Iteration with two analytic fixed points

[JmsNxn]

Again bo, not to be the nitpicker that I am. As Milnor would describe it, a polynomial is a map from \(\widehat{\mathbb{C}} \to \widehat{\mathbb{C}}\). It is not a "euclidean mapping" as Milnor describes, because it's not a transcendental entire function.

But James, really, you making up a lot of artificial conditions. Needs to be euclidian, needs to have a vicinity independent of t, etc, etc.
This all is not needed!
Look at the Karlin&McGregor paper, the class of functions they describe contains all *meromorphic functions*.
They use standard regular iteration - the vicinity *depends* on t (and explicitly state that on the 4th line of this paper).
And from *there* they conclude that the only functions inside their class of functions that can have the same regular iterations at both fixed points can only be the linear fractional functions.

Yes, there was a misread here. I thought you were arguing we could do it for this polynomial, and I was trying to say you can't--and we agree. Also, I'm mostly unconcerned with polynomials, lol; and they behave essentially the same as rational functions; that's all I was getting at.

You are absolutely correct though, I think our wires just got crossed. We are on the same page. Exact same page.

The vicinity depends on \(t\) is built in to regular function as well; which is the old timey definition of what it means, I wasn't confused by that.

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So to keep it on the same page then; can we agree that if \(f^{\circ t}(z) : D \times H \to H\) for two domains \(D\subset \mathbb{C}\) and \(H\subset \mathbb{C}\)--where \(D\) is closed under addition (is a semigroup under \(\{+\}\)); such that:

\[
\begin{align}
f^{\circ t}(f^{\circ s}(z)) &= f^{\circ t+s}(z)\\
f^{\circ 1}(z) &= f(z)\\
\end{align}
\]

Then there can only be one point \(p \in H\) such that \(f(p) = p\).

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I just want to make sure we can agree on that. Which is essentially my statement that there can't be a local iteration about two fixed points. Which has started this whole spiral. (And come to think of it, I think I learned this by way of Sheldon, but I never bothered to look into the proof, as it was just "apparent" to me as to how Schroder iterations behave. We'll have a branching problem of some kind at the other fixed point.)