# Equivalence of σ-convex hull and closed convex hull

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.

• Related: this question Apr 13, 2020 at 14:34
• Consider $\ J:=(0;1)\subseteq\Bbb R.\$ Then the sigma closure is $\ J;\$ it is not the closure, i.e. $\ [0;1].$ Apr 13, 2020 at 22:24
• @WlodAA: It seems, though, that the OP considers only compacts sets $K$. Apr 14, 2020 at 4:14
• @JochenGlueck, thank you. Apr 14, 2020 at 19:30
• 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
– erz
Apr 15, 2020 at 13:42

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 $$e(\lambda) \not\in e(\sigma\mbox{-}\mathrm{co}(K)) = \sigma\mbox{-}\mathrm{co}(e(K))$$.

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 $$\mathbb{R}^{\mathbb{N}}$$.

• 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 $K$? Apr 15, 2020 at 17:00
• @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. Apr 15, 2020 at 18:36
• @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. Apr 15, 2020 at 18:48
• @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. Apr 15, 2020 at 20:56
• 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. Apr 17, 2020 at 19:57