Laplace equation and strong maximal principal. 
Let $\Omega=\{x\in \mathbb{R}^n:||x||>1\}$ and $u\in C^2(\bar\Omega)$, $\triangle u=0$ in $\Omega$ and suppose that $\lim_{||x||\rightarrow \infty} u(x)=0$. P rove that
  $$\max_{\bar\Omega} |u|=\max_{\partial \Omega} |u|$$

I have resolved a very similar problem before. But the hypothesis of been bounded was crucial. I do not know how to begin. Somebody can give any hint. Why do the maximum exist?
Thanks!
 A: This is a Liouville type theorem, test $\Delta u=0$ with $u^+=\max\{0,u\}$ over the annulus $\Omega_R=\{x:1<\|x\|\lt R\}$ and we get:
$$-\int_{\Omega_R} u^+\Delta u\,dx=0.$$ An application of integration by parts gives us
$$\int_{\Omega_R}\nabla u^+\cdot\nabla u\,dx=\int_{\Omega_R}|\nabla u^+|^2\,dx=\int_{\partial\Omega_R}u^+\nabla u\cdot n\,ds.$$
assume that the maximum is not attained at the boundary ($\|x\|=1$), and we see that $\nabla u\cdot n$ is non positive, and since $u^+$ is positive, we have that $\int_{\|x\|=1}u^+\nabla u\cdot n\,ds\le 0$, now we can take the limit as $R\to\infty$ to obtain:
$$\int_{\Omega}|\nabla u^+|^2\,dx=\lim_{R\to\infty}\int_{\Omega_R}|\nabla u^+|^2\,dx\le 0.$$
(Noting that $\|u\|\to 0$ as $\|x\|\to\infty\Rightarrow\int_{\|x\|=R}u^+\nabla u\cdot n\,ds\to 0$.)
Thus $u^+=Const$, do the same but testing with $u^-=\max\{-u,0\}$, we obtain $u^-=Const$, and we can write $|u|=u^++u^-=Const$, so $|u|=Const$. But this is a contradiction, since we have assumed the maximum is not attained at the boundary.
Thus the maximum must be attained at the boundary $\{\|x\|=1\}=\partial\Omega$.
Also note that we have the equality $\int_{\Omega_R}\nabla u^+\cdot\nabla u\,dx=\int_{\Omega_R}|\nabla u^+|^2\,dx$, because if $u<0$, then $u^+=0$.
