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SAGA constant-step linear contraction with a memory-capped factor

Statement

Assume AC. Let n,d≥1 and let fi:Rd→R be convex, continuously differentiable, with L-Lipschitz gradients for a common L>0. Set F=n−1∑ifi, assume F is μ-strongly convex for μ>0, and let x∗ satisfy ∇F(x∗)=0. Run Finite-sum SAGA from deterministic x0 with its standard initialized table and independent uniform component draws. For any constant 0<η≤1/(10L), define Vk=∥xk−x∗∥22,Sk=1n∑i∥αik−∇fi(x∗)∥22,Lk=Vk+7η2nSk, δ=min⁡{μη,1/(7n)},q=1−δ. Then 0<δ<1, and for every k≥0, E[Lk+1∣Fk]≤qLk,E[Lk]≤qkL0,E[F(xk)−F(x∗)]≤L2qkL0. In particular η=1/(10L) is allowed independently of n, with δ=min⁡{μ/(10L),1/(7n)}. When additionally μη≤1/(7n) the factor is 1−μη; no such faster factor is asserted without that condition. These are expectation bounds for the exact SAGA recurrence, not almost-sure convergence or finite-oracle success claims.

Facts & Assumptions

Given: The functions, x∗, SAGA initialization and independent uniform sampling in this Statement, with AC for conditional-expectation classes. Finite-time moments exist because for each finite k the algorithm has finitely many index histories and each gradient evaluation is finite.

[F1]

Finite-sum SAGA: xk+1=xk−ηvk, then the sampled table entry is replaced by ∇fIk(xk) using the pre-update iterate.

[F2]

SAGA conditional estimator, variance, and table recurrence: conditionally on the history, E[vk∣Fk]=∇F(xk), E[∥vk∥22∣Fk]≤6L(F(xk)−F(x∗))+6Sk, and E[Sk+1∣Fk]≤(1−1/n)Sk+(2L/n)(F(xk)−F(x∗)).

[F3]

Gradient of a finite sum is the finite sum of gradients: ∇F(w)=n−1∑i∇fi(w).

[F4]

Descent inequality for a Lipschitz gradient: each fi satisfies its L-smooth upper descent inequality.

[F5]

Strong convexity makes a differentiable gradient strongly monotone: F(y)≥F(x)+⟨∇F(x),y−x⟩+(μ/2)∥y−x∥22.

[F6]

The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn: the Euclidean inner product is bilinear and ∥z∥22=⟨z,z⟩.

[F7]

Lipschitz map, α-Hölder map for rational 0<α≤1, and contraction: an L-Lipschitz gradient satisfies ∥∇fi(x)−∇fi(y)∥2≤L∥x−y∥2.

[F8]

Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation: ∣⟨z,w⟩∣≤∥z∥2∥w∥2; applying Cauchy–Schwarz to the finite vector of component norms bounds an average by its root-mean-square norm.

[F9]

Basic algebra and order properties of conditional expectation: conditional expectation preserves order and E[E[X∣Fk]]=E[X] for integrable X.

[F10]

Finite sums and finite products, by recursion: the finite average is the recursively defined finite sum scaled by 1/n.

[F11]

Laws of finite sums and finite products: finite sums are additive, obey scaling, and are monotone termwise.

[A1]

The Axiom of Choice: AC is assumed by the conditional-expectation interface [F2].

Proof

1.1F3F4F5F11givenalgebra

Put Δk=F(xk)−F(x∗) and ek=xk−x∗. By [F5] at xk with y=x∗, ⟨∇F(xk),ek⟩≥Δk+(μ/2)Vk. By [F5] at x∗, Δk≥(μ/2)Vk≥0. Averaging [F4] at x∗ and using [F3], [F11] and ∇F(x∗)=0 also gives Δk≤(L/2)Vk.

1.2F3F5F7F8F10F11givenalgebra

The average gradient is L-Lipschitz: by [F3], [F7], [F8], [F10] and [F11], ∥∇F(x)−∇F(y)∥22≤n−1∑i∥∇fi(x)−∇fi(y)∥22≤L2∥x−y∥22. The strong-monotonicity consequence of [F5] and Cauchy–Schwarz [F8] imply μ≤L, since d≥1 allows x≠y. Thus 0<μη≤1/10 and 0<δ≤1/(7n)≤1/7<1.

2.1A1F1F2F6step 1.1givenalgebra

Expand [F1] and condition on the current history. Unbiasedness and the second-moment bound in [F2], followed by the gap-retaining inner-product estimate in step 1.1, give E[Vk+1∣Fk]≤Vk−2η⟨∇F(xk),ek⟩+η2(6LΔk+6Sk)≤(1−μη)Vk+(−2η+6Lη2)Δk+6η2Sk. Finite-time moments exist as stated in the Given paragraph, so all conditional quantities are integrable.

3.1A1F2step 1.1step 2.1givenalgebra

Multiply the table recurrence in [F2] by 7η2n and add step 2.1. This yields E[Lk+1∣Fk]≤(1−μη)Vk+(−2η+20Lη2)Δk+(7n−1)η2Sk. Since η≤1/(10L), the gap coefficient is nonpositive; since Δk≥0, it can be dropped.

4.1step 1.2step 3.1givenalgebra

By definition δ≤μη and δ≤1/(7n). Hence (1−μη)Vk≤(1−δ)Vk and (7n−1)η2Sk=7η2n(1−1/(7n))Sk≤7η2n(1−δ)Sk. Apply these two inequalities to step 3.1 to obtain E[Lk+1∣Fk]≤qLk. The positivity from step 1.2 gives q∈(0,1).

5.1A1F9F11step 1.1step 4.1giveninductionalgebra∎

Apply the tower property and order preservation in [F9] to step 4.1. Induction gives E[Lk]≤qkL0, beginning with equality at k=0. Step 1.1 and Vk≤Lk give the objective bound after taking expectations. Substituting η=1/(10L) gives the advertised constant-step factor; if μη≤1/(7n), the minimum defining δ is μη.

Remarks

The expectation estimate is not a total-runtime guarantee and does not bound the cost of an individual component-gradient evaluation. At η=1/(10L), 1/δ=max⁡{10L/μ,7n} describes the geometric-rate scale; it is not a total runtime bound.

The separate memory-scaled theorem and the standard-initialization counterexample remain unchanged. The counterexample refutes omitting the memory cap from the faster factor, not the present capped rate. This local proof retains the gap term supplied by strong convexity instead of discarding it before combining iterate and table recurrences.

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