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.



A unique equilibrium state which does not have Gibbs property
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**](http://dx.doi.org/10.1007/BF01762666), Math. Syst. Theory 8(1974), 193-202 (1975). [ZBL0299.54031](https://zbmath.org/?q=an:0299.54031).


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**](http://dx.doi.org/10.2307/1998528), Trans. Am. Math. Soc. 228, 223-241 (1977). [ZBL0355.28010](https://zbmath.org/?q=an:0355.28010).


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**](http://dx.doi.org/10.1112/jlms/jds054), J. Lond. Math. Soc., II. Ser. 87, No. 2, 401-427 (2013). [ZBL1276.37023](https://zbmath.org/?q=an:1276.37023).


