Structure of Iso-Symmetric Operators

Abstract

For a Hilbert space operator $T\in B\left(\mathcal{H}\right)$, let $L_T$ and $R_T\in B\left(B\left(\mathcal{H}\right)\right)$ denote, respectively, the operators of left multiplication and right multiplication by T. For positive integers m and n, let ${▵}_{T^{\ast },T}^m\left(I\right)={(L_{T^{\ast }}{R}_T-I)}^m\left(I\right)$ and ${\delta }_{T^{\ast },T}^n\left(I\right)={(L_{T^{\ast }}-R_T)}^m\left(I\right)$. The operator T is said to be $(m,n)$-isosymmetric if ${▵}_{T^{\ast },T}^m\left({\delta }_{T^{\ast },T}^n\left(I\right)\right)=0$. Power bounded $(m,n)$-isosymmetric operators $T\in B\left(\mathcal{H}\right)$ have an upper triangular matrix representation $T=\left(\begin{array}{cc}{T}_1& T_3\\ 0& T_2\end{array}\right)\in B({\mathcal{H}}_1\oplus {\mathcal{H}}_2)$ such that $T_1\in B\left({\mathcal{H}}_1\right)$ is a $C_{0.}$-operator which satisfies ${\delta }_{T_1^{\ast },T_1}^n\left(I{|}_{{\mathcal{H}}_1}\right)=0$ and $T_2\in B\left({\mathcal{H}}_2\right)$ is a $C_{1.}$-operator which satisfies $A{T}_2=(V_u\oplus V_b){|}_{{\mathcal{H}}_2}A$, $A={lim}_{t\to \infty }{T}_2^{\ast t}{T}_2^t$, $V_u$ is a unitary and $V_b$ is a bilateral shift. If, in particular, T is cohyponormal, then T is the direct sum of a unitary with a $C_{00}$-contraction.

1. Introduction

Let $B\left(\mathcal{H}\right)$ denote the algebra of operators, i.e., bounded linear transformations, on an infinite dimensional complex Hilbert space $\mathcal{H}$ into itself. Let $\mathbb{C}$ denote the complex plane and $\overline{z}$ the conjugate of $z\in \mathbb{C}$. For a given polynomial $f\left(z\right)={\sum }_{i,j}{c}_{ij}{\overline{z}}^i{z}^j$ on $\mathbb{C}$ and an operator $T\in B\left(\mathcal{H}\right)$, define $f\left(T\right)$ by $f\left(T\right)={\sum }_{i,j}{c}_{ij}{T}^{\ast i}{T}^j$. Then T is said to be a (hereditary) root of f if $f\left(T\right)=0$. An operator $T\in B\left(\mathcal{H}\right)$ is n-selfadjoint for some positive integer n if T is a root of the polynomial $f\left(z\right)={(\overline{z}-z)}^n$, equivalently, if

$${\delta }_{T^{\ast },T}^n\left(I\right)=\sum _{j=0}^n{(-1)}^j\left(\begin{array}{c}n\\ j\end{array}\right){T^{\ast }}^{(n-j)}{T}^j=0,$$

and T is m-isometric for some positive integer m if it is a root of the polynomial $f\left(z\right)={(\overline{z}z-1)}^m$, equivalently, if

$${▵}_{T^{\ast },T}^m\left(I\right)=\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){T^{\ast }}^{(m-j)}{T}^{m-j}=0.$$

The classes consisting of n-selfadjoint and m-isometric operators have been studied extensively by a large number of authors in the recent past (see list of references for further references).

The development of the theory of m-selfadjoint operators in infinite dimensional Hilbert spaces was motivated by the seminal work of Helton [1], who observed an unexpected, intimate connection with differential equations, in particular conjugate point theory and disconjugacy. McCullough and Rodman [2] in their consideration of algebraic and spectral properties of n-symmetric operators remark [2] (p. 419), that the authors of [1,3,4] were certainly aware of the fact that every 2-symmetric operator is 1-symmetric, even though they do not explicitly state so. More generally, McCullough and Rodman [2] (Theorem 3.1) state that the techniques of Helton [1] lead to a possible proof of the more general result that “$2n$-symmetric operators are $(2n-1)$-symmetric”. The class of m-symmetric operators was introduced by Agler [3] and studied in a series of papers by Agler and Stankus [5,6,7]; properties of m-isometric operators, amongst them the spectral picture, strict m-isometries, perturbation by commuting nilpotents and the product of m-isometries, have since been studied by a large number of authors, amongst them Bayart [8], Bermudez et al. [9,10,11], Botelho and Jamison [12], Duggal et al. [13,14,15], and Gu et al. [16,17,18]. The (hereditary) roots of the polynomial ${(\overline{z}z-1)}^m{(\overline{z}-z)}^n=0$ have been called $(m,n)$-isosymmetric operators; thus T is $(m,n)$-isosymmetric if and only if

$$\begin{array}{ccc}\hfill {▵}_{T^{\ast },T}^m\left({\delta }_{T^{\ast },T}^n\left(I\right)\right)& =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){T^{\ast }}^{(m-j)}\left(\sum _{k=0}^n{(-1)}^k\left(\begin{array}{c}n\\ k\end{array}\right){T^{\ast }}^{(n-k)}{T}^k\right)T^{m-j}\hfill \\ & =& {\delta }_{T^{\ast },T}^n\left({▵}_{T^{\ast },T}^m\left(I\right)\right)\hfill \\ & =& \sum _{k=0}^n{(-1)}^k\left(\begin{array}{c}n\\ k\end{array}\right){T^{\ast }}^{(n-k)}\left(\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){T^{\ast }}^{(m-j)}{T}^{m-j}\right)T^k\hfill \\ & =& 0.\hfill \end{array}$$

Examples of $(m,n)$-isosymmetric operators occur naturally. Thus, every isometric operator $T\in B\left(\mathcal{H}\right)$ is $(1,1)$-isosymmetric. Indeed, if $T\in B\left(\mathcal{H}\right)$ is m-isometric, or n-symmetric, then T is $(m,n)$-isosymmetric. A study of this class of operators has been carried out by Stankus [19,20], and Gu and Stankus [18], amongst others.

For an operator $T\in B\left(\mathcal{H}\right)$, define the operators $L_T$ and $R_T\in B\left(B\left(\mathcal{H}\right)\right)$ of left multiplication and (respectively) right multiplication by T by

$$L_T\left(X\right)=TX,R_T\left(X\right)=XT.$$

Then T is n-symmetric, respectively, m-isometric, if and only if

$${(L_{T^{\ast }}-R_T)}^n\left(I\right)={\delta }_{T^{\ast },T}^n\left(I\right)=0,,\mathrm{respectively}{(L_{T^{\ast }}{R}_T-I)}^m\left(I\right)={▵}_{T^{\ast },T}^m\left(I\right)=0$$

and T is $(m,n)$-isosymmetric if and only if

$${(L_{T^{\ast }}{R}_T-I)}^m\left({(L_{T^{\ast }}-R_T)}^n\left(I\right)\right)={▵}_{T^{\ast },T}^m\left({\delta }_{T^{\ast },T}^n\left(I\right)\right)=0.$$

Trivially, ${\delta }_{T^{\ast },T}^n\left(I\right)=0$ if and only if ${\delta }_{{(T-\lambda )}^{\ast },T-\lambda }^n\left(I\right)=0$ for all $\lambda \in \mathtt{C}$, and if $\lambda \in \mathbb{R}$ is such that $\lambda \notin \sigma \left(T\right)$, then

$$\begin{array}{ccc}\hfill {\delta }_{T^{\ast },T}^n\left(I\right)=0& ⟺& {\delta }_{{(T-\lambda )}^{\ast },T-\lambda }^n\left(I\right)=0\hfill \\ & ⟺& R_{T-\lambda }^{-n}{\delta }_{{(T-\lambda )}^{\ast },T-\lambda }^n\left(I\right)=0\hfill \\ & ⟺& {▵}_{T^{\ast }-\lambda ,{(T-\lambda )}^{-1}}^n\left(I\right)=0\hfill \\ & ⟺& L_{T^{\ast }-\lambda }^{-n}{\delta }_{{(T-\lambda )}^{\ast },T-\lambda }^n\left(I\right)=0\hfill \\ & ⟺& {▵}_{{(T^{\ast }-\lambda )}^{-1},T-\lambda }^n\left(I\right)=0.\hfill \end{array}$$

In this note, we exploit relationships of this type, using little more than some basic properties of elementary operators, to give a formal, simple proof of the result that $2n$-symmetric operators are $(2n-1)$-symmetric. The case $n=1$ of this result is of some interest, more so for the reason that 2-symmetric operators are cohyponormal. Cohyponormal $(m,n)$-isosymmetric operators have a particularly simple structure: they are the direct sum of a unitary operator and a $C_{00}$-contraction (where either of the components may be absent). The cohyponormality condition is redundant in the case in which ($n=2$ and) ${\delta }_{T^{\ast },T}^2\left(I\right)\ge 0$; if also $m=2$, then ${▵}_{T^{\ast },T}^2\left({\delta }_{T^{\ast },T}^2\left(I\right)\right)\ge 0$ is sufficient to guarantee T is the direct sum of a unitary operator and a $C_{00}$-contraction. For hyponormal, more generally normaloid, $(m,n)$-isosymmetric T, T is a contraction, hence power bounded. Power bounded $(m,n)$-isosymmetric operators T have an upper triangular matrix representation $T\in B({\mathcal{H}}_1\oplus {\mathcal{H}}_2)$ such that the $(1,1)$-entry is a $C_{0.}$-operator $T_1$ satisfying ${\delta }_{T_1^{\ast },T_1}^n\left(I{|}_{{\mathcal{H}}_1}\right)=0$ and the $(2,2)$-entry $T_2$ satisfies $A{T}_2=(V_u\oplus V_b){|}_{{\mathcal{H}}_2}A$ for an injective positive operator $A\in B\left({\mathcal{H}}_2\right)$ (defined by $A={lim}_{t⟶\infty }{T}_2^{\ast t}{T}_2^t$), unitary $V_u$ and a bilateral shift $V_b$.

We introduce our notation/terminology, along with some complementary results, in the following section, Section 3 is devoted to considering $2n$-symmetric and related operators, and our Section 4 considers the structure of cohyponormal and power bounded $(m,n)$-isosymmetric operators.

2. Some Complementary Results

In the following, $⟨.,.⟩$ will denote the inner product on $\mathcal{H}$. We shall denote the approximate point spectrum and the spectrum of an operator $T\in B\left(\mathcal{H}\right)$ by ${\sigma }_a\left(T\right)$ and $\sigma \left(T\right)$, respectively. We shall denote the open unit disc in the complex plane $\mathbb{C}$ by $\mathbb{D}$ and the boundary of the unit disc in $\mathbb{C}$ by $\partial \mathbb{D}$. The operator T is power bounded if there exists a scalar $M>0$ such that

$$\underset{n\in \mathbb{N}}{sup}\parallel T^n\parallel \le M.$$

It is clear from the definition that if $T\in B\left(\mathcal{H}\right)$ is power bounded, then $T^{\ast }$ is power bounded, the spectral radius

$$r\left(T\right)=\underset{n\to \infty }{lim}\parallel T^n{\parallel }^{\frac{1}{n}}\le 1$$

and the spectrum $\sigma \left(T\right)$ of T satisfies $\sigma \left(T\right)\subseteq \overline{\mathbb{D}}$ ($=\{\lambda \in \mathbb{C}:|\lambda |\le 1\}$). The operator T is a $C_{0.}$, respectively, $C_{1.}$, operator if

$$\underset{n\to \infty }{lim}\parallel T^{n}x\parallel =0\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}x\in \mathcal{H},$$

$$\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{l}\mathrm{y},\underset{n\in \mathtt{N}}{inf}\parallel T^{n}x\parallel >0\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l}0\ne x\in \mathcal{H};$$

$T\in C_{.0}$ (resp., $T\in C_{.1}$) if $T^{\ast }\in C_{0.}$ (resp., $T^{\ast }\in C_{1.}$) and $T\in C_{\alpha \beta }$ if $T\in C_{\alpha .}\cap C_{.\beta }$ ($\alpha ,\beta =0,1$). It is well known [21] that every power bounded operator $T\in B\left(\mathcal{H}\right)$ has an upper triangular matrix representation

$$T=\left(\begin{array}{cc}{T}_1& T_3\hfill \\ 0& T_2\hfill \end{array}\right)\in B({\mathcal{H}}_1\oplus {\mathcal{H}}_2)$$

for some decomposition $\mathcal{H}={\mathcal{H}}_1\oplus {\mathcal{H}}_2$ of $\mathcal{H}$ such that $T_1\in C_{0.}$ and $T_2\in C_{1.}$. Recall that every isometry $V\in B\left(\mathcal{H}\right)$ has a direct sum decomposition

$$V=V_c\oplus V_u\in B({\mathcal{H}}_c\oplus {\mathcal{H}}_u),V_c\in C_{10}\mathrm{and}{V}_u\in C_{11}$$

into its completely non-unitary (i.e., unilateral shift) and unitary parts [22]. Hyponormal contractions T, i.e., contractions $T\in B\left(\mathcal{H}\right)$ such that $T{T}^{\ast }\le T^{\ast }T$, are known to have $C_{.0}$ cnu (=completely non-unitary) parts [23].

The following result from [24] will be used in some of our argument below.

Theorem 1.

If $A,B\in B\left(\mathcal{H}\right)$, then the following statements are pairwise equivalent.

(i)

$\mathrm{ran}\left(\mathrm{A}\right)\subseteq \mathrm{ran}\left(\mathrm{B}\right)$.

(ii)

There is a $\mu \ge 0$ such that $A{A}^{\ast }\le {\mu }^{2}B{B}^{\ast }$.

(iii)

There is an operator $C\in B\left(\mathcal{H}\right)$ such that $A=BC$.

Furthermore, if these conditions hold, then the operator C may be chosen so that (a) ${\parallel C\parallel }^2=\inf \{\lambda :A{A}^{\ast }\le \lambda B{B}^{\ast }\}$; (b) $\ker \left(A\right)=\ker \left(C\right)$; (c) $\mathrm{ran}\left(C\right)\subseteq \ker {\left(B\right)}^{\perp }$.

A pair of operators $A,B\in B\left(\mathcal{H}\right)$ satisfies the Putnam–Fuglede (commutativity) property if ${\delta }_{A,B}^{-1}\left(0\right)\subseteq {\delta }_{A^{\ast },B^{\ast }}^{-1}\left(0\right)$. It is easily seen that if $A,B$ satisfy the Putnam–Fuglede property and ${\delta }_{A,B}\left(X\right)=0$, then $\overline{X\left(\mathcal{H}\right)}$ reduces A, ${ker}^{\perp }\left(X\right)$ reduces B, and ${A|}_{\overline{X\left(\mathcal{H}\right)}}$ and ${B|}_{{ker}^{\perp }\left(X\right)}$ are unitarily equivalent normal operators. Normal operators satisfy the Putnam–Fuglede property [25]. Indeed, more is true. An asymmetric version of the Putnam–Fuglede property holds for a variety of classes of Hilbert space operators [26], amongst them hyponormal pairs A and $B^{\ast }\in B\left(\mathcal{H}\right)$: if $A,B^{\ast }$ are hyponormal operators, then ${\delta }_{A,B^{\ast }}^{-1}\left(0\right)\subseteq {\delta }_{A^{\ast },B}^{-1}\left(0\right)$. Even more interestingly:

Theorem 2

([26]). If $A,B^{\ast }\in B\left(\mathcal{H}\right)$ are hyponormal operators and n is some positive integer, then

$${\delta }_{A,B^{\ast }}^{n^{-1}}\left(0\right)={\delta }_{A,B^{\ast }}^{-1}\left(0\right)\subseteq {\delta }_{A^{\ast },B}^{-1}\left(0\right).$$

3. $\mathit{n}$-Symmetric Operators for $\mathit{n}$ Even

We start by proving that n-symmetric operators for n even are $(n-1)$-symmetric. This property of n-symmetric operators is stated in [2] (Theorem 3.4) without a proof (but with the remark that a proof can be given using the techniques of [1]). Our proof below uses little more than some well understood properties of elementary operators of left and right multiplication.

Theorem 3.

If $T\in B\left(\mathcal{H}\right)$ is n-symmetric for some positive even integer n, then T is $(n-1)$-symmetric.

Proof. 

A straightforward argument shows that ${\sigma }_a\left(T\right)\subset \mathbb{R}$ for n-symmetric operator T. Hence $\sigma \left(T\right)\subset \mathbb{R}$, and there exists a non-zero real number $\lambda \notin \sigma \left(T\right)$. Since

$${\delta }_{T^{\ast },T}^n\left(I\right)=0⟺{\delta }_{T^{\ast }-\mu ,T-\mu }^n\left(I\right)=0$$

for all real $\mu$, we have

$${\delta }_{T^{\ast },T}^n\left(I\right)=0⟺R_{T-\lambda }^{-n}{\delta }_{T^{\ast }-\lambda ,T-\lambda }^n\left(I\right)=0⟺{▵}_{T^{\ast }-\lambda ,{(T-\lambda )}^{-1}}^n\left(I\right)=0.$$

It is easily seen (use an induction argument) that

$$\begin{array}{ccc}\hfill {▵}_{A,B}^n\left(I\right)={(L_A{R}_B-I)}^n\left(I\right)=0& ⟺& {\left(L_A{R}_B\right)}^n\left(I\right)-\sum _{j=0}^{n-1}\left(\begin{array}{c}n\\ j\end{array}\right){▵}_{A,B}^j\left(I\right)=0\hfill \\ & ⟺& {\left(L_A{R}_B\right)}^n\left(I\right)=\sum _{j=0}^{n-1}\left(\begin{array}{c}n\\ j\end{array}\right){▵}_{A,B}^j\left(I\right)\hfill \end{array}$$

for all operators $A,B\in B\left(\mathcal{H}\right)$. Hence, given ${▵}_{A,B}^n\left(I\right)=0$,

$$\begin{array}{ccc}\hfill {\left(L_A{R}_B\right)}^{n+1}\left(I\right)& =& \sum _{j=0}^{n-1}\left(\begin{array}{c}n\\ j\end{array}\right)L_A{R}_B{▵}_{A,B}^j\left(I\right)\hfill \\ & =& \sum _{j=0}^{n-1}\left(\begin{array}{c}n\\ j\end{array}\right){▵}_{A,B}^{j+1}\left(I\right)+\sum _{j=0}^{n-1}\left(\begin{array}{c}n\\ j\end{array}\right){▵}_{A,B}^j\left(I\right)\hfill \\ & =& \left(\begin{array}{c}n\\ n-1\end{array}\right){▵}_{A,B}^n\left(I\right)+\sum _{j=0}^{n-1}\left(\begin{array}{c}n+1\\ j\end{array}\right){▵}_{A,B}^j\left(I\right)\hfill \\ & =& \sum _{j=0}^{n-1}\left(\begin{array}{c}n+1\\ j\end{array}\right){▵}_{A,B}^j\left(I\right)\hfill \\ & =& \left(\begin{array}{c}n+1\\ n-1\end{array}\right){▵}_{A,B}^{n-1}\left(I\right)+\sum _{j=0}^{n-2}\left(\begin{array}{c}n+1\\ j\end{array}\right){▵}_{A,B}^j\left(I\right),\hfill \end{array}$$

and by an induction argument that

$$\begin{array}{c}\hfill {\left(L_A{R}_B\right)}^t\left(I\right)=\left(\begin{array}{c}t\\ n-1\end{array}\right){▵}_{A,B}^{n-1}\left(I\right)+\sum _{j=0}^{n-2}\left(\begin{array}{c}t\\ j\end{array}\right){▵}_{A,B}^j\left(I\right)\end{array}$$

(1)

for all $A,B\in B\left(\mathcal{H}\right)$ and integers $t\ge n$. Translating to the operator ${\delta }_{T^{\ast },T}^n\left(I\right)={▵}_{T^{\ast }-\lambda ,{(T-\lambda )}^{-1}}^n$ $\left(I\right)=0$, we have

$${\left(L_{T^{\ast }-\lambda }{R}_{T-\lambda }^{-1}\right)}^t\left(I\right)=\left(\begin{array}{c}t\\ n-1\end{array}\right){▵}_{T^{\ast }-\lambda ,{(T-\lambda )}^{-1}}^{n-1}\left(I\right)+\sum _{j=0}^{n-2}\left(\begin{array}{c}t\\ j\end{array}\right){▵}_{T^{\ast }-\lambda ,{(T-\lambda )}^{-1}}^j\left(I\right)$$

for all $t\ge n$ and real $\lambda \notin \sigma \left(T\right)$. Trivially,

$$\begin{array}{ccc}\hfill {▵}_{A,B}^n\left(I\right)=L_A{R}_B{▵}_{A,B}^{n-1}\left(I\right)-{▵}_{A,B}^{n-1}\left(I\right)=0& ⟹& L_A{R}_B{▵}_{A,B}^{n-1}\left(I\right)={▵}_{A,B}^{n-1}\left(I\right)\hfill \\ & ⟹& \cdots \hfill \\ & ⟹& {\left(L_A{R}_B\right)}^t{▵}_{A,B}^{n-1}\left(I\right)={▵}_{A,B}^{n-1}\left(I\right)\hfill \end{array}$$

for all $A,B\in B\left(\mathcal{H}\right)$ and integers $t\ge 1$. Hence

$$\begin{array}{ccc}& & I=\left(\begin{array}{c}t\\ n-1\end{array}\right){▵}_{T^{\ast }-\lambda ,{(T-\lambda )}^{-1}}^{n-1}\left(I\right)+\sum _{j=0}^{n-2}\left(\begin{array}{c}t\\ j\end{array}\right){\left(L_{T^{\ast }-\lambda }^{-1}{R}_{T-\lambda }\right)}^t{▵}_{T^{\ast }-\lambda ,{(T-\lambda )}^{-1}}^j\left(I\right)\hfill \\ & ⟹& 0\le {\parallel x\parallel }^2=\left(\begin{array}{c}t\\ n-1\end{array}\right)⟨{▵}_{T^{\ast }-\lambda ,{(T-\lambda )}^{-1}}^{n-1}\left(I\right)x,x⟩\hfill \\ & & +\sum _{j=0}^{n-2}\left(\begin{array}{c}t\\ j\end{array}\right)⟨{▵}_{T^{\ast }-\lambda ,{(T-\lambda )}^{-1}}^j\left(I\right){(T-\lambda )}^{t}x,{(T-\lambda )}^{-t}x⟩\hfill \end{array}$$

for all $x\in \mathcal{H}$ and integers $t\ge 1$. Letting $t\to \infty$, and observing that $\left(\begin{array}{c}t\\ n-1\end{array}\right)$ is of the order of $t^{n-1}$ and $\left(\begin{array}{c}t\\ j\end{array}\right)$, $0\le j\le n-2$, is of the order of $t^{n-2}$ as $t\to \infty$,

$$0\le ⟨{▵}_{T^{\ast }-\lambda ,{(T-\lambda )}^{-1}}^{n-1}\left(I\right)x,x⟩$$

for all $x\in \mathcal{H}$. Conclusion:

$${▵}_{T^{\ast }-\lambda ,{(T-\lambda )}^{-1}}^{n-1}\left(I\right)\ge 0.$$

Equivalently,

$$\begin{array}{ccc}\hfill 0& \le & {▵}_{T^{\ast }-\lambda ,{(T-\lambda )}^{-1}}^{n-1}\left(I\right)\hfill \\ & =& R_{T-\lambda }^{-n+1}{\delta }_{T^{\ast }-\lambda ,T-\lambda }^{n-1}\left(I\right)\hfill \\ & =& {\left({▵}_{T^{\ast }-\lambda ,{(T-\lambda )}^{-1}}^{n-1}\left(I\right)\right)}^{\ast }\hfill \\ & =& {▵}_{{(T^{\ast }-\lambda )}^{-1},T-\lambda }^{n-1}\left(I\right)\hfill \\ & =& {(-1)}^{n-1}{L}_{T^{\ast }-\lambda }^{-n+1}{\delta }_{T^{\ast }-\lambda ,T-\lambda }^{n-1}\left(I\right).\hfill \end{array}$$

Thus

$$L_{T^{\ast }-\lambda }^{n-1}{R}_{T-\lambda }^{-n+1}{\delta }_{T^{\ast }-\lambda ,T-\lambda }^{n-1}\left(I\right)={(-1)}^{n-1}{\delta }_{T^{\ast }-\lambda ,T-\lambda }^{n-1}\left(I\right).$$

Since ${\delta }_{T^{\ast }-\lambda ,T-\lambda }^n\left(I\right)=0$ implies ${\left(L_{T^{\ast }-\lambda }{R}_{T-\lambda }^{-1}\right)}^{n-1}{\delta }_{T^{\ast }-\lambda ,T-\lambda }^{n-1}\left(I\right)={\delta }_{T^{\ast }-\lambda ,T-\lambda }^{n-1}\left(I\right)$, and the integer n is even,

$${\delta }_{T^{\ast }-\lambda ,T-\lambda }^{n-1}\left(I\right)=-{\delta }_{T^{\ast }-\lambda ,T-\lambda }^{n-1}\left(I\right)⟺{\delta }_{T^{\ast }-\lambda ,T-\lambda }^{n-1}\left(I\right)={\delta }_{T^{\ast },T}^{n-1}\left(I\right)=0.$$

This completes the proof. □

It is immediate from Theorem 3 that 2-symmetric $B\left(\mathcal{H}\right)$ operators are symmetric. A proof of this of a different flavour and (in some respects) of interest in itself may be given as follows.

Corollary 1

([2]). A 2-symmetric $B\left(\mathcal{H}\right)$ operator is self-adjoint.

Proof. 

For operators $T\in B\left(\mathcal{H}\right)$,

$$0\le {\left({\delta }_{T^{\ast },T}\left(I\right)\right)}^{\ast }\left({\delta }_{T^{\ast },T}\left(I\right)\right)=T^{\ast }T+T{T}^{\ast }-T^2-T^{\ast 2}.$$

If also T is 2-symmetric, then

$${\delta }_{T^{\ast },T}^2\left(I\right)=T^{\ast 2}-2T^{\ast }T+T^2=0.$$

Hence

$${\delta }_{T^{\ast },T}^2\left(I\right)=0\le {\left({\delta }_{T^{\ast },T}\left(I\right)\right)}^{\ast }\left({\delta }_{T^{\ast },T}\left(I\right)\right)⟹T^{\ast }T\le T{T}^{\ast },$$

i.e., $T^{\ast }$ is hyponormal. Set ${\delta }_{T^{\ast },T}\left(I\right)=X$; then T is 2-symmetric if and only if

$${\delta }_{T^{\ast },T}\left(X\right)=T^{\ast }X-XT=0.$$

Applying the Putnam–Fuglede commutativity theorem for hyponormal operators, we have

$$T^{\ast }X-XT=0⟹TX-X{T}^{\ast }=0⟺T^{\ast 2}-2T{T}^{\ast }+T^2=0.$$

Already $T^{\ast 2}-2T^{\ast }T+T^2=0$. Hence $T^{\ast }T=T{T}^{\ast }$, i.e., T is normal. However, then

$${\delta }_{T^{\ast },T}^2\left(I\right)=0⟺{\delta }_{T^{\ast },T}\left(I\right)=0$$

(see Theorem 2). Hence $T^{\ast }=T$. □

The argument of the proof of Corollary 1 is suggestive of an interesting proof of a well known result on invertible 2-isometries [8].

Corollary 2.

Invertible 2-isometric $B\left(\mathcal{H}\right)$ operators are unitary.

Proof. 

The operator ${▵}_{T^{\ast },T}\left(I\right)\in B\left(\mathcal{H}\right)$ being self-adjoint,

$${\left({▵}_{T\ast ,T}\left(I\right)\right)}^2={\left(T^{\ast }T\right)}^2-2T^{\ast }T+I\ge 0.$$

Since ${▵}_{T^{\ast },T}^2\left(I\right)=T^{\ast 2}{T}^2-2T^{\ast }T+I=0$ and T is invertible, we have

$$T^{\ast 2}{T}^2\le {\left(T^{\ast }T\right)}^2⟺T^{\ast }T\le T{T}^{\ast },$$

i.e., $T^{\ast }$ is invertible hyponormal (with a hyponormal inverse ${T^{\ast }}^{-1}$). We have

$${▵}_{T^{\ast },T}^2\left(I\right)=0⟺{\delta }_{T^{\ast },T^{-1}}^2\left(I\right)=0.$$

Putnam–Fuglede commutativity theorem for hyponormal operators applies and we conclude that

$$\begin{array}{c}\hfill {\delta }_{T^{\ast },T^{-1}}^2\left(I\right)=0⟺{\delta }_{T^{\ast },T^{-1}}\left(I\right)=0⟺T^{\ast }T=T{T}^{\ast }=I,\end{array}$$

i.e., T is unitary. □

A generalised version of Corollary 2 is known to hold: if ${▵}_{T^{\ast },T}^m\left(I\right)=0$ for an invertible $T\in B\left(\mathcal{H}\right)$ and an even positive integer m, then ${▵}_{T^{\ast },T}^{m-1}\left(I\right)=0$ [8] (Proposition 2.4). Here the pair $(T^{\ast },T)$ may be replaced by the pair $(T^{\ast },T^{-1})$.

Corollary 3.

If ${▵}_{T^{\ast },T^{-1}}^m\left(I\right)=0$ for an invertible $T\in B\left(\mathcal{H}\right)$ and even positive integer m, then ${▵}_{T^{\ast },T^{-1}}^{m-1}\left(I\right)=0$.

Proof. 

The proof is an application of Theorem 3. The hypothesis ${▵}_{T^{\ast },T}^m\left(I\right)=0$ implies

$$\begin{array}{ccc}\hfill L_{T^{\ast }}^{-m}{▵}_{T^{\ast },T^{-1}}^m\left(I\right)={(-1)}^m{\delta }_{{T^{\ast }}^{-1},T^{-1}}^m\left(I\right)=0& ⟺& {\delta }_{{T^{\ast }}^{-1},T^{-1}}^m\left(I\right)=0\hfill \\ & ⟹& {\delta }_{{T^{\ast }}^{-1},T^{-1}}^{m-1}\left(I\right)=0\hfill \\ & ⟺& L_{T^{\ast }}^{m-1}{\delta }_{{T^{\ast }}^{-1},T^{-1}}^{m-1}\left(I\right)=0\hfill \\ & ⟺& {▵}_{T^{\ast },T^{-1}}^{m-1}\left(I\right)=0.\hfill \end{array}$$

This completes the proof. □

Yet another generalisation of Corollary 2 is obtained upon considering operators $T\in B\left(\mathcal{H}\right)$ such that $T\in (m,X)$-isometric, i.e., operators $T\in B\left(\mathcal{H}\right)$ satisfying ${▵}_{T^{\ast },T}^m\left(X\right)={\sum }_{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)T^{{\ast }^{m-j}}X{T}^{m-j}=0$, for some positive operator $X\in B\left(\mathcal{H}\right)$. For such operators T, it is clear from the argument leading to equality (1) that

$$0\le {\left(L_{T^{\ast }}{R}_T\right)}^t\left(X\right)=\left(\begin{array}{c}t\\ m-1\end{array}\right){▵}_{T^{\ast },T}^{m-1}\left(X\right)+\sum _{j=0}^{m-2}\left(\begin{array}{c}t\\ j\end{array}\right){▵}_{T^{\ast },T}^j\left(X\right)$$

for all integers $t\ge m$. Letting $t\to \infty$, one obtains

$$\begin{array}{c}\hfill 0\le {▵}_{T^{\ast },T}^{m-1}\left(X\right).\end{array}$$

(2)

Proposition 1.

If $T\in B\left(\mathcal{H}\right)$ is an invertible $(m,X)$-isometric operator for some positive operator $X\in B\left(\mathcal{H}\right)$, then $T\in (m-1,X)$-isometric.

Proof. 

T being invertible

$${▵}_{T^{\ast },T}^m\left(X\right)={(L_{T^{\ast }}{R}_T-I)}^m\left(X\right)={(-1)}^m{\left(L_{T^{\ast }}{R}_T\right)}^m{\left({\left(L_{T^{\ast }}{R}_T\right)}^{-1}-I\right)}^m\left(X\right)$$

and this since $T\in (m,X)$-isometric implies ${▵}_{{T^{\ast }}^{-1},T^{-1}}^m\left(X\right)=0$. Arguing as above, we have

$$0\le {▵}_{{T^{\ast }}^{-1},T^{-1}}^{m-1}\left(X\right)={(-1)}^{m-1}{\left(L_{T^{\ast }}{R}_T\right)}^{-m+1}{▵}_{T^{\ast },T}^{m-1}\left(X\right)⟹{▵}_{T^{\ast },T}^{m-1}\left(X\right)\le 0.$$

Combining with inequality (2), we obtain the required equality. □

Remark 1. (i) In the presence of the hyponormality hypothesis on T (or $T^{\ast }$), the hypothesis that T is 2 -symmetric is not necessary. Indeed, hyponormal n-symmetric operators T are self-adjoint. This is seen as follows. A straightforward argument shows ${\sigma }_a\left(T\right)\subset \mathbb{R}$; hence $\sigma \left(T\right)\subset \mathbb{R}$. Since hyponormal operators with spectrum in $\mathbb{R}$ are self-adjoint [27], T is self-adjoint.

(ii) It is known that hyponormal m-isometric operators are isometric [28]. The following argument shows that a cohyponormal m-isometric operator is unitary. If T is m-isometric, then ${\sigma }_a\left(T\right)$ is a subset of the boundary of the unit disc in $\mathbb{C}$. Hence T is a contraction and therefore isometric [28] (Proposition 2.6). The proof now follows, since a cohyponormal isometry is necessarily unitary.

4. Structure of $(\mathit{m},\mathit{n})$-Isosymmetric Operators

In this section, we consider the structure of power bounded $(m,n)$-isosymmetric operators. We start, however, by considering cohyponormal $(m,n)$-isosymmetric operators. It is seen that such operators T have a particulary simple structure: T is the direct sum of a unitary operator with a $C_{00}$-contraction satisfying $T\in (1,1)$-isosymmetric.

By the definition of the approximate point spectrum of an operator, if a $\lambda \in {\sigma }_a\left(T\right)$, then there exists a sequence of unit vectors $\left\{x_t\right\}\subseteq \mathcal{H}$ such that ${lim}_{t\to \infty }\parallel (T-\lambda )x_t\parallel =0$. Hence, if $T\in (m,n)$-isosymmetric and $\lambda \in {\sigma }_a\left(T\right)$, then

$$\begin{array}{ccc}\hfill 0& =& \underset{t\to \infty }{lim}\sum _{j=0}^n{(-1)}^j\left(\begin{array}{c}n\\ j\end{array}\right)\sum _{k=0}^m{(-1)}^k\left(\begin{array}{c}m\\ k\end{array}\right)⟨T^{m+j-k}{x}_t,T^{m+n-j-k}{x}_t⟩\hfill \\ & =& \sum _{j=0}^n{(-1)}^j\left(\begin{array}{c}n\\ j\end{array}\right){\overline{\lambda }}^{(n-j)}{\lambda }^j\sum _{k=0}^m{(-1)}^k\left(\begin{array}{c}m\\ k\end{array}\right){\left|\lambda \right|}^{2(m-k)}\hfill \\ & =& {(\overline{\lambda }-\lambda )}^n{(1-|\lambda |}^2{)}^m\hfill \\ \hfill ⟹& & {\sigma }_a\left(T\right)\subseteq \partial \mathbb{D}\cup \mathbb{R}and\sigma \left(T\right)\subseteq \overline{\mathbb{D}}\cup \mathbb{R}.\hfill \end{array}$$

Recall that an operator $T\in B\left(\mathcal{H}\right)$ is normaloid if $\parallel T\parallel$ equals the spectral radius $r\left(T\right)={lim}_{t\to \infty }{\parallel T^t\parallel }^{\frac{1}{t}}$ of T. Hyponormal operators are normaloid.

Theorem 4.

(a)If $T\in B\left(\mathcal{H}\right)$ is cohyponormal, then the following statements are mutually equivalent.

(i)

${▵}_{T^{\ast },T}^m\left({\delta }_{T^{\ast },T}^n\left(I\right)\right)=0$ for some positive integers $m,n$.

(ii)

${▵}_{T^{\ast },T}\left({\delta }_{T^{\ast },T}\left(I\right)\right)=0$.

(iii)

T is the direct sum of a unitary with a selfadjoint $C_{00}$-contraction.

(b) If $T\in B\left(\mathcal{H}\right)$ is an invertible operator and m is a positive even integer such that ${▵}_{T^{\ast },T}^m\left({\delta }_{T^{\ast },T}^n\left(I\right)\right)$$\ge 0$ and ${\delta }_{T^{\ast },T}^n\left(I\right)\ge 0$, or, ${▵}_{T^{\ast },T}^m\left({\delta }_{T^{\ast },T}^n\left(I\right)\right)\le 0$ and ${\delta }_{T^{\ast },T}^n\left(I\right)\le 0$, for some positive integer n, then ${▵}_{T^{\ast },T}^m\left({\delta }_{T^{\ast },T}^n\left(I\right)\right)=0$.

Proof. (a) $\left(iii\right)⟹\left(ii\right)⟹\left(i\right)$. If we let $T=T_u\oplus T_c\in B({\mathcal{H}}_1\oplus {\mathcal{H}}_2)$, $T_u$ unitary and $T_c$ a $C_{00}$-contraction such that $T_c^{\ast }=T_c$, then

$$\begin{array}{ccc}\hfill {▵}_{T^{\ast },T}\left({\delta }_{T^{\ast },T}\left(I\right)\right)& =& {▵}_{T^{\ast },T}\left((T_u^{\ast }-T_u)\oplus 0\right)\hfill \\ & =& \left(0\oplus {▵}_{T_c^{\ast },T_c}\right)\left((T_u^{\ast }-T_u)\oplus 0\right)\hfill \\ & =& 0\hfill \end{array}$$

and

$${▵}_{T^{\ast },T}^m\left({\delta }_{T^{\ast },T}^n\left(I\right)\right)={▵}_{T^{\ast },T}^{m-1}\left[{\delta }_{T^{\ast },T}^{n-1}\left({▵}_{T^{\ast },T}\left({\delta }_{T^{\ast },T}\left(I\right)\right)\right)\right]=0.$$

$\left(i\right)⟹\left(iii\right)$. In view of our observation on the spectrum of operators $T\in B\left(\mathcal{H}\right)$ satisfying the equality of $\left(i\right)$, the hypothesis $T^{\ast }$ is hyponormal implies $T^{\ast }$, hence T, is a contraction. Decompose T into its normal and pure (i.e., completely non-normal) parts by $T=T_1\oplus T_2\in B({\mathcal{H}}_1\oplus {\mathcal{H}}_2)$. Then $T_2$ is a cnu (= completely non-unitary) $C_{0.}$-contraction. The hypothesis

$${▵}_{T^{\ast },T}^m\left({\delta }_{T^{\ast },T}^n\left(I\right)\right)=0⟺{\oplus }_{i=1}^2{▵}_{T_i^{\ast },T_i}^m\left({\delta }_{T_i^{\ast },T_i}^n\left(I_i\right)\right)=0,$$

where $I_i$ is the identity of $B\left({\mathcal{H}}_i\right)$. Since

$${▵}_{T_i^{\ast },T_i}^m\left({\delta }_{T_i^{\ast },T_i}^n\left(I_i\right)\right)=0⟺{\delta }_{T_i^{\ast },T_i}^n\left({▵}_{T_i^{\ast },T_i}^m\left(I\right)\right)=0,$$

if we let ${▵}_{T_i^{\ast },T_i}^m\left(I_i\right)=X_i$ and apply Theorem 2 to ${\delta }_{T_i^{\ast },T_i}^n\left(X_i\right)=0$, then

$${\delta }_{T_i^{\ast },T_i}\left(X_i\right)={\delta }_{T_i^{\ast },T_i}\left({▵}_{T_i^{\ast },T_i}^m\left(I_i\right)\right)={▵}_{T_i^{\ast },T_i}^m\left({\delta }_{T_i^{\ast },T_i}\left(I_i\right)\right)=0.$$

Choose $i=2$. Set ${▵}_{T_2^{\ast },T_2}^{m-1}\left({\delta }_{T_2^{\ast },T_2}\left(I_2\right)\right)=Y_{m-1}$ and consider ${▵}_{T_2^{\ast },T_2}\left(Y_{m-1}\right)$. Since

$${▵}_{T_2^{\ast },T_2}\left(Y_{m-1}\right)=0⟹T_2^{\ast }{Y}_{m-1}{T}_2=Y_{m-1}⟹\cdots ⟹T_2^{\ast t}{Y}_{m-1}{T}_2^t=Y_{m-1}$$

for all positive integers t,

$$|⟨Y_{m-1}x,x⟩|=\left|⟨Y_{m-1}{T}_2^{t}x,T_2^{t}x⟩\right|\le \parallel Y_{m-1}\parallel {∥T_2^{t}x∥}^2$$

for all $x\in {\mathcal{H}}_2$. Since $T_2$ is a $C_{0.}$-contraction, letting $t\to \infty$, we have

$$|⟨Y_{m-1}x,x⟩|=0\mathrm{for}\mathrm{all}x\in {\mathcal{H}}_2.$$

Hence $Y_{m-1}={▵}_{T_2^{\ast },T_2}^{m-1}\left({\delta }_{T_{,}^{\ast }{T}_2}\left(I_2\right)\right)=0$. Repeating the argument, considering ${▵}_{T_2^{\ast },T_2}\left(Y_{m-2}\right)$ and ${▵}_{T_2^{\ast },T_2}\left(Y_{m-3}\right)$ etc., it follows that

$$Y_1={▵}_{T_2^{\ast },T_2}\left({\delta }_{T_2^{\ast },T_2}\left(I_2\right)\right)=0⟹Y_0={\delta }_{T_2^{\ast },T_2}\left(I_2\right)=0.$$

Thus, $T_2\in C_{00}$ is a selfadjoint contraction.

Considering next the case $i=1$, the normal contraction $T_1$ is the direct sum of a unitary and a cnu contraction. Let

$$T_1=T_{11}\oplus T_{12}\in B({\mathcal{H}}_{11}\oplus {\mathcal{H}}_{12}),T_{11}\mathrm{unitary}\mathrm{and}{T}_{12}\mathrm{cnu}.$$

Then

$${▵}_{T_1^{\ast },T_1}^n\left({\delta }_{T_1^{\ast },T_1}\left(I\right)\right)={\oplus }_{j=1}^2{▵}_{T_{1j}^{\ast },T_{1j}}^n\left({\delta }_{T_{1j}^{\ast },T_{1j}}\left(I_{1j}\right)\right)=0,$$

where $I_{1j}$ is the identity of $B\left({\mathcal{H}}_{1j}\right)$. Since $T_{11}$ is unitary,

$$\begin{array}{ccc}\hfill {▵}_{T_{11}^{\ast },T_{11}}^n\left({\delta }_{T_{11}^{\ast },T_{11}}\left(I_{11}\right)\right)=0& ⟺& {\delta }_{T_{11}^{\ast },T_{11}^{-1}}^n\left({\delta }_{T_{11}^{\ast },T_{11}}\left(I_{11}\right)\right)=0\hfill \\ & ⟺& {\delta }_{T_{11}^{\ast },T_{11}^{-1}}\left({\delta }_{T_{11}^{\ast },T_{11}}\left(I_{11}\right)\right)=0\hfill \\ & ⟺& {▵}_{T_{11}^{\ast },T_{11}}\left({\delta }_{T_{11}^{\ast },T_{11}}\left(I_{11}\right)\right)=0.\hfill \end{array}$$

The operator $T_{12}$ being a normal cnu-contraction is a $C_{00}$-contraction. Arguing as above, this implies

$${▵}_{T_{12}^{\ast },T_{12}}^n\left({\delta }_{T_{12}^{\ast },T_{12}}\left(I_{12}\right)\right)=0⟹{\delta }_{T_{12}^{\ast },T_{12}}\left(I_{12}\right)=0,$$

i.e., $T_{12}\in C_{00}$-contraction is selfadjoint. To complete the proof, define $T_u$ and $T_c$ by $T_u=T_{11}$ and $T_c=T_{12}\oplus T_2$.

(b) We prove that either of the hypotheses implies equality $\left(i\right)$ of part a. The proof in both the cases being almost the same, simply substitute $-X$ for X in the argument below, we consider the case ${▵}_{T^{\ast },T}^m\left({\delta }_{T^{\ast },T}^n\left(I\right)\right)\ge 0$ and ${\delta }_{T^{\ast },T}^n\left(I\right)\ge 0$. Let ${\delta }_{T^{\ast },T}^n\left(I\right)=X$; then

$$0\le {▵}_{T^{\ast },T}^m\left(X\right)⟺0\le {\left(L_{T^{\ast }}{R}_T\right)}^m\left(X\right)-\sum _{j=0}^{m-1}\left(\begin{array}{c}m\\ j\end{array}\right){▵}_{T^{\ast },T}^j\left(X\right)$$

implies

$$\begin{array}{ccc}\hfill 0& \le & {\left(L_{T^{\ast }}{R}_T\right)}^{m+1}\left(X\right)-\sum _{j=0}^{m-1}\left(\begin{array}{c}m\\ j\end{array}\right)L_{T^{\ast }}{R}_T{▵}_{T^{\ast },T}^j\left(X\right)\hfill \\ & =& {\left(L_{T^{\ast }}{R}_T\right)}^{m+1}\left(X\right)-\left(\begin{array}{c}m+1\\ m-1\end{array}\right){▵}_{T^{\ast },T}^{m-1}\left(X\right)-\sum _{j=0}^{m-2}\left(\begin{array}{c}m+1\\ j\end{array}\right){▵}_{T^{\ast },T}^j\left(X\right)\hfill \end{array}$$

and this (using an induction argument as in the proof of $\left(1\right)$) implies

$$\begin{array}{c}\hfill 0\le {\left(L_{T^{\ast }}{R}_T\right)}^t\left(X\right)-\left(\begin{array}{c}t\\ m-1\end{array}\right){▵}_{T^{\ast },T}^{m-1}\left(X\right)+\sum _{j=0}^{m-2}\left(\begin{array}{c}t\\ j\end{array}\right){▵}_{T^{\ast },T}^j\left(X\right)\end{array}$$

(3)

for all integers $t\ge m$. Thus

$$\begin{array}{ccc}\hfill ⟨{▵}_{T^{\ast },T}^{m-1}\left(X\right)x,x⟩& \le & \frac{1}{\left(\begin{array}{c}t\\ m-1\end{array}\right)}\left[⟨\left\{{\left(L_{T^{\ast }}{R}_T\right)}^t\left(X\right)+\sum _{j=0}^{m-2}\left(\begin{array}{c}t\\ j\end{array}\right){▵}_{T^{\ast },T}^j\left(X\right)\right\}x,x⟩\right]\hfill \\ & =& \frac{1}{\left(\begin{array}{c}t\\ m-1\end{array}\right)}\left[{∥X^{\frac{1}{2}}{T}^{t}x∥}^2+⟨\sum _{j=0}^{m-2}\left(\begin{array}{c}t\\ j\end{array}\right){▵}_{T^{\ast },T}^j\left(X\right)x,x⟩\right]\hfill \end{array}$$

for all $x\in \mathcal{H}$. Since $\left(\begin{array}{c}t\\ m-1\end{array}\right)$ is of the order of $t^{m-1}$ and $\left(\begin{array}{c}t\\ j\end{array}\right)$ is of the order of $t^{m-2}$ (for $0\le j\le m-2$) as $t\to \infty$, letting $t\to \infty$ we have

$$⟨{▵}_{T^{\ast },T}^{m-1}\left(X\right)x,x⟩\le 0\mathrm{for}\mathrm{all}x\in \mathcal{H}⟹{▵}_{T^{\ast },T}^{m-1}\left(X\right)\le 0.$$

The invertibility of T implies

$${▵}_{T^{\ast },T}^m\left(X\right)={(-1)}^m{\left(L_{T^{\ast }}{R}_T\right)}^m{▵}_{{T^{\ast }}^{-1},T^{-1}}^m\left(X\right)$$

and hence since m is even

$${\left(L_{T^{\ast }}{R}_T\right)}^m{▵}_{{T^{\ast }}^{-1},T^{-1}}^m\left(X\right)=0⟺{▵}_{{T^{\ast }}^{-1},T^{-1}}^m\left(X\right)=0.$$

Arguing as above, we conclude

$$\begin{array}{ccc}\hfill 0\le {▵}_{{T^{\ast }}^{-1},T^{-1}}^{m-1}\left(X\right)={\left(L_{T^{\ast }}^{-1}{R}_T^{-1}\right)}^{m-1}{(-1)}^{m-1}{▵}_{T^{\ast },T}^{m-1}\left(X\right)& ⟹& 0\le {(-1)}^{m-1}{▵}_{T^{\ast },T}^{m-1}\left(X\right)\hfill \\ & ⟺& 0\ge {▵}_{T^{\ast },T}^{m-1}\left(X\right).\hfill \end{array}$$

Hence

$${▵}_{T^{\ast },T}^{m-1}\left(X\right)=0⟹{▵}_{T^{\ast },T}^m\left(X\right)=0⟺{▵}_{T^{\ast },T}^m\left({\delta }_{T^{\ast },T}^n\left(I\right)\right)=0$$

and the proof is complete. □

The hypothesis $T^{\ast }$ is hyponormal is redundant in the case in which $n=2$ and ${\delta }_{T^{\ast },T}^2\left(I\right)\ge 0$. (For then ${\delta }_{T^{\ast },T}^2\left(I\right)\ge 0$ and $({\delta }_{T^{\ast },T}{\left(I\right)}^{\ast }\left({\delta }_{T^{\ast },T}\left(I\right)\right)\ge 0$ imply $T{T}^{\ast }\ge T^{\ast }T$.) Furthermore, if also $m=2$, then the hypothesis T is invertible may be dispensed with in Theorem 4(b).

Theorem 5.

If ${\delta }_{T^{\ast },T}^2\left(I\right)$ and ${▵}_{T^{\ast },T}^2\left({\delta }_{T^{\ast },T}^2\left(I\right)\right)$ are both greater than or equal to 0, then ${▵}_{T^{\ast },T}\left({\delta }_{T^{\ast },T}\left(I\right)\right)=0$ and T is the direct sum of a unitary with a $C_{00}$-contraction.

Proof. 

The cohyponormality of T implies T is a contraction, hence has a direct sum decomposition

$$T=T_u\oplus T_c\in B({\mathcal{H}}_1\oplus {\mathcal{H}}_2),T_u=T{{|}_{{\mathcal{H}}_1}\mathrm{unitary}\mathrm{and}{T}_c=T|}_{{\mathcal{H}}_2}\mathrm{a}\mathrm{cnu}{C}_{0.}-\mathrm{contraction}.$$

If we let

$$X={\delta }_{T^{\ast },T}^2\left(I\right)={\delta }_{T_u^{\ast },T_u}^2\left(I_1\right)\oplus {\delta }_{T_c^{\ast },T_c}^2\left(I_2\right)=X_1\oplus X_2\in B({\mathcal{H}}_1\oplus {\mathcal{H}}_2),I_i{=I|}_{{\mathcal{H}}_i},i=1,2,$$

then $X_i\ge 0$ for $i=1,2$ and

$${▵}_{T^{\ast },T}^2\left(X\right)={▵}_{T_u^{\ast },T_u}^2\left(X_1\right)\oplus {▵}_{T_c^{\ast },T_c}^2\left(X_2\right)\ge 0⟺{▵}_{T_u^{\ast },T_u}^2\left(X_1\right)\ge 0,{▵}_{T_c^{\ast },T_c}^2\left(X_2\right)\ge 0.$$

The operator $T_u$ being unitary, Theorem 4(b) implies

$${▵}_{T_u^{\ast },T_u}^2\left(X_1\right)\ge 0⟺{▵}_{T_u^{\ast },T_u}\left({\delta }_{T_u^{\ast },T_u}\left(I_1\right)\right)=0.$$

Consider now the operator ${▵}_{T_c^{\ast },T_c}^2\left(X_2\right)={▵}_{T_c^{\ast },T_c}\left(X_{21}\right)\ge 0$; $X_{21}={▵}_{T_c^{\ast },T_c}\left(X_2\right)$. We have

$${▵}_{T_c^{\ast },T_c}\left(X_{21}\right)\ge 0⟹T_c^{\ast }{X}_{21}{T}_c\ge X_{21}⟹T_c^{\ast 2}{X}_{21}{T}_c^2\ge X_{21}⟹\cdots ⟹T_c^{\ast t}{X}_{21}{T}_c^t\ge X_{21}$$

for all positive integers t. Hence

$$⟨X_{21}x,x⟩\le ⟨T_c^{\ast t}{X}_{21}{T}_c^{t}x,x⟩\le ∥X_{21}∥{∥T_c^{t}x∥}^2$$

for all $x\in {\mathcal{H}}_2$. Letting $t\to \infty$, this implies

$$⟨X_{21}x,x⟩\le \underset{t\to \infty }{lim}∥X_{21}∥{∥T_c^{t}x∥}^2=0$$

for all $x\in {\mathcal{H}}_2$. Hence

$$X_{21}={▵}_{T_c^{\ast },T_c}\left({\delta }_{T_c^{\ast },T_c}^2\left(I_2\right)\right)={\delta }_{T_c^{\ast },T_c}^2\left({▵}_{T_c^{\ast },T_c}\left(I_2\right)\right)=0.$$

The operator $T_C^{\ast }$ being hyponormal, it follows from an application of Theorem 2 that

$${\delta }_{T_c^{\ast },T_c}\left({▵}_{T_c^{\ast },T_c}\left(I_2\right)\right)={▵}_{T_c^{\ast },T_c}\left({\delta }_{T_c^{\ast },T_c}\left(I_2\right)\right)=0.$$

This completes the proof. □

A result similar to that of Theorem 4 does not hold for hyponormal T. For example, if T is the forward unilateral shift $T(x_1,x_2,x_3,\cdots )=(0,x_1,x_2,\cdots )$, then ${▵}_{T^{\ast },T}^m\left({\delta }_{T^{\ast },T}^n\left(I\right)\right)=0$ for all positive integers $m,n$. However, hyponormal T is neither unitary nor self-adjoint nor a direct sum of the two. If T is hyponormal and satisfies ${▵}_{T^{\ast },T}^m\left({\delta }_{T^{\ast },T}^n\left(I\right)\right)=0$, then T is a contraction, hence power bounded. For power bounded operators $T\in B\left(\mathcal{H}\right)$ satisfying ${▵}_{T^{\ast },T}^m\left({\delta }_{T^{\ast },T}^n\left(I\right)\right)=0$, Theorem 4 has the following analogue.

Theorem 6.

If a power bounded operator $T\in B\left(\mathcal{H}\right)$ satisfies ${▵}_{T^{\ast },T}^m\left({\delta }_{T^{\ast },T}^n\left(I\right)\right)=0$ for some positive integers m and n, then:

(i)

${▵}_{T^{\ast },T}\left({\delta }_{T^{\ast },T}^n\left(I\right)\right)=0$;

(ii)

there exist decompositions $\mathcal{H}={\mathcal{H}}_1\oplus {\mathcal{H}}_2={\mathcal{H}}_1\oplus ({\mathcal{H}}_{21}\oplus {\mathcal{H}}_{22})$, a Hilbert space $\mathcal{K}={\mathcal{H}}_{22}\oplus (\mathcal{K}\ominus {\mathcal{H}}_{22})$ and operators $T_1\in B\left({\mathcal{H}}_1\right)$, $T_2\in B\left({\mathcal{H}}_2\right)$, $T_3\in B({\mathcal{H}}_2,{\mathcal{H}}_1)$, $V_u\in B\left({\mathcal{H}}_{21}\right)$, $V_c\in B\left(H_{22}\right)$, $V_b=\left(\begin{array}{cc}{V}_c& Z\hfill \\ 0& Y\hfill \end{array}\right)\in B\left(\mathcal{K}\right)$ (for some operators $Z\in B(\mathcal{K}\ominus {\mathcal{H}}_{22},{\mathcal{H}}_{22})$, $Y\in B(\mathcal{K}\ominus {\mathcal{H}}_{22})$) such that $T=\left(\begin{array}{cc}{T}_1& T_3\hfill \\ 0& T_2\hfill \end{array}\right)\in B({\mathcal{H}}_1\oplus {\mathcal{H}}_2)$, $T_1\in C_{0.}$ satisfies ${\delta }_{T_1^{\ast },T_1}^n\left(I{|}_{{\mathcal{H}}_1}\right)=0$, $V_u$ is unitary, $V_c$ is a unilateral shift, $V_b$ is a bilateral shift, the positive operator ${lim}_{t\to \infty }{T}_2^{\ast t}{T}_2^t=A$ is injective and $A{T}_2=(V_u\oplus V_b){|}_{{\mathcal{H}}_2}A$.

We remark here that either of the components in Theorem 6, as also in Theorem 4, may be missing.

Proof. 

If we set ${\delta }_{T^{\ast },T}^n\left(I\right)=X$, then ${▵}_{T^{\ast },T}^m\left(X\right)=0$ and

$${\left(L_{T^{\ast }}{R}_T\right)}^t\left(X\right)=\left(\begin{array}{c}t\\ m-1\end{array}\right){▵}_{T^{\ast },T}^{m-1}\left(X\right)+\sum _{j=0}^{m-2}\left(\begin{array}{c}t\\ j\end{array}\right){▵}_{T^{\ast },T}^j\left(X\right)$$

for all integers $t\ge m$ (see the proof of Theorem 4(b) above). The operator T being power bounded, there exists a real number $M>0$ such that $\left|\right|T^t\left|\right|\le M$ for all integers $t>0$. We have

$$\begin{array}{c}\hfill ∥{▵}_{T^{\ast },T}^{m-1}\left(X\right)∥\le \underset{t\to \infty }{lim}\frac{1}{\left(\begin{array}{c}t\\ m-1\end{array}\right)}\left[∥{\left(L_{T^{\ast }}{R}_T\right)}^t\left(X\right)∥+∥\sum _{j=0}^{m-2}\left(\begin{array}{c}t\\ j\end{array}\right){▵}_{T^{\ast },T}^j\left(X\right)∥\right]=0.\end{array}$$

Hence

$${▵}_{T^{\ast },T}^{m-1}\left(X\right)=0.$$

Repeating the argument a finite number of time, we conclude

$${▵}_{T^{\ast },T}\left(X\right)=0\left(⟺{▵}_{T^{\ast },T}\left({\delta }_{T^{\ast },T}^n\left(I\right)\right)={\delta }_{T^{\ast },T}^n\left({▵}_{T^{\ast },T}\left(I\right)\right)=0\right).$$

Recall [21], that the power bounded operator T has an upper triangular matrix representation

$$\begin{array}{c}\hfill T=\left(\begin{array}{cc}{T}_1& T_3\hfill \\ 0& T_2\hfill \end{array}\right)\in B({\mathcal{H}}_1\oplus {\mathcal{H}}_2),\end{array}$$

(4)

where $T_1\in C_{0.}$ and $T_2\in C_{1.}$. Evidently,

$${▵}_{T^{\ast },T}\left(X\right)=0⟹{▵}_{T_1^{\ast },T_1}\left({\delta }_{T_1^{\ast },T_1}^n\left(I_1\right)\right)=0,I_1{=I|}_{{\mathcal{H}}_1}.$$

Set ${\delta }_{T_1^{\ast },T_1}^n\left(I_1\right)=X_1$. Then

$$\begin{array}{c}\hfill {▵}_{T_1^{\ast },T_1}\left(X_1\right)=0⟺T_1^{\ast }{X}_1{T}_1=X_1⟹T_1^{\ast 2}{X}_1{T}_1^2=X_1⟹\cdots ⟹T_1^{\ast t}{X}_1{T}_1^t=X_1\end{array}$$

for all integers $t\ge 0$. Since $T_1\in C_{0.}$, for every $x\in {\mathcal{H}}_1$,

$$\begin{array}{c}\hfill \left|⟨X_{1}x,x⟩\right|=\underset{t\to \infty }{lim}\left|⟨T_1^{\ast t}{X}_1{T}_1^{t}x,x⟩\right|\le \underset{t\to \infty }{lim}\parallel X_1\parallel {∥T_1^{t}x∥}^2=0.\end{array}$$

Hence

$$X_1={\delta }_{T_1^{\ast },T_1}^n\left(I_1\right)=0.$$

Consider now the power bounded operator $T_2$. Since $T_2\in C_{1.}$, $T_2$ is injective and

$$\underset{t\to \infty }{lim}{T}_2^{\ast t}{T}_2^t=A$$

exists and is a positive injective operator which satisfies

$$T_2^{\ast }A{T}_2=A$$

Ref. [29] (Theorem 5.1). An application of Theorem 1 implies the existence of an isometry $V\in B\left({\mathcal{H}}_2\right)$ satisfying

$$A^{\frac{1}{2}}{T}_2=V{A}^{\frac{1}{2}}.$$

Since every isometry is part of a unitary, there exists a decomposition ${\mathcal{H}}_2={\mathcal{H}}_{21}\oplus {\mathcal{H}}_{22}$, a Hilbert space $\mathcal{K}={\mathcal{H}}_{22}\oplus (\mathcal{K}\ominus {\mathcal{H}}_{22})$ and a unitary

$$W=\left(\begin{array}{ccc}Vu& 0\hfill & 0\\ 0& V_c\hfill & Y\\ 0& 0\hfill & Z\end{array}\right)\in B({\mathcal{H}}_{21}\oplus {\mathcal{H}}_{22}\oplus (\mathcal{K}\ominus {\mathcal{H}}_{22})),$$

$Y\in B(\mathcal{K}\ominus {\mathcal{H}}_{22},{\mathcal{H}}_{22})$ and $Z\in B(\mathcal{K}\ominus {\mathcal{H}}_{22})$ some operators, such that $V_u$ is unitary, $V_c$ is a unilateral shift, $\left(\begin{array}{cc}{V}_c& Y\hfill \\ 0& Z\hfill \end{array}\right)$ is a bilateral shift and ${V=W|}_{{\mathcal{H}}_2}$ [22] (Lemma 5.7, Page 82). Evidently $A^{\frac{1}{2}}{T}_2{=W|}_{{\mathcal{H}}_2}{A}^{\frac{1}{2}}$. □

If $n=2$ in the preceding theorem, then ${\delta }_{T_1^{\ast },T_1}\left(I_1\right)=0$ and the operator $T_1$ is a selfadjoint $C_{00}$-operator. Furthermore, if the normal parts of the operator T reduce T, then $T_3=0$.

An operator $S\in B\left(\mathcal{H}\right)$ is paranormal if ${\parallel Sx\parallel }^2\le \parallel S^{2}x\parallel$ for all unit vectors $x\in \mathcal{H}$. Hyponormal operators are paranormal, paranormal operators are normaloid, the restriction of a paranormal operator to an invaraint subspace is again paranormal [25] and ${\delta }_{S,V^{\ast }}^{-1}\left(0\right)\subseteq {\delta }_{S^{\ast },V}^{-1}\left(0\right)$ for paranormal S and isometric $V\in B\left(\mathcal{H}\right)$ [30] (p. 316). Hence if the operator $T^{\ast }$ of Theorem 4 is paranormal, then ${\delta }_{V,T_2}\left(A^{\frac{1}{2}}\right)=0$ implies ${\delta }_{V^{\ast },T_2^{\ast }}\left(A^{\frac{1}{2}}\right)=0$. Consequently, $T_2$ is unitary and (since $T^{\ast }$ is necessarily a contraction and the unitary parts of a contraction reduce the contraction) $T_3=0$ in representation $\left(4\right)$ of T. Thus, $T=T_1\oplus T_2$, ${\delta }_{T_1^{\ast },T_1}^n\left(I_1\right)=0$ and $T_2$ is unitary. If we now assume $n=2$ in Theorem 4, then we have the following generalisation of a result of Stankus [19] (Proposition 5.22).

Corollary 4.

If ${▵}_{T^{\ast },T}^m\left({\delta }_{T^{\ast },T}^2\left(I\right)\right)=0$ for some paranormal operator $T^{\ast }\in B\left(\mathcal{H}\right)$ and integer $m\ge 1$, then T is the direct sum of a selfadjoint operator with a unitary.

Proof. 

As seen above $T=T_1\oplus T_2$, where $T_2$ is unitary and ${\delta }_{T_1^{\ast },T_1}^2\left(I_1\right)=0$. Since ${\delta }_{T_1^{\ast },T_1}^2\left(I_1\right)=0$ if and only if ${\delta }_{T_1^{\ast },T_1}\left(I_1\right)=0$, the proof follows. □