Generalized bilateral shifts of higher multiplicity

Munmun Hazarika can be contacted at: munmun@tezu.ernet.in

Declarations: The authors have no competing interests to declare that are relevant to the content of this article.

Purpose

Since weighted shifts play a vital role in linear dynamics so in 2021 Chan & Sanders considered weighted shift operators on $ℓ^{p} (Z)$ that are not hypercyclic and proved that under certain conditions they can be factorized as products of hypercyclic shifts. The purpose of this article is to extend their results to operator weighted shifts on $ℓ$ ²(K) by using the concept of generalized shift of higher multiplicity developed herein.

Design/methodology/approach

Traditionally, an operator T is considered as a shift on a space $H$ if there is some canonical basis for $H$ such that T shifts every basis vector to the immediate next basis vector, maybe with some weight attached to it. This clearly indicates that an operator which is a shift with respect to a basis for $H$ may not remain a shift if the basis is changed. This motivates the definition of a generalized shift operator. We begin with the case of multiplicity one and then extend it to higher finite multiplicity. We then develop the idea so that we can frame conditions under which a bilateral shift can be factorized as product of hypercyclic generalized shifts.

Findings

A generalized bilateral backward weighted shift (GBBWS) is defined in terms of a bijection on $Z$ and it is shown that if there are two bijections σ and ρ which generate the same generalized shift, then there exists a unique c such that σ(i) = ρ(i + c) for all integers i. We determine conditions under which the direct sum of generalized shifts is again a generalized shift. We also show that for a uniformly bounded sequence of invertible diagonal operators {A_i} on a separable complex Hilbert space K of finite dimension, if W is bilateral backward weighted shift on ℓ²(K) with weight sequence {A_i} then there exists hypercyclic GBBWS T and P on ℓ²(K) such that W = TP.

Originality/value

The idea of generalized shift of multiplicity one was introduced by Chan & Sanders in 2018. However, we have developed the idea further, particularly extending it to the case of shifts of higher multiplicity. The factorization of shifts as product of hypercyclic shifts was also done by them. We have extended their result to higher dimension, and our approach to the problem is completely different from theirs.

1. Introduction

Consider an infinite dimensional complex Hilbert space $H$ ⁠. Recall that a bounded linear operator T on $H$ is said to be cyclic if $\exists h \in H$ such that {p(T)h: p a polynomial} is dense in $H$ ⁠; T is said to be hypercyclic if $\exists x \in H$ such that Orbit{Tⁿx: n ≥ 0} is dense in $H$ ⁠. The study of hypercyclicity is specific to infinite dimensional spaces because no operator on a finite dimensional space is ever hypercyclic. The first examples of hypercyclic operators constructed on a Hilbert space were multiples λB with |λ| > 1, where B is the unweighted unilateral backward shift [1].

In [2] Wu proved that even if an operator is not cyclic it can still be expressed as the sum of cyclic operators. In Ref. [3] Grivaux showed that a bounded linear operator T can be expressed as the sum of two hypercyclic operators. Again in Ref. [2] Wu proved that even when T is not cyclic it can be factorized as product of cyclic operators. Hence it is natural to explore operator factorization in the stronger setting of hypercyclicity. Since weighted shifts play a vital role in linear dynamics (refer [4]) so in Refs. [5, 6] Chan & Sanders considered weighted shift operators on $ℓ^{p} (Z)$ that are not hypercyclic and proved that under certain conditions they can be factorized as products of hypercyclic shifts. In this article we extend their results to operator weighted shifts on ℓ²(K) by using the concept of generalized shift of higher multiplicity developed in Sections 3 and 4.

2. Preliminaries

Shift operators on separable Hilbert and Banach spaces have always been a subject of interest since the first systematic study done by Shields [7]. One of the reasons for this continued exploration and interest is because they provide a fertile domain for examples and counter examples. Traditionally, an operator T is considered as a shift on a space H if there is some canonical basis for H such that T shifts every basis vector to the immediate next basis vector, maybe with some weight attached to it. This clearly indicates that an operator which is a shift with respect to a basis for H may not remain a shift if the basis is changed. Given a space H, it is common practice to identify a cannonical basis $B$ of H and throughout the particular study the action of all linear operators on H are considered with respect to this arbitrary fixed basis $B$ ⁠.

For example, let $H = ℓ^{2} (C)$ ⁠, the space of all two way scalar sequences which are square summable. Let ${e_{n}}_{n \in Z}$ be an orthonormal basis for H. Given a bounded sequence of non-zero scalars ${α_{n}}_{n \in Z}$ ⁠, an operator T is a shift on H if $T e_{n} = α_{n} e_{n + 1} \forall n \in Z$ ⁠. In this case T is called bilateral forward weighted shift (abbreviated as BFWS) with weight sequence {α_n}. If $T e_{n} = α_{n - 1} e_{n - 1} \forall n \in Z$ ⁠, then T is called bilateral backward weighted shift (or BBWS) with weight sequence {α_n}.

Now consider the operator T on H such that $T e_{n} = \{\begin{cases} e_{0}, & if n = 2 \\ e_{1}, & if n = 0 \\ e_{- 1}, & if n = 1 \\ e_{n - 1}, & otherwise. \end{cases}$

Clearly, T is not a shift on H with respect to basis ${e_{n}}_{n \in Z}$ ⁠. However, if we re-arrange the basis vectors and consider the basis ${f_{n}}_{n \in Z}$ where f₁ = e₀, f₀ = e₁ and f_n = e_n for n ≠ 0, 1, then $T f_{n} = f_{n - 1} \forall n \in Z$ ⁠, and so T is a shift on H with respect to basis ${f_{n}}_{n \in Z}$ ⁠. In this case, T is called a generalized shift on H with respect to basis ${e_{n}}_{n \in Z}$ ⁠.

Thus, given a basis ${e_{n}}_{n \in Z}$ of H if there exists a bijection σ on $Z$ such that an operator T on H is a shift with respect to basis ${e_{σ (n)}}_{n \in Z}$ then T is called a generalized shift on H with respect to bijection σ. References to generalized shifts on $ℓ^{2} (C)$ can be found in Refs. [5, 6].

In this article, we consider generalized shifts of multiplicity higher than 1. For this we consider a separable complex Hilbert space K and define ℓ²(K) to be the space of all two way sequences ${x_{n}}_{n \in Z}$ in K such that $\sum_{n \in Z} ‖ x_{n} ‖^{2} < \infty$ ⁠. In Section 3, we define generalized bilateral backward weighted shifts (GBBWS) on ℓ²(K) and prove the following:

If T is a GBBWS on ℓ²(K) with respect to the bijection σ then there exists infinitely many bijections on $Z$ distinct from σ, such that T is a GBBWS with respect to each of these bijections. However, there exists only one such bijection ρ for which ρ(0) = 0, and in this sense the bijection corresponding to GBBWS T is unique.
Every GBBWS on ℓ²(K) is similar to the BBWS on ℓ²(K) with same sequence of weights.

In Section 4, we consider ℓ²(K₁) ⊕ ℓ²(K₂) for separable complex Hilbert spaces K₁ and K₂. We show that if T₁, T₂ are GBBWS on ℓ²(K₁), ℓ²(K₂) respectively then T₁ ⊕ T₂ is again a GBBWS. Conversely, if T is a GBBWS on ℓ²(K₁ ⊕ K₂) with weight sequence ${A_{n}}_{n \in Z}$ then T = T₁ ⊕ T₂, where T₁, T₂ are GBBWS on ℓ²(K₁), ℓ²(K₂) respectively provided A_n = B_n ⊕ C_n, where B_n, C_n are non zero bounded linear operators on K₁, K₂ respectively.

The main objective of the paper is addressed in Section 5. Here we apply the results developed in Sections 3 and 4 to provided conditions under which a BBWS W on ℓ²(K) can be factorized as W = AB where A and B are hypercyclic GBBWS.

3. Generalized bilateral shifts

Let $C$ denote the complex space and $Z$ denote the set of integers. $ℓ^{2} (C)$ denotes the space of all two way square summable sequences ${α_{n}}_{n \in Z}$ in $C$ ⁠. If ${α_{n}}_{n \in Z}$ is in $ℓ^{2} (C)$ then it is denoted as (…, α₋₁, [α₀], α₁, …) where [.] denotes the 0^th position of the sequence with respect to index set $Z$ ⁠. Thus,

ℓ^{2} (C) = \{x = (\dots, α_{- 1}, [α_{0}], α_{1}, \dots) : α_{i} \in C, \sum_{i \in Z} {| α_{i} |}^{2} < \infty

For x = (…, α₋₁, [α₀], α₁, …) and y = (…, y₋₁, [y₀], y₁, …) in $ℓ^{2} (C)$ ⁠, define $〈 x, y 〉 = \sum_{i \in Z} x_{i} {\bar{y}}_{i}$ ⁠. Then $ℓ^{2} (C)$ is a Hilbert space with respect to this inner product.

If e_i denotes the sequence (…, δ_−1,i, [δ_0,i], δ_1,i, …) where $δ_{j, i} = \{\begin{cases} 1, & if j = i \\ 0, & otherwise \end{cases}$ then ${e_{i}}_{i \in Z}$ is an orthonormal basis for $ℓ^{2} (C)$ and x = (…, α₋₁, [α₀], α₁, …) in $ℓ^{2} (C)$ can be written as $x = \sum_{i \in Z} α_{i} e_{i}$ ⁠.

Let K be a separable complex Hilbert space and ℓ²(K) denote the space of all two way square summable sequences ${x_{n}}_{n \in Z}$ in K. If ${x_{n}}_{n \in Z}$ is in ℓ²(K), we denote it as (…, x₋₁, [x₀], x₁, …) so that $ℓ^{2} (K) ≔ {x = (\dots, x_{- 1}, [x_{0}], x_{1}, \dots) : x_{i} \in K, \sum_{i \in Z} ‖ x_{i} ‖^{2} < \infty}$ ⁠.

For x = (…, x₋₁, [x₀], x₁, …) and y = (…, y₋₁, [y₀], y₁, …) in ℓ²(K), we define x + y = (…, x₋₁ + y₋₁, [x₀ + y₀], x₁ + y₁, …) and λx = (…, λx₋₁, [λx₀], λx₁, …) for $λ \in C$ ⁠. Also, $〈 x, y 〉 = \sum_{i \in Z} 〈 x_{i}, y_{i} 〉$ and ℓ²(K) is a separable Hilbert space with respect to these operations.

For t ∈ K, we follow the convention of denoting the sequence (…, 0, [t], 0, …) as te₀. In fact, for $i \in Z$ ⁠, if x = (…, x₋₁, [x₀], x₁, …) ∈ ℓ²(K) with x_i = t and x_j = 0 for j ≠ i, then we denote x as te_i. Therefore, x = (…, x₋₁, [x₀], x₁, …) ∈ ℓ²(K) can be written as $\sum_{i \in Z} x_{i} e_{i}$ ⁠.

Definition 1.

Let ${A_{n}}_{n \in Z}$ be a sequence of non-zero bounded linear operators on K such that sup_n‖A_n‖ < ∞. Let W: ℓ²(K) → ℓ²(K) be defined as follows:

W (\dots, x_{- 1}, [x_{0}], x_{1}, \dots) = (\dots, A_{- 1} x_{0}, [A_{0} x_{1}], A_{1} x_{2}, \dots) .

Equivalently, $W (\sum_{i \in Z} x_{i} e_{i}) = \sum_{i \in Z} A_{i - 1} x_{i} e_{i - 1}$ ⁠. Then W is called bilateral backward weighted shift (BBWS) on ℓ²(K) with weight sequence ${A_{n}}_{n \in Z}$ ⁠.

Similarly, if for each $x = \sum_{i \in Z} x_{i} e_{i} \in ℓ^{2} (K)$ ⁠, we have $W (x_{i} e_{i}) = A_{i} x_{i} e_{i + 1} \forall i \in Z$ ⁠, then W is called bilateral forward weighted shift (BFWS) on ℓ²(K) with weight sequence ${A_{n}}_{n \in Z}$ ⁠.

If dim(K) = 1, then ℓ²(K) is isomorphic to $ℓ^{2} (C)$ and is therefore considered as identical with $ℓ^{2} (C)$ ⁠. Let σ be a bijection on $Z$ ⁠. For y = (…, y₋₁, [y₀], y₁, …) in ℓ²(K), let z_j≔y_σ(j). Then {y_j} and {z_j} are same as sets but different as sequences since the order of the elements in {y_j} have been changed in the sequence {z_j}. We use the notation ${(\dots, z_{- 1}, [z_{0}], z_{1}, \dots)}_{σ}$ to understand that $\sum_{j \in Z} z_{j} e_{σ (j)} \in ℓ^{2} (K)$ ⁠. Let us define ℓ²(K)_σ as follows:

ℓ^{2} {(K)}_{σ} = {z = {(\dots, z_{- 1}, [z_{0}], z_{1}, \dots)}_{σ} : z_{i} \in K, \sum_{i \in Z} z_{i} e_{σ (i)} \in ℓ^{2} (K)} .

So, if $z = {(\dots, z_{- 1}, [z_{0}], z_{1}, \dots)}_{σ} \in ℓ^{2} {(K)}_{σ}$ then

z = \sum_{i \in Z} z_{i} e_{σ (i)} = \sum_{i \in Z} z_{σ^{- 1} (i)} e_{i} = (\dots, z_{σ^{- 1} (- 1)}, [z_{σ^{- 1} (0)}], z_{σ^{- 1} (1)}, \dots) \in ℓ^{2} (K) .

Similarly, if y = (…, y₋₁, [y₀], y₁, …) ∈ ℓ²(K) then

y = \sum_{i \in Z} y_{i} e_{i} = \sum_{i \in Z} y_{σ (i)} e_{σ (i)} = {(\dots, y_{σ (- 1)}, [y_{σ (0)}], y_{σ (1)}, \dots)}_{σ} \in ℓ^{2} {(K)}_{σ}

Definition 2.

For a uniformly bounded sequence ${A_{n}}_{n \in Z}$ of non-zero operators on K, T: ℓ²(K)_σ → ℓ²(K)_σ is said to be a BBWS with weight sequence ${A_{n}}_{n \in Z}$ if for each $z = {(\dots, z_{- 1}, [z_{0}], z_{1}, \dots)}_{σ} \in ℓ^{2} {(K)}_{σ}, T {(\dots, z_{- 1}, [z_{0}], z_{1}, \dots)}_{σ} = {(\dots, A_{- 1} z_{0}, [A_{0} z_{1}], A_{1} z_{2}, \dots)}_{σ}$ ⁠.

That is, T (\sum_{i \in Z} z_{i} e_{σ (i)}) = \sum_{i \in Z} A_{i - 1} z_{i} e_{σ (i - 1)}

Definition 3.

An operator T on ℓ²(K) is said to be a generalized bilateral backward weighted shift (GBBWS) if there exists a bijection σ on $Z$ and a uniformly bounded sequence ${A_{n}}_{n \in Z}$ of non-zero operators on K such that T is BBWS on ℓ²(K)_σ with weight sequence ${A_{n}}_{n \in Z}$ ⁠. In this case, we say that T is a GBBWS on ℓ²(K) with respect to bijection σ and weight sequence ${A_{n}}_{n \in Z}$ ⁠. Thus, for $x = \sum_{i \in Z} x_{i} e_{i} \in ℓ^{2} (K)$ ,

T {(\dots, x_{σ (- 1)}, [x_{σ (0)}], x_{σ (1)}, \dots)}_{σ} = {(\dots, A_{- 1} x_{σ (0)}, [A_{0} x_{σ (1)}], A_{1} x_{σ (2)}, \dots)}_{σ}

or equivalently, $T (x_{σ (i)} e_{σ (i)}) = A_{i - 1} x_{σ (i)} e_{σ (i - 1)} \forall i \in Z$ ⁠.

Example 1.

Let T on ℓ²(K) be defined as follows:

For $x = \sum_{i \in Z} x_{i} e_{i} \in l^{2} (K), Tx = (\dots, x_{- 1}, x_{0}, [x_{2}], x_{3}, x_{1}, x_{4}, x_{5}, \dots) .$

Equivalently,

T x_{i} e_{i} = \{\begin{cases} x_{i} e_{i - 1}, & if i ⩽ 0 or i ⩾ 4 \\ x_{i} e_{i + 1}, & if i = 1 \\ x_{i} e_{i - 2}, & if 2 ⩽ i ⩽ 3 . \end{cases}

Figure 1 below gives a pictoral representation of the action of T on x = (…, x₋₁, [x₀], x₁, …).

Consider the bijection σ on $Z$ given by:

σ (i) = \{\begin{cases} i, & if i ⩽ 0 or i ⩾ 3 \\ 2, & if i = 1 \\ 1, & if i = 2 . \end{cases}

It can easily be verified that $T x_{σ (i)} e_{σ (i)} = x_{σ (i)} e_{σ (i - 1)} \forall i \in Z$ where $x = \sum_{i \in Z} x_{i} e_{i} \in ℓ^{2} (K)$ ⁠. Thus, T is a GBBWS on ℓ²(K) with respect to bijection σ and weight sequence ${A_{n}}_{n \in Z}$ , where each A_n is the identity operator on K.

Proposition 1.

Let ${D_{n}}_{n \in Z}$ be a uniformly bounded sequence of non-zero operators on K. Let T be an operator on ℓ²(K) such that for each $x = \sum_{i \in Z} x_{i} e_{i} \in ℓ^{2} (K), T (x_{i} e_{i}) = D_{i - 1} x_{i} e_{τ (i)}$ , where τ is a bijection on $Z$ ⁠. If T is a GBBWS on ℓ²(K) with respect to bijection σ and weight sequence ${A_{n}}_{n \in Z}$ then (τ◦σ)(i) = σ(i − 1) and $A_{i - 1} = D_{σ (i) - 1} \forall i \in Z$ ⁠.

Proof. For $x = \sum_{i \in Z} x_{i} e_{i} \in ℓ^{2} (K)$ ⁠, we have, T(x_σ(i)e_σ(i)) = A_i−1x_σ(i)e_σ(i−1).

Also, T(x_σ(i)e_σ(i)) = D_σ(i)−1x_σ(i)e_τ(σ(i)).

Thus, $A_{i - 1} x_{σ (i)} e_{σ (i - 1)} = D_{σ (i) - 1} x_{σ (i)} e_{τ ◦ σ (i)} \forall i \in Z$ which implies that

$σ (i - 1) = (τ ◦ σ) (i) \forall i \in Z$ and $A_{i - 1} x_{σ (i)} = D_{σ (i) - 1} x_{σ (i)} \forall i \in Z$

As x_σ(i) is an arbitrary element of K so we must have A_i−1 = D_σ(i)−1 ∀ i. □

Using the above proposition, we will give an example of an operator T on ℓ²(K) such that T is not a GBBWS.

Example 2.

Let T on ℓ²(K) be defined as follows:

T (\dots, x_{- 1}, [x_{0}], x_{1}, \dots) ≔ (\dots, x_{0}, [x_{3}], x_{2}, x_{1}, x_{4}, \dots) .

Figure 2 below gives a pictoral representation of the action of T on x ∈ ℓ²(K).

We will show that T is not a GBBWS on ℓ²(K).

Let if possible there exists a bijection σ on $Z$ and a weight sequence ${A_{n}}_{n \in Z}$ of operators on K such that T is a GBBWS on ℓ²(K) with respect to σ and weight sequence ${A_{n}}_{n \in Z}$ ⁠.

From the definition of T, we have Tx_ie_i = x_ie_τ(i), where $τ (i) ≔ \{\begin{cases} 2, & if i = 1, \\ 0, & if i = 3, \\ i - 1, & otherwise. \end{cases}$

So, by P1, we should have each A_n to be the identity operator on K, and $(τ ◦ σ) (i) = σ (i - 1) \forall i \in Z$ ⁠.

Claim 1.

If τ◦σ(i) = σ(i − 1) ∀i then $∄ j \in Z$ such that σ(j) = 1.

To establish our claim, suppose $\exists j_{0} \in Z$ such that σ(j₀) = 1.

Then σ(j₀ − 1) = (τ◦σ)(j₀) = τ(σ(j₀)) = τ(1) = 2

and σ(j₀ − 2) = (τ◦σ)(j₀ − 1) = τ(σ(j₀ − 1)) = τ(2) = 1.

Therefore, σ(j₀) = 1 = σ(j₀ − 2), and since σ is injective, we must have j₀ = j₀ − 2, a contradiction. Thus, our claim is established.

Now, by Claim1, σ is not surjective which contradicts the fact that σ is a bijection on $Z$ ⁠. Hence, ∄ any bijection σ on $Z$ such that T is a GBBWS on ℓ²(K) with respect to σ.

Example 3.

Consider the operator given in Example1 where it was shown that ∃ a bijection σ on $Z$ such that T is GBBWS on ℓ²(K) with respect to σ.

Consider the bijection φ on $Z$ given by $φ (i) = \{\begin{cases} i + 4, & if i ⩽ - 4 or i ⩾ - 1 \\ 2, & if i = - 3 \\ 1, & if i = - 2 . \end{cases}$

It follows easily that T is a GBBWS on ℓ²(K) with respect to bijection φ.

Remark 1.

From examples 1 and 3, we can conclude that for a bounded linear operator T on ℓ²(K) there may exist different bijections σ and φ on $Z$ such that T is GBBWS on ℓ²(K) with respect to both σ and φ. In other words, σ corresponding to T is not unique.
In examples 1 and 3, we observe that $φ (i) = σ (i + 4) \forall i \in Z$ ⁠. This observation leads us to the next few results.

Theorem 2.

Let T be a GBBWS on ℓ²(K) with respect to bijection σ and weight sequence ${A_{n}}_{n \in Z}$ ⁠. Let $c \in Z$ and $φ (i) ≔ σ (i + c) \forall i \in Z$ ⁠. Then T is a GBBWS on ℓ²(K) with respect to bijection φ and weight sequence {B_n} where $B_{i} = A_{i + c} \forall i \in Z$ ⁠.

Proof. For $x = \sum_{i \in Z} x_{i} e_{i} \in ℓ^{2} (K)$ we have $T x_{σ (i)} e_{σ (i)} = A_{i - 1} x_{σ (i)} e_{σ (i - 1)} \forall i \in Z$ ⁠. Therefore $T x_{σ (i + c)} e_{σ (i + c)} = A_{i + c - 1} x_{σ (i + c)} e_{σ (i + c - 1)} \forall i \in Z$ which implies that Tx_φ(i)e_φ(i) = A_i₊_c−1x_φ(i)e_φ(i−1). Let B_i≔A_i₊_c so that B_i−1 = A_i₊_c−1. Thus, we have $T x_{φ (i)} e_{φ (i)} = B_{i - 1} x_{φ (i)} e_{φ (i - 1)} \forall i \in Z$ ⁠, and so T is GBBWS on ℓ²(K) with respect to φ and weight sequence ${B_{n}}_{n \in Z}$ ⁠. □

Theorem 3.

Let T be a bounded linear operator on ℓ²(K) such that T is GBBWS on ℓ²(K) with respect to bijection σ and weight sequence ${A_{n}}_{n \in Z}$ and T is also a GBBWS on ℓ²(K) with respect to bijection φ and weight sequence ${B_{n}}_{n \in Z}$ ⁠. Then there exists a unique $c \in Z$ such that $φ (i) = σ (i + c) \forall i \in Z$ ⁠. Also, $B_{i} = A_{i + c} \forall i \in Z$ ⁠.

Proof. Existence:

Step 1: We show that if $m, n \in Z$ such that σ(m) = φ(n) then σ(m − λ) = φ(n − λ) for all positive integers λ. For this, let $x = \sum_{i \in Z} x_{i} e_{i} \in ℓ^{2} (K)$ and suppose σ(m) = φ(n) = i₀. Then $T x_{i_{0}} e_{i_{0}} = T x_{σ (m)} e_{σ (m)} = A_{m - 1} x_{σ (m)} e_{σ (m - 1)}$ and $T x_{i_{0}} e_{i_{0}} = T x_{φ (n)} e_{φ (n)} = B_{n - 1} x_{φ (n)} e_{φ (n - 1)}$ ⁠. Thus, $A_{m - 1} x_{i_{0}} e_{σ (i - 1)} = B_{n - 1} x_{i_{0}} e_{φ (n - 1)}$ which implies that σ(m − 1) = φ(n − 1).

Therefore σ(m) = φ(n) ⇒ σ(m − 1) = φ(n − 1). Applying this argument repeatedly we get σ(m − λ) = φ(n − λ) for every positive integer λ, and Step1 is proved.

As $1 \in Z$ and σ and φ are surjective, so $\exists m_{0}, n_{0} \in Z$ such that σ(m₀) = 1 = φ(n₀). Together with the conclusion of Step1 we can say that ∃ integers m₀ and n₀ such that σ(m₀ − λ) = φ(n₀ − λ) for all non negative integers λ.

Step 2: To show σ(m₀ + λ) = φ(n₀ + λ) for all positive integers λ.

Let $S_{1} ≔ {σ (m_{0} - λ) : λ \in N \cup {0}}$ and $S_{2} = {φ (n_{0} - λ) : λ \in N \cup {0}}$ ⁠.

Since, σ(m₀ − λ) = φ(n₀ − λ) so S₁ = S₂.

Now, let λ be a positive integer and j₀ = σ(m₀ + λ). Since φ is surjective $\exists n_{1} \in Z$ such that φ(n₁) = j₀. Also j₀∉S₁ implies j₀∉S₂ and so n₁ > n₀.

Thus, φ(m₀ + λ) = φ(n₁) and so, by applying the conclusion of Step1, we get σ(m₀) = φ(n₁ − λ). This together with the fact that σ(m₀) = φ(n₀) gives φ(n₀) = φ(n₁ − λ). Since φ is injective, so this implies that n₀ = n₁ − λ, or equivalently, n₁ = n₀ + λ. Thus, σ(m₀ + λ) = j₀ = φ(n₀ + λ), and Step2 is established.

Thus, there exist integers m, n such that $σ (m + λ) = φ (n + λ) \forall λ \in Z$ ⁠. If c ≔ m − n then for $i \in Z$ ⁠, we have σ(i + c) = σ(m + (i − n)) = φ(n + (i − n)) = φ(i).

Uniqueness:

Suppose there exists $c_{1}, c_{2} \in Z$ such that $φ (i) = σ (i + c_{1}) = σ (i + c_{2}) \forall i \in Z$ ⁠. Then by injectivity of σ, we must have c₁ = c₂.

Step 3: To show $B_{i} = A_{i + c} \forall i \in Z$ ⁠. Since for $x = \sum_{i \in Z} x_{i} e_{i} \in ℓ^{2} (K)$ ⁠, we have

Tx_φ(i)e_φ(i) = B_i−1x_φ(i)e_φ(i−1) = B_i−1x_σ(i₊_c)e_σ(i₊_c−1) and

Tx_φ(i)e_φ(i) = Tx_σ(i₊_c)e_σ(i₊_c) = A_i₊_c−1x_σ(i₊_c)e_σ(i₊_c−1),

therefore we must have B_i−1 = A_i₊_c−1, or equivalently, $B_{i} = A_{i + c} \forall i \in Z$ ⁠. □

Remark 2.

In view of theorems 2 and 3, given a GBBWS T on ℓ²(K) with respect to σ we can always choose σ(0) = 0 without any loss of generality. If we follow this convention then we can say that σ corresponding to T is unique. Henceforth, in the sequel we will always assume σ(0) = 0 so that σ corresponding to T is unique.

Theorem 4.

Let T be a GBBWS on ℓ²(K) with respect to bijection σ and weight sequence ${A_{n}}_{n \in Z}$ ⁠. Then T is similar to the BBWS on ℓ²(K) with weight sequence ${A_{n}}_{n \in Z}$ ⁠.

Proof. Define P, Q: ℓ²(K) → ℓ²(K) as follows: for $x = \sum_{i \in Z} x_{i} e_{i} \in ℓ^{2} (K)$ ⁠,

$P x ≔ \sum_{i \in Z} x_{i} e_{σ (i)} and Q x ≔ \sum_{i \in Z} x_{i} e_{σ^{- 1} (i)}$ ⁠.

Therefore, PQ and QP are both identity on ℓ²(K)and so P, Q are invertible and Q = P⁻¹. As T is a BBWS on ℓ²(K)_σ, so for $y = {(\dots, y_{- 1}, [y_{0}], y_{1}, \dots)}_{σ} \in ℓ^{2} {(K)}_{σ}, T {(\dots, y_{- 1}, [y_{0}], y_{1}, \dots)}_{σ} = {(\dots, A_{- 1} y_{0}, [A_{0} y_{1}], A_{1} y_{2}, \dots)}_{σ}$ ⁠. Now for $x = (\dots, x_{- 1}, [x_{0}], x_{1}, \dots) \in ℓ^{2} (K),$

Q T P x = Q T P (\dots, x_{- 1}, [x_{0}], x_{1}, \dots) = Q T {(\dots, x_{- 1}, [x_{0}], x_{1}, \dots)}_{σ} = Q {(\dots, A_{- 1} x_{0}, [A_{0} x_{1}], A_{1} x_{2}, \dots)}_{σ} = (\dots, A_{- 1} x_{0}, [A_{0} x_{1}], A_{1} x_{2}, \dots) = W (\dots, x_{- 1}, [x_{0}], x_{1}, \dots) = W x

Thus, W = P⁻¹TP ⇒ T is similar to W. □

We can apply Theorem 4 to show that a given operator is not a GBBWS on ℓ²(K).

Example 4.

For simplicity, we consider dim(K) = 1. So ℓ²(K) is replaced with $ℓ^{2} (C)$ ⁠. From [Ref. 7], we know that the bilateral backward shift W on $ℓ^{2} (C)$ have no eigen values, or the point spectrum of W is empty.

Now consider T as defined in Example 2, where Te₁ = e₂ and Te₂ = e₁. So for $x = e_{1} + e_{2} \in ℓ^{2} (C)$ , we have Tx = x and so 1 is an eigen value of T. Thus, the point spectrum of T is non-empty.

But if two operators are similar then they must have the same point spectrum. Thus, T is not similar to W and so by Theorem 4, T cannot be a GBBWS on $ℓ^{2} (C)$ ⁠.

We end the section with the following question:

“Suppose we have two separable complex Hilbert spaces K₁ and K₂ such that K = K₁ ⊕ K₂. If T is a GBBWS on ℓ²(K) then can we express T as T₁ ⊕ T₂ where T₁ and T₂ are GBBWS on ℓ²(K₁) and ℓ²(K₂) respectively?”

In the next section, we address this and other related questions.

4. Direct sum of generalized bilateral shifts

Let K = K₁ ⊕ K₂. Then for a sequence of elements {x_i} in K, we get sequences {y_i} and {z_i} in K₁ and K₂ respectively where x_i = y_i ⊕ z_i. Also we define ‖x_i‖² = ‖y_i‖² + ‖z_i‖² so that {x_i} is square summable if and only if {y_i} and {z_i} are square summable.

Following the notation introduced in Section 3, if x = (…, x₋₁, [x₀], x₁, …) ∈ ℓ²(K) where K = K₁ ⊕ K₂, we have the representation $x = \sum_{i \in Z} x_{i} e_{i}$ ⁠. But each x_i = y_i ⊕ z_i where y_ie_i ∈ ℓ²(K₁), z_ie_i ∈ ℓ²(K₂). So, in this case x_ie_i actually has to be interpreted as y_ie_i ⊕ z_ie_i. To avoid confusion, we introduce a new notation in case K = K₁ ⊕ K₂. For each $i \in Z$ ⁠, define f_i = e_i ⊕ e_i and for α ⊕ β ∈ K₁ ⊕ K₂, let (α ⊕ β)f_i = αe_i ⊕ βe_i. So, x = (…, x₋₁, [x₀], x₁, …) ∈ ℓ²(K₁ ⊕ K₂) will now be represented as $x = \sum_{i \in Z} x_{i} f_{i} = \sum_{i \in Z} (y_{i} \oplus z_{i}) f_{i} = \sum_{i \in Z} (y_{i} e_{i} \oplus z_{i} e_{i})$ ⁠. Also, note that if H₁ and H₂ are two Hilbert spaces then for x ⊕ y and ξ ⊕ η in H₁ ⊕ H₂, addition in H₁ ⊕ H₂ is conventionally defined as (x ⊕ y) + (ξ ⊕ η) = (x + ξ) ⊕ (y + η).

Following the same rule in ℓ²(K₁) ⊕ ℓ²(K₂), we have $x = \sum_{i \in Z} (y_{i} e_{i} \oplus z_{i} e_{i}) = (\sum_{i \in Z} y_{i} e_{i}) \oplus (\sum_{i \in Z} z_{i} e_{i})$ ⁠, i.e., x = y ⊕ z where $y = \sum_{i \in Z} y_{i} e_{i} \in ℓ^{2} (K_{1})$ and $z = \sum_{i \in Z} z_{i} e_{i} \in ℓ^{2} (K_{2})$ ⁠. Thus, we can define ℓ²(K₁ ⊕ K₂) as $\{(\dots, x_{- 1}, [x_{0}], x_{1}, \dots) : x_{i} = y_{i} \oplus z_{i} \in K_{1} \oplus K_{2}, \sum_{i \in Z} y_{i} e_{i} \in ℓ^{2} (K_{1}), \sum_{i \in Z} z_{i} e_{i} \in ℓ^{2} (K_{2})\}$ so that ℓ²(K₁ ⊕ K₂) = ℓ²(K₁) ⊕ ℓ²(K₂). Equivalently, for $y = \sum_{i \in Z} y_{i} e_{i} \in ℓ^{2} (K_{1})$ and $z = \sum_{i \in Z} z_{i} e_{i} \in ℓ^{2} (K_{2})$ ⁠, we have $y \oplus z = \sum_{i \in Z} (y_{i} \oplus z_{i}) f_{i} \in ℓ^{2} (K_{1} \oplus K_{2})$ ⁠.

Definition 4.

Let K = K₁ ⊕ K₂ and {A_n} be a uniformly bounded sequence of non-zero operators on K. Linear operator W: ℓ²(K₁ ⊕ K₂) → ℓ²(K₁ ⊕ K₂) is called bilateral backward weighted shift (BBWS) on ℓ²(K₁ ⊕ K₂) with weight sequence {A_n} if $W (\sum_{i \in Z} x_{i} f_{i}) = \sum_{i \in Z} A_{i - 1} x_{i} f_{i - 1}$ ⁠.

Theorem 5.

Let W be BBWS on ℓ²(K₁ ⊕ K₂) with weight sequence ${A_{n}}_{n \in Z}$ , where A_n = B_n ⊕ C_n, B_n, C_n non zero bounded linear operators on K₁, K₂ respectively and sup_n‖A_n‖ < ∞. Then W = W₁ ⊕ W₂, where W₁ is BBWS on ℓ²(K₁) with weight sequence ${B_{n}}_{n \in Z}$ and W₂ is BBWS on ℓ²(K₂) with weight sequence ${C_{n}}_{n \in Z}$ ⁠.

Proof. Let $x = \sum_{i \in Z} x_{i} f_{i} \in ℓ^{2} (K_{1}) \oplus ℓ^{2} (K_{2})$ ⁠. Then x_i = y_i ⊕ z_i ∈ K₁ ⊕ K₂, and

W x = W (\sum_{i \in Z} x_{i} f_{i}) = \sum_{i \in Z} A_{i - 1} x_{i} f_{i - 1} = \sum_{i \in Z} (B_{i - 1} y_{i} e_{i - 1} \oplus C_{i - 1} z_{i} e_{i - 1}) = (\sum_{i \in Z} B_{i - 1} y_{i} e_{i - 1}) \oplus (\sum_{i \in Z} C_{i - 1} z_{i} e_{i - 1})

(1)

As $y = \sum_{i \in Z} y_{i} e_{i} \in ℓ^{2} (K_{1})$ so $W_{1} y = \sum_{i \in Z} B_{i - 1} y_{i} e_{i - 1}$ ⁠, and as $z = \sum_{i \in Z} y_{i} e_{i} \in ℓ^{2} (K_{2})$ so $W_{2} z = \sum_{i \in Z} C_{i - 1} z_{i} e_{i - 1}$ ⁠. Observe that y ⊕ z ∈ ℓ²(K₁) ⊕ ℓ²(K₂) and so (W₁ ⊕ W₂)(y ⊕ z) ≔ W₁y ⊕ W₂z. Using this in (1) we get Wx = W₁y ⊕ W₂z = (W₁ ⊕ W₂)(y ⊕ z) = (W₁ ⊕ W₂)x. Thus, W = W₁ ⊕ W₂. □

Theorem 6.

If W₁ and W₂ are BBWS on ℓ²(K₁) and ℓ²(K₂) with weight sequences {B_n} and {C_n} respectively, then W ≔ W₁ ⊕ W₂ is BBWS on ℓ²(K₁ ⊕ K₂) with weight sequence {A_n}, where A_n ≔ B_n ⊕ C_n.

Proof. Let $x = \sum_{i \in Z} x_{i} e_{i} \in ℓ^{2} (K_{1} \oplus K_{2})$ ⁠. Then x_i = y_i ⊕ z_i ∈ K₁ ⊕ K₂. Also $y = \sum_{i \in Z} y_{i} e_{i} \in ℓ^{2} (K_{1}), z = \sum_{i \in Z} z_{i} e_{i} \in ℓ^{2} (K_{2})$ and x = y ⊕ z ∈ ℓ²(K₁ ⊕ K₂).

Therefore,

W x = (W_{1} \oplus W_{2}) (y \oplus z) = W_{1} y \oplus W_{2} z = (\sum_{i \in Z} B_{i - 1} y_{i} e_{i - 1}) \oplus (\sum_{i \in Z} C_{i - 1} z_{i} e_{i - 1}) = \sum_{i \in Z} A_{i - 1} x_{i} f_{i - 1}

Thus, W is BBWS on ℓ²(K₁ ⊕ K₂) with weight sequence ${A_{n}}_{n \in Z}$ ⁠. □

Now for K = K₁ ⊕ K₂ let W be a GBBWS on ℓ²(K) with bijection σ and weight sequence ${A_{n}}_{n \in Z}$ ⁠, where ${A_{n}}_{n \in Z}$ is a sequence of uniformly bounded non-zero operators on K. Then for each $x = \sum_{i \in Z} x_{i} f_{i} \in ℓ^{2} (K)$ ⁠, $W {(\dots, x_{σ (- 1)}, [x_{σ (0)}], x_{σ (1)}, \dots)}_{σ} = {(\dots, A_{- 1} x_{σ (0)}, [A_{0} x_{σ (1)}], A_{1} x_{σ (2)}, \dots)}_{σ}$ i.e., $W (\sum_{i \in Z} x_{σ (i)} f_{σ (i)}) = \sum_{i \in Z} A_{i - 1} x_{σ (i)} f_{σ (i - 1)}$ where f_σ(i) = e_σ(i) ⊕ e_σ(i). Let us assume that each A_n = B_n ⊕ C_n, where B_n and C_n are non zero bounded linear operators on K₁ and K₂ respectively. Also since x_i ∈ K₁ ⊕ K₂ so there exists y_i ∈ K₁ and z_i ∈ K₂ such that x_i = y_i ⊕ z_i. Thus, x = y ⊕ z, where $y = \sum_{i \in Z} y_{i} e_{i} \in ℓ^{2} (K_{1})$ and $z = \sum_{i \in Z} z_{i} e_{i} \in ℓ^{2} (K_{2})$ ⁠.

Let W₁ be GBBWS on ℓ²(K₁) with respect to bijection σ and weight sequence {B_i}, W₂ be GBBWS on ℓ²(K₂) with respect to bijection σ and weight sequence {C_i}. Then $W_{1} y = \sum_{i \in Z} B_{i - 1} y_{σ (i)} e_{σ (i - 1)}$ and $W_{2} z = \sum_{i \in Z} C_{i - 1} z_{σ (i)} e_{σ (i - 1)}$ ⁠.

∴ we have

W x = \sum_{i \in Z} A_{i - 1} x_{σ (i)} f_{σ (i - 1)} = \sum_{i \in Z} (B_{i - 1} y_{σ (i)} \oplus C_{i - 1} z_{σ (i)}) f_{σ (i - 1)} = \sum_{i \in Z} (B_{i - 1} y_{σ (i)} e_{σ (i - 1)} \oplus C_{i - 1} z_{σ (i)} e_{σ (i - 1)}) = (\sum_{i \in Z} B_{i - 1} y_{σ (i)} e_{σ (i - 1)}) \oplus (\sum_{i \in Z} C_{i - 1} z_{σ (i)} e_{σ (i - 1)}) = W_{1} y \oplus W_{2} z = (W_{1} \oplus W_{2}) (y \oplus z) .

Thus, Wx = (W₁ ⊕ W₂)(y ⊕ z) = (W₁ ⊕ W₂)x. So, we can represent W as W₁ ⊕ W₂, where W₁ and W₂ are GBBWS on ℓ²(K₁) and ℓ²(K₂) respectively both with respect to bijection σ. We can summarize this in the following theorem.

Theorem 7.

Let W be a GBBWS on ℓ²(K₁ ⊕ K₂) with respect to bijection σ and weight sequence ${A_{n}}_{n \in Z}$ where A_n = B_n ⊕ C_n with B_n, C_n being non-zero bounded linear operators on K₁ and K₂ respectively. Then W = W₁ ⊕ W₂ where W₁ is GBBWS on ℓ²(K₁) with respect to bijection σ and weight sequence ${B_{n}}_{n \in Z}$ and W₂ is GBBWS on ℓ²(K₂) with respect to bijection σ and weight sequence ${C_{n}}_{n \in Z}$ ⁠.

Let us consider the converse of the above result by taking GBBWS W₁ and W₂ on ℓ²(K₁) and ℓ²(K₂) respectively and defining W = W₁ ⊕ W₂ on ℓ²(K₁) ⊕ ℓ²(K₂) = ℓ²(K₁ ⊕ K₂). As W₁ is a GBBWS on ℓ²(K₁), so there exists a bijection σ on $Z$ and a weight sequence ${B_{n}}_{n \in Z}$ of non-zero operators such that for each $y = \sum_{i \in Z} y_{i} e_{i} \in ℓ^{2} (K_{1}), W_{1} (\sum_{i \in Z} y_{σ (i)} e_{σ (i)}) = \sum_{i \in Z} B_{i - 1} y_{σ (i)} e_{σ (i - 1)}$ ⁠. Similarly, W₂ is a GBBWS on ℓ²(K₂) and so there exists a bijection ρ on $Z$ and weight sequence ${C_{n}}_{n \in Z}$ of non-zero operators such that for each $z = \sum_{i \in Z} z_{i} e_{i} \in ℓ^{2} (K_{2}), W_{2} (\sum_{i \in Z} y_{ρ (i)} e_{ρ (i)}) = \sum_{i \in Z} C_{i - 1} y_{ρ (i)} e_{ρ (i - 1)}$ ⁠.

As it is not necessary that ρ = σ, hence the steps that proved Theorem 7 cannot be reversed. To address this issue, we proceed to introduce a new bijection τ to represent the pair (σ, φ). For this, let $τ : Z \to Z$ be defined as τ(σ(i)) = φ(i), i.e. τ = φ◦σ⁻¹. Then τ is a bijection on $Z$ ⁠. We define ${(f_{τ})}_{i} = e_{σ (i)} \oplus e_{φ (i)}$ ⁠. For α ⊕ β ∈ K₁ ⊕ K₂, we set $(α \oplus β) {(f_{τ})}_{i} = α e_{σ (i)} \oplus β e_{φ (i)}$ ⁠. Thus for each x_i = y_i ⊕ z_i ∈ K₁ ⊕ K₂ we use the representation $x = {(\dots, x_{- 1}, [x_{0}], x_{1}, \dots)}_{τ}$ to understand

x = \sum_{i \in Z} x_{i} {(f_{τ})}_{i} = \sum_{i \in Z} (y_{i} e_{σ (i)} \oplus z_{i} e_{φ (i)}) = (\sum_{i \in Z} y_{i} e_{σ (i)}) \oplus (\sum_{i \in Z} z_{i} e_{φ (i)}) = {(\dots, y_{- 1}, [y_{0}], y_{1}, \dots)}_{σ} \oplus {(\dots, z_{- 1}, [z_{0}], z_{1}, \dots)}_{φ} \in ℓ^{2} {(K_{1})}_{σ} \oplus ℓ^{2} {(K_{2})}_{φ}

Thus, we can define $ℓ^{2} {(K_{1} \oplus K_{2})}_{τ} = {{(\dots, x_{- 1}, [x_{0}], x_{1}, \dots)}_{τ} : x_{i} = y_{i} \oplus z_{i} \in K_{1} \oplus K_{2}, \sum_{i \in Z} y_{i} e_{σ (i)} \in ℓ^{2} (K_{1}), \sum_{i \in Z} z_{i} e_{φ (i)} \in ℓ^{2} (K_{2})}$ ⁠, so that $ℓ^{2} {(K_{1} \oplus K_{2})}_{τ} = ℓ^{2} {(K_{1})}_{σ} \oplus ℓ^{2} {(K_{2})}_{φ}$ ⁠. Equivalently, for $y = {(\dots, y_{- 1}, [y_{0}], y_{1}, \dots)}_{σ} \in ℓ^{2} {(K_{1})}_{σ}$ and $z = {(\dots, z_{- 1}, [z_{0}], z_{1}, \dots)}_{φ} \in ℓ^{2} {(K_{2})}_{φ}$ ⁠, we have $y \oplus z = \sum_{i \in Z} (y_{i} \oplus z_{i}) {(f_{τ})}_{i} \in ℓ^{2} {(K_{1} \oplus K_{2})}_{τ}$ ⁠.

Remark 3.

(i) Note that when $φ (i) = i = σ (i) \forall i \in Z$ ⁠, i.e. φ and σ are the identity bijections on $Z$ ⁠, then ${(f_{τ})}_{i} = f_{i}$ ⁠.

(ii) If σ = φ, not necessarily identity on $Z$ ⁠, then

${(f_{τ})}_{i} = e_{σ (i)} \oplus e_{φ (i)} = e_{σ (i)} \oplus e_{σ (i)} = f_{σ (i)}$ ⁠.

(iii) If $x = \sum_{i \in Z} (y_{i} \oplus z_{i}) f_{i} = (\dots, y_{- 1} \oplus z_{- 1}, [y_{0} \oplus z_{0}], y_{1} \oplus z_{1}, \dots) \in ℓ^{2} (K_{1} \oplus K_{2})$ then
$x = (\sum_{i \in Z} y_{i} e_{i}) \oplus (\sum_{i \in Z} z_{i} e_{i}) = (\sum_{i \in Z} y_{σ (i)} e_{σ (i)}) \oplus (\sum_{i \in Z} z_{ρ (i)} e_{ρ (i)}) = \sum_{i \in Z} (y_{σ (i)} \oplus z_{φ (i)}) {(f_{τ})}_{i} \in ℓ^{2} {(K_{1} \oplus K_{2})}_{τ}$

Notational Summary:

ℓ²(K₁ ⊕ K₂) = ℓ²(K₁) ⊕ ℓ²(K₂) where f_i = e_i ⊕ e_i, and if x = ∑_i(y_i ⊕ z_i)f_i ∈ ℓ²(K₁ ⊕ K₂) then x = y ⊕ z where y = ∑_iy_ie_i ∈ ℓ²(K₁), z = ∑_iz_ie_i ∈ ℓ²(K₂).
For bijections σ and ρ let τ≔ρ◦σ⁻¹.

Then $ℓ^{2} {(K_{1} \oplus K_{2})}_{τ} = ℓ^{2} {(K_{1})}_{σ} \oplus ℓ^{2} {(K_{2})}_{ρ}$ ⁠. Also ${(f_{τ})}_{i} ≔ e_{σ (i)} \oplus e_{ρ (i)}$ ⁠, and if x = ∑_i(y_i ⊕ z_i)f_i ∈ ℓ²(K₁ ⊕ K₂) then $x = \sum_{i \in Z} {(x_{τ})}_{i} {(f_{τ})}_{i} = {(\dots, {(x_{τ})}_{- 1}, [{(x_{τ})}_{0}], {(x_{τ})}_{1}, \dots)}_{τ} \in ℓ^{2} {(K_{1} \oplus K_{2})}_{τ}$ where ${(x_{τ})}_{i} = y_{σ (i)} \oplus z_{ρ (i)}$ ⁠.

Definition 5.

A bounded linear operator T on ℓ²(K₁ ⊕ K₂) is said to be a generalized bilateral backward weighted shift (GBBWS) with respect to bijection τ and weight sequence {A_n} if the following conditions hold:

There exist bijections σ and ρ on $Z$ such that τ = ρ◦σ⁻¹ and $ℓ^{2} {(K_{1} \oplus K_{2})}_{τ} = ℓ^{2} {(K_{1})}_{σ} \oplus ℓ^{2} {(K_{2})}_{ρ}$ ⁠.
For $x = \sum_{i \in Z} (y_{i} \oplus z_{i}) f_{i} \in ℓ^{2} (K_{1} \oplus K_{2})$ , we have $T (\sum_{i \in Z} {(x_{τ})}_{i} {(f_{τ})}_{i}) = \sum_{i \in Z} A_{i - 1} {(x_{τ})}_{i} {(f_{τ})}_{i - 1}$ , where ${(x_{τ})}_{i} = y_{σ (i)} \oplus z_{ρ (i)}$ , ${(f_{τ})}_{i} = e_{σ (i)} \oplus e_{ρ (i)}$ ⁠.

Theorem 8.

Let T₁ be GBBWS on ℓ²(K₁) with respect to bijection σ and weight sequence ${B_{n}}_{n \in Z}$ , and T₂ be GBBWS on ℓ²(K₂) with respect to bijection ρ and weight sequence ${C_{n}}_{n \in Z}$ ⁠. If A_n = B_n ⊕ C_n and τ≔ρ◦σ⁻¹ then T = T₁ ⊕ T₂ is GBBWS on ℓ²(K₁ ⊕ K₂) with respect to bijection τ and weight sequence ${A_{n}}_{n \in Z}$ ⁠.

Proof. Let T ≔ T₁ ⊕ T₂ on ℓ²(K₁) ⊕ ℓ²(K₂) = ℓ²(K₁ ⊕ K₂) and $x = \sum_{i \in Z} x_{i} f_{i} \in ℓ^{2} (K_{1} \oplus K_{2})$ ⁠. Then x_i = y_i ⊕ z_i ∈ K₁ ⊕ K₂ and x = y ⊕ z ∈ ℓ²(K₁) ⊕ ℓ²(K₂), where $y = \sum_{i \in Z} y_{i} e_{i}$ and $z = \sum_{i \in Z} z_{i} e_{i}$ ⁠.

Thus $x = \sum_{i \in Z} {(x_{τ})}_{i} {(f_{τ})}_{i} \in ℓ^{2} {(K_{1} \oplus K_{2})}_{τ}$ ⁠, where ${(x_{τ})}_{i} ≔ y_{σ (i)} \oplus z_{ρ (i)}$ ⁠.

Also $T_{1} y = T_{1} (\sum_{i \in Z} y_{σ (i)} e_{σ (i)}) = \sum_{i \in Z} B_{i - 1} y_{σ (i)} e_{σ (i - 1)}$ and

$T_{2} z = T_{2} (\sum_{i \in Z} z_{ρ (i)} e_{ρ (i)}) = \sum_{i \in Z} C_{i - 1} z_{ρ (i)} e_{ρ (i - 1)}$ ⁠.

Let A_n ≔ B_n ⊕ C_n on K₁ ⊕ K₂. Then for $x = \sum_{i \in Z} x_{i} f_{i} \in ℓ^{2} (K_{1} \oplus K_{2})$ ⁠, it follows that $T (\sum_{i \in Z} {(x_{τ})}_{i} {(f_{τ})}_{i}) = \sum_{i \in Z} A_{i - 1} {(x_{τ})}_{i} {(f_{τ})}_{i - 1}$ ⁠. Therefore, T is a GBBWS on ℓ²(K₁ ⊕ K₂) with respect to bijection τ and weight sequence ${A_{n}}_{n \in Z}$ ⁠. □

Finally, the following result gives an affirmative answer to the question raised at the end of Section 3.

Theorem 9.

Let K = K₁ ⊕ K₂ and A_n = B_n ⊕ C_n where B_n and C_n are non-zero bounded linear operators on K₁ and K₂ respectively. Also, let sup_n‖A_n‖ < ∞. If T is a GBBWS on ℓ²(K₁ ⊕ K₂) with respect to bijection τ and weight sequence ${A_{n}}_{n \in Z}$ then the following holds:

There exist bijections σ and φ on $Z$ such that τ = φ◦σ⁻¹,
∃ GBBWS T₁ on ℓ²(K₁) with respect to bijection σ and weight sequence ${B_{n}}_{n \in Z}$ and GBBWS T₂ on ℓ²(K₂) with respect to bijection φ and weight sequence ${C_{n}}_{n \in Z}$ such that T = T₁ ⊕ T₂.

Proof. Let σ be a bijection on $Z$ and φ = τ◦σ. Then $ℓ^{2} {(K_{1} \oplus K_{2})}_{τ} = ℓ^{2} {(K_{1})}_{σ} \oplus ℓ^{2} {(K_{2})}_{φ}$ ⁠. Also for each $x = \sum_{i \in Z} (y_{i} \oplus z_{i}) f_{i} \in ℓ^{2} (K_{1} \oplus K_{2})$ ⁠,

$T x = T (\sum_{i \in Z} {(x_{τ})}_{i} {(f_{τ})}_{i})$ ⁠, [where ${(x_{τ})}_{i} = y_{σ (i)} \oplus z_{φ (i)}$ ]

$= \sum_{i \in Z} A_{i - 1} {(x_{τ})}_{i} {(f_{τ})}_{i - 1}$ ]

$= (\sum_{i \in Z} B_{i - 1} y_{σ (i)} e_{σ (i - 1)}) \oplus (\sum_{i \in Z} C_{i - 1} z_{φ (i)} e_{φ (i - 1)})$ ⁠.

Now as x = y ⊕ z where $y = \sum_{i \in Z} y_{i} e_{i} \in ℓ^{2} (K_{1}), z = \sum_{i \in Z} z_{i} e_{i} \in ℓ^{2} (K_{2})$ ⁠,

therefore $T_{1} y = T_{1} (\sum_{i \in Z} y_{σ (i)} e_{σ (i)}) = \sum_{i \in Z} B_{i - 1} y_{σ (i)} e_{σ (i - 1)}$ and

$T_{2} z = T_{2} (\sum_{i \in Z} z_{φ (i)} e_{φ (i)}) = \sum_{i \in Z} C_{i - 1} z_{φ (i)} e_{φ (i - 1)}$ ⁠. Thus we get

Tx = T₁y ⊕ T₂z = (T₁ ⊕ T₂)x which implies that T = T₁ ⊕ T₂. □

Theorem 10.

Let K = K₁ ⊕ K₂ and A_n = B_n ⊕ C_n where B_n and C_n are non-zero bounded linear operators on K₁ and K₂ respectively. If T is a GBBWS on ℓ²(K₁ ⊕ K₂) with respect to bijection τ and weight sequence ${A_{n}}_{n \in Z}$ then T is similar to the BBWS W on ℓ²(K₁ ⊕ K₂) with weight sequence ${A_{n}}_{n \in Z}$ ⁠.

Proof. By Theorem 9, ∃ bijections σ and φ on $Z$ such that τ = φ◦σ⁻¹ and T = T₁ ⊕ T₂ where T₁ is GBBWS on ℓ²(K₁) with respect to bijection σ and weight sequence ${B_{n}}_{n \in Z}$ ⁠, and T₂ is GBBWS on ℓ²(K₂) with respect to bijection φ and weight sequence ${C_{n}}_{n \in Z}$ ⁠.

Again by Theorem 4, T₁ is similar to BBWS W₁ on ℓ²(K₁) with weight sequence ${B_{n}}_{n \in Z}$ ⁠, and T₂ is similar to BBWS W₂ on ℓ²(K₂) with weight sequence ${C_{n}}_{n \in Z}$ ⁠. So, T₁ ⊕ T₂ is similar to W₁ ⊕ W₂. By Theorem 6, W = W₁ ⊕ W₂ is BBWS on ℓ²(K₁ ⊕ K₂).

Thus, T is similar to BBWS W on ℓ²(K₁ ⊕ K₂) with weight sequence ${A_{n}}_{n \in Z}$ ⁠. □

5. Generalized shifts and hypercyclicity

Weighted shift operators play a vital role in linear dynamics (refer [4]). Beginning with Kitai [8], several studies have been conducted to determine necessary and sufficient conditions for hypercyclicity of weighted shift operators, refer [9–13]. Very often a weighted shift which is not hypercyclic turns out to be Cesaro-hypercyclic and vice versa, refer [14–16]. With weighted shifts serving as a testing ground for hypercyclicity, it is natural to examine hypercyclic factorization with this class of operators.

In Theorem 2.4 [6], it was shown that if T is a BBWS on $ℓ^{2} (C)$ then there exists GBBWS A and B such that A and B are hypercyclic and T = AB. Using the results from Section 4, we will show that this result by Chan & Sanders [6] can be extended to BBWS of any finite higher multiplicity but with some restrictions on the operator weights. We begin by considering BBWS of multiplicity 2, and for this we consider $K = C \oplus C$ ⁠. Also for easy reference Theorem 2.4 [6] is stated below.

Theorem 11.

[6] Let 1 ≤ p < ∞. If $T : ℓ^{p} (Z) \to ℓ^{p} (Z)$ is a bilateral weighted shift, then there are hypercyclic bilateral weighted shifts $A, B : ℓ^{p} (Z) \to ℓ^{p} (Z)$ such that T = AB. Furthermore, for any ϵ > 0, the weighted shifts A and B can be chosen so that their norms ‖A‖, ‖B‖ are no larger than (1 + ϵ) max{1, ‖T‖}.

Theorem 12.

Let K be a complex Hilbert space of dimension 2 and {A_n} be a uniformly bounded sequence of invertible diagonal operators on K. If W is BBWS on ℓ²(K) with weight sequence ${A_{n}}_{n \in Z}$ , then there exists hypercyclic GBBWS T and P on ℓ²(K) such that W = TP.

Proof. We consider K = K₁ ⊕ K₂, where $K_{1} = K_{2} = C$ and A_n = B_n ⊕ C_n where B_n and C_n are non zero scalar multiples of identity operator on $C$ ⁠. By Theorem 5, we have W = W₁ ⊕ W₂ where W₁ is BBWS on ℓ²(K₁) with weight sequence ${B_{n}}_{n \in Z}$ and W₂ is BBWS on ℓ²(K₂) with weight sequence ${C_{n}}_{n \in Z}$ ⁠.

Since $K_{1} = K_{2} = C$ so by Theorem 11, there are hypercyclic GBBWS T₁, P₁ on ℓ²(K₁) and T₂, P₂ on ℓ²(K₂) such that W₁ = T₁P₁ and W₂ = T₂P₂. Also for any ϵ > 0, the weighted shifts T₁, T₂ and P₁, P₂ can be chosen so that

‖ T_{1} ‖, ‖ P_{1} ‖ \leq (1 + ϵ) \max {1, ‖ W_{1} ‖} and ‖ T_{2} ‖, ‖ P_{2} ‖ \leq (1 + ϵ) \max {1, ‖ W_{2} ‖}

(2)

For i = 1, 2, let T_i be GBBWS on ℓ²(K_i) with respect to bijection σ_i and weight sequence ${D_{n}^{(i)}}_{n \in Z}$ ⁠, and let P_i be GBBWS on ℓ²(K_i) with respect to bijection φ_i and weight sequence ${Q_{n}^{(i)}}_{n \in Z}$ ⁠. Let $τ_{1} = φ_{1} ◦ σ_{1}^{- 1}$ and $τ_{2} = φ_{2} ◦ σ_{2}^{- 1}$ ⁠. Also let $D_{n} = D_{n}^{(1)} \oplus D_{n}^{(2)}$ and $Q_{n} = Q_{n}^{(1)} \oplus Q_{n}^{(2)}$ ⁠. Then by Theorem 8, T = T₁ ⊕ T₂ is GBBWS on ℓ²(K₁ ⊕ K₂) with respect to bijection τ₁ and weight sequence ${D_{n}}_{n \in Z}$ ⁠. Similarly, P = P₁ ⊕ P₂ is GBBWS on ℓ²(K₁ ⊕ K₂) with respect to bijection τ₂ and weight sequence ${Q_{n}}_{n \in Z}$ ⁠.

First we show that T is hypercyclic. For this let L₁ and L₂ be BBWS on $ℓ^{2} (C)$ with weight sequence ${D_{n}^{(1)}}$ and ${D_{n}^{(2)}}$ respectively. Then by Theorem 4, L₁ is similar to T₁, and L₂ is similar to T₂. As mentioned in Ref. [4], hypercyclicity is preserved under similarity, and as T₁, T₂ are hypercyclic so L₁, L₂ are also hypercyclic. This together with (1) implies that L₁ ⊕ L₂ must also be hypercyclic, by [12, Theorem 2.5]. Let L ≔ L₁ ⊕ L₂. Then L is a hypercyclic BBWS on ℓ²(K) with weight sequence {D_n}. Since by Theorem 10, T is similar to L, so T is also hypercyclic.

By a similar argument it follows that P is hypercyclic.

Claim: (T₁ ⊕ T₂)(P₁ ⊕ P₂) = T₁P₁ ⊕ T₂P₂.

To establish the claim, let x = y ⊕ z ∈ ℓ²(K₁) ⊕ ℓ²(K₂).

Then

\begin{array}{l} (T_{1} \oplus T_{2}) (P_{1} \oplus P_{2}) x & = (T_{1} \oplus T_{2}) (P_{1} y \oplus P_{2} z) \\ = T_{1} (P_{1} y) \oplus T_{2} (P_{2} z) \\ = (T_{1} P_{1}) y \oplus (T_{2} P_{2}) z \\ = (T_{1} P_{1} \oplus T_{2} P_{2}) (y \oplus z) \\ = (T_{1} P_{1} \oplus T_{2} P_{2}) x . \end{array}

Therefore (T₁ ⊕ T₂)(P₁ ⊕ P₂) = T₁P₁ ⊕ T₂P₂, and the claim is established.

So, we have W = W₁ ⊕ W₂ = T_IP₁ ⊕ T₂P₂ = (T₁ ⊕ T₂)(P₁ ⊕ P₂) = TP, where T and P are hypercyclic GBBWS on ℓ²(K). □

Now suppose BBWS W with weight sequence {A_n} is of multiplicity 3. Then we can consider K = K₁ ⊕ K₂, where dim(K₁) = 1 and dim(K₂) = 2. If A_n = B_n ⊕ C_n on K₁ ⊕ K₂ then we have W = W₁ ⊕ W₂ where W₁, W₂ are BBWS on ℓ²(K₁) and ℓ²(K₂) respectively. We will then have W₁ = T₁P₁ by Theorem 2.4 [6]. We can also apply Theorem 12 and have W₂ = T₂P₂, but for this C_n must be an invertible diagonal operator on K₂. In other words, we need to assume that each A_n is an invertible diagonal operator on K with respect to some common basis. This leads us to the following result:

Theorem 13.

Let K be a separable complex Hilbert space of finite dimension n, and {A_i} be a uniformly bounded sequence of invertible diagonal operators on K. If W is BBWS on ℓ²(K) with weight sequence {A_i} then there exists hypercyclic GBBWS T and P on ℓ²(K) such that W = TP.

Proof. We prove the result by induction on n.

By Theorem 2.4 [6], the result is true for n = 1. Suppose the result is true for n. We will show that it is true for n + 1.

For this we first express K as K = K₁ ⊕ K₂ where dim(K₁) = 1 and dim(K₂) = n. Since A_n is invertible diagonal, we can express A_n as A_n = B_n ⊕ C_n where B_n, C_n are invertible diagonal operators on K₁, K₂ respectively. So by Theorem 5, W = W₁ ⊕ W₂ where W₁, W₂ are BBWS on ℓ²(K₁), ℓ²(K₂) with weight sequences {B_n}, {C_n} respectively.

So by our assumption we have W₁ = T₁P₁ and W₂ = T₂P₂ where T_i, P_i are hypercyclic GBBWS on ℓ²(K_i) for each i = 1, 2. Let T ≔ T₁ ⊕ T₂ and P ≔ P₁ ⊕ P₂. Then following an argument similar to what was done in Theorem 12, we can conclude that W = TP where T and P are hypercyclic GBBWS on ℓ²(K).

Thus, by induction on n, the conclusion of the theorem follows. □

References

Rolewicz

On orbits of elements

Studia Math

1969

;

(

. doi:

Generalized bilateral shifts of higher multiplicity

1. Introduction

2. Preliminaries

3. Generalized bilateral shifts

4. Direct sum of generalized bilateral shifts

5. Generalized shifts and hypercyclicity

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