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.

  • 3
    $\begingroup$ 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


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.

  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.