The space of temperate distributions $S'(\mathbb{R}^d)$ is often equipped with the weak-$\ast$ or with the strong topology. When defining the notion of a probability measure on $S'(\mathbb{R}^d)$, this makes no difference since the corresponding Borel $\sigma$-algebras are the same: see for instance this article by Becnel.

However, when it comes to weak convergence of probability measures, the topology cannot be ignored. Even though $S'(\mathbb{R}^d)$ is not a Polish space, I am using the standard definition: $\mu_n\rightarrow \mu$ weakly if for all bounded continuous functions $F:S'(\mathbb{R}^d)\rightarrow \mathbb{R}$ one has $$ \lim_{n\rightarrow\infty} \int F\ d\mu_n\ =\ \int F\ d\mu\ . $$ My question is: does weak convergence for the weak-$\ast$ topology imply weak convergence for the strong topology?

The reason I ask is that the strong topology seems to be the preferred one in the literature on topological vector spaces. On the other hand, the weak-$\ast$ topology is very nice as far as probability theory goes, in particular Prokhorov's Theorem and Levy's Continuity Theorem hold for it.


1 Answer 1


Just after posting this MO answer Generalization of Lévy's continuity theorem for nuclear spaces I went and looked again at Fernique's remarkable article and he answered my question above in the affirmative as Corollary 1 of Theorem III.6.5.


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