# Infinite convex combinations in a Banach space

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.  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.

$$*$$

 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.

• 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). Feb 21, 2011 at 10:50
• @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. Feb 21, 2011 at 22:16
• 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. Feb 21, 2011 at 22:22
• Good point. I've added a remark, just to explain why I become interested in the subject. Feb 22, 2011 at 7:28
• 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. Feb 22, 2011 at 14:55

Possibly I misunderstood your question, but I do not see a hope for a suitable characterization because of the following example (inspired by the comment of Bill Johnson): Consider a set $A$ in $\ell_2^2$ ($2$-dimensional Euclidean space) consisting of the union of the open unit ball and some subset of the unit sphere. The set $A$ is always $\sigma$-convex. On the other hand $A$ is (this can be shown directly) homeomorphic to a subset $O$ consisting of the union of the open unit ball of $\ell_1^2$ and a suitable subset of the sphere. The set $O$ is $\sigma$-convex (and even convex) in some exceptional cases only. Unless the set $A$ contains at most one point on the boundary, the homeomorphism can be modified in such a way that the (modified) $O$ is not $\sigma$-convex.