2
$\begingroup$

I know if $f\in L^2(\mathbb R)$ is two times continuously differentiable, then we must have that the Fourier transform is integrable. Is there any more relaxed condition than this? For example if $f$ is continuously differentiable and $f^\prime $ is of bounded variation, will it imply that $f^\prime$ is integrable. Assume that all the functions are compactly supported.

$\endgroup$
2
  • 2
    $\begingroup$ No, for example the FT of step function has decay $1/x$. (I assume that a hat is missing and you're asking if a BV function has integrable FT.) $\endgroup$ Commented Jun 27, 2020 at 20:20
  • $\begingroup$ When f is of bounded variation the Fourier sine transform F^s(f) and the Fourier cosine transform F^c(f) -which form the Fourier transform of a function f := F^f(f)= F^c(f) -i F^s(f) have different behavior. Respectively, F^s(f) is not (HK-)integrable while F^c(f) is. $\endgroup$ Commented Jan 11 at 19:30

1 Answer 1

1
$\begingroup$

If $f'$ is of bounded variation, then $\hat{f}$ will be integrable. To see this, note that by your assumptions, $f''$ (in the sense of tempered distributions) is a finite complex-valued measure, so that $\widehat{f''}$ is bounded, meaning that $|\hat{f}(\xi)|\lesssim 1/(1+|\xi|^2)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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