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Problem

Given Hilbert spaces $\mathcal{H}$ and $\mathcal{K}$.

Consider a closed operator: $$A:\mathcal{D}(A)\subseteq\mathcal{H}\to\mathcal{K}:\quad A=A^{**}$$

Polar decompose: $$A=J|A|:\quad J^*J=1_{\overline{\mathcal{R}|A|}}$$

Note that one has: $$\overline{\mathcal{R}|A|}=\overline{\mathcal{R}A^*}\quad\overline{\mathcal{R}|A^*|}=\overline{\mathcal{R}A}$$

Then identity holds: $$|A^*|:=\sqrt{AA^*}=J|A|J^*$$

Especially that gives: $$A^*=|A|J^*=J^*|A^*|$$

How can I prove this?

References

For construction see Square Root

For the relation see: Ranges

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freishahiri
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1 Answers1

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Finally I got it!! :D

Reducibility

It acts on the range: $$J^*J|A|=1_\overline{\mathcal{R}|A|}|A|=|A|$$

As it is bounded: $$J^*J\in\mathcal{B}(\mathcal{H},\mathcal{H}):\quad(J^*J|A|)^*=|A|J^*J$$

So one obtains: $$|A|J^*J=(J^*J|A|)^*=|A|^*=|A|$$

Concluding reducibility.

Positivity

Regard the selfadjoint: $$S:=J|A|J^*=(J|A|J^*)^*=S^*$$

Note that it is: $$\mathcal{D}|A|\subseteq\mathcal{D}|A|^{1/2}:\quad|A|=|A|^{1/2}|A|^{1/2}$$

Numerical range: $$\langle J|A|J^*\psi,\psi\rangle=\langle|A|^{1/2}J^*\psi|A|^{1/2}J^*\psi\rangle\geq0$$

But for selfadjoints: $$S=S^*:\quad\langle\sigma(S)\rangle=\overline{\mathcal{W}(S)}$$

Concluding positivity.

Identity

As it is bounded: $$J\in\mathcal{B}(\mathcal{H},\mathcal{K}):\quad(J|A|)^*=|A|J^*$$

So one obtains: $$|A^*|^2=AA^*=J|A|\cdot|A|J^*=J|A|J^*J|A|J^*=S^2$$

As both are positive: $$|A^*|^2=S^2\implies|A^*|=S$$

Concluding identity.

Relation

As it is bounded: $$J\in\mathcal{B}(\mathcal{H},\mathcal{K}):\quad (J|A|)^*=|A|J^*$$

So one obtains: $$A^*=|A|J^*=J^*J|A|J^*=J^*|A^*|$$

Concluding relation.

freishahiri
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