# Functions of bounded variation from ideal perspective

Jacek Gulgowski  Institute of Mathematics
Faculty of Mathematics, Physics and Informatics
University of Gdańsk
ul. Wita Stwosza 57
80-308 Gdańsk
Poland<EMAIL_ADDRESS>,  Adam Kwela  Institute of Mathematics
Faculty of Mathematics
Physics and Informatics
University of Gdańsk
ul. Wita Stwosza 57
80-308 Gdańsk
Poland<EMAIL_ADDRESS>https://mat.ug.edu.pl/ akwela and  Jacek Tryba
Institute of Mathematics
Faculty of Mathematics, Physics and Informatics
University of Gdańsk
ul. Wita Stwosza 57
80-308 Gdańsk
Poland<EMAIL_ADDRESS>
###### Abstract.

We present a unified approach to two classes of Banach spaces defined with the
aid of variations: Waterman spaces and Chanturia classes. Our method is based
on some ideas coming from the theory of ideals on the set of natural numbers.

##  1\. Introduction

The concept of the variation of the function was introduced in 1881 by Camille
Jordan and found plenty of applications and generalizations since that time.
When we look closely at the definition of the variation of the function we can
see its tight relation to the question of (un)boundedness of the series. This
becomes even more evident, when we look at some of the generalizations of
Jordan’s definition, namely the $\Lambda$-variation introduced by Waterman in
1972 (see [23]) and Chanturia classes introduced in 1974 (see [6]).

The question of (un)boundedness of the series of real numbers naturally
appears also in the studies of the ideals on the set of natural numbers, with
the paper [10] as the very recent example of this perspective and the concept
of the summable ideal (introduced below), which is a very basic notion in the
theory of ideals on the set of natural numbers.

Looking at these two separate threads in the realm of mathematics it appeared
to be very appealing to us to join them: to look at different spaces of
functions of bounded variation from the perspective of the theory of ideals
defined on the set of natural numbers. The additional inspiration arrived from
the recent paper [4] by Borodulin-Nadzieja and Farkas, who showed that the
concept of ideals introduced by l.s.c. submeasure on $\mathbb{N}$ (see
definitions below) naturally defines certain Banach sequence spaces. On the
other hand, these sequence spaces may be used as a natural foundation for the
definition of spaces of functions of some type of bounded variation, with a
concept of variation generalizing many different attitudes (especially
Waterman variation and Chanturia classes).

In this paper we define the concept of the variation of the function defined
on a compact real interval originating from the l.s.c submeasure $\phi$
defined on $\mathbb{N}$ (this is laid out in Section 3). In the Section 4 we
study the inclusions between different spaces and corresponding relations
between ideals generated by the submeasure, in a general setting. Then, in
Section 5 we show that for simple density ideals we recreate the Chanturia
classes; while in Section 6 we show that summable ideals correspond directly
to the concept of Waterman $\Lambda$-variation. It appears that in these two
special cases the inclusions between the spaces of functions of bounded
variation may be nicely described by the relation between corresponding ideals
(in terms of inclusion or Katětov order). One of these results also leads us
to a new characterization of Katětov order between summable ideals.

##  2\. Preliminaries

###  2.1. Basics about sequence spaces

By $\mathbb{R}^{\mathbb{N}}$ we will denote the family of all real-valued
sequences.

We will refer to several standard Banach sequence spaces. Here we will present
a notation, and basic properties, which will be used in the sequel. First of
all, in the examples listed below we set a sequence
$x=(x_{n})_{n\in\mathbb{N}}\in\mathbb{R}^{\mathbb{N}}$ of real numbers.

  * •

$\ell_{\infty}$ denotes the space of all bounded sequences equipped with the
supremum norm $\left\|x\right\|_{\infty}=\sup_{n\in\mathbb{N}}|x_{n}|$;

  * •

$c_{0}$ denotes the subspace of $l_{\infty}$ consisting of all sequences such
that $\lim_{n\to+\infty}x_{n}=0$.

  * •

$\ell_{1}$ denotes the space of such sequences that
$\left\|x\right\|_{\ell_{1}}=\sum_{n\in\mathbb{N}}|x_{n}|<+\infty$.

###  2.2. Basics about $\Lambda BV$ spaces

Let us assume that $A=(a_{n})_{n\in\mathbb{N}}$ is such nonincreasing sequence
of positive real numbers that $\sum_{n=1}^{\infty}a_{n}=+\infty$. We call such
sequence _a Waterman sequence_. If additionally $\lim_{n\to+\infty}a_{n}=0$,
we say that the sequence $A$ is a _proper Waterman sequence_.

######  Remark 2.1.

In many sources the Waterman sequence is defined in a form
$(\frac{1}{\lambda_{n}})_{n\in\mathbb{N}}$ where
$(\lambda_{n})_{n\in\mathbb{N}}$ is nondecreasing and such that
$\sum_{n=1}^{\infty}\frac{1}{\lambda_{n}}=+\infty$. Then if
$\lim_{n\to+\infty}\lambda_{n}=+\infty$ we have a proper Waterman sequence. Of
course by putting $a_{n}=\frac{1}{\lambda_{n}}$ we can see that the two
definitions are essentially identical.

Let us denote the unit interval by $I=[0,1]$. Moreover, by ${\mathcal{P}}_{I}$
we denote the set of all sequences of nonoverlapping, closed subintervals
$\\{I_{1},I_{2},\ldots,I_{N},\ldots\\}$ of $I$. The intervals may be
degenerate, i.e. it may happen that $I_{n}$ consists of only one point.

######  Definition 2.2.

Let $A=(a_{n})_{n\in\mathbb{N}}$ be a Waterman sequence and let $x\colon
I\to\mathbb{R}$. We say that $x$ is of bounded $A$-variation if there exists a
positive constant $M$ such that for any sequence of nonoverlapping
subintervals $\\{I_{1},I_{2},I_{3},\ldots,\\}\in{\mathcal{P}}_{I}$, the
following inequality holds

$\sum_{n=1}^{+\infty}a_{n}\lvert x(I_{n})\rvert\leq M,$

where $I_{n}=[s_{n},t_{n}]$ and $|x(I_{n})|=|x(t_{n})-x(s_{n})|$. The supremum
of the above sums, taken over the family ${\mathcal{P}}_{I}$ of all sequences
of nonoverlapping subintervals of $I$, is called the $A$-variation of $x$ and
it is denoted by $\operatorname{var}_{A}(x)$.

######  Remark 2.3.

The special case of a sequence constantly equal to $1$ corresponds to the
classical Jordan variation of a function $x$, which will later be denoted by
$\operatorname{var}(x)$.

This concept was introduced by Waterman in [23]. Since then the functions of
bounded $A$-variation were intensively studied by many authors – for an
overview we refer to [1].

It is worth to mention that there are many equivalent ways to express that the
function $x\colon I\to\mathbb{R}$ is of bounded $A$-variation (cf. [24,
Theorem 1, p. 34] and [5, Proposition 1]), but we will not go into the details
here.

The space of all functions defined on the interval $I$ and of bounded
$A$-variation, endowed with the norm $\left\|x\right\|_{\Lambda
BV}{\mathrel{\mathop{:}}=}\lvert x(0)\rvert+\operatorname{var}_{A}(x)$ forms a
Banach space ${ABV}(I)$ (see [24, Section 3]).

The spaces ${ABV}(I)$ are proper subspaces of the space $B(I)$ of all bounded
functions $x\colon I\to\mathbb{R}$. The space $B(I)$ is equipped with the
standard supremum norm

$\left\|x\right\|_{\infty}=\sup_{t\in I}|x(t)|.$

###  2.3. Basics about ideals

######  Definition 2.4.

A family $\mathcal{I}\subseteq\mathcal{P}(\mathbb{N})$ is called an _ideal_ if

  * •

$\mathbb{N}\notin\mathcal{I}$,

  * •

if $F\subseteq\mathbb{N}$ is finite, then $F\in\mathcal{I}$,

  * •

if $C\in\mathcal{I}$ and $D\subseteq C$, then $D\in\mathcal{I}$,

  * •

if $C,D\in\mathcal{I}$, then $C\cup D\in\mathcal{I}$.

######  Definition 2.5.

An ideal is called a _summable ideal_ if it is of the form

$\mathcal{I}_{A}=\left\\{C\subseteq\mathbb{N}:\sum_{n\in
C}a_{n}<\infty\right\\},$

for some sequence of positive real numbers $A=(a_{n})_{n\in\mathbb{N}}$ such
that $\sum_{n=1}^{\infty}a_{n}=\infty$.

######  Remark 2.6.

Note that in the definition of summable ideals we do not require that $A$ is
nonincreasing. However, in our paper we will only consider summable ideals
given by nonincreasing sequences.

By $\mathrm{Fin}$ we denote the smallest ideal, i.e., the one consisting only
of all finite subsets of $\mathbb{N}$. Note that $\mathrm{Fin}$ is a summable
ideal (given by the sequence $(a_{n})$ constantly equal to $1$).

######  Definition 2.7.

If $\mathcal{I}$ and $\mathcal{J}$ are ideals then we say that _$\mathcal{I}$
is below $\mathcal{J}$ in the Katětov order_ and write
$\mathcal{I}\leq_{K}\mathcal{J}$ whenever there is a function
$f:\mathbb{N}\to\mathbb{N}$ such that $f^{-1}[C]\in\mathcal{J}$ for every
$C\in\mathcal{I}$.

Note that actually, despite its name, Katětov order is only a pre-order, not a
partial order (it is not antisymetric). Katětov order was introduced in the
1970s in papers [12] and [13] by M. Katětov.

We say that an ideal $\mathcal{I}$ is _tall_ if for every infinite
$C\subseteq\mathbb{N}$ there is an infinite $D\subseteq C$ such that
$D\in\mathcal{I}$. It is easy to see that $\mathcal{I}$ is not tall if and
only if $\mathcal{I}\leq_{K}\mathrm{Fin}$. Consequently, all non-tall ideals
are $\leq_{K}$-equivalent (i.e., $\mathcal{I}\leq_{K}\mathcal{J}$ and
$\mathcal{J}\leq_{K}\mathcal{I}$ for any two non-tall ideals $\mathcal{I}$ and
$\mathcal{J}$). If $A=(a_{n})_{n\in\mathbb{N}}$ is a sequence of positive real
numbers such that $\sum_{n=1}^{\infty}a_{n}=\infty$, then $\mathcal{I}_{A}$ is
tall if and only if $\lim_{n\to\infty}a_{n}=0$.

An ideal $\mathcal{I}$ is a _P-ideal_ if for every sequence
$(A_{n})_{n\in\mathbb{N}}$ of elements of $\mathcal{I}$ there is
$A\in\mathcal{I}$ such that $A_{n}\setminus A$ is finite for all
$n\in\mathbb{N}$. It is easy to verify that all summable ideals are P-ideals.

##  3\. Submeasures and objects induced by them

###  3.1. Submeasures

A function $\phi:\mathcal{P}(\mathbb{N})\to[0,\infty]$ is called a
_submeasure_ if $\phi(\emptyset)=0$, $\phi(\\{n\\})<\infty$ for every
$n\in\mathbb{N}$, and

$\phi(C)\leq\phi(C\cup D)\leq\phi(C)+\phi(D)$

for all $C,D\subseteq\mathbb{N}$. A submeasure $\phi$ is _lower
semicontinuous_ (lsc, in short) if
$\phi(C)=\lim_{n\to\infty}\phi(C\cap\\{1,2,\ldots,n\\})$ for each
$C\subseteq\mathbb{N}$.

An lsc submeasure $\phi$ is _non-pathological_ , if

$\phi(C)=\sup\\{\mu(C):\ \mu\text{ is a measure such that }\mu\leq\phi\\},$

for all $C\subseteq\mathbb{N}$. Not every lsc submeasure is non-pathological –
see [7, Section 1.9], [9, Theorem 4.12], [16, Section 6.2], [17] or [22,
Theorem 4.7] for such examples.

Given an lsc submeasure $\phi$, let $\mathcal{M}_{\phi}$ be the family of all
measures $\mu$ on $\mathbb{N}$ such that $\mu\leq\phi$. Then, by definition,
$\phi$ is non-pathological if and only if
$\phi(C)=\sup\\{\mu(C):\mu\in\mathcal{M}_{\phi}\\}$ for all
$C\subseteq\mathbb{N}$.

###  3.2. Ideals induced by submeasures

By identifying subsets of $\mathbb{N}$ with their characteristic functions, we
can treat ideals as subsets of the Cantor space $\\{0,1\\}^{\mathbb{N}}$.
Mazur in [18, Lemma 1.2] proved that an ideal is $\bf{F_{\sigma}}$ if and only
if it is of the form:

$\mathrm{Fin}(\phi)=\left\\{C\subseteq\mathbb{N}:\ \phi(C)<\infty\right\\}$

for some lower semicontinuous submeasure $\phi$ such that
$\mathbb{N}\notin\mathrm{Fin}(\phi)$ (see also [8, Theorem 1.2.5]).

Solecki in [21, Theorem 3.1] showed that an ideal is an analytic P-ideal if
and only if it is of the form:

$\mathrm{Exh}(\phi)=\left\\{C\subseteq\mathbb{N}:\
\lim_{n\to\infty}\phi(C\setminus\\{1,2,\ldots,n\\})=0\right\\}$

for some lower semicontinuous submeasure $\phi$ such that
$\mathbb{N}\notin\mathrm{Exh}(\phi)$ (see also [8, Theorem 1.2.5]).

It is easy to see that $\mathrm{Exh}(\phi)\subseteq\mathrm{Fin}(\phi)$ for
every lsc submeasure $\phi$. Moreover, for every lsc submeasure $\phi$ we can
find an lsc submeasure $\phi^{\prime}$ such that
$\mathrm{Fin}(\phi)=\mathrm{Fin}(\phi^{\prime})$,
$\mathrm{Exh}(\phi)=\mathrm{Exh}(\phi^{\prime})$ and additionally
$\phi^{\prime}(\\{k\\})>0$ for all $k\in\mathbb{N}$ (it suffices, for
instance, to put $\phi^{\prime}(C)=\phi(C)+\sum_{n\in C}\frac{1}{2^{n}}$ for
all $C\subseteq\mathbb{N}$).

Note that every summable ideal is of the form $\mathrm{Exh}(\phi_{A})$ as well
as of the form $\mathrm{Fin}(\phi_{A})$, where $\phi_{A}(C)=\sum_{n\in
C}a_{n}$. For more examples of ideals induced by submeasures see [8, Example
1.2.3].

###  3.3. Banach spaces of real sequences

######  Definition 3.1.

Let $\phi$ be a non-pathological lsc submeasure. Define a function
$\hat{\phi}:\mathbb{R}^{\mathbb{N}}\to[0,\infty]$ by:

$\hat{\phi}(x)=\sup\left\\{\sum_{n\in\mathbb{N}}\mu(\\{n\\})|x_{n}|:\mu\in\mathcal{M}_{\phi}\right\\}$

for all $x=(x_{n})_{n\in\mathbb{N}}\in\mathbb{R}^{\mathbb{N}}$. Define also:

$\mathrm{FIN}(\phi)=\left\\{x\in\mathbb{R}^{\mathbb{N}}:\hat{\phi}(x)<\infty\right\\};$
$\mathrm{EXH}(\phi)=\left\\{x\in\mathbb{R}^{\mathbb{N}}:\lim_{n\to\infty}\hat{\phi}(x\cdot\chi_{\\{n,n+1,\ldots\\}})=0\right\\}.$

Moreover, let $\mathrm{mFIN}(\phi)=\mathrm{FIN}(\phi)\cap\mathrm{Mon}$ and
$\mathrm{mEXH}(\phi)=\mathrm{EXH}(\phi)\cap\mathrm{Mon}$, where

$\mathrm{Mon}=\\{(x_{n})_{n\in\mathbb{N}}\in\mathbb{R}^{\mathbb{N}}:|x_{n+1}|\leq|x_{n}|\text{
for all }n\in\mathbb{N}\\}.$

Spaces $\mathrm{FIN}(\phi)$ and $\mathrm{EXH}(\phi)$ were introduced in [4,
Section 5]. Note that
$\hat{\phi}(x)=\lim_{n\to\infty}\hat{\phi}(x\cdot\chi_{\\{1,2,\ldots,n\\}})$
for every $x\in\mathbb{R}^{\mathbb{N}}$ (see [4, Proposition 5.3]).

######  Example 3.2 ([4, Examples 5.6 and 5.7]).

  * •

Consider the submeasure given by:

$\phi(C)=\begin{cases}1,&\text{if }C\neq\emptyset,\\\ 0,&\text{if
}C=\emptyset,\end{cases}$

for every $C\subseteq\mathbb{N}$. Then $\mathrm{FIN}(\phi)=\ell_{\infty}$ and
$\mathrm{EXH}(\phi)=c_{0}$.

  * •

Consider the submeasure given by $\phi(C)=|C|$ for every
$C\subseteq\mathbb{N}$. Then $\mathrm{FIN}(\phi)=\mathrm{EXH}(\phi)=\ell_{1}$.

######  Remark 3.3.

Notice that for every $x=(x_{n})_{n\in\mathbb{N}}\in\mathbb{R}^{\mathbb{N}}$
and $k\in\mathbb{N}$ we have
$\hat{\phi}(x\cdot\chi_{\\{k\\}})=\phi(\\{k\\})\cdot|x_{k}|$. Therefore, for
any $A\subseteq\mathbb{N}$ we have

$\sup_{k\in
A}\left(\phi(\\{k\\})\cdot|x_{k}|\right)\leq\hat{\phi}(x\cdot\chi_{A})\leq\sup_{k\in
A}|x_{k}|\cdot\sum_{k\in A}\phi(\\{k\\})$

######  Remark 3.4.

Observe that $C\in\mathrm{Fin}(\phi)$ if and only if
$\chi_{C}\in\mathrm{FIN}(\phi)$, where $\chi_{C}$ denotes the characteristic
function of $C$. Similar equivalence holds for $\mathrm{Exh}(\phi)$ and
$\mathrm{EXH}(\phi)$.

######  Proposition 3.5.

[4, Propositions 5.1 and 5.3] Suppose that $\phi$ is a non-pathological lsc
submeasure. Then $\mathrm{FIN}(\phi)$ and $\mathrm{EXH}(\phi)$ are Banach
spaces normed by $\hat{\phi}$. Moreover,
$\mathrm{EXH}(\phi)\subseteq\mathrm{FIN}(\phi)$.

######  Proposition 3.6.

Let $\phi$ be a non-pathological lsc submeasure such that $\phi(\\{k\\})>0$
for all $k\in\mathbb{N}$. Then $\mathrm{mFIN}(\phi)$ is a closed subspace of
$\mathrm{FIN}(\phi)$ and $\mathrm{mEXH}(\phi)$ is a closed subspace of
$\mathrm{EXH}(\phi)$. In particular, $\mathrm{mFIN}(\phi)$ and
$\mathrm{mEXH}(\phi)$ are Banach spaces normed by $\hat{\phi}$. Moreover,
$\mathrm{mEXH}(\phi)\subseteq\mathrm{mFIN}(\phi)$.

###### Proof.

The inclusion $\mathrm{mEXH}(\phi)\subseteq\mathrm{mFIN}(\phi)$ follows from
Proposition 3.5.

Actually, it suffices to show that $\mathrm{Mon}$ is closed in
$\mathrm{FIN}(\phi)$ and in $\mathrm{EXH}(\phi)$. Let
$x=(x_{n})\in\mathbb{R}^{\mathbb{N}}\setminus\mathrm{Mon}$. Then there is
$n\in\mathbb{N}$ such that $|x_{n+1}|>|x_{n}|$. Define
$r=\frac{|x_{n+1}|-|x_{n}|}{3}\min\\{\hat{\phi}(e_{n}),\hat{\phi}(e_{n+1})\\}$,
where $e_{n}\in\mathbb{R}^{\mathbb{N}}$ is the sequence given by:

$(e_{n})_{i}=\begin{cases}1,&\text{if }i=n,\\\
0,&\text{otherwise.}\end{cases}$

Note that $r>0$ since $\hat{\phi}(e_{k})=\phi(\\{k\\})>0$ for all
$k\in\mathbb{N}$.

We claim that if $y\in\mathbb{R}^{\mathbb{N}}$ is such that
$\hat{\phi}(x-y)<r$, then $r\notin\mathrm{Mon}$ (which shows that
$\mathrm{Mon}$ is closed in $\mathrm{FIN}(\phi)$ and in $\mathrm{EXH}(\phi)$).
Indeed, if $\hat{\phi}(x-y)<r$, then:

$|x_{n}-y_{n}|\hat{\phi}(e_{n})\leq\hat{\phi}(x-y)<r\leq\frac{|x_{n+1}|-|x_{n}|}{3}\hat{\phi}(e_{n}),$

so $|x_{n}-y_{n}|<\frac{|x_{n+1}|-|x_{n}|}{3}$. Similarly,
$|x_{n+1}-y_{n+1}|<\frac{|x_{n+1}|-|x_{n}|}{3}$. Hence,
$|y_{n+1}|-|y_{n}|>\frac{|x_{n+1}|-|x_{n}|}{3}$ and we get that
$|y_{n+1}|>|y_{n}|$, which shows $y\notin\mathrm{Mon}$. ∎

######  Proposition 3.7.

[4, Theorem 5.4] The following are equivalent for any non-pathological lsc
submeasure $\phi$:

  * •

$\mathrm{Exh}(\phi)=\mathrm{Fin}(\phi)$;

  * •

$\mathrm{EXH}(\phi)=\mathrm{FIN}(\phi)$;

  * •

$\mathrm{FIN}(\phi)$ is separable.

######  Remark 3.8.

Obviously, if $\mathrm{EXH}(\phi)=\mathrm{FIN}(\phi)$, then also
$\mathrm{mEXH}(\phi)=\mathrm{mFIN}(\phi)$.

###  3.4. Variations

######  Definition 3.9.

Let $\phi$ be a non-pathological lsc submeasure. For
$J=(J_{n})\in\mathcal{P}_{I}$ denote by $x(J)$ the sequence $(|x(J_{n})|)$.
Define:

$\mathrm{BV}(\phi)=\left\\{x\in
B(I):\sup_{J\in\mathcal{P}_{I}}\hat{\phi}(x(J))<\infty\right\\}.$

######  Remark 3.10.

The requirement $x\in B(I)$ may be removed. Indeed, assume that there exists
such a sequence $(t_{n})\subseteq I$ that $|x(t_{n})|\to+\infty$. Fix any
$k\in\mathbb{N}$ such that $\phi(\\{k\\})>0$. Then if we take for each
$n\in\mathbb{N}$ any $J^{n}=(J^{n}_{i})\in\mathcal{P}_{I}$ such that
$J^{n}_{k}=[0,t_{n}]$ we have

$\sup_{J\in\mathcal{P}_{I}}\hat{\phi}(x(J))\geq|x(t_{n})-x(0)|\phi(\\{k\\})\to+\infty,$

which means that the condition
$\sup_{J\in\mathcal{P}_{I}}\hat{\phi}(x(J))<\infty$ will not be satisfied
anyway.

######  Proposition 3.11.

Let $\phi$ be a non-pathological lsc submeasure. Then $\mathrm{BV}(\phi)$ is a
Banach space normed by

$\|x\|_{\phi}=|x(0)|+\sup_{J\in\mathcal{P}_{I}}\hat{\phi}(x(J)),$

for all $x\in\mathrm{BV}(\phi)$.

###### Proof.

Let us assume that $x\in B(I)$ is such that $x\neq 0$ and $x(0)=0$. Then,
there exists such $t\in I$ that $x(t)\neq 0$. Let us assume that
$\phi(\\{k\\})>0$ for certain $k\in\mathbb{N}$ and let $J\in\mathcal{P}_{I}$
be such sequence of intervals that $J_{k}=[0,t]$. Then

$\hat{\phi}(|x(J)|)\geq|x(t)-x(0)|\phi(\\{k\\})>0.$

Let $c\in\mathbb{R}$ be any constant. The condition $\hat{\phi}(|c\cdot
x(J)|)=c\hat{\phi}(|x(J)|)$ is obvious for any $J\in\mathcal{P}_{I}$.
Similarly the triangle inequality

$\hat{\phi}(|(x+y)(J)|)\leq\hat{\phi}(|x(J)|)+\hat{\phi}(|y(J)|)$

for any functions $x,y\colon I\to\mathbb{R}$. Passing to the supremum for
$J\in{\mathcal{P}}_{I}$ keeps these conditions.

Now it remains to prove that the space is complete. The proof will be a
standard one. Let us take a Cauchy sequence
$(x_{n})\subseteq\mathrm{BV}(\phi)$. Let us fix $\varepsilon>0$ and take such
$n,m\geq N$ that

$|x_{n}(0)-x_{m}(0)|+\sup_{J\in\mathcal{P}_{I}}\hat{\phi}((x_{n}-x_{m})(J))\leq\varepsilon.$

First of all, let us observe that it means that the sequence $(x_{n}(0))$ is a
real Cauchy sequence so it converges to some real number $x(0)$. We can also
observe that the sequence $x_{n}(t)$ converges to $x(t)$ for any $t\in I$.
Indeed, fix any $t\in I$ and as before, let us assume that $\phi(\\{k\\})>0$
for certain $k\in\mathbb{N}$. Let $J\in\mathcal{P}_{I}$ be such sequence of
intervals that $J_{k}=[0,t]$. Then

$|(x_{n}-x_{m})(J_{k})|\phi(\\{k\\})\leq\varepsilon$

and

$|x_{n}(t)-x_{m}(t)|\leq\frac{1}{\phi(\\{k\\})}\varepsilon+|x_{n}(0)-x_{m}(0)|,$

which eventually proves that $(x_{n}(t))$ is a Cauchy sequence, so it
converges to some $x(t)$.

Now, as we have the pointwise limit $x(t)$, we are going to show that $x\in
BV(\phi)$ and that $\|x_{n}-x\|_{\phi}\to 0$ as $n\to+\infty$. Let us take any
$\mu\in\mathcal{M}_{\phi}$, any natural number $K\in\mathbb{N}$ and any
sequence of intervals $J\in{\mathcal{P}}_{I}$. Then we have

$\sum_{k=1}^{K}|(x_{n}-x_{m})(J_{k})|\mu(\\{k\\})\leq\varepsilon$

and we may pass to the limit with $m\to+\infty$ (as we know the sequence
$x_{m}(t))$ converges pointwise to $x(t)$). This gives us

$\sum_{k=1}^{K}|(x_{n}-x)(J_{k})|\mu(\\{k\\})\leq\varepsilon.$

Since the inequality holds for all $K$, $\mu$ and $J$ we can see that

$\sup_{J\in\mathcal{P}_{I}}\hat{\phi}((x_{n}-x)(J))\leq\varepsilon.$

To show that $x\in\mathrm{BV}(\phi)$ it is enough to take any fixed
$x_{n}\in\mathrm{BV}(\phi)$ such that

$\sup_{J\in\mathcal{P}_{I}}\hat{\phi}((x_{n}-x)(J))\leq 1$

and refer to seminorm properties to see that

$\sup_{J\in\mathcal{P}_{I}}\hat{\phi}(x(J))-\sup_{J\in\mathcal{P}_{I}}\hat{\phi}(x_{n}(J))\leq\sup_{J\in\mathcal{P}_{I}}\hat{\phi}((x_{n}-x)(J))\leq
1$

∎

######  Proposition 3.12.

Let $\phi$ be a non-pathological lsc submeasure.

  * (a)

If the sequence $\phi(\\{k\\})$ is unbounded, then $BV(\phi)$ reduces to the
space of constant functions.

  * (b)

If the sequence $\phi(\\{k\\})$ is bounded, then the space $\mathrm{BV}(\phi)$
contains the space of functions of bounded classical Jordan variation as a
subset.

###### Proof.

(a): If $x\in B(I)$ is such that $|x(s)-x(t)|=a>0$ for some $[s,t]\in(0,1)$,
then taking for each $k\in\mathbb{N}$ such
$J^{k}=(J^{k}_{n})_{n\in\mathbb{N}}\in{\mathcal{P}}_{I}$ that
$J^{k}_{k}=[s,t]$ we get

$\hat{\phi}(x(J))\geq a\phi(\\{k\\}).$

Thus the variation $\sup_{J\in\mathcal{P}_{I}}\hat{\phi}(|x(J)|)$ is
unbounded.

(b): Let us take any $x\in\mathrm{BV}(I)$ and any $J\in{\mathcal{P}}_{I}$.
Then

$\hat{\phi}(|x(J)|)\leq\sum_{k\in\mathbb{N}}\phi(\\{k\\})|x(J_{k})|)\leq
M\operatorname{var}(x).$

The last inequality actually shows that the space $\mathrm{BV}(I)$ is
continuously embedded in $\mathrm{BV}(\phi)$. ∎

######  Remark 3.13.

Let us observe that if $\phi(\mathbb{N})<+\infty$, then $BV(\phi)=B(I)$. To
see this let us take any bounded function $x\in B(I)$ such that $|x(t)|\leq M$
for all $t\in I$. Then for any interval $J\subset I$ we get $|x(J)|\leq 2M$
and for any $\mu\in{\mathcal{M}}_{\phi}$ and any $(J_{k})\in{\mathcal{P}}_{I}$
we have

$\sum_{k\in N}|x(J_{k})|\mu\\{k\\}\leq 2M\mu(\mathbb{N})$

leading to $\hat{\phi}(x)\leq 2M\phi(\mathbb{N})<+\infty.$

######  Proposition 3.14.

Let $\phi$ be a non-pathological lsc submeasure such that
$\phi(\mathbb{N})=+\infty$. Then the space $BV(\phi)$ is a subset of the space
of all bounded regulated functions (i.e. bounded functions having finite left
and right limit in every point of their domain).

###### Proof.

Assume, contrary to our claim, that there exist monotone sequences
$(t_{n})\subseteq I$ and $(s_{n})\subseteq I$ converging to $a\in I$ from the
same side and such that $x(s_{n})-x(t_{m})\geq\delta>0$ for all
$n,m\in\mathbb{N}$. Then, taking subsequences if necessary, we have the
sequence of nonoverlapping intervals $I_{n}=[s_{n},t_{n}]\subset I$ such that
$|x(I_{n})|\geq\delta$. Then for any $\mu\in{\mathcal{M}}_{\phi}$ we have

$\sum_{n\in\mathbb{N}}|x(I_{n})|\mu(\\{n\\})\geq\delta\mu(\mathbb{N})$

so

$\hat{\phi}(|x(I_{n})|)\geq\delta\phi(\mathbb{N})=+\infty.$

∎

######  Remark 3.15.

For any non-pathological lsc submeasure $\phi$ and any permutation
$\pi\colon\mathbb{N}\to\mathbb{N}$, the function
$\psi:\mathcal{P}(\mathbb{N})\to[0,\infty]$ given by

$\psi(C)=\phi(\pi[C]),$

for all $C\subseteq\mathbb{N}$, also is a non-pathological lsc submeasure.
What is more, we can easily see that $\mathrm{BV}(\phi)=\mathrm{BV}(\psi)$.

######  Proposition 3.16.

Let $\phi$ and $\psi$ be two non-pathological submeasures. If there exists
$M>0$ such that for every $A\subseteq\mathbb{N}$ we have
$|\phi(A)-\psi(A)|\leq M$ then $\mathrm{BV}(\phi)=\mathrm{BV}(\psi)$.

###### Proof.

Suppose to the contrary that there exists
$x\in\mathrm{BV}(\psi)\setminus\mathrm{BV}(\phi)$. Then $x$ is bounded by some
$N>0$ and there exists $M_{1}>0$ such that for every $J\in\mathcal{P}_{I}$ we
have $\hat{\psi}(x(J))\leq M_{1}$.

On the other hand, there exists a measure $\mu\leq\phi$, $J\in\mathcal{P}_{I}$
and $k\in\mathbb{N}$ such that $\sum_{n\leq
k}\mu(\\{n\\})|x(J_{n})|>2NM+2N+M_{1}$. Meanwhile, there exists a measure
$\nu\leq\psi$ such that $\sum_{n\leq k}\nu(\\{n\\})=\psi(\\{1,\ldots,k\\})$.
Clearly, we have $\sum_{n\leq k}\nu(\\{n\\})|x(J_{n})|\leq M_{1}$. Therefore,
we obtain

$\sum_{n\leq
k}(\mu(\\{n\\})-\nu(\\{n\\}))|x(J_{n})|>2NM+2N+M_{1}-M_{1}=2N(M+1).$

Since $|x(J_{n})|\leq 2N$ for every $n$, it follows that

$\sum_{n\leq k}(\mu(\\{n\\})-\nu(\\{n\\}))\geq\frac{\sum_{n\leq
k}(\mu(\\{n\\})-\nu(\\{n\\}))|x(J_{n})|}{2N}\geq M+1.$

Therefore,

$\phi(\\{1,\ldots,k\\})-\psi(\\{1,\ldots,k\\})\geq\sum_{n\leq
k}(\mu(\\{n\\})-\nu(\\{n\\}))\geq M+1>M,$

a contradiction. ∎

######  Corollary 3.17.

For every non-pathological submeasure $\phi$ there exists a non-pathological
submeasure $\psi$ such that $\mathrm{BV}(\phi)=\mathrm{BV}(\psi)$ and the
sequence $(\psi(\\{n\\}))$ does not tend to zero.

###### Proof.

Define the function $\delta:\mathcal{P}(\mathbb{N})\to[0,\infty]$ by

$\delta(A)=\begin{cases}0,&\text{if }A=\emptyset,\\\ 1,&\text{if
}A\not=\emptyset.\end{cases}$

Clearly, $\delta$ is a non-pathological submeasure. Define the function
$\psi:\mathcal{P}(\mathbb{N})\to[0,\infty]$ by
$\psi(A)=\max\\{\phi(A),\delta(A)\\}$ for every $A\subseteq\mathbb{N}$. Then
$\psi$ is non-pathological submeasure as a maximum of two non-pathological
submeasures. It is also clear that $(\psi(\\{n\\}))$ does not tend to zero
because $\psi(\\{n\\})\geq\delta(\\{n\\})=1$ for every $n\in\mathbb{N}$.
Moreover, since $|\phi(A)-\psi(A)|\leq 1$ for every $A\subseteq\mathbb{N}$, by
Proposition 3.16 we obtain $\mathrm{BV}(\phi)=\mathrm{BV}(\psi)$. ∎

###  3.5. Basic results

######  Proposition 3.18.

For any non-pathological lsc submeasure $\phi$ the following are equivalent :

  * (a)

$\phi(\mathbb{N})<\infty$;

  * (b)

$\mathrm{Fin}(\phi)=\mathcal{P}(\mathbb{N})$;

  * (c)

$\mathrm{FIN}(\phi)\supseteq\ell_{\infty}$;

  * (d)

$\mathrm{mFIN}(\phi)=\mathrm{Mon}$;

  * (e)

$\mathrm{mFIN}(\phi)\not\subseteq c_{0}$;

  * (f)

every bounded function $x:[0,1]\to\mathbb{R}$ belongs to $\mathrm{BV}(\phi)$.

###### Proof.

(a)$\implies$(c): If $x\in\ell_{\infty}$, then there is $M>0$ such that
$|x_{n}|\leq M$ for all $n\in\mathbb{N}$. Hence, if
$\hat{1}\in\mathbb{R}^{\mathbb{N}}$ is the infinite sequence constantly equal
to $1$, then

$\hat{\phi}(x)\leq\hat{\phi}(M\cdot\hat{1})=M\hat{\phi}(\hat{1})=M\phi(\mathbb{N})<\infty.$

Thus $x\in\mathrm{FIN}(\phi)$.

(c)$\implies$(d): If $\mathrm{FIN}(\phi)\supseteq\ell_{\infty}$, then

$\mathrm{Mon}\supseteq\mathrm{mFIN}(\phi)=\mathrm{FIN}(\phi)\cap\mathrm{Mon}\supseteq\ell_{\infty}\cap\mathrm{Mon}=\mathrm{Mon}.$

(d)$\implies$(e): The sequence constantly equal to one belongs to
$\mathrm{Mon}=\mathrm{mFIN}(\phi)$, but not to $c_{0}$.

(e)$\implies$(b): Since $\mathrm{mFIN}(\phi)\not\subseteq c_{0}$, there is
some $x\in\mathrm{mFIN}(\phi)\setminus c_{0}$. Since
$\mathrm{mFIN}(\phi)\subseteq\mathrm{Mon}$, $|x|$ converges to some $l>0$ and
$|x_{n}|\geq l$ for all $n\in\mathbb{N}$. Thus, if $\hat{1}$ denotes the
infinite sequence constantly equal to $1$, then
$\phi(\mathbb{N})=\hat{\phi}(\hat{1})=\frac{1}{l}\hat{\phi}(l\cdot\hat{1})\leq\frac{1}{l}\hat{\phi}(x)<\infty$,
so $\mathbb{N}\in\mathrm{Fin}(\phi)$ and consequently
$\mathrm{Fin}(\phi)=\mathcal{P}(\mathbb{N})$.

(b)$\implies$(f): If $x:[0,1]\to\mathbb{R}$ is bounded by some $M>0$, then
$|x(J)|$ is bounded by $2M$ for every interval $J\subseteq I$. Hence, since
$\mathbb{N}\in\mathrm{Fin}(\phi)$, we have
$\sup_{J\in\mathcal{P}_{I}}\hat{\phi}(|x(J)|)\leq 2M\phi(\mathbb{N})<\infty$.

(f)$\implies$(a): Consider the Dirichlet function $x_{D}:[0,1]\to\mathbb{R}$
given by:

$x_{D}(t)=\begin{cases}1,&\text{if }t\in\mathbb{Q}\cap[0,1];\\\
0,&\text{otherwise.}\end{cases}$

Then $x_{D}$ is bounded, so it belongs to $\mathrm{BV}(\phi)$. Note also that
there is $J\in\mathcal{P}_{I}$ such that $x_{D}(J)$ is constantly equal to
$1$. Thus, $\phi(\mathbb{N})=\hat{\phi}(x_{D}(J))<\infty$. ∎

######  Proposition 3.19.

The following are equivalent for any non-pathological lsc submeasure $\phi$:

  * (a)

$\lim_{n}\phi(\\{n,n+1,\ldots\\})=0$;

  * (b)

$\mathrm{Exh}(\phi)=\mathcal{P}(\mathbb{N})$;

  * (c)

$\mathrm{EXH}(\phi)\supseteq\ell_{\infty}$;

  * (d)

$\mathrm{mEXH}(\phi)=\mathrm{Mon}$;

  * (e)

$\mathrm{mEXH}(\phi)\not\subseteq c_{0}$.

###### Proof.

(a)$\implies$(c): If $x\in\ell_{\infty}$, then there is $M>0$ such that
$|x_{n}|\leq M$ for all $n\in\mathbb{N}$. Hence, if
$\hat{1}\in\mathbb{R}^{\mathbb{N}}$ is the infinite sequence constantly equal
to $1$, then

$\hat{\phi}(x\cdot\chi_{\\{n,n+1,\ldots\\}})\leq\hat{\phi}(M\cdot\hat{1}\cdot\chi_{\\{n,n+1,\ldots\\}})=M\phi(\\{n,n+1,\ldots\\})\to
0.$

Thus $x\in\mathrm{EXH}(\phi)$.

(c)$\implies$(d): If $\mathrm{EXH}(\phi)\supseteq\ell_{\infty}$, then

$\mathrm{Mon}\supseteq\mathrm{mEXH}(\phi)=\mathrm{EXH}(\phi)\cap\mathrm{Mon}\supseteq\ell_{\infty}\cap\mathrm{Mon}=\mathrm{Mon}.$

(d)$\implies$(e): The sequence constantly equal to one is in
$\mathrm{Mon}=\mathrm{mEXH}(\phi)$, but not in $c_{0}$.

(e)$\implies$(b): Since $\mathrm{mEXH}(\phi)\not\subseteq c_{0}$, there is
some $x\in\mathrm{mEXH}(\phi)\setminus c_{0}$. Since
$\mathrm{mEXH}(\phi)\subseteq\mathrm{Mon}$, $x$ converges to some $l>0$ and
$x_{n}\geq l$ for all $n\in\mathbb{N}$. Thus, if $\hat{1}$ denotes the
infinite sequence constantly equal to $1$, then
$\phi(\mathbb{N}\setminus\\{1,2,\ldots,n-1\\})=\phi(\\{n,n+1,\ldots\\})=\hat{\phi}(\hat{1}\cdot\chi_{\\{n,n+1,\ldots\\}})=\frac{1}{l}\hat{\phi}(l\cdot\hat{1}\cdot\chi_{\\{n,n+1,\ldots\\}})\leq\frac{1}{l}\hat{\phi}(x\cdot\chi_{\\{n,n+1,\ldots\\}})\to
0$, so $\mathbb{N}\in\mathrm{Exh}(\phi)$ and consequently
$\mathrm{Exh}(\phi)=\mathcal{P}(\mathbb{N})$.

(b)$\implies$(a): Since $\mathrm{Exh}(\phi)=\mathcal{P}(\mathbb{N})$,
$\mathbb{N}\in\mathrm{Exh}(\phi)$, which means that
$\lim_{n}\phi(\\{n,n+1,\ldots\\})=0$. ∎

##  4\. Inclusions in the general case

###  4.1. Two orders

Actually, $\hat{\phi}$ is a function defined on infinite sequences of reals.
However, for simplicity, we will sometimes write $\hat{\phi}(x)$ even for
finite sequences $x$ – in such cases we mean $\hat{\phi}(x^{\frown}0)$, where
$x^{\frown}0$ is the infnite sequence starting with $x$ and followed by zeros.

######  Definition 4.1.

Let $\phi_{1}$ and $\phi_{2}$ be two non-pathological lsc submeasures. We
write:

  * •

$\phi_{2}\preceq\phi_{1}$ if there is $M>0$ such that $\hat{\phi}_{2}(x)\leq
M\hat{\phi}_{1}(x)$ for every finite sequence
$x\in\bigcup_{n\in\mathbb{N}}\mathbb{R}^{n}$;

  * •

$\phi_{2}\preceq_{m}\phi_{1}$ if there is $M>0$ such that
$\hat{\phi}_{2}(x)\leq M\hat{\phi}_{1}(x)$ for every non-increasing finite
sequence $x\in\bigcup_{n\in\mathbb{N}}\mathbb{R}^{n}$.

###  4.2. Ideals

######  Proposition 4.2.

Let $\phi_{1}$ and $\phi_{2}$ be two non-pathological lsc submeasures.

  * (a)

If $\phi_{2}\preceq\phi_{1}$ then
$\mathrm{Exh}(\phi_{1})\subseteq\mathrm{Exh}(\phi_{2})$ and
$\mathrm{Fin}(\phi_{1})\subseteq\mathrm{Fin}(\phi_{2})$

  * (b)

If either $\mathrm{Exh}(\phi_{1})\subseteq\mathrm{Exh}(\phi_{2})$ or
$\mathrm{Fin}(\phi_{1})\subseteq\mathrm{Fin}(\phi_{2})$, then
$\mathrm{Exh}(\phi_{1})\subseteq\mathrm{Fin}(\phi_{2})$.

  * (c)

$\mathrm{Exh}(\phi_{1})\subseteq\mathrm{Fin}(\phi_{2})$ if and only if

$\exists_{M>0}\ \forall_{F\in\mathrm{Fin}}\ \phi_{1}(F)>\frac{1}{M}\text{ or
}\phi_{2}(F)<M.$

###### Proof.

(a): This is clear with the use of Remark 3.4.

(b): This follows from the inclusions
$\mathrm{Exh}(\phi_{1})\subseteq\mathrm{Fin}(\phi_{1})$ and
$\mathrm{Exh}(\phi_{2})\subseteq\mathrm{Fin}(\phi_{2})$ (see Proposition 3.5).

(c): Suppose first that there is $M>0$ such that for all $F\in\mathrm{Fin}$
either $\phi_{1}(F)>\frac{1}{M}$ or $\phi_{2}(F)<M$. Let
$C\in\mathrm{Exh}(\phi_{1})$. Then there is $k\in\mathbb{N}$ such that
$\phi_{1}(C\setminus\\{1,2,\ldots,k\\})<\frac{1}{M}$. Hence, if $F\subseteq
C\setminus\\{1,2,\ldots,k\\}$ is finite, then $\phi_{1}(F)<\frac{1}{M}$ and
$\phi_{2}(F)<M$ (by our assumption). Thus, since $\phi_{2}$ is lsc, we have:

$\displaystyle\phi_{2}(C)$
$\displaystyle\leq\phi_{2}(C\cap\\{1,2,\ldots,k\\})+\phi_{2}(C\setminus\\{1,2,\ldots,k\\})$
$\displaystyle\leq\left(\sum_{i\leq k}\phi_{2}(\\{i\\})\right)+M<\infty.$

Suppose now that for each $n\in\mathbb{N}$ there is $F_{n}\in\mathrm{Fin}$
such that $\phi_{1}(F_{n})\leq\frac{1}{2^{n}}$ and $\phi_{2}(F_{n})\geq
2^{n}$. Then clearly
$C=\bigcup_{n\in\mathbb{N}}F_{n}\notin\mathrm{Fin}(\phi_{2})$. On the other
hand, we will show that $C\in\mathrm{Exh}(\phi_{1})$. Given any
$\varepsilon>0$ there is $n_{0}\in\mathbb{N}$ such that
$\frac{1}{2^{n_{0}}}<\varepsilon$. Find $k_{0}\in\mathbb{N}$ such that
$F_{1}\cup\ldots\cup F_{n_{0}}\subseteq[1,k_{0}]$ and observe that for each
$k>k_{0}$ we have:

$\displaystyle\phi_{1}(C\setminus\\{1,2,\ldots,k\\})$
$\displaystyle\leq\phi_{1}(C\setminus\\{1,2,\ldots,k_{0}\\})$
$\displaystyle\leq\sum_{i=n_{0}+1}^{\infty}\phi_{1}(F_{i})\leq\sum_{i=n_{0}+1}^{\infty}\frac{1}{2^{i}}=\frac{1}{2^{n_{0}}}<\varepsilon.$

∎

######  Remark 4.3.

Observe that there are non-pathological lsc submeasures $\phi_{1}$ and
$\phi_{2}$ such that $\mathrm{Exh}(\phi_{1})\subseteq\mathrm{Exh}(\phi_{2})$,
but $\mathrm{Fin}(\phi_{1})\not\subseteq\mathrm{Fin}(\phi_{2})$. Indeed, this
is true for

$\phi_{1}(C)=\begin{cases}1,&\text{if }C\neq\emptyset,\\\ 0,&\text{if
}C=\emptyset,\end{cases}$

and $\phi_{2}(C)=\sum_{i\in C}\frac{1}{i}$, since in this case
$\mathrm{Exh}(\phi_{1})=\mathrm{Fin}\subseteq\mathcal{I}_{1/n}=\mathrm{Exh}(\phi_{2})$
and
$\mathrm{Fin}(\phi_{1})=\mathcal{P}(\mathbb{N})\not\subseteq\mathcal{I}_{1/n}=\mathrm{Fin}(\phi_{2})$.

On the other hand, there are also non-pathological lsc submeasures $\psi_{1}$
and $\psi_{2}$ such that
$\mathrm{Exh}(\psi_{1})\not\subseteq\mathrm{Exh}(\psi_{2})$, but
$\mathrm{Fin}(\psi_{1})\subseteq\mathrm{Fin}(\psi_{2})$. This is the case for
$\psi_{1}=\phi_{2}$ and $\psi_{2}=\phi_{1}$ (where $\phi_{1}$ and $\phi_{2}$
are as in the previous paragraph).

###  4.3. Spaces of real sequences

######  Lemma 4.4.

Let $\phi_{1}$ and $\phi_{2}$ be two non-pathological lsc submeasures and
assume that $\phi_{1}(\\{i\\})>0$ and $\phi_{2}(\\{i\\})>0$ for every
$i\in\mathbb{N}$. Then for every $m\in\mathbb{N}$ there is $L>0$ such that
$\hat{\phi}_{2}(y)\leq L\hat{\phi}_{1}(y)$ for every $y\in\mathbb{R}^{m}$.

###### Proof.

Define:

$L=\frac{\sum_{i\leq m}\phi_{2}(\\{i\\})}{\min_{i\leq m}\phi_{1}(\\{i\\})}.$

Then $L>0$. Fix any $y\in\mathbb{R}^{m}$. If $y$ is constantly equal to zero,
then $\hat{\phi}_{2}(y)=0\leq L\hat{\phi}_{1}(y)$ and we are done. Otherwise,
find $r=\max_{i\leq m}|y_{i}|$ and let $j\leq m$ be such that $r=|y_{j}|$.
Note that:

$\displaystyle\hat{\phi}_{2}\left(\frac{y}{r}\right)$
$\displaystyle\leq\sum_{i\leq
m}\phi_{2}(\\{i\\})\frac{|y_{i}|}{r}\leq\sum_{i\leq
m}\phi_{2}(\\{i\\})=L\min_{i\leq m}\phi_{1}(\\{i\\})$ $\displaystyle\leq
L\frac{|y_{j}|}{r}\phi_{1}(\\{j\\})\leq
L\hat{\phi}_{1}\left(\frac{y}{r}\right).$

Hence, after multiplying by $r$ we get that $\hat{\phi}_{2}(y)\leq
L\hat{\phi}_{1}(y)$. ∎

######  Theorem 4.5.

Let $\phi_{1}$ and $\phi_{2}$ be two non-pathological lsc submeasures and
assume that $\phi_{1}(\\{i\\})>0$ and $\phi_{2}(\\{i\\})>0$ for every
$i\in\mathbb{N}$. The following are equivalent:

  * (a)

$\phi_{2}\preceq\phi_{1}$;

  * (b)

$\mathrm{EXH}(\phi_{1})\subseteq\mathrm{EXH}(\phi_{2})$;

  * (c)

$\mathrm{FIN}(\phi_{1})\subseteq\mathrm{FIN}(\phi_{2})$;

  * (d)

$\mathrm{EXH}(\phi_{1})\subseteq\mathrm{FIN}(\phi_{2})$;

###### Proof.

(a)$\implies$(b) and (a)$\implies$(c) are clear, since
$\hat{\phi}(x)=\lim_{n\to\infty}\hat{\phi}(x\cdot\chi_{\\{1,2,\ldots,n\\}})$
for every $x\in\mathbb{R}^{\mathbb{N}}$.

(b)$\implies$(d) and (c)$\implies$(d) follow from
$\mathrm{EXH}(\phi_{1})\subseteq\mathrm{FIN}(\phi_{1})$ and
$\mathrm{EXH}(\phi_{2})\subseteq\mathrm{FIN}(\phi_{2})$ (see Proposition 3.5).

(d)$\implies$(a): Suppose that $\phi_{2}\not\preceq\phi_{1}$, i.e., for every
$n\in\mathbb{N}$ there is a finite real sequence $z$ such that
$\hat{\phi}_{2}(z)>2^{2n+1}\hat{\phi}_{1}(z)$.

We will recursively construct sequences
$(n_{k}),(m_{k}),(L_{k})\subseteq\mathbb{N}$ and
$(x_{k})\subseteq\bigcup_{n\in\mathbb{N}}\mathbb{R}^{n}$ such that for each
$k\in\mathbb{N}$:

  * (i)

$n_{1}=1$ and $n_{k+1}>n_{k}$;

  * (ii)

$2^{2n_{k+1}}>L_{k}$;

  * (iii)

$\hat{\phi}_{2}(y)\leq L_{k}\hat{\phi}_{1}(y)$ for every
$y\in\mathbb{R}^{m_{k}}$;

  * (iv)

$m_{k}$ is the length of $x_{k}$;

  * (v)

$\hat{\phi}_{1}(x_{k}\chi_{\\{m_{k-1}+1,m_{k-1}+2,\ldots,m_{k}\\}})=\frac{1}{2^{n_{k}}}$;

  * (vi)

$\hat{\phi}_{2}(x_{k}\chi_{\\{m_{k-1}+1,m_{k-1}+2,\ldots,m_{k}\\}})>2^{n_{k}}$.

Start with $n_{1}=1$ (so that item (i) is met; item (ii) is empty in this
case), using our assumption we find some finite real sequence $z_{1}$ such
that $\hat{\phi}_{2}(z_{1})>2^{2n_{1}+1}\hat{\phi}_{1}(z_{1})$ and put
$x_{1}=\frac{z_{1}}{2^{n_{1}}\hat{\phi}_{1}(z_{1})}$. Let $m_{1}$ be the
length of $x_{1}$. Note that items (v) and (vi) are satisfied, since:

$\hat{\phi}_{2}(x_{1})=\frac{\hat{\phi}_{2}(z_{1})}{2^{n_{1}}\hat{\phi}_{1}(z_{1})}>\frac{2^{2n_{1}+1}\hat{\phi}_{1}(z_{1})}{2^{n_{1}}\hat{\phi}_{1}(z_{1})}=2^{n_{1}+1}>2^{n_{1}}.$

Using Lemma 4.4, we can find $L_{1}\in\mathbb{N}$ satisfying item (iii).

If $n_{i},m_{i},L_{i}$ and $x_{i}$ for all $i\leq k$ are already defined, we
can find $n_{k+1}$ such that items (i) and (ii) are met. By our assumption,
there is some finite real sequence $z_{k+1}$ such that
$\hat{\phi}_{2}(z_{k+1})>2^{2n_{k+1}+1}\hat{\phi}_{1}(z_{k+1})$. Let $m_{k+1}$
be the length of $z_{k+1}$. Observe that $m_{k+1}>m_{k}$ (by items (ii) and
(iii)). Moreover,
$\hat{\phi}_{2}(z_{k+1}\cdot\chi_{\mathbb{N}\setminus\\{1,2,\ldots,m_{k}\\}})>\left(2^{2n_{k+1}+1}-L_{k}\right)\hat{\phi}_{1}(z_{k+1}\cdot\chi_{\mathbb{N}\setminus\\{1,2,\ldots,m_{k}\\}})>2^{2n_{k+1}}\hat{\phi}_{1}(z_{k+1}\cdot\chi_{\mathbb{N}\setminus\\{1,2,\ldots,m_{k}\\}})$.
Indeed, the second inequality follows from item (ii) and if the first one
would be false, using item (iii) we should get:

$\displaystyle\hat{\phi}_{2}(z_{k+1})$
$\displaystyle\leq\hat{\phi}_{2}(z_{k+1}\cdot\chi_{\\{1,2,\ldots,m_{k}\\}})+\hat{\phi}_{2}(z_{k+1}\cdot\chi_{\mathbb{N}\setminus\\{1,2,\ldots,m_{k}\\}})$
$\displaystyle\leq
L_{k}\hat{\phi}_{1}(z_{k+1}\cdot\chi_{\\{1,2,\ldots,m_{k}\\}})+\left(2^{2n_{k+1}+1}-L_{k}\right)\hat{\phi}_{1}(z_{k+1}\cdot\chi_{\mathbb{N}\setminus\\{1,2,\ldots,m_{k}\\}})$
$\displaystyle\leq\left(L_{k}+\left(2^{2n_{k+1}+1}-L_{k}\right)\right)\hat{\phi}_{1}(z_{k+1})=2^{2n_{k+1}+1}\hat{\phi}_{1}(z_{k+1}),$

which contradicts the choice of $z_{k+1}$. Put:

$(x_{k+1})_{i}=\begin{cases}(x_{k})_{i},&\text{if }i\leq m_{k},\\\
\frac{(z_{k+1})_{i}}{2^{n_{k+1}}\hat{\phi}_{1}(z_{k+1}\cdot\chi_{\mathbb{N}\setminus\\{1,2,\ldots,m_{k}\\}})},&\text{if
}m_{k}<i\leq m_{k+1}.\end{cases}$

Then we have:

$\hat{\phi}_{2}(x_{k+1}\chi_{\\{m_{k}+1,m_{k}+2,\ldots,m_{k+1}\\}})=\frac{\hat{\phi}_{2}(z_{k+1}\chi_{\\{m_{k}+1,m_{k}+2,\ldots,m_{k+1}\\}})}{2^{n_{k+1}}\hat{\phi}_{1}(z_{k+1}\cdot\chi_{\mathbb{N}\setminus\\{1,2,\ldots,m_{k}\\}})}>2^{n_{k+1}}.$

Using Lemma 4.4, find $L_{k+1}$ satisfying item (iii) and observe that all
conditions are met.

Once the construction is completed, define $x=\bigcup_{k\in\mathbb{N}}x_{k}$.
We need to show that
$x\in\mathrm{EXH}(\phi_{1})\setminus\mathrm{FIN}(\phi_{2})$.

The fact that $x\notin\mathrm{FIN}(\phi_{2})$ follows from the observation
that for each $k\in\mathbb{N}$ we have
$\hat{\phi}_{2}(x)\geq\hat{\phi}_{2}(x_{k}\chi_{\\{m_{k-1}+1,m_{k-1}+2,\ldots,m_{k}\\}})>2^{n_{k+1}}$
(by item (vi)).

On the other hand, $x\in\mathrm{EXH}(\phi_{1})$ follows from:

$\hat{\phi}_{1}(x\cdot\chi_{\mathbb{N}\setminus\\{1,2,\ldots,m_{k}\\}})\leq\sum_{i>k}\hat{\phi}_{1}(x_{i}\chi_{\\{m_{i-1}+1,m_{i-1}+2,\ldots,m_{i}\\}})=\sum_{i>k}\frac{1}{2^{n_{i}}}\leq\frac{1}{2^{k}}.$

∎

###  4.4. Variations

######  Theorem 4.6.

Let $\phi_{1}$ and $\phi_{2}$ be two non-pathological lsc submeasures and
assume that $\phi_{1}(\\{i\\})>0$ and $\phi_{2}(\\{i\\})>0$ for every
$i\in\mathbb{N}$. The following are equivalent:

  * (a)

$\phi_{2}\preceq_{m}\phi_{1}$;

  * (b)

$\mathrm{mEXH}(\phi_{1})\subseteq\mathrm{mEXH}(\phi_{2})$;

  * (c)

$\mathrm{mFIN}(\phi_{1})\subseteq\mathrm{mFIN}(\phi_{2})$;

  * (d)

$\mathrm{mEXH}(\phi_{1})\subseteq\mathrm{mFIN}(\phi_{2})$.

Moreover, assuming that $\phi_{1}$ and $\phi_{2}$ satisfy:

$\forall_{j=1,2}\forall_{x,y\in\mathbb{R}^{\mathbb{N}}}\left(\left(\forall_{n\in\mathbb{N}}\sum_{i=1}^{n}|x_{i}|\leq\sum_{i=1}^{n}|y_{i}|\right)\implies\hat{\phi}_{j}(x)\leq\hat{\phi}_{j}(y)\right),$
(1)

the above conditions are also equivalent to the following one:

  * (e)

$\mathrm{BV}(\phi_{1})\subseteq\mathrm{BV}(\phi_{2})$.

###### Proof.

Firstly, we will show equivalence of items (a), (c) and (d). Secondly, we will
show the implications (b)$\implies$(d) and (a)$\implies$(b). Lastly, we will
deal with item (e) by showing (a)$\implies$(e) and (e)$\implies$(c).

(a)$\implies$(c): Straightforward, since
$\hat{\phi}(x)=\lim_{n\to\infty}\hat{\phi}(x\cdot\chi_{\\{1,2,\ldots,n\\}})$
for every $x\in\mathbb{R}^{\mathbb{N}}$.

(c)$\implies$(d): This follows from the inclusion
$\mathrm{mEXH}(\phi_{1})\subseteq\mathrm{mFIN}(\phi_{1})$ (see Proposition
3.6).

(d)$\implies$(a): Suppose that $\phi_{2}\not\preceq_{m}\phi_{1}$, i.e., for
every $n\in\mathbb{N}$ there is a finite nonincreasing real sequence $z$ such
that $\hat{\phi}_{2}(z)>2^{2n+1}\hat{\phi}_{1}(z)$.

We will recursively construct sequences
$(n_{k}),(m_{k}),(L_{k})\subseteq\mathbb{N}$ and
$(x_{k})\subseteq\bigcup_{n\in\mathbb{N}}\mathbb{R}^{\mathbb{N}}$ such that
for each $k\in\mathbb{N}$:

  * (i)

$n_{1}=1$ and $n_{k+1}>n_{k}$;

  * (ii)

$2^{2n_{k+1}}>L_{k}$;

  * (iii)

$\hat{\phi}_{2}(y)\leq L_{k}\hat{\phi}_{1}(y)$ for every
$y\in\mathbb{R}^{m_{k}}$;

  * (iv)

$m_{k}$ is the length of $x_{k}$;

  * (v)

$\hat{\phi}_{1}(x_{k}\chi_{\\{m_{k-1}+1,m_{k-1}+2,\ldots,m_{k}\\}})=\frac{1}{2^{n_{k}}}$;

  * (vi)

$\hat{\phi}_{2}(x_{k}\chi_{\\{m_{k-1}+1,m_{k-1}+2,\ldots,m_{k}\\}})>2^{n_{k}}$;

  * (vii)

$\frac{1}{2^{n_{k+1}}}<\hat{\phi}_{1}(e_{m_{k}+1}x_{k}(m_{k}))$ (here
$e_{i}\in\mathbb{R}^{\mathbb{N}}$ is the sequence having $1$ on $i$th
coordinate and zeros on all other coordinates);

  * (viii)

$x_{k}\in\mathrm{Mon}$.

Note that items (i)-(vi) are exactly the same as in the proof of the
implication (d)$\implies$(a) in Theorem 4.5. Hence, we will omit some details.

Start with $n_{1}=1$ (note that item (vii) is empty in the case of $k=1$),
find some finite nonincreasing real sequence $z_{1}$ such that
$\hat{\phi}_{2}(z_{1})>2^{2n_{1}+1}\hat{\phi}_{1}(z_{1})$ and put
$x_{1}=\frac{z_{1}}{2^{n_{1}}\hat{\phi}_{1}(z_{1})}$. Note that item (viii) is
satisfied, since $z_{1}$ is nonincreasing. Let $m_{1}$ be the length of
$x_{1}$. Then items (v) and (vi) are satisfied for the same reason as in the
proof of Theorem 4.5. Moreover, using Lemma 4.4, we find $L_{1}\in\mathbb{N}$
as in (iii).

If $n_{i},m_{i},L_{i}$ and $x_{i}$ for all $i\leq k$ are already defined, we
can find $n_{k+1}$ such that items (i), (ii) and (vii) are met. There is some
finite nonincreasing real sequence $z_{k+1}$ such that
$\hat{\phi}_{2}(z_{k+1})>2^{2n_{k+1}+1}\hat{\phi}_{1}(z_{k+1})$. Let $m_{k+1}$
be the length of $z_{k+1}$. Similarly as in the proof of Theorem 4.5, we have
$m_{k+1}>m_{k}$ and
$\hat{\phi}_{2}(z_{k+1}\cdot\chi_{\mathbb{N}\setminus\\{1,2,\ldots,m_{k}\\}})>2^{2n_{k+1}}\hat{\phi}_{1}(z_{k+1}\cdot\chi_{\mathbb{N}\setminus\\{1,2,\ldots,m_{k}\\}})$.
Put:

$(x_{k+1})_{i}=\begin{cases}(x_{k})_{i},&\text{if }i\leq m_{k},\\\
\frac{(z_{k+1})_{i}}{2^{n_{k+1}}\hat{\phi}_{1}(z_{k+1}\cdot\chi_{\mathbb{N}\setminus\\{1,2,\ldots,m_{k}\\}})},&\text{if
}m_{k}<i\leq m_{k+1}.\end{cases}$

Then items (iv), (v) and (vi) are met. We will show that (viii) is satisfied.
Observe that $x_{k+1}\restriction\\{1,\ldots,m_{k}\\}$ is nonincreasing by
(viii) applied to $k$ and $x_{k+1}\restriction\\{m_{k}+1,m_{k}+2,\ldots\\}$ is
nonincreasing, since $z_{k+1}$ is. Thus, it suffices to check that
$|x_{k+1}(m_{k})|\geq|x_{k+1}(m_{k}+1)|$ and the latter follows from:

$\displaystyle|x_{k+1}(m_{k}+1)|\hat{\phi}_{1}(e_{m_{k}+1})$
$\displaystyle=\hat{\phi}_{1}(|x_{k+1}(m_{k}+1)|e_{m_{k}+1})$
$\displaystyle\leq\hat{\phi}_{1}(x_{k+1}\chi_{\\{m_{k}+1,m_{k}+2,\ldots,m_{k+1}\\}})=\frac{1}{2^{n_{k+1}}}$
$\displaystyle<\hat{\phi}_{1}(e_{m_{k}+1}x_{k}(m_{k}))=|x_{k+1}(m_{k})|\hat{\phi}_{1}(e_{m_{k}+1}).$

To end the recursion step, use Lemma 4.4 to find $L_{k+1}$ as in (iii).

Once the construction is completed, define $x=\bigcup_{k\in\mathbb{N}}x_{k}$.
Then $x\in\mathrm{Mon}$ follows from (viii) and
$x\in\mathrm{EXH}(\phi_{1})\setminus\mathrm{FIN}(\phi_{2})$ can be shown in
the same way as in the proof of Theorem 4.5.

(b)$\implies$(d): This follows from the inclusion
$\mathrm{mEXH}(\phi_{2})\subseteq\mathrm{mFIN}(\phi_{2})$ (see Proposition
3.6).

(a)$\implies$(b): Assume that
$\mathrm{mEXH}(\phi_{1})\not\subseteq\mathrm{mEXH}(\phi_{2})$ and take any
$x\in\mathrm{mEXH}(\phi_{1})\setminus\mathrm{mEXH}(\phi_{2})$. Then
$x\in\mathrm{Mon}\setminus\mathrm{mEXH}(\phi_{2})$, thus
$\mathrm{mFIN}(\phi_{2})\subseteq c_{0}$ (by Proposition 3.19). By the fact
that $\mathrm{mEXH}(\phi_{1})\subseteq\mathrm{mFIN}(\phi_{1})$ (Proposition
3.6), we have $\hat{\phi_{1}}(x)<\infty$, thus, without losing generality, we
may assume that $x$ is non-negative, non-increasing and $\hat{\phi_{1}}(x)=1$.

There are two possible cases: either $x\not\in\mathrm{mFIN}(\phi_{2})$ or
$x\in\mathrm{mFIN}(\phi_{2})$.

In the case that $x\not\in\mathrm{mFIN}(\phi_{2})$, we have
$\hat{\phi_{2}}(x)=\infty$ and $\hat{\phi_{1}}(x)=1$. Since
$\hat{\phi_{2}}(x)=\lim_{n\to\infty}\hat{\phi_{2}}(x\cdot\chi_{\\{1,2,\ldots,n\\}})$
and
$\hat{\phi_{1}}\left(x\cdot\chi_{\\{1,2,\ldots,n\\}}\right)\leq\hat{\phi_{1}}\left(x\right)=1$
for all $n$, we obtain:

$\lim_{n\to\infty}\frac{\hat{\phi_{2}}\left(x\cdot\chi_{\\{1,2,\ldots,n\\}}\right)}{\hat{\phi_{1}}\left(x\cdot\chi_{\\{1,2,\ldots,n\\}}\right)}\geq\lim_{n\to\infty}\hat{\phi_{2}}\left(x\cdot\chi_{\\{1,2,\ldots,n\\}}\right)=\infty,$

hence $\phi_{2}\not\preceq_{m}\phi_{1}$.

In the case that $x\in\mathrm{mFIN}(\phi_{2})$, we know that $x\in c_{0}$, as
$\mathrm{mFIN}(\phi_{2})\subseteq c_{0}$. Take $\alpha>0$ such that
$\lim_{n\to\infty}\hat{\phi_{2}}(x\cdot\chi_{\\{n,n+1,\ldots\\}})=\alpha$
(such $\alpha$ exists, as $\hat{\phi_{2}}(x\cdot\chi_{\\{n,n+1,\ldots\\}})$ is
a non-increasing sequence in $[0,\hat{\phi_{2}}(x)]$ and
$x\notin\mathrm{mEXH}(\phi_{2})$). Define recursively the sequence $(n_{k})$
in such way that for all $k\in\mathbb{N}$ we get
$\hat{\phi_{1}}(x\cdot\chi_{\\{n_{k}+1,n_{k}+2,\ldots\\}})\leq 1/k$ and

$\frac{x(n_{k+1})}{x(n_{k})}\leq\frac{1}{k}.$

This is possible, because $\lim_{n\to\infty}x(n)=0$, $x$ is non-increasing and
$\lim_{n\to\infty}\hat{\phi_{1}}(x\cdot\chi_{\\{n,n+1,\ldots\\}})=0$.

Now, for any $k\in\mathbb{N}$, we can define the sequence $y_{k}$ by
$y_{k}(n)=x(n_{k+1})$ for $n\leq n_{k+1}$ and $y_{k}(n)=x(n)$ otherwise.
Clearly, $y_{k}$ is non-increasing as $x$ is non-increasing. Moreover, taking
$m_{k}$ such that
$\hat{\phi_{2}}\left(x\cdot\chi_{\\{n_{k+1}+1,n_{k+1}+2,\ldots,m_{k}\\}}\right)\geq\alpha/2$,
we may notice that:

$\frac{\hat{\phi_{2}}\left(y_{k}\cdot\chi_{\\{1,2,\ldots,m_{k}\\}}\right)}{\hat{\phi_{1}}\left(y_{k}\cdot\chi_{\\{1,2,\ldots,m_{k}\\}}\right)}\geq\frac{\hat{\phi_{2}}\left(y_{k}\cdot\chi_{\\{n_{k+1}+1,n_{k+1}+2,\ldots,m_{k}\\}}\right)}{\hat{\phi_{1}}\left(y_{k}\cdot\chi_{\\{1,2\ldots,n_{k}\\}}\right)+\hat{\phi_{1}}\left(y_{k}\cdot\chi_{\\{n_{k}+1,n_{k}+2,\ldots,m_{k}\\}}\right)}\geq$
$\geq\frac{\hat{\phi_{2}}\left(x\cdot\chi_{\\{n_{k+1}+1,n_{k+1}+2,\ldots,m_{k}\\}}\right)}{\frac{x(n_{k+1})}{x(n_{k})}\hat{\phi_{1}}\left(x\cdot\chi_{\\{1,2\ldots,n_{k}\\}}\right)+\hat{\phi_{1}}\left(x\cdot\chi_{\\{n_{k}+1,n_{k}+2,\ldots,m_{k}\\}}\right)}\geq\frac{\alpha/2}{\frac{1}{k}+\frac{1}{k}}=\frac{k\alpha}{4},$

hence $\phi_{2}\not\preceq_{m}\phi_{1}$.

(a)$\implies$(e): Assume that $\phi_{2}\preceq_{m}\phi_{1}$, i.e. there is
$M>0$ such that $\phi_{2}(y)\leq M\phi_{1}(y)$for every non-increasing finite
sequence $y$. Let $x\in\mathrm{BV}(\phi_{1})$ and fix any
$\hat{J}=(\hat{J}_{n})\in\mathcal{P}_{I}$. If $\phi_{1}(\mathbb{N})<+\infty$,
then $BV(\phi_{1})=BV(\phi_{2})=B(I)$ (by Proposition 3.18), so we may assume
that $\phi_{1}(\mathbb{N})=+\infty$. Then we have $x(\hat{J}_{n})\in c_{0}$
and we can apply the procedure described below. Let
$\pi:\mathbb{N}\to\mathbb{N}$ be such that
$|x(\hat{J}_{\pi(1)})|=\sup\\{|x(\hat{J}_{n})|:n\in\mathbb{N}\\}$ and
$|x(\hat{J}_{\pi(k+1)})|=\sup\\{|x(\hat{J}_{n})|:n\in\mathbb{N}\setminus\\{\pi(1),\ldots,\pi(k)\\}\\}$
for all $k\in\mathbb{N}$. Then
$J^{\star}=(\hat{J}_{\pi(n)})\in\mathcal{P}_{I}$ and
$x(J^{\star})\in\mathrm{Mon}$. Hence, by the condition imposed on $\phi_{2}$,
we have:

$\hat{\phi}_{2}(x(\hat{J}))\leq\hat{\phi}_{2}(x(J^{\star}))\leq
M\hat{\phi}_{1}(x(J^{\star}))\leq
M\sup_{J\in\mathcal{P}_{I}}\hat{\phi}_{1}(x(J)).$

Since $\hat{J}$ was arbitrary, we get that

$\sup_{J\in\mathcal{P}_{I}}\hat{\phi}_{2}(x(J))\leq
M\sup_{J\in\mathcal{J}_{I}}\hat{\phi}_{1}(x(J)),$

so $x\in\mathrm{BV}(\phi_{2})$.

(e)$\implies$(c): We will show that the condition
$\mathrm{mFIN}(\phi_{1})\not\subseteq\mathrm{mFIN}(\phi_{2})$ implies
$\mathrm{BV}(\phi_{1})\not\subseteq\mathrm{BV}(\phi_{2})$.

Assume first that $\mathrm{mFIN}(\phi_{1})=\mathrm{Mon}$. Since
$\mathrm{mFIN}(\phi_{1})\not\subseteq\mathrm{mFIN}(\phi_{2})$,
$\mathrm{mFIN}(\phi_{2})\neq\mathrm{Mon}$. Then Proposition 3.18 gives us
$\mathrm{BV}(\phi_{1})=B(I)\not\subseteq\mathrm{BV}(\phi_{2})$.

Assume now that $\mathrm{mFIN}(\phi_{1})\not=\mathrm{Mon}$. Fix
$x=(x_{n})\in\mathrm{mFIN}(\phi_{1})\setminus\mathrm{mFIN}(\phi_{2})$. Let
$f:[0,1]\to\mathbb{R}$ be a piecewise linear function such that $f(1)=0$,
$f(0)=\sum_{n=1}^{\infty}(-1)^{n+1}|x_{n}|$ and
$f(\frac{1}{2^{k}})=\sum_{n=1}^{k}(-1)^{n+1}|x_{n}|$ for all $k\in\mathbb{N}$.
This function is well-defined as $x\in c_{0}$ (by Proposition 3.18)..

Observe that $f\notin\mathrm{BV}(\phi_{2})$, since for the sequence of
intervals $\hat{J}=(\hat{J}_{n})\in\mathcal{J}_{I}$ given by
$\hat{J}_{1}=[\frac{1}{2},1]$ and
$\hat{J}_{k+1}=[\frac{1}{2^{k+1}},\frac{1}{2^{k}}]$ for all $k\in\mathbb{N}$,
we have $|f(\hat{J}_{k})|=|x_{k}|$. Thus
$\sup_{J\in\mathcal{J}_{I}}\hat{\phi}_{2}(f(J))\geq\hat{\phi}_{2}(f(\hat{J}))=\hat{\phi}_{2}(x)=\infty$.
On the other hand, for each $J\in\mathcal{J}_{I}$ we have:

$\forall_{n\in\mathbb{N}}\sum_{i=1}^{n}|f(J_{i})|\leq\sum_{i=1}^{n}|f(\hat{J}_{i})|.$

This is actually a simple observation, which may either be proved directly or
deduced from the general observation (see [20, Proposition 1.1]) stating that
when selecting the ends of the intervals we should select points of varying
monotonicity to get higher value of the sum.

Hence, by the assumption imposed on $\phi_{1}$, we have
$\sup_{J\in\mathcal{J}_{I}}\hat{\phi}_{1}(f(J))=\hat{\phi}_{1}(f(\hat{J}))=\hat{\phi}_{1}(x)<\infty$.
∎

##  5\. Particular case I: simple density ideals

In this Section we are interested in lsc submeasures of the form
$\phi_{g}(C)=\sup_{n\in\mathbb{N}}\frac{|C\cap\\{1,2,\ldots,n\\}|}{g(n)}$,
where $g:\mathbb{N}\to\mathbb{N}$ satisfies the following conditions:

  * (a)

$(g(n))$ is nondecreasing;

  * (b)

$\lim_{n}g(n)=\infty$;

  * (c)

$\frac{n}{g(n)}$ does not tend to zero;

  * (d)

$\frac{n}{g(n)}$ is nondecresing.

Ideals of the form $\mathrm{Exh}(\phi_{g})$ (for functions $g$ satisfying all
the above requirements except the last one) have been extensively studied in
[3], [14] and [15]. For that reason, we decided to write the last item
separately despite the fact that (c) follows from (d). We will denote by
$\mathcal{G}$ the set of all functions $g$ satisfying conditions (a)-(d). Note
that in this case
$\hat{\phi}_{g}(x)=\sup_{n\in\mathbb{N}}\frac{\sum_{i=1}^{n}|x_{i}|}{g(n)}$.

In 1974 Chanturia introduced the concept of the modulus of variation of the
bounded function (see [6]), which for $x\colon I\to\mathbb{R}$ is given as a
sequence

$v(x,n)=\sup_{{\mathcal{P}}_{n}}\sum_{k=1}^{n}|x(I_{k})|,$

where ${\mathcal{P}}_{n}$ denotes the set of all $n$-element families of
nonoverlapping intervals of $I$. In the mentioned paper [6] the Author
introduced set of functions $V[g]$, for a given sequence
$g\colon\mathbb{N}\to\mathbb{R}$, as a family of those functions for which
$v(x,n)=O(g(n))$. These classes are now called Chanturia (or Chanturiya)
classes in literature. One of the statements (see Theorem 1 in the mentioned
paper) was that the necessary and sufficient condition for a sequence to be
$v(x,n)$ for some function $x$ is that it is nondecreasing and concave. These
classes were studied since then in many papers, mainly in relation to a
convergence of Fourier series and relations to other families of functions of
bounded variation (see especially the relation between Chanturia classes and
Waterman spaces given by Avdispahić in [2]).

The sequences defining Chanturia classes are defined as real valued sequences
but as we can see we may redefine any real-valued sequence $h(n)$ to a
sequence $g(n)=\lceil h(n)\rceil$, which has values in natural numbers and the
same asymptotics as $n\to+\infty$. One more important observation for
nondecreasing and concave sequences $h(n)$ as considered in a context of
Chanturia classes is that they are such that $\frac{n}{h(n)}$ are
nondecreasing (as required by the definition of a family $\mathcal{G}$ above).
Without the loss of generality we may assume that $h(0)=0$ and then from being
concave we may deduce that for each $k\in\\{1,...,n-1\\}$ we have

$h(k)=h\left(\frac{k}{n}n+\left(1-\frac{k}{n}\right)0\right)\geq\frac{k}{n}h(n)+\left(1-\frac{k}{n}\right)h(0)=\frac{k}{n}h(n),$

which gives

$\frac{h(k)}{k}\geq\frac{h(n)}{n},$

as desired.

######  Theorem 5.1.

The following are equivalent for every $g,h\in\mathcal{G}$:

  * (a)

$g(n)=O(h(n))$, i.e., there is $\eta>0$ such that $\frac{g(n)}{h(n)}\leq\eta$
for all $n\in N$;

  * (b)

$\phi_{h}\preceq\phi_{g}$;

  * (c)

$\phi_{h}\preceq_{m}\phi_{g}$;

  * (d)

$\mathrm{Fin}(\phi_{g})\subseteq\mathrm{Fin}(\phi_{h})$;

  * (e)

$\mathrm{Exh}(\phi_{g})\subseteq\mathrm{Exh}(\phi_{h})$;

  * (f)

$\mathrm{Exh}(\phi_{g})\subseteq\mathrm{Fin}(\phi_{h})$;

  * (g)

$\mathrm{FIN}(\phi_{g})\subseteq\mathrm{FIN}(\phi_{h})$;

  * (h)

$\mathrm{EXH}(\phi_{g})\subseteq\mathrm{EXH}(\phi_{h})$;

  * (i)

$\mathrm{EXH}(\phi_{g})\subseteq\mathrm{FIN}(\phi_{h})$;

  * (j)

$\mathrm{mFIN}(\phi_{g})\subseteq\mathrm{mFIN}(\phi_{h})$;

  * (k)

$\mathrm{mEXH}(\phi_{g})\subseteq\mathrm{mEXH}(\phi_{h})$;

  * (l)

$\mathrm{mEXH}(\phi_{g})\subseteq\mathrm{mFIN}(\phi_{h})$;

  * (m)

$\mathrm{BV}(\phi_{g})\subseteq\mathrm{BV}(\phi_{h})$.

###### Proof.

(a)$\implies$(b): We claim that $\phi_{h}(x)\leq\eta\phi_{g}(x)$ for every
$x\in\mathbb{R}^{\mathbb{N}}$. Indeed, for every $n\in\mathbb{N}$ we have:

$\frac{\sum_{i=1}^{n}|x_{i}|}{h(n)}\leq\eta\frac{\sum_{i=1}^{n}|x_{i}|}{g(n)}.$

Items (b), (g), (h) and (i) are equivalent thanks to Theorem 4.5.

The (b)$\implies$(c) is obvious.

Items (c), (j), (k), (l) and (m) are equivalent thanks to Theorem 4.6, since
for any $g\in\mathcal{G}$ and any $x,y\in\mathbb{R}^{\mathbb{N}}$ such that
$\sum_{i=1}^{n}|x_{i}|\leq\sum_{i=1}^{n}|y_{i}|$ for all $n\in\mathbb{N}$, it
is easy to see that $\phi_{g}(x)\leq\phi_{g}(y)$.

By Proposition 4.2, (b)$\implies$(d), (b)$\implies$(e), (d)$\implies$(f) and
(e)$\implies$(f).

Therefore, we only need to show (f)$\implies$(a) and (c)$\implies$(a).

(c)$\implies$(a): Assume that (a) does not hold. We need to show that (c) does
not hold, i.e., for every $M>0$ there is a finite nonincreasing sequence $x$
such that $\phi_{h}(x)>M\phi_{g}(x)$.

Fix $M>0$. Since (a) does not hold, there is $n\in\mathbb{N}$ such that
$\frac{g(n)}{h(n)}>M$. Define $x=(x_{i})\in\mathbb{R}^{\mathbb{N}}$ by:

$x_{i}=\begin{cases}\frac{g(n)}{n},&\text{if }i\leq n,\\\
0,&\text{otherwise}.\end{cases}$

Then $x$ is nonincreasing and

$\phi_{h}(x)\geq\frac{\sum_{i=1}^{n}|x_{i}|}{h(n)}=\frac{g(n)}{h(n)}>M.$

Now we will show that $\phi_{g}(x)\leq 1$, which will finish the proof. Fix
$m\in\mathbb{N}$. There are three possibilities:

  * •

If $m=n$, then $\frac{\sum_{i=1}^{n}|x_{i}|}{g(m)}=1$.

  * •

If $m>n$, then $\frac{\sum_{i=1}^{n}|x_{i}|}{g(m)}=\frac{g(n)}{g(m)}\leq 1$,
since $g$ is nondecreasing.

  * •

If $m<n$, then

$\frac{\sum_{i=1}^{n}|x_{i}|}{g(m)}=\frac{\frac{g(n)}{n}}{\frac{g(m)}{m}}\leq
1,$

since $\frac{n}{g(n)}$ is nondecresing.

(f)$\implies$(a): Assume that (a) does not hold. We will use Proposition
4.2(c) to show that (f) does not hold.

Fix any $k\in\mathbb{N}$. We are looking for $F\in\mathrm{Fin}$ such that
$\phi_{g}(F)\leq\frac{1}{2^{k}}$ and $\phi_{h}(F)\geq 2^{k}$. Let $\delta>1$
be such that $\frac{1}{g(1)}\geq\frac{1}{\delta}$. Note that
$\frac{i}{g(i)}\geq\frac{1}{\delta}$ for all $i\in\mathbb{N}$ (by the fact
that $\frac{i}{g(i)}$ is nondecresing).

Since (a) does not hold, there is $n\in\mathbb{N}$ such that
$\frac{g(n)}{h(n)}>2^{k}2^{k+1}\delta$. Actually, there are infinitely many
such $n$ (given one such $n$ we can always find $n^{\prime}\in\mathbb{N}$ with
$\frac{g(n^{\prime})}{h(n^{\prime})}>\frac{g(n)}{h(n)}>2^{k}2^{k+1}\delta$).
Thus, without loss of generality we may assume that $n$ is big enough to
guarantee that $\frac{1}{g(n)}<\frac{1}{2^{k}\delta}<\frac{1}{2^{k}}$ (since
$\lim_{n}g(n)=\infty$).

Find $j\in\mathbb{N}$ such that $2^{j+1}\delta\geq g(n)>2^{j}\delta$ and note
that $j\geq k$ (as $\frac{1}{g(n)}<\frac{1}{2^{k}\delta}$). Moreover, since
$\frac{n}{g(n)}\geq\frac{1}{\delta}$, we have
$n\geq\frac{g(n)}{\delta}>2^{j}$. Hence, we can find $a\leq n$ such that
$n-a=2^{j-k}$.

Define $F=\\{n-a+1,n-a+2,\ldots,n\\}$. Then

$\phi_{h}(F)\geq\frac{n-a}{h(n)}>\frac{n-a}{g(n)}2^{k}2^{k+1}\delta>\frac{2^{j-k}}{2^{j+1}\delta}2^{k}2^{k+1}\delta=2^{k}.$

In order to finish the proof, we need to show that

$\phi_{g}(F)=\sup_{m\in\mathbb{N}}\frac{|F\cap\\{1,\ldots,m\\}|}{g(m)}\leq\frac{1}{2^{k}}.$

Fix $m\in\mathbb{N}$. There are four possibilities:

  * •

If $m=n$, then
$\frac{|F\cap\\{1,\ldots,m\\}|}{g(m)}=\frac{n-a}{g(n)}<\frac{2^{j-k}}{2^{j}\delta}<\frac{1}{2^{k}}$.

  * •

If $m>n$, then
$\frac{|F\cap\\{1,\ldots,m\\}|}{g(m)}=\frac{n-a}{g(m)}\leq\frac{n-a}{g(n)}<\frac{1}{2^{k}}$,
since $g$ is nondecreasing.

  * •

If $m\leq a$, then $\frac{|F\cap\\{1,\ldots,m\\}|}{g(m)}=0$.

  * •

If $a<m<n$, then
$\frac{|F\cap\\{1,\ldots,m\\}|}{g(m)}=\frac{m-a}{g(m)}=\frac{m}{g(m)}-\frac{a}{g(m)}\leq\frac{n}{g(n)}-\frac{a}{g(n)}=\frac{n-a}{g(n)}<\frac{1}{2^{k}}$,
since $\frac{n}{g(n)}$ and $g$ are nondecresing.

∎

##  6\. Particular case II: summable ideals and $\Lambda BV$ spaces

In this Section we are interested in lsc submeasures of the form
$\phi_{A}(C)=\sum_{n\in C}a_{n}$, where $A=(a_{n})$ is a Waterman sequence
(i.e., $A$ is a nonincreasing sequence of positive real numbers such that
$\sum_{n\in\mathbb{N}}a_{n}=\infty$). Note that in this case $\phi$ is
actually a measure and $\hat{\phi}_{A}(x)=\sum_{n\in\mathbb{N}}a_{i}|x_{i}|$.
In such case:

  * •

$\mathrm{Fin}(\phi_{A})=\mathrm{Exh}(\phi_{A})$ is the summable ideal
$\mathcal{I}_{A}$ given by the sequence $A$ (see Definition 2.5 and [8,
Example 1.2.3]);

  * •

$\mathrm{FIN}(\phi_{A})=\mathrm{EXH}(\phi_{A})$ and
$\mathrm{mFIN}(\phi_{A})=\mathrm{mEXH}(\phi_{A})$ (by the previous item,
Proposition 3.7 and Remark 3.8);

  * •

$\mathrm{BV}(\phi_{A})$ is the space ${ABV}$ of functions of bounded
$A$-variation.

We will need the following result.

######  Theorem 6.1 ([11, Theorem 4.1]).

Assume we have two Waterman sequences $A=(a_{n})_{n\in\mathbb{N}}$ and
$B=(b_{n})_{n\in\mathbb{N}}$. The following statements are equivalent:

  1. (a)

$\mathcal{I}_{A}\leq_{K}\mathcal{I}_{B}$;

  2. (b)

There exist positive numbers $m,M$ and a partition of $\mathbb{N}$ into
consecutive nonempty finite intervals $(I_{n})_{n\in\mathbb{N}}$ such that

$ma_{n}\leq\sum_{i\in I_{n}}b_{i}\leq Ma_{n}$

for every $n\in\mathbb{N}$;

  3. (c)

There exists $M>0$ such that

$\sum_{i=1}^{k}b_{i}\geq M\sum_{i=1}^{l}a_{i}\Rightarrow b_{k}\leq Ma_{l}$

for all $k,l\in\mathbb{N}$.

######  Theorem 6.2.

Assume we have two Waterman sequences $A=(a_{n})_{n\in\mathbb{N}}$ and
$B=(b_{n})_{n\in\mathbb{N}}$. The following statements are equivalent:

  1. (a)

$\sum_{i=1}^{n}b_{i}=O(\sum_{i=1}^{n}a_{i})$, i.e., there is $\eta>0$ such
that

$\frac{\sum_{i=1}^{n}b_{i}}{\sum_{i=1}^{n}a_{i}}\leq\eta$

for all $n\in N$;

  2. (b)

$\mathcal{I}_{A}\leq_{K}\mathcal{I}_{B}$;

  3. (c)

$\phi_{B}\preceq_{m}\phi_{A}$;

  4. (d)

${ABV}\subseteq{BBV}$;

  5. (e)

$\mathrm{mEXH}(\phi_{A})\subseteq\mathrm{mEXH}(\phi_{B})$.

  6. (f)

$\mathrm{mFIN}(\phi_{A})\subseteq\mathrm{mFIN}(\phi_{B})$;

  7. (g)

$\mathrm{mEXH}(\phi_{A})\subseteq\mathrm{mFIN}(\phi_{B})$;

###### Proof.

The equivalence of items (c)-(g) follows from Theorem 4.6, since given any
Waterman sequence $A=(a_{n})_{n\in\mathbb{N}}$ and any
$x,y\in\mathbb{R}^{\mathbb{N}}$ such that
$\sum_{i=1}^{n}|x_{i}|\leq\sum_{i=1}^{n}|y_{i}|$ for all $n\in\mathbb{N}$,
using the fact that $(a_{n})$ is nonincreasing we have:

$\sum_{k=1}^{n}a_{k}|x_{k}|=\sum_{k=1}^{n-1}\left(\sum_{i=1}^{k}|x_{i}|\right)(a_{k}-a_{k+1})+a_{n}\sum_{i=1}^{n}|x_{i}|\leq$
$\leq\sum_{k=1}^{n-1}\left(\sum_{i=1}^{k}|y_{i}|\right)(a_{k}-a_{k+1})+a_{n}\sum_{i=1}^{n}|y_{i}|=\sum_{k=1}^{n}a_{k}|y_{k}|.$

Hence, $\phi_{A}(x)\leq\phi_{A}(y)$.

The equivalence of items (a) and (d) is proved in [19, Theorem 3], so it
remains to show that (a) and (b) are equivalent.

$(b)\Rightarrow(a)$: By the previous theorem, we can find $M>0$ and a
partition of $\mathbb{N}$ into consecutive nonempty intervals
$(I_{n})_{n\in\mathbb{N}}$ such that for all $n\in\mathbb{N}$ we have

$\frac{\sum_{i\in I_{n}}b_{i}}{a_{n}}\leq M,$

thus

$\frac{\sum_{i=1}^{\max I_{n}}b_{i}}{\sum_{i=1}^{n}a_{i}}\leq M.$

Let us put $\eta=M$ and take $n\in\mathbb{N}$. Then there exists
$j\in\mathbb{N}$ such that $n\in I_{j}$. It follows that $n\geq j$, thus

$\frac{\sum_{i=1}^{n}b_{i}}{\sum_{i=1}^{n}a_{i}}\leq\frac{\sum_{i=1}^{\max
I_{j}}b_{i}}{\sum_{i=1}^{j}a_{i}}\leq M=\eta.$

$(a)\Rightarrow(b)$: Suppose that $\mathcal{I}_{A}\not\leq_{K}\mathcal{I}_{B}$
and fix $\eta>0$. Let $M=2\eta$. By the previous theorem, we can find
$k,l\in\mathbb{N}$ such that $\sum_{i=1}^{k}b_{i}\geq M\sum_{i=1}^{l}a_{i}$
and $b_{k}>Ma_{l}$. We have two cases.

If $k\leq l$ then

$\frac{\sum_{i=1}^{k}b_{i}}{\sum_{i=1}^{k}a_{i}}\geq\frac{\sum_{i=1}^{k}b_{i}}{\sum_{i=1}^{l}a_{i}}\geq
M>\eta.$

If $k>l$ then

$\sum_{i=l+1}^{k}a_{i}\leq\sum_{i=l+1}^{k}a_{l}<\sum_{i=l+1}^{k}\frac{b_{k}}{M}\leq\sum_{i=l+1}^{k}\frac{b_{i}}{M}\leq\sum_{i=1}^{k}\frac{b_{i}}{M},$

thus

$\sum_{i=1}^{k}a_{i}=\sum_{i=1}^{l}a_{i}+\sum_{i=l+1}^{k}a_{i}<\frac{\sum_{i=1}^{k}b_{i}}{M}+\sum_{i=1}^{k}\frac{b_{i}}{M}=\frac{2\sum_{i=1}^{k}{b_{i}}}{M},$

hence

$\frac{\sum_{i=1}^{k}b_{i}}{\sum_{i=1}^{k}a_{i}}>\frac{M}{2}=\eta.$

Therefore, in both cases condition (a) does not hold. ∎

##  7\. Comparison of two particular cases

The next theorem is an interesting comment to the paper [2] giving different
inclusions between Chanturia classes and Waterman spaces: this is a general
proof that many of the inclusions given there are strict (these, for which the
assumption on monotonicity of $(g(n+1)-g(n))$ is satisfied).

######  Theorem 7.1.

Let $g\in\mathcal{G}$ be such that $(g(n+1)-g(n))$ monotonically tends to $0$.
Then $\mathrm{BV}(\phi_{g})$ is not equal to any ${ABV}$.

###### Proof.

Let $A=(a_{n})$ be a Waterman sequence. We will obtain the thesis by showing
that $\mathrm{mFIN}(\phi_{g})\not=\mathrm{mFIN}(\phi_{A})$ and applying
Theorem 4.6.

Define the sequence $x=(x_{n})$ by the formula $\sum_{i=1}^{n}x_{n}=g(n)$.
Clearly, $x$ is nonincreasing, tends to $0$ and belongs to
$\mathrm{mFIN}(\phi_{g})$ as $\hat{\phi}_{g}(x)=1$. We have two cases:

In the first case we assume that $\sum_{n=1}^{\infty}a_{n}x_{n}=\infty$. Then
$x\in\mathrm{mFIN}(\phi_{g})\setminus\mathrm{mFIN}(\phi_{A})$.

In the second case we assume that $\sum_{n=1}^{\infty}a_{n}x_{n}<\infty$.
Notice that then for every $k\in\mathbb{N}$ we have
$\sum_{n=1}^{\infty}a_{n}(k\cdot
x_{n})=k\cdot\sum_{n=1}^{\infty}a_{n}x_{n}<\infty$. We will show that
$\mathrm{mFIN}(\phi_{A})\setminus\mathrm{mFIN}(\phi_{g})\not=\emptyset$.

We will now define recursively sequences $(n_{i})$ and $(m_{i})$ of natural
numbers. We put as $n_{1}$ the smallest $N$ such that
$2\cdot\sum_{n>N}a_{n}x_{n}<\frac{1}{4}$. Next, we put as $m_{1}$ the smallest
$N>n_{1}$ such that $2x_{N}\leq x_{n_{1}}$. We can find such because
$\lim_{n\to\infty}x_{n}=0$.

Now, suppose we have already defined $n_{i}$ and $m_{i}$ for some
$i\in\mathbb{N}$. Then we define $n_{i+1}$ as the smallest $N>m_{i}$ such that

$(i+2)\sum_{n>N}a_{n}x_{n}<\frac{1}{2^{i+2}}\text{ and
}(i+1)\sum_{n=m_{i}}^{N}x_{n}\geq\frac{i+1}{2}g(N).$

We can find such $N$, since $\lim_{j\to\infty}\sum_{n>j}a_{n}x_{n}=0$ and
$(i+1)\sum_{n=1}^{j}x_{n}=(i+1)g(j)$, which tends to infinity, thus

$\lim_{j\to\infty}\frac{(i+1)\sum_{n=m_{i}}^{j}x_{n}}{g(j)}=i+1.$

Next, we define $m_{i+1}$ as the smallest $N>n_{i+1}$ such that
$(i+2)x_{N}\leq(i+1)x_{n_{i+1}}$.

We can now proceed to defining the sequence $y=(y_{n})$ such that
$y\in\mathrm{mFIN}(\phi_{A})\setminus\mathrm{mFIN}(\phi_{g})$, which will end
the proof. We put

$y_{n}=\begin{cases}x_{n},&\text{if }n\leq n_{1},\\\ (i+1)x_{m_{i}},&\text{if
}n_{i}<n<m_{i}\text{ for some }i,\\\ (i+1)x_{n},&\text{if }m_{i}\leq n\leq
n_{i+1}\text{ for some }i.\end{cases}$

First, notice that $y$ is nonincreasing as $(x_{n})$ is nonincreasing and
$(i+1)x_{m_{i}}\leq i\cdot x_{n_{i}}$ for every $i\in\mathbb{N}$. Next,
observe that $y\not\in\mathrm{mFIN}(\phi_{g})$ as

$\hat{\phi}_{g}(y)\geq\frac{\sum_{n=m_{i}}^{n_{i+1}}(i+1)x_{n}}{g(n_{i+1})}\geq\frac{i+1}{2}$

for every $i\in\mathbb{N}$. Finally, we obtain $y\in\mathrm{mFIN}(\phi_{A})$
by the fact that

$\sum_{n=1}^{\infty}a_{n}y_{n}\leq\sum_{n\leq
n_{1}}a_{n}x_{n}+\sum_{i=1}^{\infty}\sum_{n=n_{i}+1}^{n_{i+1}}a_{n}(i+1)x_{n}\leq$
$\leq\sum_{n\leq
n_{1}}a_{n}x_{n}+\sum_{i=1}^{\infty}\sum_{n>n_{i}}(i+1)a_{n}x_{n}<\sum_{n\leq
n_{1}}a_{n}x_{n}+\sum_{i=1}^{\infty}\frac{1}{2^{i+1}}=$ $=\sum_{n\leq
n_{1}}a_{n}x_{n}+\frac{1}{2}<\infty.$

∎

######  Example 7.2.

The sequence $g(n)=\sqrt{n}$ belongs to $\mathcal{G}$ and is such that
$g(n+1)-g(n)$ monotonically tends to $0$. Therefore, $\mathrm{BV}(\phi_{g})$
is not equal to any ${ABV}$.

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