# proof of classification of separable Hilbert spaces

The strategy will be to show that any separable^{}, infinite
dimensional Hilbert space^{} $H$ is equivalent^{} to ${\mathrm{\ell}}^{2}$, where
${\mathrm{\ell}}^{2}$ is the space of all square summable sequences. Then it
will follow that any two separable, infinite dimensional Hilbert
spaces, being equivalent to the same space, are equivalent to each
other.

Since $H$ is separable, there exists a countable^{} dense subset $S$ of
$H$. Choose an enumeration of the elements of $S$ as ${s}_{0},{s}_{1},{s}_{2},\mathrm{\dots}$. By the Gram-Schmidt orthonormalization procedure, one
can exhibit an orthonormal set^{} ${e}_{0},{e}_{1},{e}_{2},\mathrm{\dots}$ such that
each ${e}_{i}$ is a finite linear combination^{} of the ${s}_{i}$’s.

Next, we will demonstrate that Hilbert space spanned by the ${e}_{i}$’s is in fact the whole space $H$. Let $v$ be any element of $H$. Since $S$ is dense in $H$, for every integer $n$, there exists an integer ${m}_{n}$ such that

$$\parallel v-{s}_{{m}_{n}}\parallel \le {2}^{-n}$$ |

The sequence $({s}_{{m}_{0}},{s}_{{m}_{1}},{s}_{{m}_{2}},\mathrm{\dots})$ is a Cauchy
sequence^{} because

$$\parallel {s}_{{m}_{i}}-{s}_{{m}_{j}}\parallel \le \parallel {s}_{{m}_{i}}-v\parallel +\parallel v-{s}_{{m}_{j}}\parallel \le {2}^{-i}+{2}^{-j}$$ |

Hence the limit of this sequence must lie in the Hilbert space spanned by
$\{{s}_{0},{s}_{1},{s}_{2},\mathrm{\dots}\}$, which is the same as the Hilbert space
spanned by $\{{e}_{0},{e}_{1},{e}_{2},\mathrm{\dots}\}$. Thus, $\{{e}_{0},{e}_{1},{e}_{2},\mathrm{\dots}\}$ is an orthonormal basis^{} for $H$.

To any $v\in H$ associate the sequence $U(v)=(\u27e8v,{s}_{0}\u27e9,\u27e8v,{s}_{1}\u27e9,\u27e8v,{s}_{2}\u27e9,\mathrm{\dots})$. That this sequence lies in ${\mathrm{\ell}}^{2}$ follows from the generalized Parseval equality

$${\parallel v\parallel}^{2}=\sum _{k=0}^{\mathrm{\infty}}\u27e8v,{s}_{k}\u27e9$$ |

which also shows that ${\parallel U(v)\parallel}_{{\mathrm{\ell}}^{2}}={\parallel v\parallel}_{H}$. On the other hand, let $({w}_{0},{w}_{1},{w}_{2},\mathrm{\dots})$ be an element of ${\mathrm{\ell}}^{2}$. Then, by definition, the sequence of partial sums $({w}_{0}^{2},{w}_{0}^{2}+{w}_{1}^{2},{w}_{0}^{2}+{w}_{1}^{2}+{w}_{2}^{2},\mathrm{\dots})$ is a Cauchy sequence. Since

$${\parallel \sum _{i=0}^{m}{w}_{i}{e}_{i}-\sum _{i=0}^{n}{w}_{i}{e}_{i}\parallel}^{2}=\sum _{i=0}^{m}{w}_{i}^{2}-\sum _{i=0}^{n}{w}_{i}^{2}$$ |

if $m>n$, the sequence of partial sums of ${\sum}_{k=0}^{\mathrm{\infty}}{w}_{i}{e}_{i}$
is also a Cauchy sequence, so ${\sum}_{k=0}^{\mathrm{\infty}}{w}_{i}{e}_{i}$ converges^{}
and its limit lies in $H$. Hence the operator $U$ is invertible^{} and
is an isometry between $H$ and ${\mathrm{\ell}}^{2}$.

Title | proof of classification of separable Hilbert spaces |
---|---|

Canonical name | ProofOfClassificationOfSeparableHilbertSpaces |

Date of creation | 2013-03-22 14:34:11 |

Last modified on | 2013-03-22 14:34:11 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 5 |

Author | rspuzio (6075) |

Entry type | Proof |

Classification | msc 46C15 |

Related topic | VonNeumannAlgebra |