Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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.

The joint relaxed objective controls Bregman and range errors

Statement

Let 1≤p<∞. Let X,Y,Z be real Banach spaces, X0⊆X a closed linear subspace, D⊆X a domain containing X0 and U, and x0∈U∩X0. Let F:D→Y, K∈L(Z,Y), and r:D→Z satisfy F(x)−F(x0)=Kr(x) on D, with r∣U:U→U~ bijective and r Lipschitz with constant Lr on D. Let P:X→X be a bounded linear projection such that P∣X0=0, (I−P)(X)⊆X0, and P(x0)=P(x†)=0 for an exact solution x†∈U∩X0; put y:=F(x†). Assume the projected points (I−P)r−1(r^) belong to D∩X0 for all admitted r^∈U~.

Let Rˇ:X0→[0,+∞] be proper and convex, choose ξ†∈∂Rˇ(x†)⊆X0∗, and suppose that for some R0≥Rˇ(x†), b∈[0,1), and continuous nondecreasing concave ψ:[0,∞)→[0,∞) with ψ(0)=0, the variational source inequality

−⟨ξ†,xˇ−x†⟩≤bDRˇξ†(xˇ,x†)+ψ(∥F(xˇ)−F(x†)∥Yp)

holds whenever xˇ∈D∩X0 and Rˇ(xˇ)≤R0, where DRˇξ†(xˇ,x†):=Rˇ(xˇ)−Rˇ(x†)−⟨ξ†,xˇ−x†⟩. Set RP=0 in the joint objective. Let Q:U~×U~→[0,∞] satisfy Q(a,b)=Q(b,a) and Q(a,b)=0 exactly when a=b, and assume there is a constant CQ≥0 such that

∥P(r−1(a))−P(r−1(b))∥X≤CQQ(a,b)(a,b∈U~).

For δ>0, let yδ∈Y satisfy ∥yδ−y∥Y≤δ. Let α,β>0, Cη≥0, and let (r^δ,xδ)∈U~×U be an ηδ-minimizer of the joint objective, meaning

Jα,βδ(r^δ,xδ)≤Jα,βδ(r^,x)+ηδ((r^,x)∈U~×U),0≤ηδ≤Cηδp.

Assume xˇδ:=(I−P)r−1(r^δ) lies in the stated VSC sublevel. Define

sδ:=21−p∥K(r^δ−r(x†))∥Yp+βQ(r(xδ),r^δ)p+∥Pxδ∥Xp,eδ:=DRˇξ†(xˇδ,x†),

and

C∗:=22p−2max⁡{1,∥K∥pLrp,∥K∥pLrpCQpβ},dδ:=(2+Cη)δp.

Then

sδ+(1−b)αeδ≤dδ+αψ(C∗sδ).

Facts & Assumptions

Given: The Banach spaces, range factorization, projection, regularizer, VSC, compatibility inequality, exact datum, noise bound, and approximate minimizer in the Statement.

[A1]

The candidate projected point xˇδ belongs to D∩X0 and satisfies Rˇ(xˇδ)≤R0, so the VSC applies to it.

[A2]

The noisy datum and approximate-minimizer tolerance satisfy the bounds in the Statement.

[A3]

The projection-coupling bound holds for all range variables in the Statement.

[F1]

Differential range invariance for Banach-space maps: F(x)−F(x0)=Kr(x) for every x in the range-factorization domain.

[F2]

Joint and frozen-Newton objectives for range-invariant relaxations: with RP=0, the joint objective is ∥Kr^+F(x0)−yδ∥Yp+αRˇ((I−P)r−1(r^))+βQ(r(x),r^)p+∥Px∥Xp.

[F3]

Banach-space convex subgradients and dual pairings: ξ†∈∂Rˇ(x†) means Rˇ(z)≥Rˇ(x†)+⟨ξ†,z−x†⟩ for every z∈X0.

[F4]

Bregman distances for convex regularizers: DRˇξ†(z,x†)=Rˇ(z)−Rˇ(x†)−⟨ξ†,z−x†⟩.

[F5]

Variational source conditions on a regularizer sublevel: on the named regularizer sublevel, −⟨ξ†,z−x†⟩≤bDRˇξ†(z,x†)+ψ(∥F(z)−F(x†)∥Yp).

Proof

1.1A2F1F2givenalgebra

Compare the approximate-minimum inequality with (r^,x)=(r(x†),x†). By [F1], Kr(x†)+F(x0)=F(x†)=y, while Q(r(x†),r(x†))=0 and Px†=0. Since r−1(r(x†))=x† and P(x†)=0, the comparison regularizer is Rˇ(x†), so its objective is at most δp+αRˇ(x†). Writing the estimated data residual as K(r^δ−r(x†))+(y−yδ), the triangle inequality gives ∥K(r^δ−r(x†))∥≤∥Kr^δ+F(x0)−yδ∥+δ. For p=1 the power inequality is equality; for p>1, f(t)=tp has f′′(t)=p(p−1)tp−2≥0 for t>0 and is continuous at zero, so it is convex on [0,∞); midpoint convexity gives (a+b)p≤2p−1(ap+bp) for a,b≥0. Rearranging yields 21−p∥K(r^δ−r(x†))∥p≤∥Kr^δ+F(x0)−yδ∥p+δp. Combining this with the comparison and ηδ≤Cηδp gives sδ+α(Rˇ(xˇδ)−Rˇ(x†))≤dδ.

2.1A1F3F4F5step 1.1algebra

By [F3], ξ† is a subgradient at x†, and [F4] gives Rˇ(xˇδ)−Rˇ(x†)=eδ+⟨ξ†,xˇδ−x†⟩. The VSC [F5] therefore implies Rˇ(xˇδ)−Rˇ(x†)≥(1−b)eδ−ψ(∥F(xˇδ)−F(x†)∥p). Substituting this lower bound in step 1.1 yields sδ+(1−b)αeδ≤dδ+αψ(∥F(xˇδ)−F(x†)∥p).

2.2A3F1step 1.1algebra

Put uδ=r−1(r^δ), so xˇδ=(I−P)uδ. Since r−1(r(xδ))=xδ, assumption [A3] and the triangle inequality give ∥Puδ∥≤∥Pxδ∥+CQQ(r(xδ),r^δ). Lipschitz continuity of r gives ∥r(xˇδ)−r^δ∥=∥r((I−P)uδ)−r(uδ)∥≤Lr∥Puδ∥. Applying [F1] at xˇδ and x†, then applying (a+b)p≤2p−1(ap+bp) twice, yields ∥F(xˇδ)−F(x†)∥p≤2p−1∥K(r^δ−r(x†))∥p+22p−2∥K∥pLrp(∥Pxδ∥p+CQpQ(r(xδ),r^δ)p)≤C∗sδ. Monotonicity of ψ gives ψ(∥F(xˇδ)−F(x†)∥p)≤ψ(C∗sδ).

3.1step 2.1step 2.2algebra∎

Substitute the last bound from step 2.2 into the inequality from step 2.1. This gives sδ+(1−b)αeδ≤dδ+αψ(C∗sδ), as claimed.

Depends on

Used by

Dependency tree · two levels

11 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