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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: enter image description here

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 $\{u = 0\}$?

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