Finite Blaschke product and proper maps on the unit disc

Question

I have to present tomorrow on an introductory section in several complex variables about proper maps, and they gloss over a fact that seems important to me, but I do not know how to prove it.

Suppose $f: \Bbb D \to \Bbb D$ is an analytic proper map. Prove that $f$ is a finite Blaschke product.

That is, prove that $$f(z) = e^{i \theta} \prod_{j=1}^{k} {{z-a_j} \over {1- \overline a_jz}}$$

where, $\theta$ is real, and $a_j \in \Bbb D$.

They mention that to show this, you should consider the fiber of the origin. I'm not sure what to do with this information, though.

Daniel Fischer · Accepted Answer · 2015-05-17T19:44:34.240

13

Since $f\colon \mathbb{D} \to \mathbb{D}$ is proper, for every $r < 1$, the preimage $f^{-1}(\overline{D_r(0)})$ of the closed disk with radius $r$ is compact, thus contained in $D_\rho(0)$ for some $\rho < 1$. Hence we have

$$\lim_{\lvert z\rvert \to 1} \lvert f(z)\rvert = 1.\tag{1}$$

In particular, $f$ is not constant. So the fibre above the origin, the zero set of $f$ is compact and discrete, hence finite.

Now let $g$ be the product of the Blaschke factors for the zeros of $f$ (according to multiplicity) and consider the function

$$h\colon z \mapsto \frac{f(z)}{g(z)}.$$

Then $h$ is a zero-free holomorphic function on $\mathbb{D}$, and since finite Blaschke products have constant modulus $1$ on the unit circle, we have

$$\lim_{\lvert z\rvert\to 1} \lvert h(z)\rvert = 1.\tag{2}$$

The minimum modulus principle implies that $h$ is a constant of modulus $1$, i.e. $h(z) = e^{i\theta}$ for some $\theta\in\mathbb{R}$.

edited May 17 '15 at 19:44

answered Sep 16 '14 at 19:33

Daniel Fischer

206,697

Ok...I knew that fibers were compact, but I have never heard of Blaschke factors, so this function $g$ is new to me. – Johnny Apple Sep 16 '14 at 19:34
Very nice proof! – Bombyx mori Sep 16 '14 at 19:35
The $\frac{z-a_j}{1-\overline{a_j}z}$ are often called Blaschke factors, @JohnnyApple, because they are factors of Blaschke products. – Daniel Fischer Sep 16 '14 at 19:37
So, where have we used compactness of the fibers here? – Johnny Apple Sep 16 '14 at 19:37
@JohnnyApple A compact and discrete set is $______$. We need that to deduce the desired representation. – Daniel Fischer Sep 16 '14 at 19:39
Ahhhh of course. We needed a finite product. Thanks! – Johnny Apple Sep 16 '14 at 19:40
@JohnnyApple And we need that the modulus of $g$ is $1$ on the unit circle to deduce that the constant has modulus $1$. – Daniel Fischer Sep 16 '14 at 19:42
One more thing... Why is this fiber discrete? – Johnny Apple Sep 16 '14 at 20:06
@JohnnyApple Because the fibres of non-constant holomorphic functions on a domain are discrete. A proper holomorphic map on the unit disk cannot be constant (because ...). – Daniel Fischer Sep 16 '14 at 20:15
I don't know. I'm learning basic complex analysis and several variables simultaneously here so I don't know some basic facts like I should. – Johnny Apple Sep 16 '14 at 20:33
If $f\colon \mathbb{D}\to\mathbb{C}$ is constant, say $f(z) \equiv c$, is the fibre $f^{-1}(c)$ compact? – Daniel Fischer Sep 16 '14 at 20:37
Oh, of course not. I had forgotten momentarily that the disc is open exclusively. I'm still in topology mode. – Johnny Apple Sep 16 '14 at 20:42
Where are you using the fact that $h$ is zero free? – Johnny Apple Sep 17 '14 at 15:31
@JohnnyApple That implies that we cannot have a $z_0 \in \mathbb{D}$ with $$\lvert h(z_0)\rvert < \min \left{ \lvert h(\zeta)\rvert : \lvert \zeta\rvert = \rho\right}$$ for any $\lvert z_0\rvert < \rho < 1$, for otherwise $h$ would have a zero in ${ z : \lvert z\rvert < \rho}$, the minimum modulus principle says. And thus we must have $\lvert h(z)\rvert = \lim\limits_{\lvert \zeta\rvert\to 1} \lvert h(\zeta)\rvert = 1$ for all $z\in\mathbb{D}$. – Daniel Fischer Sep 17 '14 at 15:35
How does you got limit $ |f(z)|$ is 1 as $ |z|$ goes to 1 – Sushil Mar 23 '18 at 14:57
@Sushil Because $f$ is proper. So for every $r < 1$ the preimage $f^{-1}({z : \lvert z\rvert \leqslant r})$ is a compact subset of $\mathbb{D}$. – Daniel Fischer Mar 23 '18 at 15:08
yes this I got. But unable to see from it, how we imply that – Sushil Mar 23 '18 at 15:29
1

@Sushil Thus $$\rho(r) := \sup { \lvert z\rvert : \lvert f(z)\rvert \leqslant r} < 1,,$$ and that means $\lvert z\rvert > \rho \implies \lvert f(z)\rvert > r$. – Daniel Fischer Mar 23 '18 at 15:35

Finite Blaschke product and proper maps on the unit disc

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