Cayley transform is well-defined

Consider a Hilbert space $H$ and write $\mathscr{L}_h$ for the set of self-adjoint operators on $H$. The Cayley-transform is defined to be the map $$\mathscr{L}_h \to U(H): x \mapsto (x-i)(x+i)^{-1}$$

I want to show that this map is well-defined, i.e. that $x+i$ is invertible in $B(H)$ and that $(x-i)(x+i)^{-1}$ is a unitary.

I proceeded as follows:

The $C^*$-algebra $C^*(x,1)$ generated by $x$ and $1$ is abelian and unital, so it is $*$-isomorphic to $C(X)$ for some compact Hausdorff space $X$. Then the statement becomes trivial because the function $x + i$ does not take the value $0$ (because $x$ is real-valued) and the calculation $$\overline{\left(\frac{x-i}{x+i}\right)} \frac{x-i}{x+i}= \frac{x^2+1}{x^2+1}=1$$ shows that $(x-i)(x+i)^{-1}$ is unitary in $C(X)$ and thus also in $C^*(x,1)$ and thus also in $B(H)$, as desired.

Would this be a correct way to show this? It is just a quick sanity check.

Solution 1:

Your argument is fine. But you don't really need functional calculus here.

Since $x$ is selfadjoint, $\sigma(x)\subset\mathbb R$. Then $\sigma(x+i)\subset\mathbb R+i$. In particular, $0\not\in\sigma(x+i)$ and so $x+i$ is invertible. Same argument works for $x-i$.

Because $x+i$ commutes with $x-i$, so does $(x+i)^{-1}$. Then $$ [(x-i)(x+i)^{-1}]^*(x-i)(x+i)^{-1}=(x+i)(x-i)^{-1}(x-i)(x+i)^{-1}=(x+i)(x+i)^{-1}=1. $$