ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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.

A no-sinks sandwich on the real line

Example

On $V = R$ , let $p (x) : = \max {x, 0}$ and $L (x) : = x / 2$ . Then $p$ is an everywhere-finite no-sinks sublinear function and $L$ satisfies $- p (- x) \leq L (x) \leq p (x)$ for every real $x$ .

Facts & Assumptions

[A1]

The example's definitions: $p (x) = \max {x, 0}$ and $L (x) = x / 2$ on the real vector space $R$ .

[F1]

Infinite-valued no-sinks Hahn–Banach sandwich and conditional offsets: the sandwich theorem assumes $p (0) = 0$ , positive homogeneity, subadditivity, and symmetric finiteness of $p (x), p (- x)$ .

Verification

Given: [A1] and [F1].

1.1A1F1algebra

For $x = 0$ , $p (x) = 0$ . For $t > 0$ , $p (t x) = \max {t x, 0} = t \max {x, 0} = t p (x)$ . Also $p (x + y) \leq p (x) + p (y)$ because the right side is nonnegative and at least $x + y$ . Every value is finite, so $p (x) < + \infty$ iff $p (- x) < + \infty$ . Thus all assumptions in [F1] hold.

2.1A1F1step 1.1algebra∎

The map $L (x) = x / 2$ is real-linear. If $x \geq 0$ , then $p (x) = x$ , $p (- x) = 0$ , and $0 \leq x / 2 \leq x$ . If $x \leq 0$ , then $p (x) = 0$ , $p (- x) = - x$ , and $x \leq x / 2 \leq 0$ . These two cases include $x = 0$ and give $- p (- x) \leq L (x) \leq p (x)$ for all $x$ .

Depends on

Dependency tree · 2 levels

Sources

MIT 18.102, Introduction to Functional Analysis, complete lecture notes