Wirtinger's inequality for functions

Remarks

Proof idea: expand $f$ in its Fourier series $\sum_k c_k e^{i k x}$. The mean zero condition gives $c_0 = 0$, and $f'$ has coefficients $i k c_k$. By Parseval's identity the two sides are $2\pi \sum_{k \ne 0} \abs{c_k}^2$ and $2\pi \sum_{k \ne 0} k^2 \abs{c_k}^2$, and $k^2 \ge 1$ with equality only for $k = \pm 1$.

The mean zero condition is needed: the constant function $f = 1$ gives $2\pi \le 0$, which is false.

A related form holds on an interval with zero boundary values: if $f \colon [0, \pi] \to \R$ is continuously differentiable with $f(0) = f(\pi) = 0$, then $\int_0^\pi f^2 \le \int_0^\pi f'^2$.

Going deeper

For a student

This is the simplest Poincaré-type inequality, and the constant $1$ is the smallest nonzero eigenvalue of $-d^2/dx^2$ on the circle of length $2\pi$. It gives a short proof of the isoperimetric inequality in the plane, by applying it to the coordinates of a closed curve parametrized by arc length.