Rudin Proof of Liouville Theorem (Complex A.)

  • Thread starter Bachelier
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  • #1
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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## ?

Thanks
 

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  • #2
Infrared
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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.
 
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  • #3
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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.
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}}##
 
  • #4
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Thm 10.22
 

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