Convex function on closed bounded Interval Implies Lipschitz counterexample

Question

I am considering a convex function $f:[a,b] \rightarrow \mathbb{R}$ and have been asked to show that $f$ is absolutely continuous on $[a,b]$. I've attempted to use the Chordal Slope Theorem to show that $f$ is Lipschitz on $[a,b]$, but have only been able to show that $f$ is Lipschitz for $[c,d] \subseteq (a,b)$. I'm beginning to think that the problem as stated is false, but can't think of a counterexample. I appreciate any guidance.

Answer 1

A convex function on $[0,1]$ that is not continuous (and therefore not absolutely continuous) is $$ f(x) = \cases{1 & if $x=0$\cr 0 & otherwise}$$

Answer 2

$f(x)=-\sqrt x$ on $[0,1]$ is a counter-example. It is not Lipschitz because its derivative on $(0,1)$ is unbounded.

It can be shown that if $f$ is convex and continuous on $[a,b]$ the $f'(x+)$ exist for $a<x<b$, it is integrable and and $f(x)=f(a)+\int_a ^{x} f'(t+)dt$ for all $x$. Hence $f$ is absolutely continuous.