# Localised time-periodic solutions of discrete nonlinear Klein-Gordon systems
with convex on-site potentials

Dirk Hennig

###### Abstract

Abstract: The existence of nonzero localised periodic solutions for general
one-dimensional discrete nonlinear Klein-Gordon systems with convex on-site
potentials is proved. The existence problem of localised solutions is
expressed in terms of a fixed point equation for an operator on some
appropriate function space which is solved by means of Schauder’s Fixed Point
Theorem.

Non-uniform structures in nonlinear discrete lattice systems in the form of
spatially localised and time-periodically varying states, referred to as
breather solutions, have attracted considerable interest. Rigorous results
regarding proofs of existence and stability of intrinsic localised time-
periodic modes (also known as breathers) were provided in MacKay -b12 .
Several methods and techniques were used such as continuation of (trivially
localised) solutions from the anti-integrable of zero couplings for
sufficiently weak couplings, center manifold reduction methods, Lyapunov-
Schmidt decomposition and continuum approximation techniques, variational
approaches, perturbative arguments for the characteristic exponents of the
Floquet monodromy matrices and normal form theorems and Nekoroshev-type
proofs. With this work we present a comparatively concise existence proof for
breathers in general discrete nonlinear Klein-Gordon systems with convex on-
site potential.

We discuss the existence of localised time-periodic solutions of general
discrete nonlinear Klein-Gordon systems on infinite lattices given by the
following set of coupled oscillator equations

$\frac{d^{2}q_{n}}{dt^{2}}=\kappa\,[q_{n+1}-2q_{n}+q_{n-1}]-V^{\prime}(q_{n}),\,\,\,n\in{\mathbb{Z}},$
(1)

where the $q_{n}(t)$ describe the coordinate of the $n$th oscillator at time
$t$ and the prime ′ in the last term on the r.h.s. of (1) stands for the
derivative with respect to $q_{n}$. Each oscillator interacts with its
neighbours to the left and right and the strength of the interaction is
determined by the value of the parameter $\kappa$. This system has a
Hamiltonian structure related to the energy

$H=\sum_{n\in{\mathbb{Z}}}\left(\frac{1}{2}p_{n}^{2}+V(q_{n})+\frac{\kappa}{2}(q_{n+1}-q_{n})^{2}\right),$
(2)

and it is time-reversible with respect to the involution $p\mapsto-p$.

Assume that the following condition on the on-site potential $V(x)$ holds:

A: $V:\,\,{\mathbb{R}}\,\rightarrow\,{\mathbb{R}}$ is even, $V(x)=V(-x)$, non-
negative and at least twice continuously differentiable and uniformly convex
which is characterised by the following properties: There exists two constants
$0<\underline{K}<\overline{K}<\infty$ such that

$\underline{K}\leq V^{\prime\prime}(x)\leq\overline{K},\,\,\,\forall
x\in{\mathbb{R}}.$ (3)

Then, one has

$\frac{1}{2}\underline{K}\,x^{2}\leq
V(x)\leq\frac{1}{2}\overline{K}\,x^{2},\,\,\,|V^{\prime}(x)|\leq\overline{K}|x|,\,\,\,\forall
x\in{\mathbb{R}}.$ (4)

Further assume that

$V(0)=V^{\prime}(0)=0,\,\,\,V^{\prime\prime}(0)=1.$ (5)

That is, the unique equilibrium at $x=0$ is a global minimum of $V(x)$ and
$V^{\prime}(x)$ can be expressed as

$V^{\prime}(x)=x+W^{\prime}(x),$ (6)

with non-negative $W(x)$ and
$\operatorname{sign}(V^{\prime}(x))=\operatorname{sign}(W^{\prime}(x))$.
Hence,

$V(x)=\frac{1}{2}x^{2}+W(x).$ (7)

The solutions of the system obtained by linearising equations (1) around the
equilibrium $q_{n}=0$ are superpositions of plane wave solutions (phonons)

$q_{n}(t)=\exp(i(kn-\omega t)),$ (8)

with frequencies

$\omega^{2}(k)=1+4\kappa\sin^{2}\left(\frac{k}{2}\right),\,\,\,k\in[-\pi,\pi].$
(9)

These (extended) states disperse. Therefore the frequency $\Omega$ of a
localised time-periodic solution must satisfy the non-resonance condition
$\Omega\neq|\omega(k)|/m$ for any integer $m\geq 1$. This requires
$\Omega^{2}>1+4\kappa$ as a necessary condition for the existence of localised
time-periodic solutions of system (1).

In the following we prove the existence of localised periodic solutions of
system (1) on infinite lattices.

To this end, some appropriate function spaces are introduced on which the
original problem is presented as a fixed point problem for a corresponding
operator. Utilising Schauder’s Fixed Point Theorem we establish the existence
of localised solutions in a similar vein to the approach in SIAM  treating the
existence of travelling wave solutions in finite arrays of coupled Josephson
junction systems.

With concern to the existence of localised periodic solutions to system (1) we
state:

Theorem: Let A hold. Then there exist nonzero sequences
$q\equiv\\{q_{n}\\}_{n\in{\mathbb{Z}}}\in H^{2}([0,T];l^{2})$ with zero time
average $\int_{0}^{T}q_{n}(t)dt=0,$ that are localised time-periodic solutions
of system (1) with period $T=\frac{2\pi}{\Omega}$ provided

$\Omega^{2}>1+4\kappa.$ (10)

For the proof of the assertions of the Theorem the strong version of
Schauder’s Fixed Point Theorem is used Schauder : Let $G$ be a closed convex
subset of a Banach space and $f$ a continuous map of $G$ into a compact subset
of $G$. Then $f$ has a fixed point.

Proof: For the following considerations Eq. (1) is suitably re-written as:

$\ddot{q}_{n}+q_{n}-\kappa[q_{n+1}-2q_{n}+q_{n-1}]=-W^{\prime}(q_{n})$ (11)

Thus only the r.h.s. of (11) features terms nonlinear in $q$. Ultimately, (11)
shall be re-formulated as a fixed point equation in $q$.

We introduce appropriate function spaces. Denote by

${X}^{0}=\left\\{q\in
L^{2}_{per}((0,T);{l}^{2})\,|\,\int_{0}^{T}q_{n}(t)dt=0,\,\,\,n\in{\mathbb{Z}}\right\\}$
(12)

the space of $T-$periodic square integrable functions of time with zero time
average assuming values in $l^{2}$. $X^{0}$ is a closed convex subset of
$L^{2}_{per}((0,T);{l}^{2})$. We introduce further the Sobolev space
$X^{2}=\\{q\in H^{2}_{per}((0,T);{l}^{2})\\}$ containing $T-$periodic
functions of time assuming values in $l^{2}$ which, together with their weak
derivative $D^{\alpha\leq 2}$ up to order $2$, are in $X^{0}$. The following
norms are used:

$\displaystyle\parallel q\parallel_{X^{0}}^{2}$ $\displaystyle=$
$\displaystyle\frac{1}{T}\int_{0}^{T}\,\parallel
q(t)\parallel^{2}_{{l}^{2}}dt=\sum_{m\in{\mathbb{Z}}\setminus\\{0\\}}\frac{1}{2\pi}\int_{0}^{2\pi}\left|\hat{\tilde{q}}_{k,m}\right|^{2}dk$
(13) $\displaystyle\parallel q\parallel_{X^{2}}^{2}$ $\displaystyle=$
$\displaystyle\frac{1}{T}\int_{0}^{T}\,\left(\parallel
q(t)\parallel^{2}_{{l}^{2}}+\parallel D{q}(t)\parallel^{2}_{{l}^{2}}+\parallel
D^{2}{q}(t)\parallel^{2}_{{l}^{2}}\right)dt=\sum_{m\in{\mathbb{Z}}\setminus\\{0\\}}\left(1+(m\Omega)^{2}+(m\Omega)^{4}\right)\frac{1}{2\pi}\int_{0}^{2\pi}|\hat{\tilde{q}}_{m,k}|^{2}dk,$
(14)

where $\hat{q}_{n,k}$ determines the temporal Fourier-transform:

$\displaystyle q_{n}(t)$ $\displaystyle=$
$\displaystyle\sum_{m\in{\mathbb{Z}}\setminus\\{0\\}}\hat{q}_{n,m}\exp\left(i\Omega
mt\right),\,t\in[0,T],\,\,\,\,\,\,{\hat{q}}_{n,m}=\frac{1}{T}\int_{0}^{T}q_{n}(t)\exp(-i\Omega
mt)dt,$ (15) $\displaystyle\hat{q}_{n,m}$ $\displaystyle=$
$\displaystyle\overline{\hat{q}}_{n,-m},$ (16)

and $\tilde{q}_{k}$ is the spatial Fourier-transform:

$\displaystyle q_{n}(t)$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi}\int_{0}^{2\pi}\tilde{q}_{k}(t)\exp(ikn),\,\,\,\,\,\tilde{q}_{k}(t)=\sum_{n}q_{n}(t)\exp(-i\,k\,n),$
(17) $\displaystyle\tilde{q}_{k}$ $\displaystyle=$
$\displaystyle\overline{\tilde{q}}_{-k},$ (18)

of $q_{n}(t)$, respectively, and $\overline{u}$ denotes the conjugate of $u$.

The Hilbert space of square-summable sequences, ${l}^{2}(\mathbb{Z})$, is
endowed with the standard scalar product:

$\left<q,p\right>:=\sum_{n}\,q_{n}\,p_{n}\,,\qquad\parallel
q\parallel^{2}_{l^{2}}:=\left<q,q\right>\,.$ (19)

For later use we note that $X^{2}$ is compactly embedded in $X^{0}$
($X^{2}\Subset X^{0}$).

We relate the l.h.s. of (11) to the linear mapping:
$M\,:\,X^{2}\,\rightarrow\,X^{0}$:

$M(u_{n})=\ddot{u}_{n}+{u}_{n}-\kappa[u_{n+1}-2u_{n}+u_{n-1}].$ (20)

Using the relations for the temporal and spatial Fourier-transforms (16),(17),
one obtains the representation

$u_{n}(t)=\sum_{m\in{\mathbb{Z}}\setminus\\{0\\}}\frac{1}{2\pi}\int_{0}^{2\pi}{\hat{\tilde{u}}}_{k,m}\exp(ikn)dk\exp(i\Omega
mt).$ (21)

Applying the operator $M$ to the Fourier elements $\exp(ikn)\exp(i\,\Omega
mt)$ in (21), results in

$M\exp(ikn)\exp(i\,\Omega mt)=\nu_{m}(k)\exp(ikn)\exp(i\,\Omega mt),$ (22)

where

$\nu_{m}(k)=-\Omega^{2}\,m^{2}+1+4\,\kappa\sin^{2}\left(\frac{k}{2}\right).$
(23)

As the hypothesis (10) guarantees that $\nu_{m}(k)\neq 0$,
$\forall\,\,m\in{\mathbb{Z}}$ and $\forall\,\,k\in[0,2\pi]$, the mapping $M$
possesses an inverse obeying $M^{-1}\exp(i\,\Omega
lt)=(1/\nu_{l})\exp(i\,\Omega lt)$ and for the norm of the linear operator
$M^{-1}:\,\,X^{0}\rightarrow X^{2}$ one derives using the hypothesis (10) the
upper bound:

$\displaystyle\parallel M^{-1}\parallel_{X^{0},X^{2}}$ $\displaystyle=$
$\displaystyle\sup_{0\neq q\in X^{0}}\frac{\parallel
M^{-1}\,q\parallel_{X^{2}}}{\parallel q\parallel_{X^{0}}}$ (24)
$\displaystyle=$ $\displaystyle\sup_{0\neq q\in
X^{0}}\frac{\left(\int_{0}^{T}\,\left[\parallel
M^{-1}q(t)\parallel^{2}_{{l}^{2}}+\parallel
D{M^{-1}q(t)}\parallel^{2}_{{l}^{2}}+\parallel
D^{2}{M^{-1}q(t)}\parallel^{2}_{{l}^{2}}\right]dt\right)^{1/2}}{\parallel
q\parallel_{X^{0}}}$ $\displaystyle=$ $\displaystyle\sup_{0\neq q\in
X^{0}}\frac{\left(\sum_{m^{\prime}}\frac{1}{2\pi}\int_{0}^{2\pi}\left|\frac{\hat{\tilde{q}}_{k,m}}{\nu_{m}(k)}\right|^{2}dk+\sum_{m^{\prime}}\frac{1}{2\pi}\int_{0}^{2\pi}\left|\frac{i\Omega
m\hat{\tilde{q}}_{k,m}}{\nu_{m}(k)}\right|^{2}dk+\sum_{m^{\prime}}\frac{1}{2\pi}\int_{0}^{2\pi}\left|\frac{(i\Omega
m)^{2}\hat{\tilde{q}}_{k,m}}{\nu_{m}(k)}\right|^{2}dk\right)^{1/2}}{\parallel
q\parallel_{X^{0}}}$ $\displaystyle\leq$
$\displaystyle\sup_{m\in{\mathbb{Z}}\setminus\\{0\\}}\sup_{k\in[0,2\pi]}{\frac{1}{|\nu_{m}(k)|}}\cdot\frac{\left(\sum_{m^{\prime}}\left(1+(\Omega\,m)^{2}+(\Omega\,m)^{4}\right)\frac{1}{2\pi}\int_{0}^{2\pi}\left|\hat{\tilde{q}}_{k,m}\right|^{2}dk\right)^{1/2}}{\parallel
q\parallel_{X^{0}}}$ $\displaystyle\leq$
$\displaystyle\frac{1}{\Omega^{2}-(1+4\,\kappa)}<\infty,$

verifying boundedness of $M^{-1}$, and we used the notation
$\sum_{l^{\prime}}=\sum_{l\in{\mathbb{Z}}\setminus\\{0\\}}$.

Furthermore associated with the r.h.s. of (11) we introduce the nonlinear
operator $N\,:\,X^{0}\,\rightarrow\,X^{0}$, as

$(N(u))_{n}=-W^{\prime}(u_{n}).$ (25)

In order to show that the operator $N$ is continuous on $X^{0}$ we prove that
$N$ is Frechet differentiable at any $u$ with bounded derivative. In detail we
have

$N^{\prime}(u):\,h\in X^{0}\,\mapsto N^{\prime}(u)[h]=-W^{\prime\prime}(u)h\in
X^{0},$ (26)

so that

$\displaystyle\parallel N^{\prime}(u)[h]\parallel_{X^{0}}^{2}$
$\displaystyle=$
$\displaystyle\int_{0}^{T}\sum_{n\in\mathbb{Z}}\left|\,-W^{\prime\prime}(u_{n}(t))\,h_{n}(t)\right|^{2}\,dt$
(27) $\displaystyle\leq$
$\displaystyle\int_{0}^{T}\sum_{n\in\mathbb{Z}}|1-\,V^{\prime\prime}(u_{n}(t))|^{2}\,|h_{n}(t)|^{2}\,dt$
$\displaystyle\leq$ $\displaystyle(1+\overline{K}^{2})\,\parallel
h\parallel^{2}_{X^{0}},$

proving the (uniform) boundedness of the differential. Hence,

$\parallel
N^{\prime}(u)\parallel_{{\cal{L}}(X^{0},X^{0})}\leq\sqrt{1+\overline{K}^{2}},$
(28)

viz. the derivative is bounded and for the range of $N$ in $X^{0}$ we get the
bound

$\displaystyle\parallel N(u)\parallel_{X^{0}}^{2}$ $\displaystyle=$
$\displaystyle\int_{0}^{T}\sum_{n\in\mathbb{Z}}|\,-W^{\prime}(u_{n}(t))|^{2}\,dt$
(29) $\displaystyle\leq$
$\displaystyle\int_{0}^{T}\sum_{n\in\mathbb{Z}}|\,u_{n}(t)-V^{\prime}(u_{n}(t))|^{2}\,dt$
$\displaystyle\leq$
$\displaystyle(1+\overline{K}^{2})\int_{0}^{T}\sum_{n\in\mathbb{Z}}|u_{n}(t)|^{2}\,dt\leq(1+\overline{K}^{2})\parallel
u\parallel^{2}_{X^{0}}.$

Finally, we express the problem (11) as a fixed point equation in terms of a
mapping $X^{0}\,\rightarrow\,X^{2}\Subset X^{0}$:

$q=M^{-1}\circ N(q)\equiv S(q).$ (30)

We verify that the range of $S$ is contained in a compact subset of $X^{0}$.
Representing $N(q)\in X^{0}$ in terms of its spatial and temporal Fourier-
transforms as

$(N(q))_{n}(t)=\sum_{m\in{\mathbb{Z}}\setminus\\{0\\}}\frac{1}{2\pi}\int_{0}^{2\pi}\hat{\tilde{N}}_{k,m}\exp(ikn)dk\exp\left(i\Omega
mt\right),$ (31)

the Fourier coefficients of $M^{-1}(N(q))$ fulfill $\forall q\in X^{0}$,
$\forall k\in[0,2\pi],\,\,\,\forall m\in{\mathbb{Z}}\setminus\\{0\\}$

$\left|\frac{\hat{{\tilde{N}}}_{k,m}}{-\Omega^{2}\,m^{2}+1+4\,\kappa\sin^{2}\left(\frac{k}{2}\right)}\right|^{2}\leq\frac{\sum_{m\in{\mathbb{Z}}\setminus\\{0\\}}\frac{1}{2\pi}\int_{0}^{2\pi}\left|\hat{{\tilde{N}}}_{k,m}\right|^{2}dk}{m^{4}[\Omega^{2}-(1+4\,\kappa)]^{2}}=\frac{\parallel
N(q)\parallel_{X^{0}}^{2}}{m^{4}[\Omega^{2}-(1+4\,\kappa)]^{2}}.$ (32)

Exploiting the coercivity of the conserved total energy $E=H$ in (2) and the
expression (7) one can bound the $l^{2}$-norm of
$q=\\{q_{n}\\}_{n\in{\mathbb{Z}}}$ as follows $\parallel
q(t)\parallel_{l^{2}}^{2}\leq 2E$ for every $t\in[0,T]$, so that

$\parallel{q}\parallel_{X^{0}}^{2}\leq 2E,$ (33)

from which, using (29), we deduce

$\parallel N(q)\parallel_{X^{0}}^{2}\leq 2(1+\overline{K}^{2})E.$ (34)

Hence, from (32) and (34) one concludes that $S$ maps $X^{0}$ into the subset

$Y^{0}=\left\\{p=\left\\{p\right\\}_{n\in{\mathbb{Z}}}\in
X^{0},\,p_{n}(t)=\sum_{m\in{\mathbb{Z}}\setminus\\{0\\}}\frac{1}{2\pi}\int_{0}^{2\pi}\hat{{\tilde{p}}}_{k,m}\,\exp(ikn)dk\exp\left(i\Omega
mt\right)\,\left|\,|\hat{{\tilde{p}}}_{k,m}|\leq\frac{M}{m^{2}}\right.\right\\},$
(35)

which is, for every

$M=\frac{\sqrt{2(1+\overline{K}^{2})E}}{\Omega^{2}-(1+4\,\kappa)}>0$ (36)

compact in $X^{0}$.

That is, the operator $S$ maps closed convex subsets $X^{0}$ of
$L_{per}^{2}((0,T);l^{2})$ into compact subsets $Y^{0}$ of $X^{0}$.

Clearly $S$ is continuous on $X^{0}$ as the relations (24) and (28) establish
that its constituents $M^{-1}$ and $N$ are continuous. Schauder’s fixed point
theorem implies then that the fixed point equation $q=S(q)$ has at least one
solution. Moreover, by the hypothesis (10) the values of the frequency of
oscillations $\Omega$ lie above the upper edge of the continuous (phonon)
spectrum determined by $1+4\kappa$. Thus the corresponding solutions have to
be anharmonic which necessitates nonzero nonlinearity. Only the latter
facilitates amplitude-depending tuning of the frequency of oscillations so
that the breather frequency $\Omega$ lies outside of the phonon spectrum. Thus
it must hold that $\parallel
N(q)\parallel_{Y^{0}}=\parallel-W^{\prime}(q)\parallel_{Y^{0}}\not\equiv 0$,
which is by A satisfied if and only if $q\not\equiv 0$. That is, the fixed
point equation (30) possesses a nonzero solution and the proof is finished.

$\square$

Remarks: (i) As the obtained time-periodic $H^{2}$ fixed-point-solutions are
by Sobolev embeddings $C^{1}$ in time and since the operator $q\mapsto
W^{\prime}(q)$ maps $C^{1}$ into itself, one concludes from Eq. (11) that
$\ddot{q}\in C^{1}$ so that the solutions are in fact $C^{3}$ in time and
hence, are classical solutions.

(ii) The localised solutions on the infinite lattice ${\mathbb{Z}}$ are
represented by (infinite) square-summable sequences, viz. decay of the states
for $|n|\rightarrow\infty$ takes place in the sense of the $l^{2}$ norm. This
approach allows for general localised patterns (e.g. multi-site breathers).
Regarding so-called single-site breathers, their exponential decay can be
directly incorporated in our existence proof by using an exponentially
weighted norm instead of the standard one in (19) and the corresponding
existence proof proceeds in the same manner as the general one above. (See
also the relevant discussion in MacKay .)

In summary, we have proven the existence of nonzero localised periodic
solutions for general Klein-Gordon systems with convex on-site potentials on
the lattice ${\mathbb{Z}}$. The existence problem has been reformulated as a
fixed point problem for an operator on a function space which is solved with
the help of Schauder’s Fixed Point Theorem. The method employed in this paper
can be straightforwardly extended to treat also the general Klein-Gordon
systems on lattices of higher dimension.

## References

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# Some results on Higher orders quasi-isometries

Sid Ahmed Ould Ahmed Mahmoud , Adel SADDI and Khadija Gherairi  Sid Ahmed Ould
Ahmed Mahmoud Mathematics Department, College of Science, Jouf
University,Sakaka P.O.Box 2014. Saudi Arabia<EMAIL_ADDRESS>Adel SADDI
Mathematics Department, College of Science,Gabes University. Tunisia
<EMAIL_ADDRESS>Khadija Gherairi Laboratory of Mathematics and
Applications, College of Sciencs,Gabes University. Tunisia
<EMAIL_ADDRESS>
(Date: _November 3-2018_)

###### 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 ${\mathcal{H}}$. 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.

###### Key words and phrases:

$m$-isometry, strict $m$-isometry $n$-quasi-$m$-isometry.

###### 2010 Mathematics Subject Classification:

Primary 47B99; Secondary 47A05.

##  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

$\sum_{0\leq k\leq m}(-1)^{m-k}\binom{m}{k}T^{\ast k}T^{k}=0.$ (1.1)

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

$\sum_{0\leq k\leq
m}(-1)^{m-k}\binom{m}{k}\|T^{k}x\|^{2}=0\quad\forall\;x\in{\mathcal{H}},$
(1.2)

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

$\displaystyle T\bigg{(}\sum_{0\leq k\leq
m}(-1)^{k}\binom{m}{k}T^{*m-k}T^{m-k}\bigg{)}=0,$ (1.3)

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

$\displaystyle T^{q}\bigg{(}\sum_{0\leq k\leq
m}(-1)^{k}\binom{m}{k}T^{*m-k}T^{m-k}\bigg{)}=0,$ (1.4)

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

$\displaystyle T^{*m+1}T^{m+1}-T^{*m}T^{m}=0.$ (1.5)

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

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

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

$\displaystyle T^{*n}\bigg{(}\sum_{0\leq k\leq
m}(-1)^{k}\binom{m}{k}T^{*m-k}T^{m-k}\bigg{)}T^{n}=0,$ (1.6)

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

$\beta_{m}(T):=\sum_{0\leq k\leq m}(-1)^{m-k}\binom{m}{k}T^{\ast k}T^{k},$
(2.1) $\beta_{m,\;n}(T):=T^{\ast n}\bigg{(}\sum_{0\leq k\leq
m}(-1)^{m-k}\binom{m}{k}T^{\ast k}T^{k}\bigg{)}T^{n}$ (2.2)

and

$\Delta_{m,\;n}(T,x):=\sum_{0\leq k\leq
m}(-1)^{m-k}\binom{m}{k}\|T^{k+n}x\|^{2},\quad\;\;\;\;x\in{\mathcal{H}}.$
(2.3)

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

$\Delta_{m,\;n}(T,x)=0\quad\forall\;\;x\in{\mathcal{H}}.$

######  Lemma 2.1.

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

$\sum_{0\leq k\leq
m}(-1)^{m-k}\binom{m}{k}\|T^{k}x\|^{2}=0\quad\forall\;\;x\in\overline{{\mathcal{R}}(T^{n})},$
(2.4)

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

$\sum_{0\leq k\leq m}(-1)^{m-k}\binom{m}{k}a_{j+k}=0\;\;\;\text{for}\;\;j\geq
0$

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

$\beta_{m}(T_{1}):=\sum_{0\leq k\leq
m}(-1)^{m-k}\binom{m}{k}T_{1}^{*k}T_{1}^{k}=0\;\;\text{and}\;\;T_{3}^{n}=0.$

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

$\displaystyle T^{*n}\bigg{(}\sum_{0\leq k\leq
m}(-1)^{m-k}\binom{m}{k}T^{*k}T^{k}\bigg{)}T^{n}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}T_{1}&T_{2}\\\ 0&T_{3}\\\
\end{array}\right)^{*n}\bigg{\\{}(-1)^{m}I+\sum_{1\leq k\leq
m}(-1)^{m-k}\binom{m}{k}\left(\begin{array}[]{cc}T_{1}&T_{2}\\\ 0&T_{3}\\\
\end{array}\right)^{*k}\left(\begin{array}[]{cc}T_{1}&T_{2}\\\ 0&T_{3}\\\
\end{array}\right)^{k}\bigg{\\}}$
$\displaystyle\times\left(\begin{array}[]{cc}T_{1}&T_{2}\\\ 0&T_{3}\\\
\end{array}\right)^{n}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}T_{1}^{*n}&0\\\
\displaystyle\sum_{0\leq j\leq
n-1}T_{3}^{*n-1-j}T_{2}^{*}T_{1}^{*j}&T_{3}^{*n}\\\ \end{array}\right)$
$\displaystyle\times\bigg{\\{}(-1)^{m}I+\sum_{1\leq k\leq
m}(-1)^{k}\binom{m}{k}\left(\begin{array}[]{cc}T_{1}^{*k}&0\\\
\displaystyle\sum_{0\leq j\leq
k-1}T_{3}^{*k-1-j}T_{2}^{*}T_{1}^{*j}C_{1}&T_{3}^{*k}\\\
\end{array}\right)\left(\begin{array}[]{cc}T_{1}^{k}&\displaystyle\sum_{j=0}^{k-1}T_{1}^{j}T_{2}T_{3}^{k-1-j}\\\
0&T_{3}^{k}\\\ \end{array}\right)\bigg{\\}}$
$\displaystyle\times\left(\begin{array}[]{cc}T_{1}^{n}&\displaystyle\sum_{j=0}^{n-1}T_{1}^{j}T_{2}T_{3}^{n-1-j}\\\
0&T_{3}^{n}\\\ \end{array}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}T_{1}^{*n}&0\\\
\displaystyle\sum_{0\leq j\leq n-1}T_{3}^{*n-1-j}T_{2}^{*}T_{1}^{*j}&0\\\
\end{array}\right)$
$\displaystyle\times\bigg{\\{}\left(\begin{array}[]{cc}\Lambda_{m}(T_{1})&C\\\
\\\ D&B\\\
\end{array}\right)\bigg{\\}}\times\left(\begin{array}[]{cc}T_{1}^{n}&\displaystyle\sum_{j=0}^{n-1}T_{1}^{j}T_{2}T_{3}^{n-1-j}\\\
0&0\\\ \end{array}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}T_{1}^{\ast
n}\Lambda_{m}(T_{1})T_{1}^{n}&0\\\ \\\ 0&0\\\ \end{array}\right)$
$\displaystyle=$ $\displaystyle 0.$

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

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

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]

$\sigma(T)=\sigma(T_{1})\cup\sigma(T_{3})=\sigma(T_{1})\cup\\{0\\}.$

$(ii)$ 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

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

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

$Te_{1}=2e_{2},\;Te_{2}=3e_{3}\;\;\;\;\text{and}\;\;\;T_{k}=e_{k+1}\;\;\;\;\text{for}\;\;k\geq
3.$

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

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

we deduce that

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

This means that

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

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

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

Thus we have

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

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

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

Using the inclusion

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

we get

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

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

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

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

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

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

$P\bigg{(}\sum_{0\leq k\leq m}(-1)^{m-k}\binom{m}{k}\big{(}T^{r}\big{)}^{\ast
k}\big{(}T^{r}\big{)}^{k}\bigg{)}P=0.$

That is

$\sum_{0\leq k\leq m}(-1)^{m-k}\binom{m}{k}\big{(}T_{1}^{r}\big{)}^{\ast
k}\big{(}T_{1}^{r}\big{)}^{k}=0.$

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

$\displaystyle\langle\big{(}T_{3}^{r}\big{)}^{n}x_{2},x_{2}\rangle$
$\displaystyle=$
$\displaystyle\langle\big{(}T^{r}\big{)}^{n}(I-P)x,(I-P)x\rangle$
$\displaystyle=$ $\displaystyle\langle(I-P)x,\big{(}T^{r}\big{)}^{\ast
n}(I-P)x\rangle=0.$

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

$\|T^{k}\xi\|\leq C\|\xi\|.$

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

$[T,S]=TS-ST.$

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

$\displaystyle\beta_{m+l-1,\;n_{0}}(TS)$ $\displaystyle=$
$\displaystyle\big{(}TS\big{)}^{*(n_{0})}\beta_{m+l-1}(TS)\big{(}TS\big{)}^{n_{0}}$
$\displaystyle=$
$\displaystyle(T^{*})^{n_{0}}(S^{*})^{n_{0}}\,\Lambda_{m+l-1}(TS)\,(T)^{n_{0}}(S)^{n_{0}}$
$\displaystyle=$
$\displaystyle(T^{*})^{n_{0}}(S^{*})^{n_{0}}\bigg{(}\sum_{0\leq j\leq
m+l-1}\binom{m+l-1}{j}{T^{\ast}}^{j}\beta_{m+l-1-j}(T)\,T^{j}\beta_{j}(S)\bigg{)}(T)^{n_{0}}(S)^{n_{0}}$
$\displaystyle=$ $\displaystyle\bigg{(}\sum_{0\leq j\leq
m+l-1}\binom{m+l-1}{j}\,{T^{\ast}}^{j}\underbrace{\big{(}T^{*}\big{)}^{n_{0}}\beta_{m+l-1-j}(T)T^{n_{0}}}\,T^{j}\underbrace{\big{(}S^{*}\big{)}^{n_{0}}\beta_{j}(S)S^{n_{0}}}\bigg{)}.$

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

$T\otimes S=(T\otimes I)(I\otimes S)=(I\otimes S)(T\otimes I)$

it follows that

$[T\otimes I,I\otimes S]=[T\otimes I,(I\otimes S)^{*}]=0.$

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

$\beta_{q}\big{(}T+Q\big{)}=\sum_{0\leq k\leq q}\sum_{0\leq j\leq
q-k}\binom{q}{k}\binom{q-k}{j}\big{(}T^{*}+Q^{*}\big{)}^{k}Q^{*j}\beta_{q-k-j}(T)T^{j}Q^{k}.$

In fact, note that

$\displaystyle\beta_{m+2p-2,\;\alpha}\big{(}T+Q\big{)}$ $\displaystyle=$
$\displaystyle\big{(}T+Q\big{)}^{*\alpha}\beta_{m+2p-2}\big{(}T+Q\big{)}\big{(}T+Q\big{)}^{\alpha}$
$\displaystyle=$ $\displaystyle\bigg{(}\sum_{0\leq
r\leq\alpha}\binom{\alpha}{r}T^{*(\alpha-r)}Q^{*r}\bigg{)}\bigg{(}\sum_{0\leq
k\leq q}\sum_{0\leq j\leq
q-k}\binom{q}{k}\binom{q-k}{j}\big{(}T^{*}+Q^{*}\big{)}^{k}Q^{*j}\beta_{q-k-j}(T)T^{j}Q^{k}\bigg{)}$
$\displaystyle\times\bigg{(}\sum_{0\leq
r\leq\alpha}\binom{\alpha}{r}T^{\alpha-r}Q^{r}\bigg{)}.$

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

$\big{(}T^{*}+Q^{*}\big{)}^{k}Q^{*j}\beta_{q-k-j}(T)T^{j}Q^{k}=0.$

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

$q-k-j=m+2p-2-k-j\geq m+2p-2-(p-1)-(p-1)=m$

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

$T^{*(n+p-r)}\beta_{q-k-j}(T)T^{n+p-r}=0\;\;\;\text{for}\;\;r=0,\cdots,p$

and

$T^{*(n+p-r)}Q^{*r}\beta_{q-k-j}(T)T^{n+p-r}Q^{r}=0\;\;\;\text{for}\;\;r=p+1,\cdots,n+p.$

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

$S=\left(\begin{array}[]{cccc}T_{1}&\alpha_{1}I&0&\cdots\\\
0&\ddots&\ddots&\ddots\\\ \ddots&\ddots&\ddots&\alpha_{d-1}I\\\
0&\ddots&0&T_{d}\\\
\end{array}\right)\;\text{on}\;\;{\mathcal{H}}^{(d)}:={\mathcal{H}}\oplus\cdots\oplus{\mathcal{H}}$

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

$T=\left(\begin{array}[]{cccc}T_{1}&0&0&\cdots\\\ 0&\ddots&\ddots&\ddots\\\
\ddots&\ddots&\ddots&0\\\ 0&\ddots&0&T_{d}\\\
\end{array}\right)\;\;\text{and}\;\;Q=\left(\begin{array}[]{cccc}0&\alpha_{1}I&0&\cdots\\\
0&\ddots&\ddots&\ddots\\\ \ddots&\ddots&\ddots&\alpha_{d-1}I\\\
0&\ddots&0&0\\\ \end{array}\right)\;\;\text{for}\;\;j=1,\cdots,d.$

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

$\displaystyle\lim_{k\to\infty}\|T_{k}-T\|=0.$

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

$\displaystyle\|\beta_{m,\;n}(T)\|=\|\beta_{m,\;n}(T_{k})-\beta_{m,\;n}(T)\|$
$\displaystyle=$ $\displaystyle\|T_{k}^{*n}\bigg{(}\sum_{0\leq j\leq
m}(-1)^{m-j}\binom{m}{j}T_{k}^{*j}T_{k}^{j}\bigg{)}T_{k}^{n}-T^{*n}\bigg{(}\sum_{0\leq
j\leq m}(-1)^{m-j}\binom{m}{j}T^{*j}T^{j}\bigg{)}T^{n}\|$ $\displaystyle\leq$
$\displaystyle\|T_{k}^{*n}\bigg{(}\sum_{0\leq j\leq
m}(-1)^{m-j}\binom{m}{j}T_{k}^{*j}T_{k}^{j}\bigg{)}T_{k}^{n}-T_{k}^{*n}\bigg{(}\sum_{0\leq
j\leq m}(-1)^{m-j}\binom{m}{j}T_{k}^{*j}T^{j}\bigg{)}T^{n}\|$ $\displaystyle+$
$\displaystyle\|T_{k}^{*n}\bigg{(}\sum_{0\leq j\leq
m}(-1)^{m-j}\binom{m}{j}T_{k}^{*j}T^{j}\bigg{)}T^{n}-T^{*n}\bigg{(}\sum_{0\leq
j\leq m}(-1)^{m-j}\binom{m}{j}T^{*j}T^{j}\bigg{)}T^{n}\|$ $\displaystyle\leq$
$\displaystyle\|\sum_{0\leq j\leq
m}(-1)^{m-j}\binom{m}{j}T_{k}^{*n+j}\bigg{(}T_{k}^{j}T_{k}^{n}-T^{j}T^{n}\bigg{)}\|$
$\displaystyle+\|\sum_{0\leq j\leq
m}(-1)^{m-j}\bigg{(}T_{k}^{*n+j}-T^{*n+j}\bigg{)}T^{j}T^{n}\|$ (2.13)
$\displaystyle\leq$ $\displaystyle\sum_{0\leq j\leq
m}\binom{m}{j}\|T_{k}^{*}\|^{n+j}\|T_{k}^{j}T_{k}^{n}-T^{j}T^{n}\|+\sum_{0\leq
j\leq m}\binom{m}{k}\|T_{k}^{*n+j}-T^{n+j}\|\|T\|^{j+n}.\ \ \ \ \ \ $

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:

$\left\\{\begin{array}[]{ll}Te_{1}=ae_{2}&a\neq\sqrt{2}\\\
Te_{p}=\sqrt{\frac{p+1}{p}}e_{p+1}&(p=2,3,...)\end{array}\right.$

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
$z=(z_{1},...,z_{q}):$

$\big{(}z_{1}+\cdots+z_{q})^{k}=\displaystyle\sum_{p_{1}+p_{2}+\cdots+p_{q}=k}\binom{k}{p_{1},p_{2},\cdots,p_{q}}z_{1}^{p_{1}}z_{2}^{p_{2}}.\cdots
z_{q}^{p_{q}}.$

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

$\sum_{p_{1}+p_{2}+\cdots+p_{q}=k}\binom{k}{p_{1},p_{2},\cdots,p_{q}}=q^{k}.$

######  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

$\Delta_{m,\;n}(T,x)=\Delta_{m-1,\;n}(T,Tx)-\Delta_{m-1,\;n}(T,x).$ (3.1)

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

$\Delta_{m-1,\;n}(T,T^{k}x)=\Delta_{m-1,\;n}(T,x).$ (3.2)

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

$\Delta_{m,\;n}(T^{k},x)=\sum_{p_{1}+\cdots+p_{k}=m}\binom{m}{p_{1},\cdots,p_{k}}\Delta_{m,\;n}\big{(}T,T^{(0.p_{1}+1.p_{2}+\cdots+(k-1)p_{k})+(k-1)n}x\big{)}.$
(3.3)

###### Proof.

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

$\displaystyle\Delta_{m,\;n}(T,x)$ $\displaystyle=$ $\displaystyle\sum_{0\leq
k\leq m}(-1)^{m-k}\binom{m}{k}\|T^{k+n}x\|^{2}$ $\displaystyle=$
$\displaystyle(-1)^{m}\|T^{n}x\|^{2}+\sum_{1\leq k\leq
m}(-1)^{m-1-k}\binom{m}{k}\|T^{k+n}x\|^{2}+\|T^{m+n}x\|^{2}$ $\displaystyle=$
$\displaystyle(-1)^{m}\|T^{n}x\|^{2}+\sum_{1\leq k\leq
m}(-1)^{m-1-k}\bigg{(}\binom{m-1}{k}+\binom{m-1}{k-1}\bigg{)}\|T^{k+n}x\|^{2}+\|T^{m+n}x\|^{2}$
$\displaystyle=$ $\displaystyle\sum_{0\leq k\leq
m-1}(-1)^{m-1-k}\binom{m-1}{k}\|T^{k+n}Tx\|^{2}-\sum_{0\leq k\leq
m-1}(-1)^{m-1-k}\binom{m-1}{k}\|T^{k+n}x\|^{2}$ $\displaystyle=$
$\displaystyle\Delta_{m-1,\;n}(T,Tx)-\Delta_{m-1,n}(T,x).$

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

$\Delta_{m-1,\;n}(T,Tx)=\Delta_{m-1,\;n}(T,x).$

Hence,

$\Delta_{m-1,\;n}(T,T^{k}x)=\Delta_{m-1,\;n}(T,T^{k-1}x)=\cdots=\Delta_{m-1,\;n}(T,x).$

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

$\Delta_{m,\;n}(T^{k},x)=\sum_{0\leq j\leq
m}(-1)^{m-j}\binom{m}{j}\|\big{(}T^{k}\big{)}^{j+n}x\|^{2}$ (3.4)

and

$\displaystyle\sum_{p_{1}+\cdots+p_{k}=m}\binom{m}{p_{1},\cdots,p_{k}}\Delta_{m,\;n}\big{(}T,T^{(0.p_{1}+1.p_{2}+\cdots+(k-1)p_{k})+(k-1)n}x\big{)}$
$\displaystyle=$
$\displaystyle\sum_{p_{1}+\cdots+p_{k}=m}\binom{m}{p_{1},\cdots,p_{k}}\bigg{(}\sum_{0\leq
j\leq
m}(-1)^{m-j}\binom{m}{j}\|T^{j+kn}T^{(0.p_{1}+1.p_{2}+\cdots+(k-1)p_{k})}x\|^{2}\bigg{)}.$

Thus we need to prove the following combinatorial identity:

$\displaystyle\sum_{0\leq j\leq m}(-1)^{m-j}\binom{m}{j}(z^{k})^{j+n}$
$\displaystyle=$
$\displaystyle\sum_{p_{1}+\cdots+p_{k}=m}\binom{m}{p_{1},\cdots,p_{k}}\bigg{(}\sum_{0\leq
j\leq
m}(-1)^{m-j}\binom{m}{j}z^{j+kn}z^{(0.p_{1}+1.p_{2}+\cdots+(k-1)p_{k})}\bigg{)}.$

In fact, observe that

$\sum_{0\leq j\leq
m}(-1)^{m-j}\binom{m}{j}(z^{k})^{j+n}=\big{(}z^{k}-1\big{)}^{m}z^{kn}$

and

$\displaystyle\sum_{p_{1}+\cdots+p_{k}=m}\binom{m}{p_{1},\cdots,p_{k}}\bigg{(}\sum_{0\leq
j\leq
m}(-1)^{m-j}\binom{m}{j}z^{j+kn}z^{(0.p_{1}+1.p_{2}+\cdots+(k-1)p_{k})}\bigg{)}$
$\displaystyle=$ $\displaystyle
z^{kn}\sum_{p_{1}+\cdots+p_{k}=m}\binom{m}{p_{1},\cdots,p_{k}}\bigg{(}\sum_{0\leq
j\leq
m}(-1)^{m-j}\binom{m}{j}z^{j}\bigg{)}z^{(0.p_{1}+1.p_{2}+\cdots+(k-1)p_{k})}.$

By applying the multinomial formula in reverse order, we have

$\displaystyle\sum_{p_{1}+\cdots+p_{k}=m}\binom{m}{p_{1},\cdots,p_{k}}\bigg{(}\sum_{0\leq
j\leq
m}(-1)^{m-j}\binom{m}{j}z^{j}\bigg{)}z^{(0.p_{1}+1.p_{2}+\cdots+(k-1)p_{k})}$
$\displaystyle=$
$\displaystyle\big{(}z-1\big{)}^{m}\sum_{p_{1}+\cdots+p_{k}=m}\binom{m}{p_{1},\cdots,p_{k}}z^{(0.p_{1}+1.p_{2}+\cdots+(k-1)p_{k})}$
$\displaystyle=$
$\displaystyle\big{(}z-1\big{)}^{m}\big{(}z^{k-1}+z^{k-2}+\cdots+z+1)^{m}=\big{(}z^{k}-1)^{m}.$

∎

######  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

$\Delta_{m-1,\;n}(T^{k},x)=k^{m-1}\Delta_{m-1,\;n}(T,x).$

###### 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

$\displaystyle\Delta_{m-1,n}(T^{k},x)$ $\displaystyle=$
$\displaystyle\sum_{p_{1}+\cdots+p_{k}=m-1}\binom{m-1}{p_{1},\cdots,p_{k}}\Delta_{m-1,n}(T,T^{(0.p_{1}+1.p_{2}+\cdots+(k-1)p_{k})+(k-1)n}x)$
$\displaystyle=$
$\displaystyle\sum_{p_{1}+\cdots+p_{k}=m-1}\binom{m-1}{p_{1},\cdots,p_{k}}\Delta_{m-1,n}(T,x)$
$\displaystyle=$ $\displaystyle k^{m-1}\Delta_{m-1,\;n}(T,x).$

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

$\sum_{0\leq k\leq h+1}(-1)^{h+1-k}a_{k+j}=0$

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

$\sum_{0\leq k\leq m}(-1)^{m-k}\binom{m}{k}a_{r(k+n)+j}=0\;\;\quad\text{for
all}\;\;j\geq 0.$

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

$\sum_{0\leq k\leq l}(-1)^{l-k}\binom{l}{k}a_{s(k+n)+j}=0\;\;\quad\text{for
all}\;\;j\geq 0.$

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

$\big{(}T^{*}\big{)}^{n+l-1}\beta_{m-1}(T)T^{n+l-1}S^{*n}\beta_{l-1}(S)S^{n}\not=0.$

###### 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

$\beta_{q}(TS)=\sum_{0\leq k\leq
q}\binom{q}{k}T^{*k}\beta_{q-k}(T)T^{k}\beta_{k}(S).$

From which it follows that

$\displaystyle\beta_{m+l-2,\;n}(TS)$ $\displaystyle=$
$\displaystyle\big{(}TS)^{n}\beta_{m+l-2}(TS)\big{(}TS\big{)}^{n}$
$\displaystyle=$
$\displaystyle\big{(}T^{*}\big{)}^{n}\big{(}S^{*}\big{)}^{n}\bigg{(}\sum_{0\leq
k\leq
m+l-2}\binom{m+l-2}{k}T^{*k}\beta_{m+l-2-k}(T)T^{k}\beta_{k}(S)\bigg{)}T^{n}S^{n}$
$\displaystyle=$ $\displaystyle\sum_{0\leq k\leq
m+l-2}\binom{m+l-2}{k}T^{*k}\big{(}T^{*}\big{)}^{n}\beta_{m+l-2-k}T^{n}T^{k}\big{(}S^{*}\big{)}^{n}\beta_{k}(S)S^{n}$
$\displaystyle=$ $\displaystyle\sum_{0\leq k\leq
l-2}\binom{m+l-2}{k}T^{*k}\big{(}T^{*}\big{)}^{n}\beta_{m+l-2-k}T^{n}T^{k}\big{(}S^{*}\big{)}^{n}\beta_{k}(S)S^{n}$
$\displaystyle+\binom{m+l-2}{l-1}\big{(}T^{*}\big{)}^{n+l-l}\beta_{m-1}(T)T^{n+l-1}\big{(}S^{*}\big{)}^{n}\beta_{l-1}S^{n}$
$\displaystyle+\sum_{l\leq k\leq
m+l-2}\binom{m+l-2}{k}T^{*k}\big{(}T^{*}\big{)}^{n}\beta_{m+l-2-k}T^{n}T^{k}\big{(}S^{*}\big{)}^{n}\beta_{k}(S)S^{n}.$

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

$\big{(}T^{*}\big{)}^{n+l-l}\beta_{m-1}(T)T^{n+l-1}\big{(}S^{*}\big{)}^{n}\beta_{l-1}(S)S^{n}=0.$

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

$\beta_{q}(T\otimes S)=\sum_{0\leq k\leq
q}\binom{q}{k}T^{*k}\beta_{q-k}(T)T^{k}\otimes\beta_{k}(S).$

By calculations we have

$\displaystyle\beta_{m+l-2,\;n}(T\otimes S)$ $\displaystyle=$
$\displaystyle\big{(}T\otimes S\big{)}^{*n}\beta_{m+l-2}(T\otimes
S)\big{(}T\otimes S\big{)}^{n}$ $\displaystyle=$
$\displaystyle\big{(}T^{*n}\otimes S^{*n}\big{)}\bigg{(}\sum_{0\leq k\leq
m+l-2}\binom{m+l-2}{k}T^{*k}\beta_{m+l-2-k}(T)T^{k}\otimes\beta_{k}(S)\bigg{)}\big{(}T^{n}\otimes
S^{n}\big{)}$ $\displaystyle=$ $\displaystyle\sum_{0\leq k\leq
m+l-2}\binom{m+l-2}{k}T^{*k}T^{*n}\beta_{m+l-2-k}(T)T^{n}T^{k}\otimes
S^{*n}\beta_{k}(S)S^{n}.$

A similar arguments as in the proof of Theorem 3.3 give

$\beta_{m+l-2,\;n}(T\otimes
S)=\big{(}T^{*}\big{)}^{n+l-l}\beta_{m-1}(T)T^{n+l-1}\otimes\big{(}S^{*}\big{)}^{n}\beta_{l-1}(S)S^{n}.$

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

$\\{\;\beta_{k,\;n}(T),k=0,1,\;\cdots,m-1\;\\}$

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

$\displaystyle\beta_{m,\;n}(T)=T^{*n}\beta_{m}(T)T^{n}$ $\displaystyle=$
$\displaystyle T^{*n}\bigg{(}T^{*}\beta_{m-1}(T)T-\beta_{m-1}(T)\bigg{)}T^{n}$
$\displaystyle=$ $\displaystyle
T^{*n+1}\beta_{m-1}(T)T^{n+1}-T^{*n}\beta_{m-1}(T)T^{n}.$

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

$\sum_{0\leq k\leq m-1}\lambda_{k}\beta_{k,\;n}(T)=0$

or equivalently

$\sum_{0\leq k\leq m-1}\lambda_{k}T^{*n}\beta_{k}T^{n}(T)T^{n}=0.$

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

$\sum_{0\leq k\leq
m-1}\lambda_{k}\bigg{(}T^{*n+1}\beta_{k}(T)T^{n+1}-T^{*n}\beta_{k}(T)T^{n}\bigg{)}=\sum_{0\leq
k\leq m-1}\lambda_{k}\beta_{k+1,\;n}(T)=0$

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

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

By continuing this process we obtain

$\sum_{0\leq k\leq m-1}\lambda_{k}\beta_{k+j,\;n}(T)=0\;\;\;\text{for
all}\;\;j\in\mathbb{N}.$

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. ∎

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