Weak Topology on Spaces of Test Functions: A Simple Question With the usual notation of distribution theory, let $(\phi_j)_{j=1}^{\infty}$ be a sequence in $\mathscr{D}(\mathbb{R}^n)$ such that for every $T \in \mathscr{D'}(\mathbb{R}^n)$, we have $T(\phi_j) \rightarrow 0$ (which is to say that $\phi_j \rightarrow 0$ in the weak topology of $\mathscr{D}(\mathbb{R}^n)$). Does we have $\phi_j \rightarrow 0$ in the original topology of $\mathscr{D}(\mathbb{R}^n)$?
Analogously, let $(\phi_j)_{j=1}^{\infty}$ be a sequence in $\mathscr{S}(\mathbb{R}^n)$ such that for every $T \in \mathscr{S'}(\mathbb{R}^n)$, we have $T(\phi_j) \rightarrow 0$ (which is to say that $\phi_j \rightarrow 0$ in the weak topology of $\mathscr{S}(\mathbb{R}^n)$). Does we have $\phi_j \rightarrow 0$ in the original topology of $\mathscr{S}(\mathbb{R}^n)$?
I think the answers to these two questions are negative, but I could not find any counter-example.
Thank you very much for your attention.

Theorem: Let $E$ be an locally convex space, where every bounded set is relative compact (a so-called Semi-Montel-space). On a bounded subset $B$ of $E$ the weak topology $\sigma(E,E')$ and the original topology coincide.
Proof: Let $C$ be the closure of $B$ (with respect to the original topology). Then $C$ is a compact topological hausdorff space, so its topology is minimal, i.e. does not admit a strictly coarser hausdorff topology. As the weak topology is coarser than the original topology, they must coincide on $B\subseteq C$.
As $\mathcal{D}$ and $\mathcal{S}$ are Montel spaces you may apply the theorem on $B=\{\phi_j; j\in\mathbb{N}\}\cup \{0\}$ for your specific question.
