# proof of existence and uniqueness of best approximations

- Proof of the theorem on existence and uniqueness of - ( entry (http://planetmath.org/BestApproximationInInnerProductSpaces))

Existence : Without loss of generality we can suppose $x=0$ (we could simply translate by $-x$ the set $A$).

Let $d=d(0,A)=inf\{\parallel a\parallel :a\in A\}$ be the distance of $A$ to the origin. By defintion of infimum^{} there exists a sequence $({a}_{n})$ in $A$ such that

$$\parallel {a}_{n}\parallel \u27f6d$$ |

Let us see that $({a}_{n})$ is a Cauchy sequence^{}. By the parallelogram law we have

$${\parallel \frac{{a}_{n}-{a}_{m}}{2}\parallel}^{2}+{\parallel \frac{{a}_{n}+{a}_{m}}{2}\parallel}^{2}=\frac{1}{2}{\parallel {a}_{n}\parallel}^{2}+\frac{1}{2}{\parallel {a}_{m}\parallel}^{2}$$ |

i.e.

$${\parallel \frac{{a}_{n}-{a}_{m}}{2}\parallel}^{2}=\frac{1}{2}{\parallel {a}_{n}\parallel}^{2}+\frac{1}{2}{\parallel {a}_{m}\parallel}^{2}-{\parallel \frac{{a}_{n}+{a}_{m}}{2}\parallel}^{2}$$ |

As $A$ is convex, $\frac{{a}_{n}+{a}_{m}}{2}}\in A$, and therefore

$$\parallel \frac{{a}_{n}+{a}_{m}}{2}\parallel \ge d$$ |

So we see that

$${\parallel \frac{{a}_{n}-{a}_{m}}{2}\parallel}^{2}\le \frac{1}{2}{\parallel {a}_{n}\parallel}^{2}+\frac{1}{2}{\parallel {a}_{m}\parallel}^{2}-{d}^{2}\u27f60\mathit{\hspace{1em}\hspace{0.5em}\u2006}\text{when}m,n\to \mathrm{\infty}$$ |

which means that $\parallel {a}_{n}-{a}_{m}\parallel \u27f60$ when $m,n\to \mathrm{\infty}$, i.e. $({a}_{n})$ is a Cauchy sequence.

Since $A$ is complete^{} (http://planetmath.org/Complete), ${a}_{n}\u27f6{a}_{0}$ for some ${a}_{0}\in A$.

As ${a}_{0}\in A$ its norm must be $\parallel {a}_{0}\parallel \ge d$. But also

$$\parallel {a}_{0}\parallel \le \parallel {a}_{0}-{a}_{n}\parallel +\parallel {a}_{n}\parallel \u27f6d$$ |

which shows that $\parallel {a}_{0}\parallel =d$. We have thus proven the existence of best approximations (http://planetmath.org/BestApproximationInInnerProductSpaces).

Uniqueness : Suppose there were ${a}_{0},{b}_{0}\in A$ such that $\parallel {a}_{0}\parallel =\parallel {b}_{0}\parallel =d$. Then, by the parallelogram law

$${\parallel \frac{{a}_{0}-{b}_{0}}{2}\parallel}^{2}+{\parallel \frac{{a}_{n}+{b}_{0}}{2}\parallel}^{2}=\frac{1}{2}{\parallel {a}_{0}\parallel}^{2}+\frac{1}{2}{\parallel {b}_{0}\parallel}^{2}={d}^{2}$$ |

If ${a}_{0}-{b}_{0}\ne 0$ then we would have $$, which is contradiction^{} since $\frac{{a}_{0}+{b}_{0}}{2}}\in A$ ($A$ is convex).

Therefore ${a}_{0}={b}_{0}$, which proves the uniqueness of the . $\mathrm{\square}$

Title | proof of existence and uniqueness of best approximations |
---|---|

Canonical name | ProofOfExistenceAndUniquenessOfBestApproximations |

Date of creation | 2013-03-22 17:32:22 |

Last modified on | 2013-03-22 17:32:22 |

Owner | asteroid (17536) |

Last modified by | asteroid (17536) |

Numerical id | 6 |

Author | asteroid (17536) |

Entry type | Proof |

Classification | msc 49J27 |

Classification | msc 46N10 |

Classification | msc 46C05 |

Classification | msc 41A65 |

Classification | msc 41A52 |

Classification | msc 41A50 |