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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).

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  • $\begingroup$ 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., $\psi u \in \mathscr{S}'$"? $\endgroup$
    – md2perpe
    Commented Jun 29, 2020 at 21:15
  • $\begingroup$ Yes, thanks. I've missed a '. $\endgroup$ Commented Jun 29, 2020 at 21:37
  • $\begingroup$ But also at the end, I think that it should be "i.e. $\psi u \in \mathscr{S}'$." $\endgroup$
    – md2perpe
    Commented Jun 29, 2020 at 21:41

1 Answer 1

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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.

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    $\begingroup$ 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. $\endgroup$ Commented Jun 30, 2020 at 18:38

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