Mixing $C^r$ maps of the interval without maximal measure

TL;DR

This paper constructs a $C^r$ interval map that is topologically mixing but lacks a measure of maximal entropy, demonstrating the necessity of high smoothness for such measures.

Contribution

It provides a $C^r$ example of a mixing map without a maximal measure, highlighting the importance of smoothness in entropy measure existence.

Findings

Constructed a $C^r$ map with no maximal measure

Showed smoothness is crucial for measure existence

Computed the local entropy of the example

Abstract

We construct a $C^r$ transformation of the interval (or the torus) which is topologically mixing but has no invariant measure of maximal entropy. Whereas the assumption of $C^{\infty}$ ensures existence of maximal measures for an interval map, it shows we cannot weaken the smoothness assumption. We also compute the local entropy of the example.

Equations

$$f_r(x)={\lambda }^{-nr}\left[x_n+(y_n-x_n)s_n\left(M_n\frac{x-x_n}{y_n-x_n}\right)\right].$$

$$|f^{mp+k}(J)|\ge {\mu }^m\alpha C(a-z_0{)}^{\alpha -1}|J|.$$

$${\mu }^m\ge \frac{z_2-z_1}{a^{\prime }-z_1}=\frac{z_2-z_1}{C(a-z_0{)}^{\alpha }},$$

$$|f^{mp+k}(J)|\ge \frac{\alpha |z_2-z_1|}{|a-z_0|}|J|\ge \frac{\alpha |z_2-z_1|}{\delta }|J|\ge {\mu }_0|J|.$$

$$2\left[\frac{{\lambda }^n}{2n^2}\right]-1\ge \frac{{\lambda }^n}{2n^2}+\left(\frac{{\lambda }^n}{2n^2}-3\right)\ge \frac{{\lambda }^n}{2n^2}$$

$$f_r(x)={\lambda }^{-nr}\left[x_n+(y_n-x_n)s_n\left(M_n\frac{x-x_n}{y_n-x_n}\right)\right].$$

$$f_r^{(k)}(x)=\frac{{\lambda }^{-nr}{M}_n^k}{(y_n-x_n{)}^{k-1}}{s}_n^{(k)}\left(M_n\frac{x-x_n}{y_n-x_n}\right)\text{mbox}forx\in [x_n,y_n],$$

$$|f_r^{(k)}(x)|\le {\lambda }^{-n(r-k)}{n}^{-2}{2}^{k-1}{A}_k.$$

$$|f_r^{(k)}(x)|\to 0\text{mbox}whenx\to 1,x\in \bigcup _{n\ge 1}[x_n,y_n].$$

$$w_n=y_{n+1}+\frac{n+2}{2n(n+1{)}^2{M}_{n+1}{k}_{n+1}}.$$

$$f_r(w_n)={\lambda }^{-(n+1)r}{y}_{n+1}+{\lambda }^{-(n+1)r}\frac{n+2}{2n(n+1{)}^2}={\lambda }^{-(n+1)r}\left(1+\frac{1}{n}\right),$$

$$f_r^{\prime }(x_n)=2{\lambda }^{-(n-1)r},f_r^{\prime }(w_n)=2{\lambda }^{-nr},\text{mbox}and{f}_r^{(k)}(x_n)=f_r^{(k)}(w_n)=0\text{mbox}fork\ge 2.$$

$$h_n={\lambda }^{-nr}\left(x_n-{\lambda }^{-r}{y}_{n+1}-{\lambda }^{-r}\frac{n+2}{2n(n+1{)}^2}\right)\le 2{\lambda }^{-nr}.$$

$$\frac{n+2}{2n(n+1{)}^2}=\frac{3}{8}\text{mbox}forn=1$$

$$\frac{n+2}{2n(n+1{)}^2}=\frac{n^2+2n}{2n^2(n+1{)}^2}\le \frac{1}{2n^2}\le \frac{1}{8}\text{mbox}forn\ge 2.$$

$$h_n\ge {\lambda }^{-nr}(1-2{\lambda }^{-r})\ge \frac{6}{7}{\lambda }^{-nr}.$$

$$l_n$$

$$\frac{n+2}{2n(n+1{)}^2}\ge \frac{1}{2(n+1{)}^2}$$

$$\frac{6}{7}{\lambda }^{-nr}\le h_n\le 2{\lambda }^{-nr}\text{mbox}and\frac{1}{4(n+1{)}^2}\le l_n\le \frac{3}{8}.$$

$${\phi }_n(x)=h_n^{-1}[f_r(w_n+l_{n}x)-f_r(w_n)].$$

$${\phi }_n^{\prime }(1)=2h_n^{-1}{l}_n{\lambda }^{-(n-1)r},{\phi }_n^{\prime }(0)=2h_n^{-1}{l}_n{\lambda }^{-nr},$$

$${\phi }_n^{(k)}(0)={\phi }_n^{(k)}(1)=0\text{mbox}fork\ge 2,$$

$$\forall k\ge 1,\exists B_k,\forall n\ge 1,\Vert {\phi }_n^{(k)}{\Vert }_{\infty }\le B_k$$

$$\forall x\in [0,1],\frac{2}{3}{\phi }_n^{\prime }(0)\le {\phi }_n^{\prime }(x)\le {\lambda }^r.$$

$$f_r^{(k)}(x)=h_n{l}_n^{-k}{\phi }_n^{(k)}\left(\frac{x-w_n}{l_n}\right)\le {\lambda }^{-nr}(n+1{)}^{2k}{2}^{2k+1}{B}_k\text{mbox}for{w}_n\le x\le x_n,$$

$$f_r^{(k)}(x)\to 0\text{mbox}whenx\to 1,x\in \bigcup _{n\ge 1}[y_{n+1},x_n].$$

$$h_n{l}_n^{-1}\le \frac{8(n+1{)}^2}{{\lambda }^{nr}}\le 1\text{mbox}byLemma\text{ref}lem:lambda-n(i).$$

$$(f_r^{n+1}{)}^{\prime }(x)\ge \min \{M_n{k}_n,M_n/2\}\ge 2\text{mbox}forallx\in \left[y_n-\frac{(y_n-x_n)(1-\delta )}{M_n},y_n\right].$$

$$f_r^{n+1}\left(y_n-\frac{(y_n-x_n)(1-\delta )}{M_n}\right)<x_n.$$

$${\left|\frac{M_{n}t}{y_n-x_n}\right|}^2\le \frac{y_n-z_n}{C(y_n-x_n)}\le \frac{1}{C{M}_n{\lambda }^r}\le {\delta }^2.$$

Full text

Introduction

We are interested in topological dynamical systems on the interval, that is systems of the form $f\colon I\to I$ where $f$ is at least continuous and $I$ is a compact interval. One can wonder whether such a system has maximal measures, i.e. invariant measures of maximal entropy.

Hofbauer [15], [16] studied piecewise monotone maps, i.e. interval maps with a finite number of monotone continuous pieces (the whole map is not necessarily continuous). He proved in this case that the system admits a non zero finite number of maximal measures if its topological entropy is positive, and transitivity implies intrinsic ergodicity, that is existence of a unique maximal measure. For this purpose, he built a Markov chain which is isomorphic modulo “small sets” with the first system. Buzzi [9] generalized the construction of the Markov extension to any continuous interval map. He showed that the same conclusions as in the piecewise monotone case hold for $C^{\infty}$ maps.

One can wonder if these results are still valid under a weaker regularity assumption, at least in the mixing case. Actually, if a topological dynamical system is expansive and satisfies the specification property, then it has a unique maximal measure (Bowen [6], [7]). Specification is a strong property on periodic points, which must closely follow arbitrary pieces of orbits (see e.g. [11] for more details). In the particular case of continuous interval maps, the system is never expansive, but the mixing property implies the specification property (this result is due to Blokh [4], see [10] for the proof). More recently, Ruelle [20] worked on positively expansive maps satisfying specification.

In fact, transitivity is not much weaker than mixing since for any transitive continuous interval map $f\colon I\to I$ either the map is mixing or there exist two subintervals $J,K$ such that $J\cup K=I$, $J\cap K$ is reduced to a single point, $f(J)=K$, $f(K)=J$ and $f^{2}|_{J},f^{2}|_{K}$ are mixing [2, p59]. We also recall that the topological entropy of any transitive continuous interval map is positive (it is greater than or equal to $\frac{\log 2}{2}$ [3], see [1] for the proof) and, if in addition the map is Lipschitz, it is finite (this classical result appears in the proof of Proposition 2.3).

Gurevich and Zargaryan [12] built a continuous interval map with finite entropy which is transitive (in fact mixing) and has no maximal measure. This map has countably many intervals of monotonicity. The authors asked is this example can be made smooth on the whole interval. Actually it cannot: the end points [math] and $1$ are fixed points and the map is not monotone in a neighbourhood of [math] and $1$; on the other hand it is not hard to see that a $C^{1}$ transitive interval map must have non zero derivatives at fixed points, hence it is monotone near these points.

In [9, Appendix A] Buzzi built a $C^{r}$ interval map which has no transitive component of maximal entropy, hence it has no maximal measure. He also sketched without details the construction of a $C^{r}$ interval map with positive entropy which admits no maximal measure and which is transitive after restriction to its unique transitive component (which may be a Cantor set). His proof of non existence of any maximal measure relies on a result of Salama [21] whose proof turned out to be false (see Theorem 2.3 and Errata in [22]). Nevertheless Buzzi’s proof can be modified – using extension graphs instead of subgraphs, as we do in Subsection 2.3 – so as to be based on another theorem of Salama.

The aim of this article is to build for any integer $r\geq 1$ a $C^{r}$ mixing interval map which has no maximal measure. Transitivity instead of mixing would be enough, yet it is not more difficult to prove directly the mixing property. This family of examples is inspired by Buzzi’s [9], the important addition is that the system is transitive on the whole interval. Non existence of maximal measure prevents the metric entropy from being an upper semi-continuous map on the set of invariant measures. This is to be put in parallel with the result of Misiurewicz and Szlenk [17], which shows that the topological entropy, considered as a map on the set of $C^{r}$ interval transformations, is not upper semi-continuous for the $C^{r}$ topology.

In Section 1, we define for any $r\geq 1$ a $C^{r}$ transformation of the interval $[0,4]$ which is topologically mixing. In fact it is $C^{\infty}$ everywhere except at one point. The map $f_{r}$ is made of a countable number of monotone pieces and is Markov with respect to a countable partition. Moreover, it can also be seen as a $C^{r}$ transformation of the torus by identifying the two end points. In the next section, we study the Markov chain associated with $f_{r}$ and we conclude it has no maximal measure, thanks to results of Gurevič [13], [14] and Salama [22]. As there is an isomorphism modulo countable sets between the two systems, the interval map has no maximal measure either. In Section 3, we compute the local entropy of our examples. Buzzi [9] showed that this quantity bounds the defect in upper-semicontinuity and he gave an estimate of it depending on the differential order and the spectral radius of the derivative. Our examples show these bounds are sharp since the two are realized. Moreover, it also equals the topological entropy. It may be of some importance: we conjecture that the Markov extension admits a maximal measure when the topological entropy is strictly greater than the local entropy.

In addition to the problem of existence of maximal measure, one can ask the question of uniqueness of such a measure. Recently, Buzzi [8] proved that, if the interval transformation is $C^{1+\alpha}$ (i.e. the map is $C^{1}$ and its derivative is $\alpha$-Hölder), then there is no measure of positive entropy on the non Markov part of the system. Since a transitive Markov chain admits at most one maximal measure, a transitive $C^{1+\alpha}$ transformation has a unique maximal measure if it exists. For transitive non smooth interval maps we still do not know if several maximal measures can exist. It would imply that the topological entropy of the critical points would be equal to the topological entropy of the whole map.

I am indebted to Jérôme Buzzi for many discussions which have led to the ideas of this paper.

1 Construction and proof of mixing property

In this section, we construct a family of $C^{r}$ maps $f_{r}\colon I\to I$ for $r\geq 1$, where $I=[0,4]$. We first give a general idea of their aspect (see Figure 2). Then we give some lemmas which will be useful to prove the mixing property. Finally, we define $f_{r}$ by pieces and check some properties at each step. At the end of the section, the maps $f_{r}$ are totally defined and are proved to be mixing.

1.1 General description

Let $\lambda\geq 14$ ($\log\lambda$ will be the entropy of $f_{r}$). The map $f_{r}$ is increasing on $[0,1/2]$ and decreasing on $[1/2,1]$. Moreover, $f_{r}(x)=\lambda^{r}x$ for $0\leq x\leq\frac{5}{2}\lambda^{-r},\;f_{r}(0)=f_{r}(1)=0,\;f_{r}(1/2)=4$.

Let $x_{n}=1+\frac{1}{n}$ and $y_{n}=x_{n}+\frac{1}{2n^{2}}$ for every $n\geq 1$, and let $M_{n}$ be a sequence of odd numbers with $(\log M_{n})/n\longrightarrow\log\lambda$. We choose a family of $C^{\infty}$ maps $s_{n}\colon[0,M_{n}]\to[-1,1]$ such that $s_{n}$ is nearly 2-periodic and has $M_{n}$ oscillations; $s_{n}(0)=0$ and $s_{n}(M_{n})=1$ (see Figure 1).

Then we define $f_{r}$ on $[x_{n},y_{n}]$ by

[TABLE]

In this way, $f_{r}(x_{n})=\lambda^{-nr}x_{n}$, $f_{r}(y_{n})=\lambda^{-nr}y_{n}$ and $f_{r}$ oscillates $M_{n}$ times between $x_{n}$ and $y_{n}$ like $s_{n}$. It is worth mentioning that $x_{n}$ and $y_{n}$ are periodic points with period $n+1$, because $f_{r}$ is linear of slope $\lambda^{r}$ on $[0,y_{1}\lambda^{-r}]$.

On $[y_{n+1},x_{n}]$, $f_{r}$ is increasing.

Finally, $f_{r}$ is increasing on $[y_{1},4]$, with $f_{r}(4)=4$. Figure 2 gives a general idea of $f_{r}$.

The map $f_{r}$ will be built to be mixing and $C^{r}$ on $[0,4]$, and $\|f_{r}^{\prime}\|_{\infty}=\lambda^{r}$. Furthermore, the minimum of $s_{n}$ will be chosen such that $f_{r}(x)=\lambda^{-nr}y_{n+1}$ if $x$ is a local minimum of $f_{r}$ in $]x_{n},y_{n}[$ in order to obtain a Markov map.

This brief description is sufficient to build the Markov chain associated with $f_{r}$ and prove that $f_{r}$ has no maximal measure, which is done in Section 2. The rest of this section, which may be skipped a first reading, is devoted to prove that maps satisfying these properties do exist.

1.2 Method for the proof of mixing property

We recall the definition of mixing for a topological dynamical system.

**Definition 1.1  **Let $T\colon X\to X$ be a continuous map where $X$ is a compact metric space. The system is (topologically) mixing if for every non empty open sets $U$ and $V$, there exists $N\geq 0$ such that for every $n\geq N,\>T^{-n}U\cap V\not=\emptyset$.

In our case, we will show that for any non degenerate subinterval $J\subset I$, there exists $n\geq 0$ such that $f_{r}^{n}(J)=I$. So $f_{r}^{k}(J)=I$ for every $k\geq n$ and the system is mixing. For this, we will show that, for some constant $\mu_{0}>1$, any non degenerate subinterval $J$ satisfies one of the two following conditions:

(1)

$\exists k\geq 0$ such that $|f_{r}^{k}(J)|\geq\mu_{0}|J|$, where $|J|$ denotes the length of $J$,

or

(2)

$\exists k\geq 0,\exists n\geq 1$ such that either $0\in f_{r}^{k}(J)$ or $\mbox{\rm Int}\left(f_{r}^{k}(J)\right)$ contains $x_{n}$ or $y_{n}$.

Then it will be enough to show that for any non degenerate subinterval $J$ containing [math] or $x_{n}$ or $y_{n}$, there is a $k$ such that $f_{r}^{k}(J)=I$.

Lemma 1.2 says that an interval near a suitable extremum satisfies (1) or (2). Lemma 1.2, which is trivial, says how an interval containing a repelling periodic point behaves.

*Lemma 1.2  ** * Let $f\colon I\to I$ where $I$ is a compact interval and let $z_{0}$ be a periodic point of period $p$. Assume $(f^{p})^{\prime}(x)\geq\mu>1$ for every $x\in[z_{0},z_{1}]$. Then for every $x>z_{0}$ there exists $n\geq 0$ such that $f^{n}([z_{0},x])\supset[z_{0},z_{1}]$.

*Lemma 1.3  ** * Let $f\colon I\to I$ be a $C^{r}$ map where $I$ is a compact interval and let $z_{0}$ be an extremum such that $z_{1}=f^{k}(z_{0})$ is a periodic point of period $p$. Suppose $f^{k}(x)=z_{1}+C(x-z_{0})^{\alpha}$ for $|x-z_{0}|\leq\delta$, with $C\not=0$ and $\alpha$ an even integer. Let $z_{2}=f^{k}(z_{0}-\delta)=f^{k}(z_{0}+\delta)$. Suppose $\displaystyle f^{p}|_{[z_{1},z_{2}]}$ is linear of slope $\mu>1$, and $\frac{\alpha|z_{2}-z_{1}|}{\delta}\geq\mu_{0}$. Then for every non degenerate interval $J\subset[z_{0}-\delta,z_{0}+\delta]$, there exists $n\geq 0$ for which one of the following cases holds:

1. (i)

$|f^{n}(J)|\geq\mu_{0}|J|$. 2. (ii)

$z_{2}\in\mbox{\rm Int}\left(f^{n}(J)\right)$.

Proof:

Let $J=[a,b]$ be an interval in $[z_{0}-\delta,z_{0}+\delta]$ with $a<b$. If $z_{0}\in J$ then $f^{k}(J)=[z_{1},y]$ for some $y$. The hypotheses imply that $f^{p}(z_{2})>z_{2}$, hence $z_{2}$ cannot be an end point of $I$ and one can choose $1<\mu^{\prime}<\mu$ and $z_{3}>z_{2}$ such that $(f^{p})^{\prime}(x)>\mu^{\prime}$ for all $x\in[z_{1},z_{3}]$. According to Lemma 1.2 there exists $n$ such that $f^{n}(J)\supset[z_{1},z_{3}]$, thus $z_{2}\in\mbox{\rm Int}\left(f^{n}(J)\right)$, which is (ii).

Now assume that $z_{0}\not\in J$. We restrict to the case $C>0$ and $z_{0}<a<b\leq z_{0}+\delta$. Let $J^{\prime}=f^{k}(J)=[a^{\prime},b^{\prime}]\subset]z_{1},z_{2}]$ and $g=f^{p}$. The point $z_{1}$ is fixed for $g$ and $g$ is linear of slope $\mu>1$ on $[z_{1},z_{2}]$, so the map $g$ can be iterated on $J^{\prime}$ as long as $g^{m}(b^{\prime})\leq z_{2}$. Let $m$ be the first integer satisfying $g^{m}(b^{\prime})>z_{2}$. Then there are two cases:

• •

$g^{m}(a^{\prime})<z_{2}<g^{m}(b^{\prime})$, which implies (ii) with $n=mp+k$.

• •

$z_{2}\leq g^{m}(a^{\prime})<g^{m}(b^{\prime})$.

In this case, as $(f^{k})^{\prime}$ is positive and increasing on $[z_{0},z_{0}+\delta]$ one gets $|J^{\prime}|\geq\alpha C(a-z_{0})^{\alpha-1}|J|$ and

[TABLE]

But $g^{m}(a^{\prime})-z_{1}=\mu^{m}(a^{\prime}-z_{1})\geq z_{2}-z_{1}$, so

[TABLE]

and

[TABLE]

We add a lemma which will be useful for some estimates.

*Lemma 1.4  ** * Let $\lambda\geq 8$ and $[\cdot]$ refer to the entire part of a number. Then for all $n\geq 1$:

1. (i)

$\displaystyle\frac{\lambda^{n}}{n^{2}}\geq\lambda$. 2. (ii)

$\displaystyle\frac{\lambda^{n}}{2n^{2}}\leq 2\left[\frac{\lambda^{n}}{2n^{2}}\right]-1\leq\frac{\lambda^{n}}{n^{2}}$. 3. (iii)

$\displaystyle 2\left[\frac{\lambda^{n}}{2n^{2}}\right]-1\geq\lambda-3$.

Proof:

(i) is obtained by studying the function $x\mapsto\lambda^{x-1}-x^{2}$.

For the first inequality of (ii), we write

[TABLE]

thanks to (i). The second inequality is obvious.

(iii) comes from $\displaystyle 2\left[\frac{\lambda^{n}}{2n^{2}}\right]-1\geq\frac{\lambda^{n}}{n^{2}}-3$ and from (i).

1.3 Construction of $f_{r}$ on $[1,y_{1}]$

Recall that $\lambda\geq 14$, $f_{r}(1)=0$, $x_{n}=1+\frac{1}{n}$ and $y_{n}=x_{n}+\frac{1}{2n^{2}}\,$ for $n\geq 1$; in particular $y_{1}=\frac{5}{2}$. In this subsection, we define $f_{r}$ on $[1,y_{1}]$ with more details. For this purpose, we define $f_{r}$ on each $[x_{n},y_{n}]$ and then on each $[y_{n+1},x_{n}]$. At each step, we check that the various pieces can be glued together in a $C^{\infty}$ way and $|f_{r}^{\prime}(x)|\leq\lambda^{r}$ for $x\in[1,y_{1}]$. In addition, we show that $f_{r}$ is $C^{r}$ on the right of $1$. Finally, we focus on the mixing property. The map $f_{r}$ is not totally defined yet, but at this stage we only need to know that $f_{r}(x)=\lambda^{r}x$ for $0\leq x\leq\frac{5}{2}\lambda^{-r}$ and $f_{r}(\frac{1}{2})=4$ in order to prove that any non degenerate subinterval of $[1,y_{1}]$ satisfies (1) or (2) with $\mu_{0}=\frac{4}{3}$. Then we show that for an open interval $J$ containing $x_{n}$ or $y_{n}$ there is a $k$ satisfying $f_{r}^{k}(J)=[0,4]$.

1.3.1 On the subintervals $[x_{n},y_{n}]$

Set \displaystyle M_{n}=2\left[\frac{\lambda^{n}}{2n^{2}}\right]-1\mbox{ (where [\cdot] denotes the entire part)}, $\displaystyle m_{n}=1-\frac{1}{(n+1)^{2}}$, $\delta=\lambda^{-r}$, $C=\frac{1}{4\delta^{2}}$ and $k_{n}=\frac{2\lambda^{r}}{M_{n}}.$

First, we choose a sequence of $C^{\infty}$ functions $s_{n}\colon[0,M_{n}]\to[-m_{n},1]$ satisfying:

(3)

$s_{n}(0)=0,\>s_{n}(M_{n})=1$, $s_{n}$ is increasing on each $[2k,2k+1]\>(0\leq k\leq(M_{n}-1)/2)$, $s_{n}$ is decreasing on each $[2k+1,2k+2]\>(0\leq k\leq(M_{n}-3)/2)$.

(4)

$s_{n}(x)=1-C(x-a)^{2}$ for $|x-a|\leq\delta$ if $a$ is a local maximum of $s_{n},\,a\not=M_{n}$, and $s_{n}(x)=-m_{n}+C(x-b)^{2}$ for $|x-b|\leq\delta$ if $b$ is a local minimum of $s_{n},\;b\not=0$.

(5)

$s_{n}(x)=k_{n}(x-M_{n})+1$ for $M_{n}-\delta\leq x\leq M_{n}$.

(6)

$s_{n}(x)=k_{n}x$ for $x\in[0,\delta]$.

(7)

$\forall k\geq 1,\>\exists A_{k},\>\forall n\geq 1,\>\|s_{n}^{(k)}\|_{\infty}\leq A_{k}$.

(8)

$\|s_{n}^{\prime}\|_{\infty}\leq\lambda^{r}$ and $|s_{n}^{\prime}(x)|\geq\min\{1/2,k_{n}\}$ if $|x-d|\geq\delta$ for all local extrema $d\in]0,M_{n}[$.

Property (7) can be fulfilled because $m_{n}$ and $k_{n}$ are bounded ($3/4\leq m_{n}\leq 1$, $k_{n}\leq\lambda^{r}$) and the maps $s_{n}$ have a 2-periodic looking.

If $d$ is a local extremum in $]0,M_{n}[$, then $|s_{n}(d-\delta)-s_{n}(d)|=|s_{n}(d+\delta)-s_{n}(d)|=1/4$; moreover $|s_{n}(\delta)-s_{n}(0)|\leq 1/4$ and $|s_{n}(M_{n}-\delta)-s_{n}(M_{n})|\leq 1/4$. Thus if $d$ and $d^{\prime}$ are two successive extrema in $[0,M_{n}]$ the absolute value of the average slope between $d+\delta$ and $d^{\prime}-\delta$ is at least $\displaystyle\frac{m_{n}+1/2}{1-2\delta}>\frac{1}{2}$ and is less that $2$. Since $|s_{n}^{\prime}(d+\delta)|=|s_{n}^{\prime}(d-\delta)|=\frac{\lambda^{r}}{2}$ for any extremum $d\in]0,M_{n}[$, Property (8) can be fulfilled.

Secondly, recall that $f_{r}$ is defined for $x\in[x_{n},y_{n}]$ by

[TABLE]

Now, we look at the $C^{r}$ character of $f_{r}$ near $1$. The definition of $f_{r}$ gives

[TABLE]

where $f_{r}^{(k)}(x_{n})$ and $f_{r}^{(k)}(y_{n})$ are to be understood as left (resp. right) derivatives at this stage.

Since $M_{n}\leq\frac{\lambda^{n}}{n^{2}}$, Property (7) leads to

[TABLE]

One has $A_{1}=\lambda^{r}$ by (8), thus $\,|f^{\prime}(x)|\leq\lambda^{r}$. Moreover, for $0\leq k\leq r$,

[TABLE]

Notice that the main factor in this estimate is $\lambda^{-n(r-k)}$. If $k>r$, the $k$-th derivative $f_{r}^{(k)}$ does not tend to zero any longer and it can be shown that $f_{r}$ cannot be $C^{r+1}$ at point $1$.

As $f_{r}(x)=\lambda^{r}x$ for $x\in[0,y_{1}\lambda^{-r}]$, the $(n+1)$-th iterate of the map on $[x_{n},y_{n}]$ is given by $f_{r}^{n+1}(x)=\lambda^{nr}f_{r}(x)$.

Notice that $m_{n}$ is chosen such that $\min\{f_{r}^{n+1}(x)\colon x\in[x_{n},y_{n}]\}=y_{n+1}$. Moreover $f_{r}^{n+1}(x_{n})=x_{n}$ and $f_{r}^{n+1}(y_{n})=y_{n}$.

We sum up the previous results in two lemmas, the first one is about derivatives and the second summarizes the behaviour of $f_{r}$ on $[x_{n},y_{n}]$.

**Lemma 1.5  ** **

• •

$|f_{r}^{\prime}(x)|\leq\lambda^{r}$* for $x\in[x_{n},y_{n}]$.*

• •

\displaystyle\lim_{\mbox{\tiny\begin{tabular}[]{c}{\scriptsizex\to 1}\\ x\in\cup_{n\geq 1}[x_{n},y_{n}]\end{tabular}}}f_{r}^{(k)}(x)=0\mbox{ for }\,0\leq k\leq r.**

*Lemma 1.6  ** * Let $\displaystyle t_{i}^{n}=x_{n}+\frac{i(y_{n}-x_{n})}{M_{n}}$ for $i=0,\cdots,M_{n}$. Then

• •

$f_{r}$* is monotone on $[t_{i-1}^{n},t_{i}^{n}]$, $1\leq i\leq M_{n}$.*

• •

$f_{r}(t_{i}^{n})=\lambda^{-nr}y_{n+1}$* if $i$ is even, $i\not=0$, and $f_{r}(x_{n})=\lambda^{-nr}x_{n}$.*

• •

$f_{r}(t_{i}^{n})=\lambda^{-nr}y_{n}$* if $i$ is odd.*

1.3.2 On the subintervals $[y_{n+1},x_{n}]$

We define

[TABLE]

We have $w_{n}\in]y_{n+1},x_{n}[$. On $[y_{n+1},w_{n}]$, we define $f_{r}$ to be affine of slope $\lambda^{-(n+1)r}M_{n+1}k_{n+1}$ (recall that $f_{r}(y_{n+1})=\lambda^{-(n+1)r}y_{n+1}$ is already defined). Because of this definition $f_{r}$ is affine (thus $C^{\infty}$) in a neighbourhood of $y_{n+1}$. Moreover

[TABLE]

so $f_{r}(w_{n})=\lambda^{-(n+1)r}x_{n}$ and $f_{r}^{n+2}(w_{n})=x_{n}$. As we are going to extend $f_{r}$ in a $C^{\infty}$ way on $[w_{n},x_{n}]$, we will have

[TABLE]

Set $h_{n}=f_{r}(x_{n})-f_{r}(w_{n})$ and $l_{n}=x_{n}-w_{n}$. We compute upper and lower bounds for $h_{n}$ and $l_{n}$. First

[TABLE]

We have

[TABLE]

and

[TABLE]

Since $x_{n}\geq 1,y_{n+1}\leq y_{2}=\frac{13}{8}$ and $\frac{n+2}{2n(n+1)^{2}}\leq\frac{3}{8}$ for all $n\geq 1$ one gets

[TABLE]

For $l_{n}$ one has

[TABLE]

As $\frac{n+2}{2n(n+1)^{2}}\leq\frac{3}{8}$, one gets $l_{n}\leq\frac{3}{8}$ too. Moreover

[TABLE]

and $\frac{1}{2\lambda^{r}}\leq\frac{1}{2}$ thus $l_{n}\geq\frac{1}{4(n+1)^{2}}$. Finally we obtain the inequalities

[TABLE]

We normalize $f_{r}$ on $[w_{n},x_{n}]$ as follows: we define $\varphi_{n}\colon[0,1]\to[0,1]$ by

[TABLE]

The aim of this normalization is to check that the sequence $\varphi_{n}$ can be chosen with uniformly bounded $k$-th derivatives then to come back to $f_{r}$ and show that $f_{r}$ is $C^{r}$ at the right of $1$. We want to have

[TABLE]

and

[TABLE]

thus $\varphi_{n}^{\prime}(1)\leq\frac{7}{8}\lambda^{r}$ and $\varphi_{n}^{\prime}(0)\leq\frac{7}{8}$. Consequently, it is possible to build a sequence of functions $\varphi_{n}$ satisfying these conditions and the following additional conditions:

[TABLE]

and

[TABLE]

By definition of $\varphi_{n}$, the derivatives of $f_{r}$ are given by

[TABLE]

hence for every $k\geq 0$

[TABLE]

Moreover, $\frac{4}{3}\lambda^{-nr}\leq f_{r}^{\prime}(x)\leq\lambda^{r}h_{n}l_{n}^{-1}$ for every $x\in[w_{n},x_{n}]$ and

[TABLE]

The next lemma recalls the behaviour of $f_{r}$ on $[y_{n+1},x_{n}]$.

**Lemma 1.7  ** **

• •

$\frac{4}{3}\lambda^{-nr}\leq f_{r}^{\prime}(x)\leq\lambda^{r}$* for $x\in[y_{n+1},x_{n}]$.*

• •

$f_{r}(w_{n})=\lambda^{-(n+1)r}x_{n}$.

• •

\displaystyle\lim_{\mbox{\tiny\begin{tabular}[]{c}{\scriptsizex\to 1}\\ x\in\bigcup_{n\geq 1}[y_{n+1},x_{n}]\end{tabular}}}f_{r}^{(k)}(x)=0\mbox{ for }\,0\leq k\leq r.**

1.3.3 Beginning of the proof of the mixing property

We show that any non degenerate subinterval $J\subset[1,y_{1}]$ satisfies (1) or (2) with $\mu_{0}=\frac{4}{3}$. It is sufficient to consider $J\subset[x_{n},y_{n}]$ or $J\subset[y_{n+1},x_{n}]$.

First, we look at $[y_{n+1},x_{n}]$. For $x\in[y_{n+1},x_{n}]$, $f_{r}^{n+1}(x)=\lambda^{nr}f_{r}(x)$ and the derivative of $f_{r}$ satisfies $f_{r}^{\prime}(x)\geq\frac{4}{3}\lambda^{-nr}$ by Lemma 1.3.2, so $|f_{r}^{n+1}(J)|\geq\frac{4}{3}|J|$ if $J\subset[y_{n+1},x_{n}]$.

Now, we focus on $[x_{n},y_{n}]$. According to Property (8), $s_{n}^{\prime}(x)\geq\min\{k_{n},1/2\}$ for all $x\in[M_{n}-1+\delta,M_{n}]$ thus

[TABLE]

Because of Property (4), $s_{n}(M_{n}-1+\delta)=-m_{n}+1/4<0$, thus

[TABLE]

Let $\displaystyle t_{n}=\frac{y_{n}-x_{n}}{\lambda^{r}M_{n}}$, then according to Lemma 1.2, there exists an integer $\alpha$ such that $f_{r}^{(n+1)\alpha}([y_{n}-t_{n},y_{n}])\supset[x_{n},y_{n}]$, so there exists $z\in[y_{n}-t_{n},y_{n}[$ with $f_{r}^{(n+1)\alpha}(z)=x_{n}$. Because of the choice of $t_{n}$ and Property (5), $f_{r}^{n+1}$ is affine of slope $k_{n}M_{n}=2\lambda^{r}$ on $[y_{n}-t_{n},y_{n}]$. Let $k\geq 0$ be the maximal integer $i$ such that $\lambda^{ri}(y_{n}-z)\leq t_{n}$. Then $z_{n}=y_{n}-\lambda^{rk}(y_{n}-z)$ belongs to $\left[y_{n}-t_{n},y_{n}-\frac{t_{n}}{2\lambda^{r}}\right]$ and $f_{r}^{(n+1)\alpha_{n}}(z_{n})=x_{n}$ if $\alpha_{n}=\alpha+k$.

Set $\displaystyle\delta_{n}=\sqrt{\frac{(y_{n}-z_{n})(y_{n}-x_{n})}{CM_{n}^{2}}}$, and let $a$ be a local maximum of $f_{r}$ on $]x_{n},y_{n}[$. If $|t|\leq\delta_{n}$, then

[TABLE]

Now we check the hypotheses of Lemma 1.2 for the extremum $a$:

• •

$f_{r}^{n+1}(a)=y_{n}$ and $f_{r}^{n+1}(y_{n})=y_{n}$.

• •

$\displaystyle f_{r}^{n+1}(a+t)=y_{n}-\frac{C(M_{n}t)^{2}}{y_{n}-x_{n}}$ if $|t|\leq\delta_{n}$ (because of Property (4)).

• •

$f_{r}^{n+1}(a-\delta_{n})=f_{r}^{n+1}(a+\delta_{n})=z_{n}$.

• •

$f_{r}^{n+1}$ is linear on $[z_{n},y_{n}]$, with a slope $k_{n}M_{n}\geq 2$.

• •

$\displaystyle\frac{2|z_{n}-y_{n}|}{\delta_{n}}=2\sqrt{\frac{CM_{n}^{2}(y_{n}-z_{n})}{y_{n}-x_{n}}}\geq 2\sqrt{\frac{CM_{n}^{2}t_{n}}{2\lambda^{r}(y_{n}-x_{n})}}=2\sqrt{\frac{M_{n}}{8}}$

and the last quantity is greater than $2$ because $M_{n}\geq\lambda-3$ by Lemma 1.2 (iii) and $\lambda\geq 14$.

Consequently, we can apply Lemma 1.2 at this maximum: for any non degenerate subinterval $J\subset[a-\delta_{n},a+\delta_{n}]$, there exists $k$ such that either $z_{n}\in\mbox{\rm Int}\left(f_{r}^{k}(J)\right)$ or $|f_{r}^{k}(J)|\geq 2|J|$. Since $f_{r}^{(n+1)\alpha_{n}}(z_{n})=x_{n}$ and $f_{r}^{(n+1)\alpha_{n}}$ is a local homeomorphism in a neighbourhood of $z_{n}$, if $z_{n}\in\mbox{\rm Int}\left(f_{r}^{k}(J)\right)$ then $x_{n}\in\mbox{\rm Int}\left(f_{r}^{k^{\prime}}(J)\right)$ with $k^{\prime}=k+(n+1)\alpha_{n}$.

Set $\displaystyle\delta_{n}^{\prime}=\sqrt{\frac{(w_{n}-y_{n+1})(y_{n}-x_{n})}{CM_{n}^{2}}}$ and let $b$ be a local minimum of $f_{r}$ on $]x_{n},y_{n}[$. If $|t|\leq\delta_{n}^{\prime}$, then

[TABLE]

We check the hypotheses of Lemma 1.2 for the extremum b:

• •

$f_{r}^{n+1}(b)=y_{n+1}$ and $f_{r}^{n+2}(y_{n+1})=y_{n+1}$.

• •

$\displaystyle f_{r}^{n+1}(b+t)=y_{n+1}+\frac{C(M_{n}t)^{2}}{y_{n}-x_{n}}$ if $|t|\leq\delta_{n}^{\prime}$ (because of Property (4)).

• •

$f_{r}^{n+1}(b-\delta_{n}^{\prime})=f_{r}^{n+1}(b+\delta_{n}^{\prime})=w_{n}$ and $f_{r}^{n+2}(w_{n})=x_{n}$.

• •

$f_{r}^{n+2}$ is linear on $[y_{n+1},w_{n}]$ of slope $M_{n+1}k_{n+1}\geq 2$.

• •

$\displaystyle\frac{2|y_{n+1}-w_{n}|}{\delta_{n}^{\prime}}\geq 2$.

To prove the last point, define

[TABLE]

One has $M_{n}\geq\lambda-3$ (Lemma 1.2 (iii)), $\lambda\geq 14$ and

[TABLE]

thus $C_{n}\geq\frac{14\times 11^{2}}{16}>1$.

Hence we can apply Lemma 1.2 to this extremum: for any non degenerate subinterval $J\subset[b-\delta_{n}^{\prime},b+\delta_{n}^{\prime}]$, there exists $k$ such that either $x_{n}\in\mbox{\rm Int}\left(f_{r}^{k}(J)\right)$ or $|f_{r}^{k}(J)|\geq 2|J|$.

If $\displaystyle|x-d|\geq\delta|y_{n}-x_{n}|/M_{n}$ for all local extrema $d\in]x_{n},y_{n}[$, then $|(f_{r}^{n+1})^{\prime}(x)|\geq\min\{2\lambda^{r},M_{n}/2\}\geq 2$ according to Property (8). If $a\in]x_{n},y_{n}[$ is a local maximum and $\displaystyle\delta_{n}\leq|x-a|\leq\frac{\delta|y_{n}-x_{n}|}{M_{n}}$, then

[TABLE]

If $b\in]x_{n},y_{n}[$ is a local minimum and $\displaystyle\delta_{n}^{\prime}\leq|x-b|\leq\frac{\delta|y_{n}-x_{n}|}{M_{n}}$, then

[TABLE]

Consequently, $|(f_{r}^{n+1})^{\prime}(x)|\geq 2$ if for all local maxima $a$, $|x-a|\geq\delta_{n}$ and for all local minima $b$, $|x-b|\geq\delta^{\prime}_{n}$.

Finally, if $J$ is a non degenerate subinterval of $[x_{n},y_{n}]$, there exists $k$ such that either $|f_{r}^{k}(J)|\geq 2|J|$ or $\mbox{\rm Int}\left(f_{r}^{k}(J)\right)$ contains $x_{n}$. Together with the previous result on $[y_{n+1},x_{n}]$ it gives:

*Lemma 1.8  ** * If $J$ is a non degenerate subinterval of $[1,y_{1}]$, there exist $k\geq 0$ and $n\geq 1$ such that either $|f_{r}^{k}(J)|\geq\frac{4}{3}|J|$ or $x_{n}\in\mbox{\rm Int}\left(f_{r}^{k}(J)\right)$ or $y_{n}\in\mbox{\rm Int}\left(f_{r}^{k}(J)\right)$.

The point $x_{n}$ is periodic of period $n+1$, and $(f_{r}^{n+1})^{\prime}(x)\geq 2$ for $x_{n}\leq x\leq x_{n}+\frac{y_{n}-x_{n}}{2M_{n}}$. In this situation, we can apply Lemma 1.2. For any interval $J=[x_{n},x]$ with $x>x_{n}$ there exists $k$ such that $f_{r}^{k}(J)\supset[x_{n},x_{n}+\frac{y_{n}-x_{n}}{2M_{n}}]$. But

[TABLE]

Hence $f_{r}^{k+2(n+1)}(J)\supset[x_{n},y_{n}]$.

We do the same thing for $y_{n}$: for any interval $J=[y,y_{n}]$ with $y<y_{n}$ there exists $k$ such that $f_{r}^{k}(J)\supset[x_{n},y_{n}]$.

Moreover

[TABLE]

so $f_{r}^{2(n+1)+1}([x_{n},y_{n}])=[0,4]$. This leads to the next lemma.

*Lemma 1.9  ** * If $J$ is an open subinterval with $x_{n}\in J$ or $y_{n}\in J$, then there exists $k\geq 0$ such that $f_{r}^{k}(J)=[0,4]$.

1.4 Construction of $f_{r}$ on $[0,1]$ and $[y_{1},4]$ and end of the

proof of the mixing property

Recall that $f_{r}(x)=\lambda^{r}x$ for $0\leq x\leq\frac{5}{2}\lambda^{-r}$ and $\delta=\lambda^{-r}$. We define $f_{r}$ near the points $1/2,\>1$ and $4$ as follows:

• •

$f_{r}(x)=4-C_{0}(x-1/2)^{2}$ for $|x-1/2|\leq\delta$, with $C_{0}=\frac{3}{2}\delta^{-1}$.

• •

$f_{r}(x)=C_{1}(x-1)^{\alpha_{1}}$ for $1-\delta\leq x\leq 1$, with $\alpha_{1}=2r$ and $C_{1}=\delta^{1-\alpha_{1}}$.

• •

$f(x)=4+\lambda^{r}(x-4)$ for $4-\frac{3}{2}\delta\leq x\leq 4$.

The definition of $f_{r}$ on the left of $1$, together with Lemmas 1.3.1 and 1.3.2, leads to the next lemma.

Lemma 1.10  ** $f_{r}$ is $C^{r}$ in a neighbourhood of $1$.

Now we complete the map such that the pieces are glued together in a $C^{\infty}$ way (except at $1$ where $f_{r}$ is only $C^{r}$). As $f_{r}^{\prime}(1/2-\delta)=3$ and

[TABLE]

the map can be chosen such that $3/2\leq f_{r}^{\prime}(x)\leq\lambda^{r}$ for every $x\in[\frac{5}{2}\delta,\frac{1}{2}-\delta]$. In the same way, it is possible to have $-\lambda^{r}\leq f_{r}^{\prime}(x)\leq-3/2$ for every $x\in[1/2+\delta,1-\delta]$ because $f_{r}^{\prime}(1/2+\delta)=-3$, $f_{r}^{\prime}(1-\delta)=-2r$ and

[TABLE]

Finally, $f_{r}^{\prime}(y_{1})=2$ because of the earlier construction of $f_{r}$ on $[x_{1},y_{1}]$ (see parag. 1.3.1) and

[TABLE]

Hence it is possible to have $\frac{3}{2}\leq f_{r}^{\prime}(x)\leq\lambda^{r}$ for $y_{1}\leq x\leq 4$.

Consequently, $\frac{3}{2}\leq|f_{r}^{\prime}(x)|\leq\lambda^{r}$ if $x\in[0,\frac{1}{2}-\delta]\cup[\frac{1}{2}+\delta,1-\delta]\cup[y_{1},4]$.

A quick check shows that Lemma 1.2 can be applied to the two extrema $1/2$ and $1$ (we apply it only to the left of $1$). For $z_{0}=1$, the repulsive periodic point is $z_{1}=0$, the interval $[z_{1},z_{2}]$ is $[0,\lambda^{-r}]$, and the growth factor is $\frac{\alpha_{0}\delta}{\delta}=2r$. For $z_{0}=1/2$, the repulsive periodic point is $z_{1}=4$, the interval $[z_{1},z_{2}]$ is $[4-\frac{3}{2}\lambda^{-r},4]$, and the growth factor is $\frac{2\delta}{\frac{3}{2}\lambda^{-r}}=3$.

Since $f_{r}^{2}(\lambda^{-r})=0$ and $f_{r}(4-\frac{3}{2}\lambda^{-r})=y_{1}$, for any non degenerate interval $J\subset[0,1]\cup[y_{1},4]$ there exists $k$ such that either $|f_{r}^{k}(J)|\geq\frac{3}{2}|J|$ or $f_{r}^{k}(J)$ contains one of the points $0,4,y_{1}$.

*Lemma 1.11  ** * If $J$ is a non degenerate subinterval of $[0,1]\cup[y_{1},4]$, there exists $k\geq 0$ such that either $|f_{r}^{k}(J)|\geq\frac{3}{2}|J|$ or $0\in f_{r}^{k}(J)$ or $4\in f_{r}^{k}(J)$ or $y_{1}\in\mbox{\rm Int}\left(f_{r}^{k}(J)\right)$.

Since $f_{r}^{2}([0,\lambda^{-r}])=[0,4]$ and $f_{r}^{3}([4-\frac{3}{2}\lambda^{-r},4])=f_{r}^{2}([y_{1},4])=[0,4]$, applying Lemma 1.2 we obtain the next lemma.

*Lemma 1.12  ** * If $J$ is a non degenerate subinterval containing either [math] or $4$, then there exists $k\geq 0$ such that $f_{r}^{k}(J)=[0,4]$.

The construction of $f_{r}\colon[0,4]\to[0,4]$ is now finished. The map is $C^{r}$ on $[0,4]$ (and is $C^{\infty}$ on $[0,4]\backslash\{1\}$), and $\|f_{r}^{\prime}\|_{\infty}=\lambda^{r}$. Furthermore, if we put together Lemmas 1.3.3, 1.3.3, 1.4 and 1.4, we see that for any non degenerate subinterval $J\subset[0,4]$, there exists $k\geq 0$ such $f_{r}^{k}(J)=[0,4]$.

*Proposition 1.13  ** * $f_{r}\colon I\to I$ is $C^{r}$, mixing and $\|f_{r}^{\prime}\|_{\infty}=\lambda^{r}$.

**Remark 1.14  **If we identify the two end points [math] and $4$, the map $f_{r}$ can be seen as a mixing $C^{r}$ map on the torus, since $f_{r}^{(k)}(0)=f_{r}^{(k)}(4)$ for every $k\geq 1$.

2 Markov chain associated with $f_{r}$

We show that $f_{r}$ is a Markov map for a suitable countable partition. The associated Markov chain reflects almost all topological properties of the system $(I,f_{r})$.

2.1 Definition of the graph

We explicit the Markov partition $V_{r}$ and the associated graph $G_{r}$.

Let $t^{n}_{0}=x_{n}<t^{n}_{1}<\cdots<t^{n}_{M_{n}}=y_{n}$ the local extrema of $f_{r}$ on $[x_{n},y_{n}]$. Let

[TABLE]

The elements of $V_{r}$ have pairwise disjoint interior and their union is $]0,4]$. We check that the map $f_{r}$ is monotone on each element of $V_{r}$ and if $J\in V_{r}$ then $f_{r}(J)$ is a union of elements of $V_{r}\cup\{0\}$.

• •

By Lemma 1.3.1, $f_{r}$ is monotone on $[t_{i-1}^{n},t_{i}^{n}]$, $f_{r}([t_{0}^{n},t_{1}^{n}])=[\lambda^{-nr}x_{n},\lambda^{-nr}y_{n}]$ and $f_{r}([t_{i-1}^{n},t_{i}^{n}])=[\lambda^{-nr}y_{n+1},\lambda^{-nr}x_{n}]\cup[\lambda^{-nr}x_{n},\lambda^{-nr}y_{n}]$ if $2\leq i\leq M_{n}$.

• •

By Lemmas 1.3.1 and 1.3.2, $f_{r}$ is increasing on $[y_{n+1},x_{n}]$ for all $n\geq 1$ and

$\displaystyle f_{r}([y_{n+1},x_{n}])=[\lambda^{-(n+1)r}y_{n+1},\lambda^{-nr}x_{n}]\\ =[\lambda^{-(n+1)r}y_{n+1},\lambda^{-nr}]\cup\bigcup_{k\geq n}[\lambda^{-nr}x_{k+1},\lambda^{-nr}y_{k+1}]\cup[\lambda^{-nr}y_{k+1},\lambda^{-nr}x_{k}].$

• •

Since $f_{r}(x)=\lambda^{r}x$ for $x\in[0,\lambda^{-r}y_{1}]$ we have

– $f_{r}([\lambda^{-kr}x_{n},\lambda^{-kr}y_{n}])=[\lambda^{-(k-1)r}x_{n},\lambda^{-(k-1)r}y_{n}]$ for $1\leq k\leq n$ and this interval is an element of $V_{r}$ except $\displaystyle[x_{n},y_{n}]=\bigcup_{i=1}^{M_{n}}[t_{i-1}^{n},t_{i}^{n}]$ which is a union of elements of $V_{r}$.

– $f_{r}([\lambda^{-kr}y_{n+1},\lambda^{-kr}x_{n}])=[\lambda^{-(k-1)r}y_{n+1},\lambda^{-(k-1)r}x_{n}]$ for $1\leq k\leq n$.

– $\displaystyle f_{r}([\lambda^{-(n+1)r}y_{n+1},\lambda^{-nr}])=[\lambda^{-nr}y_{n+1},\lambda^{-(n-1)r}]$

   $\displaystyle=[\lambda^{-nr}y_{n+1},\lambda^{-nr}x_{n}]\cup[\lambda^{-nr}x_{n},\lambda^{-nr}y_{n}]\cup[\lambda^{-nr}y_{n},\lambda^{-(n-1)r}]$ for $n\geq 1$.

• •

$f_{r}$ is monotone on $[0,1/2]$, $[1/2,1]$ and $[y_{1},4]$ (see Subsection 1.4) and

– $f_{r}([\lambda^{-r}y_{1},1/2])=[y_{1},4]$.

– $\displaystyle f_{r}([1/2,1])=[0,4]=\{0\}\cup\bigcup_{J\in V_{r}}J$.

– $\displaystyle f_{r}([y_{1},4])=[\lambda^{-r}y_{1},4]$

   \displaystyle=[\lambda^{-r}y_{1},1/2]\cup[1/2,1]\cup[y_{1},4]\cup\bigcup_{n\geq 1}[y_{n+1},x_{n}]\cup\bigcup_{\mbox{\scriptsize\begin{tabular}[]{c}1\leq n$\\$1\leq i\leq M_{n}\end{tabular}}}[t_{i-1}^{n},t_{i}^{n}].

We define the directed graph $G_{r}$ as follows: the set of vertices of $G_{r}$ is $V_{r}$ and there is an arrow from $J$ to $K$ if and only if $K\subset f_{r}(J)$. The decomposition above of $f_{r}(J)$ into elements of $V_{r}$ for all $J\in V_{r}$ gives an exhaustive description of the arrows in $G_{r}$.

Notice that the graphs $G_{r}$ are identical for all $r\geq 1$. The only difference is the name of the vertices, corresponding to the partition of $f_{r}$.

2.2 Isomorphism between $f_{r}$ and the Markov chain

Let $\Gamma_{r}^{+}$ be the set of all one-sided infinite sequences $(D_{n})_{n\geq 0}$ such that $D_{n}\in V_{r}$ and $D_{n}\to D_{n+1}$ $\forall n\in\mathbb{N}$, and let $\Gamma_{r}$ be the set of all two-sided infinite sequences $(D_{n})_{n\in\mathbb{Z}}$. We write $\sigma$ for the shift transformation in both spaces. $(\Gamma_{r},\sigma)$ is called the Markov chain associated with $f_{r}$. As the systems $(\Gamma_{r},\sigma)$ are isomorphic for all $r\geq 1$, we just write $(\Gamma,\sigma)$ when we want to talk about one of them without referring to the partition associated with $f_{r}$.

We are going to build an isomorphism modulo countable sets between $(I,f_{r})$ and $(\Gamma_{r}^{+},\sigma)$, that is a map $\phi_{r}\colon I\backslash{\cal N}_{r}\longrightarrow\Gamma_{r}^{+}\backslash{\cal M}_{r}$ where ${\cal N}_{r},\>{\cal M}_{r}$ are countable sets, $\phi_{r}$ is bijective bimeasurable (in fact bicontinuous) and $\phi_{r}\circ f_{r}=\sigma\circ\phi_{r}$.

Define

[TABLE]

and let ${\cal N}_{r}=\bigcup_{n\geq 0}f_{r}^{-n}({\cal P}_{r})$ which is countable. We have $f_{r}({\cal N}_{r})={\cal N}_{r}$ and $f_{r}(I\backslash{\cal N}_{r})=I\backslash{\cal N}_{r}$. If $x\in I\backslash{\cal P}_{r}$ then there is a unique $D\in V_{r}$ such that $x\in D$ (in fact $x\in\mbox{\rm Int}\left(D\right)$). Hence if $x\in I\backslash{\cal N}_{r}$, for every $n\geq 0$ there is a unique $D_{n}\in V$ such that $f_{r}^{n}(x)\in D_{n}$. Moreover $(D_{n})_{n\geq 0}\in\Gamma_{r}^{+}$. We define

[TABLE]

This application satisfies $\phi_{r}\circ f_{r}(x)=\sigma\circ\phi_{r}(x)$.

For any $(D_{n})_{n\geq 0}\in\Gamma_{r}^{+}$, the set $J=\bigcap_{n\geq 0}f_{r}^{-n}(D_{n})$ is a compact interval because $f_{r}$ is monotone on each $D_{n}$. The map $f_{r}$ is mixing (Proposition 1.4) and $f_{r}^{n}(J)\subset D_{n}$, hence $J$ is necessarily reduced to a single point $\{x\}$. We define

[TABLE]

Let ${\cal M}_{r}=\psi_{r}^{-1}({\cal N}_{r})$. The application $\psi_{r}$, restricted to $\Gamma_{r}^{+}\backslash{\cal M}_{r}$, is the inverse of $\phi_{r}$. Moreover, both $\phi_{r}$ and $\psi_{r}$ are continuous. Indeed, choose $x_{0}\in I\backslash{\cal N}_{r}$ and write $(D_{n})_{n\geq 0}=\phi_{r}(x_{0})$ and $J_{n}=\bigcap_{k=0}^{n}f_{r}^{-k}(D_{k})$. The diameters of the compact intervals $J_{n}$ tend to [math], the point $x_{0}$ belongs to $\mbox{\rm Int}\left(J_{n}\right)$ for every $n$, and for every $x\in J_{n}\backslash{\cal N}_{r}$ the sequence $\phi_{r}(x)$ begins with $(D_{0},\cdots,D_{n})$. Hence $\phi_{r}$ is continuous. Inversely, fix $\displaystyle\gamma_{0}=(D_{n})_{n\geq 0}\in\Gamma_{r}^{+}\backslash{\cal M}_{r}$, then for every sequence $\gamma\in\Gamma_{r}^{+}\backslash{\cal M}_{r}$ beginning with $(D_{0},\cdots,D_{n})$ the point $\psi_{r}(\gamma)$ belongs to $J_{n}$ which is an arbitrarily small neighbourhood of $\psi_{r}(\gamma_{0})$. Hence $\psi_{r}$ is continuous too.

Now, we are going to show that ${\cal M}_{r}$ is countable. It is sufficient to show that $\psi_{r}^{-1}(x)$ is finite for any $x\in{\cal N}_{r}$. For any $y\in I$ there are at most two elements of $V_{r}$ containing $y$. Let $x\in{\cal N}_{r}$. If there is a $k$ such that $f_{r}^{k}(x)=0$ then $\psi_{r}^{-1}(x)=\emptyset$. If there is a $k$ such that $f_{r}^{k}(x)=4$ then $\psi^{-1}(x)$ is finite because $\psi^{-1}(4)$ contains only the constant sequence of symbol $[y_{1},4]$. Otherwise there exist $k,\,n$ such that $f_{r}^{k}(x)=x_{n}$ or $f_{r}^{k}(x)=y_{n}$. Thus it is sufficient to focus on the points $x_{n}$ and $y_{n}$.

We begin with $x_{n}$. The intervals $C_{0}=[y_{n+1},x_{n}]$ and $D_{0}=[x_{n},t^{n}_{1}]$ are the only two elements of $V_{r}$ containing $x_{n}$. If we try to build $\displaystyle(C_{k})_{k\geq 0}$ and $\displaystyle(D_{k})_{k\geq 0}$ which are elements of $\psi_{r}^{-1}(x_{n})$, we see that there are only two possibilities, which are cycles, namely:

• •

$C_{0}=[y_{n+1},x_{n}]\to C_{1}=[\lambda^{-nr}y_{n+1},\lambda^{-nr}x_{n}]\to\cdots\to C_{n+1}=C_{0}\to\cdots$

• •

$D_{0}=[x_{n},t^{n}_{1}]\to D_{1}=[\lambda^{-nr}x_{n},\lambda^{-nr}y_{n}]\to\cdots\to D_{n+1}=D_{0}\to\cdots$

Hence, $\mbox{\rm Card}\left(\psi_{r}^{-1}(x_{n})\right)=2$.

The situation is the same for $y_{n},n\geq 2$, with two slightly different cycles, namely:

• •

$C_{0}=[t^{n}_{M_{n}-1},y_{n}]\to[\lambda^{-nr}x_{n},\lambda^{-nr}y_{n}]\to\cdots\to C_{n+1}=C_{0}\to\cdots$

• •

$D_{0}=[y_{n},x_{n-1}]\to[\lambda^{-nr}y_{n},\lambda^{-(n-1)r}]\to\cdots\to D_{n+1}=D_{0}\to\cdots$

A quick look at the map $f_{r}$ gives the last two cycles for $y_{1}$.

Consequently, $\mbox{\rm Card}\left(\psi_{r}^{-1}(x)\right)<+\infty$ for every $x\in{\cal N}_{r}$, ${\cal M}_{r}$ is countable, and the map $\phi_{r}\colon I\backslash{\cal N}_{r}\longrightarrow\Gamma_{r}^{+}\backslash{\cal M}_{r}$ is an isomorphism modulo countable sets.

$\phi_{r}$ transforms any invariant measure that does not charge ${\cal N}_{r}$ into an invariant measure that does not charge ${\cal M}_{r}$, and inversely. A measure supported by ${\cal N}_{r}$ or ${\cal M}_{r}$ is of zero entropy and the metric entropy $\mu\mapsto h_{\mu}$ is affine (see e.g. [11]), thus $h_{top}(f_{r})=h(\Gamma_{r}^{+},\sigma)$, where

[TABLE]

and $\phi_{r}$ establishes a bijection between the sets of maximal measures.

On the other hand, $h(\Gamma_{r}^{+},\sigma)=h(\Gamma_{r},\sigma)$ and there is a bijection between the maximal measures of $(\Gamma_{r}^{+},\sigma)$ and those of $(\Gamma_{r},\sigma)$, because the latter is the natural extension of the former (see e.g. [19]). Recall that all systems $(\Gamma_{r},\sigma)$ are identical and $(\Gamma,\sigma)$ represents equally one of them. Hence the question of existence of maximal measure for $(I,f_{r})$ can be studied by looking at $(\Gamma,\sigma)$.

*Proposition 2.1  ** * $h_{top}(f_{r})=h(\Gamma,\sigma)$ and $(I,f_{r})$ admits a maximal measure if and only if $(\Gamma,\sigma)$ admits one.

2.3 Non existence of maximal measure

Following the terminology of Vere-Jones [23] a transitive Markov chain is either transient, positive recurrent or null recurrent. According to a result of Gurevič [14], a transitive Markov chain has a maximal measure if and only if its graph is positive recurrent. We do not give the definitions of transience, positive recurrence and null recurrence because we will only need a criterion due to Salama (Theorem 2.1(i) in [22]), which is stated below.

If $H$ is a (strongly) connected directed graph and $(\Gamma_{H},\sigma)$ is the associated Markov chain, i.e. the set of all sequences $\displaystyle(h_{n})_{n\in\mathbb{Z}}$ with $h_{n}\to h_{n+1}$ in $H$, we define $h(H)=h(\Gamma_{H},\sigma)=\sup\{h_{\mu}\colon\mu\;\sigma\mbox{-invariant probability on }\Gamma_{H}\}$.

*Theorem 2.2  ** * (Gurevič) Let $H$ be a connected directed graph and $(\Gamma_{H},\sigma)$ be the associated Markov chain. If its entropy $h(H)$ is finite then $(\Gamma_{H},\sigma)$ admits a maximal measure if and only if $H$ is positive recurrent. In this case, the measure is unique.

*Theorem 2.3  ** * (Salama) Let $H$ be a connected directed graph. If there exists a graph $H^{\prime}$ such that $H\subseteq_{\mbox{\rm{\scriptsize/} \hskip 0.70004pt}}H^{\prime}$ and $h(H)=h(H^{\prime})$ then $H$ is transient.

Next, we compute $h(G_{r})$ then we show that $G_{r}$ is transient, which is enough to conclude that $f_{r}$ has no maximal measure by Proposition 2.2. As all graphs $G_{r}$ are identical, it is sufficient to focus on $G_{1}$.

*Proposition 2.4  ** * $h_{top}(f_{r})=h(G_{r})=\log\lambda$.

Proof:

It is already known that $h_{top}(f_{r})=h(G_{r})=h(G_{1})$ by Proposition 2.2.

A subset $E\subset I$ is called $(n,\varepsilon)$-separated for $f_{1}$ if for any two distinct points $x,y$ in $E$ there exists $k$, $0\leq k<n$, with $|f_{1}^{k}(x)-f_{1}^{k}(y)|>\varepsilon$. Let $s_{n}(f_{1},\varepsilon)$ be the maximal cardinality of an $(n,\varepsilon)$-separated set. Then the topological entropy of $f_{1}$ is given by the following formula (see e.g. [11]):

[TABLE]

Let $E$ be an $(n,\varepsilon)$-separated set of $I$ of maximal cardinality. As $\|f_{1}^{\prime}\|_{\infty}=\lambda$ (Proposition 1.4), we have $|f_{1}(x)-f_{1}(y)|\leq$ $\lambda|x-y|$ for all $x,y\in I$. If $x,y$ are two distinct points of $E$, there exists $k<n$ such that $|f_{1}^{k}(x)-f_{1}^{k}(y)|>\varepsilon$. But $|f_{1}^{k}(x)-f_{1}^{k}(y)|\leq\lambda^{n}|x-y|$, hence $|x-y|\geq\lambda^{-n}\varepsilon$ and

[TABLE]

Consequently, $h_{top}(f_{1})=h(G_{1})\leq\log\lambda$.

Now, let $H_{n}\subset G_{1}$ be the subgraph whose vertices are:

[TABLE]

The edges of $H_{n}$ are all possible edges of $G_{1}$ between two vertices, namely:

• •

$[t^{n}_{i-1},t^{n}_{i}]\to[\lambda^{-n}x_{n},\lambda^{-n}y_{n}]$ for $1\leq i\leq M_{n}$,

• •

$[\lambda^{-k}x_{n},\lambda^{-k}y_{n}]\to[\lambda^{-k+1}x_{n},\lambda^{-k+1}y_{n}]$ for $2\leq k\leq n$,

• •

$[\lambda^{-1}x_{n},\lambda^{-1}y_{n}]\to[t^{n}_{i-1},t^{n}_{i}]$ for $1\leq i\leq M_{n}$.

The graph $H_{n}$ is represented in Figure 3.

The system $(H_{n},\sigma^{n+1})$ is a full shift on $M_{n}$ symbols, plus $n$ fixed points, thus $h(H_{n},\sigma^{n+1})=\log M_{n}$ (see e.g. [11, p111]) and $\displaystyle h(H_{n})=\frac{\log M_{n}}{n+1}$.

By definition of $M_{n}$,

[TABLE]

As $H_{n}$ is a subgraph of $G_{1}$, $h(H_{n})\leq h(G_{1})$. Therefore $h(G_{1})=\log\lambda$.

*Proposition 2.5  ** * The graph $G_{1}$ is transient.

Proof:

We are going to build a Markov map $g$, very similar to $f_{1}$, such that $\|g^{\prime}\|_{\infty}\leq\lambda$ and the Markov graph $H$ associated with $g$ expands strictly $G_{1}$. Suppose $g$ is already built. The same argument as in the proof of Proposition 2.3 shows that $h(H)\leq\log\|g^{\prime}\|_{\infty}\leq\log\lambda$. As $G_{1}\subset H$ we have $h(H)\geq h(G_{1})$, thus $h(H)=h(G_{1})$ by Proposition 2.3. This is enough to conclude that $G_{1}$ is transient by Theorem 2.3.

The map $g\colon I\to I$ is defined as $g(x)=f_{1}(x)$ for all $x\in I\setminus[x_{2},y_{2}]$. Let

[TABLE]

and choose $\widetilde{s}_{2}\colon[0,\widetilde{M}_{2}]\to[-m_{2},1]$ satisfying Properties (3)-(8) except that $M_{2}$ and $k_{2}$ are replaced respectively by $\widetilde{M}_{2}$ and $\widetilde{k}_{2}$. Then we define $g$ on $[x_{2},y_{2}]$ by

[TABLE]

By Properties (5) and (6), $g^{\prime}(x_{2})=g^{\prime}(y_{2})=\lambda^{-2}\widetilde{M}_{2}\widetilde{k}_{2}=2\lambda^{-1}$, thus $g^{\prime}(x_{2})=f_{1}^{\prime}(x_{2})$, $g^{\prime}(y_{2})=f_{1}^{\prime}(y_{2})$ and $g$ is $C^{1}$. Moreover for all $x\in[x_{2},y_{2}]$,

[TABLE]

thus $|g^{\prime}(x)|<\lambda$ because $\widetilde{M}_{2}=M_{2}+2=2\left[\frac{\lambda}{8}\right]+1<\lambda^{2}$. Since $\|f_{1}^{\prime}\|_{\infty}=\lambda$ by Proposition 1.4, one concludes that $\|g^{\prime}\|_{\infty}\leq\lambda$

Define the Markov graph $H$ associated with $g$ as in Subsection 2.1, and denote by $W$ the set of vertices of $H$. Compared to $V_{1}$, $W$ has two additional vertices because $f_{1}$ has $M_{2}$ monotone pieces between $x_{2}$ and $y_{2}$ and $g$ has $M_{2}+2$. If

[TABLE]

for $0\leq i\leq M_{2}+2$ then it is not hard to check that the graph $G_{1}$ is equal to $H$ deprived of the vertices $[\widetilde{t}_{M_{2}},\widetilde{t}_{M_{2}+1}]$ and $[\widetilde{t}_{M_{2}+1},\widetilde{t}_{M_{2}+2}]$ and all the edges that begin or end at one of them. Consequently $G_{1}\subseteq_{\mbox{\rm{\scriptsize/} \hskip 0.70004pt}}H$, which ends the proof.

**Remark 2.6  ** We can see intuitively what happens for an $f_{r}$-invariant measure when its entropy tends to $\log\lambda$. On each finite subgraph $H_{n}$, there is a measure of entropy $\frac{\log M_{n}}{n+1}$. This measure has a corresponding measure $\mu_{n}$ on the interval, the support of which is contained in $\bigcup_{k=0}^{n}[\lambda^{-kr}x_{n},\lambda^{-kr}y_{n}]$ (in fact, the support of $\mu_{n}$ is exactly the Cantor set of all points which never escape from that set). We have of course $h_{\mu_{n}}(f_{r})\to\log\lambda$. But if we consider what happens near [math], we see that $\mu_{n}$ converges to $\delta_{0}$, the Dirac measure at [math], whose entropy is null.

3 Local entropy

We recall first some definitions due to Bowen [5] and then we define the local entropy. There exist different definitions of local entropy, we give here that of Buzzi [9].

**Definition 3.1  **Let $T\colon X\to X$ be a continuous map on a compact metric space $X$.

The Bowen ball of radius $r$ and order $n$, centered at $x$ is $B_{n}(x,r)=\{y\in X\colon d(T^{k}(y),T^{k}(x))<r,$ $\forall\,k=0,\cdots,n-1\}$.

An $(\varepsilon,n)$-separated set of $Y\subset X$ is a subset $E\subset Y$ such that $\forall\,y\not=y^{\prime}$ in $E$, $\exists\;0\leq k<n$, $d(T^{k}(y),T^{k}(y^{\prime}))>\varepsilon$. The maximal cardinality of an $(\varepsilon,n)$-separated set of $Y$ is denoted by $s_{n}(T,\varepsilon,Y)$.

**Definition 3.2  **The local entropy of $T$, $h_{loc}(T)$, is defined as

[TABLE]

**Remark 3.3  **An $(\varepsilon,n)$-cover of $Y\subset X$ is a subset $S\subset X$ such that $\displaystyle Y\subset\bigcup_{x\in S}B_{n}(x,\varepsilon)$. Some people use $(\varepsilon,n)$-covers instead of $(\varepsilon,n)$-separated sets: it leads to the same definition of the local entropy.

Local entropy is interesting because it bounds the defect of upper semicontinuity of the metric entropy $\mu\mapsto h_{\mu}(f)$. On a compact Riemannian m-dimensional manifold, local entropy itself is bounded by $\frac{m\log R(f)}{r}$, where $R(f)$ is the spectral radius of the differential and $r$ is the differential order. These results are stated by Buzzi [9] and follow works of Yomdin [24] and Newhouse [18]. In particular, they directly imply that a $C^{\infty}$ map on a compact Riemannian manifold always has a maximal measure (this result can be found in Newhouse’s work [18]). These results are given in the next two theorems, the second one is stated for interval maps only.

**Theorem 3.4  ** Let $T\colon X\to X$ be a continuous map on a compact metric space. Assume that $\mu_{n}$ is a sequence of $T$-invariant measures on $X$, converging to a measure $\mu$. Then

[TABLE]

*Theorem 3.5  ** * Let $f\colon I\to I$ be a $C^{r}$ map on a compact interval $I$, $r\geq 1$, and let $\displaystyle R(f)=\inf_{k\geq 1}\sqrt[k]{\|(f^{k})^{\prime}\|_{\infty}}$. Then the local entropy satisfies

[TABLE]

In our family of examples, the local entropy can be computed explicitly.

**Proposition 3.6  ** For every $n\geq 1$ the local entropy of $f_{r}$ is

[TABLE]

Proof:

The map $f_{r}$ is such that $\|f_{r}^{\prime}\|_{\infty}\leq\lambda^{r}$ (Proposition 1.4) and [math] is a fixed point with $f_{r}^{\prime}(0)=\lambda^{r}$. Hence $R(f_{r})=\lambda^{r}$ and

[TABLE]

according to Theorem 3.

We are going to show the reverse inequality.

Fix $\varepsilon>0$ and choose $n$ such that $\frac{1}{2n^{2}}<\varepsilon$. Put $\delta_{0}=\frac{1}{2n^{2}M_{n}}$. If $x\in[x_{n},y_{n}]$ satisfies $f^{n+1}(x)\in[x_{n},y_{n}]$ then $|f^{i}(x)-f^{i}(x_{n})|<\varepsilon$ for $0\leq i\leq n+1$. We write $I_{i}=[t^{n}_{i-1},t^{n}_{i}]$ for $1\leq i\leq M_{n}$. The length of each $I_{i}$ is $\delta_{0}$.

Choose a finite sequence $\omega=(\omega_{0},\cdots,\omega_{p-1})$ with $1\leq\omega_{i}\leq M_{n}$. Thanks to the isomorphism between $(I,f_{r})$ and its Markov extension (Section 2), there is a point $x_{\omega}\in[x_{n},y_{n}]$ with $f^{(n+1)i}(x_{\omega})\in I_{\omega_{i}}$ for $0\leq i\leq p-1$. Consider the set $E_{n,p}=\{x_{\omega}\colon\omega=(\omega_{0},\cdots,\omega_{p-1}),\omega_{i}\mbox{ odd}\}$. The cardinality of $E_{n,p}$ is

[TABLE]

by Lemma 1.2 (ii). If $x\in E_{n,p}$ then $|f^{k}(x_{n})-f^{k}(x)|<\varepsilon$ for $0\leq k<(n+1)p$. Moreover, if $x_{\omega},x_{\omega^{\prime}}$ are two distinct elements of $E_{n,p}$, then there exists $0\leq i\leq p-1$ with $|\omega_{i}-\omega^{\prime}_{i}|\geq 2$, hence $|f^{(n+1)i}(x_{\omega})-f^{(n+1)i}(x_{\omega^{\prime}})|\geq\delta_{0}$. Consequently, $E_{n,p}$ is an $((n+1)p,\delta,B_{(n+1)p}(x_{n},\varepsilon))$-separated set for every $\delta<\delta_{0}$, and

[TABLE]

This computation shows that the bound $\frac{\log R(f)}{r}$ is a sharp one to estimate the local entropy. Moreover, we remarked (Remark 3) that there exists a sequence of measures $\mu_{n}$ converging to the Dirac measure $\delta_{0}$, with $h_{\mu_{n}}(f_{r})\to h_{top}(f_{r})$. Hence, the local entropy is exactly the defect of upper semicontinuity of the metric entropy in this case.