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 $\mathbb{R}^{n}$?



A property of weakly harmonic functions
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
$$
|u(x)|\leq\frac1{|B(x,r)|}\int\_{B(x,r)}|u|\leq \frac1{|B(x,r)|}\|u\|\_{L^p(B(x,r))}|B(x,r)|^{1-1/p}\leq C r^{-n/p},
$$
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 $u\equiv0$.


