Schwarz function if $|f(x)|\leq\frac{C}{(1+|x|^2)^N}$ for all $x$?

Question

Knowing that $f:\mathbb{R}^n\longrightarrow\mathbb{C}$ is a measurable function such that for all $N\in\mathbb{N}$, there exists a constant $C=C_N$ such that $|f(x)|\leq\frac{C}{(1+|x|^2)^N}$ for all $x\in\mathbb{R}^n$, it is possible to prove that $f$ is a Schwartz function, ie. $f\in\mathcal{S}(\mathbb{R}^n)$? I know that the condition for being a Schwartz function is almost exactly the one given above, just with the derivatives $D^{\alpha}f(x)$ instead of $f(x)$ on the left hand side. How can I deduce the inequality with the derivatives from the one with the original function? Thanks for your help!

Answer 1

There is no way to estimate derivatives. Here is an example.

Let $f = e^{-x^2} sin(e^{x^2})$ (function of only one variable). This function satisfies inequalities $|f(x)|\leq\frac{C}{(1+|x|^2)^N}$ for all $N$ with some constants $C$ (that depend on $N$). Now take the derivative $$ \frac{\partial f}{\partial x} = - 2 x e^{-x^2} sin(e^{x^2}) + 2 x e^{x^2}e^{-x^2}cos(e^{x^2}) = - 2 x e^{-x^2} sin(e^{x^2}) + 2x cos(e^{x^2})$$ So, derivative of $f$ isn't even bouded (because of second summand). Therefore $f$ does not belong to Schwartz space.