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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.

Edit: See my answer.

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This answer could also be an edit to the post, but as I am quite sure that this is the answer I am looking for my question, I can happily close the question before wasting anyone's time more than I already have.

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.

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  • $\begingroup$ You're not wasting anybody's time by asking a good question! Don't fear that. +1 to both contributions. Give me like half an hour, and I'll respond to this question-answer pair. I'll also MathJax the image in the question so that it can be visually accessible to a larger audience. $\endgroup$ Oct 28, 2022 at 13:41
  • $\begingroup$ 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. $\endgroup$ Oct 28, 2022 at 14:34
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    $\begingroup$ @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? $\endgroup$ Oct 28, 2022 at 14:58
  • $\begingroup$ 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. $\endgroup$ Oct 28, 2022 at 15:58

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