Let $\Omega\subset \Bbb R^d$ be open. The space $BV(\Omega)$ consists in functions $u\in L^1(\Omega)$ with bounded variation, i.e. $|u|_{BV(\Omega) }<\infty$ where \begin{align}\label{eq:bounded-variation} |u|_{BV(\Omega) }:= \sup\Big\lbrace \int_{\Omega} u(x) \operatorname{div} \phi(x) d x:~\phi \in C_c^\infty(\Omega, \mathbb{R}^d),~\|\phi\|_{L^\infty(\Omega)}\leq 1\Big\rbrace. \end{align}

The vector $\nabla u =(\Lambda_1, \Lambda_2,\cdots, \Lambda_d)$ can be seen as a vector valued Radon measure on $\Omega$ such that \begin{align*} \int_\Omega u(x)\frac{\partial \varphi}{\partial x_i}(x) d x= -\int_\Omega \varphi(x)d \Lambda_i(x), \quad \text{for all }\quad \varphi\in C_c^\infty(\Omega),~~i=1,\cdots, d. \end{align*} In particular if $u\in W^{1,1}(\Omega)$ then $u\in BV(\Omega)$, $\partial_{x_i} u(x)d x= d\Lambda_i(x)$ and $ |u|_{BV(\Omega)}\asymp\|\nabla u\|_{L^1(\Omega)}.$

Let us recall that $\Omega \subset \mathbb{R}^d$ is called an $W^{1,p}$-extension (resp. $BV$-extension) domain if there exists a linear operator $E:W^{1,p}(\Omega)\to W^{1,p}(\mathbb{R}^d)$ (resp. $E: BV(\Omega)\to BV(\mathbb{R}^d)$) and a constant $C: = C(\Omega, d)$ such that \begin{align*} Eu\mid_{\Omega} &= u \qquad\hbox{and} \qquad \|Eu\|_{W^{1,p}(\mathbb{R}^d)}\leq C \|u\|_{W^{1,p}(\Omega)} \qquad\text{for all}\quad u \in W^{1,p}(\Omega)\\ (\text{resp.}\quad Eu\mid_{\Omega}& = u \qquad\hbox{and} \qquad \|Eu\|_{BV(\mathbb{R}^d)}\leq C \|u\|_{BV(\Omega)} \qquad\quad\, \text{for all}\quad u \in BV(\Omega)). \end{align*}

Question:

1-If $\Omega$ is an $BV$-extension domain do we have $|EU|_{BV(\Bbb R^d)}(\partial\Omega)=0?$

2-If $\Omega$ is an $W^{1,p}$-extension domain do we have $|\nabla EU|_{W^{1,p}(\Bbb R^d)}(\partial\Omega)=0?$

Intuitively, the second question is more likely to hold if we assume in addition that the boundary $\partial \Omega$ has zero Lebesgue measure. Since $\nabla Eu$ is still a function when $Eu\in W^{1,p}(\Bbb R^d)$.

Or, does the fact that $\Omega$ is an extension domain implies that $|\partial\Omega|=0$?

Let $\Omega\subset \Bbb R^d$ be open. The space $BV(\Omega)$ consists in functions $u\in L^1(\Omega)$ with bounded variation, i.e. $|u|_{BV(\Omega) }<\infty$ where \begin{align}\label{eq:bounded-variation} |u|_{BV(\Omega) }:= \sup\Big\lbrace \int_{\Omega} u(x) \operatorname{div} \phi(x) d x:~\phi \in C_c^\infty(\Omega, \mathbb{R}^d),~\|\phi\|_{L^\infty(\Omega)}\leq 1\Big\rbrace. \end{align}

The vector $\nabla u =(\Lambda_1, \Lambda_2,\cdots, \Lambda_d)$ can be seen as a vector valued Radon measure on $\Omega$ such that \begin{align*} \int_\Omega u(x)\frac{\partial \varphi}{\partial x_i}(x) d x= -\int_\Omega \varphi(x)d \Lambda_i(x), \quad \text{for all }\quad \varphi\in C_c^\infty(\Omega),~~i=1,\cdots, d. \end{align*} In particular if $u\in W^{1,1}(\Omega)$ then $u\in BV(\Omega)$, $\partial_{x_i} u(x)d x= d\Lambda_i(x)$ and $ |u|_{BV(\Omega)}\asymp\|\nabla u\|_{L^1(\Omega)}.$

Let us recall that $\Omega \subset \mathbb{R}^d$ is called an $W^{1,p}$-extension (resp. $BV$-extension) domain if there exists a linear operator $E:W^{1,p}(\Omega)\to W^{1,p}(\mathbb{R}^d)$ (resp. $E: BV(\Omega)\to BV(\mathbb{R}^d)$) and a constant $C: = C(\Omega, d)$ such that \begin{align*} Eu\mid_{\Omega} &= u \quad\hbox{and} \quad \|Eu\|_{W^{1,p}(\mathbb{R}^d)}\leq C \|u\|_{W^{1,p}(\Omega)} \quad\text{for all}\quad u \in W^{1,p}(\Omega)\\ (\text{resp.}\quad Eu\mid_{\Omega}& = u \quad\hbox{and} \quad \|Eu\|_{BV(\mathbb{R}^d)}\leq C \|u\|_{BV(\Omega)} \quad\, \text{for all}\quad u \in BV(\Omega)). \end{align*}

Question:

1-If $\Omega$ is an $BV$-extension domain do we have $|EU|_{BV(\Bbb R^d)}(\partial\Omega)=0?$

2-If $\Omega$ is an $W^{1,p}$-extension domain do we have $|\nabla EU|_{W^{1,p}(\Bbb R^d)}(\partial\Omega)=0?$

Intuitively, the second question is more likely to hold if we assume in addition that the boundary $\partial \Omega$ has zero Lebesgue measure. Since $\nabla Eu$ is still a function when $Eu\in W^{1,p}(\Bbb R^d)$.

Or, does the fact that $\Omega$ is an extension domain implies that $|\partial\Omega|=0$?

Let $\Omega\subset \Bbb R^d$ be open. The space $BV(\Omega)$ consists in functions $u\in L^1(\Omega)$ with bounded variation, i.e. $|u|_{BV(\Omega) }<\infty$ where \begin{align}\label{eq:bounded-variation} |u|_{BV(\Omega) }:= \sup\Big\lbrace \int_{\Omega} u(x) \operatorname{div} \phi(x) d x:~\phi \in C_c^\infty(\Omega, \mathbb{R}^d),~\|\phi\|_{L^\infty(\Omega)}\leq 1\Big\rbrace. \end{align}

The vector $\nabla u =(\Lambda_1, \Lambda_2,\cdots, \Lambda_d)$ can be seen as a vector valued Radon measure on $\Omega$ such that \begin{align*} \int_\Omega u(x)\frac{\partial \varphi}{\partial x_i}(x) d x= -\int_\Omega \varphi(x)d \Lambda_i(x), \quad \text{for all }\quad \varphi\in C_c^\infty(\Omega),~~i=1,\cdots, d. \end{align*} In particular if $u\in W^{1,1}(\Omega)$ then $u\in BV(\Omega)$, $\partial_{x_i} u(x)d x= d\Lambda_i(x)$ and $$ |u|_{BV(\Omega)}\leq \|\nabla u\|_{L^1(\Omega)}\leq d^2 |u|_{BV(\Omega)}.$$

Let us recall that $\Omega \subset \mathbb{R}^d$ is called an $W^{1,p}$-extension (resp. $BV$-extension) domain if there exists a linear operator $E:W^{1,p}(\Omega)\to W^{1,p}(\mathbb{R}^d)$ (resp. $E: BV(\Omega)\to BV(\mathbb{R}^d)$) and a constant $C: = C(\Omega, d)$ such that \begin{align*} Eu\mid_{\Omega} &= u \qquad\hbox{and} \qquad \|Eu\|_{W^{1,p}(\mathbb{R}^d)}\leq C \|u\|_{W^{1,p}(\Omega)} \qquad\text{for all}\quad u \in W^{1,p}(\Omega)\\ (\text{resp.}\quad Eu\mid_{\Omega}& = u \qquad\hbox{and} \qquad \|Eu\|_{BV(\mathbb{R}^d)}\leq C \|u\|_{BV(\Omega)} \qquad\quad\, \text{for all}\quad u \in BV(\Omega)). \end{align*}

Question:

1-If $\Omega$ is an $BV$-extension domain do we have $|EU|_{BV(\Bbb R^d)}(\partial\Omega)=0?$

2-If $\Omega$ is an $W^{1,p}$-extension domain do we have $|\nabla EU|_{W^{1,p}(\Bbb R^d)}(\partial\Omega)=0?$

Intuitively, the second question is more likely to hold if we assume in addition that the boundary $\partial \Omega$ has zero Lebesgue measure. Since $\nabla Eu$ is still a function when $Eu\in W^{1,p}(\Bbb R^d)$.

Or, does the fact that $\Omega$ is an extension domain implies that $|\partial\Omega|=0$?

Let $\Omega\subset \Bbb R^d$ be open. The space $BV(\Omega)$ consists in functions $u\in L^1(\Omega)$ with bounded variation, i.e. $|u|_{BV(\Omega) }<\infty$ where \begin{align}\label{eq:bounded-variation} |u|_{BV(\Omega) }:= \sup\Big\lbrace \int_{\Omega} u(x) \operatorname{div} \phi(x) d x:~\phi \in C_c^\infty(\Omega, \mathbb{R}^d),~\|\phi\|_{L^\infty(\Omega)}\leq 1\Big\rbrace. \end{align}

The vector $\nabla u =(\Lambda_1, \Lambda_2,\cdots, \Lambda_d)$ can be seen as a vector valued Radon measure on $\Omega$ such that \begin{align*} \int_\Omega u(x)\frac{\partial \varphi}{\partial x_i}(x) d x= -\int_\Omega \varphi(x)d \Lambda_i(x), \quad \text{for all }\quad \varphi\in C_c^\infty(\Omega),~~i=1,\cdots, d. \end{align*} In particular if $u\in W^{1,1}(\Omega)$ then $u\in BV(\Omega)$, $\partial_{x_i} u(x)d x= d\Lambda_i(x)$ and $ |u|_{BV(\Omega)}\asymp\|\nabla u\|_{L^1(\Omega)}.$

Let us recall that $\Omega \subset \mathbb{R}^d$ is called an $W^{1,p}$-extension (resp. $BV$-extension) domain if there exists a linear operator $E:W^{1,p}(\Omega)\to W^{1,p}(\mathbb{R}^d)$ (resp. $E: BV(\Omega)\to BV(\mathbb{R}^d)$) and a constant $C: = C(\Omega, d)$ such that \begin{align*} Eu\mid_{\Omega} &= u \qquad\hbox{and} \qquad \|Eu\|_{W^{1,p}(\mathbb{R}^d)}\leq C \|u\|_{W^{1,p}(\Omega)} \qquad\text{for all}\quad u \in W^{1,p}(\Omega)\\ (\text{resp.}\quad Eu\mid_{\Omega}& = u \qquad\hbox{and} \qquad \|Eu\|_{BV(\mathbb{R}^d)}\leq C \|u\|_{BV(\Omega)} \qquad\quad\, \text{for all}\quad u \in BV(\Omega)). \end{align*}

Question:

1-If $\Omega$ is an $BV$-extension domain do we have $|EU|_{BV(\Bbb R^d)}(\partial\Omega)=0?$

2-If $\Omega$ is an $W^{1,p}$-extension domain do we have $|\nabla EU|_{W^{1,p}(\Bbb R^d)}(\partial\Omega)=0?$

Intuitively, the second question is more likely to hold if we assume in addition that the boundary $\partial \Omega$ has zero Lebesgue measure. Since $\nabla Eu$ is still a function when $Eu\in W^{1,p}(\Bbb R^d)$.

Or, does the fact that $\Omega$ is an extension domain implies that $|\partial\Omega|=0$?