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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 $L^{\infty}(\Omega)$

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 $L^{\infty}(\Omega)$

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 $L^{\infty}(\Omega)$

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 $L^{\infty}(\Omega)$

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 $\mathcal{H}^{n-1}$ 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 $L^{\infty}(\Omega)$

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 $L^{\infty}(\Omega)$