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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A Hilbert space with a given orthonormal basis is $ℓ^{2}$ of the index set

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $(e_{i})_{i \in I}$ be an orthonormal basis of a real or complex Hilbert space $H$ , that is, a complete orthonormal family (Orthonormal families, complete orthonormal systems and Hilbert bases), and define the Fourier coefficient map

$Φ : H \to ℓ^{2} (I, F), Φ (x) : = (⟨ x, e_{i} ⟩)_{i \in I}$ with values in the space of Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ . Then $Φ$ is a linear bijection satisfying $∥ Φ (x) ∥_{2} = ∥ x ∥, ⟨ Φ (x), Φ (y) ⟩_{ℓ^{2}} = ⟨ x, y ⟩ for all x, y \in H .$

In particular $ℓ^{2} (I, F)$ is complete, hence a Hilbert space, with the inner product of Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ .

Facts & Assumptions

[A1]

If an orthonormal family is complete, then Parseval's identity holds: $\sum_{i \in I} ∣ ⟨ x, e_{i} ⟩ ∣^{2} = ∥ x ∥^{2}$ for every $x \in H$ (Parseval equivalences for an orthonormal family).

[A2]

Bessel's inequality $\sum_{i \in I} ∣ ⟨ x, e_{i} ⟩ ∣^{2} \leq ∥ x ∥^{2}$ shows that $Φ (x)$ lies in $ℓ^{2} (I, F)$ , and $Φ$ is linear because the inner product is linear in its first argument (The Bessel inequality for an arbitrary orthonormal family, Real and complex inner-product spaces and their induced length).

[A3]

If $a \in ℓ^{2} (I, F)$ , then the finite-subset net $\sum_{i \in F} a_{i} e_{i}$ converges to a limit $s$ with $⟨ s, e_{j} ⟩ = a_{j}$ for every $j$ (Square-summable orthogonal families have norm-convergent finite sums).

[A4]

For finite $F$ and $x, y \in H$ , $⟨ P_{F} x, P_{F} y ⟩ = \sum_{i \in F} ⟨ x, e_{i} ⟩ \overline{⟨ y, e_{i} ⟩}$ , where $P_{F} x = \sum_{i \in F} ⟨ x, e_{i} ⟩ e_{i}$ ; and if $z_{F} \to z$ in $H$ then $⟨ z_{F}, v ⟩ \to ⟨ z, v ⟩$ (The finite Bessel inequality and best approximation by a finite orthonormal family, Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[A5]

$ℓ^{2} (I, F)$ is an inner-product space with norm $∥ \cdot ∥_{2}$ and pairing $⟨ a, b ⟩ = \sum_{i \in I} a_{i} \overline{b_{i}}$ , and $H$ is complete for its norm (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ , Hilbert space).

[A6]

Under completeness the finite-subset net of Fourier partial sums $P_{F} x$ converges to $x$ (Fourier expansion in a Hilbert space).

Proof

technique · direct

Given: Countable Choice, an orthonormal basis $(e_{i})_{i \in I}$ of $H$ , and the coefficient map $Φ$ .

1.1A1A2

The map $Φ$ takes values in $ℓ^{2} (I, F)$ and is linear, and Parseval's identity holds for every $x$ because the family is complete; hence $∥ Φ (x) ∥_{2} = ∥ x ∥$ for every $x$ , and $Φ (x) = 0$ forces $∥ x ∥ = 0$ , so $x = 0$ by definiteness of the norm. Thus $Φ$ is linear, norm preserving and injective.

1.2A3

The finite-subset net $\sum_{i \in F} a_{i} e_{i}$ converges for every $a \in ℓ^{2} (I, F)$ , and its limit $s$ has $⟨ s, e_{j} ⟩ = a_{j}$ for every $j$ ; hence $Φ (s) = a$ and $Φ$ is surjective.

1.3A4A5A6

For all $x, y \in H$ the inner products are preserved: for every finite $F$ orthonormality gives $⟨ P_{F} x, P_{F} y ⟩ = \sum_{i \in F} ⟨ x, e_{i} ⟩ \overline{⟨ y, e_{i} ⟩}$ , both sides converge along the finite-subset net, the left to $⟨ x, y ⟩$ by continuity of the pairing and convergence of the partial sums to $x$ and $y$ , the right to the $ℓ^{2}$ pairing of $Φ (x)$ and $Φ (y)$ by the definition of the sum of a scalar family; limits being unique, $⟨ Φ (x), Φ (y) ⟩_{ℓ^{2}} = ⟨ x, y ⟩$ .

2.1step 1.1step 1.2A5

Consequently $ℓ^{2} (I, F)$ is complete: if $(a^{(n)})$ is a Cauchy sequence in $ℓ^{2} (I, F)$ , then the vectors $x_{n} : = Φ^{- 1} (a^{(n)})$ form a Cauchy sequence in $H$ by norm preservation, hence converge to some $x \in H$ ; then $∥ a^{(n)} - Φ (x) ∥_{2} = ∥ x_{n} - x ∥ \to 0$ , so the sequence converges to $Φ (x)$ .

3.1step 1.1step 1.2step 1.3step 2.1∎

Steps 1.1 to 1.3 show that $Φ$ is a linear bijection preserving norms and inner products, and step 2.1 shows that $ℓ^{2} (I, F)$ is complete; hence $H$ and $ℓ^{2} (I, F)$ are isometrically isomorphic Hilbert spaces and $ℓ^{2} (I, F)$ is itself a Hilbert space.

A Hilbert space with a given orthonormal basis is $ℓ^{2}$ of the index set

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

A Hilbert space with a given orthonormal basis is ℓ2ℓ2 of the index set

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

A Hilbert space with a given orthonormal basis is $ℓ^{2}$ of the index set