1
$\begingroup$

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.)

$\endgroup$
1
$\begingroup$

$f$ 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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – danielle Oct 9 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 at 14:36
  • 1
    $\begingroup$ Oh, exactly! A continuous function on a closed interval must attain a maximum value somewhere. Thank you! $\endgroup$ – danielle Oct 9 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.