Boundary Nevanlinna Pick interpolation problems for generalized Schur functions

Transcription

1 , Boundary Nevanlinna Pick interpolation problems for generalized Schur functions Vladimir Bolotnikov and Alexander Kheifets Abstract. Three boundary Nevanlinna-Pick interpolation problems at finitely many points are formulated for generalized Schur functions. For each problem, the set of all solutions is parametrized in terms of a linear fractional transformation with a Schur class parameter.. Introduction The Schur class S of complex-valued analytic functions mapping the unit disk D into the closed unit disk D can be characterized in terms of positive kernels as follows: a function w belongs to S if and only if the kernel K w (z, ζ) := w(ζ)w(z) ζz (.) is positive definite on D (in formulas: K w 0), i.e., if and only if the Hermitian matrix n K w (z j, z i ) n i,j= = w(z i )w(z j ) (.2) z i z j i,j= is positive semidefinite for every choice of an integer n and of n points z,..., z n D. The significance of this characterization for interpolation theory is that it gives the necessity part in the Nevanlinna-Pick interpolation theorem: given points z,...,z n D and w,..., w n C, there exists w S with w(z j ) = w j for wiw j =,...,n if and only if the associated Pick matrix P = j z iz j is positive semidefinite. There are at least two obstacles to get an immediate boundary analogue of the latter result just upon sending the points z,...,z n in (.2) to the unit circle T. Firstly, the boundary nontangential (equivalently, radial) limits w(t) := lim z t w(z) (.3)

2 2 Vladimir Bolotnikov and Alexander Kheifets exist at almost every (but not every) point t on T. Secondly, although the nontangential limits d w (t) := lim z t w(z) 2 z 2 0 (t T) (.4) exist at every t T, they can be infinite. However, if d w (t) <, then it is readily seen that the limit (.3) exists and is unimodular. Then we can pass to limits in (.2) to get the necessity part of the following interpolation result: Given points t,..., t n T and numbers w,..., w n and γ,..., γ n such that there exists w S with w i = and γ i 0 for i =,...,n, (.5) w(t i ) = w i and d w (t i ) γ i for i =,...,n (.6) if and only if the associated Pick matrix P = P ij n i,j= with the entries P ij = is positive semidefinite. w i w j t i t j for i j γ i for i = j (.7) This result in turn, suggests the following well known boundary Nevanlinna Pick interpolation problem. Problem.. Given points t,...,t n T and numbers w,..., w n, γ,...,γ n as in (.5) and such that the Pick matrix P defined in (.7) is positive semidefinite, find all functions w S satisfying interpolation conditions (.6). Note that assumptions (.5) and P 0 are not restrictive since they are necessary for the problem to have a solution. The boundary Nevanlinna Pick interpolation problem was worked out using quite different approaches: the method of fundamental matrix inequalities 2, the recursive Schur algorithm 7, the Grassmannian approach 3, via realization theory 2, and via unitary extensions of partially defined isometries,. If P is singular, then Problem. has a unique solution which is a finite Blaschke product of degree r = rankp. If P is positive definite, Problem. has infinitely many solutions that can be described in terms of a linear fractional transformation with a free Schur class parameter. Note that a similar problem with equality sign in the second series of conditions in (.6) was considered in 9, 9, 6:

3 Boundary Nevanlinna Pick problem 3 Problem.2. Given the data as in Problem., find all functions w S such that w(t i ) = w i and d w (t i ) = γ i for i =,...,n (.8) The solvability criteria for this modified problem is also given in terms of the Pick matrix (.7) but it is more subtle: condition P 0 is necessary (not sufficient, in general) for the Problem.2 to have a solution while the condition P > 0 is sufficient. The objective of this paper is to study the above problems in the setting of generalized Schur functions. A function w is called a generalized Schur function if it is of the form w(z) = S(z) B(z), (.9) for some Schur function S S and a finite Blaschke product B. Without loss of generality we can (and will) assume that S and B in representation (.9) have no common zeroes. For a fixed integer κ 0, we denote by S κ the class of generalized Schur functions with κ poles inside D, i.e., the class of functions of the form (.9) with a Blaschke product B of degree κ. Thus, S κ is a class of functions w such that. w is meromorphic in D and has κ poles inside D counted with multiplicities. 2. w is bounded on an annulus {z : ρ < z < } for some ρ (0, ). 3. Boundary nontangential limits w(t) := lim z t w(z) exist and satisfy w(t) for almost all t T. It is clear that the class S 0 coincides with the classical Schur class. The class S κ can be characterized alternatively (and sometimes this characterization is taken as the definition of the class) as the set of functions w meromorphic on D and such that the kernel K w (z, ζ) defined in (.) has κ negative squares on D ρ(w) (ρ(w) stands for the domain of analyticity of w); in formulas: sq (K w ) = κ. The last equality means that for every choice of an integer n and of n points z,..., z n D ρ(w), the Hermitian matrix (.9) has at most κ negative eigenvalues: sq n w(z i )w(z j ) κ, (.0) z i z j i,j=

4 4 Vladimir Bolotnikov and Alexander Kheifets and for at least one such choice it has exactly κ negative eigenvalues counted with multiplicities. In what follows, we will say w has κ negative squares rather than the kernel K w has κ negative squares. Due to representation (.9) and in view of the quite simple structure of finite Blaschke products, most of the results concerning the boundary behavior of generalized Schur functions can be derived from the corresponding classical results for the Schur class functions. For example, the nontangential boundary limit d w (t) (defined in (.4)) exists for every t T and satisfies d w (t) > (not necessarily nonnegative, in contrast to the definite case). Indeed, if w is of the form (.9), then w(z) 2 ( S(z) 2 z 2 = B(z) 2 z 2 ) B(z) 2 z 2. (.) Passing to the limits as z tends to t T in the latter equality and taking into account that B(t) =, we get d w (t) = d S (t) d B (t) >, since d w (t 0 ) 0 and d B (t) <. Furthermore, as in the definite case, if d w (t) <, then the nontangential limit (.3) exists and is unimodular. Now we formulate indefinite analogues of Problems. and.2. The data set for these problems will consist of n points t,..., t n on T, n unimodular numbers w,...,w n and n real numbers γ,..., γ n : t i T, w i =, γ i R (i =,..., n). (.2) As in the definite case, we associate to the interpolation data (.2) the Pick matrix P via the formula (.7) which is still Hermitian (since γ j R), but not positive semidefinite, in general. Let κ be the number of its negative eigenvalues: where P = P ij n i,j= and P ij = κ := sq P, (.3) w i w j t i t j for i j, γ j for i = j. (.4) The next problem is an indefinite analogue of Problem.2 and it coincides with Problem.2 if κ = 0.

5 Boundary Nevanlinna Pick problem 5 Problem.3. Given the data set (.2), find all functions w S κ (with κ defined in (.3)) such that and d w (t i ) := lim z ti w(z) 2 z 2 = γ i (i =,...,n) (.5) w(t i ) := lim z ti w(z) = w i (i =,...,n). (.6) The analogue of Problem. is: Problem.4. Given the data set (.2), find all functions w S κ (with κ defined in (.3)) such that d w (t i ) γ i and w(t i ) = w i (i =,...,n). (.7) Interpolation conditions for the two above problems are clear: existence of the nontangential limits d w (t i ) s implies existence of the nontangential limits w(t i ) s; upon prescribing the values of these limits (or upon prescribing upper bounds for d w (t i ) s) we come up with interpolation conditions (.5) (.7). The choice (.3) for the index of S κ should be explained in some more detail. Remark.5. If a generalized Schur function w satisfies interpolation conditions (.7), then it has at least κ = sq P negative squares. Indeed, if w is a generalized Schur function of the class S eκ and t,..., t n are distinct points on T such that d w (t i ) < for i =,...,n, then the nontangential boundary limits w(t i ) s exist (and are unimodular) and one can pass to the limit in (.0) (as t i z i for i =,...,n) to conclude that the Hermitian matrix P w (t,...,t n ) = P w ij n i,j= satisfies with Pij w = w(t i )w(t j ) t i t j for i j d w (t i ) for i = j (.8) sq P w (t,..., t n ) κ. (.9)

6 6 Vladimir Bolotnikov and Alexander Kheifets If w meets conditions (.6), then the nondiagonal entries in the matrices P w (t,..., t n ) and P coincide which clearly follows from the definitions (.4) and (.8). It follows from the same definitions that P P w (t,...,t n ) = γ d w (t ) γ n d w (t n ) and thus, conditions (.5) and the first series of conditions in (.7) can be written equivalently in the matrix form as P w (t,..., t n ) = P and P w (t,..., t n ) P, (.20) respectively. Each one of the two last relations implies, in view of (.9) that sq P κ. Thus, the latter condition is necessary for existence of a function w of the class S eκ satisfying interpolation conditions (.7) (or (.5) and (.6)). The choice (.3) means that we are concerned about generalized Schur functions with the minimally possible negative index. Problems.3 and.4 are indefinite analogues of Problems.2 and., respectively. Now we introduce another boundary interpolation problem that does not appear in the context of classical Schur functions. Problem.6. Given the data set (.2), find all functions w S κ for some κ κ = sq P such that conditions (.7) are satisfied at all but κ κ points t,...,t n. In other words, a solution w to the last problem is allowed to have less then κ negative squares and to omit some of interpolation conditions (but not too many of them). The significance of Problem.6 will be explained in the next section. 2. Main results The purpose of the paper is to obtain parametrizations of solution sets S 3, S 4 and S 6 for Problems.3,.4 and.6, respectively. First we note that S 3 S 4 S 6 and S 4 = S 6 S κ. (2.) Inclusions in (2.) are self-evident. If w is a solution of Problems.6 with κ = κ, then κ κ = 0 which means that conditions (.7) are satisfied at all points

7 Boundary Nevanlinna Pick problem 7 t,..., t n and thus, w S 4. Thus, S 4 S 6 S κ. The reverse inclusion is evident, since S 4 S κ. Note also that if κ = 0, then Problems.4 and.6 are equivalent: S 4 = S 6. It turns out that in the indefinite setting (i.e., when κ > 0), Problem.6 plays the same role as Problem.4 does in the classical setting: it always has a solution and, in the indetereminate case, the solution set S 6 admits a linear fractional parametrization with the free Schur class parameter. The case when P is singular, is relatively simple: Theorem 2.. Let P be singular. Then Problem.6 has a unique solution w which is the ratio of two finite Blaschke products with no common zeroes and such that w(z) = B (z) B 2 (z) deg B + deg B 2 = rankp. Furthermore, if deg B 2 = κ, then w is also a solution of Problem.4. The proof will be given in Section 7. Now we turn to a more interesting case when P is not singular. In this case, we pick an arbitrary point µ T\{t,...,t n } and introduce the 2 2 matrix valued function Θ (z) Θ Θ(z) = 2 (z) (2.2) Θ 2 (z) Θ 22 (z) C = I 2 + (z µ) (zi E n T) P (I n µt ) C E where T = t..., E =..., C = w... w n. (2.3) t n Note that the Pick matrix P defined in (.4) satisfies the following identity P T PT = E E C C. (2.4) Indeed, equality of nondiagonal entries in (2.4) follows from the definition (.8) of P, whereas diagonal entries in both sides of (2.4) are zeroes. Identity (2.4) and all its ingredients will play an important role in the subsequent analysis.

8 8 Vladimir Bolotnikov and Alexander Kheifets The function Θ defined in (2.2) is rational and has simple poles at t,..., t n. Note some extra properties of Θ. Let J be a signature matrix defined as 0 J =. (2.5) 0 It turns out that Θ is J-unitary on the unit circle, i.e., that Θ(t)JΘ(t) = J for every t T ρ(θ) (2.6) and the kernel K Θ,J (z, ζ) := J Θ(z)JΘ(ζ) z ζ has κ = sq P negative squares on D: (2.7) sq K Θ,J = κ. (2.8) We shall use the symbol W κ for the class of 2 2 meromorphic functions satisfying conditions (2.6) and (2.8). It is well known that for every function Θ W κ, the linear fractional transformation T Θ : E Θ E + Θ 2 Θ 2 E + Θ 22 (2.9) is well defined for every Schur class function E and maps S 0 into κ κ S κ. This map is not onto and the question about its range is of certain interest. If Θ is of the form (2.2), the range of the transformation (2.9) is S 6 : Theorem 2.2. Let P, T, E and C be defined as in (.4) and (2.3) and let w be a function meromorphic on D. If P is invertible, then w is a solution of Problem.6 if and only if it is of the form for some Schur function E S 0. w(z) = T Θ E(z) := Θ (z)e(z) + Θ 2 (z) Θ 2 (z)e(z) + Θ 22 (z), (2.0) It is not difficult to show that every rational function Θ from the class W κ with simple poles at t,...,t n T and normalized to I 2 at µ T, is necessarily of the form (2.2) for some row vector C C n with unimodular entries, with E as in (2.3) and with a Hermitian invertible matrix P having κ negative squares and being subject to the Stein identity (2.4). Thus, Theorem 2.2 clarifies the interpolation meaning of the range of a linear fractional transformation based on a rational function Θ of the class W κ with simple poles on the boundary of the unit disk.

9 Boundary Nevanlinna Pick problem 9 The necessity part in Theorem 2.2 will be obtained in Section 3 using an appropriate adaptation of the V. P. Potapov s method of the Fundamental Matrix Inequality (FMI) to the context of generalized Schur functions. The proof of the sufficiency part rests on Theorems 2.3 and 2.5 which are of certain independent interest. To formulate these theorems, let us introduce the numbers c,..., c n and ẽ,...,ẽ n by c i := lim z t i (z t i )Θ 2 (z) and ẽ i := lim z t i (z t i )Θ 22 (z) (i =,...,n) (2.) (for notational convenience we will write sometimes a rather than a for a C). It turns out c i = ẽ i 0 (see Lemma 3. below for the proof) and therefore the following numbers are unimodular: η i := c i ẽ i = ẽ i c i = lim z ti Θ 22 (z) Θ 2 (z) (i =,...,n) (2.2) η i = (i =,...,n). (2.3) Furthermore let p ii stand for the i-th diagonal entry of the matrix P, the inverse of the Pick matrix. It is self-evident that for a fixed i, any function E S 0 satisfies exactly one of the following six conditions: C : The function E fails to have a nontangential boundary limit η i at t i. C 2 : E(t i ) := lim z ti E(z) = η i and d E (t i ) := E(z) 2 z 2 =. (2.4) C 3 : E(t i ) = η i and p ii ẽ i 2 < d E(t i ) <. (2.5) C 4 : E(t i ) = η i and 0 d E (t i ) < p ii ẽ i 2. (2.6) C 5 : E(t i ) = η i and d E (t i ) = p ii > 0. ẽ i 2 (2.7) C 6 : E(t i ) = η i and d E (t i ) = p ii = 0. (2.8) Note that condition C means that either the nontangential boundary limit E(t i ) := lim z t i E(z) fails to exist or it exists and is not equal to η i. Let us denote by C 4 6 the disjunction of conditions C 4, C 5 and C 6 : C 4 6 : E(t i ) = η i and d E (t i ) p ii ẽ i 2. (2.9) The next theorem gives a classification of interpolation conditions that are or are not satisfied by a function w of the form (2.0) in terms of the corresponding parameter E.

10 0 Vladimir Bolotnikov and Alexander Kheifets Theorem 2.3. Let the Pick matrix P be invertible, let E be a Schur class function, let Θ be given by (2.2), let w = T Θ E and let t i be an interpolation node.. The nontangential boundary limits d w (t i ) and w(t i ) exist and are subject to d w (t i ) = γ i and w(t i ) = w i if and only if the parameter E meets either condition C or C The nontangential boundary limits d w (t i ) and w(t i ) exist and are subject to d w (t i ) < γ i and w(t i ) = w i if and only if the parameter E meets condition C The nontangential boundary limits d w (t i ) and w(t i ) exist and are subject to γ i < d w (t i ) < and w(t i ) = w i. if and only if the parameter E meets condition C If E meets C 5, then w is subject to one of the following: (a) The limit w(t i ) fails to exist. (b) The limit w(t i ) exists and w(t i ) w i. (c) w(t i ) = w i and d w (t i ) =. 5. If E meets C 6, then w is the ratio of two finite Blaschke products, d w (t i ) < and w(t i ) w i. We note an immediate consequence of the last theorem. Corollary 2.4. A function w = T Θ E meets the i-th interpolation conditions for Problem.4: d w (t i ) γ i and w(t i ) = w i if and only if the corresponding parameter E S 0 meets the condition C 3 := C C 2 C 3 at t i. Note that Problem.3 was considered in 2 for rational generalized Schur functions. It was shown (2, Theorem 2..2) that all rational solutions of Problem.3 are parametrized by the formula (2.0) when E varies over the set of all rational Schur functions such that (in the current terminology) E(t i ) η i for i =,...,n.

11 Boundary Nevanlinna Pick problem Note that if E is a rational Schur function admitting a unimodular value E(t 0 ) at a boundary point t 0 T, then the limit d w (t 0 ) always exists and equals t 0 E (t 0 )E(t 0 ). The latter follows from the converse Carathéodory-Julia theorem (see e.g., 8, 20): E(z) 2 d w (t 0 ) := lim z t0 z 2 = E(z)E(t 0 ) lim z t0 z t 0 = E(t 0 ) E(z) lim E(t 0) z t0 t 0 z t 0 = t 0 E (t 0 )E(t 0 ) <. Thus, a Schur function E cannot satisfy condition C 2 at a boundary point t i therefore, Statement () in Theorem 2.3 recovers Theorem 2..2 in 2. The same conclusion can be done when E is not rational but still analytic at t i. In the case when E is not rational and admits the nontangential boundary limit E(t i ) = η i, the situation is more subtle: Statement () shows that even in this case (if the convergence of E(z) to E(t i ) is not too fast), the function w = TE may satisfy interpolation conditions (.5), (.6). The next theorem concerns the number of negative squares of the function w = T Θ E. Theorem 2.5. If the Pick matrix P is invertible and has κ negative eigenvalues, then a Schur function E S 0 may satisfy conditions C 4 6 at at most κ interpolation nodes. Furthermore, if E meets conditions C 4 6 at exactly l ( κ) interpolation nodes, then the function w = T Θ E belongs to the class S κ l. Corollary 2.4 and Theorem 2.5 imply the sufficiency part in Theorem 2.2. Indeed, any Schur function E satisfies either conditions C 4 6 or C 3 at every interpolation node t i (i =,...,n). Let E meet conditions C 4 6 at t i,...,t il and C 3 at other n l interpolation nodes t j,..., t jn l. Then, by Corollary 2.4, the function w = T Θ E satisfies interpolation conditions (.7) for i {j,...,j n l } and fails to satisfy at least one of these conditions at the remaining l interpolation nodes. On the other hand, w has exactly κ l negative squares, by Theorem 2.5. Thus, for every E S 0, the function w = T Θ E solves Problem.6. Note also that Theorems 2.2 and 2.5 lead to parametrizations of solution sets for Problems.3 and.4. Indeed, by inclusions (2.), every solution w to Problem.3 (or to Problem.4) is also of the form (2.0) for some E S 0. Thus, there is a chance to describe the solution sets S 3 and S 4 by appropriate selections of

12 2 Vladimir Bolotnikov and Alexander Kheifets the parameter E in (2.0). Theorem 2.5 indicates how these selections have to be made. Theorem 2.6. A function w of the form (2.0) is a solution to Problem.3 if and only if the corresponding parameter E S 0 satisfies either condition C or C 2 for every i {,..., n}. Theorem 2.7. A function w of the form (2.0) is a solution to Problem.4 if and only if the corresponding parameter E S 0 either fails to have a nontangential boundary limit η i at t i or E(t i ) = η i and d E (t i ) > p ii ẽ i 2 for every i =,...,n (in other words, E meets one of conditions C, C 2, C 3 at each interpolation node t i ). As a consequence of Theorems 2.2 and 2.7 we get curious necessary and sufficient conditions (in terms of the interpolation data (.2)) for Problems.4 and.6 to be equivalent (that is, to have the same solution sets). Corollary 2.8. Problems.4 and.6 are equivalent if and only if all the diagonal entries of the inverse P of the Pick matrix are positive. Indeed, in this case, all the conditions in Theorem 2.7 are fulfilled for every E S 0 and every i {,..., n} and formula (2.6) gives a free Schur class parameter description of all solutions w of Problem.4. In the course of the proof of Theorem 2.5 we will discuss the following related question: given indices i,..., i l {,...,n}, does there exist a parameter E S 0 satisfying conditions C 4 6 at t i,...,t il? Due to Theorems 2.2 and 2.3, this question can be posed equivalently: does there exist a solution w to Problem.6 that misses interpolation conditions at t i,...,t il (Theorem 2.5 claims that if such a function exists, it belongs to the class S κ l ). The question admits a simple answer in terms of a certain submatrix of P = p ij n i,j=, the inverse of the Pick matrix. Theorem 2.9. There exists a parameter E satisfying conditions C 4 6 at t i,...,t il if and only if the l l matrix P := p iα,i β l α,β=

13 Boundary Nevanlinna Pick problem 3 is negative semidefinite. Moreover, if P is negative definite, then there are infinitely many such parameters. If P is negative semidefinite (singular), then there is only one such parameter, which is a Blaschke product of degree r = rank P. Note that all the results announced above have their counterparts in the context of the regular Nevanlinna-Pick problem with all the interpolation nodes inside the unit disk 5 The paper is organized as follows: Section 3 contains some needed auxiliary results which can be found (probably in a different form) in many sources and are included for the sake of completeness. In Section 4 we prove the necessity part in Theorem 2.2 (see Remark 4.4). In Section 5 we prove Theorem 2.3. In Section 6 we present the proofs of Theorems 2.9 and 2.5 and complete the proof of Theorem 2.2 (see Remark 6.2). The proof of Theorem 2. is contained in Section 7; some illustrative numerical examples are presented in Section Some preliminaries In this section we present some auxiliary results needed in the sequel. We have already mentioned the Stein identity P T PT = E E C C (3.) satisfied by the Pick matrix P constructed in (.4) from the interpolation data. Most of the facts recalled in this section rely on this identity rather than on the special form (2.3) of matrices T, E and C. Lemma 3.. Let T, E and C be defined as in (2.3), let P defined in (.4) be invertible and let µ be a point on T \ {t,...,t n }. Then. The row vectors Ẽ = ẽ... ẽ n and C = c... c n (3.2) defined by C Ẽ C = E (µi T) P (I µt ) (3.3) satisfy the Stein identity P TP T = Ẽ Ẽ C C. (3.4)

14 4 Vladimir Bolotnikov and Alexander Kheifets 2. The numbers c i and ẽ i are subject to ẽ i = c i 0 for i =,...,n. (3.5) 3. The nondiagonal entries p ij of P are given by p ij = ẽ iẽj c i c j t i t j (i j). (3.6) Proof: Under the assumption that P is invertible, identity (3.4) turns out to be equivalent to (3.). Indeed, by (3.3) and (3.), Ẽ Ẽ C C = (I µt)p ( µi T ) E E C C (µi T) P (I µt ) = (I µt)p ( µi T ) P T PT(µI T) P (I µt ) = (I µt)p (I µt ) P + PT(µI T) P (I µt ) = (I µt)p + µtp (I µt ) = P TP T. Let P = p ij n i,j=. Due to (3.2) and (2.3), equality of the ij-th entries in (3.4) can be displayed as p ij t i t j p ij = ẽ iẽ j c i c j (3.7) and implies (3.6) if i j. Letting i = j in (3.7) and taking into account that t i =, we get ẽ i = c i for i =,...,n. It remains to show that ẽ i and c i do not vanish. To this end let us assume that ẽ i = c i = 0. (3.8) Let e i be the i-th column of the identity matrix I n. Multiplying (3.4) by e i on the right we get or equivalently, since T e i = t i e i, P e i TP T e i = Ẽ ẽ i C c i = 0 (I t i T)P e i = 0. Since the points t,...,t n are distinct, all the diagonal entries but the i-th in the diagonal matrix I t i T are not zeroes; therefore, it follows from the last equality that all the entries in the vector P e i but the i-th entry are zeroes. Thus, P e i = αe i (3.9)

15 Boundary Nevanlinna Pick problem 5 for some α C and, since P is not singular, it follows that α 0. Now we compare the i-th columns in the equality (3.3) (i.e., we multiply both parts in (3.3) by e i on the right). For the left hand side we have, due to assumption (3.8), C ci 0 e i = =. Ẽ ẽ i 0 For the right hand side, we have, due to (3.9) and (2.3), C (µi T) E P (I µt )e i = α µt i C wi e µ t i E i = αt i By (3.3), the right hand side expressions in the two last equalities must be the same, which is not the case. The obtained contradiction completes the proof of (3.5). Remark 3.2. The numbers ẽ i and c i introduced in (3.2), (3.3) coincide with those in (2.). For the proof we first note that the formula (2.2) for Θ can be written, on account of (3.3), as C Θ(z) = I 2 + (z µ) (zi E n T) (µi n T) C Ẽ and then, since lim z t i (z t i )(zi T) = e i e i and e i (µi T) = (µ t i ) e i. (3.0) (recall that e i is the i-th column of the identity matrix I n ), we have C lim (z t i )Θ(z) = lim (z µ) e z t i z ti E i e i (µi T) C Ẽ C = e E i e i C Ẽ wi = c i ẽ i. (3.) Comparing the bottom entries in the latter equality we get (2.). In the rest of the section we recall some needed results concerning the function Θ introduced in (2.2). These results are well known in a more general situation C when T, C and E are matrices such that the pair (, T) is observable: E Ker j 0 C E T j = {0}, (3.2)

16 6 Vladimir Bolotnikov and Alexander Kheifets and P is an invertible Hermitian matrix satisfying the Stein identity (3.) (see e.g., 2). Note that the matrices defined in (2.3) satisfy a stronger condition: KerET j = {0}. (3.3) j 0 KerCT j = j 0 Remark 3.3. Under the above assumptions, the function Θ defined via formula (2.2) belongs to the class W κ with κ = sq P. Proof: The desired membership follows from the formula C K Θ,J (z, ζ) = (zi T) P ( ζi T ) C E E (3.4) for the kernel K Θ,J defined in (2.7). The calculation is straightforward and relies on the Stein identity (3.) only (see e.g., 2). It follows from (3.4) that Θ is J -unitary on T (that is, satisfies condition (2.6)) and that sq K Θ,J sq P = κ. Condition (3.2) guarantees that in fact sq K Θ,J = κ (see 2). Remark 3.4. Since Θ is J-unitary on T it holds, by the symmetry principle, that Θ(z) = JΘ(/ z) J, which together with formula (2.2) leads us to Θ(z) C = I 2 (z µ) (µi T) E P (I zt ) C E. (3.5) Besides (3.4) we will need realization formulas for two related kernels. Verification of these fomulas (3.6) and (3.7) is also straightforward and is based on the Stein identities (3.) and (3.4), respectively. Remark 3.5. Let Θ be defined as in (2.2). The following identities hold for every choice of z, ζ {t,...,t n }: Θ(ζ) JΘ(z) J z ζ J Θ(ζ) JΘ(z) z ζ = = C E C Ẽ (I ζt) P (I zt ) C E, (3.6) ( ζi T ) P(zI T) C Ẽ. (3.7)

17 Boundary Nevanlinna Pick problem 7 Let us consider conformal partitionings P P P = 2, P P 2 P = 22 P P2 P 2 E = E E 2, C = C C 2, Ẽ = P22, T = T 0, (3.8) 0 T 2 Ẽ Ẽ 2, C = C C2 (3.9) where P 22, P 22, T 2 C l l and E 2, C 2, Ẽ2, C 2 C l. Note that these decompositions contain one restrictive assumption: it is assumed that the matrix T is block diagonal. Lemma 3.6. Let us assume that P is invertible and let sq P = κ κ. Then P 22 is invertible, sq P22 = κ κ and the functions Θ () C (z) = I 2 + (z µ) (zi T E ) P (I µt ) C E (3.20) and Θ (2) (z) = I 2 + (z µ) C2 Ẽ 2 (I µt 2 ) P 22 (zi T 2) C 2 Ẽ 2 (3.2) belong to W κ and W κ κ, respectively. Furthermore, the function Θ defined in (2.2) admits a factorization Θ(z) = Θ () (z) Θ (2) (z). (3.22) Proof: The first statement follows by standard Schur complement arguments: since P and P are invertible, the matrix P 22 P 2 P P 2 (the Schur complement of P in P) is invertible and has κ κ negative eigenvalues. Since the block P 22 in P equals (P 22 P 2 P P 2), it also has κ κ negative eigenvalues. Realization formulas K Θ (),J(z, ζ) = R(z)P R(ζ) and K eθ (2),J (z, ζ) = R(z) P 22 R(ζ), (3.23) where we have set for short C R(z) = (zi T ), E R(z) = C2 Ẽ 2 (I µt 2 ) P 22 (zi T 2), are established exactly as in Remark 3.3 and rely on the Stein identities P T P T = E E C C and P 22 T 2 P 22 T 2 = Ẽ 2Ẽ2 C 2 C 2 (3.24)

18 8 Vladimir Bolotnikov and Alexander Kheifets which hold true, being parts of identities (3.) and (3.4). Formulas (3.23) guarantee that the rational functions Θ () and Θ (2) are J-unitary on T and moreover, that sq K Θ (),J sq P = κ and sq K eθ (2),J sq P 22 = κ κ. (3.25) Assuming that the factorization formula (3.22) is already proved, we have and thus, K Θ,J (z, ζ) = K Θ (),J(z, ζ) + Θ () (z)k eθ (2),J (z, ζ)θ() (ζ) κ = sq K Θ,J sq K Θ (),J + sq K eθ (2),J which together with inequalities (3.25) imply sq K Θ (),J = κ and sq K eθ (2),J = κ κ. It remains to prove (3.22). Making use of the well known equality P P = 0 P + P 2 P P2 P (3.26) we conclude from (3.3) that C2 C = Ẽ 2 E C = E (µi n T) P (I n µt 0 ) Il (µi n T) P P 2 This last relation allows us to rewrite (3.2) as Θ (2) C (z) = I 2 +(z µ) (µi T) E P P 2 P 22 (I l µt2 ). (3.27) (zi T 2 ) C 2 Ẽ 2. (3.28) Now we substitute (3.26) into the formula (2.2) defining Θ and take into account (3.20) and (3.27) to get C Θ(z) = Θ () (z) + (z µ) E P 2 P = Θ () C (z) + (z µ) E (zi n T) P P 2 (I n µt ) C E (µi T 2 ) C 2 Ẽ 2 (zi n T) P P 2. P 22

19 Boundary Nevanlinna Pick problem 9 Thus, (3.22) is equivalent to Θ (2) (z) = I 2 + (z µ)θ () (z) C (zi E n T) P P 2 (µi T 2 ) C 2 Ẽ 2. Comparing the last relation with (3.28) we conclude that to complete the proof it suffices to show that C Θ () (z) E C = E (zi n T) P P 2 (µi T) P P 2 (µi T 2 ) (zi T 2 ). (3.29) The explicit formula for Θ () (z) can be obtained similarly to (3.5): Θ () (z) C = I 2 (z µ) (µi T E ) P (I zt ) C E. (3.30) Next, comparing the top block entries in the Stein identity (3.) we get, due to decompositions (3.8) and (3.9), P P 2 T P P 2 T = E E C C which, being multiplied by (I zt ) on the left and by (zi T) on the right, leads us to (I zt ) (E E C C) (zi T) = (I zt ) T P P 2 + P P 2 (zi T). (3.3)

20 20 Vladimir Bolotnikov and Alexander Kheifets Upon making use of (3.29) and (3.3) we have C Θ () (z) (zi E n T) P P 2 C = (zi T) P P 2 E C +(z µ) C = E +(z µ) C = C = E E (µi T ) I (zi T ) P C E (µi T ) P P C2 2 + P P 2 (zi T) P P 2 E 2 (zi T 2 ) (µi T E ) ( P P 2(zI T 2 ) (zi T ) P P 2 P C2 2(µI T 2 ) + (zi T E 2 ) 2 (µi T) P P 2 (zi T 2 ) which proves (3.29) and therefore, completes the proof of the lemma. Remark 3.7. The case when l = in Lemma 3.6 will be of special interest. In this case, P 22 = γ n, P22 = p nn, T 2 = t n, C 2 = w n, E 2 =, C2 = c n, Ẽ 2 = ẽ n. Then the formula (3.2) for Θ (2) simplifies to Θ (2) (z) = I 2 + z µ ( µ t n )(z t n ) cn ẽ n p nn c n ẽ n ). (3.32) 4. Fundamental Matrix Inequality In this section we characterize the solution set S 6 of Problem.6 in terms of certain Hermitian kernel. We start with some simple observations. Proposition 4.. Let K(z, ζ) be a Hermitian kernel defined on Ω C and with sq K = κ. Then. For every choice of an integer p, of a Hermitian p p matrix A and of a p vector valued function B, A B(z) sq B(ζ) κ + p. K(z, ζ)

21 Boundary Nevanlinna Pick problem 2 2. If λ,..., λ p are points in Ω and if then A = K(λ j, λ i ) p i,j= and B(z) = K(z, λ ). K(z, λ p ), (4.) A B(z) sq B(ζ) = κ. (4.2) K(z, ζ) Proof: For the proof of the first statement we have to show that for every integer m and every choice of points z,...,z m Ω, the block matrix m A B(zj ) M = B(z i ) K(z j, z i ) i,j= (4.3) has at most κ + p negative eigenvalues. It is easily seen that M contains m block identical rows of the form A B(z ) A B(z 2 )... A B(z n ). Deleting all these rows but one and deleting also the corresponding columns, we come up with the (m + p) (m + p) matrix A B(z )... B(z m ) B(z ) K(z, z )... K(z, z m ) M =... B(z m ) K(z m, z )... K(z m, z m ) having the same number of positive and negative eigenvalues as M. The bottom m m principal submatrix of M has at most κ negative eigenvalues since sq K = κ. Since M is Hermitian, we have by the Cauchy s interlacing theorem (see e.g., 4, p. 59), that sq M κ + p. Thus, sq M κ + p which completes the proof of Statement. If A and B are of the form (4.), then the matrix M in (4.3) is of the form K(ζ j, ζ i ) m+pm i,j= where all the points ζ i live in Ω. Since sq K = κ, it follows that sq M κ for every choice of z,..., z m in Ω which means that the kernel A B(z) B(ζ) has at most κ negative squares on Ω. But it has at least κ K(z, ζ) negative squares since it contains the kernel K(z, ζ) as a principal block. Thus, (4.2) follows. Theorem 4.2. Let P, T, E and C be defined as in (.4) and (2.3), let w be a function meromorphic on D and let the kernel K w be defined as in (.). Then w

22 22 Vladimir Bolotnikov and Alexander Kheifets is a solution of Problem.6 if and only if the kernel P (I zt K w (z, ζ) := ) (E C w(z)) (E w(ζ) C)(I ζt) K w (z, ζ) has κ negative squares on D ρ(w): (4.4) sq K w (z, ζ) = κ. (4.5) Proof of the necessity part: Let w be a solution of Problem.6, i.e., let w belong to the class S κ κ κ interpolation nodes. for some κ κ and satisfy conditions (.7) at all but First we consider the case when w S κ. Then w satisfies all the conditions (.7) (i.e., w is also a solution to Problem.4). Furthermore, sq K w = κ and by the second statement in Proposition 4., the kernel K w (z, z )... K w (z n, z ) K w (z, z ) K () (z, ζ) :=... K w (z, z n )... K w (z n, z n ) K w (z, z n ) K w (z, ζ)... K w (z n, ζ) K w (z, ζ) (4.6) has κ negative squares on D ρ(w) for every choice of points z,..., z n D ρ(w). Since the limits d w (t i ) and w(t i ) = w i exist for i =,...,n, it follows that K w (z j, z i ) n i,j= = w(zi ) w(z j ) z i z j n i,j= (by definition (.8) of the matrix P w (t,..., t n )) and also P w (t,..., t n ) (4.7) K w (z i, ζ) = w(ζ) w(z i ) ζz i w(ζ) w i ζt i (i =,...,n). Note that by the structure (2.3) of the matrices T, E and C, (E w(ζ) C)(I ζt) w(ζ) w w(ζ) w n = ζt... ζt n which, being combined with the previous relation, gives Kw (z, ζ)... K w (z n, ζ) (E w(ζ) C)(I ζt). (4.8) Now we take the limit in (4.6) as z i t i for i =,...,n; on account of (4.7) and (4.8), the limit kernel has the form K (2) P (z, ζ) := w (t,..., t n ) (I zt ) (E C w(z)) (E w(ζ) C)(I ζt). K w (z, ζ)

23 Boundary Nevanlinna Pick problem 23 Since K (2) is the limit of a family of kernels each of which has κ negative squares, sq K (2) κ. It remains to note that the kernel K w defined in (4.4) is expressed in terms of K (2) as K w (z, ζ) = K (2) (z, ζ) + P P w (t,...,t n ) and since the second term on the right hand side is positive semidefinite (due to the first series of conditions in (.7); see also (.20)), sq K w sq K (2) κ. On the other hand, since K w contains the kernel K w as a principal submatrix, sq K w sq K w = κ which eventually leads us to (4.5). Note that in this part of the proof we have not used the fact that sq P = κ. Now we turn to the general case: let w S κ for some κ κ and let conditions (.7) be fulfilled at all but l := κ κ interpolation nodes t i s. We may assume without loss of generality that conditions (.7) are satisfied at t i for i =,...,n l: d w (t i ) γ i and w(t i ) = w i (i =,...,n l). (4.9) Let us consider conformal partitionings (3.8), (3.9) for matrices P, T, C and E and let us set for short so that F i (z) = (I zt i ) (E i C i w(z)) (i =, 2) (4.0) F (z) = (I zt F 2 (z) ) (E C w(z)). (4.) The matrix P is the Pick matrix of the truncated interpolation problem with the data t i, w i, γ i (i =,..., n l) and with interpolation conditions (4.9). By the first part of the proof, the kernel K w (z, ζ) := P F (z) F (ζ) K w (z, ζ) (4.2) has κ negative squares on D ρ(w). Now we apply the first statement in Proposition 4. to K(z, ζ) = K w (z, ζ), B(z) = P 2 F 2 (z) and A = P 22 (4.3) to conclude that P22 B(z) sq B(ζ) sq Kw (z, ζ) Kw + l = κ + (κ κ ) = κ. (4.4)

24 24 Vladimir Bolotnikov and Alexander Kheifets By (4.3) and (4.2), the latter kernel equals P22 B(z) B(ζ) = P 22 P 2 F 2 (z) P 2 P F (z). Kw (z, ζ) F 2 (ζ) F (ζ) K w (z, ζ) Now it follows from (4.4) and (4.2) that K w (z, ζ) = U P22 B(z) B(ζ) U, where U = Kw (z, ζ) 0 I n l 0 I l which, on account of (4.4), implies that sq K w κ. Finally, since K w contains P as a principal submatrix, sq K w sq P = κ which now implies (4.5) and completes the proof of the necessity part of the theorem. The proof of the sufficiency part will be given in Sections 6 and 7 (see Remarks 6.3 and 7.3 there). In the case when P is invertible, all the functions satisfying (4.5) can be described in terms of a linear fractional transformation. Theorem 4.3. Let the Pick matrix P be invertible and let Θ = Θ ij be the 2 2 matrix valued function defined in (2.2). A function w meromorphic on D is subject to FMI (4.5) if and only if it is of the form for some Schur function E S 0. w(z) = T Θ E := Θ (z)e(z) + Θ 2 (z) Θ 2 (z)e(z) + Θ 22 (z) (4.5) Proof: The proof is about the same as in the definite case. Let S be the Schur complement of P in the kernel K w defined in (4.4): S(z, ζ) := K w (z, ζ) (E w(ζ) C)(I ζt) P (I zt ) (E C w(z)). Obvious equalities K w (z, ζ) := w(ζ) w(z) ζz = w(ζ) J where J is the matrix introduced in (2.5), and E w(ζ) C = w(ζ) J allows us to represent S in the form S(z, ζ) = w(ζ) { J z ζ + C E C E (I zt ) C E } w(z) w(z) (I ζt) P

25 Boundary Nevanlinna Pick problem 25 or, on account of identity (3.6), as S(z, ζ) = w(ζ) By the standard Schur complement argument, Θ(ζ) JΘ(z) z ζ sq K w = sq P + sq S w(z). which implies, since sq P = κ, that (4.5) holds if and only if the kernel S is positive definite on ρ(w) D: w(ζ) Θ(ζ) JΘ(z) z ζ w(z) 0. (4.6) It remains to show that (4.6) holds if and only if w is of the form (4.5). To show the only if part, let us consider meromorphic functions u and v defined by u(z) w(z) := Θ(z) v(z). (4.7) Then inequality (4.6) can be written in terms of these functions as u(ζ) v(ζ) J u(z) ζz = v(ζ) v(z) u(ζ) u(z) v(z) ζz 0. (4.8) As it follows from definition (4.7), u and v are analytic on ρ(w) D. Moreover, v(z) 0 for every z ρ(w) D. (4.9) Indeed, assuming that v(ξ) = 0 at some point ξ D, we conclude from (4.8) that u(ξ) = 0 and then (4.7) implies that det Θ(ξ) = 0 which is a contradiction. Due to (4.9), we can introduce the meromorphic function E(z) = u(z) v(z) which is analytic on ρ(w) D. Writing (4.8) in terms of E as v(ζ) E(ζ) E(z) ζz we then take advantage of (4.9) to conclude that E(ζ) E(z) ζz v(z) 0 (z, ζ ρ(w) D), 0 (z, ζ ρ(w) D). (4.20) The latter means that E is (after an analytic continuation to the all of D) a Schur function. Finally, it follows from (4.7) that w u Θ u + Θ = Θ = 2 v v Θ 2 u + Θ 22 v

26 26 Vladimir Bolotnikov and Alexander Kheifets which in turn implies w = Θ u + Θ 2 v Θ 2 u + Θ 22 v = Θ E + Θ 2 Θ 2 E + Θ 22 = T Θ E. Now let E be a Schur function. Then the function V (z) = Θ 2 (z)e(z) + Θ 22 (z) does not vanish identically. Indeed, since Θ is rational and Θ(µ) = I 2, it follows that Θ 22 (z) and Θ 2 (z) 0 if z is close enough to µ. Since E(z) everywhere in D, the function V does not vanish on U δ = {z D : z µ < δ} if δ is small enough. Thus, formula (4.5) makes sense and can be written equivalently as Then it is readily seen that E(ζ) E(z) ζz w(z) = Θ(z) E(z) = E(ζ) J ζz = V (ζ) V (z) w(ζ) V (z) E(z) Θ(ζ) JΘ(z) z ζ w(z) for z, ζ ρ(w) D. Since E is a Schur function, the latter kernel is positive on ρ(w) D and since V 0, (4.6) follows. Remark 4.4. Combining Theorems 4.2 and 4.3 we get the necessity part in Theorem 2.2. Indeed, by the necessity part in Theorem 4.2, any solution w of Problem.6 satisfies (4.5); then by Theorem 4.3, w = T Θ E for some E S 0. In the case when κ = 0, Theorem 4.2 was established in 2. Theorem 4.5. Let the Pick matrix P be positive semidefinite. Then a function w defined on D is a solution to Problem. (i.e., belongs to the Schur class S 0 and meets conditions (.6)) if and only if where K w (z, ζ) is the kernel defined in (4.4). K w (z, ζ) 0 (z, w D) (4.2) Under the a priori assumption that w is a Schur function, condition (4.2) can be replaced by a seemingly weaker matrix inequality K w (z, z) 0 for every z D

27 Boundary Nevanlinna Pick problem 27 which is known in interpolation theory as a Fundamental Matrix Inequality (FMI) of V. P. Potapov. We will follow this terminology and will consider relation (4.5) as an indefinite analogue of V. P. Potapov s FMI. It is appropriate to note that a variation of the Potapov s method was first applied to the Nevanlinna-Pick problem (with finitely many interpolation nodes inside the unit disk) for generalized Schur functions in 0. We conclude this section with another theorem concerning the classical case which will be useful for the subsequent analysis. Theorem 4.6. () If the Pick matrix P is positive definite then all the solutions w to Problem. are parametrized by the formula (2.0) with the coefficient matrix Θ defined as in (2.2) with E being a free Schur class parameter. (2) If P is positive semidefinite and singular, then Problem. has a unique solution w which is a Blaschke product of degree r = rankp. Furthermore, this unique solution can be represented as w(z) = x (I zt 2 ) E x (I zt 2 ) C (4.22) where T, C and E are defined as in (2.3) and where x is any nonzero vector such that Px = 0. These results are well known and has been established using different methods in, 2, 3, 2,. In regard to methods used in the present paper, note that the first statement follows immediately from Theorems 4.5 and 4.3. This demonstrates how the Potapov s method works in the definite case (and this is exactly how the result was established in 2). The second statement also can be derived from Theorem 4.5: if w solves Problem., then the kernel K w (z, ζ) defined in (4.4) is x positive definite. Multiplying it by the vector on the right and by its adjoint on the left we come to the positive definite kernel x Px x (I zt ) (E C w(z)) (E w(ζ) C)(I ζt) 0. x K w (z, ζ) Thus, for every x 0 such that Px = 0, we also have x (I zt ) (E C w(z)) 0. Solving the latter identity for w we arrive at formula (4.22). The numerator and the denominator in (4.22) do not vanish identically due to conditions (3.3). Since x can be chosen so that n rankp its coordinates are zeros, the rational

28 28 Vladimir Bolotnikov and Alexander Kheifets function w is of McMillan degree r = rank P. Due to the Stein identity (3.), w is inner and therefore, it is a finite Blachke product of degree r. 5. Parameters and interpolation conditions In this section we prove Theorem 2.3. It will be done in several steps formulated as separate theorems. In what follows, U E and V E will stand for the functions U E (z) = Θ (z)e(z) + Θ 2 (z), V E (z) = Θ 2 (z)e(z) + Θ 22 (z) (5.) for a fixed Schur function E, so that UE (z) E(z) = Θ(z) V E (z) and (2.0) takes the form (5.2) w(z) := T Θ E = U E(z) V E (z). (5.3) Substituting (3.0) into (5.2) and setting ( Ψ(z) = (zi T) Ẽ C ) E(z) for short, we get (5.4) U E (z) = E(z) (z µ)c(µi T) Ψ(z), (5.5) V E (z) = (z µ)e(µi T) Ψ(z). (5.6) Furthermore, for w of the form (5.3), we have w(ζ) w(z) ζz = V E (ζ) V E (z) VE(ζ) V E (z) U E (ζ) U E (z). (5.7) ζz Note that V E (ζ) V E (z) U E (ζ) U E (z) = U E (ζ) V E (ζ) UE (z) J V E (z) = E(ζ) Θ(ζ) E(z) JΘ(z) = E(ζ) E(z) + ( ζz)ψ(ζ) PΨ(z), where the second equality follows from (5.2), and the third equality is a consequence of (3.7) and definition (5.4) of Ψ. Now (5.7) takes the form w(ζ) ( w(z) E(ζ) ) E(z) ζz = V E (ζ) V E (z) ζz + Ψ(ζ) PΨ(z). (5.8)

29 Boundary Nevanlinna Pick problem 29 Remark 5.. Equality (5.8) implies that for every E S 0 and Θ W κ, the function w = T Θ E belongs to the generalized Schur class S κ for some κ κ. Indeed, it follows from (5.8) that sq K w sq K E + sq P = 0 + κ. Upon evaluating (5.8) at ζ = z we get w(z) 2 z 2 = ( ) E(z) 2 V E (z) 2 z 2 + Ψ(z) PΨ(z) (5.9) and realize that boundary values of w(t i ) and d w (t i ) can be calculated from asymptotic formulas for Ψ, U E, V E and E as z tends to one of the interpolation nodes t i. These asymptotic relations are presented in the next lemma. Lemma 5.2. Let E be a Schur function, let Ψ, U E and V E be defined as in (5.4), (5.5) and (5.6), respectively, and let t i be an interpolation node. Then the following asymptotic relations hold as z tends to t i nontangentially: (z t i )Ψ(z) = e i (ẽ i c i E(z)) + O( z t i ), (5.0) (z t i )U E (z) = w i (ẽ i c i E(z)) + O( z t i ), (5.) (z t i )V E (z) = (ẽ i c i E(z)) + O( z t i ). (5.2) Proof: Recall that e i be the i-th column in the identity matrix I n. Since (z t i )(zi T) = e i e i + O( z t i ) as z t i, and since E(z) is uniformly bounded on D, we have by (5.4), ( (z t i )Ψ(z) = (z t i )(zi T) Ẽ C ) E(z) = e i e (Ẽ i C E(z)) + O( z t i ) which proves (5.0), since e i C = c i and e i Ẽ = ẽ i by (3.2). Now we plug in the asymptotic relation (5.0) into the formulas (5.5) and (5.0) for U E and V E and make use of evident equalities C(µI T) e i = w i µ t i and E(µI T) e i = µ t i (5.3)

30 30 Vladimir Bolotnikov and Alexander Kheifets to get (5.) and (5.2): (z t i )U E (z) = (z t i )E(z) (z t i )(z µ)c(µi T) Ψ(z) = (µ z)c(µi T) e i (ẽ i c i E(z)) + O( z t i ) = µ z µ t i w i (ẽ i c i E(z)) + O( z t i ) = w i (ẽ i c i E(z)) + O( z t i ), (z t i )V E (z) = (z t i ) (z t i )(z µ)e(µi T) Ψ(z) = (µ z)e(µi T) e i (ẽ i c i E(z)) + O( z t i ) = (ẽ i c i E(z)) + O( z t i ). Lemma 5.3. Let w S κ, let t 0 T, and let us assume that the limit w(r j t 0 ) 2 d := lim j rj 2 < (5.4) exists and is finite for some sequence of numbers r j (0, ) such that lim j r j =. Then the nontangential limits d w (t 0 ) and w(t 0 ) (defined as in (.3) and (.4)) exist and moreover d w (t 0 ) = d and w(t 0 ) =. (5.5) Proof: Since w is a generalized Schur function, it admits the Krein-Langer representation (.9) and identity (.) holds at every point z D. In particular, ( ) w(r j t 0 ) 2 S(r j t 0 ) 2 rj 2 = B(r j t 0 ) 2 rj 2 B(r jt 0 ) 2 rj 2. (5.6) Since B is a finite Blaschke product, it is analytic at t 0 and the limit d B (t 0 ) := B(z) 2 lim z t 0 z 2 exists and is finite. Assumption (5.4) implies therefore that the limit S(r j t 0 ) 2 lim j rj 2 = d + d B (t 0 ) exists and is finite. Since S S 0, we then conclude by the Carathéodory-Julia theorem (see e.g., 7, 8, 20) that the nontangential limits d S (t 0 ) and S(t 0 ) exist and moreover, d S (t 0 ) = d + d B (t 0 ) and S(t 0 ) =. (5.7) Now we pass to limits in (.9) and (.) as z tends to t 0 nontangentially to get w(t 0 ) := lim z t0 w(z) = S(t 0) B(t 0 ) and d w (t 0 ) := lim z t0 w(z) 2 z 2 = d S (t 0 ) d B (t 0 )

31 Boundary Nevanlinna Pick problem 3 and relations (5.7) imply now (5.5) and complete the proof. Theorem 5.4. If E S 0 meets condition C at t i (i.e., the nontangential boundary limit lim E(z) is not equal to η i = ẽ i z ti c i T Θ E is subject to or fails to exist), then the function w = w(z) 2 lim w(z) = w i and lim z t i z ti z 2 = γ i. (5.8) Proof: By the assumption of the theorem, there exists ε > 0 and a sequence of points {r α t i } α= tending to t i radially (0 < r α < and r α ) such that ẽ i c i E(r αt i ) ε for every α. (5.9) Since e i Pe i = γ i by the definition (.4) of P, it follows from (5.0) that Furthermore, relation z t i 2 Ψ(z) PΨ(z) = ẽ i c i E(z) 2 γ i + O( z t i ). z t i 2 V E (z) 2 = ẽ i c i E(z) 2 + O( z t i ) is a consequence of (5.2) and, since E is uniformly bounded on D, it is clear that lim z t i 2 E(z) 2 z t i z 2 = 0. Now we substitute the three last relations into (5.9) and let z = r α t i t i ; due to (5.9) we have E(z) 2 w(z) 2 z t i 2 z lim z=r αt i t i z 2 = lim 2 + z t i 2 Ψ(z) PΨ(z) z=r αt i t i z t i 2 V E (z) 2 = 0 + γ i = γ i. Since w is a generalized Schur function (by Remark 5.), we can apply Lemma 5.3 to conclude that the nontangential limit d w (t i ) exists and equals γ i. This proves the second relation in (5.8). Furthermore, by (5.) and (5.2) and in view of (5.9), (z t i )U E (z) lim w(z) = lim z=r αt i t i z ti (z t i )V E (z) = w i. (5.20) Again by Lemma 5.3, the nontangential limit w(t i ) exists; therefore, it is equal to the subsequential limit (5.20), that is, to w i. This proves the first relation in (5.8) and completes the proof of the theorem.

32 32 Vladimir Bolotnikov and Alexander Kheifets The next step will will be to handle condition C 2 (see (2.4)). We need an auxiliary result. Lemma 5.5. Let t 0 T and let E be a Schur function such that Then E(z) 2 lim E(z) = E 0 ( E 0 = ) and lim z t 0 z t0 z 2 =. (5.2) E(z) 2 lim z t 0 z 2 z t 0 E(z) E 0 2 = 0 and lim z t0 z t 0 E(z) E 0 = 0. (5.22) and thus, Proof: Since E 0 =, we have 2Re( E(z)E 0 ) = ( E(z)E 0 ) + ( E 0 E(z)) = E(z)E E 0 2 E(z) 2 E(z) 2 E(z) E 0 = E(z)E 0 Re ( E(z)E 0 ) 2 Furthermore, for every z in the Stoltz domain it holds that Γ a (t 0 ) = {z D : z t 0 < a( z )}, a >, z 2 z t 0 z z t 0 > a, ( E(z) 2 ). (5.23) which together with (5.23) leads us to E(z) E 0 z t 0 2 E(z) 2 = z t 0 2 E(z) 2 z 2 z 2 z t 0 > 2a E(z) 2 z 2 which is equivalent to E(z) 2 z 2 z t 0 E(z) E 0 2a. (5.24) Note that the denominator E(z) E 0 in the latter inequality does not vanish: assuming that E(z 0 ) = E 0 at some point z 0 D, we would have by the maximum modulus principle (since E 0 = ) that E(z) E 0 which would contradict the second assumption in (5.2). Finally, by this latter assumption, d E (t 0 ) = and relations (5.22) follow immediately from (5.24).

33 Boundary Nevanlinna Pick problem 33 Theorem 5.6. Let E S 0 meet condition C 2 at t i : lim E(z) = η i = ẽ i E(z) 2 z t i c and lim z ti i z 2 =. (5.25) Then the function w = T Θ E is subject to relations (5.8). Proof: Let for short and note that i (z) := ẽ i c i E(z) t i z i (z) 0 (z D). (5.26) To see this we argue as in the proof of the previous lemma: assuming that E(z 0 ) = η i at some point z 0 D, we would have by the maximum modulus principle (since η i = ) that E(z) η i which would contradict the second assumption in (5.25). Furthermore, since η i = and due to assumptions (5.25), we can apply Lemma 5.5 (with E 0 = η i and t 0 = t i ) to conclude that and E(z) 2 lim z t 0 z 2 i (z) 2 = 0 (5.27) lim i (z) = 0. (5.28) z t 0 Now we divide both parts in asymptotic relations (5.0) (5.2) by (ẽ i c i E(z)) and write the obtained equalities in terms of i as i (z) Ψ(z) = e i + i (z) O(), i (z) U E (z) = w i + i (z) O(), i (z) V E (z) = + i (z) O(). By (5.28), the following nontangential limits exist lim i (z) Ψ(z) = e i, z t i lim i (z) U E (z) = w i, z ti and we use these limits along with (5.27) to pass to limits in (5.9): lim i (z) V E (z) = z ti w(z) 2 i (z) 2 E(z) 2 z lim z t i z 2 = lim 2 + i (z) 2 Ψ(z) PΨ(z) z ti i (z) 2 V E (z) 2 Finally, = 0 + e i Pe i lim z t i w(z) = lim z ti which completes the proof. = γ i. i (z) U E (z) i (z) V E (z) = w i = w i,