Denote by $H^s(\mathbb{R}^n)$ the Sobolev space on $\mathbb{R}^n$ of order $s \in \mathbb{R}$ and recall that we have $H^s(\mathbb{R}^n)^\ast \cong H^{-s}(\mathbb{R}^n)$ for the dual space of $H^s(\mathbb{R}^n)$. Furthermore, let $$H^\infty(\mathbb{R}^n) := \bigcap_{s \in \mathbb{N}} H^s(\mathbb{R}^n)$$ equipped with the natural Frechet topology, and let $$H^{-\infty}(R^n) := \bigcup_{s \in \mathbb{N}} H^{-s}(\mathbb{R}^n)$$ equipped with the weak-$^\ast$-topology that it gets as the dual of $H^\infty(\mathbb{R}^n)$.
A continuous operator $T\colon H^{-\infty}(\mathbb{R}^n) \to H^\infty(\mathbb{R}^n)$ is called a smoothing operator. Since both $H^{-\infty}(\mathbb{R}^n)$ and $H^\infty(\mathbb{R}^n)$ are locally convex spaces, this means that for all $s \in \mathbb{N}$ there are finitely many functions $f_1, \ldots, f_{N_s} \in H^\infty(\mathbb{R}^n)$ and an $M_s > 0$ such that $$\|T \varphi\|_{H^s} \le M_s \cdot \max_{1 \le j \le N_s} \varphi(f_j)$$ for all $\varphi \in H^{-\infty}(\mathbb{R}^n)$.
From this we can conclude that $T$ is a bounded operator $H^{-r}(\mathbb{R}^n) \to H^s(\mathbb{R}^n)$ for all $r,s \in \mathbb{N}$.
Does the converse hold, i.e., if $T\colon H^{-r}(\mathbb{R}^n) \to H^s(\mathbb{R}^n)$ is bounded for all $r,s \in \mathbb{N}$, does $T$ define a continuous operator $H^{-\infty}(\mathbb{R}^n) \to H^\infty(\mathbb{R}^n)$?