If the derivative of a function is lipschitz,,,does it mean that the function itself is also lipschitz? Any proof for that?
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closed as too localized by Bill Johnson, Anthony Quas, Michael Renardy, Noah Stein, Mark Meckes Sep 21 '12 at 15:37This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question. 


No. $f(x)=x^2$ on the whole real line is not Lipschitz, but the derivative $f'(x)=2x$ is. However, if the function is defined on a bounded interval, then the statement is true. Indeed, if $f'$ is Lipschitz, then it is continuous and thus bounded, and a function with bounded derivative is Lipschitz. 

