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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.

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  • $\begingroup$ 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)? $\endgroup$
    – Alexandre Eremenko
    Commented Aug 17, 2015 at 2:48
  • $\begingroup$ 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? $\endgroup$
    – Alexandre Eremenko
    Commented Aug 17, 2015 at 2:51
  • $\begingroup$ 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 $\xi_0$. $\endgroup$
    – user78249
    Commented Aug 17, 2015 at 16:44
  • $\begingroup$ @AlexandreEremenko, see the same question and my brief answer at math.stackexchange.com/questions/1397468/… 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: mathoverflow.net/questions/45608/… $\endgroup$
    – Will Jagy
    Commented Aug 18, 2015 at 17:05
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    $\begingroup$ @Will Jagy: I apologize for confusion. I thought I am answering to the person who asked the question. $\endgroup$
    – Alexandre Eremenko
    Commented Aug 19, 2015 at 3:14

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