Is the function $ f $ here Lipschitz?

Question

Assume that $ f:\mathbb{R}^n\to\mathbb{R} $ is a continuous function and $ E\subset\mathbb{R}^n $ is a closed set with zero Lebesgue measure, i.e. $ m(E)=0 $. Suppose that $ f\in C^1(\mathbb{R}^n\backslash E) $, $ f(x)=0 $ for any $ x\in E $ and there exists a constant $ C>0 $ such that $ |Df(x)|\leq C $ for any $ x\in\mathbb{R}^n\backslash E $. The question is that if $ f $ is a Lipschitz function.

Here I want to use the characterization $ W^{1,\infty}(\mathbb{R}^n)=\operatorname{Lip}(\mathbb{R}^n) $ but do not know how to go on. Can you give me some hints or references?

Answer 1

Since you require $f$ to be constant on $E$ you can more or less do the usual Lipschitz proof. Let $x,y\in \Bbb R^n$ and let $x(t) = (1-t)x+ty$. Now distinguish between two cases:

1. $x(t)\notin E$ for any $t\in[0,1]$. Then $f(x)-f(y) = \int_0^1 D_{x(t)}f\cdot (y-x) \,dt$ and the bound $\|D_{x(t)}f\|\leq C$ implies $|f(x)-f(y)|\leq C\|x-y\|$.
2. There is a $t\in[0,1]$ so that $x(t)\in E$. In this case let $t^*=\inf\{ t\in[0,1]\mid x(t)\in E\}$. Note that $t^*=0$ is possible. At any rate $x(t)\notin E$ for $t\in(0,t^*)$, which gives: $$f(x)-0=f(x)-f(x(t^*)) = \int_0^{t^*} D_{x(t)}f\cdot (y-x)\,dt$$ and you get $|f(x)|\leq C\,t^*\|x-y\|$. Similarly let $t^{**}=\sup\{t\in[0,1]\mid x(t)\in E\}$ and the same calculation implies $|f(y)| \leq C(1-t^{**})\|x-y\|$. In total: $$|f(x)-f(y)|\leq |f(x)|+|f(y)|\leq C \|x-y\| (1-t^{**}+t^*)\leq C\|x-y\|$$ since $t^{**}\geq t^*$.