A Supercyclic Weighted Shift Has Cyclic Square

A Supercyclic Weighted Shift Has Cyclic Square

Musa Siddig¹, Amani Elseid Abuzeid², Abdalgadir Albushra³, and Shazli Mohammed⁴

¹University of Kordofan, Faculty of  Science, Department of  Mathematics, Sudan

²Department of Mathematics, College of Aldaier, Jazan University, Saudi Arabia

³Sudan University of Science and Technology College of Education, Department of Mathematics, Sudan

⁴University of Bahri, Colloge of Applied & Industrial Sciences ,Department of  Mathematics, Sudan

DOI: https://dx.doi.org/10.47772/IJRISS.2025.909000594

Received: 27 June 2025; Accepted: 08 July 2025; Published: 21 October 2025

ABSTRACT 

It is demonstrated that for a weighted abounded bilateral shift  acting on when supercyclicity of  weak supercyclicity of   cyclicity of  and cyclicity of    are equivalent.

A new sufficient condition for cyclicity of  a weighted bilateral shift is proved, which implies, in particular, that any compact weighted bilateral shift is cyclic.

Keywords: Cyclicity, supercyclicity, Hypercyclicity, Quasisimilarity, Weighted bilateral shifts, Banach space, bounded linear operator.

INTRODUCTION

All vector spaces in this article are assumed to be over the field  of complex numbers,  is the set of whole numbers, the set of positive whole numbers , and the set of non-negative numbers.

As is customary, symbol  is the space of continuous linear functions on  and  denotes the space of bounded linear operators on a Banach space .

For  and  represents the bounded linear operator on if  or if   described using the standard framework  by

If furthermore  ,  the weighted bilateral shift with the weight sequence  is called for each operator   . We have the un weighted bilateral shift in this specific instance  .

Remember that if there exists  such that  is dense in , then a bounded linear operator  on a Banach space  is said to be cyclicis referred to as supercyclic if it has a dense interior .  is likewise referred to as hypercyclic if there  such that there is a dense orbit in  Lastly, if the density is necessary in relation to the weak topology,  it is referred to as weakly supercyclic or weakly hypercyclic. For further information on hypercyclicity and supercyclicity, we consult surveys [8, 9, 12]. Weakly supercyclic operators have an appealing quality in that all of their powers are cyclic and again weakly supercyclic . Ansari [2] demonstrated this result for norm supercyclicity, and the same proof holds for weak supercyclicity.

Weighted bilateral shifts  cyclicity characteristics have been thoroughly researched. Salas [15, 16] described the hypercyclicity and supercyclicity of weighted bilateral shifts in terms of the weight sequences.The following simpler equivalent form of the Salas criteria is admissible, as was noted in [19, Proposition 5.1] .

Theorem S For  a weighted bilateral shift  can only be considered hypercyclic if and when any 

and   it is only Supercyclic if and only when any 

where

for  with   

However, it turns out that the cyclicity of a weighted bilateral shift is a far more nuanced matter, see, for example, see [10, 11, 14, 18]. It is important to note that cyclicity of a weighted bilateral shift depends on , in contrast to hyper- or supercyclicity. The un weighted bilateral shift, for example, is non-cyclic on and cyclic on  A weighted bilateral shift might have several sufficient and necessary criteria for cyclicity ; for example, Herrero’s studies[10, 11] provide evidence of this.

One of the most important of these requirements is that a weighted bilateral shift  on for is non-cyclic if its adjoint has a non-empty point spectrum. The weighted bilateral shift with the weight sequence   for  and  for   with  is implied to be non-cyclic for any Beauzamy’s [5] initial instance  of a non-cyclic weighted bilateral shift has exactly this form.

Remember that the Supercyclicity Criterion [12] states that  a bounded linear operator on a Banach space  is said to satisfy the if there exist a strictly growing sequence  of positive integers, dense subsets  and of   and an a map such that , for every  , and as  for any  and  In [12], the next two results are demonstrated.

Theorem SC.  A supercyclic operator is one that satisfies the Supercyclicity  Criterion.

Theorem MS. If and only if a weighted bilateral shift on  for  or on  meets supercyclicity Criterion, it can be considered Supercyclic.

The final theorem is not that mysterious. All that is required is to consider  the space of sequences with finite support, being the opposite of the limitation of   to  and utilize the theorem to identify a suitable order . Also take note that [3, 6, 17, 19] examined the weak hypercyclicity of weighted bilateral shifts. It is demonstrated in [19] that for  any weighted bilateral shift on  , it is either weakly supercyclic or supercyclic. We expand on this duality.

Theorem ( The following statements are equivalent, assuming   and  be a weighted bilateral shift on   :

  fulfills the Supercyclicity requirement;

  is supercyclic;

 is weakly supercyclic;

 is cyclic;

 there is   for which is cyclic;

 for any  , is cyclic.

We emphasize that there a weakly supercyclic non-supercyclic weighted bilateral shift on   each  as demonstrated in   We can observe that ( (does not imply  when   ) since powers of a weakly supercyclic operator are cyclic. This observation and equivalency of  and  for , allow us to derive the following consequence right away.

Corollary If and only if  is cyclic, a weighted bilateral shift  occurring on  with is supercyclic. However, there is a non-supercyclic weighted bilateral shift on  for each , with powers for all of them. It is also important to remember that weak supercyclicity of weighted bilateral shifts  in the necessary condition invariably results in weak supercyclicity, thus, cyclicity of   Conversely, non-supercyclic operators  created in have the non-cyclic property. In  a sufficient condition for a weighted bilateral shift to be unicellular (and therefore cyclic) is given. This result together with Theorem S imply that there are cyclic non-supercyclic weighted bilateral shifts on for and on. As a result, the requirement  in is crucial. Which relationships between the criteria apply for any bounded linear operator on a separable Banach space may be easily from the Theorem’s proof below.

Additionally, we will demonstrate that the inference  holds true for any weighted bilateral shift with . However, general operators are not satisfied with the final implication. The Volterra operator    acting on  , for example , both satisfies  and does not fulfill .

In conclusion, we will demonstrate an additional necessary condition for the cyclicity of a weighted bilateral shift. It is not consistent with any known sufficient condition, even the most current one that Abakumov, Atzmon and Grivaux  have published.

Theorem . Assume that    is a finite series of complex numbers that not zero,  for   and   for  , the definition of the numbers  is found in  Additionally, suppose that  there a sub multiplicative sequence  of positive numbers such that   and

  as well. In the event that the sequence  is not a part of  , where

  , the weighted bilateral shift  is cyclic 

This extremely complex outcome fails to provide a description of cyclicity for bilateral shifts that are weighted. The weight sequence criteria, for example, excide compact weighted bilateral shift. The following theorem is applicable to a greater range of weight sequences, although it becomes a weaker assertion when used with weight that mee Theorem’s  requirements.

Theorem  Let  be a finite series of complex numbers that are not zero,  such that for any ,

The weighted bilateral shift  is hence cyclic for .

By substituting  into  we get the following corollary right away.

Corollary (1.4). Let be a bounded series of complex numbers that are non-zero, such that

The weighted bilateral shift  is hence cyclic for .

Since  each quasinilpotent weighted bilateral shift for each   implies that the spectral radius formula satisfies It should be noted that a compact weighted bilateral shift is inherently quasinilpotent. Therefore, the following corollary is accurate.

Corollary   A weighted bilateral shift that is quasinilpotent is cyclic. Specifically, any compact weighted bilateral shift has a cyclic structure.

If we do  the following corollary follows right away.

Corollary  Let   be a finite series complex numbers that are not zero, for which there exists  such that

  for every  .                                  (6)

After then, the weighted bilateral shift  is period.

Example Assume that  and    is a series of positive numbers such that as  It is from corollary (4.1.10) that the weighted bilateral shift  is cyclic for  . Conversely, Theorem  is only relevant if  . Also take note that by Theorem    is non-supercyclic if   and supercyclic if   .

Proof of Theorem  We begin with these three simple, well- known, yet elegant observations.

Lemma   Allow   and be Banach spaces, and  such that satisfies the existence of a bounded linear operator  with dense range  Consequently, cyclicity of   implies cyclicity of  .

Proof. Remember that   for every  . Thus, for every cyclic vector  for   , there is a cyclic vector for  .

 . Lemma  holds true even when cyclicity is substituted with hypercyclicity, supercyclicity, weak hypercyclicity or weak supercyclicity, as demonstrated by the same argument.

Lemma  Assuming that  is a Banach space, and  the operator  acting on it is non-cyclic.

Proof.  Let be distinct from zero. Consequently, the non-zero continuous linear functional  on  is defined by  . We’ve got,

Therefore , the kernel of a non-zero continuous linear functional contains orbit of any non-zero vector under   Consequently , is not cyclic.

Corollary  Suppose that  is a Banach space and  that there is a bounded linear operator   with dense range satisfies   In that case, the operator   acting on  is not cyclical.

Proof.  Given that  we have Let’s say that is cyclic.

Lemma  suggests that is cyclic since   is bounded and has dense range, , which is not feasible based on Lemma (2.2).

Lemma. On a Banach space ,  let   and  be bounded linear operator with dense range that  is cyclic. Additionally, let . Next, the operator functioning

on is cyclic.

Proof. For    Let   be a cyclic vector. The space of polynomials on one variable with complex coefficients is, then  dense in . The spaces are crowded in since  has a dense range. Verifying that  is a cyclic vector is sufficient for  Let   be the orbit’s closed linear span under   and  Then

Thus, includes the vectors of the shape forand 

Since  is closed and   dense in B, we may conclude that

 for.

Finally, the matrix is invertible since its determinant is a Vander Monde type. The latter matrix is invertible, which indicates that the union of   for  spans  . As a result   is a cyclic vector for .

For weighted bilateral shifts, the final lemma can be phrased more elegantly. Remember, if  for any  the weighted bilateral shifts  and then are isometrically similar for each Indeed, take the sequence defined as .

For   and  for  Then  for each  and thus the diagonal operator  it acts on the basic vectors using the formula  for  an invertible isometry. That is easy to check. That is,  and are isometrically comparable. Any   is particularly similar to if  and . This observation together with the previous lemma, leads to the following consequence.

Corollary ).  Let us consider  a weighted bilateral shift that  is cyclic. Then  is cyclical.

Proof. According to lemma (2.4), the operator is cyclic, where   Based on the preceding observation, it follows that  is similar to   for   As a result, the direct sum of   copies of is cyclic, implying that  is cyclic.

The following lemma establishes a sufficient condition for a direct sum of two weighted bilateral shifts to be non-cyclic.

Lemma  Considera bounded succession of non-zero complex numbers,and

Assume there exists such that, where

Then is non-cyclical.

Proof. For brevity   if  and   Consider the bilateral sequence  described by

if  and  if .

It is easy to confirm that  for each 

Since then , we have . Let  be defined by the formula. Based on the definition of  we have  

The Hölder inequality allows us to define a bounded linear operator   on the canonical basis   It is simple to prove, by computing the values of the operators on the basic vectors  that  where is the bounded linear operator  defined as for  .

Assume  it is cyclical. Since  has dense range, Lemma (2.1) implies that  is cyclic, this is impossible, according to Lemma (2.2), if  then  and if  then. In any case,is the direct sum of one operator and its dual.

The following corollary is the special case   of the preceding lemma.

Corollary  Consider  a bounded sequence of non-zero complex numbers, and  if ,  if  Assume that there exists such that , as stated in (8). Thenis not cyclic.

To show the next proposition, we use Lemma  in the instance

Proposition  Consider  a bounded series of non zero complex numbers. Then  is supercyclic if and only if    it is cyclic, where  

Proof.  According to theorem , supercyclicity of  does not depend on  Specifically,  is supercyclic if and only if ′ is supercyclic. So, without losing generality, we can assume that If is not supercyclic,  it satisfies the supercyclicity Criterion, according to the theorem . Since an operator  satisfies the supercyclicity Criterion if and only if  it does, we have that  fulfills the supercyclicity Criterion and  is hence cyclic. Since   is tightly constantly in if  and into  we can see that cyclicity  entails cyclicity  . Thus,  it is cyclic.

Assume that is not supercyclic. The presence of  such that  is not satisfied is implied by the theorem . Then  where is defined as in . It is easy to observe  that

   By Lemma ,  is not cyclic.

Proposition (2.9). Let us consider   a bounded linear operator on a separable Banach space. Then the conditions  of Theorem (1.1), are connected in the following manner:

Proof. By Theorem  implies  The implication follows from the same theorem and the fact that  satisfies the supercyclicity Criterion if and only if  does. Since the powers of a weakly supercyclic operator are weakly supercyclic, we see that  implies  The implications

and are trivial.

Proof of Theorem . According to Corollary (2.5)  implies  By Theorem ,  implies  Taking into account Proposition  we see that it suffices to show that  implies 

If  is cyclic on  then, since  it is cyclic on By Proposition ,  is supercyclic, which  implies 

Proof of Theorem (1.3). Refer to [8, p. 348–349] for general finding on universal families.. Let us consider a family of continuous maps from a complete metric space  to a separable metric space. If and only if the set  is dense in , then the set  of universal elements for   is dense in 

The following theorem can be obtained by directly applying this result to the family

whereis a bounded linear operator on a Banach space.

Theorem DC . Define  a bounded linear operator and a separable Banach space, respectively. If and only if the set is dense in , then the set of cyclic vectors for   is dense in

When a bounded linear operator acts on a subset of a Banach space  and is dense in , we say that the subset  if  is cyclic. The following refinement is allowed by Theorem .

Corollary (3.1). Letbe a bounded linear operator,   a separable Banach space, and  two cyclic subsets of . Additionally, assume that the dual operator’s  point spectrum  has an empty interior. Then the set of cyclic vectors for is thus dense in  if and only if   and  there exist  and   such that for any

Proof. The part that states ’only if’ follows directly from Theorem DC. The ’if’ portion still needs to be proven. Evidently

As a result, for any   and  we have  and for that reson  Given that every  is closed and  is cyclic  we have   for every  . Let   be the collection of polynomials  with all of zeros in  Then  provides a wide range for any Specifically, we observe that the set

is thick  for all purposes   and  Lastly, given that our data

is dense  for every  and 

Using the concept of  and cyclicity of  for  , we obtain  Since   has empty interior

The final two visualizations suggest that the collection is dense in

There is density in  the set of cyclic vectors for  by Theorem DC.

We’ll use the corollary mentioned above for weighted bilateral shifts. The sentence that follows is an  example of a corollary 

Corollary (3.2). Let be a sequence of elements of  such that  is dense in  and

  for each  , and let   be a bounded linear operator on a Banach space. If and only if there exists  and  such that  and  and for any    and any , then the set of cyclic vectors for  is dense 

 The condition ’only if’ is a superfluous outcome of Theorem DC. The ’if’ portion still needs to be proven. Let  Given that  each   of the things we have

 is dense in 

Thus,   a cyclic set is  Allow . Then  and  in some cases . If , then a constant  such that   where  exists. Specifically,   and  for  regarding any  If  and   when the presumptions are met, there is  and  so that  and   There is still corollary (3.1) to apply.

Proposition (3.3). Let    be a sequence of elements of  such that  is dense in  , and let   be a bounded linear operator on a Banach space  and for each 

Moreover, suppose that

Then  contains a dense collection of cyclic vectors.

Proof. Suppose  and  are such that We examine and discuss a polynomial  defined by for any  and  ,

It is simple to verify that    by using the fact that  for each and the summation formula for a finite geometric progression. Therefore

Assume  for the moment that. Next

The last two shows provide us with

Therefore, it can be determined that and   such that the inequality above’s  right hand side does not exceed . In this instance  Given that Corollary implies that possesses a dense set of cyclic vectors, it follows that  from the definition of .

Proof of Theorem . We have previously stated that  for any  the weighted bilateral shifts  and  are isometrically comparable for each  Therefore, we can presume, without losing generality, that for each

Allow  where if   and   if  Seeing that   for each  Clearlyis dense is simple. Since is now   is equal to  it is still necessary to apply Proposition 

Final thoughts

Another dichotomy for weighted bilateral shifts is presented in [20], which is also worth mentioning. In particular, if there   is a weighted bilateral shift on  with ,then it is either ,   weakly convergent to zero or Supercyclic as    for each non-zero  Similarly, weighted bilateral shifts on succeed and weighted bilateral shifts on   fail. We want to emphasize that the following issue is yet unresolved.

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