# If $f$ is absolutely continuous then $\sum|\hat f (n) | < \infty$

Let $f$ be a $2\pi$ periodic function on $[0,2\pi]$. If $f$ is absolutely continuous, is it true that the sum of its Fourier coefficient converges absolutely $$S(t)=\sum_{n=-\infty}^{\infty} | \hat f (n) | < \infty \text{ ?}$$ If so how do we proove it.

My understanding, which I think is wrong, is that the derivative $f'$ is in $L^2$ (how do we justify that?). But then how do I bound $S(t)$?

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## 1 Answer

The statement is not true. A counterexample, due to Zygmund, is $$f(x) = \sum_{n=2}^\infty \frac{1}{n \ln n} \sin nx.$$ It is easy to see that its Fourier coefficients do not converge absolutely. It is not quite so easy to see (but true) that $f$ is absolutely continuous. If you want absolute convergence of the Fourier coefficients, you usually need some additional conditions on the modulus of continuity.

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