HYPERCYCLICITY OF OPERATORS THAT $\lambda$-COMMUTE WITH THE HARDY BACKWARD SHIFT

Abstract

An operator T acting on a separable complex Banach space $\mathcal {B}$ is said to be hypercyclic if there exists $f\in \mathcal {B}$ such that the orbit $\{T^n f:\ n\in \mathbb {N}\}$ is dense in $\mathcal {B}$. Godefroy and Shapiro (J. Funct. Anal., 98(2):229–269, 1991) characterized those elements, which are hypercyclic, in the commutant of the Hardy backward shift. In this paper, we study some dynamical properties of operators X that $\lambda$-commute with the Hardy backward shift B, that is, $BX=\lambda XB$.

Introduction

A bounded linear operator T, defined on a complex separable Banach space X, is said to be hypercyclic if there exists a vector $x\in X$ such that the orbit: $\mathcal {O}(x,T):=\{T^n x:\ n\in \mathbb {N}\}$ is dense in X. In this case, the vector x is called a hypercyclic vector for T.

The main ancestors of this paper are [7, 14]. In 1969, Rolewicz [14] showed the first example of hypercyclic operator defined on a Banach space. If we consider $T=\lambda B$ where B is the backward shift operator defined on the sequences spaces $\ell _p$, Rolewicz proved that $T=\lambda B$ is hypercyclic if and only if $|\lambda |>1$. Rolewicz’s findings were united in 1991 by G. Godefroy and J. H. Shapiro (see [7]), who demonstrated that each non-scalar operator that commutes with B defined on the Hardy space $H^2(\mathbb {D})$ is hypercyclic if and only if the interior of its punctual spectrum intersects the unit circle. Godefroy and Shapiro’s seminal work points to the idea that the hypercyclic properties of a given operator are somehow transferred to the commutator of the operator. Surprisingly, for the Bergman backward shift the commutant hypercyclicity problem is much more delicate (see [4]).

On the other hand, we found in the literature the $\lambda$-commutant notion, that is, two operators are $\lambda$-commuting if they commute up to a complex factor $\lambda$. The term $\lambda$-commuting was introduced by Conway and Prajitura [6], and since this relation is not symmetric, a more precise terminology was introduced: a complex number $\lambda$ is said to be an extended eigenvalue of an operator T if there exists an operator $X\ne 0$ (later called extended $\lambda$-eigenoperator of T) such that $TX=\lambda XT$.

Let us remark that although we have defined a hypercyclic operator on a Banach space, this property is not exclusive for Banach space theorists. In fact, the first examples of hypercyclic operator were discovered a century ago defined on the space of entire functions endowed with the compact-open topology (see [9] for more historical notes). Recent research is focused on the attempt to see how the hypercyclic properties are transferred to the $\lambda$-commutant of a given operator. For instance, some dynamics properties of the extended eigenoperators of the differentiation operator defined on the space of entire functions were studied in [2, 8, 11]. In this paper, we wish to explore the dynamical properties of extended eigenoperators of the backward shift operator defined on the Hardy space $H^2(\mathbb {D})$. We see that in Banach spaces, the existence of the uniform norm of operators makes it difficult for hypercyclic properties to be transferred to the $\lambda$-commutant. In strong contrast, in Fréchet spaces this transfer seems to be much easier.

Let us recall that a Banach space operator T is called supercyclic if there exists a vector $x\in X$ such that the set of scalar multiples of $\mathcal {O}(x,T)$:

$$\begin{aligned} \mathbb {C}.\mathcal {O}(x,T):=\{\mu T^nx:\ \mu \in \mathbb {C},\ n\in \mathbb {N}\} \end{aligned}$$

is dense in X. And let us recall that a linear operator $\mathcal {B}$ on a Banach space X is said to be a generalized backward shift if it satisfies the following conditions:

• (1) The kernel of $\mathcal {B}$ is one-dimensional.

• (2) $\bigcup \{\ker \mathcal {B}^n:\ n = 0, 1, 2,...\}$ is dense in X.

We will see that other dynamics properties, such as supercyclicity, that is transferred very well to the commutant, have some serious difficulties to be transferred to the $\lambda$-commutant. For instance, Müller [12], solving a longstanding question posed by Godefroy and Shapiro in [7], proved that any non-scalar operator commuting with a generalized backward shift is supercyclic. However, this result is no longer true for the $\lambda$-commutant of the backward shift operator defined on the Hardy space $H^2(\mathbb {D})$.

The paper is structured as follows. In Section 2, we introduce the main tools that we will use throughout the paper. We will use the Hypercyclicity Criterion, that is, a sufficient condition for hypercyclicity. Next, in order to find enough intuition to address our research, we will review the result of Godefroy and Shapiro that characterizes which elements of the commutant of B are hypercyclic. Specifically, in Godefroy and Shapiro’s result, we will see a dichotomy on the orbits of operators T that commute with the operator B: the operator T is either hypercyclic or the orbits of T or $T^{-1}$ are bounded. In general, something similar seems to happen in the case of operators that $\lambda$-commute with B.

The cornerstone in Section 3 is a factorization of the extended $\lambda$-eigenoperators of the backward shift. This factorization complements the results obtained by Petrovic in [13] and it is the main tool that will be used in the next sections. When $|\lambda |=1$ using the ideas of [10], we get that each extended $\lambda$-eigenoperator of the backward shift factorizes as $R_\lambda \phi (B)$ when $R_\lambda f(z)=f(\lambda z)$ is the dilation operator and $\phi (B)$ is an element of the commutant of B, that is, the adjoint of a multiplier on $H^2(\mathbb {D})$. Surprisingly enough, when $|\lambda |<1$ the extended $\lambda$-eigenoperators have the form $R_\lambda \phi (B)$ when $\phi$ is an element of $H^2(\mathbb {D})$.

In Section 4, we will study the hypercyclicity of an extended $\lambda$-eigenoperator of the backward shift. Firstly, we show that if $R_\lambda \phi (B)$ is hypercyclic, then $\phi (B)$ must be hypercyclic. In strong contrast with the results in [2], the converse is not true, a hypercyclic operator $\phi (B)$ in the commutant of B in general does not induce a hypercyclic extended $\lambda$-eigenoperator. The problem depends on the intrinsic geometry of $\phi (\mathbb {D})$. However, when $\lambda$ is a root of the unit, we can get a full characterization in terms of the properties of $\phi$. When $\lambda$ is an irrational rotation, the problem is connected to the study of the dynamical properties of a sequence of functions. A characterization of the hypercyclicity of $R_\lambda \phi (B)$ in terms of the geometry of $\phi (\mathbb {D})$ seems to be a challenging problem.

In Section 5, we will show that if $T=R_\lambda \phi (B)$ is an extended $\lambda$-eigenoperator of the backward shift operator with $|\lambda |<1$, then T is supercyclic if and only if $\phi (0)=0$. As byproduct, we obtain that Müller’s result is no longer true for elements in the $\lambda$-commutant of the backward shift operator. The paper is closed with a brief section with open questions and further directions.

Some tools

We will use the following version of the Hypercyclicity Criterion formulated by Bès and Peris in [3]:

Theorem 2.1

Let T be an operator on an F-space X satisfying the following conditions: there exist $X_0$ and $Y_0$ dense subsets of X, a sequence $(n_k)$ of non-negative integers, and (not necessarily continuous) mappings $S_{n_k}: Y_0 \rightarrow X$ so that:

• (1) $T^{n_k} \rightarrow 0$ pointwise on $X_0$;

• (2) $S_{n_k} \rightarrow 0$ pointwise on $Y_0$;

• (3) $T^{n_k}S_{n_k} \rightarrow Id_{Y_0}$ pointwise on $Y_0$.

Then the operator T is hypercyclic.

Specifically, we will use the following spectral sufficient condition discovered by Godefroy and Shapiro [7].

Theorem 2.2

(Godefroy-Shapiro) Let T be a bounded linear operator defined on a Banach space X. If the eigenspaces $\text {ker}(T-\lambda I)$ with $|\lambda |>1$ span a dense subspace of X as well as the eigenspaces $\text {ker}(T-\lambda I)$ with $|\lambda |<1$, then T is hypercyclic.

Let us denote by $k_a(z)=\frac{1}{1-\overline{a}z}$ the reproducing kernels on $H^2(\mathbb {D})$ and $M_g$ denotes the analytic Toeplitz operator with symbol $g\in H^\infty (\mathbb {D})$. It is well known that $M_g^\star k_a(z)=\overline{g(a)} k_{a}(z)$. As usual, we denote by $\overline{g}(z)=\overline{g(\overline{z})}\in H^\infty (\mathbb {D})$. For convenience, we will denote the elements of the commutant of B, by $\phi (B)$, with $\phi \in H^\infty (\mathbb {D})$.

The spectrum of $M_{g}^{\star }$ is usually big, in fact $M_{\overline{\phi }}^{\star } k_a(z)= \phi (a) k_a(z)$. Thus, using Theorem 2.2, if $\phi (\mathbb {D})$ meets the unit circle, then $M_{\overline{\phi }}^\star$ is hypercyclic. In other cases, that is, if $\phi (\mathbb {D})\subset \mathbb {D}$ (respectively $\phi (\mathbb {D})\subset \mathbb {C}\setminus \overline{\mathbb {D}}$), then the orbit is bounded $\Vert (M_{\overline{\phi }}^{\star })^nf\Vert \le M$ (respectively the orbit of f under $(M_{\overline{\phi }}^{\star })^{-1}$ is bounded). This dichotomy that appears in this result will reappear in the results of this work, and it will be central in the discussion.

Let X be a topological space and $(\phi _k)$ a sequence of continuous mappings on X. A dynamical system $(X,(\phi _k))$ is topologically transitive if for any non-empty open sets $U,V\subset X$ there exists a positive integer n such that $\phi _n(U)\cap V\ne \emptyset$.

Theorem 2.3

(Birkhoff’s transitivity theorem) If X is a second countable, complete metric space, then topological transitivity implies that there is a dense set of points x in X with dense orbit $\{\phi _k(x)\}_{k\ge 1}$.

The following version of Montel’s theorem in practice can be seen as a type of Birkhoff’s transitivity theorem.

Theorem 2.4

(Montel’s Theorem) Let us suppose that $\mathcal {F}$ is a family of meromorphic functions defined on an open subset D. If $z_0\in D$ is such that $\mathcal {F}$ is not normal at $z_0$ and $z_0\in U\subset D$, then

$$\begin{aligned} \underset{{f\in \mathcal {F}}}{\bigcup } \;f(U) \end{aligned}$$

is dense for any non-empty neighborhood U of $z_0$.

Factorization of extended $\lambda$-eigenoperators of the backward shift

Assume that X is an extended $\lambda$-eigenoperator of the backward shift operator B on $H^2(\mathbb {D})$. Let us discard a trivial case: $\lambda =0$. An easy check shows that if $\lambda =0$, then $X1=c_0\ne 0$ and $Xz^n=0$ for any $n\ge 1$. That is, X is a projection over the constant functions and, therefore, it is not hypercyclic.

As a consequence of a result by Petrovic [13], when $|\lambda |>1$ then there is no extended $\lambda$-eigenoperators. That is, the extended spectrum of B is the closed unit disk. Next, we state a result that factorizes X when it exists.

Theorem 3.1

Assume that $\lambda \in \overline{\mathbb {D}}$ and X is an extended $\lambda$-eigenoperator of B. Then,

• (1) If $|\lambda |=1$, then X is an extended $\lambda$-eigenoperator of B if and only if $X=R_\lambda \phi (B)$ with $\phi \in H^\infty (\mathbb {D})$.

• (2) If $|\lambda |<1$ then X is an extended $\lambda$-eigenoperator of B if and only if $X=R_\lambda \phi (B)$ with $\phi \in H^2(\mathbb {D})$.

Proof

To show (1), assume that if $X=R_\lambda \phi (B)$ when $\phi \in H^\infty (\mathbb {D})$. Since $R_\lambda$ is an extended $\lambda$-eigenoperator of B, an easy check shows that X is an extended $\lambda$-eigenoperator of B.

Conversely, let X be an extended $\lambda$-eigenoperator of B. Since $R_{\lambda ^{-1}}$ is an extended $1/\lambda$-eigenoperator of B, then $BR_{1\over \lambda }= {1\over \lambda } R_{1\over \lambda }B,$ hence

$$\begin{aligned} (R_{1\over \lambda }X)B= {1 \over \lambda } R_{1 \over \lambda }BX ={1 \over \lambda } {\lambda } BR_{1 \over \lambda }X= B(R_{1 \over \lambda }X), \end{aligned}$$

that is $B(R_{1\over \lambda }X)$ commutes with B. So, there exists $\phi \in H^\infty (B)$ such that $R_{1 \over \lambda }X = \phi (B)$, therefore, $X= R_{ \lambda } \phi (B)$ as we desired.

To show (2), we will use a result by Petrovic (see [13]). Specifically, Petrovic proved that if X is an extended $\lambda$-eigenoperator of B ($|\lambda |<1$), then

$$\begin{aligned} X=R_\lambda \phi (B) \end{aligned}$$

where $\phi (z)=\sum _{k=0}^\infty c_n z^n$ is a non-zero formal power series. Taking adjoints, we get that $X^\star =M_{\overline{\phi }} R_{\overline{\lambda }}$ where $M_{\overline{\phi }}$ could be an unbounded multiplication operator. Let us observe that $X^\star 1=\overline{\phi }\in H^2(\mathbb {D})$ which proves the first implication.

Conversely, assume that $\phi \in H^2(\mathbb {D)}$, and $|\lambda |<1$. We have to show that $M_{\overline{\phi }} R_{\overline{\lambda }}$ is bounded. Indeed, for any $f\in H^2(\mathbb {D})$, since $|\lambda |<1$ the map $R_{\overline{\lambda }}f$ sends $H^2(\mathbb {D})$ on $H^\infty (\mathbb {D})$. Thus, since $R_{\overline{\lambda }}f\in H^\infty (\mathbb {D})$, we get

$$\begin{aligned} \Vert M_{\overline{\phi }} R_{\overline{\lambda }} f\Vert _2^2\le \Vert \phi \Vert _2^2 \max _{|w|=|\lambda |}|f(w)|^2. \end{aligned}$$

(1)

By using the maximum modulus principle, we get $\max _{|w|=|\lambda |}|f(w)|^2=|f(w_0)|^2$ for some $|w_0|=|\lambda |$. Using the subharmonicity of $|f|^2$, we get that there exists a constant $C>0$ such that for any $r>0$ such that $|\lambda |<r<1$, we get that

$$\begin{aligned} |f(w_0)|^2\le C \frac{1}{2\pi }\int _{0}^{2\pi } |f(re^{i\theta }|^2\,d\theta . \end{aligned}$$

(2)

Thus, substituting inequality Eq. 2 into inequality Eq. 1, we get

$$\begin{aligned} \Vert M_{\overline{\phi }} R_{\overline{\lambda }} f\Vert _2^2\le C \Vert \phi \Vert _2^2 \Vert f\Vert _2^2, \end{aligned}$$

which gives the desired result. $\square$

Hypercyclicity of extended $\lambda$-eigenoperators of the backward shift

In this section, we study when an extended $\lambda$-eigenoperator of B is hypercyclic.

First of all, we will see some basics consequences from the theory of hypercyclic operators. Next, we will study some sufficient conditions for hypercyclicity.

Hypercyclicity: basic consequences

It is known that for Banach space operators, extended $\lambda$-eigenoperators of a given operator A are never hypercyclic provided $|\lambda |<1$ (see [2, Proposition 3.1]). Thus, we get:

Corollary 4.1

Assume that T is an extended $\lambda$-eigenoperator of B with $|\lambda |<1$, then T is not hypercyclic.

The following example provides some intuition about the last statement in our special case.

Example 4.2

If we consider Rolewicz operator 2B, the point spectrum of 2B is big. If we consider $T=R_\lambda (2B)$ with $|\lambda |<1$, we get that $R_\lambda$ is compact, therefore, T is compact. Hence T cannot be hypercyclic.

We stress here that $T=R_\lambda \phi (B)$ with $|\lambda |<1$ is not hypercyclic even if $\phi (B)$ is unbounded. However, T could be supercyclic. In fact, in Section 5, we characterize when $R_\lambda \phi (B)$ is supercyclic in the case $|\lambda |<1$.

Let us assume that $|\lambda |=1$. In such a case, our extended $\lambda$-eigenoperators factorize as $T=R_\lambda \phi (B)$, with $\phi \in H^\infty (\mathbb {D})$. Hence, if T is hypercyclic, then $\phi (B)$ is hypercyclic too.

Proposition 4.3

Assume that $T=R_\lambda \phi (B)$ with $|\lambda |=1$ and $\phi \in H^\infty (\mathbb {D})$. If $\phi (\mathbb {D}) \cap \partial \mathbb {D}=\emptyset$ then T is not hypercyclic.

Proof

Indeed, if $\phi (\mathbb {D})\subset \mathbb {D}$, then since $\Vert T\Vert \le \Vert R_\lambda \Vert \Vert \phi \Vert _{\infty }$, we get that T is a contraction, therefore, T is not hypercyclic.

On the other hand, if $\phi (\mathbb {D})\subset \overline{\mathbb {D}}^c$, then we see that T is invertible and $\Vert T^{-1}\Vert \le 1$, therefore, $T^{-1}$ is not hypercyclic. Hence T is not hypercyclic. $\square$

Assume that $T=R_\lambda \phi (B)$, with $|\lambda |=1$ and $\phi (B)$ is hypercyclic. In strong contrast with the results obtained in [2], if $\phi (B)$ is hypercyclic, then we cannot guarantee that T is hypercyclic too. The next example provides a hypercyclic operator $\phi (B)$, with $\phi \in H^\infty (\mathbb {D})$, such that for some $\lambda$ with $|\lambda |=1$, the extended $\lambda$-eigenoperator $R_\lambda \phi (B)$ is not hypercyclic.

Example 4.4

Let us consider $\lambda =i$ the imaginary unit and the maps

$$\begin{aligned} \phi _0(z)=\frac{1}{2}z+1-\frac{1}{10}. \end{aligned}$$

and

$$\begin{aligned} \phi _1(z)=\frac{1}{2}z+1-\frac{1}{100}. \end{aligned}$$

Clearly, $\phi _0(\mathbb {D})$ is the disk centered at $1-\frac{1}{10}$ with radius 1/2 and $\phi _1(\mathbb {D})$ is the disk centered at $1-\frac{1}{100}$ with radius 1/2 (see Fig. 1).

Fig. 1

The image $\phi _0(\mathbb {D})$ and $\phi _1(\mathbb {D})$

Thus, $\phi _0(\mathbb {D})\cap \partial \mathbb {D}\ne \emptyset$ and $\phi _1(\mathbb {D})\cap \partial \mathbb {D}\ne \emptyset$, therefore, $\phi _0(B)$ and $\phi _1(B)$ are hypercyclic.

Let us consider the operators $T_0=R_{i}\phi _0(B)$ and $T_1=R_{i}\phi _1(B)$, and the powers $T^4_0$ and $T_1^4$. Since $i^4=1$, we obtain that $T_0^4$ and $T_1^4$ commute with B. Specifically, $T_0^4=\phi _0(B)\phi _0(iB)\phi _0(-iB)\phi _0(-B)$ and $T_1^4=\phi _1(B)\phi _1(iB)\phi _1(-iB)\phi _1(-B)$ Let us denote $\Phi _0(z)=\phi _0(z)\phi _0(iz)\phi _0(-iz)\phi _0(-z)$ and $\Phi _1(z)=\phi _1(z)\phi _1(iz)\phi _1(-iz)\phi _1(-z)$.

Now, using Matlab computations, we see that $\Phi _0(\mathbb {D})\subset \mathbb {D}$ and $\Phi _1(\mathbb {D}) \cap \partial \mathbb {D}\ne \emptyset$ (see Fig. 2).

Fig. 2

The image $\Phi _0(\mathbb {D})$ and $\Phi _1(\mathbb {D})$

Therefore, $T_0^4$ is not hypercyclic and $T_1^4$ is hypercyclic. Hence, $T_0$ is not hypercyclic, but $T_1$ is hypercyclic by Ansari’s result [1].

From the above considerations, we see that the hypercyclicity of $R_\lambda \phi (B)$ does not depend generally on intersection of the point spectrum of $\phi (B)$ at $\partial \mathbb {D}$, but it clearly depends on the shape of the point spectrum of $\phi (B)$. This shape is intimately connected to the limits of the sequence of sets $\Phi _{n}(\mathbb {D})$ when $\Phi _{n}(z)=\phi (z)\phi (\lambda z)\cdots \phi (\lambda ^{n-1}z)$. Specifically, if the sequence $\Phi _{n}(\mathbb {D})$ is separated from $\infty$ or from 0, then T is not hypercyclic.

Proposition 4.5

Assume that $\lambda \in \partial \mathbb {D}$ and $T=R_{\lambda }\phi (B)$ is an extended $\lambda$-eigenoperator of B.

• (1) If $\Vert \Phi _{n}\Vert _\infty <M$ for all n, then T is not hypercyclic.

• (2) If there exists $c>0$ such that $c\le Inf_{z\in \mathbb {D}}|\Phi _n(z)|$ for all $n\ge n_0$ for some $n_0$, then $T=R_{\lambda }\phi (B)$ is not hypercyclic.

Proof

To show (1), let us see that

$$\begin{aligned} \Vert (R_{\lambda }\phi (B))^n\Vert= & {} |1\cdot \lambda \cdots \lambda ^{n-1}| \Vert R_\lambda ^n\Vert \Vert \Phi _n(B)\Vert \\\le & {} \Vert \Phi _n\Vert _{\infty }\le M, \end{aligned}$$

that is $T=R_{\lambda }\phi (B)$ is power bounded, and, therefore, T is not hypercyclic.

For (2), let us observe that $T^n$ is invertible for $n\ge n_0$. Since T is hypercyclic if and only if $T^n$ is hypercyclic, we get that T is hypercyclic if and only if $T^{-n}$ is hypercyclic. However, an easy check shows that $\Vert T^{-n}\Vert \le \frac{1}{c}$ for all $n\ge n_0$. Therefore, $T^{-n}$ with $n\ge n_0$ is not hypercyclic. Hence T is not hypercyclic. $\square$

However, for values $\lambda \in \partial \mathbb {D}$ which are roots of the unity, a characterization can be obtained using Ansari and Godefroy-Shapiro results.

Proposition 4.6

Assume that $T=R_\lambda \phi (B)$ is an extended $\lambda$-eigenoperator of B and $\lambda ^n=1$ for some positive integer n. If we denote $\Phi (z)=\phi (z)\phi (\lambda z)\cdots \phi (\lambda ^{n-1}z)$, then T is hypercyclic if and only if $\Phi (\mathbb {D})\cap \partial \mathbb {D}\ne \emptyset .$

Proof

Let us see that $T^n=R_\lambda \phi (B)\cdots R_{\lambda }\phi (B)$, and since $R_\lambda$ is an extended $\lambda$-eigenoperator of B, we have $\phi (B)R_\lambda =\phi (\lambda B)R_\lambda$. Thus, $T^n=(R_\lambda )^n \phi (\lambda ^{n-1}B)\phi (\lambda ^{n-2}B)\cdots \phi (B)$. Now, since $\lambda ^n=1$, we get $(R_\lambda )^n=R_{\lambda ^n}=I$. Therefore, since $T^n=\Phi (B)$ is an element of the commutant of B, using Godefroy-Shapiro characterization, $T^n$ is hypercyclic if and only if $\Phi (\mathbb {D})\cap \partial \mathbb {D}\ne \emptyset$. Hence, using Ansari’s result, we get that T is hypercyclic if and only if $\Phi (\mathbb {D})\cap \partial \mathbb {D}\ne \emptyset$ as we desired. $\square$

To simplify the notation, set $\alpha =\overline{\lambda }$ and $\omega =1/\overline{\lambda }$. Thus, the existence of hypercyclic extended $\lambda$-eigenoperators will depend on the dynamics of the sequence of analytic functions

$$\begin{aligned} \Psi _n(z)=\overline{\phi }(z) \cdot \overline{\phi }(\alpha z)\cdots \overline{\phi }(\alpha ^{n-1}z) \end{aligned}$$

and

$$\begin{aligned} \Omega _n(z)=\overline{\phi }(\omega z)\cdots \overline{\phi }(\omega ^{n}z). \end{aligned}$$

In fact, the dynamics of both sequences are related by the formula $\Omega _n(z)=\Psi _n(\omega ^nz)$. The reader keeps in mind that $\Psi _n(z)\rightarrow \infty$ (resp. 0) if and only if $\overline{\Psi _n(z})\rightarrow \infty$ (resp. 0), and the same property is satisfied for the sequence of mappings $\Omega _n(z)$. The next result sheds light on this direction.

Proposition 4.7

Assume that $\lambda \in \partial \mathbb {D}$ is an irrational rotation. If there exists a sequence $(n_k)\subset \mathbb {N}$ such that the subsets

$$\begin{aligned} C_0=\{z\in \mathbb {D}\,:\,\Psi _{n_k}(z)\rightarrow 0\} \end{aligned}$$

and

$$\begin{aligned} C_1=\{z\in \mathbb {D}\,:\, \Omega _{n_k}(z)\rightarrow \infty \} \end{aligned}$$

have both an accumulation point on $\mathbb {D}$, then $T=R_\lambda \phi (B)$ is hypercyclic on $H^2(\mathbb {D})$.

Proof

Indeed, we can see that the Hypercyclicity Criterion is satisfied by selecting the dense subsets

$$\begin{aligned} X_0=\mathrm{{linear~span}}\{k_{z_0}(z)\,:\,z_0\in C_0\} \end{aligned}$$

and

$$\begin{aligned} Y_0=\mathrm{{linear~span}}\{k_{z_0}(z)\,:\,z_0\in C_1\}. \end{aligned}$$

Both subsets $X_0, Y_0$ are dense and

$$\begin{aligned} T^{n_k}k_{z_0}(z)=\overline{\Psi _{n_k}(z_0)}k_{\alpha ^{n_k}z_0}\rightarrow 0 \end{aligned}$$

pointwise on $X_0$. On the other hand, if $a\in C_1$, then $\overline{\phi }(\omega ^n a)\ne 0$, we can define

$$\begin{aligned} S k_a(z)=\frac{1}{\overline{\overline{\phi }(\omega a)}} k_{\omega a}(z) \end{aligned}$$

then S is well defined on $Y_0$ and $S^{n_k}$ converges pointwise to 0 on $Y_0$. Therefore, by appying the Hypercyclicity Criterion, we get that T is hypercyclic as we desired. $\square$

The above result is a natural way to apply the Hypercyclicity Criterion. However, the hypothesis is very restrictive and next, we will get some weakening.

Proposition 4.8

Assume that $\lambda \in \partial \mathbb {D}$ is an irrational rotation and $\phi \in H^\infty (\mathbb {D})$.

• (1) If there exist $z_0\in \mathbb {D}$ and a subsequence $(n_k)\subset \mathbb {N}$ such that $\Psi _{n_k}(z_0)\rightarrow 0$, then $T^{n_k}$ converges pointwise to zero on a dense subset.

• (2) If there exists $z_1\in \mathbb {D}$ such that $\Omega _{n_k}(z_1)\rightarrow \infty$ and there exists $M>0$ such that $|\phi (\omega ^{n}z_0)|>M>0$ for all $n\ge 1$, then the right inverse $Sk_a=\frac{1}{\overline{\overline{\phi }(\omega a)}}k_{\omega a}(z)$ is well defined on a dense subset $Y_0$ and $S^{n_k}$ converges pointwise to zero on $Y_0$.

Proof

Indeed, if $\overline{\phi }(\alpha ^{n_0}z_0)=0$ for some $n_0\in \mathbb {N}$, then let us take

$$\begin{aligned} X_{0}=\mathrm{{linear~span}} \{k_{\omega ^nz_0}(z)\,\,:\,\,n\in \mathbb {N}\} \end{aligned}$$

Since $\omega$ is an irrational rotation, the subset $\{\omega ^nz_0\}$ has an accumulation point in $\mathbb {D}$, which implies that $X_0$ is dense. Let us fix $k_0\in \mathbb {N}$, if $n>n_0+k_0$, then

$$\begin{aligned} T^n k_{\omega ^{k_0}z_0}(z)=0. \end{aligned}$$

That is, $T^{n}$ converges pointwise to zero on $X_0$ for the full sequence of natural numbers. In particular $T^{n_k}$ converges pointwise to zero on $X_0$.

If $\overline{\phi }(\alpha ^{n}z_0)\ne 0$ for all $n\in \mathbb {N}$, then we consider the subset

$$\begin{aligned} X_{0}=\mathrm{{linear~span}}\{k_{\alpha ^n z_0}(z)\,:\,n\ge 1\}. \end{aligned}$$

Again, since $\alpha$ is an irrational rotation, the subset $\{\alpha ^n z_0\}$ has an accumulation point in $\mathbb {D}$, therefore, the subset $X_0$ is dense. On the other hand,

$$\begin{aligned} T^{n_k}k_{\alpha ^{n_0}z_0} (z)= & {} \overline{\overline{\phi }(\alpha ^{n_0}z_0)\cdots \overline{\phi }( \alpha ^{n_0+n_{k}-1}z_0)} k_{\alpha ^{n_0+n_k}z_0}(z)\\= & {} \frac{\overline{\Phi _{n_k}(z_0)}\overline{\overline{\phi }(\alpha ^{n_k}z_0)\cdots \overline{\phi }(\alpha ^{n_0+n_k-1}z_0)} }{\overline{\overline{\phi }(z_0)\cdots \overline{\phi }(\alpha ^{n_0-1}z_0)}} k_{\alpha ^{n_0+n_k}z_0}(z). \end{aligned}$$

Thus,

$$\begin{aligned} \Vert T^{n_k}k_{\alpha ^{n_0}z_0} (z)\Vert \le C |\Phi _{n_k}(z_0)|\Vert \phi \Vert _{\infty }^{n_0}. \end{aligned}$$

We obtain again that $T^{n_k}$ converges pointwise to zero on $X_0$ as $k\rightarrow \infty$. This fact proves (1).

To show (2), we consider the following subset

$$\begin{aligned} Y_{0}=\mathrm{{linear~span}}\{k_{\omega ^nz_1}(z)\,:\,n\ge 1\} \end{aligned}$$

which is clearly dense because the sequence $\omega ^nz_1$ has an accumulation point in $\mathbb {D}$. Next, we will see that on $Y_0$ the following map

$$\begin{aligned} Sk_a(z)=\frac{1}{\overline{\overline{\phi }(\omega a)}} k_{\omega a}(z) \end{aligned}$$

is well defined. Indeed, since $\Omega _{n_k}(z_1)\rightarrow \infty$, we get that $\overline{\phi }(\omega ^nz_1)\ne 0$ for all n.

On the other hand,

$$\begin{aligned} S^{n_k}k_{\omega ^{m_0}z_1}= & {} \frac{1}{\overline{\overline{\phi }(\omega ^{m_0+1}z_1)}\cdots \overline{\overline{\phi }(\omega ^{m_k+m_0}z_1)}}k_{\omega ^{m_k+m_0}z_1}(z)\\= & {} \frac{\overline{\overline{\phi }(\omega z_1)}\cdots \overline{\overline{\phi }(\omega ^{m_0}z_1)}}{\overline{\Omega _{n_k}(z_1)} \overline{\overline{\phi }(\omega ^{m_k+1}z_1)}\cdots \overline{\overline{\phi }(\omega ^{m_k+m_0}z_1)}}. \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert S^{n_k}k_{\omega ^{m_0}z_1}\Vert \le \frac{\Vert \varphi \Vert _\infty ^{m_0}}{M^{m_0}|\Omega _{n_k}(z_1)|}\rightarrow 0 \end{aligned}$$

as $k\rightarrow \infty$, as we wanted. $\square$

From the above results, we now wish to obtain some geometrical sufficient conditions to guarantee hypercyclicity. Since we will apply the Hypercyclicity Criterion, we will look for sufficient conditions to obtain separately the conditions of the Hypercyclicity Criterion. We will say that T satisfies the condition $X_0$ if T satisfies the condition (1) of the Hypercyclicity Criterion. And we will say that T satisfies the condition $Y_0$ if there exists a dense subset $Y_0$ and a sequence of mappings $S_{n_k}$ such that the condition (2) and (3) of the Hypercyclicity Criterion are satisfied for these mappings.

Proposition 4.9

Assume that $T=R_\lambda \phi (B)$ is an extended $\lambda$-eigenoperator of B. Then

• (1) If $\phi (z_0)=0$, then T satisfies the condition $X_0$ in the Hypercyclicity Criterion for the full sequence of natural numbers.

• (2) If $|\phi (0)|<1$, then T satisfies the condition $X_0$ for the full sequence of natural numbers.

• (3) If $|\phi (0)|>1$, then T satisfies the condition $Y_0$ for the full sequence of natural numbers.

• (4) If $|\phi (0)|=1$ and $\lambda$ is a root of the unity, then T satisfies conditions $X_0$ and $Y_0$ for the full sequence $\mathbb {N}$.

Proof

Condition (1) appears in the proof of Proposition 4.8. If $z_0=0$ and $\phi (0)=0$, then it is easy to verify that T has dense generalized kernel. If $z_0\ne 0$, since $0=\phi (z_0)=\overline{\phi }(\overline{z_0})$ considering the following dense subset:

$$\begin{aligned} X_0=\mathrm{{linear~span}}\{ k_{\omega ^n\overline{z_0}}(z)\,:\,n\ge 1\}, \end{aligned}$$

an easy check proves that T converges pointwise to 0 on $X_0$.

To show (2), by continuity there exists $\delta >0$ such that $|\phi (z_0)|<1$ for every $z_0\in D(0,\delta )$. The result follows by considering the following dense subset:

$$\begin{aligned} X_0=\text {linear~span} \{k_{\overline{z_0}}(z)\,\,:\,\,z_0\in D(0,\delta )\}. \end{aligned}$$

In a similar way in the case (3), there exists $\delta >0$ such that for each $z_0\in D(0,\delta )$, $|\phi (z_0)|>1+\varepsilon$ for some $\varepsilon >0$. Now, condition (3) follows by considering:

$$\begin{aligned} Y_0=\text {linear~span}\{k_{\overline{z_0}}(z)\,;\,z_0\in D(0,\delta )\}. \end{aligned}$$

To show (4), we use the open mapping theorem. We know that if we consider $T^n$, then $T^n=\Phi (B)$ and $\Phi (z)=\phi (z)\cdot \phi (\lambda z)\cdots \phi (\lambda ^{n-1}z)$ is analytic on $\mathbb {D}$. Since $|\Phi (0)|=1$ and $\Phi (0)$ is an interior point of the image $\Phi (\mathbb {D})$, we get that $\Phi (\mathbb {D})\cap \partial \mathbb {D}\ne \emptyset$; therefore, by Godefroy and Shapiro’s result, we get that T satisfies the Hypercyclicity Criterion for the full sequence of natural numbers. $\square$

Some sufficient conditions of hypercyclicity of the elements of the $\lambda$-commutant of the Hardy backward shift

In this subsection, we will analyze the case $\lambda \in \partial \mathbb {D}$ and $\lambda$ an irrational rotation.

Now, let us select some of the ideas used in Proposition 6.1 in [2].

Clearly, we can select $r_n<1$ such that $r_n\rightarrow 1$ and $\varepsilon _n>0$ such that for all z satisfying $r_n-\varepsilon _n<|z|<r_n+\varepsilon _n$, we can get $\phi (z)\ne 0$. Let us denote by $C_n=\{z:\,r_n-\varepsilon _n<|z|<r_n+\varepsilon _n\}$.

Lemma 4.10

Let us fix an increasing subsequence $\{n_k\}\subset \mathbb {N}$, and let us assume that for any $z_0$, satisfying $|z_0|=r_n$, the family $\mathcal {G}=\{\Omega _{n_k}\}$ is normal at $z_0$ and pointwise bounded on $z_0$. Then $\mathcal {G}$ is uniformly bounded on compact subsets of $\mathbb {D}$.

Proof

Indeed, let us fix $z_0\in C_n$ with $|z_0|=r_n$. By assumption there exists $\varepsilon >0$ such that $B_0=D(z_0,\varepsilon )\subset C_n$ and $\mathcal {G}$ is uniformly bounded on $B_0$. Let us show that $\mathcal {G}$ is uniformly bounded on $|z|\le r_n$. Indeed, since $\omega$ is an irrational rotation, by compactness there is an integer $n_0$ such that

$$\begin{aligned} \partial D(0,|z_0|)\subset B_0\cup \omega B_0\cup \cdots \cup \omega ^{n_0}B_0. \end{aligned}$$

Since $A=B_0\cup \cdots \cup \omega ^{n_0}B_0$ is strictly contained in $\{z :\, r_0<|z|<r_1 \}$ and $\phi$ does not vanish on A, let $C=\frac{\max \{|\phi (z)|: |z|=r_n\}}{\min \{|\phi (z)|: |z|=r_n\}}>0$.

By using the maximum modulus principle and by showing that $\mathcal {G}$ is uniformly bounded on $A=B_0\cup \cdots \cup \omega ^{n_0}B_0$, we get that $\mathcal {G}$ is uniformly bounded on $D(0,|z_0|)$.

Indeed, since $\mathcal {G}$ is uniformly bounded on $B_0$, let M such that $|\Omega _{n_k}(z)|<M$ for all $z\in B_0$ and $\Omega _{n_k}\in \mathcal {G}$. Let us show that $\mathcal {G}$ is uniformly bounded on each $\omega ^lB_0$, $1\le l\le n_0$. Indeed, each element in $\omega ^lB_0$ has the form $\omega ^lz$ with $z\in B_0$. Thus,

$$\begin{aligned} |\Omega _{n_k}(\omega ^l z)|= & {} |\overline{\phi }(\omega ^{l+1}z)\cdots \overline{\phi }(\omega ^{l+n_k}z)|\\= & {} \frac{|\Omega _{n_k}(z)| |\overline{\phi }(\omega ^{n_k+1}z)\cdots \overline{\phi }(\omega ^{n_k+l}z)|}{|\overline{\phi }(\omega z)\cdots \overline{\phi }(\omega ^l z)|} \\\le & {} M C^l, \end{aligned}$$

for all $\omega ^l z\in \omega ^l B_0$ and $\Omega _{n_k}\in \mathcal {G}$. Therefore, $\mathcal {G}$ is uniformly bounded on $D(0,|z_0|)$. Since this argument is for any $r_n$ converging to 1, we get that $\mathcal {G}$ is uniformly bounded on compact subsets of $\mathbb {D}$ as we wanted to show. $\square$

Theorem 4.11

Set $\lambda \in \partial \mathbb {D}$ an irrational rotation. If there exist a sequence $(n_k)$, a point $z_0\in \mathbb {D}$ such that $|\Omega _{n_k}(z_0)|>c>0$ and a point $z_1\in \mathbb {D}$ such that $\Psi _{n_k}(z_1)\rightarrow 0$, then $T=R_\lambda \phi (B)$ is hypercyclic.

Proof

First of all, if $\phi$ vanishes on $\mathbb {D}$, then by applying Proposition 4.9, we get that T satisfies condition $X_0$ for the full sequence of natural numbers.

Let us assume that $\phi$ does not vanish on $\mathbb {D}$. Using the same trick as in Lemma 4.10, we can get that $\Psi _{n_k}(\alpha ^l z_1)\rightarrow 0$. Indeed, if $|z_1|=r$, then we denote by $C_r$ an annulus like in Lemma 4.10. And let us denote $C=\frac{\max \{|\phi (z)|: z\in C_r\}}{\min \{|\phi (z)|: z\in C_r\}}>0$. Then,

$$\begin{aligned} |\Psi _{n_k}(\alpha ^l z_1)|= & {} |\overline{\phi }(\alpha ^{l}z_1)\cdots \overline{\phi }(\alpha ^{l+n_k-1}z_1)|\\= & {} \frac{|\Psi _{n_k}(z_1)| |\overline{\phi }(\alpha ^{n_k}z_1)\cdots \overline{\phi }(\alpha ^{n_k+l-1}z_1)|}{|\overline{\phi }(z_1)\cdots \overline{\phi }(\alpha ^{l-1} z_1)|} \\\le & {} |\Psi _{n_k}(z_1)| C^l\rightarrow 0. \end{aligned}$$

Then, by applying Proposition 4.8 (1), the operator $T=R_\lambda \phi (B)$ satisfies condition $X_0$ of the Hypercyclicity Criterion for the sequence of natural numbers $(n_k)$. Therefore, in both cases, we obtain that T satisfies condition $X_0$ for the sequence $(n_k)$.

We set $\mathcal {G}=\{\Omega _{n_k}\}$. Let us consider the annulus $C_n$ of the previous lemma. Then, we have two possibilities:

Case 1. The family $\mathcal {G}$ is normal at no point $z\in C_{n_0}$ for some $n_0$. Then, by using Montel’s Theorem (Theorem 2.4), we have that $\bigcup _{l}\Omega _{n_l} (C_{n_0})$ is dense in $\mathbb {C}$. Since $C_{n_0}$ is homemorphic to a complete metric space, using Birkhoff’s transitivity Theorem (Theorem 2.3) there exists $z_2\in C_{n_0}$ such that $\{\Omega _{n_l}(z_2)\}_{l\ge 1}$ is dense in $\mathbb {C}$. In particular, there exists a subsequence $\{r_k\}\subset \{n_k\}$ such that $\Omega _{r_k}(z_2)\rightarrow \infty$. On the other hand, since $z_2\in C_{n_0}$, we obtain that $|\phi (\omega ^nz_2)|\ge \min _{C_{n_0}}|\phi (z)|>0$. Thus, the conditions of Proposition 4.8 (2) are satisfied. Hence, T satisfies the condition $Y_0$ of the Hypercyclicity Criterion for the subsequence $(r_k)$.

As a consequence, the conditions $X_0$ and $Y_0$ or the Hypercyclicity Criterion are satisfied for the sequence $(r_k)$, therefore, T is hypercyclic.

Case 2. For each $n_0$ there exists a point $z_2$ such that $\mathcal {G}$ is normal at $z_2$. We have two possibilities again. If the orbit $\{\Omega _{n_l}(z_2)\}_{l\ge 1}$ is unbounded, then there exists a subsequence $(r_k)\subset \{n_k\}$ such that $\Omega _{r_k}(z_2)\rightarrow \infty$. In a similar way, using Proposition 4.8 (2), we obtain that T satisfies the condition $Y_0$ of the Hypercyclicity Criterion for the sequence $(r_k)$. Thus, T satisfies the Hypercyclicity Criterion for the sequence $(r_k)$, which we wanted to prove.

Now, we suppose that for any subsequence $(n_k)$ the family $\mathcal {G}=\{\Omega _{n_k}\}$ is normal at $z_0$ and $\{\Omega _{n_k}(z_0)\}$ is bounded, then by applying Lemma 4.10, we get that $\Omega _{n_k}(z)$ is uniformly bounded on compact subsets of $\mathbb {D}$. Therefore, by Montel’s theorem, there exist a subsequence $(r_k)$ and an analytic function on $\mathbb {D}$, g, such that $\Omega _{r_k}\rightarrow g$ uniformly on compact subsets of $\mathbb {D}$.

Since $\Omega _n(z)=\Psi _n(\omega ^nz)$, we obtain also that $\Psi _{n_l}(z)$ is uniformly bounded on compact subsets of $\mathbb {D}$. Moreover, extracting a subsequence if it is necessary, there exist a subsequence $\{r_k\}\subset \{n_k\}$, an analytic function $g\in H(\mathbb {D})$ and $\mu \in \partial \mathbb {D}$ such that $\Omega _{r_k}(z)\rightarrow g(z)$ and $\Psi _{r_k}(z)\rightarrow g(\mu z)$ uniformly on compact subsets of $\mathbb {D}$.

Since $\Psi _{n_k}(\alpha ^l z_1)\rightarrow 0$ for all $l\ge 1$, we get that $g=0$. On the other hand, since $|\Omega _{n_k}(z_0)|>c$, we get that $g(z_0)\ne 0$. Thus, we have a contradiction. Therefore, we do not have the situation that $\Omega _{n_k}$ is uniformly bounded on compact subsets of $\mathbb {D}$, and, therefore, T is hypercyclic. $\square$

Corollary 4.12

Assume that $\lambda \in \partial \mathbb {D}$ an irrational rotation. If $|\phi (0)|\ge 1$ and $\phi$ has a zero on $\mathbb {D}$, then $R_\lambda \phi (B)$ is hypercyclic.

Proof

Indeed, if $\varphi (a)=0$ for some $a\in \mathbb {D}$, then by applying Proposition 4.9 (1), we get that $T=R_\lambda \phi (B)$ satisfies condition $X_0$ in the Hypercyclicity Criterion for the full sequence of natural numbers. If $|\phi (0)|\ge 1$, then $|\Omega _n(0)|=|\phi (0)|^n\ge 1$, therefore, by applying Theorem 4.11, we get that T is hypercyclic. $\square$

Moreover, we can obtain the same result by replacing the existence of a zero of $\phi$ by pointwise convergence to zero for some subsequence.

Corollary 4.13

Assume that $\lambda \in \partial \mathbb {D}$ is an irrational rotation. If $|\phi (0)|\ge 1$ and there exists a subsequence $(n_k)$ such that $\Psi _{n_k}(z_0)\rightarrow 0$ for some $z_0\in \mathbb {D}$, then $R_\lambda \phi (B)$ is hypercyclic.

Example 4.14

For $0< p<1$ let us consider the family of automorphisms $\varphi _p(z)$ defined by

$$\begin{aligned} \varphi _p(z)=\frac{p-z}{1-pz}. \end{aligned}$$

And we consider the family of bounded analytic functions $\psi _p(z)=\varphi _p(z)+1-p$. We see that $\psi _p(0)=1$ and $\phi _p(z)$ vanished on $\frac{2p-1}{p^2-p+1}\in \mathbb {D}$. Thus, according to Corollary 4.13, we get that $R_\lambda \psi _{p}(B)$ is hypercyclic for all $0< p<1$ and for all $\lambda \in \partial \mathbb {D}$.

If $\lambda$ is an irrational rotation and $|\phi (0)|<1$ in many cases, we can obtain supercyclicity.

Corollary 4.15

Set $\lambda \in \partial \mathbb {D}$ an irrational rotation. Assume that $|\phi (0)|<1$ and $\phi$ has a zero on $\mathbb {D}$. If T is not hypercyclic, then T is supercyclic.

Proof

Indeed, it is sufficient to choose a constant $c>0$ such that $|c\phi (0)|\ge 1$. Then, $T=R_\lambda c\phi (B)$, is an extended $\lambda$-eigenoperator of B which satisfies the hypothesis of Theorem 4.11. Therefore, cT is hypercyclic, which implies that T is supercyclic. $\square$

We will separate the case $T=R_\lambda \phi (B)$ invertible and will now proceed to analyze the question according to values of $\phi$ at the origin. In the next result, we will remove from the discussion the case in which the sequences $\{\Omega _n(z)\}$ or $\{\Psi _n(z)\}$ are not uniformly bounded on compact subsets of $\mathbb {D}$.

Theorem 4.16

Assume that $\lambda \in \partial \mathbb {D}$ an irrational rotation:

• (1) If $\{\Omega _n\}$ is not uniformly bounded on compact subsets and $|\phi (0)|<1$ for some $z_1\in \mathbb {D}$, then T is hypercyclic

• (2) If $\{1/\Phi _n\}$ is not uniformly bounded on compact subsets and $|\phi (0)|>1$, then T is hypercyclic.

Proof

Let us show 1). Firstly, if $|\phi (0)|<1$, by applying Proposition 4.9 (2), we get that T satisfies the condition $X_0$ of the Hypercyclicity Criterion for the full sequence of natural numbers.

Since $\{\Omega _n\}$ is not uniformly bounded on compact subsets of $\mathbb {D}$, then we have two possibilities:

(a) The family $\mathcal {G}=\{\Omega _n\}$ is normal at no point of $\mathbb {D}$. In this case by applying Theorems 2.3 and 2.4, there is a complex number $z_1\in \mathbb {D}$ with $\mathcal {G}$-orbit dense. In particular, we can select a subsequence $\{n_k\}$ such that $\Omega _{n_k}(z_1)\rightarrow \infty$. Moreover, using the same trick as in Lemma 4.10, we can get that $\Omega _{n_k}(\alpha ^l z_1)\rightarrow \infty$. Therefore, the condition $Y_0$ of the Hypercyclicity Criterion is satisfied for the subsequence $(n_k)$. As a consequence T satisfies the Hypercyclicity Criterion for the sequence $(n_k)$.

(b) There exists a point $z_1\in \mathbb {D}$ such that $\Omega _{n_k}(z_1)\rightarrow \infty$. In such a case, we get again that T satisfies the Hypercyclicity Criterion for the sequence $(n_k)$, and therefore T is hypercyclic as we wanted to show.

Part 2) runs similarly. Again, condition $|\phi (0)|>1$ asserts that condition $Y_0$ is satisfied for the full sequence of natural numbers.

If $\mathcal {G}'=\{(1/\Psi _n)\}$ is not uniformly bounded on compact subsets, then we have two possibilities. (a) The family $\{(1/\Psi _n)\}$ is normal at no point of $\mathbb {D}$. In this case by applying Theorems 2.3 and 2.4, there is a complex number $z_1\in \mathbb {D}$ with $\mathcal {G}'$-orbit dense. In particular, we can select a subsequence $\{n_k\}$ such that $(1/\Psi _{n_k})(z_1)\rightarrow \infty$. And b), there exists a point $z_1\in \mathbb {D}$ such that $(1/\Psi _{n_k})(z_1)\rightarrow \infty$.

In both cases, we can guarantee that $\Psi _{n_k}(z_1)\rightarrow 0$, and as before there exists a dense subset $X_0$ such that $T^{n_k}x_0\rightarrow 0$ for all $x_0\in X_0$. Therefore, the Hypercyclicity Criterion is satisfied, hence T is hypercyclic. $\square$

Remark 4.17

Let us observe that we can improve Theorem 4.16 in such a form: If there exists a subsequence $(n_k)$ such that $\Psi _{n_k}(z_1)\rightarrow 0$ for some $z_1\in \mathbb {D}$ and the sequence $\{\Omega _{n_k}\}$ is not uniformly bounded on compact subsets, then T is hypercyclic.

Analogously, we can prove that if for some subsequence $(n_k)$, $\Omega _{n_k}(z_1)\rightarrow \infty$ for some $z_1\in \mathbb {D}$ and $1/\Psi _{n_k}$ is not uniformly bounded on compact subsets, then T is hypercyclic.

Supercyclicity of extended $\lambda$-eigenoperators of the backward shift

In this section, we study if an operator that $\lambda$-commutes with the backward shift is supercyclic. For element of the commutant this result is true. Moreover, solving a question posed by Godefroy and Shapiro, V. Müller was able to prove that any non-scalar operator that commutes with a generalized backward shift is supercyclic [12]. Let us recall that a bounded linear operator $\mathcal {B}$ on a Banach space X is a generalized backward shift if it satisfies the following conditions:

• (1) The kernel of $\mathcal {B}$ is one-dimensional.

• (2) $\bigcup \{\ker \mathcal {B}^n:\ n = 0, 1, 2,...\}$ is dense in X.

For more information about generalized backward shift see [7].

It is natural to consider an operator T that $\lambda$-commutes with a generalized backward shift $\mathcal {B}$ and to try to see if T is supercyclic. We see that this property is no longer true for all elements of the $\lambda$-commutant of $\mathcal {B}$. Nonetheless, the result is true for many elements in the $\lambda$-commutant.

Proposition 5.1

Assume that $\mathcal {B}$ is a generalized backward shift on a Banach space X, and A is an extended $\lambda$-eigenoperator of $\mathcal {B}$. If $\text {ker}(A)\supset \text {ker}(\mathcal {B})$, then A is supercyclic.

Proof

Following to Godefroy-Shapiro, if $\mathcal {B}$ is a generalized backward shift, then there exists a sequence of vectors $(x_k)_{n\ge 0}$ such that $\mathcal {B}x_{n}=x_{n-1}$ for $n\ge 1$ and $\mathcal {B}x_0=0$. Moreover, if A commutes with $\mathcal {B}$, then the matrix of A relative to the basis $(x_k)_{k\ge 0}$ is upper triangular and it is constant on each superdiagonal. That is, A can be represented as a formal power series of $\mathcal {B}$.

In a similar way, following [13], if A is an extended $\lambda$-eigenoperator of $\mathcal {B}$, then A has the following matrix representation with respect to the basis $(x_k)_{k\ge 0}$. Namely, the $(p+1)$-superdiagonal of A has the form $(c_p\lambda ^n)_{n\ge 0}$, that is, $Ax_n=c_nx_0+c_{n-1}\lambda x_1+\cdots +c_0\lambda ^{n}x_n$.

Thus, if p is the first p for which $c_p\ne 0$, then A can be expressed as $A=R_{\lambda } \mathcal {B}^pA_p$ where $A_p$ is the formal power series $\phi (\mathcal {B})=c_pI+c_{p+1}\mathcal {B}+c_{p+2}\mathcal {B}^2+\cdots$ and $R_\lambda x_n=\lambda ^nx_n$. Moreover, if $\text {ker}(A)\supset \text {ker}(\mathcal {B})$, then $p\ge 1$.

Now, we mimic the proof of Proposition 3.6 in [7]. If we denote by

$$\begin{aligned} Y_{n+1}=\mathrm{{linear~span}}\{x_0,x_1,\cdots , x_n\} \end{aligned}$$

for $n\ge 0$, then $Y_{n+1}$ is invariant under $A_p$ and $A_p$ is invertible on $Y_{n+1}$. Therefore, $A_p$ is invertible on $Y=\text {linear span}\{x_n:\,n\ge 0\}$. We consider

$$\begin{aligned} C=A_p^{-1} F^p R_{1/\lambda }, \end{aligned}$$

where F is the generalized forward shift with respect to the basis $(x_k)_{k\ge 0}$. Clearly, C maps $Y_n$ into $Y_{n+p}$ and $AC=I$ on Y. Let us denote by $\sigma (n)$ the norm of C restricted to $Y_n$, then if $y\in Y_k$, then

$$\begin{aligned} \Vert C^ny\Vert= & {} \Vert CC^{n-1}y\Vert \\\le & {} \sigma (k+(n-1)p) \Vert C^{n-1}y\Vert \\\le & {} \sigma (k+(n-1)p)\sigma (k+(n-2)p)\cdots \sigma (k)\Vert y\Vert \\\le & {} (\sigma (k+(n-1)p))^n \Vert y\Vert . \end{aligned}$$

By considering $r_n=n (\sigma (n+(n-1)p))^n$ and the sequences of operators $T_n=r_nA^n$ and $S_n=\frac{1}{r_n}C^n$ we see that $T_n$ acting on vectors of $Y_n$ is eventually zero, therefore, $T_n$ converges pointwise to zero on Y. On the other hand, for each $y\in Y_k$, we get that $T_nS_n=I$ on Y and $\Vert S_ny\Vert \le \frac{1}{n}\Vert y\Vert$. Thus all requirements of the Hypercyclicity Criterion are satisfied for $T_n=r_nA^n$, which implies that A is supercyclic as we wanted to prove. $\square$

Remark 5.2

Surprisingly enough, dropping the hypothesis $\text {ker}(A)\supset \text {ker}(\mathcal {B})$, we can not assert that an operator A that $\lambda$-commutes with a generalized backward shift is supercyclic. In fact, we can find examples of extended $\lambda$-eigenoperators of $\mathcal {B}$ which are not supercyclic. That is, there is not a result analogous to Müller’s result for operators in the $\lambda$-commutant of a generalized backward shift. For example, following Chan and Shapiro (see [5]), we consider a comparison entire function $\gamma (z)=\sum _{n=0}^\infty \gamma _nz^n$, $\gamma _n>0$ and $\frac{n\gamma _{n}}{\gamma _{n-1}}$ bounded. On such considerations the differentiation operator D is bounded on the Hilbert space $E^2(\gamma )$ of entire functions $\sum a_kz^k$ satisfying $\sum _{k=0} |a_k|^2 \gamma _n^{-2}<\infty$. Moreover, clearly D is a generalized backward shift on $E^2(\gamma )$. If we consider $R_\lambda z^n=\lambda ^nz^n$, then $R_\lambda (I+D)$ with $|\lambda |=1$ and $\lambda$ a root of unity is hypercyclic on $E^2(\gamma )$. However, if $|\lambda |<1$ by the results in [8], we can obtain that $R_\lambda (I+D)$ is not supercyclic on $H(\mathbb {C})$ endowed with the compact-open topology and, therefore, $R_\lambda (I+D)$ is not supercyclic on $E^2(\gamma )$ for $|\lambda |<1$.

Now, we reduce our study to the supercyclicity of extended $\lambda$-eigenoperators of B, defined on the Hardy space $H^2(\mathbb {D})$. If T is an extended eigenoperator of B, then $T=R_\lambda \phi (B)$, where $\phi (B)$ is a formal power series.

If $\phi (0)=0$ and $|\lambda |< 1$, then $T=R_\lambda \phi (B)$ by applying Proposition 5.1, we get that T is supercyclic.

If $\phi (0)\ne 0$, we consider the operator $R_\lambda \frac{1}{\phi (0)}\phi (B)=R_\lambda \psi (B)$. If $\lambda$ is a root of the unity, since $\psi (0)=1$, by applying Proposition 4.9 (4), then $R_{\lambda }\psi (B)$ is hypercyclic, therefore, T is supercyclic.

Now, let us see that if $\phi (0)\ne 0$ and $|\lambda |<1$, then $T=R_\lambda \phi (B)$ is not supercyclic on $H^2(\mathbb {D})$.

Lemma 5.3

Assume that $\phi \in H^2(\mathbb {D})$ and $\phi (0)=1$. If $|\lambda |<1$, then the sequence $\Phi _n(z)=\phi (z)\phi (\lambda z)\cdots \phi (\lambda ^{n-1}z)\in H^2(\mathbb {D})$ for all n and there exists $h\in H^2(\mathbb {D}){\setminus } \{0\}$ such that $\Vert \Phi _n-h\Vert _2\rightarrow 0$.

Proof

Indeed, since $|\lambda |<1$, then $\phi (\lambda z)\cdots \phi (\lambda ^{n-1}z)\in H^\infty (\mathbb {D})$ for all $n\ge 1$

$$\begin{aligned} \Vert \Phi _n(z)\Vert _2=\Vert M_{\phi (\lambda z)\cdots \phi (\lambda ^{n-1}z)}\phi \Vert _2 \le \Vert \phi (\lambda z)\cdots \phi (\lambda ^{n-1}z)\Vert _\infty \Vert \phi \Vert _2. \end{aligned}$$

Hence, $\Phi _n\in H^2(\mathbb {D})$ for all $n\ge 1$. Now, let us see that the infinite product $\prod _{n\ge 1}\phi (\lambda ^{n-1}z)$ is convergent uniformly on compact subsets to some $h\in H(\mathbb {D})$. Since $\Phi _n(0)=1$ for all n, then $h\ne 0$.

Indeed, let us define $\phi _0(z)=\phi (z)-1$ and let us denote $C=\Vert \phi _0\Vert _2$:

$$\begin{aligned} |\phi ( z)\cdots \phi (\lambda ^{n-1}z)|\le & {} \frac{\Vert \Phi _n\Vert _2}{\sqrt{1-|z|^2}}\\\le & {} \frac{\Vert \phi (\lambda z)\cdots \phi (\lambda ^{n-1}z)\Vert _\infty \Vert \phi \Vert _2}{\sqrt{1-|z|^2}}\\\le & {} \frac{\prod _{n\ge 1} \Vert \phi (\lambda ^{n-1}z)\Vert _{2}}{{\sqrt{1-|z|^2}}}\\\le & {} \frac{\prod _{n\ge 1} (1+|\lambda |^{n-1}C)}{{\sqrt{1-|z|^2}}}, \end{aligned}$$

Since $|\lambda |<1$, $\sum _{n\ge 1}C|\lambda |^{n-1}$ is convergent, which gives that the infinite product $\Phi _n(z)$ converges uniformly on compact subsets to some $h\in H(\mathbb {D})$. Finally, a similar computation yields that $\Phi _n(z)\in H^2(\mathbb {D})$ is a Cauchy sequence. Indeed, if $n>m$, then

$$\begin{aligned} \Vert \Phi _n(z)-\Phi _m(z)\Vert _2= & {} \left\| \prod _{k=1}^{n} \phi (\lambda ^{k-1}z)-\prod _{k=1}^{m}\phi (\lambda ^{k-1}z)\right\| _2 \\\le & {} \left\| \prod _{k=2}^{m}\phi (\lambda ^{k-1}z)\right\| _\infty \Vert \phi \Vert _2 \left\| \prod _{k=m+1}^{n} \phi (\lambda ^{k-1}z)-1\right\| _\infty \end{aligned}$$

The results follow if we show that $h_m(z)=\prod _{n=m+1}^\infty \phi (\lambda ^{k-1}z)$ converges uniformly to 1. Since $h_m(z)$ is the remainder of a convergent infinite product, we get that $h_{m-1}(z)\rightarrow 1$ uniformly on compact subsets of $\mathbb {D}$, hence

$$\begin{aligned} \sup _{|z|=|\lambda |} |h_{m-1}(z)|=\sup _{|z|=1} |h_m(z)|\rightarrow 1 \end{aligned}$$

as we wanted. Therefore, $\Vert \Phi _n-h\Vert _2\rightarrow 0$ as $n\rightarrow \infty$ as we desired. $\square$

Theorem 5.4

Assume that $|\lambda |<1$ and $T=R_\lambda \phi (B)$ is an extended $\lambda$-eigenoperator of B. If $\phi (0)\ne 0$, then T is not supercyclic.

Proof

Indeed, let us denote by $H(\mathbb {D})$ the space of analytic functions in the unit disk provided with the uniform convergence on compact subsets of the unit disk.

Since $H^2(\mathbb {D})\hookrightarrow H(\mathbb {D})$, if T is hypercyclic or supercyclic on $H^2(\mathbb {D})$, then the sequence of operators $T^n: H^2(\mathbb {D})\rightarrow H(\mathbb {D})$ is hypercyclic or supercyclic on $H(\mathbb {D})$. Let us remark that the operator $T=R_\lambda \phi (B)$ could be not continuous on $H(\mathbb {D})$, but we do not need such requirements.

We will see that the projective orbit $\{\lambda T^n(f):\,\lambda \in \mathbb {D}, n\in \mathbb {N}\}$ of some vector $f\in H^2(\mathbb {D})$ is not dense in $H(\mathbb {D})$ endowed with the compact open topology.

Without loss of generality, we can suppose that $a_0=\phi (0)=1$. Let us turn our attention to the following function:

$$\begin{aligned} f_0(z)=\frac{1}{1-z}=\sum _{k=0}^\infty z^k \in H(\mathbb {D}). \end{aligned}$$

Clearly, $f_0\in H(\mathbb {D})$ and $f_0$ is an eigenvector of B associated to the eigenvalue 1, that is, $Bf_0=f_0$.

By contradiction, suppose that there exists a sequence of complex numbers $(\lambda _n)$ and a function $f\in H^2(\mathbb{D})$ such that the sequence of functions $(\lambda _n T^nf)$ is dense in $H(\mathbb {D})$ endowed with the compact-open topology. In particular, there will exist a sequence $(n_k)$ such that $\lambda _{n_k}T^{n_k}f\rightarrow f_0$ uniformly on compact subsets of $\mathbb {D}$.

Since B is continuous on $H(\mathbb {D})$ and $Bf_0=f_0$, then

$$\begin{aligned} B^m(\lambda _{n_k}T^{n_k}f)\rightarrow f_0=B^mf_0 \end{aligned}$$

uniformly on compact subsets of $\mathbb {D}$. Hence, the quotient:

$$\begin{aligned} \frac{\lambda _{n_k}B^mT^{n_k}f(z)}{\lambda _{n_k}(T^{n_k}f)(0)}=\frac{B^mT^{n_k}f(z)}{(T^{n_k}f)(0)}\rightarrow f_0(z) \end{aligned}$$

(3)

uniformly on compact subsets of $\mathbb {D}$.

   We will show that there exists $m\in \mathbb {N}$ such that for any $z\in \mathbb {D}$ the quotients:

$$\begin{aligned} \frac{B^mT^{n}f(z)}{(T^{n}f)(0)}\rightarrow 0 \end{aligned}$$

as $n\rightarrow \infty$, obtaining a contradiction.

Indeed, since $B^mT=\lambda ^m TB^m$, we get

$$\begin{aligned} B^mT^nf=\lambda ^{nm}T^nB^mf. \end{aligned}$$

By hypothesis $f\in H^2(\mathbb {D})$ and $T^nB^m$ is bounded on $H^2(\mathbb {D})$, therefore, $T^nB^mf\in H^2(\mathbb {D})$. We can suppose without loss of generality that $\Vert f\Vert =1$. Hence, since $\Vert B\Vert =1$, then

$$\begin{aligned} |\lambda ^{nm}T^nB^mf(z)|\le & {} \frac{\Vert T^nB^mf\Vert }{\sqrt{1-|z|^2}} \end{aligned}$$

(4)

$$\begin{aligned}\le & {} \frac{|\lambda |^{nm}\Vert T\Vert ^n}{\sqrt{1-|z|^2}}. \end{aligned}$$

(5)

Here $\Vert T\Vert$ denotes the uniform norm of T as linear operator on the space $H^2(\mathbb {D})$.

Let us obtain inferior estimates of the sequence $|(T^{n}f)(0)|$. Recall that $T^n=R_\lambda \phi (B)\cdots R_\lambda \phi (B)=R_{\lambda }^n\phi (B)\cdots \phi (\lambda ^{n-1}B)$. On the other hand, since $|\lambda |<1$ the sequence $\Phi _n\rightarrow h\ne 0$ on $H^2(\mathbb {D})$. Since the set of supercyclic vectors is dense, we can suppose without loss of generality that $\langle f,h\rangle \ne 0$.

Thus,

$$\begin{aligned} |(T^{n}f)(0)|= & {} |\langle T^{n}f,1 \rangle | \\= & {} |\langle f, M_{\overline{\phi }(z)\cdots \overline{\phi }(\lambda ^{n-1}z)} R_{\lambda ^n}^\star 1\rangle | \\= & {} |\langle f,\overline{\phi }(z)\cdots \overline{\phi }(\lambda ^{n-1}z)\rangle |\\\rightarrow & {} \langle f,h \rangle \ne 0 \end{aligned}$$

as $n\rightarrow \infty$. Using this fact, the Eqs. 3, 5, and by taking m such that $|\lambda |^m \cdot \Vert T\Vert <1$, we get

$$\begin{aligned} \frac{\lambda _{n_k}B^mT^{n_k}f(z)}{\lambda _{n_k}(T^{n_k}f)(0)}= & {} \frac{B^mT^{n_k}f(z)}{(T^{n_k}f)(0)}\\\le & {} \frac{\frac{|\lambda |^{nm}\Vert T\Vert ^n}{\sqrt{1-|z|^2}}}{|(T^{n}f)(0)|}\rightarrow 0 \end{aligned}$$

as $n\rightarrow \infty$. Therefore, f cannot be supercyclic for T as we desired to prove. $\square$

Concluding remarks

The results we have obtained here especially Theorems 4.11 and 4.16, indicate there is a dichotomy theorem waiting to be proved that establishes the elements of the $\lambda$-commutant of B are either hypercyclic or they have quite regular orbits (that is, either they converge to zero or the orbits of the inverse operator converge to zero).

Our results on the hypercyclicity problem of the element of the $\lambda$-commutant are as follows. $R_\lambda \phi (B)$ is connected with the shape of the spectrum of $\phi (B)$ and it remains a mysterious characterization the hypercyclicity of $R_\lambda \phi (B)$ in terms of the geometry of the spectrum of $\phi (B)$.

In any case, we see that it is particularly striking when we are dealing the $\lambda$-commutant hypercyclicity problem for Banach space operators. We see that the existence of a uniform norm of the operator makes the transference of the hypercyclicity difficult. In strong contrast, in Fréchet spaces such transference is easier (see [2, 11]).