Conxex combinations of max and min Is the following true?
$$α \left( \max_{p\in P}\int g\mathrm dp\right)+\left (1-\alpha \right ) \left(
\max_{q\in Q}\int g\mathrm dq \right )=\max_{z\in\left (\alpha P+\left(1-\alpha \right )Q \right)}\int g\mathrm dz$$
where $P,Q$ are weak* closed convex set of probability measures, $\alph\in(0,1)$ and $g$ is a bounded and measurable function. 
Is it true if I substitute the max with the min function?
It seems to me that the answer to both my questions is YES.
Thank you in advance to let me know your opinions.
 A: Denote by $\def\Prob{\mathop{\rm Prob}}\Prob(X)$ the set of all probiability measures on $X$, which is a weak$^*$ compact convex subset of the dual of the bounded measurable functions on $X$. Given a bounded measurable $g \colon X \to \mathbb R$, the map $\phi_g \colon \Prob(X) \to \mathbb R$, $\phi_g(p) = \int_X g \, dp$ is an affine weak$^*$ continuous functional. 
Now, as $P$ and $Q$ are weak$^*$ compact, the maxima exist, let $p \in P$, $q \in Q$, and $r \in (1-\alpha)P + \alpha Q$, such that 
$$ \phi_g(p) = \max_P \phi_g, \quad \phi_g(q) = \max_Q \phi_g, \quad \phi_g(r) = \max_{(1-\alpha)P + \alpha Q} \phi_g $$
Then $(1-\alpha)p + \alpha q \in (1-\alpha)P +\alpha Q$, hence 
$$ \phi_g\bigl((1-\alpha)p + \alpha q\bigr) \le \phi_g(r) $$
which gives, as $\phi_g$ is affine
$$ (1-\alpha)\phi_g(p) + \alpha \phi_g(q) \le \phi_g(r) $$
On the other hand, for some $p' \in P$, $q'\in Q$, we have $(1-\alpha)p' + \alpha q' = r$, hence
$$ \phi_g(r) = (1-\alpha)\phi_g(p') + \alpha\phi_g(q') \le (1-\alpha)\phi_g(p) + \alpha \phi_g(q) $$
which gives 
$$ (1-\alpha)\max_P \phi_g + \alpha \max_Q\phi_g = \max_{(1-\alpha)P+\alpha Q} \phi_g $$
as wished. The same formula for $\min$ can be proved along the same lines (or by replacing $g$ with $-g$ an noting that $\max {\phi_{-g}}= -\min \phi_g $.
