Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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  • 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.
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A unit VSC coefficient need not force Bregman convergence

Statement refuted

For X=Y=R with the usual norm and pairing, D(F)=R, F(x)=0, R(x)=max⁡{0,−x}, x†=0, y=0, ξ†=−1∈∂R(0), R0=1, p=2, and ψ(t)=t, the boundary source inequality with coefficient β1=1

−ξ†(x−x†)≤DRξ†(x,x†)+ψ ⁣(∣F(x)−F(x†)∣2)(R(x)≤R0),

holds for every x with R(x)≤R0, where DRξ†(x,0):=R(x)−R(0)−ξ†x. For every δ∈(0,1], set yδ=0, α(δ)=δ, and Jδ(x):=12∣F(x)−yδ∣2+α(δ)R(x). A global minimizer means a point xδ with Jδ(xδ)≤Jδ(z) for every z∈R. The false claim is that every family of such global minimizers satisfies DRξ†(xδ,0)=O(δ) as δ↓0, meaning DRξ†(xδ,0)≤Cδ for some fixed C and all sufficiently small δ>0.

Facts & Assumptions

Given: The scalar model and source inequality in the Statement.

[F1]

Banach-space convex subgradients and dual pairings: ξ∈∂R(x) exactly when R(z)≥R(x)+⟨ξ,z−x⟩ for every z.

[F2]

Bregman distances for convex regularizers: DRξ(x,x†)=R(x)−R(x†)−⟨ξ,x−x†⟩.

[F3]

Variational source conditions on a regularizer sublevel: the rate theorem's VSC requires a coefficient strictly below one; this item tests the same source inequality at its excluded boundary value one.

Counterexample

1.1givenF1F2F3algebra

The function R is finite and nonnegative, hence proper. For u,v∈R and t∈[0,1], R((1−t)u+tv)=max⁡{(1−t)0+t0,(1−t)(−u)+t(−v)}≤(1−t)max⁡{0,−u}+tmax⁡{0,−v}, so R is convex. For every z∈R, R(z)≥−z=R(0)+(−1)(z−0), so [F1] gives ξ†=−1∈∂R(0). By [F2], DR−1(x,0)=R(x)+x, which equals x for x≥0 and 0 for x<0. Since F is zero, the source inequality has left side x and right side DR−1(x,0); it holds for all x, hence on R(x)≤1, with boundary coefficient β1=1.

2.1givenstep 1.1algebra∎

For every δ>0, Jδ(x)=δR(x)≥0, and it equals zero for every x≥0. Thus xδ=1 is a global minimizer for every δ, but DR−1(1,0)=1. No fixed constant bounds 1 by Cδ for all sufficiently small δ, so the claimed rate fails.

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