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<record version="5" id="3047">
 <title>Schwarz lemma</title>
 <name>SchwarzLemma</name>
 <created>2002-06-05 11:33:25</created>
 <modified>2006-09-13 22:18:32</modified>
 <type>Theorem</type>
 <creator id="127" name="Koro"/>
 <author id="127" name="Koro"/>
 <author id="2760" name="yark"/>
 <author id="338" name="ariels"/>
 <classification>
	<category scheme="msc" code="30C80"/>
 </classification>
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 <content>Let $\Delta = \{z: \size{z} &lt; 1\}$ be the open unit disk in the complex plane $\Complex$.  Let $f\colon\Delta\to\Delta$ be a holomorphic function with $f(0)=0$.  
Then $\size{f(z)} \le \size{z}$ for all $z\in\Delta$, and $\size{f'(0)} \le 1$.  If the equality $\size{f(z)}=\size{z}$ holds for \emph{any} $z\ne 0$ or $\size{f'(0)}=1$, then $f$ is a rotation: $f(z)=az$ with $\size{a}=1$.

This lemma is less celebrated than the bigger guns (such as the Riemann mapping theorem, which it helps prove); however, it is one of the simplest results capturing the ``rigidity'' of holomorphic functions.  No \PMlinkescapetext{similar} result exists for real functions, of course.</content>
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