A problem that involves a supremum

Question

We let $[a,b]\subset \mathbb{R}$ and $f:[a,b]\to \mathbb{R}$. How do we prove that $$\sup \left|\sum_{i=1}^n a_i[f(x_i)-f(x_{i-1})]\right|=\sup\sum_{i=1}^n|f(x_i)-f(x_{i-1})|$$ where $a_i\in [-1,1]$ and the supremum is taken all divisions $$a=x_0<x_1<x_2<\dots<x_n=b$$ of $[a,b]$. The left hand side is $\le$ to the right hand side by the triangle inequality. May I ask a tip (or solution as I wish) for proving the reverse inequality?

Answer 1

Tip: put $a_i=\operatorname{sign} [f(x_i)-f(x_{i-1})]$.