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Abstract and Applied Analysis
Volume 2012, Article ID 629178, 22 pages
http://dx.doi.org/10.1155/2012/629178
Research Article

Perturbation Bound of the Group Inverse and the Generalized Schur Complement in Banach Algebra

1Faculty of Science, Guangxi University for Nationalities and Guangxi Key Laboratory of Hybrid Computational and IC Design Analysis, Nanning 530006, China
2College of Mathematics and Computer Science, Guangxi University for Nationalities, Nanning 530006, China
3Department of Computer Science, Fudan University, Shanghai 200433, China

Received 5 April 2012; Revised 26 June 2012; Accepted 11 July 2012

Academic Editor: Patricia J. Y. Wong

Copyright © 2012 Xiaoji Liu et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We investigate the relative perturbation bound of the group inverse and also consider the perturbation bound of the generalized Schur complement in a Banach algebra.

1. Introduction

Let 𝒜 denote a Banach algebra with unit 1. The symbols 𝒜1, 𝒜𝐷, 𝒜𝑑, 𝒜𝑔, 𝒜nil, 𝒜qnil, and 𝒜 stand for the sets of all invertible, Drazin invertible, generalized Drazin invertible, group invertible, nilpotent, quasinilpotent, and idempotent elements of a Banach algebra 𝒜, respectively.

Some definitions will be given in the following.

Letting 𝑎𝒜𝐷, there is an unique element 𝑥𝒜 such that 𝑎𝑘+1𝑥=𝑎𝑘,𝑥𝑎𝑥=𝑥,𝑎𝑥=𝑥𝑎.(1.1) Then 𝑥 is called the Drazin inverse of 𝑎, denoted by 𝑎𝐷. The smallest nonnegative integer 𝑘 which satisfies (1.1) is called the index of 𝑎, denoted by Ind(𝑎)=𝑘. If Ind(𝑎)1, then 𝑎𝐷=𝑎(or𝑎𝑔).

Let 𝑎𝒜, if the conditions (1.1) are replaced by 𝑎𝑥𝑎=𝑎,𝑥𝑎𝑥=𝑥,𝑎𝑥=𝑥𝑎.(1.2) Then 𝑥 is called the group inverse of 𝑎, denoted by 𝑥=𝑎. If the conditions (1.1) are replaced by 𝑥𝑎𝑥=𝑥,𝑎𝑥=𝑥𝑎,𝑎(1𝑎𝑥)isquasinilpotent.(1.3) Then 𝑥 is called the generalized Drazin inverse of 𝑎, denoted by 𝑥=𝑎𝑑.

Some notations of the Schur complement are given in the following.

For a 2×2 block complex matrix 𝑀 is defined as 𝑀=𝐴𝐵𝐶𝐷,(1.4) where 𝐴𝐶𝑚×𝑚, 𝐷𝐶𝑝×𝑝, 𝐵𝐶𝑚×𝑝, and 𝐶𝐶𝑝×𝑚. If 𝐴 is nonsingular, then the classical Schur complement of 𝐴 in 𝑀 is given as follows (see [1]): 𝑆=𝐷𝐶𝐴1𝐵.(1.5)

In [2], Benítez and Thome considered the expression 𝐴𝑁=+𝐴𝐵𝑆𝐶𝐴𝐴𝐵𝑆𝑆𝐶𝐴𝑆,(1.6) and 𝑁 is called the generalized Schur form of the matrix 𝑀 given in (1.4) being 𝑆=𝐷𝐶𝐴𝐵 for some fixed generalized inverses 𝐴𝐴{1}, 𝑆𝑆{1}, where 𝑆 is called generalized Schur complement of 𝐴 in 𝑀. In [2, Theorem 2], Benítez and Thome investigated the expression of the group inverse of 𝑀 in (1.4) by the generalized Schur complement, where (1.5) is replaced by 𝑆=𝐷𝐶𝐴𝐵.(1.7) Similar results also were given by Sheng and Chen in [3, Theorem 3.2]. The Drazin inverse of a 2 × 2 block complex square matrix in (1.4) with a singular generalized Schur complement was considered in [46], where 𝑆=𝐴𝐶𝐴𝐷𝐷.(1.8) For the expression of a 2 × 2 block operator matrix was investigated by Deng and Wei in [7].

Some notations for the block matrix form of a given element 𝑎𝒜 are introduced in [8]. Let 𝑎𝒜 and 𝑠𝒜 (see [8, Chapter VII]) which denotes the set of all idempotent elements in 𝒜. Then we write 𝑎=𝑠𝑎𝑠+𝑠𝑎(1𝑠)+(1𝑠)𝑎𝑠+(1𝑠)𝑎(1𝑠)(1.9) and use the notations 𝑎11=𝑠𝑎𝑠,𝑎12=𝑠𝑎(1𝑠),𝑎21=(1𝑠)𝑎𝑠,𝑎22=(1𝑠)𝑎(1𝑠).(1.10) For a representation of arbitrary element 𝑎𝒜 is given as the following matrix form: (=𝑎𝑎=𝑠𝑎𝑠𝑠𝑎(1𝑠)1𝑠)𝑎𝑠(1𝑠)𝑎(1𝑠)11𝑎12𝑎21𝑎22𝑠.(1.11)

In this paper, we will consider some results on the relative perturbation bounds of group inverse and also give the perturbation bounds of the generalized Schur complement of an element 𝑎𝒜 under some certain conditions in a Banach algebra.

2. Perturbation Bound of (𝑎+𝑏)𝑎𝑑 in Banach Algebra

In recent years, perturbation theory for the Drazin inverse of a given matrix 𝐴𝐶𝑛×𝑛 and its applications have been considered in [920]. In [12], Yimin and Guorong gave a perturbation result for the Drazin inverse under condition (𝒲) (see [12] for details). In [8, Ch 5], Djordjević and Rakočević extended the perturbation bound of Yimin and Guorong [12] to Banach algebra. In [13], Wei had discussed the upper perturbation bound of 𝐵𝐴/𝐴 with 𝐵=𝐴+𝐸 and had answered the question of Campbell and Meyer [21] when Ind(𝐴)=1. In [14], Wei and Wu presented the perturbation upper bounds of 𝐵𝐷𝐴𝐷/𝐴𝐷 under the weaker condition corerank𝐵=corerank𝐴 and completely answered the question of Campbell and Meyer in [21]. In [16], Wei derived a relative perturbation upper bound of 𝐵𝐴𝐷/𝐴𝐷 by Jordan canonical of 𝐴. In [5], Li gave sharper upper bounds for 𝐵𝐴𝐷 under weaker conditions: rank(𝐵)=rank(𝐴𝑘) and 𝐴𝐷𝐸<1/(1+𝐴𝐷𝐴). In [17], Wei et al. derived constructive perturbation bound of the Drazin inverse of a square matrix by using a technique proposed by Stewart and based on perturbation theory for invariant subspaces. In [18], Xu et al. gave some upper bounds for 𝐵𝐷𝐴𝐷/𝐴𝐷 only under the condition that 𝐵 is a stable perturbation of 𝐴. In [22], González and Koliha investigated the perturbation of the Drazin inverse of a closed linear operator and derived explicit bounds for the perturbations under certain restrictions on the perturbing operators. In [23], González and Vélez-Cerrada analyzed the perturbation of the Drazin inverse and also gave explicit upper bounds of 𝐵𝐴𝐷 and 𝐵𝐵𝐴𝐴𝐷 and obtained a result on the continuity of the group inverse for operators on Banach space.

In this section, we will investigate the relative perturbation bound of the group inverse in Banach algebra.

At first, we will give some concepts and lemmas as follows.

For 𝑎𝒜𝑑, let 𝑝=𝑎𝑎𝑑 and 𝑝𝒜 (see [24]): 𝑎𝑎=100𝑎2𝑝,𝑎𝑑=𝑎11000𝑝,𝑎𝜋=1𝑝=0001𝑝𝑝,(2.1) where 𝑎1𝑝𝒜𝑝 is invertible and 𝑎2(1𝑝)𝒜(1𝑝) is quasinilpotent.

For any 𝑎𝒜𝑑, we write 𝜎(𝑎), 𝜌(𝑎), and 𝑟(𝑎) for the spectrum, the resolvent set, and the spectral radius of 𝑎, respectively. For 𝜆𝜌(𝑎) and let 𝑅(𝜆,𝑎)=(𝜆𝑎)1. If 0 is an isolated point of 𝜎(𝑎), then the spectral idempotent corresponding to the set {0} is defined by 𝑎𝜋=12𝜋𝑖𝛾𝑅(𝜆,𝑎)𝑑𝜆,(2.2) where 𝛾 is a small circle surrounding 0 and separating 0 from 𝜎(𝑎)/{0}.

Some lemmas will be useful for the following proof in this paper.

Lemma 2.1 (see [24, Theorem  2.3]). Let 𝑥,𝑦𝒜, and let 𝑝𝒜. Assume that 𝑥=𝑎𝑐0𝑏𝑝,𝑦=𝑏0𝑐𝑎1𝑝.(2.3)(i)If 𝑎(𝑝𝒜𝑝)𝑑 and 𝑏((1𝑝)𝒜(1𝑝))𝑑, then 𝑥,𝑦𝒜𝑑 and 𝑥𝑑=𝑎𝑑𝑢0𝑏𝑑𝑝,𝑦𝑑=𝑏𝑑0𝑢𝑎𝑑1𝑝,(2.4) where 𝑢=𝑛=0(𝑎𝑑)𝑛+2𝑐𝑏𝑛𝑏𝜋+𝑛=0𝑎𝜋𝑎𝑛𝑐(𝑏𝑑)𝑛+2𝑎𝑑𝑐𝑏𝑑. (ii)If 𝑥𝒜𝑑 and 𝑎(𝑝𝒜𝑝)𝑑, then 𝑏[(1𝑝)𝒜(1𝑝)]𝑑 and 𝑥𝑑 is given by (2.4).

Lemma 2.2 (see [24, Corollary  3.4]). If 𝑎,𝑏𝒜 are generalized Drazin invertible, 𝑏 is quasinilpotent, and 𝑎𝑏=0, then 𝑎+𝑏 is generalized Drazin invertible and (𝑎+𝑏)𝑑=𝑛=0𝑏𝑛𝑎𝑑𝑛+1.(2.5)

The following lemma is a generalization of [25, Theorem 1].

Lemma 2.3. Let 𝑎,𝑏𝒜𝑑 such that 𝑎𝑏=𝑏𝑎. Then 𝑎+𝑏𝒜𝑑 if and only if 1+𝑎𝑑𝑏𝒜𝑑. In this case (𝑎+𝑏)𝑑=𝑎𝑑1+𝑎𝑑𝑏𝑑𝑏𝑏𝑑+𝑛=0𝑏𝜋(𝑏)𝑛𝑎𝑑𝑛+1+𝑛=0𝑏𝑑𝑛+1(𝑎)𝑛𝑎𝜋.(2.6)

Now we will state a lemma for the representation of the group inverse of an element 𝑎𝒜𝑑 with 2×2 block form in Banach algebra (see [26, Theorem 7.7.7] and [23, Theorem 2.2.] which were established for a finite dimensional case and partitioned operators matrix, resp.).

Lemma 2.4. Let 𝑧𝒜, and it has the block matrix form as 𝑧=𝑧1𝑧12𝑧21𝑧2𝑝, where 𝑝𝒜 is an idempotent element, 𝑧1 is invertible in 𝑝𝒜𝑝, and 𝑧2=𝑧21𝑧11𝑧12. Let 𝛿=𝑝+𝑧11𝑧12𝑧21𝑧11. Then 𝑧 is group invertible if and only if 𝛿 is an invertible element in 𝑝𝒜𝑝. In this case 𝑧=(𝛿𝑧1𝛿)1(𝛿𝑧1𝛿)1𝑧11𝑧12𝑧21𝑧11(𝛿𝑧1𝛿)1𝑧21𝑧11(𝛿𝑧1𝛿)1𝑧11𝑧12𝑝,𝑧𝜋=𝑝𝛿1𝛿1𝑧11𝑧12𝑧21𝑧11𝛿11𝑝𝑧21𝑧11𝛿1𝑧11𝑧12𝑝.(2.7)

Let 𝑏𝒜 be a perturbation element of 𝑎. According to (2.1), we obtain 𝑏𝑏=1𝑏12𝑏21𝑏2𝑝𝑎,𝑎+𝑏=1+𝑏1𝑏12𝑏21𝑎2+𝑏2𝑝,(2.8) where 𝑝=𝑎𝑎𝑑.

Theorem 2.5. Let 𝑎𝒜𝑑 and 𝑏𝒜 be a perturbation element of 𝑎, 𝑎 and 𝑎+𝑏 which are defined as (2.1) and (2.8), respectively. If 𝑎𝑑𝑏𝑎𝑎𝑑<1, then 𝑎1+𝑏1 is invertible in subalgebra 𝑝𝒜𝑝. Furthermore let 𝑎2+𝑏2=𝑏21(𝑎1+𝑏1)1𝑏12 and 𝛿=𝑝+[𝑝(𝑎+𝑏)𝑝]𝑑𝑏(1𝑝)𝑏[𝑝(𝑎+𝑏)𝑝]𝑑𝑝𝒜𝑝. Then 𝑎+𝑏 is group invertible if and only if 𝛿𝑝𝒜𝑝 is invertible and 𝛿 is invertible if and only if [𝑝(𝑎+𝑏)𝑝]2+𝑝𝑏(1𝑝)𝑏𝑝𝑝𝒜𝑝 is invertible. In this case, (𝑎+𝑏)𝑎𝑑𝑎𝑑𝑇21𝑎111𝑏1𝑏12+𝑏21+𝑇21𝑎113𝑎111𝑏14𝑏12𝑏21+𝑇1𝑇2+𝑇2+𝑎11𝑏1𝑎111𝑏1𝑇1𝑇1,(2.9) where 𝑎1=𝑝𝑎𝑝,𝑏1=𝑝𝑏𝑝,𝑏12=𝑝𝑏(1𝑝),𝑏21𝑇=(1𝑝)𝑏𝑝,1=𝑎1+𝑏12𝑎1+𝑏12+𝑏12𝑏211,𝑇2=𝑎11𝑏12𝑏21𝑎111𝑏1.(2.10)

Proof. Let 𝑝=𝑎𝑎𝑑. Then 𝑎, 𝑎𝑑, 𝑏 and 𝑎+𝑏 have the matrix form as (2.1) and (2.8), respectively, where 𝑎1 is invertible in 𝑝𝒜𝑝 and 𝑎2 is quasinilpotent in (1𝑝)𝒜(1𝑝).
It follows from the hypothesis 𝑎𝑑𝑏𝑎𝑎𝑑<1 that 𝑎11𝑏1<1. Thus, it implies that 𝑝+𝑎11𝑏1𝑝𝒜𝑝 is invertible. It is easy to see that (𝑎1+𝑏1)1=(𝑝+𝑎11𝑏1)1𝑎11. Let 𝛿=𝑝+[𝑝(𝑎+𝑏)𝑝]𝑑𝑏(1𝑝)𝑏[𝑝(𝑎+𝑏)𝑝]𝑑𝑝𝒜𝑝; that is, we have 𝛿=𝑝+(𝑎1+𝑏1)1𝑏12𝑏21(𝑎1+𝑏1)1. Therefore, we have 𝑎𝛿=1+𝑏11𝑎1+𝑏1𝑝𝑎1+𝑏1+𝑏12𝑏21𝑎1+𝑏11=𝑎1+𝑏11𝑎1+𝑏12+𝑏12𝑏21𝑎1+𝑏11=[]𝑝(𝑎+𝑏)𝑝𝑑[]𝑝(𝑎+𝑏)𝑝2[]+𝑝𝑏(1𝑝)𝑏𝑝𝑝(𝑎+𝑏)𝑝𝑑.(2.11) From the previous equations, we get that 𝛿 is invertible if and only if [𝑝(𝑎+𝑏)𝑝]2+𝑝𝑏(1𝑝)𝑏𝑝 is invertible. Since 𝑎2+𝑏2=𝑏21(𝑎1+𝑏1)1𝑏12 and by Lemma 2.4, we obtain that 𝑎+𝑏 is group invertible if and only if 𝛿𝑝𝒜𝑝 is invertible.
In the following, we consider the upper bound of (𝑎+𝑏)𝑎𝑑/𝑎𝑑.
Applying Lemma 2.4, we obtain (𝑎+𝑏)=𝑎𝜂𝜂1+𝑏11𝑏12𝑏21𝑎1+𝑏11𝜂𝑏21𝑎1+𝑏11𝜂𝑎1+𝑏11𝑏12𝑝,(2.12) where 𝜂=(𝛿(𝑎1+𝑏1)𝛿)1.
Note that 𝜂𝑎11=𝛿𝑎1+𝑏11𝛿1𝑎11=𝛿1𝑎1+𝑏11𝛿1𝑎11=𝛿1𝑎11𝛿1+𝛿1𝑛=1𝑎11𝑏1𝑛𝑎11𝛿1𝑎11=𝛿1𝑎11𝛿𝑎11𝛿𝛿1+𝛿1𝑛=1𝑎11𝑏1𝑛𝑎11𝛿1=𝛿1𝑎11(𝑝+𝜃)𝑎11𝛿(𝑝+𝜃)1+𝛿1𝑛=1𝑎11𝑏1𝑛𝑎11𝛿1=𝛿1𝑎11𝑝𝑎11𝑝+𝜃𝑎11𝑝+𝑝𝑎11𝜃+𝜃𝑎11𝜃𝛿1+𝛿1𝑛=1𝑎11𝑏1𝑛𝑎11𝛿1=𝛿1𝜃𝑎11𝑝+𝑝𝑎11𝜃+𝜃𝑎11𝜃𝛿1+𝛿1𝑛=1𝑎11𝑏1𝑛𝑎11𝛿1=𝛿1𝜃𝑎11𝑝+𝛿𝑎11𝜃𝛿1+𝛿1𝑛=1𝑎11𝑏1𝑛𝑎11𝛿1=𝛿1𝜃𝑎11𝛿1𝑎11𝜃𝛿1+𝛿1𝑛=1𝑎11𝑏1𝑛𝑎11𝛿1,𝛿(2.13)1=𝑎𝑝+1+𝑏11𝑏12𝑏21𝑎1+𝑏111𝑎1+𝑏12𝑎1+𝑏12+𝑏12𝑏211=𝑇1,(2.14) where 𝜃=(𝑎1+𝑏1)1𝑏12𝑏21(𝑎1+𝑏1)1.
It shows from 𝑎𝑑𝑏𝑎𝑎𝑑<1 (i.e., 𝑎11𝑏1<1) that 𝑎𝜃=1+𝑏11𝑏12𝑏21𝑎1+𝑏11𝑎11𝑏12𝑏21𝑎111𝑏1=𝑇2.(2.15)
From (2.13), (2.14), (2.15), and by 𝑎11𝑏1<1, we obtain that 𝜂𝑎11=𝛿1𝜃𝑎11𝛿1𝑎11𝜃𝛿1+𝛿1𝑛=1𝑎11𝑏1𝑛𝑎11𝛿1𝛿1𝜃𝑎11𝛿1+𝑎11𝜃𝛿1+𝛿1𝑛=1𝑎11𝑏1𝑛𝑎11𝛿1𝑇1𝑇2+𝑇2+𝑎11𝑏1𝑎111𝑏1𝑇1𝑇1𝑎11.(2.16)
It follows from (2.12) that (𝑎+𝑏)𝑎𝑑=𝜂𝑎11𝜂𝑎1+𝑏11𝑏12𝑏21𝑎1+𝑏11𝜂𝑏21𝑎1+𝑏11𝜂𝑎1+𝑏11𝑏12𝑝.(2.17)
Therefore, according to (2.14), (2.15), and (2.16), we obtain (𝑎+𝑏)𝑎𝑑=𝜂𝑎11𝜂(𝑎1+𝑏1)1𝑏12𝑏21𝑎1+𝑏11𝜂𝑏21𝑎1+𝑏11𝜂𝑎1+𝑏11𝑏12𝑝𝜂𝑎11+𝜂𝑎1+𝑏11𝑏12+𝑏21𝑎1+𝑏11𝜂+𝑏21𝑎1+𝑏11𝜂𝑎1+𝑏11𝑏12𝑎11𝑇21𝑎111𝑏1𝑏12+𝑏21+𝑇21𝑎11𝑎111𝑏14𝑏12𝑏21+𝑇1𝑇2+𝑇2+𝑎11𝑏1𝑎111𝑏1𝑇1𝑇1𝑎11.(2.18) Since 𝑎11=𝑎𝑑 and by (2.18), it is easy to see that the conclusion holds.
Thus, we complete the proof.

Let 𝐴,𝐸𝐵(𝑋) be both bounded linear operators with 𝐵=𝐴+𝐸 on Banach space, where 𝑋 denotes Banach space. If 𝐴𝐷𝐸+𝐵𝜋𝐴𝜋<1 is satisfied, (it implies that 𝐴𝐷𝐸<1 and 𝐴𝐷𝐸𝐴𝐴𝐷<1), then we have the remark.

Remark 2.6 (see [23, Theorem  4.2]). Let 𝐴,𝐵𝐵(𝑋) be Drazin invertible and group invertible, respectively. If 𝐴𝐷𝐸+𝐵𝜋𝐴𝜋<1, then 𝐵#𝐴𝐷𝐴𝐷𝐴𝐷𝐸+2𝐵𝜋𝐴𝜋𝐴1𝐷𝐸𝐵𝜋𝐴𝜋.(2.19)

Let 𝑎=𝑎1𝑎2 and 𝑎𝑑=[𝑎1]𝑝1=𝑎#1; if we put 𝛿𝑎=𝑏+𝑎2, then 1+𝛿𝑎𝑎 is invertible in 𝒜 when 𝑎𝑑𝑏𝑎𝑎𝑑<1. From the Proposition 2.2 (5) of [20], we have 𝑎2+𝑏2=𝑏21(𝑎1+𝑏1)1𝑏12 when (𝑎1+𝛿𝑎)𝒜(1𝑎𝑎𝑑)𝒜={0} for [𝑎1]𝑝[𝑎1]𝑝1=𝑝=𝑎𝑎𝑑. Therefore, for 𝑎#1𝑏+𝑎2<(1+1𝑎𝑎𝑑)1, we arrive at [20, Theorem 4.2]. In fact, the following remark implies that Theorem 2.5 improves the upper bound of (𝑎+𝑏)#𝑎# of [20, Theorem 4.2].

Remark 2.7 (see [20, Theorem  4.2]). Let 𝑎𝐺(𝒜) and let 𝑎=𝑎+𝛿𝑎𝒜 with 𝒦#𝜖𝑎<(1+𝑎𝜋). Assume that 𝑎𝒜(1𝑎𝑎#)𝒜={0}. Then 𝑎𝐺(𝒜) and 𝑎#𝑎#(1+2𝑎𝜋)𝑎#𝒦#(𝑎)𝜖𝑎1(1+𝑎𝜋)𝑎#𝒦#(𝑎)𝜖𝑎2,(2.20) where 𝒦#=𝑎𝑎# and 𝜖𝑎=𝛿𝑎𝑎1.

Theorem 2.8. Let 𝑎,𝑏𝒜 be generalized Drazin invertible and satisfy the conditions 𝑎𝑑𝑏𝑎𝑎𝑑<1,𝑎𝜋𝑏𝑎=0.(2.21) Then (𝑎+𝑏) exists if and only if 𝑎𝜋(𝑎+𝑏) is group invertible. In this case, (𝑎+𝑏)𝑎𝑑𝑎𝑑𝑎𝑑𝑎1𝑑𝑏2𝑏+𝑏𝑎𝜋𝑛=0𝑎𝑛+1𝑏𝑑𝑛+1+𝑎𝑑𝑎1𝑑𝑏2𝑏𝑎𝜋𝑏+𝑏𝑎𝜋𝑎1𝑑𝑏𝑛=0𝑎𝑛𝑏𝑑𝑛+1+𝑎𝜋𝑎𝑑1𝑛=0𝑎𝑛𝑏𝑑𝑛+1+𝑎𝑑𝑏𝑎1𝑑𝑏.(2.22)

Proof. Since 𝑎𝑑 exists, 𝑎𝑑 is defined as (2.1). Let 𝑏 have the block matrix form as 𝑏𝑏=1𝑏3𝑏4𝑏2𝑝.(2.23)
Applying the condition 𝑎𝜋𝑏𝑎=0, we have 𝑏2𝑎2=0 and 𝑎𝜋𝑏𝑎𝑎𝑑𝑏𝑎=004𝑎10𝑝=0.(2.24) It follows from (2.24) that 𝑎1𝑝𝒜𝑝 is invertible, 𝑏4=0, and 𝑏𝑏=1𝑏30𝑏2𝑝.(2.25) Combining (2.1) and (2.25), we obtain 𝑎𝑎+𝑏=1+𝑏1𝑏30𝑎2+𝑏2𝑝.(2.26)
The condition 𝑎𝑑𝑏𝑎𝑎𝑑<1 implies 𝑎11𝑏1<1 in the subalgebra 𝑝𝒜𝑝. Therefore, we conclude that 𝑎1+𝑏1𝑝𝒜𝑝 is invertible and Ind(𝑎1+𝑏1)=0. According to (2.26) and by Lemma 2.1, one observes that (𝑎+𝑏)𝑑 exists if and only if (𝑎2+𝑏2)𝑑 also. Thus, (𝑎+𝑏) exists if and only if 𝑎𝜋(𝑎+𝑏) is group invertible.
If 𝑎𝜋(𝑎+𝑏) is group invertible and by Lemma 2.1, we obtain (𝑎+𝑏)=𝑎1+𝑏11𝑥0𝑎2+𝑏2𝑝,(2.27) where 𝑥=(𝑎1+𝑏1)2𝑏3(𝑎2+𝑏2)𝜋(𝑎1+𝑏1)1𝑏3(𝑎2+𝑏2).
Since 𝑏2𝑎2=0 and 𝑎2 is quasinilpotent, by Lemma 2.2, we obtain 𝑎2+𝑏2=𝑛=0𝑎𝑛2𝑏𝑑2𝑛+1=𝑎𝜋𝑛=0𝑎𝑛𝑏𝑑𝑛+1.(2.28)
From 𝑎𝑑𝑏𝑎𝑎𝑑<1, one easily has 𝑎1+𝑏110=𝑛=0𝑎11𝑏1𝑛𝑎110=𝑛=0𝑎𝑑𝑏𝑛𝑎𝑑=𝑎111+𝑏1𝑎1110=𝑎𝑑𝑛=0𝑏𝑎𝑑𝑛.(2.29) It follows from (2.27) and (2.29) that 𝑎𝑥=1+𝑏12𝑏3𝑎2+𝑏2𝜋𝑎1+𝑏11𝑏3𝑎2+𝑏2=𝑛=0𝑎𝑑𝑏𝑛𝑎𝑑2𝑏𝑏𝑎𝜋𝑛=0𝑎𝑛+1𝑏𝑑𝑛+1+𝑛=0𝑎𝑑𝑏𝑛𝑎𝑑2𝑏𝑎𝜋𝑏𝑛=0𝑎𝑑𝑏𝑛𝑎𝑑𝑏𝑎𝜋𝑛=0𝑎𝑛𝑏𝑑𝑛+1.(2.30)
Combining (2.27), (2.28), and (2.29), we obtain (𝑎+𝑏)=𝑛=0𝑎11𝑏1𝑎11𝑛𝑥0𝑛=0𝑎𝑛2𝑏𝑑2𝑛+1𝑝=𝑛=0𝑎𝑑𝑏𝑎𝑑𝑛2𝑏𝑏𝑎𝜋𝑛=0𝑎𝑛+1𝑏𝑑𝑛+1+𝑛=0𝑎𝑑𝑏𝑛𝑎𝑑2𝑏𝑎𝜋𝑏𝑛=0𝑎𝑑𝑏𝑛𝑎𝑑𝑏𝑎𝜋𝑛=0𝑎𝑛𝑏𝑑𝑛+1+𝑛=0𝑎𝑑𝑏𝑛𝑎𝑑+𝑎𝜋𝑛=0𝑎𝑛𝑏𝑑𝑛+1.(2.31)
From (2.31), we derive (𝑎+𝑏)𝑎𝑑=𝑛=0𝑎𝑑𝑏𝑛𝑎𝑑2𝑏𝑏𝑎𝜋𝑛=0𝑎𝑛+1𝑏𝑑𝑛+1+𝑛=0𝑎𝑑𝑏𝑛𝑎𝑑2𝑏𝑎𝜋𝑏𝑛=0𝑎𝑑𝑏𝑛𝑎𝑑𝑏𝑎𝜋𝑛=0𝑎𝑛𝑏𝑑𝑛+1+𝑎𝜋𝑛=0𝑎𝑛𝑏𝑑𝑛+1+𝑛=1𝑎𝑑𝑏𝑛𝑎𝑑.(2.32) Moreover, by (2.32) we get (𝑎+𝑏)𝑎𝑑𝑎𝑑𝑎1𝑑𝑏2𝑏+𝑏𝑎𝜋𝑛=0𝑎𝑛+1𝑏𝑑𝑛+1+𝑎𝑑𝑎1𝑑𝑏2𝑏𝑎𝜋𝑎𝑏+𝑑𝑏𝑎𝜋𝑎1𝑑𝑏𝑛=0𝑎𝑛𝑏𝑑𝑛+1+𝑎𝜋𝑛=0𝑎𝑛𝑏𝑑𝑛+1+𝑎𝑑𝑎𝑑𝑏𝑎1𝑑𝑏.(2.33)
Finally, from (2.33) we easily finish the proof.

Corollary 2.9. Let 𝑎𝒜𝑔 and let 𝑏𝒜𝑑. If 𝑎,𝑏 satisfy the conditions 𝑎𝑏𝑎𝑎<1,𝑎𝜋𝑏𝑎=0,(2.34) then (𝑎+𝑏) exists if and only if 𝑎𝜋𝑏 is group invertible. In this case, (𝑎+𝑏)𝑎𝑎𝑏(𝑎𝜋𝑏)𝜋𝑎𝑎1𝑏2+𝑏𝑎𝜋𝑏𝑎1𝑏+𝑎𝜋𝑏𝑎+𝑎𝑏𝑎1𝑏.(2.35)

The conditions of Theorem 2.8  𝑎𝑑𝑏𝑎𝑎𝑑<1,𝑎𝜋𝑏𝑎=0 are weaker than the conditions (𝒲) (see [12, Theorem 3.2] for finite dimensional cases and [8, Theorem 5.3.2 and Corollary 5.3.3] for Banach algebra). According to 𝑎𝜋𝑏𝑎=0, we obtain that (2.26) holds. However, in view of (𝒲), we have 𝑎𝑎+𝑏=1+𝑏100𝑎2𝑝.(2.36) Thus, by the conditions (𝒲), we know that 𝑎 and 𝑎+𝑏 have the same Drazin invertible property (see [12, Theorem 3.1]). Thus, if 𝑎 is group invertible, then (𝑎+𝑏) is group invertible. It is easy to see that 𝑎𝑑𝑏𝑎𝑎𝑑<1, 𝑎𝜋𝑏𝑎=0 are weaker than the conditions (𝒲). From [8, Theorem 5.3.2 and Corollary 5.3.3], we easily state the following remark.

Remark 2.10. Let 𝑎𝒜𝑔 and let 𝑏𝒜𝑑. If 𝑎,𝑏 satisfy the condition (𝒲) 𝑎𝑏𝑎𝑎<1,𝑏=𝑎𝑎𝑏𝑎𝑎,(2.37) then 𝑎+𝑏 is group invertible and (𝑎+𝑏)𝑎𝑎𝑎𝑏𝑎1𝑏.(2.38)

Theorem 2.11. Let 𝑎,𝑏𝒜 be generalized Drazin invertible and satisfy the conditions 𝑎max𝑑𝑏𝑎𝑎𝑑,𝑎𝜋𝑎𝑎𝜋𝑏𝑑<1,𝑎𝜋𝑏𝑎=0.(2.39) Then (𝑎+𝑏) exists if and only if 𝑎𝜋(𝑎+𝑏) is group invertible. In this case, (𝑎+𝑏)𝑎𝑑𝑎𝑑𝑎𝑑𝑎1𝑑𝑏2𝑏+𝑏𝑎𝜋𝑏𝑎𝑑𝑏1𝑎𝑑+𝑎𝑑𝑎1𝑑𝑏2𝑏𝑎𝜋𝑏+𝑏𝑎𝜋𝑎1𝑑𝑏𝑏𝑎𝑑𝑏1𝑎𝑑+𝑎𝑑1𝑎𝜋𝑏𝑑𝑏1𝑎𝑑+𝑎𝑑𝑏𝑎1𝑑𝑏.(2.40)

Proof. The notations are taken as Theorem 2.8, and the rest of proof of theorem is similar to Theorem 2.8. Now, we only consider the perturbation of 𝑎2+𝑏2. From (2.28) and the first condition of (2.39), we have 𝑎2𝑏𝑑2<1 and 𝑎2+𝑏2=𝑛=0𝑎𝑛2𝑏𝑑2𝑛+1𝑎𝜋𝑏𝑑𝑏1𝑎𝑑.(2.41) Thus, from (2.41) we completed the proof.

Theorem 2.12. Let 𝑎,𝑏𝒜 be generalized Drazin invertible and satisfy the conditions 𝑎𝑑𝑏𝑎𝑎𝑑<1,𝑎𝜋𝑏𝑎=𝑎𝑏𝑎𝜋.(2.42) Then (𝑎+𝑏) exists if and only if 𝑎𝜋(𝑎+𝑏) is group invertible. In this case, (𝑎+𝑏)𝑎𝑑𝑎𝑑𝑎𝑑𝑏𝑎1𝑑𝑏+𝑎𝜋𝑎𝑑1𝑛=0𝑏𝑑𝑛+1𝑎𝑛.(2.43)

Proof. Letting 𝑝=𝑎𝑎𝑑, and it is similar to Theorem 2.8, we obtain that 𝑎, 𝑎𝑑, and 𝑝 have the matrix forms as (2.1). Here 𝑏 is taken as (2.23) in the proof of Theorem 2.8. The condition 𝑎𝜋𝑏𝑎=𝑎𝑏𝑎𝜋 implies that 𝑎𝜋𝑏𝑎=0001𝑝𝑝𝑏1𝑏3𝑏4𝑏2𝑝𝑎100𝑎2𝑝=𝑏004𝑎1𝑏2𝑎2𝑝,𝑎𝑏𝑎𝜋=𝑎100𝑎2𝑝𝑏1𝑏3𝑏4𝑏2𝑝0001𝑝𝑝=0𝑎1𝑏30𝑎2𝑏2𝑝.(2.44) Thus, according to (2.44), we obtain 𝑏4𝑎1=0, 𝑎1𝑏3=0 and 𝑎2𝑏2=𝑏2𝑎2. Because 𝑎1 is invertible in subalgebra 𝑝𝒜𝑝, we have 𝑏3=𝑏4=0. Thus, 𝑏, 𝑎+𝑏 have the matrix forms as follows: 𝑏𝑏=100𝑏2𝑝𝑎,𝑎+𝑏=1+𝑏100𝑎2+𝑏2𝑝.(2.45)
It follows from the condition 𝑎𝑑𝑏𝑎𝑎𝑑<1 that 𝑎11𝑏1<1. Thus, it shows from 𝑎11𝑏1<1 that 𝑎1+𝑏1 is invertible in subalgebra 𝑝𝒜𝑝. Therefore, easily we observe that 𝑎+𝑏 is Drazin invertible if and only if 𝑎2+𝑏2(1𝑝)𝒜(1𝑝) is Drazin invertible. That is, (𝑎+𝑏) exists if and only if 𝑎𝜋(𝑎+𝑏) is group invertible.
In the following, we will consider the perturbation of 𝑎2.
Let 𝑎2+𝑏2 be group invertible. The condition 𝑎𝜋𝑏𝑎=𝑎𝑏𝑎𝜋 implies that 𝑎2𝑏2=𝑏2𝑎2 holds. Since 𝑎2 is quasinilpotent in subalgebra (1𝑝)𝒜(1𝑝) and by Lemma 2.3, we get 𝑎2+𝑏2=𝑛=0𝑏𝑑2𝑛+1𝑎2𝑛=𝑎𝜋𝑛=0𝑏𝑑𝑛+1(𝑎)𝑛.(2.46)
By virtue of 𝑎11𝑏1<1, we get that 𝑎𝑎𝑑(𝑎+𝑏)1=𝑎1+𝑏11=𝑛=0𝑎11𝑏1𝑛𝑎11=𝑛=0𝑎𝑑𝑏𝑛𝑎𝑑.(2.47)
It follows from (2.46) and (2.47) that 𝑎𝑎𝑑(𝑎+𝑏)1𝑎𝑎𝑑𝑎𝑝1=𝑛=1𝑎𝑑𝑏𝑛𝑎𝑑𝑎𝑑𝑎𝑑𝑏𝑎1𝑑𝑏,[𝑎𝜋](𝑎+𝑏)𝑑=𝑎2+𝑏2𝑎𝜋𝑛=0𝑏𝑑𝑛+1𝑎𝑛.(2.48)
Next, according to (2.48), we obtain (𝑎+𝑏)𝑎𝑑𝑎𝑑𝑎𝑑𝑏𝑎1𝑑𝑏+𝑎𝜋𝑛=1𝑏𝑑𝑛+1𝑎𝑛.(2.49)
Finally, using (2.49) the proof is finished.

Corollary 2.13. Let 𝑎𝒜𝑔 and let 𝑏𝒜𝑑. If 𝑎,𝑏 satisfy the conditions 𝑎𝑏𝑎𝑎<1,𝑎𝜋𝑏𝑎=𝑎𝑏𝑎𝜋.(2.50) Then (𝑎+𝑏) exists if and only if 𝑎𝜋𝑏 is group invertible. In this case, (𝑎+𝑏)𝑎𝑎𝑎𝑏𝑎1𝑏+𝑎𝜋𝑏𝑑𝑎.(2.51)

Let 𝐴,𝐸𝐶𝑛×𝑛 with 𝐵=𝐴+𝐸, and let 𝐴=𝑃1𝐴100𝐴2𝑃,𝐸=𝑃1𝐸1𝐸12𝐸21𝐸2𝑃.(2.52) If 𝐵𝜋=𝐴𝜋 (see [10, Theorem 2.1] ), then 𝐵=𝑃1𝐵100𝐵2𝑃,𝐴+𝐸=𝑃1𝐴1+𝐸100𝐴2+𝐸2𝑃,(2.53) where 𝐵1 is invertible and 𝐵2=𝐴2 is quasinilpotent (it follows that 𝐸2=0). It follows from (2.53) that 𝐴𝜋=𝐵𝜋 implies that 𝐴𝜋𝐵𝐴=𝐴𝐵𝐴𝜋, (i.e., 𝐴𝜋𝐸𝐴=𝐴𝐸𝐴𝜋). If 𝐴 is group invertible, then 𝐵 is group invertible and 𝐵=𝑃1𝐵11000𝑃,(2.54) where 𝐵1=𝐴1+𝐸1.

By virtue of 𝐴𝜋=𝐵𝜋 and 𝐴𝐷(𝐵𝐴)<1 (see [10]), we give the following remark.

Remark 2.14 (see [10, Theorem  3.1]). Let 𝐴,𝐵𝐶𝑑×𝑑 with 𝐴𝜋=𝐵𝜋. Then 𝐴𝐷𝐴1+𝐷𝐵(𝐵𝐴)𝐷.(2.55) If 𝐴𝐷(𝐵𝐴)<1, then 𝐵𝐷𝐴𝐷𝐴1𝐷,𝐵(𝐵𝐴)𝐷𝐴𝐷𝐴𝐷𝐴𝐷(𝐵𝐴)𝐴1𝐷(.𝐵𝐴)(2.56)

Theorem 2.15. Let 𝑎,𝑏𝒜 be generalized Drazin invertible and satisfy the conditions 𝑎max𝜋𝑎𝑏𝑑,𝑎𝑑𝑏𝑎𝑎𝑑<1,𝑎𝜋𝑏𝑎=𝑎𝑏𝑎𝜋.(2.57) Then (𝑎+𝑏) exists if and only if 𝑎𝜋(𝑎+𝑏) is group invertible. In this case, (𝑎+𝑏)𝑎𝑑𝑎𝑑𝑎𝑑𝑏𝑎1𝑑𝑏+𝑎𝜋𝑏𝑑𝑎𝑑𝑏1𝑑𝑎.(2.58)

Proof. Similarly to Theorem 2.12, we have that the formulas (2.45) hold. The details will be omitted. In the following we only give the simple proof.
By the condition 𝑎max𝜋𝑎𝑏𝑑,𝑎𝑑𝑏𝑎𝑎𝑑<1,(2.59) it shows that 𝑎11𝑏1<1 and 𝑎2𝑏𝑑2<1. Thus, the first result shows that 𝑎1+𝑏1𝑝𝒜𝑝 is invertible. In view of Lemma 2.1, one concludes that 𝑎+𝑏 is Drazin invertible if and only if 𝑎2+𝑏2 is Drazin invertible. That is, (𝑎+𝑏) exists if and only if 𝑎𝜋(𝑎+𝑏) is group invertible.
In the following, we consider the perturbation of 𝑎2.
After application of the hypothesis 𝑎𝜋𝑏𝑎=𝑎𝑏𝑎𝜋, we find that 𝑎2𝑏2=𝑏2𝑎2. It follows from Theorem 2.12 and Lemma 2.3 that 𝑎2+𝑏2=𝑛=0𝑏𝑑2𝑛+1𝑎2𝑛=𝑎𝜋𝑛=0(1)𝑛𝑏𝑑𝑛+1𝑎𝑛.(2.60)
It follows from the condition 𝑎2𝑏𝑑2<1 and 𝜎𝑎2𝑏𝑑2𝑏{0}=𝜎𝑑2𝑎2{0}.(2.61) It implies that 𝑏𝑑2𝑎2<1 and 𝑎2+𝑏2𝑎𝜋𝑏𝑑𝑏1𝑑𝑎.(2.62)
Therefore, combining (2.49) with (2.62), we have (𝑎+𝑏)𝑎𝑑𝑎𝑑𝑎𝑑𝑏𝑎1𝑑𝑏+𝑎𝜋𝑏𝑑𝑏1𝑑𝑎.(2.63) Thus, by (2.63), we complete the proof.

3. Perturbation Bound of the Generalized Schur Complement

The perturbation bounds of the Schur complement are investigated in [2931]. In [29] Stewart gave perturbation bounds for the Schur complement of a positive definite matrix in a positive semidefinite matrix. In [30] Wei and Wang generalized the results in [29] and enrich the perturbation theory for the Schur complement. In [31] the authors derived some new norm upper bounds for Schur complements of a positive semidefinite operator matrix. In this section, we consider the perturbation bounds of the generalized Schur complement in Banach algebra.

Some notations of the generalized Schur complement over Banach algebra will be stated in the following.

Let 𝑎𝒜, and let it be written in the form as follows: 𝑎=𝑎11+𝑎12+𝑎21+𝑎22.(3.1) It has the following matrix form: 𝑎=11𝑎12𝑎21𝑎22𝑠,(3.2) where 𝑠𝒜 is idempotent element in 𝒜 and 𝑎𝑖𝑗 is taken as (1.10).

The formulas (1.5) and (1.6) are written in Banach algebra, respectively: 𝑠1=𝑎22𝑎21𝑎111𝑎12,𝑎=11+𝑎11𝑎12𝑠1𝑎21𝑎11𝑎11𝑎12𝑠1𝑠1𝑎21𝑎11𝑠1𝑠.(3.3) Similarly, the generalized Schur complement in (1.7) and (1.8) is defined in the following over Banach algebra, respectively: 𝑠1=𝑎22𝑎21𝑎11𝑎12,𝑠1=𝑎22𝑎21𝑎𝑑11𝑎12,(3.4) where 𝑠1 denotes the generalized Schur complement of 𝑎11 in .

Theorem 3.1. Let be given as (3.2) let and 𝑎=11+Δ𝑎11𝑎12+Δ𝑎12𝑎21+Δ𝑎21𝑎22+Δ𝑎22𝑠=𝑎11𝑎12𝑎21𝑎22𝑠(3.5) be perturbed version of , and the following conditions are satisfied: Δ𝑎11𝑎𝜖11,Δ𝑎12𝑎𝜖12,Δ𝑎21𝑎𝜖21,Δ𝑎22𝑎𝜖22,(3.6) where 𝜖>0. If 𝑎11, Δ𝑎11 and 𝑎11 satisfy the conditions of Theorem 2.8, then 𝑠1𝑠1𝑎𝜖22+𝑎21𝑎12𝜃2𝜖+𝜖2+𝑎𝜋11𝜂2𝑎+𝜖𝑑11𝑎11𝜂1𝑎+𝜖21𝑎11𝑎12𝜂21+𝑎𝜋11𝜂1𝜂2+𝜖+𝜖2𝑎𝜋11𝑎11𝜂21𝜂2,(3.7) where 𝜂1=𝑎𝑑11𝑎1𝑑11Δ𝑎11,𝜂2=𝑛=0𝑎𝑛+111Δ𝑎11𝑑𝑛+1,𝑎(3.8)𝜃=𝜖11𝜂21𝑎1+𝜋11𝜂2𝑎+𝜖11𝑎𝜋11𝜂1𝜂21+𝜖𝜂1𝑎11+𝑎𝜋11𝜂2𝑎+𝜖𝑑11𝑎11𝜂1,(3.9) and  𝑠1 and 𝑠1 are Schur complement of 𝑎11 in and Schur complement of  𝑎11 in , respectively.

Proof. Since 𝑎11, Δ𝑎11 and 𝑎11 satisfy Theorem 2.8, according to (2.31), we obtain 𝑎11=𝑛=0𝑎𝑑11Δ𝑎11𝑛𝑎𝑑112Δ𝑎11Δ𝑎11𝑎𝜋11𝑛=0𝑎𝑛+111Δ𝑎11𝑑𝑛+1+𝑛=0𝑎𝑑11Δ𝑎11𝑛𝑎𝑑112Δ𝑎11𝑎𝜋11Δ𝑎11𝑛=0𝑎𝑑11Δ𝑎11𝑛𝑎𝑑11Δ𝑎11𝑎𝜋11𝑛=0𝑎𝑛11Δ𝑎11𝑑𝑛+1+𝑛=0𝑎𝑑11Δ𝑎11𝑛𝑎𝑑11+𝑎𝜋11𝑛=0𝑎𝑛11Δ𝑎11𝑑𝑛+1.(3.10) Therefore, it is easy to see that 𝑠1=𝑎22+Δ𝑎22𝑎21+Δ𝑎21𝑎11𝑎12+Δ𝑎12=𝑎22+Δ𝑎22𝑎21𝑎11𝑎12Δ𝑎21𝑎11𝑎12𝑎21𝑎11Δ𝑎12Δ𝑎21𝑎11Δ𝑎12=𝑠1+Δ𝑎22Δ𝑎21𝑎11𝑎12𝑎21𝑎11Δ𝑎12Δ𝑎21𝑎11Δ𝑎12𝑎21𝑎𝜋11𝑛=0𝑎𝑛11Δ𝑎11𝑑𝑛+1𝑎12+𝑎21𝑛=1𝑎𝑑11Δ𝑎11𝑛𝑎𝑑11𝑎12+𝑎21𝑛=0𝑎𝑑11Δ𝑎11𝑛𝑎𝑑112Δ𝑎111𝑎𝜋11𝑎11+Δ𝑎11𝑛=0𝑎𝑛11Δ𝑎11𝑑𝑛+1𝑎12𝑎21𝑛=0𝑎𝑑11Δ𝑎11𝑛𝑎𝑑11Δ𝑎11𝑎𝜋11𝑛=0𝑎𝑛11Δ𝑎11𝑑𝑛+1𝑎12,𝑎11𝑎𝜖11𝑎𝑑11𝑎1𝑑11Δ𝑎112𝑎1+𝜋11𝑛=0𝑎𝑛+111Δ𝑎11𝑑𝑛+1+𝜖2𝑎𝑑11𝑎1𝑑11Δ𝑎112𝑎𝜋11𝑎112+𝜖𝑎𝑑11𝑎11𝑎𝜋11𝑎1𝑑11Δ𝑎11𝑛=0𝑎𝑛11Δ𝑎11𝑑𝑛+1+𝑎𝜋11𝑛=0𝑎𝑛11Δ𝑎11𝑑𝑛+1+𝜖𝑎𝑑112𝑎11𝑎1𝑑11Δ𝑎11.(3.11)
From (3.11) and by the conditions (3.6), we obtain𝑠1𝑠1Δ𝑎22+𝜃2𝜖+𝜖2𝑎21𝑎12+𝑎21𝑎12𝑎𝜋11𝑛=0𝑎𝑛11Δ𝑎11𝑑𝑛+1+𝜖𝑎𝑑112𝑎11𝑎1𝑑11Δ𝑎11𝑎+𝜖21𝑎11𝑎12𝑎𝑑11𝑎1𝑑11Δ𝑎112𝑎1+(1+𝜖)11𝑛=0𝑎𝑛11Δ𝑎11𝑑𝑛+1𝑎+𝜖21𝑎12𝑎𝑑11𝑎11𝑎𝜋11𝑎1𝑑11Δ𝑎11𝑛=0𝑎𝑛11Δ𝑎11𝑑𝑛+1,(3.12) where 𝑎𝜃=𝜖11𝑎𝑑11𝑎1𝑑11Δ𝑎112𝑎1+𝜋11𝑛=0𝑎𝑛+111Δ𝑎11𝑑𝑛+1+𝜖𝑎𝑑11𝑎11𝑎𝜋11𝑎1𝑑11Δ𝑎11𝜖𝑎1+𝑑11𝑎11𝑎𝜋11𝑎1𝑑11Δ𝑎11𝑛=0𝑎𝑛11Δ𝑎11𝑑𝑛+1+𝑎𝜋11𝑛=0𝑎𝑛11Δ𝑎𝑑11𝑛+1+𝜖𝑎𝑑112𝑎11𝑎1𝑑11Δ𝑎11.(3.13) Thus, we finish the proof.

Similar to Theorem 3.1. It follows from the proof of Theorem 2.11 that the results are given as follow.

Theorem 3.2. Let and be taken as Theorem 3.1, and let the relations in (3.6) be satisfied, where 𝜖>0. If 𝑎11, Δ𝑎11 and  𝑎11 satisfy the conditions of Theorem 2.11, then 𝑠1𝑎𝑠𝜖22+𝑎21𝑎12𝜃2𝜖+𝜖2+𝑎𝜋11𝜂2𝑎+𝜖𝑑11𝑎11𝜂1𝑎+𝜖21𝑎11𝑎12𝜂21+𝑎𝜋11𝜂1𝜂2+𝜖+𝜖2𝑎𝜋11𝑎11𝜂21𝜂2,(3.14) where 𝜂1,𝜃,𝑠1 and 𝑠1 are taken as Theorem 3.1.

Theorem 3.3. Let and be taken as Theorem 3.1, and let the relations in (3.6) be satisfied, where 𝜖>0. If 𝑎11, Δ𝑎11 and 𝑎11 satisfy the conditions of Theorem 2.12, then 𝑠1𝑎𝑠𝜖22+1+2𝜖+𝜖2𝛿1𝑎+𝜖𝑑112𝑎11𝑎12𝑎2