Let $\beta\mathbb N$ is the set of ultrafilters on $\mathbb N$ and $\mathscr F\in\beta\mathbb N$. Assume that $l\_{\mathscr F}\in\big(\ell^\infty(\mathbb N)\big)^{\!\*}$ is the functional which assigns to a bounded sequence $\{a\_n\}\_{n\in\mathbb N}\subset\mathbb C$ its limit, with respect to the ultrafilter $\mathscr F$. Set 
$$
X=\mathrm{span}\,\big\{l\_{\mathscr F}: \mathscr F\in\beta\mathbb N\big\}.
$$
It is relatively straightforward to show that $X$ is dense in $\big(\ell^\infty(\mathbb N)\big)^{\!\*}$, in the weak-$\*$ topology.


My question is whether $X$ is also dense in the dual-norm topology.



Description of $\big(\ell^\infty(\mathbb N)\big)^{\!*}$ via ultrafilters
Identify $\mathbb{N}$ with $\mathbb{Q}\cap[0,1]$ via a bijection, and consider the subspace $C([0,1])\subset\ell^\infty(\mathbb{N})$ of sequences which extend to a continuous function on $[0,1]$. When we restrict each $l\_{\mathscr{F}}$ to $C([0,1])$, we get evaluation at some point of $[0,1]$ (namely, the limit of $\mathscr{F}$ considered as an ultrafilter on $\mathbb{Q}\cap[0,1]$). The span of these is not dense in $C([0,1])^\*$, since (for instance) the closure of the span does not contain the functional given by integration with respect to Lebesgue measure (given finitely many points of $[0,1]$, you can find a continuous function of norm $1$ with Lebesgue integral arbitrarily close to $1$ but which vanishes at all of the points). Extending this Lebesgue measure functional from $C([0,1])$ to all of $\ell^\infty(\mathbb{N})$ by Hahn-Banach, we get an element of $\ell^\infty(\mathbb{N})^\*$ which is not in the closure of $X$.


