Let $T:\Sigma \rightarrow \Sigma$ be a topologically mixing subshift of finite type and let $f:\Sigma \rightarrow \mathbb{R}$ be a continuous functions over $(T, \Sigma)$. Assume that there is a unique equilibrium measure $\mu$ for $f$ because of some reason.

$\textit{Question}:$ Does $\mu$ necessarily have Gibbs property?

I guess the answer is no, but I can't find a reference.


1 Answer 1


The measure $\mu$ does not necessarily have the Gibbs property. In fact, it has the Gibbs property if and only if $f$ has the Bowen property: $\sup_n \sup \{ |S_n f(x) - S_n f(y)| : x_1 \dots x_n = y_1 \dots y_n \} < \infty$. Every such $f$ has a unique equilibrium measure, but there are some potentials without the Bowen property that still have unique equilibrium measures.

$\mu$ Gibbs iff $f$ Bowen. The Gibbs property requires that there be $K>0$ such that for every $x\in \Sigma$ we have $$ K^{-1}\leq \frac{\mu[x_1\dots x_n]}{e^{-nP(f) + S_nf(x)}} \leq K. $$ Given $x,y \in \Sigma$ with $x_1\dots x_n = y_1 \dots y_n$, the only quantity in the corresponding inequalities that can vary is $S_n f$, and comparing them gives $$ K^{-2} \leq e^{S_n f(x) - S_n f(y)} \leq K^2. $$ Thus $|S_n f(x) - S_n f(y)| \leq 2\log K$, which proves the Bowen property. The other direction is classical; see

Bowen, Rufus, Some systems with unique equilibrium states, Math. Syst. Theory 8(1974), 193-202 (1975). ZBL0299.54031.

which gives a more general result (expansive systems with specification, which includes mixing SFTs).

An example of a non-Bowen potential that has a unique equilibrium state.

Hofbauer, Franz, Examples for the nonuniqueness of the equilibrium state, Trans. Am. Math. Soc. 228, 223-241 (1977). ZBL0355.28010.

The example there is the full shift on two symbols 0,1, and the potential is $f(x) = a_k$ whenever $x = 1^k 0\dots$, where $a_k$ is a sequence of real numbers converging to $0$. (Also $f(1^\infty) = 0$.) Writing $s_k = a_0 + \cdots + a_k$, the table on page 239 of that paper is useful. The potential $f$ has the Bowen property iff $\sum a_k$ converges, but there are examples where $\sum a_k$ diverges and $f$ still has a unique equilibrium measure.

It is often the case that unique equilibrium measures, including the ones in Hofbauer's paper, satisfy a "non-uniform" Gibbs property: see

Climenhaga, Vaughn; Thompson, Daniel J., Equilibrium states beyond specification and the Bowen property, J. Lond. Math. Soc., II. Ser. 87, No. 2, 401-427 (2013). ZBL1276.37023.

  • $\begingroup$ Thank you very much for your great answer. $\endgroup$
    – Adam
    Sep 9, 2020 at 9:04

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