This is just an extension of my previous question [Tightness of probabilty distributions](https://mathoverflow.net/questions/41259/tightness-of-probabilty-distributions)
Let $\mathcal{P}(\mathbb{N})$ be the set of all PMF's on $\mathbb{N}=\{1,2,\dots \}$. Let $E$ be a convex subset of $\mathcal{P}(\mathbb{N})$ and $Q\notin E$. Let $\alpha>1$ and $\beta=\frac{1}{\alpha}$.
Let us suppose that $s:=\displaystyle \sup\_{P' \in E'}\sum P'(x)^{\beta}Q'(x)^{1-\beta} >0$, where $P'(x)=\frac{P(x)^{\alpha}}{\sum P(x)^{\alpha}}$,and $E'=\{P':P\in E\}$.
My problem is to find a convex $E$ such that $E'$ is closed (with respect to the total variation metric) but $s$ is not attained in $E'$.


By making use of the example given by Bill Bradley in the above mentioned thread, I have the following very close counterexample.


Let $Q=(1,0,0,0,...)$ and let $R\_n$, for $n=2,3,...,$ as $R\_n(1)=\frac{1}{2}-\frac{1}{n}, R\_n(n)=\frac{1}{2}+\frac{1}{n}$ and $R\_n(x)=0$ for all other values and let $E=$ Convex hull of $\{R\_n\}$.


As you can see $s=1$ and is not attained in $E'$ (actually attained at $(1, 0, 0,\dots)$ which is not in $E'$.) This is what I would be happy with.
But the problem here is that $E'$ is not closed also, because if we take $P\_n=\frac{1}{n} \sum\_{i=2}^{n+1}R\_i$, then $P\_n'\to (1, 0, 0, \dots)\notin E'$. But I want $E'$ to be closed. 


Can one somehow change the things here a bit and get a counterexample ( i.e., $E$ convex, $E'$ closed but $s$ is not attained)?
Or may the result be true?



Showing non-attainment of supremum
I think that in your assumption, the supremum is actually attained.


Consider the set 
$$\hat E:=\{tp \ :\ t\geq0 \ , \quad p\in E\ \} \cap\bar B(0,1;\ell^\alpha).$$
Since $E$ is convex, $\hat E$ is convex too (here $\bar B(0,1;\ell^\alpha)$ denotes the closed unit ball of the sequence space $\ell^\alpha$).


Moreover, we are going to show that the assumption that $E'$ is closed in $\ell^1,$ implies that $\hat E$ is a closed bounded subset of the reflexive space $\ell^\alpha$, thus weakly compact. Indeed, let $u$ belong to the $\ell^\alpha$ norm closure of $\hat E.$ So, there exists a sequence $t\_j\geq0,$ and a sequence $p\_j\in E,$ such that $u \_j:=t \_j\\ p\_j$ converges to $u$ in $\ell^\alpha.$ If $u=0$ then $u\in \hat E$ and there's nothing to prove; otherwise we have (for large $j$) that $p \_j / \| p \_j \| \_\alpha = u \_j / \| u \_j \| \_\alpha $ which converges in $\ell^\alpha$ to $u/\|u\| \_\alpha.$ Hence $p'\_j:=\big(p \_j/\|p \_j \| \_\alpha\big)^\alpha$ converges in $\ell^1$ to $\big(u/\|u \| \_\alpha\big)^\alpha, $ showing that the latter belongs to $E'$, which is $\|\cdot\| \_1$-closed. This implies that for some $p\in E,$ $u$ has the form $\frac{\|u\| \_\alpha}{\|p\| \_\alpha} p,$ so is in $\hat E$.


Now consider $v:=\big(q/\|q\| \_\alpha\big)^{\alpha-1}.$ It is a norm-one element of the dual space $\ell^{\alpha'},$ and your optimization problem can be rewritten as
$$s:=\sup \_{p\in E}\big(\frac{p}{\|p\| \_\alpha}\cdot v \big) =\sup \_{u\in\hat E} (u\cdot v),$$
that is attained by the weak compactness of $\hat E.$


