Let $\mathcal{P}(\mathbb{N})$ be the set of all probability mass functions on $\mathbb{N}=\{1,2,\dots \}$. Let $E$ be a closed(with respect to pointwise convergence, or equivalently the total variation metric) subset of $\mathcal{P}(\mathbb{N})$ and $Q\notin E$. Let $0<\beta<1$.
Now $\displaystyle \sum\_{x\in N} P(x)^{\beta}Q(x)^{1-\beta}\le 1$ for any $P$ and $Q$ by Holder's inequality. Let us suppose that $0< s:=\displaystyle \sup\_{P\in E}\sum P(x)^{\beta}Q(x)^{1-\beta}$.


Let $\{P\_n\}$ be a sequence in $E$ such that $\sum P\_n(x)^{\beta}Q(x)^{1-\beta}\to s$


Now my goal is to examine whether $\{P\_n\}$ have a convergent subsequence converging to a true probability distribution. By the diagonal argument I can always extract a convergent subsequence, but the limit need not be a probability distribution; it could be a defective probability distribution. But in this problem can one somehow argue that we can extract a convergent subsequence converging to a probability distribution?


I am also thinking of showing that $P\_n$ are tight, because in the examples which I have, mass cannot escape to infinity.
I hope, if we show tightness, we can extract a convergent subsequence converging to a true probability distribution.



Tightness of probabilty distributions
I think this conjecture is false, that is, there does not necessarily exist a subsequence that converges to a true probability distribution. Consider the following situation:


Let $Q=(1,0,0,0,...)$, i.e. the probability distribution with all mass at $x=1$.


Define the distribution $R\_n$, 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.


Let $E$ be the set of all $R\_n$, for $n\geq 2$. Note that $E$ is closed (because any distribution not in $E$ can be separated from it by a sufficiently small $\epsilon$ ball) and that $s=(1/2)^\beta>0$.


Next, we argue that any convergent distributions $P\_n$ contained in $E$ can be viewed as a subsequence of the $R\_n$. For any $R\_n$, $$\sum\_{x=1}^\infty R\_n^\beta (x) Q^{1-\beta}(x)=\left(\frac{1}{2}-\frac{1}{n}\right)^\beta$$
As $n$ increases, this value increases monotonically to $s$. Therefore, if we have any set of points $P\_n$ in $E$ such that $$\lim\_{n\rightarrow \infty} \sum P\_n^\beta (x) Q^{1-\beta}(x) =s$$
then $P\_n$ has a convergent subsequence that is a subsequence of the $R\_n$.


Finally, if we consider any subsequence of $R\_n$, it does not converge to a probability distribution since half of its probability wafts off to infinity. (A formal proof is straightforward.)


