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A minimising sequence need not converge strongly

Statement refuted

Counterexample. Assume the Axiom of Choice (The Axiom of Choice). Let H=ℓ2(N,R), let B={u∈H:∥u∥≤1} be its closed unit ball and let I(u)=∑j≥01j+1uj2(u∈H). Then I is nonnegative, convex and continuous, I(0)=0, and uk:=ek is a minimising sequence for inf⁡BI=0 with I(ek)=1/(k+1)→0; moreover ek⇀0∈B and 0 is the unique minimiser of I on B, but ∥ek−0∥=1 for every k, so no subsequence of the minimising sequence converges strongly to the minimiser. The compactness recovered in the direct method is therefore genuinely weak compactness (The direct method in a reflexive Banach space).

Facts & Assumptions

Given: The Axiom of Choice; the real Hilbert space H=ℓ2(N,R) (Square-summable families on an arbitrary index set and the space ℓ2(I)), its coordinate vectors ek, namely the family that is 1 at k and 0 elsewhere, its closed unit ball B={u∈H:∥u∥≤1}, and the functional I(u)=∑j≥0uj2/(j+1). The counting-measure dictionary ℓp is the Lp space of counting measure identifies H with real L2(#), which is reflexive (and thus Banach) by Reflexivity of Lp for one less p less infinity under Countable Choice, supplied here by AC; the series pairing is its inner product.

[F1]

The series defining I converges absolutely for every u∈H because 0≤uj2/(j+1)≤uj2 and ∑juj2=∥u∥2<∞; moreover ∣I(u+v)−I(u)∣≤2∥u∥∥v∥+∥v∥2, so I is continuous, and I is convex and nonnegative (Square-summable families on an arbitrary index set and the space ℓ2(I)).

[F2]

Each coordinate vector satisfies ek∈H, ∥ek∥=1 and ek⇀0: by the duality of ℓp and ℓq every bounded linear functional on ℓ2 is Λ(a)=∑kakbk for a unique b∈ℓ2 (Counting measure specializes the representation theorem to ℓp and ℓq), so Λ(ek)=bk, and bk→0 because a square-summable family has small tails, that is, for every ε>0 there is a finite F with ∑k∉F∣bk∣2<ε (Square-summable families on an arbitrary index set and the space ℓ2(I)), whence ∣bk∣2<ε for every k beyond all elements of F; weak convergence means convergence against every bounded linear functional (Weak convergence of nets and sequences).

[F3]

In the direct method the compactness recovered is weak sequential compactness: a norm-bounded sequence in a reflexive Banach space has a weakly convergent subsequence (A bounded sequence in a reflexive Banach space has a weakly convergent subsequence), and the limit need not be a strong limit (The direct method in a reflexive Banach space, Proper, coercive and weakly lower semicontinuous extended-real functionals).

Counterexample

technique · direct computation along the standard basis
1.1F1algebra

I is nonnegative, convex and continuous. Nonnegativity and convexity are immediate from the formula. For continuity, expanding the squares gives I(u+v)−I(u)=∑j(2ujvj+vj2)/(j+1), and by Cauchy-Schwarz, ∑j∣2ujvj∣/(j+1)≤2(∑juj2)1/2(∑jvj2)1/2=2∥u∥∥v∥ while ∑jvj2/(j+1)≤∥v∥2; hence ∣I(u+v)−I(u)∣≤2∥u∥∥v∥+∥v∥2→0 as v→0.

1.2F1F2algebra

The minimising sequence. The point 0∈B has I(0)=0, and I≥0, so inf⁡BI=0. By [F2] the vectors ek lie in B and satisfy I(ek)=1/(k+1)→0, so (ek) is a minimising sequence for inf⁡BI.

2.1step 1.1algebra

The unique minimiser. If u∈B has I(u)=0, then uj2/(j+1)=0 for every j, hence uj=0 for every j and u=0. So 0 is the unique minimiser of I on B.

3.1F2step 2.1

No strong convergence of the minimising sequence. By [F2] one has ek⇀0 with ∥ek∥=1 for every k, so ∥ek−0∥=1 does not tend to 0; a fortiori no subsequence of (ek) converges in norm to 0, the unique minimiser of step 2.1.

4.1F2F3step 1.2step 3.1∎

Conclusion. The bounded minimising sequence (ek) converges weakly to 0 by [F2], but no subsequence converges in norm to the unique minimiser by step 3.1. Thus the weak subsequence conclusion described in [F3], under that theorem's stated principles, cannot be upgraded to strong convergence of minimising sequences. No additional extraction is needed for this explicit witness.

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