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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.

Davide Giraudo
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bond
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  • Do you know that there is an option to thank people who help you by accepting their answers? If not, I seriously recommend you to go through the faq. –  Dec 25 '11 at 22:16
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    It's possible that $f(z)$ is not a real number for some $z$, so the inequality doesn't make sense (you probably missed $|\cdot |$). – Davide Giraudo Dec 25 '11 at 22:25
  • I believe you imitate the proof of the Shwartz lemma. Divide f(z) by all of the factors on the right. What does the maximum modulus principle give you? – Potato Dec 25 '11 at 22:32
  • yes, thanks, I've edited it. – bond Dec 25 '11 at 22:32
  • Also note that the function you get when you do what I said will be analytic, because all singularities will be removable. – Potato Dec 25 '11 at 22:45
  • Does that hint help, bond? I can write up a formal answer if needed. – Potato Dec 25 '11 at 22:48
  • 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. – bond Dec 27 '11 at 14:22

1 Answers1

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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 $D$.

For $|z| < r < 1$ we have by the maximum modulus principle $$ \frac{|f(z)|}{|B(z)|} \le \max_{\theta} \frac{1}{|B(re^{i\theta})|} \overset{r\to 1}\longrightarrow 1 $$ 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.

Martin R
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Ehsan M. Kermani
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    I don't think you really need to assume that $f$ extends to $\overline D$ continuously for your argument to work. Couldn't just remark that (by the maximum modulus principle) for $|z|<r<1$: $$\frac{|f(z)|}{|B(z)|} \le \max_{\theta} \frac{1}{|B(re^{i\theta})|} \overset{r\to 1}\longrightarrow 1$$ so $|f(z)|\le |B(z)|$? – Sam Dec 26 '11 at 03:00
  • @Sam: That's a good point. It can be improved in this way. – Ehsan M. Kermani Dec 27 '11 at 04:20
  • It's look good but Where have you used the fact of $f(z_k)=0$? – bond Dec 27 '11 at 14:23
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    @bond: The zeros of $B(z)$ are precisely $z_k$'s, that's why $g$ is holomorphic on $D.$ – Ehsan M. Kermani Dec 27 '11 at 19:55
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    I have taken the liberty to update the answer according to @Sam's comment, so that it works without additional conditions. – Martin R May 29 '20 at 01:39