Does $\max_{x\in X}E[Y\mid X=x] = \max_{P_X}\sum_{x\in X}P_X(x)E[Y\mid X=x]$? Does this expression hold? $$\max_{x\in X}E[Y\mid X=x] = \max_{P_X}\sum_{x\in X} P_X(x)E[Y\mid X=x]$$
Left side says: max over all available $x$ for the expected value of $Y$ given $X=x$.
Right side says: maximize over probabilities distribution of $X$ 
I can't seem to figure it out.
I think that ifyou wan't to max over probablity distribution on the right hand side you would eventually maximize over $E[Y\mid X=x]$ and say you get that maximization is for $x=x'$ then you would simply give $P_X(x')=1$ and then the expressions are equal.
Am I right? How can you prove this?
Thanks 
 A: Let me call the set in which $X$ takes values $\mathcal X$ and define $h(x) = E(Y\mid X = x)$ for cleaner notation [I take it as given that such a $h$ exists].
Your question statement seems to imply that $h$ attains its supremum on $\mathcal X$: $h(x^\star) = \sup_{x\in \mathcal X}h(x)$ for some $x^\star \in \mathcal X$.
Now for any variable $X$ with distribution $P_X$ on $\mathcal X$:
$$ E(h(X)) = \int_{\mathcal X} h(x)P_X(dx) \leq \int_{\mathcal X}h(x^\star)P_X(dx) = h(x^\star).$$ 
But if $X$ takes the value $x^\star$ with probability one, then the inequality is an equality.
A: If state space is discrete (as your notation implies) the $\max$ exists (e.g. because the state space is finite) then the expression holds almost trivially- your argument says the RHS attains the LHS.  Moreover, it cannot exceed the left because it is less than
$$
\max_{x\in\mathcal{S}} \mathbb{E}[Y | X = x] \max_{p \in \mathcal{P}(\mathcal{S}) }\sum_{x\in\mathcal{S}} p(x),
$$
which is the LHS.  (As @Alex R. said your original notation is bizarre and I took the liberty of denoting by $\mathcal{P}(\mathcal{S})$ a suitable space of probability measures on the state space $\mathcal{S}$.
If you are actually talking about $\sup$ (still assuming the space is discrete) then you can take a sequence of $x^\star_n$  such that 
$$
\mathbb{E}[Y | X = x^\star_n] \to \sup_{x\in\mathcal{S}} \mathbb{E}[Y | X = x]
$$
and the argument still works.
Moreover, if your space is continuous and your space $\mathcal{P}{(\mathcal{S})}$ includes discrete measures, again the same argument works but you need integrals instead of the sums etc.  If you only want continuous distributions then you need to work a little harder to approximate $\delta$-measures.
So basically, your inequality works with $\max$ replaced by $\sup$.
