Rudin Proof of Liouville Theorem (Complex A.)

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Discussion Overview

The discussion revolves around the proof of Theorem 10.23 from Rudin's work, specifically addressing why the coefficients ##c_n## must equal zero for all ##n > 0## in the context of Liouville's Theorem in complex analysis.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions the necessity of ##c_n = 0## for all ##n > 0##, seeking clarification on the proof.
  • Another participant argues that if any ##|c_n| > 0## for ##n > 0##, it leads to a contradiction by allowing the sum to exceed any individual term, thus violating the theorem's conditions.
  • A later reply acknowledges a misunderstanding related to Theorem 22, which is noted as a different version of Gauss's Mean Value Theorem, contributing to the initial confusion.
  • There is a correction regarding the expression for ##r^{2n}##, suggesting it should be ##\frac{M^2}{|c_n|^2}## instead of ##\frac{M}{|c_n|^2}##.

Areas of Agreement / Disagreement

The discussion includes competing views on the implications of the coefficients ##c_n##, with no consensus reached on the necessity of ##c_n = 0## for all ##n > 0##.

Contextual Notes

Participants reference different theorems and their implications, indicating potential limitations in understanding the relationships between them. The discussion reflects a reliance on specific definitions and conditions outlined in the theorems.

Bachelier
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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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Note that it says FOR ALL r. If I had a |c_n| > 0 for n>0 then I could let r^{2n}=\frac{M}{|c_n|^2} 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:
HS-Scientist said:
Note that it says FOR ALL r. If I had a |c_n| > 0 for n>0 then I could let r^{2n}=\frac{M}{|c_n|^2} 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}}##
 
Thm 10.22
 

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