I don't know if you can see your answer at this link, but it should be viewable to users with at least 10000 points. Here is the text of the question and answer, separated by a line segment:
This is just an extension of my previous question http://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?
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 $\hat E.$
]]>Someone with access to moderator tools maybe able to address your question.
]]>I think I'm in agreement with the other moderators in saying that we're quite pleased to see that by moderating heavily in the early days, we've managed to ensure that the "aggregate attitude of its participants" agrees pretty well with our own ideas. :-) After all, MO was started for entirely selfish reasons -- Anton wanted a community of people eager to answer his questions!
Nevertheless, I acknowledge your request for moderate moderation.
]]>Splitting an account in two doesn't look like a good idea since it would confuse some users and encourage others to have several accounts as well, which goes against the net etiquette.
]]>If a question is deleted, what happens to the rep? As an example, consider this question which I've just voted to close. Within the hour of it being asked, it got an answer which was accepted. The answerer now has 16 rep, enough to start voting. The asker has 3 rep, not so much. If this question just gets closed, I presume that these reps stay as they are. This might be a case for deletion.
]]>More generally, since there's no-one yet with 10000 rep, the only people with access to the lists are the moderators. As someone with 3000 rep, I have most of the powers of a moderator but don't have the ability to selectively use those abilities because I don't have access to the lists. So I can't quickly scan through a list of questions that others have thought worthy of being closed and decide whether or not to add my voice to the list. I have to do it the hard way by going through every question. I'm not going to do that as it just takes too long. So maybe while there aren't any (or many) people with at least 10000 rep, perhaps we should have a thread here so that when someone votes to close a question then they mention it here and others with 3000 rep can go and see if they agree. Also, it can be used by those who don't have enough rep to vote to close but who would like to bring something to the attention of those who do (rather than just the moderators).
]]>can I do this somehow?
Well, there's always a back-up option of starting a thread on meta!
]]>http://mathoverflow.net/questions/3471/are-the-asymptotics-of-fourier-coefficients-to-periodic-solutions-of-ode-known
http://mathoverflow.net/questions/1828/is-an-nth-root-of-unity-a-square
http://mathoverflow.net/questions/6718/how-many-vertices-do-i-need-before-i-get-n-independent-edges
http://mathoverflow.net/questions/6895/unbounded-countable-subset
This weekend brought a flood of dripping, blatant homework questions. If we could agree on a standard for things that are atrociously homework, I'd be fine with deleting them and all their answers.
]]>At the moment, our policy seems to be: 1 for all posts which seem to be good faith or HW, 3 for "trolling" (spam/abusive) posts, but we haven't carefully formulated one. I'm starting this read in hopes we will.
At this point, I have to wonder if we shouldn't start being more severe and deleting more questions.
]]>