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Does anyone know any sufficient conditions (and necessary would be great too if possible) on $f$ such that the following is satisfied:

$|f(x)-f(y)| \leq C|x-y|$, where $C$ is a constant, $f(x): \mathbb{R} \rightarrow \mathbb{R}$ and $x,y, \in \mathbb{R}$

Jessica
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  • This is precisely what it means for $f$ to be "Lipschitz continuous": http://en.wikipedia.org/wiki/Lipschitz_continuity – BaronVT Mar 26 '15 at 15:34
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    well this is called a Lipschitz function – Surb Mar 26 '15 at 15:34
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    ah thanks you two – Jessica Mar 26 '15 at 15:35
  • In particular, if $f$ is continuously differentiable and has bounded derivative, then it is Lipschitz. – Clement C. Mar 26 '15 at 15:48
  • A useful sufficient condition is that $f$ is differentiable and $|f'(x)| \le C$. This is a corollary of the mean value theorem. – Qiaochu Yuan Mar 26 '15 at 16:54
  • @QiaochuYuan continuously differentiable, no? – Clement C. Mar 26 '15 at 17:07
  • @Clement: no, I don't think so. The mean value theorem doesn't require this hypothesis. – Qiaochu Yuan Mar 26 '15 at 17:10
  • @QiaochuYuan see e.g. http://math.stackexchange.com/questions/114633/if-f-is-differentiable-on-a-b-then-it-is-also-lipschitz-on-it – Clement C. Mar 26 '15 at 17:14
  • @Clement: my claim is not that if $f$ is differentiable then it is Lipschitz. It is that if $f$ is differentiable and its derivative is bounded in absolute value by $C$, then it is Lipschitz with Lipschitz constant $C$. That, again, is a corollary of the mean value theorem, and it does not require the derivative to be continuous. – Qiaochu Yuan Mar 26 '15 at 17:32
  • You're right, indeed -- sorry, I took a stupid shortcut there. (if it it continuously differentiable, then it does imply it is bounded, but the important part is only the boundedness). – Clement C. Mar 26 '15 at 17:33

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