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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?

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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 $\hat E.$

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I met TeX troubles :-( I hope you can read it. –  Pietro Majer Oct 16 '10 at 10:26
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Try inserting some spaces. Especially before undrescores –  Piero D'Ancona Oct 16 '10 at 10:57
    
Oh, Piero! :-) –  Pietro Majer Oct 16 '10 at 17:29
    
Thank you very much Pietro Majer. I have two questions on your solution. 1. How does $p_j/\|pj\|_{\alpha}$ converges to $\frac{u}{\|u\|_{\alpha}}inl^{\alpha}$ imply $(p_j/\|p_j\|_{\alpha})^{\alpha}$ converges to $(u/\|u\|_{\alpha})^{\alpha}$ in $l^1$? 2.In the last line,how arethosetwo supremums equal? Wehave $\{\frac{p}{\|p\|_{\alpha}}:p\in E\}\subset \hat{E}$ But these two sets are not equal right? –  Ashok Oct 17 '10 at 5:17
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1. that's a general fact about $L^p$ spaces: if $fj$ is a sequence in $L^p(X,\mu)$ converging to $f$, then $|fj|^p$ converges to $|f|^p$ in $L^1(X,\mu).$ (Sketch of Proof: up to extracting a subsequence we may assume $fj$ converges a.e. and is dominated in $L^p$; then apply dom.conv.thm). 2. Yes, they are different: the latter set is just the set of all $tu$ with $0\le t\le 1$ and $u$ in the former set. So what has been added does not affect the value of the supremum, for the reason that $q(x)\ge 0, \forall x$. The reason for introducing $\hat E$ is just convexity. –  Pietro Majer Oct 17 '10 at 6:04

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