LemmaStatement: Literature-sourcedProof: 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.

Uniform error bound for Bernstein approximation of a Lipschitz function

Statement

Let $L \geq 0$ , let $f : [0, 1] \to R$ satisfy $∣ f (x) - f (y) ∣ \leq L ∣ x - y ∣$ , and let $N \geq 1$ be an integer. The Bernstein polynomial $B_{N} f (x) : = \sum_{j = 0}^{N} f (j / N) (\binom{N}{j}) x^{j} (1 - x)^{N - j}$ satisfies $\sup_{x \in [0, 1]} ∣ B_{N} f (x) - f (x) ∣ \leq \frac{L}{2 \sqrt{N}} .$

Statement references

Lorentz, Chapter 1: Bernstein polynomials and binomial moments

Sullivan, §5.1: Bernstein expectation representation

Facts & Assumptions

Given: A real number $L \geq 0$ , an $L$ -Lipschitz function $f : [0, 1] \to R$ , an integer $N \geq 1$ , and $x \in [0, 1]$ .

Proof

1.1givenconstructalgebra

Take independent Bernoulli variables $X_{1}, \dots, X_{N}$ with success probability $x$ , and put $S = \sum_{i} X_{i}$ . The binomial formula gives $B_{N} f (x) = E f (S / N)$ , so the Lipschitz hypothesis yields $∣ B_{N} f (x) - f (x) ∣ \leq L E ∣ S / N - x ∣$ .

2.1step 1.1algebra∎

Independence gives $E S = \sum_{i = 1}^{N} E X_{i} = N x$ and $Var (S) = \sum_{i = 1}^{N} Var (X_{i}) = N x (1 - x)$ , because each Bernoulli variable has variance $x (1 - x)$ . Therefore Cauchy–Schwarz yields $E ∣ S / N - x ∣ \leq \sqrt{E (S / N - x)^{2}} = \sqrt{x (1 - x) / N} \leq 1 / (2 \sqrt{N})$ . The estimate holds for every $x$ , proving the uniform bound.

Used by

A single high-degree gate defeats the stated small-k separation Counterexample
The unrestricted all-parameter depth separation is false False statement
Depth separation with explicit semi-algebraic crossing budgets Theorem

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

George G. Lorentz, Bernstein Polynomials, Chapter 1 (standard reference, not scraped)
Michael J. Sullivan, Deep Learning Theory, §5.1 (standard reference, not scraped)