LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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 McShane formula extends a 1-Lipschitz function from a finite subset

Statement

Let $(X, d)$ be a metric space, let $F = {x_{1}, \dots, x_{m}} \subseteq X$ be finite and nonempty with distinct ordered points, and let $f_{1}, \dots, f_{m} \in R$ satisfy $∣ f_{i} - f_{j} ∣ \leq d (x_{i}, x_{j})$ for every $i, j$ . Define $\tilde{f} (x) : = \min_{1 \leq i \leq m} (f_{i} + d (x, x_{i})) (x \in X) .$ Then $\tilde{f}$ is 1-Lipschitz on $X$ and $\tilde{f} (x_{i}) = f_{i}$ for every $i$ .

Facts & Assumptions

Given: The metric space, finite set, and real values in the Statement.

[F1]

A metric is symmetric and satisfies the triangle inequality. Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric

[F2]

Every nonempty finite set of reals has an attained minimum. Every nonempty finite set of reals has a maximum and a minimum

Proof

technique · direct

1.1givenF1F2

For each $k$ , every term in the defining minimum at $x_{k}$ satisfies $f_{i} + d (x_{k}, x_{i}) \geq f_{k}$ , because $f_{k} - f_{i} \leq d (x_{k}, x_{i})$ by [F1] and the hypothesis. The term $i = k$ equals $f_{k}$ , so $\tilde{f} (x_{k}) = f_{k}$ .

2.1givenF1F2algebra∎

For any $x, y \in X$ and every $i$ , [F1] gives $d (x, x_{i}) \leq d (x, y) + d (y, x_{i})$ . Thus $f_{i} + d (x, x_{i}) \leq d (x, y) + f_{i} + d (y, x_{i})$ for every $i$ ; taking finite minima yields $\tilde{f} (x) \leq d (x, y) + \tilde{f} (y)$ . Interchanging $x, y$ and using symmetry gives $\tilde{f} (y) \leq d (x, y) + \tilde{f} (x)$ , hence $∣ \tilde{f} (x) - \tilde{f} (y) ∣ \leq d (x, y)$ .

Depends on

Used by

Dependency tree · two levels

9 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

Dmitriy Panchenko, Lecture Notes on Probability Theory, §4.4 (standard reference, not scraped)
Filippo Santambrogio, Introduction to Optimal Transport Theory, §1.3, PDF pp. 5–6 (standard reference, not scraped)