8
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

12
$\begingroup$

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.

$\endgroup$
6
  • 2
    $\begingroup$ 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
  • 4
    $\begingroup$ @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
  • 1
    $\begingroup$ Ah, yes such sets do exist. Thank you for your answer! $\endgroup$
    – Nate River
    Commented Apr 7, 2022 at 5:30
  • 1
    $\begingroup$ @NikWeaver : Thank you for your very kind words. $\endgroup$ Commented Apr 7, 2022 at 13:24
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.