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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)
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The projection onto a closed convex set is characterised by a variational inequality

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)). Let H be a real Hilbert space (Hilbert space), let K⊆H be nonempty, closed and convex (Convex sets and continuous real-hyperplane separation in a normed space), let x∈H, and let PKx denote the unique nearest point of K to x, whose existence and uniqueness are supplied by Projection onto a nonempty closed convex set. Then for every u∈K one has u=PKx⟺⟨u−x, v−u⟩≥0for every v∈K.

Facts & Assumptions

Given: A real Hilbert space H, a nonempty closed convex set K⊆H, a vector x∈H and a point u∈K; PKx is the unique nearest point of K to x.

[A1]

The Axiom of Countable Choice (ACω): Countable Choice is the selection principle consumed by the existence-and-uniqueness theorem for nearest points.

[F1]

Projection onto a nonempty closed convex set: under Countable Choice, a nonempty closed convex subset K of a real or complex Hilbert space has exactly one nearest point to x; hence PKx is well defined and u=PKx holds exactly when ∥x−u∥≤∥x−v∥ for every v∈K.

[F2]

Variational characterisation of the nearest point: for p∈K, p is the nearest point of K to x if and only if Re⁡⟨x−p, v−p⟩≤0 for every v∈K.

[F3]

Real and complex inner-product spaces and their induced length: the inner product of a real inner product space is real-valued, so Re⁡⟨a,b⟩=⟨a,b⟩=⟨b,a⟩ and ⟨a,b⟩=−⟨−a,b⟩ for all vectors a,b; in particular u−x=−(x−u).

Proof

technique · direct

Given: A real Hilbert space H, a nonempty closed convex set K⊆H, vectors x∈H and u∈K, and the unique nearest point PKx of K to x.

1.1A1F1

By [F1] the point u equals PKx if and only if u is a nearest point of K to x, that is, ∥x−u∥≤∥x−v∥ for every v∈K; here Countable Choice enters through the existence and uniqueness of the nearest point recorded in [A1].

1.2F2

By [F2] the point u is a nearest point of K to x if and only if Re⁡⟨x−u, v−u⟩≤0 for every v∈K.

2.1step 1.1step 1.2F3algebra∎

Since the inner product is real-valued [F3], Re⁡⟨x−u,v−u⟩=⟨x−u,v−u⟩=−⟨u−x,v−u⟩, so the inequality of step 1.2 is equivalent to ⟨u−x,v−u⟩≥0 for every v∈K; combining this with step 1.1 gives u=PKx if and only if ⟨u−x,v−u⟩≥0 for every v∈K, which is the asserted characterisation.

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