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.
Best Answer
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, 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.