On subspace convex-cyclic operators

TL;DR

This paper introduces the concept of subspace convex-cyclicity for operators on infinite-dimensional Hilbert spaces, providing conditions for such operators and exploring their relation to invariant subspace problems.

Contribution

It defines subspace convex-cyclicity, establishes sufficient conditions for transitive operators to be subspace convex-cyclic, and presents a counterexample illustrating the concept's nuances.

Findings

Provided a sufficient condition for subspace convex-cyclicity

Related subspace convex-cyclicity to a specialized Kitai criterion

Constructed a counterexample of a non-transitive subspace convex-cyclic operator

Abstract

Let $\mathcal{H}$ be an infinite dimensional real or complex separable Hilbert space. We introduce a special type of a bounded linear operator $T$ and its important relation with invariant subspace problem on $\mathcal{H}$: operator $T$ is said to be is subspace convex-cyclic for a subspace $\mathcal{M}$, if there exists a vector whose orbit under $T$ intersects the subspace $\mathcal{M}$ in a relatively dense set. We give the sufficient condition for a subspace convex-cyclic transitive operator $T$ to be subspace convex-cyclic. We also give a special type of Kitai criterion related to invariant subspaces which implies subspace convex-cyclicity. We conclude showing a counterexample of a subspace convex-cyclic operator which is not subspace convex-cyclic transitive.

Equations

$$\widehat{Orb(T,x)}$$

$$=\left\{P(T)x:P(T):=a_0+a_{1}T+a_2{T}^2+\cdots +a_n{T}^n,n\in \mathbb{N},\sum _{i=0}^n{a}_i=1\right\}.$$

$$\widehat{Orb}(T\oplus I,m)$$

$$=\left\{P(T)x\oplus 0:P\text{is convex polynomial}\right\}$$

$$\subseteq \mathcal{H}\oplus \{0\}=\mathcal{M}.$$

$$CoC(T,\mathcal{M})=\bigcap _{j=1}^{\infty }\bigcup _{P\in \mathcal{P}}P(T{)}^{-1}({\mathcal{B}}_j),$$

$$P(T)(W)\subset \mathcal{M}.$$

$$\Vert (v+x_k)-v\Vert =\Vert x_k\Vert <ϵ,$$

$$\Vert P_k(T)(v+x_k)-u\Vert \le \Vert P_k(T)(v)\Vert +\Vert P_k(T)(x_k)-u\Vert <\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ,$$

$$\underset{j\to \infty }{\lim }\left(j{\xi }_j+\sum _{i=j+1}^{\infty }{\xi }_i\right)=0.$$

$$\Vert P_{k_j}(T)(x)-y_j\Vert$$

$$\le \Vert P_{k_j}(T)(x_j)-y_j\Vert +\sum _{i=1}^{j-1}\Vert P_{k_j}(T)(x_i)\Vert +\sum _{i=j+1}^{\infty }\Vert P_{k_j}(T)(x_i)\Vert .$$

$$\le j{\xi }_j+\sum _{i=j+1}^{\infty }{\xi }_i.$$

$$X=\bigcup _{j\ge 1}\{S^{N_i}{y}_j:i\ge j\}.$$

$$X\subset \bigcap _{j=1}^{\infty }{P}_{N_j}(T{)}^{-1}(\mathcal{M}).$$

$$p_{N_l}(T)(x)$$

$$=\sum _{\lambda =0}^{N_{i-1}}{a}_{\lambda }{T}^{\lambda }x+a_{N_i}{T}^{N_i}x+\sum _{\lambda =N_{i+1}}^{N_l}{a}_{\lambda }{T}^{\lambda }x$$

$$=\sum _{\lambda =0}^{N_{i-1}}{a}_{\lambda }{T}^{\lambda }x+a_{N_i}{y}_i+0\text{Since T is backward shift operator and x=SNiyj}$$

$$=\sum _{\lambda =0}^{N_{l-1}}{a}_{\lambda }{S}^{N_i-\lambda }{y}_j+a_{N_i}{y}_j.$$

$$p_{N_l}(T)(x)$$

$$=\sum _{\lambda =0}^{N_l}{a}_{\lambda }{S}^{-\lambda }x=\sum _{\lambda =0}^{N_l}{a}_{\lambda }{S}^{N_i-\lambda }{y}_j$$

$$p_{N_l}(T)(x)$$

$$=\sum _{\lambda =0}^{N_{l-1}}{a}_{\lambda }{T}^{\lambda }x+a_{N_l}{T}^{N_l}x$$

$$=\sum _{\lambda =0}^{N_{l-1}}{a}_{\lambda }{S}^{N_i\lambda }(y_j)+a_{N_l}{S}^{N_i-N_l}(y_j)$$

$$=\sum _{\lambda =0}^{N_{l-1}}{a}_{\lambda }{S}^{N_i-\lambda }(y_j)+a_{N_l}(y_j).$$

$$P_m(T)e_{n_{j+1}}$$

$$=\sum _{\lambda =0}^m{a}_{\lambda }{e}_{n_{j+1}-\lambda }\in Lin\{e_r:n_j\le r\le m_j\}\subseteq \mathcal{M}.$$

$$\mathcal{M}=\overline{\bigcup _{k\ge }{\mathcal{M}}_k}.$$

$$L_{k+1}\subset Lin\{e_j:n_{k+1}\le j\le \sum _{i=0}^{k+1}{n}_i\}.$$

$$n_{k+1}>2\sum _{i=0}^k{n}_i,$$

$$n_1$$

$$n_2$$

$$n_3$$

$$n_4$$

$$⋮$$

$$So,$$

$$L_0$$

$$L_1$$

$${\mathcal{M}}_1$$

$$L_2$$

Full text

keywords: ergodic dynamical systems, convex-cyclic operators; Kitai criterion; convex-cyclic transitive operators

MSC 2010: 47A16, 37A25.

1 Introduction

Ergodic dynamical systems seem to be of interest for a few decades, with an increasing number of papers appearing lately (see, for example, [1, 2, 3, 4, 5]), a large number of them concerning convex-cyclic operators.

A bounded linear operator $T$ on an infinite dimensional separable Hilbert space is convex-cyclic(see [13]) if there exists a vector $x$ in $\mathcal{H}$ such that $\widehat{Orb(T,x)}=\{P(T)x:P\,\text{is a}\text{ convex polynomial}\}$ is dense in $\mathcal{H}$ and the vector $x$ is said to be convex-cyclic vector for $T$. A bounded linear operator $T$ is said to be cyclic if there exists a vector $x$ in $\mathcal{H}$ such that the linear span of the orbit $[T,x]=span\{T^{n}x:n\in\mathbb{N}\}$ is dense in $\mathcal{H}$ and $x$ is called cyclic vector. If the orbit $Orb(T,x)=\{T^{n}x:n\in\mathbb{N}\}$ itself is dense in $\mathcal{H}$ without of linear span, then $T$ is called hypercyclic and $x$ is called hypercyclic vector. The operator $T$ is said to be supercyclic if the cone generated by $Orb(T,x)$ i.e.$\mathbb{C}Orb(T,x)=\{\lambda T^{n}x:\lambda\in\mathbb{C}\,\text{and}\,n\in\mathbb{N}\}$ is dense in $\mathcal{H}$ and $x$ is called supercyclic vector[11], [12]. In [7] it is mentioned that between a set and its linear span there is a convex hull, from this we get that every hypercyclic operator is convex-cyclic and every convex-cyclic operator is cyclic.

In this work we want to modify the notions given above and introduce the new concept of subspace convex-cyclic operator. In our case the orbit of subspace hypercyclic, subspace supercyclic and subspace convex-cyclic under hypercyclic, supercyclic and convex-cyclic operator respectively, intersected with a given selected subspace is dense in that subspace.

The paper is organized as follows. First we define the concept of subspace convex-cyclic operators and construct an example of subspace convex-cyclic operator which does not need to be convex-cyclic operator(Example 1). Next we prove that being subspace convex-cyclic transitive implies being subspace convex-cyclic operator. Then we show that a “subspace convex-cyclic criterion” holds. To this end we find an interesting relation between our new operator and invariant subspaces. We also show, by giving a proper example, that for being subspace convex-cyclic being transitive and fulfilling the criterion are not necessary conditions. In the end we present some open questions concerning subspace cyclic operators.

2 Definition and Examples

Let $\mathcal{H}$ be an infinite dimensional real or complex separable Hilbert space. Whenever we talk about a subspace $\mathcal{M}$ of $\mathcal{H}$ we will assume that $\mathcal{M}$ is closed topologically. And let $\textbf{B}(\mathcal{H})$ be the algebra of all linear bounded operators on $\mathcal{H}$. We start with our main definition.

Definition 1**.**

Let $T\in\textbf{B}(\mathcal{H})$ and let $\mathcal{M}$ be a non-zero subspace of $\mathcal{H}.$ We say that $T$ is subspace convex-cyclic operator, if there exist $x\in\mathcal{H}$ such that $\widehat{Orb(T,x)}\cap\mathcal{M}$ is dense in $\mathcal{M}$, where

[TABLE]

Such a vector $x$ is said to be a subspace convex-cyclic vector.

We will write $\mathcal{M}$ convex-cyclic instead of subspace convex-cyclic. Moreover, let us define $CoC(T,\mathcal{M}):=\{x\in\mathcal{H}:\widehat{Orb(T,x)}\cap\mathcal{M}\text{\, is dense in }\mathcal{M}\}$   as the set of all subspace convex-cyclic vectors for $\mathcal{M}$.

Remark 1**.**

Note that $\mathcal{M}$ can be any non-empty subset, convex or not.

Example 1**.**

*Let $T$ be a convex-cyclic operator on $\mathcal{H}$ and $I$ be the identity operator on $\mathcal{H}$. Then $T\oplus I:\mathcal{H}\oplus\mathcal{H}\to\mathcal{H}\oplus\mathcal{H}$ is subspace convex-cyclic operator for subspace $\mathcal{M}=\mathcal{H}\oplus\{0\}$ with subspace convex-cyclic vector $x\oplus 0.$

In fact, since $T$ is convex-cyclic operator on $\mathcal{H}$, so there exist $x\in\mathcal{H}$, such that $\widehat{Orb(T,x)}$ is dense in $\mathcal{H}$. Now we can consider the $T\oplus I:\mathcal{H}\oplus\mathcal{H}\to\mathcal{H}\oplus\mathcal{H}.$

Let $\mathcal{M}:=\mathcal{H}\oplus\{0\}\subseteq\mathcal{H}$ and there exist $m:=x\oplus 0$ such that*

[TABLE]

And since $\widehat{Orb(T,x)}$ is dense in $\mathcal{H}$, then $\widehat{Orb}(T\oplus I,(x\oplus 0))\cap\mathcal{H}\oplus\{0\}$ is dense in $\mathcal{H}\oplus\{0\}=\mathcal{M}$ so we get that $T\oplus I$ is a subspace convex-cyclic operator.

Remark 2**.**

The above example shows that if the operator $T$ is subspace convex-cyclic then $T$ does not need to be convex-cyclic. For clarifing that let us recall the following Propositions 1 and 2, from [13] and [9] respectively.

Proposition 1** (see [13]).**

Let $T:X\to X$ be an operator. If $T$ is convex-cyclic, then

$||T||>1$, 2. 2.

$\sup\{||T^{n}||:n\geq 1\}=+\infty$, 3. 3.

$\sup\{||T^{*n}\Lambda||:n\in\mathbb{N}\}=+\infty,\>$* for every $\Lambda\neq 0\>in\>X^{*}$.*

Proposition 2** (see [9]).**

Let $S:X\to X$ and $T:Y\to Y$ be operators. If $S\oplus T$ is hypercyclic then so are $S$ and $T$.

As we mentioned before, every hypercyclic operator is convex-cyclic, so Proposition 2 is of a great usage here.

Remark 3**.**

Clearly $T\oplus I$ is not convex-cyclic operator. In fact, assume that $T\oplus I$ is convex-cyclic on $\mathcal{H}\oplus\mathcal{H},$ then by Proposition 2 the identity operator must be convex-cyclic on $\{0\}$, which is impossible, because the norm of identity operator is equal to one, and by Proposition 1 we get a contradiction.

3 Subspace Convex-Cyclic Transitive Operators

In this section we define subspace convex-cyclic transitive operators, and we will show that they will be subspace convex-cyclic operators. First we state the classical equivalence of topological transitivity[6] and [9], also convex-cyclic operators [7].

Definition 2**.**

Let $T\in\textbf{B}(\mathcal{H})$ and let $\mathcal{M}$ be a non-zero subspace of $\mathcal{H}$. We say that $T$ is $\mathcal{M}$ convex-cyclic transitive with respect to $\mathcal{M}$ if for all non-empty sets $U\subset\mathcal{M}$ and $V\subset\mathcal{M}$, both are relatively open, there exist a convex polynomial $P$ such that $U\cap P(T)(V)\neq\phi$   or   $P(T)^{-1}(U)\cap V\neq\phi$ contains a relatively open non-empty subset of $\mathcal{M}$.

We use the ideas from [6, 9, 12] changing them to work for convex polynomial spans and generalizing them, obtaining the following.

Theorem 1**.**

Let $T\in\textbf{B}(\mathcal{H})$ and let $\mathcal{M}$ be non-zero subspace of $\mathcal{H}$. Then

[TABLE]

where $\mathcal{P}$ is the collection of all convex polynomials and $\{\mathcal{B}_{j}\}$ is a countable open basis for relative topology $\mathcal{M}$ as a subspace of $\mathcal{H}$.

Proof.

Observe that $x\in\bigcap\limits_{j=1}^{\infty}\bigcup\limits_{P\in\mathcal{P}}P(T)^{-1}(B_{j})$, if and only if, for all $j\in\mathbb{N}$, there exist a convex polynomials $P$ such that $x\in P(T)^{-1}(B_{j})$ which implies $P(T)(x)\in B_{j}$. But since $\{B_{j}\}$ is a basis for the relatively topology of $\mathcal{M}$, this occurs if and only if $\widehat{Orb(T,x)}\cap\mathcal{M}$ is dense in $\mathcal{M}$, that is $x\in CoC(T,\mathcal{M})$. ∎

Lemma 1**.**

Let $T\in\textbf{B}(\mathcal{H})$ and let $\mathcal{M}$ be a non-zero subspace of $\mathcal{H}$. Then the following are equivalent:

$T$* is $\mathcal{M}$ convex-cyclic transitive with respect to $\mathcal{M}$.* 2. 2.

for each relatively open subsets $U\;and\;V\;of\;\mathcal{M}$, there exist $P\in\mathcal{P}$ such that $P(T)^{-1}(U)\cap V$ is relatively open subset in $\mathcal{M}$. Where $\mathcal{P}$ is the set of all convex polynomials. 3. 3.

for each relatively open subsets $U\;and\;V\;of\;\mathcal{M}$, there exist $P\in\mathcal{P}$ such that $P(T)^{-1}(U)\cap V\neq\phi$ and $P(T)(\mathcal{M})\subset\mathcal{M}$.

Proof.

$(3)\Rightarrow(2)$ Since $P(T):\mathcal{M}\to\mathcal{M}$ is continuous and we know that $U$ is relatively open in $\mathcal{M}$, then $P(T)^{-1}(U)$ is also relatively open in $\mathcal{M}$. Now, if we take any $V|_{open}\subset\mathcal{M}$, then let $W:=P(T)^{-1}(U)\cap V$, which is open and $W\subset\mathcal{M}.$

$(2)\Rightarrow(1)$ Since for each relatively open subsets $U$ and $V$,  $P(T)^{-1}(U)\cap V$ is relatively open subset in $\mathcal{M}$, so $P(T)^{-1}(U)\cap V\neq\phi;$ now let $W:=P(T)^{-1}(U)\cap V,$ then $\;W|_{open}\subset\mathcal{M}.$

$(1)\Rightarrow(3)$ By definition of $\mathcal{M}$ convex-cyclic transitive, there exist $U$ and $V$ relatively open subsets in $\mathcal{M}$ such that $W:=P(T)^{-1}(U)\cap V\neq\phi,\;$and this set $W$ is relatively open in $\mathcal{M}$, and $W\subset P(T)^{-1}(U)$. Then $P(T)(W)\subset U\;and\;U\subset\mathcal{M}$, so we get that

[TABLE]

Let $x\in\mathcal{M}$, we must show that $P(T)(\mathcal{M})\subset\mathcal{M}$. Take $w_{0}\in W,$ since $W$ is relatively open in $\mathcal{M}$ and $x\in\mathcal{M}$ so there exist $r>0$ such that $w_{0}+rx\in W$.

But $P(T)(W)\subseteq\mathcal{M},$ that is $P(T)(w_{0}+rx)=P(T)(w_{0})+rP(T)x\in\mathcal{M}$, so $P(T)(w_{0})\in\mathcal{M}$ and $\mathcal{M}$ is subspace thus $r^{-1}\left(-P(T)(w_{0})+P(T)(w_{0})+rP(T)(x)\right)\in\mathcal{M}$, that is $P(T)(x)\in\mathcal{M}$. This is true for any $x\in\mathcal{M},$ hence for $P(T)(x)\in\mathcal{M}$, that is $P(T)(\mathcal{M})\subseteq\mathcal{M}.$ ∎

Theorem 2**.**

Let $T\in\textbf{B}(\mathcal{H})$ and let $\mathcal{M}$ be non-zero subspace of $\mathcal{H}$. If $T$ is $\mathcal{M}$ convex-cyclic transitive, then $T$ is $\mathcal{M}$ convex-cyclic.

Proof.

It is a direct consequence of proofs of Lemma 1 and Theorem 1 ∎

Remark 4**.**

It is natural to ask if the converse of Theorem 2 is true or not. We will answer this question later in Proposition 3.

4 Subspace Convex-Cyclic Criterion

This section is devoted to introduce a type of Kitai’s criterion [10], which is a sufficient criterion for an operator to be $\mathcal{M}$ convex-cyclic. Also we will relate it with invariant subspaces and we will see that the converse of Theorem 2 in general is not true.

Theorem 3**.**

Let $T\in\textbf{B}(\mathcal{H})$ and let $\mathcal{M}$ be a non-zero subspace of $\mathcal{H}$. Assume that there exist $X$ and $Y$, dense subsets of $\mathcal{M}$ such that for every $x\in X$ and $y\in Y$ there exist a sequence $\{P_{k}\}_{k\geq 1}$ of convex polynomials such that

$P_{k}(T)x\to 0,\>\forall x\in X$, 2. 2.

for each $y\in Y$, there exists a sequence $\{x_{k}\}$ in $\mathcal{M}$ such that $x_{k}\to 0$ and $P_{k}(T)x_{k}\to y,$ 3. 3.

$\mathcal{M}$* is an invariant subspace for $P_{k}(T)$ for all $k\geq 0$.*

Then $T$ is $\mathcal{M}$ convex-cyclic operator.

Proof.

To prove that $T$ is $\mathcal{M}$ convex-cyclic operator we will use Lemma 1 and Theorem 2. Let $U$ and $V$ be non-empty relatively open subsets of $\mathcal{M}$. We will show that there exists $k\geq 0$ such that $P_{k}(T)(U)\cap V\neq\phi$. Since $X$ and $Y$ are dense in $\mathcal{M}$, there exists $v\in X\cap V$ and $u\in Y\cap U$. Furthermore, since $U$ and $V$ are relatively open, there exists $\epsilon>0$ such that the $\mathcal{M}$-ball centered at $v$ of radius $\epsilon$ is contained in $V$ and the $\mathcal{M}$-ball centered at $u$ of radius $\epsilon$ is contained in $U$. By hypothesis, given these $v\in X$ and $u\in Y$ , one can choose $k$ large enough such that there exists $x_{k}\in\mathcal{M}$ with

$\|P_{k}(T)v\|<\frac{\epsilon}{2}$,  $\|x_{k}\|<\epsilon$  and  $\|P_{k}(T)x_{k}-u\|<\frac{\epsilon}{2}.$ We have:

Since $v\in\mathcal{M}$ and $x_{k}\in\mathcal{M}$, it follows that $v+x_{k}\in\mathcal{M}$. Also, since

[TABLE]

it follows that $v+x_{k}$ is in $\mathcal{M}-ball$ centered at $v$ of radius $\epsilon$ and hence $v+x_{k}\in V.$ 2. 2.

Since $v$ and $x_{k}$ are in $\mathcal{M}$ and $\mathcal{M}$ is invariant under $P_{k}(T),$ it follows that $P_{k}(T)(v+x_{k})\in\mathcal{M}.$ Also

[TABLE]

and hence

$P_{k}(T)(v+x_{k})$ is in the $\mathcal{M}-ball$ centered at $u$ of radius $\epsilon$ and thus $P_{k}(T)(v+x_{k})\in U$.

So by steps (1) and (2), $T$ is $\mathcal{M}$ convex-cyclic transitive and by Theorem 2 we get that $v+x_{k}\in P_{k}(T)^{-1}(U)\cap V$, that is $P_{k}(T)^{-1}(U)\cap V\neq\phi$ which means $T$ is $\mathcal{M}$ convex-cyclic operator. ∎

As we mentioned before, most of the papers related to the topic of convex-cyclic operators depend on the Kitai’s [10] criterion, which does not contain an analogue of our condition 3. We clarify the need of this additional condition in details in Example 3.

Theorem 4**.**

Let $T\in\textbf{B}(\mathcal{H})$ and let $\mathcal{M}$ be a non-zero subspace of $\mathcal{H}$. Assume there exist $X$ and $Y$, subsets of $\mathcal{M}$ where just $Y$ is dense in $\mathcal{M}$ such that for every $x\in X$ and $y\in Y$ there exist a sequence $\{P_{k}\}_{k\geq 0}$ of convex polynomials such that

$P_{k}(T)x\to 0,\>\forall x\in X$** 2. 2.

for each $y\in Y$, there exists a sequence $\{x_{k}\}$ in $\mathcal{M}$ such that $x_{k}\to 0$ and $P_{k}(T)x_{k}\to y,$ 3. 3.

$X\subset\bigcap\limits_{k=1}^{\infty}P_{k}(T)^{-1}(\mathcal{M})$.

Then $T$ is $\mathcal{M}$ convex-cyclic operator.

Proof.

We will use the idea from [11], namely let $\{\xi_{j}\}^{\infty}_{j=1}$ be a sequence of positive number such that

[TABLE]

In fact, from condition (1) for all $\lambda_{j}>0,\quad\|P_{k_{j}}(T)(x)\|<\lambda_{j}$,

and from condition (2) for all $\epsilon_{j}>0\quad\|P_{k_{j}}(T)(x_{j})-y_{j}\|<\epsilon_{j}$.

So, we can define a sequence of positive numbers $\{\xi_{j}\}^{\infty}_{j=1}$ as follows:

$\xi_{i}=\lambda_{i}\quad for\;i=1,2,\cdots,j.$ And $\xi_{i}=\epsilon_{i}\quad for\;i=j+1,\cdots$ such that

[TABLE]

Since $\mathcal{H}$ is separable, we can assume that $Y=\{y_{j}\}^{\infty}_{j=1}$ for some sequence $\{y_{j}\}^{\infty}_{j=1}.$ We can construct a sequence $\{x_{j}\}^{\infty}_{j=1}\subset X$ and $\{k_{j}\}^{\infty}_{j=1}$ of $\{k\}^{\infty}_{j=1}$ by induction. Let $x_{1}\in X$ and $k_{1}$ be such that $\|x_{1}\|+\|P_{k_{1}}(T)(x_{1})-y_{1}\|<\xi_{1}.$ for each $j$ choose $k_{j}$ and $x_{j}\in X$ such that $\|x_{j}\|+\|P_{k_{j}}(T)(x_{i})\|+\|P_{k_{i}}(T)(x_{j})\|+\|P_{k_{j}}(T)(x_{j})-y_{j}\|<\xi_{j}$  for all $i<j.$ Since $\sum\limits_{i=1}^{\infty}\|x_{i}\|<\sum\limits_{i=1}^{\infty}\xi_{i},$ we can let $x=\sum\limits_{i=1}^{\infty}x_{i},$ and $x$ is well defined.

From condition (3) for every $j$ we have $P_{k_{j}}(T)(x)\in\mathcal{M}$ and

[TABLE]

Thus $\lim\limits_{j\to\infty}\|P_{k_{j}}(T)(x)-y_{j}\|=0$, that is there exists $x\in X\subseteq\mathcal{M}$ such that $\widehat{Orb(T,x)}\cap\mathcal{M}$ is dense in $\mathcal{M}$. Then $T$ is $\mathcal{M}$ convex-cyclic operator. ∎

Remark 5**.**

Notice that in Theorem 4 the set $X$ is not dense in $\mathcal{M}$.

Similar to Theorem 3, in previous theorem conditions $(1)$ and $(2)$ are not sufficient for $T$ to be $\mathcal{M}$ convex-cyclic operator, which we show in Example 3 in the next section.

5 Examples

Before we start this section111All examples in $\ell^{p}(\mathbb{N})$ space are based on [14]., we need some notions. Let $(n_{k})_{k=1}^{\infty}$ and $(m_{k})_{k=1}^{\infty}$ be increasing sequences of of positive integers such that $n_{k}<m_{k}<n_{k+1}\,$ for all $k.$ In $\ell^{p},\,p\geq 1,$ denote by $\{e_{n}\}_{n=1}^{\infty}$ the canonical basis for $\ell^{p},$ let $B$, $Be_{n}=e_{n-1},$ be the backward shift operator. Consider the closed linear subspace $\mathcal{M}$ generated by the set $\{e_{j}:n_{k}\leq j\leq m_{k},\;k\geq 1\}$. As an example case we could set $n_{k}=\{1,3,9,27,81,\cdots\}$ and $m_{k}=\{2,4,5,6,7,8,10,11,\cdots,26,\cdots,80,\cdots\}$, then we would obtain $\mathcal{M}:=\{e_{1},e_{2},e_{3},\cdots\}=\{e_{j}:j\in\mathbb{N}\}.$

The following two lemmas show that being $\mathcal{M}$ convex-cyclic operator does not imply $\mathcal{M}$ convex-cyclic transitivity; we have used similar arguments as in [12].

Lemma 2**.**

If $\sup_{k\geq 1}(m_{k}-n_{k})=\infty,$  then $T=2B$ is $\mathcal{M}$ convex-cyclic operator.

Proof.

Let $Y=(y_{j})_{j=1}^{\infty}\subset c_{00}\cap\mathcal{M}$ be a dense subset of $\mathcal{M}$. Since $\sup_{k\geq 1}(m_{k}-n_{k})=\infty,$  for $y_{1}$ there exist $k_{1}$ and $N_{1}$ such that   $|y_{1}|<n_{k_{1}}<N_{1}<N_{1}+|y_{1}|<m_{k_{1}}.$ By induction, it is easy to see that there exist increasing sequences $(N_{j})_{j=1}^{\infty}$  $(k_{j})_{j=1}^{\infty}$   such that for every fixed $j>1$ we have

$|y_{j}|<n_{k_{j}}<N_{j}<N_{j}+|y_{i}|<m_{k_{j}}.$  for all   $1\leq i\leq j,$ 2. 2.

$N_{j}-N_{i}>n_{k_{j}}$  for all   $1\leq i<j,$

Let

[TABLE]

It is clear that $X\subset\mathcal{M}.$ To verify that  $T,X,Y$  and $(N_{j})_{j=1}^{\infty}\,$ satisfy all conditions of Theorem 4, hence that $T$ is $\mathcal{M}$ convex-cyclic operator. Depending on the Example 3 conditions 1 and 2 are holds. It is enough to check that

[TABLE]

Let $x=S^{N_{i}}y_{j}\in X,$ where $i\geq j.$ For every $l$ consider $P_{N_{l}}(T)(x).$ If $l>i,$ so

[TABLE]

Since $N_{i}-\lambda\geq N_{i}-N_{i-1}>n_{k_{i}}$ and $N_{i}-N_{i-1}<N_{i}-\lambda+|y_{j}|<N_{i}+|y_{j}|<m_{k_{i}}$.

So $n_{k_{i}}<N_{i}-\lambda<m_{m_{k}}$.

Hence $p_{N_{l}}(T)(x)=\sum_{\lambda=0}^{N_{l-1}}a_{\lambda}S^{N_{i}-\lambda}y_{j}+a_{N_{i}}y_{j}\in Lin\{e_{r}:n_{k_{i}}\leq r\leq m_{k_{i}}\}+\mathcal{M}\subseteq\mathcal{M}.$ That is $p_{N_{l}}(T)(x)\in\mathcal{M},$ hence $x\in p_{N_{l}}(T)^{-1}\left(\mathcal{M}\right).$

If $l<i,$ then

[TABLE]

But $N_{i}-\lambda\geq N_{i}-N_{l}>n_{k_{i}},$ and $N_{i}-N_{l}\leq N_{i}-\lambda+|y_{l}|<N_{i}+|y_{l}|<m_{k_{i}}$.

So $n_{k_{i}}<N_{i}-\lambda<m_{k_{i}}$

Hence $p_{N_{l}}(T)(x)=\sum_{\lambda=0}^{N_{l}}a_{\lambda}S^{N_{i}-\lambda}y_{j}\in Lin\{e_{r}:n_{k_{i}}\leq r\leq m_{k_{i}}\}\subset\mathcal{M},$ hence $x\in p_{N_{l}}(T)^{-1}\left(\mathcal{M}\right).$

If $l=i,$ then

[TABLE]

Since $N_{i}-\lambda\geq N_{i}-N_{l-1}>N_{k_{i}}$ and $N_{i}-N_{l-1}\leq N_{i}-\lambda+|y_{l}|<N_{i}+|y_{l}|<m_{k_{i}}$.

So $n_{k_{i}}<N_{i}-\lambda<m_{k_{i}}$.

Hence $p_{N_{l}}(T)(x)=\sum_{\lambda=0}^{N_{l-1}}a_{\lambda}S^{N_{i}-\lambda}(y_{j})+a_{N_{l}}(y_{j})Lin\{e_{r}:n_{k_{i}}\leq r\leq m_{k_{i}}\}+\mathcal{M}\subset\mathcal{M}.$

That is $x\in p_{N_{l}}(T)^{-1}\left(\mathcal{M}\right).$

Consequently, $T$ satisfies all conditions in Theorem 4, so $T$ is $\mathcal{M}$ convex-cyclic operator. ∎

Lemma 3**.**

If $\sup_{k\geq 1}(n_{k+1}-m_{k})=\infty,$  then $T=2B$ is not $\mathcal{M}$ convex-cyclic transitive.

Proof.

Suppose that $T=2B$ is $\mathcal{M}$ convex-cyclic transitive. Let $U$ and $V$ be two non-empty open subsets of $\mathcal{M}$ and suppose that there exists a positive number $m$ such that $U\cap P_{m}(T)^{-1}V$ contains an open subset $W$ of $\mathcal{M}$. For $x\in W$, there exist $\epsilon>0$ such that if $y\in\mathcal{M}$ and $||x-y||<\epsilon$; since $W$ is an open, and then $y\in W$.

Since $\sup_{k\geq 1}(n_{k+1}-m_{k})=\infty,$ then $\exists j\in\mathbb{Z}^{+}\;\text{such that}\quad n_{j+1}-m_{j}>m.$

Since $||e_{n_{j+1}}||\leq 1$   and $\frac{1}{2}\epsilon<\epsilon$,   $||\frac{1}{2}\epsilon\;e_{n_{j+1}}||\leq\epsilon.1$

So $||x-(x+\frac{1}{2}\epsilon\;e_{n_{j+1}})||<\epsilon.1$

Now, consider $y:=x+e_{n_{j+1}}\;\frac{1}{2}\epsilon\in\mathcal{M}$ and $||x-y||<\epsilon$ since $y\in\mathcal{M}$ so $y\in W\subset U\cap P_{m}(T)^{-1}V$, and $y\in P_{m}(T)^{-1}V$  which means $P_{m}(T)(y)\in V$  and

$P_{m}(T)(y)=P_{m}(T)\left(x+\frac{1}{2}\epsilon P_{m}(T)e_{n_{j+1}}\right)=P_{m}(T)(x)+\frac{1}{2}\epsilon P_{m}(T)e_{n_{j+1}}\in V\subset\mathcal{M}$

Then we get that $P_{m}(T)e_{n_{j+1}}\in\mathcal{M}$   which is contradiction, since

[TABLE]

That is $n_{j+1}-\lambda\leq m_{j}$ and $n_{j+1}-m\leq n_{j+1}-\lambda\leq m_{j}$. Which is contradicts the fact that $n_{j+1}-m>m_{j}$. ∎

Proposition 3**.**

If $\sup_{k\geq 1}(m_{k}-n_{k})=\sup_{k\geq 1}(n_{k+1}-m_{k})=\infty,$  then $T=2B$ is $\mathcal{M}$ convex-cyclic operator. But $T$ is not $\mathcal{M}$ convex-cyclic transitive.

Example 2**.**

Let $S$ be a linear operator on $\ell^{p},\,p\geq 1,$ defined as $Se_{n}=\frac{1}{2}e_{n+1}.$ Let $\{n_{k}\}_{k=1}^{\infty}$ be an increasing sequence of positive integer numbers such that $n_{0}=0$ and $n_{k+1}>2\sum_{i=1}^{k}n_{i}.$ Let $L_{0}=\mathcal{M}_{0}=Lin\{e_{0}\}$ be the linear space generated by $e_{0},$ $L_{1}=S^{n_{1}}\mathcal{M}_{0},$   $\mathcal{M}_{1}=\mathcal{M}_{0}\oplus L_{1},$ and in general, let $L_{k+1}=S^{n_{k+1}}\mathcal{M}_{k},$   $\mathcal{M}_{k+1}=\mathcal{M}_{k}\oplus L_{k+1}.$ Define

[TABLE]

It is easy to show by induction that $\mathcal{M}_{k}\subset Lin\{e_{j}:j\leq\sum_{i=0}^{k}n_{i}\},$ and

[TABLE]

*Thus if $x\in\mathcal{M},$ then $x(j)=0$  for all $\sum\limits_{i=0}^{k}n_{i}<j<n_{k+1}.$

As a special case, for more clarify $n_{0}=0$ and since*

[TABLE]

So we can suppose that.

[TABLE]

*Also notice that $e_{2}\notin\mathcal{M}_{2}$, and $e_{2},e_{5},e_{6},e_{7},e_{8},e_{11}\notin\mathcal{M}_{3}.$ This can be represented by the the following condition

if $x\in\mathcal{M}$ and $\mathcal{M}=\overline{\bigcup_{k\geq 0}\mathcal{M}_{k}}$, then

for $k=1,$   $x(j)=0$   for $\sum\limits_{i=0}^{k}n_{i}=1<j<3=n_{k+1}$

for $k=2,$   $x(j)=0$   for $\sum\limits_{i=0}^{k}n_{i}=4<j<9=n_{k+1}$.

Thus we can define $\mathcal{M}_{k},\mathcal{M},$ and $L_{k}$ as above, which satisfy all above conditions of Theorem 4.*

Example 3**.**

Let $B$ be an operator defined as above. Then the operator $T=2B$ satisfies the first two conditions of Theorem 4, but $T$ is not $\mathcal{M}$ convex-cyclic operator.

Proof.

Let $Y=\left(y_{j}\right)_{j=1}^{\infty}\subset c_{00}\bigcap\mathcal{M}$ be a dense sequence in $\mathcal{M}$. Then there exists an increasing sequence $\left(k_{j}\right)_{j=1}^{\infty}$ such that $y_{i}\in\mathcal{M}_{k_{j}},i\leq j.$

Let $X:=\bigcup\limits_{j\geq 1}\{S^{n_{k_{i}}}y_{j}\;:\;i\geq j\;\}$

Now we will verify that $T$ with $X,Y$ and $\left(P_{n_{k_{i}}}\right)_{i=1}^{\infty}$; sequence of convex polynomial, satisfies the first two conditions of Theorem 4.

1. First:

Let $x\in X$ be an arbitrary element. Then we will show that $P_{n_{k_{r}}}(T)x\to 0.$ Since $x\in X,$ then there exist $j\geq 1$ such that $x:=S^{n_{k}}y_{j}\;$ for $i\geq j,$ we choose $n_{k_{i}}$ large enough such that $\dfrac{1}{2^{n_{k_{i}}}}\to 0.$

[TABLE] 2. Second:

For each $y\in Y\subset c_{00}\bigcap\mathcal{M},$ we must show that there exists a sequence $\left(x_{k_{i}}\right)_{i=1}^{\infty}$ in $X:=\bigcup\limits_{j\geq 1}\{S^{n_{k_{i}}}y_{j}\;:\;i\geq j\;\}$ such that.

$x_{k_{i}}(y_{j})\to 0$ and $P_{n_{k_{i}}}(T)x_{k_{i}}\to y.$

Since $y\in Y=\left(y_{j}\right)^{\infty}_{j=1},$ then there exist $j\geq 1$ such that $y=y_{j}$ and $y_{j}\in c_{00}\bigcap\mathcal{M}.$ So there exist $k_{i}$ such that $y_{j}\in\mathcal{M}_{k_{i}},$ for $j\leq i.$ But $\mathcal{M}_{k_{i}}=\mathcal{M}_{k_{i}-1}\oplus L_{k_{i}},$ so there exist $x_{k_{i}}\in L_{k_{i}}$ such that $y_{j}=y_{j-1}+x_{k_{i}},$ and $L_{k_{i}}=S^{n_{k_{i}}}(\mathcal{M}_{k_{i}-1})$ and $i\geq j$ so $x_{k_{i}}\in X.$ Hence, the existence of the sequence $\left(x_{k_{i}}\right)_{i=1}^{\infty}$ in $X$ is done.

$x_{k_{i}}=S^{n_{k_{i}}}(y_{j})=\frac{1}{2^{n_{k_{i}}}}y_{j+n_{k_{i}}}$ as $k_{i}\to\infty$ and $(y_{j})\in c_{00}$ then $n_{k_{i}}\to\infty$ and $x_{k_{i}}\to 0.$

To show $P_{n_{k_{i}}}(T)x_{k_{i}}\to y.$ Since

[TABLE]

So the first two conditions of Theorem 4 satisfied.

It remains two show that $T$ is not $\mathcal{M}$ convex-cyclic operator. Suppose that $T$ is $\mathcal{M}$ convex-cyclic operator with $\mathcal{M}$ convex-cyclic vector $x\in\mathcal{M}$. For an arbitrary $k_{i}$, there exists $m$ such that $P_{n_{k_{i}}}(T)x\in\mathcal{M},$ where

[TABLE]

One can choose $l_{i}>k_{i}$, such that

[TABLE]

this implies that $\sum_{\mu=0}^{l_{i}}n_{\mu}\leq\sum_{\mu=0}^{l_{i}}n_{\mu}+m<n_{l_{i}+1}-m$ so

[TABLE]

From Equation (1) we get that $-m+\sum_{\mu=0}^{k_{i}}n_{\mu}$ has negative value, so

[TABLE]

Equation (4) gives us $\sum_{\mu=0}^{l_{i}}n_{\mu}+m<n_{l_{i}+1}<n_{l_{i}+1}+m$ so there exists some positive integer $r$ such that

[TABLE]

that is $\sum_{\mu=0}^{l_{i}}n_{\mu}\leq r-m\leq n_{l_{i}+1}.$ Now from Remark 5 if $x\in\mathcal{M},$ then $x(r-m)=0$ for all $\sum_{\mu=0}^{l_{i}}n_{\mu}\leq r-m\leq n_{l_{i}+1}$, so for $r$ satisfies Equation (5),

[TABLE]

By construction of $\mathcal{M},$ it is easy two see that $\mathcal{M}\bigoplus_{q=0}^{\infty}S^{n_{k_{q}}}\mathcal{M}_{l_{1}}$ for some increasing sequence $\left(n_{k_{q}}\right)_{q=0}^{\infty}$ such that $n_{k_{0}}=0$ and $n_{k_{q}+1}-n_{q}-\sum_{\mu=0}^{l_{i}}n_{\mu}>2m,\quad q\geq 0.$

Now, for all  $n_{k_{q}}\leq r\leq n_{k_{q}}+\sum_{\mu=0}^{k_{i}}n_{\mu}\quad q\geq 1.$

[TABLE]

where $0\leq r-\lambda\leq\sum_{\mu=0}^{{k_{i}}}n_{\mu}=\kappa$   for all $\lambda=0,1,\cdots,n_{k_{q}}$.

So we get for some $x\in X,P_{n_{k_{i}}}(T)x\notin\mathcal{M}$ i.e. $x\in X$ but $x\notin P_{n_{k_{i}}}(T)^{-1}(M).$ ∎

To explain the above proof numerically let $n_{k}=\{0,1,3,9,27,\cdots\}$,  $e_{4}\in X$ since $2^{4}e_{4}=S^{3}(e_{1})$ as $3\geq 1$. We have

[TABLE]

Example 4**.**

Let $\lambda\in\mathbb{C}$ such that $|\lambda|>1$, and consider $T:=\lambda B$ where $B$ is the backward shift on $\ell^{2}.$ Let $\mathcal{M}$ be the subspace of $\ell^{2}$ consisting of all sequences with zeros on the even entries: $\mathcal{M}=\left\{\{a_{n}\}_{n=0}^{\infty}\in\ell^{2}:a_{2k}=0\;\text{for all}\;k\right\}.$ Then $T$ is $\mathcal{M}$ convex-cyclic operator for $\mathcal{M}$.

Proof.

We will apply Theorem 3 to give an alternative proof. Let $X=Y$ be the subsets of $\mathcal{M}$ consisting of all finite sequences; i.e., those sequences that only have a finite number of non-zero entries: this clearly is a dense subset of $\mathcal{M}$. Let $P_{k}(T):=T^{2k}$ where $\{P_{k}\}_{k>1}$ is a sequence of convex polynomial.

Now let us check that conditions $(1),(2)$ and $(3)$ of Theorem 3 hold.

Let $x\in X$. Since $x$ only has finitely many non-zero entries, $P_{k}(T)x$ will be zero eventually for $k$ large enough. Thus $(1)$ holds. Let $y\in Y$ and define $x_{k}:=\frac{1}{\lambda^{2k}}S^{2k}y,$ where$S$ is the forward shift operator on $\ell^{2}$. Each $x_{k}$ is in $\mathcal{M}$ since the even entries of $y$ are shifted by $S^{2k}$ into the even entries of $x_{k}$. We have

$||x_{k}||=\frac{1}{|\lambda|^{2k}}||y||$, and thus it follows that $x_{k}\to 0,$ since $|\lambda|>1.$ Also, because

$P_{k}(T)(x_{k}):=T^{2k}(x_{k})=(\lambda B)^{2k}(x_{k})=(\lambda B)^{2k}\frac{1}{\lambda^{2k}}S^{2k}y=y,$ we have that condition $(2)$ holds.

The fact that condition $(3)$ holds follows from the fact that if a vector has a zero on all even positions then it will also have a zero entry on all even positions after the application of the backward shift any even number of times. So the $\mathcal{M}$ convex-cyclic operator of $T$ now follows. ∎

Remark 6**.**

There is a relation between $\mathcal{M}$ convex-cyclic operator and invariant subspace, as we see in Theorem 1 that $\mathcal{M}$ is invariant for $P_{k}(T)$ for all $k$, also in Example 1 is invariant for $P_{k}(T)$ whenever $k=1$. But the converse is not true, i.e. if $P_{k}(T)$ for all $k$ is $\mathcal{M}$ convex-cyclic operator then it does not need to $\mathcal{M}$ to be invariant under $P_{k}(T)$, as in Example 4 when $k=1$. But subspace $\mathcal{M}$ is invariant under $\mathcal{M}$ convex-cyclic operator for $T^{2k}$.

6 Some questions

We end our paper with some open questions on $\mathcal{M}$ convex-cyclic operators.

As Bourdon and Feldman [8] proved that for hypercyclic operator that somewhere dense orbits are everywhere dense so it is natural to ask:

Open question 1. As any orbit of $\mathcal{M}$ convex-cyclic operator in $\mathcal{M}$ is somewhere dense does it imply it to be everywhere dense in $\mathcal{M}$?

Open question 2. Does there exist $\mathcal{M}$ convex-cyclic operator for $\mathcal{M}$ such that we do not have $P_{k}(T)(M)\subset M$ for any $k$?

H. Rezaei [13] defined convex-cyclic operator and characterized completely such operators on finite dimensional vector spaces, we can also ask:

Open question 3. Is there any $\mathcal{M}$ convex-cyclic operator on finite dimensional vector spaces?

Open question 4. If $T:\mathcal{H}\to\mathcal{H}$ is $\mathcal{M}$ convex-cyclic operator over $\mathcal{M}$ on Hilbert space $\mathcal{H}$, then is $T^{m}$ $\mathcal{M}$ convex-cyclic operator for every integer $m>1$?