Cayley transform of a self-adjoint operator

Assume Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Let $T$ be a self-adjoint operator on $H$. By Resolvent of a self-adjoint operator: nonreal resolvents and the estimate the points $\pm i$ lie in $\rho (T)$, so $T+i$ and $T-i$ are bijections of $D(T)$ onto $H$ with bounded inverses, and $C_T:=(T-i)(T+i{)}^{-1}\in \mathcal{B}(H)$ is a bounded everywhere defined operator, the Cayley transform of $T$. In the resolvent convention $R_T(z)=(z-T{)}^{-1}$ of Resolvent and spectrum of an unbounded operator one has $(T+i{)}^{-1}=-R_T(-i),C_T=I+2i{R}_T(-i)=I-2i(T+i{)}^{-1}\mathrm{.}$

Its properties, with proofs. Writing $C:=C_T$:

1. $C$ is isometric. For $x\in H$ put $y:=(T+i{)}^{-1}x$, so that $x=(T+i)y$ with $y\in D(T)$; then $Cx=(T-i)y$, and the symmetry computation $\mathrm{\parallel }(T\mp i)y{\mathrm{\parallel }}^2=\mathrm{\parallel }Ty{\mathrm{\parallel }}^2+\mathrm{\parallel }y{\mathrm{\parallel }}^2$ of Resolvent of a self-adjoint operator: nonreal resolvents and the estimate gives $\mathrm{\parallel }Cx\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$.
2. $C$ is unitary, with $C^{-1}=C^{\ast }=(T+i)(T-i{)}^{-1}$. Using $C=I-2i(T+i{)}^{-1}$ and the adjoint rule $((T+i{)}^{-1}{)}^{\ast }=((T+i{)}^{\ast }{)}^{-1}=(T-i{)}^{-1}$ for the self-adjoint $T$ one gets $C^{\ast }=I+2i(T-i{)}^{-1}=(T+i)(T-i{)}^{-1}$; the elementary resolvent identity then gives $C^{\ast }C=C{C}^{\ast }=I$, as follows also from applying the computation of item 1 to $C$ and to $C^{\ast }$ and using that a surjective isometry of $H$ onto $H$ is unitary (A bounded linear operator between normed spaces). Concretely $C^{\ast }Cx=x$ for $x\in D(T)$ by direct substitution, and both sides are continuous.
3. $\ker (I-C)=\{0\}$, and $I-C=2i(T+i{)}^{-1}$, $I+C=2T(T+i{)}^{-1}$ as maps on $H$. Indeed $I-C=2i(T+i{)}^{-1}$ is injective with inverse $(2i{)}^{-1}(T+i)$, while $(I+C)(T+i)x=(T+i)x+(T-i)x=2Tx$ for $x\in D(T)$.
4. Domain recovery. $\mathrm{ran}(I-C)=D(T)$: by item 3, $\mathrm{ran}(I-C)=\mathrm{ran}(2i(T+i{)}^{-1})=D(T)$.