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Rudin Proof of Liouville Theorem (Complex A.)

  1. Apr 20, 2013 #1
    Please see attached.

    I am talking about Thm. 10.23 proof.

    Why is it that ##c_n## must be equal to ##0, \ \forall n>0## ?


    Attached Files:

    Last edited: Apr 20, 2013
  2. jcsd
  3. Apr 20, 2013 #2
    Note that it says FOR ALL r. If I had a [itex] |c_n| > 0 [/itex] for n>0 then I could let [itex] r^{2n}=\frac{M}{|c_n|^2} [/itex] so the n-th term would be equal to M and the sum would be at least as large as M (all the terms are non-negative if I choose r this way so the sum is at least as large as any individual term). This is a contradiction.
    Last edited: Apr 20, 2013
  4. Apr 21, 2013 #3
    Thank you for the answer. Yeah I get it.
    It was my misunderstanding of theorem 22 that caused the confusion.
    Theorem 22 is a different version of Gauss's Mean Value Thm.

    BTW I think you meant to say take:

    ##r^{2n} = \Large{\frac{M^2}{|c_n|^2}}##
  5. Apr 21, 2013 #4
    Thm 10.22

    Attached Files:

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