# Clarification between solution by Distribution, a.e., and functional minimization

In page 20-22 of this book, its is shown that there exists a solution to $$\Delta u = f \chi_{(u>0)}$$ by two steps:

1. Showing that $$J(u)$$ has a minimizer on $$K$$:$$J(v) = \int_D (|\nabla v|^2 + 2fv)\,dx \\K = \{v \in W^{1,2}(D):v-g \in W^{1,2}_0(D),v \geq 0 \text{ a.e. in }D \subset \mathbb{R}^n\}$$ In this step they show the minimizer, call it $$u$$, is actually in$$W^{2,p}_{loc}$$
2. Then they show $$u$$ satisfies $$\Delta u = f \chi_{(u>0)}$$, in the sense of distributions. They say this amounts to verifying that $$\Delta u = f \chi_{(u>0)}$$ a.e. in $$D$$.

They go on to show $$\Delta u = f \chi_{(u>0)}$$ a.e. in $$D$$ by saying the minimizing sequence in part 1: $$u_k \rightarrow u$$ is a sequence in $$C^{1,\alpha}_{loc}$$ for some $$p$$ large enough by Sobolev embedding. Then:

My questions:

1. Doesn't proving step 1 amount to showing that $$u$$ satisfies $$\Delta u = f \chi_{(u>0)}$$, in the sense of distributions? After all, we obtain the minimization problem by multiplying $$\Delta u = f \chi_{(u>0)}$$ by test functions.

2. How does showing showing that $$u$$ satisfies $$\Delta u = f \chi_{(u>0)}$$, in the sense of distributions imply $$\Delta u = f \chi_{(u>0)}$$ a.e. in $$D$$?

3. How does local uniform convergence imply $$\Delta u = f$$ a.e. on the set $$\{u > 0\}$$?

4. $$u \in W^{2,p}_{loc}$$ implies $$\Delta u \in L^p_{loc}$$. From here are they saying: $$\Delta u \in L^p_{loc}$$ are defined a.e. so $$\Delta u = 0$$ a.e. on $$\{u = 0\}$$?