Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)
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Weak closedness keeps the direct-method limit admissible

Statement

Let X be a normed space and let K⊆X be weakly sequentially closed (Weak convergence of nets and sequences). If (uj)⊆K and uj⇀u, then u∈K. In particular, under the Axiom of Choice every nonempty convex norm-closed K has this property, by A norm-closed convex set is weakly sequentially closed.

Facts & Assumptions

Given: A normed space X and a subset K⊆X that is weakly sequentially closed: every sequence (uj)⊆K with uj⇀u satisfies u∈K (Weak convergence of nets and sequences).

[F1]

A set K is weakly sequentially closed when it contains the weak limit of every weakly convergent sequence contained in it; the relation uj⇀u denotes convergence in the weak topology σ(X,X∗) (Weak convergence of nets and sequences).

[F2]

Under the Axiom of Choice, a convex subset of a real or complex normed space that is closed in the norm topology is closed in the weak topology σ(X,X∗), hence weakly sequentially closed (A norm-closed convex set is weakly sequentially closed).

Proof

technique · direct, unpacking the definition and quoting the closed-convex lemma
1.1F1given

First assertion. Let (uj)⊆K with uj⇀u. By the definition [F1] of weak sequential closedness of K recorded in the hypothesis, u∈K; this is exactly the first sentence of the statement.

2.1F1F2step 1.1∎

Second assertion. Assume additionally that the Axiom of Choice holds and that K is nonempty, convex and norm closed. By [F2] the set K is weakly closed, and a weakly closed set is in particular weakly sequentially closed: if (uj)⊆K and uj⇀u, then u lies in the weak closure of K, which is K. Hence such a K satisfies the hypothesis of step 1.1 and contains every weak limit of its weakly convergent sequences.

Depends on

Used by

Dependency tree · two levels

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Sources