Nonseperable Hilbert space: Explicit ONB?

Question

Every Hilbert space admits an ONB by axiom of choice.
For separable Hilbert spaces this can in fact be constructed by Gram-Schmidt.
For nonseparable Hilbert spaces there can be no general construction by induction.
However is there Hilbert spaces where one still knows some ONB explicitely?

Answer 1

One can artificially construct such a thing, but whether there is any non-separable Hilbert space that is actually of some interest in which one can do such a thing is a more challenging question.

Here's an example: Look at the set of all functions $f:\mathbb R\to\mathbb C$ such that $$ \sum_{x\in\mathbb R} |f(x)|^2 = \sup\left\{\sum_{x\in S} |f(x)|^2 : S\subseteq\mathbb R,\ S\text{ is finite} \right\} < \infty. $$ For every such function there are at most countably many $x\in\mathbb R$ such that $f(x)\ne0$.

Let the inner product be $$ \langle f,g\rangle= \sum_{x\in\mathbb R} f(x)\overline{g(x)} $$ where $\overline{c}=\text{complex conjugate of }c$.

Here is an orthonormal basis: $$ \{ f_x : x\in\mathbb R\}, \text{ where }f_x(y) = \begin{cases} 1 & \text{if }x=y, \\ 0 & \text{otherwise}. \end{cases} $$