I'm given a statement:
For a function $f$ in the Schwartz Space $S(\mathbb{R})$, the Fourier transform $\hat{f}(y) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i xy} dx$ is absolutely convergent.
and the explanation:
Since $f\in S(\mathbb{R})$, $f(x) = O(|x|^{-k})$, in particular for $|x|\geq 1$. So $\hat{f}(y)$ is an absolutely convergent integral.
How is this argument is sufficient? I understand that for $|x|\geq 1$, the integral would be absolutely convergent by the comparison test with $O(|x|^{-k})$, but how can we rule out the part of integral within $(-1, 1)$?
Thank you.
Also the definition of the Schwartz space is below.
Schwartz Space $S(\mathbb{R})$ is the space of all infinitely differentiable functions $f:\mathbb{R}\to\mathbb{C}$ such that for every $j, k\in\mathbb{Z}^+$, we have $f^{(j)}(x) = O_{k, j} (|x|^{-k})$. ($O$ is the big Oh notation.)
Best Answer
$f$ is continuous on $|x|\leq 1$, so $\displaystyle\int_{|x|\leq 1}|f(x)|dx\leq\max_{|x|\leq 1}|f(x)|v_{n}$, the unit volume of the unit ball.