If $f$ is Lipschitz and $g \in C^\infty_c(\mathbb{R})$, is $g \circ f$ Lipschitz?

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be Lipschitz continuous and $g \in C^\infty_c(\mathbb{R})$ (i.e. smooth with compact support). Is the composition $g \circ f$ Lipschitz?

I have tried proving this myself but I'm not sure how to relate $g(f(x))$ to $g(f(y))$ to show that $$|g(f(x)) - g(f(y))| \leq C |x-y|.$$

Answer 1

If $f \in C^1_c(\mathbb{R})$, then $f^\prime$ is continuous with compact support, so $|f^\prime| $ attains its maximum, $L_f$, say. By the mean value theorem, for given $x, y$ there is $\xi\in (x,y)$ such that

$$f(x)- f(y) = f^\prime(\xi)(x-y)$$ so that $$|f(x)- f(y)| \le |f^\prime(\xi)||x-y| \le L_f |x-y|$$ Since every $C^\infty_c$ function $f$ is in $C^1_c$, every $C^\infty_c$ function is Lipschitz continuous for some constant $L = L_f$ (which, of course, depends on $f$).

If now $g\in C^\infty_c$ and $f$ is Lipschitz with Lipschitz constant $L_f$, then this implies $$ |g(f(x))- g(f(y))|\le L_g |f(x) - f(y)|\le L_g L_f |x-y|$$