A rapidly decreasing function is a measurable function $f:\mathbb{R}^d:\to\mathbb{C}$ such that $|x|^n f(x)$ is bounded for every non-negative integer $n$. A Schwartz function is a smooth function $f:\mathbb{R}^d\to\mathbb{C}$ such that all derivatives $\partial_{x_1}^{n_1}\cdots \partial_{x_d}^{n_d}f$ are rapidly decreasing.
I need to find a function that is smooth rapidly decreasing but not Schwartz, I can only think of the exponential function $e^{-|x|^2}$ which is Schwartz.
Best Answer
I believe the function $$f(x)=e^{-x}\sin(e^{x})$$ should work well because this is definitely rapidly decreasing, but its first derivative is $$f'(x)=-e^{-x}\sin(e^{x})+\cos(e^{x})$$ which is not decreasing to $0$ as $x\to\infty$ at all.