Given a Schwartz function, is it always possible to write it as a product of two Schwartz function?

Fix any $f\in\mathcal{S}_x(\mathbb{R}^d\to\mathbb{C})$, i.e. $f$ is a Schwartz function from $\mathbb{R}^d$ to $\mathbb{C}$. Is it always possible to find $g,h\in\mathcal{S}_x(\mathbb{R}^d\to\mathbb{C})$ such that $f(x)=g(x)h(x)$ for all $x\in\mathbb{R}^d$?

The square root seems a good choice, but I find this problem. So I am not sure what I am supposed to do to solve or disprove this problem.

Thank you!

The answer is yes.

It is Lemma 1 from the article by K. Miyazaki, Distinguished elements in a space of distributions. J. Sci. Hiroshima Univ. Ser. A 24 (1960), 527–533.

This also done in H. Petzeltová and P. Vrbová, Factorization in the algebra of rapidly decreasing functions on $R_n$. Comment. Math. Univ. Carolin. 19 (1978), no. 3, 489–499 as well as in J. Voigt, Factorization in some Fréchet algebras of differentiable functions. Studia Math. 77 (1984), no. 4, 333–348.

Finally, there is a pedagogical account in the note "Weil-Schwartz envelopes for rapidly decreasing functions" by Paul Garrett.