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Let $S$ be the Schwartz space and $S'$ be its dual space. If a linear operator $T:S'\rightarrow S'$ is weakly* sequentially continuous, is it also weakly* continuous?

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Yes. Since boundedness is determined by sequences, $T$ maps bounded (= equicontinuous) sets of $S'$ to bounded sets. Since the Schwartz space $S$ is a Frechet-Schwartz space (i.e., a projective limit of Banach spaces with compact linking maps), either by results of Grothendieck or a theorem of Laurent Schwartz, $S'$ endowed with the strong topology $\beta(S',S)$ (of uniform convergence on bounded subsets of $S$) is bornological and hence $T$ is continuous as a map on $(S',\beta(S',S))$. Then, the transposed $T^t$ is a continuous map on $S''=S$ (Frechet-Schwartz spaces are reflexive) and thus, $T=T^{tt}$ is weak$^*$-continuous on $S$.

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