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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 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:

  1. Is $\mathcal{L}^{*}$ also continuous when $\mathscr{S}'$ is equipped with the strong topology?

  2. 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.

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  • $\begingroup$ I think that you have missed a prime sign at the end of $\mathcal{L}^{*} : \mathscr{S}' \to \mathscr{S}$. $\endgroup$
    – md2perpe
    Aug 7, 2018 at 21:30
  • $\begingroup$ What is the strong topology on $\mathscr S'$? $\endgroup$
    – md2perpe
    Aug 7, 2018 at 21:32
  • $\begingroup$ Corrected, thanks. The strong topology is a notion I actually have some trouble understanding its implications. It is the topology induced by the family of seminorms definied by $p_{B}(T) = \sup_{\varphi \in B}|\langle T , \varphi \rangle | $, whit $B$ describing the bounded sets of $\mathscr{S}$. Thus, a net of tempered distributions $(T_{n})_{n}$ is convergent to $0$ in the strong topology if it converges to $0$ uniformly over bounded sets of $\mathscr{S}$. $\endgroup$
    – CarrizoV
    Aug 7, 2018 at 22:15
  • $\begingroup$ 1 is definitely true but I don't know about 2, I would suspect not. The right topology to use for $\mathscr{S}'$ is the strong topology. The weak-$\ast$ is quite pathological. $\endgroup$ Aug 8, 2018 at 14:05

1 Answer 1

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  1. Yes: $$\sup_{\|T\|\leq1}{\|\mathcal{L}^*T\|}=\sup_{\|T\|\leq1,\|\phi\|\leq1}{(\mathcal{L}^*T)(\phi)}=\sup_{\|T\|\leq1,\|\phi\|\leq1}{T(\mathcal{L}\phi)}\leq\|\mathcal{L}\|$$
  2. Also yes; see this MSE question or Exercise 3.10 in Brezis, Functional Analysis, Sobolev Spaces, and PDE (2011).
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