# Clarkson inequality

The *Clarkson inequality* says that for all $f,g\in {L}^{p}$, for $$ we have:

$${\parallel \frac{f+g}{2}\parallel}_{p}^{p}+{\parallel \frac{f-g}{2}\parallel}_{p}^{p}\le \frac{1}{2}\left({\parallel f\parallel}_{p}^{p}+{\parallel g\parallel}_{p}^{p}\right).$$ |

The inequality^{} can be used to prove that ${L}^{p}$ space is uniformly convex for $$.

Remark. If $$, then the Clarkson inequality becomes:

$${\parallel \frac{f+g}{2}\parallel}_{p}^{q}+{\parallel \frac{f-g}{2}\parallel}_{p}^{q}\le {\left(\frac{1}{2}{\parallel f\parallel}_{p}^{p}+\frac{1}{2}{\parallel g\parallel}_{p}^{p}\right)}^{\frac{1}{p-1}}$$ |

.

for functions $f,g\in {L}^{p}$, where $q=\frac{p}{p-1}$.

Title | Clarkson inequality |
---|---|

Canonical name | ClarksonInequality |

Date of creation | 2013-03-22 16:04:59 |

Last modified on | 2013-03-22 16:04:59 |

Owner | georgiosl (7242) |

Last modified by | georgiosl (7242) |

Numerical id | 13 |

Author | georgiosl (7242) |

Entry type | Theorem |

Classification | msc 28A25 |