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Let $\Omega\subset\mathbb{R}^2$ a open and bounded set with smooth boundary and $\phi:\Omega\to\mathbb{R}$ a smooth function such that:

$\bullet$ $\phi^{-1}(0)\neq\emptyset$;

$\bullet$ $\nabla\phi(x)\neq 0$ on a neighborhood $W\subset\Omega$ of the curve $\phi^{-1}(0)$.

WLOG we can assume that $W=\{x\in\mathbb{R}^2\ |\ |\phi(x)|<\varepsilon_0\}\subset\Omega$.
How can we prove that:

$$\lim\limits_{\varepsilon\to 0^+} \mathcal{H}^1\big (\{x\in\Omega\ |\ \phi(x)=\varepsilon\}\big )=\mathcal{H}^1\big (\{x\in\Omega\ |\ \phi(x)=0\}\big )\;\; ?$$

Here $\mathcal{H}^1$ denotes the Hausdorff 1-dimensional measure.

In the article of L.Modica -The gradient theory of phase transitions and the minimal interface criterion that can be found here: https://www.math.cmu.edu/~tblass/CNA-PIRE/Modica1987.pdf this property is proved only for the signed distance function (see Lemma 3, at page 8), but from more examples that I take it seems to be valid for many other level functions.

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    $\begingroup$ This follows from the implicit function theorem, no? $\endgroup$
    – Leo Moos
    Jul 23, 2021 at 14:45
  • $\begingroup$ I do not see how. Can you give some details please? $\endgroup$
    – Bogdan
    Jul 23, 2021 at 14:47

2 Answers 2

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As Leo Moos suggested in the comments, in any dimension $d$, this is a simple consequence of the implicit function theorem.

The implicit function theorem implies that every point $x\in\phi^{-1}(0)$ has a neighborhood $\Omega_x$ that is a diffeomorphic image $\varphi_x(Q_{\delta_x})$ of a box $Q_{\delta_x}=\{|y_i|\leq\delta_x,\;1\leq i \leq d\}$, such that $\phi(\varphi(y))=|\nabla \phi(x)|\cdot y_1$, and such that $\varphi_x(0)=x$ and $D\varphi_x(0)$ is a rotation. (Details: assume wlog that $x=0$, and, by rotating, that $\nabla \phi (0)/|\nabla \phi (0)|=e_1,$ the first basis vector. Consider the function $g:\mathbb{R}^d\to \mathbb{R}^d$ defined by $g(x)=(\phi(x)/|\nabla \phi (0)|,x_2,\dots,x_d)$. Then $Dg(0)$ is the identity, and $\varphi_x$ is the inverse $g^{-1}$ provided by the inverse function theorem.) Given $\varepsilon>0$, by choosing smaller $\delta_x$ if necessary, we can ensure that the restriction of $\varphi_x$ to the leaves $\{y_1=h\}$ distorts the $(d-1)$-area by no more than $1+\varepsilon$, i. e., in $Q_{\delta_x}$, $$1-\varepsilon<\left(\det_{2\leq i,j\leq d}\left(\partial_{y_i}\varphi\cdot\partial_{y_j}\varphi\right)\right)^\frac12\leq 1+\varepsilon.$$

By compactness, we can choose a finite cover $\Omega_i=\Omega_{x_i}$, $i=1,\dots,n$ of $\phi^{-1}(0)$. Then, for $\epsilon_0>0$ small enough, we have $W_{\epsilon_0}:=\{x:|\phi(x)|\leq\epsilon_0\}\subset \cup_{i=1}^n\Omega_i$. Put $\Omega_0=\Omega\setminus W_{\epsilon_0}$, then $\Omega_0,\Omega_1,\dots,\Omega_n$ is a finite cover of $\Omega$, and we can pick a partition of unity $f_0,\dots,f_n$ subordinate to that cover. We have for $|h|<\epsilon_0$, $$ \mathcal{H}^{d-1}(\phi^{-1}(h))=\sum_{i=1}^n\int f_i d\mathcal{H}^{d-1}(\phi^{-1}(h)), $$ hence, by changing the variable, $$ (1-\varepsilon)\sum_{i=1}^n\int_{y_1=h} f_i\circ\varphi_i\leq \mathcal{H}^{d-1}(\phi^{-1}(h))\leq (1+\varepsilon)\sum_{i=1}^n\int_{y_1=h} f_i\circ\varphi_i. $$ As $h\to 0$, the terms in the right-hand side tend to $$ (1+\varepsilon)\int_{y_1=0}f_i\circ\varphi_i\leq (1+\varepsilon)^2\int f_i d\mathcal{H}^{d-1}(\phi^{-1}(0)), $$ and similarly for the LHS. This means that $$ (1-\varepsilon)^2\mathcal{H}^{d-1}(\phi^{-1}(0))\leq\liminf_{h\to 0} \mathcal{H}^{d-1}(\phi^{-1}(h)), $$ $$ \limsup_{h\to 0} \mathcal{H}^{d-1}(\phi^{-1}(h))\leq (1+\varepsilon)^2\mathcal{H}^{d-1}(\phi^{-1}(0)), $$ and since $\varepsilon>0$ is arbitrary, we are done.

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    $\begingroup$ @Bogdan, $f=max\{f,0\}-(-min\{f,0\})$, so it suffices to prove the result for positive f, for which the proof extends verbatim. $\endgroup$
    – Kostya_I
    Jul 28, 2021 at 20:54
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    $\begingroup$ @Bogdan, indeed it would be probably more accurate to say "inverse function theorem" - for me these are minor variants of the same thing. I added some details. $\endgroup$
    – Kostya_I
    Aug 1, 2021 at 9:22
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    $\begingroup$ @Bogdan, I am not sure I understand the question. I fix the point $x$ and then rotate the coordinate axes so that the first basis vector points in the direction of the gradient of $\nabla \phi(x)$, the second one is any vector orthogonal to it, etc. $\endgroup$
    – Kostya_I
    Aug 16, 2021 at 16:08
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    $\begingroup$ @Bogdan, in these new coordinates, the gradient of $\phi$ has coordinates $(\frac{\partial \phi}{\partial x_1},0,\dots,0)$, so $\frac{\partial \phi}{\partial x_1}$ cannot be zero because we assume the gradient does not vanish. $\endgroup$
    – Kostya_I
    Aug 16, 2021 at 18:18
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    $\begingroup$ @Bogdan, sure, but this is the derivative with respect to the first coordinate in the original coordinates, not the new ones. $\endgroup$
    – Kostya_I
    Aug 16, 2021 at 18:35
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Not a complete answer, but just some ideas that are too long for a comment.

Note: Here we use $\mathcal H^n, \mathcal L^n$ to denote the Hausdorff and Lebesgue measure respectively.

Let $\Omega \subset \mathbb R^n$ be an open bounded domain, and let $f: \Omega \to \mathbb R$ be a $C^1$ function such that $\nabla f(x) \neq 0$ for all $x \in \Omega$,

Define $g: \mathbb R\to \mathbb R$ by $g(y) := \mathcal L^n \{x \in \mathbb \Omega \ | \ f(x) \leq y\}$.

Then an application of the coarea formula gives us the following:

Lemma: $g$ is differentiable at $y$ if and only if $h(x) := \mathbb 1_{\{f(x) = y\}}(x) \frac{1}{|\nabla f(x)|}$ is $\mathcal H^{n-1}$-integrable, and we have

$$g’(y) = \int_{\Omega} \mathbb 1_{\{f(x) = y\}}(x) \frac{1}{|\nabla f(x)|}.$$

Now in the problem given, by compactness we may assume wlog that $\nabla \phi$ is bounded below on $\Omega$ by some absolute constant $c > 0$. Since $\phi$ is smooth, the level sets $E_y := \{\phi(x) = y\}$ are smooth submanifolds of $\Omega$, in particular they have finite $\mathcal H^1$ measure.

Thus $\mathbb 1_{\{\phi(x) = y\}}(x) \frac{1}{|\nabla \phi(x)|}$ is $\mathcal H^1$ integrable, and so, taking $f = \phi$ in the lemma, we have that $g$ is differentiable for all $-\varepsilon < y < \varepsilon$.

Now by the uniform bound on $|\nabla f|$, $g’(y)$ is comparable to $\mathcal H^1 (E_y)$ up to a scalar constant. After some epsilon chasing, we will be done if we can show that $g$ is in addition continuously differentiable. However I am having trouble showing this.

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