# Cartan Theorem Note 2

As we can see from Cartan Theorem can be proved in $\mathbb{C}^1$ case by using Schwarz Lemma and Riemann Mapping Theorem. However, this method cannot be applied to $\mathbb{C}^n$.

We shall prove it in a technical way.

Proof:

Because $\Omega$ is a bounded domain, then there exists $0\textless r\textless R\textless\infty$, such that:

$B(0,r)\subset\Omega\subset B(0,R)$

then we have the expansion of $f$ as:

$f(z)=z+a_2 z^2+a_3 z^3+\dots$

assume that the first nonzero $a_k$ is from $a_m$, then

$f(z)=z+\sum_{s=m}^{\infty}a_s z^s$

where $s\ge 2$.

Consider $F(z)=f\circ f$, then

$F(z)=f(f(z))=f(z)+\sum_{s=m}^{\infty}a_s (f(z))^s$

$=z+\sum_{s=m}^{\infty} a_s z^s+\sum_{s=m}^{\infty} a_s (z+\sum_{j=m}^{\infty}a_s z^j)^s$

By induction,

$G_k(z)=f^{(k)}(z)=f\circ f\circ f \cdots \circ f(z)=z+k a_m z^m+\dots$, then

$G_k(e^{i\theta}z)=e^{i\theta}z+k a_m e^{im\theta} z^m+\dots$,

multiply with $e^{-im\theta}$ and integrate it.

$\displaystyle\frac{1}{2\pi}\int_0^{2\pi}G_k(e^{i\theta}z)e^{-im\theta}\mathrm{d}\theta=k a_m$.

Since $f:\Omega\rightarrow \Omega$, then $G_k:\Omega\rightarrow\Omega$,

thus $|G_k(e^{i\theta}z)|\le R$, for any $k\in \mathbb{N}$.

where $\theta\in (0,2\pi)$, $z\in B(0,r)$.

thus $|k a_m|\le R$, i.e. $\displaystyle|a_m|\le\frac{k}{R}$, which means $a_m=0$.

Therefore $a_s=0$ for all $s\ge 2$, and $F(z)=z$.

However this is still for $\mathbb{C}^1$ case, for higher dimensional cases, we will talk over it later.

Remark:

In last post for Cartan Theorem, we required that the domain should be connected, while for this post, we do not have to make this assumption.