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.

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 $\sup_{\operatorname{sco}(M)} f = \sup_M f$.
