How to prove continuity of a concave, non-decreasing $f : [0, 1] \to [0, 1]$. Let $f : [0, 1] \to [0, 1]$ be non-decreasing, $f(0) = 0$ and $f(1) = 1$. In a paper which I have read, concavity for this $f$ is defined as follows: $f$ is called concave, if for each $q \in (0,1]$, there are reals $a_q, b_q$ and a line $l_q(x) = a_q x + b_q$ such that $l_q(q) = f(q)$ and $l_q(p) \ge f(p)$ for all $p \in (0, 1]$. Such an $f$ is called convex, if $-f$ is concave.
Now there are made three statements, and I do not quite see how to prove them with the given definition. Let $f : [0, 1] \to [0, 1]$ be non-decreasing.


*

*If $f$ is concave, then $f$ is continuous on $(0, 1]$ and a jump of f can only occur at $0$.

*If $f$ is convex, then $f$ is continuous on $[0, 1)$ and a jump of $f$ can only occur at $1$.

*If $f$ is concave or convex and does not have a jump, then $f$ is absolutely continuous.


I have managed to prove in (1) that $f$ is right-continuous, but I still need to prove the left-continuity and struggle with that.
Does (2) follow directly from statement (1) as $-f$ is concave?
About (3), I could already prove for the case that $f$ is concave, $f$ is absolutely continuous on $(0, 1]$ but I do not know how to show that this is the case for the entire interval $[0, 1]$. I suppose that in the scenario that $f$ is convex one uses the statement for $-f$.
Any help is appreciated. Thanks!
 A: Let us prove point number 1. The idea is that the graph of the function is, at any point, locked between the constant function and the line given by concavity, so it has no chance but to be continuous. Here a formal proof without looking at the boundary points:
Let $q \in ]0,1[$, and let $(x_n)_{n \in \mathbb{N}}$ be a sequence that converges to $q$ from above (i.e. $x_n \geq q \ \forall n \in \mathbb{N}$). We show that $\lim_{n \to \infty} f(x_n) = f(q)$. 
Observe that, because $f$ is non-decreasing, $f(q) \leq f(x_n) \ \forall n$, and also by concavity, $f(x_n) \leq l_q(x_n) = a_qx_n + b_q$. This together implies
$$ \lim_{n \to \infty} f(q) \leq  \lim_{n \to \infty} f(x_n) \leq  \lim_{n \to \infty} a_q x_n + b_q $$
$$ \Longleftrightarrow  f(q) \leq \lim_{n \to \infty} f(x_n) \leq a_q q + b_q = f(q).$$
$$\Rightarrow \lim_{n \to \infty} f(x_n) = f(q). $$
An analogous argument would give the same for a sequence converging to $q$ from below and thus, putting both steps together, we have continuity of $f$ at $q$.
A: A more tractable way of posing the condition is to say that for every $q\in (0,1]$ there is $a_q\geq 0$  so that for all $x\in [0,1]$:
  $$ f(x) \leq f(q) + a_q (x-q) $$
When $x\in (0,1]$ as well we may exchange $x,q$ to get:
$$ f(q)\leq f(x) + a_x (q-x)$$
Therefore,
 $$ a_x(x-q) \leq f(x)-f(q) \leq a_q(x-q)$$
We conclude two things: $0<q<x\leq 1  \Rightarrow 0\leq a_x\leq a_q$. And $|f(x)-f(q)|\leq a_q|x-q|$. 
Pick now $\delta>0$. There is some $L=a_\delta\geq 0$ associated. Then for $x,y\in [\delta,1]$ we have: $0\leq a_x,a_y\leq L$ so that the above inequality applied to $x,y$ yields:
  $$ |f(x)-f(y)| \leq L|x-y|$$
So $f$ is $L=a_\delta$ Lipschitz on $[\delta,1]$.
This being true for all $\delta>0$ we conclude that $f$ is continuous on $(0,1]$.
For point (2) you may look at the function $g(t)=-f(1-t)$ which verifies the conditions of being non-decreasing, concave, whence continuous on $(0,1]$. Thus $f$ is continuous on $[0,1)$.
For (3) when $f$ is continuous at zero: Let $\epsilon>0$ and pick $\delta_0>0$ so that $0\leq f(\delta_0)-f(0)< \epsilon/2$. Now let $L=a_{\delta_0}$ be the Lipschitz constant for $f$ on $[\delta_0,1]$ and set $\delta=\min\{\delta_0,\frac{\epsilon}{2L}\}$. You verify that if $[x_i,y_i]$, $1\leq i\leq N$ is a collection of disjoint intervals of total length at most $\delta$, then the sum of the lengths of their images does not exceed $\epsilon$, so $f$ is abs cont.
More generally, you don't need monotonicity for this. $f$ continuous on $[0,1]$ and Lipschitz continuous on $[\delta,1]$ for every $0<\delta<1$ implies abs cont. on $[0,1]$. 
