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The Liouville's theorem states that if $u$ is a non negative, subharmonic function, $L^\infty(\mathbf{R}^n)$, then $u$ is constant ($n\leq2$). Someone knows a counter example if $n>2$, or where can I find this Liouville's theorem?

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Consider the fundamental solution $u(x) = ||x||^{2 - n}$ of Laplace's equation which is harmonic in $\mathbb{R}^n \setminus \{0\}$ and take $v(x) = \max\{-u, -1\} + 1$. This is an example of a continuous bounded non-negative and non-constant subharmonic function. If you want a smooth example, you can take a smooth compactly supported $\rho : \mathbb{R}^n \rightarrow \mathbb{R}$ with $0 \leq \rho \leq 1$ and $\int_{\mathbb{R}^n} \rho = 1$ and use the fundamental solution to construct a solution to $\Delta u = \rho$ given by $$ u(y) = \int_{\mathbb{R}^n} \frac{\rho(x)}{||x - y||^{n-2}} dy. $$ You can check directly that this is bounded.

If you assume that $u$ is harmonic, then the theorem is also true for $n > 2$. The proof, which is an immediate consequence of the mean value equality, can be found here.

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The theorem is only valid for $n>2$, if the solution is harmonic. Unfortunatly in this case, we have a subharmonic solution. – José Carlos Nov 24 '12 at 20:22
It is always good to read the question... thanks! – levap Nov 24 '12 at 20:29
Do you have any ideia of a counter example for subharmonic functions? – José Carlos Nov 24 '12 at 20:39
Or, where can I find this theorem to subharmonic functions when $n\leq2$? – José Carlos Nov 24 '12 at 20:39
You can see a proof here from which I also took the smooth counter example idea. – levap Nov 24 '12 at 21:05

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