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My question is rather simple and I hope someone has some sort of an answer. I am looking for a simple yes or no answer, and a reference if anyone has one.

We have a holomorphic function $f$ defined on some infinite subset of $\mathbb{C}$ and sends to this set. If $f(z_0) = z_0$ and $|f'(z_0)| > 1$ does the following limit construct the Koenigs function $\Psi$? Such that $\Psi(f(z)) = f'(z_0) \Psi(z)$ for $z$ in a sufficiently small enough neighborhood of $z_0$.

$$\lim_{n\to\infty} \frac{f^{\circ n}(z) - z_0}{f'(z_0)^n}$$

I'm well aware for the attracting case this is true (when $0 < |f'(z_0)| < 1$), but I am unsure of the repelling case. I have seen it mentioned that this still holds, but I don't trust the source.

Perhaps someone has a proof this happens, or can link to a proof. Or simply give me a reason this doesn't happen. I'm at my wits end on how to prove or disprove this.

Thanks a whole bunch. I hope somebody can straighten this out for me.

Myshkin
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  • Holomorphic functions are defined on open sets. Is your "infinite set" open? Or $f$ is defined in a neighborhood of your set, and sends only the set into itself (not the whole neighborhood)? – Alexandre Eremenko Aug 17 '15 at 02:48
  • Also, for most open sets, there is no holomorphic function with a repelling fixed point inside, by the Schwarz lemma. Can you give an example of a function and a set for which you would like to prove this? – Alexandre Eremenko Aug 17 '15 at 02:51
  • Thank you thank you, these are all things that I know. I am aware on simply connected and sufficiently well behaved sets there is no repelling fixed point. My infinite set is open, it is the immediate basin about a fixed point of a different entire function $\phi$--which happens to have an infinite basin. So that $f : I \to I$, fixes $\xi_0$, and $I$ is an immediate basin of attraction that contains infinity of a different entire function $\phi$ that also fixes $\xi_0$. –  Aug 17 '15 at 16:44
  • @AlexandreEremenko, see the same question and my brief answer at http://math.stackexchange.com/questions/1397468/constructing-the-koenigs-function-about-a-repelling-fixed-point/1397472#1397472 The chapter in Milnor, in the third edition, is chapter 10, Parabolic Fixed Points: The Leau-Fatou Flower. However, the condition here that $|f'(x_0)| > 1$ simplifies matters, we just take a local inverse function, and we do not need the full Ecalle machinery. I'm quite proud of figuring out how to do that, see: http://mathoverflow.net/questions/45608/formal-power-series-convergence/46765#46765 – Will Jagy Aug 18 '15 at 17:05
  • @Wil Jagy: I agree with your answer there (45608) that "all you can do is to to consider the inverse function". But I do not understand your present question, until you state more specifically what kind of "infinite subset" you mean. – Alexandre Eremenko Aug 18 '15 at 23:05
  • @AlexandreEremenko, I see, I think you are referring to the language "infinite subset" used in the question above by james.nixon. I don't know what that means either. – Will Jagy Aug 18 '15 at 23:53
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    @Will Jagy: I apologize for confusion. I thought I am answering to the person who asked the question. – Alexandre Eremenko Aug 19 '15 at 03:14
  • @james.nixon This is absolutely standard - although as Alex points out, some of the hypotheses in your question are rather odd. (If your function sends an open subset of the plane to itself that omits more than two points, then it cannot have a repelling fixed point in that set.) Since the construction of the Koenigs function is a local one, it does not matter whether you consider attracting or repelling points; one is obtained from the other by taking inverses. Of course, the formula for the limit is exactly the one you would expect. – Lasse Rempe Aug 21 '15 at 21:42
  • @LasseRempe-Gillen So you are saying under strict enough conditions on the set that contains infinity and how $\phi$ behaves on it this limit will construct the Koenigs function? –  Aug 28 '15 at 19:59
  • @james.nixon I do not understand this question. What I am saying is that, if a function is holomorphic near a repelling fixed point $z_0$, then its linearising function can be constructed by iterating backwards $n$ times (using the branch fixing $z_0$), multiply by the $n$-th power of the multiplier, and take the limit as $n\to\infty$. – Lasse Rempe Sep 01 '15 at 08:38
  • The limit you write in the question doesn't make any sense since the function $f^{\circ n}$ need not even be defined on a neighbourhood of the orbit. Even if it was, the limit would not exist, much less give the Koenigs function, for any interesting map. – Lasse Rempe Sep 01 '15 at 08:41

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