If the derivative of a function is lipschitz,,,does it mean that the function itself is also lipschitz? Any proof for that?
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.