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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.

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1 Answer 1

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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 $f(x_1)$.

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  • $\begingroup$ Ah, very nice example! I did not consider a construction like this. $\endgroup$
    – Nate River
    Nov 26, 2021 at 10:29

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