# Stability Theory for Nullity and Deficiency of Linear Relations

Authors: Silas Luliro Kito; Gerald Wanjala
Journal: Abstract and Applied Analysis
 (2021)

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## Abstract

LetA and B be two closed linear relations acting between two Banach spaces X and Y, and let λ be a complex number. We study the stability of the nullity and deficiency of A when it is perturbed by λB. In particular, we show the existence of a constant ρ>0 for which both the nullity and deficiency of A remain stable under perturbation by λB for all λ inside the disk λ<ρ.


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## Body

## 1. Introduction

For purposes of introduction, we shall consider bounded linear operatorsA and B with domain X and range in Y. As usual, let NA and RA denote the null space and range of A respectively. The dimensions of NA and Y/RA are called the nullity and the deficiency of A respectively and denoted by αA and βA. It is well known that αA and βA have some kind of stability when A is subjected to some kind of perturbation (see for example [1]). More precisely, αA and βA are unchanged when A is perturbed by some bounded linear operator B under certain prescribed conditions. This stability can be described in the form
(1)αA−B−βA−B=αA−βA.Another convenient way of describing this stability is to put it in the form(2)αA−B=αA,βA−B=βA.The stability concept described here is very useful in studying eigenvalue problems of the formAx=λBx and A∗y=λB∗y, where A∗ denotes the adjoint operator.This paper deals with the stability theory for nullity and deficiency of linear relations, and it can be seen as a generalization of the classical theory for the corresponding quantities for linear operators. The theory and exposition developed here goes along the lines of the classical texts on the perturbation theory for linear operators (see for example [1, 2]), but in a more general setting. Some stability theorems for multivalued linear operators or what we refer to here as linear relations have been considered in [3] and more recently in [4]. In either of these cases, the perturbing multivalued linear operator B does not vary with the varying λ as the case we consider here.


## 2. Preliminaries

### 2.1. Relations on Sets

In this section, we introduce some notation and consider some basic concepts concerning relations on sets. LetU and V be two nonempty sets. By a relationT from U to V, we mean a mapping whose domain DT is a nonempty subset of U and, taking values in 2V\∅, the collection of all nonempty subsets of V. Such a mapping T is also referred to as a multivalued operator or at times as a set-valued function. If T maps the elements of its domain to singletons, then T is said to be a single valued mapping or operator. Let T be a relation from U to V, and let Tu denote the image of an element u∈U under T. If we define Tu=∅ for u∈U and u∉DT, then the domain DT of T is given by
(3)DT=u∈U:Tu≠∅.Denote byRU,V the class of all relations from U to V. If T belongs to RU,V, the graph of T, which we denote by GT, is the subset of U×V defined by
(4)GT=u,v∈U×V:u∈DT,v∈Tu.A relationT∈RU,V is uniquely determined by its graph, and conversely, any nonempty subset of U×V uniquely determines a relation T∈RU,V.For a relationT∈RU,V, we define its inverse T−1 as the relation from V to U whose graph GT−1 is given by
(5)GT−1=v,u∈V×U:u,v∈GT.LetT∈RU,V. Given a subset M of U, we define the image of M, TM to be
(6)TM=⋃‍Tm:m∈M∩DT.With this notation we define the range ofT by
(7)RT≔TU.LetN be a nonempty subset of V. The definition of T−1 given in (5) above implies that
(8)T−1N=u∈DT:N∩Tu≠∅.If in particularv∈RT, then
(9)T−1v=u∈DT:v∈Tu.For a detailed study of relations, we refer to [3, 5–8], and [9].


### 2.2. Linear Relations

LetX and Y be linear spaces over a field K=ℝ (or ℂ), and let T∈RX,Y. We say that T is a linear relation or a multivalued linear operator if for all x,z∈DT and any nonzero scalar α we have
(1)
Tx+Tz=Tx+z(2)
αTx=TαxThe equalities in items (1) and (2) above are understood to be set equalities. These two conditions indirectly imply that the domain of a linear relation is a linear subspace. The class of linear relations inRX,Y will be denoted by LRX,Y. If X=Y, then we denote LRX,X by LRX. We say that T is a linear relation in X if T∈LRX. We shall use the term operator to refer to a single valued linear operator while a multivalue linear operator will be generally referred to as a linear relation.IfX and Y are normed linear spaces, we say that T∈LRX,Y is closed if its graph GT is a closed subspace of X×Y. The collection of all such T will be denoted by CLRX,Y.We conclude this section with the following theorems which are taken from [3].Theorem 1.
LetT∈RX,Y. The following properties are equivalent.
(i)
T is a linear relation(ii)
GT is a linear subspace of X×Y(iii)
T−1 is a linear relation(iv)
GT−1 is a linear subspace of Y×XCorollary 2.
LetT∈RX,Y.
(i)
Then,T is a linear relation if and only if
(10)Tαx1+βx2=αTx1+βTx2holds for all x1,x2∈DT and some nonzero scalars α and β(ii)
IfT is a linear relation, then T0 and T−10 are linear subspacesFor a linear relationT, the subspace T−10 is called the null space (or kernel) of T and is denoted by NT.Theorem 3.
LetT be a linear relation in a linear space X, and let x∈DT. Then, y∈Tx if and only if
(11)Tx=T0+y.Theorem3 shows that T is single valued if and only if T0=0.Theorem 4.
LetT∈RX,Y. Then, T is a linear relation if and only if for all x1,x2∈DT and all scalars α and β(12)αTx1+βTx2⊂Tαx1+βx2.Theorem 5.
LetT∈LRX,Y. Then,
(a)
TM+N=TM=TN for M⊂X and N⊂DT(b)
TT−1M=M∩RT+T0 for M⊂Y(c)
T−1TM=M∩DT+T−10 for M⊂X


### 2.3. Normed Linear Relations

LetX be a normed linear space. By BX, we shall mean the set
(13)BX≔x∈X:∣x∣≤1.For a closed linear subspaceE of X, we denote by QE the natural quotient map with domain X and null space E. For T∈LRX,Y, we shall denote QT0¯ by QT. It is well known that for T∈LRX,Y, the operator QTT is single valued (see [3]).ForT∈LRX,Y, we set Tx=QTTx for x∈DT and T=QTT. Note that these notions do not define a norm since nonzero relations can have zero norm.Lemma 6.
LetA,B∈CLRX,Y be such that DB⊃DA and B0⊂A0. If x1,x2∈DA are such that Ax1∩Bx2≠∅, then Ax1−Bx2⊂A0.Proof.
Letz∈Ax1∩Bx2. Since QA and QB are single valued, we see that
(14)QAAx1−Bx2=QAAx1−QABx2=z~−z~=0^.
Hence,Ax1−Bx2∈A0.☐The following lemma is proved in [3].Lemma 7.
The following properties are equivalent for a linear relationA.
(i)
A is closed(ii)
QAA is closed and A0 is closedLemma 8.
(a)
LetT∈LRX,Y be bounded. Then, Tx≤Tx(b)
ForS,T∈LRX,Y with DS⊂DT and T0⊂S0, we have(15)Sx+Tx≥Sx−Tx.Proof.
(a)
From ([3], II.1.6), we have T=supx∈BDTTx so that(16)T=supx∈DT1∥x∥Tx,T≥1xTx,x∈DT.
The inequalityTx≥Tx, for all x∈DT, then follows from ([3], II.1.5).
(b)
SinceT0⊂S0, we see that S+T0=S0+T0=S0 since S0 is a subspace (linear subset). For x∈DS, let s∈Sx and let t∈Tx. Then, s+t∈S+Tx=Sx+Tx, and so by ([3], II.1.4), we get(17)Sx+Tx=dists+t,S+T0=dists+t,S0≥dists,S0−distt,S0≥dists,S0−distt,T0=Sx−Tx.☐LetX be a normed space. By X′, we denote the norm dual of X, that is, the space of all continuous linear functionals x′ defined on X, with norm
(18)x′=infλ:x,x′≤λx,forallx∈X,where x,x′≔x′x denotes the action of x′∈X′ on x∈X. If M⊂X and N⊂X′, we write M⊥ and N⊤ to mean
(19)M⊥≔x′∈X′:x,x′=0,forallx∈M,N⊤≔x∈X:x,x′=0,forallx′∈N.LetT be a linear relation with DT⊂X and RT⊂Y. We define the adjoint T′ of T by
(20)GT′≔G−T−1⊥⊂Y′×X′,where
(21)y,x,y′,x′=x,x′+y,y′.This means that(22)y′,x′∈GT′ifandonlyify,y′−x,x′=0,forallx,y∈GT.From (22), we see that y′y=x′x for all y∈Tx,x∈DT. Hence,
(23)x′∈T′y′,ifandonlyify′Tx=x′x,forallx∈DT.This means thatx′ is an extension of y′Tx, and therefore, the adjoint T′ can be characterized as follows:
(24)GT′=y′,x′∈Y′×X′suchthatx′isanextensionofy′T.Please note thatT′∈CLRY′,X′ (see [3], III.1.2).Lemma 9 ([3], III.1.4).
LetT be a closed linear relation. Then,
(1)
NT′=RT⊥(2)
T′0=DT⊥(3)
NT=RT′⊤(4)
T0=DT′⊤Remark 10.
IfT and S are closed linear relations with DT⊂DS and S0⊂T0, then S′0⊂T′0 by Lemma 9(2).


## 2.1. Relations on Sets

In this section, we introduce some notation and consider some basic concepts concerning relations on sets. LetU and V be two nonempty sets. By a relationT from U to V, we mean a mapping whose domain DT is a nonempty subset of U and, taking values in 2V\∅, the collection of all nonempty subsets of V. Such a mapping T is also referred to as a multivalued operator or at times as a set-valued function. If T maps the elements of its domain to singletons, then T is said to be a single valued mapping or operator. Let T be a relation from U to V, and let Tu denote the image of an element u∈U under T. If we define Tu=∅ for u∈U and u∉DT, then the domain DT of T is given by
(3)DT=u∈U:Tu≠∅.Denote byRU,V the class of all relations from U to V. If T belongs to RU,V, the graph of T, which we denote by GT, is the subset of U×V defined by
(4)GT=u,v∈U×V:u∈DT,v∈Tu.A relationT∈RU,V is uniquely determined by its graph, and conversely, any nonempty subset of U×V uniquely determines a relation T∈RU,V.For a relationT∈RU,V, we define its inverse T−1 as the relation from V to U whose graph GT−1 is given by
(5)GT−1=v,u∈V×U:u,v∈GT.LetT∈RU,V. Given a subset M of U, we define the image of M, TM to be
(6)TM=⋃‍Tm:m∈M∩DT.With this notation we define the range ofT by
(7)RT≔TU.LetN be a nonempty subset of V. The definition of T−1 given in (5) above implies that
(8)T−1N=u∈DT:N∩Tu≠∅.If in particularv∈RT, then
(9)T−1v=u∈DT:v∈Tu.For a detailed study of relations, we refer to [3, 5–8], and [9].


## 2.2. Linear Relations

LetX and Y be linear spaces over a field K=ℝ (or ℂ), and let T∈RX,Y. We say that T is a linear relation or a multivalued linear operator if for all x,z∈DT and any nonzero scalar α we have
(1)
Tx+Tz=Tx+z(2)
αTx=TαxThe equalities in items (1) and (2) above are understood to be set equalities. These two conditions indirectly imply that the domain of a linear relation is a linear subspace. The class of linear relations inRX,Y will be denoted by LRX,Y. If X=Y, then we denote LRX,X by LRX. We say that T is a linear relation in X if T∈LRX. We shall use the term operator to refer to a single valued linear operator while a multivalue linear operator will be generally referred to as a linear relation.IfX and Y are normed linear spaces, we say that T∈LRX,Y is closed if its graph GT is a closed subspace of X×Y. The collection of all such T will be denoted by CLRX,Y.We conclude this section with the following theorems which are taken from [3].Theorem 1.
LetT∈RX,Y. The following properties are equivalent.
(i)
T is a linear relation(ii)
GT is a linear subspace of X×Y(iii)
T−1 is a linear relation(iv)
GT−1 is a linear subspace of Y×XCorollary 2.
LetT∈RX,Y.
(i)
Then,T is a linear relation if and only if
(10)Tαx1+βx2=αTx1+βTx2holds for all x1,x2∈DT and some nonzero scalars α and β(ii)
IfT is a linear relation, then T0 and T−10 are linear subspacesFor a linear relationT, the subspace T−10 is called the null space (or kernel) of T and is denoted by NT.Theorem 3.
LetT be a linear relation in a linear space X, and let x∈DT. Then, y∈Tx if and only if
(11)Tx=T0+y.Theorem3 shows that T is single valued if and only if T0=0.Theorem 4.
LetT∈RX,Y. Then, T is a linear relation if and only if for all x1,x2∈DT and all scalars α and β(12)αTx1+βTx2⊂Tαx1+βx2.Theorem 5.
LetT∈LRX,Y. Then,
(a)
TM+N=TM=TN for M⊂X and N⊂DT(b)
TT−1M=M∩RT+T0 for M⊂Y(c)
T−1TM=M∩DT+T−10 for M⊂X


## 2.3. Normed Linear Relations

LetX be a normed linear space. By BX, we shall mean the set
(13)BX≔x∈X:∣x∣≤1.For a closed linear subspaceE of X, we denote by QE the natural quotient map with domain X and null space E. For T∈LRX,Y, we shall denote QT0¯ by QT. It is well known that for T∈LRX,Y, the operator QTT is single valued (see [3]).ForT∈LRX,Y, we set Tx=QTTx for x∈DT and T=QTT. Note that these notions do not define a norm since nonzero relations can have zero norm.Lemma 6.
LetA,B∈CLRX,Y be such that DB⊃DA and B0⊂A0. If x1,x2∈DA are such that Ax1∩Bx2≠∅, then Ax1−Bx2⊂A0.Proof.
Letz∈Ax1∩Bx2. Since QA and QB are single valued, we see that
(14)QAAx1−Bx2=QAAx1−QABx2=z~−z~=0^.
Hence,Ax1−Bx2∈A0.☐The following lemma is proved in [3].Lemma 7.
The following properties are equivalent for a linear relationA.
(i)
A is closed(ii)
QAA is closed and A0 is closedLemma 8.
(a)
LetT∈LRX,Y be bounded. Then, Tx≤Tx(b)
ForS,T∈LRX,Y with DS⊂DT and T0⊂S0, we have(15)Sx+Tx≥Sx−Tx.Proof.
(a)
From ([3], II.1.6), we have T=supx∈BDTTx so that(16)T=supx∈DT1∥x∥Tx,T≥1xTx,x∈DT.
The inequalityTx≥Tx, for all x∈DT, then follows from ([3], II.1.5).
(b)
SinceT0⊂S0, we see that S+T0=S0+T0=S0 since S0 is a subspace (linear subset). For x∈DS, let s∈Sx and let t∈Tx. Then, s+t∈S+Tx=Sx+Tx, and so by ([3], II.1.4), we get(17)Sx+Tx=dists+t,S+T0=dists+t,S0≥dists,S0−distt,S0≥dists,S0−distt,T0=Sx−Tx.☐LetX be a normed space. By X′, we denote the norm dual of X, that is, the space of all continuous linear functionals x′ defined on X, with norm
(18)x′=infλ:x,x′≤λx,forallx∈X,where x,x′≔x′x denotes the action of x′∈X′ on x∈X. If M⊂X and N⊂X′, we write M⊥ and N⊤ to mean
(19)M⊥≔x′∈X′:x,x′=0,forallx∈M,N⊤≔x∈X:x,x′=0,forallx′∈N.LetT be a linear relation with DT⊂X and RT⊂Y. We define the adjoint T′ of T by
(20)GT′≔G−T−1⊥⊂Y′×X′,where
(21)y,x,y′,x′=x,x′+y,y′.This means that(22)y′,x′∈GT′ifandonlyify,y′−x,x′=0,forallx,y∈GT.From (22), we see that y′y=x′x for all y∈Tx,x∈DT. Hence,
(23)x′∈T′y′,ifandonlyify′Tx=x′x,forallx∈DT.This means thatx′ is an extension of y′Tx, and therefore, the adjoint T′ can be characterized as follows:
(24)GT′=y′,x′∈Y′×X′suchthatx′isanextensionofy′T.Please note thatT′∈CLRY′,X′ (see [3], III.1.2).Lemma 9 ([3], III.1.4).
LetT be a closed linear relation. Then,
(1)
NT′=RT⊥(2)
T′0=DT⊥(3)
NT=RT′⊤(4)
T0=DT′⊤Remark 10.
IfT and S are closed linear relations with DT⊂DS and S0⊂T0, then S′0⊂T′0 by Lemma 9(2).


## 3. Lower Bound of a Closed Linear relation

Consider a closed linear relationA on a Banach space X, and let NA denote the null space of A which is closed since A is closed. Since NA⊂DA, a coset x~∈X~=X/NA which contains a point of x∈DA consists entirely of points of DA. To see that this is the case, let x~∈X~ and let x,y∈x~ with x∈DA. Then, y−x∈NA⊂DA and the linearity of DA implies that y=x+y−x∈DA. Let D~ denote the collection of all such cosets x~. On setting,
(25)Ax~≔QAAx,forx~∈D~,we define a linear operator A:X~⟶X^, where X^≔X/A0. To see that (25) is well defined, let x,y∈x~. Then, x−y∈NA, and therefore,
(26)0∈A0=Ax−y=Ax−Ay.We see from (26) that Ax∩Ay≠∅. So, let u∈Ax∩Ay. Then,
(27)Ax=A0+u=Ay,so that QAAx=QAAy. We have
(28)DA=D~,RA=RQAA,NA=0~.Remark 11.
SinceA0⊂RA, we also have that a coset x^∈X^ that contains a point of RA consists entirely of element of RA. To see that this is the case, let x^ be a coset in X^ and let u,v∈x^ with u∈RA. Then, v−u∈A0⊂RA. The linearity of RA implies that v=u+v−u∈RA.Lemma 12.
The linear operatorA defined by (25) is closed.Proof.
Letx~n be a sequence in D~ such that x~n⟶x~∈X~, and let QAAxn be a sequence in RA such that QAAxn⟶y^∈X^. Let xn∈x~n and x∈x~. Since x~n⟶x~, we see that distxn−x,NA⟶0. This means that xn−x converges to some element of NA, say,
(29)xn−x⟶u∈NA.
From (29), we see that xn⟶x+u=w∈x~.
SinceQAAxn⟶y^∈X^, that is, z^n⟶y^, we see that distzn−y,A0⟶0 as n⟶∞ and so zn⟶y+v=z∈y^ for some v∈A0 (where zn∈Axn for each n∈ℕ). The closedness of A implies that w∈DA and z∈Aw. Hence, x~∈D~ and Ax~=QAAx=y^, showing that A is closed.☐We see thatA−1 is single valued since A−10^=0~. We now introduce the quantity γA called the lower bound of the linear relation A. By definition,
(30)γA=1A−1,with the understanding that γA=0 if A−1 is unbounded and that γA=∞ if A−1=0. It follows from (30) that
(31)γA=supγ∈ℝ:Ax≥γx~=distx,NA,∀x∈DA.Note thatγA=∞ if and only if Ax=A0 for all x∈DA. In order for (31) to hold even for this case, one should stipulate that ∞×0=0. Obviously, γA=γA.Please note that characterization (31) implies that if γA=0 then the domain of A cannot consist of the zero element alone.The fact thatγA=∞ if and only if Ax=A0 for all x∈DA leads to Lemma 13 (see also [3], Proposition II.2.2).Lemma 13.
ForA∈CLRX,Y, we have
(32)γA=∞,ifDA⊂NA,infAxx~:x∈DA&x∉NA,otherwise.Remark 14.
A bounded linear operatorT is closed if and only if DT is closed.Proof.
Suppose thatun⟶u with un∈DT. The boundedness of T implies that Tun is a Cauchy sequence and therefore converges, say Tun⟶v. The closedness of T implies that u∈DT and Tu=v. This shows that DT is closed.IfS is a closed linear relation from X to Y, the graph of S, GS is a closed subset of X×Y. Sometimes, it is convenient to regard it as a subset of Y×X. More precisely, let G′S be the linear subset of Y×X consisting of all pairs of the form v,u, where u∈DS and v∈Su. We shall call G′S the inverse graph of S. As in the case of the graph GS, G′S is closed if and only S−1 is closed. Clearly, GS=G′S−1. Thus, S−1 is closed if and only S is closed.Lemma 15.
IfA is a closed linear relation in a Banach space X, then RA is closed if and only if γA>0.Proof.
By definition,γA>0 if and only if A−1 is bounded (where A is the operator defined in (25)), and this is true if and only if DA−1=RA=RQAA is closed (we use the fact that A−1 is closed because A is closed, and then apply Remark (14)).
Now, assume thatγA>0 and let yn be a convergent sequence in RA with
(33)yn⟶y.
SinceQA is a bounded linear operator, the sequenceQAyn is a Cauchy sequence in RQAA and therefore converges to a point z^∈RQAA⊂X^=X/A0 since RQAA is closed. We see that distyn−z,A0⟶0 as n⟶∞ so that yn−z⟶v for some v∈A0, that is,
(34)yn⟶z+v∈z^.
SinceA0⊂RA, a coset x^∈X^ that contains a point of RA consists entirely of element of RA. To see that this is the case, let x^ be a coset in X^ and let u,v∈x^ with u∈RA. Then, v−u∈A0⊂RA. The linearity of RA implies that v=u+v−u∈RA.
We see from (33) and (34) that y∈z^ and that y∈RA since z∈RA and y∈z^. This shows that RA is closed.
On the other hand, assume thatRA is closed. Since A−1 is closed (since A is closed), it is enough, by the closed graph theorem, to show that DA−1=RA=RQAA is closed. So, assume that z^n is a sequence in RQAA such that z^n⟶z^∈X^. Then, distzn−z,A0⟶0 as n⟶∞. Hence, there exists an element w∈A0 such that zn⟶z+w∈z^. The closedness of RA implies that z+w∈RA so that z^∈RQAA.Please see ([3], III.5.3) for another proof of Lemma 15.For the definition of continuity and openness of a linear relationT mentioned in Lemmas 16 and 17, please refer to [3].Lemma 16 ([3], II.3.2, III.1.3, III.1.5, III.4.6).
LetS,T∈LRX,Y. Then,
(a)
T is continuous if and only if T<∞(b)
λT′=λT′forλ≠0(c)
T is open if and only if γT>0(d)
IfDS⊃DT and S<∞ then T+S′=T′+S′Lemma 17([3], III.4.6).
(a)
T is continuous if and only if DT′=T0⊥(b)
T is open if and only if RT′=NT⊥(c)
IfT is continuous, then T′=T(d)
IfT is open, then γT=γT′


## 4. The Gap between Closed Linear Manifolds and Their Dimensions

LetZ be a Banach space, and let L be a closed subspaces of Z. We denote by SL the unit sphere of L, that is, SL≔u∈L:u=1. For any two closed linear manifolds M and N of Z with M≠0, define the gap between M and N, denoted by δM,N to be
(35)δM,N≔supu∈SMdistu,N,and set δM,N=0 if M=0. δM,N can also be characterized as the smallest number δ for which
(36)distu,N≤δu,forallu∈M.It can be seen from the definition that0≤δM,N≤1.See [1] for Lemma 18.Lemma 18.
LetM and N be linear manifolds in a Banach space Z. If dimM>dimN, then there exists an x∈M such that
(37)distx,N=x>0.Lemma18 can be expressed in the language of the quotient space as follows.Lemma 19.
LetM and N be linear manifolds in a Banach space Z. If dimM>dimN, then there exists an x∈M such that
(38)x~=x>0,x~∈X~≔X/NNisclosedsincedimN<∞.Lemma20 is a direct consequence of the preceding one.Lemma 20.
Ifx~<x for every none zero x∈M, where x~∈X~=X/N, then dimM≤dimN.See ([1], Page 200) and [2] for Lemmas 21 and 22 respectively.Lemma 21.
LetM and N be closed linear manifolds of a Banach space Z. If δM,N<1, then dimM≤dimN.Lemma 22.
Letx be an element of a normed linear space X, and let M and N be closed linear subspaces of X. Consider the quotient space X~≔X/N, and let x~ denote the quotient class of x. For any ε>0, there exists x0∈x~ such that
(39)distx0,M≥1−ε1−δM,N1+δM,Nx0.


## 5. The QuantityνA:B

LetX and Y be two linear spaces and let A,B∈LRX,Y with B0⊂A0. For n∈ℕ, let Mn and Nn be the linear manifolds of X and Mn′ and Nn′ be the linear manifolds of Y′ defined inductively as follows:
(40)M0=X,Mn=B−1AMn−1,forn=1,2,⋯,(41)N1=A−10,Nn=A−1BNn−1,forn=2,3,⋯,(42)M0′=Y′,Mn′=B′−1A′Mn−1′,forn=1,2,⋯,(43)N1′=A′−10,Nn′=A′−1B′Nn−1′,forn=2,3,⋯.IfMk⊃Mk+1, then AMk⊃AMk+1, and therefore,
(44)Mk+1=B−1AMk⊃B−1AMk+1=Mk+2.SinceM0=X⊃DB⊃M1, we conclude by induction that
(45)M0⊃M1⊃M2⊃⋯⊃NB.Similarly,(46)N1⊂N2⊂N3⊂⋯⊂DA.Note that(47)N1=NA.Lemma 23.
Letn be a positive integer. The following first n conditions are equivalent to one another, and they in turn imply that condition κ holds.
1N1⊂Mn
2N2⊂Mn−1
⋮
nNn⊂M1,
κANk+1∩BNk≠∅,Nk⊂DB,fork=1,2,⋯,n.Proof.
First, we prove the equivalence of the conditions1 to n. For each r=1,2,⋯,n−1, r implies r+1. In fact if Nr⊂Mn−r+1, then (44), (45), and (47) imply that Nr+1=A−1BNr⊂A−1BMn−r+1⊂A−1AMn−r+B0⊂A−1AMn−r+A0=A−1AMn−r+ANA=A−1AMn−r+NA⊂Mn−r+NA+A−10=Mn−r+A−10=Mn−r+N1⊂Mn−r+Nr⊂Mn−r+Mn−r+1=Mn−r.
Conversely,r+1 implies r. In fact, if Nr+1⊂Mn−r, then
(48)Nr⊂Nr+1⊂Mn−r=B−1AMn−r−1,so that each x∈Nr has the property that there exists a z∈Bx such that z∈Ay for some y∈Mn−r−1. Then, y∈A−1BNr=Nr+1⊂Mn−r and so x∈B−1AMn−r=Mn−r+1 This proves that Nr⊂Mn−r+1.
Next, we prove thatn implies κ. So, suppose that n is satisfied. Then, Nk⊂Nn⊂M1=B−1AX⊂DB for k<n, so that for each x∈Nk, there exists a z∈Bx such that z∈Ay for some y∈X. Then, y∈A−1BNk=Nk+1 and so ANk+1∩BNk≠∅.☐IfN1⊂Mn, then N1⊂Mn′, for all n′<n since Mn is a nonincreasing sequence. We denote by νA:B the smallest number n for which the condition N1⊂Mn (or any one of the other equivalent conditions) is not satisfied. We set νA:B=∞ if there is no such n. This is the case if for example A−10⊂B−10.Lemma 24.
LetX and Y be Banach spaces and let A,B∈CLRX,Y with DA=DB=X. Then,
(49)Mn′⊂BNn⊥,Nn′⊂AMn−1⊥,forn=1,2,⋯Proof.
First, we show that (49) holds for n=1. To begin with, let y′∈M1′ and let x∈DB∩N1. Then, by definition, y′∈B′−1A′Y′ and x∈A−10∩DB. Hence, there exists an element x′∈A′Y′∩RB′ such that y′,x′∈GB′. Since x′∈A′Y′, there exists an element f′∈DA′⊂Y′ such that f′,x′∈GA′. Since x,0∈GA, (22) implies that f′0=x′x so that x′x=0. So, for y∈Bx, y′y=x′x=0, showing the y′∈BN1⊥.
The second inclusion follows from (see Lemma9(1)).
(50)N1′=NA′=RA⊥=AM0⊥
We shall therefore assume that (49) has been proved for n=k and prove it for n=k+1. So, let g′∈Mk+1′ and let z∈DB∩Nk+1. Then, g′∈B′−1A′Mk′ and z∈A−1BNk∩DB. Hence, there exists an element h′∈A′Mk′ such that g′,h′∈GB′. Since h′∈A′Mk′, it follows that there exists an element l′∈Mk′ such that l′,h′∈GA′. The fact that z∈Nk+1 means that there is an element w∈BNk such that z,w∈GA. This means that l′w=h′z and h′z=0 since l∈BNk⊥. So, for u∈Bz, g′u=h′z=0 meaning that g′∈BNk+1⊥ and that Mk+1′⊂BNk+1⊥. This proves the first inclusion in (49). The second inclusion can be proved in a similar way.☐Lemma 25.
LetA∈CLRX,Y. For every f′∈NA⊥, there exists g′∈Y′ such that g′y=f′x, for all y∈Ax, and all x∈DA.Proof.
Define a linear functionalg′ on Y′ by setting g′y=f′x for all y∈Ax and all x∈DA. Then, g′ is defined on RA and is bounded. To show that g′ is indeed bounded, we first note that for y∈Ax,
(51)g′y=f′x≤fx,and consider the quotient space X~≔X/NA. Let x1∈x~. Then, x−x1=u for some u∈NA so that fx=fx1. This equality means that x in (51) can be replaced with x1, for any x1∈x~ without changing the inequality. This therefore means that
(52)g′y≤f′x~≤f′γA−1Ax=f′γA−1QAy≤f′γA−1QAy,that is, g′ is bounded on RA. The Hahn-Banach extension theorem implies that g′ can be extended to the whole of Y′ without changing its bound.☐Remark 26.
Lemma25 above implies that NA⊥⊂RA′ and that NA⊥=RA′ by Lemma 9(3).Lemma 27.
LetA,B∈CLRX,Y with DA=DB=X, RA closed and B bounded. If B0⊂A0, then
(53)M1′=BN1⊥(54)νA′:B′=νA:BProof.
Letf′∈BN1⊥=BA−10⊥. Since B0⊂BN1, Lemma 16a together with Lemma 17 (a) imply that f′∈DB′. So, let g′∈B′f′, that is, f′,g′∈GB′. This means that for x∈N1 and y∈Bx, g′x=f′y=0, which shows that g′∈N1⊥=NA⊥ and therefore g′∈RA′ and so g′∈RA′ by Remark 26. It follows that f′∈B′−1A′Y′=M1′. This shows that BN1⊥⊂M1′. Equality (53) then is followed by (49). To prove the second equality, let v=νA:B. Then, N1⊂Mn for all n<v. Since Mn=B−1AMn−1, we see that
(55)BN1⊂BMn⊂AMn−1+B0⊂AMn−1+A0=AMn−1,where the last equality follows from the fact that A0⊂AMn−1 and AMn−1 is a linear space. We see from (55) that AMn−1⊥⊂BN1⊥. It then follows from (49) and (53) that Nn′⊂M1′. This means that v′=vA′:B′>n and that v′≥v.
To prove the opposite inequality, letn<v′. Then, we have N1′⊂Mn′. If follows from Lemmas 9(1), (47), and (49) that AX⊥⊂BNn⊥. Since RA=AX is closed, this implies that BNn⊂AX. Since DB=X, we see that Nn⊂Nn+B0⊂B−1AX=M1. This shows that v>n and therefore v≥v′.☐


## 6. Nullity and Deficiency

In this section, we study the behaviour of the nullity and deficiency for linear relations under some perturbations. ForA∈LRX,Y, the nullity αA and the deficiency βA are defined by
(56)αA≔dimNA,βA≔dimY/RA.Lemma 28 ([3], III.7.2).
LetT be a closed linear relation with γT>0. Then αT′=βT.LetX and Y be Banach spaces, and let A be a closed linear relation with DA⊂X and RA⊂Y. Let n∈ℕ∪∞ be such that for any ε>0 there exists an n-dimensional closed linear subset Nε of NA such that
(57)Ax≤εx,forallx∈Nε,while this is not true if n is replaced by a larger number. In such a case, we set α′A≔n and define β′A to be
(58)β′A≔α′A′.Lemmas29 and 30 show that α′A is defined for every closed linear relation A.Lemma 29.
Assume that for everyε>0 and any closed linear subset M of X of finite codimension; there is an x∈M∩DA such that x=1 and Ax≤ε, then α′A=∞.Proof.
We have to show that for eachε>0, there exists an infinite dimensional closed linear subset Nε⊂DA with the property (57). First, we construct two sequences xn∈DA and fn∈X′ such that
(59)xn=1,fn=1,fnxn=1,fkxn=0,k=1,2,⋯,n−1,Axn≤3−nε,n∈ℕ.
Forn=1, the result holds by ([1], III-Corollary 1.24). Suppose that xn,fk have been constructed for k=1,2,⋯,n−1. Then, xn and fn can be constructed in he following way. Let M⊂X be the collection of all x∈X such that fkx=0,k=1,2,⋯,n−1. Since M is a closed linear subset of X with finite codimension (dim M⊥≤n−1 and use codim M=dim M⊥), there is an xn∈M∩DA such that xn=1 and Axn≤3−nε. For this xn, there exists an fn∈X′ such that fn=1 and fnxn=1 (see [1], III-Corollary 1.24). It follows from (59) that the xn are linearly independent so that Mε′≔spanx1,x2,⋯ is infinite dimensional. Each x∈Mε′ has the form
(60)x=ξ1x1+ξ2x2+⋯+ξnxn,for some positive integer n. Hence, for k=1,2,⋯,n,
(61)fkx=ξ1fkx1+ξ2fkx2+⋯+ξk−1fkxk−1+ξk.
We show that the coefficientsξk satisfy the inequality
(62)ξk≤2k−1x,k=1,2,⋯,n.
Fork=1, this is clear from (59) and (61). If we assume that (62) has been proved for k<j, we see from (61) that
(63)ξj≤fjx+ξ1fjx1+⋯+ξj−1fjxj−1≤x+∣ξ1∣+∣ξ2∣+⋯+∣ξj−1∣≤x+x+2x+⋯+2j−2x=x2+21+2+22+⋯+2j−1=2j−1x.
It follows from (59), (61), and (62) that
(64)Ax≤ξ1Ax1+⋯+ξnAxn≤13+232+2233+⋯+2n−13nεx≤εx.
Letu∈M¯ε′, and let un be a sequence in Mε′ such that un⟶u. The boundedness of QAA on Mε′ implies that QAAxn is a Cauchy sequence in Y~≔Y/A0 and therefore converges, say QAAxn⟶v~∈Y~. This means that distxn−v,A0⟶0 as n⟶∞, that is, xn−v⟶z∈A0 for some z∈A0. In other words, xn⟶v+z=w∈v~. The closedeness of A implies that x∈DA and w∈Ax. Hence, QAA is defined and bounded on the closure of Mε′ with the same bound.☐Lemma 30.
IfA is a closed linear relation with closed range(that is, γA>0, then α′A=αA and β′A=βA.Proof.
By Lemma16, γA>0 implies γA′>0 while Lemma 28 implies that αA′=βA. In view of (58), it is enough to show that α′A=αA. It is clear that α′A≥αA. Now suppose that there exists a closed linear manifold Nε with dimNε>αA=dimNA and with property (57). Pick x∈Nε such that x~=x=1 where x~∈X~≔X/NA (this is possible by ([2], Lemma 241). For this x, Ax≥γA on the one hand and Ax≤ε on the other hand, leading to the inequality γA≤ε. In other words, there is no Nε with dimNε>αA=dimNA for ε<γA. This proves that α′A≤αA and that α′A=αA. The second equality follows from (58) and Lemma 28.☐Lemma 31.
LetT∈CLRX with nonclosed range (that is, γT=0), then
(65)α′T=∞.Proof.
LetM be any closed linear manifold of X with finite codimension, and let QT be denoted by Q. Consider the mapping T:X/M⟶QTX/QTM defined by setting Tx~=QTx~. Then, T is clearly well defined and linear. It is well defined since
(66)Tx+v~=QTx+v~=QTx+QTv~=QTx~=Tx~,for any v∈M. It follows that QTX/QTM is a finite dimensional space since M has finite codimension. ([1], III-Lemma 1.9) implies that QTX is a closed subset of Y^≔Y/T0 if QTM is a closed subspace of the same space. This would mean that TX is a closed subset of Y. To see why this is true, let yn be a convergent sequence in TX with yn⟶y∈Y. Then, Qyn is a Cauchy sequence in Y^ and therefore converges to some point z^∈QTX. In other words, yn−z⟶w∈T0, so that yn⟶z+w∈z^. The uniqueness of the limit implies that y=z+w∈z^ and that y∈RT since z∈RT and every coset that contains and element of RT consists entirely of elements of RT. Next, we show that if TM is closed then QTM is closed. So, assume that TM is closed and let z^ be a sequence in QTM that converges to an element z^∈Y^. Then, zn−z⟶v∈T0 and so zn⟶z+v∈z^. The closedness of TM implies that z+v∈TM and that z^∈QTM.
The contradiction thatTX is both open and closed means that QTM is not closed and that TM is not closed and therefore γTM=0. Hence, there exists, for any ε>0, an x∈M∩DT such that x=1 and Tx≤εx~≤εx=ε, where x~∈X~=X/NT. This shows that the conditions of Lemma 29 are satisfied and therefore α′T=∞.☐Theorem 32.
LetX and Y be Banach spaces, and let A be a closed linear relation with DA⊂X, having closed range RA⊂Y, and with αA finite. Let B be a closed bounded linear relation such that DB⊃DA, B0⊂A0, and
(67)B<γA.
Then, the linear relationA+B is closed and has closed range. Moreover,
(68)αA+B≤αA,βA+B≤βA.Proof.
Letxn be a sequence in DA such that xn⟶x∈X, and let yn be a sequence in RA+B such that yn⟶y∈Y, where yn=un+vn with un∈Axn and vn∈Bxn for each n∈ℕ. In other words,
(69)un+vn⟶y.
Note that (67) implies that QBBxn is a Cauchy sequence in Y~≔Y/B0 and therefore converges to a point of Y~, say QBBxn⟶v~∈Y~. Hence, distvn−v,B0⟶0 as n⟶0, that is, vn−v⟶z for some z∈B0. Hence, vn⟶v+z∈v~. The closedness of B implies that x∈DB and v+z∈Bx. Hence, y=y−v−z+v+z∈Ax+Bx and so A+B is closed.
To complete the proof, it is enough to show that(70)α′A+B≤αA,β′A+B≤β′A,and then apply Lemma 31 to conclude that A+B has closed range and Lemma 30 to establish the inequalities in the theorem since α′A+B≥αA+B by definition and β′A+B≥αA+B by (58) and Lemma 28.
To prove (70), suppose that for a given ε>0 there exists a closed linear manifold Nε⊂DA+B=DA such that
(71)A+Bx≤εx,foreveryx∈Nε.
It then follows from (71) and Lemma 8 that
(72)B+εx≥Bx+A+Bx≥Bx+Ax−Bx≥Ax≥γAx~,where x~∈X~≔X/NA. If we pick ε such that 0<ε<γA−B, we see from (72) that x~<x for all nonzero x∈DA. It therefore follows from Lemma 20 that
(73)dimNε≤dimNA=αA,which means that α′A+B≤αA.
To prove the second inequality, we note that Lemma16 together with Lemma 17 implies that B′=B, γA′=γA, and A+B′=A′+B′. It therefore follows that B′≤γA′. Applying what has been proved above to the pair A′,B′, we see that
(74)β′A+B=α′A+B′=α′A′+B′≤αA′=βA,where the last equality follows from Lemma 28.☐Lemma 33.
LetX and Y be Banach spaces and let T be a closed linear relation with DT⊂X and RT⊂Y. Set
(75)xDT≔x+Tx,x∈DT.
Then,DT becomes a Banach space if ∥·∥DT is chosen as the norm.Proof.
That∥·∥DT defines a norm on DT is clear. To prove completeness, assume that xn is a Cauchy sequence in DT. Then, xn and QTTxn are Cauchy sequences in X and Y~=YT0, respectively, and therefore converge, say xn⟶x∈X and QTTxn⟶u~∈Y~. Let un∈Txn for each n∈ℕ. Then, u~n⟶u~ and so distun−u,T0⟶0 as n⟶∞, that is, un−u⟶v∈T0. We therefore see that un⟶u+v=s∈u~. The closedness of T implies that x∈DT and that s∈Tx. Now,
(76)xn−xDT=xn−x+QTTxn−x=xn−x+QTun−QTs=xn−x+u~n−u~⟶0asn⟶∞.
This shows thatDT is complete.☐LetX and Y be Banach spaces, and let A,B∈CLRX,Y be such that DA⊂DB and B0⊂A0. In Theorem 34, we write BxA to mean the quantity QABx. The quantities AxA and BxB are defined in a similar wayTheorem 34.
LetX and Y be Banach spaces, and let A be a closed linear relation with DA⊂X and with closed range RA⊂Y. Let B be a closed linear relation such that DA⊂DB⊂X, RB⊂Y, B0⊂A0, and
(77)BxB≤σx+τAxA,∀x∈DA,where σ and τ are nonnegative constants such that
(78)σ+τγA<γA.
Then, the linear relationA+B is closed and has closed range. If αA<∞, then
(79)αA+B≤αA,βA+B≤βA.Proof.
Letxn be a sequence in DA such that xn⟶x∈X, and let yn be a sequence in RA+B such that yn⟶y∈Y, where yn=un+vn with un∈Axn and vn∈Bxn for each n∈ℕ. Note that (77) implies that
(80)AxA−BxB≥1−τAxA−σx.
SinceBxB=QBBxB≥QABxA, we see that
(81)QAAxA−QABxA≥1−τAx−σx and that
(82)QAAx+QABxA≥1−τQAAx−σx.
Inequality (82) and the linearity of QA imply that
(83)QAun+vnA≥1−τQAun−σxn,so that
(84)yn=un+vn≥1−τQAun−σxn.
It therefore follows that form,n∈ℕ,
(85)yn−ym≥1−τQAun−QAum−σxn−xm.
Since1−τ>0 by (78) and both xn and yn are Cauchy sequences, it follows by (85) that QAun is a Cauchy sequence and therefore converges, say
(86)u~n⟶u~,where we denote QAun by u~n in Y/A0. The convergence in (86) implies that distun−u,A0⟶0 as n⟶∞. This means that un−u converges to an element of A0¯=A0, say un−u⟶z∈A0. This means that un⟶z−u=s. The closedness of A implies that x∈DA and s∈Ax. Since un⟶s, we see that QAun=QAs. Applying (77) to xn−x, we see that QBBxn⟶QBBx, that is, distvn−v,B0⟶0 as n⟶∞, v∈Bx. This shows that vn−v converges to an element say w of B0, that is, vn⟶w−v=r∈Bx since Bx=B0+v. Hence, y=s+r∈A+Bx, showing that A+B is closed.
We introduce a norm onDA by
(87)xD˘≔σ+εx+τ+εAx≥εx,for some arbitrary but fixed positive constant ε. Note that the space DA becomes a Banach space by Lemma 33, which we denote by D˘. We now regard A and B as linear relations with DA=DB=D˘ and denote them by A˘ and B˘ respectively. Since xD˘=σ+εx+τ+εAx>σx+τAx≥Bx for every x∈D˘ and B˘≔supx∈BD˘B˘x, we see that B˘≤1. From Ax≤τ=ε−1xD˘ and the definition of A˘, we also see that A˘≤τ+ε−1.
It is clear thatRA˘=RA is closed and that
(88)αA˘=αA,βA˘=βA,αA˘+B˘=αA+B,βA˘+B˘=βA+B.
Please note thatγA˘=γA if γA=∞. In order to relate γA˘ to γA in the other case, we recall that in this case,
(89)γA˘=infA˘xx~D˘:x∈D˘,x∉NA˘=infAxx~D˘:x∈D˘,x∉NA˘, where x~∈X~≔X/NA.
But(90)x~D˘=infz∈NAx−zD˘=infz∈NAσ+εx−z+τ+εAx−z=σ+εx~+τ+εAx,where we have used the linearity of the natural quotient map and the fact that Az=A0.
Hence,(91)γA˘=infAxσ+εx~+τ+εAx:x∈DA,x∉NA=γAσ+ε+τ+εγA,where we have used the fact that ft=t/α+t is an increasing function for any constant α.
In view of (78), we can make γA˘>1 by choosing ε small enough. Since B˘≤1, we can apply Theorem 32 to the pair A˘,B˘ with the result that RA˘+B˘=RA+B is closed and (68) holds with A,B replaced with A˘,B˘. The result then follows (88).☐


## 7. Stability Theorems

Consider an eigenvalue problem of the form(92)Ax=λB,where A and B are linear operators from X to Y and the associated problem
(93)A∗f′=λB∗f′,where the adjoints A∗ and B∗ exist. The null space NA−λB of the linear operator A−λB is the solution set of the eigenvalue problem (92). Similarly, NA∗−λB∗=RA−λB⊥ is the solution set of the eigenvalue problem (93). In studying the above eigenvalue problems, one therefore gets interested in the behaviour of NA−λB and NA∗−λB∗.In the setting of linear relations, the eigenvalue problems (92) and (93) can be formulated as
(94)Ax∩λBx≠∅,(95)A′x′∩λB′x′≠∅,where A,B∈LRX,Y. Conditions (94) and (93) are equivalent to
(96)A−λBx=A−λB0,(97)A′−λB′x′=A′−λB′0,respectively.As before, the solution sets of (96) and (97) are NA−λB and NA′−λB′=RA−λB⊥, respectively. In this last section, we study the stability of the dimensions of the null spaces of A−λB and A′−λB′ as λ varies in some specified subset of the complex plane.Theorem 35.
LetX and Y be Banach spaces and let A,B∈CLRX,Y be such that A has closed range, DB⊃DA, B0⊂A0, and
(98)Bx≤σx+τAx,foreveryx∈DA,where σ and τ are nonnegative constants. Then A−λB is closed for λ<γA/σ+τγA, and if RA\A0≠∅, then γA−λB<∞ for λ≥γA/σ+τγA.Proof.
It follows from Theorem34 that A−λB is closed if λ<γA/σ+τγA.
IfγA−λB=∞, then A−λBx=A−λB0=A0. The fact that A−λBx=A−λB0 for every x∈DA−λB=DA implies that Ax∩λBx≠∅ for every x∈DA. Since B0⊂A0, it follows that Ax≤λBx for every x∈DA and therefore
(99)Ax≤λBx≤λσx+τAx,so that
(100)1−λτAx≤σλx.
SinceRA≠A0, we see that there exists at least one x~ in X~=X/NA with x~≠0. Inequality (100) implies that
(101)γAx~≤Ax≤σλx/1−λτ.
Sincex can vary freely in x~, we conclude that γA≤σλ/1−λτ and that λ≥γA/σ+τγA.☐Theorem 36.
LetX and Y be Banach spaces, and let A,B∈CLRX,Y be such that A has closed range, DB⊃DA, B0⊂A0, and
(102)Bx≤σx+τAx,foreveryx∈DA,where σ and τ are nonnegative constants. If νA:B=∞, then
(103)δNA,NA−λB≤σλγA−λσ+τγA.Proof.
LetNk be as defined in (41) and consider a sequence zk with the following properties:
(104)zk∈Nk,Azk+1∩Bzk≠∅,ξzk+1≤Azk+1,k=1,2,⋯,where ξ is a positive constant. We show that for each z∈NA and ξ<γA, there is a sequence zk that satisfies (104) such that z=z1. We set z=z1 and construct zk by induction. Suppose z1,z2,⋯xk have been constructed with properties (104). Since zk∈Nk⊂M1=B−1AX, there exists a zk+1∈DA such that Azk+1∩Bzk≠∅. Since γAz~k+1≤Azk+1 and zk+1 can be replaced by any other element of z~k+1, we can choose zk+1 such that ξzk+1≤Azk+1. Since Azk+1∩Bzk≠∅, we see that zk+1∈A−1BNn=Nk+1. This completes the induction process.
SinceAzk+1∩Bzk≠∅ and A0⊃B0, we see that
(105)Azk+1≤Bzk≤σzk+τAzk.
Fork=1, (105) gives Az2≤Bz1≤σz1 since z1∈NA. For k≥2, (104) implies that
(106)Azk+1≤Bzk≤σzk+τAzk≤σξ−1+τAzk≤σξ−1+τ2Azk−1≤⋯≤σξ−1+τk−1Az2=ξ−k−1σ+ξτk−1Az2≤σξ−k−1σ+ξτk−1z1.
We also see from (104) and (106) that
(107)zk+1≤σξ−kσ+ξτk−1z1,k=1,2,⋯
The bounds in (106) and (107) imply that the series
(108)uλ=∑k=1∞λk−1zk,λA=∑k=1∞λkQAAzk+1,λB=∑k=1∞λk−1QBBzk,λBA=∑k=1∞λk−1QABzkare absolutely convergent for λ<ξ/σ+ξτ. The convergence of the last series follows from the fact that QABzk≤QBBzk since B0⊂A0.
Letunλ, λnA, λnB, and λnBA denote the sequences of the partial sums of the above series in that order. Then, for each n, unλ∈DA and λnA∈Y~≔Y/A0. Furthermore, unλ⟶uλ and λnA⟶λA. Since QAA is closed by Lemma 7, we see that uλ∈DQAA=DA and that
(109)QAAuλ=λA=∑k=1∞λkQAAzk+1.
SinceAzk+1∩Bzk≠∅, a similar argument shows that
(110)QABuλ=λBA=∑k=1∞λk−1QABzk=∑k=1∞λkQAAzk+1=λA.
One also obtains the equalityQBBuλ=λB=∑k=1∞λkQBBzk using the closedness of B.
From (109) and (110), we see that
(111)QAAuλ−λBuλ=0~,and so uλ∈NA−λB.
Furthermore,(112)uλ−z1≤∑k=1∞‍λk−1zk≤σλξ−λσ+τξz1.
Since there is such auλ∈NA−λB for every z−z1∈NA, we conclude that
(113)δNA,NA−λB≤σλγA−λσ+τγA.☐We observe that ifαA<∞ then Theorem 34 can be used to conclude that A−λB has closed range if λ<γA/σ+τγA. However, this conclusion is not possible if no restriction is imposed on αA. This case is considered in Lemma 37.Lemma 37.
LetA and B be as in Theorem 36 with νA:B=∞. Then, A−λB has closed range for λ<γA/3σ+τγA.Proof.
In the present case, letx∈X and set y=x−u for any u∈NA−λB. Lemma 22 implies that for any ε>0,
(114)y~=disty,NA≥1−δNA,NA−λB1+δNA,NA−λB1−εy.
Suppose thatx∈DA=DA−λB, and let δ≔δNA:NA−λB. Since A−λBu=A−λB0=A0, we see that,
(115)A−λBx=A−λBy≥Ay−λBy≥Ay−λσy+τAy=1−τλAy−σλy≥1−τλγAy~−σλy≥1−τλγA1−δ1+δ1−εy−σλyby115≥γA−2σ+τγAλ1−εy−σλyby104=γA−2σ+τγAλ1−ε−σλy.
LetX^ denote the quotient space X/NA−λB. Since x−y=u∈NA−λB, we see that y≥y^=x^, and therefore, (115) implies that
(116)A−λBx≥γA−2σ+τγAλ1−ε−σλx^.
Lettingε⟶0 in (116) leads to the inequality
(117)A−λBx≥γA−2σ+τγAλ−σλx^,from which we conclude that
(118)γA−λB≥γA−3σ+τγAλ.
It therefore follows thatγA−λB>0, and therefore, RA−λB is closed if λ<γA/3σ+τγA.☐Finally, we establish the stability of both the nullity and deficiency ofA−λB for λ inside the disk λ<ρ for some constant ρ.Theorem 38.
LetX and Y be Banach spaces, and let A,B∈CLRX,Y be such that A has closed range, DB⊃DA, B0⊂A0, and
(119)Bx≤σx+τAx,foreveryx∈DA,where σ and τ are nonnegative constants. If νA:B=∞, then αA−λB and βA−λB are constants for all λ for which λ<γA/3σ+τγA.Proof.
Letu∈NA−λB. Then, Au∩λBu≠∅ and we see from (101) that
(120)u~≤σλu1−λτγA.
Sinceu~=distu,NA, we see from characterization (36) that
(121)δNA−λB,NA≤σλ1−λτγA.
Sinceσλ/1−λτγA<1 if λ<γA/σ+τγA, Lemma 21 implies that
(122)αA−λB≤αA,forλ<γAσ+τγA.
The reverse inequality follows from Theorem36 by noting that the right-hand side of (103) is less than one if λ<γA/2σ+τγA. We therefore conclude by Lemma 21 that αA≤αA−λB if λ<γA/2σ+τγA. Combined with (122), we conclude that
(123)αA=αA−λB,forλ<γA2σ+τγA.
To show thatβA−λ=βA, we make use of the linear relations A˘ and B˘ as defined in the proof of Theorem 34. Since A˘ is bounded, Lemmas 16 (c), 17 (d), and 15 imply that RA˘′ has closed range. Since B˘0′⊂A˘0′ by Remark 10 and νA˘′:B˘′=∞ by Lemma 27, all the assumptions of Theorem 38 are satisfied by the pair A˘′ and B˘′. Since B˘′=B˘<1 by Lemmas 16 (a) and 17 (c), it follows from (123) that
(124)αA˘′−λB˘′=αA˘′,forλ<γA˘′2B˘′.
SinceA˘−λB˘′=A˘′−λB˘′ by Lemma 16 (b) and (d) and A˘−λB˘ has closed range (since A−λB has closed range), it follows from (88), Lemma 28, and (124) that
(125)βA−λB=βA˘−λB˘=αA˘′−λB˘′=αA˘′=βB˘=βB.☐Theorem38 remains true if we replace the requirement νA:B=∞ with B−10⊂A−10.



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