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Separable reflexive space has separable dual

Statement

Assume the Axiom of Countable Choice ${A C}_{ω}$ and the relative Hahn–Banach principle HB. If a real or complex Banach space $X$ is reflexive and norm separable, then its continuous dual $X^{*}$ is norm separable.

Facts & Assumptions

Given: ${A C}_{ω}$ , HB, and a real or complex separable reflexive Banach space $X$ .

[F1]

A space is separable precisely when it has an at most countable dense subset (Separability: the existence of an at most countable dense subset).

[F2]

Reflexivity says that the canonical map $J_{X} : X \to X^{* *}$ is surjective (Reflexivity is surjectivity of the canonical map); under the assumed relative HB principle it is also an isometric embedding (Relative Hahn–Banach makes the canonical bidual map an isometry).

[F3]

Under ${A C}_{ω}$ and HB, a real or complex normed space whose continuous dual is norm separable is itself norm separable (Separable dual implies separable primal).

Proof

technique · transport a dense set through the canonical isometry and apply the preceding theorem to the dual space

1.1givenF1F2

By [F1], fix an at most countable norm-dense set $D \subseteq X$ . Its image $J_{X} [D]$ is at most countable: the restriction of the injective map $J_{X}$ is a bijection from $D$ onto that image.

2.1step 1.1F1F2

The image $J_{X} [D]$ is norm dense in $X^{* *}$ . Indeed, for $Φ \in X^{* *}$ and $ε > 0$ , surjectivity in [F2] gives $x \in X$ with $Φ = J_{X} x$ , and density of $D$ gives $d \in D$ with $∥ x - d ∥ < ε$ ; the isometry in [F2] then gives $∥ Φ - J_{X} d ∥ = ∥ J_{X} (x - d) ∥ = ∥ x - d ∥ < ε$ . Thus $X^{* *}$ is norm separable by [F1].

3.1givenstep 2.1F3∎

Apply [F3] to the normed space $Y = X^{*}$ . Its continuous dual is $Y^{*} = X^{* *}$ , which is separable by step 2.1, so $X^{*}$ is norm separable. No new selection or separation is made here: ${A C}_{ω}$ and HB are used exactly through [F3].

Source notes

Brezis proves the same implication by identifying $X$ with $X^{* *}$ and applying the separable-dual theorem to $X^{*}$ . Reflexivity is essential: Brezis's Remark 19 records $L^{1}$ as separable with nonseparable dual $L^{\infty}$ ; that warning is source context and is not used as a supplier in the proof above.

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Sources

Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Corollary 3.27 (standard reference, not scraped)
Bühler–Salamon, Functional Analysis, Theorem 2.73(ii) (standard reference, not scraped)