Eberlein-Smulian and Weakly Convergent Subsequences

Compactness and Sequences in the Weak Topology

In normed spaces, boundedness alone does not generally produce a convergent subsequence. Weak compactness supplies a useful substitute: it guarantees subsequences that converge when tested against every continuous linear functional. The Eberlein–Šmulian theorem makes this relationship precise for Banach spaces, where weak compactness and weak sequential compactness are equivalent.

Let \(X\) be a Banach space over the real or complex scalars. Its weak topology is the topology in which a net \(x_\alpha\) converges to \(x\) exactly when \(\Lambda(x_\alpha)\to\Lambda(x)\) for every \(\Lambda\in X^*\), the space of continuous linear functionals on \(X\). A sequence \(x_n\) converges weakly to \(x\), written \(x_n\rightharpoonup x\), when this same test holds for the sequence. A subset \(K\subseteq X\) is weakly compact if it is compact in the weak topology; it is weakly sequentially compact if every sequence in \(K\) has a subsequence converging weakly to an element of \(K\).

Theorem (Eberlein–Šmulian): If \(X\) is a Banach space and \(K\subseteq X\), then \(K\) is weakly compact if and only if \(K\) is weakly sequentially compact. The theorem concerns compactness in the weak topology, not compactness in the norm topology. This is a deep standard theorem; its proof is beyond this tutorial.

The equivalence is significant because compactness is a topological property, whereas subsequences are often the practical way to obtain limits. In general topological spaces, sequential compactness need not be equivalent to compactness. The Banach-space weak topology has the special property asserted by Eberlein–Šmulian.

Bounded Sequences in Reflexive Banach Spaces

A Banach space \(X\) is reflexive if its canonical map into its bidual \(X^{**}\), given by \(J(x)(\Lambda)=\Lambda(x)\), is onto. We use the standard reflexivity criterion established in earlier material: the closed unit ball of a reflexive Banach space is weakly compact. By scaling, every norm-closed ball is weakly compact as well. Combining this fact with Eberlein–Šmulian gives the subsequence principle below.

Corollary: Let \(X\) be a reflexive Banach space, and let \((x_n)\) be a norm-bounded sequence in \(X\). Then there are indices \(n_1<n_2<\cdots\) and \(x\in X\) such that \(x_{n_j}\rightharpoonup x\).

Proof. Since the sequence is norm-bounded, there is a finite \(C\ge0\) such that \(\|x_n\|\le C\) for every \(n\). If \(C=0\), then \(\|x_n\|=0\), so \(x_n=0\) for every \(n\); the whole sequence converges weakly to \(0\). If \(C>0\), let \(K=\{x\in X:\|x\|\le C\}\). This is the closed ball of radius \(C\), and it is weakly compact because \(X\) is reflexive. Every \(x_n\) belongs to \(K\). The Eberlein–Šmulian theorem therefore says that \(K\) is weakly sequentially compact, so a subsequence \(x_{n_j}\) converges weakly to some \(x\in K\). In particular, the limit belongs to \(X\). \(\square\)

For measure spaces, take any measure space \((X,\mathcal{M},\mu)\). When \(1<p<\infty\), \(L^p(\mu)\) is a reflexive Banach space (with functions identified when they agree \(\mu\)-a.e.). Consequently, every norm-bounded sequence in \(L^p(\mu)\) has a subsequence weakly convergent in \(L^p(\mu)\). No finiteness or \(\sigma\)-finiteness assumption on \(\mu\) is required for this reflexivity conclusion.

Corollary for \(L^p\): If \(1<p<\infty\), \((X,\mathcal{M},\mu)\) is a measure space, and \(\sup_n\|f_n\|_p<\infty\), then there are a subsequence \((f_{n_j})\) and \(f\in L^p(\mu)\) such that \(f_{n_j}\rightharpoonup f\). Equivalently, \(\int f_{n_j}g\,d\mu\to\int fg\,d\mu\) for every \(g\in L^q(\mu)\), where \(q=p/(p-1)\), in the real case; in the complex case the corresponding dual pairing includes complex conjugation according to convention.

Proof. The hypothesis \(1<p<\infty\) makes \(L^p(\mu)\) reflexive, and its norm boundedness hypothesis places the sequence in a closed ball of finite radius. Apply the preceding corollary to the Banach space \(L^p(\mu)\). The equivalent integral formulation follows from the standard \(L^p\) duality theorem: every continuous linear functional on \(L^p(\mu)\) is represented by an element of \(L^q(\mu)\). \(\square\)

Worked Applications

Worked Example: A Norm-Convergent Sequence of Constants

On \([0,1]\) with Lebesgue measure \(m\), set \(f_n(x)=\frac{n}{n+1}\) for every \(x\in[0,1]\), with \(1<p<\infty\). Find its limit and verify weak convergence.

The constant function \(1\) has \(L^p\) norm \(\left(\int_0^1 1^p\,dm\right)^{1/p}=1\). Hence

\|f_n-1\|_p =\left(\int_0^1\left|\frac{n}{n+1}-1\right|^p\,dm\right)^{1/p} =\left(\int_0^1\left(\frac{1}{n+1}\right)^p\,dm\right)^{1/p} =\frac{1}{n+1}.

This tends to zero, so \(f_n\to1\) in \(L^p\). For any \(\Lambda\in (L^p)^*\), the definition of the operator norm gives \(|\Lambda(f_n)-\Lambda(1)|=|\Lambda(f_n-1)|\le\|\Lambda\|/(n+1)\), which tends to zero. Thus \(f_n\rightharpoonup1\). Check: at \(n=1\), the norm difference is \(1/2\); at \(n=3\), it is \(1/4\), agreeing with the formula \(1/(n+1)\).

Worked Example: The Unit Vectors in \(\ell^p\)

Let \(\mathbb{N}\) have counting measure, so \(L^p(\mathbb{N})=\ell^p\). For \(1<p<\infty\), let \(e_n\) be the sequence with value \(1\) at index \(n\) and \(0\) elsewhere. Show that \(e_n\rightharpoonup0\), although the sequence does not converge to \(0\) in norm.

Directly from the sum defining the norm, \(\|e_n\|_p=(\sum_{k=1}^{\infty}|e_n(k)|^p)^{1/p}=1\). Write \(q=p/(p-1)\). By the standard duality \((\ell^p)^*=\ell^q\), every continuous linear functional has the form \(\Lambda_a(x)=\sum_{k=1}^{\infty}a_kx_k\) for some \(a\in\ell^q\), with the usual conjugation adjustment in the complex case. Then \(\Lambda_a(e_n)=a_n\). Since \(\sum_k|a_k|^q<\infty\), its terms satisfy \(a_n\to0\): otherwise some \(\varepsilon>0\) would have \(|a_n|\ge\varepsilon\) for infinitely many \(n\), forcing the series of \(q\)-th powers to diverge. Thus \(\Lambda_a(e_n)\to0\) for every \(\Lambda_a\), proving weak convergence to \(0\).

Check: the sequence is bounded because every \(\|e_n\|_p=1\), and its weak limit \(0\) has norm \(0\). Nevertheless, \(\|e_n-0\|_p=1\) for every \(n\), so the sequence does not converge in norm.

Worked Example: Normalized Indicators on Disjoint Intervals

On \(\mathbb{R}\) with Lebesgue measure \(m\), let \(E_n=[n,n+1)\) for \(n\ge1\), and set \(f_n=\chi_{E_n}\). For \(1<p<\infty\), verify weak convergence to zero in \(L^p(m)\).

The intervals are pairwise disjoint and each has measure \(1\). Therefore \(\|f_n\|_p=(\int_{\mathbb{R}}\chi_{E_n}^p\,dm)^{1/p}=m(E_n)^{1/p}=1\); in particular the sequence is bounded. Let \(g\in L^q(m)\), where \(q=p/(p-1)\). Since the intervals are disjoint, countable additivity of the integral for the nonnegative function \(|g|^q\) gives \(\sum_{n=1}^{\infty}\int_{E_n}|g|^q\,dm\le\int_{\mathbb{R}}|g|^q\,dm<\infty\). The terms of this convergent nonnegative series tend to zero. Hölder's inequality on each \(E_n\) now gives

\left|\int_{\mathbb{R}} f_ng\,dm\right| =\left|\int_{E_n}g\,dm\right| \le m(E_n)^{1/p}\left(\int_{E_n}|g|^q\,dm\right)^{1/q} =\left(\int_{E_n}|g|^q\,dm\right)^{1/q}\longrightarrow0.

Thus every dual functional tends to zero on \(f_n\), so \(f_n\rightharpoonup0\) in \(L^p(m)\). Check: taking \(g=\chi_{E_k}\), the pairing is \(1\) when \(n=k\) and \(0\) when \(n\ne k\), consistent with convergence to zero as \(n\to\infty\) for each fixed \(k\).

What the Subsequence Conclusion Does—and Does Not—Say

The extraction corollary guarantees existence, not a formula for the indices or the weak limit. For a general bounded sequence in a reflexive \(L^p\) space, Eberlein–Šmulian may be the tool that supplies a weakly convergent subsequence without identifying it explicitly. The limit is an element of the same \(L^p\) space, and weak convergence means convergence under every continuous linear functional, not pointwise convergence of representatives.

A frequent mistake is to infer norm convergence from weak convergence. The unit-vector example gives a concrete correction. A mistaken calculation might treat the weakly convergent sequence \(e_n\rightharpoonup0\) as if it also had \(\|e_n-0\|_p\to0\). In fact, the defining norm calculation gives \(\|e_n-0\|_p=(|1|^p)^{1/p}=1\) for each \(n\). Weak convergence tests functionals one at a time; it does not require the norm of the difference to vanish.

Another important limit is the range \(1<p<\infty\). The reflexivity premise is essential to the \(L^p\) conclusion as stated; at the endpoints \(p=1\) and \(p=\infty\), one cannot invoke reflexivity of \(L^p\) to infer weak subsequence extraction from boundedness. The relevant compactness questions at those endpoints require additional hypotheses or different topologies.

Check Your Understanding

For each question, distinguish convergence in the weak topology from convergence in norm.

State the Eberlein–Šmulian theorem for a subset of a Banach space.
Why does a bounded sequence in a reflexive Banach space lie in a weakly compact set?
For \(1<p<\infty\), what hypotheses on a sequence in \(L^p(\mu)\) guarantee a weakly convergent subsequence?
For the unit vectors \(e_n\in\ell^p\), compute \(\|e_n\|_p\) and explain why \(e_n\rightharpoonup0\) does not imply norm convergence.
In the disjoint-interval example, where is the assumption \(g\in L^q(m)\) used to show the pairings tend to zero?