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I am working on this problem which appears to be a preliminary exam question in UTexas.

Assume that $f$ is analytic outside the closed unit disk and $ |f(z)| <1$ in this region. Prove that $|f'(2)| \leq \frac{1}{3}$

I can prove that $|f'(2)| < 1$ but the upper bound $1/3$ is very hard to get. Any help would be appreciated.

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Let $g(z):=f(1/z)$ then $g$ is holomorphic in $\mathbb{D}=\{z\in\mathbb{C}:|z|<1\}$ and $g( \mathbb{D})\subseteq \mathbb{D}$.

Then we can apply the Schwarz–Pick theorem (a variant of Schwarz lemma): for all $z\in \mathbb{D}$, $$|g'(z)|\leq \frac{1-|g(z)|^2}{1-|z|^2}.$$ Since $g'(z)=f'(1/z)\cdot(-1/z^2)$, it follows that $$|f'(1/z)|\leq \frac{(1-|f(1/z)|^2)\cdot|z|^2}{1-|z|^2}.$$ Now for $z=1/2$, we finally obtain $$|f'(2)|\leq \frac{(1-|f(2)|^2)\cdot|1/2|^2}{1-|1/2|^2}=\frac{1-|f(2)|^2}{3}\leq \frac{1}{3}.$$

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  • $\begingroup$ It's the first time I see Schwarz–Pick theorem, it is indeed helpful. Thank you. $\endgroup$ – iamvegan Aug 16 '16 at 20:07
  • $\begingroup$ @iamvegan Usually the Schwarz–Pick theorem is not covered by standard courses in complex analysis. However it is nice result about the rigidity of holomorphic functions. $\endgroup$ – Robert Z Aug 16 '16 at 20:15
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    $\begingroup$ Very interesting (I upvote). A little thought: if this is the only way to answer to a "preliminary entrance exam question to UTexas", what extraordinary level have its students ! $\endgroup$ – Jean Marie Aug 16 '16 at 23:34

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