This is just an extension of my previous question 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?