ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge 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.

Uniformly bounded cell reconstructions need not have a strong local limit

Example

For $k \geq 1$ , let $h_{k} = 2^{- k}$ and define the bounded cellwise functions $V_{k} (x, t) = {\begin{cases} (- 1)^{⌊ x / h_{k} ⌋}, & 0 \leq x < 1, \\ 0, & x \notin [0, 1), \end{cases} 0 < t < 1.$ Then $∣ V_{k} ∣ \leq 1$ , but $(V_{k})$ has no strongly $L_{l o c}^{1}$ -convergent subsequence. These are a bounded sequence of cell reconstructions illustrating why a compactness premise is explicit in the weak-limit theorem; they are not asserted to arise from one fixed initial datum or from a stable update.

Facts & Assumptions

Given: The functions $V_{k}$ in the Statement and the strong-convergence hypothesis of Conservative finite-volume strong-limit theorem.

[F1]

The weak-limit theorem assumes strong $L_{l o c}^{1}$ convergence of the reconstructions; a uniform bound is not stated as a replacement for it. (Conservative finite-volume strong-limit theorem)

Verification

1.1given

Use the compact rectangle $K = [0, 1] \times [1 / 4, 3 / 4]$ inside the space-time domain. Each $V_{k}$ takes only the values $- 1, 1$ on $[0, 1) \times (0, 1)$ and zero outside, so $∣ V_{k} ∣ \leq 1$ and $∥ V_{k} ∥_{L^{1} (K)} = 1 / 2$ .

1.2givenalgebra

If $k < ℓ$ , every $k$ -cell in $[0, 1)$ is the union of $2^{ℓ - k}$ finer $ℓ$ -cells, an even number. The signs of $V_{ℓ}$ alternate on those finer cells, so their integral on each coarse cell is zero; since $V_{k}$ is constant on that coarse cell, $\int_{K} V_{k} V_{ℓ} = 0$ .

2.1step 1.1step 1.2algebra

For signs $a, b \in {- 1, 1}$ , $∣ a - b ∣ = 1 - a b$ . Thus for $k \neq ℓ$ , $∥ V_{k} - V_{ℓ} ∥_{L^{1} (K)} = 1 / 2 - \int_{K} V_{k} V_{ℓ} = 1 / 2.$ Every subsequence is therefore non-Cauchy in $L^{1} (K)$ , so no subsequence can converge strongly in $L_{l o c}^{1}$ .

3.1F1step 2.1∎

This example supplies only a bounded sequence without strong compactness; it makes no finite-volume recurrence claim. Therefore the convergence premise in [F1] remains a separate hypothesis.

Depends on

Conservative finite-volume strong-limit theorem

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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