 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/27h/c/solution

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

Regard the same triangular functions as cutoffs in the frequency variable. They satisfy

0 \leq θ_{n} (ξ) ↑ 1.

Since

f \geq 0

, monotone convergence gives

∥ f ∥_{1} = lim_{n \to \infty} \int_{R} θ_{n} (ξ) f (ξ) d ξ .

Fubini's theorem and evenness of

θ_{n}

give

\int θ_{n} (ξ) f (ξ) d ξ = \int f (x) θ_{n} (x) d x .

Part (b) shows that

θ_{n} \geq 0

. Moreover, Fourier inversion at zero yields

\int_{R} θ_{n} (x) d x = 2 π θ_{n} (0) = 2 π .

Consequently

0 \leq \int θ_{n} f = \int f θ_{n} \leq ∥ f ∥_{\infty} \int θ_{n} = 2 π ∥ f ∥_{\infty} .

Passing to the limit proves

∥ f ∥_{L^{1}} \leq 2 π ∥ f ∥_{L^{\infty}} .

Thus one may take the universal constant $α = 2 π$ .

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