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Fall 2013 - Problem 6

Cauchy estimates

Let $U \subseteq \C$ be a bounded open set with $0 \in U$ , and $\func{f}{U}{\C}$ be a holomorphic function with $f\p{U} \subseteq U$ and $f\p{0} = 0$ . Show that $\abs{f'\p{0}} \leq 1$ .

Hint: Consider the iterations $f^n \coloneqq \underbrace{f \circ \cdots \circ f}_{n \text{ times}}$ of $f$ .

Solution.

Suppose otherwise, and that $\abs{f'\p{0}} > 1$ . First, notice that

\p{f^n}'\p{0} = f'\p{f^n\p{0}}\p{f^n}'\p{0} = f'\p{0}\p{f^n}'\p{0}.

By induction, we see that $\p{f^n}'\p{0} = \p{f'\p{0}}^n$ .

Let $R > 0$ be big enough so that $U \subseteq B\p{0, R}$ . Thus, $\abs{f\p{z}} \leq R$ for all $z \in U$ . Let $\cl{B\p{0, r}} \subseteq U$ . Since $f^n$ is holomorphic, we have the Cauchy estimate

\abs{\p{f^n}'\p{0}} \leq \frac{R}{r} \implies \abs{f'\p{0}}^n \leq \frac{R}{r}.

Thus, $\abs{f'\p{0}} \leq 1$ or else we may send $n \to \infty$ to see $\frac{R}{r} = \infty$ , which is impossible.