TITLE: Wirtinger Inequality: $\int_0^1 f'^2 \geq \pi^2 \int_0^1 f^2 $
QUESTION [5 upvotes]: I have a  basic question
Let $f$ be a $\mathcal{C}^1$  map  from  $[0,1]$ into $\mathbb{R}$ vanishing in $0$ and $1$. I want to prove that 
$$\int_0^1 f'^2 \geq \pi^2 \int_0^1 f^2 $$
and to see when we have equality.

REPLY [2 votes]: If $f(0)=f(1)=$, then $f$ can be expressed as a sine series, i.e.,
$$
f(x)=\sum_{n=1}^\infty a_n\sin n\pi x.
$$
Then
$$
f'(x)=\sum_{n=1}^\infty n\pi a_n\cos n\pi x,
$$
and 
$$
\int_0^1 \big(f(x)\big)^2\,dx=\cdots=\frac{1}{2}\sum_{n=1}^\infty a_n^2,
$$
while
$$
\int_0^1 \big(f'(x)\big)^2\,dx=\cdots=\frac{1}{2}\sum_{n=1}^\infty \pi^2 n^2a_n^2.
$$
Thus
$$
\int_0^1 \big(f'(x)\big)^2 \ge \pi^2\int_0^1 \big(f(x)\big)^2.
$$
Equality for $f(x)=\sin\pi x$.