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.
Best Answer
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.