# Parseval equality

## 0.1 Parseval’s Equality

Theorem. – If $\{{e}_{j}:j\in J\}$ is an orthonormal basis of an Hilbert space^{} $H$, then for every $x\in H$ the following equality holds:

$${\parallel x\parallel}^{2}=\sum _{j\in J}{|\u27e8x,{e}_{j}\u27e9|}^{2}.$$ |

The above theorem is a more sophisticated form of Bessel’s inequality^{} (where the inequality is in fact an equality). The difference is that for Bessel’s inequality it is only required that the set $\{{e}_{j}:j\in J\}$ is an orthonormal set, not necessarily an orthonormal basis.

## 0.2 Parseval’s Theorem

Applying Parseval’s equality on the Hilbert space ${L}^{2}([-\pi ,\pi ])$ (http://planetmath.org/LpSpace), of square integrable functions^{} on the interval $[-\pi ,\pi ]$, with the orthonormal basis consisting of trigonometric functions^{}, we obtain

Theorem (Parseval’s theorem). – Let $f$ be a Riemann square integrable function from $[-\pi ,\pi ]$ to $\mathbb{R}$. The following equality holds

$$\frac{1}{\pi}{\int}_{-\pi}^{\pi}{f}^{2}(x)\mathit{d}x=\frac{{({a}_{0}^{f})}^{2}}{2}+\sum _{k=1}^{\mathrm{\infty}}[{({a}_{k}^{f})}^{2}+{({b}_{k}^{f})}^{2}],$$ |

where ${a}_{0}^{f}$, ${a}_{k}^{f}$, ${b}_{k}^{f}$ are the Fourier coefficients of the function $f$.

The function $f$ can be a Lebesgue-integrable function, if we use the Lebesgue integral^{} in place of the Riemann integral.

Title | Parseval equality |

Canonical name | ParsevalEquality |

Date of creation | 2013-03-22 13:57:10 |

Last modified on | 2013-03-22 13:57:10 |

Owner | asteroid (17536) |

Last modified by | asteroid (17536) |

Numerical id | 11 |

Author | asteroid (17536) |

Entry type | Theorem |

Classification | msc 42B05 |

Synonym | Parseval equation |

Synonym | Parseval identity |

Synonym | Lyapunov equation |

Related topic | BesselInequality |

Related topic | ValueOfTheRiemannZetaFunctionAtS2 |

Defines | Parseval theorem |