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ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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Exact truncated Landweber and a learned finite-depth update

Example

Assume Countable Choice. Let X=Y=R, let Ax=ax for a fixed a>0, and let the exact datum be y=ax† with x†≠0 and noise bound δ=0. Choose a discrepancy factor τ>1, N≥1, and 0<ωa2<1. The exact Landweber prefix starts at x0=0 and uses xk+1=xk+ωa(y−axk) through its finite budget N. For a learned finite-depth architecture with zero initialization, identity update blocks, and layer parameters ω~0,…,ω~N−1, the updates are x~k+1=x~k+ω~ka(y−ax~k). If every ω~k=ω, the two depth-N outputs coincide; otherwise the learned output is the finite composition determined by its own layer parameters.

Facts & Assumptions

Given: Countable Choice, a>0, x†≠0, y=ax†, noise bound δ=0, discrepancy factor τ>1, N≥1, 0<ωa2<1, and the stated scalar exact and learned recurrences.

[F1]

Landweber iteration with discrepancy and budget statuses: under Countable Choice, the path starts at x0=0, uses xk+1=xk+ωA∗(y−Axk), and, with supplied discrepancy factor τ>1, the finite procedure returns at its first residual test ∥y−Axk∥≤τδ or at budget N.

[F2]

Unrolled regularization architecture: the deployed unrolled map is a finite composition of its declared update maps with fixed layer parameters.

[F3]

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

[A1]

The Axiom of Countable Choice (ACω): the Countable Choice axiom is assumed for the published Landweber interface; the scalar recurrence calculation below makes no additional choice.

Verification

technique · direct
1.1A1F1F2F3givenalgebra

The scalar map A:R→R, x↦ax, is nonzero and compact because a>0 and the space is finite-dimensional. For every x, ∥Ax∥2=a∣x∣, so ∥A∥=a. For every u,v∈R, ⟨Au,v⟩=auv=⟨u,av⟩, hence A∗ is also multiplication by a. With [A1], 0<ωa2<1 implies the supplier's condition 0<ω<2/∥A∥2. The initialization and scalar update formulas are continuous, hence Borel by [F3], so they satisfy [F2]'s estimator-domain condition. Together with the supplied τ>1 and noise bound δ=0, these verify every input condition of [F1], which gives the exact recurrence and its finite-budget stopping rule for this instance.

1.2F1givenalgebra

Substituting y=ax† gives xk+1−x†=(1−ωa2)(xk−x†). Starting from x0=0, induction gives xk=x†[1−(1−ωa2)k] for every 0≤k≤N, in particular xN=x†[1−(1−ωa2)N]. Hence ∣y−axk∣=a∣x†∣(1−ωa2)k>0 for every such finite k. Since τδ=0, the discrepancy test fails even at k=N, so no earlier discrepancy stop occurs and the procedure returns this value at budget exhaustion.

2.1F2F3givenstep 1.2algebra∎

For the learned architecture, x~0=0 and its displayed recurrence give x~k+1−x†=(1−ω~ka2)(x~k−x†). Starting with x~0−x†=−x†, repeated substitution yields x~N−x†=−x†∏k=0N−1(1−ω~ka2), or x~N=x†[1−∏k=0N−1(1−ω~ka2)]. The fixed-parameter updates are continuous and Borel by [F3], so this is an admissible finite point map. If all layer parameters equal ω, the formula is the exact Landweber value in step 1.2; for other parameters it is the finite composition specified by [F2].

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