CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-09

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Relative Hahn–Banach makes the canonical bidual map an isometry

Statement

Assume HB. For every real or complex normed space $X$ , the canonical scalar-linear map $J_{X} : X \to X^{* *}$ , given by $J_{X} (x) (f) = f (x)$ , satisfies $∥ J_{X} x ∥ = ∥ x ∥ (x \in X) .$ It preserves distances and is injective. Surjectivity is not claimed.

Facts & Assumptions

[F1]

Under HB, each nonzero $x$ has a functional $f$ with $∥ f ∥ = 1$ and $f (x) = ∥ x ∥$ (Relative dual norming, point separation, and recovery of the norm).

[F2]

Evaluation defines a scalar-linear $J_{X} : X \to X^{* *}$ with $∥ J_{X} x ∥ \leq ∥ x ∥$ (Evaluation defines a bounded scalar-linear map into the bidual).

Proof

Given: HB, a normed real or complex space $X$ , and the evaluation map $J_{X}$ .

1.1givenF2

By the evaluation construction, $J_{X}$ is scalar-linear and $∥ J_{X} x ∥ \leq ∥ x ∥$ for every $x$ . In particular $J_{X} 0 = 0$ and equality of the norms holds at zero.

2.1step 1.1F1F2

For $x \neq 0$ , HB norming gives $f \in X^{*}$ with $∥ f ∥ = 1$ and $f (x) = ∥ x ∥$ . The bidual norm is the supremum over the dual unit ball, which contains this $f$ , so $∥ J_{X} x ∥ \geq ∣ J_{X} x (f) ∣ = ∣ f (x) ∣ = ∥ x ∥$ . Combining with step 1.1 proves equality at every $x$ .

3.1step 1.1step 2.1algebra∎

For $x, y \in X$ , linearity and the established equality give $∥ J_{X} x - J_{X} y ∥ = ∥ J_{X} (x - y) ∥ = ∥ x - y ∥$ . If $J_{X} x = J_{X} y$ , the left side is zero, so definiteness of the norm gives $x = y$ .

Source notes

Brezis §1.3, pp.8–9, first displayed isometry calculation; Teschl Theorem 4.20, pp.115–116.

Relative Hahn–Banach makes the canonical bidual map an isometry

Statement

Facts & Assumptions

Proof

Source notes

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Sources