Let $M$ be a compact Riemannian manifold and $\Sigma\subset M$ a closed submanifold. Given $x\in M$ we define the distance function to $\Sigma$ by $$d_\Sigma(x):=\inf\{d(x,y):y\in \Sigma\},$$ where $d$ is the metric on $M$. Of course, in a small tubular neighborhood of $\Sigma$ the function $d_\Sigma$ will be smooth. Rather, my questions have to do with global properties of $d_\Sigma$.
Since $\Sigma$ is a closed subset of $M$, it is not hard to prove, using the triangle inequality, that $d_\Sigma$ is a Lipschitz-continuous function with respect to the metric $d$, with Lipschitz constant $1$. In fact, $d_\Sigma \in W^{1,\infty}(M)$ (see Section 5.8 in Evans' PDE book) and it is differentiable a.e. on $M$ by Rademacher's Theorem.
My first question is the following:
- If $M=\mathbb{R}^n$ then $d_\Sigma$ is a solution to the Eikonal equation, i.e. $\|\nabla d_\Sigma\|=1$ a.e. Is this also true for a general manifold $M$?
My second question is related to the behavior of $d_\Sigma$ when we vary the set $\Sigma$.
Suppose $\Sigma_t$ are closed submanifolds of $M$ that vary continously in the Hausdorff distance $d_H$, with respect to $t$. Remember that $d_H$ is a metric in the set of compact subsets of $M$. In particular we have the triangle inequality $$d(x,\Sigma_t)\leq d(x,\Sigma_s) + d_H(\Sigma_s,\Sigma_t).$$ This implies that the functions $d(\cdot,\Sigma_t)$ form a continuous curve in $L^\infty(M)$.
- Is it also true that $d(\cdot,\Sigma_t)$ is a continuous curve in $W^{1,\infty}(M)$? i.e. does it's gradients vary continuously? If not, would it be continuous (perhaps under extra assumptions) in a less regular $L^p$-norm, e.g. $W^{1,2}(M)$?
Best Answer
Question 1. Yes sure and the same proof should work.
Question 2. The answer is "yes" in $W^{1,p}$ for any $p<\infty$ and "no" in $W^{1,\infty}$.
$W^{1,1}$: Note that $$d_{\mathrm{H}}(\Sigma_t,\Sigma_s)<\varepsilon \ \ \iff\ \ \sup_x\{\,|d_{\Sigma_t}(x)-d_{\Sigma_s}(x)|\,\}<\varepsilon.$$
So it is sufficient to show the following:
Assume contrary. Then for some fixed $\varepsilon>0$ and any $t>0$ there is a set $C_t$ of measure $>\varepsilon$ such that $|\nabla_x f_t-\nabla_x f_0|>\varepsilon$ for any $x\in C_t$. Move (almost) along $\nabla f_0$ and integraite both $df_0$ and $df_1$. You get a positive lower bound on $\sup |f_t(z)-f_0(z)|$. The later contradicts that $\sup_x|f_t(x)-f_0(x)|\to 0$.
$W^{1,p}$: All this proves that $f_t$ is continuous in $W^{1,1}$. Since $\nabla f_t$ is bounded, you get continuity in $W^{1,p}$ for any $p<\infty$.
$W^{1,\infty}$: There is no continuity in $W^{1,\infty}$; this can be seen for $M=\mathbb R$ and $f_t(x)=|x-t|$. (The same works for compact $M$)