If $\psi,\phi \in \mathcal{S}(\mathbb{R}^n)$ then I know that the product $\psi\phi \in \mathcal{S}(\mathbb{R}^n)$ is also in the Schwartzspace.

Now I was wondering if $\psi\in \mathcal{S}(\mathbb{R}^n)$ but $\phi \notin \mathcal{S}(\mathbb{R}^n)$ if it is possible for the product $\psi\phi \in \mathcal{S}(\mathbb{R}^n)$ to be in the Schwartz space.

I am not sure if this is true, however I think that the multiplication with a non-smooth $\phi$ will always give a non-smooth function back and thus $\psi\phi \notin \mathcal{S}(\mathbb{R}^n)$ How do I go about showing this?

  • $\begingroup$ Let $\psi$ have compact support, and $\phi$ have disjoint support to $\psi$ but also highly irregular. $\endgroup$ – Lord Shark the Unknown Aug 9 at 15:23
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    $\begingroup$ If you take $\phi \equiv 1$ then $\psi\phi\in \mathcal{S}(\mathbb{R}^n)$ although $\phi \not\in \mathcal{S}(\mathbb{R}^n)$. $\endgroup$ – md2perpe Aug 9 at 16:35

$\forall \psi \in S(\Bbb{R}^n), \psi f \in S(\Bbb{R}^n)$ iff $f$ is smooth and each of its derivative has at most polynomial growth.

The proof is that for $\phi \in C^\infty_c$ and $ h \in C^0$ rapidly decreasing then $\phi \ast h$ is Schwartz, thus if $f$ has a more than polynomial growth we can take $h(x) = (\sup_{|y|< |x|} |f(y)|)^{-1/2}$ which is rapidly decreasing so $\psi = \phi \ast h$ is Schwartz and $\psi f$ isn't rapidly decreasing. For the growth of the derivatives of $f$ it is slightly more complicated, we need to find when a Schwartz function is the $k$-th derivative of a Schwartz function and modify $\phi \ast h$ accordingly (substracting a Schwartz function supported on strips $x_i \in [a,b]$ chosen such that the line integrals vanish).

But there are other interesting cases : if $\psi \in S(\Bbb{R}),\psi(0) = 0$ and $f(x) = \frac1x$ then $\psi f$ is Schwartz,

with $\psi(x) = e^{-x^2}, f(x) = e^x$ then $\psi f$ is Schwartz.

  • $\begingroup$ and therefore say $\phi \in \psi \in S(\mathbb{R})$ and $f(x) = \exp(-x)$ and thus $f(x) \in \psi \in S(\mathbb{R})$ since each of f(x) derivatives has more than polynomial growth then $ \psi f \text { is not Schwartz }$? $\endgroup$ – James Aug 10 at 5:59
  • $\begingroup$ +1 but I will just add that the space of such functions $f$ is an important (multiplier) space studied by Schwartz usually denoted by $\mathcal{O}_M$. $\endgroup$ – Abdelmalek Abdesselam Aug 20 at 17:53

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