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Liouville's theorem - (probably) easy question

  1. May 2, 2012 #1
    1. The problem statement, all variables and given/known data

    If f is an entire function and |f(z)|\leq C|z|^(1/2) for all complex numbers z, where C is a positive constant, show that f is constant.

    2. Relevant equations

    All bounded and entire functions are constant.

    3. The attempt at a solution

    I'm 99% sure this can be easily proven using Liouville's theorem, I'm just having trouble proving that f is bounded above by a constant. What should I do with the |z|^(1/2) term?

    Thanks for the help!
     
  2. jcsd
  3. May 2, 2012 #2

    Dick

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    You can't prove it directly with Liouville's theorem, since f(z) isn't bounded. You could look at how Liouville's theorem is proved and adapt the proof.
     
  4. May 2, 2012 #3
    Would that involve looking at the Cauchy differentiation formula and using the maximum*length principle? Is it because f is entire we know we can use CDF? In other words, do we know there exists a z_0 inside a simple closed curve gamma such that CDF holds because f is entire?
     
  5. May 2, 2012 #4

    Dick

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    Yes, it would. If |f(z)| is bounded you can use the Cauchy integral formula to show all of the ak for k>0 in the power series expansion are equal to zero. |f(z)|<C|z^(1/2)| is also good enough. Basically the same proof.
     
    Last edited: May 2, 2012
  6. May 3, 2012 #5
    Ah, got it! Thank you so much!
     
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