Alphabeta Math
TheoremStatement: 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.

Proximal bundle accumulation under vanishing aggregate-model error

Statement

Let F:Rn→(−∞,+∞] be proper, closed, and convex. Suppose C=dom⁡F is nonempty, closed, and convex, and F∣C is continuous. Consider an infinite full-memory continuation of Finite-memory-free proximal bundle method with serious and null steps with fixed t>0 and ms∈(0,1), centers xk∈C, finite affine models Mk≤F on C touching their centers (Mk(xk)=F(xk)), and exact candidates zk=arg⁡min⁡y∈C(Mk(y)+∥y−xk∥222t). Assume every predicted decrease Pk=F(xk)−Mk(zk) is positive, each trial uses the stated serious/null acceptance test, the initial sublevel set K={x∈C:F(x)≤F(x0)} is compact, and the aggregate-model errors ek=F(zk)−Mk(zk) satisfy ek→0. Then every accumulation point, meaning a limit of a subsequence in the Euclidean metric, is a global minimizer of F. At least one accumulation point exists. The vanishing-error premise is an additional assumption; full-memory cutting planes alone are not asserted to imply it.

Facts & Assumptions

[F1]

Each candidate is unique, and the exact proximal point pk=prox⁡tF(xk) exists, is unique, and the proximal map is nonexpansive (The proximal bundle model subproblem has a unique minimizer, Proximal mapping, Existence, resolvent characterization, and firm nonexpansiveness of the proximal map).

[F2]

The model error is nonnegative and ∥zk−pk∥22≤4tek (Vanishing aggregate-model error controls the proximal candidate).

[F3]

If 0∈∂F(x), then F(y)≥F(x) for every y by the extended-real subgradient definition (Subdifferentials of proper extended-real convex functions).

Proof

1.1givenF4algebra

A serious step has F(xk)−F(zk)≥msPk>0; a null step leaves the center fixed. Thus centers remain in the initial sublevel set K, and F(xk) is nonincreasing and bounded below there by [F4]. The decreases dk=F(xk)−F(xk+1)≥0 have bounded partial sums, since ∑k<qdk=F(x0)−F(xq). If dk failed to tend to zero, some positive ϵ would be exceeded by infinitely many terms, making these partial sums unbounded; hence dk→0.

1.2F1F2F4givenalgebra

Let xkj→xˉ be any accumulation subsequence, which exists for at least one subsequence by [F4]. For any chosen convergent subsequence, compactness gives a further subsequence converging to a point of K; the uniqueness-of-limits fact in [F4] identifies that point with xˉ, so xˉ∈K. Applying compactness to the whole center sequence also supplies at least one accumulation subsequence. By nonexpansiveness in [F1], pkj=prox⁡tF(xkj)→p=prox⁡tF(xˉ). By [F2] and ekj→0, ∥zkj−pkj∥2→0, so zkj→p. Since F∣C is continuous and all these points lie in C, F(xkj)→F(xˉ) and F(zkj)→F(p).

2.1F1F3step 1.2algebraassume-contra

Suppose for contradiction that xˉ is not a global minimizer. Then p≠xˉ: otherwise the resolvent equivalence in [F1] gives 0=(xˉ−p)/t∈∂F(xˉ), and [F3] would make xˉ globally minimizing. By proximal minimality, F(p)+∥p−xˉ∥22/(2t)≤F(xˉ); therefore δ:=F(xˉ)−F(p)>0.

3.1givenstep 1.1step 1.2step 2.1algebradischarge-contradiction∎

Along the chosen subsequence, Dj:=F(xkj)−F(zkj)→δ by step 1.2, while Pkj=F(xkj)−Mkj(zkj)=Dj+ekj. Hence Dj−msPkj=(1−ms)Dj−msekj→(1−ms)δ>0. For all sufficiently large j the step is serious and dkj=Dj≥δ/2, contradicting dk→0 from step 1.1. Thus every accumulation point is a global minimizer.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

92 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