I was trying to understand Koenigs Linearization Theorem given in J.Milnor (2006) Dynamics in One Complex Variable: Introductory Lectures and I got stuck in the following polynomial property:

Consider the holomorphic function $f(z) = \lambda z + a_2z^2 + a_3z^3 + \dots$. Then, it has been proven that $\left(\frac{f^n(z)}{\lambda^n}\right)_n$ converges uniformly to $\phi$ (also holomporhic) in some neighbourhood of $z=0$ denoted by U. Here is where it arises my doubt, so I will write precisely what is in the book:

$|\phi(z) - z|$ is less than or equal to some constant times $|z^2|$ . Therefore $\phi$ has derivative $\phi'(0) = 1$.

Questions (the main question is (b)):

(a) If I'm not wrong, the first statement is proven by the Taylor's Theorem. Not sure about it.

(b) What is the general property that implies the second statement from the first one?

Thanks in advance.


1 Answer 1


With regard to (b): $|\phi(z) - z| < C |z|^2$ in a neighborhood of $z=0$ implies $\phi(0) = 0$. For $z\ne 0$ $$ \left|\frac{\phi(z)}{z} - 1 \right| < C |z| $$ then implies $$ \phi'(z) = \lim_{z\to 0} \frac{\phi(z)}{z} = 1 \, . $$

  • $\begingroup$ Thank you for your help Martin. Could you help me to formalize the computation of the constant $C$? My problem is that I try to find it using Taylor's Theorem in $|f^n(z)/\lambda^n - z|$ ($\leq C_n|z^2|$) and then calculate the limit both sides, but my $C_n$ depends on $n$ and $z$... $\endgroup$
    – dudas
    Commented Feb 23, 2018 at 14:26
  • $\begingroup$ @dudas: Unfortunately I do not have a answer to that part at present. $\endgroup$
    – Martin R
    Commented Feb 24, 2018 at 17:28
  • $\begingroup$ Don't worry. Thank you anyway! $\endgroup$
    – dudas
    Commented Feb 24, 2018 at 23:54

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