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Integral operator identity involving square root of an operator

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Integral operator identity involving square root of an operator

I would be most thankful if you could help me prove the following operator identity. Let AAA and BBB be two completely continuous Hermitian operatorsPlanetmathPlanetmath on a Hilbert space HHH, such that AAA and BBB do not commute. If AAA is a positive operator, then show that

x,(AB-BA)x=1π0x,(B1A+ω-1A+ωB)xωdωxABBAx1πsuperscriptsubscript0xB1Aω1AωBxωdω\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 zzz and z0subscriptz0z_{0} real and positive numbers one has

z-z0=1π0(1z0+ω-1z+ω)ωdωzsubscriptz01πsuperscriptsubscript01subscriptz0ω1zωωdω\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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