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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.

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

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  • $\begingroup$ -1: it is not true that every continuous function on an interval is Lipchitz. Try $f(x)=\sqrt x$. $\endgroup$ – Martin Argerami Feb 24 '16 at 20:29
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    $\begingroup$ The derivative of f is continuous, not f itself $\endgroup$ – Javier Feb 24 '16 at 20:30
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    $\begingroup$ Replacing $f$ by $f^\prime$ after "Since" would fix the answer, yet a reference to the theorem used would be nice. $\endgroup$ – Clement C. Feb 24 '16 at 20:46
  • $\begingroup$ @MartinArgerami Not all function but the derivative is bounded $\endgroup$ – user312943 Feb 24 '16 at 21:18
  • $\begingroup$ @BenS. What I wrote is not Lipschitz for $\;f\;$ ? $\endgroup$ – user312943 Feb 24 '16 at 21:19

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