Theorem 1

— Let E be a hilbertian space which is a hilbertian sum of a family $(E_i)_{i \in I}$ of closed vector subspaces of E. There exists an isomorphism $f$ and only one, from E onto the external hilbertian sum $\bigoplus_{i \in I} E_i = F$ of the family $(E_i)$ such that, for all $i \in I$, the restriction of $f$ to E is the canonical mapping $f_i$ from $E_i$ into F.

Let $S \subset F$ be the « algebraic » direct sum of the $E_i$, and let $g$ be the linear mapping $(x_i)_{i \in I} \mapsto \sum_{i \in I} x_i$ from S into E. We shall show that $g$ is an isomorphism from the prehilbertian space S onto the (prehilbertian) subspace $g(S)$ of E, generated by the union of the $E_i$ : for, for two elements $x = (x_i)_{i \in I}$, $y = (y_i)_{i \in I}$, we have

But if $i \neq k$, $\langle x_i|y_k \rangle = 0$ by hypothesis, hence

this proves our assertion. Since S is dense in F and $g(S)$ dense in E, the isomorphism $g$ extends to an isomorphism $\overline{g}$ from F onto E (V, p. 8, cor.). It is clear that the inverse isomorphism $f$ of $\overline{g}$ is the required mapping; its uniqueness follows from the fact that the closed subspace of E generated by the union of the $E_i$ is E itself.

When E is the hilbertian sum of a family $(E_i)_{i \in I}$ of subspaces, we shall often identify E with the external hilbertian sum F of the $E_i$ by means of the isomorphism $f$. If the set I is finite, saying that E is the hilbertian sum of the family $(E_i)_{i \in I}$ means that the $E_i$ are two by two orthogonal and that the vector space E is the direct sum of the family $(E_i)_{i \in I}$ of subspaces.