Technische Universität Wien. Habilitationsschrift GENERALIZED RESOLVENTS IN DEGENERATED INNER PRODUCT SPACES AND APPLICATIONS

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1 Technische Universität Wien Habilitationsschrift GENERALIZED RESOLVENTS IN DEGENERATED INNER PRODUCT SPACES AND APPLICATIONS Dipl.-Ing. Dr.techn. Harald WORACEK Wien, im Februar 1998

2 Inhalt In dieser Schrift sind die folgenden Artikel zusammengefaßt: H.Woracek: Multiple point interpolation in Nevanlinna-classes, Integral Equations and Operator Theory 28 (1997), M.Kaltenbäck, H.Woracek: The Krein formula for generalized resolvents in degenerated inner product spaces, Monatshefte für Mathematik. M.Kaltenbäck, H.Woracek: On extensions of hermitian functions with a finite number of negative squares, Journal of Operator Theory. M.Kaltenbäck, H.Woracek: On representations of matrix valued Nevanlinna functions by u-resolvents, Mathematische Nachrichten. Sie enthalten eine Verallgemeinerung der Kreinschen Formel zur Beschreibung aller verallgemeinerten Resolventen eines symmetrischen Operators, sowie einige Anwendungen derselben auf Interpolations- bzw. Fortsetzungsprobleme der klassischen Analysis.

3 Integr. equ. oper. theory 28 (1997), X/97/ Birkhäuser Verlag, Basel 1997 MULTIPLE POINT INTERPOLATION IN NEVANLINNA CLASSES Harald Woracek Abstract A Nevanlinna-Pick type interpolation problem for generalized Nevanlinna functions is considered. We prescribe the values of the function and its derivatives up to a certain order at finitely many points of the upper half plane. An operator theoretic approach is used to parametrize the solutions of this interpolation problem by means of selfadjoint extensions of a certain symmetry. 1 Introduction A function f is said to be of Nevanlinna class Nν π if f is meromorphic in the open upper half plane C + and the so called Nevanlinna kernel N f (z, w) = f(z) f(w) z w for z, w (f) has π positive and ν negative squares. Here (f) denotes the domain of analyticity of f in C + and π and ν are nonnegative integers or. More explicitly this means that for each number n N and points z 1,..., z n (f) the quadratic form n N f (z i, z j )ξ i ξ j (1.1) i,j=1 has at most π positive (ν negative) squares, and that for some choice of n and z 1,..., z n this upper bound is attained. If there is no upper bound for the number of positive (negative) squares of the forms (1.1) put π = (ν = ). We consider only such classes N π ν where at least one index is finite. Denote by N ν the union N ν = π=0 N π ν where ν is finite. 1

4 Multiple point interpolation 2 A multiple point interpolation problem is a problem where not only values for the function itself, but also for its derivatives up to a certain order are prescribed. The interpolation data thus consist of numbers n N, k 1,..., k n N 0 and points z 1,..., z n C + and w ij C for j = 0,..., k i and i = 1,...,n. A solution is a function f satisfying f (j) (z i ) = w ij for j = 0,...,k i and i = 1,...,n. (1.2) In this note we develop an operator theoretic method to describe the solutions of the interpolation problem (1.2) which are contained in some Nevanlinna class. This generalizes the well known operator method of Krein and Langer concerning simple point interpolation which has been developed in [KL2] and [NK]. In the classical case the so called Pick matrix plays an important role. We introduce in Section 2 a generalized Pick matrix for multiple point interpolation problems. Furthermore we recall the operator representation of an N ν -function f and compute the derivatives of f by means of this representation. This motivates the structure of the generalized Pick matrix. In Section 3 a model space, i.e. an inner product space H and a symmetric operator S, is associated to the given data and investigated. After these considerations we are in position to prove the main results (Theorem 4.1, Proposition 4.2 and Proposition 4.3) of this note which is done in Section 4. They establish a bijective correspondence between the solutions of the multiple point interpolation problem and the selfadjoint extensions of the above mentioned symmetric operator. More precisely, our result can be formulated as follows (here e 10 is a certain element of H): The formula f S(z) = w 10 + (z z 1 )[(I + (z z 1 )( S z) 1 )e 10, e 10 ]. (1.3) establishes a bijective correspondence between the solutions of the interpolation problem (1.2) within N π ν and the selfadjoint extensions S of S acting in a Pontryagin space P H with positive (negative) index π (ν), which contain z 1,...,z n in their resolvent set and are e 10 -minimal. In the proof some results of [KL1], [KL2] and [W2] are used. Similar results for the simple point case and some generalizations to interpolation problems by matrix valued functions can be found e.g. in [B], [BH], [DGK], [G], [NK] or [W2]. In the definite case also some generalizations of Nevanlinna- Pick type problems to the case of so-called Nevanlinna-pairs are given in [ABDS1], [ABDS2] and [Br1]. In the classical case one can give a parametrization of the solutions of (1.2) by means of a fractional linear transformation involving a parameter function, e.g. by starting from a formula like (1.3) and using the theory of so-called resolvent matrices (see e.g. [KL3]), i.e. apply Krein s formula for the generalized resolvents of a symmetry. Since in our case the space H is in general degenerated this theory cannot be applied. For some partial results in this direction (concerning different types of interpolation problems) see [AD], [Br1], [Br2] and [LW].

5 Multiple point interpolation 3 We would like to remark that interpolation by rational functions is also covered by the theory developed here, as the set of (real) rational functions equals the union of all sets Nν π with both indices finite (see e.g. [W1]). The generalized Pick matrix is in this context sometimes replaced by the so called generalized Löwner matrix which has a quite similar structure. Of course rational interpolation problems are well studied, see e.g. [AA], [ABKW] or [BK]. For further approaches to rational interpolation problems and an extensive coverage of the literature we refer to [BGR]. We will use some results concerning indefinite inner product spaces and their linear operators which can be found e.g. in [Bo] or [IKL]. 2 Derivatives of N ν -functions Let f N ν, then it is well known (see e.g. [KL2]) that f has a representation by means of a selfadjoint relation S in a Pontryagin space P with negative index ν. This representation can be obtained as follows: Denote by (f) the domain of analyticity of f in C 0 (= C + C ) and let H f be the inner product space consisting of all formal sums z (f) x ze z where all but finitely many coefficients vanish, endowed with the inner product given by [e z, e w ] = f(z) f(w) z w for z, w (f), z w, [e z, e z ] = f (z) for z (f). Then P is the Pontryagin space obtained from H f by completion of the factor space H f /H f, where H f denotes the isotropic part of H f. Let à be the operator in H f with domain domã = { z (f) x ze z H f z (f) x z = 0} which acts as Ã( z (f) x ze z ) = z (f) zx ze z. Then S is obtained from à by factorization and closure, and the relation f(z) = f(z 0 ) + (z z 0 )[(I + (z z 0 )( S z) 1 )e z0, e z0 ] (2.1) holds where z 0 is a fixed point of (f). Here the resolvent set of S coincides with the domain of analyticity of f. Note that the function f (and thus also its derivatives) is real, i.e. f(z) = f(z). This can be proved by a straightforward computation. Denote by s(z) the expression s(z) = (I + (z z 0 )( S z) 1 )e z0. From (2.1) we can compute the derivatives of f by means of derivatives of s(z): Lemma 2.1. Let f N ν, and let S be such that the representation (2.1) holds. Then for k 1 we have f (k) (z) = (z z 0 )[s (k) (z), e z0 ] + k[s (k 1) (z), e z0 ].

6 Multiple point interpolation 4 The proof of Lemma 2.1 is immediate by an easy computation. Suppose now we are given finitely many points z 1,..., z n C + (f) and numbers k 1,..., k n N 0, and we are interested in the values of f (k) (z i ) for k = 0,...,k i and i = 1,...,n. We may assume that z 0 = z 1. As then s(z 1 ) = e z1 = e z0, Lemma 2.1 shows that the information about the values f (k) (z i ) is contained in the subspace G = s (k) (z i ) k = 0,...,k i, i = 1,...,n of P. To determine the inner product [.,.] on G explicitly we will use the following relation. Lemma 2.2. The Nevanlinna kernel of f can be computed via s(z): f(z) f(w) z w = [s(z), s(w)]. Proof : We will repeatedly use the formula g(z)h(z) g(w)h(w) z w = g(z) g(w) h(z) h(w) h(z) + g(w). z w z w Note that the function f defined by (2.1) is real, i.e. f(z) = f(z). Thus f(z) f(w) z w = f(z) f(w) z w = (z z 0)[s(z), e z0 ] (w z 0 )[s(w), e z0 ] z w = As [ s(z) s(w) = [s(z), e z0 ] + (w z 0 )[ z w, e z 0 ]. s(z) s(w) z w, e z 0 ] = (w z 0)(w z) 1 e z0 (z z 0 )(z z) 1 e z0 w z = we find f(z) f(w) z w = (w z) 1 e z0 + (z z 0 ) (w z) 1 e z0 (z z) 1 e z0 w z = (w z) 1 e z0 + (z z 0 )(w z) 1 (z z) 1 e z0, = [e z0, e z0 ] + [(z z 0 )(z z) 1 e z0, e z0 ] + (w z 0 )[(w z) 1 e z0, e z0 ]+ On the other hand we have +(w z 0 )[(z z 0 )(w z) 1 (z z) 1 e z0, e z0 ]. (2.2) [s(z), s(w)] = [e z0 + (z z 0 )(z z) 1 e z0, e z0 + (w z 0 )(w z) 1 e z0 ] = =

7 Multiple point interpolation 5 = [e z0, e z0 ] + [(z z 0 )(z z) 1 e z0, e z0 ] + [e z0, (w z 0 )(w z) 1 e z0 ]+ +[(z z 0 )(z z) 1 e z0, (w z 0 )(w z) 1 e z0 ]. (2.3) Obviously the right hand sides of (2.2) and (2.3) coincide. From the above lemma we find that [s (k) (z i ), s (l) (z j )] = k z k l w l ( ) f(z) f(w) (2.4) z w z=zi w=z j holds. Thus the information about the values of f and its derivatives at the points z i is contained in the generalized Pick matrix associated with the points z 1,..., z n P = (P ij ) n j, (2.5) where P ij is a block of size (k j + 1) (k i + 1) with entries ( ) p ij lk = k l f(z) f(w) z k w l for l = 0,...,k j, k = 0,...,k i. (2.6) z w z=zi w=z j In the case k 1 =... = k n = 0 we obtain the classical Pick matrix. The following lemma gives some information about the action of S on G. Lemma 2.3. We have n (i) x i s(z i ) dom S whenever n x i = 0, and n n ( x i s(z i ), x i z i s(z i )) S. (ii) s (k) (z i ) dom S whenever k 1, and Proof : Note that and that (s (k) (z i ), z i s (k) (z i ) + ks (k 1) (z i )) S. n n x i s(z i ) = ( x i ) e 10 + }{{} =0 n x i (z i z 1 )( S z i ) 1 e 10, (( S z i ) 1 e 10, e 10 + z i ( S z i ) 1 e 10 ) S.

8 Multiple point interpolation 6 Thus n ( x i (z i z 1 )( S n n z i ) 1 e 10, x i (z i z 1 )e 10 + x i (z i z 1 )z i ( S z i ) 1 e 10 ) S which proves (i), as n x i z 1 e 10 = 0 and thus the right hand side equals n x i z i s(z i ). As s (k) (z i ) = (z i z 1 )k!( S z i ) (k+1) e 10 + k!( S z i ) k e 10 we find s (k) dom S and, as ks (k 1) (z i ) = (z i z 1 )k!( S z i ) k e 10 + k!( S z i ) (k 1) e 10 ) holds: (ks (k 1) (z i ), s (k) (z i )) ( S z i ) 1 which yields the assertion. 3 An inner product space connected to the interpolation data Consider now a multiple point interpolation problem, i.e. let n N, k 1,..., k n N 0, z 1,...,z n C k and w ik C for k = 0,...,k i and i = 1,..., n be given. From this data we can buildt the generalized Pick matrix P associated with the points z 1,..., z n. Lemma 3.1. The entries p ij lk (l = 0,...,k j, k = 0,...,k i ; i, j = 1,...,n) of the generalized Pick matrix associated with the points z 1,...,z n are given explicitly as k ( ) k p ij lk = ( 1) k h (k + l h)! w ih h (z i z j ) k+l+1 h + h=0 Proof : As the entries p ij lk equal k z k formula ( ) k l f(z) f(w) z k w l = z w + l h=0 l h=0 ( l f(z) f(w) w l z w k h=0 ( ) l ( 1) l h (k + l h)! w jh. (3.1) h (z j z i ) k+l+1 h ) z=zi w=z j it suffices to proof the ( ) k f(z) (h)( 1)k h (k + l h)! h (z w) k+l+1 h + ( ) l f(w) h (h) ( 1)l h (k + l h)! (w z) k+l+1 h. (3.2)

9 Multiple point interpolation 7 In order to prove (3.2) use induction on l and k. For l = k = 0 the assertion is clearly true. As (3.2) is symmetric with respect to l and k we only have to do the inductive step for one of l and k, say, k k + 1, i.e. we have to differentiate the right hand side of (3.1) with respect to z. We then obtain k h=0 = ( ) k ( 1) k h (k+l h)![f(z) (h+1) 1 (k + l h) +f(z)(i) h (z w) k+l+1 h (z w) + h=0 k+l+2 h ]+ k ( ) l f(w) (h) ( 1) l h 1 (k + l h)!(k + l + 1 h) h (w z) k+l+2 h = k ( ) ( ) k k [ + ]( 1) (k+1) h ((k+1)+l h)!f(z) (h) 1 h 1 h 1 (z w) + }{{} (k+1)+l+1 h h=1 =( k+1 h ) ( ) ( ) k + l!f(z) (k+1) 1 k k (z w) }{{} l+1 + ( 1) k+1 1 ((k+1)+l)!f(z) 0 (z w) + }{{} (k+1)+l+1 h =( k+1) ( k+1 0 ) k ( ) l + f(w) (h) ( 1) l h 1 ((k + 1) + l h)! h (w z) (k+1)+l+1 h h=0 which proves the assertion. Taking the subspace G as a model we define an inner product space associated with the interpolation data. Definition 3.2. Let H be the inner product space of all formal sums H = { x ik e ik x ik C} k=0,...,k i,...,n endowed with the inner product defined by [e ik, e jl ] = p ij lk. Remark 3.3. If we identify the element k=0,...,k i,...,n x ik e ik with the column vector (x 10,..., x 1k1 ; x 20,...,...,x nkn ) T, (3.3) then the Gram matrix of [.,.] on H is the generalized Pick matrix associated with z 1,..., z n.

10 Multiple point interpolation 8 Note that, as the elements e ik are per definitionem linearly independent we have dim H = n (k i + 1). In our model G in general only the inequality dim G n (k i + 1) holds. We define an operator S acting in H (compare Lemma 2.3). Definition 3.4. Let S be the operator with domain which acts as j=1 doms = { k=0,...,k i,...,n x ik e ik n x i0 = 0} = n n = x j0 e j0, e ik x j0 = 0, k = 1,...,k i, i = 1,...,n, S( j=1 n x i0 e i0 ) = and, for k = 1,...,k i, i = 1,...,n n z i x i0 e i0 (3.4) Se ik = z i e ik + ke i,k 1. (3.5) Remark 3.5. If we again identify the element k=0,...,k i,...,n vector (3.3), S admits a representation as a block matrix x ik e ik with the column S (S ij ) n i,j=1 (3.6) where the blocks S ij are of size (k j + 1) (k i + 1), the off-diagonal blocks contain only zeros and the diagonal blocks are of the form z i S ii =. for i = 1,...,n... ki 0 z i Before we investigate some properties of S we give another lemma. Lemma 3.6. The entries of the generalized Pick matrix satisfy the following relations (i, j {1,..., n}): (i) (z i z j )p ij lk = kpij l,k 1 + lpij l 1,k for l, k 1,

11 Multiple point interpolation 9 (ii) (z i z j )p ij l0 = w jl+lp ij l 1,0 for l 1 and similarly (z i z j )p ij 0k = kpij w ik for k 1, (iii) (z i z j )p ij 00 = w i0 w j0. 0,k 1 + Proof : We use the explicit formula for p ij lk given in Lemma 3.1. Then we find k ( ) (z i z j )p ij k lk = ( 1) k h (k + l h)! w ih h (z i z j ) k+l h lp ij h=0 h=0 l h=0 h=0 ( ) l ( 1) l h (k + l h)! w jh h (z j z i ) k+l h, k ( ) k l 1,k = ( 1) k h (k + l 1 h)! l 1 ( ) l ( 1) l h (k + l 1 h)! l w ih h (z i z j ) k+l h l w jh h (z j z i ) k+l h and k 1 kp ij l,k 1 = ( ) k k h h=0 w ih ( 1) k h (k + l 1 h)! (z i z j ) k+l h Comparing the coefficients of the term w ih ( 1) k h (z i z j ) k+l h l h=0 ( ) l ( 1) l h (k + l 1 h)! w jh h (z j z i ) k+l h. for h = 0,...,k 1 in the first sums of the above right hand sides we find that we have to prove ( ) ( ) ( ) k k 1 k (k + l 1 h)!(l + k ) = (k + l h)!. h h h This relation is proved by direct computation. The coefficients of w jh ( 1) l h (z j z i ) k+l h of the second sums on the right hand sides give (for h = 0,...,l 1) ( ) ( ) ( ) l l 1 l (k + l h)! = (k + l 1 h)!(l + k ) h h h which is also proved by direct computation. For h = k the coefficients equal obviously. To prove the second assertion of the lemma we compute (see Lemma 3.1) (z i z j )p ij l0 = l! w i0 l ( ) l (z i z j ) l ( 1) l h (l h)! w jh h (z j z i ) l h h=0

12 Multiple point interpolation 10 and lp ij l 1,0 l 1 w i0 = l(l 1)! (z i z j ) l l ( l 1 h h=0 ) w jh ( 1) l h (l 1 h)! (z j z i ) l h. Putting these expression together finishes the proof. The last assertion is an immediate consequence of the definition of p ij 00. Proposition 3.7. The operator S is symmetric and has no eigenvectors. Proof : The matrix (3.6) representing S (or, more exactly, the extension of S given by (3.6)) obviously has the eigenvalues z i with corresponding eigenvectors e i0 (i = 1,...,n). But these are exactly the elements excluded by the choice of doms. Thus S has no nonzero eigenvectors. To prove that S is symmetric we have to show that [Sx, y] = [x, Sy] (3.7) for x, y doms. First consider the case x = e ik and y = e jl where k, l 1. Then and similarly [Sx, y] = [z i e ik + ke i,k 1, e jl ] = z i [e ik, e jl ]+ +k[e i,k 1, e jl ] = z 1 p ij lk + kpij l,k 1, [x, Sy] = z j p ij lk + lpij l 1,k. Lemma 3.6 thus implies (3.7). If x = n x i0 e i0 where n x i0 = 0 and y = e jl with l 1 we find and [Sx, y] = [x, Sy] = = n x i0 [z i e i0, e jl ] = n x i0 e i0 n x i0 [e i0, z j e jl + le j,l 1 ] = Thus we find, again using Lemma 3.6 [Sx, y] [x, Sy] = n n x i0 z j p ij l0 + l x i0 p ij l 1,0. n x i0 ((z i z j )p ij l0 + lpij l 1,0 }{{} ). (3.8) = f(z j) (l)

13 Multiple point interpolation 11 The right hand side of (3.8) equals 0, as n x i0 = 0. Finally consider the case that x = n x i0 e i0 and y = n y j0 e j0 where n x i0 = n y i0 = 0. We have = j=1 [Sx, y] [x, Sy] = n i,j=1 n n x i0 f(z i ) y j0 j=1 j=1 j=1 x i0 y j0 (z i z j )p ij 00 = n n y j0 f(z j ) x i0 = 0. Lemma 3.8. For each i {1,..., n} and k {0,...,k i } we have e 10 ran(s z i ) (k+1). In fact for i 1 (S z i ) (k+1) e 10 = holds, and for i = 1 we have k h=0 ( 1) k h h!(z i z 1 ) k h+1 e ih + ( 1)k+1 (z i z 1 ) k+1 e 10 (3.9) (S z i ) (k+1) e 10 = 1 k! e 1k. Proof : To prove the first assertion use induction on k. If k = 0 (3.9) follows immediately from the definition (3.4). Then we compute ( S z i ) 1 ( S z i ) (k+1) e 10 = = k+1 h=0 k h=0 ( 1) (k+1) h h!(z i z 1 ) (k+1) h+1 e ih + where we used (3.4) and (3.5). The second assertion is clear from (3.5). ( 1) k h h!(z i z 1 ) k h+1 1 h + 1 e ih+ ( 1)k+1 (z i z 1 ) k+1 e i0 e 10 z i z 1 = ( 1)(k+1)+1 (z i z 1 ) (k+1)+1 e 10 = ( S z i ) (k+2) e 10 4 Correspondence of solutions and extensions In this section we will prove the main results of the present note which establish a connection between the solutions of the interpolation problem (1.2) and the

14 Multiple point interpolation 12 selfadjoint extensions of S acting in some Pontryagin space extending H. These results follow similar as in the classical case (see [KL2], [NK], [W2]). But first we have to recall some notions. We call an extension S of S acting in a Pontryaginspace P e 10 -minimal if ( S z) 1 e 10 z ( S) = P. We call two extensions S 1 and S 2 acting in Pontryaginspaces P 1 and P 2, respectively e 10 -unitary equivalent if there is a unitary operator U : P 1 P 2 which satisfies Ue 10 = e 10 (remember that both, P 1 and P 2, are extensions of H) and S 1 = U 1 S2 U. Theorem 4.1. Let n N, z 1,...,z n C +, k 1,...,k n N 0 and w ik C for k = 0,...,k i, i = 1,...,n be given. Assume that the generalized Pick matrix is regular. Then the solutions of the multiple point interpolation problem f (k) (z i ) = w ik for k = 0,...,k i and i = 1,...,n within a Nevanlinna class correspond to the selfadjoint (relational) extensions S of the symmetric operator S which operate in a Pontryagin space P extending H and contain the points z 1,..., z n in their resolvent set. This connection is established by the formula S f S(z) = w 10 + (z z 1 )[(I + (z z 1 )( S z) 1 )e 10, e 10 ]. Proof : Assume that f is a solution of the considered problem, then the results of Section 2 show that there exist S and P extending S and H, respectively and satisfy the required conditions. The fact that z 1,...,z n ( S) follows similar as in [W2]. Suppose conversely that S extends S and that z 1,...,z n ( S). In order to compute f (k) (z i ) we have to compute (use similar notation as in Section 2) s (k) (z i ) for k = 0,...,k i. In case k = 0 we find from Lemma 3.8 (i 1) (I + (z i z 1 )( S z) 1 )e 10 = e 10 + (z i z 1 ) e i0 e 10 z 1 z 1 = e i0 (the case i = 1 is obvious) whereas for k 1 a straightforward computation again using Lemma 3.8 implies Thus (iii) of Lemma 3.6 shows k z k (I + (z z 1)( S z) 1 ) z=zi e 10 = e ik. f(z i ) = w 10 + (z i z 1 )[e i0, e 10 ] = w i0, }{{} =p i1 00

15 Multiple point interpolation 13 and (ii) of Lemma 3.6 together with Lemma 2.1 shows (k = 1,...,k i ) f (k) (z i ) = (z i z 1 )[s (k) (z i ), e 10 ] +k [s (k 1) (z i ), e 10 ] = w ik. }{{}}{{} =p i1 0k =p i0 0,k 1 Proposition 4.2. The correspondence S f S between extensions and solutions becomes bijective if we consider only e 10 -minimal extensions of S and do not distinguish between e 10 -unitary equivalent extensions. Proof : It is proved in [W2] that two minimal extensions induce the same function if and only if they are e 10 -unitary equivalent. As we can restrict our attention to e 10 -minimal extensions in any case, the assertion is clear. There is also a connection between the indices of the space P and the indices of the Nevanlinna class f S belongs to. Although the following result is well known (see e.g. [KL2] or [W2]) and does not use the fact that H P we include it, especially with view to Remark 4.4. Proposition 4.3. If S acts in a Pontryagin space with index (π, ν) then the corresponding solution f S is in a class Nν π with π π and ν ν. If we assume that S is minimal the full number of positive and negative squares is attained, i.e. π = π and ν = ν. Remark 4.4. Assume that the generalized Pick matrix has π 0 positive and ν 0 negative eigenvalues and is regular. Then there exist solutions of (1.2) in those classes N π ν where π π 0 and ν ν 0. On the other hand it is obvious that in classes N π ν where π < π or ν < ν no solutions can exist. References [ABDS1] D.Alpay, P.Bruinsma, A.Dijksma, H.de Snoo: Interpolation problems, extensions of symmetric operators and reproducing kernel spaces I, Oper. Theory Adv. Appl. 50 (1991), [ABDS2] D.Alpay, P.Bruinsma, A.Dijksma, H.de Snoo: Interpolation problems, extensions of symmetric operators and reproducing kernel spaces II, Integral Equations Operator Theory 14 (1991), [AD] D.Alpay, H.Dym: On applications of reproducing kernel spaces to the Schur algorithm and rational J-unitary factorization, Oper. Theory Adv. Appl. 18 (1986),

16 Multiple point interpolation 14 [AA] A.C.Antoulas, B.D.O.Anderson: On the scalar rational interpolation problem, IMA J. Math. Control Inform. 3 (1986), [ABKW] A.C.Antoulas, J.A.Ball, J.Kang, J.C.Willems: On the solution of the minimal rational interpolation problem, Linear Algebra Appl. 137/138 (1990), [B] J.A.Ball: Interpolations problems of Pick-Nevanlinna and Löwner Type for meromorphic matrix functions, Integral Equations Operator Theory 6 (1983), [BGR] J.A.Ball, I.Gohberg, L.Rodman: Interpolation of rational matrix functions, Birkhäuser Verlag, Basel [BH] J.A.Ball, J.W.Helton: Interpolation problems of Pick-Nevanlinna and Loewner types for meromorphic matrix functions: parametrization of the set of all solutions, Integral Equations Operator Theory 9 (1986), [BK] J.A.Ball, J.Kim: Stability and McMillan degree for rational matrix interpolants, Linear Algebra Appl. 196 (1994), [Bo] J.Bognar: Indefinite inner product spaces, Springer Verlag, Berlin [Br1] P.Bruinsma: Interpolation for Schur and Nevanlinna pairs, Doctoral Dissertation, University of Groningen [Br2] P.Bruinsma: Degenerate interpolation problems for Nevanlinna pairs, Indag. Math. (N.S.) 2(2) (1991), [DGK] Ph.Delsarte, Y.Genin, Y.Kamp: The Nevanlinna-Pick problem for matrix valued functions, SIAM J. Appl. Math. 36 (1979), [G] L.B.Golinskii: On one generalization of the matrix Nevanlinna-Pick problem, Izv. Akad. Nauk Armenii Mat. 18 (1983), [IKL] I.S.Iohvidov, M.G.Krein, H.Langer: Spectral theory in spaces with an indefinite metric, Akademie Verlag, Berlin [KL1] M.G.Krein, H.Langer: Über die Q-Funktion eines Π-hermiteschen Operators im Raume Π κ, Acta Sci. Math. (Szeged) 34 (1973), [KL2] M.G.Krein, H.Langer: Über einige Fortsetzungsprobleme, die eng mit der Theorie hermitescher Operatoren im Raume Π κ zusammenhängen.i. Einige Funktionenklassen und ihre Darstellungen, Math. Nachr. 77 (1977), [KL3] M.G.Krein, H.Langer: Über einige Fortsetzungsprobleme, die eng mit der Theorie hermitescher Operatoren im Raume Π κ zusammenhängen.ii. Verallgemeinerte Resolventen, u-resolventen und ganze Operatoren, J. Funct. Anal. 30 (1978), [LW] H.Langer, H.Woracek: Resolvents of symmetric operators. The degenerated Nevanlinna-Pick problem, to appear in Oper. Theory Adv. Appl.

17 Multiple point interpolation 15 [NK] B.Sz.Nagy, A.Korany: Operatortheoretische Behandlung und Verallgemeinerung eines Problemkreises in der komplexen Funktionentheorie, Acta Math. 100 (1958), [W1] H.Woracek: Nevanlinna-Pick interpolation: The degenerated case, to appear in Linear Algebra Appl. [W2] H.Woracek: An operator theoretic approach to degenerated Nevanlinna-Pick interpolation, Math. Nachr. 176 (1995), Institut für Analysis, Technische Mathematik und Versicherungsmathematik TU Wien Wiedner Hauptstr. 8-10/114.1 A-1040 Wien AUSTRIA AMS Classification Numbers: 47A57, 30E05, 47B50

18 to appear in Monatsh. Math Springer Verlag, Wien THE KREIN FORMULA FOR GENERALIZED RESOLVENTS IN DEGENERATED INNER PRODUCT SPACES Michael Kaltenbäck, Harald Woracek Abstract Let S be a symmetric operator with defect index (1,1) in a Pontryagin space H. The Krein formula establishes a bijective correspondence between the generalized resolvents of S and the set of Nevanlinna functions as parameters. We give an analogue of the Krein formula in the case that H is a degenerated inner product space. The set of parameters is determined by a kernel condition. These results are applied to some classical interpolation problems with singular data. 1 Introduction Let H be a, possibly degenerated, inner product space such that H/(H H ) is a Pontryagin space, and let S be a symmetric operator in H with defect index (1, 1). If P is a Pontryagin space containing H as a singular subspace, and A is a selfadjoint relation in P extending S we call the set of analytic functions [(A z) 1 x, y], x, y H (1.1) a generalized resolvent of S. Note that, if H is nondegenerated, there exists an orthogonal projection P of P onto H, hence a generalized resolvent (1.1) can be viewed as an operator valued function P(A z) 1 H. If H is degenerated such a projection does not exist and we have to use the form (1.1). Denote in the following by. the linear span of the elements between the brackets. If A is H-minimal in the sense that H, (A z) 1 H z (A) = P, 16

19 Krein s formula 17 the corresponding generalized resolvent is called minimal. In this case, if P has index κ of negativity (κ = ind P), the generalized resolvent is said to have index κ. If τ is a complex valued meromorphic function, denote by (τ) its domain of holomorphy. For κ N 0 let the set N κ of generalized Nevanlinna functions τ be defined as follows: τ is meromorphic in C \ R, τ(z) = τ(z) for z (τ), and the Nevanlinna kernel (z, w (τ)) N τ (z, w) = { τ(z) τ(w) z w τ (z), z w, z = w has κ negative squares. If H is nondegenerated, i.e. a Pontryagin space, Krein s formula (see [KL1] or [DLS]) [(A z) 1 x, y] = [(A z) 1 1 x, y] [x, χ(z)] [χ(z), y], (1.2) τ(z) + q(z) establishes a bijective correspondence between the minimal generalized resolvents of index κ of S and the set N κ κ0 (κ 0 = ind H) of parameters τ. Here A H 2 is a fixed selfadjoint extension of S, χ(z) are certain defect elements of S and q(z) is a corresponding Q-function, which means that the relation N q (z, w) = [χ(z), χ(w)] holds. In this paper we consider the case that H is actually degenerated, dim H = > 0. It turns out that the relation (1.2) still holds, where A, χ(z) and q(z) have a similar meaning. However, the parameter τ does not run through the Nevanlinna class N κ κ0, but through a different set of functions K κ κ 0 which is defined as follows: Definition 1.1. For ν, N 0, denote by Kν the set of all complex valued functions τ(z), meromorphic in C \ R, which satisfy τ(z) = τ(z) for z (τ), and are such that the maximal number of the negative squares of the quadratic forms (m N 0, z 1,..., z m (τ)) Q(ξ 1,..., ξ m ; η 0,...,η 1 ) = i,j=1 1 N τ (z i, z j )ξ i ξ j + k=0 R ( zi k ) ξ i η k (1.3) is ν. Note that K 0 ν = N ν. In 2 we collect some results on the defect spaces of a symmetric relation S. Our analogue of Krein s formula (1.2) is proved in 3. This formula enables us to treat some interpolation problems with singular data. We consider the extension problem of a hermitian function with a finite number of negative squares (4), and the indefinite Nevanlinna-Pick interpolation problem (Section 5) in the case of

20 Krein s formula 18 degenerated datas. This means for each problem that there exists a unique solution with minimal index of negativity. Our parametrization of generalized resolvents leads to a parametrization of the solutions with higher index of negativity. We use the notation and some results on symmetric and selfadjoint relations in Pontryagin spaces provided in [DS]. For elementary facts concerning the geometry of Pontryagin spaces we refer to [IKL]. 2 Defect elements Let H be a degenerated inner product space and assume that H = H r [ +]H, (2.1) where 1 dim H = < and where H r is a Pontryagin space. The condition (2.1) ensures that H can be embedded canonically into a Pontryagin space P c with negative index ind P c = ind H + dim H. To see this define P c = H r [ +](H +H 1 ), (2.2) where H 1 is a neutral space and H and H 1 are skewly linked (see [IKL]). A linear relation S H 2 P 2 c is called closed if it is closed in the topology of P 2 c. Let S H 2 be a closed symmetric relation. It is shown similar as in the classical (nondegenerated) situation that the number codim ran S z is constant on the upper (lower) half plane with possible exception of finitely many points. Hence we may speak of the defect index of the symmetry S in the usual sense (see [DS]). Throughout this paper we assume that S is a closed symmetric relation with defect index (1, 1) which satisfies the following regularity conditions: For each h H S ( h h ) = {0}, (2.3) and for some z C + and some z C rans z + H = H. (2.4) Remark 2.1. Condition (2.4) implies that the relation S/H acting in the Pontryagin space H/H is selfadjoint and has nonempty resolvent set. This follows from the considerations in [DS], in particular from Proposition 4.4 and Theorem 4.6 with its corollary. It also follows that (2.4) holds for all z (S/H ). Proposition 2.2. The relation S can be decomposed (as a subspace of H 2 ) as S = S 0 +S 1, where S 0 = S (H ) 2 and S 1 is a closed symmetric relation such that

21 Krein s formula 19 rans 1 z is nondegenerated for z (S/H ). There exists a basis {h 0,...,h 1 } of H, such that S 0 = 0 (H ) 2 if = 1, and S 0 = (h i ; h i+1 ) i = 0,..., 2 (2.5) otherwise. Proof : Let S 1 = S (S 0 ) ( ), where the orthogonal complement has to be understood with respect to a definite inner product on P 2 c. Then S 1 is symmetric, satisfies S = S 0 +S 1 and is closed. Choose a decomposition (2.1) and denote by P the projection of H onto H r with kernel H. Define S P = (P P)S, then we have H r = H/H and S P = S/H. Since S 0 = S kerp P we have S 1 kerp P = {0}, and therefore the restriction (P P) S1 maps S 1 bijectively onto S P. Hence there exists an inverse mapping Φ : S P S 1. Clearly Φ I maps S P into (H ) 2. We show that rans 1 z H = {0} if z (S/H ). Assume the opposite. Then there exists a pair (a; b) S 1, such that b za 0 and b za H. We have (a; b) = Φ(x; y) for some element (x; y) S P. If we put (x ; y ) = (Φ I)(x; y), we have (x ; y ) (H ) 2 and b za = (y zx) + (y zx ). Since the left hand side as well as the second term on the right hand side of the above relation are elements of H, y zx H r, and H r H = {0}, we find y zx = 0. As z (S/H ) = (S P ), this shows that x = y = 0, hence a = b = 0. Since rans 0 z H and rans z = rans 0 z + rans 1 z, we find due to (2.4) and Remark 2.1, for z (S/H ) rans 1 z +H = rans z + H = H. This shows that rans 1 z is nondegenerated for z (S/H ). It remains to show that S 0 is of the form (2.5). The condition (2.3) asserts that S 0 has no eigenvalue, in particular S 0 is (the graph of) an operator. Since S has defect index (1, 1), and S 0 acts in a -dimensional space we find that dimrans 0 z = 1 and hence dim doms 0 = 1. Consider the domain of S 2 0. Since S 0domS 2 0 = doms 0 rans 0 we have either dimdoms 2 0 = dimdoms 0 or dimdoms 2 0 = dimdoms 0 1. If dimdoms0 2 = dimdoms 0, i.e. doms0 2 = doms 0, this space is an invariant subspace of S 0. Hence S 0 has a nonzero eigenvector in doms0 2, a contradiction to condition (2.3). We find (by an inductive procedure) that H doms 0 doms doms 1 0 {0},

22 Krein s formula 20 and that at each step the dimension decreases by 1. Let h 0 H be such that doms0 1 = h 0 and put h i = S0 ih 0, i = 1,..., 1. A straightforward argument shows that {h 0,..., h 1 } is a basis of H. Clearly S 0 = (h i ; h i+1 ) i = 0,..., 2. Corollary 2.3. The element h 0 is uniquely determined (up to constant multiples) by the properties h 0 doms 1 H and S l h 0 H, l = 0,..., 1. Proof : Assume that h doms 1 H and S l h H for l = 0,..., 1. Write, with respect to the basis {h 0,..., h 1 } of H constructed in Proposition 2.2, h = η k h k, η m 0 k=0 and assume on the contrary that m > 0. Then S m h = η k S m h k = k=0 hence Sh 1 H, a contradiction. m 1 k=0 η k h k+ m + η m Sh 1, Corollary 2.4. Let k {0,..., 1} and z 0. Then z σ(s/h ) if and only if h k rans z. Proof : This follows from Remark 2.1, together with the fact that for z 0 h k +rans 0 z = H. Due to the fact that S has defect index (1, 1) we have for z (S/H ) rans z Pc = H + χ where χ is an element of rans z Pc \ H. Choose χ(z 0 ) rans z 0 Pc \ H. If A is a selfadjoint extension of S in Pc, denote by χ(z) the element χ(z) = (I + (z z 0 )(A z) 1 )χ(z 0 ).

23 Krein s formula 21 Lemma 2.5. The relation rans z Pc = H + χ(z) holds for z (A ) (S/H ) with possible exception of an isolated set. Proof : Similar as in the classical case (see e.g. [KL1]) we find that χ(z) rans z. It remains to show that χ(z) H. Choose a basis {h 0,...,h 1 } of H. Since H = (H ) Pc, we have χ(z) H if and only if [χ(z), h i ] = 0 for i = 0,..., 1. The element χ(z) depends analytically on z (A ) and χ(z 0 ) H, hence for some i the function [χ(z), h i ] does not vanish identically. Its zeros are therefore isolated in (A ), hence the points z with χ(z) H are also isolated in (A ). Proposition 2.6. There exists a selfadjoint extension A P 2 c of S with (A ), such that for z (A ) (S/H ) (A z) 1 H H, (A z) 1 h 0 = 0 (2.6) and [χ(z), h i ] = z i, i = 0,..., 1, (2.7) holds. Proof : Consider the relation S = S + (0; h 0 ). Then S is symmetric and rans z = H for all z (S/H ). We show that kers z = {0} if z (S/H ): Assume that h kers z then, since rans z = H (see Proposition 2.2) and rans z = kers z, we find h H. The relation S decomposes according to Proposition 2.2 as S = S 0 +S 1 + (0; h 0 ). Since S 1 (H ) 2 = {0}, we may write, with respect to this decomposition 2 1 (h; zh) = ( λ i h i ; λ i 1 h i ) + (0; λh 0 ). i=0 Hence 1 2 λh 0 + λ i 1 h i = z λ i h i, i=0

24 Krein s formula 22 which implies that h = 0. Note that the resolvent set of the relation S 0 + (0; h 0 ) is C. Now choose z 0 (S/H ) and consider the Cayley transform C of S. Then C is an isometric operator in P c, and codimdomc = codimranc =. Hence C may be extended to a unitary operator U. The inverse Cayley transform A of U is a selfadjoint relation extending S. Since (U) we find (A ), and for z (A ) (S/H ) we have (A z) 1 H = (A z) 1 rans z H. To define defect elements choose z 0 (A ) (S/H ), and let P c be decomposed as P c = rans 1 z 0 [ +](H +H ), (2.8) where H and H are skewly linked. Let {h 0,...,h 1 } be a basis of H which is skewly linked to the basis {h 0,...,h 1 } of H given by Proposition 2.2, i.e. let [h i, h j ] = δ ij. A short computation shows that we may choose χ(z 0 ) = h 0 + z 0h z 1 0 h 1. Then χ(z) = (I + (z z 0 )(A z) 1 )χ(z 0 ) are defect elements (see Lemma 2.5). We compute (i = 0,..., 1) [χ(z), h i ] = [(I + (z z 0 )(A z) 1 )χ(z 0 ), h i ] = = [χ(z 0 ), h i ] + (z z 0 )[χ(z 0 ), ( S z) A zh i ]. Since A extends S we have (A z) 1 h i = (S z) 1 h i. A straightforward computation shows that (S z) 1 h i = z i 1 h 0 + z i 2 h h i 1, i = 1,..., 1. Moreover (S z) 1 h 0 = 0, hence [χ(z), h i ] = z i 0 + (z z 0)(z i 1 + z i 2 z z i 1 0 ) = z i, i = 0,..., 1. Corollary 2.7. For any set M (A ) (S/H ) with M we have H rans z = {0}. z M

25 Krein s formula 23 Proof : Let h = 1 k=0 η k h k and assume that h z M rans z. Then [h, χ(z)] = 1 k=0 η k z k = 0, z M. As a nonzero polynomial of degree 1 has at most 1 different zeros, we find η 0 =... = η 1 = 0, i.e. h = 0. Remark 2.8. Note that each selfadjoint extension of S satisfying (2.6) induces the same generalized resolvent of S. For if π denotes the canonical projection of H onto H/H, we have for x, y H and z (A ) (S/H ) [(A z) 1 x, y] = [(S/H z) 1 πx, πy]. If χ(z) are defect elements as in Proposition 2.6 we define q(z) by q(z) q(w) z w = [χ(z), χ(w)]. (2.9) The function q is determined by (2.9) up to a real constant (see [KL1]). In order to determine the generalized resolvents of S, the defect elements χ(z) play (similar as in the classical situation) an important role. Proposition 2.9. Let A P 2 be a selfadjoint extension of S acting in a Pontryagin space P H, with (A). Assume that the relation A is chosen according to Proposition 2.6. Then there exists a function Ψ(z) analytic on (A) (A ) (S/H ), with Ψ(z) = Ψ(z) such that for x, y H [(A z) 1 x, y] = [(A z) 1 x, y] + [x, χ(z)]ψ(z)[χ(z), y]. (2.10) In fact Ψ(z) = [(A z) 1 h 0, h 0 ]. Proof : We will define Ψ(z) for z (A) (A ) (S/H ). As for such z we have [χ(z), h 0 ] = 1, we can decompose the space H as H = rans z + h 0. Due to (2.7) we can write x = x z + [x, χ(z)]h 0 and y = y z + [y, χ(z)]h 0, with x z rans z and y z rans z. Since (A z) 1 x = (A z) 1 x for x rans z,

26 Krein s formula 24 we find ] [((A z) 1 (A z) 1 )x, y] = [((A z) 1 (A z) 1 )[x, χ(z)]h 0, [y, χ(z)]h 0 = ] = [x, χ(z)] [((A z) 1 (A z) 1 )h 0, h 0 [χ(z), y]. Due to (2.6) we have [(A z) 1 h 0, h 0 ] = 0, hence the assertion follows. 3 Parametrization of the generalized resolvents In this section we will prove the main result of this paper, which is the following Theorem 3.1. Let S be a closed symmetric relation in the degenerated inner product space H, assume that S has defect index (1, 1) and satisfies (2.3) and (2.4). Let κ 0 = ind H and = dim H > 0. Then for z (A) (S/H ) and x, y H the relation [(A z) 1 x, y] = [(S/H z) 1 1 πx, πy] [x, χ(z)] [χ(z), y] (3.1) τ(z) + q(z) establishes a bijective correspondence between the generalized resolvents of S of index κ and the set (K κ κ 0 \ { q}) of parameters τ. Here χ(z) are the defect elements of S chosen according to Proposition 2.6, q(z) is as in (2.9) and by π we denote the canonical projection of H onto H/H. The unique generalized resolvent [(S/H z) 1 πx, πy] corresponds to the parameter τ =. Remark 3.2. If we set σ(z) = 1 τ(z) for τ 0, the quadratic form (1.3) rewrites i,j=1 1 N σ (z i, z j )ξ i ξ j + k=0 R ( zi k ) σ(z i )ξ i η k Note that, since the quadratic form (1.3) is the same for each constant function τ(z) = λ, λ R, the case τ = 0 is also covered. By the results of [KW2] we find that τ K κ κ 0 if and only if σ = 1 τ has a representation σ(z) = [(B z) 1 h, h], where B is a selfadjoint relation in a Pontryagin space with negative index κ κ 0, which extends a shift operator S 0 with ( 1)-dimensional domain, acting in a - dimensional neutral space L, where h = doms0 1, and where B is L-minimal. In particular, for κ κ 0 the sets Kκ κ 0 contain infinitely many elements.

27 Krein s formula 25 We also see that the set K κ κ 0 is empty, if κ κ 0 <. This corresponds to the fact that the negative index of a Pontryagin space extending H must be at least κ 0 +. Theorem 3.1 is proved in several steps. Lemma 3.3. Let A be a selfadjoint extension of S and assume that the generalized resolvent [(A z) 1 x, y] does not coincide with [(S/H z) 1 πx, πy]. Then there exists a H-minimal extension A 1 of S with [(A z) 1 x, y] = [(A 1 z) 1 x, y], z (A) (A 1 ), x, y H. Proof : Consider the subspace M = H, (A z) 1 H z (A). The relation A induces a selfadjoint relation A 1 in the Pontryagin space P 1 = M/M which is H-minimal. If M H = {0}, we can assume that H P 1. Then A 1 extends S and clearly [(A 1 z) 1 x, y] = [(A z) 1 x, y], x, y H. If h M H, h 0, then h H and Corollary 2.7 implies rans z + h = H for all but finitely many z (A ) (S/H ). Hence [(A z) 1 x, y] = [(S/H z) 1 πx, πy]. Assume now that a selfadjoint extension A of S is given, where A P 2 and (A), and put R z = (A z) 1. If Ψ = 0, then τ =. Otherwise we may assume due to Lemma 3.3 that A is H-minimal. Let A be chosen according to Proposition 2.6 and put R z = (A z) 1. We define the operator valued function R z by (x P c) where Ψ(z) is given by Proposition 2.9. Remark 3.4. For x, y H we have R zx = R zx + [x, χ(z)]ψ(z)χ(z), (3.2) [R z x, y] = [R zx, y]. (3.3)

28 Krein s formula 26 If either x or y does not belong to H, this relation need not hold. 1 Let a function τ be defined by Ψ(z) = τ(z)+q(z), i.e. let τ(z) = 1 Ψ(z) q(z). Note that, since Ψ, we have τ q. Denote by P H the projection of P c onto H with kernel H 1. Let m N 0, z 1,...,z m (A) (S/H ) and a, a 1,...,a m, b, b 1,...,b m H. Consider the expression U = i,j=1 ( ) [R zi a i, R zj b j ] [R z i a i, R z j b j ] + Then U can be written in two different ways. Lemma 3.5. With the above notation we have and [R zi a i, b] + [a, R zj b j ]. U = [a + (R zi P H R z i )a i, b + (R zj P H R z j )b j ], (3.4) U = j=1 j=1 Ψ(z i )[a i, χ(z i )]N τ (z i, z j )[χ(z j ), b j ]Ψ(z j )+ i,j=1 + Ψ(z i )[a i, χ(z i )][χ(z i ), b] + [a, χ(z j )][χ(z j ), b j ]Ψ(z j ). (3.5) Proof : First we show that the relation (3.4) holds. Note that rani P H = H 1 is a neutral subspace. Due to (3.3) we have for x H, y P c We compute = U = i,j=1 i,j=1 j=1 [(R z R z)x, y] = [(R z R z)x, (I P H )y]. ( [(R zi R z i )a i, (R zj R z j )b j ] + [(R zi R z i )a i, R z j b j ]+ ) + [R z i a i, (R zj R z j )b j ] + [R zi a i, b] + [a, R zj b j ] = ( [(R zi R z i )a i, (R zj R z j )b j ] + [(R zi R z i )a i, (I P H )R z j b j ]+ j=1

29 Krein s formula 27 ) + [(I P H )R z i a i, (R zj R z j )b j ] + j=1 = i,j=1 [(R zi P H R z i )a i, b]+ [a, (R zj P H R z j )b j ] = [(R zi P H R z i )a i, (R zj P H R z j )b j ] + [(R zi P H R z i )a i, b]+ + [a, (R zj P H R z j )b j ] = [a + (R zi P H R z i )a i, b + (R zj P H R z j )b j ]. Hence (3.4) is proved. On the other hand use the definition (3.2) and the relation (2.10) to compute U. We find, due to the resolvent identity which holds for R z (recall that a i, b j H ) [R zi a i, R zj b j ] = [ R z i R zj z i z j j=1 j=1 a i, b j ] = [ R z i R z j a i, b j ]+ z i z j and + [a i, χ(z i )]Ψ(z i )[χ(z i ), b j ] [a i, χ(z j )]Ψ(z j )[χ(z j ), b j ] z i z j, [R z i a i, R z j b j ] = [R z i a i, R z j b j ] + [R z i a i, [b j, χ(z j )]Ψ(z j )χ(z j )]+ +[[a i, χ(z i )]Ψ(z i )χ(z i ), R z j b j ] + [[a i, χ(z i )]Ψ(z i )χ(z i ), [b j, χ(z j )]Ψ(z j )χ(z j )]]. Using the resolvent identity for R z and the relations we find R z i χ(z j ) = χ(z i) χ(z j ) z i z j, R z j χ(z i ) = χ(z j) χ(z i ) z j z i, [R z i a i, R z j b j ] = [ R z i R z j z i z j a i, b j ]+ +[χ(z j ), b j ]Ψ(z j )[a i, χ(z i) χ(z j ) z i z j ] + [a i, χ(z i )]Ψ(z i )[ χ(z j) χ(z i ) z j z i, b j ]+ Due to (2.6) we have +[a i, χ(z i )]Ψ(z i )[χ(z i ), χ(z j )]Ψ(z j )[χ(z j ), b j ]. [R z i a i, b] = [a, R z j b j ] = 0, hence the last two terms in the definition of U compute as [R zi a i, b] = [a i, χ(z i )]Ψ(z i )[χ(z i ), b] [a, R zj b j ] = [a, χ(z j )]Ψ(z j )[χ(z j ), b j ].

30 Krein s formula 28 We find U = ( ) Ψ(z i ) Ψ(z j ) [a i, χ(z i )] Ψ(z i )[χ(z i ), χ(z j )]Ψ(z j ) [χ(z j ), b j ]+ z i z j i,j=1 + [a i, χ(z i )]Ψ(z i )[χ(z i ), b] + [a, χ(z j )]Ψ(z j )[χ(z j ), b j ]. 1 If we substitute Ψ(z) = τ(z)+q(z), where q(z) is defined by (2.9), we obtain ( τ(zi ) + q(z i ) τ(z j ) q(z j ) U = [a i, χ(z i )]Ψ(z i ) z i,j=1 i z j ) N q (z i, z j ) Ψ(z j )[χ(z j ), b j ] + [a i, χ(z i )]Ψ(z i )[χ(z i ), b] + [a, χ(z j )]Ψ(z j )[χ(z j ), b j ], j=1 hence (3.5) follows. j=1 Lemma 3.6. For any nonzero element h H, we have H, (R z P H R z )h z (A) (S/H ) = H r. Proof : Since (R z P H R z)h H r, the linear space L = H, (R z P H R z )h z (A) is dense in H r if and only if the linear space L 1 = H, (R z P H R z)h z (A) is dense in P. Since for all z C \ R, with possible exception of finitely many, the relation rans z + h = H holds, and since R z rans z H, we find that L 1 = H, R z h z (A) = H, R z H z (A). By the assumption that A is H-minimal, the last space is dense in P. If we put in Lemma 3.5 a = b = 1 k=0 η k h k, a i = b i = ξ i h 0,

31 Krein s formula 29 we obtain with (2.7) U = Ψ(z i )ξ i N τ (z i, z j )ξ j Ψ(z j ) + i,j=1 1 k=0 Ψ(z i )ξ i zi k η k + z k j η k ξ j Ψ(z j ). Due to Lemma 3.6 this quadratic form in the variables ξ 1,...,ξ m ; η 0,..., η 1 has ind H r = κ κ 0 negative squares. After the change of variables Ψ(z i )ξ i ξ i, 2η k η k and use of the fact that Ψ(z) vanishes only on an isolated set, we obtain the form (1.3), hence τ(z) K κ κ 0. We have shown that the generalized resolvent [R z x, y] has the representation (3.1) with the parameter τ K κ κ 0. It remains to prove the converse implication of Theorem 3.1. Let A and χ(z) be as in Proposition 2.6 and assume that a parameter τ K κ κ 0, τ q, is given. We will construct a selfadjoint extension A of S, such that for x, y H [(A z) 1 x, y] = [(S/H z) 1 1 πx, πy] [x, χ(z)] [χ(z), y]. (3.6) τ(z) + q(z) 1 Put ψ(z) = τ(z)+q(z). Let H be decomposed as in (2.1) and consider the linear space L = H r [ +](H + f z z C \ R, τ(z) + q(z) 0 ), where f z are formal elements, equipped with the inner product k=0 [x, y] L = [x, y] H, x, y H r, 1 1 [ η k h k + ξ i f zi, η k h k k=0 [ The τ K κ κ 0 implies x, = k=0 j=1 ξ j f z j ] L = j=1 ψ(z i )ξ i N τ (z i, z j )ξ j ψ(z j)+ i,j=1 m ψ(z i )ξ i zi k η k + z k j η k ξ j ψ(z j), (3.7) 1 k=0 η k h k + j=1 ] ξ i f zi L = 0, x H r. ind L = ind H r + (κ κ 0 ) = κ.

32 Krein s formula 30 Hence we obtain from L by factorization with respect to its isotropic part and completion a Pontryagin space P with ind P = κ: P = L/L. Lemma 3.7. We have H P. Proof : Since H r is nondegenerated it suffices to show that H L = {0}. Assume that 1 η k h k L. Choose disjoint numbers z 1,..., z (ψ), such k=0 that ψ(z i ) 0 for i = 1,...,. Then, for arbitrary numbers ξ i, we have 1 0 = [ η k h k, k=0 Since the matrix (z k i ),..., k=0,..., 1 0. ξ i f zi ] = 1 k=0 = (η 0,...,η 1 ) ( k z ) i,..., k=0,..., 1 z k i η k ξ i ψ(z i ) = ψ(z 1 )ξ 1. ψ(z )ξ. is regular this relation implies η 0 =... = η 1 = The space P c can be viewed as a subspace of P: P c = H r [ +](H +H ) with H H + f z z C \ R, τ(z) + q(z) 0. Let R z be defined by (x P c ) Note that, due to Proposition 2.6, Define elements Clearly P = H + e z z (ψ). R z x = R zx + [x, χ(z)]ψ(z)χ(z). R zh 0 = ψ(z)χ(z). (3.8) e z = f z + P H R zh 0. Lemma 3.8. The inner product of expressions involving elements e z is given by Proof : We compute [e z, e w ] = N ψ (z, w), z, w (ψ), [e z, x] = [χ(z), x]ψ(z), z (ψ), x H. [e z, e w ] = [f z, f w ] + [f z, P H R w h 0] + [P H R z h 0, f w ] + +[P H R z h 0, P H R w h 0]. (3.9)

33 Krein s formula 31 Denote by {h 0,...,h 1 } a basis of H skewly linked to {h 0,...,h 1 }. Since [χ(z), χ(w)] = N q (z, w), H is a neutral subspace, and (3.8) holds, the last term on the right hand side of (3.9) computes as [P H R zh 0, P H R wh 0 ] = ψ(z)(n q (z, w) (z k [h k, χ(w)] + w k [χ(z), h k]))ψ(w). 1 k=0 The definition (3.7) of the inner product [.,.] shows that We find [P H χ(z), f w ] = 1 k=0 w k [χ(z), h k ]ψ(w). [e z, e w ] = ψ(z)(n τ (z, w) + N q (z, w)) ψ(w) = N ψ (z, w). Let x H be decomposed as x = x r + 1 and (3.8), we find [e z, x] = [f z, x] + [P H R z h 0, x] = 1 k=0 k=0 η k h k with x r H r. Then, due to (3.7) z k ψ(z)η k + ψ(z)[χ(z), x r ] = ψ(z)[χ(z), x]. Let A P 2 be defined as A = S, (e z ; h 0 + ze z ) z (ψ). Lemma 3.9. The relation A is selfadjoint and has a nonempty resolvent set. In fact (A) (ψ) (S/H ). Moreover, A is H-minimal. Proof : We first show that A is symmetric. To see this it suffices to note that for (a; b) S and z, w (ψ) the relations [e z, h 0 + we w ] [h 0 + ze z, e w ] = ψ(z) + wn ψ (z, w) ψ(w) zn ψ (z, w) = 0, [e z, b] [h 0 + ze z, a] = [χ(z), b]ψ(z) z[χ(z), a]ψ(z) = = [χ(z), b za]ψ(z) = 0 hold. It remains to prove that rana z is dense in P if z (ψ) (S/H ), since then, by the same argument as in Remark 2.1 (using [DS]), the assertion follows.

34 Krein s formula 32 Let z (ψ) (S/H ), then (e z ; h 0 ) A z (3.10) ( e z e w z w ; e w) A z, w z. Since rans z rana z and rans z + h 0 = H, we obtain Hence rana z is dense in P. The relation (3.10) shows that In particular A is H-minimal. H, e w w (ψ) \ {z} rana z. H, (A z) 1 h 0 z (A) = P. By Proposition 2.9 we have 1 Ψ(z) = [R z h 0, h 0 ] = [e z, h 0 ] = ψ(z) = τ(z) + q(z). Hence the generalized resolvent corresponding to the relation A has the representation (3.6) with the prescribed parameter τ. All assertions of Theorem 3.1 are proved. In the classical case dim H = 0, the canonical extensions of S, i.e. those acting the space H itself, correspond to the parameter functions τ(z) = t R. The space P c is the minimal Pontryagin space extending H. Hence we will refer to an extension A Pc 2 of S as a canonical extension. The parameters corresponding to such canonical extensions are of more complicated structure. However, they can be identified by making use of Remark 3.2. The following statement deals with operator extensions (A(0) = {0}) of S. For proper relations a similar argument can be applied if the spectrum is transformed conveniently. Remark Let B be a minimal selfadjoint operator in the space C 2, endowed with the inner product induced by the Gram matrix ( ) 0 I J =. I 0 Assume that B extends a shift operator with ( 1)-dimensional domain as in Remark 3.2. Then, with respect to a convenient basis, the operator B has the