Integration of a function in Schwartz space

Question

How to prove the following: If $f(x)$ belongs to Schwartz space then integral from $x$ to infinity of $f(x)$ also belongs to Schwartz space?

Answer 1

Let $F(x) = \int_x^\infty f(t)\,dt$. In order to have $F(-\infty)=0$, we must assume $\int_{\mathbb R}f(t)\,dt=0$. (As user8268 remarked.) Under this assumption, $F$ is indeed a test function in the Schwartz space. The smoothness is clear from $F'=-f$. The decay at $+\infty$ follows from $$|F(x)| \le \int_x^\infty C(n)\, t^{-n}\,dt = O(x^{1-n})$$ where $n$ can be arbitrarily large (the first inequality holds because $f$ is a test function). The decay at $-\infty$ follows similarly, by using $F(x) = -\int_{-\infty}^x f(t)\,dt$: $$|F(x)| \le \int_{-\infty}^x C(n)\, |t|^{-n}\,dt = O(|x|^{1-n})$$