# 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!

• I think that this should be accomplishable. Where $f$ is bounded away from zero the square root works. For zeros I would suggest a combination of the function itself and the constant $1$ function. Then you need to glue square root and original function respectively constant $1$ and original function together via a suitable partition of unity or so. However you have to be careful since when the function has to much zeros (like one with compact support) you cannot do it in that way but take the product of the function with itself. – Sebastian Bechtel Mar 9 '17 at 10:33
• Another suggestion would be Fourier transform. Then you would need something like a convolution unit in the Schwartz space, but not sure whether something like this exists. Sadly delta is not a regular tempered distribution :P – Sebastian Bechtel Mar 9 '17 at 10:34
• @SebastianBechtel, one could argue that such a convolution unit would equal the Dirac delta measure (as a tempered distribution), and therefore could not be an $L^{1}$ function. In particular, it could not be in the Schwarz space. – fourierwho Jul 9 '18 at 23:22

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.