Check Lipschitz condition

Question

My mathematical analysis professor gave this exercise to us: check that the function $$f(x)=\frac{x}{1+|x|}$$ verifies the Lipschitz condition globally. Can someone help me to understand how I can reach this result?

Answer 1

You can calculate that $ f'(x) = \frac{1}{1+|x|^2} $. Now fix $ x,y \in \mathbb{R} $ then by the mean value theorem there is a $ \xi $ between $ x $ and $y $ with $ \frac{f(x) - f(y)}{x-y} = f'(\xi) = \frac{1}{1+|\xi|^2} $. This implies

$$ | f(x) - f(y) | \le \sup_{\xi \in \mathbb{R}} {|f'(\xi)|} \cdot | x-y | \le |x-y|$$