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Let $\Delta = \{z: \size{z} < 1\}$ be the open unit disk in the complex plane $\Complex$. Let $f\colon\Delta\to\Delta$ be a holomorphic function with f(0)=0.
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Let $\Delta = \{z: \size{z} < 1\}$ be the open unit disk in the complex plane $\Complex$. Let $f:\Delta\to\Delta$ be a holomorphic function with f(0)=0.
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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 $\size{a}=1$.
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Then $\size{f(z)} \le \size{z}$ for all $z\in\Delta$, and $\size{f'(0)} \le 1$. If 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 $\size{a}=1$.
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| 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 similar result exists for real functions, of course. |
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 similar result exists for real functions, of course. |
