0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 3
    $\begingroup$ 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
5
$\begingroup$

$\forall \psi \in S(\Bbb{R}^n), \psi f \in S(\Bbb{R}^n)$ 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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.