TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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.

Hilbert spaces are reflexive by Riesz representation

Statement

Assume the Axiom of Countable Choice. Every complete real or complex inner-product space $H$ , with its inner-product norm, is reflexive.

Facts & Assumptions

[A1]

Countable Choice selects one member from each countable family of nonempty sets (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L1]

The inner product is linear in its first argument and conjugate-linear in its second, and its norm is the square root of the diagonal pairing (Real and complex inner product spaces, with the inner product linear in the first argument, The norm $∥ v ∥ = \sqrt{⟨ v, v ⟩}$ induced by a real or complex inner product).

[L2]

Cauchy--Schwarz bounds inner products by products of norms (Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors), and the parallelogram identity holds (Pythagoras, the parallelogram identity, and the real and complex polarisation identities).

[L3]

Nonempty real sets bounded below have infima, characterized by points arbitrarily close from above (Every nonempty set bounded below has an infimum, Epsilon characterisation of the infimum).

[L4]

Completeness for the inner-product norm is the Banach condition (Banach space). The continuous dual uses the operator norm (The dual space X^* of a normed space and its dual norm) and is Banach because its scalar target is Banach (If (Y) is Banach then (\mathcal B(X,Y)) is Banach).

[L5]

Reflexivity means surjectivity of the canonical evaluation map $J_{H} : H \to H^{* *}$ (Reflexivity is surjectivity of the canonical map).

Proof

technique · direct

Given: Countable Choice and a complete real or complex inner-product space $H$ .

1.1

Set up the Riesz representation problem. Let $φ \in H^{*}$ . If $φ = 0$ , then $φ (x) = ⟨ x, 0 ⟩$ for every $x$ . Suppose $φ \neq 0$ and put $M = {x \in H : φ (x) = 1}$ . This is a nonempty closed affine set. The nonempty set of its norms is bounded below, so let $d = \inf_{x \in M} ∥ x ∥$ . Since $1 = ∣ φ (x) ∣ \leq ∥ φ ∥ ∥ x ∥$ on $M$ , one has $d \geq 1 / ∥ φ ∥ > 0$ .

givenL1L2L3

2.1

Select and control a norm-minimizing sequence. For every $n \geq 1$ , [L3] makes $M_{n} = {x \in M : ∥ x ∥ < d + 1 / n}$ nonempty. Use [A1] exactly here to select $y_{n} \in M_{n}$ for all $n$ . Since $(y_{n} + y_{m}) / 2 \in M$ , [L2] gives

A1L2L3step 1.1choose

$∥ y_{n} - y_{m} ∥^{2} \leq 2 (d + 1 / n)^{2} + 2 (d + 1 / m)^{2} - 4 d^{2} .$

The right side tends to zero as $m, n \to \infty$ , so $(y_{n})$ is Cauchy.

3.1

Obtain the unique minimum. Completeness gives $y_{n} \to y \in H$ . Continuity of $φ$ gives $φ (y) = 1$ , so $y \in M$ , while norm continuity gives $∥ y ∥ = d$ . Thus $y$ realizes the positive minimum of the norm on $M$ .

L4step 1.1step 2.1

4.1

Derive Riesz representation with the linear-first convention. If $z \in \ker φ$ , then $y + t z \in M$ for every scalar $t$ , and minimality gives

L1L2step 3.1

$d^{2} \leq ∥ y + t z ∥^{2} = d^{2} + 2 Re ⁣ (t ⟨ z, y ⟩) + ∣ t ∣^{2} ∥ z ∥^{2} .$

If $⟨ z, y ⟩ \neq 0$ , choosing the scalar phase of a sufficiently small $t$ to make the middle term negative contradicts this inequality. Therefore $⟨ z, y ⟩ = 0$ . For arbitrary $x \in H$ , the vector $z = x - φ (x) y$ lies in $\ker φ$ , and linearity in the first argument now gives $⟨ x, y ⟩ = φ (x) ∥ y ∥^{2}$ . Hence

$φ (x) = ⟨ x, R φ ⟩, R φ : = y / ∥ y ∥^{2} .$

Together with $R 0 = 0$ , this represents every functional. Uniqueness follows by evaluating the difference of two representing vectors at that same difference. Cauchy--Schwarz and the unit vector in the representing direction give $∥ R φ ∥ = ∥ φ ∥$ .

5.1

Put the transported Hilbert structure on the dual. Define $C : H \to H^{*}$ by $(C x) (u) = ⟨ u, x ⟩$ . Step 4.1 says that $C$ is onto with inverse $R$ , and [L1] shows that both are conjugate-linear in the complex case and linear in the real case. They are isometries. Define on $H^{*}$

L1L4step 4.1construct

$⟨ φ, ψ ⟩_{*} : = ⟨ R ψ, R φ ⟩_{H} .$

The reversed order and the two conjugate-linear occurrences make this inner product linear in $φ$ , conjugate-linear in $ψ$ , and positive definite; its norm is the existing dual norm. By [L4], $H^{*}$ is complete for that norm, so it too is a Hilbert space.

6.1

Identify every bidual functional with canonical evaluation. Apply the representation proved in steps 1.1--4.1 to the Hilbert space $H^{*}$ . For $Φ \in H^{* *}$ there is $w \in H^{*}$ with $Φ (φ) = ⟨ φ, w ⟩_{*}$ for every $φ \in H^{*}$ . Put $x = R w \in H$ . Since $φ (u) = ⟨ u, R φ ⟩_{H}$ , the definition in step 5.1 gives

L1L5step 4.1step 5.1

$Φ (φ) = ⟨ R w, R φ ⟩_{H} = ⟨ x, R φ ⟩_{H} = φ (x) = (J_{H} x) (φ) .$

Thus $Φ = J_{H} x$ , so $J_{H}$ is surjective.

7.1

Conclude reflexivity and record all boundaries. [A1, L5, step 4.1, step 6.1] Surjectivity in step 6.1 is reflexivity by [L5]. If $H = {0}$ , then both $H^{*}$ and $H^{* *}$ are zero and the canonical map is onto. The zero functional was separated before division, while nonzero $φ$ gives $d > 0$ , so every quotient is defined. The real case has trivial conjugation; step 5.1 tracks both conjugations in the complex case. Countable Choice is used only to select the minimizing sequence in step 2.1 (and again when the same proved representation is applied to $H^{*}$ ), not for any basis or uncountable family.

A1L5step 4.1step 6.1∎

Depends on

Used by

Hilbert spaces have the Radon--Nikodym property Example

Dependency tree · two levels

38 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (standard reference, not scraped)