A detail in one step in a theorem from a paper of Brezis and Merle

Question

In paper [1] Brezis and Merle prove theorem 3 by using the following fact. Let $w_n=u_n-v_n$, $\Delta w_n=0$ on $\Omega$ (a bounded domain in $\mathbb{R^2}$) and $w_n^+$ is bounded in $L^{\infty}_\mathrm{loc}(\Omega)$. Then by Harnack's principle either

1. a subsequence $w_{n_k}$ is bounded in $L_\mathrm{loc}^{\infty}(\Omega)$

2. $w_n$ converges uniformly to $-\infty$ on compact subsets of $\Omega$.

Here $v_n$ is a solution of the boundary value problem $$ \begin{cases} -\Delta v_n=V_n e^{u_n} &\text{in }\Omega\\ \quad\: v_n=0 &\text{on }\partial\Omega \end{cases}$$ while $u_n$ solves $-\Delta u_n= V_n e^{u_n}$.

Since there is no boundary condition on $u_n$, I cannot say anything about the increasing or decreasing of the sequence $w_n$, while this would be required for the application of Harnack's principle as stated in most of the references I'aware of.
Any help would be appreciated: in particular I hope to get a hint regarding the use of the customary Harnack's principle in this case in order to prove the above claim.

Reference

[1] Haïm Brezis, Frank Merle, "Uniform estimates and blow up behavior of $-\Delta u=Ve^u$ in two dimensions", Communications in Partial Differential Equations 16, No. 8-9, 1223-1253 (1991), MR1132783, Zbl 0746.35006.

Answer 1

This follows from the mean value theorem. Assume that (up to a subsequence) $w_n(x_n) \geq -B$ with $(x_n) \in K$ (a compact subset of $\Omega$). If $x_n \to x_0 \in K$ and $B(x_n,r) \in \Omega$ for every $n$, then $\int_{B(x_n,r)} w_n \geq -B$ and then $$\int_{B(x_n,r)} w_n^- \leq B+Ar^N, \quad \int_{B(x_n,r)} |w_n| \leq B+2Ar^N$$ (here $w_n^+ \leq A$) and $\int_{B(x_0,\frac r2)} |w_n| \leq B+2Ar^N$ . The mean value property again yields $|w_n(x)| \leq 4^N (Br^{-N}+2A)$ if $x \in B(x_0, r/4)$. Let $G$ be a connected compact set contained in $\Omega$ and containing $x_0$ and $E$ the points $x \in G$ having a neighborhood where $(w_n)$ is bounded. Then $x_0 \in E$ and $E$ is open in $G$. Let $(x_k) \subset E$ converge to $z \in G$. Then $|w_n(x_k)| \leq B_k$ and, as above, $|w_n|$ is uniformly bounded in a neighborhood of $x_k$ which depends only on the distance of $x_k$ from $\partial \Omega$. If $k$ is sufficiently large, this neighborhood contains $z$, hence $E$ is closed in $G$ and $E=G$. Now, it suffices to cover $G$ with a finite number open sets where $(w_n)$ is bounded.