Multi-variable Schwartz functions can be controlled by one variable Schwartz functions

Question

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!

Answer 1

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 $f_i(x_i)$.

Answer 2

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 $$|f(x)|= \prod_{j=1}^n |f(x)|^{1/n}< \prod_{j=1}^n f_j(x_j)$$