 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/23f/b/solution

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

φ (t) = t^{p}

gives

∥ M f ∥_{p}^{p} = p \int_{0}^{\infty} s^{p - 1} λ ({M f > s}) d s .

For each

s > 0

, split

f = f_{0} + f_{1}, f_{0} = f 1_{{∣ f ∣ > s /2}}, f_{1} = f 1_{{∣ f ∣ \leq s /2}} .

Since

M f_{1} \leq s /2

and the maximal operator is sublinear,

{M f > s} \subseteq {M f_{0} > s /2} .

The weak

(1, 1)

λ ({M f > s}) \leq \frac{2 C _{1}}{s} \int_{{∣ f ∣ > s /2}} ∣ f (x) ∣ d x .

Substitution into the distribution formula and another application of Tonelli theorem yield

∥ M f ∥_{p}^{p} \leq 2 p C_{1} \int_{R^{n}} ∣ f (x) ∣ \int_{0}^{2∣ f (x) ∣} s^{p - 2} d s d x = \frac{2 ^{p} p C _{1}}{p - 1} ∥ f ∥_{p}^{p} .

∥ M f ∥_{p} \leq (\frac{2 ^{p} p C _{1}}{p - 1})^{1/ p} ∥ f ∥_{p} .

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