# Convergence in $\sigma(\mathcal{E}',\mathcal{E})$ versus $\beta(\mathcal{E}',\mathcal{E})$

Let $$\mathcal{E}'(\mathbb{R})$$ be the space of all compactly supported distributions on $$\mathbb{R}$$. Suppose that $$(T_n)$$ is a sequence in $$\mathcal{E}'(\mathbb{R})$$ that converges to $$T$$ in the weak topology $$\sigma(\mathcal{E}',\mathcal{E})$$ on $$\mathcal{E}'(\mathbb{R})$$. Does this imply that $$(T_n)$$ converges to $$T$$ in the strong dual topology $$\beta(\mathcal{E}',\mathcal{E})$$ too?

The answer is yes. First, since $$\newcommand{E}{\mathcal{E}}\E$$ is a Fréchet space, it is barrelled, and so any $$\sigma(\E',\E)$$-bounded subset of $$\E'$$ is equicontinuous, and therefore bounded in any dual topology. Convergent sequences (including their limits) are compact sets, and therefore bounded. So each $$\sigma(\E',\E)$$-convergent sequence is a closed, $$\beta(\E',\E)$$-bounded set.
To finish the proof, recall that $$\E'$$ with the $$\beta(\E',\E)$$ topology is a complete nuclear space, and that in such spaces closed bounded subsets are compact. It follows that if $$B \subseteq \E'$$ is bounded, the identity mapping $$(B, \beta(\E',\E)) \rightarrow (B,\sigma(\E',\E))$$ is a continuous bijection of compact Hausdorff spaces, and therefore a homeomorphism. By taking $$B$$ to be a $$\sigma(\E',\E)$$-convergent sequence, we see it must also be a $$\beta(\E',\E)$$-convergent sequence.