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

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.

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.