Convex function and weak convergence of measures

Question

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 $\mu$?

Answer 1

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.