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An R-only proximal coordinate rate fails for nonsmooth penalties

Statement refuted

The claim that the uniform randomized proximal update for every proper closed convex separable penalty satisfies E[F(xk)−F∗]≤2nLmax⁡R2k+4 under only a bounded initial sublevel radius R and positive coordinate majorizers Li is false without an error term or other dependence on the penalty.

Witness

Let n=2, f(x)=0, g1(u)=g2(u)=4∣u∣, L1=L2=Lmax⁡=1, and x0=(1,1). The objective is F(x)=4(∣x1∣+∣x2∣) and F∗=0 at x∗=(0,0). The initial sublevel is the closed diamond ∣x1∣+∣x2∣≤2, whose radius about x∗ is R=2.

Facts & Assumptions

[F1]

Randomized proximal coordinate descent with certified inner solves: for selected block i the exact update minimizes ⟨∇if(x),u−xi⟩+(Mi/2)∣u−xi∣2+gi(u).

[F2]

The one-coordinate absolute-value quadratic minimizer: a2u2−bu+τ∣u∣ has unique minimizer Sτ(b)/a when a,τ>0.

[F3]

Proper closed convex extended-real functions on Rn: a block penalty is proper, closed and convex when its effective domain is nonempty, its epigraph is closed, and it satisfies the segment inequality.

[F4]

Absolute value in an ordered field, Basic properties of the absolute value and The triangle inequality: absolute value is nonnegative, positively homogeneous, and subadditive.

[F5]

The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn and Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation: the two-coordinate Euclidean norm is a norm and its triangle inequality gives ∥(x1,x2)∥2≤∣x1∣+∣x2∣ by decomposing into coordinate vectors.

[A1]

The proximal coordinate algorithm uses common model majorizers M1=M2=1, samples uniformly, and runs at least one update with no stopping tolerance.

Proof

1.1A1F1F3F4F5givenalgebra

The assumptions hold: f=0 is convex and continuously differentiable and has zero gradient, so Li=Mi=1 is a valid positive block-gradient bound and majorizer. Each gi(u)=4∣u∣ is finite everywhere and hence proper; [F4] gives convexity; and epi⁡gi={(u,t):t≥4u, t≥−4u} is an intersection of closed half-planes, so it is closed in the epigraph convention [F3]. Thus F(x)=4(∣x1∣+∣x2∣) and its minimum is F∗=0 at x∗=(0,0). The initial value is F(x0)=8 and its sublevel is ∣x1∣+∣x2∣≤2. There ∥x−x∗∥2=x12+x22≤∣x1∣+∣x2∣≤2 by [F4]--[F5], with equality at (2,0); hence R=2.

2.1F1F2step 1.1algebra

For either selected coordinate the subproblem at its current value 1 is 4∣u∣+12(u−1)2, which differs by a constant from 12u2−u+4∣u∣. By [F2] its unique minimizer is S4(1)=0. The other coordinate remains 1, so after one update F(x1)=4 for either draw and E[F(x1)−F∗]=4.

3.1step 1.1step 2.1algebra∎

The proposed upper bound at k=1 is 2⋅2⋅1⋅22/(1+4)=16/5<4, contradicting step 2.1. Therefore the claimed penalty-independent composite bound is false.

Remarks

The controlling plan's O(1/k) rate is for the smooth coordinate-gradient branch gi=0. This witness does not refute that branch.

Depends on

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