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For $\phi\in C^1(\mathbb{R}^N)$ with $$\omega_{\phi}=\{x\in\mathbb{R}^N\ |\ \phi(x)>0\}$$ being a bounded set with $\nabla\phi (x)\neq 0,\ \forall\ x\in\phi^{-1}(0)=\partial\omega_{\phi}\neq \emptyset$ consider the signed distance function:

$$ d_{\phi}:\mathbb{R}^N\to\mathbb{R},\ d_{\phi}(x)=\begin{cases} -\text{dist}(x,\partial\omega_{\phi}),\ x\in\omega_{\phi} \\ 0,\ x\in\partial\omega_{\phi} \\ \text{dist}(x,\partial\omega_{\phi}),\ x\in\mathbb{R}^N\setminus\overline{\omega}_{\phi}\end{cases}$$

I have some technical questions that seem to be true about $d_{\phi}$.

Let's take a point $x_0\in\mathbb{R}^N$ with a unique $y(0)\in\partial\omega_{\phi}$ such that: $|x_0-y|=\text{dist}(x_0,\partial\omega_{\phi})$ and any $\psi\in C^1(\mathbb{R}^N)$ with $|\psi(x)|<M,\forall\ x\in\mathbb{R}^N$.

For $|\theta|$ sufficiently small we associate to $x_0$ a set denoted $Y(\theta)$ such that $\forall y\in Y(\theta)\subseteq\mathbb{R}^N$ we have $\phi(y)+\theta\psi(y)=0$ and:

$$d_{\phi+\theta\psi}(x_0)=|x_0-y|.$$

How can we prove that:

(1) There is a selection for the multifunction $Y:[0;\theta_0)\rightrightarrows \phi^{-1}(0)$ denoted $y=y(\theta)$ which is continuous and differentiable for $\theta\in [0,\theta_0)$.

Note that we cannot apply here Michael's Selection Theorem because in general $Y(\theta)$ is not a convex set. (https://en.wikipedia.org/wiki/Michael_selection_theorem)

(2) $\lim\limits_{\theta\to 0} d_{\phi+\theta\psi}(x_0)=d_{\phi}(x_0)$

(3) Exists $\lim\limits_{\theta\to 0}\dfrac{d_{\phi+\theta\psi}(x_0)-d_{\phi}(x_0)}{\theta}$

I tried a lot to prove the continuity but I did not succeed.

It is well known (from KKT conditions) that $y(0)$ and $y(\theta)$ satisfy:

$$\begin{cases} \phi(y(0))=0,\ \phi(y(\theta))+\theta\psi(y(\theta))=0 \\ \lambda\nabla \phi (y(0))=y(0)-x_0,\ \lambda_{\theta}\Big (\nabla\phi(y(\theta))+\theta\psi(y(\theta))\Big )=y(\theta)-x_0\end{cases} $$

I proved easily that for $\theta$ sufficiently small $\nabla\phi+\theta\nabla\psi\neq 0$ on $(\phi+\theta\psi)^{-1}(0)\neq \emptyset$. Also note that if we prove $(1)$ then the answer to $(2)$ follows immediately and $(3)$ can be computed from KKT conditions given the result: $-\dfrac{\psi(y(0))}{|\nabla\phi(y(0))|}$.

Maybe the following sources will help:

Set of points with a unique closest point in a compact set

Concavity near the boundary of the distance function

Finally, are there some references that treats the signed distance function with the level set method (not with a shape derivative approach, but a functional approach)?

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Your assertions are wrong. Assume that in a neighboorhood of $(0,0)$ we have $$ \phi(x)=(x_1-1)^2-1+x_2^2-x_2^4\\ \psi(x)=x_2^2 $$ Then for $x_0=(1,0)$ the unique minimum is $y_0=(0,0)$. However for $\theta>0$ there are multiple minima regardless of how small $\theta$ is.

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    $\begingroup$ Nice example! However we can take a selection of that multifunction that associates to $x_0$ the points on $\phi^{-1}(0)$ which realizes the minimum distance between $x_0$ and $\phi^{-1}(0)$ that is both continuous and differentiable w.r.t. $\theta$ (in your example we could take $y>0$). I don't think that all my assertions are wrong. Somehow I could calculate the limit in $(4)$ being $-\dfrac{\psi(y(0))}{|\nabla\phi(y(0))|}$, but the details are a bit messy (even on the continuity condition) which is true in general. $\endgroup$
    – Bogdan
    Commented Dec 31, 2020 at 17:20
  • $\begingroup$ What assumptions are wrong? $\endgroup$
    – Pietro Majer
    Commented Jan 4, 2021 at 0:40
  • $\begingroup$ I have edited my post. Now I think there are none. $\endgroup$
    – Bogdan
    Commented Jan 4, 2021 at 6:02

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