proof of classification of separable Hilbert spaces

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

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, onecan exhibit an orthonormal set $e_0,e_1,e_2,\mathrm{\dots }$ such thateach $e_i$ is a finite linear combination of the $s_i$’s.

Next, we will demonstrate that Hilbert space spanned by the $e_i$’sis 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 aninteger $m_n$ 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 Cauchysequence 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 spacespanned 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 $U(v)=(⟨v,s_0⟩,⟨v,s_1⟩,⟨v,s_2⟩,\mathrm{\dots })$.That this sequence lies in ${\mathrm{\ell }}^2$ follows from the generalized Parseval equality

$${\parallel v\parallel }^2=\sum _{k=0}^{\mathrm{\infty }}⟨v,s_k⟩$$

which also shows that ${\parallel U(v)\parallel }_{{\mathrm{\ell }}^2}={\parallel v\parallel }_H$. On the otherhand, 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 ${\sum }_{k=0}^{\mathrm{\infty }}{w}_i{e}_i$ convergesand its limit lies in $H$. Hence the operator $U$ is invertible andis an isometry between $H$ and ${\mathrm{\ell }}^2$.