When sequentially continuous linear functional is continuous?

Question

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?

Answer 1

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.

Answer 2

One can use the theory of Silva spaces as proposed in janacek's answer. However, I think that things are easier, here. Let $E$ be any reflexive Frechet space (in our situation, $E=C^\infty(X)$ which is in fact even nuclear and thus a Schwartz space). The question then is if every weak* sequentially continuous linear functional $F:E'\to\mathbb C$ is weak* continuous.

To prove this we first note that $F$ is bounded on all equicontinuous sets (because they are weak* compact by Alaoglu). Since at the same time every $\beta(E',E)$-bounded set is equicontinuous, $F$ is bounded on $\beta(E',E)$-bounded sets which implies $\beta(E',E)$-continuity since reflexive Frechet spaces are distinguished (meaning that the strong dual is bornological). This is probably due to Grothendieck. Finally, $F\in (E'\beta(E',E))' = E$ because of reflexivity and therefore, $F$ is weak*-continuous.

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By the way, the positive answer to your question also holds for $\mathscr D'(X)$ for e.g. open subsets of some $\mathbb R^d$ because this space is again bornological and reflexive (but it is not a Silva space).

Answer 3

Thanks for the two previous answers. Both are very interesting. Meantime I have found another general result which seems to be particularly convenient for my purposes.

Theorem. Let $E$ be a separable complete locally convex space. Then any sequentially continuous (in the weak* topology) linear functional on $E'$ is weak* continuous.

See Schaefer's book "Topological vector spaces", Ch. IV, $\S$ 6, Corollary 3.

In our situation $E=C^\infty(X)$ is obviously complete and separable when $X$ is a compact manifold. Furthermore the above theorem is applicable when $E=\mathcal{D}(X)$ is the space of infinitely smooth compactly supported functions on $X$ when $X$ is a countable at infinity manifold (i.e. $X$ is representable as a countable union of compact subsets). Indeed $\mathcal{D}(X)$ is a strict inductive limit of a sequence of complete locally convex (even Frechet) spaces which is complete by Ch. II, $\S$ 7, (6.6) of Schaefer's book. The separability of $\mathcal{D}(X)$ is obvious.