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Consider $\mathscr{S}$ the Schwartz space of rapidly decreasing complex smooth functions over $\mathbb{R}^{d}$, equipped with its usual metric topology, and $\mathscr{S}'$ its topological dual (the space of tempered distributions). I have a few questions concerning the reflexivity of the space $\mathscr{S}$ when $\mathscr{S}'$ is equipped with the weak-$*$ topology.

The first question: Does it hold that $(\mathscr{S}')' = \mathscr{S}$ in an algebraic sense?. To be precise, if $L : \mathscr{S}' \to \mathbb{C}$ is a linear functional, continuous with the weak-$*$ topology on $\mathscr{S}'$, is there a unique function $\varphi \in \mathscr{S}$ such that $$L(T) = \langle T , \varphi \rangle, \quad \forall T \in \mathscr{S}'. $$ According to classical litterature in Theory of Distributions (for example, Shcwartz's book Théorie des Distributions), this is true when $\mathscr{S}'$ is equipped with the strong topology, although I have trouble understanding the notion of this topology (for instance, what does it means for a subset of $\mathscr{S}$ to be bounded? Is it just the classical definition on metric spaces?). I would like to know if it also holds for the weak-$*$ topology.

The second question is if the weak topology on $\mathscr{S}$ is equivalent to the metric one. To be precise, if $(\varphi_{n})_{n \in \mathbb{N}}$ is a sequence of functions in $\mathscr{S}$ such that $ \langle T , \varphi_{n} \rangle \to 0 $ for all $T \in \mathscr{S}'$, does it hold that $\varphi_{n} \to 0 $ in the sense of the metric of the Schwartz space?

Thank you very much for your help.

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  • $\begingroup$ For the first question the answer is yes because if the map $L$ is continuous for the weak-$\ast$ topology on the domain, it will be continuous for the strong topology. The second question is interesting. I don't know the answer right off the bat. The best way to approach it, I think, is to use the isomorphism of Schwartz space with the space of sequences with fast decay, see e.g. aip.scitation.org/doi/abs/10.1063/1.1665472 $\endgroup$ – Abdelmalek Abdesselam Aug 9 '18 at 15:27

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