Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-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.

Quasi-orthogonality for constant convection on a fine initial mesh

Statement

Assume Countable Choice, as in the zero-boundary Sobolev convention.

Let TH≤Th be nested conforming meshes and let UH,Uh be exact P1 Galerkin solutions for the constant-convection model of Residual indicators and data oscillation for P1 AFEM. Write ej=u−Uj and d=Uh−UH. For b=0, ∥∇eh∥22=∥∇eH∥22−∥∇d∥22. For constant b, assume the explicit dual-pickup bound ∥eh∥L2≤CDh0s∥∇eh∥L2 for the nested sequence, with s>0, where h0=max⁡K∈T0diam⁡K, and put μ=CDh0s∣b∣<1. Then ∥∇eh∥22≤(1−μ)−1∥∇eH∥22−∥∇d∥22. The dual-pickup bound is an additional hypothesis. It is not asserted for every polygon or mesh.

Facts & Assumptions

[A1]

Countable Choice is assumed in the cited Sobolev and reference-simplex suppliers (The Axiom of Countable Choice (ACω)).

Given: Countable Choice, nested conforming P1 spaces, exact Galerkin solutions, constant b, and in the nonsymmetric case the stated dual-pickup bound.

[F1]

Galerkin orthogonality says a(u−Uh,vh)=0 for every vh∈V(Th); nestedness puts d=Uh−UH in the fine space (Galerkin orthogonality for conforming approximations).

[F2]

H01 is the closure of compactly supported smooth functions in H1; constant-vector integration by parts on those functions extends by continuity to this closure (Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms).

Proof

1.1A1F1givenalgebra

If b=0, [F1] with vh=d gives (∇eh,∇d)=0. Since eH=eh+d, expanding the squared norm yields ∥∇eH∥22=∥∇eh∥22+∥∇d∥22. Rearrangement is the symmetric identity.

1.2A1F1F2givenalgebra

For constant b, [F1] gives (∇eh,∇d)+(b⋅∇eh,d)=0. By [F2], (b⋅∇eh,d)=−(b⋅∇d,eh). Expanding eH=eh+d therefore gives ∥∇eH∥22=∥∇eh∥22+∥∇d∥22+2(b⋅∇d,eh). The sign in this identity follows directly from the preceding orthogonality equation.

2.1A1step 1.2givenalgebra∎

By the assumed dual-pickup estimate and Cauchy–Schwarz, ∣2(b⋅∇d,eh)∣≤2∣b∣∥∇d∥2∥eh∥2≤2μ∥∇d∥2∥∇eh∥2≤μ(∥∇d∥22+∥∇eh∥22). Using this in step 1.2 gives (1−μ)∥∇eh∥22≤∥∇eH∥22−(1−μ)∥∇d∥22. Divide by 1−μ>0 to obtain the stated estimate.

Depends on

Used by

Dependency tree · two levels

25 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