Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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Projection onto a nonempty closed convex set

Statement

Assume the Axiom of Countable Choice. Let H be a real or complex Hilbert space, let C⊆H be nonempty, closed and convex, and let x∈H. Then there is exactly one point p∈C with ∥x−p∥=inf⁡c∈C∥x−c∥, that is, a unique nearest point of C to x.

Facts & Assumptions

[A1]

A Hilbert space is an inner-product space complete for its induced norm: every Cauchy sequence converges to a point of the space (Hilbert space).

[A2]

Every nonempty real set bounded below has an infimum, characterised by points arbitrarily close from above (Every nonempty set bounded below has an infimum, Epsilon characterisation of the infimum).

[A3]

Countable Choice selects a point from each set of a countable family of nonempty sets (The Axiom of Countable Choice (ACω)).

[A4]

A minimizing sequence in a nonempty convex set is Cauchy (A minimizing sequence in a convex set is Cauchy).

[A5]

C is convex and d≤∥x−c∥ for every c∈C (Convex sets and continuous real-hyperplane separation in a normed space, Greatest lower bound (infimum)). A closed set equals its closure, and a point lies in the closure whenever every ball about it meets the set (The closure of a nonempty A is {x:d(x,A)=0}, equals A together with its limit points, and is the smallest closed superset); limits in a metric space are unique (A sequence in a metric space has at most one limit).

[A6]

Norm distance is continuous: ∣∥u∥−∥v∥∣≤∥u−v∥ (The reverse triangle inequality in a normed space).

[A7]

Convergence of a sequence of reals to L means that for every ε>0 the terms are eventually within ε of L, and for every ε>0 some 1/n is below ε (Limits and Cauchy sequences of reals, For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε).

[A8]

Every inner-product norm satisfies the parallelogram law (The parallelogram law).

Proof

technique · direct

Given: Countable Choice, a real or complex Hilbert space H, a nonempty closed convex set C⊆H and a vector x∈H.

1.1A2

The set D={∥x−c∥:c∈C} is nonempty and bounded below by 0, so d=inf⁡D exists and for every n some c∈C has ∥x−c∥<d+1/(n+1) by the epsilon characterisation of the infimum.

2.1step 1.1A3

Countable Choice selects for every n a point cn∈C with ∥x−cn∥<d+1/(n+1), the sets being nonempty by step 1.1.

3.1step 2.1A1A4A5A7

Then 0≤∥x−cn∥−d<1/(n+1) for every n, so ∥x−cn∥→d by the Archimedean reciprocal bound; hence (cn) is Cauchy by the minimizing-sequence lemma, completeness of H gives a limit p∈H, and closedness of C places p in C.

4.1step 3.1A5A6

Moreover ∥x−p∥=d: norm continuity along the limit gives ∣∥x−p∥−∥x−cn∥∣≤∥p−cn∥→0, and a limit of the sequence ∥x−cn∥ is unique.

5.1step 4.1A5A8algebra∎

If p,q∈C both satisfy ∥x−p∥=∥x−q∥=d, then the midpoint 12(p+q) lies in C by convexity, so ∥x−12(p+q)∥≥d, and the parallelogram law gives ∥p−q∥2=2∥x−p∥2+2∥x−q∥2−4∥x−12(p+q)∥2≤4d2−4d2=0, whence p=q: the nearest point is unique.

Depends on

Dependency tree · 2 levels

Sources