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CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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Zero training loss can fail under an operator shift

Statement refuted

Let X=Y=R and consider the one-step learned update Fθ(a,y)=θay on the declared domain of all scalar operator/data pairs. For a Borel point rule g and a joint input-target probability law P, define squared risk by RP(g)=∫∣g(a,y)−x†∣2 P(d((a,y),x†)). Let PS be the probability law assigning mass one to ((a,y),x†)=((1,1),1), and let Qt assign mass one to ((a,y),x†)=((t,t),1) for each shift t>1. Define the training risk LS(θ)=RPS(Fθ) and the test risk RQt(θ)=RQt(Fθ). The forward residual is rt=∣tFθ(t,t)−t∣. The assertion refuted is that some finite constant C bounds both rt and RQt(θ) for every t>1 whenever LS(θ)=0.

Facts & Assumptions

Given: A scalar one-step architecture with initialization x0=0, identity update block, trainable layer step θ, training sample (1,1,1), and test laws Qt for t>1.

[F1]

Unrolled regularization architecture: for one layer, the update applies its declared block to the data-consistency-corrected input, and the deployed output is the finite one-step composition.

[F2]

Learned inverse estimators and reconstruction targets: a point reconstruction rule is Borel, and a finite unrolled learned update uses a Borel initialization, Borel layer maps, and has a point-valued output.

[F3]

Inverse reconstruction risk under train and test laws: a deterministic Borel point rule has risk equal to its extended nonnegative loss integral under a joint input-target law.

[F4]

A continuous map has Borel preimages of Borel sets: a continuous map between topological spaces has Borel preimages of Borel sets.

Counterexample

technique · direct

Given: The data in the Statement refuted.

1.1F1F2F4givenconstructalgebra

Choose the initialization Iθ(a,y)=0, identity block Bθ0((a,y),v)=v, and layer parameter τ0=θ. By [F1], this is an admissible one-step architecture and Fθ(a,y)=θa(y−a⋅0)=θay. This polynomial map is continuous and hence Borel by [F4], so it satisfies the learned point-rule interface [F2].

2.1F3step 1.1givenalgebra

At the training point, Fθ(1,1)=θ. The law PS assigns mass one to that point, so [F3] gives LS(θ)=RPS(Fθ)=∣θ−1∣2. It is zero exactly at θ=1.

3.1step 2.1givenalgebra

For that zero-loss parameter and any t>1, F1(t,t)=t2, so rt=∣t t2−t∣=t(t2−1).

4.1F3step 2.1step 3.1algebra∎

The test law Qt assigns mass one to its displayed point, so [F3] gives RQt(1)=∣t2−1∣2=(t2−1)2. Both rt and RQt(1) tend to +∞ as t→+∞. For every finite C, choosing t sufficiently large violates at least one claimed bound (in fact both), while the training risk remains zero by step 2.1. Therefore the refuted assertion is false.

Remark

In the separate empirical InverseBench study, Figure 5(a) and §4.3 report that most tested plug-and-play diffusion methods on black-hole digit inputs exhibit a ring feature characteristic of the diffusion prior. Figure 5(b) shows inverse-scattering sources with nine cells while the prior was trained on images with one to six cells; §4.3 presents this case as one where the methods recover effectively when the source is close to the prior. These observations concern those methods and datasets only; they are not premises in the scalar proof above. The cited diffusion survey describes certain posterior-sampling procedures as asymptotically exact only in method-specific infinite-computation limits and notes that practical scaling remains unresolved; this sampling discussion also supplies no premise to the counterexample.

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