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.
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asked Jun 27, 2020 at 20:15
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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.) $\endgroup$– Christian RemlingCommented Jun 27, 2020 at 20:20
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$\begingroup$ 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. $\endgroup$– Juan H. ArredondoCommented Jan 11 at 19:30
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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 $|\hat{f}(\xi)|\lesssim 1/(1+|\xi|^2)$.
