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.

  • $\begingroup$ What does your question have to do with the Bernoulli measure? $\endgroup$
    – R W
    Oct 25, 2018 at 22:06
  • $\begingroup$ 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. $\endgroup$ Oct 26, 2018 at 2:32
  • $\begingroup$ The link is math.uvic.ca/faculty/aquas/papers/paper05.pdf $\endgroup$ Oct 26, 2018 at 3:05
  • $\begingroup$ @RW i want to work measure of maximal entropy and in case Bernoulli,we can easily write $H(\mu)=-\sum p_{i} log p_{i}$ $\endgroup$
    – Michal
    Oct 26, 2018 at 9:37
  • $\begingroup$ Many thanks for your answer, professor @AnthonyQuas $\endgroup$
    – Michal
    Oct 26, 2018 at 9:39

1 Answer 1


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 $f_n\to f$.

  • $\begingroup$ Yes,i don't assume continuity potential. $\endgroup$
    – Michal
    Oct 27, 2018 at 13:54
  • $\begingroup$ 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? $\endgroup$
    – Michal
    Oct 27, 2018 at 14:00
  • $\begingroup$ I am not sure what you mean by "continuity potential". Do you mean the assumption $f_n\to f$? $\endgroup$
    – Algernon
    Oct 27, 2018 at 18:25
  • $\begingroup$ yes. I mean exactly that one. $\endgroup$
    – Michal
    Oct 27, 2018 at 18:36
  • 1
    $\begingroup$ 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, $|p(f)-p(g)|\leq||f-g||$. $\endgroup$
    – Algernon
    Oct 27, 2018 at 19:53

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