Return to Answer

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 $L^p({\mathbb R}^n)$ is zero.

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, hence identically $0$. 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 $u\equiv0$.

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 $u\equiv0$.

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 $L^p({\mathbb R}^n)$ is zero.

Another way is to note that weakly harmonic functions are smooth (in fact analytic), hence harmonic. In particular, you do not need to talk about "almost everywhere". Then use Liouville's theorem.

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 $L^p({\mathbb R}^n)$ is zero.

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, hence identically $0$. 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 $u\equiv0$.

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 $L^p({\mathbb R}^n)$ is zero. Also, you do not need to talk about "almost everywhere" because weakly harmonic functions are smooth (in fact analytic).

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 $L^p({\mathbb R}^n)$ is zero.

Another way is to note that weakly harmonic functions are smooth (in fact analytic), hence harmonic. In particular, you do not need to talk about "almost everywhere". Then use Liouville's theorem.