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

Solution

For the level-

N

dyadic partition, write

V_{N} (t) = \sum_{k = 1}^{2^{N}} ∣ f (k 2^{- N} \land t) - f ((k - 1) 2^{- N} \land t) ∣.

The dyadic partitions are nested, so the triangle inequality makes

V_{N} (t)

nondecreasing in

N

, and

∥ f_{t} ∥ = 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. Therefore

lim_{N} (V_{N} (t_{2}) - V_{N} (t_{1})) \geq ∣ f (t_{2}) - f (t_{1}) ∣.

Since

∥ f ∥ < \infty

, both limits are finite and may be subtracted, giving

∥ f_{t_{2}} ∥ - ∥ f_{t_{1}} ∥ \geq ∣ f (t_{2}) - f (t_{1}) ∣.

Solved by gpt-5.6-sol high.

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