# Showing that $|f(z)| \leq \prod \limits_{k=1}^n \left|\frac{z-z_k}{1-\overline{z_k}z} \right|$

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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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. –  user21436 Dec 25 '11 at 22:16
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

Let $f$ be non-constant, and continuously extends 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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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 3:00
@Sam: That's a good point. It can be improved in this way. –  Ehsan M. Kermani Dec 27 '11 at 4:20
It's look good but Where have you used the fact of $f(z_k)=0$? –  bond Dec 27 '11 at 14:23
@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