A convex function with its related increasing function Let $I$ be an open interval, $f:I\rightarrow \mathbb{R}$ be a convex function and $g:I\rightarrow \mathbb{R}$
increasing. Then, $g(x^-), g(x^+)$, $f'_-(x)$ and $f'_+(x)$ exist, for all $x\in I$.
Now, assume that the following inequalities hold
\begin{equation}
g(x^-)\leq f'_-(x)\leq g(x)\leq f'_+(x)\leq g(x^+)\;\; ;\;\; x \in I.
 \end{equation}
For example, the functions $f(x)=|x|$ and $g(x)=\mbox{sgn}(x)$  satisfy all the above conditions ($I=\mathbb{R}$).
Putting $I_g:=\{x\in I: g\mbox{ is continuous at } x\}$ and $I_f:=\{x\in I: f\mbox{ is differentiable at } x\}$
we have $I_g\subseteq I_f$ and $f'(x)=g(x)$  on $I_f$.
Now, we are looking for such functions $f$ and $g$ such that $I_f\neq I_g$.
Also, some examples such that $g(x^+)\neq f'_+(x)$ but  $g(x^-)= f'_-(x)$, for some $x\in I$
(and similarly $g(x^-)\neq f'_-(x)$ but  $g(x^+)=f'_+(x)$, for some $x\in I$).
Thanks in advance.
 A: The condition
$$
g(x^-)\leq f'_-(x)\leq g(x)\leq f'_+(x)\leq g(x^+)
$$
for all $x \in I$ implies that in fact
$$
\tag{*}
 f'_+(x) = g(x^+) \text{ and } f'_-(x) = g(x^-)
$$
for all $x \in I$, so that $f$ is differentiable exactly at the points where $g$ is continuous, i.e. the sets $I_f$ and $I_g$ are identical.
Proof:  For fixed $x < y$ consider the function
$$
 h(t) = f(t) - \frac{f(y)-f(x)}{y-x}(t-x)
$$
which has the left derivative
$$
h'_-(t) = f'_-(t) - \frac{f(y)-f(x)}{y-x} \, .
$$
We have $h(x) = h(y)$ so that $h$ attains its minimum on the interval $[x, y]$ at some point $t_0$ with $x < t_0 \le y$. Then $h'_-(t_0) \le 0$ so that
$$
 \frac{f(y)-f(x)}{y-x} = f'_-(t_0) - h'_-(t_0) \ge  f'_-(t_0) \ge g(t_0^-) \ge g(x^+)  \, .
$$
Taking the limit $y \to x^+$ we conclude that $f'_+(x) \ge g(x^+)$. This proves the first equality in $(*)$, the second one can be proved in the same way.
Remarks:

*

*The essential idea in the proof is that $ g(x^+) \le  g(y^-) \le f'_-(y) $ for $x < y$ and that a lower bound on the left derivative implies a lower bound on the right derivative, compare Infimum of right derivative and infimum of left derivative are equal?.

*For the desired examples you can choose $g = f'_-$ or $g= f'_+$.

