= Partition example of total variation
{title2=$V_0^4(f)=11,\quad f(4)-f(0)=1$}
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> give
$$
\boxed{V_0^4(f)=2+3+4+2=11.}
$$
This 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 $11$.
\Image[/total-variation-partition-refinement-and-jordan.png]
{title=A piecewise linear function has net change 1 and total variation 11; refinements reveal hidden movement, while accumulated rises and falls give its Jordan decomposition}
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.
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