 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/23g/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For

f_{z} (y) = f (yz)

, the substitution

u = yz

gives

∥ f_{z} ∥_{p}^{p} = z^{- 1} ∥ f ∥_{p}^{p}, ∥ f_{z} ∥_{p} = z^{- 1/ p} ∥ f ∥_{p} .

Homogeneity of

K

and the substitution

x = yz

give

T f (y) = \int_{0}^{\infty} K (yz, y) f (yz) y d z = \int_{0}^{\infty} K (z, 1) f_{z} (y) d z .

Apply part (a), or equivalently the Minkowski integral inequality, to this integral and use the preceding scaling identity:

∥ T f ∥_{p} \leq \int_{0}^{\infty} ∣ K (z, 1) ∣ ∥ f_{z} ∥_{p} d z = ∥ f ∥_{p} \int_{0}^{\infty} ∣ K (z, 1) ∣ z^{- 1/ p} d z = ∥ f ∥_{p} .

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