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A maximal-monotone resolvent is everywhere-defined and firmly nonexpansive

Statement

Let A be maximal monotone on E=Rn and let ρ>0. For every y∈E there is a unique x∈E with y−x∈ρA(x). Thus the resolvent from Resolvent of a monotone relation is an everywhere-defined map and, for all y,y′∈E, ∥JρAy−JρAy′∥22≤⟨JρAy−JρAy′,y−y′⟩. This is firm nonexpansiveness; it also implies the map is 1-Lipschitz.

Facts & Assumptions

Given: A maximal-monotone relation A and ρ>0.

[A1]

The relation A is maximal monotone and ρ>0.

[F1]

Monotone and maximal-monotone relations on Euclidean space: monotonicity means ⟨x−x′,a−a′⟩≥0 for graph pairs, and maximality excludes any proper monotone graph extension.

[F2]

Resolvent of a monotone relation: x∈JρA(y) iff y=x+ρa for some a∈A(x).

[F3]

One-point extension of a nonexpansive map in finite dimensions: a 1-Lipschitz map on any subset of E admits a value at each added point satisfying all 1-Lipschitz inequalities.

Proof

1.1A1F1constructalgebra

Transform every graph pair (x,a)∈A into p=x+ρa and q=x−ρa. If two graph pairs are used, then ∥q−q′∥22≤∥p−p′∥22, because the difference of the right and left sides is 4ρ⟨x−x′,a−a′⟩≥0 by [F1]. Also, if p=p′, this inequality forces q=q′. Thus the rule T(p)=q is a well-defined 1-Lipschitz map on its range D={x+ρa:(x,a)∈A}.

2.1A1F1F2F3step 1.1algebra

Fix arbitrary y∈E. By [F3], extend T at y by some q satisfying ∥q−(x−ρa)∥2≤∥y−(x+ρa)∥2 for every (x,a)∈A. Set x~=(y+q)/2 and a~=(y−q)/(2ρ). Expanding the squared inequality shows ⟨x~−x,a~−a⟩≥0 for every graph pair, so adjoining (x~,a~) preserves monotonicity. By maximality in [F1], it already belongs to A. Hence y=x~+ρa~, proving existence. This also covers an initially empty graph: [F3] gives the extension and maximality rules out the empty graph.

3.1A1F1F2step 2.1algebra

If x,x′ both satisfy y−x∈ρA(x) and y−x′∈ρA(x′), monotonicity gives 0≤⟨x−x′,(y−x)−(y−x′)⟩=−∥x−x′∥22. Thus x=x′, so the resolvent is single-valued.

4.1A1F1F2F4step 3.1algebra∎

For inputs y,y′, write x=JρAy, x′=JρAy′. Monotonicity applied to (x,(y−x)/ρ) and (x′,(y′−x′)/ρ) gives ∥x−x′∥22≤⟨x−x′,y−y′⟩. Cauchy–Schwarz [F4] then gives ∥x−x′∥2≤∥y−y′∥2 when x≠x′, and the inequality is immediate when x=x′. This proves firm nonexpansiveness and the Lipschitz consequence.

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