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.



On functions with strict Lipschitz constant
I guess it suffices to give an example for $n = 1$. If $f: \mathbb{R} \to \mathbb{R}$ is an example then $g(x\_1, \ldots, x\_n) = f(x\_1)$ will be an example for any $n \geq 1$.


All we need is a measurable set $A \subseteq \mathbb{R}$ such that both $A$ and its complement have positive measure in every interval. See [here](https://math.stackexchange.com/questions/57317/construction-of-a-borel-set-with-positive-but-not-full-measure-in-each-interval), for example. Then define
$$
f(x) = \int\_0^x 1\_A = \begin{cases}
m(A \cap [0,x])&x \geq 0\cr
-m(A\cap [x,0])&x < 0
\end{cases}.
$$ It should be clear that 1 is a strict Lipschitz constant, but $f'(x) = 1$ at every Lebesgue point of $A$, so the derivative is $1$ on a measure dense set.


