The strong matrix Stieltjes moment problem
aa r X i v : . [ m a t h . F A ] J un The strong matrix Stieltjes moment problem.
A.E. Choque Rivero, S.M. Zagorodnyuk
In this paper we analyze the following problem: to find a non-decreasingmatrix function M ( x ) = ( m k,l ( x )) N − k,l =0 , on R + = [0 , + ∞ ), M (0) = 0, whichis left-continuous on (0 , + ∞ ), and such that Z R + x n dM ( x ) = S n , n ∈ Z , (1)where { S n } n ∈ Z is a prescribed sequence of Hermitian ( N × N ) complexmatrices (moments), N ∈ N . This problem is said to be a strong matrixStieltjes moment problem . The problem is said to be determinate ifit has a unique solution and indeterminate in the opposite case.The scalar ( N = 1) strong Stieltjes moment problem (in a slightly differ-ent setting) was introduced in 1980 by Jones, Thron and Waadeland [1].Necessary and sufficient conditions for the existence of a solution with aninfinite number of points of increase and for the uniqueness of such a solutionwere established [1, Theorem 6.3]. Also necessary and sufficient conditionsfor the existence of a unique solution with a finite number of points of in-crease were obtained [1, Theorem 5.2]. The approach of Jones, Thron andWaadeland’s investigation was made through the study of special continuedfractions related to the moments.In 1995, Nj˚astad described some classes of solutions of the scalar strongStieltjes moment problem [2],[3]. He used properties of some associatedLaurent polynomials.In 1996, Kats and Nudelman obtained necessary and sufficient conditions forthe existence of a solution of the scalar strong Stieltjes moment problem [4,Theorem 1.1]. The degenerate case was studied in full: in this case thesolution is unique, given explicitly and it has a finite number of pointsof increase. In the non-degenerate case, conditions for the determinacywere given and the unique solution was presented. In the (non-degenerate)indeterminate case a Nevanlinna-type parameterization for all solutions ofthe scalar strong Stieltjes moment problem was obtained [4, Theorem 4.1].Canonical solutions and Weyl-type lunes were studied, as well. Kats andNudelman used the results of Krein on the semi-infinite string theory.Various other results on the scalar strong Stieltjes moment problem can befound in papers [5],[6],[7],[8],[9] (see also References therein).1he moment problem (1) where the half-axis R + is replaced by the wholeaxis R is said to be the strong matrix Hamburger moment problem .The scalar ( N = 1) strong matrix Hamburger moment problem has beenintensively studied since 1980-th, see a survey [7], a recent paper [10] andReferences therein. For the matrix case, see papers [11],[12] and papers citedthere.The aim of our present investigation is threefold. Firstly, we obtainnecessary and sufficient conditions for the solvability of the strong matrixStieltjes moment problem (1). Consider the following block matrices con-structed by moments:Γ n = ( S i + j ) ni,j = − n = S − n . . . S − n . . . S ... ... ... S − n . . . S . . . S n ... ... ... S . . . S n . . . S n , (2) e Γ n = ( S i + j +1 ) ni,j = − n = S − n +1 . . . S − n +1 . . . S ... ... ... S − n +1 . . . S . . . S n +1 ... ... ... S . . . S n +1 . . . S n +1 , n ∈ Z . (3)We shall prove that conditionsΓ n ≥ , e Γ n ≥ n = 0 , , , ..., (4)are necessary and sufficient for the solvability of the moment problem (1).Secondly, we obtain an analytic description of all solutions of the momentproblem (1). We shall use an abstract operator approach similar to the ”pureoperator” approach of Sz¨okefalvi-Nagy and Koranyi to the Nevanlinna-Pickinterpolation problem, see [13],[14]. We shall need some properties of gener-alized Π-resolvents of non-negative operators and generalized sc -resolventsof Hermitian contractions, established by Krein and Ovcharenko in [15],[16].As a by-product, we present a description of generalized Π-resolvents of anon-negative operator which does not use improper elements or relations asit was done in the original work of Krein [17] and in the paper of Derkachand Malamud [18]. Here we adapt some ideas from [19] of Chumakin whostudied generalized resolvents of isometric operators.2hirdly, we obtain necessary and sufficient conditions for the strong ma-trix Stieltjes moment problem to be determinate. Notations.
As usual, we denote by R , C , N , Z , Z + , the sets of realnumbers, complex numbers, positive integers, integers and non-negativeintegers, respectively. The space of n -dimensional complex vectors a =( a , a , . . . , a n − ), we denote by C n , n ∈ N . If a ∈ C n , then a ∗ means thecomplex conjugate vector. By P L we denote the space of all complex Laurentpolynomials, i.e. functions P bk = a α k x k , a, b ∈ Z : a ≤ b , α k ∈ C .Let M ( x ) be a left-continuous non-decreasing matrix function M ( x ) =( m k,l ( x )) N − k,l =0 on R , M ( −∞ ) = 0, and τ M ( x ) := P N − k =0 m k,k ( x ); Ψ( x ) =( dm k,l /dτ M ) N − k,l =0 . By L ( M ) we denote a set (of classes of equivalence) ofvector-valued functions f : R → C N , f = ( f , f , . . . , f N − ), such that (see,e.g., [20]) k f k L ( M ) := Z R f ( x )Ψ( x ) f ∗ ( x ) dτ M ( x ) < ∞ . The space L ( M ) is a Hilbert space with a scalar product( f, g ) L ( M ) := Z R f ( x )Ψ( x ) g ∗ ( x ) dτ M ( x ) , f, g ∈ L ( M ) . We denote ~e k = ( δ ,k , δ ,k , ..., δ N − ,k ), 0 ≤ k ≤ N −
1, where δ j,k is theKronecker delta.If H is a Hilbert space then ( · , · ) H and k · k H mean the scalar productand the norm in H , respectively. Indices may be omitted in obvious cases.For a linear operator A in H , we denote by D ( A ) its domain, by R ( A ) itsrange, by Ker A its kernel, and A ∗ means the adjoint operator if it exists.If A is invertible then A − means its inverse. A means the closure of theoperator, if the operator is closable. If A is self-adjoint, by R z ( A ) we denotethe resolvent of A , z ∈ C \ R . If A is bounded then k A k denotes its norm.For an arbitrary set of elements { x n } n ∈ Z in H , we denote by Lin { x n } n ∈ Z and span { x n } n ∈ Z the linear span and the closed linear span (in the norm of H ), respectively. For a set M ⊆ H we denote by M the closure of M in thenorm of H . By E H we denote the identity operator in H , i.e. E H x = x , x ∈ H . If H is a subspace of H , then P H = P HH is an operator of theorthogonal projection on H in H . By [ H ] we denote a set of all boundedlinear operators A in H , D ( A ) = H . 3 The solvability of the strong matrix Stieltjes mo-ment problem.
In this section we are going to establish the following theorem.
Theorem 1.
Let the strong matrix Stieltjes moment problem (1) with a setof moments { S n } n ∈ Z be given. The moment problem has a solution if andonly if conditions (4) are satisfied.Proof. Necessity. Let the strong matrix Stieltjes moment problem (1) has asolution M ( x ). Choose an arbitrary vector function f ( x ) = P nk = − n P N − j =0 f j,k x k ~e j , f j,k ∈ C . This function belongs to L ( M ) and0 ≤ Z R f ( x ) x s dM ( x ) f ∗ ( x ) = n X k,r = − n N − X j,l =0 f j,k f l,r Z R + x k + r + s ~e j dM ( x ) ~e ∗ l = n X k,r = − n N − X j,l =0 f j,k ~e j S k + r + s f l,r ~e ∗ l = n X k,r = − n ( f ,k , f ,k , ..., f N − ,k ) S k + r + s ∗ ( f ,r , f ,r , ..., f N − ,r ) ∗ = (cid:26) v Γ n v ∗ , s = 0 v e Γ n v ∗ , s = 1 , where v = ( f , − n , f , − n , ..., f N − , − n , f , − n +1 , f , − n +1 , ..., f N − , − n +1 , ...,f ,n , f ,n , ..., f N − ,n ). Here we make use of the rules for the multiplicationof block matrices. Sufficiency.
Let the strong matrix Stieltjes moment problem (1) begiven and conditions (4) be satisfied. Let S j = ( S j ; k,l ) N − k,l =0 , S j ; k,l ∈ C , j ∈ Z . Consider the following infinite block matrix:Γ = ( S i + j ) ∞ i,j = −∞ = ... ... ... . . . S − n . . . S − n . . . S . . .. . . ... ... ... . . .. . . S − n . . . S . . . S n . . .. . . ... ... ... . . .. . . S . . . S n . . . S n . . . ... ... ... , (5)where the element in the box corresponds to the indices i = j = 0.We assume that the left upper entry of the element in the box stands inrow 0, column 0. Let us numerate rows (columns) in the increasing order to4he bottom (respectively to the right). Then we numerate rows (columns) inthe decreasing order to the top (respectively to the left). Thus, the matrixΓ may be viewed as a numerical matrix: Γ = ( γ k,l ) ∞ k,l = −∞ , γ k,l ∈ C . Observethat the following equalities hold γ rN + j,tN + n = S r + t ; j,n , r, t ∈ Z , ≤ j, n ≤ N − . (6)From conditions (4) it easily follows that( γ k,l ) rk,l = − r ≥ , ( γ k + N,l ) rk,l = − r ≥ , ∀ r ∈ Z + . (7)The first inequality in the latter relation implies that there exist a Hilbertspace H and a set of elements { x n } n ∈ Z in H such that( x n , x m ) H = γ n,m , n, m ∈ Z , (8)and span { x n } n ∈ Z = H , see Lemma in [14, p. 177]. The latter fact is wellknown and goes back to a paper of Gelfand, Naimark [21].By (6) we get γ a + N,b = γ a,b + N , a, b ∈ Z . (9)Set L = Lin { x n } n ∈ Z . Choose an arbitrary element x ∈ L . Let x = P ∞ k = −∞ α k x k , x = P ∞ k = −∞ β k x k , where α k , β k ∈ C . Here only a finitenumber of coefficients α k , β k are non-zero. In what follows, this will be as-sumed in analogous situations with elements of the linear span.
By (8),(9)we may write ∞ X k = −∞ α k x k + N , x l ! = ∞ X k = −∞ α k γ k + N,l = ∞ X k = −∞ α k γ k,l + N == ∞ X k = −∞ α k ( x k , x l + N ) = ∞ X k = −∞ α k x k , x l + N ! = ( x, x l + N ) , l ∈ Z . Similarly we conclude that (cid:0)P ∞ k = −∞ β k x k + N , x l (cid:1) = ( x, x l + N ), l ∈ Z . Since L = H , we get P ∞ k = −∞ α k x k + N = P ∞ k = −∞ β k x k + N .Set A x = ∞ X k =0 α k x k + N , x ∈ L, x = ∞ X k = −∞ α k x k , α k ∈ C . (10)The above considerations ensure us that the operator A is defined correctly.Choose arbitrary elements x, y ∈ L , x = P ∞ k = −∞ α k x k , y = P ∞ n = −∞ β n x n , α k , β n ∈ C , and write( A x, y ) H = ∞ X k = −∞ α k x k + N , ∞ X n = −∞ β n x n ! H = ∞ X k,n = −∞ α k β n ( x k + N , x n ) H =5 ∞ X k,n = −∞ α k β n ( x k , x n + N ) H = ∞ X k = −∞ α k x k , ∞ X n = −∞ β n x n + N ! H = ( x, A y ) H . Moreover, we have( A x, x ) H = ∞ X k,n = −∞ α k β n ( x k + N , x n ) H = ∞ X k,n = −∞ α k β n γ k + N,n ≥ . (11)Thus, the operator A is a non-negative symmetric operator in H . Set A = A . The operator A always has a non-negative self-adjoint extension e A in a Hilbert space e H ⊇ H [15, Theorem 7, p.450] . We may assumethat Ker e A = { } . In the opposite case, since Ker e A ⊥ R ( e A ), R ( e A ) ⊇ L , weconclude that Ker e A ⊥ H . Therefore the operator e A , restricted to e H ⊖ Ker e A ,also will be a self-adjoint extension of the operator A , with a null kernel.Let { e E λ } λ ∈ R be the left-continuous orthogonal resolution of unity of theoperator e A . By the induction argument it is easy to check that x rN + j = A r x j , r ∈ Z , ≤ j ≤ N − . By (6),(8) we may write S r ; j,n = γ rN + j,n = ( x rN + j , x n ) H = ( A r x j , x n ) H = ( e A r x j , x n ) e H = Z R + λ r d ( e E λ x j , x n ) e H = Z R + λ r d (cid:16) P e HH e E λ x j , x n (cid:17) H , ≤ n ≤ N − . Therefore we get S r = Z R + λ r d f M ( λ ) , r ∈ Z , (12)where f M ( λ ) := (cid:16)(cid:16) P e HH e E λ x j , x n (cid:17) H (cid:17) N − j,n =0 . Therefore the matrix function f M ( λ ) is a solution of the moment problem (1) (From the properties of theorthogonal resolution of unity it easily follows that f M ( λ ) is left-continuouson (0 , + ∞ ), non-decreasing and f M (0) = 0). Let A be an arbitrary closed Hermitian operator in a Hilbert space H , D ( A ) ⊆ H . Let b A be an arbitrary self-adjoint extension of A in a Hilbert6pace b H ⊇ H . Denote by { b E λ } λ ∈ R its orthogonal resolution of unity. Recallthat an operator-valued function R z = P b HH R z ( b A ) is said to be a generalizedresolvent of A , z ∈ C \ R . A function E λ = P b HH b E λ , λ ∈ R , is said to be a spectral function of A . There exists a bijective correspondence betweengeneralized resolvents and left-continuous (or normalized in some other way)spectral functions established by the following relation ([22]):( R z f, g ) H = Z R λ − z d ( E λ f, g ) H , f, g ∈ H, z ∈ C \ R . (13)If the operator A is densely defined symmetric and non-negative ( A ≥ b A is self-adjoint and non-negative, then the correspond-ing generalized resolvent R z and the spectral function E λ are said to be a generalized Π -resolvent and a Π -spectral function of A . Relation (13)establishes a bijective correspondence between generalized Π-resolvents andleft-continuous Π-spectral functions.If the operator A is a Hermitian contraction ( k A k ≤ b A is a self-adjoint contraction, then the corresponding generalized resolvent R z and the spectral function E λ are said to be a generalized sc -resolvent and a sc -spectral function of A . Relation (13) establishes a bijective corre-spondence between generalized sc -resolvents and left-continuous sc -spectralfunctions, as well.If a generalized Π-resolvent (a generalized sc -resolvent) is generated byan extension inside H , i.e. b H = H , then it is said to be a canonical Π -resolvent (respectively a canonical sc -resolvent ).Firstly, we shall obtain a description of solutions of the strong matrixStieltjes moment problem by virtue of Π-spectral functions. Theorem 2.
Let the strong matrix Stieltjes moment problem (1) be givenand condition (4) be satisfied. Suppose that the operator A = A in a Hilbertspace H is constructed for the moment problem by (10) and the precedingprocedure. All solutions of the moment problem have the following form M ( λ ) = ( m k,j ( λ )) N − k,j =0 , m k,j ( λ ) = ( E λ x k , x j ) H , (14) where E λ is a left-continuous Π -spectral function of the operator A . Onthe other hand, each left-continuous Π -spectral function of the operator A generates by (14) a solution of the moment problem. Moreover, the corre-spondence between all left-continuous Π -spectral functions of the operator A and all solutions of the moment problem is bijective. roof. Let E λ be an arbitrary Π-spectral function of the operator A . Itcorresponds to a self-adjoint operator e A ⊇ A in a Hilbert space e H ⊇ H .Then we repeat considerations after (11) to obtain that M ( λ ), given by (14),is a solution of the moment problem (1).Let c M ( x ) = ( b m k,l ( x )) N − k,l =0 be an arbitrary solution of the moment prob-lem (1). Consider the space L ( c M ) and let Q be the operator of multipli-cation by an independent variable in L ( c M ). A set (of classes of equiva-lence) of functions f ∈ L ( c M ) such that (the corresponding class includes) f = ( f , f , . . . , f N − ), f ∈ P L , we denote by P L ( c M ). Set L L ( c M ) := P L ( c M ).For an arbitrary vector Laurent polynomial f = ( f , f , . . . , f N − ), f j ∈ P L , there exists a unique representation of the following form: f ( x ) = N − X k =0 ∞ X j = −∞ α k,j x j ~e k , α k,j ∈ C , (15)where all but finite number of coefficients α k,j are zero. Choose anothervector Laurent polynomial g with a representation g ( x ) = N − X l =0 ∞ X r = −∞ β l,r x r ~e l , β l,r ∈ C . (16)We may write( f, g ) L ( c M ) = N − X k,l =0 ∞ X j,r = −∞ α k,j β l,r Z R + x j + r ~e k d c M ( x ) ~e ∗ l = N − X k,l =0 ∞ X j,r = −∞ α k,j β l,r Z R + x j + r d b m k,l ( x ) = N − X k,l =0 ∞ X j,r = −∞ α k,j β l,r S j + r ; k,l . (17)On the other hand, we have ∞ X j = −∞ N − X k =0 α k,j x jN + k , ∞ X r = −∞ N − X l =0 β l,r x rN + l H = N − X k,l =0 ∞ X j,r = −∞ α k,j β l,r ∗ ( x jN + k , x rN + l ) H = N − X k,l =0 ∞ X j,r = −∞ α k,j β l,r γ jN + k,rN + l = N − X k,l =0 ∞ X j,r = −∞ α k,j β l,r S j + r ; k,l . (18)8y (17),(18) we get( f, g ) L ( c M ) = ∞ X j = −∞ N − X k =0 α k,j x jN + k , ∞ X r = −∞ N − X l =0 β l,r x rN + l H . (19)Set V f = ∞ X j = −∞ N − X k =0 α k,j x jN + k , (20)for a vector Laurent polynomial f ( x ) = P N − k =0 P ∞ j = −∞ α k,j x j ~e k . If f , g are vector Laurent polynomials with representations (15),(16), such that k f − g k L ( c M ) = 0, then by (19) we may write k V f − V g k H = ( V ( f − g ) , V ( f − g )) H = ( f − g, f − g ) L ( c M ) = k f − g k L ( c M ) = 0 . Thus, V is correctly defined as an operator from P ( c M ) to H . Relation (19)shows that V is an isometric transformation from P L ( c M ) on L . We extend V by continuity to an isometric transformation from L L ( c M ) on H . Observethat V x j ~e k = x jN + k , j ∈ Z ; 0 ≤ k ≤ N − . (21)Let L ( c M ) := L ( c M ) ⊖ L L ( c M ), and U := V ⊕ E L ( c M ) . The operator U isan isometric transformation from L ( c M ) on H ⊕ L ( c M ) =: b H . Set b A := U QU − . The operator b A is a self-adjoint operator in b H . Let { b E λ } λ ∈ R be its left-continuous orthogonal resolution of unity. Notice that U QU − x jN + k = V QV − x jN + k = V Qx j ~e k = V x j +1 ~e k = x ( j +1) N + k == x jN + k + N = Ax jN + k , j ∈ Z ; 0 ≤ k ≤ N − . By linearity we get:
U QU − x = Ax , x ∈ L , and therefore b A ⊇ A . Choosean arbitrary z ∈ C \ R and write Z R + λ − z d ( b E λ x k , x j ) b H = (cid:18)Z R + λ − z d b E λ x k , x j (cid:19) b H = (cid:18) U − Z R + λ − z d b E λ x k , U − x j (cid:19) L ( c M ) (cid:18)Z R + λ − z dU − b E λ U~e k , ~e j (cid:19) L ( c M ) = (cid:18)Z R + λ − z dE λ ~e k , ~e j (cid:19) L ( c M ) = (cid:0) ( Q − z ) − ~e k , ~e j (cid:1) L ( c M ) = Z R + λ − z ~e k d c M ( λ ) ~e j = Z R + λ − z d b m k,j ( λ ) , where E λ is a left-continuous orthogonal resolution of unity of the operators Q . By the Stieltjes-Perron inversion formula (e.g. [23]) we conclude that b m k,j ( λ ) = ( P b HH b E λ x k , x j ) H , λ ∈ R . Thus, c M is generated by a Π-spectral function of A .Let us check that an arbitrary element u ∈ L can be represented in thefollowing form u = u z + u , u z ∈ H z , u ∈ L N , (22)where L N := Lin { x n } N − n =0 , H z := ( A − zE H ) D ( A ). Let u = P ∞ k = −∞ c k x k , c k ∈ C , and choose a number z ∈ C \ R . Suppose that c k = 0, if k ≤ r or k ≥ R , where r ≤ − R ≥ N + 1. Set d k := 0, if k ≤ r or k ≥ R − N . Thenwe set d k := 1 z ( d k − N − c k ) , k = r + 1 , ..., − d k − N := zd k + c k , k = R − , R − , ..., N. Set v := P ∞ k = −∞ d k x k ∈ L . Then we directly calculate that( A − zE H ) v − u = N − X k =0 ( d k − N − zd k − c k ) x k , and relation (22) holds. From the latter equality it easily follows that thedeficiency index of A is equal to ( n, n ), 0 ≤ n ≤ N .Let us check that different left-continuous Π-spectral functions of theoperator A generate different solutions of the moment problem (1). Supposethat two different left-continuous Π-spectral functions generate the samesolution of the moment problem. This means that there exist two self-adjointoperators A j ⊇ A , in Hilbert spaces H j ⊇ H , such that P H H E ,λ = P H H E ,λ ,and( P H H E ,λ x k , x j ) H = ( P H H E ,λ x k , x j ) H , ≤ k, j ≤ N − , λ ∈ R , where { E n,λ } λ ∈ R are orthogonal left-continuous resolutions of unity of op-erators A n , n = 1 ,
2. By the linearity we get( P H H E ,λ x, y ) H = ( P H H E ,λ x, y ) H , x, y ∈ L N , λ ∈ R . (23)10et R n,λ := P H n H R λ ( A n ), n = 1 ,
2. By (23),(13) we get( R ,λ x, y ) H = ( R ,λ x, y ) H , x, y ∈ L N , λ ∈ C \ R . (24)Since R z ( A j )( A − zE H ) x = ( A j − zE H j ) − ( A j − zE H j ) x = x, x ∈ L = D ( A ) , then R z ( A ) u = R z ( A ) u ∈ H , u ∈ H z ; R ,z u = R ,z u, u ∈ H z , z ∈ C \ R . (25)We may write( R n,z x, u ) H = ( R z ( A n ) x, u ) H n = ( x, R z ( A n ) u ) H n = ( x, R n,z u ) H , where x ∈ L N , u ∈ H z , n = 1 ,
2, and therefore( R ,z x, u ) H = ( R ,z x, u ) H , x ∈ L N , u ∈ H z . (26)By (22) an arbitrary element y ∈ L can be represented in the following form y = y z + y ′ , y z ∈ H z , y ′ ∈ L N . Using (24) and (26) we obtain( R ,z x, y ) H = ( R ,z x, y z + y ′ ) H = ( R ,z x, y z + y ′ ) H = ( R ,z x, y ) H , where x ∈ L N , y ∈ L . Since L = H , we obtain R ,z x = R ,z x, x ∈ L N , z ∈ C \ R . (27)For arbitrary x ∈ L , x = x z + x ′ , x z ∈ H z , x ′ ∈ L N , using relations (25),(27)we get R ,z x = R ,z ( x z + x ′ ) = R ,z ( x z + x ′ ) = R ,z x, x ∈ L, z ∈ C \ R , and therefore R ,z = R ,z , z ∈ C \ R . By (13) this means that the corre-sponding Π-spectral functions coincide. The obtained contradiction com-pletes the proof.We shall use some known important facts about sc-resolvents, see [16].Let B be an arbitrary Hermitian contraction in a Hilbert space H . Set D = D ( B ), R = H ⊖ D . A set of all self-adjoint contractive extensionsof B inside H , we denote by B H ( B ). A set of all self-adjoint contractiveextensions of B in a Hilbert space e H ⊇ H , we denote by B e H ( B ). ByKrein’s theorem [15, Theorem 2, p. 440], there always exists a self-adjoint11xtension b B of the operator B in H with the norm k B k . Therefore theset B H ( B ) is non-empty. There are the ”minimal” element B µ and the”maximal” element B M in this set, such that B H ( B ) coincides with theoperator segment B µ ≤ e B ≤ B M . (28)In the case B µ = B M the set B H ( B ) consists of a unique element. This caseis said to be determinate . The case B µ = B M is called indeterminate .The case B µ x = B M x , x ∈ R\{ } , is said to be completely indeter-minate . The indeterminate case can be always reduced to the completelyindeterminate. If R = { x ∈ R : B µ x = B M x } , we may set B e x = Bx, x ∈ D ; B e x = B µ x, x ∈ R . (29)The sets of generalized sc-resolvents for B and for B e coincide ([16, p. 1039]).Elements of B H ( B ) are canonical (i.e. inside H ) extensions of B and theirresolvents are said to be canonical sc-resolvents of B . On the other hand,elements of B e H ( B ) for all possible e H ⊇ H generate generalized sc-resolventsof B (here the space e H is not fixed). The set of all generalized sc-resolventswe denote by R c ( B ). Set C = B M − B µ , (30) Q µ ( z ) = (cid:16) C R µz C + E H (cid:17)(cid:12)(cid:12)(cid:12) R , z ∈ C \ [ − , , (31)where R µz = ( B µ − zE H ) − .An operator-valued function k ( z ) with values in [ R ] belongs to the class R R [ − ,
1] if1) k ( z ) is analytic in z ∈ C \ [ − ,
1] andIm k ( z )Im z ≤ , z ∈ C : Im z = 0;2) For z ∈ R \ [ − , k ( z ) is a self-adjoint non-negative contraction.Notice that functions from the class R R [ − ,
1] admit a special integral rep-resentation, see [16].
Theorem 3. ([16, p. 1053]). Let B be a Hermitian contraction in a Hilbertspace H with D ( B ) = D ; R = H ⊖ D . Suppose that for B it takes placethe completely indeterminate case and the corresponding operator C , as anoperator in R , has an inverse in [ R ] . Then the following equality: e R cz = R µz − R µz C k ( z ) ( E R + ( Q µ ( z ) − E R ) k ( z )) − C R µz , (32)12 here k ( z ) ∈ R R [ − , , e R cz ∈ R c ( B ) , establishes a bijective correspondencebetween the set R R [ − , and the set R c ( B ) .Moreover, the canonical resolvents correspond in (32) to the constantfunctions k ( z ) ≡ K , K ∈ [0 , E R ] . Let A be an arbitrary non-negative symmetric operator in a Hilbertspace H , D ( A ) = H . We are going to obtain a formula for the generalizedΠ-resolvents of A , by virtue of Theorem 3. Set T = ( E H − A )( E H + A ) − = − E H +2( E H + A ) − , D ( T ) = ( A + E H ) D ( A ) . (33)Then A = ( E H − T )( E H + T ) − = − E H +2( E H + T ) − , D ( A ) = ( T + E H ) D ( T ) . (34)The latter transformations were introduced and intensively studied by Krein [15].The operator T is a Hermitian contraction in H . In fact, for an arbitrary h = ( A + E H ) f , f ∈ D ( A ) we may write k T h k H = k ( − E H + 2( E H + A ) − )( A + E H ) f k H = k − Af + f k H = k Af k H + k f k H − Af, f ) H ≤ k Af k H + k f k H + 2( Af, f ) H = k h k H . Let e A ⊇ A be a non-negative self-adjoint extension of A in a Hilbert space e H ⊇ H . Then the operator e T = ( E e H − e A )( E e H + e A ) − = − E e H +2( E e H + e A ) − , D ( e T ) = ( e A + E e H ) D ( e A ) , (35)is a self-adjoint contraction e T ⊇ T in e H , and e A = ( E e H − e T )( E e H + e T ) − = − E e H +2( E e H + e T ) − , D ( e A ) = ( e T + E e H ) D ( e T ) . (36)Consider the following fractional linear transformation: z = 1 − λ λ = − λ ; λ = 1 − z z = − z . (37)Choose an arbitrary z ∈ C \ R and set λ := − z z . Observe that λ ∈ C \ R .Then R z ( e T ) = ( e T − zE e H ) − = (cid:18) − E e H + 2( E e H + e A ) − − − λ λ E e H (cid:19) − = (cid:18) ( − λ ( E e H + e A )( E e H + e A ) − + 2( E e H + e A ) − (cid:19) − (cid:18)(cid:18) λ λ E e H −
21 + λ e A (cid:19) ( E e H + e A ) − (cid:19) − = − λ + 12 (cid:16)(cid:16) e A − λE e H (cid:17) ( E e H + e A ) − (cid:17) − = − λ + 12 ( E e H + e A )( e A − λE e H ) − = − ( λ + 1) e A − λE e H ) − − λ + 12 E e H = − ( λ + 1) R λ ( e A ) − λ + 12 E e H . Therefore R λ ( e A ) = − λ + 1) R − λ λ ( e T ) − λ + 1 E e H , ∀ λ ∈ C \ R . (38)Applying the orthogonal projection on H , we get R λ ( A ) = − λ + 1) R − λ λ ( T ) − λ + 1 E H , ∀ λ ∈ C \ R . (39)Here R λ ( A ) is the generalized Π-resolvent corresponding to e A , and R z ( T )is the generalized sc -resolvent corresponding to e T . Thus, an arbitrary gen-eralized Π-resolvent of A can be constructed by a generalized sc -resolventof T by relation (39).On the other hand, choose an arbitrary sc -resolvent R ′ z ( T ) of T . It cor-responds to a self-adjoint contractive extension b T ⊇ T in a Hilbert space b H ⊇ H . Observe thatKer( E b H + b T ) ⊥ R ( E b H + b T ) ⊇ R ( E H + T ) = D ( A ) , and therefore Ker( E b H + b T ) ⊥ H . We may assume that H := Ker( E b H + b T ) = { } , since in the opposite case one may consider the operator b T restrictedto b H ⊖ H ⊇ H . Then we set b A = ( E b H − b T )( E b H + b T ) − = − E b H +2( E b H + b T ) − , D ( b A ) = ( b T + E b H ) D ( b T ) . (40)The operator b A is densely defined since b A ⊇ A , and it is self-adjoint. Foran arbitrary u ∈ D ( b T ) we may write( b A ( b T + E b H ) u, ( b T + E b H ) u ) b H = ( − b T u + u, b T u + u ) b H = k u k b H − k b T u k b H ≥ . Thus, the operator b A is non-negative. Observe that b T = ( E b H − b A )( E b H + b A ) − = − E b H + 2( E b H + b A ) − . (41)14epeating the considerations after relation (37), we obtain that R ′ λ ( A ) = − λ + 1) R ′ − λ λ ( T ) − λ + 1 E H , ∀ λ ∈ C \ R , (42)gives a generalized Π-resolvent of A (corresponding to b A ).Consequently, the relation (39) establishes a bijective correspondence be-tween the set of all sc -resolvents of T and the set of all Π-resolvents of A . Itis not hard to see that the canonical sc -resolvents are related to the canonicalΠ-resolvents.They say that for the operator A it takes place a completely inde-terminate case , if for the corresponding operator T it takes place thecompletely indeterminate case [24].It is known that all self-adjoint contractive extensions of T are extensionsof the extended operator T e defined by (29) [16, Theorem 1.4]. Set A e = ( E H − T e )( E H + T e ) − = − E H +2( E H + T e ) − , D ( A e ) = ( T e + E H ) D ( T e ) . (43)It is easily seen that the above operator e A is an extension of A e . Thereforethe sets of generalized Π-resolvents for A and for A e coincide. Theorem 4.
Let A be a non-negative symmetric operator in a Hilbert space H , D ( A ) = H . Suppose that for A it takes place the completely indetermi-nate case. Let T be given by (33); D = D ( T ) , R = H ⊖ D . Suppose that thecorresponding operator C = T M − T µ , as an operator in R , has an inversein [ R ] . Then the following equality: R λ ( A ) = − λ + 1) R µ − λ λ − λ + 1 E H + 2( λ + 1) R µ − λ λ C k ( λ ) ( E R + ( Q µ ( λ ) − E R ) k ( λ )) − C R µ − λ λ , (44) where Q µ ( λ ) = Q µ (cid:16) − λ λ (cid:17) , k ( λ ) = k (cid:16) − λ λ (cid:17) ; k ( · ) ∈ R R [ − , , establishesa bijective correspondence between the set R R [ − , and the set of all gen-eralized Π -resolvents of A . Here Q µ is defined by (31) for T , R µz = ( T µ − zE H ) − , and R λ ( A ) is a generalized Π -resolvent of A .Moreover, the canonical resolvents correspond in (44) to the constantfunctions k ( z ) ≡ K , K ∈ [0 , E R ] .Proof. It follows directly from the preceding considerations, formula (39)and by applying Theorem 3. 15et the strong matrix Stieltjes moment problem be given and condi-tions (4) hold. Consider an arbitrary Hilbert space H and a sequence ofelements { x n } n ∈ Z in H , such that relation (8) holds. Let A = A , where theoperator A is defined by (10). Denote L N = Lin { x k } N − k =0 . Define a lineartransformation G from C N onto L N by the following relation: G~u k = x k , k = 0 , , ..., N − , (45)where ~u k = ( δ ,k , δ ,k , ..., δ N − ,k ). Theorem 5.
Let the strong matrix Stieltjes moment problem (1) be givenand conditions (4) be satisfied. Let { x n } n ∈ Z be a sequence of elements of aHilbert space H such that relation (8) holds. Let A = A , where the operator A is defined by relation (10). Let T = − E H + 2( E H + A ) − . The followingstatements are true:1) If T µ = T M , then the moment problem (1) has a unique solution. Thissolution is given by M ( t ) = ( m j,n ( t )) N − j,n =0 , m j,n ( t ) = ( E µt x j , x n ) H , ≤ j, n ≤ N − , (46) where { E µt } is the left-continuous orthogonal resolution of unity of theoperator A µ = − E H + 2( E H + T µ ) − .2) If T µ = T M , define the extended operator T e by (29); R e = H ⊖ D ( T e ) , C = T M − T µ , and R µz = ( T µ − zE H ) − , Q µ,e ( z ) = (cid:16) C R µz C + E H (cid:17)(cid:12)(cid:12)(cid:12) R e , z ∈ C \ [ − , . An arbitrary solution M ( · ) of the moment problem canbe found by the Stieltjes-Perron inversion formula from the followingrelation Z R + t − z dM T ( t )= A ( z ) − C ( z ) k ( z )( E R e + D ( z ) k ( z )) − B ( z ) , (47) where k ( λ ) = k (cid:16) − λ λ (cid:17) , k ( z ) ∈ R R e [ − , , and on the right-handside one means the matrix of the corresponding operator in C N . Here A ( z ) , B ( z ) , C ( z ) , D ( z ) are analytic operator-valued functions given by A ( z ) = − λ + 1) G ∗ R µ − λ λ G − λ + 1 G ∗ G : C N → C N , (48) B ( z ) = C R µ − λ λ G : C N → R e , (49)16 ( z ) = 2( λ + 1) G ∗ R µ − λ λ C : R e → C N , (50) D ( z ) = Q µ,e (cid:18) − λ λ (cid:19) − E R e : R e → R e . (51) Moreover, the correspondence between all solutions of the momentproblem and k ( z ) ∈ R R e [ − , is bijective.Proof. Consider the case 1). In this case all self-adjoint contractions e T ⊇ T in a Hilbert space e H ⊇ H coincide on H with T µ , see [16, p. 1039]. Thus,the corresponding sc-spectral functions are spectral functions of the self-adjoint operator T µ , as well. However, a self-adjoint operator has a unique(normalized) spectral function. Thus, a set of sc-spectral functions of T consists of a unique element. Therefore the set of Π-resolvents of A consistsof a unique element, as well. This element is the spectral function of A µ .Consider the case 2). By Theorem 3 and relation (13) it follows that anarbitrary solution M ( t ) = ( m j,n ( t )) N − j,n =0 of the moment problem (1) can befound from the following relation: Z R + t − z dm j,n ( t ) = ( R z x j , x n ) H , ≤ j, n ≤ N − z ∈ C \ R , where R z is a generalized Π-resolvent of the operator A . Moreover, thecorrespondence between the set of all generalized Π-resolvents of A (whichis equal to the set of all generalized Π-resolvents of A e ) and the set of allsolutions of the moment problem is bijective. Notice that T µ = T µe and T M = T Me . By Theorem 4 (applied to the operator A e ) we may rewrite thelatter relation in the following form: Z R + t − z dm j,n ( t ) = (cid:18)(cid:26) − λ + 1) R µ − λ λ − λ + 1 E H + 2( λ + 1) R µ − λ λ C k ( λ ) ( E R e + ( Q µ,e ( λ ) − E R e ) k ( λ )) − C R µ − λ λ (cid:27) x j , x n (cid:19) H , where k ( λ ) = k (cid:16) − λ λ (cid:17) , k ( z ) ∈ R R e [ − , Q µ,e ( λ ) = Q µ,e (cid:16) − λ λ (cid:17) . Then Z R + t − z dm j,n ( t ) = (cid:18)(cid:26) − λ + 1) G ∗ R µ − λ λ G − λ + 1 G ∗ G + 2( λ + 1) G ∗ ∗ R µ − λ λ C k ( λ ) ( E R e + ( Q µ,e ( λ ) − E R e ) k ( λ )) − C R µ − λ λ G (cid:27) u j , u n (cid:19) C N . Introducing functions A ( z ) , B ( z ) , C ( z ) , D ( z ) by formulas (48)-(51) one easilyobtains relation (47). 17 heorem 6. Let the strong matrix Stieltjes moment problem (1) be givenand conditions (4) be satisfied. Let { x n } n ∈ Z be a sequence of elements of aHilbert space H such that relation (8) holds. Let A = A , where the operator A is defined by relation (10). The moment problem is determinate if andonly if T µ = T M , where T µ , T M are the extremal extensions of the operator T = − E H + 2( E H + A ) − .Proof. The sufficiency follows from Statement 1 of Theorem 5. The necessityfollows from Statement 2 of Theorem 5, if we take into account that the class R R e ([ − , R e >
0, has at least two different elements. In fact,from the definition of the class R R e ([ − , k ( z ) ≡
0, and k ( z ) ≡ E R e , belong to R R e ([ − , Example . Consider the moment problem (1) with N = 2 and S n = √ √ ! , n ∈ Z . In this case we haveΓ = ( S i + j ) ∞ i,j = −∞ = ( γ n,m ) ∞ n,m = −∞ , where γ k, l = γ k +1 , l +1 = 1 , γ k, l +1 = γ k +1 , l = 3 √ , k, l ∈ Z . Consider the space C and elements u , u ∈ C : u = 1 √ , , u = 1 √ , . Set x k = u , x k +1 = u , k ∈ Z . Then relation (8) holds. Define by (10) the operator A . In this case A = A = E C . Therefore the operators A and T = − E H + 2( E H + A ) − areself-adjoint and have unique spectral functions. Hence, T M = T µ , and byTheorem 6 we conclude that the moment problem has a unique solution. ByTheorem 2 it has the following form M ( λ ) = ( m k,j ( λ )) N − k,j =0 , m k,j ( λ ) = ( E λ x k , x j ) H , where E λ is the left-continuous spectral function of the operator E C . Con-sequently, the matrix function M ( t ) is equal to 0, for t ≤
1, and M ( t ) = √ √ ! , for t >
1. 18 eferences [1] Jones W. B., Thron W. J., Waadeland H., A strong Stieltjes momentproblem // Trans. Amer. Math. Soc. , n.2 (1980), 503–528.[2] Nj˚astad O., Solutions of the strong Stieltjes moment problem // Methodsand Applications of Analysis , n.3 (1995), 320–347.[3] Nj˚astad O., Extremal solutions of the strong Stieltjes moment problem// J. of Computational and Applied Mathematics , n.1-3 (1995), 309–318.[4] Kats I. S., Nudelman A. A., Strong Stieltjes moment problem // St.Petersburg Math. J. , n. 6 (1997), 931–950; Russian original: Algebrai analiz , n. 6 (1996).[5] Nj˚astad O., Nevanlinna matrices for the strong Stieltjes moment problem// J. of Computational and Applied Mathematics , n.1-2 (1998), 311–318.[6] Bonan-Hamada C. M., Jones W. B., Para-orthogonal Laurent polynomi-als and the strong Stieltjes moment problem // J. Comput. Appl. Math. , n.1-2 (1999), 175–185.[7] Jones W. B., Nj˚astad O., Orthogonal Laurent polynomials and strongmoment theory: a survey // J. Comput. Appl. Math. , n.1-2 (1999),51–91.[8] Nj˚astad O., Lens-shaped regions for Strong Stieltjes moment problems// Methods and Applications of Analysis , n.2 (1999), 195–208.[9] Nudelman A. A., Extension of the Stieltjes moment sequences to the leftand related problems of the spectral theory of inhomogeneous string //Ukrainian Mathematical Journal, , n.6 (2007), 894–906.[10] Berezansky Yu. M., Dudkin M. E., The strong Hamburger momentproblem and related direct and inverse spectral problems for blockJacobi-Laurent matrices // Methods Funct. Anal. Topology , n.3(2010), 203–241.[11] Simonov K. K., Strong matrix moment problem of Hamburger // Meth-ods Funct. Anal. Topology , n.2 (2006), 183–196.1912] Zagorodnyuk S. M., On the strong matrix Hamburger moment problem// Ukrainian Mathematical Journal, , n.4 (2010), 537–551.[13] Sz¨okefalvi-Nagy B., Koranyi A., Relations d’un probl´eme de Nevanlinnaet Pick avec la th´eorie de l’espace hilbertien // Acta Math. Acad. Sci.Hungar. (1957), 295–302.[14] Sz¨okefalvi-Nagy B., Koranyi A., Operatortheoretische Behandlung undVerallgemeinerung eines Problemkreises in der komplexen Funktionen-theorie // Acta Math., (1958), 171–202.[15] Krein M. G., The theory of self-adjoint extensions of semi-bounded Her-mitian transformations and its applications // Matematicheskiy sbornik , n.3 (1947), 431–495. (Russian).[16] Krein M. G., Ovcharenko I. E., On Q-functions and sc-resolvents ofnondensely defined Hermitian contractions // Sibirskiy matem. zhurnal, XVIII , n.5 (1977), 1032–1056. (Russian).[17] Krein M., Concerning the resolvents of an Hermitian operator with thedeficiency-index (m,m) // Doklady Akad. Nauk SSSR
LII , n.8 (1946),651–654.[18] Derkach V. A., Malamud M. M., Generalized resolvents and the bound-ary value problems for Hermitian operators with gaps // Journal ofFunctional Analysis , n.1 (1991), 1–95.[19] Chumakin M. E., On a class of the generalized resolvents of an isometricoperator // Uchenye zapiski Ul’yanovskogo pedag. instituta , 4 (1966),361–372. (Russian).[20] Malamud M. M, Malamud S. M., Operator measures in a Hilbert space// Algebra i analiz , n.3. (2003), 1–52. (Russian).[21] Gelfand I. M., Naimark M. A., On the inclusion of a normed ring intothe ring of operators in a Hilbert space // Matem. Sbornik (1943),197–213. (Russian).[22] Akhiezer N. I., Glazman I. M., Theory of linear operators in a Hilbertspace // Gos. izdat. teh.-teor. lit., Moscow, Leningrad, 1950. (Russian).[23] Akhiezer N. I. Classical moment problem // Fizmatlit., Moscow, 1961.(Russian). 2024] Krein M. G., Ovcharenko I. E., On generalized resolvents and resolventmatrices of positive Hermitian operators // Doklady AN SSSR, , n.5(1976), 1063–1066. (Russian). The strong matrix Stieltjes moment problem.