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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-6/3/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 6 / 3 / b / Solution by

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

For the normalized velocity-reset collision operator,

⟨ L f, g ⟩ = ρ (f) \overline{ρ (g)} - ⟨ f, g ⟩ = ⟨ f, Lg ⟩ .

This bounded everywhere-defined operator is self-adjoint. Put

h = f / M

. Expanding the square with the probability measure

M (v) d v

gives

\int \int ∣ h (v_{*}) - h (v) ∣^{2} M (v) M (v_{*}) d v d v_{*} = 2∥ f ∥_{H}^{2} - 2∣ ρ (f) ∣^{2} .

It follows that

⟨ L f, f ⟩ = - \frac{1}{2} \int \int ∣ h (v_{*}) - h (v) ∣^{2} M (v) M (v_{*}) d v d v_{*} \leq 0 .

Replacing the difference by

(f (v_{*}) M (v) - f (v) M (v_{*})) / (M (v) M (v_{*}))

gives the other printed integral expression. For real functions the modulus squares are ordinary squares; the modulus version also proves the complex-space statement.

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