4
$\begingroup$

While reading a paper, I encountered the following statement:

Let $K$ be a convex compact subset of a locally convex topological vector space. If $\mu \in P(K)$ is a Radon probability measure on $K$, then there is a unique point $x_\mu \in K$ such that $$\int_K fd \mu = f(x_\mu)$$ for every continuous, affine, real-valued function $f: K \to \mathbb{R}$. Moreover, the map $P(K) \to K: \mu \mapsto x_\mu$ is a surjective affine map.

I consulted "Lectures on Choquet's theorem" but I couldn't find this version in the book. Does anybody know an appropriate reference/proof?

Edit: With an affine function $f: K \to \mathbb{R}$, I mean a function with the property

$$f\left(\sum_i \lambda_i k_i\right)= \sum_i \lambda_i f(k_i)$$ when $0 \le \lambda_i \le 1$ and $\sum_i \lambda_i = 1$, i.e. $f$ preserves convex combinations.

$\endgroup$

1 Answer 1

4
$\begingroup$

The point $x_\mu$ is the barycentre of the measure $\mu$. Its existence and uniqueness is Proposition 1.1 in "Lectures on Choquet's theorem" by Phelps.

$\endgroup$
7
  • $\begingroup$ @R W But this proposition states something about $f \in E^*$ instead of affine functions! $\endgroup$
    – Andromeda
    Commented Oct 20, 2021 at 15:15
  • 1
    $\begingroup$ What's the difference between affine and linear functions? $\endgroup$
    – R W
    Commented Oct 20, 2021 at 15:21
  • $\begingroup$ @R W An affine map preserves convex combinations. A linear map is much stronger. $\endgroup$
    – Andromeda
    Commented Oct 20, 2021 at 15:23
  • 2
    $\begingroup$ Not so much - think about it $\endgroup$
    – R W
    Commented Oct 20, 2021 at 15:28
  • 2
    $\begingroup$ I presume Hahn - Banach would help $\endgroup$
    – R W
    Commented Oct 20, 2021 at 16:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .