$ n $-quasi-$ A $-$ (m, q) $-isometry on a Banach space

Khadija Gherairi; Zayd Hajjej; Haiyan Li; Hedi Regeiba; Khadija Gherairi; Zayd Hajjej; Haiyan Li; Hedi Regeiba

doi:10.3934/math.20231448

AIMS Mathematics

2023, Volume 8, Issue 12: 28308-28321. doi: 10.3934/math.20231448

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Research article Special Issues

$n$ -quasi- $A$ - $(m, q)$ -isometry on a Banach space

1.
Laboratory of Mathematics and Applications, College of Sciencs, Gabes University, Tunisia
2.
Department of Mathematics, College of Science, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia
3.
College of Mathematics and Information Science, North Minzu University, Yinchuan 750021, China

Received: 04 August 2023 Revised: 25 September 2023 Accepted: 07 October 2023 Published: 17 October 2023
MSC : 47B99, 47A05

In this paper, we introduce the class of $n$ -quasi- $A$ - $(m, q)$ -isometry operators on a Banach space $X$ , which represents a generalization of the $n$ -quasi- $(m, q)$ -isometry on a Banach space and the $n$ -quasi- $(A, m)$ -isometry on a Hilbert space. After giving some basic properties of this class of operators, we study the product and the power of such operators in this class.

Keywords:

Citation: Khadija Gherairi, Zayd Hajjej, Haiyan Li, Hedi Regeiba. $n$ -quasi- $A$ - $(m, q)$ -isometry on a Banach space[J]. AIMS Mathematics, 2023, 8(12): 28308-28321. doi: 10.3934/math.20231448

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Abstract

1. Introduction

The class of $m$ -isometry operators was introduced in $1990$ by Agler in ^[1] and was developed in 1995 by Agler and Stankas in ^[2,3,4]. A bounded linear operator $T\in\mathcal{L}(\mathcal{H})$ on a Hilbert space $\mathcal{H}$ is called an $m$ -isometry, for a positif integer $m$ (that is $m\geq 1$ ), if

$\begin{eqnarray*} \beta_m(T): = \overset{m}{\underset{k = 0}{\sum}}(-1)^k\binom{m}{k}T^{*k}T^k = 0, \end{eqnarray*}$

where $T^*$ denotes the adjoint operator of $T$ . This latter is equivalent to

$\begin{eqnarray*} \Delta_{m}(T,x): = \sum\limits_{k = 0}^m(-1)^{k}\binom{m}{k}\Vert T^{k}x\Vert^2 = 0,\ \forall\; x\in \mathcal{H}. \end{eqnarray*}$

Some generalizations of this class of operators exist in the literature, like the $(A, m)$ -isometry, which was introduced in $2012$ by Saddi and Sid Ahmed ^[9], for a positif operator $A$ , by

$\begin{eqnarray*} \beta_m(T,A): = \overset{m}{\underset{k = 0}{\sum}}(-1)^k\binom{m}{k}T^{*k}AT^k = 0 \end{eqnarray*}$

or equivalently

$\begin{eqnarray*} \Delta_{m}(T,A,x): = \sum\limits_{k = 0}^m(-1)^{k}\binom{m}{k}\Vert A^{\frac{1}{2}}T^{k}x\Vert^2 = 0,\ \forall\; x\in \mathcal{H}. \end{eqnarray*}$

We also mention the class of $n$ -quasi- $m$ -isometry on a Hilbert space defined by

$\begin{eqnarray*} \beta_{m,n}(T): = \sum\limits_{k = 0}^m(-1)^{k}\binom{m}{k}T^{*k+n}T^{k+n} = 0 \end{eqnarray*}$

or in an equivalent manner

$\begin{eqnarray*} \Delta_{m,n}(T,x): = \sum\limits_{k = 0}^m(-1)^{k}\binom{m}{k}\Vert T^{n+k}x\Vert^2 = 0,\ \forall\; x\in \mathcal{H}. \end{eqnarray*}$

For more details about these class, please see references ^[10,11].

The $n$ -quasi- $(A, m)$ -isometries (which are particular cases of $n$ -quasi- $(m, q)$ -isometries) were thoroughly studied by Agler and Stankus in a series of three papers in which the authors employed the theory of periodic distributions to derive a function theory model for m-isometrics, a disconjugacy theory for a subclass of Toeplitz operators. In addition, they introduced a class of $2$ -isometrics operators arising from a class of non stationary stochastic processes related to Brownian motion.

Recently, Sid Ahmed et al. ^[7] combined these two classes and introduced the $n$ -quasi- $(A, m)$ -isometry on a Hilbert space. Noting that all these works are in a Hilbert space $\mathcal{H}$ , a generalization of those on a Banach space $X$ was developed. For example, we can mention the work of Bayart ^[5], who introduced the $(m, q)$ -isometry, for an integer $q\geq1$ , by

$\begin{eqnarray*} \Delta_m^q(T): = \overset{m}{\underset{k = 0}{\sum}}(-1)^k\binom{m}{k}\Vert T^kx\Vert^q = 0,\ \forall\; x\in X. \end{eqnarray*}$

For $q = 2$ , the $(m, 2)$ -isometry coincide with the $m$ -isometry defined on a Hilbert space.

We recall that the $(A, m)$ -isometry was first introduced by Duggal in ^[6], for any operator $A\in \mathcal{L}(X)$ , by

$\begin{eqnarray*} \Delta_m^q(T,A,x): = \overset{m}{\underset{k = 0}{\sum}}(-1)^k\binom{m}{k}\Vert AT^kx\Vert^q = 0,\ \forall\; x\in X. \end{eqnarray*}$

In this paper, we will generalize this last class by introducing the $n$ -quasi- $A$ - $(m, q)$ -isometry on a Banach space and present some properties of this class like the product and the powers.

The main motivation for developing this work lies in the fact of knowing whether the different properties (spectral, product and power) of the $n$ - quasi- $(m, q)$ -isometry on a Banach space and the $n$ -quasi- $(A, m)$ -isometry on a Hilbert space are valid for operators in the new generalized class introduced in Definition 2.1. Note that in our study, we remove the condition of positivity on the operator $A$ , which exists in the Hilbert case.

The paper is organized as follows. In Section 2, we define our class of $n$ -quasi- $A$ - $(m, q)$ -isometry operators and present its basic properties. The power and the product of such operator belonging to this class are discussed in the last section.

2. Some basics properties

In this section, we define our new class and give its basic properties.

Given a Banach space $X$ , we denote by $\mathcal{L}(X)$ the class of all the (linear bounded) operators on $X$ . Hereafter, $I = I_X$ , $\mathcal{R}(T)$ and $\sigma_{ap}(T)$ denote the identity operator, the range and the approximate spectrum of an operator $T\in \mathcal{L}(X)$ , respectively. We define the class of $n$ -quasi- $A$ - $(m, q)$ -isometry operators by:

Definition 2.1. Let $A, T\in\mathcal{L}(X)$ . $T$ is called $n$ -quasi- $A$ - $(m, q)$ -isometry if and only if

$\begin{eqnarray*} \mathcal{Q}_{m,n}^q(T,A,x): = \sum\limits_{k = 0}^m(-1)^{m-k}\binom{m}{k}\Vert AT^{n+k}x\Vert^q = 0,\ \forall\; x\in X. \end{eqnarray*}$

Remark 2.1. Let $A, T\in\mathcal{L}(X)$ . Then $T$ is a $n$ -quasi- $A$ - $(m, q)$ -isometry if and only if $T$ is a $A$ - $(m, q)$ -isometry on $\overline{\mathcal{R}(T^n)}$ .

Indeed, $T$ is a $n$ -quasi- $A$ - $(m, q)$ -isometry if and only if

$\begin{eqnarray*} 0& = &\sum\limits_{k = 0}^m(-1)^{m-k}\binom{m}{k}\Vert AT^{n+k}x\Vert^q,\ \; \forall\; x\in X\\ & = &\sum\limits_{k = 0}^m(-1)^{m-k}\binom{m}{k}\Vert AT^{k}T^nx\Vert^q,\ \; \forall\; x\in X\\ & = &\sum\limits_{k = 0}^m(-1)^{m-k}\binom{m}{k}\Vert AT^{k}y\Vert^q,\ \;\forall\; y\in \overline{\mathcal{R}(T^n)}. \end{eqnarray*}$

In the following proposition, we give some spectral properties of the $n$ -quasi- $A$ - $(m, q)$ -isometry operators.

Proposition 2.1. Let $A, T\in \mathcal{L}(X)$ such that $T$ is a $n$ -quasi- $A$ - $(m, q)$ -isometry. If $0\notin\sigma_{ap}(A)$ , then $\sigma_{ap}(T) = \zeta(0, 1)\cup\{0\},$ where

$\zeta(0,1) = \{x\in X,\ \Vert x\Vert = 1\}.$

Proof. Let $(x_p)_p$ such that $\Vert x_p\Vert = 1$ and $\underset{p\to\infty}{\lim}(T-\lambda I)x_p = 0.$

Since $T$ is a $n$ -quasi- $A$ - $(m, q)$ -isometry, then

$\begin{eqnarray*} 0& = &\overset{m}{\underset{k = 0}{\sum}}(-1)^{m-k}\binom{m}{k}\Vert AT^{n+k}x_p\Vert^q\\ & = &\overset{m}{\underset{k = 0}{\sum}}(-1)^{m-k}\binom{m}{k}\Vert A(T^{n+k}-\lambda^{n+k})x_p+A\lambda^{n+k}x_p\Vert^q. \end{eqnarray*}$

As $\underset{p\to\infty}{\lim}(T-\lambda I)x_p = 0$ , then $\underset{p\to\infty}{\lim}(T^{n+k}-\lambda^{n+k} I)x_p = 0,$ for all $k = 0, 1, \cdots, m.$

Therefore

$\begin{eqnarray*} 0& = &\underset{p\to\infty}{\lim}\overset{m}{\underset{k = 0}{\sum}}(-1)^{m-k}\binom{m}{k}\Vert A\lambda^{n+k}x_p\Vert^q\\ & = &\vert\lambda\vert^{nq}\overset{m}{\underset{k = 0}{\sum}}(-1)^{m-k}\binom{m}{k}(\vert\lambda\vert^q)^k\underset{p\to\infty}{\lim}\Vert Ax_p\Vert^q\\ & = &\vert\lambda\vert^{nq}(\vert\lambda\vert^q-1)^m\underset{p\to\infty}{\lim}\Vert Ax_p\Vert^q. \end{eqnarray*}$

Since $0\notin\sigma_{ap}(A),$ then $\lambda = 0$ or $\vert\lambda\vert = 1.$ □

Proposition 2.2. Let $A, T\in \mathcal{L}(X)$ . If $T$ is an $n$ -quasi- $A$ - $(m, q)$ -isometry, then $T$ is a $n_1$ -quasi- $A$ - $(m, q)$ -isometry, for all $n_1\geq n$ .

Proof. Let $T$ be a $n$ -quasi- $A$ - $(m, q)$ -isometry on $X$ . By Remark 2.1, $T$ is a $A$ - $(m, q)$ -isometry on $\overline{\mathcal{R}(T^n)}$ . Since $\overline{\mathcal{R}(T^n)}\supset\overline{\mathcal{R}(T^{n_1})}$ for all $n_1\geq n$ , therefore $T$ is a $A$ - $(m, q)$ -isometry on $\overline{\mathcal{R}(T^{n_1})}$ . According to Remark 2.1, we obtain that $T$ is a $n_1$ -quasi- $A$ - $(m, q)$ -isometry, for all $n_1\geq n$ . □

In the following proposition, thanks to a suitable condition, we give the inverse sense of the Proposition 2.2.

Proposition 2.3. Let $1\leq p\leq n-1$ such that $\overline{\mathcal{R}(T^{p})} = \overline{\mathcal{R}(T^{p+1})}$ . If $T$ is a $n$ -quasi- $A$ - $(m, q)$ -isometry, then $T$ is a $p$ -quasi- $A$ - $(m, q)$ -isometry.

Proof. Thanks to the hypothesis $\overline{\mathcal{R}(T^p)} = \overline{\mathcal{R}(T^{p+1})}$ , it follows that $\overline{\mathcal{R}(T^p)} = \overline{\mathcal{R}(T^{n})}$ . Since $T$ is a $n$ -quasi- $A$ - $(m, q)$ -isometry on $X$ , then $T$ is a $A$ - $(m, q)$ -isometry on $\overline{\mathcal{R}(T^n)} = \overline{\mathcal{R}(T^p)}$ . Therefore, $T$ is a $p$ -quasi- $A$ - $(m, q)$ -isometry. □

As in the Hilbert case, we have the following result.

Proposition 2.4. Let $T\in\mathcal{L}(X)$ be a $n$ -quasi- $A$ - $(m, q)$ -isometry. Then $T$ is a $n$ -quasi- $A$ - $(\ell, q)$ -isometry for all $\ell\geq m.$

Proof.

$\begin{eqnarray*} \mathcal{Q}_{m+1,n}^q(T,A,x)& = &\sum\limits_{k = 0}^{m+1}(-1)^{m+1-k}\binom{m+1}{k}\Vert AT^{n+k}x\Vert^q\\ & = &(-1)^{m+1}\Vert AT^{n}x\Vert^q+\Vert AT^{n+m+1}x\Vert^q\\ &&+\sum\limits_{k = 1}^{m}(-1)^{m+1-k}\left[\binom{m}{k}+\binom{m}{k-1}\right] \Vert AT^{n+k}x\Vert^q\\ & = & -(-1)^{m}\Vert AT^{n}x\Vert^q-\sum\limits_{k = 1}^{m}(-1)^{m-k}\binom{m}{k}\Vert AT^{n+k}x\Vert^q\\ &&+\sum\limits_{k = 1}^{m}(-1)^{m+1-k}\binom{m}{k-1} \Vert AT^{n+k}x\Vert^q+\Vert AT^{n+m+1}x\Vert^q\\ & = & -\mathcal{Q}_{m,n}^q(T,A,x)+\sum\limits_{k = 0}^{m-1}(-1)^{m-k}\binom{m}{k} \Vert AT^{n+1+k}x\Vert^q\\ &&+\Vert AT^{n+1+m}x\Vert^q\\ & = &\mathcal{Q}_{m,n+1}^q(T,A,x)-\mathcal{Q}_{m,n}^q(T,A,x)\\ & = &0. \end{eqnarray*}$

□

Example 2.1. Let $T, A\in\mathcal{L}(X)$ , where $X = \ell^q(\mathbb{N})$ , defined by

$T\alpha_n = w_n\alpha_{n}\text{ and } A\alpha_n = \alpha_{n+1},$

where $w_n = \left(\frac{n+1}{n}\right)^{\frac{1}{q}}$ . By simple calculations, we get that $T$ is a $2$ -quasi- $A$ - $(2, q)$ -isometry but it is not a $2$ -quasi- $A$ - $(1, q)$ -isometry. Indeed, we have

$\begin{eqnarray*} \mathcal{Q}_{2,2}^q(T,A,\alpha_n)& = &\Vert AT^4\alpha_n\Vert^q-2\Vert AT^3\alpha_n\Vert^q+\Vert AT^2\alpha_n\Vert^q\\ & = &\underset{n\geq1}{\sum}\left(\vert w_nw_{n+1}w_{n+2}w_{n+4}\vert^q-2\vert w_nw_{n+1}w_{n+2}\vert^q+\vert w_nw_{n+1}\vert^q\right)\vert\alpha_{n+1}\vert^q\\ & = &\underset{n\geq1}{\sum}\left(\frac{n+4}{n}-2\frac{n+3}{n}+\frac{n+2}{n}\right)\vert\alpha_{n+1}\vert^q\\ & = &0, \end{eqnarray*}$

and

$\begin{eqnarray*} \mathcal{Q}_{1,2}^q(T,A,\alpha_n)& = &\Vert AT^3\alpha_n\Vert^q-\Vert AT^2\alpha_n\Vert^q\\ & = &\underset{n\geq1}{\sum}\left(\vert w_nw_{n+1}w_{n+2}\vert^q-\vert w_nw_{n+1}\vert^q\right)\vert\alpha_{n+1}\vert^q\\ & = &\underset{n\geq1}{\sum}\left(\frac{n+3}{n}-\frac{n+2}{n}\right)\vert\alpha_{n+1}\vert^q\\ & = &\underset{n\geq1}{\sum}\frac{\vert\alpha_{n+1}\vert^q}{n}\\ &\ne&0. \end{eqnarray*}$

Proposition 2.5. Let $A, T\in\mathcal{L}(X)$ such that $T$ is a $n$ -quasi- $A$ - $(m, q)$ -isometry. Then, for all $p\geq0$ , we have

(1) $\Vert AT^{n+p}x\Vert^q = \sum\limits_{k = 0}^{m-1}\binom{p}{k}\mathcal{Q}_{k, n}^q(T, A, x)$ .

(2) $\mathcal{Q}_{m-1, n}^q(T, A, x) = \underset{p\to\infty}{\lim}\frac{\Vert AT^{n+p}x\Vert^q}{\binom{p}{m-1}}\geq0$ .

Proof. $(1)$ By induction, we prove that, for all $p\geq0$ ,

$\begin{eqnarray*} \Vert AT^{n+p}x\Vert^q = \sum\limits_{k = 0}^{p}\binom{p}{k}\mathcal{Q}_{k,n}^q(T,A,x). \end{eqnarray*}$

For $p = 0$ , we infer that

$\begin{eqnarray*} \sum\limits_{k = 0}^{0}\binom{0}{k}\mathcal{Q}_{k,n}^q(T,A,x)& = &\mathcal{Q}_{0,n}^q(T,A,x)\\ & = &\Vert AT^{n}x\Vert^q. \end{eqnarray*}$

We suppose that $\Vert AT^{n+j}x\Vert^q = \underset{k = 0}{\overset{j}{\sum\limits}}\binom{j}{k}\mathcal{Q}_{k, n}^q(T, A, x)$ for all $j\leq p.$

We know that

$\begin{eqnarray*} & &\Vert AT^{n+p+1}x\Vert^q\\& = &\mathcal{Q}_{p+1,n}^q(T,A,x)-\sum\limits_{k = 0}^{p}(-1)^{p+1-k}\binom{p+1}{k}\Vert AT^{n+k}x\Vert^q\\ & = &\mathcal{Q}_{p+1,n}^q(T,A,x)-\sum\limits_{k = 0}^{p}(-1)^{p+1-k}\binom{p+1}{k}\sum\limits_{j = 0}^k\binom{k}{j}\mathcal{Q}_{j,n}^q(T,A,x)\\ & = &\mathcal{Q}_{p+1,n}^q(T,A,x)-\sum\limits_{j = 0}^p\mathcal{Q}_{j,n}^q(T,A,x)\sum\limits_{k = j}^{p}(-1)^{p+1-k}\binom{p+1}{k}\binom{k}{j}\\ & = &\mathcal{Q}_{p+1,n}^q(T,A,x)-\sum\limits_{j = 0}^p\binom{p+1}{j}\mathcal{Q}_{j,n}^q(T,A,x)\underset{ = -1}{\underbrace{\sum\limits_{k = j}^{p}(-1)^{p+1-k}\binom{p+1-j}{k-j}}}\\ & = &\mathcal{Q}_{p+1,n}^q(T,A,x)+\sum\limits_{j = 0}^p\binom{p+1}{j}\mathcal{Q}_{j,n}^q(T,A,x)\\ & = &\sum\limits_{j = 0}^{p+1}\binom{p+1}{j}\mathcal{Q}_{j,n}^q(T,A,x). \end{eqnarray*}$

Then, for all $p\geq0$ , we have that $\Vert AT^{n+p}x\Vert^q = \underset{k = 0}{\overset{p}{\sum\limits}}\binom{p}{k}\mathcal{Q}_{k, n}^q(T, A, x)$ .

Since $T$ is a $n$ -quasi- $A$ - $(m, q)$ -isometry on $X$ , then, by Proposition 2.4, we obtain that

$\mathcal{Q}_{k,n}^q(T,A,x) = 0,\ \text{ for all } k\geq m.$

Hence, for all $p\geq0$ , we get that

$\Vert AT^{n+p}x\Vert^q = \underset{k = 0}{\overset{m-1}{\sum\limits}}\binom{p}{k}\mathcal{Q}_{k,n}^q(T,A,x).$

$(2)$ We know, by assertion (1), that

$\begin{eqnarray*} \Vert AT^{n+p}x\Vert^q& = &\sum\limits_{k = 0}^{m-1}\binom{p}{k}\mathcal{Q}_{k,n}^q(T,A,x)\\ & = &\binom{p}{m-1}\mathcal{Q}_{m-1,n}^q(T,A,x)+\sum\limits_{k = 0}^{m-2}\binom{p}{k}\mathcal{Q}_{k,n}^q(T,A,x). \end{eqnarray*}$

Dividing both sides by $\binom{p}{m-1}\neq0$ , we see that

$\mathcal{Q}_{m-1,n}^q(T,A,x) = \frac{1}{\binom{p}{m-1}}\Vert AT^{n+p}x\Vert^q-\frac{1}{\binom{p}{m-1}}\sum\limits_{k = 0}^{m-2}\binom{p}{k}\mathcal{Q}_{k,n}^q(T,A,x).$

Upon taking the limit as $p\rightarrow \infty$ , we know that $\underset{p\to\infty}{\lim}\frac{\binom{p}{k}}{\binom{p}{m-1}} = 0$ , for all $k = 0, 1, ..., m-2$ . Therefore, since $\Vert AT^{n+p}x\Vert^q\geq0$ , it holds that

$\mathcal{Q}_{m-1,n}^q(T,A,x) = \underset{p\to\infty}{\lim}\frac{\Vert AT^{n+p}x\Vert^q}{\binom{p}{m-1}}\geq0.$

□

3. Product and powers of $n$ -quasi- $A$ - $(m, q)$ -isometry operators

In this section, we study the product and power of an $n$ -quasi- $A$ - $(m, q)$ -isometry operators.

Let $n^{(k)}$ be the (descending Pochhammer) symbol defined by:

$n^{(k)} = \left\{ \begin{array}{cc} 0&if\ n = 0,\\ 0& if\ n > 0\ and \ k > n,\\ k!\binom{n}{k}& if\ n > 0\ and\ k\leq n. \end{array} \right.$

Proposition 3.1. $T$ is a $n$ -quasi- $A$ - $(m, q)$ -isometry if and only if we have

$\Vert AT^{n+p}x\Vert = \overset{m-1}{\underset{j = 0}{\sum\limits}}(-1)^{m-j-1}\frac{p(p-1)\cdots\overbrace{(p-j)}\cdots(p-m+1)}{j!(m-j-1)!}\Vert AT^{j+n}x\Vert^q,$

for all $p\geq0$ and all $x \in X$ , where $\overbrace{(p-j)}$ denotes that the factor $(p -j)$ is omitted.

Proof. $T$ is a $n$ -quasi- $A$ - $(m, q)$ -isometry if and only if

$\begin{eqnarray*} \Vert AT^{n+p}x\Vert^q& = &\overset{m-1}{\underset{k = 0}{\sum}}\binom{p}{k}\mathcal{Q}_{k,n}^q(T,A,x)\\ & = &\overset{m-1}{\underset{k = 0}{\sum}}\binom{p}{k}\overset{k}{\underset{j = 0}{\sum}}(-1)^{k-j}\binom{k}{j}\Vert AT^{n+j}x\Vert^q\\ & = &\overset{m-1}{\underset{j = 0}{\sum}}\Vert AT^{n+j}x\Vert^q\overset{m-1}{\underset{k = j}{\sum}}(-1)^{k-j}\binom{p}{k}\binom{k}{j}. \end{eqnarray*}$

By [12, Lemma 2.3], we have

$\overset{m-1}{\underset{k = j}{\sum\limits}}(-1)^{k-j}\binom{p}{k}\binom{k}{j} = (-1)^{m-j-1}\frac{p(p-1)\cdots\overbrace{(p-j)}\cdots(p-m+1)}{j!(m-j-1)!}.$

Then,

$\Vert AT^{n+p}x\Vert = \overset{m-1}{\underset{j = 0}{\sum\limits}}(-1)^{m-j-1}\frac{p(p-1)\cdots\overbrace{(p-j)}\cdots(p-m+1)}{j!(m-j-1)!}\Vert AT^{j+n}x\Vert^q.$

□

Lemma 3.1. Let $T$ be a $n$ -quasi- $A$ - $(m, q)$ -isometry and $\ell > m\geq1$ . For all $t\in\{0, \cdots, \ell-2\}$ , we have

$\underset{j = 0}{\overset{m+\ell-1}{\sum\limits}}(-1)^{m+\ell-1-j}\binom{m+\ell-1}{j}\underset{i = 0}{\overset{t}{\prod\limits}}(j-i)\Vert AT^{j+n}x\Vert^q = 0.$

Proof. Let $t\in\{0, \cdots, \ell-2\}$ , we have

$\begin{eqnarray*} \binom{m+\ell-1}{j}\underset{i = 0}{\overset{t}{\prod}}(j-i)& = &\frac{(m+\ell-1)!}{j!(m+\ell-1-j)!}j(j-1)\cdots(j-t)\\ & = &\frac{(m+\ell-t-2)!\underset{i = 0}{\overset{t}{\prod\limits}}(m+\ell+i)}{(j-t-1)!(m+\ell-1-j)!}\\ & = &\binom{m+\ell-t-2}{j-t-1}\underset{i = 1}{\overset{t+1}{\prod}}(m+\ell-i). \end{eqnarray*}$

Then

$\begin{eqnarray*} & &\underset{j = 0}{\overset{m+\ell-1}{\sum}}(-1)^{m+\ell-1-j}\binom{m+\ell-1}{j}\underset{i = 0}{\overset{t}{\prod}}(j-i)\Vert AT^{j+n}x\Vert^q\\ & = &\underset{j = 0}{\overset{m+\ell-1}{\sum}}(-1)^{m+\ell-1-j}\binom{m+\ell-t-2}{j-t-1}\underset{i = 1}{\overset{t+1}{\prod}}(m+\ell-i)\Vert AT^{j+n}x\Vert^q\\ & = &\underset{i = 1}{\overset{t+1}{\prod}}(m+\ell-i)\left(\underset{j = t+1}{\overset{m+\ell-1}{\sum}}(-1)^{m+\ell-1-j}\binom{m+\ell-t-2}{j-t-1}\Vert AT^{j+n}x\Vert^q\right)\\ & = &\underset{i = 1}{\overset{t+1}{\prod}}(m+\ell-i)\left(\underset{j = 0}{\overset{m+\ell-t-2}{\sum}}(-1)^{m+\ell-t-2-j}\binom{m+\ell-t-2}{j}\Vert AT^{j+n}(T^{t+1}x)\Vert^q\right)\\ & = &0. \end{eqnarray*}$

□

Lemma 3.2. Let $T$ be a $n$ -quasi- $A$ - $(m, q)$ -isometry, $p\geq0$ and $\ell\geq m\geq1$ . Then, there exists a finite sequence $(a_{j, i})_{i = 0}^{m-1}$ such that

$\Vert AT^{p+n}x\Vert^q = \sum\limits_{k = 0}^{m-1}\frac{(-1)^{m-1-k}}{k!(m-k-1)!}\left[a_{j,0}+\underset{i = 1}{\overset{m-1}{\sum\limits}}a_{j,i}\underset{t = 0}{\overset{i-1}{\prod\limits}}(p-t)\right]\Vert AT^{n+k}x\Vert^q ,$

for $j = 0, 1, ..., \ell-1.$

Proof. By using ^[13], there exists a finite sequence $(a_{j, i})_{i = 0}^{m-1}$ such that

$p(p-1)\cdots\overbrace{(p-j)}\cdots(p-m+1) = a_{j,0}+\underset{i = 1}{\overset{m-1}{\sum\limits}}a_{j,i}\underset{t = 0}{\overset{i-1}{\prod\limits}}(p-t).$

Since $T$ is a $n$ -quasi- $A$ - $(m, q)$ -isometry, then by using Proposition 3.1, we obtain that

$\begin{eqnarray*} & &\Vert AT^{n+p}x\Vert^q\\ & = &\overset{m-1}{\underset{j = 0}{\sum}}\frac{(-1)^{m-j-1}}{j!(m-j-1)!}\left[p(p-1)\cdots\overbrace{(p-j)}\cdots(p-m+1)\right]\Vert AT^{j+n}x\Vert^q\\ & = &\overset{m-1}{\underset{j = 0}{\sum}}\frac{(-1)^{m-j-1}}{j!(m-j-1)!}\left[a_{j,0}+\underset{i = 1}{\overset{m-1}{\sum}}a_{j,i}\underset{t = 0}{\overset{i-1}{\prod}}(p-t)\right]\Vert AT^{j+n}x\Vert^q. \end{eqnarray*}$

□

Theorem 3.1. Let $n_1, \, n_2, \, m, \, l$ be positive integers and $T, S\in\mathcal{L}(X)$ . If $T$ is a $n_1$ -quasi- $A$ - $(m, q)$ -isometry and $S$ is a $n_2$ -quasi- $A$ - $(l, q)$ -isometry such that $ST = TS$ , then $TS$ is a $n$ -quasi- $A$ - $(m+l-1, q)$ -isometry, with $n = \max(n_1, n_2)$ .

Proof. We have

$\begin{eqnarray*} \mathcal{Q}_{m+l-1,n}^q(TS,A,x)& = &\sum\limits_{k = 0}^{m+l-1}(-1)^{m+l-1-k}\binom{m+l-1}{k}\Vert A(TS)^{n+k}x\Vert^q\\ & = &\sum\limits_{k = 0}^{m+l-1}(-1)^{m+l-1-k}\binom{m+l-1}{k}\Vert A T^{n+k}\left( S^{n+k}x\right)\Vert^q. \end{eqnarray*}$

Using Lemma 3.2, we see that

$\begin{eqnarray*} &&\mathcal{Q}_{m+l-1,n}^q(TS,A,x) = \sum\limits_{k = 0}^{m+l-1}(-1)^{m+l-1-k}\binom{m+l-1}{k}\overset{m-1}{\underset{j = 0}{\sum}}\frac{(-1)^{m-j-1}}{j!(m-j-1)!}\\ &&\times\left[a_{j,0}+\underset{i = 1}{\overset{m-1}{\sum}}a_{j,i}\underset{t = 0}{\overset{i-1}{\prod}}(p-t)\right]\Vert A T^{n+j}\left( S^{n+k}x\right)\Vert^q\\ & = &\overset{m-1}{\underset{j = 0}{\sum}}\frac{(-1)^{m-j-1}}{j!(m-j-1)!}\sum\limits_{k = 0}^{m+l-1}(-1)^{m+l-1-k}\binom{m+l-1}{k}\left[a_{j,0}+\underset{i = 1}{\overset{m-1}{\sum}}a_{j,i}\underset{t = 0}{\overset{i-1}{\prod}}(p-t)\right]\\ & &\times\Vert A T^{n+j}\left( S^{n+k}x\right)\Vert^q\\ & = &\overset{m-1}{\underset{j = 0}{\sum}}\frac{(-1)^{m-j-1}}{j!(m-j-1)!}a_{j,0}\sum\limits_{k = 0}^{m+l-1}(-1)^{m+l-1-k}\binom{m+l-1}{k}\Vert A S^{n+k}\left( T^{n+j}x\right)\Vert^q\\ &&+\overset{m-1}{\underset{j = 0}{\sum}}\frac{(-1)^{m-j-1}}{j!(m-j-1)!}\underset{i = 1}{\overset{m-1}{\sum}}a_{j,i}\sum\limits_{k = 0}^{m+l-1}(-1)^{m+l-1-k}\binom{m+l-1}{k}\\ & &\times\underset{t = 0}{\overset{i-1}{\prod}}(p-t)\Vert A S^{n+k}\left( T^{n+j}x\right)\Vert^q. \end{eqnarray*}$

Since $S$ is a $n_2$ -quasi- $A$ - $(l, q)$ -isometry, then according to Proposition 2.4, we get that $S$ is a $n$ -quasi- $A$ - $(m+l-1, q)$ -isometry. Hence,

$\overset{m-1}{\underset{j = 0}{\sum\limits}}\frac{(-1)^{m-j-1}}{j!(m-j-1)!}a_{j,0}\underset{ = 0}{\underbrace{\sum\limits_{k = 0}^{m+l-1}(-1)^{m+l-1-k}\binom{m+l-1}{k}\Vert A S^{n+k}\left( T^{n+j}x\right)\Vert^q}} = 0.$

Since $i = 1, ..., m-1$ , then, by using Lemma 3.1, we infer that

$\sum\limits_{k = 0}^{m+l-1}(-1)^{m+l-1-k}\binom{m+l-1}{k}\underset{t = 0}{\overset{i-1}{\prod\limits}}(p-t)\Vert A S^{n+k}\left( T^{n+j}x\right)\Vert^q = 0.$

Consequently, one obtains that

$\overset{m-1}{\underset{j = 0}{\sum\limits}}\frac{(-1)^{m-j-1}}{j!(m-j-1)!}\underset{i = 1}{\overset{m-1}{\sum\limits}}a_{j,i}\underset{ = 0}{\underbrace{\sum\limits_{k = 0}^{m+l-1}(-1)^{m+l-1-k}\binom{m+l-1}{k}\underset{t = 0}{\overset{i-1}{\prod\limits}}(p-t)\Vert A S^{n+k}\left( T^{n+j}x\right)\Vert^q}} = 0,$

which gives that

$\mathcal{Q}_{m+l-1,n}^q(TS,A,x) = 0.$

□

The following example shows that Theorem 3.1 is not necessarily true if $S$ and $T$ are not commuting.

Example 3.1. We consider the operators on the two dimensional $\left(\mathbb{R}^2, \Vert.\Vert_2\right).$

$T = \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right), \ \ S = \left( \begin{array}{cc} 1 & 0 \\ 1 & 1 \\ \end{array} \right)\ \mbox{and}\ A = \left( \begin{array}{cc} 1 & 1 \\ 1 & 2 \\ \end{array} \right).$

Note that $ST\neq TS$ . Moreover, by a direct computation, we show that $T$ and $S$ are quasi- $A$ - $(3, 2)$ -isometry. However neither $TS$ nor $ST$ is a quasi- $A$ - $(5, 2)$ -isometry.

Corollary 3.1. Let $n_1, \, n_2, \, m, \, l$ be positive integers and $T, S, A_1, A_2\in\mathfrak{L}(X)$ such that $TS = ST, A_1A_2 = A_2A_1, TA_1 = A_1T$ and $SA_2 = A_2S$ . If $T$ is a $n_1$ -quasi- $A_1$ - $(m, q)$ -isometry and $S$ is a $n_2$ -quasi- $A_2$ - $(l, q)$ -isometry, then $TS$ is a $n$ -quasi- $A_1A_2$ - $(m+l-1, q)$ -isometry, with $n = \max(n_1, n_2)$ .

Proof. Following the same steps as in the proof of Theorem 3.1, we can prove that

$\mathcal{Q}_{m+l-1,n}^q(TS,A_1A_2,x) = 0.$

Indeed, we have

$\begin{eqnarray*} \mathcal{Q}_{m+l-1,n}^q(TS,A_1A_2,x)& = &\sum\limits_{k = 0}^{m+l-1}(-1)^{m+l-1-k}\binom{m+l-1}{k}\Vert A_1A_2(TS)^{n+k}x\Vert^q\\ & = &\sum\limits_{k = 0}^{m+l-1}(-1)^{m+l-1-k}\binom{m+l-1}{k}\Vert A_1 T^{n+k}\left( A_2S^{n+k}x\right)\Vert^q. \end{eqnarray*}$

□

Theorem 3.2. Let $T$ be a $n$ -quasi- $A$ - $(m, q)$ -isometry. Then, for each positive integer $k$ , $T^k$ is a $n$ -quasi- $A$ - $(m, q)$ -isometry.

Proof.

$\begin{eqnarray*} \mathcal{Q}_{m,n}^q(T^k,A,x)& = &\sum\limits_{j = 0}^{m}(-1)^{m-j}\binom{m}{j}\Vert A\left(T^k\right)^{n+j}x\Vert^q\\ & = &\sum\limits_{j = 0}^{m}(-1)^{m-j}\binom{m}{j}\Vert AT^{kn+kj}x\Vert^q\\ & = &\sum\limits_{j = 0}^{m}(-1)^{m-j}\binom{m}{j}\sum\limits_{i = 0}^{m-1}\binom{kj}{i}\mathcal{Q}_{i,kn}^q(T,A,x)\\ & = &\sum\limits_{i = 0}^{m-1}\frac{1}{i!}\left[\sum\limits_{j = 0}^{m}(-1)^{m-j}\binom{m}{j}(kj)^{(i)}\right]\mathcal{Q}_{i,kn}^q(T,A,x). \end{eqnarray*}$

According to [, Lemma 1], we have $\sum\limits_{j = 0}^{m}(-1)^{m-j}\binom{m}{j}(kj)^{(i)} = 0$ for each $i = 0, 1, ..., m-1$ . It follows that $\mathcal{Q}_{m, n}^q(T^k, A, x) = 0$ . □

The converse of Theorem 3.2 is not true in general as shown in the following example.

Example 3.2. Let $A = \left(\begin{array}{ccc} 1 & 0 & 1\\ 0 & 1 & 1\\ 1 & 1 & 0 \end{array} \right).$ It is not difficult to prove that the operator $T = \left(\begin{array}{ccc} 0 & 1 & 1\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{array} \right)$ defined in $\left(\mathbb{R}^3, \Vert.\Vert_2\right)$ satisfies $T^3$ is a quasi- $A$ - $(3, 2)$ -isometry but $T$ is not a quasi- $A$ - $(3, 2)$ -isometry.

Proposition 3.2. Let $T\in\mathcal{L}(X)$ and $n_1, \, n_2, \, r, \, s, \, m, \, l$ be positive integers. If $T^r$ is a $n_1$ -quasi- $A$ - $(m, q)$ -isometry and $T^s$ is a $n_2$ -quasi- $A$ - $(l, q)$ -isometry, then $T^t$ is a $n_0$ -quasi- $A$ - $(p, q)$ -isometry, where $t$ is the greatest common divisor of $r$ and $s$ , $n_0 = \max(n_1, n_2)$ and $p = \min(m, \ l).$

Proof. Let's put $a_j = \Vert A T^{n+j}x\Vert^q, \; \forall\; j\geq0$ . Since $T^r$ is a $n_1$ -quasi- $A$ - $(m, q)$ -isometry and $T^s$ is a $n_2$ -quasi- $A$ - $(l, q)$ -isometry, then $T^r$ is a $n_0$ -quasi- $A$ - $(m, q)$ -isometry and $T^s$ is a $n_0$ -quasi- $A$ - $(l, q)$ -isometry. Hence,

$\sum\limits_{j = 0}^m(-1)^{m-j}\binom{m}{j}a_{j+n_0r} = 0\quad\mbox{and}\quad\sum\limits_{j = 0}^l(-1)^{l-j}\binom{l}{j}a_{j+n_0s} = 0.$

By [15, Lemma 3.15], we infer that

$\sum\limits_{j = 0}^p(-1)^{p-j}\binom{p}{j}a_{j+n_0t} = 0,$

which ends the proof. □

As an immediate consequence of Proposition 3.2, we have the following result.

Corollary 3.2. Let $T, A \in\mathcal{L}(X)$ and $r, s, m, n, l$ be positive integers. Then, the following properties hold.

(1) If $T$ is a $n$ -quasi- $A$ - $(m, q)$ -isometry such that $T^s$ is a $n$ -quasi- $A$ - $(1, q)$ -isometry, then $T$ is a $n$ -quasi- $A$ - $(1, q)$ -isometry.

(2) If $T^r$ and $T^{r+1}$ are a $n$ -quasi- $A$ - $(m, q)$ -isometries, then $T$ is a $n$ -quasi- $A$ - $(m, q)$ -isometry.

(3) If $T^r$ is a $n$ -quasi- $A$ - $(m, q)$ -isometry and $T^{r+1}$ is a $n$ -quasi- $A$ - $(l, q)$ -isometry with $m < l$ , then $T$ is a $n$ -quasi- $A$ - $(m, q)$ -isometry.

As an immediate consequence of Proposition 3.2, Theorem 3.1 and Corollary 3.1, we have the following result.

Corollary 3.3. Let $n_1, \, n_2, \, r, \, s, \, m, \, l$ be positive integers and $T, S, A, A_1, A_2\in\mathcal{L}(X)$ . Let $n = \max(n_1, n_2)$ . The following properties hold true.

(1) If $T$ is a $n_1$ -quasi- $A$ - $(m, q)$ -isometry and $S$ is a $n_2$ -quasi- $A$ - $(l, q)$ -isometry such that $ST = TS$ , then $T^rS^s$ is a $n$ -quasi- $A$ - $(m+l-1, q)$ -isometry.

(2) If $T$ is a $n_1$ -quasi- $A_1$ - $(m, q)$ -isometry and $S$ is a $n_2$ -quasi- $A_2$ - $(l, q)$ -isometry such that $TS = ST$ , $A_1A_2 = A_2A_1$ , $TA_1 = A_1T$ and $SA_2 = A_2S$ , then $T^rS^s$ is a $n$ -quasi- $A_1A_2$ - $(m+l-1, q)$ -isometry.

4. Conclusions

We focus on some properties of a new class of operators called $n$ -quasi- $A$ - $(m, q)$ -isometry operators. First, we give spectral properties and relationship between $n$ -quasi- $A$ - $(m, q)$ -isometry and $p$ -quasi- $A$ - $(m, q)$ -isometry. Second, the power and product of such operators have been investigated. As a future work, we can generalize our study on a metric, dislocated metric or dislocated quasi metric space (see references ^[8,16,17]).

Use of AI tools declaration

The authors declare that they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

This work is supported by Researchers Supporting Project number (RSPD2023R736), King Saud University, Riyadh, Saudi Arabia.

Conflict of interest

The authors declare that there is no conflict of interest

References

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$n$ -quasi- $A$ - $(m, q)$ -isometry on a Banach space

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n n -quasi-A A -(m,q) (m, q) -isometry on a Banach space

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$n$ -quasi- $A$ - $(m, q)$ -isometry on a Banach space

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