The distance function in the presence of an obstacle

Abstract

We study the Riemannian distance function from a fixed point (a point-wise target) of Euclidean space in the presence of a compact obstacle bounded by a smooth hypersurface. First, we show that such a function is locally semiconcave with a fractional modulus of order one half and that, near the obstacle, this regularity is optimal. Then, in the Euclidean setting, we prove that the singularities of the distance function propagate, in the sense that each singular point belongs to a nontrivial singular continuum. Finally, we investigate the lack of differentiability of the distance function when a convex obstacle is present.

Appendix A. Proof of Theorem 1.2

Let $K\subset {\mathbb {R}}^n$ be a compact set such that $k_0\notin K$ and $K\cap \partial {\mathcal {O}}\not =\emptyset$. For $x\in K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}}$ we define

$$\begin{aligned} E(x)=\inf _\gamma \int _0^1 \langle A(\gamma (t)){\dot{\gamma }}(t),{\dot{\gamma }}(t)\rangle \, dt \end{aligned}$$

where the infimum is taken w.r.t. $\gamma \in AC([0,1];\overline{{\mathbb {R}}^n\setminus {\mathcal {O}}})$ with $\gamma (0)=x$ and $\gamma (1)=k_0$. We recall that

$$\begin{aligned} d(x)=\sqrt{E(x)},\quad x\in K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}} . \end{aligned}$$

(A.59)

Equation (A.59) describes a well-known property whose proof can be found, for instance, in [22, p. 93]. Now, observe that, since $d>0$ on $K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}}$, we have that $E>0$ on $K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}}$. Furthermore, in [13, Proposition 3.9], it is shown that $E\in SC^{\frac{1}{2}}(K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}})$. Then, it suffices to prove that the square root of a positive semiconcave function is semiconcave. For this purpose, first we show that $\sqrt{E}$ is a Lipschitz function on $K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}}$. Indeed, the existence of a constant L such that

$$\begin{aligned} |E(x)-E(y)|\le L |x-y|,\qquad \forall x,y\in K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}} \end{aligned}$$

implies that, for every $x,y\in K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}}$,

$$\begin{aligned} |\sqrt{E(x)}-\sqrt{E(y)}|\le \frac{L}{\sqrt{E(x)}+\sqrt{E(y)}} |x-y|\le L'|x-y| , \end{aligned}$$

with

$$\begin{aligned} L'=\frac{L}{2\min _{x\in K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}}} \sqrt{E(x)}}. \end{aligned}$$

Furthermore, the fact that $E\in SC^{\frac{1}{2}}(K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}})$ implies that, for every $x\in K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}}$, we have that

$$\begin{aligned} E(y)-E(x)\le \langle p, y-x\rangle +C |y-x|^{3/2} \end{aligned}$$

for every $y\in K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}}$ such that $[x,y]\subset K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}}$ and for every $p\in D^+E(x)$. Then,

$$\begin{aligned}&\sqrt{E(y)}\le \sqrt{E(x)} +\frac{1}{\sqrt{E(x)}+\sqrt{E(y)}} (\langle p, y-x\rangle +C |y-x|^{3/2}) \nonumber \\&\quad \le \sqrt{E(x)} +\frac{1}{2\sqrt{E(x)}} \langle p, y-x\rangle +\left( \frac{1}{\sqrt{E(x)}+\sqrt{E(y)}} - \frac{1}{2\sqrt{E(x)}}\right) \langle p, y-x\rangle \nonumber \\&\qquad +\frac{C}{\sqrt{E(x)}+\sqrt{E(y)}} |y-x|^{3/2} \nonumber \\&\quad \le \sqrt{E(x)} +\frac{1}{2\sqrt{E(x)}} \langle p, y-x\rangle + C' |y-x|^{3/2} \end{aligned}$$

(A.60)

with

$$\begin{aligned} C'=\max _{x,y\in K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}}} \left( \frac{C}{\sqrt{E(x)}+\sqrt{E(y)}} + \frac{ L L' \sqrt{|y-x|} }{2\sqrt{E(x)} [ \sqrt{E(x)}+\sqrt{E(y)}]} \right) , \end{aligned}$$

i.e. for every $x,y\in K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}}$ such that $[x,y]\subset K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}}$ and for every $p\in D^+E(x)$

$$\begin{aligned} \sqrt{E(y)}\le \sqrt{E(x)} +\frac{1}{2\sqrt{E(x)}} \langle p, y-x\rangle + C' |y-x|^{3/2}. \end{aligned}$$

(A.61)

Now, let $x,y\in K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}}$ such that $[x,y]\subset K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}}$ and let $\lambda \in [0,1]$. Then, by (A.61), with x replaced with $\lambda x+(1-\lambda )y$, we find

$$\begin{aligned} \sqrt{E(y)}\le & {} \sqrt{E(\lambda x+(1-\lambda )y )} \nonumber \\&+\frac{1}{2\sqrt{E(\lambda x+(1-\lambda )y)}}\lambda \langle p, x-y\rangle + C'\lambda ^{3/2} |y-x|^{3/2}, \end{aligned}$$

(A.62)

with $p\in D^+d( \lambda x+(1-\lambda )y)$. Analogoulsy, we find that

$$\begin{aligned} \sqrt{E(x)}\le & {} \sqrt{E(\lambda x+(1-\lambda )y )} \nonumber \\&+\frac{1}{2\sqrt{E(\lambda x+(1-\lambda )y)}}(1-\lambda ) \langle p, y-x\rangle + C'(1-\lambda )^{3/2} |y-x|^{3/2}.\nonumber \\ \end{aligned}$$

(A.63)

Taking the convex combination of (A.62) with (A.63), we find that

$$\begin{aligned}&\lambda \sqrt{E(x)}+(1-\lambda ) \sqrt{E(y)}- \sqrt{E(\lambda x+(1-\lambda )y )} \\&\quad \le C' ((1-\lambda ) \lambda ^{3/2}+\lambda (1-\lambda )^{3/2}) |y-x|^{3/2}\le C' (1-\lambda ) \lambda |y-x|^{3/2}. \end{aligned}$$

This completes our proof.

Remark A.12

The above proof is based on the property that the square root of the positive function $E\in SC^{\frac{1}{2}}(K\cap \overline{{\mathbb {R}}^n\setminus {\mathcal {O}}})$ is itself fractionally semiconcave of order 1/2. We note that, here, known results dealing with the semiconcavity of a composition (see, e.g., [15, Proposition 2.1.12 (i)]) cannot be applied for two reasons. First, they do not address semiconcave functions on a closed domain and, second, they do not provide a control of the fractional semiconcavity modulus.