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.

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About YiMin

This is just a nerd PhD student of Math@UT Austin.