We say that a locally integrable function $u$ defined in $RR^N$ is harmonious in $\Omega\subseteq RR^N$ if it satisfies $(2.4)$ for any ball $B_r(x)$ with $x\in\Omega$. Prove that, if $u$ is harmonious in $\Omega$, then
(i) $u$ is bounded on $\Omega$;
(ii) $u$ is locally Lipschitz continuous in $\Omega$;
(iii) $u\inC^\infty(\Omega)$ and all its derivatives are harmonious;
(iv) $u$ is analytic in $\Omega$, if it is bounded on $RR^N$.
Let $\Omega\subseteqRR^N$ be a connected open set and set $\Omega_r ={y + z : y\inOmega,|z|<r }=\Omega+B_r(0)$. Let $u$ be a harmonious function in $\Omega$. If there is an $x_0\in\Omega$ is such that $u(x_0) = \text{sup}_{\Omega_r} u$, then $u$ is constant in $\Omega$.
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