# Characterization of weak convergence in $\ell_\infty$

Is there some simple characterization of weak convergence of sequences in the space $\ell_\infty$? If yes, is there some similar claim for nets?

I was only able to come up with a characterization of sequential weak convergence using limit along ultrafilters, which I will describe below. I wonder whether there is some insight into this characterization. (E.g. whether there is some simple reformulations which does not use ultralimits.) Moreover, I do not know whether at least this characterization works for nets, too.

We have the following result describing weak convergence in $C(K)$, see e.g. Corollary 3.138, p.140. (It is a consequence of Rainwater theorem and characterization of extreme points of unit ball in $C(K)$.)

Let $K$ be a compact topological space. Let $\{f_n\}$ be a bounded sequence in $C(K)$ and $f\in C(K)$. Then, if $f_n\to f$ pointwise, we have $f_n\overset{w}\to f$.

Moreover we have isometric isomorphism between $\ell_\infty$ and $C(\beta\mathbb N)$, which is described e.g. in the Wikipedia article on Stone–Čech compactification or in Chapter 15 of Carothers' book A short course on Banach space theory.. This isomorphism assigns to each bounded sequence $(x_n)$ the continuous function $\overline x$ on $\beta\mathbb N$ defined by $$\overline x(\mathscr U) = \operatorname{\mathscr U-lim} x_n,$$ where $\operatorname{\mathscr U-lim} x_n$ denotes the ultralimit of $x_n$ w.r.t the ultrafilter $\mathscr U$.

Combining the above results we get the following characterization:

Let $f^{(n)},f\in\ell_\infty$. The sequence $f^{(n)}$ converges to $f$ weakly if and only if for every ultrafilter $\mathscr U$ $$\lim_{n\to\infty} \operatorname{\mathscr U-lim} f^{(n)}= \operatorname{\mathscr U-lim} f.$$

(The above claim for principal ultrafilters is just a pointwise convergence. But in the above claim the equality is required for free ultrafilters, too.)

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Dunford and Schwartz has a characterization (IV.6.31): in $\ell_\infty$ the sequence $(f_n)$ coverges to $f$ weakly if and only if it is bounded and , together with every subsequence, converges to $f$ quasi-uniformly. 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 $m$ $$\min_{1\le i\le n}|f_{\alpha_i}(m) -f(m)|<\epsilon$$ –  David Mitra Aug 18 '12 at 19:34
Actually, the 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. –  David Mitra Aug 18 '12 at 19:38
@DavidMitra Could you post your comments as an answer. It answers the first part of my question. (And you gave me some information I did not know before.) So I'll definitely upvote your answer, I will wait a few days to see whether someone will add something about nets before accepting. (And perhaps I'll be able to modify the t.b.'s counterexample from here by myself. He has shown that a characterization similar to the one given in my question does not work for nets in $C(0,1)$. Perhaps similar trick could work in $C(\beta\mathbb N)\cong\ell_\infty$. –  Martin Sleziak Aug 20 '12 at 6:30

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 $A$.

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The $K$ here is of course nothing but $\beta S$, the Stone–Čech compactification of $S$ (here constructed as the spectrum of the $C^\ast$-algebra $B(S)$.) –  t.b. Aug 20 '12 at 18:04


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 $C(X)$.

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