Lemma 14

Let A be a star-algebra, and let $x$ and $y$ be hermitian elements of A.

a) If $x\leqslant y$, then for every element $w$ of A, we have $w^*xw\leqslant w^*yw$. In particular, if $y\geqslant 0$, we have $w^*yw\geqslant 0$;

b) Suppose that A is unital. If $0\leqslant x\leqslant y$ and if $x$ is invertible, then $y$ is invertible and $y^{-1}\leqslant x^{-1}$;

c) If $0\leqslant x\leqslant y$ then $\|x\|\leqslant \|y\|$.

Let $u= (y-x)^{1/2}$. We have $w^*yw-w^*xw=w^*u^2w= (uw)^*(uw)$, and assertion a) follows from Theorem 2.

Let us prove assertion b). Suppose first that $x= 1$. Let B be the unital *-subalgebra generated by $y$. By the Gelfand isomorphism, $y$ corresponds to a continuous function $\geqslant 1$ on the compact space $\mathsf{X}(B)$. This function is therefore invertible and its inverse is $\leqslant 1$. This implies that $y$ is invertible and $y^{-1}\leqslant 1 =x^{-1}$. In the general case, we observe that $0\leqslant 1\leqslant x^{-1/2}yx^{-1/2}$ by a), hence the preceding case implies that $z=x^{-1/2}yx^{-1/2}$ is invertible and $z^{-1}\leqslant 1$. Thus $y$ is invertible and $y^{-1}\leqslant x^{-1}$ by a) again.

To prove assertion c), one may suppose that A is unital (Proposition 3 of I, p. 105). Suppose first that $y$ is invertible. Put

$b=y$. By a), the conditions $0\leqslant x\leqslant y$ imply $0\leqslant bxb\leqslant$ $b^{-1}yb^{-1}= 1$, hence $\|b^{-1}xb^{-1}\|\leqslant 1$ by Lemma 12, c) of I, p. 116. We then have

In the general case, for every real $\varepsilon >0$, the element $y+\varepsilon$ is invertible and $0\leqslant x\leqslant y+\varepsilon$. By what precedes, we therefore have $\|x\|\leqslant \|y+\varepsilon \|$ for every real number $\varepsilon >0$, whence the result.