Proposition 3

Suppose u is continuous. Let $(e_i)_{i \in I}$ be an orthonormal basis of $E_1$ and $(f_j)_{j \in J}$ an orthonormal basis of $E_2$. Then

(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