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ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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Logarithmic index functions give log-rate Bregman convergence

Example

Let X,Y be real Banach spaces, D(F)⊆X, F:D(F)→Y, and let R:X→(−∞,+∞] be proper and convex. Choose x†∈D(F)∩dom⁡R, set y=F(x†), and choose ξ†∈∂R(x†). Fix 1≤p<∞, ν>0, 0<t0<e−(ν+1), 0≤β<1, and R0>R(x†). Define ψ(0)=0, ψ(t)=(−log⁡t)−ν for 0<t<t0, and ψ(t)=(−log⁡t0)−ν for t≥t0. Suppose

−⟨ξ†,x−x†⟩X∗,X≤βDRξ†(x,x†)+ψ ⁣(∥F(x)−F(x†)∥Yp)(x∈LR0:={x∈D(F):R(x)≤R0}),

where DRξ†(x,x†):=R(x)−R(x†)−⟨ξ†,x−x†⟩X∗,X. For data yδ∈Y with ∥yδ−y∥Y≤δ, let 0≤εδ≤Cεδp and suppose xαδ,ε∈D(F) satisfies

Jαδ(xαδ,ε)≤inf⁡z∈D(F)Jαδ(z)+εδ,Jαδ(x):=p−1∥F(x)−yδ∥Yp+αR(x).

Choose α(δ)=δp(log⁡(1/δ))ν for sufficiently small 0<δ<1, and write cp=21−p/p. For H(q):=sup⁡s≥0(ψ(s)−qs), one has H(q)∼(log⁡q)−ν as q→∞, so ρδ:=H(cp/α(δ))≍(log⁡(1/δ))−ν and α(δ)ρδ≍δp. The variational source rate theorem therefore gives

DRξ†(xαδ,ε,x†)=O((log⁡(1/δ))−ν),∥F(xαδ,ε)−yδ∥Y=O(δ),∣R(xαδ,ε)−R(x†)∣=O((log⁡(1/δ))−ν).

If also DRξ†(x,x†)≥cR∥x−x†∥Xs on LR0 for constants cR,s>0, then ∥xαδ,ε−x†∥X=O((log⁡(1/δ))−ν/s). The parameter coupling here is derived from αH(cp/α)≍δp.

Facts & Assumptions

Given: The conditions in the Example hold for sufficiently small 0<δ<1.

[F1]

Variational source conditions on a regularizer sublevel: the VSC uses a continuous, nondecreasing, concave index function vanishing at zero and a coefficient 0≤β<1 on the stated regularizer sublevel.

[F2]

A conjugate bound for residual and Bregman errors: H(q)=sup⁡s≥0(ψ(s)−qs) is the conjugate quantity used to bound Bregman and residual errors.

[F3]

Variational source conditions yield nonlinear regularization rates: if αH(cp/α)≍δp and H(cp/(2α))=O(H(cp/α)), then the theorem yields Bregman error O(H(cp/α)) and noisy-data residual O(δ); it also yields the penalty rate when ψ(C0δp)=O(H(cp/α)).

[A1]

ν>0 and 0<t0<e−(ν+1) are fixed, and α(δ)=δp(log⁡(1/δ))ν.

Verification

1.1A1F1algebra

For 0<t<t0, put L=−log⁡t. Differentiation gives ψ′(t)=ν/(tLν+1) and ψ′′(t)=ν((ν+1)−L)/(t2Lν+2)<0, since L≥−log⁡t0>ν+1. Thus ψ′ decreases on (0,t0) and has positive left limit at t0; after t0 the function is constant, so its derivative drops to zero and remains nonincreasing. Also ψ(t)→0 as t↓0. Therefore its continuous extension at zero is nondecreasing, concave on [0,∞), vanishes at zero and is positive for t>0, as required by [F1].

2.1A1step 1.1algebra

For q>ψ′(t0−), the derivative of s↦ψ(s)−qs decreases from +∞ to a negative value on (0,t0) and is −q<0 for s>t0. Hence the unique global maximizer is sq∈(0,t0) satisfying q=ν/(sqLqν+1), where Lq=−log⁡sq. This equation gives log⁡q=Lq−(ν+1)log⁡Lq+log⁡ν, so Lq→∞ and Lq/log⁡q→1. At the maximizer, qsq=νLq−ν−1; consequently H(q)=Lq−ν(1−ν/Lq)∼(log⁡q)−ν.

3.1A1F2F3step 2.1algebra∎

Set ℓδ=log⁡(1/δ). Then α(δ)→0 and qδ=cp/α(δ)→∞, while log⁡qδ=pℓδ−νlog⁡ℓδ+log⁡cp∼pℓδ. By step 2.1, ρδ=H(qδ)≍ℓδ−ν and αρδ≍δp. Replacing qδ by qδ/2 changes its logarithm by only −log⁡2, so H(cp/(2α))≍ρδ. For any fixed C0>0, eventually C0δp<t0, and ψ(C0δp)=(−log⁡(C0δp))−ν≍ℓδ−ν≍ρδ. Thus all coupling and penalty-index hypotheses in [F3] hold; applying [F3] gives the three asserted rates. If the additional Bregman lower bound holds, its combination with the Bregman rate gives ∥xαδ,ε−x†∥Xs=O(ℓδ−ν), and taking the positive s-th root proves the stated norm rate.

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