PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04

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The centered maximal operator is bounded on $L^{\infty}$

Statement

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

Let $f \in L^{\infty} (R^{n})$ . Then $M f (x) \leq ∥ f ∥_{\infty} (x \in R^{n}) .$ In particular, $M$ is of strong type $(\infty, \infty)$ with operator norm at most $1$ .

Facts & Assumptions

Given: The Axiom of Countable Choice and a function $f \in L^{\infty} (R^{n})$ .

[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 $L^{\infty}$ norm is the essential supremum. In particular, $∣ f ∣ \leq ∥ f ∥_{\infty}$ almost everywhere. (The space $L^{\infty} (μ)$ of essentially bounded measurable functions)

Proof

technique · direct

1.1

Fix $x \in R^{n}$ and $r > 0$ . By [L2], [L1, L2, given, algebra] $\int_{B (x, r)} ∣ f (y) ∣ d λ (y) \leq ∥ f ∥_{\infty} λ (B (x, r)) .$ Dividing by $λ (B (x, r))$ gives $A_{r} ∣ f ∣ (x) \leq ∥ f ∥_{\infty} .$

L1L2givenalgebra

2.1

Taking the supremum of the inequality from step 1.1 over all $r > 0$ yields [step 1.1, L1, algebra] $M f (x) \leq ∥ f ∥_{\infty}$ . Since $x$ was arbitrary, the centered maximal operator is bounded on $L^{\infty}$ with norm at most $1$ .

step 1.1L1algebra∎

Depends on

Used by

The centered maximal operator is bounded on Lᵖ(ℝⁿ) for 1<p<∞ Corollary

Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed., Corollary 6.35 (standard reference, not scraped)

The centered maximal operator is bounded on L∞L∞