Showing that a sequence in $\ell^\infty(\mathbb{N})^*$ has a weak* convergent subnet

I am currently looking at the sequence of functionals in the closed unit ball of $\ell^\infty(\mathbb{N})^*$ which are given by $e_n(x) := x_n$.

I know that the closed unit ball is weak*-compact by the Banach Alaoglu theorem and I have shown that it does not have a weak*-convergent subsequence.

I would now like to show that I can still find a convergent subnet. It is clear to me that in compact spaces, every net has a convergent subnet, but I am struggling to prove the following:

Let $N_k= \{n \in \mathbb{N}: n \ge k\}$ be a filterbasis and let $\mathcal{U}$ denote the utrafilter of the filter that is induced by the sets $N_k$. Furthermore, let $I := \{(U,u) \ | \ U \in \mathcal{U}, u \in U \}$ be a directed set with the relation $(U,u) \triangleleft (V,v)$ iff $V \subseteq U$.I am trying to view the net over the set $I$ as a subnet of the sequence $e_n$ and then I would like to show that this subnet does indeed converge in the weak*-topology.

I am not quite sure how to do this. I was thinking, since weak*-convergence is basically pointwise convergence it would be easier to look at a sequence $x \in \ell^\infty$, since the $e_n$ are just evaluating the sequences at a certain index. My idea was that any bounded sequence has a convergent subsequence, $x_{n_k}$ so I could look at the set $\{n_k :k\in \mathbb{N}\}$ and maybe use the fact that for any set $A \subseteq \mathbb{N}$ either $A \in \mathcal{U}$ or $A^C \in \mathcal{U}$, since $\mathcal{U}$ is an ultrafilter.

• Don't they tend weak star to the 0 functional? Any element of $\ell^1$ has to have its late terms tend to 0, so this sequence of functionals tends to 0 evaluating on elements of $\ell^1$, which is the definition of weak star convergence to 0. – Ashwin Trisal May 19 '18 at 20:10
• @AshwinTrisal The functionals $e_n$ are defined on $\ell ^\infty$, so I am not sure if they have to tend to zero necessarily – Jack4t3 May 19 '18 at 21:26
• $\ell^\infty(\mathbb{N})$ is isometric to $C(\beta N)$, the continuous function space of $\beta N$ in the sup-metric. So its dual is the space of measures on $\beta N$. So the limit will be a probability measure. – Henno Brandsma May 21 '18 at 8:52

Assume for simplicity that $0\leq x_n\leq 1$ for all $n$ (the proof works in general, only one needs partitions of the disk instead of the interval, which are a bit more annoying to write; or, write $x$ as a linear combination of four sequences each with values in $[0,1]$ and work with each of them).

For each $n\in\mathbb N$, define sets $$A^n_j=x^{-1}\left(\left[\frac{j-1}{2^n},\frac{j}{2^n}\right)\right),\ \ \ j=1,\ldots,2^n$$ (make $A^n_{2^n}=x^{-1}([1-2^{-n},1])$). So for each $n$ we have a partition $\bigcup_j A^n_j=[0,1]$. Since $\mathcal U$ is an ultrafilter, for each $n$ there is a single $j(n)$ with $A^n_{j(n)}\in\mathcal U$ (this follows from the property $B\in\mathcal U\iff B^c\not\in\mathcal U$).

As $A^n_{j(n)},A^{n+1}_{j(n+1)}\in\mathcal U$, it follows that $A^{n+1}_{j(n+1)}\subset A^n_{j(n)}$ (otherwise, they would be disjoint, and that would mean that their intersection, the empty set, is in $\mathcal U$) and that they are all nonempty. This implies that $$\left[\frac{j(n+1)}{2^{n+1}},\frac{j(n+1)}{2^{n+1}}\right)\subset \left[\frac{j(n)}{2^{n}},\frac{j(n)}{2^{n}}\right),\ \ \ n\in\mathbb N.$$ So we have a decreasing nest of intervals, which guarantees that there exists $$r=\lim_n\frac{j(n)}{2^n}\in[0,1].$$ Now let us show that $\lim_{I}x_{n_I}=r$. Fix $\varepsilon>0$, and $n$ with $2^n>1/\varepsilon$. For any $m\in A^n_{j(n)}$, since $r\in [(j(n)-1)/2^n,j(n)/2^n)$, we have $$|x_m-r|<\frac1{2^n}<\varepsilon.$$ And if $(A^n_{j(n)},m)\triangleleft (V,v)$, then $V\subset A^n_{j(n)}$ so $v\in A^n_{j(n)}$ and then $$\tag{*}|x_v-r|<\frac1{2^n}<\varepsilon.$$Thus the net $\{x_{(U,u)}\}_{(U,u)\in I}$ converges to $r$.

The above works for any ultrafilter, and different ultrafilters will often give different limits. If principal ultrafilters are allowed, you get $x_n$ for the ultrafilter generated by $n$; and when you use a free ultrafilter (as the one in your question) you get a proper limit. This limit will always be one of the accumulation points of the sequence, so in particular will agree with the usual limit if the sequence is convergent.

To finally address the question, given an arbitrary $x\in\ell^\infty$ the above constructs $r_x\in\mathbb C$, so now we have a map $\gamma_I:\ell^\infty\to\mathbb C$. Being a limit, it is linear (and multiplicative). Because $r_x$ is an accumulation point for $x$, we have $|r_x|\leq\|x\|_\infty$, so $\gamma_I$ is bounded. Finally, we can now see $(*)$ as $$|e_v(x)-\gamma_I(x)|<\frac1{2^n},$$ showing that $$\lim_I e_v(x)=\gamma_I(x).$$

I think I have found a result to this problem, which is actually a bit more general. It turns out, that the subnet that is constructed above converges to the probability measure $\pi_\mathcal{U}$, which satisfies $\pi_\mathcal{U}(A) = 1$ if $A \in \mathcal{U}$ and equals to $0$ otherwise.

The result can be found in Aliprantos' Book Infinite Dimensional Analysis - A Hitchhiker's Guide, Theorem 16.36