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.



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


J. Voigt: [Factorization in some Fréchet algebras of differentiable functions](https://eudml.org/doc/218540). Studia Math. 77 (1984): 333-347


