Reflexivity
This post is about reflexivity, the bidual, and the natural compactification of a Banach space. We start by describing the weak and weak-* topologies.
Weak and Weak-* Topologies
Generally, given a dual pair \((E,F)\) with bilinear pairing \(\langle e,f\rangle\), the weak topology on \(E\) relative to \(F\) is denoted \(\sigma(E,F)\) and is the coarsest topology making every map \(e \mapsto \langle e,f\rangle\), \(f \in F\), continuous. In a Banach space \(X\), the usual weak topology on \(X\) is therefore \(\sigma(X,X^*)\). But the same construction applies to many other spaces. For example, if we look at \(X^*\) and choose a subspace \(Y \subset X^{**}\), then \(\sigma(X^*,Y)\) is the weak topology on \(X^*\) generated by the functionals in \(Y\). Thus the notation \(\sigma(\cdot,\cdot)\) always records both the space carrying the topology and the family of test functionals used to define it.
For a Banach space \(X\), the canonical isometric embedding (see below) \[ J:X \to X^{**}, \qquad J(x)(f)=f(x), \] identifies each \(x \in X\) with the functional on \(X^*\) given by evaluation at \(x\). That embedded copy \(J(X) \subset X^{**}\) is exactly the predual used to define the weak-* topology on \(X^*\). Thus the weak-* topology on \(X^*\) is \[ \sigma(X^*,X)=\sigma(X^*,J(X)), \] while the full weak topology on \(X^*\) is \[ \sigma(X^*,X^{**}). \] So weak-* convergence tests a functional in \(X^*\) only against points coming from \(X\), whereas weak convergence tests against all of \(X^{**}\). When \(X\) is reflexive, the embedding is onto, so \(J(X)=X^{**}\), and the two topologies coincide. In that precise sense, reflexivity says that no extra test functionals appear in passing from \(X\) to its bidual.
More generally, if a space \(Z\) happens to be presented as a dual space \(Z = Y^*\), then its weak-* topology is \(\sigma(Z,Y)\). So weak-* is not an intrinsic topology on the underlying vector space alone; it depends on the chosen predual. A space can have more than one predual, in which case it can have more than one weak-* topology. For example, we can take \[ Y = X^{**}, \qquad Z = X^*. \] Since \(X^{**} = (X^*)^*\), the weak-* topology on \(X^{**}\) is exactly \(\sigma(X^{**},X^*)\).
Norm Closed vs. Weakly Closed
Let \(X\) be an infinite-dimensional Banach space. The unit sphere \[ S_X = \{x \in X : \|x\|=1\} \] is norm closed, since the norm map \(x \mapsto \|x\|\) is norm continuous. But \(S_X\) is not weakly closed. In fact, \[ 0 \in \overline{S}_X^{\,\sigma(X,X^*)}. \] To see why, take any basic weak neighborhood of \(0\). It has the form \[ U = \{x \in X : |f_1(x)|<\varepsilon,\dots,|f_n(x)|<\varepsilon\} \] for some \(f_1,\dots,f_n \in X^*\) and \(\varepsilon>0\). Now \[ M = \bigcap_{i=1}^n \ker f_i \] is a closed subspace of finite codimension. Since \(X\) is infinite dimensional, \(M\) is not \(\{0\}\). So we can choose \(x \in M\) with \(\|x\|=1\). Then \(x \in S_X\), and also \[ f_i(x)=0 \qquad \text{for all } i, \] hence \(x \in U\). Thus every weak neighborhood of \(0\) meets \(S_X\), so \(0\) lies in the weak closure of \(S_X\). Since \(0 \notin S_X\), the sphere is not weakly closed.
Why Reflexivity Matters
A Banach space \(X\) is reflexive when the canonical embedding into its bidual is onto. At first sight that sounds technical, but the underlying meaning is simple and important: in a reflexive space, bounded sets do not acquire new limit points when we place them in the bidual. In a non-reflexive space, they do.
That point of view gives a clean answer to a natural question. If closed bounded sets are not compact in infinite-dimensional normed spaces, where does the missing compactness go? The answer is that bounded sets become compact after we embed \(X\) into \(X^{**}\) and use the weak-* topology there. Reflexivity is exactly the condition saying that this compactification does not add anything new.
The Canonical Embedding
The canonical embedding \(J\) is linear and isometric. Indeed, \[ \|J(x)\| = \sup_{\|f\|\le 1} |J(x)(f)| \\ = \sup_{\|f\|\le 1} |f(x)| = \|x\| \] by the Hahn–Banach theorem. Hence we may identify \(X\) with the closed subspace \(J(X) \subset X^{**}\).
Reflexivity means precisely that \(J\) is onto, so that \(X\) and \(X^{**}\) are the same Banach space up to canonical identification.
Weak And Weak-* Topologies
The weak topology on \(X\) is the coarsest topology making every \(f \in X^*\) continuous. Thus \[ x_n \rightharpoonup x \quad \Longleftrightarrow \quad f(x_n) \to f(x) \text{ for every } f \in X^*. \]
The weak-* topology on \(X^{**}\) is the coarsest topology making every evaluation map at points of \(X^*\) continuous. Thus \[ \Phi_\alpha \overset{*}{\rightharpoonup} \Phi \quad \Longleftrightarrow \quad \Phi_\alpha(f) \to \Phi(f) \text{ for every } f \in X^*. \]
When we view \(X\) inside \(X^{**}\) via \(J\), the weak-* topology on \(J(X)\) is exactly the weak topology on \(X\). Indeed, \[ J(x_n)(f) = f(x_n). \] So weak convergence in \(X\) is the same thing as weak-* convergence of the embedded sequence in \(X^{**}\), provided the limit still lies in \(J(X)\). That final caveat is the whole story.
Banach–Alaoglu: Where Compactness Goes
Theorem 1 (Banach–Alaoglu) For any normed space \(Y\), the closed unit ball \[ B_{Y^*} = \{\varphi \in Y^* : \|\varphi\| \le 1\} \] is compact in the weak-* topology \(\sigma(Y^*,Y)\).
Applied with \(Y = X^*\), we learn that the closed unit ball of \(X^{**}\) is compact in the weak-* topology \(\sigma(X^{**},X^*)\).
That gives the natural compactification of bounded sets in \(X\):
- embed \(X\) into \(X^{**}\) via \(J\);
- place a bounded set \(B \subset X\) inside a ball of \(X^{**}\);
- take its weak-* closure in \(X^{**}\).
That closure is compact. Bounded sets in \(X\) become compact after we place them in \(X^{**}\) with the weak-* topology.
The proof is not hard once we know Tychonoff’s theorem, but it is not an elementary metric compactness proof. The standard proof is a lovely product-topology argument. Take \(Y\) a normed space. Each \(\varphi \in B_{Y^*}\) defines a scalar value \(\varphi(y)\) for every \(y \in Y\), and \[ |\varphi(y)| \le \|y\|. \] So \(B_{Y^*}\) sits inside the product \[ \prod_{y \in Y} \{z \in \mathbb{K}: |z| \le \|y\|\}, \] where \(\mathbb{K}=\mathbb{R}\) or \(\mathbb{C}\). Each factor is compact, and by Tychonoff the whole product is compact in the product topology. The weak-* topology is exactly the topology of pointwise convergence on \(Y\), so it is the subspace topology inherited from that product. One then checks that the set of points corresponding to linear functionals of norm at most \(1\) is closed in the product. Therefore \(B_{Y^*}\) is compact.
So the proof is conceptually straightforward:
- represent functionals by their values on points;
- place them in a product of compact scalar disks;
- invoke Tychonoff;
- observe that the dual ball is a closed subset.
The dependence on Tychonoff means the theorem is topological rather than computational.
Reflexivity Reinterpreted
Now we can state the geometric meaning of reflexivity.
Reflexive Spaces
If \(X\) is reflexive, then \(J(X)=X^{**}\). Therefore the weak-* compact balls in \(X^{**}\) are really weakly compact balls in \(X\). Equivalently, bounded (norm) closed sets in \(X\) are relatively weakly compact. (Relatively meaning we take the closure wrt the weak topology.)
So in a reflexive space, bounded sequences have weakly convergent subsequences. Infinite-dimensional norm compactness is still lost, but weak compactness survives.
Non-Reflexive Spaces
If \(X\) is not reflexive, then \(J(X)\) is a proper subspace of \(X^{**}\). A bounded sequence in \(X\) may have weak-* cluster points in \(X^{**}\) that do not belong to \(J(X)\). In that case the sequence has no weakly convergent subsequence in \(X\) itself.
That is what we miss in non-reflexive spaces. Compactness still exists, but only after we enlarge the space.
The Natural Compactification
The phrase “natural compactification” now becomes precise. For a bounded set \(B \subset X\), consider its image \(J(B) \subset X^{**}\). Since \(B\) is bounded, \(J(B)\) lies in some weak-* compact ball of \(X^{**}\). The weak-* closure \[ \overline{J(B)}^{\,\sigma(X^{**},X^*)} \] is compact. That closure is the natural compactification of \(B\).
If \(X\) is reflexive, no new points are added. If \(X\) is not reflexive, extra boundary points can appear in \[ X^{**} \setminus J(X). \]
Those are the new generalized limit points of bounded families from \(X\).
Example: A Bad Sequence In \(L^1[0,1]\)
Consider \[ f_n = n \, 1_{(0,1/n)}. \] Then \[ \|f_n\|_1 = 1 \] for every \(n\), so the sequence is bounded in \(L^1[0,1]\).
But it has no weakly convergent subsequence. To see why, suppose some subsequence \(f_{n_k}\) converges weakly to \(f \in L^1\). Then for every \(a>0\), \[ \int_0^a f_{n_k}(x)\,dx \to \int_0^a f(x)\,dx. \] Yet for all large \(k\), since \(1/n_k < a\), \[ \int_0^a f_{n_k}(x)\,dx = 1. \] Therefore \[ \int_0^a f(x)\,dx = 1 \qquad \text{for every } a>0. \] That is impossible, because absolute continuity of the integral gives \[ \int_0^a f(x)\,dx \to 0 \qquad \text{as } a \downarrow 0. \]
So \(L^1[0,1]\) is not reflexive, and this sequence exhibits the failure of weak sequential compactness. Intuitively, the mass concentrates into a narrower and narrower spike. In the bidual picture, the missing limit lives outside the canonical copy of \(L^1\).
Example: A Bad Sequence In \(C[0,1]\)
Consider \[ g_n(x) = x^n. \] Then \[ \|g_n\|_\infty = 1, \] so the sequence is bounded in \(C[0,1]\).
For each \(x \in [0,1)\) we have \(x^n \to 0\), while \(g_n(1)=1\) for all \(n\). Hence the pointwise limit is \[ g(x)= \begin{cases} 0, & 0 \le x < 1,\\ 1, & x=1, \end{cases} \] which is not continuous. Therefore no subsequence can converge uniformly, since uniform limits of continuous functions are continuous.
That only rules out norm convergence, not weak convergence, but it still gives a vivid picture of lost compactness: bounded sets in \(C[0,1]\) are far from norm compact unless we impose extra structure such as equicontinuity. In fact \(C[0,1]\) is also non-reflexive.
Example: The Standard Basis In \(\ell^1\)
The cleanest discrete example is the standard basis \(e_n\) in \(\ell^1\). Each \(e_n\) has norm \(1\), so the sequence is bounded. But it has no weakly convergent subsequence.
Indeed, weak convergence \(e_{n_k} \rightharpoonup x\) would imply coordinatewise convergence against elements of \(\ell^\infty = (\ell^1)^*\). The obstruction is that we can choose bounded test sequences in \(\ell^\infty\) that detect arbitrary behavior of the indices. There is no candidate \(x \in \ell^1\) against which all these tests can converge consistently.
What happens in the bidual? Since \(\ell^1\) embeds into \(\ell^{1**}\) and the unit ball of \(\ell^{1**}\) is weak-* compact, the sequence has weak-* cluster points there. Those cluster points are the added boundary points of the natural compactification; they do not come from actual elements of \(\ell^1\).
What We Lose In Non-Reflexive Spaces
The failure of reflexivity is not that the weak topology on \(X\) suddenly uses more test functionals than before. The weak topology on \(X\) is always defined by all of \(X^*\). The issue is instead that bounded sets in \(X\) need not be weakly compact. The losses are manifold.
Loss Of Weak Compactness
In a reflexive space, closed bounded sets are weakly compact. In a non-reflexive space they need not be.
Loss Of Weakly Convergent Subsequences
In reflexive spaces, bounded sequences admit weakly convergent subsequences. In non-reflexive spaces, a bounded sequence can drift toward a generalized limit that only exists in \(X^{**}\).
Loss Of Direct Existence Arguments
Many existence proofs in analysis take a bounded minimizing sequence, extract a weakly convergent subsequence, and pass to the limit by weak lower semicontinuity. Reflexivity is often the hypothesis that makes that strategy work. Without reflexivity, the minimizing sequence may fail to have a limit in the original space.
The Finite-Dimensional Analogy
In finite dimensions, bounded closed sets are compact. In infinite dimensions, norm compactness fails almost completely. Reflexivity says that although norm compactness is gone, a weaker but still powerful replacement survives: weak compactness of bounded closed sets. So reflexive spaces remain close in spirit to finite-dimensional spaces. Non-reflexive spaces do not.
Summary and a Useful Mental Picture
A bounded set in \(X\) does not usually have enough compactness in norm topology. Banach–Alaoglu says that after embedding into \(X^{**}\), bounded sets sit inside weak-* compact sets. Thus \(X^{**}\) supplies the missing compactification. Reflexivity says that this compactification adds no new points. Non-reflexivity says that it does.