# proof of Marty’s theorem

(i) Fix $K\subseteq\Omega$ compact. We have:

 $\displaystyle\frac{2|f^{\prime}(z)|}{1+|f(z)|^{2}}$ $\displaystyle\leq M_{K}\ \forall\ f\in\mathcal{F},z\in K$ ($*$)

Let $V$ be a region with $K=\overline{V}$ and let $\gamma\colon[a,b]\to V$ be the $C^{1}$ curves connecting the points $P,Q\in\Omega$. Then we have:

 $\displaystyle d_{\sigma}(f(P),f(Q))$ $\displaystyle=\inf_{\gamma}l_{\sigma}(f\circ\gamma)=\inf_{\gamma}\int_{a}^{b}% \|(f\circ\gamma)^{\prime}(t)\|_{\sigma,f\circ\gamma(t)}\,dt$ $\displaystyle=\inf_{\gamma}\int_{a}^{b}\frac{2|f^{\prime}(\gamma(t))|}{1+|f(% \gamma(t))|^{2}}|\gamma^{\prime}(t)|\,dt$ $\displaystyle\lx@stackrel{{\scriptstyle\leq}}{{\begin{subarray}{c}\eqref{1}% \end{subarray}}}M_{K}\inf_{\gamma}\int_{a}^{b}|\gamma^{\prime}(t)|\,dt$ $\displaystyle=M_{K}\inf_{\gamma}l(\gamma)=M_{K}|P-Q|$

Thus $f$ is Lipschitz continuous and thus $\mathcal{F}$ is equicontinuous. By the Ascoli-ArzelÃÂ  Theorem we conclude that $\mathcal{F}$ is normal.

(ii) Now assume $\mathcal{F}$ to be normal. Define:

 $\displaystyle f^{\sharp}(z)$ $\displaystyle:=\frac{2|f^{\prime}(z)|}{1+|f(z)|^{2}}$

Let $K\subseteq\Omega$ be compact. To obtain contradiction assume $\{f^{\sharp}:f\in\mathcal{F}\}$ is not uniformly bounded on $K$. But then there exists a sequence $\{f_{n}\}\subset\mathcal{F}$ such that:

 $\displaystyle\max_{z\in K}f_{n}^{\sharp}(z)$ $\displaystyle\to\infty\ (n\to\infty)$

Since $\mathcal{F}$ is normal for each $P\in\Omega$ let there be a neighbourhood $U_{P}$ such that $\{f_{n}\}$ converges normally to a meromorphic function $f$. But from $(1/f_{n})^{\sharp}=f_{n}^{\sharp}$ we see that $\{f_{n}^{\sharp}\}$ converges normally on $U_{P}$. Since $K$ is compact it can be covered by finitely many sets $U_{P}$. We conclude that $\{f_{n}^{\sharp}\}$ must be bounded on $K$ and obtain a contradiction. ∎

Title proof of Marty’s theorem ProofOfMartysTheorem 2013-03-22 18:23:11 2013-03-22 18:23:11 karstenb (16623) karstenb (16623) 4 karstenb (16623) Proof msc 30D30