Scaling in Schwartz space

Question

Suppose that $f\in \mathcal{S}(\mathbb R^n)$=(Schwartz space).

Can we find some constat $M>0$ such that $|\frac{1}{R} |x|^2 \nabla f(x/R)| \leq M$ for all $x\in \mathbb R^n$ and $M$ is independent of $R>0$?

Answer 1

No. Take $y\in\mathbb{R}^n$ such that $y\ne0$ and $\nabla f(y)\ne0$, and let $x=R\,y$. If such an inequality were possible we would have $R\,|y|^2|\nabla f(y)|\le M$ for all $R>0$.