# Hypercyclicity of operators that $\lambda$-commute with the Hardy backward
shift

Mohamed Amouch, Fernando León-Saavedra, and M.P. Romero de la Rosa  Mohamed
Amouch , University Chouaib Doukkali. Department of Mathematics, Faculty of
science Eljadida, Morocco  Fernando León-Saavedra. Universidad de Cádiz,
Departamento de Matemáticas, Spain  M.P Romero de la Rosa. Universidad de
Cádiz, Departamento de Matemáticas, Spain<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

An operator $T$ acting on a separable complex Hilbert space $H$ is said to be
hypercyclic if there exists $f\in H$ such that the orbit $\\{T^{n}f:\
n\in\mathbb{N}\\}$ is dense in $H$. Godefroy and Shapiro [7] characterized
those elements in the commutant of the Hardy backward shift which are
hypercyclic. In this paper we study some dynamics properties of operators $X$
that $\lambda$-commute with the Hardy backward shift $B$, that is, $BX=\lambda
XB$.

###### Key words and phrases:

Space of entire functions, Shift operator, Extended eigenoperators,
Hypercyclic operators.

###### 2010 Mathematics Subject Classification:

Primary 47A16, 37B20; Secondary 46E50, 46T25

##  1\. 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 [14] and [7]. In 1969 Rolewicz [14] show
up the first example hypercyclic operator defined on a Banach space. If we
considered $T=\lambda B$ 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 operators are
somehow transferred to the commutator of the operator. Suprisingly, 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 introduce 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 $X\neq 0$
(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 por 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 to try 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 was studied in [2, 8, 11]. In this note, we wish to explore the
dynamic 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
$\mathcal{O}(x,T)$:

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

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. (1)

The kernel of $\mathcal{B}$ is one dimensional.

  2. (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, it founds some serious dificulties to
be transferred to the $\lambda$-commutant. For instance, V. 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 results is not 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 $\lambda$-commutes 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 S. Petrovic in [13] and it is a 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 factorize 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. But
the converse is not true. In strong contrast the with the results in [2], the
converse is not true, a hypercyclic operator $\phi(B)$ in the commutant of $B$
in general it does not induce a hypercyclic extended $\lambda$-eigenoperator.
The problem depends on the instrinsic 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 not 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.

##  2\. Some tools

We will use the following version of the hypercyclicity Criterion formulated
by J. Bès and A. 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. (1)

$T^{n_{k}}\rightarrow 0$ pointwise on $X_{0}$;

  2. (2)

$S_{n_{k}}\rightarrow 0$ pointwise on $Y_{0}$;

  3. (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
$\overline{g}(z)=\overline{g(\overline{z})}\in H^{\infty}(\mathbb{D})$. By
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 $\phi(\mathbb{D})$ meet 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 $\|(M_{\phi}^{\star})^{n}f\|\leq M$ are bounded (respectively the orbit
of $f$ under $(M_{\phi}^{\star})^{-1}$ 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 $\phi_{n}(U)\cap V\neq\emptyset$.

######  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 $\\{\phi_{k}(x)\\}_{k\geq 1}$.

The following version of Montel’s theorem in the practice can 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

$\bigcup_{f\in\mathcal{F}}f(U)$

is dense for any non-empty neighbourhood $U$ of $z_{0}$.

##  3\. 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 show that if $\lambda=0$ then $X1=c_{0}\neq 0$ and
$Xz^{n}=0$ for any $n\geq 1$. That is, $X$ is a projection over the constant
functions and therefore it is not hypercyclic.

As a consequence of a result by S. 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 factorize $X$
when it exists.

######  Theorem 3.1.

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

  1. (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. (2)

If $|\lambda|<1$ then 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 show 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

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

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 S. Petrovic (see [13]). Specifically,
Petrovic proved that if $X$ is an extended $\lambda$-eigenoperator of $B$
($|\lambda|<1$) then

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

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
$R_{\overline{\lambda}}f$ send $H^{2}(\mathbb{D})$ on
$H^{\infty}(\mathbb{D})$. Thus, since $R_{\overline{\lambda}}f\in
H^{\infty}(\mathbb{D})$ we get:

(1)
$\|M_{\overline{\phi}}R_{\overline{\lambda}}f\|_{2}^{2}\leq\|\phi\|_{2}^{2}\max_{|w|=|\lambda|}|f(w)|^{2}$

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

(2) $|f(w_{0})|^{2}\leq
C\frac{1}{2\pi}\int_{0}^{2\pi}|f(re^{i\theta}|^{2}\,d\theta.$

Thus, using inequality (2) into inequality (1) we get

$\|M_{\overline{\phi}}R_{\overline{\lambda}}f\|_{2}^{2}\leq
C\|\phi\|_{2}^{2}\|f\|_{2}^{2},$

which gives the desired result.

∎

##  4\. Hypercyclicity of extended $\lambda$-eigenoperators of the backward
shift.

In this section we study when a $\lambda$-extended 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.

###  4.1. 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 $\phi\in
H^{\infty}(\mathbb{D})$. Then, 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
$\|T\|\leq\|R_{\lambda}\|\|\phi\|_{\infty}$ 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 $\|T^{-1}\|\leq 1$, therefore $T^{-1}$ is
not hypercyclic. Hence $T$ is not hypercyclic.∎

Assume that $T=R_{\lambda}\phi(B)$, with $|\lambda|=1$ and $\phi(B)$
hypercyclic. In strong contrast with the results obtained in [2], if $\phi(B)$
is hypercyclicity 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

$\phi_{0}(z)=\frac{1}{2}z+1-\frac{1}{10}.$

and

$\phi_{1}(z)=\frac{1}{2}z+1-\frac{1}{100}.$

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 Figure 1).

Figure 1. The image $\phi_{0}(\mathbb{D})$ and $\phi_{1}(\mathbb{D})$.

Thus $\phi_{0}(\mathbb{D})\cap\partial\mathbb{D}\neq\emptyset$ and
$\phi_{1}(\mathbb{D})\cap\partial\mathbb{D}\neq\emptyset$, therefore
$\phi_{0}(B)$ and $\phi_{1}(B)$ are hypercyclic.

Let us see 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
$\Phi_{1}(\mathbb{D})\cap\partial\mathbb{D}\neq\emptyset$ (see Figure 2).

Figure 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 of the
shape of the point spectrum of $\phi(B)$. This shape is intimatelly 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. (1)

If $\|\Phi_{n}\|_{\infty}<M$ for all $n$ then $T$ is not hypercyclic.

  2. (2)

If there exists $c>0$ such that $c\leq Inf_{z\in\mathbb{D}}|\Phi_{n}(z)|$ for
all $n\geq n_{0}$ for some $n_{0}$, then $T=R_{\lambda}\phi(B)$ is not
hypercyclic.

###### Proof.

To show (1), let us see that

$\displaystyle\|(R_{\lambda}\phi(B))^{n}\|$ $\displaystyle=$
$\displaystyle|1\cdot\lambda\cdots\lambda^{n-1}|\|R_{\lambda}^{n}\|\|\Phi_{n}(B)\|$
$\displaystyle\leq$ $\displaystyle\|\Phi_{n}\|_{\infty}\leq M,$

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 $n\geq n_{0}$. 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, and easy check
show that $\|T^{-n}\|\leq\frac{1}{c}$ for all $n\geq n_{0}$. Therefore
$T^{-n}$ with $n\geq n_{0}$, is not hypercyclic. Hence $T$ is not hypercyclic.
∎

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
$\Phi(\mathbb{D})\cap\partial\mathbb{D}\neq\emptyset.$

###### 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
$\Phi(\mathbb{D})\cap\partial\mathbb{D}\neq\emptyset$. Hence, using Ansari’s
result we get that $T$ is hypercyclic if and only if
$\Phi(\mathbb{D})\cap\partial\mathbb{D}\neq\emptyset$ as we desired.

∎

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

$\Psi_{n}(z)=\overline{\phi}(z)\cdot\overline{\phi}(\alpha
z)\cdots\overline{\phi}(\alpha^{n-1}z)$

and

$\Omega_{n}(z)=\overline{\phi}(\omega z)\cdots\overline{\phi}(\omega^{n}z).$

In fact, the dynamics of both sequences are related by the formula
$\Omega_{n}(z)=\Psi_{n}(\omega^{n}z)$. The reader keep in mind that
$\Psi_{n}(z)\to\infty$ (resp. $0$) if and only if
$\overline{\Psi_{n}(z})\to\infty$ (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

$C_{0}=\\{z\in\mathbb{D}\,:\,\Psi_{n_{k}}(z)\to 0\\}$

and

$C_{1}=\\{z\in\mathbb{D}\,:\,\Omega_{n_{k}}(z)\to\infty\\}$

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 Hypercyclicy Criterion is satisfied by selecting
the dense subsets

$X_{0}=\textrm{linear span}\\{k_{z_{0}}(z)\,:\,z_{0}\in C_{0}\\}$

and

$Y_{0}=\textrm{linear span}\\{k_{z_{0}}(z)\,:\,z_{0}\in C_{1}\\}.$

Both subsets $X_{0},Y_{0}$ are dense and

$T^{n_{k}}k_{z_{0}}(z)=\overline{\Psi_{n_{k}}(z_{0})}k_{\alpha^{n_{k}}z_{0}}\to
0$

pointwise on $X_{0}$. On the other hand, if $a\in C_{1}$ then
$\overline{\phi}(\omega^{n}a)\neq 0$ we can define

$Sk_{a}(z)=\frac{1}{\overline{\overline{\phi}(\omega a)}}k_{\omega a}(z)$

then $S$ is well defined on $Y_{0}$ and $S^{n_{k}}$ converges pointwise to $0$
on $Y_{0}$, then $T$ is hypercyclic. ∎

The above result is a natural way to apply the Hypercyclicity Criterion.
However, the hyphotesis are 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. (1)

If there exist $z_{0}\in\mathbb{D}$ and a subsequence
$(n_{k})\subset\mathbb{N}$ such that $\Psi_{n_{k}}(z_{0})\to 0$, then
$T^{n_{k}}$ converge pointwise to zero on a dense subset.

  2. (2)

If there exists $z_{1}\in\mathbb{D}$ such that
$\Omega_{n_{k}}(z_{1})\to\infty$ and there exists $m>0$ such that
$|\phi(\omega^{n}z_{0})|>M>0$ for all $n\geq 1$, 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

$X_{0}=\textrm{linear
span}\\{k_{\omega^{n}z_{0}}(z)\,\,:\,\,n\in\mathbb{N}\\}$

Since $\omega$ is an irrational rotation, the subset $\\{\omega^{n}z_{0}\\}$
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

$T^{n}k_{\omega^{k_{0}}z_{0}}(z)=0.$

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 $\overline{\phi}(\alpha^{n}z_{0})\neq 0$ for all $n\in\mathbb{N}$ then we
consider the subset

$X_{0}=\textrm{linear span}\\{k_{\alpha^{n}z_{0}}(z)\,:\,n\geq 1\\}.$

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

$\displaystyle T^{n_{k}}k_{\alpha^{n_{0}}z_{0}}(z)$ $\displaystyle=$
$\displaystyle\overline{\overline{\phi}(\alpha^{n_{0}}z_{0})\cdots\overline{\phi}(\alpha^{n_{0}+n_{k}-1}z_{0})}k_{\alpha^{n_{0}+n_{k}}z_{0}}(z)$
$\displaystyle=$
$\displaystyle\frac{\overline{\Phi_{n_{k}}(z_{0})}\overline{\overline{\phi}(\alpha^{n_{k}}z_{0})\cdots\overline{\phi}(\alpha^{n_{0}+n_{k}-1}z_{0})}}{\overline{\overline{\phi}(z_{0})\cdots\overline{\phi}(\alpha^{n_{0}-1}z_{0})}}k_{\alpha^{n_{0}+n_{k}}z_{0}}(z),$

Thus,

$\|T^{n_{k}}k_{\alpha^{n_{0}}z_{0}}(z)\|\leq
C|\Phi_{n_{k}}(z_{0})|\|\phi\|_{\infty}^{n_{0}}.$

We obtain again that $T^{n_{k}}$ converges pointwise to zero on $X_{0}$ as
$k\to\infty$. This fact proves (1).

To show (2), we consider the following subset

$Y_{0}=\textrm{linear span}\\{k_{\omega^{n}z_{1}}(z)\,:\,n\geq 1\\}$

which is clearly dense because the sequence $\omega^{n}z_{1}$ has an
accumulation point in $\mathbb{D}$. Next we will see that on $Y_{0}$ the
following map

$Sk_{a}(z)=\frac{1}{\overline{\overline{\phi}(\omega a)}}k_{\omega a}(z)$

is well defined, because $\Omega_{n_{k}}(z_{1})\to\infty$ therefore we get
that $\overline{\phi}(\omega^{n}z_{1})\neq 0$ for all $n$.

On the other hand

$\displaystyle S^{n_{k}}k_{\omega^{m_{0}}z_{1}}$ $\displaystyle=$
$\displaystyle\frac{1}{\overline{\overline{\phi}(\omega^{m_{0}+1}z_{1})}\cdots\overline{\overline{\phi}(\omega^{m_{k}+m_{0}}z_{1})}}k_{\omega^{m_{k}+m_{0}}z_{1}}(z)$
$\displaystyle=$ $\displaystyle\frac{\overline{\overline{\phi}(\omega
z_{1})}\cdots\overline{\overline{\phi}(\omega^{m_{0}}z_{1})}}{\overline{\Omega_{n_{k}}(z_{1})}\overline{\overline{\phi}(\omega^{m_{k}+1}z_{1})}\cdots\overline{\overline{\phi}(\omega^{m_{k}+m_{0}}z_{1})}},$

Therefore

$\|S^{n_{k}}k_{\omega^{m_{0}}z_{1}}\|\leq\frac{\|\varphi\|_{\infty}^{m_{0}}}{M^{m_{0}}|\Omega_{n_{k}}(z_{1})|}\to
0$

as $k\to\infty$, as we wanted. ∎

From the above results, now we wish to obtain some geometrical sufficient
conditions to guarantee hypercyclicity. Since we will apply the Hypercyclicity
Criterion, we will look sufficient conditions to obtain separately the
conditions of the Hypercyclicity Criterion. We will say that $T$ satisfies the
condition $X_{0}$ (respectively $Y_{0}$) if $T$ satisfies the condition (1)
(respectively (2)) of the Hypercyclicity Criterion.

######  Proposition 4.9.

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

  1. (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. (2)

If $|\phi(0)|<1$ then $T$ satisfies the condition $X_{0}$ is satisfied for the
full sequence of natural numbers.

  3. (3)

If $|\phi(0)|>1$ then $T$ satisfies the condition $Y_{0}$ for the full
sequence of natural numbers.

  4. (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 $z_{0}\neq 0$, since $0=\phi(z_{0})=\overline{\phi}(\overline{z_{0}})$
considering the following dense subset:

$X_{0}=\textrm{linear span}\\{k_{\omega^{n}\overline{z_{0}}}(z)\,:\,n\geq
1\\},$

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:

$X_{0}=\textrm{linear span}\\{k_{\overline{z_{0}}}(z)\,\,:\,\,z_{0}\in
D(0,\delta)\\}.$

In a similar way 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:

$Y_{0}=\textrm{linear span}\\{k_{\overline{z_{0}}}(z)\,;\,z_{0}\in
D(0,\delta)\\}.$

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 in the image $\Phi(\mathbb{D})$ we get that
$\Phi(\mathbb{D})\cap\partial\mathbb{D}\neq\emptyset$, therefore by Godefroy
and Shapiro’s result we get that $T$ satisfies the Hypercyclicity Criterion
for the full sequence of natural numbers. ∎

###  4.2. 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, can select $r_{n}<1$ such that $r_{n}\to 1$ and $\varepsilon_{n}>0$
such that for all $z$ satisfying $r_{n}-\varepsilon_{n}<|z|<r+\varepsilon_{n}$
we can get $\phi(z)\neq 0$. 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 shown that $\mathcal{G}$
is uniformly bounded on $|z|\leq r_{n}$. Indeed, since $\omega$ is an
irrational rotation, by compactness there is an integer $n_{0}$ such that

$\partial D(0,|z_{0}|)\subset B_{0}\cup\omega
B_{0}\cup\cdots\cup\omega^{n}_{0}B_{0}.$

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 modulus maximum 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 $\omega^{l}B_{0}$, $1\leq l\leq n_{0}$. Indeed each element in
$\omega^{l}B_{0}$ have the form $\omega^{l}z$ with $z\in B_{0}$. Thus,

$\displaystyle|\Omega_{n_{k}}(\omega^{l}z)|$ $\displaystyle=$
$\displaystyle|\overline{\phi}(\omega^{l+1}z)\cdots\overline{\phi}(\omega^{l+n_{k}}z)|$
$\displaystyle=$
$\displaystyle\frac{|\Omega_{n_{k}}(z)||\overline{\phi}(\omega^{n_{k}+1}z)\cdots\overline{\phi}(\omega^{n_{k}+l}z)|}{|\overline{\phi}(\omega
z)\cdots\overline{\phi}(\omega^{l}z)|}$ $\displaystyle\leq$ $\displaystyle
MC^{l},$

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. ∎

######  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
$\Psi_{n_{k}}(z_{1})\to 0$, then $T=R_{\lambda}\phi(B)$ is hypercyclic.

###### Proof.

First of all, if $\phi$ vanishes on $\mathbb{D}$, the by applying we get that
$T$ satisfies condition $X_{0}$ for the full sequence of natural numbers.

Let us assume that $\phi$ does not vanishes on $\mathbb{D}$. By hyphotesis
Using the same trick as in Lemma 4.10, we can get that
$\Psi_{n_{k}}(\alpha^{l}z_{1})\to 0$. 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:

$\displaystyle|\Psi_{n_{k}}(\alpha^{l}z_{1})|$ $\displaystyle=$
$\displaystyle|\overline{\phi}(\alpha^{l}z_{1})\cdots\overline{\phi}(\alpha^{l+n_{k}-1}z_{1})|$
$\displaystyle=$
$\displaystyle\frac{|\Psi_{n_{k}}(z_{1})||\overline{\phi}(\alpha^{n_{k}}z_{1})\cdots\overline{\phi}(\alpha^{n_{k}+l-1}z_{1})|}{|\overline{\phi}(z_{1})\cdots\overline{\phi}(\alpha^{l-1}z_{1})|}$
$\displaystyle\leq$ $\displaystyle|\Psi_{n_{k}}(z_{1})|C^{l}\to 0.$

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
$\\{\Omega_{n_{l}}(z_{2})\\}_{l\geq 1}$ is dense in $\mathbb{C}$. In
particular, there exists a subsequence $\\{r_{k}\\}\subset\\{n_{k}\\}$ such
that $\Omega_{r_{k}}(z_{2})\to\infty$. On the other hand, since $z_{2}\in
C_{n_{0}}$, we obtain that
$|\phi(\omega^{n}z_{2})|\geq\min_{C_{n_{0}}}|\phi(z)|>0$. 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 satisfy 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
$\\{\Omega_{n_{l}}(z_{2})\\}_{l\geq 1}$ is unbounded, then there exist a
subsequence $(r_{k})\subset\\{n_{k}\\}$ such that
$\Omega_{r_{k}}(z_{2})\to\infty$. 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
$\Omega_{r_{k}}\to g$ uniformly on compact subsets of $\mathbb{D}$.

Since $\Omega_{n}(z)=\Psi_{n}(\omega^{n}z)$, 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 $\Omega_{r_{k}}(z)\to
g(z)$ and $\Psi_{r_{k}}(z)\to g(\mu z)$ uniformly on compact subsets of
$\mathbb{D}$.

Since $\Psi_{n_{k}}(\alpha^{l}z_{1})\to 0$ for all $l\geq 1$, we get that
$g=0$. On the other hand, since $|\Omega_{n_{k}}(z_{0})|>c$, we get that
$g(z_{0})\neq 0$. A contradiction. Therefore, we don’t have the situation that
$\Omega_{n_{k}}$ is uniformly bounded on compact subsets of $\mathbb{D}$, and
therefore $T$ is hypercyclic. ∎

######  Corollary 4.12.

Assume that $\lambda\in\partial\mathbb{D}$ is an irrational rotation. If
$|\phi(0)|\geq 1$ 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 $|\phi(0)|\geq 1$ then $|\Omega_{n}(0)|=|\phi(0)|^{n}\geq 1$,
therefore by appying Theorem 4.11, we get that $T$ is hypercyclic. ∎

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
$|\phi(0)|\geq 1$ and there exists a subsequence $(n_{k})$ such that
$\Psi_{n_{k}}(z_{0})\to 0$ 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

$\varphi_{p}(z)=\frac{p-z}{1-pz}.$

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. If $|\phi(0)|<1$
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 $|c\phi(0)|\geq
1$. 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. ∎

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. (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. (2)

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

###### Proof.

Let us to 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
$\Omega_{n_{k}}(z_{1})\to\infty$. Moreover, using the same trick as in Lemma
4.10, we can get that $\Omega_{n_{k}}(\alpha^{l}z_{1})\to\infty$. Therefore,
the condition $Y_{0}$ of the hypercyclicity criterion is satisfies 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
$\Omega_{n_{k}}(z_{1})\to\infty$. 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) run similar. Again, condition $|\phi(0)|>1$ assert that condition
$Y_{0}$ is satisfied for the full sequence of natural numbers.

If $\mathcal{G}^{\prime}=\\{(1/\Psi_{n})\\}$ 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 $\mathcal{G}^{\prime}$-orbit dense. In particular, we can select a
subsequence $\\{n_{k}\\}$ such that $(1/\Psi_{n_{k}})(z_{1})\to\infty$. And
b), there exists a point $z_{1}\in\mathbb{D}$ such that
$(1/\Psi_{n_{k}})(z_{1})\to\infty$.

In both cases, we can guarantee that $\Psi_{n_{k}}(z_{1})\to 0$, and as before
there exists a dense subsets $X_{0}$ such that $T^{n_{k}}x_{0}\to 0$ for all
$x_{0}\in X_{0}$. Therefore the hypercyclicity criterion is satisfied, hence
$T$ is hypercyclic. ∎

######  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 $\Psi_{n_{k}}(z_{1})\to 0$ 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 $(n_{k})$,
$\Omega_{n_{k}}(z_{1})\to\infty$ for some $z_{1}\in\mathbb{D}$ and
$1/\Psi_{n_{k}}$ is not uniformly bounded on compact subsets, then $T$ is
hypercyclic.

##  5\. Supercyclicity of extended $\lambda$-eigenoperators of the backward
shift.

In this section we study if and operator that $\lambda$-commute 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. (1)

The kernel of $\mathcal{B}$ is one dimensional.

  2. (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 $\lambda$-commute with a
generalized backward shift $\mathcal{B}$ and to try to see if $T$ is
supercyclic. We see that this property is not 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 a $\lambda$-extended eigenoperator of $\mathcal{B}$. If
$\textrm{ker}(A)\supset\textrm{ker}(\mathcal{B})$ then $A$ is supercyclic.

###### Proof.

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

In a similar way, following [13], if $A$ is an $\lambda$-extended
eigenoperator of $\mathcal{B}$ then $A$ has the following matrix
representation with respect to the basis $(x_{k})_{k\geq 0}$, namelly, the
$(p+1)$ superdiagonal of $A$ has the form $(c_{p}\lambda^{n})_{n\geq 0}$. That
is, $Ax_{n}=c_{n}x_{0}+c_{n-1}\lambda x_{1}+\cdots+c_{0}\lambda^{n}x_{n}$.

Thus, if $p$ is the first $p$ for which $c_{p}\neq 0$ then $A$ can be expresed
as $A=R_{\lambda}B^{p}A_{p}$ where $A_{p}$ is the formal power series
$\phi(\mathcal{B})=c_{p}I+c_{p+1}\mathcal{B}+c_{p+2}\mathcal{B}^{2}+\cdots$
and $R_{\lambda}x_{n}=\lambda^{n}x_{n}$. Moreover, if
$\textrm{ker}(A)\supset\textrm{ker}(\mathcal{B})$ then $p\geq 1$.

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

$Y_{n+1}=\textrm{linear span}\\{x_{0},x_{1},\cdots,x_{n}\\}$

for $n\geq 0$, then $Y_{n+1}$ is invariant under $A_{p}$ and $A_{p}$ is
invertible on $Y_{n+1}$. Therefore $A_{p}$ is invertible on $Y=\textrm{linear
span}\\{x_{n}\,:\,n\geq 0\\}$. We consider

$C=A_{p}^{-1}F^{p}R_{1/\lambda},$

where $F$ is the generalized forward shift with respect to the basis
$(x_{k})_{k\geq 0}$. 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

$\displaystyle\|C^{n}y\|$ $\displaystyle=$ $\displaystyle\|CC^{n-1}y\|$
$\displaystyle\leq$ $\displaystyle\sigma(k+(n-1)p)\|C^{n-1}y\|$
$\displaystyle\leq$
$\displaystyle\sigma(k+(n-1)p)\sigma(k+(n-2)p)\cdots\sigma(k)\|y\|$
$\displaystyle\leq$ $\displaystyle(\sigma(k+(n-1)p))^{n}\|y\|.$

By considering $r_{n}=n(\sigma(n+(n-1)p))^{n}$ and the sequences of operators
$T_{n}=r_{n}A^{n}$ 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
$T_{n}S_{n}=I$ on $Y$ and $\|S_{n}y\|\leq\frac{1}{n}\|y\|$. Thus all
requeriments of the hypercyclicity criterion are satisfied for
$T_{n}=r_{n}A^{n}$, which implies that $A$ is supercyclic as we wanted to
prove. ∎

######  Remark 5.2.

Suprisingly enough, dropping the hypothesis
$\textrm{ker}(A)\supset\textrm{ker}(\mathcal{B})$, we can not assert that an
operator $A$ that $\lambda$-commute 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
$\gamma(z)=\sum_{n=0}^{\infty}\gamma_{n}z^{n}$, $\gamma_{n}>0$ and
$\frac{n\gamma_{n}}{\gamma_{n-1}}$ bounded. On such considerations the
differentiation operator $D$ is bounded on the the Hilbert space
$E^{2}(\gamma)$ of entire functions $\sum a_{k}z^{k}$ 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
$R_{\lambda}z^{n}=\lambda^{n}z^{n}$, then $R_{\lambda}(I+D)$ with
$|\lambda|=1$ and $\lambda$ a root of unity, is hypercyclic on $E^{\gamma}$.
However if $|\lambda|<1$ by the results in [8] we can obtain that
$R_{\lambda}(I+D)$ 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 $\phi(0)\neq 0$, 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 $\phi(0)\neq 0$ 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 $\|\Phi_{n}-h\|_{2}\to 0$.

###### Proof.

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

$\|\Phi_{n}(z)\|_{2}=\|M_{\phi(\lambda
z)\cdots\phi(\lambda^{n-1}z)}\phi\|_{2}\leq\|\phi(\lambda
z)\cdots\phi(\lambda^{n-1}z)\|_{\infty}\|\phi\|_{2}.$

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

Indeed, set $\phi(z)=1+\phi_{0}$ and $C=\|\phi_{0}\|_{2}$:

$\displaystyle|\phi(z)\cdots\phi(\lambda^{n-1}z)|$ $\displaystyle\leq$
$\displaystyle\frac{\|\Phi_{n}\|_{2}}{\sqrt{1-|z|^{2}}}$ $\displaystyle\leq$
$\displaystyle\frac{\|\phi(\lambda
z)\cdots\phi(\lambda^{n-1}z)\|_{\infty}\|\phi\|_{2}}{\sqrt{1-|z|^{2}}}$
$\displaystyle\leq$ $\displaystyle\frac{\prod_{n\geq
1}\|\phi(\lambda^{n-1}z)\|_{2}}{{\sqrt{1-|z|^{2}}}}$ $\displaystyle\leq$
$\displaystyle\frac{\prod_{n\geq 1}(1+|\lambda|^{n-1}C)}{{\sqrt{1-|z|^{2}}}},$

Since $|\lambda|<1$, $\sum_{n\geq 1}C|\lambda|^{n-1}$ 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

$\displaystyle\|\Phi_{n}(z)-\Phi_{m}(z)\|_{2}$ $\displaystyle=$
$\displaystyle\left\|\prod_{k=1}^{n}\phi(\lambda^{k-1}z)-\prod_{k=1}^{m}\phi(\lambda^{k-1}z)\right\|_{2}$
$\displaystyle\leq$
$\displaystyle\left\|\prod_{k=2}^{m}\phi(\lambda^{k-1}z)\right\|_{\infty}\|\phi\|_{2}\left\|\prod_{k=m+1}^{n}\phi(\lambda^{k-1}z)-1\right\|_{\infty}$

The results follows 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 $h_{m-1}(z)\to 1$ uniformly on compact subsets of $\mathbb{D}$, hence

$\sup_{|z|=|\lambda|}|h_{m-1}(z)|=\sup_{|z|=1}|h_{m}(z)|\to 1$

as we wanted. Therefore $\|\Phi_{n}-h\|_{2}\to 0$ as $n\to\infty$ as we
desired.

∎

######  Theorem 5.4.

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

###### Proof.

Indeed, let us denote by $H(\mathbb{D})$ the space of analytic funcions 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
$T^{n}:H^{2}(\mathbb{D})\to H(\mathbb{D})$ 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 don’t need such requeriments.

We will see that the projective orbit $\\{\lambda
T^{n}(f)\,:\,\lambda\in\mathbb{D},n\in\mathbb{N}\\}$ of some vector $f\in
H^{2}$ 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 turn our
attention to the following function:

$f_{0}(z)=\frac{1}{1-z}=\sum_{k=0}^{\infty}z^{k}\in H(\mathbb{D}).$

Clearly $f_{0}\in H(\mathbb{D})$ and $f_{0}$ is an eigenfunction 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 $f\in H^{2}$ such that the sequence of
functions $(\lambda_{n}T^{n}f)$ is dense in $H(\mathbb{D})$ endowed with the
compact-open topology. In particular it will exists a sequence $(n_{k})$ such
that $\lambda_{n_{k}}T^{n_{k}}f\to f_{0}$ uniformly on compact subsets of
$\mathbb{D}$.

Since $B$ is continuos on $H(\mathbb{D})$ and $Bf_{0}=f_{0}$ then

$B^{m}(\lambda_{n_{k}}T^{n_{k}}f)\to f_{0}=B^{m}f_{0}$

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

(3)
$\frac{\lambda_{n_{k}}BT^{n_{k}}f(z)}{\lambda_{n_{k}}(T^{n_{k}}f)(0)}=\frac{BT^{n_{k}}f(z)}{(T^{n_{k}}f)(0)}\to
f_{0}(z)$

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:

$\frac{B^{m}T^{n}f(z)}{(T^{n}f)(0)}\to 0$

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

Indeed, since $B^{m}T=\lambda^{m}TB^{m}$ we get

$B^{m}T^{n}f=\lambda^{nm}T^{n}B^{m}f.$

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

(4) $\displaystyle|\lambda^{nm}T^{n}B^{m}f(z)|$ $\displaystyle\leq$
$\displaystyle\frac{\|T^{n}B^{m}f\|}{\sqrt{1-|z|^{2}}}$ (5)
$\displaystyle\leq$
$\displaystyle\frac{|\lambda|^{nm}\|T\|^{n}}{\sqrt{1-|z|^{2}}}.$

Here $\|T\|$ 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
$T^{n}=\mathbb{R}_{\lambda}\phi(B)\cdots
R_{\lambda}\phi(B)=R_{\lambda}^{n}\phi(B)\cdots\phi(\lambda^{n-1}B)$. On the
other hand, since $|\lambda|<1$ the sequence $\Phi_{n}\to h\neq 0$ on
$H^{2}(\mathbb{D})$. Since the set of supercyclic vectors is dense, we can
suppose without loss of generality that $\langle f,h\rangle\neq 0$.

Thus:

$\displaystyle|(T^{n}f)(0)|$ $\displaystyle=$ $\displaystyle|\langle
T^{n}f,1\rangle|$ $\displaystyle=$ $\displaystyle|\langle
f,M_{\overline{\phi}(z)\cdots\overline{\phi}(\lambda^{n-1}z)}R_{\lambda^{n}}^{\star}1\rangle|$
$\displaystyle=$ $\displaystyle|\langle
f,\overline{\phi}(z)\cdots\overline{\phi}(\lambda^{n-1}z)\rangle|$
$\displaystyle\to$ $\displaystyle\langle f,h\rangle\neq 0$

as $n\to\infty$. Using this fact, the equations (3), (5), and by taking $m$
such that $|\lambda|^{m}<\|T\|$ we get:

$\displaystyle\frac{\lambda_{n_{k}}BT^{n_{k}}f(z)}{\lambda_{n_{k}}(T^{n_{k}}f)(0)}$
$\displaystyle=$ $\displaystyle\frac{BT^{n_{k}}f(z)}{(T^{n_{k}}f)(0)}$
$\displaystyle\leq$
$\displaystyle\frac{\frac{|\lambda|^{nm}\|T\|^{n}}{\sqrt{1-|z|^{2}}}}{|(T^{n}f)(0)|}\to
0$

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

##  6\. Concluding remarks

The results we have obtained here especially Theorem 4.11 and Theorem 4.16,
indicate there is a dichotomy theorem waiting to be proved that establishs the
elements of the $\lambda$-commutant of $B$ are either hypercyclic or they have
orbits quite regulars (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: $R_{\lambda}\phi(B)$ is connected with the shape of the
spectrum of $\phi(B)$ and it remains a mysterious a characterization of
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 make difficult the
transference of the hypercyclicity. In strong contrast, in Fréchet spaces such
transference is easier (see [2, 11]).

Conflicts of Interest: The authors declare no conflict of interest.

Funding:  The second and the third author are supported by Junta de Andalucía,
Consejería de Universidad, Investigación e Innovación: ProyExcel_00780
”Operator Theory: An interdisciplinary approach.”

Ethical Conduct:  This article is original, it has not been previously
published and it has not been simultaneously submitted for evaluation to
another journal. All authors have contributed substantially to the article
without omission of any person.

Data Availability Statements: Not applicable.

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