From Weak-Star Compactness to Weak Compactness
Compactness can turn boundedness into a convergence guarantee, but the topology matters. In a reflexive Banach space, closed balls are compact in the weak topology. This gives a useful conclusion for bounded nets: they have weakly convergent subnets. It also supplies a compactness method for norm-closed convex constraint sets in \(L^p\), such as sets defined by a norm bound and a continuous linear equation.
There is an important qualification. A norm-closed bounded set need not be weakly compact if it is not convex. The valid general statement is that a weakly closed bounded set is weakly compact. A norm-closed convex set is weakly closed by the Hahn–Banach separation theorem, so closed bounded convex sets are a particularly useful case.
Let \((X,\mathcal M,\mu)\) be a \(\sigma\)-finite measure space, and let \(1<p<\infty\), with conjugate exponent \(q=p'\), so \(1/p+1/q=1\). We work with real or complex \(L^p(\mu)\), whose elements are equivalence classes of measurable functions equal \(\mu\)-a.e., with finite norm \(\|f\|_p=(\int |f|^p\,d\mu)^{1/p}\). The weak topology on \(L^p(\mu)\) is the topology in which \(f_\alpha\) converges to \(f\) exactly when \(\Lambda(f_\alpha)\to\Lambda(f)\) for every bounded linear functional \(\Lambda\) on \(L^p(\mu)\). By \(L^p\)-duality, these functionals are represented by \(g\in L^q(\mu)\), using \(\Lambda_g(f)=\int f\overline{g}\,d\mu\) in the complex case. The reflexivity theorem for \(L^p\), \(1<p<\infty\), says that the canonical map \(J:L^p(\mu)\to L^p(\mu)^{**}\), defined by \(J(f)(\Lambda)=\Lambda(f)\), is onto.
The Weak Compactness Theorem
Proof. First consider the closed ball \(B_R=\{f\in L^p(\mu):\|f\|_p\le R\}\), for \(R\ge0\). The Banach–Alaoglu theorem says that the closed ball of radius \(R\) in \(L^p(\mu)^{**}\) is compact in the weak-star topology \(\sigma(L^p{}^{**},L^p{}^*)\). Reflexivity makes \(J\) onto, and \(J\) is an isometry, so it maps \(B_R\) onto that bidual ball. Moreover, the weak-star topology on the bidual, pulled back by \(J\), is exactly the weak topology on \(L^p(\mu)\): for every \(\Lambda\in L^p(\mu)^*\), the evaluation of \(J(f)\) at \(\Lambda\) is \(J(f)(\Lambda)=\Lambda(f)\). Thus \(B_R\) is weakly compact.
If \(C\) is bounded, then \(C\subseteq B_R\) for some \(R\). If \(C\) is weakly closed in \(L^p(\mu)\), it is also closed as a subset of \(B_R\) with its relative weak topology. A closed subset of a compact space is compact, so \(C\) is weakly compact.
It remains to justify the convex-set formulation. Let \(C\) be norm closed and convex, and let \(f\notin C\). The Hahn–Banach separation theorem gives a bounded linear functional and a real number that strictly separates \(f\) from \(C\). The resulting strict inequality defines a weakly open neighborhood of \(f\) disjoint from \(C\). Therefore every point outside \(C\) has a weakly open neighborhood outside \(C\), which means that \(C\) is weakly closed. The first part of the theorem now applies. \(\square\)
Worked Example: A Norm Ball with a Linear Constraint
Let \(1<p<\infty\), take Lebesgue measure \(m\) on \([0,1]\), and work in the real space \(L^p([0,1],m)\). Consider \(K=\{f:\|f\|_p\le2,\ \int_0^1 f\,dm=1\}\). We check the hypotheses and conclude that \(K\) is weakly compact.
The function \(f(x)=1\) belongs to \(L^p([0,1],m)\), and \(\|1\|_p=(\int_0^1 1\,dm)^{1/p}=1\). Also \(\int_0^1 1\,dm=1\), so \(K\) is nonempty. It is bounded because each of its elements has norm at most \(2\). It is convex: if \(f,h\in K\) and \(0\le t\le1\), then \(\int_0^1(tf+(1-t)h)\,dm=t+(1-t)=1\), and the norm triangle inequality gives \(\|tf+(1-t)h\|_p\le t\|f\|_p+(1-t)\|h\|_p\le2\).
For closedness, the norm ball is norm closed and the map \(f\mapsto\int_0^1 f\,dm\) is a bounded linear functional: Hölder’s inequality gives \(|\int_0^1 f\,dm|\le\|f\|_p\|1\|_q=\|f\|_p\), since \(\|1\|_q=(\int_0^1 1\,dm)^{1/q}=1\). Thus the condition that this integral equal \(1\) defines a norm-closed set. The set \(K\) is norm closed and convex, hence weakly closed by separation. The theorem proves that \(K\) is weakly compact. Check: the constant function \(1\) satisfies both constraints, with norm \(1\le2\) and integral \(1\).
Bounded Nets Have Weakly Convergent Subnets
A net is a family \((x_i)_{i\in I}\) indexed by a directed set \(I\): any two indices have a common upper bound. A subnet is a reindexed net whose indices move eventually beyond every original index. The compactness fact needed here is that every net in a compact space has a convergent subnet. For completeness, the fact follows from the finite-intersection characterization of compactness. The closed sets \(F_i=\overline{\{x_j:j\ge i\}}\) have the finite-intersection property: for finitely many indices, choose a common upper bound \(i_0\); then \(F_{i_0}\) lies in each of the corresponding \(F_i\). Compactness gives a point \(x\in\bigcap_iF_i\), so every neighborhood of \(x\) meets every tail of the net. This makes \(x\) a cluster point, and the directed set of choices of a tail together with a neighborhood of \(x\) constructs a subnet converging to \(x\).
Proof. If the net \((f_i)\) is bounded, there is an \(R\ge0\) such that \(\|f_i\|_p\le R\) for every index \(i\). Its values lie in the weakly compact ball \(B_R\). The compactness fact just described gives a subnet converging weakly to some \(f\in B_R\). If all \(f_i\) lie in a weakly closed set \(C\), then the subnet lies in \(C\), and its limit lies in \(C\) because \(C\) is weakly closed. \(\square\)
Worked Example: Extracting a Weakly Convergent Subnet
On \([0,1]\) with Lebesgue measure, take the net \(f_n=(-1)^n1\) in real \(L^p([0,1],m)\), indexed by the positive integers in their usual order. For every \(n\), \(\|f_n\|_p=(\int_0^1|(-1)^n|^p\,dm)^{1/p}=1\), so the net is bounded. The indices \(n=2k\), \(k\ge1\), form a subnet. At every point, \(f_{2k}=1\), and therefore for every \(g\in L^q([0,1],m)\),
The right side is finite by Hölder’s inequality, since \(|\int_0^1g\,dm|\le\|g\|_q\|1\|_p=\|g\|_q\), with \(\|1\|_p=1\). Thus this subnet converges weakly to the constant function \(1\). Check: its limit has norm \((\int_0^1 1^p\,dm)^{1/p}=1\), consistent with the bound on the subnet.
Why Norm-Closedness Alone Is Not Enough
Weak compactness forces weak closedness because the weak topology is Hausdorff: bounded linear functionals separate distinct elements of a normed space. In contrast, norm-closedness alone does not imply weak closedness. The following example shows why the convexity qualification in the theorem matters.
Worked Example: A Closed Bounded Set That Is Not Weakly Compact
Take counting measure on \(\mathbb N=\{1,2,\ldots\}\), so \(L^p\) is the sequence space \(\ell^p\). Let \(e_n\) be the sequence whose \(n\)-th entry is \(1\) and whose other entries are \(0\), and set \(A=\{e_n:n\ge1\}\). Each vector has norm \(\|e_n\|_p=1\), so \(A\) is bounded. For distinct \(n,m\), the two nonzero entries of \(e_n-e_m\) have absolute value \(1\); hence \(\|e_n-e_m\|_p=(1^p+1^p)^{1/p}=2^{1/p}\).
This fixed positive distance shows that \(A\) is norm closed. Indeed, any convergent sequence of elements of \(A\) must be Cauchy. Once the distance between different elements is \(2^{1/p}\), the Cauchy condition with tolerance \(2^{1/p}/2\) forces all sufficiently late terms to be the same \(e_n\), so the limit belongs to \(A\).
But \(e_n\) converges weakly to \(0\). By \(L^p\)-duality, any bounded linear functional on \(\ell^p\) is represented by some \(a\in\ell^q\), and its value at \(e_n\) is the \(n\)-th coordinate of \(a\) (or its complex conjugate, depending on the pairing convention). Since \(\sum_{n=1}^{\infty}|a_n|^q<\infty\), we have \(a_n\to0\): otherwise some \(\varepsilon>0\) would satisfy \(|a_n|\ge\varepsilon\) for infinitely many \(n\), forcing the series to diverge. Hence every bounded linear functional tends to \(0\) on \(e_n\), which is precisely weak convergence to \(0\). Since \(0\notin A\), \(A\) is not weakly closed and therefore cannot be weakly compact. Check: for any two distinct indices, the computed distance is \(2^{1/p}>0\); thus no sequence of distinct \(e_n\)'s can converge in norm, while the weak limit is the omitted zero vector.
Using the Compactness Conclusion
When applying the theorem, separate three questions:
- Boundedness: is the set contained in some norm ball?
- Weak closedness: is it weakly closed directly, or does it follow from being norm closed and convex?
- Reflexivity: is the space \(L^p(\mu)\) with \(1<p<\infty\), under the stated duality hypotheses?
If all three conditions hold, weak compactness follows. The conclusion is compactness in the weak topology, not in the norm topology; it does not assert that a bounded net has a norm-convergent subnet. Nor does it claim that an arbitrary norm-closed bounded nonconvex set is weakly compact. The practical gain is the subnet conclusion: weak compactness supplies a limit point for bounded nets without requiring the net itself to converge.
Check Your Understanding
For each question, identify the topology and check every hypothesis before using the theorem.
- Which part of the proof turns weak-star compactness of a bidual ball into weak compactness of a ball in \(L^p(\mu)\)?
- Why does a norm-closed convex subset of \(L^p(\mu)\) have to be weakly closed?
- For \(K=\{f\in L^p([0,1]):\|f\|_p\le1,\ \int_0^1 f\,dm=0\}\), verify that \(K\) is nonempty, bounded and norm closed, and state the theorem’s conclusion.
- In \(\ell^p\), why does every \(a\in\ell^q\) satisfy \(a_n\to0\), and how does this establish weak convergence of \(e_n\) to \(0\)?
- What compactness conclusion is available for a bounded net in a reflexive \(L^p\) space, and why does it concern a subnet rather than necessarily the original net?