Show that $\lim_{|y|\to\infty}|y|^{n-2}u(y)>0$

Question

For $n\ge 3$,Let $u\in C^2(R^n) $, $\Delta u\le 0, u>0$ in $R^n$ ,show that $\lim_{|y|\to\infty}|y|^{n-2}u(y)>0$.

I consider the maximum principle,but I don't know how to deal with.

Answer 1

I don't know if the limit necessarily exists, but you can at least show that $\liminf_{y\to \infty} |y|^{n-2}u(y)>0$. I've sketched the argument below.

Recall the (scaled) fundamental solution $G(x) := |x|^{2-n}$ is harmonic away from $x=0$. Hence $u-cG$ is superharmonic ($\Delta (u-cG)\leq 0$) away from the origin for any constant $c$. Choose a constant $c>0$ small enough so that that $u(x) - cG(x)\geq 0$ for all $|x|=1$. This is possible because $u>0$ and $u$ is continuous, so $u$ assumes a positive minimum on the set $\{x\, : \, |x|=1\}$.

Now consider the annulus $A_R=\{x \, : \, 1 \leq |x| \leq R\}$, where $R>1$. By the maximum principle, $u-cG$ attains its minimum value over $A_R$ on the boundary, either $|x|=1$ or $|x|=R$. For $|x|=1$ we have $u(x)-cG(x)\geq 0$. For $|x|=R$ we have

$$u(x) - cG(x) \geq -cR^{2-n}.$$

Therefore, for all $x \in A_R$ we have

$$u(x) \geq cG(x) -cR^{2-n}.$$

Send $R\to \infty$ to find that $u\geq cG$ on the domain $|x|\geq 1$. Therefore

$$\liminf_{y\to \infty} |y|^{n-2}u(y) \geq c>0.$$