The dual of ${\mathrm{\ell }}^{\mathrm{\infty }}$ is $ba(2^{\mathbb{N}})$, the finitely additive charges

The dual of ${\mathrm{\ell }}^{\mathrm{\infty }}$ is isometrically isomorphic to $ba(2^{\mathbb{N}})$, the space of bounded finitely additive signed measures (charges) on the full power set of $\mathbb{N}$, normed by total variation; the pairing sends $\mu$ to the functional $x\mapsto \int xd\mu$ defined by the finitely additive integral.

Under this identification ${\mathrm{\ell }}^1$ is exactly the subspace of countably additive charges, a proper norm closed subspace of $({\mathrm{\ell }}^{\mathrm{\infty }}{)}^{\ast }$. In particular $({\mathrm{\ell }}^1{)}^{\ast }={\mathrm{\ell }}^{\mathrm{\infty }}$ but $({\mathrm{\ell }}^{\mathrm{\infty }}{)}^{\ast }\mathrm{\ne }{\mathrm{\ell }}^1$, so ${\mathrm{\ell }}^1$ and ${\mathrm{\ell }}^{\mathrm{\infty }}$ are not reflexive.

Not proved in this library. Recorded with a citation; both the finitely additive integral and the duality are outside the current stack.

What would prove it. Given $\phi \in ({\mathrm{\ell }}^{\mathrm{\infty }}{)}^{\ast }$, define $\mu (A)=\phi (1_A)$ for $A\subseteq \mathbb{N}$; finite additivity is linearity, boundedness of the total variation is boundedness of $\phi$, and the two operations invert each other because simple functions are dense in ${\mathrm{\ell }}^{\mathrm{\infty }}$ in the supremum norm.

Why it matters here. It is where the naive pattern "the dual of a sequence space is a sequence space" breaks. Every element of $({\mathrm{\ell }}^{\mathrm{\infty }}{)}^{\ast }\setminus {\mathrm{\ell }}^1$ is produced by an extension argument and never by a formula, the Banach limits of Banach limits ‡ being the standard examples, so the failure of reflexivity for ${\mathrm{\ell }}^1$ is inseparable from the choice discussion in The set-theoretic cost of Hahn-Banach ‡.