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The Hardy-Littlewood maximal operator is not strong type $(1, 1)$

Statement refuted

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

The centered Hardy-Littlewood maximal operator maps $L^{1} (R^{n})$ to $L^{1} (R^{n})$ .

More strongly, if $f \in L^{1} (R^{n})$ satisfies $∥ f ∥_{1} > 0$ , then $M f \notin L^{1} (R^{n})$ .

Facts & Assumptions

Given: The Axiom of Countable Choice and a function $f \in L^{1} (R^{n})$ with $∥ f ∥_{1} > 0$ .

[L1]

The centered maximal function is $M f (x) = \sup_{r > 0} \frac{1}{λ (B (x, r))} \int_{B (x, r)} ∣ f (y) ∣ d λ (y) .$ (The centered and uncentered Hardy-Littlewood maximal functions)

[L2]

The $L^{1}$ norm is $∥ f ∥_{1} = \int_{R^{n}} ∣ f ∣ d λ .$ (The class $L^{1} (μ)$ of integrable functions)

[L3]

Lebesgue measure is sigma-finite, and every bounded measurable set has finite measure. (Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure)

[L4]

Lebesgue measure on balls scales like $r^{n}$ under dilation. (For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it)

Counterexample

technique · direct

1.1

Since $∥ f ∥_{1} > 0$ , the integral in [L2] is positive. Hence there is [L2, L3, given, choose, algebra] $ε > 0$ such that the measurable set $E : = {y \in R^{n} : ∣ f (y) ∣ > ε}$ has positive measure. Because $R^{n} = ⋃_{m \geq 1} B (0, m)$ and measure is countably subadditive, some $R \geq 1$ satisfies $0 < λ (F) < \infty, F : = E \cap B (0, R),$ the finiteness coming from [L3].

L2L3givenchoosealgebra

2.1

Let $x \in R^{n}$ with $∥ x ∥_{2} \geq 2 R$ . If $y \in F$ , then [step 1.1, L1, L4, algebra] $∥ y - x ∥_{2} \leq ∥ y ∥_{2} + ∥ x ∥_{2} \leq R + ∥ x ∥_{2} \leq \frac{3}{2} ∥ x ∥_{2},$ so $F \subseteq B (x, \frac{3}{2} ∥ x ∥_{2})$ . Since $∣ f ∣ \geq ε$ on $F$ , [L1] gives $M f (x) \geq \frac{1}{λ (B (x, \frac{3}{2} ∥ x ∥_{2}))} \int_{F} ∣ f (y) ∣ d λ (y) \geq \frac{ε λ (F)}{λ (B (0, 1)) (\frac{3}{2} ∥ x ∥_{2})^{n}} = \frac{C}{∥ x ∥_{2}^{n}}$ for a positive constant $C$ .

step 1.1L1L4algebra

3.1

For each integer $k \geq 1$ , set [step 2.1, L4, algebra] $A_{k} : = B (0, 2^{k + 1} R) ∖ B (0, 2^{k} R) .$ On $A_{k}$ one has $∥ x ∥_{2} \leq 2^{k + 1} R$ , so step 2.1 yields $M f (x) \geq \frac{C}{(2^{k + 1} R)^{n}} (x \in A_{k}) .$ Therefore $\int_{A_{k}} M f d λ \geq \frac{C}{(2^{k + 1} R)^{n}} λ (A_{k}) .$ Using [L4] again, $λ (A_{k}) = λ (B (0, 1)) ((2^{k + 1} R)^{n} - (2^{k} R)^{n}),$ so the right-hand side is a positive constant independent of $k$ . Since the annuli $A_{k}$ are pairwise disjoint, the integral of $M f$ over $⋃_{k \geq 1} A_{k}$ diverges.

step 2.1L4algebra

4.1

Thus $M f \notin L^{1} (R^{n})$ whenever $∥ f ∥_{1} > 0$ , so the [step 3.1] strong type $(1, 1)$ claim is false.

step 3.1∎

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Sources

Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed., Exercise 22 after Theorem 3.22 (standard reference, not scraped)
G. H. Hardy and J. E. Littlewood, A maximal theorem with function-theoretic applications, Section I and Theorem 14 (standard reference, not scraped)

The Hardy-Littlewood maximal operator is not strong type (1,1)(1,1)