Multi-variable Schwartz functions can be controlled by one variable Schwartz functions I met the following concrete problem in my study, which I have no idea to write down a precise proof. So let $f(x_1,...,x_n)$ be a Schwartz function defined on $\mathbb{R}^n$. I want to show that there exsits Schwartz functions $f_1,...,f_2$ on $\mathbb{R}$ such that for any $x=(x_1,...,x_n)\in \mathbb{R}^n$,
$$f(x_1,...,x_n)<f_1(x_1)f_2(x_2)...f_n(x_n).$$
So it suffices to deal with the case $n=2$ and we may assume $f$ is non-negative. Then my natural idea is: for $f(x_1,x_2)$, I fix $x_1$ and let $x_2$ run through $\mathbb{R}$, and I take supremum:
$$f_1(x_1):=\sup_{x_2\in \mathbb{R}}~ (f(x_1,x_2))^{1/2}.$$
OK, so this is quite natural. But then I have to prove that such construction is Schwartz: smooth and has its required estimation. Now I have no idea how to overcome the technical difficulty here.
Or is there any pure existence proof for the problem? Any hint or suggestion would be welcome. Thanks in advance!
 A: Not sure it's the fastest way, but here is a solution following the famous motto: Be wise, discretize!.
If this was about discrete Schwartz functions $a:\mathbb{Z}^n\rightarrow \mathbb{R}$ of rapid decay,
then the OP's idea works totally fine.
Now take a 1-dimensional smooth partition of unity
$$
1=\sum_{\ell\in\mathbb{Z}}\phi(x-\ell)
$$
for all $x\in\mathbb{R}$. Here $\phi$ is a bump function, say with values in $[0,1]$, infinitely differentiable and with compact support, e.g., inside $[-R,R]$.
From this one can get an $n$-dimensional partition of unity by tensorization
$$
1=\sum_{k\in\mathbb{Z}^n}\psi(x-k)
$$
for all $x=(x_1,\ldots,x_n)\in\mathbb{R}^n$, where we defined
$$
\psi(x)=\phi(x_1)\cdots\phi(x_n)\ .
$$
Inserting the partition of unity, we get
$$
f(x)=f(x)\times 1=\sum_{k\in\mathbb{Z}^n}f(x)\psi(x-k)
\le \sum_{k\in\mathbb{Z}^n}a(k)\psi(x-k)
$$
where
$$
a(k)=\sup\limits_{x\in k+[-R,R]^n}|f(x)|\ . 
$$
Using the OP's trick, it is easy to see there exists $b_i:\mathbb{Z}\rightarrow\mathbb[0,\infty)$ of rapid decay, such that
$$
a(k)\le b_1(k_1)\cdots b_n(k_n)\ .
$$
Thus
$$
f(x)\le \sum_{k\in\mathbb{Z}^n}b_1(k_1)\cdots b_n(k_n)\psi(x-k)
=f_1(x_1)\cdots f_n(x_n)
$$
where for all $i$,
$$
f_i(x_i)=\sum_{\ell\in\mathbb{Z}}b_i(\ell)\phi(x_i-\ell)
$$
which is a Schwartz function.
If having $<$ instead of $\le$ is essential, then add $e^{-x_i^2}$ to $f_i(x_i)$.
A: Let $$h_j(t) = \sup_{x\in \Bbb{R}^n, |x_j|\in (|t|-1,|t|+1)} |f(x)|^{1/n}$$ Take $\phi\in C^\infty_c(-1,1),\phi\ge 0,\int \phi=1$.
$h_j$ is rapidly decreasing so $$f_j =e^{-t^2}+2 h_j \ast \phi \in S(\Bbb{R})$$
Finally
$$|f(x)|= \prod_{j=1}^n |f(x)|^{1/n}< \prod_{j=1}^n f_j(x_j)$$
