Parallel Sum of Bounded Operators with Closed Ranges

Abstract

Let $\mathcal{H}$ be a separable infinite dimensional complex Hilbert space and $\mathcal{B}\left(\mathcal{H}\right)$ be the set of all bounded linear operators on $\mathcal{H}$. In this paper, we present several conditions under which the distributive law of the parallel sum is valid. It is proved that the parallel sum for positive operators with closed ranges is continued at 0. For $A,B\in \mathcal{B}\left(\mathcal{H}\right)$ with closed ranges, it is proved that $A\overline{\le }B$ if and only if A and $B-A$ are parallel summable with the parallel sum $A:(B-A)=0$, where the symbol “$\overline{\le }$” denotes the minus partial order.

1. Induction

For $A\in \mathcal{B}\left(\mathcal{H}\right)$, let $A^{*}$, $A^{†}$, $\mathcal{R}\left(A\right)$, and $\mathcal{N}\left(A\right)$ be the adjoint, Moore–Penrose inverse, range, and null space of A, respectively. An operator $A\in \mathcal{B}\left(\mathcal{H}\right)$ is said to be self-adjoint and positive if $A^{*}=A$ and $⟨Ax,x⟩\ge 0$ for all $x\in \mathcal{H}$, respectively, and the set of all self-adjoint and positive operators are denoted by ${\mathcal{B}}^{\mathcal{S}}\left(\mathcal{H}\right)$, and ${\mathcal{B}}^{+}\left(\mathcal{H}\right)$, respectively. It is known that for a bounded operator $A\in \mathcal{B}\left(\mathcal{H}\right)$, $A^{†}$ exists if and only if $\mathcal{R}\left(A\right)$ is closed. The capital “I” denotes the identity on the corresponding subspace, and the symbol “⊕” means the orthogonal direct sum of two subspaces. $P_{\mathcal{M}}$ denotes the orthogonal projection onto $\mathcal{M}$.

The parallel sum originally arose in an attempt to generalize a network synthesis procedure of Duffin [1] and has already been studied in the scalar case by Erickson [2]. Later, Anderson and Duffin defined the parallel sum operation on Hermitian positive semi-definite matrices and investigated its most important properties in [3]. For the $n\times n$ complex Hermitian positive semi-definite matrices A and B, the parallel sum of A and B is defined by

$$A:B=A{(A+B)}^{†}B,$$

which can be proved to be equal to $A{(A+B)}^{-}B$ for any $\left\{1\right\}$-inverse ${(A+B)}^{-}$ of $A+B$. In [4], Mitra and Odell generalized the definition of the parallel sum to the non-square matrices A and B of the same size such that

$$\mathcal{R}\left(A\right)\subseteq \mathcal{R}(A+B)\mathrm{and}\mathcal{R}\left(A^{*}\right)\subseteq \mathcal{R}(A^{*}+B^{*}).$$

In [5], Antezana, Corach, and Stojanoff extended parallel summability for bounded linear operators between two different Hilbert spaces and defined the parallel sum of two bounded operators in terms of shorted operators (see [6] and Definition 5.1 in [5]). Recently, the properties of parallel sum for adjointable operators on Hilbert $C^{*}$-modules have been researched by Luo, Song, and Xu [7].

The purpose of this paper is to generalize some known results of the parallel sum for matrices, such as, Lemma 9, Theorem 6, and Corollary 2 to bounded operators with closed ranges on infinite dimensional Hilbert spaces. In order to obtain the main results, some lemmas and the definition of parallel summablity for bounded operators are introduced in the following section.

The following basic lemmas concerning the ranges of linear operators are given by Douglas [8].

Lemma 1 ([8], Theorem 1). 

Let $A,B\in \mathcal{B}\left(\mathcal{H}\right)$. Then, the following statements are equivalent:

(i) 

$\mathcal{R}\left(A\right)\subseteq \mathcal{R}\left(B\right)$;

(ii) 

$A{A}^{*}\le kB{B}^{*}$ for some $k>0$;

(iii) 

There exists a bounded operator C such that $A=BC$.

Moreover, if one of (i)–(iii) is valid, then there exists a unique operator $C_0\in \mathcal{B}\left(\mathcal{H}\right)$ such that $B{C}_0=A$ with $\mathcal{R}\left(C_0\right)\subseteq \mathcal{N}{\left(B\right)}^{\perp }$, $\mathcal{N}\left(C_0\right)=\mathcal{N}\left(A\right)$, and $\parallel C_0{\parallel }^2=inf\{\mu :A{A}^{*}\le \mu B{B}^{*}\}$. $C_0$ is called the Douglas reduced solution of the operator equation $BX=A$.

Lemma 2 ([9], Remark 1.1). 

Let $T\in \mathcal{B}\left(\mathcal{H}\right)$. Then, the closeness of any one of the following sets implies the closeness of the remaining three sets:

$$\mathcal{R}\left(T\right),\mathcal{R}\left(T^{*}\right),\mathcal{R}\left(T{T}^{*}\right),\mathit{and}\mathcal{R}\left(T^{*}T\right).$$

Moreover, $\mathcal{R}\left(T\right)$ is closed if and only if $\mathcal{R}\left(T\right)=\mathcal{R}\left(T{T}^{*}\right)$ and $\mathcal{R}\left(T^{*}\right)=\mathcal{R}\left(T^{*}T\right)$.

The next Lemma is connected with the reverse order law of bounded operators, which is useful in calculation of the parallel sum.

Lemma 3 ([10,11,12]). 

Let $A,B$, and $AB$ be closed range operators on $\mathcal{H}$. Then, ${\left(AB\right)}^{†}=B^{†}{A}^{†}$ if and only if $\mathcal{R}\left(A^{*}AB\right)\subseteq \mathcal{R}\left(B\right)$ and $\mathcal{R}\left(B{B}^{*}{A}^{*}\right)\subseteq \mathcal{R}\left(A^{*}\right)$.

Lemma 4 ([13], Theorem 2.2). 

If $A,B\in \mathcal{B}\left(\mathcal{H}\right)$, then

$$\mathcal{R}\left(A\right)+\mathcal{R}\left(B\right)=\mathcal{R}({(A{A}^{*}+B{B}^{*})}^{\frac{1}{2}}).$$

Lemma 5 ([13], Theorem 2.2). 

If $A\in \mathcal{B}\left(\mathcal{H}\right)$ is a positive operator, then $\overline{\mathcal{R}\left(A\right)}=\overline{\mathcal{R}\left(A^{\frac{1}{2}}\right)}$, where $\overline{\mathcal{K}}$ denotes the closure of a subset $\mathcal{K}\subseteq \mathcal{H}$.

In the following part, we introduce the definition of the parallel sum for bounded operators on infinite dimensional Hilbert spaces.

Definition 1 ([5], Definition 5.1). 

Let $A,B\in \mathcal{B}\left(\mathcal{H}\right)$. We say A and B are weakly parallel summable if the next range inclusions hold:

(i) 

$\mathcal{R}\left(A\right)\subseteq \mathcal{R}\left(\right|A^{*}+B^{*}{|}^{\frac{1}{2}})$ and $\mathcal{R}\left(B\right)\subseteq \mathcal{R}\left(\right|A^{*}+B^{*}{|}^{\frac{1}{2}})$.

(ii) 

$\mathcal{R}(A^{*})\subseteq {\mathcal{R}(|A+B|}^{\frac{1}{2}})$ and $\mathcal{R}(B^{*})\subseteq {\mathcal{R}(|A+B|}^{\frac{1}{2}})$. In this case, the parallel sum of A and B, denoted by $A:B\in \mathcal{B}\left(\mathcal{H}\right)$, is

$$\left(\begin{array}{cc}A:B& 0\\ 0& 0\end{array}\right)={\left(\begin{array}{cc}A& A\\ A& A+B\end{array}\right)}_{{/}_{(\mathcal{H}\oplus \{0\},\mathcal{H}\oplus \{0\left\}\right)}},$$

(1)

where ${\mathsf{\Lambda }}_{{/}_{(\mathcal{S},\mathcal{T})}}$ denotes the shorted operator of Λ with respect to closed subspaces $\mathcal{S}$, $\mathcal{T}\in \mathcal{H}$.

Antezana et al. gave out the definition of shorted operators in ([5], Definition 4.1). In ([5], Corollary 5.4). The authors proved that if bounded operators A and B on $\mathcal{H}$ are weakly parallel summable, then $A:B=B:A$.

Definition 2 ([5], Definition 5.8). 

Let $A,B\in \mathcal{B}\left(\mathcal{H}\right)$. If all the following conditions hold:

(i) 

$\mathcal{R}\left(A\right)\subseteq \mathcal{R}(A+B)$;

(ii) 

$\mathcal{R}\left(A^{*}\right)\subseteq \mathcal{R}(A^{*}+B^{*})$;

(iii) 

$\mathcal{R}\left(B\right)\subseteq \mathcal{R}(A+B)$;

(iv) 

$\mathcal{R}\left(B^{*}\right)\subseteq \mathcal{R}(A^{*}+B^{*})$, then A and B are parallel summable.

Note that the parallel summability could imply the weakly parallel summability of operators on infinite dimensional Hilbert spaces, and, thus, this parallel sum is also denoted by (1). If A and $B\in \mathcal{B}\left(\mathcal{H}\right)$ are weakly parallel summable with $\mathcal{R}(A+B)$ closed, it follows from Lemmas 4 and 5 that

$$\mathcal{R}\left(\right|A^{*}+B^{*}{|}^{\frac{1}{2}})=\mathcal{R}(|A^{*}+B^{*}\left|\right)=\mathcal{R}(A+B)$$

and

$${\mathcal{R}\left(\right|A+B|}^{\frac{1}{2}})=\mathcal{R}(|A+B|)=\mathcal{R}(A^{*}+B^{*}),$$

and, thus, A and B are parallel summable. In this case, the parallel sum is written as

$$A:B=A{(A+B)}^{†}B,$$

(2)

where $T^{†}$ denotes the Moore–Penrose inverse of T (see [14]). By computation, it is easy to see that $A:B=A-A{(A+B)}^{†}A=B-B{(A+B)}^{†}B=B:A$.

In the following part, we shall use Equation (2) to study the parallel sum of closed range operators on $\mathcal{B}\left(\mathcal{H}\right)$. Note that $A:B$ is well-defined, meaning that A and B are weakly parallel summable, and, thus, they are parallel summable in the condition that $\mathcal{R}(A+B)$ is closed.

Remark 1.

Since $\mathcal{R}\left(A\right)\subseteq \mathcal{R}(A+B)$ if and only if $\mathcal{R}\left(B\right)\subseteq \mathcal{R}(A+B)$ (i.e., $\mathcal{R}\left(A^{*}\right)\subseteq \mathcal{R}(A^{*}+B^{*})$ if and only if $\mathcal{R}\left(B^{*}\right)\subseteq \mathcal{R}(A^{*}+B^{*})$), then the following statements are equivalent:

(i) 

$P_{\mathcal{R}(A+B)}A=A=A{P}_{\mathcal{R}(A^{*}+B^{*})}$;

(ii) 

$P_{\mathcal{R}(A+B)}B=B=B{P}_{\mathcal{R}(A^{*}+B^{*})}$;

(iii) 

A and B are parallel summable;

(iv) 

B and A are parallel summable;

(v) 

$A^{*}$ and $B^{*}$ are parallel summable;

(vi) 

$B^{*}$ and $A^{*}$ are parallel summable.

Remark 2.

Note that the condition $P_{\mathcal{R}(A+B)}A=A{P}_{\mathcal{R}(A^{*}+B^{*})}$ (resp. $P_{\mathcal{R}(A+B)}B=B{P}_{\mathcal{R}(A^{*}+B^{*})})$ does not imply that A and B are parallel summable. For example, put $A=\left(\begin{array}{cc}-I& 0\\ 0& I\end{array}\right)\in \mathcal{B}(\mathcal{H}\oplus \mathcal{H})$ and $B=\left(\begin{array}{cc}I& 0\\ 0& I\end{array}\right)\in \mathcal{B}(\mathcal{H}\oplus \mathcal{H})$. Clearly, $P_{\mathcal{R}(A+B)}A=A{P}_{\mathcal{R}(A^{*}+B^{*})}$ and $P_{\mathcal{R}(A+B)}B=B{P}_{\mathcal{R}(A^{*}+B^{*})}$, while $\mathcal{R}\left(A\right)⊈\mathcal{R}(A+B)$ and $\mathcal{R}\left(B\right)⊈\mathcal{R}(A+B)$. Thus, A and B are not parallel summable.

It should be noted that any two positive operators on $\mathcal{H}$ are weakly parallel summable from ([15], Theorem 2). Furthermore, for positive operators A and $B\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ that have closed ranges, it is evident that A and B are parallel summable. Therefore, if A and $B\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$, then $A:B$ is always well-defined. In the following part, we give out the details.

Lemma 6 [6], Theorem 3). 

( Let $A\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ and $\mathcal{S}\subseteq \mathcal{H}$ a closed subspace. Let the operator matrix of A is written as

$$A=\left(\begin{array}{cc}{A}_{11}& A_{12}\\ A_{21}& A_{22}\end{array}\right):\left(\begin{array}{c}\mathcal{S}\\ {\mathcal{S}}^{\perp }\end{array}\right),$$

where $A_{11}:\mathcal{S}\to \mathcal{S}$, $A_{12}:{\mathcal{S}}^{\perp }\to \mathcal{S}$, $A_{21}:\mathcal{S}\to {\mathcal{S}}^{\perp }$ and $A_{22}:{\mathcal{S}}^{\perp }\to {\mathcal{S}}^{\perp }$. Then, $\mathcal{R}\left(A_{21}\right)\subseteq \mathcal{R}(A_{22}^{\frac{1}{2}})$.

Lemma 7.

Let $A,B\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ with the operators involved having closed ranges. Then, A and B are parallel summable.

Proof. 

Let $\mathsf{\Lambda }=\left(\begin{array}{cc}A& A\\ A& A+B\end{array}\right)$. Clearly, $\mathsf{\Lambda }\ge 0$. It follows from Lemmas 5 and 6 that

$$\mathcal{R}\left(A\right)\subseteq \mathcal{R}\left(\right|A+B{|}^{\frac{1}{2}})=\mathcal{R}(A+B).$$

Thus, according to Definition 2, we obtain that A and B are parallel summable. □

2. Distributive Law of Parallel Sum

Throughout this paper, it is assumed that A and B are non-zero. In fact, if $A=0$ or $B=0$, it is evident that $A:B=0$ by Equation (2). In ([3], Theorem 2), the authors showed that if A, $B\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ are positive operators, and $M\in \mathcal{B}\left(\mathcal{H}\right)$ is a bounded operator, then $M^{*}(A:B)M\le \left(M^{*}AM\right):\left(M^{*}BM\right)$. In ([12], Theorem 5), the authors studied similar results for bounded operators. In what follows, we present an equality with regard to operators $M^{-1}AM:M^{-1}BM$ and $M^{-1}(A:B)M$ under suitable restrictions on the ranges.

Theorem 1.

Let $A,B\in \mathcal{B}\left(\mathcal{H}\right)$ be parallel summable with $\mathcal{R}(A+B)$ closed, and let $M\in \mathcal{B}\left(\mathcal{H}\right)$ be invertible. If the following conditions hold:

(i) 

$M^{-1}AM$ and $M^{-1}BM$ are parallel summable with $\mathcal{R}\left(M^{-1}(A+B)M\right)$ closed;

(ii) 

$\mathcal{R}\left(M{M}^{*}{(A+B)}^{*}\right)\subseteq \mathcal{R}(A^{*}+B^{*})$ and $\mathcal{R}\left(M^{*-1}{M}^{-1}(A+B)M\right)\subseteq \mathcal{R}\left((A+B)M\right)$, then

$$M^{-1}(A:B)M=\left(M^{-1}AM\right):\left(M^{-1}BM\right).$$

Proof. 

It follows from Lemma 3, conditions $\mathcal{R}\left(M{M}^{*}{(A+B)}^{*}\right)\subseteq \mathcal{R}(A^{*}+B^{*})$ and $\mathcal{R}(M^{*-1}$$M^{-1}(A+B)M)\subseteq \mathcal{R}\left((A+B)M\right)$ that ${\left[M^{-1}(A+B)M\right]}^{†}=M^{-1}{(A+B)}^{†}M$. Therefore, it leads to that

$$\begin{array}{ccc}& & M^{-1}AM:M^{-1}BM\hfill \\ & =& M^{-1}AM{\left[M^{-1}(A+B)M\right]}^{†}{M}^{-1}BM\hfill \\ & =& M^{-1}AM{M}^{-1}{(A+B)}^{†}M{M}^{-1}BM\hfill \\ & =& M^{-1}(A:B)M,\hfill \end{array}$$

which completes the proof. □

Theorem 2.

Let $A,B\in \mathcal{B}\left(\mathcal{H}\right)$ be such that $A:B$ is well-defined with $\mathcal{R}(A+B)$ closed. Let $M\in \mathcal{B}\left(\mathcal{H}\right)$ be such that $A_0=M^{*}AM$ and $B_0=M^{*}BM$ with $\mathcal{R}(A_0+B_0)$ closed. If $B^{*}{(A^{*}+B^{*})}^{†}A=A^{*}{(A^{*}+B^{*})}^{†}B$, $\mathcal{R}\left(M\right)\subseteq \mathcal{R}(A^{*}+B^{*})$ and $\mathcal{R}\left(M^{*}\right)=\mathcal{R}(A_0+B_0)=\mathcal{R}(A_0^{*}+B_0^{*})$, then $A_0$ and $B_0$ are parallel summable with

$$M^{*}(A:B)M+T^{*}AT+T^{*}BT=M^{*}AM:M^{*}BM,$$

where $T=M^{*}{B}^{*}{(A^{*}+B^{*})}^{†}BM-M^{*}B{(A+B)}^{†}BM$.

Proof. 

It follows from

$$\mathcal{R}\left(A_0\right)=\mathcal{R}\left(M^{*}AM\right)\subseteq \mathcal{R}\left(M^{*}\right)\subseteq \mathcal{R}(A_0+B_0),$$

and

$$\mathcal{R}\left(A_0^{*}\right)=\mathcal{R}\left(M^{*}{A}^{*}M\right)\subseteq \mathcal{R}\left(M^{*}\right)\subseteq \mathcal{R}(A_0^{*}+B_0^{*})$$

that the operator $A_0:B_0$ is well-defined. Let $X_0={(A+B)}^{†}BM$ and $Y_0={(A+B)}^{†}AM$. Then, $X_0+Y_0=M$. By Equation (2), one has that

$$A{X}_0=B{Y}_0=(A:B)M=(B:A)M$$

and

$$X_0^{*}A{X}_0+Y_0^{*}B{Y}_0=X_0^{*}(A:B)M+Y_0^{*}(B:A)M=M^{*}(A:B)M.$$

For every $X\in \mathcal{B}\left(\mathcal{H}\right)$, there is a $T\in \mathcal{B}\left(\mathcal{H}\right)$ such that $X=X_0+T$. Put $Y=Y_0-T$. Then, $X+Y=X_0+Y_0=M$. Since $B^{*}{(A^{*}+B^{*})}^{†}A=A^{*}{(A^{*}+B^{*})}^{†}B$, we obtain $X_0^{*}AT=Y_0^{*}BT$. Hence, one has

$$\begin{array}{ccc}& & X^{*}AX+Y^{*}BY\hfill \\ & =& {(X_0+T)}^{*}A(X_0+T)+{(Y_0-T)}^{*}B(Y_0-Y)\hfill \\ \hfill & =& X_0^{*}A{X}_0+Y_0^{*}B{Y}_0+T^{*}AT+T^{*}BT+T^{*}A{X}_0-T^{*}B{Y}_0+X_0^{*}AT-Y_0^{*}BT\hfill \\ \hfill & =& M^{*}(A:B)M+T^{*}AT+T^{*}BT.\hfill \end{array}$$

Now, choose that $X=M{(A_0^{*}+B_0^{*})}^{†}{B}_0^{*}$ and $Y=M{(A_0^{*}+B_0^{*})}^{†}{A}_0^{*}$. Then,

$$T=X-X_0=M{(A_0^{*}+B_0^{*})}^{†}{B}_0^{*}-(A+B^{†})BM.$$

Hence, it leads to that

$$\begin{array}{ccc}& & Y_0-T\hfill \\ & =& {(A+B)}^{†}AM+{(A+B)}^{†}BM-M{(A_0^{*}+B_0^{*})}^{†}{B}_0^{*}\hfill \\ & =& {(A+B)}^{†}(A+B)M-M{(A_0^{*}+B_0^{*})}^{†}{B}_0^{*}\hfill \\ \hfill & =& M-M{(A_0^{*}+B_0^{*})}^{†}{B}_0^{*}\hfill \\ \hfill & =& M^{*}{(A_0^{*}+B_0^{*})}^{†}{A}_0^{*}\hfill \\ \hfill & =& Y.\hfill \end{array}$$

Moreover, it follows from $\mathcal{R}\left(M^{*}\right)\subseteq \mathcal{R}(A_0+B_0)$ that $X+Y=M{(A_0^{*}+B_0^{*})}^{†}(A_0^{*}+B_0^{*})=M$. Furthermore, combining with the conditions that $B^{*}{(A^{*}+B^{*})}^{†}A=A^{*}{(A^{*}+B^{*})}^{†}B$, $\mathcal{R}\left(B^{*}\right)\subseteq \mathcal{R}(A_0^{*}+B_0^{*}),$ and $\mathcal{R}\left(M\right)\subseteq \mathcal{R}(A+B)$, we have that

$$\begin{array}{ccc}& & T^{*}AT+T^{*}BT\hfill \\ \hfill & =& -[M^{*}{B}^{*}{(A^{*}+B^{*})}^{†}AM{(A_0^{*}+B_0^{*})}^{†}{B}_0^{*}+M^{*}{B}^{*}{(A^{*}+B^{*})}^{†}BM{(A_0^{*}+B_0^{*})}^{†}{B}_0^{*}]\hfill \\ \hfill & & +[B_0{(A_0+B_0)}^{†}{M}^{*}AM{(A_0^{*}+B_0^{*})}^{†}{B}_0^{*}+B_0{(A_0+B_0)}^{†}{M}^{*}BM{(A_0^{*}+B_0^{*})}^{†}{B}_0^{*}]\hfill \\ \hfill & & -[B_0{(A_0+B_0)}^{†}{M}^{*}A{(A+B)}^{†}BM+B_0{(A_0+B_0)}^{†}{M}^{*}B{(A+B)}^{†}BM]\hfill \\ \hfill & & +[M^{*}{B}^{*}{(A^{*}+B^{*})}^{†}A{(A+B)}^{†}BM+M^{*}{B}^{*}{(A^{*}+B^{*})}^{†}B{(A+B)}^{†}BM]\hfill \\ \hfill & =& -M^{*}BM{(A_0^{*}+B_0^{*})}^{†}{B}_0+B_0{(A_0^{*}+B_0^{*})}^{†}{B}_0^{*}\hfill \\ & & -B_0{(A_0+B_0)}^{†}{M}^{*}BM+M^{*}{B}^{*}{(A^{*}+B^{*})}^{†}BM\hfill \\ \hfill & =& -B_0{(A_0+B_0)}^{†}{B}_0+M^{*}{B}^{*}{(A^{*}+B^{*})}^{†}BM.\hfill \end{array}$$

Hence, we obtain that

$$\begin{array}{ccc}& & M^{*}(A:B)M+T^{*}AT+T^{*}BT\hfill \\ \hfill & =& M^{*}(A:B)M+M^{*}{B}^{*}{(A^{*}+B^{*})}^{†}BM-B_0{(A_0+B_0)}^{†}{B}_0\hfill \\ \hfill & =& B_0-B_0{(A_0+B_0)}^{†}{B}_0\hfill \\ \hfill & =& A_0:B_0,\hfill \end{array}$$

which completes the proof. □

By Theorem 2, we obtain the following observation:

Remark 3.

If $A,B\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$, and this satisfies the conditions in Theorem 2, then

$$M^{*}(A:B)M\le M^{*}AM:M^{*}BM.$$

Moreover, if $M^{*}{B}^{*}{(A^{*}+B^{*})}^{†}BM=M^{*}B{(A+B)}^{†}BM$, it holds that $M^{*}(A:B)M=M^{*}AM:$$M^{*}BM$.

The following result presents several conditions under which the equality $(A:B)C=AC:BC$ holds. In ([16], Theorem 3.4), the authors also give out some conditions under which the right distributive law of parallel sum operation holds.

Theorem 3.

Let $A,B$ and $C\in \mathcal{B}\left(\mathcal{H}\right)$ be such that $A:B$ is well-defined with $\mathcal{R}(A+B)$ and $\mathcal{R}\left(C\right)$ closed. Then, the parallel sum $AC:BC$ is well-defined and

$$(A:B)C=AC:BC,$$

if one of the following conditions holds:

(i) 

$\mathcal{R}(A^{*}+B^{*})\subseteq \mathcal{R}\left(C\right)$ and $\mathcal{R}\left(C{C}^{*}{(A+B)}^{*}\right)\subseteq \mathcal{R}(A^{*}+B^{*})$;

(ii) 

$\mathcal{R}(A^{*}+B^{*})\subseteq \mathcal{R}\left(C\right)$ and $P_{\mathcal{R}(A^{*}+B^{*})}C=C{P}_{\mathcal{R}(A^{*}+B^{*})}$;

(iii) 

$\mathcal{R}(A^{*}+B^{*})\subseteq \mathcal{R}\left(C\right)$, $AC=CA$, $C{A}^{*}=A^{*}C$, $BC=CB,$ and $C{B}^{*}=B^{*}C$.

Proof. 

(i) 

Since $\mathcal{R}(A+B)$ is closed, it follows from Lemma 2 that $\mathcal{R}\left((A+B)\right(A^{*}$$+B^{*}\left)\right)=\mathcal{R}(A+B)$ is closed. Since $\mathcal{R}(A^{*}+B^{*})\subseteq \mathcal{R}\left(C\right)$, one has that

$$\begin{array}{ccc}\hfill \mathcal{R}(A+B)& \supseteq & \mathcal{R}\left(\right(A+B\left)C\right)\hfill \\ & \supseteq & (A+B)\mathcal{R}(A^{*}+B^{*})\hfill \\ & =& \mathcal{R}\left((A+B){(A+B)}^{*}\right)\hfill \\ & =& \mathcal{R}(A+B),\hfill \end{array}$$

which implies that $\mathcal{R}(AC+BC)=\mathcal{R}(A+B)$ closed.

It follows from

$$\mathcal{R}\left(AC\right)\subseteq \mathcal{R}\left(A\right)\subseteq \mathcal{R}(A+B)=\mathcal{R}(AC+BC)$$

and

$$\mathcal{R}\left({\left(AC\right)}^{*}\right)=\mathcal{R}\left(C^{*}{A}^{*}\right)=C^{*}\mathcal{R}\left(A^{*}\right)\subseteq C^{*}\mathcal{R}(A^{*}+B^{*})=\mathcal{R}\left({(AC+BC)}^{*}\right)$$

that operators $AC$ and $BC$ are parallel summable. Combining Lemma 3, $\mathcal{R}\left(\right(A$${+B)}^{*}(A+B)C)\subseteq \mathcal{R}(A^{*}+B^{*})\subseteq \mathcal{R}\left(C\right)$ and $\mathcal{R}\left(C{C}^{*}{(A+B)}^{*}\right)\subseteq \mathcal{R}(A^{*}+B^{*})$, we obtain ${\left[(A+B)C\right]}^{†}=C^{†}{(A+B)}^{†}$.

Noting that $C{C}^{†}{(A+B)}^{†}=P_{\mathcal{R}\left(C\right)}{(A+B)}^{†}={(A+B)}^{†}$, one obtains that

$$\begin{array}{ccc}\hfill (A:B)C& =& A{(A+B)}^{†}BC\hfill \\ & =& AC{(AC+BC)}^{†}BC\hfill \\ & =& AC{C}^{†}{(A+B)}^{†}BC\hfill \\ & =& AC:BC,\hfill \end{array}$$

which completes the proof of item (i).

(ii) 

If $P_{\mathcal{R}(A^{*}+B^{*})}C=C{P}_{\mathcal{R}(A^{*}+B^{*})}$, one has that

$$\begin{array}{ccc}\hfill \mathcal{R}\left(C{C}^{*}{(A+B)}^{*}\right)& =& C{C}^{*}\mathcal{R}\left(P_{\mathcal{R}(A^{*}+B^{*})}\right)\hfill \\ & =& C\mathcal{R}\left(C^{*}{P}_{\mathcal{R}(A^{*}+B^{*})}\right)\hfill \\ & =& C\mathcal{R}\left(P_{\mathcal{R}(A^{*}+B^{*})}{C}^{*}\right)\hfill \\ & =& \mathcal{R}\left(P_{\mathcal{R}(A^{*}+B^{*})}C{C}^{*}\right)\hfill \\ & \subseteq & \mathcal{R}(A^{*}+B^{*}),\hfill \end{array}$$

which leads to that ${\left[(A+B)C\right]}^{†}=C^{†}{(A+B)}^{†}$. The remaining part is similar to the proof of item (i), and, here, we omit it.

(iii) 

If $AC=CA$ and $BC=CB$, then

$$\mathcal{R}\left(AC\right)=\mathcal{R}\left(CA\right)\subseteq \mathcal{R}(CA+CB)=\mathcal{R}(AC+BC)$$

and

$$\mathcal{R}\left({\left(AC\right)}^{*}\right)=\mathcal{R}\left(C^{*}{A}^{*}\right)\subseteq \mathcal{R}(C^{*}{A}^{*}+C^{*}{B}^{*})=\mathcal{R}\left({(AC+BC)}^{*}\right).$$

Thus, it implies that $AC$ and $BC$ are parallel summable.

It follows from Lemma 3, $\mathcal{R}(A^{*}+B^{*})\subseteq \mathcal{R}\left(C\right)$, $C{A}^{*}=A^{*}C$, and $C{B}^{*}=B^{*}C$ that

$$\mathcal{R}\left({(A+B)}^{*}(A+B)C\right)\subseteq \mathcal{R}(A^{*}+B^{*})\subseteq \mathcal{R}\left(C\right)$$

and

$$\mathcal{R}\left(C{C}^{*}{(A+B)}^{*}\right)\subseteq \mathcal{R}(A^{*}+B^{*}).$$

Therefore, we obtain ${\left[(A+B)C\right]}^{†}=C^{†}{(A+B)}^{†}$. As a result, it holds that

$$\begin{array}{ccc}\hfill (A:B)C& =& A{(A+B)}^{†}BC\hfill \\ & =& AC{C}^{†}{(A+B)}^{†}BC\hfill \\ & =& AC{(AC+BC)}^{†}BC\hfill \\ & =& AC:BC,\hfill \end{array}$$

which completes the proof of item (iii). □

Remark 4.

It should be noted that in Theorem 3, one observes that if A and $B\in \mathcal{B}\left(\mathcal{H}\right)$ are parallel summable and the bounded operator $C\in \mathcal{B}\left(\mathcal{H}\right)$ satisfies $\mathcal{R}(A^{*}+B^{*})\subseteq \mathcal{R}\left(C\right)$, then $\mathcal{R}\left(AC\right)\subseteq \mathcal{R}\left(A\right)\subseteq \mathcal{R}(A+B)=\mathcal{R}(AC+BC)$. Combining the fact that $\mathcal{R}\left({\left(AC\right)}^{*}\right)=\mathcal{R}\left(C^{*}{A}^{*}\right)=C^{*}\mathcal{R}\left(A^{*}\right)\subseteq C^{*}\mathcal{R}(A^{*}+B^{*})=\mathcal{R}\left({(AC+BC)}^{*}\right)$, one can easily see that $AC$ and $BC$ are parallel summable.

According to Theorem 3, it is easy to obtain the following corollary.

Corollary 1.

Let A, B, and $C\in \mathcal{B}\left(\mathcal{H}\right)$ be such that $A:B$ is well-defined with $\mathcal{R}(A+B)$ and $\mathcal{R}\left(C\right)$ closed. Then, operators $CA$ and $CB$ are parallel summable and

$$CA:CB=C(A:B),$$

if one of the following conditions holds:

(i) 

$\mathcal{R}(A+B)\subseteq \mathcal{R}\left(C^{*}\right)$ and $\mathcal{R}\left(C^{*}C(A+B)\right)\subseteq \mathcal{R}(A+B)$;

(ii) 

$\mathcal{R}(A+B)\subseteq \mathcal{R}\left(C^{*}\right)$ and $P_{\mathcal{R}(A+B)}C=C{P}_{\mathcal{R}(A+B)}$.

Lastly, the following theorem presents the conditions under which the distributive law of the parallel sum is valid with respect to the usual operator product.

Theorem 4.

Let A and $B\in \mathcal{B}\left(\mathcal{H}\right)$ be such that $A:B$ is well-defined with $\mathcal{R}(A+B)$ closed. If the closed range operator $C\in \mathcal{B}\left(\mathcal{H}\right)$ satisfies that $\mathcal{R}\left(A^{*}C\right)\subseteq \mathcal{R}(A^{*}C+B^{*}C)$ and $\mathcal{R}\left(C{C}^{*}(A+B)C\right)\subseteq \mathcal{R}(A+B)\subseteq \mathcal{R}\left(C\right)=\mathcal{R}(A^{*}+B^{*})$, then $C^{*}AC:C^{*}BC$ is well-defined and

$$C^{*}(A:B)C=C^{*}AC:C^{*}BC.$$

Proof. 

According to Remark 4, we know that $\mathcal{R}\left(AC\right)\subseteq \mathcal{R}(AC+BC)$ and thus $\mathcal{R}\left(C^{*}AC\right)\subseteq \mathcal{R}(C^{*}AC+C^{*}BC)$. It follows from $\mathcal{R}\left(A^{*}C\right)\subseteq \mathcal{R}(A^{*}C+B^{*}C)$ that

$$\mathcal{R}\left({\left(C^{*}AC\right)}^{*}\right)\subseteq \mathcal{R}\left({(C^{*}AC+C^{*}BC)}^{*}\right).$$

Hence, operators $C^{*}AC$ and $C^{*}BC$ are parallel summable. By Theorem 3(i), one has that

$$\mathcal{R}(AC+BC)=\mathcal{R}(A+B),$$

with the ranges closed. It follows from $\mathcal{R}(A^{*}+B^{*})\subseteq \mathcal{R}\left(C\right)$ that

$$\mathcal{R}\left({(A+B)}^{*}C\right)=\mathcal{R}\left(P_{\mathcal{R}(A^{*}+B^{*})}C\right)=\mathcal{R}\left(C\right),$$

with the operators ranges closed. Therefore, we have that $\mathcal{R}\left(C^{*}(A+B)C\right)=\mathcal{R}\left(C^{*}\right(A$$+B\left)\right)$, with the operator ranges closed because of

$$\begin{array}{ccc}\hfill \mathcal{R}\left(C^{*}(A+B)\right)& \supseteq & \mathcal{R}\left(C^{*}(A+B)C\right)\hfill \\ \hfill & =& C^{*}(A+B)\mathcal{R}\left(C\right)\hfill \\ \hfill & \supseteq & C^{*}(A+B)\mathcal{R}(A^{*}+B^{*})\hfill \\ \hfill & =& C^{*}\mathcal{R}(A+B)\hfill \\ \hfill & =& \mathcal{R}\left(C^{*}(A+B)\right).\hfill \end{array}$$

Since $\mathcal{R}\left(C{C}^{*}(A+B)C\right)\subseteq \mathcal{R}(A+B)=\mathcal{R}\left((A+B)C\right)$ and $\mathcal{R}((A+B)C{C}^{*}(A$${+B)}^{*}C)\subseteq \mathcal{R}(A+B)\subseteq \mathcal{R}\left(C\right)$, by Lemma 3, one obtains ${\left[C^{*}(A+B)C\right]}^{†}=\left[\right(A$$+{B\left)C\right]}^{†}{\left(C^{*}\right)}^{†}$. Hence, one obtains that

$$\begin{array}{ccc}\hfill \left(C^{*}AC\right):\left(C^{*}BC\right)& =& C^{*}AC{C}^{†}{(A+B)}^{†}{\left(C^{*}\right)}^{†}BC\hfill \\ & =& C^{*}A{(A+B)}^{†}BC\hfill \\ & =& C^{*}(A:B)C,\hfill \end{array}$$

which completes the proof. □

3. Continuity of the Parallel Sum

In ([6], Theorem 25), the authors obtained that $\parallel A:B\parallel \le \parallel A\parallel :\parallel B\parallel$, where A and B are $n\times n$ complex Hermitian positive semi-definite matrices. In the following section, we study the continuity of the parallel sum of positive operators in infinite dimensional Hilbert spaces. In the following section, we present some lemmas about the properties of parallel sums.

Lemma 8.

Let A and $B\in \mathcal{B}\left(\mathcal{H}\right)$ be parallel summable with $\mathcal{R}(A+B)$ closed. Then, for every $z\in \mathcal{R}(A^{*}+B^{*})$, let $x={(A+B)}^{†}Bz$ and $y={(A+B)}^{†}Az$, we have that

$$⟨A:Bz,z⟩=⟨Ax,x⟩+⟨By,y⟩.$$

Proof. 

Let $z\in \mathcal{R}(A^{*}+B^{*})$, $x={(A+B)}^{†}Bz$ and $y={(A+B)}^{†}Az$. Then

$$Ax=By=A:Bz$$

and

$$x+y={(A+B)}^{†}(A+B)z=z.$$

Therefore,

$$\begin{array}{ccc}\hfill ⟨A:Bz,z⟩& =& ⟨A:Bz,x⟩+⟨A:Bz,y⟩\hfill \\ & =& ⟨Ax,x⟩+⟨By,y⟩,\hfill \end{array}$$

which completes the proof. □

Lemma 9 (see [16], Theorem 3.3) for matrices cases). 

Let A and $B\in \mathcal{B}\left(\mathcal{H}\right)$ be parallel summable with $\mathcal{R}\left(A\right)$, $\mathcal{R}\left(B\right)$ and $\mathcal{R}(A+B)$ closed. Then

$$\mathcal{R}(A:B)=\mathcal{R}\left(A\right)\cap \mathcal{R}\left(B\right)\subseteq \mathcal{R}(A+B),$$

and, for every $x\in \mathcal{H}$, there exist $y=A:Bx$ and $z=A^{*}:B^{*}x$ such that

$$⟨A:Bx,x⟩=⟨(A^{†}+B^{†})y,z⟩.$$

Proof. 

Because A and B are parallel summable and due to Remark 1, it follows that operators $A^{*}$ and $B^{*}$ are parallel summbale. Moreover, it follows from equality (2) that

$$A^{*}:B^{*}=A^{*}{(A^{*}+B^{*})}^{†}{B}^{*}={(A:B)}^{*}.$$

By (2) again, it is clear that $\mathcal{R}(A:B)\subseteq \mathcal{R}\left(A\right)\cap \mathcal{R}\left(B\right)$. Let $x\in \mathcal{R}\left(A\right)\cap \mathcal{R}\left(B\right)$. Then, we have that

$$\begin{array}{ccc}& & (A:B)(A^{†}+B^{†})x\hfill \\ \hfill & =& A{(A+B)}^{†}B{B}^{†}x+B{(A+B)}^{†}A{A}^{†}x\hfill \\ \hfill & =& (A+B){(A+B)}^{†}x\hfill \\ \hfill & =& x,\hfill \end{array}$$

which implies that $\mathcal{R}\left(A\right)\cap \mathcal{R}\left(B\right)=\mathcal{R}(A:B)$.

For every $x\in \mathcal{H}$, let $y=A:Bx$ and $z=A^{*}:B^{*}x$. It holds that

$$\begin{array}{ccc}& & ⟨A:Bx,x⟩\hfill \\ \hfill & =& ⟨(A:B)(A^{†}+B^{†})y,x⟩\hfill \\ \hfill & =& ⟨(A^{†}+B^{†})y,A^{*}:B^{*}x⟩\hfill \\ \hfill & =& ⟨(A^{†}+B^{†})y,z⟩,\hfill \end{array}$$

which completes the proof. □

In ([13], Theorem 4.2), the author refined the parallel sum for the positive operators on infinite dimensional spaces and then proved that $A:B$ is also positive whenever A and B are both positive. Here, we are in the position to present several properties of positive operators with closed ranges in the sequel.

Theorem 5.

Let $A,B\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ with the operators involved having closed ranges. Then, $A:B$ is a positive operator on $\mathcal{H}$.

Proof. 

Assume that A and $B\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ have closed ranges. It follows from Lemmas 4 and 5 that $\mathcal{R}\left(A\right)+\mathcal{R}\left(B\right)=\mathcal{R}(A+B)$ is closed. Combining with Definition 2 and the facts that

$$\mathcal{R}\left(A\right)=\mathcal{R}\left[\right(A+B)-B]\subseteq \mathcal{R}(A+B)+\mathcal{R}\left(B\right)=\mathcal{R}\left(A\right)+\mathcal{R}\left(B\right)$$

and

$$\mathcal{R}\left(B\right)=\mathcal{R}\left[\right(A+B)-A]\subseteq \mathcal{R}(A+B)+\mathcal{R}\left(A\right)=\mathcal{R}\left(A\right)+\mathcal{R}\left(B\right),$$

we conclude that A and B are parallel summable. Moreover, it follows from Lemma 9 that $A:B$ is a positive operator, which completes the proof. □

Theorem 6 (see ([7], Theorem 3.1) for matrices cases). 

Let $A,B\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ with the operators involved having closed ranges. Then, for every $x,y$, and $z\in \mathcal{H}$ such that $x+y=z$, it holds that

$$⟨A:Bz,z⟩\le ⟨Ax,x⟩+⟨By,y⟩.$$

Proof. 

It follows from Lemma 7 that A and B are parallel summable. For every $z\in \mathcal{H}$, and $x+y=z$, let $x^1=P_{\mathcal{R}(A+B)}x$, $y^1=P_{\mathcal{R}(A+B)}y$, and $z^1=P_{\mathcal{R}(A+B)}z$. Then, let

$$x_0^1={(A+B)}^{†}B{z}^1\mathrm{and}{y}_0^1={(A+B)}^{†}A{z}^1.$$

Since $x_0^1+y_0^1=z^1=x^1+y^1$, we may write $x^1=x_0^1+t$ and $y^1=y_0^1-t$. Then

$$⟨A{x}^1,x^1⟩=⟨A{x}_0^1,x_0^1⟩+2\mathrm{Re}⟨A{x}_0^1,t⟩+⟨At,t⟩$$

and

$$⟨B{y}^1,y^1⟩=⟨B{y}_0^1,y_0^1⟩-2\mathrm{Re}⟨A{y}_0^1,t⟩+⟨Bt,t⟩.$$

Since $A{x}_0^1=B{y}_0^1$, it follows from Lemma 8 that

$$\begin{array}{ccc}\hfill ⟨A{x}^1,x^1⟩+⟨B{y}^1,y^1⟩& =& ⟨A{x}_0^1,x_0^1⟩+⟨B{y}_0^1,y_0^1⟩+⟨At,t⟩+⟨Bt,t⟩\hfill \\ & \ge & ⟨A:B{z}^1,z^1⟩.\hfill \end{array}$$

(3)

However, $⟨A{x}^1,x^1⟩=⟨A{P}_{\mathcal{R}(A+B)}x,P_{\mathcal{R}(A+B)}x⟩=⟨Ax,x⟩$ and similarly for $y^1$, $z^1$. The result then follows from (3). □

The next remark shows a similar result of Lemma 8.

Remark 5.

Let $A,B\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ with $\mathcal{R}(A+B)$ closed. Then, for every $z\in \mathcal{H}$, there exist an x and y that satisfy $z=x+y$ such that

$$⟨A:Bz,z⟩=⟨Ax,x⟩+⟨By,y⟩.$$

It follows from Lemma 7 that A and B are parallel summable. For every $z\in \mathcal{H}$, and $x+y=z$, let $x^1=P_{\mathcal{R}(A+B)}x$, $y^1=P_{\mathcal{R}(A+B)}y$, and $z^1=P_{\mathcal{R}(A+B)}z$. It follows from Lemma 8 that

$$\begin{array}{ccc}\hfill ⟨A:B{z}^1,z^1⟩& =& ⟨A{x}^1,x^1⟩+⟨B{y}^1,y^1⟩.\hfill \end{array}$$

(4)

Note that

$$⟨A:B{z}^1,z^1⟩=⟨(A:B)P_{\mathcal{R}(A+B)}z,P_{\mathcal{R}(A+B)}z⟩=⟨A:Bz,z⟩$$

and

$$⟨A{x}^1,x^1⟩=⟨A{P}_{\mathcal{R}(A+B)}x,P_{\mathcal{R}(A+B)}x⟩=⟨Ax,x⟩,$$

and similarly for $y^1$. The result then follows from (3). □

Corollary 2 (see ([17], Lemma 18) for matrices case). 

Let $A,B,C,$ and $D\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ with the operators involved having closed ranges. Then

$$(A+B):(C+D)\ge A:C+B:D.$$

Proof. 

Note that operators $(A+B):(C+D)$, $A:C,$ and $B:D$ are well-defined. It follows from Remark 5 that for $x\in \mathcal{H}$, there exist an x and $y\in \mathcal{H}$ that satisfy $z=x+y$ such that

$$\begin{array}{ccc}\hfill ⟨(A+B):(C+D)z,z⟩& =& ⟨(A+B)x,x⟩+⟨(C+D)y,y⟩\hfill \\ & =& ⟨Ax,x⟩+⟨Bx,x⟩+⟨Cy,y⟩+⟨Dy,y⟩.\hfill \end{array}$$

(5)

It follows from Lemma 8 that

$$⟨Ax,x⟩+⟨Cy,y⟩\ge ⟨A:Cz,z⟩$$

and

$$⟨Bx,x⟩+⟨Dy,y⟩\ge ⟨B:Dz,z⟩.$$

Then, by combining with (5), we obtain the result. □

Given the two operators A and $B\in {\mathcal{B}}^{\mathcal{S}}\left(\mathcal{H}\right)$, $A\le B$ means that $B-A\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ (this is called the usual or Löwner order). The following corollary presents the monotonicity of parallel sum operation (see ([16], Lemma 4.1(vi)).

Corollary 3.

Let $A,B,$ and $C\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ with the operators involved having closed ranges. Then, $A\ge B$ implies that

$$A:C\ge B:C.$$

Proof. 

Note that operators $A:C$ and $B:C$ are well-defined. Let $A-B=D$. Then, D is a positive operator, and it follows from Corollary 2 that

$$A:C=(B+D):(C+0)\ge B:C+D:0=B:C,$$

which completes the proof. □

Theorem 7.

Let A and $B\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ with $\mathcal{R}\left(A\right)$, $\mathcal{R}\left(B\right)$ and $\mathcal{R}(A+B)$ closed. Then

$$\parallel A:B\parallel \le \parallel A\parallel :\parallel B\parallel .$$

Proof. 

Since A is a non-zero positive operator, for every $x\in \mathcal{H}$, one has that

$$⟨(A-\parallel A\parallel I)Ax,x⟩\le 0,$$

which implies that

$$⟨Ax,Ax⟩\le \parallel A\parallel ⟨Ax,x⟩.$$

(6)

For any $\epsilon >0$, there is an $x_0(\ne 0)\in \mathcal{H}$ such that

$$\frac{⟨A{x}_0,A{x}_0⟩}{⟨A{x}_0,x_0⟩}\ge \parallel A\parallel -\epsilon .$$

Similarly, it follows from Theorem 2 that $A:B$ is positive and

$$\frac{⟨A:Bx,A:Bx⟩}{⟨A:Bx,x⟩}\ge \parallel A:B\parallel -\epsilon .$$

Let $y=A:Bx$. Then $(A:B)(A^{†}+B^{†})y=y$, and

$$⟨A:Bx,x⟩=⟨y,x⟩=⟨(A:B)(A^{†}+B^{†})y,x⟩=⟨(A^{†}+B^{†})y,A:Bx⟩=⟨A^{†}y,y⟩+⟨B^{†}y,y⟩.$$

Since $y\in \mathcal{R}(A:B)$, let $y=Au=Bv$, where $u,v\in \mathcal{H}$. Therefore, it leads to that

$$⟨A^{†}y,y⟩=⟨u,Au⟩\mathrm{and}⟨B^{†}y,y⟩=⟨v,Bv⟩.$$

Hence,

$$\frac{⟨u,Au⟩}{⟨Au,Au⟩}+\frac{⟨v,Av⟩}{⟨Bv,Bv⟩}=\frac{⟨A:Bx,x⟩}{⟨A:Bx,A:Bx⟩}$$

and then, from (6), it concludes that

$$\frac{1}{\parallel A\parallel }+\frac{1}{\parallel B\parallel }\le \frac{1}{\parallel A:B\parallel -\epsilon }\mathrm{or}\parallel A\parallel :\parallel B\parallel \ge \parallel A:B\parallel -\epsilon ,$$

where $\epsilon$ is arbitrary. □

Lemma 10.

Let A and $B\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ with the operators involved having closed ranges. If $A\le B$, then

$$(A^{†}-B^{†})P_{\mathcal{R}\left(A\right)}=B^{†}(B-A)A^{†}.$$

Proof. 

It follows from $\mathcal{R}\left(A\right)\subseteq \mathcal{R}\left(B\right)$ that

$$\begin{array}{ccc}\hfill B^{†}(B-A)A^{†}& =& B^{†}B{A}^{†}-B^{†}A{A}^{†}\hfill \\ & =& P_{\mathcal{R}\left(B\right)}{A}^{†}-B^{†}{P}_{\mathcal{R}\left(A\right)}\hfill \\ & =& A^{†}-B^{†}{P}_{\mathcal{R}\left(A\right)}\hfill \\ & =& (A^{†}-B^{†})P_{\mathcal{R}\left(A\right)},\hfill \end{array}$$

which completes the proof. □

Lemma 11.

Let A and $B\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ with the operators involved having closed ranges. Then,

$$P_{\mathcal{R}(A+B)}[I-{(A+B)}^{†}B]={(A+B)}^{†}A.$$

Proof. 

It follows from ${(A+B)}^{†}A+{(A+B)}^{†}B={(A+B)}^{†}(A+B)=P_{\mathcal{R}(A+B)}$ that

$$\begin{array}{ccc}\hfill P_{\mathcal{R}(A+B)}[I-{(A+B)}^{†}B]& =& P_{\mathcal{R}(A+B)}-P_{\mathcal{R}(A+B)}{(A+B)}^{†}B\hfill \\ & =& P_{\mathcal{R}(A+B)}-{(A+B)}^{†}B\hfill \\ & =& {(A+B)}^{†}A,\hfill \end{array}$$

which completes the proof. □

Theorem 1 expressed the continuity of the parallel sum at 0, while the next theorem applies at another point. For the Hermitian positive semi-definite matrices, the authors investigated it in ([6], Theorem 28).

Theorem 8.

Let $A,B,$ and $X\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ with the operators involved having closed ranges. If $G=A:(B+X)-A:B$, then $G=A{C}^{†}(C:X)C^{†}A$ is positive and $\parallel G\parallel \le \parallel C^{†}{A\parallel }^2\parallel X\parallel$, where $C=A+B$.

Proof. 

Let $C=A+B$. Note that

$$\begin{array}{ccc}\hfill G& =& A{(C+X)}^{†}(B+X)-A{C}^{†}B\hfill \\ & =& A[{(C+X)}^{†}-C^{†}]B+A{(C+X)}^{†}X.\hfill \end{array}$$

It follows from Lemma 10 and $P_{\mathcal{R}\left(C\right)}B=B$ that

$$G=-A{(C+X)}^{†}X{C}^{†}B+A{(C+X)}^{†}X.$$

Since the positive operators A and B are parallel summable, according to Remark 4, one has

$$P_{\mathcal{R}\left(C\right)}A=A=A{P}_{\mathcal{R}\left(C\right)}\mathrm{and}{P}_{\mathcal{R}\left(C\right)}B=B=B{P}_{\mathcal{R}\left(C\right)}.$$

So, by the definition of G, we obtain that $P_{\mathcal{R}\left(C\right)}G=G=G{P}_{\mathcal{R}\left(C\right)}$ is positive, and then

$$\begin{array}{ccc}\hfill G& =& G{P}_{\mathcal{R}\left(C\right)}\hfill \\ & =& -A{(C+X)}^{†}X{C}^{†}B{P}_{\mathcal{R}\left(C\right)}+A{(C+X)}^{†}X{P}_{\mathcal{R}\left(C\right)}\hfill \\ & =& A{(C+X)}^{†}X{P}_{\mathcal{R}\left(C\right)}(I-C^{†}B).\hfill \end{array}$$

Hence, it follows from Lemma 11 that

$$\begin{array}{ccc}\hfill G& =& A{(C+X)}^{†}X{C}^{†}A\hfill \\ & =& A{C}^{†}C{(C+X)}^{†}X{C}^{†}A\hfill \\ & =& A{C}^{†}(C:X)C^{†}A.\hfill \end{array}$$

Noting that G is positive, then so is $G:X$. Using Theorem 1, we obtain that

$$\parallel G\parallel \le \parallel A{C}^{†}\parallel \parallel C:X\parallel \parallel C^{†}A\parallel \le \parallel A{C}^{†}{\parallel }^2\parallel X\parallel ,$$

which completes the proof. □

Lemma 12.

Let $A,B,$ and $X\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ with the operators involved having closed ranges. Then

$$\begin{array}{ccc}& & 2(A+X):(B+X)+(A+B):\left(2X\right)\hfill \\ & =& 2(A+B+X):X+A:(B+2X)+(A+2X):B.\hfill \end{array}$$

Proof. 

Let $A+B+2X=D$. Then, by computation both sides equal to

$$2A{D}^{†}B+4A{D}^{†}X+2X{D}^{†}B+2B{D}^{†}X+2X{D}^{†}X,$$

which leads to the result. □

Lemma 13.

Let $A,B,$ and $X\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ with the operators involved having closed ranges. Then

$$H=(A+X):(B+X)-A:B-X:X$$

is positive and, for $C=A+B$, it holds that

$$2H=A{C}^{†}(C:2X)C^{†}A+B{C}^{†}(C:2X)C^{†}B-\frac{1}{2}C:2X,$$

and

$$\parallel H\parallel \le 2(\parallel C^{†}A{\parallel }^2+\parallel C^{†}B{\parallel }^2)\parallel X\parallel .$$

By Lemma 12, one has that

$$2H=A:(B+2X)-A:B+B:(A+2X)-B:A+2X(A+B+X)-2X:X-(A+B):2X.$$

Proof. 

It follows from Theorem 8 that

$$2H=A{C}^{†}(C:2X)C^{†}A+B{C}^{†}(C:2X)C^{†}B+2X{\left(2X\right)}^{†}(C:X){\left(2X\right)}^{†}X-C:\left(2X\right),$$

which implies that

$$2H=A{C}^{†}(C:2X)C^{†}A+B{C}^{†}(C:2X)C^{†}B-\frac{1}{2}C:2X.$$

By Corollary 2, we know that H is a positive operator. In view of $C:2X\ge 0$, it leads to

$$2H\le A{C}^{†}(C:2X)C^{†}A+B{C}^{†}(C:2X)C^{†}B$$

and, as in Theorem 8, we obtain that

$$\parallel H\parallel \le \parallel A{C}^{†}{\parallel }^2\parallel 2X\parallel +\parallel B{C}^{†}{\parallel }^2\parallel 2X\parallel ,$$

which completes the proof. □

In ([6], Theorem 31), the authors obtained a perturbation estimation for the parallel sum of $n\times n$ complex Hermitian positive semi-definite matrices. The following theorem shows that the result obtained in ([6], Theorem 31) is also true for positive operators in infinite dimensional Hilbert spaces. Perturbation estimation for the parallel sum in Hilbert $C^{*}$-modules is studied in ([7], Theorem 3.4).

Theorem 9.

Let A, B, X, and $Y\in {\mathcal{B}}^{+}\left(\mathcal{H}\right)$ with the operators involved having closed ranges. Then

$$\parallel (A+X):(B+Y)-A:B\parallel \le 2(\parallel {(A+B)}^{†}A{\parallel }^2+\parallel {(A+B)}^{†}B{\parallel }^2+\frac{1}{4})\parallel X+Y\parallel .$$

Proof. 

It follows from Corollary 3 that

$$(A+X):(B+Y)-A:B\le (A+X+Y):(B+X+Y)-A:B.$$

Then, using Lemma 13, we obtain that

$$\begin{array}{ccc}& & \parallel (A+X):(B+Y)-A:B\parallel \hfill \\ & \le & \parallel (A+X+Y):(B+X+Y)-A:B-(X+Y):(X+Y)\parallel \hfill \\ & \le & {2(\parallel (A+B)}^{†}{A\parallel }^2{+\parallel (A+B)}^{†}{B\parallel }^2)\parallel X+Y\parallel +\parallel \frac{1}{2}(X+Y)\parallel \hfill \\ & \le & 2(\parallel {(A+B)}^{†}A{\parallel }^2+\parallel {(A+B)}^{†}B{\parallel }^2+\frac{1}{4})\parallel X+Y\parallel ,\hfill \end{array}$$

which completes the proof. □

4. A Characterization of the Minus Partial Order through Parallel Sum

The minus partial order was defined by Hartwig [18], weakening the conditions of the star partial order defined by Drazin [19]. In ([4], Theorem 2.1), the author characterized the minus partial order of $m\times n$ complex matrices by the parallel sum. In what follows, we present several similar results of bounded operators on $\mathcal{H}$. Now, we are in the position to introduce the definition of the minus partial for closed range operators and then give several characterizations of the minus partial order. Later, we present the relationship between the minus partial order and parallel sum.

Definition 3.

Let A and $B\in \mathcal{B}\left(\mathcal{H}\right)$ with closed ranges. We write $A\overline{\le }B$ if

$$A^{-}A=A^{-}B\mathit{and}A{A}^{-}=B{A}^{-},$$

(7)

where $A^{-}$ is some g-inverse of A. The relation “$\overline{\le }$" is called the minus partial order.

Let $A\in \mathcal{B}\left(\mathcal{H}\right)$ have a closed range. Then, as an operator from $\mathcal{R}\left(A^{*}\right)\oplus \mathcal{N}\left(A\right)$ into $\mathcal{R}\left(A\right)\oplus \mathcal{N}\left(A^{*}\right)$, it has an operator matrix block representation as

$$A=A_{11}\oplus 0,\mathrm{where}{A}_{11}\in \mathcal{B}(\mathcal{R}\left(A^{*}\right),\mathcal{R}\left(A\right))$$

(8)

is invertible. Let $G\in \mathcal{B}\left(\mathcal{H}\right)$ be a g-inverse of A and have the operator matrix form as

$$G=\left(\begin{array}{cc}{G}_{11}& G_{12}\\ G_{21}& G_{22}\end{array}\right):\left(\begin{array}{c}\mathcal{R}\left(A\right)\\ \mathcal{N}\left(A^{*}\right)\end{array}\right)⟶\left(\begin{array}{c}\mathcal{R}\left(A^{*}\right)\\ \mathcal{N}\left(A\right)\end{array}\right).$$

(9)

It follows from $AGA=A$ that $G_{11}=A_{11}^{-1}$. Thus, it holds that

$$\begin{array}{c}\hfill \left\{A^{-}\right\}=\{\left(\begin{array}{cc}{A}_{11}^{-1}& G_{12}\\ G_{21}& G_{22}\end{array}\right),where{G}_{12}\in \mathcal{B}(\mathcal{N}\left(A^{*}\right),\mathcal{R}\left(A^{*}\right)),G_{21}\in \mathcal{B}(\mathcal{R}\left(A\right),\mathcal{R}\left(A^{*}\right))\\ \hfill and{G}_{22}\in \mathcal{B}(\mathcal{N}\left(A^{*}\right),\mathcal{N}\left(A\right))\mathrm{are}\mathrm{arbitrary}\}.\end{array}$$

(10)

Let $B\in \mathcal{B}\left(\mathcal{H}\right)$ with a closed range have the corresponding operator matrix block form as follows:

$$B=\left(\begin{array}{cc}{B}_{11}& B_{12}\\ B_{21}& B_{22}\end{array}\right):\left(\begin{array}{c}\mathcal{R}\left(A^{*}\right)\\ \mathcal{N}\left(A\right)\end{array}\right)⟶\left(\begin{array}{c}\mathcal{R}\left(A\right)\\ \mathcal{N}\left(A^{*}\right)\end{array}\right).$$

(11)

In the sequel, we give out the representation of $B\in \mathcal{B}\left(\mathcal{H}\right)$ which satisfies $A\overline{\le }B$.

Theorem 10.

Let A and $B\in \mathcal{B}\left(\mathcal{H}\right)$ have closed ranges. $A\overline{\le }B$ if and only if

$$B=A+(I-AG)B(I-GA)$$

for some g-inverse G of A.

Proof. 

Assume that $A\overline{\le }B$. Let G be a g-inverse of A with the same operator matrix form as (9), and $A,B$ have the same operator matrix forms as (8) and (11), respectively. It follows from (7) that

$$\left(\begin{array}{cc}{A}_{11}^{-1}& G_{12}\\ G_{21}& G_{22}\end{array}\right)\left(\begin{array}{cc}{A}_{11}& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}{A}_{11}^{-1}& G_{12}\\ G_{21}& G_{22}\end{array}\right)\left(\begin{array}{cc}{B}_{11}& B_{12}\\ B_{21}& B_{22}\end{array}\right)$$

(12)

and

$$\left(\begin{array}{cc}{A}_{11}& 0\\ 0& 0\end{array}\right)\left(\begin{array}{cc}{A}_{11}^{-1}& G_{12}\\ G_{21}& G_{22}\end{array}\right)=\left(\begin{array}{cc}{B}_{11}& B_{12}\\ B_{21}& B_{22}\end{array}\right)\left(\begin{array}{cc}{A}_{11}^{-1}& G_{12}\\ G_{21}& G_{22}\end{array}\right).$$

(13)

It follows from (12) that

$$\left\{\begin{array}{cc}{A}_{11}^{-1}{B}_{11}+G_{12}{B}_{21}=I,\hfill & \left(\mathrm{a}\right)\hfill \\ G_{21}{B}_{11}+G_{22}{B}_{21}=G_{21}{A}_{11},\hfill & \left(\mathrm{b}\right)\hfill \\ A_{11}^{-1}{B}_{12}+G_{12}{B}_{22}=0,\hfill & \left(\mathrm{c}\right)\hfill \\ G_{21}{B}_{12}+G_{22}{B}_{22}=0.\hfill & \left(\mathrm{d}\right)\hfill \end{array}\right.$$

(14)

Similarly, it follows from (13) that

$$\left\{\begin{array}{cc}{B}_{11}{A}_{11}^{-1}+B_{12}{G}_{21}=I,\hfill & \left(\mathrm{a}\right)\hfill \\ B_{11}{G}_{12}+B_{12}{G}_{22}=A_{11}{G}_{12},\hfill & \left(\mathrm{b}\right)\hfill \\ B_{21}{A}_{11}^{-1}+B_{22}{G}_{21}=0,\hfill & \left(\mathrm{c}\right)\hfill \\ B_{21}{G}_{12}+B_{22}{G}_{22}=0.\hfill & \left(\mathrm{d}\right)\hfill \end{array}\right.$$

(15)

It follows from (b) of (14) that

$$B_{21}=-B_{22}{G}_{21}{A}_{11},$$

(16)

where $G_{21}\in \mathcal{B}(\mathcal{R}\left(A\right),\mathcal{N}\left(A^{*}\right))$ is arbitrary. Similarly, from (b) of (15), we obtain

$$B_{12}=-A_{11}{G}_{12}{B}_{22},$$

(17)

where $G_{12}\in \mathcal{B}(\mathcal{N}\left(A^{*}\right),\mathcal{R}\left(A^{*}\right))$ is arbitrary. Thus, it follows from (a) of (15) and (17) (resp. (a) of (14) and (16)) that

$$B_{11}=A_{11}+A_{11}{G}_{12}{B}_{22}{G}_{21}{A}_{11}.$$

As a result,

$$\begin{array}{ccc}\hfill B& =& \left(\begin{array}{cc}{A}_{11}+A_{11}{G}_{12}{B}_{22}{G}_{21}{A}_{11}& -A_{11}{G}_{12}{B}_{22}\\ -B_{22}{G}_{21}{A}_{11}& B_{22}\end{array}\right)\hfill \\ & =& \left(\begin{array}{cc}{A}_{11}& 0\\ 0& 0\end{array}\right)+\left(\begin{array}{cc}{A}_{11}{G}_{12}{B}_{22}{G}_{21}{A}_{11}& -A_{11}{G}_{12}{B}_{22}\\ -B_{22}{G}_{21}{A}_{11}& B_{22}\end{array}\right)\hfill \\ & =& A+\left(\begin{array}{cc}0& -A_{11}{G}_{12}\\ 0& I\end{array}\right)\left(\begin{array}{cc}{B}_{11}& B_{12}\\ B_{21}& B_{22}\end{array}\right)\left(\begin{array}{cc}0& 0\\ -G_{21}{A}_{11}& I\end{array}\right)\hfill \\ & =& A+(I-AG)B(I-GA),\hfill \end{array}$$

where G is some g-inverse of A.

Conversely, let G be a reflexive generalized inverse of A. Then, it leads to $GA=GB$ and $AG=BG$, which completes the proof. □

Theorem 11.

Let A and $B\in \mathcal{B}\left(\mathcal{H}\right)$ have closed ranges. Then, $A\overline{\le }B$ if and only if there exist projections P and $Q\in \mathcal{B}\left(\mathcal{H}\right)$ with $\mathcal{R}\left(P^{*}\right)=\mathcal{R}\left(A^{*}\right)$ and $\mathcal{R}\left(Q\right)=\mathcal{R}\left(A\right)$, respectively, such that

$$A=BP=QB=QBP.$$

Proof. 

Assume that $A\overline{\le }B$. It follows from Theorem 10 that $B=A+(I-AG)B(I-GA)$ for some g-inverse G of A. It follows from $AGA=A$ that $AG$ is a projection on $\mathcal{R}\left(A\right)$. In fact, by Lemma 1, we have that

$$\mathcal{R}\left(AG\right)\subseteq \mathcal{R}\left(A\right)=\mathcal{R}\left(AGA\right)\subseteq \mathcal{R}\left(AG\right),$$

which leads to

$$\mathcal{R}\left(AG\right)=\mathcal{R}\left(A\right).$$

Thus, $I-AG$ is a projection on $\mathcal{N}\left(A^{*}\right)$. Let $Q\in \mathcal{B}\left(\mathcal{H}\right)$ be a projection on $\mathcal{R}\left(A\right)$. Then, we have that

$$QB=QA+Q(I-AG)B(I-GA)=QA=A.$$

Similarly, ${(I-GA)}^{*}$ is a projection on $\mathcal{N}\left(A\right)$. Now take the adjoint of both sides for equation $B=A+(I-AG)B(I-GA)$. Similar to the proof above, let $P^{*}\in \mathcal{B}\left(\mathcal{H}\right)$ be a projection on $\mathcal{R}\left(A^{*}\right)$; then, we have that

$$P^{*}{B}^{*}=P^{*}{A}^{*}+P^{*}{(I-GA)}^{*}{B}^{*}{(I-AG)}^{*}=P^{*}{A}^{*}=A^{*},$$

that is, $BP=A.$ Additionally, it concludes that $QBP=AP=A$.

Conversely, let A and B have operator matrix forms (8) and (12), respectively. Let P and Q have operator matrix forms as follows:

$$Q=\left(\begin{array}{cc}I& Q_0\\ 0& 0\end{array}\right):\left(\begin{array}{c}\mathcal{R}\left(A\right)\\ \mathcal{N}\left(A^{*}\right)\end{array}\right)⟶\left(\begin{array}{c}\mathcal{R}\left(A\right)\\ \mathcal{N}\left(A^{*}\right)\end{array}\right)$$

and

$$P=\left(\begin{array}{cc}I& 0\\ P_0& 0\end{array}\right):\left(\begin{array}{c}\mathcal{R}\left(A^{*}\right)\\ \mathcal{N}\left(A\right)\end{array}\right)⟶\left(\begin{array}{c}\mathcal{R}\left(A^{*}\right)\\ \mathcal{N}\left(A\right)\end{array}\right),$$

respectively. From $A=QB$, one obtains that

$$\left(\begin{array}{cc}{A}_{11}& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}I& Q_0\\ 0& 0\end{array}\right)\left(\begin{array}{cc}{B}_{11}& B_{12}\\ B_{21}& B_{22}\end{array}\right)⟹\left\{\begin{array}{c}{A}_{11}=B_{11}+Q_0{B}_{21},\hfill \\ B_{12}+Q_0{B}_{22}=0.\hfill \end{array}\right.$$

From $A=BP$, one obtains

$$\left(\begin{array}{cc}{A}_{11}& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}{B}_{11}& B_{12}\\ B_{21}& B_{22}\end{array}\right)\left(\begin{array}{cc}I& 0\\ P_0& 0\end{array}\right)⟹\left\{\begin{array}{c}{A}_{11}=B_{11}+B_{12}{P}_0,\hfill \\ B_{21}+B_{22}{P}_0=0.\hfill \end{array}\right.$$

It holds that

$$B_{11}=A_{11}+Q_0{B}_{22}{P}_0,B_{12}=-Q_0{B}_{22},B_{21}=-B_{22}{P}_0.$$

Take

$$\begin{array}{ccc}\hfill G& =& \left(\begin{array}{cc}{A}_{11}^{-1}& A_{11}^{-1}{Q}_0\\ P_0{A}_{11}^{-1}& P_0{A}_{11}^{-1}{Q}_0\end{array}\right).\hfill \end{array}$$

Then it is easy to check that G is a g-inverse of A and

$$AG=BG\mathrm{and}GA=GB.$$

Therefore, we obtain that $A\overline{\le }B$, which completes the proof. □

The following corollary is immediate from Theorem 10 and Lemma 1.

Corollary 4.

Let A and $B\in \mathcal{B}\left(\mathcal{H}\right)$ have closed ranges. If $A\overline{\le }B$, then

$$\mathcal{R}\left(A\right)\subseteq \mathcal{R}\left(B\right)\mathit{and}\mathcal{R}\left(A^{*}\right)\subseteq \mathcal{R}\left(B^{*}\right).$$

In fact, it follows from ([5], Corollary 4.14) that the result in Corollary 4 also holds for general bounded operators on $\mathcal{B}\left(\mathcal{H}\right)$.

Definition 4.

Let A and $B\in \mathcal{B}\left(\mathcal{H}\right)$ have closed ranges. Then $A\overline{\le }B$ if and only if

$$A\mathit{and}B-A\mathit{are}\mathit{parallel}\mathit{summable},\mathit{and}\mathit{the}\mathit{parallel}\mathit{sum}A:(B-A)=0.$$

Proof. 

It follows from $A\overline{\le }B$ and Corollary 4 that

$$\mathcal{R}\left(A\right)\subseteq \mathcal{R}(A+B-A)=\mathcal{R}\left(B\right)\mathrm{and}\mathcal{R}\left(A^{*}\right)\subseteq \mathcal{R}\left({(A+B-A)}^{*}\right)=\mathcal{R}\left(B^{*}\right),$$

which implies that A and $B-A$ are parallel summable according to Definition 2.

Then, it holds that

$$A:(B-A)=A{(A+B-A)}^{†}(B-A)=A{B}^{†}(B-A)=A{B}^{†}B-A{B}^{†}A.$$

(18)

It follows from $\mathcal{R}\left(A^{*}\right)\subseteq \mathcal{R}\left(B^{*}\right)$ that

$$A=A{B}^{†}B.$$

(19)

It follows from Theorem 10 and the definition of the Moore-Penrose inverse that

$$A{B}^{†}A=QB{B}^{†}BP=QBP=A,$$

(20)

where projections $Q,P\in \mathcal{B}\left(\mathcal{H}\right)$ satisfy $\mathcal{R}\left(Q\right)=\mathcal{R}\left(A\right)$, $\mathcal{R}\left(P^{*}\right)=\mathcal{R}\left(A^{*}\right)$, respectively. Combining Equations (18)–(20), we obtain that $A:(B-A)=0$.

For the converse, assume that A and $B-A$ are parallel summable. Then, it follows from Definition 2 that

$$\mathcal{R}\left(A\right)\subseteq \mathcal{R}\left(B\right)\mathrm{and}\mathcal{R}\left(A^{*}\right)\subseteq \mathcal{R}\left(B^{*}\right).$$

Thus, it follows from $\mathcal{R}\left(A^{*}\right)\subseteq \mathcal{R}\left(B^{*}\right)$ that $A{B}^{†}B=A$. Combined with the condition that $A:(B-A)=0$, it leads to $A{B}^{†}A=A$. Let $G=B^{†}A{B}^{†}$. Then $G\in \left\{A^{-}\right\}$. In fact,

$$A{B}^{†}A{B}^{†}A=A{B}^{†}A=A.$$

By computation, we obtain that

$$GA=B^{†}A{B}^{†}A=B^{†}A=B^{†}A{B}^{†}B=GB$$

and

$$AG=A{B}^{†}A{B}^{†}=A{B}^{†}=B{B}^{†}A{B}^{†}=BG,$$

which implies that $A\overline{\le }B$. □

Remark 6.

The following counterexample shows that the condition A and $B-A$ are parallel summable alone does not imply that $A\overline{\le }B$. Consider

$$B=\left(\begin{array}{cc}I& 0\\ 0& -I\end{array}\right)\mathit{and}A=\left(\begin{array}{cc}\frac{1}{2}I& \frac{\sqrt{3}}{2}I\\ \frac{\sqrt{3}}{2}I& -\frac{1}{2}I\end{array}\right).$$

Note that A and B are self-adjoint. It is easy to see that $\mathcal{R}\left(A\right)=\mathcal{R}\left(B\right)=\mathcal{H}\oplus \mathcal{H}$, which shows that A and $B-A$ are parallel summable. Furthermore, $A^{-1}=A$. Hence, by direct computation, we obtain that

$$A^{-1}A=A^2=I\ne A^{-1}B,$$

which implies that $A\overline{\le }B$ is not valid. In fact, it holds that

$$\begin{array}{ccc}& & A:(B-A)\hfill \\ & =& A{B}^{-1}(B-A)\hfill \\ & =& \left(\begin{array}{cc}\frac{1}{2}I& \frac{\sqrt{3}}{2}I\\ \frac{\sqrt{3}}{2}I& -\frac{1}{2}I\end{array}\right)\left(\begin{array}{cc}I& 0\\ 0& -I\end{array}\right)\left(\begin{array}{cc}\frac{1}{2}I& -\frac{\sqrt{3}}{2}I\\ -\frac{\sqrt{3}}{2}I& -\frac{1}{2}I\end{array}\right)\hfill \\ & =& \left(\begin{array}{cc}I& 0\\ 0& -I\end{array}\right).\hfill \end{array}$$

5. Conclusions

The parallel sum has many applications in the fields of electrical network, statistics and control theory, and so on. We present a perturbation estimation for the parallel sum of positive operators with closed ranges, and then give a characterization of the minus partial order through parallel sum operation. These observations are generalizations of the corresponding results obtained in the matrix cases. In addition, there are still many interesting properties of the parallel sum of bounded operators on infinite dimensional Hilbert spaces, and this deserves further research.