Consider the following fragment from Takesaki's book "Theory of operator algebra I" (p82 and previous pages):
The notation $\mathscr{L}_G$ means all normal operators with spectrum contained in $G$ and similarly $\mathscr{L}_{\mathbb{C}}$ denotes the normal operators.
Why is the boxed equality true? The right hand side is the functional calculus on two elements.
I suppose it makes intuitive sense. It reminds me of the fact that composition respects classical functional calculus but I can't justify it in this case. In any case, I also think it is relevant that $u(a), v(a) \ne 1$. Can anybody formally justify why the boxed equality is true and resolve this technicality?
Best Answer
Let $u: \mathbb{C} \setminus \{-i\} \to \mathbb{C} \setminus \{1\}$ denote the fractional linear transformation $$ u(z) = \frac{z-i}{z+i}, \qquad z \neq -i. $$ It can be verified that $u$ maps the real axis to bijectively to the unit circle with $1$ excluded, the upper half plane bijectively to the interior of the unit disc, and the lower half plane with $-i$ excluded bijectively to the exterior of the unit disc. (It follows that for any self-adjoint operator $s$, the operator $u(s)$ is unitary and does not have $1$ in its spectrum. With some work, this could be verified directly from a definition of $u(s)$ as $(s-i \mathbf{1})(s+i \mathbf{1})^{-1}$ if one does not want to appeal to general facts about the functional calculus.)
A short computation shows that $u^{-1}: \mathbb{C} \setminus \{1\}$ to $\mathbb{C} \setminus \{-i\}$ is given by $$ u^{-1}(z) = \frac{z+1}{i(z-1)}, \qquad z \neq 1. $$ Note also that $\frac{w+1}{w-1} = i u^{-1}(w)$ for all $w \neq 1$.
Turning to your problem, the operators denoted by Takesaki as "$u(a)$" and "$v(a)$" respectively are literally $u(h)$ and $u(k)$ in the sense of the above definition of the function $u$ and the continuous functional calculus. Because $u^{-1}$ is given by the formula above, it follows that $(u(h)+\mathbf{1})(i(u(h)-\mathbf{1}))^{-1} = h$ and $(u(k)+\mathbf{1})(u(k)-\mathbf{1})^{-1} = ik$, again where all of this is interpreted in the sense of the continuous functional calculus.
It follows that the operator denoted by Takesaki by $g$ applied to the pair "$u(a), v(a)$", which is literally $g$ applied to the pair $u(h), u(k)$, is (from the definition of $g$) the function $f$ applied to $u^{-1}(u(h)) + i u^{-1}(u(k)) = h + ik = a$, in other words, $f(a)$.
Side note, the operator $u(s)$ is sometimes called the Cayley transform of the self adjoint operator $s$; von Neumann used the Cayley transform and knowledge of the spectral resolution of unitary operators to deduce the spectral resolutions of self adjoint operators. See, e.g., https://en.wikipedia.org/wiki/Cayley_transform