All my life, I have been operating inside separable Hilbert spaces. Everything is all dandy as any orthonormal basis is a sequence. Let $e_n$ be the orthonormal basis of the Hilbert space $H$. I have seen that if $$\sum_{n=0}^{\infty} a_n e_n=\sum_{n=0}^{\infty} b_n e_n$$ Then using continuity of $\langle e_k,\cdot \rangle$ we can show $a_n=b_n$. Great, the representation is unique. What I have not seen, is that these sums actually converge unconditionally. This is a revelation to me. This means that if $x=\sum_{n=0}^{\infty} a_n e_n=\sum_{n=0}^{\infty} b_n e_{\sigma(n)}$ then actually $a_n=b_{\sigma^{-1}(n)}$ since we can rearrange the sum on the right using the bijection $\sigma^{-1}(n)$ and then use the fact that the representation is unique.
This leads to my first question, how do you prove the Fourier expansion converges unconditionally (in the separable case)? The next question is perhaps related and more related to the actual title.
As I said, I know separable Hilbert spaces well. But only recently have I learned that if we have a non separable Hilbert space something similar happens. Let $H$ be a non-separable Hilbert space and $B$ be its (uncountable) orthonormal basis. Then $$x=\sum_{b \in b} a_b b$$ Where the convergence is unconditional. How do we even start proving this? Is there a good reference to see how this formally goes? I think I see by finite bassels inequality that $\sum_{b \in b} |a_b|^2<\|x\|^2$ but does that imply convergence? Does it even make sense to talk about uniqueness of the expansion here?I am just confused as how the set up goes here and any explanation and references would be appriciated.