Integral of Left and Right Derivative of Convex Function

Question

Suppose $f(x)$ is a convex function on $[a, b]$. How can I prove that:

$$ f(b) - f(a) = \int_{a}^{b}f'_{-}(x) dx = \int_{a}^{b}f'_{+}(x) dx $$

Answer 1

We assume that $f_{+}'$ and $f_{-}'$ are bounded, otherwise the integrals are improper and the equality does not hold (take for example $[a,b]=[0,1]$, and $f(x)=\lfloor x\rfloor$ then $f(1)-f(0)=1$ but $f_{-}'(x)=0$ on $[0,1)$, and it is $+\infty$ at $1$).

The functions $f_{+}'$ and $f_{-}'$ are monotone in $[a,b]$. If they are bounded then they are both Riemann-integrable in $[a,b]$ and $f$ is continuous in $[a,b]$.

For any $x_1,x_2\in [a,b]$ with $x_1<x_2$ then there is $t\in [x_1,x_2]$ such that $$f_{+}'(x_1)\le\frac{f(x_2)-f(x_1)}{x_2-x_1}\leq f_{+}'(t)$$ Hence, for any subdivision $\sigma$ of $[a,b]$, $a=x_0<x_1<\dots<x_{N-1}<x_N=b$, we have that $$L(f_{+}',\sigma)\leq f(b)-f(a)=\sum_{k=1}^{N}(f(x_k)-f(x_{k-1}))\leq U(f_{+}',\sigma)$$ where $U$are is the upper Riemann-sum and $L$ is the lower Riemann-sum with respect to $\sigma$. Hence $$f(b) - f(a) = \int_{a}^{b}f'_{+}(x) dx.$$ The proof of the other equality is similar.