I think we need to have a serious discussion about what the appropriate action with inappropriate posts is. As I see it, there are roughly 3 options:

1. Close the question, and do nothing else.
2. Close the question, and delete the answers.
3. Delete the question.

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.

People should be encouraged to downvote any answers to clearly inappropriate questions. If you answer the question, you don't teach the lesson. Then, such answers should be deleted.

@Ilya- As far as I know, we have never deleted a good faith answer to a question (the only deleted answer I remember was a joke about Turing's sexual orientation). The ones you pointed out just never had any answers. I'm sure you've all seen what kind of questions get closed. The ones that have been deleted were ones that were on the level of "Was Serge Lang gay?"

A question that achieves a score of, say, -10, should be automatically closed, no?

@Ben: I proposed a solution to this on another thread somewhere around here, though I don't know how well it would work: split your account into two. Have a "normal user" account (with all the accumulated rep that you have) and a "moderator" account. Use the "normal user" account for almost everything and only use the "moderator" account when you really need to. I don't know if this is possible, but if it is then it would solve that problem.

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).

I've deleted a lot of my own questions, and nothing happens. Your rep remains entirely preserved. However if you have negative rep, I don't know if it will preserve that.

I think that meant fpqc wasn't sure if you would get back any rep you lost from it being voted down, or from voting the questions or answers down.

@Andrew, I think it's a great idea to have a public list of "currently in moderation" questions.

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 post is deleted, the reputation (positive or negative) disappears, but not instantly. If there's some problem (e.g. somebody has gained enough rep to be a nuisance), a moderator can manually trigger a rep recalculation. For example, I just triggered a rep recalculation for fpqc, which brought his rep up from 281 to 286.

@davidk01

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.

Hi! I think I've recently answered a question by user Ashok... Since she/he is asking now a close question, I'd like to check the previous answer, but there's no trace of it... Well' I suppose I'd just like to know how it works about deletions of questions. Actually, the idea that my previously given answers may disappear destabilizes me a little - they may be of use to me for further references here, exercises etc. Thanks, PM.
PS: so, if possible, I'd like that that post were restored, since it may be of iterest to other people too...

PPS: thanks everybody. It seems that Ashok deleted that material "for some reason". Actually I don't get it, but it's certainly his/her right do do it, according to the MO current policy, since deletion of questions is a possible option.
Actually, I'm a bit puzzled about this: is it a good thing that one can delete a question together with other people's contributions? After thinking, I tend to think that it should be more constructive to put as a condition the permission of the administrators and of the other people involved.
(compare with the Wikipedia editing policy: "You irrevocably agree to release your contributions... etc")
cheers, pm

Your answer received one upvote before Ashok deleted the question, so my new conjecture about this software is that the person asking the question can delete as long as an answer isn't accepted. It's not clear to me why Ashok deleted the question, but I feel a little uncomfortable with the idea of using my power to undelete without more discussion. I'll send Ashok an email.

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 $\hat E.$

Pietro, Ashok just posted a revised form of the question, or a question anyway, and called your attention to it in a comment.

http://mathoverflow.net/questions/44452/weak-convergence