Consider $\mathscr{S}$ the Schwartz space of rapidly decreasing complex smooth functions over $\mathbb{R}^{d} $, equipped with its usual metric topology, and S′ its topological dual, the space of tempered distributions.
If $\mathcal{L} : \mathscr{S} \to \mathscr{S}$ is a linear and continuous operator, one can define its adjoint $\mathcal{L}^{*} : \mathscr{S}' \to \mathscr{S}'$ as the linear operator such that for every tempered distribution $T \in \mathscr{S}'$, it defines the distribution $ \mathcal{L}^{*}(T) = T \circ \mathcal{L} $, that is, $$ \langle \mathcal{L}^{*}(T) , \varphi \rangle = \langle T , \mathcal{L}(\varphi) \rangle, \quad \forall \varphi \in \mathscr{S}, T \in \mathscr{S}'. \tag{1}\label{1}$$
It is easy to prove that $\mathcal{L}^{*}$ so defined is continuous when $\mathscr{S}'$ is equipped with the weak-$*$ topology. Within this framework, I have two questions:
Is $\mathcal{L}^{*}$ also continuous when $\mathscr{S}'$ is equipped with the strong topology?
If $\mathcal{L}^{*} : \mathscr{S}' \to \mathscr{S}'$ is a linear operator which is continuous when $\mathscr{S}'$ is equipped with the weak-$*$ topology, does it exist a linear operator $\mathcal{L} : \mathscr{S} \to \mathscr{S}$, continuous with respect to the metric topology on $\mathscr{S}$, such that $\eqref{1}$ holds? (that is, does it exist a "pre-adjoint" operator?)
I've tried to work out this questions using the reflexivity property $ (\mathscr{S}')'=\mathscr{S}$, but this property is known to be truth for the strong topology on $\mathscr{S}'$ and not necessarily when using the weak-$*$ topology (I've made another post with this precise question).
Thank you for your help.
