The Abstract Interpolation Problem and Applications

Transcription

1 Holomorphic Spaces MSRI Publications Volume 33, 998 The Abstract Interpolation Problem and Applications ALEXANDER KHEIFETS Abstract A number of classical interpolation problems can be reduced to the folloing scheme One is interested in the finding Schur class operator functions (ζ) : E E 2, ith ζ D, that satisfy certain interpolation conditions The data of the problem are encoded in the Lyapunov identity D(T 2 x, T 2 y) D(T x, T y) = M x, M y M 2 x, M 2 y, here x, y are elements of a vector space X, D is a positive semidefinite quadratic form on X, T and T 2 are linear operators on X, and M, M 2 are linear operators from X to the separable Hilbert spaces E, E 2 After introducing the de Branges Rovnyak function space H associated ith, e can formulate the interpolation conditions thus: is a solution to the interpolation problem if and only if there exists a linear mapping F : X H such that F x 2 H D(x, x) and» (F T x)(t) = t(f T 2 x)(t) (t) (t)» M2 x M x for ae t ith t = The solutions turn out to be the scattering matrices of the unitary colligations that extend the isometric colligation defined by the Lyapunov identity These extensions and their scattering functions can be described using a universal extension and its scattering operator function The description formula for solutions looks like here = s 0 + s 2 ( ωs) ωs,» s s S = : N s 2 s 2 E N E 2 0 is the scattering matrix of the universal extension and ω : N N 2 is an arbitrary parameter from the Schur class The matrix S is defined essentially uniquely by the data of the problem and is called the scattering matrix of the problem Using the functional model and the Fourier representation of the universal extension one can investigate analytic properties of the scattering matrices S for classes of interpolation problems Work partly supported by MSRI under NSF grant DMS

2 352 ALEXANDER KHEIFETS Contents Lecture The Abstract Interpolation Problem 352 Lecture 2 Examples 355 Lecture 3 Solutions of the Abstract Interpolation Problem 362 Lecture 4 Description of the Solutions of the AIP 370 Acknoledgements 377 References 378 Lecture The Abstract Interpolation Problem I ill begin ith the formal setup of the Abstract Interpolation Problem, or AIP, then consider several examples and discuss the role of the AIP in their investigation The data of the AIP are encoded in the Fundamental Identity: D(T 2 x, T 2 y) D(T x, T y) = M x, M y E M 2 x, M 2 y E2, ( ) here x, y are elements of a vector space X (hich has no a priori topological structure), D is a positive-semidefinite quadratic form on X, T and T 2 are linear operators on X, E and E 2 are separable Hilbert spaces, and M, M 2 are linear mappings from X into E, E 2 De Branges Rovnyak function spaces Let be a Schur class operator function; that is, associates to each ζ in the unit disk D = {ζ : ζ } a contraction (ζ) : E E 2, varying analytically ith ζ The de Branges Rovnyak space L consists of the functions f : T E 2 E (here T = {ζ : ζ = } is the unit circle) having the form for some g L 2 (E 2 E ) We set [ /2 E2 f = g ( 2) f L E def = inf g L 2, here the infimum is taken over all g L 2 (E 2 E ) such that ( 2) is satisfied All such preimages g differ by the addition of (arbitrary) elements of Ker [, and the infimum is attained hen [ (t) g(t) Ker (t) ae on T Let π f denote the particular g that achieves the infimum Thus f 2 L = π f 2 L = 2 T (π f)(t) 2 E 2 E m(dt),

3 THE ABSTRACT INTERPOLATION PROBLEM AND APPLICATIONS 353 here m(dt) is Lebesgue measure The inner product in L is defined by f, h L = π f, π h L 2 No L is a Hilbert space As a set it is contained in L 2 It might happen that a function f L admits the representation [ (t) f(t) = (t) ǧ(t) ae, here ǧ(t) need not be in L 2 ; then, for any h L, [ /2 f, h L = ǧ(t), π L h ǧ(t), = h(t) m(dt) E 2 E 2 T We also define the space H as the subspace of L consisting of [ f2 f = ith f 2 H+(E 2 2 ) and f H (E 2 ), f here H 2 + and H 2 are the standard Hardy classes Setup of the AIP The Schur class function : E E 2 is said to be a solution of the AIP (ith the data specified above) if there exists a linear mapping F : X H such that, for all x X, the folloing conditions are satisfied: (i) F x 2 H D(x, x) (ii) tf T 2 x F T x = [ [ M2 x M x for ae t T Property (ii) is, in fact, an implicit formula for the mapping F Sometimes it defines F uniquely, sometimes not; but any mapping F that possesses (ii) is very special We ill describe all such maps in Lecture 4 If e rite [ F+ x F x =, F x D(x, x) are equivalent to the conjunc- the conditions F x H and F x 2 H tion of three conditions: (a) F + x H 2 +(E 2 ) (b) F x H 2 (E ) (c) F x 2 L D(x, x) By the definition of the inner product in L, property (c) is actually an upper bound for the average boundary values of F x My goal no is to explain hy this abstract problem is an interpolation problem Before passing to examples I ould like to consider a special case of the data (hich as considered earlier than the general case)

4 354 ALEXANDER KHEIFETS A special case Assume that the operators (ζt 2 T ) and (T 2 ζt ) are invertible for all ζ D except possibly for a discrete set Because the first components of the sides of relation (ii) above are H+ 2 functions, one can consider the analytic continuation of the relation inside the unit disk D: ζ(f + T 2 x)(ζ) (F + T x)(ζ) = M 2 x + (ζ)m x ( 3) We emphasize that the vectors M 2 x E 2 and M x E are independent of ζ Fix ζ D Because the mapping F is linear, e have Actually, for any complex number µ, ζ(f + T 2 x)(ζ) = ( F + (ζt 2 x) ) (ζ) µ(f + T 2 x)(ζ) = ( F + (µt 2 x) ) (ζ); in particular, this is true for µ = ζ We can reexpress ( 3) no as ( F+ ((ζt 2 T )x) ) (ζ) = ( M 2 + (ζ)m ) x Replacing x by (ζt 2 T ) x, e end up ith (F + x)(ζ) = ((ζ)m M 2 )(ζt 2 T ) x ( 4) One can see better no the interpolation meaning of the property F + x H 2 +: the zeros of the numerator cancel the zeros of the denominator, hich means that obeys the interpolation constraint (ζ)m x = M 2 x at certain points ζ In a similar ay, the second components of the to sides of equality (ii) can be reexpresed, under the assumptions at hand, as (F x)(ζ) = ζ(m (ζ) M 2 )(T 2 ζt ) x ( 5) The interpolation meaning of the property F x H 2 is similar to the one considered above, but no for (ζ) The meaning of property (c) ill be discussed in the examples Remark Under the assumptions of this section, condition (ii) defines F uniquely and explicitly for any solution This allos us to rite F instead of F Thus, under these assumptions, one can give a more explicit setup for the AIP: Let [ F F x = + x F x be defined by the formulas ( 4) and ( 5) The Schur class function is said to be a solution of the AIP if F possesses properties (a), (b), and (c) Generally, condition (ii) does not define F in a unique ay, but e ill see in Lecture 4 ho to describe all such mappings F

5 THE ABSTRACT INTERPOLATION PROBLEM AND APPLICATIONS 355 References for Lecture are [Katsnelson et al 987; Kheifets 988a; 988b; 990b; Kheifets and Yuditskii 994 Lecture 2 Examples Example : The Nevanlinna Pick Problem Let ζ,, ζ n, be a finite or infinite sequence of points in the unit disk D; let,, n, be a sequence of complex numbers One is interested in describing all the Schur class functions such that (ζ k ) = k The ell-knon solvability criterion is [ n k j ζ k ζ j k,j= 0 for all n We define the data of the AIP associated ith this problem Because the functions are scalar, E = E 2 = C Consider the space X that consists of all sequences x x = x n that have only a finite number of nonvanishing components No topology is assumed on X Define the sesquilinear form D on X by D(x, y) = k,j ȳ k k j ζ k ζ j x j, for x, y X The (diagonal) linear operators ζ T = T = ζ n and T 2 = X are ell defined on the space X We can no check the Fundamental Identity ( ): D(x, y) D(T x, T y) = k,j ȳ k ( k j )x j = k,j ȳ k x j k,j ȳ k k j x j = k ȳ k x j j k ȳ k k j x j j

6 356 ALEXANDER KHEIFETS By defining M x = j x j and M 2 x = j jx j, e end up ith D(x, y) D(T x, T y) = M y M x M 2 y M 2 x The products on the right-hand side are actually the standard inner product in C Consider no the AIP associated ith these data Let be a solution of this AIP This means that there exists a mapping F : X H such that conditions (i) and (ii) on page 353 hold Because (ζ T ) exists for all ζ, such that ζ < and ζ ζ j, and because ( ζt ) exists for all ζ ith ζ <, e kno that F has the folloing form (see Lecture, special case on page 354): (F + x)(ζ) = ((ζ)m M 2 )(ζ T ) x (F x)(ζ) = (M (ζ)m 2 )( ζt ) x Thus, F is defined uniquely for any solution It is easy to compute these expressions more explicitly for this example Because x ζ ζ x (ζ T ) x n = ζ ζ n x n, e obtain (F + x)(ζ) = (ζ) j ζ ζ j x j j j x j ζ ζ j = j (ζ) j ζ ζ j x j ; in the same ay (F x)(ζ) = ζ (ζ) j ζζ x j j j The function is a solution of the AIP if and only if F x H and F x 2 H D(x, x) for all x X; that is, if and only if properties (a), (b), (c) of page 353 hold One can see from the explicit formula that property (a) holds if and only if (ζ j ) = j Property (b) holds automatically To see hat property (c) means, compute the left-hand side On the boundary (for t = ), F+ x and F x look like (F + x)(t) = (t) j x j j j x j, (F x)(t) = j x j (t) j j x j Or e can put these to formulas together: [ [ (F (t) x)(t) = (t) j x j j t ζ j x j t ζ j j

7 THE ABSTRACT INTERPOLATION PROBLEM AND APPLICATIONS 357 No, F x 2 H [ F = + x F, x = j [ x j (F x)(ζ j) ζ j j x j j t ζ j x j j t ζ j = j x j L 2 = k F x, j x j L 2 j k ζ j ζ k x k = D(x, x) The latter computation depends on property (a), F+ x H+ 2 Thus, e can see that is a solution of the AIP ith these data if and only if is a solution of the Nevanlinna Pick problem For this particular problem, condition (a) actually carries all the interpolation information, (b) holds automatically, and (c) follos from (a) Moreover, e emphasize that for any solution of this problem e have the equality F x 2 H = D(x, x) for all x X, instead of inequality To continue on to the second (more general) example, I ill reformulate the first one Consider the closed linear span of the functions { } and denote by K θ H 2 the space H 2 θh 2, here θ is the Blaschke product ith zeros ζ j if they satisfy the Blaschke condition and θ 0 otherise (in this case K θ = H 2 ) We can associate the function x(t) = j x j K θ ith any x X of the first example K θ is invariant under P t, here t is the independent variable, and P is the orthogonal projection onto H 2 It is easy to see that the operator T from the first example acts as P t under this correspondence; in fact, P t j x j = j ζ j x j Let W be the operator on the space X defined by W x x n = x n x n Obviously W T = T W Under the correspondence, W x(t) = j k x j

8 358 ALEXANDER KHEIFETS A Schur class function is a solution of the Nevanlinna Pick problem if and only if W x = P x In fact, P = P (ζ j ) + P (ζ j ) = (ζ j) Observe that M x = j x j = ( x), here the last notation stands for the Fourier coefficient of index of the H 2 function x And M 2 x = ( W x) Since W x = P x for any solution of the Nevanlinna Pick problem, then W x H 2 x H 2 We can reexpress the quadratic form D as In fact, let then If x = j D(x, y) = x, ỹ H 2 W x, W y H 2 W x = x, ỹ = x j, then x = and ỹ = t ζ k ; j, ζ k ζ j, W y = W x, W y = i t ζ i, k j ζ k ζ j x 2 W x 2 = k,j x k k j ζ k ζ j x j = D(x, x) Define the operator W on K θ by W ( x) = W x This is ell defined and D 0 if and only if W is a contraction Example 2 Sarason s Problem No let θ be an arbitrary inner function (not necessarily a Blaschke product) Set K θ = H 2 θh 2 ; this space is invariant under P t Set T = P t K θ Let W be a contractive operator on K θ ith W T = T W One is interested in finding all the Schur class functions such that W x = P x for all x K θ Associate the folloing AIP data to the Sarason problem: X = K θ, T = T, T 2 =, D(x, x) = x 2 W x 2, M x = (x), M 2 x = (W x) ; here E = E 2 = C One can check the Fundamental Identity (see the beginning

9 THE ABSTRACT INTERPOLATION PROBLEM AND APPLICATIONS 359 of Lecture ) For this data the implicit definition (ii) of the mapping F yields an explicit formula for F : [ [ W x F x = ae t T, x for all x X and any solution of the AIP Thus, F is unique for any solution Then again is a solution of the AIP if and only if F x H for all x X and F x 2 H D(x, x), that is, if and only if conditions (a), (b) and (c) of page 353 hold For condition (a) e have F + x = x W x H2 + P (x W x) = 0 P x = W x for all x X In other ords, condition (a) is satisfied if solves the Sarason problem Condition (b), F x = x W x H, 2 holds automatically for any Schur class function For condition (c) e get [ [ F F x 2 L = + x W x F, = F x x x, x L 2 L 2 = x W x, x = x, x W x, x = x, x W x, P x = x, x W x, W x = D(x, x) Thus, for this data is a solution of the AIP if and only if is a solution of the Sarason problem Property (a) carries all the interpolation information, (b) holds automatically, and (c) follos from (a) And e again have the equality F x 2 H = D(x, x) for all x X and for all solutions, instead of inequality Example 2 We no associate another AIP to the same Sarason problem This AIP is best considered as being different (nonequivalent) from the one in Example 2, though they have a common set of solutions The reason is that the coefficient matrices in the description formulas for the solution sets (see the last theorem of Lecture 4) are different and the associated universal colligations (see Section 2 of Lecture 4) are nonequivalent Let θ be an inner function, and set K θ = H+ 2 θh+ 2 and Tθ x 2 = P + tx 2 for x 2 K θ Let W2 be a contractive operator on K θ that commutes ith Tθ : W 2 T θ = T θ W 2 Find all the Schur class functions such that W 2 x 2 = P + x 2 One can check that the solutions of this problem and of the one in the previous example coincide (if the operator W of Example 2 and the operator W 2 are properly connected) Consider the AIP ith the data X = K θ, T =, T 2 = Tθ, E = E 2 = C, M x 2 = (W x 2 ) 0, and M 2 x 2 = (x 2 ) 0, here the notation ( ) 0 stands for the Fourier coefficient of index 0 of an H+ 2 function As in Example 2,

10 360 ALEXANDER KHEIFETS F can be expressed explicitly and uniquely for any solution of the AIP ith this data: [ [ F x 2 x 2 = W2 x 2 No consider properties (a), (b) and (c) For (a) e have F + x 2 = x 2 W 2 x 2 H 2 +; this holds automatically for any Schur class function Property (b) becomes F x 2 = x 2 W 2 x 2 H 2 ; this holds if and only if P + x 2 = W 2 x 2; that is, if and only if solves the Sarason problem Finally, for (c) e can rite F x 2 2 L = D(x 2, x 2 ) for all x 2 X for any solution Thus, for these data property (b) carries all the interpolation information, (a) holds automatically for any Schur class, and (c) follos from (b) The equality holds for any solution F x 2 2 H = D(x 2, x 2 ) for all x 2 X Example 3 The boundary interpolation problem Property (c) dominates in this example and (a) and (b) follo The equality in condition (c) holds for some solutions but not for all of them We ill need some preliminaries A Schur class function in the unit disk D is said to have an angular derivative in the sense of Carathéodory at the point t 0 T if (ζ) has a nontangential unimodular limit 0 as ζ goes to t 0, 0 =, and has a nontangential limit 0 at t 0 (ζ) 0 ζ ζ 0 Theorem (Carathéodory) A Schur class function (ζ) has an angular derivative at t 0 T if and only if D,t0 def (ζ) 2 = lim inf ζ t 0 ζ 2 < (here ζ and ζ t 0 in an arbitrary ay) In this case 0 = D,t 0 0 (ζ) 2 ζ 2 D,t0 as ζ goes to t 0 nontangentially D,t0 vanishes if and only if is a constant of modulus t 0 and

11 THE ABSTRACT INTERPOLATION PROBLEM AND APPLICATIONS 36 Moreover, a Schur class function has an angular derivative in the sense of Carathéodory at the point t 0 T if and only if there exists a unimodular constant 0 such that (t) (t) 2 t t 0 t t 0 2 L ; that is, if and only if this function is integrable over T against Lebesgue measure In particular, this guarantees that T 0 t t 0 H 2 +, because the denominator is an outer function In that case ( (t) ) (t) 2 t t 0 t t 0 2 m(dt) = D,t0 No consider the folloing interpolation problem Let t 0 be a point of the unit circle T, let 0 be a complex number ith 0 =, and let 0 D < be a given nonnegative number One ants to describe all the Schur class functions such that (ζ) 0 as ζ t 0 (nontangentially) and D,t0 D Associate to this problem the AIP data X = C, D(x, x) = xdx, T x = t 0 x, T 2 x = x, M x = x, M 2 x = 0 x, E = E 2 = C ; then e can check the Fundamental Identity ( ) The left-hand side of the identity is and the right-hand side is xdx x t 0 Dt 0 x = 0, M x M x M 2 x M 2 x = xx x 0 0 x = 0 The mapping F of Lecture (pages ) is unique for any solution and can be computed explicitly for this data: F x = [ [ 0 t t 0 x t t 0 We can analyse no hat conditions (a), (b), and (c) tell us becomes F + x = 0 x H 2 0 t t +; that is, H 0 t t Condition (b) becomes F x = 0 x = t 0 0 x H t t 0 t t 0 t 2 0 Condition (a)

12 362 ALEXANDER KHEIFETS Hence (a) and (b) coincide Finally, condition (c) becomes [ F x 2 0 L = (F x)(t), t t 0 x m(dt) T t t 0 C 2 = x 0( 0 ) + ( 0 ) T t t 0 2 x m(dt) ( 0 )( 0 ) + = x T t t 0 2 m(dt) x ( 0 2 = x + ) 2 t t 0 t t 0 2 m(dt) x = xd,t0 x T Thus, F x 2 L D if and only if D,t 0 D; that is, is a solution of the AIP ith this data if and only if is a solution of the boundary interpolation problem A reference for Examples 2 and 2 is [Kheifets 990a References for Example 3 are [Kheifets 996; Sarason 994 Lecture 3 Solutions of the Abstract Interpolation Problem Role of the AIP In Lecture 2 e shoed ho some specific analytic problems can be included in the AIP scheme The AIP, as formulated in Lecture, is very ell adapted to this inclusion (actually it arose from the experience of treating a number of problems of this type) To include a specific problem in the scheme one has to realize (if it is possible) hat the associated data is and to prove the coincidence of to solution sets: that of the original analytic problem and that of the AIP ith the associated data The mapping F suggests an algorithm for hat exactly is to be checked to prove this coincidence In this lecture e ill slightly reformulate the AIP to adapt it better to solving the problem (no the origin of the data is unimportant to a certain extent) Thus, the AIP can be vieed as an intermediate link beteen specific interpolation problems and their resolution It has to sides: one of them looks at the specific interpolation problem and is devoted to proving the equivalence of the original problem and associated AIP, the second concerns solutions As soon as a specific problem is included in the scheme, the analysis of its solutions goes in a standard and universal ay Isometric colligation associated ith the AIP data Reformulation of the AIP Let [x stand for the equivalence class of x ith respect to the quadratic form D (The equivalence relation is defined as follos: x 0 if and only if D(x, y) = 0 for all y X, and x x 2 if and only if x x 2 0) Consider the linear space of equivalence classes { [x : x X } and define an inner product in it by def [x, [y = D(x, y)

13 THE ABSTRACT INTERPOLATION PROBLEM AND APPLICATIONS 363 This product is ell defined One can complete it and obtain a Hilbert space, hich e denote by H 0 Rerite the Fundamental Identity ( ) as D(T x, T y) + M x, M y E = D(T 2 x, T 2 y) + M 2 x, M 2 y E2 Using the notations introduced above one can reexpress this as [T x, [T y H 0 + M x, M y E = [T 2 x, [T 2 y H 0 + M 2 x, M 2 y E2 (3 ) Set and {[ } def [T x d v = Clos : x X H 0 E M x {[ } def [T2 x v = Clos : x X H 0 E 2 M 2 x Define a mapping V : d v v by the formula [ [T x V : M x [ [T2 x M 2 x (3 2) Because of (3 ), V is an isometry This implies that V is ell-defined In fact, if [ [T x [ [T x [ [T (x x ) M x = M 2 x, that is, M (x x = 0, ) then (3 ) implies that [ [T2 (x x ) M 2 (x x = 0 ) No, if x and x generate the same vector on the left-hand side of (3 2), they generate the same vector on the right-hand side of (3 2), hich shos that V is ell-defined Let be a solution of the AIP; that is, suppose there exists a mapping F : X H possessing properties (i) and (ii) of page 353 Property (i) says that F x 2 H D(x, x) [x 2 H 0 Hence, F x depends only on the equivalence class [x, not on the representative x This means that F generates a mapping G of the equivalence classes and G[x def = F x, (3 3) G[x 2 H F x 2 H [x 2 H 0 Thus G is a contraction Since { [x : x X } is dense in H 0 and G is a contraction, it can be extended to a contraction of H 0 into H Thus, any F : X H ith properties (i) and (ii) (actually up to no e used only (i)) generates a contraction G : H 0 H such that G[x = F x for all x X

14 364 ALEXANDER KHEIFETS We no try to interpret property (ii) To this end e reexpress (ii) as [ ( [ ) F T 2 x + t M 2 x = t F T x + M x, or [ ( [ ) G[T 2 x + t M 2 x = t G[T x + M x (3 4) Our further goal is to realize the latter equality as one of the form [ [ G[T2 x = A G[T x, (3 5) M 2 x M x here A : H E H E 2 is a linear operator As e kno, G maps H 0 into H, M maps H 0 into E, and M 2 maps H 0 into E 2 The relation (3 4) is of the type [ ( f + t e 2 = t f + [ ) e, (3 6) here f, f H and e E, e 2 E 2 Observe that the three subspaces {[ } { [ } e : e E, H, t e 2 : e 2 E 2 are mutually orthogonal in L (see page 352 for the definition of L ) Let P be the orthogonal projection from L onto H ; then, for any [ f2 f = L, e have [ [ [ [ P f2 f2 P f 2 = f f P + f Obviously this difference is in L It can be reritten as [ [ P f2 P+ f 2 P + f = f P f P f 2 Hence, it is in H Because f [ [ P f 2 P + f is orthogonal to H, P is really the orthogonal projection Because [ f f = + H, e can obtain from (3 6) and (3 7) [ { ( t e 2 = ( L P ) t f + = [ f [ )} e = [ t(f + (0) + (0))e 0 [ [ = t (3 7) [ P t(f + + e ) P + t(f + e ) (f +(0) + (0)e )

15 THE ABSTRACT INTERPOLATION PROBLEM AND APPLICATIONS 365 Thus No, from (3 6) and (3 8), f = t(f + That is, [ [ e ) t e 2 = t(f + ( f = t f [ e 2 = f +(0) + (0)e (3 8) [ ) f + (0) + t 0 Putting (3 8) and (3 9) together e obtain [ [ f f = A e 2 e, A = [ [ (f e ) + (0) + (0)e ) ) [ [ (0) [ A in (A ) [ [ H H A 2 A : 2 E E 2 e (3 9) here ( A in f = P tf = t f (A ) e = t [ A 2 f = f + (0) : H E 2, [ [ (0) : E H, [ ) f+ (0) : H H, 0 A 2 e = (0)e : E E 2 Thus, condition (ii) of the AIP, equivalently (3 4) or (3 5), can be expressed as [ G 0 0 E2 [ [ [T2 x G 0 = A M 2 x 0 E According to the definition (3 2) of the isometry V, [ [T2 x = V M 2 x Combining this ith (3 0) e conclude that [ G 0 0 E2 [ [T x M x [ [T x (3 0) M x V [ G 0 dv = A dv (3 ) 0 E To give (3 ) a further interpretation e digress to recall some basic facts related to unitary colligations, their characteristic functions, and functional models

16 366 ALEXANDER KHEIFETS Digression on unitary colligations, characteristic functions, and functional models Let H, E, E 2 be separable Hilbert spaces A unitary mapping A of H E onto H E 2 is said to be a unitary colligation The space H is called the state space of the colligation, E is called the input space, and E 2 is called the response space Both E and E 2 are called exterior spaces The operator-valued function (ζ) : E E 2 defined by the formula (ζ) = P E2 A( H E ζp H A) E is called the characteristic function of the colligation A Because A is unitary, (ζ) is ell defined for ζ D; is a contractive-valued analytic operatorfunction in D Using the block decomposition [ [ [ Ain A H H A = :, A 2 A 2 E the characteristic function can be reexpressed as E 2 (ζ) = A 2 + ζa 2 ( H ζa in ) A The unitary colligation A is said to be simple ith respect to the exterior spaces E and E 2 if there is no nonzero reducing subspace for A in H; that is, if there is no nonzero subspace H res H that is invariant for A and A We shall call the maximal reducing subspace for A in H the residual subspace of the colligation A Thus a unitary colligation is simple if the residual subspace is trivial Let H res H be the residual subspace of A Let H simp = H H res Then A : H simp E H simp E 2 is a simple unitary colligation, and A H res is a unitary operator on H res To unitary colligations A : H E H E 2 and A : H E H E 2 ith the same exterior spaces are said to be unitarily equivalent if there exists a unitary mapping G : H H such that [ [ G 0 G 0 A = A 0 E2 0 E Theorem To simple unitary colligations A and A are unitarily equivalent if and only if their characteristic functions coincide Let be a Schur class operator function, (ζ) : E E 2 The unitary colligation A considered above is simple and is the characteristic function of this colligation Thus any Schur class operator function is the characteristic function of a unitary colligation

17 THE ABSTRACT INTERPOLATION PROBLEM AND APPLICATIONS 367 Let A : H E H E 2 be a simple unitary colligation, ith characteristic function Then a unitary mapping G : H H that performs an equivalence beteen A and A is defined as follos: [ [ (G+ h)(ζ) PE2 A( H E ζp H A) h (Gh)(ζ) = = (G h)(ζ) ζp E A ( H E2 ζp H A ) (3 2) h The colligation A is called a functional model of A and the mapping G is called the Fourier representation of H If A is not simple, the Fourier representation G defined by the same formula (3 2) is a unitary mapping from H simp onto H and performs an equivalence beteen the simple parts of A and A It vanishes on H res : [ G 0 0 E2 A = A [ G 0 0 E Solutions of the AIP as characteristic functions We are ready to proceed ith the analysis of the AIP solutions We begin ith (3 ) Assume for simplicity that e have the equality F x 2 H = D(x, x) for all x X We noticed in Lecture 2 that for some problems this is the case for any solution In other ords, our assumption means that the map G defined in (3 3) is an isometry: Gh 0 2 H = h 0 2 H 0 Set H = H GH 0 and H = H 0 H Define a mapping G : H H by setting G H 0 = G, G H = H, and observe that G is a unitary mapping of H onto H We can rite [ G 0 V [ G 0 dv = A dv, (3 3) 0 E2 0 E instead of (3 ), because Finally, e obtain Define the operator A by d V H 0 E H E, V H 0 E 2 H E 2 [ G 0 V = 0 E2 A def = [ G 0 0 E2 [ G 0 A dv (3 4) 0 E [ G 0 A (3 5) 0 E A is a simple unitary colligation from H E onto H E 2, and is unitarily equivalent to A Hence, the characteristic function of A is By definition,

18 368 ALEXANDER KHEIFETS H 0 H and A d V = V (see (3 4)); that is, A is a unitary extension of V Thus any solution of the AIP is the characteristic function of a unitary extension A of the isometry V associated ith the data: A : H E H E 2 ith H H 0, A d V = V In general, the equality F x 2 H = D(x, x) for all x X does not hold, only the inequality F x 2 H D(x, x) for all x X Nevertheless, the conclusion is still valid, but it is more difficult to prove (A short and simple proof as recently found by J Ball and T Trent [Ball and Trent 997) Remark I ould like to emphasize one basic difference beteen the general case (inequality) and the special case (equality) considered in this section We need one more definition: a unitary extension A : H E H E 2 of an isometric colligation V : d V V, ith d V H 0 E, V H 0 E 2, H 0 H is said to be a minimal extension if A has no nonzero reducing subspace in H H 0 If an extension A is nonminimal e can discard the reducing subspace in H H 0 and end up ith a unitary colligation that has the same characteristic function and that still extends V, so e can consider minimal extensions only But a minimal extension need not be a simple colligation The absence of a reducing subspace for A in H H 0 does not mean that A has no reducing subspace in H at all By discarding such a reducing subspace, e end up ith a unitary colligation that has the same characteristic function but that no longer extends V In the case of equality, F x 2 H = D(x, x) for all x X, the corresponding minimal extension is a simple colligation, as it is equivalent to A (see (3 4)); in the case of inequality, there exists x X such that F x 2 H < D(x, x), and it is not simple A natural question arises no: does an arbitrary unitary extension A of the isometry V produce a solution of the AIP? The anser is yes, and it is easy to prove Let A : H E H E 2 be a unitary extension of V Let G be the Fourier representation associated ith the colligation A; see (3 2) Then G maps H into H, here is the characteristic function of A Thus G is a contractive operator and [ [ G 0 G 0 A = A (3 6) 0 E2 0 E Define the mapping F : X H by F x def G[x def G[x Recall that [x H 0 H Since G is a contraction, e have F x 2 H [x 2 = D(x, x); H 0

19 THE ABSTRACT INTERPOLATION PROBLEM AND APPLICATIONS 369 that is, F satisfies (i) Since A extends V, (3 6) yields [ G 0 0 E2 V [ G 0 dv = A dv 0 E Since d V H 0 E and V H 0 E 2, e can replace G ith G: [ G 0 0 E2 V [ G 0 dv = A dv 0 E As e have seen, the latter is equivalent to (ii) function of A, is a solution of the AIP Thus, the characteristic Remark Starting ith an extension A of V e do not need the unitarity of G to check that the characteristic function of A is a solution Neither the characteristic function nor the mapping G, much less F, feels the residual part Starting ith the solution in the general case e actually are given information on the simple part of the corresponding extension of V only But e have to restore the residual part also in order to obtain the extension I ill finish this lecture ith the folloing summary of the discussion Theorem Let V be the isometric colligation associated ith the AIP data V : d V V, d V H 0 E, V H 0 E 2 (no special assumptions are made on the data at present) Let A : H E H E 2 be a minimal unitary extension of V, here H H 0, so A d V = V Let (ζ) be the characteristic function of the colligation A: (ζ) def = P E2 A( H E ζp H A) E Then is a solution of the AIP The corresponding mapping F : X H is defined by F x = G[x, here (Gh)(ζ) = [ [ (G+ h)(ζ) PE2 A( H E ζp H A) h = (G h)(ζ) ζp E A ( H E2 ζp H A ) h for h H All solutions of the AIP and the corresponding mappings F that satisfy (i) and (ii) are of this form References to Lecture 3 are [Katsnelson et al 987; Kheifets 988a; 988b; 990b; Kheifets and Yuditskii 994

20 370 ALEXANDER KHEIFETS Lecture 4 Description of the Solutions of the AIP In a sense, the description of all the solutions of the AIP and the corresponding mappings F as given in the last theorem of Lecture 3 The goal of this lecture is to give a description that separates explicitly the common part (related to the data) and the free parameters The first step is the description of the unitary extensions of the given isometric colligation V in this fashion Unitary extensions of isometric colligations Let V be an isometric colligation, V : d V V ith d V H 0 E, V H 0 E 2 Let A be a minimal unitary extension of V, A : H E H E 2 ith H H 0, A d V = V Let d V and V be the orthogonal complements of d V in H 0 E and V in H 0 E 2 Let H = H H 0 Then the orthogonal complement of d V in H E is H d V and the orthogonal complement of V in H E 2 is H V Since A is a unitary operator mapping d V onto V (since A d V = V ), A has to map the orthogonal complement H d V onto the orthogonal complement H V Denote by A the restriction of A to H d V Thus, A : H d V H V is a unitary colligation Since A is a minimal extension, A is a simple colligation Conversely, take an arbitrary simple unitary colligation A ith the same exterior spaces d V and V and an arbitrary admissible state space H : Let H = H 0 H and define an extension of V by A : H d V H V (4 ) A dv = V, A H d V = A The result is a minimal unitary extension of V Thus, the free parameter of a minimal unitary extension of V is an arbitrary simple unitary colligation A ith fixed exterior spaces d V and V and arbitrary admissible state space H The ord admissible means here that there exists a unitary colligation ith this state space For example: If d V and V are finite dimensional but their dimensions are different, then H cannot be of finite dimension; if the dimensions of d V and V are equal then H can be of arbitrary dimension To give a full explanation it is enough to consider model colligations Let ω : d V V be an arbitrary Schur class operator function (contractive-valued and analytic in the unit disc) One can take as A the (model) unitary colligation [ A A ω ω = in (A ω [ [ ) H ω H ω A ω 2 A ω : 2 d V V,

21 THE ABSTRACT INTERPOLATION PROBLEM AND APPLICATIONS 37 here H ω is the de Branges Rovnyak space corresponding to the operator function ω (see page 352) and A ω in f = P ω tf ( = t f (A ω ) = t [ ω ω A ω 2 f = f +(0) : H ω V, A ω 2 = ω(0) : d V V [ [ ) ω f+ (0) ω : H ω H ω, 0 [ ω(0) : d V Hω, Let A : H d V H V and A : H d V H V be to simple unitary colligations If they are unitarily equivalent, that is, if there exists a unitary mapping G : H H such that [ [ G 0 A 0 = A G 0, V 0 d V the corresponding minimal unitary extensions of V are unitarily equivalent colligations: [ [ G 0 G 0 A = A, 0 E2 0 E ith G H = G and G H 0 = H0 We emphasize here that this is more than just equivalence of unitary colligations, as G is the identity on H 0 We ill call such extensions equivalent extensions The folloing proposition follos from the previous discussion Claim To minimal unitary extensions A and A of V are equivalent extensions if and only if the corresponding simple unitary colligations A and A are unitarily equivalent colligations The next claim is a straightforard consequence of the formulas of the last theorem of Lecture 3 Claim 2 To equivalent minimal extensions of V generate the same solution and the same mapping F : X H Thus, a solution of the AIP and a corresponding mapping F : X H represent equivalence classes of simple unitary colligations A : H d V H V Hence, the latter may serve as free parameters But e kno from the digression on page 366 that the only invariant of this equivalence class is the characteristic function of A In other ords: Claim 3 The free parameter of pairs (, F ) (consisting of a solution and a corresponding mapping F ) is an arbitrary Schur class (contractive-valued and analytic in D) operator function ω(ζ) : d V V

22 372 ALEXANDER KHEIFETS Universal unitary colligations associated ith isometric colligations In the previous section e extracted the free parameter ω that the solution and the corresponding mapping F depend on It is clear that the common part of all the extensions of the isometry V is the isometry V itself To obtain nice and explicit formulas it is convenient to associate a universal unitary colligation to the isometry V Let V be an isometric colligation, V : d V V ith d V H 0 E, V H 0 E 2, and let d V and V be the orthogonal complements of d V and V Let N be an isomorphic copy of d V (that is, there exists a unitary and surjective mapping u : d V N ) Let N 2 be an isomorphic copy of V (that is, there exists a unitary and surjective mapping u 2 : V N 2) Define a unitary colligation A 0 : H 0 E N 2 H 0 E 2 N by setting A 0 d V = V (d V H 0 E ), A 0 d V = u, A 0 N 2 = u 2 Remark A 0 is not a unitary extension of V in the sense considered earlier, because e no do not extend the state space H 0, only the exterior spaces: E N 2 instead of E and E 2 N instead of E 2 Of course, A 0 extends V but in a different sense We ill call this extension a universal extension; the name is motivated by category theory We ill see the role of this colligation belo From no on e fix the unitary mappings u and u 2, and also their images N and N 2 Also e fix the notation A for a simple unitary colligation A : H N H N 2 (this class of colligations is obviously related to the colligations of the form (4 ) defined earlier, and is denoted by the same symbol) Thus, no the equivalence classes of the simple unitary colligations A : H N H N 2 ill be free parameters of the solutions and the corresponding mappings F ; that is, the Schur class functions ω(ζ) : N N 2 are free parameters no Coupling of unitary colligations and unitary extensions of isometric colligations Let V be an isometric colligation, V : d V V ith d V H 0 E, V H 0 E 2 Let A 0 be the universal colligation associated ith V : A 0 : H 0 E N 2 H 0 E 2 N, ith A 0 d V = V, A 0 (d V ) = N, A 0 (N 2 ) = V Let A be a minimal unitary extension of V : A : H E H E 2 ith H H 0, A d V = V

23 THE ABSTRACT INTERPOLATION PROBLEM AND APPLICATIONS 373 We have seen that one can associate a simple unitary colligation A : H N H N 2 ith A, here H = H H 0 : [ [ H 0 A = 0 (A 0) (A H d V ) H 0 V 0 (A (4 2) 0) N An arbitrary simple unitary colligation A ith the exterior spaces N and N 2 arises this ay I ill no give a procedure for recovering A from A 0 (fixed colligation) and A (arbitrary parameter) Let A : H N H N 2 be an arbitrary simple unitary colligation Let h 0 e 2 = A 0 n h 0 e n 2 and [ h n 2 [ h = A n (4 3) We can choose vectors on the right-hand sides of these relations in an arbitrary ay and compute the vectors on the left-hand sides, or e can choose vectors on the left-hand sides in an arbitrary ay and compute the vectors on the right-hand sides (because A 0 and A are unitary operators) Let us consider, in addition to the relations above, to more relations n = n and n 2 = n 2, (4 4) and see hat can be chosen no in an arbitrary ay Observe that the colligation A 0 possesses the property P N A 0 N 2 = 0, because A 0 sends N 2 onto V H 0 E 2, hich is orthogonal to N This property of A 0 guarantees (although it is not necessary) that e can choose h 0, h, e in an arbitrary ay and compute h 0, h, e 2 from the system of equations (4 3), (4 4) in a unique ay, along ith n = n and n 2 = n 2 as ell Take the result of this computation as the definition of the ne linear operator A: A h h 0 e def = h h 0 e 2 (4 5) It is easy to see that A is a unitary colligation, A : H H 0 E H 0 H 0 E 2 Moreover, A extends V To see this, take an arbitrary vector [ h0 d V ; e then take [ h 0 = V e 2 and take also [ h0 e V, h = h = 0, n = n = 0, n 2 = n 2 = 0

24 374 ALEXANDER KHEIFETS This collection of vectors satisfies the system of equations (4 3), (4 4): it obviously fits (4 4) and the second equality in (4 3); it satisfies the first equality in (4 3) because A 0 d V = V ; ie, [ [ V h0 e = A 0 0 h 0 e 0 if [ h0 e d V Hence, by the definition of the colligation A, these vectors (actually, a part of them) satisfy (4 5): [ 0 [ 0 V [ h 0 = A h 0 h0 if d V e e e This means that A extends V We ill say that the colligation A is the feedback coupling of A 0 ith A (or the loading of A 0 ith A ) It is also easy to check that these to procedures the extraction of A from the extension A and the feedback coupling of A 0 ith A are mutually inverse: performing the to successively e come back to the colligation that e started ith Our next goal is to express the characteristic function and the Fourier representation G of the colligation A in terms of the characteristic functions and Fourier representations of the colligations A 0 and A But e need a digression first Digression: Unitary colligations, characteristic functions, Fourier representation, and discrete time dynamics Let A : H E H E 2 be a unitary colligation One can associate ith A the discrete-time dynamical system [ [ h(k + ) h(k) = A, (4 6) e 2 (k) e (k) here k is a nonnegative integer, h(0) is an arbitrary vector from H, and {e (k)} k=0 is an arbitrary input signal Let ζ D be the corresponding spectral parameter (complex frequency) Let h + (ζ), ẽ + (ζ), and ẽ+ 2 (ζ) be the discrete Laplace transforms of h(k), e (k), and e 2 (k), respectively; that is, h + (ζ) = ζ k h(k), k=0 ẽ + (ζ) = ζ k e (k), k=0 If the input signal {e (k)} is square summable, e (k) 2 <, k=0 ẽ + 2 (ζ) = ζ k e 2 (k) then the state evolution {h(k)} k=0 and the output signal {e 2(k)} k=0 possess the same property K=0

25 THE ABSTRACT INTERPOLATION PROBLEM AND APPLICATIONS 375 Multiplying (4 6) by ζ k and taking the summation over k, one obtains the spectral form of the dynamics equation: h + (ζ) = ( H ζa in ) h(0) + G (ζ) ẽ + (ζ) ẽ + 2 (ζ) = G +(ζ) h(0) + (ζ) ẽ + (ζ), (4 7) here (ζ) is the characteristic function of the colligation A, and G +, G are the components of the Fourier representation G (see (3 2)) One can also rerite (4 6) as A [ h(k + ) e 2 (k) = [ h(k) (4 8) e (k) No consider the negative integers k (This actually means inverting time and exchanging the roles of the input and the output) Considering the past Laplace transform, h (ζ) = k= one can rerite (4 8) as ζ k h(k), ẽ 2 (ζ) = k= ζ k e 2 (k), ẽ (ζ) = k= ζ k e (k), h (ζ) = ζ( ζa in ) h(0) + G + (ζ) ẽ 2 (ζ), ẽ (ζ) = G (ζ) h(0) + (ζ) ẽ 2 (ζ) (4 9) The second formulas in (4 7) and (4 9) are the most convenient for our purposes: ẽ + 2 (ζ) = G +(ζ)h(0) + (ζ)ẽ + (ζ) ẽ (ζ) = G (ζ)h(0) + (ζ) ẽ 2 (ζ) (4 0) Formulas describing the solutions of the AIP and the corresponding mappings F Let A be the feedback coupling of A 0 ith A Denote the characteristic function of A 0 by S(ζ) : E N 2 E 2 N Denote also the entries of S corresponding to this decomposition of the spaces as follos: [ [ [ s0 s 2 E E2 S = : s s Let the characteristic function of A be ω(ζ) : N N 2 No rite the dynamics related to A 0 and the dynamics related to A in the form (4 0) I am going to consider the + parts only no; the treatment of the parts is analogous We have ñ + 2 (ζ) = G + (ζ)h (0) + ω(ζ)ñ + (ζ), [ ẽ+ 2 (ζ) ñ + (ζ) The coupling condition is N 2 N [ = G 0 ẽ+ +(ζ)h 0 (0) + S(ζ) (ζ) ñ + 2 (ζ) ñ + = ñ +, ñ+ 2 = ñ + 2

26 376 ALEXANDER KHEIFETS Express ẽ + 2 (ζ) in terms of [ h (0) h 0 (0) and ẽ + (ζ), excluding ñ+ = ñ + and ñ+ 2 = ñ + 2 What e obtain no has to coincide (see (4 0)) ith [ ẽ + 2 (ζ) = G h (0) +(ζ) + (ζ)ẽ + h 0 (0) This leads to the formulas (ζ) = s 0 (ζ) + s 2 (ζ)ω(ζ)( N s(ζ)ω(ζ)) s (ζ) [ h (0) G + (ζ) = [ ψ(ζ)ω(ζ), E2 G 0 h 0 (0) + (ζ)h 0 (0) + ψ(ζ)g +(ζ)h (0), (4 ) here ψ(ζ) = s 2 (ζ)( N2 ω(ζ)s(ζ)) (4 2) Remark By the definition of the characteristic function, S(0) = P E2 N A 0 (E N 2 ) In particular, s(0) = P N A N 2 = 0 This guarantees that the formulas (4 ) and (4 2) make sense, since s(ζ) ζ hen ζ < by Scharz s lemma Considering the parts of (4 0) for A 0, A, and A e obtain [ h (0) G (ζ) = [ ϕ(ζ) ω(ζ), E G 0 h 0 (0) (ζ)h 0 (0) + ϕ(ζ) G (ζ)h (0), (4 3) here ϕ(ζ) = ( N s(ζ)ω(ζ)) s (ζ) (4 4) Combining the expressions (4 ) and (4 2) for G + and G e arrive at [ [ [ h ψω E2 0 0 ψ 0 G = 0 0 ϕ ω G 0 h 0 + E 0 ϕ G h (4 5) h 0 As e sa in the last theorem of Lecture 3, F x = G[x for all x X, here [x H 0 is the equivalence class generated by the quadratic form D Hence [ ψω E2 0 0 F x = 0 0 ϕ ω G 0 [x E The folloing theorem summarizes this lecture

27 THE ABSTRACT INTERPOLATION PROBLEM AND APPLICATIONS 377 Theorem Let V be the isometric colligation associated ith the AIP data (page 362) Let A 0 be the unitary colligation associated ith V (page 372) Let S be the characteristic function of A 0 : S(ζ) = P E2 N A 0 ( H0 E N 2 ζp H0 A 0 ) (E N 2 ), [ [ [ s0 (ζ) s 2 (ζ) E E2 S(ζ) = : s (ζ) s(ζ) N 2 Then the solutions of the AIP and the corresponding mappings F : X H are described by N = s 0 + s 2 ω( N sω) s [ ψω E2 0 0 F x = 0 0 ϕ ω E G 0 [x for x X here ω is an arbitrary Schur class function ω(ζ) : N N 2, for ζ <, ψ = s 2 ( N2 ωs), ϕ = ( N sω) s, [ [ G G 0 0 (ζ)h 0 = + (ζ)h 0 PE2 N G 0 = A 0 ( ζp H0 A 0 ) h 0 (ζ)h 0 ζp E N 2 A 0 ( ζp H0 A 0 ) h 0 S and G 0 depend on the data of the AIP only, hereas ω is arbitrary We can see that the parameter ω defines uniquely not only the solution but also the corresponding mapping F : X H This suggests denoting the mappings F by F ω : X H In particular cases, hen the mapping F is unique for a solution, that is, hen the F ω coincide if the corresponding solutions coincide, it can be denoted by F References to Lecture 4 are [Arov and Grossman 983; Katsnelson et al 987; Kheifets 988a; 988b; 990b; Kheifets and Yuditskii 994 Acknoledgements I am very grateful to Harry Dym for discussions and personal conversations that e had during my stay at the Weizmann Institute in Rehovot, Israel, from fall 993 through fall 996, and to that institution for its encouraging scientific atmosphere The Shabbat dinners ith Harry and Irena Dym are particularly orth mentioning I thank also the Mathematics Sciences Research Institute in Berkeley, California, for its support My participation in the Holomorphic Spaces program in fall 995 gave me a great opportunity to meet many colleagues and friends I am indebted to Don Sarason, Joe Ball, D Z Arov, Victor Katsnelson, Victor Vinnikov, Daniel Alpay, Misha Cotlar, Cora Sadosky, Stefania Marcantognini, Jim Rovnyak, Nikolas Young, Nikolai Nikolskii, Vasilii Vasyunin, Sergei

28 378 ALEXANDER KHEIFETS Treil, Vladimir Peller, Boris Freidin and many others for fruitful discussions and friendly conversations Special thanks go to Don Sarason and Silvio Levy for a lot of suggestions that improved the manuscript, and to Miriam Abraham for her perfect and careful typing of the manuscript References [Arov and Grossman 983 D Z Arov and L Z Grossman, Scattering matrices in the theory of extensions of isometric operators, Dokl Akad Nauk SSSR 270: (983), 7 20 In Russian; translation in Soviet Math Dokl 27:3 (983), [Ball and Trent 997 J Ball and T Trent, in Proceedings of IWOTA-96 (Bloomington, IN), Oper Theory Adv Appl, Birkhäuser, Basel [Katsnelson et al 987 V È Katsnel son, A Y Kheĭfets, and P M Yuditskiĭ, An abstract interpolation problem and the theory of extensions of isometric operators, pp 83 96, 46 in Operators in function spaces and problems in function theory, edited by V A Marchenko, Naukova Dumka, Kiev, 987 In Russian; translation in Topics in interpolation theory, edited by Harry Dym et al, Oper Theory Adv Appl 95, Birkhäuser, Basel, 997, pp [Kheifets 988a A Y Kheĭfets, The Parseval equality in an abstract problem of interpolation, and the coupling of open systems, I, Teor Funktsiĭ Funktsional Anal i Prilozhen 49 (988), 2 20, 26 In Russian; translation in J Sov Math 49:4 (990) 4 20 [Kheifets 988b A Y Kheĭfets, The Parseval equality in an abstract problem of interpolation, and the coupling of open systems, II, Teor Funktsiĭ Funktsional Anal i Prilozhen 50 (988), 98 03, iv In Russian; translation in J Sov Math 49:6 (990) [Kheifets 990a A Y Kheĭfets, A generalized bitangential Schur Nevanlinna Pick problem and its related Parseval equality, Teor Funktsiĭ Funktsional Anal i Prilozhen 54 (990), In Russian; translation in J Sov Math 58:4 (992), [Kheifets 990b A Y Kheĭfets, Scattering matrices and the Parseval equality in the abstract interpolation problem, PhD thesis, 990 In Russian [Kheifets 996 A Kheifets, Hamburger moment problem: Parseval equality and A- singularity, J Funct Anal 4:2 (996), [Kheifets and Yuditskii 994 A Y Kheĭfets and P M Yuditskiĭ, An analysis and extension of V P Potapov s approach to interpolation problems ith applications to the generalized bi-tangential Schur Nevanlinna Pick problem and J-inner-outer factorization, pp 33 6 in Matrix and operator valued functions, edited by I Gohberg and L A Sakhnovich, Oper Theory Adv Appl 72, Birkhäuser, Basel, 994 [Sarason 994 D Sarason, Sub-Hardy Hilbert spaces in the unit disk, University of Arkansas Lecture Notes in the Mathematical Sciences, Wiley, Ne York, 994

29 THE ABSTRACT INTERPOLATION PROBLEM AND APPLICATIONS 379 Alexander Kheifets Mathematics Division Institute for Lo-Temperature Physics and Engineering 47 Lenin Avenue Kharkov 3064 Ukraine

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