On Invertible m-isometrical Extension of an m-isometry

Abstract

We give necessary and sufficient conditions on an m-isometry to have an invertible m-isometrical extension. As particular cases, we give a useful characterization for a general m-isometrical unilateral weighted shift and for $\ell$-Jordan isometries. In particular, every $\ell$-Jordan isometry operator has an invertible $(2\ell -1)$-isometrical extension.

1 Introduction

In the last twenty years there has been an intense research activity on m-isometries. In this paper, we focus our attention on characterizing m-isometries that have an invertible extension that is also m-isometry.

The notion of m-isometric operator on a Hilbert space was introduced by J. Agler [2] and studied in detail shortly after by J. Agler and M. Stankus in three papers [4,5,6]. These publications can be considered the first ones to initiate this topic of study.

An operator $T\in L(H)$, the algebra of all bounded linear operators acting on a Hilbert space H, is called an m-isometry, for some positive integer m, if

$$\begin{aligned} \sum _{k=0}^{m} {m \atopwithdelims ()k} (-1)^{k} T^{*k}T^k=0, \end{aligned}$$

where $T^*$ denotes the adjoint operator of T. When $m=1$, we obtain an isometry. It is said that T is a strict m-isometry if either $m=1$ or T is an m-isometry with $m>1$ but it is not $(m-1)$-isometry.

As one should expect, m-isometries share many important properties with isometries. For example, the following dichotomy property: the spectrum of an m-isometry is the closed unit disc if it is not invertible or a closed subset of the unit circle if it is invertible [4]. Also, if T is an m-isometry, then T is bounded below; that is, there exists $M>0$ such that $\Vert Tx\Vert \ge M\Vert x\Vert$ for every $x\in H$.

Given an m-isometry $T \in L(H)$, we are interested in research conditions which guarantee the existence of a Hilbert space K and an operator $S \in L(K)$, which is an extension of T, such that S is an invertible m-isometry. To say that $S \in L(K)$ is an extension of $T \in L(H)$ means that K contains an isometric subspace to H, which we denote also by H, and the restriction $S_{|H}$ from H to H coincides with T.

Problem 1.1

Characterize those m-isometric operators which have an invertible m-isometrical extension.

In 1969 Douglas [13] obtained that any isometry in a Banach space has an invertible isometric extension, also valid in a Hilbert space context. So, the case $m=1$ holds. For $m\ge 2$, first immediate consideration is that m must be odd, since every invertible m-isometry with even m is an $(m-1)$-isometry by [4, Proposition 1.23].

Our problem is similar to others that arise naturally in Operator Theory and can be formulated in very general terms as follows. Given a class ${{\mathcal {C}} }$ of operators, for example defined on Hilbert spaces, and given a property P relative to those operators, we wish to characterize the operators that have an extension in the class ${{\mathcal {C}}}$ with property P.

Let $T\in L(H)$ and $S \in L(K)$ with H a closed subspace of K. Denote by $P_H$ the orthogonal projection of K onto H and by J the inclusion of H into K. It is said that

• S is a lifting of T if $P_HS=TP_H$.

• S is a dilation of T if $T^n=P_HS^nJ$, for every $n\in {\mathbb {N}}$.

Many authors have studied, for a given bounded linear operator $T \in L(H)$, some additional properties of extension, lifting, or dilation of the operator T. The following results are known and respond to these problems:

• Every contraction has an extension which in turn has a unitary lifting. Thus, every contraction has a unitary dilation. Also, a contraction has a lifting which is an isometry. See [16].

• Every isometry has a unitary extension. See [13].

• Every operator T such that the norms of its powers grow polynomially has an m-isometric lifting for some integer $m\ge 1$. This lifting can be also extended to an invertible m-isometry. See [9].

Notice that the norms of the powers of an m-isometry have a polynomial behaviour (see part (1) of Proposition 2.1). However, there are operators such that those norms have a polynomial behaviour that are not m-isometries. In [9], the authors study lifting and dilations which are m-isometries. In particular, they obtain that if T is an m-isometry, then T has an $(m+3)$-isometric lifting with other additional properties.

A special class of m-isometric operators is the $\ell$-Jordan isometries; that is, operators which are the sum of an isometry and an $\ell$-nilpotent operator which commute. It is known that every $\ell$-Jordan isometry is a strict $(2\ell -1)$-isometry, but the converse is not valid. However, every strict m-isometry on a finite dimensional Hilbert space is an $\frac{(m+1)}{2}$-Jordan isometry operator. See [3, 12, 17] for more details.

Another natural and important examples of m-isometries are certain weighted shift operators. In [1, 11], the authors obtained a characterization of weighted shift which are m-isometric.

We summarize the contents of the paper. In Sect. 2, we define a bilateral sequence of operators associated to an m-isometry that allow us to transfer important information of the m-isometry to the bilateral sequence, that it will be an important tool in the paper. In Sect. 3, we present some necessary conditions to obtain an invertible m-isometrical extension. The main results are given in Sect. 4 where we obtain characterizations for an m-isometry to have an invertible m-isometrical extension. Finally, in Sect. 5, we present particular classes of m-isometries for which one can obtain nice results. In particular, we give a useful characterization for a general m-isometrical unilateral weighted shift and for $\ell$-Jordan isometries. In particular, every $\ell$-Jordan isometry operator has an invertible $(2\ell -1)$-isometrical extension.

2 Some Previous Results

In this section, we define a bilateral sequence of operators associated to an m-isometry, that allow us to transfer important information of the m-isometry to the bilateral sequence that it will be relevant for obtaining necessary conditions for having an invertible m-isometrical extension.

Any polynomial of degree less or equal to $m-1$ is uniquely determined by its values at m distinct points. If $a_0,a_1,\dots ,a_{m-1}$ are given real (or complex) numbers, then the unique polynomial p of degree less or equal to $m-1$ satisfying $p(k)=a_k$ for all $k\in \{ 0,1,\dots ,m-1\}$ is giving by Lagrange interpolating polynomial

$$\begin{aligned} p(z)=\sum _{k=0}^{m-1}a_k \prod _{0\le j\le m-1\atop j\ne k}\frac{z-j}{k-j} . \end{aligned}$$

Note that

$$\begin{aligned} p(n)=\sum _{k=0} ^{m-1} a_k b_k(n) \end{aligned}$$

with

$$\begin{aligned} b_k(n):= \prod _{0\le j\le m-1\atop j\ne k}\frac{n-j}{k-j}= (-1)^{m-k-1} \frac{n (n-1) \ldots \widehat{(n-k)} \cdots (n-m+1)}{k!(m-k-1)!}\;\nonumber \\ \end{aligned}$$

(2.1)

where $\widehat{(n-k)}$ means that the factor $(n-k)$ is omitted.

Given $T\in L(H)$, define the bilateral sequence by

$$\begin{aligned} D_n:=\sum _{k=0}^{m-1} b_k(n) T^{*k}T^k , \end{aligned}$$

(2.2)

for every $n\in {\mathbb {Z}}$. Clearly $D_n\in L(H)$ and it is self adjoint operator for every $n\in {\mathbb {Z}}$.

Denote $p_x(k):=\langle D_kx,x\rangle$ for every $x\in H$ and $k\in {\mathbb {Z}}$.

Given $T\in L(H)$, denote $T>0$ if $\langle Tx,x \rangle >0$ for every $x\in H\setminus \{ 0\}$ and we call it strictly positive operator.

We concentrate now on the family $(D_n)_{n\in {\mathbb {Z}}}$ of operators which arise from a fixed m-isometry. Indeed, the bilateral sequence $(D_n)_{n\in {\mathbb {Z}}}$ has some interesting properties that will be important tools to solve Problem 1.1.

Proposition 2.1

Let $T\in L(H)$ be an m-isometry and $(D_n)_{n\in {\mathbb {Z}}}$ be operators defined by (2.2). Then

• (1) [11, Theorem 2.1] & [4] $D_n=T^{*n}T^n$ and $p_x(n)= \langle D_nx,x \rangle =\Vert T^nx\Vert ^2>0$ for every $x\in H{\setminus } \{ 0\}$ and $n\in {\mathbb {N}}\cup \{ 0\}$. Henceforth, there exists the square root $D_n^{1/2}$ of $D_n$, for every $n\in {\mathbb {N}} \cup \{ 0\}$.

• (2) $D_n$ is invertible for every $n\in {\mathbb {N}} \cup \{ 0\}$.

• (3) $T^{*k}D_n T^k=D_{n+k}$ for every $n\in {\mathbb {Z}}$ and $k\in {\mathbb {N}}$.

• (4) Let $y\in R(T^k)$ for some $k\in {\mathbb {N}}$. Then $p_y(-k)=\Vert x\Vert ^2$, where $y=T^kx$.

• (5) If $D_{-n}>0$ and invertible, then $D_{-k}>0$ and invertible for every $k\in \{ 1, 2, \cdots , n-1\}$.

Proof

(2) Let $n\in {\mathbb {N}}$. By [14] Theorem 2.3] & [10], Theorem 3.1] any power of T, $T^n$ is an m-isometry, so, $T^n$ is bounded below. Hence

$$\begin{aligned} \Vert D_nx\Vert \Vert x\Vert \ge |\langle D_nx, x\rangle |= \langle D_nx, x\rangle =\Vert T^nx\Vert ^2 \ge M(n)^2 \Vert x\Vert ^2 , \end{aligned}$$

where $M(n)>0$. That is, $D_n$ is bounded below. Then trivially $D_n$ is invertible since $D_n$ is self adjoint operator. (3) It is enough to prove the required equality for $k=1$. Observe that

$$\begin{aligned} p_{Tx}(n)=\Vert T^nTx\Vert ^2 = \Vert T^{n+1} x\Vert ^2 =p_x(n+1), \end{aligned}$$

for every $n\in {\mathbb {N}}$ and

$$\begin{aligned} \langle D_{n+1}x,x \rangle = p_x(n+1) =p_{Tx}(n)= \langle D_nTx, Tx\rangle =\langle T^* D_nTx,x\rangle \end{aligned}$$

for every $n\in {\mathbb {Z}}$. (4) Let $y=T^kx$ for some $k\in {\mathbb {N}}$ and $x\in H$. Then

$$\begin{aligned} p_{y}(n)= p_{T^kx}(n)=p_x (k+n) , \end{aligned}$$

for every $n\in {\mathbb {N}}$. Therefore $p_y(n)=p_x(k+n)$ for every $n\in {\mathbb {Z}}$. (5) Let $k\in \{ 1, 2, \cdots , n-1 \}$ and $x\in H{\setminus }\{ 0\}$. If $D_{-n}>0$, then by part (3),

$$\begin{aligned} \langle D_{-k}x, x \rangle = \langle T^{*n-k} D_{-n}T^{n-k}x,x \rangle =\langle D_{-n} T^{n-k}x,T^{n-k}x \rangle >0 . \end{aligned}$$

(2.3)

Since $T^{n-k}$ is bounded below and by (2.3), we have that

$$\begin{aligned} \Vert D_{-k}^{1/2} x\Vert ^2 =\Vert D^{1/2} _{-n}T^{n-k} x\Vert ^2 \ge M \Vert x\Vert ^2 . \end{aligned}$$

So, the result is obtained since $D_{-k}$ is a self adjoint operator. $\square$

We close this section by studying the bilateral sequence $(D_n)_{n\in {\mathbb {Z}}}$ associated to unilateral weighted shift which are m-isometries.

Let H be a Hilbert space with an orthonormal basis $(e_n)_{n\in {\mathbb {N}}}$. Recall that the unilateral weighted shift given by $S_we_n=w_{n} e_{n+1}$ on H, where $w_n= \displaystyle \sqrt{ \frac{p(n+1)}{p(n)}}$ with p a polynomial of degree $m-1$, is a non invertible strict m-isometry, [1]. Also

$$\begin{aligned} p_{e_j}(n)=\Vert S^n_w e_j\Vert ^2= |w_j w_{j+1} \cdots w_{n+j-1}|^2= \frac{p(j+n)}{p(j)} . \end{aligned}$$

(2.4)

The following proposition gives an explicit expression of the operator $D_n$, when T is an m-isometrical unilateral weighted shift operator.

Proposition 2.2

Let H be a Hilbert space with orthonormal basis $(e_n)_{n\in {\mathbb {N}}}$ and let $S_w\in L(H)$ be an m-isometrical unilateral weighted shift with weight sequence $w=(w_n)_{n\in {\mathbb {N}}}$. Then

• (1) $D_n$ is a diagonal operator for every $n\in {\mathbb {Z}}$, with diagonal

  $$\begin{aligned} \lambda _n(j):= \sum _{k=0}^{m-1} b_k(n) \prod _{\ell =j} ^{j+k-1} |w_{\ell } |^2 , \end{aligned}$$

  where $b_k(n)$ is giving by (2.1).

• (2) Let $n\in {\mathbb {Z}}$. The following conditions are equivalent

  • (a) $D_n$ is invertible.

  • (b) $D_n>0$.

  • (c) $\lambda _n(j)>0$ for every $j\in {\mathbb {N}}$.

Proof

(1) By [1], there exists a polynomial p of degree $m-1$, such that the weights are given by $w_n=\sqrt{\frac{p(n+1)}{p(n)}}$. So,

$$\begin{aligned} D_ne_j= & {} \sum _{k=0}^{m-1} b_k(n) S_w^{*k} S_w^k e_j = \sum _{k=0}^{m-1} b_k(n) \prod _{\ell =j}^{j+k-1} |w_\ell |^2 e_j \nonumber \\= & {} \sum _{k=0}^{m-1} b_k(n) \frac{p(j+k)}{p(j )} e_j =\lambda _n(j)e_j\,, \end{aligned}$$

(2.5)

where

$$\begin{aligned} \lambda _n(j)= \sum _{k=0}^{m-1} b_k(n) \frac{p(j+k)}{p(j )} . \end{aligned}$$

(2.6)

(2) It is immediate by (1). $\square$

In general, the converse of part (5) of Proposition 2.1 is not valid. A suitable choose of the weight sequence gives an example such that $D_{-q}>0$ and $D_{-(q+1)}$ is not positive for some $q\in {\mathbb {N}}$.

Example 2.3

Let $q\in {\mathbb {N}}$ and define $p_q(n):= (n+q)(n+q+1)$. Then $S_w$ with weight $w_n=\sqrt{\frac{p_q(n+1)}{p_q(n)}}$ is a 3-isometry and it satisfies that $D_{-n}>0$ and invertible for $n\in \{ 1, \cdots , q\}$ and $D_{-(q+1)}$ is not. In fact,

$$\begin{aligned} \lambda _{-n}(j):= \frac{p_q(j-n)}{p_q(j)} =\frac{(j+q-n)(j+q-n+1)}{(j+q)(j+q+1)} \;, \end{aligned}$$

for $n\in {\mathbb {N}}$. If $n\in \{ 1, \cdots , q\}$, then we have that $-q-1+n< -q+n < 0$. Hence, $\lambda _{-n}(j) > 0$, for every $j\in {\mathbb {N}}$. If $n=q+1$,

$$\begin{aligned} \lambda _{-(q+1)}(j) = \frac{j(j-1)}{(j+q)(j+q+1)} . \end{aligned}$$

Hence $\lambda _{-(q+1)} (1)=0$ and consequently $\langle D_{-(q+1)}e_1,e_1\rangle =0$.

3 Necessary Conditions of Having an Invertible m-isometrical Extension

In an attempt towards solution of finding necessary conditions to obtain an invertible m-isometrical extension, we draw upon an interesting connection between $D_{-1}>0$ and the invertibility of $D_{-1}$ with the existence of a particular m-isometrical extension. Notice that in the following theorem we do not obtain an invertible m-isometrical extension.

Theorem 3.1

Let $T\in L(H)$ be an m-isometry. The following statements are equivalent:

• (i) There exist a Hilbert space $K\supset H$ and an m-isometry $S\in L(K)$ such that $S_{|H}=T$ and $R(S)=H$.

• (ii) $D_{-1}> 0$ and $D_{-1}$ is invertible.

Proof

${(i)\Rightarrow (ii)}$: Let $x\in H$ and $y=S^{-1}x\in K$. For $n\in {\mathbb {Z}}$, denote

$$\begin{aligned} {\widetilde{D}}_n:= \sum _{k=0}^{m-1} b_k(n) S^{*k} S^k, \;\; \; {D}_n:= \sum _{k=0}^{m-1} b_k(n) T^{*k} T^k \end{aligned}$$

and for $n\in {\mathbb {N}}$

$$\begin{aligned} {\widetilde{p}}_x(n):=\Vert S^n x\Vert ^2, \;\; \; {p}_x(n):=\Vert T^n x\Vert ^2, \end{aligned}$$

where $b_k(n)$ is given by (2.1). Then

$$\begin{aligned} \begin{aligned} \langle {\widetilde{D}}_{-1}x, x\rangle&= \langle {\widetilde{D}}_{-1} Sy, Sy \rangle =\langle S^* {\widetilde{D}}_{-1} Sy, y\rangle =\langle {\widetilde{D}}_{0} y, y\rangle =\Vert y\Vert ^2\\&= \sum _{k=0}^{m-1} b_{k}(-1) \langle S^{*k} S^k x,x \rangle = \sum _{k=0}^{m-1} b_{k}(-1) \langle T^kx, T^k x\rangle \\&= \langle D_{-1} x, x\rangle . \end{aligned} \end{aligned}$$

Then $\langle {\widetilde{D}}_{-1} x,x \rangle =\Vert y\Vert ^2=\langle D_{-1}x,x\rangle \ge 0$ for all $x\in H$. Also

$$\begin{aligned} \Vert D_{-1}x\Vert \Vert x\Vert \ge \langle D_{-1} x,x\rangle =\Vert y\Vert ^2\ge \frac{\Vert Sy\Vert ^2}{\Vert S\Vert ^2}=\frac{\Vert x\Vert ^2}{\Vert S\Vert ^2}. \end{aligned}$$

So, $D_{-1}>0$ and bounded below. Hence $D_{-1}$ is invertible since $D_{-1}$ is self adjoint operator.

${(ii)\Rightarrow (i)}$: Consider the vector space $H\times H$ with a new seminorm

$$\begin{aligned} |||(h,h')|||:=\Vert D_{-1}^{1/2}(Th+h')\Vert \end{aligned}$$

and the subspace

$$\begin{aligned} N:=\{(h,h')\in H\times H \;\;: \;\; |||(h,h')|||=0\} . \end{aligned}$$

Let $K:=(H\times H)/N$ with the quotient norm

$$\begin{aligned} |||(h,h')+ N|||:=\Vert D_{-1}^{1/2}(Th+h')\Vert . \end{aligned}$$

Then K is a normed space. Let us prove that $|||\cdot |||$ satisfies the parallelogram law. For $u=(h,h')+N$ and $v=(g,g')+N$ in K we have

$$\begin{aligned} \begin{aligned} |||u+v|||^2+|||u-v|||^2=&\, \langle D_{-1} (Th+h' +Tg+g'),Th+h'+Tg+g'\rangle \\&+\langle D_{-1} (Th+h' -Tg-g'),Th+h'-Tg-g'\rangle \\ =&\,2\langle D_{-1} (Th+h'),Th+h'\rangle + 2\langle D_{-1} (Tg+g'),Tg+g'\rangle \\ =&\,2|||u|||^2+2|||v|||^2. \end{aligned} \end{aligned}$$

Henceforth, K is a pre-Hilbert space. The linear mapping $\phi : K \longrightarrow H$ defined by $\phi ((h,h')+N)= Th+h'$ is an isomorphism. Indeed, $\phi$ is bounded since $D_{-1}$ is an invertible operator. It is clear that $\phi$ is onto and bounded below since the square root of $D_{-1}$ is a bounded operator. Hence K is complete and so it is a Hilbert space. Moreover,

$$\begin{aligned} |||(h,0)+N|||^2= & {} \Vert D_{-1}^{1/2}(Th)\Vert ^2 =\langle D_{-1} Th,Th\rangle =\langle T^*D_{-1} Th,h\rangle \\= & {} \Vert D_0h\Vert ^2=\Vert h\Vert ^2. \end{aligned}$$

So K contains H as a subspace and we identify $h\in H$ with $(h,0)+N\in K$.

Define S on K by $\bigl ((h,h')+N\bigr ):= (Th+h',0)+N$. The operator S is well defined and bounded:

$$\begin{aligned} \begin{aligned} |||S\bigl ( (h,h')+N\bigr )|||^2&=|||(Th+h',0)+N |||^2= \Vert D_{-1}^{1/2}(T(Th+h'))\Vert ^2\\&= \langle D_{-1} (T(Th+h')), T(Th+h')\rangle = \langle D_0 (Th+h'),Th+h' \rangle \\&= \Vert Th+h'\Vert ^2\le \Vert D^{-1/2}_{-1} \Vert ^2 \Vert D_{-1}^{1/2} (Th+h')\Vert ^2 \\&= \Vert D_{-1}^{-1/2}\Vert ^2 |||(h,h')+N|||^2. \end{aligned} \end{aligned}$$

Clearly S is an extension of T. Let $h\in H$. We have identified h with $(h,0)+N\in K$ and $S((h,0)+N)=(Th,0)+N$. Also $SK=H$.

Let us prove that S is an m-isometry. Let $u=(h,h')+N\in K$ and write $y:=Th+h'\in H$. We have that $Su=(y,0)+N$, $S^ku=(T^{k-1}y,0)+N$ and $||| S^ku|||^2= \Vert D^{1/2}_{-1}(T^k y) \Vert ^2=\Vert T^{k-1} y\Vert ^2$ for $k\in {\mathbb {N}}$. So

$$\begin{aligned} \begin{aligned} \sum _{k=0}^m (-1)^k{m\atopwithdelims ()k}|||S^ku|||^2&=|||u|||^2 + \sum _{k=1}^{m} (-1)^k {m \atopwithdelims ()k} |||S^{k}u|||^2\\&= \langle D_{-1}y,y\rangle + \sum _{k=1}^m (-1)^k{m\atopwithdelims ()k}\Vert T^{k-1}y\Vert ^2\\&=\sum _{k=0}^m(-1)^k{m\atopwithdelims ()k} p_y(k-1) =0, \end{aligned} \end{aligned}$$

since $p_y$ has degree less or equal to $m-1$. Hence S is an m-isometry. $\square$

The following result gives necessary conditions of having an invertible m-isometrical extension.

Proposition 3.2

Let $T\in L(H)$ be a strict m-isometry.

• (1) If T is invertible, then $p_x(n)=\Vert T^{n} x\Vert ^2>0$ for every $x\in H\setminus \{ 0\}$ and $n\in {\mathbb {Z}}$.

• (2) If T has an invertible m-isometrical extension S, then $p_x(-k){:=}\Vert S^{-k}x\Vert ^2>0$ for every $x\in H{\setminus } \{ 0\}$ and $k\in {\mathbb {N}}$, where $p_x(n):= \Vert T^nx\Vert ^2$ for $n\in {\mathbb {N}}$. In particular, the degree of $p_x$ is even for every $x\in H\setminus \{ 0\}$.

• (3) If there exists an invertible m-isometrical extension of T, then $D_{n}>0$ and invertible operator for every $n\in {\mathbb {Z}}$.

Proof

(1) Part (3) of Proposition 2.1 yields that $T^{*n}D_{-n}T^n=D_0=I$ for $n\in {\mathbb {N}}$. So, for every $x\in H{\setminus }\{ 0\}$ and $n\in {\mathbb {N}}$,

$$\begin{aligned} p_x(-n)=\langle D_{-n}x,x\rangle =\langle T^{*-n} T^{-n}x,x\rangle =\Vert T^{-n}x\Vert ^2>0, \end{aligned}$$

since $T^{-1}$ is an m-isometry.

(2) Let $x\in H$ and $n\in {\mathbb {N}}$. Denote by

$$\begin{aligned} \begin{aligned} p_x(n):&=\langle D_nx,x \rangle :=\sum _{k=0}^{m-1} b_{k } (n) \Vert T^kx\Vert ^2\\ {\widetilde{p}}_x(n):&=\langle {\widetilde{D}}_nx,x \rangle :=\sum _{k=0}^{m-1} b_{k } (n) \Vert S^kx\Vert ^2, \end{aligned} \end{aligned}$$

where S is an invertible m-isometrical extension of T. Clearly, $p_x(n)={{\widetilde{p}}} _x(n)$ is a polynomial of degree less or equal to $m-1$. Observe that $p_x(-n)={{\widetilde{p}}}_x(-n)=\Vert S^{-n}x\Vert ^2$ for every $n\in {\mathbb {N}}$. $\square$

Remark 3.3

• (1) Observe that part (2) of the above Proposition implies that the degree of $p_x$ is even if $p_x(n)>0$ for every $n\in {\mathbb {Z}}$. Indeed, this is a different way to prove that there are no invertible strict m-isometries for even m. See also [4, Proposition 1.23].

• (2) The conditions $D_n>0$ and invertible operator for every $n\in {\mathbb {Z}}$ are not sufficient to define an invertible m-isometrical extension of T. Indeed, invertibility of $D_n$ would suffice to construct an unbounded m-isometrical extension of T with dense range.

Proposition 3.2 allow us to obtain that some m-isometries have not an invertible m-isometrical extension.

Remark 3.4

Let $T\in L(H)$ be a strict m-isometry. Denote $p_x(n):=\Vert T^nx\Vert ^2$, for $n\in {\mathbb {N}}$ and $x\in H\setminus \{ 0\}$. Then

• (1) If $m=1$, then $p_x(n)>0$ for every $x\in H{\setminus } \{0\}$ and $n\in {\mathbb {Z}}$.

• (2) If m is even, then there exist $x_0\in H$ and $n_0\in {\mathbb {Z}}$ with $n_0<0$ such that $p_{x_0} (n_0) \le 0$.

• (3) If m is odd, then it is possible that $p_x(n)>0$ for every $x\in H\setminus \{ 0\}$ and $n\in {\mathbb {Z}}$ or there exist $x_0\in H$ and $n_0\in {\mathbb {Z}}$ with $n_0<0$ such that $p_{x_0} (n_0) \le 0$.

In the following examples we present different behaviours of $p_x(n)$ with negative integer n for unilateral weighted shift.

Example 3.5

Let $p(n)= n^{m-1}$ with odd m. It is clear that $p_{e_j}(n):=\Vert S^n_we_j\Vert ^2=\displaystyle \left( \frac{j+n}{j}\right) ^{m-1}$ and $p_{e_j}(-j)=0$. So, $S_w$ can not have an invertible m-isometrical extension.

Example 3.6

Let $p(n):= \prod _{i=1}^{m-1} (mn+i)$ with odd m. It is clear that

$$\begin{aligned} p_{e_j}(n):=\Vert S^n_we_j\Vert ^2= \frac{\displaystyle \prod \nolimits _{i=1}^{m-1} (m(j+n)+i)}{\prod _{i=1}^{m-1} (mj+i)}. \end{aligned}$$

If $j\ge n$, then $p_{e_j}(-n)>0$. In other case, $p_{e_j}(-n)>0$ since $m-1$ is even. As we will see later, $S_w$ has an invertible m-isometrical extension by Theorem 5.1.

4 Characterization of Having an Invertible m-isometrical Extension

The main result of this paper is to obtain, for a fixed m-isometry, characterizations of having an invertible m-isometrical extension. In Proposition 3.2, we proved that a necessary condition is that the bilateral sequence of operators $(D_n)_{n\in {\mathbb {Z}}}$ must be strictly positive and invertible.

Now, we are in position to prove the main result.

Theorem 4.1

Let $T\in L(H)$ be an m-isometry and let $(D_n)_{n\in {\mathbb {Z}}}$ be the bilateral sequence defined by (2.2). Denote $p_x(n):=\langle D_nx,x\rangle$ for every $x\in H{\setminus } \{ 0\}$ and $n\in {\mathbb {Z}}$. The following statements are equivalent:

• (i) There exist a Hilbert space $K\supset H$ and an invertible m-isometrical operator $S\in L(K)$ such that $S_{|H}=T$.

• (ii) $p_x(j)>0$ for every $x\in H{\setminus } \{0\}$, and $j\in {\mathbb {Z}}$ and

  $$\begin{aligned} \sup \left\{ \frac{p_x(j+1)}{p_x(j)}: x\in H\setminus \{ 0\}, \;\; j\in {\mathbb {Z}}\right\} <\infty . \end{aligned}$$

  (4.7)

• (iii) $D_{n}>0$ and invertible for every $n\in {\mathbb {Z}}$, and

  $$\begin{aligned} \sup \left\{ \frac{\langle D_{-n+1}x,x\rangle }{\langle D_{-n}x,x\rangle }: x\in H,\;\; \Vert x\Vert =1, \;\; n\in {\mathbb {N}} \right\} <\infty . \end{aligned}$$

  (4.8)

Proof

${(i) \Rightarrow (ii)}$: Let $x\in H\setminus \{ 0\}$. Then

$$\begin{aligned} \Vert S^{j+1} x\Vert ^2= \Vert T^{j+1} x\Vert ^2= p_x(j+1)>0 \end{aligned}$$

for $j\in {\mathbb {Z}}$ and

$$\begin{aligned} \frac{p_x(j+1)}{p_x(j)}=\frac{\Vert S^{j+1}x\Vert ^2}{\Vert S^jx\Vert ^2}\le \Vert S\Vert ^2. \end{aligned}$$

So, we get (4.7).

${(ii) \Rightarrow (iii)}$: By parts (1) and (2) of Proposition 2.1 we have that $D_n>0$ and invertible for $n\in {\mathbb {N}}$. By hypothesis, $D_j>0$ for $j\in {\mathbb {Z}}$ since $p_x(j)=\langle D_jx,x\rangle$. Let us prove that $D_{-n}$ are bounded below for every $n\in {\mathbb {N}}$. The condition (4.7) yields that there exists $M>0$ such that

$$\begin{aligned} p_x(-n)\ge \frac{p_x(-n+1)}{M} \ge \frac{p_x(0)}{M^n}=\frac{\Vert x\Vert ^2}{M^n} \end{aligned}$$

hence

$$\begin{aligned} \Vert D^{1/2} _{-n} x\Vert ^2 \ge \frac{\Vert x\Vert ^2}{M^n}, \end{aligned}$$

for every $x\in H\setminus \{ 0\}$ and $n\in {\mathbb {N}}$. Therefore $D_{-n}$ is bounded below for $n\in {\mathbb {N}}$ and hence invertible.

It is remained to prove (4.8). Indeed, (4.8) is an immediate consequence of (4.7) using the identification $p_x(j)=\langle D_{j} x,x \rangle$ for every $x\in H{\setminus } \{ 0\}$ and $j\in {\mathbb {Z}}$.

${(iii)\Rightarrow (i)}$: Let V be the vector space of all sequences $(h_1,h_2,\dots )$ of elements of H with finite support, that is, there exists $n\in {\mathbb {N}}$ such that $h_j=0$ for $j>n$. Define a new seminorm on V by

$$\begin{aligned} |||(h_1,h_2,\dots )|||^2:=\langle D_{-n}y,y\rangle , \end{aligned}$$

where $n\in {\mathbb {N}}$ is any integer satisfying $h_j=0$ for $j>n$ and $y:=\sum _{j=1}^nT^{n-j}h_j$.

The seminorm $|||\cdot |||$ does not depend on the choice of n. Indeed, if $h_j=0$ for $j>n$, $r=n+n_0$ with $n_0\in {\mathbb {N}}$, and $y=\sum _{j=0}^nT^{n-j}h_j$, then

$$\begin{aligned} \begin{aligned}&\left\langle D_{-r} \sum _{j=1}^{r} T^{r-j }h_j, \sum _{i=1}^{r} T^{r-i }h_i\right\rangle \\&\quad = \left\langle D_{-(n+n_0)} T^{n_0} \left( \sum _{j=1}^{n+n_0} T^{n-j} h_j\right) , T^{n_0}\left( \sum _{i=1}^{n+n_0} T^{n-i }h_i\right) \right\rangle \\&\quad = \left\langle T^{*n_0} D_{-(n+n_0)} T^{n_0} \left( \sum _{j=1}^n T^{n-j} h_j\right) , \sum _{i=1}^n T^{n-i} h_i \right\rangle = \langle D_{-n}y, y \rangle \end{aligned} \end{aligned}$$

where the last equality is by part (3) of Proposition 2.1.

Let $N:=\{(h_1,h_2,\dots )\in V\;\;:\;\; |||(h_1,h_2,\dots )|||=0\}$ and let K be the completion of V/N.

Let us prove that K is a pre-Hilbert space. For that, it is enough to prove that $|||\cdot |||$ satisfies the parallelogram law. Let $u:=(h_1, h_2, \cdots ) +N, \;\; v:=(g_1, g_2, \cdots )+N \in V/N$, $n\in {\mathbb {N}}$ such that $h_j=0=g_j$ for $j>n$ and $x:= \sum _{j=1}^n T^{n-j} h_j$, $y:=\sum _{j=1}^n T^{n-j} g_j$. Then

$$\begin{aligned} \begin{aligned} |||u+v|||^2 + |||u-v|||^2&= \langle D_{-n} (x+y),x+y \rangle +\langle D_{-n} (x-y), x-y\rangle \\&= 2(|||u|||^2 + |||v|||^2). \end{aligned} \end{aligned}$$

For each $h\in H$ we have $|||(h,0,0,\dots )+N|||^2=\langle D_{-1} Th, Th \rangle =\langle D_0 h,h\rangle =\Vert h\Vert ^2$.

Let L be the closed subspace generated by $(h, 0, \cdots )+N$ with $h\in H$ and define $\phi$ on H taking values on L by $\phi (h):=(h, 0, \cdots )+ N$. Then $\Vert h\Vert ^2=||| \phi (h)|||^2$ and $R(\phi ) = L$. For each $h\in H$ we can identify h with $(h,0,\dots )+N\in K$. So, K contains H as a subspace.

Define S on V/N by $S ((h_1, h_2, \cdots )+N):=(Th_1+h_2, h_3, \cdots )+ N\in V/N$. Then the definition of S is correct and S is bounded. Indeed, let $u:=(h_1, h_2, \cdots ) + N\in V/N$, $n\in {\mathbb {N}}$ such that $h_j=0$ for $j>n$ and $y:= \sum _{j=1}^n T^{n-j} h_j$. Denote $({{\widetilde{h}}}_1, {{\widetilde{h}}}_2, \cdots ):=(Th_1+h_2, h_3, \cdots )$. Then

$$\begin{aligned} \begin{aligned} |||Su|||^2&= |||(Th_1+h_2, h_3, \cdots )+N|||^2= \langle D_{-(n-1)} {{\widetilde{y}}}, {{\widetilde{y}}} \rangle \end{aligned} \end{aligned}$$

where

$$\begin{aligned} {{\widetilde{y}}}:= \sum _{j=1}^{n-1} T^{n-1-j} {{\widetilde{h}}}_j= T^{n-1} (Th_1+h_2) + \sum _{j=2} ^{n-1} T^{n-1-j} {\widetilde{h}}_j =y \;. \end{aligned}$$

Then $|||Su|||^2=\langle D_{-(n-1)} y, y\rangle =p_y (-n+1).$ Repeating the process we have that

$$\begin{aligned} |||S^ku|||^2 = p_y (-n+k), \end{aligned}$$

for $k=0,\dots m$. Therefore

$$\begin{aligned} \sum _{k=0}^m(-1)^k{m\atopwithdelims ()k}|||S^ku|||^2= \sum _{k=0}^m(-1)^k{m\atopwithdelims ()k} p_y(-n+k)=0, \end{aligned}$$

since $p_y$ has degree less or equal to $m-1$. By continuity, S is an m-isometry.

It is easy to see that $R(S)\supset V+N$. So the range of S is dense, and consequently S is an invertible m-isometry. $\square$

Moreover, the invertible extension $S\in L(K)$ is defined uniquely (up to the unitary equivalence) if we assume that S is minimal, i.e., $K=\bigvee _{k\ge 0} S^{-k}H$.

We will prove that the converse of part (3) of Proposition 3.2 is not true in general, that is, if $D_n>0$ and invertible for $n\in {\mathbb {Z}}$ are not sufficient to have an invertible m-isometrical extension of an m-isometry. Firstly, we need a previous result on m-isometries.

Proposition 4.2

Let $(T_n)_{n\in {\mathbb {N}}}\subset L(H)$ be a uniformly bounded sequence of m-isometries. Then $T=T_1 \oplus T_2 \oplus \cdots$ is an m-isometry on $\ell ^2(H)$.

Proof

Since $(T_n)_{n\in {\mathbb {N}}}$ is a uniformly bounded, then $T=T_1 \oplus T_2 \oplus \cdots$ is well-defined on $\ell ^2(H)$.

Let $x=(x_1, x_2, \cdots )\in \ell ^2(H)$. Denote $p_{x_n}(k):=\Vert T_n^kx_n\Vert ^2$. Since $(T_n)_{n\in {\mathbb {N}}}$ is a sequence of m-isometries, then $(p_{x_n} (k))_{n\in {\mathbb {N}}}$ is a sequence of polynomials of degree less or equal to $m-1$. Fixed $k\in {\mathbb {N}}$,

$$\begin{aligned} p_x(k):=\Vert T^kx\Vert ^2= \sum _{n=1} ^\infty \Vert T^k_nx_n\Vert ^2 = \sum _{n=1}^\infty p_{x_n} (k)\; \end{aligned}$$

is a polynomial of degree less or equal to $m-1$. Hence T is an m-isometry. $\square$

It is possible to exhibit an example of m-isometry with odd m such that $D_n>0$ and invertible for every $n\in {\mathbb {Z}}$ but not fulfilling the hypothesis of Theorem 4.1. In order to simplify the presentation we include an example with a 3-isometry.

Example 4.3

Let $q_n(j):= j^2+j(2-\frac{1}{n})+1$. Let H be a Hilbert space with an orthonormal basis $(e_{n,j})_{n,j\in {\mathbb {N}}}$ and $K:=\ell ^2 (H)$. Define $T\in L(K)$ by

$$\begin{aligned} Te_{n,j}:= \sqrt{ \displaystyle \frac{q_n(j+1)}{q_n(j)}} e_{n,j+1} \end{aligned}$$

for any $n,j\in {\mathbb {N}}$. Then

• (1) T is a 3-isometry on K.

• (2) $p_x(k)>0$ for every $x\in K{\setminus }\{ 0\}$ and $k\in {\mathbb {Z}}$, where $p_x(n):=\Vert T^n x\Vert ^2$ for $n\in {\mathbb {N}}$.

• (3) $D_n>0$ and invertible for $n\in {\mathbb {Z}}$.

• (4) There is no invertible 3-isometrical extension of T.

Proof: It is clear that $q_n(j)>0$ for $n\in {\mathbb {N}}$ and $j\in {\mathbb {Z}}$.

Let $x=(x_1, x_2, \cdots )=\left( \sum _{n=1}^\infty \alpha _{n,1}e_{n,1}, \sum _{n=1}^\infty \alpha _{n,2}e_{n,2}, \cdots \right) \in K$. Then

$$\begin{aligned} T(x_1, x_2, \cdots ):= (0, T_1x_1, T_2,x_2, \cdots ), \end{aligned}$$

where

$$\begin{aligned} T_ix_i:=T_i\left( \sum _{n=1}^{\infty }\alpha _{n,i}e_{n,i}\right) = \sum _{n=1}^\infty \alpha _{n,i} w_{n,i} e_{n,i+1} \end{aligned}$$

and

$$\begin{aligned} w_{n,i}:=\sqrt{\frac{q_n(i+1)}{q_n(i)}} . \end{aligned}$$

By Proposition 4.2, the operator T is a 3-isometry, since $T_n$ is a 3-isometry for every $n\in {\mathbb {N}}$ and also $(T_n)_{n\in {\mathbb {N}}}$ is uniformly bounded, that is

$$\begin{aligned} \sup _{n\in {\mathbb {N}}} \Vert T_n\Vert \le \sup _{n,i\in {\mathbb {N}}} \sqrt{\frac{q_n(i+1)}{q_n(i)}}<M \end{aligned}$$

for some positive constant M.

Let us prove that $p_x(k)>0$ for every $x\in K{\setminus }\{ 0\}$ and $k\in {\mathbb {Z}}$. Let $x=(x_1, x_2, \cdots )=(\sum _{n=1}^\infty \alpha _{n,1}e_{n,1}, \sum _{n=1}^\infty \alpha _{n,2}e_{n,2}, \cdots )\in K{\setminus } \{ 0\}$ and $k\in {\mathbb {N}}$. Then

$$\begin{aligned} \begin{aligned} p_x(k):=\Vert T^kx\Vert ^2&= \Vert (0, \cdots , 0, T_kT_{k-1} \cdots T_1 x_1, T_{k+1}T_{k} \cdots T_2 x_2, \cdots \Vert ^2\\&= \left\| [\right\| 3]{ (0, \cdots , 0, \sum _{n=1}^\infty \alpha _{n,1} \sqrt{\frac{q_{n}(k+1)}{q_n(1)}} e_{n,k+1}, \cdots )} ^2\\&= \sum _{j=1}^\infty \left\| [\right\| 3]{ \sum _{n=1}^\infty \alpha _{n,j} \sqrt{\frac{q_{n}(k+j)}{q_n(j)}}e_{n,k+j}} ^2 \\&= \sum _{n,j=1}^\infty |\alpha _{n,j} |^2 \frac{q_n(k+j)}{q_n(j)} >0 \end{aligned} \end{aligned}$$

for $k\in {\mathbb {N}}$. Notice that

$$\begin{aligned} D_{-n}:= \frac{(n+1)(n+2)}{2} I -n(n+2)T^*T+ \frac{n(n+1)}{2} T^{*2} T^2, \end{aligned}$$

is a diagonal operator given by $D_{-n} e_{m,j}=\lambda _{-n}(k,j)e_{k,j}$ where

$$\begin{aligned} \begin{aligned} \lambda _{-n}(k,j):=&\,\frac{1}{2q_k(j)} \left( (n+1)(n+2)q_k(j)-n(n+2)q_k(j+1)+n(n+1)q_k(j+2)\right) \\ =&\, \frac{1}{2q_k(j)} \left( j^2(n^2+2n+2)+j\left( -\frac{n^2}{k} +4n^2-2\frac{n}{k} +4n -\frac{2}{k} +4\right) \right. \\&\left. -\frac{n^2}{k} +6n^2 +4n +2\right) >0 , \end{aligned} \end{aligned}$$

for $n,k,j\in {\mathbb {N}}$. So, it is immediate that $D_{-n}$ is invertible for $n\in {\mathbb {N}}$.

In order to finish the proof, let us prove that there is no invertible 3-isometrical extension of T. Taking into account that

$$\begin{aligned} \frac{p_{e_{n,1}}(-1)}{p_{e_{n,1}}(-2)}= \frac{q_n(0)}{q_n(-1)}=n, \end{aligned}$$

we have that

$$\begin{aligned} \sup \left\{ \frac{p_x(j+1)}{p_x(j)} \;:\;\; x\in K\setminus \{ 0\}, \;\; j\in {\mathbb {Z}}\right\} =\infty . \end{aligned}$$

$\Box$

5 Some Particular Cases

In this section, the goal is to study two different examples of m-isometries, the $\ell$-Jordan isometry and unilateral weighted shift that are m-isometries for some m.

In the case of unilateral weighted shift we can obtain a nice characterization of invertible m-isometrical extensions of an m-isometry, as a consequence of Theorem 4.1.

Theorem 5.1

Let H be a Hilbert space with orthonormal basis $(e_n)_{n\in {\mathbb {N}}}$ and let $S_w\in L(H)$ be an m-isometrical unilateral weighted shift associated to the weight $w:=(w_n)_{n\in {\mathbb {N}}}$. Then $S_w$ has an invertible m-isometrical extension if and only if $p_{e_1}(n)>0$ for every $n\in {\mathbb {Z}}$, where $p_{e_1} (n):=\Vert S^n_w e_1\Vert ^2$ for $n\in {\mathbb {N}}$.

Proof

If $S_w$ has an invertible m-isometrical extension S, then $p_x(n):=\Vert S^nx\Vert ^2>0$ for every $x\in H{\setminus } \{ 0\}$ and $n\in {\mathbb {Z}}$, by Proposition 3.2. Hence $p_{e_1}(n)>0$ for $n\in {\mathbb {Z}}$.

Let us prove the sufficient condition. Suppose that $p_{e_{1}}(n)>0$ for $n\in {\mathbb {Z}}$. A first consequence is that m is odd. By equality (2.4), $p_{e_1} (n)$ is a polynomial of degree $m-1$. Hence

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{ p_{e_1} (-n+1)}{p_{e_1} (-n)}=1, \end{aligned}$$

and

$$\begin{aligned} \inf \left\{ \frac{p_{e_1} (-n+1)}{p_{e_1} (-n)} \;:\;\; n\in {\mathbb {N}} \right\} >0 . \end{aligned}$$

Let K be a Hilbert space with $(e_n)_{n\in {\mathbb {Z}}}$ an orthonormal basis. Define $T_{\beta } \in L(K)$ by $T_\beta e_n=\beta _n e_{n+1}$ where $\beta _n=\sqrt{\displaystyle \frac{p_{e_1}(n)}{p_{e_1}(n-1)}}$ for $n\in {\mathbb {Z}}$. By [1, Theorem 19] we have that $T_\beta$ is an m-isometry, since $p_{e_1}(n)$ is a polynomial of degree $m-1$ by (2.4). Moreover, $T_\beta$ is an invertible extension of $S_w$ and the desired result is proved. $\square$

Remark 5.2

In the above theorem, it is possible to obtain the same information with different elements of the orthogonal basis, as a consequence of equality (2.4). Indeed, in the conditions of Theorem 5.1 the following statements are equivalent:

• (1) $S_w$ has an invertible m-isometrical extension.

• (2) $p_{e_1}(n)>0$ for $n\in {\mathbb {Z}}$.

• (3) $p_{e_j}(n)>0$ for $n\in {\mathbb {Z}}$ and some $j\in {\mathbb {N}}$.

• (4) $p_{e_j}(n)>0$ for $n\in {\mathbb {Z}}$ and $j\in {\mathbb {N}}$.

Let us obtain a first approach to $\ell$-Jordan isommetries. In the next result we obtain that any 2-Jordan isometry operator admits an invertible 3-isometric extension, as a particular case of Theorem 4.1.

Corollary 5.3

Let $T\in L(H)$ be a 2-Jordan isometry operator. Then T has an invertible 2-Jordan isometry extension.

Proof

Let T be a 2-Jordan isometry operator, that is $T=A+Q$, where A is an isometry and Q is a 2-nilpotent operator such that $AQ=QA$. By (2.2) we obtain that

$$\begin{aligned} \begin{aligned} D_{-n}&=\frac{(n+1)(n+2)}{2} I -n(n+2) T^*T+\frac{n(n+1)}{2}T^{*2}T^2 \\&= I-n(A^*Q+Q^*A) +n^2 Q^*Q . \end{aligned} \end{aligned}$$

Then

$$\begin{aligned} \langle D_{-n}x,x \rangle = \Vert x\Vert ^2-n(\langle Qx,Ax\rangle + \langle Ax, Qx\rangle ) +n^2 \Vert Qx\Vert ^2 . \end{aligned}$$

Let us prove that $\langle D_{-n} x,x\rangle >0$ for every $x\in H$ such that $\Vert x\Vert =1$ and $n\in {\mathbb {N}}$. It is enough to prove that

$$\begin{aligned} n^2\Vert Qx\Vert ^2+1> 2nRe(\langle Ax,Qx\rangle ) , \end{aligned}$$

(5.9)

where Re(z) denotes the real part of z. If $Re (\langle Ax, Qx\rangle ) \le 0$, then (5.9) is clear. Assume that $Re (\langle Ax, Qx\rangle ) > 0$. Then

$$\begin{aligned} Re (\langle Ax, Qx\rangle )=| Re (\langle Ax, Qx\rangle ) | \le |\langle Ax, Qx\rangle | \le \Vert Ax\Vert \Vert Qx\Vert \le \Vert Q\Vert . \end{aligned}$$

If $|\langle Ax, Qx \rangle |=\Vert Ax \Vert \Vert Qx\Vert$, then the vectors Ax and Qx are linearly dependent, so there exists $\lambda$ such that $Qx=\lambda Ax$. Then $\lambda =0$, since $0=\Vert Q^2x\Vert =|\lambda |^2 \Vert A^2x\Vert =|\lambda |^2$ and therefore $\Vert Qx\Vert =0$, which is an absurd with $Re (\langle Ax, Qx\rangle > 0$. If $|\langle Ax, Qx \rangle | < \Vert Ax\Vert \Vert Qx\Vert$, then

$$\begin{aligned} 2nRe(\langle Ax, Qx \rangle ) <2n \Vert Qx\Vert \le n^2 \Vert Qx\Vert ^2+1. \end{aligned}$$

So, $\langle D_{-n} x,x\rangle >0$ for every $x\in H$ such that $\Vert x\Vert =1$ and all $n\in {\mathbb {N}}$.

In order to get the result, it is enough to prove that (4.8) is bounded. Let $x\in H$ such that $\Vert x\Vert =1$ and $n\in {\mathbb {N}}$. Then

$$\begin{aligned} \begin{aligned} \frac{\langle D_{-n+1}x,x\rangle }{\langle D_{-n}x,x\rangle }&= 1+\frac{ 2Re(\langle Ax,Qx \rangle )+(-2n+1)\Vert Qx\Vert ^2}{1-2nRe(\langle Ax, Qx \rangle ) +n^2 \Vert Qx\Vert ^2}\\[1pc]&\le 1+ \left| \frac{ 2Re(\langle Ax,Qx \rangle )+(-2n+1)\Vert Qx\Vert ^2}{1-2nRe(\langle Ax, Qx \rangle ) +n^2 \Vert Qx\Vert ^2} \right| \\[1pc]&\le 1+\frac{ 2 \Vert Q\Vert + (2n-1)\Vert Q\Vert ^2}{1-2n\Vert Q\Vert -n^2 \Vert Q \Vert ^2 } \end{aligned} \end{aligned}$$

converges to zero as n tends to infinity. Hence

$$\begin{aligned} \sup \left\{ \frac{\langle D_{-n+1} x,x\rangle }{\langle D_{-n} x,x\rangle } \;\;: \;\; x\in H, \;\; \Vert x\Vert =1, \;\; n\in {\mathbb {N}} \right\} <\infty . \end{aligned}$$

$\square$

Corollary 5.4

Let $T, \; C\in L(H)$ such that $TC=CT$.

• (1) If T is an isometry, then ${\widetilde{T}}:= \left( \begin{array}{ll} T &{} C \\ 0 &{} T \end{array} \right)$ has an invertible 3-isometric extension on $K\supset H\oplus H$.

• (2) If $\lambda T$ is an isometry for some $\lambda \in {\mathbb {C}}$, then $\lambda {\widetilde{T}}=\lambda \left( \begin{array}{cc} T &{} C \\ 0 &{} T \end{array} \right)$ has an invertible 3-isometric extension on $K\supset H\oplus H$.

Proof

(1) It is clear that ${\widetilde{T}} =\left( \begin{array}{cc} T &{} 0\\ 0 &{} T\end{array} \right) +\left( \begin{array}{cc} 0 &{} C\\ 0 &{} 0\end{array} \right)$ is a 2-Jordan isometry operator. Therefore the result is consequence of Corollary 5.3.

Applying (1) to the operator $\lambda T$ we obtain (2). $\square$

A similar result of part (1) of Corollary 5.4 was obtained in [8, Corollary 4.4]. That is, if $T\in L(H)$ is a contraction and $C\in L(H)$ such that $TC=CT$, then ${\widetilde{T}}$ has a 3-isometric lifting on $K\supset H\oplus H$.

In the next theorem we can improve Corollary 5.3. Indeed, we prove that every $\ell$-Jordan isometry has an invertible $\ell$-Jordan isometry extension. The first part of our proof is based in the construction by Douglas [13], as it is presented by Laursen and Neumann in the monograph [15, Proposition 1.6,6].

Theorem 5.5

Let $T \in L(H)$ be an $\ell$-Jordan isometry. Then there exist a Hilbert space K and $S\in L(K)$, such that H is isometrically embedded in K and S is an invertible $\ell$-Jordan isometry extension of T.

Proof

As T is an $\ell$-Jordan isometry, there are an isometry $A \in L(H)$ and an $\ell$-nilpotent operator $Q \in L(H)$ such that $AQ=QA$ and $T=A+Q$.

Let $K_0$ be the linear space of all the sequences $u=(u_n)_{n\in {\mathbb {N}}}$ in H such that there is $m \in {\mathbb {N}}$ satisfying $u_{m+k} =A^{k} u_m$, for $k \in {\mathbb {N}}$. Define, for $u,v \in K_0$,

$$\begin{aligned} \langle u,v \rangle _0:= \lim _{n \rightarrow \infty } \langle u_n,v_n \rangle \;, \end{aligned}$$

being $\langle \cdot , \cdot \rangle$ the inner product on H. Note that there exists $m \in {\mathbb {N}}$ such that $\langle u_m,v_m \rangle = \langle A^k u_m, A^k v_m \rangle = \langle u_{m+k},v_{m+k} \rangle$, so the sequence $(\langle u_n,v_n \rangle )_{n\in {\mathbb {N}}}$ is eventually constant, that is, there exists $k_0\in {\mathbb {N}}$ such that $\langle u_n, v_n\rangle$ is constant for $n> k_0$. It is routine to verify what $\langle \cdot ,\cdot \rangle _0$ is a semi-inner product on $K_0$. Therefore $K_0$ is a semi pre-Hilbert space. Moreover,

$$\begin{aligned} \Vert u\Vert _0^2:= \langle u,u \rangle _0 = \lim _{n \rightarrow \infty } \langle u_n,u_n \rangle = \lim _{n \rightarrow \infty } \Vert u_n\Vert ^2 \end{aligned}$$

defines a seminorm $\Vert \cdot \Vert _0$ on $K_0$.

Let $M:= \{ u \in K_0: \langle u,u \rangle _0 = \Vert u \Vert _0^2 = 0 \}$. Then M is a closed subspace of $K_0$ and we consider the quotient space $K_0/M$. In this space are defined, for $u,v \in K_0$,

$$\begin{aligned}&\langle u+M,v+M \rangle := \langle u,v \rangle _0 \; \; \text{ and } \\&\Vert u + M\Vert ^2:= \langle u+M,u+M \rangle = \langle u,u \rangle _0 = \Vert u\Vert _0^2 \;, \end{aligned}$$

and we obtain that $K_0/M$ is a pre-Hilbert space.

Denote by K the Hilbert space what it is the completion of $K_0/M$. The operator $J \in L(H,K)$, defined by $Jx:= (A^nx)_{n\in {\mathbb {N}}}+M$ for $x \in H$, satisfies that

$$\begin{aligned} \Vert Jx \Vert = \Vert (A^nx)_{n\in {\mathbb {N}}}+M \Vert = \Vert (A^nx)_{n\in {\mathbb {N}}}\Vert _0 = \lim _{n \rightarrow \infty } \Vert A^n x\Vert = \Vert Ax\Vert = \Vert x\Vert \;, \end{aligned}$$

hence J is an isometry. So K contains an isometric copy of H. It is clear that J(H) is a closed subspace of K.

In order to define $B\in L(K)$, we define an isometry on $K_0/M$ by

$$\begin{aligned} B((u_n)_{n\in {\mathbb {N}}} +M):= (A u_n)_{n\in {\mathbb {N}}} +M \;, \end{aligned}$$

for every $(u_n)_{n\in {\mathbb {N}}} +M \in K_0/M$. Note that B is a linear isometry whose range contains $K_0/M$; in fact, given $(v_n)_{n\in {\mathbb {N}}} + M = (v_1,...,v_m,Av_m,A^2 v_m,...)+ M$, we have that

$$\begin{aligned} \begin{aligned} B((\underbrace{0,..., 0}_{m}, v_m,Av_m,A^2 v_m,...)+M)&= (\underbrace{0,..., 0}_{m},Av_m,A^2v_m,A^3 v_m,...)+M \\&= (v_1, \cdots , v_m, Av_m, A^2v_m, \cdots ) +M . \end{aligned} \end{aligned}$$

As $K_0/M$ is dense in K, we have that B can be extended to an invertible isometry defined on K. Moreover, B can be considered as an extension of A since, for $x \in H$,

$$\begin{aligned} BJx = B((A^nx)_{n\in {\mathbb {N}}} + M) = (A^{n+1}x)_{n\in {\mathbb {N}}} + M = JAx \;. \end{aligned}$$

That is, $BJ=JA$.

Define $P \in L(K)$ in the following way

$$\begin{aligned} P((u_n)_{n\in {\mathbb {N}}} + M) = (Qu_n)_{n\in {\mathbb {N}}} + M \;, \end{aligned}$$

for every $(u_n)_{n\in {\mathbb {N}}} + M \in K_0/M$. It is clear that P is an $\ell$-nilpotent. Let us prove that B and P commute. Taking into account that $AQ=QA$, we have that

$$\begin{aligned} \begin{aligned} BP ((u_n)_{n\in {\mathbb {N}}} + M )&= B ((Qu_n)_{n\in {\mathbb {N}}} + M) = (AQu_n)_{n\in {\mathbb {N}}} + M \\&= (QAu_n)_{n\in {\mathbb {N}}} + M = P ((Au_n)_{n\in {\mathbb {N}}} + M) \\&= PB ((u_n)_{n\in {\mathbb {N}}} + M) \;. \end{aligned} \end{aligned}$$

for every $(u_n)_{n \in {\mathbb {N}}} + M \in K_0/M$. Therefore, $S:=B+P\in L(K)$ is an $\ell$-Jordan isometry that extends T. Moreover, S is an invertible since $\sigma (S)=\sigma (B)$ and B is an invertible isometry. So the proof is finished. $\square$

An operator $T \in L(H)$ is a doubly $\ell$-Jordan isometry if $T=A+Q$ is an $\ell$-Jordan isometry operator such that the $\ell$-nilpotent $Q \in L(H)$ which commutes with A also commutes with $A^*$. For all scalar $\lambda$ with $|\lambda |=1$ and an $\ell$-nilpotent operator Q, we have that $\lambda I + Q$ is a doubly $\ell$-Jordan isometry.

Corollary 5.6

Let $T \in L(H)$ be a doubly $\ell$-Jordan isometry. Then there exist a Hilbert space K, such that H is isometrically embedded in K and an invertible doubly $\ell$-Jordan isometry extension $S\in L(K)$ of T.

Remark 5.7

We use the notation of the proof of Theorem 5.5.

• (1) It is easy to prove that the orthogonal subspace of J(H), $J(H)^\bot$ is the closure of the subspace of all classes

  $$\begin{aligned} (u_n)_{n\in {\mathbb {N}}} + M = (u_1,...,u_m,Au_m,A^2u_m,...) + M \in K_0/M \end{aligned}$$

  such that $u_m \in R(A^m)^\bot$.

• (2) The decomposition $K=J(H) \oplus J(H)^\bot$ gives rise to the representation of B as a operator matrix:

  $$\begin{aligned} B = \left( \begin{array}{cc} B_1 &{} B_2 \\ 0 &{} B_3 \\ \end{array} \right) \end{aligned}$$

  (5.10)

  being $B_1 \in L(J(H))$, $B_2\in L(J(H)^\bot ,J(H))$ and $B_3 \in L(J(H)^\bot )$. Notice that J(H) is a closed invariant subspace of B.

• (3) The operator P is defined by the following operator matrix, associated to the decomposition $K=J(H) \oplus J(H)^\bot$,

  $$\begin{aligned} P=\left( \begin{array}{cc} P_1 &{} P_2 \\ 0 &{} P_3 \\ \end{array} \right) \end{aligned}$$

  (5.11)

  being $P_1 \in L(J(H))$, $P_2 \in L(J(H)^\bot ,J(H))$ and $P_3 \in L(J(H)^\bot )$. Notice that J(H) is a closed invariant subspace of P.

• (4) If T is a doubly $\ell$-Jordan isometry, then $P_2=0$ in (5.11). For this purpose only it is necessary to prove that if $(u_n)_{n\in {\mathbb {N}}} + M \in J(H)^\bot$, then $P((u_n)_{n\in {\mathbb {N}}} + M ) \in J(H)^\bot$, and that $BP^*=P^*B$. In fact, given $u = (u_1,...,u_m,Au_m, A^2 u_m,...)$ such that $u_m \in R(A^m)^\bot$, we have that $Qu_m \in R(A^m)^\bot$ since, for all $x \in H$,

  $$\begin{aligned} \langle Qu_m,A^mx \rangle = \langle u_m,Q^*A^mx \rangle = \langle u_m,A^m Q^*x \rangle = 0 \;, \end{aligned}$$

  because $Q^*A= AQ^*$. Therefore $P((u_n)_{n\in {\mathbb {N}}} + M ) = (Qu_1,...,Qu_m,AQu_m, A^2Qu_m,...)+M \in J(H)^\bot$. Hence $P(J(H)^\bot ) \subset J(H)^\bot$.