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I want to know if my proof is correct. This is what I have to show: Let $u:\mathbb{R}^n\rightarrow \mathbb{R}$ be harmonic and bounded above. Show that $u$ is constant. I want to apply Harnack's inequality for non-negative harmonic functions $$\sup_{B(x_0, r)}u\leq c\inf_{B(x_0, r)} u, $$ when $B(x_0, 4r)\subset \Omega$ and where $c=3^n.$ So let $R>0$ and $x\in B(0, R)$. $u$ is bounded above so $u(y)\leq M$. Define $v=M-u$. Then $v$ is harmonic and $v\geq 0$. By Harnack's inequality we have $$v(x)\leq\sup_{B(0, R)}v\leq c\inf_{B(0, R)}v\leq cv(0).$$ Because $R$ was arbitrary we conclude that $v$ is bounded above. Since it's also bounded below, by Liouville's theorem $v$ is constant, thus $u$ is constant.

Am I right?

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  • $\begingroup$ Your statement of Harnack's inequality appears doubtful. I think the constant c depends on R. I have checked both Rudin and Conway. If c depends on R then your proof breaks down. $\endgroup$ Oct 4, 2017 at 7:35
  • $\begingroup$ @KaviRamaMurthy: notice it's only claimed for balls. By a scaling argument $c$ can be chosen independent of $r$ for balls, indeed $c$ can be chosen uniformly for any family of similar bounded domains. $\endgroup$
    – Dap
    Oct 4, 2017 at 7:54
  • $\begingroup$ @KaviRamaMurthy: that was a bit unclear: I mean in the case $R=4r$, a scaling argument shows $c$ can be chosen uniformly in $r$ (indeed $c=3^n$ as stated in the question). $\endgroup$
    – Dap
    Oct 4, 2017 at 9:02
  • $\begingroup$ For what it's worth, you could avoid the appeal to Liouville by setting $M=\sup u$ so $\inf v=0$, and picking $R$ such that $\inf_{B(0,R)} v < v(x)/c$, assuming $v(x)>0$. $\endgroup$
    – Dap
    Oct 4, 2017 at 9:02
  • $\begingroup$ @KaviRamaMurthy well, we proved that version of Harnack's inequality in class, in fact you can find similar result in Evans 's book $\endgroup$ Oct 4, 2017 at 11:32

1 Answer 1

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You are right.

I feel like I should write some more, so here's an alternate argument by generalising a proof of Liouville's theorem. Consider any two points $x,y$. Let $c$ denote the volume of $B(0,|x-y|)$. The integral of $u$ in $B(x,2|x-y|)$ is $c2^n u(x)$, and the integral of $u$ in $B(y,|x-y|)$ is $c u(y)$. This means that the average value of $u$ in $B(x,2|x-y|)\setminus B(y,|x-y|)$ is

$$\frac{2^n u(x) - u(y)}{2^n - 1} = u(y) + \frac{2^n}{2^n-1}(u(x)-u(y)).$$

In particular there exists a point $x'$ with $u(x')-u(y)\geq \frac{2^n}{2^n-1}(u(x)-u(y))$. So if we start with points $x_0,y$ with $u(x_0)>u(y)$, we can inductively pick points $x_1,x_2,x_3,\dots$ with

$$u(x_k)-u(y)\geq \left(\frac{2^n}{2^n-1}\right)^k(u(x_0)-u(y))\to+\infty.$$

Thus any non-constant harmonic function is unbounded above.

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