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.


All you can do in the repelling case is take the inverse of the function. You may not be able to write down the inverse in any sensible closed form, but there it is. In any case you get the reciprocal of the derivative and are in good shape.

I recommend D. S. Alexander, A History of Complex Dynamics, John Milnor, Dynamics in One Complex Variable (Third Edition), Kuczma, Choczewski, Ger, Iterative Functional Equations.

My best answer on this used the original function, a later one used the inverse, but it was a quadratic function so no trouble.

Oh, if the absolute value of the derivative is not $1,$ no troubles.

  • $\begingroup$ Hey, first of all thank you for your answer. I have read John Milnor's book and I know of using the inverse function to solve the koenig functions. So you're saying there's no way. Thanks a lot. Going to have to rework a proof because of this, oh well. $\endgroup$ – JmsNxn Aug 14 '15 at 22:16

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