Here is Liouville's Theorem

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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 :

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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 ?


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 $R \to \infty$.

  • $\begingroup$ Unless I'm mistaken, I don't think that inequality is done explicitly in Evans' book; for the record, it can be proved using Poisson's formula and the mean value theorem (there's a proof on Wikipedia at en.wikipedia.org/wiki/Harnack's_inequality). $\endgroup$ – Matt Rigby Sep 14 '14 at 16:09
  • $\begingroup$ The hint I was given : Look at the proof of Harnack’s inequality. Show that for any radius r, if x ∈ B(0,r), then u(x) ≤ $2^n$ u(0), and note that this bound doesn’t actually depend on r. $\endgroup$ – Peter Sep 14 '14 at 16:17
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    $\begingroup$ Following that hint, you get that $u$ is bounded, and then you can apply the unmodified Liouville theorem. $\endgroup$ – Ian Sep 14 '14 at 16:35
  • $\begingroup$ I got stuck at the part for "any radius r" , since that Harnack 's proof above in my post, r is 1/4 distance(V,dV). $\endgroup$ – Peter Sep 14 '14 at 16:53
  • $\begingroup$ That's only to make sure things are defined where they need to be; $r$ can be anything you like when it's defined on all of $\mathbb{R}^n$. $\endgroup$ – Matt Rigby Sep 14 '14 at 17:11

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.


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