Why the Dual of LInfinity Can Exceed L1

The Pairing Does Not Describe Every Functional

For the counting-measure space \((\mathbb{N},\mathcal{P}(\mathbb{N}),\#)\), where \(\mathbb{N}=\{1,2,\ldots\}\), the complex space \(L^\infty\) is the space \(\ell^\infty\) of bounded complex sequences, and \(L^1\) is \(\ell^1\), the absolutely summable complex sequences. There are no nonempty sets of counting measure zero, so these identifications are pointwise, not merely almost everywhere. Each \(a\in\ell^1\) gives a bounded linear functional on \(\ell^\infty\) by \(x\mapsto\sum_{n=1}^\infty a_nx_n\). The issue here is whether every bounded linear functional on \(\ell^\infty\) has this form. A Banach limit gives a concrete negative answer.

We first construct the functional for real sequences. We will use the real Hahn–Banach extension theorem in its dominated form: if \(p\) is a finite sublinear function on a real vector space, a linear functional on a subspace dominated by \(p\) extends to a linear functional on the whole space still dominated by \(p\). This is a standard theorem from functional analysis; its proof is not needed for the construction below.

Constructing a Banach Limit

The averages \(A_n(x)\) are real and bounded because \(|A_n(x)|\le\|x\|_\infty\). Thus \(p(x)\) is finite. We verify that \(p\) is sublinear, meaning \(p(\alpha x)=\alpha p(x)\) for \(\alpha\ge0\) and \(p(x+y)\le p(x)+p(y)\). The first equality follows from the same properties of scalar limsup. For the second, \(A_n(x+y)=A_n(x)+A_n(y)\), and the limsup of a sum of bounded real sequences is at most the sum of their limsups. This last inequality follows, for every \(\varepsilon>0\), by taking \(n\) large enough that each sequence is at most its limsup plus \(\varepsilon\); then \(A_n(x+y)\le p(x)+p(y)+2\varepsilon\), and we let \(\varepsilon\) decrease to zero.

Let \(c\) be the real vector subspace of convergent sequences. The ordinary limit functional \(L_0(x)=\lim_{n\to\infty}x_n\) is linear on \(c\). If \(x\in c\) with limit \(\ell\), then \(A_n(x)\to\ell\): for any \(\varepsilon>0\), choose \(N\) so that \(|x_k-\ell|<\varepsilon\) for \(k>N\); the contribution of the first \(N\) terms to \(A_n(x)-\ell\) tends to zero, while the remaining average has absolute value at most \(\varepsilon\). Hence \(p(x)=\ell=L_0(x)\) on \(c\), so \(L_0\le p\) there. The real Hahn–Banach theorem therefore extends \(L_0\) to a real linear functional \(L\) on all of real \(\ell^\infty\) such that \(L(x)\le p(x)\) for every real bounded sequence \(x\).

Proof. We have constructed a real linear \(L\) with \(L\le p\), and because it extends \(L_0\), it satisfies \(L(\mathbf{1})=1\). To prove positivity, let \(x_k\ge0\) for all \(k\). Then every \(A_n(-x)\le0\), so \(p(-x)\le0\). Domination gives \(-L(x)=L(-x)\le p(-x)\le0\), hence \(L(x)\ge0\).

To prove shift invariance, fix real \(x\in\ell^\infty\) and set \(y=Sx-x\). The telescoping sum gives, for every \(n\),

Since \(x\) is bounded, \(|A_n(y)|\le(\|x\|_\infty+|x_1|)/n\to0\). Consequently \(p(y)=0\) and \(p(-y)=0\). Domination applied to both sequences gives \(L(y)\le0\) and \(-L(y)=L(-y)\le0\), so \(L(y)=0\). Thus \(L(Sx)=L(x)\).

For boundedness and the norm, the coordinatewise inequalities \(-\|x\|_\infty\mathbf{1}\le x\le\|x\|_\infty\mathbf{1}\), positivity, and linearity imply \(-\|x\|_\infty\le L(x)\le\|x\|_\infty\). Hence \(|L(x)|\le\|x\|_\infty\), so \(\|L\|\le1\). Since \(L(\mathbf{1})=1\) and \(\|\mathbf{1}\|_\infty=1\), we also have \(\|L\|\ge1\); therefore \(\|L\|=1\).

For complex \(z=u+iv\), where \(u,v\) are real bounded sequences, define \(L_{\mathbb{C}}(z)=L(u)+iL(v)\). This is complex-linear: it is real-linear by construction, and \(L_{\mathbb{C}}(iz)=iL_{\mathbb{C}}(z)\), as is seen by writing \(iz=-v+iu\). To check its norm, choose \(\theta\in\mathbb{R}\) so that \(e^{-i\theta}L_{\mathbb{C}}(z)=|L_{\mathbb{C}}(z)|\) (if \(L_{\mathbb{C}}(z)=0\), the desired bound is immediate). The definition gives \(\operatorname{Re}L_{\mathbb{C}}(w)=L(\operatorname{Re}w)\) for each complex sequence \(w\). Therefore

Since \(L_{\mathbb{C}}(\mathbf{1})=1\), its norm is also at least one, hence exactly one. Its real restriction is \(L\); in particular it preserves the established positivity on real nonnegative sequences, agrees with limits of convergent complex sequences by applying the definition to real and imaginary parts, and is shift-invariant by the real result applied to those parts. This proves the theorem. \(\square\)

What the Banach Limit Computes

Why This Functional Is Not an \(\ell^1\) Pairing

For each \(j\in\mathbb{N}\), let \(e_j\) be the sequence whose \(j\)-th coordinate is one and whose other coordinates are zero. It converges to zero, so every Banach limit satisfies \(L(e_j)=0\). Suppose, for contradiction, that there were \(a=(a_j)\in\ell^1\) such that \(L(x)=\sum_{j=1}^\infty a_jx_j\) for every \(x\in\ell^\infty\). Substitution of \(x=e_j\) gives \(a_j=L(e_j)=0\) for every \(j\). Thus \(a\) is the zero sequence, and the proposed pairing would give \(L(\mathbf{1})=0\). But normalization gives \(L(\mathbf{1})=1\), a contradiction.

All sums in this argument are legitimate: if \(a\in\ell^1\) and \(x\in\ell^\infty\), then \(\sum_{j=1}^\infty|a_jx_j|\le\|x\|_\infty\sum_{j=1}^\infty|a_j|<\infty\). Hence the \(\ell^1\) pairing really does define a bounded functional, but the contradiction proves that it does not describe every bounded functional on \(\ell^\infty\). In measure-theoretic language, the pairing is integration against an \(\ell^1\) function for counting measure; it is not the full dual of \(L^\infty\) on this space.

A Common Misstep

A tempting but invalid argument is to treat \(A_n(x)\), a scalar, as though it were the sequence \(Sx\), and then apply \(L\) to \(A_n(x)\). The domain of \(L\) is sequences, so \(L(A_n(x))\) is not defined. The correct proof keeps the notations distinct: \(A_n(x)\) is a scalar average, whereas \(Sx\) is a shifted sequence. The extension is dominated directly by \(p(x)=\limsup_n A_n(x)\); applying the same domination to \(-x\) gives the useful bounds

For the left inequality, \(L(-x)\le p(-x)\) means \(-L(x)\le\limsup_n(-A_n(x))=-\liminf_n A_n(x)\). This proves the claimed bound without applying the functional to a scalar average or confusing the shift with Cesàro averaging. For convergent sequences the two bounds agree with the ordinary limit; for periodic sequences they give the same value as the explicit shift-invariance computations above.