Example in $\ell^\infty$ of a sequence that is bounded and componentwise convergent that is not weakly convergent

Question

After resolving this related question, I thought about this case but couldn't think of anything.

I have seen that, for $1 < p < \infty$, a sequence in $\ell^p$ is weakly convergent if and only if it is componentwise convergent and bounded.

Is there a counterexample for $p = \infty$? That is, what would be an example of a sequence in $\ell^\infty$ that is componentwise convergent and bounded but not weakly convergent?

I thought that maybe the same counterexample that works for $\ell^1$, namely the sequence $(e_n)_{n\in\mathbb{N}}$ of standard unit vectors, would work here, but I can't think of any functional in $(\ell^\infty)^*$ that shows weak convergence of this sequence is impossible. Certainly any functional coming from the canonical embedding of $\ell^1$ into $(\ell^\infty)^*$ can't work (i.e. won't contradict weak convergence). Indeed, if $\phi \in (\ell^\infty)^*$ is the functional corresponding to $y \in \ell^1$, then: $$ \phi(e_n) = \sum_{k =1}^\infty e_n(k)y(k) = y(n) \to 0 \quad \text{as $n\to\infty$} $$ This would be compatible with $e_n \rightharpoonup 0$.

Answer 1

Take the sequence $(x_n)$ in $l^\infty$ with $x_n=(0,\dots,0,1,1,\dots)$ (the first $n$ components are $0$). It is bounded and componentwise convergent to $0$. Thus if it is weaklyconvergent the weak limit would be $(0)$. Take a Banach limit $m \in (l^\infty)^\ast$. Then $m(x_n)=1$ $(n \in \mathbb{N})$. Thus $(x_n)$ is not weakly convergent.