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?



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


