Complex Analysis – Proof of Liouville’s Theorem for Harmonic Function

Question

In complex analysis, Liouville's theorem is that every bounded entire function is constant. To prove it, Cauchy intergral formula is used$$f(z) = \frac{1}{2\pi i}\int_C\frac{f(s)}{s-z}ds$$ where C is circle with radius of R, and $|f'(z)|\le \frac{M}{2\pi R}$ by using bounded condition($|f(z)|\le M)$ and Cauchy's differentiation formula. Then, $f'(z)$ goes to $0 $ as R goes to infinity. So, f(z) is constant.
The same method can be used in proving Liouville's theorem for harmonic function?

In Partial differential equation, Liouville's theorem is that
$u(\mathbf x)$ is constant if $u(\mathbf x)$ is harmonic function in $\Bbb R^N (N=2,3) $ and $u(\mathbf x)$ is bounded by $M>0$.

By mean value principle, $$u(\mathbf x) = \frac{1}{|\partial B_R(\mathbf x)|}\int_{\partial B_R(\mathbf x)}u(s)ds$$ Then, how can I differentiate $u(\mathbf x)$?

Answer 1

Quote (the full paper!) of A Proof of Liouville's Theorem:

Consider a bounded harmonic function on Euclidean space. Since it is harmonic, its value at any point is its average over any sphere, and hence over any ball, with the point as center. 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 the function is bounded, the averages of it over the two balls are arbitrarily close, and so the function assumes the same value at any two points. Thus a bounded harmonic function on Euclidean space is a constant.