On the Nevanlinna problem -- Characterization of all Schur-Agler class solutions affiliated with a given kernel
Tirthankar Bhattacharyya, Anindya Biswas, Vikramjeet Singh Chandel
aa r X i v : . [ m a t h . F A ] O c t On the Nevanlinna problem -Characterization of all Schur-Agler class solutionsaffiliated with a given kernel
Tirthankar Bhattacharyya ∗ Anindya Biswas † Vikramjeet Singh Chandel ‡ Abstract
Given a domain Ω in C m , and a finite set of points z , z , . . . , z n ∈ Ω and w , w , . . . , w n ∈ D (the open unit disc in the complex plane), the Pick interpolation problem asks when there isa holomorphic function f : Ω → D such that f ( z i ) = w i , ≤ i ≤ n . Pick gave a conditionon the data { z i , w i : 1 ≤ i ≤ n } for such an interpolant to exist if Ω = D . Nevanlinnacharacterized all possible functions f that interpolate the data. We generalize Nevanlinna’sresult to a domain Ω in C m admitting holomorphic test functions when the function f comesfrom the Schur-Agler class and is affiliated with a certain completely positive kernel. TheSchur class is a naturally associated Banach algebra of functions with a domain. The successof the theory lies in characterizing the Schur class interpolating functions for three domains- the bidisc, the symmetrized bidisc and the annulus - which are affiliated to given kernels. Keywords—
Nevanlinna problem, Schur-Agler class, Colligation, Test function, Severalcomplex variables
Given a solvable scalar valued interpolation problem from the unit disc into the unit disc,Nevanlinna in [13] gave a complete set of solutions. The main result of this paper, Theorem 6,is a far reaching generalization of Nevanlinna’s result. ∗ Department of Mathematics, Indian Institute of Science, Bangalore 560012, India. e-mail: [email protected] † Department of Mathematics, Indian Institute of Science, Bangalore 560012, India. e-mail: [email protected] ‡ Department of Mathematics, Indian Institute of Technology Bombay, Powai, Mumbai, 400076, India. e-mail:[email protected] .1 Test functions A collection Ψ of C -valued functions on a set Ω is called a set of test functions (see [4] and [10])if the following conditions hold:1. sup {| ψ ( x ) | : ψ ∈ Ψ } < for each x ∈ Ω .2. For each finite subset F of Ω , the collection { ψ | F : ψ ∈ Ψ } together with the constantfunction generates the algebra of all C -valued functions on F .The second condition is not essential for the development of the theory, but it makes somesituations simpler (it is excluded in [4] but not in [10]). The collection Ψ is a natural topologicalsubspace of D Ω equipped with the product topology. For every x ∈ Ω , there is an element E ( x ) in C b (Ψ) , the C ∗ -algebra of all bounded functions on Ψ , such that E ( x )( ψ ) = ψ ( x ) . Clearly, k E ( x ) k = sup ψ ∈ Ψ | ψ ( x ) | < for each x ∈ Ω . The functions E ( x ) will be used at several places inthis paper. A positive kernel k on a set Ω is a function k : Ω × Ω → C such that for any n ≥ , any n points z , z , . . . , z n in Ω and any n complex numbers c , c , . . . , c n , we have n X i =1 n X j =1 c i c j k ( z i , z j ) ≥ . If E is a Hilbert space and k : Ω × Ω → B ( E ) is a function, then k is called a positive kernel if forany n ≥ , any n points z , z , . . . , z n in Ω and any n vectors e , e , . . . , e n in E , we have n X i =1 n X j =1 h k ( z i , z j ) e j , e i i ≥ . (1)The concept of a positive kernel does not cease here. Let A and B be two C ∗ -algebras and let Γ be a function on Ω × Ω taking values in B ( A , B ) (space of all bounded linear operators from A to B ). Γ is called a completely positive kernel if n X i,j =1 b ∗ i Γ( z i , z j )( a ∗ i a j ) b j ≥ (2)for all n ≥ , a , a , . . . , a n ∈ A , b , b , . . . , b n ∈ B and z , z , . . . , z n ∈ Ω .It is well-known that a completely positive kernel has a Kolmogorov decomposition (see [3]or [4] or [6]). The following situation will occur several times in our paper. Let us consider aHilbert space Y and a collection Ψ of test functions. Take the C ∗ -algebras C b (Ψ) and B ( Y ) , aHilbert space X , a unital ∗ -representation µ : C b (Ψ) → B ( X ) and a function h : Ω → B ( X , Y ) .Then it is easy to see that Γ( x, y )( δ ) = h ( z ) µ ( δ ) h ( w ) ∗ for all z, w ∈ Ω and δ ∈ C b (Ψ)
2s a completely positive kernel. Conversely, if
Γ : Ω × Ω → B ( C b (Ψ) , B ( Y )) is a completely positive kernel, then there exists a Hilbert space X , a unital ∗ -representation µ : C b (Ψ) → B ( X ) and a function h : Ω → B ( X , Y ) such that Γ( z, w )( δ ) = h ( z ) µ ( δ ) h ( w ) ∗ for all z, w ∈ Ω and δ ∈ C b (Ψ) . (3)This is called the Kolmogorov decomposition.
For example, take a Hilbert space Y and a function ψ : Ω → C from a given collection Ψ of test functions. Since | ψ ( z ) | < for each z ∈ Ω , for any finite subset Ω ⊂ Ω we have sup z ∈ Ω | ψ ( z ) | < . Now take the map Γ ψ : Ω × Ω → B ( C b (Ψ) , B ( Y )) defined by Γ ψ ( z, w )( δ ) = δ ( ψ )1 − ψ ( z ) ψ ( w ) I Y , z, w ∈ Ω , δ ∈ C b (Ψ) . (4)Since the positivity condition (2) involves only a finite number of points from Ω , we shall haveno difficulty in checking the positivity of this map. Let X = L ∞ j =0 Y j where Y j = Y for every j . Define µ ψ : C b (Ψ) → B ( X ) by µ ψ ( δ ) = δ ( ψ ) I X and h ψ : Ω → B ( X , Y ) by h ψ ( z ) (cid:0) ⊕ ∞ j =0 y j (cid:1) = ∞ X j =0 ψ ( z ) j y j . Clearly µ ψ is a unital ∗ -representation and h ψ ( z ) is bounded linear transformation. Note that Γ ψ ( z, w )( δ ) = h ψ ( z ) µ ψ ( δ ) h ψ ( w ) ∗ . Hence by (3), Γ ψ is a completely positive kernel. Ψ -unitary Colligations Let X , U and Y be Hilbert spaces and let Ψ be a fixed set of test functions. By a Ψ -unitarycolligation, we mean a pair ( U, ρ ) where U is a unitary operator from X ⊕ U to X ⊕ Y , and ρ : C b (Ψ) → B ( X ) is a ∗ -representation. If we write U as U = (cid:18) X UX
A B Y C D (cid:19) , then we can define a bounded B ( U , Y ) valued function on Ω , given by f ( x ) = D + Cρ ( E ( x ))( I X − Aρ ( E ( x ))) − B ∀ x ∈ Ω , (5)equivalently, f ( x ) = D + C ( I X − ρ ( E ( x )) A ) − ρ ( E ( x )) B ∀ x ∈ Ω . (6)This f is called the transfer function associated with ( U, ρ ) . Since U ∗ is also a unitary, we havethat g ( x ) = D ∗ + B ∗ ( I X − ρ ( E ( x )) A ∗ ) − ρ ( E ( x )) C ∗ is the transfer function of the colligation ( U ∗ , ρ ) .3 .4 The Ψ -Schur-Agler Class Let E be a Hilbert space and Ω an abstract set. We consider a B ( E ) -valued kernel K (satisfying(1)) on Ω . For this K , there is a Hilbert space H ( K ) of E -valued functions on Ω such that spanof the set { K ( · , ω ) e : e ∈ E , ω ∈ Ω } is dense in H ( K ) and for any e ∈ E , ω ∈ Ω and h ∈ H ( K ) , we have h h, K ( · , ω ) e i H ( K ) = h h ( ω ) , e i E . Given a set of test functions Ψ on Ω , a kernel K : Ω × Ω → B ( E ) is said to be Ψ -admissible if the map M ψ , sending each element h ∈ H ( K ) to ψ · h , is a contraction on H ( K ) . We denotethe set of all B ( E ) -valued Ψ -admissible kernels by K Ψ ( E ) . For two Hilbert spaces U and Y , wesay that S : Ω → B ( U , Y ) is in H ∞ Ψ ( U , Y ) if there is a non-negative constant C such that the B ( Y ⊗ Y ) -valued function ( C I Y − S ( x ) S ( y ) ∗ ) ⊗ k ( x, y ) (7)is a positive B ( Y ⊗ Y ) -valued kernel for every k in K Ψ ( Y ) . If S is in H ∞ Ψ ( U , Y ) , then we de-note by C S the smallest C which satisfies (7). The collection of maps S ∈ H ∞ Ψ ( U , Y ) for which C S is no larger than is called the Ψ -Schur-Agler class and it is denoted by SA Ψ ( U , Y ) .The plan of the paper is as follows. In Section 2, we prove a characterization of functions inthe class SA Ψ ( U , Y ) . This is followed by Section 3 which describes the usefulness of takingholomorphic test functions. Section 4 consists of the description of an auxiliary function G .Section 5 has the main theorem. In Section 6, we give applications of our main result. SA Ψ ( U , Y ) Variants of the following theorem exist in various forms in literature, see [10] and [4] and thereferences therein. We did not find it in the form that we shall need. The most non-trivialimplication is . ⇒ . and we shall prove this since we did not find, in the literature, a proof ofit. Other implications are easy to see. Theorem 1.
Consider a function S on some subset Ω of Ω with values in B ( U , Y ) . Then the follow-ing conditions are equivalent.1. There exists an S in SA Ψ ( U , Y ) such that S | Ω = S .2. S has an Agler decomposition on Ω , that is, there exists a completely positive kernel Γ :Ω × Ω → B ( C b (Ψ) , B ( Y )) so that I Y − S ( z ) S ( w ) ∗ = Γ( z, w )(1 − E ( z ) E ( w ) ∗ ) for all z, w ∈ Ω .
3. There exists a Hilbert space X , a ∗ -representation ρ : C b (Ψ) → B ( X ) and a Ψ -unitary colliga-tion ( V, ρ ) such that writing V as V = (cid:18) X UX
A B Y C D (cid:19) , ne has S ( z ) = D + C ( I X − ρ ( E ( z )) A ) − ρ ( E ( z )) B for all z ∈ Ω . (8)
4. There exists a Hilbert space X , a ∗ -representation ρ : C b (Ψ) → B ( X ) and a Ψ -unitary colliga-tion ( W, ρ ) such that writing W as W = (cid:18) X YX A B U C D (cid:19) , one has S ( z ) ∗ = D + C ( I X − ρ ( E ( z )) ∗ A ) − ρ ( E ( z )) ∗ B for all z ∈ Ω . (9) Proof.
As mentioned before, we shall only prove that . implies . Consider an S ∈ SA Ψ ( U , Y ) and a Ψ -admissible kernel K : Ω × Ω → B ( Y ) . As is usual, denote H ( K ) = span { K ( · , w ) y : w ∈ Ω , y ∈ Y } . Define a linear transformation T ∗ on the dense subspace span { K ( · , w ) y ⊗ y : w ∈ Ω , y , y ∈ Y } by first defining T ∗ (cid:16) K ( · , w ) y ⊗ y (cid:17) = K ( · , w ) y ⊗ S ( w ) ∗ y , and then extending linearly. For w i ∈ Ω , y i , y ,i ∈ Y , ≤ i ≤ n, we have || n X i =1 K ( · , w i ) y i ⊗ y i || − || T ∗ (cid:16) n X i =1 K ( · , w i ) y i ⊗ y i (cid:17) || = n X i,j =1 *(cid:16) ( I Y − S ( w j ) S ( w i ) ∗ ) ⊗ K ( w j , w i ) (cid:17) y i ⊗ y i , y j ⊗ y j + Since S ∈ SA Ψ ( U , Y ) , the last expression is nonnegative. So T defines a contraction from H ( K ) ⊗ U to H ( K ) ⊗ Y . We need the following lemma to continue with the proof.
Lemma.
Let J : Ω × Ω → B ( Y ) be a self-adjoint function, i.e., J ( z, w ) = J ( w, z ) ∗ for all z, w ∈ Ω (see page 174 of [1]). If J ⊘ K : ( z, w ) J ( z, w ) ⊗ K ( z, w ) (10)is a positive kernel for every B ( Y ) -valued admissible kernel K , then there is a completelypositive kernel Γ : Ω × Ω → B ( C b (Ψ) , B ( Y )) such that J ( z, w ) = Γ( z, w )(1 − E ( z ) E ( w ) ∗ ) for all z, w ∈ Ω . Proof of the lemma.
We prove the result for a finite subset Ω = { w , w , . . . , w n } of Ω and apply Kurosh’s theorem.Consider the following subset of n × n self-adjoint operator matrices with entries in B ( Y ) W Ω = n(cid:16) Γ( w i , w j )(1 − E ( w i ) E ( w j ) ∗ ) (cid:17) ≤ i,j ≤ n :Γ : Ω × Ω → B ( C b (Ψ) , B ( Y )) is a completely positive kernel o W Ω ⊂ B ( Y n ) is a convex set and it is invariant under multiplication by positive realscalars in the space of n × n self-adjoint matrices. Such a set is called a wedge (see [1], page 169)).It needs to be shown that W Ω is closed in the weak ∗ -topology of B ( Y n ) . To that end, startwith a net h Γ ν ( w i , w j )(1 − E ( w i ) E ( w j ) ∗ ) i ≤ i,j ≤ n = h A ν i ≤ i,j ≤ n in W Ω and suppose that it con-verges in the weak ∗ -topology to an n × n self-adjoint matrix A = h A ij i with entries in B ( Y ) .This means that for every X = h X lp i in B ( Y n ) (the space of trace class operators on Y n ), { tr ( A ν X ) } converges to tr ( AX ) . Let u, v ∈ Y with || u || , || v || ≤ and choose X to be the opera-tor matrix which has u ⊗ v as its ( j, i ) -th entry and zeros elsewhere. Then tr ( A ν X ) = D A ν u, v E tends to tr ( AX ) = D Au, v E . We have that D Γ ν ( w i , w j )(1 − E ( w i ) E ( w j ) ∗ ) u, v E −→ D A ij u, v E . Now − E ( w i ) E ( w i ) ∗ ≥ − || E ( w i ) || > gives us that there is an ǫ > such that − E ( w i ) E ( w i ) ∗ > ǫ · for all i = 1 , , . . . , n. Hence we get that D Γ ν ( w i , w i )(1 − E ( w i ) E ( w i ) ∗ ) u, u E ≥ ǫ D Γ ν ( w i , w j ) u, u E for all i = 1 , , . . . , n. Since the left hand side converges, we can find an
M > such that sup ν D Γ ν ( w i , w i ) u, u E ≤ M for all i = 1 , , . . . , n. Also for any δ ∈ C b (Ψ) , we have D Γ ν ( w i , w i )( δδ ∗ ) u, u E ≤ || δ || D Γ ν ( w i , w i ) u, u E ≤ M || δ || . Now we need the following result which is a Cauchy-Schwarz type inequality. Since we couldnot find an exact reference, we are giving a proof of the claim.
Claim: If Γ is completely positive, δ ∈ C b (Ψ) , z, w ∈ Ω and u, v ∈ Y , then we have (cid:12)(cid:12)(cid:12)D Γ( z, w )( δδ ∗ ) u, v E(cid:12)(cid:12)(cid:12) ≤ D Γ( z, z )( δδ ∗ ) v, v ED Γ( w, w )( δδ ∗ ) u, u E . Proof of the claim:
Note that, using Kolmogorov decomposition (3), we can find a Hilbert space X , a unital ∗ -representation µ : C b (Ψ) → B ( X ) and a function h : Ω → B ( X , Y ) such that Γ( z, w )( δ ) = h ( z ) µ ( δ ) h ( w ) ∗ for all z, w ∈ Ω and δ ∈ C b (Ψ) . So we have (cid:12)(cid:12)(cid:12)D Γ( z, w )( δδ ∗ ) u, v E(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)D h ( z ) µ ( δδ ∗ ) h ( z ) ∗ u, v E(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)D µ ( δ ∗ ) h ( z ) ∗ u, µ ( δ ∗ ) h ( w ) ∗ v E(cid:12)(cid:12)(cid:12) ≤ D µ ( δ ∗ ) h ( z ) ∗ u, µ ( δ ∗ ) h ( z ) ∗ u ED µ ( δ ∗ ) h ( w ) ∗ v, µ ( δ ∗ ) h ( w ) ∗ v E . D Γ( z, z )( δδ ∗ ) v, v ED Γ( w, w )( δδ ∗ ) u, u E . Hence the claim follows. ⊓⊔ Using this result, we conclude that (cid:12)(cid:12)(cid:12)D Γ ν ( w i , w j )( δδ ∗ ) u, v E(cid:12)(cid:12)(cid:12) ≤ M || δ || for every i, j = 1 , , . . . , n. Therefore, for each δ ∈ C b (Ψ) , u, v ∈ Y and i, j = 1 , , . . . , n , the net nD Γ ν ( w i , w j )( δδ ∗ ) u, v Eo isbounded. Since Ω is finite, we get a subnet ν l such that nD Γ ν l ( w i , w j )( δδ ∗ ) u, v Eo converges tosome number depending on δ, u and v , for every i, j = 1 , , . . . , n . Define a completely positivekernel Γ : Ω × Ω → B ( C b (Ψ) , B ( Y )) by D Γ( w i , w j )( δ ) u, v E = lim l D Γ ν l ( w i , w j )( δ ) u, v E andextend it trivially to the whole set Ω × Ω . Consequently, for every u, v ∈ Y we have D Γ( w i , w j )(1 − E ( w i ) E ( w j ) ∗ ) u, v E = D A ij u, v E . This proves that W Ω is weak ∗ -closed.The n × n matrix h I Y i with each entry equal to I Y , is in W Ω . Indeed, let ψ ∈ Ψ and take Γ ψ : Ω × Ω → B ( C b (Ψ) , B ( Y )) defined by Γ ψ ( z, w )( δ ) = δ ( ψ )1 − ψ ( z ) ψ ( w ) I Y , z, w ∈ Ω . Note that sup z ∈ Ω | ψ ( z ) | < as Ω is finite. From (4), it is clear that Γ ψ is completely positive. Sowe have Γ ψ ( z, w )(1 − E ( z ) E ( w ) ∗ ) = I Y and hence we conclude that h I Y i ∈ W Ω .Also, the restriction of J (see (10)) on Ω × Ω , that is, J | Ω × Ω = J is in W Ω . If possi-ble, let J / ∈ W Ω . By Theorem . in [14], we get a weak ∗ -continuous linear functional L on B ( Y n ) whose real part is nonnegative on W Ω and strictly negative at J . We replace L ( R ) by L ( R )+ L ( R )2 , R ∈ B ( Y n ) , and denote it by L itself. Since L is weak ∗ -continuous and for any locallyconvex space X , we have ( X ∗ ; weak ∗ ) ∗ = X (see Theorem V.1.3 in [9]), we find that L is of theform L ( R ) = tr ( RC ) for some n × n self-adjoint compact C ∈ B ( Y n ) whose entries are in theideal of trace class operators on Y n .Let { e n : n ≥ } be an orthonormal basis of Y . Given a bounded operator A on Y , define its transpose to be the linear transformation on Y whose matrix entries with respect to the basisabove are h A t e j , e i i = h Ae i , e j i . It is easy to see that this defines a bounded operator. Indeed, if u ∈ Y is given by u = P u i e i , then we define u = P u i e i and then we have h A t u, v i = h Av, u i for u, v ∈ Y which on application of the Cauchy-Schwarz inequality yields boundedness.Since C obtained above is a block n × n operator matrix C = (( C ( w i , w j ))) ni,j =1 , define C t to be the block n × n operator matrix whose ( i, j ) th. entry is C t ( w i , w j ) = C ( w j , w i ) t . In otherwords, h C t ( w i , w j ) u, v i = h C ( w j , w i ) v, u i for u, v ∈ Y .7e shall show that C t is a B ( Y ) -valued positive kernel on Ω , that is, for u i ∈ Y , ≤ i ≤ n , n X i,j =1 (cid:10) C t ( w i , w j ) u j , u i (cid:11) ≥ To see this, note that n X i,j =1 (cid:10) C t ( w i , w j ) u j , u i (cid:11) = n X i,j =1 (cid:10) u i , C ( w j , w i ) ∗ u j (cid:11) = n X i,j =1 (cid:10) tr (( u i ⊗ u j ) C ji ) (cid:11) . This last quantity is tr ( DC ) , where D = ( D i,j ) = ( u i ⊗ u j ) , and hence equals L ( D ) .For any function u : Ω → Y , if ∆ ψ : Ω × Ω → B ( C b (Ψ) , B ( Y )) is the function defined by ∆ ψ ( z, w ) : δ δ ( ψ ) u ( z ) ⊗ u ( w )1 − ψ ( z ) ψ ( w ) , z, w ∈ Ω , then ∆ ψ ( z, w )( δ δ ) = δ ( ψ ) δ ( ψ )1 − ψ ( z ) ψ ( w ) u ( z ) ⊗ u ( w ) . Let δ , δ , . . . , δ n ∈ C b ( ψ ) , B , B , . . . , B n ∈ B ( Y ) , v ∈ Y and z , z , . . . , z n ∈ Ω . Then we have (cid:28)(cid:16) n X i,j =1 B ∗ i ∆ ψ ( z i , z j )( δ ∗ i δ j (cid:1) B j (cid:17) v, v (cid:29) = n X i,j =1 α i α j − ψ ( z i ) ψ ( z j ) where α i = δ i ( ψ ) (cid:10) u ( z i ) , B i ( v ) (cid:11) i = 1 , , . . . , n. The last expression is clearly non-negative. There-fore, ∆ Ψ is completely positive. Now, ∆ ψ (1 − E ( z ) E ( w ) ∗ ) = u ( z ) ⊗ u ( w ) gives that D is in W Ω .Hence, P ni,j =1 (cid:10) C t ( w i , w j ) u j , u i (cid:11) = L ( D ) ≥ . Thus, C t is a B ( Y ) -valued positive kernel on Ω .In fact, C t is admissible. To see that, consider the function Ξ ψ : Ω × Ω → B ( C b (Ψ) , B ( Y )) defined by Ξ ψ ( z, w )( δ ) = δ ( ψ ) u ( z ) ⊗ u ( w ) for ψ ∈ Ψ and u : Ω → Y . Let δ i ∈ C b (Ψ) , B i ∈ B ( Y ) and z i ∈ Ω , ≤ i ≤ n. Then for any v ∈ Y we have n X i,j =1 D(cid:16) B ∗ i Ξ ψ ( z i , z j )( δ ∗ i δ j ) B j (cid:17) v, v E = | n X i =1 δ i ( ψ ) h B i v, u ( z i ) i| ≥ . So Ξ ψ is completely positive and hence Ξ ψ ( z, w )(1 − E ( z ) E ( w ) ∗ ) | Ω × Ω ∈ W Ω . Now let u , u , . . . , u n ∈ Y . A little computation gives n X i,j =1 (cid:10) (1 − ψ ( w i ) ψ ( w j )) C t ( w i , w j ) u j , u i (cid:11) = tr (Ξ ψ C ) . But then Ξ ψ ( z, w )(1 − E ( z ) E ( w ) ∗ ) | Ω × Ω ∈ W Ω gives us tr (Ξ ψ C ) ≥ . So C t is Ψ -admissible.8y our assumption, the B ( Y ⊗ Y ) -valued function J ⊘ C t on Ω × Ω is positive. So for any u i ∈ Y , ≤ i ≤ n, we have n X i,j =1 D J ⊘ C t ( w i , w j ) u j , u i E ≥ . With { e p : p ≥ } an orthonormal basis of Y , set u i = P Nm =1 e m ⊗ e m , for all i, where N is somefixed natural number. So we have n X i,j =1 D J ⊘ C t ( w i , w j ) u j , u i E = n X i,j =1 N X p,q =1 D J ( w i , w j ) e p , e q ED C t ( w i , w j ) e p , e q E and this is nonnegative. This holds for any N ≥ . On the other hand L ( J ) = n X i,j =1 tr (cid:16) J ( w i , w j ) C ( w j , w i ) (cid:17) = n X i,j =1 ∞ X p,q =1 D J ( w i , w j ) e p , e q ED C t ( w i , w j ) e p , e q E which is nonnegative by the previous argument. But this is a contradiction since L ( J ) < .Hence J ∈ W Ω .It is easy to see that conditions in Kurosh’s theorem are satisfied (see Theorem 2.56 in [1],page 74-75 in [2]). So the finiteness condition on the set Ω can be removed.This completes the proof of the statement that under given conditions there is a completelypositive kernel Γ : Ω × Ω → B ( C b (Ψ) , B ( Y )) such that J ( z, w ) = Γ( z, w )(1 − E ( z ) E ( w ) ∗ ) for all z, w ∈ Ω . This completes the proof of the lemma. ⊓⊔ To complete the proof of the theorem, note that if S ∈ SA Ψ ( U , Y ) , then clearly ( I Y − S ( z ) S ( w ) ∗ ) ⊗ K ( z, w ) is positive for every Ψ -admissible kernel K . Hence an application of theresult above shows that there is a completely positive kernel Γ : Ω × Ω → B ( C b (Ψ) , B ( Y )) suchthat I Y − S ( z ) S ( w ) ∗ = Γ( z, w )(1 − E ( z ) E ( w ) ∗ ) for all z, w ∈ Ω . When test functions are holomorphic, strong consequences can be shown to follow. That is thepurpose of this section.There are some known cases like the bidisc ([1]), the symmetrized bidisc ([7]) and the an-nulus ([10]) where the collection of test functions has the nice property of being holomorphic.Moreover, in all these cases, we can find a point in the domain at which each all test functionsvanish. We now want our collection Ψ to have this property without changing the Ψ -Schur-Agler class. Our purpose will be served if we can find another collection Θ of holomorphic testfunctions having a common zero such that SA Ψ ( U , Y ) = SA Θ ( U , Y ) . SA Ψ ( U , Y ) is the set { S : Ω → B ( U , Y ) : ( I Y − S ( x ) S ( y ) ∗ ) ⊗ k ( x, y ) is positive for every k ∈ K Ψ ( Y ) } . So if we can show that for a collection Θ of holomorphic test functions (with our requiredproperty) the equality K Ψ ( Y ) = K Θ ( Y ) holds, then we are done. The following propositionprovides us with such a collection. Proposition 2.
Let Ψ be a collection of holomorphic test functions on a domain Ω ⊂ C m , w ∈ Ω apoint and Y a Hilbert space. Then there is a collection Θ = { θ ψ : ψ ∈ Ψ } such that the followingconditions are met.1. θ ψ ( w ) = 0 for all ψ ∈ Ψ .2. Θ is a collection of holomorphic test functions.3. K Ψ ( Y ) = K Θ ( Y ) Proof.
Fix a w ∈ Ω and for each ψ ∈ Ψ define ϕ ψ : D → D by ϕ ψ (˜ z ) = ψ ( w ) − ˜ z − ψ ( w )˜ z . Since || E ( w ) || < , ϕ ψ is non-constant. Let us define a holomorphic function θ ψ : Ω → D by θ ψ = ϕ ψ ◦ ψ . Clearly θ ψ ( w ) = 0 for all ψ ∈ Ψ .We shall show that Θ = { θ ψ : ψ ∈ Ψ } is a collection of holomorphic test functions. To provethe first defining condition of the test functions, we start with the assumption that for some z ∈ Ω , sup ψ ∈ Ψ | θ ψ ( z ) | = 1 . So there is a sequence { ψ n } ⊂ Ψ such that lim n →∞ | ϕ ψ n ◦ ψ n ( z ) | = 1 . Let f n : Ω → D denote the function ϕ ψ n ◦ ψ n . Since { f n } is a uniformly bounded collection ofholomorphic functions on Ω , by Montel’s theorem, there is a subsequence { f n l } of { f n } and aholomorphic function f : Ω → D such that f n l → f uniformly on every compact subset of Ω as l → ∞ . Clearly | f ( z ) | = 1 . Since Ω is a domain (open connected set), by maximum modulustheorem, f must be constant on Ω . But we also have f ( w ) = lim l →∞ f n l ( w ) = lim l →∞ ϕ ψ nl ◦ ψ n l ( w ) = 0 . This is a contradiction. Hence sup ψ ∈ Ψ | θ ψ ( z ) | < for each z ∈ Ω .Let F be a finite subset of Ω . Choose ψ ∈ Ψ and consider θ ψ = ϕ ψ ◦ ψ . Since F is finite and foreach z ∈ Ω , sup ψ ∈ Ψ | ψ ( z ) | < and sup ψ ∈ Ψ | θ ψ ( z ) | < , there is an ǫ ∈ (0 , such that sup F | ψ | < ǫ and sup F | θ ψ | < ǫ . Now, ϕ ψ has a power series representation which converges absolutelyand uniformly in B (0 , ǫ ) ⊂ D . So, θ ψ | F is in the closed algebra generated by { , ψ | F } . Since ψ = ϕ − ψ ◦ θ ψ , a similar argument gives us that ψ | F is in the closed algebra generated by { , θ ψ | F } .Hence the closed algebras generated by { , ψ | F : ψ ∈ Ψ } and { , θ ψ | F : ψ ∈ Ψ } coincide. Since Ψ is a collection of test functions, the closed algebra generated by { , θ ψ | F : ψ ∈ Ψ } is the algebraof all C -valued functions on F . This proves that Θ is a collection of holomorphic test functions.Now consider the Hilbert space Y . K Ψ ( Y ) and K Θ ( Y ) are the sets of all Ψ -admissible and Θ -admissible B ( Y ) -valued kernels, respectively. Let k ∈ K Ψ ( Y ) . Then for each ψ ∈ Ψ , themap M ψ : H ( k ) → H ( k ) , sending h to ψ · h is a contraction. It is a well known fact that for an10 ∈ D and a contraction T on a Hilbert space H , ( aI H − T )( I H − aT ) − is again a contraction on H (see [12], page 14). So if a = ψ ( w ) , then ( aI Y − M ψ )( I Y − aM ψ ) − : H ( k ) → H ( k ) is again a contraction. We shall show that ( aI Y − M ψ )( I Y − aM ψ ) − = M θ ψ . Choose any h ∈ H ( k ) and any k ( · , w ) y ∈ H ( k ) with w ∈ Ω and y ∈ Y . It is clear that D ( aI Y − M ψ )( I Y − aM ψ ) − h, k ( · , w ) y E H ( k ) = ( a − ψ ( w )) D ( I Y − aM ψ ) − h, k ( · , w ) y E H ( k ) . Since | a | < and M ψ is a contraction, we have an absolutely convergent series representation ( I Y − aM ψ ) − = ∞ X j =0 a j M jψ Each term of this series is a repeated application of M ψ . Hence we have D ( I Y − aM ψ ) − h, k ( · , w ) y E H ( k ) = (1 − aψ ( w )) − D h, k ( · , w ) y E H ( k ) . Since θ ψ ( w ) = a − ψ ( w )1 − aψ ( w ) , we get that D ( aI Y − M ψ )( I Y − aM ψ ) − h, k ( · , w ) y E H ( k ) = θ ψ ( w ) D h, k ( · , w ) y E H ( k ) . This proves h ( aI Y − M ψ )( I Y − aM ψ ) − i ∗ = M ∗ θ ψ and consequently M θ ψ is a contraction on H ( k ) .Thus k ∈ K Θ ( Y ) . So K Ψ ( Y ) ⊆ K Θ ( Y ) . Since ϕ − ψ = ϕ ψ and ϕ − ψ ◦ θ ψ = ψ , the same argumentused above proves K Θ ( Y ) ⊆ K Ψ ( Y ) . This completes the proof of the proposition. ⊓⊔ From now on, we shall assume without loss of generality that our collection of holomorphic testfunctions has a common zero, that is, there is a point w ∈ Ω such that E ( w ) = 0 in C b (Ψ) . We now develope a power series representation.Suppose z ∈ Ω . Then there is a polydisc P = P ( z , r ) centered at z with multi-radius r such that P ⊂ Ω . Let bP be its distinguished boundary. Since the collection Ψ consists ofholomorphic functions, for each ψ ∈ Ψ we have ψ ( z ) = X α ∈ N m (cid:16) πi ) m Z bP ψ ( ζ )( ζ − z ) α +1 dζ (cid:17) ( z − z ) α (11)where α + 1 stands for ( α + 1 , α + 1 , . . . , α m + 1) .Now we shall introduce a collection of functions in C b (Ψ) that has a great importance inthe power series development of Ψ -Schur-Agler class. For each α ∈ N m , define a function δ α : Ψ → C by δ α ( ψ ) = 1(2 πi ) m Z bP ψ ( ζ )( ζ − z ) α +1 dζ . (12)We have the following result on these δ α . Proposition 3.
With z and P as above, δ α ∈ C b (Ψ) for each α ∈ N m . roof. Clearly | δ α ( ψ ) | ≤ r α as || ψ || Ω ≤ . Since this holds for any ψ , we have || δ α || ≤ r α , that is, δ α is a bounded function on Ψ with norm less than or equal to r α . Now we shall show that δ α iscontinuous on Ψ for each α ∈ N m . Let us fix an α ∈ N m and consider a net { ψ ν : ν ∈ N } thatconverges to a ψ in Ψ . Consider the Borel probability measure dµ = dθ (2 π ) m = dθ dθ . . . dθ m (2 π ) m on [0 , π ] m and a positive ǫ . Let E ν = { θ ∈ [0 , π ] m : | ψ ν ( z + re iθ ) − ψ ( z + re iθ ) | ≥ ǫ } . Claim: lim ν µ ( E ν ) = 0 . Proof of the claim:
If not, then there is a positive number τ such that for every ν ∈ N , we can find a η ν ∈ N suchthat η ν ≥ ν and µ ( E η ν ) ≥ τ . Now { ψ η ν : ν ∈ N } is a subnet of { ψ ν : ν ∈ N } , so it convergespoint-wise to ψ on Ω . Since our collection Ψ consists of holomorphic functions, we can take theclosure of the uniformly bounded set { ψ η ν : ν ∈ N } in the compact-open topology of the spaceof all holomorphic functions on Ω and get a compact subset as a result. Hence { ψ η ν : ν ∈ N } hasa convergent subnet { ψ η ν : ν ∈ N ′ } ( N ′ ⊂ N ). Let { ψ η ν : ν ∈ N ′ } converge to some holomorphic φ : Ω → C . This convergence is uniform on each compact subset of Ω . But ψ is a point-wiselimit of the subnet { ψ η ν : ν ∈ N ′ } as well. So we have φ = ψ . Since bP is a compact subset of Ω ,we can find a ν ∈ N ′ such that sup ζ ∈ bP | ψ η ν ( ζ ) − ψ ( ζ ) | = sup θ ∈ [0 , π ] m | ψ η ν ( z + re iθ ) − ψ ( z + re iθ ) | < ǫ. So we have E η ν = ∅ . But this is a contradiction, as µ ( E η ν ) ≥ τ > . Thus we get lim ν µ ( E ν ) = 0 .Hence the claim follows. ⊓⊔ From this we conclude lim ν || ψ ν ( z + re i ( · ) ) − ψ ( z + re i ( · ) ) || L ( µ, [0 , π ] m ) = 0 (see [11] page 124, Theorem 7). Since | δ α ( ψ ν ) − δ α ( ψ ) | ≤ r α || ψ ν ( z + re i ( · ) ) − ψ ( z + re i ( · ) ) || L ( µ, [0 , π ] m ) , it follows that δ α is continuous on Ψ . So we conclude that δ α ∈ C b (Ψ) for each α . This completesthe proof. ⊓⊔ Now let us take a look at the maps E ( z ) , z ∈ Ω . The following result gives a local powerseries representation of E ( z ) . Theorem 4.
For each z ∈ Ω , there is a polydisc P ⊂⊂ Ω centered at z such that the series X α ∈ N m ( z − z ) α δ α converges absolutely and uniformly on every compact subset of P to E ( z ) . Proof.
Take P and r as above and fix a z ∈ P . Then | z − z | α || δ α || ≤ | z − z | α r α α ∈ N m . Since z is in the interior of P , we can find a q ∈ (0 , such that | z − z | α r α Let Ω be a domain in C m and let Ψ be a class of holomorphic test functions on Ω . Then foreach z ∈ Ω and S ∈ SA Ψ ( U , Y ) , there is a polydisc P ( z , ε ) ⊂⊂ Ω centered at z with multi-radius ε such that S ( z ) = X α ∈ N m ( z − z ) α a α for all z ∈ P ( z , ε ) , where a α ∈ B ( U , Y ) . Moreover, the series converges absolutely and uniformly on each compact subsetof P ( z , ε ) . ⊓⊔ We start with a given collection Ψ of test functions that vanish at the point w . So E ( w ) = 0 in C b (Ψ) . Let us prove the following lemma. Lemma. Suppose that z i ∈ Ω and B i ∈ B ( U , Y ) , ≤ i ≤ n . The interpolation problem z i B i is solvable by a function in SA ψ ( U , Y ) if and only if there is a completely positivekernel Γ : { z , z , . . . , z n } × { z , z , . . . , z n } → B ( C b (Ψ) , B ( Y )) such that I Y − B i B ∗ j = Γ( z i , z j )(1 − E ( z i ) E ( z j ) ∗ ) for all i, j = 1 , , . . . , n. (13)The proof of this lemma follows from Theorem 1. ⊓⊔ G . For a solvable problem z i B i , ≤ i ≤ n , (13) gives usa completely positive kernel Γ : { z , z , . . . , z n } × { z , z , . . . , z n } → B ( C b (Ψ) , B ( Y )) such that I Y − B i B ∗ j = Γ( z i , z j )(1 − E ( z i ) E ( z j ) ∗ ) for all i, j = 1 , , . . . , n. Using Kolmogorov decomposition (3) for Γ : { z , z , . . . , z n } × { z , z , . . . , z n } → B ( C b (Ψ) , B ( Y )) we have that there exists a Hilbert space X , a unital ∗ -representation µ : C b (Ψ) → B ( X ) anda function h : { z , z , . . . , z n } → B ( X , Y ) such that Γ( z i , z j )( δ ) = h ( z i ) µ ( δ ) h ( z j ) ∗ for all i, j ∈ { , , . . . , n } and δ ∈ C b (Ψ) . This gives us I Y − B i B ∗ j = h ( z i ) h ( z j ) ∗ − h ( z i ) µ ( E ( z i ) E ( z j ) ∗ ) h ( z j ) ∗ . So for any y, w ∈ Y , we have h y, w i − h B ∗ i y, B ∗ j w i = h h ( z i ) ∗ y, h ( z j ) ∗ w i − h µ ( E ( z i )) ∗ h ( z i ) ∗ y, µ ( E ( z j )) ∗ h ( z j ) ∗ w i . Let L = span { µ ( δ ) h ( z i ) ∗ y : 1 ≤ i ≤ n, δ ∈ C b (Ψ) , y ∈ Y } . So the last equality can be rewrittenas * (cid:18) µ ( E ( z i )) ∗ h ( z i ) ∗ yy (cid:19) , (cid:18) µ ( E ( z j )) ∗ h ( z j ) ∗ ww (cid:19) + L ⊕ Y = * (cid:18) h ( z i ) ∗ yB ∗ i y (cid:19) , (cid:18) h ( z j ) ∗ wB ∗ j w (cid:19) + L ⊕ U . Now, let N = span ( (cid:18) µ ( E ( z i )) ∗ h ( z i ) ∗ yy (cid:19) : y ∈ Y , i = 1 , , . . . , n ) and N = span ( (cid:18) h ( z i ) ∗ yB ∗ i y (cid:19) : y ∈ Y , i = 1 , , . . . , n ) . Then there is a unitary V : N → N that sends (cid:18) µ ( E ( z i )) ∗ h ( z i ) ∗ yy (cid:19) to (cid:18) h ( z i ) ∗ yB ∗ i y (cid:19) for all i and y. (14)Let M = ( L ⊕ U ) ⊖ N and M = ( L ⊕ Y ) ⊖ N . Since, N and N are unitarilyequivalent, the linear operator Q : N ⊕ M ⊕ M → N ⊕ M ⊕ M sending n ⊕ m ⊕ m to15 n ⊕ m ⊕ m is a well defined unitary operator. Since N ⊕ M ⊕ M ≃ L ⊕ Y ⊕ M and N ⊕ M ⊕ M ≃ L ⊕ U ⊕ M , we can write Q as Q = L M ⊕ YL Q Q M ⊕ U Q Q ! . (15)We consider the function G ( z ) := Q ∗ + Q ∗ ( I L − µ ( E ( z )) Q ∗ ) − µ ( E ( z )) Q ∗ , ( z ∈ Ω) . (16)Then for each z ∈ Ω , we have G ( z ) ∈ B ( M ⊕ U , M ⊕ Y ) . Moreover, by the result of Section2, we have G ∈ SA ψ ( M ⊕ U , M ⊕ Y ) . Since, Q is an extension of V , we have that Q ( n ⊕ M ⊕ M ) = V n ⊕ M ⊕ M for all n ∈ N . Hence, we can write Q (cid:18) µ ( E ( z i )) ∗ h ( z i ) ∗ yy (cid:19) = (cid:18) h ( z i ) ∗ yB ∗ i y (cid:19) for all i = 1 , , . . . , n and y ∈ Y . In other words, L M ⊕ YL Q Q M ⊕ U Q Q ! (cid:18) µ ( E ( z i )) ∗ h ( z i ) ∗ y M ⊕ y (cid:19) = (cid:18) h ( z i ) ∗ y M ⊕ B ∗ i y (cid:19) , or, Q ( µ ( E ( z i )) ∗ h ( z i ) ∗ y ) + Q ( M ⊕ y ) = h ( z i ) ∗ y and Q ( µ ( E ( z i )) ∗ h ( z i ) ∗ y ) + Q ( M ⊕ y ) = M ⊕ B ∗ i y, or, ( I L − Q µ ( E ( z i )) ∗ ) − Q ( M ⊕ y ) = h ( z i ) ∗ y and Q ( M ⊕ y ) + Q ( µ ( E ( z i )) ∗ h ( z i ) ∗ y ) = M ⊕ B ∗ i y. Combining the last two equations, we obtain Q ( M ⊕ y ) + Q µ ( E ( z i )) ∗ ( I L − Q µ ( E ( z i )) ∗ ) − Q ( M ⊕ y ) = M ⊕ B ∗ i y, and hence from the definition of G (see (16)) we get G ( z i ) ∗ ( M ⊕ y ) = M ⊕ B ∗ i y. (17)Let us write G ( z ) ∗ as G ( z ) ∗ = M YM G ( z i ) ∗ G ( z i ) ∗ U G ( z i ) ∗ G ( z i ) ∗ ! . 16o we have M YM G ( z i ) ∗ G ( z i ) ∗ U G ( z i ) ∗ G ( z i ) ∗ ! (cid:18) M y (cid:19) = (cid:18) M B ∗ i y (cid:19) which gives G ( z i ) = and G ( z i ) = B i for i = 1 , , . . . , n. This completes the construction of G .Note that for any m ∈ M , we have Q ( N ⊕ M ⊕ m ) = ( N ⊕ m ⊕ M ) . Since N ⊕ M = L ⊕ Y and N ⊕ M = L ⊕ U , we can write N ⊕ M = L ⊕ Y and N ⊕ m = l ⊕ u for some l ∈ L and u ∈ U . So using (15) we get that Q ( m ⊕ Y ) = M ⊕ u .Now we have that for any z ∈ Ω , G ( z ) ∗ := Q + Q ( I L − µ ( E ( z )) ∗ Q ) − µ ( E ( z )) ∗ Q ; and by virtue of Proposition 2, we can assume without loss of generality that there is a point w ∈ Ω such that E ( w ) = 0 . So we have G ( w ) ∗ ( m ⊕ Y ) = M ⊕ u . Since m ∈ M isarbitrary, this gives G ( w ) ≡ as an operator on M . Hence the function z 7→ || G ( z ) || from Ω to R is non-constant. Being a component of a Ψ -Schur-Agler class function G , the function G has a power series expansion by Theorem 5. So we can apply maximum modulus theoremfor Banach space valued holomorphic functions and deduce that if z ∈ Ω , then || G ( z ) || < .A proof of the maximum modulus theorem for Banach space valued holomorphic functions ofone variable can be found in ([16], page 269, Theorem . ), and this proof can easily be carriedout in our case as well.For a given set of data for a solvable interpolation problem, z i ∈ Ω and B i ∈ B ( U , Y )1 ≤ i ≤ n , we shall keep this G fixed. We start this section by noting that given a data set { z , z , . . . , z n } and { B , B , . . . , B n } wherethe B i are in B ( U , Y ) , if there are two Hilbert spaces M and M and a function G = (cid:18) G G G G (cid:19) in SA Ψ ( M ⊕ U , M ⊕ Y ) satisfying G ( z i ) = B i , G ( z i ) = 0 for all i = 1 , , . . . , n and k G ( z ) k < at all points of Ω ,then for any function t in SA Ψ ( M , M ) , the function f ( z ) = (cid:0) G + G ( I M − t G ) − t G (cid:1) ( z ) is an interpolant for the data in SA Ψ ( U , Y ) .Consider a solvable interpolation problem z i B i , where z i ∈ Ω and B i ∈ B ( U , Y ) , ≤ i ≤ n . If f ∈ SA Ψ ( U , Y ) is a solution to this interpolation problem then Theorem 1 givesus that we can find a completely positive kernel ∆ : Ω × Ω → B ( C b (Ψ) , B ( Y )) such that I Y − f ( z ) f ( w ) ∗ = ∆( z, w )(1 − E ( z ) E ( w ) ∗ ) for all z, w ∈ Ω . (18)17he restriction of ∆ to { z , z , . . . , z n } × { z , z , . . . , z n } may not agree with Γ in (13) in general.When they do, we call f an affiliated solution . To be more precise, we give a proper definition. Definition : Let z , z , . . . , z n ∈ Ω and B , B , . . . , B n ∈ B ( U , Y ) be a solvable data. Let f ∈ SA ψ ( U , Y ) be a solution. Let Γ and ∆ be as in (13) and (18), respectively. Then f is saidto be affiliated with Γ if Γ( z i , z j ) = ∆( z i , z j ) for all i, j = 1 , , . . . , n .Why does one need the concept of affiliation? Because, given a solvable interpolation prob-lem, the kernel Γ obtained in (13) may not be unique. An example can be found on page of [1]. The notion of affiliation first appeared in [5] where the authors solved the Nevanlinnaproblem for the bidisc assuming this notion. The following theorem is a generalization. Theorem 6. Let Ω be a domain in C m and let Ψ be a class of holomorphic test functions. Supposethat f ∈ SA Ψ ( U , Y ) is a solution of this interpolation problem z i B i and f is affiliated with acompletely positive kernel Γ : { z , z , . . . , z n } × { z , z , . . . , z n } → B ( C b (Ψ) , B ( Y )) . Let M , M and G be as in Section 4. Writing G as G ( z ) = (cid:18) M UM G ( z ) G ( z ) Y G ( z ) G ( z ) (cid:19) , we have f ( z ) = (cid:16) G + G ( I M − t G ) − t G (cid:17) ( z ) for some t ∈ SA Ψ ( M , M ) and for all z ∈ Ω . Note : Before we embark on the proof, we want to remark that1. without loss of generality, we can assume that all test functions vanish at a certain point w , i.e., E ( w ) = 0 because of Proposition 2,2. the inverse of I M − t G exists because of the concluding remarks of Section 4. Proof of Theorem 6. Since f is a solution and f is affiliated with Γ , we can find a completely positive kernel ∆ :Ω × Ω → B ( C b (Ψ) , B ( Y )) such that I Y − f ( z ) f ( w ) ∗ = ∆( z, w )(1 − E ( z ) E ( w ) ∗ ) for all z, w ∈ Ω (19)and Γ( z i , z j ) = ∆( z i , z j ) for all i, j = 1 , , . . . , n . Now, we know from (3) that there is a Hilbertspace X , ∗ -representation ρ : C b (Ψ) → B ( X ) and a function g : Ω → B ( X , Y ) such that ∆( z, w )( a ) = g ( z ) ρ ( a ) g ( w ) ∗ for all z, w ∈ Ω and a ∈ C b (Ψ) . (20)From these equations we can construct a unitary ˜ W : X ⊕ Y → X ⊕ U (as we did in section4) such that writing ˜ W as ˜ W = (cid:18) X YX ˜ A ˜ B U ˜ C ˜ D (cid:19) , f ( z ) ∗ = ˜ D + ˜ C ( I X − ρ ( E ( z )) ∗ ˜ A ) − ρ ( E ( z )) ∗ ˜ B for all z ∈ Ω (21)and ˜ W takes (cid:18) ρ ( E ( z )) ∗ g ( z ) ∗ yy (cid:19) to (cid:18) g ( z ) ∗ yf ( z ) ∗ y (cid:19) . Let L = span { ρ ( δ ) g ( z i ) ∗ y : y ∈ Y , δ ∈ C b (Ψ) , ≤ i ≤ n } . This is a closed subspace of X and it is reducing for ρ ( E ( z )) , for all z ∈ Ω . Recall the subspace L of X from Section 4 which was defined by L = span { µ ( δ ) h ( z i ) ∗ y : 1 ≤ i ≤ n, δ ∈ C b (Ψ) , y ∈ Y } . Now Γ( z i , z j ) = ∆( z i , z j ) for all i, j = 1 , , . . . , n, gives us g ( z i ) ρ ( δ ) g ( z j ) ∗ = h ( z i ) µ ( δ ) h ( z j ) ∗ , for all ≤ i, j ≤ n, δ ∈ C b (Ψ) . It is easy to see that the map ˜ S : L → L sending ρ ( δ ) g ( z i ) ∗ y to µ ( δ ) h ( z i ) ∗ y is a unitary.Let H = X ⊖ L and S = I H ⊕ ˜ S . Then S : X → H ⊕ L is a unitary.We define λ : C b (Ψ) → B ( H ⊕ L ) by λ ( δ ) = Sρ ( δ ) S ∗ , δ ∈ C b (Ψ) , and l : Ω → B ( H ⊕ L ⊕ Y ) by l ( z ) = g ( z ) S ∗ . Clearly, λ is a unital ∗ -representation. We have l ( z i ) ∗ y = h ( z i ) ∗ y and λ ( δ ) l ( z i ) ∗ y = µ ( δ ) h ( z i ) ∗ y for all ≤ i, j ≤ n, δ ∈ C b (Ψ) . (22)So for any δ, δ ′ ∈ C b (Ψ) , y ∈ Y and i = 1 , , . . . , n , we get λ ( δ ) (cid:0) µ ( δ ′ ) h ( z i ) ∗ y (cid:1) = µ ( δ ) (cid:0) µ ( δ ′ ) h ( z i ) ∗ y (cid:1) ,that is, λ ( δ ) | L = µ ( δ ) | L , for all δ ∈ C b (Ψ) . (23)Since L is reducing for λ ( E ( z )) for all z ∈ Ω , we get using (20) and the definitions of λ and l that I Y − f ( z ) f ( w ) ∗ = l ( z ) λ (1 − E ( z ) E ( w ) ∗ ) l ( w ) ∗ for all z, w ∈ Ω . Write W = ( S ⊕ I U ) ˜ W ( S ⊕ I Y ) ∗ . This is a unitary from H ⊕ L ⊕ Y to H ⊕ L ⊕ U that takes (cid:18) λ ( E ( z )) ∗ l ( z ) ∗ yy (cid:19) to (cid:18) l ( z ) ∗ yf ( z ) ∗ y (cid:19) for all y ∈ Y and z ∈ Ω . So writing W as W = (cid:18) H ⊕ L YH ⊕ L A B U C D (cid:19) , f ( z ) ∗ = D + Cλ ( E ( z )) ∗ ( I H ⊕ L − Aλ ( E ( z )) ∗ ) − B, for all z ∈ Ω . (24)In particular, W takes (cid:18) µ ( E ( z i )) ∗ h ( z i ) ∗ yy (cid:19) to (cid:18) h ( z i ) ∗ yB ∗ i y (cid:19) for all y ∈ Y and ≤ i ≤ n (25)because of (22).Now recall that L ⊕ Y = N ⊕ M and L ⊕ U = N ⊕ M from Section 4. So W maps H ⊕ M ⊕ N onto H ⊕ M ⊕ N and maps N onto N unitarily. So from (14) and (25) weobtain W | N = V . Hence we are allowed to write W = H ⊕ M N H ⊕ M Z N V ! (26)where Z : H ⊕ M → H ⊕ M is a unitary. Now we write Z as Z = H M H Z Z M Z Z ! (27)and take t ( z ) ∗ = Z + Z λ ( E ( z )) ∗ ( I H − Z λ ( E ( z )) ∗ ) − Z for all z ∈ Ω . (28)Clearly t ∈ SA Ψ ( M , M ) .Let us fix a z ∈ Ω and a y ∈ Y , and put u = f ( z ) ∗ y . Let k z = ( I H ⊕ L − Aλ ( E ( z )) ∗ ) − By . It isan element of H ⊕ L . A little computation gives Aλ ( E ( z )) ∗ k z + By = k z and Cλ ( E ( z )) ∗ k z + Dy = u. This can be rewritten as W (cid:18) λ ( E ( z )) ∗ k z y (cid:19) = (cid:18) k z u (cid:19) . (29)For any Hilbert space X and a closed subspace M of X , let us denote the orthogonal projec-tion of X onto M by P X → M . Let r z = P H ⊕ L → H k z . Since L is reducing for λ ( E ( z )) for all z ∈ Ω (see (22)), we have thefollowing λ ( E ( z )) ∗ r z = P H ⊕ L → H ( λ ( E ( z )) ∗ k z ) = P H ⊕ L ⊕ Y → H ( λ ( E ( z )) ∗ k z ⊕ y ) and r z = P H ⊕ L ⊕ U → H ( k z ⊕ u ) . λ ( E ( z )) ∗ k z ⊕ y ∈ H ⊕ L ⊕ Y = H ⊕ N ⊕ M and k z ⊕ u ∈ H ⊕ L ⊕ U = H ⊕ N ⊕ M , there exist n ′ i ∈ N i and m ′ i ∈ M i , i = 1 , , such that λ ( E ( z )) ∗ k z ⊕ y = λ ( E ( z )) ∗ r z ⊕ n ′ ⊕ m ′ and k z ⊕ u = r z ⊕ n ′ ⊕ m ′ . (30)So (29) gives us W ( λ ( E ( z )) ∗ r z ⊕ n ′ ⊕ m ′ ) = ( r z ⊕ n ′ ⊕ m ′ ) and using (26) we get H ⊕ M N H ⊕ M Z N V ! (cid:18) λ ( E ( z )) ∗ r z ⊕ m ′ n ′ (cid:19) = (cid:18) r z ⊕ m ′ n ′ (cid:19) . So Z ( λ ( E ( z )) ∗ r z ⊕ m ) = r z ⊕ m ′ and V n ′ = n ′ . (31)Using the decomposed form of Z with respect to its domain and range we get H M H Z Z M Z Z ! (cid:18) λ ( E ( z )) ∗ r z m ′ (cid:19) = (cid:18) r z m ′ (cid:19) . This gives us two equations from which we eliminate r z . Recalling the definition of t ( z ) ∗ (28)enables us to obtain t ( z ) ∗ m ′ = m ′ . (32)Now let q z = P H ⊕ L → L ( k z ) . So from (30) we get n ′ ⊕ m ′ = P H ⊕ N ⊕ M → N ⊕ M ( r z ⊕ n ⊕ m )= P H ⊕ L ⊕ U → L ⊕ U ( k z ⊕ u ) = q z ⊕ u and n ′ ⊕ m ′ = P H ⊕ N ⊕ M → N ⊕ M ( λ ( E ( z )) ∗ r z ⊕ n ′ ⊕ m ′ )= P H ⊕ L ⊕ Y → L ⊕ Y ( λ ( E ( z )) ∗ k z ⊕ y ) = λ ( E ( z )) ∗ q z ⊕ y. Recall the Q that was defined in (15) in Section 4. It is a unitary from N ⊕ M ⊕ M to N ⊕ M ⊕ M sending a generic element n ⊕ m ⊕ m to V n ⊕ m ⊕ m . Taking V n ′ = n ′ , t ( z ) ∗ m ′ = m ′ , n ′ ⊕ m ′ = q z ⊕ u and n ′ ⊕ m ′ = λ ( E ( z )) ∗ q z ⊕ y we see that Q sends λ ( E ( z )) ∗ q z ⊕ t ( z ) ∗ m ′ ⊕ y to q z ⊕ m ′ ⊕ u. L M ⊕ YL Q Q M ⊕ U Q Q ! (cid:18) λ ( E ( z )) ∗ q z t ( z ) ∗ m ′ ⊕ y (cid:19) = (cid:18) q z m ′ ⊕ u (cid:19) . Hence Q ( λ ( E ( z )) ∗ q z ) + Q ( t ( z ) ∗ m ′ ⊕ y ) = q z Q ( λ ( E ( z )) ∗ q z ) + Q ( t ( z ) ∗ m ′ ⊕ y ) = m ′ ⊕ u. Eliminating q z we obtain ( Q + Q λ ( E ( z )) ∗ ( I L − Q λ ( E ( z )) ∗ ) − Q )( t ( z ) ∗ m ′ ⊕ y ) = m ′ ⊕ u. Now recall that from (23) we have λ ( δ ) | L = µ ( δ ) | L for all δ ∈ C b (Ψ) . So we have (cid:16) Q + Q µ ( E ( z )) ∗ ( I L − Q µ ( E ( z )) ∗ ) − Q (cid:17) ( t ( z ) ∗ m ′ ⊕ y ) = m ′ ⊕ u. Recalling the G from (16) of Section 4, we see that the last equation is precisely G ( z ) ∗ ( t ( z ) ∗ m ′ ⊕ y ) = m ′ ⊕ u. Using the decomposition of G ( z ) ∗ with respect to its domain and range we get M YM G ( z ) ∗ G ( z ) ∗ U G ( z ) ∗ G ( z ) ∗ ! (cid:18) t ( z ) ∗ m ′ y (cid:19) = (cid:18) m ′ u (cid:19) . From this we obtain two equations. Eliminating m from those gives us u = G ( z ) ∗ y + G ( z ) ∗ t ( z ) ∗ ( I M − G ( z ) ∗ t ( z ) ∗ ) − G ( z ) ∗ y. Since we know from Section 4 that G ( z ) < for each z ∈ Ω , the right hand side is well defined.As u = f ( z ) ∗ y and, y and z are arbitrary, we have f ( z ) = G ( z ) + G ( z )( I M − t ( z ) G ( z )) − t ( z ) G ( z ) , for all z ∈ Ω . This completes the proof. The Schur class H ∞ (Ω , B ( U , Y )) of a domain Ω is the closed unit ball (in the supremum norm)of the algebra of all bounded holomorphic functions on Ω taking values in B ( U , Y ) .22 heorem 7. Suppose Ω stands for the bidisc or the symmetrized bidisc or the annulus. We con-sider two Hilbert spaces U and Y . When Ω is the annulus, we take U = Y = C . Suppose that f ∈ H ∞ (Ω , B ( U , Y )) is a solution of this interpolation problem z i B i and f is affiliated with acompletely positive kernel Γ : { z , z , . . . , z n } × { z , z , . . . , z n } → B ( C b (Ψ) , B ( Y )) . Then with M , M and G as in Section 4, we have that writing G as G ( z ) = (cid:18) M UM G ( z ) G ( z ) Y G ( z ) G ( z ) (cid:19) , one has f ( z ) = (cid:16) G + G ( I M − t G ) − t G (cid:17) ( z ) for some t ∈ H ∞ (Ω , B ( M , M )) and for all z ∈ Ω . Proof In each of these examples, there exists a certain collection of holomorphic test functions,say Ψ , which satisfies the fact that there is a point w in the domain where all test functionsvanish. We do not get into the details of writing down the test functions explicitly for thesake of brevity. While for the bidisc the collection consists of just two test functions - the co-ordinate functions z and z , for the symmetrized bidisc and the annulus, they are uncountablein number. See [7] for the symmetrized bidisc and [10] for the annulus. Now we apply ourMain Theorem. 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