# The Quasi-Linear Operator Outer Generalized Inverse with Prescribed Range and Kernel in Banach Spaces

Authors: Jianbing Cao; Yifeng Xue
Journal: Journal of Operators
 (2013)

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## Abstract

LetX and Y be Banach spaces, and let A:X→Y be a bounded linear operator. In this paper, we first define and characterize the quasi-linear operator (resp., out) generalized inverse AT,S(h) (resp., AT,S(2,h)) for the operator A, where T⊂X and S⊂Y are homogeneous subsets. Then, we further investigate the perturbation problems of the generalized inverses AT,S(h) and AT,S(2,h). The results obtained in this paper extend some well-known results for linear operator generalized inverses with prescribed range and kernel.


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## Body

## 1. Introduction and Preliminaries

LetX and Y be Banach spaces, let T:X→Y be a mapping, and let D⊂X be a subset of X. Recall from [1, 2] that a subset D in X is called to be homogeneous if for any x∈D and λ∈ℝ, we have λx∈D. If for any x∈𝒟(T) and λ∈ℝ, we have T(λx)=λT(x), then we call T as a homogeneous operator on 𝒟(T), where 𝒟(T) is the domain of T; T is called a bounded homogeneous operator if T maps every bounded set in 𝒟(T) into bounded set in Y. Denote by H(X,Y) the set of all bounded homogeneous operators from X to Y. Equipped with the usual linear operations for H(X,Y), and for T∈H(X,Y), the norm is defined by ∥T∥=sup{∥Tx∥∣∥x∥=1,x∈X}, and then similar to the space B(X,Y) of all bounded linear operators from X to Y, we can easily prove that (H(X,Y),∥·∥) is a Banach space (cf. [2, 3]). Throughout this paper, we denote by 𝒟(T), 𝒩(T), and ℛ(T) the domain, the null space, and the range of a bounded homogeneous operator T∈H(X,Y), respectively. Obviously, we have B(X,Y)⊂H(X,Y).For an operatorA∈B(X,Y), let T and S be closed subspaces of X and Y, respectively. Recall that the out inverse AT,S(2) with prescribed range T and kernel S is the unique operator G∈B(Y,X) satisfying BAB=B,ℛ(B)=T,𝒩(B)=S. It is well known that the important kinds of generalized inverses, the Moore-Penrose inverse, the Drazin inverse, the group inverse, and so on, are all generalized inverse AT,S(2) (cf. [4, 5]). Researches on the generalized inverse AT,S(2) of operators or matrices have been actively ongoing for many years (see [5–12], e. g.).LetA∈B(X,Y) and let T and S be two homogeneous subsets in X and Y, respectively. Motivated by related work on AT,S(2) in the literature mentioned above and by our own recent research papers [13, 14], in this paper, we will establish the definition of the quasi-linear operator outer generalized inverse AT,S(2,h) with prescribed range T and kernel S. We give the necessary and sufficient conditions for the existence of the generalized inverses AT,S(2,h), and we will also study the perturbation problems of the generalized inverse AT,S(2,h). Similar results on the generalized inverse AT,S(h) are also given.


## 2. Definitions and Some Characterizations ofAT,S(2,h) and AT,S(h)

We first give the concepts of quasi-additivity and quasi-linear projectors in Banach spaces, which are important for us to present the main results in this paper.Definition 1.
LetM be a subset of X and let T:X→Y be a mapping. Ones calls T as quasi-additive on M if T satisfies
(1)T(x+z)=T(x)+T(z),∀x∈X,∀z∈M.Definition 2 (see [3, 15]).
LetX be a Banach space. A mapping P:X→X is called a quasi-linear projector on X, if P satisfies the following conditions:  (1)
Pis a homogeneous operator;(2)
P is idempotent; that is, P2=P;(3)
P is quasi-additive on ℛ(P); that is, for any x∈X and any z∈ℛ(P), one has
(2)P(x+z)=P(x)+P(z)=P(x)+z.
If the mappingP only satisfies the conditions (1) and (2) above, then we call P as a homogeneous projector.From Definition2, we see that the quasi-linear projectors include all the usually used projectors, such as the linear projectors in linear spaces, the bounded linear projectors, the metric projectors in Banach spaces, and the orthogonal projectors in Hilbert spaces (cf. [3, 15, 16]). We have the following important property for quasi-linear projector.Lemma 3 (see [15, Lemma 2.5]).
IfP∈H(X,X) is a bounded quasi-linear projector on X, then the range ℛ(P) of P is a closed linear subspace of X.We should note that, although the rangeℛ(P) of a bounded quasi-linear projector P is a closed linear subspace, but in general, the kernel 𝒩(P) of a bounded quasi-linear projector P is not a (closed) subspace. Let P∈H(X) be a quasi-linear projector. Then by Lemma 3, ℛ(P) is a closed linear subspace of X and ℛ(I-P)=𝒩(P). Thus, we can define “the quasi-linearly complement” of a closed linear subspace as follows. Let V be a closed subspace of X. If there exists a bounded quasi-linear projector P on X such that V=ℛ(P), then V is said to be bounded quasi-linearly complemented in X and 𝒩(P) is the bounded quasi-linear complement of V in X. In this case, as usual, we may write X=V+˙𝒩(P), where 𝒩(P) is a homogeneous subset of X and “+˙” means that V∩𝒩(P)={0} and X=V+𝒩(P).Now we establish the definitions of some types of homogeneous (or quasi-linear) operator generalized inverses with prescribed range and kernel as follows.Definition 4.
LetA∈B(X,Y). Let T and S be homogeneous subsets in X and Y, respectively. Let B∈H(Y,X) be a bounded homogeneous operator. Consider the following equations:
(3)(1)ABA=A,(2)BAB=B,(3)ℛ(B)=T,(4)𝒩(B)=S.(i)
IfB∈H(Y,X) satisfies (2)–(4) in (3), then the operator B is called to be the homogeneous outer generalized inverse of A with prescribed range T and kernel S. We denoted it by AT,S(2,H).(ii)
IfB∈H(Y,X) satisfies all equations in (3), then the operator B is called to be the homogeneous generalized inverse of A with prescribed range T and kernel S. We denoted it by AT,S(H).(iii)
IfB∈H(Y,X) is quasi-additive on AT and satisfies (2)–(4) in (3), then the operator B is called to be the quasi-linear outer generalized inverse of A with prescribed range T and kernel S. We denoted it by AT,S(2,h).(iv)
IfB∈H(Y,X) is quasi-additive on ℛ(A) and satisfies all equations in (3), then the operator B is called to be the quasi-linear generalized inverse of A with prescribed range T and kernel S. We denoted it by AT,S(h).Remark 5.
LetA∈B(X,Y). Assume that 𝒩(A) and ℛ(A) are Chebyshev subspaces in X and Y, respectively. Then the Moore-Penrose metric generalized inverse AM of A uniquely exists (cf. [3, 17, 18]). Let
(4)S=FY-1(ℛ(A)⊥),T=FX-1((𝒩(A))⊥),
where FX:X→X* is the set-valued dual mapping of X, ℛ(A)⊥={f∈X*∣f(x)=0,∀x∈ℛ(A)} and FX-1(ℛ(A)⊥)={x∈X∣FX(x)∩ℛ(A)⊥≠∅}. Then S and T are homogeneous subsets in X and Y, respectively, and
(5)AMAAM=A,AMAAM=AM,ℛ(AM)=T,𝒩(AM)=S.
In this case, we haveAM=AT,S(H). Furthermore, if AM is quasi-additive on ℛ(A), then AM=AT,S(h). Thus, it is meaningful to study the quasi-linear operator (outer) generalized inverses with prescribed range and kernel.In this paper, we mainly study the generalized inversesAT,S(2,h) and AT,S(h). Motivated by the properties of linear operator generalized inverses with prescribed range and kernel, in the rest of this section, we will give some characterizations of the generalized inverses AT,S(2,h) and AT,S(h).Theorem 6.
LetA∈B(X,Y). Let T and S be homogeneous subsets in X and Y, respectively. Then the following statements are equivalent:  (1)
there exists someB∈H(Y,X) such that B is quasi-additive on AT and BAB=B, ℛ(B)=T, 𝒩(B)=S;(2)
𝒩(A)∩T={0}, Y=AT+˙S and T⊂X is a closed linear subspace.Proof.
(1)⇒(2) Suppose that (1) holds. PutP=BA and Q=AB. Since BAB=B, we have P=P2 and ℛ(P)=ℛ(BA)=ℛ(B)=T. In order to prove  T⊂X  is a closed linear subspace, by Lemma 3, we only need to show that P is a bounded quasi-linear projector. Let x∈X and z∈T=ℛ(BA), we get z=BAz and ABAz∈AT. Noting that B is quasi-additive on AT, then
(6)P(x+z)=BA(x+BAz)=B(Ax+ABAz)=BAx+BABAz=Px+z.
Therefore,P is a bounded quasi-linear projector and then T is a closed linear subspace in X. Now we show that 𝒩(A)∩T={0}. Let x∈𝒩(A)∩T. Since ℛ(B)=T, then Ax=0 and x=Bw for some w∈Y. So x=Bw=BABw=BAx=0. Therefore, we have 𝒩(A)∩T={0}. Next we will show that Y=AT+˙S. For any y∈Y, we have y=(IY-AB)y+ABy. Since we have proved that T is a closed linear subspace of X, we have ABy∈ℛ(Q)=ℛ(AB)=Aℛ(B)=AT. Also noting that B is quasi-additive on AT and BAB=B, we get (IY-AB)y∈𝒩(B)=S. Thus, we get Y=AT+S. From BAB=B, we can also get Q=Q2 and S=𝒩(B)=𝒩(AB)=𝒩(Q); therefore, AT∩S=∅ and then Y=AT+˙S.
(2)⇒(1) Suppose that (2) holds. Since𝒩(A)∩T={0} and T is a closed linear subspace, we see A|T:T→AT is well defined, which is also a linear bijective mapping, so A|T has a linear inverse (A|T)-1:AT→T. Since Y=AT+˙S, then for any y∈Y, it can be uniquely written as y=y1+y2 with y1∈AT and y2∈S. Now we can define a mapping B:Y→X by By=(A|T)-1y1. It is easy to see that BAB=B, ℛ(B)=T and 𝒩(B)=S by Y=AT+˙S. Since 𝒩(B)=S is a homogeneous subsets in Y, we see B∈H(Y,X) is a homogeneous operator. Now we need to prove B is quasi-additive on AT. In fact, for any y∈Y and z∈AT, since Y=AT+˙S, we can write y as y=y1+y2, where y1∈AT and y2∈S. Noting that y1+z∈AT since AT⊂X is a linear subspace, then we can check that
(7)B(y+z)=B(y1+y2+z)=B((y1+z)+y2)=(A|T)-1(y1+z)=(A|T)-1y1+(A|T)-1z=By+Bz.
Therefore,B is quasi-additive on AT.Theorem 7.
LetA∈B(X,Y). Let T and S be homogeneous subsets in X and Y, respectively. Then the following statements are equivalent.(1)
There exists someB∈H(Y,X) such that B is quasi-additive on ℛ(A) and
(8)ABA=A,BAB=B,ℛ(B)=T,𝒩(B)=S.(2)
X=𝒩(A)+˙T, Y=ℛ(A)+˙S and T⊂X is a closed linear subspace.(3)
Y=AT+˙S, 𝒩(A)∩T={0}, ℛ(A)∩S={0} and T⊂X is a closed linear subspace.Proof.
(1)⇒(2) Suppose that (1) holds, PutP=BA and Q=AB. Similar to the proof of Theorem 6, we have P=P2,  Q=Q2 and T⊂X is a closed linear subspace. Note that we also have the following relation:
(9)ℛ(BA)⊂ℛ(B)=ℛ(BAB)⊂ℛ(BA),ℛ(AB)⊂ℛ(A)=ℛ(ABA)⊂ℛ(AB),𝒩(BA)⊂𝒩(ABA)=𝒩(A)⊂𝒩(BA),𝒩(B)⊂𝒩(AB)⊂𝒩(BAB)=𝒩(B).
Thus, we get that(10)ℛ(P)=ℛ(BA)=ℛ(B)=T,ℛ(Q)=ℛ(AB)=ℛ(A),𝒩(P)=𝒩(BA)=𝒩(A),𝒩(Q)=𝒩(AB)=𝒩(B)=S.
From the above equations, we can obtainX=𝒩(A)+˙T and Y=ℛ(A)+˙S.
(2)⇒(3) IfX=𝒩(A)+˙T and Y=ℛ(A)+˙S, then it is obvious that 𝒩(A)∩T={0}, ℛ(A)∩S={0}. We only need to show AT=ℛ(A). Obviously, we have AT⊂ℛ(A). Now for any y∈ℛ(A), we have y=Ax for some x∈X. Since X=𝒩(A)+˙T, we can write x=x1+x2, where x1∈𝒩(A) and x2∈T. Thus y=Ax=Ax2∈AT. Therefore ℛ(A)⊂AT and then AT=ℛ(A).
(3)⇒(1) Suppose that (3) is true, then by Theorem6, we know that AT,S(2,h)=B exists, and then ℛ(B)=T, 𝒩(B)=S. We need to show ABA=A and B is quasi-additive on ℛ(A). Since AT,S(2,h)=B exists, then BAB=B and B is quasi-additive on AT. Noting that for any x∈X, we have ABAx∈AT. Thus, B(ABA-A)x=0 and
(11)(ABA-A)x⊂ℛ(A)∩𝒩(B)=ℛ(A)∩S={0};
that is, ABA=A. Obviously, AT⊂ℛ(A). On the converse, let y∈ℛ(A). Since Y=AT+˙S, we have y=y1+y2 with some y1∈AT and y2∈S. From AT⊂ℛ(A), we see y1∈ℛ(A). Thus, we have y2=y-y1∈ℛ(A)∩S={0}. So ℛ(A)⊂AT and then AT=ℛ(A). Hence, we obtain that B is quasi-additive on ℛ(A).We should indicate that, for Definition4, we define two different quasi-additivity in (iii) and (iv). But from Theorem 7, we see that if the generalized inverse AT,S(h) exists, then we must have ℛ(A)=AT. So there is no contradiction in our definition. The following proposition gives the uniqueness of AT,S(2,h). From this result, it is also easy to see the uniqueness of AT,S(h), which are similar to the linear (outer) generalized inverses with prescribed range and kernel.Proposition 8.
LetA∈B(X,Y). Let T and S be homogeneous subsets in X and Y, respectively. If the generalized inverse AT,S(2,h) (or AT,S(h)) exists, then it is unique.Proof.
We only need to prove this proposition forAT,S(2,h). By Definition 4, we need to prove that there is at most one bounded homogeneous operator B∈H(Y,X) such that B is quasi-additive on AT and satisfies
(12)BAB=B,ℛ(B)=T,𝒩(B)=S.
Suppose that there is anotherC∈H(Y,X) such that C is quasi-additive on AT and satisfies the equations CAC=C, ℛ(C)=T, and 𝒩(C)=S. Put Q=AB and Q′=AC. By using the equations BAB=B and CAC=C, we see that both Q and Q′ are bounded homogeneous projectors on Y and satisfy
(13)ℛ(Q)=ℛ(Q′)=AT;𝒩(Q)=𝒩(Q′)=S.
Therefore, we getQ=Q′. Thus, C=CAC=CAB. On the other hand, if we set P=BA and P′=CA, then by using the equations BAB=B and CAC=C again, and also noting that B and C are both quasi-additive on AT, similar to the proof (1)⇒(2) in Theorem 6, we see both P and P′ are bounded quasi-linear projectors and satisfy
(14)ℛ(P)=ℛ(B)=ℛ(P′)=ℛ(C)=T.
Thus,BA=P=P′P=CABA, and then B=BAB=CABAB=CAB=C.Proposition 9.
LetA∈B(X,Y). Let T and S be homogeneous subsets in X and Y, respectively. If the generalized inverse AT,S(2,h) (or AT,S(h)) exists, then T and AT are both closed linear subspaces of X.Proof.
If the generalized inverseAT,S(2,h) (or AT,S(h)) exists, then from Theorem 6 (or Theorem 7), we see T is a closed linear subspace in X. Let Q=AAT,S(2,h). Then it is easy to check that ℛ(Q)=AT. Since AT,S(2,h) is quasi-additive on AT, simple computation can show that Q=AAT,S(2,h) is a bounded quasi-linear projector. Then from Lemma 3, we get AT is also a closed linear subspace of X if AT,S(2,h) exists. If the generalized inverse AT,S(h) exists, then AT,S(2,h) exists and AT,S(2,h)=AT,S(h).Finally, in this section, we will give some representation results of the generalized inverseAT,S(2,h).Theorem 10.
LetA∈B(X,Y). Let T and S be homogeneous subsets in X and Y, respectively. Suppose that G∈H(Y,X) such that G is quasi-additive on AT, and ℛ(G)=T,𝒩(G)=S. Then the following statements are equivalent: (1)
the generalized inverseAT,S(2,h) exists;(2)
(GA)|T is invertible and T⊂X is a closed linear subspace. In this case, one has AT,S(2,h)=((GA)|T)-1G.Proof.
(1)⇒(2) Suppose that (1) holds, then by Proposition9, we get that T⊂X is a closed linear subspace. If GAx=0 for some x∈T, then x=Gz for some z∈Y. Since 𝒩(G)=S, by Theorem 6, we have Ax∈AT∩S={0}. Noting that 𝒩(A)∩T={0}, thus, we get x=0 and then (2) holds.
(2)⇒(1) LetB=((GA)|T)-1G. Obviously, we have BAB=B, and ℛ(B)=ℛ(G)=T, 𝒩(B)=𝒩(G)=S. We show that B is quasi-additive on AT. Let y∈Y and z∈AT. Since G is quasi-additive on AT, we get that ((GA)|T)-1 is a linear operator and then
(15)B(y+z)=((GA)|T)-1G(y+z)=((GA)|T)-1(Gy+Gz)=By+Bz.
By Proposition8, the uniqueness of AT,S(2,h), we can obtain AT,S(2,h)=((GA)|T)-1G.LetA∈B(X,Y). Recall that A is group invertible if there exists C∈B(Y,X) such that
(16)ACA=A,CAC=C,CA=AC.In this case, denoteC=A# as usual. For a bounded homogeneous operator W∈H(X,Y), if there exists W-∈H(Y,X) such that WW-W=W, then we call W- as a bounded homogeneous inner inverse of W or simply say W is inner regular.Theorem 11.
LetA∈B(X,Y). Let T and S be homogeneous subsets in X and Y, respectively. Suppose that G∈H(Y,X) such that G is quasi-additive on ℛ(A), and ℛ(G)=ℛ(GAG)=T, 𝒩(G)=S. If GA∈B(X,X) has the group inverse (GA)#, then AT,S(2,h) exists and, moreover, GAG∈H(Y,X) is inner regular and
(17)AT,S(2,h)=(GA)#G=G(GAG)-G.Proof.
SinceG is quasi-additive on ℛ(A), we see GA∈B(X) is well defined. Let B=(GA)#G, we show that AT,S(2,h)=B. From ℛ(G)=ℛ(GAG)=T, we get ℛ(GA)⊂ℛ(G)=ℛ(GAG)⊂ℛ(GA), and then ℛ(GA)=ℛ(G)=T. The existence of (GA)# means that
(18)𝒩(GA)∩ℛ(GA)={0},GA(GA(GA)#G-G)=0.
Therefore(19)GA(GA)#G-G∈𝒩(GA)∩ℛ(G)=𝒩(GA)∩ℛ(GA)={0}.
HenceG=GA(GA)#G. Now, we can compute as follows:
(20)BAB=(GA)#GA(GA)#G=(GA)#G=B,T=ℛ(G)=ℛ(GA(GA)#G)=ℛ((GA)#GAG)⊂ℛ((GA)#G)=ℛ(B)=ℛ((GA)#GA(GA)#G)=ℛ(GA((GA)#)2G)⊂ℛ(G)=T,S=𝒩(G)⊂𝒩((GA)#G)=𝒩(B)⊂𝒩(GA(GA)#G)=𝒩(G)=S.
Thus,ℛ(B)=T and 𝒩(B)=S. Obviously, we see B is quasi-additive on AT for AT⊂ℛ(A). So, from Definition 4 and Proposition 8, we see AT,S(2,h) exists and AT,S(2,h)=(GA)#G.
SetC=A((GA)#)2. Then it is easy to check that
(21)(GAG)C(GAG)=(GAG)A((GA)#)2(GAG)=GAG.
We getGAG is inner regular and (GAG)-=A((GA)#)2. Furthermore, we have
(22)G(GAG)-G=GCG=GA((GA)#)2G=(GA)#G=AT,S(2,h).
This completes the proof.Theorem 12.
LetA∈B(X,Y). Let T and S be homogeneous subsets in X and Y, respectively. Suppose that G∈H(Y,X) such that G is quasi-additive on ℛ(A), and ℛ(G)=ℛ(GAG)=T, 𝒩(G)=S. If GA∈B(X) has the group inverse (GA)#, then AT,S(2,h) exists and AT,S(2,h)=limλ→0,λ∉σ(-GA)(λIX+GA)-1G.Proof.
By Theorem11, we know that AT,S(2,h) exists and AT,S(2,h)=(GA)#G. Let P=AT,S(2,h)A. Noting that G is quasi-additive on ℛ(A), then P∈B(X) is a projector and P=(GA)#GA. Clearly, we see that GA,(GA)#∈B(ℛ(P)) and GA is invertible in B(ℛ(P)) with (GA|B(ℛ(P)))-1=(GA)#. Consider the decomposition X=ℛ(P)⊕ℛ(IX-P). Then we can write λIX+GA as the following matrix form:
(23)λIX+GA=[λP+GAλ(IX-P)].
From (23), we see if λ∉σ(-GA)∪{0}, then λP+GA is invertible in B(ℛ(P)). Thus, in the case, by (23) we have
(24)(λIX+GA)-1=[(λP+GA)-1λ-1(IX-P)].
From the proof of Theorem11, we know that ℛ(GA)=ℛ(G)=T. Thus, we get 𝒩(IX-P)=ℛ(P)=ℛ((GA)#GA)=T. So, by using (24), we can obtain that (λIX+GA)-1G=P(λP+GA)-1G. Now we can compute as follows:
(25)AT,S(2,h)=P(GA)#G=limλ→0λ∉σ(-GA)P(λP+GA)-1G=limλ→0λ∉σ(-GA)(λIX+GA)-1G.
This completes the proof.


## 3. Perturbations of the Generalized InversesAT,S(2,h) and AT,S(h)

In this section, we present some perturbation results for the generalized inverseAT,S(2,h) and AT,S(h). The next lemma, which is a generalized Neumman Lemma taken from [19], will be useful for us to improve perturbation bound.Lemma 13 (see [19]).
LetX be a Banach space and P∈B(X). If there exist two constants λ1,λ2∈[0,1) such that
(26)∥Px∥≤λ1∥x∥+λ2∥(I+P)x∥,(∀x∈X),
then I+P:X→X is bijective. Moreover, for any x∈X(27)1-λ11+λ2∥x∥≤∥(I+P)x∥≤1+λ11-λ2∥x∥,1-λ21+λ1∥x∥≤∥(I+P)-1x∥≤1+λ21-λ1∥x∥.LetA,δA∈B(X,Y). Suppose that T⊂X and S⊂Y are homogeneous subsets such that AT,S(2,h) exists. If AT,S(2,h) is quasi-additive on ℛ(δA), then we see that AT,S(2,h)δA is a bounded linear operator, so IX+AT,S(2,h)δA is also a bounded linear operator on X. Furthermore, if AT,S(2,h) and δA also satisfy
(28)∥AT,S(2,h)δAx∥≤λ1∥x∥1+λ2∥(IX+AT,S(2,h)δA)x∥,(∀x∈X),
where λ1,λ2∈(-1,1), then by Lemma 13, we get that IX+AT,S(2,h)δA is invertible.Lemma 14.
LetA∈H(X,Y) and B∈H(Y,X) such that A is quasi-additive on ℛ(B) and B is quasi-additive on ℛ(A); then IY+AB is invertible in H(Y,Y) if and only if IX+BA is invertible in H(X,X).Proof.
If there is aΦ∈H(Y,Y) such that (IY+AB)Φ=Φ(IY+AB)=IY, then
(29)IX=IX+BA-BA=IX+BA-S((IY+AB)Φ)A=IX+BA-((B+BAB)Φ)A111111111111111(Bquasi-additiveonℛ(A))=IX+BA-((IX+BA)SΦ)A=(IX+BA)(1X-BΦA)11111111111111(Aquasi-additiveonℛ(B)).
Similarly, we also haveIX=(IX-BΦA)(IX+BA). Thus, IX+BA is invertible on X with (IX+BA)-1=(1X-BΦA)∈H(X,X).
The converse can also be proved by using the same way as above.Lemma 15.
LetA,δA∈B(X,Y). Suppose that T⊂X and S⊂Y are homogeneous subsets such that AT,S(2,h) exists. If AT,S(2,h) is quasi-additive on ℛ(δA) and IY+δAAT,S(2,h) (resp., IX+AT,S(2,h)δA) is invertible in H(Y,Y) (resp., H(X,X)), then IX+AT,S(2,h)δA (resp., IY+δAAT,S(2,h)) is invertible in H(X,X) (resp., H(Y,Y)) and
(30)Φ=AT,S(2,h)(IY+δAAT,S(2,h))-1=(IX+AT,S(2,h)δT)-1AT,S(2,h)
is a bounded homogeneous operator with ℛ(Φ)=T and 𝒩(Φ)=S.Proof.
By Lemma14, we obtain that IY+δAAT,S(2,h) (resp., IX+AT,S(2,h)δA) is invertible in H(Y,Y) (resp., H(X,X)).
Clearly,IX+AT,S(2,h)δA is a bounded linear operator and IY+δAAT,S(2,h)∈H(Y,Y). From the equation
(31)(IX+AT,S(2,h)δA)AT,S(2,h)=AT,S(2,h)(IY+δAAT,S(2,h))
and AT,S(2,h)∈H(Y,X), we get that Φ is a bounded homogeneous operator. Finally, by using (30) and Definition 4, we can obtain that ℛ(Φ)=ℛ(AT,S(2,h))=T and 𝒩(Φ)=𝒩(AT,S(2,h))=S.Theorem 16.
LetA,δA∈B(X,Y). Suppose that T⊂X and S⊂Y are homogeneous subsets such that AT,S(2,h) exists. Put A-=A+δA. If AT,S(2,h) is quasi-additive on ℛ(δA) and satisfies (28), then A-T,S(2,h) exists and A-T,S(2,h)=(IX+AT,S(2,h)δA)-1AT,S(2,h). Moreover, one has
(32)∥A-T,S(2,h)-AT,S(2,h)∥∥AT,S(2,h)∥≤1+λ21-λ1∥AT,S(2,h)δA∥.Proof.
SinceAT,S(2,h) is quasi-additive on ℛ(δA) and satisfies (28), then from Lemma 13 and Lemma 15, we see Φ=AT,S(2,h)(IY+δAAT,S(2,h))-1=(IX+AT,S(2,h)δT)-1AT,S(2,h) is well defined with ℛ(Φ)=T and 𝒩(Φ)=S. Now, we prove that Φ is quasi-additive on A-T. Let x∈X and z∈A-T, then z=A-w=Aw+δAw for some w∈T. Since AT,S(2,h) is quasi-additive on ℛ(δA) and AT, also noting that IX+AT,S(2,h)δA is a bounded linear operator, then
(33)Φ(x+z)=(IX+AT,S(2,h)δT)-1AT,S(2,h)(x+Aw+δAw)=(IX+AT,S(2,h)δT)-1AT,S(2,h)(x+Aw)11+(IX+AT,S(2,h)δT)-1AT,S(2,h)δAw=Φx+ΦAw+ΦδAw=Φx+Φz.
We getΦ is quasi-additive on A-T. Thus, from Definition 4 and Proposition 8, we see that A-T,S(2,h) exists and A-T,S(2,h)=(IX+AT,S(2,h)δA)-1AT,S(2,h).
Finally, we show that the estimate formula is valid:(34)∥A-T,S(2,h)-AT,S(2,h)∥∥AT,S(2,h)∥1≤∥(IX+AT,S(2,h)δT)-1AT,S(2,h)-AT,S(2,h)∥∥AT,S(2,h)∥1≤∥(IX+AT,S(2,h)δT)-1[IX-(IX+AT,S(2,h)δT)]AT,S(2,h)∥∥AT,S(2,h)∥1≤∥(IX+AT,S(2,h)δT)-1∥∥AT,S(2,h)δA∥.
By using Lemma13, we get
(35)∥(IX+THδT)-1∥≤1+λ21-λ1.
Thus from (34) and (35), we can obtain
(36)∥A-T,S(2,h)-AT,S(2,h)∥∥AT,S(2,h)∥≤1+λ21-λ1∥AT,S(2,h)δA∥.
This completes the proof.Corollary 17.
LetA,δA∈B(X,Y). Suppose that T⊂X and S⊂Y are homogeneous subsets such that AT,S(2,h) exists. Put A-=A+δA. If AT,S(2,h) is quasi-additive on ℛ(δA) with ∥AT,S(2,h)δA∥<1, then A-T,S(2,h) exists and A-T,S(2,h)=(IX+AT,S(2,h)δA)-1AT,S(2,h). Moreover, one has
(37)∥A-T,S(2,h)-AT,S(2,h)∥∥AT,S(2,h)∥≤∥AT,S(2,h)δT∥1-∥AT,S(2,h)δA∥.Proof.
Obviously, this corollary follows from Theorem16 if we take λ1=∥AT,S(2,h)δT∥ and λ2=0.Lemma 18 (Neumann Lemma for quasi-linear operators).
LetA∈H(X,X). If A is quasi-additive on ℛ(A) with ∥A∥<1, then (I+A)-1 exists and
(38)(I+A)-1=∑n=0∞‍(-A)n,∥(I+A)-1∥≤11-∥A∥.Proof.
Noting thatA is quasi-additive on ℛ(A), then the proof is almost the same to the classical Neumann Lemma for bounded linear operators. Here we omit it, and we refer to [20, Lemma 4.2.6] for more information.Theorem 19.
LetA,δA∈B(X,Y). Suppose that T⊂X and S⊂Y are homogeneous subsets such that AT,S(2,h) exists. Put A-=A+δA. If AT,S(2,h) is quasi-additive on ℛ(δA) with ∥δAAT,S(2,h)∥<1, then  (1)
IY+δAAT,S(2,h) is inventible bounded homogeneous operator;(2)
A-T,S(2,h) exists and A-T,S(2,h)=AT,S(2,h)(IY+δAAT,S(2,h))-1.
Moreover, one has(39)∥A-T,S(2,h)-AT,S(2,h)∥∥AT,S(2,h)∥≤∥δAAT,S(2,h)∥1-∥δAAT,S(2,h)∥.Proof.
SinceAT,S(2,h) is quasi-additive on ℛ(δA), then ℛ(δAAT,S(2,h))⊂ℛ(δA) implies that AT,S(2,h) is quasi-additive on ℛ(δAAT,S(2,h)). Obviously, δAAT,S(2,h)∈H(Y), and it follows from Lemma 18 that IY+δAAT,S(2,h) is inventible bounded homogeneous operator. Then, by using Lemma 15, similar to the proof of Theorem 16, we can show that A-T,S(2,h) exists and A-T,S(2,h)=AT,S(2,h)(IY+δAAT,S(2,h))-1. For the estimate formula, we can compute that
(40)∥A-T,S(2,h)-AT,S(2,h)∥∥AT,S(2,h)∥1≤∥AT,S(2,h)(IY+δAAT,S(2,h))-1-AT,S(2,h)∥∥AT,S(2,h)∥1=∥[AT,S(2,h)-AT,S(2,h)(IY+δAAT,S(2,h))](IY+δAAT,S(2,h))-1∥∥AT,S(2,h)∥1≤∥δAAT,S(2,h)∥∥(IY+δAAT,S(2,h))-1∥.
Thus, our result follows from Lemma18 and (40).Finally, we present two perturbation results for the generalized inverseAT,S(h).Theorem 20.
LetA,δA∈B(X,Y). Suppose that T⊂X and S⊂Y are homogeneous subsets such that AT,S(2,h) exists. Put A-=A+δA. If AT,S(2,h) is quasi-additive on ℛ(δA) and IX+AT,S(2,h)δA is invertible in H(X,X), then A-T,S(h)=(IX+AT,S(2,h)δA)-1AT,S(2,h) if and only if ℛ(A-)∩S={0}.Proof.
IfA-T,S(h)=(IX+AT,S(2,h)δA)-1AT,S(2,h), then from Theorem 7, obviously, we have ℛ(A-)∩S={0}. Conversely, suppose that ℛ(A-)∩S={0}. First, noting that from Lemma 15, we get Φ=(IX+AT,S(2,h)δA)-1AT,S(2,h) is well defined, and ℛ(Φ)=T and 𝒩(Φ)=S. Similar to the proof of Theorem 16, we can show A-T,S(2,h) exists and A-T,S(2,h)=Φ. Noting that A-T,S(h) is unique if it exists, thus, by using Theorems 6 and 7, we can prove our theorem.Corollary 21.
LetA,δA∈B(X,Y). Suppose that T⊂X and S⊂Y are homogeneous subsets such that AT,S(2,h) exists. Put A-=A+δA. If AT,S(2,h) is quasi-additive on ℛ(δA) with ∥AT,S(2,h)δA∥<1, then A-T,S(h)=(IX+AT,S(2,h)δA)-1AT,S(2,h) if and only if ℛ(A-)∩S={0}.Proof.
By our assumption,AT,S(2,h) is quasi-additive on ℛ(δA), we get IX+AT,S(2,h)δA is a bounded linear operator, and then it is well known that IX+AT,S(2,h)δA is invertible when ∥AT,S(2,h)δA∥<1. So the corollary follows from Theorem 20.



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*Source: 204587-2013-08-04.xml*