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.

  • $\begingroup$ I'd rewrite a bit. You say "Let $f$...", thus fixing $f$. Then you ask "can there exist a function $f$". $\endgroup$ Commented Apr 7, 2022 at 0:46
  • $\begingroup$ Ah true, one second. $\endgroup$
    – Nate River
    Commented Apr 7, 2022 at 0:47

1 Answer 1


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.

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    $\begingroup$ I have been working on an answer along the same lines, but your presentation is much better than the one I was going to provide (which would include a construction of the kind of measurable set $A$ that you describe.) Upvoted. $\endgroup$ Commented Apr 7, 2022 at 4:10
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    $\begingroup$ @IosifPinelis thank you! Being praised by someone of your caliber is a very nice way to end the day. $\endgroup$
    – Nik Weaver
    Commented Apr 7, 2022 at 4:23
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    $\begingroup$ Ah, yes such sets do exist. Thank you for your answer! $\endgroup$
    – Nate River
    Commented Apr 7, 2022 at 5:30
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    $\begingroup$ @NikWeaver : Thank you for your very kind words. $\endgroup$ Commented Apr 7, 2022 at 13:24
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    $\begingroup$ @Nik Weaver By the way, by taking a set $S$ such that both it and its complement are measure dense, and integrating the function $2(\mathbf 1_S - 1)$, we get a function $f$ with $|Df| = L$ a.e.! $\endgroup$
    – Nate River
    Commented Apr 8, 2022 at 4:47

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