Lipschitz function $f^p$

Question

Let $f$ be a Lipschitz function, that is $|f(x)-f(y)|\leq L|x-y|$, and $p>1$. Further suppose that $f$ has compact support.

Does it then hold that $f^p$ is also Lipschitz?

Answer 1

Under these assumptions the answer is yes. Let's see why.

By the Mean Value Theorem (for the function $x\mapsto x^p$) we obtain

$$|f(x)^p-f(y)^p|\leq p(\max\{|f(x)|,|f(y)|\})^{p-1}|f(x)-f(y)|\\\leq Lp(\max\{|f(x)|,|f(y)|\})^{p-1}|x-y|.\ \ \ \ \ (1)$$

Lipschitz functions are continuous. Thus, $f$ is continuous and so, at its compact support $f$ is bounded with upper bound, let's say $M\geq 0.$ Since $f$ vanishes outside its compact support, $M$ is an upper bound of $f$ on the whole real line. Now $(1)$ yields

$$|f(x)^p-f(y)^p|\leq LM^{p-1}p|x-y|,\ \ \ \forall x,y\in \mathbb{R}$$

and so, $f^p$ is Lipschitz too. The new Lipschitz constant is $LM^{p-1}p.$