Proposition 15

Suppose that there exists a real hilbertian space structure on $E$ such that $H(x) = \|x\|^2$ for all $x \in E$. For $\operatorname{Tr}(Q/H)$ to be finite, it is necessary and sufficient that there exists a continuous and positive endomorphism $u$ of $E$ with finite trace, such that $Q(x) = \langle x|u(x)\rangle$ for all $x \in E$; this endomorphism $u$ is unique, and we have

for every orthonormal bases $(e_i)_{i \in I}$ of $E$.

Suppose that $\operatorname{Tr}(Q/H)$ is finite. For every $x \in E$ of norm 1, we have $H(x) = 1$, hence $Q(x) \leq \operatorname{Tr}(Q/H)$. Therefore, $Q(x) \leq \operatorname{Tr}(Q/H) \cdot \|x\|^2$ for all $x \in E$, and

by the Cauchy-Schwarz inequality. Consequently, the bilinear form $(x, y) \mapsto \langle x|y\rangle_Q$ on $E \times E$ is continuous. There exists (V, p. 16, cor. 2) a mapping $u \in \mathscr{L}(E)$ such that $\langle x|y\rangle_Q = \langle x|u(y)\rangle$. We have $\langle x|y\rangle_Q = \langle y|x\rangle_Q$ for $x, y$ in $E$, hence $u$ is hermitian; and $\langle x|u(x)\rangle = Q(x) \geq 0$, hence $u$ is positive.

Conversely, let $u$ be a continuous and positive endomorphism of $E$ such that $Q(x) = \langle x|u(x)\rangle$ for all $x \in E$. Then

which gives the uniqueness of $u$. By formula (24') (V, p. 50), we get

where $(x_1, \ldots, x_m)$ range over the set of all finite orthonormal sequences of elements of $E$. By the definition of $\operatorname{Tr}(Q/H)$, we get $\operatorname{Tr}(u) = \operatorname{Tr}(Q/H)$. Finally, for every orthonormal basis $(e_i)_{i \in I}$ of $E$, we have $\operatorname{Tr}(u) = \sum_{i \in I} \langle e_i|u(e_i)\rangle$ by formula (25) of V, p. 50, hence $\operatorname{Tr}(u) = \sum_{i \in I} Q(e_i)$.