Let $u$ be a distribution in $\mathbb{R}^{n}$ and $\Delta u$=0 in the distributional sense. In addition $u\in L^{p}(\mathbb{R}^{n})$, $p>1$, then can we conclude that $u$ is zero almost everywhere in $\mathbb{R}^{n}$?
Already the fact that $u$ is a tempered distribution and is weakly harmonic implies that $u$ is a polynomial. Then observe that the only polynomial in $L^p({\mathbb R}^n)$ with $1\leq p<\infty$ is zero, and the only polynomials in $L^\infty({\mathbb R}^n)$ are constants. Another way is to note that weakly harmonic functions are smooth (in fact analytic), hence harmonic. In particular, $u$ has pointwise values. If $p=\infty$, Liouville's theorem implies that $u$ is constant. In fact, in this case we cannot conclude that $u\equiv0$. Now assume that $1\leq p<\infty$. Let $x\in{\mathbb R}^n$, and let $B(x,r)$ denote the ball of radius $r>0$ centered at $x$. Then by the mean value property we have $$ u(x)\leq\frac1{B(x,r)}\int_{B(x,r)}u\leq \frac1{B(x,r)}\u\_{L^p(B(x,r))}B(x,r)^{11/p}\leq C r^{n/p}, $$ which, upon taking $r\to\infty$, shows that $u(x)=0$. Here $B$ denotes the volume of $B$, and we have used Hölder's inequality. Since $x$ was arbitrary, we conclude that $u\equiv0$. 

