Perimeter continuity of $BV$ sets on any sequence from $W^{1,1}$

Question

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 $\delta_{\varepsilon}(x)=H'_{\varepsilon}(x)$.

Answer 1

This seems too strong, regardless of which set $\omega \subset \Omega$ one works with. We suppose that $\omega$ has bounded perimeter, so that $\chi_\omega \in BV(\Omega)$.

Let $(f_n \mid n \in \mathbf{N})$ be a sequence of functions in $W^{1,1}(\Omega)$ with \begin{equation} \lvert f_n \rvert_{L^1} \to 0 \text{ and } V(f_n) \geq 3 \, \mathrm{Per}(\omega). \end{equation} Let moreover $g_n \in W^{1,1}(\Omega)$ be a sequence of functions so that \begin{equation} \lvert g_n - \chi_\omega \rvert_{BV} \to 0; \end{equation} for example $g_n$ can be obtained by a mollification argument. Then the two sequences combined have \begin{equation} \lvert f_n + g_n - \chi_\omega \rvert_{L^1} \to 0 \text{ but } V(f_n + g_n) \geq V(f_n) - V(g_n) \geq 3/2 \mathrm{Per}(\omega). \end{equation}

We construct Lipschitz functions $f_n$ on $\Omega$ that are independent of $y$. Specifically let, given $A > 0$, \begin{equation} f_n(x,y) =\begin{cases} Ax & \text{ on $[0,\frac{1}{2n}]$} \\ A/n - Ax & \text{ on $[\frac{1}{2n},\frac{1}{n}]$} \end{cases} \end{equation} and extend this to be $\frac{1}{n}$-periodic in $x$.

Then $\lvert f_n \rvert_{L^\infty} \leq \frac{A}{2n} \to 0$ but the total variation is constant equal to $\int_\Omega \lvert D f_n \rvert = A \lvert \Omega \rvert$. To conclude it only remains to take $A > 3 \mathrm{Per}(\omega)/\lvert \Omega \rvert$.