Proposition 1

*Let $(E_i)_{i \in I}$ be a family of hilbertian spaces, P the product vector space $\prod_{i \in I} E_i$ and E the subset of P consisting of all families $x = (x_i)_{i \in I}$ such that $\sum_{i \in I} \|x_i\|^2$ is finite.

a) E is a vector subspace of P.

b) For every $x = (x_i)_{i \in I}$ and $y = (y_i)_{i \in I}$ in E, the family $(\langle x_i|y_i \rangle)_{i \in I}$ is summable. If we put $\langle x|y \rangle = \sum_{i \in I} \langle x_i|y_i \rangle$, we define a positive separating hermitian form on E.

c) For the scalar product so defined, E is a hilbertian space; the direct sum S of the $E_i$ is dense in E.

For $x = (x_i)_{i \in I}$ and $y = (y_i)_{i \in I}$ in E, we have

hence $x + y = (x_i + y_i)_{i \in I}$ belongs to E. This proves a).

By the Cauchy-Schwarz inequality, we have

hence $\sum_{i \in I} |\langle x_i|y_i \rangle| < +\infty$. If $x \neq 0$, we have $\langle x|x \rangle = \sum_{i \in I} \|x_i\|^2 > 0$, hence assertion b) follows.

We recall that S is the subspace of P consisting of all families $x = (x_i)_{i \in I}$ such that the set of all $i \in I$ for which $x_i \neq 0$ is finite. It follows immediately that S is dense in E; hence it remains to prove that E is complete for the topology $\mathscr{T}_1$ obtained by the norm $\|x\| = \langle x|x \rangle^{1/2}$. Let $\mathscr{T}_2$ be the topology induced on E by the product topology on $\prod_{i \in I} E_i$. For every $r > 0$, let $B_r$ be the set of all $x \in E$ such that $\|x\| \leq r$. This relation implies that we have $\sum_{i \in J} \|x_i\|^2 \leq r^2$ for every finite subset J of I, and so $B_r$ is a closed subset of $\prod_{i \in I} E_i$, hence also complete. The fact that E is complete for $\mathscr{T}_1$ now follows from GT, III, § 3, No. 5, cor. 2 to prop. 10.