# Wirtinger’s inequality

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

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

Then

$${\int}_{0}^{2\pi}{f}^{\prime 2}(x)\mathit{d}x\ge {\int}_{0}^{2\pi}{f}^{2}(x)\mathit{d}x$$ | (2) |

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

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

$$f(x)=\frac{1}{2}{a}_{0}+\sum _{n\ge 1}({a}_{n}\mathrm{sin}nx+{b}_{n}\mathrm{cos}nx)$$ |

and moreover ${a}_{0}=0$ by (1). By Parseval’s identity,

$${\int}_{0}^{2\pi}{f}^{2}(x)\mathit{d}x=\sum _{n=1}^{\mathrm{\infty}}({a}_{n}^{2}+{b}_{n}^{2})$$ |

and

$${\int}_{0}^{2\pi}{f}^{\prime 2}(x)\mathit{d}x=\sum _{n=1}^{\mathrm{\infty}}{n}^{2}({a}_{n}^{2}+{b}_{n}^{2})$$ |

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 |