# Can the derivative of the distance function from a subset be zero?

Let $$M$$ be a smooth Riemannian manifold, and let $$S \subseteq M$$ be compact. Let $$d_S$$ be the distance function from $$S$$. Let $$p \in M \setminus{S}$$, and suppose that $$d_S$$ is differentiable at $$p$$. Is it possible that $$d(d_S)_p=0$$? (can $$p$$ be a critical point of $$d_S$$?)

If there exist a length-minimizing path from $$p$$ to $$S$$, then $$d(d_S)_p\neq 0$$. What happens when such a length-minimizing path does not exist?

Indeed, if $$\alpha(t)$$ is a unit speed length-minimizing path from $$p$$ to a closest point $$s(p)\in S$$, we must have $$d_S(\alpha(t))=d_S(p)-t$$. Differentiating at $$t=0$$, we get $$d(d_S)_p(\dot \alpha(0))=-1$$, so $$d(d_S)_p \neq 0$$.

Edit:

I think that the answer is positive in general. Let $$p \in M\setminus{S}$$. As observed above, it suffices to prove that there always exist a unit speed path $$\alpha(t)$$, satisfying $$d_S(\alpha(t)) = d_S(p)-t$$ for sufficiently small $$t$$.

I am quite sure that such a path must always exist; however, the proof I found is a bit cumbersome. I would be happy to see a cleaner proof, or a different approach to showing $$d(d_S)_p \neq 0$$.

• Since $S$ is compact, there must be a point $s$ of least distance. If I remember correctly, every distance is achieved by some geodesic in a complete manifold. – Chrystomath Mar 20 '19 at 16:08
• You are right. But I don't want to assume completeness... – Asaf Shachar Mar 20 '19 at 16:32

I prove that there always exist a unit speed path $$\alpha(t)$$, satisfying $$d_S(\alpha(t)) = d_S(p)-t$$ for sufficiently small $$t$$.

First, note that for every unit speed path $$\alpha(t)$$ starting from $$p$$, we have $$d_S(\alpha(t)) \ge d_S(p)-t. \tag{1}$$ Indeed, $$d_S$$ is $$1$$-Lipschitz, hence $$d_S(p)-d_S(\alpha(t)) \le d(p,\alpha(t)) \le L(\alpha|_{[0,t]})=t$$.

Now, let $$B_{\delta}(p)$$ be a normal ball around $$p$$, such that $$d_S(p)>\delta$$, and let $$S_{\delta}=\partial B_{\delta}(p)$$. Let $$x_0 \in S_{\delta}$$ be a point where $$d_S(\cdot)$$ obtains a minimal value on $$S_{\delta}$$. Let $$\alpha(t)$$ be the (unique) unit speed geodesic from $$p$$ to $$x_0$$ which lies inside $$B_{\delta}(p)$$.

We shall prove two things:

1. $$\alpha(t)$$ minimizes the distance from $$S$$ inside the sphere $$S_t$$ for every $$t<\delta$$, i.e. $$d_S(\cdot)$$ obtains a minimal value on $$S_t$$ at $$\alpha(t)$$.

2. Using $$(1)$$, we deduce that $$d_S(\alpha(t)) = d_S(p)-t$$ for sufficiently small $$t$$.

Proof of 1.:

Assume by contradiction that there exists some $$t< \delta$$, and $$x_0 \in S_t$$ such that $$d_S(x_0)< d_S(\alpha(t))$$. Let $$\beta$$ be a path from $$x_0$$ to a point $$\tilde s \in S$$ such that $$L(\beta) < d_S(\alpha(t))$$. Since we assumed $$d_S(p) > \delta$$, $$\beta$$ must intersect $$S_{\delta}$$ at some point $$y_0$$. Now, we have

$$L(\beta:x_0 \to y_0)+L(\beta:y_0 \to \tilde s)=L(\beta:x_0 \to \tilde s)=L(\beta)< d_S(\alpha(t)), \tag{2}$$

and

$$\delta-t+d_S(\alpha(\delta))\le\delta-t+d_S(y_0) \le d(x_0,y_0)+d(y_0,\tilde s) \le L(\beta:x_0 \to y_0)+L(\beta:y_0 \to \tilde s). \tag{3}$$

Combining $$(2)$$ and $$(3)$$, we deduce

$$\delta-t+d_S(\alpha(\delta)) < d_S(\alpha(t)) \Rightarrow \delta-t < d_S(\alpha(t))-d_S(\alpha(\delta)),$$

which contradicts $$d_S(\alpha(t))-d_S(\alpha(\delta)) \le d(\alpha(t),\alpha(\delta))\le L(\alpha|_{[t,\delta]})=\delta-t.$$

Proof of $$2$$:

By $$(1)$$ we know that $$\alpha(t)$$ minimizes the distance from $$S$$ inside the sphere $$S_t$$ for every $$t<\delta$$, i.e. $$d_S(\cdot)$$ obtains a minimal value on $$S_t$$ at $$\alpha(t)$$. We shall prove that $$d_S(\alpha(t)) = d_S(p)-t$$. It suffices to prove that $$d_S(\alpha(t)) \le d_S(p)-t$$. Assume otherwise; then $$d_S(\alpha(t)) > d_S(p)-t$$, so $$d_S(p) . Thus, there is a path $$\beta$$, from $$p$$ to some $$\tilde s \in S$$ such that $$L(\beta)< d_S(\alpha(t))+t$$. $$\beta$$ must intersect $$S_{t}$$ at some point $$y_0$$. (Here we use $$t< \delta). Thus, $$t+L(\beta:y_0 \to \tilde s)\le L(\beta:p \to y_0)+L(\beta:y_0 \to \tilde s)=L(\beta)< d_S(\alpha(t))+t.$$ So, we got $$d_S(y_0) \le d(y_0,\tilde s)\le L(\beta:y_0 \to \tilde s) < d_S(\alpha(t)),$$ and $$y_0,\alpha(t)$$ both are in $$S_t(p)$$ contradicting the fact that $$\alpha(t)$$ was a distance minimizer w.r.t $$S$$ in this sphere.

• If you don't assume completeness, then there need not exist a distance minimizing curve. For example, take $\mathbb{R}^2\setminus\{(1,0)\}$, $S=\{(0,0)\}$, $p=(2,0)$. – Chrystomath Mar 22 '19 at 9:29
• I agree. But this is OK-it does not contradict my claim. Even though we don't always have a minimizing path, we always have a path that for sufficiently short time, decreases the distance as quickly as possible. – Asaf Shachar Mar 22 '19 at 11:00