= Operator monotonicity of the square root
For bounded <positive operators> $0\leq A\leq B$ in the <Löwner order>, one has $A^{1/2}\leq B^{1/2}$ even if $A$ and $B$ do not commute. A useful proof uses
$$
\sqrt t=\frac1\pi\int_0^\infty\frac{t}{t+s}\,\frac{ds}{\sqrt s}\qquad(t\geq0).
$$
For $t>0$, put $s=tr^2$ 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 $(B+sI)^{-1}\leq(A+sI)^{-1}$. This follows by conjugating $B+sI\geq A+sI>0$ by $(A+sI)^{-1/2}$ and inverting its spectrum. Consequently
$$
A(A+sI)^{-1}=I-s(A+sI)^{-1}\leq I-s(B+sI)^{-1}=B(B+sI)^{-1}.
$$
Integration 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 $\|A\|s^{-3/2}$, or the corresponding bound for $B$.
Back to article page
