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?
Best Answer
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.