Lemma 3

Let u be a continuous and positive endomorphism of E, and $\mathscr{F}$ the set of all finite rank orthoprojectors on E. Then for every orthonormal basis $(e_i)_{i \in I}$ of E, we have (in $\mathbf{R}_+$) the equality

For every finite subset J of I, put $p_J = \sum_{i \in J} e_i e_i^*$; this is the orthoprojector from E onto the vector subspace generated by the vectors $e_i$, where i ranges over J. We have

hence $\tau(p_J u p_J) = \sum_{i \in J} \langle e_i | u(e_i) \rangle$. Since $p_J \in \mathscr{F}$, and so we conclude that

Let $v$ be a finite rank continuous and positive endomorphism of E and let $p \in \mathscr{F}$. By th. 2 of V, p. 23 there exists an orthonormal basis $(f_\alpha)_{\alpha \in A}$ of E and a finite subset B of A such that $(f_\alpha)_{\alpha \in B}$ is an orthonormal basis of the image of p. Then we have $p = \sum_{\alpha \in B} f_\alpha f_\alpha^*$, and so, as above, the relation $\tau(p v p) = \sum_{\alpha \in B} \langle f_\alpha | v(f_\alpha) \rangle$. By lemma 2 (V, p. 48) we have $\tau(v) = \sum_{\alpha \in A} \langle f_\alpha | v(f_\alpha) \rangle$, which gives the formula

Applying this inequality to the case where $v = p_J . u p_J$ and where J is a finite subset of I, we get For every $x \in E$, we have $p_J(x) = \sum_{i \in J} \langle e_i | x \rangle e_i$, and so $x = \lim_J p_J(x)$ with respect to the ordered directed set of finite subsets J of I. Passing to the limit over J in (23), we get and this completes the proof of lemma 3.