# Understanding a proof of the Harnack's inequality

Consider the following proof of the Harnack's inequality (taken from one of the first few pdf files that you find by Googling for a proof of Harnack's inequality):

Theorem $$4$$ : (Harnack inequality for harmonic functions). Assume $$u$$ is a non-negative solution of $$\nabla u = 0$$ in $$\Omega$$. Then for any open, connected subset $$U \subset \subset \Omega$$, we have $$\sup_U {u} \leq C \inf_{U} u$$ for some positive constant $$C$$ that depends only on $$U$$ and $$\Omega$$.

Proof : Let $$\mbox{dist}(\partial \Omega,U) = 4r$$. Let $$x \neq y \in U$$ be such that $$|x-y| \leq r$$. Then, by the mean value theorem we have \begin{align} u(x) = \def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{B_{2r}(x)} u(z)dz &= \frac{1}{\omega_n 2^nr^n} \int_{B_{2r}(x)} u(z)dz \\ & \geq \frac{1}{\omega_n 2^nr^n} \int_{B_r(y)} u(z)dz \\ &\geq \frac{1}{2^n} \avint_{B_r(y)} u(z)dz = \frac{u(y)}{2^n} \end{align}

Now , for any $$x,y \in U$$, there exists a chain of segments of length $$\leq r$$, of length $$N$$ (that depends only on $$\mbox{diam} U$$ and $$r$$) , which connects $$x$$ to $$y$$. The above estimate then implies $$u(x) \geq 2^{-nN}u(y)$$, and the proof is complete. $$\square$$

(Question:) What I do not understand is how we apply the chain of segments to conclude the lower bound coefficient $$2^{-Nn}$$? Do we reapply the mean value theorem multiple times? If so, what are the steps in the sense that from the given computations we know that $$\forall x, y \in U: |x - y| < r: u(x) \geq \frac{u(y)}{2^n}$$. But I do not see how this translates to, say, $$u(x) \geq \frac{u(y)}{\left(2^n\right)^2}$$ in the case that $$x$$ is connected to $$y$$ with two segments.

If I am not mistaken the reason for the inequality is the following: Let $$x \in U$$ be fixed and consider the chain of segments from $$x$$ to $$y$$. You can take a sequence of open balls with radius $$r$$ along the chain of segments such that each centre of an open ball is within distance $$r/2$$ from the last centre. Let $$z$$ be the centre of the second neighbourhood along the chain with $$x$$ being the first. Then $$\forall w \in B(z, r):u(z) \geq \frac{u(w)}{2^n}$$. But as $$z \in B(x, r)$$ we have that $$u(x) \geq \frac{u(z)}{2^n}$$ whence $$\forall w \in B(z, r): u(x) \geq \frac{u(w)}{\left(2^n\right)^2}$$. Induction gives the rest.
• Ok, I see the problem with your approach. Your argument is that if we take a chain of segments and then pick $z$ appropriately, then we can work by induction. However, you haven't shown that $N$ depends only on $\mbox{diam} U$ and $r$, which is crucial for the argument. It turns out that the argument you need has to be more refined than what is written. Oct 28, 2022 at 14:34
• @SarveshRavichandranIyer To avoid misunderstandings: My written answer assumes what is given in the picture I posted. What I was not sure about was how we can bound the value at the other end of the chain of segments. Maybe it could work if we take a finite open cover for $U$, since it is compact, from balls $B(x_n,r/2)$ and then apply my written answer. Namely, argue that given the fixed $x, u(x)\geq\frac{u(x_l)}{2^n}$ for some starting $x_l$, argue why the covering balls must interlace, note that you can use the prior bound in each of the covering balls and then conclude the lower bound? Oct 28, 2022 at 14:58
• Yes, you're basically right. I see that argument as quite hard in general, but I think I know a standard place where one can find it. The proof is definitely by induction, but because you don't know about the geometry of $U$ (apart from the fact that it is open and connected), you will need a very general connecting argument. I think it's harder than just having a finite open cover for a compact set, but that is because I am recollecting the argument's length from memory. Oct 28, 2022 at 15:58