Supremum norm of derivative of Lipschitz continuous function

Question

The text I am reading claims that $||h'|| \leq 1$, where the norm is the supremum norm and $h$ is a Lipschitz continuous function with Lipschitz constant 1. Since Lipschitz continuity doesn't imply differentiability everywhere, how is the supremum defined on those measure zero sets where the derivative might not exit?

Answer 1

The idea is precisely to ignore those measure zero sets. Here, I believe that $\|\cdot\|$ does not denote the supremum norm, but rather the essential supremum norm. Take $h : \mathbb{R} \to \mathbb{R}$ a $1$-Lipschitz function. As you know, by Rademacher's theorem, $h'$ is defined except on a set $S \subset \mathbb{R}$ of measure $0$. Set $$ \| h' \| := \inf \{ C \in \mathbb{R}_+ ; |h'(x)| \leq C \text{ for almost every } x \in \mathbb{R} \setminus S \}. $$ You can look up the notion of essential supremum.