$f \in C^2(\mathbb{R}^2)$ is harmonic; $f$ is in one $L^p(\mathbb{R}^2)$ $\implies$ $f$ is constant 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 $1 \le p < \infty$ ?
 A: 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.
A: I'll present the proof with $1 \le p < \infty$ since the case $p=\infty$ can be handled with Liouville's theorem.  I'll also work more generally in $\mathbb{R}^n$ for $n\ge 2$ rather than $n=2$, as you ask about.
The key is to use the mean-value property.  It shows that for $x \in \mathbb{R}^n$ and $r>0$ 
$$
u(x) = \frac{1}{\omega_n r^{n}} \int_{ B(x,r)}  u(y) dy 
$$
where $\omega_n = |B(0,1)|$.  Then we use Holder's inequality:
$$
|u(x)| \le \frac{1}{\omega_n r^n} \int_{B(x,r)} |u(y)| dy \le \frac{1}{\omega_n r^n} (\omega_n r^n)^{1/p'}  \Vert u \Vert_{L^p(B(x,r))}  \\
= \frac{1}{\omega_n^{1/p} r^{n/p}} \Vert u \Vert_{L^p(B(x,r))} 
$$
where $1/p + 1/p' =1$.  Since $u$ is such that 
$$
\Vert u \Vert_{L^p(\mathbb{R}^n)} < \infty,
$$
we plug in to see that
$$
|u(x)| \le \frac{1}{\omega_n^{1/p} r^{n/p}} \Vert u \Vert_{L^p(\mathbb{R}^n)}.
$$
Sending  $r \to \infty$ then shows that 
$$
|u(x)| \le \lim_{r \to \infty} \frac{1}{\omega_n^{1/p} r^{n/p}} \Vert u \Vert_{L^p(\mathbb{R}^n)}=0,
$$
and so $u(x) =0$ for all $x$.  Thus the only harmonic function in $L^p(\mathbb{R}^n)$ for $1 \le p < \infty$ is identically $0$.
A: Since $f\in L^p$, we have that $f$ is a tempered distribution. Notice that $\Delta f =0$. Then $|\xi|^2\widehat f =0$, which implies that $\widehat f$ has support contained in {0}. Any distribution of support in $\{0\}$ is of the form
$$\widehat f = \sum_{|\alpha|\leq d}c_\alpha \partial^{\alpha}\delta$$
where $d$ is a non negative integer and $c_\alpha \in \mathbb C$.
Taking the inverse Fourier transform we conclude that $f$ is an polynomial function.
But the only polynomial function in $L^p$ is $f=0$. 
