Let $X$ be a compact set. Given probability measures $\mu,\mu\_n: \mathcal{B}(X)\to R$. Is it possible that $$\langle f,\mu\_n\rangle \rightarrow \langle f,\mu\rangle$$
for all convex function $f:X\to R$ but $\mu\_n$ does not converge weakly to $\mu$?



Convex function and weak convergence of measures
Let $\mathscr{X}$ be a locally convex space and $X \subset \mathscr{X}$ compact.


Let $E = \{f \in C(X) : \langle f, \mu\_n \rangle \to \langle f, \mu \rangle\}$. It's easy to see that $E$ is a closed linear subspace of $C(X)$ (use the triangle inequality).


Note that for any continuous linear functional $\lambda \in \mathscr{X}^\*$, the function $f(x) = e^{\lambda(x)}$ is convex (being a convex function of a linear function). So by assumption all such functions are in $E$. Since $E$ is a linear subspace, $E$ contains the set $\mathcal{A}$ of all functions of the form
$$f(x) = \sum\_{i=1}^n a\_i e^{\lambda\_i(x)}, \quad a\_i \in \mathbb{R},\, \lambda\_i \in \mathscr{X}^\*.$$
But it's clear that $\mathcal{A}$ is an algebra which separates points (by Hahn–Banach) and contains the constants (take $\lambda\_i = 0$). So by Stone–Weierstrass, $\mathcal{A}$ is dense in $C(X)$, which means $E = C(X)$. That is to say that $\mu\_n \to \mu$ weakly.


