# every Hilbert space has an orthonormal basis

Theorem - Every Hilbert space^{} $H\beta \x89\{0\}$ has an orthonormal basis^{}.

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*Proof :* As could be expected, the proof makes use of Zornβs Lemma. Let $\mathrm{\pi \x9d\x92\u037a}$ be the set of all orthonormal sets of $H$. It is clear that $\mathrm{\pi \x9d\x92\u037a}$ is non-empty since the set $\{x\}$ is in $\mathrm{\pi \x9d\x92\u037a}$, where $x$ is an element of $H$ such that $\beta \x88\u20afx\beta \x88\u20af=1$.

The elements of $\mathrm{\pi \x9d\x92\u037a}$ can be ordered by inclusion, and each chain $\mathrm{\pi \x9d\x92\x9e}$ in $\mathrm{\pi \x9d\x92\u037a}$ has an upper bound, given by the union of all elements of $\mathrm{\pi \x9d\x92\x9e}$. Thus, Zornβs Lemma assures the existence of a maximal element $B$ in $\mathrm{\pi \x9d\x92\u037a}$. We claim that $B$ is an orthonormal basis of $H$.

It is clear that $B$ is an orthonormal set, as it belongs to $\mathrm{\pi \x9d\x92\u037a}$. It remains to see that the linear span of $B$ is dense in $H$.

Let $\stackrel{{\rm B}\u2015}{\mathrm{span}\beta \x81\u2019B}$ denote the closure of the span of $B$. Suppose $\stackrel{{\rm B}\u2015}{\mathrm{span}\beta \x81\u2019B}\beta \x89H$. By the orthogonal decomposition theorem we know that

$$H=\stackrel{{\rm B}\u2015}{\mathrm{span}\beta \x81\u2019B}\beta \x8a\x95{(\stackrel{{\rm B}\u2015}{\mathrm{span}\beta \x81\u2019B})}^{\beta \x9f\x82}$$ |

Thus, we conclude that ${(\stackrel{{\rm B}\u2015}{\mathrm{span}\beta \x81\u2019B})}^{\beta \x9f\x82}\beta \x89\{0\}$, i.e. there are elements which are orthogonal^{} (http://planetmath.org/OrthogonalVectors) to $\stackrel{{\rm B}\u2015}{\mathrm{span}\beta \x81\u2019B}$. This contradicts the maximality of $B$ since, by picking an element $y\beta \x88\x88{(\stackrel{{\rm B}\u2015}{\mathrm{span}\beta \x81\u2019B})}^{\beta \x9f\x82}$ with $\beta \x88\u20afy\beta \x88\u20af=1$, $B\beta \x88\u037a\{y\}$ would belong belong to $\mathrm{\pi \x9d\x92\u037a}$ and would be greater than $B$.

Hence, $\stackrel{{\rm B}\u2015}{\mathrm{span}\beta \x81\u2019B}=H$, and this finishes the proof. $\mathrm{\beta \x96\u2018}$

Title | every Hilbert space has an orthonormal basis |
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Canonical name | EveryHilbertSpaceHasAnOrthonormalBasis |

Date of creation | 2013-03-22 17:56:05 |

Last modified on | 2013-03-22 17:56:05 |

Owner | asteroid (17536) |

Last modified by | asteroid (17536) |

Numerical id | 4 |

Author | asteroid (17536) |

Entry type | Theorem |

Classification | msc 46C05 |