All Hilbert spaces are isometric to $l^2(E)$ - how?

Question

I was reading about Hilbert spaces and came across this line on Wikipedia:

By choosing a Hilbert basis (i.e., a maximal orthonormal subset of $L^2$ or any Hilbert space), one sees that all Hilbert spaces are isometric to $ℓ^2(E)$, where $E$ is a set with an appropriate cardinality.

My questions are:

1. What does the $E$ stands for? Is it the basis of $l^2$?
2. What is meant by "appropriate cardinality"?
3. Why is $l^2$ isometric to any other Hilbert space? Yes, the norm on $l^2$ is square root of sum of squares. But how do we know the isometry applies, even when some Hilbert spaces have elements with n coordinates, while $l^2$ has infinite coordinates? (Because elements are sequences and each sequence is infinite - has infinite "coordinates").

Thank you very much for your insights.

Answer 1

This is explained in Rudin's Real and Complex Analysis (3rd edition) Chapter 4, in the part of Orthonormal Sets. See pages 86 and 87 for more details. Recall that $$ \ell^{2}(A)=L^{2}(A,\#),$$ where $\#$ is the counting measure in $A$. In particular, $\ell^{2}(\mathbb{N})=\{y=(y_{n})_{n \in \mathbb{N}};\|y\|_{2}<\infty\}$.

1. $E$ is the set of indexes used to index a orthonormal set which is maximal (with respect to inclusion).
2. The cardinality of $E$ (?). For example: for $\mathbb{R}^{n}$ you have $|E|=n$, and for $\ell^{2}(\mathbb{N})$ you have $|E|=\aleph_{0}$.
3. Let $\{u_{\alpha}\}_{\alpha \in A}$ be a maximal orthonormal set in the Hilbert space $H$. The isomorphism is given by \begin{align*} \varphi:H &\to \ell^{2}(A), \\ x & \mapsto \hat{x}, \end{align*} where $\hat{x}(\alpha)=\langle x,u_{\alpha} \rangle.$

Answer 2

Every Hilbert space $H$ admits an orthonormal basis $\{v_i : i \in E \}$. Then $H$ is isometric to $\ell^2(E)$.

EDIT: The $E$ is dependent upon the Hilbert space. Moreover, if $E, F$ are two abstract sets, then $\ell^2(E)$ and $\ell^2(F)$ are isometric if and only if $E, F$ are of the same cardinality. The point of this question is that every Hilbert space can be represented as $\ell^2$ over an (essentially) unique set $E$.

Answer 3

A theorem by Parseval states that

Theorem: Any Hilbert $H$ space admits a maximal collection $\mathscr{E}\subset H$ of orthonormal vectors such that $H=\overline{\operatorname{span}(\mathscr{E})}$. Moreover, $x\in H$ can be expressed uniquely as $$x=\sum_{\boldsymbol{e}\in\mathscr{E}}\langle x,\boldsymbol{e}_\alpha\rangle \boldsymbol{e}_\alpha,\qquad \|x\|^2=\sum_{\boldsymbol{e}\in\mathscr{E}}|\langle x,\boldsymbol{e}\rangle|^2$$

The convergence of the sums above are in terms of nets over finite subsets of $\mathscr{E}$ (see note 1 below; also, this posting may be helpful)

Q1. For any nonempty set $R$, the space $\ell^2(R)$ is the collection of all functions $X:R\rightarrow\mathbb{C}$ such that $$\sum_{r\in R}|X(r)|^2<\infty$$ where as above, convergence is in the sense of nets. It can be seen that on $\ell^2(R)$, the map $(\cdot,\cdot):\ell^2(R)\times\ell^2(R)\rightarrow\mathbb{C}$ given by $$(X,Y)=\sum_{r\in R}X(r)\overline{Y(r)}$$ defines an inner product. The space of interest for the OP is $\ell^2(\mathscr{E})$, where $\mathscr{E}$ is a maximal orthonormal collection in the Hilbert space $H$.

Q2. For each $\boldsymbol{e}\in\mathscr{E}$, the map $E_\boldsymbol{e}:\mathscr{E}\rightarrow\mathbb{C}$ defined as $E_\boldsymbol{e}(\boldsymbol{e}')=\delta_{\boldsymbol{e},\boldsymbol{e}'}$ is clearly an element of $\ell^2(\mathscr{E})$; moreover, for any $\boldsymbol{e},\boldsymbol{d}\in\mathscr{E}$ $$(E_{\boldsymbol{e}},E_{\boldsymbol{d}})=\delta_{\boldsymbol{e},\boldsymbol{d}}$$ The dimension of $\ell^2(\mathscr{E})$ is defined as the cardinality of $\mathscr{E}$.

Q3. Appealing to Parseval's theorem and Bessel's inequality, one can se that the map $\Lambda:H\rightarrow\ell^2(\mathcal{E})$ defined as $$(\Lambda x)(\boldsymbol{e})=\langle x,\boldsymbol{e}\rangle$$ is a linear isometric isomorphism between $H$ and $\ell^2(\mathcal{E})$, that is $\Lambda$ is onto and $$(\Lambda x,\Lambda x)=\|x\|^2=\sum_{\boldsymbol{e}\in\mathscr{E}}|\langle x,\boldsymbol{e}\rangle|^2$$

Notes:

1. The potentially uncountable sums are understood in as limits of nets (instead of sequences). Order the collection $\mathcal{F}(\mathcal{E})$ of all finite subsets of $\mathcal{E}$ by inclusion. A function $\xi:\mathcal{F}(\mathcal{E})\rightarrow H$ (or net in $\mathcal{C}(\mathcal{E})$) converges to $h\in H$, iff for any $\varepsilon>0$ there is $C_\varepsilon\in\mathcal{F}(\mathcal{E})$ such that for all $D\in\mathcal{F}(\mathcal{E})$, $C_0\subset D$ implies $$|\xi(D)-h|<\varepsilon$$

2. When $H$ is separable, one can take $\mathcal{E}=\mathbb{N}$ and all the net abstract nonsense reduces to the familiar real of sequences.

3. It is possible to define $\ell^2(\mathscr{E})$ as the $L^2$ space of a measure space. Define $\mathcal{P}(\mathscr{E}$ as the collection of all subsets of $\mathscr{E}$, and $\mu$ is the counting measure on $(\mathscr{E},\mathcal{P}(\mathscr{E}))$, that is, $\mu(A)=n\in\mathbb{Z}_+$ if $A$ is finite and has $n$ elements, and $\mu(A)=\infty$ otherwise. Then, it can be shown that $\ell^2(\mathscr{E})$ is the same as $L_2(\mathscr{E},\mathcal{P}(\mathscr{E}),\mu)$.

4. It can be show that if $A$ and $B$ are sets of the same cardinality, then $\ell^2(A)$ and $\ell^2(B)$ are isometric isomorphic.