Relaxed commutant lifting and a relaxed Nehari problem: Redheffer state space formulas

Transcription

1 Relaxed commutant lifting and a relaxed Nehari problem: Redheffer state space formulas arxiv:math/ v1 mathf 3 Nov 2006 S ter Horst bstract The description of all solutions to the relaxed commutant lifting problem in terms of an underlying contraction, obtained earlier in joint work of the author with E Frazho and M Kaashoek, is transformed into a linear fractional Redheffer state space form Under certain additional conditions the coefficient functions in this representation are described explicitly in terms of the original data The main theorem is a generalization of the Redheffer description of all solutions to the classical commutant lifting problem To illustrate the result a relaxed version of the Nehari extension problem is considered, and an explicit Redheffer description of all its solutions is given, assuming that a certain truncated Hankel operator is a strict contraction The latter result is specified further for two special cases 1 Introduction The classical commutant lifting theorem, which was obtained by Sz-Nagy-Foias 11 and originated from the work of Sarason 13, has been used to solve, among other things, a large number of metric constraint interpolation and extension problems; see 5 for a recent overview In 6, extending the classical theory, a relaxed commutant lifting problem is introduced, and a particular (so-called central) solution, satisfying a maximum entropy condition, is obtained and used to solve a number of relaxed versions of the classical interpolation problems Descriptions of all solutions to the relaxed commutant lifting problem are given in 7, 10 and 8 The ones in 7 and 8 are in terms of Schur class functions, whereas 10 uses a choice sequences approach The present paper can be seen as an addition to 8, where the description of all solutions (see Theorem 11 below) is given in terms of an underlying contraction The aim of the present paper is to present a linear fractional Redheffer type description of all solutions which is explicit in terms of the original data For this description some additional conditions on the data are needed The main result (see Theorem 12 below) generalizes the corresponding result for the classical commutant lifting problem given in 5 To be more precise, recall that a lifting data set is a set = {,T,U,R,Q} consisting of five Hilbert space operators The operator is a contraction mapping H into H, the operator U on K is a minimal isometric lifting of the contraction T on H, and R and Q are operators from H 0 to H, satisfying the following constraints: T R = Q and R R Q Q (11) 1

2 Given a lifting data set as above, the relaxed commutant lifting problem is to describe all contractions B from H to K such that Π H B = and U BR = BQ (12) Here Π H is the orthogonal projection from K onto the subspace H contraction B from H into K satisfying (12) will be called a contractive interpolant for Hence the relaxed commutant lifting problem is to describe all contractive interpolants for the lifting data set Without loss of generality we can, and will, assume that U is the Sz-Nagy-Schäffer isometric lifting of T (4, Section VI3), that is, K is the direct sum of H and the Hardy space H 2 (D T ) and U = T 0 E DT D T S DT on H H 2 (D T ) (13) Here we follow the convention that for a contraction C the symbol D C denotes the positive square root of I C C, while D C stands for the closure of the range of D C s usual D C and D C are referred to as the defect operator and defect space of C, respectively Moreover, for a Hilbert space Y the unilateral shift on the Hardy space H 2 (Y) is denoted by S Y, and we write E Y for the canonical embedding of Y onto the subspace of constant functions in H 2 (Y), that is, (E Y v)(λ) = v for all λ D and each v Y Put D = (Q Q R R) 1 2 and D = D H 0 (14) Then (11) implies that Q D 2 Q = D 2 +R D 2 T R+R D 2 R (15) Indeed, this is the case because for all h H 0 we have D Qh 2 = Qh 2 Qh 2 = D h 2 + Rh 2 T Rh 2 = D h 2 + Rh 2 T Rh 2 + Rh 2 Rh 2 = D h 2 + D T Rh 2 + D Rh 2 With the lifting data set we associate a contraction ω defined by DT R ω : F D T D, ωd Q =, F = D D R QH 0 (16) The relation in (15) guarantees that ω is contractive Moreover, ω is an isometry if and only if D = 0 In terms of the operators from the lifting data set this is equivalent to R R = Q Q The first main theorem from 8 can now be formulated as follows Theorem 11 Let = {,T,U,R,Q} be a lifting data set with U the Sz-Nagy-Schäffer isometric lifting of T, and let B be an operator from H into H H 2 (D T ) Then B is a contractive interpolant for if and only if B admits a representation of the form B = ΓD : H H H 2 (D T ), (17) 2

3 where Γ is the operator mapping D into H 2 (D T ) given by ( Γd)(λ) = Π DT Z(λ)(I λπ D Z(λ)) 1 d (d D,λ D) (18) for some Schur class function Z from S(D,D T D ) with Z(λ) F = ω for each λ D Let us explain in some detail the notations that are used in the above theorem or will appear in the sequel Throughout capital calligraphic letters denote Hilbert spaces The Hilbert space direct sum of U and Y is denoted by U Y or by n operator is a bounded linear transformation acting between Hilbert spaces With L(U, Y) we denote the set of all operators from U into Y The identity operator on the space U is denotedbyi U, orjustbyi whentheunderlying spaceisclearfromthecontext Bydefinition, a subspace is a closed linear manifold Let M be a subspace of U Then U M stands for the orthogonal complement of M in U We follow the convention that the symbol Π M denotes the orthogonal projection from U onto M viewed as an operator from U to M, whereas P M stands for the orthogonal projection from U onto M acting as an operator on U Note that with this notation Π M is the canonical embedding of M into U and P M = Π M Π M n operator C on U is referred to as positive definite (notation: C > 0) if C is invertible and positive (ie, Cu,u 0 for each u U) n operator N in L(U,Y) is called left invertible if there exists an operator M in L(Y,U) such that MN = I U In that case M is called a left inverse of N Note that N is left invertible if and only if N N is positive definite By definition, the symbol H (U,Y) stands fortheclass of uniformly boundedanalytic functions on the open unit disc D with values in L(U,Y) The set H (U,Y) forms a Banach space with respect to the supremum norm In that case S(U,Y) denotes the closed unit disc in H (U,Y), and is referred to as the Schur class associated with U and Y Functions from S(U,Y) are called Schur class functions Finally, by H 2 (U,Y) we denote the class of all functions F on D with values in L(U,Y) such that for each u in U the map λ F(λ)u defines a function in the Hardy class H 2 (Y) Such a function is automatically analytic on D Obviously H (U,Y) is properly contained in H 2 (U,Y) The next theorem is the main result of the present paper Theorem 12 Let = {,T,U,R,Q} be a lifting data set with U the Sz-Nagy-Schäffer isometric lifting of T ssume that is a strict contraction and R has a left inverse Then an operator B mapping H into H H 2 (D T ) is a contractive interpolant for if and only if B admits a representation of the form B = Γ : H U Y H H 2 (D T ) where Γ is the operator mapping H into H 2 (D T ) given by, (19) (Γh)(λ) = Φ 22 (λ)h+φ 21 (λ)v(λ)(i Φ 11 (λ)v(λ)) 1 Φ 12 (λ)h (h H,λ D) (110) 3

4 for some Schur class function V from S(KerQ,D D T KerR ) Moreover, any such function V defines an operator Γ mapping H into H 2 (D T ) via (110) Here Φ 11 and Φ 21 are functions from the Schur classes S(D D T KerR,KerQ ) and S(D D T KerR,D T ), respectively, and Φ 12 and Φ 22 are functions from H 2 (H,KerQ ) and H 2 (H,D T ), respectively, and these functions are given by Φ 11 (λ) = λx 3 (I λx 1 ) 1 X 2, Φ 12 (λ) = X 3 (I λx 1 ) 1, Φ 21 (λ) = X 5 +λx 4 (I λx 1 ) 1 X 2, Φ 22 (λ) = X 4 (I λx 1 ) 1, (λ D) (111) where X 1,,X 5 are the operators defined by X 1 = R(Q D 2 Q) 1 Q D 2 on H, X 2 = R(R D 2 R) 1 J 1 2 Π D D T + D 2 Π KerR 1 2 R Π KerR : D D T KerR H, X 3 = 1 2 Q Π KerQ : H KerQ X 4 = D T R(Q D 2 Q) 1 Q D 2 : H D T, X 5 = Π DT 1 2 Π D D T : D D T KerR D T, (112) and Q on KerQ, R on KerR and on D D T are the positive definite operators defined by Q = Π KerQ D 2 Π KerQ, R = Π KerR D 2 = I +J(R D 2R) 1 J, where J = Π KerR, D D T R : H 0 D D T (113) The condition in the above theorem that is a strict contraction is equivalent to the requirement that the defect operator D of is positive definite on H Moreover, if R is left invertible, then the second condition in (11) implies that Q is also left invertible Thus the combinationofbothconditionsresults ind QandD Rbeing leftinvertible, orequivalently, Q D 2Q and R D 2R being positive definite, both on H 0 So the extra assumptions on the lifting data set in Theorem 12 imply that the operators D 2, (R D 2R) 1 and (Q D 2Q) 1 appearing in Theorem 12 are well defined When taking the zero function for V in Theorem 12 we see that (110) reduces to (Γh)(λ) = Φ 22 (λ)h for h H and λ D The solution obtained in this way is precisely the central solution given in 6 In 6, Proposition 53, it was shown that the spectral radius of X 1 in (112) is strictly less then one if R and Q are such that R λq is left invertible for each λ D Note that in this case the functions Φ 12 and Φ 22 are uniformly bounded on D, that is, Φ 12 and Φ 22 are functions from the classes H (H,KerQ ) and H (H,D T ), respectively This remark will be useful later on when we consider the relaxed Nehari extension problem The proof of Theorem 12 is given in Section 2 In this section we also derive additional properties of the operator valued functions in (111) (see Corollary 24 below) 4

5 Recall that the classical commutant lifting problem appears when in the lifting data set {,T,U,R,Q} the operator Q is an isometry and R the identity operator on H, and thus, in particular, H 0 = H Under these additional conditions Theorem 12 reduces to the first part of Theorem VI61 in 5 (see Corollary 23 below for further details) Moreover, in that case (110) provides a proper parameterization, that is, there exists a unique Schur class function V such that B is given by (19) and (110) In general, for an arbitrary lifting data set formula (110) does not provide a proper parameterization This follows from Theorem 12 in 8 To illustrate Theorem 12 we consider a relaxed version of the operator-valued Nehari extension problem Let N be a positive integer (N > 0) and let F 1,F 2, be a sequence of operators from U to Y satisfying n=1 F nu 2 < for each u U The relaxed Nehari extension problem considered in this paper is to find all sequences of operators H 0,H 1, from U to Y with the property that n=0 H nu 2 < for each u U, and such that the operator from U N into l 2 (Y) given by the operator matrix representation F 2 F 3 F (N+1) F 1 F 2 F N H 0 F 1 F (N 1) H 1 H 0 : U N l 2 (Y) (114) F 1 H0 has operator norm at most one Here U N is the Hilbert space direct sum of N copies of U, and l 2 (Y) is the Hilbert space of bilateral square summable sequences (y n ) n Z with entries in Y s usual Z stands for the set of all integers The box in (114) indicates the zero position in l 2 (Y) sequence of operators (H n ) n N from U to Y that forms a solution to the relaxed Nehari problem is referred to as an N-complementary sequence associated with (F n 1 ) n N, or just an N-complementary sequence if no confusion concerning the sequence (F n 1 ) n N can arise Here N stands for the set of nonnegative integers (ie, with zero included) The setup for this problem resembles the way relaxed versions of the Schur, Nevanlinna-Pick and Sarason interpolation problems where formulated in 6 For this relaxed Nehari problem to be solvable it is necessary that the operator given by F 3 F 4 F (N+2) F 2 F 3 F (N+1) : UN l 2 (Y) (115) F 1 F 2 F N is a contraction Here l 2 (Y) stands for the Hilbert space of all square summable sequences (,y 2,y 1 ) with entries in Y In Section 3 the relaxed Nehari extension problem is put into a relaxed commutant lifting setting, which yields that the condition that the operator in(115) is a contraction is not only 5

6 necessary but also sufficient The latter can also be seen by repeatedly applying Parrott s lemma (see Corollary IV36 in 4) We use Theorem 12 to give a Redheffer description (in Theorem 32 below) of all N-complementary sequences, under the additional assumption that the operator in (115) is a strict contraction In addition, we specify Theorem 32 for two special cases, namely, when N = 1 (see Corollary 33 below), and when F n = 0 for n = 1, 2, (see Corollary 34 below) 2 Redheffer representations and proof of the main theorem The aim in this section is to prove Theorem 12 The main tool to achieve this objective is Proposition 21 given in the next subsection This proposition gives a general scheme for rewriting the description in (18) into one of the type (110) 21 Redheffer type descriptions First we introduce some notation Let C beanoperator onu Then C is saidto bepointwise stableifc n uconverges tozeroasngoestoinfinity, foreachu U RecallthatC ispointwise stable if the spectral radius of C, denoted by r spec (C), is strictly less then 1 Next, fix a function H from H 2 (U,Y) With H we associate an operator Γ H mapping U into H 2 (Y) defined by (Γ H u)(λ) = H(λ)u (λ D,u U) (216) On the other hand, if Γ is any operator from U into H 2 (Y), then there exists a unique function G from H 2 (U,Y) such that Γ = Γ G This function is given by G(λ)u = (Γu)(λ) for u U and λ D Now assume, in addition, that H is from H (U,Y) We associate with H, in the usual way, a multiplication operator M H mapping H 2 (U) into H 2 (Y) defined by (M H f)(λ) = H(λ)f(λ) (λ D,f H 2 (U)) The operator M H is called the multiplication operator defined by H and its norm is given by M H = sup H(λ) λ D In particular, M H is a contraction if and only if H is from the Schur class S(U,Y) Proposition 21 ssume that Z is a Schur class function from S(U,Y U) given by X4 X5 Z(λ) = + V(λ) X 1 X X Y 3 : U (λ D), (217) 2 U where V is from the Schur class S(V,W) and X 1 X2 U X = X 3 0 U : V W X 4 X5 Y is a contraction (218) 6

7 Define operator-valued functions Φ 11, Φ 12, Φ 21 and Φ 22 by Φ 11 (λ) = λ X 3 (I λ X 1 ) 1 X2, Φ 12 (λ) = X 3 (I λ X 1 ) 1, Φ 21 (λ) = X 5 +λ X 4 (I λ X 1 ) 1 X2, (λ D) (219) Φ 22 (λ) = X 4 (I λ X 1 ) 1 Then Φ 11 S(W,V), Φ 21 S(W,D T ), Φ 12 H 2 (U,V), Φ 22 H 2 (U,Y) and Π Y Z(λ)(I λπ U Z(λ)) 1 = Φ 22 (λ)+ Φ 21 (λ)v(λ)(i Φ 11 (λ)v(λ)) 1 Φ12 (λ) (λ D) (220) Moreover, the operator M = M Φ11 Γ Φ12 M Φ21 Γ Φ22 : H 2 (W) U H 2 (V) H 2 (Y) (221) is a contraction which is unitary whenever X in (218) is unitary and X 1 is pointwise stable The formula on the right hand side of (220) is referred to as a linear fractional Redheffer description The term Redheffer comes from scattering theory, see Chapter XIV in 4 Indeed, let Φ 11, Φ 12, Φ 21 and Φ 22 bethefunctions in(219)where X in(218)isacontraction Consider the Redheffer scattering system g y = M Φ11 Γ Φ12 M Φ21 Γ Φ22 x u, subject to x = M V g, (222) where V is a Schur class function from S(V,W) Here u is an element from U and the vectors y, g and x are functions from the Hardy spaces H 2 (Y), H 2 (V) and H 2 (W), respectively Solving (222) we obtain that y = Γu, where Γ is the operator from U into H 2 (Y) defined by the function in the right hand side of (220), that is, ( Γu)(λ) = Φ 22 (λ)u+ Φ 21 (λ)v(λ)(i Φ 11 (λ)v(λ)) 1 Φ12 (λ)u (u U,λ D) Proof of Proposition 21 Since X is contractive, and thus X 1 is contractive, the functions in (219) are properly defined, and analytic on D Moreover, we have Π Y Z(λ) = X 4 + X 5 V(λ) X 3 and Π U Z(λ) = X 1 + X 2 V(λ) X 3 (λ D) For each λ D we then obtain that X 3 (I λπ U Z(λ)) 1 = X 3 (I λ X 1 λ X 2 V(λ) X 3 ) 1 = X 3 (I λ(i λ X 1 ) 1 X2 V(λ) X 3 ) 1 (I λ X 1 ) 1 = (I λ X 3 (I λ X 1 ) 1 X2 V(λ)) 1 X3 (I λ X 1 ) 1 = (I Φ 11 (λ)v(λ)) 1 Φ12 (λ), 7

8 and X 4 (I λπ U Z(λ)) 1 = X 4 (I λ(i λ X 1 ) 1 X2 V(λ) X 3 ) 1 (I λ X 1 ) 1 The combination of these two results gives = X 4 (I λ X 1 ) 1 +λ X 4 (I λ X 1 ) 1 X2 V(λ) X 3 (I λπ U Z(λ)) 1 = Φ 22 (λ)+( Φ 21 (λ) X 5 )V(λ)(I Φ 11 (λ)v(λ)) 1 Φ12 (λ) Π Y Z(λ)(I λπ U Z(λ)) 1 = ( X 4 + X 5 V(λ) X 3 )(I λπ U Z(λ)) 1 = Φ 22 (λ)+( Φ 21 (λ) X 5 + X 5 )V(λ)(I Φ 11 (λ)v(λ)) 1 Φ12 (λ) = Φ 22 (λ)+ Φ 21 (λ)v(λ)(i Φ 11 (λ)v(λ)) 1 Φ12 (λ) So (220) holds For the remainder of the proof we use some results from system theory The terminology corresponds to that in 5 contractive system is a quadruple Θ = {Z,B,C,D}, consisting of operators Z on a Hilbert space X, B from U to X, C from X to Y and D mapping U into Y such that the operator matrix Z B X X K Θ = : is a contraction (223) C D U Y LetΘ = {Z,B,C,D}beacontractivesystem SinceZ iscontractive, wecandefineoperatorvalued functions F Θ and G Θ on D by F Θ (λ) = D +λc(i λz) 1 B and G Θ (λ) = C(I λz) 1 (λ D) (224) Here F Θ is referred to as the transfer function for Θ From the fact that K Θ in (223) is contractive it follows that F Θ S(U,Y) and G Θ H 2 (X,Y) The operator Γ GΘ from U to H 2 (Y) is referred to as the observability operator for Θ Moreover, we have that MFΘ Γ GΘ : H 2 (U) X H 2 (Y) (225) is a contractive operator which is unitary whenever K Θ is unitary and Z is pointwise stable The statement for the case that K Θ is unitary and Z pointwise stable is obtained from Theorem III104 in5, the statement that the operator in(225) is contractive in the general case can easily be derived from Theorem III101 in 5, it also follows by specifying the result from Proposition 172 in 3 concerning time-variant systems for the time-invariant case Now put { Θ = X 1, X X3 2,, X 4 0 X5 } (226) Then the operator K Θ in (223) is equal to X in (218) So Θ is a contractive system Moreover, the functions F Θ and G Θ in (224) are given by Φ11 Φ12 F Θ = and G Θ = Φ 21 Φ 22 8

9 The proposition then follows from the theory concerning contractive systems summed up above and the observation that M = M FΘ Γ GΘ, where Θ is given by (226) Here we identify H 2 (V Y) with H 2 (V) H 2 (Y) 22 Proof of Theorem 12 Proposition 21 suggests that in order to prove Theorem 12 it suffices to show that a Schur class function Z from S(D,D T D ) with the property that Z(λ) F = ω for each λ D can be expressed as in (217), with X 1, X2, X3, X4 and X 5 the appropriate operators This is done in the next proposition under the additional assumption that is a strict contraction and R is left invertible Proposition 22 Let = {,T,U,R,Q} be a lifting data set, and let Z be a Schur class function from S(D,D T D ) ssume that is a strict contraction and R is left invertible Then Z(λ) F = ω for each λ D if and only if there exists a function V from the Schur class S(KerQ,D D T KerR ) such that Z(λ) = X 4 D 1 D X 1 D 1 + X 5 D X 2 V(λ)X 3 D 1 (λ D), (227) where X 1, X 2, X 3, X 4 and X 5 are the operators defined in (112) Finally, Z and V define each other uniquely in (227) In 8 it was already shown that a function Z from S(D,D T D ) satisfies the equality Z(λ) F = ω for each λ D if and only if there exists a Schur class function Ṽ from S(KerQ D,D ω ) such that Z(λ) = ωπ F +D ω Ṽ(λ)Π KerQ D (λ D), (228) where ω is the contraction defined in (16) Moreover, it was shown there that Z and Ṽ in (228) define each other uniquely Recall that < 1 and R being left invertible imply that D Q and D R are left invertible, or equivalently, that Q D 2Q and R D 2 R are positive definite on H 0 Then left inverses of D Q and D R are given by L D Q = (Q D Q) 1 Q D and L D R = (R D R) 1 R D, (229) respectively Note that L D Q KerQ D = 0 and L D R KerR D = 0 The proof of Proposition 22 consists of four parts In Part 1 we show that X ωπ F = 4 D 1 D X 1 9

10 Put Y = 1 2 Π D T 1 2 JL D R 0 Π KerR D : DT D D D T KerR D, where andj aretheoperatorsin(113) ThesecondpartisusedtoprovethatY Y = Dω 2 and KerY = {0} In Part 3 we show that ImD 1 Π KerQ = KerQ D and ImD 1 Π KerR = KerR D With these identities we obtain that P KerQ D = D 1 Π KerQ 1 Q Π KerQ D 1 and P KerR D = D 1 Π KerR 1 R Π KerR D 1, (230) where Q and R are the operators in (113) The results from the Parts 1 to 3 are then combined in Part 4 to complete the proof of Proposition 22 Part 1 Since D F = KerQ D, we have L D Q D F = 0, where L D Q is the left inverse of D Q given by (229) From the definition of ω in (16) we then obtain that DT R DT R(Q ωπ F = L D R D Q = D 2Q) 1 Q D X D R(Q D 2Q) 1 Q = 4 D 1 D D X (231) 1 Part 2 We first show that J intertwines the operator with Q D 2 Q(R D 2 R) 1, that is, J = Q D 2 Q(R D 2 R) 1 J (232) Toseethis, firstobservethattheidentityin(15)andthedefinitionofj implythatq D 2 Q = R D 2 R+J J With this equality we obtain that J = J (I +J(R D 2 R) 1 J ) = (I +J J(R D 2 R) 1 )J = (R D 2 R+Q D 2 Q R D 2 R)(R D 2 R) 1 J = Q D Q(R D R) 1 J So (232) holds From (232) it follows that J 1 = R D 2 R(Q D 2 Q) 1 J and (I J(Q D 2 Q) 1 J ) = J(Q D Q) 1 J = J(R D 2 R) 1 J = I Since is self adjoint, we see that the inverse of is given by 1 = I J(Q D 2 Q) 1 J (233) Moreover, we have Π DT J = D T R and L D R = (R D 2 R) 1 R D Hence L D RJ 1 JL D R = D R(R D 2 R) 1 R D 2 R(Q D 2 Q) 1 J J(R D 2 R) 1 R D = D R(Q D 2 Q) 1 J J(R D 2 R) 1 R D = D R(Q D 2 Q) 1 (Q D 2 Q R D 2 R)(R D 2 R) 1 R D = D R((R D R) 1 (Q D 2 Q) 1 )R D = P ImD R D R(Q D Q) 1 R D 10

11 For the last equality note that, since D R is left invertible, the projection on ImD R is given by P ImD R = D R(R D 2R) 1 R D = D RL D R Using the formula for 1 in (233) and the fact that P ImD R +P KerR D = I D we obtain that Y Π Y = DT Π D T 1 2 JL D R L D R J 1 2 Π KerR D 0 Π KerR D Π DT 1 = Π D Π T DT 1 JL D R L D R J 1 Π D T P KerR D +L D R J 1 JL D R ΠDT = (I J(Q D 2Q) 1 J )Π D T Π DT J(Q D 2Q) 1 R D D R(Q D 2Q) 1 J Π DT I D D R(Q D 2Q) 1 R D IDT 0 DT R = (Q D 2 0 I D D R Q) 1 R D T R D = D 2 ω To see the last identity use the first formula for ωπ F in the computation (231) and observe that L D QL D Q = (Q D 2Q) 1 Next we show that KerY = {0} Since Y KerR D = Π KerR D and Y (D D T ) D ImD R = D (D KerR D ), it suffices to show that Ker(Y D D T ) = {0} To see that this is the case observe that R D L D R = I So KerL D R = {0} Moreover, we have KerD = {0}, when D is seen as an operator on D Therefore KerY Π D D T = Ker DT 1 2 L D R J 1 2 IDT 0 ΠDT = Ker 0 L D R J 1 2 = 1 ΠDT 2 Ker J = IDT Ker D R = {0} D T Part 3 Let N be an operator from H 0 to H that has a left inverse In particular, this holds for Q and R Then D 1 Π KerN has a left inverse, and thus N = Π KerN D 2 positive definite on KerN Since D 1 (D N) = N, we see that D 1 implies that ImD 1 Π KerN = D 1 for D R in Part 2 that Π KerN is ImD N = ImN This KerN = KerN D Then we obtain in the same way as P KerN D = P ImD 1 Π KerN = D 1 Π KerN 1 N Π KerN D 1 Filling in Q and R for N gives the desired results Part 4 The identities Y Y = D ω and KerY = {0} derived in Part 2 show that there exists a unitary operator φ ω mapping D ω onto D D T KerR D such that Y φ ω = D ω Furthermore, from (230) we obtain that there exist unitary operators φ Q and φ R mapping KerQ and KerR onto KerQ D and KerR D, respectively, such that φ Q 1 2 Q Π kerq D 1 = Π KerQ D and φ R 1 2 R Π kerr D 1 = Π KerR D Now put ID D φ = T 0 ΠD D T φ ω D D 0 φ : D R Π KerR D φ ω T ω KerR 11

12 Then φ is unitary and we have X 5 Π φ D X = DT ID DT 0 2 L D R J 1 2 D 1 Π KerR φ R R ΠD D T φ ω = Y φ Π KerR D φ ω = D ω ω From the definition of φ Q we obtain that φ Q X 3 D 1 = φ Q 1 2 Q Π kerq D 1 = Π KerQ D Finally, the fact that φ Q and φ are unitary implies that an operator-valued function V is from the Schur class S(KerQ,D D T KerR D ) if and only if V(λ) = φ Ṽ(λ)φ Q (λ D) (234) for some Schur class function Ṽ from S(KerQ D,D ω ), and V and Ṽ in (234) define each other uniquely Therefore we obtain with the statement at the beginning of the proof that Z from S(D,D T D ) has Z(λ) F = ω for each λ D if and only if there exists a Schur class function V from S(KerQ,D D T KerR D ) such that Z(λ) = ωπ F +D ω φ V(λ)φ Q Π KerQ D X = 4 D 1 X D X 1 D V(λ)X D X 3 D 1 2 and V and Z define each other uniquely (λ D), Proof of Theorem 12 Let B be an operator from H into H 2 (D T ) ssume that B is given by (17) with Γ the operator defined by (18), where Z is a function from the Schur class S(D,D T D ) with Z(λ) F = ω for each λ D Let V be the Schur class function from S(KerQ,D D T KerR D ) such that Z is given by (227) Put X 1 = D X 1 D 1, X2 = D X 2, X3 = X 3 D 1, X4 = X 4 D 1 and X5 = X 5 (235) Then X 3 0 X 4 X5 X 1 X2 = = X 3 D 1 0 X 4 D 1 X 5 D X 1 D 1 D X 2 φ Q 0 ΠD F 0 0 I DT D ωπ F D ω ID 0 0 φ (236) Here φ Q and φ are the unitary operators constructed in Part 4 of the proof of Proposition 22 Since ω is a contraction, we see that the operator ΠD F 0 D KerQ : D (237) ωπ F D ω D ω D T D 12

13 is a contraction fter rearranging rows in (236) we obtain, from the fact that ψ Q and ψ are unitary, that X in (218) is a contraction Let Φ 11, Φ 12, Φ 21 and Φ 22 be the functions defined in (219) and Let Φ 11, Φ 12, Φ 21 and Φ 22 be the functions defined in (111) Using that (I λ X 1 ) 1 = D (I λx 1 ) 1 D 1 we obtain from the relations in (235) that for each λ D Φ 11 (λ) = Φ 11 (λ), Φ12 (λ) = Φ 12 (λ)d 1, Φ 21 (λ) = Φ 21 (λ) and Φ 22 (λ) = Φ 22 (λ)d 1 (238) pplying Proposition 21 we see that Φ 11 and Φ 21 are Schur class functions from S(D D T KerR,KerQ ) and S(D D T KerR,D T ), respectively, and that Φ 12 and Φ 22 are functions from H 2 (H,KerQ ) and H 2 (H,D T ), respectively Moreover, we have for each h H and each λ D that ( ΓD h)(λ) = Π DT Z(λ)(I λπ D Z(λ))D h = Φ 22 (λ)d h+ Φ 21 (λ)v(λ)(i Φ 11 (λ)v(λ)) 1 Φ12 (λ)d h = Φ 22 (λ)h+φ 21 (λ)v(λ)(i Φ 11 (λ)v(λ)) 1 Φ 12 (λ)h = (Γh)(λ), where Γ is the operator defined in (110) with V the Schur class function determined by (227) 23 Corollaries We conclude this section with two corollaries The first specifies Theorem 12 for the classical commutant lifting setting Corollary 23 Let {,T,U,R,Q} be a lifting data set where U is the Sz-Nagy-Schäffer isometric lifting of T ssume that Q is an isometry on H, R = I H and is a strict contraction Then KerR = {0}, D = {0} and the operator-valued functions Φ 11, Φ 12, Φ 21 and Φ 22 in (111) are functions from S(D T,KerQ ), H (H,KerQ ), S(D T,D T ) and H (H,D T ), respectively, and they are given by Φ 11 (λ) = λ 1 2 Q Π KerQ (I λt ) 1 D 2 D T 1 2, Φ 12 (λ) = 1 2 Q Π KerQ (I λt ) 1, Φ 21 (λ) = 1 2 D T (I λt ) 1 D 2 D T 1 2, Φ 22 (λ) = D T (I λt ) 1 T, (λ D) (239) where Q = Π KerQ D 1 Π KerQ on KerQ and T = D 2 Q Q D 2 on H, (240) = I +D T D 2 D T on D T (241) 13

14 Proof Since R = I H and R R = I H = Q Q, we obtain that KerR = {0} and D = 0 on D = {0} Clearly R is left invertible and, by assumption, is a strict contraction Moreover, we have that R λq = I H λq is (left) invertible for each λ D, because Q is contractive Hence the result of Theorem 12, and the remark in the third paragraph underneath Theorem 12 hold for this lifting data set In particular Φ 11, Φ 12, Φ 21 and Φ 22 in (111) are functions from S(D T,KerQ ), H (H,KerQ ), S(D T,D T ) and H (H,D T ), respectively Observe that J in (113) is given by J = D T and we have Q D 2 Q = Q Q Q Q = I Q Q = D 2 Q Therefore we have that X 1 = T with X 1 from (112) and T as in (240) Note that Q and given in (113) reduce to the formulas in (241) So we obtain that the operators X 1, X 2, X 3, X 4 and X 5 in (112), under the present assumptions, are X 1 = T, X 2 = D 2 D T 1 2, X 3 = 1 2 Q Π KerQ, X 4 = D T T and X 5 = 1 2 This immediately shows that the formulas for Φ 11, Φ 12 and Φ 22 in (111) are given by (239) in the classical commutant lifting setting Furthermore, we have for each λ D that Φ 21 (λ) = X 5 +λx 4 (I λx 1 ) 1 X 2 = 1 2 λd T T (I λt ) 1 D 1 D T 1 2 = (I +D T D 1 D T ) 1 2 D T (I λt ) 1 D 1 D T 1 2 = 1 2 D T (I λt ) 1 D 1 D T 1 2 = 1 2 D T (I λt ) 1 D 1 D T 1 2 So Φ 21 in (111) also reduces to its formula in (239) Corollary 24 Let = {,T,U,R,Q} be a lifting data set with U the Sz-Nagy-Schäffer isometric lifting of T ssume that is a strict contraction and R has a left inverse Let Φ 11, Φ 12, Φ 21 and Φ 22 be the functions defined by (111) and (112) Put M = 0 M Φ11 M Φ21 Γ Φ12 Γ Φ22 H : 2 (D D T KerR ) H H H 2 (KerQ ) H 2 (D T ) (242) Then M is a contraction, and M is an isometry if R R = Q Q and X 1 is pointwise stable Let X 1, X 2, X 3, X 4 and X 5 be the operators in (112) and X 1, X2, X3, X4 and X 5 the operators given by (235) Define Φ 11, Φ 12, Φ 21 and Φ 22 by (219) We saw in the proof of Theorem 12 that the results from Proposition 21 can be applied for this choice of X1, X2, X3, X4 and X 5 Then the operator M in (221) is a contraction which is unitary in case X in (218) is unitary and X 1 is pointwise stable Since X 1 and X 1 are similar (D X 1 = X 1 D and D > 0), we see 14

15 that X 1 is pointwise stable if and only if X 1 is pointwise stable Moreover, the computation in (236) and the remark below (237) show that X is unitary if and only if the operator in (237) is unitary This is precisely the case if and only if ω is an isometry In other words, X in (218) is unitary if and only if R R = Q Q We conclude that the operator M in (221) is a contraction which is unitary if X 1 is pointwise stable and R R = Q Q In terms of operators the identities in (238) become M Φ11 = M Φ11, Γ Φ12 = Γ Φ11 D 1, M Φ21 = M Φ21 and Γ Φ11 = Γ Φ11 D 1 (243) Next define a contraction by = 0 : H 2 (W) H H, where W = D D T KerR The defect operator of is given by IH D = 2 (W) 0 H : 2 (W) H 2 (W) 0 D H D Then the identities in (243) show that 0 0 M = = M Φ11 M Φ21 Γ Φ12 Γ Φ22 M Φ11 Γ Φ12 D D M Φ21 Γ Φ22 = IH 0 = MD 0 M D It is well known that the operator D is an isometry Therefore we obtain that M is a contraction which is isometric in case R R = Q Q and X 1 is pointwise stable 3 The relaxed Nehari extension problem In this section we apply Theorem 12 to obtain a description of all N-complementary sequences for the relaxed Nehari extension problem formulated in the last but one paragraph of the introduction 31 The relaxed Nehari problem in the relaxed commutant lifting setting Throughoutthissubsection N isapositiveinteger andf 1,F 2,isasequence ofoperators from U to Y satisfying n=1 F nu 2 < for each u U We refer to the operator from U N into l 2 (Y) given by F 3 F 4 F (N+2) F 2 F 3 F (N+1) : UN l 2 (Y) (344) F 1 F 2 F N 15

16 as the N-truncated Hankel operator defined by the sequence (F n 1 ) n N Our first remark is that an operator from U N into l 2 (Y) is an N-truncated Hankel operator if and only if satisfies the intertwining relation T R = Q, where T is the shift operator on l 2 (Y) given by T = 0 0 0, (345) 0 I Y I Y 0 and R and Q are the operators from U N 1 to U N defined by IU N 1 U R = : U N 1 N 1 0 and Q = : U N 1 0 U I U N 1 U U N 1 (346) Indeed, if is an operator from U N into l 2 (Y), then admits an operator matrix decomposition of the form = N N, N and hence R = N N N 1 and Q = N N N It follows that T R = Q is equivalent to kj = k 1j+1 for appropriate indices k and j, that is, T R = Q is equivalent to being an N-truncated Hankel operator We shall also need the bilateral forward shift V on l 2 (Y) which is given by V = 0 I I I on l 2 (Y) (347) Let l 2 + (Y) be the subspace of l2 (Y) consisting of the sequences that have the zero vector on all entries with strictly negative index Identifying l 2 (Y) with the subspace of l2 (Y) consisting of all sequences that have the zero vector on all entries with a non-negative index, we see that l 2 (Y) = l 2 (Y) l2 + (Y), and V partitions as V = T 0 X S : l 2 (Y) l 2 + (Y) l 2 (Y) l 2 + (Y) (348) 16

17 Here T is given by (345), the operator S is the unilateral forward shift on l 2 + (Y), and X is the operator from l 2 (Y) into l2 + (Y) given by X (,y 3,y 2,y 1 ) = (y 1,0,0,) ((,y 3,y 2,y 1 ) l 2 (Y)) (349) Since V is unitary, (348) shows that V is an isometric lifting of T s is easily seen this lifting is also minimal We are now ready to state the main result of this subsection Proposition 31 Let N be a positive integer and F 1,F 2, a sequence of operators from U to Y satisfying n=1 F nu 2 < for each u U ssume that the N-truncated Hankel operator defined by (F n 1 ) n N is a contraction Then = {,T,V,R,Q}, with T,V,R and Q defined by (345), (347) and (346), is a lifting data set, and there exists a contractive interpolant for if and only if there exists an N-complementary sequence associated with (F n 1 ) n N More precisely, an operator B from U N to l 2 (Y) is a contractive interpolant for if and only if B is of the form F 2 F 3 F (N+1) F 1 F 2 F N H 0 F 1 F (N 1) B = H 1 H 0 : U N l 2 (Y) (350) F 1 H0 with (H n ) n N an N-complementary sequence associated with (F n 1 ) n N Proof We already know that T R = Q and that V is a minimal isometric lifting of T Since R and Q are both isometries, we have R R = Q Q So the constraints (11) are fulfilled, and hence is a lifting data set Now let B be an operator from U N into l 2 (Y) Using similar arguments as in the first paragraph of this subsection we obtain that B satisfies V BR = BQ if and only if H 2 H 3 H (N+1) H 1 H 2 H N B = H 0 H 1 H (N 1) : U N l 2 (Y) (351) H 1 H 0 H (N 2) H 2 H 1 H (N 3) for some sequence of operators (H n ) n Z from U into Y satisfying n= H nu 2 < for each u U Moreover, if B is given by (351), then Π l 2 (Y) B = holds if and only if H k = F k 17

18 for k = 1, 2, 3, Henceforth, we obtain that B is a contractive interpolant for if and only if B is given by (350) with (H n ) n N an N-complementary sequence associated with (F n 1 ) n N In order to apply Theorem 12 we need to establish a relation between the contractive interpolants for the lifting data set = {,T,V,R,Q} in Proposition 31 and those for = {,T,U,R,Q}, with U being the Sz-Nagy-Schäffer isometric lifting of T For that purpose we use the Fourier transform F Y from l 2 +(Y) to the Hardy space H 2 (Y), that is, F Y is the unitary operator mapping l 2 + (Y) onto H2 (Y) defined by (F Y (y n ) n N )(λ) = λ n y n ((y n ) n N l 2 +(Y), λ D) (352) n=0 Using F Y, let Ψ be the unitary operator from l 2 (Y) = l 2 (Y) l2 + (Y) into l2 (Y) H2 (Y) given by Il 2 0 l 2 Ψ = (Y) : (Y) l 2 0 F Y l 2 + (Y) (Y) H 2 (353) (Y) Since D T = Y and the bilateral shift S on l 2 + (Y) and the bilateral shift S Y on H 2 (Y) are related via F Y S = S Y F Y, we see from (348) and (13) that Ψ intertwines the bilateral shift V with the Sz-Nagy-Schäffer isometric lifting U of T, that is, ΨV = U Ψ Moreover, we have that Ψf = f for each sequence f in l 2 (Y) Using the properties of Ψ found above, a straightforward computation shows that B is a contractive interpolant for if and only if B = Ψ B for some contractive interpolant B for Finally, observe that a sequence of operators H 0,H 1, in L(U,Y) has the property that n=0 H nu 2 < for each u U if and only if it is the sequence of Taylor coefficients at zero of a function in H 2 (U,Y) So, alternatively, we seek functions H in H 2 (U,Y) such that B in (350) is a contraction, where H n is the n th Taylor coefficient of H at zero 32 The solution to the relaxed Nehari problem In the previous subsection we saw how the solution to the relaxed commutant lifting problem can be applied to obtain all N-complementary sequences for the relaxed Nehari extension problem Two important operators in the description of all contractive interpolants in Theorem 12 are the defect operators of the contractions T and From the definition of T in (345) we immediately see that D T = Y and D T = 0 0 I Y : l 2 (Y) Y (354) The defect operator of the N-truncated Hankel operator, denoted by, is the positive square root of Λ 1,1 Λ 1,N { D 2 = I on U N n=i, where Λ i,j = F nf n if i = j, Λ N,1 Λ n=i F n F (355) n+i j if i j N,N 18

19 In what follows we assume the N-truncated Hankel operator to be a strict contraction, or equivalently, we assume that the defect operator D is positive definite We also use the entries Λ i,j in the N N operator matrix representation of D 2, Λ 1,1 Λ 1,N D 2 = on U N (356) Λ N,1 Λ N,N The fact that D 2 is positive definite implies that D 2 and Λ n,n in (356) are also positive definite for n = 1,,N In particular, this is true for n = 1 and n = N Moreover, we have that the N 1 by N 1 left upper block matrix operator of D 2 in (355) is positive definite Therefore there exists a unique solution G 1 G N 1 to the equation Λ 1,1 Λ 1,N 1 Λ N 1,1 Λ N 1,N 1 G 1 G N 1 = F 1 F N+1 (357) description of all N-complementary sequences associated with (F n 1 ) n N under the assumption that the N-truncated Hankel operator for (F n 1 ) n N is a strict contraction is given in the next theorem Theorem 32 Let N be a positive integer and F 1,F 2, a sequence of operators from U to Y satisfying n=1 F nu 2 < for each u U ssume that the N-truncated Hankel operator defined by (F n 1 ) n N is a contraction Let H be a function from H 2 (U,Y) Then the Taylor coefficients (H n ) n N of H at zero form an N-complementary sequence associated with (F n 1 ) n N if and only if there exists a Schur class function V from S(U,Y U) such that H(λ) = ˆΦ 22 (λ)+ ˆΦ 21 (λ)v(λ)(i ˆΦ 11 (λ)v(λ)) 1ˆΦ12 (λ) (λ D) (358) Moreover, for any function V from S(U,Y U) the formula (358) defines a function H from H 2 (U,Y) Here ˆΦ 11 and ˆΦ 21 are Schur class functions from S(Y U,U) and S(Y U,Y), respectively, and ˆΦ 12 and ˆΦ 22 are functions from H (U,U) and H (U,Y), respectively, and these functions are given by ˆΦ 11 (λ) = λ(λ 1,1 ) 1 2E N (I U N λt state) 1 G (I +FG ) 1 2 C 2, ˆΦ 12 (λ) = (Λ 1,1 ) 1 2 λ(λ 1,1 ) 1 2E N (I U N λt state) 1 C 1, ˆΦ 21 (λ) = (I +FG ) 1 2 FC 2 F(IU N λt state ) 1 G (I +FG ) 1 2 C 2, (359) ˆΦ 22 (λ) = F(I U N λt state ) 1 C 1, where T state on U N, E N : U U N, C 1 : U U N, C 2 : U U N, F : U N Y and 19

20 G : U N Y are the operators given by Λ 2,1 (Λ 1,1 ) 1 I 0 0 I Λ 3,1(Λ 1,1) 1 0 I T state = 0, E N = 0, Λ N,1 (Λ 1,1) 1 0 I Λ 2,1 Λ 1,N C 1 = Λ (Λ 1,1 ) 1, C 2 = (Λ N,N 2, ) 1 N,1 Λ N,N 0 F = F 1 F N and G = G 1 G N 1 0, (360) with Λ m,n for m,n = 1,,N as in (356), and G 1 G N 1 the solution to the equation (357) Furthermore, we have r spec (T state ) < 1, and the operator ˆM defined by ˆM = 0 Γ H MˆΦ 11 ΓˆΦ 12 : 2 l (Y U) 2 (Y) H U 2 (U), (361) H 2 (Y) MˆΦ21 ΓˆΦ22 is isometric, where Γ is given by Γ = F 2 F 1 : l 2 (Y) U Observe that the state operator T state in (360) is close to a companion operator To be precise, let E be the flip over operator on U N given by E(u 1,u 2,,u N ) = (u N,u N 1,,u 1 ) ((u 1,u 2,,u N ) U N ) Then E is unitary and ET state E is precisely the second companion operator corresponding to the operator-valued polynomial K(λ) = λ N 1 k=0 λk Λ N k,1, see Chapter 14 in 9 Note that the leading coefficient Λ 1,1 of K is positive definite s a computational remark, note that to obtain the solution G 1 G N 1 to the equation (357) it suffices to compute the inverse of the operator Λ N,N in (356) Indeed, with a standard Schur complement type of argument we see that the inverse of the N 1 by N 1 left upper corner of D 2 in (355) is given by Λ 1,1 Λ 1,N 1 Λ 1,N (Λ N,1 (Λ ) N,N ) 1 Λ N 1,1 Λ N 1,N 1 Λ N 1,N (Λ N,N 1 ) Furthermore, if U is finite dimensional, then all computations involve finite matrices only In this sense the relaxed Nehari problem is very different from the classical Nehari problem Proof of Theorem 32 Since R in (346) is an isometry, we have that the lifting data set = {,T,U,R,Q}, with U the Sz-Nagy-Schäffer isometric lifting of T in (345) and R and Q as in (346), has the property that is a strict contraction (by assumption) and R 20

21 is left invertible So the description of all contractive interpolants in Theorem 12 can be applied to this particular lifting data set Since also Q is an isometry, we have R R = Q Q Hence D = 0 on D = {0}, where D and D are given by (14) We already made the identification between D T and Y Moreover, we have KerQ = U {0} N 1 U N and KerR = {0} N 1 U U N So both KerQ and KerR can be identified with U In that case the projections Π KerQ and Π KerR are given by Π KerQ = I U 0 0 : U N U and Π KerR = 0 0 I U : U N U (362) In particular, Π KerQ = E N, with E N the operator in (360) Now let H be a function from H 2 (U,Y) with Taylor coefficients (H n ) n N at zero Then the operator Γ H from U into H 2 (Y) defined in (216) is given by Γ H = F Y Π l 2 + (Y) BE N = Π H 2 (Y)Ψ BE N, where B is given by (350), and F Y and Ψ are the unitary operators defined in (352) and (353), respectively ccording to Proposition 31 and the last but one paragraph of the previous subsection we have that (H n ) n N is an N-complementary sequence associated with (F n 1 ) n N if and only if Ψ B is a contractive interpolant for Moreover, all contractive interpolants for are obtained in this way Then Theorem 12 and the above analysis of the spaces KerQ, KerR, D and D T imply that (H n ) n N is an N-complementary sequence if and only if there exists a Schur class function V from S(U,Y U) such that H(λ) = Φ 22 (λ)e N +Φ 21 (λ)v(λ)(i Φ 11 (λ)v(λ)) 1 Φ 12 (λ)e N (λ D), where Φ 11, Φ 12, Φ 21 and Φ 22 are the functions defined in Theorem 12 Put for each λ D ˆΦ 11 (λ) = Φ 11 (λ), ˆΦ12 (λ) = Φ 12 (λ)e N, ˆΦ21 (λ) = Φ 21 (λ) and ˆΦ22 (λ) = Φ 22 (λ)e N (363) Then we obtain from Theorem 12 that ˆΦ 11 S(Y U,U), ˆΦ12 H 2 (U,U), ˆΦ21 S(Y U,Y) and ˆΦ 22 H 2 (U,Y) Note that ΓˆΦ 12 = Γ Φ12 E N, ΓˆΦ 22 = Γ Φ22 E N and Γ = E N Hence the operator ˆM in (361) is obtained from M in (242) after multiplication from the right by IH 2 (Y U) 0 H : 2 (Y U) H 2 (Y U) 0 E N U U N Since R R = Q Q, we obtain from Corollary 24 that M, and thus ˆM, is an isometry if X 1 in (112) is pointwise stable This proves to be the case In fact, we have that X 1 in (112) has r spec (X 1 ) < 1 To see this it suffices to show that R λq is left invertible for each λ in D See Proposition 53 in 6, and the remark in the third paragraph after Theorem 12 Indeed, we have that R λq is left invertible for each λ D In fact, a left inverse is given by I 0 0 λi I 0 : UN U N 1 λ N 2 I λi I 0 21

22 So to complete the proof it remains to show that the functions ˆΦ 11, ˆΦ 12, ˆΦ 21 and ˆΦ 22 defined in (363) can also be written as in (359), and that r spec (T state ) < 1 The later immediately shows that ˆΦ 12 and ˆΦ 22 are functions in H (U,U) and H (U,Y), respectively From (362), the definitions of Q and R in (113) and the assumption that D 2 is given by (356) we immediately obtain that Q = Λ 1,1 and R = Λ N,N Moreover, the N 1 N 1 left upper corner and N 1 N 1 right lower corner of D 2 in (355) are equal to R D 2R and Q D Q, respectively The fact that D T is given by (354) and the definition of in (115) show that D T = F, where F is given by (360) Thus D T R = FR = Y 1 Y N+1 This implies that G in (360) can also be written as G = R(R D 2R) 1 R D T Using the fact that D = 0 on D = {0}, we obtain that So we have the following identities: = I +D T R(R D 2 R) 1 R D T = I +FG Q = Λ 1,1, R = Λ N,N, = I+FG, F = D T and G = D T R(R D 2 R) 1 R Let X 1, X 2, X 3, X 4 and X 5 be the operators defined in (112) Then we immediately obtain that X 3 = (Λ 1,1 2E ) 1 N : UN U, X 4 = FX 1 : U N Y (364) X 5 = (I +FG ) 1 2Π Y : Y U Y Writing out ˆΦ 11, ˆΦ 12, ˆΦ 21 and ˆΦ 22 in terms of X 1, X 2, X 3, X 4 and X 5 with the identities for X 3, X 4 and X 5 in (364) we obtain that ˆΦ 11 (λ) = λx 3 (I λx 1 ) 1 X 2 = λ(λ 1,1) 1 2E N (I λx 1) 1 X 2, ˆΦ 12 (λ) = X 3 E N +λx 3 (I λx 1 ) 1 X 1 E N = (Λ 1,1) 1 2 +λ(λ 1,1) 1 2E N (I λx 1) 1 X 1 E N, ˆΦ 21 (λ) = X 5 +λx 4 (I λx 1 ) 1 X 2 = X 5 +λfx 1 (I λx 1 ) 1 X 2 (λ D) (365) = X 5 FX 2 +F(I λx 1 ) 1 X 2, ˆΦ 22 (λ) = X 4 (I λx 1 ) 1 E N = F(I λx 1 ) 1 X 1 E N Therefore, to complete the proof, it suffices to show that X 1 = T state and X 2 = G (I +FG ) 1 2 C 2 (366) Indeed, assume that (366) holds Then C 1 = X 1 E N and X 5 FX 2 = (I +FG ) FG (I +FG ) 1 2 FC 2 = (I +FG )(I +FG ) 1 2 FC 2 = (I +FG ) 1 2 FC 2 This computation combined with the formulas for ˆΦ 11, ˆΦ 12, ˆΦ 21 and ˆΦ 22 found in (365) show that (359) holds, and also that r spec (T state ) = r spec (X 1 ) < 1 22

23 We will now prove (366), starting with the first identity First we claim that X 1 can be written as X 1 = RQ +R(Q D 2 Q) 1 Q D 2 P KerQ To see thatthis isthe case, observe thatbothq and(q D 2Q) 1 Q D 2 areleft inverses of Q Certainly all left inverses of Q are equal on ImQ Since Q (U N ImQ) = Q KerQ = 0, we obtain that (Q D 2Q) 1 Q D 2 = Q +(Q D 2Q) 1 Q D 2P KerQ, which proves our claim Note that RQ is the backward shift on U N This implies that T state = RQ + C 1 EN Using that P KerQ = E N EN and Π KerR C 1 = 0 we obtain that the first identity in (366) holds if R C 1 = (Q D Q) 1 Q D 2E N This identity follows from the next computation: Λ 0 2,1 Q D 2 QR C 1 Λ 1,1 = Q D 2 Q = Q D 2 Λ 2,1 Λ N,1 Λ N,1 = Q D 2 D 2 E N Q D 2 E NΛ 1,1 = Q E N Q D 2 E NΛ 1,1 = Q D 2 E NΛ 1,1 Indeed, to obtain the desired equality multiply the first and the last term in the above sequence of equalities with (Q D 2Q) 1 from the left and with (Λ 1,1 ) 1 from the right To see that the identity for X 2 in (366) holds, note that X 2 can be written as X 2 = R(R D 2 R) 1 R D T 1 2 D 2 Π KerR 1 2 R : Y U U N Earlier we obtained that = I +FG and R(R D 2R) 1 R D T = G This shows that X 2 Y = G (I +FG ) 1 2 The fact that X 2 U = C 2 follows from the definition of Π KerR in (362), the formula for D 2 in (356) and the identity R = Λ N,N Thus the second identity in (366) holds 33 Special cases s an illustration we specify the result in Theorem 32 for two special cases, namely, the case that N = 1 and the case that F n = 0 for n = 1, 2, Corollary 33 Let F 1,F 2, be operators in L(U,Y) such that n=1 F nu 2 < for each u U ssume that the 1-truncated Hankel operator is a contraction Let H be a function from H 2 (U,Y) Then the Taylor coefficients (H n ) n N of H at zero form a 1- complementary sequence associated with (F n 1 ) n N if and only if there exists a Schur class function V from S(D,Y D ) such that H(λ) = Π Y V(λ)(I λπ D V(λ)) 1 D (λ D) (367) Moreover, if is a strict contraction, then D = U 23

24 Note that in Corollary 33 it is not required that the 1-truncated Hankel operator is a strict contraction We give two different proofs The first is based on Theorem 11 and Proposition 31 In the second proof, assuming < 1, we show that Theorem 32 reduces to Corollary 33, when specified for N = 1 Proof 1 (Using Theorem 11 and Proposition 31) Since N = 1, we see that = F 2 F 1 from U to l 2 (Y), and that R and Q in (346) reduce to R = Q = 0 from {0} to U Note that E N = I U, with E N the operator defined in Proposition 31 Then Proposition 31 shows that the Fourier coefficients of H form a 1-complementary sequence associated with (F n 1 ) n N if and only if Γ H = Π H 2 (Y)B for a contractive interpolant B for {,T,U,R,Q}, where U is the Sz-Nagy-Schäffer isometric lifting of the contraction T defined in (345), see also the second paragraph of the proof of Theorem 32 Since R = Q = 0, we obtain that ω = 0 and F = {0}, see (16) Therefore any function Z from S(D,Y D ) has the property that Z(λ) F = ω for each λ D Recall that D T = Y, see (354) pplying Theorem 11, we see that the Fourier coefficients of H form a 1-complementary sequence associated with (F n 1 ) n N if and only if H is given by (367), where V is from the Schur class S(D,Y D ) If is a strict contraction, then D is invertible on U In particular, this implies that D = U Proof 2 (Using Theorem 32 and assuming that < 1) ssume that the 1-truncated Hankel operator for (F n 1 ) n N is a strict contraction Let ˆΦ 11, ˆΦ12, ˆΦ21 and ˆΦ 22 be the functions defined in Theorem 32 for the case N = 1 Then the operators C 1, C 2, F and G in (360) reduce to C 1 = 0 on U, C 2 = (Λ 1,1 )1 2 on U, F = F 1 : Y U and G = 0 : Y U This implies that T state = 0 on U, E 1 = I on U and G (I +FG ) C 2 = 0 (Λ 1,1 )1 2 Moreover, observe that (Λ 1,1) 1 2 = D 1 Therefore we have for each λ D that ˆΦ 11 (λ) = λd (I λ0) 1 0 D 1 = 0 I = ΠY, ˆΦ 12 (λ) = D λd (I λ0) 1 0 = D, ˆΦ 21 (λ) = (I +0) 1 2 F 1 D 1 F 1 (I λ0) 1 0 D 1 ˆΦ 22 (λ) = F 1 (I λ0) 1 0 = 0 = I 0 = ΠU, ccording to Theorem 32 the Fourier coefficients of H form an N-complementary sequence associated with (F n 1 ) n N if and only if H(λ) = Π Y V(λ)(I +λπ U V(λ)) 1 D (λ D), for some V S(U,Y U) Note that instead of the plus in (I +Π U V(λ)) we can also take a minus since ΠY V( ) V S(U,Y U) if and only if S(U,Y U) Π U V( ) 24

25 Finally, we consider the relaxed Nehari problem with F 1 = F 2 = = 0 That is, we seek a description of all functions H from H 2 (U,Y) with Fourier coefficients (H n ) n N at zero such that the operator H H 1 H H 2 H 1 H 0 0 H N 1 H N 2 H N 3 H 0 : U N l 2 +(Y) (368) is a contraction By associating with each function H from H 2 (U,Y) the norm of the operator in (368) we induce a Banach space structure on the set H 2 (U,Y) This Banach space appears in 6, in the contexts of certain interpolation problems, and for the case that U and Y are finite dimensional, in 1 and 2 By specifying Theorem 32 for the case that F 1 = F 2 = = 0 we obtain the following description of all functions H from H 2 (U,Y) with L 0 in (368) contractive Corollary 34 Let H be a function from H 2 (U,Y) with Fourier coefficients (H n ) n N at zero Then the operator in (368) is a contraction if and only if there exists a Schur class function V from S(U,Y U) such that H(λ) = Π Y V(λ)(I λ N Π U V(λ)) 1 (λ D) (369) Corollary 34 is an operator-valued version of a result from 1 The case N = 1 appears as a corollary in 7 and is fundamental in the proof of the first main result in 8 Proof of Corollary 34 Note that the N-truncated Hankel operator for the sequence F 1 = F 2 = = 0 is the zero operator from U N to l 2 (Y) In particular, is a strict contraction So we can apply the result of Theorem 32 to this Nehari data Note that in this case both D 2 in (355) and D 2 in (356) are equal to the identity operator on U N In particular, Λ i,j = I if i = j and Λ i,j = 0 if i j, for i,j = 1,,N Moreover, the operators G 1,,G N 1 in (357) are all equal to 0 Therefore we obtain that the operators F, G, C 1 and C 2 in (360) are given by F = G = 0 from U N to Y, C 1 = 0 from U to U N and C2 = 0 0 I from U N to U Furthermore, T state in (360) reduces to 0 I 0 0 I λi λ N 1 I T state = 0, so (I λt state ) 1 0 I = λi I 0 0 I