Let $f$ be increasing on $[a,b]$ and $a < x_1 < \dotsb < x_n < b$. Show that $\sum_{k = 1}^n [f(x_k^+) - f(x_k^-)]\leq f(b^-) - f(a^+)$.
Solution 1:
First, define $x_0 = a$ and $x_{n+1}=b$. Then, for any $k=1,\ldots,n$, we have $$ f(x_k +)-f(x_k -) \leq f\bigg(\frac{{x_k + x_{k + 1} }}{2}\bigg) - f\bigg(\frac{{x_{k - 1} + x_k }}{2}\bigg) $$ (since $f$ is increasing). Thus, $$ \sum\limits_{k = 1}^n {[f(x_k + ) - f(x_k - )]} \le f\bigg(\frac{{x_n + x_{n + 1} }}{2}\bigg) - f\bigg(\frac{{x_0 + x_1 }}{2}\bigg). $$ Now, since $(x_n + x_{n + 1} )/2 = (x_n + b)/2 < b$ and $(x_0 + x_1 )/2 = (a + x_1)/2 > a$, we get $$ \sum\limits_{k = 1}^n {[f(x_k + ) - f(x_k - )]} \le f(b-) - f(a+). $$