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.



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> Is it true that $\|\nabla f\|\_{L^\infty} = [f]$?
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Thank you so much for your elaboration!



Is the Lipschitz constant of $f$ equal to $\|\nabla f\|_{L^\infty}$?
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 $\mathbb{R}^n$).


