Generalized Core Inverse with Respect to a Pair of Elements

1. Introduction

A Banach algebra $\mathcal{A}$ is called a Banach *-algebra if there exists an involution $*:x\to x^{*}$ satisfying ${(x+y)}^{*}=x^{*}+y^{*},{\left(\lambda x\right)}^{*}=\overline{\lambda }{x}^{*},{\left(xy\right)}^{*}=y^{*}{x}^{*},{\left(x^{*}\right)}^{*}=x$. An element $a\in \mathcal{A}$ has group inverse if there exists $x\in \mathcal{A}$ such that

$$x{a}^2=a,a{x}^2=x,ax=xa.$$

Such x is unique if exists, denoted by $a^{\#}$, and called the group inverse of a (see [18]). An element a in a Banach *-algebra $\mathcal{A}$ has core inverse if there exists $x\in \mathcal{A}$ such that

$$a{x}^2=x,x{a}^2=a,x{a}^2=a.$$

If such x exists, it is unique, and denote it by $a^{\overline{)\#}}$ (see [1,18]). In [12], Gao and Chen extended the core inverse and introduced the core-EP inverse (i.e., pseudo core inverse). An element $a\in \mathcal{A}$ has core-EP inverse if there exist $x\in \mathcal{A}$ and $k\in \mathbb{N}$ such that

$$a{x}^2=x,{\left(ax\right)}^{*}=ax,x{a}^{k+1}=a^k.$$

If such x exists, it is unique, and denote it by $a^{Ⓓ}$.

Recently, the authors introduced and studied generalized core inverse for an element in a Banach *-algebra. An element $a\in \mathcal{A}$ has generalized core inverse if there exists a $x\in \mathcal{A}$ such that

$$x=a{x}^2,{\left(ax\right)}^{*}=ax,\underset{n\to \infty }{lim}\left|\right|a^n-x{a}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

Such x is unique if it exists, and denote it by $a^{ⓓ}$ (see [5,6]). This new construct generalizes the core-EP inverse, extending its reach from complex matrices to operators on infinite-dimensional Hilbert spaces. Consequently, many classical properties of the core-EP inverse are now generalized to this broader algebraic setting.

An element $a\in R$ has $(b,c)$-inverse if there exists an element $x\in R$ such that

$$x\in b\mathcal{A}x\bigcap x\mathcal{A}c,xab=b,cax=c.$$

If such x exists, it is unique and is denoted by $a^{(b,c)}.$ It is evident that $x=a^{(b,c)}\mathrm{if} \mathrm{and} \mathrm{only} \mathrm{if}x=xax,x\mathcal{A}=b\mathcal{A},\mathcal{A}x=\mathcal{A}c.$ Many authors studied the $(b,c)$-inverse from very different point-views, e.g., [8,9,10,11,21,22,24,25].

In [3], the author investigated when the core inverse of a Banach element coincides with its $(b,c)$-inverse.

Definition 1.1.

An element $a\in \mathcal{A}$ has core-$(b,c)$-inverse if $a\in {\mathcal{A}}^{\overline{)\#}}$ and $a^{\overline{)\#}}=a^{(b,c)}$. We use $a^{{\overline{)\#}}_{b,c}}$ to stands for $a^{\overline{)\#}}$. The set of all $(b,c)$-core invertible elements in $\mathcal{A}$ is denoted by ${\mathcal{A}}_{b,c}^{\overline{)\#}}$.

We list several characterizations of core-$(b,c)$-inverse.

Theorem 1.2.(see [3][Theorem 2.1, Theorem 2.6 and Theorem 2.8]) Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in R_{b,c}^{\overline{)\#}}$.

(2)

$a\in R^{\overline{)\#}}$ and $aR=bR,R{a}^{*}=Rc.$

(3)

$a\in R^{(b,c)}$ and $aR=bR,R{a}^{*}=Rc.$

In particular, for $b=c$,

(4)

$a,b,ba\in R^{\overline{)\#}}$, $aR=bR,{\left(ba\right)}^{\overline{)\#}}=a^{\overline{)\#}}{b}^{\overline{)\#}}.$

(5)

$a,b\in R^{\overline{)\#}}$ and $a{a}^{\overline{)\#}}=b^{\overline{)\#}}b.$

The motivation of this paper is to study when the generalized core inverse of a Banach element coincides with its $(b,c)$-inverse. We adopt

Definition 1.3.

An element $a\in \mathcal{A}$ has generalized core-$(b,c)$-inverse if $a\in {\mathcal{A}}^{ⓓ}$ and $a^{ⓓ}=a^{(b,c)}$. We use $a_{b,c}^{ⓓ}$ to stands for $a^{ⓓ}$. The set of all generalized $(b,c)$-core invertible elements in $\mathcal{A}$ is denoted by ${\mathcal{A}}_{b,c}^{ⓓ}$.

In Section 2, this generalized inverse is examined through a novel limit-based approach. We prove that an element a has a generalized core-$(b,c)$-inverse if and only if it possesses a kind of decomposition combining the core-$(b,c)$-inverse with quasinilpotent. Some properties of the generalized core-$(b,c)$-inverse are thereby obtained.

Recall that an element $a\in \mathcal{A}$ has generalized Drazin inverse if there exists $x\in \mathcal{A}$ such that $a{x}^2=x,ax=xa,a-a^{2}x\in {\mathcal{A}}^{qnil}.$ Here, ${\mathcal{A}}^{qnil}=\{a\in \mathcal{A}\mid 1+\lambda a\in {\mathcal{A}}^{-1}\}.$ Such x is unique, if exists, and denote it by $a^d$. Evidently, $a\in \mathcal{A}$ has generalized Drazin inverse if and only if it has quasi-polar property, i.e., there exists an idempotent $p\in \mathcal{A}$ such that $ap=pa,a+p\in \mathcal{A}$ is invertible and $ap\in {\mathcal{A}}^{qnil}$. In Section 3, we present the polar-like property for the generalized core-$(b,c)$-inverse.

Finally, in Section 4, we are concerned with image-based representations for the generalized core-$(b,c)$-inverse. We characterize the generalized core-$(b,c)$-inverse of an element via the core-$(b,c)$-inverse of its generalized Drazin inverse. In particular, the relation between the generalized core-$(b,b)$-inverse and the reverse order law for the generalized core inverse are thereby obtained.

Throughout the paper, all Banach *-algebras are complex with an identity. Let ${\mathbb{C}}^{n\times n}$ be the Banach algebra of all $n\times n$ complex matrices with conjugate transpose *. $\ell \left(x\right)$ and $r\left(x\right)$ stand for the left and right annihilators of $x\in \mathcal{A}$, respectively.

2. Generalized Core-($b,c$)-Inverses

The aim of this section is to introduce the notion of the generalized core-$(b,c)$-inverse in a Banach *-algebra. We begin with

Theorem 2.1.

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in \mathcal{A}$ has generalized core-$(b,c)$-inverse.

(2)

There exist $x,y\in \mathcal{A}$ such that

$$a=x+y,x^{*}y=yx=0,x\in {\mathcal{A}}_{b,c}^{\overline{)\#}},y\in {\mathcal{A}}^{qnil}.$$

(3)

There exists $x\in {\mathcal{A}}_{b,c}^{\overline{)\#}}$ such that

$$x=a{x}^2,{\left(ax\right)}^{*}=ax,\underset{n\to \infty }{lim}\left|\right|a^n-x{a}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, $a\in {\mathcal{A}}^{ⓓ}$ and $a^{ⓓ}=a^{(b,c)}$. Then we verify that

$$\begin{array}{ccc}\hfill a^{ⓓ}& =\hfill & a^{ⓓ}\left[a{a}^{ⓓ}a\right]a^{ⓓ},\hfill \\ \hfill a^{ⓓ}& \in \hfill & b\mathcal{A}\bigcap \mathcal{A}c,\hfill \\ \hfill b& =\hfill & a^{ⓓ}\left[a{a}^{ⓓ}a\right]b,\hfill \\ \hfill c& =\hfill & c\left[a{a}^{ⓓ}a\right]a^{ⓓ}.\hfill \end{array}$$

Hence, $a^{ⓓ}={\left(a{a}^{ⓓ}a\right)}^{(b,c)}$. Thus, $a{a}^{ⓓ}a\in {\mathcal{A}}^{(b,c)}$. In view of [5][Theorem 2.1], there exist $x,y\in \mathcal{A}$ such that

$$a=x+y,x^{*}y=yx=0,x\in {\mathcal{A}}^{\overline{)\#}},y\in {\mathcal{A}}^{qnil}.$$

Moreover, we have

$$x^{\overline{)\#}}=a^{ⓓ}={\left(a{a}^{ⓓ}a\right)}^{(b,c)}=x^{(b,c)}.$$

Therefore $x\in {\mathcal{A}}_{b,c}^{\overline{)\#}}$, as desired.

$\left(2\right)\Rightarrow \left(1\right)$ By assumption, there exist $x,y\in \mathcal{A}$ such that

$$a=x+y,x^{*}y=yx=0,x\in {\mathcal{A}}_{b,c}^{\overline{)\#}},y\in {\mathcal{A}}^{qnil}.$$

Here, $x=a{a}^{\overline{)\#}}a$. Then $x\in {\mathcal{A}}^{\overline{)\#}}$. In view of [5] [Theorem 2.1], $a\in {\mathcal{A}}^{ⓓ}$ and

$$a^{ⓓ}=x^{\overline{)\#}}=x^{(b,c)}={\left(a{a}^{ⓓ}a\right)}^{(b,c)}.$$

Thus, we have

$$\begin{array}{ccc}\hfill a^{ⓓ}& =\hfill & a^{ⓓ}\left[a{a}^{ⓓ}a\right]a^{ⓓ},\hfill \\ \hfill a^{ⓓ}& \in \hfill & b\mathcal{A}\bigcap \mathcal{A}c,\hfill \\ \hfill b& =\hfill & a^{ⓓ}\left[a{a}^{ⓓ}a\right]b,\hfill \\ \hfill c& =\hfill & c\left[a{a}^{ⓓ}a\right]a^{ⓓ}.\hfill \end{array}$$

Since $a^{ⓓ}=a^{ⓓ}a{a}^{ⓓ}$, we have

$$\begin{array}{ccc}\hfill a^{ⓓ}& =\hfill & a^{ⓓ}a{a}^{ⓓ},\hfill \\ \hfill a^{ⓓ}& \in \hfill & b\mathcal{A}\bigcap \mathcal{A}c,\hfill \\ \hfill b& =\hfill & a^{ⓓ}ab,\hfill \\ \hfill c& =\hfill & ca{a}^{ⓓ}.\hfill \end{array}$$

Therefore $a^{ⓓ}=a^{(b,c)}$, as required.

$\left(2\right)\Rightarrow \left(3\right)$ By hypothesis, there exist $z,y\in \mathcal{A}$ such that

$$a=z+y,z^{*}y=yz=0,z\in {\mathcal{A}}_{b,c}^{\overline{)\#}},y\in {\mathcal{A}}^{qnil}.$$

Set $x=z_{b,c}^{\overline{)\#}}$. Then

$$x\in {\mathcal{A}}^{\overline{)\#}},x\mathcal{A}=b\mathcal{A},\mathcal{A}{x}^{*}=\mathcal{A}c.$$

This implies that

$$x^{\overline{)\#}}\in {\mathcal{A}}^{\overline{)\#}},x^{\overline{)\#}}\mathcal{A}=x\mathcal{A}=b\mathcal{A},\mathcal{A}{\left(x^{\overline{)\#}}\right)}^{*}=\mathcal{A}{x}^{*}=\mathcal{A}c.$$

By using Theorem 1.2, $x\in {\mathcal{A}}_{b,c}^{\overline{)\#}}$. One easily checks that

$$\begin{array}{ccc}\hfill ax& =\hfill & (z+y)z_{b,c}^{\overline{)\#}}=z{z}_{b,c}^{\overline{)\#}}+yz{\left(z_{b,c}^{\overline{)\#}}\right)}^2=z{z}_{b,c}^{\overline{)\#}},\hfill \\ \hfill a{x}^2& =\hfill & \left(ax\right)x=z{\left(z_{b,c}^{\overline{)\#}}\right)}^2=z_{b,c}^{\overline{)\#}}=x,\hfill \\ \hfill z_{b,c}^{\overline{)\#}}y& =\hfill & xy=xzxy=x\left(zx\right)y=x{\left(zx\right)}^{*}y\hfill \\ & =\hfill & x{x}^{*}\left(z^{*}y\right)=0.\hfill \end{array}$$

By using $z_{b,c}^{\overline{)\#}}y=0$ and [26] [Lemma 2.3], we derive that

$$axa=\left(ax\right)a=z{z}_{b,c}^{\overline{)\#}}(z+y)=z{z}_{b,c}^{\overline{)\#}}z=z.$$

Then

$$\begin{array}{c}{\left(ax\right)}^{*}={\left(z{z}_{b,c}^{\overline{)\#}}\right)}^{*}=z{z}_{b,c}^{\overline{)\#}}=ax,\\ a(1-xa)=a-axa=a-z=y\in {\mathcal{A}}^{qnil}.\end{array}$$

By using Cline’s formula, $a-x{a}^2=(1-xa)a\in {\mathcal{A}}^{qnil}$. As $yz=0$, we see that

$$(a-x{a}^2)z=[z+y-z_{b,c}^{\overline{)\#}}{(z+y)}^2]z=(z-z_{b,c}^{\overline{)\#}}{z}^2)z=0.$$

Thus we have

$$\begin{array}{ccc}\hfill \left|\right|a^n-x{a}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}& =\hfill & \left|\right|(a-x{a}^2)a^{n-1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ & =\hfill & \left|\right|(a-x{a}^2){(z+y)}^{n-1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ & =\hfill & \left|\right|(a-x{a}^2)y^{n-1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ & \le \hfill & \left|\right|a-x{a}^2{\left|\right|}^{{\displaystyle \frac{1}{n}}}\left[\right||y^{n-1}{\left|\right|}^{{\displaystyle \frac{1}{n-1}}}{]}^{1-{\displaystyle \frac{1}{n}}}.\hfill \end{array}$$

Since $y\in {\mathcal{A}}^{qnil}$, we deduce that

$$\underset{n\to \infty }{lim}\left|\right|a^n-x{a}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0,$$

as required.

$\left(3\right)\Rightarrow \left(2\right)$ By hypotheses, we have $z\in {\mathcal{A}}_{b,c}^{\overline{)\#}}$ such that

$$z=a{z}^2,{\left(az\right)}^{*}=az,\underset{n\to \infty }{lim}\left|\right|a^n-z{a}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

For any $n\in \mathbb{N}$, we have

$$\begin{array}{ccc}\hfill az& =\hfill & a\left(a{z}^2\right)=a^2{z}^2=a^2\left(a{z}^2\right)z=a^3{z}^3\hfill \\ & =\hfill & \dots =a^n{z}^n=\dots =a^{n+1}{z}^{n+1}.\hfill \end{array}$$

Hence

$$\begin{array}{ccc}\hfill \left|\right|z-zaz\left|\right|& =\hfill & \left|\right|\left(az\right)z-zaz\left|\right|\hfill \\ & =\hfill & \left|\right|\left(a^n{z}^n\right)z-z\left(a^{n+1}{z}^{n+1}\right)\left|\right|\hfill \\ & =\hfill & \left|\right|(a^n-z{a}^{n+1})z^{n+1}\left|\right|.\hfill \end{array}$$

Then

$$\begin{array}{ccc}\hfill \left|\right|z-{zaz\left|\right|}^{{\displaystyle \frac{1}{n}}}& \le \hfill & \left|\right|(a^n-z{a}^{n+1}){\left|\right|}^{{\displaystyle \frac{1}{n}}}{\left|\right|z\left|\right|}^{1+{\displaystyle \frac{1}{n}}}.\hfill \end{array}$$

We infer that

$$\underset{n\to \infty }{lim}\left|\right|z-{zaz\left|\right|}^{{\displaystyle \frac{1}{n}}}=0,$$

hence, $z=zaz$.

Set $x=aza$ and $y=a-aza.$ Then $a=x+y$. As in the proceeding discussion, we have $az=z{a}^{2}z$. We claim that x has core inverse. Evidently, we verify that

$$\begin{array}{ccc}\hfill z{x}^2& =\hfill & za\left(z{a}^{2}z\right)a=z{a}^{2}za=aza=x,\hfill \\ \hfill x{z}^2& =\hfill & aza{z}^2=a{z}^2=z,\hfill \\ \hfill {\left(xz\right)}^{*}& =\hfill & {\left(azaz\right)}^{*}={\left(az\right)}^{*}=az=\left(aza\right)z=xz.\hfill \end{array}$$

Thus, $x^{\overline{)\#}}=z$.

We verify that

$$\begin{array}{ccc}\hfill \left|\right|{(a-z{a}^2)}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n+1}}}& =\hfill & \left|\right|{(a-z{a}^2)}^n(a-z{a}^2){\left|\right|}^{{\displaystyle \frac{1}{n+1}}}\hfill \\ & =\hfill & \left|\right|{(a-z{a}^2)}^{n-1}(a-z{a}^2)a{\left|\right|}^{{\displaystyle \frac{1}{n+1}}}\hfill \\ & =\hfill & \left|\right|{(a-z{a}^2)}^{n-1}{a}^2{\left|\right|}^{{\displaystyle \frac{1}{n+1}}}\hfill \\ & ⋮\hfill & \\ & =\hfill & \left|\right|(a-z{a}^2)a^n{\left|\right|}^{{\displaystyle \frac{1}{n+1}}}\hfill \\ & \le \hfill & {\left[\left|\right|a^n-z{a}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}\right]}^{{\displaystyle \frac{n}{n+1}}}\left|\right|a^n{\left|\right|}^{{\displaystyle \frac{1}{n+1}}}.\hfill \end{array}$$

Accordingly,

$$\underset{n\to \infty }{lim}\left|\right|{(a-z{a}^2)}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n+1}}}=0.$$

This implies that $a-z{a}^2\in {\mathcal{A}}^{qnil}$. By using Cline’s formula (see [17] [Theorem 2.1]), $y=a-aza\in {\mathcal{A}}^{qnil}$. Moreover, we see that

$$\begin{array}{ccc}\hfill x^{*}y& =\hfill & {\left(aza\right)}^{*}(1-az)a=a^{*}{\left(az\right)}^{*}(1-az)a\hfill \\ & =\hfill & a^{*}{\left(az\right)}^{*}(1-az)a=0,\hfill \\ & =\hfill & a^{*}\left(az\right)(1-az)a=0,\hfill \\ \hfill yx& =\hfill & (a-aza)aza=a(a-z{a}^2)za=0.\hfill \end{array}$$

Accordingly, we have

$$z^{\overline{)\#}}={\left[x^{\overline{)\#}}\right]}^{\overline{)\#}}=x^2{x}^{\overline{)\#}}.$$

By hypothesis, we have

$$x^2{x}^{\overline{)\#}}=z^{(b,c)}.$$

Claim 1. $z\in b\mathcal{A}z\bigcap z\mathcal{A}c.$

Obviously,

$$x^2{x}^{\overline{)\#}}={\left(aza\right)}^{2}z=az{a}^{2}z=a^{2}z.$$

Thus,

$$z=\left(a^{2}z\right)z^3=\left[x^2{x}^{\overline{)\#}}\right]z^3\in b\mathcal{A}z$$

and

$$z=zaz=z^2{a}^{2}z=z^2\left[x^2{x}^{\overline{)\#}}\right]\in z\mathcal{A}c.$$

Claim 2. $zxb=b$.

We claim that

$$b=x^2{x}^{\overline{)\#}}zb={\left(aza\right)}^2{z}^{2}b=azaz=azb.$$

By hypothesis, we have

$$b=azb=a{x}^{\overline{)\#}}b=(x+y)x^{\overline{)\#}}b=x{x}^{\overline{)\#}}b.$$

Hence $x^{\overline{)\#}}xb=x^{\overline{)\#}}x\left[x{x}^{\overline{)\#}}b\right]=[\left(x^{\overline{)\#}}{x}^2\right)x^{\overline{)\#}}b=x{x}^{\overline{)\#}}b=b.$ Thus, we see that

$$\begin{array}{ccc}\hfill zab& =\hfill & x^{\overline{)\#}}(x+y)b\hfill \\ & =\hfill & x^{\overline{)\#}}x{x}^{\overline{)\#}}(x+y)b\hfill \\ & =\hfill & x^{\overline{)\#}}{\left(x{x}^{\overline{)\#}}\right)}^{*}(x+y)b\hfill \\ & =\hfill & x^{\overline{)\#}}xb=b.\hfill \end{array}$$

Therefore

$$zxb=z\left(aza\right)b=\left(zaz\right)ab=zab=b.$$

Claim 3. $cxz=c$.

We verify that

$$c=cz{x}^2{x}^{\overline{)\#}}=cz\left(aza\right)\left(aza\right)z=cz{a}^{2}z=caz.$$

Hence, $cxz=c\left(aza\right)z=ca\left(zaz\right)=caz=c.$ We infer that $z=x^{(b,c)}$, and then $x\in {\mathcal{A}}_{b,c}^{\overline{)\#}}$ and $x_{b,c}^{\overline{)\#}}=z$.

Accordingly, we have a generalized core-$(b,c)$-decomposition $a=x+y$, thus yielding the result. □

Corollary 2.2.

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in \mathcal{A}$ has generalized core-$(b,c)$-decomposition.

(2)

There exists a unique element $x\in {\mathcal{A}}_{b,c}^{\overline{)\#}}$ such that

$$x=a{x}^2,{\left(ax\right)}^{*}=ax,\underset{n\to \infty }{lim}\left|\right|a^n-x{a}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

In this case, $a_{b,c}^{ⓓ}=x$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 2.1, there exists $x\in \mathcal{A}$ such that

$$x=a{x}^2,{\left(ax\right)}^{*}=ax,\underset{n\to \infty }{lim}\left|\right|a^n-x{a}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

Then $a\in {\mathcal{A}}^{ⓓ}$ and $x=a^{ⓓ}$. Therefore x is unique by [5] [Corollary 2.2].

$\left(2\right)\Rightarrow \left(1\right)$ This is clear by Theorem 2.1. □

Corollary 2.3.

Let $a=x+y$ be the generalized core-$(b,c)$-decomposition of an element a in $\mathcal{A}$. Then

$$a_{b,c}^{ⓓ}=x_{b,c}^{\overline{)\#}}.$$

Proof. 

Let $a=x+y$ be the generalized core decomposition of $a\in \mathcal{A}$. By virtue of Theorem 2.1, $x_{b,c}^{\overline{)\#}}$ is the generalized core inverse of a, as required. □

Let ${\mathbb{C}}^{n\times n}$ be the Banach algebra of all $n\times n$ complex matrices, with conjugate transpose as the involution. For a complex $A\in {\mathbb{C}}^{n\times n}$, it follows by [5] [Corollary 3.4] that the core-EP inverse and generalized core inverse coincide with each other. Then $A_{(B,C)}^{Ⓓ}=A_{(B,C)}^{ⓓ}$.

Theorem 2.4.

Let $a\in {\mathcal{A}}_{b,c}^{ⓓ}$ and $\alpha \in {\mathcal{A}}^{ⓓ}$. Then $\alpha \in {\mathcal{A}}_{b,c}^{ⓓ}$ if and only if

(1)

$1+(\alpha -a)a^{ⓓ}\in R^{-1}$;

(2)

${\alpha }^{ⓓ}[1-a{a}^{ⓓ}]=[1-{\alpha }^{ⓓ}\alpha ]a^{ⓓ}.$

Proof. 

⟹ Since $a,\alpha \in {\mathcal{A}}_{b,c}^{ⓓ}$, we have

$$a^{ⓓ}=a^{(b,c)},{\alpha }^{ⓓ}={\alpha }^{(b,c)}.$$

By virtue of [14] [Theorem 2.19], we have

$$1+(\alpha -a)a^{ⓓ}\in R^{-1}.$$

Furthermore,

$${\alpha }^{ⓓ}=a^{ⓓ}{[1+(\alpha -a)a^{ⓓ}]}^{-1}.$$

Thus

$${\alpha }^{ⓓ}[1+(\alpha -a)a^{ⓓ}]=a^{ⓓ}.$$

This implies that

$${\alpha }^{ⓓ}[1-a{a}^{ⓓ}]+{\alpha }^{ⓓ}\alpha a^{ⓓ}=a^{ⓓ}.$$

Then

$${\alpha }^{ⓓ}[1-a{a}^{ⓓ}]=[1-{\alpha }^{ⓓ}\alpha ]a^{ⓓ}.$$

⟸ Since $1+(\alpha -a)a^{ⓓ}\in R^{-1}$, by virtue of [14] [Theorem 2.19], $\alpha$ has the $(b,c)$-inverse x, where

$$x=a^{ⓓ}{[1+(\alpha -a)a^{ⓓ}]}^{-1}.$$

By hypothesis, we have ${\alpha }^{ⓓ}[1-a{a}^{ⓓ}]=[1-{\alpha }^{ⓓ}\alpha ]a^{ⓓ}.$ This implies that

$$a^{ⓓ}{[1+(\alpha -a)a^{ⓓ}]}^{-1}={\alpha }^{ⓓ}.$$

Then ${\alpha }^{ⓓ}={\alpha }^{(b,c)}.$ Therefore $\alpha \in {\mathcal{A}}_{b,c}^{ⓓ},$ as asserted. □

Corollary 2.5.

Let $a\in {\mathcal{A}}_{b,c}^{ⓓ},j\in J\left(\mathcal{A}\right)$ and let $\alpha =a+j$. If $a^{\pi }j=0$, then $\alpha \in {\mathcal{A}}_{b,c}^{ⓓ}$ and ${\alpha }_{b,c}^{ⓓ}={(1+a_{b,c}^{ⓓ}j)}^{-1}{a}_{b,c}^{ⓓ}.$

Proof. 

Since $a\in {\mathcal{A}}_{b,c}^{ⓓ}$, by virtue of Theorem 2.1, there exist $x\in {\mathcal{A}}_{b,c}^{\overline{)\#}}$ and $y\in {\mathcal{A}}^{qnil}$ such that $a=x+y,x^{*}y=0,yx=0.$ Exactly, $x=a{a}^{ⓓ}a,y=a-a{a}^{ⓓ}a$. Then $\alpha =(x+j)+y$. Since $a^{\pi }j=0$, we see that $(1-a{a}^d)j=0$. It is easy to verify that $(1-a{a}^{ⓓ})j=j-a{a}^{ⓓ}a{a}^{d}j=j-a{a}^{d}j=0$. Hence, we have $(1-(x+y)x^{\overline{)\#}})j=(1-x{x}^{\overline{)\#}})j=0.$ In view of [16] [Theorem 3.1], $x+j\in {\mathcal{A}}^{\overline{)\#}}$ and

$${(x+j)}^{\overline{)\#}}={(1+x^{\overline{)\#}}j)}^{-1}{x}^{\overline{)\#}}.$$

Moreover, we check that $y(x+j)=yx+yj=a(1-a^{ⓓ}a)j=0$ and

$${(x+j)}^{*}y=x^{*}y+j^{*}y=j^{*}(a-a{a}^{ⓓ}a)={\left[a^{*}(1-a{a}^{ⓓ})j\right]}^{*}=0.$$

In light of [5] [Theorem 2.1], $\alpha =a+j\in {\mathcal{A}}^{ⓓ}$ and ${\alpha }^{ⓓ}={(x+j)}^{\overline{)\#}}={(1+a^{ⓓ}j)}^{-1}{a}^{ⓓ}$. Clearly, we have

$$1+(\alpha -a)a^{ⓓ}=1+j{a}^{ⓓ}\in R^{-1}.$$

One directly verifies that

$$\begin{array}{ccc}\hfill {\alpha }^{ⓓ}[1-a{a}^{ⓓ}]& =\hfill & {(1+a^{ⓓ}j)}^{-1}{a}^{ⓓ}[1-a{a}^{ⓓ}]=0,\hfill \\ \hfill \left(1-{\alpha }^{ⓓ}\alpha \right)a^{ⓓ}& =\hfill & \left(1-{(1+a^{ⓓ}j)}^{-1}{a}^{ⓓ}(a+j)\right)a^{ⓓ}\hfill \\ & =\hfill & {(1+a^{ⓓ}j)}^{-1}[(1+a^{ⓓ}j)-a^{ⓓ}(a+j)]a^{ⓓ}\hfill \\ & =\hfill & {(1+a^{ⓓ}j)}^{-1}[1-a^{ⓓ}a]a^{ⓓ}=0.\hfill \end{array}$$

Hence, ${\alpha }^{ⓓ}[1-a{a}^{ⓓ}]=[1-{\alpha }^{ⓓ}\alpha ]a^{ⓓ}.$ By virtue of Theorem 2.4, we complete the proof. □

Recall that an element a in $\mathcal{A}$ is generalized EP (i.e., a generalized EP element) if there exist $x,y\in \mathcal{A}$ such that $a=x+y,x^{*}y=yx=0,x\in \mathcal{A}\mathrm{is} \mathrm{EP},y\in {\mathcal{A}}^{qnil}$ (see [7]). We characterize the generalized EP element by the generalized core-$(b,c)$-invertibility.

Theorem 2.6.

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in \mathcal{A}$ is generalized EP.

(2)

$a\in {\mathcal{A}}_{({\left(a^d\right)}^{*},a^d)}^{ⓓ}$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ In view of [7] [Theorem 2.8], we have $a\in {\mathcal{A}}^{ⓓ}$ and $a^{ⓓ}=a^d$. We directly check that

$$\begin{array}{ccc}\hfill a^{ⓓ}& =\hfill & a^{ⓓ}\left(a{a}^{ⓓ}a\right)a^{ⓓ},\hfill \\ \hfill {\left(a^d\right)}^{*}& =\hfill & a^{ⓓ}\left(a{a}^{ⓓ}a\right){\left(a^d\right)}^{*},a^d=a^d\left(a{a}^{ⓓ}a\right)a^{ⓓ},\hfill \end{array}$$

Therefore $a\in {\mathcal{A}}_{({\left(a^d\right)}^{*},a^d)}^{ⓓ}$.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, we have

$$\begin{array}{ccc}\hfill a^{ⓓ}& =\hfill & a^{ⓓ}\left(a{a}^{ⓓ}a\right)a^{ⓓ},\hfill \\ \hfill {\left(a^d\right)}^{*}& =\hfill & a^{ⓓ}\left(a{a}^{ⓓ}a\right){\left(a^d\right)}^{*},a^d=a^d\left(a{a}^{ⓓ}a\right)a^{ⓓ},\hfill \end{array}$$

Hence, we derive

$$\begin{array}{ccc}\hfill a^{ⓓ}-a^d& =\hfill & a^n{\left(a^{ⓓ}\right)}^{n+1}-a^{d}a{a}^{ⓓ}\hfill \\ & =\hfill & [a^n-a^d{a}^{n+1}]{\left(a^{ⓓ}\right)}^{n+1}.\hfill \end{array}$$

This implies that $a^{ⓓ}=a^d$. According to [7] [Theorem 2.8], $a\in \mathcal{A}$ is generalized EP, as asserted. □

An element a in a Banach *-algebra $\mathcal{A}$ is *-DMP (i.e., *-DMP element) if there exist $m\in \mathbb{N}$ and $x\in \mathcal{A}$ such that $a{x}^2=x,{\left(ax\right)}^{*}=xa=ax,a^m=x{a}^{m+1}$ (see [13]).

Corollary 2.7.

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in \mathcal{A}$ is *-DMP.

(2)

$a\in {\mathcal{A}}_{({\left(a^D\right)}^{*},a^D)}^{Ⓓ}$.

Proof. 

As is well known, $a\in \mathcal{A}$ is *-DMP if and only $a\in \mathcal{A}$ is generalized EP and it has Drazin inverse. Therefore we complete the proof by Theorem 2.6. □

3. Polar-like Properties

We investigate polar-like properties for a generalized core-$(b,c)$-invertibility. We now derive

Theorem 3.1.

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_{b,c}^{ⓓ}$ if and only if the following two conditions hold:

(1)

$a\in {\mathcal{A}}^d$;

(2)

there exists a projection $p\in \mathcal{A}$ such that

$$a+p\in {\mathcal{A}}^{-1},1-p\in {\mathcal{A}}_{b,c}^{\overline{)\#}},pa=pap\in {\mathcal{A}}^{qnil}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Since $a\in {\mathcal{A}}_{b,c}^{ⓓ}$, by using Theorem 2.1, there exist $x,y\in \mathcal{A}$ such that

$$a=x+y,x^{*}y=yx=0,x\in {\mathcal{A}}_{b,c}^{\overline{)\#}},y\in {\mathcal{A}}^{qnil}.$$

In view of [5] [Theorem 2.5], $a\in {\mathcal{A}}^d$. By virtue of [26] [Lemma 2.3], we have

$$\begin{array}{c}x=x{x}_{b,c}^{\overline{)\#}}x,x_{b,c}^{\overline{)\#}}x{x}_{b,c}^{\overline{)\#}}=x_{b,c}^{\overline{)\#}},\\ x_{b,c}^{\overline{)\#}}{x}^2=x,x{\left(x_{b,c}^{\overline{)\#}}\right)}^2=x_{b,c}^{\overline{)\#}},{\left(x{x}_{b,c}^{\overline{)\#}}\right)}^{*}=x{x}_{b,c}^{\overline{)\#}}.\end{array}$$

Let $p=1-x{x}_{b,c}^{\overline{)\#}}$. Then $p^2=p=p^{*}$ and $px=0$. We directly check that

$$\begin{array}{c}(x+1-x{x}_{b,c}^{\overline{)\#}})(x_{b,c}^{\overline{)\#}}+1-x_{b,c}^{\overline{)\#}}x)=1\\ =(x_{b,c}^{\overline{)\#}}+1-x_{b,c}^{\overline{)\#}}x)(x+1-x{x}_{b,c}^{\overline{)\#}}).\end{array}$$

Hence,

$${(x+p)}^{-1}=x_{b,c}^{\overline{)\#}}+1-x_{b,c}^{\overline{)\#}}x.$$

Since $y(x+p)=y(x+1-x{x}_{b,c}^{\overline{)\#}})=y$, we see that $y{(x+p)}^{-1}=y\in {\mathcal{A}}^{qnil}$. By virtue of [17] [Theorem 2.1], ${(x+p)}^{-1}y\in {\mathcal{A}}^{qnil}$. Hence, $1+{(x+p)}^{-1}y\in {\mathcal{A}}^{-1}$. Therefore, we check that

$$\begin{array}{ccc}\hfill pa& =\hfill & p(x+y)=py=(1-x{x}_{b,c}^{\overline{)\#}})y\hfill \\ & =\hfill & {(1-x{x}_{b,c}^{\overline{)\#}})}^{*}y\hfill \\ & =\hfill & (1-{\left(x_{b,c}^{\overline{)\#}}\right)}^{*}{x}^{*})y\hfill \\ & =\hfill & y\in {\mathcal{A}}^{qnil},\hfill \\ \hfill pa(1-p)& =\hfill & yx{x}_{b,c}^{\overline{)\#}}=0,pa=pap,\hfill \\ \hfill a+p& =\hfill & x+y+p=(x+p)[1+{(x+p)}^{-1}y]\in {\mathcal{A}}^{-1}.\hfill \end{array}$$

Obviously, ${(1-p)}^{*}={(1-p)}^2=1-p\in {\mathcal{A}}^{\overline{)\#}}.$ Moreover, we check that

Claim 1. $(1-p)\mathcal{A}=x{x}^{\overline{)\#}}\mathcal{A}=x\mathcal{A}=b\mathcal{A}$.

Claim 2. $\mathcal{A}{(1-p)}^{*}=\mathcal{A}{\left(x{x}^{\overline{)\#}}\right)}^{*}=\mathcal{A}{x}^{*}=\mathcal{A}c$.

Therefore $1-p\in {\mathcal{A}}_{b,c}^{\overline{)\#}}$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, there exists a projection $p\in \mathcal{A}$ such that

$$u:=a+p\in {\mathcal{A}}^{-1},1-p\in {\mathcal{A}}_{b,c}^{\overline{)\#}},pa=pap\in {\mathcal{A}}^{qnil}.$$

Set $x=(1-p)a$ and $y=pa$. Then

$$\begin{array}{ccc}\hfill x^{*}y& =\hfill & {\left[(1-p)a\right]}^{*}\left(pa\right)=\left[a^{*}{(1-p)}^{*}\right]p^{*}a=a^{*}{(1-p)}^{*}{p}^{*}a=0,\hfill \\ \hfill yx& =\hfill & pa(1-p)a=\left(pap\right)(1-p)a=0,\hfill \\ \hfill y& =\hfill & pa\in {\mathcal{A}}^{qnil}.\hfill \end{array}$$

Claim 1. $x\in {\mathcal{A}}^d$.

Since $pa(1-p)=0$, we have $a=pap+(1-p)ap+(1-p)a(1-p)$, i.e., $a={\left(\begin{array}{cc}pap& 0\\ (1-p)ap& (1-p)a(1-p)\end{array}\right)}_p$. By hypothesis, $pa\in {\mathcal{A}}^{qnil}$, then $pap\in {\mathcal{A}}^{qnil}$ by [17] [Theorem 2.1]. Hence, $pap\in {\left(p\mathcal{A}p\right)}^{qnil}\subseteq {\left(p\mathcal{A}p\right)}^d.$ In view of [2] [Theorem 2.3], $(1-p)a(1-p)\in {\left((1-p)\mathcal{A}(1-p)\right)}^d\subseteq {\mathcal{A}}^d.$ By using Cline’s formula again, $x=(1-p)a\in {\mathcal{A}}^d,$ as desired.

Claim 2. $x\in {\mathcal{A}}^{\#}$. Observing that

$$\begin{array}{ccc}\hfill u^{-1}{x}^2& =\hfill & u^{-1}(1-p)a(1-p)a=u^{-1}a(1-p)a-u^{-1}\left[pa(1-p)\right]a\hfill \\ & =\hfill & u^{-1}a(1-p)a={(a+p)}^{-1}(a+p)(1-p)a=(1-p)a=x,\hfill \end{array}$$

we have $\mathcal{A}x=\mathcal{A}{x}^2$.

Write $x=z{x}^2$. Then $x=z{x}^2=z^n{x}^{n+1}=z^n(x^n-x^d{x}^{n+1})x+z^n{x}^d{x}^{n+2}$ and $x^2{x}^d=\left(z{x}^2\right)x{x}^d=z^n{x}^{n+2}{x}^d$. Hence

$$\begin{array}{ccc}\hfill \left|\right|x-x^2{x}^d{\left|\right|}^{{\displaystyle \frac{1}{n}}}& =\hfill & \left|\right|z^n(x^n-x^d{x}^{n+1})x{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ & \le \hfill & \left|\right|z^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}\left|\right|x^n-x^d{x}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}{\left|\right|x\left|\right|}^{{\displaystyle \frac{1}{n}}}.\hfill \end{array}$$

Thus

$$\underset{n\to \infty }{lim}\left|\right|x-x^2{x}^d{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

This implies that $x=x^2{x}^d$, and so $x\in {\mathcal{A}}^{\#}$.

Claim 3. $x\in {\mathcal{A}}_{b,c}^{\overline{)\#}}$.

Clearly, $x{u}^{-1}=(1-p)a{(a+p)}^{-1}=(1-p)(a+p){(a+p)}^{-1}=1-p$. Then $x{u}^{-1}x=(1-p)x=x$ and ${\left(x{u}^{-1}\right)}^{*}={(1-p)}^{*}=1-p^{*}=1-p=x{u}^{-1}$. Hence, $x\in {\mathcal{A}}^{(1,3)}$. According to [26] [Theorem 2.6], $x\in {\mathcal{A}}^{\overline{)\#}}$.

Claim 1. $x\mathcal{A}=x{u}^{-1}\mathcal{A}=(1-p)\mathcal{A}=b\mathcal{A}$.

Claim 2. $\mathcal{A}{x}^{*}=\mathcal{A}{u}^{*}{(1-p)}^{*}=\mathcal{A}{(1-p)}^{*}=\mathcal{A}c$.

In light of Theorem 1.2, $x\in {\mathcal{A}}_{b,c}^{\overline{)\#}}.$

Therefore, $a\in {\mathcal{A}}_{b,c}^{ⓓ}$ by Theorem 2.1. □

Corollary 3.2.

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_{b,c}^{ⓓ}$ if and only if the following two conditions hold:

(1)

$a\in {\mathcal{A}}^d$;

(2)

there exists a projection $p\in \ell \left(b\right)\bigcap r\left(c\right)$ such that

$$a+p\in {\mathcal{A}}^{-1},1-p\in b\mathcal{A}c,pa=pap\in {\mathcal{A}}^{qnil}.$$

Proof. 

The result follows directly from Theorem 3.1, since for a projection $p\in \mathcal{A}$, $1-p\in {\mathcal{A}}_{b,c}^{\overline{)\#}}$ if and only if $1-p\in b\mathcal{A}c$ and $p\in \ell \left(b\right)\bigcap r\left(c\right)$. □

Set

$$com{m}^{*}\left(a\right)=\{x\in \mathcal{A}\mid \left(a{x}^2\right)a=\left(xax\right)a,x\left(a^{2}x\right)=\left(axa\right)x\}$$

(the weak commutant of a). We are ready to prove:

Theorem 3.3.

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_{b,c}^{ⓓ}$ if and only if the following two conditions hold:

(1)

$a\in {\mathcal{A}}^d$;

(2)

there exists $z\in com{m}^{*}\left(a\right)\bigcap {\mathcal{A}}_{b,c}^{\overline{)\#}}$ such that

$$z=a{z}^2,{\left(az\right)}^{*}=az,a-a^{2}z\in {\mathcal{A}}^{qnil}.$$

Proof. 

⟹ Clearly, $a\in {\mathcal{A}}^d$. In view of Theorem 2.1, there exist $x,y\in \mathcal{A}$ such that

$$a=x+y,x^{*}y=yx=0,x\in {\mathcal{A}}_{b,c}^{\overline{)\#}},y\in {\mathcal{A}}^{qnil}.$$

It is easy to verify that

$$\begin{array}{ccc}\hfill x_{b,c}^{\overline{)\#}}y& =\hfill & x_{b,c}^{\overline{)\#}}x{x}_{b,c}^{\overline{)\#}}y=x_{b,c}^{\overline{)\#}}\left[x{x}_{b,c}^{\overline{)\#}}\right]y\hfill \\ & =\hfill & x_{b,c}^{\overline{)\#}}{\left(x{x}_{b,c}^{\overline{)\#}}\right)}^{*}y=x_{b,c}^{\overline{)\#}}{\left(x_{b,c}^{\overline{)\#}}\right)}^{*}\left(x^{*}y\right)=0.\hfill \end{array}$$

Set $z=x_{b,c}^{\overline{)\#}}$. Then we check that $az=(x+y)x_{b,c}^{\overline{)\#}}=x{x}_{b,c}^{\overline{)\#}}+yx{\left(x_{b,c}^{\overline{)\#}}\right)}^2=x{x}_{b,c}^{\overline{)\#}}$. Hence, ${\left(az\right)}^{*}={\left(x{x}_{b,c}^{\overline{)\#}}\right)}^{*}=x{x}_{b,c}^{\overline{)\#}}=az$. Further, we verify that Moreover, we verify that

$$\begin{array}{ccc}\hfill a^{2}z& =\hfill & {(x+y)}^2{x}_{b,c}^{\overline{)\#}}\hfill \\ & =\hfill & (x^2+xy+y^2)x_{b,c}^{\overline{)\#}}\hfill \\ & =\hfill & x^2{x}_{b,c}^{\overline{)\#}},\hfill \\ \hfill z\left(a^{2}z\right)& =\hfill & x_{b,c}^{\overline{)\#}}\left(x^2{x}_{b,c}^{\overline{)\#}}\right)\hfill \\ & =\hfill & \left(x_{b,c}^{\overline{)\#}}{x}^2\right)x_{b,c}^{\overline{)\#}}=x{x}_{b,c}^{\overline{)\#}}\hfill \\ & =\hfill & {\left(x{x}_{b,c}^{\overline{)\#}}\right)}^2=\left(aba\right)b,\hfill \\ \hfill a{z}^2& =\hfill & \left(az\right)z=\left(x{x}_{b,c}^{\overline{)\#}}\right)x_{b,c}^{\overline{)\#}}=x_{b,c}^{\overline{)\#}}=b,\hfill \\ \hfill z(1-az)& =\hfill & x_{b,c}^{\overline{)\#}}[1-x{x}_{b,c}^{\overline{)\#}}]=0,\hfill \\ \hfill a-a^{2}z& =\hfill & a(1-ab)=a(1-x{x}_{b,c}^{\overline{)\#}}).\hfill \end{array}$$

Hence, $z=zaz$, and so $a{z}^2=zaz$. This implies that $z\in com{m}^{*}\left(a\right)$. Since $(1-x{x}_{b,c}^{\overline{)\#}})a=(1-x{x}_{b,c}^{\overline{)\#}})(x+y)=y\in {\mathcal{A}}^{qnil}$, by using Cline’s formula (see [17] [Theorem 2.1]), $a-a^{2}z=a(1-x{x}_{b,c}^{\overline{)\#}})\in {\mathcal{A}}^{qnil}$.

We see that $z^{\overline{)\#}}={\left[x^{\overline{)\#}}\right]}^{\overline{)\#}}=x^2{x}^{\overline{)\#}}.$ Hence, $z\in {\mathcal{A}}^{\overline{)\#}}$.

Since $z=x^{\overline{)\#}}$, we see that $z\mathcal{A}=x\mathcal{A}$ (see [26]). In view of Theorem 1.2, $x\mathcal{A}=b\mathcal{A}$. Hence, $z\mathcal{A}=b\mathcal{A}$. Analogously, we verify that $\mathcal{A}{z}^{*}=\mathcal{A}{\left(x^{\overline{)\#}}\right)}^{*}=\mathcal{A}{x}^{*}=\mathcal{A}c$.

Therefore $z\in {\mathcal{A}}^{{\overline{)\#}}_{b,c}}$ by Theorem 1.2.

⟸ By hypothesis, $a\in {\mathcal{A}}^d$ and there exists $z\in com{m}^{*}\left(a\right)\bigcap {\mathcal{A}}_{b,c}^{\overline{)\#}}$ such that

$$z=a{z}^2,{\left(az\right)}^{*}=az,a-a^{2}z\in {\mathcal{A}}^{qnil}.$$

Hence, $z\left(a^{2}z\right)=\left(aza\right)z,a\left(z^{2}a\right)=\left(zaz\right)a,$ and so $za={\left(za\right)}^2$.

Step 1. We verify that

$$\begin{array}{ccc}\hfill (a-aza)aza& =\hfill & a^{2}za-az{a}^{2}za=a^{2}za-a\left[z\left(a^{2}z\right)\right]a\hfill \\ & =\hfill & a^{2}za-a\left[azaz\right]a=a^{2}za-a^2\left[zaza\right]=0.\hfill \end{array}$$

By induction, we have

$$\begin{array}{ccc}\hfill (a-aza)a^{n-1}& =\hfill & (a-aza)(a-aza)a^{n-2}={(a-aza)}^2{a}^{n-2}\hfill \\ & =\hfill & {(a-aza)}^2(a-aza)a^{n-3}={(a-aza)}^3{a}^{n-3}\hfill \\ & =\hfill & \cdots ={(a-aza)}^n.\hfill \end{array}$$

Since $a-a^{2}z\in {\mathcal{A}}^{qnil}$, by using Cline’s formula, $a-aza\in {\mathcal{A}}^{qnil}$. Therefore

$$\underset{n\to \infty }{lim}\left|\right|a^n-az{a}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=\underset{n\to \infty }{lim}{\left|\right|(a-aza)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

Observing that

$$\begin{array}{ccc}\hfill a^n-z{a}^{n+1}& =\hfill & [a^n-a^d{a}^{n+1}]+a^d[a^n-az{a}^n]a\hfill \\ & -\hfill & [a^n-a^d{a}^{n+1}]z^{n+1}{a}^{n+1}],\hfill \end{array}$$

we deduce that

$$\begin{array}{ccc}\hfill \left|\right|a^n-z{a}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}& \le \hfill & \left|\right|a^n-a^d{a}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}[1+||z^{n+1}{a}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}]\hfill \\ & +\hfill & \left|\right|a^d{\left|\right|}^{{\displaystyle \frac{1}{n}}}\left|\right|a^n-az{a}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}{\left|\right|a\left|\right|}^{{\displaystyle \frac{1}{n}}}.\hfill \end{array}$$

Since $\underset{n\to \infty }{lim}\left|\right|a^n-a^d{a}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=\underset{n\to \infty }{lim}\left|\right|a^n-az{a}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0$, we deduce that $\underset{n\to \infty }{lim}\left|\right|a^n-z{a}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0$. Therefore we complete the proof by Theorem 2.1. □

Lemma 3.4.

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_{b,c}^{ⓓ}$ if and only if $a\in {\mathcal{A}}^{ⓓ}$ and $a^{ⓓ}a\in b\mathcal{A}ca,a{a}^{ⓓ}\in ab\mathcal{A}c,(1-a^{ⓓ}a)b=0,c(1-a{a}^{ⓓ})=0;ca(1-a^{ⓓ}a)=0,(1-a{a}^{ⓓ})ab=0.$

Proof. 

⟹ By hypothesis, $a\in {\mathcal{A}}^{ⓓ}$ and $a^{ⓓ}={\left(a{a}^{ⓓ}a\right)}^{(b,c)}$. Then $a^{ⓓ}\in b\mathcal{A}\bigcap \mathcal{A}c$. Hence, $a^{ⓓ}a=a^{ⓓ}a{a}^{ⓓ}a\in b\mathcal{A}ca$. Likewise, we have $a{a}^{ⓓ}\in ab\mathcal{A}c.$ Moreover, we verify that $(1-a^{ⓓ}a)b=b-a^{ⓓ}\left(a{a}^{ⓓ}a\right)b=0,c(1-a{a}^{ⓓ})=c-c\left(a{a}^{ⓓ}a\right)a^{ⓓ}=0;ca(1-a^{ⓓ}a)=\left[c\left(a{a}^{ⓓ}a\right)a^{ⓓ}\right]a(1-a^{ⓓ}a)=0,(1-a{a}^{ⓓ})ab=(1-a{a}^{ⓓ})a\left[a^{ⓓ}\left(a{a}^{ⓓ}a\right)b\right]=0,$ as required.

⟸ Since $a^{ⓓ}=a^{ⓓ}a{a}^{ⓓ}$, we have $a^{ⓓ}=a^{ⓓ}\left(a{a}^{ⓓ}a\right)a^{ⓓ}$. By hypothesis, we verify that $a^{ⓓ}a{a}^{ⓓ}=a^{ⓓ}\in b\mathcal{A}\bigcap \mathcal{A}c,a^{ⓓ}\left(a{a}^{ⓓ}a\right)b=a^{ⓓ}ab=b,c\left(a{a}^{ⓓ}a\right)a^{ⓓ}=ca{a}^{ⓓ}=c.$ Therefore $a^{ⓓ}={\left(a{a}^{ⓓ}a\right)}^{(b,c)}$. This implies that $a^{ⓓ}=a^{(b,c)},$ as required. □

Let X be a complex Banach space, and let $\mathcal{B}\left(X\right)$ denote the Banach algebra of all bounded linear operators on X. For any operator $A\in \mathcal{B}\left(X\right)$, we denote its null space by $\mathcal{N}\left(A\right)=\{x\in X|Ax=0\}$ and its range by $\mathcal{R}\left(A\right)=\left\{Ax\right|x\in X\}$. We are ready to prove:

Theorem 3.5.

Let $A\in \mathcal{B}\left(X\right)$. Then $A\in {\left(\mathcal{B}\left(X\right)\right)}_{(B,C)}^{ⓓ}$ if and only if $A\in {\left(\mathcal{B}\left(X\right)\right)}^{ⓓ}$ and

$$\begin{array}{ccc}\hfill \mathcal{R}\left(A^{ⓓ}A\right)& =\hfill & \mathcal{R}\left(B\right),\mathcal{N}\left(A^{ⓓ}A\right)=\mathcal{N}\left(CA\right),\hfill \\ \hfill \mathcal{R}\left(A{A}^{ⓓ}\right)& =\hfill & \mathcal{R}\left(AB\right),\mathcal{N}\left(A{A}^{ⓓ}\right)=\mathcal{N}\left(C\right).\hfill \end{array}$$

Proof. 

⟹ In view of Lemma 3.4, $A\in {\left(\mathcal{B}\left(X\right)\right)}^{ⓓ}$ and

$$\begin{array}{c}{A}^{ⓓ}A\in B\left(\mathcal{B}\left(X\right)\right)CA,A{A}^{ⓓ}\in AB\left(\mathcal{B}\left(X\right)\right)C,\\ (I-A^{ⓓ}A)B=0,C(I-A{A}^{ⓓ})=0;\\ CA(I-A^{ⓓ}A)=0,(I-A{A}^{ⓓ})AB=0.\end{array}$$

Hence,

$$\begin{array}{ccc}\hfill \mathcal{R}\left(A^{ⓓ}A\right)& =\hfill & \mathcal{R}\left(B\right),\hfill \\ \hfill \mathcal{R}\left(A{A}^{ⓓ}\right)& =\hfill & \mathcal{R}\left(AB\right),\hfill \end{array}$$

Obviously, $\mathcal{N}\left(A^{ⓓ}A\right)\subseteq \mathcal{N}\left(CA\right)$. Since $CA=CA{A}^{ⓓ}A$, we see that $\mathcal{N}\left(CA\right)\subseteq \mathcal{N}\left(A^{ⓓ}A\right).$ This implies that

$$\mathcal{N}\left(A^{ⓓ}A\right)=\mathcal{N}\left(CA\right).$$

Clearly, we have $\mathcal{N}\left(A{A}^{ⓓ}\right)\subseteq \mathcal{N}\left(C\right)$. As $C=CA{A}^{ⓓ}$, we deduce that $\mathcal{N}\left(C\right)\subseteq \mathcal{N}\left(A{A}^{ⓓ}\right)$. Therefore $\mathcal{N}\left(A{A}^{ⓓ}\right)=\mathcal{N}\left(C\right)$, as required.

⟸ By hypothesis, $A\in {\left(\mathcal{B}\left(X\right)\right)}^{ⓓ}$.

Claim 1. $A^{ⓓ}\in B\left(\mathcal{B}\left(X\right)\right)C$. Obviously, $A^{ⓓ}=\left(A^{ⓓ}A\right)A^{ⓓ}\in \mathcal{R}\left(B\right)$; hence, $A^{ⓓ}\in B\left(\mathcal{B}\left(X\right)\right)$. On the other hand, $\mathcal{N}\left(A{A}^{ⓓ}\right)=\mathcal{N}\left(C\right)$. Thus, $\mathcal{R}\left({\left(A{A}^{ⓓ}\right)}^{*}\right)=\mathcal{R}\left(C^{*}\right)$. Hence, $\mathcal{R}\left(A{A}^{ⓓ}\right)=\mathcal{R}\left(C^{*}\right)$. Thus

$${\left(A^{ⓓ}\right)}^{*}=\left(A{A}^{ⓓ}\right){\left(A^{ⓓ}\right)}^{*}\in C^{*}\left(\mathcal{B}\left(X\right)\right).$$

Accordingly, $A^{ⓓ}\in \left(\mathcal{B}\left(X\right)\right)C$. Therefore

$$A^{ⓓ}=A^{ⓓ}A{A}^{ⓓ}\in B\left(\mathcal{B}\left(X\right)\right)C.$$

Claim 2. $A^{ⓓ}AB=B.$ Write $B=A^{ⓓ}AY$ for some $Y\in \mathcal{B}\left(X\right)$. Then $A^{ⓓ}AB=\left(A^{ⓓ}A\right)\left(A^{ⓓ}AY\right)=A^{ⓓ}AY=B$.

Claim 3. $CA{A}^{ⓓ}=C$. Since $I-A{A}^{ⓓ}\in \mathcal{N}\left(A{A}^{ⓓ}\right)$, we deduce that $I-A{A}^{ⓓ}\in \mathcal{N}\left(C\right)$. Then $CA{A}^{ⓓ}=C$.

Therefore $A^{ⓓ}=A^{(B,C)}$, as required. □

4. Representations Based on Core Inverses

The aim of this section is to establish representations of generalized core-$(b,c)$-inverses by using involved core inverses. We come now to the demonstration for which this section has been developed.

Lemma 4.1.

Let $x\in {\mathcal{A}}^{\#}$ and $y\in {\mathcal{A}}^{qnil}$. If $yx=0$, then $x+y\in {\mathcal{A}}^d$ and ${(x+y)}^d=x^{\#}.$

Proof. 

This is obvious by [4] [Lemma 15.2.2]. □

Theorem 4.2.

Let $a\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}_{b,c}^{ⓓ}$.

(2)

$a\in {\mathcal{A}}^d$ and $a^d\in {\mathcal{A}}_{b,c}^{\overline{)\#}}$.

In this case,

$$a_{b,c}^{ⓓ}={\left(a^d\right)}^2{\left(a^d\right)}_{b,c}^{\overline{)\#}}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 2.1, $a\in {\mathcal{A}}^d$ and we can find $x,y\in \mathcal{A}$ such that

$$a=x+y,x^{*}y=yx=0,x\in {\mathcal{A}}_{b,c}^{\overline{)\#}},y\in {\mathcal{A}}^{qnil}.$$

Set $z=x_{b,c}^{\overline{)\#}}$. Then $a^{ⓓ}=x^{\overline{)\#}}=x^{(b,c)}$. Moreover,

$$x\mathcal{A}=b\mathcal{A},\mathcal{A}{x}^{*}=\mathcal{A}c.$$

By virtue of Lemma 4.1, we have $a^d=x^{\#}$, and so ${\left(a^d\right)}^{\overline{)\#}}={\left(x^{\#}\right)}^{\overline{)\#}}=x^2{x}^{\overline{)\#}}$. Then we deduce that

$$a^d\mathcal{A}=x^{\#}\mathcal{A}=b\mathcal{A},\mathcal{A}{\left(a^d\right)}^{*}=\mathcal{A}{\left(x^{\#}\right)}^{*}=\mathcal{A}c.$$

In light of Theorem 1.2, $a^d\in {\mathcal{A}}_{b,c}^{\overline{)\#}}$. In this case,

$${\left(a^d\right)}_{b,c}^{\overline{)\#}}={\left(x^{\#}\right)}^{\overline{)\#}}=x^2{x}^{\overline{)\#}}.$$

Set $z={\left(a^d\right)}^2{\left(a^d\right)}_{b,c}^{\overline{)\#}}$. Then $z={\left(x^{\#}\right)}^2\left[x^2{x}^{\overline{)\#}}\right]=x^{\overline{)\#}}$. Therefore $a_{b,c}^{ⓓ}=x^{\overline{)\#}}=z$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ Set $x=a^d$. By hypothesis, we have $x^{\overline{)\#}}=x^{(b,c)}$, and so

$$x^{\overline{)\#}}\in b\mathcal{A}{x}^{\overline{)\#}}\bigcap x^{\overline{)\#}}\mathcal{A}c,x^{\overline{)\#}}xb=b,cx{x}^{\overline{)\#}}=c.$$

Then we verify that

Claim 1. $x^2{x}^{\overline{)\#}}\in b\mathcal{A}{x}^2{x}^{\overline{)\#}}\bigcap x^2{x}^{\overline{)\#}}\mathcal{A}c.$

$$\begin{array}{ccc}\hfill x^2{x}^{\overline{)\#}}& =\hfill & \left(x^2{x}^{\overline{)\#}}\right)x{x}^{\overline{)\#}}\hfill \\ & =\hfill & \left(x^2{x}^{\overline{)\#}}\right)\mathcal{A}c,\hfill \\ \hfill x^2{x}^{\overline{)\#}}& =\hfill & x{x}^{\overline{)\#}}\left[x^2{x}^{\overline{)\#}}\right]\hfill \\ & =\hfill & x^{\overline{)\#}}\left[x^2{x}^{\overline{)\#}}\right]x^2{x}^{\overline{)\#}}\hfill \\ & \in \hfill & b\mathcal{A}{x}^2{x}^{\overline{)\#}}.\hfill \end{array}$$

Claim 2. $x^2{x}^{\overline{)\#}}ab=b$.

Since $x^{\overline{)\#}}xb=b$, we see that $x{x}^{\overline{)\#}}=b$, and so $a^d{\left(a^d\right)}^{\overline{)\#}}b=b$. We infers that $a{a}^{d}b=b$. Then we derive that

$$\begin{array}{ccc}\hfill x^2{x}^{\overline{)\#}}ab& =\hfill & {\left(a^d\right)}^2{\left(a^d\right)}^{\overline{)\#}}ab\hfill \\ & =\hfill & \left[{\left(a^d\right)}^2{\left(a^d\right)}^{\overline{)\#}}a\right]\left[a^d{\left(a^d\right)}^{\overline{)\#}}b\right]\hfill \\ & =\hfill & \left[{\left(a^d\right)}^2{\left(a^d\right)}^{\overline{)\#}}{a}^d\right]\left[a{\left(a^d\right)}^{\overline{)\#}}b\right]\hfill \\ & =\hfill & {\left(a^d\right)}^{2}a{\left(a^d\right)}^{\overline{)\#}}b\hfill \\ & =\hfill & a^d{\left(a^d\right)}^{\overline{)\#}}b=b.\hfill \end{array}$$

Claim 3. $ca{x}^2{x}^{\overline{)\#}}=c$.

$$\begin{array}{ccc}\hfill ca{x}^2{x}^{\overline{)\#}}& =\hfill & ca{\left(a^d\right)}^2{\left(a^d\right)}^{\overline{)\#}}\hfill \\ & =\hfill & c{a}^d{\left(a^d\right)}^{\overline{)\#}}=c.\hfill \end{array}$$

Accordingly, $x^2{x}^{\overline{)\#}}=a^{(b,c)}$.

In view of [5] [Theorem 4.1], $a^{ⓓ}={\left(a^d\right)}^2{\left(a^d\right)}^{\overline{)\#}}=x^2{x}^{\overline{)\#}}=a^{(b,c)}$. This completes the proof. □

Corollary 4.3.

Let $a\in \mathcal{A}$. Then the following hold:

(1)

$a\in {\mathcal{A}}^{\overline{)\#}}$ if and only if $a\in {\mathcal{A}}_{(a,a^{*})}^{\overline{)\#}}$.

(2)

$a\in {\mathcal{A}}^{ⓓ}$ if and only if $a\in {\mathcal{A}}^d$ and $a\in {\mathcal{A}}_{(a^d,{\left(a^d\right)}^{*})}^{ⓓ}$.

Proof.$\left(1\right)$ This is obvious by [20] [Theorem 4.4]

$\left(2\right)$ Straightforward by Theorem 4.2 and [6] [Lemma 3.1]. □

Corollary 4.4.

Let $a,b\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}_{(b,b)}^{ⓓ}$.

(2)

$a\in {\mathcal{A}}^{ⓓ},b\in {\mathcal{A}}^{\overline{)\#}}$ and $a{a}^{ⓓ}=b^{\overline{)\#}}b.$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 4.2, $a\in {\mathcal{A}}^d$ and $a^d\in {\mathcal{A}}_{(b,b)}^{\overline{)\#}}$. In view of Theorem 1.2, $a^d,b\in {\mathcal{A}}^{\overline{)\#}}$ and $\left(a^d\right){\left(a^d\right)}^{\overline{)\#}}=b^{\overline{)\#}}b$. By Theorem 1.2 again, $a\in {\mathcal{A}}^{ⓓ}$ and $a^{ⓓ}={\left(a^d\right)}^2{\left(a^d\right)}^{\overline{)\#}}$. Therefore $a{a}^{ⓓ}=\left[a{\left(a^d\right)}^2\right]{\left(a^d\right)}^{\overline{)\#}}=a^d{\left(a^d\right)}^{\overline{)\#}}=b^{\overline{)\#}}b$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ By virtue of [5] [Theorem 4.1], $a\in {\mathcal{A}}^d$ and $a^{ⓓ}={\left(a^d\right)}^2{\left(a^d\right)}^{\overline{)\#}}$. Then $a^d{\left(a^d\right)}^{\overline{)\#}}=a{a}^{ⓓ}=b^{\overline{)\#}}b$. In light of Theorem 1.2, $a^d\in {\mathcal{A}}_{b,b}^{\overline{)\#}}$. According to Theorem 4.2, $a\in {\mathcal{A}}_{(b,b)}^{ⓓ}$. □

Theorem 4.5.

Let $a,b\in \mathcal{A}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}_{(b,b)}^{ⓓ}$.

(2)

$a\in {\mathcal{A}}^{ⓓ},b,ba\in {\mathcal{A}}^{\overline{)\#}},a^d\mathcal{A}=b\mathcal{A}$ and

$${\left(ba\right)}^{\overline{)\#}}=a^{ⓓ}{b}^{\overline{)\#}}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Since $a\in {\mathcal{A}}_{(b,b)}^{ⓓ}$, by virtue of Corollary 4.3, $a,b\in {\mathcal{A}}^{ⓓ}$. In view of Theorem 4.2, $a^d\in {\mathcal{A}}_{(b,b)}^{\overline{)\#}}$. According to [3] [Theorem 2.8], we have

$$\begin{array}{ccc}\hfill a^{ⓓ}{b}^{ⓓ}& =\hfill & \left[{\left(a^d\right)}^2{\left(a^d\right)}^{\overline{)\#}}\right]b^{\overline{)\#}}\hfill \\ & =\hfill & {\left(a^d\right)}^2{\left(b{a}^d\right)}^{\overline{)\#}}b{b}^{\overline{)\#}}\hfill \\ & =\hfill & {\left(a^d\right)}^2\left[{\left(a^d\right)}^{\overline{)\#}}{b}^{\overline{)\#}}\right]b{b}^{\overline{)\#}}\hfill \\ & =\hfill & {\left(a^d\right)}^2[{\left(a^d\right)}^{\overline{)\#}}\left[b^{\overline{)\#}}b{b}^{\overline{)\#}}\right]\hfill \\ & =\hfill & {\left(a^d\right)}^2{\left(b{a}^d\right)}^{\overline{)\#}}.\hfill \end{array}$$

We verify that $ba\left[{\left(a^d\right)}^2{\left(b{a}^d\right)}^{\overline{)\#}}\right]=\left(b{a}^d\right){\left(b{a}^d\right)}^{\overline{)\#}}$; whence ${\left(ba\left[{\left(a^d\right)}^2{\left(b{a}^d\right)}^{\overline{)\#}}\right]\right)}^{*}=\left(b{a}^d\right){\left(b{a}^d\right)}^{\overline{)\#}}=ba\left[{\left(a^d\right)}^2{\left(b{a}^d\right)}^{\overline{)\#}}\right]$. By using [3] [Corollary 2.3], we check that

$$\begin{array}{ccc}\hfill ba{\left[{\left(a^d\right)}^2{\left(b{a}^d\right)}^{\overline{)\#}}\right]}^2& =\hfill & \left(b{a}^d\right){\left(b{a}^d\right)}^{\overline{)\#}}\left[{\left(a^d\right)}^2{\left(b{a}^d\right)}^{\overline{)\#}}\right]\hfill \\ & =\hfill & \left(b{a}^d\right){\left(b{a}^d\right)}^{\overline{)\#}}\left[{\left(a^d\right)}^{\#}\right]{\left(a^d\right)}^3{\left(b{a}^d\right)}^{\overline{)\#}}]\hfill \\ & =\hfill & \left(b{a}^d\right){\left(b{a}^d\right)}^{\overline{)\#}}\left[{\left(b{a}^d\right)}^{\overline{)\#}}b\right]{\left(a^d\right)}^3{\left(b{a}^d\right)}^{\overline{)\#}}]\hfill \\ & =\hfill & \left[{\left(b{a}^d\right)}^{\overline{)\#}}b\right]{\left(a^d\right)}^3{\left(b{a}^d\right)}^{\overline{)\#}}]\hfill \\ & =\hfill & {\left[a^d\right)}^{\#}\left]{\left(a^d\right)}^3{\left(b{a}^d\right)}^{\overline{)\#}}\right]\hfill \\ & =\hfill & {\left(a^d\right)}^2{\left(b{a}^d\right)}^{\overline{)\#}}.\hfill \end{array}$$

Obviously, ${\left(a^d\right)}^{\#}{a}^{d}b=b$, and so $a{a}^{d}b=b$. This implies that $a{a}^{ⓓ}b=a{a}^{ⓓ}a{a}^{d}ba{a}^{d}b=b$. In view of [3] [Theorem 2.6], we have $a^d{\left(a^d\right)}^{\overline{)\#}}=b^{\overline{)\#}}b$, and so

$$\begin{array}{ccc}\hfill \left[{\left(a^d\right)}^2{\left(b{a}^d\right)}^{\overline{)\#}}\right]{\left(ba\right)}^2& =\hfill & \left[{\left(a^d\right)}^2{\left(a^d\right)}^{\overline{)\#}}{b}^{\overline{)\#}}\right]{\left(ba\right)}^2\hfill \\ & =\hfill & {\left(a^d\right)}^2{\left(a^d\right)}^{\overline{)\#}}\left[a^d{\left(b{a}^d\right)}^{\overline{)\#}}\right]{\left(ba\right)}^2\hfill \\ & =\hfill & a^d\left[b^{\overline{)\#}}b\right]aba\hfill \\ & =\hfill & a^d\left[a^d{\left(a^d\right)}^{\overline{)\#}}\right]aba\hfill \\ & =\hfill & a^d\left[a^d{\left(a^d\right)}^{\overline{)\#}}\right]a\left[{\left(a^d\right)}^{\#}{a}^{d}b\right]a\hfill \\ & =\hfill & a{a}^{d}ba=a{a}^d\left[{\left(a^d\right)}^{\#}{a}^{d}b\right]a=ba.\hfill \end{array}$$

This implies that ${\left(ba\right)}^{\overline{)\#}}={\left(a^d\right)}^2{\left(b{a}^d\right)}^{\overline{)\#}}$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ In view of [5] [Theorem 4.1], $a^d\in {\mathcal{A}}^{\overline{)\#}}$ and $a^{ⓓ}={\left(a^d\right)}^2{\left(a^d\right)}^{\overline{)\#}}$. We easily check that

$$\begin{array}{ccc}\hfill \left[b{a}^d\right]\left[{\left(a^d\right)}^{\overline{)\#}}{b}^{\overline{)\#}}\right]& =\hfill & ba\left[a^{ⓓ}{b}^{ⓓ}\right]=b{b}^{ⓓ},\hfill \\ \hfill {\left(\left(b{a}^d\right)\left({\left(a^d\right)}^{\overline{)\#}}{b}^{\overline{)\#}}\right)\right)}^{*}& =\hfill & \left(b{a}^d\right)\left({\left(a^d\right)}^{\overline{)\#}}{b}^{\overline{)\#}}\right).\hfill \end{array}$$

By hypothesis, $a^d\mathcal{A}=b\mathcal{A}$; hence, $a^d\in b{b}^{\overline{)\#}}\mathcal{A}\bigcap b^{\overline{)\#}}b\mathcal{A}.$ Hence, $b{b}^{ⓓ}({\left(a^d\right)}^{\overline{)\#}}=({\left(a^d\right)}^{\overline{)\#}}$. Moreover, we have $b\in {\left(a^d\right)}^{\overline{)\#}}\mathcal{A}$. Then

$$\begin{array}{ccc}\hfill \left(b{a}^d\right){\left({\left(a^d\right)}^{\overline{)\#}}{b}^{\overline{)\#}}\right)}^2& =\hfill & b{b}^{ⓓ}\left({\left(a^d\right)}^{\overline{)\#}}{b}^{\overline{)\#}}\right)\hfill \\ & =\hfill & {\left(a^d\right)}^{\overline{)\#}}{b}^{\overline{)\#}},\hfill \\ \hfill \left({\left(a^d\right)}^{\overline{)\#}}{b}^{\overline{)\#}}\right){\left(b{a}^d\right)}^2& =\hfill & {\left(a^d\right)}^{\overline{)\#}}\left[b^{\overline{)\#}}b\right]a^d)\left(b{a}^d\right)\hfill \\ & =\hfill & {\left(a^d\right)}^{\overline{)\#}}\left(a^{d}b\right)a^d\hfill \\ & =\hfill & b{a}^d.\hfill \end{array}$$

Therefore ${\left(a^d\right)}^{\overline{)\#}}{b}^{\overline{)\#}}={\left(b{a}^d\right)}^{\overline{)\#}}$, thus yielding the result by Theorem 1.2 and Theorem 4.2. □

To illustrate Theorem 4.5, we present the following numerical example.

• Example 4.6.

Let $A=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\in {\mathbb{C}}^{2\times 2}$. Then $A^{\overline{)\#}}=\left(\begin{array}{cc}{\displaystyle \frac{1}{4}}& {\displaystyle \frac{1}{4}}\\ {\displaystyle \frac{1}{4}}& {\displaystyle \frac{1}{4}}\end{array}\right).$ Let $S\in \mathcal{B}(\mathbb{C}⨁\mathbb{C})$ be the linear operator given by the $2\times 2$ matrix A.

Let $X={\ell }^2\left(\mathbb{N}\right)$ and the linear operator $B\in \mathcal{B}\left(K\right)$ is given by

$$B{e}_n={\displaystyle \frac{1}{n+1}}{e}_{n+1}(n=0,1,2,\dots ),$$

where ${\left\{e_n\right\}}_{n=0}^{\infty }$ is a standard orthogonal basis, i.e., B is defined by

$$B(x_0,x_1,x_2,\dots )=(0,{\displaystyle \frac{1}{1}}{x}_0,{\displaystyle \frac{1}{2}}{x}_1,{\displaystyle \frac{1}{3}}{x}_2,\dots ).$$

Then

$$B^k{e}_n={\displaystyle \frac{1}{n+1}}·{\displaystyle \frac{1}{n+2}}·{\displaystyle \frac{1}{n+k}}{e}_{n+k}.$$

Hence, we have $\left|\right|B^k\left|\right|={\displaystyle \frac{1}{k!}}$, and then

$$\underset{n\to \infty }{lim}\left|\right|B^k{\left|\right|}^{{\displaystyle \frac{1}{k}}}=0.$$

Thus T is quasinilpotent.

Set $H=\mathbb{C}⨁\mathbb{C}\oplus X$. Construct a block diagonal operator $T=A⨁B$, where ${T|}_{\mathbb{C}⨁\mathbb{C}}=A⨁0$, ${T|}_X=0⨁B$. Then $T\in {\left(\mathcal{B}\left(H\right)\right)}^{ⓓ}$ by [5][Theorem 2.1]. Moreover, we see that $T\in \mathcal{B}{\left(H\right)}_{L,L}^{ⓓ},$ where $L\in \mathcal{B}\left(H\right)$ given by $diag(\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right),0_{2\times 2})$. Then $T^{ⓓ}=L^{ⓓ}=diag(\left(\begin{array}{cc}{\displaystyle \frac{1}{4}}& {\displaystyle \frac{1}{4}}\\ {\displaystyle \frac{1}{4}}& {\displaystyle \frac{1}{4}}\end{array}\right),0_{2\times 2})$. We easily check that

$$T\in {\left(\mathcal{B}\left(H\right)\right)}^{ⓓ},L,TL\in {\left(\mathcal{B}\left(H\right)\right)}^{\overline{)\#}},T^d\mathcal{B}\left(H\right)=L\mathcal{B}\left(H\right)$$

and ${\left(LT\right)}^{ⓓ}=T^{ⓓ}{L}^{ⓓ}.$