Let $f$ be analytic on the unit disc $D$ and bounded in modulus by $M$ there. I want to show that $|f'(z)|\le \frac{M}{1-|z|}$ for all $z\in D$.

I want to use Schwarz's lemma here after some suitable FLTs, as in the proof of Pick's lemma, but I haven't made progress. Does anyone have an idea?

Edited: Forgot a factor of $M$ in the inequality originally.

  • $\begingroup$ Is your map univalent? $\endgroup$ – Ilies Zidane Aug 29 '12 at 16:43
  • $\begingroup$ I did not understand that condition to be a requirement for the problem, I can say. $\endgroup$ – Thelonius Aug 29 '12 at 16:46
  • $\begingroup$ Because in this case you have a better control. $\endgroup$ – Ilies Zidane Aug 29 '12 at 17:03

Use Cauchy's formula. We have that $f'(z) = \frac{1}{2\pi i}\int_{B(z,r)}{\frac{f(\zeta)}{(\zeta - z)^2}}d\zeta$, where $r$ is small enough so the ball $B(z,r)$ is contained inside the disk. By the boundedness assumption, and using the standard bounds for the integral we get that $|f'(z)| \leq M(2\pi r)\frac{1}{2\pi r^2} = \frac{M}{r}$. Now let $r \rightarrow 1-|z|$ and we are done.

  • 1
    $\begingroup$ That was very straightforward. Thanks. $\endgroup$ – Thelonius Aug 29 '12 at 16:55

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