Convexity forces the derivative to be continuous

Tools.
- If $f$ is convex and differentiable, $f'$ is non-decreasing.
- Darboux's theorem: the derivative of a differentiable function satisfies the intermediate value property, even if it is not continuous.

Proof. Suppose $f'$ is discontinuous at some point $x_0\in I$. Since $f'$ is monotone, it has one-sided limits at $x_0$, with
$$f'(x_0^-)\le f'(x_0)\le f'(x_0^+),$$
and at least one of the two inequalities is strict. Say $f'(x_0)<f'(x_0^+)$ (the other case is symmetric). Pick $y$ with $f'(x_0)<y<f'(x_0^+)$.

• For $x>x_0$ we have $f'(x)\ge f'(x_0^+)>y$ (monotone and $f'(x_0^+)$ is the infimum of the values to the right).
• For $x\le x_0$ we have $f'(x)\le f'(x_0)<y$.

So $f'$ never takes the value $y$. But Darboux's theorem, applied on $[x_0,x_1]$ with $x_1>x_0$ close enough that $f'(x_1)>y$, says that $y$, which lies between $f'(x_0)$ and $f'(x_1)$, is attained. Contradiction.

Conclusion. $f'$ is continuous on $I$.

Remark. So a convex differentiable function is automatically $C^1$. In general, derivatives can be discontinuous (see $x^2\sin\frac1x$), but never with jumps.