Let $f,g\in R(\mathbb{T})$ s.t $\hat{f}(n)n^{2\over 3}=\hat{g}(n)$ .Prove $\sum_{n=-\infty}^{\infty} |\hat{f}(n)|<\infty$.
We get $\hat{f}(n)n^{2\over 3}=\hat{g}(n) \implies \hat{f}(n)=\frac{\hat{g}(n)}{n^{2\over 3}}$ . I'm not sure how to continue or use this information.
By $$ 2ab\le a^2+b^2$$ one has $$|\hat{f}(n)|=\frac{|\hat{g}(n)|}{n^{2\over 3}}\le\frac12|\hat{g}(n)|^2+\frac12\frac{1}{n^{4\over 3}}$$ and hence $$\sum_{n=-\infty}^{\infty}|\hat{f}(n)|=|\hat{f}(0)|+\sum_{n\neq0}|\hat{f}(n)|\le|\hat{f}(0)|+\frac12\sum_{n\neq0}|\hat{g}(n)|^2+\frac12\sum_{n\neq0}\frac{1}{n^{4\over 3}}\le|\hat{f}(0)|+\frac12\sum_{n=-\infty}^{\infty}|\hat{g}(n)|^2+\frac12\sum_{n\neq0}\frac{1}{n^{4\over 3}}<|\hat{f}(0)|+\frac12\|g\|_2^2+\frac12\sum_{n\neq0}\frac{1}{n^{4\over 3}}<\infty.$$