Counterexample: you will be delighted to hear that the function defined by ( ), , is derivable on , yet its derivative is unbounded on any neighbourhood of 0.So what's my question? My question is whether the above theorem is an if-and-only-if statement or whether the implication only goes in one direction. If not, what hypotheses can we add so that it does go both ways?
In particular, is uniformly continuous and derivable on while its derivative is unbounded (but finite, that's the difference with ).
You can then define a derivable periodic function that looks like repeatedly. So that the derivative of is unbounded on any , while is uniformly continuous (it is continuous and periodic, so...).