Is $\langle T, f\rangle$ a continuous function on $S'(R)\times S(R)$?

Question

This is probably wrong, but I cannot find a counterexample.

If $T_n\rightarrow T$ (in weak topology) and $\phi_n \rightarrow \phi$ (in Schwartz space), is it true that $\langle T_n, \phi_n\rangle\rightarrow \langle T, \phi\rangle$?

Answer 1

This is true. The proof requires Banach-Steinhaus theorem.

By B-S theorem we know that if a set $B \subset \mathscr{S}'$ is weakly (i.e. pointwise) bounded then for all $\varepsilon > 0$ there exists a neighbourhood of zero $U \subset \mathscr{S}$ s.t. $|\langle T, \tau \rangle | < \varepsilon$ for all $T \in B$ and $\tau \in U$.

Now the proof is quite elementary to finish. Let $\phi_n \rightarrow \phi$ and $T_n \rightarrow T$. Then $T_n$ is a weakly bounded sequence and therefore for all $\varepsilon > 0$ there exists a neighbourhood of zero $U \subset \mathscr{S}$ s.t. $|\langle T_n, \tau \rangle | < \varepsilon$ for all $\tau \in U$. But $\phi_n - \phi$ converges to $0$ and therefore $|\langle T_n, \phi_n - \phi \rangle | < \varepsilon$ for big enough $n$. So, $\langle T_n, \phi_n \rangle - \langle T, \phi \rangle \rightarrow 0$.

Answer 2

Consider the evaluation map ${\rm eval}:\mathcal{S}'(\mathbb{R})\times \mathcal{S}(\mathbb{R})\rightarrow \mathbb{R}$, $(T,f)\mapsto \langle T,f\rangle=T(f)$ which is obviously bilinear. The correct topology to use on $\mathcal{S}'(\mathbb{R})$ is the strong topology (not the weak-star). It follows that $\mathcal{S}'(\mathbb{R})\times \mathcal{S}(\mathbb{R})$ should be given the product topology of the strong topology on $\mathcal{S}'(\mathbb{R})$ and the standard Fréchet topology on $\mathcal{S}(\mathbb{R})$. One can then ask if ${\rm eval}$ is a continuous map, in the usual sense of basic point set topology. the answer is: NO. The map is only hypocontinuous. This does not contradict the other answers which only look at the weaker notion of (joint) sequential continuity. The reason for this is that convergent sequences form bounded sets.