Show a function $f$ of bounded variation has, for a partition $-\pi = x_0 < x_1 < \dots < x_m = \pi$, the sums $\sum_{j=1}^{m} |f(x_j)-f(x_{j-1})|$ that are bounded.
My professor defined a function being of bounded variation when it can be written as the sum of a non-increasing and non-decreasing function.
I am confused on how to approach this exercise. Any help is appreciated.
Let $f\in BV[-\pi,\pi]$. According to your definition, it is equivalent to write $f=f_1-f_2$, where $f_1,f_2$ are increasing functions. Let$-\pi=x_0<x_1<\dots<x_n=\pi$ be a partition of your interval. Then $\displaystyle{\sum_{j=1}^{n}|f(x_j)-f(x_{j-1})|=\sum_{j=1}^{n}|f_1(x_j)-f_2(x_j)-f_1(x_{j-1})+f_2(x_{j-1})|\leq\sum_{j=1}^{n}|f_1(x_j)-f_1(x_{j-1})|+\sum_{j=1}^{n}|f_2(x_j)-f_2(x_{j-1})|=\sum_{j=1}^{n}f_1(x_j)-f_1(x_{j-1})+\sum_{j=1}^{n}f_2(x_j)-f_2(x_{j-1})}=f_1(\pi)-f_1(-\pi)+f_2(\pi)-f_2(-\pi):=M<\infty$.
Therefore all those sums are bounded by this constant. The inequality used is the triangular inequality.