ProofOfSchwarzLemma 2002-06-06 04:40:25 2006-10-17 08:43:08 Proof Schwarz lemma 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newcommand{\Prob}[2]{\mathbb{P}_{#1}\left\{#2\right\}} \newcommand{\norm}[1]{\left\|#1\right\|} % Some sets \newcommand{\Nats}{\mathbb{N}} \newcommand{\Ints}{\mathbb{Z}} \newcommand{\Reals}{\mathbb{R}} \newcommand{\Complex}{\mathbb{C}} \newcommand{\size}[1]{\left|#1\right|} Define $g(z)=f(z)/z$. Then $g:\Delta\to\Complex$ is a holomorphic function. The Schwarz lemma is just an application of the maximal modulus principle to $g$. For any $1>\epsilon>0$, by the maximal modulus principle $\size{g}$ must attain its maximum on the closed disk $\left\{z:\size{z}\le 1-\epsilon\right\}$ at its boundary $\left\{z:\size{z}=1-\epsilon\right\}$, say at some point $z_{\epsilon}$. But then $\size{g(z)}\le \size{g(z_{\epsilon})} \le \frac{1}{1-\epsilon}$ for any $\size{z}\le 1-\epsilon$. Taking an infinimum as $\epsilon\to0$, we see that values of $g$ are bounded: $\size{g(z)}\le 1$. Thus $\size{f(z)}\le\size{z}$. Additionally, $f'(0)=g(0)$, so we see that $\size{f'(0)}=\size{g(0)}\le 1$. This is the first part of the lemma. Now suppose, as per the premise of the second part of the lemma, that $|g(w)|=1$ for some $w\in\Delta$. For any $r>\size{w}$, it must be that $\size{g}$ attains its maximal modulus (1) \emph{inside} the disk $\left\{z:\size{z}\le r\right\}$, and it follows that $g$ must be constant inside the entire open disk $\Delta$. So $g(z)\equiv a$ for $a=g(w)$ of modulus 1, and $f(z)=az$, as required.