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I'm trying to show if $f$ defined on a closed interval has a continuous derivative throughout the interval, then it is Lipschitz continuous on the interval.

Using the Mean Value Theorem the proof almost follows trivially since we can say for $\zeta \in [a,b]$ $\implies$ $|f(x) - f(y)| = |f'(\zeta)| |x - y|$. We are nearly there, however I am unsure if the derivative being bounded follows from the MVT as well? Or is there something left to show.

Please do not supply me with a proof, a simple hint will suffice.

Javier
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    The key is the continuity of the derivative. – carmichael561 Feb 24 '16 at 20:18
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    Hint: You need to add in that the image of a compact set is compact and that a compact subset of a Euclidean space is closed and bounded. – Michael Burr Feb 24 '16 at 20:21
  • Can you imagine a continuous unbounded function on a compact set? – Marco Disce Feb 24 '16 at 21:45
  • Related: https://math.stackexchange.com/questions/795950/differentiability-and-lipschitz-continuity?rq=1, https://math.stackexchange.com/questions/476072/continuous-differentiability-implies-lipschitz-continuity – mdcq Jan 16 '18 at 17:14

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Since $\;f'\;$ is continuous in compact interval it is bounded there and you have $|f(x)-f(y)|\le M|x-y|$