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

For the level-$N$ dyadic partition, writeThe dyadic partitions are nested, so the triangle inequality makes $V_N(t)$ nondecreasing in $N$, and $\Vert f_t\Vert ={\lim }_N{V}_N(t)$.

Fix $t_1<t_2$. As the mesh tends to zero, the last dyadic point before $t_1$ approaches $t_1$. Refining from there to $t_2$, the triangle inequality says that the added variation is at least $|f(t_2)-f(t_1)|$ minus the two endpoint errors, which tend to zero by continuity. ThereforeSince $\Vert f\Vert <\infty$, both limits are finite and may be subtracted, giving

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Every dyadic partition is among the finite partitions on the right-hand side, so the displayed supremum is at least $\Vert f\Vert$. For the converse, the claim is immediate if $\Vert f\Vert =\infty$. If it is finite, apply part (a) to every interval of an arbitrary partition $0\le t_0<\cdots <t_n=1$:Summing telescopes and givesTaking the supremum proves that the dyadic definition equals the usual total variation of a function.

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If $f$ is continuously differentiable, then for every finite partition the fundamental theorem of calculus and the triangle inequality givePart (b) therefore implies $\Vert f\Vert <\infty$.

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For $k\ge 1$, setThen $t_k\downarrow 0$ and $f(t_k)=(-1{)}^k{t}_k$. Consecutive values have opposite signs, soFinite partitions containing $t_N,t_{N-1},\dots ,t_1$ therefore have variation at leastwhich diverges with $N$ by comparison with the harmonic series. Part (b) now gives $\Vert f\Vert =\infty$.

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