Semi-continuity of Oseledets subspaces
and Pesin sets with exponentially small tails

Luchezar Stoyanov111E-mail<EMAIL_ADDRESS>
Department of Mathematics and Statistics, University of Western Australia,

Crawley 6009 WA, Australia

###### Abstract

Let $f:M\longrightarrow M$ be a $C^{2}$ Axiom A diffeomorphism on a compact
Riemannian manifold. We prove that if $X$ is a basic set for $f$ and $\mu$ is
a Gibbs measure for $f$ determined by a Hölder continuous potential such that
the Oseledets subspaces $E_{i}(x)$ depend upper semi-continuously on $x\in X$
then there exists a Pesin set with exponentially small tails for $\mu$. We
then apply this result to open billiard flows in ${\mathbb{R}}^{d}$. As a
consequence we derive that the non-wandering set of such a flow admits a Pesin
set with exponentially small tails with respect to every Gibbs measure. This
is expected to lead to results on exponential mixing for open billiard flows
similar to these in [26] for contact Anosov flows.

MSC: Primary: 37D20, 28D05; Secondary: 34D08

Keywords: Lyapunov exponents, Oseledets subspaces, Pesin set, open billiard
flow

##  1 Introduction

###  1.1 Pesin sets with exponentially small tails

Let $f:M\longrightarrow M$ be a $C^{2}$ Axiom A diffeomorphism on a compact
Riemannian manifold $M$ and let $X$ be a basic set for $f$. Let $\mu$ be a
Gibbs measure on $X$ for $f$ defined by a Hölder continuous potential and let

(1.1)
$\lambda_{k}<\lambda_{k-1}<\ldots<\lambda_{\ell+1}<0<\lambda_{\ell}<\ldots<\lambda_{2}<\lambda_{1}$

be the Lyapunov spectrum of $f$ with respect to $\mu$, which is constant on a
subset ${\mathcal{L}}$ of $X$ of full measure. Denote by $E_{i}(x)$ the
Oseledets subspace of $T_{x}M$ corresponding to the Lyapunov exponent
$\lambda_{i}$.

Given $\epsilon>0$, a measurable function
$R_{\epsilon}:{\mathcal{L}}\longrightarrow(1,\infty)$ is called a Lyapunov
$\epsilon$-regularity function  if

$\frac{\|v\|}{R_{\epsilon}(x)\,e^{|n|\epsilon}}\leq\frac{\|df^{n}(x)\cdot
v\|}{e^{n\lambda_{i}}}\leq R_{\epsilon}(x)\,e^{|n|\epsilon}\|v\|\quad,\quad
v\in E_{i}(x)\;,\;n\in{\mathbb{Z}},$

for all $x\in{\mathcal{L}}$ and all $i=1,\ldots,k$ and

$e^{-\epsilon}\leq\frac{R_{\epsilon}(f(x))}{R_{\epsilon}(x)}\leq
e^{\epsilon}\quad,\quad x\in{\mathcal{L}}.$

A compact subset $P$ of ${\mathcal{L}}$ is called a Pesin set if for every
$\epsilon>0$ there exist a Lyapunov $\epsilon$-regularity function
$R_{\epsilon}$ and a constant $C>0$ such that $R_{\epsilon}(x)\leq C$ for all
$x\in P$. A Pesin set $P$ will be called a Pesin set with exponentially small
tails for $\mu$ if for every $\delta>0$ there exist constants $C,c>0$ such
that

$\mu\left(\left\\{x\in{\mathcal{L}}:\\#\\{m:0\leq m\leq n-1\;,\;f^{m}(x)\notin
P\\}\geq\delta\,n\right\\}\right)\leq C\,e^{-cn}$

for every integer $n\geq 0$. In other words, for every $n\geq 0$, removing a
set of exponentially small measure from ${\mathcal{L}}$, for the remaining
$x$, at least $(1-\delta)n$ points from the orbit $x,f(x),\ldots,f^{n-1}(x)$
belong to the Pesin set $P$.

Our first result in this paper is the following.

Theorem 1. Let $f:X\longrightarrow X$ be as above and let $\mu$ be a Gibbs
measure on $X$ determined by a Hölder continuous potential on $X$. Assume that
the Oseledets subspaces $E_{i}(x)$ depend upper semi-continuously on $x$. Then
there exists a Pesin set with exponentially small tails for $\mu$.

See Definition 2 in section 2 below for the precise meaning of upper semi-
continuity of Oseledets subspaces.

We will prove Theorem 1 in section 3 below.

###  1.2 Open billiard flows

Let $K$ be a subset of ${{\mathbb{R}}}^{d}$ ($d\geq 2$) of the form
$K=K_{1}\cup K_{2}\cup\ldots\cup K_{k_{0}},$ where $K_{i}$ are compact
strictly convex disjoint domains in ${\mathbb{R}}^{d}$ with $C^{3}$ boundaries
$\Gamma_{i}=\partial K_{i}$ and $k_{0}\geq 3$. Set
$\Omega=\overline{{{\mathbb{R}}}^{d}\setminus K}$ and $\Gamma=\partial K$. We
assume that $K$ satisfies the following (no-eclipse) condition:

${\rm(H)}\quad\quad\qquad\begin{cases}\mbox{\rm for every pair $K_{i}$,
$K_{j}$ of different connected components of $K$}\cr\mbox{\rm the convex hull
of $K_{i}\cup K_{j}$ has no common points with any other}\cr\mbox{\rm
connected component of $K$. }\end{cases}$

With this condition, the billiard flow $\phi_{t}$ defined on the sphere bundle
$S(\Omega)$ in the standard way is called an open billiard flow. It has
singularities, however its restriction to the non-wandering set $\Lambda$ has
only simple discontinuities at reflection points.

The non-wandering set $\Lambda$ for the flow $\phi_{t}$ is the set of those
$x\in S(\Omega)$ such that the trajectory $\\{\phi_{t}(x):t\in{\mathbb{R}}\\}$
is bounded. Notice that the natural projection of $\phi_{t}$ on the quotient
space $S(\Omega)/\sim$, where $\sim$ is the equivalence relation
$(q,v)\sim(p,w)$ iff $q=p$ and $v=w$ or $q=p\in\partial K$ and $v$ and $w$ are
symmetric with respect to $T_{q}(\partial K)$, is continuous. Moreover
whenever both $x$ and $\phi_{t}(x)$ are in the interior of $S(\Omega)$ and
sufficiently close to $\Lambda$, the map $y\mapsto\phi_{t}(y)$ is smooth on a
neighbourhood of $x$. It follows from results of Sinai ([22], [23]) that
$\Lambda$ is a hyperbolic set for $\phi_{t}$, and it is easily seen that there
exists an open neighbourhood $U$ of $\Lambda$ in $S(\Omega)$ such that
$\Lambda=\cap_{t\in{\mathbb{R}}}\phi_{t}(U)$. In fact, $\Lambda$ is clearly
the maximal compact $\phi_{t}$-invariant subset of $S(\Omega)$. Moreover, it
follows from the natural symbolic coding for the natural section of the flow
(the so called billiard ball map) that the periodic points are dense in
$\Lambda$ and $\phi_{t}$ is transitive on $\Lambda$ (see e.g. [13] or [24]).
Thus, $\Lambda$ is a basic set for $\phi_{t}$ and the classical theory of
hyperbolic flows applies in the case under consideration (see e.g. Bowen [6]
or part 4 in the monograph of Katok and Hasselblatt [15]; see also [9]).

Our second main result in this paper is the following.

Theorem 2. Let $f:\Lambda\longrightarrow\Lambda$ be the time-one map for the
open billiard flow on the non-wandering set ${\Lambda}$ and let $\mu$ be an
arbitrary Gibbs measure on $\Lambda$ determined by a Hölder continuous
potential on $\Lambda$. Then the Oseledets subspaces $E_{i}(x)$ depend upper
semi-continuously on $x$.

As an immediate consequence of Theorems 1 and 2 we obtain the following.

Corollary 1. Under the assumptions of Theorem 2, there exists a Pesin set with
exponentially small tails for $\mu$.

###  1.3 Exponential mixing for open billiard flows

In [26] we considered contact Anosov flows on compact Riemannian manifolds and
proved exponential mixing for Gibbs measures determined by Hölder continuous
potentials that admit Pesin sets with exponentially small tails. A similar
result is expected to be established for contact Axiom A flows on basics sets,
again for Gibbs measures admitting a Pesin set with exponentially small tails.
This, combined with Corollary 1 above, would imply exponential mixing for open
billiard flows in ${\mathbb{R}}^{d}$ ($d\geq 2$) with respect to any Gibbs
measure determined by a Hölder continuous potential. Such a result was proved
in [24] for open billiard flows in ${\mathbb{R}}^{2}$ and later in [25] for
multidimensional open billiard flows however under an additional pinching
condition.

Since the breakthrough work of Dolgopyat [10] there has been a lot of activity
in studying decay of correlations for hyperbolic and non-uniformly hyperbolic
systems, including systems with singularities. We refer the reader to the
introductions in [3] and [26] for some general information and references. We
should mention in particular the recent work by Baladi, Demers and Liverani
[3] where they established exponential decay of correlations for Sinai
billiards with finite horizon on a two-dimensional torus, thus solving a long
standing open problem.

##  2 Lyapunov exponents and Oseledets subspaces

To prove Theorem 1 we will use certain background related to the so called
Multiplicative Ergodic Theorem and some arguments from [12].

We will actually consider a more general situation222It seems an even more
general situation could be considered in a similar way. and prove a more
general result from which Theorem 1 will be easily derived.

Let $f$ be an invertible transitive subshift of finite type over a bilateral
symbol space $X$ and let $\mu$ be a Gibbs measure for $f$ determined by a
Hölder continuous potential on $X$ (see e.g. [6] or [18]). In particular,
$\mu$ is $f$-invariant and ergodic. Let $A$ be a continuous linear cocycle
over $f$ acting on a continuous ${\mathbb{R}}^{d}$-bundle $E$ over $X$. Thus,
$A(x):E(x)\longrightarrow E(f(x))$ is a linear map for each $x\in X$ and

$A^{n}(x)=A(f^{n-1}(x))\circ A(f^{n-2}(x))\circ\ldots\circ A(f(x))\circ
A(x):E(x)\longrightarrow A(f^{n}(x))$

for every integer $n\geq 1$.

Given an integer $p=1,\ldots,d$, let $\mbox{\rm Gr}_{p}({\mathbb{R}}^{d})$ be
the Grassman manifold of the linear subspaces of ${\mathbb{R}}^{d}$ of
dimension $p$ endowed with the usual distance $d(U,V)$ between subspaces
$U,V\in\mbox{\rm Gr}_{p}({\mathbb{R}}^{d})$ defined by

$d(U,V)=\max\\{|\langle u,w\rangle|:u\in U,w\in V^{\perp},\|u\|=\|w\|=1\\}.$

The following is Oseledets’ Multiplicative Ergodic Theorem stated under the
above assumptions (see e.g. [1], [5], [27] or [21] for related detailed
exposition and proofs).

Theorem 3. (Multiplicative Ergodic Theorem). There exists a subset
${\mathcal{L}}$ of $X$ with $\mu({\mathcal{L}})=1$ such that:

(a) For all $x\in{\mathcal{L}}$ there exists $\displaystyle
N(x)=\lim_{n\to\infty}(A^{n}(x)^{t}A^{n}(x))^{1/2n}.$

(b) There exist an integer $k\geq 1$ such that the operator (matrix) $N(x)$
has $k$ distinct eigenvalues $t_{1}(x)>\ldots>t_{k}(x)$ for all
$x\in{\mathcal{L}}$.

(c) There exist numbers $t_{1}>t_{2}>\ldots>t_{k}$ such that
$(t^{(n)}_{i}(x))^{1/n}\to t_{i}$ for all $x\in{\mathcal{L}}$ and all
$i=1,\ldots,k$.

(d) For all $x\in{\mathcal{L}}$ and every $j=1,\ldots,k$ the dimension
$\dim(E^{(n)}_{j}(x))=m_{j}$ is constant and there exists
$\lim_{n\to\infty}E^{(n)}_{j}(x)=E_{j}(x)$ in $\mbox{\rm
Gr}_{m_{j}}({\mathbb{R}}^{d})$.

(e) For all $x\in{\mathcal{L}}$ and every $j=1,\ldots,k$ we have

$\lim_{n\to\infty}\frac{1}{n}\log\|A^{n}(x)u\|=\log t_{j}\quad,\quad u\in
F_{j}(x)\setminus F_{j-1}(x),$

where $F_{0}(x)=\\{0\\}$ and $F_{j}(x)=E_{1}(x)\oplus\ldots\oplus E_{j}(x)$
for all $j=1,\ldots,k$.

The numbers

$\lambda_{1}=\log t_{1}>\lambda_{2}=\log t_{2}>\ldots>\lambda_{k}=\log t_{k}$

are the (distinct) Lyapunov exponents of the cocycle $A$ over $f$. Setting

$\lambda^{(n)}_{i}(x)=\frac{1}{n}\log t^{(n)}_{i}(x),$

we have $\lambda^{(n)}_{i}(x)\to\lambda_{i}$ for all $x\in{\mathcal{L}}$.

Given a finite-dimensional real inner product space $V$ with inner product
$\langle\cdot,\cdot\rangle$, consider the $r$-fold exterior product
$\Lambda^{r}V$ which is naturally identified with linear combinations of wedge
products $v_{1}\wedge v_{2}\wedge\cdots\wedge v_{r}$ of vectors in $V$ (see
e.g. Sect. 3.2.3 in [1]). A natural inner product is then defined on
$\Lambda^{r}V$ so that

$\langle u_{1}\wedge\cdots\wedge u_{r},v_{1}\wedge\cdots\wedge
v_{r}\rangle=\det(\langle u_{i},v_{j}\rangle)_{i,j=1}^{r}.$

Thus, $\|u_{1}\wedge\cdots\wedge u_{r}\|^{2}=\det(\langle
u_{i},u_{j}\rangle)_{i,j=1}^{r}$. If $L:V\longrightarrow V$ is a linear
operator, then the linear operator
$\Lambda^{r}L:\Lambda^{r}V\longrightarrow\Lambda^{r}V$ is defined by

$(\Lambda^{r}L)(v_{1}\wedge v_{2}\wedge\cdots\wedge v_{r})=Lv_{1}\wedge
Lv_{2}\wedge\cdots\wedge Lv_{r}.$

Let $\chi_{1}\geq\chi_{2}\geq\ldots\geq\chi_{d}$ be the Lyapunov exponents
$\lambda_{1},\lambda_{2},\ldots,\lambda_{k}$ of the linear cocycle $A(x)$ with
the required number of repetitions, so that for every $i=1,\ldots,d$ we have
$\chi_{i}=\lambda_{j}$ for some (unique) $j=j_{i}=1,\ldots,k$. It is well-
known that

$\log\|\Lambda^{i}A(x))\|=\chi_{1}+\chi_{2}+\ldots+\chi_{i}$

for all $i=1,\ldots,d$.

Definition 1. ([12]) We say that the cocycle $A$ has exponential large
deviations for all exponents if for any $\epsilon>0$ there exist constants
$C,c>0$ such that

$\mu\left(\left\\{x\in
X:\left|\log\|\Lambda^{i}A^{n}(x)\|-n(\chi_{1}+\ldots+\chi_{i})\right|\geq
n\epsilon\right\\}\right)\leq Ce^{-cn}$

for all $n\geq 0$ and all $i=1,\ldots,d$.

It follows from Proposition 3.3 in [12] that

$\mu\left(\left\\{x\in{\mathcal{L}}:\lambda^{(n)}_{1}(x)+\ldots+\lambda^{(n)}_{i}(x)\geq\lambda_{1}+\ldots+\lambda_{i}+\epsilon\right\\}\right)\leq
Ce^{-cn}$

for all $n\geq 1$. That is, given $i=1,\ldots,k$, for every $n\geq 1$ we have

$\lambda^{(n)}_{1}(x)+\ldots+\lambda^{(n)}_{i}(x)<\lambda_{1}+\ldots+\lambda_{i}+\epsilon$

for ‘most’ $x$.

We will prove a similar estimates from below and above for every Lyapunov
exponent $\lambda_{i}$ under a certain additional assumption.

Definition 2. We will say that the Oseledets subspaces $E_{i}(x)$ related to
$\mu$ depend upper semi-continuously on $x$ if we can choose the set
${\mathcal{L}}$ of full measure so that for every $x\in{\mathcal{L}}$, every
sequence $\\{x_{n}\\}$ in ${\mathcal{L}}$ converging to $x$, every
$i=1,\ldots,k$ and every sequence $\\{u_{n}\\}_{n=1}^{\infty}$ with $u_{n}\in
E_{i}(x_{n})$ and $\|u_{n}\|=1$ for all $n\geq 1$, if there exists
$u=\lim_{n\to\infty}u_{n}$, then $u\in E_{i}(x)$.

In the next theorem the assumption that the Oseledets spaces $E_{i}(x)$ depend
upper semi-continuously on $x$ is essential. As the counterexamples in section
A in [12] show, such a statement cannot be true without additional
assumptions.

Theorem 4. Let $(T,\mu)$ be a transitive subshift of finite type with a Gibbs
measure $\mu$ and let $A$ be an invertible continuous linear cocycle above $T$
with Lyapunov exponents $\lambda_{1}>\lambda_{2}>\ldots>\lambda_{k}$. Assume
that $A^{-1}$ is also continuous and the Oseledets subspaces $E_{i}(x)$ depend
upper semi-continuously on $x$. Then for every $\epsilon>0$ there exist
constants $C,c>0$ such that

(2.1)
$\mu\left(\left\\{x\in{\mathcal{L}}:\lambda^{(n)}_{i}(x)\geq\lambda_{i}+\epsilon\right\\}\right)\leq
Ce^{-cn}$

and

(2.2)
$\mu\left(\left\\{x\in{\mathcal{L}}:\frac{1}{n}\lambda^{(n)}_{i}(x)\leq\lambda_{i}-\epsilon\right\\}\right)\leq
Ce^{-cn}$

for all $n\geq 0$ and all $i=1,\ldots,k$.

Proof of Theorem 1.  It follows from (2.1) and (2.2) that the cocycle $A$ has
exponential large deviations for all exponents. Combining this with Theorem
1.7 in [12] yields Theorem 1.

Throughout the whole paper we will denote by $C>1$ and $c\in(0,1)$ generic
global constants that may change from place to place.

##  3 Proof of Theorem 4

Let again $\mu$ be a Gibbs measure on $X$ defined by a Hölder continuous
potential and let $\lambda_{1},\ldots,\lambda_{k}$ be the Lyapunov spectrum of
the cocycle $A$ with respect to $\mu$, constant on a measurable subset
${\mathcal{L}}$ of $X$ of full measure. Consider the corresponding invariant
decomposition

$E_{1}(x)\oplus E_{2}(x)\oplus\cdots\oplus E_{k}(x)={\mathbb{R}}^{d},$

where $E_{i}(x)$ is the Oseledets subspace corresponding to the Lyapunov
exponent $\lambda_{i}$. We assume that the Oseledets subspaces $E_{i}(x)$
depend upper semi-continuously on $x\in{\mathcal{L}}$ (see Definition 2). Set

$\widehat{E}_{i}(x)=\\{u\in E_{i}(x):\|u\|=1\\}.$

Denoting by $m_{i}$ the multiplicity of $\lambda_{i}$, we have
$\dim(E_{i}(x))=m_{i}.$

We will prove (2.1). The estimate (2.2) will be easily derived from (2.1).

Fix an $i=1,\ldots,k$. Throughout this section $i$ will stay fixed. Set

$\gamma_{n}(x)=\log\|A^{n}(x)_{|E_{i}(x)}\|\quad,\quad
x\in{\mathcal{L}}\>,\>n\geq 1.$

Clearly, $|\gamma_{n}(x)|\leq{\rm Const}\;<\infty$ for all $x\in{\mathcal{L}}$
and $\gamma_{n}$ is a sub-additive function on ${\mathcal{L}}$. We can assume
${\mathcal{L}}$ is so that

$\lim_{n\to\infty}\frac{\gamma_{n}(x)}{n}=\lambda_{i}$

for all $x\in{\mathcal{L}}$. By Kingsman’s Ergodic Theorem (see e.g. [1], [27]
or [21]) there exists the limit

(3.1)
$\lim_{n\to\infty}\frac{1}{n}\int_{X}\gamma_{n}(x)\,d\mu(x)=\inf_{n}\frac{1}{n}\int_{X}\gamma_{n}(x)\,d\mu(x)=\lambda_{i}.$

Since

$\gamma_{n}(x)=\max\left\\{\log\left\|A^{n}(x)\cdot
v\right\|:v\in\widehat{E}_{i}(x)\right\\},$

we will derive from our assumptions that $\gamma_{n}$ is upper semi-continuous
on ${\mathcal{L}}$. However it is not clear whether $\gamma_{n}$ is continuous
and whether it has a continuous extension to $X$, and so it is not immediately
possible to use Lemma 3.2333Lemma 3.2 ([12]): Let $(T,\mu)$ be a transitive
subshift of finite type with a Gibbs measure $\mu$ and let $a(n,x)$ be a
subadditive cocycle above $T$ such that $a(n,x)$ is continuous for all $n$.
Let $\chi$ be the almost sure limit of $a(n,x)/n$ and assume that
$\chi>-\infty$. Then, for any $\epsilon>0$, there exist constants $C>0$, $c>0$
such that, for all $n\geq 0$ we have $\mu\left(\\{x\in
X:a(n,x)/n\geq\chi+\epsilon\\}\right)\leq C\,e^{-cn}.$ in [12].

Lemma 1. For all $n\geq 1$, the function $\gamma_{n}$ is upper semi-continuous
on ${\mathcal{L}}$, i.e.

$U=\\{x\in{\mathcal{L}}:\gamma_{n}(x)<\omega\\}$

is an open subset of ${\mathcal{L}}$ for all $\omega\in{\mathbb{R}}$.

Proof. Let $n\geq 1$ and let $U$ be defined as above for some
$\omega\in{\mathbb{R}}$. We will prove that ${\mathcal{L}}\setminus U$ is
closed in ${\mathcal{L}}$. Let $x_{m}\to x$ in ${\mathcal{L}}$ for some
sequence $\\{x_{m}\\}\subset{\mathcal{L}}\setminus U$. Then
$\gamma_{n}(x_{m})\geq\gamma$ for all $m$. From the definition of
$\gamma_{n}$, for all $m\geq 1$ there exists $u_{m}\in\widehat{E}_{i}(x_{m})$
with $\gamma_{n}(x_{m})=\log\|A^{n}(x_{m})\cdot u_{m}\|$. Replacing
$\\{x_{m}\\}$ with an appropriate subsequence, we may assume that $u_{m}\to u$
for some $u\in{\mathbb{R}}^{d}$. Then $\|u\|=1$ and the upper semi-continuity
of the bundle $E_{i}$ (see Definition 2) implies that $u\in E_{i}(x)$. We now
get

$\omega\leq\gamma_{n}(x_{m})=\log\|A^{n}(x_{m})\cdot
u_{m}\|\to\log\|A^{n}(x)\cdot u\|\leq\gamma_{n}(x),$

Therefore $\gamma_{n}(x)\geq\omega$, so $x\in{\mathcal{L}}\setminus U$. Thus,
${\mathcal{L}}\setminus U$ is closed in ${\mathcal{L}}$ and so $U$ is open in
${\mathcal{L}}$.

Proof of Theorem 4\. Since $A$ is continuous, there exists a constant
$C_{0}>0$ such that $\|A(x)\|\leq C_{0}$ for all $x\in X$. Then
$0\leq\gamma_{n}(x)\leq nC_{0}$ for all $x\in{\mathcal{L}}$ and all $n\geq 1$.

Since $\gamma_{n}$ is bounded it follows e.g. from Theorem 2.2 in [17] that it
has an upper semi-continuous extension to $X$ (which again takes values in a
bounded interval). We assume below that for every $n\geq 1$,
$\tilde{\gamma}_{n}$ is such an upper semi-continuous extension of the
original $\gamma_{n}$ to $X$.

By (3.1), there exists $n_{0}\geq 1$ so that

$\int_{X}\frac{\tilde{\gamma}_{n}(x)}{n}\,d\mu(x)=\int_{X}\frac{\gamma_{n}(x)}{n}\,d\mu(x)<\lambda_{i}+\frac{\epsilon}{8}$

for all $n\geq n_{0}$.

Fix an $m\geq n_{0}$. It follows from Baire’s Theorem for semi-continuous
functions ([2]) and the upper semi-continuity of $\tilde{\gamma}_{m}$ that
there exists a sequence

$h_{1}\geq h_{2}\geq\ldots\geq h_{n}\geq\ldots$

of continuous functions on $X$ such that $h_{n}(x)\to\tilde{\gamma}_{m}(x)$ as
$n\to\infty$ for all $x\in X$. Now Lebesgue’s Dominated Convergence Theorem
implies

$\lim_{n\to\infty}\int_{X}\frac{h_{n}}{m}\,d\mu=\int_{X}\frac{\tilde{\gamma}_{m}}{m}\,d\mu=\int_{X}\frac{\gamma_{m}}{m}\,d\mu<\lambda_{i}+\frac{\epsilon}{8},$

therefore for sufficiently large $n$ we have

$\int_{X}\frac{h_{n}}{m}\,d\mu<\lambda_{i}+\frac{\epsilon}{8}.$

Fix such an $n\geq 1$. It follows from well-known results about approximation
of continuous functions by Lipschitz ones (see e.g. [11]) that there exists a
Lipschitz function $H$ on $X$ so that $|H(x)-h_{n}(x)|<\frac{\epsilon}{8}$ for
all $x\in X$. Set $F_{m}=H+\epsilon/8$. Then $F_{m}$ is Lipschitz and

$F_{m}\geq(h_{n}-\epsilon/8)+\epsilon/8=h_{n}\geq\tilde{\gamma}_{m}.$

In particular,

(3.2) $F_{m}(x)\geq\gamma_{m}(x)\quad,\quad x\in{\mathcal{L}}.$

Moreover

(3.3)
$\int_{X}\frac{F_{m}}{m}\,d\mu=\int_{X}\frac{H+\epsilon/8}{m}\,d\mu\leq\int_{X}\frac{h_{n}+\epsilon/4}{m}\,d\mu<\lambda_{i}+\frac{3\epsilon}{8},$

assuming $m\geq 2$.

Given arbitrary integers $n,s\geq 1$ and $x\in{\mathcal{L}}$, let
$u\in\widehat{E}_{i}(x)$ be such that

$\gamma_{n+s}(x)=\log\left\|A^{n+s}(x)\cdot u\right\|,$

that is

$\log\left\|A^{n+s}(x)\cdot u\right\|=\max\left\\{\log\left\|A^{n+s}(x)\cdot
v\right\|:v\in\widehat{E}_{i}(x)\right\\}.$

Set

$w=\frac{1}{\left\|A^{s}(x)\cdot u\right\|}\,A^{s}(x)\cdot
u\in\widehat{E}_{i}(f^{s}(x)).$

We now derive

$\displaystyle\gamma_{n+s}(x)$ $\displaystyle=$
$\displaystyle\log\left\|A^{n+s}(x)\cdot
u\right\|=\log\left\|A^{n}(f^{s}(x))\cdot(A^{s}(x)\cdot u)|\right\|$
$\displaystyle=$ $\displaystyle\log\left(\left\|A^{n}(f^{s}(x))\cdot
w\right\|\cdot\|A^{s}(x)\cdot u\|\right)$ $\displaystyle=$
$\displaystyle\log\left\|A^{n}(f^{s}(x))\cdot w\right\|+\log\|A^{s}(x)\cdot
u\|\leq\gamma_{n}(f^{s}(x))+\gamma_{s}(x).$

Thus

(3.4) $\gamma_{n+s}(x)\leq\gamma_{n}(f^{s}(x))+\gamma_{s}(x)$

for all $x\in{\mathcal{L}}$ and all integers $n,s\geq 1$.

Recall the fixed integer $m\geq n_{0}$. Clearly $\gamma_{s}(x)\leq D=mC_{0}$
for all $x\in{\mathcal{L}}$ and all $s=0,1,\ldots,m$.

Next, we will use arguments similar to the ones in the proofs of Lemmas 3.1
and 3.2 in [12].

Let $x\in{\mathcal{L}}$ and let $n\geq 1$. Fix for a moment an arbitrary
integer $j$ with $0\leq j\leq m-1$. Then $n=r+mp+j$ for some integers $p$ and
$r$ with $0\leq r\leq m-1$. Applying (3.4) inductively and using (3.2), we get

$\displaystyle\gamma_{n}(x)$ $\displaystyle=$
$\displaystyle\gamma_{(r+mp)+j}(x)\leq\gamma_{r+mp}(f^{j}(x))+\gamma_{j}(x)\leq\gamma_{(r+m(p-1))+m}(f^{j}(x))+D$
$\displaystyle\leq$
$\displaystyle\gamma_{r+m(p-1)}(f^{m+j}(x))+\gamma_{m}(f^{j}(x))+D$
$\displaystyle\leq$
$\displaystyle\gamma_{r+m(p-1)}(f^{m+j}(x))+F_{m}(f^{j}(x))+D$
$\displaystyle\leq$
$\displaystyle\gamma_{r+m(p-2)}(f^{2m+j}(x))+F_{m}(f^{m+j}(x))+F_{m}(f^{j}(x))+D\leq\ldots$
$\displaystyle\leq$
$\displaystyle\gamma_{r}(f^{mp+j}(x))+\sum_{q=0}^{p-1}F_{m}(f^{mq+j}(x))+D\leq\sum_{q=0}^{p-1}F_{m}(f^{mq+j}(x))+2D.$

Hence $\gamma_{n}(x)\leq\sum_{q=0}^{p-1}F_{m}(f^{mq+j}(x))+2D.$ Summing up
over $j=0,1,\ldots,m-1$ this gives

$m\,\gamma_{n}(x)\leq\sum_{j=0}^{m-1}\sum_{q=0}^{p-1}F_{m}(f^{mq+j}(x))+2mD\leq(F_{m})_{n}(x)+3mD.$

Thus, assuming that the integer $n_{0}\geq 1$ is chosen sufficiently large, we
have

(3.5)
$\frac{1}{n}\gamma_{n}(x)\leq\frac{1}{n}\,\left(\frac{F_{m}}{m}\right)_{n}(x)+\frac{3D}{n}<\frac{1}{n}\,\left(\frac{F_{m}}{m}\right)_{n}(x)+\frac{\epsilon}{4}$

for all $x\in{\mathcal{L}}$ and all $n\geq n_{0}$.

Set

$X_{n}=\left\\{x\in{\mathcal{L}}:\frac{1}{n}\gamma_{n}(x)\geq\lambda_{i}+\epsilon\right\\},$
$Y_{n}=\left\\{x\in{\mathcal{L}}:\frac{1}{n}\left(\frac{F_{m}}{m}\right)_{n}(x)\geq\int_{X}\frac{F_{m}}{m}\,d\mu+\frac{\epsilon}{4}\right\\}.$

Since $F_{m}$ is Lipschitz, we can now use the classical Large Deviation
Principle (see [29] or [16]) for Hölder continuous functions on $X$. It
implies that there exist constants $C,c>0$ such that

(3.6) $\mu(Y_{n})\leq C\,e^{-cn}\quad,\quad n\geq 1.$

On the other hand, $X_{n}\subset Y_{n}$ for $n\geq n_{0}$. Indeed, given
$n\geq n_{0}$ and $x\in X_{n}$, (3.5) and (3.3) imply

$\displaystyle\frac{1}{n}\,\left(\frac{F_{m}}{m}\right)_{n}(x)$
$\displaystyle\geq$
$\displaystyle\frac{1}{n}\gamma_{n}(x)-\frac{\epsilon}{4}\geq(\lambda_{i}+\epsilon)-\frac{\epsilon}{4}=\lambda_{i}+\frac{3\epsilon}{4}$
$\displaystyle>$
$\displaystyle\left(\int_{X}\frac{F_{m}}{m}\,d\mu-\frac{3\epsilon}{8}\right)+\frac{3\epsilon}{4}>\int_{X}\frac{F_{m}}{m}\,d\mu+\frac{\epsilon}{4},$

so $x\in Y_{n}$. Now (3.6) yields $\mu(X_{n})\leq Ce^{-cn}$ for all $n\geq
n_{0}$. Taking a larger constant $C$ if necessary, we get $\mu(X_{n})\leq
Ce^{-cn}$ for all $n\geq 1$. This proves (2.1).

To derive (2.2) notice that the Lyapunov exponents of the cocycle $A^{-1}$
over the inverse transformation $f^{-1}:X\longrightarrow X$ are
$-\lambda_{k}<-\lambda_{k-1}<\ldots<-\lambda_{1}$ with the same Oseledets
subspaces $E_{i}(x)$ (see e.g. section 2.6.3 in [21]). Using the analogue of
(2.1) for $f^{-1}$, it follows that there exist constants $C,c>0$ such that

$\mu\left(\left\\{x\in{\mathcal{L}}:-\frac{1}{n}\lambda^{(n)}_{i}(x)\geq-\lambda_{i}+\epsilon\right\\}\right)\leq
Ce^{-cn},$

that is (2.2) holds for all $n\geq 1$. This concludes the proof of Theorem 4.

##  4 Derivative estimates for open billiards

Let $K\subset{\mathbb{R}}^{d}$ be as in section 1.2. Set $\displaystyle
d_{0}=\min_{i\neq j}\mbox{\rm dist}(K_{i},K_{j}).$ Denote by $\mbox{\rm
pr}_{1}:T({\mathbb{R}}^{d})\longrightarrow{\mathbb{R}}^{d}$ the natural
projection $\mbox{\rm pr}_{1}(q,v)=q$. For any $x\in\Gamma=\partial K$ we will
denote by $\nu(x)$ the outward unit normal to $\Gamma$ at $x$.

Remark. Notice that the condition (H) implies the existence of a constant
$\delta_{0}>0$ depending only on the obstacle $K$ such that $\langle
u,\nu(x)\rangle\geq\delta_{0}$ for any $(x,u)\in S^{*}(\Omega)$ whose backward
and forward billiard trajectories both have common points with $\partial K$.

Since every $K_{i}$ is strictly convex, the curvature (shape) operator

$L_{x}:T_{x}(\partial K)\longrightarrow T_{x}(\partial K)\quad,\quad
L_{x}u=(\nabla_{u}\nu)(x),$

where $\nu(y)$ is the outward unit normal to $\partial K$ at $y\in\partial K$,
defines a positive definite second fundamental form $\langle L_{x}u,u\rangle$.
The normal curvature of $\partial K$ at $x$ in the direction of $u\in
T_{x}(\partial K)$ ($\|u\|=1$) is by definition $k(x,u)=\langle
L_{x}u,u\rangle$. Clearly, $\|L_{x}u\|\geq k(x,u)$.

Given $\epsilon\in(0,d_{0}/2)$, set

$S^{*}_{\epsilon}(\Omega)=\\{x=(q,v)\in S^{*}(\Omega):\mbox{\rm
dist}(q,\partial K)>\epsilon\\}\quad,\quad{\Lambda}_{\epsilon}={\Lambda}\cap
S^{*}_{\epsilon}(\Omega)\;.$

In what follows in order to avoid ambiguity and unnecessary complications we
will consider stable and unstable manifolds only for points $x$ in
$S^{*}_{\epsilon}(\Omega)$ or $\Lambda_{\epsilon}$; this will be enough for
our purposes.

Here we derive a formula which is useful in getting estimates for
$\|d\phi_{t}(x)\cdot u\|$ ($u\in E^{u}(x)$, $x\in{\Lambda}$), both from above
and below.

Fix for a moment a point $x_{0}=(q_{0},v_{0})\in{\Lambda}_{\epsilon}$. If
$\epsilon>0$ is sufficiently small, then $W^{u}_{\epsilon}(x_{0})$ has the
form (cf. [22],[23])

$W^{u}_{\epsilon}(x_{0})=\\{(x,\nu_{X}(x)):x\in X\\}$

for some smooth hypersurface $X$ in ${\mathbb{R}}^{d}$ containing the point
$q_{0}$ such that $X$ is strictly convex with respect to the unit normal field
$\nu_{X}$. Let

$B_{q}:T_{q}X\longrightarrow T_{q}X\quad,\quad
B_{q}(v)=\left(\nabla_{v}\nu_{X}\right)(q),$

be the curvature (shape) operator of $X$ at $q\in X$ defined by means of the
normal field $\nu_{X}$; then the second fundamental form $\langle
B_{q}(\xi),\xi\rangle$, $\xi\in T_{q}(X)$, is positive definite.

Given a point $q_{0}\in X$, let $\gamma(x)$ be the forward billiard trajectory
generated by $x=(q_{0},\nu_{X}(q_{0}))$. Let $q_{1}(x),q_{2}(x),\ldots$ be the
reflection points of this trajectory and let $\xi_{j}(x)\in{\mathbb{S}}^{d-1}$
be the reflected direction of $\gamma(x)$ at $q_{j}(x)$.

Set $q_{0}(x)=q_{0}$, $t_{0}(x)=0$ and denote by $t_{1}(x),t_{2}(x),\ldots$
the times of the consecutive reflections of $\gamma(x)$ at $\partial K$. Then
$t_{j}(x)=d_{0}(x)+d_{1}(x)+\ldots+d_{j-1}(x)$, where
$d_{j}(x)=\|q_{j+1}(x)-q_{j}(x)\|$, $0\leq j\;.$ Given $t\geq 0$, denote by
$u_{t}(q_{0})$ the shift of $q_{0}$ along the trajectory $\gamma(x)$ after
time $t$. Set

$X_{t}=\\{u_{t}(q):q\in X\\}\;.$

When $u_{t}(q_{0})$ is not a reflection point of $\gamma(x)$, then locally
near $u_{t}(q_{0})$, $X_{t}$ is a smooth convex $(d-1)$-dimensional surface in
${\mathbb{R}}^{d}$ with ”outward” unit normal given by the direction
$v_{t}(q_{0})$ of $\gamma(x)$ at $u_{t}(q_{0})$ (cf. [23]).

Fix for a moment $t>0$ such that $t_{m}(x_{0})<t<t_{m+1}(x_{0})$ for some
$m\geq 1$, and assume that $q_{0}(s)$, $0\leq s\leq a$, is a $C^{3}$ curve on
$X$ with $q_{0}(0)=q_{0}$ such that for every $s\in[0,a]$ we have
$t_{m}(x(s))<t<t_{m+1}(x(s))$, where $x(s)=(q_{0}(s),\nu_{X}(q_{0}(s)))$.
Assume also that $a>0$ is so small that for all $j=1,2,\ldots,m$ the
reflection points $q_{j}(s)=q_{j}(x(s))$ belong to the same boundary component
$\partial K_{i_{j}}$ for every $s\in[0,a]$.

We will now estimate $\|d\phi_{t}(x_{0})\cdot\dot{q_{0}}(0)\|$, where
$\dot{q}(0)\in T_{q_{0}}X$.

Clearly $\phi_{t}(x(s))=(p(s),v_{m}(x(s)))$, where $p(s)$, $0\leq s\leq a$, is
a $C^{3}$ curve on $X_{t}$. For brevity denote by $\gamma(s)$ the forward
billiard trajectory generated by $(q_{0}(s),\nu(q_{0}(s)))$. Let
$\xi_{j}(s)\in{\mathbb{S}}^{d-1}$ be the reflected direction of $\gamma(s)$ at
$q_{j}(s)$ and let $\varphi_{j}(s)$ be the angle between $\xi_{j}(s)$ and the
outward unit normal $\nu(q_{j}(s))$ of $\partial K$ at $q_{j}(s)$. Let
$\phi_{t}(x(s))=(u_{t}(s),v_{t}(s))$, and let $t_{j}(s)=t_{j}(x(s))$ be the
times of the consecutive reflections of the trajectory $\gamma(s)$ at
$\partial K$. Set $d_{j}(s)=d_{j}(x(s))=\|q_{j+1}(s)-q_{j}(s)\|$ ($0\leq j\leq
m-1$), $d_{m}(s)=\|p(s)-q_{m}(s)\|$, $t_{0}(s)=0$ and
$t_{m+1}(s)=t_{m}(s)+d_{m}(s)$.

Next, let $k_{0}(s)$ be the normal curvature of $X$ at $q_{0}(s)$ in the
direction of $\dot{q_{0}}(s)$. That is

$k_{0}(s)=\frac{1}{\|\dot{q}_{0}(s))\|^{2}}\,\langle
B_{q_{0}(s)}(\dot{q}_{0}(s)),\dot{q}_{0}(s)\rangle.$

Similarly, for $j>0$ let $k_{j}(s)>0$ be the normal curvature of
$X_{t_{j}(s)}=\lim_{t\searrow t_{j}(s)}X_{t}$ at $q_{j}(s)$ in the direction
$e_{j}(s)$ ($\|e_{j}(s)\|=1$) of $\lim_{t\searrow
t_{j}(s)}\frac{d}{ds^{\prime}}\left(u_{t}(s^{\prime})\right)_{|s^{\prime}=s}$,
that is

$k_{j}(s)=\langle B_{q_{j}(s)}(e_{j}(s)),e_{j}(s)\rangle.$

Set $e_{0}(s)=\dot{q}_{0}(0)/\|\dot{q}_{0}(0)\|$ and for any $j\geq 0$ define
$\ell_{j}(s)>0$ by

(4.1)
$[1+d_{j}(s)\ell_{j}(s)]^{2}=1+2d_{j}(s)k_{j}(s)+(d_{j}(s))^{2}\,\|B_{q_{j}(s)}\,(e_{j}(s))\|^{2}\;.$

Finally, set

(4.2) $\delta_{j}(s)=\frac{1}{1+d_{j}(s)\ell_{j}(s)}\>\>\>\>,\>\>\>0\leq j\leq
m\;.$

Theorem 5. For all $s\in[0,a]$ we have

(4.3)
$\|\dot{q}_{0}(s)\|=\|\dot{p}_{0}(s)\|\delta_{0}(s)\delta_{1}(s)\ldots\delta_{m}(s)\;.$

This theorem was proved in [24] in the case $d=2$ and stated without proof in
[25] in the higher dimensional case. As we mentioned in [25], the above
formula can be derived from the more general study of the evolution of
unstable fronts in multidimensional dispersing billiards in [4] (see section 5
there). See also Ch. 3 in [8]. For completeness we provide a proof in the
Appendix for all $d\geq 2$.

##  5 Proof of Theorem 2

Throughout we assume that $f:\Lambda\longrightarrow\Lambda$ is the time-one
map for the open billiard flow and $\mu$ is a Gibbs measure on $\Lambda$
determined by a Hölder continuous potential on $\Lambda$. Fix a subset
${\mathcal{L}}$ of $\Lambda$ with $\mu({\mathcal{L}})=1$ and all the
properties in Theorem 3. Let

$\lambda_{2k}=-\lambda_{1}<\lambda_{2k-1}=-\lambda_{2}<\ldots<\lambda_{k+1}=-\lambda_{k}<0<\lambda_{k}<\ldots<\lambda_{2}<\lambda_{1}$

be the Lyapunov spectrum of $df$ with respect to $\mu$. As before denote by
$E_{i}(x)$ the Oseledets subspace corresponding to the Lyapunov exponent
$\lambda_{i}$. Fix $\delta_{0}\in(0,\epsilon/10)$ so small that

$0<\delta_{0}<\frac{1}{10}\min\\{\lambda_{j}-\lambda_{j+1}:j=1,\ldots,k-1\\}.$

According to Theorem 1, it is enough to prove the following

Theorem 6. The Oseledets subspaces $E_{i}(x)$ depend upper semi-continuously
on $x\in{\mathcal{L}}$.

We will use the notation and setup in section 4, however backwards, since we
will be dealing with local stable manifolds.

Fix a point $y_{0}=(p_{0},\eta_{0})\in{\Lambda}_{\epsilon}$ for some small
$\epsilon>0$ so that $W^{s}_{\epsilon}(y_{0})$ has the form
$W^{s}_{\epsilon}(y_{0})=\\{(p,\nu_{Y}(p)):p\in Y\\}$ for some smooth
hypersurface $Y$ in ${\mathbb{R}}^{d}$ containing the point $p_{0}$ such that
$Y$ is strictly concave with respect to the unit normal field $\nu_{Y}$. Then
the shape (curvature) operator

$S_{y}:T_{p}Y\longrightarrow T_{p}Y$

of $Y$ at $p\in Y$ is negative definite at $y=(p,\nu_{Y}(p))$ with respect to
the normal field $\nu_{Y}$. Let $p_{0}(s)$, $0\leq s\leq a$, be a $C^{3}$
curve on $Y$ with $p_{0}(0)=p_{0}\in{\mathcal{L}}$ such that

(5.1) $\|\dot{p}_{0}(s)\|=1\quad,\quad s\in[0,a].$

Then $y_{0}(s)=(p_{0}(s),\nu_{Y}(p_{0}(s)))$ is a $C^{3}$ curve on
$W^{s}_{\epsilon}(y_{0})$, so $\dot{y}_{0}(0)\in E^{s}(y_{0})$.

Fix another point $z_{0}=(r_{0},\nu_{Y}(r_{0}))\in W^{s}_{\epsilon}(y_{0})$
and let $r_{0}(s)$, $0\leq s\leq a$, be another $C^{3}$ curve on $Y$ with
$r_{0}(0)=r_{0}\in{\mathcal{L}}$ such that

(5.2) $\|\dot{r}_{0}(s)\|=1\quad,\quad s\in[0,a].$

Then $z_{0}(s)=(r_{0}(s),\nu_{Y}(r_{0}(s)))$ is a $C^{3}$ curve on
$W^{s}_{\epsilon}(y_{0})$, and $\dot{z}_{0}(0)\in E^{s}(z_{0})$.

Assume that $\dot{z}_{0}(0)\in E_{i}(z_{0})$ for some $i=k+1,\ldots,2k$. Then

$\lambda_{i}=\lim_{t\to\infty}\frac{1}{t}\log\|d\phi_{t}(r_{0})\cdot\dot{r}_{0}(0)\|.$

Take $T>0$ so large that

(5.3)
$\lambda_{i}-\delta_{0}<\frac{1}{t}\log\|d\phi_{t}(r_{0})\cdot\dot{r_{0}}(0)\|<\lambda_{i}+\delta_{0}$

for all $t\geq T$.

Lemma 2. Choosing $\delta_{0}>0$ sufficiently small, $T>0$ sufficiently large,
and assuming $d(p_{0},r_{0})<\delta_{0}$ and
$\|\dot{p}_{0}(0)-\dot{r}_{0}(0)\|<\delta_{0}$, we have

(5.4)
$\lambda_{i}-2\delta_{0}<\frac{1}{t}\log\|d\phi_{t}(p_{0})\cdot\dot{p_{0}}(0)\|<\lambda_{i}+2\delta_{0}$

for all $t\geq T$. Consequently, $\dot{y}_{0}(0)\in E_{i}(y_{0})$.

Proof of Lemma 2\. Let $\gamma(y_{0}(s))$ and $\gamma(z_{0}(s))$ be the
forward billiard trajectories generated by $y_{0}(s)$ and $z_{0}(s)$. Let
$p_{1}(s),p_{2}(s),\ldots$, and $r_{1}(s),r_{2}(s),\ldots$, respectively, be
the successive reflection points of these trajectories. We will denote by
$\eta_{j}(s)\in{\mathbb{S}}^{d-1}$ the reflected direction of
$\gamma(y_{0}(s))$ at $p_{j}(s)$ and by $\zeta_{j}(s)\in{\mathbb{S}}^{d-1}$
the reflected direction of $\gamma(z_{0}(s))$ at $r_{j}(s)$. Set
$t_{0}(y_{0}(s))=\tilde{t}_{0}(z_{0}(s))=0$ and denote by
$t_{1}(s),t_{2}(s),\ldots$ the times of the consecutive reflections of the
trajectory $\gamma(y_{0}(s))$ at $\partial K$. Then
$t_{j}(s)=d_{0}(s)+d_{1}(s)+\ldots+d_{j-1}(s)$, where
$d_{j}(s)=\|p_{j+1}(s)-p_{j}(s)\|$. Similarly, denote by
$\tilde{t}_{1}(s),\tilde{t}_{2}(s),\ldots$ the times of the consecutive
reflections of the trajectory $\gamma(z_{0}(s))$ at $\partial K$. Then
$\tilde{t}_{j}(s)=\tilde{d}_{0}(s)+\tilde{d}_{1}(s)+\ldots+\tilde{d}_{j-1}(s)$,
where $\tilde{d}_{j}(s)=\|r_{j+1}(s)-r_{j}(s)\|$. Set
$y_{t}(s)=\phi_{t}(y_{0}(s))$ and $z_{t}(s)=\phi_{t}(z_{0}(s))$ for all $t\geq
0$.

Given $t>0$, denote by $Y_{t}$ the shift of the surface $Y$ along the flow
$\phi_{t}$ after time $t$. That is

$Y_{t}=\\{\mbox{\rm pr}_{1}(\phi_{t}(p,\nu_{Y}(p)):p\in Y\\},$

where $\mbox{\rm pr}_{1}(u,v)=u$ for all
$(u,v)\in{\mathbb{R}}^{d}\times{\mathbb{R}}^{d}$. When $\phi_{t}(y_{0})$ is
not a reflection point of $\gamma(y_{0})$, then locally near
$\phi_{t}(y_{0})$, $Y_{t}$ is a smooth concave $(d-1)$-dimensional surface in
${\mathbb{R}}^{d}$ with ”outward” unit normal given by the direction of
$\gamma(y_{0})$ at $\phi_{t}(y_{0})$.

Fix a (large) integer $m$ and $t>0$ such that

(5.5) $t_{m}(0)<t<t_{m+1}(0).$

Consider the surface $X=Y_{t}$ with the opposite normal
$\nu_{X}=-\nu_{Y_{t}}$. Then $X$ is strictly convex with respect to this
normal and $(q,\nu_{X}(q))\in W^{u}_{\epsilon}(x_{0})$, where
$x_{0}=(q_{0},v_{0})$ with $q_{0}=\mbox{\rm pr}_{1}(y_{t}(0))$ and
$v_{0}=-\nu_{Y_{t}}(q_{0})$. Consider the curve $q_{0}(s)=\mbox{\rm
pr}_{1}(y_{t}(s))$, $s\in[0,a]$ on $X$, and let
$x_{\tau}(s)=\phi_{\tau}(q(s),\nu_{X}(q(s))$, $0\leq\tau\leq t$. As in section
4, let $q_{1}(s),q_{2}(s),\ldots$ be the reflection points of the trajectory
$\gamma(x_{\tau}(s))$ and let $\xi_{j}(s)\in{\mathbb{S}}^{d-1}$ be the
reflected direction of $\gamma(x_{\tau}(s))$ at $q_{j}(s)$. Then
$q_{j}(s)=p_{m-j}(s)$ for all $j=1,\ldots,m$, while $\xi_{j}(s)$ and
$\eta_{m-j}(s)$ are symmetric with respect to the normal $\nu(q_{j}(s))$ to
$\partial K$ at $q_{j}(s)$. As in section 4, let

$B_{q_{j}(s)}:T_{q_{j}(s)}(X_{t_{j}(s)})\longrightarrow
T_{q_{j}(s)}(X_{t_{j}(s)})$

be the shape operator of $X_{t_{j}(s)}$ at $q_{j}(s)$. Define $\ell_{j}(s)>0$
by (4.1), where $\delta_{j}(s)$ is defined by (4.2). Then, by Theorem 5,
formula (4.3) holds, where $\|\dot{p}_{0}(s)\|=1$ by (5.1).

For the same $m$ and $t$, we will need similar objects related to the
trajectory $\gamma(y_{0}(s))$. Set
$\tilde{x}_{0}=(\tilde{q}_{0},\tilde{v}_{0})$ with $\tilde{q}_{0}=\mbox{\rm
pr}_{1}(z_{t}(0))$ and $v_{0}=-\nu_{Y_{t}}(\tilde{q}_{0})$. Consider the curve
$\tilde{q}_{0}(s)=\mbox{\rm pr}_{1}(z_{t}(s))$, $s\in[0,a]$ on $X$, and let

$\tilde{x}_{\tau}(s)=\phi_{\tau}(\tilde{q}(s),\nu_{X}(\tilde{q}(s)))=(\tilde{u}_{\tau}(s),\tilde{v}_{\tau}(s)),$

$0\leq\tau\leq t$. As in section 4, let
$\tilde{q}_{1}(s),\tilde{q}_{2}(s),\ldots$ be the reflection points of the
trajectory $\gamma(\tilde{x}_{\tau}(s))$ and let
$\tilde{\xi}_{j}(s)\in{\mathbb{S}}^{d-1}$ be the reflected direction of
$\gamma(\tilde{x}_{\tau}(s))$ at $\tilde{q}_{j}(s)$. Then
$\tilde{q}_{j}(s)=r_{m-j}(s)$ for all $j=1,\ldots,m$, while
$\tilde{\xi}_{j}(s)$ and $\zeta_{m-j}(s)$ are symmetric with respect to the
normal $\nu(\tilde{q}_{j}(s))$ to $\partial K$ at $\tilde{q}_{j}(s)$. Let

$B_{\tilde{q}_{j}(s)}:T_{\tilde{q}_{j}(s)}(X_{t_{j}(s)})\longrightarrow
T_{\tilde{q}_{j}(s)}(X_{t_{j}(s)})$

be the shape operator of $X_{t_{j}(s)}$ at $\tilde{q}_{j}(s)$. Define
$\tilde{\ell}_{j}(s)>0$ following (4.1), namely

$[1+\tilde{d}_{j}(s)\tilde{\ell}_{j}(s)]^{2}=1+2\tilde{d}_{j}(s)\tilde{k}_{j}(s)+(\tilde{d}_{j}(s))^{2}\,\|B_{\tilde{q}_{j}(s)}\,(\tilde{e}_{j}(s))\|^{2}\;,$

where $\tilde{k}_{j}(s)$ is defined analogously to $k_{j}(s)$. Namely, set

$\tilde{k}_{0}(s)=\frac{1}{\|\dot{\tilde{q}}_{0}(s))\|^{2}}\,\langle
B_{\tilde{q}_{0}(s)}(\dot{\tilde{q}}_{0}(s)),\dot{\tilde{q}}_{0}(s)\rangle,$

and for $j>0$ let $\tilde{k}_{j}(s)>0$ be the normal curvature of
$X_{\tilde{t}_{j}(s)}=\lim_{\tau\searrow\tilde{t}_{j}(s)}X_{\tau}$ at
$\tilde{q}_{j}(s)$ in the direction $\tilde{e}_{j}(s)$
($\|\tilde{e}_{j}(s)\|=1$) of
$\lim_{\tau\searrow\tilde{t}_{j}(s)}\frac{d}{ds^{\prime}}\left(\tilde{u}_{\tau}(s^{\prime})\right)_{|s^{\prime}=s}$,
that is

$\tilde{k}_{j}(s)=\langle
B_{\tilde{q}_{j}(s)}(\tilde{e}_{j}(s)),\tilde{e}_{j}(s)\rangle.$

Set $\tilde{e}_{0}(s)=\dot{\tilde{q}}_{0}(0)/\|\dot{\tilde{q}}_{0}(0)\|$ and
for any $j\geq 0$ define

(5.6)
$\tilde{\delta}_{j}(s)=\frac{1}{1+\tilde{d}_{j}(s)\tilde{\ell}_{j}(s)}\;.$

Then, by Theorem 5, the analogue of (4.3) holds, namely

(5.7)
$\|\dot{\tilde{q}}_{0}(s)\|=\|\dot{r}_{0}(s)\|\tilde{\delta}_{0}(s)\tilde{\delta}_{1}(s)\ldots\tilde{\delta}_{m}(s)=\tilde{\delta}_{0}(s)\tilde{\delta}_{1}(s)\ldots\tilde{\delta}_{m}(s)\;,$

taking into account (5.2).

As is well-known (see e.g. [23], [13], [24] or [19]) it follows from the
uniform hyperbolicity of the open billiard that there exist global constants
$C>0$ and $\rho\in(0,1)$, independent of the choice of $z_{0}$, $y_{0}$, etc.
such that

(5.8) $d(p_{j}(s),r_{j}(s))\leq C\rho^{j}\quad,\quad
d(\eta_{j}(s),\zeta_{j}(s))\leq C\rho^{j}$

for all $j\geq 0$. It is also known that the stable bundle $E^{s}(x)$ is
$\alpha$-Hölder in $x$ for some global constant $\alpha>0$ (see e.g. [14] or
[20]). That is, we can take the global constants $C>0$, $\rho\in(0,1)$ and
$\alpha>0$ so that $d(E^{s}(x),E^{s}(y))\leq C(d(x,y))^{\alpha}$ for all
$x,y\in W^{s}_{\epsilon}(z)$, $z\in\Lambda$. Considering in a natural way
$E^{s}(x)$ as a vector subspace of ${\mathbb{R}}^{2d}$, for
$d\phi_{t}(y_{0}):E^{s}(y_{0})\longrightarrow E^{s}(y_{t}(0))$ and
$d\phi_{t}(z_{0}):E^{s}(z_{0})\longrightarrow E^{s}(z_{t}(0))$ we then have
$d(E^{s}(y_{t}(s)),E^{s}(z_{t}(s))\leq{\rm Const}\;\rho^{\alpha t}$ and there
exist global constants ${\rm Const}\;>0$ and $\rho_{1}\in(0,1)$ such that

(5.9)
$\|d\phi_{t}(p_{0}(0))\cdot\dot{p}_{0}(0))-d\phi_{t}(r_{0}(0))\cdot\dot{r}_{0}(0)\|\leq{\rm
Const}\;\rho^{\alpha t}$

for all $t\geq 0$. The latter follows from well-known estimates for rate of
change of stable Jacobi fields along billiard trajectories in hyperbolic
billiards ([28]; see also Theorem 17 in [30]).

Next, it follows from (5.8) and (5.9) that we have
$d(q_{j}(s),\tilde{q}_{j}(s))\leq C\rho^{m-j}$,
$|d_{j}(s)-\tilde{d}_{j}(s)|\leq C\rho^{m-j}$ and
$\|e_{j}(s)-\tilde{e}_{j}(s)\|\leq C\rho^{\alpha(m-j)}$ for all
$j=0,1,\ldots,m$, and therefore

$|k_{j}(s)-\tilde{k}_{j}(s)|\leq
C\rho^{\alpha(m-j)}\quad,\quad|\ell_{j}(s)-\tilde{\ell}_{j}(s)|\leq
C\rho^{\alpha(m-j)}$

for all $j=0,1,\ldots,m$. Now (4.2) and (5.6) yield

$\displaystyle|\log\delta_{j}(s)-\log\tilde{\delta}_{j}(s)|$ $\displaystyle=$
$\displaystyle|\log(1+d_{j}(s)\ell_{j}(s))-\log(1+\tilde{d}_{j}(s)\tilde{\ell}_{j}(s))|$
$\displaystyle\leq$ $\displaystyle
C\,|d_{j}(s)\ell_{j}(s)-\tilde{d}_{j}(s)\tilde{\ell}_{j}(s)|\leq
C\,\rho^{\alpha(m-j)},$

so $\log\delta_{j}(s)\leq\log\tilde{\delta}_{j}(s)+C\,\rho^{\alpha(m-j)}$ for
all $j=0,1,\ldots,m$. From the latter, (4.3) and (5.7) it follows now that

$\displaystyle\frac{1}{m+1}\log\|\dot{q}_{0}(s)\|$ $\displaystyle=$
$\displaystyle\frac{1}{m+1}\sum_{j=0}^{m}\log\delta_{j}(s)\leq\frac{1}{m+1}\sum_{j=0}^{m}\left(\log\tilde{\delta}_{j}(s)+C\,\rho^{\alpha(m-j)}\right)$
$\displaystyle\leq$
$\displaystyle\frac{1}{m+1}\log\|\dot{\tilde{q}}_{0}(s)\|+\frac{C}{m+1}\,\sum_{j=0}^{m}\rho^{\alpha(m-j)}$
$\displaystyle\leq$
$\displaystyle\frac{1}{m+1}\log\|\dot{\tilde{q}}_{0}(s)\|+\frac{C^{\prime}}{m+1}$

for some global constant $C^{\prime}>0$ independent of $m$.

Assuming that $t>0$ is sufficiently large, (5.3) holds and moreover for $m$
with (5.5) we have

$\frac{1}{m+1}\log\|\dot{\tilde{q}}_{0}(s)\|\leq\frac{1}{t}\log\|d\phi_{t}(r_{0})\cdot\dot{r_{0}}(0)\|+\frac{C^{\prime\prime}}{m+1}<\lambda_{i}+\delta_{0}+\frac{C^{\prime\prime}}{m+1}$

for some global constant $C^{\prime\prime}>0$ independent of $m$ and $t$.
Thus, taking $m$ (and so $t$) sufficiently large we get

$\frac{1}{m+1}\log\|\dot{q}_{0}(s)\|\leq\lambda_{i}+\delta_{0}+\frac{C^{\prime\prime}}{m+1}.$

However, again assuming that $t$ and $m$ with (5.5) are sufficiently large, we
have

$\frac{1}{m+1}\log\|\dot{q}_{0}(s)\|\geq\frac{1}{t}\log\|d\phi_{t}(p_{0})\cdot\dot{p_{0}}(0)\|-\frac{C^{\prime\prime\prime}}{m+1}$

for some global constant $C^{\prime\prime\prime}>0$. Therefore

$\frac{1}{t}\log\|d\phi_{t}(p_{0})\cdot\dot{p_{0}}(0)\|\leq\lambda_{i}+\delta_{0}+\frac{C^{\prime\prime}}{m+1}+\frac{C^{\prime\prime\prime}}{m+1}<\lambda_{i}+2\delta_{0}$

for all sufficiently large $t$.

This proves the left-hand-side of (5.4). The other side is proved by a
symmetric argument.

Proof of Theorem 6\. This follows immediately from Lemma 2 and Definition 2.

##  6 Appendix

Here we prove Theorem 5.

To deal with Theorem 5 we will use the notation from the beginning of section
4. We will need the following simple lemma which is a more precise version of
Lemma 3.7 in [13].

Lemma 3. Let $Y$ and $Z$ be the boundaries of two disjoint compact strictly
convex domains in ${{\mathbb{R}}}^{d}$. Suppose $Y$ and $Z$ are $C^{3}$-smooth
and denote by $\nu_{Y}$ and $\nu_{Z}$ their outer unit normal fields. Given
$\eta\in Y$, let $\eta_{t}$ ($t\geq 0$) be the shift of $\eta$ along the
billiard trajectory issued from $(\eta,\nu_{Y}(\eta))$. Suppose $\eta(s)$,
$0\leq s\leq a$, is a $C^{3}$ (local) curve on $Y$ such that for any
$s\in[0,a]$ the trajectory $\\{\eta_{t}(s):t\geq 0\\}$ hits transversally $Z$
at $\eta_{\tau(s)}(s)$ and the angle $\varphi(s)$ between the reflected
trajectory $\\{\eta_{t}(s):t\geq\tau(s)\\}$ and the unit normal
$\nu_{Z}(\eta_{\tau(s)}(s))$ is bounded from above by a constant
$\varphi_{0}\in(0,\pi/2)$. Let $L_{\eta}:T_{\eta}(Y)\longrightarrow
T_{\eta}(Y)$, $L_{\eta}u=(\nabla_{u}\nu_{Y})(\eta),$ be the curvature (shape)
operator of $Y$ with respect to the normal field $\nu_{Y}$. If $k_{0}>0$ is
the normal curvature of $Y$ at $\eta(0)$ in the direction of
$u_{0}=\dot{\eta}(0)$ and $\ell_{0}>0$ is such that

$(1+\tau(0)\ell_{0})^{2}=1+2\tau(0)\,k_{0}+\tau^{2}(0)\,\|L_{\eta(0)}\,(u_{0}/\|u_{0}\|)\|^{2}\;,$

then

$\displaystyle(1+\tau(0)\ell_{0})\|\eta(s)-\eta(0)\|-C\|\eta(s)-\eta(0)\|^{2}\leq\|\eta_{\tau(0)}(s)-\eta_{\tau(0)}(0)\|$
$\displaystyle\leq$
$\displaystyle(1+\tau(0)\ell_{0})\|\eta(s)-\eta(0)\|+C\|\eta(s)-\eta(0)\|^{2}$

for $0\leq s\leq a$, where $C>0$ is a constant depending only on the minimum
and maximum of the curvatures of $Y$ and $Z$,
$M_{i}=\max_{s\in[0,a]}\|\frac{d^{i}\eta}{ds^{i}}(s)\|$ ($i=1,2,3$), the
constant $\varphi_{0}$ and $\sup_{0\leq s\leq a}\tau(s)$. Moreover, there
exists constants $0<c<C$ such that

$c\|\eta_{\tau(s)}(s)-\eta_{\tau(s)}(0)\|\leq\|\eta_{\tau(s)}(s)-\eta_{\tau(0)}(0)\|\leq
C\|\eta_{\tau(s)}(s)-\eta_{\tau(s)}(0)\|$

for all $s\in[0,a]$.

Proof of Lemma 3\. We will first deal with the case when no reflection at $Z$
is involved. Then we have to get estimates from below and above for
$\lambda(s)=\|[\eta(s)+t\nu(s)]-[\eta(0)+t\nu(0)]\|^{2}\;,$ for $t>0$ and
$s\to 0$, where $\nu(s)=\nu_{Y}(\eta(s))$.

We have $\lambda(s)=a(s)+2tb(s)+t^{2}c(s)$, where

$a(s)=\|\eta(s)-\eta(0)\|^{2}\quad,\quad
b(s)=\langle\eta(s)-\eta(0),\nu(s)-\nu(0)\rangle\quad,\quad
c(s)=\|\nu(s)-\nu(0)\|^{2}.$

Since $\dot{a}(s)=2\langle\eta(s)-\eta(0),\dot{\eta}(s)\rangle$ and
$\ddot{a}(s)=2\|\dot{\eta}(s)\|^{2}+2\langle\eta(s)-\eta(0),\ddot{\eta}(s)\rangle$,
we have $\dot{a}(0)=0$ and $\ddot{a}(0)=2\|u_{0}\|^{2}$, so
$a(s)=\|u_{0}\|^{2}s^{2}+O(s^{3})$. (The estimate here obviously involves the
constant $M_{3}$.) Consequently,
$\|\eta(s)-\eta(0)\|=\sqrt{a(s)}=\|u_{0}\|\,s+O(s^{2})$.

Similarly,
$\dot{b}(s)=\langle\dot{\eta}(s),\nu(s)-\nu(0)\rangle+\langle\eta(s)-\eta(0),\dot{\nu}(s)\rangle$
and $\ddot{b}(s)=O(s)+2\langle\dot{\eta}(s),\dot{\nu}(s)\rangle$, so
$b(0)=\dot{b}(0)=0$ and $\ddot{b}(0)=2\|u_{0}\|^{2}\,k_{0}$. Hence
$b(s)=\|u_{0}\|^{2}k_{0}\,s^{2}+O(s^{3})$.

Finally, $\dot{c}(s)=2\langle\nu(s)-\nu(0),\dot{\nu}(s)\rangle$ and
$\ddot{c}(s)=2\|\dot{\nu}(s)\|^{2}+O(s)$, so $c(0)=\dot{c}(0)=0$ and
$\ddot{c}(0)=2\|\dot{\nu}(0)\|^{2}=2\|L_{\eta(0)}u_{0}\|^{2}$. Hence
$c(s)=\|L_{\eta(0)}u_{0}\|^{2}s^{2}+O(s^{3})$. This gives

$\displaystyle\lambda(s)$ $\displaystyle=$
$\displaystyle\|u_{0}\|^{2}s^{2}+2t\|u_{0}\|^{2}k_{0}\,s^{2}+t^{2}\|L_{\eta(0)}u_{0}\|^{2}\;s^{2}+O(s^{3})$
$\displaystyle=$
$\displaystyle(\|u_{0}\|^{2}+2t\|u_{0}\|^{2}k_{0}+t^{2}\|L_{\eta(0)}u_{0}\|^{2})\,s^{2}+O(s^{3})\;.$

Consequently,

$\|[\eta(s)+t\nu(s)]-[\eta(0)+t\nu(0)]\|=\sqrt{1+2t\,k_{0}+t^{2}\|L_{\eta(0)}\,(u_{0}/\|u_{0}\|)\|^{2}}\,\|u_{0}\|\,s+O(s^{2})\;.$

As mentioned earlier, $\|\eta(s)-\eta(0)\|=\|u_{0}\|\,s+O(s^{2})$, so the term
$\|u_{0}\|\,s$ above can be replaced by $\|\eta(s)-\eta(0)\|$. Thus,

(6.1)
$\|[\eta(s)+t\nu(s)]-[\eta(0)+t\nu(0)]\|=(1+\tau(0)\ell_{0})\|\eta(s)-\eta(0)\|+O(s^{2})\;.$

$Y_{\tau}$$Z$$Y$${\mathcal{T}}$$\eta_{\tau(s)}(s)$$\tilde{\eta}_{\tau}(s)$$\eta_{\tau}(s)$$\eta_{\tau}(0)$$\eta(0)$$\eta(s)$$\zeta$

Figure 1

Next, we deal with the case when a reflection at $Z$ has occurred. It is
enough to consider the following situation. Set $\tau=\tau(0)$ and assume that
$\tau(s)\leq\tau$ for $s>0$. For $\tilde{\eta}_{\tau}(s)=\eta(s)+\tau\,\nu(s)$
it follows from (6.1) that

(6.2)
$\|\tilde{\eta}_{\tau}(s)-\tilde{\eta}_{\tau}(0)\|=(1+\tau\,\ell_{0})\|\eta(s)-\eta(0)\|+O(\|\eta(s)-\eta(0)\|^{2})\;.$

The point $\eta_{\tau}(s)$ is symmetric to $\tilde{\eta}_{\tau}(s)$ through
the tangent plane ${\mathcal{T}}=T_{\eta_{\tau(s)}(s)}Z$ to $Z$ at
$\eta_{\tau(s)}(s)$. Let $\zeta=\zeta(s)$ be the orthogonal projection of
$\eta_{\tau}(0)$ to ${\mathcal{T}}$. Since the angle of reflection
$\varphi(s)$ is bounded from above by a constant $\varphi_{0}$, it is easy to
see that $\|\zeta(s)-\eta_{\tau}(0)\|=O(s^{2})$. On the other hand,
$\zeta(s)\in{\mathcal{T}}$ and the symmetry of $\eta_{\tau}(s)$ and
$\tilde{\eta}_{\tau}(s)$ imply
$\|\eta_{\tau}(s)-\zeta(s)\|=\|\tilde{\eta}_{\tau}(s)-\zeta(s)\|$. This,
$\tilde{\eta}_{\tau}(0)=\eta_{\tau}(0)$ and (6.2) imply

$\displaystyle\|\eta_{\tau}(s)-\eta_{\tau}(0)\|$ $\displaystyle=$
$\displaystyle\|\eta_{\tau}(s)-\zeta(s)\|+O(s^{2})=\|\tilde{\eta}_{\tau}(s)-\zeta(s)\|+O(s^{2})$
$\displaystyle=$
$\displaystyle\|\tilde{\eta}_{\tau}(s)-\eta_{\tau}(0)\|+O(s^{2})=\|\tilde{\eta}_{\tau}(s)-\tilde{\eta}_{\tau}(0)\|+O(s^{2})$
$\displaystyle=$
$\displaystyle(1+\tau\,\ell_{0})\|\eta(s)-\eta(0)\|+O(\|\eta(s)-\eta(0)\|^{2})\;.$

This completes the proof of the first statement of the lemma.

The second statement in the lemma follows easily from the first.

Proof of Theorem  5\. We will be using the notation from section 4 – see the
notation just before the statement of Theorem 5.

First notice that, using Lemma 4.1 in [13] (see also the proof of the estimate
(3.64) in [7]), there exists a constant $M>0$ depending on $K$ and
$M_{i}=\max_{s\in[0,a]}\|\frac{d^{i}p}{ds^{i}}(s)\|$ ($i=1,2,3$) such that
$\max_{s\in[0,a]}\left\|\frac{d^{i}q_{j}(s)}{ds^{i}}\right\|\leq M$ for any
$i=1,2,3$ and any $j=1,2,\ldots,m$. This implies similar estimates for
$\|\frac{d^{i}u_{t_{j}(s)}(s)}{ds^{i}}\|$ for all $i=1,2,3$ and
$j=1,2,\ldots,m$.

It is enough to estimate $\|\dot{q}(s)\|$ for $s=0$. Next, fix for a moment an
arbitrary $j=1,\ldots,m$. Given $s$ close to $0$, set $s_{j}^{+}=0$ if
$t_{j}(0)\geq t_{j}(s)$ and $s_{j}^{+}=s$ otherwise. Notice that $s^{+}_{0}=0$
and $s^{+}_{m+1}=0$. (In fact $t_{0}(s)=0$ and $t_{m+1}(s)=t$ for all $s$.)
Setting $\delta_{j}=\delta_{j}(0)$ for $j=0,1,\ldots,m$ and
$\epsilon_{j}=\|u_{t_{j}(s^{+}_{j})}(s)-u_{t_{j}(s^{+}_{j})}(0)\|$ for
$j=0,1,\ldots,m$ and $\epsilon_{m+1}=\|p(s)-p(0)\|$, Lemma 3 gives

$[1+d_{j}(0)\ell_{j}(0)]\;\epsilon_{j}-C\;\epsilon_{j}^{2}\leq\epsilon_{j+1}\leq[1+d_{j}(0)\ell_{j}(0)]\;\epsilon_{j}+C\;\epsilon_{j}^{2}$

for some global constant $C>0$. Thus,
$\displaystyle\frac{\epsilon_{j}}{\delta_{j}}-C\epsilon_{j}^{2}\leq\epsilon_{j+1}\leq\frac{\epsilon_{j}}{\delta_{j}}+C\epsilon_{j}^{2}\;,$
which is equivalent to

$\displaystyle\frac{\delta_{j}}{1+C\delta_{j}\epsilon_{j}}\epsilon_{j+1}\leq\epsilon_{j}\leq\frac{\delta_{j}}{1-C\delta_{j}\epsilon_{j}}\epsilon_{j+1}\;.$

Applying these inequalities recursively, we get

$\|p(s)-p(0)\|\prod_{i=j}^{m}\frac{\delta_{i}}{1+C\delta_{i}\epsilon_{i}}\leq\epsilon_{j}\leq\|p(s)-p(0)\|\prod_{i=j}^{m}\frac{\delta_{i}}{1-C\delta_{i}\epsilon_{i}}\;$

for $s$ close to $0$. We have $\epsilon_{0}=\|q_{0}(s)-q_{0}(0)\|$, so

$\|p(s)-p(0)\|\prod_{i=0}^{m}\frac{\delta_{i}}{1+C\,\delta_{i}\epsilon_{i}}\leq\|q_{0}(s)-q_{0}(0)\|\leq\|p(s)-p(0)\|\prod_{i=0}^{m}\frac{\delta_{i}}{1-C\,\delta_{i}\epsilon_{i}}\;.$

Dividing all sides of the last row by $s$ and letting $s\to 0$, we get

$\|\dot{p}(0)\|\prod_{i=0}^{m}\delta_{i}(0)\leq\|\dot{q}_{0}(0)\|\leq\|\dot{p}(0)\|\prod_{i=0}^{m}\delta_{i}(0)\;,$

which proves (4.3).

Acknowledgments Thanks are due to Sebastien Gouëzel for his comments on
earlier drafts of a part of this paper.

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