Give an example of a Lipschitz function $f$ on $[0,\infty)$ such that its square $f^2$ is not Lipschitz.
My example is to let $f(x) = x$ for $x \in [0, \infty$)
Then $|f(x) - f(u)|$ = $|x - u|$ $\implies$ $f$ is Lipschitz.
By definition a function $f$ is Lipschitz given a set $A \subseteq \Bbb R$ and $f: A \rightarrow \Bbb R$, if $\exists$ a constant $K>0$ such that $|f(x) - f(u)| \leq K|x-u|$ $\forall x,u \in A$
Letting $f(x) = x$ then, the square of the function is $f(x) = x^2$ and
$|f(x) - f(u)|$ $=$ $|x^2 - u^2| = |x+u||x-u| > 2K|x-u|$
This example satisfies necessary conditions however isn't very exciting. I was hoping perhaps someone could suggest some different examples.