# The Cauchy dual subnormality problem for cyclic 2-isometries

## Abstract

The Cauchy dual subnormality problem asks whether the Cauchy dual operator of a 2-isometry is subnormal. Recently this problem has been solved in the negative. Here we show that it has a negative solution even in the class of cyclic 2-isometries.

## Introduction

Very recently the Cauchy dual subnormality problem was solved negatively in the class of 2-isometric operators (see ). Originally formulated for completely hyperexpansive operators (see [12, Question 2.11]), the problem entails on determining whether the Cauchy dual operator of a member of the underlying class is subnormal. It is worth mentioning that there are relatively broad subclasses of 2-isometric or 2-hyperexpansive operators for which this problem has an affirmative solution (see  and  respectively). In the present paper we show that the Cauchy dual subnormality problem has a negative solution even in the class of cyclic 2-isometric operators. The counterexample is implemented with the help of a weighted composition operator on $$L^2$$ space over a directed graph with a circuit (see Theorem 4.4).

Apart from Introduction, the paper consists of three parts. The first one gives the theoretical background on weighted composition operators on $$L^2$$ spaces needed in this paper. In the next one, we construct a concrete class of weighted composition operators on an $$L^2$$ space and characterize the subnormality of their Cauchy duals. In the last part, by specifying the weights and using Hausdorff’s moment problem technique along with subtle classical analysis, we get the required counterexample.

We will now provide the necessary concepts and facts related to the issues discussed, placing more emphasis on the Hausdorff moment problem. Given a complex Hilbert space $${\mathcal {H}},$$ we denote by $${\varvec{B}}({\mathcal {H}})$$ the $$C^*$$-algebra of all bounded linear operators on $${\mathcal {H}}.$$ Let $$T\in {\varvec{B}}({\mathcal {H}}).$$ We write |T| for $$(T^*T)^{1/2}$$ and call it the modulus of T. We say that T is cyclic if there exists a vector $$e_0,$$ called a cyclic vector of T,  such that the linear span of the set $$\{T^n e_0\}_{n=0}^{\infty }$$ is dense in $${\mathcal {H}}$$. We call Tsubnormal if there exist a complex Hilbert space $${\mathcal {K}}$$ and a normal operator $$N\in {\varvec{B}}(\mathcal K)$$ such that $${\mathcal {H}}\subseteq {\mathcal {K}}$$ (an isometric embedding) and $$Th=Nh$$ for all $$h\in {\mathcal {H}}.$$ If T is left-invertible (or equivalently T is bounded from below), then $$T^*T$$ is an invertible element of $${\varvec{B}}({\mathcal {H}})$$ and the operator $$T':=T(T^*T)^{-1}$$ is called the Cauchy dual operator of T (abbreviated to: the Cauchy dual of T). Finally, T is said to be a 2-isometry (or that T is 2-isometric) if

\begin{aligned} I - 2 T^*T + T^{*2}T^2 = 0. \end{aligned}

We refer the reader to [1,2,3, 13] and  for more information on subnormal operators, 2-isometric operators and the Cauchy dual operation, respectively.

Hereafter $${\mathbb {C}}$$ stands for the field of complex numbers. The ring of all polynomials in one complex variable z with complex coefficients is denoted by $${\mathbb {C}}[z].$$ Given a sequence $$\{\gamma _n\}_{n=0}^{\infty }$$ of complex numbers, we say that $$\gamma _n$$is a polynomial inn (of degree d) if there exists $$p\in {\mathbb {C}}[z]$$ (of degree d) such that $$\gamma _n=p(n)$$ for all $$n\in {\mathbb {Z}}_+,$$ where $${\mathbb {Z}}_+:=\{0,1,2,\ldots \}.$$ The following fact will be used later (cf. [14, Exercise 7.2]).

\begin{aligned}&\textit{If\ }\gamma _n{\textit{\ is\ a\ polynomial\ in\ }}n{\textit{\ of\ degree\ }}d,{\textit{\ then\ }}\nonumber \\&\quad \sum _{n=0}^m (-1)^n \left( {\begin{array}{c}m\\ n\end{array}}\right) \gamma _n = 0, \quad m\geqslant \max \{d+1,0\}. \end{aligned}
(1)

A sequence $$\{\gamma _n\}_{n=0}^{\infty }$$ of real numbers is called a Stieltjes moment sequence (resp. Hausdorff moment sequence) if there exists a positive Borel measure $$\mu$$ on $$[0,\infty )$$ (resp. [0, 1]) such that

\begin{aligned} \gamma _n = \int t^n d \mu (t), \quad n \in {\mathbb {Z}}_+. \end{aligned}

Clearly, any Hausdorff moment sequence is a Stieltjes moment sequence, but not conversely. Using Lebesgue’s monotone convergence theorem, one can show the following.

\begin{aligned} \textit{Any\ bounded\ Stieltjes\ moment\ sequence\ is\ a\ Hausdorff\ moment\ sequence.} \end{aligned}
(2)

Recall that Hausdorff moment sequences can be characterized as follows (see [8, Proposition 6.11]).

\begin{aligned}&\textit{ A\ sequence\ }\gamma =\{\gamma _n\}_{n=0}^{\infty }{} \textit{\ of\ real\ numbers\ is\ a\ Hausdorff\ moment\ sequence\ if\ and\ only\ if\ } \\&\quad \sum _{n=0}^{m} (-1)^n \left( {\begin{array}{c}m\\ n\end{array}}\right) \gamma _{n+j} \geqslant 0, \quad j, m \in {\mathbb {Z}}_+. \end{aligned}
(3)

In this paper, we adhere to the conventions:

\begin{aligned} \sum \nolimits _{j=m}^{n} a_j = 0\hbox { and }\prod \nolimits _{l=m}^{n} a_l =1\hbox { whenever }m > n\hbox { and whatever }a_j'\hbox {s are.} \end{aligned}
(4)

To simplify the notation, we write

\begin{aligned} \{f=0\}=\{x\in X:f(x)=0\} \quad \text {and} \quad \{f\ne 0\}=\{x\in X:f(x)\ne 0\}, \end{aligned}

whenever f is a $${\mathbb {C}}$$-valued or a $$[0,\infty ]$$-valued function on a set X.

## Weighted composition operators on $$L^2$$-spaces

Given a measure space $$(X, {\mathscr {A}}, \mu )$$, we denote by $$L^2(\mu )$$ the complex Hilbert space of all square $$\mu$$-integrable $${\mathscr {A}}$$-measurable complex functions on X endowed with the standard inner product.

### Definition 2.1

Let $$(X, {\mathscr {A}}, \mu )$$ be a $$\sigma$$-finite measure space, $$\phi :X \rightarrow X$$ be an $${\mathscr {A}}$$-measurable map and $${\mathsf {w}} :X \rightarrow {\mathbb {C}}$$ be an $${\mathscr {A}}$$-measurable function. By a weighted composition operator in $$L^2(\mu )$$ we mean a mapping $$C_{\phi , {\mathsf {w}}}:L^2(\mu ) \supseteq {\mathscr {D}}(C_{\phi , {\mathsf {w}}}) \rightarrow L^2(\mu )$$ defined by

\begin{aligned} {\mathscr {D}}(C_{\phi , {\mathsf {w}}})&= \left\{ f \in L^2(\mu ):{\mathsf {w}} \cdot (f \circ \phi ) \in L^2(\mu ) \right\} , \\ C_{\phi , {\mathsf {w}}}f&= {\mathsf {w}} \cdot (f \circ \phi ), \quad f \in {\mathscr {D}}(C_{\phi , {\mathsf {w}}}). \end{aligned}

We call $$\phi$$ and $${\mathsf {w}}$$ the symbol and the weight of $$C_{\phi , {\mathsf {w}}},$$ respectively.

As a matter of fact, $$C_{\phi , {\mathsf {w}}}$$ may not be well defined. The well-definiteness of $$C_{\phi , {\mathsf {w}}}$$ means that $${\mathsf {w}} \cdot (f \circ \phi ) = {\mathsf {w}} \cdot (g \circ \phi )$$ a.e. $$[\mu ]$$ whenever $$f, g :X \rightarrow {\mathbb {C}}$$ are $${\mathscr {A}}$$-measurable functions such that $$f=g$$ a.e. $$[\mu ]$$, $$f\in L^2(\mu )$$ and $${\mathsf {w}} \cdot (f \circ \phi ) \in L^2(\mu )$$. Below we recall several basic properties of weighted composition operators including well-definiteness (see [11, Proposition 7]).

### Proposition 2.2

Let$$(X, {\mathscr {A}}, \mu )$$be a$$\sigma$$-finite measure space, $$\phi :X \rightarrow X$$be an$${\mathscr {A}}$$-measurable map and$${\mathsf {w}} :X \rightarrow {\mathbb {C}}$$be an$${\mathscr {A}}$$-measurable function. Then$$C_{\phi , {\mathsf {w}}}$$is well defined if and only if$$\mu _{{\mathsf {w}}} \circ \phi ^{-1} \ll \mu$$, where$$\mu _{{\mathsf {w}}}$$and$$\mu _{{\mathsf {w}}} \circ \phi ^{-1}$$are measures on$${\mathscr {A}}$$defined by

\begin{aligned} \mu _{{\mathsf {w}}} (\sigma ) =\int _{\sigma } |{\mathsf {w}}|^2d\mu \quad \text {and} \quad \mu _{{\mathsf {w}}} \circ \phi ^{-1} (\sigma ) = \mu _{\mathsf {w}}(\phi ^{-1}(\sigma )) \quad \text {for } \sigma \in {\mathscr {A}}. \end{aligned}
(5)

Moreover, if$$C_{\phi , {\mathsf {w}}}$$is well defined,$${\mathsf {u}}:X \rightarrow {\mathbb {C}}$$is$${\mathscr {A}}$$-measurable and$${\mathsf {u}}={\mathsf {w}}$$a.e.$$[\mu ],$$then$$C_{\phi ,{\mathsf {u}}}$$is well defined and$$C_{\phi ,{\mathsf {u}}}=C_{\phi , {\mathsf {w}}}.$$

To avoid repetition, we distinguish the following assumption, which we will often refer to in this article.

$$(X, {\mathscr {A}}, \mu )\hbox { is a }\sigma \hbox {-finite measure space, }\phi :X \rightarrow X\hbox { is an } {\mathscr {A}}\hbox {-measurable map and }{\mathsf {w}} :X \rightarrow {\mathbb {C}}\hbox { is an } {\mathscr {A}}\hbox {-measurable function such that }\mu _{{\mathsf {w}}} \circ \phi ^{-1} \ll \mu .$$
(6)

If the condition (6) holds, then by the Radon-Nikodym theorem (see [5, Theorem 2.2.1]), there exists a unique (up to a set of $$\mu$$-measure zero) $${\mathscr {A}}$$-measurable function $$h_{\phi , {\mathsf {w}}}:X \rightarrow [0, \infty ]$$ such that

\begin{aligned} \mu _{{\mathsf {w}}} \circ \phi ^{-1}(\sigma ) = \int _{\sigma } h_{\phi , {\mathsf {w}}}d\mu , \quad \sigma \in {\mathscr {A}}. \end{aligned}
(7)

Before going further, we make an important observation.

### Lemma 2.3

Suppose that (6) holds. If$$h_{\phi , {\mathsf {w}}}> 0$$a.e.$$[\mu ],$$then

\begin{aligned} \mu (\{{\mathsf {w}}\ne 0\} \cap \{h_{\phi , {\mathsf {w}}}\circ \phi = 0\})=0. \end{aligned}
(8)

### Proof

Since $$\mu (\{h_{\phi , {\mathsf {w}}}= 0\})=0$$ and $$\{h_{\phi , {\mathsf {w}}}\circ \phi = 0\} = \phi ^{-1}(\{h_{\phi , {\mathsf {w}}}= 0\}),$$ we infer from (6) that $$\mu _{{\mathsf {w}}}(\{h_{\phi , {\mathsf {w}}}\circ \phi = 0\})=0,$$ which yields (8). $$\square$$

In view of [11, Proposition 8(v) and Theorem 18], the following is valid.

\begin{aligned} \begin{array}{l} \textit{If\ }(6) \textit{\ holds,\ then\ }C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu )){} \textit{ \ if\ and\ only\ if\ }h_{\phi , {\mathsf {w}}}\in L^{\infty }(\mu );\\ \textit{\ if\ this\ is\ the\ case,\ then\ } \Vert C_{\phi , {\mathsf {w}}}\Vert ^2 = \Vert h_{\phi , {\mathsf {w}}}\Vert _{L^{\infty }(\mu )}{} \textit{\ and\ the\ modulus\ }\\ |C_{\phi , {\mathsf {w}}}|\textit{\ of\ }C_{\phi , {\mathsf {w}}}\textit{\ equals\ the\ operator\ }M_{h^{1/2}_{\phi , {\mathsf {w}}}}{} \textit{\ of\ multiplication\ by\ }h^{1/2}_{\phi , {\mathsf {w}}}{} \textit{\ in\ }L^2(\mu ). \end{array} \end{aligned}
(9)

From now on, we assume that $$h_{\phi , {\mathsf {w}}}$$ takes finite values whenever $$C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu )).$$ This assumption is justified by (9).

### Proposition 2.4

Suppose that (6) holds and$$C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu )).$$Then

1. (i)

if$$c\in (0,\infty )$$, then$$\Vert C_{\phi , {\mathsf {w}}}f\Vert \geqslant c \Vert f\Vert$$for all$$f\in L^2(\mu )$$if and only if$$h_{\phi , {\mathsf {w}}}\geqslant c^2$$a.e. $$[\mu ],$$

2. (ii)

if$$C_{\phi , {\mathsf {w}}}$$is bounded from below, then the Cauchy dual$$C'_{\phi , {\mathsf {w}}}$$of$$C_{\phi , {\mathsf {w}}}$$equals$$C_{\phi , {\mathsf {w}}'},$$where$${\mathsf {w}}' :X \rightarrow {\mathbb {C}}$$is any$${\mathscr {A}}$$-measurable function such that

\begin{aligned} {\mathsf {w}}' = {\left\{ \begin{array}{ll} {\frac{{{\mathsf {w}}}}{h_{\phi , {{\mathsf {w}}}}\, \circ\, \phi }} &{} \text {on } \{{\mathsf {w}}\ne 0\} \cap \{h_{\phi , {\mathsf {w}}}\circ \phi >0\}, \\ 0 &{} \text {on } \{{\mathsf {w}} = 0\}; \end{array}\right. } \end{aligned}
(10)

in this case, $$h_{\phi , {\mathsf {w}}'}=\frac{{1}}{{h_{\phi , {\mathsf {w}}}}}$$a.e.$$[\mu ]$$.

### Proof

1. (i)

Adapting the proof of [9, Proposition 4], we get the statement (i).

2. (ii)

Note that by (9) and (i), there exists $$c_1,c_2 \in (0,\infty )$$ such that $$c_1 \leqslant h_{\phi , {\mathsf {w}}}\leqslant c_2$$ a.e. $$[\mu ].$$ By virtue of (9), we have

\begin{aligned} C'_{\phi , {\mathsf {w}}} = C_{\phi , {\mathsf {w}}}(|C_{\phi , {\mathsf {w}}}|^2)^{-1} = C_{\phi , {\mathsf {w}}}M^{-1}_{h_{\phi , {\mathsf {w}}}} = C_{\phi , {\mathsf {w}}}M_{g}, \end{aligned}
(11)

where $$g:X \rightarrow [0,\infty )$$ is an $${\mathscr {A}}$$-measurable function such that $$g=\frac{{1}}{{h_{\phi , {\mathsf {w}}}}}$$ on the set $$\{h_{\phi , {\mathsf {w}}}> 0\}.$$ It is easily seen that

\begin{aligned} {\mathsf {w}}' \overset{(10)}{=} {\mathsf {w}} \cdot (g \circ \phi ) \quad \text {on} \quad (\{{\mathsf {w}}\ne 0\} \cap \{h_{\phi , {\mathsf {w}}}\circ \phi >0\}) \cup \{{\mathsf {w}}= 0\}. \end{aligned}

Hence, by Lemma 2.3, $${\mathsf {w}}' = {\mathsf {w}} \cdot (g\circ \phi )$$ a.e. $$[\mu ].$$ Proposition 2.2 yields

\begin{aligned} C'_{\phi , {\mathsf {w}}}\overset{(11)}{=}C_{\phi , {\mathsf {w}} \cdot (g\, \circ\, \phi )}=C_{\phi , {\mathsf {w}}'}. \end{aligned}

Now applying [5, Theorem 1.6.12] and (7), we obtain

\begin{aligned} \mu _{{\mathsf {w}}'} \circ \phi ^{-1}(\varDelta )= \mu _{{\mathsf {w}} \cdot (g\, \circ\, \phi )} \circ \phi ^{-1}(\varDelta ) & \overset{(5)}{=} \int _X (\chi _{\varDelta } \circ \phi ) (g^2\circ \phi ) d\mu _{{\mathsf {w}}} \\&= \int _{\varDelta } g^2 h_{\phi , {\mathsf {w}}}d\mu \\&= \int _\varDelta {\frac{{1}}{{h_{\phi , {\mathsf {w}}}}}} d\mu , \quad \varDelta \in {\mathscr {A}}, \end{aligned}

which shows that $$h_{\phi , {\mathsf {w}}'}=\frac{{1}}{{h_{\phi , {\mathsf {w}}}}}$$ a.e. $$[\mu ]$$. This completes the proof.$$\square$$

Recall that (see [11, Lemma 26]) if (6) holds and $$C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu )),$$ then for any integer $$n\geqslant 1,$$ the nth power $$C^n_{\phi , {\mathsf {w}}}$$ of $$C_{\phi , {\mathsf {w}}}$$ equals $$C_{\phi ^n, {\mathsf {w}}_{[n]}},$$ where $$\phi ^n$$ denotes the n-fold composition of $$\phi$$ with itself ($$\phi ^0$$ is the identity map on X) and $${\mathsf {w}}_{[n]} :X \rightarrow {\mathbb {C}}$$ is the function given by

\begin{aligned} {\mathsf {w}}_{} = 1\text { and }{\mathsf {w}}_{[n]} = \prod _{j=0}^{n-1} {\mathsf {w}} \circ \phi ^j\text { for }n \geqslant 1. \end{aligned}
(12)

Moreover, $$h_{\phi ^0, {\mathsf {w}}_{}}=1$$ a.e. $$[\mu ]$$ and the following recurrence formula holds:

\begin{aligned} h_{\phi ^{n+1}, {\mathsf {w}}_{[n+1]}}&= h_{\phi , {\mathsf {w}}}\cdot {\mathsf {E}}_{\phi ,{\mathsf {w}}}\big (h_{\phi ^n, {\mathsf {w}}_{[n]}}\big )\circ \phi ^{-1} \text { a.e.}\,[\mu ], \quad n\in {\mathbb {Z}}_+, \end{aligned}
(13)

where $${\mathsf {E}}_{\phi ,{\mathsf {w}}}(f)$$ stands for the conditional expectation of an $${\mathscr {A}}$$-measurable function $$f:X \rightarrow [0,\infty )$$ with respect to the $$\sigma$$-algebra $$\phi ^{-1}({\mathscr {A}})$$ and the measure $$\mu _{{\mathsf {w}}}$$; we refer the reader to [11, Sect. 2.4] for the precise definitions of $${\mathsf {E}}_{\phi ,{\mathsf {w}}}(f)$$ and $${\mathsf {E}}_{\phi ,{\mathsf {w}}}(f) \circ \phi ^{-1}.$$ The above discussion and Proposition 2.4 yield the following.

### Proposition 2.5

Suppose that (6) holds, $$C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu ))$$and$$C_{\phi , {\mathsf {w}}}$$is bounded from below. Then for every$$n\in {\mathbb {Z}}_+$$, thenth power$$C'^n_{\phi , {\mathsf {w}}}$$of the Cauchy dual$$C'_{\phi , {\mathsf {w}}}$$of$$C_{\phi , {\mathsf {w}}}$$equals$$C_{\phi ^n, {\mathsf {w}}'_{[n]}},$$where$${\mathsf {w}}'_{[n]} :X \rightarrow {\mathbb {C}}$$is the function given by

\begin{aligned} {\mathsf {w}}'_{}=1 \text { and } {\mathsf {w}}'_{[n]} = \prod _{j=0}^{n-1} {\mathsf {w}}' \circ \phi ^j \text { for } n\geqslant 1, \end{aligned}
(14)

with$${\mathsf {w}}'$$as in (10). Moreover, $$h_{\phi ^0, {\mathsf {w}}'_{}}=1$$a.e.$$[\mu ]$$and

\begin{aligned} h_{\phi ^{n+1}, {\mathsf {w}}'_{[n+1]}} = {\frac{{1}}{{h_{\phi , {\mathsf {w}}}}}} \, {\mathsf {E}}_{\phi ,{\mathsf {w}}'}(h_{\phi ^{n}, {\mathsf {w}}'_{[n]}}) \circ \phi ^{-1} \text { a.e.}\, [\mu ], \quad n \in {\mathbb {Z}}_+. \end{aligned}

The subnormality of the Cauchy dual $$C_{\phi , {\mathsf {w}}}'$$ of a left-invertible bounded weighted composition operator $$C_{\phi , {\mathsf {w}}}$$ can be characterized as follows (cf. [18, Theorem 4.5]).

### Proposition 2.6

Suppose that (6) holds, $$C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu ))$$and$$C_{\phi , {\mathsf {w}}}$$is bounded from below. Then

1. (i)

$$C'_{\phi , {\mathsf {w}}}$$is subnormal if and only if$$\{h_{\phi ^n, {\mathsf {w}}'_{[n]}}(x)\}_{n=0}^{\infty }$$is a Stieltjes moment sequence for$$\mu$$-a.e.$$x \in X,$$

2. (ii)

if$$\Vert C_{\phi , {\mathsf {w}}}(f)\Vert \geqslant \Vert f\Vert$$for all$$f\in L^2(\mu )$$and$$C'_{\phi , {\mathsf {w}}}$$is subnormal, then the sequence$$\{h_{\phi ^n, {\mathsf {w}}'_{[n]}}(x)\}_{n=0}^{\infty }$$is a Hausdorff moment sequence for$$\mu$$-a.e.$$x \in X.$$

### Proof

1. (i)

Combine [11, Theorem 49] with Proposition 2.5.

2. (ii)

Suppose now that $$\Vert C_{\phi , {\mathsf {w}}}(f)\Vert \geqslant \Vert f\Vert$$ for all $$f\in L^2(\mu )$$ and $$C_{\phi , {\mathsf {w}}}'$$ is subnormal. Then $$C_{\phi , {\mathsf {w}}}'$$ is a contraction and consequently so is its nth power $$C_{\phi , {\mathsf {w}}}^{\prime n}$$ for every $$n\in {\mathbb {Z}}_+.$$ In view of (9) and Proposition 2.5, the sequence $$\{h_{\phi ^n, {\mathsf {w}}'_{[n]}}(x)\}_{n=0}^{\infty }$$ is bounded by 1 for $$\mu$$-a.e. $$x \in X.$$ Applying (i) and (2) completes the proof.$$\square$$

We conclude this section by characterizing bounded 2-isometric weighted composition operators (see [17, Lemma 2.3] for the case of composition operators).

### Proposition 2.7

Suppose that (6) holds and$$C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu )).$$Then the following conditions areequivalent:

1. (i)

$$C_{\phi , {\mathsf {w}}}$$is a 2-isometry,

2. (ii)

$$1-2h_{\phi , {\mathsf {w}}}+ h_{\phi ^2, {\mathsf {w}}_{}} = 0$$a.e.$$[\mu ],$$

3. (iii)

$$1-2h_{\phi , {\mathsf {w}}}+ h_{\phi , {\mathsf {w}}}\cdot {\mathsf {E}}_{\phi ,{\mathsf {w}}}(h_{\phi , {\mathsf {w}}}) \circ \phi ^{-1} = 0$$a.e.$$[\mu ].$$

### Proof

Since $$C^n_{\phi , {\mathsf {w}}}=C_{\phi ^n, {\mathsf {w}}_{[n]}},$$ we infer from (9) that $$C^{*n}_{\phi , {\mathsf {w}}}C^n_{\phi , {\mathsf {w}}}=M_{h_{\phi ^n, {\mathsf {w}}_{[n]}}}$$ for any $$n\in {\mathbb {Z}}_+.$$ Applying the definition of 2-isometricity, one can show that (i) and (ii) are equivalent. That (ii) and (iii) are equivalent follows from (13). $$\square$$

## A family of weighted composition operators on $$\ell ^2({\mathbb {Z}}_+)$$

In this section we concentrate on a family of weighted composition operators coming from [9, Example 42] (see also [10, Section 3.2]).

### Example 3.1

Denote by $${\mathscr {A}}$$ the power set $$2^{{\mathbb {Z}}_+}$$ of $${\mathbb {Z}}_+$$ and by $$\mu$$ the counting measure on $$2^{{\mathbb {Z}}_+}$$. Clearly, all selfmaps of $${\mathbb {Z}}_+$$ and complex functions on $${\mathbb {Z}}_+$$ are $${\mathscr {A}}$$-measurable. Note that $$({\mathbb {Z}}_+, {\mathscr {A}}, \mu )$$ is a $$\sigma$$-finite measure space. For $$n\in {\mathbb {Z}}_+,$$ we denote by $$e_n$$ the element of $$L^2(\mu )$$ given by

\begin{aligned} e_n(m) = {\left\{ \begin{array}{ll} 1 &{} \text {if } m=n, \\ 0 &{} \text {otherwise.} \end{array}\right. } \end{aligned}

Clearly, $$\{e_n\}_{n=0}^{\infty }$$ is an orthonormal basis of $$L^2(\mu ).$$ Define the map $$\phi :{\mathbb {Z}}_+ \rightarrow {\mathbb {Z}}_+$$ by

$$\phi (n) = {\left\{ \begin{array}{ll} 0 &{} n=0, \\ n-1 &{} \text {otherwise}. \end{array}\right. }$$

Let $${\mathsf {w}} :{\mathbb {Z}}_+ \rightarrow {\mathbb {C}}$$ be any function. It is obvious that $$\mu _{{\mathsf {w}}} \circ \phi ^{-1} \ll \mu$$ and so $$C_{\phi , {\mathsf {w}}}$$ is well defined (cf. Proposition 2.2). It follows from (7) that

\begin{aligned} h_{\phi , {\mathsf {w}}}(n) = \sum _{j \in \phi ^{-1}(\{n\})} |{\mathsf {w}}(j)|^2, \quad n \in {\mathbb {Z}}_+, \end{aligned}
(15)

which yields

\begin{aligned} h_{\phi , {\mathsf {w}}}(0) = \alpha _{{\mathsf {w}}} \text { and } h_{\phi , {\mathsf {w}}}(n) = |{\mathsf {w}}(n+1)|^2 \text { for } n \geqslant 1, \end{aligned}
(16)

where

\begin{aligned} \alpha _{{\mathsf {w}}} := |{\mathsf {w}}(0)|^2+ |{\mathsf {w}}(1)|^2. \end{aligned}

Combined with (9), this implies that

\begin{aligned} C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu )){}& \textit{\ if\ and\ only\ if\ }\sup _{n\geqslant 0} |{\mathsf {w}}(n)| < \infty ;\textit{\ if\ this\ is\ the\ case,\ then\ } \\ &\quad \Vert C_{\phi , {\mathsf {w}}}\Vert ^2 =\max \Big \{\alpha _{{\mathsf {w}}}, \, \sup _{n \geqslant 2}|{\mathsf {w}}(n)|^2\Big \}. \end{aligned}
(17)

Moreover, by Proposition 2.4(i), $$C_{\phi , {\mathsf {w}}}$$ is bounded from below if and only if

\begin{aligned} \min \left\{ \alpha _{{\mathsf {w}}}, \, \inf _{n \geqslant 2}|{\mathsf {w}}(n)|^2\right\} > 0. \end{aligned}
(18)

Concerning the cyclicity of $$C_{\phi , {\mathsf {w}}}$$, one can make the following observation.

\begin{aligned}& \textit{If\ }C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu )){} \textit{\ and\ }{\mathsf {w}}(n)\ne 0\textit{ \ for\ every\ }n\geqslant 1, \textit{then\ }C_{\phi , {\mathsf {w}}}\textit{\ is\ cyclic\ } \\& \quad \textit{with\ the\ cyclic\ vector\ }e_0. \end{aligned}
(19)

This can be deduced from the equality (20) below which is a direct consequence of the definition.

\begin{aligned} C_{\phi , {\mathsf {w}}}e_n&= {\left\{ \begin{array}{ll} {\mathsf {w}}(0)e_0 + {\mathsf {w}}(1)e_1 &{} \text { if } n=0, \\ {\mathsf {w}}(n+1)e_{n+1} &{} \text { if } n \geqslant 1. \end{array}\right. } \end{aligned}
(20)

Using Proposition 2.4, we can describe the Cauchy dual $$C_{\phi , {\mathsf {w}}}'$$ of a left-invertible $$C_{\phi , {\mathsf {w}}}$$ as follows:

\begin{aligned}&{ If }C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu )){} \textit{\ and\ }(18) \textit{\ holds,\ then\ }C_{\phi , {\mathsf {w}}}'=C_{\phi , {\mathsf {w}}'}, \textit{where} \\&\quad {\mathsf {w}}'(n) = {\left\{ \begin{array}{ll} {\frac{{\mathsf {w}(n)}}{{\alpha _{{\mathsf {w}}}}}} &{} \text { if\ } n=0, 1, \\ {\frac{{1}}{{\overline{{{\mathsf {w}}}(n)}}}} &{} \text { if } n \geqslant 2. \end{array}\right. } \end{aligned}
(21)

As a consequence, we get

\begin{aligned} C_{\phi , {\mathsf {w}}}' e_n = {\left\{ \begin{array}{ll} {\frac{{{\mathsf {w}}(0)}}{{\alpha _{{\mathsf {w}}}}}}e_0 + {\frac{{{\mathsf {w}}(1)}}{{\alpha _{{\mathsf {w}}}}}} e_1 &{} \text { if }n=0, \\ {\frac{{1}}{{\overline{{{\mathsf {w}}}(n+1)}}}} e_{n+1} &{} \text { if } n \geqslant 1. \end{array}\right. } \end{aligned}

$$\diamondsuit$$

Below, we give necessary and sufficient conditions for the weighted composition operator $$C_{\phi , {\mathsf {w}}}$$ from Example 3.1 to be 2-isometric. Before we do this, we define the functions $$\xi _n:[1,\infty ) \rightarrow [1,\infty )$$, where $$n\in {\mathbb {Z}}_+,$$ by

\begin{aligned} \xi _n(x) = \sqrt{{\frac{{1+ (n+1)(x^2-1)}}{{1+ n(x^2-1)}}}}, \quad x \in {[}1,\infty ), \quad n\in {\mathbb {Z}}_+. \end{aligned}
(22)

### Proposition 3.2

Let$${\mathscr {A}}$$, $$\mu$$, $$\phi$$and$${\mathsf {w}}$$be as in Example 3.1. Assume that$$C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu )).$$Then$$C_{\phi , {\mathsf {w}}}$$is a 2-isometry if and only if

\begin{aligned} |{\mathsf {w}}(2)| \ge 1, \quad |{\mathsf {w}}(n+2) |= \xi _n(|{\mathsf {w}}(2)|), \quad n \in {\mathbb {Z}}_+, \end{aligned}
(23)

and

\begin{aligned} {\left\{ \begin{array}{ll} |{\mathsf {w}}(0)|=1 &{} \text { if } {\mathsf {w}}(1)=0, \\ |{\mathsf {w}}(2)| = {\frac{{\sqrt{\alpha _{\mathsf {w}}(2-|{\mathsf {w}}(0)|^2)-1}}}{{|{\mathsf {w}}(1)|}}} &{} \text { if } {\mathsf {w}}(1) \ne 0. \end{array}\right. } \end{aligned}
(24)

Moreover, if$$C_{\phi , {\mathsf {w}}}$$is a 2-isometry, then

1. (i)

if$${\mathsf {w}}(1)\ne 0$$and either$$|{\mathsf {w}}(0)|=1$$or$$\alpha _{{\mathsf {w}}} = 1,$$then$$|{\mathsf {w}}(n)|=1$$for all$$n \geqslant 2,$$

2. (ii)

$$(\alpha _{{\mathsf {w}}}-1)(1-|{\mathsf {w}}(0)|^2) \geqslant 0.$$

### Proof

Arguing as in (15) with $$(\phi ^2,{\mathsf {w}}_{})$$ in place of $$(\phi ,{\mathsf {w}})$$ and using (12), we obtain

\begin{aligned} h_{\phi ^{2}, {\mathsf {w}}_{}} (n) = {\left\{ \begin{array}{ll} |{\mathsf {w}}(n+1)|^2 |{\mathsf {w}}(n+2)|^2 &{} \text {if } n\geqslant 1, \\ |{\mathsf {w}}(0)|^4 + |{\mathsf {w}}(0)|^2 |{\mathsf {w}}(1)|^2 + |{\mathsf {w}}(1)|^2 |{\mathsf {w}}(2)|^2 &{} \text {if } n=0. \end{array}\right. } \end{aligned}

Combined with Proposition 2.7(ii), we deduce that $$C_{\phi , {\mathsf {w}}}$$ is a 2-isometry if and only if

\begin{aligned} |{\mathsf {w}}(1)|^2|{\mathsf {w}}(2)|^2 = \alpha _{\mathsf {w}}(2-|{\mathsf {w}}(0)|^2)-1 \end{aligned}
(25)

and

\begin{aligned} 1 - 2 |{\mathsf {w}}(n+2)|^2 + |{\mathsf {w}}(n+2)|^2|{\mathsf {w}}(n+3)|^2 = 0, \quad n\in {\mathbb {Z}}_+. \end{aligned}
(26)

It is easily seen that (24) is equivalent to (25). Observe that the condition (26) holds if and only if the unilateral weighted shift on $$\ell ^2({\mathbb {Z}}_+)$$ with weights $$\{|w(n+2)|\}_{n=0}^{\infty }$$ is 2-isometric. Hence, by [19, Lemma 6.1], the condition (26) is equivalent to (23). Using $$|{\mathsf {w}}(2)| \ge 1$$ and the fact that the expression under the sign of the square root in (24) is nonnegative, we get the “moreover” part. $$\square$$

It is well known that any 2-isometry is expansive (see [20, Lemma 1]), so we can consider its Cauchy dual operator. Below, we follow the conventions (4).

### Theorem 3.3

Let$${\mathscr {A}}$$, $$\mu$$, $$\phi$$and$${\mathsf {w}}$$be as in Example 3.1. Assume that$$C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu ))$$and$$C_{\phi , {\mathsf {w}}}$$is a 2-isometry. Then the following assertions hold:

1. (i)

the Cauchy dual$$C'_{\phi , {\mathsf {w}}}$$of$$C_{\phi , {\mathsf {w}}}$$is subnormal if and only if$$\{h_{\phi ^{n}, {\mathsf {w}}'_{[n]}}(0)\}_{n=0}^{\infty }$$is a Hausdorff moment sequence,

2. (ii)

the value of$$h_{\phi ^{n}, {\mathsf {w}}'_{[n]}}$$at 0 is given by the following explicit formula

\begin{aligned} h_{\phi ^{n}, {\mathsf {w}}'_{[n]}}(0) = {\frac{{|{\mathsf {w}}(0)|^{2n}}}{{\alpha _{{\mathsf {w}}}^{2n}}}} + \sum _{j=0}^{n-1} {\frac{{|{\mathsf {w}}(0)|^{2(n-j-1)}|{\mathsf {w}}(1)|^2}}{{\alpha _{\mathsf {w}}^{2(n-j)}({1+j(|{\mathsf {w}}(2)|^2-1))}}}}, \quad n\in {\mathbb {Z}}_+. \end{aligned}
(27)

### Proof

1. (i)

The “only if” part is a direct consequence Proposition 2.6(ii). In view of Proposition 2.6(i) to prove the ”if” part, it suffices to show that the sequence $$\{h_{\phi ^{n}, {\mathsf {w}}'_{[n]}}(k)\}_{n=0}^{\infty }$$ is a Stieltjes moment sequence for every $$k\geqslant 1.$$ Fix $$k\geqslant 1.$$ Arguing as in (15) with $$(\phi ^{n}, {\mathsf {w}}'_{[n]})$$ in place of $$(\phi ,{\mathsf {w}}),$$ we verify that

\begin{aligned} h_{\phi ^{n}, {\mathsf {w}}'_{[n]}}(k) = |{\mathsf {w}}'_{[n]}(k + n)|^2 \overset{(12)}{=} \prod _{j=1}^{n} |{\mathsf {w}}'(k+j)|^2 \overset{(21)}{=} {\frac{{1}}{{\prod _{j=1}^{n} |{\mathsf {w}}(k+j)|^2}}}, \quad n\geqslant 1. \end{aligned}

Since the unilateral weighted shift on $$\ell ^2({\mathbb {Z}}_+)$$ with weights $$\{|{\mathsf {w}}(k+1+n)|\}_{n=0}^{\infty }$$ is 2-isometric (see the proof of Proposition 3.2), we deduce fromFootnote 1 [6, Remark 4] that the unilateral weighted shift on $$\ell ^2({\mathbb {Z}}_+)$$ with weights $$\big \{\frac{1}{|{\mathsf {w}}(k+1+n)|}\big \}_{n=0}^{\infty }$$ is subnormal, which by Berger-Gellar-Wallen theorem (see [15, 16]) is equivalent to the fact that the sequence $$\{h_{\phi ^{n}, {\mathsf {w}}'_{[n]}}(k)\}_{n=0}^{\infty }$$ is a Stieltjes moment sequence.

2. (ii)

Arguing as in (i), we get

\begin{aligned} h_{\phi ^{n}, {\mathsf {w}}'_{[n]}}(0)&= \sum _{j= 0}^{n} |{\mathsf {w}}'_{[n]}(j)|^2 \\& \overset{(14)}{=} \sum _{j= 0}^{n} \; \prod _{l=0}^{j-1} |{\mathsf {w}}'(\phi ^{l}(j))|^2 \prod _{l=j}^{n-1} |{\mathsf {w}}'(\phi ^l(j))|^2 \\&= |{\mathsf {w}}'(0)|^{2n} + \sum _{j= 1}^{n} |{\mathsf {w}}'(0)|^{2(n-j)} \prod _{l=1}^{j} |{\mathsf {w}}'(l)|^2 \\&\overset{(21)}{=} \ {\frac{{|{\mathsf {w}}(0)|^{2n}}}{{\alpha _{{\mathsf {w}}}^{2n}}}} + {\frac{{|{\mathsf {w}}(0)|^{2(n-1)} |{\mathsf {w}}(1)|^2}}{{\alpha _{{\mathsf {w}}}^{2n}}}} \\ &\quad+ \sum _{j= 2}^{n} {\frac{{|{\mathsf {w}}(0)|^{2(n-j)}|{\mathsf {w}}(1)|^2}}{{\alpha _{{\mathsf {w}}}^{2(n-j+1)}\prod _{l=2}^{j} |{\mathsf {w}}(l)|^2}}} \\&\overset{(23)}{=}\ {\frac{{|{\mathsf {w}}(0)|^{2n}}}{{\alpha _{{\mathsf {w}}}^{2n}}}} +{\frac{{|{\mathsf {w}}(0)|^{2(n-1)} |{\mathsf {w}}(1)|^2}}{{\alpha _{\mathsf {w}}^{2n}}}} \\ &\quad + \sum _{j= 1}^{n-1} {\frac{{|{\mathsf {w}}(0)|^{2(n-j-1)}|{\mathsf {w}}(1)|^2}}{{\alpha _{\mathsf {w}}^{2(n-j)}(1+j(|{\mathsf {w}}(2)|^2 -1))}}} \\&= {\frac{{|{\mathsf {w}}(0)|^{2n}}}{{\alpha _{\mathsf {w}}^{2n}}}} + \sum _{j= 0}^{n-1} {\frac{{|{\mathsf {w}}(0)|^{2(n-j-1)}|{\mathsf {w}}(1)|^2}}{{\alpha _{\mathsf {w}}^{2(n-j)}(1+j(|{\mathsf {w}}(2)|^2 -1))}}}, \quad n\geqslant 2. \end{aligned}

It is a matter of simple verification that (27) holds for $$n=0,1$$ as well.$$\square$$

### Corollary 3.4

Let$${\mathscr {A}}$$, $$\mu$$, $$\phi$$and$${\mathsf {w}}$$be as in Example 3.1. Assume that$$|{\mathsf {w}}(0)|=|{\mathsf {w}}(n)|=1$$for every integer$$n \ge 2.$$Then$$C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu )),$$$$C_{\phi , {\mathsf {w}}}$$is a 2-isometry and$$C_{\phi , {\mathsf {w}}}'$$is subnormal. Moreover, $$C_{\phi , {\mathsf {w}}}'$$is an isometry if and only if$$C_{\phi , {\mathsf {w}}}$$is an isometry or, equivalently, if and only if$${\mathsf {w}}(1)=0.$$

### Proof

By (17) and Proposition 3.2, $$C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu ))$$ and $$C_{\phi , {\mathsf {w}}}$$ is a 2-isometry. Using Theorem 3.3(ii), we see that if $${\mathsf {w}}(1) \ne 0,$$ then

\begin{aligned} h_{\phi ^{n}, {\mathsf {w}}'_{[n]}}(0)&= {\frac{{1}}{{\alpha _{{\mathsf {w}}}^{2n}}}} + \sum _{j=0}^{n-1} {\frac{{|{\mathsf {w}}(1)|^2}}{{\alpha _{{\mathsf {w}}}^{2(n-j)}}}} \\&= {\frac{{1}}{{\alpha _{{\mathsf {w}}}^{2n}}}} + {\frac{{|{\mathsf {w}}(1)|^2(\alpha _{{\mathsf {w}}}^{2n}-1)}}{{\alpha _{\mathsf {w}}^{2n}(\alpha _{{\mathsf {w}}}^{2}-1)}}} \\&={\frac{{|{\mathsf {w}}(1)|^2}}{{\alpha _{{\mathsf {w}}}^{2}-1}}} + {\frac{{\alpha _{{\mathsf {w}}}}}{{\alpha _{{\mathsf {w}}}+1}}} (\alpha _{{\mathsf {w}}}^{-2})^n \\&={\frac{{1}}{{2+|{\mathsf {w}}(1)|^2}}} + {\frac{{1+|{\mathsf {w}}(1)|^2}}{{2+|{\mathsf {w}}(1)|^2}}} (\alpha _{\mathsf {w}}^{-2})^n , \quad n\in {\mathbb {Z}}_+, \end{aligned}

which implies that the sequence $$\{h_{\phi ^{n}, {\mathsf {w}}'_{[n]}}(0)\}_{n=0}^{\infty }$$ is a Hausdorff moment sequence. The same is true if $${\mathsf {w}}(1)=0$$ because then by (27), $$h_{\phi ^{n}, {\mathsf {w}}'_{[n]}}(0)=1$$ for all $$n\in {\mathbb {Z}}_+.$$ Applying Theorem 3.3(i), we see that $$C_{\phi , {\mathsf {w}}}'$$ is subnormal. The first equivalence in the “moreover” part is true for arbitrary left-invertible operators, while the second is a direct consequence of [11, (2.22)] and (16). This completes the proof. $$\square$$

## Main example

The following example shows that the Cauchy dual subnormality problem has a negative solution even in the class of cyclic operators.

### Example 4.1

(Example 3.1 continued). Let $${\mathscr {A}}$$, $$\mu$$, $$\phi$$ and $${\mathsf {w}}$$ be as in Example 3.1. To achieve the main purpose of this paper, we will begin by specifying the weight $${\mathsf {w}}.$$ Let x is any positive real number and let $${\mathsf {w}}$$ be the weight function constructed as follows (for notational convenience the dependence of $${\mathsf {w}}$$ on x will not be expressed explicitly). Set

\begin{aligned} {\mathsf {w}}(0)={\frac{{1}}{{\sqrt{2}}}}, \quad {\mathsf {w}}(1)=\sqrt{\frac{1}{2} + x}. \end{aligned}
(28)

Then clearly $$\alpha _{{\mathsf {w}}}(2-{\mathsf {w}}(0)^2)-1 > 0$$ and

\begin{aligned} {\mathsf {w}}(2):= {\frac{\sqrt{{\alpha _{\mathsf {w}}(2-{\mathsf {w}}(0)^2)-1}}}{{{\mathsf {w}}(1)}}} = \sqrt{\frac{1+3x}{1+2x}} > 1. \end{aligned}
(29)

Set

\begin{aligned} {\mathsf {w}}(n+2):= \xi _n({\mathsf {w}}(2)), \quad n \geqslant 1, \end{aligned}
(30)

where the functions $$\xi _n$$ are as in (22). It follows from the definition of $${\mathsf {w}}$$ and (22) that $$\sup _{n\geqslant 0} {\mathsf {w}}(n) < \infty ,$$ so by (17), $$C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu )).$$ Using Proposition 3.2, we deduce that $$C_{\phi , {\mathsf {w}}}$$ is a 2-isometry. It follows from (27)  that

\begin{aligned} h_{\phi ^{n}, {\mathsf {w}}'_{[n]}}(0) = {\frac{{1}}{{2^n(1+x)^{2n}}}} + \sum _{j=0}^{n-1} {\frac{{1+2x}}{{2^{n-j}(1+x)^{2(n-j)}(1+j({\mathsf {w}}(2)^2-1))}}}, \quad n\in {\mathbb {Z}}_+. \end{aligned}

By (29), we have

\begin{aligned} 1+j({\mathsf {w}}(2)^2-1) = \frac{1+(j+2)x}{1+2x}, \end{aligned}

which yields

\begin{aligned} h_{\phi ^{n}, {\mathsf {w}}'_{[n]}}(0) = {\frac{{1}}{{2^n(1+x)^{2n}}}} \left( 1 + (1+2x)^2 \sum _{j=0}^{n-1} {\frac{{2^{j}(1+x)^{2j}}}{{1+(j+2)x}}} \right) , \quad n\in {\mathbb {Z}}_+. \end{aligned}
(31)

$$\diamondsuit$$

Below we will continue the necessary preparations to achieve the main goal of this paper. For every $$n\in {\mathbb {Z}}_+$$, we define the real valued function $$\omega _n$$ on $$\varOmega _n:= (-\frac{1}{n+1},\infty )$$ by

\begin{aligned} \omega _n(x) = \frac{1}{2^n(1+x)^{2n}} \left( 1 + (1+2x)^2 \sum _{j=0}^{n-1} \frac{2^{j}(1+x)^{2j}}{1+(j+2)x} \right) , \quad x \in \varOmega _n, \, n \in {\mathbb {Z}}_+. \end{aligned}

Clearly, the following holds

\begin{aligned} \omega _{n}(x) = {\frac{{1}}{{2^n(1+x)^{2n}}}} \left( 1 + (1+2x)^2 S_n(x) \right) , \quad \quad x \in \varOmega _n, \, n\in {\mathbb {Z}}_+, \end{aligned}
(32)

where

\begin{aligned} S_n(x) := \sum _{j=0}^{n-1} \frac{2^{j}(1+x)^{2j}}{1+(j+2)x}, \quad x \in \varOmega _n,\, n\in {\mathbb {Z}}_+. \end{aligned}

Set

\begin{aligned} D_m(x) = \sum _{n=0}^m (-1)^n \left( {\begin{array}{c}m\\ n\end{array}}\right) \omega _n(x), \quad x \in \varOmega _m, \, m \in {\mathbb {Z}}_+. \end{aligned}
(33)

Then

\begin{aligned} D^{(l)}_m(x) = \sum _{n=0}^m (-1)^n \left( {\begin{array}{c}m\\ n\end{array}}\right) \omega ^{(l)}_n(x), \quad x \in \varOmega _m, \, m \in {\mathbb {Z}}_+, \, l \in {\mathbb {Z}}_+, \end{aligned}
(34)

where $$D^{(l)}_m$$ (resp. $$\omega ^{(l)}_n$$) stands for the l-th derivative of $$D_m$$ (resp. $$\omega _n$$). Applying the general Leibniz rule, we see that the l-th derivative $$S_n^{(l)}$$ of $$S_n$$ is given by

\begin{aligned} S_n^{(l)}(x) = \sum _{j=0}^{n-1} {2^{j}} \sum _{k=0}^{l} \left( {\begin{array}{c}l\\ k\end{array}}\right) \Big (\frac{1}{1+(j+2)x}\Big )^{(k)} ((1+x)^{2j})^{(l-k)}, \quad x \in \varOmega _n, \, l, n \in {\mathbb {Z}}_+. \end{aligned}

In particular, for every $$n\in {\mathbb {Z}}_+$$ we have

\begin{aligned} S_n(0)&= 2^n-1, \\ S_n^{(1)}(0)&= n 2^n - 4(2^n-1), \\ S_n^{(2)}(0)&= 2n^2 2^n -10n 2^n+ 24(2^n-1), \\ S_n^{(3)}(0)&= 2 n^3 2^n - 30 n^2 2^n +100 n 2^n - 192 (2^n -1), \\ S_n^{(4)}(0)&= 8 n^4 2^n - 56 n^3 2^n + 460 n^2 2^n - 1324 n 2^n + 2208 (2^n -1). \end{aligned}
(35)

Now we compute the lth derivative $$\omega ^{(l)}_n$$ of $$\omega _n.$$

### Lemma 4.2

The following assertions are valid :

1. (i)

if$$n\in {\mathbb {Z}}_+,$$lis a positive integer andxvaries over$$\varOmega _n,$$then (see (4))

\begin{aligned} \omega ^{(l)}_n(x)&= \left( {\frac{{1}}{{(1+x)^{2n}}}} \right) ^{(l)} \left[ {\frac{{1 + (1+2x)^2 S_n(x) }}{{2^n}}}\right] \nonumber \\&\quad + l \left( {\frac{{1}}{{(1+x)^{2n}}}} \right) ^{(l-1)} \left[ {\frac{{2^2(1+2x) S_n(x) + (1+2x)^2 S_n^{(1)}(x)}}{{2^n}}} \right] \nonumber \\&\quad + \sum _{i = 2}^l \left( {\begin{array}{c}l\\ i\end{array}}\right) \left( {\frac{{1}}{{(1+x)^{2n}}}} \right) ^{(l-i)} \nonumber \\&\quad \times \left[ {\frac{{2^3 \left( {\begin{array}{c}i\\ 2\end{array}}\right) S_n^{(i-2)}(x) + 2^2 \left( {\begin{array}{c}i\\ 1\end{array}}\right) (1+2x) S_n^{(i-1)}(x) + (1+2x)^2 S_n^{(i)}(x)}}{{2^n}}} \right] , \end{aligned}
(36)
2. (ii)

if $$l \in \{ 0, 1, 2, 3\}$$ and $$m\in \{4,5,6,\ldots \},$$ then $$D_m^{(l)}(0)=0.$$

### Proof

1. (i)

Applying the general Leibniz rule twice, we get

\begin{aligned} 2^n \omega ^{(l)}_n(x)&\overset{(32)}{=} \sum _{i=0}^{l} \left( {\begin{array}{c}l\\ i\end{array}}\right) \left( {\frac{{1}}{{(1+x)^{2n}}}} \right) ^{(l-i)} \left( 1 + (1+2x)^2 S_n(x) \right) ^{(i)}\\&= \left( {\frac{{1}}{{(1+x)^{2n}}}} \right) ^{(l)} \left( 1 + (1+2x)^2 S_n(x) \right) \\&\quad + \sum _{i=1}^{l} \left( {\begin{array}{c}l\\ i\end{array}}\right) \left( {\frac{{1}}{{(1+x)^{2n}}}}\right) ^{(l-i)} \left( (1+2x)^2 S_n(x) \right) ^{(i)} \\&= \left( {\frac{{1}}{{(1+x)^{2n}}}} \right) ^{(l)} \left( 1 + (1+2x)^2 S_n(x) \right) \\&\quad + l \left( {\frac{{1}}{{(1+x)^{2n}}}} \right) ^{(l-1)}\left( 2^2(1+2x) S_n(x) + (1+2x)^2 S_n^{(1)}(x) \right) \\&\quad + \sum _{i=2}^{l} \left( {\begin{array}{c}l\\ i\end{array}}\right) \left( {\frac{{1}}{{(1+x)^{2n}}}} \right) ^{(l-i)} \sum _{m=0}^{i} \left( {\begin{array}{c}i\\ m\end{array}}\right) ((1+2x)^2)^{(m)} S_n^{(i-m)}(x), \end{aligned}

which implies that (36) holds for $$l\geqslant 2,$$$$n\in {\mathbb {Z}}_+$$ and $$x\in \varOmega _n.$$ It is a matter of routine to verify that (36) holds for $$l=1,$$$$n\in {\mathbb {Z}}_+$$ and $$x\in \varOmega _n$$ as well. This yields (i).

2. (ii)

Using (35) we verify that the factors in the square brackets appearing in (36) (the third one for $$i=2,3$$) when calculated at 0 are of the form:

\begin{aligned} &\frac{S_n(0) +1}{2^n}= p_0(n), \text { where }p_0\in {\mathbb {C}}[z]\text { is of degree }0, \nonumber \\ & \frac{4 S_n(0) + S_n^{(1)}(0)}{2^n}= {} p_1(n), \text { where }p_1\in {\mathbb {C}}[z]\text { is of degree }1, \nonumber \\ & \frac{8S_n(0) + 8S^{(1)}_n(0) + S_n^{(2)}(0)}{2^n}= p_2(n), \text { where }p_2\in {\mathbb {C}}[z]\text { is of degree }2, \nonumber \\ & \frac{24 S^{(1)}_n(0) + 12 S^{(2)}_n(0) + S_n^{(3)}(0)}{2^n}= p_3(n), \text { where }p_3\in {\mathbb {C}}[z]\text { is of degree }3. \end{aligned}
(37)

This together with (36) implies that $$\omega ^{(l)}_n(0)$$ is a polynomial in n of degree at most l whenever $$l\in \{0,1,2,3\}.$$ Therefore, (ii) is a direct consequence of (1) and (34).$$\square$$

### Lemma 4.3

For every integer $$m\geqslant 5,$$ there exists $$\varepsilon _m \in (0,\infty )$$ such that $$D_m(x) < 0$$ for every $$x\in (0,\varepsilon _m).$$

### Proof

It follows from (35) that that the third factor in the square brackets appearing in (36) for $$i=4$$ when calculated at 0 is of the form

\begin{aligned} \frac{48 S_n^{(2)}(0) + 16 S_n^{(3)}(0) + S_n^{(4)}(0)}{2^n} = p_4(n)-\frac{288}{2^n}, \quad x\in \varOmega _n, \, n\in {\mathbb {Z}}_+, \end{aligned}

where $$p_4\in {\mathbb {C}}[z]$$ is of degree 4. Combined with (37) and (36), this yields

\begin{aligned} \omega ^{(4)}_n(0)&= p_0(n) \left( \frac{1}{(1+x)^{2n}} \right) ^{(4)}\Big |_{x=0} + 4 p_1(n) \left( \frac{1}{(1+x)^{2n}} \right) ^{(3)}\Big |_{x=0} \\&\quad + 6 p_2(n) \left( \frac{1}{(1+x)^{2n}} \right) ^{(2)} \Big |_{x=0} + 4 p_3(n) \left( \frac{1}{(1+x)^{2n}} \right) ^{(1)}\Big |_{x=0} \\&\quad + \left( p_4(n)-\frac{288}{2^n}\right) \left( \frac{1}{(1+x)^{2n}} \right) ^{(0)}\Big |_{x=0}, \quad n\in {\mathbb {Z}}_+. \end{aligned}

Hence, $$\omega ^{(4)}_n(0) + \frac{288}{2^n}$$ is a polynomial in n of degree at most 4. This fact, together with (1) and (34) implies that

\begin{aligned} D_m^{(4)}(0) =-288 \sum _{n=0}^{m} (-1)^{n} \left( {\begin{array}{c}m\\ n\end{array}}\right) 2^{-n}= - \frac{288}{2^m} < 0, \quad m \ge 5. \end{aligned}

Now applying Lemma 4.2(ii) and Taylor’s theorem to $$D_m$$ (see [21, Theorem 5.15] with $$n=4$$ and $$\alpha =0$$) completes the proof. $$\square$$

Concerning Lemma  4.3, the reader is referred to Fig. 1. Now we are ready to state the main result of the paper.

### Theorem 4.4

Let$${\mathscr {A}}$$, $$\mu$$, $$\phi$$and$${\mathsf {w}}$$be as in Example 4.1 (with$${\mathsf {w}}$$given by (28)–(30)). Then there exists$$\varepsilon \in (0,\infty )$$such that for every$$x \in (0, \varepsilon ),$$$$C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu ))$$and$$C_{\phi , {\mathsf {w}}}$$is a cyclic 2-isometry such that$$C_{\phi , {\mathsf {w}}}'$$is not subnormal.

### Proof

It follows from (19) and Example 4.1 that if $$x\in (0,\infty ),$$ then $$C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu ))$$ and $$C_{\phi , {\mathsf {w}}}$$ is a cyclic 2-isometry. In turn, by Lemma 4.3, there exists $$\varepsilon \in (0,\infty )$$ such that $$D_5(x) < 0$$ for every $$x \in (0, \varepsilon ).$$ Applying (3), (33) and (31), we deduce that whenever $$x \in (0, \varepsilon ),$$$$\{h_{\phi ^{n}, {\mathsf {w}}'_{[n]}}(0)\}_{n=0}^{\infty }$$ is not a Hausdorff moment sequence and consequently, by Theorem 3.3(i), $$C_{\phi , {\mathsf {w}}}'$$ is not subnormal. $$\square$$

We conclude the paper with the following observation.

### Remark 4.5

It is worth noting that if the parameter x in Example 4.1 equals 0, then $$C_{\phi , {\mathsf {w}}}\in {\varvec{B}}(L^2(\mu ))$$ and, by (16), (28), (29) and (30) with $$x=0$$, $$h_{\phi , {\mathsf {w}}}(n) = 1$$ for all $$n\in {\mathbb {Z}}_+,$$ which implies that $$C_{\phi , {\mathsf {w}}}$$ is an isometryFootnote 2 (see [11, (2.22)]). It follows that $$C_{\phi , {\mathsf {w}}}'$$ is an isometry and, as such, is subnormal (because each isometry has a unitary extension possibly in a larger Hilbert space, see e.g., [23, Proposition I.2.3]). Therefore, Theorem 4.4 breaks down at the point $$x=0.$$$$\diamondsuit$$

1. 1.

Recall that a 2-isometry is m-isometric for every integer $$m\geqslant 2$$ (see [1, §1]), and thus by [20, Lemma 1(a)] it is completely hyperexpansive.

2. 2.

In fact, in view of (20), $$\big (C_{\phi , {\mathsf {w}}}(L^2(\mu ))\big )^{\perp } = {\mathbb {C}}\cdot (e_{0} - e_{1}),$$ so $$C_{\phi , {\mathsf {w}}}$$ is not unitary.

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Correspondence to Jan Stochel.