Let $f:\mathbb{R}\to\mathbb{C}$ which is $1$-periodic and $f\in C^1$. Also, $\int_0^1 \left| f' \right| \le 1$. Show that $\sum_{n\ne 0} \left| \hat{f}(n) \right|^2 \le \frac{1}{4\pi^2}$.
(*) $\hat{f}(n)$ is the $n$-th Fourier coefficient.
First, I'm familiar with Bessel's inequality that $\sum_{n=-\infty}^\infty \left| \hat{f}(n) \right|^2 \le \| f \| = \frac{1}{2\pi}\int_0^{2\pi}\left| f \right|^2$
We know that:
$$\left| \int_0^1 f' \right| = \left|f(1)-f(0)\right| \le \int_0^1 f' \le 1$$
But that's something we already knew because $f$ is $1$-periodic so $f(1)-f(0) = 0$.
So I'm not sure how utilize this information.
Note that $$ f(x)=\sum_{n=-\infty}^\infty \hat{f}(n)e^{2\pi inx},x\in[0,1] $$ where $$\hat{f}(n)=\int_0^{1} f(x)e^{-2\pi inx}dx. $$ Since $$ \hat{f}'(n)=\int_0^{1} f'(x)e^{-2\pi inx}dx=2\pi i n\hat{f}(n), $$ by Parsevel's Identity $$ \sum_{n=-\infty}^\infty |\hat{f}(n)|^2=\int_0^{1}|f(x)|^2dx, $$ we have \begin{eqnarray} \sum_{n\neq 0} |\hat{f}(n)|^2&=&\sum_{n\neq 0}\frac{1}{4\pi^2n^2}|\hat{f}'(n)|^2\\ &\le&\frac{1}{4\pi^2}\sum_{n\neq 0}|\hat{f}'(n)|^2\\ &\le&\frac{1}{4\pi^2} \end{eqnarray}