CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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Hilbert spaces are reflexive

Statement

Assume the Axiom of Countable Choice. Every real or complex Hilbert space $H$ is reflexive: the canonical evaluation map $J_{H} : H \to H^{* *}$ is surjective.

Facts & Assumptions

[A1]

Riesz representation: for every bounded linear functional $f$ on $H$ there is a unique $y \in H$ with $f (x) = ⟨ x, y ⟩$ for all $x$ , and $∥ f ∥ = ∥ y ∥$ ; writing $R (y) : = ⟨ \cdot, y ⟩$ defines a bijection $R : H \to H^{*}$ that is conjugate-linear, and linear in the real case (Riesz representation for Hilbert spaces, The dual space X^* of a normed space and its dual norm).

[A2]

The canonical map is $(J_{H} x) (f) = f (x)$ and does not depend on choices (The canonical evaluation map into the bidual).

[A3]

$H$ is reflexive exactly when $J_{H}$ is surjective (Reflexivity is surjectivity of the canonical map).

[A4]

The pairing is conjugate-linear in the second argument, so $⟨ z, y ⟩ = \overline{⟨ y, z ⟩}$ (Real and complex inner-product spaces and their induced length).

[A5]

Countable Choice is the hypothesis of the Riesz representation theorem used below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

Given: Countable Choice, a real or complex Hilbert space $H$ , its dual $H^{*}$ , bidual $H^{* *}$ and canonical map $J_{H}$ .

1.1A1A5

By [A1] the Riesz map $R : H \to H^{*}$ is a bijection with $R (y) (x) = ⟨ x, y ⟩$ and $∥ R (y) ∥ = ∥ y ∥$ .

2.1step 1.1A1

Let $Φ \in H^{* *}$ and define $ψ (y) = Φ (R (y))$ ; since $R$ is conjugate-linear and $Φ$ is linear, $ψ (a y) = \overline{a} ψ (y)$ and $ψ (y + y^{'}) = ψ (y) + ψ (y^{'})$ , while $∣ ψ (y) ∣ \leq ∥ Φ ∥ ∥ R (y) ∥ = ∥ Φ ∥ ∥ y ∥$ ; hence $φ : = \overline{ψ}$ is a linear functional on $H$ with $∣ φ (y) ∣ \leq ∥ Φ ∥ ∥ y ∥$ .

3.1step 2.1A1

Applying Riesz representation to $φ$ gives $z \in H$ with $φ (y) = ⟨ y, z ⟩$ for every $y$ .

4.1step 3.1A1A2A4

Then for every $y$ one has $Φ (R (y)) = \overline{φ (y)} = \overline{⟨ y, z ⟩} = ⟨ z, y ⟩ = R (y) (z) = J_{H} (z) (R (y))$ by [A4] and [A2]; since $R$ is onto $H^{*}$ , every element of $H^{*}$ has the form $R (y)$ , so $Φ = J_{H} (z)$ lies in the range of $J_{H}$ .

5.1step 1.1step 4.1A3A5∎

Thus $J_{H}$ is surjective and $H$ is reflexive; the only choice assumption is the one inherited from Riesz representation in step 1.1.

Depends on

Used by

Dependency tree · two levels

25 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

Theo Bühler and Dietmar Salamon, Functional Analysis, Theorem 5.35, p.236 (standard reference, not scraped)
Andrew Lin and Casey Rodriguez, MIT 18.102 Introduction to Functional Analysis, Theorem 184 (standard reference, not scraped)
Bruce Blackadar, Ilijas Farah and Asaf Karagila, Hilbert spaces without the Countable Axiom of Choice, Theorem 2.0.6 (standard reference, not scraped)