Complements on Hilbert spaces

a) There exists an endomorphism u of E characterized by $(x|y)_Q = (u(x)|y)_H$ for x, y in E.

b) $\operatorname{Tr}(Q/H) = \operatorname{Tr}(u)$.

c) $\operatorname{Tr}(Q/H) = \sum_{i=1}^m Q(e_i)$ for every basis $(e_1, \ldots, e_m)$ of E orthonormal for H.

a) follows from the fact that the bilinear form $(x, y) \mapsto (x|y)_H$ is non-degenerate. Every sequence in E orthonormal for H may be completed to a basis of E orthonormal for H. Consequently, $\operatorname{Tr}(Q/H)$ is the supremum of the set of numbers of the form $\sum_{i=1}^m Q(e_i)$ over all bases $(e_1, \ldots, e_m)$ of E orthonormal for H. To prove b) and c), it suffices to show that $\sum_{i=1}^m Q(e_i) = \operatorname{Tr}(u)$ for every basis of this kind. Set $a_{ij} = (u(e_i)|e_j)_H = (e_i|e_j)_Q$ for $1 \leq i, j \leq m$; then $u(e_i) = \sum_{j=1}^m a_{ij} e_j$ for $1 \leq i \leq m$, whence

Q.E.D.

for $t_1, \ldots, t_n$ real. If, moreover, the relation $\operatorname{H}(x) = 0$ implies $Q(x) = 0$ for $x \in E$, then $\operatorname{Tr}(Q/H) = \sum_{i=1}^m Q(e_i)$.

There exists a basis $(e'_1, \ldots, e'_n)$ of E orthogonal for H. One can assume the basis to be indexed in such a way that $\operatorname{H}(e'_i) > 0$ for $1 \leq i \leq m$ and $\operatorname{H}(e'_i) = 0$ for $m < i \leq n$. Then set $e_i = e'_i / \operatorname{H}(e'_i)^{1/2}$ for $1 \leq i \leq m$ and $e_i = e'_i$ for $m < i \leq n$; the relation (1) is satisfied.

Let F be the subspace of E generated by $e'_{m+1}, \ldots, e'_n$; it is the set of $x \in E$ such that $\operatorname{H}(x) = 0$. Denote by $\pi$ the canonical mapping of E onto $E_1 = E/F$. Since Q and H are zero on F, there exist two positive quadratic forms $Q_1$ and $H_1$ on $E_1$ such that $Q = Q_1 \circ \pi$ and $H = H_1 \circ \pi$. Moreover, $(\pi(e_1), \ldots, \pi(e_m))$ is a basis of $E_1$ orthonormal for $H_1$, and therefore $H_1$ is nondegenerate.

By Prop. 1 and Remark 2,

Q.E.D.

(2) Cf. TVS, V, §4, No. 7.

For every $x \in E_1$, we have $\|x\|^2 = \sum_{i \in I} (x|e_i)^2$, and similarly $\|y\|^2 = \sum_{j \in J} (y|f_j)^2$ for every $y \in E_2$. Consequently,

In particular, the number $\sum_{i \in I} \|u(e_i)\|^2$ is independent of the orthonormal basis $(e_i)_{i \in I}$ of $E_1$.

Set $t = \operatorname{Tr}(Q|H)$. For every finite subset $I'$ of $I$, by definition

whence $\sum_{i \in I} \|u(e_i)\|^2 \leq t$. Let $(e'_1, \ldots, e'_p)$ be a finite orthonormal sequence in $E$. This sequence can be completed to an orthonormal basis $(e'_\alpha)_{\alpha \in A}$ of $E_1$. Then

and, passing to the supremum over all $(e'_1, \ldots, e'_p)$, one obtains the inequality $t \leq \sum_{i \in I} \|u(e_i)\|^2$. We have thus established the equality $t = \sum_{i \in I} \|u(e_i)\|^2$.

Q.E.D.

One says that $u$ is a Hilbert–Schmidt mapping of $E_1$ into $E_2$ if the positive quadratic form $Q : x \mapsto \|u(x)\|^2$ on $E_1$ is nuclear. When this is so, one has $Q \leq \operatorname{Tr}(Q/H) \cdot H$, therefore $u$ is continuous and

Let $u : E_1 \to E_2$ be a continuous linear mapping. By Prop. 3, $u$ is a Hilbert–Schmidt mapping if and only if there exists an orthonormal basis $(e_i)_{i \in I}$ of $E_1$ such that $\sum_{i \in I} \|u(e_i)\|^2 < +\infty$. When this is so, every orthonormal basis of $E_1$ has the same property. Moreover, if $u$ is a Hilbert–Schmidt mapping, then so is its adjoint $u^*$ by virtue of the formula $\sum_{i \in I} \|u(e_i)\|^2 = \sum_{j \in J} \|u^*(f_j)\|^2$ (Prop. 3).

Exercises