1$ and $\gamma_0, \xi > 0$, so that
$\mbox{\rm diam}(\varphi(B_{\gamma}(x)))\leq C_3^n\gamma^{\xi}$
for all $x\in J$, $\varphi\in S_n$, $n\geq0$ and
$\gamma \leq\gamma_0$, provided $B_{\delta}(x)\subset\Omega_n$.
\end{lemma}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% SECTION 3
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Mixing rates for rational maps}
\noindent In this section we use the convergence properties of the
transfer operator to deduce mixing properties for rational maps which
will be sufficient to prove the main result, although as one can see
it turns out that $\mu$ is not $\phi$-mixing, since the right hand
side in lemma \ref{measure.mixing} is not independent of $n$ but
depends increases exponentially on $n$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% supremum.mixing
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lemma}\label{supremum.mixing}
Let $\kappa>1$. Then there exists a constant $C_4$ and $\sigma<1$ so that
$$
\left|\mu(A_{\varphi}\cap T^{-k-n}Q)-\mu(A_{\varphi})\mu(Q)\right|
\leq C_4\sigma^k\kappa^n\mu(Q)|g_n\varphi|_{\infty},
$$
for all $\varphi\in S_n$ ($A_{\varphi}=\varphi(J)\in{\cal A}^n$),
$k, n >0$ and $Q$ measurable.
\end{lemma}
\noindent {\bf Proof.} Let $\kappa>1$, $\varphi\in S_n$ and note that
$\hat{\cal L}^n\chi_{A_{\varphi}}=g_n\varphi$. To estimate the
H\"{o}lder norm of $g_n\varphi$ let us use lemma \ref{expansion}
according to which $|\varphi x -\varphi x'|\leq C_3^n|x-x'|^{\xi}$
for all $x, x' \in J$ and some $\xi>0$. Since $\hat{f}$ is $\beta$-H\"{o}lder
continuous we obtain for every $\beta'\in(0,\beta]$ that
\begin{eqnarray}
|g_n(\varphi x)-g_n(\varphi x')|
&=&g_n(\varphi x)\left|1-\frac{g_n(\varphi x)}{g_n(\varphi x)}\right|
\nonumber\\
&\leq&|g_n\varphi|_{\infty}\left(C_3^n|x-x'|^{\xi}\right)^{\beta'}
\nonumber\\
&\leq&|g_n\varphi|_{\infty}C_3^{\beta' n}|x-x'|^{\xi\beta'}.
\nonumber
\end{eqnarray}
Now let $\beta'>0$ be so small that $C_2^{\beta'}=\kappa$ and then
put $\gamma =\xi\beta'$. This gives
$
|g_n\varphi|_{\gamma}\leq|g_n\varphi|_{\infty}\kappa^n
$
and consequently
$$
\|g_n\varphi\|_{\gamma}\leq2\kappa^n|g_n\varphi|_{\infty}.
$$
Since
$$
\mu(A_{\varphi}\cap T^{-k-n}Q)
=\mu\left(\hat{\cal L}^n\left(\chi_{A_{\varphi}}
(\chi_Q\circ T^{k+n})\right)\right)
=\mu\left((\chi_Q\circ T^k) \,(g_n\circ\varphi)\right),
$$
we get using proposition \ref{contraction} (note:
$\mu(A_{\varphi})=\mu(g_n\varphi)$)
\begin{eqnarray}
\left|\mu((\chi_Q\circ T^k)\,(g_n\circ\varphi))-\mu(Q)\mu(A_{\varphi})\right|
&\leq&\mu\left(\chi_Q
\left|\hat{\cal L}^k(g_n\circ\varphi)-\mu(A_{\varphi})\right|\right)
\nonumber\\
&\leq&C_1\mu(Q)\sigma^k\|g_n\varphi\|_{\gamma},\nonumber
\end{eqnarray}
where $C_2$ depends on the H\"{o}lder exponent $\gamma$.
The lemma now follows with $C_4 =2C_2$. \hfill$\Box$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% supremum.mixing2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lemma}\label{supremum.mixing2}
Let $\kappa>1$. Then there exists a constant $C_5$ and $\sigma<1$ so that
$$
\left|\mu(W\cap T^{-k-n}Q)-\mu(W)\mu(Q)\right|
\leq C_5\sigma^k\kappa^n\mu(Q)\mu(W),
$$
where $W=\bigcup_jA_{\varphi_j}$ for finitely many
$\varphi_j\in S'_n$, $k, n >0$ and $Q$ measurable.
\end{lemma}
\noindent {\bf Proof.} The interiors of the $\varphi_j\in S'_n$ are
disjoint and their boundaries have measure zero.
By \cite{DU1} we have $|g_n\varphi|_{\infty}\leq c_1\inf g_n\varphi$ for some
constant $c_1$ and all $\varphi\in S'_n$. Since by assumption
$\varphi_j\in S'_n$ we obtain
$|g_n\varphi_j|_{\infty}\leq c_1\mu(A_{\varphi_j})$ for all $j$, and
therefore by lemma \ref{supremum.mixing}
\begin{eqnarray}
\left|\mu(W\cap T^{-k-n}Q)-\mu(W)\mu(Q)\right|
&\leq&\sum_j C_4\sigma^k\kappa^n\mu(Q)\mu(A_{\varphi_j})\nonumber\\
&\leq& C_5\sigma^k\kappa^n\mu(Q)\mu(W)\nonumber
\end{eqnarray}
where $C_5\leq c_1C_4$. \hfill$\Box$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% measure.mixing
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lemma}\label{measure.mixing}
There exist $\nu>1$ and $\sigma<1$ so that for all
$k, n\in{\bf N}$, $Q$ measurable and finitely many (distinct)
$A_1,\dots,A_{\ell}\in{\cal A}^n$ one has
$$
\left|\mu(W\cap T^{-k-n}Q) - \mu(W)\mu(Q)\right|
\leq C_5\sigma^k\nu^n\mu(W)\mu(Q)
$$
($C_5$ as in lemma \ref{supremum.mixing2}), where $W= \bigcup_{j=0}^{\ell}A_j$.
\end{lemma}
\noindent {\bf Proof.} Let us first prove the lemma in the case when
$W$ consists of a single atom $A_{\varphi}=\varphi(J)$ for
some $\varphi\in S_n$. Observe that there exists a
constant $\nu'>1$ (e.g.\ $\nu'=\exp(\sup \hat{f}-\inf\hat{f})$)
so that for all $n\in{\bf N}$ and $\varphi\in S_n$ one has
$$
|g_n|_{\infty}\leq\nu'^n\inf g_n\varphi,
$$
which in turn implies
$$
|g_n|_{\infty}\leq\nu'^n\inf g_n\varphi\leq\nu'^n\mu(g_n\varphi)
=\nu'^n\mu(A_{\varphi}).
$$
Now apply lemma \ref{supremum.mixing} and put $\nu=\kappa\nu'$.
For the general case we use the fact that the interiors of the
atoms of ${\cal A}^n$ are disjoint and that the boundaries have
zero measure (i.e.\ ${\cal A}^n$ is a partition for $\mu$).
Hence
\begin{eqnarray}
\left|\mu(W\cap T^{-k-n}Q) - \mu(W)\mu(Q)\right|
&\leq& \sum_{j=0}^{\ell}\left|\mu(A_j\cap T^{-k-n}Q)-\mu(A_j)\mu(Q)\right|
\nonumber\\
&\leq& \sum_{j=0}^{\ell}C_5\sigma^k\nu^n\mu(A_j)\mu(Q)\nonumber\\
&\leq&C_5\sigma^k\nu^n\mu(W)\mu(Q).\nonumber
\end{eqnarray}
\hfill$\Box$
\noindent For $r\geq1$ and (large) $N$ denote by $G_r(N)$ the $r$-vectors
$\vec{v}=(v_1,\dots,v_r)$ for which $0\leq v_1