Alphabeta Math
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 xH has a unique decomposition

x=m+n,mM,nM,

so that H=MM 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 Rexp,yp0 for every yC (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 sS} is a linear subspace, and vSS 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 (ACω)).

Proof

technique · direct

Given: Countable Choice, a Hilbert space H, a closed linear subspace MH and a vector xH.

1.1

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

A1A4
2.1

The variational inequality gives Rexp,wp0 for every wM. For uM, take w=p+u and w=pu to obtain Rexp,u=0. Over R the pairing is real-valued, so this already gives xp,u=0. Over C, also iuM, and the same real-part conclusion applied to iu gives 0=Rexp,iu=Re(ixp,u)=Imxp,u. Thus in either scalar field xp,u=0 for every uM, that is xpM.

step 1.1A1A2A3algebra
3.1

Setting m=p and n=xp gives a decomposition x=m+n with mM and nM.

step 2.1A3
4.1

If x=m1+n1=m2+n2 are two such decompositions, then v=m1m2=n2n1 lies in MM, since both M and M are linear subspaces, so v,v=0 and v=0; hence m1=m2, n1=n2, and the decomposition is unique.

step 3.1A3algebra

Depends on

Used by

Dependency tree · two levels

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Sources