# Spectral properties of the Rhaly operator on weighted null sequence spaces
and associated operator ideals

Arnab Patra  Department of mathematics, Indian Institute of Technology Bhilai,
Chhattisgarh, India 492015<EMAIL_ADDRESS>,  Jyoti Rani  Department
of mathematics, Indian Institute of Technology Bhilai, Chhattisgarh, India
492015<EMAIL_ADDRESS>and  Sanjay Kumar Mahto  Department of
mathematics, RNAR College Samastipur, LNMU Darbhanga, India
<EMAIL_ADDRESS>
###### Abstract.

In this article, a comprehensive study is made on the continuity, compactness,
and spectrum of the lower triangular terraced matrix, introduced by H. C.
Rhaly, Jr. [Houston J. Math. 15(1): 137-146, 1989], acting on the weighted
null sequence spaces with bounded, strictly positive weights. Several spectral
subdivisions such as point spectrum, residual spectrum, and continuous
spectrum are also discussed. In addition, a new class of operator ideal
$\chi_{c_{0}(r)}^{(s)}$ associated to the Rhaly operator on weighted $c_{0}$
space is defined using the concept of $s$-number and it is proved that under
certain condition, $\chi_{c_{0}(r)}^{(s)}$ forms a quasi-Banach closed
operator ideal.

###### Key words and phrases:

Rhaly operators, Infinite matrices, Fine spectrum, Sequence spaces

###### 2020 Mathematics Subject Classification:

Primary 47A10, 47B37; Secondary 46B45

##  1\. Introduction

The linear map $C$ which maps a sequence $\\{x_{n}\\}$ of real or complex
numbers to its sequence of averages $\\{\frac{x_{1}+\cdots+x_{n}}{n}\\}$ is
the well-known discrete Cesàro operator. Extensive study can be found in the
literature on the continuity and spectrum of $C$ defined over sequence spaces
such as $c_{0}$, $c$, $\ell_{p}$, $bv_{p}\ (1\leq p<\infty)$, $bv_{0}$, etc
[19, 1, 8, 13, 2, 17]. Recently Albanese et al. studied the spectral
properties of Cesàro operator defined over the weighted sequence spaces [3, 4,
5]. Several generalised version of the Cesàro operator such as $p-$Cesàro
operator [20, 7], discrete generalized Cesàro operator [21, 30, 31],
$q-$Cesàro operator [28, 10, 11], etc. are also studied. For a sequence
$a=\\{a_{n}\\}$ of real or complex numbers Rhaly [20] introduced the terraced
matrix $R_{a}$ also known as Rhaly matrix where

$R_{a}=\begin{pmatrix}a_{1}&{0}&{0}&{0}&\cdots\vspace{1.5mm}\\\
a_{2}&a_{2}&{0}&{0}&\cdots\vspace{1.5mm}\\\
a_{3}&a_{3}&a_{3}&{0}&\cdots\vspace{1.5mm}\\\
\vdots&\vdots&\vdots&\vdots&\ddots\end{pmatrix}.$

The operator represented by the above matrix is known as the Rhaly operator.
The Cesàro operator $C$ can be obtained by taking $a_{n}=\frac{1}{n}.$ Rhaly
[20] also considered the case $a_{n}=\frac{1}{n^{p}},\ p\in\mathbb{R}$ which
is the $p$-cesaro operator. Many researchers investigated the spectrum of
Rhaly operators defined over several classical sequence spaces such as $c_{0}$
[25, 27, 24], $\ell_{p}$ [26], $bv_{0}$ [29], etc. To the best of our
knowledge, no investigation is carried out so far on the spectral properties
of Rhaly operators over the weighted sequence spaces. The present work is an
attempt in this direction. The aim of this paper is to study the continuity,
compactness and spectrum of the Rhaly operator defined over the weighted null
sequence space $c_{0}(r)$ where $r=\\{r_{n}\\}$ is the weight vector. Also
using the concept of $s$-number sequences, we introduce a new class of
operator ideal $\chi_{c_{0}(r)}^{(s)}$ related to the Rhaly operator $R_{a}$
acting on weighted $c_{0}$ spaces. It is demonstrated that under some
assumptions on the sequence $a=\\{a_{n}\\}$, the class forms an quasi-Banach
closed operator ideal. Moreover, an inclusion relation among operator ideals
are also proved.

The remainder of this paper is organized as follows. Section 2 contains some
notations and some useful results. Section 3 deals with the continuity and
compactness and various spectral subdivision of Rhaly operator defined over
weighted $c_{0}$ spaces. In section 4 we introduce a new class of operator
ideal $\chi_{c_{0}(r)}^{(s)}$ and discussed few of its properties.

##  2\. Preliminaries and Notations

Throughout the article, all the infinite sequences and matrices are indexed by
the set of natural numbers $\mathbb{N}.$ Let $T:X\rightarrow Y$ be a bounded
linear operator where $X$ and $Y$ are complex Banach spaces. The range space
of $T$ and null space of $T$ are denoted by $R(T)$ and $N(T)$ respectively.
The Banach-adjoint of $T$, denoted by $T^{*}$, is a bounded linear operator
$T^{*}:Y^{*}\rightarrow X^{*}$ which is defined by

$(T^{*}\phi)(x)=\phi(Tx)\ \ \ \mbox{for all}\ \phi\in Y^{*}\ \mbox{and}\ x\in
X,$

where $X^{*}$ and $Y^{*}$ are the dual spaces of $X$ and $Y$ respectively.
$\mathcal{B}(X,Y)$ and $\mathcal{K}(X,Y)$ denote the space of all bounded
linear operators and the ideal of all compact operators form $X$ into $Y$. If
$X=Y,$ then $\mathcal{B}(X,Y)$ and $\mathcal{K}(X,Y)$ are denoted by
$\mathcal{B}(X)$ and $\mathcal{K}(X)$ respectively. For any
$T\in\mathcal{B}(X),$ the operator norm is denoted by $\|T\|$ and the norm
$\|.\|_{\infty}$ denotes the supremum norm on the sequence space $c_{0}$. For
any operator $T\in\mathcal{B}(X),$ the resolvent set of $T$ is the set of all
complex numbers $\lambda$ for which the operator $T-\lambda I$ has a bounded
inverse in $X$ where $I$ is the identity operator in $X.$ The resolvent set of
$T$ is denoted by $\rho(T,X).$ The complement of the resolvent set in the
complex plane $\mathbb{C}$ is called the spectrum of $T$ and it is denoted by
$\sigma(T,X)$. The set of points $\lambda\in\mathbb{C}$ for which $N(T-\lambda
I)\neq\\{0\\}$ is called the point spectrum of $T$ and it is denoted by
$\sigma_{p}(T,X)$. The set of points $\lambda\in\mathbb{C}$ for which
$N(T-\lambda I)=\\{0\\},$ and $\overline{R(T-\lambda I)}=X$ but ${R(T-\lambda
I)}\neq X$ is called the continuous spectrum of $T$ and denoted by
$\sigma_{c}(T,X),$ and the set of points $\lambda\in\mathbb{C}$ for which
$N(T-\lambda I)=\\{0\\}$ and $\overline{R(T-\lambda I)}\neq X$ is called the
residual spectrum of $T$ and it is denoted by $\sigma_{r}(T,X)$. The three
sets $\sigma_{p}(T,X),\sigma_{c}(T,X),\sigma_{r}(T,X)$ are disjoint and their
union is the whole spectrum $\sigma(T,X).$

Let $\mathbb{C}^{\mathbb{N}}$ denotes the space of all complex sequences. For
an infinite positive real sequence $r=\\{r_{k}\\},$ the weighted null sequence
space $c_{0}(r)$ is defined as

$c_{0}(r)=\\{\\{x_{k}\\}\in\mathbb{C}^{\mathbb{N}}:\lim_{k\to\infty}r_{k}x_{k}=0\\}$

equipped with the norm $\|x\|_{r}=\sup_{k}|x_{k}|r_{k}$. If $D_{r}$ is the
diagonal matrix with $i$-th diagonal entry $r_{i}$ then, $D_{r}$ is an
isometric isomorphism from $c_{0}(r)$ to $c_{0}$ since $x=\\{x_{k}\\}\in
c_{0}(r)$ implies $D_{r}x=\\{x_{k}r_{k}\\}\in c_{0}$, and
$\|x\|_{r}=\|D_{r}x\|_{\infty}.$ Hence $\|x\|_{r}$ defines a norm on
$c_{0}(r)$ and $(c_{0}(r),\|x\|_{r})$ is a Banach space under this norm. The
dual of $c_{0}(r)$ is linearly isometric with the weighted sequence space
$\ell_{1}(r^{-1})$ where $r^{-1}=\\{\frac{1}{r_{n}}\\}.$ Also it must be noted
that if $\inf_{k}r_{k}>0$ then $c_{0}(r)=c_{0}$ and the norms are equivalent.
Therefore we are interested in the case $\inf_{k}r_{k}=0.$

We record the following lemmas which are useful in this sequel.

######  Lemma 2.1.

[22] The matrix $A=(a_{nk})$ gives rise to a bounded linear operator
$T\in\mathcal{B}(c_{0})$ if and only if the following conditions hold,

  1. (i)

the rows of $A$ are in $\ell_{1}$ and their $\ell_{1}$ norms are bounded,

  2. (ii)

the columns of $A$ are in $c_{0}$.

The operator norm of $T$ is given by the supremum of the $\ell_{1}$ norms of
the rows.

######  Lemma 2.2.

[6, Theorem 1] For two weight vectors $r=\\{r_{n}\\}$, $s=\\{s_{n}\\}$ and an
infinite matrix $A=(a_{nk})$, the matrix $A\in\mathcal{B}(c_{0}(r),c_{0}(s))$
if and only if the following conditions hold,

  1. (i)

$\sup\limits_{n\in\mathbb{N}}s_{n}\sum\limits_{k=1}^{\infty}|\frac{a_{nk}}{r_{k}}|<\infty,$

  2. (ii)

$\lim\limits_{n\to\infty}s_{k}a_{nk}=0\ \forall\ k\in\mathbb{N}.$

Then the norm of $A$ is given by
$\|A\|_{r,s}=\sup\limits_{n\in\mathbb{N}}s_{n}\sum\limits_{k=1}^{\infty}|\frac{a_{nk}}{r_{k}}|$.

######  Lemma 2.3.

[12, p. 59] The bounded linear operator $T:X\rightarrow Y$ has dense range if
and only if $T^{*}$ is one to one.

##  3\. Boundedness and compactness of $R_{a}$

Let $r=\\{r_{n}\\}$ and $s=\\{s_{n}\\}$ are two bounded, strictly positive
weight vectors for the sequence spaces $c_{0}(r)$ and $c_{0}(s)$ respectively.
For any operator $T:c_{0}(r)\to c_{0}(s)$ the operator norm is denoted by
$\|T\|_{r,s}$ and if $T$ is defined from $c_{0}(s)$ to $c_{0}(s)$ then
$\|T\|_{r,s}=\|T\|_{s}$. Following theorem is an immediate application of
Lemma 2.2.

######  Theorem 3.1.

The Rhaly operator $R_{a}:c_{0}(r)\to c_{0}(s)$ is bounded if and only if

$\left\\{s_{n}a_{n}\sum\limits_{k=1}^{n}\frac{1}{r_{k}}\right\\}_{n\in\mathbb{N}}\in\ell_{\infty},$

and in this case, $\left\lVert
R_{a}\right\rVert_{r,s}=\sup\limits_{n}s_{n}\left|a_{n}\right|\sum\limits_{k=1}^{n}\frac{1}{r_{k}}.$

###### Proof.

It follows from Lemma 2.2 that the Rhaly operator $R_{a}$ is a bounded linear
operator from $c_{0}(r)$ to $c_{0}(s)$ if and only if

$\sup\limits_{n}s_{n}\left|a_{n}\right|\sum\limits_{k=1}^{n}\frac{1}{r_{k}}<\infty\
\mbox{and}\ \lim_{n\rightarrow\infty}a_{n}s_{n}=0.$

If
$\sup\limits_{n}s_{n}\left|a_{n}\right|\sum\limits_{k=1}^{n}\frac{1}{r_{k}}<\infty$
then there exists a positive real number $M$ such that,

$s_{n}\left|a_{n}\right|\sum\limits_{k=1}^{n}\frac{1}{r_{k}}<M\ \ \mbox{for
all}\ \ n\in\mathbb{N}.$ (3.1)

Since $r=\\{r_{k}\\}$ is a bounded sequence of strictly positive real numbers,
there exists a real number $M_{1}>0$ such that $\forall n\in\mathbb{N},$

$\sum\limits_{k=1}^{n}\frac{1}{r_{k}}>n\frac{1}{M_{1}}.$

Finally from (3.1) it follows that

$0<s_{n}\left|a_{n}\right|<\frac{MM_{1}}{n}\ \ \mbox{for all}\ \
n\in\mathbb{N},$

and consequently $\lim_{n\rightarrow\infty}a_{n}s_{n}=0.$ Hence
$\sup_{n}s_{n}\left|a_{n}\right|\sum\limits_{k=1}^{n}\frac{1}{r_{k}}<\infty$
implies $\lim_{n\rightarrow\infty}a_{n}s_{n}=0.$ Also the expression of
$\left\lVert R_{a}\right\rVert_{r,s}$ is a direct consequence of Lemma 2.2. ∎

######  Theorem 3.2.

The Rhaly operator $R_{a}:c_{0}(r)\to c_{0}(s)$ is compact if and only if

$\left\\{s_{n}a_{n}\sum\limits_{k=1}^{n}\frac{1}{r_{k}}\right\\}_{n\in\mathbb{N}}\in
c_{0}.$

###### Proof.

Let us consider the linear operator
$T_{r,s}:\mathbb{C}^{\mathbb{N}}\rightarrow\mathbb{C}^{\mathbb{N}}$ defined as

$T_{r,s}(x)=\left\\{a_{n}s_{n}\sum\limits_{k=1}^{n}\frac{x_{k}}{r_{k}}\right\\}_{n\in\mathbb{N}}.$

Then $D_{s}R_{a}=T_{r,s}D_{r}$ and consequently $R_{a}:c_{0}(r)\to c_{0}(s)$
is a compact map if and only if $T_{r,s}\in\mathcal{K}(c_{0}).$ Let
$\left\\{a_{n}s_{n}\sum\limits_{k=1}^{n}\frac{1}{r_{k}}\right\\}_{n\in\mathbb{N}}\in
c_{0}$. Define a sequence of operators $\\{T_{r,s}^{(k)}\\}_{k\in\mathbb{N}}$
where $T_{r,s}^{(k)}:c_{0}\rightarrow c_{0}$ and

$T_{r,s}^{(k)}(x)=\left\\{a_{1}s_{1}\frac{x_{1}}{r_{1}},a_{2}s_{2}\left(\frac{x_{1}}{r_{1}}+\frac{x_{2}}{r_{2}}\right),\cdots,a_{k}s_{k}\sum\limits_{i=1}^{k}\frac{x_{i}}{r_{i}},0,0,\cdots\right\\}.$

For each $k\in\mathbb{N}$ the operator $T_{r,s}^{(k)}$ is a finite rank
operator. Now

$\left\lVert(T_{r,s}-T_{r,s}^{(k)})x\right\rVert_{\infty}=\sup\limits_{n>k}\left|a_{n}\right|s_{n}\sum\limits_{i=1}^{n}\frac{x_{i}}{r_{i}}\leq\left\lVert
x\right\rVert_{\infty}\sup\limits_{n>k}\left|a_{n}\right|s_{n}\sum\limits_{i=1}^{n}\frac{1}{r_{i}}.$

Hence

$\left\lVert
T_{r,s}-T_{r,s}^{(k)}\right\rVert\leq\sup\limits_{n>k}\left|a_{n}\right|s_{n}\sum\limits_{i=1}^{n}\frac{1}{r_{i}}.$

Since $a_{n}s_{n}\sum\limits_{i=1}^{n}\frac{1}{r_{i}}\rightarrow 0$ as
$n\rightarrow\infty$, by taking $k\rightarrow\infty$ on both side of the above
relation we get

$\left\lVert T_{r,s}-T_{r,s}^{(k)}\right\rVert\rightarrow 0.$

Hence the sequence of operators $\\{T_{r,s}^{(k)}\\}$ converges to $T_{r,s}$
under the operator norm, and this implies $T_{r,s}\in\mathcal{K}(c_{0})$.

For the converse part we adopt similar approach as in Proposition 2.2 [4]. Let
$R_{a}:c_{0}(r)\rightarrow c_{0}(s)$ is a compact operator. This implies
$T_{r,s}\in\mathcal{K}(c_{0}).$ This also implies $T_{r,s}(B_{c_{0}}[0,1])$ is
a relatively compact set where $B_{c_{0}}[0,1]$ denotes the closed unit ball
in $c_{0}.$ Therefore using the result of [9, p. 15] there exists a sequence
$y=\\{y_{n}\\}\in c_{0}$ such that

$\left|(T_{r,s}(x))_{n}\right|\leq\left|y_{n}\right|\ \ \mbox{for all}\ \
n\in\mathbb{N},\ x\in B_{c_{0}}[0,1].$

Since $y\in c_{0},$ for every $\epsilon>0$ there exists a $n_{0}\in\mathbb{N}$
such that,

$\left|(T_{r,s}(x))_{n}\right|<\epsilon\ \ \mbox{for all}\ \ n>n_{0}.$

Let us consider for each $n\in\mathbb{N}$ the sequence $\\{u^{(n)}\\}$ with
$\\{u_{k}^{(n)}\\}_{k\in\mathbb{N}}\in B_{c_{0}}[0,1]$ such that for
$k\in\\{1,2,\cdots,n\\},$ $u_{k}^{(n)}=1$ and $0$ otherwise. Then

$\left|(T_{r,s}(u^{(n)}))_{n}\right|=s_{n}\left|a_{n}\right|\sum\limits_{k=1}^{n}\frac{1}{r_{k}}<\epsilon\
\ \mbox{for all}\ \ n>n_{0}.$

Hence $\left\\{a_{n}s_{n}\sum\limits_{k=1}^{n}\frac{1}{r_{k}}\right\\}_{n}\in
c_{0}$ and this proves the theorem. ∎

######  Remark 3.3.

Few important remarks on the above result are mentioned below.

  1. (a)

If $r_{n}=s_{n}=1$ for all $n\in\mathbb{N}$ then both the weighted sequence
spaces $c_{0}(r)$ and $c_{0}(s)$ reduce to the classical sequence space
$c_{0}.$ In this setting, Theorem 3.1 yields, the Rhaly operator
$R_{a}:c_{0}\rightarrow c_{0}$ is bounded if and only if
$\\{na_{n}\\}\in\ell_{\infty}.$ This result is proved by Leibowitz [14] in
Proposition 3.2.

  2. (b)

Since $|a_{n}|s_{n}\sum_{k=1}^{n}\frac{1}{s_{k}}\geq|a_{n}|,\ n\in\mathbb{N},$
a necessary condition for $R_{a}\in\mathcal{B}(c_{0}(s))$ and
$R_{a}\in\mathcal{K}(c_{0}(s))$ is that $a_{n}\in\ell_{\infty}$ and $a_{n}\in
c_{0}$ respectively.

Let $s=\\{s_{n}\\}$ be a decreasing, strictly positive sequence of real
numbers. Then

$s_{n}\left|a_{n}\right|\sum\limits_{k=1}^{n}\frac{1}{s_{k}}\leq
n\left|a_{n}\right|.$

Therefore $R_{a}\in\mathcal{B}(c_{0}(s))$ if $\\{na_{n}\\}\in\ell_{\infty}$
and $R_{a}\in\mathcal{K}(c_{0}(s))$ if $\\{na_{n}\\}\in c_{0}.$ Also let
$\\{e_{n}\\}$ be a sequence in $c_{0}(r)$ such that the $n$-th entry is one
and all other entries are zero. Then for $n\in\mathbb{N},$

$\left\lVert R_{a}e_{n}\right\rVert_{s}=\sup\limits_{k\geq
n}s_{k}\left|a_{k}\right|\geq a_{n}s_{n}=\left|a_{n}\right|\left\lVert
e_{n}\right\rVert_{s}.$

This gives the following corollary.

######  Corollary 3.4.

Let $s=\\{s_{n}\\}$ be a bounded, decreasing, strictly positive sequence of
real numbers. Then $R_{a}\in\mathcal{B}(c_{0}(s))$ if
$\\{na_{n}\\}\in\ell_{\infty}$ and $R_{a}\in\mathcal{K}(c_{0}(s))$ if
$\\{na_{n}\\}\in c_{0}.$ Also

$\sup\limits_{n}\left|a_{n}\right|\leq\left\lVert
R_{a}\right\rVert_{s}\leq\sup\limits_{n}n\left|a_{n}\right|.$

The following example implies the above corollary is merely a sufficient
condition.

######  Example 3.5.

Let for $n\in\mathbb{N},$ $a_{n}=\frac{1}{\log(n+1)}$ and
$s_{n}=\frac{1}{2^{n}}.$ Then
$s_{n}\sum_{k=1}^{n}\frac{1}{s_{k}}=\frac{2^{n+1}-2}{2^{n}},$
$n\in\mathbb{N}.$ Hence

$a_{n}s_{n}\sum\limits_{k=1}^{n}\frac{1}{s_{k}}=\frac{2^{n+1}-2}{2^{n}\log(n+1)}\to
0\mbox{ as }n\to\infty.$

Here, $\\{s_{n}\\}$ is a bounded strictly decreasing sequence and
$\\{na_{n}\\}$ is unbounded but $R_{a}\in\mathcal{K}(c_{0}(s)).$

##  4\. Spectral properties of $R_{a}$

From now onwards, let $\\{a_{n}\\}$ is a sequence of positive real numbers
such that $\lim_{n\to\infty}na_{n}$ exists finitely and equal to $\chi\neq 0$
and $S=\\{a_{n}:n\in\mathbb{N}\\}$. The following lemma ([29, Lemma 2.7], [19,
Lemma 7]) is useful in this sequel.

######  Lemma 4.1.

Let $\lambda\in\mathbb{C}\setminus S$ and $\alpha=\Re(\frac{1}{\lambda}).$
Then the following relation holds

$\prod\limits_{k=1}^{n}\left|1-\frac{a_{k}}{\lambda}\right|\simeq\frac{1}{n^{\alpha\chi}},$

where the notation $a_{n}\simeq b_{n}$ means the sequences
$\left\\{\frac{a_{n}}{b_{n}}\right\\}$ and
$\left\\{\frac{b_{n}}{a_{n}}\right\\}$ are bounded and for any
$z\in\mathbb{C},$ $\Re(z)$ denotes the real part of $z.$

Using the similar concept as [19, Lemma 7], we prove a slightly modified
result.

######  Lemma 4.2.

For a fixed $m\in\mathbb{N},$ let
$\lambda\in\mathbb{C}\setminus\\{a_{k}:k=m+1,m+2,\cdots\\}$ and
$\alpha=\Re(\frac{1}{\lambda}).$ Then the following relation holds

$\prod\limits_{k=m+1}^{n}\left|1-\frac{a_{k}}{\lambda}\right|\simeq\frac{1}{n^{\alpha\chi}}.$

###### Proof.

Let $\frac{1}{\lambda}=\alpha+i\beta$ where $\alpha,$ $\beta\in\mathbb{R}$ and
$\lim_{n\to\infty}na_{n}=\chi$ implies $a_{n}\simeq\frac{\chi}{n}.$ Imitating
the calculations of Lemma 7 in [19] and using the fact that $e^{x}\geq 1+x$
for all $x\in\mathbb{R},$ we have the following relation

$\prod\limits_{k=m+1}^{n}\left|1-\frac{a_{k}}{\lambda}\right|\leq\mathcal{O}(1)\exp\sum\limits_{k=m+1}^{n}\left(-\frac{\chi}{k}\alpha+\frac{(\alpha^{2}+\beta^{2})}{2}\frac{\chi^{2}}{k^{2}}\right).$
(4.1)

Now we use the following inequality

$\sum\limits_{k=m+1}^{n}\frac{1}{k}\geq\int\limits_{m}^{n}\frac{1}{x+1}dx=\log(n+1)-\log(m+1)\geq\log
n-\log(m+1).$

Therefore from (4.1) we deduce that

$\prod\limits_{k=m+1}^{n}\left|1-\frac{a_{k}}{\lambda}\right|\leq\frac{\mathcal{O}(1)}{n^{\chi\alpha}}.$

Also note that

$\sum\limits_{k=m+1}^{n}\frac{1}{k}\leq\int\limits_{m-1}^{n-1}\frac{1}{x+1}dx=\log
n-\log m.$

Using the above inequality, in a similar way it can be proved that

$\prod\limits_{k=m+1}^{n}\left|1-\frac{a_{k}}{\lambda}\right|^{-1}\leq\mathcal{O}(1)n^{\chi\alpha}.$

This proves the result. ∎

Next we derive the point spectrum of $R_{a}$.

######  Theorem 4.3.

Let $s=\\{s_{n}\\}$ be a bounded, strictly positive sequence such that
$R_{a}\in\mathcal{B}(c_{0}(s)).$ Then

$\sigma_{p}(R_{a},c_{0}(s))=\\{\lambda\in
S:\lim\limits_{n\to\infty}a_{n}s_{n}n^{{\alpha\chi}}=0\\}$

where $\alpha=\Re({\frac{1}{\lambda}})=\frac{1}{\lambda}.$

###### Proof.

Let us consider $R_{a}x=\lambda x$ for some $\lambda\in\mathbb{C}$ and non-
zero $x.$ This gives the system of equations

$\left.\begin{aligned} a_{1}x_{1}&=&\lambda x_{1}\\\
a_{2}(x_{1}+x_{2})&=&\lambda x_{2}\\\ &\vdots&\\\
a_{k}(x_{1}+x_{2}+\cdots+x_{k})&=&\lambda x_{k}\\\
&\vdots&\end{aligned}\right\\}.$ (4.2)

If $\lambda\notin S$ then from the above equations it follows that $x_{k}=0$
for all $k\in\mathbb{N}.$ Hence $\sigma_{p}(R_{a},c_{0}(s))\subseteq S.$. Any
element $\lambda$ of $S$ belongs to $\sigma_{p}(R_{a},c_{0}(s))$ if and only
if $R_{a}x=\lambda x$ holds for some non-zero $x=\\{x_{n}\\}\in c_{0}(s).$
Simplifying $R_{a}x=\lambda x$ we get,

$x_{n}=\prod\limits_{j=m+1}^{n}\left(\frac{\lambda a_{j-1}^{-1}}{\lambda
a_{j}^{-1}-1}\right)x_{m},\ n=m+1,m+2,\cdots,$ (4.3)

where $x_{m}$ is the first non-zero entry of $\\{x_{n}\\}.$ Now we rearrange
the right hand side of the above equation as follows;

$\displaystyle x_{n}=\prod\limits_{j=m+1}^{n}\left(\frac{\lambda
a_{j-1}^{-1}}{\lambda
a_{j}^{-1}-1}\right)x_{m}=\prod\limits_{j=m+1}^{n}\left(\frac{a_{j}a_{j-1}^{-1}}{1-\frac{a_{j}}{\lambda}}\right)x_{m}$
$\displaystyle=$
$\displaystyle\frac{\prod\limits_{j=m+1}^{n}a_{j}a_{j-1}^{-1}}{\prod\limits_{j=m+1}^{n}\left(1-\frac{a_{j}}{\lambda}\right)}x_{m}$
$\displaystyle=$
$\displaystyle\frac{\frac{a_{n}}{a_{m}}}{\prod\limits_{j=m+1}^{n}\left(1-\frac{a_{j}}{\lambda}\right)}x_{m}.$

From Lemma 4.2 it can be deduced that $\lambda\in\sigma_{p}(R_{a},c_{0}(s))$
if and only if

$\lim\limits_{n\to\infty}a_{n}s_{n}n^{\alpha\chi}=0,$

where $\alpha=\Re({\frac{1}{\lambda}})=\frac{1}{\lambda}.$

∎

######  Remark 4.4.

Here we mention few important observations related to the point spectrum.

  1. (i)

Since the weight vector $s=\\{s_{n}\\}$ is a bounded sequence, and
$a_{n}\simeq\frac{\chi}{n},$ for any $\lambda\in S$ satisfies $\lambda>\chi$
we have

$a_{n}n^{\alpha\chi}\simeq\chi n^{\alpha\chi-1}$

where $\alpha=\Re({\frac{1}{\lambda}})=\frac{1}{\lambda}.$ This proves that
$a_{n}s_{n}n^{\alpha\chi}\to 0$ as $n\to\infty$ and consequently
$\lambda\in\sigma_{p}(R_{a},c_{0}(s)).$ In this regard we have the following
inclusion relation

$\\{\lambda\in S:\lambda>\chi\\}\subseteq\sigma_{p}(R_{a},c_{0}(s)).$

  2. (ii)

Let $\lambda\in\sigma_{p}(R_{a},c_{0}(s)).$ Then for all $\mu\in S$ such that
$\lambda\leq\mu$ the following relation holds for all $n\in\mathbb{N}$

$a_{n}s_{n}n^{\frac{\chi}{\mu}}\leq a_{n}s_{n}n^{\frac{\chi}{\lambda}}.$

This shows that $\mu\in\sigma_{p}(R_{a},c_{0}(s))$ for all $\lambda\leq\mu.$

  3. (iii)

If $a_{n}=\frac{1}{n}$ and $s_{n}=1$ for all $n\in\mathbb{N}$ then $\chi=1$.
In this case $a_{n}s_{n}n^{\frac{\chi}{a_{k}}}$ does not converge to zero for
any $k\in\mathbb{N}$ and hence $\sigma_{p}(R_{a},c_{0}(s))=\emptyset.$ This
reflects the case of the point spectrum of the Cesàro matrix over $c_{0}$
space which is obtained by Reade [19, Theorem 1].

######  Theorem 4.5.

Let $s=\\{s_{n}\\}$ be a bounded strictly positive sequence such that
$R_{a}\in\mathcal{B}(c_{0}(s))$. Then the following statements hold

  1. (i)

$0\notin\sigma_{p}(R_{a}^{*},c_{0}(s)^{*}),$

  2. (ii)

$S\subseteq\sigma_{p}(R_{a}^{*},c_{0}(s)^{*}),$

  3. (iii)

Let $\lambda\in\mathbb{C}\setminus\overline{S}.$ Then
$\lambda\in\sigma_{p}(R_{a}^{*},c_{0}(s)^{*})$ if and only if

$\sum\limits_{n=1}^{\infty}\frac{1}{s_{n}n^{\alpha\chi}}<\infty,$

where $\alpha=\Re(\frac{1}{\lambda}).$

###### Proof.

(i) Let the relation $R_{a}^{*}x=0$ holds for some $x.$ This implies
$\sum_{k=n}^{\infty}a_{k}x_{k}=0$ for all $n\in\mathbb{N}$ and eventually we
have $x_{n}=0$ for all $n\in\mathbb{N}.$ This shows that
$0\notin\sigma_{p}(R_{a}^{*},c_{0}(s)^{*}).$

(ii) Consider the equation $R_{a}^{*}x=\lambda x$ for some
$\lambda\in\mathbb{C}.$ This gives the following system

$\left.\begin{aligned}
a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3}+a_{4}x_{4}+\cdots&=&\lambda x_{1}\\\
a_{2}x_{2}+a_{3}x_{3}+a_{4}x_{4}+\cdots&=&\lambda x_{2}\\\
a_{3}x_{3}+a_{4}x_{4}+\cdots&=&\lambda x_{3}\\\
&\vdots&\end{aligned}\right\\}.$ (4.4)

From the above system it follows that for all $n\in\mathbb{N}$ and $n\geq 2$

$x_{n}=\prod\limits_{j=1}^{n-1}\left(1-\frac{a_{j}}{\lambda}\right)x_{1}.$
(4.5)

If $\lambda=a_{l}$ for some $l\in\mathbb{N}$ then from the above equation, it
follows

$x_{l+1}=x_{l+2}=\cdots=0.$

Hence $\\{x_{k}\\}\in\ell_{1}(s^{-1}).$ It follows that
$S\subseteq\sigma_{p}(R_{a}^{*},c_{0}(s)^{*}).$

(iii) Now let $\lambda\in\mathbb{C}\setminus\overline{S}$. Then
$R_{a}^{*}x=\lambda x$ implies the equation (4.5) for $\\{x_{n}\\}.$ Then
using Lemma 4.1 we get for $n\in\mathbb{N},$

$|x_{n}|=\left|\prod\limits_{j=1}^{n-1}\left(1-\frac{a_{j}}{\lambda}\right)x_{1}\right|\simeq\frac{1}{n^{\alpha\chi}}\left|x_{1}\right|,$

where $\alpha=\Re(\frac{1}{\lambda}).$ Hence
$\lambda\in\sigma_{p}(R_{a}^{*},c_{0}(s)^{*})$ if and only if
$\sum_{n=1}^{\infty}\frac{1}{s_{n}n^{\alpha\chi}}<\infty.$ ∎

######  Remark 4.6.

The following facts are used to obtain the next lemma. For
$\lambda\in\mathbb{C}\setminus\\{0\\}$ and a fixed $\chi>0$ the following
statements hold

  1. (i)

$\left|\lambda-\frac{\chi}{2}\right|<\frac{\chi}{2}\mbox{ if and only if }\
\Re\left({\frac{1}{\lambda}}\right)>\frac{1}{\chi},$

  2. (ii)

$\left|\lambda-\frac{\chi}{2}\right|=\frac{\chi}{2}\mbox{ if and only if }\
\Re\left({\frac{1}{\lambda}}\right)=\frac{1}{\chi},$

  3. (iii)

$\left|\lambda-\frac{\chi}{2}\right|>\frac{\chi}{2}\mbox{ if and only if }\
\Re\left({\frac{1}{\lambda}}\right)<\frac{1}{\chi}.$

######  Lemma 4.7.

Let $s=\\{s_{n}\\}$ be a bounded strictly positive sequence such that
$R_{a}\in\mathcal{B}(c_{0}(s))$. Then

$\sigma_{p}(R_{a}^{*},c_{0}(s)^{*})\subseteq\left\\{\lambda\in\mathbb{C}:\left|\lambda-\frac{\chi}{2}\right|<\frac{\chi}{2}\right\\}\cup
S.$

###### Proof.

It is already proved that $a_{n}\in\sigma_{p}(R_{a}^{*},c_{0}(s)^{*})$ for all
$n\in\mathbb{N}.$ Let $\lambda\notin S$ such that
$\lambda\in\sigma_{p}(R_{a}^{*},c_{0}(s)^{*}).$ Then from Theorem 4.5 it
follows that $\sum_{n=1}^{\infty}\frac{1}{s_{n}n^{\alpha\chi}}<\infty,$ where
$\alpha=\Re(\frac{1}{\lambda}).$ Now

$\frac{1}{s_{n}n^{\alpha\chi}}\geq||s||_{\infty}^{-1}\frac{1}{n^{\alpha\chi}}.$

holds for all $n\in\mathbb{N}$. Hence

$\lambda\in\sigma_{p}(R_{a}^{*},c_{0}(s)^{*})\Rightarrow\sum\limits_{n=1}^{\infty}\frac{1}{n^{\alpha\chi}}<\infty\Rightarrow\alpha\chi>1,$

and consequently $\lambda$ satisfies
$\left|\lambda-\frac{\chi}{2}\right|<\frac{\chi}{2}.$ This proves the lemma. ∎

The next result provides an estimation of the spectrum of the operator
$R_{a}.$

######  Theorem 4.8.

Let $s=\\{s_{n}\\}$ be a bounded, decreasing sequence of positive real numbers
such that $R_{a}\in\mathcal{B}(c_{0}(s))$, then

$\sigma(R_{a},c_{0}(s))\subseteq\left\\{\lambda\in\mathbb{C}\setminus\bar{S}:\left|\lambda-\frac{\chi}{2}\right|\leq\frac{\chi}{2}\right\\}\cup\bar{S}.$

###### Proof.

From the inclusion relation $(ii)$ in Theorem 4.5 we have,
$S\subseteq\sigma_{p}(R_{a}^{*},c_{0}(s)^{*})$. Also it is known that for a
Banach space $X$, $\sigma_{p}(T^{*},X^{*})\subseteq\sigma(T,X).$ Therefore we
have $S\subseteq\sigma(R_{a},c_{0}(s))$ with the fact that
$\sigma(R_{a},c_{0}(s))$ is a closed subset of $\mathbb{C}.$ Hence
$\overline{S}\subset\sigma(R_{a},c_{0}(s))$. Again for
$\lambda\notin\overline{S}$ the inverse $(R_{a}-\lambda I)^{-1}$ exists and
has the matrix representation

$(R_{a}-\lambda I)^{-1}=(b_{nk})=\left\\{\begin{aligned}
&\frac{1}{a_{n}-\lambda},\ \ \ \ \ \ \ \ n=k\\\
&\frac{-a_{n}}{\lambda^{2}\prod\limits_{j=k}^{n}(1-\frac{a_{j}}{\lambda})},\ \
\ 1\leq k<n\\\ &0,\ \ \ \ \ \ \ \ \ \ \ \
\mbox{otherwise.}\end{aligned}\right.$ (4.6)

From Lemma 2.2 it follows that $(R_{a}-\lambda
I)^{-1}=(b_{nk})\in\mathcal{B}(c_{0}(s))$ if and only if both the conditions
$\sup\limits_{n\in\mathbb{N}}s_{n}\sum\limits_{k=1}^{n}\frac{|b_{nk}|}{s_{k}}<\infty,$
and $\lim\limits_{n\to\infty}s_{n}|b_{nk}|=0,\ \forall k\in\mathbb{N}$ are
satisfied. Let
$\lambda\in\left\\{\lambda\in\mathbb{C}\setminus\bar{S}:\left|\lambda-\frac{\chi}{2}\right|>\frac{\chi}{2}\right\\}$.
Then we have for $n\in\mathbb{N}$

$\displaystyle s_{n}\sum\limits_{k=1}^{n}\frac{|b_{nk}|}{s_{k}}$
$\displaystyle=$ $\displaystyle
s_{n}\sum\limits_{k=1}^{n-1}\frac{|b_{nk}|}{s_{k}}+|b_{nn}|$ (4.7)
$\displaystyle=$ $\displaystyle
s_{n}\sum\limits_{k=1}^{n-1}\frac{a_{n}}{\left|\lambda^{2}\prod\limits_{j=k}^{n}(1-\frac{a_{j}}{\lambda})\right|s_{k}}+\left|\frac{1}{a_{n}-\lambda}\right|.$

Also, since $\lim_{n\to\infty}\frac{n^{a}}{(n+1)^{a}}=1$ for all real $a$,
Lemma 4.2 yields

$\frac{D_{1}}{(n+1)^{\alpha\chi}}\leq\prod\limits_{k=1}^{n}\left|1-\frac{a_{k}}{\lambda}\right|\leq\frac{D_{2}}{(n+1)^{\alpha\chi}},$
(4.8)

for some positive constants $D_{1}$ and $D_{2}$ and Lemma 4.2 and equation
(4.8) together imply

$M_{1}a_{n}\frac{n^{\alpha\chi}}{k^{\alpha\chi}}\leq\frac{a_{n}}{\prod\limits_{j=k}^{n}\left|1-\frac{a_{j}}{\lambda}\right|}\leq
M_{2}a_{n}\frac{n^{\alpha\chi}}{k^{\alpha\chi}},$ (4.9)

for some positive constants $M_{1}$ and $M_{2}.$ Therefore using the relation
(4.9) in the equation (4.7) we obtain for $n\in\mathbb{N}$

$\displaystyle s_{n}\sum\limits_{k=1}^{n}\frac{|b_{nk}|}{s_{k}}$
$\displaystyle\leq$ $\displaystyle
M_{2}\frac{s_{n}}{|\lambda|^{2}}\sum\limits_{k=1}^{n-1}\frac{a_{n}n^{\alpha\chi}}{s_{k}k^{\alpha\chi}}+\frac{1}{d_{\lambda}},$

where $d_{\lambda}=\mbox{dist }(\lambda,\bar{S}).$ Therefore we have

$\sup_{n}a_{n}s_{n}n^{\alpha\chi}\sum\limits_{k=1}^{n-1}\frac{1}{s_{k}k^{\alpha\chi}}<\infty\Rightarrow\sup\limits_{n}s_{n}\sum\limits_{k=1}^{n}\frac{|b_{nk}|}{s_{k}}<\infty.$

As
$\lambda\in\left\\{\lambda\in\mathbb{C}\setminus\bar{S}:\left|\lambda-\frac{\chi}{2}\right|>\frac{\chi}{2}\right\\}$,
this is equivalent to $\alpha\chi<1.$ In this case for $n\in\mathbb{N},$

$\displaystyle
a_{n}n^{\alpha\chi}\sum\limits_{k=1}^{n-1}\frac{s_{n}}{s_{k}k^{\alpha\chi}}$
$\displaystyle\leq$
$\displaystyle(na_{n})n^{\alpha\chi-1}\sum\limits_{k=1}^{n-1}\frac{1}{k^{\alpha\chi}}\leq
D_{1}n^{\alpha\chi-1}\sum\limits_{k=1}^{n-1}\frac{1}{k^{\alpha\chi}}$ (4.10)

for some positive constant $D_{1}$ as $\\{na_{n}\\}$ is convergent sequence
and $\\{s_{n}\\}$ is decreasing. Also

$\sum\limits_{k=1}^{n-1}\frac{1}{k^{\alpha\chi}}\leq\int\limits_{0}^{n-1}\frac{1}{x^{\alpha\chi}}dx=\lim\limits_{\epsilon\to
0+}\int\limits_{\epsilon}^{n-1}\frac{1}{x^{\alpha\chi}}dx=\frac{(n-1)^{1-\alpha\chi}}{1-\alpha\chi}.$

The above inequality and (4.10) together imply

$\sup_{n}a_{n}s_{n}n^{\alpha\chi}\sum\limits_{k=1}^{n-1}\frac{1}{s_{k}k^{\alpha\chi}}<\infty.$

Also from (4.9) we have for each $k\in\mathbb{N}$ and large $n$

$0\leq
s_{n}|b_{nk}|=\frac{s_{n}a_{n}}{\left|\lambda^{2}\prod\limits_{j=k}^{n}(1-\frac{a_{j}}{\lambda})\right|}\leq
M_{2}a_{n}s_{n}\frac{n^{\alpha\chi}}{|\lambda|^{2}k^{\alpha\chi}}.$

Hence
$\lim_{n\to\infty}a_{n}s_{n}n^{\alpha\chi}=0\Rightarrow\lim_{n\to\infty}s_{n}|b_{nk}|=0.$

Again $0\leq a_{n}s_{n}n^{\alpha\chi}=(na_{n})s_{n}n^{\alpha\chi-1}\leq
D_{2}n^{\alpha\chi-1}$ holds for some positive constant $D_{2}$ as
$\\{na_{n}\\}$ and $\\{s_{n}\\}$ are bounded. This also proves that
$\lim\limits_{n\to\infty}a_{n}s_{n}n^{\alpha\chi}=0.$ Hence $(R_{a}-\lambda
I)^{-1}$ exists and $(R_{a}-\lambda I)^{-1}\in\mathcal{B}(c_{0}(s))$ and
consequently $\lambda\in\rho(R_{a},c_{0}(s)).$ Finally we have

$\left\\{\lambda\in\mathbb{C}\setminus\bar{S}:\left|\lambda-\frac{\chi}{2}\right|>\frac{\chi}{2}\right\\}\subseteq\rho(R_{a},c_{0}(s)).$

This proves the result.

∎

Now we summarize all the result on the spectrum and fine spectrum of the Rhaly
operator over the sequence space $c_{0}(s).$ Consider the following sets

$\displaystyle A_{1}$ $\displaystyle=$ $\displaystyle\left\\{\lambda\in
S:\lim\limits_{n\to\infty}a_{n}s_{n}n^{\alpha\chi}=0\right\\},$ $\displaystyle
A_{2}$ $\displaystyle=$
$\displaystyle\left\\{\lambda\in\mathbb{C}\setminus(S\cup\\{0\\}):\sum\limits_{n=1}^{\infty}\frac{1}{s_{n}n^{\alpha\chi}}<\infty\right\\},$

where $\alpha=\Re(\frac{1}{\lambda}).$ In this setting we have the following
result.

######  Theorem 4.9.

Let $\\{s_{n}\\}$ be a bounded and strictly positive sequence such that
$R_{a}\in\mathcal{B}(c_{0}(s))$ and $\lim_{n\to\infty}na_{n}=\chi\neq 0$ then
the following statements hold

  1. (i)

$\sigma_{p}(R_{a},c_{0}(s))=A_{1},$

  2. (ii)

$\sigma_{p}(R_{a}^{*},c_{0}(s)^{*})=A_{2}\cup S,$

  3. (iii)

$\sigma_{r}(R_{a},c_{0}(s))=(A_{2}\cup S)\setminus A_{1}=A_{2}\cup(S\setminus
A_{1}).$

In addition, if the sequence $\\{s_{n}\\}$ is a decreasing sequence then

  1. (iv)

$\sigma(R_{a},c_{0}(s))\subseteq\left\\{\lambda\in\mathbb{C}\setminus\bar{S}:\left|\lambda-\frac{\chi}{2}\right|\leq\frac{\chi}{2}\right\\}\cup\bar{S},$

  2. (v)

$\\{0\\}\subseteq\sigma_{c}(R_{a},c_{0}(s))\subseteq(\left\\{\lambda\in\mathbb{C}\setminus\bar{S}:\left|\lambda-\frac{\chi}{2}\right|\leq\frac{\chi}{2}\right\\}\cup\bar{S})\setminus(A_{2}\cup
S).$

###### Proof.

The relations (i) and (ii) are direct consequences of Theorem 4.3 and Theorem
4.5 respectively. Also from lemma 2.3 it follows that
$\sigma_{r}(R_{a},c_{0}(s))=\sigma_{p}(R_{a}^{*},c_{0}(s)^{*})\setminus\sigma_{p}(R_{a},c_{0}(s))$
which gives the expression for $\sigma_{r}(R_{a},c_{0}(s))$ in (iii). The
relation (iv) follows from Theorem 4.8. The first inclusion relation in (v)
follows from Theorem 4.3 and Theorem 4.5(i) and the other inclusion relation
in (v) follows from the fact that $\sigma_{p}(R_{a},c_{0}(s)),$
$\sigma_{r}(R_{a},c_{0}(s)),$ and $\sigma_{c}(R_{a},c_{0}(s))$ forms a
partition of $\sigma(R_{a},c_{0}(s)).$ ∎

##  5\. Operator Ideal

The theory of operator ideals holds a distinctive place in functional analysis
because of its numerous applications in spectrum theory, Banach space
geometry, the theory of eigenvalue distributions, etc. Many researchers have
developed classes of operator ideals that result from operators acting over
sequence spaces [15, 16, 23]. In this section, we introduce a new class of
operator ideal with the help of $s$-number sequences and discussed of its
properties. Let $X$, $Y$, $Z$, $X_{0}$ and $Y_{0}$ are the Banach spaces,
$\mathcal{B}$ denotes the class of all bounded linear operators between any
pair of Banach spaces, $X^{\prime}$ denotes the dual of $X$, $x^{\prime}$
denotes the continuous linear functional on $X$. The sequences
$r=\\{r_{n}\\}$, $t=\\{t_{n}\\}$ and $u=\\{u_{n}\\}$ are the weight vectors
corresponding to sequence spaces $c_{0}(r)$, $c_{0}(t)$ and $c_{0}(u)$
respectively. Let $x^{\prime}\in X^{\prime}$ and $y\in Y$ then the mapping
$(x^{\prime}\otimes y):X\to Y$ is defined by $(x^{\prime}\otimes
y)(x)=x^{\prime}(x)y,~{}x\in X$. Before proceeding, we list certain known
definitions and results which are necessary for our results.

######  Definition 5.1.

[18, p.79] A map $s:\mathcal{B}\to\mathbb{R}^{\mathbb{N}}$ which assigns every
operator $\phi\in\mathcal{B}$ to a non-negative sequence $\\{s_{n}(\phi$)} is
called a $s$-number sequence if it satisfies the following properties:

  * (i)

$\|\phi\|=s_{1}(\phi)\geq s_{2}(\phi)\geq\cdots\geq 0$ for
$\phi\in\mathcal{B}(X,Y)$,

  * (ii)

$s_{m+n-1}(\phi+\psi)\leq s_{m}(\phi)+s_{n}(\psi)$ for
$\phi,\psi\in\mathcal{B}(X,Y)$,

  * (iii)

$s_{n}(\zeta\phi\eta)\leq\|\zeta\|s_{n}(\phi)\|\eta\|$ for
$\zeta\in\mathcal{B}(Y,Y_{0})$, $\phi\in\mathcal{B}(X,Y)$,
$\eta\in\mathcal{B}(X_{0},X)$,

  * (iv)

If rank$(\phi)<n$, then $s_{n}(\phi)=0$,

  * (v)

$s_{n}(I_{2}:\ell_{2}^{(i)}\to\ell_{2}^{(i)})=1$, where $I_{2}$ denotes the
identity operator on the $i$-dimensional Hilbert space $\ell_{2}^{(i)}$.

We call $s_{n}(\phi)$ the $n$-th $s$-number of the operator $\phi$.

######  Definition 5.2.

[18, p.25] For a subset $\mathcal{A}(X,Y)$ of $\mathcal{B}(X,Y)$, the class
$\mathcal{A}=\cup_{X,Y}\mathcal{A}(X,Y)$ is said to be an operator ideal if
each component $\mathcal{A}(X,Y)$ satisfies the following conditions:

  * (i)

$x^{\prime}\otimes y\in\mathcal{A}(X,Y)$ for $x^{\prime}\in X^{\prime}$ and
$y\in Y$,

  * (ii)

$\phi+\psi\in\mathcal{A}(X,Y)$ for $\phi,\psi\in\mathcal{A}(X,Y)$,

  * (iii)

$\zeta\phi\eta\in\mathcal{A}(X_{0},Y_{0})$ for $\eta\in\mathcal{B}(X_{0},X)$,
$\phi\in\mathcal{A}(X,Y)$, $\zeta\in\mathcal{B}(Y,Y_{0})$.

######  Definition 5.3.

[18, p.25] A function $\Gamma:\mathcal{A}\to\mathbb{R}^{+}$ which assigns to
every operator $\phi\in\mathcal{A}$ a non-negative number $\Gamma(\phi)$ is
called a quasi-norm on the operator ideal $\mathcal{A}$ if it satisfies the
following properties:

  * (i)

$\Gamma(x^{\prime}\otimes y)=\|x^{\prime}\|\|y\|$ for $x^{\prime}\in
X^{\prime}$ and $y\in Y$,

  * (ii)

$\Gamma(\phi+\psi)\leq c(\Gamma(\phi)+\Gamma(\psi))$ for
$\phi,\psi\in\mathcal{A}(X,Y)$ where $c\geq 1$ is a constant,

  * (iii)

$\Gamma(\zeta\phi\eta)\leq\|\zeta\|\Gamma(\phi)\|\eta\|$ for
$\eta\in\mathcal{B}(X_{0},X),\phi\in\mathcal{A}(X,Y),\zeta\in\mathcal{B}(Y,Y_{0})$.

######  Definition 5.4.

[18, p.81] An s-number $s$ is called multiplicative if

$s_{m+n-1}(\phi\psi)\leq
s_{m}(\phi)s_{n}(\psi)~{}~{}\text{for}~{}~{}\psi\in\mathcal{L}(X,Y)~{}~{}\text{and}~{}~{}\phi\in\mathcal{L}(Y,Z).$

######  Definition 5.5.

[18, p.26] An operator ideal $\mathcal{A}$ is closed if all components
$\mathcal{A}(X,Y)$ are closed linear subsets of $\mathcal{B}(X,Y)$.

The following lemma is useful in this context.

######  Lemma 5.6.

[18, p.80] If $\phi,\psi\in\mathcal{B}(X,Y)$ then
$|s_{i}(\phi)-s_{i}(\psi)|\leq\|\phi-\psi\|,~{}~{}i\in\mathbb{N}.$

Now we define sequence space $\chi_{c_{0}(r)}$ associated to the Rhaly
operator $R_{a}$ as follows

$\chi_{c_{0}(r)}=\\{v\in\mathbb{C}^{\mathbb{N}}:R_{a}(v)\in c_{0}(r)\\}.$

Thus

$\chi_{c_{0}(r)}=\left\\{v\in\mathbb{C}^{\mathbb{N}}:\lim_{i\to\infty}\left(\sum_{j=1}^{i}a_{i}v_{j}\right)r_{i}=0\right\\}.$

An operator $\phi\in\mathcal{B}(X,Y)$ is called s-type $\chi_{c_{0}(r)}$
operator if

$\lim_{i\to\infty}\left(a_{i}\sum_{j=1}^{i}s_{j}(\phi)\right)r_{i}=0$

We denote the class of all s-type $\chi_{c_{0}(r)}$ operators by
$\chi_{c_{0}(r)}^{(s)}$ i.e.,

$\chi_{c_{0}(r)}^{(s)}=\left\\{\phi\in\mathcal{B}:\lim_{i\to\infty}\left(a_{i}\sum_{j=1}^{i}s_{j}(\phi)\right)r_{i}=0\right\\}.$

If $\phi\in\mathcal{B}(X,Y)$ then

$\chi_{c_{0}(r)}^{(s)}(X,Y)=\left\\{\phi\in\mathcal{B}(X,Y):\lim_{i\to\infty}\left(a_{i}\sum_{j=1}^{i}s_{j}(\phi)\right)r_{i}=0\right\\}.$

######  Lemma 5.7.

If $r_{n}\leq t_{n}$ for all $n$ then
$\chi_{c_{0}(t)}^{(s)}\subseteq\chi_{c_{0}(r)}^{(s)}$.

###### Proof.

Suppose $\phi\in\chi_{c_{0}(t)}^{(s)}$ then

$\lim_{i\to\infty}\left(a_{i}\sum_{j=1}^{i}s_{j}(\phi)\right)t_{i}=0$

Now,

$\displaystyle\lim_{i\to\infty}\left(a_{i}\sum_{j=1}^{i}s_{j}(\phi)\right)r_{i}\leq\lim_{i\to\infty}\left(a_{i}\sum_{j=1}^{i}s_{j}(\phi)\right)t_{i}=0$

Thus $\phi\in\chi_{c_{0}(r)}^{(s)}$. ∎

######  Theorem 5.8.

The following statements hold,

  * (i)

$\chi_{c_{0}(r)}^{(s)}$ is an operator ideal if $\\{a_{n}\\}\in c_{0}(r)$,

  * (ii)

$\chi_{c_{0}(r)}^{(s)}$ is a closed operator ideal if $\\{na_{n}\\}\in
c_{0}(r)$.

###### Proof.

(i) Let $X$ and $Y$ are any two Banach spaces. Let $x^{\prime}\in X^{\prime}$
and $y\in Y$. Then $x^{\prime}\otimes y$ is a rank one operator so
$s_{i}(x^{\prime}\otimes y)=0$ for all $i\geq 2$. Consider,

$\displaystyle\lim_{i\to\infty}\left(a_{i}\sum_{j=1}^{i}s_{j}(x^{\prime}\otimes
y)\right)r_{i}=$ $\displaystyle\lim_{i\to\infty}a_{i}s_{1}(x^{\prime}\otimes
y)r_{i}$ $\displaystyle=$ $\displaystyle\|x^{\prime}\otimes
y\|\lim_{i\to\infty}a_{i}r_{i}$ $\displaystyle=$ $\displaystyle 0.$

Thus, $(x^{\prime}\otimes y)\in\chi_{c_{0}(r)}^{(s)}(X,Y).$
Suppose $\phi,\psi\in\chi_{c_{0}(r)}^{(s)}(X,Y)$. By using properties of
$s$-number, we have

$\displaystyle\lim_{i\to\infty}a_{i}\left(\sum_{j=1}^{i}s_{j}(\phi+\psi)\right)r_{i}=$
$\displaystyle\lim_{i\to\infty}\left(a_{i}\left(\sum_{j=1}^{i}s_{2j-1}(\phi+\psi)+\sum_{j=1}^{i}s_{2j}(\phi+\psi)\right)\right)r_{i}$
$\displaystyle\leq$
$\displaystyle\lim_{i\to\infty}\left(a_{i}\left(\sum_{j=1}^{i}s_{2j-1}(\phi+\psi)+\sum_{j=1}^{i}s_{2j-1}(\phi+\psi)\right)\right)r_{i}$
$\displaystyle=$ $\displaystyle
2\lim_{l\to\infty}a_{i}\left(\sum_{j=1}^{i}s_{2j-1}(\phi+\psi)\right)r_{i}$
$\displaystyle\leq$ $\displaystyle
2a_{i}\lim_{i\to\infty}\left(\sum_{j=1}^{i}\left(s_{j}(\phi)+s_{j}(\psi)\right)\right)r_{i}$
$\displaystyle=$ $\displaystyle
2\lim_{i\to\infty}a_{i}\left(\sum_{j=1}^{i}s_{j}(\phi)\right)r_{i}+2\lim_{i\to\infty}a_{i}a_{n}\left(\sum_{j=1}^{i}s_{j}(\psi)\right)r_{i}$
$\displaystyle=$ $\displaystyle 0.$

This implies,
$\lim_{i\to\infty}a_{i}\left(\sum_{j=1}^{i}s_{j}(\phi+\psi)\right)r_{i}=0$.
Thus, $\phi+\psi\in\chi_{c_{0}(r)}^{(s)}(X,Y)$.
Suppose $\zeta\in\mathcal{B}(Y,Y_{0}),\eta\in\mathcal{B}(X_{0},X)$ and
$\phi\in\chi_{c_{0}(r)}^{(s)}(X,Y)$. Then

$\displaystyle\lim_{i\to\infty}a_{i}\left(\sum_{j=1}^{i}s_{j}(\zeta\phi\eta)\right)r_{i}=\|\zeta\|\|\eta\|\lim_{i\to\infty}a_{i}\left(\sum_{j=1}^{i}s_{j}(\phi)\right)r_{i}=0.$

Thus, $\zeta\phi\eta\in\chi_{c_{0}(r)}^{(s)}(X_{0},Y_{0})$. Hence
$\chi_{c_{0}(r)}^{(s)}$ is an operator ideal.
(ii) Let $\\{\phi_{k}\\}$ be any sequence in $\chi_{c_{0}(r)}^{(s)}(X,Y)$
converging to $\phi\in\mathcal{B}(X,Y)$ with respect to the operator norm.
Given $\epsilon>0$, we can fix $k_{0}\in\mathbb{N}$ such that

$\|\phi-\phi_{k}\|\leq\epsilon~{}~{}~{}~{}\text{for}~{}k\geq k_{0}.$

Now,

$\displaystyle\lim_{i\to\infty}\left(a_{i}\sum_{j=1}^{i}s_{j}(\phi)\right)r_{i}=$
$\displaystyle\lim_{i\to\infty}\left(a_{i}\sum_{j=1}^{i}s_{j}(\phi-\phi_{k}+\phi_{k})\right)r_{i}$
$\displaystyle\leq$
$\displaystyle\lim_{i\to\infty}\left(a_{i}\sum_{j=1}^{i}s_{1}(\phi-\phi_{k})\right)r_{i}+\lim_{i\to\infty}\left(a_{i}\sum_{j=1}^{i}s_{j}(\phi_{k})\right)r_{i}$
$\displaystyle=$
$\displaystyle\|\phi-\phi_{k}\|\lim_{i\to\infty}ia_{i}r_{i}+\lim_{i\to\infty}\left(a_{i}\sum_{j=1}^{i}s_{j}(\phi_{k})\right)r_{i}$
$\displaystyle=$ $\displaystyle 0.$

Hence, $\phi\in\chi_{c_{0}(r)}^{(s)}(X,Y)$ as
$\lim_{i\to\infty}\left(a_{i}\sum_{j=1}^{i}s_{j}(\phi)\right)r_{i}=0$. The
required result is proved.

∎

Define a mapping $Q^{(s)}:\chi_{c_{0}(r)}^{(s)}\to\mathbb{R}^{+}$ where
$\mathbb{R}^{+}$ is the set of all positive real numbers, by

$Q^{(s)}(\phi)=\sup_{i}\left|a_{i}\sum_{j=1}^{i}s_{j}(\phi)\right|r_{i}~{}~{}\text{where
$\phi\in\chi_{c_{0}(r)}^{(s)}$}.$

The next result proves that under certain condition $Q^{(s)}$ forms a quasi
norm on $\chi_{c_{0}(r)}^{(s)}$.

######  Theorem 5.9.

If $\sup_{i}|a_{i}|r_{i}=1$ then the mapping $Q^{(s)}$ is a quasi norm on the
set $\chi_{c_{0}(r)}^{(s)}$ and the operator ideal $\chi_{c_{0}(r)}^{(s)}$ is
complete under the quasi-norm $Q^{(s)}$.

###### Proof.

Let $X$ and $Y$ are any two Banach spaces and $x^{\prime}\in X^{\prime}$,$y\in
Y$. Now

$\displaystyle Q^{(s)}(x^{\prime}\otimes y)=$
$\displaystyle\sup_{i}\left|a_{i}\sum_{j=1}^{i}s_{j}(x^{\prime}\otimes
y)\right|r_{i}$ $\displaystyle=$
$\displaystyle\sup_{i}\left|a_{i}s_{1}(x^{\prime}\otimes y)\right|r_{i}$
$\displaystyle=$ $\displaystyle\|x^{\prime}\otimes y\|\sup_{i}a_{i}r_{i}$
$\displaystyle=$ $\displaystyle\|x^{\prime}\otimes y\|.$

Clearly $\|x^{\prime}\otimes y\|=\|x^{\prime}\|\|y\|$, so
$Q^{(s)}(x^{\prime}\otimes y)=\|x^{\prime}\|\|y\|$.
Suppose $\phi,\psi\in\chi_{c_{0}(r)}^{(s)}(X,Y)$ and consider

$\displaystyle Q^{(s)}(\phi+\psi)=$
$\displaystyle\sup_{i}\left|a_{i}\sum_{j=1}^{i}s_{j}(\phi+\psi)\right|r_{i}$
$\displaystyle\leq$ $\displaystyle
2\sup_{i}\left|a_{i}\sum_{j=1}^{i}(s_{j}(\phi)+s_{j}(\psi))\right|r_{i}$
$\displaystyle=$ $\displaystyle
2\sup_{i}\left|a_{i}\left(\sum_{j=1}^{i}s_{j}(\phi)\right)r_{i}+a_{i}\left(\sum_{j=1}^{i}s_{j}(\psi)\right)r_{i}\right|$
$\displaystyle\leq$ $\displaystyle
2\sup_{i}\left|a_{i}\sum_{j=1}^{i}s_{j}(\phi)\right|r_{i}+2\sup_{i}\left|a_{i}\sum_{j=1}^{i}s_{j}(\psi)\right|r_{i}$
$\displaystyle\leq$ $\displaystyle 2\left(Q^{(s)}(\phi)+Q^{(s)}(\psi)\right).$

Thus,

$Q^{(s)}(\phi+\psi)\leq 2\left(Q^{(s)}(\phi)+Q^{(s)}(\psi)\right).$

Let $\phi\in\chi_{c_{0}(r)}^{(s)}(X,Y)$ and $\zeta\in\mathcal{B}(Y,Y_{0})$,
$\eta\in\mathcal{B}(X_{0},X)$.

$\displaystyle Q^{(s)}(\zeta\phi\eta)=$
$\displaystyle\sup_{i}\left|a_{i}\sum_{j=1}^{i}s_{j}(\zeta\phi\eta)\right|r_{i}$
$\displaystyle\leq$
$\displaystyle\|\zeta\|\|\eta\|\sup_{i}\left|a_{i}\left(\sum_{j=1}^{i}s_{j}(\phi)\right)\right|r_{i}.$

This implies,

$Q^{(s)}(\zeta\phi\eta)\leq\|\zeta\|Q^{(s)}(\phi)\|\eta\|.$

Hence, $Q^{(s)}$ is quasi-norm on operator ideal $\chi_{c_{0}(r)}^{(s)}$.
Also, suppose $\phi\in\chi_{c_{0}(r)}^{(s)}(X,Y)$. We have

$\displaystyle Q^{(s)}(\phi)=$
$\displaystyle\sup_{i}\left|a_{i}\sum_{j=1}^{i}s_{j}(\phi)\right|r_{i}$
$\displaystyle\geq$ $\displaystyle\sup_{i}s_{1}(\phi)a_{i}r_{i}$
$\displaystyle=$ $\displaystyle\|\phi\|\sup_{i}a_{i}r_{i}$ $\displaystyle=$
$\displaystyle\|\phi\|.$

This implies,

$\|\phi\|\leq Q^{(s)}(\phi)\text{~{}for $\phi\in\chi_{c_{0}(r)}^{(s)}(X,Y).$}$
(5.1)

∎

Let $\\{\phi_{n}\\}$ be a Cauchy sequence in $\chi_{c_{0}(r)}^{(s)}(X,Y)$.
Then for $\epsilon>0$ there exists $k\in\mathbb{N}$ such that

$Q^{(s)}(\phi_{n}-\phi_{m})<\epsilon~{}~{}\text{for all}~{}~{}n,m\geq k.$
(5.2)

By using equation (5.1), we have

$\|\phi_{n}-\phi_{m}\|<\epsilon~{}~{}\text{for all}~{}~{}n,m\geq k.$

This implies $\\{\phi_{n}\\}$ is a Cauchy sequence in $\mathcal{B}(X,Y)$.
Since $\mathcal{B}(X,Y)$ is a Banach space, there exists
$\phi\in\mathcal{B}(X,Y)$ such that $\phi_{n}\to\phi$ as $n\to\infty$. Using
Lemma 5.6, we have

$\left|s_{i}(\phi_{n}-\phi_{m})-s_{i}(\phi-\phi_{m})\right|\leq\|\phi_{n}-\phi\|,~{}~{}i\in\mathbb{N}.$

As $\phi_{n}\to\phi$, taking $n\to\infty$ and keeping $m$ fixed we have

$s_{i}(\phi_{n}-\phi_{m})\to s_{i}(\phi-\phi_{m}),~{}~{}i\in\mathbb{N}.$ (5.3)

Now, we shall show that $\phi_{n}\to\phi$ as $n\to\infty$ in
$\chi_{c_{0}(r)}^{(s)}(X,Y)$. Again, from equation (5.2), we have

$\sup_{i}\left|a_{i}\sum_{j=1}^{i}s_{j}(\phi_{n}-\phi_{m})\right|r_{i}<\epsilon~{}~{}\text{for
all}~{}~{}m\geq k.$

Keeping $m$ fixed and letting $n\to\infty$, we obtain on using equation (5.3)

$\sup_{i}\left|a_{i}\sum_{j=1}^{i}s_{j}(\phi-\phi_{m})\right|r_{i}<\epsilon~{}~{}\text{for
all}~{}~{}n,m\geq k.$ (5.4)

which gives $Q^{(s)}(\phi-\phi_{m})<\epsilon$ for all $m\geq k$. This implies
$\phi_{m}\to\phi$ under the quasi norm $Q^{(s)}$. Again,

$\displaystyle\sum_{j=1}^{i}s_{j}(\phi)=\sum_{j=1}^{i}s_{2j-1}(\phi)+\sum_{j=1}^{i}s_{2j}(\phi)\leq$
$\displaystyle 2\sum_{j=1}^{i}s_{2j-1}(\phi)$ $\displaystyle=$ $\displaystyle
2\sum_{j=1}^{i}s_{2j-1}(\phi-\phi_{m}+\phi_{m})$ $\displaystyle\leq$
$\displaystyle
2\left(\sum_{j=1}^{i}s_{j}(\phi-\phi_{m})+\sum_{j=1}^{i}s_{j}(\phi_{m})\right).$

Thus,

$\sum_{j=1}^{i}s_{j}(\phi)\leq
2\left(\sum_{j=1}^{i}s_{j}(\phi-\phi_{m})+\sum_{j=1}^{i}s_{j}(\phi_{m})\right).$
(5.5)

The above inequality implies that

$\displaystyle\left|a_{i}\sum_{j=1}^{i}s_{j}(\phi)\right|r_{i}\leq$
$\displaystyle
2\left|a_{i}\left(\sum_{j=1}^{i}s_{j}(\phi-\phi_{m})+\sum_{j=1}^{i}s_{j}(\phi_{m})\right)\right|r_{i}$
$\displaystyle\leq$ $\displaystyle
2\left|a_{i}\sum_{j=1}^{i}s_{j}(\phi-\phi_{m})\right|r_{i}+2\left|a_{i}\sum_{j=1}^{i}s_{j}(\phi_{m})\right|r_{i}.$

By using relation (5.4), we obtain that

$\left|a_{i}\sum_{j=1}^{i}s_{j}(\phi)\right|r_{i}\leq
2\epsilon+2\left|a_{i}\sum_{j=1}^{i}s_{j}(\phi_{m})\right|r_{i}~{}~{}\forall
m\geq k.$

Taking limit on both sides, we obtain that

$\lim_{i\to\infty}\left|a_{i}\sum_{j=1}^{i}s_{j}(\phi)\right|r_{i}\leq
2\epsilon+2\lim_{i\to\infty}\left|a_{i}\sum_{j=1}^{i}s_{j}(\phi_{m})\right|r_{i}$

Then,

$\lim_{i\to\infty}\left|a_{i}\sum_{j=1}^{i}s_{j}(\phi)\right|r_{i}\leq
2\epsilon.$

Thus, $\phi\in\chi_{c_{0}(r)}^{(s)}(X,Y)$, and this implies,
$\chi_{c_{0}(r)}^{(s)}$ is complete under quasi-norm $Q^{(s)}$.

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