Improve Liouville's Theorem in Evans ' PDE

Question

Here is Liouville's Theorem

Suppose that $u \colon \mathbb{R}^n \to \mathbb{R}$ is harmonic and $u \geq 0$. Prove that $u$ is constant. (In this problem , instead of $u$ is bounded now $u \geq 0$ .)

Harnack 's inequality proof in Evans :

Can I use Harnack 's inequality and show that $u(x) < 2^n u(0)$ ? How can I create an upper bound like the above proof that can converge to 0 as r goes to infinity ?

Answer 1

If you have Harnack's inequality then you are essentially done already. Here's a way to continue:

Let $x$ be arbitrary, $\| x \| = r$, and $R>r$. Then

$$\frac{1-r/R}{1+(r/R)^{n-1}} f(0) \leq f(x) \leq \frac{1+r/R}{1-(r/R)^{n-1}} f(0).$$

Now send $R \to \infty$.

Answer 2

I would say that Evan's proof is not sharp. Actually, if $u \geq 0$ is harmonic, by the divergence theorem you can write $$ \frac{\partial u}{\partial x_i}(x_0)=\frac{n}{\omega_n R^n} \int_{\partial B_R(x_0)} u(y) \, d\Sigma. $$ Hence $$ \left| \frac{\partial u}{\partial x_i}(x_0) \right| \leq \frac{n}{R} u(x_0), $$ and you let $R \to +\infty$. This "proof" comes from the book Elliptic differential equations by Lin and Lin, AMS.