Remark 1

For every hermitian $u$ in $\mathscr{L}(E)$, put $m(u) = \inf_{\|x\|=1} \langle x|u(x)\rangle$, $M(u) = \sup_{\|x\|=1} \langle x|u(x)\rangle$. If $E$ is not just 0, $m(u)$ and $M(u)$ are finite; moreover, $M(u)$ is the smallest real number $\lambda$ such that $u \leq \lambda \cdot 1_E$ and $m(u)$ the largest real number $\mu$ such that $u \geq \mu \cdot 1_E$. Clearly we have $m(-u) = -M(u)$ and $M(-u) = -m(u)$. It is clear that

and prop. 9 (V, p. 44) implies (for $E \neq \{0\}$) that \* For another proof of this formula when $K$ is $\mathbf{C}$, see prop. 14 of TS, I, § 6, No. 8. \*