 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-4/23f/b/solution

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

Let

w = e^{- Φ}, m_{R} = min_{[- R, R]} w > 0, E_{R} (h) = \int_{- R}^{R} ∣ h^{'} ∣^{2} w .

∣ h (x) - h (y) ∣ \leq ∣ x - y ∣^{1/2} (\int_{- R}^{R} ∣ h^{'} ∣^{2})^{1/2} \leq m_{R}^{- 1/2} ∣ x - y ∣^{1/2} E_{R} (h)^{1/2} .

m_{R}^{- 1/2} E_{R} (h)^{1/2}

Z_{R} = \int_{- R}^{R} w > 0

and using the zero weighted mean, for each

x

we have

h (x) = \frac{1}{Z _{R}} \int_{- R}^{R} (h (x) - h (y)) w (y) d y .

Since

∣ x - y ∣ \leq 2 R

, the preceding estimate yields

∣ h (x) ∣ \leq (2 R)^{1/2} m_{R}^{- 1/2} E_{R} (h)^{1/2} .

Consequently the required inequality holds, for example, with

C_{R} = (1 + 2 R) m_{R}^{- 1/2} .

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