The orthonormal basis $f_n(x) = e^{2\pi i nx}$ for $L^2(\mathbb R/\mathbb Z)$ defining Fourier series

Question

I thought I understood this, but now I don't think I do.

So the functions $f_n(x) = e^{2 \pi i n x}$ form an orthonormal basis for the complex Hilbert space $L^2(\mathbb R/\mathbb Z)$. This means for any square integrable function $f: \mathbb R/\mathbb Z \rightarrow \mathbb C$, I can find unique complex numbers $c_n \in \mathbb C$ such that

$$f = \sum\limits_{n \in \mathbb Z} c_n f_n$$ Formally, all this means is that the partial sums $F_N = \sum\limits_{n=-N}^N c_nf_n$ converge in the $L^2$ norm to $f$.

But if I take a random sequence $\{c_n\}$ of complex numbers, what is the sum

$$f:=\sum\limits_{n \in \mathbb Z} c_nf_n?$$

It is supposed to be an element of $L^2(\mathbb R/\mathbb Z)$. But $f$ does not appear to define a function $\mathbb R/\mathbb Z \rightarrow \mathbb C$, or even an equivalence class of such functions (identifying those $f, g$ such that $||f-g||_2 = 0$). After all, the sum $\sum\limits_n |c_n| |f_n| = \sum\limits_n |c_n|$ need not converge absolutely.

Answer 1

It's not true that an arbitrary sequence $(c_n)$ represents an element of $L^2$. Indeed, the only sequences that represent elements of $L^2$ are those for which $\sum c_nf_n$ actually does converge with respect to the $L^2$ norm to some function. It turns out that this convergence is equivalent to the sum $\sum|c_n|^2$ converging.