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.


[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.

  • 6
    $\begingroup$ 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). $\endgroup$ Feb 21, 2011 at 10:50
  • 2
    $\begingroup$ @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. $\endgroup$ Feb 21, 2011 at 22:16
  • 3
    $\begingroup$ 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. $\endgroup$ Feb 21, 2011 at 22:22
  • $\begingroup$ Good point. I've added a remark, just to explain why I become interested in the subject. $\endgroup$ Feb 22, 2011 at 7:28
  • 3
    $\begingroup$ 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. $\endgroup$ Feb 22, 2011 at 14:55

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.