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I need to show $(\mathrm{Aut}(\mathbb{D}),\| \cdot \|_{\infty})$ is complete, where $\mathbb{D}$ is an open unit disk in the complex plane.

I know $$f\in \mathrm{Aut}(\mathbb{D})\Rightarrow f(z)=e^{i\phi}{z-\alpha\over 1-\bar{\alpha}z},-\pi<\phi\le \pi,|\alpha|<1$$

so I took $$f_n(z)=e^{i\phi}{z-\alpha_n\over 1-\bar{\alpha}_nz},-\pi<\phi\le \pi,|\alpha_n|<1$$

Say $f_n\to f$ in sup norm, $f_n$ is cauchy, then the convergence is uniform convergence right? Can now just say $$f(z)={z-\beta\over 1-\bar{\beta}z},-\pi<\phi\le \pi,|\beta|<1$$

where $\alpha_n\to\beta$? Thank you for help.

Seirios
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Myshkin
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2 Answers2

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$\mathrm{Aut}(\Bbb D)$ is a group. So if $g\in \mathrm{Aut}(\Bbb D)$ then $g^{-1}\in Aut(\Bbb D)$.

1) Show that $f_n$, which is Cauchy, has a limit say $f$, then by Weierstrass's theorem for holomorphic functions limit we have that $f$ is holomorphic.

We have that $f_n(z)=e^{iθ}φ_a(z)$.Then $f_n^{-1}(z)=e^{-iθ}φ_{-a}(z)$. Show that $f_n^{-1}\to f^{-1}$.

From 1) $f^{-1}$ is holomorphic.

Thus $f\in \mathrm{Aut}(\Bbb D)$.

Seirios
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Haha
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I am not sure if one can show in your way $|\beta| <1$ (Seems you can only get $|\beta|\leq 1$). Here's (A sketch of) another method.

(1) Show that $\mathrm{Aut}(\mathbb D) = \{ f:\mathbb D \to \mathbb D: \ f,\ f^{-1} \text{ are holomorphic }\}$

(2) If $f_n$ is a Cauchy sequence, then it has a limit $f$ which is a continuous function. Show that $f$ is holomorphic.

(3) Show that $f^{-1}$ exists and is holomorphic.

Seirios
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