Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04
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The centered maximal operator is bounded on L∞

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)).

Let f∈L∞(Rn). Then Mf(x)≤∥f∥∞(x∈Rn). In particular, M is of strong type (∞,∞) with operator norm at most 1.

Facts & Assumptions

Given: The Axiom of Countable Choice and a function f∈L∞(Rn).

[L1]

The centered maximal function is Mf(x)=sup⁡r>01λ(B(x,r))∫B(x,r)∣f(y)∣ dλ(y). (The centered and uncentered Hardy-Littlewood maximal functions)

[L2]

The L∞ norm is the essential supremum. In particular, ∣f∣≤∥f∥∞ almost everywhere. (The space L∞(μ) of essentially bounded measurable functions)

Proof

technique · direct
1.1L1L2givenalgebra

Fix x∈Rn and r>0. By [L2], [L1, L2, given, algebra] ∫B(x,r)∣f(y)∣ dλ(y)≤∥f∥∞ λ(B(x,r)). Dividing by λ(B(x,r)) gives Ar∣f∣(x)≤∥f∥∞.

2.1step 1.1L1algebra∎

Taking the supremum of the inequality from step 1.1 over all r>0 yields [step 1.1, L1, algebra] Mf(x)≤∥f∥∞. Since x was arbitrary, the centered maximal operator is bounded on L∞ with norm at most 1.

Depends on

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