Let $f$ be a holomorphic map of the open unit disc into itself. Then $$|f'(z)|\le \frac{1}{1-|z|}$$ We can write $$f(z)=\sum_{n=0}^\infty a_nz^n,\ |z|<1.$$ Also, $|f(z)|<1$. After this how I conclude that $|f'(z)|\le \frac{1}{1-|z|}$
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$|f(z)|<1$ should be placed at the beginning : it is not good mathematical writing to say "also"... as if you were reminding of something later... – Jean Marie Sep 21 '17 at 04:48
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Hint 1. Note that by the Schwarz–Pick theorem (which is an application of the maximum modulus principle), for all $z$ in the open unit disc, $$|f'(z)|\le \frac{1-|f(z)|^2}{1-|z|^2}.$$
Hint 2. By the Cauchy's integral formula, for $|z|<1$ and for all $0<r<1-|z|$, $$f'(z)=\frac{1}{2\pi i}\int_{|w-z|=r}\frac{f(w)}{(w-z)^2}dw\implies |f'(z)|\leq \frac{2\pi r}{2\pi r^2}=\frac{1}{r}.$$
Robert Z
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@Iampi It is a straightforward application of the maximum modulus principle. See the proof in the wiki page. – Robert Z Sep 21 '17 at 05:00
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But here $f(0)$ need not be $0$. And also, $|f'(z)|\le \frac{1-|f(z)|^2}{1-|z|^2}$. – I am pi Sep 21 '17 at 05:16
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