Every dyadic partition is among the finite partitions on the right-hand side, so the displayed supremum is at least $\lVert f\rVert$. For the converse, the claim is immediate if $\lVert f\rVert=\infty$. If it is finite, apply part (a) to every interval of an arbitrary partition $0\leq t_0<\cdots<t_n=1$:
$$
|f(t_k)-f(t_{k-1})|
\leq\lVert f_{t_k}\rVert-\lVert f_{t_{k-1}}\rVert.
$$
Summing telescopes and gives
$$
\sum_{k=1}^n|f(t_k)-f(t_{k-1})|
\leq\lVert f_1\rVert-\lVert f_{t_0}\rVert
\leq\lVert f\rVert.
$$
Taking the supremum proves that the dyadic definition equals the usual <total variation of a function>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 202 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-202/3/b/solution