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\bibliographystyle{plain}
\title {Multiple measures of maximal entropy and equilibrium states
for one-dimensional subshifts}
\author{Nicolai Haydn \thanks{Mathematics Department, University of Southern California,
Los Angeles, 90089-1113. Email:$<$nhaydn@mtha.usc.edu$>$.}}
\date{}
\maketitle
\begin{abstract}
\noindent We give simple examples of a one-dimensional mixing subshift
with positive
topological entropy which have two distinct measures of maximal entropy.
We also give examples of subshifts which have two mutually singular
equilibrium states for H\"{o}lder continuous functions.
\end{abstract}
\section{Introduction}
It is a well known fact that a wide class of uniformly hyperbolic
transformation that satisfy some irreducibility condition have unique
measures of maximal entropy. Irreducible subshifts have unique measures
of maximal entropy if they are for instance
of finite type or sofic (factors of shifts of finite type). More
general hyperbolic systems which satisfy specification have unique
measures of maximal entropy \cite{Bow}. In either case the systems have
a `uniform' mixing property.
Hofbauer had shown in \cite{Hof} that even subshifts of finite
type don't necessarily have unique equilibrium states. However
the potentials in the provided examples lack (of course) regularity
and one
of the equilibrium states is supported on a single fixpoint.
For a large class of H\"{o}lder continuous functions
we get here multiple equilibrium states which are supported on
embedded subshifts.
In this note we shall produce a simple example which
shows that even in the case of an irreducible subshift with positive
entropy one cannot generally expect to have a unique measure of maximal
entropy.
(It was pointed out that Krieger \cite{Kr} gave an example with a
non-unique measure of maximal entropy---although not as simple and
straighforward as this one.)
For two-dimensional subshifts of finite type, Burton and Steiff \cite{B-S}
have recently given examples whose measures of maximal entropy form
simplices that consist of more than single points. Hence:
\begin{example} There exists a subshift $\Sigma$ over a finite
alphabet with a topologically mixing shift transformation
$\sigma: \Sigma\rightarrow\Sigma$ such that $\Sigma$ carries at least
two mutually singular measures of maximal entropy.
\end{example}
\noindent An example is provided in definition 2 of the following section.
Lemmas 1, 2 and 3 serve to prove the properties claimed above.
Extending this example we are then able to produce in section 3
multiple equilibrium states for H\"{o}lder continuous functions
(see section 3 for H\"{o}lder continuity).
\begin{example} There exists a topologically mixing subshift
$\Sigma$ over a finite alphabet and a H\"{o}lder continuous
function $g:\Sigma\rightarrow\mbox{\bf R}$ such that $\Sigma$ carries at least
two mutually singular equilibirium states for $g$.
\end{example}
\section{Example} %%%%%%%%%%%%%%%%%%%% Section 2
We shall construct a subshift over the symbolset ${\cal A} = \{0,1,2,3,4\}$ and
shall refer to the symbols $\{1,2\}$ as the yellow symbols and denote
them by ${\cal Y}$. Similarly we call $\{3,4\}$ the green symbols and write
${\cal G}$ for them. Hence ${\cal A} = {\cal Y} \cup {\cal G} \cup \{0\}$.
Let us denote by $\Sigma_{\cal Y} = {\cal Y}^{\bf Z}$ the (yellow) full two-shift
over the alphabet ${\cal Y}$. Similarly we write $\Sigma_{\cal G}$
for the (green) full two-shift ${\cal G}^{\bf Z}$.
Let $\tau>0$.
The subshift $\Sigma \subset {\cal A}^{\bf Z}$ will consist of the union
of the two monochromatic shift spaces $\Sigma_{\cal Y}$ and
$\Sigma_{\cal G}$, to which we add bicoloured sequences in the
following way. A sequence in $\Sigma$ in which a word $\alpha$ of
one colour (green or yellow) is followed
by a word $\beta$ of the other colour (yellow or green) will have a string
of zeroes $\gamma$ separating the two words $\alpha$ and $\beta$.
The length $|\gamma|$ of that string of zeros is then at least,
$\tau(a + b)$, where $a = |\alpha|, b = |\beta|$ are the lengths of
the coloured words to the `left' and `right' of $\gamma$.
More precisely:
\begin{definition} The shift space $\Sigma \subset {\cal A}^{\bf Z}$
is defined as follows: We have $x \in \Sigma$ if
\noindent (i) either $x \in \Sigma_{\cal Y} \cup \Sigma_{\cal G}$, in which
case we call $x$ a {\em monochromatic} point.
\noindent (ii) or if bi-coloured blocks are of the form:
$$\dots 0y_1y_2\dots y_{a-1}y_a 0^{\lambda}g_1g_2\dots g_{b-s}g_b0\dots$$
or
$$\dots 0g_1g_2\dots g_{a-1}g_a 0^{\lambda}y_1y_2\dots y_{b-s}y_b0\dots$$
where $\lambda \geq \tau(a + b)$ and $y_i \in {\cal Y}, g_i \in {\cal G}$
($0^{\lambda}$ is a string of zeros of length $\lambda$), $\tau>0$.
\end{definition}
\noindent
If $\omega = \omega_m \omega_{m+1} \dots \omega_{m+n-1}$ is an allowed word
of some length $n$ in $\Sigma$, then we denote by $U(\omega)$ the
{\em cylinder set} $\{x \in \Sigma: x_i = \omega_i, m \leq i < m + n \}$.
The shift transformation $\sigma$ on $\Sigma$ is defined in the usual
way by $\sigma(x)_i = s_{i+1}, i \in {\bf Z}$, for $x \in \Sigma$.
Notice that the transition rules are shift invariant. Thus
$\Sigma$ is indeed a subshift which we endow with the usual topology
generated by the cylinder sets $U(\omega)$, where $\omega$
runs over all finite words in $\Sigma$.
Notice that a sequence in $\Sigma$ cannot end in an infinite one-sided
string of one colour unless the sequence already belongs to the
appropriate monochromatic subshift.
If $\tau=0$ then $\Sigma$ is a sofic system in which every string of
zeroes connects to coloured symbols of different colours,
and whose topological entropy is $\log 4$ (see lemma 2).
\begin{lemma} The shift transformation $\sigma$ on $\Sigma$ is
topologically mixing for every $\tau$.
\end{lemma}
\noindent {\bf Proof.}
We have to show that for any two finite words $\omega$
and $\eta$ there exists a number $N$ so that
$U(\omega) \cap \sigma^n(U(\eta))$ is non-empty for all $n \geq N$.
This is evidently true if $\omega$ and $\eta$ are both monochromatic
of the same colour. If the last symbol of $\eta$ and the
first symbol of $\omega$ are of different colour then consider the word
$\pi = \eta 0^{\lambda} \omega$.
Clearly, $\pi$ is an admissable word for every
$\lambda > \tau(|\eta| + |\omega|)$,
and the cylinder set
$U(\pi) \subset U(\omega) \cap \sigma^n(U(\eta))$ is non-empty for all
$n \geq N$, where $N$ can then be chosen to be equal to
$(1+\tau)(|\eta| + |\omega|)$.
If the last symbol of $\eta$ and the first symbol of $\omega$ are of
the same colour, then let us consider the word
$\pi = \eta 0^{\kappa} \varepsilon 0^{\lambda} \omega$,
where $\varepsilon$ is a symbol (or some other short word) of
the other colour. Here we have $\kappa \geq \tau(|\eta| + |\varepsilon|)$
and $\lambda \geq \tau(|\omega| + |\varepsilon|)$. Clearly
$U(\pi)$ is non-empty for those choices of $\lambda$ and $\kappa$.
Thus $U(\omega) \cap \sigma^n(U(\eta)) \not= \emptyset$
if $n \geq N = (\tau+1)(|\eta| + 2|\varepsilon| + |\omega|)$.
\hfill$\Box$
\begin{lemma} The topological entropy $\,h$ of the shift
$\sigma: \Sigma \rightarrow \Sigma$ is
$\max(\log 2,\frac{\log 4}{1+\tau})$.
\end{lemma}
\noindent {\bf Proof.}
Let us first get a lower bound on the topological entropy $h$
of $\Sigma$. For that purpose let us observe that since
$\Sigma$ contains two full two shifts, $\Sigma_{\cal Y}$ and $\Sigma_{\cal G}$,
the topological entropy of $\Sigma$ must be at least $\log 2$.
Let us now estimate the number of the remaining words of length
$n$ from above. We have two cases to consider:
\noindent (i) Monochromatic words that might also contain zeroes.
According to our rules such words begin or end with strings
of zeroes. Thus we obtain purely yellow and green words of lengths
$k = 0,\dots,n$ which on at least one side is framed by strings of zeroes.
Their number turns out to be
$2 \sum_{k=0}^n 2^k=2^{n+1}\sum_{k=0}^n 2^{-k}$.
\noindent (ii)
To estimate the number of words of length $n$ that genuinely contain
symbols of both colours, let us observe that since any such word has at
least one transition from yellow to green or vice versa, it therefore
must also contain at least $n\frac{\tau}{\tau+1}$ zeroes,
that is at most $n'=[\frac{n}{\tau+1}]$ ($[\;]$ denotes integer part)
coloured symbols (i.e.\ symbols in $\{1,2,3,4\}$).
The coloured symbols come in monochromatic blocks of
alternating colour. Denote by $P_{k,\ell}$ the number of possibilities
in which one can arrange $\ell$ symbols in $k$ blocks (separated by
the appropriate number of zeroes), where $k=1,2,\dots,\ell$.
One finds that
$$
P_{k,\ell}=\left(\begin{array}{c}\ell-1\\k-1\end{array}\right),
$$
which is the number of possibilities of picking the first element
of every block but the very first one.
The number of $n$-words
in $\Sigma$ which contain $\ell$ coloured symbols arranged
in $k\leq\ell$ monochromatic blocks of alternating colour
is $2P_{k,\ell}2^{\ell}$.
If $Q_k$ denotes the number of $n$-words in $\Sigma$ with at most
$n'$ coloured symbols in $k$ monochromatic blocks of alternating colour,
then
$$
Q_k=\sum_{\ell=k}^{n'} 2P_{k,\ell}2^{\ell}.
$$
The number of words of lengths $n$ that contain symbols of
both colours is now determined by
%%
\begin{equation}\label{coloured.words}
\sum_{k=2}^{n'}Q_k=2\sum_{\ell=2}^{n'}\sum_{k=2}^{\ell}
\left(\begin{array}{c}\ell-1\\k-1\end{array}\right)2^{\ell}
=2\sum_{\ell=2}^{n'}(2^{\ell-1}-1)2^{\ell}
=r_n2^{2n'},
\end{equation}
%%
where the constants $r_n$ and their reciprocals $1/r_n$ are
subexponential as $n$ goes to infinity.
Thus the total number of words of lengths $n$ (large enough) in the
subshift $\Sigma$ has growth rate
$$
h=\lim_{n\rightarrow\infty} \frac1n\log(2^n+2^{2n'}),
$$
which implies that $\Sigma$ has topological entropy
$h=\max(\log 2, \frac{\log4}{\tau+1})$.
In particular $\Sigma$ has topological entropy $\log 2$
if $\tau\geq 1$.
\hfill$\Box$
\begin{lemma} If $\tau\geq 1$, then there are two mutually
singular measures of maximal entropy on $\Sigma$.
\end{lemma}
\noindent {\bf Proof.}
It is well known \cite{Wal} that the measure of maximal entropy on
the full
two-shift $\{1,2\}^{\bf Z}$ is the Bernoulli measure with the
probability vector $(\frac12,\frac12)$. It's metric entropy
is $\log 2$. Let us define on $\Sigma$ a Bernoulli measure
$\mu_{\cal Y}$ which is given by the probability vector
$(0,\frac12,\frac12,0,0)$ and the measure $\mu_{\cal G}$
with the probabilities $(0,0,0,\frac12,\frac12)$.
Evidently both measures are shift invariant and have metric
entropies $\log 2$ which by lemma 2 is the topological entropy
of $\Sigma$. Hence $\mu_{\cal Y}$ and $\mu_{\cal G}$
(as well as all their linear combinations) are
(distinct) measures of maximal entropy for the subshift $\Sigma$.
\hfill$\Box$
\vspace{3mm}
\noindent In the case when $0<\tau<1$, then there is a unique
measure of maximal entropy which is given by the probability
vector $(0,\frac12,\frac12,\frac12,\frac12)$ whose entropy is
by lemma 2 equal to $\frac{\log 4}{\tau+1}$.
\section{Equilibrium states}%%%%%%%%%%%%%%%%%%%% section 3
\noindent
Let us construct now subshifts with non unique equilibrium states.
We say a function $f$ is H\"{o}lder continuous on a shift space
if there exists a $\vartheta\in(0,1)$ and a constant $C$ so that
for every $n>0$ one has
$$
\sup|f(\vec{x})-f(\vec{y})|\leq C\vartheta^n
$$
where the supremum is over all pairs of sequences
$\vec{x}=\cdots x_{-1}x_0x_1x_2\dots,
\vec{y}=\cdots y_{-1}y_0x_1y_2\dots$ in the shiftspace
for which $x_{-n}\cdots x_n=y_{-n}\cdots y_n$ and either
$x_{-n}\not=y_{-n}$ or $x_n\not=y_n$.
Let $f$ be a real valued H\"{o}lder continuous function on the full
two-shift $\Sigma_*$ (where $*={\cal G},{\cal Y}$)
so that $P_*(f)>\sup f$, where $P_*(f)$ is the pressure
of $f$ (see \cite{Wal}).
Let us define a H\"{o}lder continuous function $g$ on $\Sigma$
by putting $g(x)=f(x)$ if $x$ lies in the full two-shifts
$\Sigma_{\cal G}$ or $\Sigma_{\cal Y}$. We extend $g$ to bi-coloured
words in the following way: If $x_0=0$ then we put $g(x)=\sup f$.
If $x_0\not=0$ and $x\in\Sigma$ is a sequence which contains
symbols of both colours then let $x_{-n}\cdots x_n$ be the
longest monochromatic
string of either green or yellow symbols (that is either $x_{-n}=0$
or $x_n=0$) and pick a monochromatic point $y\in\Sigma_*$
($*={\cal G},{\cal Y}$) in the cylinder
$U(x_{-n}\cdots x_n)$. Now define $g(x)=f(y)$.
In this way $g$ is a H\"{o}lder continuous function on $\Sigma$.
Let us now prove that the pressure of $g$ (on $\Sigma$) is in fact
equal to $P_*(f)$ (on the two-symbol shiftspace) for a suitable
choice of $\tau$.
\begin{lemma}
If
$\tau\geq \frac{\log 4}{P_*(f)-\sup f} -1$, then the pressure
$P(g)$ of $g$ on $\Sigma$ is equal to $P_*(f)$.
\end{lemma}
\noindent {\bf Proof.}
Denote by $W_n$ the collection of $\Sigma$-words
$\alpha=\alpha_1\cdots \alpha_n$ of lengths $n$.
The pressure of $g$ is then given by the exponential
growth rate of the partition function
$$
Z_n=\sum_{\alpha\in W_n} e^{g^n(x(\alpha))},
$$
where $x(\alpha)$ is an arbitrary point in the cylinder set
$U(\alpha)\subset\Sigma$, and
$g^n=g+g\sigma+g\sigma^2+\cdots+g\sigma^{n-1}$ is the $n$-th
ergodic sum of $g$.
If in the above sum we restrict to monochromatic words $\alpha$ of
either colour, then we obtain the green or yellow
partition function $Z^*_n$ ($*={\cal G},{\cal Y}$)
whose growth rates are exactly $P_*(f)$.
Let us note that the growth rate of
$\sum_{k=1}^nZ^*_k$ is also $P_*(f)$. Hence, if we denote by $W'_n$
those words in $W_n$ that contain symbols of exactly one colour and
otherwise at least one zero, then it follows that growth rate of
$Z_n=\sum_{\alpha\in W'_n} e^{g^n(x(\alpha))}$
is also given by $P_*(f)$.
It thus remains to show that the exponential growth rate of
$\sum_{\alpha\in W''_n} e^{g^n(x(\alpha))}$
is $\leq P_*(f)$, where $W''_n$ denotes the genuinely bi-coloured
words in $W_n$ (i.e.\ those words that contain green and yellow symbols).
In lemma 2 equation (\ref{coloured.words}) we estimated
$$
|W''_n|\leq r_n4^{n/(\tau+1)},
$$
where the constants $r_n$ grow at most subexponentially as
$n\rightarrow\infty$. Since $\sup g=\sup f$, we obtain
$$
Z''_n=\sum_{\alpha\in W''_n} e^{g^n(x(\alpha))}
\leq |W''_n|e^{n\sup g}
\leq r_n4^{n/(\tau+1)}e^{n\sup f},
$$
whose growth rate is bounded by $\frac{\log 4}{\tau+1}+\sup f$,
which is $\leq P_*(f)$ exacty if
$\tau+1\geq \frac{\log 4}{P_*(f)-\sup f}$.
\hfill$\Box$
\begin{lemma} If $\tau\geq \frac{\log 4}{P_*(f)-\sup f} -1$,
then there are two mutually
singular equilibrium states for $g$ on $\Sigma$.
\end{lemma}
\noindent {\bf Proof.}
There is an equilibrium state for $f$ on each of the full
two-shifts $\Sigma_{\cal G}$ and $\Sigma_{\cal Y}$ both of which,
by lemma 4,
assume the pressure $P(g)=P_*(f)$ in the variational principle.
These two measures are both invariant and mutually singular.
\hfill$\Box$
\vspace{5mm}
\noindent {\bf Remark:}
This method of constructing examples can be varied and adapted
in many ways. Instead of full two-shifts one can of course
take any subshift of finite type (or other), and one
can have any number of those embedded subshifts yielding a corresponding
number of mutually singular measures of maximal entropy or
equilibria for suitable potentials. In fact it is possible to have
infinitely many embedded subshifts each of which would carry a
measure and all of them would be mutually singular. This can be done
without sacrificing the mixing property;
there still would exist transitive points.
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\bibitem{B-S} R Burton and J E Steif: Nonuniqueness of Measures
of Maximal Entropy for Subshifts of Finite Type, Ergod.\ Th.\
Dynam.\ Sys.\ 14 (1994), 213--235
\bibitem{Hof} Hofbauer: Examples for the nonuniqueness of the
equilibrium state, AMS Transactions, 228 (1977), 223--241
\bibitem{Kr} W Krieger: On the uniqueness of the equilibrium state,
Math.\ Systems Theory 8 (1974), 97--104.
\bibitem{Wal} P Walter: {\em An introduction to ergodic theory},
Springer-Verlag, GTM \#79, 1982
\end{thebibliography}
\end{document}