###### Abstract

The aim of this article is to detect the ascent and descent of weighted
composition operators on Lorentz spaces. We investigate the conditions on the
measurable transformation $T$ and the complex-valued measurable function $u$
defined on measure space $(X,\mathcal{A},\mu)$ that cause the weighted
composition operators on Lorentz space $L(p,q)$, $1<p\leq\infty$, $1\leq
q\leq\infty$ to have finite or infinite ascent (descent). We also give a
number of examples in support of our findings.

Ascent and Descent of Weighted Composition Operators

on Lorentz spaces

Gopal Datt1 and Daljeet Singh Bajaj2

1Department of Mathematics, PGDAV College, University of Delhi, Delhi-110065,
India.

2Department of Mathematics, University of Delhi, Delhi-110007, India.

E-Mail<EMAIL_ADDRESS>2<EMAIL_ADDRESS>
2020 Mathematics Subject Classification: Primary: 47B33, 47B37, 47B38;
Secondary: 46E30.
Keywords: Ascent, Descent, Weighted composition operator, Lorentz spaces.

##  1 Introduction

The theory of “Lorentz spaces” is presented in the work of G. G. Lorentz in
[13, 14]. For a pair of numbers $p$ and $q$, ($1<p<\infty,\ 1\leq q<\infty$ or
$1<p\leq\infty,\ q=\infty$), the Lorentz space is defined as the collection of
all complex-valued measurable functions $h$ defined on the $\sigma-$finite
complete measure space $(X,\mathcal{A},\mu)$ whose norm $\|h\|_{pq}<\infty$,
where the norm is defined as follows

$\displaystyle\|h\|_{pq}~{}=~{}\begin{cases}\left(\frac{q}{p}\bigintsss_{0}^{\infty}\big{(}t^{1/p}h^{**}(t)\big{)}^{q}~{}\frac{dt}{t}\right)^{1/q},&\
1<p<\infty,\ 1\leq q<\infty\\\ \underset{t>0}{\sup}\,\,\,t^{1/p}h^{**}(t),&\
1<p\leq\infty,\ q=\infty\end{cases},$

where $h^{**}$ is the average function of non-increasing rearrangement
function $h^{*}$ given by $h^{*}(t)=\inf\\{s>0:\mu\\{x\in X:|h(x)|>s\\}\leq
t\\}$, $s,~{}t\geq 0$. It is represented mathematically as $L(p,q)$. These
spaces are generalization of classical Lebesgue spaces. In $1966$, R. Hunt
[11] discussed various properties and useful tools in support to describe the
dual of Lorentz spaces. The duality problem regarding Lorentz spaces was
examined by Cwikel [5] and Cwikel and Fefferman [6, 7]. To know more about the
Lorentz spaces, we suggest the reader to read [3, 15] and the references
therein.

The study of multiplication and composition operators on different function
spaces, like $L^{p}-$spaces, Lorentz spaces, Orlicz spaces, and many more, is
the centre of attraction for several mathematicians. Many interesting
mathematical results have been developed and continue to develop in the
current period as well. A lot of literature is available about the operators,
which are formed by the product of composition and multiplication operators,
and these operators are known as weighted composition operators. Such
operators are symbolically represented as $W_{u,T}$, where $u$ is a complex-
valued measurable function and $T$ is a measurble transformation defined on
$(X,\mathcal{A},\mu)$ respectively. The available theory on weighted
composition operators involves their boundedness, compactness, closedness of
range and other mathematical results. We refer the reader to read the
following references [1, 4, 10, 12, 16] with reference therein.

The terms ascent and descent of linear operators were studied by A.E. Taylor
in his book [17] and applications of these notions in the development of
spectral study of compact operators are described in details. One of the
important theorem in his book states that “If the ascent and descent of a
linear operator $L$ defined on a normed space $(Y,\|\cdot\|_{Y})$ are finite
and equal to $k_{1}$ and $k_{2}$ respectively, then $k_{1}=k_{2}$ and $X$ can
be written as direct sum of null space of $L^{k_{1}}$ and range space of
$L^{k_{1}}$”. To look into the properties, examples and other applications,
one can refer to [2, 8, 9].

The notions of ascent and descent have been examined for composition and
weighted composition operators on $l_{p}$ spaces, $L^{p}$-spaces, Lorentz
spaces, Orlicz spaces, and several more. Motivated by the direction of study,
we discussed the properties of ascent and descent of weighted composition
operators on Lorentz sequence spaces of functions defined on the measure space
$(\mathbb{N},2^{\mathbb{N}},\mu)$, where $\mu$ denotes the counting measure
and $2^{\mathbb{N}}$ denotes the power set of natural numbers $\mathbb{N}$, in
[8]. This paper extends the study and explores it on Lorentz spaces of complex
valued functions defined over general ($\sigma$-finite) measure spaces. In
section 2, we set up the notations used in the paper and recall the definition
and important results that help us in our pursuit. In sections 3 and 4, we
discuss main results of the paper which provide classes of functions $u$ and
$T$ and compute ascent and descent of the induced weighted composition
operators $W_{u,T}$. At the end, we discuss some applications of our findings
in order to calculate the ascent and descent of weighted composition
operators.

##  2 Preliminaries

In this section, we give necessary definitions, theorems and other related
informations required to form the base for this paper. We begin with the
definition of distribution function, non-increasing rearrangement and average
function which lead to define norm on Lorentz space.

Let $(X,\mathcal{A},\mu)$ be a $\sigma$-finite complete measure space and
$h:X\mapsto\mathbb{C}$ be a measurable function. By $\mu_{h}$, we mean the
distribution function of $h$ and is given by

$\mu_{h}(s)=\mu\big{(}\\{x\in X:|h(x)|>s\\}\big{)}~{}~{}\text{ where }s\geq
0.$

With the help of $\mu_{h}$, we define the non-increasing rearrangement $h^{*}$
as

$h^{*}(t)~{}=~{}\inf\,\\{s>0:\mu_{h}(s)\leq t\\},~{}t>0.$

For $t>0$,

$h^{**}(t)~{}=~{}\frac{1}{t}\int_{0}^{t}h^{*}(s)ds.$

The set of all (equivalence classes of) measurable functions $h$ having finite
norm is known as Lorentz space. It is denoted by $L(p,q)(X,\mathcal{A},\mu)$
(or in the short form we write $L(p,q)$) where $1<p\leq\infty,~{}1\leq
q\leq\infty$. For $h\in L(p,q)$, the norm $\|h\|_{pq}$ is given by

$\displaystyle\|h\|_{pq}~{}=~{}\begin{cases}\left(\frac{q}{p}\bigintsss_{0}^{\infty}\big{(}t^{1/p}h^{**}(t)\big{)}^{q}~{}\frac{dt}{t}\right)^{1/q},&\
1<p<\infty,\ 1\leq q<\infty\\\ \underset{t>0}{\sup}\,\,\,t^{1/p}h^{**}(t),&\
1<p\leq\infty,\ q=\infty\end{cases}$

The Lorentz space $L(p,q)$ is a Banach space with respect to the norm
$\|\cdot\|_{pq}$. This norm is equivalent to quasi-norm given by

$\displaystyle\|h\|_{(pq)}~{}=~{}\begin{cases}\left(\frac{q}{p}\bigintsss_{0}^{\infty}\big{(}t^{1/p}h^{*}(t)\big{)}^{q}~{}\frac{dt}{t}\right)^{1/q},&\
1<p<\infty,\ 1\leq q<\infty\\\ \underset{t>0}{\sup}\,\,\,t^{1/p}h^{*}(t),&\
1<p\leq\infty,\ q=\infty\end{cases}.$

It is easy to note that the followings are equivalent for $h\in L(p,q)$.

  1. (i)

$\mu_{h}(0)=0$,

  2. (ii)

$h=0$ almost everywhere on $X$,

  3. (iii)

$\|h\|_{pq}=0$,

  4. (iv)

$\|h\|_{(pq)}=0$.

Further, for $p=q$, Lorentz spaces are equivalent to classical Lebesgue
spaces, that is, $L^{p}$ and $L(p,p)$ are equivalent where $1<p\leq\infty$.
For the details of Lorentz spaces, the readers are referred to [11, 13, 14].

Let $T:X\mapsto X$ be a mapping. For each $k\in\mathbb{N}\cup\\{0\\}$, we
define $T^{0}(A)=I(A)=A$, $\mu_{0}(A)=\mu(A)$ and

$\mu_{k}(A):=\mu\circ T^{-k}(A)~{}=~{}\mu\circ
T^{-(k-1)}(T^{-1}(A))~{}=~{}\mu\circ T^{-1}(T^{-(k-1)}(A))~{}~{}\text{for
}A\in\mathcal{A}.$

A mapping $T:X\mapsto X$ is said to be measurable transformation if
$T^{-1}(A)$ is measurable set whenever $A$ is measurable set. In addition, if
the condition $\mu(A)=0$ implies $\mu\circ T^{-1}(A)=0$ is satisfied then $T$
is called a non-singular measurable transformation. In this case, the measure
$\mu\circ T^{-1}$ is absolutely continuous with respect to the measure $\mu$
and we denote it as $\mu\circ T^{-1}\ll\mu$. Moreover, for every
$k\in\mathbb{N}\cup\\{0\\}$, $T^{k}:X\mapsto X$ is a non-singular measurable
tansformation. Thus, we have

$\displaystyle\cdots\ll\mu_{k+1}\ll\mu_{k}\ll~{}\mu_{k-1}\ll\cdots\ll\mu_{1}\ll\mu_{0}.$
(1)

Now, by Radon-Nikodym theorem, we get a locally integrable function
$h_{T^{k}}(\geq 0)$ on $X$ such that

$\displaystyle\mu_{k}(A)=\int_{A}h_{T^{k}}(x)\,d\mu(x),\mbox{ for all
}A\in\mathcal{A}.$ (2)

Let $T$ be a non-singular measurable transformation and $u$ be a complex-
valued measurable function on $X$. We define a mapping $W_{u,T}$ from a
Lorentz space $L(p,q)$ into the vector space of all complex-valued measurable
functions by

$W_{u,T}(f)=u\circ T\cdot f\circ T,~{}~{}f\in L(p,q).$

It can be easily verified that this mapping is a linear tansformation. If
$W_{u,T}$ is bounded with range in $L(p,q)$ then $W_{u,T}$ is known as
weighted composition operator on $L(p,q)$. On taking $T=I$ and $u=1$, we get
$W_{u,T}=M_{u}$ and $W_{u,T}=C_{T}$ which are, respectively, said to be a
multiplication and composition operator. In [1], the followings are proved:

  1. (i)

if $u$ and $h_{T}$ are essentially bounded (i.e. $u,h_{T}\in L_{\infty}(\mu)$)
then $W_{u,T}$ is bounded on $L(p,q)$,

  2. (ii)

if $W_{u,T}$ is bounded on $L(p,q)$ and $T(A_{\epsilon})\subseteq
A_{\epsilon}$ for each $\epsilon>0$, where
$A_{\epsilon}=\\{x:|u(x)|>\epsilon\\}$, then $u\in L_{\infty}(\mu)$.

Together $(i)$ and $(ii)$ imply the following.

######  Theorem 2.1.

Let $u$ be the complex valued measurable function and $T$ be the non-singular
and measurable transformation such that $h_{T}$ is essentially bounded
$(h_{T}\in L_{\infty}(\mu))$ and $T(A_{\epsilon})\subseteq A_{\epsilon}$ for
every $\epsilon>0$, where $A_{\epsilon}=\\{x:|u(x)|>\epsilon\\}$. Then, a
necessary and sufficient condition for $W_{u,T}$ to be bounded on $L(p,q)$
$1<p\leq\infty$, $1\leq q\leq\infty$ is that $u\in L_{\infty}(\mu)$.

Let $Y$ be a normed linear space and $L:Y\mapsto Y$ be a bounded operator. We
denote the null space and range space of $L$ by $\mathcal{N}(L)$ and
$\mathcal{R}(L)$ respectively. Also, note that, the symbol $\mathfrak{B}(Y)$
denotes the collection of all bounded operators on $Y$. For
$L\in\mathfrak{B}(Y)$, we define

  * •

the ascent of $L$ as

$\alpha(L):=\inf\\{k\in\mathbb{N}\cup\\{0\\}~{}|~{}\mathcal{N}(L^{k+1})=\mathcal{N}(L^{k})\\}.$

  * •

the descent of $L$ as

$\beta(L):=\inf\\{k\in\mathbb{N}\cup\\{0\\}~{}|~{}\mathcal{R}(L^{k+1})=\mathcal{R}(L^{k})\\}.$

If, in any case, the set is empty then that particular term (ascent or
descent) is equal to infinite. Several examples are given by A.E. Taylor in
his article [18].

##  3 Multiplication Operators

In this section, we discuss the ascent and descent of multiplication operator
$M_{u}$, induced by essentially bounded complex-valued function
$u:X\rightarrow\mathbb{C}$, defined on Lorentz space $L(p,q)$, $1<p\leq\infty$
and $1\leq q\leq\infty$. It is readily verified that the ascent and descent of
$M_{u}$ is equal to one, if $u=0$ almost everywhere (a.e.). We only left with
the case where $u\neq 0$ and the discussion results that if the ascent and
descent of $M_{u}$ are finite then these are either equal to zero or one.

######  Theorem 3.1.

Let $u\in L_{\infty}(\mu)$ be a function in such a way that the induce
multiplication operator $M_{u}$ on $L(p,q)$, $1<p\leq\infty$, $1\leq
q\leq\infty$ is well defined. If we assume that $u\neq 0$ on $X$, then ascent
of $M_{u}$ is either equal to zero or one.

###### Proof.

The condition $u\neq 0$ gives rise to two cases. In the first case, we assume
that $\mu(\\{x\in X~{}:~{}u(x)=0\\})=0$, that is, $u\neq 0$ a.e. In this case,
we find that $\mathcal{N}(M_{u})$, the null space of $M_{u}$, is $\\{0\\}$.
Let $f\in L(p,q)$ satisfying $M_{u}f=0$, which means that $\mu_{M_{u}f}(0)=0$,
hence we have

$\displaystyle\mu_{f}(0)$ $\displaystyle=\mu(\\{x\in X~{}|~{}|f(x)|>0\\})$
$\displaystyle=\mu(\\{x\in X~{}|~{}|u(x)|=0~{}\&~{}|f(x)|>0\\})+\mu(\\{x\in
X~{}|~{}|(M_{u}f)(x)|>0\\})$ $\displaystyle\leq\mu(\\{x\in
X~{}|~{}|u(x)|=0\\})+\mu_{M_{u}f}(0)=0,$

which implies $f=0$ almost everywhere on $X$. Thus we have
$\mathcal{N}(M_{u})=\\{0\\}$ and hence our claim $\alpha(M_{u})=0$ is
achieved.
In the second case, we assume that $\mu(\\{x\in X~{}:~{}u(x)=0\\})>0$. As the
measure $\mu$ is $\sigma-$finite, we can pick a non-zero finite measurable
subset $A$ of $\\{x\in X~{}|~{}u(x)=0\\}$. This provide a non-zero
characteristic function $\chi_{A}$ in $\mathcal{N}(M_{u})$. This results that
$\alpha(M_{u})>0$. Now, let $f\in\mathcal{N}(M_{u}^{2})$. Then
$u^{2}(x)f(x)=0$ a.e. on $X$ which implies that $\mu_{M_{u}^{2}f}(0)=0$. It
follows that

$\mu_{M_{u}f}(0)=\mu(\\{x\in X~{}|~{}|u(x)f(x)|>0\\})=\mu(\\{x\in
X~{}|~{}|u^{2}(x)f(x)|>0\\})=\mu_{M_{u}^{2}f}(0)=0.$

This shows that $f\in\mathcal{N}(M_{u})$. Hence,
$\mathcal{N}(M_{u})=\mathcal{N}(M_{u}^{2})$. This means, $\alpha(M_{u})=1$. As
a conclusion, we obtain that $\alpha(M_{u})$ is either $0$ or $1$. ∎

Now, we provide examples which illustrate the use of above theorem.

######  Example 3.2.

Consider the measure space $\left([0,1],\mathcal{A},\mu\right)$ and define
$u:[0,1]\rightarrow\mathbb{C}$ as $u(x)=x$. Then $M_{u}$ is a bounded operator
on Lorentz space $L(p,q)$, $1<p\leq\infty$, $1\leq q\leq\infty$ and $u\neq 0$
a.e. Hence by Theorem 3.1, $\alpha(M_{u})=0$.

######  Example 3.3.

Consider the measure space $\left([0,1],\mathcal{A},\mu\right)$ and define
$u(x)$ as $x$, when $\frac{1}{2}\leq x\leq 1$ and zero otherwise. Again,
$M_{u}$ is a bounded operator on Lorentz space $L(p,q)$, $1<p\leq\infty$,
$1\leq q\leq\infty$ and in this case we can say that $u\neq 0$ (but this time
it is not the case $u\neq 0$ a.e.). Hence by Theorem 3.1, $\alpha(M_{u})=1$.

In Example 3.2, we see that the ascent of multiplication operator $M_{u}$ is
equal to zero. Now, let us find its descent. Take $f\equiv 1$. Since $f$ lies
in $L(p,q)$ and $u\in L_{\infty}(\mu)$,
$g_{k}:=M_{u}^{k}f\in\mathcal{R}(M_{u}^{k})\subseteq L(p,q)$ for every $k\geq
0$. But, we can note that $g_{k}\notin\mathcal{R}(M_{u}^{k+1})$. If it is,
then we can find $h_{k}\in L(p,q)$ in such a way that
$g_{k}=M_{u}^{k+1}h_{k}$. This implies that $h_{k}(x)=1/x$ for every
$k\in\mathbb{N}\cup\\{0\\}$, where $x\in(0,1]$. This leads to a contradiction
as this $h_{k}$ does not lie in $L(p,q)$, indeed, we have
$\|h_{k}\|_{pq}=\infty$. Hence, we obtain that
$g_{k}\in\mathcal{R}(M_{u}^{k})\setminus\mathcal{R}(M_{u}^{k+1})$ for every
non-negative integer $k$. It follows that $\beta(M_{u})=\infty$.

The conclusion, that is $\beta(M_{u})=\infty$, remains the same on replacing
the measure space $\left([0,1],\mathcal{A},\mu\right)$ by
$\left((0,1),\mathcal{A},\mu\right)$ in Example 3.2. In Example 3.2, we notice
that the function $u$ is not bounded away from zero. This gives an idea to
think for the function $u$ which is bounded away from zero, and our next
result detects the descent of multiplication operator on Lorentz space induced
by such functions. Recall that, a complex valued function $u$ defined on $X$
is said to be bounded away from zero if $|u(x)|>\epsilon$ a.e. on $X$ for some
$\epsilon>0$.

######  Theorem 3.4.

If $u$ is such that it is bounded away from zero and inducing $M_{u}$ on
$L(p,q)$, $1<p\leq\infty$, $1\leq q\leq\infty$, then the descent of $M_{u}$ is
equal to zero.

###### Proof.

Let $h\in L(p,q)$. Define

$\displaystyle
g^{\prime}(x)=\left\\{\begin{array}[]{cl}\frac{h(x)}{u(x)}&\mbox{ if
}|u(x)|>\epsilon\\\ 0&\mbox{ if }|u(x)|\leq\epsilon\end{array}\right.$

Then $g^{\prime}\in L(p,q)$. Indeed, $h$ and $u$ are measurable and
$|u(x)|>\epsilon$ a.e. on $X$, for some $\epsilon>0$. Thus $g^{\prime}$ is
measurable function and, for every $s\geq 0$, we have

$\displaystyle\mu_{g^{\prime}}(s)=\mu(\\{x\in X~{}|~{}|g^{\prime}(x)|>s\\})$
$\displaystyle\leq\mu(\\{x\in
X~{}|~{}|h(x)|>s\epsilon\\})=\mu_{h}(s\epsilon).$

Thus, the non-increasing rearrangement and average function of $g^{\prime}$
satisfy that for $t>0$,

$(g^{\prime})^{*}(t)\leq\frac{1}{\epsilon}~{}h^{*}(t)\mbox{~{}~{}~{}~{} and
~{}~{}~{}~{}}(g^{\prime})^{**}(t)\leq\frac{1}{\epsilon}~{}h^{**}(t)$

respectively. As a consequence,
$\|g^{\prime}\|_{pq}\leq\frac{1}{\epsilon}\|h\|_{pq}$, $1<p\leq\infty$, $1\leq
q\leq\infty$. Since $\mu(\\{x\in X~{}|~{}M_{u}g^{\prime}(x)\neq
h(x)\\})\leq\mu(\\{x\in X~{}|~{}|u(x)|\leq\epsilon\\})=0$, so
$M_{u}g^{\prime}=h$ almost everywhere on $X$. As a result,
$h\in\mathcal{R}(M_{u})$. This shows that
$\mathcal{R}(M_{u})=L(p,q)=\mathcal{R}(M_{u}^{0})$. Thus, $\beta(M_{u})=0$. ∎

##  4 Weighted Composition Operators

Now we attempt to discuss the ascent and descent of weighted composition
operators $W_{u,T}$, which in particular, for $T\equiv I$, the identity
mapping, become multiplication operators and for $u\equiv 1$ become
composition operators. The section is divided into two subsections that
examine the ascent and descent of weighted composition operators on Lorentz
spaces, respectively. Let us define some useful notations which we use
throughout this section. For non-negative integer $k$, define $N_{0}=X$,
$N_{k}=\\{x\in X~{}|~{}u(T^{i}(x))\neq 0\text{ for all }1\leq i\leq k\\}$,
$X_{k}=\\{x\in X~{}|~{}h_{T^{k}}(x)=0\\}$ and $L(p,q)(X_{k})=\\{h\in
L(p,q):h=0\mbox{ a.e. on }X_{k}^{c}\\}$, where $X_{k}^{c}$ indicates the
complement of $X_{k}$.

###  4.1 Ascent

We devote this subsection to trace the ascent of weighted composition
operators $W_{u,T}:L(p,q)\mapsto L(p,q)$. We first recall that if mappings $u$
and $T$ are such that ($u\in L_{\infty}(\mu)$ and $T$ is non-singular with
$h_{T}\in L_{\infty}(\mu)$) they induce multiplication operator $M_{u}$ and
composition operator $C_{T}$ on $L(p,q)$ then they induce the weighted
composition operator $W_{u,T}$ on $L(p,q)$ as $W_{u,T}=C_{T}M_{u}$. But,
mappings $u$ and $T$ can induce a weighted composition operator $W_{u,T}$ on
$L(p,q)$ without being $M_{u}$ and $C_{T}$ on $L(p,q)$ are well defined as one
can verify that for $u=0$ almost everywhere (a.e.), $W_{u,T}(=0)$ is a well
defined operator on $L(p,q)$ for each choice of non-singular measurable
transformation $T$, and we can deduce that its ascent is equal to one. Now
onwards, our $u$ is always assumed to be a non-zero function, which we will
not express explicitly. In this sequel, we first observe the following.

######  Proposition 4.1.

If $u\in L_{\infty}(\mu)$ and $W_{u,T}\in\mathfrak{B}(L(p,q))$,
$1<p\leq\infty$, $1\leq q\leq\infty$, then
$L(p,q)(X_{k})\subseteq\mathcal{N}(W_{u,T}^{k})$ for each integer $k\geq 0$.

###### Proof.

For $k=0$, the result is straightforward, as $L(p,q)(X_{0})=\\{0\\}$. So, we
let $k$ be a natural number and $h$ be a member of $L(p,q)(X_{k})$. As $h=0$
a.e. on $X_{k}^{c}$, $|u(x)|$ is bounded above by $M^{*}(=\|u\|_{\infty})$ and
$T^{i}$ is non-singular for each $i\in\mathbb{N}$, we have, for $s\geq 0$,

$\displaystyle\mu_{W_{u,T}^{k}h}(s)$ $\displaystyle=\mu\left(\left\\{x\in
X~{}|~{}\left|\prod\limits_{i=1}^{k}u(T^{i}(x))h(T^{k}(x))\right|>s\right\\}\right)$
$\displaystyle\leq\mu\left(\left\\{x\in
X~{}|~{}\left|h(T^{k}(x))\right|>\frac{s}{(M^{*})^{k}}\right\\}\right)+\sum_{i=1}^{k}\mu\left(\left\\{x\in
X~{}|~{}\left|u\circ T^{i}(x)\right|>M^{*}\right\\}\right)$
$\displaystyle=\mu_{k}\left(\left\\{x\in
X~{}|~{}|h(x)|>\frac{s}{(M^{*})^{k}}\right\\}\right)+\sum_{i=1}^{k}\mu_{i}\left(\left\\{x\in
X~{}|~{}\left|u(x)\right|>M^{*}\right\\}\right)$
$\displaystyle=\int\limits_{\left\\{x\in
X~{}|~{}|h(x)|>\frac{s}{(M^{*})^{k}}\right\\}}h_{T^{k}}(x)\,d\mu(x)=0.$

As a result, $\|W_{u,T}^{k}h\|_{pq}=0$, $1<p\leq\infty$, $1\leq q\leq\infty$,
that is, $h\in\mathcal{N}(W_{u,T}^{k})$. ∎

The reverse relation in $L(p,q)(X_{k})\subseteq\mathcal{N}(W_{u,T}^{k})$ can
be obtained when $h_{T}=0$ a.e. on $(Supp(u))^{c}$, where the symbol $Supp(u)$
indicates the support of $u$ and $(Supp(u))^{c}=X\setminus Supp(u)$.

######  Proposition 4.2.

Let $u:X\mapsto\mathbb{C}$ and $T:X\mapsto X$ be the non-singular mappings
such that $W_{u,T}\in\mathfrak{B}(L(p,q))$, $1<p\leq\infty$, $1\leq
q\leq\infty$. If $h_{T}=0$ a.e. on $(Supp(u))^{c}$ then
$\mathcal{N}(W_{u,T}^{k})\subseteq L(p,q)(X_{k})$ for every non-negative
integer $k$.

###### Proof.

Evidently, the result is proved for $k=0$, as
$L(p,q)(X_{0})=\\{0\\}=\mathcal{N}(I)=\mathcal{N}(W_{u,T}^{0})$. Let $k\geq 1$
and $h\in\mathcal{N}(W_{u,T}^{k})$. Then

$\displaystyle\mu_{W_{u,T}^{k}h}(0)=\mu\big{(}\\{x\in
X~{}|~{}|(W_{u,T}^{k}h)(x)|>0\\}\big{)}=0.$

Now using the hypothesis $h_{T}=0$ a.e. on $(Supp(u))^{c}$ and equation (1),
we obtain that

$\displaystyle\mu_{h\circ T^{k}}(0)$ $\displaystyle=\mu\big{(}\\{x\in
X~{}:~{}|h\circ T^{k}(x)|>0\\}\big{)}$ $\displaystyle=\mu\bigg{(}\\{x\in
X:|h\circ T^{k}(x)|>0\\}\cap(\cap_{i=1}^{k}\\{x\in X:|u\circ
T^{i}(x)|>0\\})\bigg{)}$ $\displaystyle+\mu\bigg{(}\\{x\in X:|h\circ
T^{k}(x)|>0\\}\cap(\cup_{i=1}^{k}\\{x\in X:|u\circ T^{i}(x)|=0\\})\bigg{)}$
$\displaystyle\leq\mu_{W_{u,T}^{k}h}(0)+\sum_{i=1}^{k}\mu_{i}\big{(}\\{x\in
X:|u(x)|=0\\}\big{)}$ $\displaystyle=0.$

Thus $h\circ T^{k}=0$ almost everywhere on $X$. As a consequence, we have

$\displaystyle 0$ $\displaystyle=\mu_{k}\left(\\{x\in X~{}|~{}h(x)\neq
0\\}\right)$
$\displaystyle=\int\limits_{E}h_{T^{k}}(x)\,d\mu(x)+\int\limits_{F}h_{T^{k}}(x)\,d\mu(x)$
$\displaystyle\geq\frac{1}{n}\int_{F_{n}\cap F}d\mu=\frac{1}{n}\mu(F_{n}\cap
F)$

for each $n\in\mathbb{N}$, where $E=\\{x\in X_{k}~{}|~{}h(x)\neq 0\\}$,
$F=\\{x\in X\setminus X_{k}~{}|~{}h(x)\neq 0\\}$ and $F_{n}=\\{x\in X\setminus
X_{k}~{}|~{}h_{T^{k}}(x)>1/n\\}$. This indicates that $\mu(F_{n}\cap F)=0$ for
every $n$. As $F=\cup_{n=1}^{\infty}(F_{n}\cap F)$, we get $\mu(F)=0$.
Therefore, $h=0$ a.e. on $X\setminus X_{k}$. Thus,
$\mathcal{N}(W_{u,T}^{k})\subseteq L(p,q)(X_{k})$. ∎

With the help of next example we show that the condition $h_{T}=0$ a.e. on
$(Supp(u))^{c}$ cannot be relaxed in the Proposition 4.2. For $k=0$, we always
have $L(p,q)(X_{0})=\mathcal{N}(W_{u,T}^{0})=\mathcal{N}(I)$ and hence the
example provides the mappings $u$ and $T$ to verify that on relaxing the
condition that $h_{T}=0$ a.e. on $(Supp(u))^{c}$, we have
$\mathcal{N}(W_{u,T}^{k})\not\subseteq L(p,q)(X_{k})$ for all
$k\in\mathbb{N}$.

######  Example 4.3.

Let $(\mathbb{N},2^{\mathbb{N}},\mu)$ be the measure space. Define
$T:\mathbb{N}\rightarrow\mathbb{N}$ and $u:\mathbb{N}\rightarrow\mathbb{C}$ as

$T(n)=\left\\{\begin{array}[]{cl}1&\mbox{if }n=1,2\\\
n-1&\mbox{otherwise}\end{array}\right.~{}~{}~{}\mbox{ and
}~{}~{}~{}u(n)=\left\\{\begin{array}[]{cl}0&\mbox{if }n=1\\\
1&\mbox{otherwise}\end{array}\right..$

By simple calculations we obtain that the Radon-Nikodym derivative $h_{T^{k}}$
is given by

$h_{T^{k}}(n)=\left\\{\begin{array}[]{cl}k+1&\mbox{if }n=1\\\ 1&\mbox{if
}n\geq 2\end{array}.\right.$

Further, we see that $L(p,q)(X_{k})=\\{0\\}$ and
$\chi_{\\{1\\}}\in\mathcal{N}(W_{u,T}^{k})$ for every $k\in\mathbb{N}$.
Consequently, $\mathcal{N}(W_{u,T}^{k})\not\subseteq L(p,q)(X_{k})$ for every
natural number $k$.

As a result of the previous two propositions, we obtain that the null space of
$W_{u,T}^{k}$ is the collection of all complex valued measurable functions on
$X$ having value zero almost everywhere on $X_{k}^{c}$.

######  Proposition 4.4.

Assume that a mapping $u\in L_{\infty}(\mu)$ and a non-singular measurable
transformation $T:X\rightarrow X$ induce the weighted composition operator
$W_{u,T}$ on Lorentz space $L(p,q)$, $1<p\leq\infty$, $1\leq q\leq\infty$. If
$h_{T}=0$ a.e. on $(Supp(u))^{c}$ then
$L(p,q)(X_{k})=\mathcal{N}(W_{u,T}^{k})$ for every
$k\in\mathbb{N}\cup\\{0\\}$.

The measures $\mu$ and $\nu$ on the measureable space $(X,\mathcal{A})$ are
said to be equivalent measures if $\mu\ll\nu$ and $\nu\ll\mu$. The forthcoming
theorem generates the relationship between the equivalency of null spaces and
measures in the following form.

######  Lemma 4.5.

Let $W_{u,T}\in\mathfrak{B}(L(p,q))$, $1<p\leq\infty$, $1\leq q\leq\infty$,
where $u\in L_{\infty}(\mu)$ and $h_{T}=0$ a.e. on $(Supp(u))^{c}$. A
necessary and sufficient condition for
$\mathcal{N}(W_{u,T}^{k})=\mathcal{N}(W_{u,T}^{k+1})$ is that the measures
$\mu_{k}$ and $\mu_{k+1}$ are equivalent for every integer $k\geq 0$.

###### Proof.

Assume that $\mu_{k}\ll\mu_{k+1}\ll\mu_{k}$ for each non-negative integer $k$.
Let $h\in\mathcal{N}(W_{u,T}^{k+1})$. Then, by following the same type of
calculations as done in Proposition 4.2, we get that $h\circ T^{k+1}=0$ almost
everywhere on $X$. Hence, by our assumption, $\mu_{k}(\\{x\in X~{}:~{}h(x)\neq
0\\})=0$. This yields that

$\displaystyle\mu_{W_{u,T}^{k}h}(0)$ $\displaystyle=\mu\big{(}\\{x\in
X:|h\circ T^{k}(x)|>0\\}\cap(\cap_{i=1}^{k}\\{x\in X:|u\circ
T^{i}(x)|>0\\})\big{)}$ $\displaystyle\leq\mu_{k}(\\{x\in
X~{}:~{}|h(x)|>0\\})=0.$ (3)

This provides that $\|W_{u,T}^{k}h\|_{pq}=0$. In other words,
$h\in\mathcal{N}(W_{u,T}^{k})$. Consequently, we get
$\mathcal{N}(W_{u,T}^{k})=\mathcal{N}(W_{u,T}^{k+1})$ for all
$k\in\mathbb{N}\cup\\{0\\}$.

For the converse part, we suppose that the condition
$\mathcal{N}(W_{u,T}^{k})=\mathcal{N}(W_{u,T}^{k+1})$ holds for each non-
negative integer $k$. Let $\mu_{k+1}(A)=0$ for any measurable set $A$. Then on
replacing $h$ by $\chi_{A}$ in equation (4.1), it provides that
$\chi_{A}\in\mathcal{N}(W_{u,T}^{k+1})$, and thus,
$\chi_{A}\in\mathcal{N}(W_{u,T}^{k})$. Again, by the ideas used in
calculations in Proposition 4.2, we deduce that $\mu_{k}(A)=0$. This shows
that $\mu_{k}\ll\mu_{k+1}$. However, $\mu_{k+1}\ll\mu_{k}$ is always true.
This completes the proof. ∎

In Lemma 4.5, if the search of non-negative finite integer $k$ is over for
which $\mu_{k}\ll\mu_{k+1}\ll\mu_{k}$ then smallest such $k$ is equal to the
ascent of a bounded weighted composition operator $W_{u,T}$ on $L(p,q)$,
$1<p\leq\infty$, $1\leq q\leq\infty$. This is expressed in the following form.

######  Theorem 4.6.

Suppose that a pair of a non-singular transformation $T:X\mapsto X$ and a
mapping $u:X\mapsto\mathbb{C}$ generates a bounded operator $W_{u,T}$ on
$L(p,q)$, $1<p\leq\infty$, $1\leq q\leq\infty$. If $u\in L_{\infty}(\mu)$ and
$h_{T}=0$ a.e. on $(Supp(u))^{c}$, then $\alpha(W_{u,T})=k$ if and only if $k$
is the least non-negative integer such that $\mu_{k}$ and $\mu_{k+1}$ are
equivalent measures.

Lemma 2.1 and Theorem 2.2 of [2] discuss the ascent of composition operators
on Lorentz spaces. When we combine these results with the findings of
Proposition 4.4, Lemma 4.5 and Theorem 4.6, we finally arrive to the following
corollaries:

######  Corollary 4.7.

Let $u$ and $T$ be such that $C_{T},W_{u,T}\in\mathfrak{B}(L(p,q))$,
$1<p\leq\infty$, $1\leq q\leq\infty$. If $u\in L_{\infty}(\mu)$ and $h_{T}=0$
a.e. on $(Supp(u))^{c}$ then
$\mathcal{N}(C_{T}^{k})=L(p,q)(X_{k})=\mathcal{N}(W_{u,T}^{k})$ for all
$k\in\mathbb{N}\cup\\{0\\}$.

######  Corollary 4.8.

Let $C_{T},W_{u,T}\in\mathfrak{B}(L(p,q))$, where $u\in L_{\infty}(\mu)$ and
$h_{T}=0$ a.e. on $(Supp(u))^{c}$. Then the following are equivalent
conditions:

  1. (a)

$\mu_{k+1}\ll\mu_{k}\ll\mu_{k+1}$

  2. (b)

$\mathcal{N}(C_{T}^{k})=\mathcal{N}(C_{T}^{k+1})$

  3. (c)

$\mathcal{N}(W_{u,T}^{k})=\mathcal{N}(W_{u,T}^{k+1})$

######  Corollary 4.9.

Assume that the non-singular measurable transformation $T:X\rightarrow X$ and
the mapping $u:X\rightarrow\mathbb{C}$ generates the bounded operators $C_{T}$
and $W_{u,T}$ on $L(p,q)$, $1<p\leq\infty$, $1\leq q\leq\infty$. Further, let
$u\in L_{\infty}(\mu)$ and $h_{T}=0$ a.e. on $(Supp(u))^{c}$. Then the
following are equivalent:

  1. (a)

$\alpha(C_{T})$, the ascent of $C_{T}$, is $k$

  2. (b)

$\alpha(W_{u,T})$, the ascent of $W_{u,T}$, is $k$

  3. (c)

$k$ is the least non-negative integer such that the measures $\mu_{k}$ and
$\mu_{k+1}$ are equivalent.

Next, we discuss conditions which assure the equivalency of the measures
$\mu_{k}$ and $\mu_{k+1}$ so that the ascent of weighted composition operators
on Lorentz spaces is finite. Now we write $X_{k}^{c}\subseteq
T^{k+1}(X_{k}^{c})$ in the sense that $T^{k+1}(X_{k}^{c})$ contains almost
every element of $X_{k}^{c}$, i.e. $\mu(X_{k}^{c}\setminus
T^{k+1}(X_{k}^{c}))=0$, then we have the following.

######  Theorem 4.10.

Suppose that the pair $(u,T)$ induces an operator
$W_{u,T}\in\mathfrak{B}(L(p,q))$, $1<p\leq\infty$, $1\leq q\leq\infty$. If
$\mu\circ T(A)\leq\mu(A)$ for each $A\in\mathcal{A}$ and $X_{k}^{c}=X\setminus
X_{k}\subseteq T^{k+1}(X_{k}^{c})$ for some $k\in\mathbb{N}\cup\\{0\\}$ then
the measures $\mu_{k}$ and $\mu_{k+1}$ are equivalent. In fact,
$\mu_{m}\ll\mu_{k}\ll\mu_{m}$ for all $m\geq k$.

###### Proof.

The goal is to show that $\mu_{k}\ll\mu_{k+1}$, since $\mu_{k+1}\ll\mu_{k}$ is
evident. Let $A$ be a measurable set satisfying $\mu_{k+1}(A)=0$. Then
$\mu(A_{1})=0$, where $A_{1}=\\{x\in A~{}|~{}h_{T^{k+1}}(x)>0\\}$. Take
$A_{2}=\\{x\in A~{}|~{}h_{T^{k}}(x)>0\text{ and }h_{T^{k+1}}(x)=0\\}$. Assume
$\mu(A_{2})>0$. Then the condition $\sigma-$finite on the measure $\mu$
provides a non-zero finite measurable subset $B$ of $A_{2}$. By the given
hypothesis, we have $B=T^{k+1}(T^{-(k+1)}(B))$, as $A_{2}\subseteq X_{k}^{c}$,
which yields $\mu(B)\leq\mu(T^{-(k+1)}(B))=\int_{B}h_{T^{k+1}}(x)\,d\mu(x)=0$,
a contradiction. Hence, the assumption $\mu(A_{2})>0$ is not true. Thus, we
obtain that $\mu(A_{2})=0$. This further indicates that $\mu_{k}(A)=0$. Hence,
we are done. Further, by applying the technique of principle of mathematical
induction, we can easily see that $\mu_{m}\ll\mu_{k}\ll\mu_{m}$ for all $m\geq
k$. ∎

As the consequence of the Theorem 4.6 and Theorem 4.10, we deduce the
following.

######  Theorem 4.11.

Let $T:X\mapsto X$ be a non-singular measurable transformation and $u$ be a
complex valued measurable function defined on $X$ such that
$W_{u,T}\in\mathfrak{B}(L(p,q))$, $1<p\leq\infty$, $1\leq q\leq\infty$ and
$\mu\circ T(A)\leq\mu(A)$ for each $A\in\mathcal{A}$. Also, let the Radon-
Nikodym derivative $h_{T}=0$ a.e. on $(Supp(u))^{c}$ and $u\in
L_{\infty}(\mu)$. Then, a sufficient condition for $W_{u,T}$ to have ascent a
non-negative integer $k$ is that $k$ is the smallest integer such that
$X_{k}^{c}\subseteq T^{k+1}(X_{k}^{c})$, where $X_{k}^{c}=X\setminus X_{k}$.

###### Proof.

By Theorem 4.6 and Theorem 4.10, we get that the ascent of $W_{u,T}$ is atmost
$k$. Now assume that $k$ is the least non-negative integer satisfying
$X_{k}^{c}\subseteq T^{k+1}(X_{k}^{c})$. To show $\alpha(W_{u,T})=k$, we claim
that $k$ is the least non-negative integer such that the measures $\mu_{k}$
and $\mu_{k+1}$ are equivalent. On contrary, assume that $\mu_{p}$ and
$\mu_{p+1}$ are equivalent measures for some non-negative integer $p$ strictly
less than $k$. Using the equation (1), we have $\mu_{p}\ll\mu_{m}\ll\mu$ and
$\mu_{m}\ll\mu_{p}\ll\mu$ for every $m\geq p$. Thus, by the chain rule,

$\displaystyle h_{T^{p}}(x)$ $\displaystyle=\frac{d\mu_{p}}{d\mu_{m}}(x)\cdot
h_{T^{m}}(x),$ $\displaystyle h_{T^{m}}(x)$
$\displaystyle=\frac{d\mu_{m}}{d\mu_{p}}(x)\cdot h_{T^{p}}(x).$

As a result, we obtain $X_{p}=X_{m}$ which further implies that
$X_{p}^{c}=X_{m}^{c}$. In particular, for $m=k$ and by given hypothesis, we
obtain that $X_{p}^{c}\subseteq T^{k+1}(X_{p}^{c})$. Our target is to show
that the condition $X_{p}^{c}\subseteq T^{p+1}(X_{p}^{c})$ holds, which yields
the contradiction as $k$ is the least integer satisfying that condition.
Suppose that $\mu\big{(}X_{p}^{c}\setminus T^{p+1}(X_{p}^{c})\big{)}>0$. Since
$\mu$ is $\sigma-$finite, there exists a measurable subset $C$ of
$X_{p}^{c}\setminus T^{p+1}(X_{p}^{c})$ having non-zero finite measure. The
structure of $C$ indicates that every element $y\in C$ is of type
$T^{k+1}(x)=T^{p+1}(T^{k-p}(x))$ where $T^{k-p}(x)\in X_{p}$ and $x\in
X_{p}^{c}$, in other words, $y=T^{p+1}(z)$ for some $z\in X_{p}$. Take
$B=T^{-(p+1)}(C)\cap X_{p}$. Then $B$ is measurable set with $\mu_{p}(B)=0$.
This provides

$\chi_{C}\circ T^{p+1}(x)=\left\\{\begin{array}[]{cc}1&x\in B\\\
0&\text{elsewhere}\end{array}\right.\leq~{}\chi_{B}(x).$

Therefore, $\chi_{C}\circ T^{p+1}(T^{p+1}(x))\leq\chi_{B}\circ T^{p+1}(x)$ for
every $x\in X$. We know that for each measurable set $A\in\mathcal{A}$

$\displaystyle\|\chi_{A}\|_{pq}=\left\\{\begin{array}[]{cc}(p^{\prime})^{1/q}(\mu(A))^{1/p},&1<p<\infty,1\leq
q<\infty,\\\ (\mu(A))^{1/p},&1<p\leq\infty,q=\infty,\end{array}\right.$

where $1/p+1/p^{\prime}=1$. Finally, by using the relation
$\|\chi_{T^{-(2p+2)}(C)}\|_{pq}\leq\|\chi_{T^{-(p+1)}(B)}\|_{pq}$, we get
$\mu_{2p+2}(C)=0$, and hence, $\mu_{p}(C)=0$. This yields $\mu(C)=0$, a
contradiction. This implies that $X_{p}^{c}\subseteq T^{p+1}(X_{p}^{c})$. This
contradicts the fact that $k$ is the least non-negative integer satisfying
$X_{k}^{c}\subseteq T^{k+1}(X_{k}^{c})$. This proves that $k$ is the least
non-negative integer such that measures $\mu_{k}$ and $\mu_{k+1}$ are
equivalent. This proves that the ascent of $W_{u,T}$ is equal to $k$. ∎

Before we proceed, it is to be noted that the assumption $h_{T}=0$ a.e. on
$(Supp(u))^{c}$, in the Theorem 4.11, can not be dropped. This we can see
through the following example.

######  Example 4.12.

On the measure space $(\mathbb{N},2^{\mathbb{N}},\mu)$, if we define
$T:\mathbb{N}\rightarrow\mathbb{N}$ and $u:\mathbb{N}\rightarrow\mathbb{C}$ as

$T(n)=\left\\{\begin{array}[]{cl}1&\mbox{ if }n=1\\\ 2&\mbox{ if }n=3\\\
n-2&\mbox{ if $n$ is odd and }n\geq 5\\\ n+2&\mbox{ if $n$ is even
}\end{array}\right.~{}~{}~{}~{}~{}\mbox{ and
}~{}~{}~{}~{}u(n)=\left\\{\begin{array}[]{cl}0&\mbox{ if $n$ is odd}\\\
1&\mbox{ if $n$ is even}\end{array}\right.$

then all the conditions of Theorem 4.11 are satisfied except the condition
$h_{T}=0$ a.e. on $(Supp(u))^{c}$. Under these choices of $u$ and $T$, we
obtain that the ascent of $W_{u,T}$ is infinite, as
$\chi_{\\{1\\}}\in\mathcal{N}(W_{u,T})\setminus\\{0\\}$ and
$\chi_{\\{2k\\}}\in\mathcal{N}(W_{u,T}^{k+1})\setminus\mathcal{N}(W_{u,T}^{k})$
for $k\in\mathbb{N}$.

Now, our next result shows the existence of weighted composition operators
$W_{u,T}:L(p,q)\mapsto L(p,q)$, $1<p\leq\infty$, $1\leq q\leq\infty$, having
ascent infinity.

######  Theorem 4.13.

Let $W_{u,T}\in\mathfrak{B}(L(p,q))$, $1<p\leq\infty,1\leq q\leq\infty$. If
there is a sequence $\left<A_{k}\right>_{k=0}^{\infty}$ of measurable sets in
such a way that measure of

  1. (a)

$A_{k}$ is non-zero and finite

  2. (b)

$N_{k}\cap T^{-k}(A_{k})$ is not equal to zero

  3. (c)

$N_{k+1}\cap T^{-(k+1)}(A_{k})$ is equal to zero

for every $k\geq 0$, then the ascent of $W_{u,T}$ can not be finite.

###### Proof.

Suppose that there exists a sequence $\left<A_{k}\right>_{k=0}^{\infty}$ of
measurable sets satisfying the properties $(a),(b)$ and $(c)$. In order to
show that the ascent of $W_{u,T}$ is infinite, we aim that the characteristic
function $\chi_{A_{k}}$ lies in
$\mathcal{N}(W_{u,T}^{k+1})\setminus\mathcal{N}(W_{u,T}^{k})$ for all $k\geq
0$. For $s\geq 0$

$\displaystyle\mu_{W_{u,T}^{k+1}\chi_{A_{k}}}(s)$
$\displaystyle=\mu\left(\left\\{x\in X~{}|~{}\left|\prod_{i=1}^{k+1}u\circ
T^{i}(x)\cdot\chi_{A_{k}}\circ T^{k+1}(x)\right|>s\right\\}\right)$
$\displaystyle\leq\mu(N_{k+1}\cap T^{-(k+1)}(A_{k}))=0.$

It follows that non-increasing rearrangement of $W_{u,T}^{k+1}\chi_{A_{k}}$ is
equal to zero, and as a conclusion,
$\left\|W_{u,T}^{k+1}\chi_{A_{k}}\right\|_{pq}=0$. Thus,
$\chi_{A_{k}}\in\mathcal{N}(W_{u,T}^{k+1})$. Finally, the justification of
$\chi_{A_{k}}\not\in\mathcal{N}(W_{u,T}^{k})$ is trival. In fact, if we
consider $\chi_{A_{k}}\in\mathcal{N}(W_{u,T}^{k})$ then it contradicts the
property $(b)$, as $N_{k}\cap T^{-k}(A_{k})=\\{x\in
X:|W_{u,T}^{k}\chi_{A_{k}}(x)|>0\\}$. ∎

Our next result, the proof of which follows the lines of proof of Theorem 2.10
of [2] and hence we just state the result.

######  Theorem 4.14.

Assume that the image of measurable set under the non-singular measurable
transformation $T:X\mapsto X$ is also measurable. Then
$\alpha(W_{u,T})=\infty$ implies the existence of measurable sets sequence
$\left<A_{k}\right>_{k=0}^{\infty}$ of $X$ in such a way that for all $k\geq
0$ the following properties hold

  1. (a)

$0<\mu(A_{k})<\infty$

  2. (b)

$A_{k}\subseteq T^{k}(B\cap N_{k})$ for some $B\in\mathcal{A}$

  3. (c)

$A_{k}\not\in\left\\{T^{k+1}(D\cap N_{k+1})~{}|~{}D\in\mathcal{A}\text{ and
}\mu(D\cap N_{k+1})>0\right\\}$

###### Proof.

The proof follows from the proof of [2, Theorem 2.10] on replacing
$X_{k}^{\prime}$ by $N_{k}\cap B$, where $B=\\{x\in X:h_{k}\circ T^{k}(x)\neq
0\\}$. ∎

###  4.2 Descent

This subsection deals with results on the descent of the weighted composition
operators $W_{u,T}\in\mathfrak{B}(L(p,q))$, $1<p\leq\infty$, $1\leq
q\leq\infty$. To develop the theory on descent, we begin by introducing
necessary notations, definitions and lemmas. Following this groundwork, we
delve into the main results of this section.

######  Definition.

A measurable transformation $T:X\rightarrow X$ is known as injective almost
everywhere on $X$ if $\mathcal{E}$ is a measurable set with measure equal to
zero where $\mathcal{E}:=\\{x\in X~{}|~{}\exists~{}y\in X,y\neq x\text{ and
}T(x)=T(y)\\}$.

Let $k\in\mathbb{N}\cup\\{0\\}$ be a fixed number. Define
$\mathcal{E}_{k}:=\\{x\in T^{k}(N_{k+1})~{}|~{}\exists~{}y\in
T^{k}(N_{k+1}),y\neq x\text{ and }T(y)=T(x)\\}$. The sets $\mathcal{E}_{k}$
satisfy the relationship
$\cdots\subseteq\mathcal{E}_{k+1}\subseteq\mathcal{E}_{k}\subseteq\mathcal{E}_{k-1}\subseteq\cdots\subseteq\mathcal{E}_{1}\subseteq\mathcal{E}_{0}\subseteq\mathcal{E}$.
This can be verified using the following lemma.

######  Lemma 4.15.

For every $k\in\mathbb{N}\cup\\{0\\}$, $T^{k+1}(N_{k+2})\subseteq
T^{k}(N_{k+1})$.

###### Proof.

Let $x\in T^{k+1}(N_{k+2})$. Then we can select $y_{x}$ in $N_{k+2}$
satisfying $x=T^{k+1}(y_{x})=T^{k}(T(y_{x}))$. This results that $x\in
T^{k}(N_{k+1})$, as $u\circ T^{i}(T(y_{x}))=u\circ T^{i+1}(y_{x})\neq 0$ for
all $1\leq i\leq k+1$ because of $y_{x}\in N_{k+2}$. ∎

Corresponding to each $x\in\mathcal{E}_{k}$, we define
$A_{x}^{k}:=\\{y\in\mathcal{E}_{k}~{}|~{}T(x)=T(y)\\}$. In the next lemma, we
demonstrate that for distinct $x_{1}$ and $x_{2}$ within $\mathcal{E}_{k}$,
the sets $A_{x_{1}}^{k}$ and $A_{x_{2}}^{k}$ are either disjoint or identical.

######  Lemma 4.16.

For distinct $x_{1},x_{2}\in\mathcal{E}_{k}$, either $A_{x_{1}}^{k}\cap
A_{x_{2}}^{k}=\emptyset$ or $A_{x_{1}}^{k}=A_{x_{2}}^{k}$.

###### Proof.

If $A_{x_{1}}^{k}\cap A_{x_{2}}^{k}=\emptyset$ then we are done. Suppose that
$z\in A_{x_{1}}^{k}\cap A_{x_{2}}^{k}$. This implies $T(z)=T(x_{1})=T(x_{2})$,
and as a result, $x_{2}\in A_{x_{1}}^{k}$ and $x_{1}\in A_{x_{2}}^{k}$. Hence,
$A_{x_{1}}^{k}=A_{x_{2}}^{k}$. Indeed, if $y\in A_{x_{1}}^{k}$ then
$T(y)=T(x_{1})$ which further gives $T(y)=T(x_{2})$. Therefore, $y\in
A_{x_{2}}^{k}$. Similarly, we get the other side of containment. ∎

By the definition of injective almost everywhere on $X$, it is evident that if
the measure of $A_{x}^{k}$ is non-zero for some $x\in\mathcal{E}_{k}$ then the
mapping $T:T^{k}(N_{k+1})\rightarrow T^{k+1}(N_{k+1})~{}(\subseteq
T^{k}(N_{k+1}))$ is not injective almost everywhere on $T^{k}(N_{k+1})$.
Consequently, $T:X\rightarrow X$ is not injective almost everywhere on $X$.
These sets $A_{x}^{k}$ play an important role in ensuring the infinity descent
of $W_{u,T}$. To accomplish this, we first introduce separable measurable
sets.

######  Definition.

A measurable set $E$ of $X$ is said to be separable measurable set if we can
write $E$ as the union of two disjoint positive measurable subsets of $E$,
that is, there are $E_{1}\subseteq E$ and $E_{2}\subseteq E$ such that
$\mu(E_{1})>0$, $\mu(E_{2})>0$, $E_{1}\cap E_{2}=\emptyset$ and $E=E_{1}\cup
E_{2}$.

######  Lemma 4.17.

Let $(X,\mathcal{A},\mu)$ be a $\sigma-$finite and complete measure space.
Suppose that there exists atleast one separable subset $A_{x}^{k}$ of
$\mathcal{E}_{k}$ satisfying $\mu(N_{k+1}\cap T^{-k}(B))>0$ for all positive
measurable subset $B$ of $A_{x}^{k}$ where $k\in\mathbb{N}\cup\\{0\\}$. Then
$\beta(W_{u,T})=\infty$.

###### Proof.

Let $k\in\mathbb{N}\cup\\{0\\}$ be a fixed number. Since $A_{x}^{k}$ is
separable subset of $\mathcal{E}_{k}$, we can find two disjoint positive
measurable subsets $A_{1}$ and $A_{2}$ of $A_{x}^{k}$ in such a way that
$A_{x}^{k}=A_{1}\cup A_{2}$. Due to $\sigma-$finite, we can take
$B_{1}~{}(\subseteq A_{1})$ and $B_{2}~{}(\subseteq A_{2})$ such that
$0<\mu(B_{1}),\mu(B_{2})<\infty$, $B_{1}\cap B_{2}=\emptyset$ and
$T^{-k}(B_{1})\cap T^{-k}(B_{2})=\emptyset$. Define
$f:=\chi_{B_{1}}-\chi_{B_{2}}$. Obviously, $f\in L(p,q)$, and as a result,
$g:=W_{u,T}^{k}f\in\mathcal{R}(W_{u,T}^{k})$. Now, we assert that
$g\not\in\mathcal{R}(W_{u.T}^{k+1})$. On the contrary, we assume that
$g=W_{u,T}^{k+1}h$ for some $h\in L(p,q)$, that is, $\mu\big{(}\\{x\in
X~{}|~{}g(x)\neq W_{u,T}^{k+1}h(x)\\}\big{)}=0$. Since the measure of
$N_{k+1}\cap T^{-k}(B_{1})$ and $N_{k+1}\cap T^{-k}(B_{2})$ are positive, we
can pick $z_{1}\in N_{k+1}\cap T^{-k}(B_{1})$ and $z_{2}\in N_{k+1}\cap
T^{-k}(B_{2})$ satisfying $g(z_{1})=W_{u,T}^{k+1}h(z_{1})$ and
$g(z_{2})=W_{u,T}^{k+1}h(z_{2})$ respectively. This provides $1=u\circ
T^{k+1}(z_{1})\cdot h\circ T^{k+1}(z_{1})$ and $-1=u\circ T^{k+1}(z_{2})\cdot
h\circ T^{k+1}(z_{2})$ which further leads to a contradiction because
$T^{k}(z_{1}),T^{k}(z_{2})\in A_{x}^{k}$. ∎

######  Theorem 4.18.

Suppose that $(X,\mathcal{A},\mu)$ is a $\sigma-$finite complete measure space
and $T:X\rightarrow X$ is a non-singular measurable transformation satisfying
the property $\mu\circ T(E)\leq\mu(E)$ for every $E\in\mathcal{A}$. If there
is atleast one separable subset $A_{x}^{k}$ of $\mathcal{E}_{k}$ for every
$k\in\mathbb{N}\cup\\{0\\}$ then the descent of $W_{u,T}$ is infinity.

###### Proof.

In order to proof this we only need to show that $\mu(N_{k+1}\cap
T^{-k}(B))>0$ for all positive measurable subset $B$ of
$A_{x}^{k}~{}(\subseteq T^{k}(N_{k+1}))$ where $k\in\mathbb{N}\cup\\{0\\}$. To
attain the target we claim that $T^{k}(N_{k+1}\cap T^{-k}(B))=B$. If $x\in
T^{k}(N_{k+1}\cap T^{-k}(B))$ then $x=T^{k}(y)$ for some $y\in N_{k+1}\cap
T^{-k}(B)$. This implies $x=T^{k}(y)\in T^{k}(N_{k+1})\cap B=B$. Conversely,
if $x\in B$ then $x=T^{k}(y)$ for some $y\in N_{k+1}$. This provides $y\in
N_{k+1}\cap T^{-k}(B)$, and as a consequence, we have $x=T^{k}(y)\in
T^{k}(N_{k+1}\cap T^{-k}(B))$. Hence, $T^{k}(N_{k+1}\cap T^{-k}(B))=B$. To
complete the proof, if possible assume that $\mu(N_{k+1}\cap T^{-k}(B))=0$
whenever $\mu(B)>0$, then using the property of $T$, we achieve
$\mu(B)=\mu(T^{k}(N_{k+1}\cap T^{-k}(B)))=0$, a contradiction. Hence, we are
done with the claim, and the result follows from Lemma 4.17. ∎

In the upcoming theorem, we investigate another sufficient condition that
helps in attaining $\beta(W_{u,T})=\infty$. The Lemma 4.16 divides
$\mathcal{E}_{k}$ as union of disjoint $A_{x}^{k}$ where $x$ belongs to
$\mathcal{E}_{k}$. Now, from each disjoint $A_{x}^{k}$ we pick $x\in
A_{x}^{k}$, and we denote the collection of all such $x$ by
$\mathcal{E}_{1}^{k}$, and $\mathcal{E}_{k}\setminus\mathcal{E}_{1}^{k}$ by
$\mathcal{E}_{2}^{k}$.

######  Theorem 4.19.

Let $T$ be a non-singular measurable transformation on $\sigma-$finite and
complete measure space $(X,\mathcal{A},\mu)$ such that $\mu\circ
T(A)\leq\mu(A)$ for each $A\in\mathcal{A}$. Assume that there are measurable
subsets $A_{1}^{k}~{}(\subseteq\mathcal{E}_{1}^{k})$ and
$A_{2}^{k}~{}(\subseteq\mathcal{E}_{2}^{k})$ with non-zero finite measure
satisfying $T_{k}(A_{1}^{k})=T_{k}(A_{2}^{k})$ where
$T_{k}=T:T^{k}(N_{k+1})\rightarrow T^{k+1}(N_{k+1})$ for each
$k\in\mathbb{N}\cup\\{0\\}$. Then the operator $W_{u,T}$ on $L(p,q)$ has
infinity descent.

###### Proof.

Let $k\in\mathbb{N}\cup\\{0\\}$ be a fixed number such that the conditions
given in the statement of the theorem hold, that is, there are measurable
subsets $A_{1}^{k}~{}(\subseteq\mathcal{E}_{1}^{k})$ and
$A_{2}^{k}~{}(\subseteq\mathcal{E}_{2}^{k})$ in such a way
$0<\mu(A_{1}^{k}),\mu(A_{2}^{k})<\infty$ and
$T_{k}(A_{1}^{k})=T_{k}(A_{2}^{k})$ where $T_{k}=T:T^{k}(N_{k+1})\rightarrow
T^{k+1}(N_{k+1})$. Define $f:=\chi_{A_{1}^{k}}-\chi_{A_{2}^{k}}$ and
$g:=W_{u,T}^{k}f$. Both are non-zero and $g\in\mathcal{R}(W_{u,T}^{k})$. Now,
we show that $g$ does not belong to $\mathcal{R}(W_{u,T}^{k+1})$. On the
contrary, let us assume that $g=W_{u,T}^{k+1}h$ (a.e. on $X$) for some $h\in
L(p,q)$. Let us write

$N_{k+1}\cap T^{-k}(A_{1}^{k})=A_{11}^{k}\cup A_{12}^{k}\text{~{}~{}and
~{}~{}}N_{k+1}\cap T^{-k}(A_{2}^{k})=A_{21}^{k}\cup A_{22}^{k}$

where $A_{i1}^{k}=\\{x\in N_{k+1}\cap T^{-k}(A_{i}^{k})~{}|~{}g(x)\neq
W_{u,T}^{k+1}h(x)\\}$ and $A_{i2}^{k}=\\{x\in N_{k+1}\cap
T^{-k}(A_{i}^{k})~{}|~{}g(x)=W_{u,T}^{k+1}h(x)\\}=\\{x\in N_{k+1}\cap
T^{-k}(A_{i}^{k})~{}|~{}(-1)^{i-1}=u\circ T^{k+1}(x)h\circ T^{k+1}(x)\\}$ for
$i=1,2$. It can also be verified, using the idea of the proof of Theorem 4.18,
that the measure of $N_{k+1}\cap T^{-k}(A_{1}^{k})$ and $N_{k+1}\cap
T^{-k}(A_{2}^{k})$ are positive. Now, our aim is to show that
$T^{k+1}(A_{12}^{k})$ is a subset of $T^{k+1}(A_{21}^{k})$. Let $y=T^{k+1}(x)$
for some $x\in A_{12}^{k}$. Since $A_{12}^{k}\subseteq N_{k+1}\cap
T^{-k}(A_{1}^{k})$ and $T_{k}(A_{1}^{k})=T_{k}(A_{2}^{k})$,

$y=T^{k+1}(x)=T(T^{k}(x))=T_{k}(T^{k}(x))=T_{k}(z_{0})=T_{k}(T^{k}(z))=T^{k+1}(z)$

where $z_{0}\in A_{2}^{k}~{}\left(\subseteq T^{k}(N_{k+1})\right)$ and $z\in
N_{k+1}\cap T^{-k}(A_{2}^{k})$. More precisely, this $z$ is a member of
$A_{21}^{k}$, as if $z\in A_{22}^{k}$ then

$-1=u\circ T^{k+1}(z)h\circ T^{k+1}(z)=u\circ T^{k+1}(x)h\circ T^{k+1}(x)=1$

which is a contradiction. This gives our desired claim. In the end, we get an
contradiction to the assumption that $g=W_{u,T}^{k+1}h$ (a.e. on $X$) for some
$h\in L(p,q)$ by using the properties of $T$, that is, the properties provide

$\mu(T^{k+1}(A_{12}^{k}))\leq\mu(T^{k+1}(A_{21}^{k}))\leq\mu(A_{21}^{k})\leq\mu\bigg{(}\\{x\in
X~{}|~{}g(x)\neq W_{u,T}^{k+1}h(x)\\}\bigg{)}=0,$

and therefore,
$\mu(A_{12}^{k})\leq\mu\bigg{(}T^{-(k+1)}(T^{k+1}(A_{12}^{k}))\bigg{)}=0$. As
a result, $g$ does not belong to $\mathcal{R}(W_{u,T}^{k+1})$. Hence, the
descent of $W_{u,T}$ is infinity. ∎

Now we try to investigate the necessary conditions for the descent of
$W_{u,T}\in\mathfrak{B}(L(p,q))$ to be infinity. In this sequel, we attain the
following.

######  Theorem 4.20.

Let $(X,\mathcal{A},\mu)$ be a $\sigma-$finite complete measure space. Suppose
that $W_{u,T}\in\mathfrak{B}(L(p,q))$, $1<p\leq\infty$, $1\leq q\leq\infty$,
$u$ is bounded away from zero and $\mu(T(A))\leq\mu(A)$ for all
$A\in\mathcal{A}$. Then, a necessary condition for $\beta(W_{u,T})=\infty$ is
that the map $T_{k}=T|_{T^{k}(X)}:T^{k}(X)\rightarrow T^{k+1}(X)$ is not
injective almost everywhere on $T^{k}(X)$ for every
$k\in\mathbb{N}\cup\\{0\\}$.

###### Proof.

The proof is done by using contrapositive statement, that is, if there exists
an integer $k\geq 0$ such that $T_{k}=T|_{T^{k}(X)}:T^{k}(X)\rightarrow
T^{k+1}(X)$ is injective a.e. on $T^{k}(X)$ then the descent of $W_{u,T}$ is
finite and less than equal to $k$. Let $k\in\mathbb{N}\cup\\{0\\}$ be a fixed
number for which the map $T_{k}$ is injective a.e. on $T^{k}(X)$. So, by
definition, the measure of $\mathcal{F}_{k}$ is zero where

$\mathcal{F}_{k}:=\\{x\in T^{k}(X)~{}|~{}\exists~{}y\in T^{k}(X),y\neq x\text{
and }T_{k}(x)=T_{k}(y)\\}.$

Further, by definition of bounded away from zero, there is $\epsilon>0$
satisfying $|u(x)|>\epsilon$ a.e. on $X$. Let us take

$\mathcal{R}_{1}:=\\{x\in
X~{}|~{}|u(x)|>\epsilon\\}\text{~{}~{}and~{}~{}}\mathcal{R}_{2}:=\\{x\in
X~{}|~{}|u(x)|\leq\epsilon\\}.$

Obviously, $\mu(\mathcal{R}_{2})=0$, and we also know that
$\mathcal{R}(W_{u,T}^{k+1})\subseteq\mathcal{R}(W_{u,T}^{k})$. To demonstarte
the reverse containment, let $h\in\mathcal{R}(W_{u,T}^{k})$. Then
$h=W_{u,T}^{k}f$, $f\in L(p,q)$. Define $g:X\mapsto\mathbb{C}$ as

$g(y)=\left\\{\begin{array}[]{cl}\dfrac{f(x_{y})}{u(y)}&\mbox{ if there is a
unique }x_{y}\in
T^{-1}(\mathcal{R}_{1})\cap\bigg{(}T^{k}(X)\setminus\mathcal{F}_{k}\bigg{)}:T_{k}(x_{y})=y\\\
\\\ 0&\mbox{ otherwise }\end{array}\right..$

Clearly, $g$ is well defined and belongs to $L(p,q)$. Indeed, for $s\geq 0$,
the distribution function of $g$ is given by

$\displaystyle\mu_{g}(s)$ $\displaystyle=\mu\bigg{(}\\{y\in
X~{}|~{}|g(y)|>s\\}\bigg{)}$ $\displaystyle=\mu\bigg{(}\\{y\in
T^{k+1}(X)~{}|~{}\text{ we get }x_{y}\in
T^{-1}(\mathcal{R}_{1})\cap\bigg{(}T^{k}(X)\setminus\mathcal{F}_{k}\bigg{)}:T_{k}(x_{y})=y\text{
and }\left|\frac{f(x_{y})}{u(y)}\right|>s\\}\bigg{)}$
$\displaystyle\leq\mu\bigg{(}\\{T(x_{y})\in
T^{k+1}(X)\cap\mathcal{R}_{1}~{}\big{|}~{}|f(x_{y})|>s\epsilon\\}\bigg{)}$
$\displaystyle\leq\mu\bigg{(}\\{T(x_{y})\in
X~{}|~{}|f(x_{y})|>s\epsilon\\}\bigg{)}\leq\mu\bigg{(}\\{x_{y}\in
X~{}|~{}|f(x_{y})|>s\epsilon\\}\bigg{)}=\mu_{f}(s\epsilon).$

Thus, $g^{*}(t)\leq\frac{1}{\epsilon}~{}f^{*}(t)$ and
$g^{**}(t)\leq\frac{1}{\epsilon}~{}f^{**}(t)$, for each $t>0$. As a
consequence, $\|g\|_{pq}\leq\frac{1}{\epsilon}\|f\|_{pq}$, $1<p\leq\infty$,
$1\leq q\leq\infty$. Finally, the result follows as $f\in L(p,q)$. The target
is to show that $g$ is the pre-image of $h$ under the operator
$W_{u,T}^{k+1}$. For this purpose, we split the set $X$ as follows

$\displaystyle X$ $\displaystyle=\bigg{(}T^{-(k+1)}(\mathcal{R}_{1})\cap
T^{-k}(T^{k}(X)\setminus\mathcal{F}_{k})\bigg{)}\cup\bigg{(}T^{-(k+1)}(\mathcal{R}_{1})\cap
T^{-k}(T^{k}(X)\setminus\mathcal{F}_{k})\bigg{)}^{c}$
$\displaystyle=\bigg{(}T^{-(k+1)}(\mathcal{R}_{1})\cap
T^{-k}(T^{k}(X)\setminus\mathcal{F}_{k})\bigg{)}\cup\bigg{(}T^{-(k+1)}(\mathcal{R}_{2})\cup
T^{-k}\bigg{(}(X\setminus T^{k}(X))\cup F_{k}\bigg{)}\bigg{)}$

Now, for every $x\in\bigg{(}T^{-(k+1)}(\mathcal{R}_{1})\cap
T^{-k}(T^{k}(X)\setminus\mathcal{F}_{k})\bigg{)}$, we observe that

$\displaystyle\left(W_{u,T}^{k+1}g\right)(x)$
$\displaystyle=\prod\limits_{i=1}^{k+1}u\circ T^{i}(x)\cdot g\circ T^{k+1}(x)$
$\displaystyle=\prod\limits_{i=1}^{k}u\circ T^{i}(x)\cdot u\circ
T_{k}(T^{k}(x))\cdot g\circ T_{k}(T^{k}(x))$
$\displaystyle=\prod\limits_{i=1}^{k}u\circ T^{i}(x)\cdot f\circ T^{k}(x)$
$\displaystyle=\left(W_{u,T}^{k}f\right)(x).$

As a consequence,

$\\{x\in
X~{}|~{}\left(W_{u,T}^{k+1}g\right)(x)\neq\left(W_{u,T}^{k}f\right)(x)\\}\subseteq\bigg{(}T^{-(k+1)}(\mathcal{R}_{2})\cup
T^{-k}\bigg{(}(X\setminus T^{k}(X))\cup F_{k}\bigg{)}\bigg{)},$

and as the superset has measure zero, $W_{u,T}^{k+1}g=W_{u,T}^{k}f=h$ a.e. on
$X$. This implies that $h$ is the member of $\mathcal{R}(W_{u,T}^{k+1})$.
Thus, $\mathcal{R}(W_{u,T}^{k})=\mathcal{R}(W_{u,T}^{k+1})$, that is,
$\beta(W_{u,T})\leq k$. As stated, by contrapositive statement, we get that if
the descent of $W_{u,T}$ is not finite, then there does not exist any non-
negative integer $k$ such that $T_{k}=T|_{T^{k}(X)}:T^{k}(X)\rightarrow
T^{k+1}(X)$ is injective almost everywhere on $T^{k}(X)$. Hence the result. ∎

In the above theorem the condition $u$ is bounded away from zero cannot be
relaxed which we justify with the help of following example.

######  Example 4.21.

Consider the Lebesgue measure space $(\mathbb{R},\mathcal{A},\mu)$. It is
clearly $\sigma-$finite and complete. Define $T:\mathbb{R}\mapsto\mathbb{R}$
and $u:\mathbb{R}\mapsto\mathbb{C}$ as $T(x)=x-1$ and

$\displaystyle u(x)=\left\\{\begin{array}[]{ll}0&\mbox{ if }x\in[0,1]\\\
1&\mbox{ otherwise }\end{array}\right.$

respectively. With this pair of $(u,T)$, $W_{u,T}\in\mathfrak{B}(L(p,q))$,
$1<p\leq\infty$, $1\leq q\leq\infty$ and $\mu\circ T(E)=\mu(E)$ for all
measurable sets $E$. Moreover, the mappings $T_{k}:T^{k}(X)\rightarrow
T^{k+1}(X)$ is injective almost everywhere on $T^{k}(X)$ for each $k\geq 0$.
Finally, we show that
$\chi_{[k+1,k+2]}\in\mathcal{R}(W_{u,T}^{k})\setminus\mathcal{R}(W_{u,T}^{k+1})$.
The function $\chi_{[1,2]}$ is the pre-image of the function
$\chi_{[k+1,k+2]}$ under the mapping $W_{u,T}^{k}$. Further, if we consider
$h\in L(p,q)$ such that $\chi_{[k+1,k+2]}(x)=W_{u,T}^{k+1}h(x)$, then this
leads to a contradiction as $u(x-(k+1))=0$.

The statement of the Theorem 4.20 can be restated as follows:

######  Theorem 4.22.

Suppose that $(X,\mathcal{A},\mu)$ is a $\sigma-$finite complete measure
space, $T:X\rightarrow X$ satisfying $\mu(T(A))\leq\mu(A)$ for all
$A\in\mathcal{A}$ and $u$ is bounded away from zero. If
$k\in\mathbb{N}\cup\\{0\\}$ is the smallest number such that
$T_{k}=T|_{T^{k}(X)}:T^{k}(X)\rightarrow T^{k+1}(X)$ is injective almost
everywhere on $T^{k}(X)$ then the descent of $W_{u,T}$ is less than and equal
to $k$.

The reader can verify that the condition $u\neq 0$ a.e. on $X$ will give
$h_{T}=0$ a.e. on $(Supp(u))^{c}$. In fact,

$\displaystyle\mu(\\{x\in(Supp(u))^{c}~{}|~{}h_{T}(x)\neq 0\\})$
$\displaystyle=\mu(\\{x\in X~{}|~{}h_{T}(x)\neq 0\\}\cap\\{x\in
X~{}|~{}u(x)=0\\})$ $\displaystyle\leq\mu(\\{x\in X~{}|~{}u(x)=0\\})=0.$

We draw out here a conclusion that all the results that have been obtained
with the assumption $h_{T}=0$ a.e. on $(Supp(u))^{c}$ do hold on replacing
this condition by the condition $u\neq 0$ a.e. on $X$. Let us take another
operator defined as ${\widehat{W}}_{u,T}(h)=u\cdot h\circ T$ on $L(p,q)$,
$1<p\leq\infty$, $1\leq q\leq\infty$. If $u\in L_{\infty}(\mu)$ and $u\neq 0$
a.e. on $X$ then we can also deduce the following conclusions about
$\widehat{W}_{u,T}$ without any extra efforts.

  1. (a)

For every $k\in\mathbb{N}\cup\\{0\\}$,
$\mathcal{N}\big{(}\widehat{W}_{u,T}^{k}\big{)}=L(p,q)(X_{k})$.

  2. (b)

A necessary and sufficient condition for $\alpha(\widehat{W}_{u,T})=k$ is that
$k$ is the least non-negative integer such that $\mu_{k}$ and $\mu_{k+1}$ are
equivalent measures.

  3. (c)

Suppose that $C_{T}$, $W_{u,T}$ and
$\widehat{W}_{u,T}\in\mathfrak{B}(L(p,q))$, $1<p\leq\infty$, $1\leq
q\leq\infty$. Then
$\mathcal{N}(C_{T}^{k})=\mathcal{N}(W_{u,T}^{k})=\mathcal{N}\bigg{(}\widehat{W}_{u,T}^{k}\bigg{)}=L(p,q)(X_{k})$
for every non-negative integer $k$.

  4. (d)

If $C_{T}$, $W_{u,T}$ and $\widehat{W}_{u,T}$ are all belong to
$\mathfrak{B}(L(p,q))$, $1<p\leq\infty$, $1\leq q\leq\infty$ then
$\mu_{k}\ll\mu_{k+1}\ll\mu_{k}\Longleftrightarrow\mathcal{N}(C_{T}^{k})=\mathcal{N}(C_{T}^{k+1})\Longleftrightarrow\mathcal{N}(W_{u,T}^{k})=\mathcal{N}(W_{u,T}^{k+1})\Longleftrightarrow\mathcal{N}\bigg{(}\widehat{W}_{u,T}^{k}\bigg{)}=\mathcal{N}\bigg{(}\widehat{W}_{u,T}^{k+1}\bigg{)}$
for every $k\in\mathbb{N}\cup\\{0\\}$.

  5. (e)

If $C_{T}$, $W_{u,T}$, $\widehat{W}_{u,T}\in\mathfrak{B}(L(p,q))$,
$1<p\leq\infty$, $1\leq q\leq\infty$ then $\alpha(C_{T})=k$,
$\alpha(W_{u,T})=k$, $\alpha(\widehat{W}_{u,T})=k$ and $k$ is the least non-
negative integer satisfying $\mu_{k}\ll\mu_{k+1}\ll\mu_{k}$ are all equivalent
statements.

  6. (f)

Let $\mu\circ T(A)\leq\mu(A)$ for all $A\in\mathcal{A}$. Then a sufficient
condition for $\widehat{W}_{u,T}$ to have ascent $k$ is that $k$ is the
smallest integer from the set $\mathbb{N}\cup\\{0\\}$ in such a way that
$X_{k}^{c}\subseteq T^{k+1}(X_{k}^{c})$, where $X_{k}^{c}=X\setminus X_{k}$.

Now, we justify that the condition $u\neq 0$ a.e. on X, that is, $\mu(\\{x\in
X~{}:~{}u(x)=0\\})=0$ is essential for the results obtained above with the
help of an example.

######  Example 4.23.

On the Lebesgue measure space $([0,1],\mathcal{A},\mu)$, define $T$ and $u$ on
$[0,1]$ as $T(x)=x$ and

$\displaystyle u(x)=\left\\{\begin{array}[]{cl}0&\mbox{ if }x\in[0,1/2)\\\
x&\mbox{ if }x\in[1/2,1]\end{array}\right.$

respectively. Obviously, $\widehat{W}_{u,T}$ is a well defined weighted
composition operator and
$\chi_{[0,1/4]}\in\mathcal{N}(\widehat{W}_{u,T})\setminus\\{0\\}$. Also,
$\mu_{k}=\mu$ and $X_{k}=\emptyset$ for every $k\in\mathbb{N}\cup\\{0\\}$. As
a consequence, we get $\mu_{0}\ll\mu_{1}\ll\mu_{0}$, $X_{0}^{c}\subseteq
T(X_{0}^{c})$ and $\alpha(\widehat{W}_{u,T})\geq 1$. Hence, the statements of
part (b) and (f) above are not true in this case.

At the end, if we consider $N_{0}=X$ and $N_{k}=\\{x\in
X~{}|~{}u(T^{i}(x))\neq 0\mbox{ for all }0\leq i\leq k-1\\}$ then Theorems
4.13 and 4.14 hold for $\widehat{W}_{u,T}$. Also, Theorems 4.20 and 4.22,
under the same conditions, are true for $\widehat{W}_{u,T}$.

##  5 Examples

The goal of this section is to give some applications which support the
results achieved in the paper.

######  Example 5.1.

On the measure space $(\mathbb{N},2^{\mathbb{N}},\mu)$, define
$T:\mathbb{N}\rightarrow\mathbb{N}$ as $T(m)=2n$ if
$m\in\\{2n-1,2n\\}_{n\in\mathbb{N}}$ and $u:\mathbb{N}\rightarrow\mathbb{C}$
as $u(n)=\left\\{\begin{array}[]{cl}0&\mbox{ if $n$ is odd}\\\ 1&\mbox{ if $n$
is even}\end{array}\right.$. From the definition of $u$ and $T$, we can
conclude that $(Supp(u))^{c}=\\{2n-1:n\in\mathbb{N}\\}$ and $h_{T}$ is given
by

$h_{T}(n)=\left\\{\begin{array}[]{cl}0&\mbox{if $n$ is odd}\\\ 2&\mbox{if $n$
is even}\end{array}.\right.$

Clearly, $h_{T}=0$ on $(Supp(u))^{c}$. Further, for every $k\in\mathbb{N}$,
$T^{k}=T$, $h_{T^{k}}=h_{T}$ and $X_{k}^{c}\subseteq T^{k+1}(X_{k}^{c})$. By
Theorem 4.11, the ascent of $W_{u,T}$ is equal to $1$.

######  Example 5.2.

Consider the $\sigma-$finite complete measure space
$\left([0,1],\mathcal{A},\mu\right)$ with $\mu$ the Lebesgue measure. Define
$T:[0,1]\mapsto[0,1]$ and $u:[0,1]\mapsto\mathbb{C}$, respectively, as
$T(x)=\frac{x}{2}$ and

$\displaystyle u(x)=\left\\{\begin{array}[]{ll}x&x\mbox{ is irrational in
}[0,1]\\\ 0&x\mbox{ is rational in }[0,1]\end{array}.\right.$

With this pair of mappings, we have $W_{u,T}\in\mathfrak{B}(L(p,q))$,
$1<p\leq\infty$, $1\leq q\leq\infty$. Take
$A_{k}=\left(\left(\frac{1}{2}\right)^{k+1},\left(\frac{1}{2}\right)^{k}\right)$.
These choices give $N_{k}=\\{x\in[0,1]:x\mbox{ is irrational}\\}$ and
$T^{-k}(A_{k})=\left(\frac{1}{2},1\right)$. As a consequence, we have
$\mu(T^{-k}(A_{k})\cap N_{k})=\frac{1}{2}\neq 0$ and
$\mu(T^{-(k+1)}(A_{k})\cap N_{k+1})=\mu(\emptyset)=0$. By Theorem 4.13,
$\alpha\left(W_{u,T}\right)=\infty$.

######  Example 5.3.

Consider the $\sigma-$finite and complete measure space
$([0,\infty),\mathcal{A},\mu)$, where $\mu$ is the Lebesgue measure. Define
$T:[0,\infty)\mapsto[0,\infty)$ and $u:[0,\infty)\mapsto\mathbb{C}$ by setting

$\displaystyle T(x)=\left\\{\begin{array}[]{ll}x&\mbox{ if }x\in[0,1]\\\
x-1&\mbox{ elsewhere }\end{array}\right.$

and $u(x)=1$ respectively. With this pair of $(u,T)$,
$W_{u,T}\in\mathfrak{B}(L(p,q))$, $1<p\leq\infty$, $1\leq q\leq\infty$, and
$\mu\circ T(A)\leq\mu(A)$ for all $A\in\mathcal{A}$. Also,
$N_{k+1}=[0,\infty)$, $A_{1}^{k}=(0,1]$, $A_{2}^{k}=(1,2]$ and
$T_{k}(A_{1}^{k})=T_{k}(A_{2}^{k})$ where
$T_{k}=T|_{T^{k}(N_{k+1})}:T^{k}(N_{k+1})\rightarrow T^{k+1}(N_{k+1})$ for
every $k\in\mathbb{N}\cup\\{0\\}$. Thus, using Theorem 4.19, we get
$\beta(W_{u,T})=\infty$.

######  Example 5.4.

Let $([-1,1],\mathcal{A},\mu)$ be a $\sigma-$finite complete Lebesgue measure
space. Define $u:[-1,1]\rightarrow\mathbb{C}$ as $u\equiv 1$ and
$T:[-1,1]\rightarrow[-1,1]$ as

$T(x)=\left\\{\begin{array}[]{cl}\frac{x}{2}&\mbox{ if }0\leq x\leq 1\\\ \\\
\frac{-x}{2}&\mbox{ if }-1\leq x\leq 0\end{array}\right..$

Now, for any measurable set $A\in\mathcal{A}$, we have
$T^{-1}(A)=2\left(A\cap[0,\frac{1}{2}]\right)\cup(-2)\left(A\cap[0,\frac{1}{2}]\right)$
and $T(A)=\frac{A_{1}}{2}\cup\frac{-A_{2}}{2}$ where $A_{1}=A\cap[0,1]$ and
$A_{2}=A\cap[-1,0]$. This implies that $\mu(T^{-1}(A))\leq
2\mu\left(A\cap[0,\frac{1}{2}]\right)+2\mu\left(A\cap[0,\frac{1}{2}]\right)\leq
4\mu(A)$ and $\mu(T(A))\leq\mu(A)$. As a conclusion $h_{T}\in
L_{\infty}(\mu)$, and hence $W_{u,T}\in\mathfrak{B}(L(p,q))$. Finally, note
that, for every natural number $k$,
$T_{k}=T|_{T^{k}([-1,1])}:T^{k}([-1,1])\rightarrow T^{k+1}([-1,1])$ is
injective almost everywhere on $T^{k}([-1,1])$. By Theorem 4.22, we have
descent of $W_{u,T}$ is less than and equal to $1$. Moreover, we can see that
$\beta(W_{u,T})=1$. For this purpose one can verify that the non-zero
characteristic function $\chi_{[-1,0]}$ from $L(p,q)$ is not the member of
$\mathcal{R}(W_{u,T})$, i.e., it has no preimage under $W_{u,T}$. In fact,
suppose that $W_{u,T}h=\chi_{[-1,0]}$ for some $h\in L(p,q)$. Then
$\mu(E_{1}^{c})=\mu(E_{2}^{c})=0$ where $E_{1}:=\\{x\in[-1,0)~{}|~{}h\circ
T(x)=1\\}$, $E_{2}:=\\{x\in(0,1]~{}|~{}h\circ T(x)=0\\}$,
$E_{1}^{c}:=[-1,0)\setminus E_{1}$ and $E_{2}^{c}:=(0,1]\setminus E_{2}$.
Finally, the result follows using the fact that $T$ is non-singular,
$E_{1}\subseteq T^{-1}(T(E_{1}))$ and
$\mu(T(E_{1}))\leq\mu(T(E_{2}^{c}))\leq\mu(E_{2}^{c})=0$.

## Statements and Declarations

No funding was received for conducting this study. Authors have no competing
interest to declare that are relevant to the content of this article.

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