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The convex normal-cone operator is maximal monotone and its resolvent is projection

Statement

Let C⊆Rd be nonempty, closed, and convex, and let NC have the convention of Projected contact dynamics and explicit projection Euler step. In dimension d=0, interpret this convention directly as C=R0={0}, NC(0)={0}, and PC(0)=0. An operator A:Rd⇉Rd is monotone if ⟨u−v,x−y⟩≥0 for all u∈A(x) and v∈A(y); it is maximal monotone if its graph has no proper monotone graph extension. Then NC is maximal monotone. For every τ>0 and y∈Rd, there is a unique p with y−p∈τNC(p), and that point is PC(y). Equivalently, (I+τNC)−1=PC.

Facts & Assumptions

Given: The set C and normal-cone convention in the Statement, and a scalar τ>0.

[F1]

Scaled KKT residual and convex normal cone and Projected contact dynamics and explicit projection Euler step: for x∈C, NC(x)={n:⟨n,z−x⟩≤0 for every z∈C}; outside C, the operator is empty.

[F2]

Extended-real indicator of a set: ιC is 0 on C and +∞ outside C.

[F3]

Subdifferentials of proper extended-real convex functions: for proper convex g and x∈dom⁡g, p∈∂g(x) means g(z)≥g(x)+⟨p,z−x⟩ for every z; if x∉dom⁡g, then ∂g(x)=∅.

[F4]

Existence, resolvent characterization, and firm nonexpansiveness of the proximal map: for proper closed convex g and τ>0, u=prox⁡τg(y) iff (y−u)/τ∈∂g(u).

[F5]

The proximal map of an indicator is Euclidean projection: the indicator of this nonempty closed convex set is proper, closed, and convex, and prox⁡τιC(y)=PC(y), the unique nearest point.

Proof

1.1F1givenalgebra

Let n∈NC(x) and m∈NC(y). Testing the definition in [F1] with z=y and z=x gives ⟨n,y−x⟩≤0 and ⟨m,x−y⟩≤0. Adding and changing signs yields ⟨n−m,x−y⟩≥0. Thus NC is monotone.

1.2F1F2F3F5given

The indicator is proper, closed, and convex by [F5], and its domain is C by [F2]. At x∈C, [F2]–[F3] show p∈∂ιC(x) exactly when ⟨p,z−x⟩≤0 for every z∈C; inequalities for z∉C are automatic because ιC(z)=+∞. At x∉C, both ∂ιC(x) and NC(x) are empty. Hence ∂ιC=NC.

2.1F4F5step 1.1step 1.2givenalgebra

Given any y∈Rd, put p=PC(y)=prox⁡τιC(y) by [F5]. By [F4] and step 1.2, (y−p)/τ∈NC(p), so y−p∈τNC(p). If also y−q∈τNC(q), monotonicity [step 1.1] gives 0≤⟨(y−p−y+q)/τ,p−q⟩=−∥p−q∥22/τ, hence p=q. Therefore the resolvent exists uniquely at every y and equals PC.

3.1step 2.1givenalgebra∎

Suppose (x,v) is monotonically related to the graph of NC, so ⟨v−n,x−y⟩≥0 for every n∈NC(y). Apply step 2.1 to x+τv to obtain p and n=(x+τv−p)/τ∈NC(p). Then x−p=τ(n−v), and the relation gives 0≤⟨v−n,x−p⟩=−τ∥v−n∥22. Thus v=n and x=p, so (x,v) already lies in the graph. Every monotone extension adds no point; NC is maximal monotone.

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