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Question : Prove by use of Schwarz's lemma that every one-to-one conformal mapping of a disc onto another (or a half plane) is given by a linear fractional transformation.

I have known that there exists LFT such that it maps unit disc onto itself, but if holomorphic function $f$ is a 1-1 mapping of a disc onto another disc, can we conclude $f$ is LFT?

My try is simplifying the question as

Every one-to-one conformal mapping of a unit disc onto itself is given by a linear fractional transformation.

Am I right? Sincerely thanks for your help!

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    $\begingroup$ The Möbius transformations form a group (with composition as multiplication). Since you can map any disk to the unit disk using a Möbius transformation, your simplification of the problem is valid. $\endgroup$ – Daniel Fischer Jan 12 '18 at 15:53
  • $\begingroup$ @DanielFischer,I got it. Plus, my question is on the top, could you help me with it? Thanks! $\endgroup$ – Ferry Tau Jan 13 '18 at 15:18
  • $\begingroup$ Do you see that (and why) it suffices to show that every automorphism of the unit disk is a Möbius transformation? $\endgroup$ – Daniel Fischer Jan 13 '18 at 15:20
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Consider first the case of a biholomorphic $f \colon \mathbb{D} \to \mathbb{D}$, where $\mathbb{D}$ is the unit disk.

If we define $a = f(0)$, then the Möbius transformation

$$T_a \colon z \mapsto \frac{z - a}{1 - \overline{a}\,z}$$

is an automorphism of the unit disk, and hence $g = T_a \circ f$ is an automorphism of the unit disk that fixes $0$.

Lemma (H. A. Schwarz): Let $h \colon \mathbb{D} \to \mathbb{D}$ a holomorphic function with $h(0) = 0$. Then $\lvert h(z)\rvert \leqslant \lvert z\rvert$ for all $z \in \mathbb{D}$, and $\lvert h'(0)\rvert \leqslant 1$. If there is a $z \in \mathbb{D}\setminus \{0\}$ with $\lvert h(z)\rvert = \lvert z\rvert$ or $\lvert h'(0)\rvert = 1$, then $h(z) = cz$ for some $c$ with $\lvert c\rvert = 1$.

Since $g$ and $g^{-1}$ satisfy the assumptions of the lemma, we conclude first

$$\lvert z\rvert = \lvert g^{-1}(g(z))\rvert \leqslant \lvert g(z)\rvert \leqslant \lvert z\rvert$$

for all $z \in \mathbb{D}$. Thus the last sentence of the lemma applies and hence $g(z) = cz$, so $g$ is a Möbius transformation. But then

$$f = T_a^{-1}\circ g$$

is also a Möbius transformation since the family of Möbius transformations is closed under composition and inverses.

Now, if $U,V$ are arbitrary disks or half-planes, and $f \colon U \to V$ is biholomorphic, consider the map $g = T \circ f \circ S$, where $S$ is a Möbius transformation mapping $\mathbb{D}$ biholomorphically to $U$, and $T$ is a Möbius transformation mapping $V$ biholomorphically to $\mathbb{D}$. Then $g$ is an automorphism of the unit disk. By the first part, $g$ is a Möbius transformation, and consequently $f = T^{-1}\circ g \circ S^{-1}$ is also a Möbius transformation.

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