Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm having difficulties with a question from Complex Analysis (Gamelin). The question has been asked before, but I still have some difficulties with it. It asks to show that a function continuous on unit disk and its boundary and analytic on the unit open disk is a finite Blaschke product if its modulus on the boundary of the disk is one. The solutions I have read suggested that one divide the function by the finite Blaschke product with zeros identical to zeros of f. I first of all do not understand why f should have a finite number of zeros. Secondly, would it not follow by the maximum and minimum modulus principle that f has constant modulus on the entire disk since its maximum and minimum moduli on the boundary of the disk are one? Then, how can f acquire any zeros whatsoever? Am I misusing the max/min mod principle here?

As well, how does dividing the function by the finite Blaschke product with zeros identical to the zeros of the function help? The claim was that the function is now by the max mod principle constant (??). I still don't see how this follows.

Thank you!

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

1) If $f$ had infinitely many zeros in the disk, those would have a limit point. By continuity the value at such a limit point would be $0$ too. If $f$ has no zeros on the circle, the limit point is thus a non-isolated zero of the function. But an analytic function with a non-isolated zero is identically $0$.

2) If $f$ had no zeros in the disk and constant absolute value on the circle, it would indeed have have constant absolute value, and therefore would be constant. A function with zeros in the disk can have constant absolute value on the circle, the simplest example being $f(z) = z$.

3) After dividing by the finite Blaschke product, you get a function with no zeros in the disk, and therefore you can apply (2).

share|improve this answer
    
Thank you so much! Why is it required that the function be nowhere zero for the max/min modulus principle to apply? As well, I see that rotation is another such example, but I still do not see why the max/min mod principle does not apply here to imply a constant modulus over the disk. –  MR1992 Nov 28 '12 at 21:37
1  
@MR1992: One of the hypotheses of the minimum modulus principle is that your function be nowhere zero; its proof boils down to "apply the maximum modulus principle to the holomorphic function $1/f$." –  Micah Nov 28 '12 at 21:45
    
Thank you! Thank you! Thank you! I really appreciate your help. –  MR1992 Nov 28 '12 at 21:48
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.