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Let's say that a subset $C$ of a Banach space $X$ is $\sigma$-convex if the following property holds:

For any sequence $(x_k)_{k\ge0}$ in $C$, and for any sequence of non-negative real numbers $(\lambda_k)_ {k\ge0}$ with $\sum_{k=0}^\infty \lambda_k=1$ the series $\sum_{k=0}^\infty \lambda_k x_k$ converges to an element of $C$.

(the term $\sigma$-convex seems quite natural for this property, and indeed it is used e.g. in this 1976 paper; though I'm not certain that this is the standard current terminology).

Clearly, any $\sigma$-convex set is convex and bounded; a bounded convex set need not be $\sigma$-convex (e.g. the convex hull $\Delta$ of the orthonormal basis of $\ell^2$). A closed bounded convex set is $\sigma$-convex; and an open bounded convex set is $\sigma$-convex, too. Also, the intersection of $\sigma$-convex sets is $\sigma$-convex, and the image of a $\sigma$-convex set via a bounded linear operator is $\sigma$-convex.

Question: is there a topological characterization of those bounded convex subsets of a Banach space which are $\sigma$-convex?

Given the above mentioned facts, a reasonable conjecture could be, that a bounded convex set is $\sigma$-convex if and only in it is a Baire space. [edit] a simple counterexample in $X:=\mathbb{R}\times \ell^2 $ is $C:=(0,1]\times B\, \cup\, \{0\}\times\Delta$ where $B$ is the open unit ball of $\ell^2$, and $\Delta$ is the non-$\sigma$-convex set described above. This set is bounded, convex, and Baire, though it's not $\sigma$-convex.

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[edit] As far as I see, the interesting feature of $\sigma$-convex sets is the following "iteration lemma" (it's a piece of the Open Mapping Theorem, that in my opinion is worth to be a lemma in itself, also because its proof is repeated in several theorems).

Lemma. Let $X$ be a Banach space; $C\subset X$ $\sigma$-convex; $B\subset X$ a bounded subset, $0 < t < 1$ be such that $$B\subset C + tB \, . $$ Then $$(1-t)B\subset C \, .$$

(proof: as in the OMT: start from $b_0\in B$, represent it as $b_0=c_0+tb_1$, and iterate; one gets $(1-t)b_0$ as sum of an infinite convex combination in $C$). Curiously, this is also a characterization, in that any bounded set $C$ for which the above property holds for any bounded set $B$ (and even for just a fixed $0 < t < 1$) is indeed $\sigma$-convex.

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I'm leaving this as a comment as you ask for a topological characterisation. I don't know one, but I do know an algebraic characterisation. What you describe are totally convex spaces as described at ncatlab.org/nlab/show/totally+convex+space in particular, any such set is the image of a unit ball under a continuous map from a Banach space (indeed, an $\ell^1$-space). –  Loop Space Feb 21 '11 at 10:50
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@Andrew: You are describing the $\sigma$-absolutely convex sets. @Pietro Majer: Continuing what Andrew started, IMO the $\sigma$-convex sets are best described as the image of the positive part of the unit ball of an $\ell_1$ space. –  Bill Johnson Feb 21 '11 at 22:16
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Notice that any subset of the closed unit ball of a Hilbert space (or any strictly convex Banach space) that contains the open unit ball is $\sigma$-absolutely convex, so $\sigma$-absolutely convex sets can be complicated topologically. –  Bill Johnson Feb 21 '11 at 22:22
    
Good point. I've added a remark, just to explain why I become interested in the subject. –  Pietro Majer Feb 22 '11 at 7:28
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I should mention that the $\sigma$-convex hull of a set came up naturally in the classification of Banach spaces that have the Radon-Nikodym property through Maynard's concept of $s$-dentability. Later Davis and Phelps showed that you could use just dentability, so these days no one talks about $s$-dentability. Their paper jstor.org/pss/2040618 gives the history, including the work of Rieffel that started this line of investigation. Or see the book Vector Measures by Diestel and Uhl. –  Bill Johnson Feb 22 '11 at 14:55

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