Product of a Schwartz function and a function with polynomial bonuded derivatives is schwartz? Is true the following?: If $f\in\mathcal{S}(\mathbb{R}^n)$ and $g\in\mathcal{C}^{\infty}(\mathbb{R})$ with $|g^{k}(|x|)|\leq p_k(x)$ all $k$ where $p_k(x)$ is a polynomial then $f(x)g(|x|)$ is in $\mathcal{S}(\mathbb{R}^n)?$
 A: Yes, as long as $g(|x|)\in C^\infty(\mathbb{R}^n).$
We only need to varify the condition
$$
\sup_{x\in \mathbb{R}^n} |x^\alpha\partial^\beta(f(x)g(|x|))|<\infty,\quad \forall \alpha,\beta\in \mathbb{N}^n.
$$
We have $$|\partial^\alpha (f(x)g(|x|))|\le \sum_{\beta+\gamma=\alpha} |\partial^\beta f(x) \partial^\gamma g(|x|)|$$
Notice that $|\partial^\alpha |x||\le C_{\alpha}$ when $|x|\ge 1$, see here. So $$|\partial^\alpha g(|x|)|\le \sum_{\beta\le \alpha} C_\beta' |g^{(|\beta|)}(|x|)|\le \widetilde{C_\alpha}\sum_{\beta\le\alpha}|p_{|\beta|}(|x|)|\le \widetilde{C_\alpha}|P_\alpha(x)|$$
At last
$$
\begin{aligned}
|x^\alpha \partial^\beta(f(x)g(|x|))|&\le |x^\alpha|\sum_{\gamma\le \beta} |\partial^\gamma f(x)| \widetilde{C_{\gamma}}|P_{\gamma}(x)|\le \sum_{\gamma\le \beta} \widetilde{C_\gamma} |\widetilde{P_\gamma}(x)| |\partial^\gamma f(x)| 
\end{aligned}
$$
By $f(x)\in \mathcal{S}(\mathbb{R}^n)$, we have
$$
\sup_{x\in \mathbb{R}^n \setminus B_1} |x^\alpha\partial^\beta(f(x)g(|x|))|<\infty,\quad \forall \alpha,\beta\in \mathbb{N}^n.
$$
If $g(|x|)$ is truely smooth, so does $f(x)g(|x|)$. Then $\sup_{x\in B_1} |x^\alpha\partial^\beta(f(x)g(|x|))|<\infty$ of course. And finally we have $f(x)g(|x|)\in \mathcal{S}(\mathbb{R}^n).$ But $g(|x|)$ won't be smooth on $\mathbb{R}^n$ easily for arbitrary $g(x)\in C^\infty(\mathbb{R})$. A necessary condition is $g^{(k)}(0)=0,$ $\forall k\ge 1.$ I'm not sure whether it's sufficient.
As $\mathcal{S}(\mathbb{R}^n)\subset C^\infty(\mathbb{R}^n),$ it should be varified at first, actually. It depends on $g$.
