Lemma 5.1.1
Suppose that \(z_0\) is an attractive fixed point of \(f\text{.}\) Then, there is an \(r > 0\) such that \(f^n\) converges uniformly to the constant function \(z\to z_0\) on the disk \(D_r(z_0)\text{.}\)
Section 5.1 Linearization near an attractive fixed point
The simplest type of periodic orbit is that of an attractive fixed point. That is, we have a point \(z_0\in\mathbb C\) and an analytic function \(f\)defined in a neighborhood of \(z_0\) such that \(f(z_0)=z_0\) and \(0 < |f'(z_0)| < 1\text{.}\) In this case, the function looks like the linear function \(L(z)=\lambda z\text{,}\) where \(\lambda = f'(z_0)\text{.}\) A simpler statement, whose proof follows immediately from the definition of the derivative is that \(f\) is contractive near \(z_0\)
Proof
Choose an \(\alpha\) such that \(|f'(z_0)| < \alpha < 1\) and an \(r > 0\) such that
\begin{equation*}
\left|\frac{f(z)-f(z_0)}{z-z_0}\right| < \alpha.
\end{equation*}
Then, since \(f(z_0)=z_0\text{,}\) we have
\begin{equation*}
\left|f(z)-z_0\right| < \alpha |z-z_0|
\end{equation*}
and, by induction,
\begin{equation*}
\left|f^n(z)-z_0\right| < \alpha^n |z-z_0|.
\end{equation*}
The result follows as the statment is for all \(z\in D_r(z_0)\text{.}\)
While nice, we can make a much more precise statement. In particular, \(f\) is analytically conjugate to a linear function \(L\text{.}\) Note that the following theorem Theorem 5.1.2 is stated assuming that the fixed point is zero. The fact that this extends to any fixed point is essentially the content of exercise Exercise 5.6.1.
Theorem 5.1.2
Suppose that \(f\) is analytic at zero with \(f(0)=0\) and \(f'(0)=\lambda\text{,}\) where \(0 < |\lambda| < 1\text{.}\) Then, there is a neighborhood \(U\) of zero and an analytic function \(\varphi :U\to \mathbb{C}\) that conjugates \(L(z)=\lambda z\) to \(f\text{,}\) i.e.
\begin{equation*}
L\circ\varphi(z) = \varphi\circ f(z).
\end{equation*}
This is a tremendously important theorem with several well known proofs. We present two.