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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 $\sup_{B_{1/4}}u / u(0)$ can get very large, the estimate (from the prior step) becomes useless as $r_k \to 0$, even after I replace $u(0)$ with $\sup_{B_{1/4}} u / N$ for large enough $N$. So it's unclear to me how to use it infinitely many times. I also suspect, but am not sure, that I should use the Nash-Digiorgi-Moser theorem here. Any suggestions 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.

Edit: Here is the full outline -- and yes, I know $u$ is Holder continuous on compact subsets $K \subset B_2$, with $[u]_\alpha \le C(K)\|u\|_\infty$.

  1. Let $u: \Omega \to [0,\infty)$ be a supersolution, then show that if $B_{2R}(x_0) \subset \Omega$ and $r<R$, we have $$\inf_{B_R(x_0)} u \ge c \left(\frac{r}{R}\right)^q \inf_{B_r} u.$$

This follows from iteration of the weak Harnack inequality.

  1. Let $u:B_2 \to [0,\infty)$ be a solution. Show that $$u(0) \ge cr_k^q \inf_{B_{r_k}(x_k)}u$$ for any $x_k \in B_{1/2}$ and $r_k$ small enough. This is simple enough.
  2. As I defined the $x_{k+1}$ above, prove the inequality $$u(x_{k+1}) \ge \frac{u(x_k) - cr_k^{-q} u(0)}{1-\theta}.$$
  3. 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 $u(x_{k+1}) \ge \beta u(x_k)$. Conclude.

I have everything (including how to conclude), except for the choice of $r_k$ and $\beta$ in step $4$.

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  • $\begingroup$ Someone might know straight away what to do with this, but I think you could improve the way you've asked the question a bit to make it easier for someone to help. My first thoughts are a) What exactly do you already know about $u$ at this stage? You mention De Giorgi Nash Moser, so do you know that u is Holder continuous? And b) I got confused with use of "the preceding step, which I am led to believe is what implies the claim", "from the prior step" etc. Where does the inequality (the last displayed equation) come from? Maybe put the statements in the order in which they come in the proof? $\endgroup$ – T_M Jan 24 at 17:21
  • $\begingroup$ @T_M I have made an edit including the full proof outline (but it still needs to be peer reviewed). I do know that $u$ is Holder continuous, yes. The inequality in the last displayed equation essentially comes from rearranging terms and applying the bounds from the earlier steps, and the definition of oscillation. $\endgroup$ – David Bowman Jan 24 at 17:32

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