Lipschitz continuity proof given boundedness of a derivate (not a derivative)
Hey guys. Tough problem here (tough for me anyway...)
I can show that is of bounded variation, and therefore is differentiable almost everywhere. I don't know if that's a promising approach, however, nor even if it is, how to finish the proof.
Suppose the derivate
of a function
is bounded, where
is Lipschitz continuous.
One other possible avenue is this: The proof that a function is Lipschitz if its derivative is bounded uses the mean value theorem. Is there maybe some variation of the mean value theorem for which I could use in this case?
Any help would be much appreciated!