PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 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 and uncentered maximal functions are pointwise comparable

Statement

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

Let $f \in L_{l o c}^{1} (R^{n})$ . Then for every $x \in R^{n}$ , $M f (x) \leq M^{*} f (x) \leq 2^{n} M f (x) .$

Facts & Assumptions

Given: The Axiom of Countable Choice, a locally integrable function $f$ on $R^{n}$ , and a point $x \in R^{n}$ .

[L1]

The centered maximal function takes the supremum over balls centered at $x$ , while the uncentered maximal function takes the supremum over all balls that contain $x$ . (The centered and uncentered Hardy-Littlewood maximal functions)

[L2]

Lebesgue measure scales by $∣ c ∣^{n}$ under dilation by $c \neq 0$ . (For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it)

Proof

technique · direct

1.1

Every ball centered at $x$ is in particular a ball containing $x$ , so the [L1] supremum defining $M f (x)$ is taken over a smaller family than the one defining $M^{*} f (x)$ . Therefore $M f (x) \leq M^{*} f (x)$ .

1.2

Let $B (y, r)$ be any ball containing $x$ . If $z \in B (y, r)$ , then [L1, algebra] $∥ z - x ∥_{2} \leq ∥ z - y ∥_{2} + ∥ y - x ∥_{2} < r + r = 2 r,$ so $B (y, r) \subseteq B (x, 2 r)$ . Hence $\int_{B (y, r)} ∣ f ∣ \leq \int_{B (x, 2 r)} ∣ f ∣ d λ .$

L1algebra

2.1

Step 1.2 and [L2] give [step 1.2, L1, L2, algebra] $\frac{1}{λ (B (y, r))} \int_{B (y, r)} ∣ f ∣ \leq \frac{λ (B (x, 2 r))}{λ (B (y, r))} \cdot \frac{1}{λ (B (x, 2 r))} \int_{B (x, 2 r)} ∣ f ∣ = 2^{n} A_{2 r} ∣ f ∣ (x) \leq 2^{n} M f (x) .$ Taking the supremum over all balls $B (y, r)$ containing $x$ yields $M^{*} f (x) \leq 2^{n} M f (x)$ .

step 1.2L1L2algebra

3.1

Combining steps 1.1 and 2.1 gives the claimed comparison. [step 1.1, step 2.1]

step 1.1step 2.1∎

Depends on

Used by

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

Dependency tree · two levels

26 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., Exercise 22 (standard reference, not scraped)