Integrable Harmonic Function Let $p$ be such that $1\leq p<\infty$. Suppose $U(x,y)$ is a harmonic function in $\mathbb{R}^{2}$ that satisfies $\int\int_{\mathbb{R}^{2}}|U(x,y)|^{p}dx dy<\infty$. Show that $U(x,y)$ is constant. 
My idea is that we need to show first that $U(x,y)$ has a local maxima, and because a non constant harmonic function cannot have a local maxima, $U(x,y)$ must be constant. Unfortunately, I don't know how to prove that the function must have a local maxima. Moreover, If $U(x,y)$ is only required to be continuous (not necessarily harmonic, but still satisfies $\int\int_{\mathbb{R}^{2}}|U(x,y)|^{p}dx dy<\infty$), is it still true that U(x,y) contains a local maxima? 
Any help would be appreciated. 

Harmonic functions have mean value property: $u(x,y)=A\int_{\{0\leq r \leq 1, 0\leq \theta \leq 2\pi\}} u(x+r\cos \theta, y+r\sin \theta) r\,dr\,d\theta$ where $\frac 1A$ is the area of the unit disk. Applying Holder's inequality (writing $u$ as $u$ times $1$) we see that $u$ is actually bounded on the whole of $\mathbb R^{2}$. Hence it is a constant. [Here is the standard Complex Analysis argument to show that  bounded harmonic functions are constants:  there is an entire function $f$ with real part $u$. $e^{f}$ is then a bounded entire function because $|e^{f(z)}|=e^{\Re f(z)}$. By Liouville's Theorem $e^{f}$ is a constant from which it follows that $F$ is a  constant. So $u=\Re f$ is a constant]. 
