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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.2012-11-24
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    It is always good to read the question... thanks!2012-11-24
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    Do you have any ideia of a counter example for subharmonic functions?2012-11-24
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    Or, where can I find this theorem to subharmonic functions when $n\leq2$?2012-11-24
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    You can see a proof [here](http://www.dam.brown.edu/people/menon/am223/sol2.pdf) from which I also took the smooth counter example idea.2012-11-24
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    The first example doesn't work since you still have the singularity (and taking the minimum gives a superharmonic function instead).2012-11-24
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    Bad day... it should be ok now. Thanks.2012-11-24
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    In the first example, the function $v$ is harmonic or subharmonic? I can't see how to show that $u''>0$ in some point. And, if $v$ is harmonic, $v$ should be constant for the Liouville theorem. Could you show me that $v$ is subharmonic? (in the simple case, when $n=3$). Thank you.2012-11-26
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    $v$ 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