# Exercice on periodic function

Let $f$ be a periodic function, $\mathcal{C}^1$ on $\mathbb{R}$ such that: $$\displaystyle\int_0^{2 \pi} f(t) \, dt = 0$$

$$f(2 \pi) = f(0)$$

Prove that $$\forall t \in [0,2 \pi]: \int_0^{2 \pi} |f(t)|^2 dt \leq \int_0^{2 \pi} |f'(t)|^2 dt$$

How can we prove this please. I don't have any idea.

• You don't need the absolute value. – Git Gud Jun 9 '13 at 17:26
• @GitGud may be OP talks about complex valued function $f$ – Norbert Jun 9 '13 at 17:26
• @Norbert I considered it, but the "on $\Bbb R$" part leads me to believe everything is real. Can't be sure if she means the domain or the range. – Git Gud Jun 9 '13 at 17:28
• @GitGud "on $\mathbb{R}$" (as opposed to, I guess, "to $\mathbb{R}$") usually means that the domain is $\mathbb{R}$. – Omnomnomnom Jun 9 '13 at 17:29
• Did you want $dt$ where you wrote $dx$ in that last integral? – Michael Hardy Jun 9 '13 at 19:17

Hint: Expand $f$ in Fourier series.
• yes, the domain is $\mathbb{R}.$and this exercice is on chapitre 'Fourier series" – lili Jun 9 '13 at 17:45
• @lili you have $$f=\sum\limits_{n\in\mathbb{Z}}c_n(f)e^{int}\qquad f'=\sum\limits_{n\in\mathbb{Z}}nc_n(f)e^{int}$$ I leave it to you to verify that $$\int\limits_0^{2\pi} |f(t)|^2dt=\sum\limits_{n\in\mathbb{Z}}|c_n(f)|^2\qquad\int\limits_0^{2\pi} |f'(t)|^2dt=\sum\limits_{n\in\mathbb{Z}}n^2 |c_n(f)|^2$$ Since $\int\limits_0^{2\pi}f(t)dt=0$, then $c_0(f)$, so $$\int\limits_0^{2\pi} |f'(t)|^2dt-\int\limits_0^{2\pi} |f(t)|^2dt=\sum\limits_{n\in\mathbb{Z}\setminus\{0\}}(n^2-1)|c_n(f)|^2\geq 0$$ – Norbert Jun 9 '13 at 19:29