Okay, YES! I think I got it now. I only have part of it though, but the proof went something like this. This is only for the Schroder case, I can't remember what to do for the parabolic case--but that should take care of itself--if we stick to local iteration. There exists no local iteration about a neutral fixed points.
Assume that \(f\) is holomorphic on a domain \(H\) (not simply connected), and let's assume there are two fixed points \(p_0,p_1 \in H\), such that \(p_0\) is geometrically attracting. Let us additionally assume that \(f^{\circ t}(z)\) is holomorphic on all of \(H\), including the fixed point. Define \(\lambda = f'(p_0)\). We can define the domains:
\[
B(z) = \{q \in H\,|\,|\lambda^t| < 1,\,q = f^{\circ t}(z),\,\text{for some}\,z \in H\}\\
\]
This is a simply connected domain. And additionally, as \(t \to \infty\), we converge to \(p_0\), by the nature of the Schroder iteration. Now, we know that \(f:B(z) \to B(z)\) because \(f\) is a contraction mapping here. Therefore, for all \(z \in H\) and \(z \neq p_1\), \(p_1 \not\in B(z)\)--because it would be a simply connected domain with two fixed points. But the union of all \(B(z)\) is precisely \(H\), so \(f^{\circ t}(z)\) can't be holomorphic about all of \(H\).
Now, particularly, the assumption that might be troubling you is that we converge to \(p_0\) as \(t \to \infty\)--what if it converges somewhere else?
Well to play this out, we have to allow \(z \to J\) where \(J\) is the julia set of \(f\). The schroder coordinate change is no longer valid here, and if we could extend this value here, we've incidentally analytically continued \(\Psi^{-1}\) to a domain larger than its maximal domain; which is erroneous to say the least. I'm going to think about this more though, to further justify it for you. There should be something technical which will save me
I'll keep working on polishing this, I know it's something close to this.
And again, this says nothing about your iteration, because \(B(z)\) is no longer a domain in your case (there are poles).
Also! for a bit more intuition. Every regular iteration that is a local iteration that is a schroder iteration is periodic. So say it has period \(\ell\), then the arc \(\gamma\) connecting \(\ell\) and \(0\), induces a simply connected domain through the jordan curve \(\mu = f^{\gamma}\) about any \(z\), so there can only be one fixed point contained within the center of this jordan curve. I can't see how we can move \(z\) to include neither fixed points/the other fixed points... I'll have to think about this more.
Assume that \(f\) is holomorphic on a domain \(H\) (not simply connected), and let's assume there are two fixed points \(p_0,p_1 \in H\), such that \(p_0\) is geometrically attracting. Let us additionally assume that \(f^{\circ t}(z)\) is holomorphic on all of \(H\), including the fixed point. Define \(\lambda = f'(p_0)\). We can define the domains:
\[
B(z) = \{q \in H\,|\,|\lambda^t| < 1,\,q = f^{\circ t}(z),\,\text{for some}\,z \in H\}\\
\]
This is a simply connected domain. And additionally, as \(t \to \infty\), we converge to \(p_0\), by the nature of the Schroder iteration. Now, we know that \(f:B(z) \to B(z)\) because \(f\) is a contraction mapping here. Therefore, for all \(z \in H\) and \(z \neq p_1\), \(p_1 \not\in B(z)\)--because it would be a simply connected domain with two fixed points. But the union of all \(B(z)\) is precisely \(H\), so \(f^{\circ t}(z)\) can't be holomorphic about all of \(H\).
Now, particularly, the assumption that might be troubling you is that we converge to \(p_0\) as \(t \to \infty\)--what if it converges somewhere else?
Well to play this out, we have to allow \(z \to J\) where \(J\) is the julia set of \(f\). The schroder coordinate change is no longer valid here, and if we could extend this value here, we've incidentally analytically continued \(\Psi^{-1}\) to a domain larger than its maximal domain; which is erroneous to say the least. I'm going to think about this more though, to further justify it for you. There should be something technical which will save me
I'll keep working on polishing this, I know it's something close to this.
And again, this says nothing about your iteration, because \(B(z)\) is no longer a domain in your case (there are poles).
Also! for a bit more intuition. Every regular iteration that is a local iteration that is a schroder iteration is periodic. So say it has period \(\ell\), then the arc \(\gamma\) connecting \(\ell\) and \(0\), induces a simply connected domain through the jordan curve \(\mu = f^{\gamma}\) about any \(z\), so there can only be one fixed point contained within the center of this jordan curve. I can't see how we can move \(z\) to include neither fixed points/the other fixed points... I'll have to think about this more.
