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

In paper  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

 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.

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.