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 $O(\xi^{1})$?
1 Answer
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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 $\mu(\mathbb R)=0$.

$\begingroup$ See also mathoverflow.net/questions/98405/… . $\endgroup$– jjcaleApr 9, 2017 at 5:10