Compactness Through Evaluation Maps
Weak-star convergence describes what happens when a sequence of functionals is tested on each fixed vector. Compactness asks for a stronger, topological kind of control: every open cover of a compact set has a finite subcover. The distinction matters because compactness can provide existence results even when no norm-convergent subsequence is available.
For a normed space \(X\), the weak-star topology on \(X^*\) is the topology generated by the evaluation maps \(F\mapsto F(x)\), one for each \(x\in X\). The weak topology on \(X\) is generated by the maps \(x\mapsto F(x)\), one for each \(F\in X^*\). Thus both topologies are built from scalar-valued tests, but they live on different spaces and use different families of tests. Weak-star convergence was defined in the previous tutorial; here we use its topology to study compact sets.
The central result is the Banach–Alaoglu theorem: closed balls in \(X^*\) are compact in the weak-star topology. It does not require \(X\) to be complete. Its main ingredient is Tychonoff’s theorem, which says that an arbitrary product of compact spaces, equipped with the product topology, is compact. The product may have infinitely many coordinates; this is essential to the argument.
The Banach–Alaoglu Theorem
A functional in a bounded ball of \(X^*\) has bounded values at every fixed vector. This puts all such values into compact intervals. We can therefore represent functionals as points in a product of intervals, one coordinate for each vector of \(X\). Linearity and the norm bound then identify the functionals as a closed subset of that product.
Proof. For each \(x\in X\), let \(I_x=[-R\|x\|,R\|x\|]\), and form the product $$ P=\prod_{x\in X} I_x $$ with the product topology. Each interval is compact, including the singleton interval \(I_0=\{0\}\). By Tychonoff’s theorem, \(P\) is compact.
A point \(a\in P\) is a family of real numbers \((a_x)_{x\in X}\). Consider the subset \(A\) defined by the following equations for every \(x,y\in X\) and every real scalar \(\lambda\): $$ a_{x+y}=a_x+a_y,\qquad a_{\lambda x}=\lambda a_x. $$ Each equation specifies a closed subset of \(P\). For example, the map \(a\mapsto a_{x+y}-a_x-a_y\) is continuous because it uses only three product coordinates, and the equation sets this map equal to zero. The same reasoning applies to the homogeneity equation. Therefore \(A\), an intersection of closed sets, is closed in \(P\).
Every \(a\in A\) defines a linear functional \(F_a:X\to\mathbb{R}\) by \(F_a(x)=a_x\). Since \(a\in P\), $$ |F_a(x)|=|a_x|\leq R\|x\|\qquad(x\in X). $$ This bound shows that \(F_a\) is continuous and \(\|F_a\|\leq R\). Conversely, if \(F\in B_R^*\), then \((F(x))_{x\in X}\in P\), since \(|F(x)|\leq R\|x\|\), and its coordinates satisfy both linearity equations. Hence \(A\) consists exactly of the coordinate families associated with members of \(B_R^*\).
The coordinate map \(F\mapsto(F(x))_{x\in X}\) identifies the weak-star topology on \(B_R^*\) with the topology inherited from \(P\): both are generated by the same coordinate evaluations. Thus \(B_R^*\) is homeomorphic to the closed subset \(A\) of the compact space \(P\). A closed subset of a compact space is compact, so \(B_R^*\) is weak-star compact. This also covers \(R=0\), when the ball consists only of the zero functional. \(\square\)
The proof makes clear why boundedness alone is not the entire argument. The product provides compact ranges for all evaluations, while the closed equations enforce that the coordinates actually come from a linear functional. Without those equations, an arbitrary point of the product need not represent a functional.
Worked Example: The Dual Ball of \(\ell^1\)
Let \(X=\ell^1\), with \(\|x\|_1=\sum_{k=1}^{\infty}|x_k|\). A bounded sequence \(a=(a_k)\in\ell^\infty\) defines a continuous linear functional $$ F_a(x)=\sum_{k=1}^{\infty}a_kx_k, $$ because the series is absolutely convergent and $$ |F_a(x)|\leq\|a\|_\infty\|x\|_1. $$ In fact, every member of \((\ell^1)^*\) has this form. If \(F\in(\ell^1)^*\), set \(a_k=F(e_k)\), where \(e_k\) has value \(1\) in coordinate \(k\) and \(0\) elsewhere. Since \(\|e_k\|_1=1\), we have \(|a_k|\leq\|F\|\). For vectors with finitely many nonzero coordinates, linearity gives \(F(x)=\sum_k a_kx_k\); truncating an arbitrary \(x\in\ell^1\) and using continuity gives the same identity for every \(x\in\ell^1\).
Consequently, the unit ball of \((\ell^1)^*\) corresponds to the set of sequences with \(|a_k|\leq1\) for every \(k\). Its weak-star topology is exactly coordinatewise convergence on this ball. One direction follows by testing at \(e_k\). For the other, suppose \(a^{(n)}_k\to a_k\) for every \(k\), with all coordinates bounded in absolute value by \(1\). For \(x\in\ell^1\), fix \(m\) and estimate $$ \left|\sum_{k=1}^{\infty}(a^{(n)}_k-a_k)x_k\right| \leq\sum_{k=1}^{m}|a^{(n)}_k-a_k||x_k|+2\sum_{k>m}|x_k|. $$ The tail can be made arbitrarily small by choosing \(m\) large; for this fixed \(m\), the finite sum tends to zero. Thus \(F_{a^{(n)}}(x)\to F_a(x)\) for every \(x\in\ell^1\), which is weak-star convergence. The Banach–Alaoglu theorem says this entire dual unit ball is weak-star compact.
From Weak-Star Compactness to Weak Compactness
The canonical map \(J:X\to X^{**}\), given by \((Jx)(F)=F(x)\), is an isometry, as established in the tutorial on dual spaces. A space is reflexive when \(J\) maps \(X\) onto all of \(X^{**}\). In that case, weak tests on \(X\) correspond exactly to weak-star tests on \(X^{**}\): evaluating \(Jx\) at \(F\in X^*\) gives \(F(x)\).
Proof. Since \(J\) is an isometry and is onto when \(X\) is reflexive, it maps the closed ball of radius \(R\) in \(X\) bijectively onto the closed ball of radius \(R\) in \(X^{**}\). By the Banach–Alaoglu theorem applied to the normed space \(X^*\), that ball in \(X^{**}\) is compact in the weak-star topology.
For each \(F\in X^*\), the weak-star coordinate of \(Jx\) at \(F\) is \((Jx)(F)=F(x)\). These are precisely the evaluations that generate the weak topology on \(X\). Therefore \(J\), restricted to the ball, is a homeomorphism from the weak topology on the ball in \(X\) to the weak-star topology on the ball in \(X^{**}\). The former ball is consequently compact. \(\square\)
Worked Example: A Weakly Compact Disk in \(\mathbb{R}^2\)
Give \(X=\mathbb{R}^2\) its Euclidean norm. Every linear functional on \(X\) has the form \(F_{a,b}(u,v)=au+bv\). The two coordinate functionals show that the weak topology is at least as strong as coordinatewise, or Euclidean, convergence. Conversely, each \(F_{a,b}\) is continuous in the Euclidean topology, so the weak topology is no stronger. The two topologies therefore agree.
The closed unit ball $$ \{(u,v)\in\mathbb{R}^2:u^2+v^2\leq1\} $$ is compact by the Heine–Borel theorem, and hence is weakly compact. This finite-dimensional example is consistent with the reflexive-space theorem: every finite-dimensional normed space is reflexive. In infinite-dimensional spaces, weak compactness is a genuinely different property from compactness in the norm topology.
Compactness as an Existence Tool
A continuous real-valued function on a nonempty compact space attains its maximum and minimum. Evaluation \(F\mapsto F(x)\) is continuous in the weak-star topology by definition. Banach–Alaoglu therefore gives a useful way to produce a functional that maximizes a fixed evaluation over a weak-star closed constraint set.
Worked Example: Maximizing an Evaluation on a Constrained Dual Ball
Take \(X=C([0,1])\) with the supremum norm, and consider $$ K=\{F\in X^*:\|F\|\leq1,\ F(\mathbf{1})=1\}, $$ where \(\mathbf{1}\) is the constant function with value \(1\). This set is nonempty: evaluation at \(1\), defined by \(\delta_1(f)=f(1)\), has norm \(1\) and satisfies \(\delta_1(\mathbf{1})=1\).
The dual unit ball is weak-star compact by Banach–Alaoglu. The condition \(F(\mathbf{1})=1\) defines a weak-star closed set because \(F\mapsto F(\mathbf{1})\) is continuous. Thus \(K\) is compact. For \(g(t)=2t-1\), the function \(F\mapsto F(g)\) is weak-star continuous, so it attains a maximum on \(K\). Moreover, $$ F(g)\leq |F(g)|\leq\|F\|\,\|g\|_\infty\leq1 $$ for every \(F\in K\), since \(\|g\|_\infty=1\). The functional \(\delta_1\in K\) satisfies \(\delta_1(g)=g(1)=1\), so the maximum is exactly \(1\). The compactness argument guarantees attainment; the explicit evaluation verifies the maximizing value in this case.
Compact Does Not Always Mean Sequentially Compact
In a metric space, compactness and sequential compactness are equivalent: every sequence in a compact set has a convergent subsequence. For general topologies this equivalence can fail. Weak and weak-star topologies need not be metrizable, so a compactness theorem for a dual ball does not by itself promise that every sequence in that ball has a convergent subsequence.
Here is a product example that exhibits the issue. Let \(I=\{0,1\}^{\mathbb{N}}\), the set of all infinite binary sequences, and consider the compact product \(\{-1,1\}^{I}\). For each positive integer \(n\), define \(y_n\) by $$ y_n(i)=(-1)^{i_n}\qquad(i=(i_1,i_2,\ldots)\in I). $$ Take any subsequence \((y_{n_k})\). Choose \(i\in I\) so that \(i_{n_k}=0\) for even \(k\) and \(i_{n_k}=1\) for odd \(k\), assigning arbitrary values to the other coordinates. Then \(y_{n_k}(i)\) alternates between \(1\) and \(-1\), so it does not converge. Since convergence in a product topology requires convergence in every coordinate, this subsequence cannot converge in the product. No subsequence of \((y_n)\) converges, even though all its terms belong to a compact space.
The lesson is to keep the topology in the conclusion. Banach–Alaoglu asserts weak-star compactness, not norm compactness and not, without additional hypotheses, sequential compactness. When the weak-star topology on a particular bounded set is metrizable, compactness does imply subsequential convergence; without metrizability, the open-cover definition of compactness is the reliable one.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Which product of compact spaces is used in the proof of the Banach–Alaoglu theorem, and what are its coordinates?
- Why must the coordinate families representing functionals form a closed subset of that product?
- Does the Banach–Alaoglu theorem require \(X\) to be complete?
- How does reflexivity transfer weak-star compactness in \(X^{**}\) to weak compactness in \(X\)?
- Why is the constraint \(F(\mathbf{1})=1\) weak-star closed?
- Why does compactness alone not guarantee a convergent subsequence in an arbitrary weak or weak-star topology?