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A norm-closed convex set is weakly sequentially closed

Statement

Assume the Axiom of Choice. Let X be a real or complex normed space and let K⊆X be convex and closed in the norm topology. Then K is closed in the weak topology σ(X,X∗) (Weak topology on a normed space); in particular K is weakly sequentially closed: if (uj)⊆K and uj⇀u (Weak convergence of nets and sequences), then u∈K.

Facts & Assumptions

Given: The Axiom of Choice (The Axiom of Choice); a real or complex normed space X with dual X∗; a convex set K⊆X that is closed in the norm topology. The weak topology is σ(X,X∗) (Weak topology on a normed space) and weak sequential convergence is as in Weak convergence of nets and sequences.

[F1]

Under the Axiom of Choice, two disjoint nonempty convex sets C,K⊆X, with C closed and K compact, are strongly separated by a nonzero functional in X∗ (Strong separation of a closed and a compact convex set): there are f∈X∗, f≠0, and a positive gap sup⁡CRe⁡f<inf⁡KRe⁡f.

[F2]

The weak topology is the initial topology of the maps f:X→K, f∈X∗, hence every set {y∈X:Re⁡f(y)>α} with f∈X∗ and α∈R is weakly open, and a subset of X is weakly closed exactly when its complement is weakly open (Weak topology on a normed space).

[F3]

A sequence uj⇀u converges weakly in the sense of convergence in σ(X,X∗); a weakly closed set contains the limit of every weakly convergent sequence contained in it (Weak convergence of nets and sequences).

Proof

technique · direct, by separating an exterior point from $K$ with a weak half-space
1.1givenalgebra

Trivial case and set-up. If K=∅ then K is closed in every topology, so both assertions hold; assume henceforth K≠∅ and fix a point x∈X∖K.

2.1F1step 1.1

Strong separation of K and the singleton {x}. The sets K and {x} are nonempty and convex, K is closed in the norm topology and {x} is compact; they are disjoint because x∉K. By [F1], applied here, there are f∈X∗, f≠0, and a real number α with sup⁡KRe⁡f≤α<Re⁡f(x), the gap being the one supplied by the theorem.

3.1F2step 2.1

A weak neighbourhood of x missing K. Put U:={y∈X:Re⁡f(y)>α}. By [F2] the set U is open in σ(X,X∗); it contains x because Re⁡f(x)>α, and it is disjoint from K because every y∈K satisfies Re⁡f(y)≤sup⁡KRe⁡f≤α.

4.1step 3.1F2

K is weakly closed. Since x∈X∖K was arbitrary and step 3.1 produces for it a weak neighbourhood U⊆X∖K, the complement X∖K is weakly open; equivalently K is closed in the weak topology σ(X,X∗).

5.1F3step 4.1∎

Weak sequential closedness. Let (uj)⊆K with uj⇀u. By [F3] the convergence is convergence in σ(X,X∗), and a set closed in a topology contains the limit of every convergent sequence in it; hence u∈K.

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