# Fourier transform of a function of bounded variation

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.

• 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.) Commented Jun 27, 2020 at 20:20
• 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. Commented Jan 11 at 19:30

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 $$|\hat{f}(\xi)|\lesssim 1/(1+|\xi|^2)$$.