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.)


1 Answer 1


$f$ 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.

  • $\begingroup$ 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. $\endgroup$
    – user594480
    Oct 9, 2019 at 14:31
  • 2
    $\begingroup$ You see, when you say Schwartz function, the definition already includes the property being infinitely differentiable, and hence in particular that it must be continuous. $\endgroup$
    – user284331
    Oct 9, 2019 at 14:36
  • 1
    $\begingroup$ Oh, exactly! A continuous function on a closed interval must attain a maximum value somewhere. Thank you! $\endgroup$
    – user594480
    Oct 9, 2019 at 14:37

You must log in to answer this question.