Let $T:X\to X$ be a continuous map on a compact metric space. I'm trying to prove that for two $T$-invariant probability measures $\mu,\nu$ we have $h_{p\mu+(1-p)\nu}(T)=ph_\mu(T)+(1-p)h_\nu(T)$. Here $h_\mu(T)$ denotes the metric entropy of $T$, also known as the measure theoretic entropy.
I managed to show that we have $h_{p\mu+(1-p)\nu}(T)\geq ph_\mu(T)+(1-p)h_\nu(T)$ by defining $\Phi:[0,\infty)\to\mathbb R$ by $\Phi(x)=-x\log x$ for $x>0$ and $\Phi(0)=0$; then $\Phi$ is a concave function, hence by Jensen $$p\Phi(\mu(A))+(1-p)\Phi(\nu(A))\leq\Phi(p\mu(A)+(1-p)\nu(A))$$ for each measurable $A$. Applying this to finite measurable partitions $\alpha$ gives $H_{p\mu+(1-p)\nu}(\alpha)\geq pH_\mu(\alpha)+(1-p)H_\nu(\alpha)$. Next, we plug in $\bigvee_{i=0}^{n-1}T^{-i}\alpha$, divide by $n$ and letting $n\to\infty$ and obtain $h_{p\mu+(1-p)\nu}(\alpha,T)\geq ph_\mu(\alpha,T)+(1-p)h_\nu(\alpha,T)$. Then we find partitions $\alpha_\mu,\alpha_\nu$ such that $h_\mu(\alpha_\mu,T)\geq h_\mu(T)-\frac{\epsilon}{p}$ and $h_\mu(\alpha_\nu,T)\geq h_\nu(T)-\frac{\epsilon}{1-p}$, yielding $$h_{p\mu+(1-p)\nu}(\alpha_\mu\vee\alpha_\nu,T)\geq ph_\mu(T)+(1-p)h_\nu(T)-\epsilon;$$ passing to the supremum over all measurable partition $\alpha$ on the left hand side gives the desired inequality.
For the reverse inequality, I have no idea what to do. Any help is much appreciated!
I managed to prove the other inequality myself.
Again take $\Phi(x)=-x\log x$ and observe that $-\log x$ is decreasing. Therefore, for each measurable $A$ and $p\in(0,1)$, \begin{align*}&\phantom=\ \ \Phi(p\mu(A)+(1-p)\nu(A))\\&=-p\mu(A)\log\big(p\mu(A)+(1-p)\nu(A)\big)-(1-p)\nu(A)\log\big(p\mu(A)+(1-p)\nu(A)\big)\\&\leq -p\mu(A)\log\big(p\mu(A)\big)-(1-p)\log\big((1-p)\nu(A)\big)\\&=p\Phi(\mu(A))+(1-p)\Phi(\nu(A))-p\log(p)\mu(A)-(1-p)\log(1-p)\nu(A).\end{align*} We proceed as in the proof of the other inequality: sum over the sets in some finite measurable partition $\alpha$ to obtain $$H_{p\mu+(1-p)\nu}(\alpha)\leq pH_\mu(\alpha)+(1-p)H_\nu(\alpha)-p\log(p)-(1-p)\log(1-p).$$ Next, we plug in $\bigvee_{i=0}^{n-1}T^{-i}\alpha$, divide by $n$ and let $n\to\infty$ to obtain $$h_{p\mu+(1-p)\nu}(\alpha,T)\leq ph_\mu(\alpha,T)+(1-p)h_\nu(\alpha,T).$$ Now pass over to the supremum over all measurable partitions $\alpha$, to obtain \begin{align*}h_{p\mu+(1-p)\nu}(T)&\leq\sup_\alpha\left\{ ph_\mu(\alpha,T)+(1-p)h_\nu(\alpha,T)\right\}\\&\leq p\sup_\alpha h_\mu(\alpha,T)+(1-p)\sup_\beta h_\nu(\beta,T)\\&=ph_\mu(T)+(1-p)h_\nu(T),\end{align*} establishing the other inequality, hence finishing the proof.