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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.
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Unrolled reconstruction and finite-iteration limits

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Finite unrolled domains and outputs are Borel

Statement

Let n,m≥1, X=Rn, and let Ω be a Borel subset of Rm×n×Rm with its relative Borel sigma-algebra. Equip Ω×X with the finite product sigma-algebra B(Ω)⊗B(X). Fix an integer N≥0, real parameters τ0,…,τN−1, a Borel initialization I:Ω→X, Borel sets Dk⊆Ω×X, and Borel maps Bk:Dk→X for 0≤k<N, where each Dk carries its relative product sigma-algebra. For p=(A,y)∈Ω and x∈X, define c(p,x)=AT(y−Ax). Set E0=Ω and x0=I; recursively, for 0≤k<N, set

Ek+1:={p∈Ek:(p,xk(p)+τkc(p,xk(p)))∈Dk},xk+1(p):=Bk(p,xk(p)+τkc(p,xk(p)))(p∈Ek+1).

Then every Ek is a Borel subset of Ω and every xk:Ek→X is Borel. In particular, EN is exactly the set of inputs where all N updates are defined, and the finite output map xN:EN→X is Borel.

Facts & Assumptions

Given: The finite-dimensional Borel spaces, Borel update domains and maps, Borel initialization, fixed finite layer parameters, and the recursive definitions in the Statement.

[F1]

The product sigma-algebra and its finite iterates: the product sigma-algebra is generated by measurable rectangles.

[F2]

A measurable function between measurable spaces: a map is measurable when the preimage of each measurable codomain set is measurable.

[F3]

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

[F4]

Composition with a Borel measurable outer map preserves measurability: the composition of measurable maps between measurable spaces is measurable.

Verification

technique · induction
1.1F1F3givenalgebra

The spaces Ω and X are second-countable subspaces of finite-dimensional Euclidean spaces, so rectangles from countable bases form a countable base for Ω×X. Every open set is therefore a countable union of open rectangles and belongs to B(Ω)⊗B(X) by [F1]. Conversely, for Borel C⊆Ω and F⊆X, C×F=π1−1(C)∩π2−1(F) is Borel in the product topology because the coordinate projections are continuous and [F3] applies. Since such rectangles generate the product sigma-algebra by [F1], the product sigma-algebra is the Borel sigma-algebra of the product topology.

1.2givenbase

The base domain E0=Ω is Borel in itself, and the initial map x0=I is Borel by the Given clause.

1.3F1F2givenihassume-hypalgebra

Fix k<N and assume as the induction hypothesis that Ek is Borel and xk:Ek→X is Borel. Define ιk:Ek→Ω×X by ιk(p)=(p,xk(p)). For every measurable rectangle C×F, [F2] and the induction hypothesis give ιk−1(C×F)=(Ek∩C)∩xk−1(F) Borel in Ek. The rectangles generate the target sigma-algebra by [F1], so preimages preserve the generated sigma-algebra and ιk is Borel.

2.1F3F4step 1.1step 1.3algebra

Define Γk(A,y,x)=(A,y,x+τkAT(y−Ax)). Each coordinate of its last component is a polynomial in the matrix, data, and state coordinates, so Γk is continuous. By [F3] it is Borel; hence ψk:=Γk∘ιk is Borel by [F4].

3.1F2step 1.2step 2.1algebra

The next domain is Ek+1=ψk−1(Dk). Since Dk is Borel and ψk is Borel, [F2] makes Ek+1 Borel in Ek; because Ek is Borel in Ω, Ek+1 is Borel in Ω.

4.1F2F4step 2.1step 3.1given

The restriction ψk∣Ek+1:Ek+1→Dk is Borel: the preimage of each relative-measurable set Dk∩H is Ek+1∩ψk−1(H), which is Borel. Since Bk:Dk→X is Borel, [F4] makes xk+1=Bk∘(ψk∣Ek+1) Borel.

5.1step 1.2step 1.3step 2.1step 3.1step 4.1discharge-induction: step 1.3∎

Step 1.2 supplies the base case, and steps 1.3, 2.1, 3.1, and 4.1 prove the implication from any Borel pair (Ek,xk) to (Ek+1,xk+1). Finite induction gives the claim for every k≤N. By its recursive definition, EN contains exactly the inputs at which each successive update is defined, and step 4.1 gives Borel measurability of the final output.

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-10-22 (gpt-6.1-sol)Open item page →

Unrolled regularization architecture

Definition

Let n,m≥1 be integers, X=Rn, and Y=Rm with their standard Euclidean inner products and Borel sigma-algebras. Borel sets are members of the sigma-algebra generated by open sets, and a map is Borel if preimages of Borel sets are Borel. An instance is a pair (A,y) in a declared Borel set D⊆Rm×n×Y, where A is the measurement matrix, y is the observed data, and AT is the transpose matrix. For fixed A, the forward map is FA:X→Y, FA(x)=Ax, and the reconstruction target is the signal x itself. For x∈X, define the data-consistency correction by c(A,y,x):=AT(y−Ax).

For an integer depth N≥1, an unrolled regularization architecture consists of a Borel initialization map Iθ:D→X, parameters θ=(θ0,…,θN−1) and real step parameters τ0,…,τN−1 held fixed at deployment, and Borel update maps Bθk:Dk→X with Dk⊆D×X Borel. Set x0=Iθ(A,y) and, for k=0,…,N−1, set

xk+1=Bθk((A,y),xk+τkc(A,y,xk)).

For fixed parameters and depth, let Dθ,N be the Borel subset of D for which the initialization and all N recursive evaluations are defined. For (A,y)∈Dθ,N, write UθkA,y(x):=Bθk((A,y),x+τkc(A,y,x)). The deployed map is

Φθ,N(A,y)=(UθN−1A,y∘⋯∘Uθ0A,y)(Iθ(A,y)).

Admissibility requires this finite map to be Borel; it is a point reconstruction rule from the input pair (A,y) to the signal target in X, not a probability-distribution-valued rule. The parameters may have been selected by any training procedure, and the architecture supplies no training-law specification or convergence, data-residual, or test-risk guarantee.

An exact proximal block is permitted when, for each fixed (A,y), Bθk((A,y),v)=prox⁡αkg(v) for some αk>0, where g=gθk,(A,y):Rn→(−∞,+∞] has nonempty effective domain {u:g(u)<+∞}, satisfies g((1−λ)u+λv)≤(1−λ)g(u)+λg(v) for u,v∈Rn and 0≤λ≤1, and has closed epigraph {(u,s):g(u)≤s}. Here prox⁡αg(v) denotes the unique global minimizer u∈Rn of g(u)+∥u−v∥22/(2α). The proximal-map existence theorem supplies uniqueness under precisely these hypotheses. A general learned block is not assumed to be a proximal map or to minimize any objective.

For differentiation, fix an admissible data pair (A,y) and regard the desired trainable parameter coordinates, including any step parameters to be differentiated, as a finite-dimensional input vector η in an open parameter set; hold the other values fixed. At the differentiation point η0, require an open neighborhood V of η0 on which the actual initialization and all N deployed updates are defined for this fixed (A,y). Require the graph below to evaluate those same maps for every η∈V; a smooth extension beyond a declared Borel domain alone does not meet this requirement. Without such a neighborhood, no derivative of the deployed map is asserted. Construct a finite straight-line graph with input η, initialization node x0=Iθ(A,y), and, for each k=0,…,N−1, distinct nodes

vk=xk+τkAT(y−Axk),xk+1=Bθk((A,y),vk).

Its output nodes are the coordinates of xN. Thus each stage uses its own map Bθk and step parameter τk; no common update map or shared layer parameter is required. Represent each initialization and update by its constituent primitive nodes or by a supplied complete deterministic C1 primitive with its derivative rule; an opaque call without that map and rule is not an admissible graph primitive. Require every graph primitive to be continuously differentiable on an open neighborhood of its visited inputs. For a piecewise-C1 primitive, each visited input must lie strictly inside a declared branch region, with that branch label fixed on an open neighborhood of the input; at a branch boundary or ambiguous branch selection the automatic-differentiation procedure returns no derivative. Apply forward-mode automatic differentiation to each scalar output-coordinate graph in the desired parameter direction. The forward-mode correctness theorem then identifies the returned tangent with the directional derivative of that finite output; coordinate directions give the Jacobian. When all stages instead repeat one C1 map G(z,θ) with a common parameter, the finite fixed-point-iteration procedure is the corresponding specialization. This differentiates only the deployed finite map and adds no assertion about an infinite iteration.

For N≥3, the diagram compresses the intervening states and updates into a single arrow; when N=3, that arrow is the single update Uθ1A,y. For N=1 and N=2, the recurrence above gives the corresponding one- and two-update chains. At one fixed admissible (A,y), every arrow denotes finite function application, not a limit.

(A;y)x0=Iµ(A;y)x1=UA;yµ0(x0)xN¡1xN=©µ;N(A;y)UA;yµ0UA;yµN¡2±¢¢¢±UA;yµ1UA;yµN¡1

Definition references

PropositionStatement: AI-adaptedProof: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Finite unrolling alone gives no regularization convergence or risk guarantee

Statement

Let N≥1 and let Φθ,N be the point-valued map obtained by an initialization followed by N defined, fixed-parameter update maps on an admissible input (A,y). This finite definition does not imply that the iterates of an underlying infinite recurrence converge: even the repeated scalar update T(x)=2x is defined at every finite depth from x0=1, while Tk(x0)=2k diverges.

For the risk comparison, fix a nonempty Borel input space E⊆Dθ,N for every parameter θ under consideration, and restrict each deployed map to E. For squared reconstruction loss ℓ(t,a)=∥a−t∥22 on the signal space, let S={(Ai,yi,xi†)}i=1s be a nonempty finite training set with (Ai,yi)∈E and xi†∈X, s≥1, and let PS=s−1∑i=1sδ((Ai,yi),xi†) be its empirical joint law. The training-law risk is LS(θ)=RPS(Φθ,N), where RP(g) is the expected loss of the Borel point rule g under the joint input-target law P. At a shifted input (A,y)∈E, the forward residual is ∥AΦθ,N(A,y)−y∥2, and for a test probability law Q on E×X the test risk is RQ(θ)=RQ(Φθ,N) when this expectation is defined. No bound on either quantity follows from a bound on LS alone: for every M>0 there is a one-sample scalar training law with zero training-law risk and a shifted scalar operator with forward residual and squared test risk both greater than M.

For a recurrence zk+1=G(zk,θ), a fixed point at parameter θ is a state z∗(θ) satisfying G(z∗(θ),θ)=z∗(θ). If the initialization and update maps are continuously differentiable in their state and trainable-parameter variables on open neighborhoods of their visited inputs, automatic differentiation through the deployed graph computes the derivative of that finite map. For a piecewise-C1 primitive, assume each visited input lies strictly inside one declared branch, with its branch label fixed on an open neighborhood; at a branch boundary or ambiguous branch selection no derivative is asserted by the automatic-differentiation procedure. The derivative is that of the finite map and is not replaced by the derivative of a limiting fixed point. In particular, for G(z,θ)=qz+(1−q)θ with 0<q<1, z0(θ)=0, and θ0=0, each finite iterate has zero fixed-point residual while its parameter derivative differs from the exact fixed-point derivative.

Facts & Assumptions

Given: A depth-N architecture with all N updates defined at the selected input; for the derivative clause, a continuously differentiable initialization and recurrence in state and parameter variables on open neighborhoods and a finite number K of defined iterates, with each piecewise-C1 primitive evaluated strictly inside one declared branch whose label is fixed on an open neighborhood of every visited input; for the risk clause, a common nonempty Borel input space E⊆Dθ,N for every compared parameter, training inputs in E with signal targets in X, shifted inputs in E, and test laws on E×X.

[F1]

Unrolled regularization architecture: the deployed map is the finite composition of its initialization and the listed update maps.

[F2]

Differentiate a finite fixed-point iteration as a finite composed map: the procedure differentiates the specified finite K-step map and does not substitute the derivative of an exact fixed point.

[F3]

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

[F4]

Inverse reconstruction risk under train and test laws: a point rule's risk under a joint input-target law is the extended nonnegative loss integral, and distinct train and test laws impose no inequality between risks.

[F5]

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

[F6]

Forward mode computes the exact directional derivative of the real-arithmetic graph: forward mode returns the directional derivative of the finite scalar-output graph on a fixed strict branch.

Proof

1.1F1F3givenalgebra

Fix an admissible (A,y). By the definition of the update maps, x1=Uθ0A,y(Iθ(A,y)). If xk=Uθk−1A,y∘⋯∘Uθ0A,y(Iθ(A,y)), then applying the next defined update gives xk+1=UθkA,y∘Uθk−1A,y∘⋯∘Uθ0A,y(Iθ(A,y)). Induction through k=N−1 yields the stated finite composition. Its Borel point-rule convention is the interface in [F3].

1.2givenconstructalgebra

For the repeated scalar update T(x)=2x, induction gives T0(1)=1 and Tk+1(1)=2Tk(1)=2k+1 whenever Tk(1)=2k. Thus every finite prefix is defined, but Tk(1)=2k→+∞ as k→∞. Finite evaluability alone therefore does not imply convergence of an infinite continuation.

1.3F2F6given

Under the differentiability and strict-branch assumptions in the Given clause, [F2] forms, for each output coordinate, the finite graph of the K-step recurrence and calls forward mode in each parameter-coordinate direction. By [F6], every returned tangent is the corresponding directional derivative of that finite scalar-output graph. Assembling these entries gives the Jacobian of the finite recurrence, not the derivative of a limiting fixed point.

1.4givenalgebra

To check the zero-residual boundary, fix q∈(0,1) and set G(z,θ)=qz+(1−q)θ, θ0=0, and z0(θ)=0. At θ0, the recurrence gives zk=0 for every finite k, so the fixed-point residual zk−G(zk,θ0) is zero. The exact fixed point satisfies z∗(θ)=θ, and therefore dz∗dθ(0)=1.

1.5F1givenconstructalgebra

Use the scalar architecture with declared domain containing (A,y)=(1,1) and every (t,t) for t>1, zero initialization, identity update block, and one layer parameter τ0=θ. Choose the deployed parameter θ=1. Its one-layer formula is Φ1,1(a,y)=ay, so Φ1,1(1,1)=1.

2.1F3F4F5givenstep 1.5algebra

The map (a,y)↦ay is continuous and hence Borel by [F5], so it is a point reconstruction rule by [F3]. On the singleton training law PS=δ((1,1),1), [F4] gives LS(1)=RPS(Φ1,1)=∣1−1∣2=0.

2.2step 1.4algebra

Differentiating the finite recurrence gives e0=0 and ek+1=qek+(1−q). Induction yields eK=1−qK for each K≥1. Since 0<qK<1, eK≠1 although the finite residual in step 1.4 is zero. Thus even contraction and zero finite residual do not identify the finite-map derivative with the fixed-point derivative.

3.1F4step 1.5step 2.1algebra∎

For any shifted scalar operator A=t>1 with y=t and x†=1, the same learned parameter gives Φ1,1(t,t)=t2. Its forward residual is ∣tΦ1,1(t,t)−t∣=t(t2−1), and [F4] gives squared risk (t2−1)2 under the test law Qt=δ((t,t),1). Both tend to +∞ as t→+∞, while the training-law risk from step 2.1 remains zero. Given any M>0, choosing t sufficiently large makes both quantities exceed M, proving the claimed absence of a uniform residual or test-risk bound from training risk alone.

4 · Examples, counterexamples and false statements

None yet.

Sources