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\topmatter
 
\title
DECAY OF CORRELATIONS
\endtitle
\author
Carlangelo Liverani
\endauthor
\affil University of Rome II
\endaffil
\address
Liverani Carlangelo,
Mathematics Department,
University of Rome II, Tor Vergata,
00133 Rome, Italy.
\endaddress
\email
liverani\@mat.utovrm.it
\endemail
\date
{\sl Dedicated to Micheline Ishay}
\enddate
\abstract
This paper offers a new unified
approach for studying statistical 
properties of hyperbolic dynamical systems and their small 
random perturbations.
The approach is based on a technique that 
permits an investigation of the convergence of a semigroup of operators by
using an adapted Hilbert metric. An exponential rate of decay
of correlations is proven in several classes of examples (including
some hyperbolic billiards).
\endabstract
\thanks
\bf I would like to thank  P. Boyland, N. Chernov, L. Chierchia,
V. Donnay, G. De Martino, C. Gol\`e, 
J.L. Lebowitz, M. Lyubich, M. Rychlik, I.G. Schwarz,
S. Vaienti and especially G. Gallavotti for helpful and enlightening
discussions. In addition, I am indebted to P. Collet and, most of all, 
M. Wojtkowski for introducing me to the magical world of cones.
Finally, I thank J. Milnor, director of the Institute for Mathematical 
Sciences at Stony Brook University, where I was visiting during part 
of this work, and the Italian C.N.R.-GNFM for providing travel funds.
\endthanks
\endtopmatter
 
 
\vskip -1cm
\centerline{\bf CONTENT}
\vskip -.5cm
\roster
\item"0."  Introduction\dotfill p. \ 2
\item"1."  Operators and Invariant Cones\dotfill p. \ 3
\item"2."  One Dimensional Maps (The uniformly hyperbolic case)
	   \dotfill p. \ 6
\item"3."  One Dimensional Maps (Random perturbations)\dotfill p. 12
\item"4."  Two Dimensions (The smooth uniformly hyperbolic case)\dotfill
	   p. 15
\item"5."  Two Dimensions (The non-smooth uniformly hyperbolic case)
	   \dotfill p. 29
\item"6."  Sinai Billiard (finite horizon)\dotfill p. 40
\item"7."  Two Dimensions (Random Perturbations)\dotfill p. 55
\item" "  Appendix I (Sinai Theorem)
	   \dotfill p. 60
\item" "  Appendix II (Absolute continuity)
	   \dotfill p. 60
\item" "   References\dotfill p. 67
\endroster
 
\vfil\par\newpage 
 
\document
 
\subhead \S 0 INTRODUCTION \endsubhead
\vskip1cm
The decay of correlations or, alternatively, the rate of approach of
some initial distribution to an invariant one, is
a problem addressed in many fields. In statistical mechanics such problem
is discussed via the
study of the transfer operator (also called Ruelle-Perron-Frobenius operator)
\cite{Ru}. In the field of stochastic differential equations
it is possible to obtain general results by PDE techniques \cite{H}. For
dynamical systems one can use Markov partitions \cite{Bow}.
\par
The Markov partition is a tool whereby a hyperbolic dynamical system
can be reduced to (or, better, coded into) a statistical mechanical like 
system: the ``symbolic dynamics." 
At the moment, the main limitations of this approach 
come from the
need for uniform hyperbolicity and from the necessity to use infinite 
Markov partitions for
discontinuous systems. These limitations obstruct the extension
of this technique to more general systems, with some notable
exceptions \cite{BSC}, \cite{LSV}.
\par
As already mentioned,
in statistical mechanics the desired results are obtained
by studying the transfer operator. Given that the Perron-Frobenius operator
is a natural object in the context of dynamical systems, it seems 
plausible that some advantages could be gained by investigating 
directly the Perron-Frobenius operator,  without previously 
coding the dynamics. Moreover, adopting such point of view, the investigation 
of dynamical systems would be
more similar to that of random systems, and the study
of random perturbations of dynamical systems could be facilitated. 
\par
Results in the direction of a direct study of the Perron-Frobenius 
operator already exist, but they are limited to one 
dimensional systems; particularly relevant is the investigation 
of one-dimensional quadratic families of
maps and their random perturbations in \cite{Y},\cite{Co}, since 
it is unclear if Markov partitions could be used to study
such examples; and \cite{F}, \cite{FS2}, where one-dimensional expansive maps
are studied using an approach similar to the one proposed here.
\par           
In this paper I describe a technique, originally due to G. Birkhoff 
\cite{B1}, \cite{B2}, that permits a direct study of the
Perron-Frobenius operator, and I show that its field of applicability 
is wider than that of Markov partitions. In essence, it is possible
to construct systematically metrics (Hilbert metrics) with respect to which 
the Perron-Frobenius operator is a contraction. Such contraction allows us
to obtain the invariant measure (if not already known) by an elementary, 
and constructive, fixed point theorem, rather than by some compactness 
argument (this may please some idiosyncratic people, myself included), 
and automatically implies an exponential rate for the decay of correlations.
\par
I illustrate such an approach by applying it to several examples.
For the sake of brevity and clarity the results are not presented 
in their full generality. 
In particular, all the arguments used for two-dimensional
smooth maps can be easily extended to the $n$-dimensional case.
\par
Results concerning more general systems (notably general billiards and 
non-uniformly hyperbolic maps) will be published in separate papers.
I also hope that the present exposition will prompt others
to apply this method to the many cases where it could yield 
new results (e.g. dissipative systems, flows, etc.).
\par
The structure of the paper is as follows:
section 1 describes the Hilbert metric 
and its properties.
It is a brief review of the subject, intended to provide an  easy
reference for the reader. Section 2 shows how the technique works in 
the simplest example: a one dimensional uniformly
hyperbolic map. It also mentions other consequences that can be
obtained (e.g. Central Limit Theorem type results).
Section 3 applies the same technique to the study of random perturbations. 
As a byproduct, it is possible to explicitly estimate the change 
of the invariant measure due to random perturbations. 
In sections 4 I show how to extend the approach to the multidimensional 
case---the smooth case is treated. Section 5 is the most technical section;
in it I show that even the presence of singularities can be accommodated, 
provided they are not too wild. I do not claim to have the weakest 
possible conditions on the singularities; more work in this direction
is needed. This issue is partially explored in section 6. There I extend the
approach to prove exponential decay of correlations in some special 
billiards. The section is the only one in the paper that is not 
self contained
(it relays partially on \cite{BSC}) and cannot be read independently of the
previous sections. Unfortunately, a more detailed investigation of billiards
would require considerably more space.
Nevertheless, the present result settles a question that has been 
open for some time  notwithstanding thorough investigations (e.g., \cite{BSC},
\cite{Ch}). In addition, the numerical simulations 
(see \cite{CCG}, \cite{BD}, \cite{FM}) did not  
provide a clear cut conjecture, as is clearly
shown by recent results \cite{GG}.
Finally, sections 7 deals with random perturbations of 
multidimensional systems, and in particular shows which of 
the results holding in one dimension remain valid.
The paper includes two Appendices. The first presents a standard 
result in the theory of maps with singularities. I do not refer to the
literature because the form in which the result is needed here is non standard
and (misleadingly) appears to be stronger than the usual statement. In the
second Appendix,  I prove a result concerning the absolute continuity
of the stable foliation for discontinuous systems. More precisely, I
show  that the Jacobian of the  
canonical isomorphism between two stable
manifolds is H\"older continuous. This result is not surprising but, to my
knowledge, is not present in the literature.
\par
A final word: the reader should be aware that, 
in order to simplify the formulas, no
attempt has been made to optimize the constants involved. Many times the
value of a constant used in an inequality reflects only my predilection
for the numbers 2 and 3.
 
\vskip1cm
\subhead \S 1 OPERATORS AND INVARIANT CONVEX CONES
\endsubhead
\vskip1cm
 
This section illustrates some results in lattice 
theory originally due to
Garrett Birkhoff. For more details see \cite{B2}, and \cite{N} for a recent
overview of the field.
\par
Consider a topological vector space $\Bbb V$ with a partial 
ordering ``$\preceq$," that is a vector lattice.\footnote{We are assuming the 
partial order to be well behaved with respect to the algebraic structure:
for each $f,\,g\in\Bbb V$ $f\succeq g\Longleftrightarrow f-g\succeq 0$;
for each $f\in\Bbb V$, $\lambda\in\Bbb R^+\backslash \{0\}$ 
$f\succeq 0 \Longrightarrow
\lambda f\succeq 0$; for each $f\in \Bbb V$ $f\succeq 0$ and $f\preceq 0$ imply
$f=0$ (antisymmetry of the order relation).} 
We require the partial order to be ``continuous,"
i.e. given $\{f_n\}\in \Bbb V$ $\lim\limits_{n\to \infty}f_n=f$, if 
$f_n\succeq g$
for each $n$, then $f\succeq g$. We call such vector lattices ``integrally 
closed." \footnote{To be precise, in the literature ``integrally closed"
is used in a weaker sense. First, $\Bbb V$ does not need a topology. Second, 
it suffice that for $\{\alpha_n\}\in\Bbb R$, $\alpha_n\to\alpha$; 
$f,\,g\in\Bbb V$, if $\alpha_n f\succeq g$, then $\alpha f\succeq g$. Here we 
will ignore these and other 
subtleties: our task is limited to a brief account of 
the results relevant to the present context.}
\par
We define the closed convex cone \footnote{Here, by ``cone," we mean any
set such that, if $f$ belongs to the set, then $\lambda f$ belongs to it
as well, for each $\lambda>0$.}
$\Cal C=\{f\in\Bbb V\;|\;f\neq 0,\,f\succeq 0\}$
(hereafter, the term ``closed cone" $\Cal C$ will mean that $\Cal C\cup
\{0\}$ is closed), and the equivalence relation ``$\sim$": 
$f\sim g$ iff there exists $\lambda\in\Bbb R^+\backslash\{0\}$
such that $f=\lambda g$. If we call $\widetilde{\Cal C}$ the quotient of 
$\Cal C$ with respect to $\sim$, then $\widetilde{\Cal C}$ is a closed 
convex set. Conversely, given a closed convex cone 
$\Cal C\subset\Bbb V$, enjoying the property $\Cal C\cap -\Cal C=\emptyset$, 
we can define an order relation by
$$
f\preceq g \iff g-f\in\Cal C \cup \{0\}.
$$
Henceforth, each time that we specify a convex cone we will assume the
corresponding order relation and vice versa. The reader must therefore 
be advised that ``$\preceq$" will mean different things in different contexts.
\par
It is then possible to define a projective metric $\Theta$ 
(Hilbert metric), in $\Cal C$, by the construction:
$$
\aligned
\alpha (f,\,g)=&\sup\{\lambda\in\Bbb R^+\;|\; \lambda f\preceq g\}\\
\beta(f,\,g)=&\inf\{\mu\in\Bbb R^+\;|\;g\preceq \mu f\}\\
\Theta(f,\,g)=&\log\left[\frac{\beta(f,\,g)}{\alpha(f,\,g)}\right]
\endaligned
$$
where we take $\alpha=0$ and $\beta=\infty$ if the corresponding sets are 
empty. 
\par
The importance of the previous 
constructions is due, in our context, to the following theorem.
 
 
\proclaim{Theorem 1.1} Let $\Bbb V_1$, and $\Bbb V_2$ 
be two integrally closed vector lattices;
$T:\Bbb V_1\to \Bbb V_2$ a linear map such that 
$T(\Cal C_1)\subset \Cal C_2$, for two given closed convex 
cones $\Cal C_1\subset\Bbb V_1$ and $\Cal C_2\subset\Bbb V_2$.
Let $\Theta_i$ be the Hilbert metric corresponding to the cone
$\Cal C_i$. Setting 
$\Delta=\sup\limits_{f,\,g\in T(\Cal C_1)}\Theta_2(f,\,g)$
we have
$$
\Theta_2(Tf,\,Tg)
\leq\tanh\left(\frac \Delta 4\right)\Theta_1(f,\,g)
\qquad \forall f,\,g\in\Cal C_1
$$
($\tanh(\infty)=1$).
\endproclaim
\demo{Proof}
The proof is provided for the reader convenience.
\par
Let $f,\,g\in\Cal C_1$, on the one hand if $\alpha=0$ or 
$\beta=\infty$, then the inequality 
is obviously satisfied. On the other hand, if $\alpha\neq 0$ and 
$\beta\neq\infty$, then
$$
\Theta_1(f,\,g)=\ln\frac\beta\alpha
$$
where $\alpha f\preceq g$ and $\beta f\succeq g$, 
since $\Bbb V_1$ is integrally closed. Notice that $\alpha\geq 0$, and
$\beta\geq 0$ since $f\succeq 0$, $g\succeq 0$. If $\Delta=\infty$, then the
result follows from $\alpha Tf\preceq Tg$ and $\beta Tf\succeq Tg$.
If $\Delta<\infty$, then, by hypothesis,
$$
\Theta_2\left(T(g-\alpha f),\,T(\beta f-g)\right)\leq\Delta
$$
which means that there exist $\lambda,\,\mu\geq 0$ such that
$$
\aligned
\lambda T(g-\alpha f)&\preceq T(\beta f-g)\\
\mu T(g-\alpha f)&\succeq T(\beta f-g)
\endaligned
$$
with $\ln\frac\mu\lambda\leq\Delta$.
The previous inequalities imply
$$
\aligned
\frac{\beta+\lambda\alpha}{1+\lambda}T f&\succeq Tg\\
\frac{\mu\alpha+\beta}{1+\mu} T f&\preceq T g .
\endaligned
$$
Accordingly,
$$
\aligned
\Theta_2(Tf,\,Tg)&\leq\ln\frac{(\beta+\lambda\alpha)(1+\mu)}
{(1+\lambda)(\mu\alpha+\beta)}=\ln\frac{\e^{\Theta_1(f,\,g)}+\lambda}
{\e^{\Theta_1(f,\,g)}+\mu}-\ln\frac{1+\lambda}{1+\mu}\\
&=\int_0^{\Theta_1(f,\,g)}\frac{(\mu-\lambda)\e^\xi}{(\e^\xi +\lambda)
(\e^\xi +\mu)} d\xi\leq\Theta_1(f,\,g)\frac{1-\frac\lambda\mu}
{\left(1+\sqrt{\frac\lambda\mu}\right)^2}\\
&\leq \tanh\left(\frac\Delta 4\right)\Theta_1(f,\,g) .
\endaligned
$$
\enddemo
 
\proclaim{Remark 1.2}If $T(\Cal C_1)\subset\Cal C_2$ then it follows that
$\Theta_2(Tf,\,Tg)\leq\Theta_1(f,\,g)$. Hovewer, a uniform rate of 
contraction depends on the diameter of the image being finite.
\endproclaim
\par
In particular, if an operator maps a convex cone strictly inside itself
(in the sense that the diameter of the image is finite), 
then it is a contraction in the Hilbert metric. This implies the
existence of a ``positive" eigenfunction (provided the cone
is complete with respect to the Hilbert metric), and, 
with some additional work,
the existence of a gap in the spectrum of $T$ (see \cite{B1} for
details).
The relevance of this theorem for the study of
invariant measures and their ergodic properties is obvious.
\par
It is natural to wonder about the strength of the Hilbert metric compared
to other, more usual, metrics. While in general
it depends on the cone, it is nevertheless possible to state an interesting
result.
 
\proclaim{Lemma 1.3} Let $\|\cdot\|$ be a norm in $\Bbb V$, and
suppose that, for each $f,\,g\in\Bbb V$, 
$$
-f\preceq g\preceq f \Longrightarrow \|f\|\geq\|g\|.
$$
Then, given $f,\,g\in\Cal C\subset\Bbb V$ for which $\|f\|=\|g\|$,
$$
\|f-g\|\leq\left(\e^{\Theta(f,\,g)}-1\right)\|f\| .
$$
\endproclaim
\demo{Proof}
We know that $\Theta (f,\,g)=\ln\frac\beta\alpha$, where $\alpha f\preceq g$,
$\beta f\succeq g$. This implies that $-g\preceq 0\preceq\alpha f\preceq g$, 
i.e.
$\|g\|\geq\alpha\|f\|$, or $\alpha\leq 1$. In the same manner it follows 
that $\beta\geq 1$. Hence, 
$$
\aligned
g-f\preceq& (\beta-1)f\preceq(\beta-\alpha)f\\
g-f\succeq&(\alpha-1)f\succeq-(\beta-\alpha)f
\endaligned
$$
which implies
$$
\|g-f\|\leq(\beta-\alpha)\|f\|\leq\frac{\beta-\alpha}{\alpha}\|f\|=
\left(\e^{\Theta(f,\,g)}-1\right)\|f\| .
$$
\enddemo
 
Many vector lattices satisfy the hypothesis of Lemma 1.3 (e.g. Banach 
lattices\footnote{A Banach lattice $\Bbb V$ is defined by the property
$\|\; |f|\;\|=\|f\|$ for each 
$f\in\Bbb V$, where $|f|$ is the least 
upper bound of $f$ and $-f$. For this definition to make sense it is 
necessary to require that $\Bbb V$ is ``directed," i.e. any two elements 
have an upper bound.}); nevertheless, we will see that some important 
examples treated in this paper do not.
 
 
\vskip1cm
\subhead \S 2 ONE DIMENSIONAL MAPS (THE UNIFORMLY HYPERBOLIC CASE)
\endsubhead
\vskip1cm
 
In this section we study a simple class of uniformly hyperbolic maps
using the general approach described in
the previous section. The results discussed here are well known \cite{Ru},
but this is an ideal starting point for the task at hand.
\par
We consider a special one dimensional map (see Figure 1). 
Let $I_1,\,I_2$ be two closed intervals
such that $I_1\cup I_2=[0,\,1]$, $I_1\cap I_2=\partial I_1\cap\partial I_2$, 
and $T_i: I_i\to [0,\,1]$, one to one and 
onto, such that  $T_i\in C^{(2)}(I_i)$ and
$DT_i\geq \lambda>1$. We consider the map
\footnote{The present technique applies to a much wider class of maps
(for example any expansive Markov map);
in fact, it would be interesting to see the extent to which it can be pushed.
Nevertheless, this section has a purely propaedeutic and didactic aim; an
attempt to present the results in their full generality would be incompatible
with such purpose.}
$$
T(x)=\left\{\aligned
                  & T_1(x)\qquad\text{if }x\in I_1\\
                  & T_2(x)\qquad\text{if }x\in I_2
            \endaligned\right.
\tag 2.1
$$
 
\topinsert
\vskip3in
\hsize=4.5in
\raggedright
\noindent{\bf Figure 1}  The map.
\endinsert
 
We will see how the results in section 1 can be used in order to show
that the map (2.1) has a unique invariant measure that is 
absolutely continuous with respect to the Lebesgue measure. 
In addition, any absolutely continuous measure,
if iterated by the map, will converge to the invariant one at an 
exponential rate (implying the exponential decay of correlations).
 
Let us see how the map can be lifted to the measures.
Let $f,\,g\in L^2([0,\,1])$
$$
\int_0^1f\circ T(x)g(x)dx=\sum_{i=1}^2\int_0^1f(x)g\circ T_i^{-1}(x)
|D_xT_i^{-1}|dx=\int_0^1f\widetilde Tg
$$
The operator $\widetilde T$, which describe how the density of the measure
$g(x)dx$ is transformed under the iteration of the map $T$, is called
Perron-Frobenius operator \footnote{Other definitions
of the Perron-Frobenius operator can be found in the literature, 
but the differences with the
one adopted here are immaterial.}.
\par
The result will follow from the study of the semigroup generated by the
operator $\widetilde T$. Such a semigroup maps a cone of functions 
strictly inside itself,
and therefore it has a fixed point that corresponds to the density of the 
unique absolutely continuous invariant measure.
\par
We define the following cone \footnote{This choice is made in order to 
simplify the exposition. Essentially, we require the logarithm of the 
functions in the cone to be Lipschitz, with a Lipschitz constant less than
$a$. We could have imposed a condition on the H\"older continuity or the
variation of the functions in the cone; this would have produced an estimate
for the decay of correlations that is valid for a wider class of functions, 
and allowed us to treat maps that are $C^{(\alpha)}$, $\alpha>1$, 
but not $C^{(2)}$.}
$$
\Cal C_a=\left\{g\in C^{(0)}\;\big|\;\forall x,\,y\in[0,\,1]\;\; 
g(x)> 0;\;\frac{g(x)}{g(y)}\leq\e^{a|x-y|}\right\}.
$$
 
\proclaim{Lemma 2.1}Given $\sigma\in(\lambda^{-1},\,1]$, 
$\widetilde T\Cal C_a\subset\Cal C_{\sigma a}$ provided
$a\geq \frac D{\sigma-\lambda^{-1}}$, where
$$
D=\sup\Sb x\in[0,\,1]\\i\in\{1,\,2\}\endSb
\left|\frac{D^2_xT^{-1}_i}{D_xT^{-1}_i}\right|.
$$
\endproclaim
\demo{Proof}
Let $g\in\Cal C$, $x\leq y\in[0,\,1]$
$$
\aligned
\widetilde Tg(x)&\leq \sum_{i=1}^2g(T^{-1}_i(y))|D_x T^{-1}_i|
\e^{a|T^{-1}_i(x)-T^{-1}_i(y)|}\\
&\leq\sum_{i=1}^2g\circ T^{-1}_i(y)|D_y T^{-1}_i|
\e^{a|T^{-1}_i(x)-T^{-1}_i(y)|+\left|\ln(|D_xT^{-1}_i|)-
\ln(|D_yT^{-1}_i|)\right|}\\
&\leq\sum_{i=1}^2g\circ T^{-1}_i(y)|D_y T^{-1}_i|
\e^{a\int_x^y|D_\xi T_i^{-1}|d\xi+\int_x^y
\left|\frac{D^2_\xi T^{-1}_i}{D_\xi T^{-1}_i}\right|d\xi}\\
&\leq \wt Tg(y)\e^{a\lambda^{-1}|x-y|+D|x-y|}\leq\wt Tg(y)\e^{\sigma 
a|x-y|}.
\endaligned
$$
\enddemo
 
In order to see if we can apply Theorem 1.1, we have to derive
the Hilbert metric associated with our cone.
 
\proclaim{Lemma 2.2} Let $\Theta$ be the Hilbert metric associated to 
$\Cal C_a$. For each $f,\,g\in\Cal C_a$
$$
\Theta(f,\,g)=\ln\sup\Sb x,\,y\in[0,\,1]\\ u,\,v\in[0,\,1]\endSb
\frac{\left(\e^{a|x-y|}g(y)-g(x)\right)\left(\e^{a|u-v|}f(v)-f(u)\right)}
{\left(\e^{a|x-y|}f(y)-f(x)\right)\left(\e^{a|u-v|}g(v)-g(u)\right)}
$$
\endproclaim
\demo{Proof}
We need to compute $\alpha$ and $\beta$.
>From the definitions, it follows that $\alpha f\preceq g$ iff
$$
g(x)-\alpha f(x)\geq 0\qquad \forall x\in[0,\,1]
$$
and
$$
g(x)-\alpha f(x)\leq \e^{a|x-y|}\left(g(y)-\alpha f(y)\right)
\ \forall x,\,y\in[0,\,1].
$$
 
Accordingly:
$$
\alpha=\min\left\{\inf_{x\in[0,\,1]}\frac{g(x)}{f(x)};\;
\inf\Sb x,\,y\in[0,\,1]\\ x\neq y\endSb\frac{\e^{a|x-y|}g(y)-g(x)}
{\e^{a|x-y|}f(y)-f(x)}\right\}.
$$
However, choosing $x_0\in [0,\,1]$ such that
$$
\frac{g(x_0)}{f(x_0)}=\inf_{x\in [0,\,1]}
\frac{g(x)}{f(x)}
$$
then, for each $x\neq x_0$,
$$
\frac{\e^{a|x_0-x|}g(x_0)-g(x)}{\e^{a|x_0-y|}f(x_0)-f(x)}=
\frac{\e^{a|x_0-x|}\frac{g(x_0)}{f(x_0)}f(x_0)-\frac{g(x)}{f(x)}f(x)}
{\e^{a|x_0-y|}f(x_0)-f(x)}\leq\frac{g(x)}{f(x)}.
$$
 
This means that
$$
\alpha=\inf\Sb x,\,y\in[0,\,1]\\ x\neq y\endSb\frac{\e^{a|x-y|}g(y)-g(x)}
{\e^{a|x-y|}f(y)-f(x)}.
$$
Analogously,
$$
\beta=\sup\Sb x,\,y\in[0,\,1]\\ x\neq y\endSb\frac{\e^{a|x-y|}g(y)-g(x)}
{\e^{a|x-y|}f(y)-f(x)}.
$$
\enddemo
 
 
\proclaim{Lemma 2.3}
$$
\Delta=\text{diam}(\Cal C_{\sigma a})=2\ln\frac{1+\sigma}{1-\sigma}+
2\sigma a .
$$
\endproclaim
\demo{Proof}
For each $f,\,g\in\Cal C_{\sigma a}$ we have
$$
\aligned
\Theta(f,\,g)&=\ln\sup\Sb x,\,y\in[0,\,1]\\ u,\,v\in[0,\,1]\endSb
\frac{\left(\e^{a|x-y|}-\e^{-\sigma a|x-y|}\right)
\left(\e^{a|u-v|}-\e^{-\sigma a|u-v|}\right)g(y)f(v)}
{\left(\e^{a|x-y|}-\e^{\sigma a |x-y|}\right)
\left(\e^{a|u-v|}-\e^{\sigma a|u-v|}\right)f(y)g(v)}\\
&\leq\ln\frac{(1+\sigma)^2}{(1-\sigma)^2}\e^{2\sigma a}.
\endaligned
$$
\enddemo
 
Lemmas 2.1--2.3, together with Theorem 1.1, imply that $\forall f,\,g\in
\Cal C_a$
$$
\Theta (\wt Tf,\,\wt Tg)\leq\Lambda \Theta(f,\,g)
$$
with
$$
\Lambda=\tanh\left(\frac{\Delta}4\right).
$$
 
\proclaim{Remark 2.4}In order to have $\Delta$ as small as possible
it is convenient to choose $a=\frac D{\sigma-\lambda^{-1}}$. 
It is then possible to choose the $\sigma=\sigma_*$ that minimizes
$\Delta$. Hence, the contraction rate $\Lambda$ depends only on $\lambda$
and $D$.
\endproclaim
 
\par
Note that, for each $g_1,\,g_2\in C^{(0)}([0,\,1])$, 
$-g_1\preceq g_2\preceq g_1$
implies $-g_1(x)\leq g_2(x)\leq g_1(x)$, 
that is $|g_2(x)|\leq g_1(x)$, for each
$x\in[0,\,1]$. Consequently, for each $L^p$ norm, $\|g_2\|_p\leq \|g_1\|_p$,
and Lemma 1.3 applies.
\par
Let $g\in C^{(0)}([0,\,1])$, $\int_0^1 g=1$, and $g\in\Cal C_{a_*}$ with
$a_*=\frac D{\sigma_*-\lambda^{-1}}$. Since $\int_0^1\wt T^n g=\int_0^1
g=1$,
$$
\|\wt T^{n+m}g-\wt T^n g\|_1\leq\e^{\Theta(\wt T^{n+m}g,\,\wt T^n g)}
-1\leq\e^{\Lambda^{n-1}\Theta(\wt T(\wt T^mg),\,\wt T g)}-1
\leq \e^{\Delta\Lambda^{-1}}\Delta\Lambda^{n-1}.
$$
This means that $\{\wt T^n g\}$ is a Cauchy sequence in $L^1$; in 
addition, $\{\wt T^n g\}$ are equicontinuous. Let
$\varphi_*=\lim\limits_{n\to\infty}\wt T^ng$, clearly $\wt T \varphi_*
=\varphi_*$, and
$\varphi_*\in\Cal C_{\sigma_* a_*}$, and $\varphi_*$ is not dependent on 
$g$; i.e., $\varphi_*$ is the density of the unique invariant measure 
absolutely continuous with respect to Lebesgue.
\par
Up to now, we have shown the existence of the invariant measure, obtained
some of its regularity properties, and been 
able to estimate the decay
of correlations for all functions $f\in L^\infty([0,\,1])$, $g\in
\Cal C_{a_*}$. The next Theorem gives a slightly stronger result, and
shows how to avoid Lemma 1.3.
 
\proclaim{Theorem 2.5} There exist $K,\,r\in\Bbb R^+$ such that,
for each  $f\in L^1([0,\,1])$, $g\in C^{(1)}([0,\,1])$ with $\int_0^1g=1$
$$
\left|\int_0^1 f\circ T^n gdx-\int_0^1f\varphi_*dx\right|
\leq K\|f\|_1(\|g\|_1+r\|g'\|_\infty)\Lambda^n
$$
\endproclaim
\demo{Proof}
Consider $g\in\Cal C_{a_*}$, normalized so that $\int_0^1g=1$ (i.e.,
$g$ can be thought as the density of a measure). Then,
$$
\left|\int_0^1 f(\wt T^n g-\varphi_*)dx\right| \leq\|f\|_1
\left\|\frac{\wt T^ng}{\varphi_*}-1\right\|_\infty\|\varphi_*\|_\infty .
$$
\proclaim{Lemma 2.6}If $\Cal C_+=\{g\in C^{(0)}([0,\,1])\;|\;
g(x)\geq 0,\;\;\forall x\in [0,\,1]\}$, and $\Theta_+$ is the corresponding 
Hilbert metric, then, for each $g_1,\,g_2\in\Cal C_a$,
$$
\Theta_+(g_1,\,g_2)\leq \Theta(g_1,\,g_2)
$$
\endproclaim
\demo{Proof}
Since $\Cal C_+\supset\Cal C_a$, the identity is a map from 
$C^{(0)}([0,\,1])$ to itself that maps $\Cal C_a$ into $\Cal C_+$.
The result follows then from Theorem 1.1.
\enddemo
A simple computation yields
$$
\Theta_+(g_1,\,g_2)=\ln\sup_{x,\,y\in[0,\,1]}\frac{g_1(x)g_2(y)}
{g_1(y)g_2(x)}.
$$
Using the previous facts, and the trivial equality
$$
\frac{(\wt T^n g)(x)}{\varphi_*(x)}=\frac{\wt T^n g(x)\varphi_*(y)}
{\wt T^ng(y)\varphi_*(x)}
\frac{\wt T^n g(y)}{\varphi_*(y)},
$$
we have
$$
\e^{-\Theta_+(\wt T^ng,\,\varphi_*)} \frac{\wt T^ng(y)}{\varphi_*(y)}\leq
\frac{\wt T^ng(x)}{\varphi_*(x)}\leq
\e^{\Theta_+(\wt T^ng,\,\varphi_*)} \frac{\wt T^ng(y)}{\varphi_*(y)}
$$
for each $x,\,y\in[0,\,1]$.
Because $\int_0^1(\wt T^n g-\varphi_*)=0$, and remembering that
both functions are continuous, there exists $y_n\in[0,\,1]$:
$\wt T^n g(y_n)=\varphi_*(y_n)$.
Using the previous inequalities, with $y=y_n$, we obtain, 
for each $x\in[0,\,1]$,
$$
\e^{-\Theta_+(\wt T^ng,\,\varphi_*)} \leq\frac{\wt T^n g(x)}{\varphi_*(x)}\leq
\e^{\Theta_+(\wt T^ng,\,\varphi_*)}
$$
and
$$
\left\|\frac{\wt T^n g}{\varphi_*}-1\right\|_\infty
\leq\e^{\Theta_+(\wt T^ng,\,\varphi_*)} -1
\leq \e^{\Lambda^{n-1}\Theta(\wt T g,\,\varphi_*)}-1
\leq \e^{\Lambda^{-1}\Delta}\Delta\Lambda^{n-1} .
$$
This estimate shows that for each $f\in L^1([0,\,1])$, $g\in\Cal C_{a_*}$
$$
\aligned
&\left|\int_0^1 f\circ T^n gdx-\int_0^1f\varphi_*dx\int_0^1gdx\right|
\leq K\|f\|_1\|g\|_1\Lambda^n\\
& K=\e^{\Lambda^{-1}\Delta}\Delta\Lambda^{-1}\|\varphi_*\|_\infty .
\endaligned
$$
\par
Let us now consider $g\in C^{(0)}([0,\,1])$, $g\geq 0$, and piecewise
differentiable with $\|g'\|_\infty<\infty$.
If we define $g_\rho=g+\rho \varphi_*$, 
$\rho\in\Bbb R^+$, for each $x>y$ we have
$$
\aligned
g_\rho(x)&\leq\e^{\ln[g(x)+e^{\sigma_* a_*(x-y)}\rho \varphi_*(y)]-
\ln[g(y)+\rho \varphi_*(y)]}g_\rho(y)\\
&\leq\e^{\int_y^x\frac{|g'(\xi)+\rho\sigma_* a_*
e^{\sigma_* a_*(\xi-y)}\varphi_*(y)|}
{g(\xi)+\rho e^{\sigma_* a_*(\xi-y)}\varphi_*(y)}d\xi} g_\rho(y)
\leq \e^{\left[\frac{\|g'\|_\infty}{\varphi_*(y)}\rho^{-1}
+\sigma_* a_*\right](x-y)}g_\rho(y) .
\endaligned
$$
Choosing,
$$
\rho=\sup_{x\in[0,\,1]} \frac{\|g'\|_\infty}{\varphi_*(x)a_*(1-\sigma_*)}
$$
we have $g_\rho\in\Cal C_{a_*}$; hence, the decay of correlations for $g_\rho$ 
implies the decay of correlations for $g$.
The result for arbitrary $g\in C^{(1)}([0,\,1])$
follows since any smooth function can be written as the difference
of two piecewise differentiable positive functions.
\enddemo
 
Theorem 2.5 immediately yields
\proclaim{Corollary 2.7 (Mixing)} For each $f\in L^\infty([0,\,1])$,
$g\in L^1([0,\,1])$, $\int_0^1 g=1$, we have
$$
\lim_{n\to\infty}\left|\int_0^1\left(f\circ T^n\right) g
-\int_0^1 f \varphi_*\right|=0 .
$$
\endproclaim
\par
Another corollary of Theorem 2.5 is the so called Central Limit Theorem
\cite{Ke}.
 
\proclaim{Corollary 2.8 (CLT)} For each $f\in C^1([0,\,1])$,
$\int_0^1 f \varphi_*=0$, the distribution of the random variable
$$
\frac 1{\sqrt n}\sum_{i=1}^n f\circ T^n
$$
converges to a Gaussian law.
\endproclaim
 
 
\vskip1cm
\subhead\S 3 ONE DIMENSIONAL MAPS (RANDOM PERTURBATIONS)
\endsubhead
\vskip1cm
 
In this section we will investigate random perturbations of the example
discussed in the previous section. Again, the argument presented holds
for more general examples (at least for random perturbations of Markov maps),
but I will ignore this, since my primary goal is to present the method.
\par
We will consider two types of questions: first, establishing
results about the invariant measure and the rate of mixing; second,
estimating the difference between the invariant measure in the deterministic
case and the invariant measure in the stochastic case.
\par
Let $\phi:[0,\,1]\times\Bbb R^+\to \Bbb R^+$ be a continuous function such that
$\int_0^\infty\phi(x,\,\xi)d\xi=1$ for each $x\in[0,\,1]$, $\phi(x,\,\xi)\geq
\varphi(\xi)>0$ for each $x\in[0,\,1]$, $\varphi$ monotone decreasing,
and $\phi_+=\sup\phi<\infty$.
Letting $z_\varepsilon(x)=\varepsilon^{-1}\int_0^1\phi(x,\,
\varepsilon^{-1}|x-\xi|)d\xi$, we
define the following Markov process
$$
\wh T_\varepsilon f(x)=\frac 1{\varepsilon z_{\varepsilon}(Tx)} 
\int_0^1\phi(Tx,\,\varepsilon^{-1}|Tx-y|)f(y) dy.
$$
The corresponding P-F operator is given by
$$
\wt T_\varepsilon g(x)=\varepsilon^{-1}  
\int_0^1 z_{\varepsilon}(Ty)^{-1}\phi(Ty,\,\varepsilon^{-1}|Ty-x|)g(y) dy.
$$
We want to study the operator $\wt T_\varepsilon$.
\par
>From a purely probabilistic point of view, it makes sense to consider
the cone $\Cal C_+$ (see Lemma 2.6). Clearly, $\wt T_\varepsilon \Cal C_+ 
\subset \Cal C_+ $. In addition,
$$
\wt T_\varepsilon g(x)\geq \varphi(\varepsilon^{-1})
\|g\|_1,
$$
and, by choosing $\varepsilon$ so small that $\varepsilon
z_{\varepsilon}(x)\geq\frac 12$
for each $x\in[0,\,1]$,
$$
\wt T_\varepsilon g(x)\leq 2\phi_+\|g\|_1 .
$$
These estimates imply that the diameter $\Delta$ of 
$\wt T_\varepsilon \Cal C_+ $ is bounded by
$$
\Delta \leq 2\ln\frac{\|g\|_+}{\|g\|_-}\leq 2\ln\frac{\phi_+}
{\varphi(\varepsilon^{-1})}
$$
so we have a unique smooth invariant measure, which is mixing. 
\par
The previous proof is extremely simple, especially because we have
completely obliterated the dynamics. But we paid a heavy price: we
have very little information about the invariant measure, and the estimate
for the decay of correlations has an unreasonably bad dependence upon 
$\varepsilon$.
\par
If we want better results we have to take into account explicitly
the dynamics of the underlying deterministic system. We will do so in a
slightly different model.
\par
First we will identify $[0,\,1]$ with the unit circle $S^1$. This is done
to eliminate the boundary effects. In fact, the presence of the boundary
tends to change considerably the derivative of the density  
of the invariant measure in a
small neighborhood of the boundary, although it does not change substantially
its $C^{(0)}$ norm. This would force us to consider a more complex cone 
taking such behavior into account (of course, any cone that works in
our approach must contain the invariant density), making our life harder, 
without substantially adding to the argument. From now on, $|\cdot |$ will
mean the distance in $S^1$; nevertheless we will continue writing the set 
as $[0,\,1]$.
\par
Let us require $\phi(x,\,\xi)\geq 0$, $\phi(x,\,\xi)=0$ for each $\xi\geq 1$,
$x\in[0,\,1]$ (this hypothesis is not essential, but we use it to
simplify the argument), and
$$
\frac{\phi(x,\,\xi)}{\phi(y,\,\xi)}\leq\e^{\theta|x-y|}
$$
(a sort of weak translation invariance).
\par
This time we will use the cone $\Cal C_a$ (see beginning of \S 2).
For $x\in S^1$,
$$
\aligned
\wt T_\varepsilon g(x)&=\varepsilon^{-1}  
\int_0^1 \phi(Ts,\,\varepsilon^{-1}|Ts-x|)g(s) ds\\
&=\sum_{i=1}^2\int_0^1 \phi(s,\,\varepsilon^{-1}|s-x|)g(T_i^{-1}s)
|D_s T_i^{-1}| ds\\
&=\int_0^1 \phi(s,\,\varepsilon^{-1}|s-x|)(\wt Tg)(s)ds  .
\endaligned
$$
It is important to notice that, since we are now on a circle, we can assume,
without loss of generality, that $x$ is far away from $0$ or $1$ (which, on
the circle, are the same arbitrary point). From the previous section we
know that, for $a$ sufficiently big, there exist $\sigma<1$, such that
$\wt T g\in\Cal C_{\sigma a}$. Consequently, for each $y\in[0,\,1]$ such that
$|x-y|\ll 1$, we have
$$
\aligned
\wt T_\varepsilon g(x)&=\int_0^1 \phi(s,\,\varepsilon^{-1}|(s-x+y)-y|)  
(\wt Tg)(s)ds\\ 
&\leq \int_0^1 \phi(s-y+x,\,\varepsilon^{-1}|(s-y|)  
(\wt Tg)(s-y+x)ds\leq  \e^{(\sigma a+\theta)|x-y|}\wt T_\varepsilon g(y).
\endaligned
$$
This implies that, for $a$ large enough, there exists $\sigma' <1$ such that
$\wt T_\varepsilon\Cal C_a\subset\Cal C_{\sigma' a}$, which suffices to obtain
a unique smooth invariant measure and an exponential rate for the decay
of correlations.  Let us call $\varphi_*^\varepsilon$ the invariant measure 
associated to $\wt T_\varepsilon$, and let $\varphi_*^0=\varphi_*$ 
be the invariant measure 
associated with the deterministic map). The proof of the next theorem
is based on an argument in \cite{Ry}.
 
\proclaim{Theorem 3.1}For each stochastic process defined by a function
$\phi$ with the above mentioned properties, there exists a constant $K_0\in
\Bbb R^+$ such that
$$
\|\varphi_*-\varphi_*^\varepsilon\|_\infty\leq K_0
\varepsilon\ln \varepsilon^{-1}.
$$
\endproclaim
\demo{Proof}
We start by noticing that, for each function $g\in\Cal C_a$,
$$
\left|\wt T_\varepsilon g(x)-\wt Tg(x)\right|\leq (e^{a\ve}-1) 
\wt T_\varepsilon g(x) .
$$
Let $u(x)=(\wt T \varphi_*^\varepsilon)(x)-\varphi_*^\varepsilon (x)$.
On the one hand,
$$
\|u\|_\infty=
\|\wt T\varphi_*^\varepsilon-\wt T_\varepsilon 
\varphi_*^\varepsilon\|_\infty \leq
(\e^{a\varepsilon}-1)\|\varphi_*^\varepsilon\|_\infty \leq \e^{2a}\varepsilon,
$$
and
$$
\|u'\|_\infty\leq 2\e^{2a}.
$$
On the other hand
$$
(\Id-\wt T)(\varphi_* - \varphi_*^\varepsilon)= 
\wt T \varphi_*^\varepsilon-\varphi_*^\ve=u .
$$
Since, $\int_0^1 u=0$ Theorem 2.5 implies that 
$$
\|\wt T^n \left( \wt T u\right)\|_\infty\leq K
\Lambda^n(\|u\|_1+r\|u'\|_\infty) .
$$
Hence, we can write
$$
\varphi_*^\varepsilon-\varphi_* =\sum_{n=0}^\infty \wt T^n u.
$$
>From the latter equality we have
$$
\|\varphi_*^\varepsilon-\varphi_*\|_\infty\leq N 
\e^{2a}\varepsilon+\sum_{n=N}^\infty
\Lambda^n(\varepsilon+2r)\e^{2a}\leq N \e^{2a}\varepsilon+ 
(1-\Lambda)^{-1}\Lambda^N(\varepsilon+2r)\e^{2a}
$$
which yields the result by choosing $\Lambda^N= \varepsilon $.
\enddemo
 
As a consequence of the above results we can reach some conclusions
about the use of computers to compute Lyapunov exponents. It is commonly 
thought that the simulation of a dynamical system by a computer introduces
some randomness into the system due to roundoff errors.  This is obviously
false from a rigorous point of view (since the algorithms run by computers
are deterministic), but it is likely to be a reasonable approximation.
\par
If we consider such a model, then two natural questions are: how good 
are the
computed results, and which length of trajectories is it reasonable to
consider in the computations. 
\par
Let $\Omega=[0,\,1]^{\Bbb N}$ be the space of the trajectories of 
the random system, and
$\Bbb E$ the expectation induced by the measure;
what we can evaluate by using a computer is
$$
\lambda_n(\omega)=\frac 1n\sum_{i=0}^n\ln |D_{(\sigma^i\omega)_0}T|
$$
where $\omega\in\Omega$ is the trajectory, and 
$\sigma(\omega)_i=\omega_{i+1}$. By ergodicity,
$$
\lambda_\varepsilon 
=\lim_{n\to\infty}\lambda_n(\omega)=\int_0^1\ln|D_zT| 
\varphi_*^\varepsilon(z)dz .
$$
If $\lambda$ is the Lyapunov exponent of the deterministic system, then
$$
|\lambda- \lambda_\varepsilon|\leq K_0 \varepsilon
\ln \varepsilon^{-1}     .
$$
In addition, by the CLT,
$$
\Bbb E\left(|\lambda_n(\omega)-\lambda_\varepsilon|\right)\leq 
\frac  {\text{constant}}{\sqrt n}.
$$
Accordingly, if $\varepsilon$ is the precision with which we are carrying
out the computations, it is unreasonable to expect a precision better
than $ \varepsilon  \ln \varepsilon^{-1}$, or to hope for an 
improvement of the result by looking at trajectories
longer than  $(\varepsilon  \ln \varepsilon^{-1})^{-2}$.
 
\vskip1cm
\subhead \S 4 TWO DIMENSIONS (THE SMOOTH UNIFORMLY HYPERBOLIC CASE)
\endsubhead
\vskip1cm
 
In the rest of the paper we will see how the approach presented in section 
two can be implemented in more general situations.
\par
Clearly the first possible generalization is the multidimensional case.
For simplicity, we will confine ourselves 
to the two dimensional area--preserving
case. Hence, this section is dedicated to symplectic smooth uniformly
hyperbolic two dimensional maps.
\par
The symplectic nature of the maps implies the existence of a special
invariant measure (i.e. the symplectic volume), which simplifies
our approach. 
The higher dimensional conservative case can be treated in the same fashion.
On the contrary, in the case of dissipative systems the present technique
should still yield results, but more care would be needed 
in its application. 
\par
The main idea is that in one direction the P-F operator acts as 
in the one dimensional case, i.e. it makes functions smoother, but
in the other
direction the P-F operator increases oscillations. To overcome 
this phenomenon we integrate the functions on short curves in the latter 
direction. Under the action of the P-F operator, this is equivalent to 
averaging the functions on longer and longer curves. Hence, also in the 
second direction a regularization is taking place.\footnote{The necessity
to take averages comes from the fact that, in the multidimensional case,
measures converge to the invariant one only weakly, 
contrary to what we have seen in the one dimensional case.}
\par
The previous argument will be made precise by Lemma 4.2. The 
Hilbert metric used in this section is weaker than any $L^p$ norm
(in particular Lemma 1.3 does not hold); nonetheless, according to Theorem 
4.9, it suffices to control the decay of correlation
for H\"older continuous functions.
\par
We will start by describing the class of maps under consideration and by
recalling some standard constructions. We will then introduce the cones of 
functions used in the proof.
 
\vskip.5cm
\line{\bf The map.\hfil}
\vskip.5cm
\par
Let $\Cal M$ be a compact smooth symplectic manifold and
$T:\Cal M\to\Cal M$ a symplectomorphysm. Let there exist  a bundle 
of cones $\{\Cal C(x)\}_{x\in\Cal M}$ in the tangent bundle (i.e.,
$\Cal C(x)\subset\Cal T_x\Cal M$) strictly invariant (i.e.,
$D_xT^{-1}\Cal C(x)\subset\hbox{int }\Cal C(T^{-1}x)\cup\{0\}$) and continuous
(i.e., the ``edges" of the cones vary with continuity). Then the map $T$ is
uniformly hyperbolic and its stable and unstable foliations are continuous
(conversely, for such maps, it is easy to construct a family of 
continuous strictly invariant cones with the required properties; for more
details see \cite{LW}).
\par
It follows that there exists a smooth metric on $\Cal M$ such that,
calling $v^s(x)$, $v^u(x)$ the stable and unstable normalized vectors at $x$
$$
\aligned
\| D_xTv^s(x)\|&\leq \lambda^{-1}\\
\| D_xTv^u(x)\|&\geq \lambda
\endaligned
$$
for some $\lambda>1$. Furthermore, the Riemannian volume is equal to the 
symplectic one. We will, from now on, use this metric. In addition, we 
will suppose that 
$$
\aligned
\inf_{v\in\Cal C(x)}\|D_xT^{-1}v\|&\geq\lambda\\
\sup \Sb v,\,w\in\Cal C(x)\\ \|v\|=\|w\|=1\endSb
\| v-w\|&\leq\frac 12         .
\endaligned
$$
(This can always be achieved, eventually by choosing a smaller $\lambda$
and by replacing $\Cal C(x)$ by $D_{T^{n}x}
T^{-n}\Cal C(T^nx)$ for a large enough $n$).
\par
Further, we assume that the map enjoys the mixing 
property.\footnote{The previous assumptions
imply that the map is locally ergodic 
\cite{BG},\cite{LW}, \cite{KB}, and mixing on each ergodic component 
\cite{GO}, \cite{P}, but several ergodic components could still be present.}
In the present case the Perron-Frobenius operator 
$\widetilde T: L^1(\Cal M)\to L^1(\Cal M)$ is given by
$$
\widetilde Tg=g\circ T^{-1}.
$$
\vskip.5cm
\line{\bf Cones of functions.\hfil}
\vskip.5cm
\par
Due to the presence of both contracting and expanding directions the 
approach of the previous sections must be slightly modified.
To overcome this difficulty we will, in some sense, quotient away
(or, better yet, ``integrate away") the
unwanted direction through the introduction of suitable test functions.
This is achieved by introducing a class of closed, connected 
and differentiable curves
$$
\Gamma_\delta=\{\gamma\subset \Cal M\;|\; 
\delta\leq \hbox{length}(\gamma)\leq 2\delta;\;\forall x\in\gamma,\, 
\gamma'(x)\subset\Cal C(x);\;|\kappa_\gamma(x)|\leq H\}
$$
where $\gamma'(x)$ is the unitary tangent vector to $\gamma$ at $x$,
$\kappa_\gamma(x)$ is the curvature of $\gamma$ at $x$,
and $\delta\in(0,\,1)$.\footnote{To fix a length scale, let us assume
diam$(\M)=1$.}
On each of these curves, the measure $m_\gamma$, 
induced by the Riemannian structure (arc-length), is naturally defined.
The plan is to perform integrals on $\gamma$, with respect to a large class
of measures. We proceed to characterize 
such class by specifying the properties
of their densities with respect to $m_\gamma$.
\par
Let us define the cones of ``density" functions. Given any connected
smooth curve $\gamma$, let
$$
\widetilde{\Cal D}_a(\gamma)=\left\{f\in C^{(0)}(\gamma)\;\big|\;f> 0;\;
\frac{f(x)}{f(y)}\leq\e^{a d(x,\,y)^\nu}\right\}
$$
where $d$  means the distance along $\gamma$, and $\nu\in(0,\,\frac 12]$. 
In other words, we are considering positive 
functions whose logarithm has a bounded modulus of $\nu$-H\"older continuity.
\par
The reader will certainly notice the similarity with the cones used in the 
one dimensional case. Let us remember that
$$
\rho_\gamma(f_1,\,f_2)=
\ln\left[\sup_{x,\,y,\,u,\,v\in\gamma}\frac{\left[\e^{ad(x,\,y)^\nu}
f_1(x)-f_1(y)\right]\left[\e^{ad(u,\,v)^\nu}f_2(u)-f_2(v)\right]}
{\left[\e^{ad(x,\,y)^\nu}f_2(x)-f_2(y)\right]\left[\e^{ad(u,\,v)^\nu}
f_1(u)-f_1(v)\right]}\right]
$$
is the Hilbert metric in $\widetilde{\Cal D}_a(\gamma)$.
\par
The set of functions we will consider is
$$
\Cal D_a(\gamma)=\{f\in\widetilde{\Cal D}_a(\gamma)\;|\;\rho_\gamma(1,\,f)
\leq R\}.
$$
\par
For later purposes we also define
$$
\Cal D_+(\gamma)=\{f\in C^{(0)}(\gamma)\;|\; f\geq 0\},
$$
together with its Hilbert metric
$$
\rho_+(f_1,\,f_2)=\ln\left[\sup_{x,\,y\in\gamma}\frac{f_1(x)f_2(y)}
{f_1(y)f_2(x)}\right] .
$$
Of course, by Theorem 1.1, $\rho_+\leq\rho_\gamma$ (since the inclusion
of $\widetilde{\Cal D}_a(\gamma)$ in $\Cal D_+(\gamma)$ is a contraction).
\par
It is now possible to introduce the cone of functions used in the proof.
We introduce two fundamental quantities relative to a given function 
$g\in C^{(0)}(\Cal M)$
$$
\aligned
\tb g \tb_+=&\sup\Sb \gamma\in\Gamma_\delta\\ f\in\Cal D_a(\gamma)\endSb
\frac {\int_{\gamma}fg}{\int_{\gamma}f}\\
\tb g \tb_-=&\inf\Sb\gamma\in\Gamma_\delta\\ f\in\Cal 
D_a(\gamma)\endSb
\frac {\int_{\gamma}fg}{\int_\gamma f}  .
\endaligned
$$
(Here, as in the following, the integration is done 
with respect to the measure $m_\gamma$).
\par
The cone of functions is defined as follows:
$$
\aligned
\Cal C_{b,\,c}=&\bigg\{g\in C^{(0)}(\Cal M)\;\bigg|\;
\forall\gamma\in\Gamma_\delta,\;
\forall x\in\gamma,\, \forall f,\,f_1,\,f_2\in\Cal D_a(\gamma)
\,\,\int_\gamma gf>0;\\
&\frac {f_2(x)\int_{\gamma}gf_1}
{f_1(x)\int_\gamma gf_2}\leq \e^{b\rho_\gamma(f_1,\,f_2)};\;
\frac{\| D^u g\|_\infty}{\tb g\tb_-}\leq c
\bigg\}
\endaligned
$$
where, by $D^u$ we mean the derivative in the unstable direction.
We require that $D^ug$ is well defined, apart from, at most, countably many  
points, on each unstable
manifold (the reader may feel free to think that $g$ 
is everywhere differentiable:
we will use our weaker assumption only at the very end of Theorem 4.9, and
not in an essential way).
 
The choice of piecewise smooth functions, in the unstable direction, 
is done only to simplify the exposition.
More general functions can easily be considered by defining larger cones.
\footnote{If $\gamma$ and $\widetilde\gamma$ are two curves in 
$\Gamma_\delta$, obtained one from the other by ``translation" along the 
unstable direction and $\phi$ is the canonical isomorphism from $\widetilde 
\gamma$ to $\gamma$ along the unstable direction,
one can replace the last condition in $\Cal C_{b,c}$ by
$$
\frac{\int_\gamma gf}{\int_{\widetilde \gamma}g 
f\circ\phi}\leq\e^{c\tau^\sigma}
$$
for some $\sigma\in(0,\,1)$ and
where $\tau$ is the distance between $\gamma$ and $\widetilde \gamma$.}
As anticipated, once we integrate along the ``problematic" direction,
we can consider a cone quite similar to the previous ones.
\vskip.5cm
\line{\bf Contraction of the Hilbert metric.\hfil}
\vskip.5cm
\par
The reader may wonder about how big the cone $\Cal C_{b,\,c}$ is. The next
Lemma shows that it is far from empty.
 
\proclaim{Lemma 4.1} For each $b>1$ the cone
$\Cal C_{b,\,c}$ is a closed convex cone with non empty
interior in the $C^{(1)}$ topology.
\endproclaim
\demo{Proof}
The closure and the convexity can be easily checked by direct computation.
The third property follows since the open set 
$\{g\in C^{(1)}(\Cal M)\;|\; g>p\geq 0;\;
\|Dg\|_\infty< cp\}$ is contained in $\Cal C_{b,\,c}$. In fact,
$$
\aligned
f_2(x)\int_\gamma gf_1=&f_1(x)\int_\gamma g(y)f_2(y)
\frac{f_2(x)f_1(y)}{f_1(x)f_2(y)}dm_\gamma(y)\leq
\e^{\rho_+(f_1,\,f_2)}f_1(x)\int_\gamma gf_2\\
\leq&\e^{\rho_\gamma(f_1,\,f_2)}f_1(x)\int_\gamma gf_2 .
\endaligned
$$
\enddemo
By now, it is clear that we need the cone to be invariant under the action
of the P-F operator. Lemma 4.2 shows that this is the case.
 
\proclaim{Lemma 4.2} For $a$, $b$ and $H$ large enough, 
$\widetilde T\Cal C_{b,\,c}\subset
\Cal C_{\chi b,\,\chi c}$ for some $\chi<1$.
\endproclaim
\demo{Proof}
We check the first property first.
Let $\gamma\in\Gamma_\delta$, $x\in\gamma$ and $f_1,\,f_2\in\Cal D_a(\gamma)$
then, for each $g\in\Cal C_{b,\,c}$,
$$
\int_\gamma(\widetilde T g)f_1=\int_{\gamma}g\circ T^{-1} 
f_1=\int_{T^{-1}(\gamma)} g f_1\circ T\left|\hbox{det}
(D_xT\big|_{T^{-1}\gamma})\right|.
$$
By $D_xT\big|_{T^{-1}\gamma}$ we mean the restriction of $D_xT$ to a map
from the direction $(T^{-1}\gamma)'$ to the direction
$\gamma'$. Moreover, we use $\text{det}(\cdot)$, in spite of the fact that, 
in our case, is superfluous, to emphasize the 
multidimensional nature of the argument.
\par
\proclaim{Sub-Lemma 4.3} For each $\gamma\in\Gamma_\delta$, 
there exists a collections of curves $\{\gamma_i\}\in\Gamma_\delta$, such 
that $\gamma_i\cap\gamma_j=\partial\gamma_i\cap\partial\gamma_j$ and 
$\cup_i\gamma_i=T^{-1}\gamma$, provided $H$ has been chosen large 
enough.
\endproclaim
\demo{Proof}
We start by subdividing $T^{-1}(\gamma)$ in connected pieces
of length $\delta$ plus, at most, one piece longer than $\delta$
but shorter than $2\delta$. 
Since $\gamma'(x)\in\Cal C(x)$, and $D_xT^{-1}\Cal C(x)
\subset\Cal C(T^{-1}x)$, we have $\left(T^{-1}\gamma\right)'(T^{-1}x)\in
\Cal C(T^{-1}x)$.
We are left with the condition on the curvature.
 
Let $\wt\gamma(s)$ and $\wt\gamma_1(s_1)$ be the curves $\gamma$ and $T^{-1}
\gamma$ parametrized by arc-length. If $\eta$ is the unit vector
in the
direction $\wt\gamma_1''$, then $\langle\eta,\,DT^{-1}\gamma'\rangle=0$
and $|\kappa_{T^{-1}\gamma}(s_1)|=|\langle\eta,\,\frac d{ds_1}
\wt\gamma_1'(s_1)\rangle|$.
Since $\frac{ds_1}{ds}=
\|DT^{-1}\gamma'\|$, it follows
 
$$
\aligned
|\kappa_{T^{-1}\gamma}(s_1)|&\leq \lambda^{-1}|\langle\eta,\,\frac d{ds}
\frac{D_{\wt\gamma(s)}T^{-1}\wt\gamma'(s)}
{\|D_{\wt\gamma(s)}T^{-1}\wt\gamma'(s)\|}\rangle|\leq\lambda^{-2}
|\langle\eta,\,\frac d{ds} D_{\wt\gamma(s)}T^{-1}\wt\gamma'(s) \rangle|\\
&\leq \lambda^{-2}(c_2+
|\langle\eta,\, D_{\wt\gamma(s)}T^{-1}\wt\gamma''(s) \rangle| )
\endaligned
$$
where $c_2$ depends only on the second derivative of $T$.
To conclude the estimate, let us fix $s$ and choose explicit orthogonal
coordinates at $\wt\gamma(s)$ and $T^{-1}\wt\gamma(s)$. Namely, we take 
the axes in the directions $\wt\gamma'$, $\wt\gamma''$, and $\wt\gamma_1'$,
$\wt\gamma_1''$, respectively. In these coordinates $DT^{-1}$ has the
form
$$
DT^{-1}=\pmatrix  a&b\\
                  0&c
         \endpmatrix
$$
with $a=\|D_{\wt\gamma(s)}T^{-1}\wt\gamma'(s)\|\geq \lambda$.
In addition, since the Riemannian volume equals the symplectic one,
det$(DT^{-1})=1$, i.e., $a=c^{-1}$. Using such a representation yields
$$
|\langle\eta,\, D_{\wt\gamma(s)}T^{-1}\wt\gamma''(s) \rangle|\leq
(0\,\,1)\pmatrix a&b\\ 0& a^{-1}\endpmatrix\pmatrix 0\\ |\kappa_\gamma(s)|
\endpmatrix=  |\kappa_\gamma(s)| a^{-1} ,
$$
or
$$
|\kappa_{T^{-1}\gamma}(s_1)|\leq\lambda^{-2}(c_2+\lambda^{-1} H)\leq H .
$$
\enddemo
 
It is then natural to write
 
$$
\int_\gamma(\widetilde T g)f_1=
\sum_i \int_{\gamma_i}g f_1\circ T |\hbox{det}(DT\big|_{\gamma_i})| .
$$
To continue we need to analyze the maps $\widehat T_i:C^{(0)}(\gamma)
\to C^{(0)}(\gamma_i)$  defined by
$$
(\widehat T_if)(x)=f\circ T (x)\left|\hbox{det}(D_xT\big|_{\gamma_i})\right|
\qquad\forall x\in\gamma_i  .
\tag 4.1
$$
 
>From now on, with a slight abuse of notations, we will drop the subscript
``$i$", since the range of $\wh T$ is always clear from the context.
 
\proclaim{Sub-Lemma 4.4} $\widehat T(\widetilde{\Cal D}_a(\gamma))
\subset \widetilde{\Cal D}_{\sigma a}
(\gamma_i)$ for some $\sigma<1$, provided $a>c_3\delta^{1-\nu}
(1-\lambda^{-\nu})^{-1}$, 
where $c_3$ depends only on the system and $H$ (see definition of 
$\Gamma_\delta$).
\endproclaim
\demo{Proof}
For each $f\in \wt{\Cal D}_a(\gamma)$, $x,\,y\in\gamma_i\in\Gamma_\delta$
$$
\frac {(\widehat T f)(x)}{(\widehat Tf)(y)}\leq
\e^{a d(Tx,\,Ty)^\nu+\mu(c_2+\mu H) d(x,\,y)}\leq
\e^{(a \lambda^{-\nu}+c_3\delta^{1-\nu})d(x,\,y)^\nu} ,
$$
where $\mu$ is the maximal expansion coefficient.
\enddemo
 
\proclaim{Sub-Lemma 4.5} If $R$ is sufficiently large, then
$\widehat T(\Cal D_a(\gamma))\subset\Cal D_a(\gamma_i)$; in
addition for each $f_1,\,f_2\in\Cal D_a(\gamma)$
$$
\rho_{\gamma_i}(\widehat Tf_1,\,\widehat Tf_2)
\leq\xi\rho_{\gamma}(f_1,\,f_2)
$$
for some $\xi<1$.
\endproclaim
\demo{Proof}
The constant  function $1$ is an element 
of $\Cal D_a(\gamma_i)$; 
for each
$f\in\widetilde{\Cal D}_{\sigma a}(\gamma_i)$ we have
$$
\aligned
\rho_{\gamma_i}(1,\,f)=&\ln\left[\sup_{x,\,y,\,u,\,v\in \gamma_i}
\frac{\left[\e^{ad(x,\,y)^\nu}
-1\right]\left[\e^{ad(u,\,v)^\nu}f(u)-f(v)\right]}
{\left[\e^{ad(x,\,y)^\nu}f(x)-f(y)\right]\left[\e^{ad(u,\,v)^\nu}
-1\right]}\right]\\
&\leq \ln\left[\sup_{x,\,y,\,u,\,v\in\gamma_i}\frac{\left[\e^{ad(x,\,y)^\nu}
-1\right]\left[\e^{a(1+\sigma)d(u,\,v)^\nu}-1\right]f(v)}
{\left[\e^{a(1-\sigma)d(x,\,y)^\nu}-1\right]\left[\e^{ad(u,\,v)^\nu}
-1\right]f(y)}\right]\\
&\leq\ln\left[\frac{(1+\sigma)}
{(1-\sigma)} \e^{a\sigma 2^\nu\delta^\nu}\right]\leq R.
\endaligned
$$
Hence, $\widehat T(\Cal D_a(\gamma))\subset\Cal D_a(\gamma_i)$. In 
addition, since the diameter of 
$\widehat T_i(\widetilde{\Cal D}_a(\gamma))$ in 
$\widetilde{\Cal D}_a(\gamma_i)$ is less than
$ R$,
the second part of the result follows from Theorem 1.1.
\enddemo
 
Accordingly, 
$$
\int_\gamma\wt T gf=\sum_i\int_{\gamma_i}g\wh Tf\geq 0.
$$
Next, let $x\in\gamma$, and choose points $z_i\in T\gamma_i$;
$$
\aligned
f_2(x)\int_{\gamma}(\widetilde T g)f_1=&
\sum_i \frac{f_2(x)}{\widehat Tf_2(z_i)}\widehat Tf_2(z_i)
\int_{\gamma_i}g \widehat Tf_1\\
\leq&\sum_i\e^{b\rho_{\gamma_i}(\widehat Tf_1,\,\widehat Tf_2)}
\frac{f_2(x)}{\widehat Tf_2(z_i)}\widehat Tf_1(z_i)
\int_{\gamma_i}g \widehat Tf_2\\
\leq&\e^{\xi b\rho_\gamma(f_1,\,f_2)}
\sup_{z,\,z'\in\gamma} \frac{f_2(z)f_1(z')}{f_1(z)f_2(z')}f_1(x)\sum_i
\int_{\gamma_i}g\wh T f_2\\
\leq& \e^{\left[\xi b\rho_\gamma(f_1,\,f_2)+\rho_+(f_1,\,f_2)\right]}
f_1(x)\sum_i\int_{T\gamma_i}\wt Tg f_2\\
\leq&
\e^{(\xi b+1)\rho_\gamma(f_1,\,f_2)}f_1(x)\int_{\gamma}(\widetilde T g)f_2
\endaligned
$$
which yields the wanted result for $b$ large enough.
The third of the conditions is easily checked after noticing that
$$
\frac{\int_\gamma (\widetilde Tg) f}{\int_\gamma f}\geq
\sum_i \frac {\int_{\gamma_i} \widehat T_i f}{\int_\gamma f}\tb g\tb_-=
\tb g \tb_-
$$
implies $\tb\widetilde Tg\tb_-\geq\tb g\tb_-$.
\enddemo
We know now that the cone is mapped inside itself, yet this does not 
necessarily means that the diameter of the image if finite. To check
this last property we start with a preliminary result.
 
\proclaim{Lemma 4.6} Setting $K=\frac{(1+\chi)(1+b)}{(1-\chi)(b-1)}$, 
for each
$g\in\Cal C_{\chi b,\chi c}$,
$$
\Theta(g,\,1)\leq\ln\left[K\frac{\tb g\tb_+}{\tb g\tb_-}\right]
$$
(from now on $\Theta$ designates the Hilbert metric in $\C_{b,\,c}$).
\endproclaim
\demo{Proof}
For each $\lambda>0$, such that $g-\lambda\in 
\Cal C_{\chi b,\chi c}$, a direct computation shows
$$
\aligned
\lambda\leq& \inf\Sb\gamma\in\Gamma_\delta\\ f\in\Cal D_a(\gamma)\endSb 
\frac{\int_\gamma gf}{\int_\gamma f}=\alpha_0 ,\\
\lambda\leq& \inf\Sb\gamma\in\Gamma_\delta\\ f_1,\,f_2\in\Cal 
D_a(\gamma)\endSb 
\frac{\e^{b\rho_\gamma(f_1,\,f_2)}f_1\int_\gamma g f_2-f_2\int_\gamma g 
f_1}
{\e^{b\rho_\gamma(f_1,\,f_2)}f_1\int_\gamma f_2-f_2\int_\gamma 
f_1}=\alpha_1 ,\\
\lambda\leq&\frac{c\tb g\tb_--\|D^u g\|_\infty}c=\alpha_2 ,
\endaligned
$$
which implies $\alpha=\min\{\alpha_0,\,\alpha_1,\,\alpha_2\}$
(see \S 1 for a definition of $\alpha$ in this context). In addition,
$$
\aligned
\alpha_0=&\tb g\tb_-,\\
\alpha_1\geq& \inf\Sb\gamma\in\Gamma_\delta\\ f_1,\,f_2\in\Cal 
D_a(\gamma)\endSb
\frac{\left[\e^{b\rho_\gamma(f_1,\,f_2)}-
\e^{b\chi\rho_\gamma(f_1,\,f_2)}\right]
f_1\int_\gamma g f_2}{\left[\e^{b\rho_\gamma(f_1,\,f_2)}-
\e^{-\rho_\gamma(f_1,\,f_2)}\right]
f_1\int_\gamma f_2}\\
\geq&
\tb g\tb_-\inf_{t\in\Bbb R^+}\frac {\e^{bt}-\e^{\chi 
bt}}{\e^{bt}-\e^{-t}}
\geq\frac{(1-\chi)b}{b+1}\tb g\tb_-   ,
\endaligned
$$
and
$$
\alpha_2\geq(1-\chi) \tb g\tb_-  .
$$
This shows that $\alpha\geq \frac{(1-\chi)b}{b+1}\tb g\tb_-$.
Moreover, a similar computation shows that $\beta\leq \frac{(1+\chi)b}{b-1}
\tb g\tb_+$ and concludes 
the proof, since $\Theta(g,\,1)=\ln\frac\beta\alpha$.
\enddemo
 
It is interesting (and helpful) to notice that, if we consider the cone
$$
\Cal C_+=\left\{g\in C^{(0)}(\Cal M)\;\bigg|\; g\not\equiv 0;\;
\int_\gamma fg\geq 0\;\;
\forall \gamma\in\Gamma_\delta,\,f\in\Cal D_a(\gamma)\right\}
$$
ant its Hilbert metric $\Theta_+$, then
$$
\Theta_+(g,\, 1)=\ln\frac{\tb g\tb_+}{\tb g\tb_-}
$$
In this language, Lemma 4.6 states: for each $g\in\Cal C_{\chi b,\,\chi c}$, 
$\Theta(g,\,1)\leq \ln 
K+\Theta_+(g,\,1)$.
\par
Up to now we have let the scale $\delta$ be arbitrary. In the next step it is
instead essential to compare averages on curves having some fixed scale.
In the present context we could simply fix $\delta$ to be some appropriate
value $\delta_0$. Nevertheless, it is inspiring to note that one can
relate averages on any small scale $\delta$ with averages on some fixed scale
$\delta_0$.
\par
Let $\delta_0>0$ be fixed, define
$$
\aligned
\tb g\tb_+^0=&\sup\Sb \gamma\in\Gamma_{\delta_0}\\ f\in\Cal D_a(\gamma)\endSb
\frac{\int_\gamma gf}{\int_\gamma f}\\
\tb g\tb_-^0=&\inf\Sb \gamma\in\Gamma_{\delta_0}\\ f\in\Cal D_a(\gamma)\endSb
\frac{\int_\gamma gf}{\int_\gamma f}.
\endaligned
$$
\proclaim{Lemma 4.7}For each $\delta\leq\delta_0$, 
there exists $N_0\in\Bbb N$
such that, for each $g\in\Cal C_{b,c}$, $n\geq N_0$,
$$
\aligned
\tb\wt T^n g\tb_+&\leq\tb g\tb_+^0 \\
\tb\wt T^n g\tb_-&\geq\tb g\tb_-^0 ,
\endaligned
$$
provided $a>\frac{c_3}{1-\lambda^{-1}}\delta_0^{1-\nu}$.
\endproclaim
\demo{Proof}
For each $\gamma\in\Gamma_\delta$, $f\in\Cal D_a(\gamma)$ and
$n\geq\frac {\ln(2\delta_0\delta^{-1})}{\ln\lambda}=N_0$ we have
$|T^{-n}\gamma|>2\delta_0$, we can then divide $T^{-n}\gamma$ in curves
$\{\gamma_i\}\subset\Gamma_{\delta_0}$.   The condition on $a$ insures
that $\wh T^nf\in\Cal D_a(\gamma_i)$.
$$
\frac 1{\int_\gamma f}\int_\gamma\widetilde T^n g f=
\frac 1{\int_\gamma f}\sum_i\int_{\gamma_i}g\widehat T^n f\leq
\frac 1{\int_\gamma f}\sum_i\int_{\gamma_i}\widehat T^n f\tb g\tb_+^0=
\tb g\tb_+^0
$$
that is, $\tb\widetilde T^n g\tb_+\leq\tb g\tb_+^0$. The other inequality
is proven in the same way.
\enddemo
\vskip.4cm
\line{\bf Diameter of the image.\hfil}
\vskip.4cm
While up to now we did not use the mixing property, it will be necessary
to do so in what follows. The main implication of mixing, in our context, is
of a geometric nature: it asserts that given a finite number of 
sets of positive measure, some image of each one will intersect all the 
others. Consequently,
given two curves that intersect such sets, the images of one curve will 
eventually get close to the other.
 
 
\topinsert
\vskip3in
\hsize=4.5in
\raggedright
\noindent{\bf Figure 2} Square neighborhood.
\endinsert
 
 
\par
To make this argument precise we will construct a covering of $\Cal M$, made
up of sets that are well behaved with respect to the curves in 
$\Gamma_{\delta_0}$. The construction of the covering may seem a bit 
technical, but its main point  is to insure that each curve in 
$\Gamma_{\delta_0}$ is contained in some element of the covering, and that
the relevant geometric properties of the iterates of 
any curve can be described solely in terms of the covering.
\par
Consider $z\in\Cal M$. There exists a Cartesian system of coordinates in
a neighborhood of $z$ such that the axes are parallel to $W^s(z)$ and
$W^u(z)$, respectively. By the continuity of the cone family, there exists
a smaller neighborhood of $z$ where each curve parallel to one of the
axes is either in $\Cal C$ (in the stable direction) or in its complement
(unstable direction). Let $\wh Q(z)$ be a square (in the Cartesian 
coordinates), with center in $z$, enjoying the above properties.
\par
We can use the Cartesian structure to identify all the tangent space at
$w\in\wh Q(z)$ with the tangent space at $z$, and to use the Riemannian 
metric at $z$ to define a Cartesian metric in $\wh Q(z)$. 
We then consider some smaller
neighborhood $\wh Q_1(z)$ such that, calling $d_c$ the Cartesian distance
and $d$ the Riemannian one, $\frac 34 d\leq d_c\leq \frac 54 d$, and
such that $\bigcup_{w\in\wh Q_1(z)}\Cal C(w)\subset\{|\xi|\leq\frac 12
|\eta|\}=\wh{\Cal C}$; this means that we can consider a unique cone 
$\wh{\Cal C}$  in the neighborhood $\wh Q_1(z)$. We finally consider
the square neighborhood $Q(z)$, centered at $z$, but with size half
the size of $\wh Q_1(z)$ (see \cite{LW} for more detail on such construction). 
\par
Since $\Cal M$ is compact, there are finitely many $z_i\in\Cal M$ such that
$\bigcup_i Q(z_i)=\Cal M$.   
\par
We choose $\delta_0$ to be $\frac 14$ the size of the
smallest of the $Q(z_i)$.
\par
Given a square $G\subset\wh Q_1(z_i)$, for some $i$, of size $4\delta_0$,
it is easy to see that every curve in $\Gamma_{\delta_0}$ that intersects
a concentrical square of size $\frac {\delta_0}4$ will be completely 
contained in $G$. We will call such concentrical square the ``core" of 
$G$ (see \cite{LW, \S 1-3} for a more extensive discussion of the concept
of core and its use). We also define the ``central strip" of a square,
as a strip, parallel to $W^u(z)$, 
of size $\frac{\delta_0}4$, and containing the core.
\par
In each $\wh Q_1(z_i)$ we construct squares $G$ such that the union of their
cores covers $Q(z_i)$; the collection of such squares is a covering of
$\Cal M$ with the following properties
\roster
\item   $\bigcup_{G\in\Cal G}\text{core}(G)=\Cal M $;
\item   $\forall \gamma\in\Gamma_{\delta_0},\,G\in\Cal G$ if
        $\gamma\cap\text{core}(G)\neq\emptyset$, then $\gamma\subset G$;
\item   $\forall \gamma\in\Gamma_{\delta_0}$ there exists $G\in\Cal G$:
        $\gamma\cap\text{core}(G)\neq\emptyset$, and the part of $\gamma$
        that intersects the central strip of $G$ consists of $\gamma$ minus
        two pieces, each of length larger than $\frac {\delta_0}4$.
\endroster                                                  
We will say that $\gamma$ ``intersects properly" $G$ if the last of the
latter properties is satisfied.
\par
Such covering is the main ingredient in the following fundamental Lemma.
 
\proclaim{Lemma 4.8} Let $\delta\leq\delta_0$,
then there exist $N_*\in\Bbb N$ and $\Delta\in\Bbb R^+$
such that
$$
\hbox{diameter }(\widetilde T^{N_*}\Cal C_{b,\,c})\leq\Delta<\infty ,
$$
$N_*\leq N+N_0$, where $N$ depends
only on $\delta_0$ and $T$.
\endproclaim
\demo{Proof}
Since the map under consideration is mixing there exists
$N\in\Bbb N$ such that, for each $G,\,G'\in\Cal G$,
$$
T^{-N}\hbox{core}( G)\cap \hbox{core}(G')\neq\emptyset .
$$
Without loss of generality we can impose $N>n_1$ ($n_1$ to be chosen later).
Given any function $g\in\Cal C_{b,c}$,
we choose $\gamma_*\in\Gamma_{\delta_0}$ and $f_*\in\Cal D_a(\gamma_*)$
such that
$$
\frac{\int_{\gamma_*}gf_*}{\int_{\gamma_*}f_*}\geq\frac 12 \tb g\tb_+^0.
$$
Let us call $G_*$ an element of $\Cal G$  properly intersected by   
$\gamma_*$. 
Given any $\gamma\in\wh \Gamma_{\delta_0}$, there will be an element $G$ of
$\Cal G$ that is properly intersected by $\gamma$.
This implies that there exists $w\in\text{core}(G)\cap 
T^N\text{core}(G_*)$; by definition, 
$W^u_{\text{loc}}(w)\cap\gamma\neq\emptyset$, and consists of only one point 
$w_1$, moreover $d(T^N(w_1),\,T^N(w))\leq\lambda^{-N}\frac{\delta_0}4$.
Hence, there exists $\gamma_1\subset T^{-N} 
\gamma$ such that it is the ``translate"\footnote{By this we mean that
for each $z\in\gamma_*$ $W^u_{\text{loc}}(z)\cap\gamma_1\neq\emptyset$ 
and consists 
of a single point; in addition, for each $w\in\gamma_1$ 
$W^u_{\text{loc}}(w)\cap\gamma_*\neq\emptyset$ and consists of a single point. 
See Appendix II for more details.}, along the unstable direction, of
$\gamma_*$.
In fact, by construction, $W^u(T^{-N}w)\cap\gamma_*\neq\emptyset$,
and, if we cut $ T^{-N} \gamma$ at $T^{-N}w_1$, each of the
two resulting pieces is longer than $\lambda^{N}\frac{\delta_0}8
\geq\lambda^{n_1}\frac{\delta_0}8\geq 3\delta_0$ (where we have chosen 
$n_1$ such that $\lambda^{n_1}\geq 24$). Hence, 
$T^{-N}\gamma\backslash\gamma_1$ can be divided into pieces belonging
to $\Gamma_{\delta_0}$, and
$$
\frac 1{\int_\gamma f}\int_\gamma\widetilde T^{N} g f
\geq\frac 1{\int_\gamma f}\int_{\gamma_{1}}g\widehat T^{N}f.
$$
Let $\phi$ be the canonical isomorphism (see Appendix II) from $\gamma_*$ to
$\gamma_{1}$ and $J_{\phi}$ its Jacobian.
$$
\frac 1{\int_\gamma f}\int_\gamma\widetilde T^{N} gf\geq
\frac 1{\int_\gamma f}\int_{\gamma_*}g
\circ \phi\left[ (\widehat T^{N}f)\circ\phi\right] J_{\phi^{-1}}
$$
>From the results recalled in Appendix II it follows that, for $R$, $a$ 
large and $\delta_0$ small enough,
$\widetilde f=(\widehat T^{N}f)\circ\phi\, J_{\phi^{-1}}
\in\Cal D_a(\gamma_*)$. Hence,  for some $z\in\gamma_*$,
$$
\aligned
\frac 1{\int_\gamma f}\int_\gamma\widetilde T^{N} gf\geq&
\frac 1{\int_\gamma f}\int_{\gamma_*}g
\widetilde f-\frac{\delta_0\|D^u g\|_\infty}
{\int_\gamma f}\int_{\gamma_*}\widetilde f\\
\geq&\frac 1{\int_\gamma f}\left[\frac{\widetilde f(z)}{f_*(z)}
\int_{\gamma_*}f_*g\e^{-2bR}-c\delta_0\tb g\tb_-
\int_{\gamma_*}\widetilde f\right]\\
\geq&\frac 1{\int_\gamma f}\left[\frac{\widetilde f(z)}
{2f_*(z)}
\tb g\tb_+^0\e^{-2bR} \int_{\gamma_*} f_*-
c\delta_0\tb g\tb_-\int_{\gamma_*}\widetilde f\right]\\
\geq &\left[\frac{\e^{-2(b+1)R}}2 
\tb g\tb_+^0-c\delta_0\tb g\tb_-\right]
\frac{\int_{\gamma_{1}}\widehat T^{N} f}
{\int_\gamma f} .
\endaligned
$$
On the one hand, if $\frac{e^{2(b+1)R}}{4}\tb 
g\tb^0_+\leq c\delta_0\tb g\tb_-$, we have immediately
$$
\frac{\tb \wt T^{N+N_0}g\tb_+}{\tb \wt T^{N+N_0}g\tb_-} \leq  
\frac{4c\delta_0}{e^{2(b+1)R}}.
$$
 
On the other hand, if $\frac{e^{2(b+1)R}}{4}\tb 
g\tb^0_+\geq c\delta_0\tb g\tb_-$, we choose  an arbitrary point $\zeta\in 
T^{N}\gamma_{1}\subset\gamma$, and we have
$$
\aligned
\frac 1{\int_\gamma f}\int_\gamma\widetilde T^N gf\geq&
\frac {\int_{T^{N}\gamma_{1}}
\frac{f(y)}{f(\zeta)}dm(y)}
{\int_{\gamma} \frac{f(y')}{f(\zeta)}dm(y')}                     
\frac{\e^{-2(b+1)R}}4 \tb g\tb_+^0\\
\geq&\e^{-2^\nu a\delta_0^\nu}\frac{|T^{N}\gamma_{1}| } 
{|\gamma| }\frac{\e^{-2(b+1)R}}4\tb g\tb_+^0
\geq\e^{-2^\nu a\delta_0^\nu} \mu^{-N}\frac{\e^{-2(b+1)R}}4
\tb g\tb_+^0
\endaligned
$$
where $\|DT\|\leq \mu$.
>From the above chain of inequalities follows:
$$
\frac 1{\int_\gamma f}\int_\gamma\wt T^N gf
\geq\e^{-2^\nu a\delta_0^\nu} \mu^{-N}\frac{\e^{-2(b+1)R}}4
\tb \wt T^Ng\tb_+^0  .
$$
This last inequality yields
 
$$
\tb \widetilde T^N g\tb_-^0\geq
\frac{\e^{-2^\nu a\delta_0^\nu}\mu^{-N}\e^{-2(b+1)R}}{4}
\tb \widetilde T^Ng\tb_+^0 ,
$$
 
which implies
$$
\frac{\tb\widetilde T^{N+N_0} g\tb_+}{\tb\widetilde T^{N+N_0} g\tb_-}\leq
4\e^{2^\nu a\delta_0^\nu}\mu^N\e^{2(b+1)R}   .
$$
\enddemo
 
Thanks to Lemma 4.8 we can apply Theorem 1.1. 
\par
Summarizing, there 
exists $\Lambda_0\in(0,\,1)$ such that,
for each $g_1,\,g_2\in\Cal C_{b,\,c}$
$$
\Theta(\widetilde T^ng_1,\,\widetilde T^ng_2)\leq\Lambda_0^{\frac n{N+N_0}}
\Theta(g_1,\,g_2).
$$
Let $\Lambda=\Lambda_0^{(N+N_0)^{-1}}$.
 
\vskip.5cm
\line{\bf Decay of correlations.\hfil}
\vskip.5cm
\par
                
\proclaim{Theorem 4.9}
There exist $D>0$, and $r>0$ such that,
for each $g,\,f\in C^{(1)}(\Cal M)$
$$
\bigg|\int_{\Cal M}f\widetilde T^n g-\int_{\Cal M}f\int_{\Cal M}g\bigg|\leq 
D\|g\|_*\|f\|_*\Lambda^n
$$
with $\| h\|_*=\int_{\Cal M}| h|+r\| h'\|_\infty$.
\endproclaim
\demo{Proof}
The basic idea behind this result is to partition $\Cal M$ into
curves belonging to $\Gamma_\delta$. Unfortunately, such a simple idea it
is not trivially implementable. Of course, such partitions exist (e.g.
Markov partitions) but it is not immediately obvious which constraints
they force on $\delta$. In order to avoid the problem entirely, we content
ourselves with a much coarser partition and use a dynamical argument to
refine it.
\par
We have already constructed a covering $\{Q(z_i)\}$ of $\Cal M$. From
such covering we can obtain a partition $\Cal P=\{P_i\}$ of $\Cal M$,
such that $Q(z_i)\supset P_i$ and $\partial P_i$ is piecewise smooth.
\par
Next, for each $\delta_1<\delta$, we can partition $Q(z_i)$ into squares
of size $\delta_1$, with sides parallel to the axes. We will collect
into $\Cal P_+^i(\delta_1)$ all the squares $p$ such that $p\cap P_i
\neq \emptyset$, and into $\Cal P_-^i(\delta_1)$  all the squares $p$
such that $p\subset P_i$; let $\Cal P_+(\delta_1)=\cup_i 
\Cal P_+^i(\delta_1)$, $\Cal P_+(\delta_1)=\cup_i\Cal P_+^i(\delta_1)$.
Clearly, for each $\varepsilon>0$, there exists $\delta_1\in\Bbb R^+$
such that
$$
\aligned
\sum_{p\in\Cal P_+}m(p)&\leq 1+\varepsilon\\
\sum_{p\in\Cal P_-}m(p)&\geq 1-\varepsilon .
\endaligned
\tag 4.2
$$
This two series of ``quasi-partitions" of $\Cal M$ will be sufficient to
prove:
 
\proclaim{Lemma 4.10} If $f\in C^{(0)}(\Cal M)$, 
$f\in\wt{\Cal D}_{\sigma a/2}(\gamma)$ for
each $\gamma\in\Gamma(\delta)$, and $g\in\Cal C_{b,c}$, then
$$
\int_{\Cal M}f\tb g\tb_-\leq\int_{\Cal M}fg\leq\int_{\Cal M}f \tb g\tb_+.
$$
\endproclaim
\demo{Proof}
For each $\varepsilon>0$, choose $\delta_1$ such that (4.2) holds, then,
for each $n_1>0$,
$$
\int_{\Cal M}f g=\int_{\Cal M}\wt T^{n_1}(fg)\geq\sum_{p\in\Cal P_-}
\int_{p}\wt T^{n_1}(fg).
$$
By construction, each $p$ is naturally partitioned into curves $\gamma_s$
belonging to $\Gamma_{\delta_1}$. If $\wt h_p$ is the smooth density of
the invariant measure expressed in the local Cartesian coordinate 
$(\eta,\,\xi)$ of the square $Q(z_i)$, then, by applying Fubini Theorem,
$$
\int_{\Cal M}fg\geq\sum_{p\in\Cal P_-}\int d\xi\int_{\gamma_\xi}\wt h_p 
\wt T^{n_1}(fg)= \sum_{p\in\Cal P_-}\int d\xi\int_{T^{-n_1}\gamma_\xi}
\wh T^{n_1}(\wt h_p)fg .
$$
We choose $n_1$ large enough
such that, for each $p$, $f\wh T^{n_1}\wt h_p\in\wt{\Cal D}_{\sigma a}(\gamma)
\subset\Cal D_a(\gamma)$, and $| T^{-n_1}\gamma_{\xi}|>\delta$ for each
$\xi$, 
$$
\aligned
\int_{\Cal M}fg&\geq \sum_{p\in\Cal P_-}\int d\xi\int_{T^{-n_1}\gamma_\xi}
\wh T^{n_1}(\wt h_p)f\tb g\tb_-=
\sum_{p\in\Cal P_-}\int d\xi\int_{\gamma_\xi}  \wt h_p\wt T^{n_1}f\tb g\tb_-\\
&\geq \left(\int_{\Cal M}f-\int_{\Cal M\backslash\bigcup\limits_{p\in\Cal P_-}p}
\wt T^{n_1}f\right)\tb g\tb_-\geq\left(\int_{\Cal M}f-\varepsilon\|f\|_\infty
\right)\tb g\tb_-  .
\endaligned
$$
This yields the first inequality, since $\varepsilon$ is arbitrary.
The second inequality is proven in the same way.
\enddemo
The previous Lemma implies
$$
\int_{\Cal M}f\widetilde T^n g\leq
\frac{\tb \widetilde T^n g\tb_+}{\tb \widetilde T^n g\tb_-} 
\int_{\Cal M}f  \int_{\Cal M}g\leq\e^{\Theta (\widetilde T^ng,\,1)}
\int_{\Cal M}f  \int_{\Cal M}g  
$$
and
$$
\int_{\Cal M}f\widetilde T^n g
\geq  \e^{-\Theta (\widetilde T^ng,\,1)} \int_{\Cal M}f  \int_{\Cal M}g .
$$
The preceding inequalities shows that there exists a constant $D$ such that
the Theorem holds for the functions under consideration.
\par
Next, we will consider $f,\,g\in C^{(0)}(\Cal M)$, $f\geq 0$, $g\geq 0$,
$f$, $g$ almost everywhere differentiable with $\|f'\|_\infty<\infty$,
and $\|g'\|_\infty<\infty$; in addition, for each stable curve $\gamma$
$f$ is piecewise differentiable along $\gamma$, and $g$ is piecewise 
differentiable along each unstable manifold.
Let $f_\rho=f+\rho$, and $g_\xi=g+\xi$ ($\rho,\,\xi\in\Bbb R^+$), then
$$
\frac {f(x)}{f(y)}\leq\e^{\int_x^y\frac{\|f'\|_\infty}\rho}\leq
\e^{\frac{\|f'\|_\infty}\rho d(x,\,y)^\nu\delta^{1-\nu}}.
$$
It follows then that, choosing $\rho=\frac{2\|f'\|_\infty\delta^{1-\nu}}
{\sigma a}$, we
have $f\in\Cal D_{\frac{\sigma a}2}(\gamma)$ for each $\gamma\in\Gamma_\delta$.
We can perform a similar computation for $g_\xi$; it yields
$g_\xi\in\Cal C_{b,\,c}$, provided $\xi=\frac {\| g'\|_\infty}c$. 
This implies
$$
\aligned
\left|\int_{\Cal M}f\wt T^ng- \int_{\Cal M}  f\int_{\Cal M} g\right|&=
\left|\int_{\Cal M}f_\rho\wt T^ng_\xi- 
\int_{\Cal M}  f_\rho\int_{\Cal M} g_\xi\right|\leq
D\Lambda^n\int_{\Cal M}  f_\rho\int_{\Cal M} g_\xi\\
&\leq D\Lambda^n\left[\int_{\Cal M} f+
\delta^{1-\nu}\frac{2\|f'\|_\infty}{a\sigma}\right]
\left[ \int_{\Cal M} g+\frac{\|g'\|_\infty}c\right] .
\endaligned
$$
Again, the proof is concluded since any smooth function 
can be written as the difference
of positive functions with the above properties.
\enddemo
 
\proclaim{Remark 4.11}
An interesting feature of the proof presented in this section 
is the possibility of computing explicitly the
constants $D$ and $\Lambda$ by only looking at a finite number of 
iterations of the map. It would be interesting to perform such computations
(eventually by some numerical method) in specific examples to verify if the 
present strategy can produce realistic estimates. By contrast, the attempt 
to implement numerically the proof based on Markov partitions is by no 
means trivial \cite{Ga}.
\endproclaim
\par
             
\line{\bf Review\hfil}
\par
Let us review the strategy of the proof before going to the next section.
The proof consists, essentially, of five steps.
The first is the specification of an invariant cone of functions in terms
of smoothness in the unstable direction and averages in the stable direction.
The second is the reduction of the estimate of the diameter of the images
to the estimate of the ratio of the maximal and minimal average.
The third (not particularly essential in this context but essential in the
following) is the reduction of the estimate of the averages on the scale
$\delta$ to averages on some fixed scale $\delta_0$.
The fourth is the estimate on the averages on the scale $\delta_0$ 
by using mixing first
and the absolute continuity second.
The fifth (and last) step consists of showing that the contraction in the 
Hilbert  metric suffices to imply the exponential decay of correlation.
\par
In the next section we will carry out the same five steps in a more general
context. I advise the reader to refer to this section if some confusion
arises in the next, more technical, one.
 
\vskip1cm
\subhead \S 5 TWO DIMENSIONS (THE NON-SMOOTH UNIFORMLY HYPERBOLIC CASE)
\endsubhead
\vskip1cm
In this section we will consider a class of maps that is similar to, 
but more restrictive than, the one considered in 
\cite{LW} and \cite{KS}. 
\par
More precisely, let $\Cal M\subset \Bbb R^2$ be a connected compact set
on which the map $T$ is defined: $T:\Cal M\to\Cal M$ .
\footnote{The case in which the map is defined on finitely many
disconnected sets can be treated in the same way.} 
For simplicity, we will assume $\Cal M$ to be a square, but we could
as well require that
$\partial \Cal M$ consists of a finite number of smooth lines intersecting
transversally. We will call a set with the above more general property
a ``symplectic box" (see \cite{LW} for a general definition). The
symplectic box $\Cal M$ is
partitioned in two ways into unions of an equal number of symplectic
boxes
$$
\Cal M = \Cal M_1^+ \bigcup \dots \bigcup \Cal M_m^+
= \Cal M_1^- \bigcup \dots \bigcup \Cal M_m^-.
$$
Two boxes of one partition can overlap at most on their boundaries, i.e.,
$$
\Cal M_i^{\pm}\bigcap \Cal M_j^{\pm} \subset \partial \Cal M_i^{\pm}\bigcap
\partial \Cal M_j^{\pm} ,\ \ i,j = 1,\dots ,m.
$$
 
 
The map $T$ is defined separately on each of the symplectic boxes
$\Cal M_i^+, \ i =1,\dots,m$. It is a symplectomorphism 
(i.e. it preserves the area and it is smooth) of the interior of
each $\Cal M_i^+$ onto the interior $\Cal M_i^-, \ i =1,\dots,m$
and  a homomorphism of
$\Cal M_i^+$ onto  $\Cal M_i^-, \ i =1,\dots,m$. 
We will assume that $T$ is $C^{(2)}$ in each box up to and including the 
boundary. \footnote{This is a serious restriction: in particular,
billiards do not satisfy it. See \S 6 for a discussion of one example
that does not satisfy such restrictions.}
\par
We will say that $T$ is a (discontinuous) symplectic map of $\Cal M$.
Formally $T$ is not well defined on the set of points which belong
to the boundaries of several plus-boxes: it has several values.
We adopt the convention that the image of a subset of $\Cal M$ under $T$
contains all such values.
 
Let us introduce the singularity sets $\R^{+}$ and $\R^-$.
$$
\R^{\pm} = \{ p\in \Cal M \;|\; p
\text { belongs to at least two of the boxes }\Cal M_i^{\pm}, i =1,\dots,m\}.
$$
 
The plus-singularity set $\R^+$ is a closed subset, and $T$ is
continuous on its complement. Similarly $T^{-1}$ is continuous on the
complement of $\R^-$. 
 
It is the case that $\R^+ \bigcup \partial \Cal M$ is the union of all 
the boundaries
of the plus-boxes and $\R^- \bigcup \partial \Cal M$ is the union of all the
boundaries of the minus-boxes, i.e.,
$$
\R^{\pm} \bigcup \partial \Cal M = \bigcup_{i=1}^m\partial \Cal M_i^{\pm}.
$$
Note that most of the points in the boundary $\partial \Cal M$
of $\Cal M$ do not belong to $\R^-$ or $\R^+$.
 
Our assumptions imply that the singularity sets $\R^{\pm}$
and the union of boundaries $\bigcup_{i=1}^m\partial \Cal M_i^{\pm}$
is the finite union of smooth lines intersecting transversally.
 
We assume $T$ to be uniformly 
hyperbolic, and that at each point $x\in\partial\Cal M^{\pm}_i$ the
stable and unstable direction be uniformly 
transversal to  $\partial\Cal M^{\pm}_i$.
 
We assume that $T$ is mixing. \footnote{The observations in footnote 9
apply here as well.}
\par
In addition, we chose $\zeta\in[\sqrt 3-1,\,1]$ such that
$$
\lambda^{-\zeta}\mu^{1-\zeta}<1
\tag 5.1
$$
where $\lambda$ and $\mu$ are the minimal and maximal expansion coefficients 
of $T$, respectively (see \S 4 for a more precise definition of $\lambda$
and $\mu$). 
 
 
Next, we have a condition specific to the present context, which
does not play any role in the study of weaker statistical properties
(like mixing and ergodicity). 
We assume that there exists $n_0\in\Bbb N$, such that
$$
M(n_0)\lambda^{-\zeta n_0}<\frac 1{25},
\tag 5.2
$$
$M(n)$ being
the maximal number of smooth lines belonging to $\bigcup_{i=0}^n T^i\R^-$
that intersect at one point.\footnote{The choice of $25$ is completely
arbitrary. The point is that if $M^n\lambda^{-\zeta n}<1$, for some
$n$, then there exists some $n_0$ that satisfies (5.2).}
 
\proclaim{Remark 5.1} Note that condition 5.2 can always be fulfilled if 
$M(n)$ has a polynomial growth. But it is not necessary to be able to control
$M(n)$ for each $n$, since 5.2 is a condition concerning some fixed
power of $T$.
\endproclaim
 
\par
Henceforth, we will  let $n_0=1$, since this can always be achieved,
eventually replacing $T$ by $T^{n_0}$.
\par
Finally, we assume that there exists a family 
of cones in the tangent of $\Cal M$ (not to be
confused with the cones of functions used later). First we have a 
cone family $\Cal C_-(x)\in\Cal T_x\Cal M$, continuous on
int$(\Cal M)$. 
We assume that $\Cal C_-(x)$ contains
the stable, but not the unstable, direction, and that if 
$x\in\partial\Cal M$, then one edge of the cone is tangent to $\partial
\Cal M$. \footnote{This condition amounts to little more than 
transversality of $\partial\Cal M$ with respect to the unstable direction, 
together with uniform transversality of the stable and unstable direction 
at each point at which they are both defined. In fact,
in almost all the known examples the proof of the hyperbolicity of the
system starts exactly with such a cone family.}
\par
We define $\Cal C_+(x)$ to be the complement of $\Cal C_-(x)$. In addition, 
we will consider $\wh{\Cal C}_-(x)=DT^{-l}\Cal C_-(T^lx)$, where $l$ is 
chosen so that the width of $\wh{\Cal C}_-(x)$ is less 
than some $\epsilon\in\Bbb R^+$, and $DT^{-1}\wh{\Cal C}_-(x)\subset
\Cal C_-(T^{-1}x)$ (which implies $DT^{-l-1}\wh{\Cal C}(x)\subset\wh{\Cal C}
(T^{-l-1}x)$; i.e., $\wh{\Cal C}$ is a strictly invariant cone family for
some fixed power of $T$).
 
If the reader wishes to keep in mind a concrete example of the type of 
systems the results of this section apply to, she can consider the
following: $\Cal M=\Bbb T^2$, $T(x,\,y)=F(A(x,\,y))$
where
$$
A=\left({\aligned  1&\quad \ \ a \\
                   a&\quad 1+a^2
         \endaligned}\right)
$$
with $a\not\in\Bbb N$ (which implies that the linear part is discontinuous on
the torus); $T$ is area preserving; 
$F:\Bbb T^2\to\Bbb T^2$ is a $C^2$ function
sufficiently close to the identity so that the hyperbolicity is not spoiled (see
\cite{LW} for a discussion of the structure of the symplectic boxes 
formalism in this case).
 
\proclaim{Definition 5.2} A curve $\gamma$ is said to be ``stable (unstable)"
if it is smooth, and $\gamma'(x)\in\text{int}(\Cal C_-(x) )$
($\gamma'(x)\in\text{int}(\Cal C_+(x) )$ ) for each $x\in\gamma$.
\par
A curve $\gamma$ is said to be ``(strictly) weakly stable" if it is
connected; it consists of finitely many smooth pieces; $\gamma'(x)\in\Cal
C_-(x)$ ($\gamma'(x)\in\wh{\Cal C}_-(x)$) for each $x\in\gamma$ for which
$\gamma'(x)$ exists; and for each unstable curve $\gamma^u$ shorter than
some fixed $\delta_0$, if
$x\in\gamma^u\cap\gamma$, then we have $\gamma\bigcap\gamma^u=\{x\}$.
\endproclaim
 
\line{\bf Cones of Functions\hfil}
 
We will use the same notations of the previous section. The only difference 
is due to the presence of the discontinuities, which forces us to choose
a more complicated cone instead than $\Cal C_{bc}$.
$$
\aligned
\Gamma_\delta=&\{\gamma\subset\Cal M\;|\;\delta\leq\text{length}(\gamma)
\leq2\delta;\;\text{weakly stable};\; |\kappa_\gamma|\leq H\}\\
\wh\Gamma_\delta=&\{\gamma\subset\Cal M\;|\;\delta\leq\text{length}(\gamma)
\leq2\delta;\;\text{strictly weakly stable};\;|\kappa_\gamma|\leq H\}\\
\Cal C_{b,\,c,\,d,\,L}(\delta)=&\bigg\{g\in C^{(0)}\bigg(\Cal M\backslash
\bigcup_{n=0}^\infty T^{n}\Cal S^-\bigg)\bigcap L^1(\Cal M)
\;\bigg|\;\forall \gamma\in 
\Gamma_\delta;\;f,\,f_1,\,f_2\in\Cal D_a(\gamma);\\ 
& x\in \gamma;\; I\subset\gamma\hbox{ connected,}\;\;
\int_\gamma gf> 0;\;\frac{f_2(x)\int_\gamma f_1 g}
{f_1(x)\int_\gamma f_2 g}\leq \e^{b\rho_\gamma(f_1,\,f_2)};\\
& \frac{\|D^u g\|_\infty}{\tb g\tb_-}\leq c;\;\;
\frac{\left|\int_I gf\right|}{f(x)}\leq d\delta^{1-\zeta}|I|^\zeta\tb 
g\tb_-;\;\;\frac{\tb g\tb_+}{\tb g\tb_-}\leq L\bigg\}
\endaligned
$$
 
It was necessary to weaken the regularity requirements on $g$ since,
given $g\in C^{(0)}$, $g\circ T^{-1}$ may be discontinuous; our requirement
is the maximal invariant regularity one can ask for.
In addition, although $g$ may be discontinuous on a dense set, it makes sense
to talk about the derivative in the unstable direction. In fact, 
the local unstable manifold $W^u_{\text{loc}}(x)$ is well defined
at almost every point  
\cite{KS}, and, by definition, enjoys the property 
$W^u_{\text{loc}}(x)\cap\,\cup_{n=0}^\infty T^n\R^-=\emptyset$. Hence, we
can define $D^u_xg$ to be the derivative along $W^u_{\text{loc}}(x)$.
Finally, we have a condition concerning integrals on arbitrarily short curves:
since such pieces are unavoidable, a condition of this type seems 
necessary; the last condition turns out to be needed for technical 
reasons.
\vskip.4cm
\line{\bf Contraction of the Hilbert Metric\hfil}
\vskip.4cm
Of course, our aim is to prove the following:
 
\proclaim{Theorem 5.3} There exists $m\in\Bbb N$, 
$b,\,c,\,d,\,L\in\Bbb R^+$, $\delta_s\in\Bbb R^+$, 
and $\chi\in (0,\,1)$ such
that, for $\delta\leq \delta_s$, 
$$
\wt T^m\Cal C_{b,\,c,\,d,\,L}(\delta)\subset
\Cal C_{\chi b,\,\chi c,\,\chi d,\,\chi L}(\delta).
$$
\endproclaim
Indeed, we shall see that this is sufficient to obtain a contraction of the
Hilbert metric (Lemma 5.17). Nevertheless, the route to Theorem 5.3 
(at least the one found by the author) is not very direct. 
Several intermediate steps shall be needed.
\par
The next Lemma is analogous to Lemma 4.2. It is however less 
satisfactory: it does not provide us with an invariant cone. The condition
that fails pertains the ratio of the averages. Yet, in the previous section
we were able to control effectively this ratio only at the very end, by 
an independent argument. It is then reasonable to content oneself 
temporarily with the 
following result, and to proceed with crossed fingers.
\par
\proclaim{Lemma 5.4}There exists $N(\delta)$, $\lim_{\delta\to 0}
N(\delta)=\infty$ and $A\in\Bbb R^+$, such that, if $L>1$, $8L<d$, $b$
large enough (see 5.8), and 
$2^\nu\delta^\nu a<\ln 2$, then, for each 
$n\in\{A\ln d,\,\dots,\, N(\delta)\}$,
$$
\widetilde T^n\Cal C_{b,\,c,\,d,\,L}(\delta)\subset
\Cal C_{\chi b,\,\chi c,\,\chi d,\,\tau^2_n L}(\delta)
$$
where $\chi<1$, and $\tau^{-1}_n=1-3d\lambda^{-n}M^n\geq 
1/\sqrt 2$.
\endproclaim
\demo{Proof}
Not surprisingly all the difficulties arise because $T^{-n}\gamma$ may 
contain many, eventually very short, connected pieces. The main tool to 
overcome such problems follows:
 
 
\proclaim{Sub-Lemma 5.5}There are $\delta_0\in\Bbb R^+$
and $N:\Bbb R^+\to\Bbb N$, such that, for each $\delta\leq\delta_0$, 
$n\leq N(\delta)$, and
$\gamma\in\Gamma_\delta$, $T^{-n}\gamma$ consists, at most, of
$M^n$ pieces.
\endproclaim
\demo{Proof}
Property (5.2) implies that there exists $\delta_0$ such that
$\R^-$ intersects at most $M$ times a 
segment in the stable direction of length $\delta_0$, and 
$M\lambda^{-\zeta}<1$. Given $\delta\leq\delta_0$ and 
$\gamma\in\Gamma_\delta$, $T^{-1}\gamma$ will consists, at most of $M$
pieces. If $\mu=\|DT^{-1}\|_\infty$, then the length 
of such pieces will be, at most,
$\mu\delta$. The result follows then by choosing $N(\delta)=
\left[\frac{\ln\delta_0-\ln\delta}{\ln \mu}\right]$.
\enddemo
 
The first step consists in comparing $\tb g\tb_-$ and $\tb 
\widetilde T^n g\tb_-$, for $n\leq N(\delta)$.
Given $\gamma\in\Gamma_\delta$, $T^{-n}\gamma$ consists, 
at most, of $M^n$ pieces
$\{\wt \gamma_i\}$. We divide such pieces into two collections. The first,
$\Cal J_-$, will contain all the maximal connected pieces
of length less than $\delta$.
The second collection, $\Cal J_+$, will consist of connected pieces of 
$T^{-n}\gamma$ belonging to $\Gamma_\delta$. In addition, we require
$\left(\bigcup_{\wt\gamma\in\Cal J_-}\wt\gamma\right)\bigcup
\left(\bigcup_{\wt\gamma\in\Cal J_+}\wt\gamma\right)=T^{-n}\gamma$.
For each $\wt\gamma\in\Cal J_-$ we choose some $x_{\wt\gamma}\in\wt\gamma$.
Using such notations we can write
$$
\aligned
\frac 1{\int_\gamma f}\int_\gamma\wt T^n g f=&
\frac 1{\int_\gamma f}\int_{\bigcup\limits_{\wt\gamma\in\Cal J_+}
\wt\gamma}g\wh T^n f+
\frac 1{\int_\gamma f}\int_{\bigcup\limits_{\wt\gamma\in\Cal J_-}\wt\gamma} g 
\wh T^nf\\ 
\geq&\frac 1{\int_\gamma f}\sum_{\wt\gamma\in\Cal J_+}
\int_{\wt\gamma}g\wh T^n f-
\frac 1{\int_\gamma f}\sum_{\wt\gamma\in\Cal J_-}f(T^nx_{\wt\gamma})d 
\lambda^{-n}\delta^{1-\zeta}|\wt\gamma|^\zeta\tb g\tb_-\\
\geq&\left\{\left(1-\sum_{\wt\gamma\in\Cal J_-}
\frac{\int_{\wt\gamma}\wh T^n f}{\int_\gamma f}\right)-
dM^n\lambda^{-n}\e^{a2^\nu\delta^\nu}\right\}\tb g\tb_-\\
\geq& \left\{(1-2\lambda^{-n} M^n)-
2dM^n\lambda^{-n}\right\}\tb g\tb_- \\
\geq& \left(1-3d(M\lambda^{-1})^n\right)\tb g\tb_- .
\endaligned
$$
 
Hence, if $n\geq\frac {2\ln d}{\ln M^{-1}\lambda}>
\frac{\ln 3d}{\ln M^{-1}\lambda}$, we have
$$
\int_\gamma \wt T^ngf\geq 0.
$$
 
In addition, for $A>0$ sufficiently large, given
$n\in\{A\ln d,\,\dots,\,N(\delta)\}$ we get
$$
\tb \wt T^n g\tb_-\geq \tau^{-1}_n\tb g\tb_-\geq\frac 1{\sqrt 2}\tb g\tb_-.
\tag 5.3
$$
 
A similar computation yields
$$
\tb \wt T^n g\tb_+\leq \tau_n\tb g\tb_+.
\tag 5.4
$$
 
We are now ready to verify the remaining conditions.
We start by analyzing integrals over short curves.
 
Let $n\leq N(\delta)$ and consider $I\subset\gamma\in\Gamma_\delta$,
$I$ connected. The image $T^{-n}I$ can be divided into many connected 
pieces that are either in $\Gamma_\delta$ or are shorter than $\delta$;
we call $\Cal J_+$ the collection of the first and $\Cal J_-$ the 
collection of the second. Given $f\in\Cal D_a(\gamma)$ and $x\in I$
$$
\frac{\left|\int_I\widetilde T^n g f\right|}{f(x)}\leq
\frac 1{f(x)}\left\{\sum_{J\in\Cal J_-}\left|\int_J g\widehat T^n f\right|
+\sum_{J\in\Cal J_+}\int_J g\widehat T^n f\right\}.
$$
For each $J\in\Cal J_-$, choose $x_J\in J$
such that $\left|\hbox{det}(D_{x_J}T^n\big|_{J})\right|=
\inf\limits_{z\in J}\left|\hbox{det}(D_{z}T^n\big|_{J})\right|$, then
$$
\aligned
\frac{\left|\int_I\widetilde T^n g f\right|}{f(x)}\leq&
\frac 1{f(x)}\left\{\sum_{J\in\Cal J_-}f(T^n x_J)
\left|\hbox{det}(D_{x_J}T^n\big|_{J})\right|d|J|^\zeta
\delta^{1-\zeta}\tb g\tb_-\right.\\
&\left.
+\sum_{J\in\Cal J_+}\int_J \widehat T^n f\tb g\tb_+\right\}.
\endaligned
$$
Since,
$$
|J|\left|\det(D_{x_J}T^{n}\big|_{J})\right|\leq\int_{T^nJ}
\frac{\left|\det(D_{x_J}T^n\big|_J)\right|}
{\left|\det(D_{T^{-n}y}T^n\big|_J)\right|} dm(y)\leq|T^nJ|,
\tag 5.5
$$
it follows that
$$
\aligned
\frac{\left|\int_I\widetilde T^n g f\right|}{f(x)}\leq&
\left\{\sum_{J\in\Cal J_-} d\frac{f(T^nx_J)}{f(x)}\lambda^{(\zeta-1)n}
|T^nJ|^\zeta\delta^{1-\zeta}+L\sum_{J\in\Cal J_+}
\int_{T^nJ}\frac{f(y)}{f(x)}dm(y)\right\}\\ 
\times&\tb g\tb_-\leq d\e^{a2^\nu\delta^\nu}
\left\{\lambda^{(\zeta-1)n}\delta^{1-\zeta}
\left[\sum_{J\in\Cal J_-}|T^nJ|\right]^\zeta\left[\sum_{J\in\Cal 
J_-}1\right]^{1-\zeta}\right.\\ 
+&\left. Ld^{-1}\left|\bigcup_{J\in\Cal J_+}T^nJ\right|^\zeta
\delta^{1-\zeta}\right\}\tb g\tb_-
\endaligned
$$
which implies:
$$
\frac{\left|\int_I\widetilde T^n g f\right|}{f(x)}
\leq dK_1\delta^{1-\zeta}|I|^\zeta\tb g\tb_-
\tag 5.6
$$
where $K_1$ can be taken to be
$4\max\{(\lambda^{-1}M)^{(1-\zeta)n},\,Ld^{-1}\}$,
provided $\delta$ is sufficiently small.
\par
Choosing $\delta$ small and $A$ large enough, 
it is possible to have $K_1\leq 1/2$, 
provided that
$$
L\leq\frac d8 .
\tag 5.7
$$
Collecting 5.3 and 5.6 we get
$$
\frac{\left|\int_I\widetilde T^n g f\right|}{f(x)}\leq \chi d\delta^{1-\zeta}
|I|^\zeta\tb\wt T^n g\tb_- ,
$$
provided $n\in\{A\ln d,\,\dots,\,N(\delta)\}$.
 
Moreover,
$$
0\leq\frac{\| D^u\wt T^n g\|_\infty}{\tb\wt T^n g\tb_-}\leq\chi c.
$$
 
Finally, when we look at the ratios of different averages, difficulties again
originate from the possible existence of short pieces. The next result 
resolves these difficulties.
 
\proclaim{Sub-Lemma 5.6} For each $\gamma\in\Gamma_\delta$, $f_1,\,f_2\in
\Cal D_a(\gamma)$, $I\subset\gamma$ connected, $|I|<\delta$, $x\in I$, and
$g\in\Cal C_{b,c,d,L}(\delta)$ holds
$$
\frac{\int_I gf_1}{f_1(x)}\leq  \e^{b\rho_\gamma(f_1,\,f_2)}
\frac{\int_I gf_2}{f_2(x)}+8\e^{b\rho_\gamma(f_1,\,f_2)}\delta L
b\rho_\gamma(f_1,\,f_2)\tb g\tb_-.
$$
\endproclaim
 
\topinsert
\vskip3in
\hsize=4.5in
\raggedright
\noindent{\bf Figure 3}  Extending $\gamma$.
\endinsert
 
\demo{Proof}
We introduce a curve $\gamma_e\in\Gamma_\delta$ such that 
$\text{int}(I)\bigcap\text{int}(\gamma_e)=\emptyset$ but
$\ov\gamma=I\bigcup\gamma_e$ is connected and
$\ov\gamma\in\Gamma_\delta$.\footnote{Such a curve always exists.
The only delicate point arises if either one or both of the ends of $\gamma$ 
are closer than 
$\delta$ to $\partial\Cal M$. If one of the edges is close to $\partial M$, 
then we can clearly prolong $\gamma$ at one side, 
provided the curve $\gamma_e$ 
cannot meet a corner of $\partial\Cal M$ in that direction.
In this last case it is easy to convince oneself that $\gamma$ can
be prolonged at the other end, since $\Cal T\partial\Cal M\subset \Cal C_-$.}
In addition, given $f\in\Cal D_a(\gamma)$ we modify it to a function
$\wt f\in\Cal D_a(\ov\gamma)$ by the definition: for each $z\in\gamma_e$
we set $\wt f(z)=f(y)$, where $\{y\}=I\bigcap\gamma_e$ 
is the point in $I$ closest to $z$.
For simplicity, hereafter, we will identify $f$ with $\wt f$, since
this can be done without any ambiguity as long as we are interested 
in integrating on $I$ only.
Also, note that $\rho_{I}(f_1,\,f_2)=\rho_{\ov\gamma}(\wt f_1,\,\wt f_2)
\leq \rho_{\gamma}(f_1,\,f_2)$.
 
$$
\aligned
\frac{\int_I gf_1}{f_1(x)}=& \frac{\int_{\ov\gamma} gf_1}{f_1(x)}
-\frac{\int_{\gamma_e} gf_1}{f_1(x)}\leq
\e^{b\rho_{\ov\gamma}(f_1,\,f_2)}\frac{\int_{\ov\gamma} gf_2}{f_2(x)} 
-\e^{-b\rho_{\ov\gamma}(f_1,\,f_2)}\frac{\int_{\gamma_e} gf_2}{f_2(x)}\\
\leq& \left(\e^{b\rho_\gamma(f_1,\,f_2)}-\e^{-b\rho_\gamma(f_1,\,f_2)}\right) 
\frac{\int_{\gamma_e} gf_2}{f_2(x)}+
\e^{b\rho_\gamma(f_1,\,f_2)}\frac{\int_I gf_2}{f_2(x)}\\ 
\leq&4\delta\left(\e^{b\rho_\gamma(f_1,\,f_2)}-
\e^{-b\rho_\gamma(f_1,\,f_2)}\right)\tb g\tb_+ 
+\e^{b\rho_\gamma(f_1,\,f_2)}\frac{\int_I gf_2}{f_2(x)}\\
\leq&  \e^{b\rho_\gamma(f_1,\,f_2)}\frac{\int_I gf_2}{f_2(x)}
+8\delta\e^{b\rho_\gamma(f_1,\,f_2)}\rho_\gamma(f_1,\,f_2)bL\tb g\tb_-  ,
\endaligned
$$
where we have used the fact that the modified functions are constant on 
$\gamma_e$.
\enddemo
 
\proclaim{Remark 5.7}It is possible to assume the previous relation
in our definition of the cone. This would remove the necessity of being
able to extend a weakly stable curve, and therefore slightly weaken our
conditions on the cone family. However, I do not know of any concrete
example in which such weakening could be useful.
\endproclaim
 
 
Let $\gamma\in\Gamma_\delta$ and $\Cal J$ be
the collection of the connected pieces of $T^{-n}\gamma$. In $\Cal J_-$
we put the pieces shorter than $\delta$, and we collect the others (after
sub-dividing them in pieces belonging to $\Gamma_\delta$) in $\Cal J_+$.
$$
\aligned
\int_\gamma\wt T^ng f_1\leq&\sum_{\wt\gamma\in\Cal J} 
\frac{\wh T^n f_1(x_{\wt\gamma})} {\wh T^n f_2(x_{\wt\gamma})}
\int_{\wt\gamma} g\wh T^nf_2\e^{\xi b\rho_{\gamma}(f_1,\,f_2)}\\
+&\sum_{\wt\gamma\in\Cal J_-}\xi\wh T^n f_1(x_{\wt\gamma})
8\delta\e^{\xi b\rho_\gamma(f_1,\,f_2)}bL\rho_\gamma(f_1,\,f_2)\tb g\tb_-
\endaligned
$$
where we have used the fact that the analogues of 
Lemmas 4.4 and 4.5 hold in this case (providing us with $\xi<1$).
$$
\aligned
\frac{\int_\gamma\wt T^ng f_1}{f_1(x)}\leq
&\bigg[\sum_{\wt\gamma\in\Cal J}  
\frac{f_1(T^nx_{\wt\gamma})f_2(x)} {f_2(T^nx_{\wt\gamma})f_1(x)}
\int_{\wt\gamma}g\wh T^nf_2\\
+&16\lambda^{-n}M^n\delta bL
\rho_\gamma(f_1,\,f_2)\tb \wt T^ng\tb_-\sum_{\wt\gamma\in\Cal J_-}
\frac{f_2(x)f_1(T^nx_{\wt\gamma})}{f_1(x)}\bigg]
\frac{\e^{\xi b\rho_{\gamma}(f_1,\,f_2)}}{f_2(x)}.
\endaligned
$$
Let us call $\wt{\Cal J}_-\subset\Cal J_-$ the set of curves $\wt\gamma$
such that $\int_{\wt\gamma}g\wh T^nf_2<0$, then
$$
\aligned
\frac{\int_\gamma\wt T^ng f_1}{f_1(x)}
\leq& \bigg[\sum_{\wt\gamma\in\Cal J}\e^{\rho_\gamma(f_1,\,f_2)}
\int_{\wt\gamma}g\wh T^n f_2+\sum_{\wt\gamma\in\wt{\Cal J}_-}\left[
\e^{\rho_\gamma(f_1,\,f_2)}-\e^{-\rho_\gamma(f_1,\,f_2)}\right]
\left|\int_{\wt\gamma}g\wh T^n f_2\right|\\
+&16\lambda^{-n}M^n\delta bL\rho_\gamma(f_1,\,f_2)\tb\wt T^n g\tb_-
\frac{f_2(x)f_1(T^nx_{\wt\gamma})}{f_1(x)}\bigg]
\frac{\e^{\xi b\rho_{\gamma}(f_1,\,f_2)}}{f_2(x)}\\
\leq&\left[1+16bd\rho_\gamma(f_1,\,f_2)M^n\lambda^{-n}\right]
\e^{(\xi b+1)\rho_{\gamma}(f_1,\,f_2)}
\frac{\int_{\gamma}\wt T^ng f_2}{f_2(x)}\\ 
\leq&\e^{\left[\xi + b^{-1}+
16M^n\lambda^{-n}d\right]
b\rho_\gamma(f_1,\,f_2) }
\frac{\int_\gamma \wt T^n gf_2}{f_2(x)}\\
\leq&\e^{\chi b\rho_\gamma(f_1,\,f_2)}\frac{\int_\gamma\wt T^ngf_2}{f_2(x)}
\endaligned
$$
where we have set $b$ large enough, and
$$
b^{-1}+\xi+16(M\lambda^{-1})^nd\leq\chi<1.
\tag 5.8
$$
\enddemo 
 
Let us choose $n_*=A \ln d$, and $\delta$ so small that $N(\delta)\geq
2A\ln d$.
\vskip.4cm
\line{\bf Averages on different scales\hfil}
\vskip.4cm
Next, we need estimates that allow us to compare the average of $\wt T^ng$
on the scale $\delta$ with the average of $g$ on a 
fixed scale $\delta_0$; for the time being, the only assumption on $\delta_0$
is $a\in\left(\frac{c_4\delta_0^{1-\nu}}{1-\lambda^{-\nu}},\,
2^{-\nu}\ln 2\delta_0^{-\nu}\right)$.
 
We define
$$
\aligned
\tb g\tb_+^0=&\sup\Sb \gamma\in\wh\Gamma_{\delta_0}\\ 
f\in\Cal D_a(\gamma)\endSb \frac{\int_\gamma gf}{\int_\gamma f}\\
\tb g\tb_-^0=&\inf\Sb \gamma\in\wh\Gamma_{\delta_0}\\ 
f\in\Cal D_a(\gamma)\endSb \frac{\int_\gamma gf}{\int_\gamma f} .
\endaligned
$$
 
\proclaim{Lemma 5.8}  There exists $N_-(\delta)\leq\frac{N(\delta)
\delta^\rho}{16 d\delta_0^\rho}=N_+(\delta)$ such that,
for each $g\in\wt T^{[A\ln d]}\Cal C_{b,c,d,L}(\delta)$, 
and $n\in\{N_-(\delta),\,\dots,\,N_+(\delta)\}$ either
$$
\frac{\tb\wt T^ng\tb_+}{\tb\wt T^ng\tb_-}\leq \frac 12
\frac{\tb g\tb_+}{\tb g\tb_-},
$$
or
$$
\aligned
&\tb \wt T^ng\tb_+\leq 4\tb g\tb_+^0\\
&\tb \wt T^ng\tb_-\geq \frac 12 \tb g\tb_-^0.
\endaligned
$$
\par
\endproclaim
\demo{Proof}
We start by noticing that, although Lemma 5.4 did not provide us with
an invariant cone, we can control a huge number of images.
 
\proclaim{Sub-Lemma 5.9}There exists $\rho\in\Bbb R^+$ such that, given
$n\in\{A\ln d,\,\dots,\,\frac {N(\delta)\delta_0^\rho }
{16d\delta^\rho }\}$,
$$
\wt T^{n}\Cal C_{b,\,c,\,d,\,L}\subset\Cal C_{\chi b,\,\chi c,\,
\chi d,\, 4L}
$$
provided that $32L\leq d$. In addition, for each $g\in\Cal C_{b,c,d,L}$
$$
\aligned
\tb\wt T^ng\tb_+&\leq 2\tb g\tb_+\\
\tb\wt T^ng\tb_-&\geq \frac 12\tb g\tb_-  .
\endaligned
$$
\endproclaim
\demo{Proof}
For each $n\in\{A\ln d,\,\dots,\,\left(\frac{\delta_0}{\delta}\right)^{\rho}
\frac{N(\delta)}{16d}\}$, we write 
$n=\frac{k N(\delta)}2+m$, where we require 
$m\in\{A\ln d,\,\dots,\,N(\delta)\}$.
According to Lemma 5.4 (eq. 5.3, 5.4), for $l\in\{A\ln d,\,\dots,\,
N(\delta)\}$ 
$$\aligned
\tb\wt T^lg\tb_-&\geq (1-3d\lambda^{-l}M^l)\tb g\tb_-
=\tau_l^{-1}\tb g\tb_-\\
\tb\wt T^lg\tb_+&\leq \tau_l\tb g\tb_+
\endaligned
$$
Moreover, $A$ has been chosen such that $\tau^{-1}_{A\ln d}
\geq\frac 1{\sqrt 2}$. Setting $\tau=\tau_{N(\delta)/2}$, 
all we need to check is that
$$
\tau^{2(k-1)}\leq 2.
$$
The result follows, since 
$$
\tau^2\leq\e^{6d\lambda^{-\frac{N(\delta)}2}
M^{\frac{N(\delta)}2}},
$$
and, defining $\rho=\frac{\ln\lambda M^{-1}}{2\ln\mu}$,
$$
\tau^2 \leq\e^{6d\left(\frac\delta{\delta_0}\right)^\rho}
$$
which implies the Lemma.
\enddemo
 
The idea of the proof is to iterate $\gamma\in\Gamma_\delta$ until the
majority of the pieces produced are longer than $\delta_0$. To be more 
precise we will iterate pieces of the images in order to obtain long 
connected pieces,
but we will stop iterating them once they become longer than $\delta_0$.
\par
Let $\gamma_0=\gamma$; let $n_i\in\Bbb N$ be the first integer for which,
given $\gamma_i\subset\gamma$,
$T^{-n_i}\gamma_i$ contains a connected piece longer than $\delta_0$, 
\footnote{It is easy to convince oneself that, if $\gamma_i$ contains
an interval, then $n_i$ exists finite.} we will collect
such long pieces in $\Cal J_i$, after subdividing them in pieces belonging
to $\wh\Gamma_{\delta_0}$.\footnote{Since, $n_1$ must be larger than 
$N(\delta)$, choosing $\delta$ small enough, we can insure $N(\delta)\geq 
l$ (see the definition of $\wh \C_-$ and $\wh \Gamma$ at the beginning of 
the section).}
The sets $\gamma_i\subset\gamma$ are defined
by induction: $\gamma_{i+1}=\gamma_i\backslash
\left(\bigcup_{\wt\gamma\in\Cal J_i}T^{n_i}\wt\gamma\right)$. 
In addition, we will
collect all the connected pieces belonging to $T^{-n_i}\gamma_{i+1}$ (by 
construction such pieces are all shorter than $\delta_0$) into two groups:
$\Cal J^+_i$ will contain the pieces longer than $\delta$, and $\Cal J^-_i$
the shorter pieces; $\wt{\Cal J}_i=\Cal J^+_i\cup\Cal J^-_i$.
\par
By construction $n_{i+1}> n_i$, and $T^{-n_i}\gamma_i$ consists, at
most, of $M^{n_i}$ connected pieces. For each $i_*>0$, and $m\geq n_{i_*}$,
provided $\left(\frac{\delta}{\delta_0}\right)^{-\rho}\frac{N(\delta)}
{16 d}\geq m$,
$$
\aligned
\frac{\int_\gamma\wt T^mgf}{\int_\gamma f}&=\sum_{i=0}^{i_*}
\sum_{\wt\gamma\in\Cal J_i}\frac{\int_{\wt\gamma}\wt T^{m-n_i}g\wh T^{n_i}f}
{\int_\gamma f} +\sum_{\wt\gamma\in\wt{\Cal J}_{i_*}}
\frac{\int_{\wt\gamma}\wt T^{m-n_{i_*}}g\wh T^{n_{i_*}}f}
{\int_\gamma f}\\
&\leq \sup_{n\in\{0,\dots,m\}}\tb \wt T^n g\tb_+^0+
\frac 2{|\gamma|} \tb\wt T^{m-n_{i_*}}g\tb_+ 
\sum_{\wt\gamma\in\Cal J^+_{i_*}}|T^{n_{i_*}}\wt \gamma|\\
&+2 d (\lambda^{-1}M)^{n_{i_*}}\tb\wt T^{m-n_{i_*}}g\tb_-  .
\endaligned
$$
To conclude the argument, we have to choose $i_*$ large 
enough, so that the unwanted terms in the
previous equation are negligible but not beyond the applicability of
Sub-Lemma 5.9 $\left(\hbox{i.e., }n_{i_*}\leq 
\left(\frac{\delta}{\delta_0}\right)^{-\rho}
\frac{N(\delta)}{16 d}\right)$.
\par
Since $n_{i_*}\geq N(\delta)$,
it follows that $2d(\lambda^{-1}M)^{n_{i_*}}\leq\frac 1{8}$, provided
$\delta$ is sufficiently small. 
Let $i_*$ be the first integer $i$ for which
$\sum_{\wt\gamma\in\wt{\Cal J}_i}|T^{n_i}\wt \gamma|=|\gamma_{i+1}|
\leq \frac {|\gamma|}{128L}$. 
This means that $|\gamma_{i_*}|>\frac \delta {128L}$;
in addition, $T^{-k}\gamma_{i_*}$ will consist, at most, of $M^k$ pieces,
shorter than $\delta_0$, for each $k<n_{i_*}$. Let us call $\Cal J_k^*$ the
collection of such pieces. 
$$
\frac\delta{128 L}\leq\sum_{\wt\gamma\in{\Cal J}^*_k}\frac{|T^k\wt\gamma|} 
{|\wt\gamma|} |\wt\gamma|\leq
\sum_{\wt\gamma\in{\Cal J}^*_k}\frac{|T^k\wt\gamma|} 
{|\wt\gamma|}\delta_0\leq\left(\lambda^{-1}M\right)^k\delta_0. 
$$
The previous computation shows that, setting $k_0=\frac{\ln (128L\delta_0
\delta^{-1})}{\ln\lambda M^{-1}}$, $n_{i_*}\leq k_0+1$.
\par
Using our choice of $i_*$ and the previous estimates, we conclude that,
for each $m\in\{k_0+1,\,\dots,\,N_+(\delta)\}$,
$$
\frac{\int_\gamma\wt T^mgf}{\int_\gamma f}\leq
\sup_{n\in\{0,\dots,m\}}\tb \wt T^n g\tb_+^0
+\frac 14\tb\wt T^m g\tb_- .
$$
In the same manner one can prove:
$$
\frac{\int_\gamma\wt T^mgf}{\int_\gamma f}\geq  (1-2\lambda^{-1}M)
\inf_{n\in\{0,\dots,m\}}\tb \wt T^n g\tb_-^0
-\frac 14\tb \wt T^m g\tb_-   .
$$
\par
To obtain a useful inequality it is necessary to estimate $\tb \wt T^n 
g\tb_\pm^0$. This is done by the following.
 
\proclaim{Sub-Lemma 5.10}
For each $g\in\wt T^{[A\ln d]}\Cal C_{b,c,d,L}(\delta)$, holds
$$
\aligned
\tb\wt T^n g\tb_+^0&\leq\tb g\tb_+^0+\frac 18\tb g\tb_+\\
\tb\wt T^n g\tb_-^0&\geq\frac 34\tb g\tb_-^0 ,
\endaligned
$$
for each $n\leq N_+(\delta)$.
\endproclaim
\demo{Proof}
The proof is by induction. Let $\gamma\in\wh\Gamma_{\delta_0}$, 
if $\Cal J_1$ contains the connected pieces of  
$T^{-1}\gamma$ longer than $\delta_0$, $\Cal J_2$ the one shorter than 
$\delta_0$ but longer than $\delta$, $\Cal J_3$ the shorter pieces, and 
$x_{\widetilde\gamma}\in\widetilde\gamma$ is an arbitrary point, then
$$
\aligned
\frac{\int_\gamma\wt Tgf}{\int_\gamma f}&\leq
\sum_{\wt\gamma\in\Cal J_1}
\frac{\int_\gamma g\wh Tf}{\int_\gamma f}
+\sum_{\wt\gamma\in\Cal J_2}
\frac{\int_{\wt \gamma} \wh Tf}{\int_\gamma f}\tb g\tb_+
+\sum_{\wt\gamma\in\Cal J_3}
\frac{\wh Tf(x_{\wt\gamma})}{\int_\gamma f}\delta d\tb g\tb_-\\
&\leq\tb g\tb_+^0+(\lambda^{-1}M)\left[2+2d\frac{\delta}{\delta_0}\right]
\tb g\tb_+\leq\tb g\tb_+^0+3(\lambda^{-1}M)\tb g\tb_+
\endaligned
$$
provided $\delta$ is sufficiently small.
\par
Our induction hypothesis is
$$
\tb\wt T^ng\tb_+^0\leq\tb g\tb_+^0+3\tb g\tb_+\sum_{i=1}^n(\lambda^{-1}M)^i .
$$
We use a construction similar to the one used previously. We collect
in $\Cal J_1$ all the pieces of $T^{-1}\gamma$ longer than $\delta_0$,
in $\Cal J_2$ all the pieces $\wt\gamma$ of $T^{-2}\gamma$ longer than 
$\delta_0$, provided $T\wt\gamma$  was not contained in an element of
the collection $\Cal J_1$, and so on. The connected pieces $\wt\gamma$ of
$T^{-n}\gamma$ that are shorter than $\delta_0$, and such that 
$T^i\wt\gamma$, $i\in\{1,\,\dots,\,n\}$, was always contained in maximal 
connected  pieces shorter than $\delta_0$, are collected in two sets:
$\Cal J_+$ contains the pieces longer than $\delta$, and $\Cal J_-$
the shorter ones.
$$
\aligned
\frac{\int_\gamma\wt T^{n+1}gf}{\int_\gamma f}&\leq
\sum_{i=0}^n\sum_{\wt\gamma\in\Cal J_{i+1}}
\frac{\int_{\wt \gamma}\wt T^{n-i}g\wh T^{i+1}f}{\int_\gamma f}+
\sum_{\wt\gamma\in\Cal J_+}
\frac{\int_{\wt\gamma} g\wh T^{n+1}f}{\int_\gamma f}+
\frac{2(M\lambda^{-1})^{n+1}d\delta}{\delta_0}\tb g\tb_-\\
&\leq\sum_{i=0}^n\sum_{\wt\gamma\in\Cal J_{i+1}}
\frac{\int_{\wt\gamma}\wh T^{i+1}f}{\int_\gamma f}\left\{\tb g\tb_+^0+
3\tb g\tb_+\sum_{j=1}^n(\lambda^{-1}M)^j\right\}\\
&+3(\lambda^{-1}M)^{n+1}\tb g\tb_+
\leq \tb g\tb_+^0+3\tb g\tb_+\sum_{i=1}^{n+1}(\lambda^{-1}M)^i
\endaligned
$$
which implies
$$
\tb\wt T^n g\tb_+^0\leq\tb 
g\tb_+^0+\frac{3(\lambda^{-1}M)}{1-\lambda^{-1}M}\tb g\tb_+
\leq\tb g\tb_+^0+\frac 18\tb g\tb_+  .
$$
The second inequality is proven in a similar way. In fact,
$$
\aligned
\frac{\int_\gamma \wt Tg f}{\int_\gamma f}&\geq\sum_{\wt\gamma\in\Cal J_1}
\frac{\int_\gamma g \wh Tf}{\int_\gamma f} -2d\lambda^{-1}M
\frac\delta{\delta_0}\tb g\tb_-
\geq (1-2\lambda^{-1}M)\tb g\tb_+^0-\frac{M\lambda^{-1}}2\tb g\tb_-\\
&\geq (1-2\lambda^{-1}M)\tb g\tb_+^0-M\lambda^{-1}\tb \wt T g\tb_- ,
\endaligned
$$
since, for each function $\wt g\in\C_{b,c,d,L}$, $\tb \wt g\tb_-\leq
\tb\wt g\tb_-^0$, we have
$$
\frac{\int_\gamma \wt Tg f}{\int_\gamma f}\geq
\frac{1-2\lambda^{-1}M}{1+\lambda^{-1}M }\tb g\tb_+^0  .
$$
This time, we will use the following induction hypothesis
$$
\tb\wt T^ng\tb_-^0\geq h_n\tb g\tb_-^0
$$
where $h_1= \frac{1-2\lambda^{-1}M}{1+\lambda^{-1}M }$ and
$h_{n+1}=\frac{1-2(\lambda^{-1}M)^{n+1}}{1+(\lambda^{-1}M)^{n+1}}h_n$.
$$
\frac{\int_\gamma \wt T^{n+1}gf}{\int_\gamma f}\geq
(1-2(\lambda^{-1}M)^{n+1})h_n\tb g\tb_-^0-\frac{(\lambda^{-1}M)^{n+1}}2
\tb g\tb_- .
$$
This concludes the argument, since 
$$
h_n=\prod_{i=1}^n \frac{1-2(\lambda^{-1}M)^i}{1+(\lambda^{-1}M)^i}\geq
\e^{-4\sum_{i=1}^n(\lambda^{-1}M)^i}\geq \frac 34 .
$$
\enddemo
>From the previous Lemma we obtain immediately
$$
\frac 54\tb\wt T^n g\tb_-\geq(1-\frac 2{25})\frac 34\tb g\tb_-^0
$$
which implies the last inequality in the statement of the Lemma.
 
To conclude the argument let us consider two cases. First suppose
that
$$
\tb\wt T^m g\tb_+\leq\frac 14\tb g\tb_+ .
$$
This would immediately imply the first possibility in the statement.
Second, suppose that
$$
\tb\wt T^m g\tb_+\geq\frac 14\tb g\tb_+.
$$
The above inequality, together with the inequalities obtained before, yields
$$
\tb\wt T^m g\tb_+\leq\tb g\tb_+^0+ \frac34\tb \wt T^m g\tb_+ ,
$$
which conclude the proof.
\enddemo
 
\vskip.4cm
\line{\bf Comparison of averages \hfil}
\vskip.4cm
\par
The only task left is to compare the value of the integral on different
$\gamma$'s. Due to the presence of the singularities, this is much more
complicated that in the smooth case. Nonetheless, the basic idea is the same.
The only difference is that we will break up a curve in 
$\wh \Gamma_{\delta_0}$  into pieces belonging to $\wh \Gamma_{\delta_1}$,
with $\delta_0\geq\delta_1\geq\delta$, and we will compare such pieces.
The essential feature of this decomposition is that $\delta_1$ can be
chosen smaller than $\delta_0$, depending on the parameters
of the cone.
\par
To carry out this plan we need to introduce appropriate neighborhoods of
$\Cal S^-$.
 
\proclaim{Definition 5.11}
$$
\aligned
\Cal S^-_\varepsilon &=\{z\in \Cal M\;|\;\text{ dist}(z,\,\Cal S^-)
\leq \varepsilon\} \\
\Sigma_n &=\bigcup_{i=0}^n T^i\left(\Cal S^-_{\lambda^{-i}\delta_1}\right).
\endaligned
$$
Moreover, we will say that two curves 
$\gamma_1,\,\gamma_2\in\wh \Gamma_\varepsilon$
are the translate of each other in the unstable direction 
if each sufficiently long unstable manifold intersecting one curve 
intersects the other as well.
\endproclaim
 
We are now able to state our main comparison Lemma.
 
\proclaim{Lemma 5.12} There exists $m_*\in\Bbb N$ such that, 
if two curves $\gamma_1,\,
\gamma_2\in\wh \Gamma_{\delta_1}$  are the translate of each other 
in the unstable direction, their distance is less than $\delta_1/2$, and 
$\gamma_1\bigcap\Sigma_{m_*}=\emptyset$, then
$$
\frac 12\int_{\gamma_2}g-\frac {\e^{-bR}}{256}\tb g\tb_-
\delta_1\leq\int_{\gamma_1} g
\leq  2\int_{\gamma_2}g+\frac {\e^{-bR}}{256}\tb g\tb_-\delta_1
$$
\endproclaim
\demo{Proof}
The Lemma is a consequence of the results stated in Appendix II.
$$
\aligned
\int_{\gamma_1}g&=\int_{D(\phi)}g+\int_{\gamma_1\backslash D(\phi)}g\leq
\int_{R(\phi)}g\circ\phi^{-1}J_\phi+ D\delta_1(\lambda^{-\zeta}
\mu^{1-\zeta})^{m_*}\tb g\tb_-\\
&\leq \int_{R(\phi)}gJ_\phi+\|D^u g\|_\infty\frac{\delta_1}2 
\int_{R(\phi)}J_\phi+
D\delta_1(\lambda^{-\zeta}
\mu^{1-\zeta})^{m_*}\tb g\tb_-
\endaligned
$$
Here we have used Lemma II.2.
To use effectively the properties of $g$ it is necessary for the 
domain of integration to be connected, while $R(\phi)$ it is not. But,
from Lemma II.5 and II.6 it follows that both $\phi$ and $J_\phi$
can be extended to continuous functions defined on all $\gamma_2$, 
\footnote{For example, one can take the extension to be linear where
the functions are not defined.} and the 
extensions satisfies 
the same bounds (i.e., the extension of $\phi$ is Liptchitz and the
extension of $J_\phi$ is H\"older continuous). 
Let us use a tilde to identify such extensions.
$$
\int_{\gamma_1}g
\leq\int_{\gamma_2}g\wt J_\phi+ c\tb 
g\tb_-\delta_1^2
+2D\delta_1(\lambda^{-\zeta}
\mu^{1-\zeta})^{m_*}\tb g\tb_- .
$$
Lemma II.6 shows that for $a$ large and $\delta_1$ small enough, 
$\wt J_\phi\in\Cal D_a(\gamma_2)$, therefore
$$
\int_{\gamma_1}g\leq 
\e^{b\rho_{\gamma_2}(1,\,\wt J_\phi)}
\int_{\gamma_2}g+\left( 2D(\lambda^{-\zeta}
\mu^{1-\zeta})^{m_*}+c\delta_1 \right)\tb g\tb_-  \delta_1 .
$$
The result follows by choosing $\delta_1$ sufficiently small, and 
$m_*$ sufficiently large, with respect to $b$ and $c$.
\enddemo
To conclude this part of the argument, as we did in \S 4, we need to
construct an appropriate covering of $\Cal M$. We will use such 
covering to control  the images of a given curve.
\par 
We start with the same collection of squares $Q(z_i)$ constructed in
\S 4--``Diameter of the image".\footnote{The presence of singularities 
restrict the possible points that can be chosen as the center of a square:
only points with well defined stable and unstable direction are allowed.
Nevertheless, almost all points are available, and the squares can still
be chosen to be  of uniform size.}
As in the previous section given any square we have 
a notion of ``core," although a more stringent one. In the
previous section was called ``core of a square" a concentrical square of 
size $\frac 14$, here we will call  ``weak-core" a concentrical square
of size $\frac 1{4t}$ ($t>1$ will be chosen in Sub-Lemma 5.16), while
by ``core" of a square of size $t\varepsilon$, we mean  
the set of points in the weak-core that have a stable
and unstable manifold longer than $4\varepsilon t$. In addition, we will call
a square ``$\alpha$--connecting" (for some $\alpha<1$) 
if the measure of its core is at least 
$\alpha$ times the measure of the square.
\footnote{The definition of core differs from
the one given in \cite{LW, Definition 11.1}: there the requirement on the
size of the manifolds is absent. Nevertheless, the definition of 
$\alpha$-connecting  is essentially the same.}
\par
The key ingredient in the construction of our covering is a result very  
similar to the
``Sinai Theorem," which is the cornerstone of the proof of
ergodicity for the systems under consideration. In fact, the proof of 
the following theorem is almost indistinguishable 
from the proof of Sinai Theorem in the
simple uniformly hyperbolic case
(I refer to the version in \cite{LW, \S 3}). However, 
since the statement looks  stronger, a sketch 
of a proof is provided in Appendix I. 
 
\proclaim{Sinai Theorem 5.13}
For each square $Q$ of size $\delta_0$,and $t>1$, there exist  $\alpha_Q<1$,
$\delta_Q\leq\delta_0$, and $m_Q\in\Bbb N$   
such that, for each $\alpha\leq\alpha_Q$, 
$m\geq m_Q$, and $\delta_1\leq\delta_Q$  we can construct
a covering $\Cal G(Q,\,\delta_1,\,m)$ made of squares of size $t\delta_1$ 
with the following properties:
\roster
\item "i)" the number of squares $G$ for which 
           $\bigcup\limits_{|n|\leq m_Q}T^n\R\cap G=\emptyset$, and       
           $G$ is not $\alpha$-connecting, is less than
           $\frac 1{256} K_Q^{-1}$ ($K_Q$ being the maximal number of 
           overlaps in the covering);
\item "ii)" for each $\gamma\in\wh\Gamma_{\delta_1}$, $\gamma\subset Q$,
there exists $G\in\Cal G(Q,\,\delta_1,\,m)$ 
such that $\gamma\subset G$ and $\gamma$ intersects properly (see \S 4) $G$.
\endroster
\endproclaim
 
In other words, a square $Q$ can be covered by the weak--core of
sufficiently small squares whose overwhelming 
majority contains an abundance of stable fibers. 
\par
According to Theorem 5.13, for every square $Q\in\{Q(z_i)\}$ 
we can construct coverings $\Cal G(Q,\,\delta_1,\,m)$. 
Let us choose $m_0\geq\max\{\{m_Q\}_{Q\in\Cal Q},\,m_*\}$, 
and $\delta_1\leq\min\left\{
\{\delta_Q\}_{Q\in\Cal Q},\,\mu^{-m_0}\delta_0\right\}$, and
$\alpha=\min\{\alpha_Q\}$,
then $\Cal G=\cup_{Q\in\Cal Q}\Cal G(Q,\,\delta_1,
\,m_0)$ is the covering of $\Cal M$ that will be used in the following. 
 
We are now positioned to make the last step: estimate the ratio of the 
averages; as in \S 4, a new element, the mixing property, allows us to 
close the argument.
 
\proclaim{Lemma 5.14}There exists $B\in \Bbb R$ and $N\in\Bbb N$ 
such that, for $m\in\{N_-(\delta),\,\dots,$ $N_+(\delta)\}$, 
$$
\tb \wt T^mg\tb_+\leq B\mu^N\tb\wt T^mg\tb_- .
$$
\endproclaim
\demo{Proof}
We  have assumed the map to be
mixing. Hence, there exists $N\in\Bbb N$ such that, given any 
two $\alpha$-connecting  $G,\,G'\in\Cal G$ 
$$
T^{-N}(\text{core}(G))\cap \text{core}(G')\neq\emptyset .
$$
 
The strategy of the proof is as follows: first we will show that it is 
possible to control the average on any curve of size $\delta_0$
via the average on a shorter 
curve contained in a connecting square (Sub-Lemma 5.15). Then we will see 
that the mixing property implies that a connected piece belonging 
to some image of such shorter
curve is $\delta_1$-close to any given curve. We will then be in a position
to compare any two curves.
 
 
\proclaim{Sub-Lemma 5.15}For each $\delta\leq\delta_1$ 
small enough,
$g\in\Cal C_{b,c,d,4L}(\delta)$, $\gamma\in\wh\Gamma_{\delta_0}$, 
$f\in\Cal D_a(\gamma)$, there exists $\gamma^+,\,\gamma^-\in
\wh \Gamma_{\delta_1}$ $\left(\gamma^+,\,\gamma^-\subset\gamma\right)$, 
such that
$$
\frac 1 2\frac{\int_{\gamma^-}fg}{\int_{\gamma^-}f}\leq
\frac{\int_{\gamma}fg}{\int_{\gamma}f}\leq
\frac{\int_{\gamma^+}fg}{\int_{\gamma^+}f} +\frac 1{64}\tb g\tb_+ .
$$
In addition, both $\gamma^+$ and $\gamma^-$ intersect properly an
$\alpha$--connecting squares belonging to $\Cal G$, and are
disjoint from $\bigcup\limits_{|n|\leq m_0}T^n\R$.
\endproclaim
\demo{Proof}
Consider the set $\gamma\backslash(\bigcup\limits_{|n|\leq m_0}T^n\R)$. 
Since $\delta_1\leq \delta_0\mu^{-m_0}$, it
consists of, at most, $2M^{m_0}$ connected curves. 
We collect in $\Cal J_1$ the ones that are longer than $\beta\delta_1$; in
$\Cal J_2$ we collect the curves shorter than $\beta\delta_1$, but longer 
than
$\delta$; finally in $\Cal J_3$ we collect the shorter ones.
$$
\frac{\int_\gamma fg}{\int_\gamma f}\leq
\sum_{\wt\gamma\in\Cal J_1}\frac{\int_{\wt\gamma} fg}
{\int_\gamma f}+
\sum_{\wt\gamma\in\Cal J_2}\frac{\int_{\wt\gamma} f}
{\int_\gamma f}\tb g\tb_+ +
\sum_{\wt\gamma\in\Cal J_3}\frac{f(x_{\wt\gamma})}
{\int_\gamma f}d\delta\tb g\tb_- .
$$
Next we split $\Cal J_1$ into three parts: the curves properly contained in a 
connecting square, collected in $\Cal J_0$, after
subdividing them into curves belonging to $\wh \Gamma_{\delta_1}$; 
the curves closest to
$\bigcup\limits_{|n|\leq m_0}T^n\R$, collected in $\Cal J_4$;
and the rest, collected in $\Cal J_5$. Since we have assumed the uniform
transversality between $\R^\pm$ and the stable and unstable directions,
and since the direction of $\gamma'$, for $\gamma\in\wh\Gamma_\delta$, may 
differ from the stable direction by, at most, 
some fixed arbitrary $\epsilon$, we can assume, without
loss of generality, that the curves in $\wh \Gamma_\delta$ are uniformly
transversal to $\R^\pm$. We can therefore choose $\beta$ so large that,
if $z\in\wt\gamma\in\wh\Gamma_\delta$ and the distance of $z$ from 
$\R^\pm$, along $\gamma$, is larger than $\beta$, than $d(z,\,\R^\pm)\geq
t\delta_1$.
Hence, by construction, the curves in $\Cal J_5$ 
belong to squares that do not intersect $\bigcup\limits_{|n|\leq 
m_0}T^n\R$. The Sinai Theorem implies that 
the total length of the curves in $\Cal J_5$ must be less than 
$\frac{\delta_0}{256}$.
Among the curves in $\Cal J_0$ there exists one on which the average is
larger, let us call it $\gamma^+$, then
$$
\frac{\int_\gamma fg}{\int_\gamma f}\leq 
\frac{\int_{\gamma^+} fg}{\int_{\gamma^+} f}+
2\left(\frac 1{256}+4\beta M^{m_0}\frac{\delta_1}{\delta_0}\right)
\tb g\tb_++4dM^{m_0}\frac{\delta}{\delta_0}  \tb g\tb_-.
$$
Similarly, we have
$$
\frac{\int_\gamma fg}{\int_\gamma f}\geq\sum_{\wt\gamma\in\Cal J_0}
\frac{\int_{\wt\gamma} fg}{\int_{\wt\gamma} f}-
\sum_{\wt\gamma\in\Cal J_3}\frac{\left|\int_{\wt\gamma}gf\right|}
{\int_{\wt\gamma}f}.
$$
Let $\gamma^-$ be the curve in $\Cal J_0$ 
on which the average is smaller, then
$$
\frac{\int_\gamma fg}{\int_\gamma f}\geq
\frac 34\frac{\int_{\gamma^-} fg}{\int_{\gamma^-} f}-
4dM^{m_0}\frac{\delta}{\delta_0}  \tb g\tb_-\geq
\frac 34\frac{\int_{\gamma^-} fg}{\int_{\gamma^-} f}
-4dM^{m_0}\frac{\delta}{\delta_0} \frac{\int_\gamma fg}{\int_\gamma f}
$$
The Lemma follows by choosing $\delta$ and $\delta_1$ sufficiently small.
\enddemo
Given $g\in\Cal C_{b,c,d,4L}(\delta)$, 
choose $\gamma^*\in\wh\Gamma_{\delta_0}$,
and $f_*\in\Cal D_a(\gamma^*)$,
such that
$$
\tb g\tb_+^0\leq 2\frac{\int_{\gamma^*}g f_*}{\int_{\gamma^*} f_*} .
$$
Then, there exists $\gamma^*_+\in\wh\Gamma_{\delta_1}$, and an 
$\alpha$-connecting square $G_*\in\Cal G$ containing it, such that
$$
\tb g\tb_+^0\leq 2 \frac{\int_{\gamma^*_+} gf_*}{\int_{\gamma^*_+} f_*}+ 
\frac 1{32}\tb g\tb_+ .
$$
Furthermore, for each $g\in\Cal C_{b,c,d,4L}(\delta)$, $\gamma\in
\wh\Gamma_{\delta_0}$, and $f\in\Cal D_a(\gamma)$, there exists 
$\gamma_-\in\wh\Gamma_{\delta_1}$, and an $\alpha$-connecting square 
$G_\gamma\in\Cal G$ such that $\gamma_-\subset G_\gamma$, and
$$
\frac{\int_\gamma fg}{\int_\gamma f}\geq
\frac 12\frac{\int_{\gamma_-} fg}{\int_{\gamma_-} f}.
$$
We can finally conclude our argument.
 
 
\proclaim{Sub-Lemma 5.16} $T^{-N}\gamma_-$ contains a translate of
$\gamma^*_+$.
\endproclaim
\demo{Proof}
Let $w\in \text{core}(G_\gamma)\cap T^N\text{core}(G_*)$, then, by
definition, there exists a stable manifold $W^s$ and an unstable manifold
$W^u$ of size $a\delta_1 t$ (consequently intersecting $G_\gamma$ completely),
and $W^s\cap W^u=\{w\}$. This implies $W^u\cap\gamma_-\neq\emptyset$.
Let $y=W^u\cap\gamma_-$, and $\wt\gamma_n$ the connected component of
$T^{-n}\gamma_-\cap G_*$ containing $T^{-n}y$.
 
Since $T^{-n}W^u\cap \R^-=\emptyset$, and since $\R^-$ is uniformly 
transversal to $W^u$, there exits $\theta\in\Bbb R^+$ such that, for each
$z\in T^{-n}W^u$,
$$
d(z,\,\R^-)\geq \theta\min\{d(z,\,T^{-n}w_1),\,d(z,\,T^{-n}w_2)\}
$$
where $\{w_1,\,w_2\}=\partial(W^u\cap G_\gamma)$.
Let $w_1$ be the point of  $\partial(W^u\cap G_\gamma)$ further away from
$y$, and $\wt w_1$ the point of $\partial(W^s(T^{-N}w)\cap G_*)$ further 
away from $T^{-N}y$. Then, setting $W^s_1=W^s(T^{-N}w)\cap G_*$,
$$
\aligned
d(T^{-n}y,\,\R^-)\geq& \theta d(T^{-n}y,\,T^{-n}w_1)\\
d(T^{-N+n}y,\,\R^-)\geq & d(T^{-N+n}w,\,T^n\wt w_1)-\theta 
d(\wt\gamma_{N-n},\,T^nW^s_1).
\endaligned
$$
Next, we choose $n_*\in\{0,\,\dots,\,N\}$ such that, for the first time,
$$
d(T^{-n_*}w,\,T^{N-n_*}\wt w_1)\geq 4 d(\wt\gamma_{n_*},\,T^{N-n_*}W^s_1).
$$
Then, for each $n\leq n_*$,
$$
|\wt\gamma_n|\geq\theta d(T^{-n}y,\,T^{-n}w_1).
$$
 
In fact, the latter relation is certainly true for $n=0$, now suppose that 
it is true for $n-1$ and let $\wh \gamma_n$ be the connected component of
$\wt\gamma_{n-1}\cap\R^-$ containing $T^{-n}y$, then
$$
|\wh\gamma_n|\geq\theta d(T^{-n+1}y,\,T^{-n+1}w_1)
$$
and 
$$
|\wt\gamma_n|=|T^{-1}\wh\gamma_n|\geq\lambda^2\theta d(T^{-n}y,\, 
T^{-n}w_1).
$$
Accordingly,
$$
\aligned
|\wh\gamma_{n_*}|&\geq\theta d(T^{-n_*+1}y,\,T^{-n_*+1}w_1)\geq
\theta d(\wt\gamma_{n_*-1},\,T^{N-n_*+1}W^s_1)\\
&\geq\frac \theta 4 d(T^{-n_*+1}w,\,T^{N-n_*+1}\wt w_1)
\endaligned
$$
and
$$
|\wt\gamma_{n_*}|\geq\frac\theta 4\e^{-c_4\lambda^{-n_*}\delta_1 t}
d(T^{-n_*}w,\,T^{N-n_*}\wt w_1).
$$
where we have used the usual distortion estimate and $c_4$ is some fixed 
constant.
>From our bounds on the distance between the images of $y$ and $\R^-$ 
follows that $\wt\gamma_n\cap\R^-=\emptyset$, hence
$$
|\wt\gamma_n|\geq\frac\theta 4\e^{-c_4\lambda^{-n_*}\sum_{i=0}^{n-n_*}
\lambda^i\delta_1 t}d(T^{-n}w,\,T^{N-n}\wt w_1)
$$
and
$$
|\wt\gamma_N|\geq\frac\theta 4\e^{-c_4\delta_1 t}\delta_1 t\geq 6\delta_1
$$
where we have chosen $t$ sufficiently large ($t\geq 52\theta$) and 
$\delta_1$ sufficiently small ($\e^{-c_4\delta_1 t}\geq\frac 12$). Moreover, 
from the argument is clear that dividing $\wt\gamma_N$ at $T^{-N}y$ each 
of the two pieces is long, at least, $3\delta_1$. From this follows that 
$\wt\gamma_N$ must contain a translate of $\gamma_+^*$. 
\enddemo
 
Let $\gamma_0\subset T^{-N}\gamma_-$ be a translate of $\gamma_*^+$, 
then
$$
\aligned
\frac{\int_\gamma f\wt T^{N}g}{\int_\gamma f}&\geq
\frac 12\frac{\int_{\gamma^-}\wt T^{N}gf}{\int_{\gamma^-}f}\geq
\frac 12\frac{\int_{\gamma_0}g\wh T^{N}f}{\int_{\gamma^-}f}
\geq\frac {\e^{-bR}}{8\delta_1}\mu^{-N}\int_{\gamma_0} g\\
&\geq\frac{\e^{-bR}}{16\delta_1}\mu^{-N}\int_{\gamma^+_*} g
-\frac{\e^{-2bR}\mu^{-N}}{2048\delta_1}\tb g\tb_+\\
&\geq\frac{\e^{-2bR}}{32}\mu^{-N}
\frac{\int_{\gamma_+^*} g f_*}{\int_{\gamma_+^*}f_*}
-\frac{\e^{-2bR}\mu^{-N}}{2048\delta_1}\tb g\tb_+\\
&\geq\frac{\e^{-2bR}}{64}\mu^{-N}\left(\tb g\tb^0_+-
\frac 1{16}\tb g\tb_+ \right)
\geq  \frac {\e^{-2bR}}{64}\mu^{-N}
\left(\tb g\tb^0_+-\frac L{16}\tb g\tb_- \right) .
\endaligned
$$
 
Accordingly, 
$$
\tb\wt T^{N}g\tb_-^0\geq \frac{\e^{-2bR}\mu^{-N}}{64}
\left(\tb g\tb_+^0
-\frac L{16}\tb g\tb_-\right) .
$$
Lemma 5.8 implies that there exists  
$m\in\{N_-(\delta),\,\dots,\,N_+(\delta)\}$  such that,
calling $\wt g=\wt T^{[A\ln d]}g$, we have
$$
\tb \wt T^{m+N}\wt g\tb_-\geq\frac 12\tb\wt T^N\wt g\tb_-^0\geq
\frac{\e^{-2bR}\mu^{-N}}{128}\left(\tb\wt g\tb_+^0-
\frac L{16}\tb\wt g\tb_-\right).
$$
If $\tb\wt g\tb_+^0\leq\frac{3L}{32}\tb\wt g\tb_-$, then 
$$
\tb\wt T^m\wt g\tb_+\leq 4\tb\wt g\tb_+^0\leq\frac{3L}8\tb\wt g\tb_-
\leq\frac{3L}4\tb\wt T^m\wt g\tb_- ;
$$
otherwise
$$
\tb\wt T^{m+N}\wt g\tb_-\geq
\frac{\e^{-2bR}\mu^{-N}}{128\times12}\tb\wt T^{m+N}\wt g\tb_+ .
$$
\enddemo
 
\vskip.4cm
\line{\bf Decay of correlations\hfil}
\vskip.4cm
Lemma 5.14 was the last piece in the proof of Theorem 5.3: 
it suffices to choose
$$
L>1536 \chi^{-1}\mu^{N}\e^{2bR},
$$ 
and 
$$
m=A\ln d+N_-(\delta)+N,
$$
in order to close the argument.\footnote{At this point the reader may 
wonder if such 
a choice is possible. In fact, $N$ depends on $\delta_1$, but not on 
$\delta$. In turn, the choice of $\delta_1$ depends on $c$ and $b$, but not 
on $d$ and $L$. It is therefore possible to choose al the parameters 
involved but $L$, $d$ and $\delta$. The next relation forces then a choice 
of $L$ (and, consequently, of $d$), nonetheless we are still free to choose 
$\delta$ sufficiently small to satisfy all the necessary requirements.}
\par
We announced at the beginning of the section that Theorem 5.3 was sufficient
for our purposes; this is shown by the next Lemma.
\proclaim{Lemma 5.17} There exists $\Delta\in\Bbb  R^+$:
$$
\text{diam}(\Cal C_{\chi b,\,\chi c,\,\chi d,\,\chi L}(\delta))\leq\Delta
$$
where the diameter in computed with respect to the Hilbert metric $\Theta$
of $\Cal C_{b,\,c,\,d,\,L}(\delta)$.
\endproclaim
\demo{Proof}
This Lemma is similar to Lemma 4.5, only more computations
are needed since we are now dealing with a more complex geometry.
 
Let $g\in\Cal C_{\chi b,\,\chi b,\,\chi d,\,\chi L}$, if $\lambda\preceq 
g\preceq\mu$, then, in analogy with Lemma 4.5, we have
 
$$
\aligned
\lambda\leq& \inf\Sb\gamma\in\Gamma_\delta\\ f\in\Cal D_a(\gamma)\endSb 
\frac{\int_\gamma gf}{\int_\gamma f}=\alpha_0\\
\lambda\leq& \inf\Sb\gamma\in\Gamma_\delta\\ f_1,\,f_2\in\Cal 
D_a(\gamma)\endSb 
\frac{\e^{b\rho_\gamma(f_1,\,f_2)}f_1\int_\gamma g f_2-f_2\int_\gamma g 
f_1}
{\e^{b\rho_\gamma(f_1,\,f_2)}f_1\int_\gamma f_2-f_2\int_\gamma 
f_1}=\alpha_1\\
\lambda\leq&\frac{c\tb g\tb_--\|D^u g\|_\infty}c=\alpha_2
\endaligned
$$
 
In addition, in the present context, we have two extra conditions defining 
the cone that produce the following constraints:
 
$$
\aligned
\lambda\leq&\inf\frac{d\delta^{1-\zeta}|I|^\zeta\tb g\tb_--
\frac{|\int_I gf|}{f(x)}}{d\delta^{1-\zeta}|I|^\zeta-\frac{\int_I f}{f(x)}}
=\alpha_3\\
\text{or }&\\
\lambda\leq&\inf\frac{d\delta^{1-\zeta}|I|^\zeta\tb g\tb_-+
\frac{|\int_I gf|}{f(x)}}{d\delta^{1-\zeta}|I|^\zeta+\frac{\int_I f}{f(x)}}
=\alpha_4\\
\text{and}&\\ 
\lambda\leq&\frac{L\tb g\tb_--\tb g\tb_+}{L-1}=\alpha_5 ,
\endaligned
$$
again $\alpha=\min\{\alpha_i\}$.
>From Lemma 4.5 we know that $\min\{\alpha_0,\,\alpha_1,\,\alpha_2\}\geq
\frac{(1-\chi)b}{b+1}\tb g\tb_-$, a direct computation shows that
$$
\aligned
\alpha_3\geq&\inf \frac {(1-\chi)d\delta^{1-\zeta}|I|^\zeta}
{d\delta^{1-\zeta}|I|^\zeta-\frac12|I|}\tb g\tb_-=(1-\chi)\tb g\tb_-\\
\alpha_4\geq&\inf \frac{d\delta^{1-\zeta}|I|^\zeta}{d\delta^{1-\zeta}
|I|^\zeta+2|I|}\tb g\tb_-=\frac d{d+2}\tb g\tb_-\\
\alpha_5\geq&\frac{(1-\chi)L}{L-1}\tb g \tb_- .
\endaligned
$$
These inequalities, together with the analogous ones for $\beta_i$, show
that there exists a constant $C$, depending only on $b,\,c,\,d,\,L$, and 
$\chi$,
such that
$$
\Theta(g,\,1)\leq\ln\frac\beta\alpha\leq\ln C\frac{\tb g\tb_+}{\tb g\tb_-}
\leq\ln CL .
$$
\enddemo
 
 
\proclaim{Theorem 5.18}For each $f,\,g\in C^{(1)}(\Cal M)$, there exist
$D,\,r>0$ and $\Lambda<1$ such that
$$
\left|\int_{\Cal M} (f\circ T^n)g-\int_{\Cal M}f\int_{\Cal M}g\right|\leq
D\Lambda^n \|f\|_*\|g\|_*,
$$
where, given $h\in C^{(1)}$, $\|h\|_*=\|h\|_1+r\|h'\|_\infty$.
\endproclaim
\demo{Proof}
We start with the same coarse partition $\Cal P=\{P_i\}$ 
used in the smooth case (notice that, in the present context $P_i$ can
be chosen to be polygons).
It would be possible to perform the same dynamical argument, only the
presence of the singularities would make it much more complicated.
Yet, since in the present case it has been necessary to introduce a very 
small   scale $\delta$, nothing much is lost by placing some further 
restrictions on it.  We further refine the partitions until it consists
only of convex polygons. 
\par
We can assume, without loss of generality, that each element of the new
partition
has a bottom and a top corner, and that the line joining such corners
is completely contained in the element and it is a stable line. Consider
the partition obtained by dividing each element of such partition
into two by the above mentioned line.
\par
Each element of the final partition can be 
foliated into stable curves in many ways.
An easy way to construct a measurable partition is the following.
Let $\varphi(t)$, $t\in[-1,\,1]$, be the equation of the middle line, and
$\zeta(t)$ be the equation of the rest of the boundary; 
$ s\varphi((1-\frac s2)t)+(1-s)\zeta(t)$ for $(s,\,t)\in[0,\,1]\times 
[-1,\,1]$ is a regular partition of the element into weakly--stable curves. 
By construction,
all the weakly--stable curves so generated are longer than some 
$\delta_*\leq\delta_0$; in addition, it is possible to choose the cone 
large enough to contain the Jacobian of the corresponding change of 
variables.
Using such a partition the analogue of Lemma 4.9 follows immediately, 
and the proof of the theorem is concluded by the same argument used in 
Theorem 4.8
\enddemo
 
\vskip1cm
\subhead  \S 6  SINAI BILLIARD
\endsubhead
\vskip1cm
 
 
In the previous sections we have seen how to deal with discontinuities,
provided the map behaves nicely near them. Unfortunately, some important
examples do not satisfy such conditions. In order to have a more satisfactory
theory it would be necessary to allow the first and second derivative
of the map to diverge near a line of discontinuity.
 
A little thought about what we have done in the latter section, in
particular concerning the constant use of distortion estimates 
and the use of an upper
bound on the expansion coefficient, shows that a generalization of the
present strategy to a larger class of examples is not a trivial
matter. It would certainly be necessary to impose some conditions on
the singularities of the map but it is unclear which ones.
Nonetheless, some extensions are possible. To show how, I will briefly
discuss the chief example not covered by the previous conditions:
the Sinai Billiard.
 
The Sinai billiard consists of the motion of a point particle in a two
dimensional torus, where the motion is linear, apart from elastic reflection
by a finite number of disjoint convex scatterers. 
To fix the ideas we will assume 
the scatterers to be circles of radius one.
Here we will consider the simplest case: finite horizon, i.e., 
the scatterers are disposed in such a way that
the particle cannot travel more than a fixed distance without colliding.
In addition, to further simplify matters, we require that there are no 
trajectories with more than two consecutive tangent collisions.
 
The standard way to treat this system is to introduce a Poincar\'e map
\cite{Si}, namely the map from immediately after a collision to immediately
after the next. If we take as coordinates the distance $s$ 
of the collision point from a fixed point on the boundary of 
the scatterer, together with the angle $\varphi$
between the direction of the particle and the tangent to the boundary of the
scatterer, the phase space turns out to be the union of squares (or, better 
yet, cylinders).
More precisely  each scatterer $S_i$ will have associated with it the 
cylinder
$Q_i=S^1\times[0,\,\pi]$. The flow of the particles induces
a map $T:Q=\cup_i Q_i\to Q$ which preserves the area element 
$\sin\varphi ds\wedge d\varphi$; we also introduce the natural 
metric $\sin\varphi ds^2+d\varphi^2$.
 
Due to the existence of both a minimal and a maximal length 
between consecutive collisions, it is possible to choose a 
strictly invariant cone  family smaller
than the usual one (the convergent family). More precisely,
we can choose 
$\Cal C_-(s,\,\varphi)=\{(ds,\,d\varphi)\in\Cal T_{(s,\,\varphi)}
Q_i\;|\; 1\leq -\frac{d\varphi}{ds}\leq u_0\}$ for some appropriated $u_0$, 
depending upon the billiard.
 
The detailed derivation of the properties of the map $T$ would take 
considerable
space; in addition such properties have been discussed at length in many
articles. We will therefore refer to \cite{BSC} where the
decay of correlations for such  (and more general) 
systems is discussed at length by using
a technique akin to Markov partitions. In the aforementioned article, it is
shown that the decay of correlations is, at least, sub-exponential.
 
Setting $(s_1,\,\varphi_1)=T(s_0,\,\varphi_0)$, one can write
$$
D_{(s_0,\,\varphi_0)}T=\pmatrix
\frac{\sin\varphi_0}{\sin\varphi_1}+\frac{\tau}{\sin\varphi_1}
& \frac{\tau}{\sin\varphi_1} \\
\frac{\sin\varphi_0}{\sin\varphi_1}+\frac{\tau}{\sin\varphi_1} +1 
& \frac{\tau}{\sin\varphi_1} +1 
\endpmatrix
\tag 6.1
$$
where $\tau$ is the length, in the usual Riemannian metric, of the 
trajectory between the two collisions.
 
We will measure the length of a stable curve 
\footnote{Note that, in this section, 
we do not have the distinction between stable and weakly stable curves (or 
between $\wh\Gamma_\delta$ and $\Gamma_\delta$). In fact, thanks to the 
very explicit representation of the phase space and the cone, we can both 
use a sufficiently small cone and extend stable curves without 
difficulties.} by the semi-metric
$\sin\varphi ds$, since it is uniformly increased, and (for 
$\varphi\neq 0$ and $\varphi\neq\pi$) is equivalent to the true metric, 
when restricted to a stable curve.
If $v=(1,\,-u)$ is a tangent vector ($v\in\C_-$ iff $1\leq u\leq u_0$) at 
$x=(s,\,\varphi)\in Q$, then
it is easy to compute the expansion $\lambda_x(v)$ 
of $v$ under $D_xT^{-1}$: 
$\lambda_x(v)= 1+\frac{\tau}{\sin\varphi}(1+u)$.
Finally, we will assume that the 
obstacles are sufficiently apart, so that the minimal expansion rate
$\lambda$,
for vectors in the cone, is larger than 100.\footnote{Again, 100 is a 
completely  arbitrary number. It is chosen just in order to simplify
the proof. Using some more care, any value larger than one should be 
admissible.}
 
We adopt the same cone definitions as in \S 5, with the only remark that,
in the definition of $\Gamma_\delta$, the curvature $\kappa_\gamma$ 
is replaced by $\left|\frac{d^2\varphi}{ds^2}\right|$, and
the integral on a curve $\gamma$  is now taken with respect to the
measure $dm_\gamma= \sin\varphi ds$.
 
As already mentioned, the problems here arise due to the unboundedness of the
first derivative of the map at discontinuous points (i.e., tangent 
collisions). The key idea to tackle such a problem is found in \cite{BSC}
and consists of introducing ``homogeneity strips" around singularity
lines.
 
In the present context the singularity lines are images or preimages
of $(s,\,0)$ or $(s,\,\pi)$ (i.e., tangency lines) for the various
$Q_i$.
 
\proclaim{Definition 6.1} By ``homogeneity strip" we mean a strip between
two lines belonging to the set $\{(s,\,D n^{-\rho})\}_{n\in\Bbb N}$ 
or the set $\{(s,\,\pi-D n^{-\rho})\}_{n\in\Bbb N}$ for each square 
$Q_i$ and some $\rho>1$; 
in addition, we will call the images, $\{\R^h_n\}$, of the previous
set of lines under $T$, ``h-discontinuity lines."
\endproclaim
 
Clearly, $\Cal S^-$ (the discontinuity set of $T^{-1}$) consists of the 
images of $\{\varphi=0\}$ and $\{\varphi=\pi\}$ under $T$ and is therefore
the accumulation set of the $\{\Cal S^h_n\}$.
The importance of the above definition is illustrated by the following results:
 
\proclaim{Lemma 6.2} If $\gamma\in\Gamma_\delta$, $f\in\Cal D_a^\nu
(\gamma)$  and 
$\wt\gamma=T^{-1}\big(\gamma\cap($homogeneity strip$)\big)\neq\emptyset$, 
then $\wh Tf\in\Cal D_{\sigma a}^\nu (\wt \gamma)$, provided
$\nu<(1+\rho)^{-1}$.\footnote{Here, we call $\Cal D_a^\nu$ the cone that,
in the preceding sections, was called $D_a$. The reason being that now
$\nu$ plays a much more important role.}
\endproclaim
\demo{Proof}
The fact that all the connected pieces  $\wt\gamma\subset 
T^{-1}\gamma$, of length $\ell$,
belong to $\Gamma_\ell$ is proved exactly as Sub-Lemma 4.3, by using 
formula (6.1). The only fact to check explicitly is the condition on 
$\varphi''$.
\par
Let us use the representation $(t,\,\varphi_1(t)))$ for $\gamma$ and
$(s_0(t),\,\varphi_0(t))=T^{-1}(t,\,\varphi_1(t))$ for $\wt\gamma$.
A direct computation yields
$$
\aligned
\frac{d\varphi_0}{ds_0}(t)=&-\left[1+\frac{(1-\varphi'_1(t))\sin\varphi_0(t)}
{\sin\varphi_1(t)+\tau(1-\varphi'_1(t))}\right]\\
\frac{ds_0}{dt}(t)=&\frac{\sin\varphi_1(t)}{\sin\varphi_0(t)}+
\frac{\tau(1-\varphi_1'(t))}{\sin\varphi_0(t)},
\endaligned
$$
accordingly
$$
\left|\frac{d^2\varphi_0}{ds_0^2}\right|\leq 
|\varphi_1''|\frac{\sin\varphi_1}{[\tau(1-\varphi_1')+\sin\varphi_1]^3}+
c_3\left|\frac{d\tau}{dt}\right|\sin\varphi_0+c_4.
$$
(Through this lemma $c_i$ will designate constants that depend only on the 
system and the parameters of the cone.)
For each $t_1,\,t_2\in\Bbb R$ it is clear that
$$
|\tau(t_1)-\tau(t_2)|\leq |t_1-t_2|+|s_0(t_1)-s_0(t_2)|
$$
so
$$
\left|\frac{d\tau}{dt}\right|\leq 1+\left|\frac{ds_0}{dt}\right|.
$$
Remembering that we have assumed the existence of $\tau_-$ and $\tau_+$, 
such that $0<\tau_-\leq\tau\leq\tau_+<\infty$, follows
$$
\left|\frac{d^2\varphi_0}{ds_0^2}\right|\leq \frac 
1{27\tau_-^2}|\varphi_1''| 
+c_5,
$$
which suffice, provided $\tau_->\frac 1{\sqrt{27}}$.
 
 
In the present case the operator $\wh T $ reads
$$
\wh T f(x)=f(Tx)\lambda_{Tx}(\gamma')^{-1}.
$$
 
Next, let $x,\,y\in\wt\gamma$ contained in the $n$--th homogeneity
strip (i.e., the strip between the $(n-1)$--th and the $n$--th homogeneity 
line), and let $\lambda$ be the minimal expansion coefficient under 
$DT^{-1}$. Then, by choosing $D$ sufficiently small, follows 
$$
\frac{\wh Tf(x)}{\wh Tf(y)}\leq \e^{a\lambda^{-1}n^{-\rho\nu} d(x,\,y)^\nu}
\frac{\lambda_{Ty}(\gamma')}{\lambda_{Tx}(\gamma')}. 
$$
 
Using again the representation $(t,\,\varphi_1(t))$ for $\gamma$, and setting
$(t_1,\,\varphi_1(t_1))=Tx$ and $(t_2,\,\varphi_1(t_2))=Ty$, we can 
estimate
$$ 
\frac{\lambda_{Ty}(\gamma')}{\lambda_{Tx}(\gamma')}\leq
\e^{\left|\int_{t_1}^{t_2}\lambda_{\gamma(\xi)}(\gamma'(\xi))^{-1}
\frac d{d\xi}\lambda_{\gamma(\xi)}(\gamma'(\xi))d\xi\right| }.
$$
A direct computation (see also \cite{BSC} and remember
the bound on $\varphi''$ 
contained in the definition of $\Gamma_\delta$) shows that
$$
\left| \frac d{d\xi}\lambda_{\gamma(\xi)}(\gamma'(\xi))\right|\leq
c_6\left(\frac 1{\varphi_1(\xi)^2}+\frac 
1{\varphi_1(\xi)\sin\varphi_0(\xi)}\right).
$$
Using the above estimate yields
$$ 
\frac{\lambda_{Ty}(\gamma')}{\lambda_{Tx}(\gamma')}\leq
\e^{c_7\int_{t_1}^{t_2}\left[\frac 1{\varphi(\xi)}+\frac 
1{\sin\varphi_0(\xi)}\right]d\xi}\leq
\e^{c_8\left\{n^\rho\left[\int_{s_0(t_1)}^{s_0(t_2)}dm_{\wt\gamma}\right]^\nu
|t_1-t_2|^{1-\nu}+ \int_{s_0(t_1)}^{s_0(t_2)}\frac 1{\sin\varphi_0}
dm_{\wt\gamma}\right\}}.
$$
Since $\gamma'(\xi)\in\Cal C_-(\gamma(\xi))$, by definition, the intersection 
between $\gamma$ and the $n$--th homogeneity strip must be very short, more
precisely $|t_1-t_2|\leq\hbox{const}\times n^{-\rho-1}$. 
In addition, 
$$
\int_{s_0(t_1)}^{s_0(t_2)}\frac 1{\sin\varphi_0}dm_{\wt\gamma}=
|s_0(t_2)-s_0(t_1)|,
$$
but
$$
d(x,\,y)\geq c_9\int_0^{|s_0(t_2)-s_0(t_1)|}\xi d\xi=\frac{c_9}2
|s_0(t_2)-s_0(t_1)|^2.
$$
Consequently
$$ 
\frac{\wh Tf(x)}{\wh Tf(y)}
\leq  \e^{[\lambda^{-1}n^{-\rho\nu} 
+a^{-1}c_{10} n^{\rho-(\rho+1)(1-\nu)}+c_{10}a^{-1}d(x,\,y)^{\frac12-\nu}]
a d(x,\,y)^\nu}.
$$
\enddemo
 
The above Lemma is the counterpart of Lemmas 4.4 and 4.5. In addition, it 
shows that it is possible to control the distortion,
although on shorter and shorter curves. The next Lemma takes advantage of
the fact that the expansion coefficient is unbounded near
the singularities to show that the union of the too short pieces is
negligible. Although we will not use it directly in the following,
it illustrates well the key property on which all the further arguments
rest.
 
\proclaim{Lemma 6.3} If $\gamma\in\Gamma_\delta$, then it is possible 
to write $T^{-1}\gamma= 
\bigcup\limits_{\wt\gamma\in\Cal J_1\cup\Cal J_2}\wt \gamma$, where
$\wt\gamma\in\Cal J_1$ are connected, longer than $\delta$ and 
$T\wt\gamma$ belongs
to only one homogeneity strip. In addition, for each $\xi\in[\zeta,\,1]$,
$$
\sum_{\wt\gamma\in\Cal J_2}|T\wt\gamma|^\xi\leq c_1\delta^{\frac 32\xi}.
$$
\endproclaim
\demo{Proof} 
As we have seen above, a stable curve is expanded by 
$\lambda n^\rho$, if it falls in the $n$-th  homogeneity strip.
A maximal piece of $T^{-1}\gamma$ that crosses completely the $n$-th  
homogeneity strip has a length of at least $n^{-\rho-1}\delta_0$.
Hence, if we choose $n_0=\left(\frac{\delta_0}{\delta}
\right)^{\frac 1{\rho+1}}$ the
pieces in $\Cal J_2$ must belong to homogeneity strips with $n$ greater 
than some $n_1>n_0$.
Accordingly,
$$
\sum_{\wt\gamma\in\Cal J_2}|T\wt\gamma|^\xi
\leq\sum_{n>n_1}c_0^{-\xi}n^{-\rho\xi}|\wt\gamma|^\xi
\leq\sum_{n>n_0}c_0^{-\xi}n^{-\rho\xi}\delta^\xi .
$$
The result then follows by choosing $\rho=\frac{\zeta+2}\zeta\geq 3$.
\enddemo
 
The ideas in Lemmas 6.2 and 6.3 are all that is needed to 
prove the equivalent of Lemma 5.4.
 
\proclaim{Lemma 6.4} If $L>1$, $4L<d$, $2^\nu a\delta^\nu<\ln 2$, and 
$\rho>2$, then for each $n\in[A\ln d,\,\dots,\,N(\delta)]$
$$
\wt T^n\Cal C_{b,c,d,L}(\delta)\subset
\Cal C_{\chi b,\chi c,\chi d, \tau^2_n L}(\delta)
$$
where $\chi<1$, and $\tau_n=(1-3d(4\lambda^{-1})^n)\leq
\frac 1{\sqrt 2}$.
\endproclaim
\demo{Proof}
The proof follows faithfully the path layed down in Lemma 5.4. 
Except that more care is required in keeping track of the pieces
generated iterating a curve, since now there can be countably many.
 
In fact, given $\gamma\in\Gamma_\delta$, we will divide $\gamma$
in connected pieces belonging to the same homogeneity strip, with the
possible exception of the pieces containing the endpoints or intersecting 
$\R^-$ (in this last case it is certainly possible to readjust a homogeneity 
line, so that the corresponding breaking point on $\gamma$ coincide with 
the one determined by $\R^-$, without spoiling any of the above estimates). 
This means
that $T^{-1}\gamma$ can consist, at most, of a countable collections
of curves if $\gamma$ intersect an homogeneity region, and consists at most 
of three curves otherwise 
(since we have assumed no more than two consecutive tangencies).
Since curves that expand very much are very short, it is easy to see that 
for at least $N(\delta)\sim\ln\delta^{-1}$ iterations no curves become 
longer than some fixed $\delta_0$.
So, if we put into $\Cal J_-$ all the curves shorter than $\delta$
belonging to $T^{-n}\gamma$ and we call $\lambda(\wt \gamma)$ the minimal 
expansion  coefficient of $T^{-n}$ on $T^n\wt\gamma$ we have
$$
\aligned
\sum_{\wt\gamma\in\Cal J_-}\lambda(\wt\gamma)^{-1}&\leq
\sum_{k_1=0}^\infty\cdots
\sum_{k_n=0}^\infty
\lambda^{-n}\prod_{j=1}^n \left((k_j+1)^{-\rho}+2\right)\\
&=\lambda^{-n}\left(2+\sum_{k=0}^\infty (k+1)^{-\rho}\right)^n \\
&\leq\left(\lambda^{-1}(3+\frac 1{2^{\rho-1}(\rho-1)})\right)^n
\leq(4\lambda^{-1})^n
\endaligned
$$
which, remembering our assumptions on $\lambda$, is less than one (here we 
have used our assumption on $\rho$).
 
Since the above quantity is the only one used in Lemma 5.4 to estimate 
the influence of short pieces (there it is estimated by $\lambda^{-n}M^n$)
the same results follows in the same way.
\enddemo
 
This implies that it is possible to control a large number of iterates
of $\wt T$: the analogue of Sub-Lemma 5.9.
 
In Theorem A3.1 of \cite{BSC} it is shown that a curve in $\Gamma_\delta$
will consist overwhelmingly of pieces longer than $\delta_0$ after
a number of iterations proportional to $\ln\delta^{-1}$. Hence, we can
obtain the analogue of Lemma 5.8.
 
To obtain the wanted result we need only to be able to
compare averages on curves in $\Gamma_{\delta_0}$. This can be done 
exactly  in the same way as in \S 5, Lemma 5.14. 
In fact, the arguments used there
are completely general and depend only on the results contained the 
Appendices. But, Appendix II can be generalized
to include the present case, as it is explained at the end of the Appendix
itself.
In addition, Appendix I can be not only generalized to the present context, 
but it is also possible to ask that a weak-core contains a positive measure 
of long {\sl homogenous} stable and unstable manifolds (i.e., manifolds whose
images fall always in a unique homogeneity strip) instead that simply 
stable and unstable manifolds (see \cite{BSC, Appendix 1} for details). 
Such a generalization suffices to prove the corresponding of Sub-Lemma 
5.16, and consequently establish the like of Lemma 5.14.\footnote{In Lemma 
5.14 the maximal expansion coefficient $\mu$ appears only in two places.
First, in Lemma 5.12 where we use Lemma II.2. The analogous result for 
billiards (stated at the end of Appendix II) does not involves the maximal 
expansion (that is now $\infty$) but forces us to choose $\zeta$ 
sufficiently close to 1 as well. Second, in the final estimate of the ratio 
of the averages. Here $\mu^N$ is substituted by 
$\sup\limits_{\gamma\in\Gamma_{\delta_1}}\frac{|\gamma|}{|T^N\gamma|}
<\infty$.}
 
This concludes the discussion of the proof of the equivalent of 
Theorem 5.3. Since all the rest is equal, the exponential decay of
correlations follows.
 
 
 
 
\vskip1cm
\subhead \S 7  TWO DIMENSIONS (RANDOM PERTURBATIONS)
\endsubhead
\vskip1cm
 
In this section we will study the random perturbations of a two dimensional, 
area--preserving smooth system. 
Although the class of perturbations treated is much less general than the
one considered in \cite{Ki}, the present technique provides an easier
approach and new results concerning the dependence of the invariant measure
from the size of the perturbation.
The reader will easily see the possibility of generalizing the 
following reasoning.
 
For simplicity we will assume $\M=\Bbb T^2$.
 
We will assume that the deterministic map $T$ satisfies the hypotheses of
\S 4, moreover, calling $v^u(x),\,v^s(x)$ the unstable and stable direction
at $x$, we assume
$$
\aligned
|v^u(x)-v^u(y)|&\leq A|x-y|^\tau \\
|v^s(x)-v^s(y)|&\leq A|x-y|^\tau
\endaligned
$$
for some $\tau\in(0,\,1]$. In fact, the existence of such $\tau$ is 
quite general, since, for the case under consideration, the
stable and unstable foliations are, at least, 
H\"older continuous.
 
The random perturbation, in analogy with the one dimensional case, 
will be defined by the following Markov process.
 
Let $\phi:\Bbb T^2\times \Bbb R^+\to \Bbb R^+$ be a smooth function such that
$\phi(x,\,\xi)=0$ for each $\xi\geq 1$, $x\in\Bbb T^2$, and
$2\pi\int_0^1\phi(x,\,\xi)\xi d\xi=1$ for each $x\in\Bbb T^2$.
Then, for each $f\in L^1(\Bbb T^2)$ we define
$$
\left(\wh T_\ve f\right)(x)=\frac{1}{\pi\ve^2}\int_{\Bbb T^2}
\phi(Tx,\,\ve^{-1}|Tx-y|f(y)dy .
$$
In addition, we assume a weak form of translation invariance:
$$
\frac{\phi(x,\,\xi)}{\phi(y,\,\xi)}\leq\e^{\theta|x-y|^\tau}
$$
for each $\xi\in\Bbb R^+$, $x,\,y\in\Bbb T^2$.
 
Two types of problems are of interest:
\roster
\item"i)"  ergodic properties of the random process;
\item"ii)" distance between the invariant measure of the random and the
            deterministic processes.
\endroster
 
Our claim is that both problems can be addressed with the technique described
in this paper. The approach to the first question is reasonably 
straightforward: one chooses a cone of functions and proves contraction.
As we have seen in \S 3 various choices of cones are permissible.
One can simply choose the cone of positive function,
since it is possible to shows that, after approximately $\ln\ve^{-1}$
iterations, the kernel of the transfer operator is strictly positive 
(better, there exists constants $B_-,\,B_+$, independent of $\ve$, that
lower and upper bound it),
it follows that the correlation between any two continuous
functions $f,\,g$ decays as $\Lambda^{\frac n{\ln\ve^{-1}}}
\|f\|_\infty\|g\|_\infty$.\footnote{See \cite{Ki}  for similar results.} 
This result is both stronger (the rate of decay
it is not proportional to the derivatives of the functions) and
weaker (the rate of decay depends upon $\ve$) then the one obtained for
the deterministic case. Yet, it is likely that a cone similar to the one
used in \S 4 can yield an $\ve$-independent rate of decay for $C^{(1)}$
functions.
 
In the present paper we will not discuss point (i) any further; we will
instead concentrate on point (ii). Here it is less obvious how to obtain
results similar to the one discussed in \S 3. Nevertheless, something 
can be done. 
 
The Perron-Frobenius operator reads
$$
\wt T_\ve g(x)=\frac1{\pi\ve^2}\int_{\Bbb T^2}\phi(y,\,\ve^{-1}
|Ty-x|)g(y)dy .
$$
 
Let
$$
\aligned
\|g\|_+=&\sup_{x\in\Bbb T^2} g(x)\\
\|g\|_-=&\inf_{x\in\Bbb T^2} g(x).
\endaligned
$$
We designate by $W^s(x,\,t)$ the point obtained by translating $x$ by $t$
along its stable manifold,
we then define the cone
$$
\Cal C_{a,\,H}^\tau=\left\{ f\in C^{(1)}\;|\;  f\geq 0;\;
\frac{f(x)}{f(W^s(x,\,t))}\leq\e^{a t^\tau}\text{ for }t\leq\ve
;\; \frac{\|f\|_+}{\|f\|_-}
\leq H\right\}.
$$
 
\proclaim{Lemma 7.1}
If, for some given constant $c_1$, $\theta +c_1H+\lambda^{-1}a<a$, then 
there exists $\sigma<1$ such that 
$\wh T_\ve \Cal C_{a,\,H}^\tau \subset \Cal C_{\sigma a,\,H}^\tau $.
\endproclaim
\demo{Proof}
The only condition that is not obvious is the second one.
$$
\aligned
\wh T_\ve f(x)=&
\frac{1}{\pi\ve^2}\int_{\Bbb T^2}
\phi(Tx,\,\ve^{-1}|Tx-y|)f(y)dy\\
\leq&\frac{\e^{\theta t_1^\tau}}{\pi\ve^2}\int_{\Bbb T^2}
\phi(TW^s(x,\,t),\,\ve^{-1}|Tx-y|)f(y)dy
\endaligned
$$
where $t_1\leq\lambda^{-1}t$ is such that $W^s(Tx,\,t_1)=TW^s(x,\,t)$.
Since the direction of the stable manifold at $y$ differs from the one at
$x$ by, at most, a quantity proportional to $\ve^\tau$, it follows, setting
$Tx-y=\xi$,
$$
W^s(Tx,\,t_1)+\xi=W^s(y,\,t_1)+\Cal O(\ve^\tau t)
$$
that is
$$
|\xi|=| TW^s(x,\,t)-W^s(y,\,t_1)|+\Cal O(\ve^\tau t).
$$
Hence,
$$
\aligned
\wh T_\ve f(x)
\leq&\frac{\e^{\theta t^\tau}}{\pi\ve^2}\int_{\Bbb T^2}
\phi\left(TW^s(x,\,t),\,\frac 1{\ve}|TW^s(x,\,t)-W^s(y,\,t_1)|\right)f(y)+
\Cal O(\ve^{\tau-1}t)\|f\|_+\\
\leq&\frac{\e^{\theta t^\tau}}{\pi\ve^2}\int_{\Bbb T^2}
\phi\left(TW^s(x,\,t),\,\ve^{-1}|TW^s(x,\,t)-W^s(y,\,t_1)|\right)
f(W^s(y,\,t_1))dy\\
+&\Cal O(\ve^{\tau-1}t)H\|f\|_- .
\endaligned
$$
To conclude we need to perform the change of variable $z=W^s(y,\,t_1)$
and compute its Jacobian. 
 
Let us choose Cartesian coordinates, in a $\ve$-neighborhood of $y$, with
axes parallel to $v^s(y)$ and $v^u(y)$, respectively, and write
$y=(y_s,\,y_u)$. Then, $\frac{\partial z_s }{\partial y_s}=1$;
$\frac{\partial z_s }{\partial y_u}=\Cal O(t^\tau)$ by the conditions imposed
on the regularity of the foliations; $\frac{\partial z_u }{\partial y_s}=
\Cal O(t^\tau)$; $\frac{\partial z_u }{\partial y_u}=1+\Cal O(t^\tau)$.
 
In conclusion, the Jacobian of the change of variable is equal to
$1+\Cal O(t^\tau)$; hence
$$
\wh T_\ve f(x)\leq\left[\e^{(\theta+\lambda^{-1}a)t^\tau}+c_1 t^\tau H
\right] \wt T_\ve f(W^s(x,\,t))
$$
which proves the Lemma.
\enddemo
 
We will designate by $\wt T_0$ and $\wh T_0$ the operators corresponding
to the deterministic map. The following provides an interesting bound of 
the difference between the correlations of the random and the deterministic
processes.
 
 
\proclaim{Lemma 7.2}
For each $f,\,g\in C^{(1)}$ and $\ve$ sufficiently small, holds
$$
\left|\int_{\Bbb T^2}\wh T^n_\ve fg- \int_{\Bbb T^2}\wh T^n_0fg\right|
\leq nc_3\ve^\tau \|f\|_*\|g\|_*.
$$
where $\|\cdot\|_*$ is defined in Theorem 4.8.
\endproclaim
\demo{Proof}
Choose $f\in\C^\tau_{a,\,H}$ and $g\in\C_{b,c,d,L}$, then
 
$$
\aligned
\bigg|\int_{{\Bbb T^2}}(\wh T_\epsilon^n f)g-
\int_{{\Bbb T^2}} (\wh T_0^n f)g\bigg|&\leq
\sum_{k=0}^{n-1}\bigg| \int_{{\Bbb T^2}}\wh T_0^{n-k-1}\wh T_0 
\wh T_\epsilon^k f g-
\int_{{\Bbb T^2}}\wh T_0^{n-k-1}\wh T_\epsilon \wh T_\epsilon^k f g\bigg|\\
&\leq \sum_{k=0}^{n-1}\bigg| \int_{{\Bbb T^2}} \wh T_0 
\wh T_\epsilon^k f\wt T_0^{-n+k+1}g-
\int_{{\Bbb T^2}} \wh T_\epsilon \wh T_\epsilon^k f \wt 
T_0^{-n+k+1}g\bigg| .
\endaligned
$$
Since  $f_1=\wh T_\epsilon^k f \in \Cal C^\tau_{a,\,H}$ (Lemma 7.2)
and $g_1=\wt T_0^{-n+k+1}g\in \Cal C_{b,c,d,L}$ (Lemma 4.2)
we can write the generic term of the previous expression as
$$
\aligned
\bigg|\frac 1{\pi\ve^2}\int_{(\Bbb T^2)^2}f_1(x)&\phi(x,\,\ve^{-1}|x-Ty|)
g_1(y)- \int_{{\Bbb T^2}}T_0f_1g_1\bigg|\\
&=\bigg|\frac 1{\pi\ve^2} 
\int_{(\Bbb T^2)^2}f_1(x)\phi(x,\,\ve^{-1}|x-y|)g_1(T^{-1}y)-
\int_{{\Bbb T^2}}f_1g_1\circ T^{-1}\bigg| .
\endaligned
$$
This means that the problem is reduced to the estimate of
$$
\bigg|\frac 1{\pi\ve^2}\int_{(\Bbb T^2)^2}f(x)\phi(x,\,\ve^{-1}|x-y|)g(y)
-\int_{{\Bbb T^2}}fg\bigg|
$$
where $f,\,g$ are generic functions in the respective cones.
We introduce the following change of variable:
$$
\aligned
\xi=&x-y\\
\eta=&\alpha(x,\,y)
\endaligned
\tag 7.1
$$
where $\alpha(x,\,y)=W^s(x)\cap W^u(y)$.
 
\proclaim{Sub-Lemma 7.3} Calling $\wh J$ the Jacobian of the 
change of variable (7.1), for each $|\xi|\leq\ve$, and $\ve$ small enough,
$$
|\wh J-1|\leq c_1\ve^\tau .
$$
\endproclaim
\demo{Proof}
 
Let us fix a point $x\in\Bbb T^2$. Since the Jacobian can be computed
in any system of coordinates, we choose Cartesian coordinates with
axes parallel to the stable and unstable direction at $x$ respectively.
We can then write $x=(x_s,\,x_u)$, and the same for $y,\,\xi,\,\eta$.
We will then write the Jacobian matrix $\wt J$ in this 
coordinates and compute its determinant.
 
Clearly, $\frac {\partial \xi_l}{\partial x_l}=1$ for $l\in\{s,\,u\}$
while    $\frac {\partial \xi_l}{\partial y_l}=-1$.
By the definition of $\alpha$ follows 
$\frac {\partial \eta}{\partial x_s}=0$. The computation of
$\frac {\partial \eta_u}{\partial x_u}$ is more delicate.
The derivate, with respect to $x_u$, 
can be computed by taking the derivative along the unstable
manifold of $x$. By the absolute continuity of the unstable foliation it
follows that $\alpha(W^u(x,\,h),\,y)=W^u(\eta,\,h')$ 
with $\e^{-c_1\ve}\leq\frac{|h'|}{|h|}\leq\e^{c_1\ve}$.
Consequently, since the unstable direction at $\eta$ differs from the
unstable direction at $x$ by, at most, $\ve^\tau$ we have
$$
\aligned
\frac {\partial \eta_u}{\partial x_u}=&1+\Cal O(\ve^\tau)\\
\frac {\partial \eta_s}{\partial x_u}=&\Cal O(\ve^\tau).
\endaligned
$$
By similar arguments one can compute the remaining elements of $\wt J$,
obtaining
$$
\wt J=\pmatrix
1& 0& -1 &0\\
0& 1& 0& -1\\
0& 0& 1&0\\
0&1&0 &0
\endpmatrix+\Cal O(\ve^\tau).
$$
\enddemo
 
Using the properties of $f$ and $g$ together with the above bound
easily yields
$$
\aligned
\frac 1{\pi\ve^2}\int_{(\Bbb T^2)^2}f(x)&\phi(x,\,\ve^{-1}|x-y|)g(y)\\
=&\frac 1{\pi\ve^2} \int_{|\xi|\leq \ve}d\xi\int_{\Bbb T^2}d\eta
f(\eta)\phi(\eta,\,\ve^{-1}|\xi|)g(\eta)+\Cal O(\ve^{\tau})\|f\|_\infty 
\|g\|_\infty \\
=&\int_{\Bbb T^2}fg+ \Cal O(\ve^{\tau})\|f\|_\infty 
\|g\|_\infty .
\endaligned
$$
The result for general $f$ and $g$ follows then via the same arguments
used in Theorem 4.8.
\enddemo
 
Using the previous Lemma we can easily prove the announced estimate.
 
\proclaim{Theorem 7.4} Calling $\mu_\ve$ the invariant measure of the
random process $\wt T_\ve$, for each $f\in C^{(1)}$ 
$$
\left|\int fd\mu_\ve-\int f\right|\leq A(\ln\ve^{-1})^2\ve^\tau\|f\|_*
$$
\endproclaim
\demo{Proof}
$$
\aligned
\left|\int fd\mu_\ve-\int f\right|&\leq
\left|\int fd\mu_\ve-\int f\wt T_\ve^n 1\right| +
\left|\int f\wt T^n_\ve 1-\int f\wt T^n_0 1\right|\\
&\leq c_2\left[ n\ve^\tau+\Lambda^{-\frac{n}{\ln\ve^{-1}}}\right] 
\| f\|_*
\endaligned
$$
the Theorem follows by choosing $n=\frac\tau{\ln\Lambda}(\ln\ve^{-1})^2$.
\enddemo
 
\vskip1cm
\subhead Appendix I (Sinai Theorem)
\endsubhead
\vskip1cm
 
This appendix is dedicated to the proof of Theorem 6.13. Since the argument
is reasonably standard, we will provide only a sketch, and refer to
\cite{LW} for more details.
\par
We start by introducing a coarse partition of $Q$ via $\R_{m_Q}=
\bigcup\limits_{n=0}^{m_Q}(T^n\R^-)\cup(T^{-n}\R^+)$. Clearly $Q\backslash
\R_{m_Q}$ is partitioned into finitely many open domains $\{R_i\}$.
Since $\wt\C_-(z)=DT^{-m_Q}\C_-(T^{m_Q}z)$ and  
$\wt\C_+(z)=DT^{m_Q}\C_+(T^{-m_Q}z)$ are uniformly continuous in each 
$R_i$, it is possible to further refine the partitions and obtain a new
partition $\{R^j_i\}$ made of elements of uniform size and on which the
cone families are almost constant. Next, we choose a point $z_i^j$ in each 
$R^j_i$ such that $z^j_i\not\in\Sigma_\infty$, and we introduce in each $R^i_j$ 
a cartesian system of coordinates with axes parallel to the stable and 
unstable directions at $z^i_j$. We can then introduce in $R^i_j$ a family 
of squares of size $t\delta_1$ with sides parallel to the axes, such that 
their weak-core covers $R^i_j$. We will consider only the squares whose 
weak-core do not intersect the boundary of $R^i_j$.  
\par
The set $\Cal \R_m=\bigcup\limits_{|n|\leq m}T^n\R$ consists of
a finite collection of lines. Let us call $\Cal R_m$ the set of points
that belong to more than one line. 
By hypothesis $\Cal R_m$ is a discrete
set containing a finite number of points. We choose $\delta_1$ smaller
than $\frac t4$  the smallest distance among points in $\Cal R_m$.
If $\wt{\Cal R}_m$ is the $2t\delta_1$ neighborhood of $\Cal R_m$, then
$\R_m\backslash \wt{\Cal R}_m$ consists of a set of disjoint lines,
at least $2t\delta_1$ apart.
By choosing $\delta_1$ small enough, the total measure of the squares that
intersect $\wt{\Cal R}_m$ is less than $\frac{\delta}{512\delta_0}$.
Moreover, choosing $m_Q$ large and $\alpha$ small enough, it is 
possible to insure
that a line belonging to $\R_m\backslash\R_{m_Q}$ is not sufficient
to make a square not $\alpha$-connecting.
The result follows then by choosing $m$ sufficiently large and by the
usual estimate of the size of the tail $\Sigma_\infty\backslash\Sigma_m$.
 
\vskip1cm
\subhead Appendix II (Absolute continuity)
\endsubhead
\vskip1cm
 
This section provides some result that are not readily available in
the literature. 
\par
The main result presented here concerns the map induced by the unstable
foliation between two nearby stable curves. It is well known that such 
map is absolutely continuous, with Jacobian close to the identity, if 
the two curve are sufficiently nearby \cite{K-S}; 
I will show that the map is H\"older continuous together with its 
Jacobian. This result is well known for smooth maps \cite{P},  
\cite{M} but, to my knowledge, 
it is new for discontinuous maps, although some partial 
results along this line
exist in the case of billiards \cite{B-S-C}.
 
We will restrict ourselves to the two dimensional case, this will allow
us to have easier and more transparent proofs with respect to \cite{K-S}.
Although an extension to the multidimensional case should not be out of
the reach of the technique used here, the author feels that it would be
out of the scopes of his present efforts.
\par
We will consider the class of maps described at the beginning of \S 6.
\par
 
Consider two strictly stable
curves $\gamma_1,\,\gamma_2$, such that $d(\gamma_1,\,\gamma_2)\leq
\delta_2/2$.
 
We define $D(\phi)=\{z\in\gamma_1\;|\; W^u(z)\cap\gamma_2\neq\emptyset\}$,
were $W^u$ is the unstable manifold of the $z$ of size less or equal to
$2\delta_2$, 
\footnote{$W^u(z)=\{z\}$ if $z$ does not have an unstable manifold.} and
the map $\phi: D(\phi)\subset\gamma_1\to\gamma_2$ by
$$
\phi(x)=W^u(x)\cap\gamma_2.
$$
If $R(\phi)$ is the range of the map (and the domain of $\phi^{-1}$), then
we define a measure $\mu$ on $R(\phi)$ by 
$$
\int_{R(\phi)}fd\mu= 
\int_{D(\phi)}f\circ\phi dm.
$$
 
>From the general theory developed in \cite{K-S} follows:
 
\proclaim{Theorem II.1}The map $\phi$ is absolutely continuous.
In addition, calling $J_\phi$ its Jacobian, for each density point $z$ 
of $R(\phi)$ holds
$$
\frac{d\mu}{dm}=J_\phi(z)=\prod_{j=0}^\infty\left|\frac{D_{T^{-j}z}T
\big|_{T^{-j}\gamma_2}}
{D_{T^{-j}\phi^{-1}(z)}T\big|_{T^{-j}\gamma_1}}\right|
\leq\e^{D\delta_2}.
$$
\endproclaim
 
In order to use effectively the map $\phi$ in our context, we need three 
extra informations: an estimate of the measure of $\gamma_1\backslash 
D(\phi)$ and $\gamma_2\backslash R(\phi)$; 
an estimate of the smoothness of $\phi$; and an estimate of the
smoothness of $J_\phi$.
 
Let $\Sigma_n( c,\,\delta_2)=\cup_{i=0}^n 
T^i\Cal S^-_{c\lambda^{-  i}\delta_2}$. 
>From the construction of the unstable manifold (see \cite{K-S}) follows 
that $\gamma_1\backslash\Sigma_\infty(c,\,\delta_2)\supset D(\phi)$. 
The first estimates 
are then accomplished by the following.
 
\proclaim{Lemma II.2} There exists $D>0$ such that for each 
$\gamma\in\wh\Gamma_{\delta_2}$, and
$m\in\Bbb N$, if $\gamma\cap\Sigma_m(c,\,\delta_2)=\emptyset$, with
$\delta\leq\delta_2<\delta_0$, $c<1$, then
$$
\aligned
& m\left(\gamma\backslash D(\phi)\right)\geq 
Dc\delta_2\lambda^{-m}\\
&\frac 1{f(x)}\left| \int_{\gamma\backslash D(\phi)}gf
\right|\leq Dc\delta_2(\lambda^{-\zeta}\mu^{1-\zeta})^m
\tb g\tb_- ,
\endaligned
$$
where $m(\cdot)$ is the Lebesgue 
measure restricted to $\gamma$, and $g\in\Cal C_{b,c,d,L}(\delta)$, 
$f\in\Cal D_a(\gamma)$.
\endproclaim
\demo{Proof}
We start by recalling a simple distortion estimate.
 
\proclaim{Sub-Lemma II.3} If $I\in\wh\Gamma_{\delta_0}$, 
$J\subset I$,
and $T^n I$ is connected, then
$$
A^{-1}\frac{|I|}{|J|}\leq\frac{|T^nI|}{|T^nJ|}\leq A\frac{|I|}{|J|}.
$$
\endproclaim
\demo{Proof}
$$
\frac{|T^nI|}{|T^nJ|}=\frac{\int_{I}|D_xT^n\big|_I|}{\int_J|D_yT^n\big|_I|}
\leq \e^{c_2 \delta_0 (1-\lambda^{-1})^{-1}}\frac {|I|}{|J|}.
$$
\enddemo
 
Consider $\gamma\in\wh\Gamma_{\delta_*}$, 
$\delta_*\in [\delta_2,\,\delta_0]$, 
$c<1$. Suppose $\gamma\cap\Sigma_k(c,\,\delta_2)=\emptyset$, 
for some $k\geq 0$.
Then $T^{-k}\gamma=\gamma_0$ is connected, and 
$|\gamma_0|\geq\lambda^k\delta_2$.
 
Let $S_0=\{J_i\}$ be the collection of the, at most, $M$ connected segments
constituting $\gamma_0\cap\Sigma_0(\lambda^{-k}c,\,\delta_2)$, and
set $\wt\gamma_0=\gamma_0\backslash\cup_{J\in S_0}J$.
Consider $T^{-1}\wt\gamma_0$, it will consist, at most, of $M$ connected 
pieces. Collect in $\Cal J_1$ the pieces longer than $\delta_0$, after
subdividing them, if necessary, in pieces shorter than $2\delta_0$.
Collect the remaining pieces in $\Cal J_1^*$. For each piece in
$\Cal J_1^*$  repeat the previous construction (i.e., intersect the piece
with $\Sigma_0(\lambda^{-(k+1)}c,\,\delta_2)$, collect the 
connected pieces of the intersection
in $S_1$, consider the preimage of the rest, collect the long pieces
in $\Cal J_2$ and the remaining ones in $\Cal J_2^*$, and so forth).
 
By construction, the pieces in $\Cal J_l^*$ can be at most $M^l$ and
are all shorter than $\delta_0$. Hence, for each $\xi\in[\zeta,\,1]$,
$$
\sum_{\wt\gamma\in\Cal J_l^*} |T^{-k-l}\wt\gamma|^\xi\leq
M^l\lambda^{-(l+k)\xi}\delta_0^\xi .
$$
In addition, we can estimate the pieces that fall too close
to a singularity line during the first $l$ iterations:
$$
\sum_{i=k}^{l-1}\sum_{\wt\gamma\in S_{i-k}} |T^{-i}\wt\gamma|^\xi\leq
c^\xi\sum_{i=k}^{l-1}M^{i+1-k}\lambda^{-2i\xi}\delta_2^\xi\leq
Bc^\xi\lambda^{-2k\xi}\delta_2^\xi ,
$$
since $M\lambda^{-\zeta}<1$. Consequently, calling $\Cal J(\gamma,\,c)$
the collection of segments that either get to close to the singularity
or never grow larger than $\delta_0$, we have
$$
\sum_{\wt\gamma\in\Cal J(\gamma_0,\,c)} |\wt\gamma|^\xi\leq
B c^\xi\lambda^{-2k\xi}\delta_2^\xi .
$$
We still need to deal with the segments in the collection
$\cup_{i=k+1}^\infty T^{i}\Cal J_{i-k}$. 
 
Let us consider $\ov\gamma\in\Cal J_1$, iterating it we will obtain again 
segments that are to close to singularity, segments that 
stay short for all times, and segments that grow to size 
$\delta_0$. We will estimate the first two group while we will add the
segment belonging to the last category to the above mentioned collections of
long segments.
 
$$
\aligned
\sum_{\ov\gamma\in\Cal J_1}\sum_{\wt\gamma\in\Cal J(\ov\gamma,\,
\lambda^{-k-1}c)}|T^{k+1}\wt\gamma|^\xi &\leq
\sum_{\ov\gamma\in\Cal J_1}\sum_{\wt\gamma\in\Cal J(\ov\gamma,\,
\lambda^{-k-1}c)}\frac{|T^{k+1}\wt\gamma|^\xi}{|T^{k+1}\ov\gamma|^\xi}
|T^{k+1}\ov\gamma|^\xi\\
&\leq   A^\xi 
\sum_{\ov\gamma\in\Cal J_1}\sum_{\wt\gamma\in\Cal J(\ov\gamma,\,
\lambda^{-k-1}c)}\frac{|\wt\gamma|^\xi}{|\ov\gamma|^\xi}
|T^{k+1}\ov\gamma|^\xi\\
&\leq    A^\xi B c^\xi
\lambda^{-(k+1)\xi}\delta_2^\xi\delta_0^{-\xi}
\sum_{\ov\gamma\in\Cal J_1}|T^{k+1}\ov\gamma|^\xi\\
&\leq    A^\xi B c^\xi  
\lambda^{-(k+1)\xi}\delta_2^\xi\delta_0^{-1} \mu^{(k+1)(1-\xi)}
\sum_{\ov\gamma\in\Cal J_1}|T^{k+1}\ov\gamma|\\
&\leq   2 A^\xi B c^\xi  
\lambda^{-(k+1)\xi}\delta_2^\xi\mu^{(k+1)(1-\xi)}
\endaligned
$$
 
Still, we have not analyzed all the pieces that must be estimated.
The remaining ones are pieces $\wh\gamma$ with some preimage $T^{-j}
\wh \gamma$, $j>k$, contained in a connected component of $T^{-j}\gamma$
of size larger than $\delta_0$.
The key observation to conclude the argument is that the
last estimate does not depend from the number of pieces in $\Cal J_1$,
but only from the fact that all such pieces are longer than $\delta_0$
and that $\Cal J_1\subset T^{-k-1}\gamma$. This means that we can apply
the above estimate iteratively and obtain
$$
\sum_{I\subset\Sigma_\infty(c,\,\delta_2)}|I|^\xi\leq
\sum_{i=k}^\infty
2A^\xi B c^\xi
(\lambda^{-\xi}\mu^{(1-\xi)} )^i \delta_2^\xi 
\leq D(\lambda^{-\xi}\mu^{(1-\xi)} )^k \delta_2^\xi. 
$$
 
The Lemma easily follows from the above estimate, and from noticing that
$\gamma\backslash D(\phi)$ is a countable collection of segments, each one 
of which is contained in a segment belonging to
$\Sigma_\infty(c,\,\delta_1)$.
\enddemo
 
The first part of the wanted estimates is completed by noticing the 
symmetry between $\phi$ and $\phi^{-1}$.
 
Next, we can study the smoothness of $\phi$.
 
\proclaim{Lemma II.4} There exists $A\in\Bbb R^+$, such that 
for each $x,\,y\in\gamma_1\backslash\Sigma_\infty(c,\,\delta_2)$
$$
d(\phi(x),\,\phi(y))\leq A d(x,\,y).
$$
\endproclaim
\demo{Proof}
Let $x,\,y\in\gamma\backslash\Sigma_\infty( c,\,\delta_2)$, 
$c\leq 1$; suppose
$$
d(T^{-n}x,\,T^{-n}y)\leq d(x,\,y)^{\frac 12}
\delta_2^{\frac 12}
$$
then we have
$$
\lambda^nd(x,\,y)\leq d(x,\,y)^{\frac 12}
\delta_2^{\frac 12}
$$
that is
$$
d(T^{-n}x,\,T^{-n}y)\leq \lambda^{-n}\delta_2.
$$
The above estimate means that 
the two points cannot be on opposite sides
of $\Cal S^-$ (remember that the distance is taken along $\gamma$
or its images), otherwise one of the points would belong to $\Sigma_\infty$
contrary to the hypothesis. Accordingly, another interaction is 
possible without disconnecting the part of $\gamma$ lying between $x$ and
$y$.
This shows that it is possible to iterate the two points until
$$
d(T^{-n}x,\,T^{-n}y)\geq d(x,\,y)^{\frac 12}
\delta_2^{\frac 12}.
$$
 
 
>From the definition follows that
$$
T^{-n}\phi(z)=W^u(T^{-n}z)\cap T^{-n}\gamma_2 .
$$
 
\proclaim{Sub-Lemma II.5} If $z,\,w\in\gamma_1\backslash
\Sigma_\infty( c,\,\delta_2)$ and $z'=\phi(z)$, $w'=\phi(w)$,
then, given $m\in\Bbb N$ such that
$$
\d(T^{-m}z,\,T^{-m}w)\geq \d(z,\,w)^{\frac 12}\delta_2^{\frac 12}
$$
and $T^{-m}[z,\,w]\subset T^{-m}\gamma$ is connected,
we have
$$
\aligned
\d(T^{-m}z,\,T^{-m}z')
&\leq \rho^2\d(z,\,w)^{\frac 12}\delta_2^{\frac 12}\\
\d(T^{-m}w,\,T^{-m}w')
&\leq \rho^2\d(z,\,w)^{\frac 12}\delta_2^{\frac 12}
\endaligned
$$
for some $\rho\geq 1$.
\endproclaim
\demo{Proof}
Let us consider the rectangle $Q=[z,\,w,\,w',\,z']$. Since the cone field 
is continuous the curves $\gamma_i$ are uniformly transversal to the 
unstable direction. This implies that there exists some fixed 
$\rho\in\Bbb R^+$, such that the area of the rectangle is less than 
$\rho\d(z,\,w)\delta_2$.
Therefore, since the area of $T^{-n}Q$ will be larger than
$\rho^{-1}\d(T^{-m}z,\,T^{-m}w) \d(T^{-m}z,\,T^{-m}z')$ the result follows
by the area preserving property of $T$.
\enddemo
 
In conclusion, given $x,\,y\in\gamma_1$, $d(x,y)<\delta_2^{\frac 12}$, we
choose the first $n\in\Bbb N$ such that
$$
d(T^{-n}x,\,T^{-n}y)\geq d(x,\,y)^{\frac 12}\delta_2^{\frac 12}.
$$
Then, due to the continuity of the cone field, the fact that $T^{-n}$ is 
smooth on the rectangle determined by the points $x,\,y,\,\phi(x),\,
\phi(y)$, and uniform hyperbolicity, we have that $T^{-n}[x,\,y]$
and $[T^{-n}\phi(x),\,T^{-n}\phi(y)]$ are essentially parallel curves.
Sub-Lemma II.5 implies then that
$$
\aligned
\d(T^{-n}\phi(x),\,T^{-n}\phi(y))&\leq \d(T^{-n}x,\,T^{-n}y)
+2\rho\d(T^{-n}\gamma_1,\,T^{-n}\gamma_2)\\
&\leq\d(T^{-n}x,\,T^{-n}y)+2\rho^3\d(x,\,y)^{\frac 12}
\delta_2^{\frac 12}
\endaligned
$$
or
$$
\frac{\d(T^{-n}\phi(x),\,T^{-n}\phi(y))}{\d(T^{-n}x,\,T^{-n}y)}
\leq 1+2\rho^3
$$
so, we have
$$
B^{-1}\leq 
\frac{\d(T^{-n}\phi(x),\,T^{-n}\phi(y))}{\d(T^{-n}x,\,T^{-n}y)}
\leq B.
$$
Hence,
$$
\aligned
\frac{\d (\phi(x),\,\phi(y))}{\d(x,\,y)}&\leq
\frac{\int_{T^{-n}(\phi(x))}^{T^{-n}(\phi(y))}\left|D_zT^{n}
\big|_{T^{-n}\gamma_2}\right|}{\int_{T^{-n}(x)}^{T^{-n}(y)}\left|D_wT^{n}
\big|_{T^{-n}\gamma_1}\right|}\\
&\leq \e^{\sum_{j=0}^nc_2\delta_2\lambda^{-j}}
\frac{\d(T^{-n}\phi(x),\,T^{-n}\phi(y))}{\d(T^{-n}x,\,T^{-n}y)}
\leq A.
\endaligned
$$
\enddemo
 
Finally,
 
\proclaim{Lemma II.6}There exists $D>0$ such that, 
for each $x,\,y\in \gamma_2\backslash
\Sigma_\infty(2c,\,\delta_2)$,
holds
$$
\frac{J_\phi(x)}{J_\phi(y)}\leq\e^{D d(x,\,y)^{\frac 12}\delta_2^{\frac 12}}
$$
\endproclaim
\demo{Proof}
Again, we choose $x,\,y\in\gamma_2\backslash\Sigma_\infty(2c,\,\delta_2)$
such that $d(x,\,y)\leq\delta_2$, and $n\in\Bbb N$ such that, for the
first time,
$$
d(T^{-n}x,\,T^{-n}y) \geq d(x,\,y)^{\frac 12}\delta_2^{\frac 12}.
$$
Setting $z=\phi^{-1}(x),\,w=\phi^{-1}(y)$, we have
$$
\aligned
\frac{J_\phi(x)}{J_\phi(y)}&=
\prod_{j=0}^\infty\left|\frac{D_{T^{-j}x}T
\big|_{T^{-j}\gamma_2}}
{D_{T^{-j}\phi^{-1}(x)}T\big|_{T^{-j}\gamma_1}}\right|
\left|\frac{D_{T^{-j}\phi^{-1}y}T
\big|_{T^{-j}\gamma_1}}
{D_{T^{-j}y}T\big|_{T^{-j}\gamma_2}}\right|\\
&=\prod_{j=0}^{n-1}\left|\frac{D_{T^{-j}x}T\big|_{T^{-j}\gamma_2}}
{D_{T^{-j}(y)}T\big|_{T^{-j}\gamma_2}}\right|
\left|\frac{D_{T^{-j}\phi^{-1}y}T\big|_{T^{-j}\gamma_1}}
{D_{T^{-j}\phi^{-1}(x)}T\big|_{T^{-j}\gamma_1}}\right|\\
&\times
\prod_{j=n}^\infty\left|\frac{D_{T^{-j}(x)}T\big|_{T^{-j}\gamma_2}}
{D_{T^{-j}\phi^{-1}(x)}T\big|_{T^{-j}\gamma_1}}\right|
\left|\frac{D_{T^{-j}\phi^{-1}(y)}T\big|_{T^{-j}\gamma_1}} 
{D_{T^{-j}y}T\big|_{T^{-j}\gamma_2}}\right|\\
&\leq \e^{(1+A)\sum_{j=0}^{n-1} \lambda^{-n+j}\d(x,\,y)^{\frac 12}
\delta_2^{\frac 12}}\e^{2\sum_{j=n}^\infty \rho^2  \lambda^{n-j}
\d(x,\,y)^{\frac 12} \delta_2^{\frac 12}}\\
&\leq\e^{D \delta_2^{\frac 12}\d(x,\,y)^{\frac 12} }.
\endaligned
$$
\enddemo
 
\vskip.2cm
\noindent{\bf Billiards.}
\vskip.4cm
 
We conclude with a brief discussion on how to extend the above results to
the case of Sinai Billiards (with finite horizon and no breaking points).
 
The only real change is the necessity to better estimate $D(\phi)$.
To do so, some preliminaries are called for.
 
In \S 6 we have introduced two strictly invariant cone
fields: $\C_-=\{(1,\,-u)\;|\;u\in[1,\,u_0]\}$ and 
$\C_+=\{(1,\,u)\;|\;u\in[1,\,u_0]\}$ ($DT^{-1}\C_- \subset 
\C_- $ and $DT\C_+\subset \C_+$).
 
Moreover, if $v\in\C_-(x)$ we will call $\lambda^s(x,\,v)$ the expansion
of $v$ under $DT^{-1}$, and, for $v\in\C_+$, we  will call
$\lambda^u(x,v)$ the expansion of $v$ under $DT$ (the length of a vector
is measured in the seminorm $\sin\varphi ds$ introduced in \S 6).
Also, let us define 
$\lambda^s_n(x,\,v)=\prod\limits_{i=0}^{n-1}\lambda^s(T^{-i}x,\,
D_xT^{-i}v)$ and $\lambda^s_n(x)=\prod\limits_{i=0}^{n-1}\lambda^s(T^{-i}x,\,
0)\leq \inf\limits_{v\in\C_-}\lambda^s_n(x,\,v)$, we adopt the 
corresponding definitions for $\lambda^u_n(x,\,v)$ and $\lambda^u_n(x)$.
It is easy to check that $\lambda^u_n(x)=\lambda^s_n(T^nx)$, and that
exists $D\in\Bbb R^+$ such that $D^n\lambda^u_n(x)\geq \lambda^u_n(x,\,v)$
for each $v\in \C_+ $.
 
Within the above framework we can redefine $\Sigma_n(c,\,\delta_2)$
(our new approximation of the complement of $D(\phi)$).
 
\proclaim{Definition II.7} Let $\R_n(\delta_2)=\left\{z\in\Bbb T^2\;|\;
d(z,\,\R^-)\leq\delta_2\lambda^u_n(z)^{-\alpha}\right\}$, 
$\alpha\in(0,\,\frac 12)$,
then
$$
\Sigma_n(c,\,\delta_2)=\bigcup_{i=0}^n T^i\R_i(c\delta_2).
$$
\endproclaim
 
The next Lemma shows that the above definition it is not too restrictive.
 
\proclaim{Lemma II.8} If $\gamma_1,\,\gamma_2\in\wh\Gamma(\delta_1)$,
$\phi:\gamma_1\to\gamma_2$ is the canonical isomorphism, and their
distance is less than $\delta_2$, then 
$\gamma_1\backslash\Sigma_\infty(c,\,\delta_2)\supset D(\phi)$, provided
$\lambda^{\alpha-1}D^\alpha<1$, $c$ is chosen large enough, and $\delta_2$
is smaller than some fixed value.
\endproclaim
\demo{Proof}
All is necessary to show is that the points in
$\gamma_1\backslash\Sigma_\infty(c,\,\delta_2)$ have an unstable manifold
longer than some constant times $\delta_2$.
 
To prove such a fact, one can use the usual construction: let
$x\in\gamma_1\backslash\Sigma_\infty(c,\,\delta_2)$, $T^{-n}x=
(s_n,\,\varphi_n)$, and $\ov\gamma_n=\{(s_n+t,\,\varphi_n+t)\;|\;
t\in[-\delta_2,\,\delta_2]\}$, then it is well known that the connected
component of $T^n\ov\gamma_n$ containing $x$ 
(eventually restricted to a $\delta_2$
neighborhood of $x$) converges to $W^u(x)$.
Let us call $\wh\gamma_n$ such a connected component and let us estimate 
its size.
To do so we need first to estimate $\Lambda_-(\wh\gamma_n)=
\inf\limits_{z\in\wh\gamma_n}\lambda^s_n(z)$ and
$\Lambda_+(\wh\gamma_n)=
\sup\limits_{z\in\wh\gamma_n}\lambda^s_n(z)$.
 
Clearly, there exists $B>0$ such that 
$\frac{\Lambda_+(\wh\gamma_0)}{\Lambda_-(\wh\gamma_0)}\leq B$, 
let us suppose that
$\frac{\Lambda_+(\wh\gamma_i)}{\Lambda_-(\wh\gamma_i)}\leq B$ for each
$i<n$, then, if $y\in\wh\gamma_n$
$$
\aligned
\frac{\lambda^s_n(x)}{\lambda^s_n(y)}&\leq
\e^{c_4\sum_{i=0}^{n-1}[d(T^{-i}x,\,T^{-i}y)\Lambda_+(\wh\gamma_i)^\alpha 
D^{i\alpha}+d(T^{-i}x,\,T^{-i}y)^{\frac 12}]}\\
&\leq\e^{c_4d(x,\,y)^{\frac 12}B^\alpha
\sum_{i=0}^{n-1}\Lambda_-(\wh\gamma_i)^{\alpha-1} D^{i\alpha}}
\leq\e^{c_4B^\alpha d(x,\,y)^{\frac 12}
\sum_{i=0}^{n-1}(\lambda^{\alpha-1}D^\alpha)^i}\leq B ,
\endaligned
$$
providing we choose $\alpha$ such that $\lambda^{\alpha-1}D^\alpha<1$.
 
The latter inequality implies $B\lambda^s_n(x)\geq\Lambda_+(\wh\gamma_n)
\geq\Lambda_-(\wh\gamma_n)\geq B^{-1}\lambda^s_n(x)$,
which suffice to prove the Lemma.
\enddemo
 
The rest of the argument goes more or less as in the bounded derivative
case.
 
Lemma II.2 is proven in the same manner, apart from the fact that now we
divide $T^n\gamma$ also according to homogeneity strips, as already discussed
in \S 6. In particular, Lemma 6.2 contains the estimates necessary to prove 
the corresponding of Sub-Lemma II.3; the result follows provided one
chooses $1>\zeta>\frac 1{1+\alpha^2}$.
 
The only problem with the extension of Lemma II.4 to the present context
is in the distortion estimate. Let us discuss such issue.
 
Let $\wt\gamma_n=T^{-n}[x,\,y]\subset T^{-n}\gamma$, 
$\gamma\in\wh\Gamma_{\delta_1}$. If $\wt\gamma_i$ is
connected,
$$
d(T^{-i}x,\,T^{-i}y)\leq d(x,\,y)^{\frac 12}\delta_2^{\frac 12},
$$
and for $\Lambda_-(\wt\gamma_i)$, $\Lambda_+(\wt\gamma_i) $ (defined as above)
$$
\frac{\Lambda_+(\wt\gamma_i)}{\Lambda_-(\wt\gamma_i)}\leq B
$$
for each $i<n$, then
$$
\Lambda_-(\wt\gamma_i)^{-1}\geq d(x,\,y)^{\frac 12}\delta_2^{-\frac 12}.
$$
 
Hence, for each $z,\,w\in\wt\gamma_n$ 
$$
\frac{\lambda^u_n(z)}{\lambda^u_n(w)}\leq
\e^{c_4 \sum_{i=0}^{n-1}\Lambda_+(\wt\gamma_i)^\alpha D^{\alpha i} 
d(T^{-i}x,\,T^{-i}y)}\leq
\e^{c_4 B^\alpha \sum_{i=0}^{n-1}\Lambda_-(\wt\gamma_i)^{\alpha-1} 
D^{\alpha i}\delta_2}\leq B .
$$
 
Accordingly, we can iterate the interval $[x,\,y]$ until
$$
d(T^{-n}x,\,T^{-n}y)\geq d(x,\,y)^{\frac 12}\delta_2^{\frac 12}
$$
without disconnecting it and maintaining uniformly bounded distortion.
Lemmas II.5, II.6 follow by choosing $\alpha$ sufficiently small (although,
in this case, $J_\phi$ turns out to be H\"older continuous of order strictly
less than $\frac 12$).
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\enddocument
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