A solution to the Cauchy dual subnormality problem for 2-isometries

TL;DR

This paper demonstrates that the Cauchy dual of a 2-isometry is not necessarily subnormal by providing counterexamples, thereby resolving a long-standing open problem in operator theory.

Contribution

It provides the first counterexamples to the Cauchy dual subnormality problem for 2-isometries, clarifying conditions under which the dual operator is subnormal.

Findings

Counterexamples show the Cauchy dual of a 2-isometry can be non-subnormal.

The kernel condition is not sufficient for subnormality of the Cauchy dual.

Certain classes like quasi-Brownian isometries always have subnormal Cauchy duals.

Abstract

The Cauchy dual subnormality problem asks whether the Cauchy dual operator $T^{\prime}:=T(T^*T)^{-1}$ of a $2$-isometry $T$ is subnormal. In the present paper we show that the problem has a negative solution. The first counterexample depends heavily on a reconstruction theorem stating that if $T$ is a $2$-isometric weighted shift on a rooted directed tree with nonzero weights that satisfies the perturbed kernel condition, then $T^{\prime}$ is subnormal if and only if $T$ satisfies the (unperturbed) kernel condition. The second counterexample arises from a $2$-isometric adjacency operator of a locally finite rooted directed tree again by thorough investigations of positive solutions of the Cauchy dual subnormality problem in this context. We prove that if $T$ is a $2$-isometry satisfying the kernel condition or a quasi-Brownian isometry, then $T^{\prime}$ is subnormal. We construct a $2$-isometric adjacency operator $T$ of a rooted directed tree such that $T$ does not satisfy the kernel condition, $T$ is not a quasi-Brownian isometry and $T^{\prime}$ is subnormal.

Equations

$$B_m(T)=\sum _{k=0}^m(-1{)}^k(\genfrac{}{}{0ex}{}{m}{k}){T^{\ast }}^k{T}^k.$$

$$T^{\prime }=T(T^{\ast }T{)}^{-1}.$$

$$(T^{\prime }{)}^{\prime }=T,$$

$$T^{\ast }{T}^{\prime }=I,$$

$$T^{\prime }{T}^{\ast }\text{ is the orthogonal projection of H onto }T(\mathcal{H}),$$

$$T^{\prime \ast }{T}^{\prime }=(T^{\ast }T{)}^{-1}.$$

$$\Vert S^{n}f{\Vert }^2={\int }_{[0,\infty )}{t}^{n}d{\mu }_f(t),n=0,1,2,\dots .$$

$$L_n:=\{\begin{array}{ll}(T^n{T}^{\prime n}{)}^{\ast }& \text{if n is a nonnegative integer},\\ T^{|n|}{T}^{\prime |n|}& \text{if n is a negative integer}.\end{array}$$

$$(T^{\prime n}{)}^{\ast }=(T^n{T}^{\prime n}{)}^{\ast }{T}^{\prime n},n=0,1,2,\dots ,$$

$$T^{\ast }T(\ker T^{\ast })\subseteq \ker T^{\ast }.$$

$$|T|(\ker T^{\ast })\subseteq \ker T^{\ast }.$$

$${\nabla }_{T}T=P{\nabla }_{T}T=T^{\prime }(T^{\ast }{\nabla }_{T}T)\stackrel{(\mathrm{b})}{=}0.$$

$$T^{\prime }(T^{\ast }T-I)=T-T^{\prime }\stackrel{(\mathrm{iii})}{=}{T}^{\ast }{T}^2-T=(T^{\ast }T-I)T,$$

$${△}_T\stackrel{\text{eqref}tpr}{=}{T}^{\ast }{T}^{\prime }{△}_T\stackrel{(\mathrm{v})}{=}{T}^{\ast }{△}_{T}T,$$

$$T^{\prime }{△}_T\stackrel{\text{eqref}tpr2}{=}P{T}^{\prime }{△}_T\stackrel{(\mathrm{v})}{=}P{△}_{T}T.$$

$$(I-P){△}_{T}T={△}_{T}T-T^{\prime }{△}_T\stackrel{(\mathrm{v})}{=}0.$$

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

$$\displaystyle\big{\langle}\{g_{n}\}_{n=0}^{\infty},\{h_{n}\}_{n=0}^{\infty}\big{\rangle}=\sum_{n=0}^{\infty}\langle g_{n},h_{n}\rangle,\quad\{g_{n}\}_{n=0}^{\infty},\{h_{n}\}_{n=0}^{\infty}\in\ell^{2}_{\mathcal{M}}.$$

$$W(h_0,h_1,\dots )=(0,W_0{h}_0,W_1{h}_1,\dots ),(h_0,h_1,\dots )\in {\ell }_{\mathcal{M}}^2,$$

$$W^{\ast }(h_0,h_1,\dots )$$

$$W^{\ast }W(h_0,h_1,\dots )$$

$$W_{m,n}=\{\begin{array}{ll}{W}_{m-1}{W}_{m-2}\cdots W_n& \text{if }m>n,\\ I& \text{if }m=n.\end{array}$$

$$\displaystyle\left.\begin{gathered}W_{n}=\int_{[1,\infty)}\xi_{n}(x)E(dx),\quad n\in\mathbb{Z}_{+},\\ \begin{minipage}{266.9436pt} where $E$ is a compactly supported $\boldsymbol{B}(\mathcal{M})$-valued Borel spectral measure on the iterval $[1,\infty)$ and $\{\xi_{n}\}_{n=0}^{\infty}$ is defined by \eqref{xin}. \end{minipage}\end{gathered}\;\right\}$$

$$⟨Tf,T^{n}g⟩=⟨T^{\ast }(T^{\ast }Tf),T^{n-2}g⟩=0,f,g\in \ker T^{\ast }.$$

$$T^j(\ker T^{\ast })\perp T^k(\ker T^{\ast }),k\in \{j+1,j+2,\dots \},j\in \{0,\dots ,n\}.$$

$$T^{\ast (n-1)}{T}^{n-1}-2T^{\ast n}{T}^n+T^{\ast (n+1)}{T}^{n+1}=0,$$

$$⟨T^{n+1}f,T^{k}g⟩$$

$$\stackrel{\text{eqref}tjk-2}{=}2⟨T^{n}f,T^{k-1}g⟩-⟨T^{n-1}f,T^{k-2}g⟩$$

$$\stackrel{\text{eqref}tjk}{=}0,f,g\in \ker T^{\ast },$$

$$V(h_0\oplus h_1\oplus \dots )=(V_0{h}_0,V_1{h}_1,\dots ),h_0\oplus h_1\oplus \dots \in {\mathcal{H}}_u^{\perp }.$$

$$VA(h_0\oplus h_1\oplus \dots )$$

$$=(0,(V_1{\Lambda }_0{V}_0^{-1})V_0{h}_0,(V_2{\Lambda }_1{V}_1^{-1})V_1{h}_1,\dots )$$

$$=SV(h_0\oplus h_1\oplus \dots ),h_0\oplus h_1\oplus \dots \in {\mathcal{H}}_u^{\perp },$$

$$\Vert W_0{h}_0\Vert =\Vert (0,W_0{h}_0,0,\dots )\Vert =\Vert W(h_0,0,\dots )\Vert ⩾\Vert h_0\Vert ,h_0\in {\mathcal{M}}_0.$$

Full text

The research of the third and fourth authors was supported by the NCN (National Science Center), decision No. DEC-2013/11/B/ST1/03613.

1. Introduction

Let $\mathcal{H}$ be a (complex) Hilbert space. Denote by $\boldsymbol{B}(\mathcal{H})$ the $C^{*}$-algebra of all bonded linear operators on $\mathcal{H}$. An operator $S\in\boldsymbol{B}(\mathcal{H})$ is said to be subnormal if there exist a Hilbert space $\mathcal{K}$ containing $\mathcal{H}$ and a normal operator $N\in\boldsymbol{B}(\mathcal{K})$ such that $Nh=Sh$ for every $h\in\mathcal{H}.$ An operator $T\in\boldsymbol{B}(\mathcal{H})$ is called hyponormal if the self-commutator $T^{*}T-TT^{*}$ of $T$ is positive. We recall that every subnormal operator is hyponormal, but not conversely. For a comprehensive account on the theory of subnormal and hyponormal operators, the reader is referred to  [17].

Given a positive integer $m$, we say that an operator $T\in\boldsymbol{B}(\mathcal{H})$ is an $m$-isometry (or $m$-isometric) if $B_{m}(T)=0$, where

[TABLE]

Clearly, a $1$-isometry is an isometry. We say that $T$ is $2$-hyperexpansive (resp., completely hyperexpansive) if $B_{2}(T)\leqslant 0$ (resp., $B_{m}(T)\leqslant 0$ for all positive integers $m$). The notion of an $m$-isometric operator has been invented by Agler (see [3, p. 11]). The concept of a $2$-hyperexpansive operator goes back to Richter [35] (see also [6, Remark  2]). The notion of a completely hyperexpansive operator has been introduced by Athavale [6]. It is well-known that a $2$-isometry is $m$-isometric for every integer $m\geqslant 2$, and thus it is completely hyperexpansive (see [4, Paper I, §1]).

The Cauchy dual operator $T^{\prime}$ of a left-invertible operator $T\in\boldsymbol{B}(\mathcal{H})$ is defined  by111 Note that left-invertibility of $T$ implies invertibility of $T^{*}T$ in the algebra $\boldsymbol{B}(\mathcal{H})$.

[TABLE]

$T^{\prime}$ is also called the Cauchy dual of $T$. Recall that the range of a left-invertible operator is always closed. It is easily seen that if $T$ is left-invertible, then $T^{\prime}$ is again left-invertible and the following conditions hold*:*

[TABLE]

The notion of the Cauchy dual operator has been introduced and studied by Shimorin in the context of the wandering subspace problem for Bergman-type operators [39]. The Cauchy dual technique has been employed in [12] to prove Berger-Shaw-type theorems for $2$-hyperexpansive (or in Shimorin’s terminology, concave) operators. It is also important to note that

[TABLE]

Indeed, by [35, Lemma 1], we have $\|Tf\|\geqslant\|f\|$ for all $f\in\mathcal{H}$, which implies that $T$ is left-invertible and $\|T^{\prime}\|=\|U|T|^{-1}\|\leqslant 1$, where $T=U|T|$ is the polar decomposition of $T$. The map that sends $T$ to $T^{\prime}$ links $2$-hyperexpansive operators to hyponormal ones (see [40, Sect. 5] and [12, Theorem  2.9]). Moreover, if $T$ is a $2$-hyperexpansive operator, then $T^{\prime}$ is power hyponormal, i.e., all positive integer powers $T^{\prime n}$ of $T^{\prime}$ are hyponormal (see [13, Theorem  3.1]). What is more interesting, if $T$ is a completely hyperexpansive unilateral weighted shift, then $T^{\prime}$ is a subnormal contraction (see [6, Proposition  6]). This leads to the question originally posed in [12, Question 2.11]: is the Cauchy dual of a completely hyperexpansive operator a subnormal contraction? Here we consider the following version of this problem.

Problem**.**

Is the Cauchy dual of a $2$-isometry a subnormal contraction?

In this paper we solve the above problem and [12, Question 2.11] in the negative (see Examples  6.6 and 7.10). Since, by [13, Theorem  3.1], the Cauchy dual of a $2$-isometry is power hyponormal, we get also examples of non-subnormal power hyponormal operators (cf. [18, 19]). The considerations in [5] related to the above problem enabled the first two authors to solve negatively the problem of subnormality of module tensor product of two subnormal modules posed by Salinas in  1988.

Using (5) and [20, Corollary], we can reformulate the Cauchy dual subnormality problem as follows: is the sequence $\{T^{\prime*n}T^{\prime n}\}_{n=0}^{\infty}$ an operator Hausdorff moment sequence for a $2$-isometry $T\in\boldsymbol{B}(\mathcal{H})$? Hence, it is important to determine the exact form of the operator Hausdorff moment sequence $\{T^{\prime*n}T^{\prime n}\}_{n=0}^{\infty}$ if $T^{\prime}$ is subnormal.

It is worth mentioning that the question of subnormality of $2$-isometric operators has a simple solution. This is due to the following more general result which is a direct consequence of [35, Lemma 1]: if $T\in\boldsymbol{B}(\mathcal{H})$ is a subnormal $($or, more generally, a normaloid$)$ operator which is $2$-hyperexpansive, then $T$ is an isometry.

For future reference, we explicitly state two celebrated criteria for subnormality of bounded operators that are due to Lambert and Agler, respectively. It is worth pointing out that in view of the Hausdorff moment theorem (see [7, Theorem  4.6.11]), these two results are equivalent.

Theorem 1.1** **([31], [46, Proposition 

2.3]).

An operator $S\in\boldsymbol{B}(\mathcal{H})$ is subnormal if and only if for every $f\in\mathcal{H},$ the sequence $\{\|S^{n}f\|^{2}\}_{n=0}^{\infty}$ is a Stieltjes moment sequence, i.e., there exists a positive Borel measure $\mu_{f}$ on $[0,\infty)$ such that

[TABLE]

Theorem 1.2** ([2, Theorem  3.1]).**

An operator $S\in\boldsymbol{B}(\mathcal{H})$ is a subnormal contraction if and only if $B_{m}(S)\geqslant 0$ for all $m=1,2,3,\ldots$.

It is also of some interest to reveal the following intimate connection of the Cauchy dual subnormality problem to the theory of Toeplitz matrices.

Proposition 1.3**.**

Let $T\in\boldsymbol{B}(\mathcal{H})$ be a $2$-isometry such that for all $h\in\mathcal{H}$ and $k\in\{0,1,2,\ldots\}$, the Toeplitz matrix $[\langle{L_{j-i}h},\,{h}\rangle]_{i,j=0}^{k}$ is positive definite, where

[TABLE]

Then the Cauchy dual $T^{\prime}$ of $T$ is subnormal.

Proof.

It follows from (2) that

[TABLE]

which together with [47, Theorem  1] completes the proof. ∎

The remainder of the paper is organized as follows. In Section  2 we give a purely algebraic characterization of $2$-isometries satisfying the kernel condition (see Lemma  2.3) and provide a model for such operators built on operator valued unilateral weighted shifts (see Theorem  2.5). We also give a brief discussion of the case of cyclic analytic $2$-isometries satisfying the kernel condition (see Proposition  2.8). In Section  3, we prove that the Cauchy dual $T^{\prime}$ of a $2$-isometry $T$ satisfying the kernel condition is a subnormal contraction (see Theorem  3.3). Theorem  3.3 is not true for $m$-isometries for $m\geqslant 3$ (see Example  3.10). Section  4 deals with quasi-Brownian isometries, a subclass of $2$-isometries, which generalize Brownian isometries introduced by Agler and Stankus in [4]. Using a block operator model for this class of operators given in [32], we show that the Cauchy dual $T^{\prime}$ of a quasi-Brownian isometry $T$ is a subnormal contraction (see Theorem  4.5). We also show that a quasi-Brownian isometry satisfying the kernel condition must be an isometry (see Corollary  4.6). In Section  5, we collect selected properties of weighted shifts on directed trees that we need for further considerations. This class of operators was introduced in [28] and intensively studied since then (see e.g., [29, 23, 14, 9, 11, 33, 30]). We show, among other things, that the Cauchy dual of a left-invertible weighted shift on a directed tree is still a weighted shift on a directed tree (see Lemma  5.5). We also prove that $2$-isometric weighted shifts on rootless directed trees with nonzero weights that satisfy the kernel condition and Brownian isometric weighted shifts on rooted directed trees are automatically isometric (see Propositions  5.11 and 5.12). In Section  6, we prove a reconstruction theorem stating that if $S_{\boldsymbol{\lambda}}$ is a $2$-isometric weighted shift on a rooted directed tree with nonzero weights that satisfies the perturbed kernel condition (66) for some positive integer $k$, then $S_{\boldsymbol{\lambda}}^{\prime}$ is subnormal if and only if $S_{\boldsymbol{\lambda}}$ satisfies the kernel condition (see Theorem  6.5). The reconstruction theorem enables us to answer the Cauchy dual subnormality problem in the negative (see Example  6.6). In Section  7, we investigate the Cauchy dual subnormality problem for adjacency operators of directed trees. This class of operators plays an important role in graph theory (see [8, 22, 28]). We show that if a rooted directed tree satisfies certain degree constraints and the corresponding adjacency operator $S_{\mathbb{1}}$ is a $2$-isometry, then the Cauchy dual operator $S_{\mathbb{1}}^{\prime}$ of $S_{\mathbb{1}}$ is subnormal (Theorem  7.8). This enables us to construct a $2$-isometric adjacency operator $S_{\mathbb{1}}$ of a directed tree, which does not satisfy the kernel condition, which is not a quasi-Brownian isometry, but which has the property that $S_{\mathbb{1}}^{\prime}$ is a subnormal contraction (see Example  7.9). Finally, we show that the Cauchy dual subnormality problem has a negative solution in the class of adjacency operators of directed trees (see Example  7.10).

Now we fix notation and terminology. Let $\mathbb{Z}$, $\mathbb{R}$ and $\mathbb{C}$ stand for the sets of integers, real numbers and complex numbers, respectively. Denote by $\mathbb{N}$, $\mathbb{Z}_{+}$ and $\mathbb{R}_{+}$ the sets of positive integers, nonnegative integers and nonnegative real numbers, respectively. Given a set $X$, we write $\chi_{\varDelta}$ for the characteristic function of a subset $\varDelta$ of $X$. The $\sigma$-algebra of all Borel subsets of a topological space $X$ is denoted by ${\mathfrak{B}}(X)$. For $a\in\mathbb{R},$ $\delta_{a}$ stands for the Borel probability measure on $\mathbb{R}$ supported on $\{a\}$. In this paper, Hilbert spaces are assumed to be complex and operators are assumed to be linear. Let $\mathcal{H}$ be a Hilbert space. As usual, we denote by $\dim\mathcal{H}$ the orthogonal dimension of $\mathcal{H}$. If $f\in\mathcal{H}$, then $\langle f\rangle$ stands for the linear span of the singleton of $f$. Given another Hilbert space $\mathcal{K}$, we denote by $\boldsymbol{B}(\mathcal{H},\mathcal{K})$ the Banach space of all bounded operators from $\mathcal{H}$ to $\mathcal{K}$. The kernel and the range of an operator $T\in\boldsymbol{B}(\mathcal{H},\mathcal{K})$ are denoted by $\ker T$ and $\mathrm{ran\,}T$, respectively. We abbreviate $\boldsymbol{B}(\mathcal{H},\mathcal{H})$ to $\boldsymbol{B}(\mathcal{H})$ and regard $\boldsymbol{B}(\mathcal{H})$ as a $C^{*}$-algebra. The identity operator on $\mathcal{H}$ is denoted by $I_{\mathcal{H}}$ (or simply by $I$ if no ambiguity arises). We write $\sigma(T)$ for the spectrum of $T\in\boldsymbol{B}(\mathcal{H})$. If $S$ and $T$ are Hilbert space operators which are unitarily equivalent, then we write $S\cong T$.

Following [36], we say that an operator $T\in\boldsymbol{B}(\mathcal{H})$ is analytic if $\bigcap_{n=1}^{\infty}T^{n}(\mathcal{H})=\{0\}$. An operator $T\in\boldsymbol{B}(\mathcal{H})$ is said to be completely non-unitary (resp., pure) if there is no nonzero reducing closed vector subspace $\mathcal{L}$ of $\mathcal{H}$ such that the restriction $T|_{\mathcal{L}}$ of $T$ to $\mathcal{L}$ is a unitary (resp., a normal) operator. Clearly, every analytic operator is completely non-unitary. Recall that any operator $T\in\boldsymbol{B}(\mathcal{H})$ has a unique orthogonal decomposition $T=N\oplus R$ such that $N$ is a normal operator and $R$ is a pure operator (see [34, Corollary  1.3]). We shall refer to $N$ and $R$ as the normal and pure parts of $T$, respectively.

2. A model for $2$-isometries

satisfying the kernel condition

The goal of this section is to show that a non-unitary $2$-isometry satisfying the kernel condition is unitarily equivalent to an orthogonal sum of a unitary operator and an operator valued unilateral weighted shift (see Theorem  2.5).

We say that an operator $T\in\boldsymbol{B}(\mathcal{H})$ satisfies the kernel condition if

[TABLE]

By the square root lemma (see [43, Theorem  2.4.4]), (6) holds if and only if

[TABLE]

It is easily seen that any positive integral power of a unilateral weighted shift satisfies the kernel condition (see also the proof of Corollary  3.6). It is a routine matter to verify that weighted translation semigroups studied by Embry and Lambert in [21] consist of operators satisfying the kernel condition. Other examples of operators satisfying the kernel condition will appear in this paper when solving the Cauchy dual subnormality problem.

The kernel condition is preserved by the operation of taking the Cauchy dual.

Proposition 2.1**.**

Let $T\in\boldsymbol{B}(\mathcal{H})$ be a left-invertible operator. Then the following conditions are equivalent:

1. (i)

$T$* satisfies the kernel condition,* 2. (ii)

$T^{*}T(\ker T^{*})=\ker T^{*},$** 3. (iii)

$T^{\prime}$* satisfies the kernel condition.*

Proof.

The equivalence (i)$\Leftrightarrow$(ii) is a consequence of the following fact.

[TABLE]

This together with (1), (4) and the equation $\ker T^{\prime*}=\ker T^{*}$ yields (ii)$\Leftrightarrow$(iii). ∎

The next result, whose proof is left to the reader, shows that under some circumstances the Cauchy dual of a restriction of a left-invertible operator to its invariant subspace is equal to the restriction of the Cauchy dual operator.

Proposition 2.2**.**

Suppose that $T\in\boldsymbol{B}(\mathcal{H})$ is a left-invertible operator and $\mathcal{L}$ is a closed vector subspace of $\mathcal{H}$ such that $T(\mathcal{L})\subseteq\mathcal{L}$ and $T^{*}T(\mathcal{L})\subseteq\mathcal{L}$. Then $T|_{\mathcal{L}}$ is left-invertible, $T^{\prime}(\mathcal{L})\subseteq\mathcal{L}$, $T^{\prime*}T^{\prime}(\mathcal{L})\subseteq\mathcal{L}$ and $(T|_{\mathcal{L}})^{\prime}=T^{\prime}|_{\mathcal{L}}$. In particular, if $\mathcal{L}$ reduces $T$, then $\mathcal{L}$ and $\mathcal{L}^{\perp}$ reduce $T^{\prime}$ and $T^{\prime}=(T|_{\mathcal{L}})^{\prime}\oplus(T|_{\mathcal{L}^{\perp}})^{\prime}$.

Note that in general the assumptions of Proposition  2.2 do not imply that $\mathcal{L}$ reduces $T$. Indeed, it is enough to consider an isometric unilateral shift $V$ of multiplicity $1$ and any of its nontrivial closed invariant vector subspaces, say $\mathcal{L}$. Then $\mathcal{L}$ does not reduce $V$. This is due to the fact that $V$ is irreducible, i.e., there is no nontrivial closed vector subspace of $\mathcal{H}$ which reduces $V$ (see [38, Corollary  2, p. 63]). Recall that the Beurling theorem completely describes the lattice of all closed vector subspaces of $\mathcal{H}$ which are invariant for $V$ (see [37, Theorem  17.21]).

Now we give some purely algebraic characterizations of $2$-isometries satisfying the kernel condition.

Lemma 2.3**.**

Let $T\in\boldsymbol{B}(\mathcal{H})$ be a left-invertible operator. Then the following conditions are equivalent:

1. (i)

$T$* is a $2$-isometry such that $T^{*}T(\ker T^{*})\subseteq\ker T^{*},$* 2. (ii)

$T$* is a $2$-isometry such that $T^{*}T(\mathrm{ran\,}T)\subseteq\mathrm{ran\,}T,$* 3. (iii)

$T^{\prime}-2T+T^{*}T^{2}=0,$** 4. (iv)

$(T^{*}T^{2}T^{*}-2TT^{*}+I)T=0,$** 5. (v)

$T^{\prime}(T^{*}T-I)=(T^{*}T-I)T.$**

Observe that, by (7), Lemma  2.3 remains true if inclusions appearing in (i) and (ii) are replaced by equalities.

Proof of Lemma  2.3.

Set $\triangle_{T}=T^{*}T-I$ and $\nabla_{T}=T^{\prime}T^{\prime*}-2I+T^{*}T$. Since, by (2), $T^{\prime*}T=I$ we see that $\nabla_{T}T=T^{\prime}-2T+T^{*}T^{2}$ and $T^{*}\nabla_{T}T=I-2T^{*}T+{T^{*}}^{2}T^{2}$. This implies that

1. (a)

the equation (iii) is equivalent to $\nabla_{T}T=0,$ 2. (b)

$T$ is a $2$-isometry if and only if $T^{*}\nabla_{T}T=0,$ 3. (c)

$T$ is a $2$-isometry if and only if $T^{*}\triangle_{T}T=\triangle_{T}.$

(i)$\Leftrightarrow$(ii) This is obvious because $\mathrm{ran\,}T$ is closed.

(ii)$\Rightarrow$(iii) Assume that (ii) holds. Since by assumption $\mathrm{ran\,}(\nabla_{T}T)\subseteq\mathrm{ran\,}{T}$ and, by (3), $P:=T^{\prime}T^{*}$ is the orthogonal projection of $\mathcal{H}$ onto $\mathrm{ran\,}T$, we get

[TABLE]

Hence, by (a), the equality (iii) holds.

(iii)$\Leftrightarrow$(iv) This can be deduced from the definition of $T^{\prime}.$

(iii)$\Rightarrow$(v) If (iii) holds, then

[TABLE]

which gives (v).

(v)$\Rightarrow$(ii) Suppose (v) holds. Note that

[TABLE]

and thus, by (c), $T$ is a $2$-isometry. It suffices to check that $\triangle_{T}(\mbox{ran}~{}T)\subseteq\mbox{ran}~{}T$. As above $P=T^{\prime}T^{*}$. Since $\mathrm{ran\,}(T^{\prime})\subseteq\mathrm{ran\,}~{}T$, we have

[TABLE]

This implies that

[TABLE]

Hence $P\triangle_{T}T=\triangle_{T}T$ and thus $\triangle_{T}(\mbox{ran}~{}T)\subseteq\mbox{ran}~{}T$. This completes the proof. ∎

Below, we collect some properties (whose verifications are left to the reader) of the sequence $\{\xi_{n}\}_{n=0}^{\infty}$ of self-maps of the interval $[1,\infty)$ given by

[TABLE]

Lemma 2.4**.**

The sequence $\{\xi_{n}\}_{n=0}^{\infty}$ given by (8) has the following properties:

1. (i)

$\xi_{0}(x)=x$* for all $x\in[1,\infty)$,* 2. (ii)

$\xi_{m+n}(x)=(\xi_{m}\circ\xi_{n})(x)$* for all $x\in[1,\infty)$ and $m,n\in\mathbb{Z}_{+}$,* 3. (iii)

$\xi_{n}(1)=1$* for all $n\in\mathbb{Z}_{+}$,* 4. (iv)

$\xi_{n}(x)>\xi_{n+1}(x)>1$* for all $x\in(1,\infty)$ and $n\in\mathbb{Z}_{+}$.*

Before stating the main result of this section, we recall the definition of an operator valued unilateral weighted shift. Let $\mathcal{M}$ be a nonzero Hilbert space. Denote by $\ell^{2}_{\mathcal{M}}$ the Hilbert space of all vector sequences $\{h_{n}\}_{n=0}^{\infty}\subseteq\mathcal{M}$ such that $\sum_{n=0}^{\infty}\|h_{n}\|^{2}<\infty$ equipped with the standard inner product

[TABLE]

If $\{W_{n}\}_{n=0}^{\infty}\subseteq\boldsymbol{B}(\mathcal{M})$ is a uniformly bounded sequence of operators, then the operator $W\in\boldsymbol{B}(\ell^{2}_{\mathcal{M}})$ defined by

[TABLE]

is called an operator valued unilateral weighted shift with weights $\{W_{n}\}_{n=0}^{\infty}$. If each weight $W_{n}$ of $W$ is an invertible (resp., a positive) element of the $C^{*}$-algebra $\boldsymbol{B}(\mathcal{M})$, then we say that $W$ is an operator valued unilateral weighted shift with invertible (resp., positive) weights. Putting $\mathcal{M}=\mathbb{C}$, we arrive at the well-known notion of a unilateral weighted shift in $\ell^{2}$.

Let $W$ be as in (9). It is easy to verify that

[TABLE]

Given integers $m\geqslant n\geqslant 0$, we set (see [27, p. 409])

[TABLE]

Now we characterize non-unitary $2$-isometric operators satisfying the kernel condition. In what follows, by a unitary operator, we mean a unitary operator $U\in\boldsymbol{B}(\mathcal{K})$, where $\mathcal{K}$ is an arbitrary Hilbert space; the case $\mathcal{K}=\{0\}$ is not excluded.

Theorem 2.5**.**

If $T\in\boldsymbol{B}(\mathcal{H})$ is a non-unitary $2$-isometry, then the following conditions are equivalent:

1. (i)

$T$* satisfies the kernel condition,* 2. (ii)

$T(\ker T^{*})\perp T^{n}(\ker T^{*})$* for every integer $n\geqslant 2,$* 3. (iii)

the spaces $\{T^{n}(\ker T^{*})\}_{n=0}^{\infty}$ are mutually orthogonal, 4. (iv)

$T\cong U\oplus W$, where $U$ is a unitary operator and $W$ is an operator valued unilateral weighted shift with invertible positive weights, 5. (v)

$T\cong U\oplus W,$* where $U$ is a unitary operator and $W$ is an operator valued unilateral weighted shift on $\ell^{2}_{\mathcal{M}}$ with weights $\{W_{n}\}_{n=0}^{\infty}$ defined by*

[TABLE]

Moreover, the following hold:

1. (a)

if $T\in\boldsymbol{B}(\mathcal{H})$ is a non-unitary $2$-isometry that satisfies (i), then $\mathcal{H}_{u}:=\bigcap_{n=1}^{\infty}T^{n}(\mathcal{H})$ is a closed vector subspace of $\mathcal{H}$ reducing $T$ to a unitary operator and $T|_{\mathcal{H}_{u}^{\perp}}\cong W$, where $W$ is an operator valued unilateral weighted shift on $\ell^{2}_{\mathcal{M}}$ with weights $\{W_{n}\}_{n=0}^{\infty}$ given by (15) and $\dim\mathcal{M}=\dim\ker T^{*},$ 2. (b)

if $U$ is a unitary operator and $W$ is an operator valued unilateral weighted shift on $\ell^{2}_{\mathcal{M}}$ with weights222 Note that, in view of (18), the sequence $\{W_{n}\}_{n=0}^{\infty}\subseteq\boldsymbol{B}(\mathcal{M})$ defined by (15) is uniformly bounded, and consequently $W\in\boldsymbol{B}(\ell^{2}_{\mathcal{M}})$.* $\{W_{n}\}_{n=0}^{\infty}$ defined by (15), then $T:=U\oplus W$ is a non-unitary $2$-isometry that satisfies (i), $U$ is the normal part of $T$, $W$ is the pure part of $T$ and $\ker W^{*}=\mathcal{M}\oplus\{0\}\oplus\{0\}\oplus\ldots.$*

Proof.

Assume that $T$ is a non-unitary $2$-isometry.

(i)$\Rightarrow$(ii) Note that for every integer $n\geqslant 2$,

[TABLE]

(ii)$\Rightarrow$(iii) It suffices to show that for every integer $n\geqslant 0,$

[TABLE]

We use induction on $n$. The cases $n=0$ and $n=1$ are obvious. Suppose (16) holds for a fixed $n\geqslant 1$. Since $I-2T^{*}T+{T^{*}}^{2}T^{2}=0$ yields

[TABLE]

we deduce that for every integer $k\geqslant n+2$,

[TABLE]

which completes the induction argument and gives (iv).

(iii)$\Rightarrow$(v) By [39, Theorem  3.6] (which is also valid for inseparable Hilbert spaces), $\mathcal{H}_{u}$ is a closed vector subspace of $\mathcal{H}$ that reduces $T$ to the unitary operator $U$ and $\mathcal{H}_{u}^{\perp}=\bigvee_{n=0}^{\infty}T^{n}(\ker T^{*})$. Since $T$ is non-unitary, $\mathcal{H}_{u}^{\perp}\neq\{0\}$ and consequently $\ker T^{*}\neq\{0\}$. Hence, by the injectivity of $T$, $\mathcal{M}_{n}:=T^{n}(\ker T^{*})\neq\{0\}$ for all $n\in\mathbb{Z}_{+}$. Clearly, $A:=T|_{\mathcal{H}_{u}^{\perp}}$ is a $2$-isometry. Since the operator $T$ is bounded from below, we deduce that for every $n\in\mathbb{Z}_{+}$, $\mathcal{M}_{n}$ is a closed vector subspace of $\mathcal{H}$ and $\varLambda_{n}:=T|_{\mathcal{M}_{n}}\colon\mathcal{M}_{n}\to\mathcal{M}_{n+1}$ is a linear homeomorphism. Therefore, by [24, Problem 56], for every $n\in\mathbb{Z}_{+}$, the Hilbert spaces $\mathcal{M}_{n}$ and $\mathcal{M}_{0}$ are unitarily equivalent. Let, for $n\in\mathbb{Z}_{+}$, $V_{n}\colon\mathcal{M}_{n}\to\mathcal{M}_{0}$ be any unitary isomorphism. Since $\mathcal{H}_{u}^{\perp}=\bigoplus_{n=0}^{\infty}\mathcal{M}_{n}$, we can define the unitary isomorphism $V\colon\mathcal{H}_{u}^{\perp}\to\ell^{2}_{\mathcal{M}_{0}}$ by

[TABLE]

Let $S\in\boldsymbol{B}(\ell^{2}_{\mathcal{M}_{0}})$ be the operator valued unilateral weighted shift with (uniformly bounded) weights $\{V_{n+1}\varLambda_{n}V_{n}^{-1}\}_{n=0}^{\infty}\subseteq\boldsymbol{B}(\mathcal{M}_{0})$. Then

[TABLE]

which means that $A$ is unitarily equivalent to $S$. Since the weights of the $2$-isometry $S$ are invertible in $\boldsymbol{B}(\mathcal{M}_{0})$, we infer from [27, Corollary  2.3] that $S$ is unitarily equivalent to a $2$-isometric operator valued unilateral weighted shift $W$ on $\ell^{2}_{\mathcal{M}_{0}}$ with invertible weights $\{W_{n}\}_{n=0}^{\infty}\subseteq\boldsymbol{B}(\mathcal{M}_{0})$ such that $W_{n}\,\cdots\,W_{0}\geqslant 0$ for all integers $n\geqslant 0$. In turn, by [35, Lemma 1], $\|Wh\|\geqslant\|h\|$ for all $h\in\ell^{2}_{\mathcal{M}_{0}}$, which yields

[TABLE]

Hence $W_{0}\geqslant I$. This combined with the proof of [27, Theorem  3.3] implies that

[TABLE]

where $E$ is the spectral measure of $W_{0}$ and $\{\hat{\xi}_{n}\}_{n=0}^{\infty}$ is the sequence of self-maps of the interval $[1,\infty)$ defined recursively by

[TABLE]

Using induction, one can show that $\hat{\xi}_{n}=\xi_{n}$ for all $n\in\mathbb{Z}_{+}$, which gives (v).

The implication (v)$\Rightarrow$(iv) is obvious.

(iv)$\Rightarrow$(i) Let $W$ be as in (9). Since the weights of $W$ are invertible, we infer from (10) that $\ker W^{*}=\mathcal{M}\oplus\{0\}\oplus\{0\}\oplus\ldots.$ This combined with (11) yields $W^{*}W(\ker W^{*})=\ker W^{*}$, which gives (i).

Now we turn to the proof of the “moreover” part.

(a) This has already been done in the proof of the implication (iii)$\Rightarrow$(v).

(b) First, we show that $U$ and $W$ are the normal and pure parts of $T$, respectively. Denote by $\mathcal{K}$ the Hilbert space in which the unitary operator $U$ acts. Since $W$ is an operator valued unilateral weighted shift with invertible weights, we infer from (9) and (10) that

[TABLE]

This, together with [34, Corollary  1.3], proves our claim.

Arguing as in the proof of the implication (iv)$\Rightarrow$(i), we verify that $\ker W^{*}=\mathcal{M}\oplus\{0\}\oplus\{0\}\oplus\ldots$ and $T$ satisfies (i). Therefore, it remains to show that $W$, and consequently $T$, are $2$-isometries. Since, by the Stone-von Neumann calculus for selfadjoint operators, $\|W_{n}\|=\sup_{x\in\mathrm{supp}(E)}\xi_{n}(x)$ for every $n\geqslant 0$, where $\mathrm{supp}(E)$ denotes the closed support of $E$, we get (see Lemma  2.4)

[TABLE]

where $\eta:=\sup(\mathrm{supp}(E))$. This implies that $W\in\boldsymbol{B}(\ell^{2}_{\mathcal{M}})$ and $\|W\|\leqslant\eta$ (see [27, p. 408]). By (15) and the multiplicativity of spectral integral, we have (consult (12))

[TABLE]

This implies that for all integers $s\geqslant 0$,

[TABLE]

Hence, in view of [27, Proposition  2.5(i)], $W$ is a $2$-isometry. This completes the proof. ∎

Corollary 2.6**.**

Let $T\in\boldsymbol{B}(\mathcal{H})$ be a $2$-isometry that satisfies the kernel condition. Then the following statements are equivalent:

1. (i)

$T$* is analytic,* 2. (ii)

$T$* is completely non-unitary,* 3. (iii)

$T$* is pure,* 4. (iv)

$T$* is unitarily equivalent to an operator valued unilateral weighted shift $W$ on $\ell^{2}_{\mathcal{M}}$ with weights $\{W_{n}\}_{n=0}^{\infty}$ defined by (15), where $\mathcal{M}=\ker T^{*}$.*

Corollary 2.7**.**

Let $U$, $W,$ $E$ and $T$ be as in Theorem  2.5(b). Then $T$ is an isometry if and only if $\mathrm{supp}(E)=\{1\}.$

Proof.

If $T$ is an isometry, then $W$ is an isometry. This together with (11) and (15) implies that $W_{0}$ is positive and unitary, and so

[TABLE]

The reverse implication is obvious because, due to (15) and Lemma  2.4(iii), $W_{n}=I_{\mathcal{M}}$ for all $n\in\mathbb{Z}_{+}$. ∎

It is easily seen that if an operator $T\in\boldsymbol{B}(\mathcal{H})$ is such $\ker(T^{*}T-I)=\{0\},$ then $T$ is completely non-unitary. A converse to this implication is not true in general even in the case of non-isometric $2$-isometries satisfying the kernel condition (use (11) and Corollary  2.6 with $\dim\mathcal{M}\geqslant 2$).

We conclude this section by describing cyclic analytic $2$-isometries satisfying the kernel condition. The description itself relies on Richter’s model for cyclic analytic $2$-isometries. Namely, by [36, Theorem  5.1], a cyclic analytic $2$-isometry is unitarily equivalent to the operator $M_{z,\mu}$ of multiplication by the coordinate function $z$ on a Dirichlet-type space $\mathcal{D}(\mu)$, where $\mu$ is a finite positive Borel measure on the interval $[0,2\pi)$ (which can be identified with the finite positive Borel measure on the unit circle $\mathbb{T}=\{z\in\mathbb{C}\colon|z|=1\}$). The Hilbert space $\mathcal{D}(\mu)$ consists of all analytic function $f$ on the open unit disc $\mathbb{D}=\{z\in\mathbb{C}\colon|z|<1\}$ such that

[TABLE]

where $f^{\prime}$ stands for the derivative of $f$, $A$ denotes the normalized Lebesgue area measure on $\mathbb{D}$ and $\varphi_{\mu}$ is the positive harmonic function on $\mathbb{D}$ defined by

[TABLE]

The inner product $\langle\,\cdot\,,\mbox{-}\rangle_{\mu}$ of $\mathcal{D}(\mu)$ is given by

[TABLE]

where $\langle\,\cdot\,,\mbox{-}\rangle_{H^{2}}$ stands for the inner product of the Hardy space $H^{2}$. The induced norm of $\mathcal{D}(\mu)$ is denoted by $\|\cdot\|_{\mu}$. In this model, $2$-isometries satisfying the kernel condition can be described as follows333 This answers the question of Eva A. Gallardo-Gutiérrez asked during Workshop on Operator Theory, Complex Analysis and Applications 2016, Coimbra, Portugal, June 21-24..

Proposition 2.8**.**

Under the above assumptions, $M_{z,\mu}$ satisfies the kernel condition if and only if $\mu=\alpha\,m$ for some $\alpha\in\mathbb{R}_{+},$ where $m$ is the Lebesgue measure on $[0,2\pi).$

Proof.

Suppose $\mu=\alpha m$ for some $\alpha\in\mathbb{R}_{+}$. Since the geometric series expansion of $(1-re^{it})^{-1}$ is uniformly convergent with respect to $t$ in $\mathbb{R}$, we infer from (20) that

[TABLE]

This, together with (21) and [36, Corollary  3.8(d)], implies that the sequence $\{e_{n}\}_{n=0}^{\infty}$ defined by

[TABLE]

is an orthonormal basis of $\mathcal{D}(\mu)$. Since $M_{z,\mu}e_{n}=\sqrt{\frac{1+(n+1)\alpha}{1+n\alpha}}\,e_{n+1}$ for all $n\in\mathbb{Z}_{+}$, we deduce that $M_{z,\mu}$ is unitarily equivalent to the unilateral weighted shift with weights $\{\xi_{n}(\lambda)\}_{n=0}^{\infty}$, where $\lambda:=\sqrt{1+\alpha}$. As a consequence, $M_{z,\mu}$ satisfies the kernel condition.

Suppose now that $M_{z,\mu}$ is the operator of multiplication by the coordinate function $z$ on $\mathcal{D}(\mu)$ satisfying the kernel condition, where $\mu$ is a finite positive Borel measure on $[0,2\pi)$. Since $M_{z,\mu}$ is an analytic $2$-isometry such that $\dim\ker M_{z,\mu}^{*}=1$ (see [36, Corollary  3.8(a)]), we infer from Theorem  2.5(a) that $M_{z,\mu}$ is unitarily equivalent to a $2$-isometric unilateral weighted shift $S$ with weights $\{\xi_{n}(\lambda)\}_{n=0}^{\infty}$, where $\lambda\in[1,\infty)$ (note that the closed support of $E$ is a one-point set). In view of the previous paragraph, $S$ is unitarily equivalent to the operator $M_{z,\alpha m}$ of multiplication by the coordinate function $z$ on $\mathcal{D}(\alpha\,m)$, where $\alpha=\lambda^{2}-1\in\mathbb{R}_{+}$. Applying [36, Theorem  5.2], we conclude that $\mu=\alpha\,m$. ∎

Remark 2.9*.*

It follows from the first paragraph of the proof of Proposition  2.8 that $\mathcal{D}(m)$ is the Dirichlet space and the operator of multiplication by the coordinate function $z$ on $\mathcal{D}(m)$ (Dirichlet shift) is unitarily equivalent to the unilateral weighted shift with weights $\left\{\sqrt{\frac{n+2}{n+1}}\/\right\}_{n=0}^{\infty}.$ This means that Dirichlet shift is the fundamental example of a cyclic analytic $2$-isometry which satisfies the kernel condition. $\diamondsuit$

3. The Cauchy dual subnormality problem via the kernel condition

In this section, we answer the Cauchy dual subnormality problem in the affirmative for $2$-isometries that satisfy the kernel condition (see Theorem  3.3). We provide two proofs, the first of which depends on the model Theorem  2.5, while the second does not.

Before doing this, we recall some definitions and state two useful facts related to classical moment problems. A sequence $\boldsymbol{\gamma}=\{\gamma_{n}\}_{n=0}^{\infty}\subseteq\mathbb{R}$ is said to be a Hamburger (resp., Stieltjes, Hausdorff) moment sequence if there exists a positive Borel measure $\mu$ on $\mathbb{R}$ (resp., $\mathbb{R}_{+}$, $[0,1]$) such that $\gamma_{n}=\int t^{n}d\mu(t)$ for every $n\in\mathbb{Z}_{+}$; such a $\mu$ is called a representing measure of $\boldsymbol{\gamma}$. Note that a Hausdorff moment sequence is always determinate as a Hamburger moment sequence, i.e., it has a unique representing measure on $\mathbb{R}$. We refer the reader to [7, 42] for more information on moment problems. The following lemma describes representing measures of special rational-type Hausdorff moment sequences.

Lemma 3.1**.**

Let $a,b\in\mathbb{R}$ be such that $a+bn\neq 0$ for every $n\in\mathbb{Z}_{+}$ and let $\boldsymbol{\gamma}_{a,b}=\{\gamma_{a,b}(n)\}_{n=0}^{\infty}$ be a sequence given by $\gamma_{a,b}(n)=\frac{1}{a+bn}$ for all $n\in\mathbb{Z}_{+}$. Then $\boldsymbol{\gamma}_{a,b}$ is a Hamburger moment sequence if and only if $a>0$ and $b\geqslant 0$. If this is the case, then $\boldsymbol{\gamma}_{a,b}$ is a Hausdorff moment sequence and its unique representing measure $\mu_{a,b}$ is given by

[TABLE]

Proof.

If $\boldsymbol{\gamma}_{a,b}$ is a Hamburger moment sequence, then $\gamma_{a,b}(2n)>0$ for all $n\in\mathbb{Z}_{+},$ which implies that $a>0$ and $b\geqslant 0$. Conversely, if $a>0$ and $b\geqslant 0,$ then applying the well-known integral formula

[TABLE]

one can easily verify that $\boldsymbol{\gamma}_{a,b}$ is a Hausdorff moment sequence with a representing measure $\mu_{a,b}$. ∎

The next property of moment sequences, whose prototype appeared in [15, Note on p. 780], can be deduced from the Hamburger theorem [7, Theorem  6.2.2], the Stieltjes theorem [7, Theorem  6.2.5] and the Hausdorff theorem [7, Theorem  4.6.11].

Lemma 3.2**.**

Let $(X,\mathcal{A},\mu)$ be a measure space and $\{\gamma_{n}\}_{n=0}^{\infty}$ be a sequence of $\mathcal{A}$-measurable real valued functions on $X$. Assume that $\{\gamma_{n}(x)\}_{n=0}^{\infty}$ is a Hamburger $($resp., Stieltjes, Hausdorff$)$ moment sequence for $\mu$-almost every $x\in X$ and $\int_{X}|\gamma_{n}|d\mu<\infty$ for all $n\in\mathbb{Z}_{+}$. Then $\{\int_{X}\gamma_{n}d\mu\}_{n=0}^{\infty}$ is a Hamburger $($resp., Stieltjes, Hausdorff$)$ moment sequence.

We are now in a position to prove the main result of this section.

Theorem 3.3**.**

Let $T\in\boldsymbol{B}(\mathcal{H})$ be a $2$-isometry such that $T^{*}T(\ker T^{*})\subseteq\ker T^{*}.$ Then $T^{\prime}$ is a subnormal contraction such that $T^{\prime*}T^{\prime}(\ker T^{\prime*})\subseteq\ker T^{\prime*}$ and

[TABLE]

Proof I.

Applying Proposition  2.1, we get $T^{\prime*}T^{\prime}(\ker T^{\prime*})\subseteq\ker T^{\prime*}.$ By Proposition  2.2 and Theorem  2.5(a), it suffices to consider the case of $T=W$, where $W$ is an operator valued unilateral weighted shift on $\ell^{2}_{\mathcal{M}}$ with weights $\{W_{n}\}_{n=0}^{\infty}$ given by (15). Since the weights of $W$ are invertible, selfadjoint and commuting, we infer from (11) that

[TABLE]

which means that $W^{\prime}$ is an operator valued unilateral weighted shift on $\ell^{2}_{\mathcal{M}}$ with weights $\{(W_{n})^{-1}\}_{n=0}^{\infty}$. Thus, by the commutativity of $\{W_{n}\}_{n=0}^{\infty}$ and the inversion formula for spectral integral, we have (see (12))

[TABLE]

where $\eta:=\sup(\mathrm{supp}(E))$. This implies that

[TABLE]

Using Lemmata  3.1 and 3.2, we deduce that $\{\|(W^{\prime})_{n,0}f\|^{2}\}_{n=0}^{\infty}$ is a Stieltjes moment sequence for every $f\in\mathcal{M}$. Therefore, by [44, Corollary  3.3] (cf. [31, Theorem  3.2]), $W^{\prime}$ is a subnormal operator which, by (5), is a contraction.

It remains to prove (23). Using (9) and (10) as well as the fact that the weights of $W$ are selfadjoint, invertible and commuting, we get

[TABLE]

This implies that $W^{\prime*n}W^{\prime n}=(W^{*n}W^{n})^{-1}$ for all integers $n\geqslant 0$. Since

[TABLE]

for all $j,n\in\mathbb{Z}_{+}$, we infer from (11) and (25) that

[TABLE]

which completes Proof I of Theorem  3.3. ∎

Proof II.

By Proposition  2.1, $T^{\prime*}T^{\prime}(\ker T^{\prime*})\subseteq\ker T^{\prime*}.$ It follows from (4) and Lemma  2.3(iii) that

[TABLE]

which implies that

[TABLE]

Since $T^{*}T\geqslant I$ (see [35, Lemma 1]), it follows that $\sigma(T^{*}T)\subseteq[1,\|T\|^{2}].$ Applying (4), we get

[TABLE]

which implies that the operator $n(T^{\prime*}T^{\prime})^{-1}-(n-1)I$ is invertible in $\boldsymbol{B}(\mathcal{H})$ for every $n\in\mathbb{N}$.

Now we prove the first equality in (23) by induction on $n$. The cases $n=0,1$ are obvious. Assume that this equality holds for a fixed $n\in\mathbb{N}$. Then, by the induction hypothesis and (27), we have

[TABLE]

Multiplying by $T^{\prime*}$ and $T^{\prime}$ from the left-hand side and the right-hand side, respectively, both sides of the above equality, we get

[TABLE]

Noting that

[TABLE]

we infer from (27) that the operator $(n+1)I-nT^{\prime*}T^{\prime}$ is invertible in $\boldsymbol{B}(\mathcal{H})$ for every $n\in\mathbb{N}$. This combined with (28) yields

[TABLE]

This completes the induction argument.

Since $T$ is a $2$-isometry, we deduce by using induction on $n$ that (see also [25, Proposition  4.5])

[TABLE]

This combined with the first equality in (23) gives the second one in (23).

It follows from the first equality in (23) and the Stone-von Neumann calculus for selfadjoint operators that

[TABLE]

where $G$ is the spectral measure of $T^{*}T$. This together with Lemmata  3.1 and 3.2 implies that $\{\|T^{\prime n}f\|^{2}\}$ is a Stieltjes moment sequence for every $f\in\mathcal{H}$. By Lambert’s theorem (see Theorem  1.1), $T^{\prime}$ is a subnormal operator which, by (5), is a contraction. This completes the proof. ∎

Remark 3.4*.*

Regarding Theorem  3.3, it is worth mentioning that due to (30), for every $f\in\mathcal{H},$ the Hausdorff moment sequence $\{\|T^{\prime n}f\|^{2}\}_{n=0}^{\infty}$ comes from the Hausdorff moment sequences $\{\boldsymbol{\gamma}_{1,x-1}\colon x\in[1,\infty)\}$ appearing in Lemma  3.1 via the integration procedure described in Lemma  3.2. $\diamondsuit$

We now state a few corollaries to Theorem  3.3.

Corollary 3.5**.**

If $T\in\boldsymbol{B}(\mathcal{H})$ is a $2$-isometry satisfying the kernel condition, then the family $\{T^{*n}T^{n}\colon n\in\mathbb{Z}_{+}\}\cup\{T^{\prime*n}T^{\prime n}\colon n\in\mathbb{Z}_{+}\}$ consists of commuting selfadjoint operators.

Corollary 3.6**.**

Suppose $T\in\boldsymbol{B}(\mathcal{H})$ is a $2$-isometry satisfying the kernel condition. Then for every $n\in\mathbb{N},$ $T^{n}$ is a $2$-isometry satisfying the kernel condition and $(T^{n})^{\prime}$ is a subnormal contraction satisfying the kernel condition. In particular, this is the case for $2$-isometric unilateral weighted shifts.

Proof.

Using the fact that positive integral powers of $2$-isometries are $2$-isometries (see [25, Theorem  2.3]), Theorem 3.3 and Proposition 2.1, it suffices to prove that $T^{n}$ satisfies the kernel condition. In view of Theorem  2.5(a), we may assume without loss of generality that $T=W$, where $W$ is as in this theorem. Then $\ker W^{*}=\mathcal{M}\oplus\{0\}\oplus\{0\}\oplus\ldots.$ Using induction and the formulas (9) and (10), we deduce that for every $n\in\mathbb{N}$, $W^{n}$ satisfies the kernel condition. ∎

The next corollary is of some importance because the single equality of the form $p(T,T^{*})=0,$ where $p$ is a polynomial in two non-commuting variables of degree $5$, yields subnormality of $T.$ The reader is referred to [45, Theorem  5.4 and Proposition  7.3] for an example of an unbounded non-subnormal formally normal operator annihilated by a polynomial $p(z,\bar{z})$ of (the lowest possible) degree $3$.

Corollary 3.7**.**

Suppose $T\in\boldsymbol{B}(\mathcal{H}).$ Then

1. (i)

$T^{\prime}$* is a subnormal contraction if $T$ is left-invertible and*

[TABLE] 2. (ii)

$T$* is a subnormal contraction if $T$ is left-invertible and*

[TABLE]

Moreover, in both cases $T$ and $T^{\prime}$ satisfy the kernel condition.

Proof.

(i) Combining Proposition  2.1 and Lemma  2.3 with Theorem  3.3 yields (i) and shows that $T$ and $T^{\prime}$ satisfy the kernel condition.

(ii) Apply (i) to $T^{\prime}$ in place of $T$ and use (1). ∎

Remark 3.8*.*

A careful look at the proof of Corollary  3.7 reveals that the assertions (i) and (ii) are equivalent. In fact, a left-invertible operator $T\in\boldsymbol{B}(\mathcal{H})$ satisfies (31) (resp. (32)) if and only if $T$ (resp. $T^{\prime}$) is a $2$-isometry which satisfies the kernel condition. $\diamondsuit$

The following example shows that some classical operators on Hilbert spaces of analytic functions are closely related to Corollary  3.7.

Example 3.9**.**

For $l\in\mathbb{N},$ consider the reproducing kernel

[TABLE]

where $\mathbb{D}=\{z\in\mathbb{C}\colon|z|<1\}$ and $\overline{w}$ stands for the complex conjugate of $w.$ Let $\mathscr{H}_{l}$ denote reproducing kernel Hilbert space associated with $\kappa_{l}$ and let $M_{z,l}$ be the operator of multiplication by the coordinate function $z$ on $\mathscr{H}_{l}$. It is well-known that $M_{z,l}$ is a subnormal contraction for every $l\in\mathbb{N}$ (see [2, Section  1]). Note that $\mathscr{H}_{1}$ is the Hardy space $H^{2}$ and the operator $M_{z,1}$ (Szegö shift) is an isometry, i.e., $M^{*}_{z,1}M_{z,1}=I.$ In turn, $\mathscr{H}_{2}$ is the Bergman space. Using the standard orthonormal basis of $\mathscr{H}_{2},$ we deduce that the operator $M_{z,2}$ (Bergman shift) is unitarily equivalent to the unilateral weighted shift with weights \Big{\{}\sqrt{\frac{n+1}{n+2}}\/\Big{\}}_{n=0}^{\infty}. It is now easily seen that $M_{z,2}$ is left-invertible and satisfies the following identity

[TABLE]

This together with Remark  3.8 implies that the Cauchy dual $M^{\prime}_{z,2}$ of $M_{z,2}$ is a $2$-isometry satisfying the kernel condition (cf. Remark  2.9). $\diamondsuit$

Below we show that Theorem  3.3 is no longer true for $3$-isometries. Since $m$-isometries are $(m+1)$-isometries (see [4, p. 389]), we see that Theorem  3.3 is not true for $m$-isometries with $m\geqslant 3.$

Example 3.10**.**

Let $T$ be the unilateral weighted shift in $\ell^{2}(\mathbb{Z}_{+})$ with weights \Big{\{}\sqrt{\frac{\phi(n+1)}{\phi(n)}}\/\Big{\}}_{n=0}^{\infty}, where $\phi(n)=n^{2}+1$ for $n\in\mathbb{Z}_{+}$. It is a matter of routine to verify that $T$ is a $3$-isometry (one can also use [1, Theorem  1]). Clearly, $T$ is left-invertible and satisfies the kernel condition. Since the Cauchy dual $T^{\prime}$ of $T$ is the unilateral weighted shift with weights \Big{\{}\sqrt{\frac{\phi(n)}{\phi(n+1)}}\/\Big{\}}_{n=0}^{\infty} (see (24)), we verify easily that $T^{\prime}$ is a contraction and

[TABLE]

In view of Theorem  1.2, the Cauchy dual operator $T^{\prime}$ is not subnormal. $\diamondsuit$

4. The Cauchy dual subnormality problem for

quasi-Brownian isometries

This section deals with a class of $2$-isometries which we propose to call quasi-Brownian isometries. Our goal here is to solve the Cauchy dual subnormality problem affirmatively within this class (see Theorem  4.5). It turns out that quasi-Brownian isometries do not satisfy the kernel condition unless they are isometries (see Corollary  4.6). In Section  7, we exhibit an example of a $2$-isometry $T\in\boldsymbol{B}(\mathcal{H})$ such that $T$ does not satisfy the kernel condition, $T$ is not a quasi-Brownian isometry and the Cauchy dual operator $T^{\prime}$ is a subnormal contraction (see Example  7.9).

We say that an operator $T\in\boldsymbol{B}(\mathcal{H})$ is a quasi-Brownian isometry if $T$ is a $2$-isometry such that

[TABLE]

(Recall that $\triangle_{T}\geqslant 0$ for any $2$-isometry $T.$) In [32] such operators are called $\triangle_{T}$-regular $2$-isometries. As we see below (Corollary  4.2 and Example  4.4), the notion of a quasi-Brownian isometry generalizes that of a Brownian isometry introduced by Agler and Stankus in [4]; the latter notion arose in the study of the time shift operator on a modified Brownian motion process. Here we do not include the rather technical definition of a Brownian isometry as we do not need it. Instead, we define a Brownian isometry by using [4, Theorem  5.48]. Namely, an operator $T\in\boldsymbol{B}(\mathcal{H})$ is said to be a Brownian isometry if $T$ is a $2$-isometry such that

[TABLE]

Before proving the main result of this section, we state slightly improved versions of [32, Proposition  5.1] and [4, Proposition  5.37 and Theorem  5.48].

Theorem 4.1**.**

If $T\in\boldsymbol{B}(\mathcal{H}),$ then the following conditions are equivalent:

1. (i)

$T$* is a quasi-Brownian isometry $($resp., Brownian isometry$)$,* 2. (ii)

$T$* has the block matrix form*

[TABLE]

with respect to an orthogonal decomposition $\mathcal{H}=\mathcal{H}_{1}\oplus\mathcal{H}_{2}$ $($one of the summands may be absent$)$, where $V\in\boldsymbol{B}(\mathcal{H}_{1}),$ $E\in\boldsymbol{B}(\mathcal{H}_{2},\mathcal{H}_{1})$ and $U\in\boldsymbol{B}(\mathcal{H}_{2})$ are such that

[TABLE] 3. (iii)

$T$* is either isometric or it has the block matrix form (37) with respect to a nontrivial orthogonal decomposition $\mathcal{H}=\mathcal{H}_{1}\oplus\mathcal{H}_{2},$ where $V\in\boldsymbol{B}(\mathcal{H}_{1}),$ $E\in\boldsymbol{B}(\mathcal{H}_{2},\mathcal{H}_{1})$ and $U\in\boldsymbol{B}(\mathcal{H}_{2})$ satisfy (38) $($resp., (39)$)$ and $\ker E=\{0\}$.*

Proof.

That (i) implies (iii) follows from [32, Proposition  5.1] (resp., the proof of [4, Theorem  5.48]). Obviously, (iii) implies (ii). Finally, it is a matter of routine to show that (ii) implies (i). ∎

Corollary 4.2**.**

Every Brownian isometry is a quasi-Brownian isometry.

Remark 4.3*.*

Note that if $T\in\boldsymbol{B}(\mathcal{H})$ has the block matrix form (37), where $V\in\boldsymbol{B}(\mathcal{H}_{1}),$ $E\in\boldsymbol{B}(\mathcal{H}_{2},\mathcal{H}_{1})$ and $U\in\boldsymbol{B}(\mathcal{H}_{2})$ satisfy (38), then

[TABLE]

This means that the norm of the operator $E$ appearing in [4, Proposition  5.37] and [32, Proposition  5.1] must equal $1.$ $\diamondsuit$

The converse to Corollary  4.2 is not true.

Example 4.4**.**

Let $V\in\boldsymbol{B}(\mathcal{H}_{1}),$ $E\in\boldsymbol{B}(\mathcal{H}_{2},\mathcal{H}_{1})$ and $U\in\boldsymbol{B}(\mathcal{H}_{2})$ be isometric operators such that $U$ is not unitary and $V^{*}E=0$ (which is always possible). By Theorem  4.1, we see that the corresponding operator $T$ given by (37) is a quasi-Brownian isometry. However, $T$ is not a Brownian isometry because

[TABLE]

which, by the choice of $U$, implies that $\triangle_{T}\triangle_{T^{*}}\triangle_{T}\neq 0$. $\diamondsuit$

Now we are in a position to show that the Cauchy dual operator of a quasi-Brownian isometry is a subnormal contraction.

Theorem 4.5**.**

Suppose $T\in\boldsymbol{B}(\mathcal{H})$ is a quasi-Brownian isometry. Then $T^{\prime}$ is a subnormal contraction such that

[TABLE]

Proof.

It follows from Theorem  4.1 that $T$ has the block matrix form (37) with respect to an orthogonal decomposition $\mathcal{H}=\mathcal{H}_{1}\oplus\mathcal{H}_{2}$, where $V\in\boldsymbol{B}(\mathcal{H}_{1}),$ $E\in\boldsymbol{B}(\mathcal{H}_{2},\mathcal{H}_{1})$ and $U\in\boldsymbol{B}(\mathcal{H}_{2})$ satisfy (38). Without loss of generality, we may assume that $T$ is not an isometry, which implies that $E\neq 0$. By (38), we have

[TABLE]

Clearly, $Q$ is selfadjoint and invertible in $\boldsymbol{B}(\mathcal{H}).$ For $n\in\mathbb{Z}_{+},$ we define the rational function $r_{n}\colon[1,\infty)\to(0,\infty)$ by

[TABLE]

Since, by (41) (or by [35, Lemma 1]), $T^{*}T\geqslant I$ we see that $\sigma(T^{*}T)\subseteq[1,\|T\|^{2}].$ Applying the functional calculus (see [16, Theorem  VIII.2.6]), we deduce that (40) is equivalent  to

[TABLE]

We prove (42) by induction on $n.$ The case $n=0$ is obviously true. Suppose that (42) holds for some unspecified $n\in\mathbb{Z}_{+}.$ Using (41) and functional calculus, we get

[TABLE]

It is a matter of routine to verify that

[TABLE]

The induction hypothesis, the equalities $V^{*}E=0$ and $UQ=QU$ and the functional calculus yield

[TABLE]

which completes the induction argument.

By (5), $T^{\prime}$ is a contraction. It follows from (42) and the Stone-von Neumann calculus for selfadjoint operators that

[TABLE]

where $G$ is the spectral measure of $T^{*}T.$ Now applying Lemma  3.2, we see that $\{\|T^{\prime n}f\|^{2}\}_{n=0}^{\infty}$ is a Stieltjes moment sequence for every $f\in\mathcal{H}$. This together with Lambert’s theorem (see Theorem  1.1) completes the proof. ∎

The techniques developed in proofs of Theorems  3.3 and 4.5 give the following.

Corollary 4.6**.**

Suppose $T\in\boldsymbol{B}(\mathcal{H})$ is a quasi-Brownian isometry that satisfies the kernel condition. Then $T$ is an isometry.

Proof.

Let $G$ be the spectral measure of $T^{*}T$. Note that $0\leqslant\frac{1}{1+n(x-1)}\leqslant 1$ for all $x\in[1,\infty)$ and $n\in\mathbb{Z}_{+}$, and $\lim_{n\to\infty}\frac{1}{1+n(x-1)}=\chi_{\{1\}}(x)$ for all $x\in[1,\infty)$. Applying Lebesgue’s dominated convergence theorem to (30), we deduce that the sequence $\{T^{\prime*n}T^{\prime n}\}_{n=1}^{\infty}$ of positive operators converges to $G(\{1\})$ in the weak and consequently in the strong operator topology. A similar argument applied to (47) shows that $\{T^{\prime*n}T^{\prime n}\}_{n=1}^{\infty}$ converges to $\frac{1}{2}G(\{1\})+(I+T^{*}T)^{-1}$ in the strong operator topology. Hence $G(\{1\})=I$ and thus $T$ is an isometry. ∎

The so-called Brownian shifts introduced in [4, Definition 5.5] are examples of Brownian isometries which are not isometric, and thus by Corollary  4.6 they do not satisfy the kernel condition. In turn, using Theorem  4.1, one can show that the composition operator $C_{\phi}$ that appeared in [26, Example  4.4] (in connection with the study of $2$-hyperexpansive operators) with constant parameter sequence $\{a_{n}\}_{n=-\infty}^{\infty}$ is a non-isometric Brownian isometry which is not unitarily equivalent to a Brownian shift.

5. $2$-isometric weighted shifts

on directed trees

Here we focus our attention on $2$-isometric weighted shifts on directed trees. We refer the reader to [28, Chapters 2 and 3] for all definitions pertaining to directed trees and weighted shifts on directed trees.

Let $\mathscr{T}=(V,E)$ be a directed tree (if not stated otherwise, $V$ and $E$ stand for the sets of vertices and edges of $\mathscr{T}$ respectively). If $\mathscr{T}$ has a root, we denote it by $\mathsf{\omega}$. We write $V^{\circ}=V\setminus\{\mathsf{\omega}\}$ if $\mathscr{T}$ is rooted and $V^{\circ}=V$ otherwise. Given $W\subseteq V$ and $n\in\mathbb{Z}_{+},$ we set ${\mathsf{Chi}}^{\langle n\rangle}(W)=W$ if $n=0$ and ${\mathsf{Chi}}^{\langle n\rangle}(W)=\mathsf{Chi}({\mathsf{Chi}}^{\langle n-1\rangle}(W))$ if $n\geqslant 1$, where $\mathsf{Chi}(W)=\bigcup_{u\in W}\{v\in V\colon(u,v)\in E\}$. We also set ${{\mathsf{Des}}(W)}=\bigcup_{n=0}^{\infty}{\mathsf{Chi}}^{\langle n\rangle}(W)$. For brevity, we write $\mathsf{Chi}(v)=\mathsf{Chi}(\{v\})$, ${\mathsf{Chi}}^{\langle n\rangle}(v)={\mathsf{Chi}}^{\langle n\rangle}(\{v\})$ and ${{\mathsf{Des}}(v)}={{\mathsf{Des}}(\{v\})}$ whenever $v\in V$ and $n\in\mathbb{Z}_{+}$. A member of $\mathsf{Chi}(v)$ (resp., ${{\mathsf{Des}}(v)}$) is called a child (resp., a descendant) of $v$. For $v\in V^{\circ}$, a unique $u\in V$ such that $(u,v)\in E$ is called the parent of $v$ and denoted by $\mathsf{par}(v)$. We put $V^{\prime}=\{u\in V\colon\mathsf{Chi}(u)\neq\emptyset\}$. If $V=V^{\prime}$, we say that $\mathscr{T}$ is leafless. By the degree of a vertex $v\in V$, in notation $\deg{v}$, we understand the cardinality of $\mathsf{Chi}(v)$. A directed tree whose each vertex is of finite degree is called locally finite. If $\mathscr{T}$ is rooted, then (see [28, Corollary  2.1.5])

[TABLE]

Below we give examples of directed trees playing an essential role in this  paper.

Example 5.1**.**

(a) We begin with two classical directed trees, namely

[TABLE]

which will be denoted simply by $\mathbb{Z}_{+}$ and $\mathbb{Z},$ respectively. The directed tree $\mathbb{Z}_{+}$ is rooted, $\mathbb{Z}$ is rootless and both are leafless.

(b) Following [28, page 67], we define the directed tree $\mathscr{T}_{\eta,\kappa}=(V_{\eta,\kappa},E_{\eta,\kappa})$ by

[TABLE]

where $\eta\in\{2,3,4,\ldots\}\cup\{\infty\}$, $\kappa\in\mathbb{Z}_{+}\cup\{\infty\}$ and $J_{\iota}=\{k\in\mathbb{Z}\colon 1\leqslant k\leqslant\iota\}$ for $\iota\in\mathbb{Z}_{+}\sqcup\{\infty\}$. The directed tree $\mathscr{T}_{\eta,\kappa}$ is leafless and [math] is the only vertex of $\mathscr{T}_{\eta,\kappa}$ of degree greater than $1$. It is rooted if and only if $\kappa<\infty$,

(c) Let $l\in\{2,3,4,\ldots\}.$ We say that a directed tree $\mathscr{T}$ is a quasi-Brownian directed tree of valency $l$ (or simply a quasi-Brownian directed tree) if

[TABLE]

A quasi-Brownian directed tree of valency $l\geqslant 3$ can be defined as follows:

[TABLE]

where $X=\mathbb{Z}_{+}$ in the rooted case and $X=\mathbb{Z}$ in the rootless case. Geometrically, it is obtained by “gluing” to each $n\in X$ a copy of the directed tree $\mathscr{T}_{l-1,0}$ defined in (b). A similar construction can be performed for $l=2$. Using [28, Proposition  2.1.4] and [10, Proposition  2.2.1], one can verify that there are only two (up to graph isomorphism) quasi-Brownian directed trees of valency $l,$ one with root, the other without. $\diamondsuit$

Applying induction on $n$, we see that if $\mathscr{T}$ is a quasi-Brownian directed tree of valency $l$, then for every $u\in V$ with $\deg{u}=l$ and for every $n\in\mathbb{Z}_{+},$

1. $\bullet$

there is exactly one vertex $v\in{\mathsf{Chi}}^{\langle n\rangle}(u)$ such that $\deg{v}=l$ and the remaining vertices in ${\mathsf{Chi}}^{\langle n\rangle}(u)$ are of degree $1,$ 2. $\bullet$

${\mathsf{Chi}}^{\langle n\rangle}(u)$ consists of $1+n(l-1)$ vertices.

Obviously, quasi-Brownian directed trees are locally finite and leafless.

The following lemma characterizes rooted quasi-Brownian directed trees.

Lemma 5.2**.**

Let $\mathscr{T}$ be a rooted and leafless directed tree such that $\deg{\omega}\in\{2,3,4,\ldots\}.$ Set $l=\deg{\omega}$. Then the following conditions are equivalent:

1. (i)

$\mathscr{T}$* is a quasi-Brownian directed tree of valency $l,$* 2. (ii)

$\mathscr{T}$* satisfies (50) and (51),* 3. (iii)

$\mathscr{T}$* satisfies (49) and the following equation*

[TABLE] 4. (iv)

$\mathscr{T}$* satisfies (52) and the following condition*

[TABLE]

Proof.

The implications (i)$\Rightarrow$(ii), (iii)$\Rightarrow$(i) and (iii)$\Rightarrow$(iv) are easily seen to be true. The implication (ii)$\Rightarrow$(iii) follows from (48) by induction.

(iv)$\Rightarrow$(iii) Suppose $v\in V^{\circ}$ is such that $\deg{v}\geqslant 2.$ By [10, Proposition  2.2.1], there exists $n\in\mathbb{N}$ such that $\mathsf{par}^{n}(v)=\omega.$ It follows from (53) that $\deg{\mathsf{par}(v)}=\deg{v}.$ By induction, $\deg{v}=\deg{\omega}=l,$ which shows that $\mathscr{T}$ satisfies (49). ∎

Let $\mathscr{T}=(V,E)$ be a directed tree. In what follows $\ell^{2}(V)$ stands for the Hilbert space of square summable complex functions on $V$ equipped with the standard inner product. If $W$ is a nonempty subset of $V,$ then we regard the Hilbert space $\ell^{2}(W)$ as a closed vector subspace of $\ell^{2}(V)$ by identifying each $f\in\ell^{2}(W)$ with the function $\widetilde{f}\in\ell^{2}(V)$ which extends $f$ and vanishes on the set $V\setminus W$. Note that the set $\{e_{u}\}_{u\in V}$, where $e_{u}\in\ell^{2}(V)$ is the characteristic function of $\{u\}$, is an orthonormal basis of $\ell^{2}(V)$. Given a system $\boldsymbol{\lambda}=\{\lambda_{v}\}_{v\in V^{\circ}}\subseteq\mathbb{C}$, we define the operator $S_{\boldsymbol{\lambda}}$ in $\ell^{2}(V)$, called a weighted shift on $\mathscr{T}$ with weights $\boldsymbol{\lambda}$ (or simply a weighted shift on $\mathscr{T}$), as follows

[TABLE]

where $\mathscr{D}(S_{\boldsymbol{\lambda}})$ stands for the domain of $S_{\boldsymbol{\lambda}}$ and $\varLambda_{\mathscr{T}}\colon\mathbb{C}^{V}\to\mathbb{C}^{V}$ is defined by

[TABLE]

Now we collect some properties of weighted shifts on directed trees that are needed in this paper. We also show that weighted shifts on rooted directed trees are completely non-unitary. This is no longer true even for isometric weighted shifts on rootless directed trees.

From now on, we adopt the convention that $\sum_{v\in\emptyset}x_{v}=0$. Recall also that $\triangle_{T}=T^{*}T-I$ whenever $T\in\boldsymbol{B}(\mathcal{H})$ (see (33)).

Lemma 5.3**.**

Let $S_{\boldsymbol{\lambda}}$ be a weighted shift on $\mathscr{T}$ with weights $\boldsymbol{\lambda}=\{\lambda_{v}\}_{v\in V^{\circ}}$. Then

1. (i)

$e_{u}$* is in $\mathcal{D}(S_{\boldsymbol{\lambda}})$ if and only if $\sum_{v\in\mathsf{Chi}(u)}|\lambda_{v}|^{2}<\infty;$ if $e_{u}\in\mathscr{D}(S_{\boldsymbol{\lambda}})$, then $S_{\boldsymbol{\lambda}}e_{u}=\sum_{v\in\mathsf{Chi}(u)}\lambda_{v}e_{v}$ and $\|S_{\boldsymbol{\lambda}}e_{u}\|^{2}=\sum_{v\in\mathsf{Chi}(u)}|\lambda_{v}|^{2},$* 2. (ii)

$S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V))$* if and only if $\sup_{u\in V}\sum_{v\in\mathsf{Chi}(u)}|\lambda_{v}|^{2}<\infty;$ if this is the case, then $\|S_{\boldsymbol{\lambda}}\|^{2}=\sup_{u\in V}\|S_{\boldsymbol{\lambda}}e_{u}\|^{2}=\sup_{u\in V}\sum_{v\in\mathsf{Chi}(u)}|\lambda_{v}|^{2}.$*

Moreover, if $S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V))$, then

1. (iii)

$S_{\boldsymbol{\lambda}}^{*}e_{u}=\bar{\lambda}_{u}e_{\mathsf{par}(u)}$* if $u\in V^{\circ}$ and $S^{*}_{\lambda}e_{u}=0$ otherwise,* 2. (iv)

\ker{S_{\boldsymbol{\lambda}}^{*}}=\begin{cases}\langle e_{\mathsf{\omega}}\rangle\oplus\bigoplus_{u\in V^{\prime}}\big{(}\ell^{2}(\mathsf{Chi}(u))\ominus\langle\boldsymbol{\lambda}^{u}\rangle\big{)}&\text{if \mathscr{T} is rooted,}\\[2.15277pt] \bigoplus_{u\in V^{\prime}}\big{(}\ell^{2}(\mathsf{Chi}(u))\ominus\langle\boldsymbol{\lambda}^{u}\rangle\big{)}&\text{otherwise,}\end{cases}* *

where $\boldsymbol{\lambda}^{u}\in\ell^{2}(\mathsf{Chi}(u))$ is given by $\boldsymbol{\lambda}^{u}\colon\mathsf{Chi}(u)\ni v\to\lambda_{v}\in\mathbb{C}$,* 3. (v)

$|S_{\boldsymbol{\lambda}}|e_{u}=\|S_{\boldsymbol{\lambda}}e_{u}\|e_{u}$* for all $u\in V,$* 4. (vi)

$\triangle_{S_{\boldsymbol{\lambda}}}(e_{u})=(\|S_{\boldsymbol{\lambda}}e_{u}\|^{2}-1)e_{u}$* for every $u\in V,$* 5. (vii)

\triangle_{S_{\boldsymbol{\lambda}}^{*}}(e_{u})=\begin{cases}\big{(}\sum_{v\in\mathsf{Chi}(\mathsf{par}(u))}\lambda_{v}\bar{\lambda}_{u}e_{v}\big{)}-e_{u}&\text{if }u\in V^{\circ},\\ -e_{u}&\text{if \mathscr{T} is rooted and }u=\omega,\end{cases}** 6. (viii)

$S_{\boldsymbol{\lambda}}$* is analytic $($and thus completely non-unitary$)$ if $\mathscr{T}$ is rooted.*

Proof.

The assertions (i)-(v) follow from [28, Propositions  3.1.3, 3.1.8, 3.4.1, 3.4.3 and 3.5.1]. The assertion (vi) can be deduced from (v), while the assertion (vii) can be inferred from (i) and (iii). To prove the assertion (viii), assume that $\mathscr{T}$ is rooted. It follows from [28, Corollary  2.1.5 and Lemma 6.1.1] that

[TABLE]

where $\varOmega_{n}=\bigsqcup_{j=n}^{\infty}{\mathsf{Chi}}^{\langle n\rangle}(\mathsf{\omega})$ for $n\in\mathbb{Z}_{+}$. This implies that $\bigcap_{n=0}^{\infty}S_{\boldsymbol{\lambda}}^{n}(\ell^{2}(V))=\{0\}$, which means that $S_{\boldsymbol{\lambda}}$ is analytic and so completely non-unitary. ∎

For a weighted shift $S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V))$ on $\mathscr{T}$, we define $d_{S_{\boldsymbol{\lambda}}}\colon V\times\mathbb{Z}_{+}\to\mathbb{R}_{+}$ by

[TABLE]

We show that $S_{\boldsymbol{\lambda}}^{*n}S_{\boldsymbol{\lambda}}^{n}$ is a diagonal operator with respect to the orthonormal basis $\{e_{u}\}_{u\in V}$ with diagonal elements $\{d_{S_{\boldsymbol{\lambda}}}(u,n)\}_{u\in V}$,

Lemma 5.4**.**

Let $S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V))$ be a weighted shift on $\mathscr{T}.$ Then

[TABLE]

The function $d_{S_{\boldsymbol{\lambda}}}$ satisfies the following recurrence relation:

[TABLE]

Proof.

We will use induction on $n$. The case of $n=0$ is obvious. Assume that (55) holds for a fixed $n\in\mathbb{Z}_{+}$. Then, by Lemma  5.3, we have

[TABLE]

where ($*$) is due to the induction hypothesis. This completes the proof. ∎

Given a weighted shift $S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V))$ on $\mathscr{T}$, we  set

[TABLE]

It follows from Lemma  5.3(i) that

[TABLE]

Note that if $V_{\boldsymbol{\lambda}}^{+}=V$, then $\mathscr{T}$ is leafless (but not conversely), and if $\mathscr{T}$ is leafless and $\{\boldsymbol{\lambda}\neq 0\}=V^{\circ}$, then $V_{\boldsymbol{\lambda}}^{+}=V$ (but not conversely).

Now we show that the operation of taking Cauchy dual is an inner operation in the class of weighted shifts on directed trees.

Lemma 5.5**.**

Let $S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V))$ be a left-invertible weighted shift on $\mathscr{T}$ with weights $\boldsymbol{\lambda}=\{\lambda_{v}\}_{v\in V^{\circ}}$. Then $V_{\boldsymbol{\lambda}}^{+}=V$ and the Cauchy dual $S_{\boldsymbol{\lambda}}^{\prime}$ of $S_{\boldsymbol{\lambda}}$ is a weighted shift on $\mathscr{T}$ with weights \big{\{}\lambda_{v}\|S_{\boldsymbol{\lambda}}e_{\mathsf{par}(v)}\|^{-2}\big{\}}_{v\in V^{\circ}}.

Proof.

In view of [28, Proposition  3.4.3(iv)], $(S_{\boldsymbol{\lambda}}^{*}S_{\boldsymbol{\lambda}}f)(u)=\|S_{\boldsymbol{\lambda}}e_{u}\|^{2}f(u)$ for all $u\in V$ and $f\in\ell^{2}(V)$. Since, by the left-invertibility of $S_{\boldsymbol{\lambda}}$, $S_{\boldsymbol{\lambda}}^{*}S_{\boldsymbol{\lambda}}$ is invertible in $\boldsymbol{B}(\ell^{2}(V))$, we deduce that $V_{\boldsymbol{\lambda}}^{+}=V$. Clearly, $((S_{\boldsymbol{\lambda}}^{*}S_{\boldsymbol{\lambda}})^{-1}f)(u)=\|S_{\boldsymbol{\lambda}}e_{u}\|^{-2}f(u)$ for all $u\in V$ and $f\in\ell^{2}(V)$. This and the definition of $S_{\boldsymbol{\lambda}}$ complete the proof. ∎

The question of when a weighted shift on a directed tree satisfies the kernel condition has the following explicit answer.

Lemma 5.6**.**

Let $S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V))$ be a weighted shift on $\mathscr{T}$. Then the following conditions are equivalent:

1. (i)

$S_{\boldsymbol{\lambda}}^{*}S_{\boldsymbol{\lambda}}(\ker{S_{\boldsymbol{\lambda}}^{*}})\subseteq\ker{S_{\boldsymbol{\lambda}}^{*}}$, 2. (ii)

there exists a family $\{\alpha_{v}\}_{v\in V_{\boldsymbol{\lambda}}^{+}}\subseteq\mathbb{R}_{+}$ such that

[TABLE]

Moreover, if $\mathscr{T}$ is leafless and $S_{\boldsymbol{\lambda}}$ has nonzero weights, then (i) is equivalent to

1. (iii)

there exists a family $\{\alpha_{v}\}_{v\in V}\subseteq\mathbb{R}_{+}$ such that

[TABLE]

Proof.

Given $v\in V$, we denote by $M_{v}$ the operator in $\ell^{2}(\mathsf{Chi}(v))$ of multiplication by the function $\mathsf{Chi}(v)\ni u\mapsto\|S_{\boldsymbol{\lambda}}{e_{u}}\|^{2}\in\mathbb{R}_{+}.$ It follows from Lemma  5.3(ii) that $M_{v}\in\boldsymbol{B}(\ell^{2}(\mathsf{Chi}(v)))$. Using [28, Proposition  2.1.2] and Lemma  5.3(v) we get

[TABLE]

Hence, by Lemma  5.3(iv), the condition (i) holds if and only if

[TABLE]

Since $M_{v}=M_{v}^{*}$, (60) holds if and only if $M_{v}(\langle\boldsymbol{\lambda}^{v}\rangle)\subseteq\langle\boldsymbol{\lambda}^{v}\rangle$ for all $v\in V^{\prime}$, or equivalently, if and only if $M_{v}(\langle\boldsymbol{\lambda}^{v}\rangle)\subseteq\langle\boldsymbol{\lambda}^{v}\rangle$ for all $v\in V_{\boldsymbol{\lambda}}^{+}$ (see (58)), and the latter is equivalent to (ii).

The “moreover” part is obvious due to (58) and the equivalence (i)$\Leftrightarrow$(ii). ∎

$2$-isometric weighted shifts on directed trees can be characterized as follows.

Lemma 5.7**.**

A weighted shift $S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V))$ on $\mathscr{T}$ is a $2$-isometry if and only if either of the following two equivalent conditions holds:

[TABLE]

If $S_{\boldsymbol{\lambda}}$ is a $2$-isometry, then $\|S_{\boldsymbol{\lambda}}e_{u}\|\geqslant 1$ for all $u\in V,$ $V_{\boldsymbol{\lambda}}^{+}=V$ and $\mathscr{T}$ is leafless.

Proof.

Using (55), (56) and (57) (see Lemma  5.4), we deduce that $S_{\boldsymbol{\lambda}}$ is $2$-isometric if and only if (61) holds. By Lemma  5.3(i), the conditions (61) and (62) are equivalent (note that all series appearing in (61) and (62) are convergent).

The “moreover part” follows from [35, Lemma 1] and Lemma  5.3(viii). ∎

Remark 5.8*.*

Let $S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V))$ be a $2$-isometric weighted shift on a directed tree $\mathscr{T}$ with nonzero weights $\boldsymbol{\lambda}=\{\lambda_{v}\}_{v\in V^{\circ}}$. Since $\{\boldsymbol{\lambda}\neq 0\}=V^{\circ}$ and, by Lemma  5.7, $\mathscr{T}$ is leafless, we infer from Lemma  5.6 that $S_{\boldsymbol{\lambda}}$ satisfies the kernel condition if and only if (59) holds for some $\{\alpha_{v}\}_{v\in V}\subseteq\mathbb{R}_{+}$. $\diamondsuit$

Now we characterize $2$-isometric weighted shifts on rooted directed trees which satisfy the condition (59).

Lemma 5.9**.**

Suppose $S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V))$ is a weighted shift on a rooted directed tree $\mathscr{T}$ which satisfies the condition (59) for some $\{\alpha_{v}\}_{v\in V}\subseteq\mathbb{R}_{+}.$ Then the following conditions are equivalent:

1. (i)

$S_{\boldsymbol{\lambda}}$* is a $2$-isometry,* 2. (ii)

$1-2\|S_{\boldsymbol{\lambda}}e_{\mathsf{\omega}}\|^{2}+\alpha_{\mathsf{\omega}}^{2}\|S_{\boldsymbol{\lambda}}e_{\mathsf{\omega}}\|^{2}=0$* and $1-2\alpha^{2}_{\mathsf{par}(u)}+\alpha^{2}_{u}\alpha^{2}_{\mathsf{par}(u)}=0$ for every $u\in V^{\circ}.$*

Moreover, if $S_{\boldsymbol{\lambda}}$ is a $2$-isometry, then $($see (8)$)$

1. (iii)

$\|S_{\boldsymbol{\lambda}}e_{\mathsf{\omega}}\|\geqslant 1$* and444 This implies that $\alpha_{u}\in[1,\sqrt{2}\,)$ for all $u\in V.$ $\alpha_{u}=\xi_{n+1}(\|S_{\boldsymbol{\lambda}}e_{\mathsf{\omega}}\|)$ for all $u\in{\mathsf{Chi}}^{\langle n\rangle}(\mathsf{\omega})$ and $n\in\mathbb{Z}_{+}$,* 2. (iv)

$S_{\boldsymbol{\lambda}}$* is an isometry if and only if $\|S_{\boldsymbol{\lambda}}e_{v}\|=1$ for some $v\in V.$*

Proof.

The equivalence (i)$\Leftrightarrow$(ii) is a direct consequence of (59) and Lemmata  5.3(i) and 5.7.

To prove the “moreover” part, assume that $S_{\boldsymbol{\lambda}}$ is a $2$-isometry.

(iii) By [35, Lemma 1], $\|S_{\boldsymbol{\lambda}}e_{\mathsf{\omega}}\|\geqslant 1$. We will use induction to prove that

[TABLE]

for every $n\in\mathbb{Z}_{+}$. The case of $n=0$ follows from the first equality in (ii). Assume that (63) holds for a fixed $n\in\mathbb{Z}_{+}$. Take $u\in{\mathsf{Chi}}^{\langle n+1\rangle}(\mathsf{\omega})$. Then, by (48), $\mathsf{par}(u)\in{\mathsf{Chi}}^{\langle n\rangle}(\mathsf{\omega})$. It follows from the induction hypothesis that

[TABLE]

Using the second equation in (ii) and Lemma  2.4(ii), we get

[TABLE]

which completes the induction argument. Hence (iii) holds.

(iv) Only the “if” part needs proof. Note that by Lemma  5.7, $\mathscr{T}$ is leafless. If $\|S_{\boldsymbol{\lambda}}e_{\mathsf{\omega}}\|=1$, then by (iii) we have

[TABLE]

Hence, without loss of generality we can assume that $\|S_{\boldsymbol{\lambda}}e_{v}\|=1$ for some $v\in V^{\circ}$. Set $u=\mathsf{par}(v)$. By (48), there exists $n\in\mathbb{Z}_{+}$ such that $v\in{\mathsf{Chi}}^{\langle n+1\rangle}(\mathsf{\omega})$ and thus $u\in{\mathsf{Chi}}^{\langle n\rangle}(\mathsf{\omega})$. Then

[TABLE]

which implies that $\|S_{\boldsymbol{\lambda}}e_{\mathsf{\omega}}\|=1.$ By (48) and (iii), $\alpha_{w}=1$ for all $w\in V.$ Hence, in view of (59), $\|S_{\boldsymbol{\lambda}}e_{w}\|=1$ for all $w\in V.$ This combined with Lemma  5.3(vi) shows that $S_{\boldsymbol{\lambda}}$ is an isometry. This completes the proof. ∎

Proposition 5.10**.**

If $S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V))$ is a weighted shift on a rooted directed tree $\mathscr{T}$, then the following conditions are equivalent:

1. (i)

$S_{\boldsymbol{\lambda}}$* is a $2$-isometry satisfying the condition (59) for some $\{\alpha_{v}\}_{v\in V}\subseteq\mathbb{R}_{+},$* 2. (ii)

$\|S_{\boldsymbol{\lambda}}e_{\mathsf{\omega}}\|\geqslant 1$* and $\|S_{\boldsymbol{\lambda}}e_{v}\|=\xi_{n}(\|S_{\boldsymbol{\lambda}}e_{\mathsf{\omega}}\|)$ for all $v\in{\mathsf{Chi}}^{\langle n\rangle}(\mathsf{\omega})$ and $n\in\mathbb{Z}_{+}$.*

Proof.

The implication (i)$\Rightarrow$(ii) follows from Lemma  5.9(iii), (59) and (48). To prove the reverse implication, define $\{\alpha_{v}\}_{v\in V}\subseteq\mathbb{R}_{+}$ by $\alpha_{u}=\xi_{n+1}(\|S_{\boldsymbol{\lambda}}e_{\mathsf{\omega}}\|)$ for all $u\in{\mathsf{Chi}}^{\langle n\rangle}(\mathsf{\omega})$ and $n\in\mathbb{Z}_{+}$, and verify, using (48), that the conditions (59) and (ii) of Lemma  5.9 are satisfied. Hence, by this proposition, (i) holds. ∎

Below, we will show that $2$-isometric weighted shifts on rootless directed trees satisfying (59) must be isometric (clearly, each isometric weighted shift on a directed tree satisfies (59)). This is somehow related to [28, Theorem  7.2.1(iii)].

Proposition 5.11**.**

Let $S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V))$ be a $2$-isometric weighted shift on a rootless directed tree $\mathscr{T}$, which satisfies the condition (59) for some $\{\alpha_{v}\}_{v\in V}\subseteq\mathbb{R}_{+}.$ Then $S_{\boldsymbol{\lambda}}$ is an isometry.

Proof.

In view of Lemma  5.3(vi), it suffices to show that $\|S_{\boldsymbol{\lambda}}e_{v}\|=1$ for every $v\in V$. Fix $v\in V$. Since $\mathscr{T}$ is rootless and leafless (see Lemma  5.7), an induction argument shows that there exists a (necessarily injective) sequence $\{v_{n}\}_{n=-\infty}^{\infty}\subseteq V$ such that $v_{0}=v$ and $v_{n}=\mathsf{par}(v_{n+1})$ for all $n\in\mathbb{Z}$. Set $\beta_{n}=\|S_{\boldsymbol{\lambda}}e_{v_{n}}\|$ for $n\in\mathbb{Z}$. Clearly, $\{\beta_{n}\}_{n\in\mathbb{Z}}\subseteq[0,\|S_{\boldsymbol{\lambda}}\|]$. According to (59), (61) and Lemma  5.3(i), we have

[TABLE]

Hence, by Lemma  5.7, the bilateral weighted shift $W$ in $\ell^{2}(\mathbb{Z})$ with weights $\{\beta_{n}\}_{n\in\mathbb{Z}}$ is a $2$-isometry with dense range. Since $W$ is left-invertible, we deduce that $W$ is invertible in $\boldsymbol{B}(\ell^{2}(\mathbb{Z}))$. Therefore, by [4, Proposition  1.23] (see also [41, Remark  3.4]), $W$ is unitary. This implies that $\beta_{n}=1$ for all $n\in\mathbb{Z}$. In particular, $\|S_{\boldsymbol{\lambda}}e_{v}\|=\|S_{\boldsymbol{\lambda}}e_{v_{0}}\|=\beta_{0}=1$, which completes the proof. ∎

We conclude this section by showing that Brownian isometric weighted shifts on rooted directed trees are isometric (cf. Proposition  7.3).

Proposition 5.12**.**

Let $S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V))$ be a Brownian isometric weighted shift on a rooted directed tree $\mathscr{T}$. Then $S_{\boldsymbol{\lambda}}$ is an isometry.

Proof.

We split the proof into a few steps.

Step 1. If $u\in V$ is such that either $u=\omega$ or $u\in V^{\circ}$ and $\lambda_{u}=0,$ then $\|S_{\boldsymbol{\lambda}}e_{u}\|=1.$

Indeed, it follows from (34) and the assertions (vi) and (vii) of Lemma  5.3 that

[TABLE]

which means that $\|S_{\boldsymbol{\lambda}}e_{u}\|=1.$

Step 2. If $u\in V$ is such that $\|S_{\boldsymbol{\lambda}}e_{u}\|=1,$ then $\|S_{\boldsymbol{\lambda}}e_{v}\|=1$ for all $v\in\mathsf{Chi}(u).$

Indeed, by (61) and Lemma  5.3(i), we see that

[TABLE]

Hence, we have

[TABLE]

Since by Lemma  5.7, $\|S_{\boldsymbol{\lambda}}e_{v}\|^{2}-1\geqslant 0$ for all $v\in V,$ we infer from (65) that $\|S_{\boldsymbol{\lambda}}e_{v}\|=1$ for all $v\in\mathsf{Chi}(u)$ such that $\lambda_{v}\neq 0.$ On the other hand, if $\lambda_{v}=0$ for some $v\in\mathsf{Chi}(u),$ then by Step 1, $\|S_{\boldsymbol{\lambda}}e_{v}\|=1,$ which completes the proof of Step 2.

Finally, induction together with (48) and Steps 1 and 2 shows that $\|S_{\boldsymbol{\lambda}}e_{u}\|=1$ for all $u\in V,$ which implies that $S_{\boldsymbol{\lambda}}$ is an isometry (see Lemma  5.3(vi)). ∎

6. The Cauchy dual subnormality

problem via perturbed kernel condition

Remark  5.8 suggests considering a wider class of $2$-isometric weighted shifts on directed trees which satisfy a less restrictive condition than (59). In this section, we will discuss the question of subnormality of the Cauchy dual of a $2$-isometric weighted shift $S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V))$ on a rooted directed tree $\mathscr{T}$ for which there exist $k\in\mathbb{N}$ and a family $\{\alpha_{v}\}_{v\in{{\mathsf{Des}}({\mathsf{Chi}}^{\langle k\rangle}(\mathsf{\omega}))}}\subseteq\mathbb{R}_{+}$ such that

[TABLE]

A complete answer to this question is given in Theorem  6.5. This enables us to solve the Cauchy dual subnormality problem in the negative (see Example  6.6; see also Example  7.10 for the case of adjacency operators). For a pictorial comparison of the conditions (59) and (66) in the case of $k=1$, we refer the reader to Figure  1. The quantities $\|S_{\boldsymbol{\lambda}}e_{v}\|,$ $v\in{\mathsf{Chi}}^{\langle 2\rangle}(\omega)$, appearing therein can be calculated by using (67) and (70).

We begin by establishing an explicit formula for $d_{S_{\boldsymbol{\lambda}}^{\prime}}$ (see Lemma  5.5 and (54)), where $S_{\boldsymbol{\lambda}}$ is a $2$-isometric weighted shift on a rooted directed tree which satisfies the condition (66) for $k=1$.

Lemma 6.1**.**

Let $S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V))$ be a $2$-isometric weighted shift on a rooted directed tree $\mathscr{T}$ such that

[TABLE]

for some $\{\alpha_{v}\}_{v\in V^{\circ}}\subseteq\mathbb{R}_{+}$. Then $\mathscr{T}$ is leafless, $\|S_{\boldsymbol{\lambda}}e_{u}\|\geqslant 1$ for every $u\in V$ and

[TABLE]

Proof.

By Lemma  5.7, $\mathscr{T}$ is leafless and $\|S_{\boldsymbol{\lambda}}e_{u}\|\geqslant 1$ for all $u\in V$. Hence, by (67), the expressions appearing in (68) and (69) make sense. It follows from Lemma  5.7 and (67) that

[TABLE]

Below we shall use Lemmata  5.4 and 5.5 without explicitly mentioning them. We prove the equality (69) by induction on $n$. That it holds for $n=1$ follows from Lemma  5.3(i). Assume it holds for a fixed $n\in\mathbb{N}$. Then we have

[TABLE]

where ($*$) is due to the induction hypothesis. Hence, (69) holds.

The equality (68) will be deduced from (69). The case of $n=1$ follows from Lemma  5.3(i). Let us fix an integer $n\geqslant 2$. Then we have

[TABLE]

which completes the proof. ∎

Remark 6.2*.*

Note that the formula (69) can be derived from (23) by using the fact that the operator $T_{\boldsymbol{\lambda}}:=S_{\boldsymbol{\lambda}}|_{\mathcal{M}^{\perp}}$, where $\mathcal{M}=\langle e_{\mathsf{\omega}}\rangle,$ is a $2$-isometry which satisfies the kernel condition and the assumptions of Proposition 2.2 with $\mathcal{L}=\mathcal{M}^{\perp}$. One may refer to $S_{\boldsymbol{\lambda}}$ as a rank one $2$-isometric extension of $T_{\boldsymbol{\lambda}}.$ We will show in Example 6.6 that the Cauchy dual subnormality problem has a negative solution even for rank one $2$-isometric extensions of $2$-isometries which satisfy the kernel condition. $\diamondsuit$

Recall a criterion for a Stieltjes moment sequence to have a backward extension.

Lemma 6.3** **([28, Lemma

6.1.2]).

Let $\{\gamma_{n}\}_{n=1}^{\infty}\subseteq\mathbb{R}_{+}$ be a sequence such that $\{\gamma_{n+1}\}_{n=0}^{\infty}$ is a Stieltjes moment sequence. Set $\gamma_{0}=1$. Then the following conditions are equivalent:

1. (i)

$\{\gamma_{n}\}_{n=0}^{\infty}$* is a Stieltjes moment sequence,* 2. (ii)

there exists a representing measure $\mu$ of $\{\gamma_{n+1}\}_{n=0}^{\infty}$ concentrated on $\mathbb{R}_{+}$ such that555 We adhere to the convention that $\frac{1}{0}:=\infty$. Hence, $\int_{\mathbb{R}_{+}}\frac{1}{t}d\mu(t)<\infty$ implies $\mu(\{0\})=0$.* $\int_{\mathbb{R}_{+}}\frac{1}{t}d\mu(t)\leqslant 1$.*

If $\mu$ is as in (ii), then the positive Borel measure $\nu$ on $\mathbb{R}$ defined by

[TABLE]

is a representing measure of $\{\gamma_{n}\}_{n=0}^{\infty}$ concentrated on $\mathbb{R}_{+};$ moreover, $\nu(\{0\})=0$ if and only if $\int_{\mathbb{R}_{+}}\frac{1}{t}d\mu(t)=1$.

The next lemma is an essential ingredient of the proof of the implication (i)$\Rightarrow$(ii) of Theorem  6.5. In fact, it covers the case of $k=1$ of this implication.

Lemma 6.4**.**

Let $\mathscr{T}$ and $S_{\boldsymbol{\lambda}}$ be as in Lemma  6.1. Assume that $\lambda_{v}\neq 0$ for every $v\in\mathsf{Chi}(\mathsf{\omega})$. If the Cauchy dual $S_{\boldsymbol{\lambda}}^{\prime}$ of $S_{\boldsymbol{\lambda}}$ is subnormal, then there exists $\alpha\in\mathbb{R}_{+}$ such that $\|S_{\boldsymbol{\lambda}}e_{v}\|=\alpha$ for all $v\in\mathsf{Chi}(\mathsf{\omega}).$

Proof.

Assume that $S_{\boldsymbol{\lambda}}^{\prime}$ is subnormal. It follows from (54) applied to $S_{\boldsymbol{\lambda}}^{\prime}$ and (68) that

[TABLE]

This combined with Lemmata  3.1 and 6.1 (as well as with the Lebesgue monotone convergence theorem) implies that $\|S_{\boldsymbol{\lambda}}e_{v}\|\geqslant 1$ for all $v\in V$ and $\{\|S_{\boldsymbol{\lambda}}^{\prime(n+1)}e_{\mathsf{\omega}}\|^{2}\}_{n=0}^{\infty}$ is a Hausdorff moment sequence with a (unique) representing measure $\rho$ given by

[TABLE]

Since, by Lambert’s theorem (see Theorem  1.1), $\{\|S_{\boldsymbol{\lambda}}^{\prime n}e_{\mathsf{\omega}}\|^{2}\}_{n=0}^{\infty}$ is a Stieltjes moment sequence, we infer from Lemma  6.3 that

[TABLE]

Set $\varSigma_{1}=\{v\in\mathsf{Chi}(\mathsf{\omega})\colon\|S_{\boldsymbol{\lambda}}e_{v}\|=1\}$ and $\varSigma_{2}=\{v\in\mathsf{Chi}(\mathsf{\omega})\colon\|S_{\boldsymbol{\lambda}}e_{v}\|>1\}$. Note that $\mathsf{Chi}(\mathsf{\omega})=\varSigma_{1}\sqcup\varSigma_{2}$ (the disjoint sum). It follows from Lemma  3.1 that

[TABLE]

Since, by assumption, $\lambda_{v}\neq 0$ for all $v\in\varSigma_{2}$, the conditions (22), (73) and (74) imply that $\|S_{\boldsymbol{\lambda}}e_{v}\|^{2}<2$ for all $v\in\varSigma_{2}$, and thus for all $v\in\mathsf{Chi}(\mathsf{\omega})$. Therefore, by (22) and (74), we have

[TABLE]

This together with (73) yields

[TABLE]

Now observe that

[TABLE]

where ($*$) follows from the Cauchy-Schwarz inequality. Hence, equality holds in the Cauchy-Schwarz inequality ($*$). This means that \Big{\{}\frac{\lambda_{v}\sqrt{2-\|S_{\boldsymbol{\lambda}}e_{v}\|^{2}}}{2-\|S_{\boldsymbol{\lambda}}e_{v}\|^{2}}\Big{\}}_{v\in\mathsf{Chi}(\mathsf{\omega})} and \Big{\{}\lambda_{v}\sqrt{2-\|S_{\boldsymbol{\lambda}}e_{v}\|^{2}}\Big{\}}_{v\in\mathsf{Chi}(\mathsf{\omega})} are linearly dependent vectors in $\ell^{2}(\mathsf{Chi}(\mathsf{\omega}))$. Since the weights $\{\lambda_{v}\}_{v\in\mathsf{Chi}(\mathsf{\omega})}$ are nonzero, we deduce that there exists $\alpha\in\mathbb{R}_{+}$ such that $\|S_{\boldsymbol{\lambda}}e_{v}\|=\alpha$ for every $v\in\mathsf{Chi}(\mathsf{\omega}),$ which completes the proof. ∎

We are in a position to prove the main result of this section, which can be thought of as a reconstruction theorem.

Theorem 6.5**.**

Let $S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V))$ be a $2$-isometric weighted shift on a rooted directed tree $\mathscr{T},$ which satisfies (66) for some $k\in\mathbb{N}$ and $\{\alpha_{v}\}_{v\in{{\mathsf{Des}}({\mathsf{Chi}}^{\langle k\rangle}(\mathsf{\omega}))}}\subseteq\mathbb{R}_{+}$. Let $\lambda_{v}\neq 0$ for all $v\in\bigsqcup_{i=1}^{k}{\mathsf{Chi}}^{\langle i\rangle}(\mathsf{\omega})$. Then the following conditions are equivalent:

1. (i)

the Cauchy dual $S_{\boldsymbol{\lambda}}^{\prime}$ of $S_{\boldsymbol{\lambda}}$ is subnormal, 2. (ii)

there exists a family $\{\alpha_{v}\}_{v\in\bigsqcup_{i=0}^{k-1}{\mathsf{Chi}}^{\langle i\rangle}(\mathsf{\omega})}\subseteq\mathbb{R}_{+}$ such that

[TABLE] 3. (iii)

$S_{\boldsymbol{\lambda}}$* satisfies the condition (iii) of Lemma  5.6,* 4. (iv)

$S_{\boldsymbol{\lambda}}^{*}S_{\boldsymbol{\lambda}}(\ker{S_{\boldsymbol{\lambda}}^{*}})\subseteq\ker{S_{\boldsymbol{\lambda}}^{*}}$.

Proof.

(i)$\Rightarrow$(ii) Fix $v\in{\mathsf{Chi}}^{\langle k-1\rangle}(\mathsf{\omega})$. Note that the space $\ell^{2}({{\mathsf{Des}}(v)})$ (which is identified with the closed vector subspace $\chi_{{{\mathsf{Des}}(v)}}\cdot\ell^{2}(V)$ of $\ell^{2}(V)$) is invariant for $S_{\boldsymbol{\lambda}}$ and $S_{\boldsymbol{\lambda}}|_{\ell^{2}({{\mathsf{Des}}(v)})}$ coincides with the weighted shift $S_{\boldsymbol{\lambda}^{|v\rangle}}$ on the directed tree \mathscr{T}_{{{\mathsf{Des}}(v)}}:=\big{(}{{\mathsf{Des}}(v)},({{\mathsf{Des}}(v)}\times{{\mathsf{Des}}(v)})\cap E\big{)} with weights $\boldsymbol{\lambda}^{|v\rangle}:=\{\lambda_{u}\}_{u\in{{\mathsf{Des}}(v)}\setminus\{v\}}$ (see [28, Proposition  2.1.8] for more details). It follows from [28, Proposition  2.1.10] and the fact that $v$ is a root of $\mathscr{T}_{{{\mathsf{Des}}(v)}}$ that

[TABLE]

(The expressions $\mathsf{par}_{\mathscr{T}_{{{\mathsf{Des}}(v)}}}(\cdot)$, $\mathsf{Chi}_{\mathscr{T}_{{{\mathsf{Des}}(v)}}}(\cdot)$, ${\mathsf{Chi}}_{\mathscr{T}_{{{\mathsf{Des}}(v)}}}^{\langle 2\rangle}(\cdot)$ and ${{\mathsf{Des}}_{\mathscr{T}_{{{\mathsf{Des}}(v)}}}(\cdot)}$ are understood with respect to the directed subtree $\mathscr{T}_{{{\mathsf{Des}}(v)}}$.) This and (66) imply that $S_{\boldsymbol{\lambda}^{|v\rangle}}$ satisfies the assumptions of Lemma  6.4. Applying Lemma  5.3(v) and Proposition  2.2 to $T=S_{\boldsymbol{\lambda}}$ and $\mathcal{L}=\ell^{2}({{\mathsf{Des}}(v)})$, we deduce that the Cauchy dual $S_{\boldsymbol{\lambda}^{|v\rangle}}^{\prime}$ of $S_{\boldsymbol{\lambda}^{|v\rangle}}$ is subnormal. Hence, by Lemma  6.4, there exists $\alpha_{v}\in\mathbb{R}_{+}$ such that (cf. (76))

[TABLE]

Summarizing, we have proved that there exists a family $\{\alpha_{v}\}_{v\in{\mathsf{Chi}}^{\langle k-1\rangle}(\mathsf{\omega})}\subseteq\mathbb{R}_{+}$ such that $\|S_{\boldsymbol{\lambda}}e_{u}\|=\alpha_{\mathsf{par}(u)}$ for all $u\in\mathsf{Chi}(v)$ and $v\in{\mathsf{Chi}}^{\langle k-1\rangle}(\mathsf{\omega})$. Since, by [10, (2.2.6)], ${\mathsf{Chi}}^{\langle k\rangle}(\mathsf{\omega})=\bigsqcup_{v\in{\mathsf{Chi}}^{\langle k-1\rangle}(\mathsf{\omega})}\mathsf{Chi}(v)$, we see that $\|S_{\boldsymbol{\lambda}}e_{u}\|=\alpha_{\mathsf{par}(u)}$ for all $u\in{\mathsf{Chi}}^{\langle k\rangle}(\mathsf{\omega})$. Now, using reverse induction on $k$, we conclude that (ii) holds.

Since, in view of (48), the implication (ii)$\Rightarrow$(iii) is obvious and the implications (iii)$\Rightarrow$(iv) and (iv)$\Rightarrow$(i) are direct consequences of Lemma  5.6 and Theorem  3.3 respectively, the proof is complete. ∎

We conclude this section by answering the Cauchy dual subnormality problem in the negative. The counterexample presented below is built over the directed tree $\mathscr{T}_{2,0}$ (see Example  5.1(b)). It is easily seen that similar counterexamples can be built over any directed tree of the form $\mathscr{T}_{\eta,0}$, where $\eta\in\{2,3,4,\ldots\}\cup\{\infty\}$.

Example 6.6**.**

Let $y_{1},y_{2}\in\mathbb{R}$ be such that $1<y_{1},y_{2}<\sqrt{2}$ and $y_{1}\neq y_{2}$. Then there exist positive real numbers $x_{1}$ and $x_{2}$ such that

[TABLE]

Let $S_{\boldsymbol{\lambda}}$ be the weighted shift on $\mathscr{T}_{2,0}$ with weights $\boldsymbol{\lambda}=\{\lambda_{v}\}_{v\in V_{2,0}^{\circ}}$ defined by

[TABLE]

By Lemma  5.3(ii), $S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V_{2,0}))$ and \|S_{\boldsymbol{\lambda}}\|=\max\Big{\{}y_{1},y_{2},\sqrt{\sum_{i=1}^{2}x_{i}^{2}}\,\Big{\}}. It is a matter of routine to show that $S_{\boldsymbol{\lambda}}$ satisfies the condition (62) and thus, by Lemma  5.7, $S_{\boldsymbol{\lambda}}$ is a $2$-isometry. Moreover, by Lemma  5.3(viii), $S_{\boldsymbol{\lambda}}$ is completely non-unitary. Note that $S_{\boldsymbol{\lambda}}$ satisfies the condition (67) for $\{\alpha_{v}\}_{v\in V_{2,0}^{\circ}}\subseteq\mathbb{R}_{+}$ given by $\alpha_{i,j}=\xi_{j}(y_{i})$ for $i\in\{1,2\}$ and $j\in\{1,2,\dots\}$. Since the weights of $S_{\boldsymbol{\lambda}}$ are nonzero and $\|S_{\boldsymbol{\lambda}}e_{1,1}\|=y_{1}\neq y_{2}=\|S_{\boldsymbol{\lambda}}e_{2,1}\|$, we infer from Theorem  6.5 with $k=1$ (see also Lemma  6.4) that the Cauchy dual $S_{\boldsymbol{\lambda}}^{\prime}$ of $S_{\boldsymbol{\lambda}}$ is not subnormal. $\diamondsuit$

7. The Cauchy dual subnormality problem for

adjacency operators

In this section, we turn our attention to weighted shifts on directed trees with weights whose moduli are constant on $\mathsf{Chi}(u)$ for every vertex $u$. This class of operators contains the class of adjacency operators of directed trees; the latter class plays an important role in graph theory (see [28] for more details). We prove that the Cauchy dual of a $2$-isometric adjacency operator of a directed tree is subnormal if the directed tree satisfies certain degree constraints (see Theorem  7.8). However, as shown in Example  7.10 below, the Cauchy dual subnormality problem has a negative solution even in the class of adjacency operators of directed trees.

We begin by proving a preparatory lemma.

Lemma 7.1**.**

Let $S_{\boldsymbol{\lambda}}\in\boldsymbol{B}(\ell^{2}(V))$ be a weighted shift on a leafless and locally finite directed tree $\mathscr{T}$ such that

[TABLE]

for some $\{\beta_{u}\}_{v\in V}\subseteq(0,\infty)$. Then the following statements hold:

1. (i)

$S_{\boldsymbol{\lambda}}$* is a $2$-isometry if and only if*

[TABLE] 2. (ii)

if $S_{\boldsymbol{\lambda}}$ is left-invertible, then $d_{S_{\boldsymbol{\lambda}}^{\prime}}(u,0)=1$ for all $u\in V$ and

[TABLE]

*where $d_{S_{\boldsymbol{\lambda}}^{\prime}}$ is given by (54) with $S_{\boldsymbol{\lambda}}^{\prime}$ in place of $S_{\boldsymbol{\lambda}}$ *$($see Lemma  5.5$)$.

Proof.

It follows from Lemma  5.3(i) and (77) that

[TABLE]

The statement (i) can be straightforwardly deduced from (62), (77) and (78). In turn, the statement (ii) can be easily inferred from (56) and (57) applied to $S_{\boldsymbol{\lambda}}^{\prime}$ by using (77) and (78). ∎

By the adjacency operator of a directed tree $\mathscr{T},$ we understand the weighted shift $S_{\mathbb{1}}$ on $\mathscr{T}$ all of whose weights are equal to $1$ (see [28, p. 1] for more information). Note that in general, adjacency operators may not even be densely defined (see [28, Proposition  3.1.3]).

Below we describe some classes of $2$-isometric adjacency operators including those satisfying the kernel condition, Brownian isometries and quasi-Brownian isometries (see Proposition  7.3). The following preliminary result characterizes isometric adjacency operators.

Lemma 7.2**.**

If $S_{\mathbb{1}}$ is the adjacency operator of a directed tree $\mathscr{T},$ then the following conditions are equivalent:

1. (i)

$S_{\mathbb{1}}$* is an isometry on $\ell^{2}(V)$,* 2. (ii)

$\deg{u}=1$* for all $u\in V,$* 3. (iii)

$\mathscr{T}$* is graph isomorphic either to $\mathbb{Z}_{+}$ or to $\mathbb{Z}.$*

Proof.

That (i) and (ii) are equivalent follows from [28, Corollary  3.4.4] and (78). In turn, the equivalence (ii)$\Leftrightarrow$(iii) can be deduced from [28, Corollary  2.1.5 and Proposition  2.1.6]). ∎

Now we are in a position to describe the aforesaid classes of $2$-isometric adjacency operators (see Example  5.1 for necessary definitions).

Proposition 7.3**.**

If $S_{\mathbb{1}}\in\boldsymbol{B}(\ell^{2}(V))$ is the adjacency operator of a directed tree $\mathscr{T},$ then the following assertions are valid:

1. (i)

$S_{\mathbb{1}}$* is a $2$-isometry satisfying the kernel condition if and only if $\mathscr{T}$ is graph isomorphic either to $\mathbb{Z}_{+}$ or to $\mathbb{Z},$* 2. (ii)

if $\mathscr{T}$ is rooted, then $S_{\mathbb{1}}$ is a Brownian isometry if and only if $\mathscr{T}$ is graph isomorphic to $\mathbb{Z}_{+},$ 3. (iii)

if $\mathscr{T}$ is rooted, then $S_{\mathbb{1}}$ is a quasi-Brownian isometry if and only if either $\mathscr{T}$ is graph isomorphic to $\mathbb{Z}_{+}$ or $\mathscr{T}$ is a quasi-Brownian directed tree, 4. (iv)

if $\mathscr{T}$ is rootless, then $S_{\mathbb{1}}$ is a quasi-Brownian isometry if and only if $S_{\mathbb{1}}$ is a Brownian isometry, or equivalently, if and only if either $\mathscr{T}$ is graph isomorphic to $\mathbb{Z}$ or $\mathscr{T}$ is a quasi-Brownian directed tree.

Proof.

Since directed trees admitting $2$-isometric weighted shifts are automatically leafless, we may assume without loss of generality that $\mathscr{T}$ is leafless.

(i) It suffices to prove the “only if” part. Assume that $S_{\mathbb{1}}\in\boldsymbol{B}(\ell^{2}(V))$ is a $2$-isometry satisfying the kernel condition. First, observe that $S_{\mathbb{1}}$ satisfies the condition (59) for some $\{\alpha_{v}\}_{v\in V}\subseteq\mathbb{R}_{+}$ (see Remark  5.8). In view of Proposition  5.11 and Lemma  7.2, we can assume that $\mathscr{T}$ has a root. It follows from Lemma  5.3(i) and the implication (i)$\Rightarrow$(ii) of Proposition  5.10 that $\deg{v}=\xi_{n}(\sqrt{\deg{\mathsf{\omega}}}\,)^{2}$ for all $v\in{\mathsf{Chi}}^{\langle n\rangle}(\mathsf{\omega})$ and $n\in\mathbb{Z}_{+}$. Hence, by (48), we see that $\deg{v}=\deg{w}$ for all $v,w\in\mathsf{Chi}(u)$ and $u\in V$. This combined with Lemmata  5.3(ii) and 7.1(i) (the latter applied to $\beta_{u}\equiv 1$) implies that $\sup_{u\in V}\deg u=\|S_{\mathbb{1}}\|^{2}$ and

[TABLE]

As a consequence, we deduce that $\deg{u}=1$ for all $u\in V$, which by Lemma  7.2 implies that $\mathscr{T}$ is graph isomorphic to $\mathbb{Z}_{+}.$

Before proving the assertions (ii)-(iv), we show that

[TABLE]

Indeed, by Lemma  5.3, we see that $\triangle_{S_{\mathbb{1}}}\geqslant 0$ and

[TABLE]

Similarly

[TABLE]

Hence, $\triangle_{S_{\mathbb{1}}}S_{\mathbb{1}}=\triangle_{S_{\mathbb{1}}}^{1/2}S_{\mathbb{1}}\triangle_{S_{\mathbb{1}}}^{1/2}$ if and only if

[TABLE]

This implies (79).

(ii) This is a direct consequence of Proposition  5.12 and Lemma  7.2.

(iii) Suppose $\mathscr{T}$ is rooted. Assume $S_{\mathbb{1}}$ is a quasi-Brownian isometry. Set $l=\deg{\omega}.$ Clearly, $l<\infty.$ If $l=1,$ then in view of (48), (80) and Lemma  7.2, $\mathscr{T}$ is graph isomorphic to $\mathbb{Z}_{+}.$ If $l\geqslant 2,$ then by (79) and Lemmata  5.2 and 7.1(i), $\mathscr{T}$ is a quasi-Brownian directed tree. The converse implication is obvious in the case when $\mathscr{T}$ is graph isomorphic to $\mathbb{Z}_{+}$ (see Lemma  7.2). In turn, if $\mathscr{T}$ is a rooted quasi-Brownian directed tree, then (80) holds, and consequently $\triangle_{S_{\mathbb{1}}}S_{\mathbb{1}}=\triangle_{S_{\mathbb{1}}}^{1/2}S_{\mathbb{1}}\triangle_{S_{\mathbb{1}}}^{1/2}$, which in view of Lemmata  5.2 and 7.1(i) shows that $S_{\mathbb{1}}$ is a quasi-Brownian isometry.

(iv) Suppose $\mathscr{T}$ is rootless. Assume $S_{\mathbb{1}}$ is a quasi-Brownian isometry. By Lemma  7.2, we may assume that there exists $u_{0}\in V$ such that $l:=\deg{u_{0}}\in\{2,3,4,\ldots\}.$ Set $u_{n}=\mathsf{par}^{n}(u_{0})$ for $n\in\mathbb{N}.$ It follows from (53) (see (79)) that $\deg{u_{n}}=\deg{u_{n+1}}$ for all $n\in\mathbb{Z}_{+}.$ This implies that $\deg{u_{n}}=l$ for all $n\in\mathbb{Z}_{+}.$ Applying (79) and Lemmata  5.2 and 7.1(i), we deduce that for every $n\in\mathbb{Z}_{+},$ the rooted directed tree

[TABLE]

is a quasi-Brownian directed tree of valency $l.$ This together with [28, Proposition  2.1.6(iii)] implies that the directed tree $\mathscr{T}$ itself is a quasi-Brownian directed tree of valency $l.$

In view of Corollary  4.2, it remains to show that the adjacency operator of a rootless quasi-Brownian directed tree is a Brownian isometry. Clearly (52) holds, so by Lemma  7.1(i), $S_{\mathbb{1}}$ is a $2$-isometry. It follows from Lemma  5.3 and the definition of a quasi-Brownian directed tree that

[TABLE]

Hence $S_{\mathbb{1}}$ is a Brownian isometry. This completes the proof. ∎

Combining Lemmata  5.2 and 7.1(i) with Proposition  7.3, we get the following.

Corollary 7.4**.**

Let $S_{\mathbb{1}}\in\boldsymbol{B}(\ell^{2}(V))$ be the adjacency operator of a rooted directed tree $\mathscr{T}$ such that $\deg{\omega}\in\{2,3,4,\ldots\}$. Set $l=\deg{\omega}$. If $S_{\mathbb{1}}$ is a $2$-isometry, then the following conditions are equivalent:

1. (i)

$S_{\mathbb{1}}$* is a quasi-Brownian isometry,* 2. (ii)

$\mathscr{T}$* is a quasi-Brownian directed tree,* 3. (iii)

each vertex $u\in V^{\circ}$ has degree $1$ or $l.$

Below we concentrate on $2$-isometric adjacency operators of directed trees which satisfy certain degree constraints (see Figure 2 for an illustration of parts (i) and (ii) of Lemma  7.5).

Lemma 7.5**.**

Let $\mathscr{T}$ be a rooted, leafless and locally finite directed tree. Set $l=\deg{\mathsf{\omega}}$. Suppose $S_{\mathbb{1}}\in\boldsymbol{B}(\ell^{2}(V))$ is a $2$-isometry and $d_{S_{\mathbb{1}}^{\prime}}$ is given by (54) with $S_{\mathbb{1}}^{\prime}$ in place of $S_{\boldsymbol{\lambda}}$ $($see Lemma  5.5$)$. Then the following assertions  hold.

1. (i)

Assume that each vertex $u\in V^{\circ}$ has degree $1$ or $l$. Then

[TABLE] 2. (ii)

Assume that $l\geqslant 2$ and $l-1$ vertices of $\mathsf{Chi}(\mathsf{\omega})$ have degree $2$. Then each $u\in V$ has degree $1$, $2$ or $l$ and

[TABLE]

Proof.

Since $S_{\mathbb{1}}$ is a $2$-isometry satisfying (77) with $\beta_{u}\equiv 1$, Lemma  7.1 yields

[TABLE]

These three formulas and appropriate degree assumptions are all that are needed to prove (i) and (ii). It follows from (83) that

[TABLE]

Hence, by a simple induction argument based on (84), we get

[TABLE]

This proves the degree $1$ case of (81) and (82).

(i) Without loss of generality, we may assume that $l\geqslant 2.$ Suppose that $\deg{u}=l$. Then (83) and the assumption of (i) yield

[TABLE]

We will verify the bottom formula in (81) by using induction on $n$. The case of $n=0$ is obvious. Assume that it holds for a fixed $n\in\mathbb{Z}_{+}$. Then

[TABLE]

where ($*$) follows from (86), (85) and the induction hypothesis. This completes the proof of the assertion (i).

(ii) An induction argument based on (48) and (83) shows that666 The statement (a) holds without the degree assumption of (ii).

1. (a)

if $u\in V$ is of degree $2$, then $\mathsf{Chi}(u)$ consists of one vertex of degree $1$ and one vertex of degree $2$, 2. (b)

$\mathsf{Chi}(\mathsf{\omega})$ consists of one vertex of degree $1$ and $l-1$ vertices of degree $2$, 3. (c)

if $u\in V^{\circ}$, then $\deg{u}\in\{1,2\}$, 4. (d)

if $l\geqslant 3$, then $\mathsf{\omega}$ is the only vertex of $V$ of degree $l$.

If $\deg{u}=2$, then by (c) the directed tree \big{(}{{\mathsf{Des}}(u)},({{\mathsf{Des}}(u)}\times{{\mathsf{Des}}(u)})\cap E\big{)} satisfies the assumption of (i) for $l=2$, and thus the degree $2$ case of (82) follows from (i) by applying Proposition  2.2 (cf. the proof of Theorem  6.5). By (c) and (d), it remains to consider the case of $\deg{u}=l\geqslant 3$, i.e., $u=\mathsf{\omega}$. If $n\in\mathbb{N}$, then

[TABLE]

where ($*$) follows from (b), (85) and the degree $2$ case of (82). This completes the proof of Lemma  7.5. ∎

Remark 7.6*.*

An inspection of the proof of Lemma  7.5 reveals that

1. $\bullet$

under the assumption of (i), there are infinitely many vertices of degree $1$ and, if $l\geqslant 2$, there are infinitely many vertices of degree $l$ (cf. Corollary  7.4), 2. $\bullet$

under the assumption of (ii), there are infinitely many vertices of degree $1$ and infinitely many vertices of degree $2$; if $l\geqslant 3$, then $\mathsf{\omega}$ is the only vertex of degree $l$. $\diamondsuit$

Let us make some comments regarding the condition (83).

Remark 7.7*.*

Assume that $\mathscr{T}$ is a rooted, leafless and locally finite directed tree. If $\deg{u}=1$ or $\deg{u}=2$, then $d_{S_{\mathbb{1}}^{\prime}}(u,n)$ can be calculated explicitly. If $\deg{u}=3$, then, by (83), $\mathsf{Chi}(u)$ may consists of one vertex of degree $3$ and two vertices of degree $1$, or two vertices of degree $2$ and one vertex of degree $1$. Now we have two possibilities either the degree $3$ case appear infinitely many times, or the process stops after a finite number of steps and then the degree $1$ and the degree $2$ cases appear both infinitely many times. If $\deg{u}=4$, then, by (83) again, $\mathsf{Chi}(u)$ may consists of one vertex of degree $4$ and three vertices of degree $1$, or one vertex of degree $3$, one vertex of degree $2$ and two vertices of degree $1$, or three vertices of degree $2$ and one vertex of degree $1$, and so on. $\diamondsuit$

Now we show that the Cauchy dual subnormality problem has an affirmative solution for certain adjacency operators which, in general, do not satisfy the kernel condition (see Theorem  7.8). This leads to Hausdorff moment sequences which are structurally different from those in Section  6. We also answer the question of when such adjacency operators are Brownian or quasi-Brownian isometries.

Theorem 7.8**.**

Let $\mathscr{T}$ be a rooted, leafless and locally finite directed tree. Set $l=\deg{\mathsf{\omega}}$. Suppose that $S_{\mathbb{1}}\in\boldsymbol{B}(\ell^{2}(V))$ is a $2$-isometry and one of the following two conditions holds:

1. (i)

each vertex $u\in V^{\circ}$ has degree $1$ or $l$, 2. (ii)

$l\geqslant 2$* and $l-1$ vertices of $\mathsf{Chi}(\mathsf{\omega})$ have degree $2$.*

Then the following statements are valid:

1. ➊

the Cauchy dual $S_{\mathbb{1}}^{\prime}$ of $S_{\mathbb{1}}$ is a subnormal contraction, 2. ➋

$S_{\mathbb{1}}$* satisfies the kernel condition if and only if $l=1,$* 3. ➌

if (i) holds, then $S_{\mathbb{1}}$ is always a quasi-Brownian isometry; moreover $S_{\mathbb{1}}$ is a Brownian isometry if and only if $l=1,$ 4. ➍

if (ii) holds, then $S_{\mathbb{1}}$ is never a Brownian isometry; moreover, $S_{\mathbb{1}}$ is a quasi-Brownian isometry if and only if $l=2.$

Proof.

➊ It follows from Lemma  5.5, the formula (54) (applied to $S_{\mathbb{1}}^{\prime}$) and [28, Theorem  6.1.3] that $S_{\mathbb{1}}^{\prime}$ is subnormal if and only if

[TABLE]

If (i) holds, then (87) follows easily from (81).

Now, assume that (ii) holds. In view of (82), $\{d_{S_{\mathbb{1}}^{\prime}}(u,n)\}_{n=0}^{\infty}$ is a Stieltjes moment sequence whenever $u$ is of degree $1$ or $2$. The only nontrivial case is when $l\geqslant 3$ and $\deg{u}=l$, i.e., $u=\mathsf{\omega}$ (see the statement (d) of the proof of Lemma  7.5). Then, by (82), we have

[TABLE]

This implies that $\{d_{S_{\mathbb{1}}^{\prime}}(\mathsf{\omega},n+1)\}_{n=0}^{\infty}$ is a Stieltjes moment sequence with the representing measure $\mu=\frac{2(l-1)}{3l^{2}}\delta_{\frac{1}{4}}+\frac{l+2}{3l^{2}}\delta_{1}$. Since $l\geqslant 3$, it is easily seen that

[TABLE]

Hence, by Lemma  6.3, $\{d_{S_{\mathbb{1}}^{\prime}}(\mathsf{\omega},n)\}_{n=0}^{\infty}$ is a Stieltjes moment sequence.

➋ This is an immediate consequence of Proposition  7.3 and Lemma  7.2.

➌ This is a consequence of Lemma  7.2, Corollary  7.4 and Proposition  7.3 (cf. Proposition  5.12).

➍ Assume that (ii) holds. It follows from Proposition  7.3(ii) that $S_{\mathbb{1}}$ is never a Brownian isometry. Suppose now $S_{\mathbb{1}}$ is a quasi-Brownian isometry. Then, by Theorem  4.5, we have

[TABLE]

It follows from Lemma  5.3(v) that $S_{\mathbb{1}}^{*}S_{\mathbb{1}}$ is a diagonal operator (with respect to the orthogonal basis $\{e_{u}\}_{u\in V}$) with diagonal elements $\{\deg u\}_{u\in V}.$ Hence, (89) yields

[TABLE]

According to Lemma  7.5(ii) and (54) (with $S_{\mathbb{1}}^{\prime}$ in place of $S_{\boldsymbol{\lambda}}$), we have

[TABLE]

Comparing (90) with (91) and passing to the limit as $n\to\infty,$ we see that $l=2$ (recall that $l\geqslant 2$). Conversely, if $l=2,$ then by Lemma  7.5(ii), each $u\in V$ is of degree $1$ or $2.$ Hence, in view of Corollary  7.4, $S_{\mathbb{1}}$ is a quasi-Brownian isometry. This completes the proof. ∎

Below we construct a rooted and leafless directed tree $\mathscr{T}$ such that

[TABLE]

In fact, we show that for every integer $l\geqslant 2,$ there exists a rooted and leafless directed tree with $2$-isometric adjacency operator, which satisfies the condition (ii) of Theorem  7.8. Similar construction can be performed in the case of the condition (i) of Theorem  7.8; the resulting directed tree is either $\mathbb{Z}_{+}$ if $l=1$ or a quasi-Brownian directed tree of valency $l$ if $l\geqslant 2$.

Example 7.9**.**

Let us fix $l\in\{2,3,4,\ldots\}$. Using (48), one can construct inductively a rooted and leafless directed tree $\mathscr{T}$ with the following properties (see the right subfigure of Figure 2 in which $l=4$):

1. (i)

$\deg{\mathsf{\omega}}=l$, 2. (ii)

$\mathsf{Chi}(\mathsf{\omega})$ consists of one vertex of degree $1$ and $l-1$ vertices of degree $2$, 3. (iii)

for each vertex $u\in V^{\circ}$ of degree $1,$ $\mathsf{Chi}(u)$ consists of one vertex of degree  $1,$ 4. (iv)

for each vertex $u\in V^{\circ}$ of degree $2$, $\mathsf{Chi}(u)$ consists of one vertex of degree $1$ and one of degree $2.$

Clearly, the directed tree $\mathscr{T}$ satisfies the condition (ii) of Theorem  7.8. Let $S_{\mathbb{1}}$ be the adjacency operator of $\mathscr{T}.$ Since $\|S_{\mathbb{1}}e_{u}\|^{2}\leqslant l$ for all $u\in V$, we infer from Lemma  5.3 that $S_{\mathbb{1}}\in\boldsymbol{B}(\ell^{2}(V))$ and $\|S_{\mathbb{1}}\|^{2}=l$. It follows from (i)-(iv) that

[TABLE]

This combined with Lemma  7.1(i) implies that $S_{\mathbb{1}}$ is a $2$-isometry. If $l\geqslant 3,$ then by Theorem  7.8, the adjacency operator $S_{\mathbb{1}}$ satisfies (92). In turn, if $l=2,$ then by Theorem  7.8 again, $S_{\mathbb{1}}$ is a quasi-Brownian isometry that satisfies all but the last condition in (92). $\diamondsuit$

We conclude this section by showing that the Cauchy dual subnormality problem has a negative solution in the class of adjacency operators.

Example 7.10**.**

We begin by constructing an appropriate directed tree. Let us fix $l\in\{3,4,5,\ldots\}$. Using (48), one can construct inductively a rooted and leafless directed tree $\mathscr{T}$ with the following properties (see Figure 3):

1. (i)

$\deg{\mathsf{\omega}}=l$, 2. (ii)

$\mathsf{Chi}(\mathsf{\omega})$ consists of $l-1$ vertices of degree $1$ and one vertex, say $\psi$, of degree  $l$, 3. (iii)

$\mathsf{Chi}(\psi)$ consists of one vertex of degree $1$ and $l-1$ vertices of degree $2$, 4. (iv)

for each vertex $u\in V$ of degree $1$, $\mathsf{Chi}(u)$ consists of one vertex of degree  $1$, 5. (v)

for each vertex $u\in V$ of degree $2$, $\mathsf{Chi}(u)$ consists of one vertex of degree $1$ and one of degree $2$.

Let $S_{\mathbb{1}}$ be the adjacency operator of the directed tree $\mathscr{T}$. Exactly as in Example  7.9, we verify that $S_{\mathbb{1}}\in\boldsymbol{B}(\ell^{2}(V))$ and $\|S_{\mathbb{1}}\|^{2}=l$. It is also a routine matter to verify that the properties (i)-(v) yield

[TABLE]

This together with Lemma  7.1(i) implies that $S_{\mathbb{1}}$ is a $2$-isometry. Set \mathscr{T}_{\psi}=\big{(}{{\mathsf{Des}}(\psi)},({{\mathsf{Des}}(\psi)}\times{{\mathsf{Des}}(\psi)})\cap E\big{)}. Then $\mathscr{T}_{\psi}$ is a rooted and leafless directed tree with the root $\psi$. Recall that by Lemma  5.3, the space $\ell^{2}({{\mathsf{Des}}(\psi)})$ is invariant for $S_{\mathbb{1}}$ and $S_{\mathbb{1}}^{*}S_{\mathbb{1}}$, and that the restriction of $S_{\mathbb{1}}$ to $\ell^{2}({{\mathsf{Des}}(\psi)})$ coincides with the adjacency operator $S_{\psi,\mathbb{1}}$ of the directed tree $\mathscr{T}_{\psi}$ (cf. the proof of Theorem  6.5). Hence $S_{\psi,\mathbb{1}}\in\boldsymbol{B}(\ell^{2}({{\mathsf{Des}}(\psi)}))$ is a $2$-isometry, and by Proposition  2.2, $S_{\psi,\mathbb{1}}^{\prime}$ coincides with the restriction of $S_{\mathbb{1}}^{\prime}$ to $\ell^{2}({{\mathsf{Des}}(\psi)})$. As a consequence, we see that $d_{S_{\psi,\mathbb{1}}^{\prime}}(u,n)=d_{S_{\mathbb{1}}^{\prime}}(u,n)$ for all $u\in{{\mathsf{Des}}(\psi)}$ and $n\in\mathbb{Z}_{+}$. Applying Theorem  7.8 (the case (ii)) to the directed tree $\mathscr{T}_{\psi}$ and its adjacency operator $S_{\psi,\mathbb{1}}$ and using [28, Theorem  6.1.3], we deduce that $\{d_{S_{\mathbb{1}}^{\prime}}(\psi,n)\}_{n=0}^{\infty}$ is a Stieltjes moment sequence. A careful look at the proof of Theorem  7.8 (use (88) with $\psi$ and $S_{\psi,\mathbb{1}}^{\prime}$ in place of $\mathsf{\omega}$ and $S_{\mathbb{1}}^{\prime}$, respectively) shows that the positive Borel measure $\mu$ on $\mathbb{R}_{+}$ defined by

[TABLE]

is a unique representing measure of $\{d_{S_{\mathbb{1}}^{\prime}}(\psi,n+1)\}_{n=0}^{\infty}$. It follows from Lemma  6.3 that the positive Borel measure $\nu$ on $\mathbb{R}_{+}$ given by

[TABLE]

is a representing measure of the Stieltjes moment sequence $\{d_{S_{\mathbb{1}}^{\prime}}(\psi,n)\}_{n=0}^{\infty}$. This and (93) imply that

[TABLE]

Using (iv) and Lemma  7.1(ii), we verify that (85) holds. Hence, by applying Lemma  7.1(ii) to $u=\mathsf{\omega}$, we get

[TABLE]

where, by (94), $\rho$ is the positive Borel measure on $\mathbb{R}_{+}$ defined by

[TABLE]

This means that $\{d_{S_{\mathbb{1}}^{\prime}}(\mathsf{\omega},n+1)\}_{n=0}^{\infty}$ is a Stieltjes (in fact a Hausdorff) moment sequence with the unique representing measure $\rho$. Since $\rho(\{0\})>0$ (because $l\geqslant 3$), we infer from Lemma  6.3 that $\{d_{S_{\mathbb{1}}^{\prime}}(\mathsf{\omega},n)\}_{n=0}^{\infty}$ is not a Stieltjes moment sequence. Hence, by [28, Theorem  6.1.3], the operator $S_{\mathbb{1}}^{\prime}$ is not subnormal. $\diamondsuit$

Acknowledgement

A part of this paper was written while the second author visited Jagiellonian University in Summer of 2017. He wishes to thank the faculty and the administration of this unit for their warm hospitality.