Cartan Theorem

Under \mathbb{C}^1 case:

If \Omega is a bounded domain which contains origin,

If F:\Omega \rightarrow \Omega is analytic with F(0)=0, F'(0)=1,

then F(z)=z for any z\in \Omega.

Try to solve it by Riemann Mapping Theorem.

Scheme of Proof:

Think about that if \Omega is a disk. Then use Schwarz Lemma.

Riemann Mapping Theorem:

If G is simple connected domain of \mathbb{C}, and G\neq \mathbb{C}. Then for any point in G, there exists unique function f:G\rightarrow \mathbb{C}, such that:

  1. f is holomorphic and univalent in G .
  2. f(a)=0, f'(a)\textgreater 0.
  3. f(G)=B(0,1).

Thus, if \Omega=B(0,1), then F(0)=0, by Schwarz Lemma, |F(z)|\le |z| with |F'(0)|\le 1. The equality holds for F(z)=e^{i\theta}z.

By Riemann Mapping Theorem:

There exists a unique holomorphic and univalent function \phi:\Omega\rightarrow B(0,1), such that \phi(0)=0,\phi'(0)\textgreater 0.

Set h=\phi\circ F\circ \phi^{-1}, then h is holomorphic in B(0,1) and satisfies that h(0)=0, h'(0)=1, by Schwarz Lemma:

h:B(0,1)\rightarrow B(0,1), and h(z)=z.

Thus consider that \phi is univalent, F(\phi^{-1}(z))=\phi^{-1}(z), for any z\in B(0,1), therefore F(z)=z, for any z\in \Omega.

About YiMin

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