Show that if $u$ is harmonic in $\mathbb R^n$ and $u=o(|x|)$, then $u$ is a constant.(Hint: use the solid version of the mean value property $u(x)=\frac{1}{\omega R^n} \int _{B_{R(x)}} u(y)dy$, and estimate $|u(x)-u(x')|$.)
Can anyone tell me how to use this hint?
The key point is that for fixed $x,x'$ the volume of the symmetric difference $B(x,R)\triangle B(x',R)$ is $O(R^{n-1})$ as $R\to\infty$. Hence, the difference of integrals of $u$ over these two balls is $o(R^n)$. Dividing by the volume of $B(x,R)$ we find $u(x)-u(x') = o(1)$, which means $u(x)=u(x')$ as claimed.