Characterizing abstract Hilbert spaces

Question

Let me start with a theorem that we can prove :

If $H$ is a Hilbert space over $\mathbb{C}$ and $E$ is any set such that $|E|= dim(H)$, the dimension of $H$ being the cardinality of an orthonormal basis of $H$. Then $H$ is isomorphic to the Hilbert space $$\ell_2(E, \mathbb{C})= \{f: E \to \mathbb{C}|\{\alpha \in E| f(\alpha)\neq 0 \} \text{ is countable and } \sum_{\alpha \in E}|f(\alpha)|^2<\infty \}$$ equipped with the inner product $\langle f, g\rangle= \sum_{\alpha\in E} f(\alpha)\overline{g(\alpha)} $

Can we have a similar fact but for general measure spaces? like $L_2(E, \mathbb{C})$ for example? if we drop the countability requirement?

How big can a Hilbert space be? I am assuming if $H$ is separable then its dimension must not exceed $c=2^{\aleph_o}$ .
If $H$ is not separable, can it have a dimension bigger than $c$ ? how about arbitrarily bigger ?

This questions is important for some problems in operator theory, we are given an abstract Hilbert space $H$ and we are asked to build operators on $H$, we have some nice results from operators on $L_2$ that we want to translate in $H$, so it would be always nice to write $H\cong L_2(E,\mathbb{C})$ for some $E$ .

Answer 1

Your example, $\ell^2(E)$, produces a Hilbert space with dimension $|E|$, so any cardinal can appear as the dimension of $H$.

$H$ is separable if and only if $\dim H\leq\aleph_0$. This is simple: if you have an uncountable orthonormal basis, this is an uncountable family of elements at fixed distance $\sqrt2$ of each other; so $H$ cannot be separable. And conversely, when $\dim H<\infty$ or $\dim H=\aleph_0$, it is easy to produce a countable dense subset.

Note that $\ell^2(E)=L^2(E)$ with the counting measure, so you can always do what you want. If you want a "meaty" measure space (say, diffuse), you can start with the separable space $L^2[0,1]$ and consider $L^2[0,1]^{|E|}$. I'm not sure how straightforward or not it is to consider the product measure on an infinite Cartesian product, though. I would say that for non-separable Hilbert spaces the advantages of looking at $L^2$ would be greatly reduced as you won't have the usual things that make $L^2[0,1]$ nice (polynomials, continuous functions, etc.).

Answer 2

Let $A$ be any non-empty set and $H$ be the space of all functions $f: A \to \mathbb R$ such that $\sum_{a \in A} |f(a)|^{2}<\infty$ where $\sum_{a \in A} |f(a)|^{2}$ is defined as the the supremum of all sums of the form $ \sum\limits_{k=1}^{n}|f(a_k)|^{2}$ where $\{a_1,a_2,...,a_n\}$ runs over all finite subsets of $A$. Define $ \langle f, g \rangle =\sum_{a \in A} f(a)g(a)$. This makes $H$ a Hilbert space and the functions $\{f_a: a \in A\}$ form an orthonormal basis for $H$ where $f_a (b)=1$ if $a=b$ and $0$ otherwise. The Hilbert space dimension of this space is exactly the cardinality of $A$.