Harnack Inequality for nonnegative subsolutions to uniformly elliptic PDE I am trying to prove a Harnack inequality for a nonnegative subsolution $u \in H^1(B_2)$ to the PDE $\text{div}(A Du) \ge 0$,where $A = A(x)$ is uniformly elliptic. The proof outline I am following is from a set of notes by a professor at my university, and the key step is the following inductive scheme:
Set $x_0$ to be a point such that $$u(x_0) = \sup_{B_{(0,1/2)}} u,$$ and choose $x_k$ inductively such that $x_{k+1}$ is such that $$u(x_{k+1}) = \sup_{B(x_k, r_k)} u$$
for $r_k$ sufficiently small to be chosen in a moment.
I have all of the steps except the following: suppose $$\frac{\text{sup}_{B_{0,1/4}} u}{ u(0)}$$ is sufficiently large, then we can choose a sequence $r_k$ such that $\sum r_k <1/2$ and a $\beta>1$ such that $u(x_{k+1}) \ge \beta u(x_k)$. That this would imply the result is immediate because it would contradict the boundedness of $u$. The preceding step, which I am led to believe is what implies the claim, is the following:
$$u(x_{k+1}) \ge \frac{u(x_k) - cr_k^{-q} u(0)}{1-\theta}$$
where $c$, $q$ are absolute constants, and $1-\theta \ge \text{osc}_{B_1}u>0$ and $0<\theta \le \inf_{B_1} u$. Here $c,q>0$ are absolute constants.
I basically don't know what to do with this. Even if I assume the ratio in question gets very large, the estimate (from the prior step) becomes useless as $r_k \to 0$. So it's unclear to me how to use it infinitely many times. I have the Nash-Digiorgi Holder regularity theorem at my disposal. Any hints or references would be much appreciated! I cannot find a similar proof anywhere, and given that I have provided the details for all of the other (numerous) steps, I would like to complete it. 
 A: Was looking through my old questions and realized this was never answered. I did manage to solve this, so will post the solution for anyone curious. The proof is basically by induction.
Set $M = \displaystyle \frac{\sup_{B_{1/4}} u}{u(0)}.$
We will show that if $M$ is sufficiently large, $u$ eventually becomes unbounded, which is a contradiction. Choose $\beta_0>0$ small such that $(1-\theta)(1+\beta_0) \le 1- \beta_0 <1$. With $r_k, x_k$ defined as in the opening post, our inductive hypothesis is that $u(x_{k}) \ge M(1+\beta_0)^{k-1}u(0)$ for all $k$, which we can force (as will be seen) if $M$ is large enough. The base case is satisfied by assumption, so now we assume it holds for some $k \ge 1$. Pick $0<\delta <1$ and $a>0$ very small. We are going to set $r_k = a \delta^k$. It can be shown that
$$u(x_{j+1})\ge \frac{u(x_j) - c^{-1} r_j^{-q}u(0)}{1-\theta}$$
for $c>0, q>0$ universal, and $\theta$ related to the oscillation of $u$ as detailed in the opening post.
So, from this inequality, to prove the inductive step, it suffices to show
$$\frac{M(1+\beta_0)^{k-1} - c^{-1}a^{-q}\delta^{-kq}}{1-\theta} \ge M(1+\beta_0)^k$$
since we have assumed $u(x_k) \ge M(1+\beta_0)^{k-1}u(0)$.
This occurs if and only if
$$1 \le \frac{1}{(1-\theta)(1+\beta_0)} - \frac{1}{ca^q\delta^{kq}(1-\theta)M(1+\beta_0)^{k}}.$$
So, we choose $0<\delta <1$ such that $\delta^q(1+\beta_0)>1$, and $a>0$ small enough such that $\sum a\delta^k < 1/2.$ By our choice of $\beta_0$<
$$\frac{1}{(1-\theta)(1+\beta_0)}> \frac{1}{1-\beta_0} >1.$$
Choose $K$ large enough so that
$$\frac{1}{1-\beta_0} -1 > \frac{1}{ca^q\delta^{kq}(1-\theta)M(1+\beta_0)^{k}}$$
for all $k \ge K$ taking $M\ge 1$, as a lower bound, for instance.
Then choose $M$ large enough so that the inequality is true for $1\le k \le K$. This ensures that the inductive step goes through for each $k$.
This means that $u$ blows up on the ball $B_{1/2}$ which contradicts its boundedness. Therefore the initial assumption that the ration $\sup_{B_{1/4}} u /  u(0)$ was large was false. Harnack's inequality follows.
