advanced34 min studyLesson 786 in Mathematics

Banach–Alaoglu and weak-star compactness

Recover compactness in a weaker dual topology while keeping nets, sequences and separable preduals distinct.

On this page

What you’ll understand

  • Explain the product-space proof of dual-ball weak-star compactness.
  • State the separable-predual condition giving metrizable compact dual balls.
  • Construct a weak-star compact ball containing a sequence with no convergent subsequence.

The closed unit ball of an infinite-dimensional dual space is generally not compact in norm. Banach–Alaoglu restores compactness in the weak-star topology, where only finitely many evaluation tests appear in a basic neighborhood. General topological compactness guarantees convergent subnets rather than convergent subsequences of every sequence. A separable predual supplies the extra countable structure that turns a bounded dual ball into a compact metric space. This distinction is indispensable when a proof extracts limits from bounded functionals and when interpreting compactness as an algorithmic subsequence procedure.

Place each functional in a compact product

Let X be a real or complex normed space; completeness of X is not needed for this theorem. Every functional in its dual unit ball takes x into the closed scalar interval or disk of radius ∥x∥. Send the functional to this collection of values indexed by all x in X. Each coordinate set is compact, so Tychonoff’s theorem makes their product compact. The image consists of assignments that are linear and satisfy the stated coordinate bounds, and these conditions are closed under pointwise limits.

The map is injective because a functional is determined by all its evaluations. Its pulled-back product topology is precisely σ(X*,X), so the closed image proves weak-star compactness. This reasoning does not prove norm compactness: norm neighborhoods demand uniform control over every unit vector rather than finitely many chosen evaluations. It also explains why the theorem works over an uncountable index set. Replacing Tychonoff by a diagonal subsequence argument without an appropriate countable reduction would leave a genuine gap in the proof.

BX∗↪∏x∈X{z∈K:∣z∣≤∥x∥},f⟼(f(x))x∈X,BX∗ is σ(X∗,X)-compactB_{X^*}\hookrightarrow\prod_{x\in X}\{z\in\mathbb K:|z|\leq\|x\|\},\qquad f\longmapsto(f(x))_{x\in X},\qquad B_{X^*}\text{ is }\sigma(X^*,X)\text{-compact}
Linearity and the norm bound define a closed subset of the compact scalar product. The topology inherited through evaluation is weak-star.↔ Scroll wide equations horizontally.

Separability creates a metric on bounded balls

If X is separable, choose a countable dense set in its unit ball. On a norm-bounded subset of X*, convergence of evaluations at these vectors implies convergence at every x, because a uniform functional bound controls the approximation error. A weighted sum of the bounded evaluation discrepancies therefore gives a metric inducing weak-star topology on the dual unit ball. Compactness in this metric implies sequential compactness. Thus every bounded dual sequence has a weak-star convergent subsequence under the separable-predual assumption.

The assumption concerns X, the predual, rather than norm separability of X*. For example, ℓ¹ is separable while its dual ℓ∞ is not norm separable, yet the dual unit ball is weak-star compact and metrizable in this pairing. The metric assertion is restricted to bounded balls; it does not identify the weak-star topology on the whole dual with norm topology or claim a single countable set of tests controls unbounded functionals uniformly. In the general nonseparable setting, compactness alone provides a convergent subnet of a sequence viewed as a net, not necessarily a subsequence.

d(f,g)=∑j=1∞2−j∣f(xj)−g(xj)∣1+∣f(xj)−g(xj)∣(f,g∈BX∗, {xj} dense in BX)d(f,g)=\sum_{j=1}^\infty2^{-j}\frac{|f(x_j)-g(x_j)|}{1+|f(x_j)-g(x_j)|}\qquad(f,g\in B_{X^*},\ \{x_j\}\text{ dense in }B_X)
For separable X this metric induces weak-star topology on the dual ball. Uniform boundedness of that ball is what extends dense-set tests to every input.↔ Scroll wide equations horizontally.

Compact limits can remain far in norm

Weak-star limits preserve a dual norm bound: evaluating any fixed unit vector and passing to the limit keeps its modulus at most one. This gives a convenient way to construct bounded functionals from approximating ones. It does not make the norm differences small. Point evaluations on continuous functions illustrate the distinction: evaluation points can approach one another so every fixed continuous test agrees asymptotically, while different tests chosen for each pair still distinguish the corresponding functionals by their full dual norm.

A nonseparable predual can also make the failure of sequential compactness explicit. Give one coordinate to every subset of the natural numbers and let the nth bounded function record which subsets contain n. Any proposed subsequence can be defeated by choosing the coordinate containing exactly its even-positioned indices. The resulting scalar evaluations alternate. This does not contradict Banach–Alaoglu: its compactness is topological, and convergent subnets can have a more flexible directed index set than the natural-number subsequences that this argument excludes.

A compact dual ball without sequential compactness

Let I be the power set of N, X=ℓ¹(I), and X*=ℓ∞(I). Define fₙ(A)=1 if n∈A and zero otherwise. Show that (fₙ) lies in the weak-star compact unit ball but has no weak-star convergent subsequence.

  1. Each coefficient function has supremum norm one, so it defines a dual unit-ball functional. Banach–Alaoglu makes that ball compact for the ℓ¹(I) predual pairing.

  2. Consider any subsequence indexed by distinct increasing nⱼ. Choose the coordinate A={n₂,n₄,n₆,...}, a legitimate member of I.

  3. The predual unit vector e_A tests fₙⱼ at this coordinate. Its evaluations are zero for odd j and one for even j, so they do not converge. Weak-star convergence would require this fixed evaluation to converge.

The result

No subsequence converges weak-star, despite compactness of the containing dual ball. The predual is nonseparable, so the compact-metric subsequence conclusion is unavailable.

Explore the model

Interactive lab

Interactive lab · function

A fixed continuous test follows nearby evaluation points

Pause and Reset, then set coefficient a=1,position x=0.50. For the fixed test function t↦t², predict evaluation 0.25. For a fresh comparison Reset and set a=1,x=0.25; predict 0.0625, closer to evaluation zero at t=0. The lab computes this one quadratic test and its derivative/integral; it does not sample every continuous function or measure the dual norm between point evaluations. The compactness theorem and the norm-two calculation in the practice require separate arguments.

A fixed continuous test follows nearby evaluation pointsThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. 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 forAgreement on each fixed test is compatible with large dual norm differences when the distinguishing test changes with the pair.

Check your understanding

Which additional condition guarantees weak-star sequential compactness of a bounded dual ball by metrizability?

Try it yourself

In C[0,1]*, let δₜ evaluate a continuous function at t. Prove δ₁/ₙ converges weak-star to δ₀, but ∥δ₁/ₙ−δ₀∥=2 for every n. Explain the subsequence setting.

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

Give me a hint

Continuity handles each fixed test. For the norm lower bound choose a continuous function of norm one taking −1 at zero and +1 at 1/n.

Compare with the solution

For any fixed continuous h, h(1/n)→h(0), proving weak-star convergence. The triangle inequality bounds the dual norm difference by two. The test hₙ(t)=min(1,2nt−1) has supremum norm one and gives difference two between evaluation at 1/n and zero, attaining that bound. The space C[0,1] is separable, so its bounded dual balls are also weak-star metrizable and sequentially compact. This particular sequence already converges weak-star in full; the norm separation concerns a stronger topology.

Take it with you

  • Dual unit balls are compact in weak-star topology, even for normed preduals.
  • A separable predual makes bounded dual balls metrizable and sequentially compact.
  • General compactness gives subnets and does not imply norm convergence or sequential compactness.

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