1 Introduction and Preliminaries

Throughout the paper, \(\mathbb {H}\) denotes a separable infinite dimensional complex Hilbert space with inner product \(\langle \cdot \;|\;\cdot \rangle \) and \(\{e_n\}_{n\ge 0}\) is an orthonormal basis of \(\mathbb {H}\). A represents a nonzero (\(A\ne 0\)) positive operator and denote I the identity operator on \(\mathbb {H}\). By \(\mathcal {L}(\mathbb {H})\) we denote the Banach algebra of all linear operators on \(\mathbb {H}\). For every \(T \in \mathcal {L}(\mathbb {H})\) its range is denoted by R(T), its null space by N(T) and its adjoint by \(T^{*}\). The cone of positive (semi-definite) operators and the set of all \(T\in \mathcal {L}(\mathbb {H})\) which admit an A-adjoint are given, respectively, by

$$\begin{aligned} \mathcal {L}(\mathbb {H})^{+}:= & {} \{A\in \mathcal {L}(\mathbb {H}):\langle Au\;|\;u\rangle \ge 0,\;\forall \;u\in \mathbb {H}\},\\ \mathcal {L}_{A}(\mathbb {H}):= & {} \{T\in \mathcal {L}(\mathbb {H}):R(T^{*}A)\subset R(A)\}. \end{aligned}$$

Any \(A\in \mathcal {L}(\mathbb {H})^{+}\) defines a positive semi-definite sesquilinear form:

$$\begin{aligned} \langle \cdot \;|\;\cdot \rangle _{A}:\mathbb {H}\times \mathbb {H}\longrightarrow \mathbb {C},\;\;\langle u\;|\;v\rangle _{A}:=\langle Au\;|\;v\rangle . \end{aligned}$$

By \(\Vert \cdot \Vert _{A}\) we denote the semi-norm induced by \(\langle \cdot \;|\;\cdot \rangle _{A}\), i.e. \(\Vert u\Vert _{A}=\langle u\;|\;u\rangle ^{\frac{1}{2}}_{A}.\) Observe that \(\Vert u\Vert _{A}=0\) if and only if \(u\in N(A).\) Then \(\Vert \cdot \Vert _{A}\) is a norm if and only if A is an injective operator.

Definition 1.1

For a positive operator A, we say that \(u,\,v\in \mathbb {H}\) are A-orthogonal if

$$\begin{aligned} \langle u\;|\;v\rangle _{A}=\langle Au\;|\;v\rangle =0. \end{aligned}$$

More general, a family of vectors \((u_i )_i\) is A-orthogonal if \(\langle u_i\;|\;u_j\rangle _{A}=0\), for all \(i\ne j\).

For any \(T \in \mathcal {L}(\mathbb {H})\) and \(A\in \mathcal {L}(\mathbb {H})^+\), define

$$\begin{aligned} \Theta ^{(m)}_{A}(T) =\sum _{k=0}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\;T^{*k}AT^{k},\quad m \ge 1. \end{aligned}$$
(1)

We say that T is (A, m)-expansive if \(\Theta ^{(m)}_{A}(T)\le 0\) for some positive integer m. For more details on such a family, we refer the readers to [11].

A detailed study of m-isometries was developed by Agler and Stankus [1–3]. Recently, Sid Ahmed et al. [8] generalized the concept of those operators on a Hilbert space when an additional semi-inner product is considered. They introduced the (A, m)-isometries as a special case of (A, m)-expansive operators. In [9], we gave a detailed study concerning the behavior of the orbits of such a family. In fact, for \(m\in \mathbb {N}\), an operator \(T\in \mathcal {L}(\mathbb {H})\) is called an (A, m)-isometry if \(\Theta ^{(m)}_{A}(T)=0\), or equivalently,

$$\begin{aligned} \sum _{k=0}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\; \Vert T^{k}u\Vert ^{2}_{A}=0\quad \mathrm{{for\; all}} \;u\in \mathbb {H}. \end{aligned}$$
(2)

Remark 1.1

  1. 1.

    An (A, 1)-isometry will be called an A-isometry.

  2. 2.

    T will be said a strict (A, m)-isometry if it is an (A, m)-isometry but not an \((A,m-1)\)-isometry.

For \(T\in \mathcal {L}(\mathbb {H})\) and \(k=0,1,2,\ldots \), we consider the operator

$$\begin{aligned} \beta _{k}(T)=\frac{1}{k!}\;\Theta ^{(k)}_{A}(T). \end{aligned}$$

For \(n\ge k,\) we denote

$$\begin{aligned} n^{(k)}=\left\{ \begin{array}{ll} 1, &{} \hbox {if}\; n=0 \;\mathrm{{or}}\; k=0; \\ {n\atopwithdelims ()k}k!\;= n(n-1)(n-2)\ldots (n-k+1), &{} \hbox {otherwise.} \end{array} \right. \end{aligned}$$

Observe that \(\beta _{0}(T)=A\) and if T is an (A, m)-isometry, then \(\beta _{k}(T)=0\) for every \(k\ge m.\) Hence, according to [9] we have

$$\begin{aligned} \Vert T^{n}u\Vert ^{2}_{A}=\sum _{k=0}^{m-1}n^{(k)}\;\langle \beta _{k}(T)u\;|\; u\rangle \quad \hbox {for all}\; u\in \mathbb {H}. \end{aligned}$$
(3)

The A-covariance operator \(\Delta _{A,T}\) is defined by

$$\begin{aligned} \Delta _{A,T}:=\beta _{m-1}(T)=\frac{1}{(m-1)!}\;\Theta ^{(m-1)}_{A}(T). \end{aligned}$$
(4)

Theorem 2.1 [4] gives a useful characterization of m-isometries on a Hilbert space. The same proof works for (A, m)-isometries. In fact, we show that if T is an (A, m)-isometry then it is possible to explain the semi-norm \(\Vert T^n u\Vert _A\) of \(T^n u\) in terms of the semi-norms of the vectors \(u,Tu,\ldots , T^{m-1}u\).

Theorem 1.1

An operator \(T\in \mathcal {L}(\mathbb {H})\) is an (A, m)-isometry if and only if

$$\begin{aligned} \Vert T^{n}u\Vert ^{2}_{A}=\sum _{k=0}^{m-1}(-1)^{m-k-1}\frac{n(n-1)\ldots \overbrace{(n-k)}\ldots (n-m+1)}{k!(m-k-1)!}\Vert T^{k}u\Vert ^{2}_{A} \end{aligned}$$
(5)

for all \(n \ge 0\) and all \(u\in \mathbb {H},\) where \(\overbrace{n-k}\) denotes that the factor \((n-k)\) is omitted.

Remark 1.2

For \(k=0,\;1,\;2,\ldots ,(m-1)\), the coefficient of \(\Vert T^{k}u\Vert ^{2}_{A}\) is a polynomial in n of degree \((m-1)\).

The paper is organized as follows. In Sect. 2 we focus on unilateral weighted shifts which are (A, m)-isometries. A characterization in terms of the weight sequence is given for the forward shifts. Then, we describe the behavior of the weights for such a family. Finally, in order that the paper will be self-contained, we give a characterization for backward shifts.

Generally, an (A, m)-isometry is not an \((A,m-1)\)-isometry (see [8]). Inspired from that, the aim of Sect. 3 is the study of strict (A, m)-isometric unilateral weighted forward shifts. A characterization for a particular operator A is also given. In Sect. 4 we focus on (A, 2)-expansive (or A-concave) operators. Some properties related to unilateral weighted shifts are given.

2 (A, m)-Isometric Unilateral Weighted Shifts

The aim of this section is to give a characterization of unilateral weighted shift operators which are (A, m)-isometries in terms of their weight sequences. Before starting the study, first we recall that a unilateral weighted shift T is unitarily equivalent to a weighted shift operator with a non-negative weight sequence. So we can assume that \(w_n \ge 0\). Furthermore, if T is injective, it can be assumed that \(w_n > 0\) (see [6, 10]).

2.1 Unilateral Weighted Forward Shifts

An operator \(T\in \mathcal {L}(\mathbb {H})\) is said to be a unilateral weighted forward shift, if there exists an orthonormal basis \(\{e_n\}_{n\ge 0}\) and a sequence \(\{w_n\}_{n\ge 0}\) of complex numbers such that \(Te_n = w_n e_{n+1}\). It is well known and it is not difficult to see that T is a bounded operator if and only if the weight sequence \(\{w_n\}_{n\ge 0}\) is bounded. The iterates of T are given by \(T^0 = I\), and for \(k \ge 1\),

$$\begin{aligned} T^k e_n=\left( \prod _{i=0}^{k-1}w_{n+i}\right) e_{n+k},\quad n\ge 0. \end{aligned}$$
(6)

It was shown ([4], Proposition 3.1) that if T is an m-isometric unilateral weighted forward shift operator with weight sequence \(\{w_n\}_{n\ge 1}\), then \(w_n\ne 0\) for all \(n\ge 1.\) The characterization related to (A, m)-isometric unilateral weighted shifts is given in the following proposition.

Proposition 2.1

Let T be a unilateral weighted forward shift with weight sequence \(\{w_n\}_{n\ge 0}\). If T is an (A, m)-isometry and if there exists a nonnegative integer \(n_{0}\) such that \(e_{n_0} \not \in N(A)\), then \(w_{n_0} \ne 0\).

Proof

Assume that there exists a nonnegative integer \(n_0\) such that \(\Vert e_{n_0}\Vert _A \ne 0\). Equation (6) implies that

$$\begin{aligned} \Vert T^k e_{n_0}\Vert ^{2}_{A}=\left( \displaystyle \prod _{i=0}^{k-1}|w_{i+n_0}|^{2}\right) \Vert e_{k+n_0}\Vert ^{2}_{A},\;\;k\ge 1. \end{aligned}$$

Since \(T^0 e_{n_0}=e_{n_0}\), we obtain

$$\begin{aligned} (-1)^{m+1}\Vert e_{n_0}\Vert _{A}^{2}&=\sum _{k=1}^{m}(-1)^{m-k}\;{m\atopwithdelims ()k}\Vert T^k e_{n_0}\Vert ^{2}_{A}\\&=\sum _{k=1}^{m}(-1)^{m-k}\;{m\atopwithdelims ()k}\left( \displaystyle \prod _{i=0}^{k-1}|w_{i+n_0}|^{2}\right) \Vert e_{k+n_0}\Vert ^{2}_{A}\\&=|w_{n_0}|^{2}\Big [(-1)^{m-1}\;{m\atopwithdelims ()1}\Vert e_{n_0 +1}\Vert _{A}^{2}\\&\quad +\sum _{k=2}^{m}(-1)^{m-k}\;{m\atopwithdelims ()k} \left( \displaystyle \prod _{i=1}^{k-1}|w_{i+n_0}|^{2}\right) \Vert e_{k+n_0}\Vert ^{2}_{A}\Big ] \end{aligned}$$

which gives

$$\begin{aligned} |w_{n_0}|^{2}=\frac{(-1)^{m+1}\Vert e_{n_0}\Vert _{A}^{2}}{(-1)^{m-1}\;{m\atopwithdelims ()1}\Vert e_{n_0 +1}\Vert _{A}^{2}+\sum _{k=2}^{m}(-1)^{m-k}\;{m\atopwithdelims ()k} \left( \displaystyle \prod _{i=1}^{k-1}|w_{i+n_0}|^{2}\right) \Vert e_{k+n_0}\Vert ^{2}_{A}}. \end{aligned}$$

Thus, the proof is achieved.\(\square \)

Corollary 2.1

Let T be a unilateral weighted forward shift with weight sequence \(\{w_n\}_{n\ge 0}\). The following assertions hold true.

  1. 1.

    If T is an A-isometry and if \(e_0 \not \in N(A)\), then \(w_n \ne 0\) for all \(n\ge 0.\)

  2. 2.

    If T is an (A, 2)-isometry and \(e_{2p} \not \in N(A)\) for all \(p\ge 0\), then \(w_n \ne 0\) for all \(n\ge 0\).

Proof

  1. 1.

    If T is an A-isometry then \(\Vert Tu\Vert _A =\Vert u\Vert _A\) for all \(u\in \mathbb {H}\). For \(u=e_n\), we obtain

    $$\begin{aligned} |w_n|\Vert e_{n+1}\Vert _A =\Vert e_n\Vert _A \end{aligned}$$

    which implies that if \(e_0 \not \in N(A)\), then \(w_n \ne 0\) for all \(n\ge 0.\)

  2. 2.

    To obtain the desired claim we restrict ourselves to prove that, for each n, if \(e_{n} \not \in N(A)\) then \(w_n \ne 0\) and \(w_{n+1} \ne 0.\) Since T is an (A, 2)-isometry, \(\Vert T^2 u\Vert ^{2}_{A}-2\Vert T u\Vert ^{2}_{A}+\Vert u\Vert ^{2}_A=0\) for all \(u\in \mathbb {H}\). For \(u=e_n\), we obtain

    $$\begin{aligned} |w_n|^2 \Big \{|w_{n+1}|^2\Vert e_{n+2}\Vert ^{2}_{A}-2\Vert e_{n+1}\Vert ^{2}_{A}\Big \}=-\Vert e_{n}\Vert ^{2}_{A}. \end{aligned}$$
    (7)

    If \(e_n \not \in N(A)\), then (7) implies that \(w_n \ne 0\) and

    $$\begin{aligned} |w_{n+1}|\Vert e_{n+2}\Vert _{A}\pm \sqrt{2}\Vert e_{n+1}\Vert _{A} \ne 0. \end{aligned}$$
    (8)

    Moreover, if we assume that \(w_{n+1}=0\), then the identity (8) gives \(e_{n+1}\not \in N(A)\), and hence \(w_{n+1}\ne 0\) which is a contradiction.\(\square \)

In the following proposition we establish that if a unilateral weighted forward shift T with weight sequence \(\{w_n\}_{n\ge 0}\) is A-bounded then the sequence \(\Big \{ \frac{\Vert e_{n+1}\Vert _{A}}{\Vert e_{n}\Vert _{A}}\;w_n ,\;n\ge 0\Big \}\) is bounded.

Proposition 2.2

Let \(T\in \mathcal {L}_{A}(\mathbb {H})\) be a unilateral weighted forward shift with weight sequence \(\{w_k\}_{k\ge 0}\). Assume that \(e_k \not \in N(A)\) for all \(k\in \mathbb {N}\). Then, for all \(n\ge 1\)

$$\begin{aligned} \Vert T^n\Vert _A\ge \sup _{k\in \mathbb {N}} \left\{ \Big (\prod _{i=0}^{n-1}|w_{i+k}|\Big )\;\frac{\Vert e_{k+n}\Vert _{A}}{\Vert e_{k}\Vert _{A}}\right\} . \end{aligned}$$
(9)

Proof

Note first that if \(T\in \mathcal {L}_{A}(\mathbb {H})\), then \(T^n \in \mathcal {L}_{A}(\mathbb {H})\) for all \(n\ge 0\) and, moreover, we have

$$\begin{aligned} \Vert T^n u\Vert _A\le \Vert T^n\Vert _A\Vert u\Vert _A\quad \mathrm{{for\;all}}\; u\in \mathbb {H},\;n\ge 0. \end{aligned}$$

Let \(n\ge 1\). Since (6) holds true, we have

$$\begin{aligned} \left( \prod _{i=0}^{n-1}|w_{i+k}|\right) \Vert e_{k+n}\Vert _{A}=\Vert T^n e_{k}\Vert _A\le \Vert T^n \Vert _A\Vert e_{k}\Vert _A. \end{aligned}$$

Thus taking supremum over all \(k \in \mathbb {N}\), we obtain

$$\begin{aligned} \sup _{k\in \mathbb {N}}\left\{ \Big (\prod _{i=0}^{n-1}|w_{i+k}|\Big )\;\frac{\Vert e_{k+n}\Vert _{A}}{\Vert e_{k}\Vert _{A}}\right\} \le \Vert T^n \Vert _A. \end{aligned}$$

Remark 2.1

Equality between the two parts of (9) does not holds true for every unilateral weighted forward shift T and any positive operator A. For example, assume that \(n = 1\) and \(w_k = 1\) for all \(k \ge 0\). Let A be the operator given in the orthonormal basis \(\{e_k\}_{k\ge 0}\) by the matrix \(\{a_{jk}\}_{j,k\ge 0}\) such that \(a_{jj} = 1\) for all \(j \ge 0\), \(a_{01} = a_{10} = \frac{1}{2}\), and all other elements are equal to 0. It is easily seen that A is an invertible positive operator. Moreover, \(T\in \mathcal {L}_{A}(\mathbb {H})\). Remark that the right part of (9) is equal to 1. Indeed,

$$\begin{aligned} \Vert e_{k}\Vert ^{2}_{A}=\langle Ae_k \,|\,e_k\rangle =a_{kk} =1,\quad \mathrm{{for\;all}}\;k\ge 0. \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \Vert e_{0}-e_{1}\Vert ^{2}_{A}= & {} \langle A(e_{0}-e_{1})\,|\,(e_{0}-e_{1})\rangle \\= & {} a_{00}-2a_{01}+a_{11} = 1\\ \mathrm{{and}}\quad \Vert T(e_{0}-e_{1})\Vert ^{2}_{A}= & {} \Vert e_{1}-e_{2}\Vert ^{2}_{A}\\= & {} \langle A(e_{1}-e_{2})\,|\,(e_{1}-e_{2})\rangle \\= & {} a_{11} - 2a_{12} + a_{22} = 2. \end{aligned}$$

Consequently, \(\Vert T\Vert _A \ge \sqrt{2}.\)

Every \(u\in \mathbb {H}\) can be written as \(u=\displaystyle \sum \nolimits _{n\ge 0}\;\alpha _n e_{n}\); \(\{\alpha _n\}_{n\ge 0}\subset \mathbb {C}\). Hence, we have

$$\begin{aligned} \sum _{k=0}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\; \Vert T^{k}u\Vert ^{2}_{A}=\sum _{i,j\ge 0}\alpha _i \overline{\alpha }_j\;S^{(m)}_{i,j,A}, \end{aligned}$$

where

$$\begin{aligned} S^{(m)}_{i,j,A}=\sum _{k=0}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\;\langle AT^k e_i\,|\,T^k e_j\rangle ,\quad i,j\ge 0. \end{aligned}$$

Remark 2.2

If T is a unilateral weighted forward shift with weight sequence \(\{w_n\}_{n\ge 0}\), then we have

$$\begin{aligned} S^{(m)}_{i,j,A}=\left\{ \begin{array}{llll} (-1)^{m}\langle Ae_i\,|\,e_j\rangle +\displaystyle \sum _{k=1}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\; \Big (\displaystyle \prod _{p=0}^{k-1}w_{i+p}\overline{w}_{j+p}\Big )\langle Ae_{i+k}\,|\,e_{j+k}\rangle ,\quad (i\ne j) &{} \hbox {} \\ (-1)^{m}\Vert e_i\Vert ^{2}_{A}+\displaystyle \sum _{k=1}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\; \Big (\displaystyle \prod _{p=0}^{k-1}|w_{i+p}|^{2}\Big )\Vert e_{i+k}\Vert ^{2}_{A}:=S^{(m)}_{i,A},\quad (i= j). &{} \hbox {} \end{array} \right. \end{aligned}$$

Let us begin with the following theorem.

Theorem 2.2

Let T be a unilateral weighted forward shift with weight sequence \(\{w_n\}_{n\ge 0}\). Then T is an (A, m)-isometry if and only if \(S^{(m)}_{i,j,A}=0\) for all \(i,j\ge 0.\)

Proof

Assume that T is an (A, m)-isometry. For \(u=e_n\), (2) implies that

$$\begin{aligned} 0= & {} \sum _{k=0}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\; \Vert T^{k}e_n\Vert ^{2}_{A}\\= & {} (-1)^{m}\Vert e_n\Vert _{A}^{2}+\displaystyle \sum _{k=1}^{m}(-1)^{m-k}{m\atopwithdelims ()k}\left( \ \prod _{i=0}^{k-1}|w_{n+i}|^{2}\right) \Vert e_{n+k}\Vert _{A}^{2}. \end{aligned}$$

Hence, for all \(n\in \mathbb {N}\) we have \(S_{n,A}^{(m)}=0\). As a consequence, for all sequence \(\{\alpha _n\}_n \subset \mathbb {C}\), it yields that \(\sum _{i\ne j}\alpha _i \overline{\alpha }_j\;S^{(m)}_{i,j,A}=0\). Indeed, to prove such a claim we argue by contradiction. If there exists \(\{\alpha _n\}_n\) such that \(\displaystyle \sum \nolimits _{i\ne j}\alpha _i \overline{\alpha }_j\;S^{(m)}_{i,j,A}\ne 0,\) then if we take a nonzero \(v=\sum _{i}\;\alpha _i e_{i}\) we obtain

$$\begin{aligned} \displaystyle \sum _{k=0}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\; \Vert T^{k}v\Vert ^{2}_{A}\ne 0, \end{aligned}$$

which contradicts the fact that T is an (A, m)-isometry. Moreover, assume that there exists \(i_0 \ne j_0\) such that \(S^{(m)}_{i_0 ,j_0 ,A}\ne 0\). If we consider the sequence \(\{\beta _n\}_n\) defined by \(\beta _{i_0}\ne 0\), \(\beta _{j_0}\ne 0\) and \(\beta _{i}=0\) otherwise, then for \(s=\beta _{i_0}e_{i_0}+\beta _{j_0}e_{j_0},\) we obtain

$$\begin{aligned} \sum _{k=0}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\; \Vert T^{k}s\Vert ^{2}_{A}=\beta _{i_0} \overline{\beta _{j_0}}\;S^{(m)}_{i_0 ,j_0 ,A}\ne 0, \end{aligned}$$

which contradicts the hypothesis. Hence, \(S^{(m)}_{i,j,A}=0\) for all \(i\ne j.\)

For the converse, suppose that \(S^{(m)}_{i,j,A}=0\) for all \(i,j\ge 0\). Since every vector \(u\in \mathbb {H}\) can be written as \(u=\sum _{i}\;\alpha _i e_{i}\), it follows that

$$\begin{aligned} \sum _{k=0}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\; \Vert T^{k}u\Vert ^{2}_{A}=\sum _{i,j}\alpha _i \overline{\alpha }_j\;S^{(m)}_{i,j,A}=0. \end{aligned}$$

\(\square \)

In the following proposition we give some properties related to the coefficients \(S^{(m)}_{i,j,A}\) defined in Remark 2.2.

Proposition 2.3

Let T be an (A, m)-isometric unilateral weighted forward shift with weight sequence \(\{w_n\}_{n\ge 0}\). The following claims hold true.

  1. 1.

    If \(S_{0,A}^{(m-1)}=0\), then \(S_{n,A}^{(m-1)}=0\) for all \(-1\le n-1< \inf \{n \ge 0/w_n =0\}\), where \(\inf \{n \ge 0/w_n =0\}=+\infty \) if \(\{n \ge 0/w_n =0\}=\Phi \).

  2. 2.

    If \(e_{n}\not \in N(A)\) for all \(n\in \mathbb {N}\) and \( S_{0,A}^{(m-1)}=0\), then \(S_{n,A}^{(m-1)}=0\) for all \(n\in \mathbb {N}\).

Proof

  1. 1.

    First of all, let us remark that

    $$\begin{aligned} S_{n,A}^{(m-1)}=(m-1)!\;\langle \Delta _{A,T} e_n \;|\;e_n\rangle \quad \mathrm{{for\;all}}\;n\ge 0. \end{aligned}$$

    Since T is an (A, m)-isometry, then a simple computation shows that

    $$\begin{aligned} \triangle _{A,T}=T^* \triangle _{A,T} T . \end{aligned}$$

    Moreover,

    $$\begin{aligned} \langle \Delta _{A,T} e_n \;|\;e_n\rangle= & {} \langle T^* \triangle _{A,T} T e_n \;|\;e_n\rangle \\= & {} \langle \Delta _{A,T} Te_n \;|\;Te_n\rangle \\= & {} |w_n|^2 \;\langle \Delta _{A,T} e_{n+1} \;|\;e_{n+1}\rangle . \end{aligned}$$

    This implies that

    $$\begin{aligned} S_{n,A}^{(m-1)}=|w_n|^2\;S_{n+1,A}^{(m-1)}. \end{aligned}$$
    (10)

    As a consequence of (10), we have

    $$\begin{aligned} S_{0,A}^{(m-1)}=\prod _{i=0}^{n}|w_i|^2\;S_{n+1,A}^{(m-1)}\quad \mathrm{{for\;all}}\;n\ge 0. \end{aligned}$$

    Hence, we obtain the result.

  2. 2.

    The claim is a direct consequence of (10) and Proposition 2.1.\(\square \)

Corollary 2.3

Let T be a unilateral weighted forward shift with weight sequence \(\{w_n\}_{n\ge 0}\). Let \(H^{(m)}_{i,j,A}\) be as

$$\begin{aligned} H^{(m)}_{i,j,A}:=\left( \displaystyle \prod _{p=0}^{i}w_{p}\right) \left( \displaystyle \prod _{q=0}^{j}\overline{w}_{q}\right) S^{(m)}_{i,j,A},\quad i,j \ge 0. \end{aligned}$$
(11)

Then, we have

  1. 1.

    If T is an (A, m)-isometry, then

    1. (a)

      \(H^{(m)}_{i,A} :=H^{(m)}_{i,i,A} =0\) for all \(i\ge 0\).

    2. (b)

      \(\sum _{i, j}H^{(m)}_{i,j,A}=0\) for all \(i,j \ge 0\).

  2. 2.

    If, for all \(i,j\ge 0\) \(H^{(m)}_{i,j,A} =0\), then either T is an (A, m)-isometry or there exists \(k\ge 0\) such that \(w_k =0\).

Proof

  1. 1.

    Assuming that T is an (A, m)-isometry, (a) and (b) follow immediately from 1. and 2. of Proposition 2.3.

  2. 2.

    Suppose that for all \(i,j\ge 0\) \(H^{(m)}_{i,j,A} =0\), i.e

    $$\begin{aligned} \left( \displaystyle \prod _{p=0}^{i}w_{p}\right) \left( \displaystyle \prod _{q=0}^{j}\overline{w}_{q}\right) S^{(m)}_{i,j,A}=0 \end{aligned}$$

    which implies that either there exists \(0\le k\le \max (i,j)\) such that \(w_k=0\) or \(S^{(m)}_{i,j,A}=0\) for all \(i,j\ge 0\) and, hence, we can conclude.\(\square \)

Remark 2.3

Let \(\{f_n\}_{n\ge 1}\) be an orthonormal basis (i.e \(\Vert f_n\Vert =1\ne 0,\) \(n\ge 1)\) and \(Tf_n = w_n f_{n+1}\) for all \(n\ge 1\). Assume that \(A=I\). Then Proposition 2.1 holds, that is \(w_n\ne 0\) for all \(n\ge 1.\) Moreover, we consider

$$\begin{aligned} \widetilde{H}^{(m)}_{i,j,A}:=\left( \displaystyle \prod _{p=0}^{i-1}w_{p}\right) \left( \displaystyle \prod _{q=0}^{j-1}\overline{w}_{q}\right) S^{(m)}_{i,j,A},\quad i,j \ge 1, \end{aligned}$$

where \(w_0 :=1.\) If \(\widetilde{H}^{(m)}_{n,A} =0\) for all \(n\ge 1\), then assertion 2. of Proposition 2.3 implies that T is an m-isometry. Hence, in that case we recover Proposition 3.2 ([4]), that is T is an m-isometry if and only if

$$\begin{aligned} \displaystyle \sum _{k=0}^{m}(-1)^{m-k}\;{m\atopwithdelims ()k}\left( \displaystyle \prod _{i=0}^{n+k-1}|w_{i}|^{2}\right) =0\quad \mathrm{{for\;all}}\;n\ge 1. \end{aligned}$$

For a fixed sequence \(\{w_k\}_{k\ge 0}\), \(n\ge 0\) and \(m\ge 1\), let us denote

$$\begin{aligned} R^{(m)}_{n,A}:= & {} (-1)^{m-1}\frac{(n-1)(n-2)\ldots (n-m+1)}{(m-1)!}\;\Vert e_{0}\Vert _{A}^{2}\nonumber \\&+\sum _{k=1}^{m-1}(-1)^{m-k-1}\frac{n(n-1)\ldots \overbrace{(n-k)}\ldots (n-m+1)}{k!(m-k-1)!}\left( \displaystyle \prod _{i=0}^{k-1}|w_{i}|^{2}\right) \Vert e_{k}\Vert _{A}^{2},\nonumber \\ \end{aligned}$$
(12)

where \(\overbrace{n-k}\) denotes that the factor \((n-k)\) is omitted. Remark that \(R^{(m)}_{0,A}=\Vert e_{0}\Vert _{A}^{2}\) and for \(j =1,2,\ldots ,m - 1,\)

$$\begin{aligned} R^{(m)}_{j,A}=\left( \displaystyle \prod _{i=0}^{j-1}|w_{i}|^{2}\right) \Vert e_{j}\Vert _{A}^{2}\;\; (\ge 0). \end{aligned}$$
(13)

As it was shown in [4] for m-isometric weighted shifts, in the following result we describe the behavior of a given unilateral weighted forward shift which is (A, m)-isometric by means of his weight sequence sequence.

Theorem 2.4

Let T be a unilateral weighted forward shift with weight sequence \(\{w_n\}_{n\ge 0}\). The following claims hold true.

  1. 1.

    Assume that there exists \(0\le n_0\le m-2\) such that \(e_{j}\not \in N(A),\;0\le j\le n_0 +1\). If T is an (A, m)-isometry then

    $$\begin{aligned} |w_{i}|^{2}=\frac{R^{(m)}_{i+1,A}}{R^{(m)}_{i,A}}\frac{\Vert e_{i}\Vert _{A}^{2}}{\Vert e_{i+1}\Vert _{A}^{2}}\;(>0)\qquad \hbox {for}\;\;0\le i\le n_0. \end{aligned}$$
    (14)
  2. 2.

    Assume that there exists \(n_0\ge 0\) such that \(e_{i}\not \in N(A)\), \(\;0\le i\le n_0 +m\) and

    $$\begin{aligned} |w_{p}|^{2}=\frac{R^{(m)}_{p+1,A}}{R^{(m)}_{p,A}}\frac{\Vert e_{p}\Vert _{A}^{2}}{\Vert e_{p+1}\Vert _{A}^{2}}\;(>0)\qquad \hbox {for}\;\;0\le p\le n_0 +m-1. \end{aligned}$$
    (15)

    Then \(H^{(m)}_{j,A} =0\) for all \(1\le j\le n_0\).

Proof

  1. 1.

    Assume that T is an (A, m)-isometry. From Proposition 2.1 we obtain \(w_i \ne 0\) for any \(0\le i\le n_0 +1\). By Theorem 1.1 we have that, for every \(u\in \mathbb {H}\) and for all \(n \ge 0\),

    $$\begin{aligned} \Vert T^{n}u\Vert ^{2}_{A}=\sum _{k=0}^{m-1}(-1)^{m-k-1}\frac{n(n-1)\ldots \overbrace{(n-k)}\ldots (n-m+1)}{k!(m-k-1)!} \Vert T^{k}u\Vert ^{2}_{A}. \end{aligned}$$

    For \(u=e_0\) and for all \(n\ge 1\), we obtain

    $$\begin{aligned} \Big (\displaystyle \prod _{i=0}^{n-1}|w_{i}|^{2}\Big )\Vert e_{n}\Vert _{A}^{2}= & {} (-1)^{m-1}\frac{(n-1)(n-2)\ldots (n-m+1)}{(m-1)!}\;\Vert e_{0}\Vert _{A}^{2}\\&+\sum _{k=1}^{m-1}(-1)^{m-k-1}\frac{n(n-1)\ldots \overbrace{(n-k)}\ldots (n-m+1)}{k!(m-k-1)!}\\&\times \left( \displaystyle \prod _{i=0}^{k-1}|w_{i}|^{2}\right) \Vert e_{k}\Vert _{A}^{2}. \end{aligned}$$

    The equalities (12)–(13) give \(R^{(m)}_{k,A}\ne 0\) for all \(0\le k\le m-1\). Moreover, we have

    $$\begin{aligned} R^{(m)}_{j,A}\Vert e_{j+1}\Vert _{A}^{2}|w_{j}|^{2}=R^{(m)}_{j+1,A}\Vert e_{j}\Vert _{A}^{2}\quad \hbox {for all }\;0\le j\le n_0. \end{aligned}$$

    Consequently,

    $$\begin{aligned} |w_{j}|^{2}=\frac{R^{(m)}_{j+1,A}}{R^{(m)}_{j,A}}\frac{\Vert e_{j}\Vert _{A}^{2}}{\Vert e_{j+1}\Vert _{A}^{2}}\;(>0)\qquad (0\le j\le n_0). \end{aligned}$$
  2. 2.

    Assume that (15) is verified. First note that

    $$\begin{aligned} R^{(m)}_{j+k,A}=\left( \displaystyle \prod _{i=0}^{j+k-1}|w_{i}|^{2}\right) \Vert e_{j+k}\Vert _{A}^{2}\ne 0\quad (j\ge 1,\,k\ge 0). \end{aligned}$$

    For \(0\le j\le n_0\), we have

    $$\begin{aligned} H^{(m)}_{j,A}= & {} \left( \displaystyle \prod _{i=0}^{j}|w_{i}|^{2}\right) S^{(m)}_{j,A}\\= & {} \left( \displaystyle \prod _{i=0}^{j}|w_{i}|^{2}\right) \left[ (-1)^{m} \Vert e_j\Vert ^{2}_{A}+\displaystyle \sum _{k=1}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\;\left( \displaystyle \prod _{i=0}^{k-1}|w_{i+j}|^{2}\right) \Vert e_{j+k}\Vert ^{2}_{A}\right] \\= & {} (-1)^{m}\left( \displaystyle \prod _{i=0}^{j}|w_{i}|^{2}\right) \Vert e_j\Vert ^{2}_{A}+|w_{j}|^{2}\displaystyle \sum _{k=1}^{m}(-1)^{m-k}\;{m\atopwithdelims ()k}\left( \displaystyle \prod _{i=0}^{j+k-1}|w_{i}|^{2}\right) \\&\times \Vert e_{j+k}\Vert ^{2}_{A}. \end{aligned}$$

    Taking into account (13) and for \(1\le j\le n_0\), we obtain

    $$\begin{aligned} H^{(m)}_{j,A}= & {} |w_{j}|^{2}\Big [\displaystyle \sum _{k=0}^{m}(-1)^{m-k}\;{m\atopwithdelims ()k}R^{(m)}_{j+k,A}\Big ]\\= & {} |w_{j}|^{2}\Big \{\displaystyle \sum _{k=0}^{m}(-1)^{m-k}\;{m\atopwithdelims ()k}(-1)^{m-1}\\&\times \frac{(j+k-1)(j+k-2)\ldots (j+k-m+1)}{(m-1)!}\Vert e_{0}\Vert ^{2}_{A}\\&+\displaystyle \sum _{k=0}^{m}(-1)^{m-k}\;{m\atopwithdelims ()k}\\&\times \Big [\sum _{h=1}^{m-1}(-1)^{m-h-1}\frac{(j+k)\ldots \overbrace{(j+k-h)}\ldots (j+k-m+1)}{h!(m-h-1)!}\\&\times \displaystyle \prod _{i=0}^{h-1}|w_{i}|^{2}\Vert e_{h}\Vert _{A}^{2}\Big ]\Big \}\\= & {} |w_{j}|^{2}(A+B), \end{aligned}$$

    where,

    $$\begin{aligned} A= & {} \displaystyle \sum _{k=0}^{m}(-1)^{m-k}\;{m\atopwithdelims ()k}(-1)^{m-1}\\&\times \frac{(j+k-1)(j+k-2)\ldots (j+k-m+1)}{(m-1)!}\Vert e_{0}\Vert ^{2}_{A}\\= & {} (-1)^{m-1}m\Vert e_{0}\Vert ^{2}_{A}\\&\times \sum _{k=0}^{m}(-1)^{m-k}\frac{(j+k-1)(j+k-2)\ldots (j+k-m+1)\overbrace{(j+k-m)}}{k!(m-k)!},\\ B= & {} \displaystyle \sum _{k=0}^{m}(-1)^{m-k}\;{m\atopwithdelims ()k}\\&\times \Big [\sum _{h=1}^{m-1}(-1)^{m-h-1}\frac{(j+k)\ldots \overbrace{(j+k-h)}\ldots (j+k-m+1)}{h!(m-h-1)!}\\&\times \displaystyle \prod _{i=0}^{h-1}|w_{i}|^{2}\Vert e_{h}\Vert _{A}^{2}\Big ]\\= & {} \displaystyle \sum _{h=1}^{m-1}(-1)^{m-h-1}\frac{m!}{h!(m-h-1)!}\;\displaystyle \prod _{i=0}^{h-1}|w_{i}|^{2}\Vert e_{h}\Vert _{A}^{2}\\&\times \Big [\sum _{k=0}^{m}(-1)^{m-k}\frac{(j+k)\ldots \overbrace{(j+k-h)}\ldots (j+k-m+1)}{k!(m-k)!}\Big ].\\ \end{aligned}$$

    Taking into account equality (3.2) [4], Lemma 3.3 we obtain \(A=0\) and \(B=0\). Hence, \(H^{(m)}_{j,A}=0\) for \(1\le j\le n_0\).\(\square \)

Remark 2.4

Let T be the unilateral weighted forward shift with weight sequence \(\{w_n\}_{n\ge 0}\). From Theorem 2.4 we obtain the following characterizations:

  1. 1.

    Assume that \((e_{n})_{n\ge 0}\) is A-orthogonal, \(e_{n}\not \in N(A)\) for all \(n\ge 0\) and \(S^{(m)}_{0,A}=0\). Then, T is an (A, m)-isometry if and only if

    $$\begin{aligned} R^{(m)}_{n,A}=\left( \displaystyle \prod _{i=0}^{n-1}|w_{i}|^{2}\right) \Vert e_{n}\Vert _{A}^{2}>0\quad \hbox {for every}\;n\ge 1. \end{aligned}$$

    If \(A=I\), then we obtain a conclusion similar to that of Remark 3.5 ([4]).

  2. 2.

    Assume that there exists \(0\le n_0\le m-2\) such that \(e_{j}\not \in N(A),\;0\le j\le n_0 +1\). We have:

    1. (a)

      If T is an A-isometry, then

      $$\begin{aligned} \Vert e_{i}\Vert _{A}=|w_{i}|\,\Vert e_{i+1}\Vert _{A}\quad \hbox {for every}\; i\ge 0 . \end{aligned}$$
    2. (b)

      If T is an (A, 2)-isometry, then

      $$\begin{aligned} |w_{i}|^2= & {} \frac{(i+1)|w_{0}|^2 -i\frac{\Vert e_{0}\Vert _{A}^{2}}{\Vert e_{1}\Vert _{A}^{2}}}{i|w_{0}|^2 -(i-1)\frac{\Vert e_{0}\Vert _{A}^{2}}{\Vert e_{1}\Vert _{A}^{2}}} \;\;\frac{\Vert e_{i}\Vert _{A}^{2}}{\Vert e_{i+1}\Vert _{A}^{2}}\quad (0\le i\le n_0)\\= & {} \frac{i\Big (|w_{0}|^2 -\frac{\Vert e_{0}\Vert _{A}^{2}}{\Vert e_{1}\Vert _{A}^{2}}\Big )+|w_{0}|^2}{(i-1)\Big (|w_{0}|^2 -\frac{\Vert e_{0}\Vert _{A}^{2}}{\Vert e_{1}\Vert _{A}^{2}}\Big )+|w_{0}|^2 }\;\;\frac{\Vert e_{i}\Vert _{A}^{2}}{\Vert e_{i+1}\Vert _{A}^{2}}. \end{aligned}$$

      Moreover, observe that

      $$\begin{aligned} |w_{0}|\;\ge \; \frac{\Vert e_{0}\Vert _{A}}{\Vert e_{1}\Vert _{A}}\Longleftrightarrow |w_{i}|>0 \quad (0\le i\le n_0). \end{aligned}$$

      In particular, if \(|w_{0}|>\frac{\Vert e_{0}\Vert _{A}}{\Vert e_{1}\Vert _{A}}\), then T is a strict (A, 2)-isometry.

2.2 Unilateral Weighted Backward Shifts

A unilateral weighted backward shift \(B_w\) with weight sequence \(\{w_n\}_{n\ge 1}\) is defined by \(B_w e_n = w_n e_{n-1}\) if \(n \ge 1\) and by \(B_w e_o= 0\). The iterates of \(B_w\) are given by

$$\begin{aligned} B^{k}_{w} e_n = \left\{ \begin{array}{ll} \displaystyle \Big (\prod _{i=0}^{k-1}w_{n-i}\Big )e_{n-k} &{} \hbox {if }1\le k\le n; \\ 0 &{} \hbox {otherwise.} \end{array} \right. \end{aligned}$$
(16)

Let \(u=\displaystyle \sum _{n=0}^{\infty }\alpha _n e_n \in \mathbb {H}\). We have

$$\begin{aligned}&\sum _{k=0}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\; \Vert B^{k}_{w}u\Vert ^{2}_{A}\\= & {} \displaystyle \sum _{n=0}^{\infty }|\alpha _n|^2 \Big \{(-1)^{m}\; \Vert e_n\Vert ^{2}_{A}+\sum _{k=1}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\; \Vert B^{k}_{w}e_{n}\Vert ^{2}_{A}\Big \}\\&+\;\sum _{i\ne j}\alpha _i \overline{\alpha }_j\Big \{(-1)^{m}\langle Ae_i\,|\,e_j\rangle +\sum _{k=1}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\;\langle AB^{k}_{w}e_{i}\,|\,B^{k}_{w}e_{j}\rangle \Big \}, \end{aligned}$$

where

$$\begin{aligned} \langle AB^{k}_{w} e_i\,|\,B^{k}_{w} e_j\rangle = \left\{ \begin{array}{ll} \displaystyle \left( \prod _{p=0}^{k-1}w_{i-p}\overline{w}_{j-p}\right) \langle Ae_{i-k}\,|\,e_{j-k}\rangle &{} \hbox {if }1\le k\le i<j; \\ 0 &{} \hbox {if } k>\min (i,j). \end{array} \right. \end{aligned}$$
(17)

Unilateral weighted backward shifts cannot be m-isometric for any positive integer number m, since \(B_w\) does not satisfies Eq. (2) for the vector \(e_0\). We prove in the following result that they cannot also be (A, m)-isomeric.

Theorem 2.5

A unilateral weighted backward shift can not be an (A, m)-isometry for any positive integer m.

Proof

Let \(B_w\) be a unilateral weighted backward shift with weight sequence \(\{w_n\}_{n\ge 1}\). Assume that \(B_w\) is an (A, m)-isometry. Then \(e_n\in N(A)\) for all \(n\ge 0.\) Indeed, we use an induction argument to prove such a claim. Since \(B_w\) is an (A, m)-isometry, Eq. (2) gives

$$\begin{aligned} (-1)^{m}\; \Vert e_n\Vert ^{2}_{A}+\sum _{k=1}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\;\Vert B^{k}_{w}e_n\Vert ^{2}_{A}=0\quad \mathrm{{for \;all}}\; n\ge 0. \end{aligned}$$
(18)

Since \(B^{k}_{w} e_0 =0\) for all \(k\ge 1\), we obtain

$$\begin{aligned} 0=\sum _{k=1}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\; \Vert B^{k}_{w}e_0\Vert ^{2}_{A}=(-1)^{m+1}\; \Vert e_0\Vert ^{2}_{A}, \end{aligned}$$
(19)

which implies that \(e_0 \in N(A)\) and, hence, the property is satisfied for \(n=0.\) Assume that for \(p=0,\) 1,  2,  \(\ldots ,\) \(n-1\) (\(n\ge 1\)), \(e_p \in N(A)\) and let us show the property for the step n. Taking (17) and (18) into account, we deduce

$$\begin{aligned}&(-1)^{m}\; \Vert e_n\Vert ^{2}_{A}+\sum _{k=1}^{n}(-1)^{m-k} \;{m\atopwithdelims ()k}\;\displaystyle \left( \prod _{i=0}^{k-1}|w_{n-i}|^{2}\right) \; \Vert e_{n-k}\Vert ^{2}_{A}=0 \quad \mathrm{{if}}\;1\le n \le m,\\&(-1)^{m}\; \Vert e_n\Vert ^{2}_{A}+\sum _{k=1}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\;\displaystyle \left( \prod _{i=0}^{k-1}|w_{n-i}|^{2}\right) \; \Vert e_{n-k}\Vert ^{2}_{A}=0 \quad \mathrm{{if}}\;n\ge m+1, \end{aligned}$$

and the claim follows from that. On the other hand, every \(u\in \mathbb {H}\) can be written as \(u=\displaystyle \sum \nolimits _{i=0}^{\infty }\alpha _i e_{i}\). Hence,

$$\begin{aligned} \langle Au\,|\,u\rangle =\displaystyle \sum _{i=0}^{\infty }|\alpha _i|^{2}\Vert e_i\Vert ^{2}_{A} +\displaystyle \sum _{0\le i\ne j}\alpha _i \overline{\alpha _j}\langle Ae_i \,|\,e_j\rangle =0. \end{aligned}$$

According to Proposition 2.15, [5], we obtain \(A=0\) which is impossible.\(\square \)

3 Strict (A, m)-Isometric Unilateral Weighted Forward Shifts

Generally, an (A, m)-isometry is not an \((A,m-1)\)-isometry. Sid Ahmed et al. (Theorem 2.1 [8]) proved that if T is an (A, m)-isometry satisfying \(N(\triangle _{A,T})\) is invariant for A, then \(T|_{N(\triangle _{A,T})}\) is an \((A|_{N(\triangle _{A,T})},m-1)\)-isometry. Moreover we can see that if T is an invertible (A, m)-isometry and m is even, then it is an \((A,m-1)\)-isometry. In the following we describe when a unilateral weighted shift operator is a strict (A, m)-isometry in terms of the weights sequence. To be precise we define

$$\begin{aligned} Z_n :=|w_n|^{2}S_{n+1,A}^{(m-1)}=\displaystyle \sum _{k=0}^{m-1}(-1)^{m-1-k}\;{m-1\atopwithdelims ()k}\left( \prod _{i=0}^{k}|w_{n +i}|^{2}\right) \Vert e_{n +k+1}\Vert _{A}^{2},\;\hbox {for all}\;n\ge 0. \end{aligned}$$

Theorem 3.1

Let T be an (A, m)-isometric unilateral weighted forward shift with weight sequence \(\{w_n\}_{n\ge 0}\). If there exists \(n_0 \in \mathbb {N}\) such that \(Z_{n_0} \ne 0\) then \(w_{n_{0}}\ne 0\) and T is a strict (A, m)-isometry.

Proof

Two proofs for this theorem will be given.

First proof Since there exists \(n_0 \in \mathbb {N}\) such that \(Z_{n_0} \ne 0\), we have \(w_{n_{0}}\ne 0\) and \(S_{n_0 +1,A}^{(m-1)}\ne 0\). Hence, Theorem 2.2 allows to conclude.

Second proof Let us prove the converse. Assume that T is an (A, m)-isometry and an \((A,m-1)\)-isometry. Then, for \(u=e_n\), the identity (2) gives

$$\begin{aligned} 0= & {} (-1)^{m}\Vert e_n\Vert _{A}^{2}+\displaystyle \sum _{k=1}^{m}(-1)^{m-k}{m\atopwithdelims ()k}\left( \prod _{i=0}^{k-1}|w_{n+i}|^{2}\right) \Vert e_{n+k}\Vert _{A}^{2}, \end{aligned}$$
(20)
$$\begin{aligned} 0= & {} (-1)^{m}\Vert e_n\Vert _{A}^{2}+\displaystyle \sum _{k=1}^{m-1}(-1)^{m-k}{m-1\atopwithdelims ()k}\left( \prod _{i=0}^{k-1}|w_{n+i}|^{2}\right) \Vert e_{n+k}\Vert _{A}^{2}. \end{aligned}$$
(21)

Remarking that \({m\atopwithdelims ()k}-{m-1\atopwithdelims ()k}={m-1\atopwithdelims ()k-1}\), the identities (20) and (21) give \(Z_n =0\) for all \(n\ge 0.\) As a consequence, if there exists \(n_0 \in \mathbb {N}\) such that \(Z_{n_0} \ne 0\) then either T is an (A, m)-isometry and not an \((A,m-1)\)-isometry or T is not an (A, m)-isometry and it is an \((A,m-1)\)-isometry. The second conclusion is impossible since it is well known that every \((A,m-1)\)-isometry is an (A, m)-isometry. Hence T is a strict (A, m)-isometry.\(\square \)

We have the following result.

Theorem 3.2

Let T be a unilateral weighted forward shift with weight sequence \(\{w_n\}_{n\ge 0}\) \((w_n \ne 0\) for all \(n\in \mathbb {N})\). Assume that \((e_n )_{n\ge 0}\) is A-orthogonal. If T is a strict (A, m)-isometry then for every nonnegative integer n, \(S_{n,A}^{(m)}=0\) and \(S_{n,A}^{(m-1)}\ne 0\).

Proof

Note that if \((e_n )_{n\ge 0}\) is A-orthogonal, then \(S_{i,j,A}^{(m-1)}=0\) for all \(i\ne j\ge 0\). Suppose that T is a strict (A, m)-isometry. Then, Eq. (2) gives

$$\begin{aligned} S_{n,A}^{(m)}=(-1)^{m}\Vert e_n\Vert _{A}^{2}\!+\!\displaystyle \sum _{k=1}^{m}(-1)^{m-k}{m\atopwithdelims ()k}\left( \prod _{i=0}^{k-1}|w_{n+i}|^{2}\right) \Vert e_{n+k}\Vert _{A}^{2}=0\quad \hbox {for all}\;n\ge 0. \end{aligned}$$

Let us prove now that \(S_{n,A}^{(m-1)}\ne 0\) for all \(n\ge 0\). Assume the contrary, that \(n_0\) is the smallest non-negative integer such that \(S_{n_{0},A}^{(m-1)}=0\). Our aim is to prove that \(n_0 =0\). If \(n_0 \ge 1\) we obtain \(S_{n_{0}-1,A}^{(m-1)}\ne 0,\) which is impossible from (10). Furthermore, Proposition 2.3 yields T is an \((A,m-1)\)-isometry. Hence we obtain the desired result.\(\square \)

As an immediate corollary to Theorem 3.2, we characterize unilateral weighted forward shifts that are strictly (A, m)-isometric.

Corollary 3.3

Let T be a unilateral weighted forward shift with weight sequence \(\{w_n\}_{n\ge 0}\). Assume that \((e_n )_{n\ge 0}\) is A-orthogonal. Then, the following assertions are equivalent:

  1. 1.

    T is a strict (A, m)-isometry.

  2. 2.

    For every \(n\in \mathbb {N}\), we have

    1. (a)

      \(S_{n,A}^{(m)}=0\).

    2. (b)

      \(S_{n,A}^{(m-1)}\ne 0\).

Proof

Suppose that T is a strict (A, m)-isometry. Then Theorem 3.2 shows that 2. holds. Suppose, now, that (a) and (b) hold true. Since \((e_n )_{n \ge 0}\) is an orthonormal basis, then every \(u\in \mathbb {H}\) can be written as \(u=\displaystyle \sum \nolimits _{i=0}^{\infty }\alpha _i e_{i}\). According to (a), we have

$$\begin{aligned} \sum _{k=0}^{m}(-1)^{m-k} \;{m\atopwithdelims ()k}\; \Vert T^{k}u\Vert ^{2}_{A}=\sum _{i=0}^{\infty }|\alpha _i|^2 S_{i,A}^{(m)}=0. \end{aligned}$$

This implies that (2) holds for every \(u\in \mathbb {H}\) and hence T is an (A, m)-isometry. On the other hand, for all \(n\ge 0\),

$$\begin{aligned} \sum _{k=0}^{m-1}(-1)^{m-k} \;{m-1\atopwithdelims ()k}\; \Vert T^{k}e_n\Vert ^{2}_{A}=S_{n,A}^{(m-1)}\ne 0. \end{aligned}$$

Thus, T is not an \((A,m-1)\)-isometry.\(\square \)

We now apply Corollary 3.3 to investigate the following example.

Example 3.1

Let \(\mathbb {H}\) be a separable Hilbert space with an orthonormal basis \(\{e_n\}_{n\ge 0}\). Let T, \(A \in \mathcal {L}(\mathbb {H})\) where T is the unilateral weighted shift defined by

$$\begin{aligned} Te_n =\sqrt{\frac{n+3}{n+1}}\;e_{n+1},\quad n \ge 0 \end{aligned}$$

and A is the positive operator given by \(Ae_n=\frac{n+1 }{n+2}\;e_n\), \(n \ge 0\). It is not difficult to verify that

$$\begin{aligned} S_{n,A}^{(3)}=0, \qquad S_{n,A}^{(2)}=\frac{2}{(n+1)(n+2)},\quad n \ge 0 \end{aligned}$$

and hence T is an (A, 3)-isometry which is not an (A, 2)-isometry.

In general, (A, m)-isometries are not A-isometries. Theorem 2.3 [9] gives a first characterization of (A, m)-isometric operators which are A-isometries. In the same way, we will describe in the following result the related characterization for a family of unilateral weighted shifts.

Proposition 3.1

Let T be a unilateral weighted forward shift with weight sequence \(\{w_n\}_{n\ge 0}\) \((w_n\ne 0\) for all \(n\in \mathbb {N})\). Assume that \((e_n )_{n\ge 0}\) is A-orthogonal. If T is an (A, m)-isometry and if, for some nonzero \(u\in \mathbb {H}\), we have

$$\begin{aligned} \Vert u\Vert _{A}=\Vert T^{k}u\Vert _{A},\quad 1\le k\le m-1, \end{aligned}$$
(22)

then T is an A-isometry.

Proof

Suppose that \(\{e_n\}_{n\ge 0}\) is an orthonormal basis for \(\mathbb {H}\) and \(Te_n = w_n e_{n+1}\) for all \(n \ge 0\). Put \(u=\displaystyle \sum \nolimits _{n=0}^{\infty }\alpha _n e_{n}\). Then, (22) gives

$$\begin{aligned} \sum _{k=0}^{m-1}(-1)^{m-k-1} \;{m-1\atopwithdelims ()k}\; \Vert T^{k}u\Vert ^{2}_{A}= & {} \sum _{k=0}^{m-1}(-1)^{m-k-1} \;{m-1\atopwithdelims ()k}\; \Vert u\Vert ^{2}_{A}\\= & {} \Vert u\Vert ^{2}_{A}\left( \ \sum _{k=0}^{m-1}\;{m-1\atopwithdelims ()k}\;(-1)^{m-k-1}\right) \\= & {} \Vert u\Vert ^{2}_{A}\Big (1+(-1)\Big )^{m-1}\\= & {} 0. \end{aligned}$$

Moreover, since \((e_n )_{n\ge 0}\) is A-orthogonal, (4) implies that

$$\begin{aligned} \sum _{k=0}^{m-1}(-1)^{m-k-1} \;{m-1\atopwithdelims ()k}\; \Vert T^{k}u\Vert ^{2}_{A}= & {} \displaystyle \sum _{n=0}^{\infty }|\alpha _n|^{2}\Bigg [(-1)^{m-1}\Vert e_n\Vert _{A}^{2}\\&+\displaystyle \sum _{k=1}^{m-1}(-1)^{m-1-k}{m-1\atopwithdelims ()k}\\&\times \left( \ \prod _{i=0}^{k-1}|w_{n+i}|^{2}\right) \Vert e_{n+k}\Vert _{A}^{2}\Bigg ]\\= & {} \displaystyle \sum _{n=0}^{\infty }|\alpha _n|^{2}S_{n,A}^{(m-1)}\\= & {} \displaystyle \sum _{n=0}^{\infty }|\alpha _n|^{2}\Big ((m-1)!\langle \Delta _{A,T} e_n\,|\,e_n\rangle \Big ). \end{aligned}$$

According to Theorem 2.1 [8], \(\Delta _{A,T}\) is a positive operator (i.e \(\langle \Delta _{A,T} e_n\,|\,e_n\rangle \ge 0,\) for all \(n\ge 0\)). Since u is non-zero, \(\alpha _{n_0}\ne 0\) for some \(n_0\), then

$$\begin{aligned} S_{n_{0},A}^{(m-1)}= & {} (m-1)!\langle \Delta _{A,T} e_{n_0}\,|\,e_{n_0}\rangle =0. \end{aligned}$$

Thus, condition (b) in Corollary 3.3 does not occur and so T must be an \((A,m-1)\)- isometry. Now, applying an argument similar to the above and using Corollary 3.3, \((m-1)\) times, we conclude that T must be an A-isometry. \(\square \)

Remark 3.1

The conclusion of Proposition 3.1 is not valid, in general, for any operator T and any operator A. Indeed, let \(\mathbb {K}=\mathbb {C}^{2}\) be equipped with the norm \(\Vert (x,y)\Vert ^2=|x|^2+|y|^2\) and consider the operators

$$\begin{aligned} A=\left( \begin{array}{c@{\quad }c} 1 &{} 1 \\ 1 &{} 2 \\ \end{array} \right) ,\qquad T=\left( \begin{array}{c@{\quad }c} 1 &{} 1 \\ 0 &{} 1 \\ \end{array} \right) . \end{aligned}$$

Note that \(A\in \mathcal {L}(\mathbb {H})^{+}\) and \(T\in \mathcal {L}(\mathbb {H}).\) Moreover, by direct computation, we see that

$$\begin{aligned} \Vert (x,y)\Vert ^{2}_{A}= & {} x^2 +2xy+2y^2,\\ \Vert T(x,y)\Vert ^{2}_{A}= & {} x^2 +4xy+5y^2,\\ \Vert T^2 (x,y)\Vert ^{2}_{A}= & {} x^2 +6xy+10y^2,\\ \Vert T^3 (x,y)\Vert ^{2}_{A}= & {} x^2 +8xy+17y^2. \end{aligned}$$

Consequently,

$$\begin{aligned}&\Vert T^3 (x,y)\Vert ^{2}_{A}-3\Vert T^2 (x,y)\Vert ^{2}_{A}+3\Vert T (x,y)\Vert ^{2}_{A}-\Vert (x,y)\Vert ^{2}_{A}=0,\\&\Vert T^2 (x,y)\Vert ^{2}_{A}-2\Vert T (x,y)\Vert ^{2}_{A}+\Vert (x,y)\Vert ^{2}_{A}=2y^2\ne 0\\ \mathrm{{and}}&\Vert T (x,y)\Vert ^{2}_{A}-\Vert (x,y)\Vert ^{2}_{A}=2xy+3y^2\ne 0. \end{aligned}$$

Thus, T is an (A, 3)-isometry but is nor an (A, 2)-isometry neither an A-isometry. Furthermore, if \(u=(1,0)\) then \(\Vert u\Vert _{A}=\Vert Tu\Vert _{A}=\Vert T^2 u\Vert _{A}=1\).

4 (A, 2)-Expansive Weighted Shift Operators

An operator \(T \in \mathcal {L}(\mathbb {H})\) is said to be (A, 2)-expansive (also A-concave), if it satisfies the inequality

$$\begin{aligned} \Theta ^{(2)}_{A}(T) =T^{*2} AT^2 -2T^* AT + A\le 0 \end{aligned}$$

or, equivalently,

$$\begin{aligned} \Vert T^{2}u\Vert ^{2}_{A}-2\Vert Tu\Vert ^{2}_{A}+\Vert u\Vert ^{2}_{A}\le 0 \quad \mathrm{{for\; all}} \;u \in \mathbb {H}. \end{aligned}$$
(23)

As a matter of fact, (A, 2)-isometries and A-isometries are (A, 2)-expansive operators. In addition, it was shown in [11], Proposition 3.9 that all powers of an (A, 2)-expansive operator is also (A, 2)-expansive.

In this section we give some properties of (A, 2)-expansive operators that generalizes those described in [7].

The A-covariance operator for an (A, 2)-expansive operator is given by \(\Delta _{A,T} = T^* AT - A\). We begin with the following preliminary result.

Lemma 4.1

Let \(T \in \mathcal {L}(\mathbb {H})\). Then,

T is (A, 2)-expansive    if and only if    \(T^* \Delta _{A,T} T\le \Delta _{A,T}.\)

Proof

Let \(u\in \mathbb {H}.\) We have

$$\begin{aligned} \langle (T^* \Delta _{A,T} T-\Delta _{A,T})u,u\rangle =\Vert T^{2}u\Vert ^{2}_{A}-2\Vert Tu\Vert ^{2}_{A}+\Vert u\Vert ^{2}_{A} \end{aligned}$$

thus the claim follows from that.\(\square \)

Jung et al. proved in [11], Theorem 3.10 that the A-covariance operator for an (A, 2)-expansive operator is positive. In the following theorem we show the same claim using different approach.

Theorem 4.1

Let \(T \in \mathcal {L}(\mathbb {H})\). If T is (A, 2)-expansive, then:

  1. 1.

    \(\Delta _{A,T} \) is a positive operator.

  2. 2.

    If A is injective then T is also injective.

  3. 3.

    If T is invertible, then \(T^{-1}\) is (A, 2)-expansive.

Proof

  1. 1.

    Let \(u\in \mathbb {H}\). We have

    $$\begin{aligned} \langle \Delta _{A,T} u,u\rangle =\langle (T^* AT - A)u,u\rangle =\Vert Tu\Vert ^{2}_{A}-\Vert u\Vert ^{2}_{A}. \end{aligned}$$

    To obtain the claim let us suppose on the contrary that \(\Vert Tu_0\Vert _{A}<\Vert u_0\Vert _{A}\) for some \(u_0\in \mathbb {H}\). By using on induction argument we obtain

    $$\begin{aligned} \Vert T^n u_0\Vert ^{2}_{A}<\Vert T^{n-1} u_0\Vert ^{2}_{A}<\cdots <\Vert u_0\Vert ^{2}_{A} \end{aligned}$$

    for each positive integer number n. We deduce that the sequence \(\{\Vert T^n u_0\Vert ^{2}_{A}\}_{n\ge 0}\) is strictly decreasing, bounded and hence it is convergent. Moreover we have

    $$\begin{aligned} 0=\displaystyle \lim _{n\longrightarrow +\infty }(\Vert T^{n+1} u_0\Vert ^{2}_{A}-\Vert T^n u_0\Vert ^{2}_{A})<0 \end{aligned}$$

    which is a contradiction. Thus \(\langle \Delta _{A,T} u,u\rangle \ge 0\) for all \(u\in \mathbb {H}\).

  2. 2.

    If \(T \in \mathcal {L}(\mathbb {H})\) is (A, 2) -expansive and \(u \in N(T)\) thus \(Tu = 0\). Moreover, \(\Vert Tu\Vert ^{2}_{A}=\Vert T^2 u\Vert ^{2}_{A}=0.\) Since T is A-concave, \(\Vert u\Vert _{A}=0\) which is equivalent to the fact that \(u\in N(A)\).

  3. 3.

    By hypothesis, T is (A, 2)-expansive. Then, we have

    $$\begin{aligned} \sum _{k=0}^{2}(-1)^{k} \;{2\atopwithdelims ()k}\; \Vert T^{2-k}u\Vert ^{2}_{A}\le 0\quad \mathrm{{for\;all}}\;u\in \mathbb {H}. \end{aligned}$$
    (24)

    Replacing u by \((T^{-1})^{2}u\) in (24), we deduce that

    $$\begin{aligned} \sum _{k=0}^{2}(-1)^{k} \;{2\atopwithdelims ()k}\; \Vert T^{2-k}((T^{-1})^{2}u)\Vert ^{2}_{A}= & {} \sum _{k=0}^{2}(-1)^{k} \;{2\atopwithdelims ()2-k}\; \Vert T^{-k}u\Vert ^{2}_{A}\\= & {} \sum _{k=0}^{2}(-1)^{2-k} \;{2\atopwithdelims ()k}\; \Vert (T^{-1})^{2-k}u\Vert ^{2}_{A}\\= & {} \sum _{k=0}^{2}(-1)^{k} \;{2\atopwithdelims ()k}\; \Vert (T^{-1})^{2-k}u\Vert ^{2}_{A}. \end{aligned}$$

    \(\square \)

Proposition 4.1

Let \(T \in \mathcal {L}(\mathbb {H})\) be an A-isometry and \(S\in \mathcal {L}(\mathbb {H})\) with \(ST = TS\), then ST is (A, 2)-expansive if and only if S is (A, 2)-expansive.

Proof

Since T is an A-isometry, we have

$$\begin{aligned} \Vert T^{k}S^{k}u\Vert _{A}=\Vert S^{k}u\Vert _{A}, \quad k = 0,\; 1,\; 2. \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \sum _{k=0}^{2}(-1)^{k} \;{2\atopwithdelims ()k}\; \Vert (ST)^{2-k}u\Vert ^{2}_{A}= & {} \sum _{k=0}^{2}(-1)^{k} \;{2\atopwithdelims ()k}\; \Vert (TS)^{2-k}u\Vert ^{2}_{A}\\= & {} \sum _{k=0}^{2}(-1)^{k} \;{2\atopwithdelims ()k}\; \Vert T^{2-k}S^{2-k}u\Vert ^{2}_{A}\\= & {} \sum _{k=0}^{2}(-1)^{k} \;{2\atopwithdelims ()k}\; \Vert S^{2-k}u\Vert ^{2}_{A}\\ \end{aligned}$$

which allows us to conclude.\(\square \)

Corollary 4.2

Let \(T \in \mathcal {L}(\mathbb {H})\) be an inversible (A, 2)-expansive operator and \(S\in \mathcal {L}(\mathbb {H})\) with \(ST = TS\), then ST is (A, 2)-expansive if and only if S is (A, 2)-expansive.

Proof

It suffises to prove that T is an A-isometry. If T is an invertible (A, 2)-expansive operator, then \(T^{-1}\) is also (A, 2)-expansive. Hence, by (1.)-Theorem 4.1 \(\Delta _{A,T}\ge 0\) and

$$\begin{aligned} \Delta _{A,T^{-1}} = T^{-1*} AT^{-1} - A\ge 0. \end{aligned}$$

On the other hand,

$$\begin{aligned} \Delta _{A,T} = T^* AT - A=-T^{*}(T^{-1*} AT^{-1} - A)T\le 0, \end{aligned}$$

which implies that T is an A-isometry. Thus we complete the proof by involving Proposition 4.1.\(\square \)

Now, we specify the study to unilateral weighted shifts. We give results generalizing those described in [7].

Theorem 4.3

Let T be a unilateral weighted forward shift with weights \(\{w_n \}_{n\ge 0}\). Assume that for all \(n\in \mathbb {N}\), \(e_n \not \in N(A)\). If T is (A, 2)-expansive, then the following assertions holds.

  1. 1.

    \(S^{(2)}_{n,A}=|w_{n}|^{2}|w_{n+1}|^{2}\Vert e_{n+2}\Vert ^{2}_{A}-2|w_{n}|^{2}\Vert e_{n+1}\Vert ^{2}_{A}+\Vert e_{n}\Vert ^{2}_{A}\le 0\)   for each n;

  2. 2.

    \(\{V_n\}_n =\Big \{|w_n|\frac{\Vert e_{n+1}\Vert _{A}}{\Vert e_{n}\Vert _{A}}\Big \}_n\) is a decreasing sequence of real numbers converging to 1;

  3. 3.

    \(\frac{\Vert e_{n}\Vert _{A}}{\Vert e_{n+1}\Vert _{A}}\le |w_n|<\sqrt{2}\frac{\Vert e_{n}\Vert _{A}}{\Vert e_{n+1}\Vert _{A}},\)   for all \(n\ge 0.\)

Proof

  1. 1.

    Applying (23) for \(u=e_n\) we obtain the claim.

  2. 2.

    Let \(V_n =|w_n|\frac{\Vert e_{n+1}\Vert _{A}}{\Vert e_{n}\Vert _{A}}\). To prove the assertion let us assume the contrary that \(V_n<V_{n+1}\) for some non-negative integer n. Therefore,

    $$\begin{aligned} 0\le \Big (|w_n|^2\frac{\Vert e_{n+1}\Vert ^{2}_{A}}{\Vert e_{n}\Vert _{A}}-\Vert e_{n}\Vert _{A}\Big )^2<S^{(2)}_{n,A}\le 0, \end{aligned}$$

    which is a contradiction. Hence \(\{V_n\}_n\) is a decreasing sequence of non-negative numbers. On the other hand, by Theorem 4.1 the operator \(\Delta _{A,T}\) is positive, that is

    $$\begin{aligned} \langle \Delta _{A,T} u,u\rangle =\Vert Tu\Vert ^{2}_{A}-\Vert u\Vert ^{2}_{A}\ge 0\quad \mathrm{{for\;all}}\;u\in \mathbb {H}. \end{aligned}$$
    (25)

    Thus for \(u=e_n\) the identity (25) gives

    $$\begin{aligned} \Vert Te_n\Vert _{A}=|w_n|\Vert e_{n+1}\Vert _{A}\ge \Vert e_{n}\Vert _{A}, \end{aligned}$$

    which implies that \(V_n \ge 1\) for all \(n \ge 0.\) Since the sequence \(\{V_n\}_n\) is decreasing, it must be convergent. Let \(\textit{l}=\displaystyle \lim \nolimits _{n\longrightarrow +\infty }V_n\). Our aim now is to prove that \(l=1\). Taking into account that \(S^{(2)}_{n,A}\le 0\), it is easily seen that \(V_n\) satisfies

    $$\begin{aligned} V^{2}_{n} V^{2}_{n+1}-2V^{2}_{n}+1\le 0\quad \mathrm{{for \;all}}\;n\ge 0. \end{aligned}$$
    (26)

    It holds that

    $$\begin{aligned} \displaystyle \lim _{n\longrightarrow +\infty }(V^{2}_{n} V^{2}_{n+1}-2V^{2}_{n}+1)=(l^2 -1)^2\le 0 \end{aligned}$$

    and hence \(l=1\).

  3. 3.

    The identity (26) implies

    $$\begin{aligned} V^{2}_{n+1}-2=(V_{n+1}-\sqrt{2})(V_{n+1}+\sqrt{2})\le -\frac{1}{V^{2}_{n}}<0, \end{aligned}$$

    so \(1 \le V_n <\sqrt{2}\), for each \(n\ge 0\), which allows us to conclude.\(\square \)

Theorem 4.4

A unilateral weighted backward shift cannot be (A, 2)-expansive.

Proof

We argue by contradiction. Assume that \(B_w\) is (A, 2)-expansive. Since \(B_w\) is a unilateral weighted backward shift, \(B_w e_n =w_n e_{n-1}\) for all \(n\ge 1\) and \(B_w e_0=0\). On the other hand, (23) gives

$$\begin{aligned} \Vert B_{w} ^{2}e_n\Vert ^{2}_{A}-2\Vert B_w e_n\Vert ^{2}_{A}+\Vert e_n\Vert ^{2}_{A}\le 0\quad \mathrm{{for\;all}}\;n\ge 0. \end{aligned}$$
(27)

It is easily seen that for \(i=0,1, e_i\in N(A)\). Using (27) we prove by an induction argument that \(e_n\in N(A)\) for all \(n\ge 2\). Hence \(A=0\) which is impossible.\(\square \)