TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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Orthogonal decomposition by a closed subspace

Statement

Assume the Axiom of Countable Choice. Let $M$ be a closed linear subspace of a real or complex Hilbert space $H$ . Then every $x \in H$ has a unique decomposition

$x = m + n, m \in M, n \in M^{⊥},$

so that $H = M \oplus M^{⊥}$ as a direct sum of the subspace $M$ and its orthogonal complement.

Facts & Assumptions

[A1]

A linear subspace is convex and contains $0$ , and the nearest point of a nonempty closed convex subset of a Hilbert space exists and is unique (Linear subspace of a vector space, Projection onto a nonempty closed convex set).

[A2]

A point $p$ is the nearest point of a closed convex set $C$ to $x$ exactly when $Re ⟨ x - p, y - p ⟩ \leq 0$ for every $y \in C$ (Variational characterisation of the nearest point).

[A3]

The pairing is linear in the first argument and conjugate-linear in the second, $S^{⊥} = {v : ⟨ v, s ⟩ = 0 \forall s \in S}$ is a linear subspace, and $v \in S^{⊥} \cap S^{⊥ ⊥}$ with $⟨ v, v ⟩ = 0$ forces $v = 0$ (Orthogonality and the orthogonal complement, Real and complex inner-product spaces and their induced length).

[A4]

Countable Choice is the selection principle consumed by the nearest-point theorem (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: Countable Choice, a Hilbert space $H$ , a closed linear subspace $M \subseteq H$ and a vector $x \in H$ .

1.1A1A4

Since $M$ is a nonempty closed convex set, $x$ has a unique nearest point $p$ in $M$ .

2.1step 1.1A1A2A3algebra

The variational inequality gives $Re ⟨ x - p, w - p ⟩ \leq 0$ for every $w \in M$ . For $u \in M$ , take $w = p + u$ and $w = p - u$ to obtain $Re ⟨ x - p, u ⟩ = 0$ . Over $R$ the pairing is real-valued, so this already gives $⟨ x - p, u ⟩ = 0$ . Over $C$ , also $i u \in M$ , and the same real-part conclusion applied to $i u$ gives $0 = Re ⟨ x - p, i u ⟩ = Re (- i ⟨ x - p, u ⟩) = Im ⟨ x - p, u ⟩$ . Thus in either scalar field $⟨ x - p, u ⟩ = 0$ for every $u \in M$ , that is $x - p \in M^{⊥}$ .

3.1step 2.1A3

Setting $m = p$ and $n = x - p$ gives a decomposition $x = m + n$ with $m \in M$ and $n \in M^{⊥}$ .

4.1step 3.1A3algebra∎

If $x = m_{1} + n_{1} = m_{2} + n_{2}$ are two such decompositions, then $v = m_{1} - m_{2} = n_{2} - n_{1}$ lies in $M \cap M^{⊥}$ , since both $M$ and $M^{⊥}$ are linear subspaces, so $⟨ v, v ⟩ = 0$ and $v = 0$ ; hence $m_{1} = m_{2}$ , $n_{1} = n_{2}$ , and the decomposition is unique.

Orthogonal decomposition by a closed subspace

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources