Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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  • 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.

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Finite-horizon SGD trajectories have integrable objectives

Statement

Let d≥1, let (Ω,F,P) be a probability space, and let f:Rd→R be continuously differentiable with an L-Lipschitz gradient for some finite L≥0. Fix an integer N≥1, measurable maps x0,G0,…,GN−1:Ω→Rd, and finite real stepsizes α0,…,αN−1. Suppose E∥x0∥22<∞ and E∥Gk∥22<∞ for every k<N, and define xk+1=xk−αkGk. Then every xk and ∇f(xk) for 0≤k≤N has finite second moment, and every f(xk) has finite absolute expectation.

Facts & Assumptions

Given: The probability space, measurable finite-dimensional vector maps, continuously differentiable f, finite-horizon update and finite second-moment assumptions in the Statement.

[F1]

Descent inequality for a Lipschitz gradient: if ∇f is L-Lipschitz, then f(y)≤f(x)+⟨∇f(x),y−x⟩+(L/2)∥y−x∥22 for all x,y∈Rd.

[F3]

Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation: ∣⟨u,v⟩∣≤∥u∥2∥v∥2 and ∥u+v∥2≤∥u∥2+∥v∥2.

[F4]

The principle of mathematical induction: a property true at the first index and inherited by each successor holds at every natural index.

[F5]

Expectation of a nonnegative or integrable random variable: expectations of nonnegative measurable functions are their extended integrals; an integrable real random variable has finite absolute integral.

Proof

1.1F2F3F5givenalgebra

By [F3] and 2ab≤a2+b2, ∥u+v∥22≤2∥u∥22+2∥v∥22. Thus E∥xk+1∥22≤2E∥xk∥22+2αk2E∥Gk∥22 for every k<N. The right side is finite whenever xk has finite second moment, using the expectation convention [F5].

2.1F4F5step 1.1giveninduction

The initial second moment is finite by the Statement, and step 1.1 carries finiteness from xk to xk+1. Finite induction using [F4] therefore gives E∥xk∥22<∞ for all k≤N.

3.1F1F3F5step 2.1givenalgebra

The Lipschitz-gradient condition [F1] and [F3] give ∥∇f(xk)∥2≤∥∇f(0)∥2+L∥xk∥2, so ∥∇f(xk)∥22≤2∥∇f(0)∥22+2L2∥xk∥22. Step 2.1 makes the right side integrable; hence every stated gradient has finite second moment.

4.1F1F3F5step 2.1givenalgebra∎

Apply [F1] to (0,xk) and (xk,0). The upper bound gives f(xk)≤f(0)+∥∇f(0)∥2∥xk∥2+(L/2)∥xk∥22. The lower bound and the Lipschitz-gradient condition give f(xk)≥f(0)−∥∇f(0)∥2∥xk∥2−(3L/2)∥xk∥22. Thus [F3] and 2ab≤a2+b2 imply ∣f(xk)∣≤∣f(0)∣+12∥∇f(0)∥22+(12+3L2)∥xk∥22. The right side has finite expectation by step 2.1, so E∣f(xk)∣<∞ for every k≤N.

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