 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-67/2/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

The difference between the two cumulative sums of Fourier partial sums is

(n + m) σ_{n + m} (f) - n σ_{n} (f) = \sum_{j = n}^{n + m - 1} s_{j} (f) .

v_{n, m} (f) = \frac{n + m}{m} σ_{n + m} (f) - \frac{n}{m} σ_{n} (f) .

∥ v_{n, m} (f) ∥_{\infty} \leq \frac{n + m}{m} ∥ σ_{n + m} (f) ∥_{\infty} + \frac{n}{m} ∥ σ_{n} (f) ∥_{\infty} \leq (1 + \frac{2 n}{m}) ∥ f ∥_{\infty} .

This proves the requested operator norm bound. When

n = 0

the identity reduces to

v_{0, m} = σ_{m}

and the bound is one.

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