††thanks: The work of the first, third and fourth authors was supported by the
Theoretical Physics and Mathematics Advancement Foundation “BASIS”. The work
of the second authors was supported by RSF, grant no. 24-21-00220.

# Decomposition of the set of Banach limits into discrete and continuous
subsets

Nikolai Avdeev  Faculty of Mathematics, Voronezh State University, Voronezh,
394018, Russia<EMAIL_ADDRESS>Evgenii Semenov  Faculty of Mathematics,
Voronezh State University, Voronezh, 394018, Russia<EMAIL_ADDRESS>Alexandr Usachev  School of Mathematics and Statistics, Central South
University, Hunan, 410085, China<EMAIL_ADDRESS>Roman Zvolinskii
Faculty of Mathematics, Voronezh State University, Voronezh, 394018, Russia
<EMAIL_ADDRESS>To Professor Fedor Sukochev on the occasion of his
60-th birthday

(Date: January 1, 2004)

###### Abstract.

The aim of this work is to describe subsets of Banach limits in terms of a
certain functional characteristic. We compute radii and cardinalities for some
of these subsets.

###### Key words and phrases:

Banach limits, functional characteristic, invariant Banach limits, cardinality

###### 1991 Mathematics Subject Classification:

Primary 46B45; Secondary 47B37

##  1\. Introduction and preliminaries

Undoubtedly, singular traces is one of the Fedor Sukochev areas of interest. A
second edition of the “Singular traces” book (co-authored by him) further
supports this statement. Singular traces have a fairly long history of
interconnection with Banach limits. In the late 90s F. Sukochev with coauthors
employed Banach limits in the construction of Dixmier traces on Lorentz spaces
[1]. In its neatest form this relation was explained in [2], where it was
proved that the set of Banach limits and that of positive normalised traces on
the weak-trace class ideal (of compact operators on a Hilbert space) are
isometrically isomorphic. In this paper we present new results on the
structure of Banach limits.

Denote by $\ell_{\infty}$ the set of all bounded sequences
$x=(x_{1},x_{2},\ldots)$ equipped with the uniform norm

$\|x\|_{\ell_{\infty}}=\sup_{n\in\mathbb{N}}|x_{n}|$

and the usual partial order. Here $\mathbb{N}$ denotes the set of natural
numbers. By $\ell^{*}_{\infty}$ we denote the dual of $\ell_{\infty}$.

######  Definition 1.1.

A linear functional $B\in\ell^{*}_{\infty}$ is said to be a Banach limit if it
satisfies the following conditions:

  1. (i)

$B\geq 0$, that is $Bx\geq 0$ for all $x\in\ell_{\infty}$ such that $x\geq 0$;

  2. (ii)

$B\mathrm{1\negthickspace I}=1$, where $\mathrm{1\negthickspace
I}=(1,1,\dots)$;

  3. (iii)

$BTx=Bx$ for all $x\in\ell_{\infty}$, where $T$ is the left-shift operator,
i.e.,

$T(x_{1},x_{2},x_{3},\dots)=(x_{2},x_{3},x_{4},\dots).$

The existence of Banach limits was first announced by S. Mazur [3]. The proof
using the Hahn-Banach theorem was first presented in the book by S. Banach
[4]. From the definition, it follows that

$\underset{n\to\infty}{\lim\inf}\,x_{n}\leq
Bx\leq\underset{n\to\infty}{\lim\sup}\,x_{n}$

for all $x\in\ell_{\infty}$, and consequently,

$Bx=\lim_{n\to\infty}x_{n}$

for any convergent sequence. Furthermore, $\|B\|_{\ell^{*}_{\infty}}=1$ for
any $B\in\mathfrak{B}$, where $\mathfrak{B}$ denotes the set of Banach limits.
It is easy to see that $\mathfrak{B}$ is a closed convex set on the unit
sphere of the space $\ell^{*}_{\infty}$. G. G. Lorentz proved that for given
$x\in\ell_{\infty}$ and $\lambda\in\mathbb{R}^{1}$, the identity $Bx=\lambda$
holds for all $B\in\mathfrak{B}$ if and only if

$\lim_{n\to\infty}\frac{1}{n}\sum_{k=m+1}^{m+n}x_{k}=\lambda$

uniformly for $m\in\mathbb{N}$. In this case, the sequence
$x=(x_{1},x_{2},\ldots)$ is said to be almost convergent to
$\lambda\in\mathbb{R}^{1}$ [5]. The set of all sequences almost converging to
$\lambda\in\mathbb{R}^{1}$ is denoted by $ac_{\lambda}$.

For a bounded linear operator $H$ on $\ell_{\infty}$, a Banach limit
$B\in\mathfrak{B}$ is said to be $H$-invariant if $Bx=BHx$ for all
$x\in\ell_{\infty}$. Invariant Banach limits were first considered in the
paper by R. Agnew and A. Morse [6]. Later, W. Eberlein proved the existence of
Banach limits invariant under regular Hausdorff transformations [7]. This
approach was further developed in [8]. If the operator $H$ satisfies the
following conditions:

  1. (1)

$H\geq 0$ and $H\mathrm{1\negthickspace I}=\mathrm{1\negthickspace I}$,

  2. (2)

$Hc_{0}\subset c_{0}$,

  3. (3)

$\underset{j\to\infty}{\lim\sup}\,(A(I-T)x)_{j}\geq 0$ for all
$x\in\ell_{\infty}$, $A\in
R(H)=\mathrm{conv}\left\\{H^{k},\,k\in\mathbb{N}\right\\}$,

then there exists a $H$-invariant Banach limit. We denote the set of all such
Banach limits by $\mathfrak{B}(H)$. It can be shown that $\mathfrak{B}(H)$ is
a closed convex subset of $\mathfrak{B}$. The Cesàro operator

$(Cx)_{n}=\frac{1}{n}\sum_{k=1}^{n}x_{k},\,n\in\mathbb{N}$

and dilation operators

$(\sigma_{n}x)=(\underbrace{x_{1},x_{1},\ldots,x_{1}}_{n},\underbrace{x_{2},x_{2},\ldots,x_{2}}_{n},\ldots),\,n\in\mathbb{N}$

satisfy conditions 1-3. Therefore, the sets $\mathfrak{B}(C)$ and
$\mathfrak{B}(\sigma_{n})$ are non-empty for any $n\in\mathbb{N}.$

It is well-known that the space $\ell_{\infty}$ is lattice isometric to the
space $C(\beta\mathbb{N})$ of continuous functions on the Stone-Čech
compactification of the set $\mathbb{N}$, and therefore, it is an $AM$-space
with unity [9, Theorem 12.28]. Consequently, its dual $\ell_{\infty}^{*}$ is
an $AL$-space. Thus, it is lattice isometric to the space $L_{1}(\Omega)$ for
some set $\Omega$ with measure $\mu$ [9, Theorem 12.26]. Let
$\Omega=\Omega_{d}\cup\Omega_{c}$ be a partition of the set $\Omega$ into
measurable subsets such that the measure $\mu$ restricted to $\Omega_{d}$
(resp., to $\Omega_{c}$) is discrete (resp., continuous). Then,

$\ell^{*}_{\infty}\approx L_{1}(\Omega_{d})\oplus L_{1}(\Omega_{c}).$

The kernel of $N(T^{*}-I)$ of the operator $T^{*}-I$ is a Riesz space in
$\ell_{\infty}^{*}$ and, thus, it is an $AL$-space itself. Therefore, a
similar representation holds

$N(T^{*}-I)\approx L_{1}(\Omega_{d}^{1})\oplus L_{1}(\Omega_{c}^{1}).$

The space $N(T^{*}-I)$ consists of all linear functionals on $\ell_{\infty}$
that are invariant under the shift operator. Therefore, the positive part of
its unit sphere $S^{+}_{N(T^{*}-I)}$ is nothing but the set $\mathfrak{B}$ of
Banach limits. Thus, the set $\mathfrak{B}$ can be decomposed into a direct
sum:

$\mathfrak{B}=\mathfrak{B}_{d}\oplus\mathfrak{B}_{c},$

where $\mathfrak{B}_{d}\approx S^{+}_{L_{1}(\Omega_{d}^{1})}$ and
$\mathfrak{B}_{c}\approx S^{+}_{L_{1}(\Omega_{c}^{1})}$. Sets
$\mathfrak{B}_{d}$ and $\mathfrak{B}_{c}$ are referred to as the discrete and
continuous subsets of $\mathfrak{B}$, resp. For any $B\in\mathfrak{B}$, there
exist $B_{1}\in\mathfrak{B}_{d}$, $B_{2}\in\mathfrak{B}_{c}$, and
$\lambda\in[0,1]$ such that:

$B=(1-\lambda)B_{1}+\lambda B_{2}.$

According to the Krein-Milman theorem:

$\mathfrak{B}=\overline{\operatorname{conv}}\operatorname{ext}\mathfrak{B},$

where $\operatorname{ext}\mathfrak{B}$ is the set of extreme points of
$\mathfrak{B}$, and the closure of the convex hull is taken in the weak∗
topology. For any distinct
$B_{1},B_{2},\ldots\in\operatorname{ext}\mathfrak{B}$, and
$\lambda_{1},\lambda_{2},\ldots\in\ell_{1}$ the following equality holds:

$\left\|\sum_{k=1}^{\infty}\lambda_{k}B_{k}\right\|_{\ell^{*}_{\infty}}=\sum_{k=1}^{\infty}|\lambda_{k}|.$

The main properties of the set $\mathfrak{B}$ are discussed in the surveys
[11, 12, 13], as well as in the monograph [14]. Some of the results in this
article were previously announced in [15].

##  2\. A functional characteristic

For every $B\in\mathfrak{B}$ let us consider a function defined on $[0,1]$ by
setting:

$\gamma(B,t)=B\Bigg{(}\bigcup^{\infty}_{n=1}[2^{n},2^{n+t})\Bigg{)},$

where $\bigcup^{\infty}_{n=1}[2^{n},2^{n+t})$ stands for a sequence $x_{k}$
equal to $1$ for $2^{n}\leq k<2^{n+t}$, $k\in\mathbb{N}$ and $0$ otherwise.
This function first appeared in [16].

Let $\Gamma$ be the set of all non-decreasing functions $f$ on $[0,1]$ such
that $f(0)=0$ and $f(1)=1.$ For every $f\in\Gamma$, there exists
$B\in\mathfrak{B}$ such that $\gamma(B,\cdot)=f$ [16, Lemma 1]. For each
$f\in\Gamma$, define

$\mathfrak{B}(f)=\\{B\in\mathfrak{B}:\,\gamma(B,\cdot)=f\\}.$

It is easy to see that $\mathfrak{B}(f)$ is a weak∗-closed convex subset of
$\mathfrak{B}$.

######  Theorem 2.1.

A Banach limit $B$ belongs to the set $\mathfrak{B}_{c}$ if and only if the
function $\gamma(B,\cdot)$ is continuous.

###### Proof.

Suppose, to the contrary, that $B\in\mathfrak{B}_{c}$ and the function
$\gamma(B,\cdot)$ is discontinuous. Since it is non-decreasing, it has a jump
$a>0$ at some point $s\in[0,1]$. For the sake of clarity, assume that
$s\in(0,1)$ and that $\gamma(B,\cdot)$ is continuous from the right at $s$.
Then

$\gamma(B,t)=af(t)+(1-a)g(t),$

where

$f(t)=\left\\{\begin{array}[]{cl}0,&0\leq t<s\\\ 1,&s\leq t\leq
1.\end{array}\right.$

We shall show that

$\operatorname{ext}\mathfrak{B}(f)=\operatorname{ext}\mathfrak{B}\cap\mathfrak{B}(f).$

To this end, it is sufficient to show that

$\operatorname{ext}\mathfrak{B}(f)\subset\operatorname{ext}\mathfrak{B}.$
(2.1)

Suppose $B_{0}=\frac{1}{2}(B_{1}+B_{2})$, where
$B_{0}\in\operatorname{ext}\mathfrak{B}(f)$, and $B_{1},B_{2}\in\mathfrak{B}$.
From this, we have
$f(t)=\gamma(B_{0},t)=\frac{1}{2}\big{(}\gamma(B_{1},t)+\gamma(B_{2},t)\big{)}$.
Since $\gamma(B_{0},t)$ is an extreme point of $\Gamma$, it follows that
$\gamma(B_{0},t)=\gamma(B_{1},t)=\gamma(B_{2},t)$. This implies that $B_{1}$
and $B_{2}$ are in $\mathfrak{B}(f)$. Since
$B_{0}\in\operatorname{ext}\mathfrak{B}(f)$, it follows that
$B_{0}\in\operatorname{ext}\mathfrak{B}$, which proves (2.1).

From (2.1), it follows that

$\mathfrak{B}(f)\subset\overline{\operatorname{conv}}\operatorname{ext}\mathfrak{B}=\mathfrak{B}_{d},$

where the closure is taken in the $\ell_{\infty}$ norm. In particular, $B_{0}$
belongs to a convex combination of a finite or countable number of
$B_{3},B_{4},\ldots\in\operatorname{ext}\mathfrak{B}$. Then, for a
sufficiently small $\alpha>0$, we have $\|B_{0}-\alpha
B_{3}\|_{\ell^{*}_{\infty}}<1$. From the assumption that
$B_{0}\in\mathfrak{B}_{c}$, it follows that $B_{0}\perp\mathfrak{B}_{d}$ [17,
Theorem 25], and $\|B_{0}-\alpha B_{3}\|_{\ell^{*}_{\infty}}=1+|\alpha|$ for
all $\alpha$. The obtained contradiction proves that $B\in\mathfrak{B}_{c}$
implies the continuity of the function $\gamma(B,\cdot)$.

Conversely, let $\gamma(B,\cdot)$ be continuous. It follows from [17, Theorem
25], that a Banach limit $B$ belongs to the set $\mathfrak{B}_{c}$ if and only
if the function $t\mapsto B\\{k\in\mathbb{N}:x_{k}\leq t\\}$ is continuous for
some sequence $x\in\ell_{\infty}$.

Set $x_{1}=2$ and $x_{k}=\log_{2}k-n$ for $2^{n}\leq k<2^{n+1}$,
$n=1,2,\dots$. Then

$\\{k\in\mathbb{N}:x_{k}\leq t\\}=\bigcup_{n\geq
1}\\{k\in\mathbb{N}:\log_{2}k-n\leq t\\}=\bigcup_{n\geq 1}[2^{n},2^{n+t}].$

Since the sets $\bigcup_{n\geq 1}[2^{n},2^{n+t})$ and $\bigcup_{n\geq
1}[2^{n},2^{n+t}]$ differ only by the set $\\{2^{n+t}\\}_{n=1}^{\infty}$ (the
characteristic function of which almost converges to zero), for all
$t\in[0,1]$ we have $\gamma(B,t)=B\\{k\in\mathbb{N}:x_{k}\leq t\\}$ for the
chosen sequence $x\in\ell_{\infty}$. The claim follows from the continuity of
the function $\gamma(B,\cdot)$. ∎

######  Theorem 2.2.

A Banach limit $B$ belongs to the set $\mathfrak{B}_{d}$ if and only if the
function $\gamma(B,\cdot)$ belongs to the closed convex hull of functions
$\varphi_{s}$, where

$\varphi_{s}(t)=\left\\{\begin{array}[]{cc}0,&0\leq t<s\\\ 1,&s\leq t\leq
1\end{array}\right.\text{ or }\left\\{\begin{array}[]{cl}0,&0\leq t\leq s\\\
1,&s<t\leq 1.\end{array}\right.$

($0<s\leq 1$ in the first case and $0\leq s<1$ in the second one).

###### Proof.

If $B\in\mathfrak{B}_{d}$, then

$B=\sum_{i=1}^{\infty}\alpha_{i}B_{i}$

for some $B_{i}\in\operatorname{ext\mathfrak{B}}$, $\alpha_{i}\geq 0$,
$\sum_{i=1}^{\infty}\alpha_{i}=1$. Hence,

$\gamma(B,t)=\Bigg{(}\sum_{i=1}^{\infty}\alpha_{i}B_{i}\Bigg{)}\Bigg{(}\bigcup^{\infty}_{n=1}[2^{n},2^{n+t})\Bigg{)}=\sum_{i=1}^{\infty}\alpha_{i}B_{i}\Bigg{(}\bigcup^{\infty}_{n=1}[2^{n},2^{n+t})\Bigg{)}.$

By [18, Corollary 29] we have that
$B_{i}\big{(}\bigcup^{\infty}_{n=1}[2^{n},2^{n+t})\big{)}=0$ or $1$ for every
$i=1,2,\ldots$ and for every $t\in[0,1]$. Thus,
$B_{i}\big{(}\bigcup^{\infty}_{n=1}[2^{n},2^{n+t})\big{)}=\varphi_{s_{i}}(t)$
and

$\gamma(B,t)=\sum_{i=1}^{\infty}\alpha_{i}\varphi_{s_{i}}(t).$

This proves the first part of Theorem.

As it was proved in the first part of the proof of Theorem 2.1, if the
equality $\gamma(B,t)=\varphi_{s}(t)$ holds for some $s\in[0,1]$ and all
$t\in[0,1]$, then $B\in\mathfrak{B}_{d}$. Let $\gamma(B,\cdot)$ be a convex
combination of a finite number of functions $\varphi_{s_{i}}$, $1\leq i\leq
m$. Then $B$ is a convex combination of finite number of Banach limits from
$\mathfrak{B}_{d}$. Since $\mathfrak{B}_{d}$ is convex, it follows that $B$
belongs to $\mathfrak{B}_{d}$. If $\gamma(B,\cdot)$ is a convex combination of
a countable number of functions $\varphi_{s_{i}}$, then
$B\in\mathfrak{B}_{d}$, since the set $\mathfrak{B}_{d}$ is closed. ∎

Thus, for every $B\in\mathfrak{B}_{d}$, the function $\gamma(B,\cdot)$ has at
most countable number of discontinuities. At each such point, the function
$\gamma(B,\cdot)$ may be discontinuous from the left, from the right, or both.

For any $B\in\mathfrak{B}$, the function $\gamma(B,\cdot)$ is non-decreasing.
Since a non-decreasing function is differentiable almost everywhere, then it
makes sense to speak about the derivative of the function $\gamma(B,\cdot)$.

######  Theorem 2.3.

If $B\in\mathfrak{B}_{d}$, then $\gamma(B,t)^{\prime}=0$ for almost every
$t\in[0,1]$.

###### Proof.

The assertion holds when $B$ is a convex combination of a finite number of
extreme points of $\mathfrak{B}$, that is

$B=\sum_{i=1}^{m}\lambda_{i}B_{i},\ \lambda_{i}>0,\
\sum_{i=1}^{m}\lambda_{i}=1,\ B_{i}\in\operatorname{ext}\mathfrak{B}.$

Using this statement for an arbitrary $B\in\mathfrak{B}_{d}$, we have

$\frac{d\gamma}{dt}(B,t)=\frac{d\gamma}{dt}\bigg{(}\sum_{i=m}^{\infty}\lambda_{i}B_{i},t\Bigg{)}$

up to a finite number of points and for any $m\in\mathbb{N}$. By [19, Chapter
8, Section 2, Theorem 5], the inequalities

$\displaystyle
0\leq\int\limits_{0}^{1}\gamma\Bigg{(}\sum_{i=m}^{\infty}\lambda_{i}B_{i},t\Bigg{)}^{\prime}dt$
$\displaystyle\leq\gamma\Bigg{(}\sum_{i=m}^{\infty}\lambda_{i}B_{i},1\Bigg{)}-\gamma\Bigg{(}\sum_{i=m}^{\infty}\lambda_{i}B_{i},0\Bigg{)}$
$\displaystyle\leq\left\|\sum_{i=m}^{\infty}\lambda_{i}B_{i}\right\|_{\ell^{*}_{\infty}}\leq\sum_{i=m}^{\infty}\lambda_{i}$

hold for each $m\in\mathbb{N}$. Therefore,

$\int\limits_{0}^{1}\gamma(B,t)^{\prime}dt=0.$

Since the function $\gamma(B,\cdot)$ is non-negative, we conclude that
$\gamma(B,t)^{\prime}=0$ almost everywhere. ∎

The converse of Theorem 2.3 does not hold. Recall that a continuous, non-
decreasing, and non-constant function $h$ is called singular if $h^{\prime}=0$
almost everywhere. The most well-known example of such a function is the
Cantor function. If $h(0)=0$, $h(1)=1$, and $h$ is monotone on $[0,1]$, then
by [16, Proposition 2], there exists $B\in\mathfrak{B}$ such that
$\gamma(B,t)=h(t).$ In particular, it can be additionally assumed that $h$ is
singular. In this case, Theorem 2.1 implies that $B\in\mathfrak{B}_{c}$. Thus,
we the following result holds:

######  Corollary 2.4.

If $B\in\mathfrak{B}$ and $\gamma(B,\cdot)$ is a singular function, then
$B\in\mathfrak{B}_{c}$.

######  Remark 2.5.

For every $a>1$ let us introduce a function

$\gamma_{a}(B,t)=B\left(\bigcup_{n=1}^{\infty}[a^{n},a^{n+1})\right),\
t\in[0,1].$

Theorems 2.1-2.3 and Corollary 2.4 hold for the function $\gamma_{a}(B,\cdot)$
as well. It follows from [16, Proposition 20] that $\gamma_{a}(B,t)=t$ for
every $a>1$ and $t\in[0,1]$, provided that $B\in\mathfrak{B}(C)$.

##  3\. Cesàro invariant Banach limits

In this section, we will establish several new properties of the set
$\mathfrak{B}(C)$ of Banach limits invariant under the Cesàro operator. As it
was mentioned in Introduction, this set is non-empty. However, not every
Banach limit is Cesàro invariant. On the other hand every Banach limit is
trivially invariant under the operator $A$ if $A-I:\ell_{\infty}\to ac_{0}$.
It is worth noting that for any Banach limit, there exists a nontrivial
operator $G_{B}:\ell_{\infty}\to\ell_{\infty}$ such that
$\mathfrak{B}(G_{B})=\\{B\\}$. Indeed, one can take
$G_{B}x:=Bx\cdot\mathrm{1\negthickspace I}.$

Let $\\{n_{k}\\}^{\infty}_{k=1}$ be a strictly increasing sequence of natural
numbers. We denote by $V$ the set of sequences of the form:

$x=(x_{i})=\left\\{\begin{array}[]{cr}1,&n_{2k}\leq i<n_{2k+1}\\\
0,&n_{2k+1}\leq i<n_{2k+2}\end{array}\right.,\ i,k\in\mathbb{N},$ (3.1)

where $\\{n_{k}\\}^{\infty}_{k=1}$ satisfies the condition:

$\lim_{k\to\infty}(n_{k+1}-n_{k})=\infty.$

By $V_{0}$ we denote is a subset of $V$ where a stronger condition:

$\lim_{k\to\infty}n_{k+1}/n_{k}=\infty$ (3.2)

holds.

######  Lemma 3.1.

If $x\in V_{0}$, then

$\\{Bx:\,B\in\mathfrak{B}(C)\\}=[0,1].$

###### Proof.

The assumption (3.2) implies

$\displaystyle\underset{j\to\infty}{\lim\inf}\,(C^{m}x)_{j}=0,\quad\underset{j\to\infty}{\lim\sup}\,(C^{m}x)_{j}=1$

for every $m\in\mathbb{N}$. It was proved in [20], that

$\\{By:\,B\in\mathfrak{B}(C)\\}=\left[\lim_{m\to\infty}\underset{j\to\infty}{\lim\inf}\,(C^{m}y)_{j},\lim_{m\to\infty}\underset{j\to\infty}{\lim\sup}\,(C^{m}y)_{j}\right].$

The assertion follows from these three equalities. ∎

The result in the opposite direction can be found in [8, Theorem 15]. Setting
$n_{k}=2^{k}$ for $k\in\mathbb{N}$ in the notations of (3.1), we obtain

$\\{Bx:\,B\in\mathfrak{B}(C)\\}=\\{1/2\\}.$

This lemma also shows that the assumption (3.2) cannot be replaced by a weaker
one:

$\lim_{k\to\infty}(n_{k+1}-n_{k})=\infty.$

Therefore, $V_{0}\subsetneq V$. Also, if $x\in V$, then the Sucheston theorem
[21] implies

$\\{Bx:\,B\in\mathfrak{B}\\}=[0,1].$

The following result is a particular form of [18, Lemma 13].

######  Lemma 3.2.

For every $B\in\mathfrak{B}$

$0,1\in\\{Bx:\,x\in V_{0}\\}.$

######  Corollary 3.3.

The set $\\{Bx:\,x\in V_{0}\\}$ is finite or countable for every
$B\in\mathfrak{B}_{d}$ and $x\in V_{0}$.

It follows from [13], that both diameter and radius of the set $\mathfrak{B}$
in $\ell^{*}_{\infty}$ equal $2$, i.e.,

$d\big{(}\mathfrak{B},\ell^{*}_{\infty}\big{)}:=\sup_{x,y\in\mathfrak{B}}\|x-y\|_{\ell^{*}_{\infty}}=2;\quad
r\big{(}\mathfrak{B},\ell^{*}_{\infty}\big{)}:=\inf_{x\in\mathfrak{B}}\sup_{y\in\mathfrak{B}}\|x-y\|_{\ell^{*}_{\infty}}=2.$

Furthermore,
$d\big{(}\mathfrak{B}(C),\ell^{*}_{\infty}\big{)}=d\big{(}\mathfrak{B}(\sigma_{n}),\ell^{*}_{\infty}\big{)}=2$
[20].

For every $B\in\mathfrak{B}(C)$ one has $\gamma(B,t)=t$ for every $t\in[0,1]$.
Hence, $\mathfrak{B}(C)\subset\mathfrak{B}(f_{1})$, where $f_{1}(t)=t$ for
every $t\in[0,1]$. Thus,
$d\big{(}\mathfrak{B}(f_{1}),\ell^{*}_{\infty}\big{)}=2$. It was shown in [16,
Theorem 25] that the inclusion $\mathfrak{B}(C)\subset\mathfrak{B}(f_{1})$ is
proper, that is there exists
$B\in\mathfrak{B}(f_{1})\setminus\mathfrak{B}(C)$.

The following result shows that the radius of the set $\mathfrak{B}(C)$ is
also equal to 2.

######  Theorem 3.4.

$r\big{(}\mathfrak{B}(C),\ell^{*}_{\infty}\big{)}=2$.

###### Proof.

Let $B_{1}\in\mathfrak{B}(C)$. According to Lemma 3.2, you can find a sequence
$x\in V_{0}$ such that $B_{1}x=0$. Due to Lemma 3.1, there exists
$B_{2}\in\mathfrak{B}(C)$ such that $B_{2}x=1$. Then, $(B_{2}-B_{1})x=1-0=1$.

The sequence $y=2x-\mathrm{1\negthickspace I}$ obviously has the property that
all of its coordinates are $\pm 1$. Therefore, $\|y\|_{\ell_{\infty}}=1$, and

$(B_{2}-B_{1})y=(B_{2}-B_{1})(2x-\mathrm{1\negthickspace I})=2-0=2.$

Hence,

$r\big{(}\mathfrak{B}(C),\ell^{*}_{\infty}\big{)}\geqslant\|B_{2}-B_{1}\|_{\ell^{*}_{\infty}}\geqslant
2.$

The converse inequality is obvious. ∎

Now we shall apply the properties of the set $\mathfrak{B}(C)$, obtained in
Lemma 3.1, to study the set $\mathfrak{B}_{c}$, the continuous part of the set
$\mathfrak{B}$.

######  Lemma 3.5.

For every $x\in V_{0}$ we have

$\\{Bx:\,B\in\mathfrak{B}_{c}\\}=[0,1].$

###### Proof.

It was proved in [22] that sets $\operatorname{ext}\mathfrak{B}$ and
$\mathfrak{B}(C)$ are orthogonal. Therefore, $\mathfrak{B}(C)$ is contained in
the orthogonal complement of $\mathfrak{B}_{d}$, i.e.,
$\mathfrak{B}(C)\subset\mathfrak{B}_{c}$. Applying Lemma 3.1, we obtain:

$[0,1]=\\{Bx:\,B\in\mathfrak{B}(C)\\}\subset\\{Bx:\,B\in\mathfrak{B}_{c}\\}.$

The converse inclusion is evident. ∎

Further, we aim to compute the radius of the set $\mathfrak{B}_{c}$. Before we
proceed to the result we explain why Theorem 3.4 cannot be applied.

It is obvious from the definition that $d(M,\ell^{*}_{\infty})\leqslant
d(N,\ell^{*}_{\infty})$ for every subsets $M,N\in\ell_{\infty}^{*}$ such that
$M\subseteq N$. It turns out that a similar statement does not hold for radii
of sets, in general. We shall need the definition of relative radius of a set
$M$ in a normed space $E$, which is defined as follows:

$r_{E}(M)=\inf_{x\in E}\sup_{y\in M}\|x-y\|_{E}.$

######  Proposition 3.6.

Let $E$ be a normed space. The inequality $r(M,E)\leq r(N,E)$ holds for every
convex bounded set $M\subset E$ and every set $N\subset E$ such that $M\subset
N$ if and only if $E$ is a 2-dimensional or Hilbert space.

###### Proof.

For every $N\subset E$ such that $M\subseteq N$ one has

$\inf_{x\in N}\sup_{y\in N}\|x-y\|_{E}\geq\inf_{x\in N}\sup_{y\in
M}\|x-y\|_{E}\geq\inf_{x\in E}\sup_{y\in M}\|x-y\|_{E}.$

Thus,

$r(N,E)\geq r_{E}(M).$ (3.3)

It was proved in [23] that $E$ is a 2-dimensional or a Hilbert space if and
only if $r(M,E)=r_{E}(M)$ for every convex bounded set $M\in E$.

Let $E$ be a 2-dimensional or a Hilbert space. The assertion follows from
(3.3) and the above mentioned result.

Conversely, suppose to the contrary that there exists a convex bounded set
$M\in E$ and a set $N\in E$ such that $M\subseteq N$ and $r(M,E)>r(N,E)$.
Using (3.3), we obtain that $r(M,E)>r_{E}(M).$ The assertion follows from
[23]. ∎

######  Theorem 3.7.

$r(\mathfrak{B}_{c},\ell^{*}_{\infty})=2$.

###### Proof.

The proof follows the same idea as that of Theorem 3.4. Let
$B_{1}\in\mathfrak{B}_{c}$. By Lemma 3.2, there exists a sequence $x\in V_{0}$
such that $B_{1}x=0$. Using Lemma 3.5, we conclude that there exists
$B_{2}\in\mathfrak{B}_{c}$ such that $B_{2}x=1$. Then $(B_{2}-B_{1})x=1$. For
$y=2x-\mathrm{1\negthickspace I}$, we have $\|y\|_{\ell_{\infty}}=1$, and

$(B_{2}-B_{1})y=(B_{2}-B_{1})(2x-\mathrm{1\negthickspace I})=2.$

Hence, $r(\mathfrak{B}_{c},\ell^{*}_{\infty})=2$. ∎

##  4\. Some cardinalities

C. Chou proved that the cardinality of the set ${\rm ext}\\!\ \mathfrak{B}$ is
$2^{\mathfrak{c}}$ [10]. This implies that ${\rm card}\\!\
(\mathfrak{B}_{d})=2^{\mathfrak{c}}$. In [16, Corollary 13] it was proved that
for every $s\in[0,1]$ the set of all $B\in{\rm ext}\\!\ \mathfrak{B}$ such
that $\gamma(B,\cdot)=\chi_{[s,1]}$ has cardinality $2^{\mathfrak{c}}$. This
splits the set ${\rm ext}\\!\ \mathfrak{B}$ (and, so, the set
$\mathfrak{B}_{c}$) into continually many disjoint sets of maximal
cardinality.

Since the cardinality of $\mathfrak{B}(C)$ is maximal [24, Section 4] and
$\mathfrak{B}(C)\subset\mathfrak{B}_{c}$, it follows that $\mathfrak{B}_{c}$
is of maximal cardinality as well. However, the set $\mathfrak{B}(C)$ rather
appears to be a small subset of of $\mathfrak{B}_{c}$, since for every
$B\in\mathfrak{B}(C)$ we have $\gamma(B,t)=t,0\leq t\leq 1$, whereas by
Theorem 2.1 the set $\\{\gamma(B,\cdot):B\in\mathfrak{B}_{c}\\}$ is much more
diverse. In this section we address the question of how big is the set
$\mathfrak{B}_{c}\setminus\mathfrak{B}(C)$. We show that for every absolutely
continuous function $f\in\Gamma$ with Hölder continuous derivative, the set of
all $B\in\mathfrak{B}$ with $\gamma(B,\cdot)=f$ has cardinality
$2^{\mathfrak{c}}$. Hence, the set $\mathfrak{B}_{c}\setminus\mathfrak{B}(C)$
splits into continually many disjoint sets of maximal cardinality.

We will use the following (particular form of the) result proved in [20, Lemma
2].

######  Lemma 4.1.

Let $D$ and $M$ be convex subsets of $\ell_{\infty}^{*}$ (equipped with weak∗
topology) and let $M$ be compact. Suppose that there exist mappings
$f_{1},f_{2}:\ell_{\infty}^{*}\to\ell_{\infty}^{*}$ satisfying the following
conditions:

  1. (i)

$f_{2}$ is affine and continuous;

  2. (ii)

The following inclusion holds:

${\rm ext}\\!\ D\subseteq f_{1}^{-1}(M)\cap(f_{2}\circ
f_{1}-I)^{-1}(0)\text{;}$

  3. (iii)

The following inclusion holds: $f_{2}(M)\subseteq D$.

Then the inequality

${\rm card}\\!\ ({\rm ext}\\!\ M)\geqslant{\rm card}\\!\ ({\rm ext}\\!\ D)$

holds.

Denote by $EL$ the set of all extended limits on $\ell_{\infty}$. For every
$f\in\Gamma$ and $A\subset EL$ define the set

$A(f):=\left\\{\omega\in A:\gamma(\omega,\cdot)=f\right\\}.$

For every $n\in\mathbb{N}$, setting

$a_{n,m}=\frac{1}{\log 2}\sum_{i=2^{n}+1}^{2^{n}+m}\frac{1}{i},\
m=1,\dots,2^{n},$

we obtain a partition

$0=a_{n,0}<a_{n,1}<a_{n,2}<\dots<a_{n,2^{n}}<a_{n,2^{n}+1}=1.$

of the interval $[0,1]$ into $2^{n}+1$ intervals of length at most

$\max\\{a_{n,1},1-a_{n,2^{n}}\\}=o(1/n),\ n\to\infty.$ (4.1)

The choice of such a peculiar partition is justified in Lemma 4.2 below.

Fix an absolutely continuous $f\in\Gamma$ and define the operator $Q$ on
$\ell_{\infty}$ by setting $(Qx)_{1}=x_{1}$, $(Qx)_{2}=x_{2}$ and

$(Qx)_{n+2}=\sum_{k=2^{n}+1}^{2^{n+1}}w_{n,k}x_{k},\ n=1,2,\dots,$

where positive weights are defined as follows:

$w_{n,k}=f^{\prime}(a_{n,k-(2^{n}+1)})\cdot\frac{1}{k\log 2}.$

For every $0<t<1$ define

$(y_{t})_{k}=\begin{cases}1,2^{n}\leqslant k<2^{n+t}\\\ 0,2^{n+t}\leqslant
k<2^{n+1}\end{cases},\ k,n\in\mathbb{N}.$

Note, that $\gamma(B,t)=B(y_{t}).$

######  Lemma 4.2.

For an absolutely continuous function $f\in\Gamma$ and every $0\leqslant
t\leqslant 1$ we have $(Qy_{t})_{n}\to f(t)$, $n\to\infty$.

###### Proof.

For $t=0$ we have $y_{t}\equiv 0$ and, so, $(Qy_{t})_{n}\to 0=f(0)$. Fix
$0<t\leqslant 1$. For every $n\in\mathbb{N}$, setting

$\bar{a}_{n,m}=a_{n,m},\ m=0,\dots,\lfloor 2^{n+t}\rfloor-2^{n},\ \text{and},\
\bar{a}_{n,\lfloor 2^{n+t}\rfloor-2^{n}+1}=t$

we obtain a partition of the interval $[0,t]$ into $\lfloor
2^{n+t}\rfloor-2^{n}+1$ intervals of length at most

$\max\\{\bar{a}_{n,1},t-\bar{a}_{n,\lfloor 2^{n+t}\rfloor-2^{n}}\\}=o(1/n),\
n\to\infty.$

Thus,

$\displaystyle(Qy_{t})_{n+2}$
$\displaystyle=\sum_{k=2^{n}+1}^{2^{n+1}}w_{n,k}(y_{t})_{k}$
$\displaystyle=\sum_{k=2^{n}+1}^{\lfloor
2^{n+t}\rfloor}f^{\prime}(a_{n,k-(2^{n}+1)})\cdot\frac{1}{k\log
2}\to\int_{0}^{t}f^{\prime}(s)\,ds=f(t),$

as an integral sum (since $f$ is absolutely continuous and $f(0)=0$). ∎

Similarly,

$(Q\text{1I})_{n+2}=\frac{1}{\log
2}\sum_{k=2^{n}+1}^{2^{n+1}}f^{\prime}(a_{n,k-(2^{n}+1)})\cdot\frac{1}{k}\to\int_{0}^{1}f^{\prime}(s)\,ds=f(1)=1,$

as an integral sum.

Consider operator $R$ on $\ell_{\infty}$ given by

$(Rx)_{1}=x_{1},(Rx)_{2}=x_{2}\ \text{and}\ (Rx)_{k}=x_{n+2},\
2^{n}+1\leqslant k\leqslant 2^{n+1},\ n=1,2,\dots.$

######  Proposition 4.3.

Let $f\in\Gamma$ be an absolutely continuous function. Let $A$ be a weak∗
compact convex subset of $EL$, such that operators $Q^{*}$ and $R^{*}$ map $A$
into itself. We have

${\rm card}\\!\ ({\rm ext}\\!\ A(f))\geqslant{\rm card}\\!\ ({\rm ext}\\!\
A).$

###### Proof.

Since $A$ is weak∗ compact and convex, it is easy to check that $A(f)$ is
weak∗ compact convex subset of $A$.

We apply Lemma 4.1 for $M=A(f)$, $D=A$, $f_{1}=Q^{*}$ and $f_{2}=R^{*}$. The
condition (i) of Lemma 4.1 is clearly satisfied. Further, we have

$\displaystyle\left|(QRx)_{n+2}-x_{n+2}\right|$
$\displaystyle=\left|\left(\sum_{k=2^{n}+1}^{2^{n+1}}f^{\prime}(a_{n,k-(2^{n}+1)})\cdot\frac{1}{k\log
2}\right)x_{n+2}-x_{n+2}\right|$
$\displaystyle\leq\left|\sum_{k=2^{n}+1}^{2^{n+1}}f^{\prime}(a_{n,k-(2^{n}+1)})\cdot\frac{1}{k\log
2}-1\right|\|x\|$
$\displaystyle\mathop{\rightarrow}\limits_{n\to\infty}\left|\int_{0}^{1}f^{\prime}(s)\,ds-1\right|\|x\|=\left|f(1)-f(0)-1\right|\|x\|=0,$

since $f\in\Gamma$. Thus, $QR-I:\ell_{\infty}\to c_{0}$.

For every $\omega\in A$, we have $\omega\circ Q\in A$. Moreover,

$\gamma(\omega\circ Q,t)=\gamma(Qy_{t})=f(t),$

by Lemma 4.2. Hence, $\omega\circ Q\in A(f)$. Also, $\omega(QR-I)=0$ for every
$\omega\in A$. Therefore, condition (ii) is satisfied. Condition (iii) is
satisfied by assumption. The assertion follows from Lemma 4.1. ∎

######  Corollary 4.4.

The cardinality of $EL(f)$ equal to $2^{\mathfrak{c}}$ for every absolutely
continuous $f\in\Gamma$.

###### Proof.

It is easy to see that $Q^{*}$ and $R^{*}$ map $EL$ into itself. The result
follows from the fact that ${\rm card}\\!\ ({\rm ext}\\!\
EL)=2^{\mathfrak{c}}.$ ∎

######  Corollary 4.5.

The cardinality of $\mathfrak{B}(f)$ equal to $2^{\mathfrak{c}}$ for every
absolutely continuous $f\in\Gamma$ with Hölder continuous derivative.

###### Proof.

The result follows from Lemma 4.6 below and the fact that the cardinality of
${\rm ext}\\!\ \mathfrak{B}$ is $2^{\mathfrak{c}}.$ ∎

######  Lemma 4.6.

Operator $R^{*}$ maps $\mathfrak{B}$ into itself. If $f\in\Gamma$ is
absolutely continuous with Hölder continuous derivative, then the operator
$Q^{*}$ maps $\mathfrak{B}$ into itself.

###### Proof.

Let $B\in\mathfrak{B}$. It is clear that $B\circ Q$ and $B\circ R$ are
positive and normalised.

1\. For every $n\in\mathbb{N}$ consider

$(Q(x-Tx))_{n+2}=\sum_{k=2^{n}+1}^{2^{n+1}}w_{n,k}(x_{k}-x_{k+1})$
$=w_{n,2^{n}+1}x_{2^{n}+1}+\sum_{k=2^{n}+2}^{2^{n+1}-1}x_{k}(w_{n,k}-w_{n,k-1})-w_{n,2^{n+1}-1}x_{2^{n+1}}.$

Since $f^{\prime}$ is Hölder continuous, then it is bounded on $[0,1]$. Hence,

$w_{n,2^{n}+1}=\frac{f^{\prime}(a_{n,0})}{2^{n}+1}\to 0\ \text{and}\
w_{n,2^{n+1}-1}=\frac{f^{\prime}(a_{n,2^{n}-2})}{2^{n+1}-1}\to 0,\
n\to\infty.$

Further,

$\displaystyle\sum_{k=2^{n}+2}^{2^{n+1}-1}$
$\displaystyle|w_{n,k}-w_{n,k-1}|=\sum_{k=2^{n}+2}^{2^{n+1}-1}\left|\frac{f^{\prime}(a_{n,k-(2^{n}+1)})}{k}-\frac{f^{\prime}(a_{n,k-1-(2^{n}+1)})}{k-1}\right|$
$\displaystyle\leqslant\sum_{k=2^{n}+2}^{2^{n+1}-1}\frac{1}{k-1}\left|f^{\prime}(a_{n,k-(2^{n}+1)})-f^{\prime}(a_{n,k-1-(2^{n}+1)})\right|$
$\displaystyle+\sum_{k=2^{n}+2}^{2^{n+1}-1}\frac{\left|f^{\prime}(a_{n,k-(2^{n}+1)})\right|}{k(k-1)}$

Since $f^{\prime}$ is Hölder continuous, then

$\displaystyle\left|f^{\prime}(a_{n,k-(2^{n}+1)})-f^{\prime}(a_{n,k-1-(2^{n}+1)})\right|$
$\displaystyle\leq C|a_{n,k-(2^{n}+1)}-a_{n,k-1-(2^{n}+1)}|^{\alpha}$
$\displaystyle=o(n^{-\alpha}),\ n\to\infty$

by (4.1). Thus,

$\sum_{k=2^{n}+2}^{2^{n+1}-1}|w_{n,k}-w_{n,k-1}|\leqslant
o(n^{-\alpha})\sum_{k=2^{n}+2}^{2^{n+1}-1}\frac{1}{k-1}+O\left(\sum_{k=2^{n}+2}^{2^{n+1}-1}\frac{1}{k(k-1)}\right),$

which tends to zero as $n\to\infty$. Hence, $Q(x-Tx)\in c_{0}.$ Thus, $(B\circ
Q)(x-Tx)=0$ for every $B\in\mathfrak{B}$. Thus, $B\circ Q\in\mathfrak{B}$ for
every $B\in\mathfrak{B}$.

2\. For every $x\in\ell_{\infty}$ we have

$\displaystyle R(x-Tx)$
$\displaystyle=(x_{1}-x_{2};0,x_{2}-x_{3};0,0,0,x_{3}-x_{4};\dots)$
$\displaystyle=\sum_{n=0}^{\infty}(x_{n+1}-x_{n+2})\chi_{\\{2^{n+1}-1\\}}\in
ac_{0}.$

Thus, $(B\circ R)(x-Tx)=0$ and $B\circ R\in\mathfrak{B}$ for every
$B\in\mathfrak{B}$. ∎

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