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Statement: $\sum_{n\in \mathbb Z} \frac {|\hat f(n)|}{|n|} < \infty$ for all $f \in L^1(\Bbb{T})$

Now I can prove this is true for all $f \in L^p(\Bbb{T})$ where p $\geq$ 2.

That is because $$\sum_{n=1}^\infty \frac{|\widehat{f}(n)|}{n} \le \Big( \sum_{k=1}^\infty \frac{1}{k^2}\Big)^{1/2} \Big( \sum_{n=1}^\infty |\widehat{f}(n)|^2 \Big)^{1/2} \le \frac{\pi}{\sqrt{6}} \Big( \int_0^1 |f(x)|^2 \, \mathrm{d} x \Big)^{1/2}$$ using Parseval. Thus for $f \in L^2(\Bbb{T})$ the statement holds, and since for finite measure spaces $L^q \subseteq L^p$ for q $\geq$ p, thus the statement holds for all $f \in L^p(\Bbb{T})$ where p $\geq$ 2.

However, my guess is that this is not true for all $f \in L^1(\Bbb{T})$, that there are $f \in L^1(\Bbb{T})$\ $L^2(\Bbb{T})$ such that $\sum_{n\in \mathbb Z} \frac {|\hat f(n)|}{|n|} = \infty$. This thought comes from that fact that there exists examples like Kolmogorov where it diverges everywhere. What would be a simple counter example for the statement?

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$f(x) = \frac{1}{|x|\log^2 |x|}$ for $|x|\leq\frac12$ and exetend $1$-periodically everywhere else. $$ \sum_{n\ge 1} \frac{\hat{f}(n)}n = -\lim_{r\to 1}\int_{-1/2}^{1/2} f(x)\log(1-r e^{-2i\pi x})dx=\infty$$

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  • $\begingroup$ Is $f\in L_1(\mathbb{S}^1)$?. $\endgroup$ – Oliver Diaz Jun 22 '20 at 21:10
  • $\begingroup$ Ah I missed you said $x=1$, tks $\endgroup$ – reuns Jun 22 '20 at 21:47

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