LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Metric projection onto a closed convex set satisfies the variational inequality

Statement

Let $n \geq 1$ , let $C \subseteq R^{n}$ be nonempty, closed, and convex, let $x \in R^{n}$ , and let $p \in C$ . Then $p$ is the nearest point of $C$ to $x$ if and only if

$⟨ x - p, z - p ⟩ \leq 0 (z \in C) .$

Facts & Assumptions

Given: The Euclidean inner product The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and convexity of $C$ A convex subset of $R^{m}$ contains every line segment between two of its points.

[L1]

There is a unique $p \in C$ such that $∥ x - p ∥_{2} \leq ∥ x - z ∥_{2}$ for every $z \in C$ (Every point has a unique nearest point in a nonempty closed Euclidean convex set).

Proof

technique · direct

1.1

For the forward implication, let $p$ be the nearest point and take $z \in C$ . For $0 < t \leq 1$ , convexity puts $p + t (z - p)$ in $C$ . Comparing its squared distance with the minimum in [L1] and expanding gives $2 ⟨ x - p, z - p ⟩ \leq t ∥ z - p ∥_{2}^{2} .$ If the left inner product were positive, a sufficiently small $t > 0$ would violate this inequality, so it is nonpositive.

L1givenalgebra

2.1

For the reverse implication, suppose the displayed variational inequality holds. Expanding gives $∥ x - z ∥_{2}^{2} = ∥ x - p ∥_{2}^{2} + ∥ z - p ∥_{2}^{2} - 2 ⟨ x - p, z - p ⟩ \geq ∥ x - p ∥_{2}^{2} .$ Thus $p$ satisfies the nearest-point condition of [L1].

L1algebra∎

Depends on

Used by

A point outside a nonempty closed convex set is strictly separated from it Theorem

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

S. Boyd and L. Vandenberghe, Convex Optimization, §2.5.1 (standard reference, not scraped)
D. Bertsekas, MIT 6.253 Convex Analysis and Optimization, Lecture 6 (standard reference, not scraped)