Let $f$ be $1$-periodic and continuous. Suppose that $\hat{f}(n) = -\hat{f}(-n)\geq 0$ holds for all $n \in \mathbb{Z}$, where the $\hat{f}(n)$ denotes the $nth$ Fourier coefficient of $f$. Prove that
$$\sum^{\infty}_{n=1} \frac{\hat{f} (n)}{n} < \infty.$$
I am trying to do a proof by contradiction as I was not able to see a direct approach. Therefore, suppose that the series does not converge and hence that given any $R>0$, the is some $N \in \mathbb{N}$, such that
$$\sum^{N}_{n=1} \frac{\hat{f} (n)}{n} > R.$$
By assumption this would mean that
$$\sum^{N}_{n=-N} |\hat{f}(n)| \geq \sum^{N}_{n=-N} \left| \frac{\hat{f} (n)}{n} \right |> R.$$
From here I think I could obtain a contradiction to Parseval's Theorem, which is clear if the $\hat{f}(n) \geq 1$, but not clear to me otherwise.
Could you provide some feedback on the above approach, especially whether it goes in the right direction, and if not could you please give a hint as to how to properly approach the problem? Thank you in advance.
Best Answer
Why we don't use just the Cauchy-Schwarz inequality and Parseval's Theorem as follows? $$\sum_{n=1}^\infty \frac{|\widehat{f}(n)|}{n} \le \Big( \sum_{k=1}^\infty \frac{1}{k^2}\Big)^{1/2} \Big( \sum_{n=1}^\infty |\widehat{f}(n)|^2 \Big)^{1/2} \le \frac{\pi}{\sqrt{6}} \Big( \int_0^1 |f(x)|^2 \, \mathrm{d} x \Big)^{1/2} $$ We don't even need that $\widehat{f}(n) = - \widehat{f}(-n) \ge 0$: The series is absolute convergent.