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.