The following is taken from an exam:
$f:[a,b]\rightarrow\mathbb{R}$ is convex implies $f$ is absolutely continuous (recall $f'$ exists a.e.)
One has local Lipschitz-ness by convexity, but how to show absolute continuity without global Lipschitz-ness?
Convex functions on the real line are expressible as integrals of one-sided derivatives. The ration $k(x,y)=\frac{f(y)-f(x)}{y-x}$ is increasing in $y$ on $[x,b]$. Hence, the right-hand derivative $D_+f(x)$ exists for all $x\in [a,b]$. In a similar way we conclude that the left-hand derivative $D_-f(x)$ exists for all $x\in[a,b]$, and that $D_-f(x)\leq D_+f(x)$.
Hence the set of points for which is $D_-f(x)< D_+f(x)$ is countable.
If $f$ is convex on $[a,b]$, then then both $D_+f(x)$ and $D_-f(x)$ are integrable with respect to Lebesgue measure on $[a,b]$, and $f(x)=f(a)+ \int_a^x D_+f(t) dt=f(a)+ \int_a^x D_-f(t) dt$.
More generally, suppose D is an increasing, real-valued function defined (at least) on $[a,b)$. Define $g(x) := \int^x_a D(t)dt$, for $a \leq x \leq b$. (Possibly $g(b) =\infty$.) Then $g$ is convex.
In the book Roberts, Varberg, Convex functions, on pp.9-10 is proved that
If $f:(a,b)\rightarrow R$ is continuous and convex then f is absolutely continuous on each $[c,d]\subset (a,b)$.
$f$ is continuous since it is absolutely continuous. Moreover, $f'$ exists almost everwhere, and is increasing. Thus $f'' \geq 0$ whence $f$ is convex.
Notice that continuity is important since you could move a point on the graph of a convex function upward or downward rendering it nonconvex.
The assertion is not true as stated. In fact, the function $$f(x) = \begin{cases}1 & \text{if } x = 0 \text{ or } x = 1 \\ 0 & \text{else}\end{cases}$$ is convex on $[0,1]$, but not even continuous.
However, if we add continuity (which may only be violated at the end points), the assertion is true, here is a (brief) sketch of a possible proof.
Let $\varepsilon > 0$ be given. Then, by continuity at the end points and convexity of $f$ (which implies some monotonicy around the end points), there is $\delta > 0$, such that the sum of function value differences on $[a, a+\delta]$ and $[b-\delta,b]$ is less than $\varepsilon$ if the subintervals sum to less than $\delta$. On the remaining interval you can argue by Lipschitz continuity.
This is a supplement to the answer by @gerw. We show that $f : [a, b]\rightarrow\mathbb{R}$ is a continuous convex function, then $f$ is monotone in small neighborhoods of the endpoints (which can then be used to prove that $f$ is absolutely continuous). We work with the endpoint $b$. First, if $f_{-}'(b)\leq 0$ then clearly for any $a\leq x < y\leq b$ we have $$\frac{f(y)-f(x)}{y-x}\leq f_{-}'(b)\leq 0\implies f(y)\leq f(x).$$ On the other hand, if $f_{-}'(b) > 0$, then let $M > 0$ and $x_{0} < b$ such that $$\frac{f(b)-f(x_0)}{b-x_{0}}\geq M.$$ Since $f$ is continuous at $b$, we may pick $0 < \delta < b-x_{0}$ small enough so that $|f(b)-f(z)|\leq M(b-x_{0})/2$ for $z\in [b-\delta, b]$. If $b-\delta\leq x < y\leq b$, then using the convexity of $f$ we get \begin{align}\frac{f(y)-f(x)}{y-x}&\geq\frac{f(x)-f(x_{0})}{x-x_{0}}\\&=\frac{f(x)-f(b)}{x-x_{0}}+\frac{f(b)-f(x_{0})}{b-x_{0}}\cdot\frac{b-x_{0}}{x-x_{0}}\\&\geq \frac{-M(b-x_{0})/2}{x-x_{0}}+\frac{M(b-x_{0})}{x-x_{0}}\\ & > 0.\end{align} Thus, $f$ is monotonically increasing on $[b-\delta, b]$.
