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.

Advertisements

DOODLE SOMETH

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s