# Holomorphic Maps on D(0,1)

1. Mar 7, 2012

### Bacle2

Hi, I know this one is not too hard, but I've been stuck for a while:

Say f is holomorphic and non-constant on the closed unit disc D(0,1),

and |f|=1 on the boundary of the disk (so that, e.g., by the MVT,

f maps the disk into itself) . Is it the case that f maps

the disk _onto_ itself?

I have thought of trying to show that the integral:

∫_D (f'(z)dz/(f(z)-a ) is non-zero , for a in the interior of D.

i.e., the winding number of f(z) about any point on the disk is

non-zero. But I can't see how to show this. Any Ideas?

I am trying to use the fact that if f is analytic, then f is a finite product of Blaschke

factors , but it still does not add up. Any ideas?

2. Mar 8, 2012

### mathwonk

It suffices to show that f is not constant on the boundary of the disk I believe. But a non constant map is an open map, so.....? I guess then the image of the closed disc is both closed and open in the closed disc, hence everything?????

Think this through, is it right? does it need amplification?

3. Mar 8, 2012

### Bacle2

I don't see how it follows that f(D(0,1)) is both open and closed; I agree if I could show this, it would be a proof, by connectedness.

4. Mar 15, 2012

### lavinia

Where in the interior of the disk could f(z) be outside of the disk?

5. Mar 16, 2012

### Bacle2

You have to use fact that f(z) is analytic ; otherwise points inside can be mapped outside.

6. Mar 16, 2012

### Bacle2

Besides, the more difficult part , as I see it, is how to show that the boundary does wind about every point inside. This has to see with non-constant analytic maps being open, but I cannot see how to show that the winding number about every point inside is non-zero.

7. Mar 17, 2012

### lavinia

A non-constant holomorphic function can not have it's maximum in the interior of the unit disk,

8. Mar 20, 2012

### Bacle2

Yes, that is true, but how does it help?

9. Mar 20, 2012

### lavinia

So if its norm is 1 on the boundary then all of its values throughout the unit disc must lie on the unit circle.

But then it is a map from a 2 dimensional domain onto a 1 manifold so the Jacobian must be everywhere singular.
But The Jacobian is multiplication by a complex number and so must be zero.

I think .....

10. Mar 20, 2012

### Bacle2

I don't see how so; we have a map from the unit disk to itself, both of which are 1-D. Moreover, what you say is if f'(z)≠0 , if I nderstood you well.

11. Mar 21, 2012

### lavinia

The modulus of the function achieves its maximum at the boundary.
So the norm of f is less than or equal to 1 on the closed disk. On the interior it is less that 1 unless it is a constant.

You were right to correct me. I was assuming that f is non zero in which case you can assume the Minimum Modulus Principle. In that case, the map must be a constant. My mistake.

Last edited: Mar 21, 2012
12. Mar 22, 2012

### morphism

What can you say about the boundary points of the image? This should help you prove that it's open and closed.

13. Mar 22, 2012

### Bacle2

Yes, thanks, but I'm trying to help someone who must somehow use ( instructions in the qual. exam prep. exercise), the winding number. So all I can think is showing that the image of the boundary of the disk winds around every point in the disk. All I can think with those tools are Blaschke products, but I cannot see how .

14. Mar 23, 2012

### Bacle2

Thanks, all. My apologies for not being more clear. Unfortunately, for the May Quals. my friend is taking, one must follow the instructions more closely than the September exams.

15. Mar 23, 2012

### Citan Uzuki

As you noted, for any curve γ, $\int_\gamma \frac{f'(z)}{f(z)-a} \mathrm{d}z$ is simply 2πi times the winding number of $f \circ \gamma$ around the point a. But the winding number is constant on each connected component of the complement of the image of $f \circ \gamma$. Now, you are given that |f(z)| = 1 for |z| = 1. So if γ is the curve that goes once around the unit circle, the image of $f \circ \gamma$ is contained in the unit circle. Now, if a and b are any two points in D(0, 1), they can be connected by a straight line that does not intersect the unit circle, so they must be in the same component of the complement of the image of $f \circ \gamma$. Ergo $\int_\gamma \frac{f'(z)}{f(z)-a} \mathrm{d}z = \int_\gamma \frac{f'(z)}{f(z)-b} \mathrm{d}z$. So you only have to show that this integral is nonzero at at least one point in the unit disk, and it follows that it is nonzero at all points.

16. Mar 25, 2012

### Bacle2

Nice Job: continuous, integer-valued functions are locally-constant. Good one, Citan.

Last edited: Mar 25, 2012