# Is every Schwartz function the product of two Schwartz functions?

A Schwartz function on $$\mathbb R^d$$ is a $$C^\infty$$ function, such that all differentials of order $$k \ge 0$$ decay faster than any polynomial. They include the class $$C^\infty_c(\mathbb R^d)$$ of compactly supported, smooth functions.

I would like two know, if for every Schwartz function $$f$$, there are Schwartz functions $$g,h$$ such that $$f(x)=g(x)h(x)$$ for all $$x \in \mathbb R^d$$. If $$f \in C^\infty_c$$, we can choose $$g=f$$ and $$h$$ as a cutoff function, such that $$f(x) \neq 0 \implies h(x)=1$$ to get such a (trivial) representation. For a general Schwartz function, I was unable to find a construction or a counter example.

• I think a similar idea works (but haven't worked out the details): define $g$ piecewise as $g=x^n f$, where you increase $n$ as $x\to\infty$, and of course you need to glue together these pieces smoothly. Feb 16, 2022 at 18:59

Yes, such a decomposition exists. More general given a compact set $$B\subset\mathcal{S}(\mathbb{R}^{n})$$ there is a function $$\varphi\in\mathcal{S}(\mathbb{R}^{n})$$ and a compact set $$C\subset\mathcal{S}(\mathbb{R}^{n})$$ with $$B=\varphi C$$. This property is called the compact strong factorisation property by J. Voigt. Details can be found in the following paper: