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 $\mathbb{R}^{d}$.


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.

  • $\begingroup$ So you want $z$ and $r$ to be fixed? Why? $\endgroup$ – ThiKu Jul 30 '16 at 14:38
  • $\begingroup$ If $\partial D$ is unbounded, I think it is difficult to deal with $d(x,\partial D)$. So I cut $\partial D$ by closed ball. $\endgroup$ – sharpe Jul 30 '16 at 15:39

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. $x\in \mathbb{R}^d\setminus K$.


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