Syndetic Sensitivity and Mean Sensitivity for Linear Operators

Abstract

We study syndetic sensitivity and mean sensitivity for linear dynamical systems. For the syndetic sensitivity aspect, we obtain some properties of syndetic sensitivity for adjoint operators and left multiplication operators. We also show that there exists a linear dynamical system $(X\times Y,T\times S)$ such that $(X\times Y,T\times S)$ is cofinitely sensitive but $(X,T)$ and $(Y,S)$ are not syndetically sensitive. For the mean sensitivity aspect, we show that if $(Y,S)$ is sensitive and not mean sensitive, where Y is a complex Banach space, the spectrum of T meets the unit circle. We also obtain some results regarding mean sensitive perturbations.

1. Introduction

By a linear dynamical system (l.d.s.) $(Y,S)$, we mean a Banach space Y and a bounded linear operator $S:Y\to Y$. Throughout the manuscript, we let $0_Y$ be the zero element of the Banach space Y. We denote by I the identity operator. The collection of all positive integers (nonnegative integers, real numbers, respectively) is denoted by $\mathbb{N}$ (${\mathbb{Z}}_{+}$, $\mathbb{R}$, respectively).

Ruelle [1] first used the notion of sensitivity. According to the works by Guckenheimer [2], Auslander and Yorke [3] introduced the notion of sensitivity on the compact metric space. The sensitivity of linear dynamics has been discussed in [4,5,6,7]. Recall that an l.d.s. $(Y,S)$ is sensitive if there is $\delta >0$ such that, for any $y\in Y$ and any neighborhood W of y, there are $x\in W$ and $v\in \mathbb{N}$ with $\left|\right|S^{v}x-S^{v}y\left|\right|>\delta$. Let $\delta >0$. For any nonempty open subset $W\subset Y$, define

$$\begin{array}{c}\hfill {\mathit{N}}_{\mathit{S}}(\mathit{W},\delta )=\left\{\mathit{v}\in {\mathbb{Z}}_{+} : \mathit{there} \mathit{exist} {\mathit{y}}_1,{\mathit{y}}_2\in \mathit{W} \mathit{with} \left|\right|{\mathit{S}}^{\mathit{v}}{\mathit{y}}_1-{\mathit{S}}^{\mathit{v}}{\mathit{y}}_2\left|\right|>\delta \right\}.\end{array}$$

Then, it can be verified that an l.d.s. $(Y,S)$ is sensitive if and only if there is $\delta >0$ such that $N_S(W,\delta )\ne \varnothing$ for any nonempty open subsets $W\subset X$.

Before proceeding, let us recall some notions related to the families. Denote by $\mathcal{Q}=\mathcal{Q}\left({\mathbb{Z}}_{+}\right)$ the collection of all subsets of ${\mathbb{Z}}_{+}$. A subset $\mathcal{S}$ of $\mathcal{Q}$ is a family if $E_1\subset E_2$ and $E_1\in \mathcal{S}$ imply $E_2\in \mathcal{S}$. A subset $S=\left\{a_1<a_2<\cdots \right\}\subset {\mathbb{Z}}_{+}$ is

• thick if, for each $u\in \mathbb{N}$, there exists $a_u\in {\mathbb{Z}}_{+}$ such that $\{a_u,a_u+1,\cdots ,a_u+u\}\subset S$;
• syndetic if there exists $M\in {\mathbb{Z}}_{+}$ such that $a_{u+1}-a_u\le M$ for any $u\in \mathbb{N}$.

Moothathu [8] introduced three stronger forms of sensitivity: cofinite sensitivity, multi-sensitivity and syndetic sensitivity. Subsequently, $\mathcal{F}$-sensitivity for some families was studied in [9,10,11,12,13]. We adapt the notions of syndetic sensitivity and cofinite sensitivity for the linear systems. An l.d.s. $(Y,S)$ is called syndetically sensitive (cofinitely sensitive, respectively) if there exists $\delta >0$ such that $N_S(W,\delta )$ is syndetic (cofinite, respectively) for any nonempty open subset $W\subset Y$. Inspired by [8], we have the following query: Is there a sensitive l.d.s. $(Y,S)$ that is not syndetically sensitive? The answer is yes (see Example 1).

Let $(Y,S)$ be an l.d.s. The collection of all continuous linear functionals on Y is denoted by $B(Y,\mathbb{F})$. Notice that $B(Y,\mathbb{F})$ is the dual space of Y and is denoted by $Y^{*}$. If $y^{*}\in Y^{*}$, then we write $y^{*}\left(y\right)=\langle y,y^{*}\rangle ,y\in Y.$ Let $S^{*}:Y^{*}\to Y^{*}$ be defined by $S^{*}{y}^{*}=y^{*}\circ S$ for any $y^{*}\in Y^{*}$. Then, $S^{*}$ is called the adjoint of S and $(Y^{*},S^{*})$ is an l.d.s. (see, for instance, ([5] Appendix A)).

Inspired by the approach in ([5], Chapter 10), and [14,15], the operator S induces a bounded operator $L_S:\mathcal{K}\to \mathcal{K}$ defined by

$$\begin{array}{c}\hfill L_{S}T=ST\mathit{for} \mathit{each}T\in \mathcal{K},\end{array}$$

where $\mathcal{K}=\overline{span\{<·,y^{*}>y:y^{*}\in Y^{*},y\in Y\}}$ is endowed with the operator norm. It can be verified that $(\mathcal{K},L_S)$ is an l.d.s.

There are many different notions of sensitivity on the compact space, such as mean sensitivity [16], diam-mean sensitivity [17] and mean n-sensitivity [18]. Let $(Y,S)$ be an l.d.s. For every $y\in Y$ and any $\epsilon >0$, define

$$\begin{array}{c}\hfill B(y,\epsilon )=\{x\in Y:||x-y||<\epsilon \}.\end{array}$$

If there is $\delta >0$ such that, for any $y\in Y$ and any $\epsilon >0$, there is $x\in B(y,\epsilon )$ satisfying $\underset{v\to \infty }{limsup}\frac{1}{v}{\displaystyle \sum _{u=0}^{v-1}}∥S^{u}x-S^{u}y∥>\delta$, we say that $(Y,S)$ is mean sensitive. If there exists $\delta >0$ such that for any neighborhood W of Y satisfying $\underset{v\to \infty }{limsup}\frac{1}{v}{\displaystyle \sum _{u=0}^{v-1}}diam\left(S^{u}W\right)>\delta$, we say that $(Y,S)$ is diam-mean sensitive. Let $b\ge 2$. We say that an l.d.s. $(Y,S)$ is mean b-sensitive if there exists $\delta >0$ such that, for any nonempty open subset $W\subset Y$, there exist b pairwise distinct points $y_1,\cdots ,y_b\in W$ with $\underset{v\to \infty }{limsup}\frac{1}{v}{\displaystyle \sum _{u=0}^{v-1}}\underset{1\le c\ne d\le b}{min}\left|\right|S^u{y}_c-S^u{y}_d\left|\right|>\delta$. Naturally, we have the following query: Is there a mean sensitive l.d.s. that is not mean b-sensitive? The answer is no (see Theorem 6).

Matache [19] showed that if $(Y,S)$ is hypercyclic, then $\sigma \left(S\right)$ has a nonvoid intersection with the unit circle. We have established that $\sigma \left(S\right)$ intersects the unit circle for the sensitive l.d.s $(Y,S)$, where $(Y,S)$ is not mean sensitive and Y is a complex Banach space (Theorem 7), and that there is a sensitive l.d.s. $(Y,S)$ that is neither hypercyclic nor mean sensitive (Example 2).

The paper is organized as follows. In Section 2, we recall some results in linear dynamics, which will be used later. In Section 3, we study the adjoint operator. In Section 4, we study the left multiplication operators. In Section 5, we show that there exists an l.d.s. such that

• $(X\times Y,T\times S)$ is cofinitely sensitive;
• $(Y,S)$ is not syndetically sensitive; and
• $(X,T)$ is not syndetically sensitive (Example 1).

In Section 6, we study the mean sensitive system. Let $Y=l^p,1\le p<\infty$, or $c_0$. We prove that $(Y,I+B_{\omega })$ and $(Y,I+F_{\omega })$ are mean sensitive (see Propositions 1, 2 and 4). In Section 7, we study the spectrum property of linear dynamical systems.

2. Preliminaries

In this section, we recall some results in linear dynamics, used in the later discussion.

Let $Z_1$ and $Z_2$ be Banach spaces over $\mathbb{F}$. The map $S:Z_1\to Z_2$ is called a linear operator if $S(\alpha z_1+\beta z_2)=\alpha S{z}_1+\beta S{z}_2$ for any $\alpha ,\beta \in \mathbb{F}$ and any $z_1,z_2\in Z_1$. The collection of all bounded linear operators $S:Z_1\to Z_2$ is denoted by $B(Z_1,Z_2)$. A linear operator $S:Z_1\to Z_2$ is bounded if there is a positive constant M satisfying $\left|\right|S{z}_1\left|\right|\le M\left|\right|z_1\left|\right|$ for any $z_1\in Z_1$. A linear operator $S:Z_1\to Z_2$ is continuous if and only if $S:Z_1\to Z_2$ is bounded.

Let $(Y,S)$ be an l.d.s. Recall that the number $\left|\right|S\left|\right|=\underset{y\in Y,y\ne 0_Y}{sup}\frac{\left|\right|Sy\left|\right|}{\left|\right|y\left|\right|}$ is called the norm of the operator S, and $\left|\right|S\left|\right|=\underset{\left|\right|y\left|\right|=1}{sup}\left|\right|Sy\left|\right|=\underset{\left|\right|y\left|\right|\le 1}{sup}\left|\right|Sy\left|\right|$ (see, for instance, [20]). For $y\in Y$, we call $orb(y,S)=\{y,Sy,S^{2}y,\cdots \}$ the orbit of y under S. An l.d.s $(Y,S)$ is hypercyclic if there is some $y\in Y$ such that $\overline{orb(y,S)}=Y$.

Let $W_1$$W_2$ be the subspaces of Y if $Y=W_1+W_2$ and $W_1\cap W_2=\left\{0_Y\right\}$. Then, we say that Y is the direct sum of the subspaces $W_1$ and $W_2$, and write $Y=W_1\oplus W_2$ (see, for instance, [21], p. 68).

Let $\mathcal{N}\left(S\right)=\{y\in Y:Sy=0_Y\}$, $\mathcal{R}\left(S\right)=\{y\in Y:Sx=y\mathit{forsome}x\in Y\}$. We define the resolvent set of S by $\rho \left(S\right)=\{\lambda \in \mathbb{C}:\mathcal{N}(S-\lambda I)=0_Y\mathit{and}\mathcal{R}(S-\lambda I)=Y\}$ and the spectrum of S by $\sigma \left(S\right)=\mathbb{C}\setminus \rho \left(S\right).$

We state some theorems that will be used in the following.

Theorem 1

([20], Corollary 4.5.2). If y is a non-zero element of a Banach space Y, then there is $y^{*}\in Y^{*}$ such that $\left|\right|y^{*}\left|\right|=1$ and $y^{*}\left(y\right)=<y,y^{*}>=\left|\right|y\left|\right|$.

Theorem 2.

Let $(Y,S)$ be an l.d.s., where Y is a complex Banach space. There exist closed subsets ${\sigma }_1,{\sigma }_2$ such that $\sigma \left(S\right)={\sigma }_1\cup {\sigma }_2$ and ${\sigma }_1,{\sigma }_2$ are disjoint. Then, $Y=M_1\oplus M_2$, where $M_1$ and $M_2$ are S-invariant closed subspaces, $\sigma \left(S{|}_{M_1}\right)={\sigma }_1$ and $\sigma \left(S{|}_{M_2}\right)={\sigma }_2$ (see, for instance, [5], Theorem B.9).

Theorem 3.

Let X, Y be Banach spaces and ${\left\{T_u\right\}}_{u\in \mathbb{N}}\subset B(X,Y)$. If, for every $x\in X$, $\underset{u\in \mathbb{N}}{sup}\left|\right|T_{u}x\left|\right|<\infty$, then $\underset{u\in \mathbb{N}}{sup}\left|\right|T_u\left|\right|<\infty$ (see, for instance, [20], Theorem 4.3.1).

3. Adjoint Operator

In this section, we study the adjoint operator.

Theorem 4.

Let $(Y,S)$ be an l.d.s. Then, $(Y,S)$ is syndetically sensitive if and only if $(Y^{*},S^{*})$ is syndetically sensitive.

Proof. 

Necessity. Assume that $(Y,S)$ is syndetically sensitive; then, there is $\delta >0$ such that $N_S(B(0_Y,\epsilon ),\delta )$ is syndetic for every $\epsilon >0$. Let $\epsilon >0$ and take $v\in N_S(B(0_Y,\epsilon ),\delta )$. Then, there exists $x_v\in B(0_Y,\epsilon )$ such that $\left|\right|S^v{x}_v\left|\right|>\frac{\delta }{2}$. Note that $S^v{x}_v\ne 0_Y$. By Theorem 1, there exists $x_v^{*}\in Y^{*}$ such that $\left|\right|x_v^{*}\left|\right|=1$ and $x_v^{*}\left(S^v{x}_v\right)=<S^v{x}_v,x_v^{*}>=\left|\right|S^v{x}_v\left|\right|$. Set $y_v^{*}=\left|\right|x_v\left|\right|x_v^{*}$. Then, $\left|\right|y_v^{*}\left|\right|=\left|\right|x_v\left|\right|<\epsilon$ and

$$\begin{array}{cc}\hfill \left|\right|{\left(S^{*}\right)}^v{y}_v^{*}\left|\right|& =\left|\right|y_v^{*}\circ S^v\left|\right|\ge \frac{|y_v^{*}\circ S^v\left(x_v\right)|}{\left|\right|x_v\left|\right|}\hfill \\ \hfill & =\frac{\left|\right|x_v\left|\right|·|x_v^{*}\left(S^n{x}_v\right)|}{\left|\right|x_v\left|\right|}\hfill \\ & =x_v^{*}\left(S^v{x}_v\right)>\frac{\delta }{2},\hfill \end{array}$$

which implies that $v\in N_{S^{*}}\left(B(0_{Y^{*}},\epsilon ),\frac{\delta }{2}\right)$. Hence, $N_S(B(0_Y,\epsilon ),\delta )\subset N_{S^{*}}\left(B(0_{Y^{*}},\epsilon ),\frac{\delta }{2}\right)$, and so $N_{S^{*}}\left(B(0_{Y^{*}},\epsilon ),\frac{\delta }{2}\right)$ is syndetic. Let $x^{*}\in Y^{*}$. By linearity, $N_{S^{*}}\left(B(x^{*},\epsilon ),\frac{\delta }{2}\right)=N_{S^{*}}\left(B(0_{Y^{*}},\epsilon ),\frac{\delta }{2}\right)$, and then $N_{S^{*}}\left(B(x^{*},\epsilon ),\frac{\delta }{2}\right)$ is syndetic. Since $x^{*}\in Y^{*}$ and $\epsilon >0$ are arbitrary, we find that $(Y^{*},S^{*})$ is syndetically sensitive.

Sufficiency. Suppose that $(Y^{*},S^{*})$ is syndetically sensitive; then, there exists $\delta >0$ satisfying that $N_{S^{*}}(B(0_{Y^{*}},\epsilon ),\delta )$ is syndetic for any $\epsilon >0$. Let $\epsilon >0$ and take $v\in N_{S^{*}}(B(0_{Y^{*}},\epsilon ),\delta )$. Then, there exists $x^{*}\in B(0_{Y^{*}},\epsilon )$ such that $\left|\right|{\left(S^{*}\right)}^v{x}^{*}\left|\right|>\frac{\delta }{2}$. Since $\left|\right|{\left(S^{*}\right)}^v{x}^{*}\left|\right|=\underset{\left|\right|x\left|\right|=1}{sup}\left|\right|{\left(S^{*}\right)}^v{x}^{*}\left(x\right)\left|\right|$, there exists $x_v\in Y$ with $\left|\right|x_v\left|\right|=1$ such that $|{\left(S^{*}\right)}^v{x}^{*}\left(x_v\right)|>\frac{\delta }{2}$. Set $y_v=\left|\right|x^{*}\left|\right|x_v$. Then, $\left|\right|y_v\left|\right|=\left|\right|x^{*}\left|\right|<\epsilon$ and

$$\begin{array}{cc}\hfill \left|\right|S^v{y}_v\left|\right|& =\left|\right|x^{*}\left|\right|·\left|\right|S^v{x}_v\left|\right|\ge |x^{*}\left(S^v{x}_v\right)|\hfill \\ & =|{\left(S^{*}\right)}^v{x}^{*}\left(x_v\right)|>\frac{\delta }{2},\hfill \end{array}$$

which implies that $v\in N_S\left(B(0_Y,\epsilon ),\frac{\delta }{2}\right)$. Hence, $N_{S^{*}}(B(0_{Y^{*}},\epsilon ),\delta )\subset N_S\left(B(0_Y,\epsilon ),\frac{\delta }{2}\right)$, and so $N_S(B(0_Y,\epsilon ),\frac{\delta }{2})$ is syndetic. Let $y\in Y$ be arbitrary. By linearity, $N_S\left(B(y,\epsilon ),\frac{\delta }{2}\right)$ is syndetic, and so $(Y,S)$ is syndetically sensitive. □

Similar to Theorem 4, we have the following.

Remark 1.

Let $(Y,S)$ be an l.d.s. Then, the cofinite sensitivities of $(Y,S)$ and $(Y^{*},S^{*})$ are equivalent properties.

Corollary 1.

Let $(Y,S)$ be an l.d.s. Then, $(Y,S)$ is diam-mean sensitive if and only if $(Y^{*},S^{*})$ is diam-mean sensitive.

Proof. 

Necessity. Let $\epsilon >0$ and $v\in \mathbb{N}$. Then, $diam\left(S^{v}B(0_Y,\epsilon )\right)>0$ by the diam-mean sensitivity of $(Y,S)$, and so there exists $x_v\in B(0_Y,\epsilon )$ with $x_v\ne 0_Y$ such that $\left|\right|S^v{x}_v\left|\right|>\frac{diam\left(S^{v}B(0_Y,\epsilon )\right)}{3}>0$. Similarly to the proof of the necessity of Theorem 4, there is $y_v^{*}\in B(0_{Y^{*}},\epsilon )$ with $\left|\right|{\left(S^{*}\right)}^v{y}_v^{*}\left|\right|>\frac{diam\left(S^{v}B(0_Y,\epsilon )\right)}{3}$. This implies that $diam\left({\left(S^{*}\right)}^{v}B(0_{Y^{*}},\epsilon )\right)>\frac{diam\left(S^{v}B(0_Y,\epsilon )\right)}{3}$. By linearity, one has $(Y^{*},S^{*})$ is diam-mean sensitive.

Sufficiency. Let $\epsilon >0$ and $v\in \mathbb{N}$. Then, $diam\left({\left(S^{*}\right)}^{v}B(0_{Y^{*}},\epsilon )\right)>0$ by the diam-mean sensitivity of $(Y^{*},S^{*})$, and so there is $A_v\in B(0_{Y^{*}},\epsilon )$ such that $\left|\right|{\left(S^{*}\right)}^v{A}_v\left|\right|>\frac{diam\left({\left(S^{*}\right)}^{v}B(0_{Y^{*}},\epsilon )\right)}{3}$. Similarly to the proof of the sufficiency of Theorem 4, there exists $y_v\in B(0_Y,\epsilon )$ such that $\left|\right|S^v{y}_v\left|\right|>\frac{diam{\left(S^{*}\right)}^{v}B(0_{Y^{*}},\epsilon )}{3}$. This implies that $diam\left(S^{v}B(0_Y,\epsilon )\right)>\frac{diam{\left(S^{*}\right)}^{v}B(0_{Y^{*}},\epsilon )}{3}$. Let $y\in Y$. By linearity,

$$\begin{array}{cc}\hfill \underset{w\to \infty }{lim\; sup}\frac{1}{w}{\displaystyle \sum _{u=0}^{w-1}}diam\left(S^{u}B(y,\epsilon )\right)& >\frac{1}{3}\underset{w\to \infty }{lim\; sup}\frac{1}{w}{\displaystyle \sum _{u=0}^{w-1}}diam\left({\left(S^{*}\right)}^{u}B(0_{Y^{*}},\epsilon )\right).\hfill \end{array}$$

Since $y\in Y$ and $\epsilon >0$ are arbitrary, we find that $(Y,S)$ is diam-mean sensitive. □

4. Left Multiplication Operators

In this section, we study the left multiplication operators.

Theorem 5.

Let $(Y,S)$ be an l.d.s. Then, $(Y,S)$ is syndetically sensitive if and only if $(\mathcal{K},L_S)$ is syndetically sensitive.

Proof. 

Necessity. Since $(Y,S)$ is syndetically sensitive, there is $\delta >0$ satisfying that $N_S(B(0_Y,\epsilon ),\delta )$ is syndetic for any $\epsilon >0$. Let $\epsilon >0$ and take $v\in N_S(B(0_Y,\epsilon ),\delta )$. Then, there exists $y_v\in B(0_Y,\epsilon )$ such that $\left|\right|S^v{y}_v\left|\right|>\frac{\delta }{2}$. Note that $y_v\ne 0_Y$. By Theorem 1, there exists $y_v^{*}\in Y^{*}$ such that $∥y_v^{*}∥=1$ and $y_v^{*}\left(y_v\right)=<y_v,y_v^{*}>=∥y_v∥$. Let $T_v:Y\to Y$ be defined by $T_{v}y=<y,y_v^{*}>y_v$ for any $y\in Y$. Then,

$$\begin{array}{cc}\hfill \left|\right|T_v\left|\right|& =\underset{\left|\right|y\left|\right|\le 1}{sup}\left|\right|T_{v}y\left|\right|=\underset{\left|\right|y\left|\right|\le 1}{sup}\left|\right|<y,y_v^{*}>y_v\left|\right|\hfill \\ & \le \underset{\left|\right|y\left|\right|\le 1}{sup}\left|\right|y\left|\right|·\left|\right|y_v^{*}\left|\right||y_v\left|\right|\le \left|\right|y_v\left|\right|<\epsilon \hfill \end{array}$$

and

$$\begin{array}{cc}\hfill ∥S^v{T}_v∥\ge \frac{∥S^v{T}_v\left(y_v\right)∥}{∥y_v∥}& =\frac{∥S^v{y}_v<y_v,y_v^{*}>∥}{∥y_n∥}\hfill \\ & =∥S^v{y}_v∥>\frac{\delta }{2},\hfill \end{array}$$

which implies that $v\in N_{L_S}\left(B(0_{\mathcal{K}},\epsilon ),\frac{\delta }{2}\right)$. Thus, $N_S(B(0_Y,\epsilon ),\delta )\subset N_{L_S}\left(B(0_{\mathcal{K}},\epsilon ),\frac{\delta }{2}\right)$, and so $N_{L_S}\left(B(0_{\mathcal{K}},\epsilon ),\frac{\delta }{2}\right)$ is syndetic. Let $G\in \mathcal{K}$. By linearity, $N_{L_S}\left(B(0_{\mathcal{K}},\epsilon ),\frac{\delta }{2}\right)=N_{L_S}\left(B(G,\epsilon ),\frac{\delta }{2}\right)$, and so $N_{L_S}\left(B(G,\epsilon ),\frac{\delta }{2}\right)$ is syndetic. This implies that $(\mathcal{K},L_S)$ is syndetically sensitive.

Sufficiency. Since $(\mathcal{K},L_S)$ is syndetically sensitive, there is $\delta >0$ such that $N_{L_S}(B(0_{\mathcal{K}},\epsilon ),\delta )$ is syndetic for any $\epsilon >0$. Let $\epsilon >0$ and take $v\in N_{L_S}(B(0_{\mathcal{K}},\epsilon ),\delta )$. Then, there exists $A_v\in B(0_{\mathcal{K}},\epsilon )$ such that $\left|\right|\left(L_S^v\right)\left(A_v\right)\left|\right|>\frac{\delta }{2}$. Since $\left|\right|\left(L_S^v\right)\left(A_v\right)\left|\right|=\left|\right|S^v{A}_v\left|\right|=\underset{\left|\right|y\left|\right|=1}{sup}\left|\right|S^n\left(A_{n}y\right)\left|\right|$, there exists $y_v\in Y$ such that

• $\left|\right|y_v\left|\right|=1$;
• $\left|\right|S^v{A}_v\left(y_v\right)\left|\right|>\frac{\delta }{2}$;
• $\left|\right|A_v{y}_v\left|\right|\le \left|\right|A_v\left|\right|·\left|\right|y_v\left|\right|<\epsilon$.

This implies that $v\in N_S\left(B(0_Y,\epsilon ),\frac{\delta }{2}\right)$. Thus,

$$\begin{array}{c}\hfill N_{L_S}(B(0_{\mathcal{K}},\epsilon ),\delta )\subset N_S\left(B(0_Y,\epsilon ),\frac{\delta }{2}\right),\end{array}$$

and so $N_S\left(B(0_Y,\epsilon ),\frac{\delta }{2}\right)$ is syndetic. By linearity, $(Y,S)$ is syndetically sensitive. □

Similar to Theorem 5, we have the following.

Remark 2.

Let $(Y,S)$ be an l.d.s. Then, the cofinite sensitivities of $(Y,S)$ and $(\mathcal{K},L_S)$ are equivalent properties.

Corollary 2.

Let $(Y,S)$ be an l.d.s. Then, $(Y,S)$ is diam-mean sensitive if and only if $(\mathcal{K},L_S)$ is diam-mean sensitive.

Proof. 

Necessity. Let $\epsilon >0$ and $v\in \mathbb{N}$. Then, $diam\left(S^{v}B(0_Y,\epsilon )\right)>0$ by the diam-mean sensitivity of $(Y,S)$, and so there exists $y_v\in B(0_Y,\epsilon )$ with $y_v\ne 0_Y$ such that $\left|\right|S^v{y}_v\left|\right|>\frac{diam\left(S^{v}B(0_Y,\epsilon )\right)}{3}>0$. Similarly to the proof of the necessity of Theorem 5, there exists $T_v\in B(0_{\mathcal{K}},\epsilon )$ such that $\left|\right|S^v{T}_v\left|\right|\ge \left|\right|S^v{y}_v\left|\right|>\frac{diam\left(S^{v}B(0_Y,\epsilon )\right)}{3}$. This implies that $diam\left(S^{v}B(0_{\mathcal{K}},\epsilon )\right)\ge \frac{diam\left(S^{v}B(0_Y,\epsilon )\right)}{3}$. By linearity, we find that $(\mathcal{K},L_S)$ is diam-mean sensitive.

Sufficiency. Let $\epsilon >0$ and $v\in \mathbb{N}$. Then, $diam\left({\left(L_S\right)}^{v}B(0_{\mathcal{K}},\epsilon )\right)>0$ by the diam-mean sensitivity of $(\mathcal{K},L_S)$, and so there is $A_v\in B(0_{\mathcal{K}},\epsilon )$ such that $\left|\right|{\left(L_S\right)}^v{A}_v\left|\right|>\frac{diam\left({\left(L_S\right)}^{v}B(0_{\mathcal{K}},\epsilon )\right)}{3}$. Similarly to the proof of the sufficiency of Theorem 5, there exists $y_v\in B(0_Y,\epsilon )$ such that $\left|\right|S^v{y}_v\left|\right|>\frac{diam\left({\left(L_S\right)}^{v}B(0_{\mathcal{K}},\epsilon )\right)}{3}$. This implies that $diam\left(S^{v}B(0_Y,\epsilon )\right)>\frac{diam\left({\left(L_S\right)}^{v}B(0_{\mathcal{K}},\epsilon )\right)}{3}$. Let $y\in Y$. Then,

$$\begin{array}{cc}\hfill \underset{w\to \infty }{limsup}\frac{1}{w}{\displaystyle \sum _{u=0}^{w-1}}diam\left(S^{u}B(y,\epsilon )\right)& \ge \underset{w\to \infty }{limsup}\frac{1}{w}{\displaystyle \sum _{u=0}^{w-1}}\frac{diam\left({\left(L_S\right)}^{u}B(0_{\mathcal{K}},\epsilon )\right)}{3}.\hfill \end{array}$$

Thus, $(Y,S)$ is diam-mean sensitive. □

5. Sensitivity but Not Cofinite Sensitivity

In this section, we show that there exists an l.d.s. $(X\times Y,T\times S)$ such that

• $(X\times Y,T\times S)$ is cofinitely sensitive;
• $(X,T)$ and $(Y,S)$ are not syndetically sensitive.

Lemma 1

([5] Exercise 2.3.1). An l.d.s. $(Y,S)$ is sensitive if and only if $\underset{v\in \mathbb{N}}{sup}∥S^v∥=\infty$.

Example 1.

There is an l.d.s. $(X\times Y,T\times S)$ such that

1. 

$(X,T)$ is sensitive;

2. 

$(X,T)$ is not syndetically sensitive;

3. 

$(Y,S)$ is sensitive;

4. 

$(Y,S)$ is not syndetically sensitive;

5. 

$(X\times Y,T\times S)$ is cofinitely sensitive.

Proof. 

Let $A_1=\underset{a_1 \mathit{times}}{\underbrace{\left(1,\frac{2}{1},\cdots ,\frac{a_1}{a_1-1}\right)}}$, $B_1=\underset{1 \mathit{times}}{\underbrace{\left(\frac{1}{a_1}\right)}}$. If $A_1,\cdots ,A_t$ and $B_1,\cdots ,B_t$ are well defined for $t\ge 1$, then we set

$A_{t+1}=\underset{a_{t+1} \mathit{times}}{\underbrace{\left(\frac{1+{\displaystyle \sum _{u=1}^t}(a_u+u)}{{\displaystyle \sum _{u=1}^t}(a_u+u)},\frac{2+{\displaystyle \sum _{u=1}^t}(a_u+u)}{1+{\displaystyle \sum _{u=1}^t}(a_u+u)},\cdots ,\frac{a_{t+1}+{\displaystyle \sum _{u=1}^t}(a_u+u)}{a_{t+1}-1+{\displaystyle \sum _{u=1}^{t-1}}(a_u+u)}\right)}}$ and

$B_{t+1}=\underset{t+1 \mathit{times}}{\underbrace{\left(\frac{{\displaystyle \sum _{u=1}^t}(a_u+u)}{a_{t+1}+{\displaystyle \sum _{u=1}^t}(a_u+u)}\right)}}$.

Put

$$\begin{array}{c}\hfill \omega ={\left({\omega }_u\right)}_{u\in \mathbb{N}}=(A_1{B}_1{A}_2{B}_2\cdots ).\end{array}$$

Let $C_1=\underset{c_1 \mathit{times}}{\underbrace{\left(1,\frac{2}{1},\cdots ,\frac{c_1}{c_1-1}\right)}}$, $D_1=\underset{1 \mathit{times}}{\underbrace{\left(\frac{1}{c_1}\right)}}$. If $C_1,\cdots ,C_t$ and $D_1,\cdots ,D_t$ are well defined for $t\ge 1$, then we set

$C_{t+1}=\underset{c_{t+1} \mathit{times}}{\underbrace{\left(\frac{1+{\displaystyle \sum _{u=1}^t}(c_u+u)}{{\displaystyle \sum _{u=1}^t}(c_u+u)},\frac{2+{\displaystyle \sum _{u=1}^t}(c_u+u)}{1+{\displaystyle \sum _{u=1}^t}(c_u+u)},\cdots ,\frac{c_{t+1}+{\displaystyle \sum _{u=1}^t}(c_u+u)}{c_{t+1}-1+{\displaystyle \sum _{u=1}^t}(c_u+u)}\right)}}$ and

$D_{t+1}=\underset{t+1 \mathit{times}}{\underbrace{\left(\frac{{\displaystyle \sum _{u=1}^t}(c_u+u)}{c_{t+1}+{\displaystyle \sum _{u=1}^t}(c_u+u)}\right)}}$.

Put

$$\begin{array}{c}\hfill \mu ={\left({\mu }_v\right)}_{v\in \mathbb{N}}=(C_1{D}_1{C}_2{D}_2\cdots ).\end{array}$$

The sequences ${\left\{a_u\right\}}_{u\in \mathbb{N}},{\left\{c_u\right\}}_{u\in \mathbb{N}}$ satisfy the following conditions:

• $a_1>1$, $c_1>2a_1$, $\frac{a_1+c_1+1}{a_1+1}>2a_1$;
• $\frac{a_1+a_2+1}{a_1+1}>2c_1$, $\frac{c_1+a_2+1}{c_1+1}>2c_1$;
• $\frac{a_{v+1}+{\displaystyle \sum _{u=1}^v}(a_u+u)}{{\displaystyle \sum _{u=1}^v}(a_u+u)}>2\frac{c_v+{\displaystyle \sum _{u=1}^{v-1}}(c_u+u)}{{\displaystyle \sum _{u=1}^{v-1}}(c_u+u)}$, $\frac{a_{v+1}+{\displaystyle \sum _{u=1}^v}(c_u+u)}{{\displaystyle \sum _{u=1}^v}(c_u+u)}>2c_v$ for each $v\ge 2$;
• $\frac{c_{v+1}+{\displaystyle \sum _{u=1}^v}(c_u+u)}{{\displaystyle \sum _{u=1}^v}(c_u+u)}>2\frac{a_{v+1}+{\displaystyle \sum _{u=1}^v}(a_u+u)}{{\displaystyle \sum _{u=1}^v}(a_u+u)}$, $\frac{c_{v+1}+{\displaystyle \sum _{u=1}^{v+1}}(a_u+u)}{{\displaystyle \sum _{u=1}^{v+1}}(a_u+u)}>2a_{v+1}$ for each $v\ge 1$.

Let $X=Y=\left\{x={\left(x_u\right)}_{u\in \mathbb{N}}\in {\mathbb{R}}^{\mathbb{N}}:{\displaystyle \sum _{u=1}^{\infty }}|x_u|<\infty \right\}$ with the norm $\left|\right|x\left|\right|={\displaystyle \sum _{u=1}^{\infty }}|x_u|$.

Let $T:X\to X$ be defined by

$$\begin{array}{c}\hfill T(x_1,x_2,x_3,\cdots )=(0,{\omega }_1{x}_1,{\omega }_2{x}_2,{\omega }_3{x}_3,\cdots )\end{array}$$

(1)

for any $(x_1,x_2,x_3,\cdots )\in X$.

Let $S:Y\to Y$ be defined by

$$\begin{array}{c}\hfill S(y_1,y_2,y_3,\cdots )=(0,{\mu }_1{y}_1,{\mu }_2{y}_2,{\mu }_3{y}_3,\cdots )\end{array}$$

(2)

for any $(y_1,y_2,y_3,\cdots )\in Y$.

Now, let us check that the l.d.s. $(X,T)$ has the properties via Claims 4 and 5. We need firstly the following three claims. □

Claim 1. 

For any $v\ge 3$, we have $a_{v+1}>2^{v+1}{a}_1$.

Proof of Claim 1. 

Let $v\ge 3$. By the construction of ${\left\{a_u\right\}}_{u\in \mathbb{N}}$, ${\left\{c_u\right\}}_{u\in \mathbb{N}}$, one has

$$\begin{array}{c}\hfill \frac{a_{v+1}+{\displaystyle \sum _{u=1}^v}(a_u+u)}{{\displaystyle \sum _{u=1}^v}(a_u+u)}>2\frac{c_v+{\displaystyle \sum _{u=1}^{v-1}}(c_u+u)}{{\displaystyle \sum _{u=1}^{v-1}}(c_u+u)}>2^2\frac{a_v+{\displaystyle \sum _{u=1}^{v-1}}(a_u+u)}{{\displaystyle \sum _{u=1}^{v-1}}(a_u+u)}.\end{array}$$

and so one has

$$\begin{array}{cc}\hfill \frac{a_{v+1}+{\displaystyle \sum _{u=1}^v}(a_u+u)}{{\displaystyle \sum _{u=1}^v}(a_u+u)}& >{\left(2^2\right)}^{v-1}\frac{a_1+a_2+1}{a_1+1}\hfill \\ & >{\left(2^2\right)}^{v-1}2c_1>2^{2v}{a}_1(\mathit{by} \mathit{the} \mathit{construction} \mathit{of} {\left\{{\mathit{a}}_{\mathit{u}}\right\}}_{\mathit{u}\in \mathbb{N}},{\left\{{\mathit{c}}_{\mathit{u}}\right\}}_{\mathit{u}\in \mathbb{N}}).\hfill \end{array}$$

(3)

Thus,

$$\begin{array}{cc}\hfill a_{v+1}& >2^{2v}{a}_1\left({\displaystyle \sum _{u=1}^v}(a_u+u)\right)-{\displaystyle \sum _{u=1}^v}(a_u+u)\hfill \\ & =\left(2^{2v}{a}_1-1\right){\displaystyle \sum _{u=1}^v}(a_u+u)>2^{v+1}{a}_1.\hfill \end{array}$$

 □

Claim 2. 

Let $t\ge 1$. Then, ${\displaystyle \prod _{v=r}^{{\displaystyle \sum _{u=1}^t}(a_u+u)}}|{\omega }_v|\le 1$ for any $r\le {\displaystyle \sum _{u=1}^t}(a_u+u)$.

Proof of Claim 2. 

By the construction of ${\left({\omega }_u\right)}_{u\in \mathbb{N}}$, one has that

$$\begin{array}{c}\hfill {\displaystyle \prod _{v=1+{\displaystyle \sum _{u=1}^t}(a_u+u)}^{{\displaystyle \sum _{u=1}^{t+1}}(a_u+u)}}|{\omega }_v|\le 1\mathit{for} \mathit{any}t\ge 1\mathit{and}{\displaystyle \prod _{v=1}^{a_1+1}}|{\omega }_v|=1.\end{array}$$

(4)

Let $t=1$. Then, ${\displaystyle \prod _{v=r}^{{\displaystyle \sum _{u=1}^t}(a_u+u)}}|{\omega }_v|={\displaystyle \prod _{v=r}^{a_1+1}}|{\omega }_v|\le 1$ for any $r\le a_1+1$ by the construction of ${\left({\omega }_u\right)}_{u\in \mathbb{N}}$.

Now, let $t>1$ and take $r\le {\displaystyle \sum _{u=1}^t}(a_u+u)$. Consider the following three cases.

• Case 1: If $1\le r\le a_1+1$, then

  $$\begin{array}{c}\hfill {\displaystyle \prod _{v=r}^{{\displaystyle \sum _{u=1}^t}(a_u+u)}}|{\omega }_v|\le 1\left(\mathit{by}\left(4\right)\right).\end{array}$$
• Case 2: If ${\displaystyle \sum _{u=1}^p}(a_u+u)<r\le {\displaystyle \sum _{u=1}^{p+1}}(a_u+u),1\le p\le t-2$, then

  $$\begin{array}{c}\hfill {\displaystyle \prod _{v=r}^{{\displaystyle \sum _{u=1}^t}(a_u+u)}}|{\omega }_v|\le {\displaystyle \prod _{v=r}^{{\displaystyle \sum _{u=1}^{p+1}}(a_u+u)}}|{\omega }_v|\le 1\left(\mathit{by}\left(4\right)\right).\end{array}$$
• Case 3: If ${\displaystyle \sum _{u=1}^{t-1}}(a_u+u)<r\le {\displaystyle \sum _{u=1}^t}(a_u+u)$, then ${\displaystyle \prod _{v=r}^{{\displaystyle \sum _{u=1}^t}(a_u+u)}}|{\omega }_v|\le 1$ by the construction of ${\left({\omega }_u\right)}_{u\in \mathbb{N}}$. In summary, ${\displaystyle \prod _{v=r}^{{\displaystyle \sum _{u=1}^t}(a_u+u)}}|{\omega }_v|\le 1$ for any $r\le {\displaystyle \sum _{u=1}^t}(a_u+u)$ and any $t\ge 1$.

 □

Claim 3. 

Let $s>1$. For any $a_s+1+{\displaystyle \sum _{u=1}^{s-1}}(a_u+u)\le t\le {\displaystyle \sum _{u=1}^s}(a_u+u)$ and $l\ge 1$, one has ${\displaystyle \prod _{v=l}^{l+t-1}}|{\omega }_v|\le 2$.

Proof of Claim 3. 

Let $s>1$ and take $a_s+1+{\displaystyle \sum _{u=1}^{s-1}}(a_u+u)\le t\le {\displaystyle \sum _{u=1}^s}(a_u+u)$, $l\ge 1$. Then, $l+t-1\ge a_s+1+{\displaystyle \sum _{u=1}^{s-1}}(a_u+u)$. Consider the following three cases.

• Case 1: If $a_s+1+{\displaystyle \sum _{u=1}^{s-1}}(a_u+u)\le l+t-1\le {\displaystyle \sum _{u=1}^s}(a_u+u)$, then $l\le s$, and so

$$\begin{array}{cc}\hfill {\displaystyle \prod _{v=l}^{l+t-1}}|{\omega }_v|& =\left({\displaystyle \prod _{v=l}^{{\displaystyle \sum _{u=1}^{s-1}}(a_u+u)}}|{\omega }_v|\right)\left({\displaystyle \prod _{v=1+{\displaystyle \sum _{u=1}^{s-1}}(a_u+u)}^{l+t-1}}|{\omega }_v|\right)\hfill \\ & \le {\displaystyle \prod _{v=l}^{{\displaystyle \sum _{u=1}^{s-1}}(a_u+u)}}|{\omega }_v|\le 1(\mathit{by} \mathit{Claim} 2\mathit{and} \mathit{by} \mathit{the} \mathit{construction} \mathit{of}{\left({\omega }_u\right)}_{u\in \mathbb{N}}).\hfill \end{array}$$

• Case 2: If $l+t-1>{\displaystyle \sum _{u=1}^s}(a_u+u)$ and $l\le {\displaystyle \sum _{u=1}^s}(a_u+u)$, then

$$\begin{array}{cc}\hfill {\displaystyle \prod _{v=l}^{l+t-1}}|{\omega }_v|& =\left({\displaystyle \prod _{v=l}^{{\displaystyle \sum _{u=1}^s}(a_u+u)}}|{\omega }_v|\right)\left({\displaystyle \prod _{v=1+{\displaystyle \sum _{u=1}^s}(a_u+u)}^{l+t-1}}|{\omega }_v|\right)\left(\mathit{by} \mathit{Claim} 2\right)\hfill \\ & \le {\displaystyle \prod _{v=1+{\displaystyle \sum _{u=1}^s}(a_u+u)}^{l+t-1}}|{\omega }_v|\hfill \\ & \le {\displaystyle \prod _{v=0}^{\left({\displaystyle \sum _{u=1}^s}(a_u+u)\right)-1}}\frac{v+1+{\displaystyle \sum _{u=1}^s}(a_u+u)}{v+{\displaystyle \sum _{u=1}^s}(a_u+u)}=2\left(\mathit{observe} \mathit{that}|{\omega }_v|\le \frac{v}{v-1}\mathit{for} \mathit{any}v>1\right).\hfill \end{array}$$

• Case 3: If $l+t-1>{\displaystyle \sum _{u=1}^s}(a_u+u)$ and $l>{\displaystyle \sum _{u=1}^s}(a_u+u)$, then

$$\begin{array}{c}\hfill {\displaystyle \prod _{v=l}^{l+t-1}}|{\omega }_v|\le {\displaystyle \prod _{v=0}^{\left({\displaystyle \sum _{u=1}^s}(a_u+u)\right)-1}}\frac{v+1+{\displaystyle \sum _{u=1}^s}(a_u+u)}{v+{\displaystyle \sum _{u=1}^s}(a_u+u)}=2\end{array}$$

by the construction of ${\left({\omega }_u\right)}_{u\in \mathbb{N}}$.

In summary, ${\displaystyle \prod _{v=l}^{l+t-1}}|{\omega }_v|\le 2$ for any $a_s+1+{\displaystyle \sum _{u=1}^{s-1}}(a_u+u)\le t\le {\displaystyle \sum _{u=1}^s}(a_u+u)$ and $l\ge 1$, where $s>1$. □

Claim 4. 

$(X,T)$is sensitive.

Proof of Claim 4. 

Let $s>4$ and take $x_s=(0,\cdots ,0,\underset{\stackrel{\uparrow }{1+{\displaystyle \sum _{u=1}^s}(a_u+u)}}{1},0,\cdots )\in X$. Then, $\left|\right|x_s\left|\right|=1$ and

$$\begin{array}{c}\hfill \left|\right|T^{a_{s+1}}{x}_s\left|\right|=\frac{a_{s+1}+{\displaystyle \sum _{u=1}^s}(a_u+u)}{{\displaystyle \sum _{u=1}^s}(a_u+u)}>2^{2s}{a}_1\left(\mathit{by}\left(3\right)\mathit{and}\left(1\right)\right).\end{array}$$

and so $\left|\right|T^{a_{s+1}}\left|\right|=\underset{\left|\right|x\left|\right|=1}{sup}\left|\right|T^{a_{s+1}}x\left|\right|\ge \left|\right|T^{a_{s+1}}{x}_s\left|\right|>2^{2s}{a}_1$. This implies that $\underset{v\in \mathbb{N}}{sup}\left|\right|T^v\left|\right|=\infty$. By Lemma 1, $(X,T)$ is sensitive. □

Claim 5. 

$(X,T)$is not syndetically sensitive.

Proof of Claim 5. 

Assume that $(X,T)$ is syndetically sensitive. Then, there exists $\delta >0$ such that $N_T\left(B(0_X,\frac{\delta }{12}),\delta \right)$ is syndetic. Let $s>1$, $a_s+1+{\displaystyle \sum _{u=1}^{s-1}}(a_u+u)\le v\le {\displaystyle \sum _{u=1}^s}(a_u+u)$ and take $x={\left(x_u\right)}_{u\in \mathbb{N}}\in B(0_X,\frac{\delta }{12})$. By (1),

$$\begin{array}{cc}\hfill \left|\right|T^{v}x\left|\right|& ={\displaystyle \sum _{u=1}^{\infty }}\left({\displaystyle \prod _{r=u}^{u+v-1}}|{\omega }_r|\right)\left|x_u\right|\hfill \\ & \le 2{\displaystyle \sum _{u=1}^{\infty }}|x_u|=2\left|\right|x\left|\right|<\frac{\delta }{6},\hfill \end{array}$$

and so, for any $x_1,x_2\in B(0_X,\frac{\delta }{12})$,

$$\begin{array}{c}\hfill \left|\right|T^v{x}_1-T^v{x}_2\left|\right|<\left|\right|T^v{x}_1\left|\right|+\left|\right|T^v{x}_2\left|\right|<\frac{\delta }{3}.\end{array}$$

This implies that ${\left\{a_s+1+{\displaystyle \sum _{u=1}^{s-1}}(a_u+u),\cdots ,{\displaystyle \sum _{u=1}^s}(a_u+u)\right\}}_{s\ge 2}\subset \mathbb{N}\setminus N_T\left(B(0_X,\frac{\delta }{12}),\delta \right)$. In other words, $\mathbb{N}\setminus N_T\left(B(0_X,\frac{\delta }{12}),\delta \right)$ is thick. Thus, $N_T(B(0_X,\frac{\delta }{12}),\delta )$ is not syndetic. This is a contradiction. □

We will check that $(Y,S)$ has the required properties via Claims 9 and 10. We need firstly the following three claims.

Claim 6. 

For any $v\ge 2$, we have $c_{v+1}>2^{v+1}{a}_1$.

Proof of Claim 6. 

Let $v\ge 2$. By the construction of ${\left\{a_u\right\}}_{u\in \mathbb{N}},{\left\{c_u\right\}}_{u\in \mathbb{N}}$, one has

$$\begin{array}{c}\hfill \frac{c_{v+1}+{\displaystyle \sum _{u=1}^v}(c_u+u)}{{\displaystyle \sum _{u=1}^v}(c_u+u)}>2\frac{a_{v+1}+{\displaystyle \sum _{u=1}^v}(a_u+u)}{{\displaystyle \sum _{u=1}^v}(a_u+u)}>2^2\frac{c_v+{\displaystyle \sum _{u=1}^{v-1}}(c_u+u)}{{\displaystyle \sum _{u=1}^{v-1}}(c_u+u)},\end{array}$$

and so

$$\begin{array}{cc}\hfill \frac{c_{v+1}+{\displaystyle \sum _{u=1}^v}(c_u+u)}{{\displaystyle \sum _{u=1}^v}(c_u+u)}& >{\left(2^2\right)}^{v-1}\frac{c_2+c_1+1}{c_1+1}\hfill \\ & >{\left(2^2\right)}^{v-1}2\frac{a_1+a_2+1}{a_1+1}\hfill \\ & >2^{2v}{c}_1>2^{2v}{a}_1(\mathit{by} \mathit{the} \mathit{construction} \mathit{of} {\left\{{\mathit{a}}_{\mathit{u}}\right\}}_{\mathit{u}\in \mathbb{N}},{\left\{{\mathit{c}}_{\mathit{u}}\right\}}_{\mathit{u}\in \mathbb{N}}).\hfill \end{array}$$

(5)

Thus,

$$\begin{array}{c}\hfill c_{v+1}>\left(2^{2v}{a}_1-1\right){\displaystyle \sum _{u=1}^v}(c_u+u)>2^{v+1}{a}_1.\end{array}$$

 □

Claim 7. 

Let $t\ge 1$. Then, ${\displaystyle \prod _{v=r}^{{\displaystyle \sum _{u=1}^t}(c_u+u)}}|{\mu }_v|\le 1$ for any $r\le {\displaystyle \sum _{u=1}^t}(c_u+u)$.

Proof of Claim 7. 

By the construction of ${\left({\mu }_v\right)}_{v\in \mathbb{N}}$, one has that

$$\begin{array}{c}\hfill {\displaystyle \prod _{v=1+{\displaystyle \sum _{u=1}^t}(c_u+u)}^{{\displaystyle \sum _{u=1}^{t+1}}(c_u+u)}}|{\mu }_v|\le 1\mathit{for} \mathit{any}t\ge 1\mathit{and}{\displaystyle \prod _{v=1}^{m_1+1}}|{\mu }_v|=1.\end{array}$$

(6)

Let $t=1$. Then, ${\displaystyle \prod _{v=r}^{{\displaystyle \sum _{u=1}^t}(c_u+u)}}|{\mu }_v|={\displaystyle \prod _{v=r}^{c_1+1}}|{\mu }_v|\le 1$ for any $r\le c_1+1$ by the construction of ${\left({\mu }_v\right)}_{v\in \mathbb{N}}$.

Now, let $t>1$ and take $r\le {\displaystyle \sum _{u=1}^t}(c_u+u)$. Consider the following three cases.

• Case 1: If $1\le r\le c_1+1$, then

  $$\begin{array}{c}\hfill {\displaystyle \prod _{v=r}^{{\displaystyle \sum _{u=1}^t}(c_u+u)}}|{\mu }_v|\le {\displaystyle \prod _{v=r}^{c_1+1}}|{\mu }_v|\le 1\left(\mathit{by}\left(6\right)\right).\end{array}$$
• Case 2: If ${\displaystyle \sum _{u=1}^p}(c_u+u)<r\le {\displaystyle \sum _{u=1}^{p+1}}(c_u+u),1\le p\le t-2$, then

  $$\begin{array}{c}\hfill {\displaystyle \prod _{v=r}^{{\displaystyle \sum _{u=1}^t}(c_u+u)}}|{\mu }_v|\le {\displaystyle \prod _{v=r}^{{\displaystyle \sum _{u=1}^{p+1}}(c_u+u)}}|{\mu }_v|\le 1\left(\mathit{by}\left(6\right)\right).\end{array}$$
• Case 3: If ${\displaystyle \sum _{u=1}^{t-1}}(c_u+u)<v\le {\displaystyle \sum _{u=1}^t}(c_u+u)$, then ${\displaystyle \prod _{v=r}^{{\displaystyle \sum _{u=1}^t}(c_u+u)}}|{\mu }_v|\le 1$ by the construction of ${\left({\mu }_v\right)}_{v\in \mathbb{N}}$. In summary, ${\displaystyle \prod _{v=r}^{{\displaystyle \sum _{u=1}^t}(c_u+u)}}|{\mu }_v|\le 1$ for any $r\le {\displaystyle \sum _{u=1}^t}(c_u+u)$ and $t\ge 1$.

 □

Claim 8. 

Let $s>1$. For any $c_s+1+{\displaystyle \sum _{u=1}^{s-1}}(c_u+u)\le t\le {\displaystyle \sum _{u=1}^s}(c_u+u)$ and $l\ge 1$, one has ${\displaystyle \prod _{v=l}^{l+t-1}}|{\mu }_v|\le 2$.

Proof of Claim 8. 

Let $s>1$ and take $c_s+1+{\displaystyle \sum _{u=1}^{s-1}}(c_u+u)\le t\le {\displaystyle \sum _{u=1}^s}(c_u+u)$, $l\ge 1$. Then, $l+t-1\ge c_s+1+{\displaystyle \sum _{u=1}^{s-1}}(c_u+u)$. Consider the following three situations.

• Case 1: If $c_s+1+{\displaystyle \sum _{u=1}^{s-1}}(c_u+u)\le l+t-1\le {\displaystyle \sum _{u=1}^s}(c_u+u)$, then $l\le s$, and so

$$\begin{array}{cc}\hfill {\displaystyle \prod _{v=l}^{l+t-1}}|{\mu }_v|& =\left({\displaystyle \prod _{v=l}^{{\displaystyle \sum _{u=1}^{s-1}}(c_u+u)}}|{\mu }_v|\right)\left({\displaystyle \prod _{v=1+{\displaystyle \sum _{u=1}^{s-1}}(c_u+u)}^{l+t-1}}|{\mu }_v|\right)\hfill \\ & \le {\displaystyle \prod _{v=l}^{{\displaystyle \sum _{u=1}^{s-1}}(c_u+u)}}|{\mu }_v|\le 1(\mathit{by} \mathit{Claim} 7\mathit{and} \mathit{by} \mathit{the} \mathit{construction} \mathit{of}{\left({\mu }_v\right)}_{v\in \mathbb{N}}).\hfill \end{array}$$

• Case 2: If $l+t-1>{\displaystyle \sum _{u=1}^s}(c_u+u)$ and $l\le {\displaystyle \sum _{u=1}^s}(c_u+u)$, then

$$\begin{array}{cc}\hfill {\displaystyle \prod _{v=l}^{l+t-1}}|{\mu }_v|& =\left({\displaystyle \prod _{v=l}^{{\displaystyle \sum _{u=1}^s}(c_u+u)}}|{\mu }_v|\right)\left({\displaystyle \prod _{v=1+{\displaystyle \sum _{u=1}^s}(c_u+u)}^{l+t-1}}|{\mu }_v|\right)\left(\mathit{by} \mathit{Claim} 7\right)\hfill \\ & \le {\displaystyle \prod _{v=1+{\displaystyle \sum _{u=1}^s}(c_u+u)}^{l+t-1}}|{\mu }_v|\hfill \\ & \le {\displaystyle \prod _{v=0}^{\left({\displaystyle \sum _{u=1}^s}(c_u+u)\right)-1}}\frac{v+1+{\displaystyle \sum _{u=1}^s}(c_u+u)}{v+{\displaystyle \sum _{u=1}^s}(c_u+u)}=2\left(\mathit{observe} \mathit{that}|{\mu }_v|\le \frac{v}{v-1}\mathit{for} \mathit{any}v>1\right).\hfill \end{array}$$

• Case 3: If $l+t-1>{\displaystyle \sum _{u=1}^s}(c_u+u)$ and $l>{\displaystyle \sum _{u=1}^s}(c_u+u)$, then

$$\begin{array}{c}\hfill {\displaystyle \prod _{v=l}^{l+t-1}}|{\mu }_v|\le {\displaystyle \prod _{v=0}^{\left({\displaystyle \sum _{u=1}^s}(c_u+u)\right)-1}}\frac{v+1+{\displaystyle \sum _{u=1}^s}(c_u+u)}{v+{\displaystyle \sum _{u=1}^s}(c_u+u)}=2\end{array}$$

by the construction of ${\left({\mu }_v\right)}_{v\in \mathbb{N}}$.

In summary, ${\displaystyle \prod _{v=l}^{l+t-1}}|{\mu }_v|\le 2$ for any $c_s+1+{\displaystyle \sum _{u=1}^{s-1}}(c_u+u)\le t\le {\displaystyle \sum _{u=1}^s}(c_u+u)$ and $l\ge 1$, where $s>1$. □

Claim 9. 

$(Y,S)$ is sensitive.

Proof of Claim 9. 

Let $s>4$ and take $y_s=(0,\cdots ,0,\underset{\stackrel{\uparrow }{1+{\displaystyle \sum _{u=1}^s}(c_u+u)}}{1},0,\cdots )\in X$. Then, $\left|\right|y_s\left|\right|=1$ and

$$\begin{array}{c}\hfill \left|\right|S^{c_{s+1}}{y}_s\left|\right|=\frac{c_{s+1}+{\displaystyle \sum _{u=1}^s}(c_u+u)}{{\displaystyle \sum _{u=1}^s}(c_u+u)}>2^{2s}{a}_1\left(\mathit{by}\left(5\right)\mathit{and}\left(2\right)\right).\end{array}$$

and so $\left|\right|S^{c_{s+1}}\left|\right|=\underset{\left|\right|y\left|\right|=1}{sup}\left|\right|S^{c_{s+1}}y\left|\right|\ge \left|\right|S^{c_{s+1}}{y}_s\left|\right|>2^{2s}{a}_1$. This implies that $\underset{v\in \mathbb{N}}{sup}\left|\right|T^v\left|\right|=\infty$. Thus, $(Y,S)$ is sensitive by Lemma 1. □

Claim 10. 

$(Y,S)$ is not syndetically sensitive.

Proof of Claim 10. 

Assume that $(Y,S)$ is syndetically sensitive. Then, there exists $\delta >0$ such that $N_S\left(B(0_Y,\frac{\delta }{12}),\delta \right)$ is syndetic. Let $s>1$, $c_s+1+{\displaystyle \sum _{u=1}^{s-1}}(c_u+u)\le v\le {\displaystyle \sum _{u=1}^s}(c_u+u)$ and take $y={\left(y_u\right)}_{u\in \mathbb{N}}\in B(0_Y,\frac{\delta }{12})$. By (2),

$$\begin{array}{cc}\hfill \left|\right|S^{v}y\left|\right|& ={\displaystyle \sum _{u=1}^{\infty }}\left({\displaystyle \prod _{r=u}^{u+v-1}}|{\mu }_r|\right)\left|y_u\right|\hfill \\ & \le 2{\displaystyle \sum _{u=1}^{\infty }}|y_u|=2\left|\right|y\left|\right|<\frac{\delta }{6},\hfill \end{array}$$

and so, for any $y_1,y_2\in B(0_Y,\frac{\delta }{12})$,

$$\begin{array}{c}\hfill \left|\right|S^v{y}_1-S^v{y}_2\left|\right|<\left|\right|S^v{y}_1\left|\right|+\left|\right|S^v{y}_2\left|\right|<\frac{\delta }{3}.\end{array}$$

This implies that ${\left\{c_s+1+{\displaystyle \sum _{u=1}^{s-1}}(c_u+u),\cdots ,{\displaystyle \sum _{u=1}^s}(c_u+u)\right\}}_{s\ge 2}\subset \mathbb{N}\setminus N_S\left(B(0_Y,\frac{\delta }{12}),\delta \right)$. In other words, $\mathbb{N}\setminus N_S\left(B(0_Y,\frac{\delta }{12}),\delta \right)$ is thick. Thus, $N_S(B(0_Y,\frac{\delta }{12}),\delta )$ is nor syndetic. This is a contradiction. □

Claim 11. 

$(X\times Y,T\times S)$ is cofinitely sensitive.

Proof of Claim 11. 

Let $\epsilon >0$. Then, there is $N>4$ such that $\frac{1}{2^v{a}_1}<\epsilon$ for all $v\ge N$. Now, we show that $[a_{N+1},\infty )\subset N_{T\times S}\left(B((0_X,0_Y),\epsilon ),\frac{1}{2}\right)$. By the construction of ${\left\{a_u\right\}}_{u\in \mathbb{N}},{\left\{c_u\right\}}_{u\in \mathbb{N}}$, one has $[a_{N+1},\infty )=\left({\displaystyle \bigcup _{u\ge N}}[a_{u+1},c_{u+1}]\right)\cup \left({\displaystyle \bigcup _{u\ge N}}[c_{u+1},a_{u+2}]\right)$.

Let $s\ge N$, $a_{s+1}\le v\le c_{s+1}$ and take $y_v=(0,\cdots ,0,\underset{\stackrel{\uparrow }{1+{\displaystyle \sum _{u=1}^s}(c_u+u)}}{\frac{1}{2^N{a}_1}},0,\cdots )\in Y$. Then, $\left|\right|y_v\left|\right|=\frac{1}{2^N{a}_1}<\epsilon$ and

$$\begin{array}{cc}\hfill \left|\right|S^v{y}_v\left|\right|& =\left({\displaystyle \prod _{r=1+{\displaystyle \sum _{u=1}^s}(c_u+u)}^{v+{\displaystyle \sum _{u=1}^s}(c_u+u)}}|{\mu }_r|\right)\frac{1}{2^N{a}_1}\left(\mathit{by}\left(2\right)\right)\hfill \\ & =\frac{v+{\displaystyle \sum _{u=1}^s}(c_u+u)}{{\displaystyle \sum _{u=1}^s}(c_u+u)}\frac{1}{2^N{a}_1}\hfill \\ & \ge \frac{a_{s+1}+{\displaystyle \sum _{u=1}^s}(c_u+u)}{{\displaystyle \sum _{u=1}^s}(c_u+u)}\frac{1}{2^N{a}_1}\hfill \\ & >\frac{2c_s}{2^N{a}_1}(\mathit{by} \mathit{the} \mathit{construction} \mathit{of}{\left\{a_u\right\}}_{u\in \mathbb{N}},{\left\{c_u\right\}}_{u\in \mathbb{N}})\hfill \\ & >\frac{2^s{a}_1}{2^N{a}_1}\ge 1\left(\mathit{by} \mathit{Claim}6\right).\hfill \end{array}$$

This implies that ${\displaystyle {\displaystyle \bigcup _{u\ge N}}}[a_{u+1},c_{u+1}]\subset N_S(B(0_Y,\epsilon ),1)$. Obviously,

$$\begin{array}{c}\hfill N_S(B(0_Y,\epsilon ),1)\subset N_{T\times S}\left(B((0_X,0_Y),\epsilon ),\frac{1}{2}\right),\end{array}$$

and so

$$\begin{array}{c}\hfill \bigcup _{u\ge N}[a_{u+1},c_{u+1}]\subset N_{T\times S}\left(B((0_X,0_Y),\epsilon ),\frac{1}{2}\right).\end{array}$$

(7)

Now, let $s\ge N$, $c_{s+1}\le v\le a_{s+2}$ and take $x_v=(0,\cdots ,0,\underset{\stackrel{\uparrow }{1+{\displaystyle \sum _{u=1}^s}(c_u+u)}}{\frac{1}{2^N{a}_1}},0,\cdots )\in X$. Then, $\left|\right|x_v\left|\right|=\frac{1}{2^N{a}_1}<\epsilon$ and

$$\begin{array}{cc}\hfill \left|\right|T^v{x}_v\left|\right|& =\frac{v+{\displaystyle \sum _{u=1}^{s+1}}(a_u+u)}{{\displaystyle \sum _{u=1}^{s+1}}(a_u+u)}\frac{1}{2^N{a}_1}\left(\mathit{by}\left(1\right)\right)\hfill \\ & \ge \frac{c_{s+1}+{\displaystyle \sum _{u=1}^{s+1}}(a_u+u)}{{\displaystyle \sum _{u=1}^{s+1}}(a_u+u)}\frac{1}{2^N{a}_1}\hfill \\ & >\frac{2a_{s+1}}{2^N{a}_1}(\mathit{by} \mathit{the} \mathit{construction} \mathit{of}{\left\{a_u\right\}}_{u\in \mathbb{N}},{\left\{c_u\right\}}_{u\in \mathbb{N}})\hfill \\ & >\frac{2^{s+1}{a}_1}{2^N{a}_1}>1\left(\mathit{by} \mathit{Claim}1\right).\hfill \end{array}$$

This implies that ${\displaystyle {\displaystyle \bigcup _{u\ge N}}}[c_{u+1},a_{u+2}]\subset N_T(B(0_X,\epsilon ),1)$. Observe that

$$\begin{array}{c}\hfill N_T(B(0_X,\epsilon ),1)\subset N_{T\times S}\left(B((0_X,0_Y),\epsilon ),\frac{1}{2}\right),\end{array}$$

and so ${\displaystyle \bigcup _{u\ge N}}[c_{u+1},a_{u+2}]\subset N_{T\times S}\left(B((0_X,0_Y),\epsilon ),\frac{1}{2}\right)$; hence,

$$\begin{array}{c}\hfill [a_{N+1},\infty )\subset N_{T\times S}\left(B((0_X,0_Y),\epsilon ),\frac{1}{2}\right)\end{array}$$

by (7). Let $(x,y)\in X\times Y$ be arbitrary. By linearity,

$$\begin{array}{c}\hfill N_{T\times S}\left(B((0_X,0_Y),\epsilon ),\frac{1}{2}\right)=N_{T\times S}\left(B((x,y),\epsilon ),\frac{1}{2}\right).\end{array}$$

This implies that $(X\times Y,T\times S)$ is cofinitely sensitive. □

6. Mean Sensitivity

In this section, we study mean sensitive systems. We obtain some results regarding mean sensitive perturbations.

Recall that $(Y,S)$ is absolutely Cesàro bounded if there exists a constant $C>0$ such that $\underset{v\in \mathbb{N}}{sup}\frac{1}{v}{\displaystyle \sum _{u=1}^v}∥S^{u}y∥\le C∥y∥$ for all $y\in Y$.

Theorem 6.

An l.d.s. $(Y,S)$ is mean sensitive if and only if $(Y,S)$ is mean b-sensitive.

Proof. 

Necessity. Since $(Y,S)$ is mean sensitive, one knows that $(Y,S)$ is not absolutely Cesáro bounded, and so there exists $y_0\in Y$ such that $\underset{v\in \mathbb{N}}{sup}\frac{1}{v}{\displaystyle \sum _{u=1}^v}\left|\right|S^u{y}_0\left|\right|=\infty$ by [22], Theorem 4; hence, $\underset{v\in \mathbb{N}}{sup}\frac{1}{v}{\displaystyle \sum _{u=0}^{v-1}}\left|\right|S^u{y}_0\left|\right|=\infty$. Let $b\ge 2$ and take $\epsilon >0$. Then, there exists a strictly increasing sequence ${\left\{m_v\right\}}_{v\in \mathbb{N}}$ with $\frac{1}{m_v}{\displaystyle \sum _{u=0}^{m_v-1}}∥S^u{y}_0∥>\frac{b(b+1)\left|\right|y_0\left|\right|}{\epsilon }$. Since

$$\begin{array}{cc}\hfill \underset{2\le c\ne d\le b+1}{min}∥S^u\left(\frac{y_0}{\left|\right|y_0\left|\right|}\frac{\epsilon }{c}\right)-S^u\left(\frac{y_0}{\left|\right|y_0\left|\right|}\frac{\epsilon }{d}\right)∥& =\frac{\epsilon }{\left|\right|y_0\left|\right|}∥S^u{y}_0∥\underset{2\le c\ne d\le b+1}{min}\left|\frac{1}{c}-\frac{1}{d}\right|\hfill \\ & \ge \frac{\epsilon }{\left|\right|y_0\left|\right|}∥S^u{y}_0∥\frac{1}{b(b+1)}\hfill \end{array}$$

for every $u\ge 0$, one has

$$\begin{array}{c}\hfill \frac{1}{m_v}{\displaystyle \sum _{u=0}^{m_v-1}}\underset{2\le c\ne d\le b+1}{min}∥S^u\left(\frac{y_0}{\left|\right|y_0\left|\right|}\frac{\epsilon }{c}\right)-S^u\left(\frac{y_0}{\left|\right|y_0\left|\right|}\frac{\epsilon }{d}\right)∥>1\end{array}$$

for every $v\in \mathbb{N}$. Let $y\in Y$. By linearity, $y+\frac{y_0}{\left|\right|y_0\left|\right|}\frac{\epsilon }{u}\in B(y,\epsilon )$ for each $2\le u\le b+1$ and

$$\begin{array}{cc}& \frac{1}{m_v}{\displaystyle \sum _{u=0}^{m_v-1}}\underset{2\le c\ne d\le b+1}{min}∥S^u\left(y+\frac{y_0}{\left|\right|y_0\left|\right|}\frac{\epsilon }{c}\right)-S^u\left(y+\frac{y_0}{\left|\right|y_0\left|\right|}\frac{\epsilon }{d}\right)∥>1\hfill \end{array}$$

for every $v\in \mathbb{N}$, which implies that $(Y,S)$ is mean b-sensitive.

Sufficiency. The proof is trivial. □

Inspired by the approaches in [23], Theorem 3.3, [5,24], Corollary 8.3, we obtain the following result.

Let $1\le p<\infty$. Let $l^p=\left\{y={\left(y_u\right)}_{u\in \mathbb{N}}\in {\mathbb{F}}^{\mathbb{N}}:{\displaystyle \sum _{u=1}^{\infty }}|y_u{|}^p<\infty \right\}$ with the norm $\left|\right|y\left|\right|={\left({\displaystyle \sum _{u=1}^{\infty }}|y_u{|}^p\right)}^{1/\phantom{1p}p}$ and let $c_0=\{y={\left(y_u\right)}_{u\in \mathbb{N}}\in {\mathbb{F}}^{\mathbb{N}}:\underset{u\to \infty }{lim}{y}_u=0\}$ with the norm $\left|\right|y\left|\right|=\underset{u\in \mathbb{N}}{sup}\left|y_u\right|$. Then, $l^p$ and $c_0$ are Banach spaces.

Let $\omega ={\left({\omega }_u\right)}_{u\in \mathbb{N}}$ with $\underset{u\in \mathbb{N}}{sup}\left|{\omega }_u\right|<\infty$. Let $B_{\omega }:Y\to Y$ be defined by

$$\begin{array}{c}\hfill B_{\omega }(s_1,s_2,s_3,\cdots )=({\omega }_2{s}_2,{\omega }_3{s}_3,{\omega }_4{s}_4,\cdots )\end{array}$$

(8)

for all $(s_1,s_2,s_3,\cdots )\in Y$.

Proposition 1.

Let $Y=l^p,1\le p<\infty$, and let $\omega ={\left({\omega }_u\right)}_{u\in \mathbb{N}}$ with $\underset{u\in \mathbb{N}}{sup}\left|{\omega }_u\right|<\infty$. Then, $(Y,I+B_{\omega })$ is mean sensitive.

Proof. 

Let $\epsilon >0$. Then, there is $N>0$ satisfying $\frac{1}{v}<\epsilon$ for every $v\ge N$. Take $y_{\epsilon }=\left(0,\frac{1}{N},0,0,\cdots \right)\in Y$. Then, $\left|\right|y_{\epsilon }\left|\right|=\frac{1}{N}<\epsilon$ and

$$\begin{array}{cc}\hfill ∥{(I+B_{\omega })}^v\left(y_{\epsilon }\right)∥& =∥\left({\displaystyle \sum _{u=0}^v}\left(\begin{array}{c}v\\ u\end{array}\right)B_{\omega }^u\right)\left(y_{\epsilon }\right)∥\left(\mathit{by}\left(8\right)\right)\hfill \\ & >1\hfill \end{array}$$

for any $v>\frac{N}{|{\omega }_2|}$, which implies that there is $M>0$ satisfying $\frac{1}{v}{\displaystyle \sum _{u=0}^{v-1}}\left|\right|{(I+B_{\omega })}^u{y}_{\epsilon }\left|\right|>\frac{1}{2}$ for any $v\ge M$. By linearity, one finds that $(Y,I+B_{\omega })$ is mean sensitive. □

Let $\omega ={\left({\omega }_u\right)}_{u\in \mathbb{N}}$ with $\underset{u\in \mathbb{N}}{sup}\left|{\omega }_u\right|<\infty$. Let $F_{\omega }:Y\to Y$ be defined by

$$\begin{array}{c}\hfill F_{\omega }(s_0,s_1,s_2,\cdots )=(0,{\omega }_0{s}_0,{\omega }_1{s}_1,{\omega }_2{s}_2,\cdots )\end{array}$$

(9)

for all $(s_0,s_1,s_2,\cdots )\in Y$ (see [5], p. 98).

Proposition 2.

Let $Y=l^p$, $1\le p<\infty$, and let $\omega ={\left({\omega }_u\right)}_{u\in \mathbb{N}}$ with $\underset{u\in \mathbb{N}}{sup}\left|{\omega }_u\right|<\infty$. Then, $(Y,I+F_{\omega })$ is mean sensitive.

Proof. 

Let $\epsilon >0$. There exists $L>0$ such that $\frac{1}{v}<\epsilon$ for any $v\ge L$. Take $y_{\epsilon }=(\frac{1}{L},0,0,\cdots )\in Y$. Then, $\left|\right|y_{\epsilon }\left|\right|=\frac{1}{L}<\epsilon$. By (9),

$$\begin{array}{c}\hfill F_{\omega }^v\left(y_{\epsilon }\right)=\left(\underset{v \mathit{times}}{\underbrace{0,\cdots ,0}},\frac{{\displaystyle \prod _{u=1}^v}|{\omega }_u|}{L},0,\cdots \right),\end{array}$$

and so

$$\begin{array}{cc}\hfill ∥{(I+F_{\omega })}^v\left(y_{\epsilon }\right)∥& =∥\left({\displaystyle \sum _{u=0}^v}\left(\begin{array}{c}v\\ u\end{array}\right)F_{\omega }^u\right)\left(y_{\epsilon }\right)∥\hfill \\ & >1,\hfill \end{array}$$

which implies that there is $M>0$ satisfying $\frac{1}{v}{\displaystyle \sum _{u=0}^{v-1}}\left|\right|{(I+F_{\omega })}^u{y}_{\epsilon }\left|\right|>\frac{1}{2}$ for any $v\ge M$. Thus, $(Y,I+F_{\omega })$ is mean sensitive. □

Proposition 3.

Let $Y=l^p$, $1\le p<\infty$, and let $\omega ={\left({\omega }_u\right)}_{u\in \mathbb{N}}$ with $\underset{u\in \mathbb{N}}{sup}\left|{\omega }_u\right|<\infty$. Then, $(Y,I+F_{\omega })$ is not hypercyclic.

Proof. 

Let $x_1=(1,0,0,\cdots )\in Y$. Now, we show that for any $y={\left(y_u\right)}_{u\in \mathbb{N}}\in B\left(x_1,\frac{1}{4}\right)$, $orb(y,I+F_{\omega })$ is not dense in Y. Since $y={\left(y_u\right)}_{u\in \mathbb{N}}\in B(x_1,\frac{1}{4})$, one has $|y_1|>\frac{1}{2}$ and $|y_2|\le {\left(|y_1-1{|}^p+|y_2{|}^p+{\displaystyle \sum _{u=3}^{\infty }}|y_u{|}^p\right)}^{1/\phantom{1p}p}=\left|\right|y-x_1\left|\right|<\frac{1}{4}$. In fact, if $|y_1|\le \frac{1}{2}$, then

$$\begin{array}{cc}\hfill \left|\right|y-x_1\left|\right|& ={\left(|y_1-1{|}^p+{\displaystyle \sum _{u=2}^{\infty }}|y_u{|}^p\right)}^{1/\phantom{1p}p}\hfill \\ & \ge |y_1-1|\ge ||y_1|-1|\ge \frac{1}{2}.\hfill \end{array}$$

This is a contradiction.

By (9),

$$\begin{array}{cc}\hfill \left|\right|{(I+F_{\omega })}^{v}y\left|\right|& =∥\left({\displaystyle \sum _{u=0}^v}\left(\begin{array}{c}v\\ u\end{array}\right)F_{\omega }^u\right)y∥=\left|\right|(y_1,y_2+v{\omega }_1{y}_1,\cdots )\left|\right|\hfill \\ & \ge |y_2+v{\omega }_1{y}_1|\ge |v{\omega }_1{y}_1|-|y_2|>\frac{|v{\omega }_1|}{2}-\frac{1}{4}\hfill \end{array}$$

for any $v>\frac{1}{|{\omega }_1|}$, which implies that $\underset{v\to \infty }{lim}\left|\right|{(I+F_{\omega })}^{v}y\left|\right|=\infty$. Thus, $(Y,I+F_{\omega })$ is not hypercyclic. □

Similar to Propositions 1–3, we have the following.

Proposition 4.

Let $Y=c_0$ and let $\omega ={\left({\omega }_u\right)}_{u\in \mathbb{N}}$, satisfying $\underset{u\in \mathbb{N}}{sup}\left|{\omega }_u\right|<\infty$. Then,

1. 

$(Y,I+F_{\omega })$ is not hypercyclic;

2. 

$(Y,I+F_{\omega })$ is mean sensitive;

3. 

$(Y,I+B_{\omega })$ is mean sensitive.

7. Spectrum Property

In this section, we study the spectrum property for sensitive operators.

Lemma 2

([5], Lemma 5.2). Let $(Y,S)$ be an l.d.s, where Y is a complex Banach space. Let $s>0$. Then, one has the following:

1. 

if $\sigma \left(S\right)\subset \{w\in \mathbb{C}:|w|<s\}$, then there exist $0<\eta <s$ and $W>0$ such that $\left|\right|S^{v}y\left|\right|\le W{(s-\eta )}^v\left|\right|y\left|\right|$ for any $y\in Y$ and any $v\in \mathbb{N}$;

2. 

if $\sigma \left(S\right)\subset \{w\in \mathbb{C}:|w|>s\}$, then there exist $\eta >0$ and $W>0$ such that $\left|\right|S^{v}y\left|\right|\ge W{(s+\eta )}^v\left|\right|y\left|\right|$ for any $y\in Y$ and any $v\in \mathbb{N}$.

Theorem 7.

If a sensitive l.d.s. $(Y,S)$ is not mean sensitive, then $\sigma \left(S\right)\cap \mathbb{S}\ne \varnothing$, where $\mathbb{S}=\{w\in \mathbb{C}:|w|=1\}$ and Y is a complex Banach space.

Proof. 

Assume that $\sigma \left(S\right)$ does not intersect the unit circle.

If $\sigma \left(S\right)\subset \left\{w\in \mathbb{C}:\left|w\right|<1\right\}$, then there exist $0<{\epsilon }_1<1$ and $M_1>0$ satisfying $\left|\right|S^{v}y\left|\right|\le M_1{(1-{\epsilon }_1)}^v\left|\right|y\left|\right|$ for every $y\in Y$ and any $v\in \mathbb{N}$ by Lemma 2. This implies that, for any $y\in Y$, $\underset{v\in \mathbb{N}}{sup}\left|\right|S^{v}y\left|\right|<\infty$. This is a contradiction. In fact, since $(Y,S)$ is sensitive, by Lemma 1$\underset{v\in \mathbb{N}}{sup}∥S^v∥=\infty$, and then by Theorem 3, there is $y_0\in Y$ satisfying $\underset{v\in \mathbb{N}}{sup}\left|\right|S^v{y}_0\left|\right|=\infty$.

If $\sigma \left(S\right)\subset \{w\in \mathbb{C}:|w|>1\}$, then there exist ${\epsilon }_2>0$ and $M_2>0$ satisfying $\left|\right|S^{v}y\left|\right|\ge M_2{(1+{\epsilon }_2)}^v\left|\right|y\left|\right|$ for every $y\in Y$ and any $v\in \mathbb{N}$ by Lemma 2. This implies that $\underset{v\to \infty }{lim}\left|\right|S^{v}y\left|\right|=\infty$ for any $y\in Y$ with $y\ne 0_Y$. Thus, $\underset{v\to \infty }{lim}\frac{1}{v}{\displaystyle \sum _{u=0}^{v-1}}\left|\right|S^{u}y\left|\right|=\infty$ for any $y\in Y$ with $y\ne 0_Y$. Take $y_0\ne 0_Y$. Then, $\underset{v\to \infty }{lim}\frac{1}{v}{\displaystyle \sum _{u=0}^{v-1}}\left|\right|S^u{y}_0\left|\right|=\infty$. Let $\epsilon >0$. There is $N_{\epsilon }>0$ satisfying $\frac{1}{v}{\displaystyle \sum _{u=0}^{v-1}}∥S^u\left(\frac{y_0}{\left|\right|y_0\left|\right|}\frac{\epsilon }{2}\right)∥>1$ for any $v\ge N_{\epsilon }$. By linearity, one finds that $(Y,S)$ is mean sensitive. This is a contradiction.

Set ${\sigma }_1=\sigma \left(S\right)\cap \{w\in \mathbb{C}:|w|<1\}$ and ${\sigma }_2=\sigma \left(S\right)\cap \{w\in \mathbb{C}:|w|>1\}$. Then,

• ${\sigma }_1$ and ${\sigma }_2$ are closed and disjoint;
• $\sigma \left(T\right)={\sigma }_1\cup {\sigma }_2$.

By Theorem 2, there exist $M_1$ and $M_2$ such that

• $M_1$ and $M_2$ are nontrivial S-invariant closed subspaces;
• $Y=M_1\oplus M_2$;
• $\sigma \left(S{|}_{M_1}\right)={\sigma }_1$;
• $\sigma \left(S{|}_{M_2}\right)={\sigma }_2$.

Note that $\sigma \left(S{|}_{M_2}\right)={\sigma }_2$. Similarly, we have that $(M_2,S{|}_{M_2})$ is mean sensitive by Lemma 2. Thus, there is $\delta >0$ such that, for any $\epsilon >0$, there exists $y\in B(0_Y,\epsilon )$ such that $\underset{v\to \infty }{limsup}\frac{1}{v}{\displaystyle \sum _{u=0}^{v-1}}\left|\right|S^{i}y\left|\right|>\delta$. This implies that $(Y,S)$ is mean sensitive. This is a contradiction. In summary, $\sigma \left(S\right)\cap \mathbb{S}\ne \varnothing$, where $\mathbb{S}=\{w\in \mathbb{C}:|w|=1\}$. □

Recall that a linear dynamical system $(Y,S)$ is Li–Yorke sensitive if there is $\delta >0$ such that, for any $y\in Y$ and any $\epsilon >0$, there exists $x\in B(y,\epsilon )$ with $\underset{v\to \infty }{limsup}\left|\right|S^{v}x-S^{v}y\left|\right|>\delta$ and $\underset{v\to \infty }{liminf}\left|\right|S^{v}x-S^{v}y\left|\right|=0$.

Example 2.

There is a non hypercyclic, sensitive l.d.s. $(Y,S)$ that is not mean sensitive.

Proof. 

Let $L_1=\left(\underset{l_1-1 \mathit{times}}{\underbrace{1,1,\cdots ,1}},{\left(\frac{1}{l_1-1}\right)}^{1-{\epsilon }_0}\right)$,

$L_2={\left({\left(\frac{v}{v-1}\right)}^{1-{\epsilon }_0}\right)}_{l_1+1\le v\le {\displaystyle \sum _{u=1}^2}{l}_u}=\left({\left(\frac{l_1+1}{l_1}\right)}^{1-{\epsilon }_0},{\left(\frac{l_1+2}{l_1+1}\right)}^{1-{\epsilon }_0},\cdots ,{\left(\frac{{\displaystyle \sum _{u=1}^2}{l}_u}{\left({\displaystyle \sum _{u=1}^2}{l}_u\right)-1}\right)}^{1-{\epsilon }_0}\right)$.

If $L_1,L_2,\cdots ,L_{2t-1},L_{2t}$ are are well defined for $t\ge 1$, then we set

$L_{2t+1}=\left(\underset{l_{2t+1}-1 \mathit{times}}{\underbrace{1,1,\cdots ,1}},{\left(\frac{1}{l_{2t+1}-1}\right)}^{1-{\epsilon }_0}\right)$ and

$$\begin{array}{cc}\hfill L_{2(t+1)}& ={\left({\left(\frac{v}{v-1}\right)}^{1-{\epsilon }_0}\right)}_{\left({\displaystyle \sum _{u=1}^{2t+1}}{l}_u\right)+1\le v\le {\displaystyle \sum _{u=1}^{2(t+1)}}{l}_u}.\hfill \end{array}$$

In order to obtain the desired properties, we further require ${\left\{l_t\right\}}_{t\in \mathbb{N}}\subset \mathbb{N}$, ${\epsilon }_0$ to satisfy the following conditions:

• $l_1>4,l_t>l_{t-1}$;
• $l_{2t}=(l_{2t-1}-2){\displaystyle \sum _{u=1}^{2t-1}}{l}_u$;
• $l_{2t+1}=l_{2t}+1$;
• ${\epsilon }_0\in (0,\frac{1}{5})$.

Put

$$\begin{array}{cc}\hfill \omega & ={\left({\omega }_u\right)}_{u\in \mathbb{N}}=(L_1{L}_2{L}_3{L}_4\dots ).\hfill \end{array}$$

Let $Y=\left\{y=(y_1,y_2,\cdots ,y_u,\cdots )\in {\mathbb{C}}^{\mathbb{N}}:{\displaystyle \sum _{u=1}^{\infty }}|y_u|<\infty \right\}$ with the norm $\left|\right|y\left|\right|={\displaystyle \sum _{u=1}^{\infty }}|y_u|$. Let $S:Y\to Y$ be defined by

$$\begin{array}{c}\hfill S(y_1,y_2,y_3,\cdots )=({\omega }_2{y}_2,{\omega }_3{y}_3,\cdots )\end{array}$$

for any $(y_1,y_2,y_3,\cdots )\in Y$.

Now, we show that $(Y,S)$ is not hypercyclic. Note that

$$\begin{array}{cc}\hfill {\displaystyle \prod _{v=1}^{{\displaystyle \sum _{u=1}^{2t}}{l}_u}}|{\omega }_v|& ={\displaystyle \prod _{v=1}^t}{\left(\frac{{\displaystyle \sum _{u=1}^{2v}}{l}_u}{(l_{2v-1}-1){\displaystyle \sum _{u=1}^{2v-1}}{l}_u}\right)}^{1-{\epsilon }_0}\hfill \\ & =1.\hfill \end{array}$$

for all $t\ge 1$.

Let $t\in \mathbb{N}$. Then, there exists $k_t\in \mathbb{N}$ such that ${\displaystyle \prod _{v=1}^t}|{\omega }_v|\le {\displaystyle \prod _{v=1}^{{\displaystyle \sum _{u=1}^{2k_t}}{l}_u}}|{\omega }_v|=1$ by the construction of $\omega$. Thus, $\underset{t\in \mathbb{N}}{sup}\left\{{\displaystyle \prod _{v=1}^t}|{\omega }_v|\right\}=1$, and so $(Y,S)$ is not hypercyclic by [5], Example 4.9(a).

Define $B_{{\omega }_0}:Y\to Y$ by

$$\begin{array}{c}\hfill B_{{\omega }_0}\left(y_1,y_2,y_3,\cdots \right)=\left({\omega }_2^0{y}_2,{\omega }_3^0{y}_3,{\omega }_4^0{y}_4,\cdots \right)\end{array}$$

for any $(y_1,y_2,y_3,\cdots )\in Y$, where ${\omega }_0={\left({\omega }_u^0\right)}_{u\in \mathbb{N}}$, ${\omega }_u^0={\left(\frac{u}{u-1}\right)}^{1-{\epsilon }_0}$ for any $u\ge 2$.

Let $y={\left(y_u\right)}_{u\in \mathbb{N}}\in Y$ and any $v\in \mathbb{N}$. Then,

$$\begin{array}{c}\hfill S^{v}y=\left(\left({\displaystyle \prod _{u=2}^{v+1}}{\omega }_u\right)y_{v+1},\left({\displaystyle \prod _{u=3}^{v+2}}{\omega }_u\right)y_{v+2},\cdots ,\left({\displaystyle \prod _{u=t+1}^{v+t}}{\omega }_u\right)y_{v+t},\cdots \right),\end{array}$$

and so

$$\begin{array}{cc}\hfill ∥S^{v}y∥& ={\displaystyle \sum _{r=v+1}^{\infty }}\left({\displaystyle \prod _{u=r-v+1}^r}|{\omega }_u|\right)\left|y_r\right|\hfill \\ & \le {\displaystyle \sum _{r=v+1}^{\infty }}\left({\displaystyle \prod _{u=r-v+1}^r}|{\omega }_u^0|\right)\left|y_r\right|=∥B_{{\omega }_0}^{v}x∥(\mathit{observe} \mathit{that}{\omega }_u\le {\omega }_u^0\mathit{for} \mathit{any}u\ge 1).\hfill \end{array}$$

Since $\left(Y,B_{{\omega }_0}\right)$ is absolutely Cesàro bounded as in [25], Example 23(A), there is $C>0$ satisfying $\underset{v\in \mathbb{N}}{sup}\frac{1}{v}{\displaystyle \sum _{u=1}^v}∥B_{{\omega }_0}^{u}y∥\le C\left|\right|y\left|\right|$ for all $y\in Y$. This implies that $\underset{v\in \mathbb{N}}{sup}\frac{1}{v}{\displaystyle \sum _{u=0}^{v-1}}\left|\right|S^{u}y\left|\right|\le (C+1)\left|\right|y\left|\right|$ for any $y\in Y$. Thus, $(Y,S)$ is not mean sensitive.

Since

$$\begin{array}{c}\hfill {\displaystyle \prod _{v=1+{\displaystyle \sum _{u=1}^{2t-1}}{l}_u}^{{\displaystyle \sum _{u=1}^{2t}}{l}_u}}|{\omega }_v|={(l_{2t-1}-1)}^{1-{\epsilon }_0}.\end{array}$$

for any $t\ge 1$, one has that $(Y,S)$ is Li–Yorke sensitive by [26], Theorem 5 and [7], Theorem 1, and then $(Y,S)$ is sensitive. □

8. Conclusions

In this study, we focus on syndetic sensitivity and diam-mean sensitivity for linear operators, addressing a gap existing in the literature. We obtain some properties regarding syndetic sensitivity and diam-mean sensitivty for adjoint operators and left multiplication operators. We also show that there exists an l.d.s. $(X\times Y,T\times S)$ such that

• $(X\times Y,T\times S)$ is cofinitely sensitive;
• $(X,T)$ and $(Y,S)$ are not syndetically sensitive.

However, the following question remains unresolved.

Question 1.

Is there a syndetically sensitive l.d.s. $(Y,S)$ that is not cofinitely sensitive?

For the mean sensitivity aspect, we obtain some results regarding mean sensitive perturbations; we also obtain a spectrum property for the sensitive systems. If an l.d.s. $(Y,S)$ is mean sensitive, then $(Y,S)$ is not absolutely Cesàro bounded. In the future, we intend to explore which conditions imply that an l.d.s. $(Y,S)$ is absolutely Cesàro bounded.