Let $H$ be a Hilbert space and $\{e_k:k\in \mathbb{Z}\}$ an orthonormal set. Prove that the set is a Hilbert basis if and only if Parseval's identity is true.
The direct theorem is almost immediate: Write $x=\sum_i\langle x,e_i\rangle e_i$ and $y=\sum_j\langle y,e_j\rangle e_j$ then $\langle x,y\rangle =\langle \sum_i\langle x,e_i\rangle e_i,\sum_j\langle y,e_j\rangle e_j\rangle =\sum_i\sum_j\langle x,e_i\rangle \overline{\langle y,e_j\rangle }\langle e_i,e_j\rangle =\sum_n \langle x,e_n\rangle \overline{\langle y,e_j\rangle }$ which is Parseval's identity.
But how can I prove the reciprocal? I have to show that the expansion $x=\sum_nx_ne_n$ is unique, I thought that I could suppose the existence of another expansion $x=\sum_ny_ne_n$ and then apply Parseval's identity show that the right hand side is not $\langle x,x\rangle$.
The right hand side would be $\sum_n\langle \sum_ix_ie_i\rangle \langle\sum y_je_j,e_j\rangle=\sum_n\sum_i\sum_j x_i\overline{y_j}\langle e_i,e_n\rangle\langle e_n,e_j\rangle=\sum_nx_i\overline{y_j}$ which is not $\langle x,x\rangle=\sum_nx^2_n$ because the assumption was that the expansion was different, then the assumption that there are two expansions for a vector when Parseval's identity is true is absurd.