Corollary 4

Let F be a Hilbert space. Let $u\in \mathscr{L}(E; F)$ be a compact linear mapping. Let $B = (e_i)_{i\in I}$ be an orthonormal basis of the initial space Ker($u$)$^{\circ}$ of $u$, $C = (f_i)_{i\in I}$ an orthonormal family in F, and $(\alpha_i)_{i\in I}$ a family in $(\mathbf{R}^*_+)^I$ such that $u(e_i) =\alpha_if_i$ for every $i\in I$ (cor. 2 of IV, p. 150).

The endomorphism $|u|$ of E is of finite trace if and only if the family $(\alpha_i)_{i\in I}$ is summable. One then has $u\in \mathscr{L}_2(E; F)$ and

(8) Tr($|u|$) $=\sum_{i\in I}\alpha_i,\|u\|^2_2=$ Tr($u^*u$) $=\sum_{i\in I}\alpha^2_i$,

and in particular $\|u\|_2\leqslant$ Tr($|u|$).

The family of non-zero eigenvalues of $|u|$ is $(\alpha_i)_{i\in I}$, hence $|u|$ is of finite trace if and only if the family $(\alpha_i)_{i\in I}$ is summable (Prop. 12). If this is so, the family $(\alpha^2_i)$ of non-zero eigenvalues of $u^*u=|u|^2$ is summable, and formulas (8) result from loc. cit.