Can someone help me by checking my solution. Is there a shorter More elegant solution ?(i'm almost sure you can some how express $f$'s Fourier series using $|\hat{g}|^2$ + constant, i saw someone do it in few easy manipulations but was never able to reproduce it myself)
$f,g \in R(T)$ (integrable periodic functions with period $2\pi$) such that $\hat{f} \cdot n^{2/3} = \hat{g}$ prove that $f$'s Fourier series converges absolutely.
$$(S_{N}f)(t) = \sum\limits_{n=-N}^N \hat{g}(n) \cdot \frac{e^{int}}{n^{2/3}}$$
$$|\hat{g'(n)}|^2 = n^2|\hat{g(n)}|^2$$
$$\sum n^2|\hat{g}(n)|^2 < \infty$$
$$|\hat{g(n)}| = |\hat{g(n)}| \cdot n \cdot \frac{1}{n} \le \frac{1}{2} \cdot(|\hat{g(n)}|^2n^2 + \frac{1}{n^2})\cdot n$$
the last Inequality is just means Inequality. Now finally
$$\sum|\hat{g(n)}| \le \frac{1}{2}\cdot \sum n^2|\hat{g(n)}|^2 + |\hat{g(0)}| + \frac{\pi^2}{6} < \infty$$
so
$$\sum\limits_{n=-\infty}^{\infty}|\hat{g}(n)| < \infty$$
moreover $\lim\limits_{n \rightarrow \infty}|\frac{e^{int}}{n^{2/3}}| = 0$ now we can use dirichlet's test for convergence of functions series to finish the proof.
Is it correct ?
I'm almost sure that you can solve this by expressing $f$'s Fourier series with $(\hat{g}(n))^2$ and then using parseval's theorem to prove that the series converges .
It is enough to show that $\sum|\hat f(n)|<\infty$. $$ \sum|\hat f(n)|=\sum n^{-2/3}|\hat g(n)|\le\Bigl(\sum n^{-4/3}\Bigr)^{1/2}\Bigl(\sum|\hat g(n)|^2\Bigr)^{1/2}<\infty $$ since $4/3>1$ and $g$ is Riemann-integrable, and hence in $L^2$.