m-Generalized RightWeighted Group Inverse in Banach Algebras

1. Introduction

An involution on a Banach algebra $\mathcal{A}$ is an anti-automorphism squaring to the identity map 1. A Banach algebra $\mathcal{A}$ with an involution * is a Banach *-algebra. The involution * is proper when $x^{*}x=0⟹x=0$ for all $x\in \mathcal{A}$; for instance, this is always true in $C^{*}$-algebras. In this paper, all Banach algebras are assumed to be complex and equipped with a proper involution *.

An element $a\in R$ has weak group inverse if there exist $x\in R$ and $n\in N$ such that

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

The weak group inverse was extensively studied by many authors from different point of views (see [13,20,22,23,24]).

Recently, the weak group inverse was extended to the m-weak group inverse. An element $a\in R$ has m-weak group inverse if there exist $x\in R$ and $n\in N$ such that

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

Such an x is unique if it exists, and is denoted by $a^{W_m}$. Various expressions and applications of the m-weak group inverse are known in the literature, e.g., [9,10,14,15,17,25].

Recall that $a\in {\mathcal{A}}^d$ (i.e., generalized right Drazin inverse) if there exists $x\in \mathcal{A}$ such that

$$a{x}^2=x,ax=xa,a-axa\in {\mathcal{A}}^{qnil}.$$

Here,

$${\mathcal{A}}^{qnil}=\{x\in \mathcal{A}\mid \underset{n\to \infty }{lim}\Vert x^n{\Vert }^{{\displaystyle \frac{1}{n}}}=0\}.$$

In [2], the authors introduced a new generalized as a natural extension of the m-weak group inverse. An element $a\in \mathcal{A}$ has m-generalized group inverse if and only if $a\in {\mathcal{A}}^d$ and there exists $x\in \mathcal{A}$ such that

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

Such an x is unique if it exists, and is denoted by $a^{g_m}$. Many properties of m-weak group inverse are extended to the wider case such as linear operator on an infinitely dimensional Hilbert space (see [16]).

The one-sided generalized Drazin inverse was recently introduced. Consequently, many properties of the generalized Drazin inverse have been extended to this broader setting. An element $a\in \mathcal{A}$ possesses a generalized right Drazin inverse if there exists $x\in \mathcal{A}$ such that

$$a{x}^2=x,a^{2}x=axa,a-axa\in {\mathcal{A}}^{qnil}.$$

The set of all generalized right Drazin invertible elements in $\mathcal{A}$ is denoted by ${\mathcal{A}}^d$. Characterizations of this generalized inverse are thoroughly explored in [18,21].

The motivation of this paper is to introduce a new generalized inverse as a natural one-sided version of the (weak) generalized group inverses mentioned above.

Definition 1.1.

An element $a\in \mathcal{A}$ has generalized right w-group inverse if $aw\in {\mathcal{A}}_r^d$ and there exist $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left({\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^{2}x={\left({\left(aw\right)}_r^d\right)}^{*}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

The preceding x is denoted by $a_r^{w,g}$. The set of all generalized right w-group invertible elements in $\mathcal{A}$ is denoted by ${\mathcal{A}}_r^{w,g}$.

In Section 2, we investigate the generalized right w-group inverse that are essential for our investigation. In Section 3, we introduce a new generalized inverse based on the generalized right w-group inverse.

Definition 1.2.

An element $a\in \mathcal{A}$ has m-generalized right w-group inverse if $aw\in {\mathcal{A}}_r^d$ and there exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left[\left(aw\right){\left(aw\right)}_r^d\right]}^{*}{\left(aw\right)}^{m+1}x={\left[\left(aw\right){\left(aw\right)}_r^d\right]}^{*}{\left(aw\right)}^{m-1}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

The preceding x is called an m-generalized right w-group inverse of a. We denote it by $a_r^{w,g_m}$.

We prove that $a\in {\mathcal{A}}_r^{w,g}$ if and only if ${\left(aw\right)}^{m-1}a\in {\mathcal{A}}_r^g$. The representations of the generalized right w-group inverse are also presented.

To characterize the m-generalized right w-group inverse, we introduce the one-sided w-group inverse, which can be regarded as the one-sided version of the weighted group inverse in [3,8,15].

Definition 1.3.

An element $a\in \mathcal{A}$ has right w-group inverse if there exist $x\in \mathcal{A}$ such that

$$a{\left(wx\right)}^2=x,{\left(aw\right)}^{2}x=awxwa=a.$$

We denote the preceding x by $a_r^{w,\#}$. The set of all right w-group invertible elements in $\mathcal{A}$ is denoted by ${\mathcal{A}}_r^{w,\#}$.

In Section 4, we establish the fundamental properties of the m-generalized right w-group inverse. We prove that $a\in {\mathcal{A}}_r^{w,\#}$ if and only if a has m-generalized right w-group decomposition, i.e., there exist $x,y\in \mathcal{A}$ such that

$$a=x+y,x^{*}y=0,ywx=0,x\in {\mathcal{A}}_r^{w,\#},y\in {\mathcal{A}}_w^{qnil}.$$

Here,

$${\mathcal{A}}_w^{qnil}=\{x\in \mathcal{A}\mid xw\in {\mathcal{A}}^{qnil}\}.$$

The polar like property of the m-generalized right w-group inverse is thereby investigated.

Finally, in Section 5, we explore a special case of the m-generalized right w-group inverse.

Definition 1.4.

An element $a\in \mathcal{A}$ has m-weak right w-group inverse provide that there exist $x\in R$ and $n\in N$ such that

$$x=a{\left(wx\right)}^2,{\left({\left(aw\right)}^n\right)}^{*}{\left(aw\right)}^{m+1}x={\left({\left(aw\right)}^n\right)}^{*}{a}^{m-1}a,{\left(aw\right)}^n=\left(aw\right)xw{\left(aw\right)}^n.$$

The preceding x is denoted by $a_w^{w,w_m}$.

We prove that an element a belongs to ${\mathcal{A}}_r^{w,W_m}$ if and only if $aw\in {\mathcal{A}}_r^D$ and there exists an $x\in \mathcal{A}$ such that

$$x=a{\left(wx\right)}^2,{\left[{\left(aw\right)}^n\right]}^{*}{\left(aw\right)}^{m+1}x={\left[{\left(aw\right)}^n\right]}^{*}{\left(aw\right)}^{m-1}a,{\left(aw\right)}^n=\left(aw\right)\left(xw\right){\left(aw\right)}^n$$

for some $n\in N$. This result extends the main characterization of the m-weak group inverse to the one-side case.

2. Generalized Right w-Group Inverse

The purpose of this section is to introduce a new generalized inverse which is a natural generalization of (weak) group inverse in a *-Banach algebra. We begin by

Lemma 2.1.

Let $a\in {\mathcal{A}}_r^{w,\#}$. Then $aw,wa\in {\mathcal{A}}_r^{\#}$. In this case,

$$a_r^{w,\#}=a{\left[{\left(wa\right)}_r^{\#}\right]}^2.$$

Proof. 

Let $x=a_r^{w,\#}$. Then

$$a{\left(wx\right)}^2=x,{\left(aw\right)}^{2}x=awxwa=a.$$

Then

$$a{\left(wx\right)}^2=x,{\left(aw\right)}^{2}x=awxwa=a.$$

Hence,

$$aw({\left(xw\right)}^2=xw,{\left(aw\right)}^{2}xw=awxwaw=aw.$$

Hence, $aw\in {\mathcal{A}}_r^{\#}$ and ${\left(aw\right)}_r^{\#}=xw$. Also we check that

$$wa{\left(wx\right)}^2=wx,{\left(wa\right)}^{2}wx=wawxwa=wa.$$

Thus $wa\in {\mathcal{A}}_r^{\#}$ and ${\left(wa\right)}_r^{\#}=wx$. Therefore $x=a{\left(wx\right)}^2=a{\left[{\left(wa\right)}_r^{\#}\right]}^2,$ as required. □

Theorem 2.2.

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

(1)

$a\in \mathcal{A}$ has generalized right w-group decomposition.

(2)

$aw\in {\mathcal{A}}_r^d$ and there exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^{2}x={\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, a has the generalized right w-group decomposition $a=a_1+a_2$. Let $x={\left(a_1\right)}_r^{w,\#}$. By virtue of Lemma 2.1, we have

$$\begin{array}{ccc}\hfill awx& =\hfill & (a_{1}w+a_{2}w){\left(a_1\right)}_r^{w,\#}=a_{1}w{\left(a_1\right)}_r^{w,\#}+\left[a_{2}w{a}_1\right]{\left[{\left(w{a}_1\right)}_r^{\#}\right]}^2=a_{1}w{\left(a_1\right)}_r^{w,\#},\hfill \\ \hfill a{\left(wx\right)}^2& =\hfill & a_1{\left[w{\left(a_1\right)}_r^{w,\#}\right]}^2={\left(a_1\right)}_r^{w,\#}=x,\hfill \end{array}$$

Since $a_{2}w{a}_1=0$, we have

$$\begin{array}{cc}& aw-\left(aw\right)\left(xw\right)\left(aw\right)\hfill \\ \hfill =& (a_{1}w+a_{2}w)-\left[a_{1}w{\left(a_1\right)}_r^{w,\#}w\right](a_1+a_2)w=[1-a_{1}w{\left(a_1\right)}_r^{w,\#}w]a_{2}w,\hfill \end{array}$$

and then

$$\begin{array}{cc}& {\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ \hfill =& {\left|\right|[aw-\left(aw\right)\left(xw\right)\left(aw\right)]\left(aw\right)}^{n-1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ \hfill =& \left|\right|[1-a_{1}w{\left(a_1\right)}_r^{w,\#}]a_{2}w{\left(aw\right)}^{n-1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ \hfill =& \left|\right|[1-a_{1}w{\left(a_1\right)}_r^{w,\#}w]{\left(a_{2}w\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ \hfill \le & \left|\right|1-a_{1}w{\left(a_1\right)}_r^{w,\#}{w\left|\right|}^{{\displaystyle \frac{1}{n}}}\left|\right|{\left(a_{2}w\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}.\hfill \end{array}$$

Since $a_2\in {\mathcal{A}}_w^{qnil}$, we have $\underset{n\to \infty }{lim}\left|\right|{\left(a_{2}w\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0$. Therefore $\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$

Obviously, $aw=a_{1}w+a_{2}w$ with $\left(a_{2}w\right)\left(a_{1}w\right)=0$ and $a_{2}w\in {\mathcal{A}}^{qnil}$. Since $a_1\in {\mathcal{A}}_r^{\#,w}$, by virtue of Lemma Lemma 2.1, $a_{1}w\in {\mathcal{A}}_r^{\#}$. Then we have $aw\in {\mathcal{A}}_r^d$ and

$${\left(aw\right)}_r^d={\left(a_{1}w\right)}_r^{\#}+\sum _{n=1}^{\infty }{\left({\left(a_{1}w\right)}_r^{\#}\right)}^{n+1}{\left(a_{2}w\right)}^n.$$

Hence, $\left(aw\right){\left(aw\right)}_r^d=\left(a_{1}w\right){\left(a_{1}w\right)}_r^{\#}+\sum _{n=1}^{\infty }{\left({\left(a_{1}w\right)}_r^{\#}\right)}^n{\left(a_{2}w\right)}^n.$

As $a_1^{*}{a}_2=0$, we deduce that

$$\begin{array}{cc}& {\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}{a}_2\hfill \\ \hfill =& {\left(\left(a_{1}w\right){\left(a_{1}w\right)}_r^{\#}\right)}^{*}{a}_2+\sum _{n=1}^{\infty }{\left[{\left({\left(a_{1}w\right)}_r^{\#}\right)}^n{\left(a_{2}w\right)}^n\right]}^{*}{a}_2\hfill \\ \hfill =& {\left[{\left(a_{1}w\right)}_r^{\#}\right]}^{*}\left({\left(a_{1}w\right)}^{*}{a}_2\right)+\sum _{n=1}^{\infty }{\left[{\left({\left(a_{1}w\right)}_r^{\#}\right)}^{n+1}{\left(a_{2}w\right)}^n\right]}^{*}\left({\left(a_{1}w\right)}^{*}{a}_2\right)\hfill \\ \hfill =& 0;\hfill \end{array}$$

hence, ${\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}{a}_1={\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}a$. Therefore we have

$$\begin{array}{ccc}\hfill {\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^{2}x& =\hfill & {\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}(a_{1}w+a_{2}w)(a_{1}w+a_{2}w){\left(a_1\right)}_r^{w,\#}\hfill \\ & =\hfill & {\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}{\left(a_{1}w\right)}^2{\left(a_1\right)}_r^{w,\#}={\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}{a}_1\hfill \\ & =\hfill & {\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}a.\hfill \end{array}$$

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

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^{2}x={\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

Then

$$\begin{array}{ccc}\hfill x& =\hfill & \left(aw\right)x\left(wx\right)=\left(aw\right)\left[\left(aw\right)x\left(wx\right)\right]\left(wx\right)={\left(aw\right)}^{2}x{\left(wx\right)}^2\hfill \\ & =\hfill & \dots ={\left(aw\right)}^{n-1}x{\left(wx\right)}^{n-1}.\hfill \end{array}$$

Let $a_1={\left(aw\right)}^{2}x$ and $a_2=a-{\left(aw\right)}^{2}x$. Then we check that

$$\begin{array}{ccc}\hfill \left|\right|a_{2}w{a}_1\left|\right|& =\hfill & \left|\right|[a-{\left(aw\right)}^{2}x]w{\left(aw\right)}^{2}x\left|\right|\hfill \\ & =\hfill & \left|\right|[{\left(aw\right)}^2-{\left(aw\right)}^{2}xw\left(aw\right)]\left(aw\right)x\left|\right|\hfill \\ & =\hfill & {\left|\right|\left(aw\right)}^2[1-xw\left(aw\right)]\left(aw\right)x\left|\right|\hfill \\ & =\hfill & {\left|\right|\left(aw\right)}^2[1-xw\left(aw\right)]{\left(aw\right)}^{n}x{\left(wx\right)}^{n-1}\left|\right|\hfill \\ & \le \hfill & {\left|\right|aw\left|\right|\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|\left|\right|\left(aw\right)x\left(wx\right)}^{n-1}\left|\right|.\hfill \end{array}$$

This implies that

$$\left|\right|a_{2}w{a}_1{\left|\right|}^{{\displaystyle \frac{1}{n}}}\le {\left|\right|aw\left|\right|}^{{\displaystyle \frac{1}{n}}}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}{\left|\right|\left(aw\right)x\left(wx\right)}^{n-1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}.$$

By hypothesis, we have $\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0,$ and then

$$\underset{n\to \infty }{lim}\left|\right|a_{2}w{a}_1{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

Accordingly, $a_{2}w{a}_1=0$.

Since $aw-\left(aw\right){\left(aw\right)}_r^d\left(aw\right)\in {\mathcal{A}}^{qnil}$, we see that $1-{\lambda }^{*}[aw-\left(aw\right){\left(aw\right)}_r^d\left(aw\right)]\in {\mathcal{A}}^{-1}$; hence, $1-\lambda {[aw-\left(aw\right){\left(aw\right)}_r^d\left(aw\right)]}^{*}\in {\mathcal{A}}^{-1}$. Then ${[aw-\left(aw\right){\left(aw\right)}_r^d\left(aw\right)]}^{*}\in {\mathcal{A}}^{qnil}$. This implies that

$$\underset{n\to \infty }{lim}\left|\right|{\left({[aw-\left(aw\right){\left(aw\right)}_r^d\left(aw\right)]}^{*}\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

It is easy to verify that $(a-a{a}_r^{d}a)a{a}_r^{d}a=a^2{a}_r^{d}a-a{a}_r^d{a}^2{a}_r^{d}a=a^2{a}_r^{d}a-a^3{\left(a_r^d\right)}^{2}a=0$, and then

$$\begin{array}{ccc}\hfill {(a-a{a}_r^{d}a)}^n& =\hfill & {(a-a{a}_r^{d}a)}^{n-1}(a-a{a}_r^{d}a)={(a-a{a}_r^{d}a)}^{n-1}a\hfill \\ & =\hfill & {(a-a{a}_r^{d}a)}^{n-2}{a}^2=\dots =(a-a{a}_r^{d}a)a^{n-1}=a^n-a{a}_r^d{a}^n.\hfill \end{array}$$

Thus we have

$$\underset{n\to \infty }{lim}\left|\right|{({\left(aw\right)}^n-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^n)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=\underset{n\to \infty }{lim}\left|\right|{\left({[aw-\left(aw\right){\left(aw\right)}_r^d\left(aw\right)]}^{*}\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

Observing that

$$\begin{array}{ccc}\hfill {\left(aw\right)}^{2}x& =\hfill & {\left(aw\right)}^2\left[{\left(aw\right)}^{n-1}x{\left(wx\right)}^{n-1}\right]={\left(aw\right)}^{n+1}x{\left(wx\right)}^{n-1}\hfill \\ & =\hfill & [{\left(aw\right)}^{n+1}-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^{n+1}]x{\left(wx\right)}^{n-1}+\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^{n+1}{]x\left(wx\right)}^{n-1},\hfill \end{array}$$

we verify that

$$\begin{array}{ccc}\hfill \left|\right|a_1^{*}{a}_2\left|\right|& =\hfill & \left|\right|{\left({\left(aw\right)}^{2}x\right)}^{*}(a-{\left(aw\right)}^{2}x)\left|\right|\hfill \\ & \le \hfill & \left|\right|{({\left(aw\right)}^{n+1}-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^{n+1})}^{*}\left|\right|\left|\right|{\left(x{\left(wx\right)}^{n-1}\right)}^{*}\left|\right|\left|\right|a-{\left(aw\right)}^{2}x\left|\right|\hfill \\ & & +\left|\right|\left({\left(aw\right)}^{n+1}\right]x{\left(wx\right)}^{n-1}{)}^{*}\left|\right|\left|\right|{\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}(a-{\left(aw\right)}^{2}x)\left|\right|\hfill \\ & =\hfill & \left|\right|{({\left(aw\right)}^n-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^n)}^{*}{\left|\right|\left|\right|\left(aw\right)}^{*}\left|\right|\left|\right|{\left(x{\left(wx\right)}^{n-1}\right)}^{*}\left|\right|\left|\right|a-{\left(aw\right)}^{2}x\left|\right|.\hfill \end{array}$$

Then

$$\begin{array}{ccc}\hfill \left|\right|a_1^{*}{a}_2{\left|\right|}^{{\displaystyle \frac{1}{n}}}& \le \hfill & \left|\right|{\left({\left(aw\right)}^n-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^n\right)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{n}}}{\left|\right|\left(aw\right)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{n}}}\left|\right|{\left(x{\left(wx\right)}^{n-1}\right)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ & & \left|\right|a-{\left(aw\right)}^{2}x{\left|\right|}^{{\displaystyle \frac{1}{n}}}.\hfill \end{array}$$

Therefore $\underset{n\to \infty }{lim}\left|\right|a_1^{*}{a}_2{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0$, and so $a_1^{*}{a}_2=0$. Let $m\in N$. Since $x={\left(aw\right)}^{n-1}x{\left(wx\right)}^{n-1}$, we see that

$$\begin{array}{cc}& {\left|\right|\left(aw\right)}^{m}x-awxw{\left(aw\right)}^{m}x\left|\right|\hfill \\ \hfill =& \left|\right|\left[(aw-\left(aw\right)xw\left(aw\right)]{\left(aw\right)}^{m-1}\right)x\left|\right|\hfill \\ \hfill =& {\left|\right|[aw-\left(aw\right)xw\left(aw\right)]\left(aw\right)}^{m-1}{)\left(aw\right)}^{n-1}x{\left(wx\right)}^{n-1}\left|\right|\hfill \\ \hfill =& \left|\right|[{\left(aw\right)}^n-\left(aw\right)xw{\left(aw\right)}^n]{\left(aw\right)}^{m-1}{)x\left(wx\right)}^{n-1}\left|\right|\hfill \\ \hfill \le & {\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|\left|\right|\left(aw\right)}^{m-1}{)x\left(wx\right)}^{n-1}\left|\right|.\hfill \end{array}$$

Hence, $\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^{m}x-awxw{\left(aw\right)}^{m}x{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0$. Thus $awxw{\left(aw\right)}^{m}x={\left(aw\right)}^{m}x$. Accordingly, we deduce that

$$\begin{array}{ccc}\hfill a_{1}wx& =\hfill & {\left(aw\right)}^{2}xwx=aw\left[a{\left(wx\right)}^2\right]=awx,\hfill \\ \hfill a_1{\left(wx\right)}^2& =\hfill & \left[a_{1}wx\right]wx=a{\left(wx\right)}^2=x,\hfill \\ \hfill {\left(a_{1}w\right)}^{2}x& =\hfill & \left(a_{1}w\right)\left[a_{1}wx\right]=\left[{\left(aw\right)}^{2}x\right]wawx=aw\left[awxwaw\right]x={\left(aw\right)}^{2}x=a_1,\hfill \\ \hfill a_{1}wxw{a}_1& =\hfill & \left[a_{1}wx\right]\left[w{\left(aw\right)}^{2}x\right]=awxw{\left(aw\right)}^{2}x={\left(aw\right)}^{2}x=a_1.\hfill \end{array}$$

Thus, $a_1\in {\mathcal{A}}_r^{w,\#}$.

We verify that $[aw-{\left(aw\right)}^{2}xw]{\left(aw\right)}^{2}xw=aw[{\left(aw\right)}^{2}x-\left(aw\right)\left(xw\right){\left(aw\right)}^{2}x]w=0.$ Then

$$\begin{array}{cc}& {\left(aw-{\left(aw\right)}^{2}xw\right)}^n\hfill \\ \hfill =& {\left(aw-{\left(aw\right)}^{2}xw\right)}^{n-1}\left(aw-{\left(aw\right)}^{2}xw\right)\hfill \\ \hfill =& {\left(aw-{\left(aw\right)}^{2}xw\right)}^{n-1}\left(aw\right)={\left(aw-{\left(aw\right)}^{2}xw\right)}^{n-2}\left(aw-{\left(aw\right)}^{2}xw\right)\left(aw\right)\hfill \\ \hfill =& {\left(aw-{\left(aw\right)}^{2}xw\right)}^{n-1}{\left(aw\right)}^2=\dots =\left(aw-{\left(aw\right)}^{2}xw\right){\left(aw\right)}^{n-1}\hfill \\ \hfill =& {\left(aw\right)}^n-{\left(aw\right)}^{2}xw{\left(aw\right)}^{n-1}=aw[{\left(aw\right)}^{n-1}-\left(aw\right)\left(xw\right){\left(aw\right)}^{n-1}].\hfill \end{array}$$

Hence

$$\left|\right|{\left(aw-{\left(aw\right)}^{2}xw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}\le {\left|\right|aw\left|\right|}^{{\displaystyle \frac{1}{n}}}{\left({\left|\right|\left(aw\right)}^{n-1}-\left(aw\right)\left(xw\right){\left(aw\right)}^{n-1}{\left|\right|}^{{\displaystyle \frac{1}{n-1}}}\right)}^{1-{\displaystyle \frac{1}{n}}}.$$

Accordingly,

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

This implies that $a_{2}w=aw-{\left(aw\right)}^{2}xw\in {\mathcal{A}}^{qnil}$, i.e., $a_2\in {\mathcal{A}}_w^{qnil}$. Therefore we have a generalized right w-group inverse $a=a_1+a_2$, as asserted. □

We denote x in Theorem 2.2 by $a_r^{w,g}$, and call it a generalized right w-group inverse of a. ${\mathcal{A}}_r^{w,g}$ denotes the sets of all generalized w-group invertible elements in $\mathcal{A}$.

Corollary 2.3.

Let $a\in {\mathcal{A}}_r^{w,g}$. Then

$${\left[{\left(aw\right)}^{*}{\left(aw\right)}^2{a}_r^{w,g}w\right]}^{*}={\left(aw\right)}^{*}{\left(aw\right)}^2{a}_r^{w,g}w.$$

Proof. 

By hypothesis, a has generalized right w-group decomposition $a=a_1+a_2$. Let $x={\left(a_1\right)}_r^{w,\#}$. By virtue of Theorem 2.2, we have $aw\in {\mathcal{A}}_r^d$ and

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^{2}x={\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

Then we verify that

$$\begin{array}{ccc}\hfill {\left(aw\right)}^2\left(xw\right)& =\hfill & {(a_{1}w+a_{2}w)}^2{\left(a_1\right)}_r^{w,\#}w={\left(a_{1}w\right)}^2{\left(a_1\right)}_r^{w,\#}w=a_{1}w,\hfill \\ \hfill {\left(aw\right)}^{*}{\left(aw\right)}^2\left(xw\right)& =\hfill & {(a_{1}w+a_{2}w)}^{*}{a}_{1}w={\left(a_{1}w\right)}^{*}{a}_{1}w.\hfill \end{array}$$

Accordingly, ${\left[{\left(aw\right)}^{*}{\left(aw\right)}^2\left(xw\right)\right]}^{*}={\left[{\left(a_{1}w\right)}^{*}{a}_{1}w\right]}^{*}={\left(a_{1}w\right)}^{*}{a}_{1}w={\left(aw\right)}^{*}{\left(aw\right)}^2\left(xw\right),$ as required. □

Corollary 2.4.

Let $a\in \mathcal{A}$ and $w\in {\mathcal{A}}^{-1}$. Then the following are equivalent:

(1)

$a\in {\mathcal{A}}_r^{w,g}$.

(2)

$aw\in {\mathcal{A}}_r^g$.

(3)

There exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left[{\left(aw\right)}^{*}{\left(aw\right)}^{2}xw\right]}^{*}={\left(aw\right)}^{*}{\left(aw\right)}^{2}xw,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

In this case,

$$a_r^{w,g}={\left(aw\right)}_r^g{w}^{-1}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Since $w\in {\mathcal{A}}^{-1}$, we directly check that ${\left(aw\right)}_r^g=a_r^{w,g}w.$

$\left(2\right)\Rightarrow \left(1\right)$ We easily verify that $a_r^{w,g}={\left(aw\right)}_r^g{w}^{-1}.$

$\left(1\right)\Rightarrow \left(3\right)$ This is obvious by Corollary 2.3.

$\left(3\right)\Rightarrow \left(1\right)$ By hypothesis, there exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left[{\left(aw\right)}^{*}{\left(aw\right)}^{2}xw\right]}^{*}={\left(aw\right)}^{*}{\left(aw\right)}^{2}xw,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

Let $a_1={\left(aw\right)}^{2}x$ and $a_2=a-{\left(aw\right)}^{2}x$. As is the proof of Theorem 2.2, we verify that $a_{2}w{a}_1=0,{\left(a_1\right)}_r^{w,\#}=x$ and $x={\left(aw\right)}^{n-1}x{\left(wx\right)}^{n-1}$. It is easy to check that

$$\begin{array}{ccc}\hfill \left|\right|x-\left(xw\right)\left(awx\right)\left|\right|& =\hfill & \left|\right|[{\left(aw\right)}^{n-1}-xwaw{\left(aw\right)}^{n-1}]x{\left(wx\right)}^{n-1}\left|\right|\hfill \\ & =\hfill & {\left|\right|\left(aw\right)}^{n-1}-xw{\left(aw\right)}^n{\left|\right|\left|\right|x\left(wx\right)}^{n-1}\left|\right|.\hfill \end{array}$$

By hypothesis, we have

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

Hence,

$$\underset{n\to \infty }{lim}\left|\right|x-{\left(xw\right)\left(awx\right)\left|\right|}^{{\displaystyle \frac{1}{n-1}}}=0.$$

This implies that $awx=\left(aw\right)\left(xw\right)\left(aw\right)x$. We derive that

$$\begin{array}{ccc}\hfill w^{*}{a}_1^{*}{a}_{2}w& =\hfill & w^{*}{\left[{\left(aw\right)}^{2}x\right]}^{*}[a-{\left(aw\right)}^{2}x]w\hfill \\ & =\hfill & {\left[{\left(aw\right)}^{2}xw\right]}^{*}aw[1-\left(aw\right)xw]\hfill \\ & =\hfill & {\left[{\left(aw\right)}^{*}{\left(aw\right)}^{2}xw\right]}^{*}[1-\left(aw\right)xw]\hfill \\ & =\hfill & \left[{\left(aw\right)}^{*}{\left(aw\right)}^{2}xw\right][1-\left(aw\right)xw]\hfill \\ & =\hfill & {\left(aw\right)}^{*}{\left(aw\right)}^{2}xw-{\left(aw\right)}^{*}{\left(aw\right)}^{2}xw\left(aw\right)xw\hfill \\ & =\hfill & {\left(aw\right)}^{*}{\left(aw\right)}^{2}xw-{\left(aw\right)}^{*}\left(aw\right)\left(awxwawx\right)w\hfill \\ & =\hfill & {\left(aw\right)}^{*}{\left(aw\right)}^{2}xw-{\left(aw\right)}^{*}{\left(aw\right)}^{2}xw=0.\hfill \end{array}$$

As $w\in {\mathcal{A}}^{-1}$, we get $w^{*}\in {\mathcal{A}}^{-1}$; whence, $a_1^{*}{a}_2=0$.

By hypothesis, there exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}xw=\left(aw\right){\left(xw\right)}^2,{\left[{\left(aw\right)}^{*}{\left(aw\right)}^{2}xw\right]}^{*}={\left(aw\right)}^{*}{\left(aw\right)}^{2}xw,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(xw\right){\left(aw\right)}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

Then $xw={\left(aw\right)}_r^g$, and so we have $aw-{\left(aw\right)}^2\left(xw\right)\in {\mathcal{A}}^{qnil}$. Hence $a_2=a-{\left(aw\right)}^{2}x\in {\mathcal{A}}_w^{qnil}$. Accordingly, a has generalized right w-group decomposition $a=a_1+a_2$, thus yielding the result. □

Example 2.5.

The space ${\ell }^2\left(N\right)$ is a Hilbert space consisting of all square-summable infinite sequences of complex numbers. Let $H={\ell }^2\left(N\right)⨁{\ell }^2\left(N\right)$ be the Hilbert space of the sum of ${\ell }^2\left(N\right)$ and itself. Let $\sigma$ be defined on ${\ell }^2\left(N\right)$ by:

$$\sigma (x_1,x_2,x_3,\dots )=(0,0,1^{{\displaystyle \frac{1}{1}}}{x}_1,2^{{\displaystyle \frac{1}{2}}}{x}_2,3^{{\displaystyle \frac{1}{3}}}{x}_3,\dots );$$

$\tau$ is defined on ${\ell }^2\left(N\right)$ by:

$$\tau (x_1,x_2,x_3,\dots )=(3^{{\displaystyle \frac{1}{3}}}{x}_3,4^{{\displaystyle \frac{1}{4}}}{x}_4,5^{{\displaystyle \frac{1}{5}}}{x}_5,\dots )$$

and u is defined on ${\ell }^2\left(N\right)$ by:

$$u(x_1,x_2,x_3,\dots )=(x_1/1^{{\displaystyle \frac{1}{1}}},x_2/2^{{\displaystyle \frac{1}{2}}},x_3/3^{{\displaystyle \frac{1}{3}}},x_4/4^{{\displaystyle \frac{1}{4}}},\dots ).$$

We easily check that

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim}\left|\right|{\left(x_n\right)}^2\left|\right|/\left|\right|{(x_n/\sqrt[n]{n})}^2\left|\right|& =\hfill & \underset{n\to \infty }{lim}{\left(\sqrt[n]{n}\right)}^2=1,\hfill \\ \hfill \underset{n\to \infty }{lim}\left|\right|{\left(x_n\right)}^2\left|\right|/\left|\right|{\left(\sqrt[n]{n}{x}_n\right)}^2\left|\right|& =\hfill & \underset{n\to \infty }{lim}{\left({\displaystyle \frac{1}{\sqrt[n]{n}}}\right)}^2=1.\hfill \end{array}$$

Then $\sum _{n=1}^{\infty }{\left(\sqrt[n]{n}{x}_n\right)}^2$ and $\sum _{n=1}^{\infty }{\left(x_n/\sqrt[n]{n}\right)}^2$ converge. Hence, $\sigma ,\tau$ and u are well defined.

Let $\delta$ be defined on ${\ell }^2\left(N\right)$ by:

$$\delta (x_1,x_2,x_3,\dots )=({\displaystyle \frac{1}{3}}{x}_2,{\displaystyle \frac{1}{4}}{x}_3,{\displaystyle \frac{1}{5}}{x}_4,\dots ).$$

Let $e_n=({\underbrace{0,0,\dots ,0,1,}}_{n}0,0,\dots )$. Then $\delta \left(e_n\right)={\displaystyle \frac{1}{n+2}}{e}_{n+1}$. One directly checks that

$${\delta }^k\left(e_n\right)=\prod _{j=0}^{k-1}{\displaystyle \frac{1}{n+j+2}}{e}_{n+k}={\displaystyle \frac{(n+1)!}{(n+k+1)!}}{e}_{n+k}.$$

Then we derive that

$$\left|\right|{\delta }^k\left|\right|={\displaystyle \frac{1}{(k+1)!}}.$$

By virtue of Stirling’s formula, we have $(k+1)!\sim \sqrt{2\pi (k+1)}{\left({\displaystyle \frac{k+1}{e}}\right)}^{k+1}$, and then

$$\underset{k\to \infty }{lim}\left|\right|{\delta }^k{\left|\right|}^{{\displaystyle \frac{1}{k}}}=\underset{k\to \infty }{lim}{\displaystyle \frac{e^{1+\frac{1}{k}}}{{\left(2\pi \right)}^{\frac{1}{k}}{\left[{(k+1)}^{\frac{1}{k+1}}\right]}^{1+\frac{1}{k}}{(k+1)}^{1+\frac{1}{k}}}}=0.$$

Thus $\delta$ is quasinilpotent, and then so is ${\delta }^2$.

Let $T=\tau \oplus {\delta }^2$ be the direct sum of $\tau$ and ${\delta }^2$, $W=u\oplus I$ be the direct sum of u and the identical operator I, $S=\sigma \oplus 0$ be the direct sum of $\sigma$ and 0. Then $T,W$ and S are operators on H. We see that $T=\alpha +\beta$, where $\alpha =\tau \oplus 0$ and $\beta =0\oplus {\gamma }^2$.

Claim 1. $\alpha$ has right W-group inverse. We directly check that $\tau u\sigma u=u\tau u\sigma =I.$ Then $\alpha WSW=W\alpha WS=I\oplus 0$. Hence,

$$\begin{array}{ccc}\hfill \alpha {\left(WS\right)}^2& =\hfill & \left[\alpha WSW\right]S=S,\hfill \\ \hfill \alpha WSW\alpha & =\hfill & \left[\alpha WSW\right]\alpha =\alpha ,\hfill \\ \hfill {\left(\alpha W\right)}^{2}S& =\hfill & \alpha \left[W\alpha WS\right]=\alpha .\hfill \end{array}$$

Then $S={\alpha }_r^{W,\#}$.

Claim 2. $W\beta$ is quasinilpotent. Since $W\beta =0\oplus {\delta }^2$, we see that $W\beta$ is quasinilpotent.

Claim 3. ${\alpha }^{*}\beta =\beta W\alpha =0$. This is obvious.

Therefore T has a generalized right W-group inverse by Theorem 2.2.

3. m-Generalized Right w-Group Inverse

This section introduces a new generalized inverse, which serves as a natural generalization of the (weak) group inverse within the context of *-Banach algebras. Here, we explore the fundamental properties of the m-generalized right w-group inverse for elements in a Banach *-algebra.

Lemma 3.1.

Let $a\in \mathcal{A}$ and $n\in N$. Then $a^n\in {\mathcal{A}}_r^d$ and ${\left(a^n\right)}_r^d={\left(a_r^d\right)}^n$.

Proof. 

Set $x={\left(a_r^d\right)}^n$. Then we verify that

$$\begin{array}{ccc}\hfill a^n{x}^2& =\hfill & a^n{\left(a_r^d\right)}^n{\left(a_r^d\right)}^n=a{a}_r^d{\left(a_r^d\right)}^n=a{\left(a_r^d\right)}^{n+1}=x,\hfill \\ \hfill {\left(a^n\right)}^{2}x& =\hfill & {\left(a^n\right)}^2{\left(a_r^d\right)}^n=a^n\left[a^n{\left(a_r^d\right)}^n\right]=a^{n+1}{a}_r^d\hfill \\ & =\hfill & a{a}_r^d{a}^n=a^n{\left(a_r^d\right)}^n{a}^n=a^{n}x{a}^n,\hfill \\ \hfill a^n-a^{n}x{a}^n& =\hfill & a^n-a^n{\left(a_r^d\right)}^n{a}^n=a^n-a{a}_r^d{a}^n=[a-a{a}_r^{d}a]a^{n-1}.\hfill \end{array}$$

Since $[a-a{a}_r^{d}a]a^{n-1}=a^n-a{a}_r^d{a}^n=a^n-a^n{a}_r^{d}a=a^{n-1}[a-a{a}_r^{d}a]$ and $a-a{a}_r^{d}a\in {\mathcal{A}}^{qnil}$, it follows by [26] that $[a-a{a}_r^{d}a]a^{n-1}\in {\mathcal{A}}^{qnil}$. Hence, $a^n-a^{n}x{a}^n\in {\mathcal{A}}^{qnil}$. Therefore ${\left(a^n\right)}_r^d={\left(a_r^d\right)}^n$. □

Theorem 3.2.

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_r^{w,g_m}$ if and only if $aw\in {\mathcal{A}}_r^d$ and ${\left(aw\right)}^{m-1}a\in {\mathcal{A}}_r^{w,g}$. In this case,

$$a_r^{w,g_m}={\left(aw\right)}^{m-1}{\left({\left(aw\right)}^{m-1}a\right)}_r^{w,g}.$$

Proof. 

⟹ Since $a\in {\mathcal{A}}_r^{w,g_m}$, then $aw\in {\mathcal{A}}_r^d$ and there exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left[\left(aw\right){\left(aw\right)}_r^d\right]}^{*}{\left(aw\right)}^{m+1}x={\left[\left(aw\right){\left(aw\right)}_r^d\right]}^{*}{\left(aw\right)}^{m-1}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

In view of Lemma 3.1, $\left[{\left(aw\right)}^{m-1}a\right]w={\left(aw\right)}^m\in {\mathcal{A}}_r^d$. Let $y={\left(xw\right)}^{m-1}x$. Then we verify that

$$\begin{array}{cc}& {\left(aw\right)}^{m-1}a{\left(wy\right)}^2\hfill \\ \hfill =& {\left(aw\right)}^{m-1}a{\left(wx\right)}^{2m}={\left(aw\right)}^{m-1}\left[a{\left(wx\right)}^2\right]{\left(wx\right)}^{2m-2}\hfill \\ \hfill =& {\left(aw\right)}^{m-1}x{\left(wx\right)}^{2m-2}={\left(aw\right)}^{m-2}\left[a{\left(wx\right)}^2\right]{\left(wx\right)}^{2m-3}={\left(aw\right)}^{m-2}x{\left(wx\right)}^{2m-3}\hfill \\ \hfill =& \dots ={\left(aw\right)}^{m-m}x{\left(wx\right)}^{2m-(m+1)}=x{\left(wx\right)}^{m-1}={\left(xw\right)}^{m-1}x=y,\hfill \\ & {\left({\left({\left(aw\right)}^{m-1}aw\right)}_r^d\right)}^{*}{\left({\left(aw\right)}^{m-1}aw\right)}^{2}y\hfill \\ \hfill =& {\left({\left({\left(aw\right)}^m\right)}_r^d\right)}^{*}{\left(aw\right)}^{2m}{\left(xw\right)}^{m-1}x\hfill \\ \hfill =& {\left({\left({\left(aw\right)}^m\right)}_r^d\right)}^{*}{\left(aw\right)}^{2m}{\left(xw\right)}^{m-1}x\hfill \\ \hfill =& {\left({\left({\left(aw\right)}^m\right)}_r^d\right)}^{*}{\left(aw\right)}^{2m-1}\left[a{\left(wx\right)}^2\right]w{\left(xw\right)}^{m-3}x\hfill \\ \hfill =& {\left({\left({\left(aw\right)}^m\right)}_r^d\right)}^{*}{\left(aw\right)}^{2m-1}{\left(xw\right)}^{m-2}x\hfill \\ \hfill =& \dots ={\left({\left({\left(aw\right)}^m\right)}_r^d\right)}^{*}{\left(aw\right)}^{2m-(m-1)}{\left(xw\right)}^{m-m}x\hfill \\ \hfill =& {\left({\left({\left(aw\right)}^m\right)}_r^d\right)}^{*}{\left[{\left(aw\right)}_r^d\right]}^{*}{\left(aw\right)}^{m+1}x\hfill \\ \hfill =& {\left({\left({\left(aw\right)}^{m-1}\right)}_r^d\right)}^{*}{\left[{\left(aw\right)}_r^d\right]}^{*}{\left(aw\right)}^{m-1}a\hfill \\ \hfill =& {\left({\left({\left(aw\right)}^m\right)}_r^d\right)}^{*}{\left(aw\right)}^{m-1}a,\hfill \end{array}$$

Moreover, we have

$$\begin{array}{cc}& {\left({\left(aw\right)}^{m-1}aw\right)}^n-\left({\left(aw\right)}^{m-1}aw\right)\left(yw\right){\left({\left(aw\right)}^{m-1}aw\right)}^n\hfill \\ \hfill =& {\left({\left(aw\right)}^m\right)}^n-{\left(aw\right)}^m{\left(xw\right)}^{m-1}xw{\left({\left(aw\right)}^m\right)}^n\hfill \\ \hfill =& {\left({\left(aw\right)}^m\right)}^n{-aw)}^m{\left(xw\right)}^m{\left({\left(aw\right)}^m\right)}^n\hfill \\ \hfill =& {\left({\left(aw\right)}^m\right)}^n-\left(aw\right)\left(xw\right){\left({\left(aw\right)}^m\right)}^n\hfill \\ \hfill =& [{\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n]{\left[{\left(aw\right)}^n\right]}^{m-1}.\hfill \end{array}$$

Hence, we have

$$\begin{array}{cc}& \left|\right|{\left({\left(aw\right)}^{m-1}aw\right)}^n-\left({\left(aw\right)}^{m-1}aw\right)\left(yw\right){\left({\left(aw\right)}^{m-1}aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ \hfill \le & {\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}{\left|\right|aw\left|\right|}^{m-1}.\hfill \end{array}$$

Therefore

$$\underset{n\to \infty }{lim}\left|\right|{\left({\left(aw\right)}^{m-1}aw\right)}^n-\left({\left(aw\right)}^{m-1}aw\right)\left(yw\right){\left({\left(aw\right)}^{m-1}aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

By virtue of Theorem 2.2, ${\left(aw\right)}^{m-1}a\in {\mathcal{A}}_r^{w,g}$. Therefore ${\left({\left(aw\right)}^{m-1}a\right)}_r^{w,g}=y={\left(xw\right)}^{m-1}x$, as required.

⟸ By hypothesis, $aw\in {\mathcal{A}}_r^d$. In view of Lemma 3.1, we may assume that ${\left(a^m\right)}_r^d={\left(a_r^d\right)}^m$. Hance, we have ${\left(aw\right)}^m{\left({\left(aw\right)}^m\right)}_r^d={\left(aw\right)}^m{\left[{\left(aw\right)}_r^d\right]}^m=\left(aw\right){\left(aw\right)}_r^d$. Moreover, we can find a $y\in \mathcal{A}$ such that

$$\begin{array}{c}y={\left(aw\right)}^{m-1}a{\left(wy\right)}^2,{\left({\left(aw\right)}^m{\left({\left(aw\right)}^m\right)}_r^d\right)}^{*}{\left(aw\right)}^{2m}y={\left({\left(aw\right)}^m{\left({\left(aw\right)}^m\right)}_r^d\right)}^{*}{\left(aw\right)}^{m-1}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^{mn}-{\left(aw\right)}^m\left(yw\right){\left(aw\right)}^{mn}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

Then we verify that

$$\begin{array}{ccc}\hfill y& =\hfill & {\left(aw\right)}^{m}y\left(wy\right)={\left(aw\right)}^m\left[{\left(aw\right)}^{m}y\left(wy\right)\right]\left(wy\right)={\left(aw\right)}^{2m}y{\left(wy\right)}^2\hfill \\ & =\hfill & \dots ={\left(aw\right)}^{(n-1)m}y{\left(wy\right)}^{n-1}={\left(aw\right)}^{mn}y{\left(wy\right)}^n.\hfill \end{array}$$

Take $x={\left(aw\right)}^{m-1}y$. Then

$$\begin{array}{ccc}\hfill \left|\right|[1-\left(aw\right){\left(aw\right)}_r^d]y{\left|\right|}^{{\displaystyle \frac{1}{mn}}}& =\hfill & \left|\right|[1-\left(aw\right){\left(aw\right)}_r^d]{\left(aw\right)}^{mn}y{\left(wy\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{mn}}}\hfill \\ & \le \hfill & {\left|\right|\left(aw\right)}^{mn}-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^{mn}{\left|\right|}^{{\displaystyle \frac{1}{mn}}}{\left|\right|y\left(wy\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{mn}}}.\hfill \end{array}$$

Hence, $\left|\right|[1-\left(aw\right){\left(aw\right)}_r^d]y{\left|\right|}^{{\displaystyle \frac{1}{mn}}}=0$. This implies that $[1-\left(aw\right){\left(aw\right)}_r^d]y=0$; and so $y=\left(aw\right){\left(aw\right)}_r^{d}y$. Then we verify that

$$\begin{array}{ccc}\hfill a{\left(wx\right)}^2& =\hfill & a\left[w{\left(aw\right)}^{m-1}y\right]\left[w{\left(aw\right)}^{m-1}y\right]={\left(aw\right)}^{m}yw\left[{\left(aw\right)}^{m-1}y\right]\hfill \\ & =\hfill & {\left(aw\right)}^{m}yw{\left(aw\right)}^{m-1}\left[\left(aw\right){\left(aw\right)}_r^{d}y\right]\hfill \\ & =\hfill & {\left(aw\right)}^m\left[yw{\left(aw\right)}^m\right]{\left(aw\right)}_r^{d}y\hfill \\ & =\hfill & {\left(aw\right)}^m\left[\left(yw\right)\left({\left(aw\right)}^{m-1}aw\right)\right]{\left(aw\right)}_r^{d}y\hfill \\ & =\hfill & {\left(aw\right)}^m{\left(aw\right)}_r^{d}y={\left(aw\right)}^{m-1}y=x,\hfill \\ \hfill {\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^{m+1}x& =\hfill & {\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^{m+1}{\left(aw\right)}^{m-1}y\hfill \\ & =\hfill & {\left({\left(aw\right)}^m{\left({\left(aw\right)}^m\right)}_r^d\right)}^{*}{\left(aw\right)}^{2m}y\hfill \\ & =\hfill & {\left({\left(aw\right)}^m{\left({\left(aw\right)}^m\right)}_r^d\right)}^{*}{\left(aw\right)}^{m-1}a\hfill \\ & =\hfill & {\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^{m-1}a.\hfill \end{array}$$

Obviously, we have

$$\begin{array}{ccc}\hfill {\left(aw\right)}^{m(n+1)}-\left(aw\right)\left(xw\right){\left(aw\right)}^{m(n+1)}& =\hfill & {\left(aw\right)}^{mn+m}-\left(aw\right)xw{\left(aw\right)}^{m(n+1)}\hfill \\ & =\hfill & {\left(aw\right)}^m{\left(aw\right)}^{mn}-\left({\left(aw\right)}^{m}yw\right){\left(aw\right)}^{m(n+1)}\hfill \\ & =\hfill & {[{\left({\left(aw\right)}^m\right)}^n-{\left(aw\right)}^m\left(yw\right){\left(aw\right)}^m)}^n{]\left(aw\right)}^m.\hfill \end{array}$$

Hence,

$$\begin{array}{cc}& {\left|\right|\left(aw\right)}^{m(n+1)}-\left(aw\right)\left(xw\right){\left(aw\right)}^{m(n+1)}{\left|\right|}^{{\displaystyle \frac{1}{m(n+1)}}}\hfill \\ \hfill \le & \left|\right|{\left({\left(aw\right)}^m\right)}^n-\left(aw\right)\left(yw\right){\left({\left(aw\right)}^m\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{m(n+1)}}}{\left|\right|\left(aw\right)}^m{\left|\right|}^{{\displaystyle \frac{1}{m(n+1)}}}\hfill \\ \hfill =& {\left|\right|\left(aw\right)}^m{\left|\right|}^{{\displaystyle \frac{1}{m(n+1)}}}\left|\right|{\left({\left(aw\right)}^m\right)}^n-\left(aw\right)\left(yw\right){\left({\left(aw\right)}^m\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}.\hfill \end{array}$$

Thus, we derive

$$\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^{m(n+1)}-\left(aw\right)\left(xw\right){\left(aw\right)}^{m(n+1)}{\left|\right|}^{{\displaystyle \frac{1}{m(n+1)}}}=0.$$

Therefore $a_r^{g_m,w}=x=a^{m-1}y={\left(aw\right)}^{m-1}{\left({\left(aw\right)}^{m-1}a\right)}_r^{w,g}$, as asserted. □

Corollary .3.

Let $a\in {\mathcal{A}}_r^{w,g_m}$. Then ${\left({\left(aw\right)}^{m-1}a\right)}_r^{w,g}={\left(a_r^{w,g_m}w\right)}^{m-1}{a}_r^{w,g_m}.$

Proof. 

This is obvious by the proof of Theorem 3.2. □

An element $a\in \mathcal{A}$ has $(m,w)$-generalized group inverse if $a\in {\mathcal{A}}^{d,w}$ and there exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left[{\left(aw\right)}^d\right]}^{*}{\left(aw\right)}^{m+1}x={\left[{\left(aw\right)}^d\right]}^{*}{\left(aw\right)}^{m-1}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-xw{\left(aw\right)}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

The preceding x is unique, and is called the $(m,w)$-generalized group inverse of a. We denote it by $a_w^{g_m}$. Many elementary properties of the $(m,w)$-generalized group inverse are established in [5]. We are ready to prove:

Theorem 3.4.

An element $a\in \mathcal{A}$ has $(m,w)$-generalized group inverse if and only if $aw\in {\mathcal{A}}^d$ and $a\in \mathcal{A}$ has an m-generalized right w-group inverse.

Proof. 

⟹ This is obvious.

⟸ Set $x=a_r^{w,g_m}$. Then we have $aw\in {\mathcal{A}}_r^d$ and there exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^{m+1}x={\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}{a}^{m-1},\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

Set $y={\left(aw\right)}_r^d$. Then

$$\left(aw\right)y^2=y,{\left(aw\right)}^{2}y=\left(aw\right)y\left(aw\right),aw-\left(aw\right)y\left(aw\right)\in {\mathcal{A}}^{qnil}.$$

Set $z={\left(aw\right)}^d$. Then

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

Hence, ${\left(aw\right)}^n-z{\left(aw\right)}^{n+1}={[aw-z{\left(aw\right)}^{2}z]}^n$. Therefore $\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-z{\left(aw\right)}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$

Hence, $y=\left(aw\right)y^2={\left(aw\right)}^n{y}^{n+1}$. On the other hand, we have $z=\left(aw\right)z^2={\left(aw\right)}^n{z}^{n+1}.$ Clearly, $\left(aw\right)y{\left(aw\right)}^{2}y=\left(aw\right)y\left[\left(aw\right)y\left(aw\right)\right]=\left[\left(aw\right)y\left(aw\right)\right]y\left(aw\right)={\left(aw\right)}^2{y}^2\left(aw\right)=\left(aw\right)y\left(aw\right)$; hence,

$${\left(aw\right)}^{2}y[aw-{\left(aw\right)}^{2}y]={\left(aw\right)}^{2}y\left(aw\right)-{\left(aw\right)}^{2}y{\left(aw\right)}^{2}y=0.$$

Thus,

$${[aw-{\left(aw\right)}^{2}y]}^2=[aw-{\left(aw\right)}^{2}y][aw-{\left(aw\right)}^{2}y]=\left(aw\right)[aw-{\left(aw\right)}^{2}y].$$

By induction, we have ${[aw-{\left(aw\right)}^{2}y]}^n={\left(aw\right)}^n[aw-{\left(aw\right)}^{2}y]$. Since $aw-\left(aw\right)y\left(aw\right)\in {\mathcal{A}}^{qnil}$, by using Cline’s formula (see [11]), $aw-{\left(aw\right)}^{2}y\in {\mathcal{A}}^{qnil}$. Hence

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

For any $n\in N$, we have

$$\begin{array}{ccc}\hfill z^{n+1}{\left(aw\right)}^{n+1}y& =\hfill & z^{n+1}{\left(aw\right)}^{n+1}\left[{\left(aw\right)}^n{y}^{n+1}\right]\hfill \\ & =\hfill & z^{n+1}{\left(aw\right)}^{2n+1}{y}^{n+1}\hfill \\ & =\hfill & \left[z{\left(aw\right)}^{n+1}\right]y^{n+1}\hfill \\ & =\hfill & {\left(aw\right)}^n{y}^{n+1}=y,\hfill \\ \hfill z^{n+1}{\left(aw\right)}^n& =\hfill & \left(aw\right)z^2=z.\hfill \end{array}$$

Then

$$\left|\right|y-{z\left|\right|}^{{\displaystyle \frac{1}{n}}}\le \left|\right|y-{z\left|\right|}^{1+{\displaystyle \frac{1}{n}}}{\left|\right|\left(aw\right)}^n-{\left(aw\right)}^{n+1}y{\left|\right|}^{{\displaystyle \frac{1}{n}}}.$$

This implies that

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

whence, $\left|\right|y-z\left|\right|=0$. Thus $y=z$.

Since ${\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^{m+1}x={\left(\left(aw\right){\left(aw\right)}_r^d\right)}^{*}{a}^{m-1}$, we have

$$\begin{array}{ccc}\hfill {\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{m-1}x& =\hfill & {\left(\left(aw\right){\left(aw\right)}^d\right)}^{*}\left[{\left(\left(aw\right){\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{m-1}x\right]\hfill \\ & =\hfill & {\left(\left(aw\right){\left(aw\right)}^d\right)}^{*}\left[{\left({\left(aw\right)}^d\right)}^{*}{a}^{m-1}\right]={\left({\left(aw\right)}^d\right)}^{*}{a}^{m-1}.\hfill \end{array}$$

Let $t=aw{\left(aw\right)}^{d}x$. One directly verifies that

$$\begin{array}{ccc}\hfill \left|\right|awx-awt\left|\right|& =\hfill & \left|\right|[aw-{\left(aw\right)}^2{\left(aw\right)}^d]x\left|\right|\hfill \\ & =\hfill & \left|\right|[aw-{\left(aw\right)}^2{\left(aw\right)}^d]{\left(aw\right)}^{n-1}x{\left(aw\right)}^{n-1}\left|\right|\hfill \\ & =\hfill & \left|\right|{[aw-{\left(aw\right)}^2{\left(aw\right)}^d]}^{n}x{\left(aw\right)}^{n-1}\left|\right|\hfill \\ & \le \hfill & \left|\right|{[aw-{\left(aw\right)}^2{\left(aw\right)}^d]}^n{\left|\right|\left|\right|x\left(aw\right)}^{n-1}\left|\right|,\hfill \end{array}$$

Since $\underset{n\to \infty }{lim}\left|\right|{[aw-{\left(aw\right)}^2{\left(aw\right)}^d]}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0,$ we deduce that

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

This implies that $awx=awt$.

By hypothesis, $x=a{\left(wx\right)}^2={\left(aw\right)}^{n}x{\left(wx\right)}^n$. Hence,

$$\begin{array}{cc}& \left|\right|x-aw{\left(aw\right)}^{d}x{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ \hfill =& {\left|\right|\left(aw\right)}^{n}x{\left(wx\right)}^n-{\left(aw\right)}^d{\left(aw\right)}^{n+1}x{\left(wx\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ \hfill \le & {\left|\right|\left(aw\right)}^n-{\left(aw\right)}^d{\left(aw\right)}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}{\left|\right|x\left|\right|}^{{\displaystyle \frac{1}{n}}}{\left|\right|wx\left|\right|}^{{\displaystyle \frac{1}{n}}},\hfill \end{array}$$

This implies that $\underset{n\to \infty }{lim}\left|\right|x-aw{\left(aw\right)}^{d}x{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0,$ and then $aw{\left(aw\right)}^{d}x=x$. Therefore

$$\begin{array}{ccc}\hfill t& =\hfill & aw{\left(aw\right)}^{d}x=aw{\left(aw\right)}^{d}a{\left(wx\right)}^2={\left(aw\right)}^2{\left(aw\right)}^{d}xwx\hfill \\ & =\hfill & {\left(aw\right)}^2{\left(aw\right)}^{d}xw\left[aw{\left(aw\right)}^{d}x\right]=awtwt=a{\left(wt\right)}^2\hfill \\ \hfill {\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{m+1}t& =\hfill & {\left({\left(aw\right)}^d\right)}^{*}{\left(aw\right)}^{m+1}x={\left({\left(aw\right)}^d\right)}^{*}{a}^{m-1}.\hfill \end{array}$$

Moreover, we derive that

$$\begin{array}{cc}& {\left|\right|\left(aw\right)}^n-tw{\left(aw\right)}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ \hfill =& \left|\right|[{\left(aw\right)}^n-{\left(aw\right)}^{n+1}{\left(aw\right)}^d]+[{\left(aw\right)}^d{\left(aw\right)}^{n}aw-{\left(aw\right)}^d\left(aw\right)\left(xw\right){\left(aw\right)}^{n}aw]{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ \hfill \le & {\left|\right|\left(aw\right)}^n-{\left(aw\right)}^{n+1}{\left(aw\right)}^d{\left|\right|}^{{\displaystyle \frac{1}{n}}}+{\left|\right|\left(aw\right)}^d{\left|\right|}^{{\displaystyle \frac{1}{n}}}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}{\left|\right|aw\left|\right|}^{{\displaystyle \frac{1}{n}}},\hfill \end{array}$$

we have

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

Therefore $a\in \mathcal{A}$ has $(m,w)$-generalized group inverse. □

Corollary 3.5.

An element $a\in \mathcal{A}$ has a generalized w-group inverse if and only if $a\in {\mathcal{A}}^d$ and $a\in \mathcal{A}$ has a generalized right w-group inverse.

Proof. 

This is obvious by choosing $m=1$ in Theorem 3.4. □

Example 3.6.

Let $H={\ell }^2\left(N\right)⨁{\ell }^2\left(N\right)$. Construct $T,W,\tau$ and $\delta$ as in Example 2.5. Let $\eta$ be defined on ${\ell }^2\left(N\right)$ by:

$$\eta (x_1,x_2,x_3,\dots )=(2^{{\displaystyle \frac{1}{2}}}{x}_2,3^{{\displaystyle \frac{1}{3}}}{x}_3,4^{{\displaystyle \frac{1}{4}}}{x}_4,\dots )$$

and u is defined on ${\ell }^2\left(N\right)$ by:

$$u(x_1,x_2,x_3,\dots )=(x_1/1^{{\displaystyle \frac{1}{1}}},x_2/2^{{\displaystyle \frac{1}{2}}},x_3/3^{{\displaystyle \frac{1}{3}}},x_4/4^{{\displaystyle \frac{1}{4}}},\dots ).$$

Let $\theta$ be the right shift operator defined on ${\ell }^2\left(N\right)$ by:

$$\theta (x_1,x_2,x_3,\dots )=({\displaystyle \frac{1}{3\times 4}}{x}_3,{\displaystyle \frac{1}{4\times 5}}{x}_4,{\displaystyle \frac{1}{5\times 6}}{x}_5,\dots ).$$

As the proof in Example 2.5, we easily see that $\theta$ is well defined.

We easily see that $\tau =\eta u\eta$ and $\theta ={\delta }^2$. Let $A=\eta \oplus \theta$ and $W=u\oplus I$. Then $AWA=T$. In view of Example 2.5, T has generalized right W-group inverse. Therefore A has 2-generalized right W-group inverse by Theorem 3.2.

Clearly, $AW=\eta u\oplus \theta$, where $\eta u(x_1,x_2,x_3,\dots )=(x_1,x_2,x_3,\dots )$ for any $(x_1,x_2,x_3,\dots )\in {\ell }^2\left(N\right)$. Then $\eta u\in \mathcal{B}\left({\ell }^2\left(N\right)\right)$ has no generalized Deazrin inverse. Hence, $AW\in \mathcal{B}\left(H\right)$ has no generalized Drazin inverse. Therefore A has not 2-generalized W-group inverse by Theorem 3.4.

4. Algebraic Properties

The aim of this section is to investigate the algebraic properties of the m-generalized right w-group inverse. Next, we establish the relationship between the m-generalized right w-group inverse and the corresponding element decomposition.

Theorem 4.1.

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

(1)

$a\in {\mathcal{A}}_r^{w,g_m}$.

(2)

a has m-generalized right w-group decomposition, i.e., there exist $x,y\in \mathcal{A}$ such that

$$a=x+y,x^{*}{\left(aw\right)}^{m-1}y=ywx=0,x\in {\mathcal{A}}_r^{w,\#},y\in {\mathcal{A}}_w^{qnil}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, there exists $x\in \mathcal{A}$ such that

$$\begin{array}{c}x=a{\left(wx\right)}^2,{\left[\left(aw\right){\left(aw\right)}_r^d\right]}^{*}{\left(aw\right)}^{m+1}x={\left[\left(aw\right){\left(aw\right)}_r^d\right]}^{*}{\left(aw\right)}^{m-1}a,\\ \underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-awxw{\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\end{array}$$

Let $a_1={\left(aw\right)}^{2}x$ and $a_2=a-{\left(aw\right)}^{2}x$. Then we check that

$$\begin{array}{ccc}\hfill \left|\right|a_{2}w{a}_1\left|\right|& =\hfill & \left|\right|[a-{\left(aw\right)}^{2}x]w{\left(aw\right)}^{2}x\left|\right|={\left|\right|\left(aw\right)}^{3}x-{\left(aw\right)}^{2}x\left[{\left(wa\right)}^2\left(wx\right)\right]\left|\right|\hfill \\ & =\hfill & {\left|\right|\left(aw\right)}^{3}x-{\left(aw\right)}^2{x[\left(wa\right)}^{k+1}{\left(wx\right)}^k\left|\right|\hfill \\ & =\hfill & {\left|\right|a\left(wa\right)}^2{\left(wx\right)-a\left(wa\right)\left(wx\right)[\left(wa\right)}^{k+1}{\left(wx\right)}^k\left|\right|\hfill \\ & \le \hfill & {\left|\right|aw\left|\right|\left|\right|\left(aw\right)}^k-\left(aw\right)xw{\left(aw\right)}^k\left]\right|\left|\right||\left[a{\left(wx\right)}^k\right]\left|\right|.\hfill \end{array}$$

Since $\underset{k\to \infty }{lim}{\left|\right|\left(aw\right)}^k-\left(aw\right)xw{\left(aw\right)}^k{\left|\right|}^{{\displaystyle \frac{1}{k}}}=0$, we deduce that $\underset{k\to \infty }{lim}\left|\right|a_{2}w{a}_1{\left|\right|}^{{\displaystyle \frac{1}{k}}}=0$, and then $a_{2}w{a}_1=0$.

Moreover, we have

$$\begin{array}{cc}& \left|\right|a_1^{*}{\left(aw\right)}^{m-1}{a}_2\left|\right|\hfill \\ \hfill =& \left|\right|{\left({\left(aw\right)}^{2}x\right)}^{*}{\left(aw\right)}^{m-1}\left(a-{\left(aw\right)}^{2}x\right)\left|\right|\hfill \\ \hfill =& \left|\right|{\left({\left(aw\right)}^{2}x\right)}^{*}{\left(aw\right)}^{m-1}a-{\left({\left(aw\right)}^{2}x\right)}^{*}{\left(aw\right)}^{m+1}x\left|\right|\hfill \\ \hfill =& \left|\right|{\left({\left(aw\right)}^{2}x\right)}^{*}{\left(aw\right)}^{m-1}a-{\left({\left(aw\right)}^k{\left(xw\right)}^{k-2}x\right)}^{*}{\left(aw\right)}^{m+1}x\left|\right|\hfill \\ \hfill =& \left|\right|{\left({\left(aw\right)}^{2}x\right)}^{*}{\left(aw\right)}^{m-1}a-{\left([{\left(aw\right)}^k-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^k]{\left(xw\right)}^{k-2}x\right)}^{*}{\left(aw\right)}^{m+1}x\hfill \\ \hfill -& {\left(\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^k{\left(xw\right)}^{k-2}x\right)}^{*}{\left(aw\right)}^{m+1}x\left|\right|\hfill \\ \hfill \le & \left|\right|{({\left(aw\right)}^k-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^k)}^{*}\left|\right|{\left({\left(xw\right)}^{k-2}x\right)}^{*}{\left|\right|\left|\right|\left(aw\right)}^{m+1}x\left|\right|\hfill \\ \hfill +& \left|\right|{\left({\left(aw\right)}^{2}x\right)}^{*}{\left(aw\right)}^{m-1}a-{\left[\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^k{\left(xw\right)}^{k-2}x\right]}^{*}{\left(aw\right)}^{m+1}x\left|\right|\hfill \\ \hfill \le & \left|\right|{({\left(aw\right)}^k-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^k)}^{*}\left|\right|{\left({\left(xw\right)}^{k-2}x\right)}^{*}{\left|\right|\left|\right|\left(aw\right)}^{m+1}x\left|\right|\hfill \\ \hfill +& \left|\right|{\left({\left(aw\right)}^{2}x\right)}^{*}{\left(aw\right)}^{m-1}a-{\left[{\left(aw\right)}^k{\left(xw\right)}^{k-2}x\right]}^{*}{\left[\left(aw\right){\left(aw\right)}_r^d\right]}^{*}{\left(aw\right)}^{m+1}x\left|\right|\hfill \\ \hfill \le & \left|\right|{({\left(aw\right)}^k-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^k)}^{*}\left|\right|{\left({\left(xw\right)}^{k-2}x\right)}^{*}{\left|\right|\left|\right|\left(aw\right)}^{m+1}x\left|\right|\hfill \\ \hfill +& \left|\right|{\left({\left(aw\right)}^{2}x\right)}^{*}{\left(aw\right)}^{m-1}a-{\left[{\left(aw\right)}^k{\left(xw\right)}^{k-2}x\right]}^{*}{\left[\left(aw\right){\left(aw\right)}_r^d\right]}^{*}a\left|\right|\hfill \end{array}$$

$$\begin{array}{cc}\hfill \le & \left|\right|{({\left(aw\right)}^k-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^k)}^{*}\left|\right|{\left({\left(xw\right)}^{k-2}x\right)}^{*}{\left|\right|\left|\right|\left(aw\right)}^{m+1}x\left|\right|\hfill \\ \hfill +& \left|\right|{\left({\left(aw\right)}^{2}x\right)}^{*}{\left(aw\right)}^{m-1}a-{\left[\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^k{\left(xw\right)}^{k-2}x\right]}^{*}a\left|\right|\hfill \\ \hfill \le & \left|\right|{({\left(aw\right)}^k-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^k)}^{*}\left|\right|{\left({\left(xw\right)}^{k-2}x\right)}^{*}{\left|\right|\left|\right|\left(aw\right)}^{m+1}x\left|\right|\hfill \\ \hfill +& \left|\right|{\left({\left(aw\right)}^k{\left(xw\right)}^{k-2}x\right)}^{*}a-{\left[\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^k{\left(xw\right)}^{k-2}x\right]}^{*}a\left|\right|\hfill \\ \hfill \le & \left|\right|{({\left(aw\right)}^k-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^k)}^{*}\left|\right|{\left({\left(xw\right)}^{k-2}x\right)}^{*}{\left|\right|\left|\right|\left(aw\right)}^{m+1}x\left|\right|\hfill \\ \hfill +& \left|\right|{\left([{\left(aw\right)}^k-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^k]{\left(xw\right)}^{k-2}x\right)}^{*}a\left|\right|\hfill \\ \hfill \le & \left|\right|{({\left(aw\right)}^k-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^k)}^{*}\left|\right|{\left({\left(xw\right)}^{k-2}x\right)}^{*}{\left|\right|\left|\right|\left(aw\right)}^{m+1}x\left|\right|\hfill \\ \hfill +& \left|\right|{\left({\left(aw\right)}^k-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^k\right)}^{*}\left|\right|\left|\right|{\left({\left(xw\right)}^{k-2}x\right)}^{*}\left|\right|\left|\right|a\left|\right|.\hfill \end{array}$$

Then

$$\begin{array}{cc}& \left|\right|a_1^{*}{\left(aw\right)}^{m-1}{a}_2{\left|\right|}^{{\displaystyle \frac{1}{k}}}\hfill \\ \hfill \le & \left|\right|{({\left(aw\right)}^k-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^k)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{k}}}\left|\right|{\left({\left(xw\right)}^{k-2}x\right)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{k}}}{\left|\right|\left(aw\right)}^{m+1}x{\left|\right|}^{{\displaystyle \frac{1}{k}}}\hfill \\ \hfill +& \left|\right|{({\left(aw\right)}^k-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^k)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{k}}}\left|\right|{\left({\left(xw\right)}^{k-2}x\right)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{k}}}{\left|\right|a\left|\right|}^{{\displaystyle \frac{1}{k}}}.\hfill \end{array}$$

As in the proof of Theorem 2.2, we have $\underset{k\to \infty }{lim}\left|\right|{({\left(aw\right)}^k-\left(aw\right){\left(aw\right)}_r^d{\left(aw\right)}^k)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{k}}}=0$. Then $\underset{k\to \infty }{lim}\left|\right|a_1^{*}{\left(aw\right)}^{m-1}{a}_2{\left|\right|}^{{\displaystyle \frac{1}{k}}}=0$, and then $a_1^{*}{\left(aw\right)}^{m-1}{a}_2=0$. We further verify that

$$\begin{array}{ccc}\hfill [aw-(aw\left)\right(xw\left)\right(aw\left)\right]awx& =& [aw-\left(aw\right)\left(xw\right)\left(aw\right)]{\left(aw\right)}^{2}x\left(wx\right)\hfill \\ & =& [aw-\left(aw\right)\left(xw\right)\left(aw\right)]{\left(aw\right)}^{3}x{\left(wx\right)}^2\hfill \\ & ⋮& \\ & =& [{\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n]x{\left(wx\right)}^{n-2}.\hfill \end{array}$$

Therefore

$${\left|\right|[aw-\left(aw\right)\left(xw\right)\left(aw\right)]awx\left|\right|}^{{\displaystyle \frac{1}{n}}}\le {\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}{\left|\right|x\left(wx\right)}^{n-2}{\left|\right|}^{{\displaystyle \frac{1}{n}}}.$$

Since $\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(xw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0,$ we deduce that

$$\underset{n\to \infty }{lim}{\left|\right|[aw-\left(aw\right)\left(xw\right)\left(aw\right)]awx\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

This implies that $[aw-(aw\left)\right(xw\left)\right(aw\left)\right]\left(aw\right)x=0$, i.e., ${\left(aw\right)}^{2}x=\left(aw\right)\left(xw\right){\left(aw\right)}^{2}x$. Likewise, ${\left(aw\right)}^{2}xw\left(aw\right)x={\left(aw\right)}^{2}x$. Then we check that

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

Accordingly,

$$\underset{n\to \infty }{lim}{\left|\right|[aw-\left(aw\right)\left(xw\right)\left(aw\right)]}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n+1}}}=0;$$

hence, $aw-\left(aw\right)\left(xw\right)\left(aw\right)\in {\mathcal{A}}^{qnil}$. By using Cline’s formula (see [11]), $y=a-{\left(aw\right)}^{2}x\in {\mathcal{A}}_w^{qnil}$. Moreover, we check that

$$\begin{array}{ccc}\hfill a_1{\left(wx\right)}^2& =\hfill & \left[{\left(aw\right)}^{2}x\right]{\left(wx\right)}^2=aw\left[a{\left(wx\right)}^2\right]wx=a{\left(wx\right)}^2=x,\hfill \\ \hfill {\left(a_{1}w\right)}^{2}x& =\hfill & {\left(aw\right)}^{2}xwaw\left[a{\left(wx\right)}^2\right]=a_1,\hfill \\ \hfill a_{1}wxw{a}_1& =\hfill & {\left(aw\right)}^{2}xwxw{\left(aw\right)}^{2}x=aw\left[a{\left(wx\right)}^2\right]w{\left(aw\right)}^{2}x\hfill \\ & =\hfill & \left(aw\right)\left(xw\right){\left(aw\right)}^{2}x={\left(aw\right)}^{2}x=a_1.\hfill \end{array}$$

Hence ${\left(a_1\right)}_r^{w,\#}=x$. Hence, we have a m-generalized right w-group decomposition $a=a_1+a_2$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, a has m-generalized right w-group decomposition $a=a_1+a_2$. Then ${\left(a_1\right)}^{*}{\left(aw\right)}^{m-1}{a}_2=0$ and $a_{2}w{a}_1=0$. Let $x={\left(a_1\right)}_r^{w,\#}$. Then we verify that $a{\left(wx\right)}^2=(a_1+a_2){\left(w{\left(a_1\right)}_r^{w,\#}\right)}^2=a_1{\left(w{\left(a_1\right)}_r^{w,\#}\right)}^2={\left(a_1\right)}_r^{w,\#}=x$.

Since $a_{2}w{a}_1=0$, we have

$$\begin{array}{cc}& {\left(aw\right)}^2-\left(aw\right)\left(xw\right){\left(aw\right)}^2\hfill \\ \hfill =& \left(aw\right)[(a_1+a_2)w-\left({\left(a_1\right)}_r^{w,\#}w{a}_1+{\left(a_1\right)}_r^{w,\#}w{a}_2\right)w(a_1+a_2)w]\hfill \\ \hfill =& \left(aw\right)[a_{1}w+a_{2}w-{\left(a_1\right)}_r^{w,\#}w{a}_{1}w(a_1+a_2)w-{\left(a_1\right)}_r^{w,\#}w{a}_{2}w(a_1+a_2)w]\hfill \\ \hfill =& \left(aw\right)[a_{1}w+a_{2}w-{\left(a_1\right)}_r^{w,\#}w{\left(a_{1}w\right)}^2-{\left(a_1\right)}_r^{w,\#}w{a}_{1}w{a}_{2}w-{\left(a_1\right)}_r^{w,\#}w{\left(a_{2}w\right)}^2]\hfill \\ \hfill =& \left(aw\right)[1-{\left(a_1\right)}_r^{w,\#}w{a}_{1}w-{\left(a_1\right)}_r^{w,\#}w{a}_{2}w]\left(a_{2}w\right),\hfill \end{array}$$

and so

$$\begin{array}{cc}& {\left|\right|\left(aw\right)}^{n+1}-\left(aw\right)\left(xw\right){\left(aw\right)}^{n+1}{\left|\right|}^{{\displaystyle \frac{1}{n+1}}}\hfill \\ \hfill =& \left|\right|[{\left(aw\right)}^2-\left(aw\right)\left(xw\right){\left(aw\right)}^2){\left(aw\right)}^{n-1}|{|}^{{\displaystyle \frac{1}{n+1}}}\hfill \\ \hfill =& \left|\right|\left(aw\right)[1-{\left(a_1\right)}_r^{w,\#}w{a}_{1}w-{\left(a_1\right)}_r^{w,\#}w{a}_{2}w]\left(a_{2}w\right){(a_{1}w+a_{2}w)}^{n-1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ \hfill \le & {\left|\right|aw\left|\right|}^{{\displaystyle \frac{1}{n+1}}}\left|\right|1-{\left(a_1\right)}_r^{w,\#}w{a}_{1}w-{\left(a_1\right)}_r^{w,\#}w{a}_2{w\left|\right|}^{{\displaystyle \frac{1}{n+1}}}\left|\right|{\left(a_{2}w\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n+1}}}.\hfill \end{array}$$

Since $a_2\in {\mathcal{A}}_w^{qnil}$, we have $\underset{n\to \infty }{lim}\left|\right|{\left(a_{2}w\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n+1}}}=\underset{n\to \infty }{lim}\left[\right||{\left(a_{2}w\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}{]}^{{\displaystyle \frac{n}{n+1}}}=0$, and then

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

Since $\left(a_{2}w\right)\left(a_{1}w\right)=0,a_{1}w\in {\mathcal{A}}_r^{\#}$ and $a_{2}w\in {\mathcal{A}}^{\#}$, it follows that $aw\in {\mathcal{A}}_r^d$ and ${\left(aw\right)}_r^d={\left(a_{1}w\right)}_r^{\#}+\sum _{n=1}^{\infty }{\left({\left(a_{1}w\right)}_r^{\#}\right)}^{n+1}{\left(a_{2}w\right)}^n.$ Then

$$\begin{array}{cc}& ({\left(aw\right)}_r^d{)}^{*}{\left(aw\right)}^{m-1}{a}_2\hfill \\ \hfill =& {\left({\left(aw\right)}^{\#}\right)}^{*}{\left(aw\right)}^{m-1}{a}_2+\sum _{n=1}^{\infty }{\left({\left(a_{1}w\right)}_r^{\#}{)}^{n+1}{\left(a_{2}w\right)}^n\right)}^{*}{\left(aw\right)}^{m-1}{a}_2\hfill \\ \hfill =& {\left({\left({\left(a_{1}w\right)}_r^{\#}\right)}^2\right)}^{*}\left(w^{*}{\left(a_1\right)}^{*}{\left(aw\right)}^{m-1}{a}_2\right)+\sum _{n=1}^{\infty }{\left({\left({\left(a_{1}w\right)}_r^{\#}\right)}^{n+2}{\left(a_{2}w\right)}^n\right)}^{*}\left(w^{*}{\left(a_1\right)}^{*}{a}_2\right)\hfill \\ \hfill =& 0;\hfill \end{array}$$

hence, ${\left({\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^{m-1}{a}_1={\left({\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^{m-1}a$. Accordingly, we have

$$\begin{array}{ccc}\hfill {\left({\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^{m+1}x& =\hfill & {\left({\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^m\left(aw\right){\left(a_1\right)}_r^{w,\#}\hfill \\ & =\hfill & {\left({\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^m(a_1+a_2)w{\left(a_1\right)}_r^{w,\#}\hfill \\ & =\hfill & {\left({\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^{m+1}{\left(a_1\right)}_r^{w,\#}\hfill \\ & =\hfill & {\left({\left(aw\right)}_r^d\right)}^{*}{\left(a_{1}w\right)}^{m+1}{\left(a_1\right)}_r^{w,\#}\hfill \\ & =\hfill & {\left({\left(aw\right)}_r^d\right)}^{*}{\left(a_{1}w\right)}^{m-1}{a}_1\hfill \\ & =\hfill & {\left({\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^{m-1}{a}_1\hfill \\ & =\hfill & {\left({\left(aw\right)}_r^d\right)}^{*}{\left(aw\right)}^{m-1}a,\hfill \end{array}$$

as required. □

Corollary 4.2.

Let $a\in {\mathcal{A}}_r^{w,g}$. Then $aw{a}_r^{w,g}={\left(aw\right)}^{n-1}a{\left(w{a}_r^{w,g}\right)}^n$ for any $n\in N$.

Proof. 

These are obvious by the proof of Theorem 4.1. □

Corollary 4.3.

Let $a\in {\mathcal{A}}_r^{w,g_m}$ and $b\in {\mathcal{A}}_w^{qnil}$. If $a^{*}{\left(aw\right)}^{m-1}b=0$ and $bwa=0$, then $a+b\in {\mathcal{A}}_r^{w,g_m}$. In this case,

$${(a+b)}_r^{w,g_m}=a_r^{w,g_m}.$$

Proof. 

Since $a\in {\mathcal{A}}_r^{w,g_m}$, by virtue of Theorem 4.1, there exist $x\in {\mathcal{A}}_r^{w,\#}$ and $y\in {\mathcal{A}}_w^{qnil}$ such that $a=x+y,x^{*}{\left(aw\right)}^{m-1}y=0,ywx=0$. As in the proof of Theorem 4.1, $x={\left(aw\right)}^2{a}_r^{w,g_m}$ and $y=a-{\left(aw\right)}^2{a}_r^{w,g_m}$. Then $a=x+(y+b)$. Since $bwy=bw[a-{\left(aw\right)}^2{a}_r^{g_m}]=0$, it follows by [1] that $y+b\in {\mathcal{A}}_w^{qnil}$. Obviously, $x^{*}{\left(aw\right)}^{m-1}(y+b)=x^{*}{\left(aw\right)}^{m-1}y+{\left[{\left(aw\right)}^2{a}_r^{w,g_m}\right]}^{*}{\left(aw\right)}^{m-1}b=0$. In light of Theorem 4.1, $a+b\in {\mathcal{A}}_r^{w,g_m}$. In this case,

$${(a+b)}_r^{w,g_m}=x_r^{w,\#}=a_r^{w,g},$$

as required. □

Corollary 4.4.

Let $M=\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$ with $a,d\in {\mathcal{A}}_r^{w,g_m}$. If $a^{*}{\left(aw\right)}^{m-1}b=0,bwd=0$, then $M\in M_2{\left(\mathcal{A}\right)}_r^{w,g_m}$.

Proof. 

Write $M=P+Q$, where $P=\left(\begin{array}{cc}a& 0\\ 0& d\end{array}\right),Q=\left(\begin{array}{cc}0& b\\ 0& 0\end{array}\right).$ Then $P\in M_2{\left(\mathcal{A}\right)}_r^{w,g_m}$ and $P_r^{w,g_m}=\left(\begin{array}{cc}{a}_r^{w,g_m}& 0\\ 0& d_r^{w,g_m}\end{array}\right)$. Obviously, $Q^2=0$. BY hypothesis, we verify that

$$Q\left(w{I}_2\right)P=0,P^{*}{\left(Mw{I}_2\right)}^{m-1}Q=0.$$

Therefore $M\in M_2{\left(\mathcal{A}\right)}_r^{w,g_m}$ by Corollary 4.3. □

We come now to present the polar-like property for the m-generalized right w-group inverse.

Theorem 4.5.

Let $a\in {\mathcal{A}}_r^{w,g_m}$. Then there exists an idempotent $p\in \mathcal{A}$ such that

$$\begin{array}{c}aw+p\in {\mathcal{A}}_r^{-1},paw=pawp\in {\mathcal{A}}^{qnil},\\ {\left({\left({\left(aw\right)}^m\right)}^{*}{\left(aw\right)}^{m}p\right)}^{*}={\left(aw\right)}^m{)}^{*}{\left(aw\right)}^{m}p.\end{array}$$

Proof. 

By virtue of Theorem 4.1, there exist $z,y\in \mathcal{A}$ such that

$$a=z+y,z^{*}{\left(aw\right)}^{m-1}y=ywz=0,z\in {\mathcal{A}}_r^{w,\#},y\in {\mathcal{A}}_w^{qnil}.$$

Set $x=z_r^{w,\#}$. The we check that

$$\begin{array}{ccc}\hfill awx& =\hfill & (z+y)w{z}_r^{w,\#}=zw{z}_r^{w,\#},\hfill \\ \hfill a{\left(wx\right)}^2& =\hfill & \left(awx\right)wx=z{\left(w{z}_r^{w,\#}\right)}^2=z_r^{w,\#}=x.\hfill \end{array}$$

Since $ywz=0$, we see that ${\left(aw\right)}^{k}z={\left(aw\right)}^{k-1}(y+z)wz={\left(aw\right)}^{k-1}\left(zw\right)z={\left(aw\right)}^{k-2}(y+z)wzwz={\left(aw\right)}^{k-2}{\left(zw\right)}^{2}z\dots ={\left(zw\right)}^{k}z$. Since $z^{*}{\left(aw\right)}^{m-1}y=0$, we derive that

$$\begin{array}{ccc}\hfill {\left({\left(aw\right)}^m\right)}^{*}{\left(zw\right)}^m& =\hfill & w^{*}{a}^{*}{\left({\left(aw\right)}^{m-1}\right)}^{*}{\left(zw\right)}^m=w^{*}{\left[z^{*}{\left(aw\right)}^{m-1}(y+z)\right]}^{*}{\left(wz\right)}^{m-1}w\hfill \\ & =\hfill & w^{*}{\left[z^{*}{\left(aw\right)}^{m-1}z\right]}^{*}{\left(wz\right)}^{m-1w}=w^{*}{\left[z^{*}{\left(zw\right)}^{m-1}z\right]}^{*}{\left(wz\right)}^{m-1}w\hfill \\ & =\hfill & {\left[z^{*}{\left(zw\right)}^{m-1}zw\right]}^{*}{\left(wz\right)}^{m-1}w={\left[{\left(zw\right)}^m\right]}^{*}z{\left(wz\right)}^{m-1}w\hfill \\ & =\hfill & {\left[{\left(zw\right)}^m\right]}^{*}{\left(zw\right)}^m.\hfill \end{array}$$

Hence, we have

$$\begin{array}{cc}& {\left({\left(aw\right)}^m\right)}^{*}{\left(aw\right)}^{m+1}xw\hfill \\ \hfill =& {\left({\left(aw\right)}^m\right)}^{*}{\left(aw\right)}^m\left(awxw\right)={\left({\left(aw\right)}^m\right)}^{*}{\left(aw\right)}^m\left(zw{z}_r^{w,\#}w\right)\hfill \\ \hfill =& {\left({\left(aw\right)}^m\right)}^{*}\left({\left(aw\right)}^{m}zw\right)z_r^{w,\#}w={\left({\left(aw\right)}^m\right)}^{*}\left({\left(zw\right)}^{m}zw\right)z_r^{w,\#}w\hfill \\ \hfill =& {\left[{\left(aw\right)}^m\right)}^{*}{\left(zw\right)}^m]zw{z}_r^{w,\#}w={\left[{\left(zw\right)}^m\right]}^{*}{\left(zw\right)}^{m+1}{z}_r^{w,\#}w\hfill \\ \hfill =& {\left[{\left(zw\right)}^m\right]}^{*}{\left(zw\right)}^{m-1}\left[{\left(zw\right)}^2{z}_r^{w,\#}\right]w\hfill \\ \hfill =& {\left[{\left(zw\right)}^m\right]}^{*}{\left(zw\right)}^{m-1}zw\hfill \\ \hfill =& {\left({\left(zw\right)}^m\right)}^{*}{\left(zw\right)}^m.\hfill \end{array}$$

Therefore

$$\begin{array}{ccc}\hfill {\left[{\left({\left(aw\right)}^m\right)}^{*}{\left(aw\right)}^{m+1}xw\right]}^{*}& =\hfill & {\left[{\left({\left(zw\right)}^m\right)}^{*}{\left(zw\right)}^m\right]}^{*}={\left({\left(zw\right)}^m\right)}^{*}{\left(zw\right)}^m\hfill \\ & =\hfill & {\left({\left(aw\right)}^m\right)}^{*}{\left(aw\right)}^{m+1}xw.\hfill \end{array}$$

Let $p=1-zw{z}_r^{w,\#}w$. Then $p=1-awxw=p^2\in \mathcal{A}$. Since $aw-awzw{z}^{\#}w=aw(1-zw{z}_r^{w,\#}w)=(z+y)w(1-zw{z}_r^{w,\#}w)=yw\in {\mathcal{A}}^{qnil}$, we have $awp=aw(1-zw{z}_r^{w,\#}w)\in {\mathcal{A}}^{qnil}$. Accordingly, we have

$$\begin{array}{ccc}\hfill {\left[{\left({\left(aw\right)}^m\right)}^{*}{\left(aw\right)}^{m}p\right]}^{*}& =\hfill & ({\left({\left(aw\right)}^m\right)}^{*}{\left(aw\right)}^m-{\left[{\left({\left(aw\right)}^m\right)}^{*}{\left(aw\right)}^{m+1}xw\right]}^{*}\hfill \\ & =\hfill & {\left({\left(aw\right)}^m\right)}^{*}{\left(aw\right)}^m-{\left({\left(aw\right)}^m\right)}^{*}{\left(aw\right)}^{m+1}xw\hfill \\ & =\hfill & {\left(aw\right)}^m{)}^{*}{\left(aw\right)}^{m}p.\hfill \end{array}$$

Since $paw(1-p)=(1-zw{z}_r^{w,\#}w)(z+y)wzw{z}_r^{w,\#}w=(1-zw{z}_r^{w,\#}w){\left(zw\right)}^2{z}_r^{w,\#}w$$=0$, we see that $paw=pawp$. Obviously, we have

$$\begin{array}{cc}& [zw+1-zw{z}_r^{w,\#}w][z_r^{w,\#}w+1-zw{z}_r^{w,\#}w]\hfill \\ \hfill =& zw{z}_r^{w,\#}w+zw(1-zw{z}_r^{w,\#}w)+(1-zw{z}_r^{w,\#}w)z_r^{w,\#}w+1-zw{z}_r^{w,\#}w=1.\hfill \end{array}$$

Since $yw(z_r^{w,\#}w+1-zw{z}_r^{w,\#}w)=yw\in {\mathcal{A}}^{qnil}$, it follows by Cline’s formula that $(z_r^{w,\#}w+1-zw{z}_r^{w,\#}w)yw\in {\mathcal{A}}^{qnil}$. Hence $1+(z^{\#}w+1-zw{z}^{\#}w)yw\in {\mathcal{A}}^{-1}$. This implies that

$$\begin{array}{ccc}\hfill aw+p& =\hfill & zw+yw+1-zw{z}_r^{w,\#}w\hfill \\ & =\hfill & (zw+1-zw{z}_r^{w,\#}w)[1+{(zw+1-zw{z}_r^{w,\#}w)}^{-1}yw]\hfill \\ & \in \hfill & {\mathcal{A}}_r^{-1},\hfill \end{array}$$

as desired. □

Corollary 4.6.

Let $a\in {\mathcal{A}}_r^{w,g}$. Then there exists an idempotent $p\in \mathcal{A}$ such that

$$\begin{array}{c}aw+p\in {\mathcal{A}}_r^{-1},paw=pawp\in {\mathcal{A}}^{qnil},\\ {\left({\left(aw\right)}^{*}\left(aw\right)p\right)}^{*}={\left(aw\right)}^{*}\left(aw\right)p.\end{array}$$

Proof. 

This is obvious by choosing $m=1$ in Theorem 4.5. □

5. m-Weak Right w-Group Inverse

We now proceed to examine the interrelation between the m-weak right w-group inverse and the m-generalized right w-group inverse.

Theorem 5.1.

Let $a,w\in \mathcal{A}$. Then $a\in {\mathcal{A}}_r^{w,w_m}$ if and only if $aw\in {\mathcal{A}}_r^D$ and $a\in {\mathcal{A}}_r^{w,g_m}$. In this case, $a_r^{w,w_m}=a_r^{w,g_m}$.

Proof. 

One direction is obvious. Conversely, assume that $aw\in {\mathcal{A}}_r^D$ and $a\in {\mathcal{A}}_r^{w,g_m}$. Let $x=a_r^{w,g_m}$. Then there exists $x\in \mathcal{A}$ such that

$$x=a{\left(wx\right)}^2,{\left[{\left(aw\right)}_r^d\right]}^{*}{\left(aw\right)}^{m+1}x={\left[{\left(aw\right)}_r^d\right]}^{*}{a}^{m-1}a,\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-awxw{\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

Since $a\in {\mathcal{A}}^{D,w}$, we can find some $y\in \mathcal{A}$ such that

$$yw{\left(aw\right)}^{k+1}={\left(aw\right)}^k,ywawy=y,awy=ywa$$

for some $k\in N$. Set $z={\left(aw\right)}^{k}y{\left(wx\right)}^k$. Then we verify that

$$\begin{array}{ccc}\hfill awz& =\hfill & ({\left(aw\right)}^{k+1}yw{\left(xw\right)}^{k-1}x=\left[\left(yw\right){\left(aw\right)}^{k+1}\right]{\left(xw\right)}^{k-1}x\hfill \\ & =\hfill & {\left(aw\right)}^k{\left(xw\right)}^{k-1}xw={\left(aw\right)}^k{\left(xw\right)}^{k-1}x={\left(aw\right)}^{2}xwx.\hfill \end{array}$$

Claim 1. $z=a{\left(wz\right)}^2$.

$$\begin{array}{ccc}\hfill a{\left(wz\right)}^2& =\hfill & \left(awz\right)wz=\left[{\left(aw\right)}^{2}xwx\right]w{\left(aw\right)}^{k}y{\left(wx\right)}^k\hfill \\ & =\hfill & \left[{\left(aw\right)}^2{\left(xw\right)}^2\right]{\left(aw\right)}^{k}y{\left(wx\right)}^k\hfill \\ & =\hfill & \left[\left(aw\right)\left(xw\right){\left(aw\right)}^k\right]y{\left(wx\right)}^k\hfill \\ & =\hfill & \left[\left(aw\right){\left(aw\right)}^{k-1}\right]y{\left(wx\right)}^k\hfill \\ & =\hfill & {\left(aw\right)}^{k}y{\left(wx\right)}^k=z.\hfill \end{array}$$

Claim 2. ${\left[{\left(aw\right)}_r^d\right]}^{*}{\left(aw\right)}^{m+1}z={\left[{\left(aw\right)}_r^d\right]}^{*}{\left(aw\right)}^{m-1}a.$

$$\begin{array}{ccc}\hfill {\left[{\left(aw\right)}_r^d\right]}^{*}{\left(aw\right)}^{m+1}z& =\hfill & {\left[{\left(aw\right)}_r^d\right]}^{*}\left[{\left(aw\right)}^{k+m+1}y\right]{\left(wx\right)}^k={\left[{\left(aw\right)}_r^d\right]}^{*}\left[y{\left(wa\right)}^{k+m+1}\right]{\left(wx\right)}^k\hfill \\ & =\hfill & {\left[{\left(aw\right)}_r^d\right]}^{*}\left[\left(yw\right)a{\left(wa\right)}^{k+m+1}w\right]x{\left(wx\right)}^{k-1}\hfill \\ & =\hfill & {\left[{\left(aw\right)}_r^d\right]}^{*}\left[\left(yw\right){\left(aw\right)}^{k+m+1}\right]awx{\left(wx\right)}^{k-1}\hfill \\ & =\hfill & {\left[{\left(aw\right)}_r^d\right]}^{*}{\left(aw\right)}^{k+m+1}x{\left(wx\right)}^{k-1}\hfill \\ & =\hfill & {\left[{\left(aw\right)}_r^d\right]}^{*}a{\left(wa\right)}^{k+m-1}{\left(wx\right)}^k\hfill \\ & =\hfill & {\left[{\left(aw\right)}_r^d\right]}^{*}a{\left(wa\right)}^m\left(wx\right)\hfill \\ & =\hfill & {\left[{\left(aw\right)}_r^d\right]}^{*}{\left(aw\right)}^{m+1}x\hfill \\ & =\hfill & {\left[{\left(aw\right)}_r^d\right]}^{*}{\left(aw\right)}^{m-1}a.\hfill \end{array}$$

Claim 3. ${\left(aw\right)}^k=\left(zw\right){\left(aw\right)}^{k+1}.$

Since $\left(aw\right)\left(yw\right)=\left(yw\right)\left(aw\right)$, we see that

$$\begin{array}{ccc}\hfill {\left(aw\right)}^k-\left(zw\right){\left(aw\right)}^{k+1}& =\hfill & {\left(aw\right)}^k-{\left(aw\right)}^{k}yw{\left(xw\right)}^k{\left(aw\right)}^{k+1}\hfill \\ & =\hfill & {\left(aw\right)}^k-\left(yw\right){\left(aw\right)}^k{\left(xw\right)}^k{\left(aw\right)}^{k+1}\hfill \\ & =\hfill & {\left(aw\right)}^k-\left(yw\right)\left(aw\right)\left(xw\right){\left(aw\right)}^{k+1}\hfill \\ & =\hfill & {\left(aw\right)}^k-\left(yw\right)\left[{\left(aw\right)}^{k+1}{\left(xw\right)}^{k+1}\right]{\left(aw\right)}^{k+1}\hfill \\ & =\hfill & {\left(aw\right)}^k-\left[\left(yw\right){\left(aw\right)}^{k+1}\right]{\left(xw\right)}^{k+1}{\left(aw\right)}^{k+1}\hfill \\ & =\hfill & {\left(aw\right)}^k-{\left(aw\right)}^k{\left(xw\right)}^{k+1}{\left(aw\right)}^{k+1}\hfill \\ & =\hfill & {\left(aw\right)}^k-\left(aw\right){\left(xw\right)}^2{\left(aw\right)}^{k+1}\hfill \\ & =\hfill & {\left(aw\right)}^k-\left[a{\left(wx\right)}^2\right]w{\left(aw\right)}^{k+1}\hfill \\ & =\hfill & {\left(aw\right)}^k-xw{\left(aw\right)}^{k+1}\hfill \\ & =\hfill & {\left(aw\right)}^{k+n}{\left(yw\right)}^n-xw{\left(aw\right)}^{k+n+1}{\left(yw\right)}^n.\hfill \end{array}$$

Then

$${\left|\right|\left(aw\right)}^k-\left(zw\right){\left(aw\right)}^{k+1}\left|\right|\le {\left|\right|\left(aw\right)}^{k+n}-xw{\left(aw\right)}^{k+n+1}{\left|\right|\left|\right|yw\left|\right|}^n.$$

Since $\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^{k+n}-xw{\left(aw\right)}^{k+n+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0,$ we see that

$$\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^k-\left(zw\right){\left(aw\right)}^{k+1}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

Hence, ${\left(aw\right)}^k=\left(zw\right){\left(aw\right)}^{k+1}$. Therefore $a\in {\mathcal{A}}_w^{w_m}$. □

Corollary 5.2.

Let $a,w\in \mathcal{A}$. Then $a\in {\mathcal{A}}_r^{w_m}$ if and only if $aw\in {\mathcal{A}}_r^D\bigcap R_r^{g_m}$. In this case, $a_r^{w_m}=a_r^{g_m}$.

Proof. 

This is obvious by Theorem 5.1. □

The following example illustrates that m-weak right w-group inverse is different from the $(m,w)$-weak group inverse.

Example 5.3.

Let V be a countably generated infinite-dimensional vector space over the complex field C, and let $\{x_1,x_2,x_3,\dots \}$ be a basis of V. Let $\sigma :V\to V$ be a shift operator given by $\sigma \left(x_1\right)=0$ and $\sigma \left(x_{i+1}\right)=x_i$ for all $i\in \mathbb{N}$. Then σ has $(m,1)$-generalized group inverse, while it has no any $(m,1)$-weak group inverse for any $m\in N$.

Proof. 

Define the matrix A given by

$$\begin{array}{ccc}\hfill \sigma (x_1,x_2,x_3,\dots )& =\hfill & (0,x_1,x_2,x_3,\dots )\hfill \\ & =\hfill & (x_1,x_2,x_3,\dots )A.\hfill \end{array}$$

The linear shift operator $\sigma$ can be regarded as an element in a Banach *-algebra of complex matrix, with conjugate transpose * as the involution. Then $\sigma$ is quasinilpotent as all its eigenvalues are zero. Hence, it has $(m,1)$-generalized group inverse and ${\sigma }_1^{g_m}=0$. For any $k\in N$, ${\sigma }^k(x_1,x_2,x_3,\dots )=(0,0,\dots ,0,x_1,x_2,\dots )$; whence, $\sigma$ is not nilpotent. If $\sigma$ has $(m,1)$-weak group inverse; it has Drazin inverse by [5]. Then there exists some $n\in N$ such that ${\sigma }^n={\sigma }^{n+1}\eta$ and $\sigma \eta =\eta \sigma$ for a linear operator $\eta$. Hence ${\sigma }^n(1-\sigma \eta )=0$. As $\sigma$ is quasinilpotent and $\sigma \eta =\eta \sigma$, we have $1-\sigma \eta$ is invertible, and the ${\sigma }^n=0$. This gives a contradiction. Therefore $\sigma$ has no $(m,1)$-weak group inverse, as asserted. □

Theorem 5.4.

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

(1)

$a_r^{w,w_m}$.

(2)

$aw\in {\mathcal{A}}_r^D$ and ${\left(aw\right)}^{m-1}a\in {\mathcal{A}}_r^{w,W}$.

(3)

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

$$a=x+y,x^{*}{\left(aw\right)}^{m-1}y=ywx=0,x\in {\mathcal{A}}_r^{w,\#},y\in {\mathcal{A}}_w^{nil}.$$

Proof.$\left(1\right)\iff \left(2\right)$ This is proved by Theorem 3.2 and Theorem 5.1.

$\left(1\right)\Rightarrow \left(3\right)$ In view of Theorem 4.1, we have $x,y\in \mathcal{A}$ such that

$$a=x+y,x^{*}{\left(aw\right)}^{m-1}y=ywx=0,x\in {\mathcal{A}}_r^{w,\#},y\in {\mathcal{A}}_w^{nil}.$$

. Here, $y=a-{\left(aw\right)}^2{a}_r^{w,w_m}$. Clearly, ${\left(aw\right)}^n=\left(aw\right)\left(a_r^{w,w_m}w\right){\left(aw\right)}^n$ for some $n\in N$. Since $a_r^{w,w_m}w={\left(aw\right)}^n{\left(a_r^{w,w_m}w\right)}^n$, we verify that

$$\begin{array}{cc}& [aw-\left(aw\right)\left(a_r^{w,w_m}\right)w\left(aw\right)]\left(aw\right)\left(a_r^{w,w_m}w\right)\left(aw\right)\hfill \\ \hfill =& [aw-\left(aw\right)\left(a_r^{w,w_m}\right)w\left(aw\right)]{\left(aw\right)}^n{\left(a_r^{w,w_m}w\right)}^n\left(aw\right)\hfill \\ \hfill =& 0;\hfill \end{array}$$

hence,

$${[aw-\left(aw\right)\left(a_r^{w,w_m}w\right)\left(aw\right)]}^n=[aw-\left(aw\right)\left(a_r^{w,w_m}w\right)]{\left(aw\right)}^n=0.$$

Therefore $yw=aw[1-\left(aw\right)\left(a_r^{w,w_m}w\right)]\in {\mathcal{A}}^{nil}$, as desired

$\left(3\right)\Rightarrow \left(1\right)$ By hypothesis, there exist $x,y\in \mathcal{A}$ such that

$$a=x+y,x^{*}{\left(aw\right)}^{m-1}y=ywx=0,x\in {\mathcal{A}}_r^{w,\#},y\in {\mathcal{A}}_w^{nil}.$$

By virtue of Theorem 4.1, $a\in {\mathcal{A}}_r^{w,g_m}$. Moreover, $aw=xw+yw$ with $xw\in {\mathcal{A}}_r^D$ and $yw\in {\mathcal{A}}^{nil}$. Since $\left(yw\right)\left(xw\right)=0$, we have $aw\in {\mathcal{A}}_r^D$. Therefore $a_r^{w,w_m}$ by Theorem 5.1. □

Lemma 5.5.

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_r^{w,W_m}$ if and only if

(1)

$aw\in {\mathcal{A}}_r^D$;

(2)

There exists $x\in \mathcal{A}$ such that

$$x=a{\left(wx\right)}^2,{\left[{\left(aw\right)}^n\right]}^{*}{\left(aw\right)}^{m+1}x={\left[{\left(aw\right)}^n\right]}^{*}{\left(aw\right)}^{m-1}a,{\left(aw\right)}^n=\left(aw\right)\left(xw\right){\left(aw\right)}^n$$

for some $n\in N$.

Proof. 

⟹ Set $x=a_r^{w,W_m}$. Then $x=a{\left(wx\right)}^2,{\left[\left(aw\right){\left(aw\right)}_r^D\right]}^{*}{\left(aw\right)}^{m+1}x={\left[\left(aw\right){\left(aw\right)}_r^D\right]}^{*}{\left(aw\right)}^{m-1}a,{\left(aw\right)}^k=\left(aw\right)\left(xw\right){\left(aw\right)}^k$ for some $k\in N$. Assume that $\left(aw\right){\left(aw\right)}_r^D{\left(aw\right)}^l={\left(aw\right)}^l$. Let $n=max\{k,l\}$. Then ${\left[{\left(aw\right)}^n\right]}^{*}{\left(aw\right)}^{m+1}x={\left[{\left(aw\right)}^n\right]}^{*}{\left(aw\right)}^{m-1}a$, as required.

⟸ By hypothesis, there exists $x\in \mathcal{A}$ such that

$$x=a{\left(wx\right)}^2,{\left[{\left(aw\right)}^n\right]}^{*}{\left(aw\right)}^{m+1}x=\left[{\left(aw\right)}^n\right]{\left(aw\right)}^{m-1}a,{\left(aw\right)}^n=\left(aw\right)\left(xw\right){\left(aw\right)}^n$$

for some $n\in N$. Since $aw\in {\mathcal{A}}_r^D$, we have ${\left(aw\right)}^n{\left[{\left(aw\right)}_r^D\right]}^n=\left(aw\right){\left(aw\right)}_r^D$. Hence ${\left[\left(aw\right){\left(aw\right)}_r^D\right]}^{*}{\left(aw\right)}^{m+1}x={\left[\left(aw\right){\left(aw\right)}_r^D\right]}^{*}{\left(aw\right)}^{m-1}a$. Therefore $a\in {\mathcal{A}}_r^{w,W_m}$, as asserted. □

Theorem 5.6.

Let $a\in \mathcal{A}$ and $w\in {\mathcal{A}}^{-1}$. Then $a\in {\mathcal{A}}_r^{w,W_m}$ if and only if

(1)

$aw\in {\mathcal{A}}_r^D$;

(2)

There exists $x\in \mathcal{A}$ such that $x=a{\left(wx\right)}^2,{\left[{\left({\left(aw\right)}^m\right)}^{*}{\left(aw\right)}^{m+1}xw\right]}^{*}={\left({\left(aw\right)}^m\right)}^{*}{\left(aw\right)}^{m+1}xw,{\left(aw\right)}^n=\left(aw\right)\left(xw\right){\left(aw\right)}^n$ for some $n\in N$.

Proof. 

⟹ Set $x=a_r^{w,W_m}$. By virtue of Lemma 5.5, there exists $x\in \mathcal{A}$ such that

$$x=a{\left(wx\right)}^2,{\left[{\left(aw\right)}^n\right]}^{*}{\left(aw\right)}^{m+1}x=\left[{\left(aw\right)}^n\right]{\left(aw\right)}^{m-1}a,{\left(aw\right)}^n=\left(aw\right)\left(xw\right){\left(aw\right)}^n$$

for some $n\in N$. Therefore we have

$$\begin{array}{ccc}\hfill {\left[{\left(aw\right)}^m\right]}^{*}{\left(aw\right)}^{m+1}xw& =\hfill & {\left[{\left(aw\right)}^m\right]}^{*}{\left(aw\right)}^{m+1}\left[{\left(aw\right)}^n{\left(xw\right)}^{n+1}\right]\hfill \\ & =\hfill & {\left[{\left(aw\right)}^m\right]}^{*}\left[{\left(aw\right)}^{m+n}{\left(xw\right)}^{n+1}\right]\hfill \\ & =\hfill & {\left({\left[{\left(aw\right)}^{m+n}{\left(xw\right)}^{n+1}\right]}^{*}{\left(aw\right)}^m\right)}^{*}\hfill \\ & =\hfill & {\left({\left[{\left(aw\right)}^m{\left(xw\right)}^{n+1}\right]}^{*}{\left[{\left(aw\right)}^n\right]}^{*}{\left(aw\right)}^{m-1}aw\right)}^{*}\hfill \\ & =\hfill & {\left({\left[{\left(aw\right)}^m{\left(xw\right)}^{n+1}\right]}^{*}{\left[{\left(aw\right)}^n\right]}^{*}{\left(aw\right)}^{m+1}xw\right)}^{*}\hfill \\ & =\hfill & {\left({\left[{\left(aw\right)}^{m+n}{\left(xw\right)}^{n+1}\right]}^{*}{\left(aw\right)}^{m+1}xw\right)}^{*}\hfill \\ & =\hfill & {\left({\left[{\left(aw\right)}^{m+1}xw\right]}^{*}{\left(aw\right)}^{m+1}xw\right)}^{*}\hfill \\ & =\hfill & {\left[{\left(aw\right)}^{m+1}xw\right]}^{*}{\left(aw\right)}^{m+1}xw.\hfill \end{array}$$

Therefore ${\left[{\left({\left(aw\right)}^m\right)}^{*}{\left(aw\right)}^{m+1}xw\right]}^{*}={\left({\left(aw\right)}^m\right)}^{*}{\left(aw\right)}^{m+1}xw$, as desired.

⟸ By hypothesis, there exists some $z\in \mathcal{A}$ such that such that $x=a{\left(wx\right)}^2,{\left[{\left({\left(aw\right)}^m\right)}^{*}{\left(aw\right)}^{m+1}xw\right]}^{*}={\left({\left(aw\right)}^m\right)}^{*}{\left(aw\right)}^{m+1}xw,{\left(aw\right)}^n=\left(aw\right)\left(xw\right){\left(aw\right)}^n$ for some $n\in N$. Then we verify that

$$\begin{array}{ccc}\hfill {\left[{\left(aw\right)}^n\right]}^{*}{\left(aw\right)}^{m+1}x& =\hfill & {\left[\left(aw\right)\left(xw\right){\left(aw\right)}^n\right]}^{*}{\left(aw\right)}^{m+1}x\hfill \\ & =\hfill & {\left[{\left(aw\right)}^m{\left(xw\right)}^m{\left(aw\right)}^n\right]}^{*}{\left(aw\right)}^{m+1}x\hfill \\ & =\hfill & {\left[{\left(xw\right)}^m{\left(aw\right)}^n\right]}^{*}\left({\left[{\left(aw\right)}^m\right]}^{*}{\left(aw\right)}^{m+1}xw\right)w^{-1}\hfill \\ & =\hfill & {\left[{\left(xw\right)}^m{\left(aw\right)}^n\right]}^{*}{\left({\left[{\left(aw\right)}^m\right]}^{*}{\left(aw\right)}^{m+1}xw\right)}^{*}{w}^{-1}\hfill \\ & =\hfill & {\left({\left[{\left(aw\right)}^m\right]}^{*}{\left(aw\right)}^{m+1}xw{\left(xw\right)}^m{\left(aw\right)}^n\right)}^{*}{w}^{-1}\hfill \\ & =\hfill & {\left({\left(aw\right)}^{m+1}{\left(xw\right)}^{m+1}{\left(aw\right)}^n\right)}^{*}{\left(aw\right)}^m{w}^{-1}\hfill \\ & =\hfill & {\left(\left(aw\right)\left(xw\right){\left(aw\right)}^n\right)}^{*}{\left(aw\right)}^m{w}^{-1}\hfill \\ & =\hfill & \left[{\left(aw\right)}^n\right]{\left(aw\right)}^{m-1}a.\hfill \end{array}$$

By virtue of Lemma 5.5, $a\in {\mathcal{A}}_r^{w,W_m}$. □

Corollary 5.7.

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_r^{W_m}$ if and only if

(1)

$a\in {\mathcal{A}}_r^D$;

(2)

There exists $x\in \mathcal{A}$ such that

$$x=a{x}^2,{\left[{\left(a^m\right)}^{*}{a}^{m+1}x\right]}^{*}={\left(a^m\right)}^{*}{a}^{m+1}x,a^n=ax{a}^n$$

for some $n\in N$.

Proof. 

This is completed by choosing $w=1$ in Theorem 5.6. □

Remark 5.8.

The m-generalized left weighted group inverse can be defined by a similar way. The $(m,w)$-group inverse can be characterized by combining m-generalized left and right weighted group inverses.