Wirtinger’s inequality

Theorem: Let $f:ℝ\to ℝ$ be a periodic function of period $2\pi$, which is continuous and has a continuous derivative throughout $ℝ$, and such that

$${\int }_0^{2\pi }f(x)=0.$$ (1)

Then

$${\int }_0^{2\pi }{f}^{\prime 2}(x)𝑑x\ge {\int }_0^{2\pi }{f}^2(x)𝑑x$$ (2)

with equality if and only if $f(x)=a\cos x+b\sin x$ for some $a$ and $b$ (or equivalently $f(x)=c\sin (x+d)$ for some $c$ and $d$).

Proof: Since Dirichlet’s conditions are met, we can write

$$f(x)=\frac{1}{2}{a}_0+\sum _{n\ge 1}(a_n\sin nx+b_n\cos nx)$$

and moreover $a_0=0$ by (1). By Parseval’s identity,

$${\int }_0^{2\pi }{f}^2(x)𝑑x=\sum _{n=1}^{\mathrm{\infty }}(a_n^2+b_n^2)$$

and

$${\int }_0^{2\pi }{f}^{\prime 2}(x)𝑑x=\sum _{n=1}^{\mathrm{\infty }}{n}^2(a_n^2+b_n^2)$$

and since the summands are all $\ge 0$, we get (2), with equality if and only if $a_n=b_n=0$ for all $n\ge 2$.

Hurwitz used Wirtinger’s inequality in his tidy 1904 proof of the isoperimetric inequality.

Title	Wirtinger’s inequality
Canonical name	WirtingersInequality
Date of creation	2013-03-22 14:02:38
Last modified on	2013-03-22 14:02:38
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	9
Author	rspuzio (6075)
Entry type	Theorem
Classification	msc 42B05
Synonym	Wirtinger inequality