Koenigs function

Remarks

The limit exists, uniformly near $0$, and $\varphi$ is holomorphic with $\varphi(0) = 0$ and $\varphi'(0) = 1$. It solves Schröder's equation $\varphi \circ f = \lambda \varphi$, since replacing $z$ by $f(z)$ shifts the index $n$ by one. It is the unique solution with these normalisations, by the Koenigs linearization theorem.

Example: $f(z) = z / (2 - z)$ on the unit disc, with $\lambda = f'(0) = 1/2$. For $\varphi(z) = z / (1 - z)$ one has $\varphi(f(z)) = \frac{z/(2-z)}{(2 - 2z)/(2 - z)} = \frac{z}{2(1 - z)} = \frac{1}{2} \varphi(z)$. So $\varphi(f^n(z)) = 2^{-n} \varphi(z)$, and since $f^n(z) \to 0$ and $\varphi(w) / w \to 1$ as $w \to 0$, the limit of $2^n f^n(z)$ is $\varphi(z)$.

Non-example: for $f(z) = z^2$ one has $\lambda = 0$, so the quotient $f^n(z) / \lambda^n$ is not defined, and $f$ has no Koenigs function.

When $f$ is a univalent holomorphic map of the unit disc into itself with $f(0) = 0$ and $0 < \abs{f'(0)} < 1$, the limit exists on the whole disc and $\varphi$ is univalent there.