Characterization of weak convergence in $\ell_\infty$
In Dunford and Schwartz' Linear Operators Part 1: General Theory (DS) there is a characterization of weakly convergent sequences, appearing in item IV.6.31:
DS IV.6.31: In $\ell_\infty$ the sequence $(f_n)$ converges to $f$ weakly if and only if it is bounded and, together with every subsequence, converges to $f$ quasi-uniformly.
In the above, "quasi-uniformly convergent" means: 1) pointwise convergence, and 2) for every $n_0$ and every $\epsilon>0$ there exists a finite number of indices $\alpha_1,\ldots,\alpha_n\ge n_0$ such that for each $m$ $$\min_{1\le i\le n}|f_{\alpha_i}(m)−f(m)|<ϵ.$$
Actually, this characterization is for more general $B(S)$ spaces. Here $S$ is an arbitrary set and $B(S)$ is the set of all bounded scalar-valued functions on $S$ equipped with the sup norm.
I'm not entirely sure that this is different in spirit from your characterization. Here is an outline of the proof of IV.6.31:
In Dunford-Schwartz IV.6.19-20, the space $B(S)$ is identified (with the aid of the Stone-Weierstrass Theorem) with a certain $C(K)$ space with $K$ compact (namely $K$ is the set of non-zero, continuous, multiplicative continuous functionals in the closed unit sphere of $\cal U^*$, where $\cal U$ is $B(S)$ regarded as an algebra). Weak convergence of sequences in $B(S)$ is identified with weak convergence of the corresponding sequence in $C(K)$. Moreover, in the identification of $B(S)$ with $C(K)$, $S$ is identified as a dense subset of $K$. The result IV.6.31 follows from your characterization of weak sequential convergence in a $C(K)$ space and item IV.6.30 in DS:
DS IV.6.30: Let $A$ be a dense subset of a compact Hausdorff space $S$, and suppose that a sequence $\{f_n\}$ of continuous functions converges at every point of $A$ to a continuous limit $f_0$. Then $\{f_n\}$ converges to $f_0$ at every point of $S$ if and only if $\{f_n\}$ and every subsequence of $\{f_n\}$ converges to $f_0$ quasi-uniformly on $A$.
$\newcommand{\Fin}{\mathrm{Fin}}\newcommand{\UU}{\mathscr U}{\newcommand{\Ulim}{\operatorname{\UU}-\lim}}\newcommand{\N}{\mathbb N}\newcommand{\Uilim}{\operatorname{\UU_i}-\lim}$I will try to show here that the the characterization via ultrafilters, which is described in the text of my question, does not work for nets. I tried to modify t.b.'s answer to a related question of mine, where he shown me that $C(0,1)$ does not have similar property.
I will need the following facts about ultrafilters on $\N$ and ultralimits of sequences:
- If $A_1\cup\dots\cup A_k=\N$ and $\UU$ is an ultrafilter, then one of the sets $A_1,\ldots,A_n$ belongs to$\UU$.
- For any ultrafilter $\mathscr U$ and any $A\subseteq\mathbb N$ we have $$\Ulim \chi_A= \begin{cases} 1 & A\in\UU, \\ 0 & A\notin\UU. \end{cases} $$ (where $\chi_A$ denotes the characteristic function of the set $A$.)
- There exists a linear functional $L\in\ell_\infty^*$ such that for every sequence $x$ which converges in Cesaro mean, the value of $L(x)$ is the same as the value of Cesaro mean. For example, we can get such a fixing one free ultrafilter $\UU$ and putting $$L \colon x \mapsto \Ulim \frac{x_1+\dots+x_n}n.$$
Now let $\Fin$ bee the set of all finite subsets of $\beta\N$; i.e. finite sets of ultrafilters on $\N$. Clearly, $(\Fin,\subseteq)$ is a directed set.
Let $\UU_1,\ldots,\UU_n$ be ultrafilters. Let us define set $A_1,\dots,A_{2n}$ as the sets of the form $A_k=\{2un+k; u\in\N\}$.
Now we choose half of them to be $B_1,\dots,B_n$ in the following way: Since $A_1\cup \dots \cup A_{2n}=\N$, one of these sets belong to $\UU_1$. This will be $B_1$.
Suppose $B_1,\ldots,B_{k-1}$ are already chosen. There is at least one of the sets $A_1,\dots,A_n$ which belongs to $\UU_k$. If this set was already chosen in one of the preceding steps, then $B_k$ can be taken arbitrary of the remaining sets. If not, this set will be $B_k$.
Now put $B:=B_1\cup\dots\cup B_n$ and $x=\chi_B$. Notice that $B\in\UU_i$ for $i=1,\dots,n$. So we have $\Uilim x=1$ for each $i$.
It is also easy to see that the Cesaro mean is $\lim\limits_{n\to\infty} \frac{x_1+\dots+x_n}n=\frac12$ and thus $L(x)=\frac12$.
In this way we have constructed a bounded sequence $x_F$ for any finite set $F$ of ultrafilters. So we have a net $(x_F)_{F\in\Fin}$.
If we fix arbitrary ultrafilter $\UU$, we get $\Ulim x_F=1$ for every $F\supseteq\{\UU\}$. So the net $(\Ulim x_F)_{F\in\Fin}$ converges to $1=\Ulim \chi_{\N}$. (Basically $\chi_{\N}$ is just a fancy name for the constant sequence $(1,1,1,\dots)$.)
But each $L(\chi_F)$ is equal to $1/2$ and $L(\chi_{\N})=1$, so this net does not converge to $\chi_{\N}$ weakly.
As I have mentioned, my construction is similar to this t.b.'s answer. In both cases we work with some compact space $X$. (Here $X=\beta\N$ and in his answer $X=[0,1]$.) We are trying to find a functional $\varphi\in C(K)^*$ together with the functions $f_F\in C(X)$ for each finite subset $F\subseteq X$ such that:
- $f_F|_F=1$, i.e., $f_F(x)=1$ for each $x\in F$;
- $\varphi(f_F)=\frac12$ for each finite set $F\subseteq X$.
If we find such system of functions, the same counterexample can be used for the space $C(X)$.