7
$\begingroup$

I am trying to show the following two statements are true:

(1) For any nonempty set $\Omega\subset\mathbb{R}^n$, the set $B$ consisting of points $y\notin\Omega$ where there is not a unique closest point $x\in\partial\Omega$ for each $y$ has $\mathcal{H}^n$-measure 0.

(2) This implies that the number of points $a\in A\subset \mathbb{R}^2$ such that $\mathcal{H}^{1}(\vec{n}(a)\cap B)>0$ is countable, where $\vec{n}(a)$ is the normal to $a\in A$ and $A$ is $C^1$ (i.e. it is locally the graph of a $C^1$ function from $\mathbb{R}$ to $\mathbb{R}$. More specifically, $A$ is a $C^1$ closed embedded 1-d submanifold in $\mathbb{R}^2$).

For (1), I would like help proving only this related fact: if the distance function $f$ to $\Omega$ is differentiable at a point $y\notin\Omega$, then there exists a unique closest point $x\in\partial\Omega$ to $y$.

I thought that I could get at this by saying that the graph of $f$ has a sharp corner wherever there is more than one closest point (and hence is not differentiable there). But, I am told this is not always true.

For (2), I thought that, since we're in $\mathbb{R}^2$, if we have an uncountable number of segments with positive $\mathcal{H}^{1}$ measure, then $B$ would have to have positive $\mathcal{H}^{2}$ measure since $B$ contains these segments. I'm told this is not true either. Please assist.

$\endgroup$
6
  • 1
    $\begingroup$ (2) is unclear. How is $A$ related to $\Omega$? You say it is the image of a function $f:\mathbb{R}\to\mathbb{R}$. The image is a segment in $\mathbb{R}$ which is, I guess, not what you want. Did you mean, a graph? $\endgroup$ Apr 30, 2018 at 14:00
  • $\begingroup$ $A$ is a more specific set, whereas $\Omega$ is any set. Sorry, I did mean graph, not image. More specifically, $A$ is a $C^1$ closed embedded 1-d submanifold in $\mathbb{R}^2$. $\endgroup$
    – L P
    May 1, 2018 at 2:14
  • $\begingroup$ Please edit the question to clarify that $A$ is a $C^1$ submanifold. $\endgroup$ May 1, 2018 at 2:16
  • $\begingroup$ Added. Thanks. Do you have a hint for showing (2)? $\endgroup$
    – L P
    May 2, 2018 at 7:42
  • $\begingroup$ I am thinking about it, but I expect that there is a counterexample, but until I have one, the question remains open. By the way: do you want $\Omega$ to be open, closed, compact, any? $\endgroup$ May 2, 2018 at 13:57

3 Answers 3

7
$\begingroup$

On the "differentiability implies uniqueness" issue: Yes, it is true. Indeed:

If the distance $f(y)$ from $\Omega$ to a point $y\notin \Omega$ is realized at $x\in \overline\Omega$ (a priori not unique), that is $f(y)=\|x-y\|$, then $x$ also realizes the distance from $\Omega$ to any point in the line segment $[x,y]$: for any $x'\in\overline \Omega$, $0\le t\le 1$ and $z:=y+t(x-y)$, by the triangle inequality,

$$\|z-x'\|\ge\|y-x'\|-\|y-z\|\ge f(y)-t\|x-y\|=\|z-x\|$$ so that $$f(z)=\|z-x\|=f(y)- {x-y\over\|x-y\|}\cdot(z-y).$$

If moreover the distance is differentiable at $y$, comparing the latter with $$f(z)=f(y)+\nabla f(y)\cdot (z-y) + o(\|z-y\|),\qquad (z\to y)$$ we have that the gradient at $y$ is $\nabla f(y)=-{x-y\over\|x-y\|}$. So $x$ can be recovered from the value at $y$ of the distance and of its gradient, $$x=y-f(y)\nabla f(y)$$ and it is therefore the unique point of minimum distance.

$\endgroup$
6
  • $\begingroup$ You could explain why the gradient is equal to $\frac{x-y}{\Vert x-y\Vert}$ since for some readers it might be not obvious. I could add the explanation, but I do not want to alter your answer. $\endgroup$ Apr 30, 2018 at 15:01
  • $\begingroup$ Thank you. I've added the details and fixed a typo $\endgroup$ Apr 30, 2018 at 17:14
  • $\begingroup$ @Pietro Majer: Sorry for the late question, but it seems to me that you prove $\nabla f(y)=\cdots$ only after product with $z-y$, that is, in one particular direction. What am I missing? $\endgroup$
    – abx
    Nov 24, 2020 at 9:23
  • $\begingroup$ Just this: the differential of $f$ at $y$ is uniquely determined by the definition, i e. the first order expansion. $\endgroup$ Nov 24, 2020 at 12:04
  • $\begingroup$ @Pietro Majer: What I was trying to say is that knowing the product of $\nabla f(y)$ with one vector (namely $z-y$) is not enough to determine it. You have to use that $\lVert \nabla f(y) \rVert\leq 1$ because $f$ is 1-lipschitzienne, and Cauchy-Schwarz. $\endgroup$
    – abx
    Nov 25, 2020 at 20:33
5
$\begingroup$

For (2), I thought that, since we're in $\mathbb{R}^2$, if we have an uncountable number of segments with positive $\mathcal{H}^1$ measure, then $B$ would have to have positive $\mathcal{H}^2$ measure since B contains these segments. I'm told this is not true either. Please assist.

There are quite surprising counterexamples. Besicovitch constructed a planar set of measure zero that contains segments in every direction. Such sets are called Besicovitch or Kakeya sets.

Nikodym constructed a set $N\subset [0,1]\times [0,1]$ of measure $1$ such that for every $x\in N$ there is a line $\ell_x$ such that $\ell_x\cap N=\{ x\}$. That is this line intersects $N$ at one point only. Note that the complement of the set $N$ in $[0,1]\times [0,1]$ has measure zero so the union of segments $$ [0,1]^2\cap \bigcup_{x\in N} \ell_x\setminus \{x\} $$ has measure zero.

(1) For any nonempty set $\Omega\subset\mathbb{R}^n$, the set $B$ consisting of points $y\not\in\Omega$ where there is not a unique closest point $x\in \partial\Omega$ for each $y$ has $\mathcal{H}^n$-measure $0$.

The distance function $d(x)={\rm dist}\, (x,\partial\Omega)$ is $1$-Lipschitz and hence differentiable a.e. (Rademacher theorem). As pointed by Pietro Majer in his answer, the set $B$ is contained in the set of points there $d$ is not differentiable. Hence $B$ has measure zero.

$\endgroup$
0
$\begingroup$

One can directly show that $B$ has no interior points.

For $p \in \mathbb{R}^n$ define $c(p):=\{q\in \overline{\Omega}: ||p-q||=dist(p,\overline{\Omega})\}$.

Then $B=\{q\in\mathbb{R}^n: \#c(p)>1\}$ has no interior point.

Proof: Assume $x$ is an interior point of $B$, then $R:=dist(x,\overline{\Omega})>0$ (otherwise $c(x)=\{x\}$). Now choose some point $p\in c(x)$ and a real $\epsilon>0$, such that the open ball $B_\epsilon(x)\subset B$.

Choose a point $y\in [xp]\cap B_\epsilon(x)$, where $[xp]$ is the straight line connecting $x$ and $p$, and define $r:=||x-y||$.

It follows that $p\in c(y)$ because $||y-p||=R-r$. Otherwise there would be a $z\in c(y)$ with $||y-z||<R-r$ which implies $||x-z||\leq ||x-y|| +||y-z||<R$, such that $z$ would be a point in $\overline{\Omega}$ that is closer to $x$ than the points in $c(x)$, which is a contradiction.

Because $\#c(y)>1$ (by definition $y\in B$), there is a point $w\in c(y)$ with $w\not= p$ such that $||y-w||=R-r$. One now sees that this point $w \in \overline{\Omega}$ is too close to $x$, because the vectors $x-y$ and $w-y$ have lengths $r$ and $R-r$ but are not collinear (because $w\not= p$), such that one has the strict triangle inequality $$||x-w|| < ||x-y|| +||y-w||=R.$$ This contradicts the assumption $dist(x,\overline{\Omega})=R$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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