Let $C^\infty(X)$ denote the space of infinitely smooth functions on a compact manifold $X$ (at the beginning one may assume that $X$ is a circle, though I need a more general case). Let $\mathcal{D}(X)$ be the space of Schwartz distributions equipped with the $w^*$-topology, namely the weak topology induced by the natural pairing $C^\infty(X)\times \mathcal{D}(X)\to \mathbb{C}$.
Let $F\colon \mathcal{D}(X)\to \mathbb{C}$ be a linear functional which is sequentially continuous in the weak topology, namely it maps weakly convergent sequences to convergent ones.
Question: Is $F$ continuous in the weak topology?
Best Answer
It is continuous even for the strong topology. This follows from general locally convex space theory, in particular from the fact that the space of smooth functions is a nuclear Fréchet space. Its dual is a nuclear Silva space, i.e., an inductive limit of a sequence of Banach spaces with nuclear connecting mappings. The background theory can be found in many places, e.g., in the monograph of Gottfried Köthe on topological vector spaces. Note that in this context, weak and strong convergence for sequences coincide. Also the conclusion holds not just for linear functionals but for ANY mapping, linear or not. This was proved by Sebastião e Silva in the $50$'s.