# Local convexity of the topology of weak convergence of probability measures

Let $X$ be a Polish or standard Borel space, and $\mathcal P(X)$ be the space of all Borel probability measures on $X$ endowed with the topology of weak convergence. I am thinking of using Choquet-Bishop-de Leeuw theorem which requires working on locally convex topological spaces. I thus wonder whether $\mathcal P(X)$ can be considered as a subspace of a locally convex topological space, or shall I instead work with compact subsets of $\mathcal P(X)$ in some norm topology?

## 1 Answer

Weak convergence is defined by the criterion $$\int f(x) d\mu_n \to \int f(x)\,d\mu$$ for all $f \in C_b(X)$.

That gives you $\mathcal{P}(X)$ as a subset of the closed unit ball of $C_b(X)^\ast$ (which can be canonically identified with the space of Borel measures with finite variation on the Stone-Čech compactification $\beta X$ of $X$ if $X$ is a completely regular [Hausdorff] space), and the weak topology on $\mathcal{P}(X)$ is the subspace topology induced by the weak$^\ast$ topology $\sigma(C_b(X)^\ast,C_b(X))$ on $C_b(X)^\ast$, which is locally convex (and Hausdorff). Weak and weak$^\ast$ topologies are always locally convex topologies, since they are induced by the seminorms $x \mapsto \lvert \langle \varphi,x\rangle\rvert$, where $x\in E$ and $\varphi\in E^\ast$ or $x\in E^\ast$ and $\varphi \in E$.

• $C_b(X)$ is a Banach space with the supremum norm. So its dual (the space of regular Borel measures with finite variation on the Stone-Čech compactification $\beta X$) is a Banach space too, and the norm is the total variation of the measure. Thus the space of probability measures on $X$ is a subset of the unit sphere, hence a fortiori of the closed unit ball, of $C_b(X)^\ast$. – Daniel Fischer Jul 23 '14 at 14:15
• Right. And that's precisely the one you want. (However, it is the weak$^\ast$ topology. I blame the heat for that mistake.) – Daniel Fischer Jul 23 '14 at 14:24
• Just added that as a remark. The weak and weak$^\ast$ topologies are induced by seminorms, hence locally convex. – Daniel Fischer Jul 23 '14 at 14:31
• Thanks, Daniel, that's clear now. I hope it's fine if I clean some comments as they're now redundant. – Ilya Jul 23 '14 at 14:31