ExampleConstruction: Literature-sourcedVerification: 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 function of $1_{[0, 1]}$ on $R$

Example

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

Let $f = 1_{[0, 1]}$ on $R$ . Then the centered maximal function is $M f (x) = {\begin{cases} \frac{1}{2 (1 - x)}, & x \leq 0, \\ 1, & 0 < x < 1, \\ \frac{1}{2 x}, & x \geq 1. \end{cases}$ In particular $M f (0) = M f (1) = 1 / 2$ and $M f (x) \sim (2 ∣ x ∣)^{- 1}$ at infinity.

Facts & Assumptions

Given: The Axiom of Countable Choice and the function $f = 1_{[0, 1]}$ on $R$ .

[L1]

The centered maximal function is the supremum of normalized averages of $∣ f ∣$ over centered intervals. (The centered and uncentered Hardy-Littlewood maximal functions)

Verification

technique · direct

1.1L1givenalgebra

Let $x > 1$ , and let $I_{r} = [x - r, x + r]$ with $r > 0$ . If $r \leq x - 1$ , then [L1, given, algebra] $I_{r} \cap [0, 1] = \emptyset$ and the average is $0$ . If $x - 1 < r < x$ , then $I_{r} \cap [0, 1] = [x - r, 1]$ , so the average is $\frac{r - x + 1}{2 r} = \frac{1}{2} - \frac{x - 1}{2 r},$ which increases with $r$ . If $r \geq x$ , then $I_{r}$ contains $[0, 1]$ , so the average is $1 / (2 r)$ , which decreases with $r$ . The maximum is therefore attained at $r = x$ , with value $1 / (2 x)$ .

1.2L1givenchoosealgebra

If $0 < x < 1$ , choose $r < \min {x, 1 - x}$ ; then $[x - r, x + r] \subseteq [0, 1]$ , so [L1, given, choose, algebra] the average equals $1$ . Since no average of an indicator can exceed $1$ , one has $M f (x) = 1$ on $(0, 1)$ .

2.1step 1.1algebra

The same calculation with the reflected interval shows that for $x < 0$ the [step 1.1, algebra] maximum is attained at $r = 1 - x$ and equals $1 / (2 (1 - x))$ . At the endpoints this gives $M f (0) = M f (1) = 1 / 2$ .

3.1step 1.1step 1.2step 2.1∎

Steps 1.1, 1.2, and 2.1 give the displayed formula.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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

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

The centered maximal function of 1[0,1]1[0,1]​ on RR