CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The centered maximal operator is bounded on $L^{p} (R^{n})$ for $1 < p < \infty$

Statement

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

Let $1 < p < \infty$ . Then there is a constant $C_{n, p}$ such that every $f \in L^{p} (R^{n})$ satisfies $∥ M f ∥_{p} \leq C_{n, p} ∥ f ∥_{p} .$ By comparison, the same is true for the uncentered maximal function.

Facts & Assumptions

Given: The Axiom of Countable Choice, an exponent $1 < p < \infty$ , and a function $f \in L^{p} (R^{n})$ .

[L1]

The centered maximal operator is of weak type $(1, 1)$ with constant $5^{n}$ . (The centered Hardy-Littlewood maximal operator is weak type $(1, 1)$ )

[L2]

The centered maximal operator is of strong type $(\infty, \infty)$ with operator norm at most $1$ . (The centered maximal operator is bounded on $L^{\infty}$ )

[L3]

A sublinear operator of weak type $(1, 1)$ and strong type $(\infty, \infty)$ is of strong type $(p, p)$ for every $1 < p < \infty$ . (Marcinkiewicz interpolation from weak $(1, 1)$ and strong $(\infty, \infty)$ )

[L4]

One has $M f \leq M^{*} f \leq 2^{n} M f$ pointwise. (The centered and uncentered maximal functions are pointwise comparable)

Proof

technique · direct

1.1

The maximal operator is sublinear by definition of supremum and absolute [L1, L2, L3, given, algebra] values. Apply [L3] with the constants from [L1] and [L2]. This yields a constant $C_{n, p}$ such that $∥ M f ∥_{p} \leq C_{n, p} ∥ f ∥_{p} .$

L1L2L3givenalgebra

2.1

Step 1.1 proves the centered estimate. Then [L4] gives [step 1.1, L4, algebra] $∥ M^{*} f ∥_{p} \leq 2^{n} ∥ M f ∥_{p} \leq 2^{n} C_{n, p} ∥ f ∥_{p},$ so the uncentered estimate follows as well.

step 1.1L4algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

20 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)
G. H. Hardy and J. E. Littlewood, A maximal theorem with function-theoretic applications, Theorem 13 (standard reference, not scraped)

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