Some results on higher orders quasi-isometries

TL;DR

This paper explores the properties of n-quasi-m-isometric operators, a generalization of m-isometric operators, on infinite complex Hilbert spaces, extending previous work and providing new insights into their characteristics.

Contribution

It introduces new properties of n-quasi-m-isometric operators, expanding the understanding of their structure and behavior beyond prior studies.

Findings

Established additional properties of n-quasi-m-isometric operators

Extended the theory of m-isometric operators to the quasi-m setting

Contributed to the mathematical understanding of operator classes on Hilbert spaces

Abstract

The purpose of the present paper is to pursue further study of a class of linear bounded operators, known as n-quasi-m-isometric operators acting on an infinite complex separable Hilbert space H. This generalizes the class of m-isometric operators on Hilbert space introduced by Agler and Stankus in [1]. The class of n-quasi-m-isometric operators was defined by S. Mecheri and T. Prasad in [18] , where they have given some of their properties. Based, mainly, on [2], [3], [5] and [9], we contribute some other properties of such operators.

Equations

$$\sum _{0\le k\le m}(-1{)}^{m-k}(\genfrac{}{}{0ex}{}{m}{k})T^{\ast k}{T}^k=0.$$

$$\sum _{0\le k\le m}(-1{)}^{m-k}(\genfrac{}{}{0ex}{}{m}{k})\Vert T^{k}x{\Vert }^2=0\forall x\in \mathcal{H},$$

$$\displaystyle T\bigg{(}\sum_{0\leq k\leq m}(-1)^{k}\binom{m}{k}T^{*m-k}T^{m-k}\bigg{)}=0,$$

$$\displaystyle T^{q}\bigg{(}\sum_{0\leq k\leq m}(-1)^{k}\binom{m}{k}T^{*m-k}T^{m-k}\bigg{)}=0,$$

$$T^{\ast m+1}{T}^{m+1}-T^{\ast m}{T}^m=0.$$

$$T^{*m}\bigg{(}T^{*}T-I\bigg{)}T^{m}=0.$$

$$\displaystyle T^{*n}\bigg{(}\sum_{0\leq k\leq m}(-1)^{k}\binom{m}{k}T^{*m-k}T^{m-k}\bigg{)}T^{n}=0,$$

$${\beta }_m(T):=\sum _{0\le k\le m}(-1{)}^{m-k}(\genfrac{}{}{0ex}{}{m}{k})T^{\ast k}{T}^k,$$

$$\beta_{m,\;n}(T):=T^{\ast n}\bigg{(}\sum_{0\leq k\leq m}(-1)^{m-k}\binom{m}{k}T^{\ast k}T^{k}\bigg{)}T^{n}$$

$${\Delta }_{m,n}(T,x):=\sum _{0\le k\le m}(-1{)}^{m-k}(\genfrac{}{}{0ex}{}{m}{k})\Vert T^{k+n}x{\Vert }^2,x\in \mathcal{H}.$$

$${\Delta }_{m,n}(T,x)=0\forall x\in \mathcal{H}.$$

$$\sum _{0\le k\le m}(-1{)}^{m-k}(\genfrac{}{}{0ex}{}{m}{k})\Vert T^{k}x{\Vert }^2=0\forall x\in \overline{\mathcal{R}(T^n)},$$

$$\sum _{0\le k\le m}(-1{)}^{m-k}(\genfrac{}{}{0ex}{}{m}{k})a_{j+k}=0\text{for}j\ge 0$$

$${\beta }_m(T_1):=\sum _{0\le k\le m}(-1{)}^{m-k}(\genfrac{}{}{0ex}{}{m}{k})T_1^{\ast k}{T}_1^k=0\text{and}{T}_3^n=0.$$

$$\displaystyle T^{*n}\bigg{(}\sum_{0\leq k\leq m}(-1)^{m-k}\binom{m}{k}T^{*k}T^{k}\bigg{)}T^{n}$$

$$=$$

$$T=\left(\begin{array}[]{ccc}T_{1}&T_{2}\\ 0&T_{3}\end{array}\right)\;\;\text{on}\;\;{\mathcal{H}}=\overline{{\mathcal{R}}(T^{n})}\oplus{\mathcal{N}}(T^{*n}),$$

$$\sigma (T)=\sigma (T_1)\cup \sigma (T_3)=\sigma (T_1)\cup \{0\}.$$

$$T=\left(\begin{array}[]{ccc}T_{1}&T_{2}\\ 0&T_{3}\end{array}\right)=\left(\begin{array}[]{ccc}I&A\\ 0&I\end{array}\right)^{-1}\left(\begin{array}[]{ccc}T_{1}&0\\ 0&T_{3}\end{array}\right)\left(\begin{array}[]{ccc}I&A\\ 0&I\end{array}\right).$$

$$T{e}_1=2e_2,T{e}_2=3e_3\text{and}{T}_k=e_{k+1}\text{for}k\ge 3.$$

$$\displaystyle T^{*n}\Bigg{(}\sum_{0\leq k\leq m}(-1)^{m-k}\binom{m}{k}T^{*k}T^{k}\Bigg{)}T^{n}=0,$$

$$\displaystyle T^{*}\Bigg{(}\sum_{0\leq k\leq m}(-1)^{m-k}\binom{m}{k}T^{*k}T^{k}\Bigg{)}T^{n}=0.$$

$$\displaystyle T^{*n}\Bigg{(}\sum_{0\leq k\leq m}(-1)^{m-k}\binom{m}{k}\bigg{(}T^{*k}T^{k}\bigg{)}^{*}\Bigg{)}T=0.$$

$$\displaystyle T^{*}\Bigg{(}\sum_{0\leq k\leq m}(-1)^{m-k}\binom{m}{k}\bigg{(}T^{*k}T^{k}\bigg{)}^{*}\Bigg{)}T=0.$$

$$\displaystyle T^{*}\Bigg{(}\sum_{0\leq k\leq m}(-1)^{m-k}\binom{m}{k}T^{*k}T^{k}\Bigg{)}T=0.$$

$$<{\beta }_m(T^k)x|x>=0,\forall x\in \overline{\mathcal{R}(T^n)}.$$

$$\overline{\mathcal{R}((T^k{)}^n)}\subset \overline{\mathcal{R}(T^n)},$$

$$<{\beta }_m(T^k)x|x>=0,\forall x\in \overline{\mathcal{R}((T^k{)}^n)}.$$

$$T^{k}=\left(\begin{array}[]{ccc}T_{1}^{k}&\displaystyle\sum_{j=0}^{k-1}T_{1}^{j}T_{2}T_{3}^{k-1-j}\\ 0&T_{3}^{k}\end{array}\right),$$

$$\left(\begin{array}[]{cc}T_{1}^{r}&0\\ 0&0\\ \end{array}\right)=T^{r}P=PT^{r}P.$$

Full text

1. Introduction

Throughout this paper $\mathbb{N}$ denotes the set of non negative integers, ${\mathcal{H}}$ stands for an infinite separable complex Hilbert space with inner product $\langle.\;|\;.\rangle$, ${\mathcal{L}}({\mathcal{H}})$ is the Banach algebra of all bounded linear operators on ${\mathcal{H}}$ and $I=I_{\mathcal{H}}$ the identity operator. For every $T\in{\mathcal{B}}({\mathcal{H}})$ we denote by $\mathcal{R}(T)$, ${\mathcal{N}}(T)$ and $T^{*}$ the range, the null space and the adjoint of $T$ respectively. A closed subspace ${\mathcal{M}}\subset{\mathcal{H}}$ is invariant for $T$ (or $T$-invariant) if $T({\mathcal{M}})\subset{\mathcal{M}}$. As usual, the orthogonal complement and the closure of ${\mathcal{M}}$ are denoted ${\mathcal{M}}^{\perp}$ and $\overline{\mathcal{M}}$ respectively. We denote by $P_{\mathcal{M}}$ the orthogonal projection on ${\mathcal{M}}$.

Some of the most important subclasses of the algebra of all bounded linear operators acting on a Hilbert space, are the classes of partial isometries and quasi-isometries. An operator $T\in{\mathcal{L}}({\mathcal{H}})$ is said to be an isometry if $T^{*}T=I$, a partial isometry if $TT^{*}T=T$ and quasi-isometry if $T^{*2}T^{2}=T^{*}T.$

In recent years these classes has been generalized, in some sense, to the larger sets of operators so-called $m$-isometries, $m$-partial isometries and $n$-quasi-isometries. An operator $T\in{\mathcal{L}}({\mathcal{H}})$ is said to be

$(1)$ $m$-isometric operator for some integer $m\geq 1$ if it satisfies the operator equation

[TABLE]

It is immediate that $T$ is $m$-isometric operator if and only if

[TABLE]

$(2)$ $m$-partial-isometry for some integer $m\geq 1$ if

[TABLE]

$(3)$ $(m,q)$-partial isometry (or $q$-partial-$m$-isometry) if

[TABLE]

$(4)$  $m$-quasi-isometry for some integer $n\geq 1$ if

[TABLE]

It is immediate that $T$ is $m$-quasi-isometric if and only if

[TABLE]

Here $\displaystyle\binom{m}{k}$ is the binomial coefficient. In [1], J. Agler and M. Stankus initiated the study of operators $T$ that satisfy the identity (1.1). In [24], A. Saddi and O. A. M. Sid Ahmed studied operator $T$ which satisfies $(\ref{(1.3)})$. This concept was later generalized to the operators satisfying $(\ref{(1.4)})$, was defined by O. A. M. Sid Ahmed [17]. The study of operators satisfying $(\ref{(1.5)})$ was introduced and study by L. Suciu in [25]. The $1$-quasi-isometries are shortly called quasi-isometries, such operators being firstly studied in [21] and [22]. .

Recently, S. Mecheri and T. Prasad [18] introduced the class of $n$-quasi-$m$-isometric operators which generalizes the class of $m$-isometric operators and $n$-quasi-isometries. For positive integers $m$ and $n$, an operator $T\in{\mathcal{L}}({\mathcal{H}})$ is said to be an $n$-quasi-$m$-isometric operator if

[TABLE]

After an introduction on the subject and some connection with known facts in this context, the results of the paper are briefly described. In section two, we give a matrix characterization of $n$-quasi-$m$-isometries by using the decomposition ${\mathcal{H}}=\overline{T^{n}({\mathcal{H}})}\oplus{T^{*}}^{-n}(0)$. Several properties are proved by exploiting the special kind of operator matrix representation associated with such operators. In the course of our investigation, we find some properties of $m$-isometries which are retained by $n$-quasi-$m$-isometries. In particular, we show that if $T\in{\mathcal{L}}({\mathcal{H}})$ is an $n$-quasi-isometry then its power is an $n$-quasi-isometry. If $T$ and $S$ are doubly commuting such that $T$ is an $n_{1}$-quasi-$m$-isometry and $S$ is an $n_{2}$-quasi-$l$-isometry, then $TS$ is a $n_{0}=\max\{n_{1},n_{2}\}$-quasi-$(m+l-1)$-isometry. It has also been proved that the sum of an $n$-quasi-$m$-isometry and a commuting nilpotent operator of degree $p$ is a $2\max\{n,p\}$-quasi-$(m+2p-2)$-isometry. In section three, we recall the definition of $n$-quasi strict-$m$-isometries and we give some of their properties which are similar to those of $n$-quasi-$m$-isometries.

2. Some properties of $n$-quasi-$m$-isometric operators

In this section, we study some further properties of $n$-quasi-$m$-isometries. First, we will start with the following notations.

For $T\in{\mathcal{L}}({\mathcal{H}})$, we set

[TABLE]

[TABLE]

and

[TABLE]

Observe that $T$ is an $n$-quasi-$m$-isometric operator if and only if $\beta_{m,\;n}(T)=0$ or equivalently if

[TABLE]

Lemma 2.1**.**

Let $T\in{\mathcal{L}}({\mathcal{H}})$, then $T$ is an $n$-quasi-$m$-isometric operator if and only if

[TABLE]

or $T$ is an $m$-isometric operator on $\overline{{\mathcal{R}}(T^{n})}.$

Proof.

Obvious. ∎

Let $\mathbb{Z}$ denote the set of integers and $\mathbb{Z}_{+}$ denote the set of nonnegative integers.

Lemma 2.2**.**

([13, Lemma 5.4]) Let \big{(}a_{j}\big{)}_{j\in\mathbb{Z}_{+}} be a sequence of real numbers. Then

[TABLE]

if and only if there exists a polynomial $P$ of degree less than or equal to $m-1$ such that $a_{j}=P(j)$. In this case $P$ is the unique polynomial interpolating $\{(j,a_{j})\}$, $0\leq j\leq m-1.$

Proposition 2.1**.**

Let $T\in{\mathcal{L}}({\mathcal{H}})$ and $x\in{\mathcal{H}}$. We set $a_{k+n}:=\|T^{n+k}x\|^{2}$. Then $T$ is an $n$-quasi-$m$-isometry if and only if for each $x\in{\mathcal{H}}$, there exists a polynomial $P$ of degree less than or equal to $m-1$, such that $a_{k}=P(k)$ for $n\in\mathbb{Z}_{+}$.

Proof.

The proof is a consequence of Lemma 2.1, Lemma 2.2 and [13, Theorem 5.5]. ∎

In [18], S. Mecheri and T. Prasad studied the matrix representation of $n$-quasi-$m$-isometric operator with respect to the direct sum of $\overline{{\mathcal{R}}(T^{n})}$ and its orthogonal complement. In the following we give an equivalent condition for $T$ to be $n$-quasi-$m$-isometric operator. Using this result we obtained several important properties of this class of operators.

Theorem 2.1**.**

Let $T\in{\mathcal{L}}({\mathcal{H}})$ such that $\overline{{\mathcal{R}}(T^{n})}\not={\mathcal{H}}$, then the following statements are equivalent.

$(1)$* $T$ is an $n$-quasi-$m$-isometric operator.*

$(2)$* T=\left(\begin{array}[]{cc}T_{1}&T_{2}\\ 0&T_{3}\\ \end{array}\right) on ${\mathcal{H}}=\overline{{\mathcal{R}}(T^{n})}\oplus{\mathcal{N}}(T^{*n})$, where $T_{1}$ is an $m$-isometric operator and $T_{3}^{n}=0$.*

Proof.

$(1)\Rightarrow(2)$, follows by [18, Lemma 2.1].

$(2)\Rightarrow(1)$ Suppose that T=\left(\begin{array}[]{cc}T_{1}&T_{2}\\ 0&T_{3}\\ \end{array}\right) onto ${\mathcal{H}}=\overline{{\mathcal{R}}(T^{n})}\oplus{\mathcal{N}}(T^{*n})$ , with

[TABLE]

Since T^{k}=\left(\begin{array}[]{cc}T_{1}^{k}&\displaystyle\sum_{0\leq j\leq k-1}T_{1}^{j}T_{2}T_{3}^{k-1-j}\\ 0&T_{3}^{k}\\ \end{array}\right) we have

[TABLE]

[TABLE]

Therefore $\beta_{m,\;n}(T)=0.$ Thus $T$ is an $n$-quasi-$m$-isometric operator. ∎

For $T\in{\mathcal{L}}({\mathcal{H}})$, we denote by $\sigma(T)$, $\sigma_{ap}(T)$ and $\sigma_{p}(T)$ respectively the spectrum, the approximate point spectrum and the point spectrum of $T$.

Corollary 2.1**.**

$T\in{\mathcal{L}}({\mathcal{H}})$* be an $n$-quasi-$m$-isometric operator. The following statements hold.*

• (i)

$\sigma(T)=\sigma(T_{1})\cup\{0\}$* where $T_{1}=T_{/\overline{{\mathcal{R}}(T^{n})}}$.*

• (ii)

$T_{1}$* is bounded below.*

• (iii)

If $\mu\in\sigma_{ap}(T)\setminus\{0\}$ then ${\overline{\mu}}\in\sigma_{ap}(T^{*})$. In particular, if $\mu\in\sigma_{p}(T)\setminus\{0\}$ then ${\overline{\mu}}\in\sigma_{p}(T^{*})$.

Proof.

$(i)$  Since $T$ is an $n$-quasi-$m$-isometric operator, it follows by Theorem 2.1 that

[TABLE]

where $T_{1}$ is an $m$-isometric operator and $T_{3}^{n}=0$. From [15, Corollary 7], it follows that $\sigma(T)\cup W=\sigma(T_{1})\cup\sigma(T_{3})$, where $W$ is the union of certain of the holes in $\sigma(T)$ which is a subset of $\sigma(T_{1})\cap\sigma(T_{3})$. Further $\sigma{(T_{3})}=\{0\}$ and $\sigma(T_{1})\cap\sigma(T_{3})$ has no interior points. So we have by [15, Corollary 8]

[TABLE]

$(ii)$ By [1, Lemma 1.21], it is well known that the approximate spectrum of $T_{1}$ lies in unit circle. Hence $0\notin\sigma_{ap}(T_{1})$. Consequently, $T_{1}$ is bounded from below.

$(iii)$ The proof follows from [1, Theorem 2.2]. ∎

Recall that two operators $T\in{\mathcal{L}}({\mathcal{H}})$ and $S\in{\mathcal{L}}({\mathcal{H}})$ are similar if there exists an invertible operator $X\in{\mathcal{L}}({\mathcal{H}})$ such that $XT=SX$ (i.e; $T=X^{-1}SX$ or $S=XTX^{-1}$).

Corollary 2.2**.**

Let $T\in{\mathcal{L}}({\mathcal{H}})$ be an $n$-quasi-$m$-isometric operator. If $T_{1}=T_{/\overline{{\mathcal{R}}(T^{n})}}$ is invertible, then $T$ is similar to a direct sum of a $m$-isometric operator and a nilpotent operator.

Proof.

By Theorem 2.1 we write the matrix representation of $T$ on ${\mathcal{H}}=\overline{{\mathcal{R}}(T^{n})}\oplus{\mathcal{N}}(T^{*n})$ as follows T=\left(\begin{array}[]{ccc}T_{1}&T_{2}\\ 0&T_{3}\end{array}\right) where $T_{1}=T_{/\overline{{\mathcal{R}}(T^{n})}}$ is an $m$-isometric operator and $T_{3}^{n}=0$. Since $T_{1}$ is invertible, we have $\sigma(T_{1})\cap\sigma(T_{3})=\emptyset$. Then there exists an operator $A$ such that $T_{1}A-AT_{3}=T_{2}$ by [23]. Hence

[TABLE]

The desired result follows from Theorem 2.1. ∎

Clearly that every $n$-quasi-$m$-isometric operator is an $(n+1)$-quasi-$m$-isometric operator. In [19, Theorem 2.4], the authors S. Mechri and S. M. Patel proved that if $T$ is a quasi-$2$-isometry, then $T$ is quasi-$m$-isometry for all $m\geq 2.$ In the following corollary, we give a generalization that every $n$-quasi-$m$-isometric operator is an $n$-quasi-$k$-isometric operator for $k\geq m$.

Corollary 2.3**.**

Let $T\in{\mathcal{L}}({\mathcal{H}})$. If $T$ is an $n$-quasi-$m$-isometric operator, then $T$ is an $n$-quasi-$k$-isometric operator for every positive integer $k\geq m$.

Proof.

If ${\mathcal{R}}(T^{n})$ is dense, then $T$ is an $m$-isometric operator. Hence $T$ is a $k$-isometric operator for every positive integer $k\geq m$.

If ${\mathcal{R}}(T^{n})$ is not dense, by Theorem 2.1 we write the matrix representation of T on

${\mathcal{H}}=\overline{{\mathcal{R}}(T^{n})}\oplus{\mathcal{N}}(T^{*n})$ as follows T=\left(\begin{array}[]{ccc}T_{1}&T_{2}\\ 0&T_{3}\end{array}\right) where $T_{1}=T_{/\overline{{\mathcal{R}}(T^{n})}}$ is an $m$-isometric operator and $T_{3}^{n}=0$. Obviously, $T_{1}$ is a $k$-isometric operator for every integer $k\geq m$. The conclusion follows from the statement 2. of Theorem 2.1.

∎

We consider the following example of $n$-quasi-$m$-isometry map, which is not a quasi-$m$-isometry.

Example 2.1**.**

Let \big{(}{e_{k}}\big{)}_{k\in{\mathbb{N}}} be an orthonormal basis of ${\mathcal{H}}$. Define $T\in{\mathcal{L}}({\mathcal{H}})$ as follows

[TABLE]

Then by a straightforward calculation, one can show that $T$ is a $2$-quasi-$2$-isometry but it is not a quasi-$2$-isometry.

In the following theorem, we give a sufficient condition on an $n$-quasi-$m$-isometric operator for $n\geq 2$ to be a quasi-$m$-isometric operator.

Theorem 2.2**.**

Let $T\in\mathcal{L}(\mathcal{H})$ be an $n$-quasi-$m$-isometry for $n\geq 2$. If $\mathcal{N}(T^{*})=\mathcal{N}(T^{*2})$, then $T$ is a quasi-$m$-isometry.

Proof.

Two different proofs will be given.

First proof. Under the assumption $\mathcal{N}(T^{*})=\mathcal{N}(T^{*2})$, it follows that $\mathcal{N}(T^{*})=\mathcal{N}(T^{*n})$. Since $T$ is an $n$-quasi-$m$-isometry, we have

[TABLE]

we deduce that

[TABLE]

This means that

[TABLE]

Using again the condition ${\mathcal{N}}(T^{*})={\mathcal{N}}(T^{*n})$, we obtain

[TABLE]

Thus we have

[TABLE]

Hence $T$ is a quasi-$m$-isometric operator.

Second proof. By the assumption that ${\mathcal{N}}(T^{*})={\mathcal{N}}(T^{*2})$, we have ${\mathcal{N}}(T^{*})={\mathcal{N}}(T^{*n})$ and therefore $\overline{{\mathcal{R}}(T)}=\overline{{\mathcal{R}}(T^{n})}$. Now Since $T$ is an $n$-quasi-$m$-isometry, it follows in view of Lemma 2.1 that $T$ is an $m$-isometry on $\overline{{\mathcal{R}}(T^{n})}=\overline{{\mathcal{R}}(T)}$. This means that $T$ is an $m$-isometry on $\overline{{\mathcal{R}}(T)}$. Again by applying Lemma 2.1, we obtain that $T$ is an quasi-$m$-isometry as required. ∎

Remark 2.1*.*

The following example shows that Theorem 2.2 is not necessarily true if ${\mathcal{N}}(T^{*})\not={\mathcal{N}}(T^{*2})$.

Example 2.2**.**

Consider the operator T=\left(\begin{array}[]{cc}0&1\\ 0&0\\ \end{array}\right) acting on the two dimensional Hilbert space $\mathbb{C}^{2}$. Then by a straightforward calculation, one can show that $T$ is a $2$-quasi-isometry but it is not a quasi-isometry. However ${\mathcal{N}}(T^{*})\not={\mathcal{N}}(T^{*2})$

Patel [20, Theorem 2.1], proved that any power of a $(2,2)$-isometry is again a $(2,2)$-isometry. T. Bermúndez et al. [2, Theorem 3.1] proved that any power of $(m,p)$-isometry is an $(m,p)$-isometry. Later S. Mechri and S. M. Patel [19] gave a partial generalization to quasi-$2$-isometry. The following theorem shows that any power of an $n$-quasi-$m$-isometry is an $n$-quasi-$m$-isometry.

Theorem 2.3**.**

Let $T\in{\mathcal{L}}({\mathcal{H}})$. Suppose $T$ is an $n$-quasi-$m$-isometric operator. Then $T^{k}$ is $n$-quasi-$m$-isometric operator for any $k\in\mathbb{N}$.

Proof.

Two different proofs of this statement will be given.

First proof. Suppose that $T$ is $n$-quasi-$m$-isometric operator. By the statement in Lemma 2.1, $T$ is $m$-isometric on $\overline{{\mathcal{R}}(T^{n})}$. Therefore, in view of [2, Theorem 3.1], the operator $T^{k}$ is an $m$-isometric on $\overline{{\mathcal{R}}(T^{n})}$. Thus

[TABLE]

Using the inclusion

[TABLE]

we get

[TABLE]

Hence $T^{k}$ is a $m$-isometric on $\overline{{\mathcal{R}}((T^{k})^{n})}$. This shows, by the statement of Lemma 2.1, that $T^{k}$ is $n$-quasi-$m$-isometric.

Second proof. If ${\mathcal{R}}(T^{n})$ is dense, then $T$ is an $m$-isometric operator and hence $T^{k}$ is an $m$-isometric operator (by [2, Theorem 3.1]). If ${\mathcal{R}}(T^{n})$ is not dense, by Theorem 2.1 we write the matrix representation of $T$ on ${\mathcal{H}}=\overline{{\mathcal{R}}(T^{n})}\oplus{\mathcal{N}}(T^{*n})$ as follows T=\left(\begin{array}[]{ccc}T_{1}&T_{2}\\ 0&T_{3}\end{array}\right) where $T_{1}=T_{|\overline{{\mathcal{R}}(T^{n})}}$ is an $m$-isometric and $T_{3}^{n}=0.$ We notice that

[TABLE]

where $T_{1}^{k}$ is an $m$-isometric and $(T_{3}^{k})^{n}=0.$ Hence $T^{k}$ is an $n$-quasi-$m$-isometric operator by Theorem 2.1. ∎

Remark 2.2*.*

The converse of Theorem 2.3 in not true in general as it shows in the following example.

Example 2.3**.**

It is not difficult to prove that the operator T:=\left(\begin{array}[]{cc}-1&-1\\ 3&2\\ \end{array}\right) defined in $\mathbb{C}^{2}$ with the euclidean norm satisfies $T^{3}$ is a quasi-$3$-isometry but $T$ is not a quasi-$3$-isometry.

The following theorem generalize [2, Theorem 3.6].

Theorem 2.4**.**

Let $T\in{\mathcal{L}}({\mathcal{H}})$ and $r,s,m,n,l$ be positive integers. If $T^{r}$ is an $n$-quasi-$m$-isometry and $T^{s}$ is an $n$-quasi-$l$-isometry, then $T^{q}$ is an $n$-quasi-$p$-isometry, where $q$ is the greatest common divisor of $r$ and $s$, and $p$ is the minimum of $m$ and $l$.

Proof.

Consider the matrix representation of T with respect to the decomposition

${\mathcal{H}}=\overline{{\mathcal{R}}(T^{n})}\oplus{\mathcal{N}}(T^{*n})$ as follows T=\left(\begin{array}[]{ccc}T_{1}&T_{2}\\ 0&T_{3}\end{array}\right) where $T_{1}=T_{/\overline{{\mathcal{R}}(T^{n})}}$.

We have T^{r}=\left(\begin{array}[]{cc}T_{1}^{r}&\displaystyle\sum_{j=0}^{r-1}T_{1}^{j}T_{2}T_{3}^{r-1-j}\\ 0&T_{3}^{r}\\ \end{array}\right). Since $T^{r}$ is an $n$-quasi-$m$-isometry, we need to prove that $T_{1}^{r}$ is an $m$-isometry and \big{(}T_{3}^{r}\big{)}^{n}=0.

In fact, Let $P=P_{\overline{{\mathcal{R}}(T^{n})}}$ be the projection onto $\overline{{\mathcal{R}}(T^{n})}$. Then

[TABLE]

Since $T^{r}$ is an $n$-quasi-$m$-isometry, then we have

[TABLE]

That is

[TABLE]

Hence, $T_{1}$ is an $m$-isometry.

On the other hand, let $x=x_{1}+x_{2}\in{\mathcal{H}}=\overline{{\mathcal{R}}(T^{n})}\oplus{\mathcal{N}}(T^{*n}).$ A simple computation shows that

[TABLE]

So, \big{(}T_{3}^{r}\big{)}^{n}=0.

Analogously, as T^{s}=\left(\begin{array}[]{cc}T_{1}^{s}&\displaystyle\sum_{j=0}^{s-1}T_{1}^{j}T_{2}T_{3}^{s-1-j}\\ 0&T_{3}^{s}\\ \end{array}\right) is an $n$-quasi-$l$-isometry by similar arguments we can conclude that $T_{1}^{s}$ is an $l$-isometry and \big{(}T_{3}^{s}\big{)}^{n}=0.

Now, we obtained that $T_{1}^{r}$ is an $m$-isometry and $T_{1}^{s}$ is an $l$-isometry. By [2, Theorem 3.6], it follows that $T_{1}^{q}$ is a $p=\min(m,l)$-isometry. Moreover we have \big{(}T_{3}^{q}\big{)}^{n}=0.

Consequently, T^{q}=\left(\begin{array}[]{cc}T_{1}^{q}&\displaystyle\sum_{j=0}^{q-1}T_{1}^{j}T_{2}T_{3}^{q-1-j}\\ 0&T_{3}^{q}\\ \end{array}\right) is an $n$-quasi-$p$-isometry by Theorem 2.1. The proof is completed.

∎

The following corollary shows that if we assume that two suitable different powers of $T$ are $n$-quasi-$m$-isometries, then we obtain that $T$ is a $n$-quasi-$m$-isometry.

Corollary 2.4**.**

Let $T\in{\mathcal{L}}({\mathcal{H}})$ and $r,s,m,n,l$ be positive integers. The following properties hold.

$(1)$* If $T$ is an $n$-quasi-$m$-isometry such that $T^{s}$ is an $n$-quasi-isometry, then $T$ is*

an $n$-quasi-isometry.

$(2)$* If $T^{r}$ and $T^{r+1}$ are $n$-quasi-$m$-isometries, then so is $T$.*

$(3)$* If $T^{r}$ is an $n$-quasi-$m$-isometry and $T^{r+1}$ is an $n$-quasi-$l$-isometry with $m<l$,*

then $T$ is an $n$-quasi-$m$-isometry.

Proof.

The proof is an immediate consequence of Theorem 3.1. ∎

Recall that an operator $T\in{\mathcal{L}}({\mathcal{H}})$ is said to be power bounded, if $\displaystyle\sup_{k}\|T^{k}\|<\infty$ or equivalently, there exists $C>0$ such that for every $k$ and every $\xi\in{\mathcal{H}},$ one has

[TABLE]

In [8, Theorem 2], it was proved that every power bounded $m$-isometry operator is an isometry. The following theorem extend this result to $n$-quasi-$m$-isometry.

Theorem 2.5**.**

If $T\in{\mathcal{L}}({\mathcal{H}})$ is an $n$-quasi-$m$-isometric operator which is power bounded, Then $T$ is an $n$-quasi-isometry.

Proof.

We consider the following two cases:

Case 1: If $\overline{{\mathcal{R}}(T^{n})}$ is dense, then $T$ is an $m$-isometric operator which is power bounded, thus $T$ is a isometry by [8, Theorem 2]. It follows that $T$ is an $n$-quasi-isometry.

Case 2: If $\overline{{\mathcal{R}}(T^{n})}$ is not dense. By Theorem 2.1 we write the matrix representation of $T$ on ${\mathcal{H}}=\overline{{\mathcal{R}}(T^{n})}\oplus{mathcalN}(T^{*n})$ as follows T=\left(\begin{array}[]{ccc}T_{1}&T_{2}\\ 0&T_{3}\end{array}\right) where $T_{1}=T|\overline{{\mathcal{R}}(T^{n})}$ is an $m$-isometric operator and $T_{3}^{n}=0.$ By taking into account that $T$ is power bounded, it is easily to check that $T_{1}$ is power bounded. From which we deduce that $T_{1}$ is an isometry. The result follows by applying the statement $(2)$ of Theorem 2.1 ∎

Recall that for two operators $T,S$ in ${\mathcal{L}}({\mathcal{H}})$, the commutator $[T,S]$ is defined to be

[TABLE]

A pair of operators $(T,S)\in{\mathcal{L}}({\mathcal{H}})^{2}$ is said to be a doubly commuting pair if $(T,S)$ satisfies $TS=ST$ and $T^{*}S=ST^{*}$ or equivalently $[T,S]=[T,S^{*}]=0.$

In [16, Theorem 2.2], it has proved that if $T$ and $S$ are commuting bounded linear operators on a Banach space such that $T$ is a $2$-isometry and $S$ is an $m$-isometry, then $ST$ is an $(m+1)$-isometry. This result was improved in [3, Theorem 3.3] as follows: if $TS=ST$, $T$ is an $(m,p)$-isometry and $S$ is an $(l,p)$-isometry, then $ST$ is an $(m+l-1,p)$-isometry. It is natural to ask whether the product of two $n$-quasi-$m$-isometries is $n$-quasi-$m$-isometry. The following theorem gives an affirmative answer under suitable conditions.

Theorem 2.6**.**

Let $S$ and $T$ be in ${\mathcal{L}}({\mathcal{H}})$ are doubly commuting operators and let $m,l,n_{1},n_{2}$ be positive integers. If $T$ is an $n_{1}$-quasi-$m$-isometry and $S$ is an $n_{2}$-quasi-$l$-isometry, then $TS$ is a $n_{0}=\max\{n_{1},n_{2}\}$-quasi-$(m+l-1)$-isometry.

Proof.

Under the assumption that $T$ and $S$ are doubly commuting, it follows that $[T^{*},S^{*}]=[T,S]=[T,S^{*}]=0.$ By taking into account [10, Lemma 12] we obtain that

[TABLE]

Since $S$ is an $n_{2}$-quasi-$l$-isometry, it follows by Corollary 2.3 that \big{(}S^{*}\big{)}^{n_{0}}\beta_{j}(S)S^{n_{0}}=0 for $j\geq l$. On the other hand, we have if $j\leq l-1$, then $m+l-1-j\geq m+l-1-l+1=m$, and so \big{(}T^{*}\big{)}^{n_{0}}\beta_{k+m-1-j}(T)T^{n_{0}}=0 by the fact that $T$ is an $n_{1}$-quasi-$m$-isometry and Corollary 2.3. Therefore $TS$ is an $n_{0}=\max\{n_{1},n_{2}\}$-quasi-$(m+l-1)$-isometry. This completes the proof. ∎

The following example shows that Theorem 2.6 is not necessarily true if $S,T$ are not doubly commuting.

Example 2.4**.**

We consider the operators T=\left(\begin{array}[]{cc}1&1\\ 0&1\\ \end{array}\right)\ \mbox{and}\ S=\left(\begin{array}[]{cc}2&1\\ -1&0\\ \end{array}\right) on the two dimensional Hilbert space $\mathbb{C}^{2}.$ Note that $ST\neq TS$. Moreover, by a direct computation, we show that $T$ is a quasi-$3$-isometry and $S$ is a $2$-quasi-$3$-isometry. However neither $TS$ nor $ST$ is a $2$-quasi-$5$-isometry.

Corollary 2.5**.**

Let $T,S\in{\mathcal{L}}({\mathcal{H}})$ are doubly commuting operators such that $T$ is an $n_{1}$-quasi-$m$-isometry and $S$ is an $n_{2}$-quasi-$l$-isometry, then $T^{p}S^{q}$ is a $\max\{n_{1},\;n_{2}\}$-quasi-$(m+l-1)$-isometry for all positive integers $p$ and $q$.

Proof.

Since $T$ and $S$ are doubly commuting, then $T^{p}$ and $S^{q}$ are doubly commuting. By Theorem 2.3 we know that $T^{p}$ is an $n_{1}$-quasi-$m$-isometry and $S^{q}$ is an $n_{2}$-quasi-$l$-isometry. Now by applying Theorem 2.6, we get that $T^{p}S^{q}$ is a $\max\{n_{1},\;n_{2}\}$-quasi-$(m+l-1)$-isometry. This completes the proof. ∎

Let ${\mathcal{H}}\overline{\otimes}{\mathcal{H}}$ denote the completion, endowed with a reasonable uniform cross-norm, of the algebraic tensor product ${\mathcal{H}}\otimes{\mathcal{H}}$ of ${\mathcal{H}}$ and ${\mathcal{H}}$. For $T\in{\mathcal{L}}({\mathcal{H}})$ and $S\in{\mathcal{L}}({\mathcal{H}})$ , $T\otimes S\in{\mathcal{L}}({\mathcal{H}}\overline{\otimes}{\mathcal{H}})$ denote the tensor product operator defined by $T$ and $S$.

In the following proposition we prove that the tensor product of an $n_{1}$-quasi-$m$-isometric operator with an $n_{2}$-quasi-$l$-isometric operator is a $\max\{n_{1},n_{2}\}$-quasi-$(m+l-1)$-isometric operator. This proposition generalizes [9, Theorem 2.10].

Proposition 2.2**.**

If $T\in{\mathcal{L}}({\mathcal{H}})$ is an $n_{1}$-quasi-$m$-isometry and $S\in{\mathcal{L}}({\mathcal{H}})$ is an $n_{2}$-quasi-$l$-isometry, then $T\otimes S$ is a $\max\{n_{1},n_{2}\}$-quasi-$(m+l-1)$-isometric.

Proof.

Observe that an operator $T\in{\mathcal{L}}({\mathcal{H}})$ is $n$-quasi $m$-isometric if and only if $T\otimes I$ and $I\otimes T$ are $n$-quasi-$m$-isometry. In view of the fact that

[TABLE]

it follows that

[TABLE]

Now $T\otimes I$ is an $n_{1}$-quasi-$m$-isometry and $I\otimes S$ is an $n_{2}$-quasi-$m$-isometry such that $T\otimes I$ and $I\otimes S$ are doubly commuting operators. By applying Theorem 2.6 we obtain that $(T\otimes I)(I\otimes S)$ is a $\max\{n_{1},n_{2}\}$-quasi-$(m+l-1)$-isometric operator. Hence $T\otimes S$ is an $\max\{n_{1},n_{2}\}$-quasi-$(m+l-1)$-isometric as required. ∎

The following corollary is an immediate consequence of Theorem 2.3 and Proposition 2.2. We omitted its proof.

Corollary 2.6**.**

If $T\in{\mathcal{L}}({\mathcal{H}})$ is an $n_{1}$-quasi-$m$-isometry and $S\in{\mathcal{L}}({\mathcal{H}})$ is an $n_{2}$-quasi-$l$-isometry, then $T^{p}\otimes S^{q}$ is a $\max\{n_{1},n_{2}\}$-quasi-$(m+l-1)$-isometry.

It was proved in [4, Thoerem 2.2] that if $T\in{\mathcal{L}}({\mathcal{H}})$ is an isometry and $Q\in{\mathcal{L}}({\mathcal{H}})$ is a nilpotent operator of order $p$ such that $TQ=QT$, then $T+Q$-is a strict $(2p-1)$-isometry. Later T. Bermúdez et al. [5] gave a partial generalization to $m$-isometry, that is if $T$ is an $m$-isometry with $m>1$, $Q$ is a nilpotent operator with order $p$, and $TQ=QT$, then $T+Q$ in a $(m+2p-2)$-isometry. Recently, C. Gu. and M. Stankus [14] gave a generalization to $m$-isometry, that is if $T$ is an $m$-isometry with $m>1$, $Q$ is a nilpotent operator with order $p$, and $TQ=QT$, then $T+Q$ is a strict $(m+2p-2)$-isometry. The following Theorem states the corresponding partial generalization to the sum of an $n$-quasi-$m$-isometry and a nilpotent operator.

Theorem 2.7**.**

Let $T,Q\in{\mathcal{L}}({\mathcal{H}})$ such that $T$ commutes with $Q$. If $T$ is an $n$-quasi-$m$-isometry and $Q$ is a nilpotent operator of order $p$, then $T+Q$ is a $(n+p)$-quasi-$(m+2p-2)$-isometry.

Proof.

We need to show that \beta_{m+2p-2,\;\alpha}\big{(}T+Q\big{)}=0. Set $q=m+2p-2$ and $\alpha=n+p$, by [14, Lemma 1] we have

[TABLE]

In fact, note that

[TABLE]

Now observe that if $k\geq p$ or $j\geq p$ then $Q^{*k}=0$ or $Q^{*j}=0$ and hence

[TABLE]

However , if $k<p$ and $j<p$, we obtain

[TABLE]

and using the fact that $T$ is an $n$-quasi-$m$-isometry, we get

[TABLE]

and

[TABLE]

Combining the above arguments we obtain \beta_{m+2p-2,\;n+p}\big{(}T+Q\big{)}=0. ∎

Remark 2.3*.*

A simple example shows that the commuting condition of $T$ and $Q$ can not be removed from the above theorem.

Example 2.5**.**

Let T=\left(\begin{array}[]{cc}-5&0\\ 0&-1\\ \end{array}\right) and Q=\left(\begin{array}[]{cc}0&1\\ 0&0\\ \end{array}\right). Then T is a quasi-$3$-isometry and $Q^{2}=0$. Set $S=T+Q$, by direct calculation we show that $T$ is not $5$-quasi-$5$-isometry.

Corollary 2.7**.**

Let $T\in{\mathcal{L}}({\mathcal{H}})$ be an $n$-quasi-$m$-isometry and $Q\in{\mathcal{L}}({\mathcal{H}})$ be a nilpotent operator of order $p$. Then $T\otimes I+I\otimes Q$ is a $(n+p)$-quasi-$(m+2p-2)$-isometry.

Proof.

We note that $T\otimes I\in{\mathcal{L}}({\mathcal{H}}\overline{\otimes}{\mathcal{H}})$ is an $n$-quasi-$m$-isometry and $I\otimes Q\in{\mathcal{L}}({\mathcal{H}}\overline{\otimes}{\mathcal{H}})$ is a nilpotent of order $p$. Moreover $(T\otimes I)(I\otimes Q)=(I\otimes Q)(T\otimes I)$. ∎

Corollary 2.8**.**

Let $T_{j}\in{\mathcal{L}}({\mathcal{H}})$ be an $n_{j}$-quasi-$m_{j}$-isometry for $j=1,\cdots,d$. Define

[TABLE]

where $\alpha_{j}$ is a complex number for each $j=1,\cdots,d-1$ and ${\mathcal{H}}^{(d)}$ is the sum of $d$-copies of ${\mathcal{H}}$, then ${S}$ is a $(n+d)$-quasi-$(m+2d-2)$-isometry where $n=\displaystyle\max_{1\leq j\leq d}\{n_{j},\}$ and $m=\displaystyle\max_{1\leq j\leq d}\{m_{j},\}.$

Proof.

Consider the two operators

[TABLE]

It is easy to see that ${S}={T}+{Q}$. Observing that ${T}$ is an $n$-quasi $m$-isometry tuple (by Corollary 2.3) where $n=\displaystyle\max_{1\leq j\leq d}\{n_{j}\}$ and $m=\displaystyle\max_{1\leq j\leq d}\{m_{j}\}$. Therefore $Q$ is a $d$-nilpotent operator. According to Theorem 2.7, ${S}$ is a $(n+d)$-quasi-$(m+2d-2)$-isometry.. ∎

The following theorem shows that the class of $n$-quasi-$m$-isometry is a closed subset of ${\mathcal{L}}({\mathcal{H}})$ equipped with the uniform operator (norm) topology.

Theorem 2.8**.**

Let $T\in{\mathcal{L}}({\mathcal{H}}).$ If $(T_{k})_{k}$ is a sequence of $n$-quasi-$m$-isometry such that $\displaystyle\lim_{k\to\infty}\|T_{k}-T\|=0,$ then $T$ is also $n$-quasi-$m$-isometry.

Proof.

Suppose that $(T_{k})_{k}$ is a sequence of $n$-quasi-$m$-isometric operators such that

[TABLE]

Since for every positive integer $k$. $T_{k}$ is an $n$-quasi-$m$-isometry, we have $\beta_{m,n}(T)=0$. It follows that

[TABLE]

[TABLE]

Since the product of operators is sequentially continuous in the strong topology, one concludes that $T_{k}^{j}T_{k}^{n}$, $T_{k}^{n+j}$ converge strongly to $T^{j}T^{n}$ and $T^{n+j}$ respectively for $j=0,1,...,m$. Hence the limiting case of $(\ref{norm})$ shows that $T$ belongs to the class of $n$-quasi-$m$-isometric operators. ∎

3. $n$-quasi strict-$m$-isometries

In this section we introduce and study some properties of the class of $n$-quasi strict- $m$-isometric operators.

Recall that an operator $T\in{\mathcal{L}}({\mathcal{H}})$ is said to be a strict $m$-isometry if $T$ is an $m$-isometry but it is not an $(m-1)$-isometry.

Definition 3.1**.**

We say that $T\in{\mathcal{L}}({\mathcal{H}})$ is a $n$-quasi strict $m$-isometry if $T$ is an $n$-quasi-$m$-isometry, but $T$ is not an $n$-quasi-$(m-1$)-isometry.

Example 3.1**.**

([19]) Let \big{(}e_{k}\big{)}_{k\geq 1} be an orthonormal basis of a Hilbert space ${\mathcal{H}}$. Consider an operator $T\in{\mathcal{L}}({\mathcal{H}})$ defined by:

[TABLE]

A direct calculation shows that $T$ is a quasi-$2$-isometric, but not a quasi-isometric. Therefore $T$ is a quasi strict-$2$-isometry.

Example 3.2**.**

Consider the operator $T\in{\mathcal{L}}(\mathbb{C}^{2})$ given by T=\left(\begin{array}[]{cc}1&1\\ 0&1\\ \end{array}\right) who is quasi-$3$-isometric but is not quasi-$2$-isometric. Hence $T$ is a quasi strict-$3$-isometry.

Remark 3.1*.*

It is proved in Corollary 2.3 that an $n$-quasi-$m$-isometric operator is $n$-quasi-$k$-isometric operator for all integers $k\geq m.$ Hence if an $T\in{\mathcal{L}}({\mathcal{H}})$ is a strict $n$-quasi-$m$-isometry, then it is not a $n$-quasi-$k$-isometry for all integers $1\leq k<m.$

Recall that the multinomial coefficients is given by $\displaystyle\binom{k}{p_{1},\cdots,p_{k}}=\frac{k!}{p_{1}!p_{2}!\cdots p_{k}!}$ where $k$ and $p_{1},\cdots,p_{k}$ are nonnegative integers subject to $k=p_{1}+p_{2}+\cdots+p_{k}$.

We will use the following formula for commuting variables $z=(z_{1},...,z_{q}):$

[TABLE]

In particular, if $z_{1}=\cdots=z_{q}=1,$ we have

[TABLE]

Proposition 3.1**.**

Let $T\in{\mathcal{L}}({\mathcal{H}})$, the following statements hold.

$(1)$* If $m$ is a positive integer and $x\in{\mathcal{H}}$, then*

[TABLE]

In particular, if $T$ is an $n$-quasi-$m$-isometry, then for every positive integer $k$ one has

[TABLE]

$(2)$* If $k$ is a positive integer and $x\in{\mathcal{H}}$, then*

[TABLE]

Proof.

For $m\geq 1$ and $x\in{\mathcal{H}}$ we have

[TABLE]

Hence $(\ref{3.1})$ is proved.

If we assume that $T$ is an $n$-quasi-$m$-isometry, then $\Delta_{m,\;n}(T,x)=0$ and so that

[TABLE]

Hence,

[TABLE]

$(2)$ From $(2.3)$ we have

[TABLE]

and

[TABLE]

Thus we need to prove the following combinatorial identity:

[TABLE]

In fact, observe that

[TABLE]

and

[TABLE]

By applying the multinomial formula in reverse order, we have

[TABLE]

∎

Theorem 3.1**.**

If $T\in{\mathcal{L}}({\mathcal{H}})$ is a $n$-quasi strict $m$-isometry, then for any positive integer $k$, $T^{k}$ is a $n$-quasi strict $m$-isometry. Furthermore

[TABLE]

Proof.

Since $T$ is a $n$-quasi strict $m$-isometry, by Theorem 2.3, $T^{k}$ is an $n$-quasi-$m$-isometry. Furthermore, by (3.2) and (3.3) we get

[TABLE]

Consequently, $T^{k}$ is a quasi strict-$m$-isometry. ∎

Remark 3.2*.*

The converse of the above theorem is not true in general. In fact, by Theorem 3.1 if $T^{r}$ and $T^{s}$ are $n$-quasi-$m$-isometries for two coprime positive integers $r$ and $s$, then $T$ is an $n$-quasi-$m$-isometry.

Recall that a sequence $(a_{j})_{j\geq 0}$ in a group $G$ is an arithmetic progression of order $h$ if

[TABLE]

for any $j\geq 0.$ An arithmetic progression of order $h$ is of strict order $h$ if $h=0$ or if $h\geq 1$ and it is not of order $h-1.$ We refer the interested reader to [6] for complete details.

Lemma 3.1**.**

([6, Theorem 4.1]) Let $a=(a_{j})_{j\geq 0}$ be a numerical sequence. Suppose that $(a_{cj})_{j\geq}$ is an arithmetic progression of strict order $h$ and $(a_{dj})_{j\geq 0}$ is an arithmetic progressions of strict order $k\geq 0$, for $c,d\geq 1$ and $h,k\geq 0.$ Then $(a_{ej})_{j\geq 0}$ is an arithmetic progression of strict order $l$, being $e$ the greatest common divisor of $c$ and $d$, and $l$ the minimum of $h$ and $k$.

Theorem 3.2**.**

Let $T\in{\mathcal{L}}({\mathcal{H}})$ and $r,s,n,m,l$ be positive integers. If $T^{r}$ is a $n$-quasi strict $m$-isometry and $T^{s}$ is an $n$-quasi strict $l$-isometry, then $T^{q}$ is an $n$-quasi-$p$-isometry, where $q$ is the greatest common divisor of $r$ and $s$, and $p$ is the minimum of $m$ and $l.$

Proof.

Fix $x\in{\mathcal{H}}$ and set $a_{j}:=\|T^{j}x\|^{2}$ for $j=1,2,\cdots$. Since $T^{r}$ is a $n$-quasi strict $m$-isometry, it follows that $(a_{r(j+n)})_{j\geq 0}$ is an arithmetic progression of strict order $m-1$ satisfies the recursive equation

[TABLE]

Analogously, as $T^{s}$ is a $n$-quasi strict-$l$-isometry, it follows that $(a_{s{(j+n)}})_{j\geq 0}$ is an arithmetic progression of strict order $l-1$ satisfies the recursive equation

[TABLE]

Applying Lemma 3.1 it results that $(a_{q{(j+n)}})_{j\geq 0}$ is an arithmetic progression of strict order $p-1$, so $T^{q}$ is an $n$-quasi-strict $p$-isometry, where $q=gcd(r,s)$ and $p=\min\{m,l\}$. ∎

The following corollary shows that if we assume that two suitable different powers of $T$ are $n$-quasi strict-$m$-isometries, then we obtain that $T$ is a $n$-quasi strict-$m$-isometry.

Corollary 3.1**.**

Let $T\in{\mathcal{L}}({\mathcal{H}})$ and $r,s,m,n,l$ be positive integers. The following properties hold.

$(1)$* If $T$ is an $n$-quasi strict-$m$-isometry such that $T^{s}$ is an $n$-quasi strict-isometry,*

then $T$ is an $n$-quasi strict-isometry.

$(2)$* If $T^{r}$ and $T^{r+1}$ are $n$-quasi strict-$m$-isometries, then so is $T$.*

$(3)$* If $T^{r}$ is an $n$-quasi strict-$m$-isometry and $T^{r+1}$ is an $n$-quasi strict-$l$-isometry*

with $m<l$, then $T$ is an $n$-quasi strict-$m$-isometry.

Proof.

The proof is an immediate consequence of Theorem 3.2. ∎

Theorem 3.3**.**

Let $T\in{\mathcal{L}}({\mathcal{H}})$ and $S\in{\mathcal{L}}({\mathcal{H}})$ be doubly commuting operators. If $T$ is an $n$-quasi strict $m$-isometry and $S$ is an $n$-quasi strict-$l$-isometry, then $TS$ is a $n$-quasi strict-$(m+l-1)$-isometry if and only if

[TABLE]

Proof.

In view of Theorem 2.7, it is obvious that $\beta_{m+l-1,\;n}(TS)=0$. On the other hand, since $T$ and $S$ are doubly commuting operators, it follows by [14, Corollary 3.9] that

[TABLE]

From which it follows that

[TABLE]

If $k\in[0,l-2]$, then $(m+l-2-k)\in[m,m+l-2]$ and hence $T^{*n}\beta_{m+l-2-k}(T)T^{n}=0$ by Corollary 2.3. If $k\geq l$, then $S^{n}\beta_{k}(S)S^{n}=0$ also in view of Corollary 2.3.

Consequently, $TS$ is a $n$-quasi strict-$(m+l-1)$-isometry if and only if

[TABLE]

Hence the proof is finished. ∎

Theorem 3.4**.**

Let $T\in{\mathcal{L}}({\mathcal{H}})$ and $S\in{\mathcal{L}}({\mathcal{H}})$. If $T$ is a $n$-quasi strict-$m$-isometry and $S$ is a $n$-quasi strict-$l$-isometry, then $T\otimes S$ on ${\mathcal{H}}\overline{\otimes}{\mathcal{H}}$ is a $n$-quasi strict-$(m+l-$1)-isometry.

Proof.

In view of [CG1, Corollary 3.10], it follows that

[TABLE]

By calculations we have

[TABLE]

A similar arguments as in the proof of Theorem 3.3 give

[TABLE]

This means that $T\otimes S$ is a $n$-quasi strict-$(m+l-1)$-isometry as required. ∎

In [7, Theorem 3.1] it has been proved that if $T\in{\mathcal{L}}({\mathcal{H}})$ is a strict $m$-isometry, then the list of operators $\{\;T^{*k}T^{k},k=0,1,\;\cdots m-1\;\}$ is linearly independent which is equivalent to that $\{\;\beta_{k}(T),k=0,1,\;\cdots,m-1\;\}$ is linearly independent.

In the following proposition we extend this result to $n$-quasi strict-$m$-isometry as follows.

Proposition 3.2**.**

If $T\in{\mathcal{L}}({\mathcal{H}})$ is an $n$-quasi strict $m$-isometry, then the list of operators

[TABLE]

is linearly independent.

Proof.

The outline of the proof is inspired from [12].

It was observed in [11] that $\beta_{k}(T)=T^{*}\beta_{k-1}(T)T-\beta_{k-1}(T)$   for all $k\geq 1$.

We will just write

[TABLE]

Now assume that for some complex numbers $\lambda_{k},$

[TABLE]

or equivalently

[TABLE]

Multiplying the above equation on the left and right by $T^{*}$ and $T$ and subtracting two equations, we have

[TABLE]

By applying the same procedure to the equation $\displaystyle\sum_{0\leq k\leq m-1}\lambda_{k}\beta_{k+1,\;n}(T)=0$ we get

[TABLE]

By continuing this process we obtain

[TABLE]

Since that every $n$-quasi-$m$-isometric operator is $n$-quasi-$k$-isometric operator for all $k\geq m$ (Corollary 2.3 ) we have the following cases:

For $j=m-1$, $\displaystyle\sum_{0\leq k\leq m-1}\lambda_{k}\beta_{k+j,\;n}(T)=0\Rightarrow\lambda_{0}\beta_{m-1,\;n}(T)=0,$ so $\lambda_{0}=0.$

For $j=m-2$, $\displaystyle\sum_{0\leq k\leq m-1}\lambda_{k}\beta_{k+j,\;n}(T)=0\Rightarrow\lambda_{1}\beta_{m-1,\;n}(T)=0,$ so $\lambda_{1}=0.$

Continuing this process we see that all $\lambda_{k}=0$ for $k=2,\;\cdots,\;m-1$. Hence the result is proved. ∎