I have a question about the derivative of a distance function.


Let $D \subset \mathbb{R}^{d}$ be a connected and unbounded open subset with smooth boundary. $B(z,r)$ denotes the **closed (not open)** ball of radius $r>0$ centered at $z \in \bar{D}$. We define the following distance function $F$ on $\mathbb{R}^{d}$:
\begin{equation\*}
F: \mathbb{R}^{d} \ni x \mapsto d(x,\partial D \cap B(z,r)) \in \mathbb{R}.
\end{equation\*}
This function is differentiable in a.e. sense since it is $1$-Lipschitz continuous (Rademacher's theorem).
Note that $\{F=0\}= \partial D \cap B(z,r)$ holds,
since $\partial D \cap B(z,r)$ is a compact subset of $\mathbb{R}^{d}$.


**Question**


I think that $|\nabla F|>0$ a.e. does not hold in general. But can we show the following assertion?


**There exists $\varepsilon>0$ such that $|\nabla F|>0 \text{ a.e. on } \{F < \varepsilon\}$**


If you know related results, please let me know.



A lower estimate of the derivative of a distance function
Let $K$ be any closed set in $\mathbb{R}^d$. In your case, $K= \partial D \cap B(x,r)$ but it doesn't matter. Let $F = d(\cdot, K)$.


Let $x$ be a point of differentiability of $F$ not in $K$. Let $y\in K$ be a point with $d(x,y) = d(x,K)$.


Then $F$ decreases linearly at speed $1$ along the line segment from $x$ to $y$. Hence $|\nabla F|\geq 1$. Since $F$ is $1$-Lipschitz this is an equality for a.e. $x\in \mathbb{R}^d\setminus K$. 


