Holomorphic Maps on D(0,1)


by Bacle2
Tags: holomorphic, maps
Bacle2
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#1
Mar7-12, 10:43 PM
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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?

Thanks in Advance.
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mathwonk
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#2
Mar8-12, 12:42 AM
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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?
Bacle2
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#3
Mar8-12, 12:18 PM
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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.

lavinia
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#4
Mar15-12, 09:46 PM
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Holomorphic Maps on D(0,1)


Where in the interior of the disk could f(z) be outside of the disk?
Bacle2
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#5
Mar16-12, 06:47 PM
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You have to use fact that f(z) is analytic ; otherwise points inside can be mapped outside.
Bacle2
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#6
Mar16-12, 07:28 PM
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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.
lavinia
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#7
Mar17-12, 07:23 AM
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A non-constant holomorphic function can not have it's maximum in the interior of the unit disk,
Bacle2
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#8
Mar20-12, 04:14 PM
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Yes, that is true, but how does it help?
lavinia
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#9
Mar20-12, 05:13 PM
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Quote Quote by Bacle2 View Post
Yes, that is true, but how does it help?
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 .....
Bacle2
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#10
Mar20-12, 10:55 PM
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Quote Quote by lavinia View Post
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 .....
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.
lavinia
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#11
Mar21-12, 07:42 AM
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Quote Quote by Bacle2 View Post
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.
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.
morphism
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#12
Mar22-12, 09:00 PM
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What can you say about the boundary points of the image? This should help you prove that it's open and closed.
Bacle2
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#13
Mar22-12, 09:28 PM
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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 .
Bacle2
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#14
Mar23-12, 05:30 PM
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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.
Citan Uzuki
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#15
Mar23-12, 10:08 PM
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As you noted, for any curve γ, [itex]\int_\gamma \frac{f'(z)}{f(z)-a} \mathrm{d}z[/itex] is simply 2πi times the winding number of [itex]f \circ \gamma[/itex] around the point a. But the winding number is constant on each connected component of the complement of the image of [itex]f \circ \gamma[/itex]. 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 [itex]f \circ \gamma[/itex] 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 [itex]f \circ \gamma[/itex]. Ergo [itex]\int_\gamma \frac{f'(z)}{f(z)-a} \mathrm{d}z = \int_\gamma \frac{f'(z)}{f(z)-b} \mathrm{d}z[/itex]. 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.
Bacle2
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#16
Mar25-12, 10:28 PM
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Nice Job: continuous, integer-valued functions are locally-constant. Good one, Citan.


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