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Working with the level set method introduced by Osher & Sethian in shape optimization I came across a simple question that I did not succeed to prove. It mainly asserts that the perimeter of the zero level set is continuous with respect to the level set function. The continuity of the area is true (see here Lebesgue measure of a neighbourhood of a curve). It is well known that the perimeter is a lower semicontinuous functional, so one inequality is checked.

Here is my question:

Let $\Omega\subset\mathbb{R}^2$ be a bounded set with smooth boundary, $\phi_n:\overline\Omega\to\mathbb{R},\ n\geq 1$ be a sequence of functions such that:

$\bullet\ \phi_n\to \phi\ \text{in}\ C^1(\overline{\Omega})$ i.e. $\lim\limits_{n\to\infty} \sup_{x\in\overline{\Omega}}|\phi_n(x)-\phi(x)|+\sup_{x\in\overline{\Omega}} |\nabla\phi_n(x)-\nabla\phi(x)|=0$

$\bullet$ $\nabla\phi_n\neq 0$ on $\phi_n=0$ and $\nabla\phi\neq 0$ on $\phi=0$

$\bullet$ $\{\phi=0\}\Subset\Omega$

Then how can we prove that:

$$\lim\limits_{n\to\infty} \int_{\phi_n=0} 1 \ d\sigma=\int_{\phi=0} 1\ d\sigma$$

?

For the continuity of the perimeter on level sets with respect to the level you can see other posts here: Continuity of Hausdorff measure on level sets, Continuity of surface integrals on level sets

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As it is written it's not true, by trivial reasons: take $\Omega\subset \mathbb{R}^2$ the unit open disk, $\phi(x):=1-|x|^2$ and $\phi_n(x):=\phi(x)+\frac1n$ for $x\in\overline\Omega$. So $\{x\in\overline\Omega:\phi(x)=0\}$ is the unit circle and $\{x\in\overline\Omega:\phi_n(x)=0\}$ is empty. The measure of interior zeros is neither continuous: if $\psi_n(x):=\phi(x)-\frac1n$ then $\{x\in \Omega:\phi(x)=0\}$ is empty and $\{x\in \Omega:\psi_n(x)=0\}$ is a circle of radius $1-o(1).$

You need to add the condition $\phi\neq0$ on $\partial\Omega$, then it is true, just parametrizing the zero set of $\phi_n$ by $N\ge0$ closed curves $\{\gamma_{j,n}:\mathbb{S}^1\to\Omega\}_{1\le j\le N}$ converging in $C^1$ respectively to $N$ curves $\gamma_{j,n}:\mathbb{S}^1\to\Omega$, that parametrize the zero set of $\phi$. Here $N\ge0$ is the number of connected components of the zero set of $\phi$ and $\phi_n$, which is finite and definitively constant wrto $n$. Then the length of each component $\int_{\mathbb{S}^1}|\dot\gamma_{j,n}(s)|ds$ passes to the limit.

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  • $\begingroup$ Indeed. I have edited my post. Thanks for your solution! I think there are some typos in your argument. You probably wanted to say that $\gamma_{j,n}\to \gamma_j$ in $C^1$ and at the last row the integrand should not be squared. Am I right? Also, how can it be proved that $\phi=0$ is composed by a finite number of closed curves and that $\phi_n=0$ has the same number of curves? Is it that simple? $\endgroup$
    – Bogdan
    Aug 18 '21 at 11:57
  • $\begingroup$ And what kind of results assures that one can find $C^1$ $\gamma$-s having $C^1$ $\phi$ and the convergence of $\gamma$'s still remains $C^1$? $\endgroup$
    – Bogdan
    Aug 18 '21 at 12:11
  • $\begingroup$ Yes, sure, typo fixed $\endgroup$ Aug 18 '21 at 12:13
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    $\begingroup$ The zero set $\{\phi=0\}$ is a compact 1 manifold, hence a finite union of disjoint circles; it has a $C^1$ tubular nbd, the zero sets $\{\phi_n=0\}$ are (eventually, for $n$ large) sections of it , converging to the zero section, which is $\{\phi=0\}$. $\endgroup$ Aug 18 '21 at 12:29

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