# proof of Hilbert space is uniformly convex space

We prove that in fact an inner product space^{} is uniformly convex.
Let $\u03f5>0$, $u,v\in H$ such that $\parallel u\parallel \le 1$, $\parallel v\parallel \le 1$, $\parallel u-v\parallel \ge \u03f5$. Put $\delta =1-\frac{1}{2}\sqrt{4-{\u03f5}^{2}}$.
Then $\delta >0$ and by the parallelogram law^{}

${\parallel u+v\parallel}^{2}$ | $=$ | ${\parallel u+v\parallel}^{2}+{\parallel u-v\parallel}^{2}-{\parallel u-v\parallel}^{2}$ | ||

$=$ | $2{\parallel u\parallel}^{2}+2{\parallel v\parallel}^{2}-{\parallel u-v\parallel}^{2}$ | |||

$\le $ | $4-{\u03f5}^{2}$ | |||

$=$ | $4{(1-\delta )}^{2}.$ |

Hence, $\parallel \frac{u+v}{2}\parallel \le 1-\delta $.

Since a Hilbert space^{} is an inner product space,
a Hilbert space the conditions of a uniformly convex space.

Title | proof of Hilbert space is uniformly convex space |
---|---|

Canonical name | ProofOfHilbertSpaceIsUniformlyConvexSpace |

Date of creation | 2013-03-22 15:20:48 |

Last modified on | 2013-03-22 15:20:48 |

Owner | Mathprof (13753) |

Last modified by | Mathprof (13753) |

Numerical id | 16 |

Author | Mathprof (13753) |

Entry type | Proof |

Classification | msc 46C15 |

Classification | msc 46H05 |