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I need some help with this problem:

Let $f\colon D \to D$ analytic and $f(z_1)=0, f(z_2)=0, \ldots, f(z_n)=0$ where $z_1, z_2, \ldots, z_n \in D= \{z:|z|<1\}$. I want to show that $|f(z)| \leq \prod_{k=1}^n \left| \frac{z-z_k}{1-\overline{z_k}\, z} \right|$ for all $z \in D$.

It seems that I need to use Schwarz-Pick Lemma but it seems that the problem doesn't satisfy the conditions. Another lemma that I can use is that of Lindelöf saying: Let $f:D \to D$ analytic, then $|f(z)|\leq \frac{|f(0)|+|z|}{1+|f(0)| \cdot |z|}$ for all $ z \in D$.

It seems to be an easy problem but I couldn't succeed in solving it.

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    Yes,your hint is the base of the answer below. But I didn't understand how the fact that $f(z_k)=0$ assist the answer.2011-12-27

1 Answers 1

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Let $f$ be non-constant, and continuously extend to $\overline{D}$ (the closed unit disc.)

Let $B(z)=\prod_{k=1}^n \frac{z-z_k}{1-\overline{z_k}z}.$ Note that $|B(z)|=1$ for $|z|=1.$ Define $g(z):=f(z)/B(z).$ Now, $g$ is a holomorphic map on $D$. By maximum modulus principle, $|g(z)|$ attains its maximum value on the boundary $|z|=1.$ Therefore, $|g(z)| \leq 1$ for $|z| \leq 1.$ Hence,

$|f(z)| \leq |B(z)|= \prod_{k=1}^n \left|\frac{z-z_k}{1-\overline{z_k}z} \right|.$

See the Blaschke Product as well.

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    @bond: The zeros of $B(z)$ are precisely $z_k$'s, that's why $g$ is holomorphic on $D.$2019-03-28