restricted of the entropy function is concave and upper semi continuous Let $T:X \to X$ be a topologically mixing subshifts of finite type and $f:X\to \mathbb{R}$ be a continuous function. It is well known that the entropy function $\mu \mapsto h_{\mu}(T)$ is concave. I want to show that the following function is concave
$$ H(t)=\sup\{h_{\mu}(T), \mu \in \mathcal{M}(X, T), \int f d\mu=t\},$$
where $\mathcal{M}(X, T)$ is the space of invariant measure.
My attempt: for every $\mu_{1}, \mu_{2} \in \mathcal{M}(X, T)$, let $\mu=\lambda \mu_{1}-(1-\lambda)\mu_{2}$ for some $\lambda \in (0,1)$.
Since the entropy is concave, $h_{\mu}(T)\geq \lambda h_{\mu_{1}}(T)+(1-
\lambda)h_{
\mu_{2}}(T).$
Let $a=
\int f d\mu_{1}$ and $\int f d\mu_{2}=b.$
$$\sup \{h_{\mu}(T): \mu \in \mathcal{M}(X,T), \int f d\mu=\lambda a+(1-\lambda)b\}\geq \sup \{\lambda h_{\mu_{1}}(T)+(1-
\lambda)h_{
\mu_{2}}(T): \mu \in \mathcal{M}(X,T), \int f d\mu=\lambda a+(1-\lambda)b\}$$
I don't know how to get the result.
Edit: Is $H(t)$ upper semi continuous?
 A: You wrote all the right ingredients in your post.
You want to prove that $H$ is concave in $t$, so you want to prove that for any $s\in [0,1]$ and $a,b\in \mathbb{R}$ you have $H((1-s)a + sb) \geq (1-s) H(a) + s H(b)$.
For a real number $l\in \mathbb{R}$ let $\mathcal{M}_l$ be the set of invariant probability measures $\mu$ such that $\int fd\mu = l$, this set could be empty.
Fix $a,b \in \mathbb{R}$ and $s\in [0,1]$. Since $H(t)$ is defined by a sup, you can take two sequences of probability measures,  $(\mu^a_n)_{n\in \mathbb{N}}$ and $(\mu^b_n)_{n\in \mathbb{N}}$ such that for each $n\in \mathbb{N}$, $\mu^l_n \in \mathcal{M}_l$ and  $H(l) = \lim_{n\to +\infty} h_{\mu^l_n}(T)$, for $l=a,b$.
Define $\nu^s_n = (1-s) \mu^a_n + s \mu_n^b$. Since $\int f d\nu^s_n = (1-s) a + sb$, we have that $\nu^s_n \in \mathcal{M}_{(1-s) a + s b}$ for every $n\in \mathbb{N}$.  Also $h_{\nu^s_n}(T) = (1 -s) h_{\mu^a_n}(T) + s h_{\mu^b_n}(T)$. In particular, since $H(t)$ is defined by the supremum,
$$
H((1-s)a + sb) \geq \lim_{n\to +\infty} h_{\nu^s_n}(T) = \lim_{n\to + \infty} (1 - s ) h_{\mu^a_n}(T) + s h_{\mu^b_n}(T) = (1-s)H(a) + sH(b).
$$
