LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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.

Small coordinate-by-coordinate increments stay inside a Euclidean ball and telescope the total increment

Statement

Let $B (a, r) \subseteq U \subseteq R^{m}$ and let $∥ h ∥_{2} < r$ . Define $p_{0} = a$ and $p_{j} = a + \sum_{i < j} h_{i} e_{i}$ for $1 \leq j \leq m$ . Then every $p_{j}$ lies in $B (a, r)$ , $p_{m} = a + h$ , and for every map $f$ on $U$ ,

$f (a + h) - f (a) = \sum_{j < m} (f (p_{j + 1}) - f (p_{j})) .$

Facts & Assumptions

Given: The displayed ball, vector $h$ , and coordinate-prefix points $p_{j}$ .

[L1]

The coordinate list of $x \in F^{n}$ with respect to the ordered basis $e$ is its ordinary coordinate list (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[L2]

The Euclidean norm of $x$ is $∥ x ∥_{2} = \sqrt{\sum_{i < n} x_{i}^{2}}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

Proof

technique · direct

1.1L1L2

The standard-basis coordinate formula makes $p_{m} = a + h$ , while $p_{j} - a$ has coordinates $h_{i}$ for $i < j$ and $0$ otherwise.

2.1step 1.1L2algebra

Hence $∥ p_{j} - a ∥_{2}^{2} = \sum_{i < j} h_{i}^{2} \leq \sum_{i < m} h_{i}^{2} = ∥ h ∥_{2}^{2} < r^{2}$ , so every prefix point is in $B (a, r)$ .

3.1step 1.1step 2.1∎

Summing $f (p_{j + 1}) - f (p_{j})$ cancels all intermediate values and leaves $f (p_{m}) - f (p_{0}) = f (a + h) - f (a)$ .

Depends on

Dependency tree · 2 levels

Sources

J. Lebl, Basic Analysis I, §8.4