tl;dr: When is the closed convex hull of a set $K$ equal to the set of "continuous" convex combinations of $K$?

I am essentially asking for the most general, infinite-dimensional analogue of this related question.

Update: I forgot to specicy that $K$ is compact. As @GeraldEdgar points out below, for noncompact $K$, the answer is trivially "no".

Suppose $K\subset E$ where $E$ is a topological vector space (as far as I can tell, this is the most general kind of space for which this question makes sense). Obviously we can define the closed convex hull $\overline{\text{conv} K}$ of $K$ as usual. Now consider the set $$ K^* = \{ \int_K x\,d\mu(x) : \mu \in\mathcal{P}(K)\}, $$ where $\mathcal{P}(K)$ is the set of (say, Borel) probability measures over $K$ and integral here is to be understood in the weak (Pettis) sense.

I would like to know when $\overline{\text{conv} K} = K^*$. If $E$ is finite-dimensional, there is equality. What are the most general assumptions on $E$ and $K$ for which this equality continues to hold?

(For the curious, the inspiration for this question came from trying to understand when $K^*$ is compact.)

  • 2
    $\begingroup$ A preliminary important question is what you mean by the integral, here. Bochner? Pettis? Or is it somehow part of the question (e.g., you want some kind of integral for which there will be equality in the most general setting possible)? $\endgroup$
    – Gro-Tsen
    Aug 12, 2020 at 0:23
  • $\begingroup$ I was thinking of Lebesgue integration. If there is any reason to consider more exotic integrals, I would be interested to learn more. $\endgroup$
    – user163625
    Aug 12, 2020 at 3:57
  • $\begingroup$ I don't know what the “Lebesgue integral” means in an infinite-dimensional vector space — is what I was trying to point out. The Bochner and Pettis integrals are way to extend it to this context. $\endgroup$
    – Gro-Tsen
    Aug 12, 2020 at 9:50
  • $\begingroup$ Now I see the confusion. I am used to the convention where such integrals are generally understood as weak integrals, which I suppose is the same as the "Pettis" integral. $\endgroup$
    – user163625
    Aug 13, 2020 at 15:02

1 Answer 1


No. Even in one dimension. Say $K$ is the open interval $(0,1)$. Show $0 \notin K^*$. Let $\mu$ be a probability measure with support contained in $(0,1)$. Indeed, $$ r(\mu) := \int_K x\,d\mu(x) $$ is the integral of a positive function. That is, $x > 0$ a.e. So $\int_K x\,d\mu(x) > 0$. Similarly $1 \notin K^*$.

In a locally convex topological vector space $E$, if there is any extreme point of $M = \overline{\text{conv} K}$ that does not already belong to $K$, then it also does not belong to $K^*$. So what if $K$ is the set $\text{ex}\; M$ of extreme points of a closed convex bounded set $M$? Can we recover $M$ as $K^*$?

A very nice little book that discusses this situation is

Phelps, Robert R., Lectures on Choquet’s theorem, Lecture Notes in Mathematics. 1757. Berlin: Springer. 124 p. (2001). ZBL0997.46005.

Choquet's theorem tells us roughly that every point of a compact convex set $M$ is of the form $r(\mu)$ for some probability measure concentrated on the set $\text{ex}\; M$ of extreme points of $M$.

My first publication to attract any notice was this one, where there is a generalization of Choquet's theorem to certain closed bounded noncompact sets $M$.

Edgar, G. A., A noncompact Choquet theorem, Proc. Am. Math. Soc. 49, 354-358 (1975). ZBL0273.46012.

  • $\begingroup$ Oops. $K$ should be compact, otherwise this trivially fails even in finite dimensions, as you point out. $\endgroup$
    – user163625
    Aug 12, 2020 at 1:34

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