advanced32 min studyLesson 787 in Mathematics

Reflexivity and weak sequential compactness

Use the canonical bidual map and Eberlein–Šmulian to extract weak subsequences without assuming a separable ambient space.

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What you’ll understand

  • Define reflexivity as surjectivity of the canonical evaluation embedding.
  • State the weak compactness and sequential compactness characterizations correctly.
  • Use weak lower semicontinuity in a constrained norm minimization.

Reflexivity identifies a Banach space with all the functionals on its dual through the canonical evaluation map. This turns dual weak-star compactness into weak compactness of the original unit ball. Eberlein–Šmulian then gives a surprising sequential characterization specific to Banach-space weak topology, without requiring the whole space to be separable. These facts support existence arguments for minimizers, but they do not turn weak convergence into norm convergence or provide Hilbert-space orthogonality in every reflexive space.

The canonical map must be onto

For a Banach space X, the canonical map J sends x to the functional on X* that evaluates f at x. Hahn–Banach shows it is an isometry. Reflexivity means J is surjective, not merely that some abstract isomorphism between X and X** happens to exist. Under this canonical identification, weak topology on X corresponds to weak-star topology on X**. Banach–Alaoglu therefore gives weak compactness of the closed unit ball when X is reflexive. The converse also holds, using density of the canonical unit ball in the bidual unit ball and compactness.

Hilbert spaces are reflexive, and so are ℓᵖ spaces for 1<p<∞. The endpoints c₀ and ℓ¹ are not reflexive, despite being familiar complete separable spaces. A closed linear subspace of a reflexive Banach space is reflexive, as is a quotient by a closed subspace. Reflexivity also passes between a Banach space and its dual. These permanence results concern canonical dual structure; they do not say that a reflexive space admits an inner product norm, an orthogonal basis or a bounded projection onto every closed subspace.

J:X⟶X∗∗,Jx(f)=f(x),X reflexive  ⟺  J(X)=X∗∗  ⟺  BX is weakly compactJ:X\longrightarrow X^{**},\quad Jx(f)=f(x),\qquad X\text{ reflexive}\iff J(X)=X^{**}\iff B_X\text{ is weakly compact}
The equivalence assumes X is Banach and uses its canonical embedding. Weak compactness refers to σ(X,X*), not norm topology.↔ Scroll wide equations horizontally.

Eberlein–Šmulian restores a sequential test

For a subset K of a Banach space, weak compactness is equivalent to weak sequential compactness: every sequence in K has a weakly convergent subsequence whose limit remains in K. In the relatively compact version, the subsequential limits may lie in its weak closure. This is the Eberlein–Šmulian theorem. It is a special theorem about weak topology, not a general equivalence for arbitrary compact spaces. In particular, it does not contradict the preceding nonseparable dual-ball example, which used weak-star rather than weak topology.

Combining the theorem with reflexivity shows that every bounded sequence in a reflexive Banach space has a weakly convergent subsequence. No separability assumption on the entire space is necessary. The closed span of a given sequence is separable, which helps explain the countable structure available inside this problem, but the full equivalence is more than an unsupported diagonal argument. The weak limit can have a smaller norm: coordinate unit vectors in ℓ² converge weakly to zero while all have norm one. Thus a weakly compact unit ball need not be norm compact.

K weakly compact  ⟺  ∀(xn)⊆K ∃xnj⇀x∈K,xn⇀x ⟹ ∥x∥≤lim inf⁡n∥xn∥K\text{ weakly compact}\iff\forall(x_n)\subseteq K\ \exists x_{n_j}\rightharpoonup x\in K,\qquad x_n\rightharpoonup x\ \Longrightarrow\ \|x\|\leq\liminf_n\|x_n\|
The first equivalence is Eberlein–Šmulian for subsets of a Banach space. The second is weak lower semicontinuity of the norm.↔ Scroll wide equations horizontally.

Weak limits preserve convex constraints

A norm-closed convex set is weakly closed by Hahn–Banach separation. In a reflexive Banach space, intersecting such a set with a closed bounded ball yields a weakly compact set. To minimize the norm over a nonempty closed convex set, take a minimizing sequence, which is bounded once one finite competitor is fixed. Extract a weak subsequence, retain feasibility by weak closedness, and use the lower-semicontinuity inequality to obtain a minimizer. Uniqueness requires more geometry, such as strict convexity of the norm, and does not follow from reflexivity alone.

More general direct-method arguments replace the norm by a weakly lower-semicontinuous objective whose sublevel sets supply boundedness. Coercivity is one common sufficient condition for that boundedness. Convex norm-lower-semicontinuous objectives are weakly lower semicontinuous, but an arbitrary norm-continuous nonconvex objective need not have this property. The method therefore separates three tasks: produce a bounded minimizing sequence, keep its weak limit inside the feasible set, and pass the objective inequality to the limit. Reflexivity supplies compactness rather than performing all three tasks by itself.

A minimum-norm solution in an infinite-dimensional Hilbert space

In real ℓ², minimize ∥x∥₂ subject to 3x₁+4x₂=10. Find the minimum and show why the remaining infinitely many coordinates do not help.

  1. The constraint defines a nonempty norm-closed affine set and hence a weakly closed convex set. Reflexivity and weak lower semicontinuity ensure existence of a minimum.

  2. Cauchy–Schwarz gives 10≤√(3²+4²)√(x₁²+x₂²)≤5∥x∥₂, so every feasible vector has norm at least two.

  3. Choose x=(6/5,8/5,0,0,...). The constraint becomes 18/5+32/5=10 and its norm is two. Equality forces the first two coordinates to align with (3,4) and all other coordinates to vanish.

The result

The minimum is two and the unique minimizer is (6/5,8/5,0,...). Uniqueness here uses Hilbert norm geometry, beyond reflexive existence.

Explore the model

Interactive lab

Interactive lab · vectors

Bounded vectors need not be close to each other

Pause and Reset, then set angle=0°,magnitude=1. For a fresh comparison Reset and set angle=90°,magnitude=1. Predict norm one in both cases and Euclidean distance √2 between the two displayed vectors. This illustrates that a norm bound alone is no Cauchy estimate. In this finite plane closed bounded sets are norm compact; the infinite-dimensional ℓ² example instead needs weak compactness and all dual tests. Playback does not generate an arbitrary weakly convergent sequence.

Bounded vectors need not be close to each otherPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forWeak subsequence existence and norm-Cauchy behavior are different conclusions, even when all vectors satisfy the same norm bound.

Check your understanding

What does Eberlein–Šmulian permit for a bounded sequence in any reflexive Banach space?

Try it yourself

In c₀, let u⁽ⁿ⁾ have its first n entries equal to one and the rest zero. Prove that this bounded sequence has no weakly convergent subsequence and deduce nonreflexivity of c₀.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Any subsequence with indices tending to infinity has each fixed coordinate eventually equal to one. Coordinate evaluation is a bounded functional on c₀.

Compare with the solution

Every u⁽ⁿ⁾ has supremum norm one. If a subsequence converged weakly to some x in c₀, evaluating its kth coordinate would force xₖ=1 for every k. The constant-one sequence is not in c₀, so such a limit cannot exist. Thus the unit ball is not weakly sequentially compact. Eberlein–Šmulian and the reflexivity characterization show c₀ is not reflexive. The same sequence can have a weak-star limit when placed in a different dual space, which changes both the space and its topology.

Take it with you

  • Reflexivity is surjectivity of the canonical map into the bidual.
  • Eberlein–Šmulian supplies weak subsequences in bounded reflexive sets without ambient separability.
  • Weak compactness and lower semicontinuity support existence, while uniqueness needs extra geometry.

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