\begin{document}
\title{The H\"older continuity of Lyapunov exponents for a class of Cos-type quasiperiodic Schr\"odinger cocycles} \maketitle
\begin{center}
\author{
    Jiahao Xu\footnote{Chern Institute of Mathematics and LPMC, Nankai University, Tianjin 300071, China,  178881559@qq.com},\ \  Lingrui Ge \footnote{Department of Mathematics, Nanjing University, Nanjing, China,  824404328@qq.com},\ \  Yiqian Wang\footnote{The corresponding author.}\ \footnote{Department of Mathematics,
Nanjing University, Nanjing, China, yiqianw@nju.edu.cn}
  }
\end{center}

\fancyhead{}

\fancyhead[CO]{\small Lyapunov Exponents for a class of $C^2$ quasiperiodic Schr\"odinger cocycles}

\fancyhead[CE]{Jiahao Xu, Lingrui Ge and Yiqian Wang}

\begin{abstract} In this paper we obtain exact $\frac{1}{2}$-H\"older continuity of the
Lyapunov exponents for quasi-periodic Sch\"odinger cocycles with $C^2$ cos-type potentials,
large coupling constants, and fixed Diophantine frequency. Moreover, we prove the locally Lipschitz
continuity of the Lyapunov exponent for a full measure spectral set. Furthermore, for any given
$r$ between $\frac{1}{2}$ to $1$, we can find some energy on the spectrum and on which
Lyapunov exponent is exactly $r$-H\"older continuous.
\end{abstract}

\section{\textbf{Introduction}}
In this paper we consider the Schr\"odinger
operators on $l^2(\Z)\ni u=(u_n)_{n\in\Z}:$
\begin{equation}\label{1.1} (H_{\alpha,\lambda v,x}u)_n=u_{n+1}+u_{n-1}+\lambda v(x+n\alpha)u_n.\end{equation}
Here $v\in C^r(\R/\Z,\R),$ $r\in\N\cup\{\infty,\omega\}$ is the $potential,$ $\lambda\in\R~coupling~constant,$ $x\in\R/\Z~phase,$ and $\alpha\in\R/\Z~frequency.$ Sometimes we may leave $\alpha,\lambda v$ in $H_{\alpha,\lambda v,x}$ implicit. Let $\Sigma(H_x)$ be the spectrum of the operator. As is well known that \begin{equation} \label{spec} \Sigma(H_x)\subset [-2+|\lambda|\inf v,2+|\lambda|\sup v]:=\textit{I}.\end{equation}

Moreover, for irrational $\alpha$, $\Sigma(H_x)$ is phase-independent. Let $\Sigma_{\alpha,\lambda v}$ denote the common spectrum in this case.

Consider the eigenvalue equation $$H_xu=Eu.$$ Then so-called Sch\"odinger cocycle map $A^{(E-\lambda v)}: \R/\Z \rightarrow SL(2,\R)$ is given by

\begin{equation}\label{sm} A^{(E-\lambda v)}(x)=
\begin{pmatrix}
E-\lambda v(x)& -1\\
1& 0
\end{pmatrix},\end{equation}
Then $(\alpha,A^{(E-\lambda v)})$ defines a family of dynamical systems on $\R/\Z \times \R^2,$ which is given by

$$(x,\omega) \rightarrow (x+\alpha,A^{(E-\lambda v)}(x)\omega),$$
and is called Sch\"odinger cocycle. The $n$th iteration of dynamics is denoted by
$$(\alpha,A^{(E-\lambda v)})^n=(n\alpha,A_n^{(E-\lambda v)}).$$
Thus $$A_n(x)^{(E-\lambda v)}=A^{(E-\lambda v)}(x+(n-1)\alpha)\cdots A^{(E-\lambda v)}(x),n\geq 1, A_0^{(E-\lambda v)}=Id.$$
For $n\geq 1$, $$A_{-n}^{(E-\lambda v)}(x)=A_n^{(E-\lambda v)}(x-n\alpha)^{-1}.$$

Let $u\in \C^{\Z}$ be a solution of equation $H_{\lambda,x}u=Eu$
(note $u$ is not necessary in $l^2(\Z)$); then the relation between
the cocycle and the operator is given by
\[
A_n^{(E-\lambda v)}(x){u_0\choose u_{-1}}={u_{n}\choose u_{n-1}}.
\]

The Lyapunov Exponent $L(E,\lambda)$ of this cocycle is defined as $$\lim\limits_{n\rightarrow\infty}\frac{1}{n}\int_{\R/\Z}\ln\|A_n^{(E-\lambda v)}(x)\|dx
=\inf_n\frac{1}{n}\int_{\R/\Z}\ln\|A_n^{(E-\lambda v)}(x)\|dx\geq 0.$$ The limit
exists and is equal to the infimum since $\{\int_{\R/\Z}\ln\|A_n^{(E-\lambda v)}(x)\|
dx\}_{n\geq1}$ is a subadditive sequence. If in addition $T$ is
$\mu$-ergodic, then by Kingman's subadditive ergodic theorem we also
have
$$L(E, \lambda)=\lim\limits_{n\rightarrow\infty}\frac{1}{n}\ln\|A_n^{(E-\lambda v)}(x)\|$$
for almost every $x$.


In the last twenty years, a large amount of paper was delicated to study the regularity of Lyapunov exponent for quasi-periodic Schr\"odinger operators. Eventually, they found that the regularity of LE depends sensitively on the arithmetic property of frequency and the regularity of the potential.

For rational frequency or generic irrational frequency (extremely Liouvillean), LE is not H\"older continuous \cite{aj}.

The study of the regularity of LE starts from  G-S[00-Ann]. They proved that in the positive Lyapunov regime, if the potential is analytic and the frequency is a strong Diophantine number, Lyapunov exponent is H\"older continuous. Later, Bourgain \cite{bourgain1} proved that for AMO with the potential $2\lambda \cos x$,\ $|\lambda|\gg 1$  and Diophantine frequency, LE is $\frac 12-\epsilon$ H\"older continuous. It is generalized by G-S[08-GAFA] to the result that for arbitrary analytic potential near a trigonometric polynomial of degree $k$, it holds that LE is $\frac{1}{2k}-\epsilon$ H\"older continuous provided the frequency is Diophantine and the LE is positive. In the zero Lyapunov exponent regime, Puig\cite{Puig06} proved that for AMO with the potential $2\lambda \cos x, 0<|\lambda| \ll 1$ and Diophantine frequency, LE is locally $\frac 12$ H\"older continuous at end points of spectral gaps and cannot be better. Later, it was proved by Amor \cite{amor} that in the perturbative regime, if the frequency is Diophantine, then LE is $\frac 12$-H\"older continuous. Amor's result was extended by Avila and Jitomirskaya \cite{aj} to the non-perturbative regime and they also proved $\frac 12$ H\"older continuous for $\lambda\not=0, 1$ and all Diophantine frequencies. Recently A-J's result was further extended by Leguil-You-Zhao-Zhou \cite{LYZZ} to general subcritical potential and Diophantine frequency. In contrast, there is no $\frac 12$ H\"older continuous result on a general analytic potential in the positive LE regime.


If the potential is not analytic, generally LE is not continuous with respect to the potential in $C^k$ $(k=0,1,\cdots,\infty)$ topology, see [WY]. Then Wang and Zhang [WZ] provided the first regularity result for a finitely differential potential. They proved that for $C^2$ cos-like (Morse) potential with a large coupling, LE is weak-H\"older continuous. Subsequently it was improved to be H\"older continuous with a H\"older continuous exponent strictly less than $\frac 12$ by Liang, Wang and You [LWY].

  It can be seen that in [WZ] and [LWY], although the potential is only $C^2$, the cos-like condition is strong enough to make up for the deficiency of the lower regularity. Hence it is reasonable to expect that LE is $\frac 12$ H\"older continuous for all $C^2$ cos-like potential with a large coupling. This is the aim of this paper.


\section{\textbf{Main results}}

In this paper, from now on, for fixed $h\geq 2$ we say $v$ is $C^{h}-cos-type$ if $v$ satisfy the following conditions, which was first considered by Sinai \cite{sinai}:  $v
\in C^{h}(\R/\Z,\R);$ $\frac{dv}{dx}=0$ at exactly two points, one is minimal and the other maximal, which are denoted by $z_1$ and $z_2$. Moreover, these two extremals are non-degenerate, that is, $\frac{d^2v}{dx^2}(z_j)\neq0$ for $j=1,2.$

Fix two positive constants $\tau,\gamma.$ We say $\alpha$ satisfies a $Diophantine$ condition $DC_{\tau,\gamma}$ if\

$$\vert\alpha-\frac{p}{q}\vert\geq\frac{\gamma}{|q|^{\tau}}$$
for all $p,q\in\Z$ with $q\neq0.$

It is a standard result that for any $\tau>2,$
$$DC_{\tau}:=\bigcup\limits_{\gamma>0} DC_{\tau,\gamma}$$
is of full Lebesgue measure. We fix  $\tau>2$ and  $\alpha\in DC_{\tau}.$

For any $t_0\in \R$, we say a function $f$ is H\"older continuous at $t_0$ with a H\"older exponent $0<r\leq 1$, if there exists some constant $C>0$ and some nontrivial interval I centered by $t_0$  such that
$$|f(t_0)-f(t)|<C|t-t_0|^{r},~for~any~t~\in~I.$$
If in addition there also exists $c>0$ such that $$|f(t_0)-f(t)|>c|t-t_0|^{r},~for~any~t~\in~I,$$ we say  $f$ is exactly H\"older continuous at $t_0$ with a H\"older exponent $0<r\leq 1$.


\begin{theorem}\label{Th1} Let $\alpha$ be as above and $v$ be $C^{2+r}-cos-type$ with some fixed $r>0.$ Consider the Schr\"odinger cocycle with potential $v$ and coupling constant $\lambda$. Let $L(E,\lambda)$ be the associated Lyapunov exponents. Then there exists a $\lambda_1=\lambda_1(\alpha,v)>0 $ such that for any fixed $\lambda>\lambda_1$, the following hold true:

 (1) $L(\cdot, \lambda)$ is $\frac{1}{2}$-H\"older continuous on any compact interval I of E. That is,
$$|L(E, \lambda)-L(E',\lambda)|<C|E-E'|^{\frac{1}{2}},for~any~E, E'\in I$$
where $C>0$ depends on $\alpha,v,\lambda, I.$
\\
(2) All ends points of spectrum gaps consists of a dense set in the spectrum and for each such point $E$, $L(\cdot)$ is exactly $\frac{1}{2}$ H\"older continuous at $E$.
\\
(3)  There exists  a subset F of $\Sigma_{\alpha,\lambda v}$ with full measure such that for any $E'\in F$, $L(\cdot)$ is Lipschitz at $E'$.
\\
(4) For any fixed $\frac{1}{2}\leq\beta<1$, there exists some point $E''\in \Sigma_{\alpha,\lambda v}$ such that $L(\cdot)$ is exactly $\beta$-H\"older continuous at $E''$.
\\


\end{theorem}
\noindent\textbf{Remark(a) on Theorem \ref{Th1}:}Due to some technical problems, we consider the condition $v\in C^{2+r}$ instead of $v\in C^2.$ However, we think this condition is necessary, this is,
we can construct some $v\in C^2$ such that the H\"older constant of $LE$ is strictly less that $\frac{1}{2}.$ Besides, otherwise the first and the second conclusion, the remaining results all hold true in the condition $v\in C^2.$


\noindent\textbf{Remark(b) on Theorem \ref{Th1}:}
 Conclusion (1) in Theorem \ref{Th1} implies that $\frac{1}{2}$ H\"older continuity cannot be improved. Moreover, the set of E where the Lyapunov exponent possesses exactly $\frac{1}{2}$ H\"older continuity is small in the sense of measure but is large in the sense of topology.

 \noindent\textbf{Remark(c) on Theorem \ref{Th1}:} The H\"older continuity of LE for Schr\"odinger cocycles is also expected to play important roles in studying Cantor spectrum, typical localization length,
phase transition, etc., for quasi-periodic Schr\"odinger operators.


\noindent\textbf{Remark(d) on Theorem \ref{Th1}:} For Theorem \ref{Th1}, a related result on analytic cases was recently obtained by \cite{KXZ}.


The idea for the proof is as follows. Previous results on $\frac 12$ H\"older continuous regularity for LE with small potentials[Puig, AJ] showed that the lowest regularity for LE should be achieved exactly at the end points of spectral gaps. Thus  we first study exact local $\frac 12-$ H\"older continuity of LE at the end points of spectral gaps. It is based on a sharp estimate on the derivative of finite LE, which is not available in previous related works. Then we study local regularity of LE at any spectral point by estimating the approximation speed for it by end points of spectral gaps, which is based on the fact that the set of EP is dense in the spectrum by [WZ2]. We will show that for almost every spectral point, the approximation speed is slow enough such that LE is Lipschitz there. For those spectral points which is approximated by end points of spectral gaps with a relatively fast speed, LE possesses a regularity between $\frac 12$- H\"older and Lipschitz. Finally we will show that LE is globally $\frac 12$- H\"older continuous and cannot be better.


\subsection{\bf{Remarks on the regularity of Lyapunov exponents}}

Much work has been devoted to the regularity properties of Lyapunov exponents (LE) and integrated density of states (IDS) as well. Here we focus on the regularity of LE.

Bourgain and Goldstein\cite{bourgaingoldstein} established The first LDT for real analytic potentials with a $Diophantine$ frequency  to obtain Anderson Localization. Then Goldstein-Schlag \cite{goldsteinschlag} obtained some sharp version of large deviation theorem (LDT)  with a strong $Diophantine$ frequency and developed a powerful tool, the Avalanche Principle, by which they proved H\"older  continuity of $L(E)$ in the regime of positive LE. Then similar results were obtained for all Diophantine and some Liouvillean frequencies by \cite{YZ} and \cite{Han-Zhang}. Other type of base dynamics on which regularity of LE of analytic Schr\"odinger operators holds true include a shift or skew-shift of a higher
dimensional torus by Bourgain-Goldstein-Schlag \cite{bourgaingoldsteinschlag}, doubling map  and Anosov diffeomorphism by Bourgain-Schlag \cite{bourgainschlag}.

Jitomirskaya-Koslover-Schulteis \cite{jitomirskaya} get the continuity of LE with respect to potentials for a class of
analytic quasiperiodic $M(2, C)$ cocycles which is applicable to general quasi-periodic
Jacobi matrices or orthogonal polynomials on the unit circle in various parameters.
Jitomirskaya-Marx \cite{jitomirskaya1} later extended it to all (including singular) $M(2, C)$ cocycles.


An arithmetic version of large deviations and inductive scheme were developed by Bourgain and Jitomirskaya  in \cite{{bourgainjitomirskaya}} allowing to obtain joint continuity of LE for $\mathrm{SL}(2,\mathbb{C})$ cocycles, in frequency and cocycle map, at any irrational frequencies. This result has been crucial in many important developments, such as the proof of the Ten Martini problem \cite{avilajitomirskaya}, Avila's global theory of one-frequency cocycles \cite{avila1}. It was extended to multi-frequency case by Bourgain \cite{bourgain2} and to general $\mathrm{M}(2,\mathbb{C})$ case by Jitomirskaya and Marx  \cite{jitomirskaya2}.

H\"older continuity for $GL(d, C)$ cocycles, $d \geq 2,$ was recently obtained in Schlag \cite{schlag} and
Duarte-Klein \cite{duarteklein}.


All these results on regularity of LE
have been obtained by LDT and AP.

 Without use of LDT or Avalanche principle, H\"older continuity for $M(d, C)$ cocycles was given by Avila-Jitomirskaya-Sadel \cite{avilajitomirskayasadel}.


For lower regularity case, \cite{klein} proved the  H\"older continuity for a class of Gevrey potentials.
More recently, \cite{wz1} provides the first positive result on the continuity of LE and weak H\"older continuity of IDS on E for $C^2$ cos-type potentials. Later this weak H\"older continuity result for Diophantine frequencies was improved to be  $\gamma$-H\"older continuity of $L(E)$ for Liouville frequencies by Liang-Wang-You \cite{LWY}, where $\gamma>0$ is a small constant. For other related results, one can refer to \cite{avilakrikorian,Jiro-Mavi1,Jiro-Mavi2}.

There are many negative results on the positivity and continuity of LE for non-analytic
cases. It is well known that in $C^0$-topology, discontinuity of LE holds true at every
non-uniformly hyperbolic cocycle, see \cite{furman}, \cite{furstenberg}. Moreover, motivated by Ma\~ne \cite{mane1} and \cite{mane2},
Bochi \cite{bje} and \cite{boc} proved that with an ergodic base system, any non-uniformly hyperbolic
$SL(2, R)$-cocycle can be approximated by cocycles with zero LE in the $C^0$ topology.
Wang-You \cite{wy1} constructed examples to show
that LE can be discontinuous even in the space of $C^{\infty}$ Schr\"odinger cocycles. Recently,
Wang-You \cite{wy2}  improved the result in \cite{wy1} by showing that in $C^r $topology, $1\leq r\leq +\infty,$ there exists Schr\"odinger cocycles with a positive LE that can be approximated by
ones with zero LE.
Jitomirskaya-Marx \cite{jitomirskaya2} constructed examples showing that LE of $M(2, C)$ cocycles is
discontinuous in $C^{\infty}$ topology.


$$$$
The remaining part of the paper is as follows. In Section 3, we state some basic preparation which include some technical lemmas from \cite{wz1} without proof. In section 4, we present the structure of the spectrum and prove part of Theorem 1. In the last several sections, we focus on the remaining part of Theorem 1. The letters C and c will denote absolute constants.
\section{\bf{Preparation}}

In this section, we present some technical lemmas. The proof of them can be found in \cite{wz1}. In the following, each
$SL(2,\R)$-matrix is assumed to possess a norm strictly larger than 1.
For $\theta \in \R/\Z,$ let
$$
R_{\theta}=\begin{pmatrix}
\cos{2\pi\theta}&-\sin{2\pi\theta}\\
\sin{2\pi\theta}&\cos{2\pi\theta}
\end{pmatrix}
\in SO(2,\R)
$$

Define the map
$$s:SL(2,\R)\rightarrow\R{\mathbb{P}}^1=\R/(\pi\Z)$$
so that $s(A)$ is the most contraction direction of $A\in SL(2,\R).$ That is, for a unit vector $\hat{s}(A)\in s(A)$, it holds that $\|A\cdot\hat{s}(A)\|=\|A\|^{-1}.$ Abusing the notation a little, let
$$u:SL(2,\R)\rightarrow \R{\mathbb{P}}^1=\R/(\pi\Z)$$
be determined by $u(A)=s(A^{-1})$ and $\hat{u}(A)\in u(A)$. Then for $A\in SL(2,R)$, it is clear that
$$
A=R_u\cdot\begin{pmatrix}
\|A\|&0\\
0&\|A\|^{-1}
\end{pmatrix}
\cdot R_{\frac{\pi}{2}-s},
$$
where $s,u\in[0,\pi)$ are angles correspond to the directions $s(A),u(A)\in \R/(\pi\Z).$

The following technical lemma provide an equivalent form of the cocycle map \eqref{sm}, which turns to be more convenient.

\begin{lemma}\label{lemma4} Let $J$ be any compact interval. Let $\lambda\ge \lambda_0=\lambda_0(v)\gg 1$. For $x\in\R/\Z$ and $t\in J \ here\ t\ is \frac{E}{\lambda}$, define the following cocycles map:
\begin{equation}\label{EC}
A(x,t)=\Lambda(x,t)\cdot R_{\phi(x,t)}\triangleq
\begin{pmatrix}
\lambda(x,t)& 0\\
0& \lambda^{-1}(x,t)
\end{pmatrix}
\cdot\left({\sqrt{(t-v(x))^2+1}}^{-1}
\begin{pmatrix}
t-v(x)& -1\\
1& t-v(x)
\end{pmatrix}\right)
\end{equation}
where $\cot\phi(x,t)=t-v(x)$. Assume
\begin{equation}\label{oc}
\lambda(x,t)>\lambda,\left|\frac{\partial^m\lambda(x,t)}{\partial x^m}\right|+\left|\frac{\partial^m\lambda(x,t)}{\partial t^m}\right|+\left|\frac{\partial^2\lambda(x,t)}{\partial x\partial t}\right|<C\lambda,~m=1,2,
\end{equation}
where $C$ depends only on $v$. Then to prove Theorem 1, it suffices to consider the cocycle map \eqref{EC}.

\end{lemma}


Consider the product of two $SL(2,\R)$ matrices $E_1$ and $E_2$. We say they are in a $nonresonance$ case if
$$\vert s(E_2)-u(E_1) \vert^{-1}\ll \min\{\|E_1\|,\|E_2\|\}$$
Otherwise, we say they are in a $resonance$ case.

\noindent The following lemma gives the estimates of $nonresonance$ cases.

\begin{lemma}\label{lm5}Let $$E(x)=E_1(x)E_2(x).$$
Define
$$e_1=\|E_1\|, e_2=\|E_2\|, e_3=\|E\|,\ e_0=\min\{e_1,e_2\}.$$
Assume
$$0<e_0^{-1}\ll\eta\ll1.$$
Suppose that
$$e_1\ll e_2^{C}~or~e_2\ll e_1^{C},$$
and for
$$\theta(x)=s(E_2(x))-u(E_1(x))$$
and each $x\in J,$ it holds that
$$\vert \frac{d^m e_j(x)}{dx^m} \vert <Ce_j^{1+m\eta}, \vert \frac{d^m \theta(x)}{dx^m} \vert <Ce_0^{\eta},\quad j,m=1,2,$$
where $C$ depends only on $e_0$. Then for $m=1,2,$ we have
$$\vert \frac{d^m e_3(x)}{dx^m} \vert <e_3^{1+m\eta+2 Cm\eta},$$
and
$$\vert \frac{d^m s(E(x))}{dx^m} \vert <e_1^{2+2\eta}, \vert \frac{d^m u(E(x)-E_1(x))}{dx^m} \vert <e_2^{-\frac{3}{2}}\quad {\rm for\ } e_1\ll e_2^{C},$$
or
$$\vert \frac{d^m u(E(x))}{dx^m} \vert <e_2^{2+2\eta}, \vert \frac{d^m s(E(x)-E_2(x))}{dx^m} \vert <e_1^{-\frac{3}{2}}\quad {\rm for\ } e_2\ll e_1^{C}.$$

\end{lemma}

We can generalize the concept of nonresonance to the product of many matrices. More precisely, we consider a sequence of map $E^{(l)}\in C^2(J,SL(2,\R))$ for $0\leq l< n.$
Let$$s^{(l)}=s(E^{(l)}),\ u^{(l)}=u(E^{(l)}),\ \lambda_l=\|E^{(l)}\|,\ \Lambda^{(l)}=\begin{pmatrix}
\lambda_l&0\\
0&\lambda_l^{-1}
\end{pmatrix}.$$
By the polar decomposition theorem, we have
$$E^{(l)}=R_{u{(l)}}\Lambda^{(l)}R_{\frac{\pi}{2}-s{(l)}}.$$
Set
$$E_k(x)=E^{(0)}(x)E^{(1)}(x)\cdots E^{(k)}(x),\ 0\leq k\leq n.$$
Let
$$s_k=s(E_k),\ u_k=u(E_k),\ l_k=\|E_k\|,\ L_k=\begin{pmatrix}
l_k&0\\
0&l_k^{-1}
\end{pmatrix}.$$
Again by the polar decomposition theorem, we have
$$E_k=R_{u_k}L_kR_{\frac{\pi}{2}-s_k}.$$
Let $I\subseteq \R/\Z$ be any compact interval.

Then we have

\begin{lemma}
Let
$$\lambda'=\min_{0\leq l <n}\{\lambda_l\}\gg \eta^{-1}\gg 1.$$
We further assume that for $x\in I,~m=1,2$ and $0\leq l<n,$
$$\vert \frac{d^m\lambda_l}{dx^m}\vert <C\lambda_l^{1+m\eta},~\vert \frac{d^m s^{(l)}}{dx^m}\vert,~\vert \frac{d^m u{(l)}}{dx^m}\vert<C\lambda'^{\eta},~\vert s^{(l)}-u^{(l-1)}\vert>C\lambda'^{-\eta},$$where $C$ depends only on $\lambda'$.
Then we have
$$\|u^{(n-1)}-u_n\|_{C^2}<C\lambda_{n-1}^{-2+5\eta},~\|s^{(0)}-s_n\|_{C^2}<\lambda_0^{-2+5\eta},
$$ $$\vert\frac{d^m l_n}{dx^m}\vert<C\lambda_l^{1+m\eta},~m=1,2,$$ $$
l_n\geq C\prod_{0\leq k<n} \lambda_k\cdot \prod_{0\leq l<n} \vert s^{(l)}-u^{(l-1)}\vert .$$


\end{lemma}
\vskip 0.3cm


\begin{remark} It is not difficult to find that if we consider the functions $\lambda,\theta,e_1,e_2~and~e_3$ with respect to $t$ (i.e.$\frac{E}{\lambda}$), we still obtain similar conclusions to the above two lemmas. Actually it is clear that $x$ and $t$ play similar roles in Lemma \ref{lemma4}.
\end{remark}

The following key lemma is essentially from \cite{wz1}.
\begin{lemma}\label{lemma8}Let $\lambda_1,\lambda_2>1~and~\theta \in C^2(\R/(2\pi\Z),J)$. Matrix $A$ is defined as following:
$$A:=
\begin{pmatrix}
\lambda_1&0\\
0&{\lambda_1}^{-1}
\end{pmatrix}\begin{pmatrix}
\cos{\theta}&-\sin{\theta}\\
\sin{\theta}&\cos{\theta}
\end{pmatrix} \begin{pmatrix}
\lambda_2&0\\
0&{\lambda_2}^{-1}
\end{pmatrix}.$$

Denote $W(\lambda_1,\lambda_2\theta):=\frac{|\cot\theta|}{\cot^2\theta+(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})}.$
Then, we have
$$\frac{\partial\|A\|}{\partial\theta}\cdot\frac{1}{\|A\|}=\frac{sgn(\theta)(1-\lambda_2^{-4})(1-\lambda_1^{-4})}{\sqrt{(1-\frac{1}{\lambda_1^4\lambda_2^4})^2\cot^2\theta+(\frac{1}{\lambda_1^4}-\frac{1}{\lambda_2^4})^2\tan^2\theta+2(1+\frac{1}{\lambda_1^4\lambda_2^4})(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})-\frac{8}{\lambda_1^4\lambda_2^4}}};$$
$$(1-\frac{6}{\lambda_1^4}-\frac{6}{\lambda_2^4})W(\lambda_1,\lambda_2\theta)\leq \left\vert\frac{\partial\|A\|}{\partial\theta}\cdot\frac{1}{\|A\|}\right\vert\leq(1+\frac{6}{\lambda_1^4}+\frac{6}{\lambda_2^4})\left(\frac{\tan^2\theta\sqrt{\frac{1}{\lambda^4_1}+\frac{1}{\lambda^4_2}}}{2}+W(\lambda_1,\lambda_2\theta)\right);$$

\begin{align*}\frac{\partial\|A\|}{\partial\lambda_1}\cdot\frac{1}{\|A\|}&=\frac{sgn(\theta)\cdot{\lambda_1^{-1}}((1-\frac{1}{\lambda_1^4\lambda_2^4})\cot\theta+(\frac{1}{\lambda_2^4}-\frac{1}{\lambda_1^4})\tan\theta)}{\sqrt{(1-\frac{1}{\lambda_1^4\lambda_2^4})^2\cot^2\theta+(\frac{1}{\lambda_1^4}-\frac{1}{\lambda_2^4})^2\tan^2\theta+2(1+\frac{1}{\lambda_1^4\lambda_2^4})(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})-\frac{8}{\lambda_1^4\lambda_2^4}}}
,
\end{align*}
which implies $\left\vert\frac{\partial\|A\|}{\partial\lambda_1}\cdot\frac{1}{\|A\|}\right\vert\leq \frac{1}{\lambda_1};$
\begin{align*}\frac{\partial\|A\|}{\partial\lambda_2}\cdot\frac{1}{\|A\|}&=\frac{sgn(\theta)\cdot{\lambda_2}^{-1}((1-\frac{1}{\lambda_1^4\lambda_2^4})\cot\theta+(\frac{1}{\lambda_1^4}-\frac{1}{\lambda_2^4})\tan\theta)}{\sqrt{(1-\frac{1}{\lambda_1^4\lambda_2^4})^2\cot^2\theta+(\frac{1}{\lambda_1^4}-\frac{1}{\lambda_2^4})^2\tan^2\theta+2(1+\frac{1}{\lambda_1^4\lambda_2^4})(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})-\frac{8}{\lambda_1^4\lambda_2^4}}},
\end{align*}
which implies $\left\vert\frac{\partial\|A\|}{\partial\lambda_2}\cdot\frac{1}{\|A\|}\right\vert\leq \frac{1}{\lambda_2}.$\label{keylemma}
\end{lemma}
\begin{proof}It follows from a straight direct calculation that
$$\|A\|^2+\|A\|^{-2}=(\lambda_1^2\lambda_2^2+\lambda_1^{-2}\lambda_2^{-2})\cos^2\theta+(\lambda_1^2\lambda_2^{-2}+\lambda_1^{-2}\lambda_2^{2})\sin^2\theta.$$
Taking derivatives with respect to $\theta$ for both sides of the above equation, we have
$$2\|A\|\frac{\partial\|A\|}{\partial\theta}-2\frac{\partial\|A\|}{\partial\theta}\|A\|^{-3}=(\lambda_1^2\lambda_2^{-2}+\lambda_1^{-2}\lambda_2^{2}-(\lambda_1^2\lambda_2^2+\lambda_1^{-2}\lambda_2^{-2}))\sin2\theta.$$
Thus,
$$\frac{\partial\|A\|}{\partial\theta}\cdot{\|A\|}^{-1}=\frac{(\lambda_1^2\lambda_2^{-2}+\lambda_1^{-2}\lambda_2^{2}-(\lambda_1^2\lambda_2^2+\lambda_1^{-2}\lambda_2^{-2}))\sin2\theta}{2\sqrt{(\|A\|^2+\|A\|^{-2})^2-4}}.$$

$$=\frac{(\lambda_1^2\lambda_2^{-2}+\lambda_1^{-2}\lambda_2^{2}-(\lambda_1^2\lambda_2^2+\lambda_1^{-2}\lambda_2^{-2}))\sin2\theta}{2\sqrt{((\lambda_1^2\lambda_2^2+\lambda_1^{-2}\lambda_2^{-2})\cos^2\theta+(\lambda_1^2\lambda_2^{-2}+\lambda_1^{-2}\lambda_2^{2})\sin^2\theta)^2-4}}$$
 Equivalently, we have$$ \frac{\partial\|A\|}{\partial\theta}\cdot\frac{1}{\|A\|}=\frac{sgn(\theta)(1-\lambda_2^{-4})(1-\lambda_1^{-4})}
 {\sqrt{(1-\frac{1}{\lambda_1^4\lambda_2^4})^2\cot^2\theta+(\frac{1}{\lambda_1^4}-\frac{1}
 {\lambda_2^4})^2\tan^2\theta+2(1+\frac{1}{\lambda_1^4\lambda_2^4})(\frac{1}{\lambda_1^4}+
 \frac{1}{\lambda_2^4})-\frac{8}{\lambda_1^4\lambda_2^4}}}.$$
On one hand, it implies \begin{align*}
\left\vert\frac{\partial\|A\|}{\partial\theta}\cdot\frac{1}{\|A\|}\right\vert &\geq\frac{(1-\frac{1}{\lambda_2^4})(1-\frac{1}{\lambda_1^4})}{\sqrt{1+\frac{1}{\lambda_1^4\lambda_2^4}}}\cdot\frac{1}{\sqrt{\cot^2\theta+(\frac{1}{\lambda_1^4}-\frac{1}{\lambda_2^4})^2\tan^2\theta+\frac{2}{\lambda_1^4}+\frac{2}{\lambda_2^4}}}\\
&\geq (1-\frac{6}{\lambda_2^4}-\frac{6}{\lambda_1^4})\frac{1}{\sqrt{\cot^2\theta+(\frac{1}{\lambda_1^4}-\frac{1}{\lambda_2^4})^2\tan^2\theta+\frac{2}{\lambda_1^4}+\frac{2}{\lambda_2^4}}}\\
&\geq (1-\frac{6}{\lambda_2^4}-\frac{6}{\lambda_1^4})\left(\frac{1}{\sqrt{\cot^2\theta+(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})^2\tan^2\theta+\frac{2}{\lambda_1^4}+\frac{2}{\lambda_2^4}}}\right)\\
&= (1-\frac{6}{\lambda_1^4}-\frac{6}{\lambda_2^4})\left(\frac{|\cot\theta|}{\cot^2\theta+(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})}\right).
\end{align*}

On the other hand,\begin{align*}
\left\vert\frac{\partial\|A\|}{\partial\theta}\cdot\frac{1}{\|A\|}\right\vert &\leq  \frac{1}{\sqrt{(1-\frac{1}{\lambda_1^4\lambda_2^4})^2\cot^2\theta+(\frac{1}{\lambda_1^4}-\frac{1}{\lambda_2^4})^2\tan^2\theta+2(1+\frac{1}{\lambda_1^4\lambda_2^4})(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})-\frac{8}{\lambda_1^4\lambda_2^4}}}\\
&\leq \frac{1}{1-\frac{1}{\lambda_1^4}-\frac{1}{\lambda_2^4}}\frac{1}{\sqrt{\cot^2\theta+(\frac{1}{\lambda_1^4}-\frac{1}{\lambda_2^4})^2\tan^2\theta+\frac{2}{\lambda_1^4}+\frac{2}{\lambda_2^4}}}\\
&\leq (1+\frac{6}{\lambda_2^4}+\frac{6}{\lambda_1^4})\frac{1}{\sqrt{\cot^2\theta+(\frac{1}{\lambda_1^4}-\frac{1}{\lambda_2^4})^2\tan^2\theta+\frac{2}{\lambda_1^4}+\frac{2}{\lambda_2^4}}}\\
&\leq (1+\frac{6}{\lambda_2^4}+\frac{6}{\lambda_1^4})\left(\frac{1}{\sqrt{\cot^2\theta+(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})^2\tan^2\theta+\frac{2}{\lambda_1^4}+\frac{2}{\lambda_2^4}}}\right.\\
&\left.\frac{1}{\sqrt{\cot^2\theta+(\frac{1}{\lambda_1^4}-\frac{1}{\lambda_2^4})^2\tan^2\theta+\frac{2}{\lambda_1^4}+\frac{2}{\lambda_2^4}}}-\frac{1}{\sqrt{\cot^2\theta+(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})^2\tan^2\theta+\frac{2}{\lambda_1^4}+\frac{2}{\lambda_2^4}}}\right)\\
&\leq (1+\frac{6}{\lambda_1^4}+\frac{6}{\lambda_2^4})\left(\frac{\frac{4}{\lambda_1^4\lambda_2^4}\tan^2\theta}{2\sqrt{\frac{1}{\lambda^4_1}+\frac{1}{\lambda^4_2}}(\frac{1}{\lambda^4_1}+\frac{1}{\lambda^4_2})}+\frac{|\cot\theta|}{\cot^2\theta+(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})}\right).
\\&\leq (1+\frac{6}{\lambda_1^4}+\frac{6}{\lambda_2^4})\left(\frac{\tan^2\theta\sqrt{\frac{1}{\lambda^4_1}+\frac{1}{\lambda^4_2}}}{2}+\frac{|\cot\theta|}{\cot^2\theta+(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})}\right)\end{align*}

The next two inequalities are clearly followed from a direct calculation.
\end{proof}


In the resonance case, $s(E_1E_2)$ (or $u(E_1E_2)$) may differ greatly from  $s(E_2)$ (or $u(E_1)$). To describe the angle functions, we give the following definitions.

\begin{definition} Let $B(x,r)\subset \R/\Z$ be the ball centered at $x\in\R/\Z$ with a radius of $r$. For a connected interval $J\subset\R/\Z$ and a constant $a\in(0,1]$, let $aJ$ be the subinterval of $J$ with the same center and whose length is $a|J|.$ Let $I=B(0,r),~1\ll r^{-1}\ll l_0.$ Let $f\in C^2(I,\R P^1).$
\begin{enumerate}
\item \textbf{f is of type I}, if we have the following:$\|f\|_{C^2}<C$ and $f(x)=0$ has only one solution, say $x_0$, which is contained in $\frac{I}{3}$;
$\frac{df}{dx}=0$ has at most one solution on $I$ while $\vert \frac{df}{dx}\vert>r^2$ for all $x\in B(x_0,\frac{r}{2})$;
let $J\subset I$ be the subinterval such that $\frac{df}{dx}(J)\frac{df}{dx}(x_0)\leq 0,$ then $|f(x)|>cr^3$ for all $x\in J.$
    Let $I_+$ denotes the case $\frac{df}{dx}(x_0)>0$ and $I_-$ for $\frac{df}{dx}(x_0)<0.$
    \item \textbf{f is of type II}, if we have the following: $\|f\|_{C^2}<C$ and $f(x)=0$ has at most two solutions which are in $\frac{I}{2}$;
    $\frac{df}{dx}=0$ has one solution which is contained in $\frac{I}{2}$;
    $f(x)=0$ has one solution if and only if it is the $x$ such that $\frac{df}{dx}=0$;
    $\vert\frac{d^2 f}{dx^2} \vert > c$  whenever $\vert\frac{df}{dx}\vert<r^2.$
        \item \textbf{f is of type III}, if for $l:I\rightarrow \R^+$ such that
            $$l(x)>l_0\gg1,~\frac{d^ml(x)}{dx^m} < l(x)^{1+\beta},~x\in I, m=1,2,$$
            $$f(x)=\arctan(l^2\tan f_1(x))-\frac{\pi}{2}+f_2(x).$$

            Here, either $f_1$ is of type $I_+$ and $f_2$ is of type $I_-$, or $f_1$ is of type $I_-$ and $f_2$ is of type $I_+$.
\end{enumerate}
\end{definition}

\begin{remark} The graphs of type I and II are easy to be understood. The function of type III, as one can see, can divided to $\arctan(l^2\tan f_1(x))$, which is similar to a pulse function, and $-f_2(x)$, which is exactly of type I. One can directly calculate that if $x$ locates far from the zero of $f_1$, the first part is small enough to be neglected, and therefore $f\approx -f_2$. If $x$ is near the zero of $f_1$, the first part has a drastic change from $-\pi$ to 0, which leads to a bifurcation of no zero, one zero, and two zeros of $f$. For more details, see \cite{wz1}.
\end{remark}

\begin{lemma}Let $f:I\rightarrow \R\mathbb{P}^1$ be of type III and be defined as above. Let $r^2\leq \eta_j\leq r^{-2},$ $0\leq j\leq 4,$  be some constant. Then
$$|x_1|<Cl_0^{-\frac{3}{4}},|x_2-d|<Cl_0^{-\frac{3}{4}}.$$
In particular, if $f(x_1)=f(x_2)=0,$ then
$$0<x_1\leq x_2<d.$$
If $f(x_1)=f(x_2)\neq 0,$ then
$$x_1=x_2.$$

Then we consider the following two different cases:
\begin{enumerate}
\item  $d<\frac{r}{3}$: then there exist two distinct points $x_3, x_4 \in B(x_1,\eta_0l_0^{-1})$ such that
    $$\frac{df}{ds}(x_j)=0~for~j=3,4.$$
    Here we set $x_4$ that $x_1\leq x_4\leq x_2.$ Then $x_3$ is a local minimum with
    \begin{equation}\label{2.10}f(x_3)>\eta_1l_1^{-1}-\pi\end{equation}
    See Fig2 for positions of $x_j$, $j=1,2,3,4.$ Moreover, it holds that
    $$\left\{\begin{aligned}\vert\frac{d^2f}{dx^2}(x)\vert>c& ~whenever~\vert\frac{df}{dx}(x)\vert\leq r^2~for~x\in B(X,\frac{r}{6});& \\|f(x)|>cr^3,& ~for~all~x\notin B(X,\frac{r}{6}).& \end{aligned}\right.$$

\item $d\geq\frac{r}{3}:$ then $\frac{df}{dx}=0$ may have one or two solutions , among which the one between $x_1$ and $x_2$ always exists. In other words, there might exist $x_3$ or not while $x_4$ always exists. If $x_3$ exists, then \eqref{2.10} still holds. In any case, it always holds that
        $$\left\{\begin{aligned}|f(x)|>cr^3,~x\notin B(x_1,Cl_0^{-\frac{1}{4}})\cup B(x_2,\frac{r}{4});& &\\\|f-f_2\|_{C^2}<Cl_0^{-\frac{3}{2}},~x\in B(x_2,\frac{r}{4}).& \end{aligned}\right.$$
        \end{enumerate}
Finally, we have the following bifurcation as $d$ varies. There is a $d'$ with $l_1^{-1}<d'\eta_2<l_0^{-1}$ such that:
\begin{enumerate}
\item if $d>d'$, then $f(x)=0$ has two solutions;
\item if$d=d'$, then $f(x)=0$ has exactly one tangential solution. In other words, $x_1=x_2=x_4$ and $f(x_4)=0;$
\item  $0\leq d<d',$ then $f(x)\neq 0$ for all $x\in I$. Moreover, combining \eqref{2.10}, we have
    $$\min_{x\in I}|f(x)|>-\eta_3l_1^{-1}+\eta_4d.$$


    \end{enumerate}
\end{lemma}

No matter which type of function is, it satisfies the following non-degenerate property.

\begin{lemma} Let $f:I\rightarrow \R/\Z$ be of type I,II, or III. Define
$$X=\{x\in I:|f(x)|=\min_{y\in I}|f(y)|\}=\left\{\begin{aligned}\{x_0\},~ & ~f ~is ~of ~I,\\ \{x_1,x_2\}, & ~f ~is ~of ~type ~II ~or~ III.\end{aligned}\right.$$
In case $f$ is of type III, we further assume
$$d:=|x_1-x_2|<\frac{r}{3}.$$
Then for any $r'\in (0,r),$ we have
$$|f(x)|>cr'^3,~x\notin B(X,r').$$
For the case that $f$ is of type III, we have the same estimate for $Cl^{-\frac{1}{4}}<r'<r$ if $d\geq \frac{r}{3}.$
\end{lemma}


\subsubsection{\bf{Induction theorem}}
From now on, let $A=A(x,t)$ be as in \eqref{oc}. Abusing the notation a little bit, for $n\geq 1,$ we define
$$s_n(x,t)=s[A_n(x,t)],~u_n(x,t)=s[A_{-n}(x,t)].$$
We call $s_n(respectively,u_n)$ the $n-$ step $stable$(respectively,$unstable$) direction.

Obviously, we have that $u_1(x,t)=0$ and
$$s_1(x,t)=\frac{\pi}{2}-\phi(x,t)=\frac{\pi}{2}-\cot^{-1}[t-v(x)]=\tan^{-1}[t-v(x)].$$
We define $g_1(x,t)=s_1(x,t)-u_1(x,t).$ Thus, it clearly holds that
\begin{equation}\label{g1} g_1(x,t)=\tan^{-1}[t-v(x)].\end{equation}
To consider the H\"older regularity of $L(t)$, for $\lambda>\lambda_0\gg 1$, we may restrict $t$ to the following interval:
$$t\in J:=[\inf v-\frac{2}{\lambda},\sup v+\frac{2}{\lambda}].$$
It is due to the fact that if $t_0\notin J$, then $(\alpha,A(\cdot,t_0))\in \emph{UH},$  which implies that $L(t)$ possesses the same regularity as that of $v$. See \cite{z1} for details.

Here $\emph{UH}$ is defined as the following:

\begin{definition} We say a cocycle $(\beta,B)$ is uniformly hyperbolic ($\emph{UH}$) if there exist two functions $\bar{s}$, $\bar{u}$ :$\R/\Z\rightarrow \R\mathbb{P}^1$ such that $\bar{s}(x)\oplus\bar{u}(x)=\R^2$ (here we also consider $\bar{s}(x)$ and $\bar{u}(x)$ as one dimensional subspace of $\R^2$)
 and the following hold:
\begin{enumerate}
\item $\bar{B}(x)\cdot \bar{s}(x)=\bar{s}(x+\beta),$ $\bar{B}(x)\cdot \bar{u}(x)=\bar{u}(x+\beta).$ In other words, $\bar{s}$ and $\bar{u}$ are $\bar{B}-invariant.$
    \item There exist $c>0$ and $\rho>1$ such that for any unit vectors $\vec{w}^s\in \bar{s}(x)$ and $\vec{w}^u\in \bar{u}(x)$, it holds that
        $$\|B_n(x)w^s\|, \|B_{-n}(x)w^u\|<c\rho^{-n}$$
        for all $n\geq 1$ and for all $x\in\R/\Z.$ Here we also consider $\bar{s}(x)$ and $\bar{u}(x)$ as one dimensional subspace of $\R^2.$
        \end{enumerate}
\end{definition}

Set $I_0=\R/\Z$ for all $t\in J.$ Recall $\{\frac{p_n}{q_n}\}_{n\geq1}$ are the continued fraction approximants of $\alpha.$ Fix a  large $N=N(v).$ Then at step 1, we have the following.
\begin{enumerate}
\item First step $critical~points$:
$$C_1(t)=\{c_{1,1}(t),c_{1,2}(t)\}$$
with $c_{1,1}(t),c_{1,2}(t)\in I_0$ minimizing $\{|g_1(x,t)|,~x\in I_0\}$ for each $t\in J.$
\item First step $critical~interval$:
$$I_{1,j}(t)=\{x:|x-c_{1,j}(t)|\leq \frac{1}{2q_N^{2\tau}}\}~and~I_1(t)=I_{1,1}(t)\cup I_{1,2}(t).$$
\item First step $return~times:$
$$q_N\leq r_1^{\pm}(x,t):I_1(t)\rightarrow \Z^+$$
are the first return times (back to $I_1(t)$) after time $q_N-1.$ Here $r_1^+(x,t)$ is the forward return time and $r_1^+(x,t)$ the backward return time. Let $r_1(t)=\min{r_1^+(t),r_1^-(t)}$ with $r_1^{\pm}(t)=\min_{x\in I_1(t)}r_1^{\pm}(x,t).$
\item The second step angle $g_2:$
$$g_2(x,t)=s_{r_1(t)}(x,t)-u_{r_1(t)}(x,t):D_1\rightarrow\R\mathbb{P}^1,$$
where we define
$$D_1:=\{(x,t):x\in I_1(t),t\in J\}.$$
\end{enumerate}

It is easy to see that for each $t\in J$, $g_1(\cdot,t)=\tan^{-1}(t-v(\cdot))$ is either of type I or II. And for the second step, there may exist some $t\in J$, $g_2(\cdot,t)$ is of type III. Moreover, for any $n\geq 2$ and any $t\in J$, \cite{wz1} tells us that $g_n(\cdot,t)$ belongs to one of these three types.

More precisely, \cite{wz1} proved the following conclusion by induction. Assume that for $i\geq 1,$  the following objects are well defined:
\begin{enumerate}
\item $i$th step $critical~points:$
$$C_i(t)=\{c_{i,1}(t),c_{i,2}(t)\}$$
with $c_{i,j}(t)\in I_{i-1,j}(t)$ minimizing $\{|g_i(x,t)|,~x\in I_{i-1,j}(t)\}.$ More precise description of $C_i(t)$ will be given in the following theorem.
\item $i$th step $critical~interval:$
$$I_{i,j}(t)=\{x:|x-c_{i,j}(t)|\leq \frac{1}{2^iq_{N+i-1}^{2\tau}}\}~and~I_i(t)=I_{i,1}(t)\cup I_{i,2}(t).$$
\item $i$th step $return~times:$
$$q_{N+i-1}\leq r_i^{\pm}(x,t):I_i(t)\rightarrow \Z^+$$
are the first return times (back to $I_i(t)$) after time $q_{N+i-1}-1.$ Here $r_i^+(x,t)$ is the forward return time and $r_i^-(x,t)$ backward. Let $r_i(t)=\min{r_i^+(t),r_i^-(t)}$ with $r_i^{\pm}(t)=\min_{x\in I_i(t)} r_i^{\pm}(x,t).$
\item $i+1-th$ step angle $g_{i+1}:$
$$g_{i+1}(x,t)=s_{r_i(t)}(x,t)-u_{r_i(t)}(x,t):D_i\rightarrow \R\mathbb{P}^1,$$
where we define
$$D_i:=\{(x,t):x\in I_i(t),t\in J\}.$$
\end{enumerate}

The next theorem, which is from \cite{wz1}'s induction theorem, shows the precise description of the several important quantities mentioned above.

\begin{theorem}\label{theorem12} For any $\epsilon>0,$ there exists a $\lambda_0=\lambda_0(v,\alpha,\epsilon)>0$ such that for all $\lambda>\lambda_0,$ the following holds for each $i\geq 2.$
\begin{enumerate}
\item For each $t\in J,$ $g_i(\cdot,t):I_{i-1}(t)\rightarrow \R\mathbb{P}^1$ is of type I, II or III, which are denoted by case $(1)_I,~(i)_{II}~and~(i)_{III},$ respectively. In case $(1)_I$ and $(1)_{II}$, as functions on $I_{i-1}(t)$, it holds that
    \begin{equation}\label{jiaoducha} \|g_i-g_{i-1}\|_{C^2}\leq C\lambda^{-\frac{3}{2}r_{i-1}},\end{equation}
    where $C$ depends only on $\lambda$ and $v$. Moreover, we have the following.
    \begin{enumerate}
    \item In case $(i)_I$, $I_{i-1,1}(t)\cap I_{i-1,2}(t)=\emptyset.$ Moreover, if $g_i(\cdot,t)$ is of type $I_+$ on $I_{i-1,1}(t),$ then it is of type $I_-$ on $I_{i-1,2}(t),$ and vice versa.
        In addition, $c_{i,j}(t)$ is the only point such that $g_i(x,t)=0$ on $I_{i-1,j}(t).$
        \item  In case $(i)_{II}$, $I_{i-1,1}(t)\cap I_{i-1,2}(t)\neq \emptyset$ and $g_i$ is of type II as a function on the connected interval $I_{i-1}(t).$ Moreover, $c_{i,j}(t)$ is the only point minimizing $|g_i(x,t)|$ on $I_{i-1,j}(t)$ {\rm(}note that it is possible that $c_{i,1}(t)=c_{i,2}(t)${\rm)}.
            \item In case $(i)_{III}$, $I_{i-1,1}(t)\cap I_{i-1,2}(t)=\emptyset.$ We have multiple points minimizing $|g_i(x,t)|$ on each $I_{i-1,j}(t)$ for $j=1,2.$ Furthermore,   $c_{i,j}(t)$ can be defined as the minimal point of $g_i:I_{i-1,j}(t)\rightarrow \R\mathbb{P}^1$ that corresponds to
            $x_2~ of ~in~Lemma 10$ for case $d>d'$.

            \end{enumerate}

            \item For each $i\geq 1$ and $t\in J,$ it holds that
            \begin{equation}\label{cwc} |c_{i-1,j}(t)-c_{i,j}(t)|<C\lambda^{-\frac{3}{4}r_{i-2}},j=1,2;\end{equation}
            \item For all $x\in I_{i-1}(t)$ and $m=1,2,$ it holds that
            \begin{equation}\label{mzz}\|A_{\pm r_{i-1}^{\pm}(x,t)}(x,t)\|>\lambda^{(1-\epsilon)r_{i-1}^{\pm}(x,t)}\geq\lambda^{(1-\epsilon)q_{N+i-2}}
            \end{equation}
            and
            \begin{equation}\label{dkz} \frac{\partial^m(\|A_{\pm r_{i-1}^{\pm}(x,t)}(x,t)\|)}{\partial v^m}<\|A_{\pm r_{i-1}^{\pm}(x,t)}(x,t)\|^{1+\epsilon},v=x~or~t.\end{equation}
            \item In case $(i)_{III}$, there exists a unique $k$ such that $1\leq |k|<q_{N+i-2}$ and
                $$I_{i-1,2}(t)\cap(I_{i-1,1}(t)+k\alpha)\neq\emptyset.$$
                Moreover, there exist points $d_{i,j}(t)\in I_{i-1,j}(t)$ such that
                $$g_i(d_{i,j}(t),t)=g_i(c_{i,j}(t),t),~j=1,2,$$
                and the following hold.
                \begin{enumerate}
                \item If $|g_i(c_{i,j}(t),t)|>c\lambda^{-\frac{1}{10}r_{i-1}},$ $j=1~or~2,$ then so are $|g_i(c_{i,j'}(t),t)|$ for $j'\neq j$ and $|g_{i+1}(c_{i,j}(t),t)|~for~j=1~and~2;$
                    \item If $|g_i(c_{i,j}(t),t)|<C\lambda^{-\frac{1}{10}r_{i-1}},$ $j=1~or~j~=2,$ then
                        \begin{equation}\label{gz}\|c_{i,1}(t)+k\alpha-d_{i,2}(t)\|_{\R/\Z},~\|c_{i,2}(t)+k\alpha-d_{i,1}(t)\|_{\R/\Z}<C\lambda^{-\frac{1}{30}r_{i-1}},\end{equation}
                        where $\|\cdot\|_{\R/\Z}$ denotes the distance to the nearest integer.
                        \end{enumerate}
                               \end{enumerate}
            \end{theorem}
 \begin{remark}
 \eqref{gz} implies that as $i$-th step $critical~points,$ $d_{i,1}(t)$ essentially is $ -k-iteration$ of $c_{i,2}(t)$ while $d_{i,2}(t)$ essentially is the $k$-iteration of $c_{i,1}(t)$ under $x\longmapsto x+\alpha$ on $\R/\Z.$ Thus, we can change the notation from
                        $$d_{i,1}(t)~to~c_{i,2}^{-k}(t),~d_{i,2}(t)~to~c_{i,1}^k(t).$$
 \end{remark}

\begin{remark} In Theorem \ref{theorem12}, the size of the $'critical~interval'$ $I_{i,j},$ is chosen to be $2^{-i}q_{N+i-1}^{-2\tau}.$ However, there is no any difference if we set $|I_{i,j}|=q_{N+i-1}^{-C}$ instead for any large $C>0$ depending on the choice of $\lambda.$
\vskip 0.324213cm
In Theorem \ref{theorem12}, all the derivatives of $g_i(x,t)$ and $\|A_{r_i}(x,t)\|$ are on the variable $x.$ However, all the necessary technical lemmas in \cite{wz1} (lemma 2-5) can be directly applied to the derivative estimates of $g_i$ as a two-variable function in $(x,t).$


\end{remark}


\section{\bf{The resonance and the classification of the spectrum}}
In this section, we consider the classification of the spectrum according to the occurrence of resonance in the iteration process.

 From \cite{wz2}, we know the spectrum is a cantor set. In particular, the endpoints of spectral gaps are dense in the spectrum.

 For any fixed $t\in \frac{1}{\lambda}\Sigma,$ we write $c_{n,1}(t)-c_{n,2}(t)$ as $d_{n}(t)$ in brief.


First we give the important definitions.
\begin{definition}~\\Denote $\mathcal{F}_{C,n,k}\triangleq\left\{t \in \frac{1}{\lambda}\Sigma| |d_n(t)-k\alpha|>\frac{1}{q_{N+n-1}^C}\right\}$ and $\mathcal{E}_{C,n,k}\triangleq\left\{t \in \frac{1}{\lambda}\Sigma| |d_n(t)-k\alpha|\leq C\lambda^{-\frac{r_{N+n-1}}{10}}\right\}.$ Then we define£º
\begin{enumerate}
\item $\Sigma_1:=\bigcup\limits_{l=1}^{\infty}\bigcap\limits_{n\geq l}\bigcap\limits_{|k|\leq q_{N+n-1}}\mathcal{F}_{C,n,k}$, which is \textbf{finitely-resonant} condition, \textbf{$FR$} in brief.

\item $\Sigma_2:=\frac{1}{\lambda}\Sigma-\Sigma_1$, which is \textbf{infinitely-resonant} condition, \textbf{$IR$} in brief.

\item $\Sigma_3:=\bigcup\limits_{k\in \Z}\bigcap\limits_{n\geq N(|k|)} \mathcal{E}_{C,n,k}$.
\end{enumerate}
where $N(|k|)$ satisfies $q_{N+N(|k|)-2}<|k|\leq q_{N+N(|k|)-1}.$
\end{definition}
\begin{remark} Here the choice of constant $C$ is flexible (it is noteworthy that the constant $C$ only determines the choice of $\lambda$ in the iteration process), for instance, we can take $C=100\tau^{100}$. Anyway we can select constants according to specific needs in the later proof process. Similarly the choice of $-\frac{1}{10}$ before $r_{n-1}$ is also nonessential. This point is very clear in the next lemma. In fact, $\Sigma_3$ is the endpoints of gaps, thus this will be proved soon.
\end{remark}

It holds directly by definition that
$$\Sigma_1\bigcup\Sigma_2=\Sigma, \Sigma_1\bigcap\Sigma_2=\emptyset.$$
\vskip0.2cm
We claim that
Now we prove $$\Sigma_3\subset\Sigma_2.$$
\vskip0.2cm
In fact, for any fixed $k_0\in \Z$ and $t\in \bigcap\limits_{n\geq N(|k_0|)} \mathcal{E}_{C,n,k_0},$ by the help of $\mathcal{E}_{C,n,k_0}\subset \mathcal{F}^c_{C,n,k_0}(\triangleq \frac{1}{\lambda}\Sigma-\mathcal{F}_{C,n,k_0}) ,$ we have the following containment.
 $$\bigcap\limits_{n\geq N(|k_0|)}\mathcal{E}_{C,n,k_0}\subset \bigcap\limits_{n\geq N(|k_0|)}\mathcal{F}^c_{C,n,k_0}\subset \bigcup\limits_{n\geq N(|k_0|)}\mathcal{F}^c_{C,n,k_0}\subset \bigcup\limits_{n\geq N(|k_0|)}\bigcup\limits_{|k_0|\leq q_N+n-1}\mathcal{F}^c_{C,n,k_0}\subset \Sigma_2 .$$it hold from the  $\mathcal{E}_{C,n,k}\subset \frac{1}{\lambda}\Sigma-\mathcal{F}_{C,n,k},$ which implies $\Sigma_3\subset\Sigma_2.$

It's clear that the definition of $\Sigma_3$ directly implies $$\lim\limits_{n\rightarrow\infty}d_n(t)=k\alpha.$$ In fact, the converse is also right.
\begin{lemma}
For any fixed $t\in \frac{1}{\lambda}\Sigma,$ $$If~there ~exists ~some ~k\in\Z ~such ~that ~\lim\limits_{n\rightarrow\infty}d_{n}(t)=k\alpha, ~then~ t\in \Sigma_3.$$
\end{lemma}
\begin{proof}
We will directly prove that $$t\in \bigcap\limits_{n\geq N(|k|)} \mathcal{E}_{C,n,k}.$$ If it is not true, then we have there exists some $n'\geq N(|k|)$ such that $t\in \frac{1}{\lambda}\Sigma-\mathcal{E}_{C,n',k}.$ Then we have \begin{equation}\label{lm15.1}|d_{n'}(t)-k\alpha|> \lambda^{-\frac{r_{N+n'-2}}{10}}.\end{equation} By the help of Induction Theorem of \cite{wz1}, $$|d_{n}(t)-d_{n-1}(t)|\leq \lambda^{-\frac{1}{2}r_{N+n-2}}$$ for $n\geq 2$ and $t\in \frac{1}{\lambda}\Sigma.$ Combine it with \eqref{lm15.1}, we have $$\inf_{m\geq n'}|d_{m}(t)-k\alpha|\geq \lambda^{-\frac{r_{N+n'-2}}{100}}.$$ Then a conflict is formed with the fact $\lim\limits_{n\rightarrow\infty}d_{n}(t)=k\alpha.$


 

 
 
 
 
 
 
 
 

\end{proof}

So, we essentially have $$\Sigma_3=\left\{t\in\frac{1}{\lambda}\Sigma\vert\exists k\in\Z,s.t.\lim\limits_{n\rightarrow\infty}(d_{n}(t))=k\alpha\right\}.$$

The following several lemmas are useful for our further proof and they will show the important fact :
When we move the parameter $t$ in a small neighborhood, the image of the angle function $g_n(t)$ also move the same (a constant multiple) distance.
\begin{lemma} Fixed $t\in \mathcal{I},$
for any $g_n$ being \textbf{type II} or \textbf{III}, there exist some constant $1>c>0$ dependent on $v$ and $\alpha$ such that \begin{equation}\label{lm17-main}c^{-1}>|\frac{\partial g_n(x,t)}{\partial t}|>c, \end{equation} for any $x\in \mathcal{J}_n\triangleq[\tilde{c}_n(t),\tilde{c}_n(t)+\frac{1}{q^C_{N+n-1}}]$(or $x\in [\tilde{c}_n(t)-\frac{1}{q^C_{N+n-1}},\tilde{c}_n(t)]$) with $c_n(t)\in \mathcal{J}_n$, where $\tilde{c}_n(t):=\{x\in I_n| \frac{\partial g_n(t)}{\partial x}=0\}$ and $c_n$ is the essential zero point of $g_n(t).$
\vskip0.1cm
Meanwhile, if $g_n$ is \textbf{type I}, \eqref{lm17-main} holds for any $x\in I_n(t).$\label{lmg}
\end{lemma}

\begin{remark}In fact, take $x=\tilde{c}_n$, it's easy to see that the first part of the lemma directly implies the estimate for the length of gaps. From the proof, it's easy to check that
$$\lambda^{3\kappa(t)}>|\frac{\partial g_n(x,t)}{\partial t}|$$ for $g_n$ being any type and $x\in I_n(t),$ where $\kappa(t)=\max\{l| g_l(t)~is~Type~I~and~g_{l+1}~is~Type~III\}.$
\end{remark}

\begin{proof} We process the proof by induction.

\textbf{Case $0\rightarrow 1:$}

For $n=0$ it is clear that
$$ 1\geq |\frac{\partial g_1(t,x)}{\partial t}|=|\frac{-1}{(t-v)^2+1}|>\frac{1}{(\sup |v|+3)^2+1}:=c_1, \forall x\in I_1$$ while $sgn(\frac{\partial g_{1,1}(t,x)}{\partial t})=sgn(\frac{\partial g_{1,2}(t,x)}{\partial t})~on~I_1.$

Denote $\delta_n=c_1^{-1}\lambda^{-q_{N+n-2}},$ $n\in \N,$ and $q_{N-1}=1$.
For $n=1,$ three cases may occur.
\begin{enumerate}
\item $g_2$ is of type I.

Due to the $ |\frac{\partial g_2(t,x)}{\partial t}-\frac{\partial g_1(t,x)}{\partial t}|_{C^1(I_n)}\leq \lambda^{-\frac{3}{2}}$, we have $$1+\lambda^{-1}\geq |\frac{\partial g_2(t,x)}{\partial t}|\geq  c_1-\lambda^{-\frac{3}{2}}\geq c_1(1-\delta_1).$$

\item $g_2$ is of type II.

It's similar to the first case.

\item $g_2$ is of type III.

There exits some $1\leq k(1)\leq q_{N}$ such that $|c_{1,1}-c_{2,1}-k(1)\alpha|\leq C\lambda^{-k(1)},$ thus $g_2=\arctan \lambda^2_{k(1)}\tan \tilde{g}_{1,1}-\frac{\pi}{2}+\tilde{g}_{1,2},$ where $\lambda_{k(1)}>\lambda^{\frac{9}{10}k(1)}$, $\tilde{g}_{1,1},\tilde{g}_{1,2}$ are all Type I satisfying
$$1+\lambda^{-1}\geq|\frac{\partial \tilde{g}_{1,1}}{\partial t}|>c_1(1-\delta_1)$$ and $$1+\lambda^{-1}\geq |\frac{\partial \tilde{g}_{1,2}}{\partial t}|>c_1(1-\delta_1)$$ with $sgn(\frac{\partial \tilde{g}_{1,1}}{\partial t})=sgn(\frac{\partial \tilde{g}_{1,1}}{\partial t}).$

\vskip0.3cm
 Therefore (for convenience, we write ``$\frac{\partial}{\partial t}$" as ``$'$"), we have the lower bound as follow.
\begin{align*}&\geq |\frac{\lambda^2_{k(1)}\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda^4_{k(1)}\tan^2\tilde{g}_{1,1}}+\tilde{g}_{1,2}'|+|\frac{2\lambda_{k(1)}\lambda'_{k(1)}\tan \tilde{g}_{1,1}}{1+\lambda^4_{k(1)}\tan^2 \tilde{g}_{1,1}}|\\&\geq|\frac{\partial g_2(t,x)}{\partial t}|\\&=|\frac{2\lambda_k\lambda'_{k(1)}\tan \tilde{g}_{1,1}+\lambda_{k(1)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda_{k(1)}^4\tan^2 \tilde{g}_{1,1}}+\tilde{g}_{1,2}'|\\
&\geq |\frac{\lambda_{k(1)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda_{k(1)}^4\tan^2 \tilde{g}_{1,1}}+\tilde{g}_{1,2}'|-|\frac{2\lambda_{k(1)}\lambda_{k(1)}'\tan \tilde{g}_{1,1}}{1+\lambda_{k(1)}^4\tan^2 \tilde{g}_{1,1}}|\\&\geq c_1(1-\delta_1)(1+0.5\lambda_{k(1)}^{-2})-2\lambda_1^{-\frac{3}{2}}\\&\geq c_1(1-\delta_1)-\lambda_{k(1)}^{-\frac{3}{2}}\geq c_1(1-2\delta_1).
\end{align*}\end{enumerate}

For the upper bound, note that if $|\tilde{g}_{1,1}|>\lambda_{k(1)}^{-\frac{3}{2}},$ then we can rewrite $g_2$ as $$-\tilde{C}\lambda_{k(1)}^{-2}(\frac{\partial \tilde{g}_{1,1}}{\partial x}(c'_{1,1})+o(\frac{1}{q^C_{N}}))\tilde{g}_{1,1})^{-1}+d_1-(\frac{\partial\tilde{g}_{1,1}}{\partial x}(c'_{1,2})+o(\frac{1}{q^C_{N}}))\tilde{g}_{1,1},$$
where $\tilde{C}$ is a absolute constant.
Therefore, it follows from the estimate in \cite{wz1} that $$g_2=-C\lambda_{k(1)}^{-2}(c^1_1(x-c'_{1,1}))^{-1}+d_1-c^1_2(x+k(1)\alpha-c'_{1,2}),$$ where $$c^1_i=\frac{\partial g_1}{\partial x}(c_{1,i})+\delta'_1(x,t)\leq \frac{1}{q_N},i=1,2$$ with $\|\delta'_1(x,t)\|_{C^2}\leq \frac{1}{q^C_{N}}.$

By a direct calculation, we have $\tilde{c}_{2,i}=c'_{1,1}+(\sqrt{c^1_1\cdot c^1_2})^{-1}\lambda_{k(1)}^{-1},i=1,2.$
\vskip0.3cm
Hence, for any $x\in [\tilde{c}_2(t),\tilde{c}_2(t)+\frac{1}{q^C_{N+n-1}}],$ it holds that
\begin{equation}\begin{array}{ll}&|\frac{\partial g_2(x,t)}{\partial t}|\\&=|\frac{2\lambda_{k(1)}\lambda_{k(1)}'\tan \tilde{g}_{1,1}+\lambda_{k(1)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda_{k(1)}^4\tan^2 \tilde{g}_{1,1}}+\tilde{g}_{1,2}'|\\
&\leq |\frac{\lambda_{k(1)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda_{k(1)}^4\tan^2 \tilde{g}_{1,1}}+\tilde{g}_{1,2}'|+|\frac{2\lambda_{k(1)}\lambda_{k(1)}'\tan \tilde{g}_{1,1}}{1+\lambda_{k(1)}^4\tan^2 \tilde{g}_{1,1}}|\\&=|\frac{\lambda_{k(1)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda_{k(1)}^4\tan^2 \tilde{g}_{1,1}}+\tilde{g}_{1,2}'|+o(\lambda_{k(1)}^{-1.5})\\&\leq|\lambda_{k(1)}^{-2}\sin^{-2}\tilde{g}_{1,1}g_{1,1}'+g_{1,2}'|+o(\lambda_{k(1)}^{-1.5})\\&\leq
\left[\lambda_{k(1)}^{-2}\sin^{-2}(c'_{1,1}+((\sqrt{c^1_1\cdot c^1_2})^{-1}+\tilde{\delta}_1)\lambda_{k(1)}^{-1})\cdot g_{1,1}'\right]+g_{1,2}'\\&\leq 2(2 \sqrt{c^1_1\cdot c^2_1})^{2}(1+\lambda^{-1})+1+\lambda^{-1}\leq 1+\lambda^{-1}+\frac{4}{q_N}\end{array}\end{equation}
\iffalse
Note that, similar to the estimate for $\frac{\partial g_2}{\partial t},$ we have\begin{align*}&|\frac{\partial g_2(t,x)}{\partial x}|\\&=|\frac{2\lambda_{k(1)}\lambda_{k(1)}'\tan \tilde{g}_{1,1}+\lambda_{k(1)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda_{k(1)}^4\tan^2 \tilde{g}_{1,1}}+\tilde{g}_{1,2}'|\\
&\leq |\frac{\lambda_{k(1)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda_{k(1)}^4\tan^2 \tilde{g}_{1,1}}+\tilde{g}_{1,2}'|+|\frac{2\lambda_{k(1)}\lambda_{k(1)}'\tan \tilde{g}_{1,1}}{1+\lambda_{k(1)}^4\tan^2 \tilde{g}_{1,1}}|\\&=|\frac{\lambda_{k(1)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda_{k(1)}^4\tan^2 \tilde{g}_{1,1}}+\tilde{g}_{1,2}'|+o(\lambda_{k(1)}^{-1.5})\\&\leq|\lambda_{k(1)}^{-2}\sin^{-2}\tilde{g}_{1,1}g_{1,1}'+g_{1,2}'|+o(\lambda_{k(1)}^{-1.5}).
\end{align*}\
And for the new zero points $c_{2,1}$ and $c_{2,2}$, it follows from the calculation above and the facts $$sgn(\frac{\partial g_2}{\partial x}(c_{2,i}))\cdot sgn(\frac{\partial \tilde{g}_{1,i}}{\partial x}(c'_{1,i}))\geq 0,i=1,2;$$
$$sgn(\frac{\partial \tilde{g}_{1,1}}{\partial x}(c'_{1,1}))\cdot sgn(\frac{\partial \tilde{g}_{1,2}}{\partial x}(c'_{1,2}))\leq 0,$$that

$$|\frac{\partial g_2}{\partial x}(c_{2,i})|\leq \tilde{g}'_{1,i}\leq c^1_1+o(\lambda^{-1.5}),i=1,2.$$
\fi
Therefore, we obtain $$\frac{1}{q_{N}}+1\geq |\frac{\partial g_2(x,t)}{\partial t}|>c_1-2\delta_1$$ for any $y\in [\tilde{c}_2(t),\tilde{c}_2(t)+\frac{1}{q^C_{N+n-1}}].$

\textbf{Case $n\rightarrow n+1:$}


Suppose for all $2\leq l\leq n$, when $g_l$ is type I, we have obtained $$1+\sum_{i=1}^{l-1}\frac{1}{q_{N+i-1}}\geq |\frac{\partial g_l(x,t)}{\partial t}|>(c_1-2\sum_{i=1}^{l-1}\delta_i)$$ for any $x\in I_l(t);$
when $g_l$ is type II or III, we have  $$1+\sum_{i=1}^{l-1}q^{-1}_{N+i-1}\geq|\frac{\partial g_{l}(x,t)}{\partial t}|>c_1-2\sum_{i=1}^{l-1}\delta_i,$$ for $x\in [\tilde{c}_l(t),\tilde{c}_l(t)+\frac{1}{q_{N+l-1}}].$


For the case $n+1$, similarly we consider three cases as follows.

\begin{enumerate}
\item $g_{n+1}$ is of type I.

Due to the $ |\frac{\partial g_{n+1}(x,t)}{\partial t}-\frac{\partial g_{n}(x,t)}{\partial t}|_{C^1(I_n)}\leq \lambda^{-\frac{3}{2}q_{N+n-2}}$, we have $$1+\sum_{i=1}^{n}q^{-1}_{N+i-1}\geq$$ $$ 1+\sum_{i=1}^{n-1}q^{-1}_{N+i-1}+\lambda^{-\frac{3}{2}q_{N+n-2}}\geq |\frac{\partial g_{n+1}(x,t)}{\partial t}|\geq  (c_1-2\sum_{i=1}^{n-1}\delta_i)-\lambda^{-\frac{3}{2}q_{N+n-2}}\geq c_1(1-2\sum_{i=1}^{n}\delta_i)$$ for any $x\in I_{n+1}(t).$

\item $g_{n+1}$ is of type II.

It's similar to the first case.

\item $g_{n+1}$ is of type III.

There exits some $k(n)\ll \lambda^{ck(n)}\leq k(n+1)\leq q_{N+n}$ such that $|c_{n,1}-c_{n,2}-k(n+1)\alpha|\leq C\lambda^{-k(n+1)},$ thus $g_{n+1}=\arctan \lambda_1^2\tan \tilde{g}_{n,1}-\frac{\pi}{2}+\tilde{g}_{n,2},$ where $\lambda_n>\lambda^{\frac{9}{10}k(n)}$, $g_{n,1},g_{n,2}$ are all Type I satisfying $$1+\sum_{i=1}^{n-1}q^{-1}_{N+i-1}\geq\frac{\partial \tilde{g}_{n,1}}{\partial t}>c_1(1-\sum_{i=1}^{n-1}\delta_i)$$ and $$1+\sum_{i=1}^{n-1}q^{-1}_{N+i-1}\geq \frac{\partial \tilde{g}_{n,2}}{\partial t}>c_1(1-\sum_{i=1}^{n-1}\delta_i)$$ with $sgn(\tilde{g}_{n,1})=sgn(\tilde{g}_{n,2}).$
\vskip0.3cm
Therefore (for convenience, we write ``$\frac{\partial}{\partial t}$" as ``$'$"), we have the lower bound as follow
\begin{equation}\begin{array}{ll}&|\frac{\partial g_n(t,x)}{\partial t}|\\&=|\frac{2\lambda_{k(n+1)}\lambda_{k(n+1)}'\tan \tilde{g}_{n,1}+\lambda_{k(n+1)}^2\frac{\tilde{g}_{n,1}'}{\cos^2 \tilde{g}_{n,1}}}{1+\lambda_{k(n+1)}^4\tan \tilde{g}_{n,1}}+\tilde{g}_{n,2}'|\\
&\geq |\frac{\lambda_{k(n+1)}^2\frac{\tilde{g}_{n,1}'}{\cos^2 \tilde{g}_{n,1}}}{1+\lambda_{k(n+1)}^4\tan \tilde{g}_{n,1}}+\tilde{g}_{n,2}'|-|\frac{2\lambda_{k(n+1)}\lambda_{k(n+1)}'\tan \tilde{g}_{n,1}}{1+\lambda_{k(n+1)}^4\tan \tilde{g}_{n,1}}|\\&\geq c_1(1-2\sum_{i=1}^{n-1}{\delta_i})(1+0.5\lambda_{k(n+1)}^{-2})-2\lambda_{k(n+1)}^{-\frac{3}{2}}\\&\geq c_1(1-2\sum_{i=1}^{n-1}{\delta_i})-\lambda_{k(n+1)}^{-\frac{3}{2}}\geq c_1(1-2\sum_{i=1}^{n}{\delta_i}).
\end{array}\end{equation}\end{enumerate}
\vskip0.3cm

For the upper bound, note that if $|\tilde{g}_{n,1}|>\lambda_{k(n+1)}^{-\frac{3}{2}},$ then we can rewrite $g_{n+1}$ as $$\tilde{C}\lambda_{k(n+1)}^{-2}(\frac{\partial \tilde{g}_{n,1}}{\partial x}(c'_{n,1})+o(\frac{1}{q^C_{N}}))\tilde{g}_{n,1})^{-1}+d_1-(\frac{\partial\tilde{g}_{n,1}}{\partial x}(c'_{n,2})+o(\frac{1}{q^C_{N}}))\tilde{g}_{n,2},$$
where $\tilde{C}$ is an absolute constant.
Therefore, it follows from the estimate in \cite{wz1} that $$g_{n+1}=C_1\lambda_{k(n+1)}'^{-2}(c^n_1(x-c'_{n,1}))^{-1}+d_1-c^n_2(x+k(n+1)\alpha-c'_{n,2}),$$ where \begin{equation}\label{lm17-xishu}c^n_i=\frac{\partial g_n}{\partial x}(c_{n,i})+\sum_{i=1}^{n}\delta'_i(x,t)\leq \frac{1}{q^C_{N+n-1}} ,i=1,2\end{equation} with $\|\delta'_n(x,t)\|_{C^2}\leq \frac{1}{q^C_{N+n-1}}.$

By a direct calculation, we have $\tilde{c}_{n+1,i}=c'_{n,1}+\tilde{C}^{\frac{1}{2}}((\sqrt{c^n_1\cdot c^n_2})^{-1})\lambda_{k(n+1)}^{-1},i=1,2.$

\vskip0.3cm
Hence, for any $x\in [\tilde{c}_{n+1}(t),\tilde{c}_{n+1}(t)+\frac{1}{q^C_{N+n-1}}]$ we have\begin{align*}&|\frac{\partial g_{n+1}(x,t)}{\partial t}|\\&=|\frac{2\lambda_{k(n+1)}\lambda'_{k(n+1)}\tan \tilde{g}_{n,1}+\lambda_{k(n+1)}^2\frac{\tilde{g}_{n,1}'}{\cos^2 \tilde{g}_{n,1}}}{1+\lambda_{k(n+1)}^4\tan^2 \tilde{g}_{n,1}}+\tilde{g}_{n,2}'|\\
&\leq |\frac{\lambda_{k(n+1)}^2\frac{\tilde{g}_{n,1}'}{\cos^2 \tilde{g}_{n,1}}}{1+\lambda_{k(n+1)}^4\tan^2 \tilde{g}_{n,1}}+\tilde{g}_{n,2}'|+|\frac{2\lambda_{k(n+1)}\lambda_{k(n+1)}\tan \tilde{g}_{n,1}}{1+\lambda_{k(n+1)}^4\tan^2 \tilde{g}_{n,1}}|\\&=|\frac{\lambda_{k(n+1)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{n,1}}}{1+\lambda_{k(n+1)}^4\tan^2 \tilde{g}_{n,1}}+\tilde{g}_{n,2}'|+o(\lambda_{k(n+1)}^{-1.5})\\&\leq|\lambda_{k(n+1)}^{-2}\sin^{-2}\tilde{g}_{n,1}g_{n,1}'+g_{n,2}'|+o(\lambda_{k(n+1)}^{-1.5}).
\\&\leq \left\vert\lambda_{k(n+1)}^{-2}\sin^{-2}|c'_{n,1}+((\sqrt{c^n_1\cdot c^n_2})^{-1})\lambda_{k(n+1)}^{-1}|g_{n,1}'+g_{n,2}'\right\vert+o(\lambda_{k(n+1)}^{-1.5})\\&\leq (2(\sqrt{c^n_1\cdot c^n_2})^{-1})^{-2}(1+\sum_{i=1}^{n-1}q^{-1}_{N+i-1})+1+\sum_{i=1}^{n-1}q^{-1}_{N+i-1}\\&\leq 1+\sum_{i=1}^{n}q^{-1}_{N+i-1},\end{align*}
where the last inequality follows from \eqref{lm17-xishu}.
\iffalse
And for the new zero points $c_{{n+1},1}$ and $c_{{n+1},2}$, it follows from the calculation above and the facts $$sgn(\frac{\partial g_{n+1}}{\partial x}(c_{{n+1},i}))\cdot sgn(\frac{\partial \tilde{g}_{n,i}}{\partial x}(c'_{n,i}))\geq 0,i=1,2;$$
$$sgn(\frac{\partial \tilde{g}_{n,1}}{\partial x}(c'_{n,1}))\cdot sgn(\frac{\partial \tilde{g}_{n,2}}{\partial x}(c'_{n,2}))\leq 0,$$that

$$|\frac{\partial g_{n+1}}{\partial x}(c_{{n+1},i})|\leq \tilde{g}'_{n,i}\leq c^1_1+o(\lambda^{-1.5}),i=1,2.$$
\fi
Therefore, we obtain \begin{align*}&1+\sum_{i=1}^{n}q^{-1}_{N+n-1}\\&>|\frac{\partial g_{n+1}(\tilde{c}_{n+1,i}(t),t)}{\partial t}|\\&>c_1(1-2\sum_{i=1}^{n}{\delta_i}),i=1,2.\end{align*}
\vskip0.3cm
Finally, due to the Diophantine condition, for $N$ large enough we have $1+\sum_{i=1}^{+\infty}q^{-1}_{N+i-1}\leq 2\leq c^{-1},$ where $c=\frac{1}{2}c_1,$ which implies \eqref{lm17-main}.


\end{proof}

\iffalse
\begin{lemma}
For $g_n$ being any types, we have
$$C\lambda^{4\kappa(n)}\geq|\frac{\partial g_n(t,x)}{\partial t}|>c,$$ $\forall x\in I_n(t),$ where $c,C$ only depends on $\lambda$ and $\kappa(n)=\max\{l\leq n| g_l~is~type~I~and~g_{l+1}~is~type~III~or~II\}$.

Moreover, for $g_n$ being \textbf{type II} or \textbf{III}, it holds that $$2^nq^C_{N+n-1}>|\frac{\partial g_n(\tilde{c}_n(t),t)}{\partial t}|>c,$$ where $\tilde{c}_n(t):=\{x\in I_n| \frac{\partial g_n(t)}{\partial x}=0\}$ and $c,C$ only depend on $v.$\label{lmg}
\end{lemma}

\begin{remark}In fact, the second part of the lemma directly implies the estimate for the length of gaps.
\end{remark}

\begin{proof} We process the proof by induction.

\textbf{Case $0\rightarrow 1:$}

For $n=0$ it is clear that
$$ 1\geq |\frac{\partial g_1(t,x)}{\partial t}|=|\frac{-1}{(t-v)^2+1}|>\frac{1}{(\sup |v|+3)^2+1}:=c_1, \forall x\in I_1$$ while $sgn(\frac{\partial g_{1,1}(t,x)}{\partial t})=sgn(\frac{\partial g_{1,2}(t,x)}{\partial t})~on~I_1.$

Denote $\delta_n=c_1^{-1}\lambda^{-q_{N+n-2}},$ $n\in \N,$ and $q_{N-1}=1$.
For $n=1,$ three cases may occur.
\begin{enumerate}
\item $g_2$ is of type I.

Due to the $ |\frac{\partial g_2(t,x)}{\partial t}-\frac{\partial g_1(t,x)}{\partial t}|_{C^1(I_n)}\leq \lambda^{-\frac{3}{2}}$, we have $$2c_1\geq |\frac{\partial g_2(t,x)}{\partial t}|\geq  c_1-\lambda^{-\frac{3}{2}}\geq c_1(1-\delta_1).$$

\item $g_2$ is of type II.

It's similar to the first case.

\item $g_2$ is of type III.

There exits some $1\leq k(N)\leq q_{N}$ such that $|c_{1,1}-c_{2,1}-k(N)\alpha|\leq C\lambda^{-k(N)},$ thus $g_2=\arctan \lambda_{k^2(N)}\tan \tilde{g}_{1,1}-\frac{\pi}{2}+\tilde{g}_{1,2},$ where $\lambda_{k(N)}>\lambda^{\frac{9}{10}k(N)}$, $\tilde{g}_{1,1},\tilde{g}_{1,2}$ are all Type I satisfying
$$2c_2\geq|\frac{\partial \tilde{g}_{1,1}}{\partial t}|>c_1(1-\delta_1)$$ and $$2c_1\geq |\frac{\partial \tilde{g}_{1,2}}{\partial t}|>c_1(1-\delta_1)$$ with $sgn(\frac{\partial \tilde{g}_{1,1}}{\partial t})=sgn(\frac{\partial \tilde{g}_{1,1}}{\partial t}).$

\vskip0.3cm
 Therefore (for convenience, we write ``$\frac{\partial}{\partial t}$" as ``$'$"),
\begin{align*}&2c_1\lambda^3_{k(N)}\\&\geq |\frac{\lambda^2_{k(N)}\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda^4_{k(N)}\tan^2\tilde{g}_{1,1}}+\tilde{g}_{1,2}'|+|\frac{2\lambda_{k(N)}\lambda'_{k(N)}\tan \tilde{g}_{1,1}}{1+\lambda^4_{k(N)}\tan^2 \tilde{g}_{1,1}}|\\&\geq|\frac{\partial g_2(t,x)}{\partial t}|\\&=|\frac{2\lambda_k\lambda'_{k(N)}\tan \tilde{g}_{1,1}+\lambda_{k(N)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda_{k(N)}^4\tan^2 \tilde{g}_{1,1}}+\tilde{g}_{1,2}'|\\
&\geq |\frac{\lambda_{k(N)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda_{k(N)}^4\tan^2 \tilde{g}_{1,1}}+\tilde{g}_{1,2}'|-|\frac{2\lambda_{k(N)}\lambda_{k(N)}'\tan \tilde{g}_{1,1}}{1+\lambda_{k(N)}^4\tan^2 \tilde{g}_{1,1}}|\\&\geq c_1(1-\delta_1)(1+0.5\lambda_{k(N)}^{-2})-2\lambda_1^{-\frac{3}{2}}\\&\geq c_1(1-\delta_1)-\lambda_{k(N)}^{-\frac{3}{2}}\geq c_1(1-2\delta_1).
\end{align*}\end{enumerate}

In particular, note that if $|\tilde{g}_{1,1}|>C\lambda_{k(N)}^{-\frac{3}{4}},$ then we can rewrite $g_2$ as $$-\tilde{C}\lambda_{k(N)}^{-2}(\frac{\partial \tilde{g}_{1,1}}{\partial x}(c'_{1,1})+o(\frac{1}{q^C_{N}}))\tilde{g}_{1,1})^{-1}+d_1-(\frac{\partial\tilde{g}_{1,1}}{\partial x}(c'_{1,2})+o(\frac{1}{q^C_{N}}))\tilde{g}_{1,1},$$
where $\tilde{C}$ is a absolute constant.
Therefore, it follows from the estimate in \cite{wz1} that $$g_2=-C\lambda_{k(N)}^{-2}(c^1_1(x-c'_{1,1}))^{-1}+d_1-c^1_2(x+k\alpha-c'_{1,2}),$$ where $c^1_i=\frac{\partial g_1}{\partial x}(c_{1,i})+\delta'_1(x,t),i=1,2$ with $\|\delta'_1(x,t)\|_{C^2}\leq \frac{1}{q^C_{N}}.$

By a direct calculation, we have $\tilde{c}_{2,i}=c'_{1,1}+(\sqrt{c^1_1\cdot c^1_2}+\tilde{\delta}_1)\lambda_{k(N)}^{-1},i=1,2,$ where $\|\ \tilde{\delta}_1(x,t)\|_{C^2}\leq \frac{1}{q^C_{N}}.$
\vskip0.3cm
Hence,
\begin{equation}\begin{array}{ll}&|\frac{\partial g_2(\tilde{c}_2(t),t)}{\partial t}|\\&=|\frac{2\lambda_{k(N)}\lambda_{k(N)}'\tan \tilde{g}_{1,1}+\lambda_{k(N)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda_{k(N)}^4\tan^2 \tilde{g}_{1,1}}+\tilde{g}_{1,2}'|\\
&\leq |\frac{\lambda_{k(N)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda_{k(N)}^4\tan^2 \tilde{g}_{1,1}}+\tilde{g}_{1,2}'|+|\frac{2\lambda_{k(N)}\lambda_{k(N)}'\tan \tilde{g}_{1,1}}{1+\lambda_{k(N)}^4\tan^2 \tilde{g}_{1,1}}|\\&=|\frac{\lambda_{k(N)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda_{k(N)}^4\tan^2 \tilde{g}_{1,1}}+\tilde{g}_{1,2}'|+o(\lambda_{k(N)}^{-1.5})\\&\leq|\lambda_{k(N)}^{-2}\sin^{-2}\tilde{g}_{1,1}g_{1,1}'+g_{1,2}'|+o(\lambda_{k(N)}^{-1.5})\\&\leq
\lambda_{k(N)}^{-2}\sin^{-2}(c'_{1,1}+(\sqrt{c^1_1\cdot c^1_2}+\tilde{\delta}_1)\lambda_{k(N)}^{-1})\cdot g_{1,1}'+g_{1,2}'\\&\leq 2(2 \sqrt{c^1_1\cdot c^2_1})^{-2}c_1+2c_2\leq q_{N}\end{array}\end{equation}
\iffalse
Note that, similar to the estimate for $\frac{\partial g_2}{\partial t},$ we have\begin{align*}&|\frac{\partial g_2(t,x)}{\partial x}|\\&=|\frac{2\lambda_{k(N)}\lambda_{k(N)}'\tan \tilde{g}_{1,1}+\lambda_{k(N)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda_{k(N)}^4\tan^2 \tilde{g}_{1,1}}+\tilde{g}_{1,2}'|\\
&\leq |\frac{\lambda_{k(N)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda_{k(N)}^4\tan^2 \tilde{g}_{1,1}}+\tilde{g}_{1,2}'|+|\frac{2\lambda_{k(N)}\lambda_{k(N)}'\tan \tilde{g}_{1,1}}{1+\lambda_{k(N)}^4\tan^2 \tilde{g}_{1,1}}|\\&=|\frac{\lambda_{k(N)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{1,1}}}{1+\lambda_{k(N)}^4\tan^2 \tilde{g}_{1,1}}+\tilde{g}_{1,2}'|+o(\lambda_{k(N)}^{-1.5})\\&\leq|\lambda_{k(N)}^{-2}\sin^{-2}\tilde{g}_{1,1}g_{1,1}'+g_{1,2}'|+o(\lambda_{k(N)}^{-1.5}).
\end{align*}\
And for the new zero points $c_{2,1}$ and $c_{2,2}$, it follows from the calculation above and the facts $$sgn(\frac{\partial g_2}{\partial x}(c_{2,i}))\cdot sgn(\frac{\partial \tilde{g}_{1,i}}{\partial x}(c'_{1,i}))\geq 0,i=1,2;$$
$$sgn(\frac{\partial \tilde{g}_{1,1}}{\partial x}(c'_{1,1}))\cdot sgn(\frac{\partial \tilde{g}_{1,2}}{\partial x}(c'_{1,2}))\leq 0,$$that

$$|\frac{\partial g_2}{\partial x}(c_{2,i})|\leq \tilde{g}'_{1,i}\leq c^1_1+o(\lambda^{-1.5}),i=1,2.$$
\fi
Therefore, we obtain $$q_{N}\geq |\frac{\partial g_2(\tilde{c}_2(t),t)}{\partial t}|>c_1-2\delta_1.$$

\textbf{Case $n\rightarrow n+1:$}


Suppose for all $l\leq n$, we have obtained $$C\lambda_{\kappa(l)}^{3}\geq |\frac{\partial g_l(t,x)}{\partial t}|>(c_1-2\sum_{i=1}^{l-1}\delta_i),$$
while  $$2^lq^C_{N+l-1}\geq|\frac{\partial g_{l}(\tilde{c}_l(t),t)}{\partial t}|>c_1-2\sum_{i=1}^{l-1}\delta_i,$$ and $g_l$ is of type II or III


For the case $n+1$, similarly we consider three cases as follows.

\begin{enumerate}
\item $g_{n+1}$ is of type I.

Due to the $ |\frac{\partial g_{n+1}(t,x)}{\partial t}-\frac{\partial g_{n}(t,x)}{\partial t}|_{C^1(I_n)}\leq \lambda^{-\frac{3}{2}q_{N+n-2}}$, we have $$C\lambda_{\kappa(n)}^{3}\geq C\lambda_{\kappa(n)}^{3}+\lambda^{-\frac{3}{2}q_{N+n-2}}\geq |\frac{\partial g_{n+1}(t,x)}{\partial t}|\geq  (c_1-2\sum_{i=1}^{n-1}\delta_i)-\lambda^{-\frac{3}{2}q_{N+n-2}}\geq c_1(1-2\sum_{i=1}^{n}\delta_i).$$

\item $g_{n+1}$ is of type II.

It's similar to the first case.

\item $g_{n+1}$ is of type III.

There exits some $k(n)\ll \lambda^{ck(n)}\leq k(n+1)\leq q_{N+n-1}$ such that $|c_{n,1}-c_{n,2}-k(n+1)\alpha|\leq C\lambda^{-k(n+1)},$ thus $g_{n+1}=\arctan \lambda_1^2\tan \tilde{g}_{n,1}-\frac{\pi}{2}+\tilde{g}_{n,2},$ where $\lambda_n>\lambda^{\frac{9}{10}k(n)}$, $g_{n,1},g_{n,2}$ are all Type I satisfying $$2^nq^C_{N+n-1}\geq C\lambda^3_{\kappa(n)}\geq\frac{\partial \tilde{g}_{n,1}}{\partial t}>c_1(1-\sum_{i=1}^{n-1}\delta_i)$$ and $$2^nq^C_{N+n-1}\geq C\lambda^3_{\kappa(n)}\geq \frac{\partial \tilde{g}_{n,2}}{\partial t}>c_1(1-\sum_{i=1}^{n-1}\delta_i)$$ with $sgn(\tilde{g}_{n,1})=sgn(\tilde{g}_{n,2}).$ Therefore,
\begin{equation}\begin{array}{ll}&C\lambda^3_{k(n+1)}\geq |\frac{\partial g_n(t,x)}{\partial t}|\\&=|\frac{2\lambda_{k(n+1)}\lambda_{k(n+1)}'\tan \tilde{g}_{n,1}+\lambda_{k(n+1)}^2\frac{\tilde{g}_{n,1}'}{\cos^2 \tilde{g}_{n,1}}}{1+\lambda_{k(n+1)}^4\tan \tilde{g}_{n,1}}+\tilde{g}_{n,2}'|\\
&\geq |\frac{\lambda_{k(n+1)}^2\frac{\tilde{g}_{n,1}'}{\cos^2 \tilde{g}_{n,1}}}{1+\lambda_{k(n+1)}^4\tan \tilde{g}_{n,1}}+\tilde{g}_{n,2}'|-|\frac{2\lambda_{k(n+1)}\lambda_{k(n+1)}'\tan \tilde{g}_{n,1}}{1+\lambda_{k(n+1)}^4\tan \tilde{g}_{n,1}}|\\&\geq c_1(1-2\sum_{i=1}^{n-1}{\delta_i})(1+0.5\lambda_{k(n+1)}^{-2})-2\lambda_{k(n+1)}^{-\frac{3}{2}}\\&\geq c_1(1-2\sum_{i=1}^{n-1}{\delta_i})-\lambda_{k(n+1)}^{-\frac{3}{2}}\geq c_1(1-2\sum_{i=1}^{n}{\delta_i}).
\end{array}\end{equation}\end{enumerate}

In particular, note that for $|\tilde{g}_{n,1}|>C\lambda_{k(n+1)}'^{-\frac{3}{4}},$ we can rewrite $g_{n+1}$ as $$\tilde{C}\lambda_{k(n+1)}'^{-2}(\frac{\partial \tilde{g}_{n,1}}{\partial x}(c'_{n,1})+o(\frac{1}{q^C_{N}}))\tilde{g}_{n,1})^{-1}+d_1-(\frac{\partial\tilde{g}_{n,1}}{\partial x}(c'_{n,2})+o(\frac{1}{q^C_{N}}))\tilde{g}_{n,2},$$
where $\tilde{C}$ is an absolute constant.
Therefore, it follows from the estimate in \cite{wz1} that $$g_{n+1}=C_1\lambda_{k(n+1)}'^{-2}(c^n_1(x-c'_{n,1}))^{-1}+d_1-c^n_2(x+k'\alpha-c'_{n,2}),$$ where \begin{equation}\label{lm17-xishu}c^n_i=\frac{\partial g_n}{\partial x}(c_{n,i})+\sum_{i=1}^{n}\delta'_i(x,t)\leq \frac{1}{q^C_{N+n-1}}<1 ,i=1,2\end{equation} with $\|\delta'_n(x,t)\|_{C^2}\leq \frac{1}{q^C_{N+n-1}}.$

By a direct calculation, we have $\tilde{c}_{n+1,i}=c'_{n,1}+((\sqrt{c^n_1\cdot c^n_2})^{-1})\lambda_{k(n+1)}^{-1},i=1,2.$

\vskip0.3cm
Hence, we have\begin{align*}&|\frac{\partial g_{n+1}(\tilde{c}_{n+1}(t),t)}{\partial t}|\\&=|\frac{2\lambda_{k(n+1)}\lambda'_{k(n+1)}\tan \tilde{g}_{n,1}+\lambda_{k(n+1)}^2\frac{\tilde{g}_{n,1}'}{\cos^2 \tilde{g}_{n,1}}}{1+\lambda_{k(n+1)}^4\tan^2 \tilde{g}_{n,1}}+\tilde{g}_{n,2}'|\\
&\leq |\frac{\lambda_{k(n+1)}^2\frac{\tilde{g}_{n,1}'}{\cos^2 \tilde{g}_{n,1}}}{1+\lambda_{k(n+1)}^4\tan^2 \tilde{g}_{n,1}}+\tilde{g}_{n,2}'|+|\frac{2\lambda_{k(n+1)}\lambda_{k(n+1)}\tan \tilde{g}_{n,1}}{1+\lambda_{k(n+1)}^4\tan^2 \tilde{g}_{n,1}}|\\&=|\frac{\lambda_{k(n+1)}^2\frac{\tilde{g}_{1,1}'}{\cos^2 \tilde{g}_{n,1}}}{1+\lambda_{k(n+1)}^4\tan^2 \tilde{g}_{n,1}}+\tilde{g}_{n,2}'|+o(\lambda_{k(n+1)}^{-1.5})\\&\leq|\lambda_{k(n+1)}^{-2}\sin^{-2}\tilde{g}_{n,1}g_{n,1}'+g_{n,2}'|+o(\lambda_{k(n+1)}^{-1.5}).
\\&\leq \left\vert\lambda_{k(n+1)}^{-2}\sin^{-2}|c'_{n,1}+((\sqrt{c^n_1\cdot c^n_2})^{-1})\lambda_{k(n+1)}^{-1}|g_{n,1}'+g_{n,2}'\right\vert+o(\lambda_{k(n+1)}^{-1.5})\\&\leq (2(\sqrt{c^n_1\cdot c^n_2})^{-1})^{-2}2^nq^C_{N+n-1}+2^nq^C_{N+n-1}\\&\leq 2^{n+1}q^C_{N+n},\end{align*}
where the last inequality follows from \eqref{lm17-xishu}.
\iffalse
And for the new zero points $c_{{n+1},1}$ and $c_{{n+1},2}$, it follows from the calculation above and the facts $$sgn(\frac{\partial g_{n+1}}{\partial x}(c_{{n+1},i}))\cdot sgn(\frac{\partial \tilde{g}_{n,i}}{\partial x}(c'_{n,i}))\geq 0,i=1,2;$$
$$sgn(\frac{\partial \tilde{g}_{n,1}}{\partial x}(c'_{n,1}))\cdot sgn(\frac{\partial \tilde{g}_{n,2}}{\partial x}(c'_{n,2}))\leq 0,$$that

$$|\frac{\partial g_{n+1}}{\partial x}(c_{{n+1},i})|\leq \tilde{g}'_{n,i}\leq c^1_1+o(\lambda^{-1.5}),i=1,2.$$
\fi
Therefore, we obtain \begin{align*}&2^{n+1}q^C_{N+n}\\&>|\frac{\partial g_{n+1}(\tilde{c}_{n+1,i}(t),t)}{\partial t}|\\&>c_1(1-2\sum_{i=1}^{n}{\delta_i}),i=1,2.\end{align*}

Let $c=\frac{1}{2}c_1$ for large $\lambda$ and by the notation $\lambda^3_{\kappa(n)}\leq \lambda^{4k(n)}$, the lemma is proved.


\end{proof}
\fi

The next lemma shows that $d_i$ has the similar local property to $g_i.$
\begin{lemma} \label{lmg'}Fix $t\in \mathcal{I}$ and let $d_i(t)=c_{i,1}(t)-c_{i,2}(t),i\in \N^+,$ then there exist $c(\alpha, v)>0$ and $C(\alpha, v, \lambda)>0$ such that
\begin{enumerate}
\item if $g_i(t)$ is \textbf{type I}, it holds that
$$C(\alpha,v, \lambda)q^{2C(\alpha, v, \lambda)}_{N+i-1}\geq \vert\frac{d d_i}{dt}(t')\vert >c(\alpha, v);$$

\item if $g_i(t)$ is of \textbf{type II or III}, it holds that
$$\vert\frac{d d_i}{dt}(t')\vert >c(\alpha, v), \forall t'\in [t-\frac{1}{q^{10C(\alpha, v, \lambda)}_{N+i-1}}, t+\frac{1}{q^{10C(\alpha, v, \lambda)}_{N+i-1}}]\bigcap\{t| \min_{I_i} |g_i|=0\}.$$
\label{lm14}
\end{enumerate}\end{lemma}
\begin{proof}For the first case.

On one hand, according to Lemma \ref{lmg}, we obtain $$c(\alpha, v)\leq |\frac{\partial g_i(x,t)}{\partial t}|\leq c^{-1}(\alpha, v)$$ for any $x\in I_i(t).$

On the other hand, since $g_i$ is of  type I, we clearly have (for sufficiently large $\lambda$) \begin{equation}\label{lm19-1}1\geq |\frac{\partial g_{i}}{\partial x}|\geq q^{-C(\alpha, v,\lambda)}_{N+i-1}, \forall x\in I_i.\end{equation}

Then it follows from the Implicit Function Theorem that $$c(\alpha, v)\leq \vert\frac{d c_{i,j}(t)}{dt}\vert=\vert\frac{\partial g_i(c_{i,j}(t),t)}{\partial t}\slash\frac{\partial g_i(c_{i,j}(t),t)}{\partial x}\vert\leq c^{-1}(\alpha,v)q_{N+i-1}^{C(\alpha, v, \lambda)},$$ where $0<c(\alpha, v)<1.$

Moreover, we have
\begin{equation}\begin{array}{ll}
\frac{d c_{i,1}(t)}{dt}\cdot\frac{d c_{i,2}(t)}{dt}&=(\frac{\partial g_i(c_{i,j}(t),t)}{\partial t}\slash\frac{\partial g_i(c_{i,j}(t),t)}{\partial x})\cdot(\frac{\partial g_i(c_{i,j}(t),t)}{\partial t}\slash\frac{\partial g_i(c_{i,j}(t),t)}{\partial x})\\
&<0
\end{array}\end{equation}
Thus, it holds that
$$2C(\alpha, v, \lambda)q^{2C(\alpha, v, \lambda)}_{N+i-1}\geq \vert\frac{d d_i}{dt}(t)\vert=\vert\frac{dc_{i,1}(t)}{dt}-\frac{dc_{i,2}(t)}{dt}\vert>2c(\alpha, v).$$

The second case can be similarly handled with the help of Lemma \ref{lmg}, the left inequality of \eqref{lm19-1} and the Implicit Function Theorem.


\end{proof}

Now, we are ready to show the `leanness' of $IR$ in the sense of measure.

\begin{lemma}$Leb(IR)=0.$
\label{lm15}
\end{lemma}
\begin{proof} Denote $$B(n,k):=\{t| |d_n(t)-k\alpha|\geq \frac{1}{q^{1000C}_{N+n-1}}\};$$
$$P_1(n)=\{t|\min|g_n(t)|=0\} ~and~ P_2(n)=\{t|\min|g_n(t)|>0\};$$
$$\bigcup\limits_{k\leq q_{N+n-1}}B^c(n,k):=C(n).$$
Clearly, $$\Sigma_1=\bigcup\limits_{l\geq1}\bigcap\limits_{n\geq l}\bigcap\limits_{k\leq q_{N+n-1}}B(n,k);$$ $$\Sigma_2=\bigcap\limits_{l\geq1}\bigcup\limits_{n\geq l}\bigcup\limits_{k\leq q_{N+n-1}}B^c(n,k)=\bigcap\limits_{l\geq1}\bigcup\limits_{n\geq l}C(n);$$
By Borel Contelli Lemma, it is suffice to prove $\sum_{n=1}^{+\infty} Leb(C(n))<+\infty.$

Clearly, $g_n$ is of type III, by the help of Lemma \ref{lmg} and Lemma \ref{lmg'}, it's not difficult to see:

\textbf{a:~}If $\min|g_n(t)|=0$, when we continuously move $t$ to $t'$ with $\min|g_n(y)|=0$ for any $y$ between $t$ and $t',$  the difference between the critical points $c_{n,1}$ and $c_{n,2}$ is at least $c(\alpha, v)|t-t'|,$ that is
$$\left|d_n(t)-d_n(t')\right|\geq c(\alpha, v)\cdot|t-t'|.$$ Therefore, for any $|t'-t|\geq \frac{1}{q^{750C}_{N+n-1}}$ we obtain $$\left|d_n(t')-k\alpha\right|\geq c(\alpha, v)\frac{1}{q^{750C}_{N+n-1}}-\frac{1}{q^{1000C}_{N+n-1}}\geq \frac{1}{q^{1000C}_{N+n-1}},$$ which implies
$$Leb\left\{C(n)\bigcap P_2(n)\right\}=Leb\left\{\{t\big\vert ~~|d_n(t)-k\alpha|< \frac{1}{q^{1000C}_{N+n-1}}\}\bigcap P_2(n)\right\}\leq \frac{1}{q^{750C}_{N+n-1}}.$$

\textbf{b:~}If $\min|g_n(t)|>0$, then according to Lemma \ref{lmg}, we are allowed to continuously move $t$ until $t''$ such that $\min|g_n(t'')|=0$ with $|t-t''|\leq c(\alpha, v)\lambda^{-\frac{1}{2}r_n}\ll \frac{1}{q^{500C}_{N+n-1}}.$

Thus, combining \textbf{a} and \textbf{b} we have \begin{equation}\begin{array}{ll}&Leb\{B^c(n,k)\}=Leb\{B^c(n,k)\bigcap P_1(n)\}+Leb\{B^c(n,k)\bigcap P_2(n)\}\\&\leq \frac{1}{q^{500C}_{N+n-1}}+\frac{1}{q^{500C}_{N+n-1}}\\&\leq \frac{1}{q^{50C}_{N+n-1}}.\end{array}\end{equation}
Therefore $$\sum_{n=1}^{+\infty} Leb(C(n))\leq \sum_{n=1}\sum_{k\leq q_{N+n-1}} Leb\{B^c(n,k)\leq \sum_{n=1}\frac{1}{q^{10C}_{N+n-1}}<\infty. $$

\end{proof}

Let $EP:=\{all~the~endpoints~of~spectrum~gaps\}~and~LP:=\Sigma-EP.$ Clearly, we obtain that $LP$ is a full measure set in $\Sigma$.


when $g_n$ is of type II or III, let $$\tilde{c}_{n,1}(t):=\{x\in I_n| \frac{d g_n}{d x}(x)=0 ~and~x~is~between~c'_{n,2}~and~c_{n,1}\}.$$
Similarly, let $$\tilde{c}_{n,2}(t):=\{x\in I_n| \frac{d g_n}{d x}(x)=0 ~and~x~is~between~c'_{n,1}~and~c_{n,2}\}.$$

The following lemma states the important property of $t\in EP,$ which results in $\frac{1}{2}$-H\"older continuity.

\begin{lemma} The following are equivalent.

1:$t\in EP$;

2:For $t\in \frac{1}{\lambda}\Sigma,$ there exists some $N(t)$ such that $g_n(t)$~always~is~of~type~II~or~III~for~$n\geq N(t);$

3:For $t\in \frac{1}{\lambda}\Sigma,$ there exists some $N(t)$ such that $|g_n(t,\tilde{c}_{n,1}(t))|<\lambda^{-\frac{3}{2}r_{N+n-2}}$ for $n\geq N(t).$\label{lm16}
\end{lemma}

First, we prove the following lemma.
\begin{lemma}For any fixed $t\in J$ if $g_n(t)$ is of type I for some $n\in \N$, it holds that $[t-\lambda^{-0.5q_{N+n(t)-1}},t+\lambda^{-0.5q_{N+n(t)-1}}]:=\tilde{I}(n(t))\bigcap\Sigma\neq \emptyset.$\label{fr}
\end{lemma}
\begin{proof} Clearly, it follows from \cite{young} that $\lim\limits_{n\rightarrow \infty}\min\limits_{I_n}g_n(t)=0$ implies $t\in \Sigma.$ Thus, without loss of generality, we assume that there exists some $N(t)>n(t)$ such that $g_{N}$ is of type III and $g_i(t),$ $n(t)\leq i \leq N-1,$ are of type I. We denote $t_0=t$ and consider the following two conditions.
\begin{enumerate}
\item $|g_{N}(\tilde{c}_N)|\leq \lambda^{-\frac{3}{2}r_{N-1}}.$

By lemma \ref{lmg}, there exists some $t_1\in [t-\lambda^{-\frac{1}{2}r_{N-1}},t+\lambda^{-\frac{1}{2}r_{N-1}}]$ such that $\min\limits_{I_N}g_N=0$ and $|g_{N}(\tilde{c}_N)|=\lambda^{-\frac{3}{2}r_{N-1}}$ (This can been considered as putting the minimum of $g_N$ down or up such that $g_N$ across the horizontal axis with the minimum of $g_N$ being the fixed distance.)

\item $|g_{N}(\tilde{c}_N)|\geq \lambda^{-\frac{3}{2}r_{N-1}}.$
\begin{enumerate}
\item[a:] If $\min |g_N|>0$, it follows from \cite{wz1} that $t\in \emph{UH}$ and $\min |g_N|\leq C\lambda_N^{-1}\leq \lambda^{-\frac{2}{3}r_{N-1}}.$ Again by lemma \ref{lmg}, there exists some $t_1\in [t-\lambda^{-\frac{1}{2}r_{N-1}},t+\lambda^{-\frac{1}{2}r_{N-1}}]$ such that $\min\limits_{I_N}g_N=0$ and $|g_{N}(\tilde{c}_N)|=\lambda^{-\frac{3}{2}r_{N-1}}.$

\item[b:] If $\min |g_N|=0,$ we denote $t_1=t.$
\end{enumerate}
\end{enumerate}
Therefore, we can find some $t_1\in [t-\lambda^{-\frac{1}{2}r_{N-1}},t+\lambda^{-\frac{1}{2}r_{N-1}}]$ such that $|g_{N}(t_1)(\tilde{c}_N)|\geq \lambda^{-\frac{3}{2}r_{N-1}}$ and $\min |g_N|=0.$

Since $$\sum_{0\leq i\leq +\infty} \|g_{N+i}-g_{N+i+1}\|<C\lambda^{-\frac{7}{4}r_{N-1}}\ll \lambda^{-\frac{3}{2}r_{N-1}},$$ there exits some $N'>N$ such that $g_{N'}$ is of type I and $g_i$, $N\leq i \leq N'-1,$ are of type III. Clearly, we have $q_{N+N'-1}^{-C}\leq |I_{N'}|\leq  \lambda^{-\frac{1}{2}r_{N-1}}$, thus \begin{equation}\label{gngj}\lambda^{-\frac{1}{2}r_{N'}}\leq \lambda^{-\frac{1}{2}q_{N+N'-1}}\leq \lambda^{-\frac{1}{2}\lambda^{cr_{N-1}}}\leq \lambda^{-\frac{1}{2}r^2_{N-1}}.\end{equation}

Therefore, it's not difficult to see that if we start from $g_{N'}$, by the same process as above we can find some $t_2\in [t-\lambda^{-\frac{1}{2}r^2_{N-1}},t+\lambda^{-\frac{1}{2}r^2_{N-1}}]$ such that $|g_{N'}(t_2)(\tilde{c}_{N'})|\geq \lambda^{-\frac{3}{2}r_{N(t_2)-1}}$ and $\min |g_N'|=0.$ Meanwhile $|g_{N'}(t_2)(\tilde{c}_{N'})|\geq (1-e_1)\lambda^{-\frac{3}{2}r_{N-1}}$ and $\min |g_N'|=0$ with $e_1=\lambda^{-\frac{1}{3}r_{N-1}}.$

Repeat this process. We can find the sequences $t_i$ and $N(t_i)$ such that $g_{N(t_i)}$ is of type I and $|g_{N(t_i)}(t_i)(\tilde{c}_{N(t_i)})|\geq \lambda^{-\frac{3}{2}r_{N(t_i)-1}}$, $\min |g_{N(t_i)}|=0.$ Meanwhile for each $j<i,$ we have $|g_{N(t_j)}(t_i)(\tilde{c}_{N(t_i)})|\geq (1-\sum_{i=1}^{i-1}e_i)\lambda^{-\frac{3}{2}r_{N(t_j)-1}}$ and $\min |g_N(t_j)(t_i)|=0$ with $e_i=\lambda^{-\frac{1}{3}r^{i}_{N-1}}.$

Denote $t_0=\lim\limits_{i\rightarrow \infty} t_i.$ Clearly, $t_0\in \tilde{I}(n(t)).$
Finally, we find a sequence $g_{N_i}(t_0)$ such that $\lim\limits_{i\rightarrow \infty} \min |g_{N_i}(t_0)|=0,$ thus $t_0\in \Sigma$ as require.

\end{proof}

\begin{proof}[\textbf{The proof of lemma \ref{lm16}}]

$1\rightarrow2:$Note that if there exists some subsequence $g_{n_i}(t)$ being of type I. By lemma\ref{lmg}, it holds that for any $t^{-}_i=t-\frac{1}{q^C_{N+n(i)-1}},$  $g_{n_i}(t^{-}_i)$ is of type I. Similarly, for any $t^{+}_i=t+\frac{1}{q^C_{N+n(i)-1}},$  $g_{n_i}(t^{+}_i)$ is of type I. Therefore, it follows from lemma \ref{fr} that there exists some spectral point $\tilde{t}^{\pm}_i\in B(t^{\pm}_i,\lambda^{-\frac{1}{2}r_{n_i-1}}).$ Hence at each side of $t,$ there exists two sequence $\tilde{t}^{+}_i$ and $\tilde{t}^{-}_i$, which come from the spectrum, such that $\tilde{t}^{\pm}_i\rightarrow t.$ This is a contradiction since for any endpoint, there must be no spectral point on one side.

Therefore, there exists some $N$ such that $g_n(t)$ are of type II or III $\forall n\geq N.$

$2\rightarrow3:$, note that if $|g_m(\tilde{c}_{m,1})|\geq C\lambda^{-\frac{3}{2}r_{m-1}}$ for some $m>N,$ only the following two cases might occur.
\begin{enumerate}
\item $\min |g_m(\tilde{c}_{m,1})|=0.$

Since \eqref{gngj} holds, there must exist some $N'>m$ such that $g_{N'}$ is of type I, this contradicts the fact $g_n(t)$ are of type II or III $\forall n\geq N.$

\item $\min |g_m(\tilde{c}_{m,1})|>0.$

Again by \eqref{gngj}, it follows that $t\in \Sigma^c,$ which also contradicts  the fact $t\in \Sigma.$
\end{enumerate}

$3\rightarrow1:$For any fixed $t$ satisfies $3$, by the help of Lemma \ref{lmg}, we have a uniformly (respect with $n$ and $t$) lower bound $$|\frac{\partial g_n(x,t)}{\partial t}|>c$$ for any $x\in \mathcal{J}_n\triangleq[\tilde{c}_n(t),\tilde{c}_n(t)+\frac{1}{q^C_{N+n-1}}]$(or $x\in [\tilde{c}_n(t)-\frac{1}{q^C_{N+n-1}},\tilde{c}_n(t)]$) $\in$ $I_n(t)$. Thus for fixed large $n_0>\lambda^{\lambda^{N(t)}}$ and $t'$ satisfied $q_{N+n_0-1}^{-1000\tau^{1000}}<|t'-t|<q_{N+n_0-2}^{-1000\tau^{1000}},$ we have
$$|\frac{\partial g_{n_0}(\tilde{c_{n_0}}(t'),t')}{\partial t}-\frac{\partial g_{n_0}(\tilde{c_{n_0}}(t),t)}{\partial t}|\geq c|t-t'|\geq q_{N+n_0-1}^{-2000\tau^{1000}}.$$ Combine this with the help of \cite{wz1}, we have that for $n\geq n_0$, \begin{equation}\label{panbie}\min\{g_n(t')\}\geq c|t'-t|-\sum_{i=n_0}^{+\infty}\lambda^{-\frac{3}{2}r_i}\geq q_{N+n_0-1}^{-3000\tau^{1000}}\triangleq C(n_0)>0,\end{equation} which implies $\mathcal{UH}$. And according the definition of type III, it's clear that \eqref{panbie} holds for only one of the following two case. $(1)[t-q_{N+n_0-2}^{-1000\tau^{1000}},t-q_{N+n_0-1}^{-1000\tau^{1000}})$ or $(2)(t+q_{N+n_0-1}^{-1000\tau^{1000}},t+q_{N+n_0-2}^{-1000\tau^{1000}}].$  WLOG, we assume $(1)$ holds true, which implies $[t-q_{N+n_0-2}^{-1000\tau^{1000}},t-q_{N+n_0-1}^{-1000\tau^{1000}})\subset \frac{1}{\lambda}\Sigma^c.$ Hence ,$[t-q_{N+n_0-2}^{-1000\tau^{1000}},t)= \bigcup\limits_{l\geq n_0}[t-q_{N+l-2}^{-1000\tau^{1000}},t-q_{N+l-1}^{-1000\tau^{1000}})\subset \frac{1}{\lambda}\Sigma^c.$ Therefore, $t\in EP.$
\end{proof}


At the end of the section, we prove one of the important property of $EP.$
\begin{lemma}$\Sigma_3=EP.$\label{lemdd}
\end{lemma}
\begin{proof}Combine Lemma \ref{lm16} with the definition of $\Sigma_3,$ we obtain that for $t\in \Sigma_3$ there exists $N(t)$ such that $g_n$ is of type III for each $n\geq N(t),$ which implies $EP\supseteq\Sigma_3.$ For the other direction, we only need to prove $\Sigma^c_3\subseteq EP^c.$ Lemma \ref{lm16} shows that $\Sigma_1\cap EP=\emptyset.$ Since $\Sigma_3\subset \Sigma_2,$ it's enough to show $(\Sigma_2-\Sigma_3)\cap EP=\emptyset.$ Again by the definition of $\Sigma_3$, we can find a sequence $0<k_1<k_2<\cdots<k_n<\cdots\rightarrow\infty$ such that $$\vert d_{N(k_i)}-k_i\alpha\vert\geq \lambda^{-\frac{1}{2}r_{k_i-1}}.$$ Thus there must exists some type I cases between each $k_i$ and $k_{i+1}.$ Again by Lemma \ref{lm16} we finish the proof.
\end{proof}


\section{Indicator function for Measuring the Intensity of Resonance}

In this section, we introduce a function to measure the intensity of resonance.

let
\begin{equation}\label{defbeta}\beta(t)\triangleq\min\{\liminf_{n\rightarrow+\infty}(\frac{1}{2}+\frac{\log\|A_{k_n}\|}{2\log|c_{n,1}(t)-c_{n,2}(t)-k_n(t)\alpha|^{-1}}), 1\}.\end{equation}

Here $k_n$ denote the \textbf{resonance-distance}, which is defined as follows.
$$k_n:=\min\limits_{k}\left\{k\vert I_{n,1}+k\alpha\cap I_{n,2}\neq \emptyset\right\},$$
$n\geq 1.$

It's easy to find that either there exists some $k\geq 0$ and $N>0$ such that $k_n=k$ for any $n\geq N,$ or $k_n\rightarrow +\infty.$ In fact, for any $t\in EP,$ $t$ satisfies the first case.
With slight offense to the definition, we denote
$\left\| A_{k_n}\right\|:=\left\| A_{k_n}(x)\right\|$ for a fixed $x\in I_n.$ It's essentially well defined since lemma \ref{key2} implies $$\left|\frac{d\left\|A_{k_n}\right\|}{dx}\right|\leq \left\|A_{k_n}(x)\right\| q_{N+n-1}^C$$ for $x\in I_n$ and $$q_{N+n-1}\ll \lambda^{\frac{1}{2}k_n}\leq \left\|A_{k_n}\right\| \leq \lambda^{k_n},$$ which means $\left\|A_{k_n}(x)\right\|$ is almost a constant on $I_n.$
It's easy to check that $$\frac{1}{2}\leq \beta(t)\leq 1.$$
Besides, it's worth noting that $\beta(t)$ represents the intensity of resonance. One can check that the definition is equivalent to the following one.
\begin{lemma}\label{beta}$$\beta(t)\triangleq\min\{\liminf_{k\rightarrow+\infty}(\frac{1}{2}+\frac{k\log\lambda_{\infty}}{2\left|\log\left|c_{\infty,1}-c_{\infty,2}-k\alpha\right|\right|}),1\}.$$\end{lemma}
\begin{proof}\begin{enumerate}
\item[1:] If there exist a subsequence $n_j$ such that $\left|c_{n_j,1}(t)-c_{n_j,2}(t)-k_{n_j}(t)\alpha\right|\leq \lambda^{-\frac{1}{10}r_{n_j}},$ then by the help of the Induction Theorem of \cite{wz1}, it's clear that $\left|c_{\infty,1}(t)-c_{\infty,2}(t)-k_{n_j}(t)\alpha\right|\leq \lambda^{-\frac{1}{20}r_{n_j}}.$ Thus, it follows from the fact $\|A_{k_{n_j}}\|\le \lambda^{k_{n_j}}\le \lambda^{\frac{1}{20}r_{n_j}^{\frac{9}{10}}}$ and $\lambda^{1-\sum_{i=N}^{\infty}\frac{\log q_{i+1}}{q_i}}\le \lambda_{\infty}\le \lambda$ that
    $$\lim_{j\rightarrow+\infty}\frac{\log\|A_{k_{n_j}}\|}{-\log\left|c_{{n_j},1}(t)-c_{{n_j},2}(t)-k_{n_j}\alpha\right|}=\lim_{j\rightarrow+\infty}\frac{k_{n_j}\log\lambda_{\infty}}{-\log\left|c_{\infty,1}(t)-c_{\infty,2}(t)-k_{n_j}\alpha\right|}=0.$$
    Note that the definition shows $\beta(t)\geq \frac{1}{2},$ then it's not difficult to see that
    $$\beta(t)=\min\{\liminf_{k\rightarrow+\infty}(\frac{1}{2}+\frac{k\log\lambda_{\infty}}{2\left|\log\left|c_{\infty,1}-c_{\infty,2}-k\alpha\right|\right|}),1\}$$
    $$=\min\{\liminf_{n\rightarrow+\infty}(\frac{1}{2}+\frac{\log\|A_{k_n}\|}{2\left|\log\left|c_{n,1}-c_{n,2}-k_n\alpha\right|\right|}),1\}=\frac{1}{2}.$$


\item[2:] If for all $n\in \N,$ $\left|c_{n,1}(t)-c_{n,2}(t)-k_n(t)\alpha\right|> \lambda^{-\frac{1}{10}r_{n}},$ then it again follows from the Induction Theorem of \cite{wz1} that $$\left|c_{{n},1}(t)-c_{{n},2}(t)-k_{n}\alpha\right|-\lambda^{-\frac{3}{2}r_n}\le \left|c_{{\infty},1}(t)-c_{{\infty},2}(t)-k_{n}\alpha\right|\le \left|c_{{n},1}(t)-c_{{n},2}(t)-k_{n}\alpha\right|+\lambda^{-\frac{3}{2}r_n},$$
    which implies
    $$1-\lambda^{-r_n}\le \frac{\left|c_{{\infty},1}(t)-c_{{\infty},2}(t)-k_{n}\alpha\right|}{\left|c_{{n},1}(t)-c_{{n},2}(t)-k_{n}\alpha\right|} \le 1+\lambda^{-r_n}.$$

    Combine this with $|\log\lambda_n-\log\lambda_{\infty}|\le \sum_{i=n}^{\infty}\frac{\log q_{N+i}}{q_{N+i-1}}\triangleq e_n,$ it holds that
    \begin{equation}\label{beta-1} (1-e_n)(1-\lambda^{-r_n})\le \left|\frac{\log\|A_{k_n}\|}{k_n\log\lambda_{\infty}}\frac{\log\left|c_{{\infty},1}(t)-c_{{\infty},2}(t)-k_{n}\alpha\right|^{-1}}{\log\left|c_{n,1}(t)-c_{n,2}(t)-k_{n}\alpha\right|^{-1}}\right|\le
    (1+e_n)(1+\lambda^{-r_n}).\end{equation}

    Therefore, on one hand, for any given $t\in\Sigma,$ if we assume there exists a sequence $k_n$ such that \eqref{defbeta} convergence to the limit inferior($\beta(t)$), then with the same sequence we have
    $$\min\{\liminf_{n\rightarrow+\infty}\frac{1}{2}+\frac{k_n\log\lambda_{\infty}}{2\left|\log\left|c_{\infty,1}-c_{\infty,2}-k_n\alpha\right|\right|},1\}=\beta(t),$$
    which implies
    $$\min\{\liminf_{k\rightarrow+\infty}\frac{1}{2}+\frac{k\log\lambda_{\infty}}{2\left|\log\left|c_{\infty,1}-c_{\infty,2}-k\alpha\right|\right|},1\}\leq \beta(t).$$

    On the other hand, the definition of $k_n$ shows that
    $|c_{\infty,1}-c_{\infty,2}-j\alpha|\geq \frac{1}{10}|I_j|-\|A_{k_n}\|^{-1}-\lambda^{-\frac{3}{2}r_{n}}\geq \lambda^{-\frac{1}{20}k_{n-1}}\geq \lambda^{-\frac{1}{20}j},$ for $k_{n-1}\leq j \leq k_n.$ Thus $|c_{\infty,1}-c_{\infty,2}-k\alpha|\geq \lambda^{-\frac{1}{20}k},$ for each $k\in \N,$ which directly implies
    $$\min\{\liminf_{k\rightarrow+\infty}\frac{1}{2}+\frac{k\log\lambda_{\infty}}{2\left|\log\left|c_{\infty,1}-c_{\infty,2}-k\alpha\right|\right|},1\}\geq \beta(t).$$


\end{enumerate}In conclusion we obtain \eqref{beta}.
\end{proof}

It is easy to see that If $t\in EP,$ then $\beta(t)=\frac{1}{2},$ if $t\in FR,$ then $\beta(t)=1.$
\section{{ Regularity of LE at the end points of spectral gaps}}
In this section, we prove the exact $\frac{1}{2}$-H\"older continuity of Lyapunov exponent $L(E)$ for $E\in EP$.
That is, for each end point $E_0$ of spectral gaps, there exists a $\delta=\delta(E_0)>0$ and two positive constants $\tilde{c}=\tilde{c}(E_0), \tilde{c}'=\tilde{c}'(E_0)$ such that for each $E\in(E_0-\delta, E_0+\delta)$, it holds that $\tilde{c} |E-E_0|^{\frac{1}{2}}\le|L(E)-L(E_0)|\le \tilde{c}' |E-E_0|^{\frac{1}{2}}$.
\vskip 0.3cm

The $\frac{1}{2}$-H\"older continuity is proved by improving the estimates in a recent work \cite{LWY}. Based on a Large Deviation Theorem (LDT) of the form
$$Leb\{x\in \R/{\Z} | \frac{1}{i}\log\|A_i(x)\|\geq \frac{9}{10}\log\lambda\}<\lambda^{-ci},\quad 0<c<1/2,$$
   the $c-$H\"older continuity of Lyapunov exponent was obtained in  \cite{LWY}. The idea for the proof of \cite{LWY} is as follows.


Since $L_{N_0}(E)=\frac1{N_0}\int\log\|A_{N_0}(x,E)\|dx$, from the facts that $\|A_{N_0}(x,E)\|\ge 1$ and $|\partial_E\|A_{N_0}(x,E)\||\le C^{N_0}$ with $C>1$ depending only on $A(x)$, we have
$$
|L'_{N_0}(E)|=\left|\frac{1}{N_0}\int\frac{\partial_E\|A_{N_0}(x,E)\|}{\|A_{N_0}(x,E)\|}dx\right|\le C^{N_0}.
$$
With the help of LDT and the Avalanche Principle, for each pair $(E_0, E)$, there exist  $c>0, C'>C$ and (large) $N_0$ satisfying $|E-E_0|\thickapprox C'^{-N_0}$ such that
$$
|L(E)-L(E_0)|\le |L_{N_0}(E)-L_{N_0}(E_0)|+|L_{2N_0}(E)-L_{2N_0}(E_0)|+2e^{-cN_0}.$$


It follows that
$$
|L(E)-L(E_0)|\le 2C^{N_0}|E-E_0|+2e^{-cN_0}\le |E-E_0|^{c'}

$$
with $c'\thickapprox \min \{\log_{C'}\frac{C'}{C}, c\cdot\log_{C'}e\}>0$.

 Note that the estimate on the upper bound for $L'_{N_0}(E)$ in the above argument is too bad. In fact, in \cite{LWY} it is estimated by
$\left\|\frac{\partial_E\|A_{N_0}(x,E)\|}{\|A_{N_0}(x,E)\|}\right\|_{\mathcal{C}^0(\mathbb{S}^1)}\approx C^{N_0}$. However, the upper bound for $L'_{N_0}(E)$ only depends on $\left\|\frac{\partial_E\|A_{N_0}(x,E)\|}{\|A_{N_0}(x,E)\|}\right\|_{\mathcal{L}^1(\mathbb{S}^1)}$, which
should be much less than $\|\cdot\|{\mathcal{C}^0(\mathbb{S}^1)}$. The reason lies in the observation that the set of bad $x$ satisfying $\left|\frac{\partial_E\|A_{N_0}(x,E)\|}{\|A_{N_0}(x,E)\|}\right|\approx C^{N_0}$ is of a small measure depending on $N_0$. By a complicated computation, we will be able to obtain a sharp upper bound estimate on $L'_{N_0}(E)$ which is sufficient to prove $\frac{1}{2}$-H\"older continuity.

On the other hand, for different $E$, the measure of bad $x$ is of different order. More precisely, the measure of the bad set for end points of spectral gaps is larger than the one for each point in some set $FR$ of full measure. Thus for an end point of spectral gaps we can obtain a lower bound on $L'_{N_0}(E)$ which implies the regularity cannot be better than
$\frac{1}{2}$-H\"older continuity, while for each point in the set $FR$ we can obtain the Lipschitz continuity.


\vskip 0.4cm
From now on, the scale of the relevant quantities in \cite{wz1}  will be slightly modified as follows.
For any $n\geq 0.$
Let $I_n:=B(c_{n,1},\frac{1}{2^n q_{N+n-1}^{8\tau}})\bigcup B(c_{n,2},\frac{1}{2^n q_{N+n-1}^{8\tau}})$.
Define $q_{N+n-1}^2<r_n^{\pm}:I_n\rightarrow \Z^{+}$ to be the smallest positive number $j$ such that $j>q_{N+n-1}^2$ and $T^{\pm j}x\in I_n$ for $x\in I_n$.
Note that the choice of``$8\tau$'' is nonessential, and any large constant $C$ is adequate. It's not difficult to see the different choices only affect on the error function between the two Iteraton steps  in \cite{wz1}.


For our purpose, LDT in \cite{LWY} is needed. For the convenience of readers, we provide a simple proof of it in Appendix A.4.
\begin{theorem}\label{Th18} Let $v$ and $\alpha$ be as in Theorem 1. Then there exists $\lambda_1=\lambda(v,\alpha),i_0=i_0(\alpha)\in\Z^+$ and $0<c<1,$ such that for each $\lambda>\lambda_1$ and each $i\geq i_0,$ it holds that
$$Leb\{x\in \R/{\Z} | \frac{1}{i}\log\|A_i(x)\|\geq \frac{9}{10}\log\lambda\}<\lambda^{-ci}.$$\label{ldt}
\end{theorem}
\begin{remark} The version of LDT is not sharp but it is enough for our purpose.
\end{remark}


     The following lemma is a corollary of Avalanche Principle of \cite{goldsteinschlag}, see \cite{bourgainjitomirskaya}.

     \begin{lemma} Let $E^{(1)}$, $\cdots$, $E^{(n)}$ be a finite sequence in $SL(2,\R)$ satisfying the following conditions
     \begin{equation}\label{gs1}
     \min\limits_{1\leq j\leq n}\left\|E^{(j)}\right\|\geq \mu\geq n
     \end{equation}
     and
     \begin{equation}\label{gs2}
     \min\limits_{1\leq j\leq n}\left| \log\left\|E^{(j+1)}\right\|+\log\left\|E^{(j)}\right\|-\log\left\|E^{(j+1)}E^{(j)}\right\|\right|<\frac{1}{2}\log\mu,
     \end{equation}
     then
     \begin{equation}\label{gs3}
     \left|\log\left\|E^{(n)}\dots E^{(1)}\right\|+\sum_{2}^{n-1}\log\left\|E^{(j)}\right\|-\sum_{1}^{n-1}\log\left\|E^{(j+1)}E^{(j)}\right\|\right|\leq C\frac{n}{\mu}.
     \end{equation}\label{gs}
     \end{lemma}

With the help of LDT and Avalanche Principle, we have
     \begin{lemma}\label{lm27} Let $v,\alpha~and~\lambda$ be as in Theorem \ref{Th18}. Then for all large $n\in \Z^+$ and for all $E\in [\lambda\inf v -2,\lambda\sup v +2],$ it holds that
     $$\big\vert L_n(E)+L(E)-2L_{2n}(E)\big\vert<\lambda^{-\frac{c}{2}n}.$$\label{ap}
     \end{lemma}
     \begin{proof} For any $l\in \N,$ Let
$$n= [\lambda^{\frac{c}{2}l}],E^{(j)}=A_l(x+(j-1)l\alpha).$$

From lemma~\ref{ldt}, there is an exceptional set $\Omega\subset\R/\Z$ with $Leb(\Omega)<\lambda^{-\frac{c}{2}l}$ such that if $x\notin \Omega,$ and $j=0,\cdots,n$ then
$$\frac{9}{10}\log\lambda<\frac{1}{l}\log\|A_l(x+jl\alpha)\|,\frac{1}{2l}\log\|A_{2l}(x+jl\alpha)\|<\log\lambda.$$
Hence for $x\notin\Omega,$
$$\vert\log\|E^{(j+1)}\|+\log\|E^{(j)}\|-\log\|E^{(j+1)}E^{(j)}\|\vert<\frac{1}{5}\log\lambda<\frac{\frac{4}{5}}{2}\log\lambda.$$
Thus taking $\mu=\lambda^{\frac{4}{5}l},$
conditions \eqref{gs1} and \eqref{gs2} of lemma \ref{gs} are clearly fullfilled. For $x\notin\Omega$ the conclusion \eqref{gs3} is that
$$\vert\log\|A_{ln}(x)\|+\sum_{2\leq j\leq n-1}\log\|A_l(x+(j-1)l\alpha)\|-\sum_{1\leq j\leq n-1}\log\|A_{2l}(x+(j-1)l\alpha)\|\vert\leq C \frac{n}{\mu}.$$

Divide the above inequality by $ln$ and integrate it in $x\in \R/\Z.$ Splitting the integration as $(\R/\Z)-\Omega$ and $\Omega$, we get
$$\vert L_{ln}(E)+\frac{n-2}{n}L_l(E)-\frac{2(n-1)}{n}L_{2l}(E)\vert<C(\mu^{-1}l^{-1}+Leb(\Omega))<C\lambda^{\frac{c}{3}l}.$$
Note here $C$ depends on $\lambda.$ Hence we obtain
\begin{equation}\label{ap1} \vert L_{ln}(E)+L_l(E)-2L_{2l}(E)\vert<C \lambda^{-\frac{c}{2}l}.
\end{equation}
Applying \eqref{ap1} to $n=n_1=l$ large enough and $\log n_2\sim n_1$ yields
\begin{equation}\label{ap2}
\vert L_{n_2}(E)+L_{n_1}(E)-2L_{2n_1}(E)\vert <C\lambda^{-\frac{c}{2}n_1}.
\end{equation}
Applying \eqref{ap1} to $n=n_1=l$ and $2n_2$ yields
$$\vert L_{2n_2}(E)+L_{n_1}(E)-2L_{2n_1}(E)\vert<C \lambda^{-\frac{c}{2}n_1}.$$

Therefore, we get
\begin{equation}\label{ap3}
\vert L_{2n_2}(E)-L_{n_2}(E)\vert<C \lambda^{-\frac{c}{2}n_1}.
\end{equation}
Clearly, in \eqref{ap2} and \eqref{ap3}, we may replace $n_1$ and $n_2$ by any $n_s$ and $\log n_{s+1}\sim n_s.$ Thus we obtain
\begin{align*}
\vert L(E)-L_{n_2}(E)\vert&\leq \sum_{s\geq 2}\vert L_{n_{s+1}}(E)-L_{n_s}(E)\vert\\
&\leq 2\sum_{s\geq 2}(\vert L_{2n_s}(E)-L_{n_s}(E)\vert+C\lambda^{\frac{c}{2}n_s})\\
&\leq 4 \sum_{s\geq 1} C\lambda^{-\frac{c}{2}n_s}\\
&<C\lambda^{-\frac{c}{2}n}.
\end{align*}

Thus, replacing $L_{n_2}(E)$ by $L(E)$ in \eqref{ap2} yields lemma \ref{ap}.

\end{proof}

For any fixed $n$ and $x$, we can always assume that $r_i(t,x)\equiv r_i(t',x)$ for any $i\leq n$ when $|t-t'|$ small enough, which means $r_i(x,t)$ is locally independent on $t$ for $i\leq n$.
(This holds from continuity of $c_i(t)$ on some small neighborhood and the fact: $\vert x+r_i(x,t)\alpha-c_i\vert\leq 0.5|I_i|$ and $\vert x+s\alpha-c_i(t)\vert> 0.5|I_i|,\forall 0\leq s<r_i(x,t)$ implying $\vert x+r_i(x,t)\alpha-c_i(t')\vert\leq 0.5|I_i|+\epsilon(t')$ and $\vert x+s\alpha-c_i(t')\vert> 0.5|I_i|-\epsilon(t'),\forall 0\leq s<r_i(x,t)$ with $t'-t$ being small enough, and this allows us  to denote $r_i(x,t)=r_i(x)$ for $i\leq n$ in the next proofs since the derivative represent the property on any small neighborhood) In the same manner, for any fixed $\hat{t}$ and $x_0$,  we assume $r_i(x,\hat{t})\equiv r_i(x_0,\hat{t})$, when we estimate $\frac{1}{\|A_{r_n^+}(x,\hat{t})\|}\big\vert \frac{d\|A_{r_n^+}(x,\hat{t})\|}{d x}\big\vert(x_0).$
\vskip 0.3cm
   Now we are at a position to estimate $\frac{1}{{\|A_{n}(x,E)}\|}\frac{\partial\|A_{n}(x,E)\|}{\partial E}$.
   Let $L=L(n)$ be determined by $q_{N+L(n)-1}^4\leq n\leq q_{N+L(n)}^4$ and for $\lambda^{-\frac{c}{10}q_{N+L(n)-1}}\leq \vert E_1-E_0\vert \leq \lambda^{-\frac{c}{10}q_{N+L(n)-2}},$ let $n'=[\frac{-\log|E_1-E_0|}{\log\lambda}]+R_{N_0}^2$ with $N_0$ some constant determined later. Denote $
       \mathbb{T}=S_1\bigcup S_2 \bigcup S_3,$ where
     $$
     \begin{array}{ll}
     S_1&=\{x\in \mathbb{T}| T^{l}(x)\not\in I_{L+1}\ ~for\ any\ ~ 0\leq l\leq n \},\\
      \\
      S_2&=\{x\in \mathbb{T}|~there\ exists\ ~ \frac{1}{2}n'\leq l\leq n-\frac{1}{2}n' ~such\ that\ ~ T^{l}(x)\in I_{L+1}\ \},\\
      \\
      S_3&=\{x\not\in S_2|~there\ exists\ ~ 0\leq l<\frac{1}{2}n'\ ~or~\ n-\frac{1}{2}n'< l\leq n ~such\ that\ ~ T^{l}(x)\in I_{L+1}\ \}.
      \end{array}
      $$


    We first deal with the case $x\in S_1$. For this purpose, we need the following two lemmas. Recall that $r_n^{\pm}(x)=\min\{l\geq 1| x\pm l\alpha\in I_n\}.$

     \begin{lemma}\label{lemma22} For any $t\in FR$ or $t\in EP$, and $x\in\R/\Z$ there exists some constant $\bar{N}(t)$ and $C(t,\alpha,v)$, such that for $n\geq \bar{N}(t)$ we have
     $$\frac{1}{\|A_{r_n^+}(x,t)\|}\big\vert \frac{\partial\|A_{r_n^+}(x,t)\|}{\partial t}\big\vert \leq\max\{r_n^+(x), q_{N+n-1}\}^{C(t,\alpha,v)}.$$

     Moreover, for any $t\in EP,$ which is labeled by $k\in N,$ let $\tilde{N}(k)$ satisfy $q^2_{N+\tilde{N}(k)-2}\leq k\leq  q^2_{N+\tilde{N}(k)-1}.$ There exist some constant $C'(\alpha,v)$ (independent on $t$) such that
$$\int_{I_{\tilde{N}(k),1}}\vert\frac{1}{\|A_{r_{\tilde{N}(k)}}(x,t)\|}\frac{\partial\|A_{r_{\tilde{N}(k)}(x,t)}\|}{\partial t} dx\vert \leq r^{C'}_{\tilde{N}(k)}.$$


\label{key2}
     \end{lemma}
     \begin{proof} It follows from the definition of $FR$ that there exists some integer $L$ such that the series $\{g_s\}_{s\geq L}$ are of type $I$, which implies $$c(L)^{-1}\geq \big\vert \frac{d g_s(x)}{dx}\big\vert\bigg\vert_{I_{s-1}} >c(L),$$ where $c(L)$ can be chosen as $|I_L|^{100}$.

     Next, we'll prove the lemma by induction.

     For $n=L$, let $C_0=\lambda^{2R_L(t)}$, where $R_L(t):=\max_{x\in\R/\Z}\{r_n^+(x),r_n^-(x)\}$.
     Clearly, for any $x\in\R/\Z$
     we have
     $$\big\vert \frac{\partial\|A_{r_L^+}(x,t)\|}{\partial t}\big\vert\leq \|A_{r_L^+}\|^{1+\epsilon}\leq \lambda^{2r_L^+(x)}\leq \lambda^{2R_L(t)}\leq C_0(t).$$

     Let $s_n:=[\frac{50\tau^2\log q_{N+n-1}}{\log\lambda}]$.
     \vskip0.2cm
     Assume that for $n=k$, we have
     $$\frac{1}{\|A_{r_k^+}(x,t)\|}\big\vert \frac{\partial\|A_{r_k^+}(x,t)\|}{\partial t}\big\vert\leq r_k^+(C_0+2^{2k+1}q_{N+k-1}^{50\tau^2})$$ for any $x\in\R/\Z.$
     \vskip0.2cm
     Consider the condition $n=k+1$.  If $r_{k+1}^+(x)=r_{k}^+(x)$, then we need do nothing.
     Otherwise , let $0\leq j_m\leq r_{k+1}^+,$ $0\leq m\leq p$ be the return times such that
     $$ j_{m+1}-j_m\geq q_{N+k-1}^2,~1\leq m\leq p-1~and~x+j_m\alpha\in I_{k}\backslash I_{k+1},~ 1\leq m\leq p-1,$$
     where $j_0=0$, $j_p=r_{k+1}^+$ and $p>0$. Now consider the sequence

    $$ A_{j_1}(x),A_{j_2-j_1}(x+j_1\alpha),\cdots,A_{j_p-j_{p-1}}(x)(x+j_{p-1}\alpha).$$

    From the definition of return time, we have $$\vert g_{k+1}(x+j_m\alpha)\vert\geq 2^{-2k-2}q_{N+L}^{-16\tau}\leq 2^{-2k-2}q_{N+L}^{-200\tau}$$ for $1\leq m\leq p-1.$


   Let $g_{k+1}':=\frac{\pi}{2}-s_{r_k}+u(A_{j_1}(x)).$
    Then we need to consider the following two cases.
    \begin{enumerate}
    \item $j_1>s_{k}.$
    \vskip 0.3cm
    Then it holds that
    \begin{align*}\|g_{k+1}'-g_{k+1}\|&\leq \|u(A_{j_1}(x)-u_{r_k^-}\|\\&\leq C\|A_{j_1}(x)\|^{-\frac{3}{2}}\\&\leq C\lambda^{-\frac{3}{2}(1-\epsilon)j_1}\\&\leq C\lambda^{-\frac{3}{2}(1-\epsilon)s_k}\\&\leq C\lambda^{-s_k}\\&\leq Cq_{N+k-1}^{-50\tau^2},
    \end{align*}
    which implies
    $$\vert g_{k+1}'(x+j_1\alpha)\vert\geq 2^{-2k-2}q_{N+k}^{-16\tau}-Cq_{N+k-1}^{-50\tau^2}\geq 2^{-2k-3} q_{N+k}^{-16\tau}\geq 2^{-2k-2}q_{N+k}^{-200\tau^2}.$$


   \noindent Thus by lemma \ref{keylemma}, it follows from induction hypothesis that
    \begin{align*}\frac{1}{\|A_{r_{k+1}^+}(x,t)\|}\big\vert \frac{\partial\|A_{r_{k+1}^+}(x,t)\|}{\partial t}\big\vert&\leq \sum\limits_{0\leq m\leq p-1} \frac{1}{\|A_{j_{m+1}-j_m}(x,t)\|}\big\vert \frac{\partial\|A_{j_{m+1}-j_m}(x,t)\|}{\partial t}\big\vert+\\&(\sum\limits_{2\leq m\leq p-1} 2\tan(g_{k+1}(x+j_m\alpha))+2\tan(g_{k+1}'(x+j_1\alpha)))\\&\leq  \sum\limits_{0\leq m\leq p-1}(C_0(j_{m+1}-j_m)+2^{2k+1}q_{N+k-1}^{200\tau^2})+C(p-2)2^{2k+2}q_{N+k}^{16\tau}\\&+2^{2k+2}q_{N+k}^{200\tau}\\&\leq r_{k+1}^+(C_0+2^{2k+1}q_{N+k-1}^{200\tau^2}+2^{2k+2}q_{N+k}^{200\tau^2})\\& \leq r_{k+1}^+(C_0+2^{2k+3}q_{N+k}^{200\tau^2}).
    \end{align*}

    \item $j_1\leq s_k$.

   \vskip 0.2cm

  \noindent Then also by lemma \ref{keylemma} and induction hypothesis, it holds that
    \begin{align*}
    \frac{1}{\|A_{r_{k+1}^+}(x,t)\|}\big\vert \frac{\partial\|A_{r_{k+1}^+}(x,t)\|}{\partial t}\big\vert &\leq \sum\limits_{1\leq m\leq p-1} \frac{1}{\|A_{j_{m+1}-j_m}(x,t)\|}\big\vert \frac{\partial\|A_{j_{m+1}-j_m}(x,t)\|}{\partial t}\big\vert\\&+\frac{1}{\|A_{j_1}(x,t)\|}\big\vert \frac{\partial\|A_{j_1}(x,t)\|}{\partial t}\big\vert\\&+(\sum\limits_{2\leq m\leq p-1} 2\tan(g_{k+1}(x+j_m\alpha))+2\tan(g_{k+1}'(x+j_1\alpha)))\\&\leq \sum\limits_{1\leq m\leq p-1}(C_0(j_{m+1}-j_m)+2^{2k+1}q_{N+k-1}^{200\tau^2})\\&+\|A_{j_1}\|^3+(p-2)2^{2k+2}q_{N+k}^{200\tau^2}+\|A_{j_1}\|^{\epsilon}\\&\leq
    \sum\limits_{1\leq m\leq p-1}(C_0(j_{m+1}-j_m)+2^{2k+1}q_{N+k-1}^{200\tau^2})\\&+q_{N+k}^{150\tau^2}+(p-2)2^{2k+2}q_{N+k}^{200\tau^2}+q_{N+k-1}^{200\tau^2}\\
    &\leq r_{k+1}^+(C_0+2^{2k+1}q_{N+k-1}^{200\tau^2}+2^{2k+2}q_{N+k}^{200\tau^2})\\&\leq r_{k+1}^+(C_0+2^{2k+3}q_{N+k}^{200\tau^2}).
    \end{align*}

    \end{enumerate}

    Thus, for any $n>L$ and $x\in\R/\Z$ we obtain
    \begin{align*}\frac{1}{\|A_{r_n^+}(x,t)\|}\big\vert \frac{\partial\|A_{r_n^+}(x,t)\|}{\partial t}\big\vert
    &\leq r_n^+(C_0+2^{2n+1}q_{N+n-1}^{200\tau^2})\\&\leq C(t,\alpha,v) \max\{r_n^+(x),q_{N+n-1}\}^{200\tau^2}\\&\leq \max\{r_n^+(x),q_{N+n-1}\}^{C(t,\alpha,v)}
    \end{align*}
    for proper constant $C$.

    For the case $t\in EP$, it's easy to see that the proof is quite similar as the the condition $t\in FR$. In fact, for any fixed $t\in EP$, it follows from lemma \ref{lm16} that for some sufficient large $N_0$, we have $g_t$ is always of type III(or II)  for $t\geq N_0$ and $|g_t(\tilde{c}_{t,1})|<C\lambda^{-\frac{3}{2}r_{t-1}}$ for any $t\geq N_0$. Thus for $n$ sufficiently large, one can check that $g_t$ is locally similar as type II. So the remain part of the proof is obvious if we note that $\theta$ between two matrices we've divided still has the property as we want since the second derivative of $g_t(x)$ have an uniform lower bound.

\

\


    For the remaining part, note that $g_{\tilde{N}}$ is of type I and $I_{\tilde{N},1}+k\alpha=I_{\tilde{N},2}.$ We rewrite $A_{r_{\tilde{N}}}$ as $A_{r_{\tilde{N}}-k}A_k=R_{u(A_{r_{\tilde{N}}-k})}diag\{\|A_{r_{\tilde{N}}-k}\|,\|A_{r_{\tilde{N}}-k}\|^{-1}\}R_{\theta_k}diag\{\|A_{r_{k}}\|,\|A_{r_{k}}\|^{-1}\}R_{\frac{\pi}{2}-s(A_{k})}.$
    According to Lemma 16, we have $$|\frac{\partial \theta_k}{\partial t}|\leq c^{-1}$$ with some absolute constant $c$. Therefore, by the help of Lemma \ref{lemma8}, we have
    \begin{align*}|\frac{1}{\|A_{r_{\tilde{N}}}\|}\frac{\partial\|A_{r_{\tilde{N}}}\|}{\partial t}|&\leq |\frac{1}{\|A_{r_{\tilde{N}}-k}\|}\frac{\partial\|A_{r_{\tilde{N}}-k}\|}{\partial t}|+|\frac{1}{\|A_{k}\|}\frac{\partial\|A_k\|}{\partial t}|+\vert\frac{2}{\sqrt{\cot^2\theta_k+\frac{2}{\|A_{r_{\tilde{N}-k}\|^4}}+\frac{2}{\|A_k\|^4}}}\frac{\partial \theta_k}{\partial t}|\\&\leq (r_{\tilde{N}}-k)^C+k^C+c^{-1}|\frac{2}{\sqrt{\cot^2\theta_k+\frac{2}{\|A_{r_{\tilde{N}-k}\|^4}}+\frac{2}{\|A_k\|^4}}}|\\&\leq r^C_{\tilde{N}}|\frac{2C'}{\sqrt{(x-c_{\tilde{N},2})^2{+\frac{2}{\|A_{r_{\tilde{N}-k}\|^4}}+\frac{2}{\|A_k\|^4}}}}|\\&\leq r^C_{\tilde{N}}|\frac{1}{\sqrt{(x-c_{\tilde{N},2})^2{+\frac{1}{\lambda^{4k}}}}}|
    \end{align*}

Therefore,
\begin{align*}\vert\int_{I_{\tilde{N}(k),1}}\frac{1}{\|A_{r_{\tilde{N}(k)}}(x,t)\|}\frac{\partial\|A_{r_{\tilde{N}(k)}(x,t)}\|}{\partial t} dx\vert&\leq \int_{I_{\tilde{N}(k),1}}r^C_{\tilde{N}}|\frac{1}{\sqrt{(x-c_{\tilde{N},2})^2{+\frac{1}{\lambda^{4k}}}}}|dx\\&\leq \int_{0}^{\frac{1}{2^{\tilde{N}}q^{8\tau}_{N+\tilde{N}-1}}}r^C_{\tilde{N}}|\frac{1}{\sqrt{x^2{+\frac{1}{\lambda^{4k}}}}}|dx\\&\leq Cr^C_{\tilde{N}}k\log \lambda\\&\leq r^C_{\tilde{N}}.
\end{align*}


     \end{proof}

     The next lemma is absolutely similar to the first part of the lemma above. So we omit the proof of it.

     \begin{lemma}For any fixed $\hat{t}\in FR$ or $\hat{t}\in EP$, $x_0\in\R/\Z,$ and sufficiently large $n,$ there exists some constant $C$ depending on $v,$ $\alpha$ and $t$, such that
     $$\frac{1}{\|A_{-r_n^-}(x,t)\|}\big\vert \frac{\partial\|A_{-r_n^-}(x,t)\|}{\partial t}(\hat{t})\big\vert+\frac{1}{\|A_{-r_n^-}(x,t)\|}\big\vert \frac{\partial\|A_{-r_n^-}(x,t)\|}{\partial x}\big\vert(x_0)\leq\max\{-r_n^-(x_0), q_{N+n-1}\}^{C(v,\alpha)}.$$
     \end{lemma}

\


     Now we  deal with the case $x\in S_1$. In fact, the following Lemma immediately implies what we desire.
     \begin{corollary}For any $t\in FR$ or $t\in EP$, sufficiently large $n,$ $q_{N+n-1}^4\leq m\leq q_{N+n}^4$ and  $x\in\{x\in\R/\Z\vert T^l(x)\notin I_{n+1}~for~any~0\leq l\leq m\}.$ There exists some constant $C$ depending on $v,$ $\alpha$ and $t$, such that
     $$\frac{1}{\|A_m(x,t)\|}\big\vert \frac{\partial\|A_{m}(x,t)\|}{\partial t}\big\vert \leq \max\{m,q_{N+n-1}\}^{C(\alpha,v,t)}.$$\label{useful}
     \end{corollary}
     \begin{proof} Consider a fixed $x$ satisfying the requirement as above and sufficiently large $n$. Without loss of generality, we assume that there exists some $0\leq l\leq m$ such that $T^{l}(x)\in I_n.$
     Consider matrices
     $$ A_{j_1}(x),A_{j_2-j_1}(x+j_1\alpha),\cdots,A_{j_p-j_{p-1}}(x)(x+j_{p-1}\alpha),A_{m-j_p}(x)(x+j_{p}\alpha)$$

     where $0\leq j_k\leq m,$ $0\leq k\leq p+1$ denote the times such that
     $$ j_{k+1}-j_k\geq q_{N+n-1}^2,~1\leq k\leq p-1~and~x+j_k\alpha\in I_{n}\backslash I_{n+1},~ 1\leq k\leq p-1,$$
     with $j_0=0$ and $j_p+1=m$.

     Let $g_{k+1}':=\frac{\pi}{2}-s_{r_n}+u(A_{j_1}(x))$ and $g_{k+1}'':=\frac{\pi}{2}-s(A_{m-j_p})+u_{r_n}.$
     \begin{enumerate}
     \item
     If $j_1,m-j_p>s(n)$, it is clear that
     \begin{align*}\|g_{n+1}'-g_{k+1}\|&\leq \|u(A_{j_1}(x)-u_{r_n^-}\|\\&\leq C\|A_{j_1}(x)\|^{-\frac{3}{2}}\\&\leq C\lambda^{-\frac{3}{2}(1-\epsilon)j_1}\\&\leq C\lambda^{-\frac{3}{2}(1-\epsilon)s_n}\\&\leq C\lambda^{-s_k}\\&\leq Cq_{N+n-1}^{-50\tau^2},
     \end{align*}
     which implies
     $$\vert g_{n+1}'(x+j_1\alpha)\vert\geq 2^{-2n-2}q_{N+n}^{-16\tau}-Cq_{N+n-1}^{-50\tau^2}\geq 2^{-2n-3} q_{N+n}^{-16\tau}\geq 2^{-2k-2}q_{N+n}^{-200\tau^2}.$$

     Similarly, we have
     \begin{align*}\|g_{n+1}''-g_{n+1}\|&\leq \|s(A_{m-j_p}(x)-s_{r_n^+}\|\\&\leq C\|A_{m-j_p}(x)\|^{-\frac{3}{2}}\\&\leq C\lambda^{-\frac{3}{2}(1-\epsilon)(m-j_p)}\\&\leq C\lambda^{-\frac{3}{2}(1-\epsilon)s_n}\\&\leq C\lambda^{-s_n}\\&\leq Cq_{N+n-1}^{-50\tau^2},
     \end{align*}
     which implies
     $$\vert g_{n+1}''(x+j_1\alpha)\vert\geq 2^{-2n-2}q_{N+n}^{-16\tau}-Cq_{N+n-1}^{-50\tau^2}\geq 2^{-2n-3} q_{N+n}^{-16\tau}\geq 2^{-2n-2}q_{N+n}^{-200\tau^2}.$$

\noindent\hskip -0.2cm Thus by lemma \ref{keylemma}, it follows from induction hypothesis that
    \begin{align*}\frac{1}{\|A_{m}(x,t)\|}\big\vert \frac{\partial\|A_{m}(x,t)\|}{\partial t}\big\vert&\leq \sum\limits_{0\leq k\leq p} \frac{1}{\|A_{j_{k+1}-j_k}(x,t)\|}\big\vert \frac{\partial\|A_{j_{k+1}-j_k}(x,t)\|}{\partial t}\big\vert+\\&(\sum\limits_{2\leq k\leq p} 2\tan(g_{n+1}(x+j_m\alpha))+2\tan(g_{n+1}'(x+j_1\alpha))+2\tan(g_{n+1}''(x+j_p\alpha)))\\&\leq  \sum\limits_{0\leq k\leq p-1}((j_{k+1}-j_k)^C)+Cp2^{2n+2}q_{N+n}^{16\tau}\\&\leq m^C+mq_{N+n}^C\\& \leq\max\{m,q_{N+n-1}\}^C,
    \end{align*}
    for proper constant C.
     \item If $j_1\leq s(n)$ and $i-j_p> s(n)$.
     Straightly calculation shows that
     \begin{align*}
     \frac{1}{\|A_{m}(x,t)\|}\big\vert \frac{\partial\|A_{m}(x,t)\|}{\partial t}\big\vert&\leq \sum\limits_{1\leq k\leq p} \frac{1}{\|A_{j_{k+1}-j_k}(x,t)\|}\big\vert \frac{\partial\|A_{j_{k+1}-j_k}(x,t)\|}{\partial t}\big\vert+\frac{1}{\|A_{j_1}(x,t)\|}\big\vert \frac{\partial\|A_{j_1}(x,t)\|}{\partial t}\big\vert\\&+(\sum\limits_{2\leq m\leq p} 2\tan(g_{n+1}(x+j_m\alpha))+2\tan(g_{n+1}'(x+j_1\alpha))+2\tan(g_{n+1}''(x+j_p\alpha)))\\&\leq \sum\limits_{1\leq k\leq p}((j_{m+1}-j_m)^C)+\|A_{j_1}\|^3+(p-1)2^{2k+2}q_{N+n}^{200\tau^2}+\|A_{j_1}\|^{\epsilon}\\&\leq m^C+mq_{N+n}^C\\&\leq \max\{m,q_{N+n-1}\}^C,
     \end{align*}
     for some proper constant $C.$


     \end{enumerate}

     The cases $j_1\leq s(n)$ and $i-j_p\leq s(n)$ and $j_1> s(n)$ and $i-j_p\leq s(n)$ is quite similar as above. Thus we finish the proof.

     \end{proof}

     When we use Induction Theorem of \cite{wz1}, it's noteworthy that we do not require the orbit of $x$ starting from and ending at critical interval $I_n,$ the following lemma shows the fact.

     \begin{lemma}\label{lmwz0} For any fixed $t,$ $n\in \N,$ $q^4_{N+n-2}<l_1,l_2\leq q^4_{N+n-1},$ $x\in I_n,$ denote $\min\{l_1,r^+_n(x)\}=m^n_1,$ $\min\{l_2,r^-_n(x)\}=m^n_2$ and $s(A_{m^n_1})-u(A_{m^n_2})\triangleq g_{m^n_1,m^n_2,n},$ where $r_n^+$ is the first returning time to $I_n.$ The following holds true,
     $$\|A_{m^n_i}\|\geq \lambda^{(1-\epsilon)m^n_i},$$ where $\epsilon=\sum_{i=1}^{\infty}\frac{\log q_{N+i}}{q_{N+i-1}},$ $i=1,2;$
     $$\|g_{m^n_1,m^n_2,n}-(s(A_{r^+_n(x)})-u(A_{r^-_n(x)}))\|_{C^2}\leq \lambda^{-\frac{1}{2}r_{n-1}}.$$
     \end{lemma}
     The proof of Lemma \ref{lmwz0} is located in Appendix A.5.

\


     The following lemma is in preparation for later proofs.
     It is a stronger version of the estimate for the angle of lemma 3 in \cite{wz1}.

     
     
     
     
     

     \begin{lemma}\label{yol}For fixed $t\in FR,$ $n\in \N$ and $x\in I_n$ satisfying $g_n$ and $g_{n-1}$ are same type (Type I, II, or III with the same resonance), let $s^n_{i}$ be all the returning times (second returning time for type III) for $x$ to $I_{n-1},$ and we denote $\|A_{s^n_{i}-s^n_{i-1}}\|\triangleq \lambda_i$ and $\theta_j=s(A_{s^n_{j+1}-s^n_{j}})-u(A_{s^n_{j}-s^n_{j-1}})$ for $i=1,2,\cdots,t_n-1,j=1,2,\cdots,t_n-2.$ Then the following holds true.
     $$\frac{d\|A_{r_n}\|}{dx}\frac{1}{\|A_{r_n}\|}\leq \sum_{j=1}^{t_n-1}\frac{\lambda'_j}{\lambda_j}+\sum_{j=1}^{t_n-2}|\tan\theta_j\theta'_j|;$$
     $$\frac{d^2\|A_{r_n}\|}{dx^2}\frac{1}{\|A_{r_n}\|}\leq C\left(\sum\limits_{1\leq k\neq j\leq t_n-1}\frac{\lambda'_j\lambda'_k}{\lambda_j\lambda_k}+\sum_{j=1}^{t_n-1}(\frac{\lambda''_j}{\lambda_j}+|\theta_j''\tan\theta_j|)+\sum_{j,k=1}^{t_n-1}|\theta'_j\theta'_k|\right).$$
     \end{lemma}
     \begin{proof} It follows from lemma \ref{lm5} that \begin{equation}\label{L11}\|A_{r_n}\|\precsim_2 (\prod_{i=1}^{t_n-1}\lambda_i)(\prod_{i=1}^{t_n-2}|\cos\theta_i|)\end{equation} and \begin{equation}\label{2}\|A_{r_n}\|\geq (\prod_{i=1}^{t_n-1}\lambda_i)(\prod_{i=1}^{t_n-2}|\cos\theta_i|).\end{equation} Then it holds from \eqref{L11} and the direct calculation, we have
     $$\frac{d\|A_{r_n}\|}{dx}\leq \sum_{j=1}^{t_n-1}\left(\lambda'_j\prod_{i\neq j}\lambda_j\prod_{i=1}^{t_n-2}|\cos\theta_j|\right)+\sum_{j=1}^{t_n-2}\left(|\sin\theta_j\theta'_j|\prod_{i=1}^{t_n-1}\lambda_j\prod_{i\neq j}|\cos\theta_j|\right);$$
     \begin{align*}&\frac{d^2\|A_{r_n}\|}{dx^2}\leq \sum_{j=1}^{t_n-1}\left(\lambda''_j\prod_{i\neq j}\lambda_j\prod_{i=1}^{t_n-2}|\cos\theta_j|\right)+\sum_{1\leq j\neq k\leq t_n-1}\left(\lambda'_j\lambda'_k\prod_{i\neq j,k}\lambda_j\prod_{i=1}^{t_n-2}|\cos\theta_j|\right)\\&+\sum_{j=1}^{t_n-1}\sum_{k=1}^{t_n-2}\left(\lambda'_j|\sin\theta_k\theta_k'|\prod_{i\neq j}\lambda_j\prod_{i\neq k}|\cos\theta_k|\right) +\sum_{j=1}^{t_n-2}\left((|\sin\theta_j\theta''_j|+|\cos\theta_j\theta'_j|)\prod_{i=1}^{t_n-1}\lambda_j\prod_{i\neq j}|\cos\theta_j|\right)\\&+ \sum_{j=1}^{t_n-1}\sum_{k=1}^{t_n-2}\left(|\sin\theta_j\sin\theta_k\theta'_j\theta_k'|\prod_{i\neq j}\lambda_j\prod_{i\neq j,k}|\cos\theta_k|\right)
     \end{align*}
     Combine this with \eqref{2}, we obatain the conclusion.
     \end{proof}
     \begin{remark} \label{yolr}It's not difficult to see that lemma \ref{yol} holds for each $\|A_{s^n_{i}}\|,$ $i=1,2,\cdots,t_n-1.$
     \end{remark}

     \begin{lemma}\label{yolwz}For any fixed $t\in EP~or~FR,$ $m\leq n\in \N$ and $x_m\in I_m$ satisfying $r^+_{m}(x_m)\geq [\frac{C\log q_{N+n-1}}{\log \lambda}],$ and $r^+_{n}(x_m)\geq q_{N+n-1},$ where $r_i^+$ denote the first returning time to $I_i, i\in \N^+$ and $C$ can be chosen as $1000\tau^{1000}\cdot \kappa(\alpha)$ with $\kappa(\alpha)>1$ satisfying $\sum_{i=1}^{|I|^{-\kappa(\alpha)}}Leb\{I\bigcap \left(I+i\cdot\alpha\right)\}>0, \forall~open~interval~I\subset \R.$ Then we have
     \begin{equation}\label{1.1.1}\|A_{r^+_n(x_m)}\|\geq \lambda^{(1-\epsilon)r^{+}_n(x_m)};\end{equation}
     \begin{equation}\label{1.1.2}|s(A_{r^+_n(x_m)})-s(A_{r^+_m(x_m)})|_{C^2}\leq \lambda^{-1.5r^+_m(x_m)}.\end{equation}
     \end{lemma}
     \begin{remark}~The existence of $\kappa$ is a standard result for Diophantine translation on torus, one can see the Lemma 6 of \cite{aviladamanikzhang} for a simple proof. The lower bound of
     $r^+_m(x_m)$ implies that there are no other strong resonance between $m$-th step and $n$-th step.
     \end{remark}
     We put the proof at Appendix A.6.

    \


     \begin{lemma}\label{lmwz} For any fixed $t\in EP$ or $FR,$ $l\in\N$ large and $x\in I_{L(l)},$ the following holds true.
     $$\frac{d\|A_l\|}{dx}+\frac{d^2\|A_l\|}{dx^2}\leq \|A_l\|l^{C(\alpha,t)},$$
     for some proper constant $C$ depending on $\alpha$ and $t,$ where $L(l)=\{m| q^2_{N+m-2}\leq l \leq q^2_{N+m-1}\}.$
     \end{lemma}
     \begin{proof}We process it by induction. Note that for large $N_0$, it's enough to consider $t\in FR$ since there essentially exist only one critical interval in the case $t\in EP.$ $N_0$ can be chosen as $N_0\triangleq \min\{m| g_n~is~Type~I~for~n\geq m\}$ (or $N_0\triangleq \min\{m| g_n~is~Type~III~for~n\geq m\}$ in the case $t\in EP.$)

     \begin{enumerate}\item[\textbf{Starting Step $N_0$:}] Denote $\lambda'_0\triangleq \max\{\frac{d\|A_{k}(x)\|}{dx}\vert k\leq r_{N_0}(x),x\in I_{N_0}\}$ and $\lambda''_0\triangleq \max\{\frac{d^2\|A_{k}(x)\|}{dx^2}\vert k\leq r_{N_0}(x),x\in I_{N_0}\},$ $\lambda_0\triangleq\max\{\lambda'_0,\lambda''_0\}.$
     \item[\textbf{Step $N_0+1:$}] For any $q^2_{N+N_0-1}\leq l \leq q^2_{N+N_0},$ we write $A_{l}$ as $A_{s^1_{t_1}-s^1_{t_1-1}}\cdots A_{s^1_{2}-s^1_{1}}A_{s^1_{1}},$ where $s^1_{i}$ are all the returning timings for $x$ to $I_{N_0}.$ Note that $l \leq q^2_{N+N_0}$ implies $x$ doesn't return back to $I_{N_0+1}.$  We denote $M_{1}\triangleq[100\tau\frac{\log q_{N+N_0}}{\log\lambda}]$ and $\theta_{s^1_{i}}\triangleq s(A_{s^1_{i+1}-s^1_{i}})-u(A_{s^1_{i}}),$ $i=0,1,\cdots,t_1-1.$
         \begin{enumerate}\item[1:$s^1_{t_1}-s^1_{t_1-1}\leq M_{1}$:] On one hand, $$\|A_{s^1_{t_1}-s^1_{t_1-1}}\|\leq C\lambda^{s^1_{t_1}-s^1_{t_1-1}}\leq q^{150\tau}_{N+N_0}.$$ On the other hand, $$\|A_{s^1_{t_1-1}}\|\geq \lambda^{(1-\epsilon)s^1_{t_1-1}}=\lambda^{(1-\epsilon)(l-(s^1_{t_1}-s^1_{t_1-1}))}\geq \lambda^{(1-\epsilon)(q^2_{N+N_0}-M_1)}\gg q^{150\tau}_{N+N_0}.$$

         We write $A_l=A_{s^1_{t_1}-s^1_{t_1-1}}A_{s^1_{t_1-1}}.$ It follows from lemma \ref{lm5} that $$\|A_l\|\precsim_2 \|A_{s^1_{t_1-1}}\|\|A_{s^1_{t_1}-s^1_{t_1-1}}\|^{-1}|\sin(\frac{\pi}{2}-\theta_{s^1_{t_1-1}})|\triangleq\lambda_1\lambda_2^{-1}|\sin\theta|.$$ Therefore, $$|\frac{d\|A_l\|}{dx}|\leq \lambda'_1\lambda_2^{-1}|\sin\theta|+\lambda_1\lambda_2^{-2}\lambda'_2|\sin\theta|+\lambda_1\lambda_2^{-1}|\cos\theta\theta'|;$$
         $$|\frac{d^2\|A_l\|}{dx^2}|\leq 2(\lambda''_1\lambda_2^{-1}|\sin\theta|+\lambda'_1\lambda'_2\lambda^{-2}_2|\sin\theta|+\lambda'_1\lambda^{-1}_2|\cos\theta\theta'|+\lambda_1\lambda^{-2}_2\lambda'_2|\theta'\cdot\cos\theta|+$$ $$\lambda_1\lambda^{-3}_2(\lambda'_2)^2|\sin\theta|+\lambda_1\lambda^{-2}_2\lambda''_2|\sin\theta|+\lambda_1\lambda^{-1}_2(|\sin\theta|(\theta')^2+|\cos\theta\theta''|).$$

         Note that $\|A_l\| \thickapprox \sqrt{\lambda^2_1\lambda^2_2|\cos^2\theta|+\lambda^2_1\lambda^{-2}_2|\sin\theta|}\geq \lambda_1\lambda^{-1}_2|\sin\theta|.$ Thus, by direct calculation, we have
         $$\frac{1}{\|A_l\|}|\frac{d\|A_l\|}{dx}|\leq \lambda'_1\lambda^{-1}_1+\lambda_2\lambda_2^{-1}+|\cot\theta \theta'|;$$
         $$\frac{1}{\|A_l\|}|\frac{d^2\|A_l\|}{dx^2}|\leq \lambda''_1\lambda^{-1}_1+(\lambda'_1\lambda^{-1}_1)(\lambda'_2\lambda^{-1}_2)+(\lambda'_1\lambda^{-1}_1)|\cot\theta\theta'|+(\lambda'_2\lambda^{-1}_2)|\cot\theta\theta'|+(\lambda'_2\lambda^{-1}_2)^2$$ $$+ \lambda''_2\lambda^{-1}_2+|\theta'|^2+|\cot\theta\theta''|$$

         Note that either $t\in FR$ or $t\in EP,$ we have \begin{equation}\label{jdg}\cot\theta\leq 1;|\theta|_{C^2}\leq C(t),\end{equation} where $C$ only depends on $t.$ Therefore, the following hold true.
         \begin{equation}\label{1.1}|\frac{1}{\|A_l\|}|\frac{d\|A_l\|}{dx}|\leq C(\lambda'_1\lambda^{-1}_1+\lambda'_2\lambda^{-1}_2)\leq C\lambda_0+\lambda'_2\lambda^{-1}_2\end{equation}
         \begin{equation}\label{1.2}|\frac{1}{\|A_l\|}|\frac{d^2\|A_l\|}{dx^2}|\leq C(\lambda'_1\lambda^{-1}_1+\lambda'_2\lambda^{-1}_2+(\lambda'_1\lambda^{-1}_1)(\lambda'_2\lambda^{-1}_2)+\lambda''_1\lambda^{-1}_1+\lambda''_2\lambda^{-1}_2)\leq C\lambda_0+\lambda'_2\lambda^{-1}_2+\lambda''_2\lambda^{-1}_2\end{equation}

         We denote $\|A_{s^1_{i}-s^1_{i-1}}\|\triangleq \lambda_{2,i-1}$ and $s(A_{s^1_{j+1}-s^1_{j}})-u(A_{s^1_{j}-s^1_{j-1}})$ for $i,j=1,2,\cdots,t_n-2.$ It follows from remark \ref{yolr} and \eqref{jdg} that $$\frac{\lambda'_2}{\lambda_2}\leq C\sum_{i=1}^{t_n-2}\frac{\lambda'_{2,i}}{\lambda'_{2,i}}+\sum_{i=1}^{t_n-3}|\tan\theta_i\theta_i'|\leq C[(t_n-2)\lambda_0+(t_n-3)q^{24\tau}_{N+N_0}]\leq C(t_n-2)(\lambda_0+q^{24\tau}_{N+N_0});$$
         \begin{align*}&\frac{\lambda''_2}{\lambda_2}\\&\leq C\left(\sum_{1\leq i\neq k\leq t_n-2}(\frac{\lambda'_{2,i}}{\lambda'_{2,i}})(\frac{\lambda'_{2,k}}{\lambda'_{2,k}})+\sum_{i=1}^{t_n-2}\frac{\lambda''_{2,i}}{\lambda_{2,i}}\right)+\sum_{i=1}^{t_n-3}|\tan\theta_i\theta_i''|+\sum_{i,j}|\theta_i'\theta_j'|\\&\leq C(t_n-2)^2\lambda^2_0+C(t_n-2)^2+C(t_n-3)q^{24\tau}_{N+N_0} \leq C(t_n-2)^2(\lambda^2_0+q^{24\tau}_{N+N_0}).\end{align*}

         Combine these with \eqref{1.1} and \eqref{1.2}, we have $$|\frac{1}{\|A_l\|}|\frac{d\|A_l\|}{dx}|\leq C(t_n-1)(\lambda_0+q^{24\tau}_{N+N_0})\leq C(t_n-1)(\lambda_0+q^{30\tau}_{N+N_0});$$
         $$|\frac{1}{\|A_l\|}|\frac{d^2\|A_l\|}{dx^2}|\leq C(t_n-1)^2(\lambda^2_0+q^{24\tau}_{N+N_0})\leq C(t_n-1)^2(\lambda^2_0+q^{30\tau}_{N+N_0}).$$
         \item[2:$s^1_{t_1}-s^1_{t_1-1}> M_{1}$:] We use the same notation as the previous case. We focus on $A_{s^1_{t_1}-s^1_{t_1-1}}.$ Note that there must exist some $0\leq p\leq N_0,$ such that $r_p\geq [\frac{C\log q_{N+N_0}}{\log \lambda}]$($C$ is from the Lemma \ref{yolwz}). Without loss of generality, we assume that $r_{N_0-1}(x)\geq [\frac{C\log q_{N+N_0}}{\log \lambda}].$ We denote $A_{s^1_{t_1}-s^1_{t_1-1}}A_{s^1_{t_1-1}}\triangleq A_{r_{N_0-1}}A'A_{s^1_{t_1-1}}.$ It follows from lemma \ref{yolwz} that $\|A'A_{s^1_{t_1-1}}\|\geq \lambda^{(1-\epsilon)(l-r_{N_0-1})}$ and $\|A'\|\geq \lambda^{(1-\epsilon)(s^1_{t_1}-s^1_{t_1-1}-r_{N_0-1})}.$ Therefore, similar to the previous case, we denote $\tilde{\lambda}\triangleq\|A'A_{s^1_{t_1-1}}\|,$ $\tilde{\lambda_1}\triangleq\|A'\|,$ $\tilde{\lambda_2}\triangleq\|A_{s^1_{t_1-1}}\|$ and $\tilde{\theta}\triangleq \frac{\pi}{2}-(s(A')-u(A_{s^1_{t_1-1}}).$ It's not difficult to see that $$\tilde{\lambda}\precsim_2\tilde{\lambda_1}\tilde{\lambda_2}|\cos\tilde{\theta}|;$$ $$\frac{1}{\|A_l\|}(\frac{d\|A_l\|}{dx})\leq \frac{\tilde{\lambda}'}{\tilde{\lambda}}+C\lambda_0;$$ $$\frac{1}{\|A_l\|}(\frac{d^2\|A_l\|}{dx^2})\leq \frac{\tilde{\lambda}''}{\tilde{\lambda}}+C\lambda_0;$$ It follows from lemma \ref{yolwz} that $|\tilde{\theta}(x)-g_{N_0}|\leq \|A'\|^{-1.5}\leq \frac{C}{q^{150\tau}_{N+N_0}}.$ Thus \begin{equation}\label{jddgj}|\cos\tilde{\theta}(x)|\geq \frac{C}{q^{24\tau}_{N+N_0}}-\frac{C}{q^{150\tau}_{N+N_0}}\geq \frac{1}{q^{30\tau}_{N+N_0}}.\end{equation} By lemma \ref{yol}, it's clear that $$\frac{\tilde{\lambda'}}{\tilde{\lambda}}\leq C\frac{\tilde{\lambda_1'}}{\tilde{\lambda_1}}+C\frac{\tilde{\lambda_2'}}{\tilde{\lambda_2}}+|\tan\tilde{\theta}\tilde{\theta'}|;$$ $$\frac{\tilde{\lambda''}}{\tilde{\lambda}}\leq C\frac{\tilde{\lambda_1'}}{\tilde{\lambda_1}}+C\frac{\tilde{\lambda_2'}}{\tilde{\lambda_2}}+|\tan\tilde{\theta}\tilde{\theta''}|+\sum\frac{\tilde{\lambda_1'}}{\tilde{\lambda_1}}\frac{\tilde{\lambda_2'}}{\tilde{\lambda_2}}
             $$ $$+C\frac{\tilde{\lambda_1''}}{\tilde{\lambda_1}}+C\frac{\tilde{\lambda_2''}}{\tilde{\lambda_2}}+\theta'^2.$$ Note that the estimate for $\tilde{\lambda_2}$ has been done in the previous case. Combine this with \eqref{jddgj}, we have $$\frac{\tilde{\lambda'}}{\tilde{\lambda}}\leq C(t_n-1)(\lambda_0+q^{24\tau}_{N+N_0})+C\lambda_0+q^{30\tau}_{N+N_0}\leq C(t_n-1)(\lambda_0+q^{30\tau}_{N+N_0});$$ $$\frac{\tilde{\lambda''}}{\tilde{\lambda}}\leq C(t_n-1)^2(\lambda^2_0+q^{24\tau}_{N+N_0})+C\lambda_0+q^{30\tau}_{N+N_0}\leq C(t_n-1)^2(\lambda^2_0+q^{30\tau}_{N+N_0}).$$ Therefore, $$|\frac{1}{\|A_l\|}|\frac{d\|A_l\|}{dx}|\leq C(t_n-1)(\lambda_0+q^{30\tau}_{N+N_0});$$
         $$|\frac{1}{\|A_l\|}|\frac{d^2\|A_l\|}{dx^2}|\leq C(t_n-1)^2(\lambda^2_0+q^{30\tau}_{N+N_0}).$$

         \end{enumerate}

         In conclusion, $|\frac{1}{\|A_l\|}|\frac{d\|A_l\|}{dx}|+|\frac{1}{\|A_l\|}|\frac{d^2\|A_l\|}{dx^2}|\leq C(t_n-1)^2(\lambda_0+q^{30\tau}_{N+N_0})\leq Cl^2(\lambda_0+q^{30\tau}_{N+N_0}).$

         Assume that for any $x\in \R/\Z$ and $p>N_0,$ the following holds: $$\frac{d\|A_{r_{N+p}(x)}\|}{dx}+\frac{d^2\|A_{r_{N+p}(x)}\|}{dx^2}\leq Cr^2_{N+p}(x)(\lambda_0+\sum_{i=N_0}^{p-1}q^{30\tau}_{N+i}).$$

         \item[\textbf{Step $N_0+p+1:$}]For any $q^2_{N+N_0+p-1}\leq l \leq q^2_{N+N_0+p},$ we write $A_{l}$ as $A_{s^{p}_{t_{p}}-s^{p}_{t_{p}-1}}\cdots A_{s^{p}_{2}-s^{p}_{1}}A_{s^{p}_{1}},$ where $s^{p}_{i}$ are all the returning timings for $x$ to $I_{N_0+p+1}.$ Note that $l \leq q^2_{N+N_0+p}$ implies $x$ doesn't return back to $I_{N_0+p+1}.$  We denote $M_{p}\triangleq[100\tau\frac{\log q_{N+N_0+p}}{\log\lambda}]$ and $\theta_{s^p_{i}}\triangleq s(A_{s^p_{i+1}-s^p_{i}})-u(A_{s^p_{i}}),$ $i=0,1,\cdots,t_p-1.$
             \begin{enumerate}\item[1:$s^p_{t_p}-s^p_{t_p-1}\leq M_{p}$:]On one hand, $$\|A_{s^p_{t_p}-s^p_{t_p-1}}\|\leq C\lambda^{s^p_{t_p}-s^p_{t_p-1}}\leq q^{150\tau}_{N+N_0+p}.$$ On the other hand, $$\|A_{s^p_{t_p-1}}\|\geq \lambda^{(1-\epsilon)s^p_{t_p-1}}=\lambda^{(1-\epsilon)(l-(s^p_{t_p}-s^p_{t_p-1}))}\geq \lambda^{(1-\epsilon)(q^2_{N+N_0+p}-M_p)}\gg q^{150\tau}_{N+N_0+p}.$$

         We write $A_l=A_{s^p_{t_p}-s^p_{t_p-1}}A_{s^p_{t_p-1}}.$ It follows from lemma \ref{lm5} that $$\|A_l\|\precsim_2 \|A_{s^p_{t_p-1}}\|\|A_{s^p_{t_p}-s^p_{t_p-1}}\|^{-1}|\sin(\frac{\pi}{2}-\theta_{s^p_{t_p-1}})|\triangleq\bar{\lambda}_1\bar{\lambda}_2^{-1}|\sin\bar{\theta}|.$$ Therefore, $$|\frac{d\|A_l\|}{dx}|\leq \bar{\lambda'}_1\bar{\lambda_2}^{-1}|\sin\bar{\theta}|+\bar{\lambda}_1\bar{\lambda}_2^{-2}\bar{\lambda'}_2|\sin\bar{\theta}|+\bar{\lambda}_1\bar{\lambda}_2^{-1}|\cos\bar{\theta}\bar{\theta'}|;$$
         $$|\frac{d^2\|A_l\|}{dx^2}|\leq 2(\bar{\lambda''}_1\bar{\lambda}_2^{-1}|\sin\bar{\theta}|+\bar{\lambda'}_1\bar{\lambda'}_2\bar{\lambda}^{-2}_2|\sin\bar{\theta}|+\bar{\lambda'}_1\bar{\lambda}^{-1}_2|\cos\bar{\theta}\bar{\theta'}|+\bar{\lambda}_1\bar{\lambda}^{-2}_2\bar{\lambda'}_2|\cos\bar{\theta}\bar{\theta'}|+$$ $$\bar{\lambda}_1\bar{\lambda}^{-3}_2(\bar{\lambda'}_2)^2|\sin\bar{\theta}|+\bar{\lambda}_1\bar{\lambda}^{-2}_2\bar{\lambda''}_2|\sin\bar{\theta}|+\bar{\lambda}_1\bar{\lambda}^{-1}_2(|\sin\bar{\theta}|(\bar{\theta'})^2+|\cos\bar{\theta}\bar{\theta''})|.$$

         Note that $\|A_l\| \thickapprox \sqrt{\bar{\lambda}^2_1\bar{\lambda}^2_2\cos^2\bar{\theta}+\bar{\lambda}^2_1\bar{\lambda}^{-2}_2|\sin\bar{\theta}}|\geq \bar{\lambda}_1\bar{\lambda}^{-1}_2|\sin\bar{\theta}|.$ Thus, by direct calculation, we have
         $$\frac{1}{\|A_l\|}|\frac{d\|A_l\|}{dx}|\leq \bar{\lambda'}_1\bar{\lambda}^{-1}_1+\bar{\lambda}_2\bar{\lambda}_2^{-1}+|\cot\bar{\theta}\bar{\theta'}|;$$
         $$\frac{1}{\|A_l\|}|\frac{d^2\|A_l\|}{dx^2}|\leq \bar{\lambda''}_1\bar{\lambda}^{-1}_1+(\bar{\lambda'}_1\bar{\lambda}^{-1}_1)(\bar{\lambda'}_2\bar{\lambda}^{-1}_2)+(\bar{\lambda'}_1\bar{\lambda}^{-1}_1)|\cot\bar{\theta}\bar{\theta'}|+(\bar{\lambda'}_2\bar{\lambda}^{-1}_2)|\cot\bar{\theta}\bar{\theta'}|+(\bar{\lambda'}_2\bar{\lambda}^{-1}_2)^2$$ $$+ \bar{\lambda''}_2\bar{\lambda}^{-1}_2+(\bar{\theta'})^2+|\cot\bar{\theta}\bar{\theta''}|$$

         Note that either $t\in FR$ or $t\in EP,$ we have \begin{equation}\label{jdgp}|\cot\bar{\theta}|\leq 1;|\bar{\theta}|_{C^2}\leq C(t),\end{equation} where $C$ only depends on $t.$ Therefore, the following hold true.
         \begin{equation}\label{1.1p}|\frac{1}{\|A_l\|}|\frac{d\|A_l\|}{dx}|\leq C(\bar{\lambda'}_1\bar{\lambda}^{-1}_1+\bar{\lambda'}_2\bar{\lambda}^{-1}_2)\leq C\lambda_0+\bar{\lambda'}_2\bar{\lambda}^{-1}_2\end{equation}
         \begin{equation}\label{1.2p}|\frac{1}{\|A_l\|}|\frac{d^2\|A_l\|}{dx^2}|\leq C(\bar{\lambda'}_1\bar{\lambda}^{-1}_1+\bar{\lambda'}_2\bar{\lambda}^{-1}_2+(\bar{\lambda'}_1\bar{\lambda}^{-1}_1)(\bar{\lambda'}_2\bar{\lambda}^{-1}_2)+\bar{\lambda''}_1\bar{\lambda}^{-1}_1+\bar{\lambda''}_2\bar{\lambda}^{-1}_2)\leq C\lambda_0+\bar{\lambda'}_2\bar{\lambda}^{-1}_2+\bar{\lambda''}_2\bar{\lambda}^{-1}_2\end{equation}

         We denote $\|A_{s^1_{i}-s^1_{i-1}}\|\triangleq \bar{\lambda}_{2,i-1}$ and $s(A_{s^1_{j+1}-s^1_{j}})-u(A_{s^1_{j}-s^1_{j-1}})$ for $i,j=1,2,\cdots,t_p-2.$ It follows from remark \ref{yolr},\eqref{jdgp} and Induction hypothesis that \begin{align*}&\frac{\bar{\lambda'}_2}{\lambda_2}\\&\leq C\sum_{i=1}^{t_p-2}\frac{\bar{\lambda'}_{2,i}}{\bar{\lambda}_{2,i}}+\sum_{i=1}^{t_p-3}|\tan\bar{\theta}_i\bar{\theta}_i'|\\&\leq C\sum_{i=0}^{t_p-1}(s^p_{i+1}-s^p_i)(\lambda_0+\sum_{i=0}^{p-1} q^{30\tau}_{N+N_0+i})+(t_p-2)q^{24\tau}_{N+N_0+p}\\&\leq Cl(\lambda_0+\sum_{i=0}^p q^{30\tau}_{N+N_0+i});\end{align*}
         \begin{align*}&\frac{\bar{\lambda''}_2}{\bar{\lambda}_2}\\&\leq C\left(\sum_{1\leq i\neq k\leq t_p-2}(\frac{\bar{\lambda'}_{2,i}}{\bar{\lambda}_{2,i}})(\frac{\bar{\lambda'}_{2,k}}{\bar{\lambda}_{2,k}})+\sum_{i=1}^{t_p-2}\frac{\bar{\lambda''}_{2,i}}{\bar{\lambda}_{2,i}}\right)+\sum_{i=1}^{t_p-3}|\tan\bar{\theta}_i\bar{\theta}_i''|+\sum_{i,j}|\bar{\theta}_i'\bar{\theta}_j'|\\&\leq Cl^2(\lambda_0+\sum_{i=0}^{p-1} q^{30\tau}_{N+N_0+i})^2+C(t_p-2)^2+C(t_p-3)q^{24\tau}_{N+N_0+p} \leq C(t_p-2)^2(\lambda^2_0+\sum_{i=0}^{p} q^{30\tau}_{N+N_0+i}).\end{align*}

         Combine these with \eqref{1.1p} and \eqref{1.2p}, we have $$|\frac{1}{\|A_l\|}|\frac{d\|A_l\|}{dx}|\leq C(t_p-1)(\lambda_0+\sum_{i=0}^p q^{30\tau}_{N+N_0+i});$$
         $$|\frac{1}{\|A_l\|}|\frac{d^2\|A_l\|}{dx^2}|\leq C(t_p-1)^2(\lambda^2_0+\sum_{i=0}^p q^{30\tau}_{N+N_0+i}).$$
         \item[2:$s^p_{t_p}-s^p_{t_p-1}> M_{p}$:] We use the same notation as the previous case. We focus on $A_{s^p_{t_p}-s^p_{t_p-1}}.$ Note that there must exist some $0\leq u\leq N_0,$ such that $r_u(x)\geq [\frac{C\log q_{N+N_0+p}}{\log \lambda}]$($C$ is from the Lemma \ref{yolwz}). Without loss of generality, we assume that $r_{N_0+p-1}(x)\geq [\frac{C\log q_{N+N_0+p}}{\log \lambda}]$. We denote $A_{s^p_{t_p}-s^p_{t_p-1}}A_{s^p_{t_p-1}}\triangleq A_{r_{N_0+p-1}}A'A_{s^p_{t_p-1}}.$ It follows from lemma \ref{yolwz} that $\|A'A_{s^p_{t_p-1}}\|\geq \lambda^{(1-\epsilon)(l-r_{N_0+p-1})}$ and $\|A'\|\geq \lambda^{(1-\epsilon)(s^p_{t_p}-s^p_{t_p-1}-r_{N_0+p-1})}.$ Therefore, similar to the previous case, we denote $\tilde{\lambda}\triangleq\|A'A_{s^p_{t_p-1}}\|,$ $\tilde{\lambda_1}\triangleq\|A'\|,$ $\tilde{\lambda_2}\triangleq\|A_{s^p_{t_p-1}}\|$ and $\tilde{\theta}\triangleq \frac{\pi}{2}-(s(A')-u(A_{s^p_{t_p-1}}).$ It's not difficult to see that $$\tilde{\lambda}\precsim_2\tilde{\lambda_1}\tilde{\lambda_2}\cos\tilde{\theta};$$ $$\frac{1}{\|A_l\|}(\frac{d\|A_l\|}{dx})\leq \frac{\tilde{\lambda}'}{\tilde{\lambda}}+C\lambda_0;$$ $$\frac{1}{\|A_l\|}(\frac{d^2\|A_l\|}{dx^2})\leq \frac{\tilde{\lambda}''}{\tilde{\lambda}}+C\lambda_0;$$ It follows from lemma \ref{yolwz} that $|\tilde{\theta}(x)-g_{N_0+p-1}|\leq \|A'\|^{-1.5}\leq \frac{C}{q^{150\tau}_{N+N_0+p}}.$ Thus \begin{equation}\label{jddgjp}|\cos\tilde{\theta}(x)|\geq \frac{C}{q^{24\tau}_{N+N_0+p}}-\frac{C}{q^{150\tau}_{N+N_0+p}}\geq \frac{1}{q^{30\tau}_{N+N_0+p}}.\end{equation} By lemma \ref{yol}, it's clear that $$\frac{\tilde{\lambda'}}{\tilde{\lambda}}\leq C\frac{\tilde{\lambda_1'}}{\tilde{\lambda_1}}+C\frac{\tilde{\lambda_2'}}{\tilde{\lambda_2}}+|\tan\tilde{\theta}\tilde{\theta'}|;$$ $$\frac{\tilde{\lambda''}}{\tilde{\lambda}}\leq C\frac{\tilde{\lambda_1'}}{\tilde{\lambda_1}}+C\frac{\tilde{\lambda_2'}}{\tilde{\lambda_2}}+|\tan\tilde{\theta}\tilde{\theta''}|+\sum\frac{\tilde{\lambda_1'}}{\tilde{\lambda_1}}\frac{\tilde{\lambda_2'}}{\tilde{\lambda_2}}
             $$ $$+C\frac{\tilde{\lambda_1''}}{\tilde{\lambda_1}}+C\frac{\tilde{\lambda_2''}}{\tilde{\lambda_2}}+\theta'^2.$$ Note that the estimate for $\tilde{\lambda_2}$ has been done in the previous case. Combine this with \eqref{jddgjp}, we have $$\frac{\tilde{\lambda'}}{\tilde{\lambda}}\leq Cl(\lambda_0+\sum_{i=0}^p q^{30\tau}_{N+N_0+i})+C\lambda_0+q^{30\tau}_{N+N_0+p}\leq Cl(\lambda_0+\sum_{i=0}^p q^{30\tau}_{N+N_0+i});$$ $$\frac{\tilde{\lambda''}}{\tilde{\lambda}}\leq Cl^2(\lambda^2_0+\sum_{i=0}^p q^{30\tau}_{N+N_0})+C\lambda_0+q^{30\tau}_{N+N_0+p}\leq Cl^2(\lambda^2_0+\sum_{i=0}^p q^{30\tau}_{N+N_0+i}).$$ Therefore, $$|\frac{1}{\|A_l\|}|\frac{d\|A_l\|}{dx}|\leq Cl(\lambda_0+\sum_{i=0}^{p}q^{30\tau}_{N+N_0+i});$$
         $$|\frac{1}{\|A_l\|}|\frac{d^2\|A_l\|}{dx^2}|\leq Cl^2(\lambda^2_0+\sum_{i=0}^p q^{30\tau}_{N+N_0+i}).$$
             \end{enumerate}

             In summary, for any fixed $l,$ we have
             $$\frac{1}{\|A_l\|}(\frac{d\|A_l\|}{dx}+\frac{d^2\|A_l\|}{dx^2})\leq Cl^2(\lambda^2_0+\sum_{i=0}^{L(l)-N_0} q^{30\tau}_{N+N_0+i})\leq Cl^2L(l)q^{30\tau}_{N+L(l)}\leq Cl^{100\tau^6}.$$

     \end{enumerate}

     \end{proof}

     \begin{lemma}\label{lm2222}Let $l$ be as above, the following estimate holds true. \begin{equation}\left|s(A_l)-s(A_{r_{L(l)}})\right|_{C^2}+\left|u(A_l)-u(A_{r_{L(l)}})\right|_{C^2}\leq \|A_l\|^{-2}l^{C(\alpha,v)},     \end{equation} for some proper constant C depending on $\alpha$ and $v.$
     \end{lemma}
     \begin{proof}It's an easy extension of the lemma 3 of \cite{wz1}. We denote $M_{l}\triangleq[100\tau\frac{\log q_{N+L(l)}}{\log\lambda}].$ Note that there must exist some $N'$ such that $r_{N'}\leq M(l).$ Without loss of generality, we assume that $r_{N+L(l)-1}\leq M(l)$ and rewrite $A_l$ as $A_2A_1\triangleq A_{l-r_{N+L(l)-1}}A_{r_{N+L(l)-1}},$ where $\|A_2\|\geq \lambda^{\frac{1}{2}l}\gg q^{100\tau}_{N+L(l)}\geq \|A_1\|.$ It follows from lemma \ref{lm5} that
       $$|u(A_l)-u(A_2)|_{C^2}\leq \|A_1\|^2\|A_2\|^{-2}= (\|A_1\|\|A_2\|)^{-2}\|A_1\|^4\leq \|A_l\|^{-2}q^{100\tau}_{N+L(l)}\leq \|A_l\|^{-2}l^{500\tau^6}.$$

       Thus, it's enough to estimate $|u(A_2)-u_{r_{N+L(l)}}|_{C^2}.$
       Let $\|A_2\|=e_1$ and the angle between $A_2=A_{r_{N+L(l)-1}}$ and $A_{r_{N+L(l)}-r_{N+L(l)-1}}$ is $\theta_l$. Thus $|\theta_l-\frac{\pi}{2}|\geq |I_{N+L(l)}|\geq q^{8\tau}_{N+L(l)}\geq l^C$ for some constant C. The direct calculation and lemma \ref{lmwz} show that \begin{equation}\left|s(A_l)-s(A_{r_{L(l)}})\right|\leq |\arctan(e_1^2 \cot\theta)|\leq e_1^{-2}l^C;\end{equation}
     \begin{equation}\left|\frac{d s(A_l)}{dx}-\frac{d s(A_{r_{L(l)}})}{dx}\right|\leq C\left|\frac{2e_1'}{e_1^3\cot\theta}\right|+C\left|\frac{\theta'}{e_1^2}\right|+C\left|\frac{\theta'}{e_1^2\cot^2\theta}\right| \leq e_1^{-2}l^C;\end{equation}
     \begin{equation}\left|\frac{d^2s(A_l)}{dx^2}-\frac{d^2s(A_{r_{L(l)}})}{dx^2}\right|\leq C\left|\frac{2e_1''}{e_1^3\cot\theta}\right|+\left|\frac{4e_1'\theta'}{e_1^3\cot^2\theta}\right|+\left|\frac{6(e_1')^2}{e_1^4\cot\theta}\right|+\left|\frac{2(\theta')^2}{e_1^2\cot^3\theta}\right|+\left|\frac{2\theta''}{e_1^2\cot^2\theta}\right|\leq e_1^{-2}l^C,\end{equation} which complete the proof.


     \end{proof}


In the next two lemmas we deal with the case $x\in S_2$, which separately implies the $\frac{1}{2}-$H\"older continuity on the two subinterval centered by $t.$ We denote $k(t)$ the largest resonance-distance of $t\in EP.$
     \begin{lemma}\label{lemma25}
     Let $n'$ $A_l, A_{n-l}$ and $\theta_1$ be defined as in the beginning of the section and $c$ be as in lemma \ref{ap}. For any $t_0\in EP$ and $\epsilon>0,$ there exists some $N'=N'(t_0),$ and absolute constant $\lambda^{-\frac{1}{2}k(t_0)}<c_0''<\lambda^{-\frac{1}{2}(1-\epsilon)k(t_0)}$ such that
     for $n\geq N',x\in S_2$
     and $t_0-\lambda^{-\frac{c}{10}q_{N+M(n)-2}}\le t_1\le t_0-\lambda^{-\frac{c}{10}q_{N+M(n)-1}}$, we have
\begin{equation}\label{t1<t2}
\left|\int_{T^{l}x\in I_{L+1}} \frac{d\theta_l}{dt}\frac{d\|A_n\|}{d\theta_l}\frac{1}{\|A_n\|} dx\right|\leq c_0''\sqrt{t_1-t_0}^{-\frac{1}{2}}.
\end{equation}


\end{lemma}\begin{proof}

By a straight calculation, it holds that
$$\frac{1}{\|A_n\|}\frac{d\|A_n\|}{dt}=  \frac{1}{\|A_{n}\|}\frac{d\|A_{n}\|}{d\|A_{n-l}\|}\frac{d\|A_{n-l}\|}{dt}+ \frac{1}{\|A_{n}\|}\frac{d\|A_{n}\|}{d\|A_{l}\|}\frac{d\|A_l\|}{dt}+\frac{d\theta_l}{dt}\frac{1}{\|A_n\|}\frac{d\|A_n\|}{d\theta}$$
 By the second part of Lemma \ref{lemma8}, it holds that $\left|\frac{1}{\|A_{n}\|}\frac{d\|A_{n}\|}{d\|A_{n-l}\|}\right|\le \frac{1}{\|A_{n-l}\|} $. Subsequently by Lemma \ref{lemma22}, it holds that
$$\big\vert \int_{T^{l}x\in I_{L+1}} \frac{1}{\|A_{n}\|}\frac{d\|A_{n}\|}{d\|A_{n-l}\|}\frac{d\|A_{n-l}\|}{dt}\big\vert \le \int_{T^{l}x\in I_{L+1}}\big\vert \frac{1}{\|A_{n-l}\|}\frac{d\|A_{n-l}\|}{dt}\big\vert\leq k^C(t_0)\leq N'^C(t_0);$$
Similarly, we have $$\big\vert \int_{T^{l}x\in I_{L+1}}\frac{1}{\|A_{n}\|}\frac{d\|A_{n}\|}{d\|A_{l}\|}\frac{d\|A_l\|}{dt}\big\vert \le \int_{T^{l}x\in I_{L+1}}\big\vert \frac{1}{\|A_{l}\|}\frac{d\|A_{l}\|}{dt}\big\vert\leq k^C(t_0)\leq N'^C(t_0),$$ where $k$ is the resonance distance of $g_{N'}.$
\vskip 0.2cm

Next we estimate $\frac{d\theta_l}{dt}\frac{1}{\|A_n\|}\frac{d\|A_n\|}{d\theta}$.
  Fixed any $t_0\in EP$. By lemma \ref{lm16}, it holds that for some sufficient large $N_0$, we have $g_i$ is always of type III(or II)  and $|g_i(\tilde{c}_{t,1})|<C\lambda^{-\frac{3}{2}r_{t-1}},\ |\frac{\p^2\theta(x,t_0)}{\partial x^2}|\ge c>0$ for any $i\geq N_0$.
    It implies that there exists some $N$ such that for $n\geq N$ we have $M(n)\geq R_{N_0}^{10}$. Moreover, due to the nonexistence of resonance for the case $t_0\in EP$, it holds that
     \begin{equation}\label{second}\|g_i-g_{i+1}\|_{\mathcal{C}^2(I_{i+1})}\leq \lambda^{-\frac{3}{2}r_{i-1}}\ll\frac{1}{q_{N+i-1}}\end{equation}$i\geq N_0$
for $i\geq N_0$.
   \vskip 0.1cm
Without loss of generality, we can always assume that $\frac{\p^2\theta(x,t_0)}{\partial x^2}>0,$ for any $x\in I_M.$ Besides, we only need to consider one of the critical intervals. For instance, we consider $I_{M,1}$ and we still write it as $I_{M}.$
\vskip 0.3cm

The inequality $\lambda^{-\frac{c}{10}q_{N+M(n)-1}}\leq \vert t_0-t_1\vert \leq \lambda^{-\frac{c}{10}q_{N+M(n)-2}}$ is equivalent to
$$t_0+\lambda^{-\frac{c}{10}q_{N+M-1}}<t_1<t_0+\lambda^{-\frac{c}{10}q_{N+M-2}}$$ or
$$t_0-\lambda^{-\frac{c}{10}q_{N+M-2}}<t_1<t_0-\lambda^{-\frac{c}{10}q_{N+M-1}}.$$
Let $t_0-t_1:=\hat{\epsilon_1}$. It's obvious that $n'=o(n).$
Firstly, we consider the case $\hat{\epsilon_1}<0$.
It follows from lemma \ref{lm2222} that
\begin{equation}\label{wc}\|\theta_l-g_{M+1}\|_{C^2(I_M)}\leq\|u(A_l)-u_{r_M}\|_{C^2(I_M)}+\|s(A_{n-l}-s_{r_M})\|_{C^2(I_M)}\leq\lambda^{R_{N_0}^2}\hat{\epsilon_1},
\end{equation}
for any $\frac{1}{2}n'\leq l\leq n-\frac{1}{2}n'.$

Let $\tilde{x_l}:=\{y\in I_{M+1}| \frac{\p\theta_l}{\p x}(y)=0\}.$ It is well defined due to the fact that $g_{i}, \ i\gg 1$ is of type III (hence  $\{y\in I_{M+1}| \frac{\p g_{M+1}}{\p x}(y)=0\}$ is well defined) and \eqref{wc}.

We claim that there exist constants $0<c<C, 0<c'<C'$ and $d>0$ independent of $x, t, l$ and $n$ such that \begin{equation}\label{jd}c(T^lx-\tilde{x_l})^2-C'\hat{\epsilon_1}\le \theta_l(t_1,x)\le C(T^lx-\tilde{x_l})^2-c'\hat{\epsilon_1},\qquad |T^lx-\tilde{x_l}|\le d.\end{equation}


In fact, if we denote $J_{t_0,n}=[t_0-\lambda^{-\frac{c}{10}q_{N+M-1}},t_0+\lambda^{-\frac{c}{10}q_{N+M-1}}]$ and $I_{\tilde{x_l},n}=[\tilde{x_l}-2^{-M}q^{-8\tau}_{N+M(n)-1},\tilde{x_l}+2^{-M}q^{-8\tau}_{N+M(n)-1}],$ then by Taylor expand $\theta_l(x, t)\in C^{2+h}(I_{\tilde{x_l},n}\times J_{t_0},n)$ at $(\tilde{x_l},t_0)$, we have
\begin{equation}\begin{array}{ll}&\theta_l(x,t)=\theta_l(\tilde{x_l},t_0)+(x-\tilde{x_l})\frac{\partial\theta_l(\tilde{x_l},t_0)}{\partial x}+(t-t_0)\frac{\partial\theta_l(\tilde{x_l},t_0)}{\partial t}+\frac{1}{2}(x-\tilde{x_l})^2\frac{\partial^2\theta_l(\tilde{x_l},t_0)}{\partial x^2}+\\&\frac{1}{2}(t-t_0)^2\frac{\partial^2\theta_l(\tilde{x_l},t_0)}{\partial t^2}+(t-t_0)(x-\tilde{x_l})\frac{\partial^2\theta_l(\tilde{x_l},t_0)}{\partial x\partial t}+o((x-\tilde{x_l})^{2+h}+(t-t_0)^{2+h}).\end{array}\end{equation}

Note that $\frac{\partial\theta_l(\tilde{x_l},t_0)}{\partial x}=0,$ which follows from the definition of $\tilde{x_l}.$

Thus, \begin{equation}\label{taile}\theta_l(x,t)=\theta_l(\tilde{x_l},t_0)+[C(\tilde{x_l},t_0)+o(x-\tilde{x_l})^h](x-\tilde{x_l})^2+[D(x,t,\tilde{x_l},t_0)+o(t-t_0)^{1+h}](t-t_0),\end{equation} where $$C(\tilde{x_l},t_0)=\frac{1}{2}\frac{\partial^2\theta_l(\tilde{x_l},t_0)}{\partial x^2};$$
$$ D(x,t,\tilde{x_l},t_0)=\frac{\partial\theta_l(\tilde{x_l},t_0)}{\partial t}+(x-\tilde{x_l})\frac{\partial\theta_l(\tilde{x_l},t_0)}{\partial t\partial x}+\frac{1}{2}(t-t_0)\frac{\partial^2\theta_l(\tilde{x_l},t_0)}{\partial^2 t}.$$

Clearly, it holds from Lemma 17 that $\frac{\partial\theta_l(\tilde{x_l},t_0)}{\partial t}>c$ with some absolute constant $c\geq \frac{1}{10}.$ And by directly calculation and the choice of $n$, we have
$$ \frac{\partial^2\theta_l(\tilde{x_l},t_0)}{\partial^2 t},\frac{\partial^2\theta_l(\tilde{x_l},t_0)}{\partial^2 x},\frac{\partial^2\theta_l(\tilde{x_l},t_0)}{\partial t\partial x}\sim \lambda^{k(t_0)}\ll q_{N+M(n)-1}\ll |I_{\tilde{x_l}}|+|J_{t_0}|.$$ Therefore, \begin{equation}\label{C}C(\tilde{x_l},t_0)\sim \frac{1}{2}\lambda^{k(t_0))}\geq 1 \gg |I_{\tilde{x_l},n}|^h;\end{equation} \begin{equation}\label{D}D(x,t,\tilde{x_l},t_0)\sim \frac{\partial\theta_l(\tilde{x_l},t_0)}{\partial t}> c\geq \frac{1}{10} \gg (|I_{\tilde{x_l},n}|+|J_{t_0}|)\cdot \lambda^{k(t_0)}+|J_{t_0}|^{1+h}.\end{equation} We denote $c_l\triangleq\frac{\p^2\theta_l}{\p x^2}(\tilde{x_l}, t_0)$ and $\delta\triangleq\theta_l(\tilde{x_l}, t_0)$. Hence, combine \eqref{taile},\eqref{C} with \eqref{D}, we obtain
\begin{equation}\label{taile'}(c_l-|I_{\tilde{x_l},n}|^{\frac{1}{2}h}|)(x-\tilde{x_l})^2-\lambda^{R_{N_0}}\hat{\epsilon_1}+\delta\leq \theta_l(t,x) \leq (c_l+|I_{\tilde{x_l},n}|^{\frac{1}{2}h}|)(x-\tilde{x_l})^2-c\hat{\epsilon_1}+\delta.\end{equation}
Then \eqref{jd} is followed by $|\delta|\leq\lambda^{-q_{N+L-1}^2}\ll\hat{\epsilon_1}.$

\vskip0.4cm

Let $\lambda_1:=\|A_l\|,\lambda_2:=\|A_{n-l}\|$.
Then by Lemma \ref{keylemma} and \eqref{jd}, it holds that
\begin{equation}\label{kl} \begin{array}{ll}
&\left|\frac{d \theta_l}{dt}\right|\frac{\cot\theta_l}{(\cot\theta_l)^2+\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4}}\\&\leq \left\vert\frac{d\theta_l}{dt}\frac{1}{\|A_n\|}\frac{d\|A_n\|}{d\theta}\right\vert \\&\triangleq \left|\frac{d \theta_l}{dt}\right| |\hat{\theta}_l(T^lx)|\\&\le
\left|\frac{d \theta_l}{dt}\right|(1+O(\lambda^{-4}_1+\lambda^{-4}_2))\left(\frac{\cot\theta_l}{(\cot\theta_l)^2+\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4}}+\frac{\sqrt{\lambda^{-4}_1+\lambda^{-4}_2}}{2}\tan^2\theta_l\right).\\

\end{array}\end{equation}


Now we estimate $\int_{T^{l}x\in I_{L+1}}|\hat{\theta}_l(T^lx)|dx=\int_{x\in I_{L+1}}|\hat{\theta}_l(x)|dx
=\int_{x\in I_{L+1,1}}+\int_{x\in I_{L+1,2}}$. We only consider the case $x\in I_{L+1,1}$ and the other one is similar. It's worthy noting that, $I_{L+1,1}$ consist of two main parts: on the first part,
the image of $g_{L+1}$ is very steep and rapidly cross $[-\pi,\pi],$ and on the second part, the image of $g_{L+1}$ is seriously flat and slowly move away from the extremum point. Due to the difference between the two cases, we have to separately consider them.
\vskip0.3cm
By \cite{wz1}, we can write $\theta_l$ as $\arctan (\|A_k\|^2\tan\theta_1) -\frac{\pi}{2}+\theta_2$(we omit the dependence on $t_0$ for $k$), where $\theta_1$ and $\theta_2$ are of type I before the resonance with two zero points $c_{\tilde{N},1}$ and $c_{\tilde{N},2}$. Denote $\{c_{L+1,1},c'_{L+1,1},c_{L+1,2},c'_{L+1,2}\}=\{x\in I_{L+1,1}\bigcup I_{L+1,2}| \theta_l(x)=0\}$. Clearly, without loss of generality, we assume that $$c_{L+1,2}-k\alpha>c'_{L+1,2}>c_{L+1,1}>c_{\tilde{N},1};$$ $$c_{\tilde{N},2}>c_{L+1,2}>c'_{L+1,1}>c_{L+1,1}+k\alpha,$$ where $$|c_{\tilde{N},i}-c_{L+1,i}|+|c_{\tilde{N},i}\pm k\alpha-c'_{L+1,i}|\leq C_1\|A_k\|^{-1},$$ for $i=1,2$ ; $$C_2\|A_{r_{\tilde{N}}}\|^{-\frac{1}{2}}>|c_{\tilde{N},i}\pm k\alpha-c_{\tilde{N},i}|$$ for $i=1,2;$ and $$|c'_{L+1,i}-c_{L+1,j}|\leq C_3\|A_k\|^{-1}$$ for $i\neq j, i,j=1,2.$
We denote $C''=\{C\in \R |g_{L+1}(c_{L+1,2}+C\|A_k\|^{-1})=\max\{g_{L+1}\}\}$ and $I':=[c_{L+1,2}+C''\|A_k\|^{-1},c_{L+1,2}+\frac{1}{2^{L+1}q_{N+L+1-1}^{8\tau}}].$
\vskip0.3cm
Next we separately consider the case $x\in I'$ and $x\in I_{L+1,1}-I'.$
\vskip0.3cm
\textbf{A:The case $x\in I'$:}
For $x\in I'$, it follows from lemma \ref{lemma22} and $r_{L+1}^{C}\geq |I_{k'}|^{-C}\geq |\frac{d\theta_i}{dx}|_{I_{L+1,i}}>|I_{k'}|^{C}\geq r_{L+1}^{-C},$ where $k'=\max\{l<L+1| g_{l}~is~of~type~I\}, $ that
\begin{equation}\label{1}
\end{equation}\begin{align*}|\frac{\partial \theta_k}{\partial t}|&\leq |\frac{2\|A_k\|\|A_k\|'\tan\theta_1+\|A_k\|^2(1+\tan^2\theta)\theta_1'}{1+\|A_k\|^4\tan^2\theta_1}|+|\theta_2'|\\&\leq |\frac{r^C_{L+1}}{\|A_k\|^2\tan\theta_1}|+|\frac{r^C_{L+1}}{\|A_k\|^2\tan^2\theta_1}|+r^C_{L+1}\\&\leq \frac{r^C_{L+1}}{\|A_k\|^2}+r^C_{L+1}+r^C_{L+1}\leq k^C.
\end{align*}
Therefore, it's sufficient to consider the following simplified form, that is
\begin{equation}\label{jff}\int_{0}^{\frac{1}{2^{L+1}q_{N+L}^{8\tau}}}k^Csgn(\theta_l)|\hat{\theta}_l(T^lx)|dx.\end{equation}

We denote $x^*\triangleq\{x>\tilde{x}|\theta_l(x)=\frac{\pi}{2}\},$ $\hat{c}\triangleq \theta_l''(x^*)$, $\hat{I}\triangleq[\epsilon^8_1,\epsilon^{\frac{1}{8}}_1]$, $c^2_m=\min\{\theta_l''(x)|x\in \hat{I}\},$ $c^2_M=\max\{\theta_l''(x)|x\in \hat{I}\}.$
It's clear that $x^*=\frac{\epsilon_1}{c^*}$ for some constant $c^*$, while $|c_M-c_m|\leq \epsilon_1^{\frac{1}{100}h}.$ Let $I^*_{L+1}\triangleq[0,\frac{1}{2^{L+1}q_{N+L}^{8\tau}}]$ and $I^*_m=[\frac{\sqrt{\epsilon^2_1-\epsilon^2_2}}{c_m},\frac{\sqrt{\epsilon^2_1+\epsilon^2_2}}{c_m}],$ where $\epsilon_2^2\triangleq\sqrt{\lambda^{-4}_1+\lambda^{-4}_2}.$

We write \eqref{jff} as $$\eqref{jff}=\int_{I^*_m}Csgn(\theta_l)|\hat{\theta}_l(T^lx)|dx+\int_{\hat{I}-I^*_m}Csgn(\theta_l)|\hat{\theta}_l(T^lx)|dx+\int_{I^*_{L+1}-\hat{I}}Csgn(\theta_l)|\hat{\theta}_l(T^lx)|dx$$
$$\triangleq R_1+R_2+R_3$$

For $R_1$, note that $\cot^2\theta_l+(\lambda^{-4}_1-\lambda^{-4}_2)^2\tan^2\theta_l\geq 2|\lambda^{-4}_1-\lambda^{-4}_2|.$ Therefore, it follows from the direct calculation that
\begin{equation}\label{1r}\end{equation}\begin{align*}&R_1=\\&\int_{I^*_m}Csgn(\theta_l)|\hat{\theta}_l(T^lx)|dx\\&\leq \int_{I^*_m}\frac{C}{\sqrt{\cot^2\theta_l+(\lambda^{-4}_1-\lambda^{-4}_2)^2\tan^2\theta_l+\lambda^{-4}_1+\lambda^{-4}_2}}\\
&\leq Leb\{I_m^*\}\cdot C\min\{\lambda_1^2,\lambda_2^2\}\\&\leq (\frac{\sqrt{\epsilon^2_1+\epsilon^2_2}}{c_m}-\frac{\sqrt{\epsilon^2_1-\epsilon^2_2}}{c_m})\cdot C\min\{\lambda_1^2,\lambda_2^2\}\\&\leq
C\frac{\sqrt{\lambda_1^4+\lambda_2^4}}{\epsilon_1}\min\{\lambda_1^2,\lambda_2^2\}\\&\leq C\epsilon^{-1}_1
.\end{align*}

For $R_2$, note that \begin{equation}\label{dddz}|\cot\theta_l|\geq |c^2_mx^2-(\frac{c^m}{c^*})^2\epsilon^2_1|\geq \epsilon^2_2=\sqrt{\lambda^{-4}_1+\lambda^{-4}_2}\geq \sqrt{|\lambda^{-4}_1-\lambda^{-4}_2|}.\end{equation} For the function $C^0(\R)\backepsilon f\triangleq\frac{1}{\sqrt{x^2+(\lambda^{-4}_1-\lambda^{-4}_2)^2x^{-2}+\lambda^{-4}_1+\lambda^{-4}_2}},$ it is clear that $f$ is nondecreasing on $[\sqrt{|\lambda^{-4}_1-\lambda^{-4}_2|},+\infty).$

Therefore, \begin{equation}\begin{array}{ll}&R_2=\\&\int_{\hat{I}-I^*_m}Csgn(\theta_l)|\hat{\theta}_l(T^lx)|dx\\&\leq (1+O(\lambda^{-6}_1+\lambda^{-6}_2))\int_{\frac{\sqrt{\epsilon_1^2+\epsilon_2^2}}{c_m}}^{\epsilon^{\frac{1}{8}}_1}\left(\frac{\epsilon^2_2}{2}\tan^2\theta_l+\frac{|\cot\theta_l|}{\cot^2\theta_l+\epsilon^4_2}\right)dx
\\&-(1-O(\lambda^{-6}_1+\lambda^{-6}_2))\int_{\epsilon^8_1}^{\frac{\sqrt{\epsilon_1^2-\epsilon_2^2}}{c_m}}\left(\frac{|\cot\theta_l|}{\cot^2\theta_l+\epsilon^4_2}\right)dx
\\&\leq (1+O(\lambda^{-6}_1+\lambda^{-6}_2))\int_{\frac{\sqrt{\epsilon_1^2+\epsilon_2^2}}{c_m}}^{\epsilon^{\frac{1}{8}}_1}\left(\frac{\epsilon^2_2}{2c^4_m}\frac{1}{(x^2-(\frac{\epsilon_1}{c^*})^2)^2}+\frac{1}{c^2_m}\frac{|x^2-(\frac{\epsilon_1}{c^*})^2|}{(x^2-(\frac{\epsilon_1}{c^*})^2)^2+\frac{\epsilon^4_2}{c^4_m}}\right)dx
\\&-(1-O(\lambda^{-6}_1+\lambda^{-6}_2))\int_{\epsilon^8_1}^{\frac{\sqrt{\epsilon_1^2-\epsilon_2^2}}{c_m}}\left(\frac{1}{c^2_m}\frac{|x^2-(\frac{\epsilon_1}{c_m})^2|}{(x^2-(\frac{\epsilon_1}{c_m})^2)^2+\frac{\epsilon^4_2}{c^4_m}}\right)dx
\\&\triangleq K_1-K_2
\end{array}\end{equation}

By denoting $$F_1(\epsilon_1,\epsilon_2)\triangleq\frac{\sqrt{\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}}}{\sqrt{\epsilon_1^4+\epsilon_2^4}};$$  $$F_2(\frac{\epsilon_1}{c_m},\frac{\epsilon_2}{c_m})\triangleq2\log|\frac{\sqrt{2}(\frac{\epsilon_2}{c_m})^2}{\frac{\epsilon_1^2-\epsilon_2^2}{c^2_m}+\sqrt{2}\sqrt{(\frac{\epsilon_1}{c_m})^2+\sqrt{(\frac{\epsilon_1}{c_m})^4+(\frac{\epsilon_2}{c_m})^4}}\frac{\sqrt{\epsilon_1^2-\epsilon_2^2}}{c_m}+\sqrt{(\frac{\epsilon_1}{c^*})^4+(\frac{\epsilon_2}{c_m})^4}}|$$
 and a direct calculation, we have $$K_1-K_2\leq $$ $$\frac{\sqrt{2}}{8c^2_m}\left[\bar{C}F_1(\frac{\epsilon_1}{c^*},\frac{\epsilon_2}{c_m})(F_2(\frac{\epsilon_1}{c^*},\frac{\epsilon_2}{c_m})+\log\sqrt{2})-\b{C}F_1(\frac{\epsilon}{c_m},\frac{\epsilon_2^2}{c_m})F_2(\frac{\epsilon}{c_m},\frac{\epsilon_2^2}{c_m})\right]+C\epsilon^{-1},$$ where $(1+O(\lambda^{-6}_1+\lambda^{-6}_2))=\bar{C}$ and $(1-O(\lambda^{-6}_1+\lambda^{-6}_2))=\b{C}.$
One can check the details in Appendix A.3.

By a direct calculation, the following estimates hold:
$$|\bar{C}-\b{C}|\leq O(\lambda^{-6}_1+\lambda^{-6}_2);|F_1(\frac{\epsilon_1}{c^*},\frac{\epsilon_2}{c_m})-F_1(\frac{\epsilon_1}{c_m},\frac{\epsilon_2}{c_m})|\leq C(1-\frac{c_m}{c^*})\epsilon_1^{-1};$$
$$|F_2(\frac{\epsilon_1}{c^*},\frac{\epsilon_2}{c_m})-F_2(\frac{\epsilon_1}{c_m},\frac{\epsilon_2}{c_m})|\leq 1+\frac{c_m}{c^*};$$
$$|F_1(\frac{\epsilon_1}{c_m},\frac{\epsilon_2}{c_m})F_2(\frac{\epsilon_1}{c_m},\frac{\epsilon_2}{c_m})|\leq C\epsilon^{-1}_1\log\epsilon^{-1}_2;$$
$$|\bar{C}F_2(\frac{\epsilon_1}{c_m},\frac{\epsilon_2}{c_m})|\leq C\log\epsilon^{-1}_2;|\bar{C}F_1(\frac{\epsilon_1}{c^*},\frac{\epsilon_2}{c_m})|\leq C\epsilon_1^{-1}.$$

Hence, it follows from the inequality $$|a_1a_2a_3-b_1b_2b_3|\leq |a_1-b_1||b_2b_3|+|a_1b_3||a_2-b_2|+|a_1a_2||a_3-b_3|$$ that
\begin{equation}\label{bdsgj}K_1-K_2\leq O(\lambda^{-6}_1+\lambda^{-6}_2)C\epsilon^{-1}_1\log\epsilon^{-1}_2+C(1-\frac{c_m}{c^*})\epsilon_1^{-1}\log\epsilon^{-1}_2+(1+\frac{c_m}{c^*})C\epsilon^{-1}_1+C\epsilon^{-1},\end{equation} which implies
\begin{equation}\label{2r}R_2\leq C\epsilon_1^{-1}.\end{equation}

Combine \eqref{bdsgj} with the fact $|1-\frac{c_m}{c^*}|\leq \epsilon_1^{\frac{h}{100}}\ll \log\epsilon^{-1}_2$ and $\log\epsilon^{-1}_2\ll \min\{\lambda_1,\lambda_2\}^{-1}$, we have $K_1-K_2\leq C\epsilon^{-1}_1,$

which implies $R_2\leq C\epsilon_1^{-1}.$

For $R_3$, by the help of the process above, we have
\begin{align*}&\int_{I^*_{L+1}-\hat{I}}Csgn(\theta_l)|\hat{\theta}_l(T^lx)|dx\\&\leq 2\int_{\epsilon^{\frac{1}{8}}_1}^{\frac{1}{2^{L+1}q^{8\tau}_{N+L}}}\left(\frac{\epsilon^2_2}{2}\tan^2\theta_l+\frac{|\cot\theta_l|}{\cot^2\theta_l+\epsilon^4_2}\right)dx
\\&+2\int_{0}^{\epsilon^8_1}\left(\frac{|\cot\theta_l|}{\cot^2\theta_l+\epsilon^4_2}\right)dx
\\&\leq C\int_{\epsilon^{\frac{1}{8}}_1}^{\frac{1}{2^{L+1}q^{8\tau}_{N+L}}}\left(\frac{\epsilon^2_2}{2c^4_m}\frac{1}{(x^2-(\frac{\epsilon_1}{c^*})^2)^2}+\frac{1}{c^2_m}\frac{|x^2-(\frac{\epsilon_1}{c^*})^2|}{(x^2-(\frac{\epsilon_1}{c^*})^2)^2+\frac{\epsilon^4_2}{c^4_m}}\right)dx
\\&+C\int_{0}^{\epsilon^8_1}\left(\frac{\epsilon^2_2}{2c^4_m}\frac{1}{(x^2-(\frac{\epsilon_1}{c^*})^2)^2}+\frac{1}{c^2_m}\frac{|x^2-(\frac{\epsilon_1}{c^*})^2|}{(x^2-(\frac{\epsilon_1}{c^*})^2)^2+\frac{\epsilon^4_2}{c^4_m}}\right)dx
\\&\triangleq\bar{K}_1+\bar{K}_2.\end{align*}

It follows from the process of the estimate for $R_2$ that $\hat{K_1}\leq C\epsilon^{-1}_1|\log(1+C\epsilon^{-2}_1q^C_{N+L})+\log(1+\epsilon_1^{\frac{15}{8}})|+C\epsilon_1^{-1}$ and $\hat{K}_2\leq C\epsilon_1^{-1}|\log(1+\epsilon_1^6)|+C\epsilon^{-1}_1.$ Therefore, \begin{equation}\label{3r}R_3\leq C\epsilon^{-1}_1.\end{equation}

In summary, \eqref{1r},\eqref{2r} and \eqref{3r} imply
$$ \eqref{jff}\leq c_0''\hat{\epsilon_1}^{-\frac{1}{2}}$$
with some positive constant $c_0''\sim \lambda^{-\frac{1}{2}k(t_0)}$.
\vskip0.3cm

\


\textbf{B:The case $x\in I_{L+1}-I'$:}
For the second part, we divide it into three parts. For simplification, denote $\|A_k\|=l^k$, $I_{L+1,2}-I':=I_1\bigcup I_2\bigcup I_3,$
where $$I_1=(I_{L+1,2}-I')\bigcap[c_{L+1,1}+k\alpha-\frac{1}{2^{L+1}q^{8\tau}_{N+L+1-1}},c_{L+1,1}+k\alpha-Cl^{-2k}],$$ where the image of the angle function is located at $[-\pi+Cl^{-k},-c_4);$ $$I_2=[c_{L+1,1}+k\alpha-C'l^{-2k},c_{L+1,1}+k\alpha+C'l^{-2k}],$$ where the image of the angle function is located at $[-c_4, -c_5];$ $$I_3=[c_{L+1,1}+k\alpha+C'l^{-2k},c_{L+1,1}+k\alpha+C''l^{-k}],$$ where the image of the angle function is located at $[-c_5,maximum]$

Note that for $x\in I_3,$ we may write $\theta_k$ as $c_1l^{-2k}x^{-1}+x-c_2l^{-k}$ and $$|\frac{\partial \theta_k}{\partial t}|\leq c_3r^C_{L+1}l^{-2k}x^{-2}.$$ Hence by lemma 18, we have:
\begin{align*}\int_{I_3}The~majority~of~the~integral&\leq \int_{I_3}\frac{|\frac{\partial \theta_k}{\partial t}|}{\sqrt{(c_1l^{-2k}x^{-1}+x-c_2l^{-k})^2+\epsilon'}}dx\\&\leq \int_{I_3}\frac{c_3r^C_{L+1}l^{-2k}x^{-2}}{c_1l^{-2k}x^{-1}}dx\\&\leq \int_{C'l^{-2k}}^{C''l^{-k}}Cx^{-1}dx\leq C\log l\leq r^C_{L+1}.
\end{align*}

For the integral on $I_2.$ Note that $\theta_k>c_3$ on $I_2$ and $|\frac{\partial \theta_k}{\partial t}|\leq r^C_{L+1}l^{2k}.$ Thus, it's easy to see that
\begin{align*}\int_{I_2}The~majority~of~the~integral&\leq \int_{I_2}\frac{|\frac{\partial \theta_k}{\partial t}|}{\sqrt{(\cot \theta_k)^2+\epsilon'}}dx\\&\leq |I_2|c_3^{-1}r^C_{L+1}l^{2k}\\&\leq 2C'l^{-2k}c_3^{-1}r^C_{L+1}l^{2k}\\&\leq r^C_{L+1}.
\end{align*}

For the remaining $I_1$, note that we can write $\theta_k$ as $c_1l^{-2k}x^{-1}+x+c_4l^{-k}$ since the graph is always at a certain distance (about $l^{-k}$) from the horizontal axis. By direct calculation, we have:\begin{align*}\int_{I_1}The~majority~of~the~integral&\leq \int_{I_1}\frac{|\frac{\partial \theta_k}{\partial t}|}{\sqrt{(c_1l^{-2k}x^{-1}+x+c_4l^{-k})^2+\epsilon'}}dx\\&\leq \int_{I_1}\frac{r^C_{L+1}l^{-2k}x^{-2}}{c_1l^{-2k}x^{-1}+x+c_4l^{-k}}dx\\&\leq \int_{I_1}\frac{r^C_{L+1}}{c_4l^k x^2+l^{2k} x^3+c_1x}dx\\&\leq \int_{I_1}C\frac{r^C_{L+1}}{x+l^{2k}x^3+l^k x^2}dx\\&\leq r^C_{L+1}[\left.\frac{\log(l^{2k}x^2+l^k x+1)}{2}+\log(x)-\frac{\arctan(\frac{2l^k x+1}{\sqrt{3}})}{\sqrt{3}}\right|_{C''l^{-k}}^{\frac{1}{2^{L+1}q^{8\tau}_{N+L+1-1}}}]\\&\leq r^C_{L+1}C\log l\leq r^C_{L+1}.
\end{align*}

Therefore, we have $$\eqref{t1<t2}\leq r^C_{L+1}\|A_{k_j}\|^{-\frac{1}{2}}\hat{\epsilon_1}^{-\frac{1}{2}}+r^C_{L+1}\leq 2r^C_{L+1}\|A_{k_j}\|^{-\frac{1}{2}}\hat{\epsilon_1}^{-\frac{1}{2}}\leq c_0''\hat{\epsilon_1}^{-\frac{1}{2}}$$ with $\lambda^{-\frac{1}{2}k(t_0)}\leq c_0''\leq \lambda^{-(1-\epsilon)\frac{1}{2}k(t_0)}$ which is what we desire. Furthermore, it's not difficult to see that: the later the resonance occurs, the better (smaller) the 1/2-H\"older constant should be obtained.


\end{proof}

\begin{lemma}\label{lemma25-1}
   Let $n'$ $A_l, A_{n-l}$ and $\theta_1$ be defined as in Lemma \ref{lemma25} and $c$ be as in lemma \ref{ap}. For any fixed $t_0\in EP$ and $\epsilon>0,$ there exists some $N=N(t_0),$ and absolute constants $\lambda^{-\frac{1}{2}k(t_0)}\leq c_1'',c_2''\leq \lambda^{-\frac{1}{2}(1-\epsilon)k(t_0)}$ with $0<c''_1<c''_2\le c''_1+ |t_0-t_1|^h$ such that for $n\geq N,x\in S_2$ and $t_0+\lambda^{-\frac{c}{10}q_{N+L(n)-1}}\le t_1\le t_0+\lambda^{-\frac{c}{10}q_{N+L(n)-2}}$, we have
\begin{equation}\label{jf}
 c''_1\sqrt{t_1-t_0}^{-\frac{1}{2}}\leq \left|\int_{T^{l}x\in I_{L+1}} \frac{d\theta_l}{dt}\frac{d\|A_n\|}{d\theta_l}\frac{1}{\|A_n\|} dx\right|\leq c''_2\sqrt{t_1-t_0}^{-\frac{1}{2}}.
\end{equation}

\end{lemma}
\begin{proof}
For $$t_0+\lambda^{-\frac{c}{10}q_{N+L-1}}<t_1<t_0+\lambda^{-\frac{c}{10}q_{N+L-2}},$$ all the inferences are similar as above.
In fact, if we let $\epsilon_1:=t_1-t_0$, with the same method we can obtain that
$$\theta_l(t_1,x)\approx c(x-\tilde{x_l})^2+\hat{\epsilon_1}$$ for $\frac{1}{2}n'\leq l\leq n-\frac{1}{2}n'.$

Then Lemma \ref{keylemma} implies that
\begin{equation}\label{jfff'}\begin{array}{ll}
&\int_{T^lx\in I_{L+1}} \frac{1}{\|A_n\|}\frac{d\|A_n\|}{d\theta}dx\\
\\
&=\int_{T^lx\in I_{L+1}} \frac{(sgn(x-c_{L+1})(sgn(x-c_{L+1}'))(1-\lambda_2^{-4})(1-\lambda_1^{-4})}
{\sqrt{(1-\frac{1}{\lambda_1^4\lambda_2^4})^2(cx^2+\epsilon_1)^2+(\frac{1}{\lambda_1^4}-\frac{1}
{\lambda_2^4})^2(cx^2+\epsilon_1)^{-2}+2(1+\frac{1}{\lambda_1^4\lambda_2^4})
(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})-\frac{8}{\lambda_1^4\lambda_2^4}}}.
\end{array}
\end{equation}
Since $1\ll\epsilon_1^{-\frac12}\ll\min\{\lambda_1, \lambda_2\}$,  we obtain that
$$
\begin{array}{ll}
&\frac{(sgn(x-c_{L+1})(sgn(x-c_{L+1}'))(1-\lambda_2^{-4})(1-\lambda_1^{-4})}
{\sqrt{(1-\frac{1}{\lambda_1^4\lambda_2^4})^2(cx^2+\epsilon_1)^2+(\frac{1}{\lambda_1^4}-\frac{1}
{\lambda_2^4})^2(cx^2+\epsilon_1)^{-2}+2(1+\frac{1}{\lambda_1^4\lambda_2^4})
(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})-\frac{8}{\lambda_1^4\lambda_2^4}}}\\
\\
&\thickapprox \frac{(sgn(x-c_{L+1})(sgn(x-c_{L+1}'))}{(cx^2+\hat{\epsilon_1})^2+o(\hat{\epsilon_1}^2)}
\end{array}
$$
Note that $sgn(x-c_{L+1})\equiv 1~or~-1$ since the minimum(or maximum) of $g_{L}$ has been far away from horizontal axis.
Thus with the help of Appendix A1, we have
\begin{equation}\label{jfff}
\int_{T^lx\in I_{L+1}} \frac{1}{\|A_n\|}\frac{d\|A_n\|}{d\theta}dx\thickapprox c''\hat{\epsilon_1}^{-\frac{1}{2}}.
\end{equation}


More precisely, there exist some $c_1''$ and $c_2''$ satisfying $0<c''_1<c''_2\le c''_1+ |t_0-t_1|^h$ such that $$c_1''\hat{\epsilon_1}^{-\frac{1}{2}}\leq \eqref{jfff} \leq c_2''\hat{\epsilon_1}^{-\frac{1}{2}}.$$


\end{proof}

\


Finally we deal with the case $x\in S_3$. For $1\leq l \leq \frac{1}{2}n'$ or $n-\frac{1}{2}n'\leq l \leq n,$ $\|\theta_l-g_{L+1}\|$ may be too large such that $\delta=\theta_l(\tilde{x_l})$ is larger than $\epsilon_1$ and the argument in the proof of Lemma \ref{lemma25-1} does not work. For this case, we will prove the following result:
\begin{lemma}\label{lemma26}  Let $n'$ $A_l, A_{n-l}$ and $\theta_1$ be defined as in Lemma \ref{lemma25} and the constant $c$ is from lemma \ref{ap}. For any $t_0\in EP$ and $\epsilon>0,$ there exists some $N=N(t_0)$ and absolute constant $\lambda^{-\frac{1}{2}k(t_0)}\leq c_3''\leq \lambda^{-\frac{1}{2}(1-\epsilon)k(t_0)}$ such that  for $n\geq N,x\in S_3$ and $\lambda^{-\frac{c}{10}q_{N+L(n)-1}}\leq \vert t_1-t_0\vert \leq \lambda^{-\frac{c}{10}q_{N+L(n)-2}},$ we have \begin{equation}\label{small-or-large-l}\left|\int_{T^{l}x\in I_{L+1}} \frac{d\theta_l}{dt}\frac{d\|A_n\|}{d\theta_l}\frac{1}{\|A_n\|} dx\right|\le c_3''n^{\frac13}|t_1-t_0|^{-\frac12}.\end{equation}
\end{lemma}
\begin{proof}
For $1\leq l \leq \frac{1}{2}n'$ or $n-\frac{1}{2}n'\leq l \leq n,$  we only consider the case $t_0-\lambda^{-\frac{c}{10}q_{N+L(n)-2}}\le t_1\le t_0-\lambda^{-\frac{c}{10}q_{N+L(n)-1}}$. And for $$t_0+\lambda^{-\frac{c}{10}q_{N+L-1}}<t_1<t_0+\lambda^{-\frac{c}{10}q_{N+L-2}},$$ all the inferences are similar as above.

Let
$$\delta_1:=\delta-\epsilon_1,$$ we have the following estimate.

By lemma \ref{keylemma} and Appendix A1, we have
\begin{align*}
&\int_{T^{l}x\in I_{L+1}} \frac{d\theta_l}{dt}\frac{d\|A_n\|}{d\theta_l}\frac{1}{\|A_n\|} dx \\&\leq \int_{-\frac{1}{2^{L+1}q_{N+L}^{8\tau}}}^{\frac{1}{2^{L+1}q_{N+L}^{8\tau}}} \frac{(sgn(x-c_{L+1})(sgn(x-c_{L+1}'))C}{\sqrt{(cx^2-\delta_1)^2+(\frac{1}{\lambda_1^4}-\frac{1}{\lambda_2^4})^2(cx^2-\delta_1)^{-2}+2(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})}} dx\\&\leq \int_{-\frac{1}{2^{L+1}q_{N+L}^{8\tau}}}^{\frac{1}{2^{L+1}q_{N+L}^{8\tau}}} \frac{(sgn(x-c_{L+1})(sgn(x-c_{L+1}'))C}{\sqrt{(cx^2-\delta_1)^2+2(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})}} dx\\&\leq C\min\{2\lambda_0,\delta_1^{-\frac{1}{2}}\}\\&\leq C\min\{\lambda_0,\delta_1^{-\frac{1}{2}}\}\\&\leq C\min\{\lambda_0,(\lambda_0^{-2}{[}\frac{\log\lambda_0}{\log \lambda}{]}^{C_\tau}-\epsilon_1)^{-\frac{1}{2}}\}\\&\leq C{[}\frac{\log\lambda_0}{\log \lambda}{]}^{\frac{C_\tau}{2}}\sqrt{\epsilon_1}^{-1}\\&\leq Cq^{\frac{C_\tau}{2}}_{N+L(n)-1}\epsilon_1^{-0.5}\\&\leq C n^{\frac{1}{3}}\epsilon_1^{-0.5}
\end{align*}
where $\lambda_0=\min\{\lambda_1,\lambda_2\}.$
See the details of the third inequality in A.1 and the fifth inequality is due to $|\delta|\leq\max\{\lambda_1^{-2},\lambda_2^{-2}\}{[}\frac{\log\lambda_0}{\log \lambda}{]}^{C_\tau}\leq \lambda_0^{-2}{[}\frac{\log\lambda_0}{\log \lambda}{]}^{C_\tau}$ from lemma \ref{lm2222}. For the seventh inequality, we need slightly change the scale of $n$ and corresponding $I_{L(n)}$. Whereas, we won't specifically change it and just admit it for convenience.

\end{proof}


 \begin{corollary} Let $c$ be as in lemma \ref{ap}. For any $t_0\in EP,$ there exists some $N=N(t_0),c'''_1~and~c'''_2$ independent on $t_0$ satisfying $0<c'''_1\le c'''_2\le c'''_1+|t_0-t_1|^{h}$ such that
     for $n\geq N$
     it holds that
           \begin{equation}\label{lm-one-side}|L_n'(t_1)|\leq c_2'''|t_0-t_1|^{-\frac{1}{2}}\end{equation} for $\lambda^{-\frac{c}{10}q_{N+L(n)-1}}\leq \vert t_0-t_1\vert \leq \lambda^{-\frac{c}{10}q_{N+L(n)-2}},$
     and
          \begin{equation}\label{lm}c_1'''|t_1-t_0|^{-\frac{1}{2}}\leq |L_n'(t_1)|\leq c_2'''|t_1-t_0|^{-\frac{1}{2}}\end{equation} for  $\lambda^{-\frac{c}{10}q_{N+L(n)-1}}\leq  t_1-t_0 \leq \lambda^{-\frac{c}{10}q_{N+L(n)-2}},$ where $L(n)$ is determined by $q_{N+L(n)-1}^4\leq n\leq q_{N+L(n)}^4$.
     Both $c'''_1$ and $c'''_2$ independent  on $t_0$.
     \end{corollary}

     \begin{proof}

     For any fixed $n\geq N$, since $|I_{L(n)+1}|\le \frac{1}{2q_{N+L_n}^{2\tau}}$, from the Diophantine condition on $\alpha$ and the definition of $L(n)$, for $0\leq l\leq n$ we have that $\|T^{l}(0)\|\ge \frac{\gamma}{n^{\tau}}>\frac{\gamma}{q_{N+L(n)}^{4\tau}}\gg \frac{1}{2^Lq_{N+L(n)}^{8\tau}}= |I_{L(n)+1}|$. It implies that there exists at most one $0\leq l\leq n$ such that $T^{l}(x)\in I_{L(n)+1}.$

       Recall that $[\frac{-\log|t_1-t_0|}{\log\lambda}]+R_{N_0}^2=n'$ and $
       \mathbb{T}=S_1\bigcup S_2 \bigcup S_3,$ where
     $$
     \begin{array}{ll}
     S_1&=\{x\in \mathbb{T}| T^{l}(x)\not\in I_{L+1}\ ~for\ any\ ~ 0\leq l\leq n \},\\
      \\
      S_2&=\{x\in \mathbb{T}|~there\ exists\ ~ \frac{1}{2}n'\leq l\leq n-\frac{1}{2}n' ~such\ that\ ~ T^{l}(x)\in I_{L+1}\ \},\\
      \\
      S_3&=\{x\in \mathbb{T}|~there\ exists\ ~ 0\leq l<\frac{1}{2}n'\ ~or~\ n-\frac{1}{2}n'< l\leq n ~such\ that\ ~ T^{l}(x)\in I_{L+1}\ \}.
      \end{array}
      $$

     \begin{enumerate}
     \item For $x\in S_1$, since there is no $0\leq l<n$ such that $T^{l}x\in I_{L+1},$ by corollary \ref{useful}, it holds that
     $$\frac{1}{\|A_n(x,t)\|}\big\vert \frac{\partial\|A_{n}(x,t)\|}{\partial t}\big\vert \leq \max\{n,q_{N+L-1}\}^C\leq n^C,$$ which implies
     $$\frac{1}{n}\bigg\vert \int_{x\in S_1} \frac{1}{\|A_n\|}\frac{d\|A_n\|}{dt} dx\bigg\vert\ll \epsilon_1^{-\frac12}.$$

     \item For $x\in S_2$, there exists some $\frac{1}{2}n'\leq l\leq n-\frac{1}{2}n'$ such that $T^{l}x\in I_{L+1}.$ For this case, Lemma \ref{lemma25} and  \ref{lemma25-1} hold true. Thus it holds that
         $$\frac{c_1}{n}(n-o(n^{\frac13})){\epsilon_1}^{-\frac{1}{2}}\le\frac{1}{n}\bigg\vert \int_{x\in S_2} \frac{1}{\|A_n\|}\frac{d\|A_n\|}{dt} dx\bigg\vert\le \frac{c_2}{n}(n-o(n^{\frac13})){\epsilon_1}^{-\frac{1}{2}},$$

      \item For $x\in S_3$, there exists some $0\leq l<\frac{1}{2}n',n-\frac{1}{2}n'< l$ such that $T^{l}x\in I_{L+1}.$ For this case, Lemma \ref{lemma26} holds true. Thus it holds from $n'^{3.5}\leq n$ that
         $$\frac{1}{n}\bigg\vert \int_{x\in S_3} \frac{1}{\|A_n\|}\frac{d\|A_n\|}{dt} dx\bigg\vert= \frac{o(n^{\frac{1}{3}})}{n}n^{\frac13}{\epsilon_1}^{-\frac{1}{2}}.$$
          \end{enumerate}


Thus on one hand for $\lambda^{-\frac{c}{10}q_{N+L(n)-1}}\leq \vert t_0-t_1\vert \leq \lambda^{-\frac{c}{10}q_{N+L(n)-2}},$ it holds that
\begin{align*} |L_n'(t)| &=\bigg\vert \int_{\R/\Z}\frac{1}{n} \frac{1}{\|A_n\|}\frac{d\|A_n\|}{dt} dx\bigg\vert\\&\leq \bigg\vert\int_{S_1} \bigg\vert+\bigg\vert\int_{S_2} \bigg\vert+\bigg\vert\int_{S_3} \bigg\vert\\&\leq o(\epsilon_1^{-\frac{1}{2}})+\frac{c_2}{n}(n-o(n^{\frac13})){\epsilon_1}^{-\frac{1}{2}}+
\frac{o(n^{\frac13})}{n}n^{\frac13}{\epsilon_1}^{-\frac{1}{2}}\\&\leq c_2'''\epsilon_1^{-\frac{1}{2}}.
\end{align*}
Thus we obtain \eqref{lm-one-side}.

On the other hand, for $\lambda^{-\frac{c}{10}q_{N+L(n)-1}}\leq  t_1-t_0 \leq \lambda^{-\frac{c}{10}q_{N+L(n)-2}},$ by a similar calculation, we obtain
\begin{align*}\begin{array}{ll} |L_n'(t)| &=\bigg\vert \int_{\R/\Z}\frac{1}{n} \frac{1}{\|A_n\|}\frac{d\|A_n\|}{dt} dx\bigg\vert\\
\\
&\geq \bigg\vert\int_{S_2} \bigg\vert-\bigg\vert\int_{S_1} \bigg\vert-\bigg\vert\int_{S_3} \bigg\vert\\
\\&\geq \frac{c_2}{n}(n-o(n^{\frac13})){\epsilon_1}^{-\frac{1}{2}}-o(\epsilon_1^{-\frac{1}{2}})-
\frac{o(n^{\frac13})}{n}n^{\frac13}{\epsilon_1}^{-\frac{1}{2}}
\\
\\
&\geq c_1'''\epsilon_1^{-\frac{1}{2}},\end{array}
\end{align*}
with
$0<c_1'''\le c_2'''\le c_1'''+|t_1-t_0|^h$, which together with \eqref{lm-one-side} implies \eqref{lm}.


     
     \end{proof}

Now we are ready to proof part of theorem 1.

Fix any $\tilde{t} \in EP$ and let $L_0$ sufficiently large.
Choose the interval $$I_{\tilde{t}}:=[\tilde{t}-\lambda^{-\frac{c}{10}q_{N+L_0}},\tilde{t}+\lambda^{-\frac{c}{10}q_{N+L_0}}]$$ and denote $I^{+}_{\tilde{t}}:=[\tilde{t},\tilde{t}+\lambda^{-\frac{c}{10}q_{N+L_0}}];$ $I^{-}_{\tilde{t}}:=[\tilde{t}-\lambda^{-\frac{c}{10}q_{N+L_0}},\tilde{t}].$

Note that either $$L_k'(t)>0~or<0~for~any~t~\in~I_{\tilde{t}}~and~q_{N+l+n}^4\leq k\leq q_{N+l+n+1}^4.$$

Without loss of generality, we assume that $$L_k'(t)>0~for~any~t~\in~I_{\tilde{t}}~and~q_{N+l+n}^4\leq k\leq q_{N+l+n+1}^4.$$

Then for any fixed $t'\in I^{-}_{\tilde{t}}$ (the case $I^{+}_{\tilde{t}}$ is similar), there exists some $0<l\leq L_0$ such that
$$\lambda^{-\frac{c}{10}q_{N+l-1}}\leq \tilde{t}-t'\leq \lambda^{-\frac{c}{10}q_{N+l}}.$$

For $n\in \N^+$, let $t_n:=\tilde{t}-\lambda^{-\frac{c}{10}q_{N+l+n-1}}$ and $t_0:=t'$.


It follows from lemma \ref{ap} that for $i\in N$
\begin{align*}
L(t_i)-L(t_{i+1})&\leq 2(L_{2q_{N+l+i+1}^4}(t_{i+1})-L_{2q_{N+l+i+1}^4}(t_{i}))+(L_{q_{N+l+i+1}^4}(t_{i+1})-L_{q_{N+l+i+1}^4}(t_{i}))\\&+\big\vert L_{q_{N+l+i+1}^4}(t_{i})+L(t_i)-2L_{2q_{N+l+i+1}^4}(t_{i})\big\vert\\&+\big\vert L_{q_{N+l+i+1}^4}(t_{i+1})+L(t_{i+1})-2L_{2q_{N+l+i+1}^4}(t_{i+1})\big\vert\\&\leq 3c_2(\sqrt{t_i-\tilde{t}}-\sqrt{t_{i+1}-\tilde{t}})+2\lambda^{-\frac{c}{2}q_{N+l+i+1}^4}
\end{align*}

Similarly, we have
\begin{align*}
L(t_i)-L(t_{i+1})&\geq 2(L_{2q_{N+l+i+1}^4}(t_{i+1})-L_{2q_{N+l+i+1}^4}(t_{i}))-(L_{q_{N+l+i+1}^4}(t_{i+1})-L_{q_{N+l+i+1}^4}(t_{i}))\\&-\big\vert L_{q_{N+l+i+1}^4}(t_{i})+L(t_i)-2L_{2q_{N+l+i+1}^4}(t_{i})\big\vert\\&-\big\vert L_{q_{N+l+i+1}^4}(t_{i+1})+L(t_{i+1})-2L_{2q_{N+l+i+1}^4}(t_{i+1})\big\vert\\&\geq (2c_2-c_1)(\sqrt{t_i-\tilde{t}}-\sqrt{t_{i+1}-\tilde{t}})-2\lambda^{-\frac{c}{2}q_{N+l+i+1}^4}
\end{align*}

Thus, for arbitrary $k\in \N^+,$
\begin{align*}
\vert L(t')-L(\tilde{t})\vert &\geq \vert\sum_{0\leq i\leq k-1}(L(t_i)-L(t_{i+1}))\vert-\vert L(t_k)-L(\tilde{t})\vert\\&\geq (2c_2-c_1)\sum_{0\leq i\leq k-1}(\sqrt{t_i-\tilde{t}}-\sqrt{t_{i+1}-\tilde{t}})-\sum_{0\leq i\leq k-1}2\lambda^{-\frac{c}{2}q_{N+l+i+1}^4}-\vert L(t_k)-L(\tilde{t})\vert\\&\geq (2c_2-c_1)\sqrt{t'-\tilde{t}}-\sum_{0\leq i\leq +\infty}2\lambda^{-\frac{c}{2}q_{N+l+i+1}^4}\\&\geq(2c_2-c_1)(t'-\tilde{t})^{\frac{1}{2}}-\lambda^{-\frac{c}{2}q_{N+l+1}}\\&\geq \tilde{c}(t'-\tilde{t})^{\frac{1}{2}}
\end{align*}

and

\begin{align*}
\vert L(t')-L(\tilde{t})\vert &\leq \vert\sum_{0\leq i\leq k-1}(L(t_i)-L(t_{i+1}))\vert+\vert L(t_k)-L(\tilde{t})\vert\\&\leq 3c_2\sum_{0\leq i\leq k-1}(\sqrt{t_i-\tilde{t}}-\sqrt{t_{i+1}-\tilde{t}})+\sum_{0\leq i\leq k-1}2\lambda^{-\frac{c}{2}q_{N+l+i+1}^4}+\vert L(t_k)-L(\tilde{t})\vert\\&\leq 3c_2\sqrt{t'-\tilde{t}}+\sum_{0\leq i\leq +\infty}2\lambda^{-\frac{c}{2}q_{N+l+i+1}^4}\\&\leq3c_2(t'-\tilde{t})^{\frac{1}{2}}+\lambda^{-\frac{c}{2}q_{N+l+1}}\\&\leq \tilde{c}'(t'-\tilde{t})^{\frac{1}{2}}
\end{align*}
for proper constant $\lambda^{-\frac{1}{2}k(t_0)}\leq \tilde{c}<\tilde{c}'<\lambda^{-\frac{1}{2}(1-\epsilon)k(t_0)}.$
The third inequality is followed by the continuity of $L(t)$ and the arbitrary choice of $k$.

In conclusion, we have proved that for any $E\in EP$, $L(\cdot)$ is exactly local $\frac{1}{2}$-H\"older continuous.


\section{The proof of  local Lipschitz continuity}
In  this section we prove  local Lipschitz continuity of $L(E)$ on a full measure set $FR$, which is different from exact $\frac{1}{2}$-H\"older continuity of $L(E)$ on $EP$. The reason why the regularities for these two cases are different lies in the fact that the measure of the set of `bad' $x$ for the former is much less than the one for the latter. It can be found from the degeneration  of the function $g(x,\cdot)$ of $t\in EP$ and the nondegeneration of the one of $t\in FR$, respectively.

Roughly speaking, the  proof for $(1-\epsilon)-$H\"older continuity of $L(t)$ for any $\epsilon>0$ and $t\in FR$ is similar to the one for  $\frac{1}{2}-$H\"older continuity of $L(t)$ for $t\in EP$. However, it is quite different to improve the regularity from
$(1-\epsilon)-$H\"older continuity to Lipschitz continuity.

\vskip 0.3cm
The following result is the key lemma for this section, which shows that $L_n'$ is  uniformly bounded in some local sense when $n$ is sufficiently large.

First, we denote $Y(n)\triangleq\{l|q^2_{N+l-1}\leq n \leq q^2_{N+l}\}.$ $L_0(t)\triangleq\max\{n| g_n(t)~is~of~type~III\}$

\begin{lemma}\label{lemma29} For any fixed $t\in FR,$ there exists some $\lambda_0>0$ and $n_0>q^4_{N+L_0(t)-1}$, such that
     $$\left| L_n'(t')\right|\leq q_{N+L_0(t)-1}^{300\tau^2}$$
     for $n\geq n_0$ , $\lambda\geq\lambda_0$ and $|t'-t|\leq \lambda^{-\frac{c}{2}q_{N+Y(n)}},$ \label{final}
     \end{lemma}
     \begin{remark} The upper bound of $\left| L_n'(t')\right|$ is independent of the scale of $|I_{L_0}|$ and the best estimate is only dependent on $t$ and $\lambda.$
     \end{remark}
     \vskip 0.4cm
\noindent {\it The proof of the Lipschitz continuity for $t\in FR$ by Lemma \ref{lemma29}.}
\begin{proof}
Fix any $t\in FR.$ By Lemma \ref{lemma29}, there exists $n_0$ sufficiently large such that $$\left| L_n'(t)\right|\leq q_{N+L_0(t)-1}^{300\tau^2}$$
for any $n\geq n_0.$  Let $I_0:=[t-\lambda^{-\frac{c}{2}q_{N+Y(n_0)-1}},t+\lambda^{-\frac{c}{2}q_{N+Y(n_0)-1}}].$
Together with Lemma \ref{ap} and the definition of $Y(n)$, for any $\lambda^{-\frac{c}{2}q_{N+Y(n)-1}}\leq |t-t'|\leq\lambda^{-\frac{c}{2}q_{N+Y(n)}},$ it holds that
\begin{align*}
\vert L(t)-L(t')\vert &\leq \vert L_n(t)-L_n(t')\vert+\vert L_{2n}(t)-L_{2n}(t')\vert+\vert L_n(t)+L(t)-2L_{2n}(t)\vert\\&+\vert L_n(t')+L(t')-2L_{2n}(t')\vert\\&\leq C\vert t-t'\vert +2\lambda^{-\frac{c}{2}n}\\&\leq C\vert t-t'\vert+\lambda^{-\frac{c}{2}q^2_{N+L(n)-1}}\\&\leq C\vert t-t'\vert+\lambda^{-\frac{c}{2}q_{N+L(n)-1}}\\&\leq (C+1)\vert t-t'\vert\\&\leq C'\vert t-t' \vert.
\end{align*}


It is clear that as $n$ increases from $n_0$ to $+\infty$, $t'$ goes through $I_0.$
Thus we prove the local Lipschitz continuity of $L(\cdot)$ at t on $I_0$ with the constant $C'$.  \end{proof}
\vskip 0.3cm

We are at the position to prove Lemma \ref{final}.
From the definition, for any $t_0\in FR$, we have that $g_{N+j}(t_0)$ is of type I for all $j\geq L_0.$

Note that for any $t\in [\lambda^{-\frac{c}{2}q_{N+l-1}}+t_0,\lambda^{-\frac{c}{2}q_{N+l}}+t_0],$
$g_l(t)$ is still of type I for any $l> L_0,$ which is followed by the definition of $FR.$ And it follows from Lemma \ref{lmg} that
$$q_{N+L_0-1}^{-C}\leq \left|\frac{\partial g_l(t,x)}{\partial t}\right|\leq q_{N+L_0-1}^{C},$$
where $C$ can be chosen as $100\tau^{100}.$

\vskip0.3cm
Assume
$A_n(x)=A_{n-l}(x)A_l(x)$ and denote \begin{equation}\label{lambda12}\|A_{n-l}(x)\|:=\lambda_1,~\|A_l(x)\|:=\lambda_2,~and~\theta_l(x):=\frac{\pi}{2}-s(A_{n-l}(x))+u(A_l(x)).\end{equation}
We need the two following lemmas later.
\begin{lemma}\label{dAdlambda}

\begin{enumerate}
\item It holds that
$$\frac{1}{\|A_n\|}\frac{d\|A_n\|}{d\lambda_i}<C\lambda_i^{-1},\quad i=1,\ 2.$$


\item If \begin{equation}\label{tj1}\lambda_2^2\leq \lambda_1(resp.\lambda_1^2\leq\lambda_2),\end{equation}
then we have
\begin{equation}\label{tj3}\frac{1}{\|A_n\|}\frac{d\|A_n\|}{d\lambda_1}
=\lambda_1^{-1}+o(\lambda_1^{-1})\quad  (resp.\frac{1}{\|A_n\|}\frac{d\|A_n\|}{d\lambda_2}
=\lambda_2^{-1}+o(\lambda_2^{-1})).\end{equation}


\item If \begin{equation}\label{tj2}|\theta_l|\geq \lambda_m^{-\eta}, ~where~ \lambda_m:=\min\{\lambda_1,\lambda_2\}~ and ~\eta\ll 1,\end{equation}
Then we have

\begin{equation}\label{tj4}\frac{1}{\|A_n\|}\frac{d\|A_n\|}{d\lambda_1}
=\lambda_1^{-1}+o(\lambda_1^{-1}),\quad \frac{1}{\|A_n\|}\frac{d\|A_n\|}{d\lambda_2}=\lambda_2^{-1}+o(\lambda_2^{-1}).\end{equation}


\end{enumerate}


\end{lemma}
\begin{proof}
Without loss of generality, we only consider the estimates on $\frac{1}{\|A_n\|}\frac{d\|A_n\|}{d\lambda_1}$. From Lemma \ref{lemma8}, we have

\begin{align*}
&\frac{1}{\|A_n\|}\frac{\p\|A_n\|}{\p\lambda_1}\\
= & \frac{sgn(\theta)\cdot\lambda_1^{-1}\left((1-\frac{1}{\lambda_1^4\lambda_2^4})\cot\theta_l+(\frac{1}{\lambda_2^4}-\frac{1}{\lambda_1^4})\tan\theta_l\right)}{\sqrt{(1-\frac{1}{\lambda_1^4\lambda_2^4})^2
\cot^2\theta_l+(\frac{1}{\lambda_1^4}-\frac{1}{\lambda_2^4})^2\tan^2\theta_l+2(1+\frac{1}{\lambda_1^4\lambda_2^4})
(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})-\frac{8}{\lambda_1^4\lambda_2^4}}}\\
=&\lambda_1^{-1}\frac{a_1}{\sqrt{a_1^2+a_2^2}},
\end{align*}
where $a_1=(1-\frac{1}{\lambda_1^4\lambda_2^4})\cot\theta_l+(\frac{1}{\lambda_2^4}-\frac{1}{\lambda_1^4})\tan\theta_l$
and $a_2=\frac{2}{\lambda_1^2}(1-\frac{1}{\lambda_2^4})$.

Thus it is easy to obtain $\left|\frac{1}{\|A_n\|}\frac{d\|A_n\|}{d\lambda_1}\right|<C\lambda_1^{-1}.$

Assume the condition (\ref{tj1}) holds, then $\frac{1}{\lambda_1^4}\ll\frac{1}{\lambda_2^4}$, which implies
two terms of $a_1$ share the same sign. Thus
$$
a_1=\cot\theta_l+\frac{1}{\lambda_2^4}\tan\theta_l+o(a_1)>\frac{1}{2\lambda_2^2}.
$$
Since $a_2=\frac{2}{\lambda_1^2}+o(a_2)$, with the help of (\ref{tj1}), we have $a_2\ll|a_1|$.
Assume the condition (\ref{tj2}) holds, then $$\left|(1-\frac{1}{\lambda_1^4\lambda_2^4})\cot\theta_l\right|\gg
\max\{\left|(\frac{1}{\lambda_2^4}-\frac{1}{\lambda_1^4})\tan\theta_l\right|,\ a_2\}.$$ Thus (\ref{tj4}) holds true.
\end{proof}


\vskip 0.3cm
For any $x\in \mathbb{T}$, let $0\le j_1<j_2<\cdots<j_s\le n$ be all the times for $x$ returning to $I_{L},$ where $s$ depends on $x$, $E$.

Let $\hat{I}_{L+1}=B(C_{L+1},\frac{1}{q_{N+L}^{1000C\tau}})$ and define $
{\Omega_{L,1}}=\{x\in \mathbb{T}| T^kx\not\in \hat{I}_{L+1},\ 0\le k\le n\}$ and $\Omega_{L,2}=\mathbb{T}\backslash \Omega_{L,1}.
$
From the Diophantine condition,  there exists at most one $0\le l\le n$ such that $T^{l}x\in \hat{I}_{L+1}$ (for simplicity we only consider one interval of $\hat{I}_{L+1}$ and  still write it as $\hat{I}_{L+1}$).


It is not difficult to see that for the measure of $\Omega_{L,2}$, we have $mes(\hat{I}_{L+1})\le \frac{2}{q_{N+L}^{1000C\tau}}$, which implies
\begin{equation}\label{measure-omega-2}
mes(\Omega_{L,2})\le \frac{2n}{q_{N+L}^{1000C\tau}}.
\end{equation}The following lemma shows that  the contribution of the `bad' set  $\Omega_{L,2}$ for  $L_n'$ can be ignored.

\begin{lemma}\label{Integral-L2}
It holds that
\begin{equation}\label{integral-L2}\frac1n\left|\int_{\Omega_{L,2}} \frac{1}{\|A_n\|}\frac{\partial\|A_n\|}{\partial t}dx\right|\le 2q_{N+L}^{-500C\tau}+C \frac{\log q_{N+L+1}}{q_{N+L}}.\end{equation}
\end{lemma}
Before proving Lemma \ref{Integral-L2}, we need some notations. Consider two matrices $A$ and $B$. Assume by the singular value decomposition it holds that
 $BA=R_{\phi_l}\cdot diag(\|B\|, \|B\|^{-1})\cdot R_{\phi_c}\cdot diag(\|A\|, \|A\|^{-1})\cdot R_{\phi_r}.$ Then we denote
 $\theta(A,B)=\phi_c$.
Moreover, we define $\hat{I}_i=\{x\in \mathbb{S}^1|dist(x, C_i)\le \frac{1}{q_{N+i}^{1000C\tau}}\}$, where $C_i$ is the critical point set at the scale i and dist is the distance. To prove Lemma \ref{Integral-L2}, we need the following estimate.

\begin{lemma}\label{patheta} For $x\in I_i$ with $L_1\le i\le L-1$, let $\hat{A}_i(x,t)=A_{{r^+_i}}(x,t)\cdot A_{{-r^-_i}}(T^{-r^-_i}x,t),$
$\lambda_1=\|A_{{r^-_i}}(T^{-r^-_i}x,t)\|$, $\lambda_2=\|A_{{r^+_i}}(x,t)\|$,
 and $\theta_l=\theta(A_{{r^+_i}}(x,t), A_{{-r^-_i}}(T^{-r^-_i}x,t))$.
Then it holds that
$$\left|\int_{\hat{I}_i\backslash \hat{I}_{i+1}}\frac{1}{\|\hat{A}_i\|}\frac{\partial \|\hat{A}_i\|}{\partial \theta_l}\frac{\partial \theta_l}{\partial t}dx\right|\le
\frac{C}{q_{N+i}}\log \frac{q_{N+i}}{ q_{N+i-1}}
$$
and
$$\left|\int_{\hat{I}_L}\frac{1}{\|\hat{A}_L\|}\frac{\partial \|\hat{A}_L\|}{\partial \theta_l}\frac{\partial \theta_l}{\partial t}dx\right|\le
\frac{C}{q_{N+L}}\log \frac{q_{N+L}}{ q_{N+L-1}}.
$$
\end{lemma}
\begin{proof}
From Lemma \ref{lemma8}, we have \begin{align*}\frac{1}{\|\hat{A}_i\|}\frac{\partial \|\hat{A}_i\|}{\partial \theta_l}=
\frac{sgn(\theta_l)(1-\lambda_2^{-4})(1-\lambda_1^{-4})}{\sqrt{(1-\frac{1}{\lambda_1^4\lambda_2^4})^2\cot^2\theta_l+
(\frac{1}{\lambda_1^4}-\frac{1}{\lambda_2^4})^2\tan^2\theta_l+2(1+\frac{1}{\lambda_1^4\lambda_2^4})(\frac{1}{\lambda_1^4}
+\frac{1}{\lambda_2^4})-\frac{8}{\lambda_1^4\lambda_2^4}}}.
\end{align*}
\vskip0.3cm

Denote $K(i)=\max \{s<i| g_{s-1}~is~type~I~and~g_s~is~type~III\},$
then, it follows from lemma \ref{lmg} that
\begin{enumerate}
\item[O1:] $sgn(\frac{\partial \|\hat{A}_i\|}{\partial \theta_l})=sgn(\theta_l);$
\item[O2:] $M_1(i)\leq \left|\frac{\partial \theta_l(x,t)}{\partial t}\right|\leq M_2(i)\leq \lambda^{K(i)}\ll q_{N+i}, $ for any $x\in \hat{I}_{i}$ and $|t-t_0|\leq \lambda^{-\frac{c}{2}q_{N+i}},$ where $M_1$ and $M_2$ are some proper constants with $M_2-M_1\leq Cq_{N+i}^{-1000\tau}.$ In fact, for any $(x,t),(x',t')\in I_i\times J_t(\triangleq [t_0-\lambda^{-\frac{c}{2}q_{N+i}},t_0+\lambda^{-\frac{c}{2}q_{N+i}}]),$ by Lagrange Mean Value Theorem, we have
    \begin{equation}\begin{array}{ll}&\left|\frac{\partial\theta_l(x,t)}{\partial t}-\frac{\partial\theta_l(x',t')}{\partial t}\right|\\&\leq \left|(x-x')\frac{\partial^2 \theta_l(x+(x'-x)q,t+(t'-t)q)}{\partial t\partial x}+(t-t_0)\frac{\partial^2 \theta_l(x+(x'-x)q,t+(t'-t)q)}{\partial t^2}\right|\\&\leq \sqrt{(\frac{\partial^2 \theta_l(x+(x'-x)q,t+(t'-t)q)}{\partial t\partial x})^2+(\frac{\partial^2 \theta_l(x+(x'-x)q,t+(t'-t)q)}{\partial t^2})^2}\sqrt{(x-x')^2+(t-t_0)^2}.
    \end{array}\end{equation}

    Direct calculation shows that $\frac{\partial^2 \theta_l(x,t)}{\partial t^2},\frac{\partial^2 \theta_l(x,t)}{\partial t\partial x}\sim \lambda^{K(i)}\ll q_{N+i}.$ Therefore, $\left|\frac{\partial\theta_l(x,t)}{\partial t}-\frac{\partial\theta_l(x',t')}{\partial t}\right|\leq \lambda^{K(i)}|\hat{I_i}|,$ which implies O2.
\item[O3:] $C_1(i)\leq \left|\frac{\partial \theta_l(x,t)}{\partial x}\right|\leq C_2(i)\leq \lambda^{k(i)}\ll q_{N+i}, $ for any $x\in \hat{I}_{i+}$ and $|t-t_0|\leq \lambda^{-\frac{c}{2}q_{N+i}},$ where $C_1$ and $C_2$ are some proper constants with $C_2-C_1\leq Cq_{N+i}^{-1000\tau}.$ The proof is similar to O2.
\item[O4:] For any $x,y\in \hat{I}_{i},i=1,2$ $|\frac{\lambda_i(x)}{\lambda_i(y)}|\leq 1+\frac{e}{q^{900C\tau}_{N+i}}.$ In fact, it follows from lemma \ref{lmwz} that
\begin{align*}&\vert\log \lambda_j(x)-\log\lambda_j(y)\vert \leq \int_{\hat{I}_{i+1}}|\frac{d\log\lambda_j}{dx}|dx\\
&\leq \int_{\hat{I}_{i}}|\frac{1}{\lambda_j}\frac{d \lambda_j}{dx}|dx\\&\leq |\hat{I}_{i+1}|\cdot \max\{l,q^4_{N+i}-l,q_{N+i}\}^C\\&\leq \frac{2}{q^{1000C\tau}_{N+i}}\cdot q^{4C}_{N+i}\\&\leq \frac{1}{q^{900C\tau}_{N+i}}.
\end{align*}

Therefore, $|\frac{\lambda_i(x)}{\lambda_i(y)}|\leq \exp^{\frac{1}{q^{900C\tau}_{N+i}}}\leq 1+\frac{e}{q^{900C\tau}_{N+i}}.$
\item[O5:] For any $x\in \hat{I}_{i+1}$ and $j=1,2,$ we have $q^{1000C\tau}_{N+i+1}\ll \lambda^{\frac{1}{2}r_i} \leq \lambda_j\leq \lambda^{2R_i}.$ (Recall that $R_i=\max\{r^{+}_i,r^{-}_i\}$)\end{enumerate}

 O2 and O3 in the above show that in the domain under consideration, the change of $\theta_l$ is much slower than that of $\frac{1}{\|\hat{A}_i\|}\frac{\partial \|\hat{A}_i\|}{\partial \theta_l}$ such that it can roughly be regarded as a constant. O4 tells us that the difference between $\lambda_i(x)$ and $\lambda_i(y)$ is small and it implies that $\lambda^{-}_i\leq \lambda_i(x)\leq \lambda^{+}_i,$ where $(1\pm \frac{1}{q^{800C\tau}_{N+L}})\lambda_i(0)\triangleq \lambda^{\pm}_i.$

\textbf{The first inequality:}It follows from Lemma \ref{keylemma}, O1 and O5 that
\begin{equation}\label{lips'}\end{equation}
\begin{align*}&\left|\int_{\hat{I}_i\backslash \hat{I}_{i+1}}\frac{1}{\|\hat{A}_i\|}\frac{\partial \|\hat{A}_i\|}{\partial \theta_l}\frac{\partial \theta_l}{\partial t}dx\right|\\&\leq \left|\int_{c_{L+1}+\frac{1}{q_{N+i+1}^{1000C\tau}}}^{c_{L+1}+\frac{1}{q_{N+i}^{1000C\tau}}}\frac{(1+O(\frac{2}{\lambda^{2r_i}}))sgn(\theta_l)M_2}{\sqrt{\cot^2\theta_l+\frac{2}{\lambda^{8R_i}}}}dx\right.
\\&+\left.\int_{c_{L+1}-\frac{1}{q_{N+i}^{1000C\tau}}}^{c_{L+1}-\frac{1}{q_{N+i+1}^{1000C\tau}}}\frac{(1-O(\frac{2}{\lambda^{2r_i}}))sgn(\theta_l))M_1\cot\theta_l}{\cot^2\theta_l+
\frac{2}{\lambda^{2r_i}}}dx\right|.
\end{align*}

By O2 and O3, we have

\begin{align*}\eqref{lips}&
\leq \left|\int_{\frac{1}{q_{N+i+1}^{1000C\tau}}}^{\frac{1}{q_{N+i}^{1000C\tau}}}M_2\frac{1}{\sqrt{C_1^2x^2+\frac{2}{\lambda^{8R_i}}}}dx\right.\\
&-\left.\int_{\frac{1}{q_{N+i+1}^{1000C\tau}}}^{\frac{1}{q_{N+i}^{1000C\tau}}}M_1\frac{C_1x}{C_2^2x^2+\frac{2}{\lambda^{2r_i}}}dx\right|+\\&\left|O(\frac{2}{\lambda^{2r_i}})\int_{0 }^{\frac{1}{q_{N+L}^{1000C\tau}}}2M_2\frac{1}{\sqrt{C_1^2x^2+\frac{2}{\lambda^{8R_i}}}}dx\right|.\end{align*}

If we denote $\delta_{i}\triangleq\frac{1}{q_{N+i}^{1000C\tau}},$ then it's clear that
\begin{align*}&\eqref{lips'} \leq \left|\frac{M_2}{C_1}\ln\dfrac{\left|\sqrt{C^2_1\lambda^{8R_i}\delta_i^2+2}+C_1\lambda^{4R_i}\delta_i\right|}{\left|\sqrt{C^2_1\lambda^{8R_i}\delta_{i+1}^2+2}+C_1\lambda^{4R_i}\delta_{i+1}\right|}-\frac{C_1M_1}{C_2^2}\ln\dfrac{\left|\sqrt{C_2^2\lambda^{2r_i}\delta_i^2+2}\right|}{\left|\sqrt{C_2^2\lambda^{2r_i}{\delta_{i+1}^2+2}}\right|}\right|
\\&\triangleq \left|\frac{M_2}{C_1}\log |U_1(C_1,\lambda,R_i,\delta_i,\delta_{i+1})|-\frac{C_1M_1}{C_2^2}\log |U_2(C_2,\lambda,r_i,\delta_i,\delta_{i+1})\right|,\end{align*}
where $U_1=\log|\frac{\delta_i}{\delta_{i+1}}+o(\lambda^{-3R_i})|$ and $U_2=\log|\frac{\delta_i}{\delta_{i+1}}+o(\lambda^{-\frac{1}{3}r_i})|.$

Thus \begin{align*}&\eqref{lips'} \leq |\frac{M_2}{C_1}-\frac{C_1M_1}{C_2^2}|\max\{\log|U_1|,\log|U_2|\}+|\log|\frac{U_1}{U_2}||\max\{\frac{M_2}{C_1},\frac{C_1M_1}{C_2^2}\}\\&\leq (|\frac{M_2}{C_1}-\frac{M_1}{C_2}|+\frac{M_1}{C_2}|1-\frac{C_1}{C_2}|)\log|\delta_{i+1}|+C\log|\frac{\frac{\delta_i}{\delta_{i+1}}+o(\lambda^{-3R_i})}{\frac{\delta_i}{\delta_{i+1}}+o(\lambda^{-\frac{1}{3}r_i})}|
\\&\leq C\frac{\log q_{N+i+1}}{q^{8\tau}_{N+i}}+o(\lambda^{-\frac{1}{10}r_i})\\&\leq \frac{C}{q_{N+i}}\log \frac{q_{N+i}}{ q_{N+i-1}}\end{align*}

\textbf{The second inequality:}
With the help of lemma 8 and O1, we have
\begin{equation}\label{lips}\end{equation}
\begin{align*}&\left|\int_{\hat{I}_L}\frac{1}{\|\hat{A}_i\|}\frac{\partial \|\hat{A}_i\|}{\partial \theta_l}\frac{\partial \theta_l}{\partial t}dx\right|\\&\leq \left|\int_{c_{L+1}}^{c_{L+1}+\frac{1}{q_{N+L}^{1000C\tau}}}\frac{(1+O(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4}))sgn(\theta_l))M_2}{\sqrt{\cot^2\theta_l+(\frac{1}{\lambda_1^4}-
\frac{1}{\lambda_2^4})^2\tan^2\theta_l+\frac{2}{\lambda_1^4}+
\frac{2}{\lambda_2^4}}}dx\right.\\&+\left.\int_{c_{L+1}-\frac{1}{q_{N+L}^{1000C\tau}}}^{c_{L+1}}\frac{(1-O(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4}))sgn(\theta_l))M_1}{\sqrt{\cot^2\theta_l+(\frac{1}{\lambda_1^4}-
\frac{1}{\lambda_2^4})^2\tan^2\theta_l+\frac{2}{\lambda_1^4}+
\frac{2}{\lambda_2^4}}}dx\right|.
\end{align*}
By O2 and O3, we have
\begin{align*}\eqref{lips}&
\leq \left|\int_{0}^{\frac{1}{q_{N+L}^{1000C\tau}}}M_2\frac{1}{\sqrt{C_1^2x^2+(\frac{1}{\lambda_1^4}-\frac{1}{\lambda_2^4})^2C_2^{-2}x^{-2}+\frac{2}{\lambda_1^4}+\frac{2}{\lambda_2^4}}}dx\right.\\
&-\left.\int_{0}^{\frac{1}{q_{N+L}^{1000C\tau}}}M_1\frac{1}{\sqrt{C_2^2x^2+(\frac{1}{\lambda_1^4}-\frac{1}{\lambda_2^4})^2C_1^{-2}x^{-2}+\frac{2}{\lambda_1^4}+\frac{2}{\lambda_2^4}}}dx\right|+\\&\left|\int_{0 }^{\frac{1}{q_{N+L}^{1000C\tau}}}2M_2\frac{O(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})}{\sqrt{C_1^2x^2+\frac{2}{\lambda_1^4}+\frac{2}{\lambda_2^4}}}dx\right|.\end{align*}
By O4, we only need to consider the following two cases:
\begin{enumerate}
\item[\textbf{CASE1:}] dist$\{[\lambda^{-}_1,\lambda^{+}_1],[\lambda^{-}_2,\lambda^{+}_2]\}\leq \frac{1}{q^{600C\tau}_{N+L}}(\lambda^{+}_1+\lambda^{+}_2).$

In this case, note that \begin{align*}&0\leq |\frac{1}{\lambda^4_1(x)}-\frac{1}{\lambda^4_2(x)}|=\frac{|\lambda_1-\lambda_2|(\lambda_1+\lambda_2)(\lambda_1^2+\lambda_2^2)|}{\lambda_1^4\lambda_2^4}\\&\leq \frac{\max\{|\lambda^{+}_1-\lambda^{-}_2|,|\lambda^{+}_2-\lambda^{-}_1|\}(\lambda^{+}_1+\lambda^{+}_2)((\lambda^{+}_1)^2+(\lambda^{+}_2)^2)|}{(\lambda^{-}_1)^4(\lambda^{-}_2)^4}
\\&\leq \frac{1}{q^{700C\tau}_{N+L}}\frac{(\lambda^{+}_1+\lambda^{+}_2)^4}{(\lambda^{-}_1)^4(\lambda^{-}_2)^4}\\
&\leq \frac{1}{q^{700C\tau}_{N+L}}\frac{(1+\frac{1}{q^{800C\tau}_{N+L}})^4(\lambda^{-}_1+\lambda^{-}_2)^4}{(\lambda^{-}_1)^4(\lambda^{-}_2)^4}\\&\leq \frac{2^4}{q^{700C\tau}_{N+L}}\frac{(2^4-1)((\lambda^{-}_1)^4+(\lambda^{-}_2)^4)}{(\lambda^{-}_1)^4(\lambda^{-}_2)^4}\\&\leq \frac{1}{q^{600C\tau}_{N+L}}\frac{(\lambda^{-}_1)^4+(\lambda^{-}_2)^4}{(\lambda^{-}_1)^4(\lambda^{-}_2)^4} .\end{align*}

Hence,
\begin{align*}\eqref{lips}&
\leq \left|\int_{0}^{\frac{1}{q_{N+L}^{1000C\tau}}}M_2\frac{1}{\sqrt{C_1^2x^2+\frac{2}{(\lambda^{+}_1)^4}+\frac{2}{(\lambda^{+}_2)^4}}}dx\right.\\
&-\left.\int_{0}^{\frac{1}{q_{N+L}^{1000C\tau}}}M_1\frac{1}{\sqrt{C_2^2x^2+(\frac{1}{q^{600C\tau}_{N+L}}\frac{(\lambda^{-}_1+\lambda^{-}_2)^4}{(\lambda^{-}_1)^4(\lambda^{-}_2)^4})^2C_1^{-2}x^{-2}+\frac{2}{(\lambda^{-}_1)^4}+\frac{2}{(\lambda^{-}_2)^4}}}dx\right|+\\&\left|\int_{0 }^{\frac{1}{q_{N+L}^{1000C\tau}}}2M_2\frac{O(\frac{1}{(\lambda^{-}_1)^4}+\frac{1}{(\lambda^{-}_2)^4})}{\sqrt{C_1^2x^2+\frac{2}{(\lambda^{+}_1)^4}+\frac{2}{(\lambda^{+}_2)^4}}}dx\right|.\end{align*}
By a direct calculation, see \textbf{A.2}, we have

\begin{align*}\eqref{lips}&\leq
\left.\left(\left|\frac{M_2}{2C_1}\log S_{L,+}(x, C_1,\lambda^{+}_1,\lambda^{+}_2)
-\frac{M_1}{2C_2}\log S_{L,-}(x,C_2,C_1,\lambda^{-}_1,\lambda^{-}_2)\right|\right)\right|_{0}^{\frac{1}{q_{N+L}^{1000C\tau}}} \\&+O(\frac{1}{(\lambda^{-}_1)^4}+\frac{1}{(\lambda^{-}_2)^4})\left.\frac{2M_2}{C_1}\log\left(\sqrt{(\lambda^{+})_1^4(\lambda^{+})_2^4C_1^2x^2+2((\lambda^{+})_1^4+(\lambda^{+})_2^4)}+C_1(\lambda^{+})_1^2(\lambda^{+})_2^2x\right)\right|_{0}^{\frac{1}{q_{N+L}^{1000C\tau}}}, \end{align*}
where
\begin{align*}&S_{L,+}(x, C_1,\lambda^{+}_1,\lambda^{+}_2)\\&=\log\left|\sqrt{(\lambda^{+}_1)^8(\lambda^{+}_2)^8{C_1}^4x^4+2C_1^2(\lambda^{+}_1)^4(\lambda^{+}_2)^4((\lambda^{+}_2)^4+2(\lambda^{+}_2)^4)x^2}+(\lambda^{+}_2)^4(\lambda^{+}_2)^4C_1^2x^2+(\lambda^{+}_1)^4+(\lambda^{+}_2)^4 \right|;\end{align*}

\begin{align*}
&S_{L,-}(x,C_1,C_2,\lambda^{-}_1,\lambda^{-}_2)\\&=\log\left|\sqrt{(\lambda^{-}_1)^8(\lambda^{-}_2)^8C_2^4x^4+2C_2^2(\lambda^{-}_1)^4(\lambda^{-}_2)^4((\lambda^{-}_1)^4+(\lambda^{-}_2)^4)x^2+\frac{C_2^2}{C_1^2}\frac{1}{q^{1200C\tau}_{N+L}}((\lambda^{-}_1)^4+(\lambda^{-}_2)^4)^2}\right.\\&+\left.((\lambda^{-}_1)^4(\lambda^{-}_2)^4C_2^2x^2+(\lambda^{-}_1)^4+(\lambda^{-}_2)^4)\right|.
\end{align*}

If we denote $\delta_{L}\triangleq\frac{1}{q_{N+L}^{1000C\tau}},$ then it's clear that
\begin{equation*}\label{dengs}
\begin{array}{ll}\eqref{lips}&\le \left|\frac{M_2}{2C_1}\log \frac{S_{L,+}(\delta_L)}{S_{L,+}(0)}-\frac{M_1}{2C_2}\log \frac{S_{L,-}(\delta_L)}{S_{L,-}(0)}\right|
+o((\max\{(\lambda^{-}_1)^{-3},(\lambda^{-}_2)^{-3}\})(\triangleq o(\lambda_m^{-3})).\end{array}\end{equation*}
It follows from lemma \ref{lmg} that
\begin{equation}\label{e1}\min\{C_1,C_2\}\geq c_1, \min\{M_1,M_2\}\geq c_2,\end{equation}
where $c_1\sim |I_{L_1}|$ and $c_2$ is an absolutely constant from lemma \ref{lmg}.

By a direct calculation, we have
\begin{equation}\label{e2}\max\left\{\left|\frac{S_{L,+}(\delta_L)}{S_{L,+}(0)}\right|, \left|\log \frac{S_{L,-}(\delta_L)}{S_{L,-}(0)}\right|\right\}\leq C\log((\lambda^{+}_1)^4(\lambda^{+}_2)^4)\leq Cq^4_{N+L}\log\lambda.\end{equation}

On the other hand, from the fact $\lambda_m\gg \delta_q$ and O4 we have
\begin{equation}\label{e3}\begin{array}{ll}&\left|\log \frac{S_{L,+}(\delta_L)}{S_{L,+}(0)}-\log \frac{S_{L,-}(\delta_L)}{S_{L,-}(0)}\right|\\&=\left|\log \frac{S_{L,+}(\delta_L)}{S_{L,-}(\delta_L)}+\log \frac{S_{L,-}(0)}{S_{L,+}(0)}\right|\\
\\
&\leq \log\left|\frac{(\lambda^{+}_1)^4(\lambda^{+}_2)^4C_1^2+o(\lambda_m^{-3})}{(\lambda^{-}_1)^4(\lambda^{-}_2)^4C_2^2+o(\lambda_m^{-3})}\right|+\log\left|\frac{(1+\frac{C_2}{C_1q^{600C\tau}_{N+L}})((\lambda^{-}_1)^4+(\lambda^{-}_2)^4)}{(\lambda^{+}_1)^4+(\lambda^{+}_2)^4)}\right|
\\
&\leq \left|\log \frac{C_1^2}{C_2^2(1+\frac{1}{q^{900C\tau}})^8}+o(\lambda_m^{-3})\right|+\log\left|(1+\frac{C_2}{C_1q^{600C\tau}_{N+L}})(1+\frac{1}{q^{900C\tau}_{N+L}})^4\right|\\
\\&\leq (1+\frac{1}{q^{700C\tau}_{N+L}})^8\frac{|C_1-C_2|}{C_2}+o(\lambda_m^{-2.5})+\frac{5C_2}{C_1q^{500C\tau}_{N+L}}\leq \frac{1}{c_1}\frac{1}{q^{8\tau}_{N+i}}.
\end{array}.\end{equation}
In the above, we use the inequality $\log(1+x)\leq x$ for $x\geq 0.$

Hence, combining \eqref{e1},\eqref{e2}, and \eqref{e3}, we have
\begin{align*}
&\eqref{lips}\leq |\frac{M_2}{C_1}-\frac{M_1}{C_2}|\max\left\{\log \left|\frac{S_{L,+}(\delta_L)}{S_{L,+}(0)}\right|,\log \left|\frac{S_{L,-}(\delta_L)}{S_{L,-}(0)}\right|\right\}\\
\\
&+\max\{|\frac{M_2}{C_1}|,|\frac{M_1}{C_2}|\}\left|\log \frac{S_{L,+}(\delta_L)}{S_{L,+}(0)}-\log \frac{S_{L,-}(\delta_L)}{S_{L,-}(0)}\right|+o(\lambda_m^{-3}) \\ &\leq C\left|\frac{M_2}{C_1}-\frac{M_1}{C_2}\right|q^4_{N+L}\log\lambda+\frac{M_1}{c_1}\cdot\frac{1}{c_1}\frac{1}{q^{8\tau}_{N+i}}\\&\leq
\frac{C}{C_1C_2}[|M_1-M_2|C_2+|C_1-C_2|M_1]q^4_{N+L}\log\lambda+C\frac{1}{q^{8\tau}_{N+L}}\\&\leq C\frac{q^4_{N+L}}{q_{N+L}^{8\tau}}\\&\leq \frac{C}{q_{N+L}} \frac{\log q_{N+L}}{q_{N+L-1}}.
\end{align*}

\item[\textbf{CASE2:}]dist$\{[\lambda^{-}_1,\lambda^{+}_1],[\lambda^{-}_2,\lambda^{+}_2]\}> \frac{1}{q^{600C\tau}_{N+L}}(\lambda^{+}_1+\lambda^{+}_2).$

Without loss of generality, we assume $\lambda^{-}_1>\lambda^{+}_2,$ which implies $\lambda_1(x)>\lambda_2(x)$ for any $x\in \hat{I}_L$ ,$\frac{\lambda^{+}_2}{\lambda^{+}_1}\leq 1-\frac{1}{q^{600C\tau}_{N+L}}$ and $\frac{\lambda^{-}_2}{\lambda^{-}_1}\leq 1-\frac{1}{q^{600C\tau}_{N+L}}$

It follows from O4 that \begin{align*}&\lambda_1(x)-\lambda_2(x)\\&>dist\{[\lambda^{-}_1,\lambda^{+}_1],[\lambda^{-}_2,\lambda^{+}_2]\}\\&> \frac{1}{q^{600C\tau}_{N+L}}(\lambda^{+}_1+\lambda^{+}_2)\\&=\frac{1}{q^{600C\tau}_{N+L}}(1+\frac{1}{q^{900C\tau}_{N+L}})(\lambda_1(0)+\lambda_2(0))\\&>\frac{1}{q^{600C\tau}_{N+L}}(1+\frac{1}{q^{900C\tau}_{N+L}})(1-\frac{1}{q^{900C\tau}_{N+L}})(\lambda_1(x)+\lambda_2(x))>\frac{1}{q^{590C\tau}_{N+L}}(\lambda_1(x)+\lambda_2(x)).\end{align*}

Hence, \begin{align*}&|\frac{1}{\lambda_2(x)}-\frac{1}{\lambda_1(x)}|\\&\geq |\frac{1}{\lambda^{+}_2}-\frac{1}{\lambda^{-}_1}|\\&= |\frac{1}{\lambda^{+}_2}-\frac{1}{\lambda^{+}_1}|\frac{1-(\frac{\lambda^{+}_2}{\lambda^{-}_1})^4}{1-(\frac{\lambda^{+}_2}{\lambda^{+}_1})^4}= |\frac{1}{\lambda^{+}_2}-\frac{1}{\lambda^{+}_1}|(1-\frac{(\frac{\lambda^{+}_2}{\lambda^{-}_1})^4-(\frac{\lambda^{+}_2}{\lambda^{+}_1})^4}{1-(\frac{\lambda^{+}_2}{\lambda^{+}_1})^4})=|\frac{1}{\lambda^{+}_2}-\frac{1}{\lambda^{+}_1}|(1-\frac{(\frac{\lambda^{+}_2}{(1-\frac{1}{q^{800C\tau}_{N+L}})\lambda_1(0)})^4-(\frac{\lambda^{+}_2}{\lambda^{+}_1})^4}{1-(\frac{\lambda^{+}_2}{\lambda^{+}_1})^4})
\\& \geq |\frac{1}{\lambda^{+}_2}-\frac{1}{\lambda^{+}_1}|(1-\frac{(\frac{1+\frac{1}{q^{800C\tau}_{N+L}}}{1-\frac{1}{q^{800C\tau}_{N+L}}})^4(\frac{\lambda^{+}_2}{\lambda^{+}_1})^4-(\frac{\lambda^{+}_2}{\lambda^{+}_1})^4}{1-(\frac{\lambda^{+}_2}{\lambda^{+}_1})^4})\\&\geq |\frac{1}{\lambda^{+}_2}-\frac{1}{\lambda^{+}_1}|(1-\frac{((1+\frac{1}{q^{790C\tau}_{N+L}})-1)\cdot 1}{\frac{1}{q^{600C\tau}_{N+L}}})\geq (1-\frac{1}{q^{100C\tau}_{N+L}})|\frac{1}{\lambda^{+}_2}-\frac{1}{\lambda^{+}_1}|
\end{align*}\begin{equation}\label{lambda1}\end{equation}

Similarly, we have \begin{align*}&|\frac{1}{\lambda_2(x)}-\frac{1}{\lambda_1(x)}|\\&\leq |\frac{1}{\lambda^{-}_2}-\frac{1}{\lambda^{-}_1}|(1+\frac{1}{q^{100C\tau}_{N+L}})
\end{align*}\begin{equation}\label{lambda2}\end{equation}

With the help of \eqref{lambda1} and \eqref{lambda2}, we have
\begin{align*}\eqref{lips}&
\leq \left|\int_{0}^{\frac{1}{q_{N+L}^{1000C\tau}}}M_2\frac{1}{\sqrt{C_1^2x^2+(1-\frac{1}{q^{100C\tau}_{N+L}})^2(\frac{1}{(\lambda^{+}_1)^4}-\frac{1}{(\lambda^{+}_2)^4})^2C_2^{-2}x^{-2}+\frac{2}{(\lambda^{+}_1)^4}+\frac{2}{(\lambda^{+}_2)^4}}}dx\right.\\
&-\left.\int_{0}^{\frac{1}{q_{N+L}^{1000C\tau}}}M_1\frac{1}{\sqrt{C_2^2x^2+(1+\frac{1}{q^{100C\tau}_{N+L}})^2(\frac{1}{(\lambda^{-}_1)^4}-\frac{1}{(\lambda^{-})_2^4})^2C_1^{-2}x^{-2}+\frac{2}{(\lambda^{-}_1)^4}+\frac{2}{(\lambda^{-})_2^4}}}dx\right|+o(\lambda_m^{-3}).\end{align*}

By a direct calculation, we have

\begin{align*}\eqref{lips}&\leq
\left.\left(\left|\frac{M_2}{2C_1}\log S_L(x, C_1,C_2,\lambda^{+}_1,\lambda^{+}_2)
-\frac{M_1}{2C_2}\log S_L(x,C_2,C_1,\lambda^{-}_1,\lambda^{-}_2)\right|\right)\right|_{0}^{\frac{1}{q_{N+L}^{1000C\tau}}} \\&+o(\lambda_m^{-3}), \end{align*}
where
\begin{align*}&S_L(x, C_1,C_2,\lambda^{+}_1,\lambda^{+}_2)\\&=\log\left|\sqrt{(\lambda^{+}_1)^8(\lambda^{+}_2)^8{C_1}^4x^4+2C_1^2(\lambda^{+}_1)^4(\lambda^{+}_2)^4((\lambda^{+}_2)^4+2(\lambda^{+}_2)^4)x^2+(1-\frac{1}{q^{100C\tau}_{N+L}})^2((\lambda^{+}_1)^4-(\lambda^{+}_2)^4)^2}\right.\\&+\left.(\lambda^{+}_2)^4(\lambda^{+}_2)^4C_1^2x^2+(\lambda^{+}_1)^4+(\lambda^{+}_2)^4 \right|;\end{align*}

By a direct calculation, we have
\begin{equation}\label{e4}\max\left\{\left|\frac{S_L(\delta_L,C_1,C_2,\lambda^{+}_1,\lambda^{+}_2)}{S_L(0,C_1,C_2,\lambda^{+}_1,\lambda^{+}_2)}\right|, \left|\log \frac{S_L(\delta_L,C_2,C_1,\lambda^{-}_1,\lambda^{-}_2)}{S_L(0,C_2,C_1,\lambda^{+}_1,\lambda^{+}_2)}\right|\right\}\leq C\log((\lambda^{+}_1)^4(\lambda^{+}_2)^4)\leq Cq^4_{N+L}\log\lambda.\end{equation}

It holds from direct calculation that the following two facts holds:
\begin{enumerate}
\item[\textbf{F1:}] For $x_1,x_2,y_1,y_2\in \R^+$ satisfying $\frac{x_1}{x_2}\geq \frac{y_1}{y_2}$ and $y_2\geq x_2$, the following inequality holds true:
$$\frac{x_1+y_1}{x_2+y_2}\leq \frac{1}{2}(\frac{x_1}{x_2}+\frac{y_1}{y_2}).$$
\item[\textbf{F2:}] For $a_1,a_2,b_1,b_2\in \R^+$ satisfying $a_2\neq b_2$ and $\frac{a_1}{a_2}=\frac{b_1}{b_2},$ the following inequality holds true:
$$\frac{a_1-b_1}{a_2-b_2}=\frac{a_1+a_2}{b_1+b_2}.$$
\end{enumerate}

Similar to CASE1, it follows from the fact $\lambda_m\gg \delta_q$ and O4 we have
\begin{equation}\label{e5}\begin{array}{ll}&\left|\log \frac{S_L(\delta_L,C_1,C_2,\lambda^{+}_1,\lambda^{+}_2)}{S_L(0,C_1,C_2,\lambda^{+}_1,\lambda^{+}_2)}-\log \frac{S_L(\delta_L,C_2,C_1,\lambda^{-}_1,\lambda^{-}_2)}{S_L(0,C_2,C_1,\lambda^{-}_1,\lambda^{-}_2)}\right|\\&=\left|\log \frac{S_L(\delta_L,C_1,C_2,\lambda^{+}_1,\lambda^{+}_2)}{S_L(\delta_L,C_2,C_1,\lambda^{-}_1,\lambda^{-}_2)}+\log \frac{S_L(0,C_2,C_1,\lambda^{-}_1,\lambda^{-}_2)}{S_L(0,C_1,C_2,\lambda^{+}_1,\lambda^{+}_2)}\right|\\
\\
&\leq \log\left|\frac{(\lambda^{+}_1)^4(\lambda^{+}_2)^4C_1+o(\lambda_m^{-3})}{(\lambda^{-}_1)^4(\lambda^{-}_2)^4C_2+o(\lambda_m^{-3})}\right|+\log\left|\frac{C_2(1+\frac{1}{q^{100C\tau}})|(\lambda^{-}_1)^4-(\lambda^{-}_2)^4|+C_1((\lambda^{-}_1)^4+(\lambda^{-}_2)^4)}{C_1(1-\frac{1}{q^{100C\tau}})|(\lambda^{+}_1)^4-(\lambda^{+}_2)^4|+C_2((\lambda^{+}_1)^4+(\lambda^{+}_2)^4)}\right|
\\&\leq \log\left|\frac{(\lambda^{+}_1)^4(\lambda^{+}_2)^4C_1+o(\lambda_m^{-3})}{(\lambda^{-}_1)^4(\lambda^{-}_2)^4C_2+o(\lambda_m^{-3})}\right|+\log\frac{1}{2}\left(\frac{C_2(1+\frac{1}{q^{100C\tau}})|(\lambda^{-}_1)^4-(\lambda^{-}_2)^4|}{C_1(1-\frac{1}{q^{100C\tau}})|(\lambda^{+}_1)^4-(\lambda^{+}_2)^4|}+\left|\frac{C_1((\lambda^{-}_1)^4+(\lambda^{-}_2)^4)}{C_2((\lambda^{+}_1)^4+(\lambda^{+}_2)^4)}\right|\right)
\\
&\leq \left|\log \frac{C_1^2}{C_2^2(1+\frac{1}{q^{900C\tau}})^8}+o(\lambda_m^{-3})\right|+\log\frac{1}{2}\left|(1+\frac{1}{c_1 q^{8\tau}_{N+L}})(1+\frac{1}{q^{90C\tau}_{N+L}})(1+\frac{10}{q^{800C\tau}_{N+L}})+(1+\frac{1}{c_1q^{8\tau}_{N+L}})(1+\frac{10}{q^{800C\tau}_{N+L}})\right|\\
\\&\leq (1+\frac{1}{q^{700C\tau}_{N+L}})^8\frac{|C_1-C_2|}{C_2}+o(\lambda_m^{-2.5})+\frac{1}{c_1q^{8\tau}_{N+L}}\leq \frac{2}{c_1}\frac{1}{q^{8\tau}_{N+i}},
\end{array}\end{equation}
where the third inequality is due to \textbf{F1} and \textbf{F2}.

Therefore, it follows from \eqref{e1}, \eqref{e4} and \eqref{e5} that
\begin{align*}
&\eqref{lips}\leq |\frac{M_2}{C_1}-\frac{M_1}{C_2}|\max\left\{\log \left|\frac{S_L(\delta_L,C_1,C_2,\lambda^{+}_1,\lambda^{+}_2)}{S_{L,+}(0,C_1,C_2,\lambda^{+}_1,\lambda^{+}_2)}\right|,\log \left|\frac{S_L(\delta_L,,C_2,C_1,\lambda^{-}_1,\lambda^{-}_2)}{S_L(0,C_2,C_1,\lambda^{-}_1,\lambda^{-}_2)}\right|\right\}\\
\\
&+\max\{|\frac{M_2}{C_1}|,|\frac{M_1}{C_2}|\}\left|\log \left|\frac{S_L(\delta_L,C_1,C_2,\lambda^{+}_1,\lambda^{+}_2)}{S_{L,+}(0,C_1,C_2,\lambda^{+}_1,\lambda^{+}_2)}\right|-\log \left|\frac{S_L(\delta_L,,C_2,C_1,\lambda^{-}_1,\lambda^{-}_2)}{S_L(0,C_2,C_1,\lambda^{-}_1,\lambda^{-}_2)}\right|\right|+o(\lambda_m^{-3}) \\ &\leq C\left|\frac{M_2}{C_1}-\frac{M_1}{C_2}\right|q^4_{N+L}\log\lambda+\frac{M_1}{c_1}\cdot\frac{1}{c_1}\frac{1}{q^{8\tau}_{N+i}}\\&\leq
\frac{C}{C_1C_2}[|M_1-M_2|C_2+|C_1-C_2|M_1]q^4_{N+L}\log\lambda+C\frac{1}{q^{8\tau}_{N+L}}\\&\leq C\frac{q^4_{N+L}}{q_{N+L}^{8\tau}}\\&\leq \frac{C}{q_{N+L}} \frac{\log q_{N+L}}{q_{N+L-1}}.
\end{align*}

\end{enumerate}

In summary,
$$\left|\int_{\hat{I}_L}\frac{1}{\|\hat{A}_i\|}\frac{\partial \|\hat{A}_i\|}{\partial \theta_l}\frac{\partial \theta_l}{\partial t}dx\right|\le
\frac{C}{q_{N+L}}\log \frac{q_{N+L}}{ q_{N+L-1}}.
$$
\end{proof}
\vskip 0.4cm


{\it Proof of Lemma \ref{Integral-L2}}

For $x\in \Omega_{L,2}$, assume $l$ be the unique time such that $T^lx\in \hat{I}_{L+1}$. Let $\lambda_1, \lambda_2$ and $\theta_l$ be defined as in (\ref{lambda12}).  Recall that
\begin{equation}\label{decomposition-pa}
\frac{1}{\|A_n\|}\frac{\partial\|A_n\|}{\partial t}=\frac{1}{\|A_n\|}\frac{\partial \|A_n\|}{\partial \lambda_1}\frac{\partial \lambda_1}{\partial t}+
\frac{1}{\|A_{n}\|}\frac{\partial \|A_{n}\|}{\partial \lambda_2}\frac{\partial \lambda_2}{\partial t}+\frac{1}{\|A_n\|}\frac{\partial \|A_n\|}{\partial \theta_l}\frac{\partial \theta_l}{\partial t}.\end{equation}
With the help of  Lemma \ref{dAdlambda} and (\ref{measure-omega-2}), we have \begin{align*}&\left|\int_{\Omega_{L,2}}\frac{1}{\|A_{n}\|}\frac{\partial \|A_{n}\|}{\partial \lambda_2}\frac{\partial \lambda_2}{\partial t}dx\right|\\&\leq
\frac1n\int_{\Omega_{L,2}}\left\vert\frac{1}{\lambda_2}\frac{\partial \lambda_2}{\partial t}\right\vert dx\\& \leq \frac1n Leb(\Omega_{L,2})\cdot q_{N+L}^C\\& \leq q_{N+L}^{-500C\tau}.
\end{align*}
The second inequality is  followed by Lemma\ref{lemma22} as $l=r_{L+1}^+(x)$ and $\|A_{l}\|=\lambda_2.$ We have a similar estimate on $\int_{\Omega_{L,2}}\frac{1}{\|A_{n}\|}\frac{\partial \|A_{n}\|}{\partial \lambda_1}\frac{\partial \lambda_1}{\partial t}dx$.

From Lemma \ref{patheta}, we have
$$\frac1n\left|\int_{\Omega_{L,2}}\frac{1}{\|A_n\|}\frac{\partial \|A_n\|}{\partial \theta_l}\frac{\partial \theta_l}{\partial t}dx\right|\le \frac{C}{q_{N+L}} \frac{\log q_{N+L+1}}{\log q_{N+L}}.
$$

In conclusion, we obtain (\ref{integral-L2}).

\qed
\vskip 0.52cm

For $x\in \Omega_{L,1}$, let $n=j_s$, $0=j_0$ and
$
A_n(x)=A_{j_s-j_{s-1}}(x)A_{j_{s-1}-j_{s-2}}(x)\cdots A_{j_2}(x).
$
Note that $j_{i}-j_{i-1}\ge q_{N+L}$ with $2\le i\le s$.

Let $\lambda_2=\|A_{j_2}(x)\|$,\ $\lambda_{i-1}=A_{j_{i-1}-j_{i-2}}(x)$ for $4\le i\le s$ and
$\lambda_s=A_{n-j_{s-1}}(x)$. Define $\theta_i=\theta(A_{ j_{i}-j_{i-1}}, \ A_{ n-j_{i}})$, $2\le i\le s-1$ and $\hat{A}_{n-j_i}(x)=\Pi_{k=j_i+1}^n A(T^kx)$.
Then by repeatedly applying (\ref{decomposition-pa}), we have \begin{equation}\label{decomposition}\begin{array}{ll}
&\frac{1}{\|A_n(x,t)\|}\frac{\partial\|A_n(x,t)\|}{\partial t}=\sum\limits_{i=2}\limits^{s-1} \frac{\partial \hat{A}_{n-j_i}}{\partial \theta_i}\frac{d\theta_i}{dt}+\sum\limits_{i=2}\limits^{s} \frac{1}{\lambda_i}\frac{d\lambda_i}{dt}.
\end{array}
\end{equation}
The following lemma shows that outside of $\Omega_{L,2}$,   $\frac{1}{\|A_n\|}\frac{\partial\|A_n\|}{\partial t}$ can roughly be decomposed into the sum of $\frac{1}{\lambda_i}\frac{\partial\lambda_i}{\partial t}.$

\vskip 1.3cm
\begin{lemma}\label{iteration-Lip}
It holds that
\begin{equation}\label{A_n} \frac{1}{\|A_n\|}\frac{\partial\|A_n\|}{\partial t}=\sum_{i=2}^{s}\frac{1}{\lambda_i}\frac{\partial\lambda_i}{\partial t}+F_{L},\end{equation}
where $F_{L}$ satisfies
\begin{equation}\label{F1}
\frac1n\left|\int_{\Omega_{L,1}}F_{L}dx\right|\le  \lambda^{-\frac{1}{100}q_{N+L-1}}+\frac{1}{100}q_{N+L-1}-\frac{1}{100}q_{N+L-2}.
\end{equation}

\end{lemma}
\begin{proof}

Assume $x\in \Omega_{L,1}$. We write $A_n=\hat{A}_{n-j_2}\cdot A_{j_2}$ and define $\lambda_1=A_{j_2},\ \lambda_2=\hat{A}_{n-j_2}$ and $\theta_l=\theta(A_{j_2}, \hat{A}_{n-j_2})$. From the definition of $\Omega_{L,1}$, we have


 \begin{equation}\label{dec-ll}|\theta_l(x,t)|\ge \frac{1}{q_{N+L}^{1000C
\tau}}\gg \lambda^{-\frac{1}{100}q_{N+L-1}}\gtrsim (\min\{\|\hat{A}_{n-j_2}\|, \|A_{j_2}\|\})^{-\frac{1}{100}}.$$
Thus \eqref{tj2} holds true with $\eta=1/100$. Then  from Lemma \ref{dAdlambda} we have  $$
\begin{array}{ll}&\left|\frac{1}{\|A_n\|}\frac{\partial \|A_n\|}{\partial \|A_{j_2}\|}\frac{\partial \|A_{j_2}\|}{\partial t}
-\frac{1}{\|A_{j_2}\|}\frac{\partial \|A_{j_2}\|}{\partial t}\right|\le \lambda^{-\frac{1}{100}q_{N+L-1}},\\
\\
& \left|\frac{1}{\|A_n\|}\frac{\partial \|A_n\|}{\partial \|A_{n-j_2}\|}\frac{\partial \|A_{n-j_2}\|}{\partial t}
-\frac{1}{\|A_{n-j_2}\|}\frac{\partial \|A_{n-j_2}\|}{\partial t}\right|\le \lambda^{-\frac{1}{100}q_{N+L-1}}.\end{array}\end{equation}


With the help of Lemma \ref{keylemma} and a direct computation, we have
\begin{equation}\label{error-dec-theta}
\left|\frac{1}{\|A_n\|}\frac{\partial \|A_n\|}{\partial \theta_l}\frac{\partial \theta_l}{\partial t}-
\frac{1}{\|A_{j_3}\|}\frac{\partial \|A_{j_3}\|}{\partial \theta(A_{j_3-j_2}, A_{j_2})}\frac{\partial \theta(A_{j_3-j_2}, A_{j_2})}{\partial t}\right|\le \lambda^{-\frac{1}{100}q_{N+L-1}}.
\end{equation}
Subsequently, we have
$$
\left|\frac{1}{\|A_{{r_{N+L}^+}+{r_{N+L}^-}}\|}\frac{\partial \|A_{{r_{N+L}^+}+{r_{N+L}^-}}\|}{\partial \theta(A_{{r_{N+L}^+}},A_{{r_{N+L}^-}})}\frac{\partial \theta(A_{{r_{N+L}^+}},A_{{r_{N+L}^-}})}{\partial t}\right.$$ $$\left.-
\frac{1}{\|A_{j_3}\|}\frac{\partial \|A_{j_3}\|}{\partial \theta(A_{j_3-j_2}, A_{j_2})}\frac{\partial \theta(A_{j_3-j_2}, A_{j_2})}{\partial t}\right|\le \lambda^{-\frac{1}{100}q_{N+L-1}}.
$$
Thus to consider $\int_{T^{-l}(I_{N+L}\backslash\hat{I}_{N+L})}\frac{1}{\|A_n\|}\frac{\partial \|A_n\|}{\partial \theta_l}\frac{\partial \theta_l}{\partial t}dx$, it is sufficient to consider \begin{equation}\label{pa-theta_L}\int_{T^{-l}(I_{N+L}\backslash\hat{I}_{N+L})}\frac{1}{\|A_{{r_{N+L}^+}+{r_{N+L}^-}}(x)\|}\frac{\partial \|A_{{r_{N+L}^+}+{r_{N+L}^-}}(x)\|}{\partial \theta(A_{{r_{N+L}^+}}(x),A_{{r_{N+L}^-}}(x))}\frac{\partial \theta(A_{{r_{N+L}^+}}(x),A_{{r_{N+L}^-}}(x))}{\partial t}\end{equation} with $x\in \Omega_{L,1}$ or equivalently $T^lx\in I_{L+1}\backslash \hat{I}_{L+1}$ for some $0\le l\le n$.


Similarly, we write $\hat{A}_{n-j_2}=\hat{A}_{n-j_3}\cdot A_{j_3-j_2}$ and define  $\lambda_1=A_{j_3-j_2},\ \lambda_2=\hat{A}_{n-j_3}$ and $\theta_l=\theta(A_{j_3-j_2}, \hat{A}_{n-j_3})$.
Then we obtain
 $$\begin{array}{ll}
&\frac{1}{\|A_{n-j_2}\|}\frac{\partial \|A_{n-j_2}\|}{\partial \|A_{j_3-j_2}\|}\frac{\partial \|A_{j_3-j_2}\|}{\partial t}
\approx\frac{1}{\|A_{j_3-j_2}\|}\frac{\partial \|A_{j_3-j_2}\|}{\partial t},\\
\\
&\quad \frac{1}{\|A_{n-j_2}\|}\frac{\partial \|A_{n-j_2}\|}{\partial \|A_{n-j_3}\|}\frac{\partial \|A_{n-j_3}\|}{\partial t}
\approx\frac{1}{\|A_{n-j_3}\|}\frac{\partial \|A_{n-j_3}\|}{\partial t},\\
\\
&
\left|\frac{1}{\|A_{n-j_2}\|}\frac{\partial \|A_{n-j_2}\|}{\partial \theta_l}\frac{\partial \theta_l}{\partial t}-
\frac{1}{\|A_{{r_{N+L}^+}+{r_{N+L}^-}}\|}\frac{\partial \|A_{{r_{N+L}^+}+{r_{N+L}^-}}\|}{\partial \theta(A_{{r_{N+L}^+}},A_{{r_{N+L}^-}})}\frac{\partial \theta(A_{{r_{N+L}^+}},A_{{r_{N+L}^-}})}{\partial t}\right|\le \lambda^{-\frac{1}{100}q_{N+L-1}}.
\end{array}$$
The last inequality implies  (\ref{pa-theta}) is applicable to estimate $\int_{T^{-l}(I_{N+L}\backslash\hat{I}_{N+L})}\frac{1}{\|A_{n-j_2}\|}\frac{\partial \|A_{n-j_2}\|}{\partial \theta_l}\frac{\partial \theta_l}{\partial t}dx$.


Inductively, we can prove that
$$
 \frac{1}{\|A_n\|}\frac{\partial\|A_n\|}{\partial t}=\sum_{i=2}^{s}\frac{1}{\|A_{j_{i+1}-j_i}\|}\frac{\partial\|A_{j_{i+1}-j_i}\|}{\partial t}+F_{L}
$$
with $F_{L}$ satisfying $$\begin{array}{ll}&\left|\int_{\Omega_{L,1}}F_{L}dx\right|\le n\lambda^{-\frac{1}{100}q_{N+L-1}}+\\
\\
& \sum_{0\le l\le n}\left|\int_{T^{-l}(I_{N+L}\backslash\hat{I}_{N+L})}\frac{1}{\|A_{{r_{N+L}^+}+{r_{N+L}^-}}(x)\|}\frac{\partial \|A_{{r_{N+L}^+}+T^{r_{N+L}^-}}(x)\|}{\partial \theta(A_{{r_{N+L}^+}}(x),A_{{r_{N+L}^-}}(x))}\frac{\partial \theta(A_{{r_{N+L}^+}}(x),A_{{r_{N+L}^-}}(x))}{\partial t}\right|
\\
\\
&\le n(\lambda^{-\frac{1}{100}q_{N+L-1}}+\frac{1}{100}q_{N+L-1}-\frac{1}{100}q_{N+L-2}).\end{array}$$
The first inequality comes from (\ref{dec-ll}) and (\ref{error-dec-theta}) and the last one comes from (\ref{pa-theta}).


\end{proof}

In Lemma \ref{iteration-Lip}, $A_n(x)$ is decomposed into a product of matrices according to the returning times of $x$ with respect to $I_L$. In the following, we will repeat the procedure of the decomposition of the product of matrices in a similar method until the original matrix $A_n$ is decomposed into matrices with respect to the returning times of $x$ with respect to $I_{L_0}$.


\begin{lemma}\label{iteration-Lip1}
For any $L_0\le i\le L$, the following holds true. Let $j_{i,k},\ 1\le k\le s_i$ be all the returning times between $r_i$
and $n-r_i$ for $x$ under $T$ with respect to $I_i$. Define $\lambda_{i,k}=\|A_{j_{i,k}-j_{i,k-1}}(T^{j_{i,k-1}}(x))\|$. Then we have
\begin{equation}\label{itera-Lip}
\frac{1}{\|A_n(x)\|}\frac{\partial\|A_n(x)\|}{\partial t}=\sum_{k=1}^{s_i}\frac{1}{\lambda_{i,k}}\frac{\partial\lambda_{i,k}}{\partial t}+F_{i}
\end{equation}
where $F_i$ satisfies
\begin{equation}\label{itera-Lip1}\frac1n\left|\int_{\Omega_{L,1}}F_idx\right|\le \frac{1}{100}q_{N+L-1}-\frac{1}{100}q_{N+i}+\sum_{k=i}^L \lambda^{-\frac{1}{100}q_{N+i-1}}.
\end{equation}
\end{lemma}
\begin{proof}
The case $i=L$ is just Lemma \ref{iteration-Lip}. Inductively, assume the conclusion holds true for $i, \cdots, L$.
We will prove (\ref{itera-Lip}) holds true for the case $i-1$.

 Consider $\frac{1}{\lambda_{i,k}}\frac{\partial\lambda_{i,k}}{\partial t}$. Let $0<j_{i,k,0}< j_{i,k,1}<j_{i,k,2}<\cdots\le j_{i,k,s(i,k)}=j_{i,k}-j_{i,k-1}$ be all the returning times of $T^{j_{i,k-1}}(x)$ with respect to $I_{i-1}$ under T.
Define $\lambda_{i,k,u}=\|A_{j_{i,k,u}-j_{i,k,u-1}}(T^{j_{i,k,u-1}}x)\|,\ 1\le u\le s(i,k)-1$. Note that $T^{j_{i,k,u}}(x)\in I_{i-1}\backslash I_{i}$ for $1\le u\le s(i,k)-1$. Thus following the argument in Lemma \ref{iteration-Lip}, it holds that
 $$
 \frac{1}{\lambda_{i,k}}\frac{\partial\lambda_{i,k}}{\partial t}=\sum_{u=1}^{s(i,k)-1}\frac{1}{\lambda_{i,k,u}}\frac{\partial\lambda_{i,k,u}}{\partial t}+F_{i,k},
$$
which together with the inductive assumption implies that
$$\frac{1}{\|A_n(x)\|}\frac{\partial\|A_n(x)\|}{\partial t}=\sum_{k=1}^{s_{i-1}}(\frac{1}{\lambda_{i-1,k}}\frac{\partial\lambda_{i-1,k}}{\partial t}+F_{i,k})+F_i:=\sum_{k=1}^{s_{i-1}}\frac{1}{\lambda_{i-1,k}}\frac{\partial\lambda_{i-1,k}}{\partial t}+\tilde{F}_{i}+F_{i},
$$
where $s_{i-1}$ and $\lambda_{i-1,k}$ are defined similar as above; moreover, we have
$$\begin{array}{ll}
&\left|\frac1n\int_{\Omega_{L,1}}{\tilde F}_{i} dx\right|\\
&\le \left|\int_{I_{i-1}\backslash I_i}
\frac{1}{\|A_{{r_{N+i}^+}+{r_{N+i}^-}}(x)\|}
\frac{\partial \|A_{{r_{N+i}^+}+{r_{N+i}^-}}(x)\|}{\partial \theta(A_{{r_{N+i}^+}}(x),A_{{r_{N+i}^-}}(x))}
\frac{\partial \theta(A_{{r_{N+i}^+}}(x),A_{{r_{N+i}^-}}(x))}{\partial t}dx
\right|.
\end{array}
$$
Let $F_{i-1}=\tilde{F}_{i}+F_{i}$. Thus from (\ref{itera-Lip1}), we obtain
$$\begin{array}{ll}
&\frac1n\left|\int_{\Omega_{L,1}}F_{i-1} dx\right|\\
&\le \left|\int_{I_{i-1}\backslash I_i} \frac{1}{\|A_{{r_{N+i}^+}+{r_{N+i}^-}}(x)\|}\frac{\partial \|A_{{r_{N+i}^+}+{r_{N+i}^-}}(x)\|}{\partial \theta(A_{{r_{N+i}^+}}(x),A_{{r_{N+i}^-}}(x))}\frac{\partial \theta(A_{{r_{N+i}^+}}(x),A_{{r_{N+i}^-}}(x))}{\partial t}dx\right|\\
\\
&+\frac{1}{100}q_{N+L-1}-\frac{1}{100}q_{N+i}\\
\\
&\le \frac{1}{100}q_{N+i}-\frac{1}{100}q_{N+i-1}+\frac{1}{100}q_{N+L-1}-\frac{1}{100}q_{N+i}\le \frac{1}{100}q_{N+L-1}-\frac{1}{100}q_{N+i-1}.\end{array}
$$


\end{proof}


In the opposite direction, we have the same conclusion and similarly we replace $s$ with $u$.


\vskip 0.3cm
\noindent{\it The proof of Lemma \ref{final}}

Clearly, for any given $L_0\leq j \leq L-1,$ we have $\Omega_{j,1}\subseteq \Omega_{j+1,1}$ and $\Omega_{j,2}\supseteq \Omega_{j+1,2}.$ And $\T$ can be divided into $\Omega_{L_0,1},$ $\Omega_{L_0,2}\bigcap \Omega_{L_0+1,1}$, $\Omega_{L_0+1,2}\bigcap \Omega_{L_0+2,1},$ $\cdots,$ $\Omega_{L-1,2}\bigcap\Omega_{L,1}$ and $\Omega_{L,2}.$

On one hand, with the help of Lemma \ref{lemma22}, it's clear that
\begin{equation}\label{pa-theta}
\left|\int_{\Omega_{L_0,1}}\frac{1}{\|A_{{r_{N+L}^+}+{r_{N+L}^-}}(x)\|}\frac{\partial \|A_{{r_{N+L}^+}+{r_{N+L}^-}}(x)\|}{\partial \theta(A_{{r_{N+L}^+}}(x),A_{{r_{N+L}^-}}(x))}\frac{\partial \theta(A_{{r_{N+L}^+}}(x),A_{{r_{N+L}^-}}(x))}{\partial t}dx\right|\le |I_{L_0}|^{-3}\le q_{N+L_0-1}^{50\tau}.
\end{equation}

On the other hand, by the help of Lemma \ref{iteration-Lip}, we have
$$\frac{1}{\|A_n(x)\|}\frac{\partial\|A_n(x)\|}{\partial t}=\sum_{k=1}^{s_{L_0}}\frac{1}{\lambda_{{L_0},k}}\frac{\partial\lambda_{{L_0},k}}{\partial t}+F_{{L_0}},
$$
where
$$\begin{array}{ll}
&\frac1n\left|\int_{\Omega_{L,1}}F_{L_0}dx\right|
\\
&\le \left|\int_{ I_{L_0}} \frac{1}{\|A_{{r_{N+L}^+}+{r_{N+L}^-}}(x)\|}\frac{\partial \|A_{{r_{N+L}^+}+{r_{N+L}^-}}(x)\|}{\partial \theta(A_{{r_{N+L}^+}}(x),A_{{r_{N+L}^-}}(x))}\frac{\partial \theta(A_{{r_{N+L}^+}}(x),A_{{r_{N+L}^-}}(x))}{\partial t}dx\right|\\
\\
&+\sum_{k=L_0}^L \lambda^{-\frac{1}{100}q_{N+k-1}}.\end{array}
$$

Lemma \ref{lemma22} shows $\left|\frac{1}{\lambda_{{L_0},k}}\frac{\partial\lambda_{{L_0},k}}{\partial t}\right|\le q_{N+L_0-1}^{300\tau^2}\triangleq C(L_0),$ combine this with $S_{L_0}\le n$, we immediately have
$$
\frac{1}{n}\left|\int_{\Omega_{L_0,1}}\frac{1}{\|A_n(x)\|}\frac{\partial\|A_n(x)\|}{\partial t}dx\right|
\le nC(L_0).$$

For $\Omega_{j+1,2}\bigcap\Omega_{j+2,1}$ with $L-1\geq j\geq L_0$ $(\Omega_{L+1,1}\triangleq\emptyset),$ by directly applying Lemma \ref{integral-L2}, we have
$$\frac{1}{n}\int_{\Omega_{j+1,2}\bigcap\Omega_{j+2,1}}\frac{1}{\|A_n(x)\|}\frac{\partial\|A_n(x)\|}{\partial t}dx\le 2q_{N+j+1}^{-500C\tau}+C \frac{\log q_{N+j+2}}{q_{N+j+2}}.$$


In summary, we obtain
$$\frac{1}{n}\int_{\Omega_{j+1,2}\bigcap\Omega_{j+2,1}}\frac{1}{\|A_n(x)\|}\frac{\partial\|A_n(x)\|}{\partial t}dx\le \sum_{j=L_0}^{L}\left[2q_{N+j+1}^{-500C\tau}+C \frac{\log q_{N+j+2}}{q_{N+j+2}}\right]+C(L_0),$$ which finishes the proof.
\vskip0.3cm
All of the above work provide two things, which are as follows:
\begin{enumerate}
\item There exists a dense subset $T:=EP$ of $\Sigma_{\alpha,\lambda v}$ with a zero measure  such that for any $E\in T$, $L(\cdot)$ is exactly $\frac{1}{2}$ H\"older continuous at $E$.
\item There exists a subset $F:=FR$ of $\Sigma_{\alpha,\lambda v}$ with a full measure such that for any $E'\in F$, $L(\cdot)$ is Lipschitz at $E'$.
\end{enumerate}


\section{Regularity of LE for other $E$}
To finish the proof of Theorem 1. We need to shows $\frac{1}{2}$-H\"older continuity for all all $E\in \Sigma_{\alpha,\lambda v}$ and find  some point $E''\in \Sigma_{\alpha,\lambda v}$ such that $L(\cdot)$ is almost $\beta$-H\"older (see \eqref{def7.1}) continuous at $E''$
for any fixed $\frac{1}{2}<\beta<1.$

\begin{definition}For some fixed $\gamma\in (0,1),$ $x_0\in \R$ and $f\in C^0(\R),$ we say \begin{equation}\label{def7.1}f~ is ~almost ~\gamma-H\ddot{o}lder ~continuous~on~x_0,\end{equation} if $f$ is at least $(\gamma-\delta)$-H\"older continuous and at most $(\gamma+\delta)$-H\"older continuous on $x_0$ for any $\delta>0.$\label{def1}

We say \begin{equation}\label{def7.1}f~is~\gamma^+-H\ddot{o}lder ~continuous~on~x_0,\end{equation} if $f$ is $(\gamma+\delta)$-H\"older continuous on $x_0$ for any $\delta>0.$\label{def2}

We say \begin{equation}\label{def7.1}f~is~\gamma^--H\ddot{o}lder ~continuous~on~x_0,\end{equation} if $f$ is $(\gamma-\delta)$-H\"older continuous on $x_0$ for any $\delta>0.$\label{def3}

\end{definition}

We have the following lemma.
\begin{lemma}For any $t\in \Sigma$ and $\frac{1}{2}<\beta(t)<1,$ $L(\cdot)$ is almost $\beta(t)$-H\"older continuous at $t;$ For any $t\notin EP$ and $\beta(t)=\frac{1}{2},$ $L(\cdot)$ is exactly $\frac{1}{2}^+$-H\"older continuous at $t.$\label{lm28}
\end{lemma}
\begin{remark} The proof is divided into three parts. At the first part, we prove the lower bound of the H\"older exponent of $L(\cdot),$ which essentially follows from the previous proof of Lipshchitz continuity and $\frac{1}{2}-$H\"older continuity. The second part is to prove the upper bound of the H\"older exponent of $L(\cdot)$ and we obtain it by selecting proper sequence $t_n\rightarrow t$ such that ¡°$|L(t)-L(t_n)|$¡± being too ¡°bad¡± to overstep the upper bound we fixed. This is essentially due to the density of $EP,$ where $L(\cdot)$ is exactly $\frac{1}{2}-$ H\"older continuous. The third part, also by the density of $EP$ and the decreasing of the exponent for $\frac{1}{2}-$H\"older continuity, we obtain the exactly $\frac{1}{2}^+-$H\"older continuity.
\end{remark}
\begin{proof}[\textbf{The proof of at least $\beta(t)-\delta$}]

\

If $t\in I(\beta(t)),$ then there exists some $N,$ such that
\begin{equation}\label{tjbds}\left| c_{n,1}(t)-c_{n,2}(t)-k_n(t)\alpha\right|\geq \left\|A_{k_n}\right\|^{(-2(\beta(t))+2\delta+1)^{-1}},
\end{equation}
for $n\geq N.$

Without loss of generality, we assume $g_{N_1}$ is of type III. Due to \eqref{tjbds} and the definition of $\beta(t)$, there exist infinite $g_i$s such that $g_i$ is of type III. Assume that $\left\{i_j\right\}_{j\geq1}$ are all the time points such that $g_{i_j}$ is of type III and $g_{i_j+1}$ is of type I, and $\left\{k_j\right\}_{j\geq 1}$ are all the time points such that $g_{k_j}$ is of type I and $g_{k_j+1}$ is of type III. Thus $g_{i_j+1}, g_{i_j+2} ,\cdots, g_{k_{j+1}}$ are of type I and $g_{k_j+1}, g_{k_j+2},\cdots, g_{i_j}$ are of type III for $j\geq 1.$ Clearly, $$N_1-1=k_1<i_1<k_2<\cdots<i_n<k_{n+1}<\cdots.$$

Then we have the following lemma:
\vskip 0.3cm
\begin{lemma}\label{quasi-Lip}
 Assume for some $t$ and $l_j-1\le n\le k_{j+1}-1$ for some $j\geq 1$, $g_n$ is of type I.
Then for any fixed $\epsilon>0$ and $t'\in \left[t-q_{N+n-1}^{-1000\tau^2},t-q_{N+n}^{-1000\tau^2}\right]\bigcup\left[t+q_{N+n}^{-1000\tau^2},t+q_{N+n-1}^{-1000\tau^2}\right]$, we have
\begin{equation}\label{difference-quasi-Lip}
\left| L(t)- L(t')\right|\leq \left|t-t'\right|^{1-\epsilon}
\end{equation}
for some positive constant $\epsilon\ll 1$.
\end{lemma}
\begin{proof}
The proof is similar to the case of Lipschitz continuity. In fact, it follows from Lemmas  \ref{lemma29}, we have
$| L'_{q_{N+n+1}^2}(t)|, | L'_{2q_{N+n+1}^2}(t)|\le C(i_j)\le q_{N+i_j-1}^{300\tau^2}$

for $t'\in \left[t-q_{N+n-1}^{-1000\tau^2},t-q_{N+n}^{-1000\tau^2}\right]\bigcup\left[t+q_{N+n}^{-1000\tau^2},t+q_{N+n-1}^{-1000\tau^2}\right]$.

Note that we assume $i_j\ll k_j.$
Then it holds that
$$\begin{array}{ll}&\left| L(t)- L(t')\right|\\
&\le 2\left| L_{q_{N+n+1}^2}(t)- L_{q_{N+n+1}^2}(t')\right|+\left| L_{2q_{N+n+1}^2}(t)- L_{2q_{N+n+1}^2}(t')\right|+\lambda^{-cq_{N+n+1}^2}\\
&\le 3C(i_j)|t-t'|+\lambda^{-cq_{N+n+1}^2}\\
&\le |t-t'|^{1-\epsilon}+\lambda^{-cq_{N+n+1}^2}\\
&\le 2|t-t'|^{1-\epsilon}.
\end{array}
$$
The inequality  second to the last from the fact that  $|t-t'|\le q_{N+n-1}^{-1000\tau^2}\le q_{N+l_j-1}^{-1000\tau^2}\ll q_{N+i_j-1}^{-1000\tau^2}$. And the last one follows from $\lambda^{-cq_{N+n+1}^2}\ll q_{N+n}^{-1000\tau^2}\le |t-t'|$.
\end{proof}
\begin{remark} We rearrange the sequence of
$N_1-1=k_1<i_1<k_2<\cdots<i_n<k_{n+1}<\cdots$ into
$$N_1-1=k_1<i_1<l_1<k_2<l'_1\cdots<i_n<l_n<k_{n+1}<l'_n<\cdots$$
such that $l_j\gg i_j$, $l'_j\gg k_{j+1}$ and for any $n\in [i_j, l_{j}]$ or $n\in [k_{j+1}, l'_{j}]$, $g_n$ can be considered either as type I or as type III. Moreover,for any $n\in [l_{j}, k_{j+1}]$ or $n\in [l'_{j}, i_{j+1}]$, $g_n$ can be considered  as type I or as type III, respectively. We will apply the argument on Lipschitz continuity on $[l_j, l'_j]$ with $l_j\gg i_j$ and the one on $\frac12$ H\"older continuity on $[l'_j, l_{j+1}]$ with $l'_j\gg k_{j+1}$.
\end{remark}
\vskip 0.2cm
If $k_j\leq n\leq i_j-2$ for some $j\geq 1,$ then we claim that
$$\left| L_{q_{N+n+1}^4}(t)-L_{q_{N+n+1}^4}(t')\right|\leq C\left|t-t'\right|^{\beta(t)-\delta}.$$

Actually, as $g_{n+2}$ is of type III. There exists at most one time point $0\leq l\leq q_{N+n+1}^4$ such that $T^l(x)\in I_{n+2}.$
Consider
$$A_{q_{N+n+1}^4}(x)=A_{q_{N+n+1}^4-l}(T^lx)A_l(x).$$
Let $\lambda_1=\|A_{q_{N+n+1}^4-l}\|~and~\lambda_2=A_l.$
Similar as the proof of $\frac{1}{2}$-H\"older continuity on EP, it's not difficult to see that $\left|\frac{1}{\|A_n\|}\frac{d\|A_n\|}{dt}\right|$ is dominated by
\begin{equation}\label{kl111} \vert\frac{(1-\lambda_2^{-4})(1-\lambda_1^{-4})}{\sqrt{(1-\frac{1}{\lambda_1^4\lambda_2^4})^2(c(x-\tilde{x_l})^2-\epsilon_1)^2+(\frac{1}{\lambda_1^4}-\frac{1}{\lambda_2^4})^2(c(x-\tilde{x_l})^2-\epsilon_1)^{-2}+2(1+\frac{1}{\lambda_1^4\lambda_2^4})(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})-\frac{8}{\lambda_1^4\lambda_2^4}}}\vert.
\end{equation}

Note that $c\sim \|A_{k_j}\|$, which is followed from A.4 of the \cite{wz1}, and $\epsilon_1$ denote the distance between horizontal axis and $g_{n+2}(\tilde{c_{n+2}}).$

It is followed from lemma \ref{keylemma}, A.1 and straight calculation that
\begin{align*}
\int_{T^{l}x\in I_{n+2}} \frac{d\theta_l}{dt}\frac{d\|A_{q_{N+n+1}^4}\|}{d\theta_l}\frac{1}{\|A_{q_{N+n+1}^4}\|} dx &\leq \int_{T^{l}x\in I_{n+2}} \vert\frac{C}{\sqrt{(cx^2-\delta_1)^2+(cx^2-\delta_1)^{-2}+2(\frac{1}{\lambda_1^4}+\frac{1}{\lambda_2^4})}}\vert dx\\&\leq c^{-\frac{1}{2}}\sqrt{\epsilon_1}^{-1}\\&=\|A_{k_j}\|^{-\frac{1}{2}} \epsilon_1^{-\frac{1}{2}}.
\end{align*}
Thus $$\left| L_{q_{N+n+1}^4}'(t)\right|\leq \|A_{k_j}\|^{-\frac{1}{2}} \epsilon_1^{-\frac{1}{2}},$$
which implies
\begin{align*}
\left| L_{q_{N+n+1}^4}(t)-L_{q_{N+n+1}^4}(t_0)\right|&\leq \int L_{q_{N+n+1}^4}(t)'dt\\&\leq C\|A_{k_j}\|^{-\frac{1}{2}} \epsilon_1^{\frac{1}{2}}\\&\leq C\epsilon_1^{\frac{1}{2}+\frac{1}{2}\frac{-\log \|A_{k_j}\|}{\log \epsilon_1}}\\&\leq C\epsilon_1^{\beta(t)-\delta}
\end{align*}

Here $t_0$ denotes the $t$ such that $g_{n+2}(t,\tilde{c}_{n+2})=0$ and $\epsilon_1=\left|t_0-t\right|.$ The last inequality is followed from $t\in I(\beta(t)).$
Thus, by the help of Lemma \ref{lm27}, similarly to the case $\frac{1}{2}-$H\"older continuity, it's not difficult to obtain
$$\left|L(t)-L(t')\right|\leq C\left|t-t'\right|^{\beta(t)-\delta}$$ for each $t'\in (t_0, t-\eta(n))\bigcup(t+\eta(n),t+q^{-C}_{N+n+1})$ with some $\eta(n)\leq \lambda^{-\frac{3}{2}r_{n+1}}.$

Therefore, \begin{equation}\label{bteq}\left|L(t)-L(t')\right|\leq C\left|t_0-t'\right|^{\beta(t)-\delta},\end{equation} for any $t'\in B_n\triangleq (t_0,t-\eta(n))\bigcup(t+\eta(n),t+q^{-C}_{N+n+1}).$

Let $F_t\triangleq\bigcup\limits_{n=1}^{+\infty}B_n \bigcup\{t\}$, which is a certain interval including $t$. Hence, \eqref{bteq} holds for each $t\in F_t$.

Then the $\beta(t)-\delta$-H\"older continuity on $t$ holds.

\end{proof}
\textbf{The proof of at most $\beta(t)+\delta$:}
By the definition of $\beta(t)$, it holds that for any given $\epsilon>0,$ there exists a sequence of $n_j$ such that
$$\left| c_{n_j,1}(t)-c_{n_j,2}(t)-k_{n_j}(t)\alpha \right|< C\left\|A_{k_{n_j}}\right\|^{(-2(\beta(t)+\epsilon)+1)^{-1}}.$$
We define $t_j^0$ as the $x$ such that $g_{n_j}(x)$ has the only zero point on $I_{n_j}.$ Then, by the proof of the first part above we know that there exist some $t_j^0\rightarrow t$ such that
$$\left| L_{q_{N+n_j}^4}(t)-L_{q_{N+n_j}^4}(t_j^0)\right|\leq C\left|t-t_j^0\right|^{\gamma}$$ with $\gamma\leq \beta(t)+\epsilon$ , some proper constant $C>1$ and $\left|t-t_j^0\right|=\lambda^{-c_0 k_{n_j}}\gg \lambda^{-q_{N+n_j}^4}.$
\vskip0.3cm
Next, we choose a sequence of $t_j$ as follow:
$$t_j:=\{x| |x-t_j^0|=C^*|t-t_j^0|,sgn(x-t_j^0)sgn(t-t_j^0)=-1\},$$ where $C^*$ will be defined later. This means that $t_j$ and $t_j^0$ respectively locates in the two sides of $t_j^0$ and $t_j$ is in some spectrum gap.
Due to the $\frac{1}{2}-$H\"older continuity of LE at $t^{0}_j,$ and the exactly $\frac{1}{2}-$ H\"older continuity of LE at $t^{0-}_j$ (approximate from the semi-interval located outside the spectrum) with the H\"older exponent $\|A_{k_{n_j}}\|^{-\frac{1}{2}},$ it's clear that
\begin{equation}\begin{array}{ll}&\left| L_{q_{N+n_j}^4}(t)-L_{q_{N+n_j}^4}(t_j^0)\right|\sim C'|t-t^0_j|^{\gamma};\\& \left| L_{q_{N+n_j}^4}(t_j)-L_{q_{N+n_j}^4}(t_j^0)\right|<C''|t_j-t^0_j|^{\gamma}.\end{array}
\end{equation}


Therefore, by direct calculation, we have\begin{equation}\begin{array}{ll}&(C'(C^*)-C'')(C^*+1)|t-t_j|^{\gamma}\\&(C'(C^*)-C'')\frac{|t-t^0_j|^{\gamma}}{|t-t_j|^{\gamma}}\cdot|t-t_j|^{\gamma}\\&=C'(C^*)^{\gamma}|t-t^0_j|^{\gamma}-C''|t-t^0_j|^{\gamma}\\&=C'|t_j-t^0_j|^{\gamma}-C''|t-t^0_j|^{\gamma}\\&  \leq\bigg|\left| L_{q_{N+n_j}^4}(t_j)-L_{q_{N+n_j}^4}(t_j^0)\right|-\left| L_{q_{N+n_j}^4}(t)-L_{q_{N+n_j}^4}(t_j^0)\right|\bigg|\\& \leq \left| L_{q_{N+n_j}^4}(t)-L_{q_{N+n_j}^4}(t_j^0)\right|\\&\leq \left| L_{q_{N+n_j}^4}(t)-L_{q_{N+n_j}^4}(t_j^0)\right|+\left| L_{q_{N+n_j}^4}(t_j)-L_{q_{N+n_j}^4}(t_j^0)\right|\\& \leq C'|t_j-t^0_j|^{\gamma}+C''|t-t^0_j|^{\gamma}\\&=C'(C^*)^{\gamma}|t-t^0_j|^{\gamma}+C''|t-t^0_j|^{\gamma}\\&=(C'(C^*)+C'')\frac{|t-t^0_j|^{\gamma}}{|t-t_j|^{\gamma}}\cdot|t-t_j|^{\gamma}\\&=(C'(C^*)+C'')(C^*+1)|t-t_j|^{\gamma} \end{array}\end{equation}

Then we denote $C_1\triangleq (C'(C^*)-C'')(C^*+1)$ and $C_2\triangleq (C'(C^*)+C'')(C^*+1),$ and we choose $C^*$ large enough such that $2C_1>C_2.$(a.e. $C''>\frac{10C''}{C'}$)

Hence, we obtain
\begin{align*} \left|L(t)-L(t_j)\right|&\geq\bigg|2\left|L_{2q_{N+n_j}^4}(t)- L_{2q_{N+n_j}^4}(t_j)\right|-\left|L_{q_{N+n_j}^4}(t)-L_{q_{N+n_j}^4}(t_j)\right|\bigg|\\&-(\left|L_{2q_{N+n_j}^4}(t)-L_{q_{N+n_j}^4}(t)-L(t)\right|+\left|L_{2q_{N+n_j}^4}(t_j)-L_{q_{N+n_j}^4}(t_j)-L(t_j)\right|)\\
&\geq (2C_1-C_2)\left|t-t_j\right|^{\gamma}-\lambda^{-cq_{N+n_j}^4}\\&\geq C_3\left|t-t_j\right|^{\gamma}
\end{align*}
with $\gamma\leq \beta(t)+\epsilon.$ Which implies the lemma.
\vskip0.3cm
\textbf{The proof of exactly $\frac{1}{2}^+-$H\"older continuity:}By the definition of $\beta(t)$, for any given $\delta>0,$ there exist a sequence of $n_j\rightarrow+\infty,$ such that
\begin{equation}\label{tjbds}\left| c_{n_j,1}(t)-c_{n_j,2}(t)-k_{n_j}(t)\alpha\right|\leq \left\|A_{k_{n_j}}\right\|^{-2\delta},
\end{equation}where $g_{n_j}(t)$ is of Type III(or II), for $n_{j-1}<l<n_j,$ $g_j(t)$ is Type I.

On one hand, we can find $t_j\rightarrow t$ such that
$$|L(t)-L(t_j)|\geq c_{n_j}|t-t_j|^{\frac{1}{2}},$$ where $c_{n_j}\sim \|A_{n_j}\|^{-\frac{1}{2}},$ $|t-t_j|\sim \|A_{n_j}\|^{-\delta}.$ Hence, $$|L(t)-L(t_j)|\geq \|A_{n_j}\|^{-\frac{1}{2}-\delta}=|t-t_j|^{\frac{1}{2}+\delta},$$ which implies $L(\cdot)$ is at most $\frac{1}{2}+\delta$-H\"older on $t$¡£

On the other hand, let $M_j$ satisfying $\bigcup\limits_{j=1}^{\infty} M_j\ni t$ being an open interval, and for any $t_1\in M_j,$ the type of $g_j(t_1)$ are same (Type I, II, or III). Meanwhile, when $g_j$ is Type I on $M_j$, $L(\cdot)$ is $1-\epsilon$(for any given $\epsilon>0$) H\"older continuous on $M_j;$ when $g_j$ is of type III (or II) on $M_j$, we have $$|L(t)-L(t_j)|\leq c_{n_j}|t-t_j|^{\frac{1}{2}},$$ ÆäÖÐ$c_{n_j}\sim \|A_{n_j}\|^{-\frac{1}{2}},$ $|t-t_j|\sim \|A_{n_j}\|^{-\delta}.$ Òò´Ë, $$|L(t)-L(t_j)|\leq \|A_{n_j}\|^{-\frac{1}{2}-\delta}=|t-t_j|^{\frac{1}{2}+\delta},$$ which implies $L(\cdot)$ is at least $\frac{1}{2}+\delta$-H\"older on $t$.
$\square$


\vskip0.3cm

\textbf{proof of the $\frac{1}{2}$-H\"older continuity on $\Sigma.$} It's a obvious corollary of lemma \ref{lm28}.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$\square$

Now we are ready to result in the last problem of Theorem 1. For any given $\frac{1}{2}\leq t<1$, We'd like to find the point $t_0$ such that $L(\cdot)$ is almost $\beta$-H\"older continuous.
By lemma \ref{lm28}, it's enough to find the point $t_0$ such that $t_0\in I(\beta(t_0))$ with $\beta(t_0)=\beta.$
Moreover, it's sufficient to find $t_0$ satisfying
$$t_0\in \bigcup\limits_{k\geq 1}\bigcap\limits_{n\geq k}\left\{t\vert C_1\|A_{k_n}\|^{(-2\beta+1)^{-1}}\leq\left|c_{n,1}(t)-c_{n,2}(t)-k_n(\alpha)\right|\leq C_2\|A_{k_n}\|^{(-2\beta+1)^{-1}}\right\}$$
for some constant $C_1\leq C_2.$
We'll find it by Closed interval set theorem.

Fix $\frac{1}{2}<\beta<1$ and $t_0\in EP.$ It's clear that there exists some $N_1>0$ such that
$$\left|g_n(\tilde{c_{n}})\right|\leq C\lambda^{-\frac{3}{2}r_{n-1}}$$ for any $n\geq N_1.$

Let $\tilde{t_0}$ denotes $\left\{t|g_{N_1}(\tilde{c_{N_1},t})=\left\|A_{k_{N_1}}\right\|^{-1+\frac{2}{2\beta-1}}\right\}$, which implies $$\left|c_{N_1+1,1}(t)-c_{N_1+1,2}(t)-k_n(\alpha)\right|=\left\|A_{k_{N_1+1}}\right\|^{(1-2\beta)^{-1}}.$$

Clearly, for $\left\{g_n\right\}_{n\geq N_1}$, $g_n$ cannot be always of type III and it will change into type I at some point.
Thus, we denote $$N_2:=\min\left\{j>N_1| g_j~is~of~type~I\right\}.$$

Consider $g_{N_2}(\cdot)$. By the density of EP, there must exist some point $\tilde{t_0}'$ such that $\left\{g_n\right\}_{n\geq N_3}$ is of type III and $N_3>N_2.$
Besides, we can let $\left|\tilde{t_0}'-\tilde{t_0}\right|< \epsilon_0$, for any $\epsilon_0>0$ we want to get, such that $g_n(\tilde{t_0}')\sim g_n(\tilde{t_0})$ for any $n\leq N_2.$ For example, let $\epsilon_0:=\lambda^{-q_{N+N_2-1}^{10000}}.$ For convenience, we suppose $\tilde{t_0}=\tilde{t_0}'.$

Thus, there will exist some $k_1,k_2,\cdots,k_s$ and $m_1,m_2,\cdots,m_s$ such that
$$for~n~=k_i+1,\cdots,m_i,~g_n~is~of~type~I;$$
$$for~n~=m_i+1,\cdots,k_i,~g_n~is~of~type~III.$$
Here $m_s+1=N_3.$

Let $$\bar{j}:=\min\left\{1\leq j\leq s| g_{m_j+1}~is~of~type~III~and~\left|g_{m_j+1}(\tilde{c_{m_j+1}})\right|\leq \left\|A_{m_j+1}\right\|^{-1+\frac{2}{2\beta-1}}\right\}.$$

Now consider $g_{m_{\bar{j}+1}}.$ Let $\tilde{t_1}$ denotes $$\tilde{t_1}:=\left\{t| g_{m_{\bar{j}}+1}(\tilde{c}_{m_{\bar{j}}+1},t)=\left\|A_{k_{m_{\bar{j}}+1}}\right\|^{-1+\frac{2}{2\beta-1}}\right\}.$$

Clearly, \begin{align*}
\left|\tilde{t_1}-\tilde{t_0}\right|&\leq \lambda^{-q_{N+m_{\bar{j}}-1}^4}\\&\leq \lambda^{-\lambda^{cq_{N+m_{\bar{j}-1}-1}}}\\&\leq \lambda^{-2Cq_{N+m_{\bar{j}-1}+1}}\\&\leq \left\|A_{k_{m_{\bar{j}-1}+1}}\right\|^{-2(-1+\frac{2}{2\beta-1})}\\&= o(\left\|A_{k_{m_{\bar{j}-1}+1}}\right\|^{-(-1+\frac{2}{2\beta-1})}).
\end{align*}
The second inequality is followed from $$\lambda^{cq_{N+m_{\bar{j}-1}-1}}\sim q_{N+k_{\bar{j}}-1}^4\leq q_{N+m_{\bar{j}}-1}^4.$$

So we find some $\tilde{t_1}$ satisfies the following property:
$$\left| g_{N_1}(\tilde{c_{N_1}},\tilde{t_1})\right|=\left\|A_{k_{N_1}}\right\|^{-1+\frac{2}{2\beta-1}}+o(\left\|A_{k_{m_{\bar{j}-1}+1}}\right\|^{-(-1+\frac{2}{2\beta-1})});$$
$$\left| g_{m_{1}+1}(\tilde{c_{m_{1}+1}},\tilde{t_1})\right|>\left\|A_{k_{m_{1}+1}}\right\|^{1-2C(\beta)}-o(\left\|A_{k_{m_{\bar{j}-1}+1}}\right\|^{-(-1+\frac{2}{2\beta-1})});$$
$$\cdots$$
$$\left| g_{m_{\bar{j}}+1}(\tilde{c}_{m_{\bar{j}}+1},\tilde{t_1})\right|=\left\|A_{k_{m_{\bar{j}}+1}}\right\|^{-1+\frac{2}{2\beta-1}}.$$
This implies
$$\left|c_{N_1+1,1}(t)-c_{N_1+1,2}(t)-k_{N_1+1}(\alpha)\right|=C_{N_1}\left\|A_{k_{N_1+1}}\right\|^{(1-2\beta)^{-1}};$$
$$\left|c_{m_1+1,1}(t)-c_{m_1+1,2}(t)-k_{m_1+1}(\alpha)\right|>C{m_1}\left\|A_{k_{N_1+1}}\right\|^{(1-2\beta)^{-1}};$$
$$\cdots$$
$$\left|c_{m_{\bar{j}}+1,1}(t)-c_{m_{\bar{j}}+1,2}(t)-k_{m_{\bar{j}}+1}(\alpha)\right|=C_{m_{\bar{j}}}\left\|A_{k_{N_1+1}}\right\|^{(1-2\beta)^{-1}}.$$

Here each constant can be chosen as some number between $\frac{3}{4}$ and $\frac{5}{4}.$

Besides, for any $1\leq i\leq s,$ $g_{{k_i}+1},\cdots,g_{{m_i}}$ are of type I since $$\left\|A_{k_{m_{\bar{j}-1}+1}}\right\|^{-1+\frac{2}{2\beta-1}}\ll q_{N+l-1}$$ for $N_1\leq l\leq m_{\bar{j}}.$

Repeat the process as above, it's not difficult to obtain a sequence $\left\{t_n\right\}_{n\geq0}$ such that $$\sum\limits_{i\geq 1}\left|\tilde{t_i}-\tilde{t_{i-1}}\right|< \infty.$$ We denote $\tilde{t}$ the accumulation point.

For $\left\{g_n(\tilde{t}\right\}_{n\geq N_1}$, we have the following property:
\begin{enumerate}
\item Type I and III alternately appear.
\item There exists a sequence $\left\{n_k\right\}_{k\geq1}$ such that
$g_{n_k}$ is of type III and
$$\left|g_{n_k}(\tilde{c_{n_k}},\tilde{t})\right|=C_{n_k}\left\|A_{k_{n_k}}\right\|^{-1+\frac{2}{2\beta-1}},$$
$\frac{3}{4}<C_{n_k}<\frac{5}{4}$ for $k\geq 1.$
Thus$$\frac{3}{4}\left\|A_{k_{n_k}}\right\|^{-C(\beta)}\leq \left|c_{n_k,1}-c_{n_k,2}-k_{n_k}\alpha\right|\leq \frac{5}{4}\left\|A_{k_{n_k}}\right\|^{(1-2\beta)^{-1}}.$$
While, between $n_k$ and $n_{k+1},$ either $g_n$ is of type I or type III with $\left|g_n(\tilde{c_n})\right|\geq C\left\|A_n\right\|^{-1+\frac{2}{2\beta-1}},$ which implies
$$\left|c_{n,1}-c_{n,2}-k_n\alpha\right|\geq \left\|A_n\right\|^{(1-2\beta)^{-1}}.$$
\end{enumerate}

It's easy to check that $$\tilde{t}\in\bigcup\limits_{k\geq 1}\bigcap\limits_{n\geq k}\left\{t\vert \frac{3}{4}\|A_{k_n}\|^{(1-2\beta)^{-1}}\leq\left|c_{n,1}(t)-c_{n,2}(t)-k_n(\alpha)\right|\leq \frac{5}{4}\|A_{k_n}\|^{(1-2\beta)^{-1}}\right\},$$ which completes the proof.
\qed
\section{The $\frac{1}{2}$-H\"older continuity of LE}
In the previous section we deal with the local $\frac{1}{2}$-H\"older continuity of LE at each endpoints, local Lipschitz continuity of LE at each $FR$ and local $\beta$-H\"older continuity for other points with some $0.5<\beta<1$. In this section, we try to obtain the global $\frac{1}{2}$-H\"older continuity of LE. Actually, as one can see, the `worst' condition happen only at each $EP.$

We divide the proof into two parts.


As is know to all that, a continuous function defined on a closed interval is uniformly continuous. Similarly, we have the same property for H\"older continuity, which can be seen in the following lemma.
\vskip0.3cm


Recall that we have already obtain the following results, and all the results are under the condition $\lambda$ is large enough.
\begin{enumerate}
\item[\textbf{1}:] For any fixed $t_1<t_2$ and $k\in \N^+,$ if $g_k([t_1,t_2])$ is of type I, then we have $|L_k(t_1)-L_k(t_2)|\leq C(t_1,t_2)|t_1-t_2|\leq |t_1-t_2|^{0.9};$
\item[\textbf{2}:] Fix $t_3\in EP,$ which labeled by $l\in N$, we have $|L(t_3)-L(t)|\leq C'(t_3)|t_3-t|^{0.5}$ for any $t\in [t_3-3|I_i|,t_3+3|I_i|].$ While for some $t_3+\lambda^{-q^4_{N+i-1}}<t_1<t_2\leq t_3+3|I_i|,$ we have $|L(t_2)-L(t_1)|\leq C'(t_3)|t_2-t_1|^{0.5}.$
\item[\textbf{3}:] The $C'$ is independent on $t_3$ in the second case.
\end{enumerate}

First, we slightly change the definition of the type of the angle functions as follow.
\begin{enumerate}
\item We say the angle function $g_i(t)$, $i\in \N,$ is $essentially$ type I, if $|c_{i-1,1}-c_{i-1,2}-s\alpha|>|I_i|^{200},$ for all $s\leq q_{N+i-1}^2.$
\item We say the angle function $g_i(t)$, $i\in \N,$ is $essentially$ type III (or II), if $|c_{i-1,1}-c_{i-1,2}-s\alpha|<2|I_i|,$ for some $s\leq q_{N+i-1}^2.$
\end{enumerate}

Surely, the previous three properties also hold if we instead of the definition as above. The main motivation of it is to avoid discussing the intersected situations between type I and III.

Now, we directly prove the following lemma:

\begin{lemma}
For any given $t_1<t_2$ in the $\frac{1}{\lambda}\Sigma$ with sufficiently large $\lambda$, we have $|L(t_1)-L(t_2)|\leq C'|t_1-t_2|^{0.5},$ where $C'$ is an absolute constant.
\end{lemma}
\begin{remark} In fact, as one will see the main part has already been proved in previous lemmas' proof and the rest is only to do some simple classifications and calculations.
\end{remark}
\begin{proof}
Clearly, we can assume $\lambda^{-\frac{c}{10}q_{N+n-1}}\leq t_2-t_1\leq \lambda^{-\frac{c}{10}q_{N+n-1}},$ for some sufficiently large $n\in \N.$ We denote $K:=q_{N+n-1}^4.$ Note that for sufficiently large $\lambda$ and $n$, the following inequalities always hold:
\begin{equation}|I_n|^{10000}\gg \lambda^{-\frac{c}{10}q_{N+n-2}};
\end{equation}
\begin{equation}\lambda^{-cK}\gg \lambda^{-\frac{c}{10}q_{N+n-1}}\gg \lambda^{-q^4_{N+n-1}};
\end{equation}

We claim that only the following two cases may occur:
\begin{enumerate}
\item[\textbf{Case 1}] $g_n([t_1,t_2])$ is essentially of type I;
\item[\textbf{Case 2}] $g_n([t_1,t_2])$ is essentially of type III.
\end{enumerate}
The reason is as follow.
For the first case, if $g_n(t_1)$ is of type I, then $$|c_{n-1,1}(t_1)-c_{n-1,2}(t_1)-s\alpha|>|I_n|,$$ for all $s\leq q_{N+n-1}^2.$ Note here $|I_n|$ can be seen as a constant independent on $t$ since the error $\delta\ll |I_n|^{C}$. Combining with $$|\frac{\partial(c_{n-1,1}(t)-c_{n-1,2}(t))}{\partial t}|\leq C\leq |I_n|^{-2},$$ we obtain $$|c_{n-1,1}(t)-c_{n-1,2}(t)-s\alpha|>|I_n|^{5}>|I_n|^{200},$$ for all $s\leq q_{N+n-1}^2$ and $t\in [t_1,t_2].$ Thus the first one holds.

For the second one, we assume $g_n(t_1)$ is of type III. Clearly, there exists some $l\leq q_{N+n-1}^2$ and $s\leq n$ such that $|c_{s-1,1}(t_1)-c_{s-1,2}(t_1)-l\alpha|<\frac{4}{3}|I_n|$. Note that we can assume $g_{s-1}(t_1)$ is of type I (otherwise select the biggest one). Due to the previous case implies $g_{s-1}([t_1,t_2])$ is essentially of type I, we have $$|\frac{\partial(c_{s-1,1}(t)-c_{s-1,2}(t))}{\partial t}|\leq C\leq |I_s|^{-200},$$ which implies $|c_{s-1,1}(t)-c_{s-1,2}(t)-l\alpha|<\frac{3}{2}|I_n|$ and hence $$|c_{n-1,1}(t)-c_{n-1,2}(t)-l\alpha|< \sum_{i=s}^{n-1} \|c_{i-1,1}(t)-c_{i,1}(t)\|+\sum_{i=s}^{n-1} \|c_{i-1,2}(t)-c_{i,2}(t)\|+\frac{3}{2}|I_n|\leq 2|I_n|.$$ Therefore the second case holds.

So we only need consider such two cases as above.

For the \textbf{Case 1,} $g_n([t_1,t_2])$ is essentially of type I. This case is easy. Note that $$|L_K(t_1)-L_K(t_2)|\leq |t_1-t_2|^{0.9}\leq |t_1-t_2|^{0.5};$$ $$|L_{2K}(t_1)-L_{2K}(t_2)|\leq |t_1-t_2|^{0.9}\leq |t_1-t_2|^{0.5}.$$ Therefore \begin{align*}
\vert L(t_1)-L(t_2)\vert &\leq \vert L_K(t_1)-L_K(t_2)\vert+\vert L_{2K}(t_1)-L_{2K}(t_2)\vert+\vert L_K(t_1)+L(t_1)-2L_{2K}(t_1)\vert\\&+\vert L_K(t_2)+L(t_2)-2L_{2K}(t_2)\vert\\&\leq 5|t_1-t_2|^{0.5} +2\lambda^{-\frac{c}{2}K}\\&\leq 10|t_1-t_2|^{0.5}.
\end{align*}

For the \textbf{Case 2,} $g_n([t_1,t_2])$ is essentially of type III. We denote the nearest gap $[t_1^0,t_2^0]$ such that $g_n(t_1^0)$ or $g_n(t_2^0)$ is essentially of type III. Consider the following three cases:
\begin{enumerate}
\item $t_1^0\leq t_1, t_2^0\geq t_2.$

This is trivial because $[t_1,t_2]\subset [t_1^0,t_2^0]$ implies the uniformly hyperbolic case.

\item $t_1^0\leq t_1\leq t_2^0\leq t_2.$

Note we have already obtained the following three things:
$$|L(t_2^0)-L(t_2)|\leq C' |t_2^0-t_2|^{0.5};$$
$$|L(t_2^0)-L(t_1)|\leq C' |t_2^0-t_1|^{0.5};$$
$$x^{0.5}+y^{0.5}\leq \sqrt{2}(x+y)^{0.5}~ for ~any ~x,y\geq 0.$$

Therefore, $$|L(t_1)-L(t_2)|\leq C' |t_2-t_2^0|^{0.5}+C' |t_2^0-t_1|^{0.5}\leq 2C'|t_1-t_2|.$$

\item $t_1^0\leq t_2^0\leq t_1\leq t_2.$

If $t_1\geq t_2^0+\lambda^{-q^2_{N+n-1}},$ then by the second fact in the beginning and
$$x^{0.5}-y^{0.5}\leq (x-y)^{0.5}~ for ~any ~x>y\geq 0 ,$$ we have:
$$|L(t_1)-L(t_2)|\leq C' |t_2-t_2^0|^{0.5}-C' |t_1-t_2^0|^{0.5}\leq C'|t_1-t_2|.$$

If $t_1< t_2^0+\lambda^{-q^2_{N+n-1}}:=t',$ then we have:
\begin{align*}|L(t_1)-L(t_2)| &\leq |L(t_1)-L(t')|+|L(t_2)-L(t')|\\&\leq |L(t_2)-L(t')|+|L(t_2^0)-L(t_1)|+|L(t_2^0)-L(t')|\\&\leq C'|t_2-t'|^{0.5}+2C'|t_1-t_2^0|^{0.5}+2C'|t'-t_2^0|^{0.5}\\&\leq 10C'|t_1-t_2|,\end{align*} where the last inequality is due to the previous case for $t'=t_2$ and the fact $|t_1-t_2|\geq|t_2-t'|\geq \lambda^{-\frac{c}{10}q_{N+n-1}}-\lambda^{-q^2_{N+n-1}}\geq \lambda^{-q^2_{N+n-1}}\geq |t'-t_2^0|\geq |t'-t_1|.$


The remaining cases are similarly by symmetry.


\end{enumerate}

\end{proof}


\section{\bf{Appendix}}

\textbf{A1.} Given $\epsilon_1,\epsilon_2\ll 1,$ $1\ll q \ll \min\{\frac{1}{\epsilon_1},\frac{1}{\epsilon_2}\}$, then we have
we have
\begin{equation}\label{jf1}\int_{0}^{\frac{1}{q}} \frac{x^2-\epsilon_1^2}{(x^2-\epsilon_1^2)^2+\epsilon_2^4} dx= C_1\frac{((\epsilon_1^4+\epsilon_2^4)^{\frac{1}{2}}-\epsilon_1^2)^{\frac{1}{2}}}{(\epsilon_1^4+\epsilon_2^4)^{\frac{1}{2}}}-C_2q\frac{(\epsilon_1^4+\epsilon_2^4)^{\frac{1}{2}}+\epsilon_1^2}{(\epsilon_1^4+\epsilon_2^4)^{\frac{1}{2}}}
\end{equation}
\begin{equation}\label{jf2}\int_{0}^{\frac{1}{q}} \frac{x^2+\epsilon_1^2}{(x^2+\epsilon_1^2)^2+\epsilon_2^4} dx= C_3\frac{((\epsilon_1^4+\epsilon_2^4)^{\frac{1}{2}}+\epsilon_1^2)^{\frac{1}{2}}}{(\epsilon_1^4+\epsilon_2^4)^{\frac{1}{2}}}-C_4q\frac{(\epsilon_1^4+\epsilon_2^4)^{\frac{1}{2}}-\epsilon_1^2}{(\epsilon_1^4+\epsilon_2^4)^{\frac{1}{2}}}
\end{equation}
\begin{equation}\label{jf3}\int_{0}^{\frac{1}{q}} \frac{sgn(x^2-\epsilon_1^2)}{\sqrt{(x^2-\epsilon_1^2)^2+\epsilon_2^4}}dx\leq C_5\frac{1}{\epsilon_1}, when \epsilon_1\gg \epsilon_2;\end{equation}
\begin{equation}\label{jf4}\int_{0}^{\frac{1}{q}} \frac{1}{\sqrt{(x^2+\epsilon_1^2)^2+\epsilon_2^4}}dx= C_6\frac{1}{\epsilon_1}, when \epsilon_1\gg \epsilon_2.\end{equation}
for some proper constants $C_i>0$,$1\leq i\leq 5$.

Moreover,
\begin{equation}\label{jf3}\int_{0}^{\frac{1}{q}} \frac{x^2-\epsilon_1^2}{(x^2-\epsilon_1^2)^2+\epsilon_2^4} dx= \left\{\begin{gathered} -Cq,\epsilon_1\geq\frac{1}{q}\epsilon^{\frac{2}{3}}\\ O(1)\frac{\epsilon_2^2}{\epsilon_1^3},  \epsilon^{\frac{2}{3}}\frac{1}{q}>\epsilon_1>\epsilon_2\\ O(1)\frac{1}{\epsilon_2},\epsilon_1\leq \epsilon_2\end{gathered}\right.
\end{equation}
\begin{equation}\label{jf4}\int_{0}^{\frac{1}{q}} \frac{x^2+\epsilon_1^2}{(x^2+\epsilon_1^2)^2+\epsilon_2^4} dx= \left\{\begin{gathered} O(1)\frac{1}{\epsilon_1}, \epsilon_1>\epsilon_2\\ O(1)\frac{1}{\epsilon_2}, \epsilon_1\leq \epsilon_2\end{gathered}\right.
\end{equation}
for some constant $C>0.$


\begin{proof}Note that
\begin{equation} \frac{x^2-r}{(x^2-px+q)(x^2+px+q)}=\frac{\frac{(r+q)x}{2pq}-\frac{r}{2q}}{x^2-px+q}+\frac{-\frac{(r+q)x}{2pq}-\frac{r}{2q}}{x^2+px+q}.
\end{equation}
Thus,\begin{align*}\int\frac{x^2-r}{(x^2-px+q)(x^2+px+q)}dx&=\frac{r+q}{4pq}\log(x^2-px+q)+\frac{q-r}{2q\sqrt{4q-p^2}}\arctan\frac{2x-p}{\sqrt{4q-p^2}}\\&+\frac{-r-q}{4pq}\log(x^2+px+q)+\frac{q-r}{2q\sqrt{4q-p^2}}\arctan\frac{2x+p}{\sqrt{4q-p^2}}+C
\end{align*}.Let$$p=\sqrt{2}\sqrt{\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}},q=\sqrt{\epsilon_1^4+\epsilon_2^4},r=\epsilon_1^2,$$ by straight calculation, we have
\clearpage
\begin{align*}\int_{0}^{\frac{1}{q}} \frac{x^2-\epsilon_1^2}{(x^2-\epsilon_1^2)^2+\epsilon_2^4} dx&=\left.\frac{\sqrt{2}}{8}\frac{\sqrt{\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}}}{\sqrt{\epsilon_1^4+\epsilon_2^4}}\log(x^2-\sqrt{2}\sqrt{\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}}x+\sqrt{\epsilon_1^4+\epsilon_2^4})\right|_{0}^{\frac{1}{q}}\\&-
\left.\frac{\sqrt{2}}{8}\frac{\sqrt{\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}}}{\sqrt{\epsilon_1^4+\epsilon_2^4}}\log(x^2+\sqrt{2}\sqrt{\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}}x+\sqrt{\epsilon_1^4+\epsilon_2^4})\right|_{0}^{\frac{1}{q}}\\&+
\left.\frac{\sqrt{2}}{4}\frac{\sqrt{\sqrt{\epsilon_1^4+\epsilon_2^4}-\epsilon_1^2}}{\sqrt{\epsilon_1^4+\epsilon_2^4}}\arctan\frac{2x-\sqrt{2}\sqrt{\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}}}{\sqrt{2(\sqrt{\epsilon_1^4+\epsilon_2^4}-\epsilon_1^2)}}\right|_{0}^{\frac{1}{q}}\\&+
\left.\frac{\sqrt{2}}{4}\frac{\sqrt{\sqrt{\epsilon_1^4+\epsilon_2^4}-\epsilon_1^2}}{\sqrt{\epsilon_1^4+\epsilon_2^4}}\arctan\frac{2x+\sqrt{2}\sqrt{\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}}}{\sqrt{2(\sqrt{\epsilon_1^4+\epsilon_2^4}-\epsilon_1^2)}}\right|_{0}^{\frac{1}{q}}
\\&=\left.\frac{\sqrt{2}}{8}\frac{\sqrt{\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}}}{\sqrt{\epsilon_1^4+\epsilon_2^4}}\log(1-\frac{2\sqrt{2}\sqrt{\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}}x}{x^2+\sqrt{2}\sqrt{\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}}x+\sqrt{\epsilon_1^4+\epsilon_2^4}})\right|_{0}^{\frac{1}{q}}
\\&+\frac{\sqrt{2}}{4}\frac{\sqrt{\sqrt{\epsilon_1^4+\epsilon_2^4}-\epsilon_1^2}}{\sqrt{\epsilon_1^4+\epsilon_2^4}}(\arctan\frac{2x-\sqrt{2}\sqrt{\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}}}{\sqrt{2(\sqrt{\epsilon_1^4+\epsilon_2^4}-\epsilon_1^2)}}\\&+\left.\arctan\frac{2x+\sqrt{2}\sqrt{\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}}}{\sqrt{2(\sqrt{\epsilon_1^4+\epsilon_2^4}-\epsilon_1^2)}})\right|_{0}^{\frac{1}{q}}\\&=
\frac{\sqrt{2}}{8}\frac{\sqrt{\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}}}{\sqrt{\epsilon_1^4+\epsilon_2^4}}\log(1-O(2\sqrt{2}\sqrt{\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}}q))\\&+\frac{\sqrt{2}}{4}\frac{\sqrt{\sqrt{\epsilon_1^4+\epsilon_2^4}-\epsilon_1^2}}{\sqrt{\epsilon_1^4+\epsilon_2^4}}[(\pi-O(\frac{q\epsilon_2^2}{\sqrt{\sqrt{\epsilon_1^4+\epsilon_2^4}+\epsilon_1^2}}))\\&-(-\arctan\sqrt{\frac{\sqrt{\epsilon_1^4+\epsilon_2^4}}{\sqrt{\epsilon_1^4+\epsilon_2^4}-\epsilon_1^2}}+\arctan\sqrt{\frac{\sqrt{\epsilon_1^4+\epsilon_2^4}}{\sqrt{\epsilon_1^4+\epsilon_2^4}-\epsilon_1^2}})]\\&
=-C_2q\frac{\sqrt{\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}}}{\sqrt{\epsilon_1^4+\epsilon_2^4}}\sqrt{\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}}+C_1\frac{\sqrt{\sqrt{\epsilon_1^4+\epsilon_2^4}-\epsilon_1^2}}{\sqrt{\epsilon_1^4+\epsilon_2^4}}
\end{align*}.
which implies \eqref{jf1}. Similarly, \eqref{jf2} holds if we denote
    $$p=\sqrt{2}\sqrt{-\epsilon_1^2+\sqrt{\epsilon_1^4+\epsilon_2^4}},q=\sqrt{\epsilon_1^4+\epsilon_2^4},r=-\epsilon_1^2.$$

The \eqref{jf4} is obvious if one note the following estimate,
$$\frac{C'}{\sqrt{\epsilon_1^2+\epsilon_2^2}}= \int_{0}^{\frac{1}{q}}\frac{1}{x^2+\epsilon_1^2+\epsilon_2^2}dx\leq \int_{0}^{\frac{1}{q}}\frac{1}{\sqrt{(x^2+\epsilon_1^2)^2+\epsilon_2^4}}dx\leq\int_{0}^{\frac{1}{q}}\frac{1}{x^2+\epsilon_1^2}dx=C''\frac{1}{\epsilon_1}.$$

For the \eqref{jf3}, it's also clear if we use the following variable substitution.

$$\int_{0}^{\frac{1}{q}} \frac{sgn(x^2-\epsilon_1^2)}{\sqrt{(x^2-\epsilon_1^2)^2+\epsilon_2^4}}dx\leq\left|\int_{0}^{\sqrt{2}\epsilon_1} \frac{sgn(x^2-\epsilon_1^2)}{\sqrt{(x^2-\epsilon_1^2)^2+\epsilon_2^4}}dx\right|+\left|\int_{\sqrt{2}\epsilon_1}^{\frac{1}{q}} \frac{sgn(x^2-\epsilon_1^2)}{\sqrt{(x^2-\epsilon_1^2)^2+\epsilon_2^4}}dx\right|.$$

For the second items in the right, it's obvious by direct calculation.
$$\left|\int_{\sqrt{2}\epsilon_1}^{\frac{1}{q}} \frac{sgn(x^2-\epsilon_1^2)}{\sqrt{(x^2-\epsilon_1^2)^2+\epsilon_2^4}}dx\right|\leq \int_{\sqrt{2}\epsilon_1}^{\frac{1}{q}} \frac{1}{(x^2-\epsilon_1^2)}dx =C\frac{1}{\epsilon_1}.$$

For the first one, we have
\begin{equation}\label{jf100}\left|\int_{0}^{\sqrt{2}\epsilon_1^2} \frac{sgn(x^2-\epsilon_1^2)}{\sqrt{(x^2-\epsilon_1^2)^2+\epsilon_2^4}}dx\right|=\left|\int_{0}^{\epsilon_1^2} \frac{sgn(x^2-\epsilon_1^2)}{\sqrt{(x^2-\epsilon_1^2)^2+\epsilon_2^4}}dx+\int_{\epsilon_1^2}^{\sqrt{2}\epsilon_1^2} \frac{sgn(x^2-\epsilon_1^2)}{\sqrt{(x^2-\epsilon_1^2)^2+\epsilon_2^4}}dx\right|.\end{equation}

For the two items in the right side, separately let $\epsilon_1^2-x^2=t$ and $x^2-\epsilon_1^2=t.$

Thus for a sufficiently large $\lambda>0,$ combining with the fact $\epsilon_1\gg \epsilon_2$ we have

$$\eqref{jf100}=\int_{0}^{\epsilon_1^2}\frac{tdt}{\sqrt{t^2+\epsilon_2^4}\sqrt{\epsilon_1^4-t^2}(\sqrt{\epsilon_1^2-t}+\sqrt{\epsilon_1^2+t})}$$
$$=\int_{0}^{\frac{1}{\lambda}\epsilon_1^2}\frac{tdt}{\sqrt{t^2+\epsilon_2^4}\sqrt{\epsilon_1^4-t^2}(\sqrt{\epsilon_1^2-t}+\sqrt{\epsilon_1^2+t})}+\int_{\frac{1}{\lambda}\epsilon_1^2}^{\epsilon_1^2}\frac{tdt}{\sqrt{t^2+\epsilon_2^4}\sqrt{\epsilon_1^4-t^2}(\sqrt{\epsilon_1^2-t}+\sqrt{\epsilon_1^2+t})}$$
$$\leq \frac{\frac{1}{\lambda}\epsilon_1^2}{\sqrt{1-\frac{1}{\lambda^2}}\epsilon_1^3}+\int_{\lambda^{-1}\epsilon_1^2}^{\epsilon_1^2}\frac{1}{\sqrt{\epsilon_1^4-t^2}(\sqrt{\epsilon_1^2-t}+\sqrt{\epsilon_1^2+t})}dt$$
Let $\sqrt{\epsilon_1^2-t}=x$, we'll obtain
$$\eqref{jf100}\leq \frac{1}{2\lambda}\frac{1}{\epsilon_1}+\int_{0}^{\sqrt{1-\lambda^{-1}}\epsilon_1}\frac{1}{\sqrt{x\sqrt{2\epsilon_1^2-x^2}+2\epsilon_1^2-x^2}}dx.$$
Then let $x=\sqrt{2}\epsilon_1 \sin\theta$
$$\eqref{jf100}\leq \frac{1}{2\lambda}\frac{1}{\epsilon_1}+\frac{1}{\epsilon_1}\int_{\frac{\pi}{4}}^{\arcsin\frac{1-\lambda^{-1}}{\sqrt{2}}+\frac{\pi}{4}}\frac{1}{\sin\theta}d\theta.$$
$$\leq C\frac{1}{\epsilon_1},$$ which implies \eqref{jf3}.
\end{proof}

\textbf{A2.} For any $a,b,c,d,e\in \R$ satisfying $(b^4+a^4)d-a^4b^4c>0$, consider the following integral:
$$\int \frac{1}{\sqrt{c^2x^2+e^2(d^{-2}x^{-2})+\frac{2}{a^4}+\frac{2}{b^4}}} dx.$$

We try to calculate it out.

At first, apply linearity, we have $$={a^4b^4d}{\cdot}{\displaystyle\int}\dfrac{1}{\sqrt{a^8b^8c^2d^2x^2+\frac{a^8b^8e^2}{x^2}+\left(2a^4b^8+2a^8b^4\right)d^2}}\,\mathrm{d}x$$

Second, we substitute $u=x^2,$ it holds that
$$={\dfrac{a^4b^4d}{2}}{\displaystyle\int}\dfrac{1}{\sqrt{a^4b^4d^2\left(a^4b^4c^2u^2+2\left(b^4+a^4\right)u\right)+ea^4b^4}}\,\mathrm{d}u.$$

Next, by substituting $v=\dfrac{d\left(a^4b^4c^2u+b^4+a^4\right)}{(b^4+a^4)^2d^2-(a^4b^4e)^2c^2},$ we have
$$=\frac{1}{2c}\int\frac{1}{\sqrt{v^2-1}}dv=\frac{1}{2c}\log|\sqrt{v^2-1}+v|.$$

Eventually, it's not difficult to figure out the answer as follow:

$$\dfrac{\log\left|\sqrt{d^2c^4a^8b^8x^4+2d^2c^2a^4b^4(a^4+b^4)x^2+(a^4b^4e)^2c^2}+d\left(a^4b^4c^2x^2+b^4+a^4\right)\right|}{2c}+C$$

\vskip0.3cm

\textbf{A3.} \begin{align*}
&K_1\leq (1+O(\lambda^{-6}_1+\lambda^{-6}_2))\frac{\epsilon^2_2}{2c^2_m}\left(\left.\frac{(c^*)^3}{4\epsilon^3_1}\log|\frac{x+\frac{\epsilon_1}{c^*}}{x-\frac{\epsilon_1}{c^*}}|-\frac{x}{2(\frac{\epsilon_1}{c^*})^2(x^2-(\frac{\epsilon_1}{c^*})^2)}\right)\right|_{\frac{\sqrt{\epsilon_1^2+\epsilon_2^2}}{c_m}}^{\epsilon^{\frac{1}{8}}_1}+(1+O(\lambda^{-6}_1+\lambda^{-6}_2))
\\&\left.\frac{\sqrt{2}}{8c^2_m}\frac{\sqrt{(\frac{\epsilon_1}{c^*})^2+\sqrt{(\frac{\epsilon_1}{c^*})^4+(\frac{\epsilon_2}{c_m})^4}}}{\sqrt{(\frac{\epsilon_1}{c^*})^4+(\frac{\epsilon_2}{c_m})^4}}\log(1-\frac{2\sqrt{2}\sqrt{(\frac{\epsilon_1}{c^*})^2+\sqrt{(\frac{\epsilon_1}{c^*})^4+(\frac{\epsilon_2}{c_m})^4}}x}{x^2+\sqrt{2}\sqrt{(\frac{\epsilon_1}{c^*})^2+\sqrt{(\frac{\epsilon_1}{c^*})^4+(\frac{\epsilon_2}{c_m})^4}}x+\sqrt{(\frac{\epsilon_1}{c^*})^4+(\frac{\epsilon_2}{c_m})^4}})\right|_{\frac{\sqrt{\epsilon_1^2+\epsilon_2^2}}{c_m}}^{\epsilon^{\frac{1}{8}}_1}
\\& +\frac{\sqrt{2}}{4c^2_m}\frac{\sqrt{\sqrt{(\frac{\epsilon_1}{c^*})^4+(\frac{\epsilon_2}{c_m})^4}-(\frac{\epsilon_1}{c^*})^2}}{\sqrt{(\frac{\epsilon_1}{c^*})^4+(\frac{\epsilon_2}{c_m})^4}}(\arctan\frac{2x-\sqrt{2}\sqrt{(\frac{\epsilon_1}{c^*})^2+\sqrt{(\frac{\epsilon_1}{c^*})^4+(\frac{\epsilon_2}{c_m})^4}}}{\sqrt{2(\sqrt{(\frac{\epsilon_1}{c^*})^4+(\frac{\epsilon_2}{c_m})^4}-(\frac{\epsilon_1}{c^*})^2)}}\\&+\left.\arctan\frac{2x+\sqrt{2}\sqrt{(\frac{\epsilon_1}{c_m})^2+\sqrt{(\frac{\epsilon_1}{c^*})^4+(\frac{\epsilon_2}{c_m})^4}}}{\sqrt{2(\sqrt{(\frac{\epsilon_1}{c^*})^4+(\frac{\epsilon_2}{c_m})^4}-(\frac{\epsilon_1}{c^*})^2)}})\right|_{\frac{\sqrt{\epsilon_1^2+\epsilon_2^2}}{c_m}}^{\epsilon^{\frac{1}{8}}_1}
\\&\leq (1+O(\lambda^{-6}_1+\lambda^{-6}_2))C(\frac{\epsilon_2^2\log|\epsilon_2|}{\epsilon^3_1}+\epsilon^{-\frac{1}{8}}+\epsilon^{-1}_1)+(1+O(\lambda^{-6}_1+\lambda^{-6}_2))\cdot
\\&\frac{\sqrt{2}}{8c^2_m}\frac{\sqrt{(\frac{\epsilon_1}{c^*})^2+\sqrt{(\frac{\epsilon_1}{c^*})^4+(\frac{\epsilon_2}{c_m})^4}}}{\sqrt{(\frac{\epsilon_1}{c^*})^4+(\frac{\epsilon_2}{c_m})^4}}\cdot 2\log|\frac{(\frac{\epsilon_2}{c_m})^2}{\frac{\epsilon_1^2+\epsilon_2^2}{c^2_m}+\sqrt{2}\sqrt{(\frac{\epsilon_1}{c^*})^2+\sqrt{(\frac{\epsilon_1}{c^*})^4+(\frac{\epsilon_2}{c_m})^4}}\frac{\sqrt{\epsilon_1^2+\epsilon_2^2}}{c_m}+\sqrt{(\frac{\epsilon_1}{c^*})^4+(\frac{\epsilon_2}{c_m})^4}}|
\\&+C\epsilon^{-1}_1\end{align*}

For $K_2$, similarly we have
\begin{align*}&K_2\\&\geq \frac{\sqrt{2}}{8c^2_m}\frac{\sqrt{(\frac{\epsilon_1}{c_m})^2+\sqrt{(\frac{\epsilon_1}{c_m})^4+(\frac{\epsilon_2}{c_m})^4}}}{\sqrt{(\frac{\epsilon_1}{c_m})^4+(\frac{\epsilon_2}{c_m})^4}}\\&\left.\log(1-\frac{2\sqrt{2}\sqrt{(\frac{\epsilon_1}{c_m})^2+\sqrt{(\frac{\epsilon_1}{c_m})^4+(\frac{\epsilon_2}{c_m})^4}}x}{x^2+\sqrt{2}\sqrt{(\frac{\epsilon_1}{c_m})^2+\sqrt{(\frac{\epsilon_1}{c_m})^4+(\frac{\epsilon_2}{c_m})^4}}x+\sqrt{(\frac{\epsilon_1}{c_m})^4+(\frac{\epsilon_2}{c_m})^4}})\right|_{\epsilon^8_1}^{\frac{\sqrt{\epsilon_1^2-\epsilon_2^2}}{c_m}}
\\& +\frac{\sqrt{2}}{4c^2_m}\frac{\sqrt{\sqrt{(\frac{\epsilon_1}{c_m})^4+(\frac{\epsilon_2}{c_m})^4}-(\frac{\epsilon_1}{c_m})^2}}{\sqrt{(\frac{\epsilon_1}{c_m})^4+(\frac{\epsilon_2}{c_m})^4}}(\arctan\frac{2x-\sqrt{2}\sqrt{(\frac{\epsilon_1}{c_m})^2+\sqrt{(\frac{\epsilon_1}{c_m})^4+(\frac{\epsilon_2}{c_m})^4}}}{\sqrt{2(\sqrt{(\frac{\epsilon_1}{c_m})^4+(\frac{\epsilon_2}{c_m})^4}-(\frac{\epsilon_1}{c_m})^2)}}\\&+\left.\arctan\frac{2x+\sqrt{2}\sqrt{(\frac{\epsilon_1}{c_m})^2+\sqrt{(\frac{\epsilon_1}{c_m})^4+(\frac{\epsilon_2}{c_m})^4}}}{\sqrt{2(\sqrt{(\frac{\epsilon_1}{c_m})^4+(\frac{\epsilon_2}{c_m})^4}-(\frac{\epsilon_1}{c_m})^2)}})\right|_{\epsilon^8_1}^{\frac{\sqrt{\epsilon_1^2-\epsilon_2^2}}{c_m}}
\\&\geq (1-O(\lambda^{-6}_1+\lambda^{-6}_2))\frac{\sqrt{2}}{8}\frac{\sqrt{(\frac{\epsilon_1}{c_m})^2+\sqrt{(\frac{\epsilon_1}{c_m})^4+(\frac{\epsilon_2}{c_m})^4}}}{\sqrt{(\frac{\epsilon_1}{c_m})^4+(\frac{\epsilon_2}{c_m})^4}}\cdot \\&2\log|\frac{\sqrt{2}(\frac{\epsilon_2}{c_m})^2}{\frac{\epsilon_1^2-\epsilon_2^2}{c^2_m}+\sqrt{2}\sqrt{(\frac{\epsilon_1}{c_m})^2+\sqrt{(\frac{\epsilon_1}{c_m})^4+(\frac{\epsilon_2}{c_m})^4}}\frac{\sqrt{\epsilon_1^2-\epsilon_2^2}}{c_m}+\sqrt{(\frac{\epsilon_1}{c^*})^4+(\frac{\epsilon_2}{c_m})^4}}|
-C\epsilon^{-1}_1\end{align*}

\textbf{A.4}
\begin{proof} It is enough to prove the theorem for sufficiently large $i$. So for any fixed $i$, there exists some $L>0$ such that $q_{N+L-1}^4\leq i\leq q_{N+L}^4.$ From the Induction Theorem of \cite{wz1}, it's clear that no matter which type (I, II, or III) $g_n$ is, there always exists some (at most two) intervals $\bar{I}_{\delta}\subset I_n$ with their measure satisfying $|\bar{I}_{\delta}|\leq \lambda^{-\delta}$  such that $|sin(g_n)|\geq \lambda^{-3\delta}$ for $x\notin \bar{I}_{\delta}.$ The main problem of the proof is how to divide the matrix $A_n$ so that we can
apply the Induction Theorem in \cite{wz1}. It's worth noting that when we use the Induction Theorem it's no need starting from and ending at critical interval $I_n,$ which is followed from lemma \ref{lmwz0}.

Now we start our proof.
By Diophantine condition, we have the following three cases.
\begin{enumerate}
\item[\textbf{CASE A:}] There exists only one $0\leq l\leq i$ such that $T^lx\in I_{L+1}.$

\noindent In this case, no resonance case occur. We can directly consider $A_i(x)=A_{i-l}(T^lx)\cdot A_l(x).$  Obviously, it follows from \cite{wz1} that only the following two cases need to be considered.
 \item[\textbf{a1:}]If $\min\{i-l, l\}\geq\frac{1}{50}q_{N+L-1}^4,$ then we have $\min\{\|A_l\|,\|A_{i-l}\|\}\geq \min\{\lambda^{\frac{2}{3}l},\lambda^{\frac{2}{3}(i-l)}\}\geq \lambda^{\frac{1}{75}q^4_{N+L-1}}$. it's clear that the angle between $A_l$ and $A_{i-l}$, says $\tilde{g}_l$ is similar to the function $g_{L+1},$ which can be seen in the following estimate :
     $$\|\tilde{g}_l-g_{L+1}\|_{C^2(I_{L+1})}\leq (\min\{\|A_l\|,\|A_{i-l}\|\})^{-1.5}\leq \lambda^{\frac{1}{75}\times \frac{3}{2}q^4_{N+L-1}}\leq \lambda^{-\frac{1}{50}q^4_{N+L-1}};$$
     Thus, by direct calculation, we have
    \begin{align*}
    \|A_i(x)\|&\geq \cos(\frac{\pi}{2}-\bar{g}_{l}(T^lx))\cdot\|A_l(x)\|\cdot\|A_{i-l}(T^lx)\|\\
     &\geq\lambda^{\frac{999}{1000}(i-l)}\cdot\lambda^{-\frac{3}{500}i}\\
     &\geq \lambda^{\frac{9}{10}i}.
     \end{align*}
     for $x\notin \bigcup\limits_{0\leq j\leq i}B(c_{L+1}+j\alpha,\lambda^{-\frac{1}{500}i})\triangleq\bar{B}_1.$

  \noindent   \item[\textbf{a2:}]$i-l$ or $l$ is less than $\frac{1}{50}q_{N+L-1}^4$. Then it holds that $$\|A_i(x)\|\geq \max\{\|A_l(x)\|\cdot \|A_{i-1}(T^lx)\|^{-1},\|A_l(x)\|^{-1}\cdot \|A_{i-1}(T^lx)\|\}\geq \lambda^{\frac{9}{10}i}.$$
     \item[\textbf{CASE B:}] There exists $0\leq l_1,l_2\leq i$ such that $T^{l_1}x,T^{l_2}x\in I_{L+1}.$


     \noindent In this case, the resonance may occur and we consider $$A_i(x)=A_{i-l_2}A_{l_2-l_1}A_{l_1}.$$
     The following several subcases need to be thought about.
     \begin{enumerate}
     \item[\textbf{b1:}] $\min\{i-l_2,l_2-l_1,l_1\}>\frac{1}{50}q_{N+n-1}^4.$
     \vskip 0.cm
    \noindent In this situation, no resonance occur, hence it's absolutely same as the first case.

    For $x\notin$ $\bigcup\limits_{0\leq j_1,j_2\leq i}\bigcup\limits_{1\leq k\leq 2}\bigcup\limits_{0\leq m\leq 4} B(c_{L+1,m}+j_k\alpha,\lambda^{-\frac{1}{500}i})\triangleq \bar{B}_2,$ it is clear that
     \begin{align*}
     \|A_i(x)\|&\geq \cos(\frac{\pi}{2}-\bar{g}_{l_2}(T^{l_2}x))\cdot \cos(\frac{\pi}{2}-\bar{g}_{l_1}(T^{l_1}x))\cdot\|A_{l_1}\|\cdot\|A_{l_2-l_1}\|\cdot\|A_{i-l_2}\|\\
     &\geq \lambda^{-\frac{3}{500}i}\cdot\lambda^{-\frac{3}{500}i}\cdot\lambda^{\frac{999}{1000}(i-l_2)}\lambda^{\frac{999}{1000}(l_2-l_1)}\lambda^{\frac{999}{1000}(l_1)}\\
     &\geq\lambda^{-\frac{9}{10}i}.
     \end{align*}
          Here $c_{L+1,1},c_{L+1,2},c_{L+1,3}~and~c_{L+1,4}$ are the possible four critical points when $g_{L+1}$ is of type III. \vskip 0.3cm
     \item[\textbf{b2:}] $\min\{i-l_2,l_2-l_1,l_1\}\leq \frac{1}{50}q_{N+n-1}^4.$
     \vskip 0.2cm
     Clearly, $\max\{i-l_2,l_2-l_1,l_1\}> \frac{1}{3}i>\frac{1}{3}q^4_{N+L-1}.$ And the maximum of the three cannot be equal to the minimum of the three in this case.
     Hence by the symmetry, we only need to consider the following three cases:
     \item[\textbf{b2.1:}]If $i-l_2\leq \frac{1}{50}q_{N+n-1}^4,$ and $l_2-l_1\geq \frac{1}{3}i,$

     then either $l_1<\frac{1}{50}q^4_{N+L-1},$ which implies \begin{align*}\|A_i\|&\geq \|A_{i-l_2}\|^{-1}\cdot\|A_{l_2-l_1}\|\cdot \|A_{l_1}\|^{-1}\geq \lambda^{\frac{999}{1000}(i-(i-l_2)-l_1)-\frac{2}{50}q^4_{N+L-1}}\geq \lambda^{((\frac{999}{1000}-\frac{2}{50})-\frac{2}{50})i}\geq \lambda^{\frac{9}{10}i},\end{align*}

     or $l_1>\frac{1}{50}q_{N+L-1}^4$, which implies the angle between $A_{l_2-l_1}$ and $A_{l_1}$, says $\bar{g}_{l_1}$ is similar to $g_{L+1}$. Thus we can deal with $A_{l_2-l_1}A_{l_1}$ in
     the same way as the``no resonance" case.  By straight calculation, we have
     \begin{align*}
     \|A_i(x)\|&\geq \|A_{i-l_2}\|^{-1}\cdot \|A_{l_2-l_1}\|\cdot \|A_{l_1}\|\cdot \cos(\frac{\pi}{2}-\bar{g}_{l_1}(T^{l_1}x))\\
     &\geq \lambda^{-\frac{1}{50}q^4_{N+L-1}}\cdot\lambda^{\frac{999}{1000}(l_2-l_1)}\cdot\lambda^{-\frac{999}{1000}l_1}\cdot\lambda^{-\frac{3}{500}i}\geq \lambda^{\frac{9}{10}i}.
     \end{align*}
     for $x\notin$ $\bigcup\limits_{0\leq j\leq i}\left(B(c_{L+1,1}+j\alpha,\lambda^{-\frac{1}{500}i})\bigcup B(c_{L+1,2},\lambda^{-\frac{1}{500}i})\right)\triangleq \bar{B}_3.$
\vskip 0.2cm
     \item[\textbf{b2.2:}]If $i-l_2<\frac{1}{50}q_{N+L-1}^4$ and $l_1>\frac{1}{3}i$,

     then either $l_2-l_1\geq \frac{1}{50}q^4_{N+L-1}$, which implies the angle between $A_{l_2-l_1}$ and $A_{l_1}$, says $\bar{g}_{l_1}$ is similar to $g_{L+1}$ and we have
     \begin{align*}
     \|A_i(x)\|&\geq \|A_{i-l_2}\|^{-1}\cdot \|A_{l_2-l_1}\|\cdot \|A_{l_1}\|\cdot \cos(\frac{\pi}{2}-\bar{g}_{l_1}(T^{l_1}x))\\
     &\geq \lambda^{-\frac{1}{50}q^4_{N+L-1}}\cdot\lambda^{\frac{999}{1000}(l_2-l_1)}\cdot\lambda^{-\frac{999}{1000}l_1}\cdot\lambda^{-\frac{3}{500}i}\geq \lambda^{\frac{9}{10}i}.
     \end{align*}
     for $x\notin$ $\bar{B}_3$;

     or $l_2-l_1\leq \frac{1}{50}q^4_{N+L-1}$, which implies which implies \begin{align*}\|A_i\|&\geq \|A_{i-l_2}\|^{-1}\cdot\|A_{l_2-l_1}\|\cdot \|A_{l_1}\|^{-1}\geq \lambda^{\frac{999}{1000}(i-(i-l_2)-l_1)-\frac{2}{50}q^4_{N+L-1}}\geq \lambda^{((\frac{999}{1000}-\frac{2}{50})-\frac{2}{50})i}\geq \lambda^{\frac{9}{10}i},\end{align*}
\vskip 0.2cm
     \item[\textbf{b2.3:}]If $l_2-l_1<\frac{1}{50}q_{N+L-1}^4$ and $l_1>\frac{1}{3}i,$

      then the resonance occur and only the following two cases is possible.
\begin{enumerate}
      \item[\textbf{b2.3.1:}]$i-l_2<\frac{1}{50}q_{N+L-1}^4.$ This implies \begin{align*}\|A_i\|&\geq \|A_{i-l_2}\|^{-1}\cdot\|A_{l_2-l_1}\|^{-1}\cdot \|A_{l_1}\|\geq \lambda^{\frac{999}{1000}(i-(i-l_2)-(l_2-l_1))-\frac{2}{50}q^4_{N+L-1}}\geq \lambda^{((\frac{999}{1000}-\frac{2}{50})-\frac{2}{50})i}\geq \lambda^{\frac{9}{10}i},\end{align*}

      \item[\textbf{b2.3.2:}]$i-l_2\geq\frac{1}{50}q_{N+L-1}^4.$ In this case, by the idea of \cite{wz1}, we need to consider $A_i=A_{i-l_2}\left(A_{l_2-l_1}A_{l_1}\right)=A_{i-l_2}B_{l_2}.$ We denote $\bar{g}_{l_2}$ the angle between $A_{i-l_2}$ and $B_{l_2}.$ It's clear that $\bar{g}_{l_2}$ is similar to the $g_{L+1}$, which has four critical points $\bar{c}_1$ and $\bar{c}_2'$, which locate at $I_{L+1,1}$; and $\bar{c}_2$ and $\bar{c}_1'$, which locate at $I_{L+1,2}$. It holds from the direct calculation that
     \begin{align*}
     \|A_i(x)\|&\geq \|A_{l_2-l_1}\|^{-1}\cdot \|A_{i-l_2}\|\cdot \|A_{l_1}\|\cdot \cos(\frac{\pi}{2}-\bar{g}_{l_2}(T^{l_1}x))\\
     &\geq \lambda^{\frac{999}{1000}(i-l_2)}\left(\lambda^{\frac{999}{1000}l_1}\cdot\lambda^{-(l_2-l_1)}\lambda^{-\frac{3}{500}i}\right)\geq \lambda^{\frac{9}{10}i}.
     \end{align*}
     for $x\notin$ $$\bigcup\limits_{0\leq j\leq i}\left(B(\bar{c}_{1}+j\alpha,\lambda^{-\frac{1}{500}i})\bigcup B(\bar{c}_{2}+j\alpha,\lambda^{-\frac{1}{500}i})\bigcup B(\bar{c}_{1}'+j\alpha,\lambda^{-\frac{1}{500}i})\bigcup B(\bar{c}_{2}',\lambda^{-\frac{1}{500}i})\right)\triangleq \bar{B}_4.$$
\end{enumerate}
     \end{enumerate}
     \item[\textbf{CASE C:}] There is no $0\leq l\leq i$ such that $T^lx\in I_{L+1}$.
     It is obvious that $\|A_i\|\geq \lambda^{\frac{999}{1000}i}\geq \lambda^{\frac{9}{10}i}.$

     In conclusion, in any case for sufficient large $i$ and $\lambda$, there exists a set $K=\bigcup\limits_{i=1}^{4}\bar{B}_i$ and a constant c such that $Leb\{K\}<\lambda^{-ci}$ and $\|A_i\|\geq \lambda^{\frac{9}{10}i}$ for any $x\notin K.$
     \end{enumerate}
     \end{proof}

\textbf{A.5}
\begin{proof}We process it by induction.

     \textbf{Case $n=1\rightarrow 2$:}

     The beginning step $n=1$ is trivial. When we come to the case $n=2,$ it follows from \cite{wz1} that the following three conditions may occur.
     \begin{enumerate}\item[Type $I_1\rightarrow I_2$:] In this case, we write $A_{m^2_1}$ as $A_{s_t-s_{t-1}}\cdots A_{s_1}$ where $s_i$ are all returning times to $I_1$ of $x.$ If $t=1$, then we come back to the case $n=1.$ Thus we only need to consider $t>1.$ We focus on the following two cases:
     \begin{enumerate}
     \item[$s_t-s_{t-1}< q_{N+1}$:] Note that $s_t-s_{t-1}\ll q_{N+1}^2\ll q^4_{N+1}-q^2_{N+1}\leq s_{t-1}.$ By the help of lemma \ref{lm5}, we have $$\|A_{m^2_1}\|\geq \frac{\|A_{s_{t-1}}\|}{\|A_{s_{t}-s_{t-1}}\|}\geq \lambda^{(1-\epsilon)m^2_1};$$ $$|s(A_{m^2_1})-s(A_{s_{t-1}})|_{C^2}\leq \|A_{s_{t-1}}\|^{-1.5}.$$ Therefore $$|g_2-s(A_{m^2_1})|_{C^2}\leq |s(A_{m^2_1})-s(A_{s_{t-1}})|_{C^2}+|g_2-s(A_{s_{t-1}})|_{C^2}\leq \|A_{s_{t-1}}\|^{-1.5}+\lambda^{-1.5r_1}\leq \lambda^{-0.5r_1}.$$ Similarly, $$\|A_{m^2_2}\|\geq \lambda^{(1-\epsilon)m^2_2};$$ $$|g_2-u(A_{m^2_2})|_{C^2}\leq \lambda^{-0.5r_1}.$$ In conclusion, $|g_{m^2_1,m^2_2,2}-g_2|_{C^2}\leq \lambda^{-0.5r_1}.$
     \item[$s_t-s_{t-1}\geq q_{N+1}$:] By the definition of $m^2_1,$ it's clear that $T^{s_t}x\notin I_2,$ which implies $|T^{s_t}x-c_2|\geq \frac{1}{q^{8\tau}_{N+1}}$ and $g_2(T^{s_t}x)\geq \frac{C}{q^{24\tau}_{N+1}}.$ Denote $\tilde{g}_2\triangleq s(A_{s_{t}-s_{t-1}})-u(A_{s_{t-1}})$ and $\tilde{c}_2$ be the zero point of $\tilde{g}_2$. It follows from the conclusion of the previous step that $|\tilde{g}_2(T^{s_t}x)-g_2(T^{s_t}x)|_{C^2}\leq \lambda^{-\frac{1}{2}r_0}\ll \frac{C}{q^{24\tau}_{N+1}}.$ Therefore, it follows from lemma \ref{lm5} that $\|A_{m^2_1}\|\geq \|A_{s_{t-1}}\| \frac{C}{q^{24\tau}_{N+1}} \|A_{s_{t}-s_{t-1}}\|\geq \lambda^{(1-\epsilon)m^2_1}.$ Similarly, we have $\|A_{m^2_1}\|\geq \lambda^{(1-\epsilon)m^2_2}.$ Again from lemma \ref{lm5}, it's not difficult to see that $|g_{m^2_1,m^2_2,2}-g_2|_{C^2}\leq \lambda^{-0.5r_1}.$
     \end{enumerate}
     \item[Type $I_1 \rightarrow III_2$:] We assume that $I_{2,1}+k\alpha\cap I_{2,2}\neq \emptyset,$ for some $k\leq q^2_{N+1}.$
     \begin{enumerate}
     \item[$x\in I_{2,1}$:] In this case, note that $r_2(x)=k\leq q^2_{N+1}.$ So no resonance happen in this condition and all the estimates are quite similar to type I to type I.
     \item[$x\in I_{2,2}$:] Also no resonance happen in this case, thus it's same as the above case.
     \end{enumerate}
     \item[Type $II_1\rightarrow I_2,II_2~or~III_2$:] This is same as Type I$\rightarrow$ Type I and Type I$\rightarrow$ Type III.

     \end{enumerate}

     Now we assume that for all the case $n\leq q$ the conclusion holds.

     \textbf{Case $n=p\rightarrow p+1$:} The following conditions need be considered.
     \begin{enumerate}
     \item[Type $I_p\rightarrow I_{p+1}$:]This similar to the original case but we still give it out for completeness. We write $A_{m^{p+1}_1}$ as $A_{s_t'-s_{t'-1}}\cdots A_{s_1}$ where $s_i$ are all returning times to $I_1$ of $x.$ If $t=1$, then we need do nothing. Thus we only need to consider $t'>1.$ We focus on the following two cases:
     \begin{enumerate}
     \item[$s_t'-s_{t'-1}< q_{N+p}$:] Note that $s_t'-s_{t'-1}\ll q_{N+p}^2\ll q^4_{N+p}-q^2_{N+p}\leq s_{t'-1}.$ By the help of lemma \ref{lm5}, we have $$\|A_{m^{p+1}_1}\|\geq \frac{\|A_{s_{t'-1}}\|}{\|A_{s_{t'}-s_{t'-1}}\|}\geq \lambda^{(1-\epsilon)m^{p+1}_1};$$ $$|s(A_{m^{p+1}_1})-s(A_{s_{t'-1}})|_{C^2}\leq \|A_{s_{t'-1}}\|^{-1.5}.$$ Therefore $$|g_{p+1}-s(A_{m^{p+1}_1})|_{C^2}\leq |s(A_{m^{p+1}_1})-s(A_{s_{t'-1}})|_{C^2}+|g_{p+1}-s(A_{s_{t'-1}})|_{C^2}\leq \|A_{s_{t'-1}}\|^{-1.5}+\lambda^{-1.5r_1}\leq \lambda^{-0.5r_1}.$$ Similarly, $$\|A_{m^{p+1}_2}\|\geq \lambda^{(1-\epsilon)m^{p+1}_2};$$ $$|g_{p+1}-u(A_{m^{p+1}_2})|_{C^2}\leq \lambda^{-0.5r_{p}}.$$ In conclusion, $|g_{m^{p+1}_1,m^{p+1}_2,2}-g_{p+1}|_{C^2}\leq \lambda^{-0.5r_p}.$
     \item[$s_t-s_{t-1}\geq q_{N+p}$:] By the definition of $m^{p+1}_1,$ it's clear that $T^{s_t'}x\notin I_{p+1},$ which implies $|T^{s_t'}x-c_{p+1}|\geq \frac{1}{q^{8\tau}_{N+p}}$ and $g_{p+1}(T^{s_t}x)\geq \frac{C}{q^{24\tau}_{N+p}}.$ Denote $\tilde{g}_{p+1}\triangleq s(A_{s_{t'}-s_{t'-1}})-u(A_{s_{t'-1}})$ and $\tilde{c}_{p+1}$ be the zero point of $\tilde{g}_{p+1}$. It follows from the induction hypothesis that $|\tilde{g}_{p+1}(T^{s_t'}x)-g_{p+1}(T^{s_t'}x)|\leq \lambda^{-\frac{1}{2}r_p}\ll \frac{C}{q^{24\tau}_{N+p}}.$ Therefore, it follows from lemma \ref{lm5} that $\|A_{m^{p+1}_1}\|\geq \|A_{s_{t'-1}}\| \frac{C}{q^{24\tau}_{N+p}} \|A_{s_{t'}-s_{t'-1}}\|\geq \lambda^{(1-\epsilon)m^{p+1}_1}.$ Similarly, we have $\|A_{m^{p+1}_1}\|\geq \lambda^{(1-\epsilon)m^{p+1}_2}.$ Again from lemma \ref{lm5}, it's not difficult to see that $$|g_{m^{p+1}_1,m^{p+1}_2,2}-g_{p+1}|_{C^2}\leq \lambda^{-0.5r_p}.$$
     \end{enumerate}
     \item[Type $III_p\rightarrow I_{p+1}$:] If $\bar{t}=1,$ then we need do nothing. For $\bar{t}\geq 1$, we can also write $A_{m^{p+1}_1}$ as $A_{s_{\bar{t}}-s_{\bar{t}-1}}\cdots A_{s_1}$ where $s_i,$ $i=1,2\cdots \bar{t}-1$  are all \textbf{second} returning times to $I_p$ of $x,$ where $I_p\triangleq I_{p,1}\bigcup I_{p,2}$ and $I_{p,1}+k'\alpha \bigcap I_{p,2}\neq \emptyset$ for some $k'\leq q^2_{N+p-1}.$  And for the last returning (or not returning) time $s_{\bar{t}}-s_{\bar{t}-1},$ there are two cases should be concerned.
         \begin{enumerate}\item[$s_{\bar{t}}-s_{\bar{t}-1}<q^3_{N+p-1}$:]
          Note that $q^2_{N+p-1}\ll q^4_{N+p-1}\ll s_{\bar{t-1}}=m^{p+1}_1-(s_{\bar{t}}-s_{\bar{t}-1}).$ Thus lemma \ref{lm5} implies $$\|A_{m^{p+1}_1}\|\geq \frac{\|A_{s_{\bar{t-1}}}\|}{\|A_{s_{\bar{t}}-s_{\bar{t}-1}}\|}\geq \lambda^{(1-\epsilon)m^{p+1}_1};$$

         $$|s(A_{m^{p+1}_1})-g_{p+1}|_{C^2}\leq |g_{p+1}-s(A_{s_{\bar{t}-1}})|_{C^2}+|s(A_{s_{\bar{t}-1}})-s(A_{m^{p+1}_1})|_{C^2}\leq \lambda^{-1.5r_{p}}+\|A_{s_{\bar{t}-1}}\|^{-1.5}\leq \lambda^{-0.5r_{p}}.$$ Similarly, $|u(A_{m^{p+1}_2})-g_{p+1}|_{C^2}\leq \lambda^{-0.5r_{p}},$ which implies $|g_{m^{p+1}_1,m^{p+1}_2,2}-g_{p+1}|_{C^2}\leq \lambda^{-0.5r_p}$ as desire.
         \item[$s_{\bar{t}}-s_{\bar{t}-1}\geq q^3_{N+p-1}$:]
         In this case, note that $k'\leq q^2_{N+p-1}\leq q^3_{N+p-1}\leq s_{\bar{t}}-s_{\bar{t}-1},$ thus it either return once to $I_{p}$ or not. So the following two cases should be concerned.

         \begin{enumerate}
                  \item[\textbf{$r^{+}_{p}(T^{s_{\bar{t}-1}}x)\neq k'$:}] This implies $r^{+}_{p}(T^{s_{\bar{t}-1}}x)\geq q^4_{N+p-1}-q^2_{N+p-1},$ which means $x$ doesn't return to any critical intervals $I_{p,1}$ or $I_{p,2}.$ On one hand, the induction hypothesis tell us that $\|A_{s_{\bar{t}}-s_{\bar{t}-1}}\|\geq \lambda^{(1-\epsilon)(s_{\bar{t}}-s_{\bar{t}-1})}$ ; and for $\theta_{s_{\bar{t}-1}}\triangleq s(A_{s_{\bar{t}}-s_{\bar{t}-1}})-u(A_{s_{\bar{t}-1}})$ we have $|\theta_{s_{\bar{t}-1}}-g_{p}|_{C^2}\leq \lambda^{0.5r_{p-1}}\ll \frac{1}{q^{24\tau}_{N+p}}.$ On the other hand, it follows from the definition of $m^{p+1}_1$ that $T^{s_{\bar{t}}-s_{\bar{t}-1}}x\notin I_{p+1},$ which implies $\theta_{s_{\bar{t}-1}}(^{s_{\bar{t}}-s_{\bar{t}-1}}x)\geq \frac{1}{q^{24\tau}_{N+p}}.$ According to lemma \ref{lm5}, it's clear that $$\|A_{m^{p+1}_1}\|\geq \|A_{s_{\bar{t}-1}}\|\frac{1}{q^{24\tau}_{N+p}}\|A_{s_{\bar{t}}-s_{\bar{t}-1}}\|\geq \lambda^{(1-\epsilon)m^{p+1}_1};$$ $$|s(A_{m^{p+1}_1})-g_{p+1}|_{C^2}\leq |g_{p+1}-s(A_{s_{\bar{t}-1}})|_{C^2}+|s(A_{s_{\bar{t}-1}})-s(A_{m^{p+1}_1})|\leq \lambda^{0.5r_{p}},$$ Similarly, $$|u(A_{m^{p+1}_2})-g_{p+1}|_{C^2}\leq \lambda^{-0.5r_{p}},$$ which implies $$|g_{m^{p+1}_1,m^{p+1}_2,2}-g_{p+1}|_{C^2}\leq \lambda^{-0.5r_p}$$ as desire.
                      \item[$r^{+}_{p}(T^{s_{\bar{t}-1}}x)= k'$:] Due to $k'\leq q^2_{N+p-1}\leq q^3_{N+p-1}\leq s_{\bar{t}}-s_{\bar{t}-1},$ $x$ must return once back to $I_{p+1}.$ We are concern about the difference between $g_{p-1}$ and $\theta'\triangleq s(A_{s_{\bar{t}}-s_{\bar{t}-1}})-u(A_{s_{\bar{t}-1}}).$  It follows from \cite{wz1} that $$|g_{p}-(s(A_{s_{\bar{t}}-s_{\bar{t}-1}})-u(A_{s_{\bar{t}-1}}))|_{C^2}\leq C\lambda^{-\frac{1}{30}q^3_{N+p-1}}.$$ And by the definition of $m^{p+1}_1,$ we have
                          $T^{s_{\bar{t}-1}}x\notin I_{p+1}.$ Therefore, $|\theta'(T^{s_{\bar{t}-1}}x)|\geq \frac{1}{q^{24\tau}_{N+p}},$ and the following holds:
                          $$\|A_{m^{p+1}_1}\|\geq \|A_{s_{\bar{t}-1}}\|\frac{1}{q^{24\tau}_{N+p}}\|A_{s_{\bar{t}}-s_{\bar{t}-1}}\|\geq \lambda^{(1-\epsilon)m^{p+1}_1};$$
                          $$|s(A_{m^{p+1}_1})-g_{p+1}|_{C^2}\leq |g_{p+1}-s(A_{s_{\bar{t}-1}})|_{C^2}+|s(A_{s_{\bar{t}-1}})-s(A_{m^{p+1}_1})|\leq \lambda^{-1.5r_{p}}+\|A_{s_{\bar{t}-1}}\|^{-1.5}\leq \lambda^{-0.5r_{p}}.$$ Similarly, $$|u(A_{m^{p+1}_2})-g_{p+1}|_{C^2}\leq \lambda^{-0.5r_{p}},$$ which implies $$|g_{m^{p+1}_1,m^{p+1}_2,2}-g_{p+1}|_{C^2}\leq \lambda^{-0.5r_p}$$ as desire.

             \end{enumerate}
         \end{enumerate}

     \item[Type $III_p\rightarrow III_{p+1}$:] Without loss of generality, there exist some $k_1\leq q^2_{N+p-1}$ and $k_1\leq k_2\leq q^2_{N+p}$ such that $I_{p,1}+k_1\alpha\cap I_{p,2}\neq \emptyset$ and $I_{p+1,1}+k_2\alpha\cap I_{p+1,2}\neq \emptyset.$
         \begin{enumerate}\item[C1,$k_1=k_2:$] We denote $k_1=k_2=k''.$
         \begin{enumerate}
         \item[$x\in I_{p+1,1}$:] Note that $m^{p+1}_1\leq r^+_{p+1}=k''$ and $x\in I_{p+1,1}\in I_{p,1}.$ Thus we come back to the case $n=p.$
         \item[$x\in I_{p+1,2}$:] This is quite similar to the case  Type $III_p\rightarrow I_{p+1}$ and we omit it.
         \end{enumerate}
         \item[C2,$k_1<k_2:$] Note that either $x\in I_{p+1,1}$ or $I_{p+1,2},$ the definition of $m^{p+1}_1$ implies that $x$ come back to $I_{p+1}$ at most once. Thus we only need to deal with the each non-resonance part. By labeling each \textbf{second} returning times to $I_{p}$, we can also divide $A_{m^{p+1}_1}$ in the same way we just did in the case Type $III_p\rightarrow I_{p+1}.$ The rest is exactly the same.
         \end{enumerate}

     The following three types are similar to the case $n=0$ $\rightarrow$ $1$ and there are no new things need be present.
     $$Type~ I_p ~\rightarrow ~ III_{p+1}$$
     $$Type ~II_p ~\rightarrow ~ I_{p+1}, II_{p+1}~or~III_{p+1}.$$

     \end{enumerate}
     \end{proof}

\

\textbf{A.6}
 \begin{proof}We prove it by induction.

     For $m=n,$ the conclusion is trivial.

     For $n=m+1,$ the following several cases need to be considered.
     \begin{enumerate}\item[Type $I_m$ $\rightarrow$ $I_{m+1}$:] We write $A_{r^+_{m+1}}=A_{s^m_{t_m}-s^m_{t_m-1}}\cdots A_{s^m_1},$ where $\{s^m_{i}\}_{i=1}^{t_m}$ are all the returning times of $x$ to $I_m.$ Then the conclusion holds from \cite{wz1} or \cite{young}.
     \item[Type $I_m$ $\rightarrow$ $III_{m+1}$:] Recall that $r^+_{m+1}(x)$ is the first (not the second) returning time of $x$ to $I_{m+1}.$ Thus this case is equivalent to Type $I_m$ $\rightarrow$ $I_{m+1}.$
     \item[Type $III_m$ $\rightarrow$ $I_{m+1}$:] Assume $I_{m,1}+k_m\alpha\bigcap I_{m,2}\neq \emptyset$ and $x\in I_{m,1}.$ \begin{enumerate}\item If $r^+_{m+1}(x)=k_m,$ then $r^+_m=r^+_{m+1}$ and all the conclusion is trivial; \item If $r^+_{m+1}(x)\neq k_m,$ then we write $A_{r^+_{m+1}}=A_{s^m_{t_m}-s^m_{t_m-1}}\cdots A_{s^m_1},$ where $\{s^m_{i}\}_{i=1}^{t_m}$ are all the returning time of $x$ to $I_m.$ Note that $t_m\geq 1.$ If $t_m$ is even, then we put each twice returning time points together and the conclusion holds from \cite{wz1}; If $t_m$ is odd, then the last returning time $s^m_{t_m}-s^m_{t_m-1}=k_m\ll s^m_{t_m-1}$ and the conclusion holds from lemma \ref{lm5}.

         \end{enumerate}
     \item[Type $III_m$ $\rightarrow$ $III_{m+1}$:] This is similar to the previous case due to the definition of $r^+_{m+1}.$
     \end{enumerate}
     The left possible cases are as follow and we omit the proof since it's quite similar to the above cases which we have stated.
     $$Type II_m \rightarrow I_{m+1}, II_{m+1}, or III_{m+1}.$$

     Assume that for any $s\geq 1$ and $|n-m|\leq s,$ the conclusion hold.

     Let's focus on the case $n=m+s+1.$ We still consider the following possible cases.

     \begin{enumerate}\item[Type $I_{m+s}$ $\rightarrow$ $I_{m+s+1}$:] By the Induction hypothesis, for fixed $x_m$ from $I_{m}$ to $I_{m+s}$ and $I_{m+s}$ to $I_{m+s},$ the norms have good growth. We write $A_{r^+_{m+s+1}}=A_{s^{m+s}_{t_{m+s}}-s^{m+s}_{t_{m+s}-1}}\cdots A_{s^{m+s}_1},$ where $\{s^{m+s}_{i}\}_{i=1}^{t_{m+s}}$ are all the returning times of $x$ to $I_{m+s}.$ If $t_{m+s}=1,$ then we can see that $r^+_{m+s}=r^+_{m+s+1}>q_{N+m+s}>q_{N+m+s-1}.$ By the Induction hypothesis, we have obtained the conclusion. For $t_{m+s}\geq 2,$ the following cases should be considered.
     \begin{enumerate}\item If $s^{m+s}_1<q_{m+s},$ then $r^+_{m+s+1}\geq q^2_{m+s}\gg q_{m+s}$ and \eqref{1.1.1} holds from lemma \ref{lm5}.
     \item If $s^{m+s}_1\geq q_{m+s},$ then we have $r_{m+s}\geq q_{m+s}.$ On one hand, Induction hypothesis implies that $\|A_{s^{m+s}_1}\|\geq \lambda^{(1-\epsilon)s^{m+s}_1}$ and $|\theta_2(x_m)|\geq \frac{C}{q^{24\tau}_{N+m+s}},$ where $\theta_2\triangleq s(A_{s^{m+s}_{t_{m+s}}-s^{m+s}_{t_{m+s}-1}}\cdots A_{s^{m+s}_2})-u(A_{s^{m+s}_1}).$ On the other hand, \cite{wz1} implies $A_{s^{m+s}_{t_{m+s}}-s^{m+s}_{t_{m+s}-1}}\cdots A_{s^{m+s}_2}\geq \lambda^{(1-\epsilon)(r^+_{m+s+1}-s^{m+s}_1)}.$ Therefore,
         $$\|A_{r^+_{m+s+1}}\|\geq \lambda^{(1-\epsilon)(r^+_{m+s+1}-s^{m+s}_1)}\frac{C}{q^{24\tau}_{N+m+s}}\lambda^{(1-\epsilon)s^{m+s}_1}\geq \lambda^{(1-\epsilon)r^+_{m+s+1}}.$$
     \end{enumerate}
     \item[Type $I_{m+s}$ $\rightarrow$ $III_{m+s+1}$:] This is similar to the case Type $I_{m+s}$ $\rightarrow$ $I_{m+s+1}$.
     \item[Type $III_{m+s}$ $\rightarrow$ $I_{m+s+1}$:] In this case, we write $A_{r^+_{m+s+1}}=A_{s^{m+s}_{t_{m+s}}-s^{m+s}_{t_{m+s}-1}}\cdots A_{s^{m+s}_1},$ $\{s^{m+s}_{i}\}_{i=1}^{t_{m+s}}$ are defined as above case. If $t_{m+s}=1,$ then by the analysis in Type $I_{m+s}$ $\rightarrow$ $I_{m+s+1}$ we obtain the conclusion. For $t_{m+s}\geq 2,$ note that the short chain ( with the length $k_{m+s}\leq q^2_{N+m+s-1}$) and the long chain ( with the length $\geq q^8_{N+m+s-1}$) alternately occur. We need consider the following cases.
         \begin{enumerate}\item If $t_{m+s}\geq 3,$ then we have $r^+_{m+s+1}-s^{m+s}_1\geq q^8_{N+m+s-1}.$ We should consider the following two cases. For $s^{m+s}_1\leq q_{N+m+s-1},$ it's clear that $r^+_{m+s+1}-s^{m+s}_1\geq q^8_{N+m+s-1}\gg s^{m+s}_1$ and the conclusion holds from lemma \ref{lm5}; For $s^{m+s}_1\geq q_{N+m+s-1},$ by the help of the Induction hypothesis, we have $|u(A_{s^{m+s}_1})-u_{r_{m+s}}|_{C^2}\leq C\lambda^{-\frac{1}{2}q_{N+m+s-1}},$ which implies $\theta_1(x_m)\geq \frac{C}{q^{24\tau}_{N+m+s-1}},$ where $\theta_1\triangleq s(A_{r^+_{m+s+1}-s^{m+s}_1})-u(A_{s^{m+s}_1});$ and $\|A_{s^{m+s}_1}\|\geq \lambda^{(1-\epsilon)s^{m+s}_1}.$ Again due to lemma \ref{lm5}, we obtain the proof.
         \item If $t_{m+s}=2,$ then $r^+_{m+s}$ is the second returning time point. We rewrite $A_{r^+_{m+s+1}}$ as $A_{K_1}A_{K_2}$.
         \begin{enumerate}
         \item[$(b_1)$:] If $\min\{K_1,K_2\}\leq q^{0.5}_{N+m+s},$ then the growth of the norm is obvious. For the angle, if $K_2\leq q^{0.5}_{N+m+s},$ which implies $K_1\geq q_{N+m+s}-q^{0.5}_{N+m+s}\gg K^2_2,$ then $|u(A_{r^+_{m+s}})-u(A_{K_1})|_{C^2}\leq C\lambda^{-1.5K_1};$ if $K_1\leq q^{0.5}_{N+m+s},$ which implies $K_2\geq q_{N+m+s}-q^{0.5}_{N+m+s}\gg K^2_1,$ then we go on to divide $A_{K_2}$ into $A_{s^{m+s-1}_{t_{m+s-1}}-s^{m+s-1}_{t_{m+s-1}}}\cdots A_{s^{m+s-1}_1},$ without loss of generality, we assume $t_{m+s-1}\geq 2$ (otherwise it holds from $K_2\geq \frac{1}{2}q_{N+m+s}$ that we can go on to divide it until $t_{m+s-i}\geq 2$ for some $i\geq 2$) For the case $t_{m+s-1}\geq 3,$ it's similar to the previous case (a). For the case $t_{m+s-1}=2,$ we write $A_{r^+_{m+s+1}}\triangleq A_{K_1}A_{K_{2,1}}A_{K_{2,2}}$ and we still need to consider the following sub-cases.
             \begin{enumerate}\item[$(b_{1,1})$:] If the resonance distance $K_{m+s-1}=K_{m+s},$ then it's easy to see that $K_{2,1}\gg K_{2,2}.$ Therefore, on one hand, it follows from lemma \ref{lm5} that $$|u(A_{K_1}A_{K_{2,1}})-u(A_{K_1}A_{K_{2,1}}A_{K_{2,2}})|_{C^2}\leq \lambda^{-1.5q_{N+m+s-2}};$$ on the other hand, $$|u(A_{K_1}A_{K_{2,1}})-u_{r_{m+s}}|_{C^2}\leq \|A_{K_1}A_{K_{2,1}}\|^{-1.5}\leq \lambda^{-1.5q_{N+m+s-2}}.$$ Thus, $$|u(A_{r^+_{m+s+1}})-u_{r_{m+s}}|_{C^2}\leq \lambda^{-1.5q_{N+m+s-2}}.$$
             \item[$(b_{1,2})$:]If the resonance-distance $k_{m+s-1}\neq k_{m+s},$ then of course $k_{m+s-1}<k_{m+s}\leq q^{0.5}_{N+m+s}.$ This is impossible since $K_1>K_2$ in this case, which has contradiction with $K_2\gg K^2_1.$
                  \end{enumerate}
         \item[$(b_2)$:] If $\min\{k_1,k_2\}\geq q^{0.5}_{N+m+s},$ similar to $(b_1)$ we write $A_{r^+_{m+s+1}}\triangleq A_{K_1}A_{K_{2,1}}A_{K_{2,2}}$ and consider the following possible two cases.
             \begin{enumerate}\item[$(b_{2,1})$:] $k_{m+s-1}=k_{m+s}.$ This is same as $(b_{1,1}).$
             \item[$(b_{2,2})$:] $k_{m+s-1}\neq k_{m+s}.$ This means $k_{m+s-1}<k_{m+s}$ and we can write $A_{K_1}$ as $A_{p_u-p_{u-1}}\cdots A_{p_1},$ where $\{p_i\}_{i=1}^{u}$ are all the second returning time points to $I_{m+s-1}.$ Note that $u\geq 2,$ which means $K_1\geq q^8_{N+m+s-1}.$ Therefore, if $K_{2,2}\leq q_{N+m+s-1},$ then $K_1\gg q^2_{N+m+s-1}+q_{N+m+s-1}\geq K_{2,1}+K_{2,2}=K_2.$ It holds from lemma \ref{lm5} that the conclusion holds true. If $K_{2,2}\geq q_{N+m+s-1},$ then by the help of the Induction hypothesis it's clear that $A_{K_{2,1}}A_{K_{2,2}}$ has a good growth and $u(A_{K_1})-s(A_{K_2})\triangleq\theta(x_m)\geq \frac{C}{q^{24\tau}_{N+m+s-1}}.$ Hence, combining it with the fact $\|A_{K_1}\|\geq \lambda^{(1-\epsilon)K_1},$ lemma \ref{lm5} implies the proof.
             \end{enumerate}
         \end{enumerate}
         \item[Type $III_{m+s}$ $\rightarrow$ $III_{m+s+1}$:] This is similar to the case Type $III_{m+s}$ $\rightarrow$ $I_{m+s+1}.$
         \end{enumerate}

         The left possible cases are all similar as the above cases.
     \end{enumerate}


     \end{proof}
{"config": "arxiv", "file": "2006.03381.tex"}