# Multiplying tempered distributions by smooth functions

This is roughly exercise V.23 of Reed, Simon - Methods of Modern Mathematical Physics, vol. 1.

Let $$\psi$$ be a $$\mathcal{C}^\infty$$ function such that for every multi-index $$\alpha$$ there exist $$c_\alpha> 0$$ constant and a $$N_\alpha$$ such that

$$\|\partial^\alpha \psi(x)\| \leq c_\alpha(1 + \|x\|^2)^{N_\alpha}$$

for every $$x \in \mathbb{R}^d$$.

(i) If $$\varphi \in \mathcal{D}(\mathbb{R}^d)$$, then $$\varphi\psi \in \mathscr{S}(\mathbb{R}^d)$$.

This is clear from Leibniz' Rule. What I'm having trouble to show is parts (ii) and (iii):

(ii) If $$h$$ is a measurable function such that $$h\psi \in \mathscr{S}$$ for all $$\psi \in \mathscr{S}$$, then $$h$$ has to be smooth.

(iii) Given $$u \in \mathscr{S}'$$, putting $$(\psi u)(\varphi) = u(\psi\varphi)$$ (as expected) for every $$\varphi\in\mathscr{S}$$, then $$\psi u$$ is a tempered distribution, i.e., $$\psi u \in \mathscr{S}'$$.

I'm actually not sure on how (ii) helps me at all on showing (iii) and how to show (ii).

• Shouldn't (iii) be "Given $u \in \mathscr{S}'$, putting $(\psi u)(\varphi) = u(\psi\varphi)$ (as expected) for every $\varphi\in\mathscr{S}$, then $\psi u$ is a tempered distribution, i.e., $\psi u \in \mathscr{S}'$"? Jun 29, 2020 at 21:15
• Yes, thanks. I've missed a '. Jun 29, 2020 at 21:37
• But also at the end, I think that it should be "i.e. $\psi u \in \mathscr{S}'$." Jun 29, 2020 at 21:41

Are you sure about the statement (i)? As stated, it is trivial because for $$\varphi\in\mathscr D$$ and smooth $$\psi$$ one has $$\varphi\psi\in\mathscr D \subseteq \mathscr S$$. Perhaps you mean $$\varphi\psi\in\mathscr S$$ for all $$\varphi\in \mathscr S$$? Then the map $$\varphi\mapsto \varphi\psi$$ is continuous on $$\mathscr S$$ by the closed graph theorem and $$\psi u$$ is continuous as a composition. This solves (3).
I don't see, why (ii) should be useful. For the proof of (ii) just consider test functions $$\psi\in\mathscr D$$ which are constant on a neighbourhood of the point where you want to show smoothness.
• Yeah, it has to be. So, the argument should go as: $T_\psi(\varphi) = \psi \varphi$ for $\varphi \in \mathscr{S}$ has a closed graph, essentially because of the Leibniz Rule, the notion of convergence on the seminorms $\|\varphi\|_{\alpha,\beta} = \sup_x \|x^\beta \partial^\alpha \varphi(x)\|$ in $\mathscr{S}$ and the fact that $\psi$ has polynomial growth. Then by the Closed Graph Thm it follows easily that $T_\psi$ is continuous. Thanks. It turns out this statement has a lot of mistakes and that was the source of confusion. Jun 30, 2020 at 18:38