On the fixed points of the Ruelle operator.
Carlos Cabrera and Peter Makienko.
We discuss the relation between the existence of fixed points of the Ruelle operator, acting on different Banach spaces, with Sullivan's conjecture in holomorphic dynamics.
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\title{On the fixed points of the Ruelle operator.}
\author{Carlos Cabrera and Peter Makienko.}

\begin{document}
\begin{abstract}
We discuss the relation between the existence of fixed points of the Ruelle 
operator, acting on different Banach spaces, with Sullivan's 
conjecture in holomorphic dynamics.  
 \end{abstract}
\maketitle
\footnotetext{This work was partially supported by PAPIIT IN-102515 
and CONACYT CB2015/255633.}

\section{Introduction}

Let $Rat_d(\bar{\C})$ be the set of all rational maps on the 
Riemann sphere $\bar{\C}$ of  degree $d$. 
When $R$ is an element  in $Rat_d(\bar{\C})$, the postcritical set 
of $R$ is given by $$
P_R=\overline{\bigcup_k\bigcup_i R^i(c_k)},
$$  
where the union is taken over all critical points $c_k$ of $R$ and $i>0$. 
The \fd{Julia set} $J_R$ is the accumulation set points of all 
periodic points of $R$, with all isolated points removed. 
The map $R$ is called \fd{hyperbolic} when  $P_R\cap J_R=\emptyset$.  
The Fatou conjecture states: \textit{Hyperbolic maps are open and everywhere 
dense in $Rat_d(\bar{\C})$.}
 
Recall that a rational map $R$ is \fd{$J$-stable} if 
there is an open neighbourhood $U$ of $R$ in $Rat_d(\bar{\C})$, 
such that for $Q \in U$ there exists a homeomorphism 
$h_Q:J_R\rightarrow J_Q$, quasiconformal in the Pesin's sense (also known as 
metric quasiconformal), with 
$$Q=h_Q\circ R\circ h_Q^{-1}.$$

Due to Ma\~n\'e, Sad and Sullivan (see \cite{MSS}),  the set of $J-$stable 
maps forms an open and everywhere dense subset of $Rat_d(\bar{\C})$, even 
more, a $J-$stable map is hyperbolic if and only if there is no invariant 
Beltrami differential supported on the Julia set.
Since hyperbolic maps are $J-$stable, hence the Fatou conjecture 
becomes: \textit{The set of $J-$stable maps is precisely the set of hyperbolic 
maps.} 

An  \textit{invariant Beltrami differential}  $\mu$ is a 
$(-1,1)$-differential form locally expressed by $\mu(z)\frac{
d\bar{z}}{dz}$ and whose coefficient $\mu(z)$ is as $L_\infty$ function 
that satisfies  
$$
\mu(z)=\mu(R(z))\frac{\overline{R'(z)}}{R'(z)}.
$$ 
That is, $\mu$ is a fixed point of the Beltrami operator, as defined below, 
acting on the space $L_\infty(\bar{\C})$ with respect to the planar 
Lebesgue measure. Along this paper, whenever is clear from the context, 
we will denote both the differential and its coefficient by the same letter.

Sullivan's conjecture states: \textit{There exists an invariant Beltrami 
differential supported on the Julia set if and only if $R$ is a flexible 
Latt\'es map.} For definitions and further properties of Latt\'es 
maps, see Milnor's paper \cite{MilLat}. 

Note that if $R$ is hyperbolic 
then $R^n$ is hyperbolic and hence $J-$stable for every $n>1.$ We have the 
following statement (see \cite{CMdecomp} and the 
discussion therein): For $n>1$ the iterated map $R^n$ is $J$-stable if and only 
if $R$ is hyperbolic. Therefore the Fatou conjecture is true when one is 
considering an iterated rational map.  On the other hand, Sullivan's conjecture 
predicts not only 
the absence of fixed points 
for the Beltrami operator but also the absence of periodic points for this 
operator.
In other words, there is no eigenvalue of the Beltrami operator which is a root 
of unity. In this way, Sullivan's conjecture can be interpreted as a spectral 
problem for a semigroup of  Beltrami operators.

According to Sullivan's dictionary between Kleinian groups and 
holomorphic dynamics, rational maps correspond to finitely 
generated Kleinian groups. In this setting, we can reformulate Teichm\"uller 
theorem (see \cite{GardLakic}) in the following way:

\textit{Given a finitely generated Kleinian group $\Gamma.$ For any 
$\Gamma$-invariant Beltrami differential $\mu$ which defines a non-trivial 
quasiconformal deformation (see definition below) on the associated Riemann 
surface $S_\Gamma$, 
there exists a $\Gamma$-invariant holomorphic two form $\phi$ such that $\int_F 
\mu \phi\neq 0$ for any fundamental domain $F$ of $\Gamma$.}

In other words, the following separation principle holds: The space of 
invariant holomorphic two forms separates the space of quasiconformally 
non-trivial invariant Beltrami differentials. 

The separation principle is well-known in ergodic theory for bounded semigroups 
of linear endomorphisms of a Banach space. In fact, it is the subject of many 
ergodic theorems and is one of the oldest principles in this theory.

Due to the observations above, in this article we discuss the following 
question: 

Which consequences arise from the existence of a common 
non-trivial fixed point for a representation of the dynamics as a semigroup of 
contractions on a suitable Banach space? 

Furthermore, we will discuss what happens when these representations satisfy a 
separation principle (definitions will be given below).

To keep in line with Sullivan's conjecture we consider representations that 
arise as versions of complex pull-back or push-forward 
operators acting on either invariant subspaces $X \subset L_p(W)$, not 
necessarily closed, or spaces which are predual, dual or bidual to $X$. 
Here $W$ is a  $R$-invariant set in the Riemann sphere 
of positive Lebesgue measure and $1\leq p \leq \infty$. 
Throughout our discussion, unless otherwise stated, 
all $L_p$ spaces are taken with respect to the planar Lebesgue measure on the 
dynamical plane.  

Most results are given in terms of ergodic theory and suggest that the 
Sullivan's conjecture holds true.  

\par\medskip\noindent \textbf{Acknowledgment.}
The authors would like to thank the referee and A. Poirier  for useful comments 
and suggestions on an early draft of this paper.

\section{The Pull-back and Push-forward actions.}

In this section we give the definition of Ruelle transfer operator for 
a rational map $R$ which is a complex version of Perron-Frobenius operator. 
In this setting, perhaps, the first instance of this operator appeared in 
Ruelle's paper \cite{RuelleZeta}. 

We take the standpoint of the theory of quasiconformal deformations, which deals 
with a dual interplay between Beltrami differentials and quadratic 
differentials (i.e, between some $(-1,1)$-forms and $(2,0)$-forms). 
In that sense, there is a natural action in both spaces  induced by a degree 
$d$ rational function $R$, once we think of $R$ as a ``local change of 
variables''. 

Let $F_{m,n}$ be the space of all $(m,n)$ forms $\alpha=\phi(z)d\bar{z}^ndz^m $ 
where $\phi$ is a complex valued measurable function on $\C$. The 
\textit{pull-back} operator acting 
on $F_{m,n}$ 
is given by $$R_{*(m,n)}(\alpha)=\alpha\circ R= 
\phi(R)\overline{(R')}^n(R')^m 
d\bar{z}^ndz^m.$$ The \textit{push-forward} operator on $F_{m,n}$ is given 
by $$R^*_{(m,n)}(\alpha)=\sum \alpha(\zeta_i)$$ where the sum is taken over all 
branches $\zeta_i$ of  $R.$ Therefore we have  $R^*_{m,n}\circ R_{*\,m,n}= 
deg(R) Id.$ 

The \textit{Beltrami operator }is $B_R=R_{*(-1,1)}$. The 
operator  $|B_R|=R_{*(0,0)}$ is called the \textit{modulus} of  $B_R$ and 
satisfies 
$|B_R||\phi|=|B_R(\phi)|$ almost everywhere for every measurable function 
$\phi.$ 

The \textit{Ruelle transfer operator}, shortly the \textit{Ruelle operator}, 
is $R^*=R^*_{(2,0)}$, while the operator $|R^*|=R^*_{(1,1)}$ is called the 
\textit{modulus of Ruelle operator} and satisfies $|R^*(\phi)|\leq |R^*||\phi|$ 
almost everywhere for every measurable function $\phi.$  The modulus of Ruelle 
operator is also known as the \textit{Perron-Frobenius} operator for the map 
$R$ or as the \textit{push-forward} operator on the space of absolutely 
continuous measures. 



In coefficients, the Beltrami operator  and its modulus are defined by 
the formulas
$$B_R(\phi)=\phi(R)\frac{\overline{R'}}{R'}, \textnormal{ 
and }|B_R|(\phi)=\phi(R),$$ respectively, where $\phi$ is a measurable function 
with respect to the Lebesgue measure. 

In turn, the Ruelle operator and its modulus are given by

$$R^*(\phi)=\sum \phi(\zeta_i)(\zeta'_i)^2 \textnormal{ and } |R^*|(\phi)=\sum 
\phi(\zeta_i)|\zeta'_i|^2,$$ respectively.  Both  sums are taken over all 
branches $\zeta_i$ of $R$.
We end this section with the following simple facts. 

\begin{proposition}\label{prop.bounds}  For a Lebesgue measurable set $A$, for 
$\mu\in 
L_\infty(\C)$ and $\phi\in L_1(\C)$. 
\begin{enumerate}
 \item 
$$\int_A \mu(z) R^*(\phi(z))|dz|^2= \int_{R^{-1}(A)} 
B_R(\mu(z)) \phi(z)|dz|^2.$$
\item $$\int_A \mu(z) |R^*|(\phi(z))|dz|^2=\int_{R^{-1}(A)} 
\mu(R(z)) \phi(z)|dz|^2.$$

\item Define $R_*(\phi)=\frac{R_{*\,2,0}(\phi)}{deg(R)}=\frac{\phi(R) 
R'^{2}}{deg(R)}$, then
$$\int_A |R_*(\phi(z))| |dz|^2\leq \int _{R(A)} |\phi(z)| dz\wedge 
d\bar{z}.$$

\item If $\phi$ is a holomorphic function outside the postcritical set 
$P_R$, 
then $R^*\phi$ is also a holomorphic function outside $P_R.$
\end{enumerate}
\end{proposition}
\begin{proof} Part (1). Fix a system of branches of $R$ in the 
following way, let $\tau$ be any differentiable arc containing all critical 
values of $R.$ Let $D=\bar{\C} \setminus \tau$ then by the Monodromy Theorem 
each of the branches $\zeta_i$ of $R$ defines a holomorphic function on $D$. 
Let 
$D_i=\zeta_i(D)$, then $D_i\cap D_j=\emptyset$ for $i\neq j.$ Now we compute
$$\int_A\mu(z)R^*(\phi(z))|dz|^2
=\int_{A\cap D}\mu(z)R^*(\phi(z))|dz|^2$$
$$=\int_{A\cap D}\sum_i \mu(z)\phi(\zeta_i(z))(\zeta'_i(z)^2) dz\wedge 
d\bar{z},$$ 
after a change of variables the latter is equal to 
$$\sum_i \int_{\zeta_i(A\cap 
D)}\mu(R(z))\frac{\bar{R}'(z)}{R'(z)}\phi(z)dz\wedge 
d\bar{z}$$
$$=\int_{R^{-1}(A)}B_R(\mu(z))\phi(z)|dz|^2.$$
Similar computations render parts (2) and (3).
 
Part (4). Again by the Monodromy theorem, the Ruelle operator 
does not depend on the local choice of branches of $R.$ As every branch of $R$ 
is, outside the postcritical set, a local holomorphic function and 
$R^{-1}(\bar{\C}\setminus P_R)\subset \bar{\C}\setminus P_R$, then $R^*(\phi)$ 
is a sum of holomorphic functions, so is holomorphic. \end{proof}

If $A$ is an invariant set, that is $R(A)\subset A$, then we have that 
$R^*$ acts over $L_1(A)$ with $\|R^*\|\leq 1.$ If $A$ is completely invariant 
(i.e. $R^{-1}(A)=A$), then $B_{R}$ acts over $L_\infty(A)$ with $\|B_R\|=1$ and 
$B_R$ is dual to $R^*.$ Note that the unit ball of the space of fixed points of 
$B_R$ in $L_\infty(\C)$ parametrizes all quasiconformal deformations of the 
rational map $R.$

When  $A$ is backward invariant  (i.e. $R^{-1}(A)\subset A$) and $\phi$ 
is an integrable function on $A$, then $$\int_A |R^*(\phi)| \leq \int_A 
||R^*|(\phi)|\leq \int_A |R^*||\phi|= \int_{R^{-1}(A)}|\phi|\leq \int_A|\phi|.$$
Here, the integration is with respect to the Lebesgue measure. Hence the Ruelle 
operator is a contraction in the $L_1$ norm.

As another entry of Sullivan's dictionary, the reader can recognize Ruelle 
operator as the relative Poincar\'e Theta operator for branched coverings of 
the sphere onto itself. 

\section{The Thurston operator}\label{sec.Thurs}


Following Teichm\"uller theory ideas, an important space on which the Ruelle 
operator acts is the space of  functions that are holomorphic on certain open 
set.  

Fix an invariant compact set $K$ containing the postcritical set $P_R$.  Let  
$H(K)$ be the subspace of $L_1(\bar{\C})$ of all functions that are holomorphic 
outside of $K$ with the restricted norm of $L_1(\bar{\C})$. Then the Ruelle 
operator $R^*$ is a contractive endomorphism of $H(K).$   Take
$S_K=\C\setminus K$, and let $A(S_K)$ be the space of all 
integrable holomorphic functions on $S_K$ with the restricted measure. The 
Ruelle operator also acts as contracting endomorphism of $A(S_K)$ by the 
discussion on the previous section. 
Every element $f$ in $A(S_K)$ extends to an element in $H(K)$: Just set 
$f(z)=0$ for all $z$ in $K$. This extension gives a canonical inclusion from 
$A(S_K)$ into $H(K)$, which is an isomorphism precisely when the Lebesgue 
measure of $K$ is $0.$ 

Let $B(S_K)$ denote the associated Bergman space, that is, the space of all 
holomorphic functions $\phi$ on $S_K$ with the following $L_\infty$-norm 
$$
\|\phi\|=\sup_{z\in 
S_K}|\lambda_K^{-2}(z)\phi(z)|
$$  where $\lambda_K$ denotes the complete hyperbolic metric on $S_K.$ By the 
Bers Isomorphism theorem (see \cite[Theorem 2.1]{Krabook}), the space 
$A^*(S_K)$, the dual of $A(S_K)$, is linearly isomorphic to $B(S_K)$. 
Furthermore, there is an equivalent norm, called the \fd{Teich\-m\"uller norm}, 
on $B(S_K)$ which is the canonical supremum norm of continuous linear 
functionals on the unit sphere in $A(S_K)$.

Let $T:B(S_K)\rightarrow B(S_K)$ be the dual operator of $R^*$ up to the 
identification above. Then  $T$ is a power bounded operator which is a 
contraction in the Teichm\"uller norm. The operator $T$ is called the 
\fd{infinitesimal Thurston pull-back} operator or, for short, the \fd{Thurston 
operator.} 

Let $B_0(S_K)$ be the subspace  of all elements in $B(S_K)$ vanishing at 
infinity. In other words, $B_0(S_K)$ is the space of all $\phi$ in $B(S_K)$ 
such that $|\lambda_K^{-2} \phi(z_j)|$ converges to zero, whenever $z_j$ is a 
sequence converging to the boundary $\partial S_K.$ 

Let $A_*(S_K)$ be the subspace of $A^*(S_K)$ such that the dual space 
$(A_*(S_K))^*$ is isometrically isomorphic to $A(S_K)$ (see for instance 
Theorem 5 on page 52 of \cite{GardLakic}). The space $A_*(S_K)$ is constructed 
as follows. A sequence $\{\phi_j\}$ in $A(S_K)$ is \fd{degenerating} when 
$\|\phi_j\|=1$ for all $j$ and $\phi_j$ converges pointwise to $0$ on $S_K$. 
Then $A_*(S_K)$ is the kernel of the seminorm on $A^*(S_K)$ given by
$$\beta(l)=\sup 
|l(\phi_i)|,$$
where the supremum is taken over all degenerating sequences $\{\phi_i\}$ in 
$A(S_K).$ 

The Bers Isomorphism theorem together with Theorem 1 of \cite{BonetWolf}, 
provide a guide between the topologies and Banach structures on the spaces we 
are dealing with as follows.

\begin{equation*}
\begin{split}
B(S_K)\simeq& B_0(S_K)^{**} \overset{f^{**}}\longleftrightarrow 
A^*(S_K)\overset{g^{**}}{\longrightarrow} \ell_\infty=c_0^{**},\\
&B_0(S_K)^{*}\overset{f^*}{\longleftrightarrow} 
A(S_K)\overset{g^*}{\longleftarrow}\ell_1,\\
&B_0(S_K)\overset{f}{\longleftrightarrow} 
A_{*}(S_K)\overset{g}{\longrightarrow} 
c_0.
\end{split}
\end{equation*}


Above  $f$ is the restriction map of Bers's isomorphism over $A_*(S_K)$ 
and, by Lemma 1 and Corollary 1 of pages  259-260  in \cite{GardLakic}, 
is a surjective map onto $B_0(S_K)$. There is an isomorphism $h$ onto its image 
from $B_0(S_K)$ into $c_0$ given by Theorem 1 in \cite{BonetWolf}. The 
map $g$ is just the composition $h\circ f$. Here $\ell_\infty$, $\ell_1$ and 
$c_0$ denote the spaces of complex valued sequences that are bounded, 
absolutely summable and converging to $0$, respectively. Moreover, the $*$-weak 
topology on $A(S_K)$ induced by $A_*(S_K)$ coincides with the topology of 
pointwise convergence of bounded sequences.

Now we collect some facts about the geometry and dynamics of operators on  
$B(S_K)$ and $B_0(S_K)$. First we need some definitions. 
A Banach space $B$ is a \textit{Grothendieck space} if every $*$-weak 
convergent sequence of continuous functionals $\{l_i\}$ also converges 
in the weak topology on $B^{*}$.

Every complemented closed subspace of a Grothendieck space is again a 
Grothen\-dieck space. Clearly, every reflexive space is a Grothen\-dieck space. 

A Banach space $B$ has the \textit{Dunford-Pettis property} if every weakly 
compact operator from $B$ into any Banach space maps weakly compact sets into 
norm compact sets. As in the case of Grothendieck spaces, the Dunford-Pettis 
property is also inherited on complemented closed subspaces. 
A typical example of a Grothendieck space with the Dunford-Pettis property is 
$L_\infty(X,\mu)$ where $(X,\mu)$ is a positive measure space (see \cite{Lotz}).

A series $\sum x_n$ in a Banach space $X$ is called weakly unconditionally 
Cauchy (wuC), if for every $l\in X^*$, the series $\sum|l(x_n)|< \infty.$ A 
Banach space $X$ is said to have \textit{property (V) of Pe\l{}czy\'{n}sky} if 
every subset $K\subset X^*$ is relatively weakly compact whenever $K$ 
satisfies  $$\lim_n \sup_{l\in K}|l(x_n)|=0$$ for
every wuC series $\sum x_n$ in $X$. 

By results of functional analysis (see, for example, Corollary 3.7, page 132 in 
\cite{HarmandWWerner}), any closed subspace $Y\subset c_0$ has the property (V) 
of Pe\l{}czy\'{n}sky.

Hence we have the following 

\begin{itemize}
 \item $B_0(S_K)$ has the property (V) of  Pe\l{}czy\'{n}sky.
 \item $B(S_K)$ is a Grothendieck space with the Dunford-Pettis property.
\end{itemize}

Indeed, by results of Bonet-Wolf  in \cite{BonetWolf}, the space $B_0(S_K)$
is isomorphic to a closed subspace of $c_0.$ 

By the Bers Isomorphism theorem,  we 
have $L_\infty(S_K)\cong N\oplus B(S_K).$ So $B(S_K)$ is a complemented
subspace of a Grothendieck space with the Dunford-Pettis property.

Again, by a combination of classical results in functional analysis we have the 
following fact. 
\begin{itemize}
\item If $E:B_0(S_K)\rightarrow B_0(S_K)$ is a linear operator then either $E$ 
is compact or there exists an infinite dimensional subspace $Y$, which is 
isomorphic to $c_0$ such that $E|_Y$ is an isomorphism onto its image. 
\end{itemize}
Indeed, as noted above $B_0(S_K)$ has the property (V) of Pe\l{}czy\'{n}sky, 
then by the Lemma 3.3.A and the Theorem 3.3.B on page 128 of 
\cite{HarmandWWerner}, either $E$ is weakly compact or there exists an infinite 
dimensional subspace $Y$, which is isomorphic to $c_0$ such that $E|_Y$ is an 
isomorphism onto its image. But  if $E$ is weakly compact then 
$E^*:B^*_0(S_K)\rightarrow B_0^*(S_K)$ is also weakly compact. Let us show 
that every bounded weakly convergent sequence $\{\psi_n\} \subset A(S_K)$ 
contains a norm convergent sequence. In fact, every bounded sequence in 
$A(S_K)$ forms a normal family. Let $\{\psi_{n_k}\}$ be a subsequence of 
$\{\psi_n\}$ converging to its pointwise limit $\psi.$ Hence, $\psi_{n_k}$ is 
weakly convergent and by the Fatou lemma $\psi\in A(S_K).$ Then by the Uniform 
Integrable Convergence in Measure theorem (see Theorem 1.5.13 in 
\cite{Taomeasure}), $\psi_{n_k}$ converges to $\psi$ in norm. Therefore 
$E^*$ and, hence, $E$ are compact operators. 
 
Let us note, by using similar arguments as above and other theorems in 
functional analysis, is possible to show that in fact $B_0(S_K)$ is isomorphic 
to $c_0.$ 

We say that an invariant Beltrami differential $\mu$ defines a 
\textit{non-trivial quasiconformal deformation} if and only if 
$l_\mu(\psi)(z)=\int_{S_K} \psi(z) \mu(z) |dz|^2$ is a non-zero 
invariant 
functional on 
$A(S_K)$. In other words, $l_\mu$ induces a non-zero fixed point for $T$ on 
$B(S_K).$ 

Finally, in order to use the results of the second author from 
\cite{MakRuelle}, without cumbersome recalculations, along the paper we will 
assume that $\{0,1,\infty\}$ are fixed points of $R$.  This is always the case 
after passing to a suitable iteration of $R$ and conjugating with a M\"obius 
map. However, most of our results do not need this normalization.
We fix $K_R= P_R\cup \{0,1\}$ and set $S_R=\C\setminus K_R.$

\section{Mean ergodicity in holomorphic dynamics}\label{section4}

Given an operator $S$ on a Banach space $X$,  the 
$n$-\textit{Ces\`aro averages} of $S$ are the operators $A_n(S)$ defined for 
$x\in 
X$ 
by $$A_n(S)(x)=\frac{1}{n}\sum_{i=0}^{n-1}S^i(x).$$


An operator $S$  on a Banach space $X$ is called  \textit{mean ergodic} if $S$ 
is power bounded, that is,  satisfies $\|S^n\|\leq M$ for some number $M$ 
independent of $n$, and the Ces\`aro averages $A_n(S)(x)$ converge in  norm for 
every $x\in X$.

The topology of convergence in norm is also called the strong topology on $X.$
If $A_n(S)$ converges uniformly on the closed unit ball on $X$ (that is 
$\lim_{n,m\rightarrow \infty} \|A_n(S)-A_m(S)\|=0$), then the operator $S$ is 
called \fd{uniformly ergodic}. The following facts can be found, for example, 
in Krengel's book \cite{Krengel}.

 
\begin{enumerate}
 \item (\textbf{Separation principle)}. The operator $S$ is mean-ergodic if 
satisfies the principle of separation of points: If $x^*$ is a fixed point of 
$S^*$, then there exist  $y\in X$, a fixed point of $S$, such that 
$\langle y,x^*\rangle \neq 0.$

\item The limit $\lim_{n\rightarrow \infty} A_n(S)(x)=0$ holds if and only if 
$x\in  \overline{(Id-S)(X)}.$

\item (\textbf{Mean Ergodicity lemma}). Let $Conv(S,x)$ be the convex hull of 
the orbit of a point $x$ under $S$. Then $y$ belongs to the weak closure of 
$Conv(S,x)$ if and only if $y$ is a fixed point of $S.$ In this situation, 
$A_n(S)(x)$ converges to $y$ in norm. If $X$ is a dual space, then $y$ belongs 
to the *-weak closure of $Con(S,x)$ if and only if $y$ is a fixed point of $S.$ 
Here, a dual space is a space isometrically isomorphic to $B^*$ 
for some Banach space $B$ where the notion of $*$-weak topology is well 
defined. 

\item (\textbf{Uniform Ergodicity lemma}). A power bounded operator $S$ acting 
on a Banach space $B$ is uniformly ergodic if and only if either the subspace 
$(Id-S)B$ is  closed or, equivalently, if and only if the point $1$ 
either belongs to the resolvent set or is an isolated eigenvalue of $S$. 
If $dim(Fix(S))$ is finite then the intersection of the spectrum of $S$ with 
the unit circle consists of finitely many isolated eigenvalues. 
\end{enumerate}

Let $Hol(R)$ be the space of all rational functions $\phi$ integrable, with 
respect to the planar Lebesgue measure, with poles in the forward 
orbit of the set  $V(R)\cup \{0,1\}$ where $V(R)$ is the set of all 
critical values. Equivalently, $Hol(R)$ consists of all rational functions with 
simple poles in the forward orbit of $V(R)\cup \{0,1\}$ and a zero at infinity 
of multiplicity at least three. Note that $Hol(R)$ is a normed vector space 
with the norm  inherited from the norm of 
$L_1(\bar{\C})$. The  space  $Hol(R)$ is not complete and, by the Bers 
Approximation theorem (see \cite[Theorem 9]{GardLakic}), its completion is 
$H(K_R)$ and contains a canonical inclusion of $A(S_R)$. Hence, the completion 
of $Hol(R)$ is equal to $A(S_R)$ if and only if the Lebesgue measure of $P_R$ 
is zero.

Now we recall some facts from ergodic theory which will be used along this 
text. A positive measure set $M\subset \bar{\C}$ is called \fd{wandering} when 
the sets  $\{R^{-k}(M)\}_k$ are pairwise disjoint Lebesgue measurable sets. Let 
$D(R)$ be 
the union of all wandering sets. The set $D(R)$ is called the \fd{dissipative 
set}, the complement $C(R)=\C\setminus D(R)$  is the \fd{conservative set}. 
Similarly, a positive measure set $M\subset \bar{\C}$ is called \textit{weakly 
wandering} when the sets  $\{R^{-k}(M)\}_k$ contain infinitely many disjoint 
Lebesgue measurable sets. The weakly dissipative set $W(R)$ is the union of all 
weakly wandering sets and $SC(R)=\C\setminus W(R)$) is called the \fd{strongly 
conservative set}. 

The following proposition is a reformulation of Theorem 4.6, page 141, in 
\cite{Krengel}.
 
\begin{proposition}\label{prop.Krengel} Let $R$ be a rational map then
$$
m(R(SC(R))\triangle SC(R))=0.
$$  
Furthermore, if $SC(R)$ has positive measure there exists a non negative
integrable function $P$ which is positive on $SC(R)$ so that $P(z)dz\wedge 
d\bar{z}$ defines an invariant probability measure supported on $SC(R).$ 
Moreover, if for a given non negative measurable function $\phi$ we have that 
$\phi(z) dz \wedge d\bar{z}$ defines an invariant probability measure, then 
$supp(\phi)$ is contained in $SC(R).$
\end{proposition}

Due to the classification of Fatou  components and Proposition 
\ref{prop.Krengel},  we have  that both the conservative set $C(R)$
and the strongly conservative set $SC(R)$ intersect the Fatou set $F_R$ 
precisely at the union of all rotation domains cycles. Indeed, if a periodic 
Fatou component $D$ is not a rotation domain then $D$ consists of a union of 
wandering sets. If $D$ is a rotation domain, consider $\tau_\theta$  an 
irrational rotation on the unit disk $\triangle$, then for any annulus 
$T_r=\{r\leq |z|\leq 1\}$ the function given by $P(z)=\frac{1}{|z|^2 mod(T_r)}$ 
for $z\in T_r$ and $0$ for $z$ in $\triangle\setminus T_r$ defines a rotational 
invariant probability measure. Using a parametrization of  
$D$ and Proposition \ref{prop.Krengel}, we get that rotation domain cycles are 
conservative and strongly conservative.
Moreover, by Theorem 1.6 on \cite{Krengel} we have the following fact.

\begin{lemma}\label{lm.con.dis} [Almost everywhere convergence on the 
dissipative set] For every $f\geq 0$ in $L_1(\bar{\C})$, the series 
$$\sum_{n=0}^\infty |R^*|^n(f)$$ converges and is finite almost 
everywhere on the dissipative set.
\end{lemma}

\begin{proof}
It is enough to show that the series  $$\sum_{n=0}^\infty |R^*|^n(f)$$ 
converges on every wandering set $W$ of finite positive measure. In fact, we 
show that the series defines an integrable function. Since $W$ is wandering 
then by Proposition \ref{prop.bounds}, we have
$$\int_W \sum_{n=0}^\infty 
|R^*|^n(f)\leq \sum_{n=0}^\infty \int_{R^{-n}(W)} f=\int_{\cup 
R^{-n}(W)}f<\|f\|_1.$$
\end{proof}


Using Proposition \ref{prop.Krengel}, we reformulate results of Lyubich and 
McMullen to obtain the following lemma.

\begin{lemma}\label{lemma.dichotomy}
Let $R$ be a rational map. 
\begin{itemize}
\item Either $C(R)\cap J_R \subset P_R$ or $C(R)=\bar{\C}$.
\item Either $SC(R)\cap J_R\subset P_R$ or $SC(R)=\bar{\C}$.
\item In case $C(R)\neq P_R$ and $C(R)=\bar{\C}$, then there exist a fixed point
of Beltrami operator supported on the Julia set if and only if  $R$ is a 
flexible Latt\'es map.
\end{itemize}

\end{lemma}

\begin{proof}
By Lyubich's classical results (see \cite{LyuTypical}), either the 
accumulation set of almost every orbit in the Julia set belongs to the 
postcritical set or the Julia set is the whole sphere with conservative and  
ergodic action. If the measure of $SC(R)\setminus P_R$ is positive, then by the 
Poincar\'e Recurrence theorem, Proposition \ref{prop.Krengel}, and 
a Koebe distortion argument (see section 1.19 of \cite{L}), 
we have that $SC(R)=\bar{\C}$. The third part is  Theorem 3.17 
in McMullen's book \cite{Mc1}. 
\end{proof}

We also use the following proposition.

\begin{proposition}\label{prop.invariantmeasure} Let $R$ be a rational map and 
let $f \in L_1(\bar{\C})$ be a fixed point of the Ruelle operator $R^*$. Then 
there exist a fixed point of the Beltrami operator $\mu\in L_\infty(\bar{\C})$
such that $\int_\C f(z) \mu(z) |dz|^2 \neq 0$. Moreover we get 
$|R^*||f|=|f|$; in other 
words, $|f|$ defines an absolutely continuous finite invariant measure that 
satisfies $\frac{|f|}{f}=\mu$ almost everywhere on the support of $f.$
\end{proposition} 

\begin{proof}
This summarizes the results given in Lemma 11 and Corollary 12 in 
\cite{MakRuelle}. \end{proof}

Note that Lemma \ref{lemma.dichotomy} and Proposition 
\ref{prop.invariantmeasure} give necessary conditions for Sullivan's and 
Fatou's conjectures. 

The following lemma is a consequence of part 3 of Lemma 5  and part 1 
of Theorem 3 in \cite{MakRuelle}. For $v$ in $\C$,  define the function 
$\gamma_v$ by  $$\gamma_v(z)=\frac{v(v-1)}{z(z-1)(z-v)}.$$
Along the paper, for functions of the type of $\gamma_v$, the Ces\`aro averages 
$A_n(\gamma_v)$ will be always with respect to the Ruelle operator $R^*.$

\begin{lemma}\label{lemma.fixed.Beltrami} Let $R$ be a rational map. 
\begin{enumerate}
\item If the Fatou set $F_R$ contains a periodic attracting domain $V$, then 
there exist an invariant Beltrami differential $\mu$ supported on the grand 
orbit of $V$ and a critical value $v_0$ such that $\int_\C \mu(z) 
\gamma_{v_0}(z)|dz|^2\neq 
0$.
\item  If $\mu$ is an invariant Beltrami differential in $L_\infty(J_R)$, then 
$\mu\neq 0$ if and only if there exist a critical value $v_0$ such that 
$\int_\C \mu(z) \gamma_{v_0}(z)|dz|^2\neq 0$.
\end{enumerate}
\end{lemma}



The following theorem gives a connection between Sullivan's conjecture and 
mean ergodicity on a suitable subspace of $L_1(\C)$.

\begin{theorem}\label{th.MeanErg.1}
Let $R$ be a rational map such that $SC(R)\cap P_R$ has Lebesgue measure zero. 
Then $R$ satisfies Sullivan's conjecture if and only if $R^*$ is mean ergodic 
in $Hol(R)$ with the topology inherited from $L_1(J_R)$.
\end{theorem}

\begin{proof}
Assume that $R$ satisfies Sullivan's conjecture. Then either there is no 
invariant Beltrami differential supported on the Julia set or $R$ is a 
flexible Latt\'es map. 
If there is no invariant Beltrami differential supported on $J_R$ then 
the Beltrami operator on $L_\infty(J_R)$ does not have fixed points. Hence 
$(Id-R^*)(L_1(J_R))$ is dense in $L_1(J_R)$. Otherwise by 
the Hanh-Banach theorem, there would exist a non-zero continuous functional 
$\mathcal{L}$ on $L_1(J_R)$ such that $(Id-R^*)(L_1(J_R))\subset 
ker(\mathcal{L})$. By the Riesz Representation theorem there is $\alpha\in 
L_1(J_R)$, fixed by the Beltrami operator, representing the functional 
$\mathcal{L}$ which is a contradiction. Then $A_n(R^*)(f)$ converges to $0$ for 
all $f$ in $L_1(J_R).$ In particular, this happens when $f\in Hol(R)$. Thus 
$R^*$ is mean ergodic. 

If $R$ is a flexible Latt\'es, then since $R$ is postcritically finite, the 
space $A(S_R)$ is finite dimensional.  So $A(S_R)$ coincides with the 
subspace $Hol(R)$ and, by the Mean Ergodicity lemma, the Ruelle operator 
$R^*$ is mean ergodic.

Now assume that $R^*$ is mean ergodic in $Hol(R)$ and that there exist an 
invariant Beltrami differential supported on $J_R.$ By Lemma 
\ref{lemma.fixed.Beltrami}, every fixed point of the Beltrami operator 
supported on the Julia set defines a continuous invariant functional 
$\mathcal{L}$ on $L_1(J_R)$ which is non-zero on $Hol(R)$. Let $\phi$ in 
$Hol(R)$ such that $\mathcal{L}(\phi)\neq 0$. By mean ergodicity 
$A_n(R^*)(\phi)$ converges to some non-zero element $f$ in $L_1(J_R)$. 
By Lemma \ref{lemma.dichotomy} and Proposition \ref{prop.invariantmeasure}, 
the map $f$ is a flexible Latt\'es map. \end{proof}


Let us note that when $P_R\neq J_R$, we can consider the space 
$A(S_R)$ instead $Hol(R)$ and get the same conclusion as in the previous 
theorem. Also, after the celebrated examples of Julia sets of positive measure 
given by X. Buff and A. Cheritat of Cremer polynomials, and  A. Avila and M. 
Lyubich of infinitely renormalizable polynomials, the main conjecture is that 
the postcritical set either has measure zero or is the whole Riemann sphere. 
Thus, conjecturally, for a map $R$ with non-empty Fatou set, the condition on 
the postcritical set of the previous theorem is always fulfilled. 
On the other hand, mean ergodicity of Ruelle operator is a rather simple 
consequence of geometric conditions of $S_R$. For example, 
the Ces\`aro averages $A_n(\gamma_v)$ are weakly convergent on measurable 
subsets $Y$ of $S_R$ of finite hyperbolic area (see \cite{CabMakHyp} and 
discussion therein). Hence, the existence of fixed points for the Ruelle 
operator is the harder difficulty to extend Theorem \ref{th.MeanErg.1} in 
full generality.

In general, as the following corollary shows, the Ruelle operator for 
rational maps is not mean ergodic on $L_1(\C)$ and $A(S_R)$.

\begin{corollary}
 Let $R$ be a rational map.
\begin{enumerate}
\item If $F_R$ contains an attracting periodic component  then 
$R^*:A(S_R)\rightarrow A(S_R)$ is not mean ergodic.
 \item If $F_R$ contains a periodic non rotational component then 
$R^*:L_1(\C)\rightarrow L_1(\C)$ is not mean ergodic.


\end{enumerate}
\end{corollary}
\begin{proof}
Part (1). By contradiction. Let $V$ be an attracting domain of $F_R$ and assume 
that $R^*$ is mean ergodic on $A(S_R)$. Then by part (1) of  Lemma 
\ref{lemma.fixed.Beltrami}, there exist a Beltrami differential $\mu$, 
supported on the grand orbit of $V$, such that the functional 
$l_\mu(\phi)=\int_\C
\mu(z) 
\phi(z)|dz|^2\neq 0$ on $A(S_R)$ and $l_\mu(R^*\phi)=l_\mu( \phi)$. By the Mean 
Ergodicity lemma, there exists $\psi\in A(S_R)$ with $R^*(\psi)=\psi$ such that 
$l_\mu(\psi)\neq 0.$ This implies that the restriction $f$ of $\psi$ on the 
grand orbit of $V$ is non-zero and is a fixed point of $R^*.$ By Proposition 
\ref{prop.invariantmeasure}, the map $|f|$ defines a finite invariant measure. 
However, the grand orbit of $V$ belongs to the dissipative set. Hence 
using arguments similar to the previous theorem we can show that $f$ is $0$ 
almost everywhere on the grand orbit of $V$. This is a contradiction. 

Part (2). Under this hypothesis, it is enough to show, that the grand orbit 
of any non rotational periodic domain supports a non-zero invariant Beltrami 
differential. We conclude  using Fatou component classification and  arguments 
similar to the previous part. On part (1), we already considered the attracting 
case, so 
by the Fatou Components classification, we have to consider the cases where 
there is a Fatou component $V$ which is either parabolic or superattracting. 
In the parabolic case, let $\Phi$ be a linearisation function defined on the 
grand orbit of $V$. That is a function satisfying $\Phi(R(z))=\Phi(z)+1.$ Then 
a short computation shows that the map $\mu=\frac{\bar{\Phi}'}{\Phi'}$ gives a 
non-zero invariant Beltrami differential.
In the superattracting case, let $\Phi$ be now the B\"ottcher coordinate around 
a neighbourhood  $U$ of the superattracting  cycle. Let 
$\nu(z)=\frac{\Phi \bar{\Phi}'}{\bar{\Phi}\Phi'}$ then for $z\in U$, we have 
$\nu(R(z))\frac{\bar{R'}(z)}{R'(z)}=\nu(z)$. Using dynamics we can extend to a 
non-zero invariant Beltrami differential defined on the grand orbit of 
$U.$ \end{proof}

\textbf{Remark}. For the reasons above, in order to work with mean ergodicity 
of the Ruelle operator on $A(S_R)$ we will often assume that $R$ does not 
accept invariant Beltrami differentials defining non-trivial quasiconformal 
deformations supported on the Fatou set.  This is the same that the 
postcritical set on the Fatou set is finite and there no rotational domains. In 
other words, we assume that any non-zero invariant functional on the space 
$A(S_R)$ is induced by an invariant Beltrami differential supported on the 
Julia set. Equivalently, by the Separation principle, the Ces\`aro averages 
converges to zero with respect to the $L_1$ norm over the Fatou set. 

Later on, we will discuss the relation between the topologies on $Hol(R)$ 
induced by $L_1$ norms over the Fatou and the Julia set, respectively.

Next we give some conditions under which the Ruelle operator does not have a 
fixed point. We call an integrable function $f$ \fd{regular} if  
$\bar{\partial}f$,  in sense of distributions, is a finite complex valued 
measure.  Examples of non regular functions are given by characteristic 
functions of certain measurable sets. 

\begin{theorem}\label{thm.regularfixed}
Let $R$ be a rational map. Assume that the postcritical set $P_R$ is
such that either

\begin{itemize}
 \item the diameter of all components $D$ of $S_R$ are uniformly 
bounded away from $0$, or
 
\item $J_R\cap P_R\subset \bigcup \partial V$ for $V$ a component of  the 
Fatou set $F_R$. 
\end{itemize}

\noindent Then  $R^*$ does have a regular non-zero fixed  point (a non-trivial 
Beltrami form) if and only if $R$ is a flexible Latt\'es map.
\end{theorem}

\begin{proof}
Assume that $R$ is a flexible Latt\'es map. Then $R$ has a unique, up to 
multiplication by scalars, invariant Beltrami differential $\mu$ (see Milnor 
\cite{MilLat}).  
This differential $\mu$ defines a non-zero functional $l_\mu$ on $A(S_R)$ given 
by the pairing
$$
l_\mu(\phi)=\int_\C \phi(z) \mu(z)|dz|^2.
$$ 
Since $A(S_R)$ is finite dimensional, then $R^*$ is mean ergodic on $A(S_R)$, 
and by the Separation principle there exist a non-zero fixed point $f\in 
A(S_R)$, of the Ruelle operator, which is unique up to multiplication by 
scalars. Since $f$ is an integrable holomorphic function outside finitely many 
points on $\C$, then $f$ is rational with simple poles only. Hence the 
distributional derivative $\bar{\partial}f$ is a finite combination of Dirac 
measures supported on the poles of $f.$

The Beltrami differential $\mu$ is a unique fixed point of the 
Beltrami operator. Since the Beltrami operator $B:L_1(\C)\rightarrow L_1(\C)$
is dual to $R^*$, then by the Separation principle, the operator $R^*$ is mean 
ergodic on $L_1(\C).$ Hence $f\in L_1(\C)$ is a unique, up to scalar 
multiplication, fixed point of the Ruelle operator.

Now, let $f$ be a non-zero regular fixed point of the Ruelle operator. Then by 
Lemma \ref{lemma.dichotomy}, either $R$ is a flexible Latt\'es map or the 
support $supp(f)$ belongs to the postcritical set.
 
Without loss of generality, we can think that $P_R$ is a subset of $\C$. Let  
$\nu=\bar{\partial} f$ and put 
$$
F(z)=\int_\C \frac{\nu(t)dt}{t-z}.
$$ 
Since $\nu$ is a finite measure, then $F(z)$ is holomorphic outside $P_R.$ We 
claim that $F(z)=f(z)$ holds for Lebesgue almost every point. By Weyl's lemma 
$h(z)=F(z)-f(z)$ is an entire function. Also $F(z)$ converges to $0$ as $z$ 
tends to infinity.  Then $h(z)=0$ for almost every $z$. Hence, $F(z)$ is 
identically $0$ outside the support of $\nu$. We claim that the first 
condition implies that $\nu$ is identically $0$ on $\C$.  Recall that the 
generalized Mergelyan's theorem (see, for example, Gamelin 's paper 
\cite[Theorem 10.4]{GamelinUniAlg}) states:  \textit{if the diameters of all 
components of the complement of a compact set $K$ on the plane $\C$ are 
bounded away from $0$ then any continuous function which is holomorphic on the 
interior of $K$ is a uniform limit of rational functions with poles outside of
$K$}. Since $P_R$ satisfies the conditions of the generalized Mergelyan's 
theorem and has empty interior, then any continuous function on $P_R$ is a 
uniform limit of rational functions with poles outside $P_R.$ So for any finite 
set of complex numbers $b_i$ and points $a_i$ in $\C \setminus P_R$, for 
$r(z)=\sum \frac{b_i}{z-a_i}$ we have 
$$\int r d\nu=\sum b_i F(a_i)=0.
$$ Hence $\nu$ represents a zero functional on the space of 
continuous functions on $P_R.$ By  Riesz Representation theorem, the 
measure $\nu$ is null. Thus, $f(z)=0$ almost everywhere which is a 
contradiction.  

For the second case we follow closely the arguments of part 3 of Proposition 14 
in \cite{MakRuelle}. Let $\{Y_i\}$ be the family of all components of 
$\C\setminus P_R$, then we claim that $f(z)=0$ on $\cup_i \partial Y_i\subset 
P(R)$. Otherwise, there exist a component $Y_{i_0}$ and $E\subset \partial 
Y_{i_0}$ and $m(E)>0$ with $\int_E f\neq 0$. Since $P_R$ is compact and 
conjugating by a M\"obius map, we can assume that $\infty$ belongs to 
$Y_{i_0}$. Then the function $F_E(z)=\int_E \frac{dt\wedge \overline{dt}}{t-z}$
is a continuous function on the plane which is holomorphic outside $\partial 
Y_{i_0}.$ Again, by the generalized Mergelyan theorem $F_E$ is a uniform limit 
of rational functions with poles in $Y_{i_0}$. Hence using similar 
arguments as above we have $\int F_E(z) \bar{\partial} f(z)=0$. By Fubini's 
theorem we have   $$\int F_E(z) \bar{\partial} f(z) dz=\int 
\bar{\partial}f(z)\int_E \frac{dt\wedge \overline{dt}}{t-z}$$$$=\int_E dt\wedge 
\overline{dt}\int \frac{\bar{\partial}f(z)dz}{t-z}.$$ But, by arguments above, 
the later is equal to $-\int_E f(t)dt\wedge \overline{dt}$ which is a 
contradiction.  
Under the assumptions of part (2), we have that $f(t)=0$ almost everywhere, 
which finish the proof.  \end{proof}


On the discussion above, we saw that the convergence of Ces\`aro averages on 
subspaces of $L_1(\C)$ is tightly related to the existence of non-trivial 
invariant Beltrami differentials under some conditions. 

\begin{corollary}\label{cor.totvariation}
Let $R$ be a rational map as in  Theorem \ref{thm.regularfixed}. Suppose there 
exist a critical value $v\in V(R)$ such that the total variation 
of $\bar{\partial}A_n(\gamma_v)$ is uniformly bounded. Then  $R$ is not 
structurally stable.
\end{corollary}

\begin{proof}

Since total variation of the sequence $\bar{\partial}A_n(\gamma_v)$ is 
uniformly bounded, $\bar{\partial}A_n(\gamma_v)$ is $*$-weakly precompact 
when acting on continuous functions. Let $m_0$ be a complex valued measure 
which is a $*$-weak accumulation point of this sequence. Now 
$supp(\bar{\partial}A_n(\gamma_v))\subset P_R$, then considering 
$\bar{\partial}A_n(\gamma_v)$ as measures, by a straightforward computation 
we have $$A_n(\gamma_n)(z_0)=-\int_{\C} 
\frac{\bar{\partial}A_n(\gamma_v)(t)dt}{t-z_0}=-\int_{P_R} 
\frac{\bar{\partial}A_n(\gamma_v)(t)dt}{t-z_0}$$ for every $z_0$ outside $P_R.$ 
If $m_0\neq0$ then, as in Theorem 
\ref{thm.regularfixed}, the integral $-\int_\C \frac{m_0(t)dt}{t-z}$ is 
non-zero and is an accumulation point of $A_n(\gamma_v)$ in the pointwise 
topology and, therefore, is a 
regular non-zero fixed point. By Theorem 
\ref{thm.regularfixed}, the map $R$ is a Latt\'es map which is not 
structurally stable, this is a contradiction.
Hence $\bar{\partial}A_n(\gamma_v)$ converges to $0$ in the $*$-weak 
topology. 

Let $\mu\in L_\infty(\C)$ and define the 
potential $$F_\mu(z)=-\frac{z(z-1)}{\pi}\int  \frac{\mu(\zeta) 
|d\zeta|^2}{\zeta(\zeta -1)(\zeta-z)}.$$ This 
potential is continuous on $\C$ and satisfies $\bar{\partial}F_\mu(z)=\mu(z)$ 
in the sense of distributions. 
We claim that if $\mu$ is a fixed point of the Beltrami operator then we have 
 
$\int_\C \mu(z) \gamma_v(z)|dz|^2=0.$ Indeed, since $\int F_\mu(z) 
\bar{\partial}A_n(\gamma_v)(z)|dz|^2$ converges to $0$ and $\mu$ is 
invariant, we have 
that 
$$
\int_\C F_\mu(z)
\bar{\partial} A_n(\gamma_v)(z)|dz|^2=-\int_\C 
\bar{\partial}F_\mu(z) A_n(\gamma_v)(z)|dz|^2$$$$=-\int_\C \mu(z) 
A_n(\gamma_v)(z)|dz|^2
$$ which implies 
$$
\int_\C \mu(z) A_n(\gamma_v)(z)|dz|^2=\int_\C \mu(z) 
\gamma_v(z)|dz|^2=0
$$
as claimed. If $R$ is structurally stable, then we have a contradiction with 
Theorem 3 and part (3) of Lemma 5 in \cite{MakRuelle}.\end{proof}

 As an immediate corollary we have the following.

\begin{corollary}
Assume the Julia set $J_R$ has positive Lebesgue measure, then $J_R$  does not 
support a non-zero invariant Beltrami differential 
if and only if, for any critical value $v$, the sequence 
$\bar{\partial}A_n(\gamma_v)(\phi)$ converges to $0$ on every 
continuous function $\phi$ on $J_R$ with distributional 
derivative $\phi_{\bar{z}}\in L_\infty(J_R)$.
\end{corollary}

In the following statements we show that there are no fixed points of the 
Ruelle operator among the examples of non regular functions given above.

\begin{proposition}\label{prop.chiA}
Assume that a function $g=f+c\chi_A$, where $c$ is a constant and $\chi_A$ is 
the characteristic  function of a measurable set $A$ such that 
$A\setminus supp(f)$ has positive measure. If $g$ is a fixed point for 
the Ruelle operator then $R^*(f)=f$ and $c=0.$
\end{proposition}

\begin{proof}

If $B=A\setminus supp(f)$, then by Lemma \ref{lemma.dichotomy} and 
Proposition \ref{prop.invariantmeasure}, we have that $B\subset SC(R)$ and 
$\mu(z)=\frac{|g(z)|}{g(z)}=\mu(R(z))\frac{\overline{R'(z)}}{R'(z)}$ almost 
everywhere on $supp(g).$
Then there exist $k$ such that the measure of $B\cap R^{k}(B)$ is positive, 
and hence the set $C=B\cap (R^{k})^{-1}(B)$ has positive measure.  Moreover, 
we have $\mu=\frac{|c|}{c}$ on $B.$ If $c\neq 0$ then by invariance of $\mu$ we 
have that $(R^k)'$ is real valued on $C$ and thus $C\subset 
((R^{k})')^{-1}(\mathbb{R})$ which contradicts $m(C)>0.$ 

\end{proof}
An immediate corollary we have that a simple function can not be a fixed point 
of the Ruelle operator.

\begin{corollary}
 If $f=\sum c_i \chi_{A_i}$ where the $A_i$ are distinct measurable 
sets, then $f$ is a fixed point of the Ruelle operator if and only if $c_i=0$ 
for all $i.$
\end{corollary}


Let us note that for any closed set $K$ in $\bar{\C}$ the space $H(K)$ 
always contains a non regular function, even in the case when $K$ has zero 
Lebesgue measure. On the other hand $H(K)$ contains characteristic functions if 
and only if $K$ has positive measure.  Nevertheless, by Theorem 
\ref{thm.regularfixed}, if the postcritical set has measure zero then any fixed 
point of the Ruelle operator is necessarily a regular function. In general, we 
conjecture that any fixed point of the Ruelle operator is necessarily a regular 
function. In the last section we will discuss the existence of fixed points 
when the postcritical set has positive measure.

We endow the space $Hol(R)$ with two non-complete norms. 
The first is given by the restriction of the $L_1$ norm to the Fatou set 
and the second by restriction of that same norm to the Julia set. 
Let us call $Hol_1$ and $Hol_2$ the respective normed spaces. If the Fatou
set is empty then $Hol_1=0.$ The operator $R^*$ is a contraction on each space. 
 
Next we show that any compatibility between these two 
topologies on $Hol(R)$ gives rise to a sort of rigidity on the dynamics of 
$R$. 

\begin{proposition}\label{prop.contId}
Fix a rational map $R$ with $F_R\neq \emptyset$. 
 
\begin{enumerate}
\item If the identity map $Id:Hol_1 \to Hol_2$ is
continuous, then there are no fixed points for the 
Beltrami operator on the Julia set $J_R$.

\item If the Fatou set $F_R$ admits an invariant Beltrami differential 
defining a non-zero functional on $Hol_1$, then $Id:Hol_2 \to Hol_1$ is 
continuous if and only if $J_R$ has Lebesgue measure zero. 
\end{enumerate}
\end{proposition} 
\begin{proof}
For the first part, suppose that there is a non-zero fixed point $\mu$ 
for the Beltrami operator on the Julia  set. 
Then $\mu$ defines a non-trivial invariant continuous linear functional $l_\mu$ 
on $Hol_2$. 
Since $Id$ is a continuous operator, then the functional 
$l(\alpha)=l_\mu(Id(\alpha))$ is a continuous invariant linear 
functional on $Hol_1$ and hence extends to the completion of $Hol_1$. 
By the Hahn-Banach and Riesz Representation theorems 
there exists a function $\nu\in L_\infty(F_R)$ such that  
$$
l(\phi)=\int _{F_R} \phi(z) \nu(z) |dz|^2
$$
for all $\phi \in \overline{Hol_1}$.   
We claim that we can choose $\nu$ to be a fixed point of Beltrami operator 
supported on the Fatou set.  If $\nu$ is not fixed, consider the Ces\`aro 
averages 
$$
\frac{1}{n}\sum_{k=0}^{n-1}B^k(\nu)=\nu_n.
$$ 
If $\nu_\infty$ is a $*$-weak accumulation point of $\nu_n$, 
then $\nu_\infty$ is a fixed point of Beltrami operator supported on the Fatou 
set. We show that $\nu_\infty$ still represents $l_\mu$. In fact, letting 
$\{n_i\}$ be chosen  so that $\nu_{n_i}$ $*$-weakly converges  to $\nu_\infty$ 
and, for $\phi \in Hol(R)$, we have
$$
\int_\C \phi(z) v_\infty(z)|dz|^2=\lim_{n_i} \int_\C \phi(z) 
\nu_{n_i}(z)|dz|^2$$$$=\lim \int_\C \phi(z) 
\frac{1}{n_i}\sum_{k=0}^{n_i-1}B^k(\nu)(z)|dz|^2=\lim \int_\C 
\nu(z) \frac{1}{n_i}\sum_{k=0}^{n_i-1} (R^*)^k(\phi(z))|dz|^2$$$$=\lim 
\frac{1}{n_i}\sum_{k=0}^{n_i-1}l_\mu( (R^*)^k(\phi) =l_\mu(\phi), 
$$
and thus $\nu_\infty$ is a fixed point of the Beltrami operator supported on 
the Fatou set as claimed. In particular, as $\gamma_a\in Hol(R)$ for $a\in 
P_R,$ the continuous functions $F_{\nu_\infty}(a)=\int_\C 
\gamma_a(z)\nu_\infty(z) 
|dz|^2$ and 
$F_{\mu}(a)=\int_\C \gamma_a(z)\mu(z) |dz|^2$ coincide
on the orbit of all critical values, and hence on $P_R.$ This however 
contradicts to Theorem 3 of \cite{MakRuelle}. 

For the second part notice that if the Julia set has measure zero, 
then $Hol_2$ consists of only the $0$ function. 
So, assume that the Lebesgue measure of the Julia set is not zero. 
Let $\mu$ a fixed point of Beltrami operator with 
$$
l_\mu(\phi)=\int_{F_R} \phi(z) \mu(z)|dz|^2 
\neq 0.
$$
Applying the same arguments as in the first part, we get a 
continuous linear functional on $Hol_2$ defined by a fixed point of  
Beltrami operator now supported on the Julia set. 
This again contradicts to Theorem 3 of \cite{MakRuelle}, hence 
the Lebesgue measure of $J_R$ is zero. 
\end{proof}

Now define $X=(Id-R^*)(Hol(R))$ and let $X_1$ and $X_2$ be the 
closures of $X$ in the spaces of $Hol_1$ and $Hol_2$, respectively. 
The proof of Theorem \ref{th.MeanErg.1} shows that 
if there is no invariant Beltrami differential supported on a Julia set of 
positive measure, then we get $X_2=Hol(R)$ and hence also 
$X_1\subset X_2$. We will prove the converse in Proposition 
\ref{pr.fixedpointBelt} below. First we need a technical result. 

\begin{lemma}\label{lemma.linearfunc} 
Let $l$ be a  linear functional on $Hol(R)$. If $ X_1 \subset ker(l)$ then 
$l$ is continuous on $Hol_1$. 
\end{lemma}

\begin{proof}
Let $W$ be the finite dimensional space of all linear combinations of the 
$\gamma_v$, with $v$ a critical value of $R$. 
We will show first that $Hol_1$ equals the sum $X_1 + W$, note that $X_1\cap 
W$ may be non-zero. Indeed,  by definition, the space $Hol_1$ is the linear 
span of $\gamma_a(z)$ where $a$ is an element in the union of the forward 
orbits of all critical values. For every critical value $v$ of $R$, by Lemma 5 
of \cite{MakRuelle}, we have 
$$
R^*(\gamma_v)=\frac{1}{R'(v)}\gamma_{R(v)}+w
$$ 
for some $w$ in $W$. Since $R^*(\gamma_v)-\gamma_v\in X_1$ and 
$-w+\gamma_v \in W$, the element $\gamma_{R(v)}$ 
is a sum of elements in $X_1$ and $W$. But $X_1$ is 
invariant under $R^*$, then by an induction argument we conclude that
$\gamma_{v_n}\in X_1+ W$ for every $v_n=R^n(v)$. 
Therefore the space $X_1$ has finite codimension on $Hol_1$.  

If $X_1\subset \ker(l)$ then $l$ projects to a linear functional 
${\cal L}$ defined on the finite dimensional space $Hol_1 /X_1$. 
As $X_1$ is closed this implies that both ${\cal L}$ and $l$ are 
continuous.\end{proof}

\begin{proposition} \label{pr.fixedpointBelt}
Suppose that the Julia set $J_R\neq \bar{\C}$ and has positive Lebesgue 
measure. 
Then there is no non-zero invariant Beltrami differential supported on $J_R$ 
if and only if $X_1\subset X_2$.
\end{proposition}

\begin{proof}
If there is no invariant Beltrami differential supported on the Julia set, 
then $X_2$ is equal to the whole space $Hol(R)$ and contains $X_1$. 

On the other direction, assume $X_1\subset X_2$. As in Lemma 
\ref{lemma.linearfunc}, we have $Hol(R)=X_1+F$ for some 
finite dimensional vector space $F\subset W$. For an invariant Beltrami 
differential $\mu$ supported on the Julia set, the assignment 
$l_\mu(\phi)=\int_{J_R} \phi(z) \mu(z)|dz|^2$ defines an invariant 
linear continuous 
functional on $Hol_2$ for which we have $X_2\subset ker(l_\mu)$.   
Lemma \ref{lemma.linearfunc} shows that $l_\mu$ is 
continuous on $Hol_1$, by Riesz Representation theorem 
there is a function $\nu\subset L_\infty(F_R)$ so that 
$l_\mu(\phi)=\int_{F_R}\phi(z)  \nu(z) |dz|^2$. 
However, since $\mu$ is invariant then, using analogous arguments 
in the proof of Proposition \ref{prop.contId}, we again get a contradiction 
to Theorem 3 of \cite{MakRuelle}. \end{proof}



\section{Action on $L_p$ spaces}

If we want to extend the theory to $L_p(K)$ spaces (with $K$ completely 
invariant), we need to modify somehow the operators in an \textit{ad-hoc} 
manner. In the formulas below, $d$ will denote the degree of $R$. 
Let $p,q$ be such that $1/p+1/q=1$. 

The action of $R$ by \fd{pull-back} on $L_p$ is given 
by 
$$
R_{*p}(\phi)=\frac{1}{\sqrt[p]{d}}\phi(R)|R'|^{\frac{2}{p}}\frac{R'}{\overline 
{R'}}, 
$$
when $\phi \in L_p$. 

Similarly, the \fd{push-forward} action of $R$ on $L_q$ (for $1< q < 
\infty$)
is defined by
$$ 
R^{*}_{q}\phi(z)=\frac{1}{\sqrt[p]{d}}\sum 
\phi(\zeta_i)\frac{\zeta_i'}{\overline{\zeta_i '}}|\zeta_i'|^{\frac{2}{q}},
$$ 
where the sum is taken over all branches $\zeta_i$ of $R$, 
that is, that satisfy $R(\zeta_i(z))=z$ for all $z$. 
The constants are suitably chosen so that $R^{*}_{p}$ and $R_{*q}$ are 
mutually dual. Indeed, if $\phi\in L_p$ then for any $g\in L_q$, we have $$\int 
R_{*p}(\phi(z))\bar{g}(z)\,|dz|^2=\frac{1}{\sqrt[p]{d}}\int 
\phi(R(z))|R'(z)|^{\frac{2}{p}}\frac{\overline{R'(z)}}{R'(z)}\bar{g}(z)\,
|dz|^2$$ 
and, after changing variables, is equal to 
$$\frac{1}{\sqrt[p]{d}}\int 
\phi
\left[\sum_{i=0}^d 
\left(\bar{g}(z)|R'(z)|^{\frac{2}{p}}\frac{\overline{R'(z)}}{R'(z)}\right)
(\zeta_i(z))|\zeta'_i(z)|^2\right] |dz|^2$$ for all branches 
$\zeta_i$ of 
$R.$ By direct calculation, for every $i$ we have 
$$\left(|R'|^{\frac{p}{2}} 
\frac{\overline{R'}}{R'}\right)(\zeta_i)|\zeta'_i|^2=|\zeta'_i|^\frac{2}{q} 
\frac{\zeta'_i}{\overline{\zeta_i'}} .$$  
Hence the previous expression is equal to $$\int \phi 
\overline{\left[R^*_q(g(z))\right]}|dz|^2.$$ 
Moreover, we have $R^*_p\circ R_{*p}=Id$ on $L_p$. 

We have now two continuous family of contractions 
depending on $p >0$, which are mutually dual for 
$p\ge1$ and includes the Ruelle and Beltrami 
operators, for $p=1$ and $p=\infty$, respectively.  

The next theorem is the $L_p$ version of the action 
of Ruelle operators (compare with previous section). 
Unfortunately, the Ruelle operator on $L_p$, for $1<p< \infty$, 
does not detect whether there is an invariant Beltrami differential without 
an associated fixed point for the Ruelle operator.


\begin{theorem} 
Let $K$ be a completely invariant set of positive measure.
The operator $R^*_p$  has a fixed point in $L_p(K)$ for 
$1<p<\infty$ if and only if $R$ is a flexible Latt\'es map.
\end{theorem}
\begin{proof}
If $R^*_p$ has a fixed point then its dual $R_{*q}$ has 
a fixed point on $L_q(K)$. Given a fixed point $\psi$ of $R_{*q}$ 
then $\frac{|\psi|}{\psi}$ is an invariant Beltrami differential. 
On the other hand, the function  $f=|\psi|^q$ is integrable and 
satisfies 
$$
f(z)=\frac{f(R(z))|R(z)'|^2}{deg(R)}.
$$ 
In this case $supp(f)$ is completely invariant and, by Proposition 
\ref{prop.Krengel} and the discussion afterwards, $supp(f) \subset C(R)\cap 
J_R$.  Let $E$ be the operator $E:C(J_R)\rightarrow C(J_R)$ on the space of 
complex valued continuous functions $\phi$ on $J_R$ given by  $$E_R(\phi)= 
\frac{1}{deg(R)}\sum_i \phi(\zeta_i(z))$$ where the sum is taken over all 
branches $\zeta_i$ of $R.$ Lyubich's Theorem 5 in \cite{LyuTypical} states that 
every continuous functional invariant with respect to $E$ is induced by a 
multiple of the maximal entropy measure. 

Let $\nu(z)$ be the measure such that $d\nu(z)=f(z) \, |dz|^2$. 
Now, 
$$\int_{J_R} E_R(\phi)d\nu=\frac{1}{deg(R)}\int_{J_R} \left[\sum_i 
\phi(\zeta_i(z))\right] f(z)|dz|^2.$$
By the arguments and computations of Proposition \ref{prop.bounds}, the latter 
is equal to
$$\frac{1}{deg(R)}\int_{J_R} \phi(z)f(R(z))|R'(z)|^2|dz|^2 
=\int_{J_R}\phi d\nu.$$ Then $\nu$ is a multiple of the maximal entropy 
measure. 
By Zdunik's Theorem (see \cite{ZdunikParab}) $R$ is a postcritically finite 
rational map. Finally Lemma \ref{lemma.dichotomy} shows that $R$ is a flexible 
Latt\'es map. 

Conversely, if the map $R$ is flexible Latt\'es, then there 
exist an integrable function $f_0$ such that $f_0=\frac{f_0(R)(R')^2}{deg 
R}$ and 
$\mu(z)=\frac{\overline{f_0(z)}}{|f_0(z)|}$ is a fixed point of 
Beltrami operator. 
Therefore $\psi=|f_0|^{\frac{1}{p}}\mu$ is a fixed point for $R_{*p}$ on 
$L_p(K)$ and induces a fixed point of the Ruelle operator in $L_p(K)$. 
Since the dual operator satisfies $R^*_q\circ R_{* q}=Id$, the converse follows 
immediately.
\end{proof}
 

\section{Fixed points of bidual actions and uniform ergodicity}

In this section, we start with  the following lemma which is a summary 
of results due to H. P. Lotz in \cite{Lotz}.

\begin{lemma}\label{Lotz.lemma}
Let $B$ be a Grothendieck space with the Dunford-Pettis property and let 
$S:B\rightarrow B$ be a power bounded operator. Then
\begin{enumerate}
 \item If $S$ is mean-ergodic then $S$ is uniformly ergodic.
 \item If the space $Fix(S^*)$ of fixed points of the dual of $S$ is separable 
then $S$ is uniformly ergodic and the Ces\`aro averages uniformly converges to 
a compact projection.
\end{enumerate}
 
\end{lemma}

Part (1) is the content of Theorem 5 in \cite{Lotz}. Part (2) is the content of 
Theorem 7 in \cite{Lotz}.
 

 
\begin{theorem}\label{th.meanergosep} 
Let $R$ be a rational map such that $P_R$ is not the 
whole sphere,   the conservative set does not contain any Fatou  component and 
that $1$ belongs to the spectrum  $\sigma(R^*)$ on $A(S_R)$. 
Then the following four conditions are equivalent. 
\begin{enumerate}
\item The space $Fix(T^*)$ is separable.
\item The operator $T$ is mean ergodic.
\item The Ruelle operator $R^*$ is uniformly ergodic.
\item The map $R$ is a flexible Latt\'es map. 
\end{enumerate}
\end{theorem}
 
\begin{proof} By the discussion on Section \ref{sec.Thurs}, the space $B(S_R)$ 
is a Grothendieck space with the Dunford Pettis property. Then Lemma 
\ref{Lotz.lemma} gives the implications from (1) to (2) and from (2) to (3).

To show (3) implies (4) note that since $R^*$ is a contraction, then $R^*$ is a 
power bounded operator. By the Uniform Ergodicity lemma and the assumption,  
the value $1$ is an isolated eigenvalue of $\sigma(R^*)$. 
Then Ruelle's operator $R^*$ has a non-trivial fixed point $\phi$ in $A(S_R)$. 
Hence by Proposition \ref{prop.invariantmeasure}, the modulus $|\phi|$
defines an invariant finite measure such that the support of $|\phi|$ is $S_R$. 
By Lemma \ref{lemma.dichotomy}, the map $R$ is a flexible Latt\'es map. 

Now we show (4) implies (1). For a flexible Latt\'es map, the space $A(S_R)$ is 
finite dimensional, and hence $A^*(S_R)$ and its dual are finitely 
dimensional too, which implies (1). \end{proof}

The condition that $1$ belongs to the spectrum is necessary for the 
discussion around Sullivan's conjecture. Otherwise, the Beltrami operator does 
not have fixed points. However, in this situation there are two open questions: 

First, does there exist a rational map $R$ with infinite postcritical set such 
that the norm of $R^*$ on $A(S_R)$ is strictly smaller than 1? 

Second, is it true that if $1$ does not belong to the spectrum of $R^*$ on 
$A(S_R)$ then $R$ is postcritically finite?  

By the Uniform Ergodicity lemma, the conditions of the questions above imply 
that $R^*$ is uniformly ergodic on $A(S_R).$

The following corollary gives partial answer to these questions for the class 
of $J$-stable rational maps.  

\begin{corollary}
If $R$ is $J$-stable and the Lebesgue measure of  $P_R$ is zero, then the 
following are equivalent. 
\begin{enumerate}

\item The operator $R^*$ is uniformly ergodic. 
\item The map $R$ is hyperbolic and postcritically finite. 
\end{enumerate}
 
\end{corollary}
\begin{proof}
Condition (2) implies (1) since, in this case, $A(S_R)$ is finite dimensional. 
Now, (1) implies (2). First note that $1$ does not belong to $\sigma(R^*)$. 
Indeed, if $1$ belongs to $\sigma(R^*)$ then, by Theorem \ref{th.meanergosep}, 
$R$ is a flexible Latt\'es map which contradicts to $J-$stability.
Hence every invariant Beltrami differential supported on $S_R$ defines a
trivial quasiconformal deformation. By Lemma \ref{lemma.fixed.Beltrami}, any 
non-zero invariant Beltrami differential supported on the Julia set defines a 
non-trivial quasiconformal deformation.  Hence by Ma\~n\'e, Sad, Sullivan's 
Theorem E (see \cite{MSS}), the map $R$ is hyperbolic. Moreover, by Ma\~n\'e, 
Sad, Sullivan Theorem D, we have that $R$ is postcritically finite.\end{proof}

If the Julia set $J_R$ has positive measure and does not support non-zero 
invariant Beltrami differentials, then the action of the Ruelle operator on 
$L_1(J_R)$ is mean ergodic. In contrast with this fact, we show that the action 
of $R^*$ in $L_1(J_R)$ is uniformly ergodic only in the case when $J_R$ has 
Lebesgue measure zero. In slightly more generality, we show the following.


\begin{theorem} 
Let $R$ be a rational map and let $K$ be a completely invariant measurable set. 
Then the Ruelle operator is uniformly ergodic in $L_1(K)$ if and only if $K$ 
has Lebesgue measure zero. 
\end{theorem}

\begin{proof} 
When the measure of $K$ is $0$ then the theorem is certainly true.

Now, if the measure of $K$ is positive and $R^*$ is uniformly ergodic in 
$L_1(K)$, we claim that $R^*$ is an isomorphism in $L_1(K)$. 
It is enough to show that $R^*$ is injective since $R^*$ is surjective 
by the relation $R^*\circ R_*=Id$. 
Assume that  $\phi$ is a non-zero element in $ker(R^*)$. 
Then for all $\lambda$ such that $|\lambda|<1$, the element 
$$
\phi_\lambda=\sum_{n=0}^\infty \lambda^n R^n_*(\phi)
$$ 
satisfies $\lambda R_*(\phi_\lambda)=\phi_\lambda-\phi$. Hence 
$R^*(\phi_\lambda)=\lambda\phi_\lambda$. 
It follows that $1$ is not an isolated eigenvalue in the spectrum,  
which contradicts to the uniform ergodicity of $R^*$. 
Therefore $R^*$ is an isomorphism as we claimed.
This, in turn, implies 
$$
R_*\circ R^*=Id
$$ 
on $L_1(K)$. 
Now let $x_1$ and $x_2$ be different fixed points of $R$, and fix 
$a$ different from $x_1$ and $x_2$. 
Then the restriction of 
$\omega_a(z)=\frac{(x_1-a)(x_2-a)}{(z-x_1)(z-x_2)(z-a)}$ 
to $K$ is integrable. 
Since $\omega_a$ is rational, and the equation $R_*\circ 
R^*(\omega_a)=\omega_a$ holds almost everywhere  
on a set positive measure, then it holds on the whole Riemann sphere.  
If we take $a$ such that neither $a$, $R(a)$ nor $R^{-1}(a)$ are critical 
values or $R$, then $R^*(\omega_a)$ has a non-trivial pole on $R(a)$, 
which implies that $R_*\circ R^*(\omega)$  has poles in $R^{-1}(R(a))$ which 
are different to the set of poles of $\omega_a$. 
As this is a contradiction, then the Lebesgue measure of $K$ is zero.  
\end{proof} 

We have the following corollary.

\begin{corollary}
Let $R$ be a rational map and $K$ be a completely invariant subset of positive 
measure. Then the space $F$, of all fixed points of the bidual action of the 
Ruelle operator acting the bidual space of $L_1(K)$, is a non separable Banach 
space.
\end{corollary}

\begin{proof}
 By contradiction. If the space $F$ is separable, then by Lemma 
\ref{Lotz.lemma}, the Ruelle operator is uniformly ergodic.  Thus $K$ has 
measure $0$ by the previous theorem.  
\end{proof}

Let us observe that Thurston operator $T$ leaves $B_0(S_R)$ invariant. 
Since the Ruelle operator is continuous in the $*$-weak topology over 
$B_0(S_R)$ then $R^*$ is dual to the restriction of $T$ on $B_0(S_R)$, 
so we have $(T|_{B_0(S_R)})^{**}=T$.


Recall the facts discussed on Section \ref{sec.Thurs},
every endomorphism $Q$ on $B_0(S_R)$ is either compact or 
there exists an infinite dimensional subspace $E$ in $B_0(S_R)$ such that the 
restriction $Q|_E:E \rightarrow B_0(S_R)$ is an isomorphism onto its image. 
Let us consider $Q=Id-T$ and investigate two extremal situations: 
First when $Q:B_0(S_R)\to B_0(S_R)$ is an isomorphism onto its image, that is 
the space $E$ is the whole of $B_0(S_R)$ and, second, when $Q$ is a compact 
operator. 
Our goal is to show that  these two extremal situations imply the uniform 
ergodicity of the operator $T$ which is equivalent to the uniform ergodicity of 
$R^*$.

\begin{theorem}\label{th.Qcompact}
Assume $Q=Id-T$ is either an isomorphism onto its image or a compact operator, 
then $T$ is uniformly ergodic.  
\end{theorem}

\begin{proof}
If $Q:B_0(S_R) \to B_0(S_R)$ is an isomorphism onto its image, 
then the subspace $(Id-T)B_0(S_R)$ is closed, hence $T$ is uniformly 
ergodic by the Uniform Ergodicity lemma. 

Now assume $Q$ is compact. As a consequence of the Spectral Decomposition 
theorem, for any $\epsilon>0$ there is a splitting $B_0(S_R)=F_\epsilon \oplus 
X_\epsilon$ in which both subspaces, $F_\epsilon$ and $X_\epsilon$, are 
invariant under $Q$, with $dim(X_\epsilon)<\infty$ and such that the norm of 
$Q$ restricted to $F_\epsilon$ is less than $\epsilon$.   
As $F_\epsilon $ and $X_\epsilon$ are $Q$-invariant these are also 
$T$-invariant. 
Consider the restriction $Q|_{F_\epsilon}$. 
Since the norm on $F_\epsilon$ is small, it follows that $(Id-Q)|_{F_\epsilon}$ 
is invertible. 
The von Neumann series yield $(Id-Q)^{-1}=\sum_{n=0}^\infty Q^n$ at least on 
$F_\epsilon$. 
This means that 
$(Id-Q)^{-1}|_{F_\epsilon}$ is a compact operator 
as it is the norm limit of compact operators.
Since we have $Id-Q=T$, 
we get that $T|_{F_\epsilon}$ is a compact isomorphism, 
which implies that $F_\epsilon$ has finite dimension. 
The bottom line is that $B_0(S_R)$ is finite dimensional and 
$T$ is a contraction over a finite dimensional space, 
and as such is uniformly ergodic by the Mean Ergodicity lemma. 
\end{proof}

As an immediate corollary we have the following. 

\begin{corollary}
Under conditions of the last theorem, 
the operator $Q$ is compact if and only if $R$ is postcritically finite.
\end{corollary}

\begin{proof}
If $R$ is postcritically finite, then $B_0(S_R)$ has finite dimension.
Since $Q$ is continuous, then $Q$ is compact. 
But, if $Q$ is compact, as in the proof of the previous theorem, 
the linear space $A(S_R)$ which is dual to $B_0(S_R)$
has finite dimension. This is only possible when $P_R$ is finite. 
\end{proof}

The following is the main theorem of this section. 


\begin{theorem}\label{th.ThurstonMeanergodic}
Assume that the conservative set does not intersect the Fatou set $F_R$. Then 
Thurston operator $T$ is mean ergodic on $B_0(S_R)$. Moreover, $T$ has a non 
zero fixed point in $B_0(S_R)$ if and only if $R$ is a flexible Latt\'es map.
\end{theorem}
\begin{proof}
Assume that there is no non-zero fixed point of $R^*:B^*_0(S_R)\to B^*_0(S_R)$, 
then by arguments of the proof of Theorem \ref{th.MeanErg.1} we have that 
$(I-T)(B_0(S_R))$ is dense in $B_0(S_R)$. This implies that for $\phi\in 
B_0(S_R)$ the averages $\frac{1}{n}\sum_{i=0}^{n-1}T^i(\phi)$ converge to 
$0.$ Then $T$ is mean ergodic in $B_0(S_R)$.

If $R^*$ has a fixed point, by
Proposition \ref{prop.invariantmeasure}, Lemma \ref{lemma.dichotomy} and 
the arguments presented in the proof of Theorem \ref{th.meanergosep} 
imply that $R$ is a flexible Latt\'es map and $T$ is mean ergodic on 
$B(S_R)$ and, hence, on $B_0(S_R).$

Finally, by the Separation principle of Section \ref{section4}, a mean ergodic 
operator $T$ has a non-trivial fixed point if and only if its dual $R^*$ has a 
non trivial fixed point. By the arguments above this is the case precisely when 
$R$ is a flexible Latt\'es map.
\end{proof}

The following corollary shows that if $T$ has a fixed point in $B(S_R)$ close 
enough to $B_0(S_R)$ in the Teichm\"uller distance, with respect to the 
Teichm\"uller norm,  then $R$ is a Latt\'es map. 

\begin{corollary}
Let $R$ be as in Theorem \ref{th.ThurstonMeanergodic}. 
Let $\alpha\in B(S_R)$ with $T(\alpha)=\alpha$ and $\|\alpha\|_T=1$ where 
$\|\cdot\|_T$  is the Teichm\"uller norm. 
If 
$$
dist(\alpha,B_0(S_R))=\inf_{\phi \in B_0(S_R)}\|\alpha -\phi\|_T<1, 
$$
then $R$ is a flexible Latt\'es map.
\end{corollary}
\begin{proof}
Given the hypothesis, we can find $\phi_0\in B_0(S_R)$ subject to 
$\|\alpha-\phi_0\|_T<1$. 
By Theorem \ref{th.ThurstonMeanergodic}, $T$ is mean ergodic. 
If $T$ has a fixed point in $B_0(S_R)$ then  $R$ is a Latt\'es map. 
Otherwise the Ces\`aro averages $A_n(\phi)$ converge to $0$ 
for every $\phi$ in $B_0(S_R)$. 
In particular, we have 
$$
1=\|\alpha\|_T=\lim \|A_n(\alpha - \phi_0)\|_T \leq \|\alpha-\phi_0\|_T<1,
$$ 
which is a contradiction. 
Thus $T$ has a fixed point in $B_0(S_R)$ and $R$ is a Latt\'es map. \end{proof}

We say that $R$ is \textit{dissipative} 
if the dissipative set of $R$ is the whole Riemann sphere; in other words, the 
conservative set has Lebesgue measure zero. 

\begin{theorem}\label{th.dissipative}
Let $R$ be a dissipative map. 
Then, for any $\alpha$ in $B_0(S_R)$, 
the Thurston operator orbit $T^n(\alpha)$ converges weakly to $0$. 
\end{theorem}
 
\begin{proof}
Since the conservative set of $R$ has Lebesgue measure zero, 
by Lemma \ref{lm.con.dis},  we have that $\sum_{n} R^{*n}(\phi)(z)$ absolutely 
converges almost everywhere in $\C$ for every $\phi$ in $A(S_R)$.
 
In particular, it follows that $R^{*n}\phi$ converges to $0$ almost everywhere 
on $\C$. But  $R^{*n}\phi$ is a bounded sequence in $A(S_R)$. Therefore 
$R^{*n}\phi$ defines a normal family of holomorphic functions on $S_R$. Hence   
$R^{*n}\phi$ converges pointwise to $0.$ By duality, the $T$ orbit of any 
element in $B_0(S_R)$ weakly converges to $0$. 
\end{proof}

An operator satisfying the conclusion of the previous theorem is called 
\textit{weakly asymptotic}.  


\begin{theorem}\label{th.kukareko}
Suppose that the measure of the postcritical set $P_R$ is 
$0$ and that the image of the unit ball in $B_0(S_R)$ under $Id-T$ is closed, 
then $T$ has a fixed point in $B(S_R)$ if and only if $R$ is a Latt\'es map.
\end{theorem}
\begin{proof} 
We use Theorems 2.3 and Theorem 3.3  in \cite{FonfLinRubinov}, which state 
that, for a separable Banach space $X$ and power bounded operator 
$T$, if the image of the unit ball under $(Id-T)$ is closed then either $T$ is 
uniformly ergodic or the space $\overline{(Id-T)(X)}$ contains an isomorphic 
copy of an infinite dimensional dual Banach space. 

Now let us show that $T$ is uniformly ergodic on $B_0(S_R)$. By contradiction, 
if $T$ is not uniformly ergodic $\overline{(Id-T)(B_0(S_R))}$ contains an 
infinite dimensional dual Banach space $Y$. By Theorem 1 in \cite{BonetWolf}, 
$B_0(S_R)$ is isomorphic to a closed subspace of $c_0$. On the other hand, any 
closed infinite dimensional subspace of $c_0$ contains a closed subspace 
isomorphic to $c_0.$ Hence $Y$, contains an isomorphic copy of $c_0.$ Let $h: 
c_0\rightarrow Y$ be this isomorphism onto the image. 
Since $Y$ is a dual space, the unit ball in $Y$ is $*-$weak compact, so the 
isomorphism $h$ can be extended to a continuous linear operator $O:\ell_\infty
\rightarrow Y$.  Then $O^*:Y^*\rightarrow (\ell_\infty)^*$  is continuous with 
respect to the $*$-weak topologies. As $Y$ is separable then any bounded set of 
linear functionals contains a subsequence converging in the $*$-weak topology. 
Now $\ell_\infty$ is a Grothendieck space, then   $O^*$ is a 
weakly compact operator.  Thus $O$ and, hence,  $h$ are weakly compact. But a 
separable Banach space with weakly compact unit ball is reflexive, this 
contradicts the fact that $c_0$ is not reflexive. Therefore $T$ is uniformly 
ergodic on $B_0(S_R).$

If $T$ has a fixed point in $B(S_R)$, then $Id-T$ is not invertible in 
$B(S_R)$. Then $(Id-T):B_0(S_R)\rightarrow B_0(S_R)$ is not an isomorphism. 
Thus by the Uniform Ergodicity lemma the value $1$ is an isolated eigenvalue 
for $T$ in $B_0$. This implies that  $T$ has a non-zero fixed point in 
$B_0(S_R)$. By Lemma \ref{lemma.dichotomy}, the map $R$ is either dissipative 
or 
a flexible 
Latt\'es map. But, if $R$ is dissipative, by Theorem \ref{th.dissipative}, the 
operator $T$ does not have non-zero fixed points in $B_0(S_R)$. Hence $R$ is a 
Latt\'es map.

The reciprocal of the theorem follows from finite dimensionality of all the 
spaces involved in the Latt\'es case. \end{proof}

Let us add some comments about the closeness of the image under $(Id-T)$ of 
the unit ball. Since $B(S_R)$ is a dual space, the image 
of its unit ball under $(Id-T)$ is always closed. When the image of the unit 
ball of $B_0(S_R)$ under $(Id-T)$ is not closed, consider the space 
$$X=cl((Id-T)^{-1}(B_0(S_R))\subset B(S_R),$$ then again on $X$, the image of 
its unit ball under $(Id-T)$ is closed. But $X$ is invariant under $T$ 
and $B_0(S_R)\oplus Fix(T,B(S_R)) \subset X.$  Then we have the following 
conclusion. 
\begin{proposition} Let $X$ be as in the discussion above. 
 Assume that $X$ satisfies one of the following conditions: 
\begin{enumerate} 
 \item $X$ is separable.
\item $X$ is a Grothendieck space.
\item The operator $T:X\rightarrow X$ is mean ergodic.
\end{enumerate}
Then $T:B(S_R)\rightarrow B(S_R)$ is uniformly ergodic.
\end{proposition}

\begin{proof}
 If $X$ is separable, since the image of the unit ball under $(Id-T)$ is closed 
and $\overline{(Id-T)(X)}\subset B_0(S_R)$ then we can apply similar arguments  
of Theorem \ref{th.kukareko} to show that $T$ is uniformly ergodic.

If $X$ is a Grothendieck space, then $(Id-T):X\rightarrow B_0(S_R)$ is 
weakly compact. It follows that $(Id-T):B_0(S_R)\rightarrow B_0(S_R)$ is 
also weakly compact, which implies that the image of the unit ball in 
$B_0(S_R)$ 
is closed and Theorem \ref{th.kukareko} applies. 

Finally, if $T:X\rightarrow X$ is mean ergodic then Proposition 3.8 in 
\cite{FonfLinRubinov} states precisely that the image of the unit ball in 
$B_0(S_R)$ under $(Id-T)$ is closed and the arguments of Theorem 
\ref{th.kukareko} apply again. \end{proof}


In the next section  we analyse further properties of $X$ in a more general 
setting. 

\section{Hamilton-Krushkal sequences}

In this section, to avoid cumbersome calculations and definitions, we consider 
rational maps $R$ satisfying 
two conditions:
\begin{itemize}
\item First, there is no non-trivial quasiconformal deformation supported 
on the Fatou set.
\end{itemize} 

This condition does not impose any restriction on the study of Sullivan's 
conjecture if the Julia set is connected. In fact, when $J_R$ is 
connected, by Sullivan's Theorem, the $J-$stability component of $R$ 
contains a map $Q$ which does not admit  non-trivial quasiconformal 
deformations on the Fatou set. Moreover, $Q$ is unique, up to M\"obius 
conjugacy, only in the case when $J_Q$ does not support an  invariant Beltrami 
differential. 
\begin{itemize}
\item Second, the postcritical set $P_R$ does not support an absolutely 
continuous invariant measure with respect to the Lebesgue measure.
\end{itemize}

In other words, the measure of the intersection of $P_R$ with the strongly 
conservative set $SC(R)$ is zero. In the last section, we will discuss and give 
partial results for the case where $P_R\cap SC(R)$ has positive measure. 

Let us consider the elements $\gamma_v(z)$ with $v$ a critical value. 
Let $D=\{A_n(\gamma_{v_i})\}$ be the set of Ces\`aro averages for all 
$\gamma_{v_i}$ with $v_i$ in the critical value set $V(R)$. 
A sequence $\{\phi_k\}$ in $A^*(S_R)$ is called \fd{degenerating}, non 
normalized, if there are constants $C$ and $\epsilon>0$ with $\epsilon 
<\|\phi_k\|<C$ such that $\phi_k$ converges to $0$ pointwise almost 
everywhere. 


By Bers Isomorphism theorem, we identify $A^*(S_R)$ with the space 
$B(S_R)$ and consider the seminorm on $B(S_R)$ given by
$$
K(l)=\sup|l(\phi_k(z))|
$$
where the supremum is taken over all degenerating sequences 
$\{\phi_k\}$ in $D$. 
  

Note that a sequence in $D$ is either degenerating or precompact in norm. 
Indeed, if the sequence $\{A_n(\gamma_v)\}$ is not degenerating and does not 
converge in norm, then there exist a subsequence  which converges pointwise and 
locally uniformly to a non-zero limit which is fixed by the Ruelle operator. By 
Lemma \ref{lemma.dichotomy} and Proposition 
\ref{prop.invariantmeasure}, the map $R$ is a Latt\'es map.  By Theorem 
\ref{th.meanergosep}, the Ruelle operator $R^*$ is uniformly ergodic. Then the 
Ces\`aro averages of any function converges in norm which is a contradiction.
  
The \fd{Hamilton-Krushkal space} $HK(R)$ is the zero set of $K$.  
Since we have $K(l)\leq \|l\|$ on $B(S_R)$, the space $HK(R)$ is a closed 
subspace of $B(S_R)$.  

A subspace $Y$ of a Banach space $X$ is \fd{coseparable} when $X/Y$ is 
separable. 


\begin{theorem}\label{th.coseparable}
A rational map $R$ satisfies Sullivan's conjecture if and only if $HK(R)$ is 
coseparable in $A^*(S_R)$. 
\end{theorem}
\begin{proof}
If there is no invariant Beltrami differential supported on the Julia set 
then the action of $R^*$ on $L_1(J_R)$ is mean ergodic and the Ces\`aro 
averages converge to $0$ in $L_1(J_R)$.  Thus $R^*$ on $A(S_R)$ is mean 
ergodic. So we get $HK(R)=A^*(S_R)$ and that the quotient is separable. 
If $R$ is Latt\'es, then $A^*(S_R)$ is finite dimensional and so is $HK(R)$. 

Conversely, assume that $HK(R)$ is coseparable. Then there exist a countable 
set $\{\alpha_i\}$ of elements in $A^*(S_R)$ such that $S=\{\alpha_i\}+ 
HK(R)$ is dense in $A^*(S_R)$. By induction, and a diagonal argument, we can 
pick a sequence $\{n_k\}$ such that $\alpha_i(A_{n_k}(\gamma_v))$ converges for 
all critical values $v$.  
Since $R^*$ is a contraction and $S$ is an everywhere dense subset 
of $A^*(S_R)$, the Ces\`aro averages $A_{n_k}(\gamma_v)$ converges weakly for 
every $v$. 
By the Mean Ergodicity lemma, for all $v$, the sequence $A_{n_k}(\gamma_v)$ 
converges in norm. Let $\mu$ be a non-zero invariant Beltrami differential. 
By Lemma \ref{lemma.fixed.Beltrami} there exist a critical value $v_0$ such 
that 
$\int_\C  \mu(z) \gamma_{v_0}\neq 0$. 
Thus the limit $f_0=\lim A_{n_k}(\gamma_{v_0})$ is a non-zero fixed point of 
the Ruelle operator in $A^*(S_R)$. Since the measure of $P_R\cap SC(R)$ is 
zero, by Lemma \ref{lemma.dichotomy}, the map $R$ is a Latt\'es map. 
\end{proof}

From the arguments in the previous proof, we have the following corollary. 

\begin{corollary}\label{cor.RuellemeanErg}
If $HK(R)$ is coseparable then the Ruelle operator $R^*$ is mean ergodic on 
$A(S_R)$ equipped with the norm inherited from $L_1(J_R).$
\end{corollary}

Also, we have the following.

\begin{corollary}\label{cor.coseparable} 
The  space $HK(R)$ is coseparable if and only if 
$codim(HK(R))=0$. 
\end{corollary}


\section{Amenability and compactness}


This is the main section of this work,  our goal is to give compactness 
conditions of suitable operators under which a map $R$ satisfies the Sullivan 
conjecture. We will keep the technical assumptions given at the beginning of 
the previous section.

 For every critical value  $v$  of $R$ define the operator 
$E_v:B(S_R)\rightarrow \ell_\infty$ by the formula 
$$
E_v(\psi)=\left( \int_{\C} \lambda^{-2}(z) 
\bar{\psi}(z)A_{n}(\gamma_v)(z)|dz|^2\right )_{n=0}^{\infty}
$$

In particular,  an element $\phi \in A^*(S_R)$ belongs to $HK(R)$ 
whenever  $E_v(\phi)\in c_0$ for every critical value $v$. On the image of 
$E_v$,  the Thurston operator $T$ acts as 

$$
\hat{T}_v(E_v(\psi))= E_v(T(\psi)).
$$
This formula defines $\hat{T}_v$ as a linear endomorphism, non necessarily 
continuous, of the image of $E_v$. 

A \textit{mean} $m$ on $\ell_\infty$ is a positive linear functional which 
satisfies three conditions: 
\begin{itemize}
\item $m(1,1,1,...)=1$, 

\item  If $\sigma$ is the shift $\sigma(a_1,a_2,...)=(a_2,a_3,...)$, 
then $m(x)=m(\sigma(x))$ for any $x\in \ell_\infty$.  

\item $\liminf |a_i|\leq  |m(a_1,a_2,...)|\leq \limsup|a_i|$. 
\end{itemize}

A mean is also known as a \textit{Banach limit} on $\ell_\infty$. 

A linear operator $O:X\to X$, not necessarily continuous, defined on a linear 
subspace $X$, not necessarily closed, of $\ell_\infty$ has an 
invariant mean if there is a mean on $\ell_\infty$ with non-zero restriction to 
$X$ that satisfies $m(O(\alpha))=m(\alpha)$ for $\alpha\in X$. 
We denote by $M(O)$ the set of all invariant means for $O$. 

\begin{lemma}\label{lm.invmeanempty}
The set $M(\hat{T}_v)$ is empty if and only if 
$E_v(A^*(S_R))$ consists exclusively of sequences that converge to $0$. 
\end{lemma}
\begin{proof}
By definition means are invariant under shift and bounded by the supremum of 
the elements of the sequence, this implies that $c_0\subset ker(m)$ for every 
mean $m$. Hence, if  $E_v(A^*(S_R))\subset c_0$, then we get 
$M(\hat{T}_v)=\emptyset$. 

Conversely, if there is an element $h\in A^*(S_R)$ with $E_v(h) \in 
E_v(A^*(S_R))\setminus c_0$ then there is a subsequence $\{n_j\}$ such that 
$h(A_{n_j}(\gamma_v))$ converges to a non-zero number $a$. 
By duality this implies that $A^*_{n_j}(h)$ allows a subsequence which 
converges $*$-weakly to a non-zero element $l_0$ that satisfies $T(l_0)=l_0$. 
Set $E_v(l_0)=(l_0(\gamma_v), l_0(\gamma_v),...)$ and, 
since $E_v(A^*(S_R))$ is a subspace of $\ell_\infty$, 
we conclude that $E_v(A^*(S_R))$ contains the constant sequence $1$. 
This implies that $E_v(A^*(S_R))$ intersects the space of convergent sequences 
in a non empty set. 
On convergent sequences, the functional $l:\{c_i\}\mapsto \lim c_i$ is 
continuous. 
By Banach Limit theorem, there exist an extension ${\cal 
L}$ to all $\ell_\infty$ which is a mean. 

Next, we show that ${\cal L}$ is $\hat{T}_v$ invariant on $E_v(A^*(S_R))$.  
In fact,  we have 
$$
|T(h)(A_n(\gamma_v)) - h(A_{n+1}(\gamma_v))|\leq \frac{4\|h\|\|\gamma_v\|}{n},
$$  
for $h\in A^*(S_R)$. So the difference 
$$
|T(h)(A_n(\gamma_v)) - h(A_{n+1}(\gamma_v))|
$$ 
uniformly converges to $0$ as $n$ tends to $\infty$ on any ball of $A^*(S_R)$, 
and hence
$(\sigma -\hat{T}_v) E_v(A^*(S_R))\subset c_0.$ Thus $(\sigma -\hat{T}_v) 
E_v(A^*(S_R))$ belongs to $ker({\cal L})$. 
The invariance of ${\cal L}$ with respect to $\sigma$ implies the invariance 
of ${\cal L}$ with respect to $\hat{T}_v$. Therefore $M(\hat{T}_v)$ is non 
empty. \end{proof}

Next we show the finiteness of $M(\hat{T}_v)$ in very special cases.

\begin{theorem}\label{th.finiteBanach}
The set $M(\hat{T}_v)$ is finite if and only if $M(\hat{T}_v)$ contains at most 
one element or, equivalently,  if and only if $E_v(A^*(S_R))$ consists 
exclusively of convergent sequences. 
\end{theorem}

\begin{proof}
The first equivalence is clear after one notices that $M(\hat{T}_v)$ is convex, 
so we just worry about the second. 

If $M(\hat{T}_v)$ is empty, by Lemma \ref{lm.invmeanempty} we are done. 
Otherwise, again by the arguments on Lemma \ref{lm.invmeanempty}, if 
$E_v(A^*(S_R))\subset c$, then the only invariant mean is given by the 
restriction of the limit functional, $\{a_n\}\mapsto \lim a_n.$  
Reciprocally, we assume that there is only one invariant mean $m$ and set 
$X=\overline{E_v(A^*(S_R))}$. 
Again, by the same arguments given in Lemma \ref{lm.invmeanempty},  
the space $E_v(A^*(S_R))$ contains the element $(1,1,1,...)$.

Now we claim that if $L$ is a mean then  $L(x)=m(x)$ for all $x\in X$.
Indeed, by Lemma \ref{lm.invmeanempty}, we have that  $(\sigma -\hat{T}_v) 
E_v(A^*(S_R))\subset c_0.$ It follows that $L(\hat{T}_v(x))=L(x)$ for all $x\in 
E_v(A^*(S_R))$. Since $(1,1,1,...)\in E_v(A^*(S_R))$ then 
$L|_{E_v(A^*(S_R))}\neq 0.$ By uniqueness $L(x)=m(x)$ on $E_v(A^*(S_R))$, so 
by the continuity of $L$ and $m$ we get our claim. In particular,   
$E_v(A^*(S_R))\cap ker(m)\subset \bigcap_L ker(L)$ where the intersection is 
taken over all means $L.$ 


The space $X$ admits the decomposition $X=\C \cdot (1,1,1...) \oplus 
(ker(m)\cap X)$,  
and there is a Beltrami fixed point $\mu$ such that $E_v^{-1}(X)=\C \mu \oplus 
E_v^{-1}(ker(m))$. By Banach Limit theorem (see \cite[Theorem 4.1]{Krengel}) 
and 
the 
claim above, 
for every $h\in E_v^{-1}(ker(m))$ we get 
$$
\lim_{k\rightarrow \infty} \frac{1}{k}\sum_{j=0}^{k-1} h(A_j(\gamma_v))=0.
$$
But we have 
$$ 
\frac{1}{k}\sum_{j=0}^{k-1} 
h(A_j(\gamma_v))=h(\frac{1}{k}\sum_{j=0}^{k-1} 
A_j(\gamma_v)), 
$$ 
and thus, the sequence $f_k=\frac{1}{k}\sum_{j=0}^{k-1} A_j(\gamma_v)$ 
is a weakly convergent sequence of integrable functions. Let $\phi_0$ be a weak 
limit of $f_k$. 
By the Dunford-Pettis's theorem, the sequence $f_k$ is uniformly integrable. 
Since $A_j(\gamma_v)$ is  bounded and 
holomorphic, the sequence $f_k$ converges uniformly  on compact sets to 
$\phi_0$. Since $\mu$ is invariant, and $\int \mu \gamma_v \neq 0$ then $\int 
\mu \phi_0\neq 0.$ Now recall that the sequence  $\{A_n(\gamma_v)\}$ is either 
degenerating or converges by norm. If $\{A_n(\gamma_v)\}$ is degenerating then 
$f_k$ is also degenerating. This implies that $\phi_0=0$ which contradicts the 
existence of $\mu.$ Thus $\{A_n(\gamma_v)\}$ converges in norm. Then $E_v(h)$ 
is a convergent sequence for every $h\in A^*(S_R).$ \end{proof}

We have the following.

\begin{theorem}\label{th.Mtfinite} 

The map $R$ is satisfies Sullivan's conjecture if and only if $M(\hat{T}_v)$ 
is finite for every critical value $v$.
\end{theorem}
\begin{proof}
If $M(\hat{T}_v)$ is finite, then by Theorem \ref{th.finiteBanach} consists of 
at most one element $m_v$ which can be $0$. The space $E_v(A^*(S_R))$ 
consists only of convergent sequences for every critical value $v$. By 
definition $HK(R)$ contains  all elements $h$ in $A^*(S_R)$ such that $E_v(h)$ 
is a sequence converging to $0$ for every critical value $v.$ In other words,
$\cap_v ker(m_v(E_v(A^*(S_R))))\subset HK(R)$. But the space $\cap_v 
ker(m_v(E_v(A^*(S_R))))$  has finite codimension in $A^*(S_R)$.
 

This implies that $HK(R)$ is coseparable in $A^*(S_R)$, by Theorem 
\ref{th.coseparable}, the map $R$ satisfies Sullivan's conjecture. 

By Corollary \ref{cor.coseparable}, the space $HK(R)$ coincides with $A^*(S_R)$ 
and by Lemma \ref{lm.invmeanempty} and Theorem \ref{th.finiteBanach}, the 
converse follows. 
\end{proof}

 Now we show the following theorem.

\begin{theorem}\label{th.Evcompact} Let $R$ be a rational map without 
rotational domains. Then for any given critical value $v$ of $R$, the following 
statements are equivalent. 

\begin{enumerate}
\item The space $M(\hat{T}_v)$ is finite.
\item The restriction $E_v:B_0(S_R) \to \ell_\infty$ is weakly compact.
\item The restriction $E_v: HK(R)\rightarrow \ell_\infty$ is compact.
\item The operator $E_v:B(S_R)\rightarrow \ell_\infty$ is compact.
 
\end{enumerate}
\end{theorem}
\begin{proof}

Clearly, we have that (4) implies (3) and (3) implies (2). 

Let us show that (2) implies (4). Recall that $R$ does not admit non-trivial 
quasiconformal deformations on the Fatou set and the measure of $P_R\cap SC(R)$ 
is zero, and does not have rotational domains. Hence by Theorem 
\ref{th.ThurstonMeanergodic}, Thurston operator $T$ is mean ergodic on 
$B_0(S_R)$, so $E_v(B_0)$ consists of convergent sequences. 
Since $\ell_1$ is isometrically isomorphic to $c^*$. Then by duality, 
$E^*_v:\ell_1 \rightarrow A(S_R)=B^*_0(S_R)$ is given by 
$E^*_v(\{a_n\})=\sum_n a_n A_n(\gamma_v)$. If $E_v$ is weakly compact on 
$B_0(S_R)$ then $E^*_v$ is weakly compact on $\ell_1.$ The image of the 
canonical base of $\ell_1$ is  $\{A_n(\gamma_v)\}$. By the Mean Ergodicity 
lemma, the sequence $\{A_n(\gamma_v)\}$ is precompact in norm. Then $E^*_v$  
is a compact operator. By duality the operator $E^{**}_v:B\rightarrow 
\ell_\infty$ given by $E_v|_{B_0(S_R)}^{**}(l)=E_v(l)=\{l(A_n(\gamma_v)\}$ is 
compact.

Now let us show that (1) implies (2). By Theorem \ref{th.finiteBanach}, 
$E_v(B(S_R))$ consists of convergent sequences. In other words,  $E_v$ defines 
a continuous operator from a Grothendieck space into a separable space. Since 
the unit ball in $\ell_1$ is sequentially precompact on the $*$-weak topology. 
By definition of a Grothendieck space, we have that $E_v$ is a weakly compact 
operator on $B(S_R).$ Hence the restriction of $E_v$ on $B_0(S_R)$ is also 
weakly compact. 

Finally, let us show that (2) implies (1). By Theorem \ref{th.finiteBanach} is 
enough to show that $E_v(B(S_R))$ consists of convergent sequences. But $E_v$ 
is 
a compact operator on $B(S_R)$ and $E_v(B(S_R))=E^{**}_v(B_0(S_R)\subset 
E_v(B_0(S_R))$. By mean ergodicity of $T$ on $B_0(S_R)$, the set 
$E_v(B_0(S_R))$ consists only of convergent sequences. 
\end{proof}

 The following corollary is one of the main results of this paper. 

\begin{corollary}
A rational map $R$ satisfies Sullivan conjecture if and only if either the 
operator $Id-T$ is compact or the operator $E_v$ is compact for every critical 
value $v.$ 
\end{corollary}

\begin{proof}
If $Id-T$ is compact then, by Theorem \ref{th.Qcompact}, the operator $T$ is 
uniformly ergodic on $B(S_R)$. If $R$ admits an invariant non-zero Beltrami 
differential supported on the Julia set then $1$ belongs to 
the spectrum of $T$.  Then Theorem \ref{th.meanergosep} implies that the map 
$R$ is Latt\'es.
If $E_v$ is compact for every critical value $v$, then $R$ satisfies Sullivan's 
conjecture by Theorem \ref{th.Evcompact} and Theorem 
\ref{th.Mtfinite}. \end{proof}

Note that if $Id-T$ is compact then the operators $E_v$ are compact but 
the converse is no true. A map $R$ where $E_v\circ (Id-T)$ is not compact, 
for some critical value $v$, would serve as a counterexample to Sullivan's 
conjecture. However, we have the following observation which is one of the main 
motivations of the present work.

\begin{proposition}\label{prop.op.compact}
The operator $E_v \circ (Id-T):B(S_R) \to \ell_\infty$ is compact for every 
critical value $v$. 
\end{proposition}

\begin{proof}
A bounded sequence on $B(S_R)$ contains a subsequence which is $*$-weakly 
convergent. Let $\mu_i$ be a sequence on $B(S_R)$ with $*-$weak limit $\mu_0$. 
Define $\omega_i=E_v\circ (Id-T)(\mu_i)$ and $\omega_0=E_v\circ(Id-T)(\mu_0)$.
We will prove that $\omega_i$ converges to $\omega_0$ in norm. 

From the definition we have then 
$$
||\omega_i -\omega_0||=\sup_n \left| \int_{S_R} 
A_n(\gamma_v)(Id-T)\mu_i-A_n(\gamma_v)(Id-T)\mu_0) \right|
$$
$$
=\sup_n \left| \int_{S_R}  A_n(\gamma_v)(Id-T)(\mu_i-\mu_0)\right|.
$$ 
However, as $T$ is dual to the Ruelle operator we get  
$$
\left | \int_{S_R}  A_n(\gamma_v)(Id-T)(\mu_i-\mu_0) \right|=\left |\int_{S_R}  
(Id-R^*)A_n(\gamma_v)(\mu_i-\mu_0)\right|
\leq \frac{2\|\gamma_v\|}{n}\| \mu_i-\mu_0\|.
$$ 
Since $\|\gamma_v\|\|\mu_i-\mu_0\|$ is bounded and $\mu_i$ converges $*$-weakly 
to $\mu_0$, then  $||\omega_i -\omega_0||$ converges to $0$ as $i$ tends to 
infinity. Hence $E_v \circ (Id-T)$ is compact. \end{proof} 

In general, the compactness of the composition $E_v\circ (Id-T)$ does not imply 
the compactness of any of the factors. But this implication is true,  by 
Theorem \ref{th.Evcompact}, if and only if Sullivan's conjecture holds true. 

For every $v$, the operator $E_v$ has a canonical extension on
$L_\infty(\bar{\C})$ with the same defining formula. When 
$P_R\neq \bar{\C}$ and $SC(R)\cap P_R$ has Lebesgue measure zero,  the 
extension of $E_v$ on $L_\infty(\bar{\C})$ is compact if and only if $E_v$ is 
compact on $B(S_R).$ The extension of the operators  $E_v$ on $L_\infty$,
which we also denote $E_v$, gives a sort of ``marking" for a rational map $R$. 
Furthermore, the operators $E_v$ induce a topology on the rational maps 
as follows.

A sequence of rational maps $R_i$ converges to $R_0$ in 
$v-$sense, where $v$ is a critical value of $R_0$ if and only if  for a given 
$\mu\in L_\infty(\C)$ there exist a sequence of critical values $v_i$ of $R_i$ 
such that $E_{v_i}(\mu)\rightarrow E_v(\mu)$ in $\ell_\infty.$

When $R_i$ converges to $R_0$ on the $v$-sense we will say that $R_0$ is a 
$v-$limit of $R_i.$

\begin{proposition}
 Let $R_0$ be a $v-$limit of $R_i$ such that $M(\hat{T}_{v_i})$ if finite  for 
all critical values $v_i$ of $R_i,$ then $M(\hat{T}_v)$ is also finite.
\end{proposition}

\begin{proof}
Let $c$ be the space of convergent sequences. By Theorem \ref{th.finiteBanach}, 
it is enough to show that the image $E_v$ belongs to $c$. 
By the hypothesis, for every critical value $w$ of $R_i$ the image of $E_{w}$ 
belongs to $c$. As the space  $c$ is closed on $L_\infty$ 
then the image of $E_v$ is also a subset of $c.$\end{proof}

In other words, roughly speaking, any $v-$limit of rational maps satisfying 
Sullivan's conjecture, satisfies Sullivan's conjecture too. Moreover, in 
general, not every accumulation point, in $v-$sense, has the same degree as the 
approximating elements. Further details on this topology will be the subject of 
a forthcoming work. 
 
\section{A ``mixing'' condition.}

In this section we show that the Ruelle operator does not have fixed 
points when $R$ satisfies a kind of mixing condition over  
its strongly conservative set. 

We say that $R$ satisfies the $M$-\textit{condition} if, for an  invariant 
ergodic probability measure $\nu$ which is absolutely continuous with respect 
to the Lebesgue measure supported on the Julia set  $J_R$, $R$ satisfies the 
following two properties: 

\begin{enumerate}
\item  If $A$ and $B$ is a pair of $\nu$-measurable subsets of $SC(R)\cap J_R$, 
 
then $\lim \nu(B\cap R^{-n}(A))=\nu(A)\nu(B)$.

\item  There exist a $\nu-$measurable set $A_\nu\subset SC(R)\cap J_R$ with 
$\nu(A_\nu)>0$ such that the sequence of  functions 
$$
B^n(\chi_{SC(R)}(x))=\frac{\overline{(R^n)'(x)}}{(R^n)'(x)}
$$ 
is precompact, in the topology of convergence in measure on $A_\nu$.
\end{enumerate}
The reader might recognize in the first property the classical mixing 
condition for invariant probability measures. We will comment the second 
property at the end of this section.
If there is no invariant absolutely continuous probability measure, then the 
$M$-condition is vacuously satisfied. This is the case when the Lebesgue 
measure of the strongly conservative set is zero. 

\begin{proposition}\label{prop.converg} 
Assume $R$ satisfies the first property of the $M$-condition. 
Let $\nu$ be an invariant ergodic probability measure absolutely continuous 
with respect to Lebesgue. Let $W=supp(\nu)$ be the support of $\nu$. Then for 
every $\phi$ in $L_\infty(W,\nu)$, the sequence $$(|R^*|)^{*n}(\phi)$$ 
converges 
$*$-weakly to a constant. 
\end{proposition}
\begin{proof}
The proof follows from classical ergodic theory, for convenience to 
the reader we include it here. Let $\phi$ be a non negative function such that 
$d\nu=\phi(z) |dz|^2$. 
Since $\nu$ is an invariant probability measure we have  $|R^*|\phi=\phi$.  
Now consider the space $L_1(W,\nu)$ and the operator $S$  on $ L_1(W,\nu)$ 
given by $S(g)=\frac{1}{\phi}|R^{*}|(g\phi)$ 
with dual $S^*(\omega)=\omega(R)$ for $\omega$ in $L_1^*(W,\nu)$.  
Note that $S$ and $S^*$ are contractions in both $L_1(W,\nu)$ and 
$L_\infty(W,\nu)$. By well known ergodic theorems (see for instance Chapter 6 
of Dunford and Schwartz \cite{DunfordSchwartz}),  
both $S$ and $S^*$ are contracting mean ergodic operators on $L_p(W,\nu)$ for 
all $1 \leq p <\infty$. The first part of the $M$-condition implies that, for 
every $f$ and $g$ in $L_2(W,\nu)$,
we have 
$$
\lim \int S^n(f)\,\bar{g}d\nu=\lim \int f\,\overline{S^{*n}(g)}d\nu=
\lim \int f \,\overline{g(R^n)}d\nu=\int f d\nu \int \bar{g} d\nu.
$$

Since $\nu$ is a probability measure, we get the chain of inclusions 
$$L_\infty(W,\nu)\subset L_2(W,\nu)\subset L_1(W,\nu),$$
and $L_2(W,\nu)$ defines an everywhere dense subspace in $L_1(W,\nu)$. 
Hence the orbits of $S$ and $S^*$ converges weakly 
in $L_1(W,\nu)$ and $*$-weakly in $L_\infty(W,\nu)$, respectively. 
Let $f_0$ be an element in $L_2(W,\nu)$. 
Then the  weak limits of 
$S^n(f_0)$ and $S^{n*}(f_0)$ are fixed points for $S$ and $S^*$, respectively. 
But $\nu$ is ergodic, 
so the spaces of fixed points of $S$ and $S^*$ consists only of constants. 
The conclusion of the proposition follows from the equality 
$S^*(\mu)=\mu(R)=|R^*|^*(\mu)$. 
\end{proof}


\begin{theorem}\label{th.mixing}
Assume that $R$ is not a Latt\`es map, that satisfies the $M$-condition, and 
$P_R\neq \C$. Then there is no non-zero fixed point for the 
Ruelle 
operator in $L_1(J_R)$.   
\end{theorem}
\begin{proof} Without loss of generality we can assume that $P_R$ is bounded. 
Now, assume that there exist a non-zero fixed point $f$ in $L_1(J_R)$ for the 
Ruelle operator. 
Then by Lemma \ref{lemma.dichotomy}, Proposition \ref{prop.invariantmeasure} 
and the conditions of the theorem, the support of 
$f$ is a bounded measurable subset of the postcritical set $P_R$ and there 
exist $\mu$, a fixed point for the Beltrami operator, with 
$\mu=\frac{\bar{f}}{|f|}$ almost everywhere on $supp(f)$. 
Let $\nu$ be the invariant probability measure with density $|f|$.
For $S$ and $S^*$ as in the proof of Proposition \ref{prop.converg}, 
 consider the operator given by $$
Z(g)=\frac{1}{|f|}R^*(g|f|), 
$$ 
which defines an endomorphism of $L_1(supp(f),\nu)$. 
In this situation we have that  
$$
Z^*(\alpha)(z)=B(\alpha)(z)=\alpha(R)(z)\frac{\overline{R'}(z)}{R'(z)}
$$ 
defines an endomorphism of $L_\infty(supp(f),\nu)$. 
We obtain $Z(g)=\bar{\mu} S(\mu g)$ and $Z^*(\alpha)=\mu S^*(\bar{\mu}\alpha )$. 
By Proposition \ref{prop.converg}, the orbits of $S$ and $S^*$ weakly converge to constants;  
hence the orbits of the operators $Z$ and $Z^*$ weakly converge to scalar 
multiples of $\bar{\mu}$ and $\mu$ respectively. 
Let $c_g$ be the constant such that $Z^{*n}(g)$ weakly converge to $\mu c_g.$

Let $z_0$ be a density point of  $supp(f)$ and a continuity point of  
$\mu$. Since $supp(f)$ is a subset of the strongly conservative set $SC(R)$ and
almost every point $supp(f)$ is recurrent, we can assume that $z_0$ is 
also recurrent. 
This implies that there exist a sequence $\{n_i\}$ such that
$$
|\mu (R^{n_i}(z_0))-\mu(z_0)| \to 0.
$$ 
But $\mu$ is invariant, so the differences  
$\displaystyle{\left|\frac{\overline{(R^{n_i})'}}{(R^{n_i}})'(z_0)-1\right|}$ 
converges to $0$. 
Using together that $supp(f)$ is bounded and the $M-$condition,
we can assume that the previous sequence converges pointwise 
 almost everywhere in $A_\nu \subset supp(f)$. 
In this case we have 
$\frac{\overline{(R^{n_i})'}}{R^{n_i}}(z_0)=(Z^*)^{n_i}(\chi_{supp(f)})(z_0)$. 
By Lebesgue Dominated Convergence theorem 
$\frac{\overline{(R^{n_i})'}}{R^{n_i}}$ converge to its pointwise almost 
everywhere limit in the $L_1$ norm on $A_\nu$.  As 
norm and weak limits agree whenever exist, hence, we have  
$c_{\chi_{supp(f)}}=\frac{1}{\mu(z_0)}. $ But $c_{\chi_{supp(f)}}$ does not 
depend on the point $z_0$ and the sequence $\{n_i\}$.  
Therefore  $\mu(z)=\frac{1}{c_{\chi_{supp(f)}}}$ for almost every $z$ in 
$A_\nu$. Since $A_\nu\subset SC(R)$, then there exist a natural $k_0$ such 
that $\nu(A_\nu \cap R^{k_0}(A_\nu)>0.$ Hence for a density point $y$ of $ 
A_\nu \cap R^{k_0}(A_\nu)$ there exists a density point $x\in A_\nu$ so that 
$y=R^{k_0}(x)$ and, by invariance, we have 
$c_{\chi_{supp(f)}}(x)=\mu(x)=\mu(y)\frac{\overline{(R^{k_0})'(x)}}{(R^{
k_0 })'(x)}=c_{\chi_{supp(f)}(x)}\frac{\overline{(R^{k_0})'(x)}}{(R^{
k_0 })'(x)}$ then, again as in the proof of Proposition \ref{prop.chiA}, we 
have $A_\nu\cap (R^{k_0})^{-1}(A_\nu\cap R^{k_0}(A_\nu))\subset 
((R^{k_0})')^{-1}(\mathbb{R})$. 
But $\nu$ is absolutely continuous with respect to the Lebesgue measure and 
$\nu(A_\nu)>0$. 
This contradiction completes the proof.
\end{proof}

As an immediate corollary we have.

\begin{corollary}
If $R$ satisfies the conditions of Theorem \ref{th.mixing} and there exist a 
non-zero invariant Beltrami differential $\mu$. Then $supp(\mu)\cap SC(R)$ has 
Lebesgue measure zero.
\end{corollary}

Finally, let us comment the $M$-condition. According to M. Rees 
(see \cite{ReesErg}) the known examples of rational maps, for which the 
strongly 
conservative set has positive Lebesgue measure,  are ergodic postcritically 
finite maps $R$ with $SC(R)=\bar{\C}.$ In other words, there exists a
unique invariant absolutely continuous probability measure $\rho$ on $\bar{\C}$ 
so that for any pair $A,B$ of measurable subsets there exists a $n_0$ such that 
the Lebesgue measure of $R^{n_0}(A)\cap R^{n_0}(B)$ is positive. In ergodic 
theory, this corresponds to the fact that the operator $S$ of Proposition 
\ref{prop.converg} has strongly convergent orbits in $L_1(\C,\rho)$. The first 
part of our definition of the $M-$condition is equivalent to weak convergence of 
orbits in 
$L_1(\C,\rho)$ with respect to $|R^*|$ and this is the classical definition of 
mixing dynamical systems with respect to a non-necessarily invariant measure. 

Now, about the second property of the $M-$condition: if $\mu$ is an 
invariant Beltrami differential, then by Birkhoff theorem we have that 
$$
\frac{1}{n}\sum_{i=0}^{n-1}\bar{\mu}(R^i(z))=
\bar{\mu}(z)\left[\frac{1}{n}\sum_{
i=0}^{n-1}\left(\frac{\overline{R'^i(z)}}{R'^i(z)}\right)\right] \rightarrow 
\int \bar{\mu} d\nu.
$$ 
Here the convergence is almost everywhere and in $L_1$ norm on $supp(\nu)$, 
where $\nu$ is an invariant absolutely continuous probability measure. So the 
Ces\`aro averages $\frac{1}{n}\sum_{i=0}^{n-1} B^i(\chi_{SC(R)})$ 
converges almost everywhere to a multiple of $\mu$ on $supp(\mu)\cap 
supp(\nu)\subset SC(R).$ 
For any $j$, the sequence $\frac{1}{n}\sum_{i=0}^{n-1}
B^{j+i}(\chi_{SC(R)})$ converges almost everywhere to the same limit.

Now we claim that: \textit{if this limit exist uniformly, with respect to $j$, 
then the sequence $B^n(\chi_{SC(R)})$ converges almost everywhere.} 

In sketch, for any Banach limit $L$ consider the function 
$$\mu_L(z)=L(B^n(\chi_{SC(R)}(z))_{n=0}^\infty),$$ then by uniformity and
the Banach Limit theorem (see \cite[4.1]{Krengel}) we have 
$$\mu_L(z)=L(B^n(\chi_{SC(R)}(z))_{n=0}^\infty)=\mu(z)\int \bar{\mu}d\nu,$$
for $\nu-$almost everywhere.

Besides, a Banach limit is a positive functional. Then, by above we 
have 
$$Re \left(\mu_L(z)\right)=Re (L(B^n(\chi_{SC(R)}(z)))_{n=0}^\infty))$$$$=L(Re 
(B^n(\chi_{SC(R)}(z)))_{n=0}^\infty))=Re(\mu(z)\int \bar{\mu}d\nu).$$ 
Since  $\liminf$ and $\limsup$ 
are Banach limits, by the uniform convergence on $j$, then
$$\liminf(Re( 
B^n(\chi_{SC(R)}(z))_{n=0}^\infty))=\limsup(Re(
B^n(\chi_{SC(R)}(z))_{n=0}^\infty))$$ and analogously for $Im( 
B^n(\chi_{SC(R)}(z)))_{n=0}^\infty$. It 
follows that $B^n(\chi_{SC(R)})$ converges almost everywhere on $supp(\nu)$. 
Hence the second part of the definition of $M-$condi\-tion is formally weaker 
than almost everywhere convergence, roughly speaking the $M-$con\-dition is an 
analogue of classical mixing for the complex Perron-Frobenius operators.

Finally, next proposition is the main motivation of this section. This 
proposition follows from classical ergodic theorems and a well known fact, due 
to Sullivan, which states that a measurable set $A\subset J_R$ has zero 
Lebesgue measure whenever the iterates $R^n$ are injective on $A$ and the 
Lebesgue measure of $R^n(A)\cap R^m(A)$ is $0$ for all distinct  $m,n>0$.

\begin{proposition}
Assume that a rational map $R$ is injective on $P_R\cap J_R$ 
and the Lebesgue measure of $P_R\cap SC(R)$ is $0$. Then the Lebesgue measure 
of $P_R$ is zero if and only if $R$ is mixing on its conservative part with respect 
to the Lebesgue measure restricted on $P_R$.
\end{proposition}

\begin{proof}
If the Lebesgue measure of $P_R$ is zero then clearly $R$ is mixing on $P_R$ 
with respect to Lebesgue measure.
On the contrary, assume that the Lebesgue measure of $P_R$ is positive. Let us 
consider the dynamics of $R$ restricted on $P_R$ and let $C(P_R)$ and $D(P_R)$ 
be the conservative and the dissipative parts of this action, respectively. As 
$R$ is bijective on $P_R$, by Sullivan's lemma stated above, the Lebesgue 
measure of $D(P_R)$ is zero. Since the measure of $SC(R)\cap P_R$ is zero, then 
by Proposition \ref{prop.Krengel} there are no invariant absolutely continuous 
measures. 

Assume that $R$ is mixing on the conservative part of $P_R$. Now, we use 
Theorem 1.4 on page 255 of \cite{Krengel} which states that every positive 
contraction $E$ on a $L_1$ space has orbits strongly convergent to $0$ whenever 
$E$ has weakly convergent orbits and $E$ has no non-zero fixed points. Thus the 
orbits of $|R^*|$ in $L_1(C(R))$  converge strongly to $0.$ As $R(C(R))=C(R)$ 
then $$ 0=\lim_{n\rightarrow \infty}\int_{C(R)}|R^{*n}|\chi_{C(R)}= 
\lim_{n\rightarrow 
\infty}\int_{R^{-n}(C(R))} \chi_{C(R)}=\int \chi_{C(R)}.$$

Thus the Lebesgue measure of $C(R)$ is $0$ and hence $P_R$ has also Lebesgue 
measure  $0$. \end{proof}

Let us note that by Krengel-Sucheston result in \cite{KrenSuch}, there 
is no invertible, mixing, measure preserving, transformation on a 
$\sigma$-finite infinite space. Hence, for almost all $R$ acting bijectively on 
$P_R\cap SC(R)$ the second condition of previous proposition is fulfilled. 




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\end{document}


Ruellefixed.bbl
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