Proof of Liouville's Theorem for Harmonic functions Recall that a function $f:U (\subset \mathbb R^2) \to \mathbb R$ is called harmonic if $\Delta f=0$ where $\Delta$ is Laplacian operator.

Liouville's Theorem: Let $f$ be a harmonic and bounded function then $f$ is constant.( I know that harmonic functions have mean value property)

I know that harmonic functions have mean value property,how can i deduce Liouville's theorem using mean value property and maximum modulus principle.
 A: A very very slick proof due to Edward Nelson follows from the version of mean value that integrates over thr whole disc rather than just the circle (can be obtained by integrating the standard mean value property). Pick two points and write the value of the function as integrals over the two discs both of the same radius. Let their radius go to infinity, the symmetric difference of the two discs gets smaller and smaller in proportion to their overlap.  Since the function is bounded, the average of the function on one disc is then essentially the average of the function on the intersection of the discs.  Hence as the radius goes to infinity, the average over either disc goes to the same number and the value of the function is the same at both points. 
A: There is also a slightly fancier argument that reaches a stronger conclusion: for a tempered distribution $u$ on $\mathbb R^n$, if $\Delta u=0$ then $u$ is a polynomial (and for $u$ bounded it must be constant).
Proof: Taking Fourier transform, we obtain $r^2\cdot \widehat{u}=0$, and can conclude that $u$ is supported at $0$. Classification of such distributions (essentially the theory of Maclaurin-Taylor series) says that $\widehat{u}$ is a linear combination of derivatives of Dirac $\delta$. Thus, $u$ itself is a polynomial (annihilated by $\Delta$).
