A norm-closed dual subspace need not be weak-star closed ($c_0⊊{\ell }^{\infty }=({\ell }^1{)}^{\prime },(c_0{)}_{\circ }=0$)

The continuous dual space of ${\ell }^1$ is ${\ell }^{\infty }$ under the complex-linear coordinate pairing. Its subspace $c_0$ is norm closed: uniform approximation by a sequence with a small tail leaves a small tail. Nevertheless its preannihilator in ${\ell }^1$ is zero, because the coordinate unit vectors belong to $c_0$ and test every coordinate. Its double continuous annihilator is therefore all of ${\ell }^{\infty }$. It is weak-star dense directly as well: the finite truncations of any bounded sequence lie in $c_0$ and converge against every ${\ell }^1$ test by absolute summability. This demonstrates why norm closure cannot replace closure in the weak-star topology in the dual-side annihilator theorem.