---
abstract: 'We study a skew product with a curve of neutral points. We show that there exists a unique absolutely continuous invariant probability measure, and that the Birkhoff averages of a sufficiently smooth observable converge to a normal law or a stable law, depending on the average of the observable along the neutral curve.'
author:
- 'S. Gouëzel'
- 'Sébastien Gouëzel [^1]'
bibliography:
- 'biblio.bib'
date: November 2003
title: 'Statistical properties of a skew product with a curve of neutral points [^2]'
---

Introduction
============

Let $T:M\to M$ be a map on a compact space. While uniformly hyperbolic or uniformly expanding dynamics are well understood, problems arise when there are neutral fixed points (where the differential of $T$ has an eigenvalue equal to $1$). The one-dimensional case has been thoroughly studied, particularly when $T$ has only one neutral fixed point (see [@liverani_saussol_vaienti] and references therein). The normal form at the fixed point dictates the asymptotics of the dynamics, and in particular the speed of mixing, and the convergence of Birkhoff sums to limit laws ([@gouezel:stable]).

In this article, we study the same type of phenomenon, but in higher dimension. Contrary to [@hu:almost_hyperbolic], [@pollicott_yuri:indifferent] (where the case of isolated fixed points is considered), our models admit a whole invariant neutral curve. We show that the one-dimensional results remain essentially true.

More precisely, define a map $T_\alpha$ on $[0,1]$ by $$T_\alpha(x)=\left\{ \begin{array}{cl}
   x(1+2^\alpha x^\alpha) &\text{if }0{\leqslant}x{\leqslant}1/2
   \\
   2x-1 &\text{if }1/2<x{\leqslant}1
   \end{array}\right.$$ It has a neutral fixed point at $0$, behaving like $x(1+x^\alpha)$. To mix different such behaviors, we consider a skew product, similar to the Alves-Viana map ([@viana:multidim_attr]) but where the unimodal maps are replaced by $T_\alpha$. Let $\alpha:S^1 \to (0,1)$ be a map with minimum $\operatorname{{\alpha_{\text{min}}}}$ and maximum $\operatorname{{\alpha_{\text{max}}}}$. Assume that

1.  $\alpha$ is $C^2$.

2.  $0<\operatorname{{\alpha_{\text{min}}}}<\operatorname{{\alpha_{\text{max}}}}<1$.

3.  $\alpha$ takes the value $\operatorname{{\alpha_{\text{min}}}}$ at a unique point $x_0$, with $\alpha''(x_0)>0$.

4.  $\operatorname{{\alpha_{\text{max}}}}< \frac{3}{2}\operatorname{{\alpha_{\text{min}}}}$ (which implies $\operatorname{{\alpha_{\text{max}}}}<\operatorname{{\alpha_{\text{min}}}}+1/2$).

These conditions are for example satisfied by $\alpha(\omega)=\operatorname{{\alpha_{\text{min}}}}+{\varepsilon}(1+\sin(2\pi \omega))$ where $\operatorname{{\alpha_{\text{min}}}}\in (0,1)$ and ${\varepsilon}$ is small enough.

We define a map $T$ on $S^1 \times [0,1]$ by $$\label{definit_T}
  T(\omega,x)= (F(\omega), T_{\alpha(\omega)}(x))$$ where $F(\omega)=4\omega$.

In the following, we will generalize to this skew product the one-dimensional results on the maps $T_\alpha$. First of all, in Section \[section\_invariante\], we prove that there exists a unique absolutely continuous invariant probability measure $m$, whose density $h$ is in fact Lipschitz on every compact subset of $S^1\times (0,1]$ (Theorem \[existence\_mesure\_invariante\]). In Section \[limit\_Markov\], we prove limit theorems for abstract Markov maps (using a method essentially due to [@melbourne_torok] and recalled in Appendix \[appendice:loi\_stable\], and estimates of [@aaronson_denker] and [@gouezel:stable]). Finally, in Sections \[section\_estimee\_Xn\] and \[section:limite\], we study the limit laws of Birkhoff sums for the skew product $T$, and we obtain the convergence to a normal law or a stable law, depending on the value of $\operatorname{{\alpha_{\text{min}}}}$. We obtain the following theorem (see Theorem \[enonce\_theoreme\_limite\] for more details).

Set $$A=\frac{1}{4\left( \operatorname{{\alpha_{\text{min}}}}^{3/2}
  \sqrt{\frac{\pi}{2\alpha''(x_0)}}\right)^{1/\operatorname{{\alpha_{\text{min}}}}}}\int_{S^1\times \{1/2\}}
  h\operatorname{dLeb},$$ where $h$ is the density of the absolutely continuous invariant probability measure.

Let $f$ be a Lipschitz function on $S^1\times [0,1]$, with $\int
f{\, {\rm d}}m=0$. Write $c=\int_{S^1\times\{0\}} f \operatorname{dLeb}$ and $S_n
f=\sum_{k=0}^{n-1} f\circ T^k$. Then

-   If $\operatorname{{\alpha_{\text{min}}}}<1/2$, there exists $\sigma^2 {\geqslant}0$ such that $\frac{1}{\sqrt{n}} S_n f \to {\mathcal{N}}(0,\sigma^2)$.

-   If $\operatorname{{\alpha_{\text{min}}}}=1/2$ and $c\not=0$, then $\frac{S_n f}{\sqrt{ \frac{c^2 A }{4} n (\ln n)^2}} \to
    {\mathcal{N}}(0,1)$.

-   If $1/2<\operatorname{{\alpha_{\text{min}}}}<1$ and $c\not =0$, then $\frac{S_n f}{n^{\operatorname{{\alpha_{\text{min}}}}}\sqrt{\operatorname{{\alpha_{\text{min}}}}\ln n}} \to Z$, where the random variable $Z$ has an explicit stable distribution.

-   If $1/2{\leqslant}\operatorname{{\alpha_{\text{min}}}}<1$ and $c=0$, then there exists $\sigma^2 {\geqslant}0$ such that $\frac{1}{\sqrt{n}} S_n f \to
    {\mathcal{N}}(0,\sigma^2)$.

An interesting feature of this example is that its study involves sophisticated mixing properties of $F$, particularly a multiple decorrelation property, proved in Appendix \[appendice:pene\] using [@pene:averaging].

Theorems of [@gouezel:stable] could be used instead of the method of [@melbourne_torok] to get the limit laws. However, this elementary method is interesting in its own right, and can be generalized more easily to other settings than the results of [@gouezel:stable] (in particular to the case of more neutral fixed points).

The specific form of $F$ is of no importance at all, the results remain true when $F$ is $C^2$ with $|F'| {\geqslant}4$ (for example $F(\omega)=d\omega$ with $d {\geqslant}4$). In the same way, the only important properties of the maps $T_\alpha$ are their normal form close to $0$ and the fact that they are Markov. Finally, the hypothesis $\alpha''(x_0)\not=0$ is only useful for limit theorems, and can be replaced by: $\exists m,
\alpha^{(m)}(x_0)\not=0$ (but the normalizing factors have to be modified accordingly). For the sake of simplicity, we will restrict ourselves in what follows to the aforementioned case.

In this article, $a(n) \sim b(n)$ means that $a(n)/b(n) \to 1$ when $n\to \infty$. The integral with respect to a probability measure will sometimes be denoted by $E(\cdot)$. Finally, $\lfloor
x \rfloor$ will denote the integer part of $x$.

Invariant measure {#section_invariante}
=================

An important property of the map $T$, that will be used thoroughly in what follows, is that it is Markov: there exists a partition of the space such that every element of this partition is mapped by $T$ on a union of elements of this partition. In fact, we will consider $T_Y$ (the induced map on $Y=S^1 \times (1/2,1]$), which is also Markov, and expanding, contrary to $T$. We will apply to $T_Y$ classical results on expanding Markov maps (also called *Gibbs-Markov* maps), which we recall in the next paragraph.

Markov maps and invariant measures
----------------------------------

Let $(Y,{\mathcal{B}},m_Y)$ be a standard probability space, endowed with a bounded metric $d$. A non-singular map $T_Y$ defined on $Y$ is said to be a *Markov map* if there exists a finite or countable partition $\alpha$ of $Y$ such that $\forall a\in
\alpha$, $m_Y(a)>0$, $T_Y(a)$ is a union (mod $0$) of sets of $\alpha$, and $T_Y:a \to T_Y(a)$ is invertible. In this case, $\alpha$ is a *Markov partition* for $T_Y$.

A Markov map $T_Y$ (with a Markov partition $\alpha$) is a *Gibbs-Markov map* ([@aaronson:book]) if

1.  $T_Y$ has the big image property: $\inf_{a\in
    \alpha} m_Y(T_Y(a))>0$.

2.  \[enumere\_expansion\] There exists $\lambda>1$ such that $\forall a\in \alpha, \forall
    x,y\in a, d(T_Yx, T_Y y){\geqslant}\lambda d(x,y)$.

3.  \[enumere\_distortion\] Let $g$ be the inverse of the jacobian of $T_Y$, i.e. on a set $a\in \alpha$, $g(x)=\frac{{\, {\rm d}}m_Y}{{\, {\rm d}}\left(m_Y \circ
    (T_Y)_{|a}\right)} (x)$. Then there exists $C>0$ such that for all $a\in \alpha$, for almost all $x,y\in a$, $$\left|1-\frac{g(x)}{g(y)} \right| {\leqslant}C d(T_Yx,T_Yy).$$

This definition is slightly more general than the definition of [@aaronson:book]: the distance $d=d_\tau$ considered there is given by $d_\tau(x,y)=\tau^{s(x,y)}$ where $\tau<1$ and $s(x,y)$ is the separation time of $x$ and $y$, i.e.  $$\label{definit_separation}
  s(x,y)=\inf\{ n\in {\mathbb{N}}{\ |\ }\nexists a\in \alpha, T^n x\in a, T^n y\in
  a\}.$$

The proof of [@aaronson:book Theorem 4.7.4] still works in our context, and gives:

\[existe\_mesure\_invariante\] Let $T_Y$ be a transitive Gibbs-Markov map ($\forall a,b \in
\alpha, \exists n\in {\mathbb{N}}, m_Y(T_Y^n a \cap b)>0$) such that $\operatorname{Card}(\alpha_*)<\infty$, where $\alpha_*$ is the partition generated by the images $T_Y(a)$ for $a\in \alpha$. Then $T_Y$ is ergodic, and there exists a unique absolutely continuous (with respect to $m_Y$) invariant probability measure, denoted by $\mu_Y$.

Moreover, $\mu_Y=h m_Y$ where the density $h$ is bounded and bounded away from $0$, and Lipschitz on every set of $\alpha_*$.

Preliminary estimates
---------------------

To apply Theorem \[existe\_mesure\_invariante\], we will construct a Markov partition, and control the distortion of the inverse branches of $T_Y$.

We will write $T_\omega^n=T_{\alpha(F^{n-1}\omega)}
\circ\cdots\circ T_{\alpha(\omega)}$, whence $T^n(\omega,x)=(F^n\omega, T_\omega^n(x))$. Write also $d((\omega_1,x_1),(\omega_2,x_2))= |\omega_1-\omega_2|+|x_1-x_2|$. A point of $S^1 \times [0,1]$ will be denoted by ${\textbf{x}}=(\omega,x)$. Finally, set $\operatorname{{d_{\text{vert}}}}((\omega_1,x_1),(\omega_2,x_2))=|x_2-x_1|$.

Define $X_0(\omega)=1$, $X_1(\omega)=1/2$, and for $n{\geqslant}2$, $X_n(\omega)$ is the preimage in $[0,1/2]$ of $X_{n-1}(F\omega)$ by $T_{\alpha(\omega)}$. These $X_n$ will be useful in the construction of a Markov partition for $T$ (paragraph \[construit\_markov\]).

\[estime\_croissance\_grossiere\_Xn\] There exists $C>0$ such that $\forall n\in {\mathbb{N}}^*, \forall \omega\in
S^1$, $$\frac{1}{Cn^{1/\operatorname{{\alpha_{\text{min}}}}}} {\leqslant}X_n(\omega) {\leqslant}\frac{C}{n^{1/\operatorname{{\alpha_{\text{max}}}}}}.$$

Write $Z_1=1/2$ and $T(Z_{n+1})=Z_n$ where $T(x)=x(1+2^{\operatorname{{\alpha_{\text{max}}}}}
x^{\operatorname{{\alpha_{\text{min}}}}})$. We easily check inductively that $Z_n{\leqslant}X_n(\omega)$ for every $\omega$, since $T(x){\geqslant}T_{\alpha(\omega)}(x)$ for every $\omega$. It is thus sufficient to estimate $Z_n$ to get the minoration. As $T(x){\geqslant}x$, the sequence $Z_n$ is decreasing, and nonnegative, whence it tends to a fixed point of $T$, necessarily $0$.

We have $$\begin{aligned}
  \frac{1}{Z_n^{\operatorname{{\alpha_{\text{min}}}}}}&
  =\frac{1}{Z_{n+1}^{\operatorname{{\alpha_{\text{min}}}}}}\left(1+2^{\operatorname{{\alpha_{\text{max}}}}} Z_{n+1}^{\operatorname{{\alpha_{\text{min}}}}}\right)^{-\operatorname{{\alpha_{\text{min}}}}}
  =\frac{1}{Z_{n+1}^{\operatorname{{\alpha_{\text{min}}}}}}\left(1-\operatorname{{\alpha_{\text{min}}}}2^{\operatorname{{\alpha_{\text{max}}}}} Z_{n+1}^{\operatorname{{\alpha_{\text{min}}}}}
  +o(Z_{n+1}^{\operatorname{{\alpha_{\text{min}}}}})\right)
  \\&
  =\frac{1}{Z_{n+1}^{\operatorname{{\alpha_{\text{min}}}}}}-\operatorname{{\alpha_{\text{min}}}}2^{\operatorname{{\alpha_{\text{max}}}}}+o(1).
  \end{aligned}$$ A summation gives $\frac{1}{Z_m^{\operatorname{{\alpha_{\text{min}}}}}}\sim m \operatorname{{\alpha_{\text{min}}}}2^{\operatorname{{\alpha_{\text{max}}}}}$, whence $Z_m \sim C/m^{1/\operatorname{{\alpha_{\text{min}}}}}$, which concludes the minoration.

The majoration is similar, using a sequence $Z'_n$ with $Z'_n {\geqslant}X_n(\omega)$.

*We fix once and for all a large enough constant $D$.* The following definition is analogous to a definition of Viana ([@viana:multidim_attr]).

Let $\psi: K\to [0,1]$, where $K$ is a subinterval of $S^1$. We say that the graph of $\psi$ is an admissible curve if $\psi$ is $C^1$ with $|\psi'|{\leqslant}D$.

Let $\psi$ be an admissible curve, defined on $K$ with $|K| <1/4$, and included in $K\times [0,1/2]$ or $K\times (1/2,1]$. Then the image of $\psi$ by $T$ is still an admissible curve.

Let $(u,v)$ be a tangent vector at $(\omega,x)$ with $|v|{\leqslant}D
|u|$, we have to check that its image $(u',v')$ by $DT(\omega,x)$ still satisfies $|v'|{\leqslant}D |u'|$.

Assume first that $x{\leqslant}1/2$, whence $u'=4u$ and $v'=(1+(2x)^{\alpha(\omega)} (\alpha(\omega)+1))v+ x
\ln(2x)\alpha'(\omega)(2x)^{\alpha(\omega)}u$. As $\alpha(\omega){\leqslant}\operatorname{{\alpha_{\text{max}}}}{\leqslant}1$, we get $|v'|{\leqslant}3|v|+C|u|$ for a constant $C$ (which depends only on ${\left\| \alpha' \right\|}_\infty$). Thus, $$\frac{|v'|}{|u'|} {\leqslant}\frac{3}{4} \frac{|v|}{|u|} +\frac{C}{4}.$$ This will give $|v'|/|u'| {\leqslant}D$ if $\frac{3}{4}D +\frac{C}{4}
{\leqslant}D$, which is true if $D$ is large enough.

Assume then that $x>1/2$. Then $u'=4u$ and $v'=2v$, and there is nothing to prove.

\[controle\_rapport\_horz\_vert\] Let $(\omega_1,x_1)$ and $(\omega_2,x_2)$ be two points in $S^1\times [0,1/2]$ with $|x_1-x_2|{\leqslant}D|\omega_1-\omega_2|$ and $|\omega_1-\omega_1|{\leqslant}\frac{1}{8}$. Then their images satisfy $|x'_1-x'_2|{\leqslant}D|\omega'_1-\omega'_2|$.

Use a segment between the two points: it is an admissible curve, whence its image is still admissible.

The Markov partition {#construit_markov}
--------------------

Set $Y=S^1 \times (1/2,1]$. For ${\textbf{x}}\in Y$, set ${\varphi}_Y({\textbf{x}})=\inf\{n>0 {\ |\ }T^n({\textbf{x}})\in Y\}$: this is the first return time to $Y$, everywhere finite. The map $T_Y({\textbf{x}}):=T^{{\varphi}_Y({\textbf{x}})}({\textbf{x}})$ is the map induced by $T$ on $Y$. We will show that $T_Y$ is a Gibbs-Markov map, by constructing an appropriate Markov partition.

If $I$ is an interval of $S^1$, we will abusively write $I\times
[X_{n+1},X_n]$ for $\{(\omega,x){\ |\ }\omega \in I, x\in
[X_{n+1}(\omega), X_n(\omega)]\}$.

Set $I_n(\omega)=[X_{n+1}(\omega),X_n(\omega)]$ (or $\{\omega\}
\times [X_{n+1}(\omega),X_n(\omega)]$, depending on the context). By definition of $X_n$, $T$ maps $\{\omega\}\times I_n(\omega)$ bijectively on $\{F\omega\} \times I_{n-1}(F\omega)$. Thus, the interval $I_n(\omega)$ returns to $[1/2,1]$ in exactly $n$ steps.

Let $Y_n(\omega)$ be the preimage in $[1/2,1]$ of $X_{n-1}(F\omega)$ under $T_{\alpha(\omega)}$. Thus, the interval $J_n(\omega)= [Y_{n+1}(\omega),Y_n(\omega)]$ returns to $[1/2,1]$ in $n$ steps.

*We fix once and for all $0<{{\varepsilon}_0}<\frac{1}{8}$, small enough so that $D{{\varepsilon}_0}$ is less than the length of every interval $I_1(\omega)$*. (This condition will be useful in distortion estimates).

*Let $q$ be large enough so that $\frac{1}{4^q}<{{\varepsilon}_0}$, and consider $A_{s,n}=\left[
\frac{s}{4^{q+n}},\frac{s+1}{4^{q+n}}\right] \times J_n$, for $n\in {\mathbb{N}}^*$ and $0{\leqslant}s{\leqslant}4^{q+n}-1$*: this set is mapped by $T^n$ on $\left[\frac{s}{4^q},\frac{s+1}{4^q}\right] \times
[1/2,1]$. Let $K_0,\ldots,K_{4^q-1}$ be the sets $\left[\frac{i}{4^q},\frac{i+1}{4^q}\right] \times [1/2,1]$. Then the map $T_Y$ is an isomorphism between each $A_{s,n}$ and some $K_i$. Consequently, the map $T_Y$ is Markov for the partition $\{A_{s,n}\}$, and it has the big image property.

To apply Theorem \[existe\_mesure\_invariante\], we need expansion (for in the definition of Gibbs-Markov maps) and distortion control (for ). The expansion is given by the next proposition, and the distortion is estimated in the following paragraph.

On the intervals $[X_3(\omega),X_1(\omega)]$, the derivative of $T_{\alpha(\omega)}$ is greater than $1$, whence greater than a constant $2>\lambda>1$, independent of $\omega$.

For $(\omega_1,x_1)$ and $(\omega_2,x_2)\in S^1\times [0,1]$, set $$\label{definit_d'}
  d'((\omega_1,x_1),(\omega_2,x_2))=a|x_1-x_2|+|\omega_1-\omega_2|$$ where $a=\frac{1-\lambda/4}{D}$.

\[dilate\_markov\] On each $A_{s,n}$, the map $T^n$ is expanding by at least $\lambda$ for the distance $d'$.

For $n=1$ (the points return directly to $S^1\times [1/2,1]$), everything is linear, and the result is clear. Assume $n{\geqslant}2$.

Take $(\omega_1,x_1)$ and $(\omega_2,x_2)\in A_{s,n}$, with for example $x_2{\geqslant}x_1$. The points $(\omega_1,x_1)$ and $(\omega_2,x_1)$ return to $S^1\times [1/2,1]$ after at least $n$ iterations (by hypothesis for the first point, and the second point is under $(\omega_2,x_2)$). We can use Corollary \[controle\_rapport\_horz\_vert\] $n-1$ times, and get that in vertical distance, $\operatorname{{d_{\text{vert}}}}( T^n(\omega_1,x_1),T^n(\omega_2,x_1))
{\leqslant}D|F^n\omega_1-F^n\omega_2|$. In particular, $T_{\omega_2}^n(x_1) {\geqslant}T_{\omega_1}^n(x_1) - D {\varepsilon}_0 {\geqslant}1/2-D {\varepsilon}_0$. Thus, by definition of ${\varepsilon}_0$, $T^n(\omega_2,x_1) \in I_i(F^n \omega_2)$ for $i=0$ or $1$, whence $T^{n-1}(\omega_2,x_1) \in [X_3(F^{n-1} \omega_2),
X_1(F^{n-1}\omega_2)]$. Note that $T^{n-1}(\omega_2,x_2)$ belongs to the same interval (in fact, $T^{n-1}_{\omega_2}(x_2) \in
[X_2(F^{n-1}\omega_2), X_1(F^{n-1}\omega_2)]$). Moreover, the $T_\alpha$ are expanding, whence $\operatorname{{d_{\text{vert}}}}(T^{n-1}(\omega_2,x_1),
T^{n-1}(\omega_2,x_2)) {\geqslant}|x_1-x_2|$. We apply once more $T$, which expands at least by $\lambda$ on $[X_3(F^{n-1}\omega_2),
X_1(F^{n-1} \omega_2)]$ by definition of $\lambda$, and get $\operatorname{{d_{\text{vert}}}}(T^n(\omega_2,x_1),T^n(\omega_2,x_2)) {\geqslant}\lambda|x_1-x_2|$.

Finally, $$\begin{aligned}
  d'(T^n(\omega_1,x_1),T^n(\omega_2,x_2))&
  =a
  \operatorname{{d_{\text{vert}}}}(T^n(\omega_1,x_1),T^n(\omega_2,x_2))+|F^n\omega_1-F^n\omega_2|
  \\&
  {\geqslant}a \operatorname{{d_{\text{vert}}}}(T^n(\omega_2,x_1),T^n(\omega_2,x_2))-a \operatorname{{d_{\text{vert}}}}(
  T^n(\omega_1,x_1),T^n(\omega_2,x_1))
  \\& \hphantom{=\ }+|F^n\omega_1-F^n\omega_2|
  \\&
  {\geqslant}a \lambda|x_1-x_2| -aD
  |F^n\omega_1-F^n\omega_2|+|F^n\omega_1-F^n\omega_2|.
  \end{aligned}$$ The proposition will be proved if $(1-aD)|F^n\omega_1-F^n\omega_2|
{\geqslant}\lambda |\omega_1-\omega_2|$. But $$(1-aD)|F^n\omega_1-F^n\omega_2|
  =(1-aD)4^n |\omega_1-\omega_2|
  {\geqslant}(1-aD)4 |\omega_1-\omega_2|
  =\lambda |\omega_1-\omega_2|.$$

Distortion bounds
-----------------

\[lemme\_distortion\_deplac\_horz\] There exists a constant $E>0$ such that $$\begin{split}
  \forall n>0, \forall \omega_1,\omega_2&\in S^1 \text{ with
  }|\omega_1-\omega_2|{\leqslant}\frac{{{\varepsilon}_0}}{4^n}, \forall x_1 \in
  J_n(\omega_1) \text{ with }T_{\omega_2}^{n-1} x_1{\leqslant}1/2,\\&
  \left| \ln (T_{\omega_1}^n)'(x_1)-\ln (T_{\omega_2}^n)'(x_1)\right|
  {\leqslant}E |F^n \omega_1 -F^n \omega_2|.
  \end{split}$$

We use Corollary \[controle\_rapport\_horz\_vert\] $n-1$ times and get for $0{\leqslant}k{\leqslant}n$ that $|T_{\omega_1}^k x_1 -T_{\omega_2}^k
x_1|{\leqslant}D|F^k\omega_1-F^k\omega_2|$.

In particular, for $k=n$, $|T_{\omega_1}^n x_1|{\geqslant}1/2$, whence $|T_{\omega_2}^n x_1|{\geqslant}1/2-D{{\varepsilon}_0}$. Consequently, $T^n(\omega_2,x_1)\in I_i(F^n\omega_2)$ for some $i\in \{0,1\}$, by definition of ${{\varepsilon}_0}$. An inverse induction gives $T^k(\omega_2,x_1) \in I_{n-k+i}(F^k\omega_2)$.

For $x{\leqslant}1/2$ and $\omega\in S^1$, write $G(\omega,x)=\ln
T_{\alpha(\omega)}'(x)=
\ln\left(1+(\alpha(\omega)+1)(2x)^{\alpha(\omega)}\right)$. Then $$\frac{\partial G}{\partial x}(\omega,x)
  =\frac{(\alpha(\omega)+1)\alpha(\omega)2^{\alpha(\omega)}
  x^{\alpha(\omega)-1}}{1+(\alpha(\omega)+1)(2x)^{\alpha(\omega)}}
  {\leqslant}C x^{\operatorname{{\alpha_{\text{min}}}}-1}$$ and $$\left|\frac{\partial G}{\partial \omega}(\omega,x)\right|
  =\left|\frac{\alpha'(\omega)(2x)^{\alpha(\omega)}
  +(\alpha(\omega)+1)\alpha'(\omega)\ln(2x)(2x)^{\alpha(\omega)}}
  {1+(\alpha(\omega)+1)(2x)^{\alpha(\omega)}} \right|
  {\leqslant}C.$$

Lemma \[estime\_croissance\_grossiere\_Xn\], and the fact that $T^k(\omega_1,x_1)\in I_{n-k}(F^k\omega_1)$ and $T^k(\omega_2,x_1)
\in I_{n-k+i}(F^k\omega_2)$ with $i{\leqslant}1$, give that the second coordinates of $T^k(\omega_1,x_1)$ and $T^k(\omega_2,x_1)$ are ${\geqslant}\frac{1}{C (n-k+1)^{1/\operatorname{{\alpha_{\text{min}}}}}}$. On the set of points $(\omega,x)$ with $x{\geqslant}\frac{1}{C (n-k+1)^{1/\operatorname{{\alpha_{\text{min}}}}}}$, the estimates on the partial derivatives of $G$ show that this function is $C (n-k+1)^{1/\operatorname{{\alpha_{\text{min}}}}-1}$-Lipschitz, whence $$\begin{aligned}
  |G(T^k(\omega_1,x_1))-G(T^k(\omega_2,x_1))| &
  {\leqslant}C (n-k+1)^{1/\operatorname{{\alpha_{\text{min}}}}-1} d((T^k(\omega_1,x_1),T^k(\omega_2,x_1))
  \\&
  {\leqslant}C (n-k+1)^{1/\operatorname{{\alpha_{\text{min}}}}-1} (1+D)|F^k \omega_1-F^k \omega_2|
  \\&
  {\leqslant}C (n-k+1)^{1/\operatorname{{\alpha_{\text{min}}}}-1} (1+D) 4^k |\omega_1-\omega_2|.
  \end{aligned}$$ Finally, $$\begin{aligned}
  \left| \ln (T_{\omega_1}^n)'(x_1)-\ln (T_{\omega_2}^n)'(x_1)\right|&
  {\leqslant}\sum_{k=0}^{n-1}|G(T^k(\omega_1,x_1))-G(T^k(\omega_2,x_1))|
  \\&
  {\leqslant}C 4^n |\omega_1-\omega_2| \sum_{k=0}^{n-1} (n-k+1)^{1/\operatorname{{\alpha_{\text{min}}}}-1} 4^{k-n}
  \\&
  {\leqslant}C |F^n\omega_1-F^n\omega_2| \sum_{l=1}^\infty (l+1)^{1/\operatorname{{\alpha_{\text{min}}}}-1} 4^{-l}.
  \end{aligned}$$ The last sum is finite, which concludes the proof.

For $n{\geqslant}2$, write $J_n^+(\omega)=[Y_{n+2}(\omega),Y_n(\omega)]$. Thus, if $n{\geqslant}1$, $J_{n+1}^+(\omega)$ is the preimage of $I_n^+(F\omega)$, defined by $I_n^+(F\omega)=[X_{n+2}(F\omega),X_{n}(F\omega)]$. These intervals will appear naturally in distortion controls, since we have seen in the proof of Lemma \[lemme\_distortion\_deplac\_horz\] that, if we move away horizontally from a point of $J_n(\omega_1)$, we find a point of $J_{n+i}(\omega_2)$ for $i\in
\{0,1\}$, i.e. in $J_n^+(\omega_2)$.

\[distortion\_verticale\_bornee\] There exists a constant $C$ such that $$\forall n{\geqslant}0, \forall \omega\in S^1, \forall x,y\in
  J_n^+(\omega),\ \left|\ln (T_\omega^n)'(x)-\ln(T_\omega^n)'(y)
  \right| {\leqslant}C |T_\omega^n(x)-T_\omega^n(y)|.$$

Recall that the Schwarzian derivative of an increasing diffeomorphism $g$ of class $C^3$ is $Sg(x)=\frac{g'''(x)}{g'(x)}-\frac{3}{2}\left(\frac{g''(x)}{g'(x)}
\right)^2$. The composition of two functions with nonpositive Schwarzian derivative still has a nonpositive Schwarzian derivative.

For $\tau>0$, the Koebe principle ([@demelo_vanstrien Theorem IV.1.2]) states that, if $Sg{\leqslant}0$, and $J\subset J'$ are two intervals such that $g(J')$ contains a $\tau$-scaled neighborhood of $g(J)$ (i.e. the intervals on the left and on the right of $g(J)$ in $g(J')$ have length at least $\tau |g(J)|$), then there exists a constant $K(\tau)$ such that $$\forall x,y\in J, \left|\ln g'(x) - \ln g'(y)\right| {\leqslant}K(\tau) \frac{|x-y|}{|J|}.$$ This implies that the distortion of $g$ is bounded on $J$, whence it is possible to replace the bound on the right with $K'(\tau)
\frac{|g(x)-g(y)|}{|g(J)|}$.

In our case, if $0<\alpha<1$, the left branch of $T_\alpha$ has nonpositive Schwarzian derivative, since $T_\alpha'''<0$ and $T_\alpha'>0$. Let in particular $g$ be the composition of the left branches of $T_{\alpha(F^{n-1}\omega)},\ldots,
T_{\alpha(F\omega)}$, and of the right branch of $T_{\alpha(\omega)}$. Then, on $J_n^+$, we have $T_\omega^n=g$, and $g$ has nonpositive Schwarzian derivative.

We want to see that $\left|\ln (T_\omega^n)'(x) - \ln
(T_\omega^n)'(y)\right|{\leqslant}C|T_\omega^n(x)-T_\omega^n(y)|$. For this, we apply the Koebe principle to $J=J_n^+$ and $J'=[1/2+\delta,2]$ for $\delta$ very small. Then $g(J)=[X_2,1]$ while $g(J')$ contains $[\delta',2]$ for $\delta'>0$, arbitrarily small if $\delta$ is small enough. As the $X_{2}$ are uniformly bounded away from $0$, there exists $\tau>0$ (independent of $\omega$ and $n$) such that $g(J')$ contains a $\tau$-scaled neighborhood of $g(J)$. The Koebe principle then gives the desired result.

\[prop\_distortion\_bornee\] There exists a constant $C$ such that, for every $A_{s,n}$, for every $(\omega_1,x_1)$ and $(\omega_2,x_2)\in A_{s,n}$, $$\left| \frac{\det DT^n(\omega_1,x_1)}{\det DT^n(\omega_2,x_2)}-1
  \right|{\leqslant}C d(T^n(\omega_1,x_1),T^n(\omega_2,x_2)).$$

The matrix $DT^n(\omega,x)$ is upper triangular, with $4^n$ in the upper left corner. Thus, we have to show that $$\left| \ln(T_{\omega_1}^n)'(x_1)-\ln(T^n_{\omega_2})'(x_2)
  \right|{\leqslant}C d(T^n(\omega_1,x_1),T^n(\omega_2,x_2)).$$ Assume for example $x_2{\geqslant}x_1$, which implies that $T_{\omega_2}^k(x_1){\leqslant}1/2$ for $k=0,\ldots,n-1$. Lemma \[lemme\_distortion\_deplac\_horz\] can be applied to $x_1$, $\omega_1$ and $\omega_2$, and gives in particular that $x_1\in
J_n^+(\omega_2)$.

Write $$\begin{aligned}
  \left| \ln(T_{\omega_2}^n)'(x_2)-\ln(T^n_{\omega_1})'(x_1)
  \right|
  &{\leqslant}\left| \ln(T_{\omega_2}^n)'(x_2)-\ln(T^n_{\omega_2})'(x_1)
  \right|
  +  \left| \ln(T_{\omega_2}^n)'(x_1)-\ln(T^n_{\omega_1})'(x_1)
  \right|
  \\&
  {\leqslant}C
  d(T^n(\omega_2,x_2)),T^n(\omega_2,x_1))
  + E|F^n\omega_2-F^n \omega_1|
  \end{aligned}$$ by Lemmas \[lemme\_distortion\_deplac\_horz\] and \[distortion\_verticale\_bornee\]. For the first term, $$\begin{aligned}
  d(T^n(\omega_2,x_2),T^n(\omega_2,x_1))
  &{\leqslant}d(T^n(\omega_2,x_2),T^n(\omega_1,x_1))
  +d(T^n(\omega_1,x_1),T^n(\omega_2,x_1))
  \\&
  {\leqslant}d(T^n(\omega_2,x_2),T^n(\omega_1,x_1))
  + (D+1) |F^n \omega_1-F^n \omega_2|
  \end{aligned}$$ using admissible curves.

As $|F^n \omega_1-F^n \omega_2|{\leqslant}d(T^n(\omega_1,x_1),T^n(\omega_2,x_2))$, we get the conclusion.

Construction of the invariant measure
-------------------------------------

The previous estimates and Theorem \[existe\_mesure\_invariante\] easily give that $T_Y$ admits an invariant measure, with Lipschitz density. Inducing gives an invariant measure for $T$, whose density is Lipschitz on each set $S^1\times (X_{n+1},X_n)$. However, this does not exclude discontinuities on $S^1\times X_n$, which is not surprising since $T$ itself has a discontinuity on $S^1 \times \{1/2\}$, which will then propagate to the other $X_n$, since the measure is invariant.

However, in the one-dimensional case, Liverani, Saussol and Vaienti ([@liverani_saussol_vaienti]) have proved that the density is really continuous everywhere, since they constructed it as an element of a cone of continuous functions. This fact remains true here:

\[existence\_mesure\_invariante\] The map $T$ admits a unique absolutely continuous invariant probability measure ${\, {\rm d}}m$. Moreover, this measure is ergodic. Finally, the density $h=\frac{{\, {\rm d}}m}{\operatorname{dLeb}}$ is Lipschitz on every compact subset of $S^1 \times (0,1]$.

Consider the map $T_Y$ induced by $T$ on $Y=S^1\times (1/2,1]$. It is Markov for the partition $\alpha=\{A_{s,n}\}$, and transitive for this partition since $T_Y^2(a)=Y$ for all $a\in \alpha$. Moreover, it is expanding for $d'$ on each set of the partition (Proposition \[dilate\_markov\]) and its distortion is Lipschitz (Proposition \[prop\_distortion\_bornee\], and $d$ equivalent to $d'$).

Theorem \[existe\_mesure\_invariante\] shows that $T_Y$ admits a unique absolutely continuous invariant probability measure ${\, {\rm d}}m_Y=h \operatorname{dLeb}$, which is ergodic. Moreover, the density $h$ is Lipschitz (for the distance $d'$, whence for the usual one) on each element of the partition $\alpha_*$ generated by the sets $T_Y(a)$, i.e. on the sets $K_i$.

To construct an invariant measure for the initial map $T$, we use the classical induction process ([@aaronson:book Section 1.1.5]): let ${\varphi}_Y$ be the return time to $Y$ under $T$, then $\mu=\sum_{n=0}^\infty T_*^n(m_Y | {\varphi}_Y>n)$ is invariant. To check that the new measure has finite mass, we have to see that $\sum m_Y({\varphi}_Y>n) <\infty$. As ${\, {\rm d}}m_Y$ and $\operatorname{dLeb}$ are equivalent, we check it for $\operatorname{dLeb}$. We have $$\operatorname{Leb}({\varphi}_Y>n)=\operatorname{Leb}(S^1 \times [1/2,Y_{n+1}])=\frac{1}{2}
  \operatorname{Leb}(S^1\times [0,X_n])
  {\leqslant}\frac{1}{2} \frac{C}{n^{1/\operatorname{{\alpha_{\text{max}}}}}},$$ using Lemma \[estime\_croissance\_grossiere\_Xn\]. As $\operatorname{{\alpha_{\text{max}}}}<1$, this is summable.

We know that $h$ is Lipschitz on the sets $[\frac{s}{4^q},\frac{s+1}{4^q}]\times [1/2,1]$, we have to prove the continuity on $\{s/4^q\}\times [1/2,1]$, which is not hard: these numbers $s/4^q$ are artificial, since they depend on the arbitrary choice of a Markov partition on $S^1$. We can do the same construction using other sets than the $A_{s,n}$. For example, set $A'_{s,n}=\left[ \frac{1}{3}+\frac{s}{4^{q+n}},
\frac{1}{3}+\frac{s+1}{4^{q+n}}\right]\times J_n$, and $K'_i=\left[\frac{1}{3}+\frac{i}{4^q},
\frac{1}{3}+\frac{i+1}{4^q}\right]$. Since $1/3$ is a fixed point of $F$, the map $T_Y$ is Markov for the partition $\{A'_{s,n}\}$, and each of these sets is mapped on a set $K'_i$. Thus, the same arguments as above apply, and prove that $h$ is Lipschitz on each set $K'_i$. Since the boundaries of the sets $K_i$ and $K'_i$ are different, this shows that $h$ is in fact Lipschitz on $S^1\times
[1/2,1]$.

We show now that $h$ is Lipschitz on $S^1 \times [X_2,1]$. Note that it is slightly incorrect to say that $h$ is Lipschitz, since $h$ is defined only almost everywhere. Nevertheless, if we prove that $|h({\textbf{x}})-h({\textbf{y}})|{\leqslant}Cd({\textbf{x}},{\textbf{y}})$ for almost all ${\textbf{x}}$ and ${\textbf{y}}$, then there will exist a unique version of $h$ which is really Lipschitz. Thus, all the equalities we will write until the end of this proof will be true only almost everywhere.

Let $A_{s,n}^+ = \left[\frac{s}{4^{q+n}},
\frac{s+1}{4^{q+n}}\right]\times J_n^+$: $T^n$ is a diffeomorphism between $A_{s,n}^+$ and $K_i^+ =
\left[\frac{i}{4^q},\frac{i+1}{4^q}\right]\times [X_2,1]$. We fix some $K^+=K_i^+=I\times [X_2,1]$, and we show that $h$ is Lipschitz on $K^+$. Let $U_1,U_2,\ldots$ be the inverse branches of $T^{n_1},T^{n_2},\ldots$ whose images all coincide with $K^+$. Let $T_Y$ be the map induced by $T$ on $Y=S^1\times [1/2,1]$. Then $h\operatorname{dLeb}_{|Y}$ is invariant under $T_Y$, which means that, for each ${\textbf{x}}\in I\times [1/2,1]$, $$h({\textbf{x}})=\sum JU_j({\textbf{x}}) h(U_j {\textbf{x}})$$ where $JU_j$ is the jacobian of $U_j$.

Let $Z=S^1\times [X_2,1]$, and $T_Z$ be the map induced by $T$ on $Z$. Since $h\operatorname{dLeb}_{|Z}$ is also invariant under $T_Z$, we have the same kind of equation as above. For ${\textbf{x}}\in I\times
[X_2,1/2]$, all its preimages under $T_Z$ are in $S^1 \times
[1/2,1]$, and the invariance gives that $$h({\textbf{x}})=\sum JU_j({\textbf{x}}) h(U_j {\textbf{x}}).$$

We have shown that, for every ${\textbf{x}}\in S^1 \times [X_2,1]$, $$h({\textbf{x}})=\sum JU_j({\textbf{x}}) h(U_j {\textbf{x}}).$$ This means that $h$ is invariant under some kind of transfer operator, even though it is not a real transfer operator since the images of the maps $U_j$ are not disjoint, and since they do not cover the space. In particular, the images of the $U_j$ are included in $S^1 \times [1/2,1]$, and we already know that $h$ is Lipschitz on this set.

The bounds of the previous paragraphs still apply to the distortion of the $U_j$, and their expansion. In particular, $\left|1-\frac{JU_j({\textbf{y}})}{JU_j({\textbf{x}})}\right|{\leqslant}Cd({\textbf{x}},{\textbf{y}})$ for a constant $C$ independent of $j$, and $|h(U_j {\textbf{x}})-h(U_j {\textbf{y}})| {\leqslant}C d(U_j {\textbf{x}},U_j {\textbf{y}}){\leqslant}D d({\textbf{x}},{\textbf{y}})$ (since $h$ is Lipschitz on the image of $U_j$). Thus, $$\begin{aligned}
  |h({\textbf{x}})-h({\textbf{y}})|&
  {\leqslant}\sum |JU_j({\textbf{x}})h(U_j {\textbf{x}})-JU_j({\textbf{y}})h(U_j {\textbf{y}})|
  \\&
  {\leqslant}\sum |JU_j({\textbf{x}})| \left|1-\frac{JU_j({\textbf{y}})}{JU_j({\textbf{x}})}\right|
  |h(U_j
  {\textbf{x}})| +\sum |JU_j({\textbf{y}})| |h(U_j {\textbf{x}})-h(U_j {\textbf{y}})|
  \\&
  {\leqslant}Cd({\textbf{x}},{\textbf{y}})\sum |JU_j({\textbf{x}})| + Dd({\textbf{x}},{\textbf{y}}) \sum |JU_j({\textbf{y}})|.
  \end{aligned}$$ It remains to prove that $\sum |JU_j({\textbf{x}})|$ is bounded. The bound on distortion gives $JU_j({\textbf{x}}) \asymp \operatorname{Leb}(\operatorname{Im}U_j)$, whence $\sum JU_j({\textbf{x}}){\leqslant}C\sum \operatorname{Leb}(\operatorname{Im}U_j)$, which is finite since every point of $I\times[1/2,1]$ is in the image of at most two maps $U_j$.

We have proved that $h$ is Lipschitz on $S^1\times [X_2,1]$, except maybe on $\{\frac{s}{4^q}\}\times [X_2,1]$. As above, using another Markov partition, we exclude the possibility of discontinuities there. Thus, $h$ is Lipschitz on $S^1 \times
[X_2,1]$.

To prove that $h$ is Lipschitz on $S^1\times [X_k,1]$, we do exactly the same thing, except that we consider $[Y_{n+k},Y_n]$ instead of $J_n^+=[Y_{n+2},Y_n]$. As above, writing $U_1,U_2,\ldots$ for the inverse branches of $T^n$ defined on a set $[\frac{s}{4^{n+q}},\frac{s+1}{4^{n+q}}]\times [Y_{n+k},Y_n]$ and whose image is $K'=[\frac{i}{4^q},\frac{i+1}{4^q}] \times
[X_k,1]=I\times [X_k,1]$, we show that $h({\textbf{x}})=\sum JU_j({\textbf{x}})
h(U_j{\textbf{x}})$ for ${\textbf{x}}\in K'$. In fact, for ${\textbf{x}}\in I\times
[X_{l},X_{l-1}]$, we use the invariance of $h\operatorname{dLeb}$ under the map induced by $T$ on $S^1 \times [X_l,1]$. We conclude finally as above, using the fact that $h$ is Lipschitz on $S^1\times
[1/2,1]$, which contains the images of the $U_j$.

This concludes the proof, since every compact subset of $S^1\times(0,1]$ is contained in $S^1 \times [X_k,1]$ for large enough $k$.

Limit theorems for Markov maps {#limit_Markov}
==============================

We want to establish limit theorems for Birkhoff sums, of the form $\sum_{k=0}^{n-1} f(T^k x)$. We give in this section an abstract result, valid for a map that induces a Gibbs-Markov map on a subset of the space (which is the case of our skew product). Related limit theorems have been proved in [@gouezel:stable], but we will show here a slightly different result, which requires more control on the return time ${\varphi}$ but is more elementary, using Theorem \[thm\_probabiliste\_general\] proved in Appendix \[appendice:loi\_stable\] and inspired by results of Melbourne and Török ([@melbourne_torok]) for flows. An advantage of this new method is that, contrary to [@gouezel:stable], it can easily be extended to stable laws of index $1$.

If $Z_0,\ldots,Z_{n-1},\ldots$ are independent identically distributed random variables with zero mean, the sums $\frac{1}{B_n} \sum_{k=0}^{n-1} Z_k$ (where $B_n$ is a real sequence) converge to a nontrivial limit in essentially three cases: if $Z_k \in L^2$, there is convergence to a normal law for $B_n=\sqrt{n}$. There is also convergence to a normal law, but with a different normalization, if $P(|Z_k|>x)=x^{-2}l(x)$ with $L(x):=2\int_1^x \frac{l(u)}{u}{\, {\rm d}}u$ unbounded and slowly varying (i.e. $L:(0,\infty) \to (0,\infty)$ satisfies $\forall a
>0, \lim_{x\to \infty} L(ax)/L(x)=1$) – this is in particular true when $l$ itself is slowly varying. Finally, if $P(Z_k>x)=(c_1+o(1))x^{-p}L(x)$ and $P(Z_k<-x)=(c_2+o(1))x^{-p}L(x)$, where $L$ is slowly varying and $p\in (0,2)$, we have convergence (for a good choice of $B_n$) to a limit law called stable law. Moreover, these are the only cases where there is a convergence ([@feller:2]).

In the dynamical setting, we will prove the same kind of limit theorems, still with three possible cases: $L^2$, normal nonstandard, and stable. The normalizations will moreover be the same as in the probabilistic setting.

\[thm\_abstrait\_markov\] Let $T:X\to X$ be an ergodic transformation preserving a probability measure $m$. Assume that there exists a subset $Y$ of $X$ with $m(Y)>0$ such that the first return map $T_Y(x)=T^{{\varphi}(x)}(x)$ (where ${\varphi}(x)= \inf\{ n>0 {\ |\ }T^n(x) \in
Y\}$) is Gibbs-Markov for $m_{|Y}$, a partition $\alpha$ of $Y$ such that ${\varphi}$ is constant on each element of $\alpha$, and a distance $d$ on $Y$.

Let $f:X \to {\mathbb{R}}$ be an integrable map with $\int f=0$, such that $f_Y(y):=\sum_{n=0}^{{\varphi}(y)-1} f(T^n y)$ satisfies $$\label{condition_markov_abstrait}
  \sum_{a\in \alpha} m(a) D f_Y(a) <\infty$$ where $$D f_Y(a)=\inf\{ C >0 {\ |\ }\forall x,y\in a, |f_Y(x)-f_Y(y)| {\leqslant}C
d(x,y)\}.$$

Set $M(y)=\max_{1{\leqslant}k {\leqslant}{\varphi}(y)} \left| \sum_{j=0}^{k-1}
f\circ T^j(y) \right|$.

Then:

-   Assume that $f_Y \in L^2$ and $M \in L^2$. Assume moreover that ${\varphi}$ satisfies one of the following hypotheses:

    -   ${\varphi}\in L^2$.

    -   $m({\varphi}> x)=x^{-p} L(x)$ where $L$ is slowly varying and $p \in (1,2]$.

    Then there exists $\sigma^2 {\geqslant}0$ such that $\frac{1}{\sqrt{n}}
    S_n f \to {\mathcal{N}}(0,\sigma^2)$.

-   Assume that $m(|f_Y| > x)=x^{-2} l(x)$, with $L(x):=2\int_1^x\frac{l(u)}{u}{\, {\rm d}}u$ unbounded and slowly varying. Assume moreover that $m(M
    > x) {\leqslant}C x^{-2} l(x)$, and $m({\varphi}>x)=(c+o(1)) x^{-2}l(x)$. Let $B_n\to \infty$ satisfy $n L(B_n)=B_n^2$. Then $\frac{1}{B_n} S_n f \to {\mathcal{N}}(0,1)$.

-   Assume that $m(f_Y>x)=(c_1+o(1))x^{-p}L(x)$ and $m(f_Y<-x)=(c_2+o(1))x^{-p} L(x)$ where $L$ is a slowly varying function, $p\in (1,2)$, and $c_1,c_2{\geqslant}0$ with $c_1+c_2>0$. Assume also that $m(M>x) {\leqslant}C x^{-p} L(x)$, and $m({\varphi}>x)=(c_3+o(1)) x^{-p}L(x)$. Let $B_n\to \infty$ satisfy $n
    L(B_n)=B_n^p$. Then $\frac{1}{B_n} S_n f \to Z$ where the random variable $Z$ has a characteristic function given by $$E(e^{itZ})=e^{-c|t|^p \left( 1-i\beta \operatorname{sgn}(t) \tan
      \left(\frac{p\pi}{2} \right) \right)}$$ with $c=(c_1+c_2) \Gamma(1-p) \cos \left(\frac{p\pi}{2} \right)$ and $\beta=\frac{c_1-c_2}{c_1+c_2}$.

Note that $M(y){\leqslant}\sum_{j=0}^{{\varphi}(y)-1} |f(T^j y)|=|f|_Y(y)$. Thus, if the integrability hypotheses of the theorem are satisfied by $|f|_Y$ (which will often be the case), they are automatically satisfied by $M$.

In the second case of the theorem, when $l$ itself is slowly varying, then $L$ is automatically slowly varying.

The second case of the theorem is not the most general possible result, since one may have convergence to a normal law even when the function $l$ is not slowly varying (what really matters is that $L$ is slowly varying). The theorem can be extended without problem to this more general setting, but the result becomes more complicated to state. In the applications, the statement given in Theorem \[thm\_abstrait\_markov\] will be sufficient.

The idea is to use Theorem \[thm\_probabiliste\_general\]: we have to check all its hypotheses. We will use the notations of this theorem, and in particular write $E_Y(u)=\frac{\int_Y u {\, {\rm d}}m}{m(Y)}$.

We first treat the third case (stable law), using the results of [@aaronson_denker] (and the generalizations of [@gouezel:stable]). Let $s(x,y)$ be the separation time of $x$ and $y$ defined in , $\tau=1/\lambda$ and $d_\tau=\tau^s$ the corresponding metric. Since every iteration of $T_Y$ expands by at least $\lambda$, we get $d(x,y){\leqslant}C d_\tau(x,y)$. In particular, we can assume without loss of generality that $d=d_\tau$, which is the setting of [@aaronson_denker] and [@gouezel:stable].

Let $P$ be the transfer operator associated to $T_Y$ (i.e.defined by $\int u\cdot v\circ T_Y = \int P(u)\cdot v$), and $P_t(u)=P(e^{itf_Y} u)$. Let ${\mathcal{L}}$ be the space of bounded Lipschitz functions (i.e. such that there exists $C$ such that, $\forall a\in \alpha, \forall x,y\in a, |g(x)-g(y)|{\leqslant}C
d(x,y)$). Theorem 5.1 of [@aaronson_denker] ensures that, for small enough $t$, $P_t$ acting on ${\mathcal{L}}$ has an eigenvalue $\lambda(t)=e^{-\frac{c}{m(Y)}|t|^p \left( 1-i\beta \operatorname{sgn}(t) \tan
\left(\frac{p\pi}{2} \right) \right)L(|t|^{-1})(1+o(1))}$, the remaining part of its spectrum being contained in a disk of radius ${\leqslant}1-\delta<1$. In fact, this theorem requires that $Df_Y(a)$ is bounded, but [@gouezel:stable Theorem 3.8] shows that it remains true under the weaker assumption $\sum m(a)D
f_Y(a)<\infty$.

The slow variation of $L$ easily implies that $\lambda\left(\frac{t}{B_n} \right)^{\lfloor n m(Y) \rfloor} \to
e^{-c|t|^p \left( 1-i\beta \operatorname{sgn}(t) \tan \left(\frac{p\pi}{2}
\right) \right)}$, whence, for $g\in {\mathcal{L}}$, $$\label{converge_dans_L}
  E_Y\left(g e^{i\frac{t}{B_n}
  S^Y_{\lfloor n m(Y) \rfloor} f_Y}\right) \to E_Y(g)E(e^{itZ})$$ where the random variable $Z$ is as in the statement of the theorem (see [@aaronson_denker] or [@gouezel:stable] for more details). We can not apply this result to $g={\varphi}$, since ${\varphi}$ is not bounded. However, ${\varphi}$ is Lipschitz and integrable, whence $P{\varphi}\in {\mathcal{L}}$ ([@aaronson_denker Proposition 1.4]). Equation applied to $P{\varphi}$ gives $E_Y\left({\varphi}e^{i\frac{t}{B_n} S^Y_{\lfloor n
m(Y) \rfloor} f_Y\circ T_Y }\right) \to E(e^{itZ})$, since $E_Y(P{\varphi})=E_Y({\varphi})=1$ by Kac’s Formula. Let $k(n)$ be a sequence such that $\lfloor k(n)m(Y) \rfloor = \lfloor n m(Y) \rfloor
-1$. Since $k(n)\sim n$, the same arguments give in fact that $E_Y\left({\varphi}e^{i\frac{t}{B_n} S^Y_{\lfloor k(n)
m(Y) \rfloor} f_Y\circ T_Y }\right) \to E(e^{itZ})$, i.e.  $E_Y\left({\varphi}e^{i\frac{t}{B_n} (S^Y_{\lfloor n
m(Y) \rfloor} f_Y  - f_Y)}\right) \to E(e^{itZ})$. The difference between this term and $E_Y\left({\varphi}e^{i\frac{t}{B_n} (S^Y_{\lfloor n
m(Y) \rfloor} f_Y }\right)$ is bounded by $E_Y\left(
{\varphi}\left|e^{-i\frac{t}{B_n}
f_Y}-1\right| \right)$, which tends to $0$ by dominated convergence. Thus, $$\label{eq_presque_bonne}
  E_Y\left({\varphi}e^{i\frac{t}{B_n}
  S^Y_{\lfloor n m(Y) \rfloor} f_Y} \right)
  \to E(e^{itZ}).$$ This is . Finally, since $L$ is slowly varying, the equation $n L(B_n)=B_n^p$ implies that $\sup_{r{\leqslant}2n} \frac{B_r}{B_n}<\infty$, $\inf_{r{\geqslant}n} \frac{B_r}{B_n}>0$ (using for example [@feller:2 Corollary page 274]).

Let ${\varepsilon}>0$, we bound $m(M {\geqslant}{\varepsilon}B_n)$. $$m(M {\geqslant}{\varepsilon}B_n)
  {\leqslant}C ({\varepsilon}B_n)^{-p} L( {\varepsilon}B_n)
  =C {\varepsilon}^{-p} B_n^{-p} L(B_n)
  \frac{L({\varepsilon}B_n)}{L(B_n)}.$$ But $B_n^{-p}L(B_n)=\frac{1}{n}$ by definition of $B_n$, and $\frac{L({\varepsilon}B_n)}{L(B_n)}$ tends to $1$ since $L$ is slowly varying. Thus, $m(M {\geqslant}{\varepsilon}B_n) {\leqslant}\frac{D}{n}$, which proves .

Hypothesis \[hypothese\_3\] of Theorem \[thm\_probabiliste\_general\] is satisfied for $b=1$, according to the Birkhoff Theorem applied to ${\varphi}-E_Y({\varphi})$ (and because $T_Y$ is ergodic, which is a consequence of the ergodicity of $T$). Finally, the hypothesis on the distribution of ${\varphi}$ ensures, once again by [@aaronson_denker], that $\frac{S_{\lfloor
nm(Y)\rfloor}^Y
{\varphi}-n m(Y)E_Y({\varphi})}{B_n}$ converges in distribution. Thus, is satisfied. We can use Theorem \[thm\_probabiliste\_general\], and get that $\frac{S_n f}{B_n}
\to Z$.

The proof of the second case of Theorem \[thm\_abstrait\_markov\] is exactly the same, using [@aaronson_denker:central] instead of [@aaronson_denker] to show the convergence in distribution of $\frac{S_{\lfloor nm(Y)\rfloor} ^Y f_Y}{B_n}$ and $\frac{S_{\lfloor 
nm(Y) \rfloor} ^Y
{\varphi}-n m(Y) E_Y({\varphi})}{B_n}$.

In the first case ($f_Y\in L^2$), the proof is again identical when ${\varphi}\in L^2$, with $B_n=\sqrt{n}$, using [@guivarch-hardy] (or the remarks of [@aaronson_denker:central]). However, when $m({\varphi}>x)=x^{-p}
L(x)$, we check in a different way the hypotheses \[hypothese\_3\] and \[hypothese\_4\] of Theorem \[thm\_probabiliste\_general\]. [@aaronson_denker] ensures that, if $B'_n$ is given by $$\label{definit_bn}
  nL(B'_n)=(B'_n)^p,$$ then $\frac{S_n^Y {\varphi}-n E_Y({\varphi})}{B'_n}$ converges in distribution. Moreover, [@gouezel:stable Lemma 3.4] proves that $Pf_Y \in {\mathcal{L}}$, and has a vanishing integral. As $P$ has a spectral gap on ${\mathcal{L}}$, $P^n f_Y \to 0$ exponentially fast. In particular, $\int f_Y\circ T_Y^n \cdot f_Y=\int (P^n f)\cdot
f=O((1-\delta)^n)$ for some $0<\delta<1$. Thus, as $f_Y\in L^2$, [@vitesse_birkhoff Theorem 16] gives that, for every $b>1/2$, $\frac{1}{N^b}\sum_{k=0}^{N-1}f_Y(T_Y^k) \to 0$ almost everywhere when $N\to \infty$. In the natural extension, $\int f_Y\circ
T_Y^{-n} \cdot f_Y=\int f_Y \cdot f_Y\circ T_Y^n$ decays also exponentially fast, whence the same argument gives that $\frac{1}{|N|^b}\sum_{k=0}^{N-1}f_Y(T_Y^k) \to 0$ when $N\to
-\infty$. Thus, Hypothesis \[hypothese\_3\] of Theorem \[thm\_probabiliste\_general\] is satisfied for any $b>1/2$. Let $\kappa>0$ be very small. As $L$ is slowly varying, $L(B'_n)=O((B'_n)^\kappa)$, whence Equation gives $B'_n=O(n^{1/(p-\kappa)})$. Thus, if $b<\frac{p}{2}$, we have $B'_n=O(B_n^{1/b})$, which implies .

Asymptotic behavior of $X_n$ {#section_estimee_Xn}
============================

We return to the study of the skew product . To prove limit theorems using Theorem \[thm\_abstrait\_markov\], we will need to estimate $m({\varphi}_Y>n)$, which is directly related to the speed of convergence of $X_n$ to $0$. This section will be devoted to the proof of the following theorem:

\[estimee\_Xn\_L1\] We have $$\left(\frac{n}{\sqrt{\ln n}}\right)^{1/\operatorname{{\alpha_{\text{min}}}}} X_n
  \to \frac{1}{\left(2^{\operatorname{{\alpha_{\text{min}}}}} \operatorname{{\alpha_{\text{min}}}}^{3/2}
  \sqrt{\frac{\pi}{2\alpha''(x_0)}}\right)^{1/\operatorname{{\alpha_{\text{min}}}}}}$$ almost everywhere and in $L^1$.

\[lemme\_asympt\_Xn\] We have $$E(e^{-(\alpha-\operatorname{{\alpha_{\text{min}}}})w})\sim \sqrt{\frac{\pi}{2\alpha''(x_0)}}
  \frac{1}{\sqrt{w}} \text{
  when }w\to\infty.$$

Write $\beta=\alpha-\operatorname{{\alpha_{\text{min}}}}$, and $f(b)=\operatorname{Leb}\{ \omega {\ |\ }\beta(\omega) \in [0,b)\}$. In a neighborhood of $x_0$ (the unique point where $\alpha$ takes its minimal value $\operatorname{{\alpha_{\text{min}}}}$), $\alpha$ behaves like the parabola $\operatorname{{\alpha_{\text{min}}}}+\frac{\alpha''(x_0)}{2}(x-x_0)^2$, whence $f(b) \sim \sqrt{\frac{2}{\alpha''(x_0)}} \sqrt{b}$ when $b
\to 0$.

Writing $P_\beta$ for the distribution of $\beta$, an integration by parts gives $$\begin{aligned}
  E\left(e^{-(\alpha-\operatorname{{\alpha_{\text{min}}}})w}\right)&=\int_0^\infty e^{-b w} {\, {\rm d}}P_\beta(b)
  =w \int_0^\infty e^{-b w} f(b) {\, {\rm d}}b
  =\int_0^\infty e^{-u} f(u/w) {\, {\rm d}}u
  \\ &
  =\frac{1}{\sqrt{w}} \int_0^\infty e^{-u}
  \left(\sqrt{w}f(u/w)\right) {\, {\rm d}}u.
  \end{aligned}$$ But $e^{-u} \left(\sqrt{w}f(u/w)\right) \to e^{-u}
\sqrt{\frac{2}{\alpha''(x_0)}}\sqrt{u}$ when $w\to \infty$. There exists a constant $E$ such that $f(u){\leqslant}E \sqrt{u}$ (this is clear in a neighborhood of $0$, and elsewhere since $f$ is bounded), whence $e^{-u} \left(\sqrt{w}f(u/w)\right) {\leqslant}E
e^{-u}\sqrt{u}$ integrable. By dominated convergence, $$\int_0^\infty e^{-u}
  \left(\sqrt{w}f(u/w)\right) {\, {\rm d}}u
  \to \sqrt{\frac{2}{\alpha''(x_0)}} \int_0^\infty e^{-u} \sqrt{u}{\, {\rm d}}u
  =\sqrt{\frac{2}{\alpha''(x_0)}} \frac{\sqrt{\pi}}{2}.$$

As in Proposition \[estime\_croissance\_grossiere\_Xn\], we write $$\frac{1}{X_n(F\omega)^{\operatorname{{\alpha_{\text{min}}}}}}=\frac{1}{X_{n+1}(\omega)^{\operatorname{{\alpha_{\text{min}}}}}}-\operatorname{{\alpha_{\text{min}}}}2^{\operatorname{{\alpha_{\text{min}}}}} (2X_{n+1}(\omega))^{\alpha(\omega) -\operatorname{{\alpha_{\text{min}}}}}
  +O(X_{n+1}(\omega)^{2\alpha(\omega) -\operatorname{{\alpha_{\text{min}}}}}).$$ Proposition \[estime\_croissance\_grossiere\_Xn\] gives $$X_{n+1}(\omega)^{2\alpha(\omega)-\operatorname{{\alpha_{\text{min}}}}} {\leqslant}X_{n+1}(\omega)^{\operatorname{{\alpha_{\text{min}}}}}
 {\leqslant}\frac{C}{(n+1)^{\operatorname{{\alpha_{\text{min}}}}/\operatorname{{\alpha_{\text{max}}}}}}
  {\leqslant}\frac{C}{\sqrt{n+1}}$$ as $\operatorname{{\alpha_{\text{min}}}}/\operatorname{{\alpha_{\text{max}}}}{\geqslant}1/2$ by hypothesis. Thus, $$\frac{1}{X_{n+1}(\omega)^{\operatorname{{\alpha_{\text{min}}}}}}-\frac{1}{X_n(F\omega)^{\operatorname{{\alpha_{\text{min}}}}}}=2^{\operatorname{{\alpha_{\text{min}}}}} \operatorname{{\alpha_{\text{min}}}}(2X_{n+1}(\omega))^{\alpha-\operatorname{{\alpha_{\text{min}}}}} +O(1/\sqrt{n}).$$ Summing from $1$ to $n$, we get a constant $P$ (independent of $\omega$) such that $$\begin{aligned}
  \label{minore_1/X_n}
  \frac{1}{X_n(\omega)^{\operatorname{{\alpha_{\text{min}}}}}}{\geqslant}2^{\operatorname{{\alpha_{\text{min}}}}} \operatorname{{\alpha_{\text{min}}}}\left[\sum_{k=1}^{n} (2X_k(F^{n-k} \omega))^{\alpha(F^{n-k}\omega)-\operatorname{{\alpha_{\text{min}}}}}
  -P\sqrt{n} \right]
  \\
  \label{majore_1/X_n}
  \frac{1}{X_n(\omega)^{\operatorname{{\alpha_{\text{min}}}}}}{\leqslant}2^{\operatorname{{\alpha_{\text{min}}}}} \operatorname{{\alpha_{\text{min}}}}\left[\sum_{k=1}^n (2X_k(F^{n-k}\omega))^{\alpha(F^{n-k}\omega)-\operatorname{{\alpha_{\text{min}}}}}
  +P\sqrt{n} \right]
  \end{aligned}$$

Equation and Proposition \[estime\_croissance\_grossiere\_Xn\] imply that $$\label{definit_An}
  \frac{\sqrt{\ln n}}{n} \frac{1}{2^{\operatorname{{\alpha_{\text{min}}}}} \operatorname{{\alpha_{\text{min}}}}X_n(\omega)^{\operatorname{{\alpha_{\text{min}}}}}}
  {\geqslant}\frac{\sqrt{\ln n}}{n}\sum_{k=1}^n \left(\frac{2C^{-1}}{k^{1/\operatorname{{\alpha_{\text{min}}}}}}
  \right)^{\alpha(F^{n-k} \omega)-\operatorname{{\alpha_{\text{min}}}}}
  -P\sqrt{\frac{\ln n}{n}}=: A_n(\omega).$$

We first study the convergence of $A_n$. The functions $\alpha$ and $\alpha \circ F^{n-k}$ have the same distribution since $F$ preserve Lebesgue measure. Thus, by Lemma \[lemme\_asympt\_Xn\], $$E\left(\left(\frac{2C^{-1}}{k^{1/\operatorname{{\alpha_{\text{min}}}}}}\right)^{\alpha\circ F^{n-k}-\operatorname{{\alpha_{\text{min}}}}}
  \right)
  \sim \sqrt{\frac{\pi}{2\alpha''(x_0)}}
  \frac{1}{\sqrt{\ln (k^{1/\operatorname{{\alpha_{\text{min}}}}})-\ln(2C^{-1})}}
  \sim \sqrt{\frac{\pi \operatorname{{\alpha_{\text{min}}}}}{2\alpha''(x_0)}}\frac{1}{\sqrt{\ln k}}.$$ Summing, we get that $$E(A_n)\to C_1:=\sqrt{\frac{\pi \operatorname{{\alpha_{\text{min}}}}}{2\alpha''(x_0)}},$$ since $\sum_{k=2}^n \frac{1}{\sqrt{\ln k}} \sim \frac{n}{\sqrt{\ln
n}}$.

We will need $L^p$ estimates, for $p{\geqslant}1$. To get them, we use a result of Françoise Pène, recalled in Appendix \[appendice:pene\]. Let us denote by ${\left\| g \right\|}$ the Lipschitz norm of a function $g:S^1 \to {\mathbb{R}}$.

We define $f_k(\omega)=\left(\frac{2C^{-1}}{k^{1/\operatorname{{\alpha_{\text{min}}}}}}
\right)^{\alpha(\omega)-\operatorname{{\alpha_{\text{min}}}}}$, and $g_k=f_k-E(f_k)$. Thus, $A_n=\frac{\sqrt{\ln n}}{n}\sum_{k=1}^n f_k \circ
F^{n-k}-P\sqrt{\frac{\ln n}{n}}$. As $g'_k=\ln
\left(\frac{2C^{-1}}{k^{1/\operatorname{{\alpha_{\text{min}}}}}} \right)\alpha' f_k$, there exists a constant $L$ such that, for $k{\leqslant}n$, ${\left\| g_k \right\|} {\leqslant}L \ln
n$. As a consequence, Theorem \[thm\_borne\_Lp\_pene\] applied to $g_k/(L \ln n)$ gives $${\left\| A_n-E(A_n) \right\|}_p
  = \frac{\sqrt{\ln n}}{n} L\ln n
  {\left\| \sum_{k=1}^{n} g_k \circ F^{n-k} / (L \ln n) \right\|}_p
  {\leqslant}\frac{\sqrt{\ln n}}{n} L\ln n K_p \sqrt{n},$$ i.e. $${\left\| A_n-E(A_n) \right\|}_p {\leqslant}L_p \sqrt{\frac{\ln^3 n}{n}}.$$

This implies in particular that $A_n$ converges almost everywhere to $C_1$. Namely, if $\delta>0$, $$\operatorname{Leb}\{|A_n-E(A_n)|>\delta\}
  {\leqslant}\int \frac{|A_n-E(A_n)|^4}{\delta^4}
  {\leqslant}\frac{L_4}{\delta^4} \left(\frac{\ln^3 n}{n}\right)^{4/2}$$ which is summable, and $E(A_n)\to C_1$.

We have $$\begin{aligned}
  A_n(\omega)&
  {\geqslant}\frac{\sqrt{\ln n}}{n} \left[\sum_{k=1}^n \left( \frac{2C^{-1}}
  {k^{1/\operatorname{{\alpha_{\text{min}}}}}}\right)^{\operatorname{{\alpha_{\text{max}}}}-\operatorname{{\alpha_{\text{min}}}}} - P\sqrt{n} \right]
  {\geqslant}\frac{\sqrt{\ln n}}{n} \bigl[ K n^{2-\operatorname{{\alpha_{\text{max}}}}/\operatorname{{\alpha_{\text{min}}}}} - P
  \sqrt{n}\bigr]
  \\&
  {\geqslant}K' \frac{\sqrt{\ln n}}{n} n^{2-\operatorname{{\alpha_{\text{max}}}}/\operatorname{{\alpha_{\text{min}}}}}
  \end{aligned}$$ since $\operatorname{{\alpha_{\text{max}}}}/\operatorname{{\alpha_{\text{min}}}}<3/2$. Thus, $${\left\| \frac{1}{A_n} \right\|}_\infty {\leqslant}K'' \frac{n^{\operatorname{{\alpha_{\text{max}}}}/\operatorname{{\alpha_{\text{min}}}}-1}}{\sqrt{\ln
  n}}.$$ Note that $E(A_n)$ tends to $C_1 \not=0$, whence $\frac{1}{E(A_n)}$ is bounded. Thus, $$\begin{aligned}
  {\left\|  \frac{1}{A_n} - \frac{1}{E(A_n)} \right\|}_p &
  {\leqslant}{\left\| \frac{1}{A_n} \right\|}_\infty \frac{1}{E(A_n)}
  {\left\| A_n-E(A_n) \right\|}_p
  {\leqslant}K''' \frac{n^{\operatorname{{\alpha_{\text{max}}}}/\operatorname{{\alpha_{\text{min}}}}-1}}{\sqrt{\ln
  n}} L_p \sqrt{\frac{\ln^3 n}{n}}
  \\&
  = M_p \frac{\ln n}{n^\kappa}
  \end{aligned}$$ where $\kappa=\frac{3}{2}-\frac{\operatorname{{\alpha_{\text{max}}}}}{\operatorname{{\alpha_{\text{min}}}}}>0$. In particular, $\frac{1}{A_n}$ tends to $\frac{1}{C_1}$ in every $L^p$. Equation shows that $$\label{un_petit_label}
  \left(\frac{n}{\sqrt{\ln n}}\right)^{1/\operatorname{{\alpha_{\text{min}}}}} X_n {\leqslant}\frac{1}
  {(2^{\operatorname{{\alpha_{\text{min}}}}}\operatorname{{\alpha_{\text{min}}}}A_n )^{1/\operatorname{{\alpha_{\text{min}}}}}}.$$ The right hand side tends to $$C_2:=\frac{1}{\left(2^{\operatorname{{\alpha_{\text{min}}}}}\operatorname{{\alpha_{\text{min}}}}^{3/2}
  \sqrt{\frac{\pi}{2\alpha''(x_0)}}\right)^{1/\operatorname{{\alpha_{\text{min}}}}}}$$ in every $L^p$, and in particular in $L^1$. Thus, $$\label{eq_esperance}
  \varlimsup E\left(\left( \frac{n}{\sqrt{\ln n}}\right)^{1/\operatorname{{\alpha_{\text{min}}}}} X_n
  \right) {\leqslant}C_2.$$ Moreover, $A_n$ converges almost everywhere to $C_1$, whence yields that, almost everywhere, $$\label{eq_limsup}
  \varlimsup \left( \frac{n}{\sqrt{\ln n}}\right)^{1/\operatorname{{\alpha_{\text{min}}}}}
  X_n(\omega)
  {\leqslant}C_2.$$

Set $Q=\sup_n \left(\frac{1}{E(A_n)}\right)+1$, we estimate $\operatorname{Leb}\left\{ \frac{1}{A_n} {\geqslant}Q\right\}$. If $p{\geqslant}1$, $$\operatorname{Leb}\left\{ \frac{1}{A_n} {\geqslant}Q\right\}
  {\leqslant}\operatorname{Leb}\left\{ \left|\frac{1}{A_n} -\frac{1}{E(A_n)} \right|
  {\geqslant}1 \right\}
  {\leqslant}E\left( \left|\frac{1}{A_n} -\frac{1}{E(A_n)} \right| ^p
  \right)
  {\leqslant}\left( M_p \frac{\ln n}{n^\kappa} \right)^p.$$ In particular, choosing $p$ large enough gives $$\operatorname{Leb}\left\{ \frac{1}{A_n} {\geqslant}Q\right\}{\leqslant}\frac{M}{n^5}.$$ Setting $Q'=\frac{Q}{2^{\operatorname{{\alpha_{\text{min}}}}} \operatorname{{\alpha_{\text{min}}}}}$, thus yields that $$\operatorname{Leb}\left\{ X_n {\geqslant}\left(\frac{Q'\sqrt{\ln n}}{n}\right)^{1/\operatorname{{\alpha_{\text{min}}}}}
  \right\} {\leqslant}\frac{M}{n^5}.$$ Consequently, $U_n:=\left\{ \omega {\ |\ }\exists \sqrt{n} {\leqslant}k{\leqslant}n \text{ with } X_k(F^{n-k} \omega) {\geqslant}\left(\frac{Q'\sqrt{\ln
k}}{k}\right)^{1/\operatorname{{\alpha_{\text{min}}}}} \right\}$ has a measure at most $\sum_{\sqrt{n}}^n \frac{M}{k^5} {\leqslant}\frac{M'}{n^2}$ (since $\operatorname{Leb}$ is invariant under $F^{n-k}$). Finally, Borel-Cantelli ensures that there is a full measure subset of $S^1$ on which $\omega \not \in U_n$ for large enough $n$.

Set $$A'_n(\omega)=\frac{\sqrt{\ln n}}{n} \left[\sum_{k=1}^n
  \left(\frac{2(Q'\sqrt{\ln
  k})^{1/\operatorname{{\alpha_{\text{min}}}}}}{k^{1/\operatorname{{\alpha_{\text{min}}}}}}\right)^{\alpha(F^{n-k} \omega) -\operatorname{{\alpha_{\text{min}}}}}
  + (P+1)\sqrt{n}\right].$$ As for $A_n$, we show that $A'_n \to C_1$ in every $L^p$ and almost everywhere.

Let $\omega$ be such that $\omega \not \in U_n$ for large enough $n$, and $A'_n(\omega) \to C_1$ (these properties are true almost everywhere). Then, for large enough $n$, Equation and the fact that $X_k(F^{n-k} \omega) {\leqslant}\left(\frac{Q'\sqrt{\ln k}}{k}\right)^{1/\operatorname{{\alpha_{\text{min}}}}}$ for $\sqrt{n}{\leqslant}k{\leqslant}n$, yield that $$\begin{aligned}
  \frac{1}{2^{\operatorname{{\alpha_{\text{min}}}}} \operatorname{{\alpha_{\text{min}}}}X_n(\omega)^{\operatorname{{\alpha_{\text{min}}}}}}&
  {\leqslant}\left[\sum_{k=1}^{\sqrt{n}} 1+
  \sum_{k=\sqrt{n}}^n \left(\frac{2(Q'\sqrt{\ln
  k})^{1/\operatorname{{\alpha_{\text{min}}}}}}{k^{1/\operatorname{{\alpha_{\text{min}}}}}}\right)^{\alpha(F^{n-k}\omega) -\operatorname{{\alpha_{\text{min}}}}}
  +P \sqrt{n} \right]
  \\&
  {\leqslant}\frac{n}{\sqrt{\ln n}} A'_n(\omega)\sim \frac{n}{\sqrt{\ln n}} C_1.
  \end{aligned}$$ Thus, $$\label{eq_liminf}
  \varliminf \left( \frac{n}{\sqrt{\ln n}}\right)^{1/\operatorname{{\alpha_{\text{min}}}}}
  X_n(\omega)
  {\geqslant}C_2.$$

Equations and prove that $\left( \frac{n}{\sqrt{\ln n}}\right)^{1/\operatorname{{\alpha_{\text{min}}}}} X_n$ tends almost everywhere to $C_2$. We get the convergence in $L^1$ from the inequality and the following elementary lemma.

Let $f_n$ be nonnegative functions on a probability space, with $f_n \to f$ almost everywhere, and $\varlimsup E(f_n) {\leqslant}E(f)<\infty$. Then $f_n \to f$ in $L^1$.

Limit theorems {#section:limite}
==============

Set $$\label{definit_A}
  A=\frac{1}{4\left( \operatorname{{\alpha_{\text{min}}}}^{3/2}
  \sqrt{\frac{\pi}{2\alpha''(x_0)}}\right)^{1/\operatorname{{\alpha_{\text{min}}}}}}\int_{S^1\times \{1/2\}}
  h\operatorname{dLeb},$$ where $h$ is the density of $m$ with respect to $\operatorname{Leb}$.

In this section, we prove the following theorem:

\[enonce\_theoreme\_limite\] Let $f$ be a Hölder function on $S^1\times [0,1]$, with $\int f{\, {\rm d}}m=0$. Write $c=\int_{S^1\times\{0\}} f \operatorname{dLeb}$. Then

-   If $\operatorname{{\alpha_{\text{min}}}}<1/2$, there exists $\sigma^2 {\geqslant}0$ such that $\frac{1}{\sqrt{n}} S_n f \to {\mathcal{N}}(0,\sigma^2)$.

-   If $\operatorname{{\alpha_{\text{min}}}}=1/2$ and $c\not=0$, then $\frac{S_n f}{\sqrt{ \frac{c^2 A }{4} n (\ln n)^2}} \to
    {\mathcal{N}}(0,1)$.

-   If $1/2<\operatorname{{\alpha_{\text{min}}}}<1$ and $c\not =0$, then $\frac{S_n f}{n^{\operatorname{{\alpha_{\text{min}}}}}\sqrt{\operatorname{{\alpha_{\text{min}}}}\ln n}} \to Z$, where the random variable $Z$ has a characteristic function given by $$E(e^{itZ})
      =e^{- A
      |c|^{1/\operatorname{{\alpha_{\text{min}}}}} \Gamma(1-1/\operatorname{{\alpha_{\text{min}}}})\cos\left(\frac{\pi}{2\operatorname{{\alpha_{\text{min}}}}}\right) |t|^{1/\operatorname{{\alpha_{\text{min}}}}}
      \left( 1-i\operatorname{sgn}(ct) \tan
      \left(\frac{\pi}{2\operatorname{{\alpha_{\text{min}}}}} \right) \right)}$$

-   If $1/2{\leqslant}\operatorname{{\alpha_{\text{min}}}}<1$ and $c=0$, assume also that there exists $\gamma
    >0$ such that $|f(\omega,x)-f(\omega,0)|{\leqslant}Cx^\gamma$, with $\gamma>\operatorname{{\alpha_{\text{max}}}}\left(1-\frac{1}{2\operatorname{{\alpha_{\text{min}}}}}\right)$. Then there exists $\sigma^2 {\geqslant}0$ such that $\frac{1}{\sqrt{n}} S_n f \to
    {\mathcal{N}}(0,\sigma^2)$.

The random variable $Z$ in the third case has a stable distribution of exponent $1/\operatorname{{\alpha_{\text{min}}}}$ and parameters $A|c|^{1/\operatorname{{\alpha_{\text{min}}}}}
\Gamma(1-1/\operatorname{{\alpha_{\text{min}}}})\cos\left(\frac{\pi}{2\operatorname{{\alpha_{\text{min}}}}}\right)$ and $\operatorname{sgn}(c)$.

To prove this theorem, we will use Theorem \[thm\_abstrait\_markov\]. For this, we need a control of $m({\varphi}_Y>n)$ which comes from the asymptotic behavior of $X_n$ proved in Theorem \[estimee\_Xn\_L1\]. It will also be necessary to estimate $m(f_Y>x)$, through the study of the integrability of $f_Y$ (Lemmas \[L2\_am\_petit\] and \[lemme\_dans\_Lp\]).

In the rest of this section, $f$ will be a Hölder function on $S^1\times [0,1]$, fixed once and for all. Recall that $f_Y(y)=\sum_{k=0}^{{\varphi}_Y(y)-1} f(T^k y)$, where ${\varphi}_Y$ is the first return time to $Y=S^1\times(1/2,1]$.

Estimates on measures
---------------------

\[controle\_mesure\_phi\] We have $$m({\varphi}_Y>n) \sim \left(\frac{\sqrt{\ln n}}{n}\right)^{1/\operatorname{{\alpha_{\text{min}}}}} A$$ where $A$ is given by .

We have $$\begin{aligned}
  m({\varphi}_Y> n)&
  =\int_{S^1} \int_{1/2}^{Y_{n+1}(\omega)} h(\omega,u){\, {\rm d}}u {\, {\rm d}}\omega
  =\int_{S^1} \int_0^{X_{n}(F\omega)/2} h(\omega,1/2+u){\, {\rm d}}u {\, {\rm d}}\omega
  \\&
  =\int_{S^1} \frac{X_{n}(F\omega)}{2} h(\omega,1/2){\, {\rm d}}\omega
  +\int_{S^1} \int_0^{X_{n}(F\omega)/2} \bigl[h(\omega,1/2+u)-h(\omega,1/2)
  \bigr] {\, {\rm d}}u{\, {\rm d}}\omega
  \\&
  =I+II.
  \end{aligned}$$ As $\left(\frac{n}{\sqrt{\ln n}}\right)^{1/\operatorname{{\alpha_{\text{min}}}}} X_n(F\omega) \to
\frac{1}{\left(2^{\operatorname{{\alpha_{\text{min}}}}} \operatorname{{\alpha_{\text{min}}}}^{3/2}
  \sqrt{\frac{\pi}{2\alpha''(x_0)}}\right)^{1/\operatorname{{\alpha_{\text{min}}}}}}$ in $L^1$ and almost everywhere (Theorem \[estimee\_Xn\_L1\]) and $h(\omega,1/2)$ is bounded, we get that $I\sim
\left(\frac{\sqrt{\ln n}}{n}\right)^{1/\operatorname{{\alpha_{\text{min}}}}} A$. Moreover, for large enough $n$, $|h(\omega,1/2+u)-h(\omega,1/2)|{\leqslant}{\varepsilon}$, whence $II=o\left(\frac{\sqrt{\ln n}}{n}\right)^{1/\operatorname{{\alpha_{\text{min}}}}}$.

\[L2\_am\_petit\] If $\operatorname{{\alpha_{\text{min}}}}<1/2$, then $f_Y \in L^2(Y, {{\rm d}}m)$.

We have $$\begin{aligned}
  \int f_Y^2 {\, {\rm d}}m &
  {\leqslant}C\sum_n m({\varphi}_Y=n)  n^2
  =C \sum \bigl( m({\varphi}_Y>n-1)-m({\varphi}_Y>n)\bigr)n^2
  \\&
  {\leqslant}C\sum m({\varphi}_Y>n) n
  \end{aligned}$$ which is summable since $m({\varphi}_Y>n)\sim A \left(\frac{\sqrt{\ln
n}}{n}\right)^{1/\operatorname{{\alpha_{\text{min}}}}}$ with $1/\operatorname{{\alpha_{\text{min}}}}>2$.

\[lemme\_dans\_Lp\] Assume that $\int_{S^1\times \{0\}} f=0$. Let $\operatorname{{\alpha_{\text{max}}}}>\gamma>0$ be such that $|f(\omega,x)-f(\omega,0)| {\leqslant}C x^\gamma$. If $1<p<
\min\left( \frac{2}{\operatorname{{\alpha_{\text{min}}}}}, \frac{1}{\operatorname{{\alpha_{\text{min}}}}(1-\gamma/\operatorname{{\alpha_{\text{max}}}})} \right)$, then $f_Y \in L^p(Y,{{\rm d}}m)$.

As $h$ is bounded on $Y$, it is sufficient to prove that $f_Y \in
L^p(Y,\operatorname{dLeb})$.

Assume first that $f \equiv 0$ on $S^1 \times \{0\}$. Then, if ${\textbf{x}}=(\omega,x)$ satisfies ${\varphi}_Y({\textbf{x}})=n$, we have $f_Y({\textbf{x}})=\sum_0^{n-1} f(T^k {\textbf{x}})$. If $k {\geqslant}1$, $T_\omega^k(x){\leqslant}X_{n-k}(F^k \omega) {\leqslant}\frac{C}{(n-k)^{1/\operatorname{{\alpha_{\text{max}}}}}}$, whence $|f(T^k {\textbf{x}})|{\leqslant}\frac{C}{(n-k)^{\gamma/\operatorname{{\alpha_{\text{max}}}}}}$, and a summation yields that $|f_Y({\textbf{x}})|{\leqslant}C n^{1-\gamma/\operatorname{{\alpha_{\text{max}}}}}$. Thus, $$\begin{aligned}
  \int |f_Y|^p &
  {\leqslant}C \sum m({\varphi}_Y=n) n^{p(1-\gamma/\operatorname{{\alpha_{\text{max}}}})}
      \\&
  {\leqslant}C \sum m({\varphi}_Y>n) n^{p(1-\gamma/\operatorname{{\alpha_{\text{max}}}})-1}.
  \end{aligned}$$ As $m({\varphi}_Y>n) \sim A \left(\frac{\sqrt{\ln n}}{n}
\right)^{1/\operatorname{{\alpha_{\text{min}}}}}$, this last series is summable as soon as $$-\frac{1}{\operatorname{{\alpha_{\text{min}}}}}+p\left(1-\frac{\gamma}{\operatorname{{\alpha_{\text{max}}}}}\right)-1 <-1,$$ which is the case by assumption on $p$.

Assume now that $f$ has a vanishing integral on $S^1$. Let $g(\omega,x)=f(\omega,0)$. The function $f-g$ vanishes on $S^1
\times \{0\}$, whence $f_Y-g_Y \in L^p$ according to the first part of this proof. Consequently, it is sufficient to prove that $g_Y \in L^p$. Write $\chi(\omega)=f(\omega,0)$ and $S_n
\chi(\omega)=\sum_{k=0}^{n-1} \chi(F^k \omega)$: then $g_Y(\omega,x)=S_{{\varphi}_Y(\omega,x)} \chi(\omega)$.

Let $M_n \chi(\omega)=\max_{k{\leqslant}n}|S_k \chi(\omega)|$. Let $\delta>0$, and $l=\frac{1+\delta}{\delta}$, so that $\frac{1}{l}+\frac{1}{1+\delta}=1$. We have $$\begin{aligned}
  \int |g_Y|^p &
  =\sum_{n=0}^{\infty} \int_{S^1}
  \int_{1/2+X_{n}(F\omega)/2}^{1/2+X_{n-1}(F\omega)/2} \bigl|S_n
  \chi(\omega)\bigr|^p {\, {\rm d}}u {\, {\rm d}}\omega
  \\&
  {\leqslant}\sum_{k=1}^{\infty} \int_{S^1}
  \int_{1/2+X_{2^k}(F\omega)/2}^{1/2+X_{2^{k-1}}(F\omega)/2} |M_{2^k}
  \chi(\omega)|^p {\, {\rm d}}u {\, {\rm d}}\omega
  \\&
  {\leqslant}\sum_{k=1}^\infty \int_{S^1} X_{2^{k-1}}(F\omega)
  |M_{2^k}\chi(\omega)|^p {\, {\rm d}}\omega
  {\leqslant}\sum_{k=1}^\infty {\left\| X_{2^{k-1}}\circ F \right\|}_{1+\delta}
  {\left\| M_{2^k}\chi \right\|}_{lp}^p,
  \end{aligned}$$ where the last inequality is Hölder inequality. If $\delta$ is small enough, $lp>2$, whence Corollary \[ineg\_max\] yields that ${\left\| M_{2^k}\chi \right\|}_{lp} {\leqslant}C k^{\frac{lp-1}{lp}}\sqrt{2^k}$. Moreover, $${\left\| X_{2^{k-1}}\circ F \right\|}_{1+\delta}
  ={\left\| X_{2^{k-1}} \right\|}_{1+\delta}
  {\leqslant}\left( \int X_{2^{k-1}} \right)^{1/(1+\delta)}
  \sim C \left( \frac{\sqrt{\ln(2^{k-1})}}{2^{k-1}}
  \right)^{\frac{1}{(1+\delta) \operatorname{{\alpha_{\text{min}}}}}}$$ by Theorem \[estimee\_Xn\_L1\]. Thus, $\int|g_Y|^p<\infty$ if $\frac{1}{(1+\delta)\operatorname{{\alpha_{\text{min}}}}} > \frac{p}{2}$, and it is possible to choose $\delta$ such that this inequality is true, since $\frac{1}{\operatorname{{\alpha_{\text{min}}}}}>\frac{p}{2}$ by hypothesis.

Proof of Theorem \[enonce\_theoreme\_limite\]
---------------------------------------------

To apply Theorem \[thm\_abstrait\_markov\], we first check the condition . Let $\theta$ be the Hölder exponent of $f$. We will work with the distance $d_{
\lambda^{-\theta}}= \lambda^{-\theta s(x,y)}$. For this distance, $T_Y$ is a Gibbs-Markov map.

*Fact: if $f$ is $\theta$-Hölder on $S^1\times [0,1]$, then $$\sum m[A_{s,n}] D f_Y(A_{s,n}) <\infty.$$*

Recall that $D f_Y(A_{s,n})$ (defined in Theorem \[thm\_abstrait\_markov\]) is the best Lipschitz constant of $f_Y$ on $A_{s,n}$, here for the distance $d_{ \lambda^{-\theta}}$.

Take $(\omega_1,x_1)$ and $(\omega_2,x_2)\in A_{s,n}$ with for example $x_2{\geqslant}x_1$. This implies that $x_1\in J_n^+(\omega_2)$ and that, for $0{\leqslant}k{\leqslant}n$, $d(T^k(\omega_1,x_1),T^k(\omega_2,x_2)) {\leqslant}(1+D) |F^k \omega_1
-F^k \omega_2|$ (see the beginning of the proof of Proposition \[prop\_distortion\_bornee\]). Moreover, $d(T^k(\omega_2,x_1),T^k(\omega_2,x_2)) {\leqslant}d(T^n(\omega_2,x_1),T^n(\omega_2,x_2))$ (since, if $\omega$ is fixed, the map $T_{\alpha(\omega)}$ is expanding).

Thus, for $0{\leqslant}k{\leqslant}n$, $$\begin{aligned}
  d(T^k(\omega_1,x_1),T^k(\omega_2,x_2))&
  {\leqslant}d(T^k(\omega_1,x_1),T^k(\omega_2,x_1))
  +d(T^k(\omega_2,x_1),T^k(\omega_2,x_2))
  \\&
  {\leqslant}(1+D)|F^k \omega_1 -F^k \omega_2|
  +d(T^n(\omega_2,x_1),T^n(\omega_2,x_2))
  \\&
  {\leqslant}(1+D)|F^n \omega_1 -F^n \omega_2|
  +d(T^n(\omega_1,x_1),T^n(\omega_2,x_1))
  \\& \hphantom{=\ }
  +d(T^n(\omega_1,x_1),T^n(\omega_2,x_2))
  \\&
  {\leqslant}(1+D)|F^n \omega_1 -F^n \omega_2|
  +(1+D)|F^n \omega_1 -F^n \omega_2|
  \\& \hphantom{=\ }
  +d(T^n(\omega_1,x_1),T^n(\omega_2,x_2))
  \\&
  {\leqslant}(3+2D)d(T^n(\omega_1,x_1),T^n(\omega_2,x_2)).
  \end{aligned}$$ We deduce that $$\begin{aligned}
  |f_Y(\omega_1,x_1)-f_Y(\omega_2,x_2)|&
  {\leqslant}\sum_{k=0}^{n-1}
  |f(T^k(\omega_1,x_1))-f(T^k(\omega_2,x_2))|
  \\&
  {\leqslant}\sum_{k=0}^{n-1}C
  d(T^k(\omega_1,x_1),T^k(\omega_2,x_2))^\theta
  \\&
  {\leqslant}C' n d(T^n(\omega_1,x_1),T^n(\omega_2,x_2))^\theta.
  \end{aligned}$$ As $T_Y$ is expanding for the distance $d'$ (defined in , and equivalent to $d$), we get $$d(T^n(\omega_1,x_1),T^n(\omega_2,x_2)) {\leqslant}C
  d_{\lambda^{-1}}(T^n(\omega_1,x_1),T^n(\omega_2,x_2))=C \lambda
  d_{\lambda^{-1}}((\omega_1,x_1),(\omega_2,x_2)),$$ whence $d(T^n(\omega_1,x_1),T^n(\omega_2,x_2))^\theta {\leqslant}C
d_{\lambda^{-\theta}}((\omega_1,x_1),(\omega_2,x_2))$.

Thus, $Df_Y(A_{s,n}) {\leqslant}C n$, and $$\sum m(A_{s,n}) D f_Y(A_{s,n})
  {\leqslant}C \sum m({\varphi}_Y=n) n
  = C < +\infty,$$ by Kac’s Formula.

In the case $\operatorname{{\alpha_{\text{min}}}}<1/2$, Lemma \[L2\_am\_petit\] gives that $f_Y \in
L^2$. Moreover, $|f|_Y\in L^2$ for the same reason, and ${\varphi}\in
L^2$ (since ${\varphi}=g_Y$ for $g\equiv 1$, whence Lemma \[L2\_am\_petit\] applies also). We have already checked the condition , so we can apply (the first case of) Theorem \[thm\_abstrait\_markov\]. This yields the central limit theorem for $f$.

The second and third cases are analogous. Let us prove for example the third one, i.e. $1/2<\operatorname{{\alpha_{\text{min}}}}<1$ and $c\not=0$. Assume for example $c>0$. We estimate $m(f_Y>x)$.

*Fact: $m(f_Y>x) \sim \left(\frac{c \sqrt{\ln
x}}{x}\right)^{1/\operatorname{{\alpha_{\text{min}}}}} A$ and $m(f_Y<-x)=o\left(\frac{\sqrt{\ln
x}}{x}\right)^{1/\operatorname{{\alpha_{\text{min}}}}}$.*

We prove the estimate on $m(f_Y>x)$, the other one being similar.

Let $g\equiv c$ on $S^1 \times [0,1]$. Then $g_Y=nc$ on $[{\varphi}_Y=n]$, which implies that $m(g_Y>nc)=m({\varphi}_Y>n)\sim
\left(\frac{\sqrt{\ln n}}{n}\right)^{1/\operatorname{{\alpha_{\text{min}}}}} A$ by Lemma \[controle\_mesure\_phi\].

In the general case, consider $j=f-g$, and let us prove that $m(|j_Y|>x)=o\left(\frac{\sqrt{\ln x}}{x}\right)^{1/\operatorname{{\alpha_{\text{min}}}}}$. As $f_Y=g_Y+j_Y$, it will give $$m(g_Y>x(1+{\varepsilon})) -m(|j_Y|>x{\varepsilon}) {\leqslant}m(f_Y>x) {\leqslant}m(g_Y>x(1-{\varepsilon}))+m(|j_Y|>x{\varepsilon}),$$ which gives the conclusion. Let $\gamma>0$ with $\gamma<\min(\theta,\operatorname{{\alpha_{\text{max}}}})$ (where $\theta$ is the Hölder coefficient of $f$). Lemma \[lemme\_dans\_Lp\] gives that $j_Y\in L^p$ if $p<\min\left(\frac{2}{\operatorname{{\alpha_{\text{min}}}}},\frac{1}{\operatorname{{\alpha_{\text{min}}}}(1-\gamma/\operatorname{{\alpha_{\text{max}}}})} \right)$. We can in particular choose $p>1/\operatorname{{\alpha_{\text{min}}}}$. Then $m(|j_Y|>x) {\leqslant}\int
\left(\frac{|j_Y|}{x}\right)^p =O(x^{-p})$, which concludes the proof of the fact.

The same fact holds for ${\varphi}_Y$ and $|f|_Y$, with the same proof, whence we are in the third case of Theorem \[thm\_abstrait\_markov\]. This gives the desired result.

Assume finally that $\frac{1}{2}{\leqslant}\operatorname{{\alpha_{\text{min}}}}<1$ and that $c=0$. Under the hypotheses of the theorem, we can apply Lemma \[lemme\_dans\_Lp\] with $p=2$, and get that $f_Y \in L^2$. The proof of this lemma shows in fact that the function $M$ (defined in Theorem \[thm\_abstrait\_markov\]) is also in $L^2$. Finally, Lemma \[controle\_mesure\_phi\] shows that $m[{\varphi}_Y>x] \sim
\left(\frac{\sqrt{\ln x}}{x}\right)^{1/\operatorname{{\alpha_{\text{min}}}}}A$. We have checked all the hypotheses of the first case of Theorem \[thm\_abstrait\_markov\].

Induced maps and limit theorems {#appendice:loi_stable}
===============================

The aim of this section is to prove very general results stating that, if a function satisfies a limit theorem for an induced map, it also satisfies one for the initial map. Similar theorems have been proved in [@gouezel:stable], by spectral methods. We will describe here a more elementary method, essentially due to Melbourne and Török for flows ([@melbourne_torok]).

If $Y$ is a subset of a probability space $(X,m)$, $T:X\to X$, and $T_Y$ is the induced map on $Y$, we will write $S_n^Y
g=\sum_{k=0}^{n-1} g\circ T_Y^k$: this is the Birkhoff sum of $g$, for the transformation $T_Y$. We will also write $E_Y(g)=\frac{\int_Y g}{m[Y]}$. Finally, for $t\in {\mathbb{R}}$, $\lfloor t
\rfloor$ denotes the integer part of $t$.

\[thm\_probabiliste\_general\] Let $T:X\to X$ be an ergodic endomorphism of a probability space $(X,m)$, and $f:X\to {\mathbb{R}}$ an integrable function with vanishing integral. Let $Y\subset X$ have positive measure. For $y\in Y$, write ${\varphi}(y)=\inf\{n>0 {\ |\ }T^n(y)\in Y\}$ and $f_Y(y)=\sum_{k=0}^{{\varphi}(y)-1} f(T^k y)$, and $M(y)=\max_{1{\leqslant}k{\leqslant}{\varphi}(y)} \left|\sum_{j=0}^{k-1} f(T^j y) \right|$.

We assume the following properties:

1.  There exists a sequence $B_n\to +\infty$, with $\sup_{r {\leqslant}2n}
    \frac{B_r}{B_n} < \infty$ and $\inf_{r {\geqslant}n} \frac{B_r}{B_n}>0$, such that $(f_Y, {\varphi})$ satisfies a mixing limit theorem for the normalization $B_n$: there exists a random variable $Z$ such that, for every $t\in {\mathbb{R}}$, $$\label{limite_mixing}
      E_Y\left({\varphi}e^{it \frac{S^Y_{\lfloor n m(Y) \rfloor}  f_Y}{B_n}}\right)
      \to E_Y({\varphi}) E\left(e^{itZ}\right).$$

2.  For every ${\varepsilon}>0$, there exists $C$ such that, for any $n\in
    {\mathbb{N}}^*$, $$\label{majore_fY}
      m\{y\in Y {\ |\ }M(y) {\geqslant}{\varepsilon}B_n \} {\leqslant}\frac{C}{n}.$$

3.  \[hypothese\_3\] There exists $b>0$ such that, in the natural extension of $T_Y$, $\frac{1}{N^b} \sum_0^{N-1} f_Y(T_Y^k)$ tends almost everywhere to $0$ when $N \to \pm \infty$.

4.  \[hypothese\_4\] For every ${\varepsilon}>0$, there exists $A>0$ and $N_0$ such that, for every $n{\geqslant}N_0$, $$\label{eq_hyp4}
      m\left\{ y\in Y {\ |\ }\left| \frac{S_n^Y {\varphi}- n
      E_Y({\varphi})}{B_n^{1/b}} \right| {\geqslant}A \right\} {\leqslant}{\varepsilon}.$$

Then the function $f$ satisfies also a limit theorem: $$E\left(e^{it \frac{S_n f}{B_n}}\right)
  \to E(e^{it Z}),$$ i.e. $\frac{S_n f}{B_n}$ tends in distribution to $Z$.

The hypotheses of the theorem are tailor-made so that the following proof works, but they are in fact often satisfied in natural cases. Let us comment on these 4 hypotheses:

1.  The convergence is very often satisfied when $f_Y$ satisfies a limit theorem. Namely, the martingale proofs or spectral proofs of limit theorems automatically give this kind of convergence.

2.  If $Z_0,Z_1,\ldots$ are independent identically random variables such that $\frac{\sum_{0}^{n-1} Z_k}{B_n}$ converges in distribution to a nontrivial limit, then for all ${\varepsilon}>0$, there exists $C$ such that $P(|Z_0| {\geqslant}{\varepsilon}B_n) {\leqslant}\frac{C}{n}$: this is a consequence of the classification of the stable laws, see [@feller:2].

    Here, we are not in the independent setting, and there is no such classification. However, the same kind of results holds very often: usually, it is not hard to check in practical cases that $m(|f_Y(x)| {\geqslant}{\varepsilon}B_n) {\leqslant}\frac{C}{n}$, since $f_Y$ satisfies a limit theorem by the first assumption. Set $|f|_Y(y)=\sum_{j=0}^{{\varphi}(y)-1} |f(T^j y)|$. As $|f|_Y$ and $f_Y$ have more or less the same distribution, $|f|_Y$ satisfies also often $$\label{majore_fY'}
      \tag{\ref{majore_fY}'}
      m\{ y\in Y {\ |\ }|f|_Y(y) {\geqslant}{\varepsilon}B_n \} {\leqslant}\frac{C}{n}.$$ Since $M {\leqslant}f_Y$, implies . Thus, it will often be sufficient to check . However, is sometimes strictly weaker than , because of cancellations, which is why we have stated the theorem with .

3.  The natural extension is useful so that we can let $N$ tend to $-\infty$, and consider $T_Y^{-1}$ in the proof. Generally, Birkhoff’s Theorem yields that this assumption is satisfied for $b=1$. This is often sufficient. However, sometimes, it is important to have better estimates. It is then possible to use [@vitesse_birkhoff Theorem 16], for example: this theorem ensures that, if the correlations of $f_Y\in L^2$ decay at least as $O(1/n)$, then the hypothesis is satisfied for any $b>1/2$ (for $N \to -\infty$, use the fact that $\int f_Y \cdot f_Y \circ T_Y^n
    = \int f_Y\circ T_Y^{-n} \cdot f_Y$, and apply the result to $T_Y^{-1}$).

4.  The fourth assumption is weaker than $$\label{eq_hyp4'}
      \tag{\ref{eq_hyp4}'}
      \exists B'_n=O(B_n^{1/b}) \text{ such that }
      \frac{S_n^Y {\varphi}- nE_Y({\varphi})}{B'_n}\text{ converges in
      distribution.}$$ Moreover, ${\varphi}$ is often simpler than $f_Y$. Since $f_Y$ satisfies a limit theorem (this is more or less the first hypothesis), this is also often the case of ${\varphi}$, which implies . Thus, – and hence – are satisfied quite generally.

Without loss of generality, we can work in a tower, i.e. assume that $X=\{ (y,i) {\ |\ }y\in Y, i\in\{0,\ldots,{\varphi}(y)-1\} \}$ and that, for $i<{\varphi}(y)-1$, $T(y,i)=(y,i+1)$, while $T(y,{\varphi}(y)-1)=(T_Y(y),0)$. Namely, it is possible to build an extension of $X$ satisfying these properties, and it is equivalent to prove a limit theorem in $X$ or in this extension (see for example [@gouezel:stable Section 4.1]). Note that $E_Y({\varphi})=1/m(Y)$ by Kac’s Formula. Let $\pi$ be the projection from $X$ to $Y$, given by $\pi(y,i)=y$.

In this proof, we will write $S_t f(x)$, even when $t$ is not an integer, for $S_{\lfloor t \rfloor }f(x)$. In the same way, $T^t$ should be understood as $T^{\lfloor t \rfloor}$. We also extend $B_n$ to ${\mathbb{R}}_+$, setting $B_t:=B_{\lfloor t \rfloor}$.

As $T$ is ergodic, $T_Y$ is also ergodic ([@aaronson:book Proposition 1.5.2]). Birkhoff’s Theorem gives that $$\label{asymp_phin}
  S_n^Y {\varphi}= \frac{n}{m(Y)}  +o(n)$$ almost everywhere on $Y$. For $y\in Y$ and $N\in {\mathbb{N}}$, let $n(y,N)$ be the greatest integer $n$ such that $S_n^Y {\varphi}(y) < N$. If $y$ is such that $S_n^Y {\varphi}(y)=\frac{n}{m(Y)}+o(n)$ (which is true almost everywhere), then $n(y,N)$ is finite for every $N$, and $\frac{n(y,N)}{m(Y)} \sim N$, i.e.$$\label{renouvellement_basique}
  \frac{n(y,N)}{N m(Y)} \to 1.$$

Since $\int_X e^{it (S_N^Y f_Y)\circ \pi }=\int_Y {\varphi}e^{it S_N^Y
f_Y}$, yields that $$\label{Thm_limite_SY}
  \frac{(S^Y_{Nm(Y)} f_Y)\circ \pi}{B_N} \to Z$$ in distribution on $X$. The idea of the proof will be to see that $(S^Y_{Nm(Y)}f_Y)\circ \pi$ and $S_N f$ are close (this is not surprising, since one iteration of $T_Y$ corresponds roughly to $1/m(Y)$ iterations of $T$). This will give that $\frac{S_N
f}{B_N}$ tends to $Z$.

We write $$\begin{aligned}
    S_N f(y,i)=
  \left(S_N f(y,i)-S_N f(y,0) \right)
  &+ \left(S_N f(y,0)-S^Y_{n(y,N)}f_Y(y) \right)
  \\&
  +\left( S^Y_{n(y,N)}f_Y(y) - S^Y_{Nm(Y)}f_Y(y) \right)
  +S^Y_{ Nm(Y)}f_Y(y).
  \end{aligned}$$ The last term, equal to $\bigl(S^Y_{N m(Y)} f_Y\bigr)\circ \pi$, satisfies a limit theorem by . To conclude the proof, we will see that the three other terms, divided by $B_N$, tend to $0$ in probability.

The second and third terms depend only on $y$. Thus, the following lemma will be useful to prove that they tend to $0$ on $X$:

\[proba\_sur\_Y\_donne\_proba\_sur\_X\] Let $f_n$ be a sequence of functions on $Y$, tending to $0$ in probability on $Y$. Then $f_n \circ \pi$ tends to $0$ in probability on $X$.

Take ${\varepsilon}>0$. As $f_n \to 0$ in probability, the measure of $E_n:=\{ y\in Y {\ |\ }|f_n(y)|{\geqslant}{\varepsilon}\}$ tends to $0$. As ${\varphi}\in L^1$, dominated convergence yields that $\int_{E_n} {\varphi}\to 0$, i.e. the measure of $\pi^{-1}(E_n)$ tends to $0$. But $\pi^{-1}(E_n)$ is exactly the set where $|f_n \circ \pi|{\geqslant}{\varepsilon}$.

*Fact: $\frac{1}{B_N} \left(S_N f(y,i)-S_N f(y,0) \right)$ tends to $0$ in probability on $X$.*

Set $V_N(y)=\sum_{i=0}^{{\varphi}(y)-1} |f \circ T^N(y,i)|$ on $Y$. Then ${\left\| V_N \right\|}_{L^1(Y)} ={\left\|  f\circ T^N \right\|}_{L^1(X)}
={\left\| f \right\|}_{L^1(X)}$ since $T$ preserves the measure. Thus, $V_N/B_{N}$ tends to $0$ in $L^1(Y)$, and in probability. Lemma \[proba\_sur\_Y\_donne\_proba\_sur\_X\] yields that $\frac{1}{B_N}
V_N\circ \pi$ tends to $0$ in probability on $X$.

As $S_N f(y,i)-S_N f(y,0) =\sum_N^{N+i-1} f(T^k(y,0))
-\sum_0^{i-1} f(T^k(y,0))$, we get $|S_N f(y,i)-S_N f(y,0)| {\leqslant}V_N(y)+V_0(y)$. Thus, $\frac{1}{B_N} \left(S_N f(y,i)-S_N f(y,0)
\right)$ is bounded by a function going to $0$ in probability.

*Fact: $\frac{1}{B_N} \left(S_N f(y,0)-S_{n(y,N)}^Y
f_Y(y)\right)$ tends to $0$ in probability on $X$.*

By Lemma \[proba\_sur\_Y\_donne\_proba\_sur\_X\], it is sufficient to prove it on $Y$. We have $$\left|S_N f(y,0)-S_{n(y,N)}^Y f_Y(y)\right|
  =
  \left| \sum_{S_{n(y,N)}^Y {\varphi}(y)}^{N-1} f\circ T^k(y,0) \right|
  {\leqslant}M\left(T_Y^{n(y,N)}y\right).$$ Let $a>0$ be very small, we show that $m\left\{ y {\ |\ }M\left(T_Y^{n(y,N)}y\right) {\geqslant}a B_N \right\} \to 0$.

Let ${\varepsilon}>0$. Let $C$ be such that $m( M(y) {\geqslant}a B_n) {\leqslant}\frac{C}{n}$, by . Set $\delta=\frac{{\varepsilon}}{2Cm(Y)}$. By , for large enough $N$, $$m\left\{\left|\frac{n(y,N)}{m(Y) N} - 1 \right|
  {\geqslant}\delta\right\}
  {\leqslant}{\varepsilon}.$$ When $\left|\frac{n(y,N)}{m(Y) N} - 1 \right| {\leqslant}\delta$, the fact that $M\left(T_Y^{n(y,N)}y\right) {\geqslant}aB_N$ implies that there exists $n \in [(1-\delta) m(Y)N, (1+\delta)m(Y)N]$ such that $M(T_Y^n y) {\geqslant}aB_N$. Thus, $$m\left\{ y {\ |\ }M\left(T_Y^{n(y,N)}y\right) {\geqslant}a B_N \right\}
  {\leqslant}{\varepsilon}+
  \sum_{n=(1-\delta) m(Y)N}^{(1+\delta) m(Y)N}
  m\{M(T_Y^n y) {\geqslant}a B_N\}.$$

As $m$ is invariant by $T_Y$, we have $m\{M(T_Y^n y) {\geqslant}a B_N\}
=m\{ M {\geqslant}a B_N\} {\leqslant}\frac{C}{N}$. Thus, $$m\left\{ y {\ |\ }M\left(T_Y^{n(y,N)}y\right) {\geqslant}a B_N \right\}
  {\leqslant}{\varepsilon}+
  2\delta m(Y)N \frac{C}{N}
  =2{\varepsilon}.$$

*Fact: $\frac{1}{B_N} \left(S^Y_{n(y,N)} f_Y- S^Y_{N
m(Y)}f_Y\right)$ tends to $0$ in probability on $X$ when $N \to
\infty$.*

By Lemma \[proba\_sur\_Y\_donne\_proba\_sur\_X\], it is sufficient to prove it on $Y$. Without loss of generality, we can use the natural extension and assume that $T_Y$ is invertible.

For $n<0$, write $S^Y_n f_Y=\sum_0^{|n|-1} f_Y\circ T_Y^{-j}$. Then, setting $\nu(y,N)=n(y,N)-N m(Y)$, $$\label{definit_nu}
  S^Y_{n(y,N)} f_Y(y)- S^Y_{N m(Y)}f_Y(y)
  =S^Y_{\nu(y,N)} f_Y \left(T^{N m(Y)}(y)\right).$$ If $A>0$ and $N\in {\mathbb{N}}$, as $E_Y({\varphi})=1/m(Y)$, we get $$\begin{gathered}
   \{y{\ |\ }\nu(y,N) {\geqslant}A B_N^{1/b}\}
   =\{ n(y,N) {\geqslant}A B_N^{1/b}+ N m(Y)\}
   =\{ S^Y_{A B_N^{1/b}+Nm(Y)}{\varphi}< N\}
   \\=
   \left \{ \frac{S^Y_{A B_N^{1/b}+N m(Y)}{\varphi}- (A B_N^{1/b}+
   Nm(Y))E_Y({\varphi})}{\left(B_{AB_N^{1/b}+Nm(Y)}\right)^{1/b}} <
   -\frac{A}{m(Y)}\left(\frac{B_N}{B_{AB_N^{1/b}+Nm(Y)}}\right)^{1/b}
   \right\}.
   \end{gathered}$$ For some integer $k$, we have $N {\leqslant}2^k Nm(Y) {\leqslant}2^k
(AB_{Nm(Y)}^{1/b}+Nm(Y))$. The assumption $\sup_{r{\leqslant}2n}
\frac{B_r}{B_n} {\leqslant}C<\infty$ thus yields that $\frac{B_N}{B_{AB_{Nm(Y)}^{1/b}+Nm(Y)}} {\leqslant}C^k$. In particular, $$\{y{\ |\ }\nu(y,N) {\geqslant}A B_N^{1/b}\}
  \subset \left \{ \frac{S^Y_{A B_N^{1/b}+N m(Y)}{\varphi}- (A B_N^{1/b}+
   Nm(Y))E_Y({\varphi})}{\left(B_{AB_N^{1/b}+Nm(Y)}\right)^{1/b}} <
   -\frac{AC^{k/b}}{m(Y)}
   \right\}.$$ Consequently, if $A$ is large enough, Assumption \[hypothese\_4\] yields that $m\{y{\ |\ }\nu(y,N) {\geqslant}A B_N^{1/b}\} {\leqslant}{\varepsilon}$ for large enough $N$. We handle in the same way the set of points where $\nu(y,N) {\leqslant}-A B_N^{1/b}$, using the assumption $\inf_{r{\geqslant}n}\frac{B_r}{B_n}>0$.

We have thus proved: $$\label{convergence_nuy}
  \forall {\varepsilon}>0, \exists A>0, \exists N_0>0, \forall N {\geqslant}N_0,\
  m\{ y{\ |\ }|\nu(y,N)| {\geqslant}A B_N^{1/b} \} {\leqslant}{\varepsilon}.$$

Set $W_N(y)=\frac{1}{B_N} S_{\nu(y,N)} f_Y \left(T_Y^{ N
m(Y)}(y)\right)$, we will show that it tends to $0$ in distribution, which will conclude the proof, by . Take $a>0$, we show that $m(|W_N|>a) \to 0$ when $N\to \infty$.

Let ${\varepsilon}>0$. Assumption \[hypothese\_3\] ensures that there exists ${\widetilde}{Y}$ with $m({\widetilde}{Y}){\geqslant}m(Y)-{\varepsilon}$ and $N_1$ such that $\frac{1}{|N|^b}|S_N^Y f_Y| {\leqslant}{\varepsilon}$ on ${\widetilde}{Y}$, for every $|N|{\geqslant}N_1$. Define $Y'_N=\{y\in Y {\ |\ }|\nu(y,N)|< N_1\}$ and $Y''_N=\{y \in Y {\ |\ }|\nu(y,N)|{\geqslant}N_1\}$. We estimate first the contribution of $Y'_N$.

Set $\psi(y)=\sum_{-N_1}^{N_1-1} |f_Y\circ T_Y^j|$. Since $\psi$ is measurable, there exists a constant $C$ and a subset $Z$ of $Y$ with $m(Z){\geqslant}m(Y)-{\varepsilon}$ and $\psi {\leqslant}C$ on $Z$. Then, for $y\in Y'_N$, we have $|W_N(y)|{\leqslant}\frac{1}{B_N} \psi\left(T_Y^{N
m(Y)} y\right)$. Set $Z_N=Y'_N \cap T_Y^{- N m(Y)}(Z)$: it satisfies $m(Z_N) {\geqslant}m(Y'_N)-{\varepsilon}$. On $Z_N$, we have $|W_N|{\leqslant}\frac{C}{B_N}$, whence, for large enough $N$, $|W_N|<a$ on $Z_N$. Thus, for large enough $N$, $$m\left\{y \in Y'_N {\ |\ }|W_N(y)|{\geqslant}a\right\}
  {\leqslant}m\left\{y\in Z_N {\ |\ }|W_N(y)|{\geqslant}a\right\}+{\varepsilon}={\varepsilon}.$$

We estimate then the contribution of $Y''_N$. Set ${\widetilde}{Y}''_N=Y''_N \cap T_Y^{-N m(Y)}({\widetilde}{Y})$, satisfying $m({\widetilde}{Y}''_N) {\geqslant}m(Y''_N)-{\varepsilon}$. Thus, $$m(|W_N|{\geqslant}a) {\leqslant}m\{y\in {\widetilde}{Y}''_N {\ |\ }|W_N(y)|{\geqslant}a\}+2{\varepsilon}.$$ On ${\widetilde}{Y}''_N$, $|\nu(y,N)|{\geqslant}N_1$, whence $\frac{1}{|\nu(y,N)|^b}\left|S^Y_{\nu(y,N)}f_Y \left(T_Y^{N
m(Y)}y\right)\right| {\leqslant}{\varepsilon}$. Thus, $|W_N(y)|{\leqslant}{\varepsilon}\frac{|\nu(y,N)|^b}{B_N}={\varepsilon}\left(\frac{|\nu(y,N)|}{B_N^{1/b}}\right)^b$. Consequently, $$m(|W_N|{\geqslant}a) {\leqslant}m\left (\frac{|\nu(y,N)|}{B_N^{1/b}} {\geqslant}\left(\frac{a}{{\varepsilon}}\right)^{1/b} \right) +2{\varepsilon}.$$ Thus, if ${\varepsilon}$ is small enough, and $N$ large enough, yields that $m(|W_N|{\geqslant}a) {\leqslant}3{\varepsilon}$.

The three facts we have just proved imply that $\frac{S_N
f(y,i)}{B_N}-\frac{S_{Nm(Y)}^Y f_Y(y)}{B_N} \to 0$ in distribution on $X$. As $\frac{S_{Nm(Y)}^Y f_Y(y)}{B_N} \to Z$ in distribution on $X$, by , this concludes the proof.

Multiple decorrelations and $L^p$-boundedness {#appendice:pene}
=============================================

The following theorem has been useful in this paper:

\[thm\_borne\_Lp\_pene\] Let $F: \omega \to 4\omega$ on the circle $S^1$. Then, for every $p \in [1,\infty)$, there exists a constant $K_p$ such that, for every $n\in {\mathbb{N}}$, for every $f_0,\ldots,f_{n-1}:S^1 \to {\mathbb{R}}$ bounded by $1$, of zero average and $1$-Lipschitz, $${\left\| \sum_{k=0}^{n-1} f_k \circ F^k  \right\|}_p {\leqslant}K_p \sqrt{n}.$$

This result has essentially been proved by Françoise Pène, in a much broader context. Her proof depends on a property of multiple decorrelations, which is implied by the spectral gap of the transfer operator:

Let ${\left\| f \right\|}$ be the Lipschitz norm of the function $f$ on the circle $S^1$. Then, for every $m,m'\in {\mathbb{N}}$, there exist $C>0$ and $\delta<1$ such that, for every $N\in {\mathbb{N}}$, for every increasing sequences $(k_1,\ldots,k_m)$ and $(l_1,\ldots,l_{m'})$, for every Lipschitz functions $G_1,\ldots,G_m,H_1,\ldots, H_{m'}$, $$\label{decorr_mult}
  \left|\operatorname{Cov}\left(\prod_{i=1}^m G_i \circ F^{k_i}, \prod_{j=1}^{m'}
    H_j\circ F^{N+l_j}\right)\right|
  {\leqslant}C \left(\prod_{i=1}^m {\left\| G_i \right\|}\right) \left(\prod_{j=1}^{m'}
    {\left\| H_j \right\|} \right) \delta^{N-k_m}.$$

Here $\operatorname{Cov}(u,v)=\int uv -\int u \int v$.

Let ${\widehat}{F}$ be the transfer operator associated to $F$, and acting on Lipschitz functions. It is known that it admits a spectral gap and that its iterates are bounded, i.e. there exist constants $M>0$ and $\delta<1$ such that $\bigl\|{\widehat}{F}^n
f\bigr\|{\leqslant}M {\left\| f \right\|}$, and $\bigl\|{\widehat}{F}^n f\bigr\|{\leqslant}M\delta^n {\left\| f \right\|}$ if $\int f=0$.

We can assume that $N{\geqslant}k_m$ (otherwise, $\delta^{N-k_m}{\geqslant}1$, and the inequality becomes trivial). Then, writing ${\varphi}=\prod_{i=1}^m G_i \circ F^{k_i}$ and $\psi=\prod_{j=1}^{m'} H_j\circ F^{l_j}$, we get $$\begin{aligned}
  \left|\operatorname{Cov}({\varphi},\psi\circ F^N)\right|
  &
  =\left|
  \int \left({\varphi}-\int {\varphi}\right) \psi\circ F^N\right|
  =\left|\int {\widehat}{F}^N \left({\varphi}-\int {\varphi}\right) \psi \right|
  \\&
  {\leqslant}{\left\| {\widehat}{F}^N \left({\varphi}-\int {\varphi}\right) \right\|} {\left\| \psi \right\|}_\infty.
  \end{aligned}$$ But $$\begin{aligned}
  {\widehat}{F}^N({\varphi})&
  ={\widehat}{F}^N \left(\prod G_i^{k_i}\right)
  ={\widehat}{F}^{N-k_m} (G_m {\widehat}{F}^{k_m-k_{m-1}}(G_{m-1}
  {\widehat}{F}^{k_{m-1}-k_{m-2}}(\ldots {\widehat}{F}^{k_2-k_1} (G_1))\ldots)
  \\&
  =:{\widehat}{F}^{N-k_m}(\chi).
  \end{aligned}$$ As the iterates of ${\widehat}{F}$ are bounded on Lipschitz functions, we get a bound on the Lipschitz norm of $\chi$: ${\left\| \chi \right\|}{\leqslant}M^{m-1} \prod {\left\| G_i \right\|}$. Moreover, $\int \chi=\int {\varphi}$, whence $$\begin{aligned}
  {\left\| {\widehat}{F}^N \left({\varphi}-\int {\varphi}\right) \right\|}
  &
  ={\left\| {\widehat}{F}^{N-k_m} \left(\chi -\int \chi\right) \right\|}
  {\leqslant}M\delta^{N-k_m} {\left\| \chi-\int \chi \right\|}
  \\&
  {\leqslant}M\delta^{N-k_m} M^{m-1} \prod {\left\| G_i \right\|}.
  \end{aligned}$$

When $p$ is an even integer, Theorem \[thm\_borne\_Lp\_pene\] is then a consequence of [@pene:averaging Lemma 2.3.4]. The Hölder inequality gives the general case.

\[remarque\_pene\] The same result holds for Hölder functions instead of Lipschitz functions, with the same proof.

We will also need the following result:

\[ineg\_max\_abstraite\] Let $T$ be a measure preserving transformation on a space $X$. Let $f:X\to {\mathbb{R}}$ and $p>2$ be such that $$\exists C>0, \forall n\in {\mathbb{N}}^*, {\left\| S_n f \right\|}_p {\leqslant}C \sqrt{n}.$$ Write $M_nf(x)=\sup_{1{\leqslant}k{\leqslant}n} |S_k f(x)|$. Then there exists a constant $K$ such that $$\forall n{\geqslant}2, {\left\| M_n f \right\|}_p {\leqslant}K (\ln n)^{\frac{p-1}{p}}
  \sqrt{n}.$$

Let $n\in {\mathbb{N}}^*$. Let $k<2^n$, and write its binary decomposition $k=\sum_{j=0}^{n-1} {\varepsilon}_j 2^j$, with ${\varepsilon}_j \in
\{0,1\}$. Set $q_j=\sum_{l=j}^{n-1} {\varepsilon}_l 2^l$ (in particular, $q_0=k$ and $q_n=0$). Then $S_k f=\sum_{j=0}^{n-1}
(S_{q_j}f-S_{q_{j+1}}f)$. Consequently, the convexity inequality $(a_0+\ldots +a_{n-1})^p {\leqslant}n^{p-1} (a_0^p+\ldots+a_{n-1})^p$ gives that $$|S_k f|^p {\leqslant}n^{p-1} \sum_{j=0}^{n-1}
  |S_{q_j}f-S_{q_{j+1}}f|^p.$$ Note that $q_{j+1}$ is of the form $\lambda 2^{j+1}$ with $0{\leqslant}\lambda {\leqslant}2^{n-j-1}-1$, and $q_j$ is equal to $q_{j+1}$ or $q_{j+1}+2^j$. Thus, $$|S_k f|^p {\leqslant}n^{p-1} \sum_{j=0}^{n-1} \left(
  \sum_{\lambda=0}^{2^{n-j-1} -1} \left|S_{\lambda 2^{j+1} +2^j}f
  -S_{\lambda 2^{j+1}}f \right|^p \right).$$ The right hand term is independent of $k$, and gives a bound on $|M_{2^n -1} f|^p$. Moreover, $$\int \left|S_{\lambda 2^{j+1} +2^j}f -S_{\lambda 2^{j+1}}f
  \right|^p =\int \left|S_{2^j} f\right|^p {\leqslant}C^p \sqrt{2^j}^{p}.$$ Thus, we get $$\int |M_{2^n -1} f|^p {\leqslant}n^{p-1} \sum_{j=0}^{n-1} 2^{n-j-1} C^p 2^{pj/2}
  {\leqslant}K n^{p-1} 2^n 2^{(\frac{p}{2}-1)n}
  = K n^{p-1} \sqrt{2^n}^p.$$ For times of the form $2^n-1$, this is a bound of the form ${\left\| M_t \right\|}_p {\leqslant}K (\ln t)^{\frac{p-1}{p}} \sqrt{t}$. To get the same estimate for an arbitrary time $t$, it is sufficient to choose $n$ with $2^{n-1} {\leqslant}t<2^n$, and to note that $M_t {\leqslant}M_{2^n -1}$.

\[ineg\_max\] Let $F: \omega \to 4\omega$ on the circle $S^1$, let $\chi:S^1 \to
{\mathbb{R}}$ be a Hölder function with $0$ average, and $p>2$. Write $M_n
\chi(x)=\sup_{1{\leqslant}k{\leqslant}n} |S_k \chi(x)|$. Then there exists a constant $K$ such that $$\forall n{\geqslant}2, {\left\| M_n \chi \right\|}_p {\leqslant}K (\ln
  n)^{\frac{p-1}{p}} \sqrt{n}.$$

Theorem \[thm\_borne\_Lp\_pene\] (or rather the remark following it, for the Hölder case) shows that ${\left\| S_n \chi \right\|}_p {\leqslant}C\sqrt{n}$. Consequently, Theorem \[ineg\_max\_abstraite\] gives the conclusion.

[^1]: Département de Mathématiques et Applications, École Normale Supérieure, 45 rue d’Ulm 75005 Paris (France). e-mail `Sebastien.Gouezel@ens.fr`

[^2]: *keywords*: intermittency, countable Markov shift, central limit theorem, stable laws. *2000 Mathematics Subject Classification:* 37A50, 37C40, 60F05
