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LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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Gram control and a Taylor remainder imply one-step loss contraction

Statement

Let r∈Rn, let G be a real symmetric matrix with eigenvalues in [λ,Λ] for some 0<λ≤Λ, and suppose 0<ηΛ≤1. If an update has the form r+=(I−ηG)r+R with ∥R∥2≤ηλ∥r∥2/4, then for the quadratic loss L=12∥r∥22 and L+=12∥r+∥22,

L+≤(1−ηλ/2)L.

Facts & Assumptions

Proof

technique · bound the spectral norm of the linear contraction and then square the residual-norm bound
1.1F1F2givenalgebra

By [F1], the eigenvalues μj of G lie in [λ,Λ], so those of I−ηG lie in [1−ηΛ,1−ηλ]⊆[0,1−ηλ]. Expanding r in an orthonormal eigenbasis gives ∥(I−ηG)r∥22=∑j(1−ημj)2rj2≤(1−ηλ)2∥r∥22. The norm triangle inequality [F2] with the remainder hypothesis then yields ∥r+∥2≤(1−3ηλ/4)∥r∥2.

2.1step 1.1algebra∎

Put t=ηλ∈(0,1]. Then (1−3t/4)2≤1−t/2, since 1−t/2−(1−3t/4)2=t(1−9t/16)≥7t/16>0. Squaring the estimate in step 1.1 and multiplying by 1/2 yields L+≤(1−ηλ/2)L.

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