8
$\begingroup$

Let $f: \mathbb R^n \to \mathbb R^m$ be a non constant Lipschitz map, for $m < n$. We denote by

$$\text{Lip}(f):= \sup_{x, y \in \mathbb R^n} \frac{|f(x) - f(y)|}{|x - y|}$$

the best Lipschitz constant of $f$. Let

$$S := \left \{(x, y) \in \mathbb R^n \times \mathbb R^n \, \big |\, \frac{|f(x) - f(y)|}{|x - y|} = \text{Lip}(f) \right \}$$

denote the stretch set of $f$, the set of pairs of points on which $f$ realises its Lipschitz constant.

Question: Is it true that $\text{dim}_H (S) \leq n+ m$?

Here $\dim_H$ denotes Hausdorff dimension.

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
6
$\begingroup$

I believe this is true (maybe just if $m$ is significantly smaller than $n$?), as a result of a theorem of Ciosmak, see "Leaves decompositions in Euclidean spaces and optimal transport of vector measures". However, there is a singular set on which I am not sure what happens. So this isn't really an answer, it just zooms in on the what the essential issue is (namely, understanding the structure of the singular set of Ciosmak's leaf decomposition), but it's too long for a comment.

Let me normalize $\mathrm{Lip}(f) = 1$. Ciosmak's idea is that we should think of $f$ as a Kantorovich potential for the optimal transport of an $\mathbb R^m$-valued measure, and then he is able to generalize some of the regularity theory of the transport set in the classical ($m = 1$) theory.

A leaf of $f$ is a maximal set $S$ such that $f|_S$ is an isometry. Clearly $\dim S \leq m$. Ciosmak's theorem says that, for every leaf $S$ of $f$:

  1. $S$ is closed and convex.
  2. $f|_S$ is affine.
  3. For every leaf $S'$ of $f$ which is distinct from $S$, $S \cap S' \subseteq \partial S$ and for every $x \in S'$, $f$ is not differentiable at $x$.

Let us say that $x$ is an intersection point if there are distinct leaves $S, S'$ such that $x \in S \cap S'$. By Ciosmak's theorem and Rademacher's theorem, the set of intersection points has measure zero. It feels like maybe it should have dimension $\leq m - 1$? If so, and $n$ is large enough compared to $m$, then the intersection points can be ignored entirely.

Henceforth I shall assume that the set of intersection points has such small dimension that it can be ignored entirely. The map $x \mapsto S(x)$, which sends a non-intersection point to the unique leaf through $x$, is Borel measurable. Therefore the stretch set is contained in a Borel measurable subbundle $E \to \mathbb R^n$ of the trivial bundle $\mathbb R^n \times \mathbb R^n \to \mathbb R^n$ such that the fiber dimension of $E$ is $m$, so its Hausdorff dimension is $\leq n + m$. (The last point can be justified, for example, using the point-to-set principle.)

Improve this answer
$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.