# Taking a convex hull does not increase a supremum of a linear function

Let $X$ be a topological vector space, let $f:X\to\Bbb R$ be a continuous linear function and let $P(X)$ denote the set of all Borel probability measures on $X$. For any $M\subseteq X$ we define the strong convex hull as $$\operatorname{sco}(M) = \left\{\int y\;\mathrm d\nu: \nu\in P(X) \text{ and } \nu^*(M) =1\right\}.$$ Is that true that for any non-empty $M$ it holds that $\sup_M f = \sup_{\operatorname{sco(M)}}f$? In case the latter fact is true, I would be happy if someone can provide a reference to this.

I am not sure whether the integral is always defined over topological vector spaces, so the motivation was the case $X=P(A)$ where $A$ is a Borel space, and $P(A)$ is endowed with the topology of weak convergence. In the latter situation the integration is well-defined, but I guess that a similar result shall hold for a more general case as well. Feel free to correct me.

• By "bounded linear function", do you mean continuous, or "bounded" as in $\bigl(\forall B \subset X\bigr)(B\text{ bounded } \Rightarrow f(B)\text{ bounded})$? – Daniel Fischer Jul 8 '13 at 17:53
• I know very little about the field, but it seems you are looking for stuff in Choquet theory. – Michael Greinecker Jul 8 '13 at 19:22
• What is $\nu^*$ ? – Bunder Jul 9 '13 at 6:50
• @Bunder: an outer measure induced by the probability measure $\nu$ – Ilya Jul 9 '13 at 7:03
• @Daniel: sorry, I meant a continuous function – Ilya Jul 9 '13 at 7:50

For continuous $f$, I think - I'm not sure enough about my measure theory to be near certain - that you can argue as follows:

$M \subset \operatorname{sco}(M)$ follows by choosing $\delta_x$ for $x \in M$ as the probability measure. Hence $\sup_M f \leqslant \sup_{\operatorname{sco}(M)} f$. Thus if $\sup_M f = \infty$, you have $\sup_{\operatorname{sco}(M)} f = \sup_M f$. And if $c = \sup_M f < \infty$, for any $\varepsilon > 0$, the open half space $H_\varepsilon := f^{-1}((-\infty,\, c+\varepsilon))$ is a neighbourhood of $M$, thus

$$x_\nu := \int_X y\,d\nu = \int_{H_\varepsilon} y\,d\nu \in \overline{H_\varepsilon} \subset f^{-1}((-\infty,\, c+\varepsilon])$$

for each $\nu \in P(X)$ with $\nu^\ast(M) = 1$. Letting $\varepsilon \to 0$ then shows $f(x_\nu) \leqslant c$, hence $\sup_{\operatorname{sco}(M)} f = \sup_M f$.

• I think the idea is clear to me, yet indeed some technical details I have to check (as I said, we even don't talk here which integral do we use). I guess, you don't have an idea about a book where a proof of a similar fact can be found, do you? – Ilya Jul 9 '13 at 9:55
• I'm afraid I can't name a book where something like that is treated. I expect that one can make the argument hold for every reasonable integral, by filling in the details, but as I said, I'm not sure about it. – Daniel Fischer Jul 9 '13 at 10:11