Let $u :\mathbb{R}^2 \to \mathbb{R}$ be a smooth harmonic function ($\Delta u = 0$). Furthermore, suppose that $u \in L^p(\mathbb{R}^2)$, with $1 < p < \infty$. Does it follow that $u = 0$ everywhere?
By the theorem of Liouville, this follows immediately if $u \in L^\infty(\mathbb{R}^2)$. For $u \in L^p(\mathbb{R}^2)$, this would also follow if we knew that $u$, for example, is uniformly continuous ($u$ is necessarily unbounded, find a sequence of points $x_k$ such that $u(x_k) \to \infty$, and apply the definition of uniform continuity in a neighbourhood of those points to get a contradiction with $u \in L^p(\mathbb{R}^2)$).
The enemy seems therefore higher and narrower ``spikes'' going off to infinity. Any help would be much appreciated!
This may settle it:
By the mean value property, we have, leting $r >0$, $$ |u(x)| \leq \frac 1 {\pi r^2} \int_{B_r(x)} |u(y)| \ d y. $$ Using H\"older on the RHS, we have $$ \frac 1 {\pi r^2} \int_{B_r(x)} |u(y)| \ d y \leq \frac 1 {{\pi}^{1-1/q}} r^{-2 + 2/q} \|u \|_{L^p(\mathbb{R}^2)}. $$ Where $1/q + 1/p =1$. Hence $-2+2/q = -2/p$. Then send $r \to \infty$.