On m-Isometric and m-Symmetric Operators of Elementary Operators

Abstract

Given Hilbert space operators $A,B$, and X, let ${▵}_{A,B}$ and ${\delta }_{A,B}$ denote, respectively, the elementary operators ${▵}_{A,B}\left(X\right)=I-AXB$ and the generalised derivation ${\delta }_{A,B}\left(X\right)=AX-XB$. This paper considers the structure of operators $D_{d_1,d_2}^m\left(I\right)=0$ and $D_{d_1,d_2}^m$ compact, where m is a positive integer, $D=▵$ or $\delta$, $d_1={▵}_{A^{*},B^{*}}$ or ${\delta }_{A^{*},B^{*}}$ and $d_2={▵}_{A,B}$ or ${\delta }_{A,B}$. This is a continuation of the work performed by C. Gu for the case where ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, and the author with I.H. Kim for the cases where ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$ or ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m$ is compact, and ${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$ or ${\delta }_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m$ is compact. Operators $D_{d_1,d_2}^m\left(I\right)=0$ are examples of operators with a finite spectrum; indeed, the operators $A,B$ have at most a two-point spectrum, and if $D_{d_1,d_2}^m$ is compact, then (the non-nilpotent operators) $A,B$ are algebraic. $D_{d_1,d_2}^m\left(I\right)=0$ implies $D_{d_1,d_2}^n\left(I\right)=0$ for integers $n\ge m$: the reverse implication, however, fails. It is proved that $D_{d_1,d_2}^m\left(I\right)=0$ implies $D_{d_1,d_2}\left(I\right)=0$ if and only if of A and B (are normal and hence) satisfy a Putnam–Fuglede commutativity property.

1. Introduction

Given a complex infinite dimensional Hilbert space $(\mathcal{H};\langle .,.\rangle )$, let $B\left(\mathcal{H}\right)$ denote the algebra of operators, equivalently bounded linear transformations, on $\mathcal{H}$ into itself. For an operator $A\in B\left(\mathcal{H}\right)$, let $L_A$ and $R_A\in B\left(B\left(\mathcal{H}\right)\right)$ denote, respectively, the operators

$$L_A\left(X\right)=AX\mathrm{and}{R}_A\left(X\right)=XA$$

of left multiplication by A and right multiplication by A. For $A,B\in B\left(\mathcal{H}\right)$, the elementary operator ${▵}_{A,B}$ of length two and the generalised derivation ${\delta }_{A,B}\in B\left(B\left(\mathcal{H}\right)\right)$ are defined by

$${▵}_{A,B}\left(X\right)=(I-L_A{R}_B)\left(X\right)=X-AXB\mathrm{and}{\delta }_{A,B}\left(X\right)=(L_A-R_B)\left(X\right)=AX-XB,$$

respectively. We say that the pair $(A,B)$ of operators in $B\left(\mathcal{H}\right)\times B\left(\mathcal{H}\right)$ is m-isometric (respectively, m-symmetric) for some positive integer m if

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

$$\begin{array}{ccc}\hfill (\mathrm{resp}.,& & {\delta }_{A,B}^m\left(I\right)={(L_A-R_B)}^m\left(I\right)=\left(\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)L_A^{m-j}{R}_B^j\right)\left(I\right)\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)A^{m-j}{B}^j=0).\hfill \end{array}$$

(In the case in which the pair $(A,B)$ is the pair $(T^{*},T)$ for an operator $T\in B\left(\mathcal{H}\right)$, we shorten $(T^{*},T)$ is m-isometric (respectively, m-symmetric) to T is m-isometric (respectively, m-symmetric)). Structural properties of m-isometric and m-symmetric pairs $(A,B)$, and their connections with classical function theory, non-stationary stochastic processes, Toeplitz operators, and nilpotent perturbations of Hermitian operators (etc.), have been studied by a number of authors in the recent past: see [1,2,3,4,5,6,7,8,9,10,11] for further references.

A characterisation of m-isometric operators with a finite spectrum was carried out in [7,12,13]. An example of such an operator, first considered by Botelho and Jamison [14,15] for the cases $m=2$ and $m=3$, is the m-isometric operator pair $(L_{A^{*}}{R}_{B^{*}},L_A{R}_B)$,

$${▵}_{L_{A^{*}}{R}_{B^{*}},L_A{R}_B}^m\left(I\right)=\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)A^{*j}{A}^j{B}^j{B}^{*j}=0;A,B\in B\left(\mathcal{H}\right).$$

It is straightforward to see that if $\overline{\lambda }$ is in the approximate point spectrum ${\sigma }_a\left(B^{*}\right)$ of $B^{*}$, then ${▵}_{\overline{\lambda }{A}^{*},\lambda A}^m\left(I\right)=0$, $\sigma \left(\lambda A\right)$ is a subset of the boundary $\partial \mathtt{D}$ of the unit disc $\mathtt{D}\subset \mathtt{C}$ and $(1-\left(\overline{\lambda }{\sigma }_a\left(A^{*}\right)\right)\left(\lambda {\sigma }_a\left(A\right)\right)=0$. There exists a non-trivial scalar $\beta$, $\left|\beta \right|=1$ such that $\lambda \alpha =\beta$ and $\overline{\lambda \alpha }={\displaystyle \frac{1}{\beta }}=\overline{\beta }$ for all $\alpha \in {\sigma }_a\left(A\right)$. We assert that ${\sigma }_a\left(B^{*}\right)$ consists, at most, of two points. For if there exist non-trivial $\overline{\mu },\overline{\nu }\in {\sigma }_a\left(B^{*}\right)$, $\mu \ne \lambda \ne \nu$, then ${\sigma }_a\left(\mu A\right)={\sigma }_a\left(\lambda A\right)+{\sigma }_a\left((\mu -\lambda )A\right)$ and ${\sigma }_a\left(\nu A\right)={\sigma }_a\left(\lambda A\right)+{\sigma }_a\left((\nu -\lambda )A\right)$: since $0\notin {\sigma }_a\left(A\right)$, not both of these translates of ${\sigma }_a\left(\lambda A\right)$ are in $\partial \mathtt{D}$. This argument applies equally to ${\sigma }_a\left(A\right)$; hence $\sigma \left(A\right)$ and $\sigma \left(B\right)$ consist at most of two points. A version of the preceding argument applies to m-isometric operators ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$. Indeed Gu, [16] proves that if ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$ for some operators $A,B\in B\left(\mathcal{H}\right)$, then $A,B$ have a spectrum consisting at most of two points. An extension of this result to operators ${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$ was recently considered by Duggal and Kim [17].

For operators $A,B\in B\left(\mathcal{H}\right)$, let $D_{d_1,d_2}^m\left(I\right)$ denote the operator defined by the choices

$$D=▵\mathrm{or}\delta ,d_1={▵}_{A^{*},B^{*}}\mathrm{or}{\delta }_{A^{*},B^{*}}\mathrm{and}{d}_2={▵}_{A,B}\mathrm{or}{\delta }_{A,B}.$$

This paper considers the structure of the resulting eight operators, two of which have already been discussed in [16,17,18], to prove that the spectra $\sigma \left(A\right)$ and $\sigma \left(B\right)$ consist at best of two points. It is fairly easily seen that $D_{d_1,d_2}^m\left(I\right)=0$ implies $D_{d_1,d_2}^n\left(I\right)=0$ for all integers $n\ge m$. The reverse problem of whether $D_{d_1,d_2}^m\left(I\right)=0$ implies $D_{d_1,d_2}^n\left(I\right)=0$ for some positive integer $n<m$ does not, in general, have a positive answer. We prove that a necessary and sufficient condition for $D_{d_1,d_2}^m\left(I\right)=0$ to imply $D_{d_1,d_2}\left(I\right)=0$ is that the pairs of operators $({\alpha }_{1}A+{\beta }_1,{\alpha }_{2}B+{\beta }_2)$, ${\alpha }_i$ and ${\beta }_i$ ($i=1,2$) scalars, satisfy a Putnam–Fuglede commutativity property [19,20,21]. The paper considers also the case in which the operator $D_{d_1,d_2}^m$ is compact. Here it is seen that the operators A and B (are algebraic operators, and as such) have a finite spectrum consisting of poles (of the resolvent) of the operators.

2. Complementary Results: The Adjoint Operator ${\left(D_{d_1,d_2}^m\left(I\right)\right)}^{*}$

In the following, $A,B$ will denote Hilbert space operators such that $\sigma \left(A\right)\ne \left\{0\right\}\ne \sigma \left(B\right)$, and m will denote a positive integer. Given complex scalars ${\alpha }_i$ and ${\beta }_i$, $1\le i\le 2$, we say that the pair of operators $({\alpha }_{1}A+{\beta }_1,{\alpha }_{2}B+{\beta }_2)$ satisfies the Putnam–Fuglede commutativity property, shortened to $({\alpha }_{1}A+{\beta }_1,{\alpha }_{2}B+{\beta }_2)\in \left(PF\right)$, if

$$D_{{\alpha }_{1}A+{\beta }_1,{\alpha }_{2}B+{\beta }_2}^{-1}\left(0\right)\subseteq D_{\overline{{\alpha }_1}{A}^{*}+\overline{{\beta }_1},\overline{{\alpha }_2}{B}^{*}+\overline{{\beta }_2}}^{-1}\left(0\right);D=▵\mathrm{or}\delta .$$

It is well known, see [19,22], that if $A,B$ are normal operators, then the pair $({\alpha }_{1}A+{\beta }_1,{\alpha }_{2}B+{\beta }_2)\in \left(PF\right)$.

The ascent (respectively, descent) of A, denoted as $\mathrm{asc}\left(A\right)$ (respectively, $\mathrm{dsc}\left(A\right)$), is the least positive integer n such that

$$A^{-(n+1)}\left(0\right)\subseteq A^{-n}\left(0\right)(\mathrm{resp}.A^n\left(\mathcal{H}\right)\subseteq A^{n+1}\left(\mathcal{H}\right)).$$

Finite ascent and finite descent imply their equality [23,24]; we say that a point $\lambda$ of the spectrum $\sigma \left(A\right)$ of A is a pole (of the resolvent) of A if $\mathrm{asc}(A-\lambda )=\mathrm{dsc}(A-\lambda )$ [23,24]. (Here, and in the sequel, we write $A-\lambda$ for $A-\lambda I$.) We observe here that if $\lambda \in \sigma \left(A\right)$ is a pole of A, then necessarily $\lambda$ is an isolated point of $\sigma \left(A\right)$ (i.e., $\lambda \in \mathrm{iso}\sigma \left(A\right)$). We say that $D_{A,B}^{-1}\left(0\right)\perp D_{A,B}\left(\mathcal{H}\right)$, i.e., the kernel of the operator $D_{A,B}$ is orthogonal to the range of $D_{A,B}$ in the sense of James [25,26], if

$$\left|\right|X\left|\right|\le \left|\right|X+Y\left|\right|,\mathrm{all}X\in D_{A,B}^{-1}\left(0\right)\mathrm{and}Y\in D_{A,B}\left(B\left(\mathcal{H}\right)\right).$$

The following lemma, linking “range-kernel orthogonality” to the “ascent” of operators satisfying the $\left(PF\right)$-property, will be useful in our considerations below.

Lemma 1

([27], Proposition 2.26). If $D_{A,B}\in \left(PF\right)$, then ($D_{A,B}^{-1}\left(0\right)\perp D_{A,B}\left(\mathcal{H}\right)$, hence) $\mathrm{asc}\left(D_{A,B}\right)\le 1$.

The quasi-nilpotent part $H_0(A-\lambda )$ of an operator $A\in B\left(\mathcal{H}\right)$ at $\lambda \in \mathtt{C}$ is the set ([23], p. 119)

$$H_0(A-\lambda )=\{x\in \mathcal{H}:\underset{n\to \infty }{lim}{\left|\right|(A-\lambda )}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}\left|\right|=0\}$$

A has SVEP, the single-valued extension property, at ${\lambda }_0$ if for every open disc ${\mathtt{D}}_{{\lambda }_0}$ centred at ${\lambda }_0$ the only analytic function $f:{\mathtt{D}}_{{\lambda }_0}\to \mathcal{H}$ satisfying $(A-\lambda )f\left(\lambda \right)=0$ for all $\lambda \in {\mathtt{D}}_{{\lambda }_0}$ is the function $f\equiv 0$ [23,28]; A has SVEP if it has SVEP at all $\lambda$. Operators A with a countable spectrum have SVEP, and if $\sigma \left(A\right)=\{{\alpha }_1,\dots ,{\alpha }_p\}$ for some natural number p, then $A={\oplus }_{i=0}^n{A|}_{H_0(A-{\alpha }_i)}$ ([23], p. 303, Lemma 4.17). A necessary and sufficient condition for the points ${\alpha }_i$ to be poles of (the resolvent of) A of order $t_i$, for some positive integers $t_i$, $1\le i\le p$, is that $H_0(A-{\alpha }_i)={(A-{\alpha }_i)}^{-t_i}\left(0\right)$ [23]. Observe that if $\sigma \left(A\right)=\{{\alpha }_1,\dots ,{\alpha }_p\}$ and the points ${\alpha }_i$ are poles of A of order $t_i$, then $A={\oplus }_{i=1}^p{A|}_{{(A-{\alpha }_i)}^{-t_i}\left(0\right)}+N$ for some nilpotent operator N.

The operator A is algebraic if there exists a non-trivial polynomial $p(.)$ such that $p\left(A\right)=0$. It is well known, see for example ([23], Theorem 3.83), that A is algebraic if and only if $\sigma \left(A\right)$ is a finite set consisting of the poles of A (i.e., if and only if $\sigma \left(A\right)$ is finite and A is polaroid [23]). Recall from Boasso [29] that A is polaroid, i.e., points $\lambda \in \mathrm{iso}\sigma \left(A\right)$ are poles of A, if and only if $L_A$ and $R_A$ are polaroid. The (Banach space) operator ${\delta }_{A,B}$ is algebraic if and only if $A,B$ are algebraic. $L_A{R}_B$ algebraic does not imply A and B algebraic, only that at least one of A and B is algebraic ([27], Proposition 2.6).

A is nilpotent, more precisely, A is n-nilpotent for some positive integer n, if $A^n=0$ (and $A^{n-1}\ne 0$). Either of A and B nilpotent implies $L_A{R}_B$ nilpotent; conversely, $L_A{R}_B$ nilpotent implies, at best, that at least one of A and B is nilpotent [30]. We consider next the adjoint of the operator $D_{d_1,d_2}^m\left(I\right)$.

Start by observing that for all $X\in B\left(\mathcal{H}\right)$,

$${\left({\delta }_{A,B}\left(X\right)\right)}^{*}={\left((L_A-R_B)\left(X\right)\right)}^{*}=(R_{A^{*}}-L_{B^{*}})\left(X^{*}\right)=(-1){\delta }_{B^{*},A^{*}}\left(X^{*}\right)$$

and

$${\left({▵}_{A,B}\left(X\right)\right)}^{*}={\left((I-L_A{R}_B)\left(X\right)\right)}^{*}=(I-L_{B^{*}}{R}_{A^{*}})\left(X^{*}\right)={▵}_{B^{*},A^{*}}\left(X^{*}\right).$$

Hence,

$$\begin{array}{ccc}& & {\left({\delta }_{A,B}^n\left(X\right)\right)}^{*}={(-1)}^n{\delta }_{B^{*},A^{*}}^n\left(X^{*}\right);\hfill \\ & & {\left({▵}_{A,B}^n\left(X\right)\right)}^{*}={▵}_{B^{*},A^{*}}^n\left(X^{*}\right);\hfill \\ & & {\left({▵}_{A^{*},B^{*}}\left({▵}_{A,B}\left(X\right)\right)\right)}^{*}={▵}_{B,A}\left({\left({▵}_{A,B}\left(X\right)\right)}^{*}\right)={▵}_{B,A}\left({▵}_{B^{*},A^{*}}\left(X^{*}\right)\right);\hfill \\ & & {\left({\delta }_{A^{*},B^{*}}\left({\delta }_{A,B}\left(X\right)\right)\right)}^{*}=(-1){\delta }_{B,A}{\left({\delta }_{A,B}\left(X\right)\right)}^{*}{)=(-1)}^2{\delta }_{B,A}\left({\delta }_{B^{*},A^{*}}\left(X^{*}\right)\right);\hfill \\ & & {\left({▵}_{A^{*},B^{*}}\left({\delta }_{A,B}\right)\left(X\right))\right)}^{*}={▵}_{B,A}\left({\left({\delta }_{A,B}\left(X\right))\right)}^{*}\right)=(-1){▵}_{B,A}\left({\delta }_{B^{*},A^{*}}\left(X^{*}\right)\right)\mathrm{and}\hfill \\ & & {\left({\delta }_{A^{*},B^{*}}\left({▵}_{A,B}\left(X\right)\right)\right)}^{*}=(-1){\delta }_{B,A}\left({\left({▵}_{A,B}\left(X\right)\right)}^{*}\right)=(-1){\delta }_{B,A}\left({▵}_{B^{*},A^{*}}\left(X^{*}\right)\right).\hfill \end{array}$$

Letting

$${\nabla }_{E,F}\left(X\right)=(I+L_E{R}_F)\left(X\right)\mathrm{and}{◊}_{E,F}\left(X\right)=(L_E+R_F)\left(X\right)$$

for Banach space operators $E,F$ and X, we have

$$\begin{array}{ccc}\hfill & & {\left({▵}_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)\right)}^{*}\hfill \\ \hfill & =& {\left(\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){\delta }_{A^{*},B^{*}}^j\left({▵}_{A,B}^j\left(I\right)\right)\right)}^{*}\hfill \\ \hfill & =& \sum _{j=0}^m{(-1)}^{2j}\left(\begin{array}{c}m\\ j\end{array}\right){\delta }_{B,A}^j{\left({▵}_{A,B}^j\left(I\right)\right)}^{*}\hfill \\ \hfill & =& \sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right){\delta }_{B,A}^j\left({▵}_{B^{*},A^{*}}^j\left(I\right)\right)\hfill \\ \hfill & =& {\nabla }_{{\delta }_{B,A},{▵}_{B^{*},A^{*}}}^m\left(I\right);\hfill \end{array}$$

(1)

$$\begin{array}{ccc}\hfill & & {\left({\delta }_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)\right)}^{*}\hfill \\ \hfill & =& {\left(\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){\delta }_{A^{*},B^{*}}^{m-j}\left({▵}_{A,B}^j\left(I\right)\right)\right)}^{*}\hfill \\ \hfill & =& \sum _{j=0}^m{(-1)}^m\left(\begin{array}{c}m\\ j\end{array}\right){\delta }_{B,A}^{m-j}\left({▵}_{B^{*},A^{*}}^j\left(I\right)\right)\hfill \\ \hfill & =& {(-1)}^m{◊}_{{\delta }_{B,A},{▵}_{B^{*},A^{*}}}^m\left(I\right);\hfill \end{array}$$

(2)

$${\left({▵}_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)\right)}^{*}={\nabla }_{{▵}_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right);$$

(3)

$${\left({\delta }_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)\right)}^{*}={◊}_{{▵}_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right);$$

(4)

$${\left({▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)\right)}^{*}={▵}_{{▵}_{B,A},{▵}_{B^{*},A^{*}}}^m\left(I\right);$$

(5)

$${\left({\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)\right)}^{*}={(-1)}^m{\delta }_{{\delta }_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right)$$

(6)

$${\left({\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)\right)}^{*}={\delta }_{{▵}_{B,A},{▵}_{B^{*},A^{*}}}^m\left(I\right);$$

(7)

$${\left({▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)\right)}^{*}={▵}_{{\delta }_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right).$$

(8)

3. The Spectrum

The approximate point spectrum of an operator being a non-empty set, if the pair $(A,B)$ satisfies the operator equation $D_{d_1,d_2}^m\left(I\right)=0$, $D=▵$ or $\delta$, $d_1={▵}_{A^{*},B^{*}}$ or ${\delta }_{A^{*},B^{*}}$ and $d_2={▵}_{A,B}$ or ${\delta }_{A,B}$, then there exist non-trivial scalars $\alpha \in {\sigma }_a\left(A\right)$ and $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$. (We assume in the following that $\sigma \left(A\right)\ne \left\{0\right\}\ne \sigma \left(B^{*}\right)$; see [31] for the spectra of left and right multiplication operators.) By definition

$$\begin{array}{ccc}& & {▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){\left(L_{{\delta }_{A^{*},B^{*}}}{R}_{{\delta }_{A,B}}\right)}^j\left(I\right)\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left[\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}j\\ p\end{array}\right)L_{A^{*}}^{j-p}{R}_{B^{*}}^j\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)L_A^{j-k}{R}_B^k\right)\right]\left(I\right)\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left[\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}j\\ p\end{array}\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)\right)\right]A^{*(j-p)}{A}^{j-k}{B}^k{B}^{*p}\hfill \\ & =& C_{\sum }{\mathcal{E}}_{A^{*(j-p)}{A}^{j-k},B^k{B}^{*p}},\hfill \end{array}$$

where $C_{\sum }=C(m,j,p,k)$ denotes

$$C_{\sum }=\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left[\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}j\\ p\end{array}\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)\right)\right]$$

and ${\mathcal{E}}_{E,F}$ is the operator

$${\mathcal{E}}_{E,F}\left(X\right)=EXF.$$

A similar argument shows that

$${▵}_{{\delta }_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right)=C_{\sum }{\mathcal{E}}_{B^{j-p}{B}^{*(j-k)},A^{*k}{A}^p}.$$

If $\alpha \in {\sigma }_a\left(A\right)$ and $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$, and $\left\{x_n\right\},\left\{y_n\right\}\subseteq \mathcal{H}$ are sequences of unit vectors such that ${lim}_{n\to \infty }(B^{*}-\overline{\lambda })y_n=0={lim}_{n\to \infty }(A-\alpha )x_n$, then for all $x\in \mathcal{H}$

$$\begin{array}{ccc}& & \left[{\mathcal{E}}_{A^{*(j-p)}{A}^{j-k},B^k{B}^{*p}}(x\otimes y_n)\right]y_n\hfill \\ & =& A^{*(j-p)}{A}^{j-k}x\langle y_n,B^p{B}^{*k}{y}_n\rangle \hfill \end{array}$$

and

$$\begin{array}{ccc}& & \left[{\mathcal{E}}_{B^{j-p}{B}^{*(j-k)},A^{*k}{A}^p}(x\otimes x_n)\right]x_n\hfill \\ & =& B^{j-p}{B}^{*(j-k)}x\langle x_n,A^{*p}{A}^k{x}_n\rangle .\hfill \end{array}$$

(Here, and in the sequel, $\langle .,.\rangle$ denotes the inner product on $\mathcal{H}$). Thus, if ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, then (for all $x\in \mathcal{H}$)

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}{C}_{\sum }{A}^{*(j-p)}{A}^{(j-k)}x\langle y_n,B^p{B}^{*k}{y}_n\rangle \hfill \\ & =& C_{\sum }\left({\overline{\lambda }}^p{A}^{*(j-p)}\right)\left({\lambda }^k{A}^{j-k}\right)x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left[\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}j\\ p\end{array}\right){\overline{\lambda }}^p{A}^{*(j-p)}\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right){\lambda }^k{A}^{j-k}\right)\right]x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(A^{*}-\overline{\lambda })}^j{(A-\lambda )}^{j}x,\hfill \end{array}$$

i.e.,

$${▵}_{A^{*}-\overline{\lambda },A-\lambda }^m\left(I\right)=0.$$

(9a)

A similar argument shows that if ${▵}_{{\delta }_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right)=0$, then

$${▵}_{B-\alpha ,B^{*}-\overline{\alpha }}^m\left(I\right)=0.$$

(9b)

If ${▵}_{A^{*}-\overline{\lambda },A-\lambda }^m\left(I\right)=0={▵}_{B-\alpha ,B^{*}-\overline{\alpha }}^m\left(I\right)$, then ${\sigma }_a(A-\lambda )$ and ${\sigma }_a(B^{*}-\overline{\alpha })$ are subsets of the boundary $\partial \mathtt{D}$ of the unit disc in the complex plane $\mathtt{C}$. Considering ${\sigma }_a(A-\lambda )$, if $\overline{\mu }\in {\sigma }_a(B^{*}-\overline{\lambda })$; then ${\sigma }_a(A-\lambda -\mu )\in \partial \mathtt{D}$ and, for a $\beta \in {\sigma }_a(A-\lambda )$, $1=|{\sigma }_a(A-\lambda -\mu )|=|{\sigma }_a(A-\lambda )-\mu |=|\beta -\mu |$. Let $\mu =e^{it}\left|\mu \right|$, then $\left|\mu \right|\le 2$ and

$$\begin{array}{ccc}& & |\beta -\mu |=1⟹{\beta }^2-e^{it}\left|\mu \right|\beta +e^{2it}=0\hfill \\ & ⟺& \beta =e^{it}\left({\displaystyle \frac{\left|\mu \right|\pm i\sqrt{4-{\left|\mu \right|}^2}}{2}}\right).\hfill \end{array}$$

A similar argument shows that if $\nu =e^{i\theta }\left|\nu \right|\in {\sigma }_a(A-\alpha )$, then $\left|\nu \right|\le 2$ and, for every $\overline{\tau }\in {\sigma }_a(B^{*}-\overline{\alpha })$,

$$\begin{array}{ccc}& & |\tau -\nu |=1⟹{\tau }^2-e^{i\theta }\left|\nu \right|\tau +e^{2i\theta }=0\hfill \\ & ⟺& \tau =e^{i\theta }\left({\displaystyle \frac{\left|\nu \right|\pm i\sqrt{4-{\left|\nu \right|}^2}}{2}}\right).\hfill \end{array}$$

Spectra $\sigma \left(A\right)$ and $\sigma \left(B\right)$ consist at most of two points: for if $\overline{\mu },\overline{\eta }\in {\sigma }_a(B^{*}-\overline{\lambda })$, then ${\sigma }_a(A-\lambda -\eta )={\sigma }_a(A-\lambda -\mu )+(\mu -\eta )$, and both ${\sigma }_a(A-\lambda -\eta )$ and its non-trivial translates can not be in $\partial \mathtt{D}$. The following proposition encompasses this known result see ([16,17,18]).

Proposition 1.

Given $\alpha \in {\sigma }_a\left(A\right)$ and $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$, if

(i) 

${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, then

$${▵}_{A^{*}-\overline{\lambda },A-\lambda }^m\left(I\right)=0={▵}_{B-\alpha ,B^{*}-\overline{\alpha }}^m\left(I\right),$$

$$\sigma (A-\lambda )={\sigma }_a(A-\lambda )=\{e^{it}\left({\displaystyle \frac{\left|\mu \right|\pm i\sqrt{4-{\left|\mu \right|}^2}}{2}}\right):\overline{\mu }=e^{-it}\left|\mu \right|\in {\sigma }_a(B^{*}-\overline{\lambda }),0\le t<2\pi ,\left|\mu \right|\le 2\}$$

and

$$\sigma (B-\alpha )={\sigma }_a(B-\alpha )=\{e^{it}\left({\displaystyle \frac{\left|\mu \right|\pm i\sqrt{4-{\left|\mu \right|}^2}}{2}}\right):\mu =e^{it}\left|\mu \right|\in {\sigma }_a(A-\alpha ),0\le t<2\pi ,\left|\mu \right|\le 2\}.$$

(ii) 

${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$, then

(a) 

${\delta }_{I-\overline{\lambda }{A}^{*},I-\lambda A}^m\left(I\right)=0={\delta }_{I-\alpha B,I-\overline{\alpha }{B}^{*}}^m\left(I\right)$,

(b) 

There exist real scalars ${\beta }_1$ and ${\beta }_2$, ${\beta }_1\le {\beta }_2$, such that

$$\left\{\lambda \sigma \left(A\right)\right\}\cup \left\{\alpha \sigma \left(B\right)\right\}\subseteq [1-{\beta }_2,1-{\beta }_1],$$

(c) 

$\sigma \left(A\right)$ and $\sigma \left(B\right)$ consist at most of two points.

Proof. 

Part (i) having already been proved, we start proving part (ii) by proving ${\delta }_{I-\overline{\lambda }{A}^{*},I-\lambda A}^m\left(I\right)=0={\delta }_{I-\alpha B,I-\overline{\alpha }{B}^{*}}^m\left(I\right)$. By definition

$$\begin{array}{ccc}& & {\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=C_{{\sum }_1}{\mathcal{E}}_{A^{*p}{A}^k,B^k{B}^{*p}}\mathrm{and}\hfill \\ & & {\delta }_{{▵}_{B,A},{▵}_{B^{*},A^{*}}}^m\left(I\right)=C_{{\sum }_1}{\mathcal{E}}_{B^p{B}^{*k},A^{*k}{A}^p},\hfill \end{array}$$

where

$$C_{{\sum }_1}=C(m,j,p,k)=\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left[\sum _{p=0}^{m-j}{(-1)}^p\left(\begin{array}{c}m-j\\ p\end{array}\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)\right)\right].$$

Arguing as above, it is seen that

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}{\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)\hfill \\ & =& \underset{n\to \infty }{lim}{C}_{{\sum }_1}{A}^{*p}{A}^{k}x\langle y_n,B^k{B}^{*p}{y}_n\rangle \hfill \\ & =& C_{{\sum }_1}\left({\overline{\lambda }}^p{A}^{*p}\right)\left({\lambda }^k{A}^k\right)x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left[\sum _{p=0}^{m-j}{(-1)}^p\left(\begin{array}{c}m-j\\ p\end{array}\right){\overline{\lambda }}^p{A}^{*p}\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right){\lambda }^k{A}^k\right)\right]x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(I-\overline{\lambda }{A}^{*})}^{m-j}{(I-\lambda A)}^{j}x,\hfill \end{array}$$

for all $x\in \mathcal{H}$ and

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}{\delta }_{{▵}_{B,A},{▵}_{B^{*},A^{*}}}^m\left(I\right)\hfill \\ & =& \underset{n\to \infty }{lim}{C}_{{\sum }_1}{B}^p{B}^{*k}x\langle x_n,A^{*p}{A}^k{x}_n\rangle \hfill \\ & =& C_{{\sum }_1}\left({\alpha }^p{B}^p\right)\left({\overline{\alpha }}^k{B}^{*k}\right)x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(I-\alpha B)}^{m-j}{(I-\overline{\alpha }{A}^{*})}^{j}x\hfill \end{array}$$

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

$${\delta }_{I-\overline{\lambda }{A}^{*},I-\lambda A}^m\left(I\right)=0$$

(10a)

and

$${\delta }_{I-\alpha B,I-\overline{\alpha }{B}^{*}}^m\left(I\right)=0$$

(10b)

This proves $\left(a\right)$. To prove $\left(b\right)$, we start by observing from the spectral mapping theorem that

$$\begin{array}{ccc}& & {\sigma }_a(I-\overline{\lambda }{A}^{*})-{\sigma }_a(I-\lambda A)=(I-\overline{\lambda }{\sigma }_a\left(A^{*}\right))-(1-\lambda {\sigma }_a\left(A\right))\hfill \\ & =0& \\ & =& {\sigma }_a(I-\alpha B)-{\sigma }_a(I-\overline{\alpha }{B}^{*})=(1-\alpha {\sigma }_a\left(B\right))-(1-\overline{\alpha }{\sigma }_a\left(B^{*}\right)).\hfill \end{array}$$

Hence there exist real numbers ${\beta }_i$, $1\le i\le 2$ such that

$$1-\lambda \sigma \left(A\right)\subseteq [{\beta }_1,{\beta }_2],\mathrm{equivalently}\lambda \sigma \left(A\right)\subseteq [1-{\beta }_2,1-{\beta }_1].$$

(This inclusion being true for all $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$, we have also that $\alpha \sigma \left(B\right)\subseteq [1-{\beta }_2,1-{\beta }_1]$). To prove $\left(c\right)$, assume that there exist distinct scalars $\mu ,\nu$ such that $\overline{\mu },\overline{\nu }\in {\sigma }_a(B^{*}-\overline{\lambda })$. Then $\overline{\lambda +\mu }$ and $\overline{\lambda +\nu }\in {\sigma }_a\left(B^{*}\right)$, and hence

$$\left\{(\lambda +\mu )\sigma \left(A\right)\right\}\cup \left\{(\lambda +\nu )\sigma \left(A\right)\right\}\subseteq [1-{\beta }_2,1-{\beta }_1].$$

Recalling the hypothesis that ${\sigma }_a\left(A\right)\ne \left\{0\right\}$, we have

$$\begin{array}{ccc}& & {\sigma }_a(I-(\lambda +\nu )A)={\sigma }_a(I-(\lambda +\mu )A)+(\mu -\nu ){\sigma }_a\left(A\right)\hfill \\ & ⟹& (\lambda +\nu ){\sigma }_a\left(A\right)\subseteq [1-{\beta }_2,1-{\beta }_1]+(\mu -\nu ){\sigma }_a\left(A\right)⊈[1-{\beta }_2,1-{\beta }_1],\hfill \end{array}$$

since $1-{\beta }_1+(\mu -\nu ){\sigma }_a\left(A\right)>1-{\beta }_1$ if $(\mu -\nu ){\sigma }_a\left(A\right)>0$ and $1-{\beta }_2+(\mu -\nu ){\sigma }_a\left(A\right)<1-{\beta }_2$ if $(\mu -\nu ){\sigma }_a\left(A\right)<0$. A similar argument proves that ${\sigma }_a\left(A\right)$ consists at most of two points. □

The point 0 cannot be in both ${\sigma }_a\left(B^{*}\right)$ and ${\sigma }_a\left(A\right)$ in the case in which ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$. This follows from the following argument. If $0\in {\sigma }_a\left(B^{*}\right)$, then (see above) for all $x\in \mathcal{H}$,

$$\begin{array}{ccc}& & {▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0⟹C_{\sum }{A}^{*(j-p)}{A}^{j-k}x\underset{n\to \infty }{lim}\langle B^{*p}{y}_n,A^{*k}{y}_n\rangle \hfill \\ & =& 0\mathrm{for}\mathrm{all}p,k>0\mathrm{and}\mathrm{if}p=k=0,\mathrm{then}\hfill \\ & & \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)A^{*j}{A}^{j}x={▵}_{A^{*},A}^m\left(I\right)x=0\hfill \\ & ⟹& {▵}_{A^{*},A}^m\left(I\right)=0.\hfill \end{array}$$

Thus, A is m-isometric, and hence ${\sigma }_a\left(A\right)\subseteq \partial \mathtt{D}$. A similar argument shows that if $0\in {\sigma }_a\left(A\right)$, then $B^{*}$ is m-isometric and ${\sigma }_a\left(B^{*}\right)\subseteq \partial \mathtt{D}$. This fails for operators $A,B$ satisfying ${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$. Consider, for example, the operators $A,B$ such that $\sigma \left(A\right)=\{a{e}^{i\theta },0\}$, $\sigma \left(B\right)=\{b{e}^{-i\theta },0\}$ for some real numbers $a,b$ and $0\le \theta <2\pi$, $A=a{e}^{i\theta }{I}_1\oplus 0I_2$ and $B=b{e}^{-i\theta }{I}_1\oplus 0I_2$; $I=I_1\oplus I_2$. Then ${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=C_{{\sum }_1}(I_1\oplus 0I_2)=0$.

The following theorem proves that the conclusion $\sigma \left(A\right)$ and $\sigma \left(B\right)$ consists at most of two point holds for the remaining six choices of the operators $D,d_1$ and $d_2$ (in $D_{d_1,d_2}^m\left(I\right)=0$). For continuity, we keep the numbering above; thus, our theorem starts with case $\left(iii\right)$ and ends with case $\left(viii\right)$. We assume in the proof of the theorem that $\left\{x_n\right\}$ and $\left\{y_n\right\}$ are sequences of unit vectors in $\mathcal{H}$ such that ${lim}_{n\to \infty }(A-\alpha )x_n=0={lim}_{n\to \infty }(B^{*}-\overline{\lambda })y_n=0$. (Thus, $\alpha \in {\sigma }_a\left(A\right)$ and $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$.)

Theorem 1.

(iii) 

If ${▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$, then ${▵}_{I-\overline{\lambda }{A}^{*},I-\lambda A}^m=0={▵}_{I-\alpha B,I-\overline{\alpha }{B}^{*}}^m\left(I\right)$ and for each $\alpha \in \sigma \left(A\right)$ and $\lambda \in \sigma \left(B\right)$ there exists a θ, $0\le \theta <2\pi$, such that $\left|\alpha \lambda \right|=2(1-cos\theta )$.

(iv) 

If ${▵}_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, then ${▵}_{I-\overline{\lambda }{A}^{*},A-\lambda }^m=0={\nabla }_{I-\alpha B,B^{*}-\overline{\alpha }}^m\left(I\right)$ and there exists a non-zero scalar β such that

$$\lambda ={\displaystyle \frac{-{\left|\beta \right|}^2\pm i\sqrt{4\overline{\beta }(1-\overline{\beta })-{\left|\beta \right|}^4}}{2\overline{\beta }}}\mathrm{and}\alpha ={\displaystyle \frac{{\left|\beta \right|}^2\pm i\sqrt{4\overline{\beta }(1-\overline{\beta })-{\left|\beta \right|}^4}}{2\overline{\beta }}}.$$

(v) 

If ${▵}_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$, then ${▵}_{A^{*}-\overline{\lambda },I-\lambda A}^m=0={\nabla }_{B-\alpha ,I-\overline{\alpha }{B}^{*}}^m\left(I\right)$ and there exists a non-zero scalar β such that

$$\lambda ={\displaystyle \frac{-{\left|\beta \right|}^2\pm i\sqrt{4\beta (1-\beta )-{\left|\beta \right|}^4}}{2\beta }}\mathrm{and}\alpha ={\displaystyle \frac{{\left|\beta \right|}^2\pm i\sqrt{4\beta (1-\beta )-{\left|\beta \right|}^4}}{2\beta }}.$$

(vi) 

If ${\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, then

$${\delta }_{A^{*}-\overline{\lambda },A-\lambda }^m\left(I\right)=0={\delta }_{B-\alpha ,B^{*}-\overline{\alpha }}^m\left(I\right),$$

$0\in {\sigma }_a\left(A\right)$ (respectively, $0\in {\sigma }_a\left(B^{*}\right)$) implies $B^{*}$ (respectively, A) m-symmetric, and ${\sigma }_a\left(A\right)\subseteq \{r+\lambda :\overline{\lambda }\in {\sigma }_a\left(B^{*}\right),r\mathrm{real}\}$.

(vii) 

If ${\delta }_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$ and $A-I,B+I$ are non-nilpotent, then

$${\delta }_{I-\overline{\lambda }{A}^{*},A-\lambda }^m\left(I\right)=0={◊}_{I-\alpha B,B^{*}-\overline{\alpha }}^m\left(I\right)$$

and there exists a scalar β such that

$$\lambda ={\displaystyle \frac{-\overline{\beta }\pm i\sqrt{4(\beta -1)-{\overline{\beta }}^2}}{2}}\mathrm{and}\alpha ={\displaystyle \frac{\overline{\beta }\pm i{\sqrt{4(\beta -1)-\overline{\beta }}}^2}{2}}.$$

(viii) 

If ${\delta }_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$ and $A-I,B+I$ are non-nilpotent, then

$${\delta }_{I-\overline{\lambda }{A}^{*},A-\lambda }^m\left(I\right)=0={◊}_{I-\alpha B,B^{*}-\overline{\alpha }}^m\left(I\right)$$

and there exists a scalar β such that

$$\lambda ={\displaystyle \frac{-\beta \pm i\sqrt{4(\overline{\beta }-1)-{\beta }^2}}{2}}\mathrm{and}\alpha ={\displaystyle \frac{\beta \pm i\sqrt{4(\overline{\beta }-1)-{\beta }^2}}{2}}.$$

Furthermore, the spectra $\sigma \left(A\right)$ and $\sigma \left(B\right)$ consist at most of two points in each of the (six) cases.

Proof. 

The proof of $\left(iv\right)$ and $\left(v\right)$, and $\left(vii\right)$ and $\left(viii\right)$, is similar: we prove $\left(iii\right),\left(iv\right),\left(vi\right)$ and $\left(vii\right)$.

Case (iii). We start by proving that $I-\lambda A$ and $I-\overline{\alpha }{B}^{*}$ are m-isometric. By definition

$$\begin{array}{ccc}& & {▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left[\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}j\\ p\end{array}\right)A^{*p}\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)A^k\right)B^k{B}^{*p}\right].\hfill \end{array}$$

Hence, for all $x\in \mathcal{H}$,

$$\begin{array}{ccc}\hfill 0& =& \left[\underset{n\to \infty }{lim}{▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)(x\otimes y_n)\right]y_n\hfill \\ & =& C_{\sum }{A}^{*p}{A}^{k}x\underset{n\to \infty }{lim}\langle B^{*p}{y}_n,B^{*k}{y}_n\rangle \hfill \\ & =& C_{\sum }{\left(\overline{\lambda }{A}^{*}\right)}^p{\left(\lambda A\right)}^{k}x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(I-\overline{\lambda }{A}^{*})}^j{(I-\lambda A)}^{j}x\hfill \\ & =& {▵}_{I-\overline{\lambda }{A}^{*},I-\lambda A}^m\left(I\right)x.\hfill \end{array}$$

Consequently, ${▵}_{I-\overline{\lambda }{A}^{*},I-\lambda A}^m\left(I\right)=0$. A similar argument, working this time with

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}\left[{▵}_{{▵}_{B,A},{▵}_{B^{*},A^{*}}}^m\left(I\right)(x\otimes x_n)\right]x_n\hfill \\ & =& C_{\sum }{B}^p{B}^{*k}x\underset{n\to \infty }{lim}\langle A^p{x}_n,A^k{x}_n\rangle ,\hfill \end{array}$$

proves ${▵}_{I-\alpha B,I-\overline{\alpha }{B}^{*}}^m\left(I\right)=0$.

We claim that ${\sigma }_a\left(A\right)$ and ${\sigma }_a\left(B^{*}\right)$, and hence also $\sigma \left(A\right)$ and $\sigma \left(B\right)$, consist at most of two points. Assume to the contrary that (along with $\overline{\mu }$) $\overline{\nu }\in {\sigma }_a(B^{*}-\overline{\lambda })$; $\nu \ne \mu$. Then ${\sigma }_a(I-(\lambda +\nu )A)\subseteq \partial \mathtt{D}$ and

$${\sigma }_a(I-(\lambda +\nu )A)={\sigma }_a(I-(\lambda +\mu )A)+(\mu -\nu ){\sigma }_a\left(A\right).$$

Since ${\sigma }_a\left(A\right)\ne \left\{0\right\}$ and ${\sigma }_a(I-(\lambda +\mu )A)\subseteq \partial \mathtt{D}$, ${\sigma }_a((I-(\lambda +\nu )A)$ is a translate by a non-zero scalar of points in $\partial \mathtt{D}$, it is not a point in the boundary of $\mathtt{D}$. This being a contradiction, our claim is proved.

The operators $I-\lambda A$ and $I-\overline{\alpha }{B}^{*}$ being m-isometric, the spectra ${\sigma }_a(I-\lambda A)$, ${\sigma }_a(I-\overline{\alpha }{B}^{*})$ are subsets of $\partial \mathtt{D}$ and the operators $I-\lambda A$, $I-\overline{\alpha }{B}^{*}$ are (left invertible, hence, since their spectra lie in the boundary of the unit disc,) invertible.The spectral mapping theorem implies that

$$\sigma \left({▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}\right)=0⟹\alpha \lambda +\overline{\alpha \lambda }={\left|\alpha \lambda \right|}^2$$

for $\alpha \in \sigma \left(A\right)$ and $\lambda \in \sigma \left(B\right)$. Since $\sigma (I-\lambda A)\subseteq \left\{e^{i\theta }\right\}$, equivalently, $\lambda \alpha \subseteq \{1-e^{i\theta }\}$, for each $\alpha \in \sigma \left(A\right)$ and $\lambda \in \sigma \left(B\right)$ there exists a $0\le \theta <2\pi$ such that $\left|\alpha \lambda \right|=2(1-cos\theta )$.

Case (iv). We start by proving ${▵}_{I-\overline{\lambda }{A}^{*},A-\lambda }^m=0={\nabla }_{I-\alpha B,B^{*}-\overline{\alpha }}^m\left(I\right)$ for all $\alpha \in {\sigma }_a\left(A\right)$ and $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$. Arguing as above,

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}\left[{▵}_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)(x\otimes y_n\right]y_n,x\in \mathcal{H}\hfill \\ & =& C_{\sum }{A}^{*p}{A}^{j-k}x\underset{n\to \infty }{lim}\langle B^{*p}{y}_n,B^{*k}{y}_n\rangle \hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left(\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}j\\ p\end{array}\right){\left(\overline{\lambda }{A}^{*}\right)}^p\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right){\lambda }^k{A}^{j-k}\right)x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(I-\overline{\lambda }{A}^{*})}^j{(A-\lambda )}^{j}x\hfill \\ & ⟹& {▵}_{I-\overline{\lambda }{A}^{*},A-\lambda }^m\left(I\right)=0\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}\left[{\nabla }_{{▵}_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right)(x\otimes x_n)\right]x_n,x\in \mathcal{H}\hfill \\ & =& \sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right)\left(\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}j\\ p\end{array}\right)B^p\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)B^{*(j-k)}\right)x\underset{n\to \infty }{lim}\langle A^p{x}_n,A^k{x}_n\rangle \right)\hfill \\ & =& \sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right)\left(\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}j\\ p\end{array}\right){\left(\alpha B\right)}^p\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right){\overline{\alpha }}^k{B}^{*(j-k)}\right)x\hfill \\ & ⟹& {\nabla }_{I-\alpha B,B^{*}-\overline{\alpha }}^m\left(I\right)=0.\hfill \end{array}$$

If ${▵}_{I-\overline{\lambda }{A}^{*},A-\lambda }^m\left(I\right)=0$, then (by the spectral mapping theorem)

$$1-(1-\overline{\lambda }\overline{\alpha })(\alpha -\lambda )=0;\mathrm{all}\alpha \in {\sigma }_a\left(A\right)\mathrm{and}\overline{\lambda }\in {\sigma }_a\left(B^{*}\right).$$

Consequently, there exists a non-zero scalar $\beta$ such that

$$1-\overline{\lambda }\overline{\alpha }={\displaystyle \frac{1}{\overline{\beta }}}, \alpha -\lambda =\beta ⟹\beta {\lambda }^2+{\left|\beta \right|}^2\lambda +(1-\beta )=0$$

and this (solving for $\lambda$ and $\alpha$) implies

$$\lambda ={\displaystyle \frac{-{\left|\beta \right|}^2\pm i\sqrt{4\overline{\beta }(1-\overline{\beta })-{\left|\beta \right|}^4}}{2\beta }}\mathrm{and}\alpha ={\displaystyle \frac{{\left|\beta \right|}^2\pm i\sqrt{4\overline{\beta }(1-\overline{\beta })-{\left|\beta \right|}^4}}{2\beta }},$$

$\sigma \left(A\right)={\sigma }_a\left(A\right)$ and $\sigma \left(B\right)={\sigma }_a\left(B\right)$ consist at most of two points. (In particular, if $0\in \sigma \left(B\right)$, then $\beta =1$, $\sigma \left(A\right)=\{0,1\}$ and $\sigma \left(B\right)=\{-1,0\}$.)

Case (vi). We have

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}\left[{\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)(x\otimes y_n)\right]y_n,x\in \mathcal{H}\hfill \\ & =& C_{{\sum }_1}{A}^{*(m-j-p)}{A}^{j-k}x\underset{n\to \infty }{lim}\langle B^{*p}{y}_n,B^{*k}{y}_n\rangle \hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left(\sum _{p=0}^{m-j}{(-1)}^p\left(\begin{array}{c}m-j\\ p\end{array}\right){\overline{\lambda }}^p{A}^{*(m-j-p)}\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right){\lambda }^k{A}^{j-k}\right)x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(A^{*}-\overline{\lambda })}^{m-j}{(A-\lambda )}^{j}x\hfill \\ & ⟹& {\delta }_{A^{*}-\overline{\lambda },A-\lambda }^m\left(I\right)=0.\hfill \end{array}$$

Again,

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}\left[{\delta }_{{\delta }_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right)(x\otimes x_n)\right]x_n\hfill \\ & =& C_{{\sum }_1}{B}^{m-j-p}{B}^{*(j-k)}x\underset{n\to \infty }{lim}\langle A^p{x}_n,A^k{x}_n\rangle \hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left(\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}m-j\\ p\end{array}\right){\alpha }^p{B}^{m-j-p}\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)({\overline{\alpha }}^k{B}^{*(j-k)}\right)x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(B-\alpha )}^{m-j}{(B^{*}-\overline{\alpha })}^{j}x\hfill \\ & ⟹& {\delta }_{B-\alpha ,B^{*}-\overline{\alpha }}^m\left(I\right)=0.\hfill \end{array}$$

Thus

$${\delta }_{A^{*}-\overline{\lambda },A-\lambda }^m\left(I\right)=0={\delta }_{B-\alpha ,B^{*}-\overline{\alpha }}^m\left(I\right)$$

for all $\alpha \in {\sigma }_a\left(A\right)$ and $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$.

Evidently, $A-\lambda$ and $B^{*}-\overline{\alpha }$ are m-symmetric. Furthermore, since ${\delta }_{A^{*}-\overline{\lambda },A-\lambda }^m\left(I\right)=0$ implies $\overline{(\alpha -\lambda )}-(\alpha -\lambda )=0$, $\Im \alpha =\Im \lambda$ for all $\alpha$ and $\lambda$. Hence

$${\sigma }_a\left(A\right)\subseteq \{r+\lambda :\overline{\lambda }\in {\sigma }_a\left(B^{*}\right),r\mathrm{real}\}.$$

Trivially, if $0\in {\sigma }_a\left(A\right)$, then $B^{*}$ is m-symmetric, and if $0\in {\sigma }_a\left(B^{*}\right)$, then A is m-symmetric. The set $\left\{\sigma \right(A)-\lambda \}$ being a compact set, there exist scalars ${\beta }_i$, $1\le i\le 2$, such that (${\beta }_1\le {\beta }_2$ and) $\sigma \left(A\right)-\lambda \subseteq [{\beta }_1,{\beta }_2]$. Suppose that there exist scalars $\overline{\mu },\overline{\nu }\in {\sigma }_a(B^{*}-\overline{\lambda })$, $\mu \ne \nu$. Then ($\overline{\lambda +\mu }$ and $\overline{\lambda +\nu }\in {\sigma }_a\left(B^{*}\right)$, hence) ${\sigma }_a(A-\lambda -\mu )=\sigma (A-\lambda -\mu )$ and ${\sigma }_a((A-\lambda -\nu )=\sigma (A-\lambda -\nu )$ are subsets of $[{\beta }_1,{\beta }_2]$. Since

$$\begin{array}{ccc}& & \sigma (A-\lambda -\nu )=\sigma (A-\lambda -\mu )+(\mu -\nu ),\mathrm{where}0\ne \mu -\nu \mathrm{is}\mathrm{real},\hfill \\ & & {\beta }_1-(\mu -\nu )<{\beta }_1\mathrm{if}(\mu -\nu )>0\mathrm{and}{\beta }_2-(\mu -\nu )>{\beta }_2\mathrm{if}(\mu -\nu )>0,\hfill \end{array}$$

we have a contradiction. Conclusion: $\sigma \left(A\right)$ and $\sigma \left(B\right)$ consist, at most, of two points.

Case (vii). In this case

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}\left[{\delta }_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)(x\otimes y_n)\right]y_n,x\in \mathcal{H}\hfill \\ & =& C_{{\sum }_1}{A}^{*p}{A}^{j-k}x\underset{n\to \infty }{lim}\langle B^{*p}{y}_n,B^{*k}{y}_n\rangle \hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left(\sum _{p=0}^{m-j}{(-1)}^p\left(\begin{array}{c}m-j\\ p\end{array}\right){\left(\overline{\lambda }\right)}^p{A}^{*p}\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right){\lambda }^k{A}^{j-k}\right)x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(I-\overline{\lambda }{A}^{*})}^j{(A-\lambda )}^{j}x\hfill \\ & ⟹& {\delta }_{I-\overline{\lambda }{A}^{*},A-\lambda }^m\left(I\right)=0.\hfill \end{array}$$

Let

$$C_{{\sum }_0}=C(m,j,p,k)=\sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right)\left[\sum _{p=0}^{m-j}{(-1)}^p\left(\begin{array}{c}m-j\\ p\end{array}\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)\right)\right].$$

Then, for all $x\in \mathcal{H}$,

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}\left[{◊}_{{\delta }_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right)(x\otimes x_n)\right]x_n\hfill \\ & =& C_{{\sum }_0}{B}^p{B}^{*(j-k)}x\underset{n\to \infty }{lim}\langle A^p{x}_n,A^k{x}_n\rangle \hfill \\ & =& \sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right)\left(\sum _{p=0}^{m-j}{(-1)}^p\left(\begin{array}{c}m-j\\ p\end{array}\right){\alpha }^p{B}^p\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)({\overline{\alpha }}^k{B}^{*(j-k)}\right)x\hfill \\ & =& \sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right){(I-\overline{\lambda }{B}^{*})}^{m-j}{(B-\lambda )}^{j}x\hfill \\ & ⟹& {◊}_{I-\lambda B,B^{*}-\overline{\lambda }}^m\left(I\right)=0.\hfill \end{array}$$

This proves ${\delta }_{I-\overline{\lambda }{A}^{*},A-\lambda }^m\left(I\right)=0={◊}_{I-\alpha B,B^{*}-\overline{\alpha }}^m\left(I\right)$.

The conclusion ${\delta }_{I-\overline{\lambda }{A}^{*},A-\lambda }^m\left(I\right)=0$ implies (by the spectral mapping theorem) that $(1-\overline{\alpha }\overline{\lambda })-(\alpha -\lambda )=0$ for all non-zero $\alpha \in {\sigma }_a\left(A\right)$ and $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$. Hence there exists a non-zero scalar $\beta$ such that

$$\begin{array}{ccc}& & \alpha -\lambda =\beta =1-\overline{\alpha }\overline{\lambda }⟹{\lambda }^2+\beta \lambda +(\overline{\beta }-1)=0\hfill \\ & ⟹& \lambda ={\displaystyle \frac{-\beta \pm i\sqrt{4(\overline{\beta }-1)-{\beta }^2}}{2}}\mathrm{and}\hfill \\ & & \alpha ={\displaystyle \frac{\beta \pm i\sqrt{4(\overline{\beta }-1)-{\beta }^2}}{2}}.\hfill \end{array}$$

Evidently, ${\sigma }_a\left(A\right)=\sigma \left(A\right)$ and ${\sigma }_a\left(B\right)=\sigma \left(B\right)$ consist, at most, of two points (which, if $0\in \sigma \left(B\right)$ so that $\beta =1$, are respectively the sets $\{0,1\}$ and $\{-1,0\}$; observe that $I-A$ is m-nilpotent if $0\in \sigma \left(B\right)$ and $I+B$ is m-nilpotent if $0\in \sigma \left(A\right)$). □

The conclusions of Theorem 1 help in building a picture of the operators $A,B$ satisfying $D_{d_1,d_2}^m\left(I\right)=0$. The following examples consider cases $\left(iii\right)$, $\left(iv\right)$, $\left(vi\right)$ and $\left(vii\right)$ of the theorem.

Example 1.

0 may be in both ${\sigma }_a\left(A\right)$ and ${\sigma }_a\left(B^{*}\right)$ in the case in which ${▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$: consider, for example, the operators $A,B$ such that ${\sigma }_a\left(A\right)=\{0,{\displaystyle \frac{1-e^{-i\theta }}{a}}\}$, ${\sigma }_a\left(B^{*}\right)=\{0,a\}$, $A=0I_1\oplus {\displaystyle \frac{1-e^{-i\theta }}{a}}{I}_2$ and $B=0I_1\oplus a{I}_2$, $0\ne a$ real and $0<\theta <2\pi$. It is seen that

$$\begin{array}{ccc}& & {▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)\hfill \\ & =& C_{\sum }{\left(0I_1\oplus {\displaystyle \frac{1-e^{i\theta }}{a}}{I}_2\right)}^p{\left(0I_1\oplus {\displaystyle \frac{1-e^{-i\theta }}{a}}{I}_2\right)}^k\left(0I_1\oplus a^{p+k}{I}_2\right)\hfill \\ & =& C_{\sum }\left(0I_1\oplus {(1-e^{i\theta })}^p{(1-e^{-i\theta })}^k{I}_2\right)=\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(I_1\oplus e^{i\theta }{I}_2)}^j{(I_1\oplus e^{-i\theta }{I}_2)}^j\hfill \\ & =& 0.\hfill \end{array}$$

Example 2.

If ${▵}_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, then 0 may belong to both ${\sigma }_a\left(A\right)$ and ${\sigma }_a\left(B^{*}\right)$ (consider, for example the operators $A=I_1\oplus 0I_2$ and $B=0I_1\oplus -I_2$), or, 0 may just be in one of ${\sigma }_a\left(A\right)$ and ${\sigma }_a\left(B^{*}\right)$ (consider the operators $A=I_1\oplus (1-\sqrt{{\displaystyle \frac{3}{2}}})I_2$ and $B=0I_1\oplus (-1-\sqrt{{\displaystyle \frac{3}{2}}})I_2$).

Example 3.

In the case in which ${\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, $0\in {\sigma }_a\left(B^{*}\right)$ implies A is m-symmetric (thus $\sigma \left(A\right)$ is real) and $0\in {\sigma }_a\left(A\right)$ implies $B^{*}$ is m-symmetric (thus $\sigma \left(B\right)$ is real). If $a,b,r_1,r_2$ are real numbers, $\sigma \left(A\right)=\{a,r_1+ib\}$ and $\sigma \left(B\right)=\{0,r_2+ib\}$, then $A=a{I}_1\oplus (r_1+ib)I_2$ and $B=0I_1\oplus (r_2+ib)I_2$ is an example of a pair of operators satisfying ${\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$. Again, if we let $\beta =i$ and $A=I_1\oplus (-1-i)I_2$, $B=(1+i)I_1\oplus -I_2$ (in the proof of case $\left(vii\right)$), then ${\delta }_{A,B}\left(I\right)=-iI$, ${▵}_{A^{*},B^{*}}\left(I\right)=iI$ and ${\delta }_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}\left(I\right)=(I-L_{iI}{R}_{-iI})\left(I\right)=0$.

4. Equivalence $D_{d_1,d_2}^m\left(I\right)=0⟺D_{d_1,d_2}\left(I\right)=0$ and the Putnam–Fuglede Property

It is well known (indeed, easily proven) that if a pair $(A,B)$ of (Banach space) operators is m-isometric (similarly, m-symmetric), then it is n-isometric (respectively, n-symmetric) for all integers $n\ge m$ [5,11]. The proposition that $(A,B)$ is m-isometric (or, m-symmetric) implies it is n-isometric (respectively, n-symmetric) for some positive integer $n<m$ fails. A prime example here is that of strictly m-isometric (respectively, strictly m-symmetric) operators, where an operator A is strictly m-isometric (respectively, strictly m-symmetric) if ${▵}_{A^{*},A}^m\left(I\right)=0$ and ${▵}_{A^{*},A}^{m-1}\left(I\right)\ne 0$ (respectively, ${\delta }_{A^{*},A}^m\left(I\right)=0$ and ${\delta }_{A^{*},A}^{m-1}\left(I\right)\ne 0$). Partial results exist. Thus, if $T\in B\left(\mathcal{H}\right)$ is such that ${\delta }_{T^{*},T}^m\left(I\right)=0$ for some even positive integer m (respectively, ${▵}_{T^{*},T}^m\left(I\right)=0$ for some invertible $T\in B\left(\mathcal{H}\right)$ and a positive even integer m), then ${\delta }_{T^{*},T}^{m-1}\left(I\right)=0$ (respectively, ${▵}_{T^{*},T}^{m-1}\left(I\right)=0$); see [17], Theorem 3 and Proposition 1, respectively. Again, if T is power bounded, i.e., ${sup}_n\left|\right|T^n\left|\right|<M$ for some positive real number M, then ${▵}_{T^{*},T}^m\left(I\right)=0$ if and only if ${▵}_{T^{*},T}\left(I\right)=0$ [32]. Equivalently, if T is power bounded and ${▵}_{T^{*},T}^m\left(I\right)=0$, then T is isometric. More generally, if $S\in B\left(\mathcal{H}\right)$ is left m-left invertible by a power-bounded operator $T\in B\left(\mathcal{H}\right)$, thus ${▵}_{T,S}^m\left(I\right)=0$ and T is power bounded, then S is similar to an isometric operator (where we may choose the invertible operator effecting similarity to be a positive operator), and T is similar to $S^{*}$ ([33], Theorem 2.4). The following theorem considers the equivalence $D_{d_1,d_2}^m\left(I\right)=0$ if and only if $D_{d_1,d_2}\left(I\right)=0$ for the operators $D,d_1$ and $d_2$ of Theorem 1. It is seen that just as for the cases ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$ and ${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$ considered in [17], a necessary and sufficient condition for the required equivalence is that the operators A and B satisfy a Putnam–Fuglede commutativity property. Before, however, going on to state the theorem, we consider a few examples.

Example 4.

Given ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, let ${\sigma }_a\left(B^{*}\right)=\{0,\mu \}$ for somel non-zero real number μ. Then

$$\begin{array}{ccc}& & {\sigma }_a(A-\mu )={\sigma }_a(A-(0+\mu ))={\displaystyle \frac{\mu \pm i\sqrt{4-{\mu }^2}}{2}}\hfill \\ & ⟹& {\sigma }_a\left(A\right)=\left\{{\displaystyle \frac{-\mu \pm i\sqrt{4-{\mu }^2}}{2}}\right\}=\{e^{i\theta },-e^{-i\theta }\}\hfill \end{array}$$

for some $0\le \theta \le 2\pi$. Letting

$$A=e^{i\theta }{I}_1\oplus -e^{-\theta }{I}_2\mathrm{and}B=0I_1\oplus \mu I_2,$$

it is seen that

$$\begin{array}{ccc}& & {▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^1\left(I\right)=I-\left((e^{-i\theta }{I}_1\oplus -(e^{i\theta }+\mu )I_2)(e^{i\theta }{I}_1\oplus -(e^{-i\theta }+\mu )I_2)\right)\hfill \\ & =& I-\left(I_1\oplus (1+\mu (e^{i\theta }+e^{-i\theta }){\mu }^2)I_2\right)=0,\hfill \end{array}$$

since $e^{i\theta }+e^{-i\theta }=-\mu$. Define operators $E,F\in B\left(B\right(\mathcal{H}\left)\right)$ by $E={\delta }_{A^{*},B^{*}}$ and $F={\delta }_{A,B}$. Let $N\in B\left(\mathcal{H}\right)$ be an n-nilpotent operator, $N=N_{1}I\oplus N_2{I}_2$; let $L_{N^{*}}={\mathcal{N}}_1$ and $L_N={\mathcal{N}}_2$. Then ${\delta }_{A^{*}+N^{*},B^{*}}={\delta }_{A^{*},B^{*}}+L_{N^{*}}=E+{\mathcal{N}}_1$ and ${\delta }_{A+N,B}=F+{\mathcal{N}}_2$, where ${\mathcal{N}}_i$, $1\le i\le 2$, are n-nilpotent operators. Trivially, ${▵}_{A^{*}-\mu ,A-\mu }^1\left(I\right)=0$ and, since N is n-nilpotent, ${▵}_{A^{*}+N^{*}-\mu ,A+N-\mu }^{2n-1}\left(I\right)=0$ (see [5,34]). Similarly, ${▵}_{B+N-e^{-i\theta },B^{*}+N^{*}-e^{-i\theta }}^{2n-1}\left(I\right)=0$. However, neither of ${▵}_{E+{\mathcal{N}}_1,F+{\mathcal{N}}_2}^1\left(I\right)$, ${▵}_{A^{*}+N^{*}-\mu ,A+N-\mu }^1\left(I\right)$ and ${▵}_{B+N-e^{i\theta },B^{*}+N^{*}-e^{-i\theta }}^1\left(I\right)$ is the 0 operator.

Example 5.

For some non-zero real numbers $r,r_1,r_2$ and $0\le \theta <2\pi$, let $\sigma \left(A\right)=\{{\alpha }_1,{\alpha }_2\}=\{r_1{e}^{-i\theta },r_2{e}^{-i\theta }\}$, $\sigma \left(B\right)=\left\{\lambda \right\}=\left\{r{e}^{i\theta }\right\}$, $A=r{e}^{-i\theta }{I}_1\oplus r_2{e}^{-i\theta }{I}_2$ and $B=r{e}^{i\theta }(I_1\oplus I_2)$.Then

$${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^1\left(I\right)=(I-A^{*}{B}^{*})-(I-AB)=0.$$

Let $N=N(I_1\oplus I_2)$ be an n-nilpotent operator in $B\left(\mathcal{H}\right)$. Let ${▵}_{A^{*},B^{*}}=E$, ${▵}_{A,B}=F$, ${▵}_{A^{*}+N^{*},B^{*}}=I-L_{A^{*}}{R}_{B^{*}}-L_{N^{*}}{R}_{B^{*}}=E+{\mathcal{N}}_1$ and ${▵}_{A+N,B}=F+{\mathcal{N}}_2$. Then, ${\mathcal{N}}_i$, $1\le i\le 2$, are n-nilpotent operators such that ${\mathcal{N}}_1$ commutes with E, ${\mathcal{N}}_2$ commutes with F and see [5], ${\delta }_{E+{\mathcal{N}}_1,F+{\mathcal{N}}_2}^{2n-2}\left(I\right)=0$. Evidently,

$${\delta }_{I-\overline{\lambda }{A}^{*},I-\lambda A}^1\left(I\right)=0={\delta }_{I-\alpha B,I-\overline{\alpha }{B}^{*}}^1\left(I\right),$$

and hence

$${\delta }_{I-\overline{\lambda }(A^{*}+N^{*}),I-\lambda (A+N)}^{2n-1}\left(I\right)=0={\delta }_{I-{\alpha }_{1}B,I-\overline{{\alpha }_2}{B}^{*}}^{2n-1}\left(I\right).$$

However, neither of ${\delta }_{I-\overline{\lambda }(A^{*}+N^{*}),I-\lambda (A+N)}^1\left(I\right)$ and ${\delta }_{{▵}_{A^{*}+N^{*},B^{*}},{▵}_{A+N,B}}^1\left(I\right)$ is the 0 operator.

Observe here that the operator $A+N$ is not normal in either of the above examples.

Theorem 2.

If $A,B\in B\left(\mathcal{H}\right)$ are such that $\sigma \left(A\right)\ne \left\{0\right\}\ne \sigma \left(B\right)$, $\alpha \in {\sigma }_a\left(A\right)$ and $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$, then we have the following:

(i) 

${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$ implies ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}\left(I\right)=0$ if and only if the pairs of operators $(A^{*}-\overline{\lambda },A-\lambda )$ and $(B-\alpha ,B^{*}-\overline{\alpha })$ are $\left(PF\right)$-pairs.

(ii) 

${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$ implies ${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}\left(I\right)=0$ if and only if the pairs of operators $(I-\overline{\lambda }{A}^{*},I-\lambda A)$ and $(I-\alpha B,I-\overline{\alpha }{B}^{*})$ are $\left(PF\right)$-pairs.

(iii) 

${▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$ implies ${▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}\left(I\right)=0$ if and only if the pairs of operators $(I-\overline{\lambda }{A}^{*},I-\lambda A)$ and $(I-\alpha B,I-\overline{\alpha }{B}^{*})$ are $\left(PF\right)$-pairs.

(iv) 

${▵}_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$ implies ${▵}_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}\left(I\right)=0$ if and only if the pairs of operators $(I-\overline{\lambda }{A}^{*},A-\lambda )$ and $(I-\alpha B,\overline{\alpha }-B^{*})$ are $\left(PF\right)$-pairs.

(v) 

${▵}_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$ implies ${▵}_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}\left(I\right)=0$ if and only if the pairs of operators $(A^{*}-\overline{\lambda },I-\lambda A)$ and $(B-\alpha ,\overline{\alpha }{B}^{*}-I)$ are $\left(PF\right)$-pairs.

(vi) 

${\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$ implies ${\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}\left(I\right)=0$ if and only if the pairs of operators $(A^{*}-\overline{\lambda },A-\lambda )$ and $(B-\alpha ,B^{*}-\overline{\alpha })$ are $\left(PF\right)$-pairs.

(vii) 

${\delta }_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$ implies ${\delta }_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}\left(I\right)=0$ if and only if the pairs of operators $(A^{*}-\overline{\lambda },I-\lambda A)$ and $(B-\alpha ,\overline{\alpha }{B}^{*}-I)$ are $\left(PF\right)$-pairs.

(viii) 

${\delta }_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$ implies ${\delta }_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}\left(I\right)=0$ if and only if the pairs of operators $(I-\overline{\lambda }{A}^{*},A-\lambda )$ and $(I-\alpha B,\overline{\alpha }-B^{*})$ are $\left(PF\right)$-pairs.

Proof. 

Before going on to prove the theorem, we make a few observations (which we will use in the sequel without further reference). As seen in the proofs of Proposition 1 and Theorem 1, if $D_{d_1,d_2}^m\left(I\right)=0$ for any of the choices $D=▵$ or $\delta$, $d_1={▵}_{A^{*},B^{*}}$ or ${\delta }_{A^{*},B^{*}}$ and $d_2={▵}_{A,B}$ or ${\delta }_{A,B}$, then the spectra $\sigma \left(A\right)$ and $\sigma \left(B\right)$ consist at most of two points. Consequently, ${\sigma }_a\left(A\right)=\sigma \left(A\right)$ and ${\sigma }_a\left(B\right)=\sigma \left(B\right)$; also, if $I-\lambda A$ or $\lambda -A$, similarly $I-\lambda B$ or $\lambda -B$, is left invertible for some scalar $\lambda$, then it is invertible. Furthermore, if A (similarly, B) is normal, then the spectral points of A (respectively, B) are (simple poles of the resolvent of the operator, hence) normal eigenvalues of the operator. The operator A (respectively, B) has a direct sum decomposition $A={\alpha }_1{I}_1\oplus {\alpha }_2{I}_2$ (respectively, $B={\lambda }_1{I}_1\oplus {\lambda }_2{I}_2$), $I=I_1\oplus I_2$, and to every $x\in \mathcal{H}$ there corresponds an $\alpha ={\alpha }_1$ or ${\alpha }_2$ (respectively, $\lambda ={\lambda }_1$ or ${\lambda }_2$) such that $Ax=\alpha x$ (respectively, $Bx=\lambda x$).

The proof for almost all the cases is similar, except for differences in detail: we prove below cases $\left(i\right)$ to $\left(iii\right)$ in some detail and provide a brief outline of the proof for cases $\left(v\right)$ and $\left(vi\right)$.

$\left(i\right)$. If ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$ (implies ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}\left(I\right)=0$), then ${▵}_{A^{*}-\overline{\lambda },A-\lambda }\left(I\right)=0={▵}_{B-\alpha ,B^{*}-\overline{\alpha }}\left(I\right)$ for all $\alpha \in \sigma \left(A\right)$ and $\lambda \in \sigma \left(B\right)$ (see the proof of Proposition 1). The conclusion ${▵}_{A^{*}-\overline{\lambda },A-\lambda }\left(I\right)=0$ implies

$$A-\lambda \mathrm{is}(\mathrm{isometric},\mathrm{hence})\mathrm{left}\mathrm{invertible}⟹A-\lambda \mathrm{is}\mathrm{invertible}$$

and the conclusion ${▵}_{B-\alpha ,B^{*}-\overline{\alpha }}\left(I\right)=0$ implies

$$B-\alpha \mathrm{is}\mathrm{right}\mathrm{invertible}⟹B-\alpha \mathrm{is}\mathrm{invertible}.$$

(Indeed $A-\lambda$ and $B-\alpha$ are unitary.) Consequently,

$$\begin{array}{ccc}& & (A^{*}-\overline{\lambda })(A-\lambda )=I=(A-\lambda )(A^{*}-\overline{\lambda })⟹A\mathrm{is}\mathrm{normal}\mathrm{and}\hfill \\ & & (B-\alpha )(B^{*}-\overline{\alpha })=I=(B^{*}-\overline{\alpha })⟹B\mathrm{is}\mathrm{normal}.\hfill \end{array}$$

Hence $(A^{*}-\overline{\lambda },A-\lambda )$ and $(B^{*}-\overline{\alpha },B-\alpha )$ are $\left(PF\right)$-pairs.

To prove the sufficiency, we recall from Proposition 1 that

$${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0⟹{▵}_{A^{*}-\overline{\lambda },A-\lambda }^m\left(I\right)=0={▵}_{B-\alpha ,B^{*}-\overline{\alpha }}^m\left(I\right)$$

for all $\alpha \in \sigma \left(A\right)$ and $\lambda \in \sigma \left(B\right)$. The hypothesis on the $\left(PF\right)$-property implies (that the ascent is less than or equal to one, and hence)

$${▵}_{A^{*}-\overline{\lambda },A_{\lambda }}\left(I\right)=0={▵}_{A-\lambda ,A^{*}-\overline{\lambda }}\left(I\right)$$

and

$${▵}_{B-\alpha ,B^{*}-\overline{\alpha }}\left(I\right)=0={▵}_{B^{*}-\overline{\alpha },B-\alpha }\left(I\right).$$

Consequently, $A,B$ are normal, $A={\oplus }_{i=1}^2{\alpha }_i{I}_i$, $B={\oplus }_{i=1}^2{\lambda }_i{I}_{1i}$ ($I_1\oplus I_2=I=I_{11}\oplus I_{12}$), and for each $x\in \mathcal{H}$ there exists a $\lambda$ ($={\lambda }_1$ or ${\lambda }_2$) such that

$$\begin{array}{ccc}& & \langle {▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}\left(I\right)x,x\rangle =\langle (I-A^{*}A+A^{*}B-B^{*}A+B^{*}B)x,x\rangle \hfill \\ & =& \langle (I-A^{*}A+\lambda A^{*}-\overline{\lambda }A+{\left|\lambda \right|}^2)x,x\rangle =\langle {▵}_{A^{*}-\overline{\lambda },A-\lambda }\left(I\right)x,x\rangle \hfill \\ & =& 0.\hfill \end{array}$$

This proves the sufficiency. (Observe that we also have ${▵}_{{\delta }_{A,B},{\delta }_{A^{*},B^{*}}}\left(I\right)=0.$)

$\left(ii\right)$. If ${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$ implies ${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}\left(I\right)=0$, then ${\delta }_{I-\overline{\lambda }{A}^{*},I-\lambda A}\left(I\right)=0={\delta }_{I-\alpha B,I-\overline{\alpha }{B}^{*}}\left(I\right)$ for all $\alpha \in \sigma \left(A\right)$ and $\lambda \in \sigma \left(B\right)$. Hence $I-\lambda A$ and $I-\alpha B$ are self-adjoint. Consequently, A and B are normal, and the necessity is proved.

For sufficiency, we start by recalling ${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$ implies ${\delta }_{I-\overline{\lambda }{A}^{*},I-\lambda A}^m\left(I\right)=0={\delta }_{I-\alpha B,I-\overline{\alpha }{B}^{*}}^m\left(I\right)$ for all $\alpha \in \sigma \left(A\right)$ and $\lambda \in \sigma \left(B\right)$. The $\left(PF\right)$-property hypothesis thus implies ${\delta }_{I-\overline{\lambda }{A}^{*},I-\lambda A}\left(I\right)=0={\delta }_{I-\lambda A,I-\overline{\lambda }{A}^{*}}\left(I\right)$ and ${\delta }_{I-\alpha B,I-\overline{\alpha }{B}^{*}}\left(I\right)=0={\delta }_{I-\overline{\alpha }{B}^{*},I-\alpha B}\left(I\right)$, A and B are normal operators with direct sum representations of type $A={\oplus }_{i=1}^2{\alpha }_i{I}_i$ and $B={\oplus }_{i=1}^2{\lambda }_i{I}_{1i}$. Furthermore, for each $x\in \mathcal{H}$, there exists a $\lambda$ ($={\lambda }_1$ or ${\lambda }_2$) such that $Bx=\lambda x$. Since this implies

$$\begin{array}{ccc}& & \langle \left({\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}\left(I\right)\right)x,x\rangle =\langle \left(AB-A^{*}{B}^{*}\right)x,x\rangle \hfill \\ & =& \langle \left(\lambda A-\overline{\lambda }{A}^{*}\right)x,x\rangle =\langle {\delta }_{I-\overline{\lambda }{A}^{*},I-\lambda A}\left(I\right)x,x\rangle =0,\hfill \end{array}$$

the proof is complete. (Remark here that ${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}\left(I\right)=-{\delta }_{{▵}_{A,B},{▵}_{A^{*},B^{*}}}\left(I\right)$, and hence also ${\delta }_{{▵}_{A,B},{▵}_{A^{*},B^{*}}}\left(I\right)=0$.)

$\left(iii\right)$. We start by proving the sufficiency of the conditions. If $\alpha \in {\sigma }_a\left(A\right)$ and $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$, then (see Theorem 1)

$${▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0⟹{▵}_{I-\overline{\lambda }{A}^{*},I-\lambda A}^m=0={▵}_{I-\alpha B,I-\overline{\alpha }{B}^{*}}^m\left(I\right),$$

and hence, by the $\left(PF\right)$-property

$$\begin{array}{ccc}& & {▵}_{I-\overline{\lambda }{A}^{*},I-\lambda A}\left(I\right)=0={▵}_{I-\alpha B,I-\overline{\alpha }{B}^{*}}\left(I\right)\mathrm{and}\hfill \\ & & {▵}_{I-\lambda A,I-\overline{\lambda }{A}^{*}}\left(I\right)=0={▵}_{I-\overline{\alpha }{B}^{*},I-\alpha B}\left(I\right).\hfill \end{array}$$

Consequently,

$$\begin{array}{ccc}& & I-(I-\overline{\lambda }{A}^{*})(I-\lambda A)=0=I-(I-\lambda A)(I-\overline{\lambda }{A}^{*})\hfill \\ & ⟺& \lambda A+\overline{\lambda }{A}^{*}={\left|\lambda \right|}^2{A}^{*}A={\left|\lambda \right|}^{2}A{A}^{*}\hfill \end{array}$$

and

$$\begin{array}{ccc}& & I-(I-\alpha B)(I-\overline{\alpha }{B}^{*})=0=I-(I-\overline{\alpha }{B}^{*})(I-\alpha B)\hfill \\ & ⟺& \alpha B+\overline{\alpha }{B}^{*}={\left|\alpha \right|}^{2}B{B}^{*}={\left|\alpha \right|}^2{B}^{*}B.\hfill \end{array}$$

Thus $A,B$ are normal operators, $A={\alpha }_1{I}_1\oplus {\alpha }_2{I}_2$, $B={\lambda }_1{I}_{11}\oplus {\lambda }_2{I}_{12}$ (for some not necessarily distinct scalars ${\alpha }_i,{\lambda }_i$; $1\le i\le 2$). Furthermore, to each $x\in \mathcal{H}$, there corresponds a $\lambda (={\lambda }_1$ or ${\lambda }_2)$ such that $Bx=\lambda x$. We have

$$\langle \left({▵}_{{▵}_{A,B},{▵}_{A^{*},B^{*}}}^1\left(I\right)\right)x,x\rangle =\langle \left(\overline{\lambda }{A}^{*}+\lambda A-{\left|\lambda \right|}^2{A}^{*}A\right)x,x\rangle =0$$

for all $x\in \mathcal{H}$. Hence, also, ${▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^1\left(I\right)=0$. (Remark here that

$$\begin{array}{ccc}\hfill {▵}_{{▵}_{A^{*},B^{*}}.{▵}_{A,B}}^1\left(I\right)& =& \left(L_{A^{*}}{R}_{B^{*}}+L_A{R}_B-L_{A^{*}}{L}_A{R}_{B^{*}}{R}_B\right)\left(I\right)\hfill \\ & =& A^{*}{B}^{*}+AB--A^{*}A{B}^{*}B\hfill \\ & =& AB+A^{*}{B}^{*}-A{A}^{*}B{B}^{*}=\left(L_A{R}_B+L_{A^{*}}{R}_{B^{*}}-L_A{L}_{A^{*}}{R}_B{R}_{B^{*}}\right)\left(I\right)\hfill \\ & =& {▵}_{{▵}_{A,B},{▵}_{A^{*},B^{*}}}^1\left(I\right),\hfill \end{array}$$

and hence ${▵}_{{▵}_{A,B},{▵}_{A^{*},B^{*}}}^1\left(I\right)=0$.)

To prove the necessity of the condition, we start by observing that the hypothesis $\sigma \left(A\right)\ne \left\{0\right\}\ne \sigma \left(B\right)$ implies the existence of non-trivial $\alpha \in {\sigma }_a\left(A\right)$ and $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$. Hence, see the proof of Theorem 1(iii), ${▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^1\left(I\right)=0$ implies

$$\begin{array}{ccc}& & {▵}_{I-\overline{\lambda }{A}^{*},I-\lambda A}\left(I\right)=0,\hfill \\ & & {▵}_{I-\alpha B,I-\overline{\alpha }{B}^{*}}\left(I\right)=0,\hfill \end{array}$$

i.e., $I-\lambda A$ and $I-\overline{\lambda }{B}^{*}$ are left invertible, hence invertible. We have

$$\begin{array}{ccc}& & (I-\overline{\lambda }{A}^{*})(I-\lambda A)=I=(I-\lambda A)(I-\overline{\lambda }{A}^{*})\hfill \\ & ⟺& \overline{\lambda }{A}^{*}+\lambda A-{\left|\lambda \right|}^2{A}^{*}A=\overline{\lambda }{A}^{*}+\lambda A-{\left|\lambda \right|}^{2}A{A}^{*},\mathrm{and}\hfill \\ & & (I-\alpha B)(I-\overline{\alpha }{B}^{*})=I=(I-\overline{\alpha }{B}^{*})(I-\alpha B)\hfill \\ & ⟺& \alpha B+\overline{\alpha }{B}^{*}-{\left|\alpha \right|}^{2}B{B}^{*}=\alpha B+\overline{\alpha }{B}^{*}-{\left|\alpha \right|}^2{B}^{*}B,\hfill \end{array}$$

i.e., once again the operators A and B are normal. Hence $(I-\overline{\lambda }{A}^{*},I-\lambda A)$ and $(I-\overline{\lambda }{B}^{*},I-\lambda B)$ are $\left(PF\right)$-pairs for all $\alpha$ and $\lambda$.

An outline proof of $\left(vi\right)$ and $\left(vii\right)$. If ${\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$ implies ${\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}\left(I\right)=0$, then

$${\delta }_{A^{*}-\overline{\lambda },A-\lambda }\left(I\right)=0={\delta }_{B-\alpha ,B^{*}-\overline{\alpha }}\left(I\right)⟹A-\lambda \mathrm{and}B-\alpha \mathrm{self}-\mathrm{adjoint}.$$

Hence $A,B$ are normal and the necessity follows. For sufficiency,

$$\begin{array}{ccc}& & {\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0⟹{\delta }_{A^{*}-\overline{\lambda },A-\lambda }^m\left(I\right)=0={\delta }_{B-\alpha ,B^{*}-\overline{\alpha }}^m\left(I\right)\hfill \\ & ⟹& {\delta }_{A^{*}-\overline{\lambda },A-\lambda }\left(I\right)=0={\delta }_{B-\alpha ,B^{*}-\overline{\alpha }}\left(I\right)\hfill \\ & ⟹& A,B\mathrm{normal}⟹\langle \left({\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}\left(I\right)\right)x,x\rangle ,x\in \mathcal{H},\hfill \\ & ⟹& \langle {\delta }_{A^{*}-\overline{\lambda },A-\lambda }\left(I\right)x,x\rangle =0\hfill \\ & ⟹& {\delta }_{{\delta }_{A^{*},B^{*}}{\delta }_{A,B}}\left(I\right)=0.\hfill \end{array}$$

If ${\delta }_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$ implies ${\delta }_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}\left(I\right)=0$, then

$$\begin{array}{ccc}& & {\delta }_{I-\overline{\lambda }{A}^{*},A-\lambda }\left(I\right)=0={◊}_{I-\alpha B,B^{*}-\overline{\alpha }}\left(I\right)\hfill \\ & ⟹& (I-\overline{\lambda }{A}^{*})(\lambda -A)=(\lambda -A)(I-\overline{\lambda }{A}^{*}),(I-\alpha B)(B^{*}-\overline{\alpha })=(B^{*}-\overline{\alpha })(I-\alpha B)\hfill \\ & ⟹& A,B\mathrm{are}\mathrm{normal}.\hfill \end{array}$$

(Notice here that if $0\in \sigma \left(A\right)$, then $B=I$ and $A^{*}=A$, and if $0\in \sigma \left(B\right)$, then $A=I$ and $B=B^{*}$.) This proves the necessity. For sufficiency,

$$\begin{array}{ccc}& & {\delta }_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0⟹{\delta }_{I-\overline{\lambda }{A}^{*},A-\lambda }^m\left(I\right)=0={◊}_{I-\alpha B,B^{*}-\overline{\alpha }}^m\left(I\right)\hfill \\ & ⟹& {\delta }_{I-\overline{\lambda }{A}^{*},A-\lambda }\left(I\right)=0={◊}_{I-\alpha B,B^{*}-\overline{\alpha }}\left(I\right)\hfill \\ & ⟹& A,B\mathrm{normal}⟹\langle \left({\delta }_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}\left(I\right)\right)x,x\rangle ,x\in \mathcal{H},\hfill \\ & ⟹& \langle {\delta }_{I-\overline{\lambda }{A}^{*},A-\lambda }\left(I\right)x,x\rangle =0\hfill \\ & ⟹& {\delta }_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}\left(I\right)=0.\hfill \end{array}$$

□

Remark 1.

If ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, then Proposition 1 implies $0\notin \sigma (A-\lambda )\wedge \sigma (B^{*}-\overline{\alpha })$ for all $\alpha \in \sigma \left(A\right)$ and $\lambda \in \sigma \left(B\right)$, and $A-\lambda$ and $B^{*}-\overline{\alpha }$ are invertible m-isometries. As such, $A-\alpha$ and $B^{*}-\overline{\lambda }$ (hence, also $B-\lambda$) are generalised scalar operators (see [9,16,35] and ([17], Lemma 2.9)). Since generalised scalar operators are polaroid, and, since (for a non-constant) polynomial $p(.)$, $p\left(T\right)$ is polaroid for an operator T if and only if T is polaroid [23], $A,B$ are polaroid. If $\sigma \left(A\right)=\{{\alpha }_1,{\alpha }_2\}$ and $\sigma \left(B\right)=\{{\lambda }_1,{\lambda }_2\}$, then there exists a positive integer n (equal to the maximum of the order of the poles at the points ${\alpha }_i$ and ${\lambda }_i$, $1\le i\le 2$) and nilpotent operators $N_i$, $1\le i\le 2$, such that

$$A=\left({A|}_{{(A-{\alpha }_1)}^{-n}\left(0\right)}{\oplus A|}_{{(A-{\alpha }_2)}^{-n}\left(0\right)}\right)+N_1$$

and

$$B=\left({B|}_{{(B-{\lambda }_1)}^{-n}\left(0\right)}{\oplus B|}_{{(B-{\lambda }_2)}^{-n}\left(0\right)}\right)+N_2.$$

(Here, either of the components in the representation of A, respectively B, may be absent). This known representation of operators A and B, see [36], extends to operators $A,B$ such that ${▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$.

Proposition 2.

Let $A,B\in B\left(\mathcal{H}\right)$ be such that $\sigma \left(A\right)\ne \left\{0\right\}\ne \sigma \left(B\right)$. If ${▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$, then there exist scalars ${\alpha }_i$ and ${\lambda }_i$, nilpotents $N_i$ ($1\le i\le 2$) and a positive integer n such that

$$A=\left({A|}_{{(A-{\alpha }_1)}^{-n}\left(0\right)}{\oplus A|}_{{(A-{\alpha }_2)}^{-n}\left(0\right)}\right)+N_1$$

and

$$B=\left({B|}_{{(B-{\lambda }_1)}^{-n}\left(0\right)}{\oplus B|}_{{(B-{\lambda }_2)}^{-n}\left(0\right)}\right)+N_2.$$

(Either of the components in the representation of A, respectively B, may be absent).

Proof. 

The hypothesis ${▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$ implies $I-\lambda A$ and $I-\overline{\alpha }{B}^{*}$ are m-isometric left invertible operators, hence invertible operators (since the spectrum consists at most of two points). This implies $I-\lambda A$ and $I-\overline{\alpha }{B}^{*}$ are generalised scalar, hence polaroid, operators. Consequently, A and B are polaroid, and if $\sigma \left(A\right)=\{{\alpha }_1,{\alpha }_2\}$ and $\sigma \left(B\right)=\{{\lambda }_1,{\lambda }_2\}$, then A and B have the representation of the statement of the proposition. □

Do operators A and B satisfying the hypotheses of the remaining cases of Theorem 1 have analogous representations? We do not know the answer to this question. However, the following proposition shows that there is an answer in the affirmative for certain choices of the spectral points $\alpha$ and $\lambda$, and m (for the remaining cases). In the following proposition, the scalar $\beta$ shall refer to the scalar $\beta$ (of the corresponding case) of the statement of Theorem 1. The scalar n in the statement of the proposition is not necessarily the same in the direct sum representations of the operators A and B.

Proposition 3.

Let $A,B\in B\left(\mathcal{H}\right)$ be such that $\sigma \left(A\right)\ne \left\{0\right\}\ne \sigma \left(B\right)$.

(a) 

If either ${▵}_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, or ${▵}_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$, and either $-1\in \sigma \left(B\right)$ or $1\in \sigma \left(A\right)$ or $\left|\beta \right|=1$, then

$$\begin{array}{ccc}& & A=\left({A|}_{{(A-1)}^{-n}\left(0\right)}{\oplus A|}_{{(A+1-e^{i\theta })}^{-n}\left(0\right)}\right)+N_1\mathrm{and}\hfill \\ & & B=\left({B|}_{{(B+1)}^{-n}\left(0\right)}{\oplus B|}_{{(B-1+e^{i\theta })}^{-n}\left(0\right)}\right)+N_2\hfill \end{array}$$

for some positive integer n, nilpotents $N_i$ ($1\le i\le 2$) and $0\le \theta <2\pi$.

(b) 

Let $m\le 4$.

(i) 

If either ${\delta }_{{▵}_{A^{,}{B}^{*}},{▵}_{A,B}}^m\left(I\right)=0$ or ${\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, then then there exist scalars ${\alpha }_i$ and ${\lambda }_i$, nilpotents $N_i$ ($1\le i\le 2$) and a positive integer n such that

$$A=\left({A|}_{{(A-{\alpha }_1)}^{-n}\left(0\right)}{\oplus A|}_{{(A-{\alpha }_2)}^{-n}\left(0\right)}\right)+N_1$$

and

$$B=\left({B|}_{{(B-{\lambda }_1)}^{-n}\left(0\right)}{\oplus B|}_{{(B-{\lambda }_2)}^{-n}\left(0\right)}\right)+N_2.$$

(ii) 

If either ${\delta }_{{\delta }_{A^{,}{B}^{*}},{▵}_{A,B}}^m\left(I\right)=0$, or ${\delta }_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, the operators $A-I,B+I$ are not nilpotent and either $-1\in \sigma \left(B\right)$ (or $1\in \sigma \left(A\right)$ or $\beta \ne 1$ is real), then

$$\begin{array}{ccc}& & A=\left({A|}_{{(A-1)}^{-n}\left(0\right)}{\oplus A|}_{{(A+1-\beta )}^{-n}\left(0\right)}\right)+N_1\mathrm{and}\hfill \\ & & B=\left({B|}_{{(B+1)}^{-n}\left(0\right)}{\oplus B|}_{{(B-1+\beta )}^{-n}\left(0\right)}\right)+N_2\hfill \end{array}$$

for some positive integer n and nilpotents $N_i$ ($1\le i\le 2$).

Proof. 

(a)

If either of the hypotheses on $\sigma \left(A\right)$, $\sigma \left(B\right)$ and $\beta$, holds, then necessarily

$$\sigma \left(A\right)=\{-1+e^{i\theta },1\},\sigma \left(B\right)=\{-1,1-e^{i\theta }\}\mathrm{and}{▵}_{A^{*}+I,A+I}^m\left(I\right)=0={▵}_{I-B,I-B^{*}}^m\left(I\right)$$

(see Theorem 1). The operator $A+I$ and $I-B^{*}$ being invertible m-isometric are generalised scalar, hence polaroid. This implies $A,B$ are polaroid, hence have the representation of the statement of the proposition.

(b)

Before going on to considering the operators ${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$ and ${\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, $m\le 4$, we recall some facts. It is known, see [7,12], that an operator $T\in B\left(\mathcal{H}\right)$ is 3-symmetric if and only if it being unitarily equivalent to the restriction to a closed invariant subspace of the operator

$$S=\left[\begin{array}{cc}V& E\hfill \\ 0& V\hfill \end{array}\right]\in B(\mathcal{H}\oplus \mathcal{H}),[V,E]=0,V\mathrm{self}\mathrm{adjoint}.$$

Letting $H_0\left(p\right)$, p being a positive integer, denote the class of operators $T\in B\left(\mathcal{H}\right)$ such that

$$H_0(T-\lambda )={(T-\lambda )}^{-p}\left(0\right),\lambda \in \mathtt{C}.$$

(Recall, $H_0\left(T\right)$ denotes the quasi-nilpotent part

$$H_0\left(T\right)=\{x\in \mathcal{H}:\underset{n\to \infty }{lim}\left|\right|T^n{x\left|\right|}^{{\displaystyle \frac{1}{n}}}=0\}$$

of T ([23], p. 119)).

Normal operators are $H_0\left(1\right)$ operators ([23], Theorem 4.46). Since

$${\left|\right|(S-\lambda )}^n{(x\oplus y)\left|\right|}^{{\displaystyle \frac{1}{n}}}={\left|\right|(V-\lambda )}^n\left(x\right)+nE{(V-\lambda )}^{n-1}{y\left|\right|}^{{\displaystyle \frac{1}{n}}}+{\left|\right|(V-\lambda )}^{n}x{\left|\right|}^{{\displaystyle \frac{1}{n}}}$$

for all $x\oplus y\in \mathcal{H}\oplus \mathcal{H}$, $H_0(V-\lambda )={(V-\lambda )}^{-1}\left(0\right)$ implies

$$H_0(S-\lambda )\subseteq {(S-\lambda )}^{-2}\left(0\right).$$

The reverse inclusion ${(S-\lambda )}^{-n}\left(0\right)\subseteq H_0(S-\lambda )$ being always true,

$$H_0(S-\lambda )={(S-\lambda )}^{-2}\left(0\right),\mathrm{all}\lambda \in \mathtt{C}.$$

Property $H_0\left(p\right)$ is inherited by restrictions to closed invariant subspaces ([23], Theorem 4.36). Hence “a 3-symmetric operator $T\in H_0\left(2\right)$”, i.e., $H_0(T-\lambda )={(T-\lambda )}^{-2}\left(0\right)$ for all $\lambda \in \mathtt{C}$.

(i)

If ${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$, then ${\delta }_{I-\overline{\lambda }{A}^{*},I-\lambda A}^m\left(I\right)=0={\delta }_{I-\alpha B,I-\overline{\alpha }{B}^{*}}^m\left(I\right)$ for all $\alpha \in \sigma \left(A\right)$ and $\lambda \in \sigma \left(B\right)$. Recall from ([36], Theorem 3) that if an operator $T\in B\left(\mathcal{H}\right)$ is m-symmetric for an even positive integer m, then it is $(m-1)$-symmetric. Hence we may assume that our operators $1-\lambda A$ and $I-\overline{\alpha }{B}^{*}$ are 3-symmetric (recall $m\le 4$), and therefore $H_0\left(2\right)$ operators. Since $H_0\left(p\right)$ operators are polaroid (of index p ([23], Corollary 4.37)), $1-\lambda A$ and $I-\overline{\alpha }{B}^{*}$, hence also A and B, are polaroid operators. This implies that if $\sigma \left(A\right)=\{{\alpha }_1,{\alpha }_2\}$ and $\sigma \left(B\right)=\{{\lambda }_1,{\lambda }_2\}$, then $A,B$ have the representation of the statement of the proposition.

A similar argument applied to ${\delta }_{{\delta }_{A^{,}{B}^{*}},{\delta }_{A,B}}^m\left(I\right)=0$ implies $A-\lambda$ and $B^{*}-\overline{\alpha }$, hence A and B, are polaroid operators. Consequently, A and B have the representation of the statement of the proposition.

(ii)

If ${\delta }_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$ (or, ${\delta }_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$), then ${\delta }_{A^{*}-\lambda ,I-\lambda A}^m\left(I\right)=0={\delta }_{B-\alpha ,\overline{\alpha }{B}^{*}-I}^m\left(I\right)$ (respectively, ${\delta }_{I-\overline{\lambda }{A}^{*},A-\lambda }^m\left(I\right)=0={\delta }_{I-\alpha B,\overline{\alpha }-B^{*}}^m\left(I\right)$); see Theorem 1. Furthermore, if $\lambda =-1$ (or $\alpha =1$ or $\beta$ is real), then

$$\sigma \left(A\right)=\{-1+\beta ,1\}\mathrm{and}\sigma \left(B\right)=\{-1,1-\beta \}$$

(in both the cases). Observe here that $\beta \ne 1$: for if it were, then we would have $0\in \sigma \left(A\right)$ and $0\in \sigma \left(B\right)$, which then forces $A-1$ and $B+1$ to be nilpotents (and by assumption $\sigma \left(A\right)\ne \left\{0\right\}\ne \sigma \left(B\right)$). Evidently, if $\alpha =1$, or $\lambda =-1$, then ${\delta }_{A^{*}+I,A+I}^4\left(I\right)=0={\delta }_{B-I,B^{*}-I}^4\left(I\right)$ implies that the operators $A,B$ are polaroid and have the stated representations. □

5. The Compact Case

We consider in the following the structure of operators $A,B$ such that the operator $D_{d_1,d_2}^m$ is compact. The argument below is patterned on that used to consider the cases ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m$ and ${\delta }_{{▵}_{A^{*},B^{*}}{▵}_{A,B}}^m$ in [17]. We start by recalling our standing hypothesis that $A,B$ are not nilpotent (indeed, $\sigma \left(A\right)\ne \left\{0\right\}\ne \sigma \left(B\right)$). If $D_{d_1,d_2}^m$ is compact, then the Fredholm spectrum ${\sigma }_e\left(D_{d_1,d_2}^m\right)$ of $D_{d_1,d_2}^m$ equals 0 [20,23], and this by the spectral mapping theorem implies [37,38]

$${\sigma }_e\left(D_{d_1,d_2}\right)=({\sigma }_e\left(d_1\right)-\sigma \left(d_2\right))\cup (\sigma \left(d_1\right)-{\sigma }_e\left(d_2\right))=\left\{0\right\},D=\delta$$

and

$${\sigma }_e\left(D_{d_1,d_2}\right)=1-({\sigma }_e\left(d_1\right)\sigma \left(d_2\right)\cup \sigma \left(d_1\right){\sigma }_e\left(d_2\right))=\left\{0\right\},D=▵$$

In either case, there exists a non-zero scalar $\beta$ (not necessarily the same for the two cases) such that

$${\sigma }_e\left(d_1\right)=\sigma \left(d_1\right)=\left\{\beta \right\}={\sigma }_e\left(d_2\right)=\sigma \left(d_2\right)ifD=\delta ,\mathrm{and}$$

$${\sigma }_e\left(d_1\right)=\sigma \left(d_1\right)=\left\{\beta \right\},{\sigma }_e\left(d_2\right)=\sigma \left(d_2\right)=\left\{{\displaystyle \frac{1}{\beta }}\right\},\mathrm{if}D=▵.$$

We prove that the operators $A,B$ are algebraic.

Proposition 4.

If $D_{d_1,d_2}^m$ is compact, then $A,B$ are algebraic.

Proof. 

By definition

$$\begin{array}{ccc}& & D_{d_1,d_2}^m={\delta }_{d_1,d_2}^m=\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)L_{d_1}^{m-j}{R}_{d_2}^j,\mathrm{and}\hfill \\ & & D_{d_1,d_2}^m={▵}_{d_1,d_2}^m=\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)L_{d_1}^j{R}_{d_2}^j.\hfill \end{array}$$

We claim that the sequences

$$\begin{array}{ccc}& & {\left\{L_{d_1}^{m-j}\right\}}_{j=0}^m,{\left\{R_{d_2}^j\right\}}_{j=0}^m\mathrm{if}D=\delta \mathrm{and}\hfill \\ & & {\left\{L_{d_1}^j\right\}}_{j=0}^m,{\left\{R_{d_2}^j\right\}}_{j=0}^m\mathrm{if}D=▵\hfill \end{array}$$

are linearly dependent. Suppose, for example, that the sequence ${\left\{R_{d_2}^j\right\}}_{j=0}^m$ is linearly independent. Then, since ${\sum }_{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)L_{d_1}^{m-j}{R}_{d_2}^j$ (similarly, ${\sum }_{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)L_{d_1}^j{R}_{d_2}^j$) is compact, the sequence ${\left\{L_{d_1}^{m-j}\right\}}_{j=0}^m$ (respectively, the sequence ${\left\{L_{d_1}^j\right\}}_{j=0}^m$) is compact. In particular, $L_{d_1}^0=I$ is compact. This being a contradiction, our claim is proved. The linear dependence of the sequences implies that the operators $L_{d_1}$ and $R_{d_2}$ are algebraic operators. Since the operator $L_E$ (or, $R_E$) is algebraic if and only if E is algebraic, and since (for non-nilpotent operators $A,B$) ${\delta }_{A,B}=L_A-R_B$ and ${▵}_{A,B}=I-L_A{R}_B$ are algebraic if and only if A and B are algebraic ([30], Proposition 2.6), the operators $A,B$ are algebraic. □

If $A,B$ are algebraic, and $\sigma \left({\delta }_{A,B}\right)=\left\{\beta \right\}$ (respectively, $\sigma \left({▵}_{A,B}\right)=\left\{\beta \right\}$) for some scalar $\beta$, then there exist finite sequences $a={\left\{a_j\right\}}_{j=1}^n$ and $b={\left\{b_j\right\}}_{j=1}^n$ such that

$$\begin{array}{ccc}& & \sigma \left({\delta }_{A,B}\right)=\sigma \left(A\right)-\sigma \left(b\right)={\{a_j-b_j\}}_{j=1}^n(=a-b)\hfill \\ & & (\mathrm{resp}.,\sigma \left({▵}_{A,B}\right)=1-\sigma \left(A\right)\sigma \left(B\right)={\{1-a_j{b}_j\}}_{j=1}^n(=1-ab).\hfill \end{array}$$

The following table lists spectra $\sigma \left(D_{d_1,d_2}\right)$ for various choices of $D,d_1$ and $d_2$.

There exists finite scalar sequences sequences ${\left\{a_j\right\}}_{j=1}^n,{\left\{b_j\right\}}_{j=1}^n$ and a scalar β such that if

(Ai) 

$D=\delta$, $d_1={\delta }_{A^{*},B^{*}}$ and

(a) 

$d_2={\delta }_{A,B}$, then $\sigma \left(d_1\right)={\left\{\overline{a_j-b_j}\right\}}_{j=1}^n=\left\{\beta \right\}={\{a_j-b_j\}}_{j=1}^n=\sigma \left(d_2\right)$;

(b) 

$d_2={▵}_{A,B}$, then $\sigma \left(d_1\right)={\left\{\overline{a_j-b_j}\right\}}_{j=1}^n=\left\{\beta \right\}={\{1-a_j{b}_j\}}_{j=1}^n=\sigma \left(d_2\right)$.

(Aii) 

$D=\delta$, $d_1={▵}_{A^{*},B^{*}}$ and

(a) 

$d_2={\delta }_{A,B}$, then $\sigma \left(d_1\right)={\{1-\overline{a_j{b}_j}\}}_{j=1}^n=\left\{\beta \right\}={\{a_j-b_j\}}_{j=1}^n=\sigma \left(d_2\right)$;

(b) 

$d_2={▵}_{A,B}$, then $\sigma \left(d_1\right)={\{1-\overline{a_j{b}_j}\}}_{j=1}^n=\left\{\beta \right\}={\{1-a_j{b}_j\}}_{j=1}^n=\sigma \left(d_2\right)$.

(Bi) 

$D=▵$, $d_1={▵}_{A^{*},B^{*}}$ and

(a) 

$d_2={▵}_{A,B}$, then $\sigma \left(d_1\right)={\{1-\overline{a_j{b}_j}\}}_{j=1}^n=\left\{\beta \right\}$, $\sigma \left(d_2\right)={\{1-a_j{b}_j\}}_{j=1}^n=\left\{{\displaystyle \frac{1}{\beta }}\right\}$;

(b) 

$d_2={\delta }_{A,B}$, then $\sigma \left(d_1\right)={\{1-\overline{a_j{b}_j}\}}_{j=1}^n=\left\{\beta \right\}$, $\sigma \left(d_2\right)={\{a_j-b_j\}}_{j=0}^n=\left\{{\displaystyle \frac{1}{\beta }}\right\}$.

(Bii) 

$D=▵$, $d_1={\delta }_{A^{*},B^{*}}$ and

(a) 

$d_2={▵}_{A,B}$, then $\sigma \left(d_1\right)={\left\{\overline{a_j-b_j}\right\}}_{j=1}^n=\left\{\beta \right\}$, $\sigma \left(d_2\right)={\{1-a_j{b}_j\}}_{j=1}^n=\left\{{\displaystyle \frac{1}{\beta }}\right\}$;

(b) 

$d_2={\delta }_{A,B}$, then $\sigma \left(d_1\right)={\left\{\overline{a_j-b_j}\right\}}_{j=1}^n=\left\{\beta \right\}$, $\sigma \left(d_2\right)={\{a_j-b_j\}}_{j=1}^n=\left\{{\displaystyle \frac{1}{\beta }}\right\}$.

The following theorem characterises operators $A,B$ such that $D_{d_1,d_2}^m$ is compact. Recall from ([23], p. 128) that a point $\lambda$ is a pole of A if and only if $H_0(A-\lambda )={(A-\lambda )}^{-n}\left(0\right)$ for some positive integer n.

Theorem 3.

Let $A,B\in B\left(\mathcal{H}\right)$ be such that $\sigma \left(A\right)\ne \left\{0\right\}\ne \sigma \left(B\right)$. The operator $D_{d_1,d_2}^m$ is compact if and only if the following conditions are satisfied.

(i) There exist finite sequences $a={\left\{a_j\right\}}_{j=1}^n\in \sigma \left(A\right)$ and $b={\left\{b_j\right\}}_{j=1}^n\in \sigma \left(B\right)$, not necessarily the same for the two cases under consideration, such that

$$\begin{array}{ccc}& & \sigma \left(d_1\right)={\left\{\overline{a_j-b_j}\right\}}_{j=1}^n\mathrm{if}{d}_1={\delta }_{A^{*},B^{*}}\mathrm{and}\sigma \left(d_1\right)={\{1-\overline{a_j{b}_j}\}}_{j=1}^n\mathrm{if}{d}_1={▵}_{A^{*},B^{*}},\hfill \\ & & \sigma \left(d_2\right)={\left\{a_j-b_j\right\}}_{j=1}^n\mathrm{if}{d}_2={\delta }_{A,B}\mathrm{and}\sigma \left(d_2\right)={\{1-a_j{b}_j\}}_{j=1}^n\mathrm{if}{d}_2={▵}_{A,B}.\hfill \end{array}$$

(ii) There exist decompositions

$$\begin{array}{ccc}& & \mathcal{H}={\oplus }_{j=1}^n{H}_0(A-a_j)={\oplus }_{j=1}^n{(A-a_j)}^{-r_j}\left(0\right),\mathcal{H}={\oplus }_{j=1}^n{H}_0(B-b_j)={\oplus }_{j=1}^n{(B-b_j)}^{-s_j}\left(0\right),\hfill \\ & & A={\oplus }_{j=1}^n{A}_j={\oplus }_{j=1}^n\left(A{|}_{H_0(A-a_j)}\right),B={\oplus }_{j=1}^n{B}_j={\oplus }_{j=1}^n\left(B{|}_{H_0(B-b_j)}\right)\hfill \end{array}$$

such that

$$A-a={\oplus }_{j=1}^n(A_j-a_j)\mathrm{and}B-b={\oplus }_{j=1}^n(B_j-b_j)$$

are nilpotent operators (of order $r=max\{r_j:1\le j\le n\}$ and $s=max\{s_j:1\le j\le n\}$, respectively).

Proof. 

The necessity of condition (i) having already been seen, we prove the necessity of condition (ii). The operators $A,B$ being algebraic are polaroid (i.e., all isolated points of the spectrum are poles of the resolvent of the operator [23], p. 299). Since $\sigma \left(A\right),\sigma \left(B\right)$ are finitely countable, (i) follows.

The proof of the sufficiency of the conditions requires some argument. However, the argument being very similar in all cases, we restrict ourselves to considering cases $D=▵,d_1={\delta }_{A^{*},B^{*}},d_2={▵}_{A,B}$ and $D=\delta ,d_1={▵}_{A^{*},B^{*}},d_2={\delta }_{A,B}$. (The cases $D=▵,d_1={\delta }_{A^{*},B^{*}},d_2={\delta }_{A,B}$ and $D=\delta ,d_1={▵}_{A^{*},B^{*}},d_2={▵}_{A,B}$ are proved in [17]).

The hypotheses imply that the spectral points $a_j$ and $b_j$ are poles of $A,B$, respectively. Hence A and B, consequently also $L_A$ and $R_B$, are polaroid operators. This, see ([27], Proposition 1), implies that the operators $L_A{R}_B$ and $L_A-R_B$ are polaroid. The functional calculus for polaroid operators ([23], p. 305) now tells us that the operators ${▵}_{A,B}-\mu$ and ${\delta }_{A,B}-\mu$ (hence, also ${▵}_{A^{*},B^{*}}-\mu$ and ${\delta }_{A^{*},B^{*}}-\mu$) are polaroid for all scalars $\mu$.

Considering first the case $D=▵,d_1={\delta }_{A^{*},B^{*}}$ and $d_2={▵}_{A,B}$, we have from the above that the operators ${\delta }_{A^{,}{B}^{*}}-\beta$ and ${▵}_{A,B}-{\displaystyle \frac{1}{\beta }}$ are polaroid. Since $\sigma \left({\delta }_{A^{*},B^{*}}\right)=\left\{\beta \right\}$ and $\sigma \left({▵}_{A,B}\right)=\left\{{\displaystyle \frac{1}{\beta }}\right\}$,

$$\sigma ({\delta }_{A^{*},B^{*}}-\beta )=\left\{0\right\}=\sigma ({▵}_{A,B}-{\displaystyle \frac{1}{\beta }})$$

and the operators ${\delta }_{A^{*},B^{*}}-\beta$ and ${▵}_{A,B}-{\displaystyle \frac{1}{\beta }}$ are nilpotents of some orders r and s, respectively. Set $r+s-1=t$. Then

$$\begin{array}{ccc}& & {▵}_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}=\left(I-L_{{\delta }_{A^{*},B^{*}}}{R}_{{▵}_{A,B}}\right)\hfill \\ & =& (-1)\left(L_{{\delta }_{A^{*},B^{*}}-\beta }{R}_{{▵}_{A,B}}+L_{\beta }{R}_{{▵}_{A,B}-{\displaystyle \frac{1}{\beta }}}\right)\hfill \\ & \mathrm{and}& \\ & & {▵}_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}^t={(-1)}^t\sum _{p=0}^t\left(\begin{array}{c}t\\ p\end{array}\right)R_{{▵}_{A,B}}^{t-p}{L}_{\beta }^p{L}_{{\delta }_{A^{*},B^{*}}-\beta }^{t-p}{R}_{{▵}_{A,B}-{\displaystyle \frac{1}{\beta }}}^p\hfill \\ & =& 0,\hfill \end{array}$$

since $R_{{▵}_{A,B}-{\displaystyle \frac{1}{\beta }}}^p=0$ for all $p\ge s$, and if $p\le s-1$ then $t-p\ge t-(s-1)=r$ and this implies $L_{{\delta }_{A^{*},B^{*}}-\beta }^{t-p}=0$. Trivially, being nilpotent, ${▵}_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}$ is compact.

To complete the proof of the theorem, we observe that if $D=\delta ,d_1={▵}_{A^{*},B^{*}}$ and $d_2={\delta }_{A,B}$, then there exist positive integers r and s such that

$${\left({▵}_{A^{*},B^{*}}-\beta \right)}^r=0={\left({\delta }_{A,B}-\beta \right)}^s$$

and

$$\begin{array}{ccc}& & {\left({\delta }_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}\right)}^{r+s-1}={\left(L_{{▵}_{A^{*},B^{*}}}-R_{{\delta }_{A,B}}\right)}^{r+s-1}\hfill \\ & =& {\left(L_{{▵}_{A^{*},B^{*}}-\beta }-R_{{\delta }_{A,B}-\beta }\right)}^{r+s-1}\hfill \\ & =& \sum _{p=0}^{r+s-1}{(-1)}^p\left(\begin{array}{c}r+s-1\\ p\end{array}\right)L_{{▵}_{A^{*},B^{*}}-\beta }^{r+s-1-p}{R}_{{\delta }_{A,B}-\beta }^p\hfill \\ & =0& ,\hfill \end{array}$$

since $R_{{\delta }_{A,B}-\beta }^p=0$ for all $p\ge s$, and if $p\le s-1$, equivalently $r+s-1-p\ge r$, then $L_{{▵}_{A^{*},B^{*}}-\beta }^r=0$. □