Let $\Omega \subseteq \mathbb{R}^n$ open and $f \in C^1(\Omega)$.

Given $c >0$, take $f_c=f\mathbb{I}_{\{|f|\leq c\}}+c\mathbb{I}_{\{f\geq c\}}-c\mathbb{I}_{\{f\leq -c\}}$ ($\mathbb{I}$ is the indicator function here).

Is $f$ Lipschitz continuous on $\Omega$?

This property clearly doesn't hold if $f$ is assumed to be just continuous, as a counterexample could be given $f(x)=\sqrt{|x|}$ and $\Omega \subseteq \mathbb{R}$ containing $0$.

However when $f$ is required to be smooth things might change, as it follows that $f$ must be at least locally Lipschitz on $\Omega$.


1 Answer 1


Note that

$$||f_c(x) - f_c(y) ||\le ||f(x)-f(y)||$$


Now since $f$ is locally Lipshitz it follows that $f_c$ is (with the same local Lipshitz constants). Since you cannot expect more for $f$ (even if $-c\le f\le c$) this is (in general) all you can hope for. (But it's obvious that $f_c$ is "at least as Lipshitz" as $f$ is).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .