Remark 3

Let $E$ and $F$ be two hilbertian spaces and $v$ a linear, not necessarily continuous mapping from $E$ into $F$. Let $H(x) = \|x\|^2$ and $Q(x) = \|v(x)\|^2$ for all $x \in E$. It follows from prop. 15 that $v$ is a Hilbert-Schmidt mapping if and only if $\operatorname{Tr}(Q/H)$ is finite, and then $\operatorname{Tr}(Q/H) = \|v\|_2^2$.