# On the Lipschitz constant outside the stretch set

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.

• The set $S$ may be reduced to a single point, like for $f(x):=\arctan x$, or be empty, like for $g:=x-\arctan x$. What is $\text{Lip}(f,S)$ in these cases? (Not very relevant, as in these cases $\text{Lip}(f,\mathbb R^n\setminus S)= \text{Lip}(f,\mathbb R^n)$ ) Commented Jul 12 at 6:26
• @PietroMajer Ah, let us say that we set those cases to be $0$ by convention. Commented Jul 12 at 6:28

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.
• What about a finite partition, e.g. $\mathbb R^n=A\cup B\cup C$? I’d say $$\text{Lip}(f)=\max\big(\text{Lip}(f,A), \text{Lip}(f,B), \text{Lip}(f,C)\big),$$ but it’s not quite clear. Commented Jul 12 at 9:40
• One may consider a covering into (wlog) closed sets $A_j$, and also define, with the obvious meaning, $\lambda_{ij}:=\text{Lip}(f,A_i,A_j)$. There are a number of inequalities between these constants Commented Jul 12 at 10:38