# proof of conformal Möbius circle map theorem

Let $f$ be a conformal map from the unit disk $\Delta$ onto itself. Let $a=f(0)$. Let $g_{a}(z)=\frac{z-a}{1-\overline{a}z}$. Then $g_{a}\circ f$ is a conformal map from $\Delta$ onto itself, with $g_{a}\circ f(0)=0$. Therefore, by Schwarz’s Lemma for all $z\in\Delta$ $|g_{a}\circ f(z)|\leq|z|$.

Because $f$ is a conformal map onto $\Delta$, $f^{-1}$ is also a conformal map of $\Delta$ onto itself. $(g_{a}\circ f)^{-1}(0)=0$ so that by Schwarz’s Lemma $|(g_{a}\circ f)^{-1}(w)|\leq|w|$ for all $w\in\Delta$. Writing $w=g_{a}\circ f(z)$ this becomes $|z|\leq|g_{a}\circ f(z)|$.

Therefore, for all $z\in\Delta$ $|g_{a}\circ f(z)|=|z|$. By Schwarz’s Lemma, $g_{a}\circ f$ is a rotation. Write $g_{a}\circ f(z)=e^{i\theta}z$, or $f(z)=e^{i\theta}g_{a}^{-1}$.

Therefore, $f$ is a Möbius Transformation  .

Title proof of conformal Möbius circle map theorem ProofOfConformalMobiusCircleMapTheorem 2013-03-22 13:36:45 2013-03-22 13:36:45 brianbirgen (2180) brianbirgen (2180) 5 brianbirgen (2180) Proof msc 30E20 SchwarzLemma MobiusTransformation AutomorphismsOfUnitDisk