I know the following must be very standard, but I haven't found it in any of the functional analysis books to which I have access. Do you know where I can find a self-contained proof?:
If $G$ is a compact Hausdorff abelian topological group, the set of characters (i.e. continuous homomorphisms $G \rightarrow S^1$) form a Hilbert basis for $L^2(G)$.
From Folland's book: Let $G$ be an abelian compact Hausdorff group with measure $\mu$ so that $\mu(G) = 1$.
(1) The set of characters, $\hat{G}$, is orthonormal: If $\xi \neq \eta$ are two characters, then there exists an $x_0 \in G$ such that $\xi \eta^{-1}(x_0) \neq 1$. By a change of variable and invariance of Haar measure, $$ \int \xi \bar{\eta} = \xi \eta^{-1}(x_0) \int \xi \bar{\eta} $$ so $\int \xi \bar{\eta} = 0$.
(2) The set is a basis: If $f \in L^2(G)$ is orthogonal to all characters, $\xi$, then $$ 0 = \int f \bar{\xi} = \hat{f}(\xi), $$ so $f = 0$ by the Plancherel theorem. ($\hat{f}$ is the Fourier transform of $f$).