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:

  1. 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 $L^\infty(M)$.

  1. 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. $W^{1,2}(M)$?
  • 3
    $\begingroup$ It is always a solution to the eikonal equation. $\endgroup$
    – Ben McKay
    Commented Dec 22, 2014 at 18:21
  • 1
    $\begingroup$ Have you looked at a proof of the tubular neighbourhood theorem? You can put everything in terms of Hausdorff distance if you like. Perhaps look at the one in the Guillemin and Pollack text? It adapts to general manifolds quite directly. $\endgroup$ Commented Dec 22, 2014 at 21:57

2 Answers 2


I'm posting this sketch just for the sake of completeness. This question was already marked as answered by Anton. But I thought of this, more geometrical, argument after the discussion that answered the question in the first time. Notice that the fact that the $\Sigma$ are assumed to be submanifolds is of no relevance to this questions, the important feature is that they are compact subsets of $M$. Hence I will substitute $\Sigma$ and $\Sigma_n$ by $K$ and $K_n$ in what follows.

One important observation is the following:

Assertion If $d_K$ is differentiable at a point $x\in M$, then there exists a unique geodesic $\gamma$ starting at $x$ that minimize the distance to $K$.

To see this, observe that we can always find at least one such geodesic by compacity. Along one of such geodesics the distance to $K$ decreases linearly with time (otherwise it would contradict the fact that is minimizing), then we can differentiate along it, in particular in $t=0$ to obtain $$\gamma'(0)\cdot \nabla d_K(x) = -1.$$ But $\|\gamma'(0)\|=1$ (because is geodesic) and $\|\nabla d_K(x)\|\leq 1$ (because of the Lipschitz bound). This implies $\gamma'(0)=- \nabla d_K(x)$, therefore $\gamma$ is unique and also $\|\nabla d_K(x)\|= 1$.

This answers Question 1: The function is always a solution to the Eikonal equation.

Next, suppose we have a sequence of compact sets $K_n$ converging to $K$ in the Hausdorff distance. As Anton pointed out in his answer, since the gradients of the functions $d_{K_n}$ are all bounded, we can obtain convergence in $L^p$, for high $p$, from a weaker form of convergence. In particular, by Lebesgue $L^p$-dominated convergence, it is enough to prove pointwise convergence almost everywhere.

In almost every point, all the functions $d_{K_n}$ and $d_K$ are differentiable. By the Assertion above we have a unique minimizing geodesic $\gamma_n$ and $\gamma$ for each one, respectively. If $\gamma_n \nrightarrow \gamma$ we would find another geodesic minimizing the distance to $K$, contradicting the uniqueness. Then the geodesics converge, and therefore they velocities at $x$ too, i.e. the gradients of the functions $d_{K_n}$.

This answers Question 2 for every $1\leq p<\infty$. The case $p=\infty$, does not hold, as Anton pointed out, even for distance functions to points in $\mathbb R$.


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:

If (1) $\sup_x|f_t(x)-f_0(x)|\to 0$ as $t\to 0$, (2) $|\nabla_xf_t|\le 1$ for any $t,x$ and (3) $|\nabla_xf_0|=1$ for almost all $x$ then $$\int\limits_M|\nabla_x f_t-\nabla_x f_0|\cdot dx\to 0.$$

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 $M$)

  • $\begingroup$ Fix a chart, near $z_0$ and apply Lebesgue differentiation theorem to each component of gradient. $\endgroup$ Commented Dec 23, 2014 at 0:14
  • $\begingroup$ P.S. You can not get $W^{1,\infty}$, it does not hold even for one-point sets in $\mathbb R$. The proof gives continuity in $W^{1,1}$; since gradient is bounded, it implies continuity in $W^{1,p}$ for any finite $p$. $\endgroup$ Commented Dec 23, 2014 at 0:27
  • $\begingroup$ Lebesgue's theorem implies that at most of the points near $z_0$ the value is almost the same. $\endgroup$ Commented Dec 23, 2014 at 1:34
  • $\begingroup$ Oh, yes, you are right, one needs to be more careful, I will update the answer. $\endgroup$ Commented Dec 23, 2014 at 5:58
  • $\begingroup$ I apologize for being so picky, but I still do not find your answer satisfactory. Particularly the line that begins with "Move (almost) along...". I understand what you want to do, but for doing it in detail, several $\epsilon$'s need to be picked and I haven't being able to do it in the right way. Of course this could be my fault. Nonetheless, I will accept your answer as satisfactory because it helped me to think about another one for the specific case of the distance functions, but I encourage you to make your response more precise. Thanks the help and I will post my own answer later. $\endgroup$
    – mmen
    Commented Dec 23, 2014 at 19:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.