Lipschitz continuity of $f(x) \to x^2$
Solution 1:
The function $f(x)=x^2$ is Lipschitz continuous on compact sets. For example, on $[-L,L]$, we have $$ \begin{align} |x^2-y^2| &=|x+y|\,|x-y|\\ &\le2L|x-y| \end{align} $$ $f$ is not Lipschitz continuous on all of $\mathbb{R}$: $$ |(x+1)^2-x^2|=|2x+1|\,|(x+1)-x| $$ shows that there is no single bound that works over all $\mathbb{R}$.