Is derivative of any Lipschitz function continuous on its domain?

Question

Due to the Rademacher's theorem, we know that every Lipschitz function, $f$, on euclidean space is almost everywhere differentiable. I am interested to know whether the derivative of $f$ is continuous on almost all points in its domain. If this statement does not hold necessarily, please provide a counterexample.

Answer 1

Here's a slight modification of the classic example of a function whose derivative fails to be continuous on its domain: $$ f(x) = \begin{cases} x^2 \cos(1/x) & x \in (-1/\pi ,0) \cup(0, 1/\pi) \\ 0 & \text{otherwise} \end{cases} $$ Which satisfies $$ f'(x) = \begin{cases} 2x \cos(1/x) - \sin(1/x^2) & x \in(-1/\pi,0) \cup (0,1/\pi)\\ 0 & x = 0, x<-1/\pi, \text{ or } x > 1/\pi \end{cases} $$ Verify that $f$ is Lipschitz (e.g. using the mean value theorem), but $f'$ is not continuous at $x = 0$.

Answer 2

I apologize that I did not state my question properly. In fact, I want to know whether Aleksandrov Theorem for convex functions holds for Lipschitz functions.

Alexandrov Theorem states that if $f$ is a convex function on $\mathbf{R}^n$, then for almost $x\in\mathbf{R}^n$ there are $v$ and $A$ such that

$f(x+d)=f(x)+v^Td+\frac{1}{2}d^TAd+o(||d||^2)$.