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If $f$ is a compactly supported bounded variation function on the real line $\mathbb R$, its Fourier transform $\widehat f$ can be estimated as $|\widehat f(\xi)| = O(|\xi|^{-1})$, $|\xi|\to\infty$. This estimate is exact: for a function $f$ which is a sum of a jump and an absolutely continuous function we have $\widehat f(\xi) \sim C |\xi|^{-1}$ at infinity (due to the jump). If the function $f$ is BV and absolutely continuous, we have $|\widehat f(\xi)| = o(|\xi|^{-1})$, $|\xi| \to \infty$. But what if $f$ is continuous and BV but not absolutely continuous? Do we have $o(|\xi|^{-1})$ or the general estimate $O(|\xi|^{-1})$?

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No. There are continuous measures $\mu$ whose Fourier transform does not go to zero (for example, the Cantor measure). We can now take $f(x)=\mu(-\infty, x)$ (so $f'=\mu$ as distributions), and this function is BV and continuous and satisfies $\limsup |\xi \widehat{f}(\xi|=\limsup |\widehat{\mu}(\xi)|>0$.

This $f$ is not automatically compactly supported, but of course you can have that property too, if you take $\mu$ as a finite signed measure with $\mu(\mathbb R)=0$.

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