Recently I have read a paper Weighted Trudinger-type Inequalities written by Stephen M. Buckley and Julann O'Shea and published by Indiana University Mathematics Journal in 1999. I have some questions of pp.95-96 of this paper.

Now let me state the background of this paper. Let $\Omega\subset \mathbb{R}^n\ (n\geqslant 2)$ be a domain (i.e., open connected set). Given a point $x_0\in \Omega$. Let $\{S_i\}_{i=0}^j,\ j\geqslant 2$ be a pairwise disjoint collection of open subsets of $\Omega$ with $x_0\in S_0$. Define \begin{equation*} l_i(y)=\inf_{\lambda\in \mathcal{F}_{y,x_0}}\mathrm{len} (\lambda\cap S_i),\quad y\in \Omega,\quad 0<i<j, \end{equation*} where $\mathcal{F}_{y,x_0}$ is the set of all rectifiable curves in $\Omega$ joining $y$ and $x_0$, and $\mathrm{len}(\lambda\cap S_i)$ stands for the arc length of $\lambda$ lying in $S_i$. Then $l_i$ is Lipschitz and \begin{equation}\tag{*}\label{*} |\nabla l_i|\leqslant 1. \end{equation}

My questions are:

1. Why $l_i$ is Lipschitz?

2. Why \eqref{*} holds?

3. Why $\nabla l_i$ is supported on $S_i$?

For those questions, I am sure that $l_i$ is locally Lipschitz with constant $1$ on $\Omega$, that is, any point in $\Omega$ has a neighborhood on which $l_i$ is Lipschitz with constant $1$. Hence by Rademacher's theorem, we know that $l_i$ is differentiable a.e. in $\Omega$. But question 1 and 2 seem require $l_i$ is (global) Lipschitz with constant $1$ on $\Omega$, that is, for any $x',x''\in \Omega$, \begin{equation}\tag{**}\label{**} |l_i(x')-l_i(x'')|\leqslant |x'-x''|. \end{equation} I do not know how to prove \eqref{**}.

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Update: $x_0$ and $\{S_i\}$ come from the following property: Suppose $\Omega\subset \mathbb{R}^n\ (n\geqslant 2)$ is a domain with a distinguished point $x_0$ and $C_0>1$. We say $\Omega$ has the $C_0$-slice property with respect to $x_0$ if, for each $x\in \Omega$, there is a path $\gamma\colon [0,1]\to \Omega$, $\gamma(0)=x_0$, $\gamma(1)=x$, and a pairwise disjoint collection of open subsets $\{S_i\}_{i=0}^j, j\geqslant 2$ of $\Omega$ such that

(a) $x_0\in S_0$, $x\in S_j$, and $x_0$ and $x$ are in different connected components of $\Omega\setminus \overline{S}_i$ for all $0<i<j$.

(b) If $\lambda$ is a curve containing both $x$ and $x_0$, and $0<i<j$, then \begin{equation*} \mathrm{diam}(S_i)\leqslant C_0\mathrm{len}(\lambda\cap S_i). \end{equation*}

(c) $ \gamma\bigl([0,1])\subset \bigcup_{i=0}^j \overline{S}_i.$

(d) There exists $x_i\in \gamma_i\equiv \gamma([0,1])\cap S_i$, such that $x_0$ is as previously defined, $x_j=x$, and $B(x_j,2r_i)\subset S_i$, where $r_i=C_0^{-1}\mathrm{diam}(S_i)$. Additionally, $\delta(x)\geqslant C_0^{-1}\mathrm{diam}(S_i)$, for all $x\in \gamma_i,\ 0\leqslant i\leqslant j$.

Recently I have read a paper "Weighted Trudinger-type Inequalities" written by Stephen M. Buckley and Julann O'Shea and published by Indiana University Mathematics Journal in 1999, MR1722194, Zbl 0965.46024. I have some questions of pp. 95-96 of this paper.

Now let me state the background of this paper. Let $\Omega\subset \mathbb{R}^n\ (n\geqslant 2)$ be a domain (i.e., open connected set). Given a point $x_0\in \Omega$. Let $\{S_i\}_{i=0}^j,\ j\geqslant 2$ be a pairwise disjoint collection of open subsets of $\Omega$ with $x_0\in S_0$. Define $$ l_i(y)=\inf_{\lambda\in \mathcal{F}_{y,x_0}}\mathrm{len} (\lambda\cap S_i),\quad y\in \Omega,\quad 0<i<j, $$ where $\mathcal{F}_{y,x_0}$ is the set of all rectifiable curves in $\Omega$ joining $y$ and $x_0$, and $\mathrm{len}(\lambda\cap S_i)$ stands for the arc length of $\lambda$ lying in $S_i$. Then $l_i$ is Lipschitz and $$ \tag{$*$}\label{*} |\nabla l_i|\leqslant 1. $$

My questions are:

1. Why $l_i$ is Lipschitz?

2. Why \eqref{*} holds?

3. Why $\nabla l_i$ is supported on $S_i$?

For those questions, I am sure that $l_i$ is locally Lipschitz with constant $1$ on $\Omega$, that is, any point in $\Omega$ has a neighborhood on which $l_i$ is Lipschitz with constant $1$. Hence by Rademacher's theorem, we know that $l_i$ is differentiable a.e. in $\Omega$. But question 1 and 2 seem require $l_i$ is (global) Lipschitz with constant $1$ on $\Omega$, that is, for any $x',x''\in \Omega$, \begin{equation}\tag{${**}$}\label{**} |l_i(x')-l_i(x'')|\leqslant |x'-x''|. \end{equation} I do not know how to prove \eqref{**}.

********************************************

Update: $x_0$ and $\{S_i\}$ come from the following property: Suppose $\Omega\subset \mathbb{R}^n\ (n\geqslant 2)$ is a domain with a distinguished point $x_0$ and $C_0>1$. We say $\Omega$ has the $C_0$-slice property with respect to $x_0$ if, for each $x\in \Omega$, there is a path $\gamma\colon [0,1]\to \Omega$, $\gamma(0)=x_0$, $\gamma(1)=x$, and a pairwise disjoint collection of open subsets $\{S_i\}_{i=0}^j, j\geqslant 2$ of $\Omega$ such that

1. $x_0\in S_0$, $x\in S_j$, and $x_0$ and $x$ are in different connected components of $\Omega\setminus \overline{S}_i$ for all $0<i<j$.

2. If $\lambda$ is a curve containing both $x$ and $x_0$, and $0<i<j$, then \begin{equation*} \mathrm{diam}(S_i)\leqslant C_0\mathrm{len}(\lambda\cap S_i). \end{equation*}

3. $ \gamma\bigl([0,1])\subset \bigcup_{i=0}^j \overline{S}_i.$

4. There exists $x_i\in \gamma_i\equiv \gamma([0,1])\cap S_i$, such that $x_0$ is as previously defined, $x_j=x$, and $B(x_j,2r_i)\subset S_i$, where $r_i=C_0^{-1}\mathrm{diam}(S_i)$. Additionally, $\delta(x)\geqslant C_0^{-1}\mathrm{diam}(S_i)$, for all $x\in \gamma_i,\ 0\leqslant i\leqslant j$.

Recently I have read a paper Weighted Trudinger-type Inequalities written by Stephen M. Buckley and Julann O'Shea and published by Indiana University Mathematics Journal in 1999. I have some questions of pp.95-96 of this paper.

Now let me state the background of this paper. Let $\Omega\subset \mathbb{R}^n\ (n\geqslant 2)$ be a domain (i.e., open connected set). Given a point $x_0\in \Omega$. Let $\{S_i\}_{i=0}^j,\ j\geqslant 2$ be a pairwise disjoint collection of open subsets of $\Omega$ with $x_0\in S_0$. Define \begin{equation*} l_i(y)=\inf_{\lambda\in \mathcal{F}_{y,x_0}}\mathrm{len} (\lambda\cap S_i),\quad y\in \Omega,\quad 0<i<j, \end{equation*} where $\mathcal{F}_{y,x_0}$ is the set of all rectifiable curves in $\Omega$ joining $y$ and $x_0$, and $\mathrm{len}(\lambda\cap S_i)$ stands for the arc length of $\lambda$ lying in $S_i$. Then $l_i$ is Lipschitz and \begin{equation}\tag{*}\label{*} |\nabla l_i|\leqslant 1. \end{equation}

My questions are:

1. Why $l_i$ is Lipschitz?

2. Why \eqref{*} holds?

For those questions, I am sure that $l_i$ is locally Lipschitz with constant $1$ on $\Omega$, that is, any point in $\Omega$ has a neighborhood on which $l_i$ is Lipschitz with constant $1$. Hence by Rademacher's theorem, we know that $l_i$ is differentiable a.e. in $\Omega$. But question 1 and 2 seem require $l_i$ is (global) Lipschitz with constant $1$ on $\Omega$, that is, for any $x',x''\in \Omega$, \begin{equation}\tag{**}\label{**} |l_i(x')-l_i(x'')|\leqslant |x'-x''|. \end{equation} I do not know how to prove \eqref{**}.

********************************************

Update: $x_0$ and $\{S_i\}$ come from the following property: Suppose $\Omega\subset \mathbb{R}^n\ (n\geqslant 2)$ is a domain with a distinguished point $x_0$ and $C_0>1$. We say $\Omega$ has the $C_0$-slice property with respect to $x_0$ if, for each $x\in \Omega$, there is a path $\gamma\colon [0,1]\to \Omega$, $\gamma(0)=x_0$, $\gamma(1)=x$, and a pairwise disjoint collection of open subsets $\{S_i\}_{i=0}^j, j\geqslant 2$ of $\Omega$ such that

(a) $x_0\in S_0$, $x\in S_j$, and $x_0$ and $x$ are in different connected components of $\Omega\setminus \overline{S}_i$ for all $0<i<j$.

(b) If $\lambda$ is a curve containing both $x$ and $x_0$, and $0<i<j$, then \begin{equation*} \mathrm{diam}(S_i)\leqslant C_0\mathrm{len}(\lambda\cap S_i). \end{equation*}

(c) $ \gamma\bigl([0,1])\subset \bigcup_{i=0}^j \overline{S}_i.$

(d) There exists $x_i\in \gamma_i\equiv \gamma([0,1])\cap S_i$, such that $x_0$ is as previously defined, $x_j=x$, and $B(x_j,2r_i)\subset S_i$, where $r_i=C_0^{-1}\mathrm{diam}(S_i)$. Additionally, $\delta(x)\geqslant C_0^{-1}\mathrm{diam}(S_i)$, for all $x\in \gamma_i,\ 0\leqslant i\leqslant j$.

Recently I have read a paper Weighted Trudinger-type Inequalities written by Stephen M. Buckley and Julann O'Shea and published by Indiana University Mathematics Journal in 1999. I have some questions of pp.95-96 of this paper.

Now let me state the background of this paper. Let $\Omega\subset \mathbb{R}^n\ (n\geqslant 2)$ be a domain (i.e., open connected set). Given a point $x_0\in \Omega$. Let $\{S_i\}_{i=0}^j,\ j\geqslant 2$ be a pairwise disjoint collection of open subsets of $\Omega$ with $x_0\in S_0$. Define \begin{equation*} l_i(y)=\inf_{\lambda\in \mathcal{F}_{y,x_0}}\mathrm{len} (\lambda\cap S_i),\quad y\in \Omega,\quad 0<i<j, \end{equation*} where $\mathcal{F}_{y,x_0}$ is the set of all rectifiable curves in $\Omega$ joining $y$ and $x_0$, and $\mathrm{len}(\lambda\cap S_i)$ stands for the arc length of $\lambda$ lying in $S_i$. Then $l_i$ is Lipschitz and \begin{equation}\tag{*}\label{*} |\nabla l_i|\leqslant 1. \end{equation}

My questions are:

1. Why $l_i$ is Lipschitz?

2. Why \eqref{*} holds?

3. Why $\nabla l_i$ is supported on $S_i$?

For those questions, I am sure that $l_i$ is locally Lipschitz with constant $1$ on $\Omega$, that is, any point in $\Omega$ has a neighborhood on which $l_i$ is Lipschitz with constant $1$. Hence by Rademacher's theorem, we know that $l_i$ is differentiable a.e. in $\Omega$. But question 1 and 2 seem require $l_i$ is (global) Lipschitz with constant $1$ on $\Omega$, that is, for any $x',x''\in \Omega$, \begin{equation}\tag{**}\label{**} |l_i(x')-l_i(x'')|\leqslant |x'-x''|. \end{equation} I do not know how to prove \eqref{**}.

********************************************

Update: $x_0$ and $\{S_i\}$ come from the following property: Suppose $\Omega\subset \mathbb{R}^n\ (n\geqslant 2)$ is a domain with a distinguished point $x_0$ and $C_0>1$. We say $\Omega$ has the $C_0$-slice property with respect to $x_0$ if, for each $x\in \Omega$, there is a path $\gamma\colon [0,1]\to \Omega$, $\gamma(0)=x_0$, $\gamma(1)=x$, and a pairwise disjoint collection of open subsets $\{S_i\}_{i=0}^j, j\geqslant 2$ of $\Omega$ such that

(a) $x_0\in S_0$, $x\in S_j$, and $x_0$ and $x$ are in different connected components of $\Omega\setminus \overline{S}_i$ for all $0<i<j$.

(b) If $\lambda$ is a curve containing both $x$ and $x_0$, and $0<i<j$, then \begin{equation*} \mathrm{diam}(S_i)\leqslant C_0\mathrm{len}(\lambda\cap S_i). \end{equation*}

(c) $ \gamma\bigl([0,1])\subset \bigcup_{i=0}^j \overline{S}_i.$

(d) There exists $x_i\in \gamma_i\equiv \gamma([0,1])\cap S_i$, such that $x_0$ is as previously defined, $x_j=x$, and $B(x_j,2r_i)\subset S_i$, where $r_i=C_0^{-1}\mathrm{diam}(S_i)$. Additionally, $\delta(x)\geqslant C_0^{-1}\mathrm{diam}(S_i)$, for all $x\in \gamma_i,\ 0\leqslant i\leqslant j$.

Recently I have read a paper Weighted Trudinger-type Inequalities written by Stephen M. Buckley and Julann O'Shea and published by Indiana University Mathematics Journal in 1999. I have some questions of pp.95-96 of this paper.

Now let me state the background of this paper. Let $\Omega\subset \mathbb{R}^n\ (n\geqslant 2)$ be a domain (i.e., open connected set). Given a point $x_0\in \Omega$. Let $\{S_i\}_{i=0}^j,\ j\geqslant 2$ be a pairwise disjoint collection of open subsets of $\Omega$, such that \begin{equation*} \text{$x_0\in S_0$ and $x_0$ is in different connected components of $\Omega\setminus\overline{S_i}$ for all $0<i<j$.} \end{equation*} Define \begin{equation*} l_i(y)=\inf_{\lambda\in \mathcal{F}_{y,x_0}}\mathrm{len} (\lambda\cap S_i),\quad y\in \Omega,\quad 0<i<j, \end{equation*} where $\mathcal{F}_{y,x_0}$ is the set of all rectifiable curves in $\Omega$ joining $y$ and $x_0$, and $\mathrm{len}(\lambda\cap S_i)$ stands for the arc length of $\lambda$ lying in $S_i$. Then $l_i$ is Lipschitz and \begin{equation}\tag{*}\label{*} |\nabla l_i|\leqslant 1. \end{equation}

My questions are:

1. Why $l_i$ is Lipschitz?

2. Why \eqref{*} holds?

For those questions, I am sure that $l_i$ is locally Lipschitz with constant $1$ on $\Omega$, that is, any point in $\Omega$ has a neighborhood on which $l_i$ is Lipschitz with constant $1$. Hence by Rademacher's theorem, we know that $l_i$ is differentiable a.e. in $\Omega$. But question 1 and 2 seem require $l_i$ is (global) Lipschitz with constant $1$ on $\Omega$, that is, for any $x',x''\in \Omega$, \begin{equation}\tag{**}\label{**} |l_i(x')-l_i(x'')|\leqslant |x'-x''|. \end{equation} I do not know how to prove \eqref{**}.

Recently I have read a paper Weighted Trudinger-type Inequalities written by Stephen M. Buckley and Julann O'Shea and published by Indiana University Mathematics Journal in 1999. I have some questions of pp.95-96 of this paper.

Now let me state the background of this paper. Let $\Omega\subset \mathbb{R}^n\ (n\geqslant 2)$ be a domain (i.e., open connected set). Given a point $x_0\in \Omega$. Let $\{S_i\}_{i=0}^j,\ j\geqslant 2$ be a pairwise disjoint collection of open subsets of $\Omega$ with $x_0\in S_0$. Define \begin{equation*} l_i(y)=\inf_{\lambda\in \mathcal{F}_{y,x_0}}\mathrm{len} (\lambda\cap S_i),\quad y\in \Omega,\quad 0<i<j, \end{equation*} where $\mathcal{F}_{y,x_0}$ is the set of all rectifiable curves in $\Omega$ joining $y$ and $x_0$, and $\mathrm{len}(\lambda\cap S_i)$ stands for the arc length of $\lambda$ lying in $S_i$. Then $l_i$ is Lipschitz and \begin{equation}\tag{*}\label{*} |\nabla l_i|\leqslant 1. \end{equation}

My questions are:

1. Why $l_i$ is Lipschitz?

2. Why \eqref{*} holds?

For those questions, I am sure that $l_i$ is locally Lipschitz with constant $1$ on $\Omega$, that is, any point in $\Omega$ has a neighborhood on which $l_i$ is Lipschitz with constant $1$. Hence by Rademacher's theorem, we know that $l_i$ is differentiable a.e. in $\Omega$. But question 1 and 2 seem require $l_i$ is (global) Lipschitz with constant $1$ on $\Omega$, that is, for any $x',x''\in \Omega$, \begin{equation}\tag{**}\label{**} |l_i(x')-l_i(x'')|\leqslant |x'-x''|. \end{equation} I do not know how to prove \eqref{**}.

********************************************

Update: $x_0$ and $\{S_i\}$ come from the following property: Suppose $\Omega\subset \mathbb{R}^n\ (n\geqslant 2)$ is a domain with a distinguished point $x_0$ and $C_0>1$. We say $\Omega$ has the $C_0$-slice property with respect to $x_0$ if, for each $x\in \Omega$, there is a path $\gamma\colon [0,1]\to \Omega$, $\gamma(0)=x_0$, $\gamma(1)=x$, and a pairwise disjoint collection of open subsets $\{S_i\}_{i=0}^j, j\geqslant 2$ of $\Omega$ such that

(a) $x_0\in S_0$, $x\in S_j$, and $x_0$ and $x$ are in different connected components of $\Omega\setminus \overline{S}_i$ for all $0<i<j$.

(b) If $\lambda$ is a curve containing both $x$ and $x_0$, and $0<i<j$, then \begin{equation*} \mathrm{diam}(S_i)\leqslant C_0\mathrm{len}(\lambda\cap S_i). \end{equation*}

(c) $ \gamma\bigl([0,1])\subset \bigcup_{i=0}^j \overline{S}_i.$

(d) There exists $x_i\in \gamma_i\equiv \gamma([0,1])\cap S_i$, such that $x_0$ is as previously defined, $x_j=x$, and $B(x_j,2r_i)\subset S_i$, where $r_i=C_0^{-1}\mathrm{diam}(S_i)$. Additionally, $\delta(x)\geqslant C_0^{-1}\mathrm{diam}(S_i)$, for all $x\in \gamma_i,\ 0\leqslant i\leqslant j$.