 Partition example of total variation

Define a piecewise linear function by joining the five points $(0,0),(1,2),(2,-1),(3,3),(4,1)$. Its net change is $1$, but the increments at these knots are $2,-3,4,-2$. Their absolute values giveThis is its exact total variation, not merely one lower bound. Given any partition of an interval, insert all five knots. The triangle inequality ensures that this partition refinement cannot decrease the sum. On each resulting monotone function segment, the absolute increments telescope, so every refined sum is $11$. The knot partition of an interval itself attains $11$.

Sampling only at $0,4$ gives $1$. Sampling at $0,1,3,4$ gives $2+1+2=5$. Inserting the remaining knot at $2$ reveals the descent and rise that were hidden inside $[1,3]$, giving $11$. These are nested partitions of an interval: their sums increase because each is a partition refinement of the preceding one. Arbitrarily chosen unrelated partitions of an interval need not produce an increasing sequence of sums.