Rudin Proof of Liouville Theorem (Complex A.)

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SUMMARY

The discussion centers on the proof of Theorem 10.23 from Rudin's "Principles of Mathematical Analysis," specifically addressing why the coefficients ##c_n## must equal zero for all ##n > 0## in the context of Liouville's Theorem. The user highlights a contradiction arising from assuming ##|c_n| > 0##, leading to a sum that exceeds any individual term, which violates the theorem's conditions. The confusion stemmed from a misunderstanding of Theorem 22, which is a variant of Gauss's Mean Value Theorem, and the correct application of the relationship ##r^{2n} = \frac{M^2}{|c_n|^2}## as stated in Theorem 10.22.

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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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