac{d}{dt} \BB^{-1} \left( tx +
(1-t)y\right) dt \\
& = & \int_0^1 \left( 1 + B_{(tx + (1-t)y)} \right)^{-1} (x - y) dt
\end{eqnarray*}
the integrals being defined in the Bochner sense. Then the stated
Lipschitz condition follows from the uniform boundedness of $\| (I +
B_z)^{-1}\|$ on $\xmodz$.
\end{enumerate}
Now we are ready to prove the almost periodicity of the potentials.
\vspace{5mm}
\begin{theorem}
The quantities $\omega(n)$ given by the trace formula are almost
periodic in n.
\end{theorem}
\proof Recall the trace formula
$$
\omega(n) = \sum_i \frac{1}{2} (\tau_{2i} + \tau_{2i-1} - 2 \xi_i (n))
= \sum_i \frac{1}{2}\ell_i ~ cos^2(\theta_i(n))
$$
where we have written $\xi_i(n) = \tau_{2i-1} + \ell_i ~
sin^2(\theta_i(n))$.
So that we have for each n and k in $\BZ$ that
$$
\omega(n) - \omega(n+k) = \sum_i \frac{1}{2}\ell_i ~ \left( cos^2(\theta_i(n))
- cos^2(\theta_i(n+k)\right)
$$
Now the stated almost periodicity is immediate from the above equation
if we show that given $\epsilon > 0$ there is an N independent of n
and a sequence $m_i(n,N)$ of integers , i in $\ZZ^+$ such that
\begin{equation}
\label{ap}
sup_{n \in \BZ} \left[ \sum_{i=1}^{\infty} \ell_i \vert \theta_i (n) -
\theta_i(n+N) + \pi m_i(n,N) \vert \right] < \epsilon
\end{equation}
We shall prove this as follows. Consider the $\epsilon$ and let
$k(\epsilon)$ be such that
\begin{equation}
\label{feps}
\sum_{i=k+1}^{\infty} \ell_i < \frac{\epsilon}{2}
\end{equation}
Then , for each M we have,
\begin{eqnarray*}
\vert \omega(n) - \omega(n+M) \vert & \leq & \left[\sum_{1}^{k(\epsilon)}
+ \sum_{k(\epsilon)+1}^\infty\right]
\ell_i \left( cos^2(\theta_i(n) - cos^2(\theta_i(n+M))\right)
\\ & \leq & \sum_{i=1}^{k(\epsilon)}
\ell_i \left( cos^2(\theta_i(n) - cos^2(\theta_i(n+M))\right) +
\epsilon
\end{eqnarray*}
Now it is enough to show that there exist integers N and $m_i(n,N)$ for each i
such that
\begin{equation}
\sum_{i=1}^{k(\epsilon)} \ell_i \vert \theta(n) + \pi m_i(n,N) -
\theta(n+N) \vert < \epsilon
\end{equation}
It is clear that this will follow if we show that, there exists an N
and $\vec{m(n,N)} \in \xz$ such that,
\begin{equation}
\label{required}
\left(sup_{i=1,..,k(\epsilon)} \frac{\ell_i}{s_i}\right)\| \theta(n)
+\vec{m(n,N)} - \theta(n,N)\| < \epsilon
\end{equation}
Consider for any $\delta > 0$ an $\ell$ such that $\sum_{i=\ell}^{\infty} \pi
s_i < \delta$. Now given $\ell$ and $\delta$ it is possible to choose
integers M, $M_1,..,M_{\ell}$ such that
$$
sup_{i=1,..,\ell} \vert Mc_i - \pi M_i \vert < \delta
$$
as can be seen by a simple application of Poincare recurrence theorem.
Now we look at
\begin{equation}
\label{final}
\AA(\theta(n)) - \AA(\theta(n+M)) = \vec{y} + \vec{m}
\end{equation}
where
$$
y_i = Mc_i - \pi M_i ~ for ~ 0\leq i\leq \ell ~ and ~ y_i = Mc_i -
\pi[\frac{Mc_i}{\pi}] ~ if ~ i > \ell
$$
where $\vec{c}$ is the vector appearing in Proposition (\ref{image}).
Therefore we choose the $\delta$ in the above to be
$(\sum s_i)^{-1}(inf_{i=1,..,k}(\frac{s_i}{\ell_i}))\epsilon$,
take the corresponding M
in the equation (\ref{final}) to be N and use the Lipschitz estimate
$$
\| \theta(n) + \vec{m(n,N)} - \theta(n+N) \| \leq C
\|\AA(\theta(n)+\vec{m(n,N)}) - \AA(\theta(n+N))\| \leq C \|y\|
$$
to obtain the estimate required in equation (\ref{required}) .
%%%%%%%%%%%%%%%%%%%%%%%%%%%% end Almost Periodicity %%%%%%%%%%%%%
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