Proving a Mapping Property Between Tempered Distributions and Schwartz Functions

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?

• Asking whether $\psi:\mathcal{S}'(\mathbb{R}^n)\rightarrow \mathcal{S}(\mathbb{R}^n)$ doesn't make much sense, because in fact $\psi:\Bbb C\to\Bbb C$. Maybe what you actually had in mind was $\psi(u)(x)=\langle u,\varphi(x,\cdot)\rangle$? Or, with the notation you used, the obvious question seems to be whether $\psi\in\mathcal S$ Commented Aug 2, 2019 at 14:09
• @DavidC.Ullrich Yes, I meant to write $\psi (u)(x),$ my mistake! Commented Aug 2, 2019 at 19:55
• Now I'm not, did you show that continuous on the Schwartz space implies $|<u,\phi>| \le C_u \sum_{|r|,|s| \le N_u} \|D_y^r y^s \phi\|_\infty$, the argument is quite the same as in mathoverflow.net/a/348398/84768 Commented Dec 23, 2019 at 20:29
• @reuns Do you mean have I seen the definition of continuity for elements of $\mathcal{S}'$ in that manner. If so, I have seen that argument before awhile ago. I think what was tripping me up on your answer was that I wasn't sure of the definition of a continuous function between two locally convex topological vector spaces in terms of seminorms. I looked it up yesterday and saw that was what your answer was showing. Looking back at that argument, I see exactly what you mean. Thanks! Commented Dec 23, 2019 at 20:33

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

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

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

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

• I don't think this is correct. I haven't checked but I also think the map is not continuous if $S'$ is given its usual (wrong) topology, i.e., the weak-star one. If $S'$ has the strong topology then everything works fine. Not only $\psi$ is continuous but the map $\phi\mapsto\psi$ is a TVS isomorphism (all topologies being strong,i.e., given by uniform convergence on bounded sets). Commented Dec 30, 2019 at 13:33
• Also, this result is just a part of Fubini's Theorem for temperate distributions. The best way to prove it is to rewrite it in terms of sequence spaces, thanks to Hermite functions. Commented Dec 30, 2019 at 13:35
• @AbdelmalekAbdesselam I didn't notice, but I think you're right. The RHS of the inequality would need to be in terms of the seminorms coming from the weak* topology, I think. Also, the result should use the weak* topology on $\mathcal{S}',$ not the strong one. For reference, I was trying to fill in the details of the proof of theorem 4.1 on page 40 of math.berkeley.edu/~evans/semiclassical.pdf Commented Jan 3, 2020 at 20:41
• Why do you care, continuous for the strong topology implies continuous for the weak topology. @user269711 Commented Jan 3, 2020 at 20:49
• @reuns: Could you state a bit more completely the thm about strong continuity implies weak continuity. For what maps and spaces? Commented Jan 6, 2020 at 17:57