 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-4/22g/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Choose

A > 0

and set

M = max_{∣ x ∣ \leq A} ∣ ϕ (x) ∣.

Take

T_{1} \leq \frac{A}{m a x { M , 1 }} .

Inductively, if

∣ f_{n} (s) ∣ \leq A

[0, T_{1}]

, then

∣ f_{n + 1} (t) ∣ \leq \int_{0}^{t} ∣ ϕ (f_{n} (s)) ∣ d s \leq M T_{1} \leq A .

Since

f_{0} = 0

, every iterate is bounded by

A

. Moreover,

∣ f_{n + 1} (t) - f_{n + 1} (s) ∣ = \int_{s}^{t} ϕ (f_{n} (r)) d r \leq M ∣ t - s ∣.

Thus $(f_{n})_{n \geq 1}$ is uniformly bounded and uniformly Lipschitz, hence equicontinuous, on $[0, T_{1}]$ .

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