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There is a holomorphic function $f$ from the unit disc to itself s.t $f(0)=f(i/2)=0$. How can you prove the estimates $|f'(0)|\leq 1/2$ and $|f(-i/2)|\leq 2/5$?

Applying Schwarz's lemma we can get $|f(z)|\leq |z|$ but this gives an estimate too large. Cauchy's integral formula seems of no use since we don't know anything about other values of the function.

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I'll give you a solution for the first one. Let $p = i/2$; this will simplify the notation. Let $m(z)$ be an automorphism of the unit disc, interchanging $0$ and $p$, i.e. $$m(z) = \frac{z-p}{1-\bar p z}.$$

Since $f$ maps the disc to itself and $f(0) = f(p) = 0$, we can write $$f(z) = m(z) g(z),$$ where $g$ (is holomorphic and) maps the disc to itself, satisfying $g(0) = 0$. Schwarz' lemma on $g$ gives us that $|g'(0)| \le 1$. Differentiating the equality above, we get $$ f'(z) = m'(z)g(z) + m(z)g'(z). $$ In particular $$ |f'(0)| = |m'(0)g(0) + m(0)g'(0)| \le |m'(0)g(0)| + |m(0)g'(0)| \le |m(0)| = \frac12. $$

Added: I might as well give the second half too. Again by Schwarz' lemma on $g$, $|g(z) | \le |z|$. So from the factorization above, $$\left| f\left(-\frac{i}2\right) \right| = \left|m\left(-\frac{i}2\right) g\left(-\frac{i}2\right) \right| \le \left| \frac{-i/2 - i/2}{1 - i^2/4} \right| \cdot \left| -\frac{i}2 \right| = \frac25$$

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  • $\begingroup$ I don't get why Since f maps the disc to itself and $f(0)=f(p)=0$ we can write $f(z)=m(z)g(z)$.. $\endgroup$ – balestrav Apr 16 '12 at 19:37
  • $\begingroup$ $|m(z)| = 1$ for all $z$ on the unit circle and has $p$ as its only zero. I.e. $g = f/m$ is holmorphic on the disc and $|g|$ is bounded by $1$ on the circle. The maximum modulus theorem shows that $g$ maps the disc to itself. (For a formally correct argument, look at the disc of radius $r < 1$ and let $r \to 1$.) $\endgroup$ – mrf Apr 16 '12 at 19:44
  • $\begingroup$ Ok, thank! Is there an example proving those estmates are sharp? $\endgroup$ – balestrav Apr 16 '12 at 19:49
  • $\begingroup$ When are the conclusions in Schwarz' lemma sharp? I.e. take $f(z) = z\,m(z)$. $\endgroup$ – mrf Apr 16 '12 at 19:50

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