Square integrable harmonic function

Question

Let $u$ be harmonic function on $\mathbb{R}^n$ such that $\int_{\mathbb{R}^n} u^2 dx1...dxn < +\infty$ (1). I've got to prove that it implies $u=0$. My idea was to prove that (1) implies that $u$ is bounded on $\mathbb{R}^n$. Then it would mean that $u$ is constant (by Liouville's Theorem) and, because of condition (1), $u=0$. Other legitimate way to attack the problem would be using the fact that $u^2$ is subharmonic (since $f(u)=u^2$ is convex function, Evans, chapter 2) and trying with analogous maximum principle. I would be glad if someone provided any useful hint.

Answer 1

Hint: The subharmonic function $u^2$ satisfies the following averaging inequality: For $x\in \Bbb R^n$ and $B(x,r)$ the ball of radius $r>0$ centered at $x$, $$ u(x)^2\le{\int_{B(x,r)}u(y)^2\,dV(y)\over V(B(x,r))}, $$ where $V$ is $n$-dimensional volume.