 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-6/4/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-07

Nonnegativity and unit mass give, by the triangle inequality for the Fourier transform,

∣ f (ξ) ∣ \leq \int f = 1 = f (0)

. The finite second moment implies a finite first absolute moment, so differentiating under the Fourier integral is justified and

∥ f ∥_{\infty} \leq 1, f (0) = 1, \nabla f (0) = - i \int v f (v) d v = 0 .

The exponent convention has no

2 π

, so no such factor occurs in this derivative.

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