Integral operator identity involving square root of an operator

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# Integral operator identity involving square root of an operator

Submitted by andrew_miller on Mon, 07/01/2013 - 14:55

Forums:

I would be most thankful if you could help me prove the following operator identity. Let $A$ and $B$ be two completely continuous Hermitian operators^{} on a Hilbert space $H$, such that $A$ and $B$ do not commute. If $A$ is a positive operator, then show that

$\left\langle x,\left(\sqrt{A}B-B\sqrt{A}\right)x\right\rangle=\frac{1}{\pi}% \int_{{0}}^{{\infty}}\left\langle x,\left(B\frac{1}{A+\omega}-\frac{1}{A+% \omega}B\right)x\right\rangle\sqrt{\omega}d\omega$ |

It is known that for $z$ and $z_{0}$ real and positive numbers one has

$\sqrt{z}-\sqrt{z_{0}}=\frac{1}{\pi}\int_{{0}}^{{\infty}}\left(\frac{1}{z_{0}+% \omega}-\frac{1}{z+\omega}\right)\sqrt{\omega}d\omega$ |

Tahnk you!

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## Versions

(v1) by andrew_miller 2013-07-01