Endpoint Nonreflexivity in Sequence Spaces

Why the Endpoints Behave Differently

For \(1<p<\infty\), the usual duality theorem identifies the dual of \(L^p\) with \(L^{p'}\), and reflexivity follows by applying that identification twice. At the endpoints, the duality pattern changes. For sequence spaces on \(\mathbb N\), \(\ell^1\) has dual \(\ell^\infty\), but the dual of \(\ell^\infty\) is strictly larger than \(\ell^1\). The canonical embedding then shows that \(\ell^1\) is not reflexive. A separate argument, using the closed subspace \(c_0\), shows that \(\ell^\infty\) is not reflexive either.

Throughout the sequence-space discussion, use counting measure on \(\mathbb N=\{1,2,\ldots\}\), with the sigma-algebra of all subsets. This measure is complete and sigma-finite, but not finite. The real-valued space \(\ell^p\) consists of sequences \(x=(x_n)\) with \(\sum_n|x_n|^p<\infty\) for \(1\le p<\infty\); \(\ell^\infty\) consists of bounded sequences with norm \(\|x\|_\infty=\sup_n|x_n|\). The space \(c_0\) consists of sequences tending to zero, with the same supremum norm. These are Banach spaces, and no almost-everywhere identification changes their elements: for counting measure, a null set is empty.

The map \(J_X\) is an isometry: \(\|J_X(x)\|=\|x\|\). The inequality \(\|J_X(x)\|\le\|x\|\) follows from \(|x^*(x)|\le\|x^*\|\|x\|\). For the reverse inequality when \(x\ne0\), define a norm-one functional on the one-dimensional space spanned by \(x\), taking \(tx\) to \(t\|x\|\), and extend it to \(X\) by the Hahn–Banach theorem. The case \(x=0\) is immediate. We use the standard Hahn–Banach theorem in its norm-preserving extension form.

The Dual of \(\ell^1\) and Its Bidual

Proof. If \(a\in\ell^\infty\), then for every \(x\in\ell^1\) the series \(\sum_na_nx_n\) converges absolutely, since \(\sum_n|a_nx_n|\le\|a\|_\infty\|x\|_1\). It defines a bounded linear functional \(\Phi_a\) with \(\|\Phi_a\|\le\|a\|_\infty\). For each \(n\), let \(e_n\) be the sequence equal to \(1\) at \(n\) and \(0\) elsewhere. Since \(\|e_n\|_1=1\), \(|a_n|=|\Phi_a(e_n)|\le\|\Phi_a\|\). Taking the supremum over \(n\) proves \(\|a\|_\infty\le\|\Phi_a\|\).

Conversely, let \(\Phi\in(\ell^1)^*\) and set \(a_n=\Phi(e_n)\). Then \(|a_n|\le\|\Phi\|\) for every \(n\), so \(a\in\ell^\infty\). For any \(x\in\ell^1\), its finite truncations \(x^{(N)}=\sum_{n=1}^Nx_ne_n\) converge to \(x\) in \(\ell^1\), because \(\sum_{n>N}|x_n|\to0\). Continuity gives \(\Phi(x)=\lim_N\Phi(x^{(N)})=\lim_N\sum_{n=1}^Na_nx_n=\sum_{n=1}^\infty a_nx_n\). Thus every dual functional is of the stated form, with equality of norms as already shown. Taking the dual again gives the bidual identification. \(\square\)

Under the identification \((\ell^1)^{**}=(\ell^\infty)^*\), the canonical image of \(x\in\ell^1\) is the functional \(a\mapsto\sum_n a_nx_n\) on \(\ell^\infty\). To show that this image is not all of \((\ell^\infty)^*\), we construct another bounded functional.

Why \(\ell^\infty\) Is Not Reflexive

The preceding construction shows that \((\ell^\infty)^*\) is large, but that fact alone does not establish whether the canonical map from \(\ell^\infty\) onto its own bidual is surjective. Instead, use \(c_0\), a closed subspace of \(\ell^\infty\). First identify its dual and bidual.

Proof. Given \(a\in\ell^1\), the formula \(\Phi_a(x)=\sum_na_nx_n\) defines a bounded functional on \(c_0\), since \(|\Phi_a(x)|\le\|a\|_1\|x\|_\infty\). Its norm is \(\|a\|_1\): for each \(N\), take \(x_n=1\) when \(n\le N\) and \(a_n\ge0\), \(x_n=-1\) when \(n\le N\) and \(a_n<0\), and \(x_n=0\) when \(n>N\). This finite-support sequence is in \(c_0\), has norm at most \(1\), and gives \(\Phi_a(x)=\sum_{n=1}^N|a_n|\). Letting \(N\to\infty\) proves the lower norm bound; the upper bound was already shown.

Conversely, if \(\Phi\in c_0^*\), put \(a_n=\Phi(e_n)\). For each \(N\), choose the finite-support sequence with coordinates \(x_n=1\) if \(a_n\ge0\), \(x_n=-1\) if \(a_n<0\), for \(n\le N\), and zero otherwise. Then \(\|x\|_\infty\le1\) and \(\sum_{n=1}^N|a_n|=\Phi(x)\le\|\Phi\|\). The partial sums are bounded, so \(a\in\ell^1\). For any \(x\in c_0\), its truncations converge in supremum norm: \(\sup_{n>N}|x_n|\to0\). Hence \(\Phi(x)=\lim_N\sum_{n=1}^Na_nx_n=\sum_na_nx_n\). This proves \(c_0^*=\ell^1\) isometrically. The theorem above identifies \((\ell^1)^*\) with \(\ell^\infty\), so \(c_0^{**}=\ell^\infty\). Under these pairings, \(J_{c_0}(x)\) acts on \(a\in\ell^1\) by \(\sum_na_nx_n\), which is exactly the element \(x\) viewed in \(\ell^\infty\). \(\square\)

We also need the fact that a closed subspace of a reflexive Banach space is reflexive. Here is the argument. Let \(Y\) be a closed subspace of a reflexive Banach space \(X\), and take \(y^{**}\in Y^{**}\). Define \(\Phi\in X^{**}\) by \(\Phi(x^*)=y^{**}(x^*|_Y)\). This is bounded, so reflexivity of \(X\) gives \(x\in X\) with \(\Phi(x^*)=x^*(x)\) for all \(x^*\in X^*\). If \(x^*\) vanishes on \(Y\), then \(\Phi(x^*)=0\), hence \(x^*(x)=0\). This implies \(x\in Y\): if \(x\notin Y\), closedness gives \(d=\inf_{y\in Y}\|x-y\|>0\); the functional \(y+tx\mapsto t\) on \(Y+\operatorname{span}\{x\}\) is bounded because \(|t|d\le\|y+tx\|\), and Hahn–Banach extends it to a functional vanishing on \(Y\) but not on \(x\), a contradiction. Finally, for any \(y^*\in Y^*\), Hahn–Banach extends it to some \(x^*\in X^*\), and then \(y^{**}(y^*)=\Phi(x^*)=x^*(x)=y^*(x)\). Thus \(y^{**}=J_Y(x)\), proving \(Y\) reflexive.

The space \(c_0\) is closed in \(\ell^\infty\): a uniform limit of sequences tending to zero also tends to zero. Since \(c_0^{**}=\ell^\infty\) and its canonical image is \(c_0\), the constant-one sequence proves that \(c_0\) is not reflexive. If \(\ell^\infty\) were reflexive, its closed subspace \(c_0\) would be reflexive by the result just proved. This contradiction establishes that \(\ell^\infty\) is not reflexive.

Finite-Atom Exceptions and a Common Pitfall

Nonreflexivity here is an infinite-dimensional endpoint phenomenon, not a claim about every \(L^1\) or \(L^\infty\) space. For example, on a finite set with counting measure, both spaces are finite-dimensional and reflexive. More generally, suppose a measure space \((X,\mathcal M,\mu)\) is, modulo a null set, a union of finitely many disjoint measurable atoms \(A_1,\ldots,A_k\) of positive measure. An atom means that every measurable subset of \(A_j\) has either measure zero or complement in \(A_j\) of measure zero.

Every measurable real-valued function is constant \(\mu\)-a.e. on each atom: its rational sublevel sets inside that atom must each be null or full, which forces its value to be a single constant outside a null set. Thus an \(L^p\) function is determined by at most \(k\) constants. For \(1\le p<\infty\), a coordinate on an atom of infinite measure must be zero for the function to have finite \(p\)-th power integral; on a finite-measure atom the coordinate contributes \(|c_j|^p\mu(A_j)\). For \(p=\infty\), each coordinate is simply an essentially bounded constant. In every case the space has finite dimension, so the canonical map onto the bidual is surjective: in a finite basis, every linear functional on the dual is evaluation at a vector. The condition is that the whole space, modulo a null set, is covered by finitely many atoms; merely having finitely many atoms does not exclude a non-atomic part.

Common mistake. A tempting but false argument says that every \(\Lambda\in(\ell^\infty)^*\) is determined by \(a_n=\Lambda(e_n)\), and therefore equals \(x\mapsto\sum_na_nx_n\). For the functional constructed above, every \(a_n=0\), so this argument would conclude \(\Lambda=0\); nevertheless \(\Lambda(\mathbf1)=1\). The error is that finite linear combinations of the \(e_n\) are dense in \(c_0\), not in \(\ell^\infty\). Coordinate values determine the restriction to \(c_0\), but need not determine a functional on all of \(\ell^\infty\).