The of norm $L^2(0,2\pi)$ is an element of $\mathcal{l}^2(\mathbb{C})$

Question

Consider a square integrable function $u \in L^2(0,2\pi)$, whose Fourier expansion is given by

$$u(x) = \sum_{m\in\mathbb{Z}} \hat u_m e^{imx}$$

Is it correct stating that the functions $e^{imx}, m\in\mathbb{Z}$ are basis of the $ L^2(0,2\pi)$ space?

Moreover, consider the norm induced by the usual inner product $$ (f,g) = \int_{0}^{2\pi}fgdx$$ After a straightforward calculation we get

$$ \Vert u \Vert^2 = 2\pi\sum_{m\in\mathbb{Z}} \vert\hat u_m\vert^2 $$

This last expression is a convergent sequence with complex coefficients, meaning the norm of $u$ is an element of $\mathcal{l}^2(\mathbb{C})$. I find this a very interesting fact and I was wondering if there was any generalization for a generic domain $\Omega$ and for $L^p(\Omega)$ where $p> 2$.

EDIT: Thanks to the comments I understood that what I wrote above is actually wrong. The norm is by definition an element of $[0,\infty)$ and obviously not of $\mathcal{l}^2(\mathbb{C})$. The sequence of Fourier coefficients is indeed part of such space, but it doesn't mean the norm itself is too.

Answer 1

You are correct, $L^2(0,2\pi)$ is a separable Hilbert space, and every separable Hilbert space is isomorphic to $l^2(\mathbb{C})$. In fact, Hilbert spaces are indexed by the cardinality of any orthonormal basis in them, and any two Hilbert spaces with same dimension are isomorphic.