Alphabeta Math
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Coordinatewise minima of a separable convex composite are global minima

Statement

Let f:Rd→R be differentiable and convex, and let B1,…,Bn be a partition of the coordinate labels {1,…,d} into nonempty blocks, with n≥1. Let g(x)=∑i=1ngi(xBi) be proper, closed and convex, with each gi proper, closed and convex on its block. Set F=f+g. If F(x)<+∞ and x globally minimizes every block slice u↦F(x−i,u), where (x−i,u) means replace block Bi of x by u and keep all other blocks fixed, then −∇if(x)∈∂gi(xBi)(1≤i≤n),−∇f(x)∈∂g(x), and x is a global minimizer of F. For an extended-real convex h, p∈∂h(a) means h(z)≥h(a)+⟨p,z−a⟩ for every z.

Facts & Assumptions

Given: The finite block partition, the functions and the coordinatewise minimizing point in the Statement.

[F1]

Subdifferentials of proper extended-real convex functions: p∈∂h(a) exactly when h(z)≥h(a)+⟨p,z−a⟩ for every z in the full Euclidean space.

[F2]

Proper closed convex extended-real functions on Rn: each gi is convex and has nonempty effective domain; closed epigraph is lower semicontinuity in this convention.

[F3]

For a differentiable scalar field, Dvf(a)=⟨∇f(a),v⟩ and the unit direction of steepest ascent is the normalized gradient: for a totally differentiable scalar function, the directional derivative in direction v is ⟨∇f(x),v⟩.

[F4]

Differentiable convex functions are characterized by the gradient inequality: convex differentiability gives f(z)≥f(x)+⟨∇f(x),z−x⟩.

[F5]

Identify block label j∈{1,…,d} with standard coordinate j−1; let Ui insert a block displacement into those coordinates, and define ∇if(x) as the entries of ∇f(x) on Bi, in increasing order. By Laws of finite sums and finite products and The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, finite block sums add and the full inner product is the sum of the block inner products, so ⟨∇f(x),Uih⟩=⟨∇if(x),h⟩.

[A1]

f is differentiable and convex, g=∑igi is proper closed convex, F(x)<+∞, and x minimizes each exact one-block slice.

Proof

1.1A1F1F2F3F5givenalgebra

Fix a block i, a point zi with gi(zi)<+∞, and put h=zi−xBi. For 0<t<1, block minimality and convexity of gi give 0≤f(x+Ui(th))−f(x)+gi(xBi+th)−gi(xBi)≤f(x+Ui(th))−f(x)+t[gi(zi)−gi(xBi)]. Rearranging, dividing by t>0, and taking t↓0, [F3] and [F5] give gi(zi)−gi(xBi)≥−⟨∇f(x),Uih⟩=−⟨∇if(x),zi−xBi⟩. For gi(zi)=+∞ the same subgradient inequality is automatic. Thus −∇if(x)∈∂gi(xBi) by [F1].

2.1F1F5step 1.1algebra

Fix any z=(zBi)i. If g(z)=+∞, the subgradient inequality at z is automatic. Otherwise each gi(zBi) is finite, so summing the block inequalities from step 1.1 and using [F5] gives g(z)≥g(x)−∑i⟨∇if(x),zBi−xBi⟩=g(x)−⟨∇f(x),z−x⟩. Thus the subgradient inequality holds for every z, and [F1] gives −∇f(x)∈∂g(x).

3.1A1F4step 2.1algebra∎

By [F4], f(z)≥f(x)+⟨∇f(x),z−x⟩. Adding the inequality in step 2.1 gives F(z)≥F(x) for every z, so x is globally minimizing and all the stated stationarity inclusions hold.

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