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Let $u \in C^{\infty}(\mathbb{R}^3)$ be a harmonic function on $\mathbb{R}^3$: $\Delta u(x) = 0$ for all $x \in \mathbb{R}^3$.

Assume that $|u(x)| \leq \sqrt{|x|}$ holds for all $x$, where $x = (x_1,x_2,x_3)$. Show that $u(x) = 0$ for all $x \in \mathbb{R}^3$.

I want to apply Harnack Inequality to $u$ on $B_R(0)$ because then I can get something like $$u(x) \leq \frac{R(R+|x|)}{(R - |x|)^2}u(0) \leq \frac{R(R+|x|)}{(R - |x|)^2}\sqrt{|0|} \leq 0.$$ However this only applies when $u$ is non-negative on the ball, which is not something I am given.

Thanks.

Sebastiano
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Jamal
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    $u$ is harmonic, so each $\frac{\partial u}{\partial x^i}$ is harmonic. Now use the mean-value property, over a ball of radius $R$, on $\frac{\partial u}{\partial x^i}$, followed by the divergnce theorem to get an upper bound on the derivatives. Show this vanishes as $R\to\infty$. i.e you just run the proof of Liouville's theorem as much as possible, and you'll see that it holds in more generality than simply "bounded harmonic functions on $\Bbb{R}^n$ are constant". – peek-a-boo Nov 26 '22 at 11:29
  • @peek-a-boo would you mind showing me how you apply divergence theorem to the mean value property integral? thanks – Jamal Nov 26 '22 at 11:40
  • write out what the mean-value theorem says first. The divergence theorem implies $\int_{\Omega}\frac{\partial f}{\partial x^i},dV=\int_{\partial \Omega}f\nu_i,dA$. This is easily seen by applying the divergence theorem to the vector field $\mathbf{F}=(0,\dots, f, \dots, 0)$, with the function $f$ in the $i^{th}$ spot. – peek-a-boo Nov 26 '22 at 11:40
  • so mean value on ball gives $u_{x_i}(x) = \frac{1}{|B_R(0)|}\int_{B_R(0)} u_{x_i}(y) dy$. Then the divergence theorem gives $\frac{1}{|B_R(0)|}\int_{B_R(0)} u_{x_i}(y) dy =\frac{1}{|B_R(0)|}\int_{\partial B_R(0)} u(y) v_i dS(y)$, where $v_i$ is the $i^{th}$ component of the unit normal vector to the ball, which is $\frac{y_i}{|y|}$. So overall this becomes $\frac{1}{|B_R(0)|}\int_{\partial B_R(0)} u(y)\frac{y_i}{|y|} dS(y)$? – Jamal Nov 26 '22 at 11:51
  • You should be integrating over $B_R(x)$ not $B_R(0)$ (so the normal also needs to be modified). Anyway, the explicit formula for the normal doesn't matter. Once you put absolute values, just use $|\nu_i|\leq 1$. So, now you have an upper bound like $\frac{1}{\text{volume of $B_R$}}\int_{\partial B_R(x)}|u|,dS$. Use your assumption on $u$, and take the limit $R\to\infty$. Why do you get $0$? Therefore... – peek-a-boo Nov 26 '22 at 12:06
  • @peek-a-boo I get $u_{x_i}(x) = \frac{1}{|B_R(x)|}\int_{\partial B_R(x)} |u(y)| dS(y) \leq \frac{1}{|B_R(x)|}\int_{\partial B_R(x)} \sqrt{|y|} dS(y)$. But on $\partial B_R(x)$, we have that $|y| = R$. So the integral becomes $u_{x_i}(x) \leq \frac{1}{|B_R(x)|}\int_{\partial B_R(x)} \sqrt{R} dS(y) = \frac{1}{|B_R(x)|}\sqrt{R} \int_{\partial B_R(x)}dS(y) = \frac{1}{|B_R(x)|}\sqrt{R} |\partial B_R(x)|$. But in $\mathbb{R}^3$, we have $|B_R(x)| = 4 \pi R^3$ and $|\partial B_R(x)| = 4 \pi R^2$. So then $u_{xi}(x) \leq \frac{3 \sqrt{R}}{R}$ which goes to $0$ as $R \rightarrow \infty$. – Jamal Nov 26 '22 at 12:24
  • @peek-a-boo so then we have that each derivative of $u$ is $0$ and so $u$ is constant. But then that means $u(0) = u(x)$ for all $x$, and $|u(0)| \leq 0$, so also $|u(x)| \leq 0$ for all $x$. But does this not mean that $u$ could be a negative constant instead of just $0$? – Jamal Nov 26 '22 at 12:28
  • Almost: one small mistake is that because we're working on the the boundary of the ball $B_R(x)$, it is not true that $|y|=R$. The magnitude of $y$ will actually depend on where you are on the boundary. But this is not a big problem, because the maximum over the boundary is related to some multiple of $R$ (i.e you have an inequality, not an equality, but the inequality still has the correct order of magnitude in $R$, so the integral still decays like $1/\sqrt{R}$). – peek-a-boo Nov 26 '22 at 12:30
  • ... If $|u(0)|\leq 0$, then it is equal to 0 for obvious reasons. – peek-a-boo Nov 26 '22 at 12:31
  • oh ye, sorry that was silly of me. Thank you very much for walking me through that – Jamal Nov 26 '22 at 12:32
  • As an exercise, convince yourself this proof works in $n$-dimensions as well. – peek-a-boo Nov 26 '22 at 12:35

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