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TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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Vanishing Tikhonov bias and controlled noise amplification imply convergence

Statement

Assume Countable Choice. Let A:X→Y be compact between real or complex Hilbert spaces, x†⊥ker⁡A, and y=Ax†. For a sequence δj≥0, data yj=y+ej with ∥ej∥≤δj, and parameters λj>0, suppose λj→0,δjλj→0. Then the Tikhonov solutions xj=(A∗A+λjI)−1A∗yj converge in norm to x†. No source condition on x† is required.

Facts & Assumptions

Given: Countable Choice, compact A:X→Y, the minimum-norm solution x†⊥ker⁡A of Ax=y, noisy data yj=y+ej with ∥ej∥≤δj, and positive parameters λj satisfying the stated limits.

[F1]

Compact-operator Tikhonov bias vanishes on the minimum-norm solution: For exact data y=Ax†, the bias ∥x†−(A∗A+λI)−1A∗y∥ tends to zero as λ↓0.

[F2]

Hilbert-space Tikhonov amplifies data by at most 1/(2λ): For bounded A, Rλ=(A∗A+λI)−1A∗ satisfies ∥Rλe∥≤∥e∥/(2λ).

[F3]

Compact spectral filters, source conditions, and qualification: The regularized solution is Rλy and the parameter λ is the coefficient of the squared norm penalty.

Proof

1.1givenF1F2F3algebra

Let xλ=Rλy be the exact-data solution. Linearity gives xj−x†=(xλj−x†)+Rλjej. By [F1] and [F2], ∥xj−x†∥≤∥xλj−x†∥+δj2λj.

2.1givenstep 1.1algebraF1∎

The first term on the right tends to zero because λj→0 and [F1]; the second tends to zero by the assumed ratio. Thus ∥xj−x†∥→0.

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