Is there a function on a compact interval that is differentiable but not Lipschitz continuous?

Question

Consider a function $f:[a,b]\rightarrow \mathbb{R}$, does there exist a differentiable function that is not Lipschitz continuous?

After discussing this with friends we have come to the conclusion that none exist. However there is every chance we are wrong. If it is true that none exist how could we go about proving that? It is true that if $f$ is continuously differentiable then $f$ is Lipschitz, but what if we don't assume the derivative is continuous?

Answer 1

The map $f : [0,1] \to \mathbb{R}$, $f(0) = 0$ and $f(x) = x^{3/2} \sin(1/x)$ is differentiable on $[0,1]$ (in particular $f'(0) = \lim_{x \to 0^+} f(x)/x = 0$), but it is not Lipschitz (the derivative $f'(x)$ is unbounded).

Answer 2

If you require $f'$ to be continious then there can be no $f$ since $f'([0,1])$ is compact and indeed bounded.