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LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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Coordinate vectors form the Hilbert basis of real ell-two

Statement

Let H=ℓ2(N,R) with its canonical real inner product, and for each j∈N let ej be the sequence whose jth coordinate is 1 and whose other coordinates are 0. Then (ej)j∈N is a complete orthonormal basis of H. For every x=(xj)∈H, the truncations ∑j<Nxjej converge in H to x as N→∞ (the N=0 sum is empty and equals 0), and ∥x∥22=∑j∈N∣xj∣2.

Facts & Assumptions

Given: The real sequence Hilbert space H=ℓ2(N,R) and the coordinate vectors (ej).

[F1]

Square summability means ∑j∣xj∣2<∞, the norm squared is that sum, and finite tails of the sum can be made arbitrarily small. Square-summable families on an arbitrary index set and the space ℓ2(I)

[F2]

The canonical inner product makes real ℓ2(N) a complete real Hilbert space. The real square-summable sequence space is complete

[F3]

An orthonormal family is a Hilbert basis when its closed linear span is the whole Hilbert space. Orthonormal families, complete orthonormal systems and Hilbert bases

Proof

Given: H=ℓ2(N,R) and ej with coordinate 1 at j and 0 elsewhere.

1.1F1F2givenalgebra

For all i,j, the canonical inner product gives ⟨ei,ej⟩=δij, so the coordinate family is orthonormal.

2.1F1F3step 1.1givenalgebra∎

Let x=(xj)∈H and ε>0. By [F1], a finite set of coordinates has complement square-sum less than ε2. Every finite subset of N is contained in {j:j<N} for some N∈N, so choose N such that ∑j≥N∣xj∣2<ε2. The coordinate formula gives ∥x−∑j<Nxjej∥22=∑j≥N∣xj∣2<ε2. Thus finite linear combinations of the ej are dense in H; by [F3], the orthonormal family is a complete orthonormal basis. The same tail estimate proves the stated truncation convergence, and the norm identity is [F1].

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Sources