On the derivative of a distance function

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.

• $\text{ess sup}_{x \in \mathbb{R}^{d}}|\nabla F(x)|\le1$ holds because $F$ is $1$-Lipschitz. – Iosif Pinelis Jul 29 '16 at 18:01
• If $D$ is the complement of the closed unit ball and $r>2$, then $\nabla F(0)=0$. – Iosif Pinelis Jul 29 '16 at 18:09
• Thanks for your reply. But I don't think $\nabla F(0)=0$ implies $\text{ess inf}_{x \in \mathbb{R}^{d}}|\nabla F(x)|=0$. – sharpe Jul 29 '16 at 21:32

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.