# Example of a Schwartz function

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?

• Let $\psi$ have compact support, and $\phi$ have disjoint support to $\psi$ but also highly irregular. – Lord Shark the Unknown Aug 9 at 15:23
• If you take $\phi \equiv 1$ then $\psi\phi\in \mathcal{S}(\mathbb{R}^n)$ although $\phi \not\in \mathcal{S}(\mathbb{R}^n)$. – 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.
• 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 $\psi f \text { is not Schwartz }$? – James Aug 10 at 5:59
• +1 but I will just add that the space of such functions $f$ is an important (multiplier) space studied by Schwartz usually denoted by $\mathcal{O}_M$. – Abdelmalek Abdesselam Aug 20 at 17:53