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Holomorphic Maps on D(0,1) 
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#1
Mar712, 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 nonconstant 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 nonzero , for a in the interior of D. i.e., the winding number of f(z) about any point on the disk is nonzero. 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. 


#2
Mar812, 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? 


#3
Mar812, 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.



#4
Mar1512, 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?



#5
Mar1612, 06:47 PM

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



#6
Mar1612, 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 nonconstant analytic maps being open, but I cannot see how to show that the winding number about every point inside is nonzero.



#7
Mar1712, 07:23 AM

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



#8
Mar2012, 04:14 PM

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



#9
Mar2012, 05:13 PM

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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
Mar2012, 10:55 PM

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#11
Mar2112, 07:42 AM

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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. 


#12
Mar2212, 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.



#13
Mar2212, 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 .



#14
Mar2312, 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.



#15
Mar2312, 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.



#16
Mar2512, 10:28 PM

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Nice Job: continuous, integervalued functions are locallyconstant. Good one, Citan.



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