On Topological Numerical Transitivity, C*-Transitivity, and Generalized Numerical Transitivity

Abstract

Motivated by the concept of numerical hypercyclicity, in this paper we introduce three new notions in linear dynamics: topological numerical transitivity, generalized numerical transitivity, and $C^{\ast }$-transitivity. The first notion is defined for operators on general Banach spaces, whereas the latter two are formulated in the setting of operators on Banach algebras. The concept of $C^{\ast }$-transitivity is also applicable to operators on the space of Hilbert–Schmidt operators. We prove that these new notions are mutually different, and we also show that they differ from the standard notions in dynamics like hypercyclicity, supercyclicity, and numerical hypercyclicity. In particular, while topological numerical transitivity implies numerical hypercyclicity, as we argue in the paper, we provide examples of numerically hypercyclic and strongly numerically hypercyclic operators that are not topologically numerically transitive. We also provide examples of numerically hypercyclic operators that are neither $C^{\ast }$-transitive nor generalized numerically transitive. Also, we construct nonsupercyclic $C^{\ast }$-transitive and generalized numerically transitive operators on the $C^{\ast }$-algebra of compact operators on a separable Hilbert space, as well as $C^{\ast }$-transitive operators on the standard Hilbert $C^{\ast }$-module. In addition, we study diagonal operators on finite-dimensional Hilbert spaces and on ${\mathsf{\ell }}^p\left(\mathbb{N}\right)$, $1<p<\infty$, and we obtain complete characterizations of both numerical hypercyclicity and topological numerical transitivity in terms of coefficient-simplex criteria. As a consequence of this result, we prove that, unlike numerical and strong numerical hypercyclicity, a diagonal operator on a finite-dimensional complex Hilbert space of dimension at least two can never be topologically numerically transitive. Furthermore, we provide some sufficient conditions for diagonal operators on the standard Hilbert $C^{\ast }$-module to be $C^{\ast }$-transitive, and illustrate this result with examples. At the end of the paper, we compare topological numerical transitivity and $C^{\ast }$-transitivity with numerous standard notions in linear dynamics, like N-weak supercyclicity, weak-n supercyclicity, $ϵ$–cyclicity, Li–Yorke chaos, distributional chaos of type 3, subspace hypercyclicity, and subspace supercyclicity, and we prove that they genuinely differ from all these standard concepts in dynamics. Finally, we also introduce the notion of $C^{\ast }$-numerical transitivity, and we argue that many of the results on $C^{\ast }$-transitivity hold analogously for $C^{\ast }$-numerical transitivity.

1. Introduction

Linear dynamics of bounded operators on Banach spaces has become a major theme in modern functional analysis, at the crossroads of operator theory, topological dynamics, and approximation theory [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]. In the classical setting, an operator T acting on a topological vector space X is called hypercyclic if there exists a vector $x\in X$ whose orbit under T

$$\mathrm{Orb}(T,x):=\{T^{n}x:n\ge 1\}$$

is dense in X, whereas T is called supercyclic if there exists a vector $x\in X$ whose projective orbit under T

$${\mathrm{Orb}}_p(T,x):=\{\lambda T^{n}x:n\ge 1\lambda \in \mathbb{C}\}$$

is dense in X. Furthermore, if X and Y are general topological spaces and ${\left\{T_n\right\}}_{n\in \mathbb{N}}$ is a family of continuous mappings from X into $Y,$ we say that ${\left\{T_n\right\}}_{n\in \mathbb{N}}$ is topologically transitive if for each pair of open nonempty subsets $O_1\subseteq X$ and $O_2\subseteq Y$ there exists some $n\in \mathbb{N}$ such that $T_n\left(O_1\right)\cap O_2\ne ⌀.$ Also, we say that ${\left\{T_n\right\}}_{n\in \mathbb{N}}$ is topologically mixing if for each pair of open nonempty subsets $O_1\subseteq X$ and $O_2\subseteq Y$ there exists some $N\in \mathbb{N}$ such that $T_n\left(O_1\right)\cap O_2\ne ⌀$ for every $n\ge N.$

Since the pioneering works of Birkhoff and Rolewicz, these notions and their relatives have generated a rich theory, involving mixing, weak mixing, recurrence, universality, frequent hypercyclicity, and spectral criteria [2,3,4,5,6,7,8,9,11,13,14,15,16]. Also, we refer to [17,18,19] for some recent results on topological transitivity, ergodicity and mixing.

A complementary viewpoint arises when one observes the dynamics not through vector orbits in the ambient space, but through scalar evaluations of iterates. This philosophy is classical in numerical range theory [20,21,22], and it led Kim, Peris, and Song to introduce the notion of numerical hypercyclicity [23]. Let X be a complex Banach space and define

$$\mathrm{\Pi }\left(X\right):=\{(x,x^{\ast })\in S_X\times S_{X^{\ast }}:x^{\ast }\left(x\right)=1\}.$$

For $T\in \mathcal{B}\left(X\right)$ and $(x,x^{\ast })\in \mathrm{\Pi }\left(X\right)$, the associated numerical orbit is

$$\mathrm{NO}(T;x,x^{\ast }):=\{x^{\ast }\left(T^{n}x\right):n\ge 1\}\subseteq \mathbb{C}.$$

The operator T is called numerically hypercyclic if there exists $(x,x^{\ast })\in \mathrm{\Pi }\left(X\right)$ such that $\mathrm{NO}(T;x,x^{\ast })$ is dense in $\mathbb{C}$ [23]. Thus, the emphasis shifts from density of vector orbits in X to density of scalar observations in the complex plane.

This scalar-dynamical viewpoint is markedly different from ordinary hypercyclicity. In particular, Kim, Peris, and Song proved that every complex finite-dimensional Banach space of dimension at least two supports numerically hypercyclic operators, a phenomenon impossible in the classical hypercyclic theory [23]. In the same work, they obtained characterizations of numerically hypercyclic weighted shifts on classical sequence spaces [23].

The theory was subsequently deepened by Shkarin in a substantial way [24]. In particular, he introduced the notions of weakly numerically hypercyclic and strongly numerically hypercyclic operators, thereby clarifying the role of similarity and showing that numerical hypercyclicity is not, in general, similarity invariant [24]. His work also provided spectral criteria, finite-dimensional descriptions, examples and counterexamples, and several structural results concerning powers, scalar multiples, and concrete operator classes [24]. Altogether, these developments show that numerical hypercyclicity is a subtle scalar version of orbit complexity, parallel to classical hypercyclicity but with its own genuinely distinctive behavior [23,24,25].

The natural domain for this theory is the state space $\mathrm{\Pi }\left(X\right)$. For every $T\in \mathcal{B}\left(X\right)$ and every $n\ge 1$, one has the numerical evaluation map

$$F_n:\mathrm{\Pi }\left(X\right)\to \mathbb{C},F_n(x,x^{\ast })=x^{\ast }\left(T^{n}x\right).$$

From this perspective, ordinary numerical hypercyclicity says precisely that the sequence ${\left(F_n\right)}_{n\ge 0}$ has at least one universal point [23,24]. This observation places the subject naturally in the broader framework of universality theory and Baire-category methods [10,13,14,15]. Indeed, in classical linear dynamics, topological transitivity is closely tied to the existence of residual sets of hypercyclic vectors, and the same philosophy suggests that one should go beyond the mere existence of a single numerically hypercyclic pair and ask for a genuinely topological form of numerical orbit complexity [10,14,15].

The purpose of the present paper is to initiate the systematic study of this stronger phenomenon. Given $T\in \mathcal{B}\left(X\right)$, we say that T is topologically numerically transitive if for every pair of nonempty open sets

$$U\subset \mathrm{\Pi }\left(X\right),V\subset \mathbb{C},$$

there exists $n\ge 1$ such that

$$F_n\left(U\right)\cap V\ne ⌀.$$

Thus, the question is no longer whether there exists one exceptional pair with dense numerical orbit, but whether every nonempty open region of the state space contains a pair whose numerical orbit visits any prescribed open subset of the plane. In other words, we pass from an existence statement to a genericity statement.

This notion is motivated by several converging ideas. First, it is formally analogous to topological transitivity for operator iterates and to universality for sequences of continuous maps [10,13,14,15]. Second, it fits naturally with the Baire-category structure of the state set $\mathrm{\Pi }\left(X\right)$, where universal pairs may be organized into residual subsets rather than treated merely as isolated objects. Third, it reflects a broader trend in linear dynamics, where one seeks not only dense orbits, but generic orbit behavior, frequency properties, and robust transitivity mechanisms [8,11,14,15,16]. In this sense, topological numerical transitivity should be viewed as the Baire-generic analogue of numerical hypercyclicity.

This viewpoint fits naturally into the classical theory of universality for sequences of continuous maps [10,15]. Indeed, since $\mathrm{\Pi }\left(X\right)$ is a Baire space and $\mathbb{C}$ is second-countable, the density of ${\bigcup }_{n\ge 1}{F}_n^{-1}\left(V\right)$ in $\mathrm{\Pi }\left(X\right)$ for every nonempty open set $V\subseteq \mathbb{C}$ guarantees a dense $G_{\delta }$ set of universal numerical pairs. Thus, topological numerical transitivity strengthens ordinary numerical hypercyclicity by requiring generic rather than merely existential universality [23,24]. This distinction is also relevant when comparing the new notion with other forms of linear dynamics. Frequent hypercyclicity imposes positive lower density conditions on visiting times [11,16], whereas weak mixing is characterized by the transitivity of the corresponding product operator [12,15]. Related recurrence, mixing, and ergodic properties have also been investigated for $C_0$-semigroups [17,18]. Unlike these classical orbit-theoretic properties, topological numerical transitivity concerns the generic density of scalar numerical observations without imposing additional frequency or product-transitivity requirements.

The main abstract result of the paper shows that this interpretation is exact: under the natural topology on $\mathrm{\Pi }\left(X\right)$, topological numerical transitivity is equivalent to the fact that the set of universal pairs is a dense $G_{\delta }$-subset of $\mathrm{\Pi }\left(X\right)$. Hence, the new notion is not merely a formal variant of numerical hypercyclicity, but a genuinely stronger property, forcing residual abundance of universal pairs rather than the existence of one such pair. This also leads naturally to the converse problem: does numerical hypercyclicity imply topological numerical transitivity? As our results show, this question is already nontrivial in concrete classes such as weighted shifts and diagonal operators. We provide in this paper examples of numerically hypercyclic diagonal operators that are not topologically numerically transitive, thus giving a negative answer to the above question. Also, we give some examples of topologically numerically transitive operators. As shown in our examples, it turns out that if B is a unilateral or a bilateral shift and R is a positive number strictly greater than 1, then $RB$ is topologically numerically transitive. This shows that there exist topologically numerically transitive operators that are neither supercyclic nor weakly hypercyclic, as we argue in the paper.

After developing the general Baire-category framework, we study diagonal operators in both finite-dimensional Hilbert spaces and on ${\mathsf{\ell }}^p\left(\mathbb{N}\right)$ for $1<p<\infty$. In these classes, the problem reduces to coefficient-simplex criteria: numerical hypercyclicity becomes an existence condition for a suitable scalar exponential sum, whereas topological numerical transitivity becomes the corresponding density condition. This dichotomy provides one of the main conceptual contributions of the paper and shows that the new notion is both tractable and genuinely different from ordinary numerical hypercyclicity. In particular, we show that, unlike numerical and strong numerical hypercyclicity, no diagonal operator on a finite-dimensional complex Hilbert space of dimension at least two can be topologically numerically transitive.

Now, linear dynamics of operators on Banach algebras and their ideals constitutes an active area of research; see, for instance, refs. [26,27,28,29,30,31] and the references therein. For example, a characterization of universal and chaotic multipliers on some ideals of operators, in particular on the ideal of compact operators on a Banach space with the approximation property, has been provided in [26]; see also [27]. Dynamical properties of multipliers and elementary mappings on Schatten p-ideals of operators on a separable Hilbert space have been studied, for instance, in refs. [28,31], whereas sufficient conditions for *-supercyclicity of general bounded linear operators on $\mathcal{B}\left(H\right)$ (with H being a separable Hilbert space) have been given in [30]. Furstenberg semi-transitivity of generalized weighted shifts on the standard Hilbert C*-module and on more general non-unital Banach algebras has been studied in [29,32]. Motivated by this line of research and inspired by the concept of topological numerical transitivity, we introduce two new notions for bounded linear operators on Banach algebras and Banach $C^{\ast }$-modules, which we call generalized numerical transitivity and $C^{\ast }$-transitivity. The first notion is defined as follows:

Let $\mathcal{A}$ be a Banach algebra and let $T\in \mathcal{B}\left(\mathcal{A}\right)$. For each $n\in \mathbb{N}$, we consider the induced map

$${\Theta }_n:\mathcal{A}\times \mathcal{A}⟶\mathcal{A},{\Theta }_n(a,b)=a{T}^n\left(b\right).$$

We say that T is generalized numerically transitive if the family ${\left\{{\Theta }_n\right\}}_{n\in \mathbb{N}}$ is topologically transitive when $\mathcal{A}\times \mathcal{A}$ is equipped with the product topology.

For the second notion, we let $\mathcal{M}$ be a Banach bimodule over a $C^{\ast }$-algebra $\mathcal{C}$, and $\mathcal{K}$ be a cone in $\mathcal{M}$. We assume that X is a Banach space such that there exists a sesquilinear map

$$\phi :X\times X⟶\mathcal{M}$$

with the property that $\phi (x,x)\in \mathcal{K}$ and such that

$${\parallel x\parallel }_X={\parallel \phi (x,x)\parallel }_{\mathcal{M}}^{1/2}\mathrm{for}\mathrm{all}x\in X.$$

As we argue in the paper, $C^{\ast }$-algebras, Hilbert $C^{\ast }$-modules and the space of Hilbert–Schmidt operators on a separable Hilbert space are examples of such spaces. More precisely, in the case when X is a Hilbert $C^{\ast }$-module over a $C^{\ast }$-algebra $\mathcal{C}$, $\phi$ can be taken to be the $\mathcal{C}$-valued inner product on X, so $\mathcal{M}=\mathcal{C}$ in this case. In particular, if X is a $C^{\ast }$-algebra, then it is a Hilbert module over itself. On the other hand, if X is the space of Hilbert–Schmidt operators on a separable Hilbert space H, then it is a Banach space whose norm is induced by a $B_1\left(H\right)$-valued inner product, where $B_1\left(H\right)$ stands for the Banach space consisting of all nuclear (trace-class) operators on H and equipped with the trace norm. Hence, in this case we can let $\mathcal{M}=B_1\left(H\right)$ and $\phi$ be the $\mathcal{M}$-valued inner product on $X.$

Also, we let again $S_X$ denote the unit sphere of $X,$ and for $T\in \mathcal{B}\left(X\right)$ and $n\in \mathbb{N}$ we let

$${\tilde{\Theta }}_n:S_X\times S_X⟶\mathcal{M}$$

be given by

$${\tilde{\Theta }}_n(x,y)=\phi \left(T^{n}x,y\right).$$

Then we say that T is topologically $C^{\ast }$-transitive with respect to φ (or shortly $C^{\ast }$-transitive with respect to φ) if the sequence ${\left\{{\tilde{\Theta }}_n\right\}}_{n\in \mathbb{N}}$ is topologically transitive.

Our first result regarding these two new notions shows that, when X is a separable Banach space with the approximation property, every hypercyclic operator on the Banach algebra $K\left(X\right)$ of compact operators on X is generalized numerically transitive; see Corollary 4. However, it turns out that the converse does not hold in general. In Proposition 4, we construct $C^{\ast }$-transitive and generalized numerically transitive operators on $B_0\left(H\right)$ (the $C^{\ast }$-algebra of compact operators on a separable Hilbert space) that are not even supercyclic and, hence, not hypercyclic.

Next, in Proposition 3, we show that in the special case of operators on the $C^{\ast }$-algebra of compact operators on a Hilbert space, topological $C^{\ast }$-transitivity implies generalized numerical transitivity; however, in Proposition 5, we construct a $C^{\ast }$-transitive operator on the standard Hilbert $C^{\ast }$-module which is not generalized numerically transitive, illustrating in this way the difference between $C^{\ast }$-transitivity and generalized numerical transitivity. In this context, we would like to emphasize that it has been proved in [32] that the standard Hilbert $C^{\ast }$-module is also a Banach algebra under componentwise multiplication; hence, the concept of generalized numerical transitivity applies also in this concrete case. As a consequence of the proof of Proposition 5, we provide some sufficient conditions for diagonal operators on the standard Hilbert $C^{\ast }$-module to be $C^{\ast }$-transitive (Corollary 5 ), and we illustrate this result with concrete examples (Remark 3). Furthermore, in Proposition 7, we establish the existence of topologically numerically transitive and numerically hypercyclic operators that are not $C^{\ast }$-transitive. Finally, in the series of corollaries at the end of the paper, we show that there exist topologically numerically transitive and $C^{\ast }$-transitive operators that are neither weakly n-supercyclic nor N-weakly supercyclic for any $n,N\in \mathbb{N},$ that are not $ϵ$–cyclic for any $ϵ\in (0,1),$ that are neither Li–Yorke chaotic nor distributionally chaotic of type 3, and that are not subspace hypercyclic. Also, we show that there exist strongly numerically hypercyclic operators that are neither topologically numerically transitive nor $C^{\ast }$-transitive. At the end of the paper, we introduce the notion of $C^{\ast }$-numerical transitivity for operators on Hilbert $C^{\ast }$-modules. In Proposition 8, we construct a $C^{\ast }$-numerically transitive left multiplier on $B_0\left(H\right)$ (the $C^{\ast }$-algebra of compact operators on a separable Hilbert space), and in Remark 4, we argue that many of the results on $C^{\ast }$-transitivity hold analogously for $C^{\ast }$-numerical transitivity.

Our results show that $C^{\ast }$-transitivity, topological numerical transitivity and generalized numerical transitivity are genuinely different from the standard notions in linear and numerical dynamics, and also, in general, they differ from each other. We hope that the introduction of these concepts will motivate further investigations of their properties.

The paper is organized as follows. In Section 2, we introduce the state space $\mathrm{\Pi }\left(X\right)$, fix the norm-product topology and some other notions needed for the rest of the paper. In Section 3.1, we define topological numerical transitivity, establish a Baire-category criterion for it, and study its relation with ordinary numerical hypercyclicity. In Section 3.2, we study diagonal operators in finite-dimensional Hilbert spaces and on ${\mathsf{\ell }}^p\left(\mathbb{N}\right)$, $1<p<\infty$, obtaining complete characterizations in these settings. In Section 4, we introduce generalized numerical transitivity and $C^{\ast }$-transitivity for operators on Banach algebras and operator ideals. We compare these two new notions with each other and study their relations with many other concepts in linear dynamics. Also, we present examples showing that they are genuinely distinct from several known notions of numerical dynamics.

The main results in the paper are Proposition 1, Theorem 1, Theorem 3, Propositions 4 and 5, Corollary 5, Proposition 7, and Corollaries 6–10. In Proposition 1, we provide several equivalent conditions for topological numerical transitivity of general bounded linear operators on Banach spaces. Thanks to Theorem 1, in Corollary 3, we prove that no diagonal operator on a finite-dimensional complex Hilbert space of dimension at least two is topologically numerically transitive. In Theorem 3, we completely characterize topologically numerically transitive diagonal operators on ${\mathsf{\ell }}^p$-spaces in terms of the coefficient-simplex criteria. Thanks to this theorem and a resulting remark (Remark 1), we show in Corollary 9 that there exist strongly numerically hypercyclic operators on ${\mathsf{\ell }}^2\left(\mathbb{N}\right)$ which are not topologically numerically transitive, illustrating in this way the difference on infinite-dimensional spaces between numerical and strong numerical hypercyclicity on one side and topological numerical transitivity on the other side. Moreover, in Example 4, we construct a topologically numerically transitive operator on ${\mathsf{\ell }}^2\left(\mathbb{Z}\right)$ which is neither supercyclic nor weakly hypercyclic, whereas in Proposition 4, we construct a $C^{\ast }$-transitive nonsupercyclic left multiplier on $B_0\left(H\right).$ Furthermore, in Proposition 5, we construct a $C^{\ast }$-transitive diagonal operator on the standard Hilbert $C^{\ast }$-module which is not generalized numerically transitive, showing in this way the difference between $C^{\ast }$-transitivity and generalized numerical transitivity. Then, as a consequence of the proof of this proposition, in Corollary 5, we provide some sufficient conditions for generalized compact diagonal operators on the standard Hilbert $C^{\ast }$-module to be $C^{\ast }$-transitive. Moreover, in Proposition 6, we establish the existence of generalized numerically transitive operators on the standard Hilbert $C^{\ast }$-module. In Proposition 7, we show that topological numerical transitivity essentially differs from $C^{\ast }$-transitivity, whereas in Corollaries 6–10, we prove that topological numerical transitivity and $C^{\ast }$-transitivity differ from the standard concepts in linear dynamics, such as N-weak supercyclicity, weak-n supercyclicity, $ϵ$–cyclicity, Li–Yorke chaos, distributional chaos of type 3, and subspace hypercyclicity. In addition, in Corollary 10, we show that there exist subspace supercyclic operators that are not $C^{\ast }$-transitive. Finally, after introducing the notion of $C^{\ast }$-numerical transitivity at the end of the paper, in Proposition 8, we construct a $C^{\ast }$-numerically transitive nonsupercyclic left multiplier on $B_0\left(H\right).$

2. Preliminaries

Let X be a complex Banach space, let $X^{\ast }$ be its dual, and denote by

$$S_X:=\{x\in X:\parallel x\parallel =1\},S_{X^{\ast }}:=\{x^{\ast }\in X^{\ast }:\parallel x^{\ast }\parallel =1\}$$

the unit spheres of X and $X^{\ast }$. Following the standard framework of numerical hypercyclicity, we consider the state space

$$\mathrm{\Pi }\left(X\right):=\{(x,x^{\ast })\in S_X\times S_{X^{\ast }}:x^{\ast }\left(x\right)=1\}.$$

By the Hahn–Banach theorem, for every $x\in X,$ there exists some $x^{\ast }\in X^{\ast }\mathrm{with}\parallel x^{\ast }\parallel =1$ such that $x^{\ast }\left(x\right)=\parallel x\parallel ,$ and therefore $\mathrm{\Pi }\left(X\right)$ is nonempty.

For $T\in \mathcal{B}\left(X\right)$ and $(x,x^{\ast })\in \mathrm{\Pi }\left(X\right)$, we write

$$\mathrm{NO}(T;x,x^{\ast }):=\{x^{\ast }\left(T^{n}x\right):n\ge 1\}\subseteq \mathbb{C}$$

for the associated numerical orbit. Recall that T is numerically hypercyclic if there exists $(x,x^{\ast })\in \mathrm{\Pi }\left(X\right)$ such that $\mathrm{NO}(T;x,x^{\ast })$ is dense in $\mathbb{C}$.

Given $T\in \mathcal{B}\left(X\right)$, we shall study the family of maps

$$F_n:\mathrm{\Pi }\left(X\right)\to \mathbb{C},F_n(x,x^{\ast })=x^{\ast }\left(T^{n}x\right),n\in {\mathbb{N}}_0.$$

We equip $X\times X^{\ast }$ with the product norm

$$\parallel (x,x^{\ast }){\parallel }_{\times }:=\max \{\parallel x\parallel ,\parallel x^{\ast }\parallel \},$$

and endow $\mathrm{\Pi }\left(X\right)$ with the induced topology.

Lemma 1. 

The set $\mathrm{\Pi }\left(X\right)$ is closed in $S_X\times S_{X^{\ast }}$.

Proof. 

Consider

$$\mathrm{\Phi }:S_X\times S_{X^{\ast }}\to \mathbb{C},\mathrm{\Phi }(x,x^{\ast })=x^{\ast }\left(x\right).$$

For $(x,x^{\ast }),(y,y^{\ast })\in S_X\times S_{X^{\ast }}$, we have

$$|\mathrm{\Phi }(x,x^{\ast })-\mathrm{\Phi }(y,y^{\ast })|=|x^{\ast }\left(x\right)-y^{\ast }\left(y\right)|\le |x^{\ast }(x-y)|+|(x^{\ast }-y^{\ast })\left(y\right)|\le \parallel x-y\parallel +\parallel x^{\ast }-y^{\ast }\parallel ,$$

so $\mathrm{\Phi }$ is continuous. Since $\mathrm{\Pi }\left(X\right)={\mathrm{\Phi }}^{-1}\left(\left\{1\right\}\right),$ the claim follows. □

Lemma 2. 

The metric space $(\mathrm{\Pi }\left(X\right),\parallel ·{\parallel }_{\times })$ is complete. In particular, $\mathrm{\Pi }\left(X\right)$ is a Baire space.

Proof. 

Since X and $X^{\ast }$ are Banach spaces, the unit spheres $S_X$ and $S_{X^{\ast }}$ are complete for the induced metrics. Hence $S_X\times S_{X^{\ast }}$ is complete for ${\parallel ·\parallel }_{\times }$. By Lemma 1, $\mathrm{\Pi }\left(X\right)$ is a closed subset of $S_X\times S_{X^{\ast }}$, and therefore complete. The Baire property follows from the Baire category theorem. □

Lemma 3. 

For every $n\in \mathbb{N}$, the map

$$F_n:\mathrm{\Pi }\left(X\right)\to \mathbb{C},F_n(x,x^{\ast })=x^{\ast }\left(T^{n}x\right),$$

is continuous.

Proof. 

Let $(x,x^{\ast }),(y,y^{\ast })\in \mathrm{\Pi }\left(X\right)$. Then

$$|F_n(x,x^{\ast })-F_n(y,y^{\ast })|=|x^{\ast }\left(T^{n}x\right)-y^{\ast }\left(T^{n}y\right)|\le |x^{\ast }\left(T^n(x-y)\right)|+|(x^{\ast }-y^{\ast })\left(T^{n}y\right)|.$$

Since $\parallel x^{\ast }\parallel =1$ and $\parallel y\parallel =1$,

$$|x^{\ast }\left(T^n(x-y)\right)|\le \parallel T^n\parallel \parallel x-y\parallel ,|(x^{\ast }-y^{\ast })\left(T^{n}y\right)|\le \parallel T^n\parallel \parallel x^{\ast }-y^{\ast }\parallel .$$

Hence,

$$|F_n(x,x^{\ast })-F_n(y,y^{\ast })|\le \parallel T^n\parallel \left(\parallel x-y\parallel +\parallel x^{\ast }-y^{\ast }\parallel \right),$$

which proves continuity. □

Corollary 1. 

For every $n\in {\mathbb{N}}_0$ and every $(x,x^{\ast }),(y,y^{\ast })\in \mathrm{\Pi }\left(X\right)$,

$$|F_n(x,x^{\ast })-F_n(y,y^{\ast })|\le 2\parallel T^n\parallel \parallel (x,x^{\ast })-(y,y^{\ast }){\parallel }_{\times }.$$

In particular, each $F_n$ is Lipschitz on $\mathrm{\Pi }\left(X\right)$.

Proof. 

This follows from Lemma 3 and

$$\parallel x-y\parallel +\parallel x^{\ast }-y^{\ast }\parallel \le 2\max \{\parallel x-y\parallel ,\parallel x^{\ast }-y^{\ast }\parallel \}=2\parallel (x,x^{\ast })-(y,y^{\ast }){\parallel }_{\times }.$$

□

We end this section with the following auxiliary lemma, which we will use later in the proofs.

Lemma 4. 

Let $s,t$ be nonnegative real numbers and let $p\in (1,\infty )$. Then

$${|s-t|}^p\le |s^p-t^p|.$$

Proof. 

We may assume without loss of generality that $s\ge t$. Then $s-t=|s-t|$ and $s^p-t^p=|s^p-t^p|$, because the function $x\mapsto x^p$ is increasing on $[0,\infty )$. Set $k=s-t$, so that $k\ge 0$ and $s=t+k$.

Suppose first that $p\in \mathbb{N}$. By the binomial formula,

$$\begin{array}{cc}\hfill s^p={(t+k)}^p& ={\displaystyle \sum _{i=0}^p}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right)t^{p-i}{k}^i\hfill \\ & =t^p+k^p+{\displaystyle \sum _{i=1}^{p-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right)t^{p-i}{k}^i\hfill \\ & \ge t^p+k^p,\hfill \end{array}$$

since $t,k\ge 0$. Therefore,

$$s^p-t^p\ge k^p={(s-t)}^p.$$

Suppose now that $p\in (1,\infty )\setminus \mathbb{N}$. Write $p=n+r$, where $n\in \mathbb{N}$, $n\ge 1$, and $r\in (0,1)$. In this case,

$$\begin{array}{cc}\hfill {(t+k)}^p& ={(t+k)}^n{(t+k)}^r\hfill \\ & =(t^n+k^n){(t+k)}^r+f_n(t,k){(t+k)}^r,\hfill \end{array}$$

where

$$f_n(t,k)=\left\{\begin{array}{cc}{\displaystyle {\displaystyle \sum _{i=1}^{n-1}}\left(\genfrac{}{}{0pt}{}{n}{i}\right)t^{n-i}{k}^i,}\hfill & n\ge 2,\hfill \\ 0,\hfill & n=1.\hfill \end{array}\right.$$

In every case, $f_n(t,k)\ge 0$, since $t,k\ge 0$. Hence,

$${(t+k)}^p\ge (t^n+k^n){(t+k)}^r.$$

Moreover, the function $x\mapsto x^r$ is increasing on $[0,\infty )$, so

$$t^r\le {(t+k)}^r\mathrm{and}{k}^r\le {(t+k)}^r.$$

Consequently,

$$(t^n+k^n){(t+k)}^r\ge t^n{t}^r+k^n{k}^r=t^p+k^p.$$

Combining these inequalities gives

$${(t+k)}^p\ge (t^n+k^n){(t+k)}^r\ge t^p+k^p.$$

Thus,

$$s^p-t^p={(t+k)}^p-t^p\ge k^p={(s-t)}^p.$$

This proves the desired inequality. □

3. Topological Numerical Transitivity

In the first subsection below, we will study some general properties of topological numerical transitivity, whereas in the second subsection we will completely characterize topologically numerically transitive diagonal operators.

3.1. A Baire-Category Criterion, and Relations with Numerical Hypercyclicity

Let $T\in \mathcal{B}\left(X\right)$. We consider the family of maps

$$F_n:\mathrm{\Pi }\left(X\right)\to \mathbb{C},F_n(x,x^{\ast })=x^{\ast }\left(T^{n}x\right),n\in \mathbb{N}.$$

Definition 1. 

An operator $T\in \mathcal{B}\left(X\right)$ is called topologically numerically transitive if for every pair of nonempty open sets $U\subset \mathrm{\Pi }\left(X\right)$, $V\subset \mathbb{C},$ there exists $n\in \mathbb{N}$ such that

$$F_n\left(U\right)\cap V\ne ⌀.$$

Equivalently, for every such U and V, there exist $(x,x^{\ast })\in U$ and $n\in \mathbb{N}$ such that

$$x^{\ast }\left(T^{n}x\right)\in V.$$

Definition 2. 

For $T\in \mathcal{B}\left(X\right)$, define

$$\mathcal{U}\left(T\right):=\left\{(x,x^{\ast })\in \mathrm{\Pi }\left(X\right):\{F_n(x,x^{\ast }):n\ge 1\}\mathit{\text{is}}\mathit{\text{dense}}\mathit{\text{in}}\mathbb{C}\right\}.$$

Equivalently,

$$\mathcal{U}\left(T\right)=\left\{(x,x^{\ast })\in \mathrm{\Pi }\left(X\right):\mathrm{NO}(T;x,x^{\ast })\mathit{\text{is}}\mathit{\text{dense}}\mathit{\text{in}}\mathbb{C}\right\}.$$

Thus, $\mathcal{U}\left(T\right)$ is precisely the set of universal pairs of T, or equivalently the set of numerically hypercyclic pairs.

Fix a countable basis $\mathcal{V}=\{V_j:j\in \mathbb{N}\}$ of nonempty open subsets of $\mathbb{C}$. Then

$$\mathcal{U}\left(T\right)=\bigcap _{j\in \mathbb{N}}\bigcup _{n\ge 1}{F}_n^{-1}\left(V_j\right).$$

Indeed, $(x,x^{\ast })\in \mathrm{\Pi }\left(X\right)$ belongs to $\mathcal{U}\left(T\right)$ if and only if its numerical orbit meets every basis element $V_j$.

Now we are ready to state the main proposition in this section.

Proposition 1. 

For $T\in \mathcal{B}\left(X\right)$, the following assertions are equivalent:

(i) 

T is topologically numerically transitive;

(ii) 

for every nonempty open set $U\subset \mathrm{\Pi }\left(X\right)$ and every $j\in \mathbb{N}$, there exists $n\in \mathbb{N}$ such that $F_n\left(U\right)\cap V_j\ne ⌀;$

(iii) 

for every $j\in \mathbb{N}$, the set ${\bigcup }_{n\ge 1}{F}_n^{-1}\left(V_j\right)$ is open and dense in $\mathrm{\Pi }\left(X\right)$.

(iv) 

$\mathcal{U}\left(T\right)$ is a dense $G_{\delta }$ subset of $\mathrm{\Pi }\left(X\right)$.

(v) 

$\mathcal{U}\left(T\right)\mathit{is}\mathit{residual}\mathit{in}\mathrm{\Pi }\left(X\right).$

Proof. 

(ii)⇔(i). The implication (i)⇒(ii) is immediate. Assume now (ii) and let $U\subset \mathrm{\Pi }\left(X\right)$ and $V\subset \mathbb{C}$ be nonempty open sets. Choose $j\in \mathbb{N}$ such that $V_j\subseteq V$. Then $F_n\left(U\right)\cap V_j\ne ⌀$ for some $n\in \mathbb{N}$, hence $F_n\left(U\right)\cap V\ne ⌀$. Thus, (ii)⇒(i).

(ii)⇔(iii). Fix $j\in \mathbb{N}$. For a nonempty open set $U\subset \mathrm{\Pi }\left(X\right)$, the condition $F_n\left(U\right)\cap V_j\ne ⌀$ for some $n\in \mathbb{N}$ is equivalent to $U\cap F_n^{-1}\left(V_j\right)\ne ⌀$ for some $n\in \mathbb{N}$ and, hence, to

$$U\cap \bigcup _{n\ge 1}{F}_n^{-1}\left(V_j\right)\ne ⌀.$$

Moreover, the sets ${\bigcup }_{n\ge 1}{F}_n^{-1}\left(V_j\right)$ are open because each $F_n$ is continuous. Therefore, (ii) and (iii) are equivalent.

(iii)⇔(iv). Recall first that by Lemma 2, we have that $\mathrm{\Pi }\left(X\right)$ is a Baire space, and moreover, as argued above, we also have

$$\mathcal{U}\left(T\right)=\bigcap _{j\in \mathbb{N}}\bigcup _{n\ge 1}{F}_n^{-1}\left(V_j\right).$$

Therefore, if (iii) holds, then the set $\mathcal{U}\left(T\right)$ is a countable intersection of open dense subsets of the Baire space $\mathrm{\Pi }\left(X\right)$. Hence, $\mathcal{U}\left(T\right)$ is a dense $G_{\delta }$ subset of $\mathrm{\Pi }\left(X\right)$, so (iii)⇒(iv). Conversely, if (iv) holds, then again since $\mathcal{U}\left(T\right)={\bigcap }_{j\in \mathbb{N}}{\bigcup }_{n\ge 1}{F}_n^{-1}\left(V_j\right),$ then the density of $\mathcal{U}\left(T\right)$ implies the density of the sets ${\bigcup }_{n\ge 1}{F}_n^{-1}\left(V_j\right),$ and since the sets ${\bigcup }_{n\ge 1}{F}_n^{-1}\left(V_j\right)$ are open because each $F_n$ is continuous, we deduce (iii). Thus, (iv)⇒(iii).

(v)⇔(iv). Since $\mathrm{\Pi }\left(X\right)$ is a Baire space by Lemma 2, every residual subset of $\mathrm{\Pi }\left(X\right)$ is dense in $\mathrm{\Pi }\left(X\right)$, so (v)⇒(iv). Conversely, if (iv) holds, then, by the equivalence (ii)⇔(iii) which was shown above, (iii) also holds; hence, $\mathcal{U}\left(T\right)$ is a countable intersection of open and dense subsets of $\mathrm{\Pi }\left(X\right)$ because $\mathcal{U}\left(T\right)={\bigcap }_{j\in \mathbb{N}}{\bigcup }_{n\ge 1}{F}_n^{-1}\left(V_j\right).$ Thus, $\mathcal{U}\left(T\right)$ is residual in $\mathrm{\Pi }\left(X\right)$. □

For later use, if $V\subset \mathbb{C}$ is nonempty and open, define

$${\mathcal{O}}_T\left(V\right):=\bigcup _{n\ge 1}{F}_n^{-1}\left(V\right)=\left\{(x,x^{\ast })\in \mathrm{\Pi }\left(X\right):\exists n\ge 1\mathrm{such}\mathrm{that}{x}^{\ast }\left(T^{n}x\right)\in V\right\}.$$

Then Proposition 1 shows that T is topologically numerically transitive if and only if ${\mathcal{O}}_T\left(V_j\right)$ is dense in $\mathrm{\Pi }\left(X\right)$ for every basis element $V_j$ and $\mathcal{U}\left(T\right)={\bigcap }_{j\in \mathbb{N}}{\mathcal{O}}_T\left(V_j\right).$ Hence, topological numerical transitivity is exactly the condition ensuring that the family $\left(F_n\right)$ has a residual set of universal points in $\mathrm{\Pi }\left(X\right)$.

We now compare topological numerical transitivity with the classical notion of numerical hypercyclicity.

Corollary 2. 

Let $T\in \mathcal{B}\left(X\right)$. Then T is numerically hypercyclic if and only if $\mathcal{U}\left(T\right)\ne ⌀.$ In particular, topological numerical transitivity implies numerical hypercyclicity.

Proof. 

The first statement of the corollary is just the definition of numerical hypercyclicity, whereas the second statement follows from Proposition 1. □

Thus, numerical hypercyclicity is an existence property, whereas topological numerical transitivity is the corresponding density, equivalently residuality, property for universal pairs.

The converse implication is the first natural problem.

Problem 1. 

Does numerical hypercyclicity imply topological numerical transitivity?

Equivalently, if $\mathcal{U}\left(T\right)\ne ⌀$, must $\mathcal{U}\left(T\right)$ be dense in $\mathrm{\Pi }\left(X\right)$? By Corollary 2, this is exactly the question of whether one universal pair forces generic abundance of universal pairs.

It is also natural to compare topological numerical transitivity with strong numerical hypercyclicity in the sense of Shkarin. We recall that an operator T is said to be strongly numerically hypercyclic if every operator similar to T is numerically hypercyclic. Since every operator is similar to itself, strong numerical hypercyclicity implies numerical hypercyclicity, which naturally leads to the following question.

Problem 2. 

Does strong numerical hypercyclicity imply topological numerical transitivity?

In the next section we will give negative answers to the questions raised in Problem 1 and Problem 2. In particular, in Example 1 we will construct concrete strongly numerically hypercyclic operators on finite-dimensional spaces that are not topologically numerically transitive. Moreover, later in Section 4, in Corollary 9, we will prove that there exist strongly numerically hypercyclic diagonal operators on ${\mathsf{\ell }}^2\left(\mathbb{N}\right)$ that are not topologically numerically transitive.

3.2. Diagonal Operators: Finite-Dimensional and ${\mathsf{\ell }}^p$ Cases

Diagonal operators provide a natural class in which numerical dynamics reduce to exponential sums with nonnegative coefficients. This yields complete characterizations of both numerical hypercyclicity and topological numerical transitivity in two settings: finite-dimensional Hilbert spaces and ${\mathsf{\ell }}^p\left(\mathbb{N}\right)$ for $1<p<\infty$.

We begin with the finite-dimensional Hilbert case. Let $d\ge 2$, and let $D=\mathrm{diag}({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_d)\in \mathcal{B}\left({\mathbb{C}}^d\right),$ where ${\lambda }_1,\dots ,{\lambda }_d\in \mathbb{C}$. Denote by ${\mathrm{\Delta }}_d:=\left\{a=(a_1,\dots ,a_d)\in {[0,1]}^d:{\sum }_{j=1}^d{a}_j=1\right\}$ the standard simplex, and for $a\in {\mathrm{\Delta }}_d$ define ${\mathcal{O}}_D\left(a\right):=\left\{{\sum }_{j=1}^d{a}_j{\lambda }_j^n:n\ge 1\right\}.$ Set

$$G_D:=\left\{a\in {\mathrm{\Delta }}_d:{\mathcal{O}}_D\left(a\right)\mathrm{is}\mathrm{dense}\mathrm{in}\mathbb{C}\right\}.$$

In this section, for each $x\in {\mathbb{C}}^d,$ we let $\left|x\right|:=\left(\right|x_1|,\dots ,|x_d\left|\right).$

Theorem 1. 

Let $D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_d)\in \mathcal{B}\left({\mathbb{C}}^d\right)$. Then:

(i) 

D is numerically hypercyclic if and only if $G_D\ne ⌀$.

(ii) 

D is topologically numerically transitive if and only if $G_D$ is dense in ${\mathrm{\Delta }}_d$.

Proof. 

Since ${\mathbb{C}}^d$ is a Hilbert space, $\mathrm{\Pi }\left({\mathbb{C}}^d\right)$ is naturally identified with the unit sphere $S_{{\mathbb{C}}^d}$ because for each $x\in S_{{\mathbb{C}}^d}$, the unique state functional is $x^{\ast }=⟨·,x⟩.$ Thus, for $x=(x_1,\dots ,x_d)\in S_{{\mathbb{C}}^d}$, $⟨D^{n}x,x⟩={\sum }_{j=1}^d{\left|x_j\right|}^2{\lambda }_j^n(n\ge 1).$ Hence, the numerical orbit of x depends only on the coefficient vector $a\left(x\right):=\left(\right|x_1{|}^2,\dots ,|x_d{|}^2)\in {\mathrm{\Delta }}_d,$ and one has

$$x\in \mathcal{U}\left(D\right)\iff a\left(x\right)\in G_D.$$

This proves (i).

Assume now that $G_D$ is dense in ${\mathrm{\Delta }}_d$. Let $y=(y_1,\dots ,y_d)\in S_{{\mathbb{C}}^d}$ and $\epsilon >0$. Set

$$b:=a\left(y\right)=\left(\right|y_1{|}^2,\dots ,|y_d{|}^2)\in {\mathrm{\Delta }}_d,$$

and choose $a=(a_1,\dots ,a_d)\in G_D$ such that ${\parallel a-b\parallel }_1<{\epsilon }^2.$ Define $x=(x_1,\dots ,x_d)\in {\mathbb{C}}^d$ by

$$x_j=\left\{\begin{array}{cc}\sqrt{a_j}{\displaystyle \frac{y_j}{|y_j|}},\hfill & y_j\ne 0,\hfill \\ \sqrt{a_j},\hfill & y_j=0.\hfill \end{array}\right.$$

Then $x\in S_{{\mathbb{C}}^d}$, $a\left(x\right)=a\in G_D$ and hence $x\in \mathcal{U}\left(D\right)$. Moreover, by Lemma 4, we have

$$\left(\right|x_j|-|y_j{\left|\right)}^2\le \left||x_j{|}^2-{|y_j|}^2\right|\mathrm{for}\mathrm{each}j\in \{1,\dots ,d\}.$$

Therefore, we obtain

$${\parallel x-y\parallel }_2^2={\displaystyle \sum _{j=1}^d}\left(\right|x_j|-|y_j{\left|\right)}^2\le {\displaystyle \sum _{j=1}^d}\left||x_j{|}^2-{|y_j|}^2\right|={\parallel a-b\parallel }_1<{\epsilon }^2.$$

Thus, $\mathcal{U}\left(D\right)$ is dense in $S_{{\mathbb{C}}^d}$, hence dense in $\mathrm{\Pi }\left({\mathbb{C}}^d\right)$. By Proposition 1, D is topologically numerically transitive.

Conversely, assume that D is topologically numerically transitive. Then $\mathcal{U}\left(D\right)$ is dense in $S_{{\mathbb{C}}^d}$. Let $b=(b_1,\dots ,b_d)\in {\mathrm{\Delta }}_d$ and $\epsilon >0$, and set $y=(\sqrt{b_1},\dots ,\sqrt{b_d})\in S_{{\mathbb{C}}^d}.$ Choose $x\in \mathcal{U}\left(D\right)$ such that

$${\parallel x-y\parallel }_2<{\displaystyle \frac{\epsilon }{2}},$$

and define $a:=a\left(x\right)=\left(\right|x_1{|}^2,\dots ,|x_d{|}^2)\in G_D.$ Then

$${\parallel a-b\parallel }_1={\displaystyle \sum _{j=1}^d}\left||x_j{|}^2-{|y_j|}^2\right|\le {\left({\displaystyle \sum _{j=1}^d}\left(\right|x_j|-|y_j{\left|\right)}^2\right)}^{1/2}{\left({\displaystyle \sum _{j=1}^d}\left(\right|x_j|+|y_j{\left|\right)}^2\right)}^{1/2},$$

where in the last inequality we have applied the Cauchy–Schwarz inequality for the inner product on ${\mathbb{C}}^d$ with respect to the sequences ${\left\{\right||x_j|-|y_j|\left|\right\}}_{1\le j\le d}$ and $\left\{\right|x_j|+|y_j{\left|\right\}}_{1\le j\le d}.$ Since ${\parallel x\parallel }_2={\parallel y\parallel }_2=1$, one has ${\sum }_{j=1}^d\left(\right|x_j|+|y_j{\left|\right)}^2\le 4$ by the triangle inequality for the norm in ${\mathbb{C}}^d.$ Hence

$${\parallel a-b\parallel }_1\le 2\parallel \left|x\right|-{\left|y\right|\parallel }_2.$$

However, by the reverse inequality for the absolute values, we get

$$2\parallel \left|x\right|-{\left|y\right|\parallel }_2\le 2{\parallel x-y\parallel }_2<\epsilon ,$$

hence

$${\parallel a-b\parallel }_1<\epsilon .$$

Thus, $G_D$ is dense in ${\mathrm{\Delta }}_d$, and the proof is complete. □

As a consequence of Theorem 1, we obtain the following corollary.

Corollary 3. 

No diagonal operator on a finite-dimensional complex Hilbert space of dimension at least two is topologically numerically transitive.

Proof. 

Write

$$D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_d),R=\underset{1\le j\le d}{\max }\left|{\lambda }_j\right|.$$

If $R\le 1$, then for every $a\in {\mathrm{\Delta }}_d$ and $n\ge 1$,

$$\left|{\displaystyle \sum _{j=1}^d}{a}_j{\lambda }_j^n\right|\le {\displaystyle \sum _{j=1}^d}{a}_j{\left|{\lambda }_j\right|}^n\le 1.$$

Hence, $G_D=⌀$.

Suppose that $R>1$, and choose k with $|{\lambda }_k|=R$. For every $a\in {\mathrm{\Delta }}_d$ satisfying $a_k>1/2$,

$$\begin{array}{cc}\hfill \left|{\displaystyle \sum _{j=1}^d}{a}_j{\lambda }_j^n\right|& \ge a_k{R}^n-\sum _{j\ne k}{a}_j|{\lambda }_j{|}^n=2a_k{R}^n-{\displaystyle \sum _{j=1}^d}{a}_j{|{\lambda }_j|}^n\hfill \\ & \ge (2a_k-1)R^n⟶\infty .\hfill \end{array}$$

Consequently,

$$G_D\cap \{a\in {\mathrm{\Delta }}_d:a_k>1/2\}=⌀.$$

The excluded set is nonempty and relatively open in ${\mathrm{\Delta }}_d$, so $G_D$ is not dense. Theorem 1 completes the proof. □

We wish to give an example of a strongly numerically hypercyclic operator that is not topologically numerically transitive. In order to construct such an example, we will need the following theorem:

Theorem 2 

([24], Theorem 1.10). Let X be a finite-dimensional space. Then each of the following conditions is sufficient for the strong numeric hypercyclicity of $T\in \mathcal{B}\left(X\right)$.

(i) 

There exist ${\lambda }_1,\dots ,{\lambda }_n\in {\sigma }_p\left(T\right)$ and $c_1,\dots ,c_n\in {\mathbb{R}}_{+}$ such that $\left\{{\sum }_{j=1}^n{c}_j{\lambda }_j^k:k\in {\mathbb{Z}}_{+}\right\}$ is dense in $\mathbb{C}$.

(ii) 

There exist ${\lambda }_1,{\lambda }_2,{\lambda }_3\in {\sigma }_p\left(T\right)$ such that $|{\lambda }_1|=|{\lambda }_2|=|{\lambda }_3|>1$ and ${\displaystyle \frac{{\lambda }_1}{|{\lambda }_1|}},{\displaystyle \frac{{\lambda }_2}{|{\lambda }_2|}},{\displaystyle \frac{{\lambda }_3}{|{\lambda }_3|}}$ are independent.

Now we are ready to provide an example of a strongly numerically hypercyclic diagonal operator which is not topologically numerically transitive.

Example 1 

(A diagonal operator which is SNH but not TNT). Fix $1<R<M$, let $u,w,z\in \mathbb{T}$ be independent, and define $D=\mathrm{diag}(Ru,Rw,Rz)\in \mathcal{B}\left({\mathbb{C}}^3\right).$ Then $Ru,Rw,Rz\in {\sigma }_p\left(D\right)$ and satisfy the conditions of Theorem 2 part (ii). Hence, the operator D is strongly numerically hypercyclic. However, by Corollary 3, we deduce that D is not topologically numerically transitive.

Since every strongly numerically hypercyclic operator is numerically hypercyclic, Example 1 provides a negative answer to the questions raised in Problems 1 and 2.

We now turn to diagonal operators on ${\mathsf{\ell }}^p\left(\mathbb{N}\right)$ in the reflexive range $1<p<\infty$. The same mechanism survives, with ${\mathrm{\Delta }}_d$ replaced by the probability simplex of ${\mathsf{\ell }}^1$. Let

$$D=\mathrm{diag}({\lambda }_1,{\lambda }_2,{\lambda }_3,\dots )\in \mathcal{B}\left({\mathsf{\ell }}^p\right),\underset{n\ge 1}{\sup }\left|{\lambda }_n\right|<\infty .$$

Denote by $\mathrm{\Delta }:=\left\{a={\left(a_n\right)}_{n\ge 1}\in {\mathsf{\ell }}^1:a_n\ge 0,{\sum }_{n\ge 1}{a}_n=1\right\}$ the probability simplex, and for $a\in \mathrm{\Delta }$ define ${\mathcal{O}}_D\left(a\right):=\left\{{\sum }_{n\ge 1}{a}_n{\lambda }_n^m:m\ge 1\right\}.$ Set $G_D:=\left\{a\in \mathrm{\Delta }:{\mathcal{O}}_D\left(a\right)\mathrm{is}\mathrm{dense}\mathrm{in}\mathbb{C}\right\}.$

Theorem 3. 

Let $1<p<\infty$, let $X={\mathsf{\ell }}^p\left(\mathbb{N}\right)$, and let $D=\mathrm{diag}({\lambda }_1,{\lambda }_2,{\lambda }_3,\dots )\in \mathcal{B}\left(X\right).$ Then:

(i) 

D is numerically hypercyclic if and only if $G_D\ne ⌀$.

(ii) 

D is topologically numerically transitive if and only if $G_D$ is dense in Δ.

Proof. 

Let q be the conjugate exponent of p. Since $1<p<\infty$, every $x={\left(x_n\right)}_{n\ge 1}\in S_{{\mathsf{\ell }}^p}$ has a unique norming functional $J_p\left(x\right)\in S_{{\mathsf{\ell }}^q}$ given by

$$J_p{\left(x\right)}_n=\overline{x_n}{\left|x_n\right|}^{p-2},n\ge 1.$$

Hence, $\mathrm{\Pi }\left(X\right)$ is naturally identified with $S_X$.

For $x\in S_X$, define

$$a\left(x\right):=\left(\right|x_n{{|}^p)}_{n\ge 1}\in \mathrm{\Delta }.$$

Then

$$J_p\left(x\right)\left(D^{0}x\right)=1,$$

and for every $m\ge 1$,

$$J_p\left(x\right)\left(D^{m}x\right)=\sum _{n\ge 1}\overline{x_n}|x_n{|}^{p-2}{\lambda }_n^m{x}_n=\sum _{n\ge 1}{|x_n|}^p{\lambda }_n^m=\sum _{n\ge 1}a{\left(x\right)}_n{\lambda }_n^m.$$

Thus, the numerical orbit of x is exactly ${\mathcal{O}}_D\left(a\left(x\right)\right)$, and

$$x\in \mathcal{U}\left(D\right)\iff a\left(x\right)\in G_D.$$

This proves (i).

Assume first that $G_D$ is dense in $\mathrm{\Delta }$. Let $y\in S_X$ and $\epsilon >0$, and set

$$b:=a\left(y\right)\in \mathrm{\Delta }.$$

Choose $a\in G_D$ such that

$${\parallel a-b\parallel }_1<{\epsilon }^p,$$

and define $x\in X$ by

$$x_n=\left\{\begin{array}{cc}{a}_n^{1/p}{\displaystyle \frac{y_n}{|y_n|}},\hfill & y_n\ne 0,\hfill \\ a_n^{1/p},\hfill & y_n=0.\hfill \end{array}\right.$$

Then $x\in S_X$, $a\left(x\right)=a\in G_D$ and hence $x\in \mathcal{U}\left(D\right)$. Moreover, by Lemma 4, we have

$${\left||x_n|-|y_n|\right|}^p\le \left||x_n{|}^p-{|y_n|}^p\right|\mathrm{for}\mathrm{every}n\in \mathbb{N}.$$

This gives

$${\parallel x-y\parallel }_p^p=\sum _{n\ge 1}{\left||x_n|-|y_n|\right|}^p\le \sum _{n\ge 1}\left||x_n{|}^p-{|y_n|}^p\right|={\parallel a-b\parallel }_1<{\epsilon }^p.$$

Thus, $\mathcal{U}\left(D\right)$ is dense in $S_X$, hence dense in $\mathrm{\Pi }\left(X\right)$, and Proposition 1 yields topological numerical transitivity.

Conversely, assume that D is topologically numerically transitive. Then $\mathcal{U}\left(D\right)$ is dense in $S_X$. Let $b\in \mathrm{\Delta }$ and $\epsilon >0$, and define

$$y={\left(b_n^{1/p}\right)}_{n\ge 1}\in S_X.$$

Choose $x\in \mathcal{U}\left(D\right)$ such that

$${\parallel x-y\parallel }_p<{\displaystyle \frac{\epsilon }{2p}}.$$

Set $a:=a\left(x\right)\in G_D$. For nonnegative numbers $s,t$, it holds that

$$|s^p-t^p|\le p(s^{p-1}+t^{p-1})|s-t|.$$

Indeed, we may assume without loss of generality that $s>t$. Then, by the mean value theorem, we obtain

$${\displaystyle \frac{s^p-t^p}{s-t}}=p{c}^{p-1}$$

for some $c\in (t,s)$. Therefore,

$$|s^p-t^p|=s^p-t^p\le p{s}^{p-1}(s-t)=p{s}^{p-1}|s-t|.$$

Consequently,

$$|s^p-t^p|\le p\left(s^{p-1}+t^{p-1}\right)|s-t|.$$

Applying this coordinatewise and summing, we obtain

$${\parallel a-b\parallel }_1\le p\sum _{n\ge 1}\left(|x_n{|}^{p-1}+{|y_n|}^{p-1}\right)\left||x_n|-|y_n|\right|.$$

By applying Hölder’s inequality to the sequences

$${\left\{|x_n{|}^{p-1}\right\}}_{n\in \mathbb{N}}\in {\mathsf{\ell }}^{{\displaystyle \frac{p}{p-1}}}$$

and

$${\left\{\left||x_n|-|y_n|\right|\right\}}_{n\in \mathbb{N}}\in {\mathsf{\ell }}^p,$$

we obtain

$$\sum _n{|x_n|}^{p-1}\left||x_n|-|y_n|\right|\le {\left(\sum _n{|x_n|}^p\right)}^{{\displaystyle \frac{p-1}{p}}}{∥|x|-|y|∥}_p.$$

Since ${\parallel x\parallel }_p=1,$ it follows from the above inequality that

$$\sum _n{|x_n|}^{p-1}\left||x_n|-|y_n|\right|\le {∥|x|-|y|∥}_p.$$

Similarly,

$$\sum _n{|y_n|}^{p-1}\left||x_n|-|y_n|\right|\le {∥|x|-|y|∥}_p.$$

Hence,

$${\parallel a-b\parallel }_1\le p\sum _{n\ge 1}\left(|x_n{|}^{p-1}+{|y_n|}^{p-1}\right)\left||x_n|-|y_n|\right|$$

$$\le 2p\parallel |x|-{|y|\parallel }_p\le 2p{\parallel x-y\parallel }_p<\epsilon .$$

Thus, $G_D$ is dense in $\mathrm{\Delta }$. □

Thus, in the diagonal ${\mathsf{\ell }}^p$-setting, numerical hypercyclicity is again an existence property on the coefficient simplex, whereas topological numerical transitivity is the corresponding density property. Moreover, in the case when $p=2,$ the ${\mathsf{\ell }}^p$ case reduces to the Hilbert space case.

As a consequence of Theorem 3, we obtain the following useful obstruction remark.

Remark 1. 

A bounded diagonal operator on ${\mathsf{\ell }}^p$ cannot be topologically numerically transitive if the supremum of its eigenvalue moduli is attained. Indeed, if $(R=|{\lambda }_k|={\sup }_j|{\lambda }_j|>1),$ then, for every coefficient vector with $(a_k>1/2),$

$$\left|\sum _j{a}_j{\lambda }_j^n\right|\ge (2a_k-1)R^n.$$

If $R\le 1,$ all numerical orbits are bounded.

Thus, in particular, a compact diagonal operator on ${\mathsf{\ell }}^2$ can never be topologically numerically transitive.

Next we record some examples of numerically hypercyclic and topologically numerically transitive operators.

Example 2 

(A diagonal numerically hypercyclic operator). Choose ${\lambda }_1,{\lambda }_2\in \mathbb{C}$ such that

$$\{{\lambda }_1^m+{\lambda }_2^m:m\ge 1\}$$

is dense in $\mathbb{C}$, and set

$$D=\mathrm{diag}({\lambda }_1,{\lambda }_2,0,0,\dots )\in \mathcal{B}\left({\mathsf{\ell }}^p\right).$$

Then

$$a=\left({\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}},0,0,\dots \right)\in G_D,$$

and D is numerically hypercyclic by Theorem 3 part (i).

Example 3 

(A unilateral shift which is TNT). Let $1<p<\infty$, and let B and $\mathcal{S}$ denote the backward and forward shifts on ${\mathsf{\ell }}^p\left(\mathbb{N}\right)$:

$$B(x_1,x_2,\dots )=(x_2,x_3,\dots ),\mathcal{S}(x_1,x_2,\dots )=(0,x_1,x_2,\dots ).$$

For every $R>1$, the operator $T=RB$ is topologically numerically transitive.

Indeed, for $x\in S_{{\mathsf{\ell }}^p}$, let $J_p\left(x\right)$ denote its unique norming functional:

$$J_p\left(x\right)\left(v\right)=\sum _{j\ge 1}\overline{x_j}{\left|x_j\right|}^{p-2}{v}_j,$$

with the coefficient interpreted as zero when $x_j=0$. The map $x\mapsto (x,J_p\left(x\right))$ is a homeomorphism from $S_{{\mathsf{\ell }}^p}$ onto $\mathrm{\Pi }\left({\mathsf{\ell }}^p\right)$ for the norm-product topology.

Let $U\subseteq \mathrm{\Pi }\left({\mathsf{\ell }}^p\right)$ and $V\subseteq \mathbb{C}$ be nonempty open sets. Choose a finitely supported unit vector y such that

$$(y,J_p\left(y\right))\in U,$$

and choose $z\in V$. Let m satisfy $y_j=0$ for $j>m$.

For $n>m$, set

$$c_n={\left(1+R^{-np}{\left|z\right|}^p\right)}^{1/p},x^{\left(n\right)}={\displaystyle \frac{y+R^{-n}z{\mathcal{S}}^{n}y}{c_n}}.$$

Since the supports of y and ${\mathcal{S}}^{n}y$ are disjoint, $\parallel x^{\left(n\right)}{\parallel }_p=1.$ Moreover, $x^{\left(n\right)}\to y$ in ${\mathsf{\ell }}^p$. By the norm-continuity of $J_p$,

$$\left(x^{\left(n\right)},J_p\left(x^{\left(n\right)}\right)\right)⟶(y,J_p\left(y\right)).$$

Thus, these pairs belong to U for sufficiently large n.

Since $B^{n}y=0$ and $B^n{\mathcal{S}}^{n}y=y$,

$$T^n{x}^{\left(n\right)}={\displaystyle \frac{z}{c_n}}y.$$

On the support of y, we have $x_j^{\left(n\right)}=y_j/c_n$. Therefore,

$$\begin{array}{cc}\hfill J_p\left(x^{\left(n\right)}\right)\left(T^n{x}^{\left(n\right)}\right)& ={\displaystyle \frac{z}{c_n^p}}{\displaystyle \sum _{j=1}^m}{\left|y_j\right|}^p\hfill \\ & ={\displaystyle \frac{z}{1+R^{-np}{\left|z\right|}^p}}⟶z.\hfill \end{array}$$

Since V is open, this evaluation belongs to V for sufficiently large n. Hence

$$F_n\left(U\right)\cap V\ne ⌀$$

for some $n\ge 1$, proving that T is topologically numerically transitive.

A similar construction applies to the bilateral case, as the next example illustrates. In the next example, we will construct a topologically numerically transitive operator that is neither supercyclic nor weakly hypercyclic.

Example 4. 

[A bilateral shift which is TNT] Let $1<p<\infty$, and let B be the bilateral backward shift on ${\mathsf{\ell }}^p\left(\mathbb{Z}\right)$:

$${\left(Bx\right)}_j=x_{j+1},j\in \mathbb{Z}.$$

Set $\mathcal{S}=B^{-1}$. For every $R>1$, the operator $D=RB$ is topologically numerically transitive.

To see this, for $x\in S_{{\mathsf{\ell }}^p\left(\mathbb{Z}\right)}$, let

$$J_p\left(x\right)\left(v\right)=\sum _{j\in \mathbb{Z}}\overline{x_j}{\left|x_j\right|}^{p-2}{v}_j,$$

where the coefficient is zero when $x_j=0$. The map $x\mapsto (x,J_p\left(x\right))$ is a homeomorphism from the unit sphere onto $\mathrm{\Pi }\left({\mathsf{\ell }}^p\left(\mathbb{Z}\right)\right)$ for the norm-product topology.

Let $U\subseteq \mathrm{\Pi }\left({\mathsf{\ell }}^p\left(\mathbb{Z}\right)\right)$ and $V\subseteq \mathbb{C}$ be nonempty open sets. Choose a finitely supported unit vector y such that $(y,J_p\left(y\right))\in U$, and fix $z\in V$.

For sufficiently large n, the supports of $B^{n}y$, y, and ${\mathcal{S}}^{n}y$ are pairwise disjoint. Define

$$c_n=(1+R^{-np}{\left|z\right|}^p{)}^{1/p},x^{\left(n\right)}={\displaystyle \frac{y+R^{-n}z{\mathcal{S}}^{n}y}{c_n}}.$$

The disjoint supports give $\parallel x^{\left(n\right)}{\parallel }_p=1$. Moreover, $x^{\left(n\right)}\to y$, and the norm-continuity of $J_p$ yields

$$(x^{\left(n\right)},J_p\left(x^{\left(n\right)}\right))⟶(y,J_p\left(y\right)).$$

Thus, these pairs belong to U for sufficiently large n.

Since $B^n{\mathcal{S}}^n=I$,

$$D^n{x}^{\left(n\right)}={\displaystyle \frac{R^n{B}^{n}y+zy}{c_n}}.$$

The support of $B^{n}y$ is disjoint from that of $x^{\left(n\right)}$, so its contribution to the numerical evaluation vanishes. On the support of y, $x_j^{\left(n\right)}=y_j/c_n$. Hence,

$$J_p\left(x^{\left(n\right)}\right)\left(D^n{x}^{\left(n\right)}\right)={\displaystyle \frac{z}{c_n^p}}\sum _{j\in \mathbb{Z}}{\left|y_j\right|}^p={\displaystyle \frac{z}{1+R^{-np}{\left|z\right|}^p}}⟶z.$$

Since V is open, $F_n\left(U\right)\cap V\ne ⌀$ for sufficiently large n. Therefore, D is topologically numerically transitive. On the other hand, since the operator D is normal, it cannot be supercyclic, which follows from [Theorem 3.1] in [33], nor weakly hypercyclic, which follows from [Theorem 4.5] in [34].

We end this section with the following proposition, which we will use in the next section in order to distinguish topological numerical transitivity from other concepts and notions in linear dynamics.

Proposition 2. 

Let $H_1$ and $H_2$ be complex Hilbert spaces, and let $T_1\in B\left(H_1\right)$ be topologically numerically transitive. Then $D=T_1\oplus 0$ is topologically numerically transitive on $H=H_1\oplus H_2$.

Proof. 

Identify $\mathrm{\Pi }\left(H\right)$ with $S_H$ through the unique norming functional of each unit vector. Let $U\subseteq S_H$ and $V\subseteq \mathbb{C}$ be nonempty open sets.

Choose $y=(y_1,y_2)\in U$ with $y_1\ne 0$. This is possible because such vectors are dense in $S_H$. Put

$$r=\parallel y_1\parallel ,u=y_1/r.$$

The map

$$\mathrm{\Phi }:S_{H_1}⟶S_H,\mathrm{\Phi }\left(v\right)=(rv,y_2),$$

is continuous and satisfies $\mathrm{\Phi }\left(u\right)=y$. Consequently, $W={\mathrm{\Phi }}^{-1}\left(U\right)$ is a nonempty open subset of $S_{H_1}$.

By topological numerical transitivity of $T_1$, there exist $v\in W$ and $n\ge 1$ such that

$$⟨T_1^{n}v,v⟩\in r^{-2}V.$$

Set $x=(rv,y_2)\in U$. Since $D^n=T_1^n\oplus 0$ for $n\ge 1$,

$$⟨D^{n}x,x⟩=r^2⟨T_1^{n}v,v⟩\in V.$$

Thus, D is topologically numerically transitive. □

4. Generalized Numerical and Topological ${\mathit{C}}^{\ast }$-Transitivity in Banach Algebras

Let $\mathcal{A}$ be a Banach algebra and $T\in \mathcal{B}\left(\mathcal{A}\right)$. For each $n\in \mathbb{N}$, we let

$${\Theta }_n:\mathcal{A}\times \mathcal{A}⟶\mathcal{A}$$

be given by

$${\Theta }_n(a,b):=a{T}^n\left(b\right).$$

Since the multiplication is jointly continuous on $\mathcal{A}$ and $T^n$ is bounded, it follows that ${\Theta }_n$ is continuous for each $n\in \mathbb{N}$ when $\mathcal{A}\times \mathcal{A}$ is equipped with the product topology.

We will say that T is generalized numerically transitive if, for every pair of nonempty open subsets $\tilde{O}\subseteq \mathcal{A}\times \mathcal{A}$ and $O\subseteq \mathcal{A}$, there exists some $n\in \mathbb{N}$ such that

$${\Theta }_n\left(\tilde{O}\right)\cap O\ne ⌀.$$

In the sequel, H will denote a separable Hilbert space with an orthonormal basis ${\left\{e_j\right\}}_{j\in \mathbb{Z}},$ and for each $m\in \mathbb{N}$ we let $P_m$ be the orthogonal projection onto $\mathrm{span}\{e_{-m},\dots ,e_m\}.$ Furthermore, $(B_0\left(H\right),\parallel ·\parallel )$ will denote the $C^{\ast }$-algebra of compact operators on H equipped with the operator norm. Moreover, for each p with $1\le p<\infty$, we let $B_p\left(H\right)$ denote the corresponding Schatten p-ideal, equipped with the Schatten p-norm ${\parallel ·\parallel }_p$. Furthermore, for each $D\in B_0\left(H\right)$ and $\delta >0$, we let $B(D,\delta )$ be the $\delta$-neighborhood of D in $B_0\left(H\right)$, or, in other words, $B(D,\delta )$ is the open ball in $B_0\left(H\right)$ with center at D and radius $\delta .$

We wish to show that every hypercyclic operator on $B_0\left(H\right)$ is generalized numerically transitive. For that purpose, we first need the following auxiliary lemma.

Lemma 5. 

Let X be a Banach space and let $X_0$ be a finite-dimensional subspace of X. Suppose that $T\in \mathcal{B}(X,X_0)$. Then there exist a finite-dimensional subspace $X_1\subseteq X$, with $\dim X_1=\dim X_0$, and a bounded projection $P:X\to X_1$ such that, for every $\epsilon >0$, there exist operators ${\tilde{T}}_{\epsilon }\in \mathcal{B}(X_1,X_0)$ and ${\tilde{S}}_{\epsilon }\in \mathcal{B}(X_0,X_1)$ satisfying

$$∥T-{\tilde{T}}_{\epsilon }P∥<\epsilon ,{\tilde{T}}_{\epsilon }{\tilde{S}}_{\epsilon }=I_{X_0}.$$

Proof. 

Since $X_0$ is finite-dimensional, T is a finite-rank operator. Thus, $\mathrm{Ran}T$ is finite-dimensional and hence closed. Moreover, $\ker T$ has finite codimension in X. Hence, there exists a finite-dimensional subspace $M\subseteq X$ such that $X=M\oplus \ker T$. The restriction ${T|}_M:M\to \mathrm{Ran}T$ is therefore an isomorphism.

Since $X_0$ is finite-dimensional, choose a subspace $N\subseteq X_0$ such that $X_0=\mathrm{Ran}T\oplus N$, and set $n=\dim N$.

If $n=0$, then $\mathrm{Ran}T=X_0$. Set $X_1=M$, and let $P:X\to X_1$ be the bounded projection associated with $X=X_1\oplus \ker T$. Since T vanishes on $\ker T$, we have $T=TP$. For every $\epsilon >0$, define ${\tilde{T}}_{\epsilon }{=T|}_{X_1}$ and ${\tilde{S}}_{\epsilon }={\tilde{T}}_{\epsilon }^{-1}$. Then ${\tilde{T}}_{\epsilon }{\tilde{S}}_{\epsilon }=I_{X_0}$ and $\parallel T-{\tilde{T}}_{\epsilon }P\parallel =0<\epsilon$.

Suppose now that $n>0$. Choose linearly independent vectors $x_1,\dots ,x_n\in \ker T$ and set $Y=\mathrm{span}\{x_1,\dots ,x_n\}$. Then $\dim Y=\dim N$, so there exists an isomorphism $V:Y\to N$.

Since Y is finite-dimensional, it is complemented in $\ker T$. Thus, there exists a closed subspace $Z\subseteq \ker T$ such that $\ker T=Y\oplus Z$. Consequently,

$$X=M\oplus Y\oplus Z.$$

Set $X_1=M\oplus Y$. Then $\dim X_1=\dim \mathrm{Ran}T+\dim N=\dim X_0$.

Let $P:X\to X_1$ be the bounded projection associated with $X=X_1\oplus Z$, and let $Q:X_1\to Y$ be the projection onto Y along M. Note that $QP$ is then the bounded projection of X onto Y along $M\oplus Z$. Since V is an isomorphism from Y onto N, ${VQP|}_Y{=V|}_Y$ is an isomorphism onto N. Therefore, $VQP\ne 0,$ because $Y\ne \left\{0\right\}$. Since $VQP\ne 0$, for a given $\epsilon >0$ define ${\delta }_{\epsilon }=\epsilon /(2\parallel VQP\parallel )$ and define ${\tilde{T}}_{\epsilon }:X_1\to X_0$ by

$${\tilde{T}}_{\epsilon }(m+y)=Tm+{\delta }_{\epsilon }Vy,m\in M,y\in Y.$$

Since $X_1=M\oplus Y$, $X_0=\mathrm{Ran}T\oplus N$, ${T|}_M:M\to \mathrm{Ran}T$ is an isomorphism, and $V:Y\to N$ is an isomorphism, it follows that ${\tilde{T}}_{\epsilon }:X_1\to X_0$ is an isomorphism. Therefore, define ${\tilde{S}}_{\epsilon }={\tilde{T}}_{\epsilon }^{-1}$. Hence ${\tilde{T}}_{\epsilon }{\tilde{S}}_{\epsilon }=I_{X_0}$.

Finally, if $x=m+y+z$, with $m\in M$, $y\in Y$, and $z\in Z$, then $y,z\in \ker T$, so $Tx=Tm$, while $Px=m+y$. Hence

$$T-{\tilde{T}}_{\epsilon }P=-{\delta }_{\epsilon }VQP.$$

Consequently,

$$∥T-{\tilde{T}}_{\epsilon }P∥={\delta }_{\epsilon }\parallel VQP\parallel ={\displaystyle \frac{\epsilon }{2}}<\epsilon .$$

This completes the proof. □

We now obtain the corresponding result for compact operators on Banach spaces.

Corollary 4. 

Let X be a separable Banach space and let $T\in \mathcal{B}\left(K\right(X\left)\right)$ where $K\left(X\right)$ denotes the ideal of compact operators on X. Suppose that the set of all finite-rank operators on X is dense in $K\left(X\right)$ in the operator norm. If T is hypercyclic on $K\left(X\right)$, then T is generalized numerically transitive on $K\left(X\right)$.

Proof. 

Let $O_1,O_2,O_3$ be nonempty open subsets of $K\left(X\right)$. Since the set of all finite-rank operators on X is dense in $K\left(X\right),$ we may therefore choose nonzero finite-rank operators $D\in O_1$, $F\in O_2$, and $G\in O_3$. Choose $\delta >0$ such that $B(D,\delta )\subseteq O_1$, $B(F,\delta )\subseteq O_2$, and $B(G,\delta )\subseteq O_3$.

Set

$$E=\mathrm{Ran}D+\mathrm{Ran}F+\mathrm{Ran}G.$$

Then E is a finite-dimensional subspace of X.

Regard D as an operator from X into E. Applying Lemma 5 with $\epsilon =\delta /2$, we obtain a finite-dimensional subspace $X_1\subseteq X$, a bounded projection $P_1:X\to X_1$, and operators $\tilde{D}\in \mathcal{B}(X_1,E)$ and $\tilde{S}\in \mathcal{B}(E,X_1)$ such that

$$\parallel D-\tilde{D}{P}_1\parallel <{\displaystyle \frac{\delta }{2}},\tilde{D}\tilde{S}=I_E.$$

Define $\widehat{D}=\tilde{D}{P}_1$. Since $\widehat{D}$ has finite-dimensional range, $\widehat{D}\in K\left(X\right)$, and the preceding estimate shows that $\widehat{D}\in O_1$. Moreover, since $P_1\tilde{S}=\tilde{S}$, we have $\widehat{D}\tilde{S}=I_E$.

Now set $Y=\tilde{S}G$. Since G has finite rank, Y also has finite rank and, hence, $Y\in K\left(X\right)$. Moreover,

$$\widehat{D}Y=\widehat{D}\tilde{S}G=G.$$

Since T is hypercyclic on $K\left(X\right)$, it is topologically transitive. Therefore, there exist $B\in K\left(X\right)$ and $n\in \mathbb{N}$ such that

$$\parallel B-F\parallel <{\displaystyle \frac{\delta }{2}},\parallel T^n\left(B\right)-Y\parallel <{\displaystyle \frac{\delta }{2\parallel \widehat{D}\parallel }}.$$

Hence, $B\in O_2$. Finally,

$$\parallel \widehat{D}{T}^n\left(B\right)-G\parallel =\parallel \widehat{D}(T^n\left(B\right)-Y)\parallel \le \parallel \widehat{D}\parallel \parallel T^n\left(B\right)-Y\parallel <{\displaystyle \frac{\delta }{2}}<\delta .$$

Thus, $\widehat{D}{T}^n\left(B\right)\in O_3$. Therefore, the family $\{{\Theta }_n:n\in \mathbb{N}\}$, where ${\Theta }_n(A,B)=A{T}^n\left(B\right)$ for all A, $B\in K\left(X\right)$, is topologically transitive. Hence, T is generalized numerically transitive on $K\left(X\right)$. □

Remark 2. 

We recall that if X has the bounded approximation property, then the set of all finite-rank operators on X is dense in $K\left(X\right)$ in the operator norm. In particular, taking $X=H$, we have $K\left(H\right)=B_0\left(H\right)$, and the set of all finite-rank operators on H is dense in $B_0\left(H\right)$ in the operator norm. Therefore, every hypercyclic operator on $B_0\left(H\right)$ is generalized numerically transitive.

We will now introduce another notion of transitivity that is applicable in spaces whose norm is induced by a Banach bimodule-valued inner product. As we will see in examples, such spaces are, for instance, the space of Hilbert–Schmidt operators and Hilbert $C^{\ast }$-modules. Let $\mathcal{M}$ be a Banach bimodule over a $C^{\ast }$-algebra $\mathcal{C}$, and let $\mathcal{K}$ be a cone in $\mathcal{M}$. Suppose that X is a Banach space such that there exists a sesquilinear map

$$\phi :X\times X⟶\mathcal{M}$$

with the property that

$$\phi (x,x)\in \mathcal{K}\mathrm{for}\mathrm{all}x\in X,$$

and such that

$${\parallel x\parallel }_X={\parallel \phi (x,x)\parallel }_{\mathcal{M}}^{1/2}\mathrm{for}\mathrm{all}x\in X.$$

Let $S\left(X\right)$ denote the unit sphere of X. For $T\in \mathcal{B}\left(X\right)$ and $n\in \mathbb{N}$, let

$${\tilde{\Theta }}_n:S\left(X\right)\times S\left(X\right)⟶\mathcal{M}$$

be given by

$${\tilde{\Theta }}_n(x,y)=\phi \left(T^{n}x,y\right).$$

We say that T is topologically $C^{\ast }$-transitive with respect to φ (or shortly $C^{\ast }$-transitive with respect to φ) if the sequence ${\left\{{\tilde{\Theta }}_n\right\}}_{n\in \mathbb{N}}$ is topologically transitive when $S\left(X\right)\times S\left(X\right)$ is considered to be equipped with the standard product topology. Usually, in concrete cases, we will omit mentioning with respect to φ when it is clear which sesquilinear map $\phi$ is under consideration. Also, we will say that T is topologically $C^{\ast }$-mixing with respect to φ (or shortly $C^{\ast }$-mixing with respect to φ) if the sequence ${\left\{{\tilde{\Theta }}_n\right\}}_{n\in \mathbb{N}}$ is topologically mixing. Usually, in concrete cases we will omit mentioning with respect to φ when it is clear which sesquilinear map $\phi$ is under consideration.

Before proceeding further, let us first give some examples of such Banach spaces.

Example 5. 

Let $X=\mathcal{M}=B_0\left(H\right),$ and let $\mathcal{K}$ be the natural cone consisting of positive operators in $B_0\left(H\right)$. Define $\phi :B_0\left(H\right)\times B_0\left(H\right)⟶B_0\left(H\right)$ by $\phi (A,B)=B^{\ast }A,\mathit{for}\mathit{all}A,B\in B_0\left(H\right).$ In other words,

$$\phi (x,y)=⟨y,x⟩,x,y\in B_0\left(H\right),$$

where $⟨·,·⟩$ is the inner product on $B_0\left(H\right)$ defined in [35], [Example 1.2.2], when $B_0\left(H\right)$ is considered as a Hilbert module over $B\left(H\right).$

Example 6. 

Let $X=B_2\left(H\right),\mathcal{M}=B_1\left(H\right),$ and let $\mathcal{K}$ be the natural cone consisting of positive operators in $B_1\left(H\right)$. Define $\phi :B_2\left(H\right)\times B_2\left(H\right)⟶B_1\left(H\right)$ by $\phi (A,B)=B^{\ast }A,\mathit{for}\mathit{all}A,B\in B_2\left(H\right).$

In some special cases, topological $C^{\ast }$-transitivity implies generalized numerical transitivity, as the next result illustrates.

Proposition 3. 

If T is a topologically $C^{\ast }$-transitive operator on $B_0\left(H\right)$, then T is generalized numerically transitive on $B_0\left(H\right).$

Proof. 

Let $O_1,O_2,O_3$ be nonempty open subsets of $B_0\left(H\right)$. We wish to show that there exists some $n\in \mathbb{N}$ such that ${\Theta }_n\left((O_1\times O_2)\right)\cap O_3\ne ⌀.$ To this end, notice first that we can find some $F\in O_1\setminus \left\{0\right\},D\in O_2\setminus \left\{0\right\},$ and some $\delta >0$ such that the $\delta$-neighborhood $B(D,\delta )$ of D is contained in $O_2\setminus \left\{0\right\}$. If T is a topologically $C^{\ast }$-transitive operator on $B_0\left(H\right)$, then we can find some $n\in \mathbb{N}$ and some

$$\tilde{F}\in {\displaystyle \frac{1}{\parallel F\parallel }}{O}_1\cap S\left(B_0\left(H\right)\right),\tilde{D}\in {\displaystyle \frac{1}{\parallel D\parallel }}B(D^{\ast },\delta )\cap S\left(B_0\left(H\right)\right)$$

such that ${\tilde{D}}^{\ast }{T}^n\left(\tilde{F}\right)\in {\displaystyle \frac{1}{\parallel F\parallel \parallel D\parallel }}{O}_3.$ Hence, multiplying by $\parallel F\parallel$ and $\parallel D\parallel$ we deduce the desired result. □

While every hypercyclic operator on $B_0\left(H\right)$ is generalized numerically transitive, it turns out that the converse does not hold in general. In what follows, we will construct a generalized numerically transitive operator on $B_0\left(H\right)$ which is not even supercyclic and, hence, is not hypercyclic.

Let $\mathcal{S}$ be the bilateral forward shift on H, defined by $\mathcal{S}\left(e_j\right)=e_{j+1}$ for every $j\in \mathbb{Z}$, and set

$$U=2\mathcal{S}.$$

Then U is an invertible normal operator on H. We let $L_U$ be the corresponding left multiplier by U on $B_0\left(H\right)$, that is,

$$L_U\left(F\right)=UF,F\in B_0\left(H\right).$$

In the sequel, we will say that $L_U$ is WOT-hypercyclic if there exists some $F\in B_0\left(H\right)$ whose orbit under $L_U$ is dense in the weak operator topology of $B_0\left(H\right).$

Proposition 4. 

The operator $L_U$ is a nonsupercyclic, topologically $C^{\ast }$-transitive, and generalized numerically transitive operator on $B_0\left(H\right)$. Moreover, $L_U$ is not WOT-hypercyclic.

Proof. 

Let $O_1,O_2,O_3$ be nonempty open subsets of $B_0\left(H\right)$ such that

$$O_j\cap S\left(B_0\left(H\right)\right)\ne ⌀,\mathrm{for}\mathrm{all}j\in \{1,2\}.$$

Choose $D\in O_1\cap S\left(B_0\left(H\right)\right),F\in O_2\cap S\left(B_0\left(H\right)\right),G\in O_3,$ and choose some $\delta \in (0,1)$ such that the $\delta$-neighborhoods of $D,F,G$ are contained in $O_1,O_2,O_3$, respectively. Now we recall that the involution is an isometry. Therefore, by [Proposition 2.2.1] in [35] and by passing to the adjoints, it is not hard to see that we can find some $m\in \mathbb{N}$ such that

$$\parallel D-P_{m}D\parallel ,\parallel F-P_{m}F\parallel ,\parallel G-P_{m}G\parallel <{\displaystyle \frac{\delta }{16(1+\parallel G\parallel )}}.$$

Then

$$\parallel D-P_{m}D{P}_m\parallel \le \parallel D-P_{m}D\parallel +\parallel P_m\parallel \parallel D-D{P}_m\parallel <{\displaystyle \frac{\delta }{8}}.$$

Hence,

$$\parallel P_{m}D{P}_m\parallel \ge \parallel D\parallel -{\displaystyle \frac{\delta }{8}}=1-{\displaystyle \frac{\delta }{8}}>{\displaystyle \frac{7}{8}}>0,$$

where we have used that $\parallel D\parallel =1$ because $D\in S\left(B_0\left(H\right)\right).$

Similarly, $\parallel P_{m}G\parallel ,\parallel P_{m}F\parallel >1-{\displaystyle \frac{\delta }{16}}>0.$ Moreover, since $\parallel D\parallel =1$ and $\parallel D-P_{m}D{P}_m\parallel \le {\displaystyle \frac{\delta }{8}},$ as shown above, we obtain

$$\left|\parallel P_{m}D{P}_m\parallel -1\right|=\left|\parallel P_{m}D{P}_m\parallel -\parallel D\parallel \right|\le \parallel D-P_{m}D{P}_m\parallel \le {\displaystyle \frac{\delta }{8}},$$

hence we deduce that

$$∥D-{\displaystyle \frac{P_{m}D{P}_m}{\parallel P_{m}D{P}_m\parallel }}∥\le \parallel D-P_{m}D{P}_m\parallel +∥P_{m}D{P}_m-{\displaystyle \frac{P_{m}D{P}_m}{\parallel P_{m}D{P}_m\parallel }}∥$$

$$\le {\displaystyle \frac{\delta }{8}}+\parallel P_{m}D{P}_m\parallel \left|1-{\displaystyle \frac{1}{\parallel P_{m}D{P}_m\parallel }}\right|={\displaystyle \frac{\delta }{8}}+\left|\parallel P_{m}D{P}_m\parallel -1\right|<{\displaystyle \frac{\delta }{4}}.$$

By Lemma 5, we can choose an invertible operator $\tilde{D}\in B\left(P_m\left(H\right)\right)$ such that

$$∥{\displaystyle \frac{P_{m}D{P}_m}{\parallel P_{m}D{P}_m\parallel }}-\tilde{D}∥<{\displaystyle \frac{\delta }{16}}.$$

Consequently, $∥D-{\displaystyle \frac{\tilde{D}{P}_m}{\parallel \tilde{D}{P}_m\parallel }}∥<\delta ,$ so that ${\displaystyle \frac{\tilde{D}{P}_m}{\parallel \tilde{D}{P}_m\parallel }}\in O_1\cap S\left(B_0\left(H\right)\right).$ Again, we can find some $n_0\in \mathbb{N}$ such that, for every $n\ge n_0$,

$$P_m{U}^n{P}_m=0$$

and

$$\parallel U^{-n}{P}_m\parallel <{\displaystyle \frac{\delta }{4\parallel {\tilde{D}}^{\ast -1}\parallel \parallel {\tilde{D}}^{\ast }\parallel \parallel G\parallel (1+\parallel G\parallel )}}.$$

For each $n\ge n_0$, set

$$E_n:={\displaystyle \frac{P_{m}F}{\parallel P_{m}F\parallel }}+U^{-n}{\tilde{D}}^{\ast -1}\parallel {\tilde{D}}^{\ast }\parallel P_{m}G.$$

Since $P_m\tilde{D}{P}_m=\tilde{D}$, we also have $P_m{\tilde{D}}^{\ast }{P}_m={\tilde{D}}^{\ast }$, and therefore

$$U^{-n}{\tilde{D}}^{\ast -1}=U^{-n}{P}_m{\tilde{D}}^{\ast -1}.$$

Hence,

$$\begin{array}{cc}\hfill \parallel E_n-F\parallel & \le ∥F-{\displaystyle \frac{P_{m}F}{\parallel P_{m}F\parallel }}∥+\parallel U^{-n}{\tilde{D}}^{\ast -1}\parallel {\tilde{D}}^{\ast }\parallel P_{m}G\parallel \hfill \\ & \le \parallel F-P_{m}F\parallel +\left|\parallel P_{m}F\parallel -1\right|+\parallel U^{-n}{P}_m\parallel \parallel {\tilde{D}}^{\ast -1}\parallel \parallel {\tilde{D}}^{\ast }\parallel \parallel G\parallel \hfill \\ & <{\displaystyle \frac{3\delta }{8(1+\parallel G\parallel )}}.\hfill \end{array}$$

By the last inequality,

$$∥{\displaystyle \frac{E_n}{\parallel E_n\parallel }}-F∥\le \parallel E_n-F\parallel +\left|1-\parallel E_n\parallel \right|<{\displaystyle \frac{3\delta }{4(1+\parallel G\parallel )}}<\delta .$$

Thus, ${\displaystyle \frac{E_n}{\parallel E_n\parallel }}\in O_2\cap S\left(B_0\left(H\right)\right).$ Next, since $P_m{U}^n{P}_m=0$ for all $n\ge n_0$, by some computations we obtain

$$\begin{array}{cc}\hfill \phi \left(L_U^n\left({\displaystyle \frac{E_n}{\parallel E_n\parallel }}\right),{\displaystyle \frac{\tilde{D}}{\parallel \tilde{D}\parallel }}\right)& ={\displaystyle \frac{P_{m}G}{\parallel E_n\parallel }},\hfill \end{array}$$

where we have used $\parallel \tilde{D}\parallel =\parallel {\tilde{D}}^{\ast }\parallel$. Moreover,

$$∥G-{\displaystyle \frac{P_{m}G}{\parallel E_n\parallel }}∥\le \parallel G-P_{m}G\parallel +{\displaystyle \frac{\parallel P_{m}G\parallel }{\parallel E_n\parallel }}\left|\parallel E_n\parallel -1\right|<{\displaystyle \frac{\delta }{16}}+{\displaystyle \frac{\frac{3\delta }{8}}{1-\frac{3\delta }{8}}}.$$

Here we have used

$$\parallel G\parallel \left|\parallel E_n\parallel -1\right|\le \left|\parallel E_n\parallel -\parallel F\parallel \right|\parallel G\parallel <{\displaystyle \frac{3\delta }{8}},$$

and

$$\parallel E_n\parallel >\parallel F\parallel -{\displaystyle \frac{3\delta }{8(1+\parallel G\parallel )}}>1-{\displaystyle \frac{3\delta }{8}}.$$

Since $\delta \in (0,1)$, we get $1-{\displaystyle \frac{3\delta }{8}}>{\displaystyle \frac{5}{8}},$ and, hence,

$$∥G-{\displaystyle \frac{P_{m}G}{\parallel E_n\parallel }}∥<{\displaystyle \frac{\delta }{16}}+{\displaystyle \frac{3\delta }{5}}<\delta .$$

Therefore,

$${\tilde{\Theta }}_n\left(O_1\cap S\left(B_0\left(H\right)\right),O_2\cap S\left(B_0\left(H\right)\right)\right)\cap O_3\ne ⌀$$

for every $n\ge n_0$. This proves that $L_U$ is topologically $C^{\ast }$-transitive on $B_0\left(H\right)$. From Proposition 3 it follows that $L_U$ is generalized numerically transitive on $B_0\left(H\right)$.

It remains to show that $L_U$ is not supercyclic. Suppose, to the contrary, that $L_U$ is supercyclic, and let $F\in B_0\left(H\right)$ be a supercyclic vector for $L_U$. Since $F\ne 0$, there exists $x\in H$ such that $Fx\ne 0$.

Consider the evaluation map ${\mathrm{\Phi }}_x:B_0\left(H\right)⟶H,$ given by ${\mathrm{\Phi }}_x\left(A\right)=Ax.$ This map is continuous and surjective. To see that ${\mathrm{\Phi }}_x$ is surjective, just observe that for every $y\in H$, the rank-one operator

$$A_y\left(z\right)={\displaystyle \frac{⟨z,x⟩}{{\parallel x\parallel }^2}}y$$

belongs to $B_0\left(H\right)$ and satisfies $A_{y}x=y$. Hence, ${\mathrm{\Phi }}_x\left(A_y\right)=y,$ which proves the surjectivity of ${\mathrm{\Phi }}_x.$

To see that the map ${\mathrm{\Phi }}_x$ is continuous, just observe that if ${\left\{A_n\right\}}_n\subset B_0\left(H\right)$ converges to some $A\in B_0\left(H\right)$ in the operator norm, then

$$A_{n}x⟶Ax\mathrm{as}n\to \infty ,$$

which proves the continuity of ${\mathrm{\Phi }}_x.$ Hence, ${\mathrm{\Phi }}_x\left(A_n\right)⟶{\mathrm{\Phi }}_x\left(A\right)\mathrm{as}n\to \infty .$

Since the projective orbit of F under $L_U$ is dense in $B_0\left(H\right)$, its image under ${\mathrm{\Phi }}_x$ is dense in H. Moreover,

$${\mathrm{\Phi }}_x\left(\lambda L_U^n\left(F\right)\right)=\lambda U^n\left(Fx\right)$$

for every $\lambda \in \mathbb{C}$ and $n\in \mathbb{N}$. Thus, $Fx$ would be a supercyclic vector for U. This is impossible because $U=2\mathcal{S}$ is normal, and normal operators on complex Hilbert spaces are not supercyclic by [Theorem 3.1] in [33]. Therefore, $L_U$ is not supercyclic.

Finally, by similar arguments, it can be proved that $L_U$ cannot be WOT-hypercyclic on $B_0\left(H\right).$ Indeed, to see that $L_U$ cannot be WOT-hypercyclic, assume for contradiction that $L_U$ is WOT-hypercyclic and let $F\in B_0\left(H\right)$ be a WOT-hypercyclic vector for $L_U$. Again, since $F\ne 0,$ there exists some $x\in H$ such that $Fx\ne 0.$ Moreover, since ${\left\{U^{k}F\right\}}_k$ is dense in $B_0\left(H\right)$ in the weak operator topology, given $y,z_1,\dots ,z_m\in H\mathrm{and}\delta >0,$ we can find some $n\in \mathbb{N}$ such that

$$\left|⟨\left(U^{n}F-A_y\right)x,z_j⟩\right|<\delta$$

for each $j\in \{1,\dots ,m\},$ where, as before,

$$A_y\left(h\right)={\displaystyle \frac{⟨h,x⟩}{{\parallel x\parallel }^2}}y,h\in H.$$

Since $A_{y}x=y,$ we obtain

$$\left|⟨U^n\left(Fx\right)-y,z_j⟩\right|<\delta$$

for each $j\in \{1,\dots ,m\}.$ It follows that $Fx$ is a weakly hypercyclic vector for U, because $y,z_1,\dots ,z_m\in H$ and $\delta >0$ were chosen arbitrarily. However, since U is a normal operator, by [Theorem 4.5] in [34] we obtain that U cannot be weakly hypercyclic on H. Thus, we obtain a contradiction. Therefore, we conclude that $L_U$ is not WOT-hypercyclic. □

As we will see in the next proposition, it is not true in general that topological $C^{\ast }$-transitivity implies generalized numerical transitivity.

Let ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ denote the standard Hilbert module over $B_0\left(H\right)$ (see [Example 1.3.5] in [35]). Thus, in this special case, we let

$$X={\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right),\mathcal{M}=B_0\left(H\right),$$

and let $\phi$ be the standard $B_0\left(H\right)$-valued inner product on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ given by

$$\phi \left((x_1,x_2,\dots ),(y_1,y_2,\dots )\right)={\displaystyle \sum _{j=1}^{\infty }}{y}_j^{\ast }{x}_j.$$

In other words,

$$\phi (x,y)=⟨y,x⟩,x,y\in {\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right),$$

where $⟨·,·⟩$ is the inner product on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ defined in [35], [Example 1.3.5].

For every $x\in {\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ and $j\in \mathbb{N}$, we let $x_j$ denote its j-th component. It has been proved in [Section 4] in [32] that ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ is a Banach algebra under componentwise multiplication, i.e.,

$${\left(xy\right)}_j=x_j{y}_j,x,y\in {\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right),j\in \mathbb{N}.$$

We have the following proposition.

Proposition 5. 

There exists a topologically $C^{\ast }$-transitive operator on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ which is not generalized numerically transitive.

Proof. 

Let U be the normal operator on H as above in Proposition 4, and $\tilde{D}\in \mathcal{B}\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right)$ be given by

$$\tilde{D}(T_1,T_2,\dots )=(U{T}_1,0,0,\dots ),(T_1,T_2,\dots )\in {\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right).$$

We first show that $\tilde{D}$ is $C^{\ast }$- transitive on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$. The proof of this statement will be divided into 3 main steps:

(i)

We will first approximate open sets in ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ by elements with nonzero first coordinate.

(ii)

Then we will use the $C^{\ast }$-transitivity of the operator $L_U$ on $B_0\left(H\right)$.

(iii)

Finally, we will need to control the normalization of elements in ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ to obtain the desired approximation.

While the procedure in the first two steps is somewhat similar to that in the proof of Proposition 2, the most technical work will be needed for the third step due to the fact that Pythagoras does not generally hold (in the norm) in ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ as it holds in ${\mathsf{\ell }}_2$ because the components of elements in ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ are operators and not just scalars.

Let $O_1,O_2$ be open subsets of ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ such that

$$O_j\cap S\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right)\ne ⌀,j\in \{1,2\},$$

and let $\tilde{O}$ be a nonempty open subset of $B_0\left(H\right)$. Choose

$$x\in O_1\cap S\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right),y\in O_2\cap S\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right),$$

and $F\in \tilde{O}$. By taking $\delta >0$ sufficiently small, we may assume that the $\delta$-neighborhoods of x, y, and F are contained in $O_1$, $O_2$, and $\tilde{O}$, respectively. We may moreover assume that $x=(x_1,x_2,\dots ),y=(y_1,y_2,\dots )$ satisfy $x_1\ne 0,y_1\ne 0.$ Indeed, suppose that $x_1=0$. Choose some $T\in S\left(B_0\left(H\right)\right)$ and set

$$x^{\prime }=\left({\displaystyle \frac{\delta }{2}}T,x_2,x_3,\dots \right),x^{″}={\displaystyle \frac{x^{\prime }}{\parallel x^{\prime }\parallel }}.$$

Then $x^{\prime },x^{″}\in {\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ and

$$\parallel x-x^{″}\parallel \le \parallel x-x^{\prime }\parallel +\parallel x^{\prime }-x^{″}\parallel =\parallel x-x^{\prime }\parallel +|\parallel x^{\prime }\parallel -1|<\delta .$$

Hence

$$x^{″}\in O_1\cap S\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right),$$

and the first component of $x^{″}$ is nonzero. A similar argument applies to y. Since

$$\underset{t\to 0}{\lim }\left({\displaystyle \frac{1}{{(1-t)}^2}}-1\right)=\underset{t\to 0}{\lim }\left(1-{\displaystyle \frac{1}{{(1+t)}^2}}\right)=0,$$

we can choose $\epsilon \in \left(0,{\displaystyle \frac{\delta }{4}}\right)$ such that

$$1-{\displaystyle \frac{1}{{(1+\epsilon )}^2}},{\displaystyle \frac{1}{{(1-\epsilon )}^2}}-1<{\displaystyle \frac{\delta }{4\left({\displaystyle \frac{\parallel F\parallel }{\parallel x_1\parallel \parallel y_1\parallel }}+1\right)}}.$$

Let $V_1$ and $V_2$ be the $\epsilon$-neighborhoods of ${\displaystyle \frac{x_1}{\parallel x_1\parallel }}\mathrm{and}{\displaystyle \frac{y_1}{\parallel y_1\parallel }},$ respectively, in $B_0\left(H\right)$. Also, let $\tilde{V}$ be the $\epsilon$-neighborhood of ${\displaystyle \frac{F}{\parallel x_1\parallel \parallel y_1\parallel }}$ in $B_0\left(H\right)$. Since $L_U$ is $C^{\ast }$-transitive on $B_0\left(H\right)$ by Proposition 4, there exist some $n\in \mathbb{N}$ and

$$z\in V_1\cap S\left(B_0\left(H\right)\right),w\in V_2\cap S\left(B_0\left(H\right)\right)$$

such that $w^{\ast }{U}^{n}z\in \tilde{V}.$ Thus,

$$∥w^{\ast }{U}^{n}z-{\displaystyle \frac{F}{\parallel x_1\parallel \parallel y_1\parallel }}∥<\epsilon .$$

Since $\parallel x_1\parallel \le \parallel x\parallel =1$, we have

$$∥\parallel x_1\parallel z-x_1∥=\parallel x_1\parallel ∥z-{\displaystyle \frac{x_1}{\parallel x_1\parallel }}∥<\parallel x_1\parallel \epsilon <\epsilon .$$

Similarly, $∥\parallel y_1\parallel w-y_1∥<\epsilon .$ Let $\overline{x},\overline{y}\in {\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ be given by

$$\overline{x}=(\parallel x_1\parallel z,x_2,x_3,\dots ),\overline{y}=(\parallel y_1\parallel w,y_2,y_3,\dots ).$$

Then $\parallel x-\overline{x}\parallel <\epsilon ,\parallel y-\overline{y}\parallel <\epsilon .$ Since $z,w\in S\left(B_0\left(H\right)\right)$, both $\overline{x}$ and $\overline{y}$ are nonzero. Put $\tilde{x}={\displaystyle \frac{\overline{x}}{\parallel \overline{x}\parallel }}$ and $\tilde{y}={\displaystyle \frac{\overline{y}}{\parallel \overline{y}\parallel }}.$ Then

$$\parallel x-\tilde{x}\parallel \le \parallel x-\overline{x}\parallel +∥\overline{x}-{\displaystyle \frac{\overline{x}}{\parallel \overline{x}\parallel }}∥=\parallel x-\overline{x}\parallel +|\parallel \overline{x}\parallel -1|<2\epsilon <{\displaystyle \frac{\delta }{2}}.$$

Similarly, $\parallel y-\tilde{y}\parallel <2\epsilon <{\displaystyle \frac{\delta }{2}}.$ Consequently,

$$\tilde{x}\in O_1\cap S\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right)\mathrm{and}\tilde{y}\in O_2\cap S\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right).$$

Furthermore,

$$⟨\tilde{y},{\tilde{D}}^n\left(\tilde{x}\right)⟩={\displaystyle \frac{\parallel y_1\parallel w^{\ast }}{\parallel \overline{y}\parallel }}{U}^n{\displaystyle \frac{\parallel x_1\parallel z}{\parallel \overline{x}\parallel }}.$$

Hence

$$∥⟨\tilde{y},{\tilde{D}}^n\left(\tilde{x}\right)⟩-F∥=\parallel x_1\parallel \parallel y_1\parallel ∥{\displaystyle \frac{w^{\ast }{U}^{n}z}{\parallel \overline{y}\parallel \parallel \overline{x}\parallel }}-{\displaystyle \frac{F}{\parallel x_1\parallel \parallel y_1\parallel }}∥.$$

Since $\parallel x_1\parallel ,\parallel y_1\parallel \le 1$, it follows that

$$∥⟨\tilde{y},{\tilde{D}}^n\left(\tilde{x}\right)⟩-F∥\le ∥{\displaystyle \frac{w^{\ast }{U}^{n}z}{\parallel \overline{y}\parallel \parallel \overline{x}\parallel }}-{\displaystyle \frac{F}{\parallel x_1\parallel \parallel y_1\parallel }}∥$$

$$\le ∥{\displaystyle \frac{w^{\ast }{U}^{n}z}{\parallel \overline{y}\parallel \parallel \overline{x}\parallel }}-w^{\ast }{U}^{n}z∥+∥w^{\ast }{U}^{n}z-{\displaystyle \frac{F}{\parallel x_1\parallel \parallel y_1\parallel }}∥\le \parallel w^{\ast }{U}^{n}z\parallel \left|{\displaystyle \frac{1}{\parallel \overline{y}\parallel \parallel \overline{x}\parallel }}-1\right|+{\displaystyle \frac{\delta }{4}}.$$

Moreover,

$$∥w^{\ast }{U}^{n}z-{\displaystyle \frac{F}{\parallel x_1\parallel \parallel y_1\parallel }}∥<\epsilon <{\displaystyle \frac{\delta }{4}}<1,$$

and therefore

$$\parallel w^{\ast }{U}^{n}z\parallel \le {\displaystyle \frac{\parallel F\parallel }{\parallel x_1\parallel \parallel y_1\parallel }}+1.$$

Thus,

$$∥⟨\tilde{y},{\tilde{D}}^n\left(\tilde{x}\right)⟩-F∥\le {\displaystyle \frac{\delta }{4}}+\left({\displaystyle \frac{\parallel F\parallel }{\parallel x_1\parallel \parallel y_1\parallel }}+1\right)\left|{\displaystyle \frac{1}{\parallel \overline{y}\parallel \parallel \overline{x}\parallel }}-1\right|.$$

Now, since $\parallel x-\overline{x}\parallel <\epsilon \mathrm{and}\parallel y-\overline{y}\parallel <\epsilon ,$ we have

$$1-\epsilon =\parallel x\parallel -\epsilon \le \parallel \overline{x}\parallel \le \parallel x\parallel +\epsilon =1+\epsilon ,$$

and similarly, $1-\epsilon \le \parallel \overline{y}\parallel \le 1+\epsilon .$ Therefore, ${(1+\epsilon )}^{-2}\le (\parallel \overline{x}\parallel \parallel \overline{y}{\parallel )}^{-1}\le {(1-\epsilon )}^{-2}.$ If $(\parallel \overline{x}\parallel \parallel \overline{y}{\parallel )}^{-1}\le 1,$ then $0\le 1-(\parallel \overline{x}\parallel \parallel \overline{y}{\parallel )}^{-1}\le 1-{(1+\epsilon )}^{-2}.$ On the other hand, if $(\parallel \overline{x}\parallel \parallel \overline{y}{\parallel )}^{-1}\ge 1,$ then $0\le (\parallel \overline{x}\parallel \parallel \overline{y}{\parallel )}^{-1}-1\le {(1-\epsilon )}^{-2}-1.$ By our choice of $\epsilon$, we therefore obtain

$$\left|{\displaystyle \frac{1}{\parallel \overline{x}\parallel \parallel \overline{y}\parallel }}-1\right|<{\displaystyle \frac{\delta }{4\left({\displaystyle \frac{\parallel F\parallel }{\parallel x_1\parallel \parallel y_1\parallel }}+1\right)}}.$$

Hence,

$$∥⟨\tilde{y},{\tilde{D}}^n\left(\tilde{x}\right)⟩-F∥\le {\displaystyle \frac{\delta }{4}}+\left({\displaystyle \frac{\parallel F\parallel }{\parallel x_1\parallel \parallel y_1\parallel }}+1\right)\left|{\displaystyle \frac{1}{\parallel \overline{y}\parallel \parallel \overline{x}\parallel }}-1\right|<{\displaystyle \frac{\delta }{2}}.$$

Consequently,

$$⟨\tilde{y},{\tilde{D}}^n\left(\tilde{x}\right)⟩\in \tilde{O}.$$

Since $\tilde{x}\in O_1,\tilde{y}\in O_2,$ and $\parallel \tilde{x}\parallel =\parallel \tilde{y}\parallel =1,$ we conclude that $\tilde{D}$ is topologically $C^{\ast }$-transitive on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$.

Now we show that $\tilde{D}$ is not generalized numerically transitive on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$. However, this is obvious because the second component of ${\tilde{D}}^{n}y$ is always zero for each $n\in \mathbb{N}$ and $y\in {\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$, so ${\Theta }_n$ cannot hit an open set where the second component is close to a nonzero operator. □

The following corollary can be deduced by similar arguments as in the proof of Proposition 5.

Corollary 5. 

Let T be a generalized compact (in the sense of [Section 2.2] in [35]) operator on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ and suppose that $T=\mathrm{diag}(D_1,D_2,\dots )$ with $D_j\in B_0\left(H\right)$ for all $j\in \mathbb{N}.$ Then, given any $c\in (0,1),$ there exists some $m\in \mathbb{N}$ such that $\parallel D_j\parallel \le c$ for all $j\ge m+1.$ Furthermore, if $L_{D_j}$ is $C^{\ast }$-mixing on $B_0\left(H\right)$ for every $j\le m,$ then T is $C^{\ast }$-transitive on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$.

Proof. 

The first statement follows directly from [Proposition 2.2.1] in [35] because T is generalized compact by the assumption. Then, by utilizing this fact and by applying slight modifications of the proof of Proposition 5, we can deduce the second statement of the corollary. Indeed, let $m\in \mathbb{N}$ be such that $L_{D_1},\dots ,L_{D_m}$ are $C^{\ast }$-mixing on $B_0\left(H\right)$, and such that

$$\parallel D_j\parallel \le c\mathrm{for}\mathrm{all}j\ge m+1$$

for some $c\in (0,1)$. Let again ${\mathcal{O}}_1$ and ${\mathcal{O}}_2$ be open neighborhoods in ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ such that

$${\mathcal{O}}_k\cap S\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right)\ne ⌀,k\in \{1,2\},$$

and let $\tilde{\mathcal{O}}$ be an open nonempty subset of $B_0\left(H\right)$. By similar arguments as in the proof of Proposition 5, we may, without loss of generality, assume that there exist

$$x\in {\mathcal{O}}_1\cap S\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right),y\in {\mathcal{O}}_2\cap S\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right),$$

and $F\in \tilde{\mathcal{O}}$ such that $x_j\ne 0,y_j\ne 0\mathrm{for}\mathrm{all}j\in \{1,\dots ,m\}.$ Also, we can find some $\delta \in (0,1)$ such that the $\delta$-neighborhoods of x, y, and F are contained in ${\mathcal{O}}_1$, ${\mathcal{O}}_2$, and $\tilde{\mathcal{O}}$, respectively.

Since $c\in (0,1)$, there exists some $N\in \mathbb{N}$ such that $c^N<{\displaystyle \frac{\delta }{2}}.$ As in the proof of Proposition 5, we can find some $\epsilon \in \left(0,{\displaystyle \frac{\delta }{4}}\right)$ such that

$$1-{\displaystyle \frac{1}{{(1+\epsilon )}^2}},{\displaystyle \frac{1}{{(1-\epsilon )}^2}}-1<\underset{j\in \{1,\dots ,m\}}{\min }\left\{{\displaystyle \frac{\delta }{4\left({\displaystyle \frac{\parallel F\parallel }{m\parallel x_j\parallel \parallel y_j\parallel }}+1\right)m}}\right\}.$$

For each $j\in \{1,\dots ,m\}$, let $V_{j,1}$ and $V_{j,2}$ be the ${\displaystyle \frac{\epsilon }{m}}$-neighborhoods in $B_0\left(H\right)$ of ${\displaystyle \frac{x_j}{\parallel x_j\parallel }}\mathrm{and}{\displaystyle \frac{y_j}{\parallel y_j\parallel }},$ respectively, whereas we let ${\tilde{V}}_j$ be the ${\displaystyle \frac{\epsilon }{2m}}$-neighborhood in $B_0\left(H\right)$ of

$${\displaystyle \frac{F}{m\parallel x_j\parallel \parallel y_j\parallel }}.$$

Since $L_{D_1},\dots ,L_{D_m}$ are $C^{\ast }$-mixing on $B_0\left(H\right)$, for each $j\in \{1,\dots ,m\}$ we can find some

$$z_j\in V_{j,1}\cap S\left(B_0\left(H\right)\right),w_j\in V_{j,2}\cap S\left(B_0\left(H\right)\right),$$

and some $n\ge N$ such that $w_j^{\ast }{D}_j^n{z}_j\in {\tilde{V}}_j,$ which gives for each $j\in \{1,\dots ,m\}$

$$∥w_j^{\ast }{D}_j^n{z}_j-{\displaystyle \frac{F}{m\parallel x_j\parallel \parallel y_j\parallel }}∥<{\displaystyle \frac{\epsilon }{2m}}.$$

As in the proof of Proposition 5, since $\parallel x_j\parallel \le \parallel x\parallel =1,$ we have

$$∥\parallel x_j\parallel z_j-x_j∥<{\displaystyle \frac{\epsilon }{m}},$$

and, similarly,

$$∥\parallel y_j\parallel w_j-y_j∥<{\displaystyle \frac{\epsilon }{m}}\mathrm{for}\mathrm{each}j\in \{1,\dots ,m\}.$$

Set

$$\overline{x}=\left(\parallel x_1\parallel z_1,\dots ,\parallel x_m\parallel z_m,x_{m+1},x_{m+2},\dots \right),$$

$$\overline{y}=\left(\parallel y_1\parallel w_1,\dots ,\parallel y_m\parallel w_m,y_{m+1},y_{m+2},\dots \right),$$

and put

$$\tilde{x}={\displaystyle \frac{\overline{x}}{\parallel \overline{x}\parallel }},\tilde{y}={\displaystyle \frac{\overline{y}}{\parallel \overline{y}\parallel }}.$$

Then, as in the proof of Proposition 5, we get

$$\parallel x-\tilde{x}\parallel ,\parallel y-\tilde{y}\parallel <2\epsilon .$$

Moreover,

$$⟨\tilde{y},T^n\left(\tilde{x}\right)⟩={\displaystyle \sum _{j=1}^m}{\displaystyle \frac{\parallel y_j\parallel w_j^{\ast }}{\parallel \overline{y}\parallel }}{D}_j^n{\displaystyle \frac{\parallel x_j\parallel z_j}{\parallel \overline{x}\parallel }}+{\displaystyle \sum _{j=m+1}^{\infty }}{y}_j^{\ast }{D}_j^n{x}_j,$$

so

$$⟨\tilde{y},T^n\left(\tilde{x}\right)⟩-F={\displaystyle \sum _{j=1}^m}\parallel y_j\parallel \parallel x_j\parallel \left({\displaystyle \frac{w_j^{\ast }}{\parallel \overline{y}\parallel }}{D}_j^n{\displaystyle \frac{z_j}{\parallel \overline{x}\parallel }}-{\displaystyle \frac{F}{m\parallel x_j\parallel \parallel y_j\parallel }}\right)+{\displaystyle \sum _{j=m+1}^{\infty }}{y}_j^{\ast }{D}_j^n{x}_j.$$

Hence, we obtain

$$∥⟨\tilde{y},T^n\left(\tilde{x}\right)⟩-F∥\le {\displaystyle \sum _{j=1}^m}\parallel y_j\parallel \parallel x_j\parallel ∥{\displaystyle \frac{w_j^{\ast }}{\parallel \overline{y}\parallel }}{D}_j^n{\displaystyle \frac{z_j}{\parallel \overline{x}\parallel }}-{\displaystyle \frac{F}{m\parallel x_j\parallel \parallel y_j\parallel }}∥+∥{\displaystyle \sum _{j=m+1}^{\infty }}{y}_j^{\ast }{D}_j^n{x}_j∥$$

$$\le {\displaystyle \sum _{j=1}^m}\parallel y_j\parallel \parallel x_j\parallel ∥{\displaystyle \frac{w_j^{\ast }}{\parallel \overline{y}\parallel }}{D}_j^n{\displaystyle \frac{z_j}{\parallel \overline{x}\parallel }}-{\displaystyle \frac{F}{m\parallel x_j\parallel \parallel y_j\parallel }}∥+{∥{\displaystyle \sum _{j=m+1}^{\infty }}{y}_j^{\ast }{y}_j∥}^{1/2}{∥{\displaystyle \sum _{j=m+1}^{\infty }}{x}_j^{\ast }{D}_j^{\ast n}{D}_j^n{x}_j∥}^{1/2}$$

$$\le {\displaystyle \sum _{j=1}^m}\parallel y_j\parallel \parallel x_j\parallel ∥{\displaystyle \frac{w_j^{\ast }}{\parallel \overline{y}\parallel }}{D}_j^n{\displaystyle \frac{z_j}{\parallel \overline{x}\parallel }}-{\displaystyle \frac{F}{m\parallel x_j\parallel \parallel y_j\parallel }}∥+{∥{\displaystyle \sum _{j=m+1}^{\infty }}{x}_j^{\ast }{\parallel D_j^n\parallel }^2{x}_j∥}^{1/2},$$

where the second inequality follows from the positive monotonicity of the norm in $B_0\left(H\right)$ and the fact that ${∥{\sum }_{j=1}^{\infty }{y}_j^{\ast }{y}_j∥}^{1/2}=1.$ We also notice that for the first inequality we have applied the Cauchy–Schwarz inequality [Proposition 1.2.4, part (iii)] in [35] on the sequences ${\left\{y_j\right\}}_{j=m+1}^{\infty }\mathrm{and}{\left\{D_j^n{x}_j\right\}}_{j=m+1}^{\infty }$ to deduce

$$∥{\displaystyle \sum _{j=m+1}^{\infty }}{y}_j^{\ast }{D}_j^n{x}_j∥\le {∥{\displaystyle \sum _{j=m+1}^{\infty }}{y}_j^{\ast }{y}_j∥}^{1/2}{∥{\displaystyle \sum _{j=m+1}^{\infty }}{x}_j^{\ast }{D}_j^{\ast n}{D}_j^n{x}_j∥}^{1/2}.$$

Furthermore, we have

$${∥{\displaystyle \sum _{j=m+1}^{\infty }}{x}_j^{\ast }{\parallel D_j^n\parallel }^2{x}_j∥}^{1/2}\le c^n{∥{\displaystyle \sum _{j=m+1}^{\infty }}{x}_j^{\ast }{x}_j∥}^{1/2}<{\displaystyle \frac{\delta }{2}},$$

because

$$\parallel D_j^n\parallel \le \parallel D_j{\parallel }^n\le c^n\mathrm{for}\mathrm{all}j\ge m+1,$$

and

$$c^n{∥{\displaystyle \sum _{j=m+1}^{\infty }}{x}_j^{\ast }{x}_j∥}^{1/2}\le c^N{∥{\displaystyle \sum _{j=1}^{\infty }}{x}_j^{\ast }{x}_j∥}^{1/2}=c^N<{\displaystyle \frac{\delta }{2}},$$

so

$$∥⟨\tilde{y},T^n\left(\tilde{x}\right)⟩-F∥<{\displaystyle \sum _{j=1}^m}\parallel y_j\parallel \parallel x_j\parallel ∥{\displaystyle \frac{w_j^{\ast }}{\parallel \overline{y}\parallel }}{D}_j^n{\displaystyle \frac{z_j}{\parallel \overline{x}\parallel }}-{\displaystyle \frac{F}{m\parallel x_j\parallel \parallel y_j\parallel }}∥+{\displaystyle \frac{\delta }{2}}$$

Next, we notice that by the construction of $\overline{x}$ and $\overline{y}$ we have

$$\parallel x-\overline{x}\parallel ,\parallel y-\overline{y}\parallel <\epsilon .$$

Hence, by the same arguments as in the proof of Proposition 5 and by the choice of $\epsilon$, we obtain

$$\left|{\displaystyle \frac{1}{\parallel \overline{x}\parallel \parallel \overline{y}\parallel }}-1\right|<{\displaystyle \frac{\delta }{4\left({\displaystyle \frac{\parallel F\parallel }{m\parallel x_j\parallel \parallel y_j\parallel }}+1\right)m}}\mathrm{for}\mathrm{all}j\in \{1,\dots ,m\}.$$

Therefore, since for each $j\in \{1,\dots ,m\},$ we have

$$∥w_j^{\ast }{D}_j^n{z}_j-{\displaystyle \frac{F}{m\parallel x_j\parallel \parallel y_j\parallel }}∥<{\displaystyle \frac{\epsilon }{2m}}<\epsilon <{\displaystyle \frac{\delta }{4}}<1,$$

we obtain

$$\left|{\displaystyle \frac{1}{\parallel \overline{x}\parallel \parallel \overline{y}\parallel }}-1\right|∥w_j^{\ast }{D}_j^n{z}_j∥\le {\displaystyle \frac{\delta }{4m}}.$$

By combining these two inequalities together with the triangle inequality, we deduce

$$\begin{array}{c}\parallel x_j\parallel \parallel y_j\parallel ∥{\displaystyle \frac{w_j^{\ast }}{\parallel \overline{y}\parallel }}{D}_j^n{\displaystyle \frac{z_j}{\parallel \overline{x}\parallel }}-{\displaystyle \frac{F}{m\parallel x_j\parallel \parallel y_j\parallel }}∥\le ∥{\displaystyle \frac{w_j^{\ast }}{\parallel \overline{y}\parallel }}{D}_j^n{\displaystyle \frac{z_j}{\parallel \overline{x}\parallel }}-{\displaystyle \frac{F}{m\parallel x_j\parallel \parallel y_j\parallel }}∥\hfill \\ \hspace{1em}\hspace{1em}<{\displaystyle \frac{\delta }{4m}}+{\displaystyle \frac{\epsilon }{2m}}<{\displaystyle \frac{3\delta }{8m}}\mathrm{for}\mathrm{each}j\in \{1,\dots ,m\}.\hfill \end{array}$$

Hence, we conclude that

$$∥⟨\tilde{y},T^n\left(\tilde{x}\right)⟩-F∥\le {\displaystyle \frac{3\delta }{8}}+{\displaystyle \frac{\delta }{2}}<\delta .$$

□

In view of Remark 1, which states that a compact diagonal operator on ${\mathsf{\ell }}^2$ can never be topologically numerically transitive, Corollary 5 illustrates in this way the difference between topological numerical transitivity on ${\mathsf{\ell }}^2$ and $C^{\ast }$-transitivity on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right).$

Remark 3. 

We notice that by the proof of Proposition 4 it follows that $L_U$ is actually $C^{\ast }$-mixing and not just $C^{\ast }$-transitive. In general, the operator $L_{\lambda \mathcal{S}}$ is $C^{\ast }$-mixing on $B_0\left(H\right)$ for every scalar λ with $\left|\lambda \right|>1.$

A natural question that arises now is whether there exist generalized numerically transitive operators on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ at all. A positive answer to this question is provided in the next proposition.

Proposition 6. 

There exist generalized numerically transitive nonsupercyclic operators on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right).$

Proof. 

We notice first that we may index the sequences in ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ over $\mathbb{Z}$ instead of over $\mathbb{N}$.

Let $T\in B\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right)$ be given by

$$T\left({\left(x_j\right)}_{j\in \mathbb{Z}}\right)={\left(2x_{j-1}\right)}_{j\in \mathbb{Z}};$$

that is, the j-th coordinate of $T\left({\left(x_j\right)}_{j\in \mathbb{Z}}\right)$ is $2x_{j-1}$ for each $j\in \mathbb{Z}$.

In a similar way to that in the proof of Proposition 4, it can be shown that T is generalized numerically transitive on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$. It suffices to recall from [Section 4] in [32] that

$${\left\{{\tilde{p}}_{(\gamma ,m)}\right\}}_{(\gamma ,m)\in {\mathbb{N}}^2}$$

is both a right and a left approximate unit for ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$, where

$${\left({\tilde{p}}_{(\gamma ,m)}\right)}_j=\left\{\begin{array}{cc}{P}_m,\hfill & -\gamma \le j\le \gamma ,\hfill \\ 0,\hfill & \mathrm{otherwise},\hfill \end{array}\right.j\in \mathbb{Z},$$

and

$$\parallel {\tilde{p}}_{(\gamma ,m)}x\parallel \le \parallel x\parallel \mathrm{for}\mathrm{all}x\in {\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right),\gamma ,m\in \mathbb{N}.$$

In fact, this can also be checked by a straightforward computation.

Furthermore, by an easy application of Lemma 5, we can deduce that for every $\epsilon >0$, $\gamma ,m\in \mathbb{N}$, and $x\in {\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$, there exist

$${\tilde{x}}_{\epsilon },{\tilde{x}}_{\epsilon }^{(-1)}\in {\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$$

such that

$$\parallel {\tilde{p}}_{(\gamma ,m)}x{\tilde{p}}_{(\gamma ,m)}-{\tilde{x}}_{\epsilon }\parallel <\epsilon$$

and ${\tilde{x}}_{\epsilon }{\tilde{x}}_{\epsilon }^{(-1)}={\tilde{p}}_{(\gamma ,m)}.$ Indeed, it suffices to find, for each $x_j$ with $-\gamma \le j\le \gamma$, operators

$${\tilde{x}}_{j,\epsilon },{\tilde{x}}_{j,\epsilon }^{(-1)}\in P_m{B}_0\left(H\right)P_m$$

such that

$$\parallel P_m{x}_j{P}_m-{\tilde{x}}_{j,\epsilon }\parallel <{\displaystyle \frac{\epsilon }{2\gamma +1}}$$

and ${\tilde{x}}_{j,\epsilon }{\tilde{x}}_{j,\epsilon }^{(-1)}=P_m.$ Then let ${\tilde{x}}_{\epsilon },{\tilde{x}}_{\epsilon }^{(-1)}\in {\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ be given by

$${\left({\tilde{x}}_{\epsilon }\right)}_j=\left\{\begin{array}{cc}{\tilde{x}}_{j,\epsilon },\hfill & -\gamma \le j\le \gamma ,\hfill \\ 0,\hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

and

$${\left({\tilde{x}}_{\epsilon }^{(-1)}\right)}_j=\left\{\begin{array}{cc}{\tilde{x}}_{j,\epsilon }^{(-1)},\hfill & -\gamma \le j\le \gamma ,\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Then also

$${\tilde{x}}_{\epsilon }{\tilde{x}}_{\epsilon }^{(-1)}={\tilde{p}}_{(\gamma ,m)}.$$

Finally, by the construction of the operator T, we have

$${\tilde{p}}_{(\gamma ,m)}{T}^n\left({\tilde{p}}_{(\gamma ,m)}\right)=0\mathrm{for}\mathrm{all}n\ge 2\gamma +2,$$

and, moreover,

$$\underset{n\to \infty }{\lim }\parallel T^{-n}\left({\tilde{p}}_{(\gamma ,m)}w\right)\parallel =0\mathrm{for}\mathrm{all}w\in {\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right),\gamma ,m\in \mathbb{N}.$$

Given nonempty open subsets $O_1,O_2,O_3$ of ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ and some

$$x\in O_1\setminus \left\{0\right\},y\in O_2\setminus \left\{0\right\},z\in O_3\setminus \left\{0\right\},$$

for sufficiently large $\gamma ,m\in \mathbb{N}$ and sufficiently small $\epsilon >0$, set, for each $n\in \mathbb{N}$,

$$Y_n:={\tilde{p}}_{(\gamma ,m)}y+T^{-n}\left({\tilde{p}}_{(\gamma ,m)}{\tilde{x}}_{\epsilon }^{(-1)}{\tilde{p}}_{(\gamma ,m)}z\right).$$

By similar computations as in the proof of Proposition 4, we can show that $Y_n⟶y$ and

$${\tilde{x}}_{\epsilon }{T}^n\left(Y_n\right)⟶z(n\to \infty ),$$

and that ${\tilde{x}}_{\epsilon }\in O_1$ for sufficiently small $\epsilon >0$. Thus, T is generalized numerically transitive on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$.

On the other hand, if T were supercyclic, then it is not hard to deduce from [Section 4] in [32] that, for every $\gamma ,m\in \mathbb{N}$, we would be able to find, for each j with $-\gamma \le j\le \gamma$, sequences

$${\left\{D_{k,j}\right\}}_{k\in \mathbb{N}},{\left\{G_{k,j}\right\}}_{k\in \mathbb{N}}\subseteq B_0\left(H\right),$$

and a strictly increasing sequence ${\left\{n_k\right\}}_{k\in \mathbb{N}}\subseteq \mathbb{N}$ such that

$$\underset{k\to \infty }{\lim }\parallel D_{k,j}-P_m\parallel =\underset{k\to \infty }{\lim }\parallel G_{k,j}-P_m\parallel =0$$

and

$$\underset{k\to \infty }{\lim }\left(2^{n_k}\parallel D_{k,j}\parallel \right)\left(2^{-n_k}\parallel G_{k,j}\parallel \right)=0.$$

Equivalently,

$$\underset{k\to \infty }{\lim }\parallel D_{k,j}\parallel \parallel G_{k,j}\parallel =0.$$

This is obviously impossible, since $D_{k,j}\to P_m$ and $G_{k,j}\to P_m$ while $\parallel P_m\parallel =1$. Hence, T is not supercyclic on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$. □

The proof of Proposition 5 gives us also a tool for studying the relations between topological numerical transitivity, $C^{\ast }$-transitivity and other dynamical concepts, which will be done in the next results. Although the concept of topological $C^{\ast }$-transitivity is in a certain sense motivated by topological numerical transitivity, these two concepts are essentially different, as we show in the next proposition.

Proposition 7. 

There exists a topologically numerically transitive operator on $B_2\left(H\right)$ which is not topologically $C^{\ast }$-transitive. In particular, there exists a numerically hypercyclic operator on $B_2\left(H\right)$ which is not topologically $C^{\ast }$-transitive.

Proof. 

Write $H=H_0\oplus H_1$, where both $H_0$ and $H_1$ are infinite-dimensional separable Hilbert spaces. Let ${\left\{e_j\right\}}_{j\in \mathbb{N}}$ be an orthonormal basis for $H_0$. Then

$$⟨h_1,e_j⟩=0\mathrm{for}\mathrm{every}{h}_1\in H_1\mathrm{and}\mathrm{every}j\in \mathbb{N}.$$

Recall that $\{e_i\otimes e_j^{\ast }:i,j\in \mathbb{N}\}$ is an orthonormal system in $B_2\left(H\right)$, and let

$$V:=\overline{\mathrm{span}}\{e_i\otimes e_j^{\ast }:i,j\in \mathbb{N}\}.$$

Since $H_1$ is infinite-dimensional and $⟨h_1,e_j⟩=0$ for every $h_1\in H_1$ and every $j\in \mathbb{N}$, it is easily seen that $V^{\perp }$ is also an infinite-dimensional closed subspace of $B_2\left(H\right)$.

Let D be the topologically numerically transitive operator from Example 4 on ${\mathsf{\ell }}^2\left(\mathbb{Z}\right)\cong V$, and let P be the orthogonal projection of $B_2\left(H\right)$ onto V. By Proposition 2, the operator $D\circ P$ is topologically numerically transitive on $B_2\left(H\right)$.

Choose $x\in H_0^{\perp }$ with ${\parallel x\parallel }_H=1$, and let $\tilde{R}$ be the orthogonal projection onto $\mathrm{span}\left\{x\right\}$. Then $\tilde{R}$ is a finite-rank operator, hence it belongs to $B_1\left(H\right)$. Let

$$\tilde{U}=\left\{T\in B_1\left(H\right):{\parallel T-\tilde{R}\parallel }_1<{\displaystyle \frac{1}{2}}\right\}.$$

Then $\tilde{U}$ is a nonempty open subset of $B_1\left(H\right)$. For every $T\in \tilde{U}$, we have

$$1-{\parallel T\left(x\right)\parallel }_H\le \parallel \tilde{R}{\left(x\right)-T\left(x\right)\parallel }_H\le \parallel \tilde{R}-T\parallel \le \parallel \tilde{R}{-T\parallel }_1<{\displaystyle \frac{1}{2}}.$$

Consequently, ${\parallel T\left(x\right)\parallel }_H>{\displaystyle \frac{1}{2}}\mathrm{for}\mathrm{every}T\in \tilde{U}.$ On the other hand, since $x\in H_0^{\perp }$, for each $i,j\in \mathbb{N}$ we have

$$(e_i\otimes e_j^{\ast })\left(x\right)=⟨x,e_j⟩e_i=0.$$

Hence, every operator in V vanishes at x. Moreover, the construction of $D\circ P$ gives $\mathrm{Ran}{(D\circ P)}^n\subseteq V$ for every $n\in \mathbb{N}$. Therefore, for each $n\in \mathbb{N}$ and every $F,E\in B_2\left(H\right)$, we have

$${\tilde{\Theta }}_n(F,E)\left(x\right)=E^{\ast }\left({(D\circ P)}^n\left(F\right)\left(x\right)\right)=0.$$

Thus, ${\tilde{\Theta }}_n(F,E)\notin \tilde{U}$, because every element of $\tilde{U}$ sends x to a vector of norm greater than $1/2$. Consequently,

$${\tilde{\Theta }}_n\left(B_2\left(H\right)\times B_2\left(H\right)\right)\cap \tilde{U}=⌀$$

for every $n\in \mathbb{N},$ or in other words, ${\tilde{\Theta }}_n$ cannot hit $\tilde{U}$ for any $n\in \mathbb{N}.$ Hence, $D\circ P$ is not $C^{\ast }$-transitive on $B_2\left(H\right)$. This completes the proof. □

Recall that an operator $T\in \mathcal{B}\left(X\right),$ where X is a Banach space, is called weakly n-supercyclic if there exists an n-dimensional subspace M of X such that

$${\displaystyle \bigcup _{k=0}^{\infty }}{T}^k\left(M\right)$$

is weakly dense in X. Moreover, we recall from [36] that T is said to be N-weakly supercyclic if there exists some $x\in X$ such that, for every continuous surjective linear map $\gamma :X⟶{\mathbb{C}}^N,$ it holds that $\gamma \left({\mathrm{Orb}}_p(T,x)\right)$ is dense in ${\mathbb{C}}^N$.

Corollary 6. 

There exist $C^{\ast }$-transitive operators on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ and topologically numerically transitive operators on ${\mathsf{\ell }}^2\left(\mathbb{N}\right)$ that are neither weakly n-supercyclic nor N-weakly supercyclic, for any $n,N\in \mathbb{N}$.

Proof. 

Let $\tilde{D}$ be the operator on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ appearing in the proof of Proposition 5.

Let M be any finite-dimensional subspace of ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$. Then

$${\displaystyle \bigcup _{k=0}^{\infty }}{\tilde{D}}^k\left(M\right)\subseteq \tilde{P}\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right)\oplus (I-\tilde{P})\left(M\right),$$

where $\tilde{P}$ denotes the projection onto the first coordinate of a sequence in ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right).$ Now, $(I-\tilde{P})\left(M\right)$ is a finite-dimensional, hence closed subspace of $(I-\tilde{P})\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right).$ Therefore, there exists a complemented subspace $\tilde{M}$ such that

$$(I-\tilde{P})\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right)=(I-\tilde{P})\left(M\right)\oplus \tilde{M}.$$

Clearly, $\tilde{M}$ must be infinite-dimensional since $(I-\tilde{P})\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right)$ is infinite-dimensional. Hence, we can find some nonzero $x\in \tilde{M}$ and some $\phi \in {\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right)}^{\ast }$ such that $\phi \left(x\right)\ne 0.$ This follows from the Hahn–Banach theorem. Let $\tilde{Q}$ be the projection onto $\tilde{M}$ along $(I-\tilde{P})\left(M\right)$. Then

$$\phi \circ \tilde{Q}\circ (I-\tilde{P})\in {\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right)}^{\ast }$$

and $\left(\phi \circ \tilde{Q}\circ (I-\tilde{P})\right)\left(x\right)=\phi \left(x\right)\ne 0.$ On the other hand, $\left(\phi \circ \tilde{Q}\circ (I-\tilde{P})\right)\left(y\right)=0$ for every

$$y\in \tilde{P}\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right)\oplus (I-\tilde{P})\left(M\right).$$

Thus, in particular, $\left(\phi \circ \tilde{Q}\circ (I-\tilde{P})\right)\left(y\right)=0$ for every $y\in {\bigcup }_{k=1}^{\infty }{\tilde{D}}^k\left(M\right).$ It is therefore clear that ${\bigcup }_{k=1}^{\infty }{\tilde{D}}^k\left(M\right)$ cannot be weakly dense in ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$. Consequently, $\tilde{D}$ is not weakly n-supercyclic for any $n\in \mathbb{N}$.

We now prove that $\tilde{D}$ cannot be N-weakly supercyclic. Suppose, to the contrary, that there exists an N-weakly supercyclic vector $x_0$ for $\tilde{D}$, and let $M=\mathrm{span}\left\{x_0\right\}.$ Then, again,

$$\mathrm{Orb}(\tilde{D},x_0)\subseteq \tilde{P}\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right)\oplus (I-\tilde{P})\left(M\right).$$

Let $\tilde{M}$ be the same complemented Banach subspace of $(I-\tilde{P})\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right)$ as above, and choose linearly independent vectors $y_1,\dots ,y_N\in \tilde{M}.$ Let

$$\iota :\mathrm{span}\{y_1,\dots ,y_N\}⟶{\mathbb{C}}^N$$

be an isomorphism. Since $\mathrm{span}\{y_1,\dots ,y_N\}$ is finite-dimensional, it is complemented in $\tilde{M}$. Hence, there exists a bounded linear projection

$$\beta :\tilde{M}⟶\mathrm{span}\{y_1,\dots ,y_N\}.$$

Consider the map $\gamma :{\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)⟶{\mathbb{C}}^N$ defined by $\gamma =\iota \circ \beta \circ \tilde{Q}\circ (I-\tilde{P}).$ Then $\gamma$ is a continuous surjective linear map. However,

$$\gamma \left(\mathrm{Orb}(\tilde{D},x_0)\right)=\left\{0\right\}.$$

This contradicts the assumption that $x_0$ is an N-weakly supercyclic vector. Hence, $\tilde{D}$ cannot be N-weakly supercyclic on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$.

The proof is analogous for the operator $D\circ P$ appearing in the proof of Proposition 7. As established there, $D\circ P$ is a topologically numerically transitive operator on $B_2\left(H\right)$. By letting $D\circ P$, P, and $B_2\left(H\right)$ play, respectively, the roles of $\tilde{D}$, $\tilde{P}$, and ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ in the arguments above, we deduce that $D\circ P$ cannot be weakly n-supercyclic or N-weakly supercyclic on $B_2\left(H\right)$ for any $n,N\in \mathbb{N}$. Since $B_2\left(H\right)$ is isometrically isomorphic to ${\mathsf{\ell }}_2\left(\mathbb{N}\right)$, the second statement follows. □

Recall that for $\epsilon \in (0,1)$, an operator $T\in \mathcal{B}\left(X\right)$ is said to be $\epsilon$-cyclic if there exists some $x\in X$ such that for every $y\in X$ we can find some $z\in \overline{\mathrm{span}\left({\bigcup }_{k=1}^{\infty }{\left(T\right)}^k\left(M\right)\right)}$ with

$$\parallel z-y\parallel \le \epsilon \parallel y\parallel ,$$

where $M:=\mathrm{span}\left\{x\right\}.$

Corollary 7. 

There exist topologically numerically transitive operators on ${\mathsf{\ell }}^2\left(\mathbb{N}\right)$ and $C^{\ast }$-transitive operators on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ that are not ε-cyclic for any $\epsilon \in (0,1)$.

Proof. 

Suppose that $x\in B_2\left(H\right)$ is an $\epsilon$-cyclic vector for $D\circ P$, with $\epsilon \in (0,1),$ where D and P are as in the proof of Proposition 7. Then, again,

$$\overline{\mathrm{span}\left(\bigcup _{k=0}^{\infty }{(D\circ P)}^k\left(M\right)\right)}\subseteq V\oplus (I-P)\left(\mathrm{span}\left\{x\right\}\right).$$

Let $h\in H$ with ${\parallel h\parallel }_H=1$ and choose some $g\in {(H_0+\mathrm{span}\left\{x\left(h\right)\right\})}^{\perp }$ such that ${\parallel g\parallel }_H=1$, and let $y=g\otimes h^{\ast },$ so $y\in B_2\left(H\right)$. Then, clearly, for every $z\in \overline{\mathrm{span}\left({\bigcup }_{k=0}^{\infty }{(D\circ P)}^k\left(M\right)\right)}$, we have

$$1={\parallel g\parallel }_H\le {∥\left(y-z\right)h∥}_H,$$

because ${(D\circ P)}^n\left(x\right)\left(h\right)\in H_0+\mathrm{span}\left\{x\left(h\right)\right\}\mathrm{for}\mathrm{all}n\in {\mathbb{N}}_0.$ Consequently, $1=\parallel y\parallel \le \parallel y-z\parallel$ contradicting the fact that x is an $\epsilon$-cyclic vector for $D\circ P$.

The approach is similar in the case of the operator $\tilde{D}$ on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ from the proof of Proposition 5. In that case, we have

$$\overline{\mathrm{span}\left(\bigcup _{k=0}^{\infty }{\left(\tilde{D}\right)}^k\left(\tilde{M}\right)\right)}\subseteq \tilde{P}\left({\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)\right)\oplus (I-\tilde{P})\left(\mathrm{span}\left\{\tilde{x}\right\}\right),$$

where we let now $\tilde{x}$ be an $\epsilon$-cyclic vector for $\tilde{D}$, $\tilde{M}:=\mathrm{span}\left\{\tilde{x}\right\}$ and $\tilde{P}$ is again the projection onto the first coordinate. Write $\tilde{x}=({\tilde{x}}_1,{\tilde{x}}_2,{\tilde{x}}_3,\dots )$ and choose some $\tilde{g},\tilde{h}\in H$ with

$$\parallel \tilde{h}{\parallel }_H={\parallel \tilde{g}\parallel }_H=1,\tilde{g}\perp {\tilde{x}}_2\left(\tilde{h}\right).$$

Let $\tilde{y}=(0,\tilde{g}\otimes {\tilde{h}}^{\ast },0,0,\dots ).$ Then $\parallel \tilde{y}\parallel =1$, and it can be checked that for every $n\in \mathbb{N}$,

$$\parallel {\tilde{D}}^n\left(\tilde{x}\right)-\tilde{y}\parallel \ge {∥{\left({\tilde{D}}^n\left(\tilde{x}\right)-\tilde{y}\right)}_2\left(\tilde{h}\right)∥}_H\ge 1.$$

Hence, $\tilde{x}$ cannot be an $\epsilon$-cyclic vector for $\tilde{D}$, a contradiction. Thus, $\tilde{D}$ cannot be $\epsilon$-cyclic. □

We recall that an operator $T\in \mathcal{B}\left(X\right)$ is Li–Yorke chaotic if and only if there exists some $x\in X$ and a strictly increasing sequence $\left\{n_k\right\}\subseteq \mathbb{N}$ such that

$$\underset{k\to \infty }{\lim }\parallel T^{n_k}\left(x\right)\parallel =0\mathrm{and}\sup \{\parallel T^n\left(x\right)\parallel :n\in \mathbb{N}\}=\infty ,$$

whereas a necessary condition for T to be distributionally chaotic of type 3 is that there exists some $y\in X$ and some $\delta >0$ such that both sets $A:=\{n\in \mathbb{N}:\parallel T^{n}y\parallel <\delta \}$ and $B:=\{n\in \mathbb{N}:\parallel T^{n}y\parallel >\delta \}$ are infinite; see the proof of part b) in [Proposition 42] in [37].

Corollary 8. 

There exist topologically numerically transitive operators on ${\mathsf{\ell }}_2\left(\mathbb{N}\right)$, and $C^{\ast }$-transitive operators on $B_0\left(H\right)$, $B_2\left(H\right)$ and ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ that are neither Li–Yorke chaotic nor distributionally chaotic of type 3.

Proof. 

Note that since $U=2\mathcal{S}$, where $\mathcal{S}$ is the forward bilateral shift, for each $F\in B_0\left(H\right)$ we have $\parallel U^{n}F\parallel =2^n\parallel F\parallel ,$ and similarly for each $G\in B_2\left(H\right)$, it holds that $\parallel U^n{G\parallel }_2=2^n{\parallel G\parallel }_2,$ for all $n\in \mathbb{N}.$ Consequently, the conditions for Li–Yorke chaos and for distributional chaos of type 3 given above cannot be satisfied for the operator $L_U$.

Furthermore, if we consider the operator $\tilde{D}$ on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ from the proof of Proposition 5, then for every $x=(x_1,x_2,\dots )\in {\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ and each $n\in \mathbb{N}$, we have

$$∥{\tilde{D}}^n\left((x_1,x_2,\dots )\right)∥=2^n\parallel x_1\parallel .$$

Therefore, it is again easy to deduce that $\tilde{D}$ can neither be Li–Yorke chaotic nor distributionally chaotic of type 3.

Finally, if we consider the operator $D\circ P$ on $B_2\left(H\right)$ from the proof of Proposition 7, then for every $F\in B_2\left(H\right)$ and each $n\in \mathbb{N}$ it holds that

$${\parallel (D\circ P)}^n{F\parallel }_2=R^n{\parallel P\left(F\right)\parallel }_2.$$

Since $R>1$, it is again clear that $D\circ P$ can neither be Li–Yorke chaotic nor distributionally chaotic of type 3. □

Corollary 9. 

There exist strongly numerically hypercyclic operators on $B_2\left(H\right)$, $B_0\left(H\right)$ and ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ that are not $C^{\ast }$-transitive. Furthermore, there exist strongly numerically hypercyclic operators on ${\mathsf{\ell }}^2\left(\mathbb{N}\right)$ that are not topologically numerically transitive.

Proof. 

Consider again the orthonormal basis $\{e_i\otimes e_j^{\ast }:i,j\in \mathbb{N}\}$ for the subspace V of $B_2\left(H\right)$ from the proof of Proposition 7. Let ${\psi }_1,{\psi }_2,{\psi }_3$ be any three distinct elements of $\{e_i\otimes e_j^{\ast }:i,j\in \mathbb{N}\}$ and $\tilde{V}:=\mathrm{span}\{{\psi }_1,{\psi }_2,{\psi }_3\}.$ Then $\tilde{V}$ is a subspace of $B_0\left(H\right).$ Since $\tilde{V}$ is finite dimensional, it is closed and complemented in $B_0\left(H\right).$ Let $\widehat{P}$ denote the projection onto $\tilde{V}$ along its complement in $B_0\left(H\right).$ Suppose that T is a diagonal operator with respect to the basis $\{{\psi }_1,{\psi }_2,{\psi }_3\},$ with eigenvalues $\{{\lambda }_1,{\lambda }_2,{\lambda }_3\}$ satisfying the conditions of Theorem 2 part (ii). If $\tilde{J}$ is an isomorphism of $B_0\left(H\right)$ onto some other Banach space $X,$ then it is easily seen that ${\left.\tilde{J}(T\circ \widehat{P}){\tilde{J}}^{-1}\right|}_{\tilde{J}\left(\tilde{V}\right)}$ is the diagonal operator with respect to the basis $\left\{\tilde{J}{\psi }_1,\tilde{J}{\psi }_2,\tilde{J}{\psi }_3\right\},$ with the same eigenvalues $\{{\lambda }_1,{\lambda }_2,{\lambda }_3\}$ as T. Hence, ${\left.\tilde{J}(T\circ \widehat{P}){\tilde{J}}^{-1}\right|}_{\tilde{J}\left(\tilde{V}\right)}$ is a strongly numerically hypercyclic operator on $\tilde{J}\left(\tilde{V}\right)$ by Theorem 2 part (ii). In particular, ${\left.\tilde{J}(T\circ \widehat{P}){\tilde{J}}^{-1}\right|}_{\tilde{J}\left(\tilde{V}\right)}$ is a numerically hypercyclic operator on $\tilde{J}\left(\tilde{V}\right)$. Hence, there exist some

$$x^{\ast }\in {\left(\tilde{J}\left(\tilde{V}\right)\right)}^{\ast }$$

and some $x\in \tilde{J}\left(\tilde{V}\right)$ with $\parallel x^{\ast }\parallel =\parallel x\parallel =1,x^{\ast }\left(x\right)=1,$ such that

$$\left\{x^{\ast }\left({\left({\left.\tilde{J}(T\circ \widehat{P}){\tilde{J}}^{-1}\right|}_{\tilde{J}\left(\tilde{V}\right)}\right)}^{n}x\right):n\in \mathbb{N}\right\}$$

is dense in $\mathbb{C}$. Note that for each $n\in \mathbb{N}$, it holds that

$${\left(\tilde{J}(T\circ \widehat{P}){\tilde{J}}^{-1}\right)}^n=\tilde{J}{(T\circ \widehat{P})}^n{\tilde{J}}^{-1}=\tilde{J}(T^n\circ \widehat{P}){\tilde{J}}^{-1},$$

since $\widehat{P}$ is the projection onto $\tilde{V}$ and $T\left(\tilde{V}\right)=\tilde{V}.$ Hence,

$$\left\{x^{\ast }\left(\tilde{J}(T^n\circ \widehat{P}){\tilde{J}}^{-1}x\right):n\in \mathbb{N}\right\}$$

is dense in $\mathbb{C}$. By the Hahn–Banach extension theorem, there exists some ${\tilde{x}}^{\ast }\in X^{\ast }$ such that ${\left.{\tilde{x}}^{\ast }\right|}_{\tilde{J}\left(\tilde{V}\right)}=x^{\ast }$ and $\parallel {\tilde{x}}^{\ast }\parallel =\parallel x^{\ast }\parallel =1.$ Hence, ${\tilde{x}}^{\ast }\left(x\right)=x^{\ast }\left(x\right)=1$ and

$$\left\{{\tilde{x}}^{\ast }\left(\tilde{J}(T^n\circ \widehat{P}){\tilde{J}}^{-1}x\right):n\in \mathbb{N}\right\}$$

is dense in $\mathbb{C}$, since

$${\tilde{x}}^{\ast }\left(\tilde{J}(T^n\circ \widehat{P}){\tilde{J}}^{-1}x\right)=x^{\ast }\left(\tilde{J}(T^n\circ \widehat{P}){\tilde{J}}^{-1}x\right)$$

for each $n\in \mathbb{N}$, because

$$\tilde{J}(T^n\circ \widehat{P}){\tilde{J}}^{-1}x\in \tilde{J}\left(\tilde{V}\right)$$

for all $n\in \mathbb{N}$. Thus, $\tilde{J}(T\circ \widehat{P}){\tilde{J}}^{-1}$ is numerically hypercyclic on X. Since X and $\tilde{J}$ were chosen arbitrarily, we deduce that $T\circ \widehat{P}$ is strongly numerically hypercyclic on $B_0\left(H\right).$ The proof is analogous if, instead of $B_0\left(H\right)$, we consider $B_2\left(H\right).$ However, by analogous arguments as in the proof of Proposition 7, we can conclude that $T\circ \widehat{P}$ is not $C^{\ast }$-transitive on $B_2\left(H\right)$ or on $B_0\left(H\right)$.

Next, let $\widehat{T}$ be the operator on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ given by

$$\widehat{T}(x_1,x_2,\dots )=\left((T\circ \widehat{P})\left(x_1\right),0,0,\dots \right).$$

Then, again by the arguments above with the application of the Hahn–Banach extension theorem, we obtain that $\widehat{T}$ is a strongly numerically hypercyclic operator on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$. On the other hand, since

$$⟨z,\widehat{T}y⟩=z_1^{\ast }\left((T\circ \widehat{P})\left(y_1\right)\right)\mathrm{for}\mathrm{all}z,y\in {\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right),$$

it follows that if $\widehat{T}$ is $C^{\ast }$-transitive on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$, then $T\circ \widehat{P}$ must be $C^{\ast }$-transitive on $B_0\left(H\right)$. Since this is not the case, as we argued above, we obtain that $\widehat{T}$ is not $C^{\ast }$-transitive on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$.

In order to prove the last statement of the corollary, fix $R>1$; let $u,w,z\in \mathbb{T}$ be independent, and define

$$D=\mathrm{diag}(Ru,Rw,Rz,0,0,0,\dots )\in \mathcal{B}\left({\mathsf{\ell }}^2\right).$$

From Example 1, we know that the operator $\mathrm{diag}(Ru,Rw,Rz)$ is a strongly numerically hypercyclic operator on ${\mathbb{C}}^3,$ hence again by similar arguments as above, it can be shown that the operator D is strongly numerically hypercyclic on ${\mathsf{\ell }}^2\left(\mathbb{N}\right)$. On the other hand, by Remark 1 it follows that D is not topologically numerically transitive, which proves the last statement of the corollary. □

Recall that an operator $T\in \mathcal{B}\left(X\right)$ is called subspace hypercyclic (respectively, subspace supercyclic) if there exists a closed subspace $M\subseteq X$, with $\dim M\ge 2,$ and some $x\in X$ such that $\mathrm{Orb}(T,x)\cap M$ (respectively, ${\mathrm{Orb}}_p(T,x)\cap M$) is dense in M.

Corollary 10. 

There exist topologically numerically transitive operators on ${\mathsf{\ell }}^2\left(\mathbb{Z}\right)$ and $C^{\ast }$-transitive operators on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ that are not subspace hypercyclic. Also, there exist subspace supercyclic operators on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ that are not $C^{\ast }$-transitive.

Proof. 

Let $\tilde{D}$ be the operator on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ from Proposition 5. Then $∥{\tilde{D}}^n(x_1,x_2,\dots )∥=2^n\parallel x_1\parallel$ for all $n\in \mathbb{N}$ and $(x_1,x_2,\dots )\in {\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right).$ Consequently, the norms of the elements of the orbit are either increasing or are all equal to zero. By ref. [38] (see the comments on page 511 regarding hyponormal operators in [38]), we deduce that $\tilde{D}$ cannot be subspace hypercyclic.

By analogous arguments, the operator D from Example 4 cannot be subspace hypercyclic on ${\mathsf{\ell }}^2\left(\mathbb{Z}\right).$

Choose now some $h\in H$ with ${\parallel h\parallel }_H=1,$ and let p be the orthogonal projection onto $\mathrm{span}\left\{h\right\}.$ It is easily checked that the map

$$T:{\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)⟶{\tilde{\mathsf{\ell }}}_2\left(p{B}_0\left(H\right)p\right)$$

given by $T(x_1,x_2,\dots )=(p{x}_{1}p,p{x}_{2}p,\dots )$ is a bounded surjective linear map, and $T{|}_{{\tilde{\mathsf{\ell }}}_2\left(p{B}_0\left(H\right)p\right)}$ is the identity on ${\tilde{\mathsf{\ell }}}_2\left(p{B}_0\left(H\right)p\right).$ Let now B denote any supercyclic operator on ${\mathsf{\ell }}^2\left(\mathbb{N}\right).$ Since $p{B}_0\left(H\right)p\cong \mathbb{C},$ it follows that ${\tilde{\mathsf{\ell }}}_2\left(p{B}_0\left(H\right)p\right)$ is isometrically isomorphic to ${\mathsf{\ell }}^2\left(\mathbb{N}\right)$. Let $\tilde{T}:=BT.$ Then ${\tilde{T}}^n=B^{n}T\mathrm{for}\mathrm{all}n\in \mathbb{N}.$ Since B is supercyclic on ${\mathsf{\ell }}^2\left(\mathbb{N}\right)$, we deduce that $\tilde{T}$ is subspace supercyclic with respect to ${\tilde{\mathsf{\ell }}}_2\left(p{B}_0\left(H\right)p\right).$ However, it is straightforward to see that $⟨\tilde{y},{\tilde{T}}^n\left(\tilde{x}\right)⟩\in B_0\left(H\right)p,$ hence $⟨\tilde{y},{\tilde{T}}^n\left(\tilde{x}\right)⟩\left({\left(\mathrm{span}\left\{h\right\}\right)}^{\perp }\right)=\left\{0\right\}$ for every $\tilde{x},\tilde{y}\in {\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right)$ and $n\in \mathbb{N}.$ By similar arguments as in the proof of Proposition 7, we can deduce that $\tilde{T}$ cannot be $C^{\ast }$-transitive on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right).$ □

Corollary 10 leads naturally to the following question, which we will leave open for further research.

Question 1. 

Does there exist a topologically numerically transitive or a $C^{\ast }$-transitive operator that is not subspace supercyclic?

Let now ${\tilde{\mathrm{\Delta }}}_d\left(X\right)$ be the subset of $S\left(X\right)\times S\left(X\right)$ consisting of all pairs $(x,x)$ with $x\in S\left(X\right).$ Then, obviously, ${\tilde{\mathrm{\Delta }}}_d\left(X\right)$ is closed in $S\left(X\right)\times S\left(X\right)$. We will say that T is $C^{\ast }$-numerically transitive on X if the sequence ${\left\{\left({\tilde{\Theta }}_n\right){|}_{{\tilde{\mathrm{\Delta }}}_d\left(X\right)}\right\}}_{n\in \mathbb{N}}$ is topologically transitive.

Proposition 8. 

The operator $L_U$ is $C^{\ast }$-numerically transitive on $B_0\left(H\right)$.

Proof. 

Let $O_1,O_2,O_3$ be nonempty open subsets of $B_0\left(H\right)$ such that

$$(O_1\times O_2)\cap {\tilde{\mathrm{\Delta }}}_d\left(B_0\left(H\right)\right)\ne ⌀.$$

Then

$$(O_1\cap O_2)\cap S\left(B_0\left(H\right)\right)\ne ⌀,$$

since ${\tilde{\mathrm{\Delta }}}_d\left(B_0\left(H\right)\right)$ consists of all pairs $(T,T)$ with $T\in S\left(B_0\left(H\right)\right)$.

Choose some

$$F\in (O_1\cap O_2)\cap S\left(B_0\left(H\right)\right)\mathrm{and}G\in O_3.$$

Then there exists some $\epsilon >0$ such that the $\epsilon$-neighborhood of G is contained in $O_3$.

As in the proof of Proposition 4, we can find some $m_0\in \mathbb{N}$ such that, for every $m\ge m_0$,

$$\parallel G-P_{m}G\parallel <{\displaystyle \frac{\epsilon }{2}}.$$

Also, since the function

$$f:{\mathbb{R}}^{+}⟶{\mathbb{R}}^{+},f\left(t\right)={\displaystyle \frac{1}{t^2}},$$

is continuous at $t=1$, there exists some $\delta >0$ such that

$$|t-1|<\delta ⟹\left|{\displaystyle \frac{1}{t^2}}-1\right|<{\displaystyle \frac{\epsilon }{2(1+\parallel G\parallel )}}.$$

In fact, $\delta >0$ can be chosen sufficiently small so that the $\delta$-neighborhood of F is contained in $O_1\cap O_2$, and such that $\delta <1.$ Again, as in the proof of Proposition 4, we can find some $m\ge m_0$ such that

$$\parallel F-P_{m}F{P}_m\parallel <{\displaystyle \frac{\delta }{8}}.$$

By Lemma 5, there exists an invertible operator $\tilde{F}\in B\left(P_m\left(H\right)\right)$ such that

$$∥{\displaystyle \frac{P_{m}F{P}_m}{\parallel P_{m}F{P}_m\parallel }}-\tilde{F}∥<{\displaystyle \frac{\delta }{16}}.$$

Moreover, as in the proof of Proposition 4, we have

$$∥F-{\displaystyle \frac{P_{m}F{P}_m}{\parallel P_{m}F{P}_m\parallel }}∥<{\displaystyle \frac{\delta }{4}}.$$

Since

$$\left|\parallel \tilde{F}\parallel -1\right|\le ∥\tilde{F}-{\displaystyle \frac{P_{m}F{P}_m}{\parallel P_{m}F{P}_m\parallel }}∥<{\displaystyle \frac{\delta }{16}},$$

we obtain

$$∥{\displaystyle \frac{\tilde{F}}{\parallel \tilde{F}\parallel }}-\tilde{F}∥=\parallel \tilde{F}\parallel \left|{\displaystyle \frac{1}{\parallel \tilde{F}\parallel }}-1\right|=\left|1-\parallel \tilde{F}\parallel \right|<{\displaystyle \frac{\delta }{16}}.$$

Therefore,

$$∥F-{\displaystyle \frac{\tilde{F}}{\parallel \tilde{F}\parallel }}∥\le ∥F-{\displaystyle \frac{P_{m}F{P}_m}{\parallel P_{m}F{P}_m\parallel }}∥+∥{\displaystyle \frac{P_{m}F{P}_m}{\parallel P_{m}F{P}_m\parallel }}-\tilde{F}∥+∥\tilde{F}-{\displaystyle \frac{\tilde{F}}{\parallel \tilde{F}\parallel }}∥$$

$$<{\displaystyle \frac{\delta }{4}}+{\displaystyle \frac{\delta }{16}}+{\displaystyle \frac{\delta }{16}}={\displaystyle \frac{3\delta }{8}}.$$

Set $\overline{F}:={\displaystyle \frac{\tilde{F}}{\parallel \tilde{F}\parallel }}.$ Then $\overline{F}$ is invertible in $B\left(P_m\left(H\right)\right),$ ${\overline{F}}^{\ast }\in B\left(P_m\left(H\right)\right),\parallel {\overline{F}}^{\ast }\parallel =\parallel \overline{F}\parallel =1$ and

$$\parallel \overline{F}-F\parallel <{\displaystyle \frac{3\delta }{8}}.$$

As in the proof of Proposition 4, we can choose some $n_0\in \mathbb{N}$ such that $P_m{U}^n{P}_m=0,$ and such that

$$\parallel U^{-n}{P}_m\parallel <{\displaystyle \frac{\delta }{4(1+\parallel G\parallel +\parallel {\overline{F}}^{\ast -1}{\parallel \parallel G\parallel )}^2}}n\ge n_0,$$

In what follows, since $\overline{F}$ and thus also ${\overline{F}}^{\ast }$ are invertible in $B\left(P_m\left(H\right)\right),$ we let ${\overline{F}}^{-1}$ and ${\overline{F}}^{\ast -1}$ denote the inverse of $\overline{F}$ and ${\overline{F}}^{\ast },$ respectively, in $B\left(P_m\left(H\right)\right).$ For each $n\ge n_0$, set

$$E_n:=\overline{F}+U^{-n}{\overline{F}}^{\ast -1}{P}_{m}G=\overline{F}+U^{-n}{P}_m{\overline{F}}^{\ast -1}{P}_{m}G.$$

Note that the last equality follows from the fact that $U^{-n}{P}_m{\overline{F}}^{\ast -1}=U^{-n}{\overline{F}}^{\ast -1}$ because ${\overline{F}}^{\ast -1}\in B\left(P_m\left(H\right)\right).$

Then

$$\parallel E_n-\overline{F}\parallel \le \parallel U^{-n}{P}_m\parallel \parallel {\overline{F}}^{\ast -1}\parallel \parallel G\parallel <{\displaystyle \frac{\delta }{4}}.$$

Hence,

$$∥{\displaystyle \frac{E_n}{\parallel E_n\parallel }}-\overline{F}∥\le ∥{\displaystyle \frac{E_n}{\parallel E_n\parallel }}-E_n∥+\parallel E_n-\overline{F}\parallel$$

$$\le \left|\parallel E_n\parallel -1\right|+\parallel E_n-\overline{F}\parallel \le 2\parallel E_n-\overline{F}\parallel <{\displaystyle \frac{\delta }{2}},$$

where we have used that $\parallel \overline{F}\parallel =1.$ Consequently,

$$∥{\displaystyle \frac{E_n}{\parallel E_n\parallel }}-F∥\le ∥{\displaystyle \frac{E_n}{\parallel E_n\parallel }}-\overline{F}∥+\parallel \overline{F}-F\parallel <{\displaystyle \frac{\delta }{2}}+{\displaystyle \frac{3\delta }{8}}<\delta .$$

Thus,

$${\displaystyle \frac{E_n}{\parallel E_n\parallel }}\in O_1\cap O_2,$$

or equivalently,

$$\left({\displaystyle \frac{E_n}{\parallel E_n\parallel }},{\displaystyle \frac{E_n}{\parallel E_n\parallel }}\right)\in (O_1\times O_2)\cap {\tilde{\mathrm{\Delta }}}_d\left(B_0\left(H\right)\right),n\ge n_0.$$

On the other hand,

$${\displaystyle \frac{E_n^{\ast }}{\parallel E_n\parallel }}{U}^n\left({\displaystyle \frac{E_n}{\parallel E_n\parallel }}\right)=\left({\displaystyle \frac{{\overline{F}}^{\ast }}{\parallel E_n\parallel }}+{\displaystyle \frac{G^{\ast }{P}_m{\overline{F}}^{-1}{U}^{\ast -n}}{\parallel E_n\parallel }}\right)\left({\displaystyle \frac{U^n\overline{F}}{\parallel E_n\parallel }}+{\displaystyle \frac{{\overline{F}}^{\ast -1}{P}_{m}G}{\parallel E_n\parallel }}\right)$$

$$={\displaystyle \frac{1}{\parallel E_n{\parallel }^2}}\left({\overline{F}}^{\ast }{U}^n\overline{F}+P_{m}G+G^{\ast }{P}_m{\overline{F}}^{-1}{U}^{\ast -n}{U}^n\overline{F}+G^{\ast }{P}_m{\overline{F}}^{-1}{U}^{\ast -n}{\overline{F}}^{\ast -1}{P}_{m}G\right).$$

Since $U^{\ast -1}={\displaystyle \frac{1}{4}}U,$ we get

$$U^{\ast -n}{U}^n={\displaystyle \frac{1}{4^n}}{U}^{2n},n\in \mathbb{N}.$$

Moreover,

$$\overline{F}=P_m\overline{F}{P}_m,{\overline{F}}^{-1}=P_m{\overline{F}}^{-1}{P}_m,$$

$${\overline{F}}^{\ast }=P_m{\overline{F}}^{\ast }{P}_m,{\overline{F}}^{\ast -1}=P_m{\overline{F}}^{\ast -1}{P}_m,$$

because $\overline{F}\in B\left(P_m\left(H\right)\right)$. Since $P_m{U}^n{P}_m=0$ for all $n\ge n_0,$ we therefore obtain

$${\overline{F}}^{\ast }{U}^n\overline{F}=0,{\overline{F}}^{-1}{U}^{\ast -n}{U}^n\overline{F}={\displaystyle \frac{1}{4^n}}{\overline{F}}^{-1}{U}^{2n}\overline{F}=0.$$

for all $n\ge n_0.$ Furthermore,

$${\overline{F}}^{-1}{U}^{\ast -n}{\overline{F}}^{\ast -1}={\overline{F}}^{-1}{P}_m\left({\displaystyle \frac{1}{4^n}}{U}^n\right)P_m{\overline{F}}^{\ast -1}=0.$$

Therefore,

$${\displaystyle \frac{E_n^{\ast }}{\parallel E_n\parallel }}{U}^n{\displaystyle \frac{E_n}{\parallel E_n\parallel }}={\displaystyle \frac{P_{m}G}{\parallel E_n{\parallel }^2}},n\ge n_0.$$

On the other hand,

$$\left|\parallel E_n\parallel -1\right|\le \parallel E_n-\overline{F}\parallel <{\displaystyle \frac{\delta }{4}},n\ge n_0.$$

By the choice of $\delta >0$ at the beginning of this proof, we have

$$\left|{\displaystyle \frac{1}{\parallel E_n{\parallel }^2}}-1\right|<{\displaystyle \frac{\epsilon }{2(1+\parallel G\parallel )}}.$$

Hence,

$$∥G-{\displaystyle \frac{P_{m}G}{\parallel E_n{\parallel }^2}}∥\le \parallel G-P_{m}G\parallel +\parallel P_{m}G\parallel \left|1-{\displaystyle \frac{1}{\parallel E_n{\parallel }^2}}\right|<\epsilon .$$

Thus,

$${\displaystyle \frac{P_{m}G}{\parallel E_n{\parallel }^2}}\in O_3.$$

Consequently,

$${\tilde{\Theta }}_n\left((O_1\times O_2)\cap {\tilde{\mathrm{\Delta }}}_d\left(B_0\left(H\right)\right)\right)\cap O_3\ne ⌀,n\ge n_0.$$

This proves that $L_U$ is $C^{\ast }$-numerically transitive on $B_0\left(H\right)$. □

Remark 4. 

By letting $x=y$ and $z=w$ in the proof of Proposition 5, we can prove analogously that the operator $\tilde{D}$ from that proof is $C^{\ast }$-numerically transitive on ${\tilde{\mathsf{\ell }}}_2\left(B_0\left(H\right)\right).$ Therefore, the operator $\tilde{D}$ is used in the proofs of Corollary 6, Corollary 7, and Corollary 8. Furthermore, by exactly the same arguments as in the proof of Proposition 7, it can be shown that the operator D from that proof is not $C^{\ast }$-numerically transitive, and, similarly, the strongly numerically hypercyclic operator $T\circ \widehat{P}$ on $B_0\left(H\right)$ constructed in the proof of Corollary 9 is not $C^{\ast }$-numerically transitive. Finally, if we let $x=y$ and $z_j=w_j$ for each $j\in \{1,\dots ,m\}$ in the proof of Corollary 5, then Corollary 5 remains valid in the case when $C^{\ast }$-mixing and $C^{\ast }$-transitivity are replaced by $C^{\ast }$-numerical mixing and $C^{\ast }$-numerical transitivity, respectively.