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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.

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    $\begingroup$ 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. $\endgroup$ Feb 16, 2022 at 18:59

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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. Studia Math. 77 (1984): 333-347

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The answer is yes. For a brief review of the literature on this, see the proof of Lemma 5 Sec 3.1 of my article "A Second-Quantized Kolmogorov–Chentsov Theorem via the Operator Product Expansion".

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(I do intend to read @AbdelmalekAbdesselam's literature review on this subject!)

This sort of assertion has been known since A. Weil, at latest, and some forms are not very complicated. So, for example, yes, every Schwartz function is the product of two such. A stand-alone proof is given in my little essay https://www-users.cse.umn.edu/~garrett/m/fun/weil_schwartz_envelope.pdf

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