State and prove the polar decomposition for closed operators. My attempt:
Let $A$ be a closed operator, we know $A^*A$ is self-adjoint. Let $W=(A^*A-1)(A^*A+1)^{-1}$. Then the polar decomposition for $A=W\vert A\vert$, where $\vert A\vert=(A^*A)^{\frac{1}{2}}.$
Motivation: we know the Cayley transform is a partial isometry for closed densely defined symmetric operators.
proof: I have a proof for bounded operators, and following that in this case gives rise to the problem of domains, ranges kernels and initial spaces being not equal to each other. I am a bit lost, is it true that the initial space of $W$, $\mathcal{R}(A^*A+i)$ is not equal to $\mathcal{R}(A+i)$? Could we prove one sits densely in the other? Also i suspect we can use the fact that $\{h\oplus Ah:h\in\mathcal{D}(A^*A)\}$ is dense in the graph of $A$, but I don't see how.