Iteration with two analytic fixed points
#12
Oh yes, I'm well aware that I didn't even use the iteration. I'm trying to find the more advanced proof that I had before, but I remembered this one quickly to get started. And also, I apologize for the simply connected domain part, I mean simply connected and biholomorphic to \(\mathbb{D}\). There are only two kinds of simple connectivity, all of \(\mathbb{C}\) or \(\mathbb{D}\). So that works.

For the case \(b^z\), just note that no simply connected domain (bi holomorphic to \(\mathbb{D}\)) can have a repelling fixed point. So by nature, there's no simply connected domain that has any repelling fixed points. This part was crucial to my argument, and I'm trying to remember all the details.

The second part of the proof (which I haven't written yet), is a bit more complicated, and I'm working on remembering how it goes. The iteration part does come into play. I'll see if I can remember. But essentially the rest of the argument uses that if \(f^{\circ t}(z) : D \times A \to A\) where \(A\) is the immediate basin about the fixed point, then that is its maximal domain. This stuff is buried in old notebooks though, so I'm racking my brain trying to remember how to derive it. But it isn't very far off from Milnor, where the Schroder coordinate change has a maximal domain of convergence.
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RE: Iteration with two analytic fixed points - by JmsNxn - 08/04/2022, 08:54 PM

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