Editing Liouville's theorem (complex analysis) (section)

== Proof ==

This important theorem has several proofs.

A standard analytical proof uses the fact that [[Proof that holomorphic functions are analytic|holomorphic functions are analytic]]. 

{{Math proof|title=Proof|drop=hidden|proof=
If <math>f</math> is an entire function, it can be represented by its [[Taylor series]] about 0:

: <math>f(z) = \sum_{k=0}^\infty a_k z^k</math>

where (by [[Cauchy's integral formula]])

: <math>a_k = \frac{f^{(k)}(0)}{k!} = {1 \over 2 \pi i} \oint_{C_r} \frac{f( \zeta )}{\zeta^{k+1}}\,d\zeta</math>

and <math>C_r</math> is the circle about 0 of radius <math>r > 0</math>. Suppose <math>f</math> is bounded: i.e. there exists a constant <math>M</math> such that <math>|f(z)|\leq M</math> for all <math>z</math>. We can estimate directly

: <math>| a_k | 
\le \frac{1}{2 \pi} \oint_{C_r} \frac{ | f ( \zeta ) | }{ | \zeta |^{k+1} } \, |d\zeta|
\le \frac{1}{2 \pi} \oint_{C_r} \frac{ M }{ r^{k+1} } \, |d\zeta|
= \frac{M}{2 \pi r^{k+1}} \oint_{C_r} |d\zeta|
= \frac{M}{2 \pi r^{k+1}} 2 \pi r
= \frac{M}{r^k},</math>

where in the second inequality we have used the fact that <math>|z|=r</math> on the circle <math>C_r</math>. (This estimate is known as [[Cauchy's estimate]].) But the choice of <math>r</math> in the above is an arbitrary positive number. Therefore, letting <math>r</math> tend to infinity (we let <math>r</math> tend to infinity since <math>f</math> is analytic on the entire plane) gives <math>a_k=0</math> for all <math>k\geq 1</math>. Thus <math>f(z)=a_0</math> and this proves the theorem.
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Another proof uses the mean value property of harmonic functions.
{{Math proof|title=Proof<ref>{{cite journal |last=Nelson |first=Edward |year=1961 |title=A proof of Liouville's theorem |journal=Proceedings of the American Mathematical Society |volume=12 |issue=6 |pages=995 |doi=10.1090/S0002-9939-1961-0259149-4 |doi-access=free}}</ref>|drop=hidden|proof=
Given two points, choose two balls with the given points as centers and of equal radius. If the radius is large enough, the two balls will coincide except for an arbitrarily small proportion of their volume. Since <math>f</math> is bounded, the averages of it over the two balls are arbitrarily close, and so <math>f</math> assumes the same value at any two points.  
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The proof can be adapted to the case where the harmonic function <math>f</math> is merely bounded above or below. See [[Harmonic function#Liouville's theorem]].

Another approach to prove the theorem is

{{Math proof|title=Proof<ref>{{Cite book |last1=Gamelin|first1=Theodore W.|title= Complex Analysis |publisher=Springer |year=2004|isbn=9788181281142}}</ref>|drop=hidden|proof=
Suppose <math>|f(z)| \leq M </math> for all <math> z </math> in the complex plane, we can apply the Cauchy estimate to a disk center at any <math> z_0 </math> of any radius <math> \rho </math> to obtain: 
<math> |f'(z)| \leq \frac{M}{\rho}</math>. 

Let <math> \rho </math> tend to <math> +\infty </math>, we obtain <math> f'(z) = 0 </math>. Since This is true for all <math> z_0 </math>, <math> f(z) = 0 </math> is a constant.
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