# Does the Mean Value Theorem imply that the derivative is bounded?

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.

• The key is the continuity of the derivative. – carmichael561 Feb 24 '16 at 20:18
• 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
• – philmcole Jan 16 '18 at 17:14

Since $\;f'\;$ is continuous in compact interval it is bounded there and you have $|f(x)-f(y)|\le M|x-y|$
• -1: it is not true that every continuous function on an interval is Lipchitz. Try $f(x)=\sqrt x$. – Martin Argerami Feb 24 '16 at 20:29
• Replacing $f$ by $f^\prime$ after "Since" would fix the answer, yet a reference to the theorem used would be nice. – Clement C. Feb 24 '16 at 20:46
• @BenS. What I wrote is not Lipschitz for $\;f\;$ ? – user312943 Feb 24 '16 at 21:19