# 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.

• 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
Commented Dec 25, 2011 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 $|\cdot |$). Commented Dec 25, 2011 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? Commented Dec 25, 2011 at 22:32
• yes, thanks, I've edited it.
– bond
Commented Dec 25, 2011 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. Commented Dec 25, 2011 at 22:45

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.

• 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
Commented Dec 26, 2011 at 3:00
• @Sam: That's a good point. It can be improved in this way. Commented Dec 27, 2011 at 4:20
• It's look good but Where have you used the fact of $f(z_k)=0$?
– bond
Commented Dec 27, 2011 at 14:23
• @bond: The zeros of $B(z)$ are precisely $z_k$'s, that's why $g$ is holomorphic on $D.$ Commented Dec 27, 2011 at 19:55
• I have taken the liberty to update the answer according to @Sam's comment, so that it works without additional conditions. Commented May 29, 2020 at 1:39