If $f$ is $2 \pi$ periodic and $\int_{0}^{2 \pi} f(t) dt=0$ then $\int_{0}^{2 \pi} (f(t))^2 dt \le \int_{0}^{2 \pi} (f'(t))^2 dt$
How to prove $\int_{-\pi}^\pi (f(x))^2dx\le \int_{-\pi}^\pi (f'(x))^2dx$