1
$\begingroup$

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 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:x \mapsto d(x,\partial D \cap B(z,r)). \end{equation*} This function is differentiable in a.e. sense since it is Lipschitz continuous (Rademacher's theorem).

Can we show that the following estimate holds? \begin{equation*} \text{ess inf}_{x \in \mathbb{R}^{d}} |\nabla F(x)|>0 \end{equation*} More weakly, can we show that the following? \begin{equation*} |\nabla F(x)|>0 \text{ a.e.} \end{equation*}

If you know related results, please let me know.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ $\text{ess sup}_{x \in \mathbb{R}^{d}}|\nabla F(x)|\le1$ holds because $F$ is $1$-Lipschitz. $\endgroup$ Commented Jul 29, 2016 at 18:01
  • $\begingroup$ If $D$ is the complement of the closed unit ball and $r>2$, then $\nabla F(0)=0$. $\endgroup$ Commented Jul 29, 2016 at 18:09
  • 1
    $\begingroup$ Thanks for your reply. But I don't think $\nabla F(0)=0$ implies $\text{ess inf}_{x \in \mathbb{R}^{d}}|\nabla F(x)|=0$. $\endgroup$ Commented Jul 29, 2016 at 21:32

1 Answer 1

Reset to default
2
$\begingroup$

Let $K$ be a compact set with smooth boundary. The distance function has gradient 1 everywhere where the gradient exists. The gradient exists in any $x$ there exists a unique $y \in \partial K$ boundary point minimizing the distance $d(x,y) = d(K,x)$. The proof is simple. Take the normal at $y$ and map a neighbourhood.

Improve this answer
$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.