Schwarz Lemma and the Space of All Automorphisms on the Unit Disk

Definition: Let $\Omega$ be a non-empty open set. Any biholomorphism $f: \Omega \to \Omega$ is called an automorphism on $\Omega$ . The space of automorphisms on $\Omega$ is denoted by $\text{Aut}(\Omega)$ .

Remark: Equipped with the operation of function composition, $\text{Aut}(\Omega)$ is a group.

In this post, I would like to study the space $\text{Aut}(\mathbb{D})$ , where $\mathbb{D}$ is the open unit disk. Interesting, it turns out that $\text{Aut}(\mathbb{D})$ only consists of very restricted types of functions. To study this space, we need the following theorem (whose name contains the word lemma).

Theorem: (Schwarz Lemma) Let $f: \mathbb{D} \rightarrow \mathbb{D}$ be a holomorphic function fixing the origin, i.e. $f(0)=0$ . Then
(1) for any $z \in \mathbb{D}$ , we have $|f(z)| \leq |z|$
(2) $|f'(0)| \leq 1$
(3) If there exists some $z_0 \neq 0$ such that such that $|f(z_0)|=|z_0|$ or $|f'(0)|=1$ , then $f$ is a rotation, i.e. there exists some $\theta \in \mathbb{R}$ such that $f(z)=e^{i\theta}z$ .

Proof: Consider the function $g$ defined $g(z)=\frac{f(z)}{z}$ if $z \neq 0$ and $g(0)=f'(0)$ . It is easy to check that it is a holomorphic function on $\mathbb{D}$ . For any $1>r>0$ , on $|z|=r$ , we have

$\displaystyle |g(z)|=\frac{|f(z)|}{r} \leq \frac{1}{r}$

Then by maximum modulus principle, if $z \in \mathbb{D}$ , then for any $1>r>|z|$ , we have

$\displaystyle |g(z)|=\frac{|f(z)|}{r} \leq \frac{1}{r} \rightarrow 1$

as $r \to 1^-$ . Then we got (1) and (2) in one shot.

If equality holds as mentioned in (3), then there is a point in which the maximum modulus of $g$ is attained. By maximum modulus principle, $g$ must be a constant, and that constant must have a modulus of $1$ by checking the point where the maximum modulus is attained (the modulus is $1$ there). The result thus follows.

Q.E.D.

By the Schwarz Lemma, we obtain several important results in future analysis.

Corollary:
(1) Any biholomorphism $f: \mathbb{D} \rightarrow \mathbb{D}$ fixing the origin is a rotation.
(2) If $f: \mathbb{D} \rightarrow \mathbb{D}$ is either not injective or not surjective (i.e. $f$ is not bijective), then we must have strict inequality in (1) and (2) in Schwarz Lemma.

Proof: (2) follows from the fact that a rotation is a biholomorphism, so if $f$ is not bijective, it cannot be a rotation and so the assumption in (3) of Schwarz Lemma cannot hold.

For (1), application of Schwarz Lemma on $f$ and $f^{-1}$ shows that for any $z,w \in \mathbb{D}$ , we have

$\displaystyle |f'(0)| \leq 1; \frac{1}{|f'(0)|}=|(f^{-1})'(0)|\leq 1$

This implies $|f'(0)|=1$ , and by Schwarz Lemma again, $f$ is a rotation.

Q.E.D.

Using Schwarz Lemma, we can give a characterization of the space $\text{Aut}(\mathbb{D})$ .

Theorem:

$\displaystyle \text{Aut}(\mathbb{D})=\{f: f(z)=e^{i\theta}\frac{\alpha-z}{1-\overline{\alpha}z}, \alpha \in \mathbb{D}, \theta \in \mathbb{R}\}$

Proof: We first show that any function of the form $f(z)=e^{i\theta}\frac{\alpha-z}{1-\overline{\alpha}z}, \alpha \in \mathbb{D}, \theta \in \mathbb{R}$ is an automorphism on $\mathbb{D}$ .

Denote $\psi_{\alpha}(z)=\frac{\alpha-z}{1-\overline{\alpha}z}$ . We first show that $| \psi_{\alpha}(z)| <1$ for any $z \in \mathbb{D}$ . Indeed, $|\frac{\alpha-z}{1-\overline{\alpha}z}|<1 \iff |\alpha-z|^2<|1-\overline{\alpha}z|^2$ . Note that $|\alpha-z|^2=|\alpha|^2-2Re(\alpha z)+|z|^2$ and $|1-\overline{\alpha}z|^2=1-2Re(\overline{\alpha}z)+|\alpha z|^2=1-2Re(\alpha z)+|\alpha z|^2$ . By cancellation, it suffices to show that $|\alpha|^2+|z|^2 < 1+|\alpha z|^2$ . Since the roles of $\alpha$ and $z$ in this inequality is symmetric, WLOG, we may assume $|z| \geq |\alpha|$ . Then $|z|^2 <1$ and $|\alpha|^2 \leq |\alpha z|$ . Summing up these two inequality yields the inequality desired. Therefore, indeed $f: \mathbb{D} \to \mathbb{D}$ .

Next, we need to show that $f$ is a biholomorphism. Since rotation is a biholomorphism, it suffices to show that $\psi_\alpha$ is a biholomorphism. Indeed, direct checking shows that $\psi_\alpha$ satisfies $\psi_{\alpha} \circ \psi_\alpha (z)=z$ , which shows that $\psi_\alpha$ is its own inverse.

Next, we need to show that any $f \in \text{Aut}(\mathbb{D})$ is of the form mentioned above. Let $f(\alpha)=0$ . Then $f \circ \psi_{\alpha} =f\circ \psi_{\alpha}^{-1}$ is an automorphism on $\mathbb{D}$ fixing the origin, thus it is a rotation by previous Corollary, i.e. there exists some $\theta \in \mathbb{R}$ such that for any $z \in \mathbb{D}$ , we have $f \circ \psi_{\alpha}^{-1}(z)=e^{i\theta}z$ . Replace $z$ by $\psi_{\alpha}(z)$ to yield the result.