# continuity entropy with respect gibbs measures

Let $$X=\{0,1\}^{\mathbb{N}}$$. For simplicity I consider Bernoulli measures on $$X$$ only.

Let $$f:X\to \mathbb{R}$$ be Holder continuous. The measure $$\mu$$ is a Gibbs measure with potential $$f$$ if there are $$C>0$$ and $$P\in\mathbb{R}$$, such that for every infinite sequence $$i_1 i_2\ldots$$ and all natural $$n$$, $$C^{-1} \le \frac{\mu[i_1\ldots i_n]}{\exp(-nP+f(i)+f(\sigma i)+\cdots+f(\sigma^{n-1}i))} \le C,$$ where $$\sigma$$ is the one side shift.

we write $$h(\mu)$$ for the Kolmogorov-Sinai (metric) entropy of $$\mu$$.

$$Question$$ Consider a topology on the Gibbs measures via the Holder topology on the potentials.Why is $$h(\mu)$$ continuous with respect to this topology?

comment we know that entropy is upper semi continuous,thus we have to show that why $$h(\mu)$$ is lower semi continuous.

• What does your question have to do with the Bernoulli measure?
– R W
Oct 25, 2018 at 22:06
• See my paper with Zaqueu Coelho. We show that the Gibbs measure depends d-bar continuously on the potential; and entropy is continuous with respect to d-bar distance. Oct 26, 2018 at 2:32
• Oct 26, 2018 at 3:05
• @RW i want to work measure of maximal entropy and in case Bernoulli,we can easily write $H(\mu)=-\sum p_{i} log p_{i}$ Oct 26, 2018 at 9:37
• Many thanks for your answer, professor @AnthonyQuas Oct 26, 2018 at 9:39

Let $$\mu_1,\mu_2,\ldots$$ and $$\mu$$ be invariant measures and $$f_1,f_2,\ldots$$ and $$f$$ be continuous functions such that

1. $$f_n\to f$$ uniformly,
2. $$\mu_n\to\mu$$ weakly,
3. $$\mu_n$$ is an equilibrium measure for $$f_n$$, that is, $$h(\mu_n)-\mu_n(f_n)=p(f_n)$$, where $$p(g):=\sup_\nu[h(\nu)-\nu(g)]$$ is the topological pressure of $$g$$. [Recall: an invariant Gibbs measure for a Hölder function $$g$$ is an equilibrium measure for $$g$$.]
4. $$\mu$$ is an equilibrium measure for $$f$$. [This in fact follows from the previous three conditions.]

The topological pressure is continuous, thus, $$p(f_n)\to p(f)$$, that is, \begin{align} h(\mu_n) - \mu_n(f_n) &\to h(\mu)-\mu(f) \;. \end{align} On the other hand, $$\mu_n(f_n)\to\mu(f)$$ because \begin{align} |\mu(f)-\mu_n(f_n)| &\leq |\mu(f)-\mu_n(f)| + |\mu_n(f)-\mu_n(f_n)| \end{align} and both terms on the righthand side go to $$0$$ as $$n\to\infty$$. It follows that $$h(\mu_n)\to h(\mu)$$.

This doesn't quite answer your question (if I understand it correctly) because of the assumption $$f_n\to f$$.

• Yes,i don't assume continuity potential. Oct 27, 2018 at 13:54
• If you take a look the paper,Prof.Quas mentioned who proved it, we have continuity potential,as well.I am wondering ,is there another way for proof it? Oct 27, 2018 at 14:00
• I am not sure what you mean by "continuity potential". Do you mean the assumption $f_n\to f$? Oct 27, 2018 at 18:25
• yes. I mean exactly that one. Oct 27, 2018 at 18:36
• The Lipschitz continuity of topological pressure (w.r.t. the uniform topology) is standard (e.g., Theorem 3.4 of Ruelle's book). With the variational definition of the topological pressure I used above, the proof is pretty simple: take two continuous functions $f$ and $g$. If $\mu$ is an equilibrium measure for $f$, we get $p(f)=h(\mu)-\mu(f)$ and $p(g)\geq h(\mu)-\mu(g)$, which gives $p(f)-p(g)\leq\mu(g-f)\leq||f-g||$ and by symmetry $p(g)-p(f)\leq||f-g||$. Hence, $|p(f)-p(g)|\leq||f-g||$. Oct 27, 2018 at 19:53