Polar Decomposition: Adjoint 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
 A: 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.
