2

My question is based on this question, except for infinite dimensions. Let $E \subseteq X$ be a compact subset of a normed vector space. Consider the set $\Lambda=\{\lambda: \lambda = \int_{E} x d \mu(x)$ where $\mu$ is a probability measure on $E\}$. Apparently this is the closed convex hull of $E$ - as explained in the first answer and comments to the first answer here - which I understand to be the closure of the convex hull of $E$, i.e. the set $Cl \Gamma$ where $\Gamma=\left\{\sum_{i=1}^{k} w_{i} x_i \mid k \in \mathbb{N}, x_{i} \in E, \sum_{i=1}^{k} w_{i}=1, w_{i} \geq 0 \forall i\right\}.$ Please correct me if I'm wrong.

So, my understanding is that, in general it does not hold that $\Gamma = \Lambda$, unlike in this case, where $X$ is a Euclidean space, and, as I mentioned, in general $Cl \Gamma = \Lambda$.

My question is, which elements of $\Lambda$ are in $\Gamma$? For example, are interior points of $\Lambda$ in $\Gamma$? If yes, how do we show that, and what would even be a characterization of an interior point of $\Lambda$?

Canine360
  • 1,481
  • Isn't $\Lambda$ compact, though? In this case, the interior would be empty when $X$ is infinite dimensional. – tomasz Jul 29 '21 at 17:54
  • Well $\Lambda$ is a continuous image (expectation operator) of a compact set (the set of all Borel prob measures on the compact set $E$, which is compact under the Prokhorov metric). So in my understanding $\Lambda$ should be compact. – Canine360 Jul 29 '21 at 17:58
  • Right. So it doesn't really make sense to ask about the interior when $X$ is infinite dimensional. A compact set cannot contain an open ball in that case. Unless of course you allow $E$ to be non-compact. – tomasz Jul 29 '21 at 19:42
  • 1
    An example that may be worth thinking about: In the Hilbert space $\ell^2$, let $e_k$ be the standard unit vector with $1$ in position $k$ and $0$ in all other positions. Let $E={0}\cup{e_k/k:k=1,2,\dots}$. Then $\sum_{k=1}^\infty2^{-k}e_k/k$ is in the closure of the convex hull $\Gamma$ of $E$ (because all its partial sums are in $\Gamma$), but is not itself in $\Gamma$. – Andreas Blass Jul 29 '21 at 21:44
  • @tomasz OK I do not at all understand the part about compact sets having empty interior. Is that a property of compact sets in infinite dimensions? Could you elaborate? – Canine360 Jul 30 '21 at 00:07
  • Infinite dimensional Hausdorff TVS are not locally compact. So compact sets always have empty interior. – tomasz Jul 30 '21 at 11:44
  • So let $E$ be my metric space (cpt). $f:E\rightarrow \mathbb{R}$ is a continuous function. The set $f^{-1}((-\infty,s])$ is closed, as a inverse image of a closed set under a continuous map. Therefore it is compact, as it is a subset of the compact set $E$. It was my understanding that $f^{-1}((-\infty,s))$ is contained in its interior. How is that automatically empty? I'm just trying to understand better. – Canine360 Jul 31 '21 at 01:09
  • @tomasz Thanks once again for pointing this out to me. I have started a new question here related to this: https://math.stackexchange.com/questions/4214473/compact-subset-of-infinite-dimensional-space-has-empty-interior Would you kindly have a look? – Canine360 Aug 02 '21 at 00:19
  • 1
    @Canine360: It is not contained in the interior, because $f$ is only defined on $E$. It is an open subset of $E$, but so is all of $E$. – tomasz Aug 02 '21 at 10:40

0 Answers0