Proving a Mapping Property Between Tempered Distributions and Schwartz Functions

Question

Say I have a Schwartz function $\varphi\in\mathcal{S}(\mathbb{R}^{2n})$. Consider a map $\psi(u)(x)=\langle u,\varphi(x,\cdot)\rangle$ for $u\in\mathcal{S}'(\mathbb{R}^{n}).$ I'd like to know how to show (1) that if $\psi:\mathcal{S}'(\mathbb{R}^n)\rightarrow \mathcal{S}(\mathbb{R}^n)$ and (2) that the map $\psi$ is continuous.

For (1), I initially wrote $$|x^\alpha D^\beta \psi (u)(x)|=|\langle x^\alpha D^\beta u,\varphi(x,\cdot)\rangle|=|\langle u,x^\alpha D^\beta\varphi(x,\cdot)\rangle|$$ for any $x$, but I'm not sure if moving the $x^\alpha D^\beta$ over actually makes sense here. Typically, we do this when $x$ is the variable that the action is with respect to, which is not the case here. If this were allowed, then the fact that $\varphi$ is Schwartz would guarantee that $x^\alpha D^\beta \varphi(x,\cdot)$ is Schwartz, and this would be finite for any $x$ since $u$ is tempered. So, it comes down to simply understanding how the $x^\alpha D^\beta$ action is defined.

For (2), I tried to use the sequential criterion, but I could not show that if $u_n\rightarrow u$ in $\mathcal{S}'$ (i.e. weak$^*$/pointwise), then $\psi (u_n)\rightarrow \psi (u)$ in $\mathcal{S}$ (i.e. in every Schwartz seminorm). I think it would come from the first part, as if I could move over the $x^\alpha D^\beta,$ then I would have $$x^\alpha D^\beta(\psi(u_n)-\psi(u))=\langle (u_n-u), x^\alpha D^\beta \varphi(x,\cdot)\rangle\rightarrow 0$$ since $x^\alpha D^\beta \varphi(x,\cdot)$ is Schwartz and $\langle u_n, f\rangle \rightarrow \langle u, f\rangle$ for any $f\in\mathcal{S}.$

Alternatively, since $\mathcal{S}$ is dense in $\mathcal{S}'$ in the weak$^*$ topology, might it suffice to check these claims for $u\in\mathcal{S}$ and extend by density? It's not clear to me how that extension process would work.

EDIT: Perhaps, I can make sense of the derivative claim by showing that difference quotients converge? Also, I believe that moving that $x^\alpha$ from one side to the other is justified since, for a fixed $x$, this is as good as a constant as far as $u$ is concerned, so I can use linearity. Is that correct?

Answer 1

For $\varphi \in S(\mathbb{R}^n \times \mathbb{R}^n), u \in S'(\mathbb{R}^n)$ let $$T[u](x) = <u,\varphi(x,.)> \ \in R, \qquad (D_x^a x^b) T[u](x_0)=<u,(D_x^a x^b \varphi)(x_0,.)>$$

Since $u$ is a tempered distribution you know that for some $N_u,C_u$, for all $\phi \in S(\mathbb{R}^n)$ $$|<u,\phi>| \le C_u \sum_{|r|,|s| \le N_u} \|D_y^r y^s \phi\|_\infty$$ whence

$$\sup_{x_0} |(D_x^a x^b) T[u](x_0)|=\sup_{x_0} |<u,(D_x^a x^b \varphi)(x_0,.)>|\\ \le \sup_{x_0} C_u\sum_{|r|,|s| \le N_u} \|D_y^r y^s (D_x^a x^b \varphi)(x_0,.)\|_\infty \le C_u \sum_{|r|,|s| \le N_u} \|D_y^r y^s D_x^a x^b \varphi \|_\infty = C_u \kappa(N_u,a,b)$$

Which is quite the definition of that $u \mapsto T[u]$ is a continuous linear map $S'(\mathbb{R}^n) \to S(\mathbb{R}^n)$.