Proposition 4

Let E be a Hausdorff prehilbertian space, (e_i)_{i∈I} an orthonormal family in E, V the closed vector subspace of E generated by the e_i.

1) For every x ∈ E, we have

(Bessel’s inequality); here the set of all $i \in I$ such that $⟨e_i|x⟩ \neq 0$ is countable. Moreover, the following conditions are equivalent : a) $x \in V$; b) $\|x\|^2 = \sum_{i\in I} |⟨e_i|x⟩|^2$; c) the family $⟨e_i|x⟩.e_i$ is summable in E, and $x = \sum_{i\in I} ⟨e_i|x⟩.e_i$.

2) If V is complete, then the family of all $⟨e_i|x⟩.e_i$ is summable in E for all $x \in E$, and $\sum_{i\in I} ⟨e_i|x⟩.e_i = p_V(x), \sum_{i\in I} |⟨e_i|x⟩|^2 = \|p_V(x)\|^2$.

3) Suppose V is complete. For every family $(\lambda_i)_{i\in I}$ of scalars such that $\sum_{i\in I} |\lambda_i|^2 < +\infty$, there exists a unique point $x \in V$ such that $⟨e_i|x⟩ = \lambda_i$ for all $i \in I$. If $(\mu_i)_{i\in I}$ is a second family of scalars such that $\sum_{i\in I} |\mu_i|^2 < +\infty$, and if $y \in V$ is such that $⟨e_i|y⟩ = \mu_i$ for all $i \in I$, then $⟨x|y⟩ = \sum_{i\in I} \overline{\lambda_i} \mu_i$.