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Let $f : \mathbb R^d \to \mathbb R$ be Lipschitz and $[f] := \sup_{x,y \in \mathbb R^d; x\neq y} \frac{|f(x) - f(y)|}{|x-y|}$ its Lipschitz constant. By Rademacher theorem, $f$ is differentiable a.e., so $\nabla f$ is defined a.e.

Is it true that $\|\nabla f\|_{L^\infty} = [f]$?

Thank you so much for your elaboration!

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    $\begingroup$ @ThomasKojar From your link, I get that $\|\nabla f\|_{L^\infty} \le [f]$. For the reverse inequality, Julien's proof (in Question 2.2) requires that $f$ is differentiable everywhere. $\endgroup$
    – Akira
    Commented Jul 25, 2023 at 16:06
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    $\begingroup$ You may want to check out the reference at wiki where it is stated that "any Lipschitz function on Ω is an element of the space W1,∞(Ω)". The reference is to Evans and Gariepy. $\endgroup$
    – Mikhail Katz
    Commented Jul 25, 2023 at 16:31

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Yes, Theorem 1.41 of my book Lipschitz Algebras (second edition). But you must interpret ``$\|\nabla f\|_{L^\infty}$'' to mean $\|\,|\nabla f|\,\|_{L^\infty}$ (sup of the norm of the gradient taken in $\mathbb{R}^n$).

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    $\begingroup$ I should have specified "second edition". Edited. $\endgroup$
    – Nik Weaver
    Commented Jul 25, 2023 at 16:38
  • $\begingroup$ Hi Nik! Do you prove that via Sobolev spaces, or by a more elementary argument? $\endgroup$
    – Mikhail Katz
    Commented Jul 25, 2023 at 16:43
  • $\begingroup$ @MikhailKatz Oh, it's pretty elementary. The $\leq$ part is easy, in the opposite direction pick any two points, $p$ and $q$, draw a line joining them, and attempt to argue that $d(p,q) \leq$ the derivative of $f$ restricted to that line, in absolute value. $\endgroup$
    – Nik Weaver
    Commented Jul 25, 2023 at 19:14
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    $\begingroup$ That doesn't quite work because the derivative might not exist a.e. on every line, but you can thicken that line and integrate over a ball lying in the hyperplane perpendicular to the line joining $p$ and $q$. $\endgroup$
    – Nik Weaver
    Commented Jul 25, 2023 at 19:15
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    $\begingroup$ (I doubt this is my argument, but if not I don't remember where it's from.) $\endgroup$
    – Nik Weaver
    Commented Jul 25, 2023 at 19:16

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