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In the book of S.Osher & R.Fedkiw - Level set methods and dynamic implicit surfaces, at page 15, is stated without proof, a formula like that: $$\text{Per}_{\Omega}(\omega)=\lim_{\varepsilon\to 0} \int\limits_{\Omega}\delta_{\varepsilon}(\phi(x))\ |\nabla\phi(x)|\ dx,$$

where

• $\Omega\subset\mathbb{R}^2$ is an open and bounded set,
• $\omega=\{x\in\Omega\ |\ \phi(x)>0\}$,
• $\phi:\Omega\to\mathbb{R}$ is a smooth level set function.
• $\text{Per}_{\Omega}(\omega)$ is the perimeter of $\omega$ that lies inside $\Omega$ (the relative perimeter).
• $\delta_{\varepsilon}:\mathbb{R}\to\mathbb{R}$, is a smooth approximation of the $\delta$-Dirac (generalized) function, precisely $$ \delta_{\varepsilon}(x)=\dfrac{\varepsilon}{\pi (\varepsilon^2+x^2)}\;. $$ Now, denoting by $\bigvee_{\Omega} f$ the total variation of $f$ in $\Omega$, here is my question:

Is it true that for any $\chi_{\omega}\in BV(\Omega)$ ($\chi_\omega$ being the characteristic function of $\omega$ - so by this condition we require $\omega$ to be a set with finite perimeter), whenever $\chi_n\to \chi_{\omega}$ in the $L^1(\Omega)$ norm and $0\leq \chi_n\leq 1, \chi_n\in W^{1,1}(\Omega)$ for each $n$, we have that: $$\bigvee_{\Omega} \chi_{\omega}=\lim\limits_{n\to\infty} \bigvee_{\Omega}\chi_n(x)\;\;\boldsymbol{?}$$

Notes

• It is known that: $$\bigvee_{\Omega} f = \sup\left\{\,\int\limits_{\Omega} f(x)\text{div}(\varphi(x))\ dx\ |\ \varphi\in C^{\infty}_{c}(\Omega;\mathbb{R}^2)\ \text{and} \ \Vert\varphi\Vert_{L^{\infty}(\Omega;\mathbb{R}^2)}\leq 1\right \}$$

• It is also known that for $\chi\in W^{1,1}(\Omega)\subset BV(\Omega)$ we have that: $$\bigvee_{\Omega} \chi=\int\limits_{\Omega} |\nabla\chi(x)|\ dx.$$

• We have that $\text{Per}_{\Omega}(\omega)=\bigvee_{\Omega}\chi_{\omega}$ for any $\omega\subseteq\Omega$.

• Also, one part of the inequality is well-know (lower semicontinuity of the total variation), so: $$\bigvee_{\Omega} \chi_{\omega}\leq\liminf\limits_{n\to\infty} \bigvee_{\Omega}\chi_n(x)$$

• In the cited book, at page 10, Theoreme 1.3 there's an approximation result of the $BV$ functions with $C^{\infty}_c$ functions. Maybe it is useful.

• I tried to prove it. It is correct for many examples that I take, but I can't figure out why: intuitively I understand it but technically I'm blocked.

Motivation

I'm interested in this type of formulas because if it is indeed true we will have that: $$\text{Per}_{\Omega}(\omega)=\lim\limits_{\varepsilon\to 0}\int_{\Omega} H'_{\varepsilon}(\phi(x))|\nabla\phi(x)|\ dx, $$ for any function $H_{\varepsilon}:\mathbb{R}\to\mathbb{R},\ H_{\varepsilon}\in W^{1,1}(\mathbb{R})$ that approximates in $L^1(\Omega)$ the Heaviside function $$ H(x)=\begin{cases} 1,\ x>0 \\ 0,\ x<0\end{cases}. $$ In the above example, $$ H_{\varepsilon}(x)=\dfrac{1}{2}+\dfrac{1}{\pi}\cdot\tan^{-1}(x/\varepsilon),$$ for each $\varepsilon>0$ and $\delta_{\varepsilon}(x)=H'_{\varepsilon}(x)$.

In the book of S.Osher & R.Fedkiw - Level set methods and dynamic implicit surfaces, at page 15, is stated without proof, a formula like that: $$\text{Per}_{\Omega}(\omega)=\lim_{\varepsilon\to 0} \int\limits_{\Omega}\delta_{\varepsilon}(\phi(x))\ |\nabla\phi(x)|\ dx,$$

where

• $\Omega\subset\mathbb{R}^2$ is an open and bounded set,
• $\omega=\{x\in\Omega\ |\ \phi(x)>0\}$,
• $\phi:\Omega\to\mathbb{R}$ is a smooth level set function.
• $\text{Per}_{\Omega}(\omega)$ is the perimeter of $\omega$ that lies inside $\Omega$ (the relative perimeter).
• $\delta_{\varepsilon}:\mathbb{R}\to\mathbb{R}$, is a smooth approximation of the $\delta$-Dirac (generalized) function, precisely $$ \delta_{\varepsilon}(x)=\dfrac{\varepsilon}{\pi (\varepsilon^2+x^2)}\;. $$ Now, denoting by $\bigvee_{\Omega} f$ the total variation of $f$ in $\Omega$, here is my question:

Is it true that for any $\chi_{\omega}\in BV(\Omega)$ ($\chi_\omega$ being the characteristic function of $\omega$ - so by this condition we require $\omega$ to be a set with finite perimeter), whenever $\chi_n\to \chi_{\omega}$ in the $L^1(\Omega)$ norm and $0\leq \chi_n\leq 1, \chi_n\in W^{1,1}(\Omega)$ for each $n$, we have that: $$\bigvee_{\Omega} \chi_{\omega}=\lim\limits_{n\to\infty} \bigvee_{\Omega}\chi_n(x)\;\;\boldsymbol{?}$$

Notes

• It is known that: $$\bigvee_{\Omega} f = \sup\left\{\,\int\limits_{\Omega} f(x)\text{div}(\varphi(x))\ dx\ |\ \varphi\in C^{\infty}_{c}(\Omega;\mathbb{R}^2)\ \text{and} \ \Vert\varphi\Vert_{L^{\infty}(\Omega;\mathbb{R}^2)}\leq 1\right \}$$

• It is also known that for $\chi\in W^{1,1}(\Omega)\subset BV(\Omega)$ we have that: $$\bigvee_{\Omega} \chi=\int\limits_{\Omega} |\nabla\chi(x)|\ dx.$$

• We have that $\text{Per}_{\Omega}(\omega)=\bigvee_{\Omega}\chi_{\omega}$ for any $\omega\subseteq\Omega$.

• Also, one part of the inequality is well-know (lower semicontinuity of the total variation), so: $$\bigvee_{\Omega} \chi_{\omega}\leq\liminf\limits_{n\to\infty} \bigvee_{\Omega}\chi_n$$

• In the cited book, at page 10, Theoreme 1.3 there's an approximation result of the $BV$ functions with $C^{\infty}_c$ functions. Maybe it is useful.

• I tried to prove it. It is correct for many examples that I take, but I can't figure out why: intuitively I understand it but technically I'm blocked.

Motivation

I'm interested in this type of formulas because if it is indeed true we will have that: $$\text{Per}_{\Omega}(\omega)=\lim\limits_{\varepsilon\to 0}\int_{\Omega} H'_{\varepsilon}(\phi(x))|\nabla\phi(x)|\ dx, $$ for any function $H_{\varepsilon}:\mathbb{R}\to\mathbb{R},\ H_{\varepsilon}\in W^{1,1}(\mathbb{R})$ that approximates in $L^1(\Omega)$ the Heaviside function $$ H(x)=\begin{cases} 1,\ x>0 \\ 0,\ x<0\end{cases}. $$ In the above example, $$ H_{\varepsilon}(x)=\dfrac{1}{2}+\dfrac{1}{\pi}\cdot\tan^{-1}(x/\varepsilon),$$ for each $\varepsilon>0$ and $\delta_{\varepsilon}(x)=H'_{\varepsilon}(x)$.

In the book of S.Osher & R.Fedkiw - Level set methods and dynamic implicit surfaces, at page 15, is stated without proof, a formula like that: $$\text{Per}_{\Omega}(\omega)=\lim_{\varepsilon\to 0} \int\limits_{\Omega}\delta_{\varepsilon}(\phi(x))\ |\nabla\phi(x)|\ dx,$$

where

• $\Omega\subset\mathbb{R}^2$ is an open and bounded set,
• $\omega=\{x\in\Omega\ |\ \phi(x)>0\}$,
• $\phi:\Omega\to\mathbb{R}$ is a smooth level set function.
• $\text{Per}_{\Omega}(\omega)$ is the perimeter of $\omega$ that lies inside $\Omega$ (the relative perimeter).
• $\delta_{\varepsilon}:\mathbb{R}\to\mathbb{R}$, is a smooth approximation of the $\delta$-Dirac (generalized) function, precisely $$ \delta_{\varepsilon}(x)=\dfrac{\varepsilon}{\pi (\varepsilon^2+x^2)}\;. $$ Now, denoting by $\bigvee_{\Omega} f$ the total variation of $f$ in $\Omega$, here is my question:

Is it true that for any $\chi_{\omega}\in BV(\Omega)$ ($\chi_\omega$ being the characteristic function of $\omega$ - so by this condition we require $\omega$ to be a set with finite perimeter), whenever $\chi_n\to \chi_{\omega}$ in the $L^1(\Omega)$ norm and $0\leq \chi_n\leq 1, \chi_n\in W^{1,1}(\Omega)$ for each $n$, we have that: $$\bigvee_{\Omega} \chi_{\omega}=\lim\limits_{n\to\infty} \bigvee_{\Omega}\chi_n(x)\;\;\boldsymbol{?}$$

=======================================================================

Notes

• It is known that: $$\bigvee_{\Omega} f = \sup\left\{\int\limits_{\Omega} f(x)\text{div}(\varphi(x))\ dx\ |\ \varphi\in C^{\infty}_{c}(\Omega;\mathbb{R}^2)\ \text{and} \ \Vert\varphi\Vert_{L^{\infty}(\Omega;\mathbb{R}^2)}\leq 1\right \}$$

• It is also known that for $\chi\in W^{1,1}(\Omega)\subset BV(\Omega)$ we have that: $$\bigvee_{\Omega} \chi=\int_{\Omega} |\nabla\chi(x)|\ dx.$$

• We have that $\text{Per}_{\Omega}(\omega)=\bigvee_{\Omega}\chi_{\omega}$ for any $\omega\subseteq\Omega$.

• Also, one part of the inequality is well-know (lower semicontinuity of the total variation), so: $$\bigvee_{\Omega} \chi_{\omega}\leq\liminf\limits_{n\to\infty} \bigvee_{\Omega}\chi_n(x)$$

• In the cited book, at page 10, Theoreme 1.3 there's an approximation result of the $BV$ functions with $C^{\infty}_c$ functions. Maybe it is useful.

• I tried to prove it. It is correct for many examples that I take, but I can't figure out why: intuitively I understand it but technically I'm blocked.

Motivation

I'm interested in this type of formulas because if it is indeed true we will have that: $$\text{Per}_{\Omega}(\omega)=\lim\limits_{\varepsilon\to 0}\int_{\Omega} H'_{\varepsilon}(\phi(x))|\nabla\phi(x)|\ dx, $$ for any function $H_{\varepsilon}:\mathbb{R}\to\mathbb{R},\ H_{\varepsilon}\in W^{1,1}(\mathbb{R})$ that approximates in $L^1(\Omega)$ the Heaviside function $$ H(x)=\begin{cases} 1,\ x>0 \\ 0,\ x<0\end{cases}. $$ In the above example, $$ H_{\varepsilon}(x)=\dfrac{1}{2}+\dfrac{1}{\pi}\cdot\tan^{-1}(x/\varepsilon),$$ for each $\varepsilon>0$ and $\delta_{\varepsilon}(x)=H'_{\varepsilon}(x)$.

In the book of S.Osher & R.Fedkiw - Level set methods and dynamic implicit surfaces, at page 15, is stated without proof, a formula like that: $$\text{Per}_{\Omega}(\omega)=\lim_{\varepsilon\to 0} \int\limits_{\Omega}\delta_{\varepsilon}(\phi(x))\ |\nabla\phi(x)|\ dx,$$

where

• $\Omega\subset\mathbb{R}^2$ is an open and bounded set,
• $\omega=\{x\in\Omega\ |\ \phi(x)>0\}$,
• $\phi:\Omega\to\mathbb{R}$ is a smooth level set function.
• $\text{Per}_{\Omega}(\omega)$ is the perimeter of $\omega$ that lies inside $\Omega$ (the relative perimeter).
• $\delta_{\varepsilon}:\mathbb{R}\to\mathbb{R}$, is a smooth approximation of the $\delta$-Dirac (generalized) function, precisely $$ \delta_{\varepsilon}(x)=\dfrac{\varepsilon}{\pi (\varepsilon^2+x^2)}\;. $$ Now, denoting by $\bigvee_{\Omega} f$ the total variation of $f$ in $\Omega$, here is my question:

Is it true that for any $\chi_{\omega}\in BV(\Omega)$ ($\chi_\omega$ being the characteristic function of $\omega$ - so by this condition we require $\omega$ to be a set with finite perimeter), whenever $\chi_n\to \chi_{\omega}$ in the $L^1(\Omega)$ norm and $0\leq \chi_n\leq 1, \chi_n\in W^{1,1}(\Omega)$ for each $n$, we have that: $$\bigvee_{\Omega} \chi_{\omega}=\lim\limits_{n\to\infty} \bigvee_{\Omega}\chi_n(x)\;\;\boldsymbol{?}$$

Notes

• It is known that: $$\bigvee_{\Omega} f = \sup\left\{\,\int\limits_{\Omega} f(x)\text{div}(\varphi(x))\ dx\ |\ \varphi\in C^{\infty}_{c}(\Omega;\mathbb{R}^2)\ \text{and} \ \Vert\varphi\Vert_{L^{\infty}(\Omega;\mathbb{R}^2)}\leq 1\right \}$$

• It is also known that for $\chi\in W^{1,1}(\Omega)\subset BV(\Omega)$ we have that: $$\bigvee_{\Omega} \chi=\int\limits_{\Omega} |\nabla\chi(x)|\ dx.$$

• We have that $\text{Per}_{\Omega}(\omega)=\bigvee_{\Omega}\chi_{\omega}$ for any $\omega\subseteq\Omega$.

• Also, one part of the inequality is well-know (lower semicontinuity of the total variation), so: $$\bigvee_{\Omega} \chi_{\omega}\leq\liminf\limits_{n\to\infty} \bigvee_{\Omega}\chi_n(x)$$

• In the cited book, at page 10, Theoreme 1.3 there's an approximation result of the $BV$ functions with $C^{\infty}_c$ functions. Maybe it is useful.

• I tried to prove it. It is correct for many examples that I take, but I can't figure out why: intuitively I understand it but technically I'm blocked.

Motivation

I'm interested in this type of formulas because if it is indeed true we will have that: $$\text{Per}_{\Omega}(\omega)=\lim\limits_{\varepsilon\to 0}\int_{\Omega} H'_{\varepsilon}(\phi(x))|\nabla\phi(x)|\ dx, $$ for any function $H_{\varepsilon}:\mathbb{R}\to\mathbb{R},\ H_{\varepsilon}\in W^{1,1}(\mathbb{R})$ that approximates in $L^1(\Omega)$ the Heaviside function $$ H(x)=\begin{cases} 1,\ x>0 \\ 0,\ x<0\end{cases}. $$ In the above example, $$ H_{\varepsilon}(x)=\dfrac{1}{2}+\dfrac{1}{\pi}\cdot\tan^{-1}(x/\varepsilon),$$ for each $\varepsilon>0$ and $\delta_{\varepsilon}(x)=H'_{\varepsilon}(x)$.

In the book of S.Osher & R.Fedkiw - Level set methods and dynamic implicit surfaces, at page 15, is stated without proof, a formula like that:

$$\text{Per}_{\Omega}(\omega)=\lim\limits_{\varepsilon\to 0} \int_{\Omega}\delta_{\varepsilon}(\phi(x))\ |\nabla\phi(x)|\ dx,$$

where $\omega=\{x\in\Omega\ |\ \phi(x)>0\}$ and $\phi:\Omega\to\mathbb{R}$ is a smooth level set function. By $\text{Per}_{\Omega}(\omega)$ I denote the perimeter of $\omega$ that lies inside $\Omega$. Here $\Omega\subset\mathbb{R}^2$ is an open and bounded set.

Also $\delta_{\varepsilon}:\mathbb{R}\to\mathbb{R},\ \delta_{\varepsilon}(x)=\dfrac{\varepsilon}{\pi (\varepsilon^2+x^2)}$ is a smooth approximation of the $\delta$-Dirac (generalized) function.

My question is: Is it true that for any $\chi_{\omega}\in BV(\Omega)$ ($\chi_\omega$ being the characteristic function of $\omega$ - so by this condition we require $\omega$ to be a set with finite perimeter), whenever $\chi_n\to \chi_{\omega}$ in the $L^1(\Omega)$ norm and $0\leq \chi_n\leq 1, \chi_n\in W^{1,1}(\Omega)$ for each $n$, we have that: $$\bigvee_{\Omega} \chi_{\omega}=\lim\limits_{n\to\infty} \bigvee_{\Omega}\chi_n(x)$$ ? I denote by $\bigvee_{\Omega} f$ the total variation of $f$ in $\Omega$.

=======================================================================

$\bullet$ It is known that:

$$\bigvee_{\Omega} f = \sup\left\{\int_{\Omega} f(x)\text{div}(\varphi(x))\ dx\ |\ \varphi\in C^{\infty}_{c}(\Omega;\mathbb{R}^2)\ \text{and} \ \Vert\varphi\Vert_{L^{\infty}(\Omega;\mathbb{R}^2)}\leq 1\right \}$$

$\bullet$ It is also known that for $\chi\in W^{1,1}(\Omega)\subset BV(\Omega)$ we have that:

$$\bigvee_{\Omega} \chi=\int_{\Omega} |\nabla\chi(x)|\ dx.$$

$\bullet$ We have that $\text{Per}_{\Omega}(\omega)=\bigvee_{\Omega}\chi_{\omega}$ for any $\omega\subseteq\Omega$.

$\bullet$ One part of the inequality is well-know (lower semicontinuity of the total variation), so:

$$\bigvee_{\Omega} \chi_{\omega}\leq\liminf\limits_{n\to\infty} \bigvee_{\Omega}\chi_n(x)$$

P.S. In this course, at page 10, Theoreme 1.3 gives an approximation result of the $BV$ functions with $C^{\infty}_c$ functions. Maybe it is useful.

I tried to prove it. It is for many examples that I take, but I can't figure out why. Intuitively I understand it but technically I'm blocked.

I'm interested in this type of formulas because if it is indeed true we will have that:

$$\text{Per}_{\Omega}(\omega)=\lim\limits_{\varepsilon\to 0}\int_{\Omega} H'_{\varepsilon}(\phi(x))|\nabla\phi(x)|\ dx,$$

for any function $H_{\varepsilon}:\mathbb{R}\to\mathbb{R},\ H_{\varepsilon}\in W^{1,1}(\mathbb{R})$ that approximates in $L^1(\Omega)$ the Heaviside function $H(x)=\begin{cases} 1,\ x>0 \\ 0,\ x<0\end{cases}$. In the above example $H_{\varepsilon}(x)=\dfrac{1}{2}+\dfrac{1}{\pi}\cdot\tan^{-1}(x/\varepsilon)$, for each $\varepsilon>0$ and $\delta_{\varepsilon}(x)=H'_{\varepsilon}(x)$.

In the book of S.Osher & R.Fedkiw - Level set methods and dynamic implicit surfaces, at page 15, is stated without proof, a formula like that: $$\text{Per}_{\Omega}(\omega)=\lim_{\varepsilon\to 0} \int\limits_{\Omega}\delta_{\varepsilon}(\phi(x))\ |\nabla\phi(x)|\ dx,$$

where

• $\Omega\subset\mathbb{R}^2$ is an open and bounded set,
• $\omega=\{x\in\Omega\ |\ \phi(x)>0\}$,
• $\phi:\Omega\to\mathbb{R}$ is a smooth level set function.
• $\text{Per}_{\Omega}(\omega)$ is the perimeter of $\omega$ that lies inside $\Omega$ (the relative perimeter).
• $\delta_{\varepsilon}:\mathbb{R}\to\mathbb{R}$, is a smooth approximation of the $\delta$-Dirac (generalized) function, precisely $$ \delta_{\varepsilon}(x)=\dfrac{\varepsilon}{\pi (\varepsilon^2+x^2)}\;. $$ Now, denoting by $\bigvee_{\Omega} f$ the total variation of $f$ in $\Omega$, here is my question:

Is it true that for any $\chi_{\omega}\in BV(\Omega)$ ($\chi_\omega$ being the characteristic function of $\omega$ - so by this condition we require $\omega$ to be a set with finite perimeter), whenever $\chi_n\to \chi_{\omega}$ in the $L^1(\Omega)$ norm and $0\leq \chi_n\leq 1, \chi_n\in W^{1,1}(\Omega)$ for each $n$, we have that: $$\bigvee_{\Omega} \chi_{\omega}=\lim\limits_{n\to\infty} \bigvee_{\Omega}\chi_n(x)\;\;\boldsymbol{?}$$

=======================================================================

Notes

• It is known that: $$\bigvee_{\Omega} f = \sup\left\{\int\limits_{\Omega} f(x)\text{div}(\varphi(x))\ dx\ |\ \varphi\in C^{\infty}_{c}(\Omega;\mathbb{R}^2)\ \text{and} \ \Vert\varphi\Vert_{L^{\infty}(\Omega;\mathbb{R}^2)}\leq 1\right \}$$

• It is also known that for $\chi\in W^{1,1}(\Omega)\subset BV(\Omega)$ we have that: $$\bigvee_{\Omega} \chi=\int_{\Omega} |\nabla\chi(x)|\ dx.$$

• We have that $\text{Per}_{\Omega}(\omega)=\bigvee_{\Omega}\chi_{\omega}$ for any $\omega\subseteq\Omega$.

• Also, one part of the inequality is well-know (lower semicontinuity of the total variation), so: $$\bigvee_{\Omega} \chi_{\omega}\leq\liminf\limits_{n\to\infty} \bigvee_{\Omega}\chi_n(x)$$

• In the cited book, at page 10, Theoreme 1.3 there's an approximation result of the $BV$ functions with $C^{\infty}_c$ functions. Maybe it is useful.

• I tried to prove it. It is correct for many examples that I take, but I can't figure out why: intuitively I understand it but technically I'm blocked.

Motivation

I'm interested in this type of formulas because if it is indeed true we will have that: $$\text{Per}_{\Omega}(\omega)=\lim\limits_{\varepsilon\to 0}\int_{\Omega} H'_{\varepsilon}(\phi(x))|\nabla\phi(x)|\ dx, $$ for any function $H_{\varepsilon}:\mathbb{R}\to\mathbb{R},\ H_{\varepsilon}\in W^{1,1}(\mathbb{R})$ that approximates in $L^1(\Omega)$ the Heaviside function $$ H(x)=\begin{cases} 1,\ x>0 \\ 0,\ x<0\end{cases}. $$ In the above example, $$ H_{\varepsilon}(x)=\dfrac{1}{2}+\dfrac{1}{\pi}\cdot\tan^{-1}(x/\varepsilon),$$ for each $\varepsilon>0$ and $\delta_{\varepsilon}(x)=H'_{\varepsilon}(x)$.