 Operator monotonicity of the square root

For bounded positive operators $0\le A\le B$ in the Löwner order, one has $A^{1/2}\le B^{1/2}$ even if $A$ and $B$ do not commute. A useful proof usesFor $t>0$, put $s=t{r}^2$ to obtain $2\sqrt{t}{\int }_0^{\infty }(1+r^2{)}^{-1}dr=\pi \sqrt{t}$ before the factor $1/\pi$. The zero case is immediate. The continuous functional calculus therefore gives the corresponding operator integral.

For $s>0$, inverse order yields $(B+sI{)}^{-1}\le (A+sI{)}^{-1}$. This follows by conjugating $B+sI\ge A+sI>0$ by $(A+sI{)}^{-1/2}$ and inverting its spectrum. ConsequentlyIntegration with positive weight proves the result. The integral converges in operator norm: near zero the integrand is bounded by $s^{-1/2}$, and near infinity by $\Vert A\Vert s^{-3/2}$, or the corresponding bound for $B$.