Let $f:[a,b] \to \mathbb{R}$ a non-decreasing function. Define $F:[a,b] \to \mathbb{R}$ by $F(x) = \int_{a}^{x}f(t)\text{d}t$. In the sentences below, prove or give a counterexample:
(1) The function $F$ is differentiable in $(a,b)$.
(2) Given $x_{0} \in (a,b)$, then $F(x) - F(x_{0}) \geq w(x - x_{0})$, for any $x \in (a,b)$ and $w \in [f(x_{0}^{-}),f(x_{0}^{+})]$.Here, $f(x_{0}^{-})$ $[f(x_{0}^{+})]$ is the left [right] limit of $f$ in $x_{0}$.
(3) $F$ is convex. If $f$ is inscreasing, then $F$ is strictly convex.
(1) Is false. Some functions defined by parts, works.
(3) If (2) is true, can I show that $\frac{F(x) - F(x_{0})}{x - x_{0}}$ is non-decreasing? $F$ be convex seems reasonable.
(2) $F(x) - F(x_{0}) = \int_{x_{0}}^{x}f(t)\text{d}t$ if $x \geq x_{0}$ and $F(x) - F(x_{0}) = -\int_{x_{0}}^{x}f(t)\text{d}t$ if $x \leq x_{0}$. So, probably (2) is true, but I have no idea about how to prove.
Let $f(x)=\begin{cases} -1 \,\, \text{if} \,\, x\in [-1,0[ \\ 1 \,\, \text{if} \,\, x\in [0,1] \end{cases}$. This should cover problem $(1)$.
$(3)$ is true, just check the definition of convexity (the actual definition, not the sufficient condition with the derivative, as it is not applicable here due to the above counter-example).