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?
What is the counter example?
1 Answers
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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0$v$ is subharmonic, but not in the sense $\Delta u \geq 0$, because it is not twice differentiable at the place you "cut" it (along $u(x) \equiv 1$). It is subharmonic in the sense it satisfies the mean value inequality, or in the sense described in the [wiki page](http://en.wikipedia.org/wiki/Subharmonic_function). What you want is the second example. – 2012-11-28