We say a measurable subset $S$ of $\mathbb R^n$ is *measure dense* if for every open set $U \subset \mathbb R^n$, $U \cap S$ is of positive Lebesgue measure.

Let $n \geq 2$, and let $f: \mathbb R^n \to \mathbb R$ be a Lipschitz continuous function with *strict* Lipschitz constant $L > 0$.

That is, $|f(x) - f(y)| < L|x - y|$ for all $x \neq y$ in $\mathbb R^n$.

**Question:** Is it possible that $|Df| = L$ on a measure dense set?

*Note:* Here $Df$ denotes the total derivative of $f$, and $|\cdot|$ the operator norm of a linear map.