On the correspondence between spectra of the operator pencil A−λB and of the operator B −1 A
aa r X i v : . [ m a t h . SP ] O c t On the correspondence between spectra of the operatorpencil A − λB and of the operator B − A Ivica Nakić ∗ November 11, 2018
Abstract
This paper is concerned with the reduction of the spectral problem for symmetriclinear operator pencils to a spectral problem for the single operator. Also, a Rayleigh–Ritz–like bounds on eigenvalues of linear operator pencils are obtained.Keywords: linear operator pencil, spectrum, Krein spaces2000 Mathematics Subject Classification: 47A56, 47A10
In many areas of applied mathematics spectral problems for the operator pencil L ( λ ) = A − λB arise, where A and B are symmetric or selfadjoint operators acting in some Hilbertspace H . See, for instance, [15], [26], [16], [29] and references therein.One way of dealing with this problem is reducing it to a classical spectral problem for asingle operator. If B is bounded and boundedly invertible this can be done by introducing theoperator S = B − A . It is easy to see that the spectral problems for L and S are equivalent.Additionally, one can think of the operator S as a symmetric or selfadjoint operator in theKrein space ( H , ( B · , · )) . Here ( H , ( B · , · )) denotes the Banach space ( H , (cid:13)(cid:13) | B | / · (cid:13)(cid:13) ) , where | B | / is the positive square root of B , with the structure of the (possibly indefinite) innerproduct on this space given by ( B · , · ) .If B is still bounded, but, say, zero is a point of continuous spectrum of the operator B , the space ( H , ( B · , · )) is no longer complete, and S is no longer closed. If we drop theassumption that B is bounded, the problems are even more involved, and in the case whenzero is an eigenvalue of B , S does not exists as an operator.There exists an extensive literature on the spectral theory of operator pencils with itsapplications in diverse mathematical and physical domains. Hence, we restrict ourselves tomentioning only two classical works [3] and [25].The reduction of the spectral problem for operator pencils to the spectral problem for asingle operator is a standard procedure used in numerous applications. One of the importantsteps in the reduction is to show that there is no loss of the relevant spectral information. ∗ Department of Mathematics, University of Zagreb, 10000 Zagreb, Croatia, e-mail: [email protected]. Thiswork has been partially supported by the HRZZ project grant 9345. A and B are symmetric or selfadjointoperators in a Hilbert space H and the reduction operator is symmetric or selfadjoint ina Krein space K , its topology generated by the operator B . So the aim is to investigatethose reductions for which the symmetry of the problem is preserved. We treat various casesdepending on the properties of the operator B .We now give a brief outline of our main results.In Sections 2 and 3 we investigate the case when B is bounded and ∈ σ c ( B ) . InSection 2 we give a construction of a reduction operator in the case when A is a symmetricoperator with finite and equal deficiency indices. The main result in Section 2 is Theorem 2where we show that there exists a bijective correspondence between all selfadjoint extensionsof the operator A and all selfadjoint extensions of the reduction operator S . Section 3 isconcerned with the correspondence of the spectra. We show that there is a correspondenceof spectra and point spectra, but that, in general, there does not exists correspondence forresidual and continuous parts of spectra. We give necessary and sufficient conditions for thecorrespondence of residual, and hence also continuous, parts of spectra.In Section 4 we investigate the case when B is unbounded and ∈ σ c ( B ) . A constructionof the reduction operator follows essentially the one given in [6], but the authors there do notinvestigate the questions we are concerned with, except the case of point spectrum for quasi–uniformly positive operators. The main result in this section is Theorem 5 in which we showthat under the assumptions stated at the beginning of the section, the point spectrum of thepencil is contained in the point spectrum of the reduction operator and the spectrum of thereduction operator is contained in the spectrum of the pencil. To obtain the correspondenceof the spectra, additional assumptions are needed, and some are presented at the end of thesection.In Section 5, we apply the results from Sections 2 and 4 to obtain a Rayleigh–Ritz–likeupper bound for the eigenvalues of the reduction operator given in terms of the eigenvaluesof the matrix pair ( A V , B V ) , where A V , B V are orthogonal projections of A , B on a finitedimensional subspace. To obtain a variational characterization for the reduction operator,we use the results from [5].In Section 6 we investigate the case when is an eigenvalue of B . By the use of thetheory of linear relations we give constructions of linear relations which act as a substitutefor reduced operators. We show that similar results as in Sections 3, 4 and 5 also hold inthis case. Using a result from [28], we give some examples when a reduced operator can beconstructed. 2 .2 Basic definitions and preliminaries Let H be a Hilbert space with inner product ( · , · ) and the corresponding norm k x k =( x, x ) / . By D ( T ) we denote the domain, by R ( T ) we denote the range of the operator T , and by Ker( T ) we denote the null-space of T . We say that the operator T is symmetricif T ⊂ T ∗ and T has a dense domain. The deficiency indices of a symmetric operator T are numbers n ± = n ± ( T ) defined as n + = codim R ( T − µ ) , n − = codim R ( T − µ ) , where µ ∈ C \ R , ℑ µ > is arbitrary. By ∔ we denote the direct sum.Let L () = A − B be an operator pencil in H , where A and B are densely defined operatorsin H , such that B is A –bounded, i.e. D ( B ) ⊃ D ( A ) and k Bx k ≤ C k Ax k . Obviously, D ( L ( λ )) = D ( A ) . Definition 1.
The complex number is in the resolvent set of the pencil L , (denoted by ∈ ρ ( L ) ) if ∈ ρ ( L ()) . The spectrum of L is defined as σ ( L ) = C \ ρ ( L ) , i.e. ∈ σ ( L ) if ∈ σ ( L ()) . Analogously we define the point σ p , residual σ r , continuous σ c , and approximativespectrum σ ap .We also define the set of the points of regular type of the operator pencil L by Π( L ) = { ∈ C : ∃ C > such that k L () x k ≥ C k x k ∀ x ∈ D ( L ()) } , i.e. the set of all ∈ C such that is a point of regular type for the operator L () .We say that the vectors x , x , . . . , x k ∈ H , x = 0 , form a Jordan chain of length k + 1 for L corresponding to the eigenvalue ∈ C if L ( ) x i = Bx i − , i = 0 , , . . . , k, where x − = 0 .In the next proposition we list some properties of linear operator pencils which areessentially known. Proposition 1.
Let L () = A − B be an operator pencil in the Hilbert space H , where B is A –bounded.(i) If L has a compact resolvent for some ∈ ρ ( L ) , i.e. if L () − is compact, then L has acompact resolvent in each point of the resolvent set.(ii) If B is bounded, the set of points of regular type of L is an open set.(iii) If B is bounded and A and B are symmetric operators, the deficiency indices of thesymmetric operators A and A − B coincide, for all ∈ R .Proof. The statement (i) easily follows from the generalized resolvent equation. The state-ment (ii) is evident. For the proof of (iii) see [18, Theorem 9.3].The following lemma is known, but we give a proof for reader’s convenience.
Lemma 1.
Let H be a Hilbert space, D a dense subspace of H , and N a finite dimensionalsubspace of H . Then D ∩ N ⊥ is dense in N ⊥ . roof. Let P be the orthogonal projection onto N . Since P is continuous and its range is N , the image P ( D ) is a dense subspace of N . As N is finite dimensional, P ( D ) = N . Let D be a subspace of D such that the restriction P : D → N of P onto D is an isomorphism.Then I − P − P is continuous on H , its range is N ⊥ and ( I − P − P )( D ) ⊆ D ∩ N ⊥ . Since ( I − P − P )( D ) is dense in N ⊥ , so is D ∩ N ⊥ .Since we will mainly operate in Krein spaces, we introduce some basic notions from theKrein space theory. Let K be a vector space and let [ · , · ] be an inner product on K (notnecessarily definite). The pair ( K , [ · , · ]) is said to be a Krein space if there exists a directsum decomposition K = K + ∔ K − such that ( K ± , ± [ · , · ]) are Hilbert spaces and such that [ K + , K − ] = { } . For such a decomposition the corresponding fundamental symmetry J isa linear operator defined by J ( x + + x − ) = x + − x − , where x + ∈ K + and x − ∈ K − . Let ( · , · ) = [ J · , · ] . Then the space ( K , ( · , · )) is a Hilbert space. Its topology is independent of thechoice of K + and K − . The subspace F is said to be ortho–complemented if F ∔ F ⊥ = K ,where F ⊥ = { x : [ x, y ] = 0 ∀ y ∈ F } .The definitions of symmetric and selfadjoint operators in a Krein space are analogous tothose in Hilbert space. A selfadjoint operator A is said to be definitizable if its resolventset is nonempty and there exists a nonzero polynomial p such that [ p ( A ) x, x ] ≥ for all x ∈ D ( p ( A )) . A definitizable operator has a spectral function with finitely many criticalpoints in R ∪ {∞} . A critical point z is regular if the spectral function is bounded in aneighborhood of z . A critical point is singular if it is not regular. We say that is aneigenvalue of positive (negative) type of an operator T if [ x, x ] ≥ ( ≤ ) for all eigenvectorscorresponding to . We say that an operator T is quasi–uniformly positive (qup) if thereexists a subspace M of finite codimension in D ( T ) such that inf { [ T x, x ] : x ∈ M, k x k = 1 } > . For more details about Krein spaces see [23], [8] and [1]. Basic properties of qup operatorscan be found in [11]. A symmetricand B bounded In this section we assume:1. A is a symmetric operator on a Hilbert space H with finite and equal deficiency indices n + ( A ) = n − ( A ) =: n ,2. B be is a bounded selfadjoint operator on H with ∈ σ c ( B ) ,3. Π( L ) ∩ R = ∅ .Let E be a spectral function of operator B . We define the operator J by J = E (0 , ∞ ) − E ( −∞ , . Operator J is a bounded selfadjoint operator and J = I holds. We define | B | := J B = BJ ≥ . Now, we define a space ˆ K which consists of sequences ( x n ) , x n ∈ H in the following way ˆ K = { ( x n ) : | B | / x n converges in H} , We thank the anonymous referee for this simplified version of the proof. | B | / . We introduceequivalence relation ∼ on ˆ K : ( x n ) ∼ ( y n ) iff | B | / ( x n − y n ) → as n → ∞ , and define K as the corresponding quotient space K = ˆ K / ∼ . On K we introduce two innerproducts ( · , · ) K and [ · , · ] K as follows: (( x n ) , ( y n )) K = lim n →∞ ( | B | / x n , | B | / y n ) , [( x n ) , ( y n )] K = lim n →∞ ( J | B | / x n , | B | / y n ) . It is easy to see that ( K , ( · , · ) K ) is a Hilbert space, and that ( K , [ · , · ] K ) is a Krein space.These spaces have the norm given by k ( x n ) k K = lim n →∞ (cid:13)(cid:13) | B | / x n (cid:13)(cid:13) . If we identify x ∈ H with the sequence ( x, x, . . . ) in ˆ K , it is evident that H can be regarded as a subspace in K ,and in that case k x k K = (cid:13)(cid:13) | B | / x (cid:13)(cid:13) holds. Also, H is a dense subspace in K . Operators in H can also be regarded as operators in K . Also, J can be extended to a bounded operatorin K , and this extension we also denote with J . It is easy to see that J is a fundamentalsymmetry in K . Proposition 2.
The spaces H and K are isomorphic.Proof. We define T : K → H , T ( x n ) = lim n →∞ | B | / x n . (1)Then it is easy to see that T is a linear surjection from ( K , ( · , · ) K ) to ( H , ( · , · )) which preservesthe inner products, i.e. ( T ( x n ) , T ( y n )) = (( x n ) , ( y n )) K .In applications, the space K can usually be represented as a function space as can beseen from the following example. Example 1.
Let H = L ( a, b ) , and let w ∈ L ∞ ( a, b ) be a function which is zero on some non–empty set of measure zero. We define the operator B by ( Bf )( x ) = w ( x ) f ( x ) , f ∈ L ( a, b ) .The operator B is obviously bounded and ∈ σ c ( B ) holds.The Krein space K can be identified with the space { f : p | w ( · ) | f ∈ L ( a, b ) } (i.e. a weighted L space with the weight w , see [31, Section 8.4]) with the inner productsgiven by [ f, g ] K = Z ba w ( x ) f ( x ) g ( x )d x, ( f, g ) K = Z ba | w ( x ) | f ( x ) g ( x )d x. The operator T is given by T f = p | w ( · ) | f , and the fundamental symmetry J is given by J f = (sgn w ) f . Lemma 2.
Let D be a dense subset in H . Then D is also dense in K . roof. Let ε > , x ∈ H be arbitrary. Since R ( | B | / ) is dense in H , there exists y ∈ H such that (cid:13)(cid:13) | B | / y − x (cid:13)(cid:13) < ε , and since D is dense in H , there exists z ∈ D such that k y − z k < ε (cid:13)(cid:13) | B | / (cid:13)(cid:13) − . Now (cid:13)(cid:13)(cid:13) | B | / z − x (cid:13)(cid:13)(cid:13) ≤ (cid:13)(cid:13)(cid:13) | B | / ( z − y ) (cid:13)(cid:13)(cid:13) + (cid:13)(cid:13)(cid:13) | B | / y − x (cid:13)(cid:13)(cid:13) < ε. This implies that {| B | / x : x ∈ D } is dense in H . Now let ( x n ) ∈ K ,ε > be arbitrary. Set e x = T ( x n ) ∈ H . Then there is y ∈ H such that (cid:13)(cid:13)e x − | B | / y (cid:13)(cid:13) < ε ,which implies k ( x n ) − y k K < ε . Proposition 3.
The operator B − A is symmetric in ( K , [ · , · ] K ) .Proof. Let x, y ∈ D ( B − A ) = { x ∈ D ( A ) : Ax ∈ R ( B ) } . Then [ B − Ax, y ] K = ( J | B | / B − Ax, | B | / y ) = ( Ax, y )= ( x, Ay ) = ( J | B | / x, | B | / B − Ay ) = [ x, B − Ay ] K . Let ∈ Π( L ) ∩ R be arbitrary. Then R ( L ()) is closed. From Proposition 1 and [18, Theorem9.2] follows codim R ( L ()) = n . From Lemma 1 follows that R ( L ()) ∩ R ( B ) is dense in R ( L ()) .Note that R ( L ()) ∩ R ( B ) = { ( A − B ) x : x ∈ D ( A ) , Ax ∈ R ( B ) } = { ( A − B ) x : x ∈ D ( B − A ) } . Let
C > be such that k L () x k ≥ C k x k . Let x ∈ D ( A ) , ε > be arbitrary. Set e x = L () x .Then it exists y ∈ D ( B − A ) such that k L () y − e x k < Cε . Then Cε > k L () y − e x k = k L ()( y − x ) k ≥ C k x − y k , hence k x − y k < ε . So D ( B − A ) is dense in D ( A ) , which implies that D ( B − A ) is dense in H . Now Lemma 2 implies that D ( B − A ) is dense in K , hence B − A is a symmetric operatorin K .We will need the following three lemmas. Lemma 3.
Let ( x n ) ∈ D ( A ∗ ) be a sequence such that A ∗ x n → (2) and | B | / x n → . (3) Then x n → .Proof. From (3) it follows that Bx n → . Let λ ∈ Π( L ) ∩ R be arbitrary. Since ( A − λB ) ∗ = A ∗ − λB , we have D ( A ∗ ) = D ( A ) ∔ Ker( A ∗ − λB ) . x n = x n + y n , x n ∈ D ( A ) , y n ∈ Ker( A ∗ − λB ) . Since ( A ∗ − λB ) x n = ( A − λB ) x n ,and ( A − λB ) − is a bounded operator, it follows x n → . Hence Bx n → and By n = B ( x n − x n ) → . Since the operator B | Ker( A ∗ − λB ) : Ker( A ∗ − λB ) → B (Ker( A ∗ − λB )) acts between finite–dimensional spaces and is non–singular, we have y n → , hence x n → . Lemma 4.
We have(i) The set D ( | B | − / A ∗ ) is a core of A ∗ .(ii) The set D ( | B | − / A ) is a core of A .(iii) For each selfadjoint extension e A of A , the set D ( | B | − / e A ) is a core of e A .Proof. Let ∈ Π( L ) ∩ R be arbitrary.(i) As in the proof of Proposition 3, we have codim R ( A − B ) = n . Also ( A − B ) ∗ = A ∗ − B holds. This implies D ( A ∗ ) = D ( A ) ∔ Ker( A ∗ − B ) .Let x ∈ D ( A ∗ ) be arbitrary, and let x = x + y , x ∈ D ( A ) , y ∈ Ker( A ∗ − B ) .Since Ker( A ∗ − B ) ⊂ D ( | B | − / A ∗ ) , it is enough to find a sequence ( x n ) ∈ D ( | B | − / A ) such that x n → x and Ax n → Ax . Lemma 1 implies that R ( A − B ) ∩ R ( | B | / ) isdense in R ( A − B ) , so for ( A − B ) x there exists a sequence x n ∈ D ( | B | − / A ) such that ( A − B ) x n → ( A − B ) x . Since ( A − B ) − is bounded it follows that x n → x which implies Bx n → Bx , and consequently, Ax n → Ax .(ii) Since R ( A − λB ) ∩ R ( B ) is dense in R ( A − λB ) , for an arbitrary x ∈ D ( A ) thereexists a sequence x n ∈ D ( B − A ) such that ( A − λB ) x n → ( A − λB ) x , which implies x n → x and Ax n → Ax .(iii) Since e A is a finite–dimensional extension of A , codim R ( e A − λB ) ≤ n , and R ( e A − λB ) is closed. Hence, again from Lemma 1, R ( e A − λB ) ∩ R ( | B | / ) is dense in R ( e A − λB ) . Nowwe can proceed as in the proof of the first statement of the Lemma. Lemma 5.
The formulae ( | B | − / A ) ∗ = A ∗ | B | − / and ( B − A ) ∗ = A ∗ B − hold.Proof. Since | B | / is bounded, [31, Satz 4.19] implies ( | B | / ( | B | − / A )) ∗ = ( | B | − / A ) ∗ | B | / . (4)From Lemma 3 follows ( | B | / ( | B | − / A )) ∗ = ( A | D ( | B | − / A ) ) ∗ = ( A | D ( | B | − / A ) ) ∗ = A ∗ , (5)Now (4) and (5) imply ( | B | − / A ) ∗ | B | / = A ∗ . The second formula follows analogously.Let us define the operator S in K by D ( S ) = { x ∈ D ( A ) : Ax ∈ R ( | B | / ) } = D ( | B | − / A ) Sx = T − J | B | − / Ax, where the operator T is defined by (1). 7 roposition 4. The operator S is closed.Proof. Let ( x n ) be a sequence in D ( S ) , and let x n → x in the norm k·k K , which implies x ∼ ( x n ) . Set e x = T ( x n ) , (6)and let Sx n → y in the norm of K . It follows J | B | − / Ax n → e y := T y. (7)From (6) and (7) follows Bx n → J | B | / e x and Ax n → J | B | / e y ,respectively, which implies ( A − B ) x n → J | B | / ( e y − e x ) , (8)for all ∈ C . Especially, (8) holds for ∈ Π( L ) , which implies x n → ( A − B ) − J | B | / ( e y − e x ) =: x . Hence we found x ∈ H such that x n → x in the norm k·k and Ax n → J | B | / e y , whichimplies x ∈ D ( A ) and Ax = J | B | / e y . Hence Ax ∈ R ( | B | / ) , and J | B | − / Ax = e y , i.e. Sx = y and x ∈ D ( S ) . Proposition 5.
The adjoint of S in K is given by D ( S ∗ ) = { x ∈ D ( A ∗ ) : A ∗ x ∈ R ( | B | / ) } = D ( | B | − / A ∗ ) S ∗ x = ( y n ) , where T ( y n ) = J | B | − / A ∗ x. Proof.
Let ( y n ) , ( z n ) ∈ K be such that [ Sx, ( y n )] K = [ x, ( z n )] K , ∀ x ∈ D ( S ) . This can be written as lim n ( | B | − / Ax, | B | / y n ) = lim n ( J | B | / x, | B | / z n ) ∀ x ∈ D ( S ) . (9)Let us denote e y = T ( y n ) , e z = T ( z n ) . Then (9) reads ( | B | − / Ax, e y ) = ( x, J | B | / e z ) ∀ x ∈ D ( | B | − / A ) . From Lemma 5 follows e y ∈ D ( A ∗ | B | − / ) and A ∗ | B | − / e y = J | B | / e z. Now, since e y ∈ R ( | B | / ) there exists y ∈ H such that ( y n ) ∼ y . Hence we can take e y = | B | / y , which implies A ∗ y = J | B | − / e z . Hence J | B | − / A ∗ y = e z , which implies y ∈ D ( | B | − / A ∗ ) and T ( z n ) = J | B | − / A ∗ y .On the other hand, take an arbitrary y ∈ D ( | B | − / A ∗ ) . Then one can easily prove ( | B | − / Ax, | B | / y ) = ( | B | / x, | B | − / A ∗ y ) , ∀ x ∈ D ( S ) , hence y ∈ D ( S ∗ ) and T ( S ∗ y ) = J | B | − / A ∗ y .8 emark . If n = 0 , i.e. if A is selfadjoint, then S is also selfadjoint. Theorem 1.
The operator S is the closure of the operator B − A in K .Proof. Let us denote with S ′ the closure of the operator B − A in K .Let x ∈ D ( B − A ) , and set ( y n ) = Sx , i.e. lim n →∞ | B | / y n = J | B | − / Ax ∈ R ( | B | / ) .It follows that there exists y ∈ H such that y ∼ ( y n ) , and | B | / y = J | B | − / Ax , hence y = B − Ax . Hence we have proved B − A ⊂ S , so S ′ ⊂ S .Let ( y n ) ∈ D (( B − A ) ∗ ) be arbitrary, where the adjoint is taken with respect to thegeometry in K , and set ( z n ) = ( B − A ) ∗ x . Then [ B − Ax, ( y n )] K = [ x, ( z n )] K holds for all x ∈ D ( B − A ) . This implies ( | B | − / Ax, e y ) = ( x, J | B | / e z ) , ∀ x ∈ D ( B − A ) , (10)where e y = T ( y n ) , e z = T ( z n ) . The relation (10) can be written as ( B − Ax, J | B | / e y ) = ( x, J | B | / e z ) . (11)From Lemma 5 we know that ( B − A ) ∗ = A ∗ B − , hence (11) implies A ∗ | B | − / e y = J | B | / e z. This implies e y ∈ R ( | B | / ) , hence ( y n ) ∼ y ∈ H and A ∗ y = J | B | / e z , which implies y ∈ D ( | B | − / A ∗ ) . We have proved that ( B − A ) ∗ ⊂ S ∗ , hence S ⊂ ( B − A ) ∗∗ = S ′ .Let S be a closed symmetric operator in a Hilbert space with equal deficiency indices.It is well-known that the set D ( S ) provided with the inner product ( x, y ) D ( S ∗ ) = ( x, y ) +( S ∗ x, S ∗ y ) is a Hilbert space. Definition 2 ([27],[13],[19]) . Let S be a symmetric operator in a Hilbert space H . Acollection { Ω , Γ , Γ } , in which Ω is a Hilbert space, and Γ , Γ : D ( C ∗ ) → Ω are boundedoperators, is called a space of boundary values (SBV) for S ∗ , if1. ( S ∗ x, y ) − ( x, S ∗ y ) = (Γ x, Γ y ) Ω − (Γ x, Γ y ) Ω ∀ x, y ∈ D ( S ∗ ) ,
2. the mapping
Γ : x
7→ { Γ x, Γ x } from D ( S ∗ ) to Ω × Ω is surjective.In the case of a symmetric operator S in a Krein space we have Definition 3 ([20]) . Let S be a symmetric operator in a Krein space ( K , [ · , · ]) . A collection { Ω , b J , Γ , Γ } , in which Ω is a Hilbert space, b J is an involution in Ω , and Γ , Γ : D ( S ∗ ) → Ω are bounded operators, is called a space of boundary values for S ∗ , if1. [ S ∗ x, y ] − [ x, S ∗ y ] = (Γ x, b J Γ y ) Ω − (Γ x, b J Γ y ) Ω ∀ x, y ∈ D ( S ∗ ) ,
2. the mapping
Γ : x
7→ { Γ x, Γ x } from D ( S ∗ ) to Ω × Ω is surjective.9n the case of Hilbert space operators, it can be shown that a SBV exists for any sym-metric operator with equal deficiency indices (see [13], [19]). In the case of Krein spaceoperators, it can be shown that a SBV exists for an operator T if Π( T ) = ∅ holds, and if thedeficiency indices of T are equal (see [20]). For a fixed SBV for the operator T , there existsa bijective correspondence between the collection of the closed extensions e T and the set ofclosed relations θ in Ω : x ∈ D ( e T ) = D ( e T θ ) ⇐⇒ { Γ x, Γ x } ∈ θ ⊂ Ω × Ω . (12)An extension e T is selfadjoint if and only if θ is a selfadjoint relation, or equivalently, if thereexists a selfadjoint operator G on Ω given by the relation (cos G )Γ x − (sin G )Γ x = 0 ⇐⇒ { Γ x, Γ x } ∈ θ. For more details on the theory of SBV see [19], [20], [13] and [27].Let { Ω , Γ , Γ } be a fixed SBV for the operator A ∗ . Set e Γ i = Γ i | D ( | B | − / A ∗ ) , i = 1 , .We will show that a collection { Ω , I, e Γ , e Γ } is a SBV for the operator S ∗ . Hence, we needto show that the statements 1. and 2. from Definition 3 hold. We start with [ S ∗ x, y ] K − [ x, S ∗ y ] K = ( | B | − / A ∗ x, | B | / y ) − ( J | B | / x, J | B | − / A ∗ y )= ( A ∗ x, y ) − ( x, A ∗ y ) = (Γ x, Γ y ) Ω − (Γ x, Γ y ) Ω = ( e Γ x, e Γ y ) Ω − ( e Γ x, e Γ y ) Ω . To see that e Γ i is bounded for i = 1 , , it is sufficient to prove that there exists C > suchthat k x k D ( A ∗ ) ≤ C k x k D ( S ∗ ) ∀ x ∈ D ( S ∗ ) , and since k A ∗ x k ≤ (cid:13)(cid:13)(cid:13) | B | / | B | − / A ∗ x (cid:13)(cid:13)(cid:13) ≤ (cid:13)(cid:13)(cid:13) | B | / (cid:13)(cid:13)(cid:13) (cid:13)(cid:13)(cid:13) | B | − / A ∗ x (cid:13)(cid:13)(cid:13) , it is enough to see that there exists C > such that k x k ≤ C (cid:16)(cid:13)(cid:13)(cid:13) | B | / x (cid:13)(cid:13)(cid:13) + k A ∗ x k (cid:17) ∀ x ∈ D ( | B | − / A ∗ ) . (13)But this is implied by Lemma 3. Indeed, suppose that (13) does not hold. Then there is asequence ( x n ) ∈ D ( | B | − / A ∗ ) , k x n k = 1 , such that k x n k ≥ n (cid:16)(cid:13)(cid:13)(cid:13) | B | / x n (cid:13)(cid:13)(cid:13) + k A ∗ x n k (cid:17) , which implies A ∗ x n → and | B | / x n → , hence x n → , which is a contradiction with thestatement of Lemma 3.The density of D ( S ∗ ) in D ( A ∗ ) follows directly from Lemma 4. Our next aim is to showthat e Γ = { e Γ , e Γ } is surjective. Let f, g ∈ Ω be arbitrary. Then there exists x ∈ D ( A ∗ ) suchthat Γ x = f , Γ x = g . Since D ( S ∗ ) is dense in D ( A ∗ ) , there exists a sequence ( x n ) ∈ D ( S ∗ ) such that x n → x in the norm of D ( A ∗ ) , and since Γ i , i = 1 , are bounded, it follows that e Γ i x n = Γ i x n → Γ i x , i = 1 , , so R ( e Γ) is dense in Ω × Ω . Now, since the space Ω × Ω isfinite–dimensional, it follows R ( e Γ) = Ω × Ω . 10ence, we can parameterize the selfadjoint extensions of the operator S by the use ofa SBV for the operator A ∗ . More precisely, let e A = e A θ be an extension of the operator A generated by the closed relation θ on Ω . We define an extension e S of S in the following way: D ( e S ) = D ( | B | − / e A ) and e Sx = S ∗ x. This definition implies D ( e S ) = { x ∈ D ( e A ) : e Ax = A ∗ x ∈ R ( | B | / ) } = n x ∈ D ( A ∗ ) : A ∗ x ∈ R ( | B | / ) : { Γ x, Γ x } ∈ θ o = n x ∈ D ( S ∗ ) : { e Γ x, e Γ x } ∈ D ( e S θ ) o . On the other hand, if e S = e S θ is a closed extension of the operator S generated by a closedrelation θ on Ω , then it generates a closed extension e A θ of the operator A . Thus, we haveshown Theorem 2.
There exists a bijective correspondence between all closed extensions of theoperator A and all closed extensions of the operator S . Specifically, there exists a bijectivecorrespondence between all selfadjoint extensions of the operator A and all selfadjoint exten-sions of the operator S . Moreover, if e A is a selfadjoint extension of A , then the operator e S defined by D ( e S ) = D ( | B | − / e A ) , e Sx = ( y n ) , T ( y n ) = J | B | − / e A , is a selfadjoint extensionof S .Remark . In the case when a selfadjoint extension of A is given by the relation (cos G )Γ x − (sin G )Γ x = 0 , where G is a selfadjoint operator in Ω , the relation (cos G ) e Γ x − (sin G ) e Γ x =0 defines a selfadjoint extension of S . Example 2.
Let H = L (0 , , Af = − d f d x , D ( A ) = C (0 , , ( Bf )( x ) = xf ( x ) . Theoperators A and B are obviously symmetric, B is bounded and ∈ σ c ( B ) . One can easilysee that Π( L ) = ∅ . A SBV for A ∗ can be chosen as follows: Ω = C , Γ = {− f (0) , f (1) } , Γ = { f ′ (0) , f ′ (1) } . We have D ( A ∗ ) = W (0 , , where W (0 , is the usual Sobolev space, and D ( | B | − / A ∗ ) = (cid:26) f ∈ W (01 , ) : 1 √ x d f d x ( x ) ∈ L (0 , (cid:27) . The collection { C , I, Γ | D ( | B | − / A ∗ ) , Γ | D ( | B | − / A ∗ ) } is a SBV for S ∗ , hence all selfadjointextensions of S are parameterized by the selfadjoint relations in C by the formula (12). B In this section we have the same assumptions as in the previous section. Our aim in thissection is to show the connection between the spectra of e L and e S .Let θ be a selfadjoint relation in Ω × Ω such that ρ ( e L ) = ∅ holds, where e L () = e A θ − B = e A − λB , and let e S = e S θ be a selfadjoint extension generated by the relation θ .11 roposition 6. The operator e S is not bounded.Proof. Let us assume that e S is bounded. This implies D ( e S ) = K , hence D ( | B | − / e A ) = K ,which implies K = H . This implies that for each sequence ( x n ) ∈ H such that | B | / x n → y , there exists x ∈ H such that x ∼ ( x n ) , i.e. | B | / x = y . Hence R ( | B | / ) = H , acontradiction with the assumption ∈ σ c ( B ) .For a symmetric operator S in a Hilbert or Krein space, by π ( S ) we denote the number ofnegative squares of Hermitian sesquilinear form ( Sx, y ) , x, y ∈ D ( s ) , where ( · , · ) is the innerproduct on the Hilbert or Krein space. That is, π ( S ) is the supremum of the dimensions ofall subspaces L such that ( Sx, x ) < for all = x ∈ L . Theorem 3.
For each selfadjoint extension e A of A , and e S the corresponding extension of S , the formula π ( e S ) = π ( e A ) holds.Proof. The inequality π ( e S ) ≤ π ( e A ) is evident.We will first prove the theorem in the case π ( e S ) = 0 . In this case we have [ e Sx, x ] K ≥ for all x ∈ D ( e S ) , or, equivalently ( e Ax, x ) ≥ for all x ∈ D ( | B | − / e A ) . Let us assume that π ( e A ) > . Then there exists some x ∈ D ( e A ) such that ( e Ax, x ) < . Lemma 4 impliesthat there is a sequence ( x n ) ∈ D ( | B | − / e A ) such that x n → x and e Ax n → e Ax , hence ( e Ax n , x n ) → ( e Ax, x ) , so for n large enough we have ( e Ax n , x n ) < ; a contradiction.Now we treat the case π ( e S ) > . Let integer n and elements x , . . . , x n ∈ D ( e A ) bearbitrarily chosen. Let A be the matrix (( e Ax n , x n )) ni,j =1 , and let k be the number of neg-ative eigenvalues of A . Then, by Lemma 4 (iii), for each ε > there exist x ′ , . . . , x ′ n ∈ D ( | B | − / e A ) such that k x i − x ′ i k < ε and k e A ( x i − x ′ i ) k < ε . Since | ( e Ax i , x j ) − ( e Ax ′ i , x ′ j ) | = | ( e A ( x i − x ′ i ) , x j ) + ( e Ax ′ i , x j − x ′ j ) | << ε ( k x j k + k Ax i k ) + ε , for a sufficiently small ε > the matrix A ′ = (( e Ax ′ i , x ′ j )) ni,j =1 will have at least k negativeeigenvalues. Hence, the number of negative squares of the Hermitian form ( e Ax, y ) , x, y ∈ D ( e A ) is smaller or equals the number of negative squares of the Hermitian form [ e Sx, y ] K , x, y ∈ D ( e S ) , which finishes the proof. Theorem 4.
The spectra of e S and e L coincide. Moreover, their point spectra coincide, aswell as their eigenspaces and Jordan chains.Proof. First we will show σ p ( e L ) = σ p ( e S ) . Let ∈ σ p ( e L ) be arbitrary. Then there exists x ∈ D ( e A ) , x = 0 such that e Ax = Bx. (14)It is evident that x ∈ D ( e S ) . Multiplying (14) by J | B | − / we obtain J | B | − / e Ax = | B | / , which implies T ( e S − ) x = 0 , hence ( e S − ) x = 0 .On the other hand, for ∈ σ p ( e S ) there exists x ∈ D ( e S ) , x = 0 , such that J | B | − / e Ax = | B | / x. (15)Multiplying (15) by J | B | / we obtain e Ax = Bx , i.e. ∈ σ p ( e L ) . Hence, σ p ( e L ) = σ p ( e S ) .12he relation ( e S − ) x = x implies J | B | − / e Ax − | B | / x = | B | / x . (16)Multiplying (16) by J | B | / we obtain ( e A − B ) x = Bx , hence a Jordan chain of e S corre-sponding to an eigenvalue is also a Jordan chain of e L for the same eigenvalue . The otherdirection can be seen analogously.Let ∈ ρ ( e S ) . This implies T ( R ( e S − )) = H , i.e. R ( J | B | − / e A − | B | / ) = H . This implies R ( e A − B ) ⊇ R ( | B | / ) ⊇ R ( B ) . (17)Now, let µ ∈ ρ ( e L ) be arbitrary. From (17) follows ( e A − µB ) x = ( e A − B ) x + ( − µ ) Bx ∈ R ( e A − B ) , hence R ( e A − µB ) ⊂ R ( e A − B ) , which implies ∈ ρ ( e L ) .On the other hand, let ∈ σ ( e S ) \ σ p ( e S ) . Then there exists ( y n ) ∈ K such that ( y n ) / ∈ R ( e S − ) . (18)We denote e y = T ( y n ) . Then, the relation (18) can be written as e y / ∈ R ( J | B | − / e A − | B | / ) which implies J | B | − / e y / ∈ R (( e A − B ) | D ( | B | − / e A ) ) . (19)Let us assume that J | B | / e y ∈ R ( e A − B ) . Then there exists x ∈ D ( e A ) such that e Ax − Bx = J | B | / e y , hence e Ax ∈ R ( | B | / ) , a contradiction with (19). Hence, J | B | / e y / ∈ R ( e A − B ) and ∈ σ ( e L ) . Proposition 7.
We have(i) σ c ( e S ) ⊂ σ c ( e L ) ,(ii) σ ap ( e S ) ⊂ σ ap ( e L ) .Proof. (i) Let ∈ σ c ( e S ) . Then R ( e S − ) is dense in K . For each ( y n ) ∈ K , and for each ε > there exists x ∈ D ( e S ) such that (cid:13)(cid:13)(cid:13) ( e S − ) x − ( y n ) (cid:13)(cid:13)(cid:13) K < ε. We choose y ∈ R ( B ) arbitrarily and set e y = J | B | − / y , y ′ = | B | − / e y . For y ′ ∈ H ⊂ K there exists x ∈ D ( e S ) such that k ( e S − ) x − y ′ k K < ε k| B | / k − , or, equivalently, (cid:13)(cid:13)(cid:13) J | B | − / e Ax − λ | B | / x − e y (cid:13)(cid:13)(cid:13) < ε (cid:13)(cid:13)(cid:13) | B | / (cid:13)(cid:13)(cid:13) − . Now, (cid:13)(cid:13)(cid:13) ( e A − B ) x − y (cid:13)(cid:13)(cid:13) = (cid:13)(cid:13)(cid:13) J | B | / ( J | B | − / e Ax − | B | / x − e y ) (cid:13)(cid:13)(cid:13) < ε. R ( e A − B ) is dense in R ( B ) , which implies that R ( e A − B ) is dense in H . Since / ∈ σ p ( e L ) , ∈ σ c ( e L ) follows.(ii) Let ∈ σ ap ( e S ) . Then there exists a sequence ( x n ) ∈ D ( e S ) , k x n k K = 1 such that k ( e S − ) x n k → , or equivalently, ( J | B | − / e A − | B | / ) x n → (20)and (cid:13)(cid:13)(cid:13) | B | / x n (cid:13)(cid:13)(cid:13) = 1 . (21)Multiplying relation (20) by J | B | / we obtain ( e A − B ) x n → . From (21) we see that k x n k ≥ (cid:13)(cid:13) | B | / (cid:13)(cid:13) − . Set b x n = x n / k x n k . Then k b x n k = 1 and ( e A − B ) b x n → , hence ∈ σ ap ( e L ) . Corollary 1.
The relation σ r ( e L ) ⊂ σ r ( e S ) holds. In general, Proposition 7 (and hence also Corollary 1) cannot be improved, as will beshown in Example 3.First we give sufficient and necessary conditions for the complete correspondence of thespectra of e L and e S . Lemma 6.
The subspace R ( e A − B ) is dense in H if and only if H ∩ R ( e S − ) ⊥ = { } .Proof. = ⇒ : First we show that the subspace R ( e A − B ) is dense in H if and only if R (( e A − B ) | D ( | B | − / e A ) ) is dense in H .The case = 0 follows from Lemma 4, so we assume = 0 . First we assume that R ( e A − B ) is dense in H . Let x ∈ H , ε > be arbitrary. Then there exists x ∈ D ( e A ) such that k ( e A − B ) x − x k < ε/ . Lemma 4 implies that there exists x ′ ∈ D ( | B | − / e A ) such that k x − x ′ k < ε ||k B k and k e A ( x − x ′ ) k < ε/ . Now, (cid:13)(cid:13)(cid:13) ( e A − B ) x ′ − x (cid:13)(cid:13)(cid:13) ≤ (cid:13)(cid:13)(cid:13) ( e A − B )( x ′ − x ) + ( e A − B ) x − x (cid:13)(cid:13)(cid:13) < ε. The other direction is obvious.We showed that R (( e A − B ) | D ( | B | − / e A ) ) is dense in H , hence the relation ( x, ( e A − λB ) y ) = 0 , ∀ y ∈ D ( | B | − / e A ) implies x = 0 . Now, from ( x, ( e A − λB ) y ) = ( | B | / x, | B | − / ( e A − λB ) y ) = [ x, ( e S − λ ) y ] K our claim immediately follows.The other direction follows from the similar reasoning. Lemma 7.
The subspace R ( e S − ) is dense in K if and only if R ( | B | − / ( e A − B )) is dense in H . roof. = ⇒ : Let us assume that R ( | B | − / ( e A − B )) is not dense in H . Then there exists e x ∈ H , e x = 0 such that ( e x, | B | − / ( e A − B ) y ) = 0 ∀ y ∈ D ( | B | − / e A ) . Set ( x n ) = T − J e x . Then (( x n ) , ( e S − ) y ) K = 0 for all y ∈ D ( e S ) , which implies the statement. ⇐ = : Let us assume that R ( e S − ) is not dense in K . Then it exists ( y n ) ∈ K , ( y n ) = 0 , suchthat (( y n ) , ( e S − ) x ) K = 0 for all x ∈ D ( e S ) . Set y = T ( y n ) . Then ( e y, ( J | B | − / e A − | B | / ) x ) = 0 ∀ x ∈ D ( | B | − / e A ) , hence ( J e y, | B | − / ( e A − B ) x ) = 0 ∀ x ∈ D ( | B | − / e A ) , and the statement follows.Lemmata 6 and 7 imply the following result. Proposition 8. (i) If ∈ σ r ( e S ) , then ∈ σ c ( e L ) if and only if H ∩ R ( e S − ) ⊥ = { } .(ii) If ∈ σ c ( e L ) , then ∈ σ r ( e S ) if and only if R ( | B | − / ( e A − B )) is not dense in H . Example 3.
Let H = L (0 , . We define the operators A and B in H by: ( Af )( x ) = Z G ( x, ξ ) f ( ξ )d ξ, ( Bf )( x ) = (2 x − f ( x ) , where G ( x, ξ ) = ( x , x ≤ ξ,ξ , x ≥ ξ. Note that A is the inverse of the Dirichlet Laplacian. The operator B is bounded and ∈ σ c ( B ) , ∈ σ c ( A ) holds. Note that this implies ∈ σ c ( L ) . One can easily prove that Π( L ) ∩ R = ∅ holds.We will show that R ( | B | − / A ) is not dense in H . Indeed, constant functions are notcontained in R ( | B | − / A ) . This follows from the fact that the function x p | x − | isnot contained in R ( A ) . Now Proposition 8 (ii) implies ∈ σ r ( S ) . Remark . If π ( B ) < ∞ or e A ≥ , then σ r ( e L ) = σ r ( e S ) = ∅ , and consequently, σ c ( e L ) = σ c ( e S ) .If e S is definitizable, then σ r ( e S ) = ∅ , hence σ r ( e L ) = ∅ and σ c ( e S ) = σ c ( e L ) . Corollary 2.
Let us suppose that π ( A ) < ∞ . Then each selfadjoint extension e S of S isdefinitizable.Proof. Since deficiency indices of A are finite, π ( e A ) < ∞ holds for each selfadjoint extension e A , and Theorem 3 implies π ( e S ) < ∞ . Now, let ∈ Π( L ) ∩ R be arbitrary. It is well–knownthat there exists a selfadjoint extension e A such that ∈ ρ ( e A − B ) (see [21], [13]), whichimplies ∈ ρ ( e S ) , where e S is the corresponding operator. Also π ( e S ) = π ( e A ) < ∞ holds.Now, from [10, Proposition 1.1] follows that each selfadjoint extension of S has a nonvoidresolvent set, hence, see [23], each selfadjoint extension of S is definitizable.15 emark . If e A is qup, the proof of Lemma 3.1 from [6] implies ∈ ρ ( e L ) for = 0 , || sufficiently small. Hence ∈ ρ ( e S ) for such . From this fact and from π ( e S ) = π ( e A ) < ∞ , itfollows that e S is also qup. Corollary 3. If e L has a compact resolvent in one point (and then, from Proposition 1, inall points), and if e Ax = Bx, x = 0 = ⇒ ( Bx, x ) = 0 , (22) then e S has a discrete spectrum.Proof. See [4].
Remark . The relation (22) is equivalent with the fact that e S does not have isotropiceigenvectors, i.e. if [ x, x ] = 0 for all eigenvectors x of e S . B unbounded In this section we will treat a more general case when B is possibly unbounded. In thissection we assume:1. A and B are selfadjoint and D ( A ) ⊂ D ( B ) ,2. is not an eigenvalue of B , and3. ρ ( L ) ∩ R = ∅ .Using a spectral shift (i.e. substituting A by A − λB ), if necessarily, we can assume ∈ ρ ( A ) .The construction which we use here is essentially given in [6].Similarly as in Section 2. we introduce a space K as the completion of ( D ( | B | / ) , [ · , · ]) ,where [ · , · ] = ( J | B | / · , · ) , by the use of Cauchy sequences. As in Section 2., we set J = E (0 , ∞ ) − E ( −∞ , , where E is the spectral function of B . By T we denote the operator T : K → H , T ( x n ) = lim n | B | / x n . It is easy to see that T is an isomorphism betweenspaces H and K . As was shown in [6], A − B extends to a bounded symmetric operator R in K , and R is injective. Hence, the operator S = R − is selfadjoint and boundedly invertiblein K . Moreover, we can see that S is an extension of B − A which is densely defined in K .Indeed, since R is injective, it holds ( B − A ) − = ( B − A ) − = R, so B − A = S , where the closure is taken in K . The domain of S is given by D ( S ) = { ( x n ) ∈ K : J | B | − / Ax n is convergent in H} . For ( x n ) ∈ K we can always assume x n ∈ D ( B ) . Then, if ( y n ) = ( S − )( x n ) we have T ( y n ) = lim n J | B | − / ( A − B ) x n .If A is a qup, we can, instead of the assumption 3., make the assumption ρ ( L ) = ∅ ,since in [6, Lemma 3.1] it was proved that in this case there exists ∈ R such that L ( ) isboundedly invertible. Also, in this case, S is a qup.16 heorem 5. We have σ p ( L ) ⊂ σ p ( S ) (23) and σ ( S ) ⊂ σ ( L ) . (24) Proof.
The relation (23) is obvious.Let ∈ ρ ( L ) . Then it is easy to see that | B | / | A − B | − / and | B | / | A − B | − / areeverywhere defined bounded operators. From ( | A − B | − / | B | / ) ∗ = | B | / | A − B | − / follows ( | A − B | − / | B | / ) ∗∗ = ( | B | / | A − B | − / ) ∗ , hence | A − B | − / | B | / ⊂ ( | B | / | A − B | − / ) ∗ . (25)Since the right hand side in (25) is a bounded operator, it follows that also the left handside in (25) is a bounded operator (but, in general, not everywhere defined!). Hence, theoperator | B | / ( A − B ) − | B | / = | B | / | A − B | − / sgn( A − B ) | A − B | − / | B | / is bounded.The subspace D ( | B | / ) ∩ R ( | B | / ) is dense. Indeed, we have D ( | B | / ) ∩ R ( | B | / ) = { x = | B | / y : y ∈ D ( B ) } , and the right hand side is dense since D ( B ) is a core of | B | / . Now, let x ∈ H be arbitrary.Then there exists a sequence ( x n ) ∈ D ( | B | / ) ∩ R ( | B | / ) such that x n → J x . From theconsiderations given above it follows that | B | / ( A − B ) − | B | / x n converges to some y ∈ H .Set z n = ( A − B ) − | B | / x n . Then | B | / z n → y , hence ( z n ) ∈ K and x n = | B | − / ( A − B ) z n . Now | B | − / ( A − B ) z n → J x , which implies J | B | − / ( A − B ) z n → x , so x ∈ T ( R ( S − )) . This implies ∈ ρ ( S ) .The other inclusion σ ( L ) ⊂ σ ( S ) in general does not hold, since for large the operator L () need not be closed.If we assume ∈ ρ ( B ) , then K = D ( | B | / ) and σ p ( S ) ⊂ σ p ( L ) , hence σ p ( S ) = σ p ( L ) .Also, it can be seen that σ ap ( L ) ⊂ σ ap ( S ) . Hence, if L has only approximative spectrumthen σ ( L ) = σ ( S ) .Another kind of construction is made in [24]. There it is assumed that ∈ ρ ( B ) and D ( A ) = D ( | B | / ) . Then L () is a holomorphic function in Kato sense from K to H . For thereaders convenience, we give a definition. Definition 4.
The function T () : X → Y is holomorphic in Kato sence if there exist abounded holomorphic functions U () : Z → X and V () : Z → Y such that T () U () = V () and R ( U ()) = D ( T ()) for all ∈ C , where Z is some Banach space. An operator function T () is bounded holomorphic if it takes values in the set of bounded operators and if T () x isa holomorphic function in for all x . 17n our case, we can choose U () = B − , V () = AB − − , and Z = H .We set S = B − A , and by the use of the similar methods as in the proofs of Theorem4 and Proposition 7, it can be seen that σ ( L ) = σ ( S ) , σ ap ( L ) = σ ap ( S ) , σ c ( L ) = σ c ( S ) and σ p ( L ) = σ p ( S ) , hence we have the complete equivalence of the spectra of L and S . In this section we apply the constructions of reduced operators from Sections 2 and 4 toobtain a Rayleigh–Ritz–like upper bound for the eigenvalues of the reduction operator givenin terms of the eigenvalues of the matrix pair ( A V , B V ) , where A V , B V are orthogonalprojections of A , B on a finite dimensional subspace. Hence the bound is given in terms ofthe original data ( A, B ) , which can be important in applications.To obtain a variational characterization for the reduction operator, we need a result from[5]. Let Q be a qup operator such that all finite critical points are regular, and such thatthere are no critical points embedded in the positive continuous spectrum. For such a Q , in[5] a variational characterization of eigenvalues is obtained. More precisely, we define ± j = sup (cid:26) inf (cid:26) [ Qx, x ][ x, x ] : x ∈ M ∩ C ± ∩ D ( Q ) (cid:27) : M ∈ M j (cid:27) , (26)where C ± = { x : [ x, x ] ≷ } , and M j denotes the set of all subspaces of codimension j − .Then, if by d ± we denote the positive (negative) spectral shift of Q (for more details see[5]), a number ± j + d ± is either an eigenvalue of Q , or a point on the boundary of the essentialspectrum of Q , for all j ∈ N .If the operator e S from Section 2, or the operator S from Section 4 satisfy conditionsstated above, we can variationally characterize the eigenvalues of e S or S , respectively. If B is bounded, then we have the following variational characterization ± j ( e S ) = ± j ( e L ) = sup ( inf ( ( e Ax, x )( Bx, x ) : x ∈ M ∩ C ± ∩ D ( | B | − / e A ) ) : M ∈ M j ) . (27)In the case when B is unbounded, we have [ S ( x n ) , ( x n )] K [( x n ) , ( x n )] K = lim n ( Ax n , x n )( Bx n , x n ) , where ( x n ) ∈ D ( S ) . Since D ( | B | − / A ) is a core of the operator S , from [12, Theorem 4.5.3.]follows that in this case we have ± j ( S ) = ± j ( L ) = sup (cid:26) inf (cid:26) ( Ax, x )( Bx, x ) : x ∈ M ∩ C ± ∩ D ( | B | − / A ) (cid:27) : M ∈ M j (cid:27) . (28)Now, suppose that the operator e S from Section 2 satisfies conditions stated above. Let V ⊂ D ( | B | − / e A ) be a finite dimensional ortho–complemented subspace in K . Let dim V = N .18enote by P a orthogonal projector in K onto V . Set e A V = P e AP , B V = P BP . Let ⊥ and [ ⊥ ] denote the orthogonal complement in H and K , respectively. The set of ortho–complemented subspaces of K we denote by O . Let F be a subspace of V , with dim F = k ,and let F = span { x , . . . , x k } . Set F ′ = span { B − V x , . . . , B − V x k } . Then dim F ′ = k and x [ ⊥ ] F ′ ⇐⇒ x ⊥ F, ∀ x ∈ V. On the other hand, if x [ ⊥ ] F , then we define F ′′ = span { B V x , . . . , B V x k } and then we have x [ ⊥ ] F ⇐⇒ x ⊥ F ′′ , ∀ x ∈ V. For m ≤ N , set µ m = sup F ⊂ V dim F = m − inf x ∈ Vx ⊥ Fx ∈C + ( e A V x, x )( B V x, x ) . (29)Then, from the considerations given above, we have µ m = sup F ⊂ V dim F = m − inf x ∈ Vx [ ⊥ ] Fx ∈C + ( e Ax, x )( Bx, x )= sup y ,...,y m − ∈ V span { y ,...,y m − }∈O inf x ∈ Vx ∈ span { y ,...,y m − } [ ⊥ ] x ∈C + ( e Ax, x )( Bx, x )= sup y ,...,y m − ∈K span { y ,...,y m − }∈O inf x ∈ Vx ∈ span { P y ,...,P y m − } [ ⊥ ] x ∈C + ( e Ax, x )( Bx, x )= sup y ,...,y m − ∈K span { y ,...,y m − }∈O inf x ∈ Vx ∈ span { y ,...,y m − } [ ⊥ ] x ∈C + ( e Ax, x )( Bx, x ) ≥≥ sup y ,...,y m − ∈K span { y ,...,y m − }∈O inf x ∈ D ( | B | − / e A ) x ∈ span { y ,...,y m − } [ ⊥ ] x ∈C + ( e Ax, x )( Bx, x )= sup M ∈M j inf x ∈ D ( | B | − / e A ) x ∈ M ∩C + ( e Ax, x )( Bx, x ) . The analogous relation can be obtained for the eigenvalues of the negative type. From thefinite dimensional principle given in [7], µ m is an eigenvalue of the matrix pair ( e A V , B V ) .Hence we obtained a Rayleigh–Ritz–like upper bound for eigenvalues of e S : ± j ( e A V , B V ) ≥ ± j ( e S ) . The analogous Rayleigh–Ritz upper bound holds also for the operator S from Section 4, dueto the formula (28). 19 Zero is an eigenvalue of B If is an eigenvalue of B , then in general a reduced operator cannot be constructed. But wecan construct an reduced linear relation, a generalization of the notion of linear operator,and we can recover the results from previous sections. Under some additional assumptionswe can also construct a proper reduction operator.In this section we assume:1. A and B are selfadjoint operators,2. B is A –bounded, ∈ σ p ( B ) ,3. ρ ( A, B ) ∩ R = ∅ .The last relation implies that we can assume ∈ ρ ( A ) .Then B − A does not exist as a linear operator, but it can be introduced in terms ofthe linear relations, i.e. as a subspace of H × H (for the basic definitions see [28], [14]). Wedefine S = B − A = {{ x, y } ∈ D ( A ) × D ( B ) : Ax = By } . By S we denote the closure of S as a subspace in H × H .Let K denote the factor space D ( B ) / Ker B , and let e K denote the completion of ( K , ( B · , · )) .Evidently, e K is a Krein space, and it consists of all Cauchy sequences in D ( | B | / ) , where weidentify such two sequences ( x n ) and ( y n ) if | B | / ( x n − y n ) → . This equivalence relationwe denote by the ∼ symbol. Let [ x ] denote a class in e K represented by x .Let R ( , S ) be a resolvent of S . Then the family of linear operators R ( , S ) in H induces afamily e R ( , S ) of linear operators in the space K , since R ( , S ) Ker B ⊂ Ker B (see [24]). Wedefine the linear relation e S in e K by e S = e R (0 , S ) − .We say that is a discrete eigenvalue of the operator T if is an isolated eigenvalue offinite multiplicity of the operator T .First we treat the case when B is a bounded operator and zero is a discrete eigenvalue.Then it is easy to see that we can substitute Cauchy sequences with elements of H , and that x ∼ y if and only if x − y ∈ Ker B . The inner product on e K is given by [ · , · ] = ( B · , · ) . It iseasy to see that e S = {{ [ x ] , [ y ] } : x − A − By ∈ Ker B } , and that e S is a selfadjoint relation. Theorem 6.
The spectra of e S and L coincide. Moreover, their point spectra coincide.Proof. The relation σ p ( L ) ⊂ σ p ( e S ) is obvious. On the other hand, let ∈ σ p ( e S ) . Then there exists x ∈ H such that x − A − Bx ∈ Ker B , hence Bx − BA − Bx = 0 .There exists y = 0 such that Bx = Ay , hence I − BA − ) Bx = ( I − BA − ) Ay = Ay − By, which implies ∈ σ p ( L ) . It is easy to see that ( S − ) − = ( A − B ) − B ∀ ∈ C \ σ p ( L ) = C \ σ p ( e S ) . From this relation it is easy to obtain ρ ( L ) = ρ ( e S ) .20f B is bounded, but zero is not a discrete eigenvalue of B , it can be seen that e S = {{ [ x ] , [( y n )] } : x − A − J | B | / lim n | B | / y n ∈ Ker B } , where J is defined as in section 2, is a selfadjoint relation. Using similar techniques as inSection 2, it can be shown that σ ( e S ) = σ ( L ) and σ p ( e S ) = σ p ( L ) .This kind of procedure can be implemented also in the case when B is not bounded,using the techniques from Section 4. For instance, if all assumptions given in the beginningof the Section 4 hold, except from / ∈ σ p ( B ) , we can define e S by e S = {{ [( x n )] , [( y n )] } : ( x n ) ∼ ( A − By n ) } . Reasoning analogously as in the proof of Theorem 5, it can be shown that σ p ( L ) ⊂ σ p ( e S ) and σ ( e S ) ⊂ σ ( L ) .When is it possible to reduce the relation e S to an operator, without the loss of theinformation about spectra? To answer this question we introduce the notion of the multi–valued part T ∞ = {{ , g } ∈ T } of the relation T (see [28]). We also set T (0) = { g ∈K : { , g } ∈ T } . It is easy to see that T ∞ is ortho–complemented if and only if T (0) isortho–complemented. We have Theorem 7 ([28]) . Let T be a closed symmetric relation in a Krein space K such that T ∞ is ortho–complemented. Then T s = T ∩ ( T ∞ ) ⊥ is an operator with D ( T s ) = D ( T ) and R ( T s ) ⊂ T (0) ⊥ . We also have(i) σ p ( T ) = σ p ( T s ) ,(ii) σ ( T ) = σ ( T s ) , and(iii) T is a selfadjoint relation if and only if T s is a selfadjoint operator in T (0) ⊥ . Hence, we can reduce e S to an operator if the set e S (0) = { [( y n )] : lim n | B | / A − By n ∈ Ker B } is ortho–complemented.If, for instance, R ( B ) ∩ A Ker B = { } , or if Ker B is A –invariant, e S (0) is ortho–complemented.If d = dim Ker B < ∞ and if e S (0) is ortho–complemented, we can also obtain Rayleigh–Ritz–like upper bound for the eigenvalues of e S (0) in terms of the non-zero eigenvalues ofthe matrix pair ( A V , B V ) , where V ⊂ D ( A ) , in the case when the dimension of V is greaterthen the sum of d and the spectral shift of e S s = e S ∩ ( e S (0)) ⊥ . We omit the details. Acknowledgement : The author is very grateful to anonymous referees for extremely usefulcomments and suggestions.
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