# On Stable Perturbations of the Generalized Drazin Inverses of Closed Linear Operators in Banach Spaces

Authors: Qianglian Huang; Lanping Zhu; Xiaoru Chen; Chang Zhang
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We investigate the stable perturbation of the generalized Drazin inverses of closed linear operators in Banach spaces and obtain some new characterizations for the generalized Drazin inverses to have prescribed range and null space. As special cases of our results, we recover the perturbation theorems of Wei and Wang, Castro and Koliha, Rakocevic and Wei, Castro and Koliha and Wei.


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## Body

## 1. Introduction and Preliminaries

LetX be a Banach spaces. Let L(X), C(X), and B(X) denote the linear space of all linear operators, the homogeneous set of all densely defined closed linear operators and the Banach space of all bounded linear operators from X into itself, respectively. For any T∈L(X), we denote by D(T), N(T), and R(T) the domain, the null space, and, respectively, the range of T. The identity operator will be denoted by I.As an important extension of the conventional Drazin inverse, the generalized Drazin inverse in Banach algebra was introduced firstly by Koliha [1]. Later, this notion was extended to closed linear operators by Koliha and Tran [2]. Recall that an operator S∈B(X) is said to be the generalized Drazin inverse of T∈C(X) if R(S)⊂D(T), R(I-TS)⊂D(T) and
(1.1)ST=TSonD(T),STS=S,T(I-TS)isquasinilpotent.An operatorT∈C(X) that possesses a generalized Drazin inverse is said to be generalized Drazin invertible, and its generalized Drazin inverse is denoted by Td. From [3–5], we know that T is generalized Drazin invertible if and only if 0 is not an accumulation of σ(T). In this case, we write Tπ=I-TTd for the spectral idempotent of T.Perturbations of the conventional Drazin inverse and the generalized Drazin inverse were studied in many papers [3–13]. For example, Castro González and Koliha [3] introduced the concept of T-compatible and gave the stability theorem of generalized Drazin inverse under the inverse-compatible perturbation (Theorem 1.1). Let T∈C(X) be generalized Drazin invertible. An operator δT∈B(X) is called T-compatible if it commutes with Tπ and if the operator TπδT is quasinilpotent and commutes with T, that is,
(1.2)TπδT=δTTπ,σ(TπδT)={0},TTπδT=δTTTπ.Theorem 1.1 (see [3]).
LetT∈C(X) have the generalized Drazin inverse Td∈B(X) and δT∈B(X) be T-compatible. If I+TdδT:X→X is invertible, then T¯=T+δT is generalized Drazin invertible and T¯d=Td(I+δTTd)-1=(I+TdδT)-1Td.In this paper, we investigate the stable perturbation of the generalized Drazin inverses of closed linear operators in Banach spaces and obtain some new characterizations for the generalized Drazin inverses to have prescribed range and null space. As special cases of our results, we recover the perturbation theorems of Castro González and Koliha [3], Castro González et al. [4], Wei and Wang [6], Rakočević and Wei [7].


## 2. Main Results

First we will characterize the invertibility ofI+δTTd, which is inspired from [8].Lemma 2.1.
LetT∈C(X) have a generalized Drazin inverse Td∈B(X) and δT∈L(X) with D(T)⊂D(δT). Then the following statements are equivalent:(1)
I+δTTd:X→X is bijective;(2)
I+TdδT:D(T)→D(T) is bijective;(3)
N(T¯)∩R(Td)={0} and X=T¯R(Td)+˙N(Td),whereT¯=T+δT and +˙ denotes the algebraic direct sum.Proof.
(
1
)
⇒
(
2
). We first claim that I+TdδT:D(T)→D(T) is injective. In fact, if x∈D(T) satisfies (I+TdδT)x=0, then δT(I+TdδT)x=0, that is,
(2.1)(I+δTTd)δTx=0.
Combining it with statement (1), we obtain δTx=0 and so x=-TdδTx=0. Next, we will show that I+TdδT:D(T)→D(T) is surjective, that is, for all q∈D(T), we need to find a p∈D(T) such that (I+TdδT)p=q. Let B=Td(I+δTTd)-1, then
(2.2)(I+TdδT)BδTq=(I+TdδT)Td(I+δTTd)-1δTq=Td(I+δTTd)(I+δTTd)-1δTq=TdδTq
and (I+TdδT)(I-BδT)q=(I+TdδT)q-TdδTq=q. Set p=(I-BδT)q, then p∈D(T) and (I+TdδT)p=q. Hence, I+TdδT:D(T)→D(T) is bijective.
(
2
)
⇒
(
3
). If I+TdδT:D(T)→D(T) is bijective, let x∈N(T¯)∩R(Td), then there exists y∈X satisfying x=Tdy and T¯Tdy=T¯x=0. Hence
(2.3)(I+TdδT)Tdy=Tdy+TdδTTdy=TdTTdy+TdδTTdy=TdT¯Tdy=0.
Thus Tdy=0, that is, x=0. Therefore, N(T¯)∩R(Td)={0}. Next we will show T¯R(Td)∩N(Td)={0}. Let q∈T¯R(Td)∩N(Td), then there exists p∈X with q=T¯Tdp. Hence
(2.4)0=Tdq=TdT¯Tdp=Td(T+δT)Tdp=TdTTdp+TdδTTdp=Tdp+TdδTTdp=(I+TdδT)Tdp.
Then Tdp=0 and q=T¯Tdp=0. Finally, we prove X=T¯R(Td)+N(Td). For all x∈X, Tdx∈D(T). Since I+TdδT:D(T)→D(T) is bijective, we can find y∈D(T) such that Tdx=(I+TdδT)y, that is, Tdx=(I-TdT)y+TdT¯y. Hence
(2.5)Td(x-T¯y)=(I-TdT)y∈R(Td)∩N(TdT)={0}.
Thus y=TdTy∈R(Td) and Tdx=TdT¯y, which implies x-T¯y∈N(Td). Since x=T¯y+(x-T¯y)∈T¯R(Td)+N(Td), we obtain X=T¯R(Td)+N(Td).
(
3
)
⇒
(
1
). If N(T¯)∩R(Td)={0} and X=T¯R(Td)+˙N(Td), we first prove that I+δTTd:X→X is injective. In fact, if x∈X satisfies (I+δTTd)x=0, then
(2.6)T¯Tdx=TTdx-x∈T¯R(Td)∩N(Td)={0},
which means x=TTdx and T¯Tdx=0. Thus Tdx∈N(T¯)∩R(Td), that is, Tdx=0. Hence x=TTdx=0. Next, we will show that I+δTTd:X→X is surjective. In fact, for all y∈X, since X=T¯R(Td)+N(Td),y can be expressed by y=T¯Tdy1+y2, where y1∈X and y2∈N(Td). Hence
(2.7)(I+δTTd)(TTdy1+y2)=(TTdy1+y2)+(T¯-T)TdTTdy1=TTdy1+y2+T¯Tdy1-TTdy1=y2+T¯Tdy1=y.
Therefore, I+δTTd:X→X is bijective.Lemma 2.2.
LetT∈C(X) have the generalized Drazin inverse Td∈B(X), and let δT∈L(X) satisfy D(T)⊂D(δT) and T¯=T+δT∈C(X), then I+δTTd∈B(X). Further, if I+δTTd:X→X is bijective, then its inverse (I+δTTd)-1∈B(X) and
(2.8)B=Td(I+δTTd)-1=(I+TdδT)-1Td:X→X
is a well-defined bounded linear operator with R(B)=R(Td) and N(B)=N(Td).Proof.
SinceTd is bounded and T¯ is closed, by the Closed Operator Theorem, we know that T¯Td is bounded and so is I+δTTd=I-TTd+T¯Td.  Further, if I+δTTd:X→X is bijective, it follows from the Inverse Operator Theorem that (I+δTTd)-1∈B(X). By Lemma 2.1 and (I+TdδT)Td=Td(I+δTTd), we can see that
(2.9)B=Td(I+δTTd)-1=(I+TdδT)-1Td:X→R(Td)
is a well-defined bounded linear operator. Easily, we can obtain R(B)=R(Td) and N(B)=N(Td).The following theorem points out that the generalized Drazin inverse with the prescribed range and null space must possess the expressionTd(I+δTTd)-1.Theorem 2.3.
LetT∈C(X) have the generalized Drazin inverse Td∈B(X), and let δT∈L(X) satisfy D(T)⊂D(δT) and T¯=T+δT∈C(X). Then the following statements are equivalent:(1)
T¯ is generalized Drazin invertible with T¯π=Tπ;(2)
T¯ is generalized Drazin invertible with N(T¯d)=N(Td) and R(T¯d)=R(Td);(3)
I+δTTd:X→X is bijective and
(2.10)B=Td(I+δTTd)-1=(I+TdδT)-1Tdis the generalized Drazin inverse ofT¯;(4)
T¯ is generalized Drazin invertible with Td-T¯d=TdδTT¯d=T¯dδTTd;(5)
T¯ is generalized Drazin invertible with Td-T¯d=T¯dδTTd.Proof.
It is easy to see(1)⇒(2) and (4)⇒(5).
(
2
)
⇒
(
3
). Suppose that T¯ is generalized Drazin invertible with N(T¯d)=N(Td) and R(T¯d)=R(Td). We first show that I+δTTd:X→X is bijective. From Lemma 2.1, it suffices to show N(T¯)∩R(Td)={0} and X=T¯R(Td)+˙N(Td). Since T¯d is the generalized Drazin inverse of T¯, we can obtain N(T¯)∩R(T¯d)={0} and X=R(T¯T¯d)⊕N(T¯T¯d)=T¯R(T¯d)⊕N(T¯d). Hence
(2.11)N(T¯)∩R(Td)={0},X=T¯R(Td)⊕N(Td).
Next, by N(T¯d)=N(Td) and R(T¯d)=R(Td), we can obtain T¯d(I-TTd)=0 and (T¯dT¯-I)Td=0, that is,
(2.12)T¯d=T¯dTTd,T¯dT¯Td=Td.
Then T¯d+T¯dT¯Td-T¯dTTd=Td, that is,
(2.13)T¯d(I+δTTd)=T¯d[I+(T¯-T)Td]=Td.
Therefore, T¯d=Td(I+δTTd)-1. By Lemma 2.2, we have T¯d=(I+TdδT)-1Td.
(
3
)
⇒
(
4
). Since the generalized Drazin inverse is unique, we get
(2.14)T¯d=Td(I+δTTd)-1=(I+TdδT)-1Td.
Then T¯d(I+δTTd)=Td and (I+TdδT)T¯d=Td. Hence
(2.15)Td-T¯d=T¯dδTTd=TdδTT¯d.
(
5
)
⇒
(
1
). We only need to prove T¯T¯d=TTd. In fact, by Td-T¯d=T¯dδTTd, we obtain Td=T¯d(I+δTTd) and T¯d=(I-T¯dδT)Td. Hence
(2.16)T¯dT¯Td=T¯dT¯T¯d(I+δTTd)=T¯d(I+δTTd)=Td,T¯dTTd=(I-T¯dδT)TdTTd=(I-T¯dδT)Td=T¯d.
Thus
(2.17)T¯T¯d=T¯T¯dTTd=T¯T¯dTdT=T¯dT¯TdT=TdT=TTdonD(T).
Since D(T) is dense in X, T¯T¯d and TTd are bounded, we can get T¯T¯d=TTd and so T¯π=I-T¯T¯d=I-TTd=Tπ.Remark 2.4.
To guaranteeT¯∈C(X), a general condition of T boundedness is often assumed, see [14].Theorem 2.5.
LetT∈C(X) have the generalized Drazin inverse Td∈B(X) and δT∈L(X) satisfy D(T)⊂D(δT) and T¯=T+δT∈C(X). Assume that I+δTTd:X→X is bijective, then the following statements are equivalent:(1)
(2.18)B=Td(I+δTTd)-1=(I+TdδT)-1Tdis the generalized Drazin inverse ofT¯;(2)
T¯Td=TTdT¯Td, TdT¯=TdT¯TdT and limn→+∞∥T¯n(I-TdT)∥1/n=0;(3)
T¯TdT=TTdT¯ and limn→+∞∥T¯n(I-TdT)∥1/n=0;(4)
T¯TdT=TTdT¯ and T¯Tπ is quasinilpotent;(5)
T¯Td=TTdT¯Td, TdT¯=TdT¯TdT and T¯Tπ is quasinilpotent.In this case,T¯π=Tπ and
(2.19)∥T¯d-Td∥≤∥Td∥·∥δTTd∥·∥(I+δTTd)-1∥.Proof.
(
1
)
⇒
(
2
). If B is the generalized Drazin inverse of T¯, then by Theorem 2.3, we get T¯π=Tπ and so T¯B=TTd and BT¯=TdT. Hence T¯Td=TTd[I+(T¯-T)Td]=TTdT¯Td and TdT¯=[I+Td(T¯-T)]TdT=TdT¯TdT. Also, for all x∈D(T¯),
(2.20)[T¯(I-T¯B)]2x=[T¯(I-T¯B)T¯(I-T¯B)]x=T¯(I-T¯B)T¯(I-BT¯)x=T¯T¯(I-BT¯)(I-BT¯)x=T¯2(I-BT¯)x=T¯2(I-TdT)x.
Thus we obtain T¯(I-BT¯)x=T¯x-T¯BT¯x=(I-T¯B)T¯x∈D(T¯) and
(2.21)[T¯(I-T¯B)]3x=[T¯(I-T¯B)]2[T¯(I-T¯B)]x=[T¯(I-T¯B)]2[T¯(I-BT¯)]x=T¯2(I-BT¯)T¯(I-BT¯)x(by(2.20))=T¯2(I-T¯B)T¯(I-BT¯)x=T¯3(I-BT¯)(I-BT¯)x=T¯3(I-BT¯)x=T¯3(I-TdT)x.
Hence for all n∈N,
(2.22)[T¯(I-T¯B)]n=T¯n(I-TdT)onD(T).
Therefore limn→+∞∥T¯n(I-TdT)∥1/n=limn→+∞∥[T¯(I-T¯B)]n∥1/n=0.
(
2
)
⇒
(
1
). It follows from Lemma 2.2 that B=Td(I+δTTd)-1=(I+TdδT)-1Td is a well-defined bounded linear operator with R(B)=R(Td)⊂D(T)=D(T¯). By T¯Td=TTdT¯Td and TdT¯=TdT¯TdT, we can verify
(2.23)T¯B=T¯Td(I+δTTd)-1=TTdT¯Td(I+δTTd)-1=TTd(T+δT)Td(I+δTTd)-1=(TTd+TTdδTTd)(I+δTTd)-1=TTd(I+δTTd)(I+δTTd)-1=TTd
and by Lemma 2.1,
(2.24)BT¯=(I+TdδT)-1TdT¯=(I+TdδT)-1TdT¯TdT=(I+TdδT)-1Td(T+δT)TdT=(I+TdδT)-1(TdT+TdδT)TdT=(I+TdδT)-1(I+TdδT)TdT=TdT.
Thus T¯B=BT¯ on D(T¯). Obviously, R(I-T¯B)=R(I-TTd)⊂D(T)=D(T¯). Also, we can see
(2.25)BT¯B=Td(I+δTTd)-1T¯Td(I+δTTd)-1=Td(I+δTTd)-1(T+δT)Td(I+δTTd)-1=Td(I+δTTd)-1(TTd+δTTdTTd)(I+δTTd)-1=Td(I+δTTd)-1(I+δTTd)TTd(I+δTTd)-1=Td(I+δTTd)-1=B.
To the aim, we need show T¯(I-T¯B) is quasinilpotent. In fact, as in (1)⇒(2), we can obtain (2.22) and so
(2.26)limn→+∞∥[T¯(I-T¯B)]n∥1/n=limn→+∞∥T¯n(I-TdT)∥1/n=0.
Therefore, B is the generalized Drazin inverse of T¯.
(
2
)
⇔
(
3
). If T¯Td=TTdT¯Td and TdT¯=TdT¯TdT, then
(2.27)T¯TdT=TTdT¯TdT=TTdT¯.
Conversely, if T¯TdT=TTdT¯, then
(2.28)T¯Td=T¯TdTTd=TTdT¯Td,TdT¯=TdTTdT¯=TdT¯TdT.
(
3
)
⇔
(
4
). It follows from T¯TdT=TTdT¯ that
(2.29)(T¯Tπ)2=T¯TπT¯Tπ=T¯(I-TTd)T¯Tπ=T¯(T¯-TTdT¯)Tπ=T¯(T¯-T¯TdT)Tπ=T¯2(I-TdT)Tπ=T¯2(I-TdT)onD(T)
and then for all n∈N,
(2.30)(T¯Tπ)n=T¯n(I-TdT)onD(T).
Therefore, T¯Tπ is quasinilpotent if and only if limn→+∞∥T¯n(I-TdT)∥1/n=0.
(
4
)
⇔
(
5
). It is similar to (2)⇔(3). In this case,
(2.31)∥B-Td∥=∥Td(I+δTTd)-1-Td∥≤∥TdδTTd(I+δTTd)-1∥≤∥Td∥·∥δTTd∥·∥(I+δTTd)-1∥.Remark 2.6.
Theorems2.3 and 2.5 generalize the main results in [3–8, 10, 12, 13].As applications, we recover some known results in [3, 6, 7].Corollary 2.7 (see [3]).
LetT∈C(X) have the generalized Drazin inverse Td∈B(X) and δT∈B(X) be T-compatible. If I+TdδT:X→X is invertible, then T¯=T+δT is generalized Drazin invertible and T¯d=Td(I+δTTd)-1=(I+TdδT)-1Td.Proof.
It is easy to seeT¯∈C(X) and T¯TdT=TTdT¯. Since
(2.32)T¯Tπ=TTπ+δTTπ
is the sum of two commuting quasinilpotent operators, we know that T¯Tπ is quasinilpotent. Hence by Theorem 2.5, we can get what we desired.Corollary 2.8.
LetT∈C(X) have the generalized Drazin inverse Td∈B(X) and δT∈L(X) satisfy D(T)⊂D(δT) and T¯=T+δT∈C(X). Assume that I+δTTd:X→X is bijective and
(2.33)δT=TTdδT=δTTdT,
then B=Td(I+δTTd)-1=(I+TdδT)-1Td is the generalized Drazin inverse of T¯=T+δT.Proof.
IfδT=TTdδT=δTTdT, then T¯TdT=TTdT¯ and T¯Tπ=TTπ is quasinilpotent. From Theorem 2.5, B is the generalized Drazin inverse of T¯.Remark 2.9.
The conditionδT=TTDδT=δTTDT is called the condition (W) in [6, 7], which appear as a sufficient condition to ensure that TD(I+δTTD)-1 is the Drazin inverse of T¯.The following theorem shows that the above condition(W) is exactly the stable characterization of group inverse. Recall that an operator S∈B(X) is said to be the group inverse of T∈C(X) if S satisfies R(S)⊂D(T) and
(2.34)TST=T;STS=S,ST=TSonD(T).The group inverse ofT is unique if it exists, which is always denoted by T#.Theorem 2.10.
LetT∈C(X) have the group inverse T#∈B(X) and let δT∈L(X) satisfy D(T)⊂D(δT) and T¯=T+δT∈C(X). Assume that I+δTT#:X→X is bijective, then
(2.35)B=T#(I+δTT#)-1=(I+T#δT)-1T#
is the group inverse of T¯ if and only if
(2.36)δT=TT#δT=δTT#T.
In this case, T¯π=Tπ, R(T¯)=R(T) and N(T¯)=N(T).Proof.
IfδT=TT#δT=δTT#T, then T¯=T¯T#T=TT#T¯ and
(2.37)T¯Tπ=T¯(I-TT#)=T¯(I-T#T)=0onD(T).
Hence T¯Tπ=0. And by (4) in Theorem 2.5, we know that B is the generalized Drazin inverse of T¯ with T¯B=TT# and BT¯=T#T. Hence T¯BT¯=TT#T¯=T¯. Therefore, B is also the group inverse of T¯. Conversely, if B is the group inverse of T¯, then
(2.38)R(T¯)=R(T¯B)=R(BT¯)¯=R(T#T)¯=R(TT#)=R(T),(2.39)N(T¯)=N(BT¯)=N(T¯B)∩D(T¯)=N(TT#)∩D(T)=N(T#T)=R(I-T#T).
Hence
(2.40)T¯-TT#T¯=(I-TT#)T¯=0,(2.41)T¯-T¯T#T=T¯(I-T#T)=0.
Thus T¯=TT#T¯=T¯T#T, that is, δT=TT#δT=δTT#T.The following corollary extends the Proposition 4.2 in [3] to the case of closed linear operators. Recall that T∈C(X) is an EP (Equal Projections) operator if T has the Moore-Penrose inverse T†∈B(X) satisfying T†T=TT† on D(T).Corollary 2.11.
LetX be a Hilbert space and T∈C(X) be an EP operator. Let δT∈L(X) satisfy D(T)⊂D(δT) and T¯=T+δT∈C(X). Assume that I+δTT†:X→X is bijective, then T¯ is also an EP operator and
(2.42)B=T†(I+δTT†)-1=(I+T†δT)-1T†
is its Moore-Penrose inverse if and only if
(2.43)δT=TT†δT=δTT†T.
In this case, BT¯=T¯B on D(T¯), T¯π=Tπ, R(T¯)=R(T) and N(T¯)=N(T).Proof.
IfT¯ is an EP operator and B is its Moore-Penrose inverse, then B is also its group inverse. By Theorem 2.10, we get δT=TT†δT=δTT†T. Conversely, if δT=TT†δT=δTT†T, then by Theorem 2.10 again, B is the group inverse of T¯, T¯B=TT† and BT¯=T†T. Therefore, B is the Moore-Penrose inverse of T¯.To illustrate our results, we give an example of a closedEP operator.Example 2.12 (see [15]).
Let(2.44)L2[0,1]={f,fmeasurablecomplex-valuedon[0,1],∫[0,1]‍|f(x)|2dx<∞}
be the Hilbert space with the inner product:
(2.45)〈f,g〉=∫[0,1]‍f(x)¯g(x)dx,f,g∈L2[0,1].
Sett:[0,1]→C by
(2.46)t(x)={1x=01x0<x≤1
and define the maximal operator of multiplication T by t on L2[0,1], that is,
(2.47)Tf=tf,forf∈D(T)={f∈L2[0,1],tf∈L2[0,1]},
then T is a densely defined closed linear operator [16]. Since |t(x)|≥1, forallx∈[0,1], R(T)=L2[0,1] and T has a bounded inverse T-1:L2[0,1]→L2[0,1] defined by
(2.48)T-1g=t1g,∀g∈L2[0,1],
where
(2.49)t1(x)={1x=0,x0<x≤1.
Therefore, T is a closed operator on L2[0,1] with a bounded inverse T-1, which is both the Moore-Penrose inverse and the group inverse of T.



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