\documentclass[12pt]{article}
\usepackage{amsmath,amsthm,amsfonts,amssymb,amscd}
\usepackage{graphics}

\headsep=-1truecm

\renewcommand{\baselinestretch}{1.5}

\oddsidemargin=0pt
\evensidemargin=0pt



\textwidth=17truecm
\textheight=23truecm

\begin{document}


\theoremstyle{plain}
\newtheorem{thm}{\sc Theorem}[section]
\newtheorem{lem}[thm]{\sc Lemma}
\newtheorem*{coro}{\sc Corollary}
\newtheorem{prop}[thm]{\sc Proposition}

\theoremstyle{remark}
\newtheorem{Case}{\bf Case}[section]
\newtheorem*{rem}{\sc Remark}
\newtheorem*{exem}{\sc Example}

\theoremstyle{definition}
\newtheorem{Def}[thm]{\bf Defini\c c\~ao}
\newtheorem*{pf}{\sc Proof}

\numberwithin{equation}{section}

\newcommand{\ve}{{\varepsilon}}
\newcommand{\be}{{\beta}}
\newcommand{\la}{\lambda}
\newcommand{\teta}{{\theta}}
\newcommand{\ro}{{\rho}}
\newcommand{\ov}{\overline}
\newcommand{\re}{\mathbb R}
\newcommand{\F}{{\mathbb F}}
\newcommand{\al}{{\alpha}}
\newcommand{\vr}{{\varphi}}
\newcommand\lv{\left\vert}
\newcommand\rv{\right\vert}



\begin{center}
{\Large \bf Maximal Invariant Sets for Restrictions of
 Tent and Unimodal Maps}
\end{center}

\vskip .5in

\begin{center}
{\sc Carlos Gustavo Tamm de Araujo Moreira}
\end{center}

\vskip .1in

\begin{center}
{\large Instituto de Matem\'atica Pura e Aplicada} \\
{\sl Estrada Dona Castorina, 110} \\
{\sl 22460-320, Rio de Janeiro-RJ} \\
{E-mail: gugu@impa.br}
\end{center}

\vskip .7in

\begin{abstract}
The aim of this paper is to study the restriction
of  a tent map $f\colon I\to I$ to a subinterval $J\subset I$, and to
characterize when the maximal invariant of $f|_J$ is a countable set. We 
shall extend our results for a certain class of unimodal maps, via a 
semi-conjugacy. In the case of tent maps there is a critical case where the
maximal invariant set is a Cantor set with Hausdorff dimension zero. The
caracterization of this critical case leads to interesting arithmetical
phenomena, particularly in the case of the complete tent map.
\end{abstract}

\section{Introduction}
The classical family of  tent maps is the family $(T_\la)_{1<\la\le2}$ defined by
$$
T_\la(x) = \begin{cases} \la x, \quad\text{ if }\quad 0 \le x  \le\dfrac12\\ \la(1-x), \quad\text{ if }\quad \dfrac 12 \le x \le 1\end{cases}.
$$
 It is the simplest continuous family of unimodal maps, and gives 
topological
 models for many differentiable unimodal maps (for instance, the
complete tent map $T_2$ is conjugated to $f(x)=x^2-2$ in [-2,2] via the homeomorphism $h\colon [0,1] \to [-2,2]$,\,\,\, $h(x) = 2\cos(\pi x))$.

The aim of this work is to study restrictions of a tent map $T_\la$ to closed subintervals $J \subset [0,1]$, and characterize when the maximal invariant set $K = K_{\la,J}$ of $T_\la|_J$ is an uncountable set, when it has positive Hausdorff dimension (see definition below) and when it has positive Lebesgue measure (case in which it must be a nondegenerate interval). Our main result is that, given $\la \in (1,2]$, there is 
an explicitly given
interval $[a(\la),b(\la)] \subset [0,1]$ such that $K$ is uncountable if and only if $J$ contains $[a(\la),b(\la)]$ and it has positive Hausdorff dimension if and only if the interior of $J$ contains $[a(\la), b(\la)]$. In the case $\la=2$ we relate $a(\la)$ and $b(\la)$ with the classical Morse-Thue constant and use it to show their transcendence.

We also discuss generalizations of our results to $C^2$ unimodal maps, via the study of (semi) conjugacies between them and tent maps.

Let us finish this introduction with some definitions:

Given a metric space $X$ and a positive real number $\al$, we define the Hausdorff measure of dimension $\al$ associated to a covering $\mathcal{U} = (U_\la)_{\la\in L}$ of $X$ by bounded sets $U_\la$ by $m_\al(\mathcal{U}) = \sum_{\la\in L}(\text{diam } U_\la)^\al$, where $\text{diam  }U_\la$ denotes the diameter of $U_\la$\,, and, if we define the norm of a covering $\mathcal{U}$ by $||\mathcal{U}|| = \sup_{U\in\mathcal{U}} (\text{diam }U)$, then the Hausdorff measure of dimension $\al$ of the set $X$ is $m_\al(X) = \underset{\substack{\mathcal{U} \text{ covering of } X\\ ||\mathcal{U}||\to0}}{\liminf} \,m_\al(\mathcal{U})$.

It is not difficult to see that there is a unique $d \in [0,+\infty]$ such that if $\al > d$ then $m_\al(X)=0$ and if $\al < d$ then $m_\al(X) = +\infty$. This number $d$ is called the {\it Hausdorff dimension\/} of $X$. It is easy to see that if $X \subset \re$ then its Hausdorff dimension $d =: HD(X)$ belongs to $[0,1]$.

Given a Cantor set $K \subset \re$, a gap of $K$ is a connected component of the complement of $K$. Given a bounded gap $U$, let $U_L$ and $U_R$ the nearest gaps respectively at the left and at the right  of $U$ that are bigger than $U$, and let $I_L$ and $I_R$ be the intervals between $U_L$ and $U$ and between $U$ and $U_R$\,, respectively.  We define $\tau(U) = \min\{|I_L|/|U|,\,\, |I_R|/|U|\}$, and we define the {\it thickness\/} of $K$ by $\tau(K) = \inf\{\tau(U), U$ bounded gap of $K\}$. Its importance follows from Newhouse's gap Lemma (see [P-T]), according to which if $K_1$ and $K_2$ are Cantor sets on the real line such that $\tau(K_1)\cdot\tau(K_2) \ge 1$ then either $K_1$ is contained in a gap of $K_2$\,,\,\,\, $K_2$ is contained in a gap of $K_1$ or $K_1 \cap K_2 \ne \emptyset$. It is possible to prove that we always have $HD(K) \ge \log 2/\log\big(2+\frac{1}{\tau(K)}\big)$. In particular, if $\tau(K) > 0$ then $HD(K) > 0$. The equality holds for the affine Cantor sets $K_\al$ with $\al \in \big(0, \frac 12\big)$, which are defined as the maximal invariant set of $\vr_\al\colon [0,\al] \cup [1-\al, 1] \to [0,1]$, given by $\vr_\al(x) = \begin{cases} 1-x/\al, \quad\text{if}\quad x \in [0,\al]\\ (x-1+\al)/\al, \quad\text{if}\quad x \in [1-\al,1]\end{cases}$, for which $\tau(K_\al) = \frac{\al}{1-2\al}$ and $HD(K_\al) = -\log 2/\log \al$.


\section{The Case of Tent Maps}

Let $1<\la\le 2$. The tent map $T_\la\colon[0,1]\to[0,1]$ is defined by
$$
T_\la(x)=\begin{cases} \la x, & \text{ if } x\le \frac12 \\
\la(1-x), & \text{ if } x\ge \frac 12 \end{cases}
$$
Since $T_\la([0,1])=[0,\frac\la2]$, if $M_J:=\bigcap_{n=0}^\infty T^{-
n}(J)$ is the maximal invariant of $T_\la|_J$, we have
$M_J=M_{J\cap[0,\frac\la2]}\cup \left(J\cap T_\la^{-
1}(M_{J\cap[0,\frac\la2]})\right)$. So, we may restrict our attention to the
case $J\subset[0,\frac\la2]$. Moreover, if $x\ge\la(1-
\frac\la2)=T_\la(\frac\la2)$ and $x\in[0,\frac\la2]$ then $T_\la(x)\ge\la(1-
\frac\la2)$, so the interval $L=[\la(1-\frac\la2),\frac\la2]$ is $T_\la$-invariant, and $M_{J\cap[0,\frac\la2]}=J\cap \bigcup_{r=0}^\infty\la^{-
r}M_{J\cap L}$, and so we may restrict our attention to the case
$J\subset[\la(1-\frac\la2),\frac\la2]$.

\begin{figure}[h]
%\begin{center}
\includegraphics{fig1.eps}
%\caption{}
%\end{center}
\end{figure}

Figure 1 above shows the dynamics of $T_\la$ and the interval $L$.

Let $J=[\la(1-\frac\la2)+\ve,\frac\la2]$. If $x\in M_J$ then $x\le\frac\la2-
\frac\ve\la$, since if $x>\frac\la2-\frac\ve\la$ then $T_\la(x)<\la(1-
\frac\la2)+\ve\Rightarrow M_J=M_{[\la(1-\frac\la2)+\ve,\frac\la2-
\frac\ve\la]}$.
On the other hand, if $J=[\la(1-\frac\la2),\frac\la2-\ve]$ then,
since $[\la(1-\frac\la2)+\ve\la,\frac\la2-\ve]=J\cap T_\la(J)$,
we have $M_J\subseteq \bigcup_{r=0}^\infty \la^{-r} M_{[\la(1-\frac\la2)+
\ve\la,\frac\la2-\ve]}$, and so we may restrict our attention to the intervals
$J_\ve=[\la(1-\frac\la2)+\ve\la,\frac\la2-\ve]=:[a_\ve,b_\ve]$.
Indeed, the discussion above implies that if $J=[\la(1-\frac\la2)+\delta, \frac\la2-\eta]$, then $J \supset J_\ve$, where $\ve=\max\{\frac\delta\la,\eta\}$ and $M_J\subseteq \bigcup_{r=0}^\infty \la^{-r} M_{J_\ve}$ (see Figure 2).
$T_\la|_{J_\ve}$ is conjugated to the function $\psi_{\la,c}\colon
[0,\frac1\la] \cup [1-\frac c\la,1]$ defined by
$$
\psi_{\la,c}(x)=\begin{cases}
1-\la x, & \text{ if } x\in[0,\frac1\la]\\
\la(x-1+\frac c\la), & \text{ if } x\in[1-\frac c\la,1], \end{cases}
$$
where
\begin{align*}
c=\frac{(\frac{b_\ve}\la-a_\ve)\cdot\la}{b_\ve-a_\ve} &=
\frac{b_\ve-\la a_\ve}{b_\ve-a_\ve}=
\frac{\frac\la2-\ve-\la^2(1+\ve-\frac\la2)}{\frac\la2-\ve-\la(1+\ve-
\frac\la2)} \\
&= \frac{\la(\la-1)^2-2(\la^2+1)\ve}{\la(\la-1)-2(\la+1)\ve} \\
&= \la-1-\frac{4\ve}{\la(\la-1)-2(\la+1)\ve} < \la-1 \quad \text{ for }\quad \ve>0.
\end{align*}
by the affine conjugacy $h(x)=\frac{b_\ve-x}{b_\ve-a_\ve}$ (see Figure 3).

\begin{figure}[h]
\begin{center}
\includegraphics{fig2.eps}
%\caption{}
\end{center}
\end{figure}

We shall study the maximal invariant of $\psi_{\la,c}$ in $[0,1]$. Let
$$
\tilde\psi_{\la,c}= \begin{cases}
c-\la^2x, & \text{ if } x\in[0,\frac c{\la^2}] \\
1-\la+\la^2 x, & \text{ if } x\in [\frac{\la-1}{\la^2},\frac1\la] \\
\la(x-1+\frac c\la), & \text{ if } x\in[1-\frac c\la,1] \end{cases} =
\begin{cases} \psi_{\la,c}^2, & \text{ if } x\in[0,\frac1\la]\cap
\psi_{\la,c}^{-2}([0,1]) \\
\psi_{\la,c}, & \text{ if } x\in[1-\frac c\la,1] \end{cases}.
$$
(See Figure 4).

\begin{figure}[h]
\begin{center}
\includegraphics{fig3_4.eps}
%\caption{}
\end{center}
\end{figure}


Clearly the maximal invariants of $\tilde\psi_{\la,c}$ and $\psi_{\la,c}$ in
$[0,1]$ are equal. 

Let $\la-1> p_\la=\frac1{\la+1}$ be the fixed point of $\psi_{\la,c}$. We have the
following possibilities:

\begin{itemize}

\item[i)] $c\ge p_\la$: In this case the maximal invariant of $\psi_{\la,c}$
contains (indeed is equal to) a Cantor set. Indeed, if
$\tilde I=\left[\frac{c-p_\la}{\la^2},p_\la\right]$ 
then the maximal invariant of $\tilde\psi_{\la,c}|_{\tilde
I}$ is a regular Cantor set, with positive Hausdorff dimension and positive
thickness (see Figure 5).

\begin{figure}[h]
\begin{center}
\includegraphics{fig5_6.eps}
%\caption{}
\end{center}
\end{figure}


\item[ii)] $c<\frac{\la-1}{\la^2}$: In this case the maximal invariant of
$\psi_{\la,c}$ is countable. Indeed, in this case $[0,c]$ is
$\tilde\psi_{\la,c}$-invariant, and the maximal invariant $M_{[0,1]}$ is equal
to $\bigcup_{r=0}^\infty\tilde\psi_{\la,c}^{-r}(M_{[0,c]})$, but
$M_{[0,c]}=\left\{\frac c{\la^2+1}\right\}$, since
$\tilde\psi_{\la,c}|_{[0,c]}=c-\la^2x$, if $x\in[0,\frac c{\la^2}]$, and is not
defined otherwise (see Figure 6).

\item[iii)] $\frac{\la-1}{\la^2}\le c\le\frac1{\la+1}=p_\la$: In this case
$[0,c]$ is still $\tilde\psi_{\la,c}$-invariant, and
$M_{[0,1]}=\bigcup_{r=0}^\infty\tilde\psi_{\la,c}^{-r}(M_{[0,c]})$.
Therefore, $M_{[0,1]}$ is a Cantor set (resp. countable) iff
$M_{[0,c]}$ is a Cantor set (resp. countable). On the other hand,
$\tilde\psi_{\la,c}|_{[0,c]}$ is conjugated by $h(x)=\frac xc$ to
$\psi_{\la^2,\la^2-\frac{\la-1}c}$. We have now 
a
renormalization operator
$F(\la,c)=(\la^2,\la^2-\frac{\la-1}c)$ (see Figure 7).
\end{itemize}

\begin{figure}[h]
\begin{center}
\includegraphics{fig7.eps}
%\caption{}
\end{center}
\end{figure}


\begin{lem}
Given $\la$, there is 
a unique value of $c$, denoted by $c(\la)$, such that we can iterate infinitely
many times the renormalization operator $F$ beginning with $(\la,c)$. This value is
given by the expression $c(\la)=(\la-1)\frac{\al(\la)}{1+\al(\la)}$, where
$\al(\la)=\sum_{r=1}^\infty \frac{1}{\prod_{j=1}^r(\la^{2^j}-1)}$.
\end{lem}

\begin{rem}
 The 
characterization 
of the maximal
invariant set of $\psi_{\la,c}$ will be as follows:

\begin{itemize}

\item[$\bullet$]  $\la-1 > c>c(\la)\Rightarrow M_{[0,1]}$ is a Cantor set with positive
Hausdorff
dimension. (Since the case i) will occur after a certain number of
renormalizations)

\item[$\bullet$] $c<c(\la)\Rightarrow M_{[0,1]}$ is countable. (Since the case ii)
will occur after a certain number of renormalizations)
\end{itemize}

We shall also prove that if $c=c(\la)$ then $M_{[0,1]}$ is a Cantor set with
Hausdorff dimension zero (and so with thickness $0$).
\end{rem}

\begin{pf}
Suppose that we can iterate $F$ infinitely many times. Let
\begin{align*}
F^n(\la,c) &= (\la_n,c_n)\Rightarrow\la_{n+1}=\la_n^2, c_{n+1}=\la_n^2-
\frac{\la_n-1}{c_n} \Rightarrow \\
\la_n=\la_0^{2^n}, c_n &= \frac{\la_n-1}{\la_n^2-c_{n+1}} \Rightarrow c_0=
\frac{\la_0-1}{\la_0^2-\frac{\la_0^2-1}{\la_0^4-\frac{\la_0^4-1}{\ddots
\la_0^{2^n}-\frac{\la_0^{2^n}-1}{\la_0^{2^{n+1}}-c_{n+1}}}}}
\end{align*}
Since $c_n\in[0,1]$, $\forall n$, $\la_0^{2^n}\to\infty$ and, if $\la_n\ge 2$, $q_n(c):= \frac{\la_n-1}{\la_n^2-c}$ is $\frac12$ contracting in $[0,1]$,
$c_0$ must be equal to the generalized continued fraction
$$
\frac{\la_0-1}{\la_0^2-\frac{\la_0^2-1}{\la_0^4-\frac{\la_0^4-1}{\ddots}}}.
$$
We have
$$
x_n:= \frac{b_0}{a_1+\frac{b_1}{a_2+\frac{b_2}{\ddots +\frac{b_{n-1}}{a_n}}}}
= \frac{p_n}{q_n},
$$
where $p_n$ and $q_n$ are defined by
$$
\begin{cases}
p_{-1}=1, p_0=0, p_{n+2}=a_{n+2}p_{n+1}+b_{n+1}p_n \\
q_{-1}=0, q_0=1, q_{n+2}=a_{n+2}q_{n+1}+ b_{n+1}q_n \end{cases}
$$
This can be proved by induction:
\begin{align*}
x_{n+1} &= \frac{\left(a_n+\frac{b_n}{a_{n+1}}\right) p_{n-1}+b_{n-1}p_{n-2}}
{\left(a_n+\frac{b_n}{a_{n+1}}\right) q_{n-1}+b_{n-1}q_{n-2}} \\
&= \frac{a_{n+1}(a_np_{n-1}+b_{n-1}p_{n-2})+b_np_{n-1}}
{a_{n+1}(a_nq_{n-1}+b_{n-1}q_{n-2})+b_nq_{n-1}}=
\frac{a_{n+1}p_n+b_np_{n-1}}{a_{n+1}q_n+b_nq_{n-1}}=\frac{p_{n+1}}{q_{n+1}}.
\end{align*}
In our case, $p_n$ and $q_n$ satisfy the recurrence
$Z_{n+2}=\la_0^{2^{n+2}}Z_{n+1}-(\la_0^{2^{n+1}}-1)Z_n$, $\forall n\ge0$. Then
we have $Z_{n+2}-(\la_0^{2^{n+2}}-1)Z_{n+1}=Z_{n+1}-(\la_0^{2^{n+1}}-
1)Z_n\Rightarrow Z_n-(\la_0^{2^n}-1)Z_{n-1}$ is constant $\Rightarrow$
$Z_{n+2}=(\la_0^{2^{n+2}}-1)Z_{n+1}+c$, $\forall n \ge 0\Rightarrow Z_n=
\prod_{j=1}^n(\la_0^{2^j}-1)\cdot w_n$, where $w_n$ satisfies 
$w_{n+2}=w_{n+1}+\frac c{\prod_{j=1}^{n+2}(\la_0^{2^j}-1)}$,
$\forall n \ge 0 \Rightarrow w_n=w_1+c\sum_{r=2}^n \prod_{j=1}^r
(\frac1{\la_0^{2^j}-1})$, where $c=Z_1-(\la_0^2-1)Z_0$,
$w_1=\frac{Z_1}{\la_0^2-1}$.

Applying this results for $p_n$ and $q_n$ we obtain
$p_n=\prod_{j=1}^n(\la_0^{2^j}-1)\cdot w_n$ and  \linebreak 
$q_n=\prod_{j=1}^n(\la_0^{2^j}-1)\tilde w_n$, where
$w_n=w_1+c\sum_{r=2}^n\prod_{j=1}^r(\frac1{\la_0^{2^j}-1})$ and
$\tilde w_n=\tilde w_1+\tilde
c \sum_{r=2}^n \prod_{j=1}^r(\frac1{\la_0^{2^j}-1})$.
We have $c=p_1-(\la_0^2-1)p_0=(\la_0-1)-(\la_0^2-1)\cdot 0=\la_0-1$,
and $w_1=\frac{p_1}{\la_0^2-1}=\frac1{\la_0+1}\Rightarrow w_n=(\la_0-1)
 \sum_{r=1}^n \prod_{j=1}^r (\frac1{\la_0^{2^j}-1})$. On the other
hand, $\tilde c=q_1-(\la_0^2-1)q_0=\la_0^2-(\la_0^2-1)=1$, and
$w_1=\frac{q_1}{\la_0^2-1}=\frac{\la_0^2}{\la_0^2-1}\Rightarrow \tilde w_n=1+
 \sum_{r=1}^n \prod_{j=1}^r(\frac1{\la_0^{2^j}-1})$.
Since $\frac{p_n}{q_n}=\frac{w_n}{\tilde w_n}$, if we define
$\al(\la)= \sum_{r=1}^\infty \prod_{j=1}^r(\frac1{\la^{2^j}-1})$, we
have $c(\la)= \lim_{n\to\infty}
\frac{p_n}{q_n}=\frac{(\la-1)\al(\la)}{\al(\la)+1}$. \qed
\end{pf}

Let us consider the case $c=c(\la)$ in more detail: Let $c_{-1}=1$,
$c_0=c(\la)$, $\la_n=\la_0^{2^n}$, $c_{n+1}=\la_n^2-\frac{\la_n-1}{c_n}$,
$\forall n\ge0$ and $d_n= \prod_{r=0}^nc_r$. Then, for each $n\ge0$, the
interval $[0,d_{n-1}]$ is invariant by $\psi_{\la,c}^{2^n}$, which is conjugated
by the diffeomorphism $h_n(x)=\frac x{d_n}$ to $\psi_{\la_n,c_n}$. We can
decompose $\psi_{\la,c}^{2^n}$ in his two components
$\vr_n^{(1)}:=\psi_{\la,c}^{2^n}|_{[0,\frac{d_{n-1}}{\la_n}]}$ and
$\vr_n^{(2)}:= \psi_{\la,c}^{2^n}|_{[d_{n-1}(1-\frac{c_n}{\la_n}),d_{n-1}]}$,
which are expansive affine maps.

Let $K$ be the maximal invariant set of $\psi_{\la,c}$ in $[0,1]$. Clearly $K$
is a closed set. To prove that $K$ is a Cantor set, it is enough to prove that
$K$ has no isolated points. Let $\delta= \max_{x\in K}( \min_{y\in
K}\lv y-x \rv)$. We want to prove that $\delta=0$.

Let $K_n=K\cap[0,d_{n-1}]$. We have $K_0=K$, and $K_n=(\vr_n^{(2)})^{-
1}(K_{n+1})\cup \bigcup_{j=0}^\infty(\vr_n^{(1)})^{-j}(K_{n+1})$, that is,
$K_n$ is the union of countable many congruent or reduced copies of $K_{n+1}$,
which shows that $\delta_n\le\delta_{n+1}$, where $\delta_n:= \max_{x\in
K_n}( \min_{y\in K_n} \lv y-x \rv)$. Since $\delta_n\le d_{n-1}$, we have
$ \lim_{n\to\infty}\delta_n=0\Rightarrow\delta=\delta_0=0$, because
$\delta_0\le\delta_n$, $\forall n\in \bold N$.

Let us prove now that the Hausdorff dimension of $K$ is $0$. Let $n\in\bold N$
and define 
$$
\tilde c_\la(n)= \frac{\la-1}{\la^2-\frac{\la^2-1}{\ddots \frac{}{\la^{2^n}-
1}}}.
$$
Then, if $c=\tilde c_\la(n)$, $\psi_{\la,c}$ is $n$ times renormalizable, and
after $n$ renormalizations we obtain the set $K_n$ which is the maximal
invariant set of $\psi_{\la^{2^n},1}$, such that $\tilde K$ is the union of
countable many diffeomorphic images of $K_n$, and so has the same Hausdorff
dimension of $K_n$, where $\tilde K$ is the maximal invariant set of
$\psi_{\la,c}$ in $[0,1]$. On the other hand, $K_n$ is an affine Cantor set and
his Hausdorff dimension $h$ satisfies $(\la^{2^n})^{-h}+(\la^{2^n})^{-h}=1\Rightarrow
h=\frac{\log 2}{2^n\log\la}$ (see [P-T]). Since $c(\la)<\tilde c_\la(n)$,
$\forall n\in\bold N$, we have $HD(K)\le-\frac{\log 2}{2^n\log\la}$, $\forall
n\in\bold N\Rightarrow HD(K)=0$. Our last claims follow from the fact
that if $c_1<c_2$ then there is an affine image of $K_{\la,c_1}$ contained in
$K_{\la,c_2}$, where $K_{\la,c}$ is the maximal invariant of $\psi_{\la,c}$ in
$[0,1]$. Indeed, the restriction of $\psi_{\la,c_2}$ to
$[\ve,1-\la\ve]$ is conjugated to $\psi_{\la, \frac{c_2-\la^2\ve-\ve}{1-
(1+\la)\ve}}$ (see Figure 8). Taking $\ve=\frac{c_2-c_1}{1+\la^2-c_1(1+\la)}$ we are done.

\begin{rem} If $c>c(\la)$ then $c>\tilde c_\la(n)$ for some $n$, and so
$HD(K)\ge HD(K_n)=\frac{\log 2}{2^n\log\la}>  0$
\end{rem}

\begin{figure}[h]
\begin{center}
\includegraphics{fig8.eps}
%\caption{}
\end{center}
\end{figure}


We have proved the following

\begin{thm} Let $\la \in(1,2]$ and $[a,b]\subset[0,1]$. Let
$\ve=\max\{0,\frac\la2-b, \frac a\la-(1-\frac\la 2)\}$, and let
$c=\frac{\frac\la2-\ve-\la^2(1+\ve-\frac\la2)}{\frac\la2-\ve-\la(1+\ve-
\frac\la2)}=\la-1-\frac{4\ve}{\la(\la-1)-2(\la+1)\ve}$.

If $c(\la)=\frac{(\la-1)\al(\la)}{1+\al(\la)}$, where $\al(\la)= \sum_{r=1}^\infty
\frac{1}{ \prod_{j=1}^r(\la^{2^j}-1)}$, and if $K$ is the maximal
invariant of
$T_\la$ in $[a,b]$, then:

\begin{itemize}

\item[i)] $\la-1>c>c(\la)\Rightarrow K$ is a Cantor set with positive Hausdorff
dimension;

\item[ii)] $c=c(\la)\Rightarrow K$ is a Cantor set with Hausdorff dimension
$0$;

\item[iii)] $c<c(\la)\Rightarrow K$ is countable.
\end{itemize}

\end{thm}

\bigskip

\noindent
{\sc Remark 0:} If $c=\la-1$, which is equivalent to $\ve=0$, then $K$ is an interval.

\bigskip

\noindent
{\sc Remark 1:} Conditions i), ii) and iii) are respectively equivalent to

\begin{itemize}

\item[i')] $[a(\la), b(\la)]\subset (a,b)$;

\item[ii')] $[a(\la), b(\la)]\subset [a,b]$ and $a=a(\la)$ or $b=b(\la)$;

\item[iii')] $[a(\la), b(\la)] \not\subset [a,b]$

\end{itemize}

\noindent
where, given $\ve>0$ for which $c=c(\la),a(\la)=a_\ve=\frac{\la(1-c(\la))}{1+\la^2-c(\la)(1+\la)}$ and $b(\la)=b_\ve=\frac{\la(\la-c(\la))}{1+\la^2-c(\la)(1+\la)}$. 

\bigskip

\noindent
{\sc Remark 2:} $c=\la-1- \frac{4\ve}{\la(\la-1)-2(\la+1)\ve}$ is a strictly decreasing function of $\ve$, whose largest value is $\la-1 > c(\la)$, so for each $\la\in(1,2]$, there is an unique $\ve>0$ such that $c=c(\la)$.


\section{The Case $\la=2$:}

For $\la=2$ we have an interesting arithmetic phenomenum involving the binary
representation of $b(\la)$. Let $x\in[0,1]$ such that $T_2^n(x)\ne0$, $\forall
n\in\bold N$ (this is equivalent to $x\ne\frac K{2^n}$, $\forall K,n\in\bold
N)$. Then we can associate to $x$ the sequence $(\sigma_i)_{i\ge0}$, where
$$
\sigma_i=\begin{cases} 0, & \text{ if } T_2^i(x)\in(0,1/2)\\
1, & \text{ if } T_2^i(x)\in (1/2,1) \end{cases}.
$$
If the representation of $x$ in basis two is
$x= \sum_{j=0}^\infty\frac{\mu_j}{2^{j+1}}$ then $\sigma_0=\mu_0$ and, for
$n\ge1$, 
$$
\sigma_n=\begin{cases}
\mu_n, & \text{ if } \mu_{n-1}=0\\
1-\mu_n, & \text{ if } \mu_{n-1}=1. \end{cases}
$$
and so we can reobtain the $\mu_i$'s from the $\sigma_i$'s. The restriction of
$T_2$ to $[a(2),b(2)]$ is conjugated to $\psi_{2,c(2)}$, and $b(2)$ correspond
to $0$ by this 
conjugacy. Let us study the orbit of $0$ by $\psi_{2,c(2)}$.

Let 
$$
\tau_n:= \begin{cases}
0, & \text{ if } \psi_{2,c(2)}^n(0)\in[0,\frac12]\\
1, & \text{ if } \psi_{2,c(2)}^n(0)\in (\frac12,1] \end{cases}
$$
$$
\tilde\tau_n:= \begin{cases}
0, & \text{ if } (\psi_{2,c(2)}^2)^n(0)\in[0,\frac14)\\
1, & \text{ if } (\psi_{2,c(2)}^2)^n(0)\in [\frac14,\frac12] \end{cases}
$$
then:

\begin{itemize}

\item[i)] $\tilde\tau_n$ is well-defined, because $[0,c(2)]$ is invariant by
$\psi_{2,c(2)}^2\Rightarrow(\psi_{2,c(2)}^2)^n(0)\in[0,c(2)]\subset[0,
\frac12]$, $\forall n\ge0$;

\item[ii)] $\tilde\tau_n=0\Leftrightarrow(\tau_{2n}=0$ and $\tau_{2n+1}=1)$,
and $\tilde\tau_n=1\Leftrightarrow(\tau_{2n}=0$ and $\tau_{2n+1}=0)$;

\item[iii)] $\tilde\tau_n=\tau_n$, $\forall n\ge0$, because, by the
construction,
$\psi_{2,c(2)}^2|_{[0,c(2)]}$ is conjugated to $\psi_{4,c(4)}$, and the
combinatorial behavior of the orbit of $0$ by $\psi_{\la,c(\la)}$ does not
depend on $\la$.

\end{itemize}

The properties i), ii) and iii) allow us to obtain the sequence $\tau_n$
recursively by the rules:
$$
\tau_0=0; \quad \tau_n=0\Rightarrow\tau_{2n}=0 \text{ and } \tau_{2n+1}=1;\quad
\tau_n=1\Rightarrow\tau_{2n}=0 \text{ and } \tau_{2n+1}=0.
$$
We have $(\tau_n)= 0100010101000100\dots$ .

Since the conjugacy reverses orientation, the sequence $(\sigma_i)$
associated to $b(2)$ is given by $\sigma_i=1-\tau_i$, and the sequence $\mu_i$
is given by $\mu_0=\sigma_0=1$,
$$
\mu_{n+1}=\begin{cases}
\sigma_{n+1}, & \text{ if } \mu_n=0\\
1-\sigma_{n+1}, & \text{ if } \mu_n=1. \end{cases}
$$
So, $(\sigma_n)=1011101010111011\dots$, and 
$(\mu_n)=1101001100101101\dots$, that is, \linebreak $b(2)=0,1101001100101101\dots$ in basis 2, and 
$a(2)=2-
2b(2)=0,010110011010010\dots= \sum_{i=0}^\infty\frac{\la_i}{2^{i+1}}$,
where $\la_i=1-\mu_{i+1}$, $\forall i$.

This shows, in particular, that $a(2)$ and $b(2)$ have both 50\%
of zeros and 50\% of ones in their binary representations, but do not have
three consecutive zeros nor three consecutive ones.

It is easy to prove by induction that $b(2)=2m$, where
$$
m=0,0110100110010110\dots = \sum_{i=1}^\infty \frac{m_i}{2^i},
$$
and the digits $m_i$ of the binary representation of $m$ form the 
Morse-Thue sequence defined by $m_1=0$,
$$
m_k=0 \Rightarrow m_{2k-1}=0, \quad m_{2k}=1
$$
$$
m_k=1 \Rightarrow m_{2k-1}=1, \quad m_{2k}=0.
$$
This number $m$ is called the Morse-Thue constant, and is equal to the external
angle in the Mandelbrot set corresponding to the Feigenbaum map of the
quadratic family (see [S] pp. 279, 299; [D]).
We have $b(2)=2m=f(\frac12)$, where $f(x)= \sum_{i=0}^\infty
m_{i+1}x^i$ satisfies $f(x)=(1-x)f(x^2)+\frac{x}{1-x^2}$, and so, by the
results of [L-Po], we know that $b(2)$ (and so $a(2)$) is a transcendental number.
Since we have $b(2)=\frac{4-2c(2)}{5-3c(2)}$, where
$$
c(2)=\frac{\al(2)}{1+\al(2)},\qquad \al(2)=\sum_{r=1}^\infty\frac1
{ \prod_{j=1}^r(2^{2^j}-1)},
$$
we obtain
$$
m=\frac12 b(2)=\frac{2-c(2)}{5-3c(2)}=\frac{2+\al(2)}{5+2\al(2)},
$$
which gives an explicit expression for $m$ in terms of a series.

\section{The Case of Unimodal Maps}

In this section we consider $C^2$ unimodal maps with non-flat critical point.
These maps have no wandering intervals, by Theorem A of Chapter IV of [M-S].
Theorem B of the same chapter implies that these maps have only a finite number
of periodic attractors. In what follows a pre-periodic point is a point with finite positive orbit, i.e., a point that lands on a periodic orbit.

We may suppose by the same initial considerations that we have done for tent
maps that our unimodal maps $f\colon[0,1]\to[0,1]$ satisfy $f(c)=1$, where $c$ is the critical point, and $f(1)=0$ (the tent map $T_\la$ restricted to $L=[\la(1-\frac\la2),\frac\la2]$ satisfies this combinatorial condition after an affine conjugacy).

We have now two cases:

\begin{itemize}

\item[a)] The topological entropy of $f$ is $0$: In this case $f$ is
combinatorially equivalent to $g_a(x)=ax(1-x)$ for some $a\le c$, where
$cx(1-x)$
is the Feigenbaum map of the quadratic family (which is a full family of
unimodal maps). We have two subcases.

\begin{itemize}

\item[a.i)] $a<c$: In this case $g_a(x)$ has a finite number of periodic
points which form the non-wandering set of $g_a$. In this case, if
$J\not\supset[0,1]$, the maximal invariant set of $f|_J$ is infinite if and
only if $J$ contains a recurrent interval (that is, there is an interval
$I_0\subset I$ and an $m\in\bold N$
such that $f^m(I_0)\subset I_0$, $f^k(I_0)\subset I$, for $k=1,2,\dots,m$  and
$I_0$ is contained in the basin of
attraction of some periodic point of period $m$ or $f^m|_{I_0}=id$).

\item[a.ii)] $a=c$: In this case the closure of the orbit of the critical
point is a Cantor set $K$, and the $\omega$-limit of any point $x$
can be a periodical point if $x$ is pre-periodic or else must be equal to $K$.
Here if $J\ne[0,1]$ and $J$ does not contain any recurrent interval
then the maximal invariant of $f|_J$ must be finite (see Ex. 5.3 of Chapter II
of [M-S]).
\end{itemize}

\item[b)] The topological entropy of $f$ is strictly positive: In this case we
can use Proposition 4.3 of Chapter III of [M-S] to obtain a semi-conjugacy
between $f$ and $T_{\exp(h_t(f))}$, where $h_t(f)$ is the topological entropy
of $f$. This semi-conjugacy can collapse an interval only if it is contained
into $X$, where $X$ is the set of points which are eventually mapped into a
restrictive interval (see Section II.5 of [M-S]) or into the basin of a periodic
attractor.


Let $\la=h_t(f)$. We have the following commutative diagram:
$$
\begin{matrix}
[0,1] & \overset{f}{\longrightarrow} & [0,1] \\
h \downarrow &    & \downarrow h\\
\Big[ \lambda(1-\frac{\lambda}{2}), \frac{\lambda}{2} \Big] & \overset{T_\lambda}{\longrightarrow} & \Big[ \la(1-\frac{\la} 2),  \frac{\la} 2 \Big]
\end{matrix}
$$
$h$ is a monotonous semi-conjugacy, and if $x$ is not a pre-periodic point of
$T_\la$ then $h^{-1}(x)$ is a single point.

Let $I$ be an interval. We shall prove that the maximal invariant of $f|_I$ is
uncountable if and only if $I$ contains the orbit of some recurrent 
interval 
or the maximal invariant of $T_\la|_{h(I)}$ is uncountable.

For this, let $a(\la)$ and $b(\la)$ as before, $J=[a(\la),b(\la)]$ and $\tilde
J=h^{-1}(J)$. Since $a(\la)$ and $b(\la)$ are not pre-periodic, we have $\tilde
I\not\subset\tilde J\to h(\tilde I)\not\subset h(\tilde J)=J$. Thus, if
$T_\la|_{h(I)}$ is uncountable then $J\subset h(I)\Rightarrow K$ is contained
in the maximal invariant of $T_\la|_{h(I)}$, where $K$ is the maximal invariant
of $T_\la|_J\Rightarrow h^{-1}(K)$ is contained in the maximal invariant of
$f|_I$ (because $x\in h^{-1}(K)\Rightarrow T_\la^n(h(x))\in J$, $\forall
n\in\bold N\Rightarrow h(f^n(x))\in J$, $\forall n\in\bold N\Rightarrow
f^n(x)\in h^{-1}(J)$, $\forall n\in\bold N\Rightarrow f^n(x)\in I$, $\forall
n\in\bold N$). Therefore in this case the maximal invariant of $f|_I$ is
uncountable. If $I$ contains the orbit of some recurrent interval then clearly
the maximal invariant of $f|_I$ has non-empty interior.

If $h(I)\not\supset J$ then the turning point of $T_\la$ does not belong to the
maximal invariant of $T_\la|_{h(I)}$, and so the pre-images of a restrictive
interval are disjoint of the maximal invariant of $f|_I$. Moreover, the maximal
invariant of $T_\la|_{h(I)}$ is countable, and consists only of pre-periodic
points. Thus the maximal invariant of $f|_I$ is uncountable if and only if
there is a periodic point $p$ of period $m$ and a closed interval $I_0$, with
$p\in I_0$ such that $f^m(I_0)\subset I_0$ and $f^k(I_0)\subset I$, for $0\le
k< m$, $I_0$ being collapsed by the semi-conjugacy. 

\begin{itemize}
\item[b.i)] For unimodal maps $f$ without recurrent intervals (for maps of the quadratic family or, more generally, with negative Schwarzian derivative, it means that $f$ is non-renormalizable,
and that the critical point does not belong to the immediate basin of attraction of a periodic point), the map $h$ is a conjugacy, so the maximal invariant set of $f|_I$ is uncountable if and only if $I\supset h^{-1}([a(\la),b(\la)])$. Here $\la=h_t(f)$ and $a(\la)$ and $b(\la)$ were defined in the end of Section 2.

\end{itemize}
\end{itemize}

Under these 
hypotheses, 
if $f$ is a Collet-Eckmann map (i.e., if there are $C>0$, $r>1$ such that $|Df^n(f(C))|\ge Cr^n$ for any $n\in \mathbb N$, where $C$ is the critical point of $f$) then $h$ is a H\"older-continuous
conjugacy, and the maximal invariant set of $f|_I$ has positive Hausdorff dimension if and only if the same occurs for $T_\la|_{h(I)}$. In particular, the maximal invariant set of $f|_{h^{-1}([a(\la),b(\la)]}$ is a Cantor set with zero Hausdorff dimension. We do not know if it is true in more general settings, and in particular if we drop the hypothesis of $f$ being Collet-Eckmann. \qed

\vskip .3in

\noindent
{\bf Acknowledgement:} I would like to thank Welington de Melo for helpful
conversations about this subject, and 
to the Referee for his helpful corrections and suggestions.

\vskip .5in

\noindent
{\Large\bf References:}

\begin{itemize}

\item[{[D]}] A. Douady; Julia sets and the Mandelbrot set. In H. -O. Peitgen
and P.H. Richter: The beauty of Fractals (Springer, Berlin), 
pp. 161-173.

\item[{[L-Po]}] J.H. Loxton and A.J. Van der Poorten, Arithmetic
properties of the solutions of a class of functional equations,
J. -Reine-Angew. -Math. 330 (1982), pp. 159--172.

\item[{[M-S]}] W. de Melo and S. Van Strien, One-Dimensional Dynamics, Ergebrisse
der Mathematik und ihrer Grenzgebiete 3. Folge-Band 25, Springer-Verlag,
1991.

\item[{[P-T]}] J. Palis and F. Takens, Hyperbolicity and sensitive chaotic
dynamics at homoclinic bifurcations: fractal dimensions and infinitely many
attractors, Cambridge Univ. Press (1992).

\item[{[S]}] M. Schroeder, Fractal, Chaos, Power Laws - W.H. Freeman and
Company, New York, 1991.
\end{itemize}

\end{document}
