Function of bounded variation clarity

Question

We know that the definition is : $$\sup_{a=x_0<x_1<...<x_n=b} \sum_{k=0}^{n-1} |f(x_{k+1})-f(x_{k})|$$
but intuitively what does this mean ? I don't get why there is a sup ! With the sup we only take the biggest $|f(x_{k+1})-f(x_{k})|$ ? Or do we consider all the sum ? thanks !

Answer 1

The supremum here is taken over all possible partitions.

Where you have $a = x_0 < x_1 < \cdots < x_n = b$ you have defined a partition of $[a,b]$ where you may choose the values of $x_i$ arbitrarily for all $i$ so long as you maintain the strict ordering. The supremum says that we then need to consider all of those partitions, and the one that results in the largest sum is the one we use.

In practice we don't actually work out all partitions: we can either pick out a partition that clearly maximises the sum (for example, choosing $x_n = 1/(n\pi)$ for $f(x) = \sin(1/x)$) or we use analytical methods to bound or limit the sum regardless of the partition.