Editing Harmonic function (section)

=== Liouville's theorem ===
'''Theorem''': If {{mvar|f}} is a harmonic function defined on all of {{tmath|\R^n}} which is bounded above or bounded below, then {{mvar|f}} is constant.

(Compare [[Liouville's theorem (complex analysis)|Liouville's theorem for functions of a complex variable]]).

[[Edward Nelson]] gave a particularly short proof of this theorem for the case of bounded functions,<ref>{{cite journal |first=Edward |last=Nelson |title=A proof of Liouville's theorem |journal=[[Proceedings of the American Mathematical Society]] |year=1961 |volume=12 |issue=6 |pages=995 |doi=10.1090/S0002-9939-1961-0259149-4 |doi-access=free }}</ref> using the mean value property mentioned above: 
<blockquote>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 {{mvar|f }} is bounded, the averages of it over the two balls are arbitrarily close, and so {{mvar|f }} assumes the same value at any two points.  </blockquote>
The proof can be adapted to the case where the harmonic function {{mvar|f }} is merely bounded above or below. By adding a constant and possibly multiplying by –1, we may assume that {{mvar|f }} is non-negative. Then for any two points {{mvar|x}} and {{mvar|y}}, and any positive number {{mvar|R}}, we let <math>r=R+d(x,y).</math> We then consider the balls {{math|''B{{sub|R}}''(''x'')}} and {{math|''B{{sub|r}}''(''y'')}} where by the triangle inequality, the first ball is contained in the second.

By the averaging property and the monotonicity of the integral, we have
<math display="block">f(x)=\frac{1}{\operatorname{vol}(B_R)}\int_{B_R(x)}f(z)\, dz\leq \frac{1}{\operatorname{vol}(B_R)} \int_{B_r(y)}f(z)\, dz.</math>
(Note that since {{math|vol ''B{{sub|R}}''(''x'')}} is independent of {{mvar|x}}, we denote it merely as {{math|vol ''B{{sub|R}}''}}.) In the last expression, we may multiply and divide by {{math|vol ''B{{sub|r}}''}} and use the averaging property again, to obtain
<math display="block">f(x)\leq \frac{\operatorname{vol}(B_r)}{\operatorname{vol}(B_R)}f(y).</math>
But as <math>R\rightarrow\infty ,</math> the quantity 
<math display="block">\frac{\operatorname{vol}(B_r)}{\operatorname{vol}(B_R)} = \frac{\left(R+d(x,y)\right)^n}{R^n}</math> 
tends to 1. Thus, <math>f(x)\leq f(y).</math> The same argument with the roles of {{mvar|x}} and {{mvar|y}} reversed shows that <math>f(y)\leq f(x)</math>, so that <math>f(x) = f(y).</math>

Another proof uses the fact that given a [[Wiener process|Brownian motion]] {{mvar|B{{sub|t}}}} in {{tmath|\R^n,}} such that <math>B_0 = x_0,</math> we have <math>E[f(B_t)] = f(x_0)</math> for all {{math|''t'' ≥ 0}}. In words, it says that a harmonic function defines a [[Martingale (probability theory)|martingale]] for the Brownian motion. Then a [[Coupling (probability)|probabilistic coupling]] argument finishes the proof.<ref>{{Cite web |date=2012-01-24 |title=Probabilistic Coupling |url=https://blameitontheanalyst.wordpress.com/2012/01/24/probabilistic-coupling/ |archive-url=https://web.archive.org/web/20210508091536/https://blameitontheanalyst.wordpress.com/2012/01/24/probabilistic-coupling/ |archive-date=8 May 2021 |access-date=2022-05-26 |website=Blame It On The Analyst |language=en}}</ref>