\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{mathrsfs} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2015 (2015), No. 67, pp. 1--16.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2015 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2015/67\hfil Properties of Lyapunov exponents] {Properties of Lyapunov exponents for quasiperodic cocycles with singularities} \author[K. Tao \hfil EJDE-2015/67\hfilneg] {Kai Tao} \address{Kai Tao \newline College of Sciences, Hohai University, 1 Xikang Road, Nanjing, Jiangsu 210098, China} \email{ktao@hhu.edu.cn, tao.nju@gmail.com} \thanks{Submitted October 10, 2014. Published March 20, 2015.} \subjclass[2000]{37C55, 37F10} \keywords{Lyapunov exponent; quasiperodic cocycles; H\"older exponent} \begin{abstract} We consider the quasi-periodic cocycles $(\omega,A(x,E)): (x,v)\mapsto (x+\omega, A(x,E)v)$ with $\omega$ Diophantine. Let $M_2(\mathbb{C})$ be a normed space endowed with the matrix norm, whose elements are the $2\times 2$ matrices. Assume that $A:\mathbb{T}\times \mathscr{E}\to M_2(\mathbb{C})$ is jointly continuous, depends analytically on $x\in\mathbb{T}$ and is H\"{o}lder continuous in $E\in\mathscr{E}$, where $\mathscr{E}$ is a compact metric space and $\mathbb{T}$ is the torus. We prove that if two Lyapunov exponents are distinct at one point $E_0\in\mathscr{E}$, then these two Lyapunov exponents are H\"{o}lder continuous at any $E$ in a ball central at $E_0$. Moreover, we will give the expressions of the radius of this ball and the H\"older exponents of the two Lyapunov exponents. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} Denote by $\mathbb{T}:=\mathbb{R}/\mathbb{Z}$ the torus equipped with its Haar measure $\mu$, and $\mu(\mathbb{T})=1$. Let $M_d(\mathbb{C})$ be the set of linear operators from $\mathbb{C}^d$ to $\mathbb{C}^d$, i.e. the set of $d\times d$ complex matrices. A quasi-periodic cocycle is a pair $(\omega, A)$, where $\omega$, the irrational number, is the frequency and $A\in C^0(\mathbb{T},M_d(\mathbb{C}))$ is continuous, defined by a map $(\omega, A):(x,v)\to (x+\omega, A(x)v)$. The iterates of the cocycle are given by $(\omega, A)^N=(N\omega, A_N)$, where \[A_N(x)=\prod_{j=N-1}^0 A(x+j\omega).\] In this case, the dynamical system is ergodic and the Oseledets Theorem provides us with a sequence of Lyapunov exponents $L_1\leq L_2\leq \dots \leq L_m$, and for almost every $x\in\mathbb{T}$ there exist an invariant measurable decomposition $\mathbb{C}^d=\oplus_{j=1}^nE_x^j$, and a non decreasing surjective map $k:\{1,2,\dots,d\}\to \{1,\dots,n\} $ such that for almost every $x\in\mathbb{T}$, every $1\leq i \leq m$ and every $v\in E_x^{k_i}\backslash \{0\}$ we have $L_j=\lim_{N\to\infty}\frac{1}{N}\log \|A_N(x) v\|$. Moreover, $L_i=L_{i+1}$ if and only if $k_i=k_{i+1}$, and the subspace $E_x^j$ has dimension equal to $\sharp k^{-1}(j)$, where $\sharp k^{-1}(j)$ is the number of the elements in the set $\{i|k_i=j\}$. In the past several years, some researchers focused on the continuity of Lyapunov exponent for the Schr\"{o}dinger equation: \[ (S_{x,\omega}\phi)(n)=\phi(n+1)+\phi(n-1)+v(x+n\omega)\phi(n)=E\phi(n),\quad n\in\mathbb{Z}, \] with the cocycle \[ A(x,E)=\begin{pmatrix} v(x)-E& -1 \\ 1& 0 \end{pmatrix}. \] Goldstein and Schlag \cite{GS} proved that if $v(x)$ is analytic and $\omega$ is a Diophantine, which will mean that \begin{equation}\label{dio} \|n\omega\|\geq \frac{c(\omega)}{n(\log n)^a},\quad\forall n\geq 2 \end{equation} with $a>1$ arbitrary but fixed, the Lyapunov exponent $L(E)$ is H\"{o}lder continuous. Then You and Zhang showed the similar result with more general $\omega$ in \cite{YZ}. Observing that for the Schr\"{o}dinger cocycle, $\det A\equiv 1$ and $L_1(E)+L_2(E)=0$ for any $E$, it makes that we can only study the Lyapunov exponent defined by \begin{equation}\label{lyaexp} L(E):=\lim_{N\to\infty}\frac{1}{N}\int_{\mathbb{T}}\log \|A_N(x,E) \|dx . \end{equation} Recall that we say a function $f(E)$ is H\"{o}lder continuous, when there are nonnegative real constants C, $\alpha$, such that \[ |f(E_1)-f(E_2)|0$ such that for any $E\in (E_0-\rho, E_0+\rho)$, $L_1(E)$ and $L_2(E)$ are H\"{o}lder continuous with H\"{o}lder exponent $\alpha$. Moreover, $\alpha=c\beta$, where $c$ is a positive and small constant depending only on $A(x,E)$ but not on $\tilde{L}(E_0)$; and the $\rho$ has an expression as following: \[ \rho(A,\tilde L(E_0))=\Big[\tilde L(E_0) \exp \Big(-\frac{C_{\rho,1}}{\tilde{L}^{C_{\rho,2}}(E_0)}\Big) \Big]^{1/\beta}, \] where $C_{\rho,1}, C_{\rho,2}$ are the big constants depending only on $A(x,E)$. \end{theorem} \begin{remark}\rm The parameter space $(\mathscr{E},\mathrm{d})$ here is always being the spectral spaces of operators, such like the Schr\"{o}dinger equation \cite{GS}, the extended Happer's model \cite{JM} and the Jacobi operators \cite{T}. It is well known that the cocycle is uniform hyperbolic when the energy $E$ is not in the spectral. Thus, we assume that this parameter space is compact, as the discrete operators' spectrums are always bounded. \end{remark} The study of the H\"{o}lder exponent of the Lyapunov exponents is a hot spot in our field. For the almost Mathieu operator with the cocycle \[ A(x,E)=\begin{pmatrix} \lambda\cos x-E& -1 \\ 1& 0 \end{pmatrix}, \] Avila and Jitomirskaya proved that the Lyapunov exponent defined by \eqref{lyaexp} is H\"{o}lder continuous with H\"{o}lder exponent $1/2$, provided $\omega$ is a Diophantine number \cite{AJ}. For the general Schr\"{o}dinger operator with the cocycle \[ A(x,E)=\begin{pmatrix} v(x)-E& -1 \\ 1& 0 \end{pmatrix}, \] supposing $v(x)$ is a small perturbation of a trigonometric polynomial $v_0(x)$ of degree $k_0$ and $\tilde{L}(E)>0$, Goldstein and Schlag proved that Lyapunov exponent is H\"older continuous with H\"older exponent $\frac{1}{2k_0}-\kappa$ for any $\kappa>0$ \cite{GS1}. In this article, we study the continuity of the Lyapunov exponent from a new perspective. We want to show the relationship between the continuity of the function $E\to A(x,E)$ and the continuity of the Lyapunov exponents. \cite{WY} showed that the assumption that $E\mapsto A(x,E)$ is analytic, is necessary. Here only consider the condition that $A(x,E)$ is a $2\times 2$ matrix, but we believe that the main theorem also work if $A(x,E)\in M_d(\mathbb{C})$. This article is organized as follows. In section 2 we get the Large Deviation Theorem and some upper bound estimate by using some propositions of the subharmonic functions. In section 3 we apply Avalanche principle twice to prove the sharp large deviation theorem. The proof of the main theorem is presented in section 4. Some other common senses about analytic functions which will be applied in this article are presented as follows. Because $\mathscr{E}$ is compact, there exist $D(A)$, $C_2(A)$ and $C_{\rm max}(A)$ , such that for any $E\in\mathscr{E}$, \begin{gather*} \|\log |\det A(\cdot,E)|\|_1\leq D(A),\\ \|\log \|A(\cdot,E)\|-\frac{1}{2}\log|\det A(\cdot,E)|\|_2=C_2(A,E)\leq C_2(A) \end{gather*} and for any $x\in\mathbb{T}$ and any $E\in\mathscr{E}$, \[ \|A(x,E)\|\leq C_{\rm max}(A,E)\leq C_{\rm max}(A). \] Also, by the suppose that $E\mapsto A(x,E)$ is H\"{o}lder continuous in $x\in \mathbb{T}$ with H\"older exponent $\beta$, it is easy to see that \[ |D(E)-D(E')|\leq C_D(A)|E-E'|^{\beta}, \quad \forall E,E'\in \mathscr{E}, \] where $D(E):=\int_{\mathbb{T}}\log|\det A(x,E)|dx$. \section{Large deviation theorem and the upper bound estimate} It is obvious that we can define $L_1(E)=\lim_{N\to\infty}\frac{1}{N}\int_{\mathbb{T}}\log \|A_N(x,E)\|dx$. For easy notation, we replace $L_1(E)$ by $L(E)$. It is also convenient to replace $A_N(x,E)$ by $A_N(e(x),E)$ (with $e(x)=e^{2\pi ix}$), where $A_N(z,E)$ is analytic function in the annulus $\mathscr{A}_\rho=\{z\in\mathbb{C}:1-\rho< |z|< 1+\rho\}$ uniformly in $E\in\mathscr{E}$. Set $u_n(z,E)=\frac{1}{n}\log \|A_N(z,E)\|$, $d_N(z,E)=\frac{1}{n}\log |\det A_N(z,E)|$. Sometimes we use $u_N(z)$ or $u_N$ for short, and the same for $d_N(z,E)$. Let $L_{N,r}(E)=$, $D_r(E)=<\log |\det A(re(\cdot))>$. For $r=1$ we use notations $L_N(E)$ and $D(E)$. Note that $u_N(z)$ and $d_N(z)$ are subharmonic functions in $\mathscr{A}_\rho$ and then the following Large Deviation Theorem for the subharmonic functions applies, provided $\omega$ satisfies \eqref{dio}. \begin{theorem}[{\cite[Theorem 2.15, Remark 2.16]{T}}] \label{22006} There exists $\check{N}(A,\omega)$ such that for any $N\geq \check{N}$, any $1-\frac{\rho}{2}\leq r\leq 1+\frac{\rho}{2}$ and $\delta<1$ holds \begin{gather*} \operatorname{meas}(\{x:|u_N(re(x))-L_r |>\delta \})<\exp(-\check{c}\delta^2 N),\\ \operatorname{meas}(\{x:|d_N(re(x))-D_r |>\delta \})<\exp(-\check{c}\delta^2 N), \end{gather*} where $\check{c}=\check{c}(A)$. \end{theorem} \begin{remark}\label{29} \rm (1) For the avalanche principle in the next section, we need to define the unimodular matrix of $A_N(z,E)$: \begin{equation} \label{26001} \tilde{A}_N(z,E):=\frac{1}{|\det A_N(z,E)|^{1/2}}A_N(z,E), \end{equation} and the unimodular function of $u_N(z,E)$ \[ %\label{26002} \tilde{u}_N(z,E):=\frac{1}{N}\log\|\tilde{A}_N(z,E)\|=u_N(z,E) -\frac{1}{2N}\log|\det(A_N(z,E))|. \] By Theorem \ref{22006}, for any $N\geq \check{N}$ \begin{equation}\label{25001} \operatorname{meas}(\{x:|\tilde{u}_N(re(x)) -\tilde{L}_{N,r}(E)|>\delta\})<\exp(-\tilde{c}\delta^2 N), \end{equation} where $\tilde{L}_{N,r}(E)=<\tilde{u}_N(re(\cdot))>=L_{N,r}-\frac{D_r}{2}$. (2) In the proof of Theorem \ref{22006}, the following lemma will be necessary and apply in the later of this paper: \begin{lemma}[{\cite[Lemma 2.12]{T}}] \label{lem2.3} For any $1-\frac{\rho}{2}\leq r \leq 1+\frac{\rho}{2}$, $\delta$ and $K$, \[ \operatorname{meas}\Big\{x:|\sum_{k=1}^K u_N(re(x+k\omega))-K \tilde{L}_{N,r} | >\delta K \Big\} <\exp(-c\delta K), \] where $c=c(A,\omega)$. \end{lemma} (3) let us note that the constants $c,\tilde{c}$ here do not depend on $\delta$. In particular, one can choose here $\delta$ depending on N. \end{remark} In the next part of this section, we estimate the upper bound of the subharmonic function $u_N(z,E)$ and some other functions. Now we first review a lemma from \cite{GS1}. \begin{lemma}[{\cite[Lemma 4.1]{GS1}}] \label{lem:liprad} For any $r_1,r_2$ so that $1-\frac{\rho}{2}\leq r_1,r_2\leq 1+\frac{\rho}{2}$ one has \[ u_N(re(\cdot),E)>-L_N(E)|\leq C_{\rho}|r-1|. \] \end{lemma} \begin{lemma}\label{22010} For any $N\geq \check{N}(A)$ \[ \frac{1}{N}\log\|A_N(e(x),E)\|\leq L_N(E) +C_6\big(\frac{\log N}{N}\big)^{1/2}, \] where $C_6=C_6(A)$ and $\check{N} $ is as in Theorem \ref{22006}. \end{lemma} \begin{proof} Let $0<\delta<\frac{\rho}{4}$ be arbitrary. Note that $e(x+iy)=e^{-2\pi y}e(x)$, $1-\frac{\rho}{4}\leq e^{-2\pi y}\leq 1+\frac{\rho}{4}$, if $|y|\leq \frac{\delta}{4\pi e C'_\rho}$, where $C'_\rho=\max(1,C_\rho)$ and $e=\exp(1)$. By Lemma \ref{lem:liprad}, one has \begin{equation}\label{25019} |\langle u_N(re(\cdot))\rangle -L_N(E)| \leq \delta,\quad \text{if } |y|\leq \frac{\delta}{4\pi e C'_\rho}. \end{equation} Set \[ \mathbb{B}_y:=\{x:|u_N(e(x+iy))-L_N(E)|>2\delta\}. \] It follows from \eqref{25019} that for $|y|\leq \frac{\delta}{4\pi e C'_\rho}$, there holds \[ \mathbb{B}_y\subseteq \{x:|u_N(e(x+iy))-|>\delta\}. \] By Theorem \ref{22006} one obtains $\operatorname{meas}\mathbb{B}_y\leq \exp(-\check{c}\delta^2 N)$. The function $u_N(e(x+iy))$ is subharmonic, for $e(x+iy)\in \mathscr{A}_\rho$. Let $x_0$ be arbitrary and $y_0=0$. Then $e(x_0)\in \mathscr{A}_{\frac{\rho}{4}}$. Due to subharmonicity one has for any $t_0<\frac{\rho}{4}$, \begin{align*} u_N(e(x_0))-L_N(E) &\leq \frac{1}{\pi t_0^2}\iint_{|(x,y)-(x_0,0)|\leq t_0}[u_N(e(x+iy))-L_N]\,dx\,dy \\ &=\frac{1}{\pi t_0^2}\int_{|y|\leq t_0}\int_{|x- x_0|\leq \sqrt{t_0^2-|y|^2}}[u_N(e(x+iy))-L_N]\,dx\,dy. \end{align*} Furthermore \begin{align*} &\int_{|x- x_0|\leq \sqrt{t_0^2-|y|^2}}[u_N(e(x+iy))-L_N]dx\\ &=\Big(\int_{\{|x- x_0|\leq \sqrt{t_0^2-|y|^2}\}\bigcap \mathbb{B}_y}+\int_{\{|x- x_0|\leq \sqrt{t_0^2-|y|^2}\}\setminus \mathbb{B}_y} \Big)[u_N(e(x+iy))-L_N]dx. \end{align*} Note that \[ |u_N(e(x+iy))-L_N|\leq 2\delta,\quad \text{if } x\not\in \mathbb{B}_y \text{ and } y<\frac{\delta}{4e\pi C'_\rho}. \] So \[ \big|\int_{\{|x- x_0|\leq \sqrt{t_0^2-|y|^2}\}\setminus \mathbb{B}_y}[u_N(e(x+iy))-L_N]dx\big| \leq 2\delta\times(2\sqrt{t_0^2-|y|^2}). \] By Cauchy-Schwartz inequality, \begin{align*} &\big|\int_{\{|x-x_0|\leq \sqrt{t_0^2-|y|^2}\}\bigcap \mathbb{B}_y}[u_N(e(x+iy))-L_N]dx\big|\\ &\leq \Big(\int_0^1|u_N(e(x+iy))-L_N|^2dx\Big)^{1/2} (\operatorname{meas}\mathbb{B}_y)^{1/2}\\ &\leq C_7\exp(-\frac{\check{c}}{2}\delta^2 N). \end{align*} Set $t_0=\delta/(4e\pi C'_\rho)$; then \begin{align*} u_N(e(x))-L_N &\leq \frac{1}{\pi t_0^2}\int_{|y|\leq t_0}[C_7\exp(-\frac{\check{c}}{2}\delta^2 N)+2\delta\times(2\sqrt{t_0^2-|y|^2})]dy \\ &\leq \frac{1}{\pi t_0^2}\times C_7\exp(-\frac{\check{c}}{2}\delta^2 N)\times (2t_0)+2\delta \\ &= \frac{8e C_7C'_\rho}{\delta}\exp(-\frac{\check{c}}{2}\delta^2 N)+2\delta . \end{align*} Set $\delta=\big(\frac{C_8\log N}{N}\big)^{1/2}$, where $C_8>2/\check{c}$. Then $\exp(-\frac{\check{c}}{2}C_8\log N)<\frac{1}{N}$, and \[ u_N(e(x))\leq L_N +8e C_7C'_\rho\times(\frac{N}{C_8\log N})^{1/2} \frac{1}{N}+2(\frac{C_8\log N}{N})^{1/2} \leq L_N +C_6(\frac{\log N}{N})^{1/2}. \] \end{proof} \begin{lemma}\label{22021} Set \[ F_N(x,E):=\frac{1}{2N}\sum_{n=0}^{N-1} \big(\log|\det A(e(x+n\omega),E)|-D(E)\big). \] Then \begin{itemize} \item[(1)] For any $N$ and any $k$ holds \begin{align*} &|\log\|\tilde{A}_N(e(x+k\omega),E)\|-\log\|\tilde{A}_N(e(x),E)\|\, |\\ &\leq 2k(\log C_{\rm max}(A,E)-\frac{1}{2}D(E))-kF_k(x)-kF_k(x+N\omega), \end{align*} \item[(2)] For any $N$ and any $k\geq \check{N}$ holds \begin{align*} &|\log\|\tilde{A}_N(e(x+k\omega),E)\|-\log\|\tilde{A}_N(e(x),E)\|\,|\\ &\leq 2k\tilde{L}_k(E)+2C_6(k\log k)^{1/2}-kF_k(x)-kF_k(x+N\omega). \end{align*} \end{itemize} \end{lemma} \begin{proof} (1) Recall that any $N$, $u_N(e(x),E)\leq \log C_{\rm max}(A,E)$. Then \begin{align*} &\log\|\tilde{A}_N(e(x),E)\|\\ &= Nu_N(e(x),E)-\frac{1}{2}\sum_{n=0}^{N-1}\log |\det A(e(x+n\omega),E)| \\ &\leq N\log C_{\rm max}(A,E) -\frac{1}{2}\sum_{n=0}^{N-1}\log |\det A(e(x+n\omega),E)| \\ &= N (\log C_{\rm max}(A,E)-\frac{1}{2}D(E) )-\frac{1}{2}\sum_{n=0}^{N-1}(\log|\det A(e(x+n\omega),E)|-D(E)) \\ &= N (\log C_{\rm max}(A,E)-\frac{1}{2}D(E))-NF_N(x,E) \end{align*} One has if $\det M=1$, then $\|M\|=\|M^{-1}\|$ and \[ \tilde{A}_N(e(x+k\omega),E)\tilde{A}_k(e(x),E) =\tilde{A}_k(e(x+N\omega),E)\tilde{A}_N(e(x),E). \] Thus, \begin{align*} &|\log\|\tilde{A}_N(e(x+k\omega),E)\|-\log\|\tilde{A}_N(e(x))\|\,|\\ &\leq \log\|\tilde{A}_k(e(x),E)\|+\log\|\tilde{A}_k(e(x+N\omega),E)\| \\ &\leq 2k (\log C_{\rm max}(A,E)-\frac{1}{2}D(E) )-kF_k(x,E)-kF_k(x+N\omega,E) \end{align*} (2) From the previous lemma, we know that for any $N\geq \check{N}(A)$, \[ \frac{1}{N}\log\|A_N(e(x),E)\| \leq L_N(E) +C_6(\frac{\log N}{N})^{1/2}. \] Then the rest follows as in part (1). \end{proof} \section{Avalanche principle and the sharp large deviation theorem} For the rest of the paper, without special statement, $N\geq \check{N}$ and $N\geq \check{K}$ from now on ($\delta$ in $\check{K}$ will be defined in Lemma \ref{22012}). Furthermore, we do not use $e(x+iy)$ with $y\not=0$ and write $x$ instead of $e(x)$ in all expressions. \begin{proposition} \label{prop:AP} Let $A_1,\ldots,A_n$ be a sequence of $2\times 2$-matrices whose determinants satisfy $ %\label{eq:detsmall} \max_{1\le j\le n}|\det A_j|\le 1$. Suppose that \begin{gather*} \min_{1\le j\le n}\|A_j\|\ge\mu>n, %\label{large} \\ \max_{1\le j100\delta>0$, where $\delta<1$ is a constant not depending on $N$, and $\tilde{L}_{2N}(E)>\frac{9}{10}\tilde{L}_{N}(E)$. Let $N'=m N$, $m\in \mathbb{N}$ and $ m\leq \exp(\frac{\tilde{c}}{4}\delta^2 N) $. Then \[ |\tilde{L}_{N'}(E)+\tilde{L}_{N}(E)-2\tilde{L}_{2N}(E)| \leq \exp(-\tilde{c}'\delta^2 N)+\frac{2}{9m}L_N(E), \] where $\tilde{c}'=\tilde{c}'(A)$. If $\exp(\frac{\tilde{c}}{10}\delta^2 N) \leq m\leq \exp(\frac{\tilde{c}}{4}\delta^2 N)$, we have \[ |\tilde{L}_{N'}(E)+\tilde{L}_{N}(E)-2\tilde{L}_{2N}(E)| \leq \exp(-\hat{c}\delta^2 N), \] where $\hat{c}=\hat{c}(A)$. Furthermore, if $\tilde{L}_{N_0}(E)>100\delta>0$, $\tilde{L}_{2N_0}(E)>\frac{9}{10}\tilde{L}_{N_0}(E)$ and $\exp(-\hat{c}\delta^2N_0)\leq \delta/12$, then there exists $\tilde{N}_0=\tilde{N}_0(A,\delta,N_0)\leq (\exp(\frac{\tilde{c}}{8}\delta^2 N_0)+1)N_0$ such that for any $N\geq \tilde{N}_0$, \[ |\tilde{L}_N(E)+\tilde{L}_{N_0}(E)-2\tilde{L}_{2N_0}(E)|<\exp(-\bar{c}'\delta^2 N_0), \] where $\bar{c}'=\bar{c}'(A)$. Furthermore, \[ |\tilde{L}(E)+\tilde{L}_{N_0}(E)-2\tilde{L}_{2N_0}(E)|< \exp(-\bar{c}\delta^2 N_0), \] where $\bar{c}=\bar{c}(A)$. \end{lemma} \begin{proof} In section three in \cite{T}, we proved this lemma for Jacobi operator, which is a special $2\times 2$ analytic matrix. It is easy to see that the proof there is suitable in this general condition. \end{proof} \begin{lemma}\label{62003} Assume $\tilde{L}(E_0)>0$. There exists $\check{C}(A,\tilde{L}(E_0))$ such that with $\rho_0'=\big[\frac{\tilde{L}(E_0)}{200N} \exp\big(-\check{C}(A,\tilde{L}(E_0))N\big)\big]^{1/\beta}$, one has \[ |\tilde{L}_{N}(E_0)-\tilde{L}_{N}(E)|<\frac{\tilde{L}(E_0)}{100}, \] for any $|E-E_0|<\rho_0'(E_0,N)$ and any $N$. \end{lemma} \begin{proof} Note that \begin{equation} \label{70007} \begin{aligned} &|\,\|A_{N}(x,E_0)\|-\|A_{N}(x,E)\|\, |\\ &\leq \|A_{N}(x,E_0)-A_{N}(x,E)\| \\ &\leq \sum_{j=0}^{N-1}\big(\|A(x+(N-1)\omega,E_0)\dots A(x+(j+1)\omega,E_0)\| \\ &\quad\times \|A(x+j\omega,E_0)- A(x+j\omega,E)\|\, \|A(x+(j-1)\omega,E) \dots A(x,E)\|\big ) \\ &\leq NC_{\rm max}(A)^{N-1}\times C(A)|E_0-E|^{\beta} . \end{aligned} \end{equation} By \eqref{26001}, if $|\det A_N(x,E_0)|\leq |\det A_N(x,E)|$, one has \begin{equation} \label{70008} \begin{aligned} |\,\|\tilde{A}_{N}(x,E_0)\|-\|\tilde{A}_{N}(x,E)\|\, | &= \big|\frac{\|A_{N}(x,E_0)\|}{|\det A_N(x,E_0)|^{1/2}} -\frac{\|A_{N}(x,E)\|}{|\det A_N(x,E)|^{1/2}}\big |\\ &\leq \frac{NC_{\rm max}(A)^{N-1} C(A)|E_0-E|^{\beta}}{|\det A_N(x,E_0)|^{1/2}} \end{aligned} \end{equation} Assume for instance that $\|\tilde{A}_{N}(x,E_0)\|\geq \|\tilde{A}_{N}(x,E)\|$. Then \begin{equation} \label{60002} \begin{aligned} & |\log\|\tilde{A}_{N}(x,E_0)\|-\log \|\tilde{A}_{N}(x,E)\| \,|\\ &= \log\frac{\|\tilde{A}_{N}(x,E_0)\|}{\|\tilde{A}_{N}(x,E)\|} =\log (1+\frac{\|\tilde{A}_{N}(x,E_0)\| -\|\tilde{A}_{N}(x,E)\|}{\|\tilde{A}_{N}(x,E)\|})\\ &\leq \frac{\|\tilde{A}_{N}(x,E_0)\|- \|\tilde{A}_{N}(x,E)\|}{\|\tilde{A}_{N}(x,E)\|} \leq\|\tilde{A}_{N}(x,E_0)\|- \|\tilde{A}_{N}(x,E)\| \\ &\leq \frac{NC_{\rm max}(A)^{N-1} C(A)|E_0-E|^{\beta}}{|\det A_N(x,E_0)|^{1/2}}. \end{aligned} \end{equation} By Theorem \ref{22006}, for any $\delta$ and any $K$ \[ \operatorname{meas}\Big\{x:|\sum_{k=1}^K\log |\det A(x+k\omega)|-K\langle \log |\det A(\cdot)|\rangle | >\delta K\Big\}<\exp(-c'\delta K). \] Thus if $x\not\in\mathbb{B}_1$, $\operatorname{meas}{\mathbb{B}_1}<\exp(-c'\times\frac{800C_2(A)}{\tilde{L}(E_0)c'}N) =\exp(-\frac{800C_2(A)}{\tilde{L}(E_0)}N)$, then \begin{align} |\log |\det A_N(x,E_0)|\, | &< |\langle \log |\det A(\cdot,E_0)|\rangle |N+\frac{800C_2(A)}{\tilde{L}(E_0)c'}N \nonumber \\ &=|D(E_0)|N+\frac{800C_2(A)}{\tilde{L}(E_0)c'}N \nonumber \\ &\leq D(A)N+\frac{800C_2(A)}{\tilde{L}(E_0)c'}N =\hat{C}(A,\tilde{L}(E_0))N. \label{30010} \end{align} It is obvious that the same estimate holds in the other three conditions. So if $x\not\in\mathbb{B}_1$, $\operatorname{meas}{\mathbb{B}_1} <\exp(-\frac{800\tilde{C}(A)^{1/2}}{\tilde{L}(E_0)}N)$, then \begin{align*} |\tilde{u}_{N}(x,E_0)-\tilde{u}_{N}(x,E)| &\leq C_{\rm max}(A)^{N-1}C(A)|E_0-E|^{\beta}\exp(\hat{C}(A,\tilde{L}(E_0))N)\\ &\leq \exp\left(\check{C}(A,\tilde{L}(E_0))N\right)|E_0-E|^{\beta}. \end{align*} Set $\rho'_0=\big[\frac{\tilde{L}(E_0)}{200} \exp\left(-\check{C}(A,\tilde{L}(E_0))N\right )\big]^{1/\beta}$. Then, if $|E-E_0|\leq \rho_0'$, we have $\left|\tilde{u}_{N}(x,E_0)-\tilde{u}_{N}(x,E) \right| <\frac{\tilde{L}(E_0)}{200}$, and if $x\not\in\mathbb{B}_1,\operatorname{meas}{\mathbb{B}_1} <\exp(-\frac{800\tilde{C}(p,q)^{1/2}}{\tilde{L}(E_0)}N)$, we also have \begin{equation}\label{60012} \Big|\int_{\mathbb{T}\backslash \mathbb{B}_1}\tilde{u}_{N}(x,E_0)dx-\int_{\mathbb{T}\backslash \mathbb{B}_1}\tilde{u}_{N}(x,E)dx \Big| <\frac{\tilde{L}(E_0)}{200}. \end{equation} By the Cauchy-Schwartz inequality, \begin{align*} \big|\int_{\mathbb{B}_1}\tilde{u}_{N}dx\big| &=\big|\int_{\mathbb{T}}\tilde{u}_N1_{\mathbb{B}_1}dx\big|\\ &\leq \|\tilde{u}_N(E)\|_2 (\operatorname{meas}\mathbb{B}_1)^{1/2}\\ &\leq C_2(A)\exp(-\frac{400C_2(A)}{\tilde{L}(E_0)}N) \end{align*} for $E $ or $E_0$. As $y\exp(-\xi y)\leq \xi^{-1}$ for any $y, \xi>0$. Thus \begin{equation}\label{60013} \big|\int_{\mathbb{B}_1}\tilde{u}_{N}dx\big| \leq \frac{\tilde{L}(E_0)}{400N} \leq \frac{\tilde{L}(E_0)}{400} \end{equation} for $E$ or $E_0$. Combining \eqref{60012} with \eqref{60013}, one has \[ |\tilde{L}_{N}(E_0)-\tilde{L}_{N}(E)| <\frac{\tilde{L}(E_0)}{200}+2\frac{\tilde{L}(E_0)}{400} =\frac{\tilde{L}(E_0)}{100}. \] \end{proof} \begin{lemma}\label{62006} Assume $\tilde{L}(E_0)>0$. There exists $\rho=\rho(A,\tilde{L}(E_0))>0$ and $\tilde{N}_0=\tilde{N}_0(A,E_0)<+\infty$ such that for any $N\geq\tilde{N}_0$ and any $|E-E_0|<\rho$ \[ |\tilde{L}_{N}(E)-\tilde{L}(E)|<\frac{1}{20}\tilde{L}(E),\quad \frac{11}{10}\tilde{L}(E_0)>\tilde{L}(E)>\frac{9}{10}\tilde{L}(E_0). \] \end{lemma} \begin{proof} One has $\lim_{n\to\infty}\tilde{L}(E_0)=\tilde{L}(E_0)$. Therefore, there exists $N_0=N_0(A,E_0)$ such that the following statements hold: \begin{itemize} \item[(1)] $|\tilde{L}_n(E_0)-\tilde{L}(E_0)|<\frac{\tilde{L}(E_0)}{100}$ for $n\geq N_0(A,E_0)$, which implies that $\tilde{L}_{N_0}(E_0)-\tilde{L}_{2N_0}(E_0)<\frac{\tilde{L}(E_0)}{100}$, as $\tilde{L}(E_0)\leq \tilde{L}_{2N_0}(E_0)\leq \tilde{L}_{N_0}(E_0)$; \item[(2)] $\exp(-\hat{c}\delta^2N_0)\leq \frac{\delta}{12}$, $\exp(-\bar{c}\delta^2 N_0))<\frac{1}{50}\tilde{L}(E_0)$, $\exp(-\bar{c}'\delta^2 N_0))<\frac{1}{50}\tilde{L}(E_0)$ with $\delta=\min(\frac{1}{200}\tilde{L}(E_0),\frac{1}{2})$, where $\hat{c}$ , $\bar{c}$ and $\bar{c}'$ are as Lemma \ref{62001}. \end{itemize} Using Lemma \ref{62003} applied to $N_0$ and $2N_0$. One has for $|E-E_0|<\rho(A,\tilde{L}(E_0)):=\rho'_0(\tilde{L}(E_0),2N_0)$, \begin{equation} \label{60033} \begin{aligned} \tilde{L}_{N_0}(E) &\geq \tilde{L}(E_0)- |\tilde{L}_{N_0}(E)-\tilde{L}_{N_0}(E_0)|-|\tilde{L}_{N_0}(E_0)-\tilde{L}(E_0)|\\ &> \tilde{L}(E_0)-\frac{\tilde{L}(E_0)}{100}-\frac{\tilde{L}(E_0)}{100} =\frac{49}{50}\tilde{L}(E_0), \end{aligned} \end{equation} and \begin{equation} \label{60034} \begin{aligned} &|\tilde{L}_{N_0}(E)-\tilde{L}_{2N_0}(E)|\\ &\leq |\tilde{L}_{N_0}(E)-\tilde{L}_{N_0}(E_0)|+|\tilde{L}_{N_0}(E_0) -\tilde{L}_{2N_0}(E_0)|+|\tilde{L}_{2N_0}(E_0)-\tilde{L}_{2N_0}(E)|\\ &< \frac{\tilde{L}(E_0)}{100}+\frac{\tilde{L}(E_0)}{100}+\frac{\tilde{L}(E_0)}{100} =\frac{3}{100}\tilde{L}(E_0)<\frac{1}{10}\tilde{L}_{N_0}(E). \end{aligned} \end{equation} Thus Lemma \ref{62001} applies with $\tilde{L}_{N_0}(E)$, $\delta$, $N_0$ and $E$. Then there exists a number $\tilde{N}_0=\tilde{N}_0(A,\delta,N_0)\leq (\exp(\frac{\tilde{c}}{8}\delta^2 N_0)+1)N_0$ such that for any $N\geq \tilde{N}_0$ there holds \begin{gather} |\tilde{L}_N(E)+\tilde{L}_{N_0}(E)-2\tilde{L}_{2N_0}(E)|<\exp(-\bar{c}'\delta^2 N_0), \nonumber \\ \label{60035} |\tilde{L}(E)+\tilde{L}_{N_0}(E)-2\tilde{L}_{2N_0}(E)| < \exp(-\bar{c}\delta^2 N_0), \end{gather} where $\bar{c}'=\bar{c}'(A)$ and $\bar{c}=\bar{c}(A)$ are as in Lemma \ref{62001}. These imply \begin{equation} \label{60036} \begin{aligned} |\tilde{L}(E)-\tilde{L}_{N}(E)| &\leq\exp(-\bar{c}'\delta^2 N_0)+\exp(-\bar{c}\delta^2 N_0)\\ &< \frac{1}{50}\tilde{L}(E_0)+\frac{1}{50}\tilde{L}(E_0) =\frac{1}{25}\tilde{L}(E_0). \end{aligned} \end{equation} Combining \eqref{60033}, \eqref{60034} with \eqref{60035}, one obtains \begin{align*} &|\tilde{L}(E_0)-\tilde{L}(E)|\\ &\leq |\tilde{L}(E)+\tilde{L}_{\tilde{N}_0}(E)-2\tilde{L}_{2\tilde{N}_0}(E)| +|\tilde{L}(E_0)-\tilde{L}_{\tilde{N}_0}(E)|+2|\tilde{L}_{\tilde{N}_0}(E)- \tilde{L}_{2\tilde{N}_0}(E)| \\ &< \frac{1}{50}\tilde{L}(E_0)+\frac{1}{50}\tilde{L}(E_0) +2\frac{3}{100}\tilde{L}(E_0)=\frac{1}{10}\tilde{L}(E_0). \end{align*} It implies \begin{gather*} %\label{60038} \frac{11}{10}\tilde{L}(E_0)>\tilde{L}(E)>\frac{9}{10}\tilde{L}(E_0), \\ |\tilde{L}(E)-\tilde{L}_{N}(E)| <\frac{1}{25}\tilde{L}(E_0)<\frac{1}{25} (\frac{10}{9})\tilde{L}(E) =\frac{2}{45}\tilde{L}(E)<\frac{1}{20}\tilde{L}(E). \end{gather*} \end{proof} \begin{remark}\label{2} \rm From \eqref{30010}, $\hat{C}(A,\tilde{L}(E_0))=D(A)+\frac{800\tilde{C}(A)^{1/2}}{\tilde{L}(E_0)c'}$. It is obvious that $\hat{C}(A,\tilde{L}(E_0))$ is a positive constant, which is smaller when $\tilde{L}(E_0)$ becomes larger. Thus $\check{C}(A,\tilde{L}(E_0))=\hat{C}(A,\tilde{L}(E_0))+\log C(A)$ is a positive and monotonically increasing function of $\tilde{L}(E_0)$. Recall that \begin{equation}\label{60020} 0<\rho(A,\tilde{L}(E_0))=\big[\frac{\tilde{L}(E_0)}{200} \exp\big(-2\check{C}(A,\tilde{L}(E_0))N_0\big)\big]^{1/\beta} < 1. \end{equation} $N_0$ depends on the rate of convergence of $\tilde{L}_n(E_0)\to\tilde{L}(E_0)$ (see the proof of Lemma \ref{62006}). On the other hand, in Lemma \ref{62001}, let $m=\exp(\frac{\tilde{c}}{8}\delta^2 N)$, $N'=m N$, then \[ |\tilde{L}_{N'}-\tilde{L}_{2N'}|<2\exp(-\hat{c}\delta^2 N)<\big(\frac{1}{N'}\big)^{C_1(A)}. \] Like the induction in the proof of Lemma \ref{62001}, one gets \begin{equation}\label{60021} |\tilde{L}_{N}-\tilde{L}|<\big(\frac{1}{N}\big)^{C_2(A)}. \end{equation} Actually, in \cite{S}, the better estimate holds as \[ |\tilde{L}_{N}-\tilde{L}|<\frac{C}{N}, \] with the large constant $C$ depending on $\tilde{L}$. Finally, combining \eqref{60020}, \eqref{60021} with some assumptions in Lemma \ref{62001} and Lemma \ref{62006}, we have \[ %\label{60022} \rho(A,\tilde{L}(E_0))\simeq\Big[\tilde{L}(E_0) \exp\Big(-\frac{C_{\rho,1}}{(\tilde{L}(E_0))^{C_{\rho,2}}}\Big)\Big]^{1/\beta}, \] where $C_{\rho,1}, C_{\rho,2}$ are the big constants depending only on $A(x,E)$. \end{remark} The first part of this section shows that the uniform property that for any $E\in (E_0-\rho,E_0+\rho)$ and for the uniformly large $N$, the distance between $ \tilde{L}_{N}(E)$ and $\tilde{L}(E)$ is controlled by a uniform constant $\tilde{L}(E)$. This property is very important and will be applied repeatedly in the following part of this paper, such like the sharp Large Deviation Theorem (Lemma \ref{22012}) and the proof of the main theorem (Section 4). \begin{lemma}\label{22012} Assume $\tilde{L}(E_0)>0$. There exist $N_1(A,E_0)$ such that for any $N\geq N_1$ and any $E\in(E_0-\rho,E_0+\rho)$ holds \[ \operatorname{meas}\{x:|\tilde{u}_N(x,E)-\tilde{L}(E)| > \frac{\tilde{L}(E)}{10}\}<\exp(-c \tilde{L}(E) N), \] where constant $c$ depends only on $A$, but does not depend on $E$ or $E_0$. \end{lemma} \begin{proof} Choose $\bar{N}_0$ such that \begin{equation}\label{21118} \begin{gathered} \bar{N}_0>\max\Big(\frac{10^4C_6}{ \tilde{L}(E_0)},\max_{E\in\mathscr{E}}\frac{10^6(\log C_{\rm max}(A,E)-\frac{1}{2}D(E))}{ \tilde{L}(E_0)},40,\tilde{N}_0\Big),\\ \log \bar{N}_0<\bar{N}_0^{1/3} \end{gathered} \end{equation} where $(\log C_{\rm max}(A,E)-D(E))$ is as in Lemma \ref{22021}, $C_6$ is as in Lemma \ref{22010}, $\tilde{N}_0$ is as in Lemma \ref{62006}. Take $N\geq \bar{N}_0^3$, $K:=\frac{1}{800}N\geq \bar{N}_0$. Thus for any $E\in(E_0-\rho,E_0+\rho)$, \begin{equation} \label{21119} \tilde{L}_{K}(E)<(1+\frac{1}{20})\tilde{L}(E). \end{equation} Using Lemma \ref{22021} one obtains \begin{align*} %\label{21124} &|\tilde{u}_N(x,E)-\frac{1}{K}\sum_{k=1}^K\tilde{u}_N(x+k\omega,E)|\\ &\leq \frac{1}{KN}\Big[ \sum_{k=1}^{\bar{N}_0}2k(\log C_{\rm max}(A,E) -\frac{1}{2}D(E)) +\sum_{k=\bar{N}_0+1}^K 2 k\tilde{L}_k(E)\\ &\quad +\sum_{k=\bar{N}_0+1}^K 2C_6(k\log k)^{1/2}\Big] -\frac{1}{KN}\sum_{k=1}^K(kF_k(x)+kF_k(x+N\omega)) \\ &=I+II . \end{align*} By \eqref{21118}, \eqref{21119} and Lemma \ref{62006}, one has \begin{equation} \label{21122} \begin{aligned} I&<\frac{\bar{N}_0^2(\log C_{\rm max}(A,E)-\frac{1}{2}D(E))}{KN} +4\frac{21}{20}\tilde{L}(E)\frac{K}{N}++\frac{C_6K^{1/2}(\log K)^{1/2}}{N}\\ &< \frac{ \tilde{L}(E_0)}{320}+\frac{\tilde{L}(E)}{160}+\frac{ \tilde{L}(E_0)}{320}=\frac{\tilde{L}(E)}{80}. \end{aligned} \end{equation} If $\sum_{k=1}^KkF_k(x,E)<-\frac{KN}{160} \tilde{L}(E)$, then there is a $k$ such that $ kF_k(x,E)<-\frac{N}{160}\tilde{L}(E)$. We know that \[ \operatorname{meas}\{x:|kF_k(x)-k\langle F_k(x)\rangle |>k\delta\} <\exp(-c\delta k), \] Since $\langle F_k\rangle =0$, \begin{align*} \operatorname{meas}\{x:kF_k(x) &<-\frac{N}{160} \tilde{L}(E)\}\\ &\leq \operatorname{meas}\{x:|kF_k(x)|>\frac{N}{160} \tilde{L}(E)\}\\ & <\exp(-c\frac{N \tilde{L}(E)}{160k}k) =\exp(-c_2 N \tilde{L}(E)). \end{align*} So \begin{equation} \label{21123} \begin{aligned} &\operatorname{meas}\Big\{x:\sum_{k=1}^K kF_k(x)<-\frac{KN}{160}\tilde{L}(E)\Big\}\\ &\leq K\exp(-c_2 N \tilde{L}(E)<\exp(-c'_2 \tilde{L}(E) N), \end{aligned} \end{equation} if $N$ is large enough depending on $L(E_0)$ (see Lemma \ref{62006} and \eqref{21118}). Combining \eqref{21122} with \eqref{21123} one has \begin{equation}\label{21124} \begin{aligned} &\operatorname{meas}\Big\{x:|\tilde{u}_N(x,E)-\frac{1}{K} \sum_{k=1}^K\tilde{u}_N(x+k\omega,E)|>\frac{\tilde{L}(E)}{40}\Big\}\\ &\leq 2\exp(-c'_2 N \tilde{L}(E)) <\exp(-c'' \tilde{L}(E)N). \end{aligned} \end{equation} On the other hand, recall that by Lemma \ref{lem2.3}, for any $K$, \begin{gather*} \operatorname{meas}\Big\{x:|\sum_{k=1}^Ku_N(x+k\omega)-K| >\delta K\Big\} <\exp(-c\delta K), \\ \operatorname{meas}\Big\{x:|\sum_{k=1}^K\frac{1}{N}\log |\det A(x+k\omega)|-KD|>\delta K\Big\} <\exp(-c'\delta K). \end{gather*} By the definition of $\tilde{u}_N(x,E)$ and $\tilde{L}_N(E)$, there exists $\check{K}=\check{K}(A, E_0)$ such that for any $K>\check{K}$ holds (with $\delta:=\frac{ \tilde{L}(E)}{40}$) \[ \operatorname{meas}\Big\{x:|\sum_{k=1}^K\tilde{u}_N(x+k\omega,E) -K\langle \tilde{u}_N(\cdot,E)\rangle| >\frac{\tilde{L}(E)}{40} K\Big\} \leq \exp(-\hat{c} \tilde{L}(E) K). \] Note that if \[ \big|\frac{1}{K}\sum_{k=1}^K\tilde{u}_N(x+k\omega,E)-\langle \tilde{u}_N(\cdot,E)\rangle \big|\leq\frac{ \tilde{L}(E)}{40}, \] then \begin{align*} &|\frac{1}{K}\sum_{k=1}^K\tilde{u}_N(x+k\omega,E)-\tilde{L}(E)|\\ &\leq |\frac{1}{K}\sum_{k=1}^K\tilde{u}_N(x+k\omega,E) -\tilde{L}_N(E)|+|\tilde{L}_N(E)-\tilde{L}(E)| \\ &< \frac{1}{40}\tilde{L}(E)+\frac{1}{20} \tilde{L}(E) = \frac{3}{40} \tilde{L}(E)\,. \end{align*} Therefore \[ \operatorname{meas}\Big\{x:|\frac{1}{K} \sum_{k=1}^K\tilde{u}_N(x+k\omega,E)-\tilde{L}(E)|>\frac{3}{40} \tilde{L}(E)\Big\}<\exp(-\hat{c} \tilde{L}(E) K). \] Combining this with \eqref{21124}, there exists $N_1=N_1(A,E_0)$ such that for any $N\geq N_1$ holds \[ \operatorname{meas}\{x:|\tilde{u}_N(x,E)-\tilde{L}(E)|>\frac{ \tilde{L}(E)}{10}\} <\exp(-c_0 L(E) N), \] where $c_0$ depends only on $A$. Here we replace $c_0$ by $c$ for convenient notations. \end{proof} \begin{lemma}\label{62010} Assume $\tilde{L}(E_0)>0$. Let $N\geq N_1$ and $E\in(E_0-\rho,E_0+\rho)$ be arbitrary. Then \begin{equation} |\tilde{L}(E)+\tilde{L}_{N}(E)-2\tilde{L}_{2N}(E)|<\exp(-c \tilde{L}(E) N), \end{equation} where $c=c(A)$. \end{lemma} \begin{proof} By the sharp Large Deviation Theorem, Lemma \ref{22012}, this lemma get better conclusion than Lemma \ref{62001}, which comes from Theorem \ref{22006} and Remark \ref{29}, but the proof here is the same as the later one. \end{proof} \section{proof of the main theorem} Let $\tilde{L}(E_0)>0$. By Lemma \ref{62006} and \eqref{60036}, one has that for any $N\geq \tilde{N}_0(A,E_0)$ and $E\in(E_0-\rho,E_0+\rho)$, \begin{equation} \label{40001} \tilde{L}_N(E)\leq \tilde{L}(E)+\frac{1}{25}\tilde{L}(E_0) <\frac{11}{10}\tilde{L}(E_0)+\frac{1}{25}\tilde{L}(E_0) =\frac{57}{50}\tilde{L}(E_0). \end{equation} Let $E'\to E$ such that $|D(E)-D(E')|\leq \frac{1}{5}\tilde{L}(E_0)$. Thus, \begin{equation} \label{40002} \begin{aligned} L_N(E)-\frac{1}{2}D(E') &\leq L_N(E)+\frac{1}{10}\tilde{L}(E_0)-\frac{1}{2}D(E)\\ &=\tilde{L}_N(E) +\frac{1}{10}\tilde{L}(E_0) \leq \frac{62}{50}\tilde{L}(E_0). \end{aligned} \end{equation} Assume $\|\tilde{A}_N(x,E)\|\geq \|\tilde{A}_N(x,E')\|$ and $|\det A_N(x,E)|\leq |\det A_N(x,E)|$. By \eqref{70007}, \eqref{70008} and \eqref{60002}, \begin{equation} \label{40003} \begin{aligned} &|\log\|\tilde{A}_N(x,E)\|-\log \|\tilde{A}_N(x,E')\| \,|\\ &\leq \|\tilde{A}_N(x,E)\|- \|\tilde{A}_N(x,E')\|=\frac{\|A_N(x,E)\|}{|\det A_N(x,E)|^{1/2}} -\frac{\|A_N(x,E')\|}{|\det A_N(x,E')|^{1/2}} \\ &\leq \frac{\|A_N(x,E)\|- \|A_N(x,E')\|}{|\det A_N(x,E)|^{1/2}}\leq \frac{\|A_N(x,E)-A_N(x,E')\|}{|\det A_N(x,E)|^{1/2}} \\ &\leq C(A)|E-E'|^{\beta}\frac{\sum_{j=0}^{N-1} \|\prod_{m=1}^{N-j}A(x+(N-m)\omega,E)\|\, \|\prod_{m=j-1}^0A(x+m\omega,E')\|}{|\det A_N(x,E)|^{1/2}} \\ &= C(A)|E-E'|^{\beta}\sum_{j=0}^{N-1} \Big(\|\prod_{m=1}^{N-j}A(x+(N-m)\omega,E)\|\, \|\prod_{m=j-1}^0A(x+m\omega,E')\|\Big) \\ &\quad \times \exp(-\frac{N}{2}D(E)) \exp(-N F_N(x,E)); \end{aligned} \end{equation} see Lemma \ref{22021} for the definition of $F_N(x)$. Let $N\geq N_2$, where $N_2(A):=\max_{E_1,E_2\in\mathscr{E}}\frac{2\log C_{\rm max}(A,E_1)-D(E_2)}{L(E_0)}\tilde{N}_0\geq 2\tilde{N}_0$. Then \begin{align*} \eqref{40003} &\leq \Big(\sum_{j=1}^{\tilde{N}_0}+\sum_{j=\tilde{N}_0+1}^{N-\tilde{N}_0} +\sum_{j=N-\tilde{N}_0+1}^{N}\Big) \|\prod_{m=1}^{N-j}A(x+(N-m)\omega,E)\|\\ & \quad \times\|\prod_{m=j-1}^0A(x+m\omega,E')\| \exp(-\frac{N}{2}D) \exp(-N F_N(x,E)) C(A)|E-E'|^{\beta} \\ &:=I+ II +III . \end{align*} By Lemma \ref{22010} and Lemma \ref{22021}, one has \begin{align*} I,III &\leq \sum_{j=1}^{\tilde{N}_0}\exp\Big( L_{N-j}(E)(N-j)+C_6(N-j)^{1/2}\log^{1/2}(N-j)\\ &\quad +\log C_{\rm max}(A,E')(j-1)-\frac{N}{2}D(E)\Big) \exp(-N F_N(x,E))\times C(A)|E-E'|^{\beta} \\ &\leq \sum_{j=1}^{\tilde{N}_0} \exp\Big(\tilde{L}_{N-j}(E)(N-j)+C_6N^{1/2}\log^{1/2}N +\big (\log C_{\rm max}(A,E')\\ &\quad -\frac{1}{2}D(E)\big )(j-1) -\frac{1}{2}D(E) \Big) \exp(-N F_N(x,E))\times C(A)|E-E'|^{\beta} \\ &\leq \sum_{j=1}^{\tilde{N}_0} \exp \Big(\frac{1}{2}\tilde{L}(E_0)N+\frac{62}{50}\tilde{L}(E_0)N+C_6N^{1/2}\log^{1/2}N -\frac{1}{2}D\Big) \\ &\quad\times \exp(-N F_N(x,E))C(A)|E-E'|^{\beta}, \end{align*} and \begin{align*} II &\leq \sum_{j=\tilde{N}_0+1}^{N-\tilde{N}_0}\exp\Big( L_{N-j}(E)(N-j)+C_6(N-j)^{1/2}\log^{1/2}(N-j)+L_{j-1}(E')(j-1)\\ &\quad +C_6(j-1)^{1/2}\log^{1/2}(j-1) -\frac{N}{2}D(E)\Big) \exp(-N F_N(x,E))\times C(A)|E-E'|^{\beta} \\ &\leq \sum_{j=N-\tilde{N}_0+1}^{N}\exp\Big( \big(\log C_{\rm max}(A,E) -\frac{1}{2}D(E)\big)(N-j)+ \Big(L_{j-1}(E')\\ &\quad -\frac{1}{2}D(E)\Big)(j-1) +C_6N^{1/2}\log^{1/2}N-\frac{1}{2}D(E)\Big)\\ &\quad\times \exp(-N F_N(x,E))\times C(A)|E-E'|^{\beta} \\ &\leq \sum_{j=\tilde{N}_0+1}^{N-\tilde{N}_0} \exp\Big(\frac{62}{50}\tilde{L}(E_0)N+C_6N^{1/2}\log^{1/2}N-\frac{1}{2}D \Big) \\ &\quad\times \exp(-N F_N(x,E))C(A)|E-E'|^{\beta}. \end{align*} Combining these expressions, we have \begin{align*} \eqref{40003} &\leq \sum_{j=1}^N \exp\Big(\frac{87}{50}\tilde{L}(E_0)N +C_6N^{1/2}\log^{1/2}N-\frac{1}{2}D \Big) \exp(-N F_N(x,E)) \\ &\quad \times C(A)|E-E'|^{\beta}. \end{align*} There exists $N_3=N_3(A,E_0)$ such that for any $N\geq N_3$ there holds \[ C(A)\sum_{j=1}^N \exp\Big(\frac{87}{50}\tilde{L}(E_0)N+C_6N^{1/2}\log^{1/2}N -\frac{1}{2}D \Big) \leq \exp(2\tilde{L}(E_0)N). \] It is easy to see that there are similar processes for the other three conditions. Thus \[ \eqref{40003}\leq |E-E'|^{\beta}\exp(2\tilde{L}(E_0)N) \max\{\exp(-N F_N(x,E))\exp(-N F_N(x,E'))\}. \] Set \[ \mathbb{B}(E):=\{x:NF_N(x,E)<-N\tilde{L}(E_0)\}, \] then \begin{align*} \operatorname{meas}(\mathbb{B}(E)) &\leq \operatorname{meas}(\{x:|NF_N(x,E)-N\langle F_N(\cdot,E)\rangle| >N\tilde{L}(E_0)\})\\ &<\exp(-c\tilde{L}(E_0)N), \end{align*} since $=0$. By the Cauchy-Schwartz inequality, one has \begin{align*}%\label{40004} |\tilde{L}_N(E)-\tilde{L}_N(E')| &= \int_{\mathbb{T}\backslash(\mathbb{B}(E)\bigcup\mathbb{B}(E'))}|\tilde{u}_N(x,E) -\tilde{u}_N(x,E')|dx\\ &\quad +\int_{\mathbb{B}(E)\bigcup\mathbb{B}(E')}|\tilde{u}_N(x,E) -\tilde{u}_N(x,E')|dx \\ &< |E-E'|^{\beta} \exp (3\tilde{L}(E_0)N)+4C_2(A)\exp(-\frac{c}{2}\tilde{L}(E_0)N) . \end{align*} Let $N\geq N_4:=\max(N_1,N_2,N_3)$, where $N_1$ is as in Lemma \ref{22012}. By Lemma \ref{62010}, \begin{align*} &|\tilde{L}(E)-\tilde{L}(E')|\\ &\leq |\tilde{L}(E)+\tilde{L}_N(E)-2\tilde{L}_{2N}(E)| +|\tilde{L}(E')+\tilde{L}_N(E')-2\tilde{L}_{2N}(E')| \\ &\quad +|\tilde{L}_N(E)-\tilde{L}_N(E')|+2|\tilde{L}_{2N}(E)-\tilde{L}_{2N}(E')| \\ &< 2\exp(-c \tilde{L}(E_0)N)+3|E-E'|^{\beta} \exp (3\tilde{L}(E_0)N)+12C_2(A)\exp(-\frac{c}{2}\tilde{L}(E_0)N) \\ &< \exp(-c_7\tilde{L}(E_0)N)+3 \exp(3\tilde{L}(E_0)N)|E-E'|^{\beta}, \end{align*} where $c_7=c_7(A)$. Then when $E'\to E$, there exists $N\geq N_4$ such that \[ \exp\big(-(6+c_7)\tilde{L}(E_0)(N+1)\big)\leq |E-E'|^{\beta} \leq \exp\big(-(6+c_7)\tilde{L}(E_0)N\big). \] It implies \begin{align*} |\tilde{L}(E)-\tilde{L}(E')| &< 4\exp(-c_7\tilde{L}(E_0)N)\\ &=4\exp\big(-\frac{N}{N+1}c_7\tilde{L}(E_0)(N+1)\big) <\exp\big(-\frac{2c_7}{3}\tilde{L}(E_0)(N+1)\big) \\ &= \exp\big(-\frac{2c_7}{18+3c_7}\tilde{L}(E_0)N\big) < |E-E'|^{\frac{2c_7\times\beta}{18+3c_7}}. \end{align*} By the definition, one also has \begin{align*} %\label{main} |L_1(E)-L_1(E')|,\ |L_2(E)-L_2(E')| &\leq |\tilde{L}(E)-\tilde{L}(E')|+\frac{1}{2}|D(E)-D(E')| \\ &< |E-E'|^{\frac{2c_7\times\beta}{18+3c_7}}+\frac{C_D(A)}{2}|E-E'|^{\beta} \\ &\leq (1+\frac{C(A)}{2})|E-E'|^{\frac{2c_7\times\beta}{18+3c_7}}. \end{align*} \subsection*{Acknowledgments} The author was supported by the Fundamental Research Funds for the Central Universities (Grant 2013B01014) and by the National Nature Science Foundation of China (Grant 11326133). \begin{thebibliography}{00} \bibitem{AJ} A. 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