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Fall 2011 - Problem 5

Hardy-Littlewood maximal inequality

For an f ⁣:R→R\func{f}{\R}{\R} belonging to L1(R)L^1\p{\R}, we define the Hardy-Littlewood maximal function as follows:

(Mf)(x)≔sup⁡h>0 12h∫x−hx+h∣f(y)∣ dy.\p{Mf}\p{x} \coloneqq \sup_{h>0}\,\frac{1}{2h} \int_{x-h}^{x+h} \abs{f\p{y}} \,\diff{y}.

Prove that it has the following property: There is a constant AA such that for any λ>0\lambda > 0,

∣{x∈R | (Mf)(x)>λ}∣≤Aλ∥f∥L1\abs{\set{x \in \R \st \p{Mf}\p{x} > \lambda}} \leq \frac{A}{\lambda} \norm{f}_{L^1}

where ∣E∣\abs{E} denotes the Lebesgue measure of EE. If you use a covering lemma, you should prove it.
Solution.

See Fall 2011 - Problem 5.