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

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

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.

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.

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

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

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

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

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

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

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.