# Why is the Fourier Transform of a function in the Schwartz Space absolutely convergent?

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.)

$$f$$ 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.
• Is it possible that $f$ is unbounded on $[-1, 1]$? That $f(x) = O(|x|^{-k})$ isn't particularly helpful when $|x|\leq 1$ as $|x|^{-k}$ gets arbitrarily large when $|x|$ is close to 0.