Generalized Right Weighted Core Inverse in Banach Algebras

1. Introduction

A Banach algebra 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 a Banach *-algebra $\mathcal{A}$ has core inverse if and only if there exists $x\in \mathcal{A}$ such that

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

If such x exists, it is unique, and denote it by $a^{\#◯}$. Core inverse is extensively studied by many authors from different views, e.g., [1,6,21,25].

Let $C^{n\times n}$ be the Banach *-algebra of all $n\times n$ complex matrices with conjugate transpose * and $\mathcal{R}\left(X\right)$ represent the range space of a complex matrix X. In 2014, Prasad and Mohana extended core inverse and introduced core-EP inverse for a complex matrix (see [17]). A matrix $A\in C^{n\times n}$ has core-EP inverse X if and only if

$$XAX=X,\mathcal{R}\left(X\right)=\mathcal{R}\left(X^{*}\right)=\mathcal{R}\left(A^k\right),$$

where $k=ind\left(A\right)$ is the Drazin index of A. Such X is unique, and we denote it by $A^{†◯}$.

In 2020, Gao et al. extended the concept of the core-EP inverse and introduced the notion of weighted core-EP inverse for a complex matrix (see [9]). Let $A,W\in C^{n\times n}$ and $k=max\left\{ind\right(AW),ind(WA\left)\right\}$. The weighted core-EP inverse of A is the unique solution to the system:

$$WAWX={\left(WA\right)}^k{\left[{\left(WA\right)}^k\right]}^{†},\mathcal{R}\left(X\right)\subseteq \mathcal{R}\left({\left(AW\right)}^k\right),$$

and we denote such X by $A^{†◯,W}$. Then Mosić introduced and studied weighted core-EP inverse for a bounded linear operator between two Hilbert spaces as a generalization of the weighted core-EP inverse of a matrix (see [14]). In 2021, Mosić further extended the weighted core-EP inverse of bounded linear operators on Hilbert spaces to elements of a $C^{*}$-algebra and the weighted core-EP inverse in a $C^{*}$-algebra was characterized by means of range projections (see [15]).

Wang et al. generalized the core inverse to the right core inverse (see [18]). An element $a\in \mathcal{A}$ has right core inverse if there exist $x\in \mathcal{A}$ such that

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

If such x exists, we denote it by $a_r^{\#◯}$.

An element $a\in \mathcal{A}$ has right pseudo core if there exists a $x\in \mathcal{A}$ such that

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

for some $k\in N$ (see [18]).

In [5], the authors introduced and studied generalized right core inverse. An element $a\in \mathcal{A}$ has generalized right core inverse if there exists a $x\in \mathcal{A}$ such that

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

The preceding x is called generalized right core inverse of a and we denote it by $a_r^{d◯}$. We refer the reader more properties of right generalized inverse in [18,19,22,27].

Recently, many authors studied generalized inverse with wights (see [2,8,9,12,16,20,23,24]). In [28], Zhu et al. introduced and studied a weighted generalized inverse as a generalization of core inverse. Let $a,w\in \mathcal{A}$. An element $a\in \mathcal{A}$ is w-core invertible if there exists some $x\in \mathcal{A}$ such that

$$aw{x}^2=x,{\left(awx\right)}^{*}=awx,xawa=a.$$

Such an x is called a w-core inverse of a. Many properties of w-core inverse are presented by many authors, e.g., [26,27]. In [29], Zhu et al. extended w-core inverse and introduce right w-core inverse. An element $a\in \mathcal{A}$ has right w-core inverse if there exists $x\in \mathcal{A}$ such that

$$aw{x}^2=x,{\left(awx\right)}^{*}=awx,awxa=a.$$

Many properties of right w-core inverse are investigated in [29]. However, the right w-core inverse is unrelated to the weighted core-EP inverse mentioned above. The motivation of this paper is to introduce and study a new class of weighted generalized inverses, such that the class of weighted core-EP inverses can be viewed as a subclass of this new class.

 Definition 1.1. 

An element $a\in \mathcal{A}$ has generalized right w-weighted core inverse if there exists a $x\in \mathcal{A}$ such that

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

The preceding x is denoted by $a_r^{d◯,w}$. It is evident that for any complex matrix, the generalized right weighted core inverse coincides with the weighted core-EP inverse. Consequently, many properties of the weighted core-EP inverse naturally extend to this broader class.

In Section 2, we introduce right w-weighted core inverse for an element in a Banach *-algebra.

 Definition 1.2. 

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

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

If such x exists, we denote it by $a_r^{\#◯,w}$. The investigation explores several elementary properties of the right w-weighted core inverse, which will be utilized subsequently.

In Section 3, we characterize generalized right weighted core inverse by combining right w-weighted core inverse and quasinilpotent in a Banach *-algebra. We prove that $a\in \mathcal{A}$ has generalized right w-core inverse if and only if there exist $z,y\in \mathcal{A}$ such that

$$a=z+y,{\left(wz\right)}^{*}\left(wy\right)=ywz=0,z\in {\mathcal{A}}_r^{\#◯,w},y\in {\mathcal{A}}_w^{qnil}.$$

Here, ${\mathcal{A}}_w^{qnil}=\{x\in \mathcal{A}\mid \underset{n\to \infty }{lim}{\left|\right|\left(xw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0$. Surprisingly, we observe that the preceding condition ${\left(wz\right)}^{*}\left(wy\right)=0$ can be dropped, which add some new properties for the weighted core-EP inverse as well. The polar-like property of the generalized right weighted core inverse is thereby established.

An element $a\in \mathcal{A}$ has right wg-Drazin inverse x if there exists a $x\in \mathcal{A}$ such that

$$a{\left(wx\right)}^2=x=xwawx,aw-xw{\left(aw\right)}^2\in {\mathcal{A}}^{qnil}.$$

We denote such a x by $a_r^{d,w}$. The right wg-Drazin inverse is an one-sided version of weighted generalized Drazin inverse (see [13]). In Section 4, we characterize generalized right weighted core inverse by using the right wg-Drazin inverse. The generalized right weighted core inverse is then constructed based on the right weighted core inverse.

 Definition 1.3. 

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

$$a{\left(wx\right)}^2=x,{\left(wawx\right)}^{*}=wawx,{\left(aw\right)}^k=\left(aw\right)\left(xw\right){\left(aw\right)}^k$$

for some $k\in N$.

Finally, in Section 5, we investigate the right pseudo w-core inverse for elements in a Banach *-algebra by using the generalized right w-core inverse. This approach allows us to present many new properties of the weighted core-EP inverse.

Throughout the paper, all Banach *-algebras are complex with an identity. ${\mathcal{A}}_r^{d,w},{\mathcal{A}}_r^{D,w},{\mathcal{A}}_r^{\#◯,w}$, ${\mathcal{A}}_r^{d◯,w}$ and ${\mathcal{A}}_r^{D◯,w}$ denote the sets of all right weighted g-Drazin invertible, right weighted Drazin invertible, right weighted core invertible, generalized right weighted core invertible and generalized right pseudo w-core invertible elements in $\mathcal{A}$, respectively.

2. Right w-Weighted Core Inverse

The objective of this section is to elucidate the fundamental properties of the right w-weighted core inverse. If $a,x$ and w satisfy the equations $a=awxwa$ and ${\left(wawx\right)}^{*}=wawx$, then x is called w-weighted $(1,3)$-inverse of a and is denoted by $a_w^{(1,3)}$. We use ${\mathcal{A}}_w^{(1,3)}$ to stand for sets of all w-weighted $(1,3)$-invertible elements in $\mathcal{A}$.

 Theorem 2.1. 

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

(1)

$a\in {\mathcal{A}}_r^{\#◯,w}$.

(2)

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

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

(3)

$a\in {\mathcal{A}}_w^{(1,3)}$ and $a\mathcal{A}=awa\mathcal{A}$.

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

$$a{\left(wz\right)}^2=z,{\left(wawz\right)}^{*}=wawz,awzwa=a.$$

Set $x=zwawz$. Then we verify that

$$\begin{array}{ccc}\hfill awx& =\hfill & \left(awzwa\right)wz=awz,\hfill \\ \hfill a{\left(wx\right)}^2& =\hfill & \left[a{\left(wz\right)}^2\right]wawz=zwawz=x,\hfill \\ \hfill xwawx& =\hfill & zwawzw\left(awz\right)=zw\left(awzwa\right)wz=zwawz=x,\hfill \\ \hfill awxwa& =\hfill & \left(awz\right)wa=a,\hfill \\ \hfill {\left(wawx\right)}^{*}& =\hfill & {\left(wawz\right)}^{*}=wawz=wawx.\hfill \end{array}$$

as desired.

$\left(2\right)\Rightarrow \left(1\right)$ This is trivial.

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

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

Then $wawxwa=wa$. Hence, $a\in {\mathcal{A}}_w^{(1,3)}$.

Clearly, $awa\in a\mathcal{A}$. On the other hand, we have $a=\left(aw\right)x\left(wa\right)=aw\left[a{\left(wx\right)}^2\right]wa\in awa\mathcal{A}.$ Therefore $a\mathcal{A}=awa\mathcal{A}$.

$\left(3\right)\Rightarrow \left(1\right)$ Since $a\in {\mathcal{A}}_w^{(1,3)}$, we can find some $x\in \mathcal{A}$ such that $wawxwa=wa$ and ${\left(wawx\right)}^{*}=wawx$. Write $a=awat$ for some $t\in \mathcal{A}$. Set $x=atw{a}_w^{(1,3)}$. Then we verify that

$$\begin{array}{ccc}\hfill awx& =\hfill & \left(awat\right)w{a}_w^{(1,3)}=aw{a}_w^{(1,3)},\hfill \\ \hfill {\left(wawx\right)}^{*}& =\hfill & {\left(waw{a}_w^{(1,3)}\right)}^{*}=waw{a}_w^{(1,3)}=wawx,\hfill \\ \hfill a{\left(wx\right)}^2& =\hfill & \left(awx\right)\left(wx\right)=\left(aw{a}_w^{(1,3)}\right)w\left(atw{a}_w^{(1,3)}\right)\hfill \\ & =\hfill & \left(aw{a}_w^{(1,3)}wa\right)tw{a}_w^{(1,3)}=atw{a}_w^{(1,3)}=x,\hfill \\ \hfill awxwa& =\hfill & \left(awx\right)wa=aw{a}_w^{(1,3)}wa=a.\hfill \end{array}$$

Therefore $a\in {\mathcal{A}}_r^{\#◯,w}$. □

 Corollary 2.2. 

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

(1)

$a\in {\mathcal{A}}_r^{\#◯,w}$.

(2)

$\mathcal{A}a=\mathcal{A}{\left(wa\right)}^{*}\left(wa\right)$ and $a\mathcal{A}=awa\mathcal{A}$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By virtue of Theorem 2.1, $a\mathcal{A}=awa\mathcal{A}$. By hypothesis, $wa\in {\mathcal{A}}^{(1,3)}$. In light of [7, Lemma2.3], we have $\mathcal{A}wa=\mathcal{A}{\left(wa\right)}^{*}\left(wa\right)$. As $a=aw{a}_r^{\#◯,w}wa$, we see that $\mathcal{A}wa=\mathcal{A}a$, and then $\mathcal{A}a=\mathcal{A}{\left(wa\right)}^{*}\left(wa\right)$.

$\left(2\right)\Rightarrow \left(1\right)$ Write $a=t{\left(wa\right)}^{*}\left(wa\right)$ for a $t\in \mathcal{A}$. Then $wa=\left(wt\right){\left(wa\right)}^{*}\left(wa\right)$, and so $\left(wa\right){\left(wt\right)}^{*}=\left(wt\right){\left(wa\right)}^{*}\left(wa\right){\left(wt\right)}^{*}$. Moreover, ${\left(wa\right)}^{*}={\left(wa\right)}^{*}\left(wa\right){\left(wt\right)}^{*}$; hence, ${\left(wa\right)}^{*}\left(wa\right)={\left(wa\right)}^{*}\left(wa\right){\left(wt\right)}^{*}\left(wa\right)$. Thus, $t{\left(wa\right)}^{*}\left(wa\right)=t{\left(wa\right)}^{*}\left(wa\right){\left(wt\right)}^{*}$$\left(wa\right)$. Since $a\mathcal{A}=awa\mathcal{A}$, we can find a $z\in \mathcal{A}$ such that $a=awaz$. Then we derive

$$\begin{array}{ccc}\hfill a& =\hfill & t{\left(wa\right)}^{*}\left(wa\right)=t{\left(wa\right)}^{*}\left(wa\right){\left(wt\right)}^{*}\left(wa\right)\hfill \\ & =\hfill & a{\left(wt\right)}^{*}\left(wa\right)=awaz{\left(wt\right)}^{*}wa=awxwa,\hfill \end{array}$$

where $x=az{\left(wt\right)}^{*}$. Clearly, $wawx=wawaz{\left(wt\right)}^{*}=wa{\left(wt\right)}^{*}=\left(wt\right){\left(wa\right)}^{*}\left(wa\right)$${\left(wt\right)}^{*}$. This implies that ${\left(wawx\right)}^{*}=wawx$. Accordingly, $a\in {\mathcal{A}}_w^{(1,3)}$. This completes the proof by Theorem 2.1. □

 Theorem 2.3. 

Let $a\in {\mathcal{A}}_r^{\#◯,w}$. Then there exists a projection $p\in \mathcal{A}$ such that

$$pwa=0andwa+p\in {\mathcal{A}}_r^{-1}.$$

 Proof. 

Let $x=a_r^{\#◯,w}$. Then

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

Let $p=1-wawx$. Then $p^2=p=p^{*}\in \mathcal{A}$ and $pwa=0$. We directly check that

$$\begin{array}{cc}& [wa+1-wawx][wx+1-wawx]\hfill \\ \hfill =& \left(wa\right)\left(wx\right)+\left(wa\right)(1-wawx)+(1-wawx)\left(wx\right)+{(1-wawx)}^2\hfill \\ \hfill =& \left(wa\right)\left(wx\right)+\left(wa\right)(1-wawx)+wx-wa{\left(wx\right)}^2+(1-wawx)\hfill \\ \hfill =& 1+\left(wa\right)(1-wawx).\hfill \end{array}$$

By hypothesis, we verify that

$$[1+wa(1-wawx\left)\right][1-(wa\left)\right(1-wawx\left)\right]=1.$$

Then

$$[wa+1-wawx][wx+1-wawx][1-(wa\left)\right(1-wawx\left)\right]=1.$$

Thus, $wa+1-wawx\in \mathcal{A}$ is right invertible, as required. □

 Corollary 2.4. 

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

(1)

$a\in {\mathcal{A}}_r^{\#◯,w}$.

(2)

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

$$pwa=0andwa+p\in {\mathcal{A}}_r^{-1}.$$

(3)

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

$$pwa=0andwa+p\in {\mathcal{A}}_r^{-1}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By virtue of Theorem 2.3, here exists a projection $p\in \mathcal{A}$ such that

$$pwa=0andwa+p\in {\mathcal{A}}_r^{-1}.$$

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

$$\begin{array}{c}qwa=0andwa+q\in {\mathcal{A}}_r^{-1}.\end{array}$$

Then $(1-p)wa=wa$, and so $\ell (1-p)\subseteq \ell \left(wa\right)$. On the other hand, $wa=(1-p)(wa+p)$, and then $1-p=wa{(wa+p)}_r^{-1}$. This implies that $\ell \left(wa\right)=\ell (1-p)$. Thus, we have $\ell (1-p)=\ell \left(wa\right)$. Likewise, we have $\ell (1-p)=\ell \left(wa\right)$. It follows that $\ell (1-p)=\ell (1-q)$. This implies that $p=pq$ and $q=qp$. Accordingly, $p=pq=p^{*}{q}^{*}={\left(qp\right)}^{*}=q^{*}=q$, as required.

$\left(2\right)\Rightarrow \left(3\right)$ This is obvious.

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

$$pwa=0,(wa+p)s=1andps(1-p)=0.$$

Since $(1-p)(wa+p)=(1-p)wa=wa$, we see that $1-p=was=was(1-p)$. Since $s(1-p)=(1-p)s(1-p)\in w\mathcal{A}$, we may write $s(1-p)=wx$ for some $x\in \mathcal{A}$. Then $wawx=was(1-p)=1-p$; and so ${\left(wawx\right)}^{*}={(1-p)}^{*}=1-p=wawx$. Obviously, we have $awxwa=(1-p)wa=wa$, and so $awxwa=a$.

Since $p(wa+p)=p$, we see that $p=ps$ and $(1-p)s=s-ps=s-p.$ Thus,

$$\begin{array}{ccc}\hfill a{\left(wx\right)}^2& =\hfill & awxwx=(1-p)wx=(1-p){(wa+p)}_r^{-1}(1-p)\hfill \\ & =\hfill & [{(wa+p)}_r^{-1}-p](1-p)=w\left[w^{-1}{(wa+p)}_r^{-1}(1-p)\right]=x.\hfill \end{array}$$

As $w\in {\mathcal{A}}^{-1}$, we deduce that $a{\left(wx\right)}^2=x$. Therefore $a_r^{\#◯}=x$, as asserted. □

 Corollary 2.5. 

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

(1)

$a\in {\mathcal{A}}_r^{\#◯}$.

(2)

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

$$pa=0\in {\mathcal{A}}^{qnil},and{a}^n+p\in {\mathcal{A}}_r^{-1}.$$

 Proof. 

Straightforward by choosing $w=1$ in Theorem 2.3. □

Let $p,q\in \mathcal{A}$ be idempotents. Then for any $x\in \mathcal{A}$, we have $x=pxq+px(1-p)+(1-p)xq+(1-p)x(1-q)$. Thus x can be represented in the matrix form $x={\left(\begin{array}{cc}pxp& px(1-p)\\ (1-p)xp& (1-p)x(1-q)\end{array}\right)}_{(p,q)}$. We have at our disposal all the information necessary to prove the following.

 Theorem 2.6. 

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

(1)

$a_r^{\#◯,w}=x$.

(2)

There exist an idempotent $p\in \mathcal{A}$ and a projection $q\in \mathcal{A}$ such that

$$a={\left(\begin{array}{cc}{a}_1& a_2\\ 0& 0\end{array}\right)}_{p\times q},w={\left(\begin{array}{cc}{w}_1& w_2\\ 0& w_3\end{array}\right)}_{q\times p}$$

and x is represented as

$$x={\left(\begin{array}{cc}{\left(w_1{a}_1{w}_1\right)}_r^{-1}& x_2\\ 0& 0\end{array}\right)}_{p\times q},$$

where $w_1{a}_1{w}_1{x}_2=0,a_1{w}_1{\left(w_1{a}_1{w}_1\right)}^{-1}{w}_1{x}_2=x_2.$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ Let $p=\left(aw\right)\left(a_r^{\#◯,w}w\right)$ and $q=\left(wa\right)\left(w{a}_r^{\#◯,w}\right)$. Then we verify that

$$\begin{array}{ccc}\hfill (1-p)a& =\hfill & [1-\left(aw\right)\left(a_r^{\#◯,w}w\right)]a\hfill \\ & =\hfill & a-aw{a}_r^{\#◯,w}wa=0.\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill (1-q)wp& =\hfill & [1-\left(wa\right)\left(w{a}_r^{\#◯,w}\right)]w\left(aw\right)\left(a_r^{\#◯,w}w\right)\hfill \\ & =\hfill & w[a-aw{a}_r^{d,w})wa]w{a}_r^{d,w}w\hfill \\ & =\hfill & 0.\hfill \end{array}$$

Hence, we may write

$$a={\left(\begin{array}{cc}{a}_1& a_2\\ 0& 0\end{array}\right)}_{p\times q},w={\left(\begin{array}{cc}{w}_1& w_2\\ 0& w_3\end{array}\right)}_{q\times p}.$$

Then we have

$$aw={\left(\begin{array}{cc}{a}_1{w}_1& a_1{w}_2+a_2{w}_3\\ 0& 0\end{array}\right)}_{p\times p},wa={\left(\begin{array}{cc}{w}_1{a}_1& w_1{a}_2\\ 0& 0\end{array}\right)}_{q\times q}.$$

Since $(1-p)x=[1-\left(aw\right)\left(a_r^{d,w}w\right)]a_r^{d◯,w}=0$, we may write $x={\left(\begin{array}{cc}{x}_1& x_2\\ 0& 0\end{array}\right)}_{p\times q}.$

We compute that

$$\begin{array}{ccc}\hfill wawx& =\hfill & {\left(\begin{array}{cc}{w}_1{a}_1& w_1{a}_2\\ 0& 0\end{array}\right)}_{q\times q}{\left(\begin{array}{cc}{w}_1& w_2\\ 0& w_3\end{array}\right)}_{q\times p}{\left(\begin{array}{cc}{x}_1& x_2\\ 0& 0\end{array}\right)}_{p\times q}\hfill \\ & =\hfill & {\left(\begin{array}{cc}{w}_1{a}_1& w_1{a}_2\\ 0& 0\end{array}\right)}_{q\times q}{\left(\begin{array}{cc}{w}_1{x}_1& w_1{x}_2\\ 0& 0\end{array}\right)}_{q\times q}\hfill \\ & =\hfill & {\left(\begin{array}{cc}{w}_1{a}_1{w}_1{x}_1& w_1{a}_1{w}_1{x}_2\\ 0& 0\end{array}\right)}_{q\times p}.\hfill \end{array}$$

Since ${\left(wawx\right)}^{*}=wawx$, we have $w_1{a}_1{w}_1{x}_2=0$.

It is easy to verify that

$$\begin{array}{cc}& \left(w_1{a}_1{w}_1\right)x_1\hfill \\ \hfill =& \left[\right(wawx\left)w\right(awxw\left)\right]\left[\right(awxw\left)a\right(wawx\left)\right]\left[\right(wawx\left)w\right(awxw\left)\right]\hfill \\ & \left[\right(awxw\left)x\right(wawx\left)\right]=wawx=q.\hfill \end{array}$$

Therefore

$$x={\left(\begin{array}{cc}{\left(w_1{a}_1{w}_1\right)}_r^{-1}& x_2\\ 0& 0\end{array}\right)}_{p\times p},$$

as required.

$\left(2\right)\Rightarrow \left(1\right)$ By hypothesis, we have an idempotent $p\in \mathcal{A}$ and a projection $q\in \mathcal{A}$ such that

$$a={\left(\begin{array}{cc}{a}_1& a_2\\ 0& 0\end{array}\right)}_{p\times q},w={\left(\begin{array}{cc}{w}_1& w_2\\ 0& w_3\end{array}\right)}_{q\times p}$$

and x is represented as

$$x={\left(\begin{array}{cc}{\left(w_1{a}_1{w}_1\right)}_r^{-1}& x_2\\ 0& 0\end{array}\right)}_{p\times q},$$

where $w_1{a}_1{w}_1{x}_2=0,a_1{w}_1{\left(w_1{a}_1{w}_1\right)}^{-1}{w}_1{x}_2=x_2.$ Then we verify that

$$\begin{array}{ccc}\hfill awx& =\hfill & {\left(\begin{array}{cc}{a}_1& a_2\\ 0& 0\end{array}\right)}_{p\times q}{\left(\begin{array}{cc}{w}_1& w_2\\ 0& w_3\end{array}\right)}_{q\times p}{\left(\begin{array}{cc}{\left(w_1{a}_1{w}_1\right)}_r^{-1}& x_2\\ 0& 0\end{array}\right)}_{p\times q}\hfill \\ & =\hfill & {\left(\begin{array}{cc}{a}_1{w}_1& a_1{w}_2+a_2{w}_3\\ 0& 0\end{array}\right)}_{p\times q}{\left(\begin{array}{cc}{\left(w_1{a}_1{w}_1\right)}_r^{-1}& x_2\\ 0& 0\end{array}\right)}_{p\times q}\hfill \\ & =\hfill & {\left(\begin{array}{cc}{a}_1{w}_1{\left(w_1{a}_1{w}_1\right)}_r^{-1}& a_1{w}_1{x}_2\\ 0& 0\end{array}\right)}_{p\times q},\hfill \\ \hfill a{\left(wx\right)}^2& =\hfill & {\left(\begin{array}{cc}{a}_1{w}_1{\left(w_1{a}_1{w}_1\right)}_r^{-1}& a_1{w}_1{x}_2\\ 0& 0\end{array}\right)}_{p\times q}{\left(\begin{array}{cc}{w}_1{\left(w_1{a}_1{w}_1\right)}_r^{-1}& w_1{x}_2\\ 0& 0\end{array}\right)}_{p\times q}\hfill \\ & =\hfill & {\left(\begin{array}{cc}{a}_1{w}_1{\left(w_1{a}_1{w}_1\right)}_r^{-1}{w}_1{\left(w_1{a}_1{w}_1\right)}_r^{-1}& a_1{w}_1{\left(w_1{a}_1{w}_1\right)}_r^{-1}{w}_1{x}_2\\ 0& 0\end{array}\right)}_{p\times q}\hfill \\ \\ & =\hfill & {\left(\begin{array}{cc}{\left(w_1{a}_1{w}_1\right)}_r^{-1}& x_2\\ 0& 0\end{array}\right)}_{p\times q}.\hfill \end{array}$$

Moreover, we have

$$\begin{array}{ccc}\hfill wawx& =\hfill & {\left(\begin{array}{cc}{w}_1& w_2\\ 0& w_3\end{array}\right)}_{q\times p}{\left(\begin{array}{cc}{a}_1{w}_1{\left(w_1{a}_1{w}_1\right)}_r^{-1}& 0\\ 0& 0\end{array}\right)}_{p\times q}\hfill \\ & =\hfill & {\left(\begin{array}{cc}{w}_1{a}_1{w}_1{\left(w_1{a}_1{w}_1\right)}_r^{-1}& 0\\ 0& 0\end{array}\right)}_{q\times q}\hfill \\ & =\hfill & q=q^{*}={\left(wawx\right)}^{*},\hfill \\ \hfill awxwa& =\hfill & {\left(\begin{array}{cc}{a}_1{w}_1{\left(w_1{a}_1{w}_1\right)}_r^{-1}& a_1{w}_1{x}_2\\ 0& 0\end{array}\right)}_{p\times q}{\left(\begin{array}{cc}{w}_1{a}_1& w_1{a}_2\\ 0& 0\end{array}\right)}_{q\times q}\hfill \\ & =\hfill & {\left(\begin{array}{cc}{w}_1^{-1}& w_1^{-1}\left(w_1{a}_1{w}_1{x}_2\right)\\ 0& 0\end{array}\right)}_{p\times q}{\left(\begin{array}{cc}{w}_1{a}_1& w_1{a}_2\\ 0& 0\end{array}\right)}_{q\times q}\hfill \\ & =\hfill & a.\hfill \end{array}$$

Therefore $a\in {\mathcal{A}}^{\#◯,w}=x,$ as required. □

 Corollary 2.7. 

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

(1)

$a_r^{\#◯}=x$.

(2)

There exist an idempotent $p\in \mathcal{A}$ and a projection $q\in \mathcal{A}$ such that $a={\left(\begin{array}{cc}{a}_1& a_2\\ 0& 0\end{array}\right)}_{p\times q}$ and x is represented as

$$x={\left(\begin{array}{cc}{\left(a_1\right)}_r^{-1}& x_2\\ 0& 0\end{array}\right)}_{p\times q},$$

where $a_1{x}_2=0,a_1{\left(a_1\right)}_r^{-1}{x}_2=x_2.$

 Proof. 

We obtain the result by choosing $w=1$ in Theorem 2.6. □

3. Generalized Right w-Weighted Core Decomposition

The purpose of this section is to characterize generalized w-weighted core inverse in a Banach *-algebra by using right weight core inverse and quasinilpotent. The following theorem contains new characterizations for a generalized right w-weighted core inverse.

 Theorem 3.1. 

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

(1)

$a\in {\mathcal{A}}_r^{d◯,w}$.

(2)

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

$$a=z+y,{\left(wz\right)}^{*}\left(wy\right)=ywz=0,z\in {\mathcal{A}}_r^{\#◯,w},y\in {\mathcal{A}}_w^{qnil}.$$

(3)

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

$$a=z+y,ywz=0,z\in {\mathcal{A}}_r^{\#◯,w},y\in {\mathcal{A}}_w^{qnil}.$$

In this case, $a_r^{d◯,w}=z_r^{\#◯,w}.$

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

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

Set $z=awxwa$ and $y=a-awxwa.$ Then $a=z+y$. 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 prove that

$$\left(aw\right)xwawx=awx,\left(aw\right)xw{\left(aw\right)}^{2}x={\left(aw\right)}^{2}x.$$

Step 1. We claim that z has right w-weighted core inverse x. One easily checks that

$$\begin{array}{ccc}\hfill z{\left(wx\right)}^2& =\hfill & \left(awxwa\right){\left(wx\right)}^2=awxw\left[a{\left(wx\right)}^2\right]=\left(awxw\right)x=a{\left(wx\right)}^2=x,\hfill \\ \hfill wzwx& =\hfill & w\left(awxwa\right)wx={\left(wawx\right)}^2,\hfill \\ \hfill {\left(wzwx\right)}^{*}& =\hfill & {\left[{\left(wawx\right)}^{*}\right]}^2={\left(wawz\right)}^2=wzwx,\hfill \\ \hfill zwxwz& =\hfill & \left(awxwa\right)wxw\left(awxwa\right)=\left(awxwaw\right)xwawxwa\hfill \\ & =\hfill & \left(awxwaw\right)xwa=awxwa=z.\hfill \end{array}$$

Thus, $z\in {\mathcal{A}}_r^{\#◯,w}$ and $x=z_r^{\#◯,w}$.

Moreover, we see that

$$[aw-(aw\left)xw\right(aw\left)\right]awx=0.$$

Then we verify that

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

Accordingly,

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

We infer that $aw-\left(aw\right)xw\left(aw\right)\in {\mathcal{A}}^{qnil}$. Therefore $yw=aw-awxwaw\in {\mathcal{A}}^{qnil}$. That is, $y\in {\mathcal{A}}_w^{qnil}.$

Moreover, we verify that

$$\begin{array}{ccc}\hfill {\left(wz\right)}^{*}wy& =\hfill & {\left(wawxwa\right)}^{*}(wa-wawxwa)={\left(wa\right)}^{*}{\left(wawx\right)}^{*}(1-wawx)wa\hfill \\ & =\hfill & {\left(wa\right)}^{*}\left(wawx\right)(1-wawx)wa={\left(wa\right)}^{*}(wawxw-wawxwawxw)a\hfill \\ & =\hfill & {\left(wa\right)}^{*}[wawxw-w\left(awxwawxw\right)]a\hfill \\ & =\hfill & {\left(wa\right)}^{*}[wawxw-w\left(awxw\right)]a=0,\hfill \\ \hfill ywx& =\hfill & (a-awxwa)w\left(awxwa\right)=awawxwa-\left[awxw{\left(aw\right)}^2\right]xwa\hfill \\ & =\hfill & awawxwa-{\left(aw\right)}^{2}xwa=0,\hfill \end{array}$$

as required.

$\left(2\right)\Rightarrow \left(3\right)$ This is obvious.

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

$$a=z+y,ywz=0,z\in {\mathcal{A}}_r^{\#◯,w},y\in {\mathcal{A}}_w^{qnil}.$$

Set $x=z_r^{\#◯,w}$. Then

$$\begin{array}{ccc}\hfill z{\left(wx\right)}^2& =\hfill & x,\hfill \\ \hfill {\left(wzwx\right)}^{*}& =\hfill & wzwx,\hfill \\ \hfill \left(aw\right)xw\left(zw\right)& =\hfill & zw.\hfill \end{array}$$

It is easy to verify that

$$\begin{array}{ccc}\hfill a{\left(wx\right)}^2& =\hfill & (z+y){\left(wx\right)}^2=z{\left(wx\right)}^2=x,\hfill \\ \hfill \left(waw\right)x& =\hfill & w(z+y)wx=wzwx,\hfill \\ \hfill {\left(\left(waw\right)x\right)}^{*}& =\hfill & {\left(wzwx\right)}^{*}=wzwx=\left(waw\right)x.\hfill \end{array}$$

We verify that

$$\begin{array}{cc}& {\left(aw\right)}^2-\left(aw\right)\left(xw\right){\left(aw\right)}^2\hfill \\ \hfill =& aw(zw+yw)-\left(aw\right)(xwzw+xwyw)(zw+yw)\hfill \\ \hfill =& aw(zw+yw)-\left(aw\right)[\left(xwzw\right)zw+\left(xwzw\right)yw+xw\left(ywz\right)w+xw{\left(yw\right)}^2]\hfill \\ \hfill =& aw(zw+yw)-[awxw{\left(zw\right)}^2+(z+y)wxwzwyw+awxw{\left(yw\right)}^2]\hfill \\ \hfill =& awzw+awyw-[awzw+zwxw\left(zw\right)yw+awxw{\left(yw\right)}^2]\hfill \\ \hfill =& awzw+awyw-[awzw+zwxw\left(xw{\left(zw\right)}^2\right)yw+awxw{\left(yw\right)}^2]\hfill \\ \hfill =& awyw-z{\left(wx\right)}^{2}w{\left(zw\right)}^{2}yw-awxw{\left(yw\right)}^2]\hfill \\ \hfill =& awyw-\left[xw{\left(zw\right)}^2\right]yw-awxw{\left(yw\right)}^2]\hfill \\ \hfill =& (z+y)wyw-zwyw-(z+y)wxw{\left(yw\right)}^2]\hfill \\ \hfill =& {\left(yw\right)}^2-zwxw{\left(yw\right)}^2\hfill \\ \hfill =& (1-zwxw){\left(yw\right)}^2.\hfill \end{array}$$

Since $yw\left(aw\right)=yw(zw+yw)={\left(yw\right)}^2$, we deduce that

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

Then

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

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

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

Therefore

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

Accordingly, $a\in {\mathcal{A}}_r^{d◯,w}$, as desired. □

 Corollary 3.2. 

Let $a\in {\mathcal{A}}_r^{d◯,w}$ and $b\in {\mathcal{A}}^{qnil}$. If $bwa=0$, then $a+b\in {\mathcal{A}}_r^{d◯,w}$. In this case, ${(a+b)}_r^{d◯,w}=a_r^{d◯,w}.$

 Proof. 

Since $a\in {\mathcal{A}}_r^{d◯,w}$, by virtue of Theorem 3.1, there exist $x\in {\mathcal{A}}_r^{\#◯,w}$ and $y\in {\mathcal{A}}_w^{qnil}$ such that $a=x+y,ywx=0$. As in the proof of Theorem 3.1, $x=aw{a}_r^{d◯,w}wa$ and $y=a-aw{a}_r^{d◯,w}wa$. Then $a=x+(y+b)$. As $bwy=bw(a-aw{a}_r^{d◯,w}wa)=0$, it follows by [3, Lemma 2.4] that $(y+b)w\in {\mathcal{A}}^{qnil}$. Hence, $y+b\in {\mathcal{A}}_w^{qnil}$. Obviously, $(y+b)wx=ywx+bwx=0$. In light of Theorem 3.1, $a+b\in {\mathcal{A}}_r^{d◯,w}$. In this case, ${(a+b)}_r^{d◯,w}=x_r^{\#◯,w}=a_r^{d◯,w}.$ □

 Corollary 3.3. 

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

(1)

$a\in {\mathcal{A}}_r^{d◯}$.

(2)

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

$$a=z+y,z^{*}y=yz=0,z\in {\mathcal{A}}_r^{\#◯},y\in {\mathcal{A}}^{qnil}.$$

(3)

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

$$a=z+y,ywz=0,z\in {\mathcal{A}}_r^{\#◯},y\in {\mathcal{A}}^{qnil}.$$

In this case, $a_r^{d◯,w}=z^{\#◯}.$

 Proof. 

This is obvious by choosing $w=1$ in Theorem 3.1. □

 Corollary 3.4. 

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

(1)

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

(2)

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

$$a=z+y,z^{*}y=yz=0,z\in {\mathcal{A}}_r^{\#◯},y\in {\mathcal{A}}^{nil}.$$

(3)

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

$$a=z+y,yz=0,z\in {\mathcal{A}}_r^{\#◯},y\in {\mathcal{A}}^{nil}.$$

In this case, $a^d=z_r^{\#◯}.$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ This is proved in [5, Theorem 5.2].

$\left(2\right)\Rightarrow \left(3\right)$ This is trivial.

$\left(3\right)\Rightarrow \left(1\right)$ In view of Theorem 3.1, $a\in {\mathcal{A}}_r^d$. Since $y,z\in {\mathcal{A}}_r^D$ and $yz=0$, we directly verify that $a\in {\mathcal{A}}_r^D$. Therefore $a\in {\mathcal{A}}_r^D$ by [5, Lemma 5.1]. □

We present the following example to illustrate Theorem 3.1.

 Example 3.5. 

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:

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

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

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

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

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

Obviously, we have

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

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

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

$$\gamma (x_1,x_2,x_3,\dots )=(x_2,x_3,x_4,\dots ),$$

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

$$v(x_1,x_2,x_3,\dots )=({\displaystyle \frac{1}{3}}{x}_1,{\displaystyle \frac{1}{4}}{x}_2,{\displaystyle \frac{1}{5}}{x}_3,\dots ).$$

Since $\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^2}}$ converges, we see that v is well defined.

Let $\delta =v\gamma$. Then it is given 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

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

i.e., $\delta$ is quasinilpotent.

Let $T=\tau \oplus \gamma$ be the direct sum of $\tau$ and $\gamma$, $W=u\oplus v$ be the direct sum of u and v, $S=\sigma \oplus 0$ 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 $\beta =0\oplus \gamma$.

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

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

Then $S={\alpha }_r^{\#◯,W}$.

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

Claim 3. $\beta W\alpha =0$. This is obvious.

Therefore T has a generalized right W-core inverse by Theorem 3.1.

As it is well known, an element $a\in \mathcal{A}$ has g-Drazin inverse if and only if it is quasi-polar, i.e., there exists an idempotent $p\in \mathcal{A}$ such that $pa=ap\in {\mathcal{A}}^{qnil},a+p\in {\mathcal{A}}^{-1}$ (see [4]). For generalized w-weighted core-EP inverse, we establish the following polar-like characterization.

 Theorem 3.6. 

Let $a\in {\mathcal{A}}_r^{d◯,w}$ and $n\in N$. Then there exists a projection $p\in \mathcal{A}$ such that

$$pwa\in {\mathcal{A}}^{qnil},{\left(wa\right)}^n+p\in {\mathcal{A}}_r^{-1}and(1-p)\mathcal{A}=wa(1-p)wa\mathcal{A}.$$

 Proof. 

Since $a\in {\mathcal{A}}_r^{d◯,w}$, by using Theorem 3.1, there exist $x,y\in \mathcal{A}$ such that

$$a=x+y,{\left(wx\right)}^{*}\left(wy\right)=0,ywx=0,x\in {\mathcal{A}}_r^{\#◯,w},y\in {\mathcal{A}}_w^{qnil}.$$

Let $z=x_r^{\#◯,w}$. By virtue of Theorem 2.1, there exist $x\in \mathcal{A}$ such that

$$x{\left(wz\right)}^2=z=zwxwz,{\left(wxwz\right)}^{*}=wxwz,xwzwx=x.$$

Let $p=1-wxwz$. Then $p^2=p=p^{*},1-p\in w\mathcal{A}$ and $pwx=0$. We directly check that

$$\begin{array}{cc}& [{\left(wx\right)}^n+1-wxwz][{\left(wz\right)}^n+1-wxwz]\hfill \\ \hfill =& {\left(wx\right)}^n{\left(wz\right)}^n+{\left(wx\right)}^n(1-wxwz)+(1-wxwz){\left(wz\right)}^n+{(1-wxwz)}^2\hfill \\ \hfill =& \left(wx\right)\left(wz\right)+{\left(wx\right)}^n(1-wxwz)+[wz-wx{\left(wz\right)}^2]{\left(wz\right)}^{n-1}+(1-wxwz)\hfill \\ \hfill =& 1+{\left(wx\right)}^n(1-wxwz).\hfill \end{array}$$

Then

$$[{\left(wx\right)}^n+1-wxwz][{\left(wz\right)}^n+1-wxwz][1-{\left(wx\right)}^n(1-wxwz)]=1.$$

Hence, ${\left(wx\right)}^n+1-wxwz\in {\mathcal{A}}_r^{-1}$.

Furthermore, we check that

$$\begin{array}{ccc}\hfill yw-ywxwzw& =\hfill & yw\in {\mathcal{A}}^{qnil},\hfill \\ \hfill (w-wxwzw)y& =\hfill & (1-wxwz)wy\in {\mathcal{A}}^{qnil},\hfill \\ & & (1-wxwz)wx+(w-wxwzw)y\in {\mathcal{A}}^{qnil},\hfill \\ \hfill pwa& =\hfill & pw(x+y)=(1-wxwz)wx+pwy=pyw\in {\mathcal{A}}^{qnil},\hfill \\ \hfill wa(1-p)wa& =\hfill & w(x+y)\left(wxwz\right)wa=wxwxwz(wx+wy)\hfill \\ & =\hfill & wx{\left(wxwz\right)}^{*}(wx+wy)=wx{\left(wz\right)}^{*}[{\left(wx\right)}^{*}wx+{\left(wx\right)}^{*}wy]\hfill \\ & =\hfill & wx{\left(wz\right)}^{*}{\left(wx\right)}^{*}wx=wx\left(wxwz\right)wx={\left(wx\right)}^2\hfill \\ & =\hfill & w\left(xwzwx\right)wx=\left(wxwz\right){\left(wx\right)}^2=(1-p){\left(wx\right)}^2,\hfill \\ \hfill 1-p& =\hfill & wxwz={\left(wx\right)}^2{\left(wz\right)}^3=\left[wa(1-p)wa\right]{\left(wz\right)}^3,\hfill \\ \hfill (1-p)\mathcal{A}& =\hfill & wa(1-p)wa\mathcal{A},\hfill \\ \hfill {\left(wa\right)}^n+p& =\hfill & {(wx+wy)}^n+p={\left(wx\right)}^n+\sum _{i=1}^n{\left(wx\right)}^{n-i}{\left(wy\right)}^i+p\hfill \\ & =\hfill & [{\left(wx\right)}^n+p]+\sum _{i=1}^n{\left(wx\right)}^{n-i}{\left(wy\right)}^i\hfill \\ & =\hfill & [{\left(wx\right)}^n+p][1+{({\left(wx\right)}^n+p)}_r^{-1}\sum _{i=1}^n{\left(wx\right)}^{n-i}{\left(wy\right)}^i].\hfill \end{array}$$

as required. □

 Corollary 3.7. 

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

(1)

$a\in {\mathcal{A}}_r^{d◯,w}$.

(2)

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

$$pwa\in {\mathcal{A}}^{qnil},{\left(wa\right)}^n+p\in {\mathcal{A}}_r^{-1}and(1-p)\mathcal{A}=wa(1-p)wa\mathcal{A}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ This is obvious by Theorem 3.6.

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

$$pwa\in {\mathcal{A}}^{qnil},{\left(wa\right)}^n+p\in {\mathcal{A}}_r^{-1}and(1-p)\mathcal{A}=wa(1-p)wa\mathcal{A}.$$

Then $pwa(1-p)wa=0$.

Set $x=(1-p)wa$ and $y=pwa$. Then

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

Write $(wa+p)s=1$ and $ps(1-p)=0$. Then $(1-p)was=(1-p)(wa+p)s=1-p$, and so $(1-p)was(1-p)wa=(1-p)wa$ and $(1-p)was=(1-p)(wa+p)s=1-p$. Hence, $x=(1-p)wa\in {\mathcal{A}}^{(1,3)}$.

Write $x=ws$ and $t=a-s$. Then $a=s+t,{\left(ws\right)}^{*}\left(wt\right)=x^{*}(wa-x)=x^{*}[wa-(1-p)wa]=x^{*}y=0,tws=w^{-1}\left(wt\right)x=w^{-1}[wa-(1-p)wa]x=w^{-1}yx=0$ and $t\in {\mathcal{A}}_w^{qnil}.$

We compute that $sws=w^{-1}{x}^2=w^{-1}(1-p)wa(1-p)wa=a(1-p)wa$. Since $(1-p)\mathcal{A}=wa(1-p)wa\mathcal{A},$ we can find a $z\in \mathcal{A}$ such that $1-p=wa(1-p)waz$; hence,

$$s=w^{-1}(1-p)wa=w^{-1}\left[wa(1-p)waz\right]wa=a(1-p)waz=\left(sws\right)z;$$

hence, $s\mathcal{A}\subseteq sws\mathcal{A}$. Obviously, $sws\mathcal{A}\subseteq s\mathcal{A}$. Accordingly, $s\mathcal{A}=sws\mathcal{A}$. Moreover, we have $\mathcal{A}s=\mathcal{A}ws=\mathcal{A}x=\mathcal{A}{x}^{*}x=\mathcal{A}{\left(ws\right)}^{*}\left(ws\right)$. In light of Corollary 2.2, $s\in {\mathcal{A}}_r^{\#◯,w}$. Therefore $a\in {\mathcal{A}}_r^{d◯,w}$ by Theorem 2.1. □

 Corollary 3.8. 

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

(1)

$a\in {\mathcal{A}}_r^{d◯}$.

(2)

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

$$pa\in {\mathcal{A}}^{qnil},a^n+p\in {\mathcal{A}}_r^{-1}and(1-p)\mathcal{A}=a(1-p)a\mathcal{A}.$$

 Proof. 

Straightforward by choosing $w=1$ in Corollary 3.7. □

4. Connections to Right wg-Drazin Inverses

This section establishes the relationship between the generalized right weighted core inverse and the right wg-Drazin inverse. Let $a\in \mathcal{A}$ and let

$$\left\{a_r^{d◯,w}\right\}=\{x\in \mathcal{A}|a{\left(wx\right)}^2=x=xwawx,aw-xw{\left(aw\right)}^2\in {\mathcal{A}}^{qnil}\}.$$

Next, we derive:

 Theorem 4.1. 

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

(1)

$a\in {\mathcal{A}}_r^{d◯,w}$.

(2)

$\left\{a_r^{d,w}\right\}\bigcap {\mathcal{A}}_r^{\#◯,w}\ne \varnothing$.

In this case, $a_r^{d◯}={\left(xw\right)}^2{x}_r^{\#◯,w}$ for $x\in \left\{a_r^{d◯,w}\right\}\bigcap {\mathcal{A}}_r^{\#◯,w}.$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By virtue of Theorem 3.1, there exist $z,y\in \mathcal{A}$ such that

$$a=z+y,{\left(wz\right)}^{*}\left(wy\right)=ywz=0,z\in {\mathcal{A}}_r^{\#◯,w},y\in {\mathcal{A}}_w^{qnil}.$$

Set $x=z_r^{\#◯,w}$. Then

$$a{\left(wx\right)}^2=(z+y)wxwx=z{\left(wx\right)}^2=x.$$

Moreover, we see that $xwawx=xw(z+y)wx=xwzwx=x.$ Since $awxw=(z+y)wxw=zwxw$, we see that

$$aw-\left(aw\right)xw\left(aw\right)=aw-(z+y)wxw\left(aw\right)=(z+y)w-zwxw(z+y)w.$$

Observing that

$$\begin{array}{ccc}\hfill w(z+y)-wzwxw(z+y)& =\hfill & w(z+y)-{\left(wzwx\right)}^{*}w(z+y)\hfill \\ & =\hfill & w(z+y)-{\left(wx\right)}^{*}{\left(wz\right)}^{*}(wz+wy)\hfill \\ & =\hfill & w(z+y)-{\left(wx\right)}^{*}{\left(wz\right)}^{*}wz\hfill \\ & =\hfill & wz+wy-\left(wzwx\right)wz=wy\in {\mathcal{A}}^{qnil},\hfill \end{array}$$

by using Cline’s formula, we have

$$aw-\left(aw\right)xw\left(aw\right)\in {\mathcal{A}}^{qnil}.$$

By Cline’s formula again, we see that $aw-xw{\left(aw\right)}^2\in {\mathcal{A}}^{qnil}.$ Thus, $a\in {\mathcal{A}}_r^{d◯,w}$ and $x\in \left\{a_r^{d,w}\right\}\bigcap {\mathcal{A}}_r^{\#◯,w}$. This implies that $\left\{a_r^{d,w}\right\}\bigcap {\mathcal{A}}_r^{\#◯,w}\ne \varnothing$.

$\left(2\right)\Rightarrow \left(1\right)$ Let $z\in \left\{a_r^{d,w}\right\}\bigcap {\mathcal{A}}_r^{\#◯,w}$. Then

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

Set $x={\left(zw\right)}^2{z}_r^{\#◯,w}$. It is easy to verify that

$$\begin{array}{ccc}\hfill awx& =\hfill & aw{\left(zw\right)}^2{z}_r^{\#◯,w}=\left[a{\left(wz\right)}^2\right]w{z}_r^{\#◯,w}=zw{z}_r^{\#◯,w},\hfill \\ \hfill a{\left(wx\right)}^2& =\hfill & \left(awx\right)wx=\left(zw{z}_r^{\#◯,w}\right)w{\left(zw\right)}^2{z}_r^{\#◯,w}\hfill \\ & =\hfill & \left[zw{z}_r^{\#◯,w}wz\right]wzw{z}_r^{\#◯,w}\hfill \\ & =\hfill & {\left(zw\right)}^2{z}_r^{\#◯,w}=x,\hfill \\ \hfill wawx& =\hfill & w\left(zw{z}_r^{\#◯,w}\right)=wzw{z}_r^{\#◯,w},\hfill \\ \hfill {\left(wawx\right)}^{*}& =\hfill & {\left(wzw{z}_r^{\#◯,w}\right)}^{*}=wzw{z}_r^{\#◯,w}=wawx,\hfill \end{array}$$

Hence, we check that

$$\begin{array}{ccc}\hfill \left(zw{z}_r^{\#◯,w}wa\right)wx& =\hfill & zw{z}_r^{\#◯,w}\left(wawx\right)\hfill \\ & =\hfill & zw{z}_r^{\#◯,w}\left(wzw{z}_r^{\#◯,w}\right)\hfill \\ & =\hfill & zw{z}_r^{\#◯,w},\hfill \\ \hfill {\left(w\left(zw{z}_r^{\#◯,w}wa\right)wx\right)}^{*}& =\hfill & {\left(wzw{z}_r^{\#◯,w}\right)}^{*}\hfill \\ & =\hfill & wzw{z}_r^{\#◯,w}=w\left(zw{z}_r^{\#◯,w}wa\right)wx,\hfill \\ \hfill \left(zw{z}_r^{\#◯,w}wa\right){\left(wx\right)}^2& =\hfill & zw{z}_r^{\#◯,w}w{\left(zw\right)}^2{z}_r^{\#◯,w}\hfill \\ & =\hfill & {\left(zw\right)}^2{z}_r^{\#◯,w}=x,\hfill \\ \hfill \left(zw{z}_r^{\#◯,w}wa\right)wxw\left(zw{z}_r^{\#◯,w}wa\right)& =\hfill & zw{z}_r^{\#◯,w}w\left(zw{z}_r^{\#◯,w}wa\right)\hfill \\ & =\hfill & zw{z}_r^{\#◯,w}wa.\hfill \end{array}$$

Thus, $zw{z}_r^{\#◯,w}wa\in {\mathcal{A}}_r^{\#◯,w}$ and ${\left[zw{z}_r^{\#◯,w}wa\right]}_r^{\#◯,w}=x={\left(zw\right)}^2{z}_r^{\#◯,w}$.

Write $a=a_1+a_2$, where $a_1=zw{z}_r^{\#◯,w}wa$ and $a_2=a-zw{z}_r^{\#◯,w}wa$. We verify that

$$\begin{array}{ccc}\hfill a_{2}w{a}_1& =\hfill & [a-zw{z}_r^{\#◯,w}wa]w\left[zw{z}_r^{\#◯,w}wa\right]\hfill \\ & =\hfill & awzw{z}_r^{\#◯,w}wa-zw{z}_r^{\#◯,w}wawzw{z}_r^{\#◯,w}wa\hfill \\ & =\hfill & awzw{z}_r^{\#◯,w}wa-zw{z}_r^{\#◯,w}w{z}_r^{\#◯,w}wa\hfill \\ & =\hfill & z_r^{\#◯,w}wa-z{\left(w{z}_r^{\#◯,w}\right)}^{2}wa=0.\hfill \end{array}$$

Moreover, we check that

$$\begin{array}{cc}& (aw-zw{\left(aw\right)}^2)[1-zw{z}_r^{\#◯,w}w]\hfill \\ \hfill =& aw-zw{\left(aw\right)}^2-(1-zwaw)awzw{z}_r^{\#◯,w}w\hfill \\ \hfill =& aw-zw{\left(aw\right)}^2-(1-zwaw)awzwz{\left(w{z}_r^{\#◯,w}\right)}^{2}w\hfill \\ \hfill =& aw-zw{\left(aw\right)}^2-(1-zwaw)\left[a{\left(wz\right)}^2\right]{\left(w{z}_r^{\#◯,w}\right)}^{2}w\hfill \\ \hfill =& aw-zw{\left(aw\right)}^2-(1-zwaw)z{\left(w{z}_r^{\#◯,w}\right)}^{2}w\hfill \\ \hfill =& aw-zw{\left(aw\right)}^2\in {\mathcal{A}}^{qnil}.\hfill \end{array}$$

By using Cline’s formula (see [11, Theorem 2.1]),

$$a_2=a-zw{z}_r^{\#◯,w}wa=[1-zw{z}_r^{\#◯,w}w](a-z{a}^2)\in {\mathcal{A}}^{qnil}.$$

Thus $a=a_1+a_2$ is the generalized right weighted core decomposition of a. Accordingly,

$$a_r^{e,d}={\left(a_1\right)}_r^{\#◯,w}={\left(zw\right)}^2{z}_r^{\#◯,w},$$

as asserted. □

As an immediate consequence, we provide formulas of the weighted core-EP inverse of a complex matrix.

 Corollary 4.2. 

Let $A,W\in C^{n\times n}$. Then

$$A^{†◯,W}={\left[A^{D,W}W\right]}^2{\left(A^{D,W}\right)}^{\#◯,W},$$

where $k=max\left\{ind\right(AW),ind(WA\left)\right\}$.

 Proof. 

This is obvious by Theorem 4.1. □

 Lemma 4.3. 

Let $z\in \left\{a_r^{d,w}\right\}$. Then

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

 Proof. 

Let $x=a-awzwa$. Then $xw\in {\mathcal{A}}^{qnil}$. For any $\lambda \in C$, we have $1-\overline{\lambda }xw\in {\mathcal{A}}^{-1}$, and so $1-\lambda {\left(xw\right)}^{*}\in {\mathcal{A}}^{-1}$. Thus, ${\left(xw\right)}^{*}\in {\mathcal{A}}^{qnil}$. Obviously, $(1-awzw)awzw=0$, we see that ${(aw-awzw)}^n=(1-awzw){\left(aw\right)}^n$. Hence, we derive that

$$\begin{array}{ccc}\hfill \left|\right|{({\left(aw\right)}^n-awzw{\left(aw\right)}^n)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{n}}}& =\hfill & \left|\right|{\left({\left(aw\right)}^n\right)}^{*}{(1-awzw)}^{*}\left|\right|\hfill \\ & =\hfill & \left|\right|{\left({\left(aw\right)}^n\right)}^{*}{\left[{(1-awzw)}^n\right]}^{*}\left|\right|\hfill \\ & =\hfill & \left|\right|{\left[{(aw-awzwaw)}^n\right]}^{*}\left|\right|=\left|\right|{\left({\left(xw\right)}^{*}\right)}^n\left|\right|.\hfill \end{array}$$

Since ${\left(xw\right)}^{*}\in {\mathcal{A}}^{qnil}$, we have

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

Analogously, we prove that $\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-awzw{\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$, as asserted. □

 Lemma 4.4. 

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

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

 Proof. 

Construct $x,y$ as in the proof of Theorem 3.1. Since $(1-awxw)awxwaw=0$, we deduce that ${(aw-awxwaw)}^n=(1-awxw){\left(aw\right)}^n$. Then

$$\begin{array}{cc}& \left|\right|{({\left(aw\right)}^n-axww{\left(aw\right)}^n)}^{*}\left|\right|\hfill \\ \hfill =& \left|\right|{\left[(1-axww){\left(aw\right)}^n\right]}^{*}\left|\right|\hfill \\ \hfill =& \left|\right|{\left({\left(yw\right)}^{*}\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.\hfill \end{array}$$

Therefore

$$\underset{n\to \infty }{lim}\left|\right|{(a^n-a^n{\left(a_r^{e,d}\right)}^n{a}^n)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0,$$

as asserted. □

We are ready to prove:

 Theorem 4.5. 

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_r^{d◯,w}$ if and only if there exist $x\in \mathcal{A}$ and $z\in \left\{a_r^{d,w}\right\}$ such that

$$a{\left(wx\right)}^2=x=xwawx,x\mathcal{A}=z\mathcal{A},\mathcal{A}x=\mathcal{A}{\left(wawz\right)}^{*}.$$

In this case, $a_r^{d◯,w}=x$.

 Proof. 

⟹ Choose $x=a_r^{d◯,w}$. As in the proof of Theorem 4.1, $x=xwawx=a{\left(wx\right)}^2$. By using Theorem 4.1, we can find $z\in \left\{a_r^{d,w}\right\}\bigcap {\mathcal{A}}_r^{\#◯,w}$ such that

$$x={\left(zw\right)}^2{z}_r^{\#◯,w}.$$

Then we have

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

Obviously, $z=zw{z}_r^{\#◯,w}wz={\left(zw\right)}^2{\left(z_r^{\#◯,w}w\right)}^{2}z=xw{z}_r^{\#◯,w}wz.$ Accordingly, we have $x\mathcal{A}=z\mathcal{A}$.

Since $a{\left(wx\right)}^2=x$, we have $\left(wa\right)\left(wx\right)={\left(wa\right)}^n{\left(wx\right)}^n$, and then

$$\begin{array}{ccc}\hfill x& =\hfill & xwawx=x{\left(wawx\right)}^{*}=x{\left({\left(wa\right)}^n{\left(wx\right)}^n\right)}^{*}=x{\left({\left(wx\right)}^n\right)}^{*}{\left({\left(wa\right)}^n\right)}^{*}\hfill \\ & =\hfill & x{\left({\left(wx\right)}^n\right)}^{*}{({\left(wa\right)}^n-wawz{\left(wa\right)}^n)}^{*}+x{\left({\left(wx\right)}^n\right)}^{*}{\left({\left(wa\right)}^n\right)}^{*}{\left(wawz\right)}^{*}\hfill \\ & =\hfill & x{\left({\left(wx\right)}^n\right)}^{*}{({\left(wa\right)}^n-wawz{\left(wa\right)}^n)}^{*}+x{\left(wawx\right)}^{*}{\left(wawz\right)}^{*}.\hfill \end{array}$$

Hence,

$$\begin{array}{ccc}\hfill \left|\right|x-x{\left(wawx\right)}^{*}{\left(wawz\right)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{n}}}& =\hfill & \left|\right|x{\left({\left(wx\right)}^n\right)}^{*}{({\left(wa\right)}^n-wawz{\left(wa\right)}^n)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ & \le \hfill & \left|\right|x{\left({\left(wx\right)}^n\right)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{n}}}\left|\right|{({\left(wa\right)}^n-wawz{\left(wa\right)}^n)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{n}}}.\hfill \end{array}$$

In view of Lemma 4.3,

$$\underset{n\to \infty }{lim}\left|\right|{({\left(wa\right)}^n-wawz{\left(wa\right)}^n)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0,$$

we derive that

$$\underset{n\to \infty }{lim}\left|\right|x-x{\left(wawx\right)}^{*}{\left(wawz\right)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0;$$

hence, $x=x{\left(wawx\right)}^{*}{\left(wawz\right)}^{*}$. Then $\mathcal{A}x\subseteq \mathcal{A}{\left(wawz\right)}^{*}$.

Since $a{\left(wz\right)}^2=z$, we have ${\left(wa\right)}^n{\left(wz\right)}^n=wawz$, and then we derive that

$$\begin{array}{cc}& {\left|\right|\left(wawz\right)}^{*}-{\left(wawz\right)}^{*}wawx{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ \hfill =& \left|\right|{\left({\left(wz\right)}^n\right)}^{*}{\left({\left(wa\right)}^n\right)}^{*}-{\left({\left(wz\right)}^n\right)}^{*}{\left(wawx{\left(wa\right)}^n\right)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ \hfill =& \left|\right|{\left({\left(wz\right)}^n\right)}^{*}{({\left(wa\right)}^n-wawx{\left(wa\right)}^n)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{n}}}\hfill \\ \hfill \le & \left|\right|{\left({\left(wz\right)}^n\right)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{n}}}\left|\right|{({\left(wa\right)}^n-wawx{\left(wa\right)}^n)}^{*}{\left|\right|}^{{\displaystyle \frac{1}{n}}}.\hfill \end{array}$$

In light of Lemma 4.4, we see that

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

Then

$$\underset{n\to \infty }{lim}{\left|\right|\left(wawz\right)}^{*}-{\left(wawz\right)}^{*}wawx{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0,$$

and so ${\left(wawz\right)}^{*}={\left(wawz\right)}^{*}wawx$. Hence $\mathcal{A}{\left(wawz\right)}^{*}\subseteq \mathcal{A}x$. Therefore $\mathcal{A}x=\mathcal{A}{\left(wawz\right)}^{*}$, as required.

⟸ By hypothesis, there exist $x\in \mathcal{A}$ and $z\in \left\{a_r^{d◯,w}\right\}$ such that

$$a{\left(wx\right)}^2=x=xwawx,x\mathcal{A}=z\mathcal{A},\mathcal{A}x=\mathcal{A}{\left(wawz\right)}^{*}.$$

We claim that $a_r^{d◯,w}=x$.

Claim 1.

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

Write $z=xy$ for some $y\in \mathcal{A}$. For any $n\in N$, we have

$$\begin{array}{ccc}\hfill {\left(aw\right)}^n& =\hfill & [{\left(aw\right)}^n-\left(aw\right)\left(zw\right){\left(aw\right)}^n]+\left(aw\right)\left(zw\right){\left(aw\right)}^n,\hfill \\ \hfill awxw{\left(aw\right)}^n& =\hfill & awxw[{\left(aw\right)}^n-awzw{\left(aw\right)}^n]+awxwawzw{\left(aw\right)}^n\hfill \\ & =\hfill & awxw[{\left(aw\right)}^n-awzw{\left(aw\right)}^n]+aw\left(xwawx\right)yw{\left(aw\right)}^n\hfill \\ & =\hfill & awxw[{\left(aw\right)}^n-awzw{\left(aw\right)}^n]+awxyw{\left(aw\right)}^n\hfill \\ & =\hfill & awxw[{\left(aw\right)}^n-\left(aw\right)\left(zw\right){\left(aw\right)}^n]+awzw{\left(aw\right)}^n.\hfill \end{array}$$

Hence,

$${\left(aw\right)}^n-awxw{\left(aw\right)}^n=(1-awxw)[{\left(aw\right)}^n-\left(aw\right)\left(zw\right){\left(aw\right)}^n],$$

and so

$${\left|\right|\left(aw\right)}^n-awxw{\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}\le \left|\right|1-{awxw\left|\right|}^{{\displaystyle \frac{1}{n}}}{\left|\right|\left(aw\right)}^n-\left(aw\right)\left(zw\right){\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}},$$

we have

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

Claim 2. ${\left(wawx\right)}^{*}=wawx$.

Since $\mathcal{A}x=\mathcal{A}{\left(wawz\right)}^{*}$, we have $x^{*}\mathcal{A}=wawz\mathcal{A}$. Write $wawz=x^{*}s$ for some $s\in \mathcal{A}$. Since $xwawx=x$, we have $x^{*}{w}^{*}{a}^{*}{w}^{*}{x}^{*}=x^{*}$, and so ${\left(wawx\right)}^{*}{x}^{*}=x^{*}$. We infer that ${\left(wawx\right)}^{*}wawz={\left(wawx\right)}^{*}\left(x^{*}s\right)=\left[{\left(wawx\right)}^{*}{x}^{*}\right]s={\left(xwawx\right)}^{*}s=x^{*}s=wawz$. Since $x\mathcal{A}=z\mathcal{A}$, we can find $t\in \mathcal{A}$ such that $x=zt$. Then ${\left(wawx\right)}^{*}\left(wawx\right)={\left(wawx\right)}^{*}waw\left(zt\right)=\left[{\left(wawx\right)}^{*}wawz\right]t=\left(wawz\right)t=waw\left(zt\right)=wawx$. Hence ${\left(wawx\right)}^{*}={\left[{\left(wawx\right)}^{*}\left(wawx\right)\right]}^{*}={\left(wawx\right)}^{*}\left(wawx\right)=wawx$.

Therefore $a_r^{d◯,w}=x$, as asserted. □

 Corollary 4.6. 

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_r^{d◯}$ if and only if there exist $x\in \mathcal{A}$ and $z\in \left\{a_r^d\right\}$ such that

$$a{x}^2=x=xax,x\mathcal{A}=z\mathcal{A},\mathcal{A}x=\mathcal{A}{\left(az\right)}^{*}.$$

In this case, $a_r^d=x$.

 Proof. 

Straightforward from Theorem 4.5. □

5. Right Pesudo Weighted Core Inverses

In this section, we are concerned with right pesudo weighted core inverse in a Banach *-algebra. Let $a\in \mathcal{A}$, and let

$$\left\{a_r^{D,w}\right\}=\{x\in \mathcal{A}|a{\left(wx\right)}^2=x=xwawx,{\left(aw\right)}^k=awxw{\left(aw\right)}^{k}forsomek\in N\}.$$

Evidently, $a\in \mathcal{A}$ has the right w-Drazin inverse if and only if $\left\{a_r^{D,w}\right\}\ne \varnothing$.

 Lemma 5.1. 

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

(1)

$a\in {\mathcal{A}}_r^{d◯,w}$;

(2)

$\left\{a_r^{D,w}\right\}\ne \varnothing .$

In this case, $a_r^{D◯,w}=a_r^{d◯,w}$.

 Proof. 

⟹ This implication is obvious.

⟸ Let $x=a_r^{d◯,w}$. Then

$$a{\left(wx\right)}^2=x=xwawx,{\left(wawx\right)}^{*}=wawx,\underset{n\to \infty }{lim}{\left|\right|\left(aw\right)}^n-awxw{\left(aw\right)}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}=0.$$

Let $y\in \left\{a_r^{D,w}\right\}$. Then $a{\left(wy\right)}^2=y=ywawy,{\left(aw\right)}^k=awyw{\left(aw\right)}^{k}forsomek\in N.$ Set $z=ywawx$. Then we verify that

$$\begin{array}{ccc}\hfill wawz& =\hfill & waw\left(ywawx\right)=wawy{\left(wa\right)}^k{\left(wx\right)}^k=w\left[awyw{\left(aw\right)}^k\right]x{\left(wx\right)}^{k-1}\hfill \\ & =\hfill & w{\left(aw\right)}^{k}x{\left(wx\right)}^{k-1}=waw{\left(aw\right)}^{k-1}{\left(xw\right)}^{k-1}x\hfill \\ & =\hfill & waw\left(aw\right)\left(xw\right)x={\left(wa\right)}^2{\left(wx\right)}^2=wawx.\hfill \end{array}$$

Hence ${\left(wawz\right)}^{*}={\left(wawx\right)}^{*}=wawx=wawz$. We observe that

$$\begin{array}{ccc}\hfill awzw& =\hfill & awywawxw=awyw{\left(aw\right)}^k{\left(xw\right)}^k=awxw,\hfill \\ \hfill a{\left(wz\right)}^2& =\hfill & awzwz=awxwz=awxwywawx=awxw\left[a{\left(wy\right)}^2\right]wawx\hfill \\ & =\hfill & \left[a{\left(wy\right)}^2\right]wawx=ywawx=z,\hfill \\ \hfill zwawz& =\hfill & ywawx\left(wawz\right)=ywaw\left(xwawx\right)=ywawx=z.\hfill \end{array}$$

Moreover, we have

$$\begin{array}{ccc}\hfill {\left(aw\right)}^k-awzw{\left(aw\right)}^k& =\hfill & {\left(aw\right)}^k-awxw{\left(aw\right)}^k\hfill \\ & =\hfill & \left(aw\right)\left(ay\right){\left(aw\right)}^k-awxw\left[\left(aw\right)\left(ay\right){\left(aw\right)}^k\right]\hfill \\ & =\hfill & {\left(aw\right)}^n{\left(ay\right)}^n{\left(aw\right)}^k-awxw\left[{\left(aw\right)}^n{\left(ay\right)}^n{\left(aw\right)}^k\right]\hfill \\ & =\hfill & [{\left(aw\right)}^n-awxw{\left(aw\right)}^n]{\left(ay\right)}^n{\left(aw\right)}^k.\hfill \end{array}$$

Hence,

$${\left|\right|\left(aw\right)}^k-awzw{\left(aw\right)}^k\left|\right|\le {\left|\right|\left(aw\right)}^n-awxw{\left(aw\right)}^n{\left|\right|\left|\right|\left(ay\right)}^n{\left(aw\right)}^k\left|\right|.$$

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

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

and then ${\left(aw\right)}^k=awzw{\left(aw\right)}^k$. Therefore $a\in {\mathcal{A}}_r^{D,w}$. □

We are ready to prove:

 Theorem 5.2. 

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

(1)

$a\in {\mathcal{A}}_r^{D◯,w}.$

(2)

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

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

(3)

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

$$a=x+y,ywx=0,x\in {\mathcal{A}}_r^{\#◯,w},yw\in {\mathcal{A}}^{nil}.$$

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By virtue of Lemma 5.1, $a\in {\mathcal{A}}_r^{d◯,w}$ and $a_r^{d◯,w}=a_r^{D◯,w}$. In view of Theorem 3.1, we can find $x,y\in \mathcal{A}$ such that

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

Explicitly, we have $y=a-aw{a}_r^{d◯,w}wa=a-aw{a}_r^{D◯,w}wa$. Set $z=a_r^{D◯,w}$. Then $a{\left(wz\right)}^2=z=zwawz,{\left(wawz\right)}^{*}=wawz$ and $awaz{\left(aw\right)}^k={\left(aw\right)}^k$ for some $k\in N$. It is easy to see that

$$\begin{array}{ccc}\hfill (1-awzw)aw\left(awz\right)& =\hfill & (1-awzw)aw\left({\left(aw\right)}^{k-1}z{\left(wz\right)}^{k-2}\right)\hfill \\ & =\hfill & (1-awzw){\left(aw\right)}^{k}z{\left(wz\right)}^{k-2}=0.\hfill \end{array}$$

Moreover, we verify that

$$\begin{array}{ccc}\hfill {(aw-zw{\left(aw\right)}^2)}^{k+2}& =\hfill & (1-zwaw)aw{[aw-zw{\left(aw\right)}^2]}^k[aw-zw{\left(aw\right)}^2]\hfill \\ & =\hfill & (1-zwaw)aw{[aw-zw{\left(aw\right)}^2]}^{k-1}[aw-zw{\left(aw\right)}^2]aw\hfill \\ & =\hfill & (1-zwaw)aw{(aw-zw{\left(aw\right)}^2)}^{k-1}{\left(aw\right)}^2\hfill \\ & ⋮\hfill & \\ & =\hfill & (1-zwaw)aw(aw-zw{\left(aw\right)}^2){\left(aw\right)}^k\hfill \\ & =\hfill & (1-zwaw)({\left(aw\right)}^k-awzw{\left(aw\right)}^k)\left(aw\right)\hfill \\ & =\hfill & 0.\hfill \end{array}$$

This implies that $aw-zw{\left(aw\right)}^2\in {\mathcal{A}}^{nil}$. Therefore $yw=aw-awzwaw\in {\mathcal{A}}^{nil}$, as desired.

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

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

Since ${\mathcal{A}}^{nil}\subseteq {\mathcal{A}}^{qnil}$. In view of Theorem 3.1, $a\in {\mathcal{A}}_r^{d◯,w}$ and $z:=a_r^{d◯,w}=x_r^{\#◯,w}$. Hence, $a{\left(wz\right)}^2=z$ and ${\left(wawz\right)}^{*}=wawz$. Write ${\left(yw\right)}^k=0(k\in N)$. Then

$$\begin{array}{ccc}\hfill {\left(aw\right)}^k& =\hfill & \sum _{i=0}^k{\left(xw\right)}^i{\left(yw\right)}^{n-i}={\left(yw\right)}^k+xw{\left(yw\right)}^{k-1}+\dots +{\left(xw\right)}^{k-1}yw+{\left(xw\right)}^k\hfill \\ & =\hfill & xw{\left(yw\right)}^{k-1}+\dots +{\left(xw\right)}^{k-1}yw+{\left(xw\right)}^k.\hfill \end{array}$$

Then

$$\begin{array}{ccc}\hfill awzw{\left(aw\right)}^k& =\hfill & (xw+yw)x_r^{\#◯}wx[{\left(yw\right)}^{k-1}+\dots +{\left(xw\right)}^{k-2}yw+{\left(xw\right)}^{k-1}]\hfill \\ & =\hfill & xw[{\left(yw\right)}^{k-1}+\dots +{\left(xw\right)}^{k-2}y+{\left(xw\right)}^{k-1}]={\left(aw\right)}^k.\hfill \end{array}$$

Therefore $a\in {\mathcal{A}}_r^{D◯,w},$ as asserted. □

Recall that a has w-Drazin inverse x provided that the following identities are satisfied:

$$xwawx=x,awx=xwa,xw{\left(aw\right)}^k={\left(aw\right)}^{k+1}$$

for some $k\in N$.

 Corollary 5.3. 

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

(1)

$a\in {\mathcal{A}}^{D,w};$

(2)

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

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

(3)

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

$$a=x+y,ywx=0,x\in {\mathcal{A}}_r^{\#◯},y\in {\mathcal{A}}^{nil}.$$

 Proof. 

⟹ This is obvious by Theorem 5.2 and [5, Lemma 2].

⟸ In view of Theorem 5.2, $a\in {\mathcal{A}}_r^{D,w}$. Therefore we complete the proof by [5, Lemma 2]. □

 Theorem 5.4. 

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

(1)

$a\in {\mathcal{A}}_r^{D◯,w}.$

(2)

There exists $m\in N$ such that $a{\left(wa\right)}^{k-1}\in {\mathcal{A}}_r^{\#◯,w}$ for any $k\ge m$.

(3)

$a{\left(wa\right)}^{k-1}\in {\mathcal{A}}_r^{\#◯,w}$ for some $k\in N$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ By hypothesis, there exist $z\in \mathcal{A}$ and $m\in N$ such that

$$a{\left(wz\right)}^2=z,{\left(wawz\right)}^{*}=wawz,{\left(aw\right)}^{m-1}=\left(aw\right)\left(zw\right){\left(aw\right)}^{m-1}.$$

Let $x={\left(zw\right)}^{k-1}z(k\ge m)$. Then ${\left(aw\right)}^{k-1}=\left(aw\right)\left(zw\right){\left(aw\right)}^{k-1}$, and so

$$\begin{array}{ccc}\hfill {\left(wa\right)}^{k}wx& =\hfill & {\left(wa\right)}^{k-1}waw{\left(zw\right)}^{k-1}z\hfill \\ & =\hfill & {\left(wa\right)}^{k-1}w\left[a{\left(wz\right)}^2\right]{\left(wz\right)}^{k-2}\hfill \\ & =\hfill & {\left(wa\right)}^{k-1}wz{\left(wz\right)}^{k-2}\hfill \\ & =\hfill & {\left(wa\right)}^{k-1}{\left(wz\right)}^{k-1}=wawz,\hfill \\ \hfill \left(a{\left(wa\right)}^{k-1}\right){\left(wx\right)}^2& =\hfill & w^{-1}\left(wawz\right)\left(wx\right)=awzw{\left(zw\right)}^{k-1}z\hfill \\ & =\hfill & \left[a{\left(wz\right)}^2\right]w{\left(zw\right)}^{k-2}z\hfill \\ & =\hfill & zw{\left(zw\right)}^{k-2}z={\left(zw\right)}^{k-1}z=x,\hfill \\ \hfill w\left(a{\left(wa\right)}^{k-1}\right)wx& =\hfill & {\left(wa\right)}^{k}wx=wawz,\hfill \\ \hfill \left(w\left[a{\left(wa\right)}^{k-1}\right]wx\right)& =\hfill & w\left[a{\left(wa\right)}^{k-1}\right]wx,\hfill \\ \hfill \left(a{\left(wa\right)}^{k-1}\right)wxw\left(a{\left(wa\right)}^{k-1}\right)& =\hfill & \left(a{\left(wa\right)}^{k-1}\right)w{\left(zw\right)}^{k-1}zw\left(a{\left(wa\right)}^{k-1}\right)\hfill \\ & =\hfill & {\left(aw\right)}^k{\left(zw\right)}^k\left[a{\left(wa\right)}^{k-1}\right]=awzwa{\left(wa\right)}^{k-1}\hfill \\ & =\hfill & awzw\left[a{\left(wa\right)}^{k-1}\right]=\left[awzw{\left(aw\right)}^{k-1}\right]\hfill \\ & =\hfill & a{\left(wa\right)}^{k-1}.\hfill \end{array}$$

Therefore $a{\left(wa\right)}^{k-1}\in {\mathcal{A}}_r^{\#◯,w}$, as required.

$\left(2\right)\Rightarrow \left(3\right)$ This is trivial.

$\left(3\right)\Rightarrow \left(1\right)$ Set $x=a{\left(wa\right)}^{k-2}w{\left[a{\left(wa\right)}^{k-1}\right]}_r^{\#◯,w}$. Then we verify that

$$\begin{array}{ccc}\hfill awx& =\hfill & awa{\left(wa\right)}^{k-2}w{\left[a{\left(wa\right)}^{k-1}\right]}_r^{\#◯,w}=a{\left(wa\right)}^{k-1}w{\left[a{\left(wa\right)}^{k-1}\right]}_r^{\#◯,w},\hfill \\ \hfill {\left(wawx\right)}^{*}& =\hfill & {\left(w\left[a{\left(wa\right)}^{k-1}\right]w{\left[a{\left(wa\right)}^{k-1}\right]}_r^{\#◯,w}\right)}^{*}\hfill \\ & =\hfill & w\left[a{\left(wa\right)}^{k-1}\right]w{\left[a{\left(wa\right)}^{k-1}\right]}_r^{\#◯,w}=wawx,\hfill \\ \hfill a{\left(wx\right)}^2& =\hfill & a{\left(wa\right)}^{k-1}w{\left[a{\left(wa\right)}^{k-1}\right]}_r^{\#◯,w}{\left(wa\right)}^{k-1}w{\left[a{\left(wa\right)}^{k-1}\right]}_r^{\#◯,w}\hfill \\ & =\hfill & a{\left(wa\right)}^{k-1}w{\left[a{\left(wa\right)}^{k-1}\right]}_r^{\#◯,w}wa{\left(wa\right)}^{k-1}{\left(wa\right)}^{k-1}{\left({\left[a{\left(wa\right)}^{k-1}\right]}_r^{\#◯,w}\right)}^2\hfill \\ & =\hfill & \left[a{\left(wa\right)}^{k-1}w{\left[a{\left(wa\right)}^{k-1}\right]}_r^{\#◯,w}wa{\left(wa\right)}^{k-1}\right]{\left(wa\right)}^{k-1}{\left({\left[a{\left(wa\right)}^{k-1}\right]}_r^{\#◯,w}\right)}^2\hfill \\ & =\hfill & a{\left(wa\right)}^{k-1}{\left(wa\right)}^{k-1}{\left({\left[a{\left(wa\right)}^{k-1}\right]}_r^{\#◯,w}\right)}^2\hfill \\ & =\hfill & a{\left(wa\right)}^{k-2}w\left[a{\left(wa\right)}^{k-1}\right]{\left({\left[a{\left(wa\right)}^{k-1}\right]}_r^{\#◯,w}\right)}^2\hfill \\ & =\hfill & a{\left(wa\right)}^{k-2}w{\left[a{\left(wa\right)}^{k-1}\right]}_r^{\#◯,w}=x,\hfill \\ \hfill awxw{\left(aw\right)}^k& =\hfill & \left[a{\left(wa\right)}^{k-1}w{\left[a{\left(wa\right)}^{k-1}\right]}_r^{\#◯,w}wa{\left(wa\right)}^{k-1}\right]w\hfill \\ & =\hfill & a{\left(wa\right)}^{k-1}w={\left(aw\right)}^k.\hfill \end{array}$$

This completes the proof. □

 Corollary 5.5. 

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

(1)

$a\in {\mathcal{A}}_r^{D◯,w}.$

(2)

$a{\left(wa\right)}^{k-1}\in {\mathcal{A}}_w^{(1,3)}$ and ${\left(aw\right)}^k\mathcal{A}={\left(aw\right)}^{k+1}\mathcal{A}$ for some $k\in N$.

Proof.$\left(1\right)\Rightarrow \left(2\right)$ In view of Theorem 5.4, $a{\left(wa\right)}^{k-1}\in {\mathcal{A}}_w^{(1,3)}$. By hypothesis, there exist $x\in \mathcal{A}$ such that

$$a{\left(wx\right)}^2=x,{\left(wawx\right)}^{*}=wawx,{\left(aw\right)}^k=\left(aw\right)\left(xw\right){\left(aw\right)}^k$$

for some $k\in N$. Then ${\left(aw\right)}^k={\left(aw\right)}^{k+1}{\left(xw\right)}^{k+1}{\left(aw\right)}^k$; hence, ${\left(aw\right)}^k\mathcal{A}={\left(aw\right)}^{k+1}\mathcal{A}$, as required.

$\left(2\right)\Rightarrow \left(1\right)$ In view of Theorem 5.4, we will suffice to prove $a{\left(wa\right)}^{k-1}\in {\mathcal{A}}_r^{\#◯,w}$.

We claim that $a{\left(wa\right)}^{k-1}\mathcal{A}=a{\left(wa\right)}^{k-1}wa{\left(wa\right)}^{k-1}\mathcal{A}$. Obviously, we have $a{\left(wa\right)}^{k-1}wa{\left(wa\right)}^{k-1}\mathcal{A}\subseteq a{\left(wa\right)}^{k-1}\mathcal{A}$. Since $a{\left(wa\right)}^{k-1}\in {\mathcal{A}}_w^{(1,3)}$, we can find some $z\in \mathcal{A}$ such that $a{\left(wa\right)}^{k-1}=a{\left(wa\right)}^{k-1}wzwa{\left(wa\right)}^{k-1}$. As ${\left(aw\right)}^k\mathcal{A}={\left(aw\right)}^{k+1}\mathcal{A}$, by induction, we have ${\left(aw\right)}^k\mathcal{A}={\left(aw\right)}^{k+m}\mathcal{A}$ for any $m\in N$. Thus, we derive that

$$\begin{array}{ccc}\hfill a{\left(wa\right)}^{k-1}& =\hfill & a{\left(wa\right)}^{k-1}wzwa{\left(wa\right)}^{k-1}\hfill \\ & =\hfill & {\left(aw\right)}^{k}zwa{\left(wa\right)}^{k-1}\in {\left(aw\right)}^{2k}zwa{\left(wa\right)}^{k-1}\mathcal{A}\hfill \\ & \subseteq \hfill & {\left(aw\right)}^k{\left(aw\right)}^{k-1}aw\mathcal{A}\hfill \\ & =\hfill & a{\left(wa\right)}^{k-1}wa{\left(wa\right)}^{k-1}\mathcal{A},\hfill \end{array}$$

and then $a{\left(wa\right)}^{k-1}\mathcal{A}=a{\left(wa\right)}^{k-1}wa{\left(wa\right)}^{k-1}\mathcal{A}$. By virtue of Theorem 2.1, $a{\left(wa\right)}^{k-1}\in {\mathcal{A}}_r^{\#◯,w}$. Therefore $a\in {\mathcal{A}}_r^{D,w}$ by Theorem 5.4. □

We proceed to prove:

 Theorem 5.6. 

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}_r^{D◯,w}$ if and only if there exist a projection $p\in \mathcal{A}$ and $k\in N$ such that

$$wa+p\in {\mathcal{A}}_r^{-1},(1-p)\mathcal{A}=wa(1-p)wa\mathcal{A},{\left(pwa\right)}^k=0,a{\left(wa\right)}^{k-1}\in {\mathcal{A}}^{(1,3)}.$$

 Proof. 

⟹ In light of Theorem 3.6, there exists a projection $p\in \mathcal{A}$ such that

$$wa+p\in {\mathcal{A}}_r^{-1},pwa\in {\mathcal{A}}^{qnil},(1-p)\mathcal{A}=wa(1-p)wa\mathcal{A}.$$

Explicitly, $p=1-waw{a}_r^{D◯,w}$. As in the proof of Theorem 3.1, we check that $pwa=(1-waw{a}_r^{D◯,w})wa\in {\mathcal{A}}^{nil}$. Write ${\left(pwa\right)}^m=0$ for some $m\in N$. By using Theorem 5.4, we can find a $k\ge m$ such that $a{\left(wa\right)}^{k-1}\in {\mathcal{A}}^{(1,3)}$, as required.

⟸ By hypothesis, there exist a projection $p\in \mathcal{A}$ and $k\ge 2$ such that

$$wa+p\in {\mathcal{A}}_r^{-1},(1-p)\mathcal{A}=wa(1-p)wa\mathcal{A},{\left(pwa\right)}^{k-1}=0,a{\left(wa\right)}^{k-1}\in {\mathcal{A}}_w^{(1,3)}.$$

Then $pwa(1-p)wa=0$, and so $pwa=pwap$. Hence, $p{\left(wa\right)}^2=\left(pwa\right)wa=\left(pwap\right)wa={\left(pwa\right)}^2$. By induction, we have $p{\left(wa\right)}^{k-1}={\left(pwa\right)}^{k-1}=0$. This implies that ${\left(wa\right)}^{k-1}=(1-p){\left(wa\right)}^{k-1}$. Since $1-p\in wa(1-p)\mathcal{A}$, by induction, $1-p\in {\left(wa\right)}^m$ for any $m\ge 2$. Then ${\left(wa\right)}^{k-1}\in {\left(wa\right)}^{k+1}\mathcal{A}$. Therefore ${\left(aw\right)}^k=a{\left(wa\right)}^{k-1}w\in a{\left(wa\right)}^{k+1}\mathcal{A}\subseteq {\left(aw\right)}^{k+1}\mathcal{A}$. This implies that ${\left(aw\right)}^k\mathcal{A}={\left(aw\right)}^{k+1}\mathcal{A}$. According to Corollary 5.5, $a\in {\mathcal{A}}_r^{D,w}$. □

As an immediate consequence, we derive

 Corollary 5.7. 

Let $a\in \mathcal{A}$. Then $a\in {\mathcal{A}}^{D◯,w}$ if and only if $a\in {\mathcal{A}}^{D,w}$ and there exist a projection $p\in \mathcal{A}$ and $k\in N$ such that

$$wa+p\in {\mathcal{A}}_r^{-1},(1-p)\mathcal{A}=wa(1-p)wa\mathcal{A},{\left(pwa\right)}^k=0,a{\left(wa\right)}^{k-1}\in {\mathcal{A}}^{(1,3)}.$$

 Remark 5.8. 

Generalized weighted left core inverse can be defined dually. The corresponding results for the generalized weighted left core inverse can be established in a similar way.