LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-30

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.

First-order Hadamard factorization near a point

Statement

Let $W \subseteq R^{n}$ be open, let $a \in W$ , and let $f : W \to R$ be smooth. After shrinking to a convex neighbourhood of $a$ , there are smooth functions $g_{1}, \dots, g_{n}$ such that $f (x) - f (a) = \sum_{i = 1}^{n} (x^{i} - a^{i}) g_{i} (x)$ and $g_{i} (a) = \partial_{i} f (a)$ for each $i$ .

Facts & Assumptions

Given: A smooth function $f : W \to R$ and a point $a \in W$ .

[L1]

The Newton-Leibniz formula holds on line segments, and differentiation under the integral sign preserves smoothness on compact rectangles (Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative, Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral).

Proof

technique · direct

1.1

Shrink $W$ so that the segment $a + t (x - a)$ stays in $W$ for all $x \in W$ and $t \in [0, 1]$ . Define $g_{i} (x) : = \int_{0}^{1} \partial_{i} f (a + t (x - a)) d t$ .

L1givenconstruct

2.1

Applying the one-variable Newton-Leibniz formula from [L1] to $t \mapsto f (a + t (x - a))$ gives $f (x) - f (a) = \sum_{i} (x^{i} - a^{i}) g_{i} (x)$ .

L1step 1.1

3.1

Differentiation under the integral sign in [L1] shows each $g_{i}$ is smooth, and evaluating at $x = a$ gives $g_{i} (a) = \partial_{i} f (a)$ .

L1step 1.1∎

Depends on

Used by

Dependency tree · two levels

10 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

John M. Lee, Introduction to Smooth Manifolds (standard reference, not scraped)
Will J. Merry, Differential Geometry (standard reference, not scraped)
Nigel Hitchin, Differentiable Manifolds (standard reference, not scraped)