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Bounded analytic function

  1. Apr 19, 2010 #1
    1. The problem statement, all variables and given/known data
    Let f ba analytic function on 0< |z| < 1 and suppose |f(z)| <= 4|z|^1.1 for all 0<|z|<1.
    Prove that |f(1/2)| <= 1

    2. Relevant equations



    3. The attempt at a solution

    I tried to prove it be cauchy integral formula but I got
    |f(1/2)|< 8 r ^1.1 r<1
     
  2. jcsd
  3. Apr 20, 2010 #2
    Hint: Clearly f(0) = 0. The consider f(z)/z, Cearly this functon is analytic and is also zero at z = 0, so we can divide yet again by z. We comnclude that

    g(z) = f(z)/z^2

    is an analytic function.

    What does the Maximum Modulus theorem imply for |g(z)| for
    |z|<1 given the bounds on |f(z)|?
     
  4. Apr 20, 2010 #3
     
  5. Apr 20, 2010 #4
    Yes, I see now that the point z = 0 is excluded. I would suggest you to first solve the slightly different problem in which z = 0 is in the domain. If you do that, then it is a trivialy matter to modify the proof to take into account that z = 0 is not in the domain.
     
  6. Apr 21, 2010 #5
    I think this is not related to the maximum modulus principle since it says that f cannot achieve max. on an open set or it is a constant. If it is constant, then what is the value of this constant?
     
  7. Apr 21, 2010 #6
    Just close the boundaries at first and then assume that f(z)/z^2 assumes a maximum at the boundary. Then prove the result of this modified problem and then fix the proof by taking into account that the boundary is in fact open. E.g. you can think of placing boundaries a distance of epsilon within the region and then the analogues of the above result will hold. Since epsilon is larger than zero but arbitrary, you will recover the result.
     
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