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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?

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    $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. $\endgroup$ Sep 21, 2018 at 12:11

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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

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

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