Why is $F_\phi$ defined on the whole disk

Question

This is a question about a proof on page 97 in these lecture notes.

In exercise 13, I don't understand

On the hand, $F_\phi$ is defined on the whole open disk $D$

Why is $F_\phi$ defined on all of $D$?

Here is the exercise this question is referring to:

Answer 1

By definition, $F_\phi = \phi\circ F$. The linear map $\phi$ is defined on all of $\mathcal{A}$, so $F_\phi = \phi\circ F$ is defined wherever $F$ is defined.

So where is $F$ defined? By definition $F(\zeta) = (1 - \zeta a)^{-1}$, and this is only defined where $1 - \zeta a$ is invertible. If $\zeta = 0$, you have $1 - \zeta a = 1$ is certainly invertible. On the other hand, if $\zeta\neq 0$, then $1 -\zeta a$ is invertible $\iff \zeta^{-1} - a$ is invertible. By the definition of the spectral radius of $a$, we have that if $|\zeta^{-1}| > \mathrm{rad}_\mathcal{A}(a) = R$, then $\zeta^{-1} - a$ is invertible. Thus we have derived:

$1 - \zeta a$ is invertible if $|\zeta^{-1}|>R$, or equivalently, if $|\zeta|<1/R$. In particular, the map $F(\zeta) = (1 - \zeta a)^{-1}$ is defined on the disk $D = \{|\zeta|<1/R\}$.