# proof of Clarkson inequality

Suppose $$.

${\parallel {\displaystyle \frac{f+g}{2}}\parallel}_{p}^{p}+{\parallel {\displaystyle \frac{f-g}{2}}\parallel}_{p}^{p}$ | $=$ | $\int {\left|\frac{f+g}{2}\right|}^{p}\mathit{d}\mu}+{\displaystyle \int {\left|\frac{f-g}{2}\right|}^{p}\mathit{d}\mu$ | (1) | ||

$=$ | $\frac{1}{{2}^{p}}}\left({\displaystyle \int {\left|f+g\right|}^{p}\mathit{d}\mu}+{\displaystyle \int {\left|f-g\right|}^{p}\mathit{d}\mu}\right).$ | (2) |

By the triangle inequality^{}, we have the following two inequalities^{}

$${|f+g|}^{p}\le {|f|}^{p}+{|g|}^{p}\mathit{\hspace{1em}\hspace{1em}}\text{and}\mathit{\hspace{1em}\hspace{1em}}{|f-g|}^{p}\le {|f|}^{p}+{|g|}^{p},$$ |

and summing the two inequalities we get

$${|f+g|}^{p}+{|f-g|}^{p}\le 2({|f|}^{p}+{|g|}^{p}).$$ |

This means that expression (2) above is less than or equal to

$\frac{1}{{2}^{p-1}}}{\displaystyle \int ({|f|}^{p}+{|g|}^{p})\mathit{d}\mu}.$ | (3) |

Hence it follows that

${\parallel {\displaystyle \frac{f+g}{2}}\parallel}_{p}^{p}+{\parallel {\displaystyle \frac{f-g}{2}}\parallel}_{p}^{p}$ | $\le $ | $\frac{1}{{2}^{p-1}}}\left({\displaystyle \int {|f|}^{p}\mathit{d}\mu}+{\displaystyle \int {|g|}^{p}\mathit{d}\mu}\right)$ | ||

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

which since $p\ge 2$ directly implies the desired inequality.

Title | proof of Clarkson inequality |
---|---|

Canonical name | ProofOfClarksonInequality |

Date of creation | 2013-03-22 16:24:46 |

Last modified on | 2013-03-22 16:24:46 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 8 |

Author | CWoo (3771) |

Entry type | Proof |

Classification | msc 28A25 |