Proposition 8

Let E be a complex Hilbert space. Let $u$ be an endomorphism of E. The following conditions are equivalent:

(i) The endomorphism $u$ is positive;

(ii) The numerical image of $u$ is contained in $\mathbf{R}_+$;

(iii) The endomorphism $u$ is a positive element of the involutive algebra $\mathscr{L}(E)$ ;

(iv) There exists a hermitian element $v$ of $\mathscr{L}(E)$ such that $u=v^2$;

(v) There exists a continuous linear mapping $v$ of E into a complex Hilbert space F such that $u=v^*v$.

By EVT, V, p. 45, def. $6,u$ is positive if and only if it is hermitian and if $\langle x|u(x)\rangle \geqslant 0$ for every $x\in E$. The implication (i) $=\Rightarrow$ (ii) therefore follows from the definition of the numerical image.

(ii) $=\Rightarrow$ (iii) : the assumption implies that $u$ is hermitian (EVT, V, p. 45, and remark, p. 2) and its spectrum is contained in $\mathbf{R}_+$ (prop. 6) ; hence $u$ is a positive element of the involutive algebra $\mathscr{L}(E)$.

(iii) $=\Rightarrow$ (iv) : this is a particular case of prop. 16 of I, p. 118.

(iv) $=\Rightarrow$ (v) is immediate.

(v) $=\Rightarrow$ (i) : let F be a complex Hilbert space and let $v\in \mathscr{L}(E; F)$ be such that $u=v^*v$. Let $x\in E$. One has $\langle x|u(x)\rangle =\langle x|(v^*v)(x)\rangle =\|v(x)\|^2$, which proves that $u$ is positive.

Recall (EVT, V, p. 45, remark 1) that, for every hermitian element $u$ of $\mathscr{L}(E)$, one sets

$m(u) =$ inf$_{\|xx\in\|=1E}\langle x|u(x)\rangle =$ inf $\iota (u) =$ inf$_{x\in E-\{0\}}\frac{\langle x|u(x)\rangle}{\|x\|^2}$,

$M(u) =$ sup$_{x\in E}\langle x|u(x)\rangle =$ sup $\iota (u) =$ sup$_{x\in E-\{0\}}\frac{\langle x|u(x)\rangle}{\|x\|^2}$.

$\|x\|=1$

If $E =\{0\}$, one has $M(u) =-\infty ,m(u) = +\infty$ and $\iota (u) =\emptyset$.

Suppose $E$ nonzero; one then has $m(u)\leqslant M(u)$ and the numerical image of $u$ is an interval with endpoints $m(u)$ and $M(u)$. By Prop. 6, Sp($u$) is contained in the interval $[m(u),M(u)]$. More precisely: