Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A verified MGRIT error-matrix contraction controls error and residual

Statement

Let Atu∗=b be the fine-grid system from The all-at-once bidiagonal time operator and C/F points. By Eliminating F-points gives the exact coarse time operator, At is invertible. Let E be the exact error matrix for one fixed two-level cycle in A declared two-level MGRIT coarse-correction cycle. Choose a real-valued norm ∥⋅∥ on (Rd)N and define its induced operator norm by ∥T∥=sup⁡v≠0∥Tv∥/∥v∥. Assume ∥At∥ and ∥At−1∥ are finite for this norm. If a direct calculation or other proof establishes ∥E∥≤q<1, then after r≥0 cycles the error satisfies ∥e(r)∥≤qr∥e(0)∥, with the convention q0=1. The residual s(r)=b−Atu(r)=Ate(r) obeys ∥s(r)∥≤∥At∥qr∥e(0)∥, and for every error/residual pair, ∥s∥≤∥At∥∥e∥,∥e∥≤∥At−1∥∥s∥.

The contraction hypothesis is a property to prove for the declared operators, norm, and cycle. The MGRIT name alone supplies no such bound, and this theorem makes no multilevel convergence claim.

Facts & Assumptions

Given: A finite linear time system, the declared exact two-level cycle and a chosen product-space norm.

[F1]

A declared two-level MGRIT coarse-correction cycle: one cycle has error matrix E=Spost(I−PAc−1RAt)Spre.

[F2]

Eliminating F-points gives the exact coarse time operator: the fine-grid operator At is invertible.

[F3]

A norm on a real vector space, the induced metric, and the dictionary with the metric axioms: the chosen vector norm separates zero: ∥v∥=0 if and only if v=0.

[A1]

The directly calculated or proved induced norm satisfies ∥E∥≤q<1.

[A2]

The induced norms ∥At∥ and ∥At−1∥ are finite for the chosen product-space norm.

Proof

1.1givenF1algebra

By [F1], one cycle sends the current error to e(r+1)=Ee(r).

2.1givenF3A1step 1.1inductionalgebra

At r=0 the claimed inequality is equality. If the bound holds at r and e(r)=0, [F3] gives Ee(r)=0 and the next bound follows. If e(r)≠0, the induced-norm definition in the Statement and [A1] give ∥e(r+1)∥=∥Ee(r)∥≤∥E∥ ∥e(r)∥≤qr+1∥e(0)∥. Induction proves the error bound for every r≥0.

3.1givenF2F3A2step 2.1algebra

Since Atu∗=b, subtraction gives s(r)=At(u∗−u(r))=Ate(r). If e(r)=0, [F3] gives s(r)=0. Otherwise the induced-norm definition in the Statement and [A2] give ∥s(r)∥≤∥At∥∥e(r)∥, and step 2.1 yields the displayed residual estimate.

4.1givenF2F3A2algebra∎

For any error/residual pair, s=Ate and [F2] gives e=At−1s. If e=0, then s=0 and [F3] gives both inequalities. If e≠0, invertibility forces s≠0, so the induced-norm definition in the Statement and [A2] give both ∥s∥≤∥At∥∥e∥ and ∥e∥≤∥At−1∥∥s∥.

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