11
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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. $\endgroup$
    – user21436
    Dec 25, 2011 at 22:16
  • 3
    $\begingroup$ It's possible that $f(z)$ is not a real number for some $z$, so the inequality doesn't make sense (you probably missed $|\cdot |$). $\endgroup$ Dec 25, 2011 at 22:25
  • $\begingroup$ 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? $\endgroup$
    – Potato
    Dec 25, 2011 at 22:32
  • $\begingroup$ yes, thanks, I've edited it. $\endgroup$
    – bond
    Dec 25, 2011 at 22:32
  • $\begingroup$ Also note that the function you get when you do what I said will be analytic, because all singularities will be removable. $\endgroup$
    – Potato
    Dec 25, 2011 at 22:45

1 Answer 1

15
$\begingroup$

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.

$\endgroup$
5
  • 2
    $\begingroup$ 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 $|f(z)|\le |B(z)|$? $\endgroup$
    – Sam
    Dec 26, 2011 at 3:00
  • $\begingroup$ @Sam: That's a good point. It can be improved in this way. $\endgroup$ Dec 27, 2011 at 4:20
  • $\begingroup$ It's look good but Where have you used the fact of $f(z_k)=0$? $\endgroup$
    – bond
    Dec 27, 2011 at 14:23
  • 1
    $\begingroup$ @bond: The zeros of $B(z)$ are precisely $z_k$'s, that's why $g$ is holomorphic on $D.$ $\endgroup$ Dec 27, 2011 at 19:55
  • 1
    $\begingroup$ I have taken the liberty to update the answer according to @Sam's comment, so that it works without additional conditions. $\endgroup$
    – Martin R
    May 29, 2020 at 1:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .