Cayley Transform: well defined?

Question

Why is the Cayley backtransformation well-defined: $$A_U:=\imath(1+U)(1-U)^{-1}$$ In general $1-U$ is not invertible for example for $U=1$.

Answer 1

The Caley transform is invertible on its range:

First it follows by symmetry: $$\langle Ax,y\rangle=\langle x,Ay\rangle\Rightarrow \mathcal{N}(A+\imath)=\{0\}$$ So we have: $$A+\imath:\mathcal{D}(A)\to\mathcal{R}(A+\imath): \mathcal{N}(A+\imath)=\{0\}$$ $$A-\imath:\mathcal{D}(A)\to\mathcal{R}(A-\imath)$$ Especially: $$U_A\text{ well defined}$$

Next by formal calculations it holds: $$(1+U_A)(A+\imath)=2A$$ $$(1-U_A)(A+\imath)=2\imath\Rightarrow\mathcal{N}(\ldots)=\{0\}$$ Thus we also have: $$(1+U_A)(A+\imath):\mathcal{D}(A)\to\mathcal{D}(A):x\mapsto 2Ax$$ $$(1-U_A)(A+\imath):\mathcal{D}(A)\to\mathcal{R}:x\mapsto 2\imath x,\mathcal{N}(\ldots)=\{0\}$$ Especially: $$(1-U_A)\text{ invertible}$$

Concluding: $$A_U=(1+U_A)(1-U_A)^{-1}=(1+U_A)(A+\imath)(A+\imath)^{-1}(1-U_A)^{-1}=A$$

Note that it was crucial that the isometric operator was given by the Cayley transform of a symmetric operator; otherwise the Cayley backtransformation is ill defined in general.