Conditions for global Lipschitz continuity of composite functions

Suppose $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$, $g:\mathbb{R}^n\rightarrow\mathbb{R}^n$ and that $f$ is globally Lipschitz continuous, i.e. there exists an $L>0$ such that

$|f(x)-f(y)|\leq L|x-y|\quad \forall x,y\in\mathbb{R}^n,$

where $|\cdot|$ denotes any vector norm. Does there exists any weaker requirement on $g$ than global Lipschitz continuity that ensures that $f\circ g$ is globally Lipschitz continuous?

You want to compare $L|g(x)-g(y)|$ and $|x-y|$. The Lipschitz continuity of $g$ seems to be a rather weak assumption. –  Siminore Aug 21 '12 at 15:11
What do you mean 'Lipschitz continuity of $g$'? (Global) Lipschitz continuity of $g$ is sufficient, say $f$ has a Lipschitz constant $L_1$ and $g$ $L_2$, then $|f(g(x))-f(g(y))|\leq L_1|g(x)-g(y)|\leq L_2L_3|x-y|$. –  jkn Aug 21 '12 at 15:16
Not really (as in having weaker conditions), unless you have more conditions on $f$. If you take $f(x) = x$, then $f \circ g$ and $g$ being Lipschitz continuous are obviously synonymous. –  copper.hat Aug 21 '12 at 15:20