alternate way to show function is schwarz

Question

We know that $f (x)$ is in Schwarz space if for given $ m, k $ non negative integers Supremum of $ |x^{m } f^{(k)}(x) | $ is finite over real numbers. Is it true that $f (x)$ is Schwarz if and only if $ |x^{m } f^{(k)}(x) | $ going to 0 as $|x|$ tends to $ \infty $ for each given pair m and k ( non negative integers)

Answer 1

Yes. To see that the original property implies the second, note that if you didn't have $x^m f^{(k)}(x)$ going to $0$ at infinity, then $x^{m+1}f^{(k)}(x)$ would be unbounded. To see that the second property implies the original, note that by continuity $x^{m}f^{(k)}(x)$ is bounded on bounded intervals, and if it goes to $0$ at $\infty$, then you can make it arbitrarily small outside sufficiently large bounded intervals, getting boundedness on the entire real line.