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](https://en.wikipedia.org/wiki/Choquet_theory), 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.



Equivalence of σ-convex hull and closed convex hull
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 $e(\lambda) \not\in e(\sigma\mbox{-}\mathrm{co}(K)) = \sigma\mbox{-}\mathrm{co}(e(K))$. 




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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](https://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_theorem_(convex_hull)) 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 $\mathbb{R}^{\mathbb{N}}$.


