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\begin{document}
\title{Expanding maps of an interval with holes}
\author{H. van den Bedem\thanks{William E. Wecker Associates, Inc.
505 San Marin Drive Novato, CA 94945} $\ $ and N. Chernov\thanks{
Department of Mathematics, University of Alabama at Birmingham,
Birmingham, AL 35294}}
%E-mail: chernov@vorteb.math.uab.edu
\date{\today}
\maketitle
\begin{abstract}
We study a class of open chaotic dynamical systems. Consider an
expanding map of an interval from which a few small open
subintervals are removed (thus creating ``holes''). Almost every
point of the original interval then eventually escapes through the
holes, so there can be no absolutely continuous invariant
measures. We construct a so called conditionally invariant measure
that is equivalent to the Lebesgue measure. Our measure is unique
and naturally generates another measure, which is singular but
invariant. By this, we generalize early results by Pianigiani,
Yorke, Collet, Martinez and Schmidt, who studied similar maps
under an additional Markov assumption. We do not assume any Markov
property here and use ``bounded variation'' techniques rather than
Markov coding.
\end{abstract}
\renewcommand{\theequation}{\arabic{section}.\arabic{equation}}
\section{Introduction}
\label{secI} \setcounter{equation}{0}
Expanding maps of an interval are very popular in the theory
of chaotic dynamical systems. A map $\hat{T}\colon [0,1]\to [0,1]$
is said to be {\em expanding} if it is piecewise smooth and
its derivative satisfies the condition $\inf |\hat{T}'(x)|>1$.
Simple examples of expanding maps are:\\
(i) the so-called beta-transformations defined by $\hat{T}(x)=
\beta x$ (mod 1) with $\beta>1$,
in particular the doubling map $\hat{T}(x)=2x$ (mod 1);\\
(ii) the tent map defined by $\hat{T}(x)=2x$ for $0\leq x\leq 1/2$ and
$\hat{T}(x)=2(1-x)$ for $1/21/2$, with some $c>2$. This is
a map $\hat{T}\colon \IR\to\IR$ shown on Fig.~\ref{Fig1}. One can
easily check that the points $x\in H\colon =(c^{-1},1-c^{-1})$
will be mapped out of the interval $[0,1]$ and eventually, under
the iterations of $\hat{T}$, they will escape to $-\infty$. The
same will happen to all the points $x\in\hat{T}^{-k}H$ for all
$k\geq 1$. The set $\cup_k\hat{T}^{-k}H$ is open and dense in
$[0,1]$ and has full Lebesgue measure. In other words, almost
every point $x\in [0,1]$ eventually escapes through the
``opening'' $H$, which can be also regarded as a {\em hole} in the
interval $[0,1]$. Note that the map $\hat{T}$ was used in \cite{D}
to illustrate the concept of open dynamics and escape through
holes.
In physical applications, the process of escape of points from the
unit interval is characterized by a conditionally invariant
measure. A probability measure $\mu$ on $[0,1]$ is said to be {\em
conditionally invariant} if there exists $\lambda >0$ such that
$\hat{T}_{\ast}\mu=\lambda\mu$. Here, as usual, $\hat{T}_{\ast}$
is a dual operator acting on measures as defined by
$(\hat{T}_{\ast}\mu)(A)=\mu(\hat{T}^{-1}A)$ for all Borel sets
$A\subset [0,1]$. The constant $\lambda>0$ is called the
eigenvalue of $\mu$, see \cite{Ch85}, and $-\ln\lambda$ is called
the escape rate \cite{D,GD}.
In physical experiments, if one choses $N$ points in $[0,1]$ at
random according to the distribution $\mu$, then after $n$
iterations of $\hat{T}$ there will be $\approx \lambda^nN$ points
remaining on $[0,1]$, and they will be distributed according to
the measure $\mu$. Note that conditionally invariant measures are
analogous to quasi-stationary distributions for Markov chains with
absorbing states (or absorbing boundary conditions) - see
\cite{CMM,FKMP}.
>From the physicist point of view, only absolutely continuous
conditionally invariant measures (a.c.c.i.m.'s) are relevant,
and this is what we will study in this paper. It is
also important that an a.c.c.i.m. attracts other smooth measures,
i.e. there is a large class $\cal C$ of absolutely continuous
measures on $[0,1]$ such that for each $\nu\in\cal C$ the sequence
$c_n\,\hat{T}^n_{\ast}\mu$, where
$c_n^{-1}=(\hat{T}^n_{\ast}\nu)([0,1])$ is the normalizing factor,
converges to the a.c.c.i.m. $\mu$ (in the sense that the densities
converge uniformly) . We will refer to this as the {\em
convergence property} for $\mu$ (in the class $\cal C$).
We note that the convergence property allows to choose $N$ points in
the above mentioned physical experiment according to any distribution
in the class $\cal C$, with the same final result for large $n$.
The first results on a.c.c.i.m.'s were obtained by G.~Pianigiani
and J.~Yorke:
\begin{theorem}[\cite{PY}]
Let $I\subset [0,1]$ be a finite union of disjoint closed
intervals $I_1,\ldots,I_m$. Let $T\colon I\to [0,1]$ be a map that is
$C^2$ smooth and expanding on each interval $I_j\subset I$ and
satisfies the condition $T(I)\supset I$ and the Markov property
${\rm int}\, I\cap T(\partial I)=\emptyset$ (i.e. for each pair of
intervals $I_i,I_j$ we have either $T(I_i)\supset I_j$ or
$T(I_i)\cap {\rm int}\, I_j=\emptyset$). Then an a.c.c.i.m.
exists. If $T$ satisfies an additional transitivity assumption,
then an a.c.c.i.m. is unique in a class of absolutely continuous
measures whose densities have positive upper and lower bounds.
In that class, the a.c.c.i.m. has the convergence property. \label{tmPY}
\end{theorem}
For example,
let $T$ be the restriction of the modified tent map $\hat{T}$ shown on
Fig.~1 to the set $I\colon =[0,1]\setminus H$. Then $T$ satisfies
the conditions of Theorem~\ref{tmPY} and hence admits an a.c.c.i.m.,
which is in this case just the (normalized) Lebesgue measure on $I$.
\medskip
\noindent {\bf Remark}. One would expect that under the conditions
of Theorem~\ref{tmPY} the a.c.c.i.m. is unique (and has the
convergence property) even if we relax the requirement that the
densities be bounded away from zero. Surprisingly, this is not
true, even for the just mentioned modified tent map. Indeed,
choose any $0<\alpha<1/2$ and define a density by setting it to
$\alpha^k$ on the set $T^{-k}H$ for each $k\geq 1$. This defines
the density almost everywhere on $I$, and the corresponding
measure (after normalization) will be an a.c.c.i.m. with
eigenvalue $\lambda=2\alpha c^{-1}$. Moreover, the variation of
this density is finite (less than $4/(1-2\alpha)$). This example
shows that there is a mysterious difference between the properties
of a.c.i.m.'s for ordinary expanding maps and those of
a.c.c.i.m.'s for expanding maps with holes.
\medskip
An a.c.c.i.m. naturally generates a $\hat{T}$-invariant measure
$\bar{\mu}$ (which is singular). Its construction is described below, here
we give its physical interpretation. Suppose one choses $N$
points at random according to any distribution in the class
$\cal C$, then after $n$ iterations of $\hat{T}$ there will
be $\approx \lambda^nN$ points remaining on $[0,1]$. We take
their images after $m\ll n$ (instead of $n$) iterations of
$\hat{T}$, then those will be distributed according to the
measure $\bar{\mu}$.
\begin{theorem}[\cite{CMS1}]
Under the conditions of the previous theorem (including
transitivity) there is a unique $T$-invariant measure
associated with the a.c.c.i.m. $\mu$.
\label{tmCMS}
\end{theorem}
The purpose of this paper is to extend Theorems~\ref{tmPY} and
\ref{tmCMS} to expanding maps with holes that do not necessarily
satisfy the Markov property. In other words, we allow more generic
holes than in \cite{PY,CMS1}. Our approach is the following. We
start with an ordinary expanding map $\hat{T}$ of the unit
interval. For simplicity, we assume that it admits a mixing
a.c.i.m. whose density is bounded away from zero. Let $H$ be a
finite union of small open intervals (holes in $[0,1]$), and $T$
the restriction of $\hat{T}$ to $I=[0,1]\setminus H$. Here is our
main theorem, stated a bit loosely:
\begin{theorem}
Suppose that the total length of holes is small enough and they
are in generic position (i.e. their images do not overlap for
sufficiently many iterations of $\hat{T}$). Then the map $T\colon
I\to [0,1]$ admits an a.c.c.i.m. $\mu$, which is unique in a class
of absolutely continuous measures whose densities have bounded
variation. In that class, the a.c.c.i.m. $\mu$ has the convergence
property. The measure $\mu$ generates a unique $T$-invariant
measure $\bar{\mu}$. \label{tmmain}
\end{theorem}
The precise conditions and claims of the theorem will be specified
below.
\medskip
\noindent {\bf Remark}. We emphasize that even though
we are making a rather strong assumption on
the given expanding map $\hat{T}$ (that it has a
mixing a.c.i.m. with a strictly positive density), the holes
are allowed to be rather arbitrary, in fact,
they must be in generic position.
We note that our assumption on $\hat{T}$ is not too
restrictive either: if it fails, it is usually possible to find a
finite union of subintervals $J\colon =\cup J_i$ in $[0,1]$ and a higher
iteration of the map, $\hat{T}_1\colon =\hat{T}^k$, such that
$\hat{T}_1(J)\subset J$ and the map
$\hat{T}_1\colon J\to J$ admits an a.c.i.m. that is mixing and has
a strictly positive density on $J$. Then the problem can be reduced to the map
$\hat{T}_1$ on $J$.
\section{Notation and preliminary lemmas}
\label{secNPL} \setcounter{equation}{0}
Here we introduce our notation and collect necessary tools
for the proof of the main theorem.
Let $\hat{T}\colon I\to I$ be a piecewise $C^2$ expanding map. We denote
$$
s\colon =\sup_{x} 1/|T'(x)|<1
$$
and
$$
S\colon =\sup_{x} |T'(x)|<\infty
$$
We also put
\be
\bar{C}=\sup_{x\in I}|T''(x)|
\label{barC}
\ee
and
\be
\tilde{C}=\frac{s^2\bar{C}}{1-s}
\label{tildeC}
\ee
For each $n\geq 1$, we denote by $\hat{I}_k^n$, $k\geq 1$,
the intervals of monotonicity for the map $\hat{T}^n$.
We put $\hat{J}_k^n=\hat{T}^n\hat{I}_k^n$.
Let $\delta_n$ be the length of the smallest interval
$\hat{I}_k^n$, $k\geq 1$, that is
\be
\delta_n=\min_k m(\hat{I}^n_k)
\label{deltan}
\ee
Here and on $m$ stands for the Lebesgue measure on $[0,1]$.
\medskip
\noindent{\bf Assumption on $\hat{T}$}. We assume that the map
$\hat{T}$ preserves a mixing measure with density $h(x)$ that is
bounded away from zero: $h(x)\geq h_{\min}>0$.
\medskip
We make an important note. Under this assumption, the map
$\hat{T}$ has the following {\em covering property} \cite{L}: for
every $n\geq 1$ there is a $K(n)<\infty$ such that, for every $k$,
the set $\hat{T}^{K(n)}\hat{I}_k^n$ covers $[0,1]$, up to finitely
many points. For a proof, see C.~Liverani \cite{L}. We note that
Liverani \cite{L} proved the converse: the covering property
implies that the invariant density is bounded away from zero. It
is also interesting to note that recent results by J.~Buzzi
\cite{Bu} imply that if $\hat{T}$ is topologically mixing, then it
has the covering property.
\medskip
\noindent{\bf Holes}. We now introduce holes in the interval
$[0,1]$. Let $H= \cup_l H_l$ be a finite union of disjoint
open subintervals $H_l$ (holes). We denote by $L$ the number of
$H_l$ (holes) and by
$$
h=m(H)=\sum_{l=1}^Lm(H_l)
$$
their total length.
Let $I=[0,1]\setminus H$. Put $I^n=\cap_{i=0}^n\hat{T}^{-i}I$
and $H^n=[0,1]\setminus I^n$.
We now define the map $T$ on $I$ and its iterations.
We denote by $T^n$ the restriction of $\hat{T}^n$ to the set
$I^{n}$. Note that $T^n(I^n)\subset I$. That is, the map $T^n$
is the restriction of the map $\hat{T}$ to the set of points whose
trajectories do not enter the holes (the set $H$) at times
$0,1,\ldots,n$.
We put $I_k^n=\hat{I}_k^n\cap I^n$. Note that $T^n$ is well defined
and monotonic on each set $I_k^n$. Put $J_k^n=T^nI_k^n$. Note that
both $I_k^n$ and $J_k^n$ are finite unions of intervals.
As the reader may have noticed, we attach ``hats''
to notation related to the original map $\hat{T}$, and remove hats
when dealing with the map $T$ generated by holes.
Note that the set $H^n$ consists of points where the map $T^n$ is not
defined. We have a simple bound on its Lebesgue measure:
\be
m(H^n)\leq h\, \sum_{i=0}^n (sq)^i
\label{mHn}
\ee
where $q$ is the number of intervals $\hat{I}_k^1$ of monotonicity
of the map $\hat{T}$.
\medskip
\noindent{\bf Perron-Frobenius operator}. Since we study
absolutely continuous measures on $[0,1]$, we will need the
Perron-Frobenius operator that describes the transformation of
densities under the map $T$.
Let $f\geq 0$ be the density function of a measure $\nu$ on $I$.
The density of the measure $T_{\ast}^n\nu$ is given by
\be
(P_{T^n}f)(x)=\sum_{y\in T^{-n}x}\frac{f(y)}{|(T^n)'(y)|}
\label{PFo}
\ee
and if $T^{-n}x=\emptyset$, we set $(P_{T^n}f)(x)=0$
(here and on $(T^n)'$ stands for the derivative of the map $T^n$).
This equation defines a linear operator $P_{T^n}$ on $L^1(I)$, known as the
Perron-Frobenius operator. We will denote it simply by $P^n$. One
can easily verify that $P_{T^n}=\underbrace{P_T\circ\cdots\circ
P_T}_{n}$, so this notation is consistent.
Note that $P^nf\geq 0$ if $f\geq 0$. Also, by a simple change of variable,
$$
\| P^nf \|_1=\int_{I^n}f(x)\,dx = \nu(I^n)
$$
hence $\| P^nf\|_1\leq \| f\|_1=\nu(I)$.
This shows that, when holes are present, the Perron-Frobenius operator
decreases the norm of measures. For this reason we also consider a {\em modified}
(or {\em normalized}) Perron-Frobenius operator
\be
P^n_1f=P^nf / \| P^nf\|_1
\ee
defined only if $\| P^nf\|_1\neq 0$. Now we have $\| P^n_1f\|=1$.
The operator $P^n_1$ does preserve the norm of probability measures,
but it is not a linear operator anymore.
We study the action of the Perron-Frobenius operator on functions
of bounded variation. The variation of a function $f$ on $I$ is defined by
$$
{\rm Var}\, f = \sup\left\{\sum_{k=1}^n|f(a_k)-f(a_{k-1})|\colon\
a_0<\cdots 0$ we put
$$
E_b=\left \{ f\in L^1(I)\colon \, f\geq 0,\ \| f\|_1=1,\
{\rm Var}\, f\leq b \right\}
$$
It is straightforward to verify that $E_b$ is a compact, convex subset
of $L^1(I)$.
A crucial property of the Perron-Frobenius operator for ordinary
expanding maps is that they reduce variation of densities, i.e.
they satisfy the bound ${\rm Var}\, Pf\leq \alpha\, {\rm Var}\,
f+\beta \| f\|_1$ for some constants $0<\alpha<1$ and $\beta>0$,
see \cite{LaY,HK}. We now prove a similar property for our maps
with holes:
\begin{proposition}[Variation bound]
For every $n \geq 1$ and a nonnegative function $f$ of bounded
variation on $I$ we have
\be
{\rm Var}_{I} \left ( P^n f \right ) \leq
s^n(2L(n+1)+3)\, {\rm Var}_{I} f +
[\tilde{C}+(\delta_n-h)^{-1}]\, \int_{I} f \, dx
\label{varestimate}
\ee
(the value of $\delta_n$ is defined in (\ref{deltan})).
\label{prVar}
\end{proposition}
We first prove an auxiliary bound on distortions. Let $I^n_k$ be
one of the intervals of the set $I^n$ and $J^n_k=T^nI^n_k$. The
map $T^n$ is defined and monotonic, hence invertible, on the
interval $I^n_k$. Denote by $\psi\colon\, J^n_k\to I^n_k$ the inverse
of the map $T^n$ restricted to $I^n_k$. For any points $x,y\in
J^n_k$ we put $x'=\psi(x)$, $y'=\psi(y)$. Note that $|\psi'|\leq
s^n$. Hence, $|x'-y'|\leq s^n\cdot |x-y|$, and for the same reason
\be
|T^i(x')-T^i(y')|\leq s^{n-i}\cdot |x-y|
\label{Tix'y'}
\ee
for all $0\leq i\leq n$.
\begin{lemma}[Distortion bounds]
In the above notation
$$
\Big |\ln |(T^n)'(x')|-\ln |(T^n)'(y')|\Big | \leq \tilde{C}\cdot |x-y|
$$
\label{lmdisb}
\end{lemma}
\noindent {\em Proof}. By a simple application of the chain rule and then
(\ref{barC}) and (\ref{Tix'y'}) we have
\begin{eqnarray*}
\Big |\ln |(T^n)'(x')|-\ln |(T^n)'(y')|\, \Big | & \leq &
\sum_{i=0}^{n-1} \Big |\ln |T'(T^ix')|-|\ln T'(T^iy')|\, \Big |\\
& \leq &
\sum_{i=0}^{n-1}s\, \bar{C}\, |T^i(x')-T^i(y')|\\
& \leq &
\bar{C}(s^2+s^3+\cdots+s^{n+1})\, |x-y|
\end{eqnarray*}
This proves the lemma. $\Box$\medskip
Lemma~\ref{lmdisb} implies
\be
\left |\frac{(T^n)'(x')}{(T^n)'(y')}-1\right |\leq \tilde{C}\cdot |x-y|
\label{Tn'x'y'}
\ee
\noindent {\em Proof of Proposition~\ref{prVar}}.
Let $\hat{I}^n_k$ be an arbitrary interval of
monotonicity of the map $\hat{T}^n$. Recall that
$\hat{J}_k^n=\hat{T}^n\hat{I}_k^n$. Note that $\hat{T}^n\colon\,
\hat{I}_k^n\to \hat{J}_k^n$ is a monotonic, hence invertible, map,
and denote by $\hat{\psi}_k\colon\, \hat{J}_k^n\to \hat{I}_k^n$ the
inverse of the map $\hat{T}^n$ restricted to $\hat{I}^n_k$. Note
that $|\hat{\psi}_k'|\leq s^n$.
The set $I_k^n=\hat{I}_k^n\cap I^{n}$ consists of some number,
say $L_{n,k}$, of connected components (intervals). We claim that
\be
L_{n,k}\leq L(n+1)+1
\label{Lnk}
\ee
To prove this claim, we note that the gaps between the above
subintervals of $I_k^n$ are made by the components of
$H^n\cap\hat{I}^n_k$. The maps $\hat{T}^{i}$ for $1\leq i\leq n$
are monotonic on $\hat{I}^n_k$, thus the set
$\hat{T}^{-i}(H\cap\hat{T}^i\hat{I}_k^n)$ has no more than $L$
connected components for every $i=0,1,\ldots,n$. So, the set
$H^n\cap\hat{I}^n_k$ has no more than $L(n+1)$ connected
components. Hence (\ref{Lnk}). Denote by
$b^n_{k,1},\ldots,b^n_{k,q_k}$ the endpoints of the intervals of
which the set $I_k^n$ consists. Note that $q_k\leq 2L(n+1)+2$ by
(\ref{Lnk}).
It follows from the definition of the Perron-Frobenius operator
$P^n$ that
\begin{eqnarray}
{\rm Var} P^nf & \leq & \sum_{k} {\rm Var}_{J_k^n}
[(f\circ \hat{\psi}_k)\, |\hat{\psi}_k'|]
+ s^{n}\sum_k\sum_i f(b^n_{k,i}) \nonumber\\
& \leq & \sum_{k} {\rm Var}_{J_k^n} [(f\circ \hat{\psi}_k)\, |\hat{\psi}_k'|]
+ s^{n}(2L(n+1)+2)\sum_k \max_{I_k^n} f(x)
\label{VarIf}
\end{eqnarray}
For each $n$ and $k$ we have
\begin{eqnarray*}
{\rm Var}_{J_k^n} [(f\circ \hat{\psi}_k)\, |\hat{\psi}_k'|] &\leq&
\int_{J_k^n} \left | d[(f\circ \hat{\psi}_k)\, |\hat{\psi}_k'|] \right | \\
&\leq& \int_{J_k^n} | d(f\circ \hat{\psi}_k)|\cdot |\hat{\psi}_k'| +
\int_{J_k^n} (f\circ \hat{\psi}_k)\cdot |\hat{\psi}_k''|\, dx
\end{eqnarray*}
Lemma~\ref{lmdisb} implies that
\be
\sup_{x\in J_k^n}|\hat{\psi}_k''(x)/\hat{\psi}_k'(x)|
=\sup_{x\in J_k^n}|(\ln |\hat{\psi}_k'(x)|)'|
\leq\tilde{C}
\label{Dk}
\ee
and we obtain
\begin{eqnarray*}
{\rm Var}_{J_k^n} [(f\circ \hat{\psi}_k)\, |\hat{\psi}_k'|]
&\leq& s^{n}\,\int_{J_k^n} |d(f\circ \hat{\psi}_k)| +
\tilde{C}\int_{J_k^n} |(f\circ \hat{\psi}_k)|\, |\hat{\psi}_k'|\, dx \nonumber\\
& = & s^{n}\, {\rm Var}_{I_k^n} f +
\tilde{C}\int_{I_k^n} |f(x)|\, dx
\end{eqnarray*}
Now summing over $k$ yields
\be
\sum_{k} {\rm Var}_{J_k^n} [(f\circ \hat{\psi}_k)\, |\hat{\psi}_k'|]\leq
s^n\, {\rm Var}_{I^n} f + \tilde{C}\, \int_{I^n} |f(x)| \, dx
\label{sumkVarJkn}
\ee
To estimate the second term in (\ref{VarIf}), we note that
\begin{eqnarray}
\max_{I_k^n} f(x) &\leq& \max_{\hat{I}_k^n\setminus H} f(x) \nonumber\\
&\leq& \min_{\hat{I}_k^n\setminus H} f(x)
+{\rm Var}_{\hat{I}_k^n\setminus H}f\nonumber \\
&\leq& [m(\hat{I}_k^n\setminus H)]^{-1}\int_{\hat{I}_k^n\setminus H}f(x)\, dx
+{\rm Var}_{\hat{I}_k^n\setminus H}f
\label{maxIkn}
\end{eqnarray}
Note that $m(\hat{I}_k^n\setminus H)\geq\delta_n-h$. Now summing
over $k$ in (\ref{maxIkn}) and combining with (\ref{VarIf}) and
(\ref{sumkVarJkn}) complete the proof of Proposition~\ref{prVar}.
$\Box$ \medskip
\section{Existence and uniqueness of a.c.c.i.m.}
\label{secEU} \setcounter{equation}{0}
Fix $N\in\IN$ such that
\be
\alpha\colon= s^N(2L(N+1)+3)<1
\label{alpha}
\ee
\begin{proposition}
There exists a $b_{\min}>0$ such that for every $b_{\max}\geq b_{\min}$
there is an $h_0=h_0(b_{\min},b_{\max})>0$ such that whenever
$hb_{\min}$ be given. Obviously,
\be
\|f\|_{\infty} 1-\frac{1-\alpha}{2}=\alpha+\frac{1-\alpha}{2}
\label{PNf1}
\ee
for all $f\in E_b$, $b\leq b_{\max}$. Then an easy calculation
yields
$$
{\rm Var}\, P^N_1f \leq \|P^Nf\|_1^{-1}(\alpha\, {\rm Var}\, f+\beta)
< b
$$
for all $f\in E_b$. $\Box$ \medskip
It is easy to see that when the conditions of the above proposition
hold, the operator $P^N_1$ on $E_b$ is continuous (in the $L^1$ metric).
The Shauder-Tykhonov theorem then implies the existence of a fixed
point:
\begin{corollary}
Under the conditions of the previous proposition, the operator $P^N_1$
preserves a density $f\in E_b$. Hence, $P^Nf=\|P^Nf\|_1f$, so the density
$f$ defines an a.c.c.i.m. for the map $T^N$ with eigenvalue $\lambda=\|P^Nf\|_1$.
\label{crexist}
\end{corollary}
\noindent{\bf Remark}. Since Var$P_1^Nf**b_{\min}$ such that whenever $h c\} \geq \delta
$$
\end{lemma}
\noindent{\em Proof}. Suppose, on the contrary, that there exists
a $\tilde{c}\in (0,1)$ such that for any $\delta\in (0,1)$ we can
find $f_\delta \in E_{b_{\max}}$ for which $m\{x\colon f_\delta(x) >
\tilde{c} \} < \delta$. Choose $\delta < (1-\tilde{c})/(b_{\max}+2)$.
Since $\|f\|_{\infty} 3/4$ and the corresponding $\delta_0$ according
to the above lemma. Let $M$ denote a multiple of $N$ satisfying
$$
m(\hat{I}^{M}_k) < \frac{\delta_0}{2(2L+1)(b_{\max}+2)}
$$
for all intervals $\hat{I}^{M}_k$. Note that the set $\{x\colon
f(x) > c_0\}$ intersects at least $2(2L+1)(b_{\max}+2)$ intervals
$\hat{I}^{M}_k$.
Since the map $\hat{T}^{M}$ has the same mixing invariant
density $h$ as the map $\hat{T}$, it also has the covering
property, cf. Sect.~\ref{secNPL}. Therefore, there is a $K\geq 1$
such that the set $\hat{T}^{KM}\hat{I}^{M}_k$ covers $[0,1]$,
up to finitely many points, for every $k$. Observe that now for
every $x\in [0,1]$ (except finitely many) we have
$\hat{T}^{-KM}x \cap \hat{I}^{M}_k \neq \emptyset$ for all
intervals $\hat{I}^{M}_k$.
Recall that the Perron-Frobenius operator $P^n=P_{T^n}$ is defined
by (\ref{PFo}). If we assume $H=\emptyset$ (i.e., $I=[0,1]$), that
definition would give us the ordinary Perron-Frobenius operator
for the expanding map $\hat{T}^n$ (without holes). We denote it by
$\hat{P}^n \colon =P_{\hat{T}^n}$. It acts on $L^1([0,1])$.
For every function $f\in E_{b_{\max}}$ denote by $\hat{f}$ its
extension to $[0,1]$ obtained by setting $f$ to zero on the set
$H=[0,1]\setminus I$. Denote the space of the so defined functions
$\hat{f}$ by $\hat{E}_{b_{\max}}$. Such extension of $f$ certainly
increases its variation, but since $\|f\|_{\infty} 0$ such that if $\hat{f}\in\hat{E}_{b_{\max}}$
and $\hat{P}^{KM}\hat{f}(x) \leq \varepsilon_c$, then $\hat{f}(y)
\leq c$ for all $y \in \hat{T}^{-KM}x$.
\end{lemma}
\noindent{\em Proof}. Pick a $c < 1/4$, choose $\varepsilon_c <
cS^{-KM}$, where $S=\sup |\hat{T}'|$, and suppose $\hat{f}(\tilde
y)
> c$ for some $\tilde y \in \hat{T}^{-KM}x$. Then
$$
\hat{P}^{KM}\hat{f}(x)=
\sum_{y \in {\hat{T}}^{-KM}x}\frac{\hat{f}(y)}{|({\hat{T}}^{KM})'(y)|}
> \frac{c}{S^{KM}} + \sum_{y \in {\hat{T}}^{-KM}x \backslash
\tilde{y}}\frac{\hat{f}(y)}{|({{\hat{T}}^{KM}})'(y)|} >
\varepsilon_c
$$
$\Box$
\medskip
\begin{lemma}
There exists an $\varepsilon_0 > 0$ such that
$\inf\hat{P}^{KM}\hat{f} \geq \varepsilon_0$ for all $\hat{f} \in
\hat{E}_{b_{\max}}$. \label{lmeps1}
\end{lemma}
\noindent{\em Proof}. Fix some $0< c < 1/4$, and set
$\varepsilon_0=\varepsilon_c$ according to Lemma~\ref{epsexist}.
Suppose $\hat{P}^{KM}f(x) < \varepsilon_0$ for some $\hat{f} \in
\hat{E}_{b_{\max}}$ and $x\in [0,1]$. From Lemma~\ref{epsexist} we
obtain $\hat{f}(y) \leq c < 1/4$ for all $y \in \hat{T}^{-KM}x$.
Since $\hat{T}^{KM}$ is covering, each interval $\hat{I}^{M}_k$
contains one such point $y$. Our choice of $M$ implies that at
least $2(2L+1)(b_{\max}+2)$ of those intervals contain a point $z$
for which $\hat{f}(z) > c_0 > 3/4$. On each of the latter
intervals, the variation of $\hat{f}$ exceeds $1/2$. Thus,
$$
{\rm Var}\, \hat{f} > 2(2L+1)(b_{\max}+2)\frac 12=
(2L+1)(b_{\max}+2)
$$
which contradicts (\ref{Varhatf}). $\Box$
\medskip
Next, we will take a closer look at the holes. For technical
reasons we will choose an even higher iterate of the map,
$\hat{T}^{2KM}$. Let $x \in I$. Since $\hat{T}^{KM}$ is covering,
we may expect $x$ to have plenty of pre-images under $T^{2KM}$
(and not only under $\hat{T}^{2KM}$). However, it may happen that
too many of those preimages are ``eaten up'' by the holes, so that
not enough are left in $T^{-2KM}x$. We need to prevent this from
happening. Note that if $y\in\hat{T}^{-n}x\cap H$ for some $1\leq
n<2KM$, then $\hat{T}^{n-2KM}y\cap T^{-2KM}x=\emptyset$, i.e. all
the further preimages of the point $y$ are excluded from the set
$T^{-2KM}x$.
To ensure that sufficiently many pre-images of each $x \in I$
survive the removal of the holes, we impose the condition that the
holes are not only small, but in ``generic'' position:
\medskip
\noindent{\bf Genericity Conditions on the Holes.}
The collection of holes $H_l$, $1\leq l\leq L$,
satisfies the following assumptions:
\begin{enumerate}
\item[(G1)] $T^{-1}x \neq \emptyset$ for each $x \in I$.
\item[(G2)] The images of the holes under the maps $\hat{T}^n$,
$1\leq n\leq 2KM$ do not overlap. In particular, we assume that
the map $\hat{T}^{2KM}$ is one-to-one on each hole $H_l$ and
furthermore require that
$$
\hat{T}^{i}(H_p) \cap \hat{T}^{j}(H_q) \neq \emptyset
$$
for $1\leq i,j\leq 2KM$ if and only if $i=j$ and $p=q$.
Hence, for each $x \in I$ the set
\be
\left (\cup_{n=1}^{2KM}\hat{T}^{-n}x\right) \cap H
\label{capH}
\ee
consists of at most one point.
\end{enumerate}
\noindent{\bf Remark}. While the condition (G2) looks ``generic'',
(G1) may look more restrictive. However, if we require that
$\#\{\hat{T}^{-1}x\} \geq 2$ for each $x\in [0,1]$, then (G1)
would be ``generic'' as well. By the covering property, we can
always fulfill the above requirement by replacing $\hat{T}$ with
its higher iterate, before making holes in $[0,1]$. We do not do
that because our version of (G1) is the weakest one we need.
\medskip
\begin{lemma}
\label{plentypreimages}
Let $x \in I$. Then there exists a point
$y \in T^{-KM}x$ such that $y$ has a full set of pre-images
under $T^{KM}$; i.e. $T^{-KM}y = \hat{T}^{-KM}y$.
\end{lemma}
\noindent{\em Proof}. If $T^{-2KM}x = \hat{T}^{-2KM}x$, there
is nothing to show. If not, the set (\ref{capH}) is not empty, and
according to (G2) it consists of a single point, call it $z =
\hat{T}^{-i}x \cap H$ with some $i = 1, \ldots, 2KM$. If $i \leq
KM$, then any point $y\in T^{-KM}x$ has a full set of
preimages under $T^{KM}$ (and the set $T^{-KM}x$ is not empty
according to (G1)). If $i > KM$, then $T^{-KM}x =
\hat{T}^{-KM}x$, and we know that $\#\{\hat{T}^{-KM}x\} \geq
2$ by the covering property. Now we have that $\hat{T}^{-i+KM}y
\cap H \neq \emptyset$ for at most one point $y \in T^{-KM}x$,
so all the others will have a full set of preimages. See an
illustration to our argument in Fig.~2. $\Box$\medskip
\begin{lemma}
There exists an $\varepsilon_1 > 0$ such that $P_1^{2KM}f \geq
\varepsilon_1$ for all $f \in E_{b_{\max}}$. \label{lmeps2}
\end{lemma}
\noindent{\em Proof}. Let $x \in I$. By the previous lemma
there exists $y_1 \in T^{-KM}x$ such that $y_1$ has
a full set of pre-images under $T^{KM}$. Then,
\begin{eqnarray*}
(P_1^{2KM}f)(x) &=&
\|P^{2KM} f\|_1^{-1} \sum_{y \in T^{-2KM}x}
\frac{f(y)}{|(T^{2KM})'(y)|} \\
& \geq &
\sum_{y \in \hat{T}^{-KM}y_1}
\frac{f(y)}{|(\hat{T}^{KM})'(y_1)|\,|(\hat{T}^{KM})'(y)|} \\
& \geq &
S^{-KN0} (\hat{P}^{KM}f)(y_1)\\
& \geq &
\varepsilon_0 S^{-KM}
\end{eqnarray*}
where we used Lemma~\ref{lmeps1} at the last step.
We set $\varepsilon_1= \varepsilon_0 S^{-KM}$ and
complete the proof. $\Box$ \medskip
\begin{proposition}
\label{unique}
Assume that the holes satisfy the genericity conditions
(G1) and (G2). Then there exists a unique $f \in E_{b_{\max}}$
such that $P_1^{N}f = f$.
\end{proposition}
\noindent{\em Proof}. We only need to prove the uniqueness.
Suppose $P_1^{N}$ fixes two distinct densities $f_1$ and $f_2$ in
$E_{b_{\max}}$. Since $M$ is a multiple of $N$, we have
$f_1 = P_1^{KM}f_1 \neq P_1^{KM}f_2 = f_2$.
We distinguish two cases:
1. The densities $f_1$ and $f_2$ have equal eigenvalues, hence
$\|P^{KM}f_1\|_1 = \|P^{KM}f_2\|_1$. For $s \in \IR$ set
$f_s = sf_1 + (1-s)f_2$. Then for all $s$, $\int_{I} f_s\, dm = 1$,
and as long as $f_s \geq 0$ we have $P_1^Nf_s=f_s$, hence
$f_s\in E_{b_{\min}}$ by Remark after Corollary~\ref{crexist}.
Let $\sigma > 1$ such that $\inf_x f_{\sigma}(x) = 0$ for some
$x\in I$. Since $f_{\sigma} = \lim_{s \to \sigma} f_s$, we
have $f _{\sigma} \in E_{b_{\min}}$. Therefore,
$P_1^{2KM}f_{\sigma} \geq \varepsilon_1$ by the previous lemma.
But then $f_{\sigma} = P_1^{2KM}f_{\sigma} \geq \varepsilon_1$,
a contradiction.
2. The eigenvalues are not equal. For example, let
$\|P^{2KM}f_1\|_1 > \|P^{2KM}f_2\|_1$.
Since $f_2 > \varepsilon_1$, there exists
a $\beta > 0$ such that $\beta f_2 \geq f_1$.
Therefore, $\beta P^{2KMn}f_2 \geq P^{2KMn}f_1$,
and so $\beta \|P^{2KM}f_2\|_1^n \geq \|P^{2KM}f_1\|_1^n$
for all $n\geq 1$, a contradiction.
$\Box$ \medskip
Hence we proved the following:
\begin{theorem}
Suppose the holes satisfy the conditions
$h0$.
\section{The convergence property}
\label{secCP} \setcounter{equation}{0}
Here we prove the following:
\begin{theorem}
For any $f\in E_{b_{\max}}$ the sequence $\{P^n_1f\}$ converges to
$f_{\ast}$, as $n\to\infty$, uniformly on $I$. Furthermore, there
are constants $C>0$ and $\theta\in (0,1)$ such that
$$
\|P_1^nf - f_{\ast} \|_{\infty}\leq C\,\theta^n
$$
for all $f\in E_{b_{\max}}$. \label{tmconv}
\end{theorem}
\noindent{\em Proof}. By Lemma~\ref{lmeps2}, it is enough to prove
this fact for $f\in E_{b_{\max}}^1$, where
$$
E_{b_{\max}}^1\colon =\{f\in E_{b_{\max}}\colon\, \inf f\geq\varepsilon_1\}
$$
For $f\in E_{b_{\max}}^1$ and $n\geq 1$ denote
$$
\lambda_n(f)=\| P^{n}f\|_1
$$
Note that $\lambda_n(f_{\ast})=\lambda_{\ast}^n$.
\begin{lemma}
For all $f\in E_{b_{\max}}^1$ and $n\geq 1$
$$
00$ is a constant (independent of $f,g$ and $n$).
\end{lemma}
\noindent{\em Proof}. For brevity, we write $\|\cdot\|$ for
$\|\cdot\|_{\infty}$. By using (\ref{chi}) and (\ref{lamlam}), we have
\begin{eqnarray*}
\|P_1^{n}f-P_1^{n}g\| &=& \|P^{n}f/\lambda_n(f)-P^{n}g/\lambda_n(g)\| \\
& \leq &
\|(P^{n}f-P^{n}g)/\lambda_n(f)\| \\
& & + \|(P^{n}g)(\lambda_n(f)-\lambda_n(g))/(\lambda_n(f)\cdot\lambda_n(g))\| \\
& \leq &
C_2\,\|f-g\| \left (\lambda_{\ast}^n/\lambda_n(f) +
\|P^{n}g\|\,\lambda_{\ast}^n/(\lambda_n(f)\cdot\lambda_n(g)) \right )
\end{eqnarray*}
Now the result follows from Lemma~\ref{lmlala} and
(\ref{Pnfinfty}). $\Box$
\medskip
By this lemma, it is enough to prove that
\be
\|P_1^{nR}f-f_{\ast}\|_{\infty}\leq C\,\theta^n
\label{convR}
\ee
for any fixed $R\geq 1$ and all $f\in E^1_{b_{\max}}$ with some
$\theta<1$ and $C>0$. We do this next.
\begin{lemma}
There are constants $\kappa>0$ and $\theta_1<1$ such that for
every $f\in E^1_{b_{\max}}$\\ {\rm (a)} $f-\kappa f_{\ast}> 0$;\\
{\rm (b)} Put $f_{\kappa}=(f-\kappa f_{\ast})/\|f-\kappa
f_{\ast}\|_1$, then $P_1^Nf_{\kappa}\in E_{b_{\max}}$;\\ {\rm (c)}
$f-f_{\kappa}\leq \theta_1 f$. \end{lemma}
\noindent{\em Proof}. The property (a) holds for all $\kappa <
\varepsilon_1/(b_{\max}+2)$. For such $\kappa$, we have $\|f-\kappa
f_{\ast}\|_1=1-\kappa$, hence
$$
{\rm Var}\, f_{\kappa}\leq \frac{{\rm Var}\, f+\kappa\,{\rm
Var}\, f_{\ast}}{1-\kappa}\leq
\frac{1+\kappa}{1-\kappa}\, b_{\max}
$$
Now, by Proposition~\ref{prVar} and (\ref{alpha})-(\ref{PNf1}) we
have
$$
{\rm Var}\, P_1^Nf_{\kappa}\leq
\frac{\alpha\,\frac{1+\kappa}{1-\kappa}\,b_{\max}+\beta}{(1+\alpha)/2}0$;\\ {\rm (b)} $f_1 > 0$ and $f_1/\|f_1\|_1\in
E^1_{b_{\max}}$;\\ {\rm (c)} $f_1\leq \theta_1\, P^Rf$.
\end{corollary}
Applying the corollary to the function $f_1/\|f_1\|_1$ and
continuing in the same manner $n$ times yields
\begin{proposition}
For every $n\geq 1$ and $f\in E^1_{b_{\max}}$ there is a
decomposition $P^{nR}f=f_n+f_{n\ast}$ such that\\ {\rm (a)}
$f_{n\ast}=cf_{\ast}$ for some $c>0$;\\ {\rm (b)} $f_n > 0$
and $f_n/\|f_n\|_1\in E^1_{b_{\max}}$;\\
{\rm (c)} $f_n\leq \theta_1^n\, P^{nR}f$.
\end{proposition}
Note that $\|P^{nR}f\|_1=\lambda_{nR}(f)=O(\lambda_{\ast}^{nR})$ by
Lemma~\ref{lmlala}, hence in the decomposition
$P^{nR}f=f_n+f_{n\ast}$ we have
$$
\sup f_n \leq {\rm const}\cdot \theta_1^n\, \inf f_{n\ast}
$$
i.e. the first term $f_n$ is exponentially smaller than the second
term $f_{n\ast}$.
This proves (\ref{convR}), and hence Theorem~\ref{tmconv}. $\Box$ \medskip
\begin{proposition}
For every $f\in E_{b_{\max}}$ there is a limit
$$
B_f\colon =\lim_{n\to\infty} \lambda_n(f)/\lambda_{\ast}^n
$$
Hence,
$$
\lim_{n\to\infty}\frac{P^nf}{\lambda_{\ast}^n}=B_f\, f_{\ast}
$$
\label{prDf}
\end{proposition}
\noindent{\em Proof}. Note that for any $m\frac 12
$$
Hence $s^N<\lambda_{\ast}^N$. $\Box$ \medskip
Now choose $m\geq 1$ so that
\be
2(b_{\max}+2)(s/\lambda_{\ast})^m < \varepsilon_1/4
\label{slm}
\ee
Note that the map $T^m$, just like $T$,
has finitely many discontinuity points.
We take a tiny open interval $J\subset I$ containing $x_0$ such the map $T^m$
has at most one discontinuity point on $J$. Then $TJ=J_1\cup J_2$
where $J_1,J_2$ are two intervals (which may overlap).
For $n\geq 1$ denote by $\mu_n|_J$ the restriction of $\mu_n$
on $J$, and put $\nu_n=T_{\ast}^m(\mu_n|_J)$. Note that
\be
\liminf_{n\to\infty}\nu_n(J_1\cup J_2)= \liminf_{n\to\infty}\mu_n(J)\geq p_0
\label{p0}
\ee
for any open interval $J$ containing $x_0$.
On the other hand, note that any point $x\in J_1\cup J_2$ has
at most two preimages under $T^m$ on the interval $J$,
call them $y_1$ and $y_2$, hence
$$
\frac{d\nu_n}{dm}(x)=
\frac{f_{\ast}(y_1)}{\lambda_{\ast}^n\,|(T^m)'(y_1)|}+
\frac{f_{\ast}(y_2)}{\lambda_{\ast}^n\,|(T^m)'(y_2)|}
\leq\frac{2(b_{\max}+2)s^m}{\lambda_{\ast}^n}
$$
and by (\ref{slm}) we get
$$
\frac{d\nu_n}{dm}(x)\leq
\frac{\varepsilon_1}{4\lambda_{\ast}^{n-m}}\leq
\frac{f_{\ast}(x)}{4\lambda_{\ast}^{n-m}}
$$
Therefore,
$$
\nu_n(J_1\cup J_2)\leq\frac 14\, \mu_{n-m}(J_1\cup J_2)
$$
If the interval $J$ is small enough, then so are $J_1$
and $J_2$, and then obviously
$$
\limsup_{n\to\infty} \mu_{n-m}(J_1\cup J_2) < 3p_0
$$
which contradicts (\ref{p0}). Theorem~\ref{tminv} is proved.
$\Box$ \medskip
We conclude with some open questions.
It would be interesting to study
the properties of the measure $\bar{\mu}$. We conjecture
that it is ergodic, Bernoulli, and an equilibrium state
for the potential function $-\ln |T'(x)|$.
Under the assumptions of Theorem~\ref{tmPY}, these
properties of $\bar{\mu}$ have been proved in \cite{CMS1}, along with
a remarkable {\em escape rate formula}:
$$
\chi(\bar{\mu}) = h_{\rm KS}(\bar{\mu}) + \gamma
$$
Here $\chi(\bar{\mu})=\int_{\Lambda}\ln |T'|\,d\bar{\mu}$ is the
Lyapunov exponent, $h_{\rm KS}(\bar{\mu})$ is the Kolmogorov-Sinai
entropy, and $\gamma=-\ln\lambda$ is the escape rate of the
a.c.c.i.m. $\mu$ (recall that $\lambda=\mu(I^1)$ is its
eigenvalue). The proofs in \cite{CMS1,CMS2} were based on thermodynamic
formalism and the symbolic representation of the system by a finite
Markov chain (which existed because of the Markov property, see
Theorem~\ref{tmPY}).
In our case, no finite Markov partition exists, hence one needs to
develop a different approach. One way to do that is approximate
the holes $H$ by slightly larger holes that satisfy the Markov
property, and then use the above results. This approach was
employed in \cite{CMT1,CMT2} where Anosov diffeomorphisms with
small open holes were studied.
\medskip
\noindent{\bf Acknowledgement}. We are grateful to J.~Buzzi and
C.~Liverani for helpful discussions and referring us to their
results on the covering property of expanding maps. N.~Chernov was
partially supported by NSF grant DMS-9732728.
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\end
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