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A convex norm-lower-semicontinuous functional is weakly lower semicontinuous

Statement

Assume HB (The real dominated-extension principle as an additional hypothesis over ZF) and Countable Choice (The Axiom of Countable Choice (ACω)). Let X be a real normed space, let K⊆X be nonempty and convex (Convex and strictly convex functionals on a convex subset of a real vector space) and let I:K→(−∞,+∞] be convex and sequentially lower semicontinuous in the norm topology (Proper, coercive and weakly lower semicontinuous extended-real functionals). Then I is weakly sequentially lower semicontinuous on K: for every (uj)⊆K with uj⇀u∈K (Weak convergence of nets and sequences), I(u)≤lim inf⁡jI(uj).

Facts & Assumptions

Given: HB and Countable Choice; a real normed space X, a nonempty convex set K⊆X, and a convex functional I:K→(−∞,+∞] that is sequentially lower semicontinuous in the norm topology.

[F1]

A convex functional has convex sublevel sets: for every t∈R the set {v∈K:I(v)≤t} is convex (Convex and strictly convex functionals on a convex subset of a real vector space).

[F2]

Norm sequential lower semicontinuity means I(v)≤lim inf⁡jI(vj) whenever vj→v∈K in norm with vj∈K (Proper, coercive and weakly lower semicontinuous extended-real functionals). It gives closedness of sublevels relative to K, not necessarily in X.

[F3]

Under HB the norm and weak closures in X of any convex subset coincide (Norm closed convex iff weakly closed).

[F6]

Countable Choice selects a point from each nonempty set S∩B(u,1/m), m≥1, whenever u lies in the norm closure of S (The Axiom of Countable Choice (ACω)).

[F4]

Weak convergence uj⇀u is convergence in σ(X,X∗); in particular every subsequence of a weakly convergent sequence converges weakly to the same limit (Weak convergence of nets and sequences).

[F5]

Limit inferior: if lim inf⁡jaj<t for a sequence in (−∞,+∞] and a real t, then aj≤t for infinitely many j, so a strictly increasing sequence of indices jk with ajk≤t for all k exists (Limit superior and limit inferior of a real sequence as inf⁡nsup⁡k≥nxk and sup⁡ninf⁡k≥nxk in R‾).

Proof

technique · direct, by comparing the norm and weak closures of a convex sublevel set
1.1F5givenalgebra

Suppose uj⇀u with (uj)⊆K and u∈K, and put ℓ:=lim inf⁡jI(uj). If I(u)>ℓ, then since I(u)∈(−∞,+∞] there is a real t with ℓ<t<I(u) (if I(u) is finite take t between; if I(u)=+∞ take any real t>ℓ).

2.1F4F5step 1.1

A subsequence in the sublevel set. By [F5], applied to the sequence (I(uj)) and this t, there is a strictly increasing sequence of indices jk with I(ujk)≤t for every k. By [F4] the subsequence still satisfies ujk⇀u.

3.1F1F3step 2.1

Use the ambient closures. Put St:={v∈K:I(v)≤t}. It is nonempty by step 2.1 and convex by [F1]. Since ujk∈St and ujk⇀u, the point u lies in the weak closure of St in X. By [F3] it therefore lies in its norm closure. No ambient closedness of K or St is required.

4.1F2F6step 3.1given

Recover the relative sublevel inequality. For each integer m≥1, choose vm∈St with ∥vm−u∥<1/m, using [F6]. Then vm→u in norm, vm∈K and u∈K. Thus [F2] gives I(u)≤lim inf⁡mI(vm)≤t.

5.1step 4.1step 1.1∎

Conclusion. Step 4.1 gives I(u)≤t<I(u) by the choice of t in step 1.1, a contradiction; hence I(u)≤ℓ=lim inf⁡jI(uj). As (uj) and u were arbitrary, I is weakly sequentially lower semicontinuous on K.

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