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Let $X$ be a locally convex topological space, and let $K \subset X$ be a compact set. Recalling that the standard convex hull is defined as $$\text{co}(K) = \Big\{ \sum_{i=1}^n a_i x_i : a_i \geq 0,\, \sum_{i=1}^n a_i = 1,\, x_i \in K \Big\},$$ define the $\sigma$-convex hull as $$\sigma\text{-}\mathrm{co}(K) = \Big\{ \sum_{i=1}^\infty a_i x_i : a_i \geq 0,\, \sum_{i=1}^\infty a_i = 1,\, x_i \in K \Big\},$$ where the summation is to be understood as convergence of the sequence in the topology of $X$.

I would like to understand conditions under which $\sigma\text{-}\mathrm{co}(K)$ is exactly the closure of $\mathrm{co}(K)$. In particular, does this property hold for any separable normed space $X$, or are further constraints on $X$ (and $K$?) required?

The motivation for this question is Choquet's theorem, which allows one to write $$\overline{\mathrm{co}}(K) = \Big\{ \int x d\mu(x) : \mu \in M(K) \Big\}$$ with $M(K)$ standing for probability measures on $K$ for any compact subset $K$ in a normed space. I would like to understand the "countable" version of this theorem as presented above, but I could not find any references nor do I have an idea about how one could prove it.

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    $\begingroup$ Related: this question $\endgroup$ Apr 13, 2020 at 14:34
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    $\begingroup$ Consider $\ J:=(0;1)\subseteq\Bbb R.\ $ Then the sigma closure is $\ J;\ $ it is not the closure, i.e. $\ [0;1].$ $\endgroup$
    – Wlod AA
    Apr 13, 2020 at 22:24
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    $\begingroup$ @WlodAA: It seems, though, that the OP considers only compacts sets $K$. $\endgroup$ Apr 14, 2020 at 4:14
  • $\begingroup$ @JochenGlueck, thank you. $\endgroup$
    – Wlod AA
    Apr 14, 2020 at 19:30
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    $\begingroup$ there are exercises 1.66 and 1.67 in Fabian, Habala Hajek, Montesinos, Zizler - Banach space theory, dedicated to these notions, although they do not adress your specific question $\endgroup$
    – erz
    Apr 15, 2020 at 13:42

1 Answer 1

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Wlod AA gave a good counterexample for the case when $K$ is not required to be compact, here I give a counterexample $K$ compact, first in a locally convex space, and then for a(n infinite-dimensional) separable normed space, and (after an edit) for all infinite-dimensional Banach spaces.

There is a standard counterexample if $X$ is only required to be locally convex, which is to take $X = C([0,1])^*$ with the weak-* topology, and to take $K$ to be the set of unital ring homomorphisms $C([0,1]) \rightarrow \mathbb{R}$. Making free use of the Riesz representation theorem to consider elements of $C([0,1])^*$ as measures on $[0,1]$, the elements of $K$ are the Dirac $\delta$-measures. Now, for each element $\mu$ of $\sigma\mbox{-}\mathrm{co}(K)$, there exists a countable set $S \subseteq [0,1]$ such that $\mu([0,1]\setminus S) = 0$. However, $\overline{\mathrm{co}}(K)$ consists of $P([0,1])$, the set of all positive unital linear functionals on $C([0,1])$, i.e. all probability measures on $[0,1]$, and so Lebesgue measure is an element of $\overline{\mathrm{co}}(K) \setminus \sigma\mbox{-}\mathrm{co}(K)$.

To get this to happen in a normed space, we will use $\ell^2$, and embed $P([0,1])$ affinely and continuously into it. First, observe that we can affinely embed $P([0,1])$ into $[0,1]^{\mathbb{N}}$, getting each coordinate by evaluating at $x^n$ (including $n = 0$). This is injective because polynomials are norm dense in $C([0,1])$, and continuous by the definition of the weak-* topology. We can then embed $[0,1]^{\mathbb{N}}$ into $\ell^2$ by the mapping: $$ f(a)_n = \frac{1}{n+1}a_n $$ this is affine and continuous from the product topology on $[0,1]^\mathbb{N}$ to the norm topology on $\ell^2$ (in fact, it defines a continuous linear map from the bounded weak-* topology on $\ell^\infty$ to the norm topology on $\ell^2$). We use $e$ for the composition of these two embeddings, and it is affine and continuous on $P([0,1])$.

A continuous injective map from a compact Hausdorff space to a Hausdorff space is a homeomorphism onto its image, and as we also preserved convex combinations by making the embedding affine, we have that $\overline{\mathrm{co}}(e(K)) = e(\overline{\mathrm{co}}(K)) = e(P([0,1]))$, while, taking $\lambda$ to be the element of $P([0,1])$ defined by Lebesgue measure, $e(\lambda) \in e(P([0,1]))$, but $e(\lambda) \not\in e(\sigma\mbox{-}\mathrm{co}(K)) = \sigma\mbox{-}\mathrm{co}(e(K))$.


Added in edit:

As Bill Johnson points out, there is an injective bounded map from $\ell^2$ into any infinite-dimensional Banach space $E$. By the same argument used to transfer the example to $\ell^2$, this allows us to transfer the example to $E$.

In the other direction, the convex hull of a compact subset $K$ of a finite-dimensional space is compact (using Carathéodory's theorem we can express the convex hull of $K$ as the continuous image of the compact set $K^{d+1} \times P(d+1)$, where $d$ is the dimension. Therefore the $\sigma$-convex hull and closed convex hull of $K$ coincide.

All together, this means:

If $E$ is a Banach space, the statement "for all compact sets $K \subseteq E$, the closed convex hull equals the $\sigma$-convex hull" is equivalent to "$E$ is finite-dimensional".

There are, however, complete locally convex spaces in which every bounded set, and therefore every compact set, is contained in a finite-dimensional subspace, and for which, therefore, the $\sigma$-convex and closed convex hulls of compact sets coincide. One example is the space $\phi$ of finitely supported functions $\mathbb{N} \rightarrow \mathbb{R}$, topologized as an $\mathbb{N}$-fold locally convex coproduct of $\mathbb{R}$ with itself, or equivalently as the strong dual space of $\mathbb{R}^{\mathbb{N}}$.

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  • $\begingroup$ Thank you, I'll need to go through this carefully. Do you have any ideas about the least set of assumptions about the space $X$ to make the property hold true for any compact $K$? $\endgroup$
    – Gregory D.
    Apr 15, 2020 at 17:00
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    $\begingroup$ @GregoryD. A sufficient condition is that $X$ is finite-dimensional. Then the convex hull of a compact set is compact, so is equal to both the $\sigma$-convex and closed convex hulls (it is easy to prove this using Carathéodory's theorem. It follows that the property does hold in spaces in which every compact set is finite-dimensional, such as an infinite locally convex coproduct of $\mathbb{R}$ with itself. $\endgroup$ Apr 15, 2020 at 18:36
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    $\begingroup$ @GregoryD. It may well be that this property does not hold for any infinite-dimensional Banach space, but my knowledge of this kind of functional analysis has run out, so I hope one of the experts can help on that subject. $\endgroup$ Apr 15, 2020 at 18:48
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    $\begingroup$ @RobertFurber: Beautiful counterexamples! Interestingly, there are also infinite-dimensional compact sets $K$ for which the $\sigma$-convex hull coincides with the closure of the convex hull - for instance $K = \{e_n/n: \, n \in \mathbb{N}\} \subseteq \ell^2$, where $e_n$ denotes the $n$-th canonical unit vector. It really seems interesting how to characterize such sets $K$; but I wasn't able to come up with any ideas, yet. $\endgroup$ Apr 15, 2020 at 20:56
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    $\begingroup$ Once you have an example in $\ell_2$ you can transfer it to any infinite dimensional Banach space $X$ by taking an injective continuous linear mapping from $\ell_2$ into $X$. It is an exercise for students that such a map exists. $\endgroup$ Apr 17, 2020 at 19:57

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