#  A class of Zero Divisors and Topological Divisors of Zero in some Banach
algebras

Anurag Kumar Patel Anurag Kumar Patel
Department of Mathematics
Banaras Hindu University
Varanasi 221005, India<EMAIL_ADDRESS>and  Harish Chandra Harish
Chandra
Department of Mathematics
Banaras Hindu University
Varanasi 221005, India<EMAIL_ADDRESS>
###### Abstract.

In this paper, we establish necessary and sufficient conditions that must be
met for weighted composition operators to act as zero divisors in
$\mathcal{B}(\ell^{p}).$ We also give a necessary condition and a sufficient
condition for a composition operators to act as zero divisors in
$\mathcal{B}(L^{p}(\mu)).$ Subsequently, we characterize TDZ in $C(X)$.
Afterward, we establish that a multiplication operator $M_{h}$ in
$\mathcal{B}(C(X))$ becomes a TDZ if and only if $h$ is a TDZ in $C(X).$
Further, motivated by the definition of TDZ, we introduce notions of
polynomially TDZ and strongly TDZ and prove that every element in $C(X)$ and
in $L^{\infty}(\mu)$ is a polynomially TDZ. We then prove that a
multiplication operator $M_{h}$ in $\mathcal{B}(C(X))$ as well as in
$\mathcal{B}(L^{p}(\mu))$ is a polynomially TDZ. Lastly, we show that each
$T\in\mathcal{B}(H)$, where $H$ is a separable Hilbert space, is a strongly
TDZ.

###### Key words and phrases:

Banach algebra, Composition operators, Zero divisor, Measurable functions

###### 1991 Mathematics Subject Classification:

46E25, 47B33, 13A70, 28A20

The first author is supported by the Council of Scientific and Industrial
Research (CSIR) NET-JRF, New Delhi, India, through grant
$09/013(0891)/2019-EMR-I$

##  1\. Introduction

The notion of zero divisor[7] is well known from ring theory. In a ring
$\mathfrak{R},$ an element $z$ is termed a left $(right)$ zero divisor if
$zx=0~{}(xz=0)$ for some $0\neq x\in\mathfrak{R}$ and if it is either a left
zero divisor or a right zero divisor, we call it a zero divisor.

The notion of topological divisors of zero(TDZ) is an important generalization
in a Banach algebra, introduced by Shilov in [16] with significant later
contributions by Zelasko, Kaplansky et al. In a Banach algebra $\mathfrak{B},$
an element $z$ is called a TDZ if there exists a sequence
$\\{z_{n}\\}_{n=1}^{\infty}$ in $\mathfrak{B}$ with $\|z_{n}\|=1,$ for each
$n\geq 1$ such that either $zz_{n}\to 0$ or $z_{n}z\to 0$ as $n\to\infty$.

In [15] F. Schulz, R. Brits, M. Hasse gave a characterization of Banach
algebras in which every element is a TDZ. The purpose of the present paper is
to determine zero divisors and TDZ in some Banach algebras which do not fall
in the preceding category.

In this paper, we consistently employ the notation $\mathbb{N}$ to signify the
set of positive integers, $\mathbb{C}$ designates the set of complex numbers,
$\mathcal{B}(X)$ denotes the Banach algebra of all bounded linear operators on
$X,$ $\mathcal{R}(f)$ denotes the range of a function $f,$ $[f]$ denotes the
linear span of $f,$ $C(X)$ denotes the space of all complex-valued continuous
functions on $X$ and $Z(f)=\\{x:f(x)=0\\}$ refers to the zero set of $f$.

Let $(X,\Omega,\mu)$ be a $\sigma$-finite measure space. For $1\leq
p\leq\infty$, we abbreviate the Lebesgue space $L^{p}(X,\Omega,\mu)$ to
$L^{p}(\mu),$ and denote the $L^{p}$-norm by $\|\cdot\|_{p}.$ Note that
$L^{p}(\mu)$ is a Banach space under the $L^{p}$-norm $(1\leq p\leq\infty)$
(see [1, 3, 14]).

In the scenario where $X$ equals $\mathbb{N},$ and $\mu$ represents the
counting measure on $\mathbb{N},$ the $L^{p}(\mu)$ becomes $\ell^{p}.$

Let $(X,\Omega,\mu)$ be a $\sigma$-finite measure space and $u\in
L^{\infty}(\mu).$ Let $\phi$ be a measurable transformation on $X$ and suppose
that the measure $\mu_{\phi}$ defined by
$\mu_{\phi}(E)=\mu(\phi^{-1}(E))~{}\forall~{}E\in\Omega,$ is absolutely
continuous w.r.t. $\mu$ (we write $\mu_{\phi}\ll\mu,$ as usual). Then we can
define the linear transformation $uC_{\phi}:L^{p}(\mu)\to L^{p}(\mu)$ as

$uC_{\phi}f(x)=u(x)f(\phi(x))~{}\text{for all}~{}x\in X~{}\text{and}~{}f\in
L^{p}(\mu)$

and the multiplication operator $M_{u}:L^{p}(\mu)\to L^{p}(\mu)$ defined as

$M_{u}(f)(x)=u(x)f(x)~{}\forall~{}x\in X.$

Assumption $\mu_{\phi}\ll\mu$ is used in the well-definedness of $uC_{\phi}$.
Further, if the transformation $uC_{\phi}$ is bounded, then it is called a
weighted composition operator on $L^{p}(\mu).$ Specifically, when considering
the constant function $1$ as $u$, the associated operator $uC_{\phi}$ on
$L^{p}(\mu)$ becomes $C_{\phi}$ and is called a composition operator (see [3,
21, 23, 24]).

In [22], R. K. Singh and B. S. Komal have shown that a linear transformation
$C_{\phi}$ on $\ell^{p}~{}(1\leq p<\infty)$ defined as

$C_{\phi}(f)=\sum_{n=1}^{\infty}f(n)\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{\phi^{-1}(n)}$

is bounded if and only if $\\{|\phi^{-1}(n)|:n\in\mathbb{N}\\}$ is a bounded
set, where $|\phi^{-1}(n)|$ represents the cardinality of the set
$\phi^{-1}(n).$ Further, in [11], it is proved that a linear transformation
$uC_{\phi}$ on $L^{p}(\mu)$ is bounded if and only if
$\frac{d\mu\phi^{-1}}{d\mu}:X\setminus Z(u)\to\mathbb{C}$ is essentially
bounded, where $\frac{d\mu\phi^{-1}}{d\mu}$ represents the Radon-Nikodym
derivative.

Weighted composition operators constitute a significant and diverse class of
operators. There are several important reasons to study these operators. For
example, Lamperti [14] showed that every isometry on $L^{p}(\mu)$ spaces is a
weighted composition operator. Most recently, in [2], it is shown that
analytic m-isometries without the wandering subspace property are weighted
composition operators which can be realised as weighted shifts, thus providing
a partial answer to the problem raised by Shimorin in [[17], p. 185].

The current article is aimed at finding weighted composition operators on the
spaces $\ell^{p}$ and $L^{p}(\mu)$ as zero divisors in $\mathcal{B}(\ell^{p})$
and $\mathcal{B}(L^{p}(\mu))$ respectively. Further, we find TDZ in the Banach
algebras $C(X)$ in its full generality. Consequently, we determine condition
under which the multiplication operator $M_{h}$ is a TDZ in
$\mathcal{B}(C(X)).$ We show that every element in $C(X)$ and in
$L^{\infty}(\mu)$ is a polynomially TDZ. We also prove that every
multiplication operator $M_{h}$ is a polynomially TDZ in $\mathcal{B}(C(X))$
and $\mathcal{B}(L^{p}(\mu))$ respectively. Finally, we show that if $H$ is a
separable Hilbert space, then each $T\in\mathcal{B}(H)$ is a strongly TDZ.
Lastly, we construct a class of compact operators in $\mathcal{B}(\ell^{p})$
which are also TDZ. The paper is arranged as follows.

In section 2, we state some definitions and basic results which are relevant.
Section 3.1 is devoted for the characterization of weighted composition
operators on the $\ell^{p}$ spaces as zero divisors in $\mathcal{B}(\ell^{p})$
and in section 3.2, we characterize composition operators on the $L^{p}(\mu)$
spaces as zero divisors in $\mathcal{B}(L^{p}(\mu)).$ In section 3.3, we
completely characterize the TDZ in the Banach algebras $C(X)$ in its full
generality. Section 3.4 is devoted for the characterization of polynomially
TDZ. In this section we show that every element in $C(X)$ and
$L^{\infty}(\mu)$ is a polynomially TDZ. Further, we prove that every
multiplication operator $M_{h}$ is a polynomially TDZ in $\mathcal{B}(C(X))$
and $\mathcal{B}(L^{p}(\mu))$ respectively. In section 3.5, we show that every
TDZ is also a strongly TDZ. We also show that if $H$ is a separable Hilbert
space, then each $T\in\mathcal{B}(H)$ is a strongly TDZ. Lastly, we construct
a class of compact operators in $\mathcal{B}(\ell^{p})$ which are also TDZ.

##  2\. Preliminaries

In this section, we state few definitions and some known results, which we
shall use later.

######  Definition 2.1.

[6, 18] In a Banach algebra $\mathfrak{B}$ with identity $1,$ an element $x$
is called regular if $xy=yx=1$ for some $y\in\mathfrak{B},$ otherwise it is
called a singular element.

######  Definition 2.2.

([24]) Let $(X,\Omega,\mu)$ be a measure space, $Y$ be a topological space and
$f:X\to Y$ . An element $y\in Y$ is in the essential range of $f$ if
$\mu(f^{-1}(U))>0$ for each neighborhood $U$ of $y.$ We denote the essential
range of $f$ by ess.range$(f).$

######  Theorem 2.3.

[19, 21] Let $\phi:\mathbb{N}\to\mathbb{N}$ and $1\leq p<\infty.$ Let
$C_{\phi}$ be the composition operator on $\ell^{p}.$ Then the following
statements are true

  1. $(1)$

$C_{\phi}$ is injective $\iff$ $\phi$ is surjective;

  2. $(2)$

$C_{\phi}$ is surjective $\iff$ $\phi$ is injective;

  3. $(3)$

$C_{\phi}$ is invertible $\iff$ $\phi$ is invertible.

######  Theorem 2.4.

[19, 21] Suppose $u\in\ell^{\infty}$ with $u(n)\neq
0~{}\forall~{}n\in\mathbb{N}.$ Let $\phi:\mathbb{N}\to\mathbb{N}$ and $1\leq
p<\infty.$ Then the weighted composition operator $uC_{\phi}$ on $\ell^{p}$ is
injective if and only if $\phi$ is surjective .

######  Definition 2.5.

[10] Let $X$ be a non-empty set. We say that $u:X\to\mathbb{C}$ is bounded
away from zero if there exists $a>0$ satisfying $0<\frac{1}{|u(x)|}\leq
a~{}\forall~{}x\in X.$

######  Theorem 2.6.

[19, 21] Suppose $u\in\ell^{\infty}$ is bounded away from zero. Let
$\phi:\mathbb{N}\to\mathbb{N}$ and $1\leq p<\infty.$ Let
$uC_{\phi}:\ell^{p}\to\ell^{p}$ be the weighted composition operator. Then
$uC_{\phi}$ is surjective if and only if $\phi$ is injective.

######  Definition 2.7.

[8] Let $H$ be an infinite dimensional Hilbert space. An operator
$T\in\mathcal{B}(H)$ is called antinormal if
$d(T,\mathcal{N})=\inf_{N\in\mathcal{N}}\|T-N\|=\|T\|,$ where $\mathcal{N}$ is
the class of all normal operators in $\mathcal{B}(H).$

######  Theorem 2.8.

[9] Suppose $u\in\ell^{\infty}$ is bounded away from zero and let $\phi$ be a
bijective self map on $\mathbb{N}.$ Then $uC_{\phi}$ is not antinormal on
$\ell^{2}$.

######  Theorem 2.9.

[13] Let $h\in L^{\infty}(\mu).$ Then $h$ is a TDZ if and only if
$0\in\textrm{ess.range}(h).$

######  Theorem 2.10.

[13] Let $h\in L^{\infty}(\mu)$ and $1\leq p\leq\infty.$ Then the
multiplication operator $M_{h}$ is a TDZ in $B(L^{p}(\mu))$ if and only if $h$
is a TDZ in $L^{\infty}(\mu).$

######  Theorem 2.11.

[5] Let $f\in C[a,b].$ Then $f$ is a TDZ if and only if $Z(f)\neq\emptyset.$

##  3\. Main results

###  3.1. Weighted composition operators on $\ell^{p}~{}(1\leq p<\infty)$
spaces as zero divisors

In this section, we give a characterization of the weighted composition
operators on $\ell^{p},$ which are zero divisors in $\mathcal{B}(\ell^{p}).$
We begin with the following theorem, which shows that under a certain
condition on $\phi$ and the zero set $Z(u)$ of the weight function $u$, the
weighted composition operator $uC_{\phi}$ is a right zero divisor.

######  Theorem 3.1.

Let $u\in\ell^{\infty}$ and $Z(u)$ denote the zero set of $u.$ If
$\phi:\mathbb{N}\to\mathbb{N}$ is such that there exists $n_{0}\in\mathbb{N}$
with $\phi^{-1}(n_{0})\neq\emptyset$ and
$\phi^{-1}(n_{0})=\\{n_{1},n_{2},...,n_{k}\\}\subset Z(u)$ then $uC_{\phi}$ is
right zero divisor.

###### Proof.

Let $\phi^{-1}(n_{0})=\\{n_{1},n_{2},...,n_{k}\\}.$ Then $T$ defined as

$T(f)=\begin{cases}f,&\text{ if
}f=\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{n_{1}},\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{n_{2}},...,\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{n_{k}}\\\
0,&\text{ if }f=\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{n}\text{ and
}n\neq n_{1},n_{2},...,n_{k}\end{cases}$

linearly extends to a bounded linear operator $T$ on $\ell^{p}.$
For any
$g=\sum_{n=1}^{\infty}g_{n}\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{n}\in\ell^{p}$

$\displaystyle T(g)$
$\displaystyle=\sum_{n=1}^{\infty}g_{n}T(\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{n})$
$\displaystyle=\sum_{i=1}^{k}g_{n_{i}}\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{n_{i}}$

and

$(T\circ
uC_{\phi})(\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{n})=\begin{cases}\sum_{i=1}^{k}u(n_{i})\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{n_{i}},&\text{
if }n=n_{0}\\\ 0,&\text{ otherwise. }\end{cases}.$

Since $\phi^{-1}(n_{0})\subset Z(u),$ therefore $T\circ uC_{\phi}=0.$ Hence
$uC_{\phi}$ is right zero divisor. ∎

The following theorem gives a necessary and sufficient condition for
$uC_{\phi}$ to be a right zero divisor when $Z(u)=\emptyset.$

######  Theorem 3.2.

Let $\phi:\mathbb{N}\to\mathbb{N}$ and suppose $u\in\ell^{\infty}$ is such
that $u(n)\neq 0~{}\forall~{}n\geq 1.$ Then $uC_{\phi}$ is a right zero
divisor if and only if $\phi$ is not injective.

###### Proof.

Suppose $\phi$ is injective and $u\in\ell^{\infty}$ is such that $u(n)\neq
0~{}\forall~{}n\geq 1.$ Let $T\in\mathcal{B}(\ell^{p})$ be such that
$T(uC_{\phi})=0.$ This implies that

$\displaystyle
T(uC_{\phi})(\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{n})$
$\displaystyle=T(u\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{\phi^{-1}(n)})$
$\displaystyle=u(\phi^{-1}(n))T(\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{\phi^{-1}(n)})$
$\displaystyle=0~{}~{}\text{for each~{}}n\geq 1.$

Since $u(n)\neq 0~{}\forall~{}n\geq 1.$ Therefore

$T(\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{\phi^{-1}(n)})=0\ \forall\ n\geq
1.$

Again, as $\phi$ is injective, hence

$|\phi^{-1}(n)|\leq 1\ \forall\ n\in\mathbb{N}\text{ and
}\bigcup_{n=1}^{\infty}\phi^{-1}(n)=\mathbb{N}.$

Therefore, it follows that
$T(\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{n})=0~{}\forall~{}n\geq 1.$ Thus
$T=0.$ Hence $uC_{\phi}$ can not be a right zero divisor.
Conversely, suppose $\phi$ is not injective and $u(n)\neq 0~{}\forall~{}n\geq
1.$ Now we first prove that $uC_{\phi}$ is not surjective.
Since $\phi$ is not injective, then by Theorem 2.3, $C_{\phi}$ is not
surjective. Hence, there exists $f_{0}\in\ell^{p}$ such that
$f_{0}\notin\mathcal{R}(C_{\phi}).$ We claim that
$uf_{0}\notin\mathcal{R}(uC_{\phi}).$
For, if $uf_{0}\in\mathcal{R}(uC_{\phi}),$ then there exists
$g_{0}\in\ell^{p}$ such that $uf_{0}=uC_{\phi}g_{0}.$ This implies that

$u(n)(f_{0}(n)-(g_{0}\circ\phi)(n))=0~{}~{}\text{for each }~{}n\geq 1.$

Since $u(n)\neq 0~{}\forall~{}n\geq 1,$ hence
$f_{0}=g_{0}\circ\phi\in\mathcal{R}(C_{\phi}).$ This is a contradiction.

Thus, if $\phi$ is not injective and $u(n)\neq 0~{}\forall~{}n\geq 1,$ then
$uC_{\phi}$ is not surjective.
Hence there exists an $f_{0}\in\ell^{p}\backslash\mathcal{R}(uC_{\phi})$. Let
$M=[f_{0}].$ Since $M$ is a finite dimensional subspace of $\ell^{p},$ it
follows from Hahn-Banach theorem that there exists a closed subspace $N$ such
that $\ell^{p}=M\bigoplus N.$ Let $T$ denote the projection on $M.$ Then $T$
is a non-zero bounded linear operator on $\ell^{p}$ and
$\mathcal{R}(uC_{\phi})$ is contained in $N.$ Hence

$~{}(T\circ uC_{\phi})(f)=T(uC_{\phi}f)=0~{}\forall~{}f\in\ell^{p}.$

This implies that $uC_{\phi}$ is a right zero divisor. ∎

We now give examples of two weighted composition operators, which are right
zero divisors.

######  Example 3.3.

Define $\phi:\mathbb{N}\to\mathbb{N}$ as
$\phi(1)=1,~{}\phi(n)=n+1~{}\forall~{}n\geq 2$ and $u\in\ell^{\infty}$ as
$u(1)=0,~{}u(n)=\frac{1}{n}~{}\forall~{}n\geq 2.$ Then, by Theorem 3.1,
$uC_{\phi}$ is a right zero divisor.

######  Example 3.4.

Define $\phi:\mathbb{N}\to\mathbb{N}$ as
$\phi(1)=\phi(2)=1,~{}\phi(n)=n~{}\forall~{}n\geq 3$ and $u\in\ell^{\infty}$
as $u(n)=\frac{1}{n}~{}\forall~{}n\geq 1.$ Then, by Theorem 3.2, $uC_{\phi}$
is a right zero divisor.

######  Theorem 3.5.

Let $u\in\ell^{\infty}$ and $\phi:\mathbb{N}\to\mathbb{N}$ be such that
$uC_{\phi}$ is a bounded linear operator on $\ell^{p}.$ Then $uC_{\phi}$ is a
left zero divisor if $\phi$ is not surjective.

###### Proof.

Suppose $\phi$ is not surjective. Then there exists $n_{0}\in\mathbb{N}$ such
that $\phi^{-1}(n_{0})=\emptyset.$
Let $f_{0}=\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{n_{0}}$ and $M=[f_{0}].$
Since $M$ is a one dimensional subspace of $\ell^{p},$ then there exists a
closed subspace $N$ of $\ell^{p}$ such that $\ell^{p}=M\bigoplus N.$ Let $T$
denote the projection on $M.$ Clearly, $0\neq T$ defines a bounded linear
operator on $\ell^{p}$. Observe that for
$f=\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{n_{0}}$

$\displaystyle(uC_{\phi}\circ
T)(\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{n_{0}})$
$\displaystyle=u\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{\phi^{-1}(n_{0})}$
$\displaystyle=0$

and for $f\notin M$

$\displaystyle(uC_{\phi}\circ T)(f)$ $\displaystyle=uC_{\phi}(T(f))$
$\displaystyle=0.$

This implies that $uC_{\phi}\circ T=0.$ Hence, $uC_{\phi}$ is a left zero
divisor. ∎

######  Remark 3.6.

The following example shows that the converse of the above Theorem 3.5 is not
true.

######  Example 3.7.

Let $\phi:\mathbb{N}\to\mathbb{N}$ be defined as
$\phi(2n-1)=\phi(2n)=n~{}\forall~{}n\geq 1$ and $u\in\ell^{\infty}$ as
$u(1)=u(2)=0$ and $u(n)=1~{}\forall~{}n\geq 3.$ It is easy to see that null
space $M$ of $uC_{\phi}$ is one dimensional. Let $T$ denote the projection on
$M.$ Then $T$ is a non-zero bounded linear operator on $\ell^{p}$ and
$uC_{\phi}\circ T=0.$ Hence $uC_{\phi}$ is a left zero divisor.

######  Theorem 3.8.

Suppose $u\in\ell^{\infty}$ is such that $u(n)\neq
0~{}\forall~{}n\in\mathbb{N}$. Then $uC_{\phi}$ is a left zero divisor if and
only if $\phi$ is not surjective.

###### Proof.

Suppose $\phi$ is surjective. Since $u(n)\neq 0~{}\forall~{}n\in\mathbb{N},$
hence by Theorem 2.4, $uC_{\phi}$ is injective. Let
$T\in\mathcal{B}(\ell^{p})$ be such that $uC_{\phi}\circ T=0.$ Then

$(uC_{\phi}\circ
T)(\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{n})=uC_{\phi}(T(\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{n}))=0\text{
for each }n\geq 1.$

Now injectivity of $uC_{\phi}$ implies that
$T(\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{n})=0~{}\forall~{}n\geq 1.$
Therefore $T=0.$ Hence, $uC_{\phi}$ can not be a left zero divisor.
Conversely, suppose $\phi$ is not surjective. Then, again by Theorem 2.4,
$uC_{\phi}$ is not injective. Hence there exists $0\neq f_{0}\in\ell^{p}$ such
that $uC_{\phi}(f_{0})=0.$ Now, define a linear operator
$T:\ell^{p}\to\ell^{p}$ as follows.

$T(f)=f(1)f_{0}~{}\forall~{}f\in\ell^{p}.$

Clearly, $T\neq 0$ since
$T(\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{1})=f_{0}\neq 0.$ Also, for each
$f\in\ell^{p},$

$\|Tf\|=|f(1)|\|f_{0}\|\leq\|f\|\|f_{0}\|.$

Hence $T$ is bounded. Further, observe that for each $f\in\ell^{p},$

$uC_{\phi}(T(f))=uC_{\phi}(f(1)f_{0})=f(1)uC_{\phi}(f_{0})=0.$

Hence $uC_{\phi}\circ T=0.$ Therefore $uC_{\phi}$ is a left zero divisor. ∎

Now we give examples of two weighted composition operators successively: one
that is a left zero divisor and another that is not a left zero divisor.

######  Example 3.9.

Let $\phi:\mathbb{N}\to\mathbb{N}$ be defined as
$\phi(n)=n^{2}~{}\forall~{}n\in\mathbb{N}$ and $u\in\ell^{\infty}$ be defined
as $u(n)=\frac{1}{n}~{}\forall~{}n\geq 1.$ Then by Theorem 3.8, $uC_{\phi}$ is
a left zero divisor.

######  Example 3.10.

Let $\phi:\mathbb{N}\to\mathbb{N}$ be defined as $\phi(n)=n~{}\forall~{}n\geq
1$ and $u\in\ell^{\infty}$ be defined as $u(n)=1+\frac{1}{n}~{}\forall~{}n\geq
1.$ Then by Theorem 3.8, $uC_{\phi}$ is not a left zero divisor.

######  Theorem 3.11.

Suppose $u\in\ell^{\infty}$ is such that $u(n)\neq
0~{}\forall~{}n\in\mathbb{N}$. If $\phi$ is not invertible, then
$uC_{\phi}\in\mathcal{B}(\ell^{p})$ is a zero divisor.

###### Proof.

The proof follows from Theorem 3.2 and Theorem $\ref{anurag13}.$ ∎

######  Theorem 3.12.

Suppose $u\in\ell^{\infty}$ is bounded away from zero. Then
$uC_{\phi}\in\mathcal{B}(\ell^{p})$ is a zero divisor if and only if $\phi$ is
not invertible.

###### Proof.

If $\phi$ is invertible then, by Theorem $\ref{anurag17},$ $uC_{\phi}$ is also
invertible. Hence $uC_{\phi}$ can not be zero divisor. Conversely, if $\phi$
is not invertible, then by Theorem $\ref{Anurag34},$ $uC_{\phi}$ is a zero
divisor. ∎

The subsequent theorem establishes a connection between the antinormality of a
weighted composition operator and its characteristic of being a zero divisor.

######  Theorem 3.13.

Let $u\in\ell^{\infty}$ be bounded away from zero such that $uC_{\phi}$ is
antinormal. Then $uC_{\phi}$ is a zero divisor in $\mathcal{B}(\ell^{2}).$

###### Proof.

Let $u\in\ell^{\infty}$ be bounded away from zero. Suppose $uC_{\phi}$ is
antinormal, then by Theorem $\ref{DAntinormal},$ $\phi$ is not bijective.
Hence by Theorem 3.12, $uC_{\phi}$ is a zero divisor. ∎

Here is an example of an antinormal weighted composition operator that is also
a zero divisor.

######  Example 3.14.

Define $\phi:\mathbb{N}\to\mathbb{N}$ as
$\phi(n)=n^{2}~{}\forall~{}n\in\mathbb{N}$ and $u\in\ell^{\infty}$ as
$u(n)=1~{}\forall~{}n\geq 1.$ Then by Theorem 3.8, $uC_{\phi}$ is a left zero
divisor.

In the following theorem, we characterize another class of weighted
composition operators, which are left zero divisors.

######  Theorem 3.15.

Suppose $u\in\ell^{\infty}$ and $\phi:\mathbb{N}\to\mathbb{N}.$ If for some
$n_{0}\in\mathbb{N},~{}\phi^{-1}(n_{0})\neq\emptyset$ and
$u(m)=0~{}\forall~{}m\in\phi^{-1}(n_{0}),$ then $uC_{\phi}$ is a left zero
divisor.

###### Proof.

Let $f_{0}=\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{n_{0}}$ and $M=[f_{0}].$
Since $M$ is a one dimensional subspace of $\ell^{p},$ then there exists a
closed subspace $N$ of $\ell^{p}$ such that $\ell^{p}=M\bigoplus N.$ Let $T$
denote the projection on $M.$ Clearly, $0\neq T$ defines a bounded linear
operator on $\ell^{p}$. Observe that for
$f=\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{n_{0}}$

$\displaystyle(uC_{\phi}\circ
T)(\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{n_{0}})$
$\displaystyle=u\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{\phi^{-1}(n_{0})}$
$\displaystyle=0$

and for $f\notin M$

$\displaystyle(uC_{\phi}\circ T)(f)$ $\displaystyle=uC_{\phi}(T(f))$
$\displaystyle=0.$

This implies that $uC_{\phi}\circ T=0.$ Hence, $uC_{\phi}$ is a left zero
divisor. ∎

######  Theorem 3.16.

Let $u\in\ell^{p}$ and $\phi:\mathbb{N}\to\mathbb{N}.$ If $u(n_{0})=0$ for
some $n_{0}\in\mathbb{N},$ then $uC_{\phi}$ is a right zero divisor.

###### Proof.

Observe that for any $f\in\ell^{p},$

$\displaystyle(uC_{\phi}\circ f)(n_{0})$
$\displaystyle=u(n_{0})(f\circ\phi)(n_{0})$
$\displaystyle=0\quad(\text{since}~{}u(n_{0})=0).$

Let $g=\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{n_{0}}.$ Since for each
$f\in\ell^{p},$

$\displaystyle 0=(uC_{\phi}\circ f)(n_{0})$ $\displaystyle\neq g(n_{0})=1.$

Therefore $g\notin\mathcal{R}(C_{\phi}).$ Hence $uC_{\phi}$ is not surjective.
Consequently, using the arguments given in Theorem 3.2, $uC_{\phi}$ is a right
zero divisor. ∎

######  Theorem 3.17.

Let $u\in\ell^{\infty}$ and $\phi:\mathbb{N}\to\mathbb{N}.$ Then $uC_{\phi}$
is injective if and only if $\phi$ is surjective and for each $n\geq 1,$ there
exists $m_{n}\in\phi^{-1}(n)$ such that $u(m_{n})\neq 0.$

###### Proof.

Suppose $uC_{\phi}$ is injective. Then

$\displaystyle
uC_{\phi}(\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{n})\neq
0~{}~{}\text{for each}~{}n\geq 1$ $\displaystyle\implies
u\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{\phi^{-1}(n)}\neq
0~{}~{}\text{for each}~{}n\geq 1.$

This implies that for each $n\geq 1,~{}\phi^{-1}(n)\neq\emptyset$ and there
exists $m_{n}\in\phi^{-1}(n)$ such that $u(m_{n})\neq 0.$
Conversely, suppose for each $n\geq 1$ there exists $m_{n}\in\phi^{-1}(n)$
such that $u(m_{n})\neq 0.$
Then

$\displaystyle
u\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{\phi^{-1}(n)}\neq
0~{}~{}\text{for each}~{}n\geq 1$ $\displaystyle\implies
uC_{\phi}(\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{n})\neq
0~{}~{}\text{for each}~{}n\geq 1.$

Therefore $uC_{\phi}$ is injective. ∎

The subsequent corollary is a straightforward outcome of the aforementioned
theorem.

######  Corollary 3.18.

Let $u\in\ell^{\infty}$ and $\phi:\mathbb{N}\to\mathbb{N}.$ Then $uC_{\phi}$
is not a left zero divisor if and only if $\phi$ is surjective and for each
$n\geq 1$ there exists $m_{n}\in\phi^{-1}(n)$ such that $u(m_{n})\neq 0.$

###  3.2.  Composition operators on $L^{p}(\mu)~{}(1\leq p\leq\infty)$ spaces
as zero divisors

In this section, we study composition operators on $L^{p}(\mu)$ spaces, which
are zero divisors.

######  Theorem 3.19.

Suppose $\phi:X\to X$ is a non-singular measurable transformation such that
$\frac{d\mu\phi^{-1}}{d\mu}$ is bounded away from zero on the complement of
$Z(\frac{d\mu\phi^{-1}}{d\mu}).$ Then $C_{\phi}\in\mathcal{B}(L^{2}(\mu))$ is
surjective if and only if $\phi$ is injective almost everywhere.

###### Proof.

Let $\phi:X\to X$ be a non-singular measurable transformation such that
$\frac{d\mu\phi^{-1}}{d\mu}$ is bounded away from zero on the complement of
zero-set of $\frac{d\mu\phi^{-1}}{d\mu}.$ Then by Theorem 2.2 [20] range of
$C_{\phi}$ is closed in $L^{2}(\mu)$.
Suppose $\phi$ is injective almost everywhere. Let $E$ be any measurable set
with $0<\mu(E)<\infty.$ Then there exists a measurable set $E_{0}$ with
$\mu(E_{0})=0$ and $\mu(E)=\mu(E\cap(X-E_{0})).$
Let $F=E\cap(X-E_{0})$ and
$f=\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{\phi(F)}.$ Then
$C_{\phi}(f)=\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{F}=\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{E}.$
Thus for each measurable set $E$, with $0<\mu(E)<\infty$, there exists $f\in
L^{2}(\mu)$ such that
$C_{\phi}(f)=\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{E}.$ This proves that
$\mathcal{R}(C_{\phi})$ is dense in $L^{2}(\mu)$. Consequently, $C_{\phi}$ is
surjective.
Conversely, suppose $C_{\phi}$ is surjective. To show $\phi$ is injective
almost everywhere.
If not, then there exist two distinct measurable sets $E_{1}$ and $E_{2}$ with
$0<\mu(E_{1})<\infty,~{}0<\mu(E_{2})<\infty$ such that
$\mu(E_{1}-E_{2})>0,~{}\mu(E_{2}-E_{1})>0$ and $\phi(E_{1})=\phi(E_{2}).$ Let
$g=\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{E_{1}}\in L^{2}(\mu).$ Since
$C_{\phi}$ is surjective, hence there exists $f\in L^{2}(\mu)$ such that

$C_{\phi}(f)=\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{E_{1}}\implies
f\circ\phi=\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{E_{1}}.$

Since $\phi(E_{1})=\phi(E_{2}),$ therefore

$\displaystyle 1$ $\displaystyle=f\circ\phi(E_{1})$
$\displaystyle=f\circ\phi(E_{2})$
$\displaystyle=\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{E_{1}}(E_{2})$
$\displaystyle=0\quad on~{}E_{2}-E_{1}$

which is a contradiction because $\mu(E_{2}-E_{1})>0$. Therefore $\phi$ must
be injective almost everywhere. ∎

######  Theorem 3.20.

$C_{\phi}\in\mathcal{B}(L^{p}(\mu))$ is a left zero divisor if and only if
there exists a measurable set $E_{0}$ with $0<\mu(E_{0})<\infty$ such that
$\mu(\phi^{-1}(E_{0}))=0.$

###### Proof.

Suppose there exists a measurable set $E_{0}$ with $0<\mu(E_{0})<\infty$
satisfying $\mu(\phi^{-1}(E_{0}))=0.$ Then for $u=1,$ Theorem 3.21 implies
that $C_{\phi}$ is a left zero divisor.
Conversely, suppose there does not exist any measurable set $E_{0}$ with
$0<\mu(E_{0})<\infty$ satisfying $\mu(\phi^{-1}(E_{0}))=0.$ Then for each
measurable set $E$ with $\mu(\phi^{-1}(E))=0$ implies $\mu(E)=0.$ Hence by
Theorem LABEL:amar2, $C_{\phi}$ is injective. Let $C_{\phi}\circ T=0.$ Then
for each measurable set $E,$ injectivity of $C_{\phi}$ implies that
$T(\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{E})=0.$ This implies that $T=0.$
Hence $C_{\phi}$ can not be left zero divisor.
This completes the proof. ∎

######  Theorem 3.21.

Let $uC_{\phi}\in\mathcal{B}(L^{p}(\mu)).$ Then it is a left zero divisor if
there exists a measurable set $E_{0}$ with $0<\mu(E_{0})<\infty$ such that
$\phi^{-1}(E_{0})=A\cup B$ with $u=0$ a.e. on $A$ and $\mu(B)=0.$

###### Proof.

Suppose there exists a measurable set $E_{0}$ with $0<\mu(E_{0})<\infty$ such
that $\phi^{-1}{(E_{0})}=A\cup B,$ where $u=0$ a.e. on $A$ and $\mu(B)=0.$ Let
$f_{0}=\raisebox{0.0pt}[6.2pt][6.2pt]{$\chi$}_{E_{0}}$ and $M=[f_{0}].$ Let
$T$ denote the projection operator on $M.$ Clearly, $0\neq T$ is a bounded
linear operator on $L^{p}(\mu)$. Now for the measurable set $E_{0},$

$\displaystyle(uC_{\phi}\circ
T)(\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{E_{0}})$
$\displaystyle=uC_{\phi}(T(\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{E_{0}}))$
$\displaystyle=uC_{\phi}(\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{E_{0}})$
$\displaystyle=u\raisebox{0.0pt}[6.19998pt][6.19998pt]{$\chi$}_{\phi^{-1}(E_{0})}$
$\displaystyle=0\quad(\text{since}~{}\phi^{-1}(E_{0})=A\cup
B~{}\text{and}~{}u=0~{}\text{a.e. on}~{}A~{}\text{and}~{}\mu(B)=0)$

and for $f\notin M$

$\displaystyle(uC_{\phi}\circ T)(f)$ $\displaystyle=uC_{\phi}(T(f))$
$\displaystyle=0.$

Hence $uC_{\phi}\circ T=0.$ Therefore $uC_{\phi}$ is a left zero divisor in
$\mathcal{B}(L^{p}(\mu)).$ ∎

###  3.3. Topological divisors of zero in $C(X)$

Consider $X$ as a compact Hausdorff space, and let $C(X)$ represent the Banach
algebra consisting of all continuous complex-valued functions on X. In [4], F.
Azarpanah, D. Esmaeilvandi and A.R. Salehi characterized the regular elements
in $C(X).$ With this context, the following results have been obtained:

######  Theorem 3.22.

Let $f\in C(X).$ Then $f$ is a TDZ if and only if it is singular in $C(X).$

###### Proof.

If $f$ is a TDZ in $C(X)$ then it is clearly a singular element.
Conversely, suppose that $f\in C(X)$ is a singular. Then there exists
$x_{0}\in X$ such the $f(x_{0})=0.$ Now, by continuity of $f,$ for each $n\geq
1,$ there exists an open set $E_{n}$ containing $x_{0}$ such that

$|f(x)|<\frac{1}{n}~{}\text{for each}~{}x\in E_{n}.$

Let $A_{n}=E^{c}_{n}$ and $B_{n}=\\{x_{0}\\}.$ By Urysohn’s Lemma, there
exists a function $f_{n}:X\to[0,1]$ such that

$f_{n}(B_{n})=1~{}\text{and}~{}f_{n}(A_{n})=0.$

Clearly $\|f_{n}\|=1$ and $\|ff_{n}\|<\frac{1}{n}.$ This implies that
$\|ff_{n}\|\to 0$ as $n\to\infty.$ Hence $f$ is a TDZ. ∎

######  Theorem 3.23.

Let $h\in C(X)$ and $M_{h}:C(X)\to C(X)$ be the multiplication operator
defined as $M_{h}f=hf.$ Then $M_{h}$ is a TDZ in $\mathcal{B}(C(X))$ if and
only $h$ is a TDZ in $C(X)$.

###### Proof.

Suppose $h\in C(X)$ be a TDZ. Let $\\{h_{n}\\}_{n=1}^{\infty}$ be a sequence
in $C(X)$ with $\|h_{n}\|=1\ \forall\ n\in\mathbb{N}$ such that $\|hh_{n}\|\to
0$ as $n\to\infty.$
For each $n\geq 1,$ define $M_{h_{n}}:C(X)\to
C(X)~{}\text{by}~{}M_{h_{n}}(f)=h_{n}f.$ Observe that

$\|M_{h_{n}}\|=\|h_{n}\|=1\ \forall\
n\in\mathbb{N}~{}\text{and}~{}M_{h}M_{h_{n}}(f)=hh_{n}f.$

Then $\|M_{h}M_{h_{n}}\|=\|hh_{n}\|\to 0~{}\text{as}~{}n\to\infty.$ Therefore,
$M_{h}$ is a TDZ in $\mathcal{B}(C(X))$.
Conversely, suppose $h\in C(X)$ is not a TDZ in $C(X)$. Then $h$ is a regular
element in $C(X)$. Therefore there exists a $g\in C(X)$ such that $h\cdot
g=g\cdot h=1.$ This implies that $M_{hg}=\text{I}.$ Hence
$M_{hg}=M_{h}M_{g}=M_{g}M_{h}=\text{I}.$ This shows that $M_{h}$ is regular
element in $\mathcal{B}(C(X))$. Hence $M_{h}$ can not be a TDZ.
∎

######  Theorem 3.24.

Let $\phi:X\to X$ be a continuous map and $C_{\phi}:C(X)\to C(X)$ be the
composition operator defined as $C_{\phi}f=f\circ\phi.$ Then $C_{\phi}$ is a
bounded linear operator on $C(X)$ with $\|C_{\phi}\|=1.$
Further, the following statements hold.
(1) If $\phi$ is bijective, then $C_{\phi}$ is invertible in
$\mathfrak{B}(C(X)).$
(2) $C_{\phi}$ is injective if and only if $\phi$ is surjective.
(3) $C_{\phi}$ is surjective if and only if $\phi$ is injective.

###### Proof.

(1) If $\phi:X\to X$ is bijective and continuous, then it is a homeomorphism.
Hence $C^{-1}_{\phi}=C_{\phi^{-1}}.$
(2) Suppose $\phi$ is surjective and let $C_{\phi}f=0.$ Then
$(f\circ\phi)(x)=0~{}\forall~{}x\in X.$ Now, surjectivity of $\phi$ implies
that $f(x)=0~{}\forall~{}x\in X.$ Therefore $f=0.$ This implies $C_{\phi}$ is
injective.
Conversely, suppose $\phi$ is not surjective. Then $\phi(X)$ is a proper
closed and compact subset of $X.$ Since $X$ is completely regular, therefore
for $x_{0}\in X\setminus\phi(X)$ there exists a non-zero continuous function
$f_{0}:X\to[0,1]$ such that

$f_{0}(x_{0})=1~{}\text{and}~{}f_{0}(\phi(X))=0.$

This implies that $C_{\phi}f_{0}=0.$ Since $f_{0}\neq 0,$ therefore $C_{\phi}$
is not injective. This completes the proof.
(3) Suppose $\phi$ is injective. Let $g\in C(X)$ and $\phi^{-1}:\phi(X)\to X.$
Define $f=g\circ\phi^{-1}.$ By Tietz extension theorem, $f$ can be
continuously extended on $X.$ Clearly,

$\displaystyle C_{\phi}(f)$ $\displaystyle=f\circ\phi$
$\displaystyle=g\circ\phi^{-1}\circ\phi$ $\displaystyle=g.$

Hence $C_{\phi}$ is surjective.
Conversely, suppose $\phi$ is not injective. Then there exists two distinct
elements $x_{1},x_{2}\in X$ such that $\phi(x_{1})=\phi(x_{2})$ and a
continuous function $f_{0}:X\to[0,1]$ such that

$f_{0}(x_{1})=0,~{}f_{0}(x_{2})=1.$

We, claim that $f_{0}\notin\mathcal{R}(C_{\phi}).$ Observe that if
$g\in\mathcal{R}(C_{\phi}),$ then $g=f\circ\phi$ for some $f\in C(X).$
Then,

$\displaystyle g(x_{1})=f\circ\phi(x_{1})=f\circ\phi(x_{2})=g(x_{2}).$

This implies that $f_{0}\notin\mathcal{R}(C_{\phi}).$ Hence $C_{\phi}$ is not
surjective. ∎

###  3.4.  Polynomially topological divisors of zero

Motivated by the notion of polynomially compact operators[12], we give the the
following definition.

######  Definition 3.25.

Let $\mathfrak{B}$ be a Banach algebra. An element $z\in\mathfrak{B}$ is said
to be polynomially TDZ if $p(z)$ is a TDZ for some non-zero complex polynomial
$p$.

######  Theorem 3.26.

Every element in $C(X)$ is a polynomially TDZ.

###### Proof.

If $f\in C(X)$ is a TDZ. For the polynomial $p(x)=x,~{}f$ becomes a
polynomially TDZ.
Now, let $f\in C(X)$ be such that $f(x)\neq 0~{}\forall~{}x\in X.$ Let
$x_{0}\in X$ and $p(x)=x-f(x_{0}).$ Then, $p(f)(x_{0})=0.$ Then $p(f)$ is a
TDZ in $C(X).$ ∎

######  Theorem 3.27.

Let $X$ be a compact Hausdorff space and $h:X\to X$ be a continuous map. Let
$M_{h}:C(X)\to C(X)$ be the multiplication operator defined as $M_{h}f=hf.$
Then $M_{h}$ is a polynomially TDZ in $\mathcal{B}(C(X))$.

###### Proof.

Let $x_{0}\in X$ and $p(z)=z-h(x_{0}).$ Clearly $p(h(x))\in C(X)$ and
$p(h(x_{0}))=0.$ Hence by Theorem $\ref{CHS TDZ},~{}p(h(x))$ is a TDZ in
$C(X)$ which is also a singular element. For each $f\in C(X)$

$p(M_{h})f=(M_{h}-h(x_{0})I)f$ $=(h-h(x_{0}))f$ $=M_{h-h(x_{0})}f.$

Hence $p(M_{h})=M_{p(h)}.$ By Theorem 3.23, $p(M_{h})$ is a TDZ in
$\mathcal{B}(C(X))$. Hence $M_{h}$ is a polynomially TDZ in
$\mathcal{B}(C(X))$. ∎

######  Theorem 3.28.

Let $\mathfrak{B}$ be a Banach algebra and let $z\in\mathfrak{B}$ be the TDZ.
Then $z$ is also a polynomially TDZ.

###### Proof.

Let $z\in\mathfrak{B}$ be the TDZ in $\mathfrak{B}.$ For the polynomial
$p(x)=x,$ clearly $p(z)$ is a TDZ. Hence $z$ is a polynomially TDZ in
$\mathfrak{B}.$ ∎

######  Remark 3.29.

The converse of above theorem is not true. Let $\mathfrak{B}$ be an arbitrary
unital Banach algebra with the unit element denoted as $e$ in $\mathfrak{B}.$
Let $p(x)=x-1,$ then $p(e)=e-e=0$ which is a TDZ, while identity $e$ is not a
TDZ in $\mathfrak{B}$.

######  Theorem 3.30.

In $L^{\infty}(\mu)$ every element is a polynomially TDZ.

###### Proof.

Let $h\in L^{\infty}(\mu)$ and $\alpha\in\textrm{ess.range}(h).$ Then for the
polynomial $p(x)=x-\alpha,$ clearly $0\in\textrm{ess.range}(p(h)).$ Hence by
the Theorem 2.9, $p(h)$ is a TDZ. Therefore $h$ is a polynomially TDZ. ∎

######  Theorem 3.31.

Let $h\in L^{\infty}(\mu)$ and $1\leq p\leq\infty.$ Define the multiplication
operator $M_{h}:L^{p}(\mu)\to L^{p}(\mu)$ by $M_{h}(f)=h\cdot f=hf$. Then
$M_{h}$ is a polynomially TDZ in $\mathcal{B}(L^{p}(\mu)).$

###### Proof.

Let $h\in L^{\infty}(\mu).$ Then by Theorem 3.30 there exists a non-zero
polynomial $p$ such that $p(h)$ is a TDZ. Consequently by the Theorem 2.10,
$p(M_{h})=M_{p(h)}$ is a TDZ in $\mathcal{B}(L^{p}(\mu)).$ Hence $M_{h}$ is a
polynomially TDZ in $\mathcal{B}(L^{p}(\mu)).$ ∎

###  3.5. Strongly topological divisors of zero

The following notion is a generalization of the concept of TDZ.

######  Definition 3.32.

Let $X$ be a non-empty set. An operator $T\in\mathcal{B}(X)$ is called
strongly TDZ if there exists a sequence $\\{T_{n}\\}_{n=1}^{\infty}$ in
$\mathcal{B}(X)$ such that

  1. (1)

$\|T_{n}\|=1~{}\forall~{}n\in\mathbb{N}$

  2. (2)

and $\|T_{n}Tx\|\to 0$ or $\|TT_{n}x\|\to 0$ as $n\to 0~{}~{}\forall x\in X.$

######  Theorem 3.33.

Let $X$ be a Banach space. If $T\in\mathcal{B}(X)$ is a TDZ then $T$ is a
strongly TDZ.

###### Proof.

If $T\in\mathcal{B}(X)$ is a TDZ then there exists a sequence
$\\{T_{n}\\}_{n=1}^{\infty}$ such that
$\|T_{n}\|=1~{}\forall~{}n\in\mathbb{N}$ and $\|T_{n}T\|\to 0$ or
$\|TT_{n}\|\to 0$ as $n\to\infty.$ Let $x\in X$ then

$\|TT_{n}x\|\leq\|TT_{n}\|\|x\|\to 0~{}or~{}\|T_{n}Tx\|\leq\|T_{n}T\|\|x\|\to
0~{}as~{}n\to 0.$

Thus $TT_{n}x\to 0$ or $T_{n}Tx\to 0~{}\forall x\in X.$ Hence $T$ is a
strongly TDZ. ∎

######  Remark 3.34.

The subsequent example illustrates that the converse of the above theorem 3.33
is not true.

######  Example 3.35.

Let $T=I\in\mathcal{B}(\ell^{p})~{}(1\leq p\leq\infty)$ be the indentity
operator. For each $n\in\mathbb{N},$ define $T_{n}:\ell^{p}\to\ell^{p}$ as

$T_{n}(x)=(0,0,...,x_{n+1},0,0,...).$

Clearly $\|T_{n}\|=1~{}\forall~{}n\in\mathbb{N}$ and
$\|TT_{n}(x)\|=|x_{n+1}|\to 0$ as $n\to\infty.$ Hence $T$ is a strongly TDZ
but not a TDZ in $\mathcal{B}(\ell^{p})$.

######  Theorem 3.36.

Let $H$ be a separable Hilbert space and $T\in\mathcal{B}(H).$ Then $T$ is a
strongly TDZ in $\mathcal{B}(H)$.

###### Proof.

Let $\\{e_{n}:n\geq 1,\|e_{n}\|=1\\}$ be an orthonormal basis for $H$. Then
for each $x\in H$ there exists a unique sequence of scalars
$\\{\alpha_{n}\\}_{n=1}^{\infty}$ such that
$x=\sum_{k=1}^{\infty}\alpha_{k}e_{k}.$ For each $n\in\mathbb{N},$ define
$T_{n}:H\to H$ by

$T_{n}(x)=\sum_{k=n+1}^{\infty}\alpha_{k}e_{k}.$

Clearly $\|T_{n}(x)\|\leq\|x\|$ and $T_{n}(e_{n+1})=e_{n+1}.$ This implies
that $\|T_{n}\|=1~{}\forall~{}n\in\mathbb{N}.$ Now observe that

$\|T_{n}(x)\|=\|\sum_{k=n+1}^{\infty}\alpha_{k}e_{k}\|\to
0~{}\text{as}~{}n\to\infty.$

Hence for each $T\in\mathcal{B}(H)$ and $H\in H$

$\|TT_{n}(x)\|\leq\|T\|\|T_{n}(x)\|\to 0~{}as~{}n\to\infty.$

Therefore $T$ is a strongly TDZ in $\mathcal{B}(H)$. ∎

######  Remark 3.37.

We conjecture that in the above theorem $H$ can be replaced by any Banach
space with Schauder basis.

######  Remark 3.38.

Since the space $\ell^{p}~{}(1\leq p\leq\infty)$ has
$\\{e_{n}\\}_{n=1}^{\infty}$ as a Schauder basis, where each $e_{n}$ is
defined as $e_{n}=(0,0,...,1,0,0,...).$ Therefore by Theorem 3.36, each
$T\in\mathcal{B}(\ell^{p})~{}(1\leq p<\infty)$ is a strongly TDZ in
$\mathcal{B}(\ell^{p})$.

In the following example, we construct a class of TDZ in
$\mathcal{B}(\ell^{p})~{}(1\leq p\leq\infty).$

######  Example 3.39.

Let
$c_{0}=\\{(y_{n})_{n=1}^{\infty}:y_{n}\in\mathbb{C}~{}\forall~{}n\in\mathbb{N}~{}\text{and}~{}y_{n}\to
0~{}\text{as}~{}n\to\infty\\}.$ Let $y\in c_{0}$ and $T:\ell^{p}\to\ell^{p}$
be defined by $T(x):=(x_{1}y_{1},x_{2}y_{2},...,x_{n}y_{n},...)$. Then $T$ is
TDZ in $\mathcal{B}(\ell^{p}).$

###### Proof.

For each $n\in\mathbb{N},$ define $T_{n}:\ell^{p}\to\ell^{p}$ by

$T_{n}(x)=(0,0,...,x_{n+1}y_{n+1},x_{n+2}y_{n+2},...).$

Then $\|T_{n}\|=1$ and
$T_{n}T(x)=(0,0,...,x_{n+1}y_{n+1},x_{n+2}y_{n+2},...).$ Clearly, for each
$x\in\ell^{p}$

$\|T_{n}Tx\|\leq\sup_{k\geq n+1}|y_{k}|\|x\|.$

Since $y\in c_{0},$ then $\sup_{k\geq n+1}|y_{k}|\to 0$ as $n\to 0.$ Therefore
$\|T_{n}Tx\|\to 0$ as $n\to 0.$ Hence $T$ is a TDZ. ∎

######  Remark 3.40.

We note that the class of operators in the above Example 3.39 forms a class of
compact operators.

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