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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 bases 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.

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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 of the orthonormal bases 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.

$$*$$

 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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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 of the orthonormal bases 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.

$$*$$

 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 included 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

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