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Separable reflexive space has separable dual
Statement
Assume the Axiom of Countable Choice and the relative Hahn–Banach principle HB. If a real or complex Banach space is reflexive and norm separable, then its continuous dual is norm separable.
Facts & Assumptions
Given: , HB, and a real or complex separable reflexive Banach space .
A space is separable precisely when it has an at most countable dense subset (Separability: the existence of an at most countable dense subset).
Reflexivity says that the canonical map is a surjective isometric embedding (Reflexivity is surjectivity of the canonical map).
Under and HB, a real or complex normed space whose continuous dual is norm separable is itself norm separable (Separable dual implies separable primal).
Proof
By [F1], fix an at most countable norm-dense set . Its image is at most countable: the restriction of the injective map is a bijection from onto that image.
The image is norm dense in . Indeed, for and , surjectivity in [F2] gives with , and density of gives with ; the isometry in [F2] then gives . Thus is norm separable by [F1].
Apply [F3] to the normed space . Its continuous dual is , which is separable by step 2.1, so is norm separable. No new selection or separation is made here: and HB are used exactly through [F3].
Source notes
Brezis proves the same implication by identifying with and applying the separable-dual theorem to . Reflexivity is essential: Brezis's Remark 19 records as separable with nonseparable dual ; that warning is source context and is not used as a supplier in the proof above.
Depends on
Used by
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Dependency tree · two levels
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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)