Square-root resolvent integral in a C-star algebra ($x^{1/2}={\pi }^{-1}{\int }_0^{\infty }{t}^{-1/2}x(t1+x{)}^{-1}dt$)

For a Positive element of a C-star algebra $x$, the displayed improper integral converges in norm. Near zero its integrand has norm at most $t^{-1/2}$; near infinity it has norm at most $\Vert x\Vert t^{-3/2}$. The scalar identity follows by setting $t=s{u}^2$ for $s>0$ and using ${\int }_0^{\infty }2/(1+u^2)du=\pi$; for $s=0$ both sides vanish. The bounds are uniform on the compact spectrum of an element, so continuous functional calculus gives the operator identity.

For Positive elements of a C-star algebra $x\le y$, inversion reverses the order of strictly positive elements, so $(t1+x{)}^{-1}\ge (t1+y{)}^{-1}$ for $t>0$. Subtracting their square-root resolvent integral in a C-star algebra formulas givesThe integrand is positive, and the positive cone is closed, so the norm limit of these integrals is positive. No commutativity of $x$ and $y$ is required. In particular, $b^2\ge a^2$ for positive $a,b$ implies $b\ge a$, even though squaring itself is not generally order preserving.