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.

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    $\begingroup$ My guess is "no", because the norm closure of your set X may be viewed as the space of atomic finite measures on $\beta{\bf N}$, whereas $\ell^\infty({\bf N})^*$ is the space of all finite Radon measure on $\beta{\bf N}$, and it just seems very unlikely that all finite Radon measures on this space are atomic $\endgroup$ – Yemon Choi Feb 1 '16 at 11:42
  • $\begingroup$ @YemonChoi: Is there any literature you could suggest? $\endgroup$ – smyrlis Feb 1 '16 at 12:03
  • $\begingroup$ I can't think of anything at the moment, sorry $\endgroup$ – Yemon Choi Feb 1 '16 at 12:22

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 $X$.


Eric Wofsey's answer is very nice and simple. Nevertheless this might be interesting. Very old results of Kakutani [Concrete Representation of Abstract (M)-Spaces, Ann. of Math. 42 (1941)] show that for every compact space $K$ the bidual of $C(K)$ is isomorphic to $C(Z)$ for some compact space $Z$ containing $K$. If $X=$span$\lbrace \delta_x:x\in K\rbrace$ were norm-dense in $C(K)^*$ the Hahn-Banach theorem implies that the only continuous linear functional on $C(K)^*$ vanishing on $X$ is $0$. This means that every $f\in C(Z)$ with $f|_K=0$ is $0$ which implies $K=Z$. (Here, however, one has to check that $f$ interpreted as an element of $C(K)^*$ acs on $\delta_x$ as $f(\delta_x)=f(x)$). Applying this to $K=\beta \mathbb N$ shows that your $X$ is not norm-dense.


Both the answers of Eric Wofsey and Jochen Wengenroth are very elegant. Let me give a quite different, a more constructive one. (Although AC is used repeatedly.)

Definitions. A generalised limit is a bounded linear functional on $(\ell^\infty(\mathbb N))^*$, which assigns to every convergent sequences their limit. One such limit is called positive if it assigns non-negative values to sequences with non-negative terms. A Banach limit is a positive generalised limit $\varphi$ with the property $\varphi(S\boldsymbol{a})=\varphi(\boldsymbol{a})$, where $\boldsymbol{a}=\{a_n\}$ and $S$ is the shift operator, i.e. $S\{a_n\}=\{a_{n+1}\}$.

Facts. Positive generalised limits are exactly the generalised limits of unit norm. Limits with respect to ultrafilters are positive generalised limits. Limits with respect to ultrafilters are exactly the generalised limits which assign to sequences subsequential limits. Also, if $A_1,\ldots, A_k$ is a partition of $\mathbb N$ and $\mathscr F$ an ultrafilter, then exactly one of the $A_i$'s belongs to $\mathscr F$.

For $A\subset\mathbb N$, define as $\chi_A=\{a_n\}$, with $a_n=1$ if $n\in A$ and $a_n=0$ if $n\in A^c$. If $\mathscr F$ is an ultrafilter, then $l_{\mathscr F}(\chi_A)=1$ iff $A\in\mathscr F$. Otherwise $l_{\mathscr F}(\chi_A)=0$. Moreover, if $\mathscr F_i$, $i=1,\ldots,k$, are distinct ultrafilters, there exist disjoint $A_1,\ldots,A_k\subset\mathscr N$, such that $A_i\in\mathscr F_i$, $i=1,\ldots,k$.

The fact that $X=\mathrm{span}\{l_{\mathscr F}:\mathscr F\in\beta\mathbb N\}$ is not dense in $(\ell(\mathbb N))^*$ shall be a consequence of the lemma:

Lemma. If $\varphi$ is a Banach limit, $\mathscr F_1,\ldots,\mathscr F_k$ are distinct ultrafilters on $\mathbb N$ and $\lambda_1,\ldots,\lambda_k\in\mathbb C$, then $$\Big\|\varphi-\sum_{j=1}^k\lambda_j\,l_{\mathscr F_j}\Big\|_*=1+\sum_{j=1}^k\lvert\lambda_j\rvert.$$

Proof. Let $S_{m,r}=\{mj+r: j\in\mathbb N\}$. It's not hard to see that $\varphi(\chi_{S_{m,r}})=1/m$. Meanwhile, each $\mathscr F_i$ contains exactly one of the $S_{m,r}$'s, $r=0,\ldots,m-1$, and thus, for every $\varepsilon>0$, there exist $A_i\in\mathscr F_i$, $i=1,\ldots,k$, with $\varphi(A_i)<\varepsilon$. The $A_i$'s can be chosen to be disjoint, as the $\mathscr F_i$'s are distinct. Now let $B=(\bigcup_{i=1}^k A_i)^c$. Then $$ l_{\mathscr F_i}(\chi_{A_i})=1, \quad i=1,\ldots,k\quad\text{while}\quad 1-k\varepsilon\le\varphi(\chi_B)\le 1. $$ Define now the sequence $$ a_n=\left\{\begin{array}{ccc} 1 & \text{if} & n\in B, \\ \frac{\overline{\lambda}_i}{\lvert\lambda_i\rvert} & \text{if} & n\in A_i.\end{array}\right. $$ It's not hard to check that $\|\{a_n\}\|_\infty=1$, $\lambda_i l_{\mathscr F_i}(\{a_n\})=\lvert\lambda_i\rvert$, while $\lvert\varphi(\{a_n\})-1\rvert\le 2k\varepsilon$. Therefore $$ \Big\|\varphi(\{a_n\})-\sum_{j=1}^k\lambda_j\,l_{\mathscr F_j}(\{a_n\})\Big\|_*\ge\Big\lvert\varphi(\{a_n\})-\sum_{j=1}^k\lambda_j\,l_{\mathscr F_j}(\{a_n\})\Big\lvert\ge 1+\sum_{j=1}^k\lvert\lambda_j\rvert-2k\varepsilon. $$ And the Lemma follows.


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