Rudin Proof of Liouville Theorem (Complex A.)

[Bachelier]

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

 

[Infrared]

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.

 

[Bachelier]

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:

$r^{2n} = \Large{\frac{M^2}{|c_n|^2}}$

 

[Bachelier]

Thm 10.22

 

[blue_raver22]

for your question. In the proof of the Liouville Theorem, we are considering an entire function f(z) = ∑c_nz^n. By definition, an entire function is a function that is analytic everywhere in the complex plane, which means it has a power series representation that converges for all complex numbers z. In this case, we are considering the series ∑c_nz^n, which is a power series representation of f(z).

Now, in order for a power series to converge for all complex numbers z, there is a necessary condition that the coefficients c_n must satisfy. This condition is known as the Cauchy-Hadamard theorem, which states that the radius of convergence of the power series is given by R = limsup|c_n|^(1/n). In other words, the series will converge for |z| < R.

In the proof of the Liouville Theorem, we are assuming that the function f(z) is bounded, which means that the absolute value of f(z) is less than or equal to some constant M for all complex numbers z. This implies that the power series ∑c_nz^n must also be bounded for all complex numbers z. However, as the radius of convergence R is given by limsup|c_n|^(1/n), this means that for all n > 0, the value of c_n must be equal to 0 in order for the series to converge for all z.

In conclusion, the reason why c_n must be equal to 0 for all n > 0 is because of the necessary condition for the power series to converge for all complex numbers z. This is a crucial step in the proof of the Liouville Theorem, as it allows us to conclude that the function f(z) is a constant function, which is the main result of the theorem.