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


*

*$f_n\to f$ uniformly,

*$\mu_n\to\mu$ weakly,

*$\mu_n$ 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 $g$.]

*$\mu$ 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 $f_n\to f$.
