Interpolation by holomorphic maps from the disc to the tetrablock
aa r X i v : . [ m a t h . C V ] J a n INTERPOLATION BY HOLOMORPHIC MAPS FROM THE DISC TOTHE TETRABLOCK
HADI O. ALSHAMMARI AND ZINAIDA A. LYKOVA
Abstract.
The tetrablock is the set E = { x ∈ C : 1 − x z − x w + x zw = 0 whenever | z | ≤ , | w | ≤ } . The closure of E is denoted by E . A tetra-inner function is an analytic map x from theunit disc D to E such that, for almost all points λ of the unit circle T ,lim r ↑ x ( rλ ) exists and lies in b E , where b E denotes the distinguished boundary of E . There is a natural notion of degree of a rational tetra-inner function x ; it is simply the topological degree of the continuousmap x | T from T to b E .In this paper we give a prescription for the construction of a general rational tetra-inner function of degree n . The prescription exploits a known construction of thefinite Blaschke products of given degree which satisfy some interpolation conditionswith the aid of a Pick matrix formed from the interpolation data. It is known thatif x = ( x , x , x ) is a rational tetra-inner function of degree n , then x x − x either isidentically 0 or has precisely n zeros in the closed unit disc D , counted with multiplicity.It turns out that a natural choice of data for the construction of a rational tetra-innerfunction x = ( x , x , x ) consists of the points in D for which x x − x = 0 and thevalues of x at these points. Contents
1. Introduction 22. The phasar derivatives of Ψ ω ◦ x and Υ ω ◦ x
63. Rational tetra-inner functions and royal polynomials 114. Criteria for the solvability of the Blaschke interpolation problem 125. From the royal tetra-interpolation problem to the Blaschke interpolationproblem 166. From the Blaschke interpolation problem to the royal tetra-interpolationproblem 217. The algorithm 288. Examples 30References 34
Date : 6th January 2021.2010
Mathematics Subject Classification.
Primary 30E05, 32F45, 32A07, Secondary 93B36, 93B50.
Key words and phrases.
Blaschke product, tetrablock, inner functions, interpolation, Pick matrix,Distinguished boundary.The first author was supported by the Government of Saudi Arabia. The second author was partiallysupported by the UK Engineering and Physical Sciences Research Council grant EP/N03242X/1. Introduction
In this paper we present an algorithm for the construction of a general rational innerfunction from D to the tetrablock. The algorithm is based on a known solution of theNevanlinna-Pick interpolation problem on D . Different versions of the Nevanlinna-Pickinterpolation problem have been studied by many authors, beginning with G. Pick in1916 [29] and continuing with R. Nevanlinna in 1922 [28], and they still attract interest,because they are natural questions in function theory and because of their applicationsto engineering, particular electric networks and control theory, see [7] for some references.We should mention particularly papers of J. A. Ball and J. W. Helton [8], D. Sarason[32], D. R. Georgijevi´c [20], G.-N. Chen and Y.-J. Hu [18] and V. Bolotnikov and A.Kheifets [15], and the books of J. A. Ball, I. C. Gohberg and L. Rodman [9], and of V.Bolotnikov and H. Dym [14]. There are many further papers on this interesting topic andapplications (see, for example, [19, 33]).The closed tetrablock E is the set in C defined by E = { ( x , x , x ) ∈ C : 1 − x z − x w + x zw = 0 whenever | z | < , | w | < } . The original motivation for the study of E was an attempt to solve a µ -synthesis problem[1], which is itself motivated by basic unsolved problems in H ∞ control theory [21, 22]. Thetetrablock has attracted considerable interest in recent years. It has interesting complexgeometry [1, 23, 34, 25, 26], rich function theory [16, 5, 6] and associated operator theory[11]. The solvability of the µ -synthesis problem connected to E can be expressed in termsof the existence of rational inner functions from the open unit disc D in the complex plane C to the closure of E [16].Observe that if x : D → E is analytic then, by Fatou’s Theorem, for almost all λ ∈ T with respect to Lebesgue measure, the radial limitlim r ↑ x ( rλ )exists. We say that an analytic map x : D → E is a tetra-inner function , or alternatively,an E -inner function if, for almost all λ ∈ T with respect to Lebesgue measure,lim r ↑ x ( rλ )lies in the distinguished boundary b E of E . The distinguished boundary b E of E is home-omorphic to the solid torus D × T , which has a boundary [1]. The E -inner functionsconstitute a natural analogue (in the context of the tetrablock) of the inner functions introduced by A. Beurling [10], which play an important role in the function theory ofthe unit disc and in operator theory [17].A basic question about rational inner functions ϕ from D to D was studied by W.Blaschke [12]. Specifically, he obtained (inter alia) a formula for the general rationalinner function ϕ of degree n in terms of its zeros. Indeed, by the Argument Principle, anyrational inner function ϕ of degree n has exactly n zeros in D , counted with multiplicity.From this fact one can see that ϕ is a “finite Blaschke product”, having the form ϕ ( λ ) = c n Y j =1 λ − α j − α j λ (1.1) NTERPOLATION BY TETRA-INNER MAPS 3 for some unimodular constant c and some α , ..., α n ∈ D . The α j are the zeros of ϕ . It isevident from equation (1.1) that ϕ extends to a continuously differentiable function on D ,given by the same formula. In this paper our aim is to write down a formula analogous toequation (1.1) for the general rational E -inner function of degree n . The first question thatarises is: what data should replace the α j , the zeros of ϕ ? We have found that an effectivechoice is the set of royal nodes of the tetra-inner function, which we shall now define. Itwas shown in [5] that if x = ( x , x , x ) is a rational E -inner function of degree n then x − x x either is identically 0 or has exactly n zeros in the closed unit disc D , countedwith an appropriate notion of multiplicity. Here, the degree of a rational E -inner function x is defined to be the topological degree of the restriction of x that maps T continuouslyto b E . Since b E is homeomorphic to the solid torus D × T , which is homotopic to T , thefundamental group π ( b E ) is Z , and so the degree of x is an integer; it will be denoted bydeg( x ).The variety R E = { ( x , x , x ) ∈ E : x = x x } has an important role in the function theory of E ; it is called the royal variety of E . Forany rational tetra-inner function x = ( x , x , x ), the zeros of x x − x in D are the points λ ∈ D such that x ( λ ) ∈ R ¯ E . We call these points the royal nodes of x . If σ ∈ D is a royalnode of x , so that x ( σ ) = ( η, ˜ η, η ˜ η ) for some η, ˜ η ∈ D , then we call η, ˜ η the royal valuesof x corresponding to the royal node σ of x . In this paper, in Theorem 1.7, we give aprescription for the construction of a general rational tetra-inner function of degree n interms of its royal nodes and royal values. We shall make use of a known solution of aninterpolation problem for finite Blaschke products.The E -inner functions x such that x ( D ) ⊆ R E are simply the functions of the form( ϕ , ϕ , ϕ ϕ ) where ϕ , ϕ are inner functions. These E -inner functions behave differentlyfrom the others, and we shall often specifically exclude them from consideration.To describe our main results on the construction of a general rational tetra-inner func-tion we need to recall some definitions and results on the Blaschke interpolation problem. Definition 1.1. [3, Definition 1.2]
Let n ≥ and ≤ k ≤ n . By Blaschke interpolationdata of type ( n, k ) we mean a triple ( σ, η, ρ ) where (i) σ = ( σ , σ , ..., σ n ) is an n -tuple of distinct points of D such that σ j ∈ T for j = 1 , ..., k and σ j ∈ D for j = k + 1 , ..., n ;(ii) η = ( η , η , ..., η n ) where η j ∈ T for j = 1 , ..., k and η j ∈ D for j = k + 1 , ..., n ;(iii) ρ = ( ρ , ρ , ..., ρ k ) where ρ j > for j = 1 , ..., k . Definition 1.2. [2, Definition 7.5]
For any differentiable function f : T → C \ { } the phasar derivative of f at z = e i θ ∈ T is the derivative with respect to θ of the argumentof f (e i θ ) at z ; we denote it by Af ( z ) . Thus, if f (e i θ ) = R ( θ )e i g ( θ ) is differentiable, where g ( θ ) ∈ R and R ( θ ) >
0, then g isdifferentiable on [0 , π ) and the phasar derivative of f at z = e i θ ∈ T is equal to Af (e i θ ) = ddθ arg f (e i θ ) = g ′ ( θ ) . (1.2)We shall find it useful in the sequel. HADI O. ALSHAMMARI AND ZINAIDA A. LYKOVA
Problem 1.3. (The Blaschke interpolation problem)
For given Blaschke interpola-tion data ( σ, η, ρ ) of type ( n, k ) , find all rational inner functions ϕ on D of degree n withthe properties ϕ ( σ j ) = η j for j = 1 , ..., n (1.3) and Aϕ ( σ j ) = ρ j for j = 1 , ..., k. (1.4)Problem 1.3 has been analysed by several authors [31, 32, 8]. In the absence of thetangential conditions (1.4) the problem would be ill-posed, in that the solvability of theproblem would depend only on the interpolation conditions at nodes in D , and the condi-tions at σ , . . . , σ k would be irrelevant. With the conditions (1.4), however, the problemhas a pleasing solution. The existence of a solution of the Blaschke interpolation problemcan be characterized in terms of an associated “Pick matrix”, and all solutions ϕ areparametrized by a linear fractional expression in terms of a parameter ζ ∈ T . There arepolynomials a, b, c and d of degree at most n such that the general solution of Problem1.3 is ϕ = aζ + bcζ + d where the parameter ζ ranges over a cofinite subset of T . The polynomials a, b, c and d are unique subject to a certain normalization. Definition 1.4. [3, Definition 3.10]
Let ( σ, η, ρ ) be Blaschke interpolation data of type ( n, k ) . Suppose that Problem 1.3 is solvable. We say that ϕ = aζ + bcζ + d is a normalized linear fractional parametrization of the solutions of Problem 1.3 if (i) a, b, c, d are polynomials of degree at most n ; (ii) for all but at most k values of ζ ∈ T , the function ϕ ( λ ) = a ( λ ) ζ + b ( λ ) c ( λ ) ζ + d ( λ ) (1.5) is a solution of Problem 1.3 ; (iii) for some point τ ∈ T \{ σ , ..., σ k } , (cid:20) a ( τ ) b ( τ ) c ( τ ) d ( τ ) (cid:21) = (cid:20) (cid:21) ;(iv) every solution ϕ of Problem 1.3 has the form (1.5) for some ζ ∈ T . The data for the construction of a rational tetra-inner function x consists of the royalnodes and royal values of x . Definition 1.5.
Let n > , and ≤ k ≤ n . By royal tetra-interpolation data of type ( n, k ) we mean a four-tuple ( σ, η, ˜ η, ρ ) where (i) σ = ( σ , σ , ..., σ n ) is an n-tuple of distinct points such that σ j ∈ T for j = 1 , ..., k and σ j ∈ D for j = k + 1 , ..., n ;(ii) η = ( η , η , ..., η n ) where η j ∈ T for j = 1 , ..., k and η j ∈ D for j = k + 1 , ..., n ;(iii) ˜ η = ( ˜ η , ˜ η , ..., ˜ η n ) where ˜ η j ∈ T for j = 1 , ..., k and ˜ η j ∈ D for j = k + 1 , ..., n . NTERPOLATION BY TETRA-INNER MAPS 5 (iv) ρ = ( ρ , ρ , ..., ρ k ) where ρ j > for j = 1 , ..., k . Problem 1.6. (The royal tetra-interpolation problem)
Given royal tetra-interpolationdata ( σ, η, ˜ η, ρ ) of type ( n, k ) , find all rational E -inner functions x = ( x , x , x ) of degree n such that x ( σ j ) = ( η j , ˜ η j , η j ˜ η j ) for j = 1 , ..., n, and Ax ( σ j ) = ρ j for j = 1 , ..., k. The connection between Problems 1.6 and 1.3 can be described with the aid of a certain1-parameter family of rational functions on E that is parametrized by the unit circle T .These functions play a central role in the function theory of E (see [1, 16]). They aredefined, for ω ∈ T , by Ψ ω ( x ) = x ω − x x ω − x ω − = 0 . (1.6)Ψ ω is holomorphic on E , except at points x ∈ E where x ω − E at which it is defined into D .The main theorem of this paper is the following. Theorem 1.7.
For royal tetra-interpolation data ( σ, η, ˜ η, ρ ) of type ( n, k ) the followingtwo statements are equivalent: (i) The royal tetra-interpolation problem (Problem 1.6) with data ( σ, η, ˜ η, ρ ) admits arational E -inner interpolating function x such that x ( D ) * R E ;(ii) The Blaschke interpolation problem (Problem 1.3) with data ( σ, η, ρ ) of type ( n, k ) is solvable and there exist x ◦ , x ◦ , x ◦ ∈ C such that | x ◦ | = 1 , | x ◦ | < , | x ◦ | < , x ◦ = x ◦ x ◦ , and x ◦ c ( σ j ) + x ◦ d ( σ j ) x ◦ c ( σ j ) + d ( σ j ) = ˜ η j for j = 1 , ..., n, where a, b, c and d are the polynomials in the normalized parametrization ϕ = aζ + bcζ + d of the solution of Problem . The theorem follows from Theorems 5.1 and 6.4. The proofs of these theorems aregiven in Section 5 and Section 6 respectively. Theorem 6.4 provides us with a formula fora solution x of Problem 1.6 in terms of s , p , a, b, c and d . The formula is in terms of thepolynomials a, b, c and d computed in [3, Theorem 3.9] (see Remark 4.11). In this way wederive an explicit solution of Problem 1.6. Theorem 6.4 . Let ( σ, η, ρ ) be Blaschke interpolation data of type ( n, k ) , and let ( σ, η, ˜ η, ρ ) be royal tetra-interpolation data, where ˜ η j ∈ T , for j = 1 , ..., k , and ˜ η j ∈ D , for j = k + 1 , ..., n . Suppose that the Blaschke interpolation problem (Problem ) with thesedata is solvable and the solutions ϕ of Problem have normalized parametrization ϕ = aζ + bcζ + d . Suppose that there exist scalars x ◦ , x ◦ , x ◦ ∈ C such that | x ◦ | = 1 , | x ◦ | < , | x ◦ | < , x ◦ = x ◦ x ◦ , HADI O. ALSHAMMARI AND ZINAIDA A. LYKOVA and x ◦ c ( σ j ) + x ◦ d ( σ j ) x ◦ c ( σ j ) + d ( σ j ) = ˜ η j for j = 1 , ..., n. (1.7) Then there exists a rational tetra-inner function x = ( x , x , x ) given by, x ( λ ) = x ◦ a ( λ ) + b ( λ ) x ◦ c ( λ ) + d ( λ ) (1.8) x ( λ ) = x ◦ c ( λ ) + x ◦ d ( λ ) x ◦ c ( λ ) + d ( λ ) (1.9) x ( λ ) = x ◦ b ( λ ) + x ◦ a ( λ ) x ◦ c ( λ ) + d ( λ ) , (1.10) for λ ∈ D , such that (i) x is a solution of the royal tetra-interpolation problem with the data ( σ, η, ˜ η, ρ ) ,that is, x ( σ j ) = ( η j , ˜ η j , η j ˜ η j ) for j = 1 , ..., n, and Ax ( σ j ) = ρ j for j = 1 , ..., k, (ii) for all but finitely many ω ∈ T , the function Ψ ω ◦ x is a solution of Problem . The solution sets of the royal tetra-interpolation problem and the corresponding Blaschkeinterpolation problem admit an explicit connection in terms of the functions Ψ ω . Corollary 6.5 . Let ( σ, η, ρ ) be Blaschke interpolation data of type ( n, k ) . Suppose that x is a solution of the Problem 1.6 with data ( σ, η, ˜ η, ρ ) for some ˜ η j ∈ D , j = 1 , ..., n, andthat x ( D )
6⊆ R E . For all ω ∈ T \{ ˜ η , ..., ˜ η k } , the function ϕ = Ψ ω ◦ x is a solution of Problem 1.3 with Blaschke interpolation data ( σ, η, ρ ) . Conversely, if ϕ is a solution ofthe Blaschke interpolation problem with data ( σ, η, ρ ) , then there exists ω ∈ T such that ϕ = Ψ ω ◦ x . In [5] there is a construction of the general rational E -inner function x = ( x , x , x )of degree n , in terms of different data, namely, the royal nodes of x and the zeros of x and x . A major step in the construction in [5] is to perform a Fej´er-Riesz factorizationof a non-negative trigonometric polynomial, which requires an iterative process, whereas,in contrast, the construction of x in this paper is purely algebraic and can be carriedout entirely in rational arithmetic. The algorithm for the solution of the royal tetra-interpolation problem is presented in Section 7.The authors are grateful to Nicholas Young for some helpful suggestions.2. The phasar derivatives of Ψ ω ◦ x and Υ ω ◦ x The tetrablock was introduced in [1], and it is related to the µ Diag -synthesis problem.By [1, Theorem 2.9],
E ∩ R is the open tetrahedron with vertices (1 , , , − , − − , , −
1) and ( − , − , E . Definition 2.1. [1, Definition 2.1]
For x = ( x , x , x ) ∈ C and z ∈ C we define Ψ( z, x ) = x z − x x z − x z − = 0 , (2.1) NTERPOLATION BY TETRA-INNER MAPS 7 Υ( z, x ) = x z − x x z − x z − = 0 . (2.2)For ω ∈ T , let Ψ ω ( x ) = x ω − x x ω − x ω − = 0 . (2.3)Further we will need the following description of the tetrablock from [1]. Theorem 2.2. [1, Theorem 2.4]
For x ∈ C the following are equivalent. (i) x ∈ ¯ E ;(ii) || Ψ( ., x ) || H ∞ ≤ and if x x = x then, in addition, | x | ≤ ; (iii) || Υ( ., x ) || H ∞ ≤ and if x x = x then, in addition, | x | ≤ ; (iv) | x | − | x | + | x | + 2 | x − ¯ x x | ≤ and | x | ≤ ; (v) | x − ¯ x x | + | x − ¯ x x | ≤ − | x | and if | x | = 1 then, in addition, | x | ≤ . By [1, Theorem 2.9], ¯ E is polynomially convex. Since ¯ E is polynomially convex, thereis a smallest closed boundary b ¯ E of E , which is called the distinguished boundary of E . Theorem 2.3. [1, Theorem 7.1]
For x ∈ C the following are equivalent. (i) x = ¯ x x , | x | = 1 and | x | ≤ ; (ii) either x x = x and Ψ( ., x ) is an automorphism of D or x x = x and | x | = | x | = | x | = 1 ; (iii) x is a peak point of ¯ E ; (iv) there exists a × unitary matrix U such that x = π ( U ) where π : C × → C : U = [ u ij ] ( u , u , det U );(v) there exists a symmetric × unitary matrix U such that x = π ( U ) ; (vi) x ∈ b ¯ E ; (vii) x ∈ ¯ E and | x | = 1 . By [1, Corollary 7.2], b ¯ E is homeomorphic to D × T . Lemma 2.4.
Let ( x , x , x ) ∈ E be such that x x = x . For any ω ∈ T , | Ψ ω ( x , x , x ) | = 1 if and only if ω ( x − x x ) = 1 − | x | + | x | − | x | . Proof.
Consider ω ∈ T . | Ψ ω ( x , x , x ) | = 1 ⇔ (cid:12)(cid:12)(cid:12)(cid:12) x ω − x x ω − (cid:12)(cid:12)(cid:12)(cid:12) = 1 ⇔ | ωx − x | = | x ω − | ⇔ | x | − ωx x ) + | x | = | x | − x ω ) + 1 ⇔ | x | − | x | + | x | + 2Re( ω ( x − x x )) = 1 ⇔ ω ( x − x x )) = 1 − | x | + | x | − | x | . (2.4)Since ( x , x , x ) ∈ E , by [1, Theorem 2.4 (vii)] (see Theorem 2.2),2 | x − ¯ x x | ≤ − | x | + | x | − | x | , and | x | ≤
1. Therefore,2Re( ω ( x − x x )) ≤ | x − x x | ≤ − | x | + | x | − | x | = 2Re( ω ( x − x x )) . HADI O. ALSHAMMARI AND ZINAIDA A. LYKOVA
Thus 2Re( ω ( x − x x )) = 2 | x − x x | = 1 − | x | + | x | − | x | . Hence, for any ω ∈ T , | Ψ ω ( x , x , x ) | = 1 if and only if2 ω ( x − x x ) = 1 − | x | + | x | − | x | . (cid:3) For a rational E -inner function x = ( x , x , x ) : D → E , we consider the rationalfunctions ψ ω : D → D and υ ω : D → D given, for any ω ∈ T , by ψ ω ( λ ) = Ψ ω ◦ x ( λ ) = ωx − x x ω − λ ) , for all λ ∈ D such that x ( λ ) ω − = 0 , and υ ω ( λ ) = Υ ω ◦ x ( λ ) = x ω − x x ω − λ ) , for all λ ∈ D such that x ( λ ) ω − = 0 . In [2] we introduced the terminology of the phasar derivative Af ( z ) for any differen-tiable function f : T → C \ { } at z = e iθ ∈ T and wrote down some useful elementaryproperties of phasar derivatives, see Definition 1.2. Proposition 2.5. [2](i)
For differentiable functions ψ, ϕ : T → C \ { } and for any c ∈ C \ { } , A ( ψϕ ) = Aψ + Aϕ and A ( cψ ) = Aψ. (2.5)(ii)
For any rational inner function ϕ and for all z ∈ T , Aϕ ( z ) = z ϕ ′ ( z ) ϕ ( z ) . (2.6)(iii) If α ∈ D and B a ( z ) = z − α − ¯ αz , then AB α ( z ) = 1 − | α | | z − α | > f or z ∈ T . (iv) For any rational inner function p , Ap ( z ) > f or all z ∈ T . Let us recall that, by definition, σ ∈ T is a royal node of a tetra-inner function x =( x , x , x ) if x ( σ ) − x ( σ ) x ( σ ) = 0. Lemma 2.6.
Let x = ( x , x , x ) be a rational tetra-inner function and let σ ∈ T be aroyal node of x . Then σ is a zero of the function x − x x of multiplicity at least .Proof. If λ ∈ T , we have x ( λ ) − x ( λ ) x ( λ ) = 0 if and only if λ is a royal node of x .For λ ∈ T , since x is a tetra-inner function, by Theorem 2.3 (i), x ( λ )( x ( λ ) − x ( λ ) x ( λ )) = x ( λ ) x ( λ ) − x ( λ ) x ( λ ) x ( λ )= | x ( λ ) | − x ( λ )( x ( λ ) x ( λ )) x ( λ )= 1 − | x ( λ ) | | x ( λ ) | = 1 − | x ( λ ) | ≥ , since | x ( λ ) | ≤ T . (2.7) NTERPOLATION BY TETRA-INNER MAPS 9
For θ ∈ R , let f ( θ ) = x ( e iθ )( x ( e iθ ) − x ( e iθ ) x ( e iθ )) , and let σ = e iξ . By assumption, x ( σ ) − x ( σ ) x ( σ ) = 0, and so f ( ξ ) = 0. By inequality(2.7), the function f ( θ ) = x ( e iθ )( x ( e iθ ) − x ( e iθ ) x ( e iθ )) = 1 − | x ( e iθ ) | ≥ , and so it has a local minimum at ξ . Therefore ξ is a critical point of f , and ddθ (1 − | x ( e iθ ) | ) | ξ = 0 . Thus,0 = ddθ (1 − | x ( e iθ ) | ) | ξ = ddθ ( x ( e iθ )( x ( e iθ ) − x ( e iθ ) x ( e iθ )) | ξ = ddθ ( x ( e iθ )) | ξ ( x ( e iθ ) − x ( e iθ ) x ( e iθ )) | ξ + x ( e iθ ) | ξ ddθ ( x ( e iθ ) − x ( e iθ ) x ( e iθ )) | ξ = ddθ ( x ( e iθ )) | ξ × x ( e iξ ) (cid:20) ddθ x ( e iθ ) | ξ − ddθ ( x ( e iθ ) x ( e iθ )) | ξ (cid:21) = x ( e iξ ) (cid:20) ie iξ x ′ ( e iξ ) − ( x ( e iξ ) ie iξ x ′ ( e iξ ) + ie iξ x ′ ( e iξ ) x ( e iξ )) (cid:21) . Note that | x ( e iξ ) | = 1, hence x ′ ( σ ) = x ( σ ) x ′ ( σ ) + x ′ ( σ ) x ( σ ). Thus x ( σ ) − x ( σ ) x ( σ ) = 0 and ( x ( σ ) − x ( σ ) x ( σ )) ′ = 0 . Therefore σ is a zero of ( x − x x ) of multiplicity at least 2. (cid:3) Proposition 2.7.
Let x = ( x , x , x ) be a rational E -inner function. Let σ ∈ T be aroyal node of x . Suppose x ( σ ) = ( η, ˜ η, η ˜ η ) , ω ∈ T and ω ˜ η = 1 . Then A (Ψ ω ◦ x )( σ ) = Ax ( σ ) . Proof.
Since x is a rational E -inner function, for almost all λ ∈ T , x ( λ ) ∈ b E , and, byTheorem 2.3, for almost all λ ∈ T , x ( λ ) = x ( λ ) x ( λ ) , | x ( λ ) | = 1 and | x ( λ ) | ≤
1. ByProposition 2.5, for every z ∈ T , and every rational inner function ϕ , Aϕ ( z ) = z ϕ ′ ( z ) ϕ ( z ) . For σ ∈ T such that x ( σ ) ∈ R E and ω ˜ η = 1, A (Ψ ω ◦ x )( σ ) = A ( ωx − x )( σ ) − A ( x ω − σ )= σ ( ωx − x ) ′ ( σ )( ωx − x )( σ ) − σ ( x ω − ′ ( σ )( x ω − σ )= σω ˜ η − (cid:16) ωx ′ ( σ ) − x ′ ( σ ) η − ωx ′ ( σ ) (cid:17) . Since x ( σ ) ∈ R ¯ E , we have x ( σ ) = x ( σ ) x ( σ ), and, by Lemma 2.6, σ is a zero of x − x x of multiplicity at least 2. Thus ( x − x x ) ′ ( σ ) = 0 and x ′ ( σ ) = x ( σ ) x ′ ( σ ) + x ( σ ) x ′ ( σ ) = ηx ′ ( σ ) + ˜ ηx ′ ( σ ) . (2.8)Thus, by equation (2.8), we have A (Ψ ω ◦ x )( σ ) = σω ˜ η − (cid:16) ω ( ηx ′ ( σ ) + ˜ ηx ′ ( σ )) − x ′ ( σ ) η − ωx ′ ( σ ) (cid:17) = σω ˜ η − (cid:16) ωηx ′ ( σ ) + x ′ ( σ )( ω ˜ η − − ηωx ′ ( σ ) η (cid:17) = σ (cid:16) x ′ ( σ ) η (cid:17) = σ (cid:16) x ′ ( σ ) x ( σ ) (cid:17) = Ax ( σ ) . (cid:3) Proposition 2.8.
Let x = ( x , x , x ) be a rational E -inner function. Let σ ∈ T be aroyal node of x . Suppose x ( σ ) = ( η, ˜ η, η ˜ η ) , ω ∈ T and ωη = 1 . Then A (Υ ω ◦ x )( σ ) = Ax ( σ ) . Proof.
Since x is a rational E -inner function, then for almost all λ ∈ T , x ( λ ) ∈ b E , and, byTheorem 2.3 (i ), for almost all λ ∈ T , x = ¯ x x , | x | = 1 and | x | ≤
1. By Proposition2.5, for every z ∈ T , and every rational inner function ϕ , Aϕ ( z ) = z ϕ ′ ( z ) ϕ ( z ) . For σ ∈ T such that x ( σ ) ∈ R ¯ E , and ωη = 1, A (Υ ω ◦ x )( σ ) = A ( ωx − x )( σ ) − A ( ωx − σ )= σ ( ωx − x ) ′ ( σ )( ωx − x )( σ ) − σ ( x ω − ′ ( σ )( x ω − σ )= σωη − (cid:16) ωx ′ ( σ ) − x ′ ( σ )˜ η − ωx ′ ( σ ) (cid:17) . Since x ( σ ) ∈ R ¯ E , we have x ( σ ) = x ( σ ) x ( σ ), and, by Lemma 2.6, σ is a zero of x − x x of multiplicity at least 2. Thus ( x − x x ) ′ ( σ ) = 0 and x ′ ( σ ) = x ( σ ) x ′ ( σ ) + x ( σ ) x ′ ( σ ) = ηx ′ ( σ ) + ˜ ηx ′ ( σ ) . (2.9)Thus, by equation (2.9), we have A (Υ ω ◦ x )( σ ) = σωη − (cid:16) ω ( ηx ′ ( σ ) + ˜ ηx ′ ( σ )) − x ′ ( σ )˜ η − ωx ′ ( σ ) (cid:17) = σωη − (cid:16) ωηx ′ ( σ ) + x ′ ( σ ) ω ˜ η − x ′ ( σ ) − ˜ ηωx ′ ( σ )˜ η (cid:17) = σωη − (cid:16) x ′ ( σ )( ωη − η (cid:17) = σ (cid:16) x ′ ( σ )˜ η (cid:17) = σ (cid:16) x ′ ( σ ) x ( σ ) (cid:17) = Ax ( σ ) . (cid:3) NTERPOLATION BY TETRA-INNER MAPS 11 Rational tetra-inner functions and royal polynomials
In this section we will show how to construct rational ¯ E -inner functions with prescribedroyal nodes and values. To describe this construction we need several theorems anddefinitions from [5]. Detailed proofs of these statements are given in [5, 6].For a polynomial p of degree less than or equal to n , where n ≥
0, we define thepolynomial p ∼ n by p ∼ n ( λ ) = λ n p ( 1 λ ) . Theorem 3.1. [5, Theorem 4.15] If x = ( x , x , x ) is a rational ¯ E -inner function ofdegree n , then there exist polynomials E , E , D such that (i) deg ( E ), deg ( E ), deg (D) ≤ n , (ii) D ( λ ) = 0 on D , (iii) E ( λ ) = E ∼ n ( λ ) , for all λ ∈ T , (iv) | E i ( λ ) | ≤ | D ( λ ) | on D , i = 1 , , (v) x = E D on D , (vi) x = E D on D , (vii) x = D ∼ n D on D . Definition 3.2.
Let x = ( x , x , x ) be a rational tetra-inner function of degree n . The royal polynomial of x is R x ( λ ) = D ( λ ) D ∼ n ( λ ) − E ( λ ) E ( λ ) , where E , E , D are as in Theorem 3.1 . Remark 3.3.
For a rational tetra-inner function x , since D ( λ ) = 0 on D , the zeroes of R x are the zeroes of the function x − x x . Lemma 3.4.
Let x = ( x , x , x ) be a rational E -inner function, and let σ ∈ ¯ D be a royalnode of x . If σ ∈ T , then | x ( σ ) | = 1 and | x ( σ ) | = 1 .Proof. Since x is an E -inner function, by definition of tetra-inner functions, x ( σ ) ∈ b E for σ ∈ T . By Theorem 2.3, x ( σ ) = x ( σ ) x ( σ ), | x ( σ ) | = 1 and | x ( σ ) | ≤
1. By assumption σ is a royal node of x . Thus x ( σ ) = x ( σ ) x ( σ ), and so | x ( σ ) | = 1 and | x ( σ ) | = 1 since | x ( σ ) | = 1. (cid:3) Definition 3.5. [4, Definition 3.4]
We say that a polynomial f is n -symmetric if deg( f ) ≤ n and f ∼ n = f . Proposition 3.6. [5]
Let x be a rational E -inner function of degree n and let R x be theroyal polynomial of x . Then R x is n-symmetric and the zeros of R x on T have evenorder or infinite order. Definition 3.7.
Let x = ( x , x , x ) be a rational ¯ E -inner function such that x ( D ) * R E and let R x be the royal polynomial of x . If σ is a zero of R x of order ℓ , we define themultiplicity σ of σ (as a royal node of x ) by σ = ( ℓ if σ ∈ D ℓ if σ ∈ T . We define the type of x to be the ordered pair ( n, k ) , where n is the number of royal nodesof x that lie in ¯ D , counted with multplicity, and k is the number of royal nodes of x that liein T , counted with multiplicity. R n,k denotes the collection of rational ¯ E -inner functionsof type ( n, k ) . Definition 3.8. [5]
The degree of a rational E -inner function x , denoted by deg( x ) isdefined to be x ∗ (1) , where x ∗ : Z = π ( T ) → π ( b E ) is the homomorphism of fundamentalgroups induced by x when x is regarded as a continuous map from T to b E . Proposition 3.9. [5]
For any rational E -inner function x , deg( x ) is the degree deg( x ) (in the usual sense) of the finite Blaschke product x . Theorem 3.10. [5] If x ∈ R n,k is non-constant, then the degree of x is equal to n . Theorem 3.11. [5]
Let x be a non-constant rational E -inner function of degree n . Then,either x ( D ) ⊆ R E or x ( D ) meets R E exactly n times. Proposition 3.12.
Let x = ( x , x , x ) be a non-constant rational E -inner function andlet ω ∈ T be such that ωx ( λ ) − = 0 for all λ ∈ D . Then the rational function Ψ ω ◦ x = ωx − x x ω − has a cancellation at ζ ∈ D if and only if the following conditions aresatisfied : ζ ∈ T , ζ is a royal node of x and ω = x ( ζ ) .Proof. Let ζ ∈ T be a royal node of x such that x ( ζ ) = ( η, ˜ η, η ˜ η ). By Lemma 3.4, | η | = 1and | ˜ η | = 1. If ω = ˜ η ∈ T , then ωx ( ζ ) − x ( ζ ) = ˜ ηη ˜ η − η = | ˜ η | η − η = η − η = 0 , and x ( ζ ) ω − η ˜ η − | ˜ η | − . Thus, Ψ ω ◦ x = ωx ( λ ) − x ( λ ) x ( λ ) ω − ζ ∈ T .Conversely, by assumption Ψ ω ◦ x has a cancellation at ζ ∈ D , and so( ωx − x )( ζ ) = 0 = ( x ω − ζ ) . Therefore, x ( ζ ) ω = 1 and ωx ( ζ ) = x ( ζ ). Since x ( ζ ) ω = 1 , it implies that x ( ζ ) = ω ∈ T , so | x ( ζ ) | = 1. Since x : D → D is a rational analytic function with | x ( ζ ) | = 1, bythe maximum principle, ζ ∈ T , or x ( λ ) = ω for all λ ∈ D . By assumption, ωx ( λ ) − = 0for all λ ∈ D . Hence the function x = ω on D . Therefore, ζ ∈ T . Note ωx ( ζ ) = x ( ζ ) = ⇒ x ( ζ ) x ( ζ ) = x ( ζ )= ⇒ x ( ζ ) = x ( ζ ) x ( ζ ) . Thus, ζ ∈ T is a royal node for x , and ω = x ( ζ ). (cid:3) Criteria for the solvability of the Blaschke interpolation problem
In this section, for the convenience of the reader, we collect some known facts aboutfinite Blaschke products that we need. They may be found in several places, but the mosteconomical source for our purposes is [3], which assembles precisely the results which werequire.
NTERPOLATION BY TETRA-INNER MAPS 13
As mentioned in the Introduction, there is an extensive literature on boundary inter-polation problems. A very valuable source of information about all manner of complexinterpolation problems is the book of Ball, Gohberg and Rodman [9]. The authors of[8, 32, 15, 9, 14] make use of Krein spaces, moment theory, measure theory, reproducingkernel theory, realization theory and de Branges space theory. They obtain far-reachingresults, including generalizations to matrix-valued functions and to functions allowed tohave a limited number of poles in a disc or half-plane. See also papers of [13, 27, 30]for elementary treatments of interpolation problems. The monograph [14] by Bolotnikovand Dym is entirely devoted to boundary interpolation problems for the Schur class.They reformulate the problem within the framework of the Ukrainian school’s AbstractInterpolation Problem and solve it by means of operator theory in de Branges-Rovnyakspaces.The Blaschke interpolation Problem 1.3 as described in [3] is an algebraic variant ofthe classical Pick interpolation problem. One looks for a Blaschke product of degree n satisfying n interpolation conditions, rather than a Schur-class function as in the originalPick interpolation problem. We admit interpolation nodes both in the open unit disc andon the unit circle. There is a criterion for the solvability of the Blaschke interpolationproblem in terms of the positivity of a “Pick matrix” formed from the interpolationdata. To obtain a well-posed problem one imposes additional interpolation conditions,on phasar derivatives at the interpolation nodes on the unit circle. These bounds on thephasar derivatives appear as the diagonal entries of the Pick matrix. Definition 4.1.
The Schur class is the set of analytic functions S from D to ¯ D , S : D → ¯ D . Definition 4.2.
A function f : D → D is inner if it is an analytic map such that theradial limit lim r → − f ( rλ ) exists and belongs to T for almost all λ ∈ T with respect to Lebesgue measure. Definition 4.3.
The
Pick matrix associated with Blaschke interpolation data ( σ, η, ρ ) oftype ( n, k ) is the n × n matrix M = [ m ij ] ni,j =1 where m ij = ρ i if i = j ≤ k. − η i η j − σ i σ j otherwise . Definition 4.4.
The Pick matrix M = [ m ij ] ni,j =1 is minimally positive if M ≥ and thereis no strictly positive diagonal n × n matrix D such that M ≥ D . The following is a refinement of the Sarason Interpolation Theorem [32].
Theorem 4.5. [3, Theorem 3.3]
Let M be the Pick matrix associated with Blaschkeinterpolation data ( σ, η, ρ ) of type ( n, k ) . (i) There exists a function ϕ in the Schur class such that ϕ ( σ j ) = η j f or j = 1 , ..., n, (4.1) and the phasar derivative Aϕ ( σ j ) exists and satisfies Aϕ ( σ j ) ≤ ρ j f or j = 1 , ..., k, (4.2) if and only if M > ; (ii) if M is positive semi-definite and of rank r < n then there is a unique function ϕ in the Schur class satisfying conditions (4.1) and (4.2) above, and this function isa Blaschke product of degree r ; (iii) the unique function ϕ in statement (ii) satisfies Aϕ ( σ j ) = ρ j f or j = 1 , ..., k, (4.3) if and only if M is minimally positive. In [3] the authors described a strategy for the construction of the general solution ofthe Blaschke interpolation problem (Problem 1.3). It is to adjoin an additional boundaryinterpolation condition ϕ ( τ ) = ζ where τ ∈ T \ { σ , ..., σ k } and ζ ∈ T . This augmentedproblem has a unique solution. All the solutions of Problem 1.3 are then obtained interms of a unimodular parameter. Lemma 4.6. [3, Lemma 3 . If C is an n × n positive definite matrix, u is an n × column, ρ = h C − u, u i and the ( n + 1) × ( n + 1) matrix B is defined by B = (cid:20) C uu ∗ ρ (cid:21) , then B is positive semi-definite, rank( B ) = n and B (cid:20) − C − u (cid:21) = 0 . The Pick matrix B ζ,τ of the augmented problem is the ( n + 1) × ( n + 1) matrix, B ζ,τ = (cid:20) M u ζ,τ u ∗ ζ,τ ρ ζ,τ (cid:21) , (4.4)where ρ ζ,τ = h M − u ζ,τ , u ζ,τ i ,M is the Pick matrix associated with Problem 1.3, and u ζ,τ is the n × u ζ,τ = − ¯ η ζ − ¯ σ τ ... − ¯ η n ζ − ¯ σ n τ . (4.5) Theorem 4.7. [3, Proposition 3.6]
If the Pick matrix M associated with Problem 1.3 ispositive definite then, for any τ ∈ T \ { σ , ..., σ k } and ζ ∈ T , there is at most one solution ϕ of Problem 1.3 for which ϕ ( τ ) = ζ . The j th standard basis vector in C n will be denoted by e j . Theorem 4.8. [3, Proposition 3.7]
If the Pick matrix M for Problem is positivedefinite, if τ ∈ T \ { σ , ..., σ k } and ζ ∈ T and if h M − u ζ,τ , e j i 6 = 0 (4.6) for j = 1 , ..., k , then there is a unique solution ϕ of Problem that satisfies ϕ ( τ ) = ζ . NTERPOLATION BY TETRA-INNER MAPS 15
The exceptional set Z τ for Problem 1.3 is defined to be Z τ = { ζ ∈ T : for some j such that 1 ≤ j ≤ k, h M − u ζ,τ , e j i = 0 } (4.7)Define n × x λ and y λ for λ ∈ D \{ σ , ..., σ k } by x λ = − ¯ σ λ ... − ¯ σ n λ , y λ = ¯ η − ¯ σ λ ... ¯ η n − ¯ σ n λ , (4.8)so that u ζ,τ = x τ − ζ y τ (4.9) Theorem 4.9. [3, Proposition 3.8](i)
For any τ ∈ T \ { σ , ..., σ k } , if h x τ , M − e j i = 0 = h y τ , M − e j i for some j such that ≤ j ≤ k, then Z τ = T . (ii) There exist uncountably many τ ∈ T \ { σ , ..., σ k } such that h x τ , M − e j i = 0 = h y τ , M − e j i does not hold for any j such that ≤ j ≤ k . Moreover, for such τ , the set Z τ consists of at most k points. Theorem 4.10. [3, Theorem 3 . Let the Pick matrix M of Problem be positivedefinite, and let τ ∈ T \ { σ , ..., σ k } be such that the set Z τ = { ζ ∈ T : u ζ,τ ⊥ M − e j for some j such that ≤ j ≤ k } contains at most k points, where u ζ,τ is defined by equation (4.5) . (i) If ζ ∈ T \ Z τ , then there is a unique solution ϕ ζ of Problem that satisfies ϕ ζ ( τ ) = ζ . (ii) There exist unique polynomials a τ , b τ , c τ , and d τ of degree at most n such that (cid:20) a τ ( τ ) b τ ( τ ) c τ ( τ ) d τ ( τ ) (cid:21) = (cid:20) (cid:21) (4.10) and, for all ζ ∈ T , if ϕ is a solution of a Problem 1.3 such that ϕ ( τ ) = ζ , then ϕ ( λ ) = a τ ( λ ) ζ + b τ ( λ ) c τ ( λ ) ζ + d τ ( λ ) (4.11) for all λ ∈ D . (iii) If ˜a, ˜b, ˜c, ˜d are rational functions satisfying the equation (cid:20) ˜a ( τ ) ˜b ( τ ) ˜c ( τ ) ˜d ( τ ) (cid:21) = (cid:20) (cid:21) (4.12) and such that for three distinct points ζ in T \ Z τ , the equation a τ ( λ ) ζ + b τ ( λ ) c τ ( λ ) ζ + d τ ( λ ) = ˜a ( λ ) ζ + ˜b ( λ ) ˜c ( λ ) ζ + ˜d ( λ ) (4.13) holds for all λ ∈ D , then there exists a rational function X such that ˜a = Xa τ , ˜b= Xb τ , ˜c = Xc τ and ˜d = Xd τ . Remark 4.11.
Let ( σ, η, ρ ) be Blaschke interpolation data of type ( n, k ). Suppose thePick matrix M of this problem is positive definite. The proof of [3, Theorem 3.9] gives an explicit linear fractional parametrization of the solutions of Problem 1.3. As in Theorem4.10 choose τ ∈ T \ { σ , . . . , σ k } such that the set Z τ contains at most k points. Then anormalized linear fractional parametrization of the solution set of Problem 1.3 is ϕ = a τ ζ + b τ c τ ζ + d τ , where the polynomials a τ , b τ , c τ and d τ are defined by the equations a τ = πA, b τ = πB, c τ = πC, and d τ = πD. (4.14)Here (see [3, Theorem 3.9]), π ( λ ) = (1 − τ λ ) n Y j =1 − σ j λ − σ j τ , and A ( λ ) = −h x λ , M − x τ i + 11 − τ λ , (4.15) B ( λ ) = h x λ , M − y τ i , (4.16) C ( λ ) = −h y λ , M − x τ i , (4.17)and D ( λ ) = h y λ , M − y τ i + 11 − τ λ . (4.18)Different choices of τ yield different normalized parametrizations of the solutions of Prob-lem 1.3.[3, Theorem 3.9] tells us the following. Corollary 4.12. [3, Corollary 3.12]
Let ( σ, η, ρ ) be Blaschke interpolation data of type ( n, k ) . Suppose the Pick matrix M of this problem is positive definite. There exists anormalized linear fractional parameterization ϕ = aζ + bcζ + d of the solutions of Problem . Moreover (i) at least one of the polynomials a, b, c, d has degree n , (ii) the polynomials a, b, c, d have no common zero in C ; (iii) | c | ≤ | d | on D . From the royal tetra-interpolation problem to the Blaschkeinterpolation problem
In this section we show that for Blaschke interpolation data ( σ, η, ρ ) of type ( n, k )knowledge of a solution x of the royal tetra-interpolation problem ( σ, η, ˜ η, ρ ) for some˜ η j ∈ D allows us to construct a solution of the Blaschke interpolation problem. NTERPOLATION BY TETRA-INNER MAPS 17
Theorem 5.1.
Let x = ( x , x , x ) be a rational E -inner function of type ( n, k ) having dis-tinct royal nodes σ , σ , ..., σ n where σ , σ , ..., σ k ∈ T and σ k +1 , ..., σ n ∈ D and correspond-ing royal values η , .., η n and ˜ η , ..., ˜ η n , that is, x ( σ j ) = ( η j , ˜ η j , η j ˜ η j ) . Let ρ j = Ax ( σ j ) for j = 1 , , .., k. (1) There exists a rational inner function ϕ that solves the Blaschke interpolationProblem for ( σ, η, ρ ) , that is, such that deg ( ϕ ) = n , ϕ ( σ j ) = η j for j = 1 , ..., n (5.1) and Aϕ ( σ j ) = ρ j for j = 1 , ..., k. (5.2) Any such function ϕ is expressible in the form ϕ = Ψ ω ◦ x for some ω ∈ T . (2) There exist polynomials a, b, c, d of degree at most n such that a normalizedparametrization of the solutions of Problem 1.3 is ϕ = aζ + bcζ + d , ζ ∈ T . (3) For any polynomials a, b, c, d as in (2) , there exist x ◦ , x ◦ , x ◦ ∈ C such that | x ◦ | = 1 , | x ◦ | < , | x ◦ | < , (5.3) x ◦ = x ◦ x ◦ , (5.4) and moreover, x = x ◦ a + bx ◦ c + d (5.5) x = x ◦ c + x ◦ dx ◦ c + d (5.6) x = x ◦ b + x ◦ ax ◦ c + d . (5.7) Proof. (1) For ω ∈ T and for a given rational E -inner function x = ( x , x , x ) : D → E ,we consider the rational function ψ ω : D → D given by ψ ω ( λ ) = Ψ ω ◦ x ( λ ) = x ω − x x ω − λ ) . (5.8)Then, if ω = ˜ η , ...., ˜ η k , ψ ω ( σ j ) = x ( σ j ) ω − x ( σ j ) x ( σ j ) ω − η j ˜ η j ω − η j ˜ η j ω − η j ω ˜ η j − η j ω − η j for j = 1 , ..., n. (5.9)We claim that, for ω ∈ T \ { ˜ η , ...., ˜ η k } , the function ϕ = ψ ω is a solution of Problem 1.3.Let us check that ϕ is an inner function from D to D . For any λ ∈ T , ϕ ( λ ) = ψ ω ( λ ) = ωx ( λ ) − x ( λ ) x ( λ ) ω − . Since x is a rational E -inner function, x ( λ ) ∈ b E for almost all λ ∈ T , and, by Theorem2.3, x ( λ ) = x ( λ ) x ( λ ) and | x ( λ ) | = 1 for almost all λ ∈ T . Thus, for almost all λ ∈ T , ϕ ( λ ) = ψ ω ( λ ) = ωx ( λ ) − x ( λ ) x ( λ ) x ( λ ) ω − x ( λ )( ω − x ( λ )) ωx ( λ ) − . Hence, for λ ∈ T , | ϕ ( λ ) | = | x ( λ ) | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ω − x ( λ ) ωx ( λ ) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Since | x ( λ ) | = 1 , | ω | = 1 and | ω − x ( λ ) | = | ω − x ( λ ) | , we have, for almost all λ ∈ T , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ω ( ω − x ( λ )) ωx ( λ ) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − x ( λ ) ω − (1 − x ( λ ) ω ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 1 . Therefore, for almost all λ ∈ T , | ϕ ( λ ) | = 1. Hence ϕ is a rational inner function.By equation (5.9), ψ ω takes the required values at σ , ..., σ n . By Proposition 2.7, A (Ψ ω ◦ x )( σ j ) = Ax ( σ j ) = ρ j for j = 1 , , ..., k. (5.10)Furthermore, deg( ψ ω ) = n for ω = ˜ η , ...., ˜ η k . By Theorem 3.1, for a rational E -innerfunction x = ( x , x , x ) such that deg( x ) = n and if D is the denominator when x iswritten in its lowest terms then x and x can also be represented as rational functionsin which the denominator is D . Thereforedeg( ψ ω ) = deg( x ) − { cancellations between ωx − x and x ω − } . (5.11)By Proposition 3.12, such cancellations can occur only at the royal nodes σ j ∈ T , j =1 , ..., k , and then only when ω = x ( σ j ) = ˜ η j , j = 1 , ..., k . Hence there are no cancellationsin equation (5.11), and so deg( ψ ω ) = n .(2) Because Problem 1.3 is solvable, its Pick matrix is positive definite and so, by Theo-rem 4.10, there are polynomials a, b, c, d of degree at most n that parametrize the solutionsof Problem 1.3. Choose four particular such polynomials, as described in Theorem 4.10.By Theorem 4.9, there is τ ∈ T \ { σ , ..., σ k } such that the exceptional set Z τ for Problem1.3, which is defined as Z τ = { ζ ∈ T : for some j such that 1 ≤ j ≤ k, h M − u ζ,τ , e j i = 0 } , (5.12)consists of at most k points. Fix a τ ∈ T \ { σ , ..., σ k } such that Z τ consists of at most k points; then there exist unique polynomials a τ , b τ , c τ , d τ of degree at most n such that (cid:20) a τ ( τ ) b τ ( τ ) c τ ( τ ) d τ ( τ ) (cid:21) = (cid:20) (cid:21) (5.13)and, for all ζ ∈ T \ Z τ , the function ϕ = a τ ζ + b τ c τ ζ + d τ (5.14)is the unique solution of Problem 1.3 satisfying ϕ ( τ ) = ζ . In addition, the general 4-tupleof polynomials that parametrizes the solutions of Problem 1.3 can be written in the form( a, b, c, d ) = ( Xa τ , Xb τ , Xc τ , Xd τ ) (5.15)for some rational function X .(3) For τ ∈ T \ { σ , ..., σ k } as above, let x ◦ = x ( τ ) , x ◦ = x ( τ ) , x ◦ = x ( τ ) . Since x istetra-inner, by Theorem 2.3, | x ◦ | = 1 and x ◦ = x ◦ x ◦ . Since τ is chosen not to be a royalnode of x, | x ◦ | < , | x ◦ | <
1. Thus the relations (5.3) and (5.4) hold.
NTERPOLATION BY TETRA-INNER MAPS 19
Lemma 5.2.
Let x ◦ , x ◦ , x ◦ ∈ C be such that | x ◦ | = 1 , | x ◦ | < , | x ◦ | < , (5.16) x ◦ = x ◦ x ◦ . (5.17) Let Z τ be defined as in equation (5.12) , let τ ∈ T \ { σ , ..., σ k } be such that Z τ consists ofat most k points, and let Z ∼ τ = n ˜ η x ◦ − x ◦ x ◦ ˜ η − , ˜ η x ◦ − x ◦ x ◦ ˜ η − , ..., ˜ η k x ◦ − x ◦ x ◦ ˜ η k − o . If ζ ∈ T \ Z ∼ τ then the function ϕ = ( x ◦ x − x ) ζ + x ◦ x − x x ◦ ( x ◦ − x ) ζ + x ◦ x − x ◦ (5.18) is a solution of Problem 1.3 which satisfies ϕ ( τ ) = ζ .Proof. By equation (5.8), for any ω ∈ T , ψ ω ( τ ) = ωx ◦ − x ◦ x ◦ ω − , which is well defined since | x ◦ | <
1. We have, for ζ ∈ T ,ψ ω ( τ ) = ζ ⇔ ωx ◦ − x ◦ x ◦ ω − ζ ⇔ ω = − ζ + x ◦ x ◦ − ζ x ◦ . Hence, as long as − ζ + x ◦ x ◦ − ζ x ◦ = ˜ η , ...., ˜ η k , (5.19)the function ϕ ( λ ) = ψ ω ( λ ) = ψ − ζ + x ◦ x ◦ − ζx ◦ ( λ )= x ( λ ) x ◦ − x ( λ ) ζx ◦ − x ◦ ζ − x ( λ ) x ( λ ) x ◦ − x ( λ ) ζx ◦ − x ◦ ζ −
1= ( x ( λ ) x ◦ − x ( λ )) ζ + x ◦ x ( λ ) − x ( λ ) x ◦ ( x ◦ − x ( λ )) ζ + x ◦ x ( λ ) − x ◦ . is a solution of Problem 1.3 which satisfies ϕ ( τ ) = ζ . Condition (5.19) can also be written,for j = 1 , , ..., k, ζ = x ◦ − ˜ η j x ◦ − x ◦ ˜ η j = ˜ η j x ◦ − x ◦ x ◦ ˜ η j − ζ / ∈ Z ∼ τ . (cid:3) For ζ ∈ T \ ( Z τ ∪ Z ∼ τ ) where Z ∼ τ is defined in Lemma 5.2, we have two formulae forthe unique solution of Problem 1.3 satisfying ϕ ( τ ) = ζ , namely, the equations (5.14) and(5.18). Note that (cid:20) x ◦ x ( τ ) − x ( τ ) x ◦ x ( τ ) − x ( τ ) x ◦ x ◦ − x ( τ ) x ◦ x ( τ ) − x ◦ (cid:21) = (cid:20) x ◦ x ◦ − x ◦ x ◦ x ◦ − x ◦ x ◦ x ◦ − x ◦ x ◦ x ◦ − x ◦ (cid:21) = ( x ◦ x ◦ − x ◦ ) (cid:20) (cid:21) . Because the set Z τ ∪ Z ∼ τ is finite, the linear fractional transformations in equations (5.14)and (5.18) are equal at infinitely many points and therefore coincide. It follows from thenormalizing condition that (cid:20) a τ b τ c τ d τ (cid:21) = 1 x ◦ x ◦ − x ◦ (cid:20) x ◦ x − x x ◦ x − x x ◦ x ◦ − x x ◦ x − x ◦ (cid:21) . (5.20)Suppose that a, b, c and d are polynomials that parametrize the solutions of Problem 1.3,as in Theorem 5.1 (2). By the observation (5.15), there is a rational function X such that Xa = x ◦ x − x , (5.21) Xb = x ◦ x − x x ◦ , (5.22) Xc = x ◦ − x , (5.23) Xd = x ◦ x − x ◦ , (5.24)Let us find connections between x , x , x and the polynomials a, b, c, d . Equations (5.23)and (5.24) for x and X can be written as Xc + x = x ◦ Xd − x ◦ x = − x ◦ . (5.25)Then, the solution of the system (5.25) is X = (cid:12)(cid:12)(cid:12)(cid:12) x ◦ − x ◦ − x ◦ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) c d − x ◦ (cid:12)(cid:12)(cid:12)(cid:12) = x ◦ x ◦ − x ◦ x ◦ c + d (5.26)and x = (cid:12)(cid:12)(cid:12)(cid:12) c x ◦ d − x ◦ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) c d − x ◦ (cid:12)(cid:12)(cid:12)(cid:12) = x ◦ c + x ◦ dx ◦ c + d . (5.27)Equations (5.21) and (5.22) give us the system x ◦ x − x = Xa − x ◦ x + x ◦ x = Xb. (5.28)
NTERPOLATION BY TETRA-INNER MAPS 21
Then, the solution of the system (5.28) is x = (cid:12)(cid:12)(cid:12)(cid:12) Xa − Xb x ◦ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x ◦ − − x ◦ x ◦ (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x ◦ x ◦ a − x ◦ ax ◦ c + d − x ◦ x ◦ b − x ◦ bx ◦ c + d x ◦ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x ◦ − − x ◦ x ◦ (cid:12)(cid:12)(cid:12)(cid:12) = x ◦ a + bx ◦ c + d and x = (cid:12)(cid:12)(cid:12)(cid:12) x ◦ Xa − x ◦ Xb (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x ◦ − − x ◦ x ◦ (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x ◦ x ◦ x ◦ a − x ◦ ax ◦ c + d − x ◦ x ◦ x ◦ b − x ◦ bx ◦ c + d (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x ◦ − − x ◦ x ◦ (cid:12)(cid:12)(cid:12)(cid:12) = x ◦ b + x ◦ ax ◦ c + d . Thus x , x , x are given by equations (5.5), (5.6) and (5.7). The proof of Theorem 5.1 iscomplete. (cid:3) Note that we can also prove a result similar to Theorem 5.1, using the function Υ ω instead of Ψ ω , where Υ ω ( x , x , x ) = x ω − x x ω − , which is defined for every ( x , x , x ) in C such that x ω − = 0. In this case we supposethat ρ j = Ax ( σ j ) for j = 1 , , .., k. From the Blaschke interpolation problem to the royaltetra-interpolation problem
In this section we will prove Theorem 6.4. This theorem shows that, if Blaschke interpo-lation data ( σ, η, ρ ) of type ( n, k ) are given and the corresponding Problem 1.3 is solvable,then we can construct a solution for the royal tetra-interpolation problem ( σ, η, ˜ η, ρ ), forsome ˜ η = ( ˜ η , ..., ˜ η n ). We will start with some technical lemmas. Lemma 6.1.
Let a, b, c, d, x ◦ , x ◦ , x ◦ ∈ C , and suppose that | x ◦ | = 1 , x ◦ = x ◦ x ◦ , x ◦ c = − d and | x ◦ | < , | x ◦ | < . Let x = x ◦ c + x ◦ dx ◦ c + d . Then (1) | x | ≤ if and only if | c | ≤ | d | , and (2) | x | < if and only if | c | < | d | . Proof. (1) | x | ≤ ⇔ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x ◦ c + x ◦ dx ◦ c + d (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ⇔ | x ◦ c + x ◦ d | ≤ | x ◦ c + d | ⇔ | c | + 2Re( x ◦ c x ◦ d ) + | x ◦ | | d | ≤ | x ◦ | | c | + 2Re( x ◦ c d ) + | d | ⇔ | c | + | x | | d | − | x ◦ | | c | − | d | ≤ x ◦ = x ◦ x ◦ ) ⇔ (1 − | x ◦ | )( | c | − | d | ) ≤ ⇔ | c | ≤ | d | (since (1 − | x ◦ | ) > . (6.1)(2) The same calculation leads to | x | < ⇐⇒ | c | < | d | . (cid:3) Proposition 6.2.
Let a, b, c, d be polynomials in the variable λ and let x ◦ , x ◦ , x ◦ ∈ C satisfy x ◦ = x ◦ x ◦ and x ◦ c = − d . Let rational functions x , x , x be defined by x ( λ ) = x ◦ a ( λ ) + b ( λ ) x ◦ c ( λ ) + d ( λ ) , x ( λ ) = x ◦ c ( λ ) + x ◦ d ( λ ) x ◦ c ( λ ) + d ( λ ) , x ( λ ) = x ◦ b ( λ ) + x ◦ a ( λ ) x ◦ c ( λ ) + d ( λ ) . (6.2) and define a rational function ζ in the indeterminate ω by ζ ( ω ) = ωx ◦ − x ◦ x ◦ ω − . (6.3) Then, as rational functions in ( ω, λ ) , ωx ( λ ) − x ( λ ) x ( λ ) ω − a ( λ ) ζ ( ω ) + b ( λ ) c ( λ ) ζ ( ω ) + d ( λ ) . Proof.
Let x , x , x be defined by equations (6.2). Then ωx ( λ ) − x ( λ ) x ( λ ) ω − ωx ◦ b ( λ ) + ωx ◦ a ( λ ) x ◦ c ( λ ) + d ( λ ) − x ◦ a ( λ ) + b ( λ ) x ◦ c ( λ ) + d ( λ ) ωx ◦ c ( λ ) + ωx ◦ d ( λ ) x ◦ c ( λ ) + d ( λ ) − ωx ◦ b ( λ ) + ωx ◦ a ( λ ) − x ◦ a ( λ ) − b ( λ ) ωx ◦ c ( λ ) + ωx ◦ d ( λ ) − x ◦ c ( λ ) − d ( λ )= a ( λ )( ωx ◦ − x ◦ ) + b ( λ )( ωx ◦ − c ( λ )( ωx ◦ − x ◦ ) + d ( λ )( ωx ◦ − a ( λ ) (cid:18) ωx ◦ − x ◦ ωx ◦ − (cid:19) + b ( λ ) c ( λ ) (cid:18) ωx ◦ − x ◦ ωx ◦ − (cid:19) + d ( λ )= a ( λ ) ζ ( ω ) + b ( λ ) c ( λ ) ζ ( ω ) + d ( λ ) , where ζ ( ω ) = ωx ◦ − x ◦ ωx ◦ − (cid:3) NTERPOLATION BY TETRA-INNER MAPS 23
Proposition 6.3.
Let a, b, c, d be polynomials having no common zero in D , and satisfy-ing | c | ≤ | d | on D . Suppose that x ◦ , x ◦ , x ◦ ∈ C satisfy x ◦ c = − d , | x ◦ | = 1 , | x ◦ | < , | x ◦ | < and x ◦ = x ◦ x ◦ . Let rational functions x , x , x be defined by x ( λ ) = x ◦ a ( λ ) + b ( λ ) x ◦ c ( λ ) + d ( λ ) , x ( λ ) = x ◦ c ( λ ) + x ◦ d ( λ ) x ◦ c ( λ ) + d ( λ ) , x ( λ ) = x ◦ b ( λ ) + x ◦ a ( λ ) x ◦ c ( λ ) + d ( λ ) , (6.4) and let ψ ζ ( λ ) = a ( λ ) ζ + b ( λ ) c ( λ ) ζ + d ( λ ) . (6.5) (i) If, for all but finitely many values of λ ∈ D , | ψ ζ ( λ ) | ≤ for all but finitely many ζ ∈ T , then x ◦ c + d has no zero in D and x = ( x , x , x ) is an analytic map from D to E .(ii) If, for all but finitely many ζ ∈ T , the function ψ ζ is inner, then either x ( D ) ⊆ R E or x = ( x , x , x ) is a rational tetra-inner function.Proof. (i) Suppose there is a finite subset E of D such that, for all λ ∈ D \ E , there is afinite subset F λ of T for which the inequality (6.6) holds for all ζ ∈ T \ F λ . Let us showthat the denominator x ◦ c + d of x , x , x has no zeros in D . Suppose that α ∈ D is a zeroof ( x ◦ c + d ). Since | c | ≤ | d | on D , | x ◦ c ( α ) + d ( α ) | ≥ | d ( α ) | − | x ◦ c ( α ) |≥ | d ( α ) | − | x ◦ || d ( α ) | = (1 − | x ◦ | ) | d ( α ) | . Thus, 0 = | x ◦ c ( α ) + d ( α ) | ≥ (1 − | x ◦ | ) | d ( α ) | . Since | x ◦ | < , (1 − | x ◦ | ) = 0, and so d ( α ) = 0, Then0 = x ◦ c ( α ) + d ( α ) = x ◦ c ( α )implies that c ( α ) = 0.Pick a sequence ( α j ) in D \ E such that α j → α . For each j , for ζ ∈ T \ F λ j , we have | ψ ζ ( λ ) | ≤ D \ E . Hence, for ζ ∈ T \ ∪ j F λ j , which is to say, for all but countablymany ζ ∈ T , (cid:12)(cid:12)(cid:12)(cid:12) a ( α j ) ζ + b ( α j ) c ( α j ) ζ + d ( α j ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ . Because c ( α j ) ζ + d ( α j ) → ζ ∈ T as j → ∞ , the sameholds for a ( α j ) ζ + b ( α j ). Therefore a ( α j ) → b ( α j ) →
0. Hence a ( α ) = b ( α ) = 0.Thus a, b, c, d all vanish at α , contrary to our assumption. So x ◦ c + d has no zeros in D .Thus x , x , x defined by equations (6.4) are rational functions having no poles in D .Consider λ ∈ D \ E . By Proposition 6.2,Ψ ω ( x ( λ ) , x ( λ ) , x ( λ )) = ωx ( λ ) − x ( λ ) x ( λ ) ω − a ( λ ) ζ ( ω ) + b ( λ ) c ( λ ) ζ ( ω ) + d ( λ ) (6.7)provided that both sides are defined, that is, for all ω ∈ T \ Ω λ whereΩ λ = { ω ∈ T : ωx ( λ ) = 1 or c ( λ ) ζ ( ω ) = − d ( λ ) } . Ω λ contains at most two points. On combining the relations (6.5), (6.6) and (6.7), wededuce that, for λ ∈ D \ E, | Ψ ω ( x ( λ ) , x ( λ ) , x ( λ )) | ≤ ω ∈ T such that ω / ∈ Ω λ ∪ ζ − ( F λ ), that is, for all but finitely many ω ∈ T . By [1,Theorem 2.4] (see Theorem 2.2), ( x ( λ ) , x ( λ ) , x ( λ )) ∈ E . Since this is true for all butfinitely many λ ∈ D , and x , x , x are rational functions without poles in D , ( x , x , x )maps D into E .(ii) Let us assume that, for some finite subset F of T , the function ψ ζ is inner for all ζ ∈ T \ F . By Part (i), ( x , x , x ) is a rational analytic map from D into E and thereforeextends to a continuous map of D into E . Let λ ∈ T . By Proposition 6.2 and equation(6.5), Ψ ω ( x ( λ ) , x ( λ ) , x ( λ )) = ψ ζ ( ω ) ( λ ) (6.9)provided that both sides are defined, that is, for all ω ∈ T \ Ω λ whereΩ λ = { ω ∈ T : ωx ( λ ) = 1 or c ( λ ) ζ ( ω ) = − d ( λ ) } . Note that Ω λ contains at most two points. For ω ∈ T \ ζ − ( F ) the function ψ ζ ( ω ) is inner.Thus, for ω ∈ T \ ( ζ − ( F ) ∪ Ω λ ), | Ψ ω ( x ( λ ) , x ( λ ) , x ( λ )) | = | ψ ζ ( ω ) ( λ ) | = 1 . (6.10)Case 1. Suppose that for all λ ∈ D , x ( λ ) x ( λ ) = x ( λ ). Then, for all λ ∈ D ,Ψ ω ( x ( λ ) , x ( λ ) , x ( λ )) = ωx ( λ ) − x ( λ ) x ( λ ) ω − ωx ( λ ) x ( λ ) − x ( λ ) x ( λ ) ω − x ( λ )( ωx ( λ ) − x ( λ ) ω − x ( λ ) . Thus x ( D ) ⊆ R E .Case 2. Suppose that for some λ ∈ D , x ( λ ) x ( λ ) = x ( λ ). To prove that x = ( x , x , x )is rational E -inner function, by Theorem 2.3, we need to show that ( x , x , x )( λ ) ∈ b E for almost all λ ∈ T , that is,(1) | x ( λ ) | = 1 for almost all λ ∈ T ,(2) | x | ≤ D ,(3) x ( λ ) = x ( λ ) x ( λ ) for almost all λ ∈ T .For (2), by Lemma 6.1, we showed that | x ( λ ) | ≤ λ ∈ D . By Lemma 2.4, for any ω ∈ T and any point x = ( x , x , x ) ∈ E such that x x = x , | Ψ ω ( x , x , x ) | = 1 if and only if 2 ω ( x − x x )) = 1 − | x | + | x | − | x | . Thus, for λ ∈ T such that x ( λ ) x ( λ ) = x ( λ ) , equation (6.10) implies2 ω ( x ( λ ) − x ( λ ) x ( λ ))) = 1 − | x ( λ ) | + | x ( λ ) | − | x ( λ ) | . Hence, for λ ∈ T , if | Ψ ω ( x ( λ ) , x ( λ ) , x ( λ )) | = 1 for two distinct ω ∈ T , say ω = ω , wehave the linear system2 ω ( x ( λ ) − x ( λ ) x ( λ )) = 1 − | x ( λ ) | + | x ( λ ) | − | x ( λ ) | ω ( x ( λ ) − x ( λ ) x ( λ )) = 1 − | x ( λ ) | + | x ( λ ) | − | x ( λ ) | . (6.11) NTERPOLATION BY TETRA-INNER MAPS 25
Thus, for λ ∈ T , 2 ω ( x ( λ ) − x ( λ ) x ( λ )) − ω ( x ( λ ) − x ( λ ) x ( λ )) = 0= ⇒ ( x ( λ ) − x ( λ ) x ( λ ))( ω − ω ) = 0= ⇒ x ( λ ) = x ( λ ) x ( λ ) . (6.12)By equations (6.11), for λ ∈ T ,1 − | x ( λ ) | + | x ( λ ) | − | x ( λ ) | = 0 . (6.13)Note for λ ∈ T , since x ( λ ) = x ( λ ) x ( λ ),(6.13) holds ⇔ − | x ( λ ) | + | x ( λ ) x ( λ ) | − | x ( λ ) | = 0 ⇔ − | x ( λ ) | + | x ( λ ) | | x ( λ ) | − | x ( λ ) | = 0 ⇔ − | x ( λ ) | − | x ( λ ) | (1 − | x ( λ ) | ) = 0 ⇔ (1 − | x ( λ ) | )(1 − | x ( λ ) | ) = 0 ⇔ | x ( λ ) | = 1 or | x ( λ ) | = 1 . Case 1. If | x ( λ ) | = 1 and x ( λ ) = x ( λ ) x ( λ ), we have x ( λ ) = x ( λ ) x ( λ ) for almostall λ ∈ T . Then since x i are rational functions for i = 1 , ,
3, and x ( λ ) = x ( λ ) x ( λ ) for λ ∈ T , it imples that x ( λ ) = x ( λ ) x ( λ ) for all λ ∈ D . Then, for all λ ∈ D ,Ψ ω ( x ( λ ) , x ( λ ) , x ( λ )) = ωx ( λ ) − x ( λ ) x ( λ ) ω − ωx ( λ ) x ( λ ) − x ( λ ) x ( λ ) ω − x ( λ )( ωx ( λ ) − x ( λ ) ω − x ( λ ) . Thus x ( D ) ⊆ R E .Case 2. If for almost all λ ∈ T , | x ( λ ) | = 1, then x ( λ ) = x ( λ ) x ( λ ) = ⇒ x ( λ ) x ( λ ) = x ( λ )= ⇒ x ( λ ) = x ( λ ) x ( λ ) . Thus, for almost all λ ∈ T , | x ( λ ) | = 1 and x ( λ ) = x ( λ ) x ( λ ) that proves (1) and(3) respectively. Therefore, the point ( x ( λ ) , x ( λ ) , x ( λ )) for almost all λ ∈ T is in thedistinguished boundary b E of E . Hence x = ( x , x , x ) is a rational E -inner function inthis case. (cid:3) Theorem 6.4.
Let ( σ, η, ρ ) be Blaschke interpolation data of type ( n, k ) , and let ( σ, η, ˜ η, ρ ) be royal tetra-interpolation data of type ( n, k ) where ˜ η = (˜ η , ˜ η , ..., ˜ η n ) , ˜ η j ∈ T , j = 1 , ..., k and ˜ η j ∈ D , j = k + 1 , ..., n . Suppose that Problem with ( σ, η, ρ ) is solvable and thesolutions ϕ of Problem have normalized parametrization ϕ = aζ + bcζ + d . Suppose that there exist scalars x ◦ , x ◦ , x ◦ in C such that | x ◦ | = 1 , | x ◦ | < , | x ◦ | < , x ◦ = x ◦ x ◦ , and x ◦ c ( σ j ) + x ◦ d ( σ j ) x ◦ c ( σ j ) + d ( σ j ) = ˜ η j , j = 1 , ..., n. (6.14) Then there exists a rational tetra-inner function x = ( x , x , x ) given by x ( λ ) = x ◦ a ( λ ) + b ( λ ) x ◦ c ( λ ) + d ( λ ) (6.15) x ( λ ) = x ◦ c ( λ ) + x ◦ d ( λ ) x ◦ c ( λ ) + d ( λ ) (6.16) x ( λ ) = x ◦ b ( λ ) + x ◦ a ( λ ) x ◦ c ( λ ) + d ( λ ) , (6.17) for λ ∈ D , such that(i) x ∈ R n,k , and x is a solution of the royal tetra-interpolation problem with the data ( σ, η, ˜ η, ρ ) , that is, x ( σ j ) = ( η j , ˜ η j , η i ˜ η j ) for j = 1 , ..., n, and Ax ( σ j ) = ρ j for j = 1 , ..., k, (ii) for all but finitely many ω ∈ T , the function Ψ ω ◦ x is a solution of Problem .Proof. By Corollary 4.12 (3), | c | ≤ | d | on D . Hence (cid:12)(cid:12)(cid:12) d ( λ ) c ( λ ) (cid:12)(cid:12)(cid:12) ≥ λ ∈ D . By assumption, | x ◦ | <
1. We claim that x ◦ c = − d on D . Suppose that x ◦ c = − d = ⇒ | x ◦ c | = | d | = ⇒ | x ◦ || c | = | d | = ⇒ | x ◦ | = | d || c | , which is a contradiction since (cid:12)(cid:12)(cid:12) d ( λ ) c ( λ ) (cid:12)(cid:12)(cid:12) ≥ λ ∈ D , and | x ◦ | < D . Therefore, x ◦ c = − d on D . By Proposition 6.3, either x ( D ) ⊆ R E or x is a rational E -inner function.Because a, b, c, d are polynomials of degree at most n , the rational function x has degreeat most n .By the definition of a normalized linear fractional parametrization of the solutions ofProblem 1.3, for some point τ ∈ T \ { σ , ..., σ k } , (cid:20) a ( τ ) b ( τ ) c ( τ ) d ( τ ) (cid:21) = (cid:20) (cid:21) . NTERPOLATION BY TETRA-INNER MAPS 27
Thus it is easy to see that x ( τ ) = x ◦ a ( τ ) + b ( τ ) x ◦ c ( τ ) + d ( τ ) = x ◦ , (6.18) x ( τ ) = x ◦ c ( τ ) + x ◦ d ( τ ) x ◦ c ( τ ) + d ( τ ) = x ◦ , (6.19) x ( τ ) = x ◦ b ( τ ) + x ◦ a ( τ ) x ◦ c ( τ ) + d ( τ ) = x ◦ . (6.20)By assumption, | x ◦ | = 1 , | x ◦ | < | x ◦ | <
1, and hence x ( τ ) = x ( τ ) x ( τ ). Therefore, x ( D ) is not in the royal variety R E .By assumption, x is defined by equation (6.19). Hence x ( σ j ) = x ◦ c ( σ j ) + x ◦ d ( σ j ) x ◦ c ( σ j ) + d ( σ j ) = ˜ η j for j = 1 , ..., n. We want to show that x satisfies the interpolation conditions x ( σ j ) = ( η j , ˜ η j , η i ˜ η j ) for j = 1 , ..., n, (6.21)which is to say that σ j , j = 1 , ..., n , is a royal node of x with corresponding royal value( η j , ˜ η j ). By assumption, there is a finite set F ⊂ T such that, for all ζ ∈ T \ F , thefunction ϕ ( λ ) = ψ ζ ( λ ) = a ( λ ) ζ + b ( λ ) c ( λ ) ζ + d ( λ )is a solution of Problem 1.3, and so ψ ζ ( σ j ) = η j for j = 1 , ..., n (6.22)and Aψ ζ ( σ j ) = ρ j for j = 1 , ..., k (6.23)for all ζ ∈ T \ F . Hence, by Proposition 6.2, ψ ζ ( ω ) ( λ ) = a ( λ ) ζ ( ω ) + b ( λ ) c ( λ ) ζ ( ω ) + d ( λ ) = ωx ( λ ) − x ( λ ) ωx ( λ ) − ω ◦ x ( λ ) (6.24)whenever both sides are defined, that is, for all ω ∈ T \ Ω λ whereΩ λ = { ω ∈ T : ωx ( λ ) = 1 or c ( λ ) ζ ( ω ) = − d ( λ ) } . Note that Ω λ contains at most two points. Thus, equation (6.24) holds as rational func-tions in ( ω, λ ), where ζ ( ω ) = ωx ◦ − x ◦ x ◦ ω − . Hence, for ω ∈ T \ ( ζ − ( F ) ∪ Ω λ ), Ψ ω ◦ x is asolution of Problem 1.3, which proves statement (ii).Equation (6.24) holds for any λ ∈ D , that is, provided both denominators are nonzero,and therefore for all but at most two values of ω ∈ T . Combine equations (6.22) and(6.24) (with λ = σ j ) to infer that, for j = 1 , ..., n and for all but finitely many ω ∈ T , ωx ( σ j ) − x ( σ j ) ωx ( σ j ) − ψ ζ ( ω ) ( σ j ) = η j . Therefore, for almost all ω ∈ T and j = 1 , ..., n , ωx ( σ j ) − x ( σ j ) = η j ( ωx ( σ j ) − . (6.25) Recall that x ( σ j ) = ˜ η j for j = 1 , ..., n. Hence from equations (6.25) it follows that x ( σ j ) = η j and x ( σ j ) = η j ˜ η j , j = 1 , ..., n , and so the interpolation conditions (6.21)hold.We have already observed that x is a rational E -inner function, deg( x ) ≤ n and that x ( D ) is not in R E . Thus by Theorem 3.10, the number of royal nodes of x is equal to thedegree of x . Consequently x has at most n royal nodes. Because the points σ j , j = 1 , ..., n are royal nodes, they contain all n royal nodes of x , and so deg( x ) = n . Observe that k of the σ j lie in T ; thus x has exactly k royal nodes in T . Hence x ∈ R n,k .Next we show that Ax ( σ j ) = ρ j for j = 1 , ..., k . Fix j ∈ { , ..., k } . By Proposition 2.7,for ω ∈ T , ω ˜ η j = 1 , A (Ψ ω ◦ x )( σ j ) = Ax ( σ j ) . (6.26)In addition there exists a set Ω j containing at most one point in T such that c ( σ j ) ζ ( ω ) + d ( σ j ) = 0 for ω ∈ Ω j . Thus, if ω ∈ T \ ( { ˜ η j } ∪ Ω j ), by equation (6.24), ψ ζ ( ω ) = Ψ ω ◦ x in a neighbourhood of σ j , and so, for such ω , Aψ ζ ( ω ) ( σ j ) = A (Ψ ω ◦ x )( σ j ) . (6.27)The equations (6.26), (6.27) and (6.23) all hold for ω in a cofinite subset of T . Therefore,for ω in the intersection of these cofinite subsets, Ax ( σ j ) = A (Ψ ω ◦ x )( σ j ) = Aψ ζ ( ω ) ( σ j ) = ρ j . Thus (i) holds. (cid:3)
Corollary 6.5.
Let ( σ, η, ρ ) be Blaschke interpolation data of type ( n, k ) . Let x be asolution of Problem with data ( σ, η, ˜ η, ρ ) for some ˜ η j ∈ D , j = 1 , ..., n, and that x ( D )
6⊂ R ¯ E . For all ω ∈ T \ { ˜ η , ..., ˜ η k } , the function ϕ = Ψ ω ◦ x is a solution ofProblem with Blaschke interpolation data ( σ, η, ρ ) . Conversely, for every solution ϕ of the Blaschke interpolation problem with data ( σ, η, ρ ) , there exists ω ∈ T such that ϕ = Ψ ω ◦ x .Proof. (= ⇒ ) Consider Blaschke interpolation data ( σ, η, ρ ). If x = ( x , x , x ) is a solutionof Problem 1.6 with data ( σ, η, ˜ η, ρ ) for some ˜ η j ∈ D , j = 1 , ..., n, and x ( D )
6⊆ R ¯ E , then,by Theorem 5.1 (1), for all ω ∈ T \ { ˜ η , ..., ˜ η k } , there exists a rational function ϕ = Ψ ω ◦ x that solves the Blaschke interpolation problem with data ( σ, η, ρ ).( ⇐ =) Let ϕ be a solution of the Blaschke interpolation problem (Problem 1.3) withdata ( σ, η, ρ ) of type ( n, k ). Then, by Theorem 5.1 and Theorem 6.4 (ii), there exists ω ∈ T such that ϕ = Ψ ω ◦ x . (cid:3) The algorithm
In this section we present a concrete algorithm for the solution of the royal E -interpolationproblem.Let ( σ, η, ˜ η, ρ ) be royal interpolation data of type ( n, k ) for the tetrablock, as in Defi-nition 1.5. One can consider the associated Blaschke interpolation data ( σ, η, ρ ) of type( n, k ) . To construct a rational E -inner function x : D → E of degree n having royal nodes σ j for j = 1 , ..., n , royal values η j , ˜ η j , and phasar derivatives ρ j at σ j for j = 1 , ..., k , weproceed as follows. NTERPOLATION BY TETRA-INNER MAPS 29 (1) Consider the Pick matrix M = [ m i,j ] ni,j =1 for the data ( σ, η, ρ ). It has entries m i,j = ρ i if i = j ≤ k − η i η j − σ i σ j otherwise . (7.1)Assume that M is positive definite; otherwise the interpolation problem 1.3 is not solvable.Introduce the notation x λ = − σ λ ... − σ n λ , y λ = η − σ λ ... η n − σ n λ , (7.2)as in equations (4.8).(2) Choose a point τ ∈ T \ { σ , σ , ..., σ k } such that the set of ζ ∈ T for which h M − x τ , e j i = ζ h M − y τ , e j i for some j ∈ { , ..., n } is finite. Here e j is the j th standard basis vector in C n .(3) Let g ( λ ) = n Y j =1 − σ j λ − σ j τ , (7.3)and let polynomials a, b, c, d be defined by a ( λ ) = g ( λ )(1 − (1 − τ λ ) h x λ , M − x τ i ) , (7.4) b ( λ ) = g ( λ )(1 − τ λ ) h x λ , M − y τ i , (7.5) c ( λ ) = − g ( λ )(1 − τ λ ) h y λ , M − x τ i (7.6) d ( λ ) = g ( λ )(1 + (1 − τ λ ) h y λ , M − y τ i ) . (7.7)Observe that (cid:20) a ( τ ) b ( τ ) c ( τ ) d ( τ ) (cid:21) = (cid:20) (cid:21) . (7.8)(See Theorem 3.9 in [3]).(4) Find x ◦ , x ◦ , x ◦ ∈ C such that | x ◦ | = 1 , | x ◦ | < , | x ◦ | < , x ◦ = x ◦ x ◦ , and x ◦ c ( σ j ) + x ◦ d ( σ j ) x ◦ c ( σ j ) + d ( σ j ) = ˜ η j , j = 1 , ..., n. If there is no ( x ◦ , x ◦ , x ◦ ) satisfying these conditions, then by Theorem 6.4, the royal E -interpolation problem is not solvable. (5) If there are such ( x ◦ , x ◦ , x ◦ ) ∈ C , we define x ( λ ) = x ◦ a + bx ◦ c + d ( λ ) ,x ( λ ) = x ◦ c + x ◦ dx ◦ c + d ( λ ) ,x ( λ ) = x ◦ b + x ◦ ax ◦ c + d ( λ ) , for λ ∈ D . It is easy to see that, since the equation (7.8) is satisfied, x ( τ ) = x ◦ , x ( τ ) = x ◦ and x ( τ ) = x ◦ . Then, by Theorem 6.4, x = ( x , x , x ) is a rational E -inner function of degree at most n such that x ( σ j ) = ( η j , ˜ η j , η j ˜ η j ) for j = 1 , ..., n , and Ax ( σ j ) = ρ j for j = 1 , ..., k . Byassumption, | x ◦ | = 1 , | x ◦ | < | x ◦ | <
1, and hence x ( τ ) = x ( τ ) x ( τ ). Hence, x ( D )is not in the royal variety of the tetrablock, and the degree of x is exactly n .Now let us relate the steps of algorithm to results earlier in the paper.(1) If the royal E -interpolation problem with data ( σ, η, ˜ η, ρ ) for some ˜ η j ∈ D is solv-able, then by Theorem 5.1, the Blaschke interpolation problem with data ( σ, η, ρ )is solvable. By [3, Proposition 3.2], M > | x ◦ | = 1 , | x ◦ | < , | x ◦ | < , and x ◦ = x ◦ x ◦ are equivalent to( x ◦ , x ◦ , x ◦ ) ∈ b E and | x ◦ | < x , x and x are equations (6.15), (6.16) and (6.17) respectively.8. Examples
The following two examples describe all rational tetra-inner functions of degree 1.
Lemma 8.1.
Let σ ∈ D , and η, ˜ η ∈ D . Let m ∈ Aut( D ) be such that m ( σ ) = 0 . Supposethere exists a rational E -inner y : D → E such that y (0) = ( η, ˜ η, η ˜ η ) . Then x = y ◦ m is arational E - inner function such that x ( σ ) = ( η, ˜ η, η ˜ η ) .Proof. By assumption, the function y : D → E is such that y (0) = ( η, ˜ η, η ˜ η ). The Blaschkefactor m : D → D such that m ( z ) = z − σ − σ z moves σ to 0.Note that ( y ◦ m )( σ ) = y ( m ( σ )) = y (0) = ( η, ˜ η, η ˜ η ) . It is easy to see that the composition x = y ◦ m is a rational E - inner function, x : D → E such that x ( σ ) = ( η, ˜ η, η ˜ η ). (cid:3) Example 8.2.
Let ( σ, η, ˜ η, ρ ) be royal interpolation data of type (1 ,
0) for the tetrablock,as in Definition 1.5. Consider Problem 1.6 with the data ( σ, η, ˜ η, ρ ). We are given asingle royal node σ ∈ D and a royal value ( η, ˜ η, η ˜ η ), where η, ˜ η ∈ D , and we need to finda E -inner function x of degree 1 such that x ( σ ) = ( η, ˜ η, η ˜ η ). By composition with anautomorphism of D , we may reduce our problem to the case that σ = 0. Step 1 . Pick an arbitrary τ ∈ T . The solution set of the associated Blaschke interpo-lation problem has the normalized parametrization, given, for some ζ ∈ T , by ϕ ( λ ) = a ( λ ) ζ + b ( λ ) c ( λ ) ζ + d ( λ ) , for all λ ∈ D , NTERPOLATION BY TETRA-INNER MAPS 31 where a, b, c, d are given by equations (7.4) - (7.7), and x λ , y λ , g and M are given byequations (7.2), (7.3) and (7.1) respectively. Note that since σ = 0, g ( λ ) = 1 − σ λ − σ τ = 1, M = 1 − ηη − σ σ = 1 − | η | , x λ = 11 − λ = 1 and y λ = η − λ = η .Thus, the polynomials a, b, c and d are defined by a ( λ ) = g ( λ )(1 − (1 − τ λ ) h x λ , M − x τ i )= 1 − − τ λ − | η | = τ λ − | η | − | η | , (8.1) b ( λ ) = g ( λ )(1 − τ λ ) h x λ , M − y τ i = 1(1 − τ λ ) η − | η | = η (1 − τ λ )1 − | η | , (8.2) c ( λ ) = − g ( λ )(1 − τ λ ) h y λ , M − x τ i = − − τ λ ) η − | η | = − η (1 − τ λ )1 − | η | , (8.3) d ( λ ) = g ( λ )(1 + (1 − τ λ ) h y λ , M − y τ i )= 1(1 + (1 − τ λ ) ηη − | η | ) = 1 − | η | τ λ − | η | . (8.4) Step 2 . The next step is to determine whether there exist x ◦ , x ◦ , x ◦ ∈ C such that | x ◦ | = 1 , | x ◦ | < , | x ◦ | < , x ◦ = x ◦ x ◦ , (8.5)and x ◦ c (0) + x ◦ d (0) x ◦ c (0) + d (0) = ˜ η. (8.6)Here, a (0) = −| η | − | η | , b (0) = η − | η | , c (0) = − η − | η | , d (0) = 11 − | η | . Let x ◦ = ω for ω ∈ T . Thus x ◦ c (0) + x ◦ d (0) x ◦ c (0) + d (0) = ˜ η ⇔ x ◦ = − x ◦ η ˜ η + ˜ η + ωη. Since x ◦ = x ◦ x ◦ and x ◦ = ω , we have the system x ◦ = ωx ◦ = x ◦ ωx ◦ = − x ◦ η ˜ η + ˜ η + ωη. (8.7)For given η, ˜ η ∈ D , we want to find a solution x ◦ , x ◦ , x ◦ of the above system such that x ◦ ∈ T , | x ◦ | <
1, and | x ◦ | <
1. This is equivalent to finding ω ∈ T and | x ◦ | <
1, such thatthe equation ω ˜ η + η = x ◦ + x ◦ ω ˜ ηη (8.8)holds. Lemma 8.3.
Let η, ˜ η ∈ D . There are ω ∈ T and | x ◦ | < such that equation (8.8) holds.Proof. Choose ω ∈ T . Let ξ = ω ˜ η , and so ξ ∈ D . Let s = ξ + η and p = ξη ; then ( s, p ) isa point of the symmetrized bidisc G = { ( z + z , z z ) : | z | < , | z | < } . One can seethat β = s − ¯ sp − | p | satisfies the equation s = β + ¯ βp and the inequality | β | <
1. Hence, for ω ∈ T and for x ◦ = β , equation (8.8) holds. (cid:3) Step 3 . Therefore, for the given data, 0 → ( η, ˜ η, η ˜ η ), there is a 1-parameter family of( x ◦ , x ◦ , x ◦ ) such that equations (8.5) and (8.6) satisfied, given by x ◦ = ( ω ˜ η + η ) − (¯ ω ˜ η + ¯ η ) ω ˜ ηη − | ˜ ηη | , x ◦ = ωx ◦ , x ◦ = ω, for any ω ∈ T . Substitute these values into equations (6.15), (6.16) and (6.17) to obtainthe degree 1 rational E -inner function x = ( x , x , x ), satisfying x (0) = ( η, ˜ η, η ˜ η ), where,for λ ∈ D , x ( λ ) = x ◦ ( τ λ − | η | ) + η (1 − τ λ ) − x ◦ η (1 − τ λ ) + 1 − | η | τ λ , (8.9) x ( λ ) = − x ◦ η (1 − τ λ ) + x ◦ (1 − | η | τ λ ) − x ◦ η (1 − τ λ ) + 1 − | η | τ λ , (8.10) x ( λ ) = x ◦ η (1 − τ λ ) + x ◦ ( τ λ − | η | ) − x ◦ η (1 − τ λ ) + 1 − | η | τ λ . (8.11) Example 8.4.
Consider the case n = 1 , k = 1. Suppose σ = 1. The points η, ˜ η ∈ T and ρ > E -inner function x = ( x , x , x ) of degree 1 such that x (1) = ( η, ˜ η, η ˜ η ) and Ax (1) = ρ . Step 1.
Choose τ ∈ T \ { } . The normalized parametrization of the solution set ofthe associated Blaschke interpolation problem is given by ϕ ( λ ) = a ( λ ) ζ + b ( λ ) c ( λ ) ζ + d ( λ ) for λ ∈ D , and some ζ ∈ T , (8.12)where a, b, c, d are given by equations (7.4) - (7.7), and x λ , y λ , g and M are given byequations (7.2), (7.3) and (7.1) respectively.Note that, since σ = 1 and k = 1, g ( λ ) = 1 − σλ − στ = 1 − λ − τ , M = ρ , x λ = 11 − λ and y λ = η − λ . Therefore, polynomials a, b, c and d are defined by a ( λ ) = g ( λ ) (cid:16) − (1 − τ λ ) h x λ , M − x τ i (cid:17) = 1 − λ − τ − (1 − τ λ ) ρ | − τ | (8.13) b ( λ ) = g ( λ )(1 − τ λ ) h x λ , M − y τ i = η (1 − τ λ ) ρ | − τ | (8.14) NTERPOLATION BY TETRA-INNER MAPS 33 c ( λ ) = − g ( λ )(1 − τ λ ) h y λ , M − x τ i = − η (1 − τ λ ) ρ | − τ | (8.15) d ( λ ) = g ( λ ) (cid:16) − τ λ ) h y λ , M − y τ i (cid:17) = 1 − λ − τ + 1 − τ λρ | − τ | . (8.16) Step 2.
Next, determine if there exist x ◦ , x ◦ , x ◦ ∈ C such that | x ◦ | = 1 , | x ◦ | < , | x ◦ | < , and x ◦ = x ◦ x ◦ , (8.17)and x ◦ c (1) + x ◦ d (1) x ◦ c (1) + d (1) = ˜ η. (8.18)Here, a (1) = 1 − τρ | − τ | , b (1) = η − ητρ | − τ | , c (1) = − η + ητρ | − τ | , d (1) = 1 − τρ | − τ | . Let x ◦ = ω for ω ∈ T . Thus x ◦ c (1) + x ◦ d (1) x ◦ c (1) + d (1) = ˜ η ⇔ x ◦ = − ηx ◦ ˜ η + ˜ η + ηω. Since x ◦ = x ◦ x ◦ and x ◦ = ω , we have x ◦ = ωx ◦ = x ◦ ωx ◦ = − ηx ◦ ˜ η + ˜ η + ηω. (8.19)For given η, ˜ η ∈ T , we want to find a solution x ◦ , x ◦ , x ◦ of the above system such that | x ◦ | < , | x ◦ | <
1. Thus we want to find ω ∈ T and x ◦ ∈ C such that | x ◦ | < x ◦ + x ◦ ω ˜ ηη = ω ˜ η + η. (8.20) Lemma 8.5.
Let η, ˜ η ∈ T . There are ω ∈ T and x ◦ ∈ D such that equation (8.20) holds.Proof. Choose ω ∈ T such that ω = η ˜ η . Let ξ = ω ˜ η , and so ξ ∈ T and − ξ ¯ η = −
1. Weseek β ∈ D such that β + ¯ βξη = ξ + η ; (8.21)that is, we seek β ∈ D such that the following equation is satisfied − ξ ¯ η = 1 − β ¯ η − ¯ βη . (8.22)Let − ξ ¯ η = e iϕ where 0 ≤ ϕ < π and ϕ = π . Hence we can choose β such that1 − β ¯ η = re iϕ/ , that is, β = η (1 − re iϕ/ ). Choose r so small that | β | <
1. In the case0 ≤ ϕ/ < π/
2, one can find r > | β | <
1. In the case π/ < ϕ/ < π , onecan find r < | β | < ω ∈ T such that ω = η ˜ η and for x ◦ = β = η (1 − re iϕ/ ), equation (8.21)holds, and so the equivalent equation (8.20) holds too. (cid:3) Step 3 . Therefore, for the given data, 1 → ( η, ˜ η, η ˜ η ) and ρ >
0, there is a familyof points ( x ◦ , x ◦ , x ◦ ) such that equations (8.17) and (8.18) are satisfied. Substitution ofthese values into equations (6.15), (6.16) and (6.17) yields the degree 1 rational E -innerfunction x = ( x , x , x ) where, for λ ∈ D , x ( λ ) = x ◦ [(1 − λ ) ρ (1 − τ ) − (1 − τ λ )] + η (1 − τ λ ) x ◦ [ − η (1 − τ λ )] + ρ (1 − τ )(1 − λ ) + (1 − τ λ ) , (8.23) x ( λ ) = x ◦ [ − η (1 − τ λ )] + x ◦ [ ρ (1 − τ )(1 − λ ) + (1 − τ λ )] x ◦ [ − η (1 − τ λ )] + ρ (1 − τ )(1 − λ ) + (1 − τ λ ) , (8.24) x ( λ ) = x ◦ η (1 − τ λ ) + x ◦ [ ρ (1 − τ )(1 − λ ) − (1 − τ λ )] x ◦ [ − η (1 − τ λ )] + ρ (1 − τ )(1 − λ ) + (1 − τ λ ) . (8.25)One can check that x = ( x , x , x ) defined by equations (8.23), (8.24) and (8.25) is a E -inner function of degree 1 satisfying x (1) = ( η, ˜ η, η ˜ η ) and Ax (1) = ρ . References [1] A. A. Abouhajar, M. C. White and N. J. Young, A Schwarz lemma for a domain related to µ -synthesis. J. Geom. Anal . (4) (2007) 717-750.[2] J. Agler, Z. A. Lykova and N. J. Young, Extremal holomorphic maps and the symmetrized bidisc. Proc. Lond. Math. Soc. (4) (2013) 781-818.[3] J. Agler, Z. A. Lykova and N. J. Young, Finite Blaschke products and the construction of rationalΓ-inner functions.
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