Let $A$ be a unital $C^\ast$ algebra. Let $\varphi$ be the Gelfand transform. Let $u$ be unitary and $f = \varphi (u)$. Let $\ln: \mathbb C \setminus [0,\infty) \to \mathbb C$ denote the natural logarithm (prinicpal branch).
(1) Then why $\ln \circ f$ is well defined? $f$ is a map $\Omega (A) \to \mathbb C$, obviously (in fact, an evaluation map for some $a \in A$). It's clear that $|f(\tau)|\le 1$. But why can't $f$ map into $[-1,0)$?
(2) Then the proof I'm reading states "Since $|f(\omega)|= 1$ for all $\omega \in \Omega$....". I simply don't see why this should be true. Since $\varphi$ is an isometry, $\|f\| = \|u\| = 1$ hence $\|f\|_\infty =1$ but it doesn't imply $|f(x)|=1$, so why should $|f(\omega)|= 1$?
Let me include the relevant text here:
Note that the Gelfand transform of $u$ is evalution at $u$, so for $\omega \in \Omega$, we have $f(\omega) = \omega(u)$. As $\sigma(u) = \{ \omega(u) \mid \omega \in \Omega\}$ and $u$ is unitary, we have ${\rm ran}\, f = \sigma(u) \subseteq S^1 - \{-1\}$.