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 15:37
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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. 

