TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedverified 2026-09-26 (gpt-6-sol)

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Double annihilators give norm and weak-star closures

Statement

Let $K = R$ or $C$ . For a normed $X$ and a linear subspace $N \leq X^{*}$ , $(^{⊥} N)^{⊥} = {\overline{N}}^{σ (X^{*}, X)} .$ Consequently $N$ is weak-star closed if and only if $N = (^{⊥} N)^{⊥}$ . If $^{⊥} N = {0}$ , then $N$ is weak-star dense in $X^{*}$ .

Assume the Axiom of Choice (The Axiom of Choice) for the remaining conclusions. Then $N$ is weak-star dense in $X^{*}$ if and only if $^{⊥} N = {0}$ . For a linear subspace $M \leq X$ , the primal formula is $^{⊥} (M^{⊥}) = {\overline{M}}^{∥ \cdot ∥}$ .

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From The weak-star topology from finite evaluations, with its stated hypotheses: Let $K = R$ or $C$ . For a normed $X$ with continuous dual $X^{*}$ from def-dual-space-of-a-normed-space, the weak-star topology $σ (X^{*}, X)$ is the initial topology of all evaluations $f \mapsto f (x)$ into $K$ with its usual topology. At $f_{0}$ , a neighbourhood basis consists of $U (f_{0}; x_{1}, \dots, x_{n}; ε) = {f \in X^{*} : ∣ (f - f_{0}) (x_{j}) ∣ < ε (1 \leq j \leq n)},$ where $n$ is finite and $ε > 0$ . For $n = 0$ the set is all of $X^{*}$ . Finite intersections of inverse images of scalar open sets form the initial-topology basis; at the given point, finitely many disks can be refined using their smallest positive radius. Weak-star closure means closure in this topology, not merely sequential closure.

[F2]

From Annihilator notation and the preannihilator, with its stated hypotheses: Let $K = R$ or $C$ . For a normed $X$ and arbitrary subsets $M \subseteq X$ , $N \subseteq X^{*}$ , define $M^{⊥} = {f \in X^{*} : f (m) = 0 for all m \in M},^{⊥} N = {x \in X : f (x) = 0 for all f \in N} .$ Here $X^{*}$ is def-dual-space-of-a-normed-space. The first notation agrees with def-continuous-annihilator-of-a-subspace on $span M$ , since linearity makes vanishing on $M$ equivalent to vanishing on its span. The preannihilator lies in $X$ , not in $X^{* *}$ . Empty sets impose no conditions: $\emptyset^{⊥} = X^{*}$ and $^{⊥} \emptyset = X$ .

[F3]

From Finite evaluations separate a functional from a dual subspace, with its stated hypotheses: Let $K = R$ or $C$ . Let $X$ be normed, $N \leq X^{*}$ a linear subspace, and $x_{1}, \dots, x_{n} \in X$ with $n \geq 1$ . Define $E (f) = (f (x_{1}), \dots, f (x_{n}))$ . If $f_{0} \in X^{*}$ satisfies $E (f_{0}) \notin E (N)$ , there is $x \in span {x_{1}, \dots, x_{n}}$ such that $g (x) = 0$ for all $g \in N$ and $f_{0} (x) = 1$ .

[F4]

Under the Axiom of Choice, The annihilator detects the closure of a subspace gives for every linear subspace $M \subseteq X$ the identity $\overline{M} = ⋂_{f \in M^{⊥}} \ker f$ ; its proof uses Hahn–Banach separation.

Proof

1.1F1F2

If $x \in^{⊥} N$ , evaluation at $x$ vanishes on $N$ . Its kernel is weak-star closed by the definition of that topology. Intersecting these kernels shows ${\overline{N}}^{σ (X^{*}, X)} \subseteq (^{⊥} N)^{⊥}$ .

1.2F1

Let $f_{0}$ lie outside the weak-star closure. Choose a basic neighbourhood $U (f_{0}; x_{1}, \dots, x_{n}; ε)$ disjoint from $N$ . Since $0 \in N$ , this neighbourhood cannot have $n = 0$ . Its finite-coordinate map satisfies $E (f_{0}) \notin E (N)$ , since equality with $E (g)$ would put that $g \in N$ in $U$ .

2.1F2F3step 1.1step 1.2

Finite-evaluation separation supplies $x \in^{⊥} N$ with $f_{0} (x) = 1$ . Hence $f_{0} \notin (^{⊥} N)^{⊥}$ . Together with step 1.1, this proves equality.

3.1F2F4step 2.1

A set is closed exactly when it equals its closure, so the first equivalence follows in both directions. If $^{⊥} N = 0$ , the equality gives closure $X^{*}$ without choice. Conversely, assume AC and suppose the closure is $X^{*}$ ; then any $x \in^{⊥} N$ is annihilated by all of $X^{*}$ . The primal formula [F4] with $M = 0$ says $^{⊥} X^{*} = 0$ , so $x = 0$ .

4.1F2F4step 2.1∎

Under AC, the primal formula in the statement is exactly [F4], with the preannihilator notation unpacked. The weak-star identity itself checks the extremes $N = 0$ and $N = X^{*}$ : their closures and double annihilators are respectively $0$ and $X^{*}$ . For $X = 0$ these coincide.

Depends on

Used by

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Sources

Bühler–Salamon, Functional Analysis, Corollary 3.26(i)–(iii), p.130; Corollary 2.55, p.84 (real source; K-linear finite-coordinate proof covers C) (standard reference, not scraped)