First, the setup: $X$ is a compact set. By Riesz's representation theorem $C(X)^\*=${all Radon measures on $X$}. $K$ is a convex, closed set of probability measures. $m$ is a probability measure out of $K$.
In a paper I read, the author uses the fact that there's a continuous function $g$ s.t. $\int gdm>sup\_{\mu \in K}\{\int gd\mu\}$


So far I was able to show that $\exists \psi\in C(X)^{\*\*}$ s.t. $\psi(m)>sup\_{\mu \in K}\{\psi(\mu)\}$


And also that the unit ball of $C(X)$ is dense in the weak-\* topology in the unit ball of $C(X)^{\*\*}$, meaning that $\exists \{g\_n\}\subset B(C(X))$ s.t. $g\_n\overset{w-\*}{\longrightarrow} \psi$, where the canonical embedding is $g\_n(\mu)=\int gd\mu$.


Making the final step of showing some continuous function actually fulfills the desired inequality, has been unsuccessful for me so far. Neither could I find any book that show a similar claim.


Any help would be appreciated, especially a simple quick solution that I missed.



How to show that there's a continuous function separating convex sets of Radon measures?
Since the weak topology is Hausdorff, there is an open set containing $m$ and disjoint from $K$. So we can find $f\_1,...,f\_n$ and $\varepsilon$ such that the open set
$$\{\mu \mid \int f\_i dm -\varepsilon < \int f\_i d\mu < \int f\_i dm +\varepsilon \quad \forall \ i \}$$
is disjoint from $K$. Now project everything on ${\bf R}^n$ with the map
$\mu \mapsto (\int f\_1 d\mu,..., \int f\_n d\mu)$. 


The projections of $K$ and the open set are two disjoint convex sets that can be separated by an hyperplane $\{H=\sum a\_i x\_i +c=0\}$ with $H<0$ on the projection of $K$ and $H>0$ on the projection of the open set (which is just a small cube). Take $g=\sum a\_i f\_i +c$.


