# Characterizing abstract Hilbert spaces

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 $$E$$ .

• $\mathcal H = L^2(X, \mathcal A, \mu)\cong \mathcal l^2(A)$ where $A$ can be chosen to be orthonormal basis, no countability condition required. $\mathcal l^2$ is certainly easy to understand in many cases, while $L^2$ is also useful in others, such as spectral theorem for normal operators. Feb 23, 2022 at 6:06
• what is $X$ and the sigma-algebra $\mathcal{A}$ in $L^2$ ? Feb 23, 2022 at 15:46
• Any measure space. Feb 23, 2022 at 16:33
• I am not sure you got my point, $H$ is given as an arbitrary Hilbert space, how can you determine that it is $L_2$ Feb 23, 2022 at 20:42
• I don't know what want. Any $H\simeq l^2$ and Any $l^2$ is automatically a $L^2$ with the measure that any singleton has measure $1$. Feb 24, 2022 at 0:04

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.).

• I am looking for something with Lebesgue measure if I may Feb 23, 2022 at 20:43
• Lebesgue measure is defined for $\mathbb R^n$, for $n\in\mathbb N$. The space $L^2(\mathbb R^n)$ is separable, and thus as Hilbert spaces $L^2(\mathbb R^n)=L^2[0,1]$ for all $n$. Feb 23, 2022 at 21:12
• here is my main problem: math.stackexchange.com/questions/4389634/… Feb 23, 2022 at 21:18

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 $$A$$.

• @NotaChoice The Hilbert space $H$ described in this answer is exactly the space $\ell_2(A,\mathbb C)$ from your question. The countability assumption is automatic (in general, a square integrable function is always supported on a $\sigma$-finite set). Feb 23, 2022 at 12:12