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Quadratic skew-product, non-uniform expansion,
Lyapunov exponent, absolutely continuous invariant measure.
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%
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\title{Weakly expanding skew-products of quadratic maps}
\author{J{\'e}r{\^o}me Buzzi, Olivier Sester, Masato Tsujii}
\begin{document}
\maketitle
\begin{abstract} We consider quadratic skew-products over angle doubling of
the circle and prove that they admit positive Lyapunov exponents almost
everywhere and an absolutely continuous invariant probability measure.
This extends corresponding results of M. Viana and J.F. Alves for skew-products
over linear strongly expanding map of the circle.
\end{abstract}
\section{Introduction}
We study the ergodic properties
of the maps $F:S^1\times\R\rightarrow S^1\times\R$
$$
F(\theta,x)=(d\theta\ \mbox{mod 1},a-x^2+\alpha\sin(2\pi \theta)).
%f(\theta,x)) \mbox{ with }
%f(\theta,x)=a-x^2+\alpha\sin(2\pi \theta).
$$
Here $d\ge 2$ is an integer, $\alpha$ is a small real number and the
parameter $a\in ]1,2[$ is such that the map $f_a(x)=a-x^2$ has preperiodic (but
not periodic) critical point.
It is hoped that these maps introduced by M. Viana in \cite{Viana}
represent a first step towards the study of perturbations
of products of good unimodal maps and a conjectural `two-dimensional Jakobson's
theorem', i.e., proving that many of those perturbations admit an
absolutely continuous invariant probability measure (or a.c.i.m.).
In~\cite{Viana}, M. Viana shows that under the assumption $d\geq 16$
these maps admit two positive Lyapunov exponents at Lebesgue almost every point,
provided that $\alpha$ is sufficiently small. J.F. Alv{\`e}s
\cite{Alves} deduces from the underlying estimates the existence of an
a.c.i.m.
%Finally they together \cite{AlvesViana} prove that this measure is unique
%and has full basin and that it is stochastically stable.
One would like to weaken the uniform expansion assumptions. A first step (and
the main purpose of this paper) is to replace the assumption $d\geq
16$ by the more natural one $d\geq2$.
\begin{theorem} \label{main}
For any integer $d\geq 2$, for every sufficiently small
$\alpha>0$, $F$ has two positive Lyapunov exponents at
Lebesgue almost every point and admits an a.c.i.m. Moreover, the same
is true of any sufficiently small $C^\infty$ perturbation of $F$ (which
is not necessarily a skew product).
\end{theorem}
%
\subsection*{Comments}
%
1. One could consider much more general skew-products of the form
$(\theta,x)\mapsto (g(\theta), h(x)+\alpha\phi(\theta))$. Especially,
it wouldn't take much effort to generalize $h(x)=a-x^2$ to any
unimodal map with negative Schwarzian derivative and having critical point
preperiodic or non-recurrent. $h$ could even be replaced by a ``good''
unimodal map of the interval, that is satisfying the so called Benedicks-Carleson conditions.
Indeed, according to~\cite{BaladiBenedicksMaume} these maps satisfy the
``Building expansion'' Lemmas in \cite{Viana} so that an adaptation of the proof
of Proposition~\ref{key} to this case would be sufficient.
%
\medbreak
\noindent
2. Our method makes stringent demands on the coupling function
$\phi(\theta)=\sin(2\pi\theta)$, demands that go much beyond the
natural requirement of non-flat critical points. A likewise
unnatural-looking restriction that we make is our $C^\infty$
smoothness assumption (or $C^r$ with $r$ arbitrarily large as $h$ goes
closer to the full unimodal map).
%
\medbreak
\noindent
3. Our proof deals only with the unperturbed case - the perturbated
case is treated like in Viana's paper \cite{Viana}.
%
\medbreak
\noindent
4. In our situation, one of the exponents is given for free by the uniform
expansion of the circle map. Thus Theorem~\ref{main} deals with the vertical
Lyapunov exponent, which is the result of 'random' compositions of
perturbations of a good unimodal map. Also we are only interested in a
`dynamical strip' (the equivalent of the dynamical interval of a unimodal
map), as points outside just escape to infinity and therefore have a
trivially positive (in fact $+\infty$) vertical exponent.
\medbreak
\noindent
5. We use the same strategy of proof as Viana, with some significant
variations:
{\it a)} Horizontal expansion dominates vertical expansion in the dynamical
strip but in our case this happens not after the first, but after a (fixed)
number of iterations. This domination allows one to concentrate on nearly
horizontal curves which remain so under iterations.
{\it b)} The non-flatness of the images of nearly horizontal curves is
directly enforced by the
non-flatness of the $\sin$ function. It ensures that crossing the critical
circle $x=0$ only removes a small proportion of any horizontal curve.
In Viana's setting, one could exploit the non-flatness just by looking at
the second order derivative. Here we have to go to finite, but arbitrarily
high order as $a$ gets closer to $2$.
{\it c)} This non-flatness ensured vertical dispersion of full-width
curves in the strongly expanding case. In our situation the dispersion
only occurs on a part of these curves.
{\it d)} Finally a large deviation estimate of the depth of returns
provides all the informations needed for proving the positivity of the
vertical Lyapunov exponent as well as constructing the a.c.i.m. This part requires
no modification in our setting. The existence of the a.c.i.m. follows from
the estimates on the vertical Lyapunov exponent just by applying verbatim
\cite{Alves} and \cite{AlvesViana}.
\medbreak
\section{Preliminaries} \label{sec:prelim}
We begin with a remark on the parameter dependences. The map we
consider has three parameters $10$. Henceforth, we fix the parameters $a$ and $d$
and do not specify dependence on them. On the other hand, the parameter
$\alpha$ is not fixed, but always assumed to be sufficiently small and
dependence on it is always specified.
We write, for simplicity,
\begin{align*}
\phi(\theta)&=\sin(2\pi \theta),\\
g(\theta)&=d \theta\mod 1,\\
f(\theta,x)&=a-x^2+\alpha\phi(\theta),
\end{align*}
For $n\ge 1$, let's define $f_{n}(\theta,x)$ by
\[
F^n(\theta,x)=(g^n(\theta),f_n(\theta,x)).
\]
Notice that the functions $f$ and $f_{n}$ depend on the parameter $\alpha$ though the dependence is not made
explicit.
The map $f_a$ has a unique negative fixed point
$p_1=\frac{-1-\sqrt{1+4a}}{2}<0$ which is repelling.
If we take $\beta$ slightly smaller than $|p_{1}|$, the interval
$B=[-\beta,\beta]$ satisfies: $f_{a}(B)\subset \mathrm{int} (B)$ and $|f_{a}'|>1$ on $\R\setminus
\mathrm{int} (B)$. Then, writing $I=S^{1}\times B$, for sufficiently
small $\alpha$, we have
\begin{itemize}
\item $F(I)\subset I$,
\item $|\partial_{x}f(\theta,x)|>1$ outside of $I$.
\end{itemize}
These imply that, for any point $(\theta,x)$ on $S^{1}\times \R$,
either its orbit eventually comes into the invariant strip $I$ or the vertical Lyapunov
exponent is positive. Thus it is enough to exhibit a positive vertical
Lyapunov exponent at almost every point of $I$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Upper bounds on derivatives} \label{sec:derivatives}
Note that we always have $|\partial_x f(\theta,x)|<2|p_{1}|<4$ for $(\theta,x)\in I$.
We prove that the vertical expansion on the dynamic strip $I$
is {\it uniformly} less than $2$:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lemma}\label{derivative-x}
There exist constants $K_1<\infty$ and $R_1<2$
such that, when $\alpha$ is sufficiently small, we have
$|\d_x f_n(\theta,x)|\leq K_1R_1^n$ for all $(\theta,x)\in I$ and all $n\in\N$.
\end{lemma}
\proof
Consider maps
\[
h: [-\pi/2,\pi/2]\to [p_{1},-p_{1}],\quad \xi\mapsto p_{1}\sin \xi.
\]
and
\[
\psi=h^{-1}\circ f_{a}\circ h: [-\pi/2,\pi/2] \to [-\pi/2,\pi/2].
\]
We claim that $|\psi'(x)|<\sqrt{2p_{1}}$ on $[-\pi/2.\pi/2]$. Take a
point
$\xi\in [-\pi/2,\pi/2]$ and let
$\bar{\xi}=\psi(\xi)$. Using the relations
$a-p_1^2\sin^2\xi=p_1\sin\bar{\xi}$, $a-p_{1}^{2}=p_{1}$ and $|p_{1}|<2$,
we obtain
\begin{align*}
\left|\psi'(\xi)\right|
&=
\left|h'(\bar{\xi})\right|^{-1}
\left|f'_{a}(h(\xi))\right|
\left|h'(\xi)\right|
= \frac{|p_{1}\cos \xi|\cdot |2p_1\sin\xi|}
{|p_{1}\cos \bar{\xi}|}\\
&
= \frac{2| p_{1}^{2}\cos \xi\sin\xi|}
{\sqrt{p_{1}^{2}-\left(a-p_{1}^{2}\sin^{2}\xi\right)^{2}}}
= \frac{2| p_{1}^{2}\cos \xi\sin\xi|}
{\sqrt{p_{1}^{2}-\left(p_{1}+p_{1}^{2}\cos^{2}\xi\right)^{2}}}
\\
&= \frac{2|p_{1}\sin\xi|}{\sqrt{-2 p_{1}-p_{1}^{2}\cos^{2}\xi}}\le
\frac{2|p_{1}|}{\sqrt{2|p_{1}|}}\cdot\frac{|\sin
\xi|}{\sqrt{1-|p_{1}/2|\cos^{2}\xi}}
\\
& \le \sqrt{2|p_{1}|}\frac{|\sin\xi|}{\sqrt{1-\cos^2\xi}}
=\sqrt{2|p_{1}|}<2.
\end{align*}
Let $\sqrt
{2|p_{1}|}100 A_{0}.
\end{equation}
Consider a curve $\widehat{X}=\graph(X)$, $X: S^1\rightarrow\R$,
which is $C^{\rho}$ on $S^1\setminus \{0\}$ and continuous to the right at 0.
If $X$ satisfies:
\begin{equation}\label{admissible_condition}
|X^{(r)}(\theta)|\le 2A_{0} (2\pi/d)^{r}\alpha\qquad \mbox{for $\theta\in S^{1}$
and $1\le r\le \rho$,}
\end{equation}
then $X$ is called {\em pre-admissible}.
Also consider the sequence of Markov partitions of $S^1$ induced by $g$:
\begin{align*}
\CP_{1}&=\{[\theta_j,\theta_{j+1}[ ; 0\leq j\leq d-1\}\qquad \mbox{ where }\theta_j=j/d;\\
\CP_{n+1}&=\{
\mbox{connected components of $g^{-1}(\omega)$; $\omega\in\CP_{n}$} \}.
\end{align*}
\begin{lemma} Suppose $\alpha>0$ is sufficiently small. If
$\widehat{X}=\graph(X)$ is a pre-admissible curve
then, for any $\omega\in \CP_{n}$ with $n\ge N_{0}$, the image
$\widehat{Y}=F^{n}(\widehat{X}|_{\omega})$ is again pre-admissible.
\end{lemma}
\proof
Consider an arbitrary curve $\widehat{X}=\graph(X)$. Let $\omega$ be an arbitrary
element of $\CP_{N_{0}}$. For $0\le i\le N_{0}$, let
$\widehat{X}_{i}=\graph(X_{i})=F^{i}(\widehat{X}|_{\omega})$.
For $i\ge 0$ and $\theta\in g^{i}(\omega)$, we have the obvious relation
\begin{equation}\label{inductive_relation}
X_{i+1}(g(\theta))=f(\theta,X_{i}(\theta)).
\end{equation}
%
{\bf Claim.} Omitting obvious dependencies, we have:
\begin{equation}\label{inductive_relation_r}
X_{i+1}^{(r)}
= \frac{1}{d^r} \partial_x f \cdot X_i^{(r)}
+ \frac{1}{d^r} \partial_\theta^r f
+ Q_r(X_i^{(1)},\dots,X_i^{(r-1)})
\end{equation}
%where
%$$
%Q_r(Y_1,\dots,Y_{r-1}) = -2\sum_{j=1}^{r-1}\left({r-1 \atop
%j}\right)Y_jY_{r-j}
%$$
%and $\left({r-1 \atop j}\right)$ denotes the binomial coefficient.
%The claim is an immediate consequence of Leibniz formula.
where
$$
Q_r(Y_1,\dots,Y_{r-1}) = \sum_{j=1}^{r-1}q_{rj}Y_j Y_{r-j}
$$
and $q_{rj}$ are constants.
The claim follows from an immediate induction on $r$ as it is obvious
that the derivative of $Q_{r-1}$ w.r.t. $\theta$ is of the form claimed.
%for%$Q_{r}$ and $Q_1=0$.
%
\medbreak
Observe that the second term in equation (\ref{inductive_relation_r}) is
$\alpha d^{-r} \phi^{(r)}(X_i(\theta))$, hence it is bounded in
absolute value by $\alpha (2\pi/d)^r$.
\medbreak
Suppose that the curve $\widehat{X}$ satisfies the condition
(\ref{admissible_condition}). An easy induction using
(\ref{inductive_relation_r}) and Lemma \ref{derivative-x} give
\begin{align}
|X_{n}^{(r)}| &\le \frac1{d^{rn}} |\partial_x f_n| \cdot |X^{(r)}|
+ \sum_{i=0}^{n-1} \frac{1}{d^{ri}} |\partial_x f_i|
\cdot \frac{1}{d^r} |\partial_\theta^r f|
+ \sum_{i=0}^{n-1} \frac{1}{d^{ri}} |\partial_x f_i| \cdot |Q_r| \notag\\
&<\frac{K_{1}R_{1}^{n}}{d^{rn}} 2 A_{0}
\left(\frac{2\pi}{d}\right)^{r}\alpha
+ \sum_{i=1}^{n-1} \frac{K_{1}R_{1}^{i}}{d^{ri}}
\cdot \left(\frac{2\pi}{d}\right)^{r} \alpha
+ \sum_{i=0}^{n-1}\frac{K_{1}R_{1}^{i}}{d^{ri}}
\cdot |Q_r|. \label{Xn_estimate}
\end{align}
\
By induction on $1\le r\le\rho$, we see that there
exists a constant
$C>0$, which is independent of $\alpha$ and $\widehat{X}$, the initial pre-admissible curve,
such that
\begin{equation}\label{bounds_Xi}
|Q_{r}(X_i^{(1)}(\theta),\dots,X_i^{(r-1)}(\theta))|0
\]
for Lebesgue measure almost every $\theta\in S^{1}$.
\medbreak
The main property of this class of nearly horizontal curves is the
\begin{corollary}
For any admissible curve $\widehat{X}=\graph(X)$ and any $\omega\in
\CP_{n}$, $n\geq0$, the image
$F^{n}(\widehat{X}|_{\omega})$ is again an admissible curve.
\end{corollary}
The second fundamental property of admissible curves is that their image are
{\em non-flat}:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lemma}\label{lower-bound}
If
$\widehat{X}=\graph(X)$ is an admissible curve, then it
satisfies either
$$
|X^{(\rho-1)}(\theta)|>(\pi/d)^{\rho-1}\alpha\mbox{ \ or \ }
|X^{(\rho)}(\theta)|>(\pi/d)^{\rho}\alpha,
$$
for each $\theta\in S^{1}$.
More precisely $S^{1}$ can be divided into at most
four (in fact three) closed intervals on each of which, one of the
above two inequalities holds.
\end{lemma}
\proof
By the definition of admissible curves, there exists a pre-admissible curve $\widehat{Y}=\graph(Y)$
and $\omega\in \CP_{1}$ such that $\widehat{X}=F(\widehat{Y}|_{\omega})$.
From (\ref{inductive_relation_r}) and (\ref{bounds_Xi}), we
have
\[
X^{(r)}(g(\theta))=\frac{-2Y(\theta)Y^{(r)}(\theta)}{d^{r}}+
\alpha\frac{\phi^{(r)}(\theta)}{d^{r}}+\mathcal{O}(\alpha^{2}).
\]
Thus, for sufficiently small $\alpha$, we have
\[
|X^{(r)}(g(\theta))|>\left|\alpha\frac{\phi^{(r)}(\theta)}{d^{r}}\right|-\frac{9A_{0}
(2\pi/d)^{r}\alpha}{d^{r}}.
\]
We assume that $\theta$ belongs to the subset $\omega\cap
([\frac18,\frac38] \cup [\frac58,\frac78])$ which consists of
at most two intervals, as the other case is
similar. For the even number $r=\rho$ or $\rho-1$, we
have $|\phi^{(r)}(\theta)|= (2\pi)^{r}|\sin(2\pi\theta)|\ge
(2\pi)^{r}/\sqrt{2}$ for
$\theta\in ([\frac18,\frac38] \cup [\frac58,\frac78])$ and
\begin{align*}
\alpha^{-1}|X^{(r)}(g(\theta))|>
\left|\frac{(2\pi)^{r}}{d^{r}\sqrt{2}}\right|-\frac{9A_{0}\cdot
(2\pi)^{r}}{d^{2r}}
&=\left(\frac{2\pi}{d}\right)^{r}\left(\frac{1}{\sqrt{2}}-\frac{9A_{0}}{d^{r}}\right)
\\ &\ge \left(\frac{2\pi}{d}\right)^r \left( \frac 1{\sqrt{2}} -
\frac{9}{100}\right)
\ge
\left(\frac{\pi}{d}\right)^{r}.
\end{align*}
In the second inequality, we used the condition in the choice of $\rho$.
\cqfd %%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Critical return}
We turn now to the central fact about the returns of admissible curves to
the critical line $S^{1}\times \{0\}$, which corresponds to Viana's ``Technical Lemma 2.6'' in
\cite{Viana}.
First we recall a fact about the map $f_{a}$ from \cite[Th {III.6.3}]{dMvS}:
\begin{lemma} \label{lemma-Mis} There exist constants $\delta>0$,
$\sigma>1$ and
$\mu>0$ such that
$ |(f_{a}^{n})'(x)|\ge \mu\sigma^{n}$ if $f_{a}^{n}(x)\in
(-\delta,\delta)$.
\end{lemma}
Define constants
\begin{equation}\label{definition_Malpha}
\Ma=\left[\frac{|\log\alpha|}{\log32}\right],\qquad
\eta=\frac{\log\sigma}{8\log32}.
\end{equation}
Fix $r\geq 0$, let $J(r)=\{x\in\R,\ |x|\leq\sqrt\alpha e^{-r}\}$. Denote by
$\widehat J(r)$ the critical strip $\widehat J(r)=S^1\times J(r)$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{proposition}\label{key}
There exist $C<\infty$ and $\beta>0$ such that, for all sufficiently
small $\alpha>0$, for any admissible curve
$\widehat Y$ and any $r\geq (\frac{1}{2}-2\eta)\log(\frac{1}{\alpha})$,
\begin{equation}\label{claim_key}
m(\{\theta\in S^1|\ F^{\Ma}(\widehat{Y}(\theta))\in\widehat
J(r-2)\})\leq Ce^{-\beta r}
\end{equation}
where $m$ is the Lebesgue measure on $S^{1}$.
\end{proposition}
The main ingredients in the proof of Proposition~\ref{key} are the two Lemmas
below. The first is a general result on the size of the pre-image
of a small interval for $C^{s}$ functions whose $s$-th derivative
is bounded away from zero. The second
states that the image under $F$ of the components of an admissible
curve are well-separated
over a large subset of $S^1$.
%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lemma}\label{morse}
Let $s$ be a positive integer and let $I\subset \R$ be an interval. Suppose that
$h:I\rightarrow \R$ is a $C^{s}$ function such that
$|h^{(s)}(\theta)| \geq \delta$
for all $\theta\in I$.
Then,
\[
m(\{\theta\in I| \ | h(\theta)|\leq \varepsilon\})<2^{s+1}\left(\varepsilon/\delta\right)^{1/s}
\]
for all $\varepsilon>0$. More precisely, we have
$$
C:=\{\theta\in I| \ | h(\theta)|\leq \varepsilon\}
\subset\bigcup_{i=1}^{2^s-1}J_i
$$
where $J_i$ are intervals with
$$
m(J_i)\leq 2\Big(\frac{\varepsilon}{\delta}\Big)^{1/s}.
$$
\end{lemma}
\medbreak
Together with Lemma \ref{lower-bound}, this immediately implies:
%%%%%%%%%%%%%%%%%%%%%%%%
\begin{corollary} \label{cor-intersection}
Let $\widehat{X}$ be an admissible curve and assume that $\alpha$ is sufficiently small. Then, for
$0<\epsilon<\alpha$, we have
$$
m(\{\theta\in S^1| \widehat{X}(\theta)\in S^{1}\times (-\eps,\eps) \})
\leq C_{1}\cdot (\epsilon/\alpha)^{1/\rho}.
$$
where $C_{1}=4\cdot 2^{\rho+1}(d/\pi)$.
\end{corollary}
\noindent
{\it Proof of Lemma~\ref{morse}.} We proceed by induction on $s$. The case $s=1$ is obvious. Assume
the result is true for $s-1\geq 1.$ Let
$$
A:=\{\theta\in I| \ | h^{(s-1)}(\theta)|\leq
\delta(\varepsilon/\delta)^{1/s}\}.
$$
Then $A$ is an interval because $h^{(s-1)}$ is
monotone.
Applying the case $s=1$ to $h^{(s-1)}$ with
$\varepsilon'=\delta(\varepsilon/\delta)^{1/s}$ we deduce
$$
m(A)\leq 2 (\varepsilon/\delta)^{1/s}.
$$
The complement of $A$ is made up of at most 2 intervals denoted by $I_1$
and $I_2$.
Let $B_j=\{\theta\in I_j;\ |h(\theta)|\leq \varepsilon \}$ for $j=1,2$. As
$|h^{(s-1)}|\geq \delta(\varepsilon/\delta)^{1/s}$ on $I_j$, the induction
assumption at level $s-1$ implies that $B_j$ is contained in the union
of $2^{s-1}-1$ intervals whose length is no more than
$$
2\left( \frac{\varepsilon}{\delta(\varepsilon/\delta)^{1/s}}\right)^{1/(s-1)}
= 2\Big(\frac{\varepsilon}{\delta}\Big)^{1/s}.
$$
Since $C\subset A\cup B_1\cup B_2$, $C$ is contained in at most
$1+2(2^{s-1}-1)=2^{s}-1$ intervals.
\cqfd %%%%%%%%%%%%%%
Let $\widehat X=\graph(X)$ be an admissible curve. Denote by
$\widehat X_j=\graph (X_j)=F(\widehat X|_{[\theta_{j},\theta_{j+1}[})$
for $0\leq j\leq d-1$, the
$d$ admissible curves contained in the image of $\widehat X$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lemma}\label{disjoint-curve}
There exists a constant $\varepsilon_{0}>0$, such that, for all sufficiently
small $\alpha>0$, for any admissible curve $X$ and any $j_1\in\{0,\dots,d-1\}$,
there exists $j_2$, $0\leq j_2\leq d-1$, satisfying
$$
\{\theta\in S^1 | \ %
|\widehat X_{j_1}(\theta)-\widehat X_{j_2}(\theta)|\leq \varepsilon_{0}\alpha\}
\subset \bigcup_{i=1}^{2^{\rho+2}}J_i
$$
where $J_i$ are intervals with $m(J_i)\leq\frac{1}{40\times2^\rho
d}$.
\end{lemma}
Remark that if $d=2$ there is no much choice in $j_2$.
\proof
It follows from~(\ref{inductive_relation_r}) and (\ref{bounds_Xi}) with $i=0$
that for all
$1\le r\le \rho$, $0\leq j\leq d-1$ and
$\theta\in S^1$,
\begin{equation}\label{estimate-Xr}
X_j^{(r)}(\theta)=d^{-r}\alpha\phi^{(r)}\left(\theta/d+\theta_{j}\right)
-2d^{-r}X\left(\theta/d+\theta_{j}\right)
X^{(r)}\left(\theta/d+\theta_{j}\right)+\mathcal{O}(\alpha^2).
\end{equation}
For $0\le j_{1}(2\pi)^{r}/10.
$$
In fact, in the case $i=1$ for example, the above inequality is clear
since
$\theta/d+\theta_{j_1}\in I_{1}= [1/8,3/8]$ and
$\theta/d+\theta_{j_1}\notin (1/24,11/24)$. The other cases are similar.
Thus, using (\ref{estimate-Xr}) and
the condition in the choice of
$\rho$,
\begin{align*}
\left|X_{j_1}^{(r)}(\theta)-X_{j_2}^{(r)}(\theta)\right|&\ge
\left(\frac{2\pi}{d}\right)^{r}\frac{\alpha}{10} -4d^{-r}\cdot 2A_{0}
\left(\frac{2\pi}{d}\right)^{r}\alpha-\mathcal{O}(\alpha^2)\\
&>\left(\frac{2\pi}{d}\right)^{r}\frac{\alpha}{20}.
\end{align*}
Now let $i=0,\dots,d-1$ and $\eps_0>0$ be small (see below) and apply
Lemma \ref{morse} to the function
$X_{j_1}^{(r)}(\theta)-X_{j_2}^{(r)}(\theta)$ on each interval
$g(I_{i}\cap [\theta_{j_{1}},\theta_{j_{1}+1}])$ with
$\varepsilon=\varepsilon_{0}\alpha$ and
$\delta=(2\pi/d)^{r}\alpha/20$. We obtain
\[
\{\theta\in g(I_{i}\cap [\theta_{j_{1}}, \theta_{j_{1}+1}]) \mid
|X_{j_1}^{(r)}(\theta)-X_{j_2}^{(r)}(\theta)|\le \varepsilon_{0}\alpha\}\subset
\bigcup_{j=1}^{2^{\rho}-1}\tilde{J}_{i,j}
\]
where $\tilde{J}_{i,j}$ is an interval such that
\[
m(\tilde{J}_{i,j})\le \frac{d(20 \varepsilon_{0})^{1/r}}{\pi}.
\]
For $\varepsilon_{0}$ sufficiently small, the right hand side is smaller than
$\frac{1}{40\times2^\rho d}$.
Finally $\bigcup_{i=0,\dots,d-1}
g(I_i\cap[\theta_{j_1},\theta_{j_1+1}])= S^1$, hence to get the lemma
it is enough to take all the $d(2^\rho-1)$ intervals $\tilde{J}_{i,j}$ just constructed.
\cqfd %%%%%%%%%%%%%%
\bigskip
\noindent
{\it Proof of Proposition~\ref{key}} There are two cases to consider.
We first assume that $r\geq (\frac{1}{2}+2\eta)\log(\frac{1}{\alpha})$. Note that in this case,
\[
1/\sqrt{\alpha}\leq\exp((1+4\eta)^{-1}r)<\exp((1-\eta)r).
\]
From Corollary \ref{cor-intersection}, we have, for any admissible curve
$\widehat{X}=\graph(X)$,
\[
m(\{\theta\in S^{1}\mid \widehat{X}(\theta)\in \widehat J(r-2)
\})
\le C_{1}\left(\frac{\sqrt{\alpha}e^{-r+2}}{\alpha}\right)^{1/\rho}
<
C_{1}\left(e^{2}e^{-\eta r}\right)^{1/\rho}.
\]
Since $F^{\Ma}(\widehat{Y})$ consists of $d^{\Ma}$ admissible curves
and since
$g^{\Ma}$ is linear, this implies (\ref{claim_key}) with $\beta=\eta/\rho$.
\medskip
Henceforth we consider the case $(\frac{1}{2}-2\eta)\log(\frac{1}{\alpha})\leq
r\leq (\frac{1}{2}+2\eta)
\log(\frac{1}{\alpha})$.
Let $r_{0}(\alpha)=(\frac{1}{2}-2\eta)\log(\frac{1}{\alpha})$.
% be the smallest of such $r$.
Actually, we are going to prove the claim
(\ref{claim_key}) only for $r=r_{0}(\alpha)$,
since the claim for other $r$'s follows by
replacing $\beta$ by
$\beta(\frac{1}{2}-2\eta)/(\frac{1}{2}+2\eta)$. We may and do assume
that there exists a point
$z_{0}=(\theta_{0},x_{0})$ on the admissible curve
$\widehat{Y}$ such that $F^{\Ma}(z_{0})\in \widehat{J}(0)$, since, otherwise,
the claim (\ref{claim_key}) is trivial.
Let $x_{i}=f_{a}^{i}(x_{0})$. Consider horizontal strips
\[
S_{i}=S^{1}\times [x_{i}-5^{i}L\alpha,
x_{i}+5^{i}L\alpha], \quad i=0,1,2,\cdots,
\]
where $L=\max\{1,2A_{0} (2\pi/d)\}$. Note that these strips are defined so that the (vertical)
width $|S_{i}|$ satisfies
$$
4|S_{i}|+\alpha<|S_{i+1}| \text{ and }
|S_{i}|\le 2\cdot 5^{\Ma}L\alpha<\sqrt{\alpha}\quad \mbox{for $0\le i\le
\Ma$}
$$
provided that $\alpha$ is sufficiently small.
From the first inequality, it follows that $F(S_{i})\subset S_{i+1}$.
Since the slope of the admissible curve $\widehat{Y}$ is bounded by
$2A_{0} (2\pi/d)\alpha\le
L\alpha$, we have
$\widehat{Y}\subset S_{0}$ and hence $F^{i}(\widehat{Y})\subset S_{i}$ for $0\le
i\le \Ma$.
The strips $S_{i}$, $0\le i\le \Ma-1$, do
not meet the critical strip $\widehat{J}(0)$. Indeed, otherwise,
$S_{\Ma}$ would intersect with both of $\widehat{J}(0)$ and
$F^{\Ma-i}(\widehat{J}(0))$. But this is impossible. In fact, we can
see inductively that for $i\ge0$,
$F^{i}(\widehat{J}(0))$ is contained in the
strip $J_{i}=S^{1}\times [f_{a}^{i}(0)-5^{i}\alpha,
f_{a}^{i}(0)+5^{i}\alpha]$.
The height of $J_i$ is bounded by $\sqrt{\alpha}$ for $1\le i\le
\Ma$. Hence the distance between
$S_{i}$ and
$\widehat{J}(0)$ is larger than
\[
|f_{a}^{i}(0)|-2\sqrt{\alpha}
\ge \text{const}-2\sqrt{\alpha}>\sqrt{\alpha}>|S_{\Ma}|.
\]
From $d(S_i,S^1\times \{0\})>\sqrt{\alpha}$ and using
$\partial_x\partial_\theta f=0$ and
$|\partial^2_xf|\leq 2$, we obtain the following distortion
estimate, for all $0\leq i\leq \Ma-1$:
\begin{equation}\label{bounded-distortion}
\sum_{j=i}^{\Ma-1}\sup_{w,w'\in
S_{j}}|\log|\partial_{x}f(w)|-\log|\partial_{x}f(w')||\le
\Ma\frac{2\cdot 2\cdot 5^{\Ma}L\alpha}{\sqrt{\alpha}}<\log 2
\end{equation}
which allows us to consider that the maps $F^{\Ma-i}$ are almost linear on $S_{i}$.
Since
$|f_a^{\Ma}(x_{0})|<|f_{\Ma}(z_{0})|+|S_{\Ma}|<2\sqrt{\alpha}<\delta$ if $\alpha$ is
sufficiently small, Lemma \ref{lemma-Mis} gives
\begin{equation}\label{estimate_dxf}
|\partial_{x}f_{\Ma-i}(\theta,x)|>
\frac{1}{2}\prod_{j=0}^{\Ma-i}|f'_{a}(x_{i+j})|>\frac{\mu\sigma^{\Ma-i}}{2}
\end{equation}
for any $(\theta,x)\in S_{i}$ with $0\le i\le \Ma$.
We introduce some more constants:
\begin{itemize}
\item Pick an integer $m_0$ such that
$d^{m_0-1}\geq 80\times 2^\rho$.
\item Let $\bar\sigma=\sqrt{\sigma}>1$.
\item Fix a constant $\kappa>4^{m_{0}}$, independent of $\alpha$, so
that
\[
(2 \kappa )(\kappa\varepsilon_{0}/4-A_{0}
(2\pi/d)-4(1-{\bar\sigma}^{-1})^{-1})\ge 3
e^{2}.
\]
\end{itemize}
Denote $\lambda_{j}=|\d_x f_{\Ma-j}(F^j(z_{0}))|/\bar\sigma^{(\Ma-j)}$
for
$0\leq j\leq \Ma-1$. Note that we have, from (\ref{estimate_dxf}),
\begin{equation}\label{estimate_lambda_i}
\lambda_{j}>\mu\sigma^{\Ma-j}/2\bar\sigma^{(\Ma-j)}=(\mu/2){\bar
\sigma}^{\Ma-j}.
\end{equation}
Let
\[
0=t_1t_i+m_0$ for all $1\le i< k_{0}(\alpha)$, because,
by the definition of $\lambda_{i}$'s and
$\kappa$,
$$
4^{m_{0}}<\kappa\le\frac{1}{2}\frac{\lambda_{t_i}}{\lambda_{t_{i+1}}}\leq
|\d_x f_{t_{i+1}-t_i}(F^{t_i}(z_0))| \bar{\sigma}^{t_{i}-t_{i+1}}
<4^{t_{i+1}-t_i}.
$$
\noindent
{\bf Claim 1.} We have $k_{0}(\alpha)\geq \gamma
r_{0}(\alpha)$ for $\gamma=\eta/\log(8\kappa)$ if $\alpha$ is sufficiently small.
\proof
On the one hand, $\lambda_{t_i}\leq 2\kappa\lambda_{t_{i+1}-1}\leq 8
\kappa\lambda_{t_{i+1}}$, hence
$
\lambda_{t_1}\leq (8\kappa)^{k_{0}(\alpha)}\lambda_{t_{k_{0}(\alpha)+1}}.
$
From (\ref{estimate_lambda_i}), it follows
$$
\lambda_{t_{k_{0}(\alpha)+1}}\ge (8\kappa)^{-k_{0}(\alpha)}\lambda_{t_1}\ge (8\kappa)^{-k_{0}(\alpha)}(\mu/2)\cdot
\bar\sigma^{\Ma}.
$$
On the other hand, by definition of $k_{0}(\alpha)$,
$\lambda_{t_{k_{0}(\alpha)+1}}\leq 2 \kappa e^{-r_{0}(\alpha)}/\sqrt\alpha$.
Combining these two bounds on $\lambda_{t_{k_{0}(\alpha)+1}}$, we derive
$$
-r_{0}(\alpha)-(\log\alpha)/2\geq \Ma\log \bar\sigma-k_{0}(\alpha)\log(8\kappa)+C
$$
where $C=\log(\mu/4\kappa)$ is a constant independent of $\alpha$.
Recall the definitions (\ref{definition_Malpha}) and that $r_{0}(\alpha)\geq
(\frac{1}{2}-2\eta)\log\frac{1}{\alpha}$, thus
\begin{align*}
k_{0}(\alpha)\log(8\kappa)
& \geq r_{0}(\alpha)-\left(\frac12-4\eta\right)\log\frac{1}{\alpha}+C \geq
r_{0}(\alpha)\left(1-\frac{\frac12-4\eta}{\frac12-2\eta}\right)+C \\
&\geq r_{0}(\alpha)\left(\frac{2\eta}{\frac12-2\eta}\right)+C\geq
\eta r_{0}(\alpha),
\end{align*}
for sufficiently small $\alpha$. This proves the claim.
\cqfd %%%%%%%%%%%%%%
%
\bigskip
%
%
A sequence $\bar l=(l_1,\dots,l_{n})$ of $n$ integers in
$\{0,1,\cdots,d-1\}$ will be called an {\em $n$-word}
below. For an $n$-word, we denote by
$\omega(\bar l)$ the unique element $\omega$ of $\CP_{n}$ satisfying
$g^{i-1}(\omega)\in[\theta_{l_{i}},\theta_{l_{i}+1}[$ for $1\le i\le n$. For an
$n$-word $\bar l$ and $1\leq j\leq n$ we set
$$
\widehat Y_j(\bar l)=\graph(Y_j(\bar l,\cdot))=F^j(\widehat{ Y}|_{\omega(\bar l)}).
$$
Two $\Ma$-words
$\bar l$ and $\bar m$ are said to be {\it incompatible} if
\begin{equation}\label{eq:defincomp}
|Y_{\Ma}(\bar l,\theta)- Y_{\Ma}(\bar m,\theta)|\geq 3
e^{-r_{0}(\alpha)+2}\sqrt{\alpha} \qquad \forall \theta\in S^1.
\end{equation}
We claim that, if two $\Ma$-words
$\bar l$ and $\bar m$ are incompatible, then at most one of the curves
$\widehat{Y}_{\Ma}(\bar{l})$ or
$\widehat{Y}_{\Ma}(\bar{m})$ can meet
$\widehat{J}(r_{0}(\alpha)-2)$.
Indeed, the right hand side of the inequality (\ref{eq:defincomp}) is chosen
so that it is larger than the sum of the height of
$\widehat{J}(r_{0}(\alpha)-2)$, which is equal to $2
e^{-r_{0}(\alpha)+2}\sqrt{\alpha}$, and the vertical width of the
admissible curve $\widehat{Y}_{\Ma}(\bar{l})$, which is bounded by
$
2A_{0} (2\pi/d)\alpha< e^{-r_{0}(\alpha)+2}\sqrt{\alpha}\sim \alpha^{1-2\eta}$.
A bound on the number of mutually compatible words will give a bound
of the type sought-for.
To estimate this number, we are first
going to establish a sufficient condition for incompatibility.
Consider a $(t_{i}-1)$-word $\bar m=(m_1,\dots,m_{t_i-1})$ and an
integer $0\le m< d$. Applying
Lemma~\ref{disjoint-curve} to the curve
$\widehat{X}=\widehat{Y}^{t_{i}-1}(\bar m)$, we find an integer $0\le m'< d$ such that,
$$
\{\theta\in S^1|\ |Y_{t_i}((\bar m,m),\theta)-Y_{t_i}((\bar m,m'),\theta)|
\leq \varepsilon_{0}\alpha \}\subset \bigcup_{\ell=1}^{2^{\rho+2}}J_\ell
$$
where $J_\ell$ are intervals satisfying
$m(J_\ell)\leq\frac{1}{40\times2^\rho d}$. We denote by $E(\bar
m, m)$ such an integer as $m'$.
We write $\Omega_{*}(\bar m,m)$ for the collection of elements $\omega$ of $\CP_{m_0}$ which
meet some $J_\ell$ above.
For a given $J_\ell$,
$$
\#\{\omega\in\CP_{m_0} | \omega\cap J_\ell\not=\emptyset\}
\leq 2+\frac{m(J_\ell)}{d^{-m_0}}
\leq 2+\frac{d^{m_0}}{40\times 2^\rho d}.
$$
Hence taking the union over all $J_\ell$'s,
$$
\#\{\omega\in\CP_{m_0} | \omega\cap (\cup J_\ell)\not=\emptyset\}
\leq 2^{\rho+3}
+\frac{2^{\rho+2}d^{m_0}}{40\times 2^\rho d}\leq \frac{1}{5d}d^{m_0}.
$$
It follows that, for any $t_{i}$-word $\bar m$, we
have
\begin{equation}\label{estimate-omega}
\#\Omega_*(\bar m) := \# \Omega_*((m_1,\dots,m_{t_i-1}),m_{t_i})
\leq\frac{1}{5d}d^{m_0}.
\end{equation}
\medskip\\
%
%
%
%
\noindent
{\bf Claim 2. A sufficient condition for incompatibility:}
Let $1\leq i