TITLE: $f \in C^2(\mathbb{R}^2)$ is harmonic; $f$ is in one $L^p(\mathbb{R}^2)$ $\implies$ $f$ is constant
QUESTION [5 upvotes]: Suppose $f \in C^2(\mathbb{R}^2)$ is harmonic. Suppose that $f$ is in one $L^p(\mathbb{R}^2)$ space, with $1 \le p \le \infty$. 
How can I prove that $f$ is constant and in particular is identically $0$ if $1 \le p < \infty$ ?

REPLY [2 votes]: I'm going to add to the proof by @Glitch because I think it can be simplified. Starting with the volume version of the mean value property (which follows from the spherical MVP), we have, for any $r>0,$
$$u(x) = \frac{1}{V(B(x,r))}\int_{B(x,r)} u\,dV,$$
where $V$ is Lebesgue volume measure on $\mathbb R^n.$ This implies
$$ |u(x)| \le \frac{1}{V(B(x,r))}\int_{B(x,r)} |u|\, dV.$$
For $1\le p <\infty,$ apply Jensen's inequality to see
$$\tag 1 |u(x)|^p \le \frac{1}{V(B(x,r))}\int_{B(x,r)} |u|^p\, dV.$$
Recall this holds for any $r>0.$ As $r\to \infty,$ $V(B(x,r)) \to \infty,$ while the integral in $(1)$ is bounded above by $ \int_{\mathbb R ^n} |u|^p\, dV < \infty.$ It follows that $|u(x)|^p = 0,$ i.e., $u(x) = 0.$ Since $x$ was an arbitrary point in $\mathbb R^n,$ we see $u\equiv 0$ as desired.