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

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?

## 1 Answer

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

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