I'm getting lost in the first couple of sentences of a paper by Cafarelli & Alt. The sections are organized as follows, 1.1 Data, 1.2 The problem, 1.3 the statement and relevant part of proof.

1.1 Data

$$\Omega \subset \mathbb{R}^n$$ is an open connected (may be unbounded) set, and locally $$\partial \Omega$$ is a Lipschitz graph. $$S \subset \partial \Omega$$ is measurable and $$\mathcal{H}^{n-1}(S)>0$$. The Dirchlet data on $$S$$ are given by a non-negative function $$u^0 \in L^1_{loc}(\Omega)$$ with $$\nabla u^0 \in L^2(\Omega)$$. The given function $$Q$$ is non-negative and measurable.

1.2 The problem

Consider the convex set: $$K = \{v \in L^1_{loc}(\Omega)/\nabla v \in L^2(\Omega)\text{ and } v = u^0 \text{ on } S\}$$ Find the absolute minimum of the functional $$J(v) = \int_{\Omega}|\nabla v|^2 + \chi(\{v > 0 \}Q^2)$$ where $$v \in K$$.

1.3 Existence Statement and Proof

If $$J(u^0)<\infty$$ then there exists an absolute minimum $$u \in K$$ of the functional $$J$$.

Since $$J$$ is non-negative there is a minimal sequence $$u_k$$, $$k \in \mathbb{N}$$. Then $$\nabla u_k$$ are bounded in $$L^2(\Omega)$$, and since $$\mathcal{H}^{n-1}(S)$$ is positive $$u_k - u^0$$ are bounded in $$L^2(B_R \cap \Omega)$$ for large $$R$$. Therefore there is an $$u \in K$$, such that for a subsequence: $$\nabla u_K \rightarrow \nabla u$$ weakly in $$L^2(\Omega)$$ and $$u_k \rightarrow u$$ a.e. in $$\Omega$$. Moreover there is a function $$\gamma \in L^{\infty}(\Omega)$$ with $$0 \leq \gamma \leq 1$$ such that $$\chi(\{u_k > 0\}) \rightarrow \gamma$$ weak star in $$L^{\infty}(\Omega)$$.

My questions:

1. The authors write "Since $$\mathcal{H}^{n-1}$$ is positive $$u_k -u^0$$ are bounded in $$L^2(B_R \cap \Omega)$$ for large $$R$$". I don't follow this. What allows them to go from the Hausdorff measure of the boundary to some estimate locally on the interior? I was thinking like some divergence theorem thing but I can't be sure.

2. The authors write "$$u_k \rightarrow u\text{ a.e. in }\Omega$$". This confuses me. I know convergence in measure implies a subsequence converging a.e. But he has the $$u_k \in L^2_{loc}$$ -- so I dont know where this fact is coming from.

3. Where does the final fact come from? -- $$\chi(\{u_k > 0\}) \rightarrow \gamma$$ weak star in $$L^{\infty}(\Omega)$$

• could you give a reference to the paper? Aug 26 '19 at 13:09
• H. W. Alt & L. A. Caffarelli, Existence and regularity for a minimum problem with free boundary, J. Reine Angew. Math. 325 (1981), 105–144 Aug 26 '19 at 13:24
• 1) is the Poincare inequality: $u_k - u^0$ is zero on $S$, and $S \subseteq \partial(B_R \cap \Omega)$ for $R$ large enough, so there you have the Poincare inequality for $u_k - u^0$ on $B_R \cap \Omega$. 3) Is the Banach-Alaoglu theorem in $L^\infty(\Omega)$ since the characteristic functions are uniformly bounded in $L^\infty$ norm by $1$. Aug 26 '19 at 13:37
• and again, 2) is the application of poincare inequality Aug 26 '19 at 15:23
• Yes; I'd say 2) is a consequence of 1) plus some tedious stuff on how to transfer the subsequences converging pw. a.e. on $B_R \cap \Omega$ to the whole $\Omega$. But maybe there is a direct argument which I am not seeing. Aug 26 '19 at 15:28