On the Lipschitz constant outside the stretch set

Question

Let $f: \mathbb R^n \to \mathbb R^m$ be a Lipschitz map. We define the local Lipschitz constant $Lf$ of $f$ at $x \in \mathbb R^n$ by

$$Lf(x) := \lim_{r \to 0_+} \text{Lip}(f, B_r (x)),$$

where $\text{Lip}(f, U) := \sup_{y,z \in U} \frac{f(y) - f(z)}{y- z}$ denotes the Lipschitz constant of $f$ on the set $U$.

Define the stretch set $S$ of $f$ by

$$S := \{x \in \mathbb R^n \, | \, Lf(x) = \text{Lip}(f, \mathbb R^n)\}.$$

Roughly, the stretch set is the set on which $f$ achieves its maximal Lipschitz constant.

Question: Is it true that $\text{Lip}(f, \mathbb R^n) = \max(\text{Lip} (f, S), \text{Lip}(f, \mathbb R^n \setminus S))$?

Remark: The stretch set plays a crucial role in Thurston's "best Lipschitz maps" approach to Teichmuller theory, see here for a reference.

Answer 1

In fact $\text{Lip}(f,\mathbb R^n)=\max\big( \text{Lip}(f,A), \text{Lip}(f,A^c)\big)$ holds for every $A\subset \mathbb R^n$.

• The inequality $\text{Lip}(f,\mathbb R^n)\ge\max\big( \text{Lip}(f,A), \text{Lip}(f,A^c)\big)$ is clear.
• For the opposite inequality, let $x\in A$ and $y \in A^c$. So we have $\partial A\cap[x ,y ]\neq\emptyset$: let $z \in \partial A\cap[x ,y ]$. Since $$ \|y -x \|=\|y -z \|+\|z -x \|, $$ the quotient $\frac{f(y )-f(x )}{\|y -x \| } $ is a convex combination of $\frac{f(y )-f(z )}{\|y -z \| } $ and $\frac{f(z )-f(x )}{\|z -x \| } $ (here with the convention “$\frac00=0$” stated in comments), so
  $$ \begin{split} \frac{\|f(y )-f(x )\|}{\|y -x \| } & \le\max\left(\frac{\|f(y )-f(z )\|}{\|y -z \| }, \frac{\|f(z )-f(x )\|}{\|z -x \| } \right)\\ & \le \max\left( \text{Lip}(f,\overline{A}), \text{Lip}(f,\overline{A^c})\right)\\ & = \max\left( \text{Lip}(f,A), \text{Lip}(f,A^c)\right), \end{split}$$ and the inequality follows.