Boundary value space associated to a given Weyl function
aa r X i v : . [ m a t h . F A ] F e b Boundary value space associated to a given Weylfunction
Muhamed Borogovac,[email protected] - Actuarial Department, Canton MA 02021, USAFebruary 16, 2021
Abstract : Let S be a symmetric linear relation in the Pontyagin space ( K , [ ., . ]) andlet Π = ( H , Γ , Γ ) be the corresponding boundary triple. We prove that the correspond-ing Weyl function Q satisfies Q ∈ N κ ( H ). Conversely, for regular Q ∈ N κ ( H ), we findlinear relation S ( A , where A is representing self-adjoint linear relation of Q , and weprove that Q is the Weyl function of the relation S . We also prove ˆ A = ker Γ , whereˆ A is the representing relation of the ˆ Q := − Q − . In addition, if we assume that thederivative at infinity Q ′ ( ∞ ) := lim z →∞ zQ ( z ) is a boundedly invertible operator then weare able to decompose A , ˆ A and S + in terms of S , i.e. we express relation matrices of A , ˆ A and S + in terms of S , which is a bounded operator in this case. Key words:
Weyl Function, Operator representation; Boundary Value Space; Pontryagin space;
MSC (2010)
The following definitions of a linear relation and basic concepts related to it can befound in [1, 14]. In the sequel C and R are sets of complex and real numbers, respectively, κ denotes a non-negative integer, and H , K , M are inner product spaces.A linear relation from H into K is a (linear) subspace T of the product space H × K .If H = K , T is said to be a linear relation in K . A linear relation is closed if it is a closedsubspace. We will use the following concepts and notation for two linear relations, T and S from H into K and a linear relation U from K into M . D ( T ) := { f ∈ H| { f, g } ∈ T f or some g ∈ K} R ( T ) := { g ∈ K | { f, g } ∈ T f or some f ∈ H} ker T := { f ∈ H| { f, } ∈ T } T (0) := { g ∈ K| { , g } ∈ T } T ( f ) := { g ∈ K| { f, g } ∈ T } , ( f ∈ D ( T )) T − := {{ g, f } ∈ K × H| { f, g } ∈ T } zT := {{ f, zg } ∈ H × K| { f, g } ∈ T } , ( z ∈ C ) S + T := {{ f, g + k }| { f, g } ∈ S, { f, k } ∈ T } S ˙+ T := {{ f + h, g + k }| { f, g } ∈ S, { h, k } ∈ T } U T := {{ f, k } ∈ H × M| { f, g } ∈ T, { g, k } ∈ U f or some g ∈ K} + := {{ k, h } ∈ K × H| [ k, g ] = ( h, f ) f or all { f, g } ∈ T } T ∞ := {{ , g } ∈ T } If T (0) = { } , we say that T is single-valued linear relation, i.e. operator . The setsof closed linear relations, closed operators, and bounded operators in K are denoted by˜ C ( K ) , C ( K ) , B ( K ), respectively. Let A be a linear relation in K . We say that A is symmetric ( self-adjoint ) if it satisfies A ⊆ A + ( A = A + ). Every point α ∈ C for which { f, αf } ∈ A , with some f = 0, is called a finite eigenvalue , denoted by α ∈ σ p ( A ). Thecorresponding vectors are eigenvectors belonging to the eigenvalue α . If for some z ∈ C the relation ( A − z ) − is a bounded operator defined on the entire K , then z belongs tothe resolvent set ρ ( A ). If operator ( A − z ) − is bounded, not necessarily defined on theentire K , then z is a point of regular type of A , symbolically, z ∈ ˆ ρ ( A ).1.2 Recall that an operator valued function Q : D ( Q ) → L ( H ) belongs to the general-ized Nevanlinna class N κ ( H ) if it is meromorphic on C \ R , such that Q ( z ) ∗ = Q (¯ z ), forall points z of holomorphy of Q , and the kernel N Q ( z, w ) := Q ( z ) − Q ( w ) ∗ z − ¯ w has κ negativesquares. A generalized Nevanlinna function Q ∈ N κ ( H ) is called regular if there existsat least one point w ∈ D ( Q ) ∩ C + such that the operator Q ( w ) is boundedly invertible.We will need the following representation of generalized Nevanlinna functions. Theorem 1.1
A function Q : D ( Q ) → L ( H ) is a generalized Nevanlinna function ofsome index κ , denoted by Q ∈ N κ ( H ) , if and only if it has a representation of the form Q ( z ) = Q ( z ) ∗ + ( z − ¯ z )Γ + z (cid:16) I + ( z − z ) ( A − z ) − (cid:17) Γ z , z ∈ D ( Q ) , (1.1) where, A is a self-adjoint linear relation in some Pontryagin space ( K , [ ., . ]) of index ˜ κ ≥ κ ; Γ z : H → K is a bounded operator; z ∈ ρ ( A ) ∩ C + is a fixed point of reference.This representation can be chosen to be minimal, that is K = c.l.s. { Γ z h : z ∈ ρ ( A ) , h ∈ H} (1.2) where Γ z = (cid:16) I + ( z − z ) ( A − z ) − (cid:17) Γ z . (1.3) Realization (1.1) is minimal if and only if the negative index of the Pontryagin space ˜ κ equals κ . In that case D ( Q ) = ρ ( A ) and the triple ( K , A, Γ z ) is uniquely determined(up to isomorphism). Such operator representations were developed by M. G. Krein and H. Langer, see e.g.[10, 11] and later converted to representations in terms of linear relations (multivaluedoperators), see e.g. [7, 9].A significant part of the study is about class of the functions Q ∈ N κ ( H ) that areholomorphic at ∞ in the Banach space of bounded operators L ( H ). Such functions arecharacterized by the following lemma: Lemma 1.2 [4, Lemma 3] A function Q ∈ N κ ( H ) is holomorphic at ∞ if and only if Q ( z ) has representation Q ( z ) = Γ + ( A − z ) − Γ , z ∈ ρ ( A ) , (1.4) with bounded operator A . In this case Q ′ ( ∞ ) := lim z →∞ zQ ( z ) = − Γ + Γ , (1.5) where, the limit denotes convergence in the Banach space of bounded operators L ( H ) . γ -field Γ z by Γ = ( A − z ) − Γ z , z ∈ ρ ( A ) . (1.6) Remark 1.3 [4, Remark 1] If Q ∈ N κ ( H ) and α is a finite generalized pole of Q , thenit holds Q ( z ) = ˜ Q ( z ) + ˜ H ( z ) , where ˜ Q ( z ) = Γ + (cid:16) ˜ A − z (cid:17) − Γ ∈ N κ ( H ) , self-adjoint bounded operator ˜ A has the sameroot manifold at α as the representing relation A of Q in (1.1), and the function ˜ H ∈ N κ ( H ) is holomorphic at α . If we set Γ so that Γ + Γ becomes a boundedly invertibleoperator, then the number of negative squares may change, i.e. κ ≤ κ + κ . The decomposition in Remark 1.3 shows us the important role that representations of theform (1.4) play for every function Q ∈ N κ ( H ). That justifies our focus on the functionscharacterized by representation (1.4).1.3 The following statements from [4] will be frequently needed in this paper. Wecopy them here for the convenience of the reader. Lemma 1.4 [4, Lemma 4] Let ( H , ( ., . )) and ( K , [ ., . ]) be a Hilbert and Pontryaginspaces, respectively. Let Γ :
H → K be a bounded operator and Γ + : K → H its ad-joint operator. Assume also that Γ + Γ is a boundedly invertible operator in the Hilbertspace ( H , ( ., . )) . Then for operator P := Γ (cid:0) Γ + Γ (cid:1) − Γ + (1.7) the following statements hold:(i) P is orthogonal projection in Pontryagin space ( K , [ ., . ]) . (ii) Scalar product does not degenerate on P ( K ) = Γ ( H ) and therefore it does notdegenerate on Γ ( H ) [ ⊥ ] = ker Γ + . (iii) ker Γ + = ( I − P ) K . (iv) Pontryagin space K can be decomposed as a direct orthogonal sum of Pontryaginspaces i.e. K = ( I − P ) K [+] P K . (1.8) Theorem 1.5 [4, Theorem 3] Let Q ∈ N κ ( H ) . (i) Q is holomorphic at ∞ and Q ′ ( ∞ ) is boundedly invertible if and only if ˆ Q ( z ) = ˜Γ + (cid:16) ˜ A − z (cid:17) − ˜Γ + ˆ S + ˆ Gz, z ∈ D ( Q ) ∩ D (cid:16) ˆ Q (cid:17) , where ˜ A = ( I − P ) A | ( I − P ) K is a self-adjoint bounded operator in the Pontryaginspace ( I − P ) K , ˜Γ : H → ( I − P ) K is bounded operator, ˆ S and ˆ G are self-adjointbounded operators in the Hilbert space H , and ˆ G is boundedly invertible.(ii) In that case function Q ∈ N κ ( H ) is regular. S denotes a closed symmetric relation or operator, not necessarilydensely defined in a separable Krein or Pontryagin space ( K [ ., . ]), and S + denotes adjointlinear relation of S in ( K [ ., . ]). For definitions and notation about Boundary Value Space(BVS) for linear relation S + , see e.g. [3, 5, 6]. We will repeat those definitions here withadjusted notation. For example, Γ in [5] is denoted by Γ in [3, 6] and here, while Γ denotes the same operator in all papers. Elements of S + are denoted by ˆ f , ˆ g, . . . , wheree.g. ˆ f := (cid:18) ff ′ (cid:19) = n f, f ′ o . Let R z := ker (cid:0) S + − z (cid:1) , z ∈ ˆ ρ ( S ) , be the defect subspace of S . Thenˆ R z := (cid:26)(cid:18) f z zf z (cid:19) : f z ∈ R z (cid:27) , R := K [ − ] D ( S ) , ˆ R := (cid:26)(cid:18) f (cid:19) : f ∈ R (cid:27) . (1.9) Definition 1.6
A triple
Π = ( H , Γ , Γ ) , where H is a Hilbert space and Γ , Γ arebounded operators from S + to H , is called a Boundary Values Space (BVS) for therelation S + if the abstract Green’s identity h f ′ , g i − h f, g ′ i = (cid:16) Γ ˆ f , Γ ˆ g (cid:17) H − (cid:16) Γ ˆ f , Γ ˆ g (cid:17) H , ∀ ˆ f , ˆ g ∈ S + , (1.10) holds, and the mapping ˜Γ : ˆ f → (cid:18) Γ ˆ f Γ ˆ f (cid:19) from S + to H × H is surjective.
Note, notation ˜Γ rather than Γ is used here . That is because notation Γ :
H → K isused for operator given by (1.6).Linear relation F Π := ˜Γ ˆ R = (cid:26)(cid:18) Γ ˆ f Γ ˆ f (cid:19) : ˆ f ∈ ˆ R (cid:27) is called a forbidden relation. A linear relation θ ∈ ˜ C ( H ) is Π- admissible if θ ∩ F Π = { } .An extension ˜ S of S is called proper, if S ( ˜ S ⊆ S + . A set of proper extensionsof S is denoted by Ext S . Two proper extensions S , S ∈ Ext S are called disjoint if S ∩ S = S , and transversal if, additionally, S ˙+ S = S + , where “ ˙+” denotes a sumof subspaces, not necessarily direct.Each BVS is naturally associated with two self-adjoint extensions of S defined by S i := ker Γ i , i = 0 ,
1, i.e. it holds S i = S + i , i = 0 ,
1, see [5, p. 4425].The following questions rise:Q1: Given function Q ∈ N κ ( H ) represented by (1.1). Is it possible to find a linearrelation S ( A in the Pontryagin state space K of Q , and a boundary triple Π =( H , Γ , Γ ) for S + such that Q becomes the Weyl function of S ?Q2: Is S uniquely determined by Q ?Q3: What are the relation matrices of S , S and S + in terms of S ?Questions Q1 and Q2 and some related questions will be answered in generalterms, in Proposition 2.1. We also prove that A = S and ˆ A = S are regular extensionsof S in this case, where ˆ A is the representing operator of the invese function ˆ Q := − Q − .Questions Q3 will be answered in Theorem 2.6 for the important special case, when thefunction Q has boundedly invertible Q ′ ( ∞ ). roposition 2.1 .(a) Let S ( A , be a closed symmetric linear relation (or operator) which is not necessarilydensely defined in a Pontryagin space K of index κ satisfying K = c.l.s. {R z : z ∈ ρ ( A ) } , (2.1) A + = A , and let Q ( z ) be the Weyl function corresponding to the boundary triple Π = ( H , Γ , Γ ) .(i) Then Q ∈ N κ ( H ) , A = ker Γ , and A is the representing relation of Q in (1.1).(ii) If ˆ A denotes linear relation that represents the inverse function ˆ Q := − Q − ∈ N κ ( H ) , then ˆ A = ker Γ .(b) Conversely, let Q ∈ N κ ( H ) be a regular function given by (1.1), with representingrelation A . Then there exists a unique closed symmetric linear relation S ( A and aboundary triple Π = ( H , Γ , Γ ) , such that it holds:(i) A = ker Γ , and Q ( z ) is the Weyl function that corresponds to Π = ( H , Γ , Γ ) .(ii) ˆ A = ker Γ , where ˆ A is representing relation of ˆ Q := − Q − .(c) In both cases it holds:(i) S = A ∩ ˆ A, (ii) S + = A ˙+ ˆ A Proof. (a) (i) Existence of the boundary triple Π = ( H , Γ , Γ ) with A = ker Γ has beenproven in [5, Proposition 2.2]. Existence of the Weyl function as defined in [5, Definition2.2] has been proven in [5, Proposition 2.3]. Let us now prove Q ∈ N κ ( H ).The γ -field functions γ ( z ) : H → R z , z ∈ ρ ( A ), that we will need in the sequel, havebeen introduced in the proof of [5, Proposition 2.2]. From identity [5, (2.13)], i.e. from Q ( z ) − Q ( w ) z − w = γ ( ¯ w ) + γ ( z ) , ∀ w, z ∈ ρ ( A ) , it follows, for all h, k ∈ H : (cid:18) Q ( z ) − Q ( w ) z − w h, k (cid:19) = (cid:0) γ ( ¯ w ) + γ ( z ) h, k (cid:1) = [ γ ( z ) h, γ ( ¯ w ) k ] = [ f, g ] , f ∈ R z , g ∈ R ¯ w . Because ( K , [ ., . ]) is the Pontryagin space with negative index κ that satisfies (2.1) weconclude Q ∈ N κ ( H ).Because γ ( z )( H ) = R z , according to assumption (2.1), the minimality condition(1.2) is fulfilled. Then, according to Theorem 1.1, the representing relation, the statespace K , and the γ -field are uniquely determined (up to isomorphism). This proves (a)(i).(a)(ii) Here, we assume existence of − Q − , i.e. regularity of Q . Then, according to[12, Proposition 2.1], for Q ∈ N κ ( H ) with representing relation A , the inverse ˆ Q admitsrepresentation ˆ Q ( z ) = ˆ Q ( ¯ z ) + ( z − ¯ z ) ˆΓ + z (cid:18) I + ( z − z ) (cid:16) ˆ A − z (cid:17) − (cid:19) ˆΓ z , (2.2)5here z ∈ ρ ( A ) ∩ ρ (cid:16) ˆ A (cid:17) is an arbitrarily selected point of reference,ˆΓ z := − Γ z Q ( z ) − (2.3)and it holds (cid:16) ˆ A − z (cid:17) − = ( A − z ) − − Γ z Q ( z ) − Γ +¯ z , ∀ z ∈ ρ ( A ) ∩ ρ (cid:16) ˆ A (cid:17) . (2.4)According to [5, (2.3)], there exists a bijective correspondence between proper extensions˜ S ∈ Ext S and closed subspaces θ in H × H defined by S θ ∈ Ext S ⇔ θ := ˜Γ S θ = (cid:26)(cid:18) Γ ˆ f Γ ˆ f (cid:19) | ˆ f ∈ S θ (cid:27) ∈ ˜ C ( H ) . (2.5)Then the Krein formula( S θ − z ) − = ( A − z ) − + Γ z ( θ − Q ( z )) − Γ +¯ z (2.6)holds. Let us set S θ := ˆ A , where ˆ A is linear relation that represents the inverse functionˆ Q in representation (2.2). Then according to (2.4), the pair: ˆ A, θ = 0, satisfies (2.6).Because the correspondence defined by (2.5) is bijection it follows θ = ˜Γ ˆ A = (cid:26)(cid:18) Γ ˆ f (cid:19) | ˆ f ∈ ˆ A (cid:27) . (2.7)Therefore, ˆ A = ker Γ =: S . This proves (a)(ii).(b)(i) We assume now that Q ∈ N κ ( H ) is a regular function with representing relation A in representation (1.1). Therefore, the state space K is minimal i.e. (1.2) holds, ρ ( A ) = ∅ , and there exist the inverse ˆ Q represented by (2.2). We define closed symmetricrelation S by: S := A ∩ ˆ A Because representations (1.1) and (2.2) are uniquely determined up to isomorphism,linear relation S is uniquely determined too. This also means that A is a self-adjointextension of S , and we can apply [5, Proposition 2.2 (2)]. Therefore, there exists aboundary triple Π = ( H , Γ , Γ ) such that A = ker Γ . According to [5, Proposition2.3], there axists the Weyl function M ( z ) corresponding to the boundary triple Π =( H , Γ , Γ ). At this point all conditions of our statement (a) are fulfilled, and we canclaim M ( z ) = Q ( z ). Hence, according to (a)(i), the given generalized Nevanlinna function Q ∈ N κ ( H ) is the Weyl function corresponding to Π = ( H , Γ , Γ ). This proves (b)(i).(b)(ii) Because all conditions of part (a) of this proposition are satisfied, accordingto (a) (ii), it holds ˆ A = ker Γ . This proves (b)(ii).(c) According to [5, (2.4)] the extension ˆ A = S corresponds to operator B = 0.Because operator B is bounded, according to [5, Proposition 2.1 (2)] the relation ˆ A = S is transversal with the relation A = S . Therefore both claims S := A ∩ ˆ A and S + = A ˙+ ˆ A hold. (cid:3) Note, because wee deal with linear relation A , Proposition 2.1 (a)(i) is a generalizationof [5, Remark 2.2], and because we work with Pontryagin space K the Proposition 2.1(b)(i) is a generalization of [15, Theorem 2.2].The following, well known statements, now can be seen also as consequences of Propo-sition 2.1. Corollary 2.2
Let regular function Q ( z ) be the Weyl function associated with a sym-metric linear relation S in the Pontryagin space K , and let γ z be the γ -field associatedwith S . Then it holds: i) Gamma field associated with S is given by γ z = − γ z Q ( z ) − and resolvent is given by ( S − z ) − = ( S − z ) − + γ z γ +¯ z , z ∈ ρ ( S ) ∩ ρ ( S ) . (ii) z ∈ σ p (cid:16) ˆ A (cid:17) ⇔ ∈ σ p ( Q ( z )) ⇔ z is a generalized zero of Q .(iii) z ∈ ρ ( ˆ A ) ⇔ ∈ ρ ( Q ( z )) . Proof.
Because z ∈ ρ ( A ) ∩ ρ ( ˆ A ) was selected arbitrarily in (2.3), the claim about γ -field γ corresponding to S follows from (2.3) for z = z . Then the claim about resolventsfollows from (2.4). Well known claims (ii) and (iii) follow here from [5, Theorem 2.1],from claims (1) and (3) respectively, because ˆ A corresponds to θ = 0. (cid:3) Note, claim that follows from [5, Theorem 2.1 (2)]: z ∈ σ r ( ˆ A ) ⇔ ∈ σ r ( Q ( z )), doesnot bring any new information, because ˆ A is self-adjoint and, therefore σ r (cid:16) ˆ A (cid:17) = ∅ .2.2 In the next theorem we will deal with the matrix of a linear relation . Let K := K [+] K be a Pontryagin space with nontrivial Pontryagin subspaces K l , l = 1 ,
2, and orthogonalprojections E l : K → K l , l = 1 ,
2. For every linear relation in K = K [+] K , thefollowing four linear relations can be defined T ji := (cid:26)(cid:18) k i k ji (cid:19) : k i ∈ D ( T ) ∩ K i , k ji ∈ E j T ( k i ) (cid:27) ⊆ K i × K j , i, j = 1 , . In this notation the subscript “ i ” is associated with the domain subspace K i , the super-script “ j ” is associated with the range subspace K j . For example (cid:18) k k (cid:19) ∈ T . We willuse “[ ∗ ]” to denote adjoint relations of T ji . Therefore T ⊆ K × K ⇒ T
21 [ ∗ ] ⊆ K × K . To every linear relation T and decomposition K := K [+] K , we can assign the following relation matrix (cid:18) T T T T (cid:19) . Then it holds T = (cid:0) T + T (cid:1) ˙+ (cid:0) T + T (cid:1) , where “+” stands for operator-like addition, and “ ˙+” stands for the sum of subspaceswhich may or may not be direct. Lemma 2.3 . Let Q ∈ N κ ( H ) be given by (1.4) Q ( z ) = Γ + ( A − z ) − Γ , z ∈ D ( Q ) := ρ ( A ) , with bounded operator A and boundedly invertible Γ + Γ . Let us define linear relation B := A | ( I − P ) K ˙+ ( { } × R (Γ)) ⊆ ( I − P ) K × K . (2.8) Then z ∈ D ( Q ) ∩ D (cid:16) ˆ Q (cid:17) ⇒ z ∈ ρ ( B ) and it holds K ⊆ ( B − z ) ( I − P ) K . (2.9)7 roof. It is sufficient to prove (2.9).Assume z ∈ D ( Q ) ∩ D (cid:16) ˆ Q (cid:17) . Then, according to (2.2) and Theorem 1.5, z ∈ ρ (cid:16) ˆ A (cid:17) ifand only if z ∈ ρ (cid:16) ˜ A (cid:17) = ρ (cid:0) ( I − P ) A | ( I − P ) K (cid:1) . Therefore, for any f = ( I − P ) f + P f ∈ K there exists g ∈ ( I − P ) K , such that( I − P ) f = (cid:16) ( I − P ) A | ( I − P ) K − z ( I − P ) (cid:17) g. In addition, there obviously exists h ∈ H such that it holds:Γ h = P f − P A | ( I − P ) K ( g ) ⇒ P f = P A | ( I − P ) K ( g ) + Γ h. We will also use the identity ( I − P ) A | ( I − P ) K + P A | ( I − P ) K = A | ( I − P ) K Now we have, f = ( I − P ) f + P f = (cid:16) ( I − P ) A | ( I − P ) K − z ( I − P ) (cid:17) g + P A | ( I − P ) K ( g ) + Γ h = (cid:0) A | ( I − P ) K − z ( I − P ) (cid:1) g + Γ h ∈ ( B − ( I − P ) z ) g ⊂ ( B − z ) ( I − P ) K which proves this lemma. (cid:3) Lemma 2.4
Let Q ∈ N κ ( H ) be given by (1.4) Q ( z ) = Γ + ( A − z ) − Γ , z ∈ D ( Q ) , with bounded operator A and boundedly invertible Γ + Γ . Then the representing relation ˆ A , of ˆ Q := − Q − , satisfies ˆ A = A | ( I − P ) K ˙+ ˆ A ∞ , (2.10) where the sum ˙+ is not necessarily direct, and ˆ A ∞ := ( { } × R (Γ)) . (2.11) Proof . Because Γ + Γ is boundedly invertible, according to Lemma 1.4 scalar product [ ., . ]does not degenerate on the subspace P ( K ) = Γ( H ), where P denotes orthogonal pro-jection defined by (1.7). According to Theorem 1.5, there exists ˆ Q ( z ) := − Q ( z ) − , z ∈D ( Q ) ∩ D ( ˆ Q ). Let ˆ Q be represented by self-adjoint linear relation ˆ A in representation(2.2). Then ˆ A satisfies (2.4).Let us now observe linear relation B given by (2.8), and let us find resolvent of B ,denoted by R ( z ), which exists according to Lemma 2.3. Let us select a point z ∈ ρ ( B )and a vector f ∈ K = ( B − zI ) ( I − P ) K and let us find R ( z ) f := ( B − z ) − f. According to (2.9) there exists an element g := R ( z ) f ∈ ( I − P ) K = ker Γ + , and itholds { g, f + zg } ∈ A | ( I − P ) K ˙+ ( { } × R (Γ)) . This means that for some h ∈ H it holds f + zg = Ag + Γ h . h := − h we have Ag − zg = f + Γ h. Hence, g = ( A − z ) − f + ( A − z ) − Γ h. Since g ∈ ( I − P ) K = R (Γ) [ ⊥ ] = ker Γ + it holds0 = Γ + g = Γ + ( A − z ) − f + Γ + ( A − z ) − Γ h = Γ + ( A − z ) − f + Q ( z ) h. According to (1.6) it holds Γ +¯ z f = Γ + ( A − z ) − f. Therefore, h = − Q ( z ) − Γ +¯ z f. This gives R ( z ) f = g = ( A − z ) − f − Γ z Q ( z ) − Γ +¯ z f, which proves that formula (2.4) holds for linear relation B ⊆ K × K defined by (2.8).Therefore, ( B − z ) − = (cid:16) ˆ A − z (cid:17) − , andˆ A = B = A | ( I − P ) K ˙+ ( { } × R (Γ)) . (2.12)According to definition (2.11) of ˆ A ∞ , formula (2.10) holds for ˆ A . (cid:3) Remark 2.5
Identity (2.10) corresponds to identity [8, (3.5)]. The proof of Lemma 2.4is a repetition of the proof given in [8, (3.5)] for one-dimensional case. However, theproof of (2.10) is not complete without the statement of Lemma 2.3. Our Lemma 2.3completes also the proof of [8, (3.5)].
The following theorem answers question Q3 for an important case.
Theorem 2.6
Let Q ∈ N κ ( H ) be given by (1.4) Q ( z ) = Γ + ( A − z ) − Γ , z ∈ D ( Q ) , with bounded operator A and boundedly invertible Γ + Γ .Then:(i) There exists a closed symmetric relation S in the Pontryagin space K and BVS Π = ( H , Γ , Γ ) such that Q is the Weyl function associated to Π .(ii) Linear relation S is a bounded operator which is not densely defined in the Pon-tryagin space ( K [ ., . ]) .(iii) S = ( I − P ) A | ( I − P ) K , R = Γ ( H ) = P ( K ) , ˆ A ∞ = ˆ R , where P is projectiondefined by (1.7).(iv) S = A = (cid:18) S ( I − P ) A | P K P A | ( I − P ) K P A | P K (cid:19) , (v) S = ˆ A = S ˙[+] ˆ R , i.e. the relation matrix of S is (cid:18) ( I − P ) A | ( I − P ) K
00 ˆ A ∞ (cid:19) = (cid:18) S
00 ˆ R (cid:19) . (vi) The relation matrix of S + is (cid:18) S ( I − P ) A | P K P A | ( I − P ) K P A | P K ˙+ ˆ R (cid:19) . vii) F Π = (cid:26)(cid:18) Γ ˆ f (cid:19) | ˆ f ∈ ˆ A (cid:27) = θ . Therefore θ := ˜Γ ˆ A is not Π -admissible, F Π ⊆H × H is closed relation, and both R ( F Π ) and D ( F Π ) are closed subsets of H . Proof .(i) According to Theorm 1.5 (ii) function Q is regular. Then according to Proposi-tion 2.1 (b), there exists a closed symmetric relation S in the Pontryagin space K andboundary triple Π = ( H , Γ , Γ ) such that Q is the Weyl function corresponding to Π.(ii) Recall, A = S and Γ z = γ z . According to Proposition 2.1 (c), S ( S = A ( S + .Hence, S must be bounded symmetric operator . This proves the first claim of (ii).The second claim of (ii) will follow from (vi), where we will see that S + is multivaluedrelation.(iii), (iv), (v) Let us again observe decomposition (1.8): K [+] K := ( I − P ) K [+] P K Relation matrix of the operator S = A , with respect to this decomposition, is A = (cid:18) ( I − P ) A | ( I − P ) K ( I − P ) A | P K P A | ( I − P ) K P A | P K (cid:19) . (2.13)Let the relation matrix of S = ˆ A be ˆ A ˆ A ˆ A ˆ A ! , where ˆ A ji ⊆ K i × K j , i, j = 1 ,
2. According to Lemma 2.4 we haveˆ A (0) = R (Γ) = R ( P ) = K . (2.14)Therefore, ˆ A (0) is non-degenerate, i.e. it is ortho-complemented subspace of K . (Let usnote here that (2.14) was already proven in [4, Proposition 5]. However, we can not omitLemma 2.4 from the paper because we will also need equality (2.10).)From (2.14), according to [14, Theorem 2.4] it holdsˆ A = ˆ A s ˙[+] ˆ A ∞ , (2.15)where ˆ A s is a self-adjoint densely defined operator in ˆ A (0) [ ⊥ ] = R (Γ) [ ⊥ ] = ( I − P ) K , R (cid:16) ˆ A s (cid:17) ⊆ ( I − P ) K , and ˙[+] denotes direct and orthogonal sum of linear relations.For g ∈ ( I − P ) K , from (2.10) and (2.15) it follows (cid:0) ( I − P ) A | ( I − P ) K ( g ) [+] P A | ( I − P ) K ( g ) (cid:1) + Γ ( h ) = A s ( g ) [+] Γ ( h )for some h , h ∈ H . Obviously, there exists h ∈ H such that it holds:Γ ( h ) := P A | ( I − P ) K ( g ) + Γ ( h ) . It follows ( I − P ) A | ( I − P ) K ( g ) [+] Γ ( h ) = A s ( g ) [+] Γ ( h ) , ∀ g ∈ ( I − P ) K . Because this sum is direct and orthogonal, it follows Γ ( h ) = Γ ( h ) and more importantly A s ( g ) = ( I − P ) A | ( I − P ) K ( g ) . (2.16)10et us now prove that ˆ A ∞ = ˆ A . From (2.15) it easily follows ˆ A ∞ = ˆ A [ ∗ ] ∞ . If weassume ˆ A ∞ = ˆ A = (cid:26)(cid:18) P f (cid:19) | f ∈ K (cid:27) ⊆ ( I − P ) K × P K , then it would be ˆ A [ ∗ ] ∞ = (cid:26)(cid:18) I − P ) g (cid:19) | g ∈ K (cid:27) ⊆ P K × ( I − P ) K . Hence, ˆ A ∞ would not be self-adjoint. Therefore, it must be ˆ A ∞ = ˆ A ⊆ P K × P K .This means that the relation matrix of ˆ A isˆ A = (cid:18) ( I − P ) A | ( I − P ) K
00 ˆ A ∞ (cid:19) . (2.17)This can be written as a direct and orthogonal sum of linear relations, i.e.ˆ A = ( I − P ) A | ( I − P ) K ˙[+] ˆ A ∞ . Because
P AP ∩ A ∞ = 0, from (2.13) and (2.17), according to Proposition 2.1 (c)(i), itfollows S = A ∩ ˆ A = ( I − P ) A | ( I − P ) K . In addition, it follows D ( S ) = ( I − P ) K . Now, according to (1.9) we have Γ( H ) = R andˆ A ∞ = ˆ R . This completes proof of (iii).Now, statement (iv) follows directly from (iii) and (2.13). Statement (v) followsdirectly from (iii) and (2.17).(vi) According to Proposition 2.1 (c)(ii), S + = S ˙+ S = A ˙+ ˆ A . Then from S ˙+ S = S it follows S + = (cid:18) S ( I − P ) A | P K P A | ( I − P ) K P A | P K ˙+ ˆ A ∞ (cid:19) . Because of ˆ A ∞ ⊆ S + , S + is a multivalued relation. Therefore, linear relation S cannotbe densely defined in K , which proves the second claim in (ii).(vii) From ˆ A ∞ = (cid:26)(cid:18) h (cid:19) | h ∈ H (cid:27) = ˆ R it follows by definition that ˜Γ ˆ A ∞ is aforbidden relation. On the other hand, according to (2.7) we know θ = ˜Γ ˆ A = (cid:26)(cid:18) Γ ˆ f (cid:19) | ˆ f ∈ ˆ A (cid:27) . According to Proposition 2.1 (c)(i), S ⊆ ker ˜Γ. Now, according to (2.17) we have˜Γ ˆ A = ˜Γ S = ˜Γ (cid:16) S ˙[+] ˆ R (cid:17) = ˜Γ (cid:16) ˆ R (cid:17) = F Π = (cid:26)(cid:18) Γ ˆ f (cid:19) | ˆ f ∈ S (cid:27) . According to [5, (2.3)], θ = ˜Γ ˆ A is closed relation. Therefore F Π = θ is closed. Theremaining claims of (vii) are now obvious. (cid:3) Recall that an extension ˜ S ∈ Ext S is regular if ˜ S ˙+ ˆ R is a closed linear relation in K × K , see [5, Definition 3.1].By means of the above results, we prove the following corollary, where we learn that A = S and ˆ A = S are regular extensions of the operator S . Corollary 2.7
Let Q ∈ N κ ( H ) satisfies conditions of Theorem 2.6. Then: i) Operator A ( I − P ) is regular extensions of S .(ii) ˆ A is regular extensions of S .(iii) A is regular extensions of S . Proof. (i) Note that I − P is identity operator in the domain of ˆ A . Therefore, from (2.10) itfollows B := A ( I − P ) ˙+ ˆ R = ( ˆ A ( f ) , f ∈ ( I − P ) K , , f ∈ P K , Obviously, relation B is closed, because ˆ A is closed. By definition of regular extension,claim (i) follows.(ii) From ˆ A = S [ ∔ ] ˆ R and ˆ R ∔ ˆ R = ˆ R it follow ˆ A ∔ ˆ R = ˆ A Hence, ˆ A ∔ ˆ R is closed.Then, by definition ˆ A is regular extension of S .(iii) According to [5, (2.3)], θ = ˆΓ ˆ A is closed. According to Theorem 2.6 (vii) F Π = θ is closed relation. According to (2.7), D ( F Π ) is obviously closed. Then according to [5,Corollary 3.2 (1)], A is regular extension od S . (cid:3) The following statement is more specific than the statement that would follow from[5, Proposition 3.2] for functions with boundadly invertible Q ′ ( ∞ ). Corollary 2.8
Let Q ∈ N κ ( H ) be a Weyl function of S corresponding to BVS Π =( H , Γ , Γ ) represented by A = S .(i) Q is holomorphic at ∞ with boundedly invertible Q ′ ( ∞ ) = lim z →∞ zQ ( z ) if and onlyif ˆ Q ( z ) = ˜Γ + ( S − z ) − ˜Γ + ˆ F + ˆ Gz, z ∈ D ( Q ) ∩ D (cid:16) ˆ Q (cid:17) , where S is a self-adjoint bounded operator in the Pontryagin space R [ ⊥ ] = R (Γ) [ ⊥ ] =( I − P ) K , ˆ F and ˆ G are self-adjoint bounded operators in the Hilbert space H , and ˆ G is boundedly invertible.(ii) In that case function Q ∈ N κ ( H ) is regular function. Proof . Both statements of the corollary follow immediately form Theorem 1.5 and The-orem 2.6 (ii). (cid:3) l = − d dx with singular point at infinity In the notation used so fare S i := ker Γ i , i = 0 ,
1. Hence, in that notation S = ˆ A corresponds to relation (operator) θ = 0, according to Proposition 2.1. In the followingexample, self-adjoint operator θ is multiplication by scalar θ ∈ R . For that reason wewill occasionally use notation from [6, p. 188] which is more intuitive with respect toparametrization of extensions. In that notation, L := S , L + := S + , extension L := ˆ A = S corresponds to θ = 0, and extension L ∞ := A = S corresponds to θ = ∞ .In the following example wee will show how to use results presented in Proposition2.1 to find the solutions of the equation − y ′′ − zy = g ( x ) , (3.1)satisfying boundary conditions that correspond to the following domains D ( A ) , D ( ˆ A ) , D ( S + ) = D ( A ) ∔ D ( ˆ A ) , D ( S ) = D ( A ) ∩ D ( ˆ A ) . Example 3.1
Let the differential expression be l := − d dx on (0 , ∞ ) and let L be theminimal operator associated with l in L (0 , ∞ ) . Let D + denote the class of all functions g ( x ) ∈ L (0 , ∞ ) for which g ( x ) , g ′ ( x ) are absolutely continuous and g ′′ ( x ) ∈ L (0 , ∞ ) . • Use extensions A and ˆ A , and Proposition 2.1 (c) to find solutions of equation (3.1)that belong to each of domains D ( S + ) , D ( S ) . Also, find analytic expressions offunctions g ∈ R ( S − z ) , z ∈ ˆ ρ ( S ) . • Find gamma filed of the inverse function ˆ Q by meas of Corollary 2.2 (i). It is well known that minimal operator L = S is densely defined on L (0 , ∞ ) andthat it holds D ( L + ) = D + . The deficiency indices of the operator L are (1 , L + is introduced by H := C , Γ ˆ f := f (0) , Γ ˆ f := f ′ (0) , (cid:18) ˆ f = (cid:18) ff ′ (cid:19) ∈ L + (cid:19) . Let us select the fundamental system u ( x ) , v ( x ) of solutions of the homogeneous equation(3.1) that satisfy boundary conditions u (0 , z ) = 1 , v (0 , z ) = 0 ,u ′ (0 , z ) = 0 , v ′ (0 , z ) = 1 . Let z ∈ C \ R be a fixed number. It is easy to verify that the fundamental system u ( x ) , v ( x ) is given by: u ( x, z ) = cos ( √ zx ) , v ( x, z ) = 1 √ z sin (cid:0) √ zx (cid:1) , where we choose ( −∞ ,
0] as the branch cut for the square root function √ z . Then theWeyl solution is f z ( x ) = cos (cid:0) √ zx (cid:1) + Q ( z ) 1 √ z sin (cid:0) √ zx (cid:1) , where the function Q ( z ) is such that it holds f z ( x ) ∈ L (0 , ∞ ), see [13, p. 18].Obviously Γ ˆ f z = f z (0) = 1 , Γ ˆ f z = f ′ z (0) = Q ( z ) Γ ˆ f z . According to the definition[5, (2.12)], Q ( z ) is the Weyl function. Let us recall that parametrization of self-adjointextensions L θ of L is given by: • For θ ∈ R , D ( L θ ) = n f ∈ D + : f ′ (0) = θf (0) o . • D ( L ∞ ) := { f ∈ D + : f (0) = 0 } , i.e. L ∞ = A := ker Γ is the representing opera-tor of Q ( z ).According to Proposition 2.1, the extension ˆ A of L = S is the representing operator ofˆ Q ( z ) = − Q ( z ) . Recall D (cid:16) ˆ A (cid:17) = n f ∈ D + : f ′ (0) = 0 o .From the requirement f z ∈ L (0 , ∞ ) for Weyl solution we get expression for Q ( z ): Q ( z ) = ( i √ z, z ∈ C + , − i √ z, z ∈ C − , Q ( z ) = ( i ( √ z ) − , z ∈ C + , − i ( √ z ) − , z ∈ C − , According to Proposition 2.1 (c)(ii) we can find D ( S + ) = D ( L + ) in terms of D ( A ) and D ( ˆ A ). For simplicity, let us assume z ∈ C + , which is not loss of generality. Then f z ( x ) = cos (cid:0) √ zx (cid:1) + i √ z √ z sin (cid:0) √ zx (cid:1) = e i √ zx . Obviously, f z (0) = 1 , f ′ z (0) = i √ z .Now, we will use formula [13, (2.5)] to find solutions of equation (3.1), correspondingto extensions A = L ∞ and ˆ A = L . In [13], author uses notation with parameter s = θ ,while L θ -notation is in [5] and here. We will simply denote the solution of (3.1) that issquare integrable near zero by y θ ( x ) := c √ z sin (cid:0) √ zx (cid:1) + c cos (cid:0) √ zx (cid:1) and the solution which is square integrable near infinity, the Weyl solution, by f z ( x ).When θ = ∞ , then we deal with the extension A , and then y ∞ (0) = 0, i.e. c = 0. Hence,the solutions which are square integrable near zero, near infinity, and the correspondingWronskian, that we will substitute into [13, (2.5)], are respectively: y ∞ ( x ) = c √ z sin (cid:0) √ zx (cid:1) , f z ( x ) = e i √ zx , W ( f z , y ∞ ) = c . Then for every g ∈ L (0 , ∞ ) there exists f g ( x ) := (cid:16) ( A − z ) − g (cid:17) ( x ) =1 √ z (cid:20) e i √ zx Z x sin (cid:0) √ zx (cid:1) g ( x ) dx + sin (cid:0) √ zx (cid:1) Z ∞ x e i √ zx g ( x ) dx (cid:21) ∈ D ( A ) . Similarly, for θ = 0 we deal with ˆ A , hence y (0) = 1, i.e. c = 1. Then the squareintegrable functions near zero, near infinity, and the corresponding Wronskian are y ( x ) = cos (cid:0) √ zx (cid:1) , f z ( x ) = e i √ zx , W ( f z , y ) = − i √ z respectively. According to [13, (2.5)], for every g ∈ L (0 , ∞ ) we have ϕ g ( x ) := (cid:18)(cid:16) ˆ A − z (cid:17) − g (cid:19) ( x ) = i √ z (cid:20) e i √ zx Z x cos (cid:0) √ zx (cid:1) g ( x ) dx + cos (cid:0) √ zx (cid:1) Z ∞ x e i √ zx g ( x ) dx (cid:21) ∈ D (cid:16) ˆ A (cid:17) . It is easy to verify that functions f g , ϕ g indeed satisfy conditions f g (0) = 0 , ϕ ′ g (0) = 0i.e. f g ∈ D ( A ) , ϕ g ∈ D (cid:16) ˆ A (cid:17) , ∀ g ∈ L (0 , ∞ ) . According to Proposition 2.1 (c)(ii), for every g ∈ L (0 , ∞ ), a particular solutions of(3.1) can be expressed in the form y ( x ) = 12 ( f g ( x ) + ϕ g ( x )) ∈ D ( A ) + D (cid:16) ˆ A (cid:17) . f g , ϕ g into this expression gives y ( x ) = i √ z (cid:20) e i √ zx Z x e − i √ zx g ( x ) dx + e − i √ zx Z ∞ x e i √ zx g ( x ) dx (cid:21) ∈ D (cid:0) S + (cid:1) , (3.2)for all g ∈ L (0 , ∞ ). Hence, formula (3.2) gives a particular solution of the equation (cid:0) S + − zI (cid:1) y = g. We will now find expression for functions that belong to D ( S ) = D ( L ). According to(3.2) y ′ ( x ) = − (cid:20) e i √ zx Z x e − i √ zx g ( x ) dx − e − i √ zx Z ∞ x e i √ zx g ( x ) dx (cid:21) . (3.3)According to Proposition 2.1 (c)(i) y ∈ D ( S ) ⇔ (cid:16) y (0) = 0 ∧ y ′ (0) = 0 (cid:17) . According to (3.2) and (3.3) this is further equivalent to Z ∞ e i √ zx g ( x ) dx = 0 . (3.4)Hence, D ( S ) = D ( L ) consists of the functions y ∈ D ( S + ) given by (3.2) such that ¯ g ( x )is orthogonal to Weyl solutions f z ( x ) = e i √ zx . It is easy to see that functions g ( x ) areof the form g ( z, k, n ; x ) = ( e i ( −√ z +2 kπ ) x , x ∈ [0 , n ] , , x ∈ ( n, ∞ ) , (3.5)where z ∈ C ; k, n ∈ N ∪ { } , i.e. they satisfy condition (3.4). Obviously, functions fromthe closed linear span in L (0 , ∞ ) of functions given by (3.5) satisfy condition (3.4). Inother words:Equation (3.1) has a solution that satisfies boundary conditions y (0) = y ′ (0) = 0 ifand only if the function g ( x ) on the right-hand side of (3.1) belongs to the closed linearspan of functions (3.5). In that case g ∈ R ( S − z ) and the solution y ∈ D ( S + ) is givenby (3.2).At the end let us find gamma filed γ z corresponding to ˆ A . According to Corollary2.2, we have γ z ( h ) := − γ z (cid:16) Q ( z ) − ( h ) (cid:17) , ∀ h ∈ H . 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