Integration inequality for a periodic function

Question

Let $E:=\{f\in C^2(\Bbb R): f \text{ is } 2\pi-\text{ periodic and} \int_{-\pi}^{\pi}f(t)\,dt=\int_{-\pi}^{\pi}f(t)e^{\pm it}\,dt=0$. Show that if $f\in E$ then prove that $$\int_{-\pi}^{\pi}|f(t)|^2\,dt \le 4\int_{-\pi}^{\pi}|f''(t)|^2\,dt.$$

I'm thinking about Cauchy Schwartz inequality, but not getting this. Any hints please?

Answer 1

Parseval gives $\int |f|^2 = 2 \pi \sum_k |f_k|^2 $, $\int |f''|^2 = 2 \pi \sum_k k^2|f_k|^2 $ (since if $f(t) = \sum_k f_k e^{i kt}$, then $f''(t) = -\sum_k k^2 f_k e^{i kt}$).

Since $f_{-1}=f_0=f_1 = 0$ we have $\int |f|^2 = 2 \pi \sum_{|k|\ge 2} |f_k|^2 $ and $\int |f''|^2 = 2 \pi \sum_{|k|\ge 2} k^2|f_k|^2 \ge 2 \pi 4 \sum_{|k|\ge 2} |f_k|^2 = 4\int|f|^2$.