# Prove that $\int_0^\pi f^2\leq \int_0^\pi (f')^2$ by Wirtinger's inequality

Use Wirtinger's inequality to show that $\forall f \in C[0,\pi]$ satiwfying $f(0)=f(1)=0$ and $f'$ exists $\forall x\in [0,\pi]$, the inequality $\int_0^\pi f^2\leq \int_0^\pi (f')^2$ holds. When does the equal signs hold?

For you information (since there may be some different version of Wirtinger's inequality), Wirtinger's inequality states that:

For any $2\pi$ periodic function integrable on $[-\pi,\pi]$ s.t. $f'$ also integable on $[-\pi,\pi]$, we have $\int_{-\pi}^\pi (f(x)-a_0)^2 > \leq \int_{-\pi}^\pi (f'(x))^2$, where $a_0=\int_{-\pi}^\pi f$, and equality holds if and only if $f(x)=a_0+a_1cosx+b_1sinx$, where $a_1,b_1$ are the fourier coefficient of $f$

Could you please give some hints? The condition $f(0)=f(1)=0$ is strange.

• Umm....maybe it was a typo by my professor. But what can I can do if it was in fact $f(\pi)=0$ ? – Longitude Sep 30 '16 at 16:57
• Expand $f(x)$ and $f'(x)$ as Fourier sine/cosine series over the interval $(0,\pi)$ and exploit Parseval's theorem. Probable duplicate of math.stackexchange.com/questions/702168/… – Jack D'Aurizio Sep 30 '16 at 17:07
• Solved. Thanks for your help. – Longitude Sep 30 '16 at 17:11