Distance function to a submanifold 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.  $W^{1,2}(M)$?  

 A: 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 $\mathbb R$.
A: 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 $M$)
