# proof of Riesz’ Lemma

Let’s consider $x\in E-S$ and let $r=d(x,S)$. Recall that $r$ is the distance between $x$ and $S$: $d(x,S)=\text{inf}\{d(x,s)\text{such that}s\in S\}$. Now, $r>0$ because $S$ is closed. Next, we consider $b\in S$ such that

$$ |

This vector $b$ exists: as $$ then

$$\frac{r}{\alpha}>r$$ |

But then the definition of infimum implies there is $b\in S$ such that

$$ |

Now, define

$${x}_{\alpha}=\frac{x-b}{\parallel x-b\parallel}$$ |

Trivially,

$$\parallel {x}_{\alpha}\parallel =1$$ |

Notice that ${x}_{\alpha}\in E-S$, because if ${x}_{\alpha}\in S$ then $x-b\in S$, and so $(x-b)+b=x\in S$, an absurd. Plus, for every $s\in S$ we have

$$\parallel s-{x}_{\alpha}\parallel =\parallel s-\frac{x-b}{\parallel x-b\parallel}\parallel =\frac{1}{\parallel x-b\parallel}\cdot \parallel \parallel x-b\parallel \cdot s+b-x\parallel \ge \frac{r}{\parallel x-b\parallel}$$ |

because

$$\parallel x-b\parallel \cdot s+b\in S$$ |

But

$$\frac{r}{\parallel x-b\parallel}>\frac{\alpha}{r}\cdot r=\alpha $$ |

QED.

Title | proof of Riesz’ Lemma |
---|---|

Canonical name | ProofOfRieszLemma |

Date of creation | 2013-03-22 14:56:14 |

Last modified on | 2013-03-22 14:56:14 |

Owner | gumau (3545) |

Last modified by | gumau (3545) |

Numerical id | 6 |

Author | gumau (3545) |

Entry type | Proof |

Classification | msc 54E35 |

Classification | msc 15A03 |