Just to go deeper into this, we can be more rigorous.
Let's say that \(f(z) : \mathbb{C}_{\Re(z) >0} \to \mathbb{C}_{\Re(z) > 0}\), while additionally \(|f(z)| < M\) is bounded by some constant \(M\). Let's say that this a space of functions.
Now the solution:
\[
f^{\circ s}(1)\\
\]
Necessarily exists; since \(f\) has an attracting fixed point; and all orbits tend to this fixed point. If \(f\) takes a simply connected domain to itself, it has one attracting fixed-point. We must additionally assume here, that \(f'(z) \neq 0\), which is totally fine for hyper operators.
If I write \(f+\epsilon\), then this equally is a bounded function within the above space. So it has a fixed point, which is attracting, and also, \((f+\epsilon)^{\circ s}(1)\), is a holomorphic function. This function, \(F[f](s) : \mathbb{C}_{\Re(s) >0} \to \mathbb{C}_{\Re(s)>0}\). Additionally, this solution is just as bounded, \(< M\). Therein:
\[
\frac{(f+\epsilon)^{\circ s}(1) - f^{\circ s}(1)}{\epsilon} \to G[f](s)\\
\]
And this statement is entirely rigorous. And \(F[f]\) is analytic in \(f\).
Let's say that \(f(z) : \mathbb{C}_{\Re(z) >0} \to \mathbb{C}_{\Re(z) > 0}\), while additionally \(|f(z)| < M\) is bounded by some constant \(M\). Let's say that this a space of functions.
Now the solution:
\[
f^{\circ s}(1)\\
\]
Necessarily exists; since \(f\) has an attracting fixed point; and all orbits tend to this fixed point. If \(f\) takes a simply connected domain to itself, it has one attracting fixed-point. We must additionally assume here, that \(f'(z) \neq 0\), which is totally fine for hyper operators.
If I write \(f+\epsilon\), then this equally is a bounded function within the above space. So it has a fixed point, which is attracting, and also, \((f+\epsilon)^{\circ s}(1)\), is a holomorphic function. This function, \(F[f](s) : \mathbb{C}_{\Re(s) >0} \to \mathbb{C}_{\Re(s)>0}\). Additionally, this solution is just as bounded, \(< M\). Therein:
\[
\frac{(f+\epsilon)^{\circ s}(1) - f^{\circ s}(1)}{\epsilon} \to G[f](s)\\
\]
And this statement is entirely rigorous. And \(F[f]\) is analytic in \(f\).