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 $\mathbb R^d$.


**Question:** What is the maximal Hausdorff dimension of the set on which $f$ is differentiable and $|Df| = L$?


**Remarks:**


1. $\mathcal H^{n-1}$ is trivially achievable.
2. By sticking together a maximising sequence, we may achieve the supremal Hausdorff dimension, so we are justified in speaking of the maximum.



On the set on which $|Df|$ is maximal for Lipschitz $f$
The maximal dimension is $n$, and it can be even of positive Lebesgue measure.


For $n=1$, consider a fat cantor set $K$. Then the primitive
$$
f(x):=\int\_0^x \chi\_K(t)dt
$$
is a maximizer. Indeed, since $K$ is nowhere dense, we get $|f(x)-f(y)|< |x-y|$. On the other hand, by the fundamental theorem of calculus (i.e., some form of Lebesgue differentiation theorem) we obtain that $f$ is differentiable with derivative 1 at all density points of $K$, which form a positive measure set.


In general dimension it is sufficient to consider the same function of one coordinate, say $f(x\_1)$.


