Show Lipschitz condition for $f(x) = x+|x|$

Question

I am struggling to prove $$x+|x|$$ to be locally Lipschitz. Since it is not continuously differentiable - not differentiable at $x=0$ - it is hence not globally Lipschitz. But how do I proceed for the local case?

$$|f(x) - f(y)| \leq L|x-y|$$ $$\Rightarrow |x+|x|-y-|y||=...$$

Answer 1

$|x+|x|-y-|y||=|(x-y)+(|x|-|y|)|\le |x-y|+||x|-|y|| \le |x-y|+|x-y| = 2|x-y|$. Thus $k = 2$. The key inequality is $xy \le |xy| = |x||y|$ which leads to $||x| - |y|| \le |x-y|$ which can be shown by simply squaring both sides.

Answer 2

$f(x)=x+|x|=\max\{0,2x\}$, so it's globally Lipschitz with constant 2.