Consider a continuous and convex function $F(x):[0,1]\longrightarrow\mathbb{R}$. I am wondering if
$F(x)$ is continuously differentiable in $[0,1]$
$F(x)$ is of bounded variation in $[0,1]$
$F(x)$ is absolute continuous in $[0,1]$.
The second one is correct, due to this post Proving a convex function is of bounded variation.
However, the remaining two became mysterious to me. Royden's chapter 6 answers them if we have an open interval.
Corollary 17: Let $\varphi$ be a convex function on $(a,b)$. Then $\varphi$ is Lipschitz, and therefore absolutely continuous on each closed, bounded subinterval $[c,d]$ and $(a,b)$
Theorem 18: Let $\varphi$ be a convex function on $(a,b)$. Then $\varphi$ is differentiable except at a countable number of points.
By the Theorem 18, it is hard to believe that $F(x)$ will become differentiable in $[0,1]$. But I cannot find a counterexample. That is, a convex function that is continuous on $[0,1]$ but is not differentiable.
The Corollary 17 gives us pretty nice result, but seems like it does not apply to the closed interval. Is it possible to say that if we have $F(x)$ on $[0,1]$ is convex, then it will be convex on $(-\epsilon, 1+\epsilon)$? and then we can use Corollary 17 to conclude that it is absolutely continuous on $[0,1]\subset (-\epsilon, 1+\epsilon)$.
Thank you!