# Relationships between Perturbations of a Linear Relation and Its Operator Part

Authors: Guixin Xu; Meirong Zhang
Journal: Discrete Dynamics in Nature and Society
 (2022)

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## Abstract

In this study, we establish some relationships between perturbations of a linear relation and its operator part by constructing an operator, which is induced by two linear relations including their closedness, hermiticity, self-adjointness, various spectra, defect indices, and perturbation terms.


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## Body

## 1. Introduction

Motivated by the study for the adjoint of nondensely defined linear differential operators, the concept of linear relations, a natural generalization of linear operators, was introduced in [1]. Along with the development of operator theory, the spectral theory for linear relations has been extensively studied and has important applications to several problems (cf., [2–14]). It is worth mentioning that the spectra of linear relations may provide some useful tools for the study of certain operators, such as the maximal and minimal operators corresponding to linear continuous Hamiltonian systems or symmetric linear difference equations [12, 15], and the inverse of certain operators in the study of some Cauchy problems associated with parabolic type equations in Banach spaces [16].To the best of our knowledge, there are still many important fundamental problems of linear relations that have neither been studied nor completed. In 1961, Arens showed that every closed linear relationT in a Hilbert space can be decomposed as an operator part Ts and a purely multivalued part T∞ [17]; this decomposition provides a bridge between linear relations and operators. In 1985, Dijksma and de Snoo proved that the operator part Ts of a self-adjoint relation T is also a self-adjoint in the Hilbert space T0⊥ [18]. Later, Shi et al. established some relationships between the spectra and various spectra of a closed relation and its operator part as well [19]. Enlightened by these works, the main idea of this study was to construct a linear operator AT, which is induced by two linear relations T and A such that some perturbations of T can be consistent with its operator part Ts and the various spectra of T+A and Ts+AT are identical. Consequently, one can study some perturbation problems about linear relations by using these results and related existing results about operators.The rest of this study is organized as follows. In Section2, some preliminary and auxiliary results that will be used in the sequel are introduced. In addition, a new linear operator AT is introduced, which is induced by two linear relations T and A, and its properties are studied. In Section 3, the decomposition of a linear relation is given in terms of reducing subspaces, and relationships between a relation and its decomposition parts are established, including their closedness, hermiticity, self-adjointness, defect indices, and spectra. Using these relationships, the corresponding relationships between relation T+A and operator Ts+AT are given in case if T is closed. Further, a concept of trace class linear relations is introduced and relationships between various perturbations of a closed relation T and its operator part Ts are discussed. It is shown that if relation A is a relatively bounded or compact perturbation term of T, or A belongs to degenerate or trace class linear relations, then AT is the same perturbation term of Ts, respectively (seeTheorems 9–12).


## 2. Preliminaries

In this section, we shall recall some basic concepts, give some fundamental results on linear relations, and introduce a new linear operator induced by two relations and study its properties, which will be used in the sequent sections. This section is divided into three subsections.


### 2.1. Some Basic Concepts and Fundamental Results of Linear Relations

LetX,Y,Z,⋯ denote normed spaces over a number field K. The norm of X×Y is defined by(1)x,y=xX2+yY21/2,x∈X,y∈Y,where ⋅X and ⋅Y are the norms of the spaces X and Y, respectively, still denoted by ⋅ without any confusion. Similarly, if X and Y are inner product spaces, then the inner product of X×Y is defined by(2)x1,y1,x2,y2=x1,x2+y1,y2,x1,y1,x2,y2∈X×Y.Obviously, ifX and Y are complete, then X×Y is also complete.C and R denote the sets of complex and real numbers, respectively.Any linear subspaceT⊂X×Y is called a linear relation (briefly, relation or subspace) of X×Y. LRX,Y denotes the set of all the linear relations of X×Y. In the case where X=Y, LRX denotes LRX,Y, briefly.The domainDT, range RT, and null space NT of T are, respectively, defined by(3)DT=x∈X:x,y∈Tforsomey∈Y,RT=y∈Y:x,y∈Tforsomex∈X,NT=x∈X:x,0∈T.Furthermore, it denotes the following:(4)Tx=y∈Y:x,y∈TT−1=y,x:x,y∈T.T is said to be injective if NT=0, and surjective if RT=Y.It is evident thatT0=0 if and only if T can uniquely determine a linear operator from DT into X whose graph is T. For convenience, a linear operator (i.e., single-valued operator) from X to Y will always be identified with a relation in X×Y via its graph. In addition, NT=0 if and only if T−10=0, i.e., T is injective if and only if T−1 is a linear operator. Further, T is said to be closed if T=T¯, where T¯ is the closure of T.LetT,A∈LRX,Y and α∈K. We define(5)αT=x,αyx,y∈T,T+A=x,y+zx,y∈T,x,z∈A,T−αx,y−αxx,y∈TinthecasethatY=X.IfT∩A=0,0, we denote(6)T+˙A=x1+x2,y1+y2x1,y1∈T,x2,y2∈A.Further, in the case thatX and Y are inner product spaces, if T and A are orthogonal, that is, x,y,u,v=0 for all x,y∈T and u,v∈A, then it denotes the following: (7)T⊕A=T+˙A.Theproduct of linear relations T∈LRX,Y and A∈LRY,Z is defined by (see [17])(8)AT=x,z∈X×Zx,y∈T,y,z∈Aforsomey∈Y.Note that ifA and T are operators, then AT is also an operator.In the following, we shall briefly recall the concepts of bounded and compact relations, which were introduced in [20, 21].LetX and Y be normed spaces and T∈LRX,Y. By QT, or simply Q when there is no ambiguity about the relation T, we denote the natural quotient map from Y onto Y/T0¯. Clearly, QT is an operator [20]. Furthermore, it denotes that BX:=x∈X:x≤1.For any givenx∈DT, the norms of Tx and T are defined by(9)Tx=QTx,T=QT=supQTxx∈DT∩BX,respectively. If T<∞, then T is said to be bounded.T is said to be compact if QT is compact.It is evident thatT is compact if and only if for every bounded sequence xnn=1∞⊂DT and QTxnn=1∞ has a convergent subsequence in Y/T0¯. Moreover, if T is compact, then T is bounded [20] (Corollary V.2.3).The following results come from [20], Proposition II.1.4, and [22], Corollary 1.Lemma 1.
LetX and Y be normed spaces and T∈LRX,Y. Then,(i)
Tx=dy,T0=d0,Tx for every x∈DT and y∈Tx;(ii)
T=supTx:x∈DT∩BX;(iii)
For any givenxnn=1∞⊂DT, QTTxn⟶y as n⟶∞ for some y∈Y/T0¯ if and only if for each n≥1, there exists yn∈Txn such that yn⟶y as n⟶∞.Definition 1.
(see [19, 23, 24]). Let X be a Banach space over the complex field C and T∈LRX.(1)
The setρT:=λ∈C:T−λ−1isaboundedlinearoperatordefinedonX is called the resolvent set of T.(2)
The setσT:=C/ρT is called the spectrum of T.(3)
Forλ∈C, if there exists x,λx∈T for some x≠0, then λ is called an eigenvalue of T, while x is called an eigenvector of T with respect to the eigenvalue λ. NT−λ is called the eigensubspace of λ, and dim NT−λ is called the multiplicity of λ. Further, the set of all the eigenvalues of T is called the point spectrum of T, denoted by σpT.(4)
Theessential spectrumσeT of T is the set of those points of σT that are either accumulation points of σT or isolated eigenvalues of infinite multiplicity.(5)
The setσdT:=σT/σeT is called the discrete spectrum of T.


### 2.2. Concepts and Fundamental Properties of Self-Adjoint Linear Relations

In this subsection, we shall briefly recall the concept of self-adjoint relations and another classification of the spectrum of a self-adjoint relation by its spectral family and some fundamental properties of them.In this part,X is always assumed to be a complex Hilbert space.LetX∈LRX. The adjoint of T is defined by(10)T∗=f,g∈X2g,x=f,yforallx,y∈T.T is said to be Hermitian in X2 if T⊂T∗ and said to be self-adjoint in X2 if T=T∗.Arens introduced the following decomposition for a closed linear relationT in X2 [17]:(11)T=Ts⊕T∞,where(12)T∞=0,y∈X2:0,y∈T,Ts=T⊖T∞.It can be easily verified thatTs is an operator, and T is an operator if and only if T=Ts. Ts and T∞ are called the operator and pure multivalued parts of T, respectively. In addition, they satisfy the following properties:(13)DTs=DT,DTs¯=DT¯=T∗0⊥,RTs⊂T0⊥,T∞=0×T0.IfT is an Hermitian, it is evident that(14)DT⊂T∗0⊥⊂T0⊥.Throughout the present study, the resolvent set and spectrum ofTs mean those of Ts are restricted to T0⊥2.Lemma 2 (see [Proposition 2.1, Theorems 2.1, 2.2 and 3.4 in [19]).
LetT be a closed Hermitian relation in X2. Then, (15)Ts=T∩T0⊥2,T∞=T∩T02.Ts is a closed Hermitian operator in T0⊥, and (16)ρT=ρTs,σT=σTs,σT∞=∅,σpT=σpTs,NT−λ=NTs−λ,λ∈σpT.Further, ifT is a self-adjoint relation in X2, then(17)σeT=σeTs,σdT=σdTs.Remark 1.
It follows fromTheorems 2.1 and 2.2 in [19] that (17) also holds ifTis a closed Hermitian relation inX2.Lemma 3 (see p. 26 in [18]).
IfT is a self-adjoint relation in X2, then Ts and T∞ are self-adjoint relations in T0⊥2 and T02, respectively.Lemma 4 (see Theorem 2.5 in [19]).
LetT be an Hermitian relation in X2. Then, T is self-adjoint in X2 if and only if RT−λ=RT−λ¯=X for some λ∈C.Next, we shall briefly recall another classification of the spectrum of a self-adjoint relation by its spectral family, including continuous spectrum, singular continuous spectrum, absolutely continuous spectrum, and singular spectrum and their some properties (see [19]). The concept of spectral family of a self-adjoint relation was introduced by Coddington and Dijksma in [8].LetT be a self-adjoint relation in X2. By Lemma 3, Ts is a self-adjoint operator in T0⊥. Then, Ts has the following spectral resolution:(18)Ts=∫tdEst,where Estt∈R is the spectral family of Ts in T0⊥. The spectral family of the relation T is defined by(19)Et=Est⊕O,t∈R,where O is the zero operator defined on T0. We denote(20)ENf=∫NdEtf,N⊂R,f∈X.LetXP denote the closed linear hull of all the eigenvectors of T and XC:=XP⊥. They are called the discontinuous and continuous subspaces in X with respect to T, respectively. Further, we denote(21)XSC=f∈XCthereexistsaBorelnullsetNsuchthatENf=f,XAC:=XC⊖XSC, and XS:=XP⊕XSC, which are called the singular continuous, absolutely continuous, and singular subspaces in X with respect to T.The (spectral) discontinuous, continuous, singular continuous, absolutely continuous, and singular parts ofT are defined by(22)TP=T∩XP2,TC=T∩XC2,TSC=T∩XSC2,TAC=T∩XAC2,TS=T∩XS2,respectively.Definition 2.
(see Definition 4.1 in [19]). Let T be a self-adjoint relation in X2. The spectra of TC, TSC, TAC, and TS are called the continuous spectrum, singular continuous spectrum, absolutely continuous spectrum, and singular spectrum of T, respectively, denoted by σcT, σscT, σacT, and σsT, respectively.Lemma 5 (see Theorem 4.1 in [19]).
LetT be a self-adjoint relation in X2. Then,(23)σTP=σTsP,σrT=σrTs,r=c,ac,sc,s.


### 2.3. An Operator Induced by Two Linear Relations

In this subsection, we shall first introduce a new linear operator induced by two linear relations, which plays an important role in the present study, and then study its some properties. Further,X is assumed to be a complex Hilbert space.LetT and A be two linear relations in X2 with DT⊂DA and A0⊂T0. PT denotes the following orthogonal projection:(24)PT:X⟶T0⊥.It defines that(25)AT=PTA|DT¯∩DA.It follows fromA0⊂T0 that AT0=PTA0⊂PTT0=0. This means AT is single-valued. Further, it is evident that DT⊂DAT⊂DT¯ and RAT⊂T0⊥. Consequently, DAT¯=DT¯. Note that AT=A|DT¯∩DA in the case that T is single-valued.Proposition 1.
LetT,A∈LRX with DT⊂DA and A0⊂T0. If T is closed and DT⊂T0⊥, then(26)AT|DT=T+A−T∩T0⊥2.Proof.
We suppose thatT is closed and DT⊂T0⊥. Then, T0 is closed by Proposition II.5.3 in [20]. Consequently, X can be decomposed as(27)X=T0⊥⊕T0.
SinceA0⊂T0, we have that(28)T+A−Tx=Ax+T0=y+T0,y∈Ax,x∈DT.
Hence,T+A−T∩T0⊥2 is single-valued.
We first show thatT+A−T∩T0⊥2⊂AT|DT. For anyx,y∈T+A−T∩T0⊥2, there exist x,y1∈A and 0,y2∈T such that y=y1+y2 by (28). Let y1=y11+y12, where y11∈T0⊥ and y12∈T0. Then, y=y11+y12+y2. Note that y,y11∈T0⊥ and y12,y2∈T0. We have y=y11. Hence, ATx=PTy1=y11=y. Therefore, T+A−T∩T0⊥2⊂AT|DT.
Now, we consider the inverse. For any givenx,y∈AT|DT, there exists z,f∈X such that x,z∈A, z,y∈PT, and x,f∈T since x∈DT. It follows from (11) that x,f can be decomposed as x,f=x,f1+0,f2 with x,f1∈Ts and 0,f2∈T∞. Let z=z1+z2, where z1∈T0⊥ and z2∈T0. Then, y=z1. Since A0⊂T0, we have T0=T+A0. Hence, 0,z2∈T+A. Note that when x,f+z∈T+A, we get x,f+y∈T+A. This together with x,f∈T implies that x,y∈T+A−T. Since x∈DT⊂T0⊥ and y∈T0⊥, we havex,y∈T0⊥2. Consequently, AT|DT⊂T+A−T∩T0⊥2.
Therefore, (26) holds. This completes the proof.
The following result is a direct consequence of (14) and Proposition 1.Corollary 1.
LetT,A∈LRX with DT⊂DA and A0⊂T0. If T is closed and Hermitian, then(29)AT|DT=T+A−T∩T0⊥2.Now, we give a decomposition ofT+A in the case that T is closed. This decomposition plays an important role in the study of some properties about linear relations in the present study.Theorem 1.
LetT,A∈LRX with DT⊂DA and A0⊂T0. If T is closed, then(30)T+A=Ts+AT⊕T∞.Proof.
We suppose thatT is closed. It is evident that Ts+AT and T∞ are orthogonal.
For anyx,y∈T+A, there exists y1,y2∈X such that x,y1∈T, x,y2∈A, and y=y1+y2. It follows from (11) that x,y1=x,y11+0,y12 with x,y11∈Ts and 0,y12∈T∞. Let y2=y21+y22, where y21∈T0⊥ and y22∈T0. Then, x,y21∈AT and 0,y12+y22∈T∞. Hence, x,y=x,y11+y21+0,y12+y22 belongs to Ts+AT⊕T∞. This implies that T+A⊂Ts+AT⊕T∞.
On the other hand, for any givenx,y∈Ts+AT⊕T∞, it can be decomposed as x,y=x,y1+0,y2 with x,y1∈Ts+AT and 0,y2∈T∞⊂T. Let y1=y11+y12, where x,y11∈Ts⊂T and x,y12∈AT. Given any z∈Ax, it can be decomposed as z=z1+z2 with z1∈T0⊥ and z2∈T0, and we have that y12=z1=z−z2 and x,y11+y2−z2∈T. Consequently,(31)x,y=x,y1+y2=x,y11+y2−z2+z∈T+A,which yields that Ts+AT⊕T∞⊂T+A.
Therefore, (30) holds and the proof is complete.Corollary 2.
LetT,A∈LRX with DT⊂DA and A0⊂T0. If T and T+A are closed, then(32)T+As=Ts+AT.Proof.
Suppose thatT and T+A are closed. It follows from A0⊂T0 that T∞=T+A∞. Then, T+A=T+As⊕T∞ by (11), which together with (30) implies T+As=Ts+AT. The proof is complete.Remark 2.
In Theorem 3.1 in [25], Shi showed thatT+As=PTs+QAsunder some conditions, whereP:T0⊥⟶T+A0⊥andQ:A0⊥⟶T+A0⊥are orthogonal projections. Note thatT+A0=T0by the assumption thatA0⊂T0. Then,Pis an identity mapping fromT0⊥onto itself andQAs|DT¯∩DA=AT. Hence, Theorem 3.1 in [25] is consistent with the result in Corollary 2 in the case that A is closed. Further, Theorem 3.2 in [25] can be directly derived from Corollary 2.


## 2.1. Some Basic Concepts and Fundamental Results of Linear Relations

LetX,Y,Z,⋯ denote normed spaces over a number field K. The norm of X×Y is defined by(1)x,y=xX2+yY21/2,x∈X,y∈Y,where ⋅X and ⋅Y are the norms of the spaces X and Y, respectively, still denoted by ⋅ without any confusion. Similarly, if X and Y are inner product spaces, then the inner product of X×Y is defined by(2)x1,y1,x2,y2=x1,x2+y1,y2,x1,y1,x2,y2∈X×Y.Obviously, ifX and Y are complete, then X×Y is also complete.C and R denote the sets of complex and real numbers, respectively.Any linear subspaceT⊂X×Y is called a linear relation (briefly, relation or subspace) of X×Y. LRX,Y denotes the set of all the linear relations of X×Y. In the case where X=Y, LRX denotes LRX,Y, briefly.The domainDT, range RT, and null space NT of T are, respectively, defined by(3)DT=x∈X:x,y∈Tforsomey∈Y,RT=y∈Y:x,y∈Tforsomex∈X,NT=x∈X:x,0∈T.Furthermore, it denotes the following:(4)Tx=y∈Y:x,y∈TT−1=y,x:x,y∈T.T is said to be injective if NT=0, and surjective if RT=Y.It is evident thatT0=0 if and only if T can uniquely determine a linear operator from DT into X whose graph is T. For convenience, a linear operator (i.e., single-valued operator) from X to Y will always be identified with a relation in X×Y via its graph. In addition, NT=0 if and only if T−10=0, i.e., T is injective if and only if T−1 is a linear operator. Further, T is said to be closed if T=T¯, where T¯ is the closure of T.LetT,A∈LRX,Y and α∈K. We define(5)αT=x,αyx,y∈T,T+A=x,y+zx,y∈T,x,z∈A,T−αx,y−αxx,y∈TinthecasethatY=X.IfT∩A=0,0, we denote(6)T+˙A=x1+x2,y1+y2x1,y1∈T,x2,y2∈A.Further, in the case thatX and Y are inner product spaces, if T and A are orthogonal, that is, x,y,u,v=0 for all x,y∈T and u,v∈A, then it denotes the following: (7)T⊕A=T+˙A.Theproduct of linear relations T∈LRX,Y and A∈LRY,Z is defined by (see [17])(8)AT=x,z∈X×Zx,y∈T,y,z∈Aforsomey∈Y.Note that ifA and T are operators, then AT is also an operator.In the following, we shall briefly recall the concepts of bounded and compact relations, which were introduced in [20, 21].LetX and Y be normed spaces and T∈LRX,Y. By QT, or simply Q when there is no ambiguity about the relation T, we denote the natural quotient map from Y onto Y/T0¯. Clearly, QT is an operator [20]. Furthermore, it denotes that BX:=x∈X:x≤1.For any givenx∈DT, the norms of Tx and T are defined by(9)Tx=QTx,T=QT=supQTxx∈DT∩BX,respectively. If T<∞, then T is said to be bounded.T is said to be compact if QT is compact.It is evident thatT is compact if and only if for every bounded sequence xnn=1∞⊂DT and QTxnn=1∞ has a convergent subsequence in Y/T0¯. Moreover, if T is compact, then T is bounded [20] (Corollary V.2.3).The following results come from [20], Proposition II.1.4, and [22], Corollary 1.Lemma 1.
LetX and Y be normed spaces and T∈LRX,Y. Then,(i)
Tx=dy,T0=d0,Tx for every x∈DT and y∈Tx;(ii)
T=supTx:x∈DT∩BX;(iii)
For any givenxnn=1∞⊂DT, QTTxn⟶y as n⟶∞ for some y∈Y/T0¯ if and only if for each n≥1, there exists yn∈Txn such that yn⟶y as n⟶∞.Definition 1.
(see [19, 23, 24]). Let X be a Banach space over the complex field C and T∈LRX.(1)
The setρT:=λ∈C:T−λ−1isaboundedlinearoperatordefinedonX is called the resolvent set of T.(2)
The setσT:=C/ρT is called the spectrum of T.(3)
Forλ∈C, if there exists x,λx∈T for some x≠0, then λ is called an eigenvalue of T, while x is called an eigenvector of T with respect to the eigenvalue λ. NT−λ is called the eigensubspace of λ, and dim NT−λ is called the multiplicity of λ. Further, the set of all the eigenvalues of T is called the point spectrum of T, denoted by σpT.(4)
Theessential spectrumσeT of T is the set of those points of σT that are either accumulation points of σT or isolated eigenvalues of infinite multiplicity.(5)
The setσdT:=σT/σeT is called the discrete spectrum of T.


## 2.2. Concepts and Fundamental Properties of Self-Adjoint Linear Relations

In this subsection, we shall briefly recall the concept of self-adjoint relations and another classification of the spectrum of a self-adjoint relation by its spectral family and some fundamental properties of them.In this part,X is always assumed to be a complex Hilbert space.LetX∈LRX. The adjoint of T is defined by(10)T∗=f,g∈X2g,x=f,yforallx,y∈T.T is said to be Hermitian in X2 if T⊂T∗ and said to be self-adjoint in X2 if T=T∗.Arens introduced the following decomposition for a closed linear relationT in X2 [17]:(11)T=Ts⊕T∞,where(12)T∞=0,y∈X2:0,y∈T,Ts=T⊖T∞.It can be easily verified thatTs is an operator, and T is an operator if and only if T=Ts. Ts and T∞ are called the operator and pure multivalued parts of T, respectively. In addition, they satisfy the following properties:(13)DTs=DT,DTs¯=DT¯=T∗0⊥,RTs⊂T0⊥,T∞=0×T0.IfT is an Hermitian, it is evident that(14)DT⊂T∗0⊥⊂T0⊥.Throughout the present study, the resolvent set and spectrum ofTs mean those of Ts are restricted to T0⊥2.Lemma 2 (see [Proposition 2.1, Theorems 2.1, 2.2 and 3.4 in [19]).
LetT be a closed Hermitian relation in X2. Then, (15)Ts=T∩T0⊥2,T∞=T∩T02.Ts is a closed Hermitian operator in T0⊥, and (16)ρT=ρTs,σT=σTs,σT∞=∅,σpT=σpTs,NT−λ=NTs−λ,λ∈σpT.Further, ifT is a self-adjoint relation in X2, then(17)σeT=σeTs,σdT=σdTs.Remark 1.
It follows fromTheorems 2.1 and 2.2 in [19] that (17) also holds ifTis a closed Hermitian relation inX2.Lemma 3 (see p. 26 in [18]).
IfT is a self-adjoint relation in X2, then Ts and T∞ are self-adjoint relations in T0⊥2 and T02, respectively.Lemma 4 (see Theorem 2.5 in [19]).
LetT be an Hermitian relation in X2. Then, T is self-adjoint in X2 if and only if RT−λ=RT−λ¯=X for some λ∈C.Next, we shall briefly recall another classification of the spectrum of a self-adjoint relation by its spectral family, including continuous spectrum, singular continuous spectrum, absolutely continuous spectrum, and singular spectrum and their some properties (see [19]). The concept of spectral family of a self-adjoint relation was introduced by Coddington and Dijksma in [8].LetT be a self-adjoint relation in X2. By Lemma 3, Ts is a self-adjoint operator in T0⊥. Then, Ts has the following spectral resolution:(18)Ts=∫tdEst,where Estt∈R is the spectral family of Ts in T0⊥. The spectral family of the relation T is defined by(19)Et=Est⊕O,t∈R,where O is the zero operator defined on T0. We denote(20)ENf=∫NdEtf,N⊂R,f∈X.LetXP denote the closed linear hull of all the eigenvectors of T and XC:=XP⊥. They are called the discontinuous and continuous subspaces in X with respect to T, respectively. Further, we denote(21)XSC=f∈XCthereexistsaBorelnullsetNsuchthatENf=f,XAC:=XC⊖XSC, and XS:=XP⊕XSC, which are called the singular continuous, absolutely continuous, and singular subspaces in X with respect to T.The (spectral) discontinuous, continuous, singular continuous, absolutely continuous, and singular parts ofT are defined by(22)TP=T∩XP2,TC=T∩XC2,TSC=T∩XSC2,TAC=T∩XAC2,TS=T∩XS2,respectively.Definition 2.
(see Definition 4.1 in [19]). Let T be a self-adjoint relation in X2. The spectra of TC, TSC, TAC, and TS are called the continuous spectrum, singular continuous spectrum, absolutely continuous spectrum, and singular spectrum of T, respectively, denoted by σcT, σscT, σacT, and σsT, respectively.Lemma 5 (see Theorem 4.1 in [19]).
LetT be a self-adjoint relation in X2. Then,(23)σTP=σTsP,σrT=σrTs,r=c,ac,sc,s.


## 2.3. An Operator Induced by Two Linear Relations

In this subsection, we shall first introduce a new linear operator induced by two linear relations, which plays an important role in the present study, and then study its some properties. Further,X is assumed to be a complex Hilbert space.LetT and A be two linear relations in X2 with DT⊂DA and A0⊂T0. PT denotes the following orthogonal projection:(24)PT:X⟶T0⊥.It defines that(25)AT=PTA|DT¯∩DA.It follows fromA0⊂T0 that AT0=PTA0⊂PTT0=0. This means AT is single-valued. Further, it is evident that DT⊂DAT⊂DT¯ and RAT⊂T0⊥. Consequently, DAT¯=DT¯. Note that AT=A|DT¯∩DA in the case that T is single-valued.Proposition 1.
LetT,A∈LRX with DT⊂DA and A0⊂T0. If T is closed and DT⊂T0⊥, then(26)AT|DT=T+A−T∩T0⊥2.Proof.
We suppose thatT is closed and DT⊂T0⊥. Then, T0 is closed by Proposition II.5.3 in [20]. Consequently, X can be decomposed as(27)X=T0⊥⊕T0.
SinceA0⊂T0, we have that(28)T+A−Tx=Ax+T0=y+T0,y∈Ax,x∈DT.
Hence,T+A−T∩T0⊥2 is single-valued.
We first show thatT+A−T∩T0⊥2⊂AT|DT. For anyx,y∈T+A−T∩T0⊥2, there exist x,y1∈A and 0,y2∈T such that y=y1+y2 by (28). Let y1=y11+y12, where y11∈T0⊥ and y12∈T0. Then, y=y11+y12+y2. Note that y,y11∈T0⊥ and y12,y2∈T0. We have y=y11. Hence, ATx=PTy1=y11=y. Therefore, T+A−T∩T0⊥2⊂AT|DT.
Now, we consider the inverse. For any givenx,y∈AT|DT, there exists z,f∈X such that x,z∈A, z,y∈PT, and x,f∈T since x∈DT. It follows from (11) that x,f can be decomposed as x,f=x,f1+0,f2 with x,f1∈Ts and 0,f2∈T∞. Let z=z1+z2, where z1∈T0⊥ and z2∈T0. Then, y=z1. Since A0⊂T0, we have T0=T+A0. Hence, 0,z2∈T+A. Note that when x,f+z∈T+A, we get x,f+y∈T+A. This together with x,f∈T implies that x,y∈T+A−T. Since x∈DT⊂T0⊥ and y∈T0⊥, we havex,y∈T0⊥2. Consequently, AT|DT⊂T+A−T∩T0⊥2.
Therefore, (26) holds. This completes the proof.
The following result is a direct consequence of (14) and Proposition 1.Corollary 1.
LetT,A∈LRX with DT⊂DA and A0⊂T0. If T is closed and Hermitian, then(29)AT|DT=T+A−T∩T0⊥2.Now, we give a decomposition ofT+A in the case that T is closed. This decomposition plays an important role in the study of some properties about linear relations in the present study.Theorem 1.
LetT,A∈LRX with DT⊂DA and A0⊂T0. If T is closed, then(30)T+A=Ts+AT⊕T∞.Proof.
We suppose thatT is closed. It is evident that Ts+AT and T∞ are orthogonal.
For anyx,y∈T+A, there exists y1,y2∈X such that x,y1∈T, x,y2∈A, and y=y1+y2. It follows from (11) that x,y1=x,y11+0,y12 with x,y11∈Ts and 0,y12∈T∞. Let y2=y21+y22, where y21∈T0⊥ and y22∈T0. Then, x,y21∈AT and 0,y12+y22∈T∞. Hence, x,y=x,y11+y21+0,y12+y22 belongs to Ts+AT⊕T∞. This implies that T+A⊂Ts+AT⊕T∞.
On the other hand, for any givenx,y∈Ts+AT⊕T∞, it can be decomposed as x,y=x,y1+0,y2 with x,y1∈Ts+AT and 0,y2∈T∞⊂T. Let y1=y11+y12, where x,y11∈Ts⊂T and x,y12∈AT. Given any z∈Ax, it can be decomposed as z=z1+z2 with z1∈T0⊥ and z2∈T0, and we have that y12=z1=z−z2 and x,y11+y2−z2∈T. Consequently,(31)x,y=x,y1+y2=x,y11+y2−z2+z∈T+A,which yields that Ts+AT⊕T∞⊂T+A.
Therefore, (30) holds and the proof is complete.Corollary 2.
LetT,A∈LRX with DT⊂DA and A0⊂T0. If T and T+A are closed, then(32)T+As=Ts+AT.Proof.
Suppose thatT and T+A are closed. It follows from A0⊂T0 that T∞=T+A∞. Then, T+A=T+As⊕T∞ by (11), which together with (30) implies T+As=Ts+AT. The proof is complete.Remark 2.
In Theorem 3.1 in [25], Shi showed thatT+As=PTs+QAsunder some conditions, whereP:T0⊥⟶T+A0⊥andQ:A0⊥⟶T+A0⊥are orthogonal projections. Note thatT+A0=T0by the assumption thatA0⊂T0. Then,Pis an identity mapping fromT0⊥onto itself andQAs|DT¯∩DA=AT. Hence, Theorem 3.1 in [25] is consistent with the result in Corollary 2 in the case that A is closed. Further, Theorem 3.2 in [25] can be directly derived from Corollary 2.


## 3. Relationships between Perturbations ofT and Ts

In this section, we shall investigate the relationships between properties ofT+A and Ts+AT and the perturbation terms of T and Ts. Note that the relation T+A can be decomposed as (30), we shall consider a general decomposition, which is induced by reducing subspaces and discuss the relationships between a relation and its decomposition parts in Subsection 3.1. Using these results, the corresponding relationships between relation T+A and operator Ts+AT are given in Subsection 3.2. Further, relationships between various perturbations of closed relation T and its operator part Ts are studied in Subsection 3.3.


### 3.1. Decomposition of Relations

The following concept of reducing subspace for a linear relation in Banach spaces can be extended to the corresponding one in Hilbert spaces (cf., [18]).LetX be a Banach space. Suppose that X has the decomposition(33)X=X1∔X2,where X1 and X2 are closed subspaces of X and X1∩X2=0. Let P:X⟶X1 be the projection on X1 along with X2 and T∈LRX. We denote(34)P2T=Px,Pyx,y∈T.It is clear thatT∩X12⊂P2T.IfP2T⊂T, then X1 is called a reducing subspace of T. We also say that X1 reduces T or T is reduced by X1. In this case, one has that(35)T∩X12=P2T.It can be easily verified thatX1 reduces T if and only if X2 reduces T. Further, if X1 reduces T, then(36)T=T1∔T2,where(37)Ti=T∩Xi2,i=1,2.Further, we suppose thatX is a Hilbert space and X1⊥X2. If T is reduced by X1, then(38)T=T1⊕T2,where T1 and T2 are defined by (37).Now, we give a relationship between the closedness ofT and its decomposition parts.Proposition 2.
LetX be a Banach space, T∈LRX be reduced by X1⊂X, and Tii=1,2 be defined by (37). Then, T is a closed relation in X2 if and only if Ti is a closed relation in Xi2 for each i=1,2.Proof.
It is evident thatTi is closed in Xi2 for i=1,2 if T is closed.
We suppose thatTi is closed in Xi2 for i=1,2. Given any sequence xn,ynn=1∞⊂T with xn,yn⟶x,y as n⟶∞, there exist xn,i,yn,i∈Ti,i=1,2 such that xn,yn=xn,1,yn,1+xn,2,yn,2 by (36). Note that Xi is closed for i=1,2. Then, P:X⟶X1 is bounded by closed graph theorem. So, there exist xi,yi∈X such that xn,i,yn,i⟶xi,yi as n⟶∞ for i=1,2. Consequently, x,y=x1,y1+x2,y2 with xi,yi∈Ti since Ti is closed in Xi2 for i=1,2. Hence, x,y∈T again by (36). Therefore, T is a closed relation in X2. This completes the proof.
In the following, we shall discuss the relationships between the hermiticity and self-adjointness ofT and its decomposition parts, respectively.Proposition 3.
LetX be a Hilbert space, T∈LRX be reduced by X1⊂X, and Tii=1,2 be defined by (37) with X1⊥X2. Then, T is an Hermitian relation in X2 if and only if Ti is an Hermitian relation in Xi2 for each i=1,2.Proof.
Obviously,Ti is an Hermitian relation in Xi2 for each i=1,2 if T is Hermitian.
We suppose thatTi is an Hermitian relation in Xi2 for i=1,2. Given any x,y,f,g∈T, there exist x1,y1,f1,g1∈T1 and x2,y2,f2,g2∈T2 such that x,y=x1,y1+x2,y2 and f,g=f1,g1+f2,g2 by (38). Since Ti is Hermitian in Xi2 for i=1,2, we have that(39)f,y=f1+f2,y1+y2=f1,y1+f2,y2=g1,x1+g2,x2=g1+g2,x1+x2=g,x
Therefore,T is an Hermitian. The proof is complete.Lemma 6 (see p. 26 in [18]).
LetX be a Hilbert space and T∈LRX.(i)
LetX1 and X2 be closed subspaces in X and Ti⊂Xi2,i=1,2. If T=T1⊕T2 and X=X1⊕X2, then Xi is a reducing subspace of T, i=1,2.(ii)
IfT is a self-adjoint relation in X2 and X1 reduces T with X1⊥X2, then Ti defined by (37) is a self-adjoint relation in Xi2 for i=1,2.Proposition 4.
LetX be a Hilbert space, T∈LRX be reduced by X1⊂X, and Tii=1,2 be defined by (37) with X1⊥X2. Then, T is a self-adjoint relation in X2 if and only if Ti is a self-adjoint relation in Xi2 for each i=1,2.Proof.
The necessity directly follows from (ii) of Lemma6. Now, we consider the sufficiency. Since Tj is a self-adjoint relation in Xj2, we have RTj±i=Xj for j=1,2 by Lemma 4. So, it follows from (37) and (38) that(40)RT±i=RT1±i⊕RT2±i=X1⊕X2=X,which implies that T is a self-adjoint relation in X2 by Lemma 4 and Proposition 3. This completes the proof.
Now, we discuss the relationships between the defect indices ofT and its decomposition parts.Definition 3.
(see Definition 2.3 in [26]). Let X be a Hilbert space and T∈LRX. The subspace RT−λ⊥ is called the defect space of T and λ, and the number βT,λ:=dimRT−λ⊥ is called the defect index of T and λ.
LetX be a Hilbert space. It follows from Theorem 2.3 in [26] that βT,λ is constant in the upper and lower half-planes if T is an Hermitian relation in X2. In this case, it denotes that(41)d+T=βT,−i,d−T=βT,i,which are called the positive and negative defect indexes of T, respectively. The pair d+T,d−T is called the defect indices of T (see [26]).Proposition 5.
LetX be a Hilbert space, T∈LRX be reduced by X1⊂X, and Tii=1,2 be defined by (37) with X1⊥X2. Then,(42)βT,λ=βT1,λ+βT2,λ,λ∈C,where Ti is regarded as a relation in Xi2 for i=1,2.Proof.
Given anyλ∈C, it suffices to show that(43)RT−λ⊥=RT1−λ⊥⊕RT2−λ⊥,whereRTi−λ⊥ is the orthogonal complement of RTi−λ in Xi for i=1,2.Given anyy∈RT−λ⊥, there exist y1∈X1 and y2∈X2 such that y=y1+y2. So, for every f∈RT1−λ⊂RT−λ∩X1 and g∈RT1−λ⊂RT−λ∩X2, we have y1,f=y,f=0 and y2,g=y,g=0. Hence, yi∈RTi−λ⊥ for i=1,2. And consequently, RT−λ⊥⊂RT1−λ⊥⊕RT2−λ⊥.On the other hand, for anyy∈RT1−λ⊥⊕RT2−λ⊥, it can be decomposed as y=y1+y2 with yi∈RTi−λ⊥⊂Xi,i=1,2. For any f∈RT−λ, there exists fi∈RTi−λ⊂Xi, i=1,2 such that f=f1+f2 by (37) and (38). Hence,(44)y,f=y1+y2,f1+f2=y1,f1+y2,f2=0,which implies that y∈RT−λ⊥. Then, RT1−λ⊥⊕RT2−λ⊥⊂RT−λ⊥. Therefore, (43) holds, which yields that (42) holds and this completes the proof.The following result can be directly derived from Propositions3 and 5.Corollary 3.
LetX be a Hilbert space and T∈LRX be Hermitian and reduced by X1⊂X. Furthermore, let Tii=1,2 be defined by (37) with X1⊥X2. Then,(45)d+T=d+T1+d+T2,d−T=d−T1+d−T2,where Ti is regarded as a relation in Xi2 for i=1,2.In the following, we shall investigate the relationships among the spectra and various spectra ofT and its decomposition parts, including point spectra, essential spectra, discrete spectra, continuous spectra, singular continuous spectra, absolutely continuous spectra, and singular spectra.The following result is a generalization of self-adjoint case in Hilbert spaces Proposition 3.2 in [19].Theorem 2.
LetX be a Banach space, T∈LRX be reduced by X1⊂X, and Tii=1,2 be defined by (37). Then,(46)σT=σT1∪σT2,ρT=ρT1∩ρT2,where σTi and ρTi are the spectrum and resolvent set of Ti in Xi2 for i=1,2, respectively.Proof.
It suffices to show that the second relation in (46) holds, which implies that σT=σT1∪σT2.
We first show thatρT⊂ρT1∩ρT2. For any given λ∈ρT, RT−λ=Xand there exists a constant c>0 such that for all x,y∈T,(47)x≤cy−λx.
Obviously, (47) holds for all x,y∈T1 since T1⊂T. Now, we show that T1−λ is surjective in X1. For any given z∈X1, there exists x,y∈T such that z=y−λx since RT−λ=X. It follows from (36) that x,y can be decomposed as x,y=x1,y1+x2,y2 with xi,yi∈Ti and i=1,2. Then, z=y1−λx1+x2−λy2. Because z,x1,y1∈X1 and x2,y2∈X2, we have z=y1−λx1 and y2−λx2=0, and thus, z∈RT1−λ. Consequently, T1−λ is surjective in X1. Hence, λ∈ρT1. With a similar argument, one can show λ∈ρT2. Then, λ∈ρT1∩ρT2. Therefore, ρT⊂ρT1∩ρT2.
Next, we consider the inverse inclusion. For any givenλ∈ρT1∩ρT2, we have RTi−λ=Xi for i=1,2, and there is a constant b>0 such that(48)xi≤byi−λxi,∀xi,yi∈Ti,i=1,2.
For anyz∈X, there exist z1∈X1 and z2∈X2 such that z=z1+z2. Then, there is xi,yi∈Ti such that zi=yi−λxi for i=1,2, which implies that z=y1+y2−λx1+x2. Further, by (36), we can get that x1+x2,y1+y2∈T. Hence, z∈RT−λ, and consequently, RT−λ=X. Note that P:X⟶X1 is bounded by the closed graph theorem. There exists a constant M>0 such that(49)Px≤Mx,1−Px≤Mx∀x∈X.
For anyx′,y′∈T, it follows from (36) that x′,y′ can be uniquely decomposed as x′,y′=x1′,y1′+x1′,y2′ with xi′,yi′∈Ti and i=1,2. By utilizing (48) and (49), one can get that(50)x′≤x1′+x2′≤by1′−λx1′+y2′−λx2′=bPy′−λx′+1−Py′−λx′≤2bMy′−λx′,which yields that T−λ−1 is a bounded linear operator defined on X, and consequently, λ∈ρT. It follows that ρT1∩ρT2⊂ρT+A. Therefore, the second relation in (46) holds. The proof is complete.
Now, we discuss the relationships between the point spectra and essential spectra ofT and its decomposition parts.Theorem 3.
LetX be a Banach space, T∈LRX be reduced by X1⊂X, and Tii=1,2 be defined by (37). Then, (51)σpT=σpT1∪σpT2,NT−λ=NT1−λ∔NT2−λ,λ∈C.Proof.
It suffices to show thatNT−λ=NT1−λ∔NT2−λ for every λ∈C. It is evident that NT1−λ∔NT2−λ⊂NT−λ since Ti⊂T and i=1,2. For any f∈NT−λ, we have f,λf∈T, which can be decomposed as f,λf=f1,g1+f2,g2 by (36), where fi,gi∈Ti and i=1,2. Note that λf=g1+g2=λf1+λf2 and fi,gi∈Xi, i=1,2. One can get that gi=λfi for i=1,2, which implies that fi∈NTi−λ and i=1,2. This yields that NT−λ⊂NT1−λ∔NT2−λ. Therefore, NT−λ=NT1−λ∔NT2−λ. This completes the proof.
The following result can be easily verified by Theorems2 and 3. So, its detail proofs are omitted.Theorem 4.
LetX be a Banach space, T∈LRX be reduced by X1⊂X, and Tii=1,2 be defined by (47). Then, (52)σeT=σeT1∪σeT2,σdT=σdT1∪σdT2.
To the end of this subsection, we shall discuss the relationships between the continuous spectra, singular continuous spectra, absolutely continuous spectra, and singular spectra ofT and its decomposition parts, separately.
Suppose thatX is a Hilbert space and X1⊥X2. If T∈LRX is self-adjoint and can be decomposed as (38), then Ti is self-adjoint in Xi2 for i=1,2 by Lemma 6. Let Ti=Ti,s⊕Ti,∞, where Ti,∞=0,y∈Xi2:0,y∈Ti, Ti,s=Ti⊖Ti,∞, and i=1,2. It is evident that(53)Ts=T1,s⊕T2,s,Et=E1t⊕E2t,t∈R,where Eitt∈R is the spectral family of Ti in Xi2 for i=1,2.It follows from the second relation in (51) that(54)XP=X1,P⊕X2,P,and consequently,(55)XC=X1,C⊕X2,C,where Xi,P and Xi,C are the discontinuous and continuous subspaces in Xi with respect to Ti for i=1,2.By utilizing (53) and (55), we can get that(56)XSC=X1,SC⊕X2,SC,which together with (54) and (55) implies that(57)XAC=X1,AC⊕X2,AC,XS=X1,S⊕X2,S,where Xi,SC, Xi,AC, and Xi,S are the singular continuous, absolutely continuous, and singular subspaces in Xi with respect to Ti for i=1,2.It is derived from (37), (38), and (54)–(57) that(58)TP=T1,P⊕T2,P,TC=T1,C⊕T2,C,TSC=T1,SC⊕T2,SC,TAC=T1,AC⊕T2,AC,TS=T1,S⊕T2,S,Ti,P=TP∩Xi,P2,Ti,C=TC∩Xi,C2,Ti,SC=TSC∩Xi,SC2,Ti,AC=TAC∩Xi,AC2,Ti,S=TS∩Xi,S2,i=1,2.where Ti,P, Ti,C, Ti,SC, Ti,AC, and Ti,S are the (spectral) discontinuous, continuous, singular continuous, absolutely continuous, and singular parts of Ti in Xi2 for i=1,2, respectively.Theorem 5.
LetX be a Hilbert space and T∈LRX be self-adjoint and reduced by X1⊂X. Further, let Tii=1,2 be defined by (37) with X1⊥X2. Then,(59)σTP=σT1,P∪σT2,P,σrT=σrT1∪σrT2,r=c,ac,sc,s.where Ti is regarded as a relation in Xi2 for i=1,2.Proof.
This theorem can be directly derived from (i) of Lemma6, Theorem 2, and (54)–(58).


### 3.2. Relationships betweenT+A and Ts+AT

In this subsection, we shall study the relationships between the properties ofT+A and Ts+AT, including their closedness, hermiticity, self-adjointness, various spectra, and defect indices. Further, X is always assumed to be a complex Hilbert space in this part.Lemma 7.
LetT,A∈LRX satisfy that DT⊂DA and A0⊂T0. If T is closed and DT⊂T0⊥, then T+A is reduced by T0.Proof.
We suppose thatT is closed and DT⊂T0⊥. It follows from the assumption DT⊂DA and A0⊂T0 that(60)T∞=T∩T02=T+A∩T02.
Note that (15) holds if DT⊂T0⊥, which together with Proposition 1 and Proposition 2.1 in [21] implies that(61)Ts+AT=Ts+AT|DT=T+A−T+T∩T0⊥2=T+A∩T0⊥2.s
Therefore,T+A is reduced by T0 by Theorem 1 and (i) of Lemma 6. This completes the proof.
The following result comes from [6], and we shall give its proof for completeness.Lemma 8.
LetT∈LRX be closed. Then, T∞ is a self-adjoint relation in T0.Proof.
We suppose thatT is closed. Obviously, T∞ is closed in T0 for any 0,f,0,g∈T∞ and 0,f=g,0=0. Then, T∞ is Hermitian in T0. Note that RT∞±i=T0. Therefore, T∞ is a self-adjoint relation in T0 by Lemma 4. The proof is complete.
By Propositions 2 ‐ 4, Lemmas7 and 8, (61) and (62), one can easily get the following results.Theorem 6.
LetT,A∈LRX satisfy that DT⊂DA and A0⊂T0. If T is closed and DT⊂T0⊥, then(i)
T+A is closed if and only if Ts+AT is closed in T0⊥;(ii)
T+A is an Hermitian relation in X2 if and only if Ts+AT is an Hermitian operator in T0⊥.(iii)
T+A is a self-adjoint relation in X2 if and only if Ts+AT is a self-adjoint operator in T0⊥.
Now, we give a relationship between the spectra and various spectra ofT+A and Ts+AT.Theorem 7.
LetT,A∈LRX satisfy that DT⊂DA and A0⊂T0.(i)
IfT is closed and DT⊂T0⊥, then(62)σeT+A=σeTs+AT,σdT+A=σdTs+AT.(ii)
IfT is closed and T+A is self-adjoint, then(63)σT+AP=σTs+ATP,σrT+A=σrTs+AT,r=c,ac,sc,s,whereTs+AT is regarded as a relation in T0⊥2.Proof.
The first assertion of Theorem7 can be easily verified by (16), Theorems 1‐4, and Lemma 7. Suppose that T+A is self-adjoint. Then, DT=DT+A⊂T+A0⊥=T0⊥. And consequently, T0 reduces T+A by Lemma 7. It follows from (16), and Theorems 1, 5, and 6 that (63) holds. This completes the proof.Remark 3.
By Corollary 2and Lemmas 3 and 5, (63) holds.
The following results can be easily directly derived from (14) and Theorem 7.Corollary 4.
LetT,A∈LRX satisfy that DT⊂DA and A0⊂T0. If T is closed and Hermitian, then(64)σeT+A=σeTs+AT,σdT+A=σdTs+AT,where Ts+AT is regarded as a relation in T0⊥2.
To the end of this subsection, we shall give a relationship between the defect indices ofT+A and Ts+AT.Theorem 8.
LetT,A∈LRX satisfy that DT⊂DA and A0⊂T0. If T is closed and T+A is Hermitian, then(65)d+T+A=d+Ts+AT,d−T+A=d−Ts+AT,where Ts+AT is regarded as a relation in T0⊥2.Proof.
We suppose thatT is closed and T+A is Hermitian. Then, DT=DT+A⊂T+A0⊥=T0⊥. And consequently, T0 reduces T+A by Lemma 7. It is derived from Theorem 1, Corollary 3, and (ii) of Theorem 6 that (65) holds. Thus, the proof is complete.


### 3.3. Relationships between Perturbation Terms ofT and Ts

In this subsection, we shall discuss the relationships between the perturbation terms ofT and Ts if T is closed. Including relatively bounded and relatively compact perturbation terms, finite rank perturbation term, and trace class perturbation term.We shall first recall the concepts of relatively bounded and compact relations, which were introduced by Cross [20].LetX and Y be normed spaces, T∈LRX,Y, and XT denote the space DT,⋅T, where(66)xT=x+Tx,x∈DT.We then defineGT∈LRXT,X by GTx=x for x∈XT. GT is called the graph operator of T.Definition 4.
(see Definition VII.2.1 in [20]). Let X, Y, and Z be normed spaces, T∈LRX,Y, and A∈LRX,Z with DT⊂DA.(1)
The linear relationA is said to be T-bounded if there exist nonnegative numbers a and b such that(67)Ax≤ax+bTx,x∈DT.IfA is T-bounded, then the infimum of all numbers b≥0 for which (67) holds with a constant a≥0, is called the T-bound of A.(2)
The linear relationA is said to be T-compact (or relatively compact toT) if AGT is compact, i.e., A:XT⟶Z is compact.Lemma 9 (see Lemma 2.7 in [22]).
LetX be a Hilbert space and T∈LRX can be decomposed as (11). Then, (11) holds and(68)Tx=Tsx,x∈DT,T=Ts.Lemma 10.
LetX be a Hilbert space and T,A∈LRX satisfy that DT⊂DA and A0⊂T0. Then,(69)ATx≤Ax,x∈DT¯∩DA,AT≤A.Proof.
It suffice to show that the first relation in (69) holds. Let x∈DT¯∩DA. It follows from (i) of Lemma 1 that for any ϵ>0, there exists y∈Ax such that y<Ax+ϵ. Let y=y1+y2, where y1∈T0⊥ and y2∈T0¯. Then, ATx=y1 by (25). Consequently, ATx=y1≤y<Ax+ϵ. Hence, ATx≤Ax by the arbitrariness of ϵ. Therefore, (69) holds and this completes the proof.
By Lemmas9 and 10, one can easily get the following result.Theorem 9.
LetX be a Hilbert space and T,A∈LRX satisfy that T is closed, DT⊂DA, and A0⊂T0. If A is T-bounded with T-bound less that b, then AT is Ts-bounded with Ts-bound less than b, b≥0.
Now, we give a relationship between the relatively compact perturbation ofT and Ts.Theorem 10.
LetX be a Hilbert space and T,A∈LRX satisfy that T is closed, DT⊂DA, and A0⊂T0. If A is T-compact, then AT is Ts-compact.Proof.
It follows from (13) and Lemma 9 that XT=XTs. We suppose that A is T-compact. Then, for any given bounded sequencexnn=1∞ in XT, and QAAxnn=1∞ has a convergent subsequence QAAxnkk=1∞ in X/A0¯. So, for each k≥1, there is ynk∈Axnk such that ynkk=1∞ is convergent in X by (iii) of Lemma 1. Let ynk=ynk,1+ynk,2 with ynk,1∈T0⊥ and ynk,2∈T0. Hence, ATxnk=ynk,1, and ynk,1k=1∞ is convergent in T0⊥. This means ATxnn=1∞ has a convergent subsequence in T0⊥. Therefore, AT is Ts-compact. Thus, the proof is complete.
With a similar argument to that used in the proof of Theorem10, one can easily show the following results hold.Corollary 5.
LetX and Y be Hilbert spaces and A∈LRX,Y. Then, A is a compact relation if and only if PAA is a compact operator, where PA:Y⟶A0⊥ is an orthogonal projection.Corollary 6.
LetX be a Hilbert space and T,A∈LRX satisfy that DT⊂DA and A0⊂T0. If A is compact, then AT is compact.Proposition 6.
LetX be a Hilbert space and T,A∈LRX satisfy that DT⊂DA∩T0⊥ and A0⊂T0. Then, A|DT¯∩DA is an Hermitian relation in X2 if and only if AT is an Hermitian operator in T0⊥.Proof.
It follows from the assumptionDT⊂T0⊥ that DAT=DT¯∩DA⊂T0⊥, which means that AT is a linear operator in the Hilbert space T0⊥.
Suppose thatA|DT¯∩DA is an Hermitian relation in X2. For any f,g and h,k∈AT, there exists f,y,h,z∈A such that y=g+y′ and z=k+z′, where y′,z′∈T0¯. Note that f,h∈DAT⊂T0⊥ and the assumption that A|DT¯∩DA is Hermitian, we get(70)g,h=y,h=f,z=f,k.
Hence,AT is an Hermitian operator in T0⊥.
Now, we consider the inverse. We assume thatAT is an Hermitian operator in T0⊥. We fixed x,y,u,v∈A|DT¯∩DA. There exist y1,v1∈T0⊥ and y2,v2∈T0¯ such that y=y1+y2 and v=v1+v2. Hence, x,y1,u,v1∈AT. By the assumption that AT is Hermitian and the fact that x,u∈DAT⊂T0⊥, we get(71)y,u=y1,u=x,v1=x,v.
Therefore,A|DT¯∩DA is an Hermitian relation in X2. This completes the proof.
LetX be a Hilbert space and T∈LRX. If T is self-adjoint, then DT¯=T∗0⊥=T0⊥by Proposition III.1.4 in [20], which together with Proposition 6, can easily achieve the following result.Corollary 7.
LetX be an Hilbert space and T,A∈LRX, satisfying DT⊂DA and A0⊂T0. If T is self-adjoint, then A|DT¯∩DA is an Hermitian relation in X2 if and only if AT is a symmetric operator in T0⊥, that is, AT is a densely defined Hermitian operator in T0⊥.
Now, we introduce the concept of degenerate linear relation, which is a generation of single-valued case.Definition 5.
LetXandYbe normed spaces andA∈LRX,Y.Ais said to be degenerate ifAis bounded anddimRA<∞.
LetA∈LRX,Y be degenerate. It is evident that QAA is degenerate. Hence, QAA is compact (see p.160 in [27]), that is, A is compact.
Motivated by the definition of the norm of a linear relation, we introduce the concept of trace class linear relations.Definition 6.
LetXandYbe Hilbert spaces andA∈LRX,YwithDA=X. We say thatAbelongs to trace class relations ifPAAbelongs to trace class operators, wherePA:Y⟶A0⊥is the orthogonal projection.Theorem 11.
LetX be a Hilbert space and T,A∈LRX satisfy that DT⊂DA and A0⊂T0. If A is degenerate, then AT is also degenerate.Proof.
We suppose thatA is degenerate, that is, dimRA=m<∞ and A is bounded. Then, AT is bounded by Lemma 10. Let y1,y2,…,ym be a base of RA. They can be decomposed as yi=yi,1+yi,2 with yi,1∈T0⊥ and yi,2∈T0¯, i=1,2,…,m. Now, we show that for any given z∈RAT⊂T0⊥, it can be expressed as a linear combination of yi,1, 1≤i≤m. We then set x,y∈A with y=z+y′, where y′∈T0¯. There exist ci and i=1,2,…,m such that(72)y=∑i=1mciyi=∑i=1mciyi,1+∑i=1mciyi,2=z+y′.
Note thatz,∑i=1mciyi,1∈T0⊥ and y′,∑i=1mciyi,2∈T0¯, we can get that z=∑i=1mciyi,1, which yields that dimRAT≤m<∞. Therefore, AT is degenerate and the proof is complete.
Now, we recall a necessary and sufficient condition about trace operators (cf., Theorem 7.12 in [28]).Lemma 11.
LetT be an operator from Hilbert space X into Hilbert space Y with DT=X. Then, T belongs to trace class operators if and only if there exist sequences fnn=1∞ from X and gnn=1∞ from Y such that fn=gn=1 for each n≥1, and there is a constant sequence cnn=1∞ for which ∑n=1∞cn<∞ and T can be expressed as(73)Tx=∑n=1∞cnx,fngn,x∈X.Remark 4.
In the case thatX=Yin Lemma 11, Tbelongs to trace class operators if and only ifTcan be expressed as(74)Tx=∑n=1∞cnx,fnfn,x∈X,where fnn=1∞ form an orthonormal family of eigenvectors of T and the cn are the associated (repeated) eigenvalues with ∑n=1∞cn<∞ (cf., [27], p.543).Theorem 12.
LetX be a Hilbert space and T,A∈LRX with DA=X and DT¯=T0⊥, and A0⊂T0. If A belongs to trace class relations, then AT belongs to trace class operators in T0⊥.Proof.
The assumptionsDA=X and DT¯=T0⊥ implies that DAT=DT¯∩DA=T0⊥.
We suppose thatA belongs to trace class relations, that is, PAA belongs to trace class operators. It follows from Lemma 11 that there exist sequences fnn=1∞ and gnn=1∞ from X and cnn=1∞ from C satisfying fn=gn=1, n≥1, and ∑n=1∞cn<∞ such that(75)PAAx=∑n=1∞cnx,fngn,x∈X.
Letfn=fn,1+fn,2 and gn=gn,1+gn,2 with fn,1,gn,1∈T0⊥ and fn,2,gn,2∈T0¯ for n≥1. We shall show that AT=S|T0⊥, where S=∑n=1∞cn⋅,fn,1gn,1. Given any x∈DAT, there is y∈X such that x,y∈A. Since A0⊂T0, y can be decomposed as y=y1+y2+y3 with y1∈T0⊥, y2∈A0⊥∩T0¯, and y3∈A0¯. Then, ATx=y1 and PAAx=y1+y2. By (75) and the fact that x∈DAT=T0⊥, one can get that(76)y1+y2=∑n=1∞cnx,fn,1+fn,2gn,1+gn,2=∑n=1∞cnx,fn,1gn,1+∑n=1∞cnx,fn,1gn,2.
Note thaty1,∑n=1∞cn⋅,fn,1gn,1∈T0⊥ and y2,∑n=1∞cnx,fn,1gn,2∈T0¯, we have that ATx=y1=∑n=1∞cn⋅,fn,1gn,1. Consequently, AT⊂S. Hence, AT=S|T0⊥. Let L:=n≥1:fn,1≠0andgn,1≠0. By setting fn,1′=fn,1/fn,1 and gn,1′=gn,1/gn,1 for each n∈L, we can get (77)AT\x=∑n∈Lcnfn,1gn,1x,fn,1′gn,1′,x∈T0⊥, where ∑n∈Lcnfn,1gn,1≤∑n=1∞cn<∞. Therefore, AT belongs to trace class in T0⊥ by Lemma 11. This completes the whole proof.Remark 5.
In the present study, we construct a linear operator, which is induced by two linear relations, and then establish the relationships between the perturbation terms of a closed relation and the perturbation terms of its operator part (see Theorems9–12), and give the relationships between spectrum of a perturbed relation and spectrum of a perturbed operator (see Theorem 7). By using the results obtained in the present study, we shall deeply study stabilities of the spectra of linear relations under some perturbations in our forthcoming study, especially the invariance of the absolutely continuous spectrum of a self-adjoint linear relation under trace class perturbation.Remark 6.
Note that the constructing technique in the present study can also be applied in our previous works. For example, Theorem 5.2 in [29] can be followed by Lemma 3, Corollary 7, Theorems 6 and 9, ([29], Lemma 5.8), and ([27], Theorem V.4.3); Theorem 5.3 in [29] can be followed by Lemmas 2, 3, and 10, Corollaries 4 and 7, Theorem 10, ([29], Lemma 5.8); and ([27], Theorem V.4.10); Theorem 4.1 in [22] can be followed by Lemmas 2 and 3, Corollaries 4 and 7, Theorems 6 and 10, ([29], Lemma 5.8), and ([28], Theorem 9.9), respectively.


## 3.1. Decomposition of Relations

The following concept of reducing subspace for a linear relation in Banach spaces can be extended to the corresponding one in Hilbert spaces (cf., [18]).LetX be a Banach space. Suppose that X has the decomposition(33)X=X1∔X2,where X1 and X2 are closed subspaces of X and X1∩X2=0. Let P:X⟶X1 be the projection on X1 along with X2 and T∈LRX. We denote(34)P2T=Px,Pyx,y∈T.It is clear thatT∩X12⊂P2T.IfP2T⊂T, then X1 is called a reducing subspace of T. We also say that X1 reduces T or T is reduced by X1. In this case, one has that(35)T∩X12=P2T.It can be easily verified thatX1 reduces T if and only if X2 reduces T. Further, if X1 reduces T, then(36)T=T1∔T2,where(37)Ti=T∩Xi2,i=1,2.Further, we suppose thatX is a Hilbert space and X1⊥X2. If T is reduced by X1, then(38)T=T1⊕T2,where T1 and T2 are defined by (37).Now, we give a relationship between the closedness ofT and its decomposition parts.Proposition 2.
LetX be a Banach space, T∈LRX be reduced by X1⊂X, and Tii=1,2 be defined by (37). Then, T is a closed relation in X2 if and only if Ti is a closed relation in Xi2 for each i=1,2.Proof.
It is evident thatTi is closed in Xi2 for i=1,2 if T is closed.
We suppose thatTi is closed in Xi2 for i=1,2. Given any sequence xn,ynn=1∞⊂T with xn,yn⟶x,y as n⟶∞, there exist xn,i,yn,i∈Ti,i=1,2 such that xn,yn=xn,1,yn,1+xn,2,yn,2 by (36). Note that Xi is closed for i=1,2. Then, P:X⟶X1 is bounded by closed graph theorem. So, there exist xi,yi∈X such that xn,i,yn,i⟶xi,yi as n⟶∞ for i=1,2. Consequently, x,y=x1,y1+x2,y2 with xi,yi∈Ti since Ti is closed in Xi2 for i=1,2. Hence, x,y∈T again by (36). Therefore, T is a closed relation in X2. This completes the proof.
In the following, we shall discuss the relationships between the hermiticity and self-adjointness ofT and its decomposition parts, respectively.Proposition 3.
LetX be a Hilbert space, T∈LRX be reduced by X1⊂X, and Tii=1,2 be defined by (37) with X1⊥X2. Then, T is an Hermitian relation in X2 if and only if Ti is an Hermitian relation in Xi2 for each i=1,2.Proof.
Obviously,Ti is an Hermitian relation in Xi2 for each i=1,2 if T is Hermitian.
We suppose thatTi is an Hermitian relation in Xi2 for i=1,2. Given any x,y,f,g∈T, there exist x1,y1,f1,g1∈T1 and x2,y2,f2,g2∈T2 such that x,y=x1,y1+x2,y2 and f,g=f1,g1+f2,g2 by (38). Since Ti is Hermitian in Xi2 for i=1,2, we have that(39)f,y=f1+f2,y1+y2=f1,y1+f2,y2=g1,x1+g2,x2=g1+g2,x1+x2=g,x
Therefore,T is an Hermitian. The proof is complete.Lemma 6 (see p. 26 in [18]).
LetX be a Hilbert space and T∈LRX.(i)
LetX1 and X2 be closed subspaces in X and Ti⊂Xi2,i=1,2. If T=T1⊕T2 and X=X1⊕X2, then Xi is a reducing subspace of T, i=1,2.(ii)
IfT is a self-adjoint relation in X2 and X1 reduces T with X1⊥X2, then Ti defined by (37) is a self-adjoint relation in Xi2 for i=1,2.Proposition 4.
LetX be a Hilbert space, T∈LRX be reduced by X1⊂X, and Tii=1,2 be defined by (37) with X1⊥X2. Then, T is a self-adjoint relation in X2 if and only if Ti is a self-adjoint relation in Xi2 for each i=1,2.Proof.
The necessity directly follows from (ii) of Lemma6. Now, we consider the sufficiency. Since Tj is a self-adjoint relation in Xj2, we have RTj±i=Xj for j=1,2 by Lemma 4. So, it follows from (37) and (38) that(40)RT±i=RT1±i⊕RT2±i=X1⊕X2=X,which implies that T is a self-adjoint relation in X2 by Lemma 4 and Proposition 3. This completes the proof.
Now, we discuss the relationships between the defect indices ofT and its decomposition parts.Definition 3.
(see Definition 2.3 in [26]). Let X be a Hilbert space and T∈LRX. The subspace RT−λ⊥ is called the defect space of T and λ, and the number βT,λ:=dimRT−λ⊥ is called the defect index of T and λ.
LetX be a Hilbert space. It follows from Theorem 2.3 in [26] that βT,λ is constant in the upper and lower half-planes if T is an Hermitian relation in X2. In this case, it denotes that(41)d+T=βT,−i,d−T=βT,i,which are called the positive and negative defect indexes of T, respectively. The pair d+T,d−T is called the defect indices of T (see [26]).Proposition 5.
LetX be a Hilbert space, T∈LRX be reduced by X1⊂X, and Tii=1,2 be defined by (37) with X1⊥X2. Then,(42)βT,λ=βT1,λ+βT2,λ,λ∈C,where Ti is regarded as a relation in Xi2 for i=1,2.Proof.
Given anyλ∈C, it suffices to show that(43)RT−λ⊥=RT1−λ⊥⊕RT2−λ⊥,whereRTi−λ⊥ is the orthogonal complement of RTi−λ in Xi for i=1,2.Given anyy∈RT−λ⊥, there exist y1∈X1 and y2∈X2 such that y=y1+y2. So, for every f∈RT1−λ⊂RT−λ∩X1 and g∈RT1−λ⊂RT−λ∩X2, we have y1,f=y,f=0 and y2,g=y,g=0. Hence, yi∈RTi−λ⊥ for i=1,2. And consequently, RT−λ⊥⊂RT1−λ⊥⊕RT2−λ⊥.On the other hand, for anyy∈RT1−λ⊥⊕RT2−λ⊥, it can be decomposed as y=y1+y2 with yi∈RTi−λ⊥⊂Xi,i=1,2. For any f∈RT−λ, there exists fi∈RTi−λ⊂Xi, i=1,2 such that f=f1+f2 by (37) and (38). Hence,(44)y,f=y1+y2,f1+f2=y1,f1+y2,f2=0,which implies that y∈RT−λ⊥. Then, RT1−λ⊥⊕RT2−λ⊥⊂RT−λ⊥. Therefore, (43) holds, which yields that (42) holds and this completes the proof.The following result can be directly derived from Propositions3 and 5.Corollary 3.
LetX be a Hilbert space and T∈LRX be Hermitian and reduced by X1⊂X. Furthermore, let Tii=1,2 be defined by (37) with X1⊥X2. Then,(45)d+T=d+T1+d+T2,d−T=d−T1+d−T2,where Ti is regarded as a relation in Xi2 for i=1,2.In the following, we shall investigate the relationships among the spectra and various spectra ofT and its decomposition parts, including point spectra, essential spectra, discrete spectra, continuous spectra, singular continuous spectra, absolutely continuous spectra, and singular spectra.The following result is a generalization of self-adjoint case in Hilbert spaces Proposition 3.2 in [19].Theorem 2.
LetX be a Banach space, T∈LRX be reduced by X1⊂X, and Tii=1,2 be defined by (37). Then,(46)σT=σT1∪σT2,ρT=ρT1∩ρT2,where σTi and ρTi are the spectrum and resolvent set of Ti in Xi2 for i=1,2, respectively.Proof.
It suffices to show that the second relation in (46) holds, which implies that σT=σT1∪σT2.
We first show thatρT⊂ρT1∩ρT2. For any given λ∈ρT, RT−λ=Xand there exists a constant c>0 such that for all x,y∈T,(47)x≤cy−λx.
Obviously, (47) holds for all x,y∈T1 since T1⊂T. Now, we show that T1−λ is surjective in X1. For any given z∈X1, there exists x,y∈T such that z=y−λx since RT−λ=X. It follows from (36) that x,y can be decomposed as x,y=x1,y1+x2,y2 with xi,yi∈Ti and i=1,2. Then, z=y1−λx1+x2−λy2. Because z,x1,y1∈X1 and x2,y2∈X2, we have z=y1−λx1 and y2−λx2=0, and thus, z∈RT1−λ. Consequently, T1−λ is surjective in X1. Hence, λ∈ρT1. With a similar argument, one can show λ∈ρT2. Then, λ∈ρT1∩ρT2. Therefore, ρT⊂ρT1∩ρT2.
Next, we consider the inverse inclusion. For any givenλ∈ρT1∩ρT2, we have RTi−λ=Xi for i=1,2, and there is a constant b>0 such that(48)xi≤byi−λxi,∀xi,yi∈Ti,i=1,2.
For anyz∈X, there exist z1∈X1 and z2∈X2 such that z=z1+z2. Then, there is xi,yi∈Ti such that zi=yi−λxi for i=1,2, which implies that z=y1+y2−λx1+x2. Further, by (36), we can get that x1+x2,y1+y2∈T. Hence, z∈RT−λ, and consequently, RT−λ=X. Note that P:X⟶X1 is bounded by the closed graph theorem. There exists a constant M>0 such that(49)Px≤Mx,1−Px≤Mx∀x∈X.
For anyx′,y′∈T, it follows from (36) that x′,y′ can be uniquely decomposed as x′,y′=x1′,y1′+x1′,y2′ with xi′,yi′∈Ti and i=1,2. By utilizing (48) and (49), one can get that(50)x′≤x1′+x2′≤by1′−λx1′+y2′−λx2′=bPy′−λx′+1−Py′−λx′≤2bMy′−λx′,which yields that T−λ−1 is a bounded linear operator defined on X, and consequently, λ∈ρT. It follows that ρT1∩ρT2⊂ρT+A. Therefore, the second relation in (46) holds. The proof is complete.
Now, we discuss the relationships between the point spectra and essential spectra ofT and its decomposition parts.Theorem 3.
LetX be a Banach space, T∈LRX be reduced by X1⊂X, and Tii=1,2 be defined by (37). Then, (51)σpT=σpT1∪σpT2,NT−λ=NT1−λ∔NT2−λ,λ∈C.Proof.
It suffices to show thatNT−λ=NT1−λ∔NT2−λ for every λ∈C. It is evident that NT1−λ∔NT2−λ⊂NT−λ since Ti⊂T and i=1,2. For any f∈NT−λ, we have f,λf∈T, which can be decomposed as f,λf=f1,g1+f2,g2 by (36), where fi,gi∈Ti and i=1,2. Note that λf=g1+g2=λf1+λf2 and fi,gi∈Xi, i=1,2. One can get that gi=λfi for i=1,2, which implies that fi∈NTi−λ and i=1,2. This yields that NT−λ⊂NT1−λ∔NT2−λ. Therefore, NT−λ=NT1−λ∔NT2−λ. This completes the proof.
The following result can be easily verified by Theorems2 and 3. So, its detail proofs are omitted.Theorem 4.
LetX be a Banach space, T∈LRX be reduced by X1⊂X, and Tii=1,2 be defined by (47). Then, (52)σeT=σeT1∪σeT2,σdT=σdT1∪σdT2.
To the end of this subsection, we shall discuss the relationships between the continuous spectra, singular continuous spectra, absolutely continuous spectra, and singular spectra ofT and its decomposition parts, separately.
Suppose thatX is a Hilbert space and X1⊥X2. If T∈LRX is self-adjoint and can be decomposed as (38), then Ti is self-adjoint in Xi2 for i=1,2 by Lemma 6. Let Ti=Ti,s⊕Ti,∞, where Ti,∞=0,y∈Xi2:0,y∈Ti, Ti,s=Ti⊖Ti,∞, and i=1,2. It is evident that(53)Ts=T1,s⊕T2,s,Et=E1t⊕E2t,t∈R,where Eitt∈R is the spectral family of Ti in Xi2 for i=1,2.It follows from the second relation in (51) that(54)XP=X1,P⊕X2,P,and consequently,(55)XC=X1,C⊕X2,C,where Xi,P and Xi,C are the discontinuous and continuous subspaces in Xi with respect to Ti for i=1,2.By utilizing (53) and (55), we can get that(56)XSC=X1,SC⊕X2,SC,which together with (54) and (55) implies that(57)XAC=X1,AC⊕X2,AC,XS=X1,S⊕X2,S,where Xi,SC, Xi,AC, and Xi,S are the singular continuous, absolutely continuous, and singular subspaces in Xi with respect to Ti for i=1,2.It is derived from (37), (38), and (54)–(57) that(58)TP=T1,P⊕T2,P,TC=T1,C⊕T2,C,TSC=T1,SC⊕T2,SC,TAC=T1,AC⊕T2,AC,TS=T1,S⊕T2,S,Ti,P=TP∩Xi,P2,Ti,C=TC∩Xi,C2,Ti,SC=TSC∩Xi,SC2,Ti,AC=TAC∩Xi,AC2,Ti,S=TS∩Xi,S2,i=1,2.where Ti,P, Ti,C, Ti,SC, Ti,AC, and Ti,S are the (spectral) discontinuous, continuous, singular continuous, absolutely continuous, and singular parts of Ti in Xi2 for i=1,2, respectively.Theorem 5.
LetX be a Hilbert space and T∈LRX be self-adjoint and reduced by X1⊂X. Further, let Tii=1,2 be defined by (37) with X1⊥X2. Then,(59)σTP=σT1,P∪σT2,P,σrT=σrT1∪σrT2,r=c,ac,sc,s.where Ti is regarded as a relation in Xi2 for i=1,2.Proof.
This theorem can be directly derived from (i) of Lemma6, Theorem 2, and (54)–(58).


## 3.2. Relationships betweenT+A and Ts+AT

In this subsection, we shall study the relationships between the properties ofT+A and Ts+AT, including their closedness, hermiticity, self-adjointness, various spectra, and defect indices. Further, X is always assumed to be a complex Hilbert space in this part.Lemma 7.
LetT,A∈LRX satisfy that DT⊂DA and A0⊂T0. If T is closed and DT⊂T0⊥, then T+A is reduced by T0.Proof.
We suppose thatT is closed and DT⊂T0⊥. It follows from the assumption DT⊂DA and A0⊂T0 that(60)T∞=T∩T02=T+A∩T02.
Note that (15) holds if DT⊂T0⊥, which together with Proposition 1 and Proposition 2.1 in [21] implies that(61)Ts+AT=Ts+AT|DT=T+A−T+T∩T0⊥2=T+A∩T0⊥2.s
Therefore,T+A is reduced by T0 by Theorem 1 and (i) of Lemma 6. This completes the proof.
The following result comes from [6], and we shall give its proof for completeness.Lemma 8.
LetT∈LRX be closed. Then, T∞ is a self-adjoint relation in T0.Proof.
We suppose thatT is closed. Obviously, T∞ is closed in T0 for any 0,f,0,g∈T∞ and 0,f=g,0=0. Then, T∞ is Hermitian in T0. Note that RT∞±i=T0. Therefore, T∞ is a self-adjoint relation in T0 by Lemma 4. The proof is complete.
By Propositions 2 ‐ 4, Lemmas7 and 8, (61) and (62), one can easily get the following results.Theorem 6.
LetT,A∈LRX satisfy that DT⊂DA and A0⊂T0. If T is closed and DT⊂T0⊥, then(i)
T+A is closed if and only if Ts+AT is closed in T0⊥;(ii)
T+A is an Hermitian relation in X2 if and only if Ts+AT is an Hermitian operator in T0⊥.(iii)
T+A is a self-adjoint relation in X2 if and only if Ts+AT is a self-adjoint operator in T0⊥.
Now, we give a relationship between the spectra and various spectra ofT+A and Ts+AT.Theorem 7.
LetT,A∈LRX satisfy that DT⊂DA and A0⊂T0.(i)
IfT is closed and DT⊂T0⊥, then(62)σeT+A=σeTs+AT,σdT+A=σdTs+AT.(ii)
IfT is closed and T+A is self-adjoint, then(63)σT+AP=σTs+ATP,σrT+A=σrTs+AT,r=c,ac,sc,s,whereTs+AT is regarded as a relation in T0⊥2.Proof.
The first assertion of Theorem7 can be easily verified by (16), Theorems 1‐4, and Lemma 7. Suppose that T+A is self-adjoint. Then, DT=DT+A⊂T+A0⊥=T0⊥. And consequently, T0 reduces T+A by Lemma 7. It follows from (16), and Theorems 1, 5, and 6 that (63) holds. This completes the proof.Remark 3.
By Corollary 2and Lemmas 3 and 5, (63) holds.
The following results can be easily directly derived from (14) and Theorem 7.Corollary 4.
LetT,A∈LRX satisfy that DT⊂DA and A0⊂T0. If T is closed and Hermitian, then(64)σeT+A=σeTs+AT,σdT+A=σdTs+AT,where Ts+AT is regarded as a relation in T0⊥2.
To the end of this subsection, we shall give a relationship between the defect indices ofT+A and Ts+AT.Theorem 8.
LetT,A∈LRX satisfy that DT⊂DA and A0⊂T0. If T is closed and T+A is Hermitian, then(65)d+T+A=d+Ts+AT,d−T+A=d−Ts+AT,where Ts+AT is regarded as a relation in T0⊥2.Proof.
We suppose thatT is closed and T+A is Hermitian. Then, DT=DT+A⊂T+A0⊥=T0⊥. And consequently, T0 reduces T+A by Lemma 7. It is derived from Theorem 1, Corollary 3, and (ii) of Theorem 6 that (65) holds. Thus, the proof is complete.


## 3.3. Relationships between Perturbation Terms ofT and Ts

In this subsection, we shall discuss the relationships between the perturbation terms ofT and Ts if T is closed. Including relatively bounded and relatively compact perturbation terms, finite rank perturbation term, and trace class perturbation term.We shall first recall the concepts of relatively bounded and compact relations, which were introduced by Cross [20].LetX and Y be normed spaces, T∈LRX,Y, and XT denote the space DT,⋅T, where(66)xT=x+Tx,x∈DT.We then defineGT∈LRXT,X by GTx=x for x∈XT. GT is called the graph operator of T.Definition 4.
(see Definition VII.2.1 in [20]). Let X, Y, and Z be normed spaces, T∈LRX,Y, and A∈LRX,Z with DT⊂DA.(1)
The linear relationA is said to be T-bounded if there exist nonnegative numbers a and b such that(67)Ax≤ax+bTx,x∈DT.IfA is T-bounded, then the infimum of all numbers b≥0 for which (67) holds with a constant a≥0, is called the T-bound of A.(2)
The linear relationA is said to be T-compact (or relatively compact toT) if AGT is compact, i.e., A:XT⟶Z is compact.Lemma 9 (see Lemma 2.7 in [22]).
LetX be a Hilbert space and T∈LRX can be decomposed as (11). Then, (11) holds and(68)Tx=Tsx,x∈DT,T=Ts.Lemma 10.
LetX be a Hilbert space and T,A∈LRX satisfy that DT⊂DA and A0⊂T0. Then,(69)ATx≤Ax,x∈DT¯∩DA,AT≤A.Proof.
It suffice to show that the first relation in (69) holds. Let x∈DT¯∩DA. It follows from (i) of Lemma 1 that for any ϵ>0, there exists y∈Ax such that y<Ax+ϵ. Let y=y1+y2, where y1∈T0⊥ and y2∈T0¯. Then, ATx=y1 by (25). Consequently, ATx=y1≤y<Ax+ϵ. Hence, ATx≤Ax by the arbitrariness of ϵ. Therefore, (69) holds and this completes the proof.
By Lemmas9 and 10, one can easily get the following result.Theorem 9.
LetX be a Hilbert space and T,A∈LRX satisfy that T is closed, DT⊂DA, and A0⊂T0. If A is T-bounded with T-bound less that b, then AT is Ts-bounded with Ts-bound less than b, b≥0.
Now, we give a relationship between the relatively compact perturbation ofT and Ts.Theorem 10.
LetX be a Hilbert space and T,A∈LRX satisfy that T is closed, DT⊂DA, and A0⊂T0. If A is T-compact, then AT is Ts-compact.Proof.
It follows from (13) and Lemma 9 that XT=XTs. We suppose that A is T-compact. Then, for any given bounded sequencexnn=1∞ in XT, and QAAxnn=1∞ has a convergent subsequence QAAxnkk=1∞ in X/A0¯. So, for each k≥1, there is ynk∈Axnk such that ynkk=1∞ is convergent in X by (iii) of Lemma 1. Let ynk=ynk,1+ynk,2 with ynk,1∈T0⊥ and ynk,2∈T0. Hence, ATxnk=ynk,1, and ynk,1k=1∞ is convergent in T0⊥. This means ATxnn=1∞ has a convergent subsequence in T0⊥. Therefore, AT is Ts-compact. Thus, the proof is complete.
With a similar argument to that used in the proof of Theorem10, one can easily show the following results hold.Corollary 5.
LetX and Y be Hilbert spaces and A∈LRX,Y. Then, A is a compact relation if and only if PAA is a compact operator, where PA:Y⟶A0⊥ is an orthogonal projection.Corollary 6.
LetX be a Hilbert space and T,A∈LRX satisfy that DT⊂DA and A0⊂T0. If A is compact, then AT is compact.Proposition 6.
LetX be a Hilbert space and T,A∈LRX satisfy that DT⊂DA∩T0⊥ and A0⊂T0. Then, A|DT¯∩DA is an Hermitian relation in X2 if and only if AT is an Hermitian operator in T0⊥.Proof.
It follows from the assumptionDT⊂T0⊥ that DAT=DT¯∩DA⊂T0⊥, which means that AT is a linear operator in the Hilbert space T0⊥.
Suppose thatA|DT¯∩DA is an Hermitian relation in X2. For any f,g and h,k∈AT, there exists f,y,h,z∈A such that y=g+y′ and z=k+z′, where y′,z′∈T0¯. Note that f,h∈DAT⊂T0⊥ and the assumption that A|DT¯∩DA is Hermitian, we get(70)g,h=y,h=f,z=f,k.
Hence,AT is an Hermitian operator in T0⊥.
Now, we consider the inverse. We assume thatAT is an Hermitian operator in T0⊥. We fixed x,y,u,v∈A|DT¯∩DA. There exist y1,v1∈T0⊥ and y2,v2∈T0¯ such that y=y1+y2 and v=v1+v2. Hence, x,y1,u,v1∈AT. By the assumption that AT is Hermitian and the fact that x,u∈DAT⊂T0⊥, we get(71)y,u=y1,u=x,v1=x,v.
Therefore,A|DT¯∩DA is an Hermitian relation in X2. This completes the proof.
LetX be a Hilbert space and T∈LRX. If T is self-adjoint, then DT¯=T∗0⊥=T0⊥by Proposition III.1.4 in [20], which together with Proposition 6, can easily achieve the following result.Corollary 7.
LetX be an Hilbert space and T,A∈LRX, satisfying DT⊂DA and A0⊂T0. If T is self-adjoint, then A|DT¯∩DA is an Hermitian relation in X2 if and only if AT is a symmetric operator in T0⊥, that is, AT is a densely defined Hermitian operator in T0⊥.
Now, we introduce the concept of degenerate linear relation, which is a generation of single-valued case.Definition 5.
LetXandYbe normed spaces andA∈LRX,Y.Ais said to be degenerate ifAis bounded anddimRA<∞.
LetA∈LRX,Y be degenerate. It is evident that QAA is degenerate. Hence, QAA is compact (see p.160 in [27]), that is, A is compact.
Motivated by the definition of the norm of a linear relation, we introduce the concept of trace class linear relations.Definition 6.
LetXandYbe Hilbert spaces andA∈LRX,YwithDA=X. We say thatAbelongs to trace class relations ifPAAbelongs to trace class operators, wherePA:Y⟶A0⊥is the orthogonal projection.Theorem 11.
LetX be a Hilbert space and T,A∈LRX satisfy that DT⊂DA and A0⊂T0. If A is degenerate, then AT is also degenerate.Proof.
We suppose thatA is degenerate, that is, dimRA=m<∞ and A is bounded. Then, AT is bounded by Lemma 10. Let y1,y2,…,ym be a base of RA. They can be decomposed as yi=yi,1+yi,2 with yi,1∈T0⊥ and yi,2∈T0¯, i=1,2,…,m. Now, we show that for any given z∈RAT⊂T0⊥, it can be expressed as a linear combination of yi,1, 1≤i≤m. We then set x,y∈A with y=z+y′, where y′∈T0¯. There exist ci and i=1,2,…,m such that(72)y=∑i=1mciyi=∑i=1mciyi,1+∑i=1mciyi,2=z+y′.
Note thatz,∑i=1mciyi,1∈T0⊥ and y′,∑i=1mciyi,2∈T0¯, we can get that z=∑i=1mciyi,1, which yields that dimRAT≤m<∞. Therefore, AT is degenerate and the proof is complete.
Now, we recall a necessary and sufficient condition about trace operators (cf., Theorem 7.12 in [28]).Lemma 11.
LetT be an operator from Hilbert space X into Hilbert space Y with DT=X. Then, T belongs to trace class operators if and only if there exist sequences fnn=1∞ from X and gnn=1∞ from Y such that fn=gn=1 for each n≥1, and there is a constant sequence cnn=1∞ for which ∑n=1∞cn<∞ and T can be expressed as(73)Tx=∑n=1∞cnx,fngn,x∈X.Remark 4.
In the case thatX=Yin Lemma 11, Tbelongs to trace class operators if and only ifTcan be expressed as(74)Tx=∑n=1∞cnx,fnfn,x∈X,where fnn=1∞ form an orthonormal family of eigenvectors of T and the cn are the associated (repeated) eigenvalues with ∑n=1∞cn<∞ (cf., [27], p.543).Theorem 12.
LetX be a Hilbert space and T,A∈LRX with DA=X and DT¯=T0⊥, and A0⊂T0. If A belongs to trace class relations, then AT belongs to trace class operators in T0⊥.Proof.
The assumptionsDA=X and DT¯=T0⊥ implies that DAT=DT¯∩DA=T0⊥.
We suppose thatA belongs to trace class relations, that is, PAA belongs to trace class operators. It follows from Lemma 11 that there exist sequences fnn=1∞ and gnn=1∞ from X and cnn=1∞ from C satisfying fn=gn=1, n≥1, and ∑n=1∞cn<∞ such that(75)PAAx=∑n=1∞cnx,fngn,x∈X.
Letfn=fn,1+fn,2 and gn=gn,1+gn,2 with fn,1,gn,1∈T0⊥ and fn,2,gn,2∈T0¯ for n≥1. We shall show that AT=S|T0⊥, where S=∑n=1∞cn⋅,fn,1gn,1. Given any x∈DAT, there is y∈X such that x,y∈A. Since A0⊂T0, y can be decomposed as y=y1+y2+y3 with y1∈T0⊥, y2∈A0⊥∩T0¯, and y3∈A0¯. Then, ATx=y1 and PAAx=y1+y2. By (75) and the fact that x∈DAT=T0⊥, one can get that(76)y1+y2=∑n=1∞cnx,fn,1+fn,2gn,1+gn,2=∑n=1∞cnx,fn,1gn,1+∑n=1∞cnx,fn,1gn,2.
Note thaty1,∑n=1∞cn⋅,fn,1gn,1∈T0⊥ and y2,∑n=1∞cnx,fn,1gn,2∈T0¯, we have that ATx=y1=∑n=1∞cn⋅,fn,1gn,1. Consequently, AT⊂S. Hence, AT=S|T0⊥. Let L:=n≥1:fn,1≠0andgn,1≠0. By setting fn,1′=fn,1/fn,1 and gn,1′=gn,1/gn,1 for each n∈L, we can get (77)AT\x=∑n∈Lcnfn,1gn,1x,fn,1′gn,1′,x∈T0⊥, where ∑n∈Lcnfn,1gn,1≤∑n=1∞cn<∞. Therefore, AT belongs to trace class in T0⊥ by Lemma 11. This completes the whole proof.Remark 5.
In the present study, we construct a linear operator, which is induced by two linear relations, and then establish the relationships between the perturbation terms of a closed relation and the perturbation terms of its operator part (see Theorems9–12), and give the relationships between spectrum of a perturbed relation and spectrum of a perturbed operator (see Theorem 7). By using the results obtained in the present study, we shall deeply study stabilities of the spectra of linear relations under some perturbations in our forthcoming study, especially the invariance of the absolutely continuous spectrum of a self-adjoint linear relation under trace class perturbation.Remark 6.
Note that the constructing technique in the present study can also be applied in our previous works. For example, Theorem 5.2 in [29] can be followed by Lemma 3, Corollary 7, Theorems 6 and 9, ([29], Lemma 5.8), and ([27], Theorem V.4.3); Theorem 5.3 in [29] can be followed by Lemmas 2, 3, and 10, Corollaries 4 and 7, Theorem 10, ([29], Lemma 5.8); and ([27], Theorem V.4.10); Theorem 4.1 in [22] can be followed by Lemmas 2 and 3, Corollaries 4 and 7, Theorems 6 and 10, ([29], Lemma 5.8), and ([28], Theorem 9.9), respectively.



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