Rational tetra-inner functions and the special variety of the tetrablock
aa r X i v : . [ m a t h . C V ] J a n RATIONAL TETRA-INNER FUNCTIONS AND THE SPECIALVARIETY OF THE TETRABLOCK
OMAR M. O. ALSALHI AND ZINAIDA A. LYKOVA
Abstract.
The set E = { x ∈ C : 1 − x z − x w + x zw = 0 whenever | z | < , | w | < } . is called the tetrablock and has intriguing complex-geometric properties. It is poly-nomially convex, nonconvex and starlike about 0. It has a group of automorphismsparametrised by Aut D × Aut D × Z and its distinguished boundary b E is homeomorphicto the solid torus D × T . It has a special subvariety R E = (cid:8) ( x , x , x ) ∈ E : x x = x (cid:9) , called the royal variety of E , which is a complex geodesic of E that is invariant underall automorphisms of E . We exploit this geometry to develop an explicit and detailedstructure theory for the rational maps from the unit disc D to E that map the unitcircle T to the distinguished boundary b E of E . Such maps are called rational E -innerfunctions. We show that, for each nonconstant rational E -inner function x , either x ( D ) ⊆ R E ∩ E or x ( D ) meets R E exactly deg( x ) times. We study convex subsets ofthe set J of all rational E -inner functions and extreme points of J . Contents
1. Introduction 22. The tetrablock E µ Diag -synthesis problem 52.2. The distinguished boundary of the tetrablock 63. The symmetrised bidisc and Γ-inner functions 64. Rational E -inner functions 84.1. Relations between E -inner functions and Γ-inner functions 84.2. The degree of a rational E -inner function 134.3. Description of rational E -inner functions 154.4. Superficial E -inner functions 215. The construction of rational E -inner functions 245.1. The royal polynomial of an E -inner function 245.2. Rational E -inner functions of type ( n, k ) 266. Convex subsets of E and extremality 336.1. Convex subsets in the tetrablock 336.2. Extremality in the set of E -inner functions 35References 47 Date : 7th January 2021.2010
Mathematics Subject Classification.
Primary 32F45, 30E05, 93B36, 93B50.
Key words and phrases.
Inner functions, Tetrablock, Convexity, Extreme point, Distinguishedboundary.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 grants EP/N03242X/1. Introduction
Basic unsolved problems in H ∞ control theory led us to consider “inner mappings”from D to certain domains in C d with d >
1. For example, a special case of the problemof robust stabilization under structured uncertainty, or the µ -synthesis problem [13, 14,3, 5], leads naturally to a class of domains of which a typical member is E = { x ∈ C : 1 − x z − x w + x zw = 0 whenever | z | ≤ , | w | ≤ } the open tetrablock, as was observed by Abouhajar, White and Young in [1]. Thecomplex geometry of E was further developed in [15, 16, 17, 20] and associated operatortheory in [9, 8]. The solvability of the µ -synthesis problem connected to E can beexpressed in terms of the existence of rational inner functions from the open unit disc D in the complex plane C to the closure of E [10].Recall that a classical rational inner function is a rational map f from the unit disc D to its closure D with the property that f maps the unit circle T into itself. See [12] for asurvey of results, linking inner functions and operator theory. We denote the closure of E by E and we define a rational E -inner or ( tetra-inner ) function to be a rational analyticmap x : D → E with the property that x maps T into the distinguished boundary b E of E . Here, b E is the smallest closed subset of E on which every continuous function on b E that is analytic in E attains its maximum modulus. Rational E -inner functions havemany similarities with rational inner functions, but also have some striking differences,which stem from the fact that E has a more subtle complex geometry than D .Here are some points of difference between well-studied domains in C , such as thetridisc D and the Euclidean ball B on the one hand and E on the other. Firstly,whereas D and B are homogeneous (so that the holomorphic automorphisms of thesedomains act transitively), E is inhomogeneous [20].Secondly, the distinguished boundary of E differs markedly in its topological prop-erties from those of D and B . The distinguished boundaries of D and B are the3-dimensional torus and the 5-sphere respectively, which are smooth manifolds with-out boundary. In contrast, the distinguished boundary b E of E is homeomorphic to asolid torus D × T , which has a boundary. For a rational E -inner function x the curve x (e it ) , ≤ t < π , lies in b E and may or may not touch the edge, with consequences forthe algebraic and geometric properties of x .We call the set(1.1) R E = (cid:8) ( x , x , x ) ∈ C : x x = x (cid:9) . the royal variety of the tetrablock. The complex geodesic R E ∩ E is invariant under thegroup of biholomorphic automorphisms of E [20]. This paper shows that the variety R E plays a central role in the function theory of E . The intersection R E ∩ b E is exactly theboundary of b E , that is, { ( x , x , x x ) ∈ C : | x | = | x | = 1 } [1, Theorem 7.1], whichis homeomorphic to the 2-torus T × T .These geometric properties of E lead to dramatic phenomena in the theory of rational E -inner functions that have no analogue for classical inner functions. The first of ourmain theorems, a corollary of Theorem 5.12, is as follows. Theorem 1.1. If x is a nonconstant rational E -inner function then either x ( D ) ⊆ R E ∩ E or x ( D ) meets R E exactly deg( x ) times. Here deg( x ) is the degree of x , defined in a natural way by means of fundamentalgroups (Subsection 4.2). In Proposition 4.14 we show that, for any rational E -innerfunction x = ( x , x , x ), deg( x ) is equal to the degree deg( x ) (in the usual sense) of the ATIONAL E -INNER FUNCTIONS 3 finite Blaschke product x . The precise way of counting the number of times that x ( D )meets R E is also described in Section 5. We call the points λ ∈ D such that x ( λ ) ∈ R E the royal nodes of x and, for such λ , we call x ( λ ) a royal point of x .The second main result describes the construction of rational E -inner functions ofprescribed degree from certain interpolation data; one can think of this result as ananalogue of the expression for a finite Blaschke product in terms of its zeros. Concretely,the following result is proved in Theorem 5.17. Theorem 1.2.
Let n be a positive integer and suppose that α , ..., α k ∈ D and α , ..., α k ∈ D , where k + k = n . Suppose that σ , ..., σ n ∈ D are distinct from the points of theset { α ij , j = 1 , ..., k i , i = 1 , } ∩ T . Then there exists a rational E -inner function x = ( x , x , x ) : D → E such that (1) the zeros of x in D , repeated according to multiplicity, are α , ..., α k ; (2) the zeros of x in D , repeated according to multiplicity, are α , ..., α k ; (3) the royal nodes of x are σ , ..., σ n ∈ D , with repetition according to multiplicityof the nodes.Such a function x can be constructed as follows. Let t + > and let t ∈ C \{ } . Let R be defined by R ( λ ) = t + n Y j =1 ( λ − σ j )(1 − σ j λ ) . Let E be defined by E ( λ ) = t k Y j =1 ( λ − α j ) k Y j =1 (1 − α j λ ) . Then the following statements hold: (i)
There exists an outer function D of degree at most n such that λ − n R ( λ ) + | E ( λ ) | = | D ( λ ) | for all λ ∈ T . (ii) The function x defined by x = (cid:18) E D , E ∼ n D , D ∼ n D (cid:19) where, for any polynomial D , D ∼ n ( λ ) = λ n D (cid:0) λ (cid:1) , is a rational E -inner function such that the degree of x is equal to n and conditions (1), (2) and (3) hold. The royal polynomial R x of x is equal to R . Here the royal polynomial of x is defined as R x = (cid:2) D ∼ n D − E E ∼ n (cid:3) . The proof ofthis theorem is constructive: it gives a prescription for the construction of a 3-parameterfamily of such functions x , subject to the computation of Fej´er-Riesz factorizations ofcertain non-negative functions on the circle.In Section 6 we study convex subsets of the set J of all rational E -inner functions andextremality. It is easy to see that the set J is not convex. On the other hand, the subsetof J with a fixed inner function x is convex (Theorem 6.5). Recall that the distinguishedboundary of the tridisc D contain no line segments. Thus every inner function in theset of analytic functions Hol( D , D ) from D to D is an extreme point of Hol( D , D ).However, this property contrasts sharply with the situation in the tetrablock. We showthat whether x is an extreme point of J depends on how many royal nodes lie on T . OMAR M. O. ALSALHI AND ZINAIDA A. LYKOVA
Theorem 1.3.
Let x be a rational E -inner function and let x have n royal nodes where k of them are in T . If k ≤ n , then x is not an extreme point of J . Here we use the precise way of counting the number of royal nodes, which we introducein Section 5. In Proposition 6.21 we provide a class of extreme functions of the set J .In [2] 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 royal values of x . The algorithm [2] for the construction of x exploits a known construction of the finiteBlaschke products of given degree which satisfy some interpolation conditions with theaid of a Pick matrix formed from the interpolation data.The authors are grateful to Nicholas Young for some helpful suggestions.2. The tetrablock E Definition 2.1. [1]
The open tetrablock is the domain defined by E = { x ∈ C : 1 − x z − x w + x zw = 0 for | z | ≤ , | w | ≤ } . Despite the fact that E is not convex, its intersection with R is . It is proved in [1]that E ∩ R is the open tetrahedron with the vertices (1 , , , (1 , − , − , ( − , , − − , − , Definition 2.2.
For x = ( x , x , x ) ∈ C and z ∈ C we define Ψ( z, x ) = x z − x x z − , whenever x z = 1 . Remark 2.3.
In the case that x = x x , z ∈ C ,Ψ( z, x ) = x x z − x x z − x ( x z − x z − x . Theorem 2.4. [1, Theorem 2.2]
Let x ∈ C . The following are equivalent (1) x ∈ E ; (2) k Ψ( ., x ) k H ∞ < and if x x = x , then, in addition, | x | < | x − x x | + | x − x x | < − | x | ; (4) there exists a × matrix A = [ a ij ] such that k A k < and x = ( a , a , det( A )) ; (5) | x | < and there exist β , β ∈ C such that | β | + | β | < and x = β + β x , x = β + β x . Theorem 2.5. [1, Theorem 2.4]
Let x ∈ C . The following are equivalent (1) x ∈ E ; (2) k Ψ( ., x ) k H ∞ ≤ and if x x = x , then, in addition, | x | ≤ | x − x x | + | x − x x | ≤ − | x | and if k x | = 1 then, in addition, | x | ≤ ; (4) there exists a × matrix A = [ a ij ] such that k A k ≤ and x = ( a , a , det( A )) ; (5) | x | ≤ and there exist β , β ∈ C such that | β | + | β | ≤ and x = β + β x , x = β + β x . ATIONAL E -INNER FUNCTIONS 5 The tetrablock and the µ Diag -synthesis problem.
The tetrablock is associatedwith the µ Diag -synthesis problem from D to C × . The structured singular value in thiscase is defined by(2.1) µ Diag ( A ) = 1inf {k X k : X ∈ Diag , det( I − AX ) = 0 } , where Diag := ( (cid:20) z w (cid:21) : z, w ∈ C ) . We set µ Diag ( A ) = 0 if ( I − AX ) is non-singular for all X ∈ Diag.
Definition 2.6.
We define the map π : C × → C for a matrix A = (cid:20) a a a a (cid:21) in C × to be π ( A ) = ( a , a , det( A )) . and Σ to be Σ := { A ∈ C × : µ Diag ( A ) < } where µ Diag ( A ) is defined by equation (2.1) . Theorem 2.7. [1, Theorem 9.2]
Suppose that λ , ..., λ n ∈ D are distinct points and A k = [ a kij ] ∈ Σ are such that a k a k = det( A k ) , ≤ k ≤ n . The following conditions areequivalent. (1) There exists an analytic function F : D → Σ such that F ( λ k ) = A k , ≤ k ≤ n ; (2) There exists an analytic function ϕ : D → E such that ϕ ( λ k ) = π ( A k ) , that is, ϕ ( λ k ) = ( a k , a k , det( A k )) , k = 1 , , ..., n. In the following theorem the authors give a necessary and sufficient condition for thesolvability of a µ Diag -synthesis problem by a rational E -inner function. Theorem 2.8. [10, Theorem 1.1 and Theorem 8.1]
Let λ , .., λ n be distinct points in D and let A k = [ a kij ] ∈ C × be such that a k a k = det( A k ) , ≤ k ≤ n . Let ( x k , x k , x k ) = (cid:0) a k , a k , det( A k ) (cid:1) , ≤ k ≤ n. The following two conditions are equivalent. (1)
There exists an analytic × matrix function F in D such that F ( λ k ) = A k for k = 1 , .., n, and µ Diag ( F ( λ )) ≤ for all λ ∈ D ;(2) there exists a rational E -inner function x : D → E such that x ( λ k ) = ( x k , x k , x k ) for k = 1 , .., n. Therefore, the understanding of rational E -inner functions will be useful for such µ -synthesis problems. OMAR M. O. ALSALHI AND ZINAIDA A. LYKOVA
The distinguished boundary of the tetrablock.Theorem 2.9. [1, Theorem 2.9] E is polynomially convex. Therefore, there exists a distinguished boundary b E of E . Let A ( E ) be the algebra ofcontinuous scalar functions on E that are holomorphic on E endowed with the supremumnorm. If there is a function f ∈ A ( E ) and a point p in E such that f ( p ) = 1 and | f ( x ) | < x ∈ E \{ p } , then p ∈ b E and is called a peak point of E and the function f iscalled a peaking function for p . Theorem 2.10. [1, Theorem 7.1]
For x ∈ C the following are equivalent. (1) x = x x , | x | = 1 and | x | ≤ ; (2) either x x = x and Ψ( ., x ) is an automorphism of D or x x = x and | x | = | x | = | x | = 1 ; (3) x is a peak point of E ; (4) there exists a × unitary matrix U such that x = π ( U ) ; (5) there exists a symmetric × unitary matrix U such that x = π ( U ) ; (6) x ∈ b E ; (7) x ∈ E and | x | = 1 . Lemma 2.11.
Let x = ( x , x , x ) ∈ C . Then x ∈ b E if and only if x = x x , | x | = 1 and | x | ≤ . Proof.
By Theorem 2.10 (1), x ∈ b E ⇔ x = x x , | x | = 1 and | x | ≤ . Since | x | = 1 this implies x x = 1. Now, since x ∈ b E , x = x x , and so x = x x . Thus x x = x x x = x . Note, by Theorem 2.5, | x | ≤ x = x x , | x | = 1 and | x | ≤ x ∈ b E . Therefore, x ∈ b E if and only if x = x x , | x | = 1 and | x | ≤ . (cid:3) The symmetrised bidisc and Γ -inner functions In Section 4.1 we show that there exist useful relations between Γ-inner functions and E -inner functions. Recall the definition of the symmetrised bidisc Γ. Definition 3.1.
The symmetrised bidisc is the set G def = (cid:8) ( z + w, zw ) : | z | < , | w | < (cid:9) , and its closure is Γ def = (cid:8) ( z + w, zw ) : | z | ≤ , | w | ≤ (cid:9) . In 1995 Jim Agler and Nicholas Young started the study of the symmetrised bidiscwith the aim of solving a robust control problem in H ∞ control theory. Although,the aim has not yet been achieved, it turned out that the symmetrised bidisc has a richstructure and it has attracted the attention of specialists in the several complex variablesand in operator theory. ATIONAL E -INNER FUNCTIONS 7 We shall often use the co-ordinates ( s, p ) for points in the symmetrized bidisc G ,chosen to suggest ‘sum’ and ‘product’. The following results afford useful criteria formembership of G , of the distinguished boundary b Γ of Γ and of the topological boundary ∂ Γ of Γ [4].
Proposition 3.2. [4, Proposition 3.2]
Let ( s, p ) belong to C . Then (1) ( s, p ) belongs to G if and only if | s − sp | < − | p | ;(2) ( s, p ) belongs to Γ if and only if | s | ≤ and | s − sp | ≤ − | p | ;(3) ( s, p ) lies in b Γ if and only if | p | = 1 , | s | ≤ and s − sp = 0;(4) ( s, p ) ∈ ∂ Γ if and only if | s | ≤ and | s − sp | = 1 − | p | . Γ-inner functions were defined and studied in [4].
Definition 3.3. A Γ-inner function is an analytic function h : D → Γ such that theradial limit (3.1) lim r → − h ( rλ ) exists and belongs to b Γ for almost all λ ∈ T with respect to Lebesgue measure. By Fatou’s Theorem, the limit (3.1) exists for almost all λ ∈ T . Definition 3.4.
Let f be a polynomial of degree less than or equal to n , where n ≥ .Then we define the polynomial f ∼ n by f ∼ n ( λ ) = λ n f (1 /λ ) . The polynomial f ∨ is defined by f ∨ ( λ ) = f (cid:0) λ (cid:1) . Remark 3.5.
One can see that(1) f ∼ n ( λ ) = λ n f ( /λ ) = λ n f ∨ (1 /λ ) . (2) If f is a polynomial of degree k , then, for n ≥ k , (cid:0) f ∼ n (cid:1) ∼ n ( λ ) = f ( λ ).Algebraic and geometric aspects of rational Γ-inner functions were studied in [6]. Weare going to use some results from the paper. Proposition 3.6. [6, Proposition 2.2] If h = ( s, p ) is a rational Γ -inner function ofdegree n then there exist polynomials E and D such that (1) deg( E ) , deg( D ) ≤ n , (2) E ∼ n = E , (3) D ( λ ) = 0 on D , (4) | E ( λ ) | ≤ | D ( λ ) | on D , (5) s = ED on D , (6) p = D ∼ n D on D . OMAR M. O. ALSALHI AND ZINAIDA A. LYKOVA
Furthermore, E and D is a second pair of polynomials that satisfy (1)–(6) if and onlyif there exists a nonzero t ∈ R such that E = tE and D = tD. Conversely, if E and D are polynomials that satisfy (1), (2), (4) , D ( λ ) = 0 on D , and s and p are defined by (5) and (6) , then h = ( s, p ) is a rational Γ -inner function of degreeless than or equal to n . The royal variety R Γ of the symmetrised bidisc is R Γ = { ( s, p ) ∈ C : s = 4 p } . Definition 3.7. [6, Page 7]
Let h = ( s, p ) be a Γ -inner function of degree n . Let E and D be as in Proposition . The royal polynomial R h of h is defined by R h ( λ ) = 4 D ( λ ) D ∼ n ( λ ) − E ( λ ) . Definition 3.8. [6, Definition 3.6]
Let h be a rational Γ -inner function such that h ( D ) * R Γ ∩ Γ . Let R h be the royal polynomial of h . If σ is a zero of R h of order ℓ , wedefine the multiplicity σ of σ (as a royal node of h ) by σ = ( ℓ if σ ∈ D , ℓ if σ ∈ T . We define the type of h to be the ordered pair ( n, k ) , where n is the sum of the multiplic-ities of the royal nodes of h that lie in D , and k is the sum of the multiplicities of theroyal nodes of h that lie in T . We define R n,k Γ to be the collection of rational Γ -innerfunctions h of type ( n, k ) . Theorem 3.9. [6, Theorem 3.8] If h ∈ R n,k Γ is nonconstant then deg( h ) = n . Rational E -inner functions In this section we give a definition of the degree of a rational tetra-inner function x by means of the fundamental group π . Recall that the rational inner functions on D of degree n are exactly the finite Blaschke products of degree n . As an analogue of thisdescription of rational inner functions on D we describe all rational E -inner functionson D in Theorem 4.15. In [6], the authors describe all rational Γ-inner functions (seeProposition 3.6). We use this description and the connection between Γ-inner functionsand E -inner functions to describe all rational E -inner functions on D .4.1. Relations between E -inner functions and Γ -inner functions.Definition 4.1. An E -inner or tetra-inner function is a map f : D → E that is analyticand is such that the radial limit lim r → − f ( rλ ) exists and belongs to b E for almost all λ ∈ T with respect to Lebesgue measure. Remark 4.2.
Let x : D → E be a rational E -inner function. Since x is rational andbounded on D it has no poles in D and hence x is continuous on D . Thus one canconsider the continuous function˜ x : T → b E , where ˜ x ( λ ) = lim r → − x ( rλ ) for all λ ∈ T . In future we will use the same notation x for both continuous functions x and ˜ x . ATIONAL E -INNER FUNCTIONS 9 Lemma 4.3.
Let x = ( x , x , x ) be an E -inner function. Then (1) x ( λ ) = x ( λ ) x ( λ ) , | x ( λ ) | ≤ and | x ( λ ) | = 1 for almost all λ ∈ T ; (2) x is an inner function on D .Proof. (1) By the definition of E -inner function x ( λ ) = ( x ( λ ) , x ( λ ) , x ( λ )) ∈ b E , for almost every λ ∈ T and, by Theorem 2.10, x ( λ ) = x ( λ ) x ( λ ) , | x ( λ ) | = 1 and | x ( λ ) | ≤ λ ∈ T . (2) Since x : D → D and , for almost all λ ∈ T , | x ( λ ) | = 1 ,x is an inner function. (cid:3) Remark 4.4.
Let x = ( x , x , x ) be a rational E -inner function. By Lemma 4.3, x isan inner function on D , and so x is a finite Blaschke product.In [9] the author shows that there is a relation between points in the symmetrisedbidisc and the tetrablock as follows Lemma 4.5. [9, Lemma 3.2]
A point x = ( x , x , x ) ∈ C belongs to the tetrablock ifand only if the pair ( x + zx , zx ) is in the symmetrised bidisc G for every z ∈ T .Proof. By Proposition 3.2 (1), ( s, p ) ∈ G if and only if(4.1) | s − sp | < − | p | . Suppose that x = ( x , x , x ) ∈ E , s z = x + zx and p z = zx . | s z − s z p z | = | x + zx − ( x + zx ) zx | = | x − x x + z ( x − x x ) |≤ | x − x x | + | x − x x | , since | z | = 1 ,< − | x | = 1 − | p z | , by Theorem 2.4 . Hence ( s z , p z ) ∈ G .Conversely, let x = ( x , x , x ) ∈ C and, for z ∈ T , let(4.2) s z = x + zx and p z = zx . Suppose for all z ∈ T , we have ( s z , p z ) ∈ G . We want to show that x = ( x , x , x ) ∈ E .Let us prove that | x − x x | + | x − x x | < − | x | . By assumption for all z ∈ T , | s z − s z p z | < − | x | . By equations (4.2), we have(4.3) | x − x x + z (cid:0) x − x x (cid:1) | < − | x | , for all z ∈ T . Let z = e iθ θ ∈ (0 , π ]; w = x − x x = | w | e iθ θ ∈ (0 , π ]; w = x − x x = | w | e iθ θ ∈ (0 , π ] . Now substitute z , w and w in inequality (4.3) (cid:12)(cid:12) | w | e iθ + e iθ ( | w | e iθ ) (cid:12)(cid:12) < − | x | . This implies that (cid:12)(cid:12) | w | e iθ + | w | e i ( θ + θ ) (cid:12)(cid:12) < − | x | , for all e iθ . We can choose θ such that θ + θ = θ , that is, θ = θ − θ . Hence (cid:12)(cid:12) | w | e iθ + | w | e iθ (cid:12)(cid:12) = | e iθ | ( | w | + | w | ) = | w | + | w | = | x − x x | + | x − x x | < −| x | . By Theorem 2.4, ( x , x , x ) ∈ E . (cid:3) Lemma 4.6.
A point x = ( x , x , x ) ∈ C belongs to the closed tetrablock if and onlyif for every a ∈ D , ( ax + ax , x ) ∈ Γ .Proof. Suppose x = ( x , x , x ) ∈ E . Consider ( s a , p a ) = ( ax + ax , x ). By Proposition3.2 (2), ( s, p ) ∈ Γ if and only if | s − sp | ≤ − | p | and | s | ≤ . (4.4) | s a − s a p a | = | ax + ax − ( ax + ax ) x | = | a ( x − x x ) + a ( x − x x ) |≤ | a ( x − x x ) | + | a ( x − x x ) | , ≤ | x − x x | + | x − x x | , since | a | ≤ , ≤ − | x | , by Theorem 2.5 . (4.5)Thus, | s a − s a p a | ≤ − | x | = 1 − | p a | and | s a | = | ax + ax | ≤ . Hence ( s a , p a ) ∈ Γ.Conversely, let x = ( x , x , x ) ∈ C . Suppose for every a ∈ D , we have ( s a , p a ) ∈ Γwhere(4.6) s a = ax + ax and p a = x . By equations (4.5) and (4.6), we have(4.7) | s a − s a p a | = | a ( x − x x ) + a (cid:0) x − x x (cid:1) | ≤ − | p a | , for all a ∈ D . Take a ∈ T , then a = e iθ θ ∈ (0 , π ]; w = x − x x = | w | e iθ θ ∈ (0 , π ]; w = x − x x = | w | e iθ θ ∈ (0 , π ] . Substitute a , w and w into inequality (4.7), to get | a ( x − x x ) + a (cid:0) x − x x (cid:1) | = (cid:12)(cid:12) e iθ | w | e iθ + e − iθ | w | e iθ (cid:12)(cid:12) = (cid:12)(cid:12) | w | e i ( θ + θ ) + | w | e i ( θ − θ ) (cid:12)(cid:12) ≤ − | x | , for every θ ∈ (0 , π ]. Now choose θ = θ − θ | x − x x | + | x − x x | = | w | + | w | = | e i ( θ θ ) | (cid:0) | w | + | w | (cid:1) = (cid:12)(cid:12) | w | e i ( θ θ ) + | w | e i ( θ θ ) (cid:12)(cid:12) = (cid:12)(cid:12) | w | e i ( θ − θ + θ ) + | w | e i ( θ − θ − θ ) (cid:12)(cid:12) ≤ − | x | . Therefore x = ( x , x , x ) ∈ E . (cid:3) Lemma 4.7.
Let s, p ∈ C be such that | s | ≤ and | p | ≤ . The pair ( s, p ) belongs to Γ if and only if (cid:0) s, s, p (cid:1) ∈ E . ATIONAL E -INNER FUNCTIONS 11 Proof.
By Theorem 2.5, (cid:0) s, s, p (cid:1) ∈ E ⇔ | s − sp | + | s − sp | ≤ − | p | . Thus ( s, s, p ) ∈ E ⇔ | s − sp | ≤ − | p | ⇔ | s − sp | ≤ − | p | . By assumption | s | ≤
2, hence by Proposition 3.2 (2),( s, s, p ) ∈ E ⇔ ( s, p ) ∈ Γ . (cid:3) Lemma 4.8.
Let x = ( x , x , x ) be a rational E -inner function. Then (1) h ( λ ) = (cid:0) x ( λ ) + x ( λ ) , x ( λ ) (cid:1) , for λ ∈ D , is a rational Γ -inner function; (2) h ( λ ) = ( ix ( λ ) − ix ( λ ) , x ( λ )) , for λ ∈ D , is a rational Γ -inner function.Proof. (1) By Lemma 4.5, for all λ ∈ D , x ( λ ) ∈ E implies that (cid:0) x ( λ ) + x ( λ ) , x ( λ ) (cid:1) ∈ G . Consider h = ( s , p ) where s ( λ ) = x ( λ ) + x ( λ ) and p ( λ ) = x ( λ ) , for λ ∈ D . It is obvious that h is a rational function from D to G . By assumption, x is an E -innerfunction. Thus x ( λ ) ∈ b E for almost every λ ∈ T . By Theorem 2.10 and Lemma 2.11,for almost all λ ∈ T ,(4.8) x ( λ ) = x ( λ ) x ( λ ) , x ( λ ) = x ( λ ) x ( λ ) , | x ( λ ) | = 1 and | x ( λ ) | ≤ . It is clear that | p ( λ ) | = | x ( λ ) | = 1 for λ ∈ T , and, for almost all λ ∈ T , | s ( λ ) | = | x ( λ ) + x ( λ ) |≤ | x ( λ ) | + | x ( λ ) | ≤ . Since, for almost all λ ∈ T , x ( λ ) = x ( λ ) x ( λ ), we have s ( λ ) p ( λ ) = [ x ( λ ) + x ( λ )] x ( λ )= x ( λ ) x ( λ ) + x ( λ ) x ( λ ) , by equations (4.8),= x ( λ ) + x ( λ ) = s ( λ ) . Hence s ( λ ) = s ( λ ) p ( λ ) for almost every λ ∈ T . Therefore, by Proposition 3.2 (3), h is a rational Γ-inner function.(2) Following the same steps as (1), let h ( λ ) = ( s ( λ ) , p ( λ )) , where s ( λ ) = ix ( λ ) − ix ( λ ) and p ( λ ) = x ( λ ) , λ ∈ D . By Lemma 4.6, h is rational function from D to G . Since x is an E -inner function, x ( λ ) ∈ b E for almost all λ ∈ T . By Proposition 3.2, to prove that h is a rationalΓ-inner function we need to show that | p ( λ ) | = 1 , | s ( λ ) | ≤ s ( λ ) = s ( λ ) p ( λ ) for almost every λ ∈ T . By Theorem 2.10, for almost all λ ∈ T , | p ( λ ) | = | x ( λ ) | = 1 and | s ( λ ) | ≤ | ix ( λ ) | + | ix ( λ ) | ≤ . By Lemma 2.11, x ( λ ) = x ( λ ) x ( λ ) for almost all λ ∈ T . Hence, for almost all λ ∈ T , s ( λ ) p ( λ ) = [ ix ( λ ) − ix ( λ )] x ( λ )= i (cid:0) x ( λ ) (cid:1) x ( λ ) − i (cid:0) x ( λ ) (cid:1) x ( λ ) , by equations (4.8) , = ix ( λ ) − ix ( λ ) = s ( λ ) . Hence s ( λ ) = s ( λ ) p ( λ ) for almost every λ ∈ T . Therefore, by Proposition 3.2, h is arational Γ-inner function. (cid:3) Lemma 4.9.
Let x = ( x , x , x ) be a rational E -inner function. Then x ( λ ) = x ∨ (1 /λ ) x ( λ ) for all λ ∈ C \ { } . Proof.
By Theorem 2.10, for all λ ∈ T , x ( λ ) = x ( λ ) x ( λ ) . For λ ∈ T , we have | λ | = 1, that is, λλ = 1, and so x ( λ ) = x ∨ ( λ ) = x ∨ ( λ ) . Therefore, for all λ ∈ T , x ( λ ) = x ∨ (1 /λ ) x ( λ ) . Since x , x , x are rational functions, x ( λ ) = x ∨ (1 /λ ) x ( λ ) for all λ ∈ C \ { } . (cid:3) Proposition 4.10.
Let x = ( x , x , x ) be a rational E -inner function (1) If a ∈ C ∪ {∞} is a pole of x of multiplicity k ≥ and a is a zero of x ofmultiplicity ℓ ≥ , then a is a pole of x of multiplicity at least k − ℓ . (2) If a ∈ C ∪ {∞} is a pole of x of multiplicity k ≥ , then a is a pole of x ofmultiplicity at least k .Proof. (1) By Lemma 4.9, we have(4.9) x ( λ ) = x ∨ (1 /λ ) x ( λ ) for λ ∈ C \ { } . Since x is a rational inner function, x cannot have any pole in D . Hence | a | > | a | <
1. We know that x ∨ is analytic in D , so a cannot be a pole of x ∨ . By equation(4.9), ( λ − a ) k − ℓ − x ( λ ) = ( λ − a ) k − ℓ − x ∨ (1 /λ ) x ( λ ) . Take the limit for both sides as λ goes to a :lim λ → a ( λ − a ) k − ℓ − x ( λ ) = lim λ → a ( λ − a ) k − ℓ − x ∨ (1 /λ ) x ( λ ) . The right hand side goes to ∞ , therefore x has a pole of multiplicity at least k − l at a .Now suppose that ∞ is a pole of x of multiplicity k and 0 is a zero of x of multiplicity ℓ . By equation (4.9), for all λ ∈ C \{ } , we have x ( λ ) = x ∨ ( λ ) x ( λ ) . Multiply both sides by λ k − λ ℓ to obtain the equation(4.10) λ k − λ ℓ x ( λ ) = λ k − λ ℓ x ∨ ( λ ) x ( λ ) . ATIONAL E -INNER FUNCTIONS 13 Since x ∨ is analytic at 0 and has a zero of multiplicity ℓ > λ → x ∨ ( λ ) λ ℓ = c, where c ∈ C \{ } . Since by assumption, x ( λ ) has a pole of multiplicity k at ∞ ,lim λ → λ k − x ( λ ) = ∞ . Hence by equation (4.10), lim λ → λ k − ℓ − x ( λ ) = ∞ . It follows that x ( λ ) has a pole of multiplicity at least k − ℓ at 0. That is, x ( λ ) has apole of multiplicity at least k − ℓ at ∞ .(2) Let a ∈ C be a pole of x of multiplicity k ≥
1. Then | a | >
1. This implies | a | <
1. Therefore x ∨ is analytic at a . Nowlim λ → a ( λ − a ) k − x ( λ ) = ∞ . Thus a is a pole of x of multiplicity at least k .If ∞ is a pole of x of multiplicity k ≥
1. Then 0 is a pole of x ( λ ) of multiplicity k ,that is, lim λ → λ k − x ( λ ) = ∞ . By relation (4.10), λ k − x ( λ ) = λ k − x ∨ ( λ ) x ( λ ) . Since x ∨ is analytic at 0, 0 cannot be a pole of x ∨ and thuslim λ → x ∨ ( λ ) = x ∨ (0) . Therefore lim λ → λ k − x ( λ ) = ∞ This completes the proof that x has a pole of multiplicity at least k at ∞ . (cid:3) The degree of a rational E -inner function. Let us clarify the notion of thedegree of a rational E -inner function x . Definition 4.11.
The degree deg( x ) of a rational E -inner function x is defined to be x ∗ (1) , where x ∗ : Z = π ( T ) → π ( b E ) is the homomorphism of fundamental groupsinduced by x when x is regarded as a continuous map from T to b E . We shall adopt the convention that deg( x ) is a non-negative integer. Lemma 4.12. b E is homotopic to T and π ( b E ) = Z .Proof. The maps f : b E → T , defined by f ( x , x , x ) = x ,g : T → b E , defined by g ( z ) = (0 , , z ) , satisfy ( g ◦ f )( x , x , x ) = g (cid:0) f ( x , x , x ) (cid:1) = g ( x ) = (0 , , x )and ( f ◦ g )( z ) = f (0 , , z ) = z, that is, f ◦ g = id T . If ( x , x , x ) ∈ b E and 0 ≤ t ≤
1, then ( tx , tx , x ) ∈ b E . Let I = [0 , h : b E × I → b E , which is defined by h ( x , x , x , t ) = ( tx , tx , x ) . One can see that h ( x , x , x ,
0) = (0 x , x , x ) = (0 , , x ) = ( g ◦ f )( x , x , x ) and h ( x , x , x ,
1) = (1 x , x , x ) = ( x , x , x ) = id b E . Therefore h defines a homotopy between g ◦ f and id b E , that is, g ◦ f ≃ id b E . Hence b E is homotopically equivalent to T and it follows that π ( b E ) = π ( T ) = Z . (cid:3) Lemma 4.13.
Let B be a finite Blaschke product. Then the degree of B is equal to B ∗ (1) .Proof. Since B is a finite Blaschke product, it can be written as B ( λ ) = e iθ N Y j =1 λ − α j − α j λ , where α j ∈ D , j = 1 . . . N, and θ ∈ [0 , π ) . One can consider the map, B : T → T , and B ∗ : π ( T ) = Z → π ( T ) = Z . Now 1 ∈ π ( T ) is the homotopy class of id T and B ∗ (1) is equal to the homotopy classof B ◦ id T = B, when B is regarded as a continuous map from T to T . Therefore B ∗ (1) = n ( γ, a ), where n ( γ, a ) is the winding number of γ about a , which lies inside γ = { B ( e it ) : 0 ≤ t ≤ π } . Thus, one can see that n ( γ, a ) = 12 πi Z γ dzz − a = 12 πi Z T B ′ ( z ) dzB ( z ) − a . By the Argument Principle, [7, Theorem 18], the integral12 πi Z T B ′ ( z ) B ( z ) − a dz, is equal to the number of zeros of B in D . It is clear that B has N zeros, countingmultiplicities, and has degree N . Therefore the number of zeros of B is equal to thewinding number of γ about a , and it is equal to N . (cid:3) Proposition 4.14.
For any rational E -inner function x = ( x , x , x ) , deg( x ) is thedegree deg( x ) ( in the usual sense ) of the finite Blaschke product x .Proof. Since x is a rational E -inner function, x is an inner function, and so x is afinite Blaschke product. Two E -inner functions x = ( x , x , x ) and y = (0 , , x ) arehomotopic if there exists a continuous mapping f : T × I → b E such that f ( λ,
0) = y ( λ ) and f ( λ,
1) = x ( λ ) , λ ∈ T . Let x t ( λ ) = (cid:0) tx ( λ ) , tx ( λ ) , x ( λ ) (cid:1) for λ ∈ D and t ∈ [0 , . ATIONAL E -INNER FUNCTIONS 15 Since x ( λ ) ∈ b E , for all λ ∈ T , by Theorem 2.10 (1), x ( λ ) = x ( λ ) x ( λ ) and | x ( λ ) | = 1 . Hence for all λ ∈ T , tx ( λ ) = tx ( λ ) x ( λ ) . Therefore, x t ( λ ) = (cid:0) tx ( λ ) , tx ( λ ) , x ( λ ) (cid:1) ∈ b E for λ ∈ T . Hence x t is a homotopy between x = x and (0 , , x ) = x .It follows that the homomorphism x ∗ : π ( T ) = Z → π ( b E ) = Z coincides with ( x ) ∗ = (0 , , x ) ∗ . By Lemma 4.13, ( x ) ∗ (1) = deg x , since x is a finiteBlaschke product. Therefore (0 , , x ) ∗ (1) is the degree of the finite Blaschke product x . (cid:3) Description of rational E -inner functions.Theorem 4.15. If x = ( x , x , x ) is a rational E -inner function of degree n then thereexist polynomials E , E , D such that (1) deg( E ) , deg( E ) , deg( D ) ≤ n , (2) D ( λ ) = 0 on D , (3) x = D ∼ n D on D , (4) x = E D on D , (5) x = E D on D , (6) | E i ( λ ) | ≤ | D ( λ ) | on D , for i = 1 , , (7) E ( λ ) = E ∼ n ( λ ) , for λ ∈ D .Conversely, if E , E and D satisfy (1), (6) and (7) , D ( λ ) = 0 on D and x , x and x are defined by (3)–(5) , then x = ( x , x , x ) is a rational E -inner function of degreeat most n .Furthermore, a triple of polynomials E , E and D satisfies (1)–(7) if and only if thereexists a real number t = 0 such that E = tE , E = tE and D = tD. Proof.
By assumption x = (cid:0) x , x , x (cid:1) is a rational E -inner function. By Lemma 4.8 (1), h = ( s, p ) where s = x + x , p = x is a rational Γ-inner function. Since x : D → D is an inner function, it is a finite Blaschke product and, by [4, Corollary 6.10], it can bewritten in the form x ( λ ) = c λ k D ∼ ( n − k ) ( λ ) D ( λ ) , where | c | = 1 , ≤ k ≤ n and D is a polynomial of degree n − k such that D (0) = 1. ByProposition 3.6, there exist polynomials E, D such that(1) deg( E ) , deg( D ) ≤ n ,(2) E ∼ n = E ,(3) D ( λ ) = 0 on D ,(4) | E ( λ ) | ≤ | D ( λ ) | on D ,(5) s = ED on D ,(6) p = D ∼ n D on D . Hence(4.11) x + x = s = ED and x = p = D ∼ n D .
By Lemma 4.8 (2), h = ( s , p ), where s = ix − ix , p = x = p is a rational Γ-innerfunction. By Proposition 3.6, for h = ( s , p ), there exist polynomials G, D such that(1) deg( G ) , deg( D ) ≤ n ,(2) G ∼ n = G ,(3) D ( λ ) = 0 on D ,(4) | G ( λ ) | ≤ | D ( λ ) | on D ,(5) s = ix − ix = GD on D ,(6) p = x = D ∼ n D on D .Therefore, by (5),(4.12) x − x = − iGD . By relation (4.11),(4.13) x + x = ED .
Add equations (4.12) and (4.13) to get x = ( E − iG ) D .
Substituting x in equation (4.13) gives x = ( E + iG ) D .
Define the polynomials E and E by E = 12 ( E − iG ) , E = 12 ( E + iG ) . Since the degrees of both polynomials
E, G are at most n , deg( E ) , deg( E ) ≤ n . Thus,for λ ∈ D , x ( λ ) = E ( λ ) D ( λ ) and x ( λ ) = E ( λ ) D ( λ ) . Since x is an E -inner function, for λ ∈ D , | x ( λ ) | ≤ | x ( λ ) | ≤ , and so | E ( λ ) | ≤ | D ( λ ) | and | E ( λ ) | ≤ | D ( λ ) | . Hence | E i ( λ ) | ≤ | D ( λ ) | on D , where i = 1 ,
2. Therefore (1)–(6) of Theorem 4.15 aresatisfied.
ATIONAL E -INNER FUNCTIONS 17 By assumption, x is a rational E -inner function. Thus, for all λ ∈ T , x ( λ ) = x ( λ ) x ( λ ) ⇔ E ( λ ) D ( λ ) = E ( λ ) D ( λ ) × D ∼ n ( λ ) D ( λ ) ⇔ E ( λ ) D ( λ ) = E ( λ ) D ∨ (1 /λ ) × λ n D ∨ (1 /λ ) D ( λ ) , since D ( λ ) = D ∨ ( λ ) = D ∨ (1 /λ ) . ⇔ E ( λ ) D ( λ ) = λ n E ( λ ) D ( λ ) ⇔ E ( λ ) D ( λ ) = E ∼ n ( λ ) D ( λ ) ⇔ E ( λ ) = E ∼ n ( λ ) . (4.14)Hence E ( λ ) = E ∼ n ( λ ) for all λ ∈ T , and therefore on D . Thus (7) is proved. Let us prove the converse statement . Let E , E and D satisfy (1), (6) and (7)of Theorem 4.15 and D ( λ ) = 0 on D , and x , x , x be defined by (3)–(5), that is, x = E D , x = E D and x = D ∼ n D .
Let us show that x = ( x , x , x ) is a rational E -inner function. By Theorem 2.10, wehave to prove that x : D → E and the following conditions are satisfied.(1) | x ( λ ) | = 1 for almost all λ on T , that is, x is inner,(2) | x | ≤ D ,(3) x ( λ ) = x ( λ ) x ( λ ) for almost all λ ∈ T .(1) Firstly, if D has no zeros on the unit circle, then D and D ∼ n have no common factor.Therefore, x ( λ ) = D ∼ n ( λ ) D ( λ ) maps T to T . Hence, x is inner function anddeg( x ) = deg D ∼ n D ! = max { deg( D ∼ n ) , deg( D ) } = n. Second case: if D has the zeros a , ..., a ℓ on T then D and D ∼ n have the common factor Q ℓi =1 ( λ − a i ) and hence x = D ∼ n D is inner anddeg( x ) = deg D ∼ n D ! ≤ n − ℓ. (2) By assumption (6), | E ( λ ) | ≤ | D ( λ ) | for all λ ∈ D . This implies | E ( λ ) D ( λ ) | ≤ | x ( λ ) | ≤ E ( λ ) = E ∼ n ( λ ), for almost all λ ∈ T and by equality (4.14), x ( λ ) = x ( λ ) x ( λ ) , for almost all λ ∈ T . Let us show that x = ( x , x , x ) = (cid:18) E D , E D , D ∼ n D (cid:19) maps D to E , that is, x ( λ ) = ( x ( λ ) , x ( λ ) , x ( λ )) ∈ E for all λ ∈ D . By Theorem 2.5, for λ ∈ D , x ( λ ) ∈ E ⇔ k Ψ( ., x ( λ )) k H ∞ ≤ , where Ψ( z, x ) = x z − x x z − z ∈ D ,Ψ( z, x ) : D → C : λ → Ψ( z, x ( λ ))is analytic on D because x i , i = 1 , ,
3, are analytic functions on D , and | x ( λ ) | ≤ x ( λ ) z = 1 for all λ ∈ D . We have shown above that, for almost all λ ∈ T , x ( λ ) ∈ b E .Thus, by Theorem 2.10 (2), x ( λ ) ∈ b E if and only if Ψ( ., x ( λ )) is an automorphism of D . By the maximum principle, for all z, λ ∈ D , | Ψ( z, x ( λ )) | <
1. Thus by Theorem 2.10, x ( λ ) ∈ E for all λ ∈ D .Suppose that t is a nonzero real number and E = tE , E = tE and D = tD. Then it is clear that E , E and D satisfy (1)–(7). Conversely, let E , E and D be asecond triple that satisfies (1)–(7). Then(4.15) x = E D = E D on D , (4.16) x = E D = E D on D , (4.17) x = D ∼ n D = D ∼ n D on D . Suppose that D ( λ ) = a + a λ + ... + a k λ k where a = 0 and k ≤ n . Then D ∼ n ( λ ) = λ n D (cid:0) /λ (cid:1) = λ n (cid:18) a + a λ + ... + a k λ (cid:19) = a λ n + a λ n − + ... + a k λ n − k . Thus, for all λ ∈ D , x = D ∼ n ( λ ) D ( λ ) = λ n − k (cid:0) a λ k + a λ k − + ... + a k (cid:1) a + a λ + ... + a k λ k . Therefore, x has a zero of multiplicity ( n − k ) at 0 , has k poles in C , counting multiplic-ity, and has degree n . Hence the poles of x in { z ∈ C : | z | > } , n and k are determinedby x . Therefore, D and D have the same degree k and the same finite number of zerosin { z ∈ C : | z | > } , counting multiplicity. Thus there exists t ∈ C , t = 0 where(4.18) D = tD on D . ATIONAL E -INNER FUNCTIONS 19 By equality (4.17), for λ ∈ D x = D ∼ n D = D ∼ n D = tD ∼ n tD Thus t = t , and so, t ∈ R \{ } . By the equalities (4.15) and (4.18) x = E D = E D = E tD , on D . This implies that E = tE . By the equalities (4.16) and (4.18) x = E D = E D = E tD , on D . Thus E = tE . (cid:3) Remark 4.16.
For a fixed polynomial D of degree n , the set of polynomials E satisfyingthe conditions of Theorem 4.15 is a subset of a real vector space of dimension 2 n + 2.Hence the set of rational E -inner functions of degree n with x = D ∼ n D is a subset of a(2 n + 2)-dimensional real space of rational functions. Lemma 4.17.
Let x = ( x , x , x ) = (cid:18) E D , E D , D ∼ n D (cid:19) be a rational E -inner function. Then, for λ ∈ T , | E ( λ ) | = | E ( λ ) | , and so | x ( λ ) | = | x ( λ ) | . Proof.
By Theorem 4.15 (7), for all λ ∈ T , E ( λ ) = E ∼ n ( λ ) = λ n E (1 /λ ) . Thus, since λλ = 1, | E ( λ ) | = | λ n E (1 /λ ) | = | E (1 /λ ) | = | E ( λ ) | . Therefore, for all λ ∈ T , | x ( λ ) | = | x ( λ ) | . (cid:3) Example 4.18.
Let x = ( x , x , x ) be a rational E -inner function such that x ( λ ) = λ .Clearly, deg( x ) = deg( x ) = 1 . By Theorem 4.15, there exist polynomials E , E , D such that D ( λ ) = 1 , deg( E ) ≤ , deg( E ) ≤ ,E ( λ ) = E ∼ n ( λ ) , | E i ( λ ) | ≤ | D ( λ ) | = 1 , i = 1 , , for all λ ∈ D , and x = (cid:18) E D , E D , D ∼ D (cid:19) . Therefore the function x ( λ ) = (cid:0) a + a λ, a + a λ, λ (cid:1) is rational E -inner for a , a ∈ D such that | a + a λ | ≤ | a + a λ | ≤ λ ∈ D . In particular, one can choose a = 1 and a = 0 to get the rational E -inner function x ( λ ) = (1 , λ, λ ) . Example 4.19. E -inner functions Suppose that B × = { A ∈ C × : k A k < } . Let us construct an analytic map from theopen unit disc D to B × . Consider nonconstant inner functions ϕ, ψ ∈ H ∞ ( D ) and thediagonal matrix h ( λ ) = (cid:20) ϕ ( λ ) 00 ψ ( λ ) (cid:21) for λ ∈ D . Note k h ( λ ) k = max {| ϕ ( λ ) | , | ψ ( λ ) |} < λ ∈ D and h : D → B × is analytic.By Theorem 2.4, for all λ ∈ D , (cid:0) ϕ ( λ ) , ψ ( λ ) , det h ( λ ) (cid:1) ∈ E , and ϕ ( λ ) ψ ( λ ) = det h ( λ ). Recall that such points are called triangular points of E .However, we are seeking more interesting and general examples. To get such exampleswe make use of the singular value decomposition.Let U, V be 2 × h : D → C × defined by h = U hV maps D to B × . For example, if U = √ √ − √ √ and V = I = (cid:20) (cid:21) then U is unitary and we obtain h ( λ ) = U h ( λ ) I = 1 √ (cid:20) − (cid:21) (cid:20) ϕ ( λ ) 00 ψ ( λ ) (cid:21) (cid:20) (cid:21) = 1 √ (cid:20) ϕ ( λ ) ψ ( λ ) − ϕ ( λ ) ψ ( λ ) (cid:21) . Define x : D → E by x = π ◦ h , where π is given in Definition 2.6. Then, for λ ∈ D ,x ( λ ) = π (cid:0) h ( λ ) (cid:1) = π (cid:18) √ (cid:20) ϕ ( λ ) ψ ( λ ) − ϕ ( λ ) ψ ( λ ) (cid:21) (cid:19) = (cid:18) ϕ ( λ ) √ , ψ ( λ ) √ , ϕ ( λ ) ψ ( λ ) (cid:19) . Note that this x ( λ ) is not a triangular point unless either ϕ ( λ ) = 0 or ψ ( λ ) = 0 . ATIONAL E -INNER FUNCTIONS 21 Let us show that this function x is E -inner. By Theorem 2.10(1), for λ ∈ T , since ϕ, ψ are inner functions, x ( λ ) x ( λ ) = (cid:18) ψ ( λ ) √ (cid:19) ϕ ( λ ) ψ ( λ ) = ϕ ( λ ) √ ψ ( λ ) ψ ( λ )= ϕ ( λ ) √ | ψ ( λ ) | = ϕ ( λ ) √ x ( λ ) . Since | ψ ( λ ) | < λ ∈ D , this implies that (cid:12)(cid:12)(cid:12) ψ ( λ ) √ (cid:12)(cid:12)(cid:12) <
1. Thus | x ( λ ) | <
1. Finally, for λ ∈ T , since ϕ, ψ are inner functions, | x ( λ ) | = (cid:12)(cid:12) ϕ ( λ ) ψ ( λ ) (cid:12)(cid:12) = | ϕ ( λ ) || ψ ( λ ) | = 1 . Therefore x is an E -inner function. Remark 4.20.
In the previous example if we choose the functions ϕ and ψ to be in theSchur class but not to be inner functions then one can check that we obtain an analyticfunction x : D → E which is not an E -inner function. Proposition 4.21.
Let ( s, p ) be a Γ -inner function. Then x = (cid:0) s , s , p (cid:1) is an E -innerfunction.Proof. By Lemma 4.7, for every λ ∈ D , x ( λ ) = (cid:0) s ( λ ) , s ( λ ) , p ( λ ) (cid:1) ∈ E . It is easy tosee that x is in the set of analytic functions Hol( D , E ) from D to the tetrablock E . ByProposition 3.2 (3), for almost all λ ∈ T , | p ( λ ) | = 1 , | s ( λ ) | ≤ s ( λ ) − s ( λ ) p ( λ ) = 0 . Thus, for almost all λ ∈ T , | p ( λ ) | = 1 , s ( λ )2 = s ( λ )2 p ( λ ) and | s ( λ ) | ≤ . Hence x is E -inner. (cid:3) See [4] for many examples of Γ-inner functions.4.4.
Superficial E -inner functions. In this section, we study E -inner functions x suchthat x ( λ ) lies in the topological boundary ∂ E of E for all λ ∈ D .The topological boundary of E is denoted by ∂ E . Lemma 4.22. [1] Let x = ( x , x , x ) ∈ C . Then x ∈ ∂ E if and only if | x − x x | + | x − x x | = 1 − | x | , | x | ≤ | x | ≤ . Here we show that, for any inner function x and β , β ∈ C such that | β | + | β | = 1,the function x = ( β + β x , β + β x , x ) is E -inner and has the property that it maps D to ∂ E . Recall the definition of superficial function in the set of analytic functionsHol( D , Γ) from D to the symmetrised bidisc Γ from [4]. Definition 4.23.
An analytic function h : D → Γ is superficial if h ( D ) ⊂ ∂ Γ . One can define a similar notion for functions in Hol( D , E ). Definition 4.24.
An analytic function x : D → E is superficial if x ( D ) ⊂ ∂ E . We also consider relations between superficial Γ-inner functions and superficial E -innerfunctions. Proposition 4.25. [4, Proposition 8.3] A Γ -inner function h is superficial if and onlyif there is an ω ∈ T and an inner function p such that h = ( ωp + ω, p ) . Proposition 4.26.
An analytic function x : D → E such that x ( λ ) = ( β + β x ( λ ) , β + β x ( λ ) , x ( λ )) , λ ∈ D , where x is an inner function and | β | + | β | = 1 is E -inner and superficial.Proof. By Lemma 4.22, we need to show that, for λ ∈ D , x ( λ ) = ( β + β x ( λ ) , β + β x ( λ ) , x ( λ )) . is in ∂ E . Here x ( λ ) = β + β x ( λ ) , x ( λ ) = β + β x ( λ ) . Note, for λ ∈ D , | (cid:0) x − x x (cid:1) ( λ ) | = (cid:12)(cid:12)(cid:12)(cid:12) β + β x ( λ ) − (cid:16) β + β x ( λ ) (cid:17) x ( λ ) (cid:12)(cid:12)(cid:12)(cid:12) (4.19) = (cid:12)(cid:12)(cid:12)(cid:12) β (cid:16) − | x ( λ ) | (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) . We also have | ( x − x x )( λ ) | = (cid:12)(cid:12)(cid:12)(cid:12) β + β x ( λ ) − (cid:16) β + β x ( λ ) (cid:17) x ( λ ) (cid:12)(cid:12)(cid:12)(cid:12) (4.20) = (cid:12)(cid:12)(cid:12)(cid:12) β (cid:16) − | x ( λ ) | (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) . Note that by equations (4.19) and (4.20), for all λ ∈ D , | ( x − x x )( λ ) | + | ( x − x x )( λ ) | = (cid:12)(cid:12)(cid:12)(cid:12) β (cid:16) − | x ( λ ) | (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) β (cid:16) − | x ( λ ) | (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:16) | β | + | β | (cid:17)(cid:0) − | x ( λ ) | (cid:1) = 1 − | x ( λ ) | . By Theorem 2.5 and Lemma 4.22, for λ ∈ D , the point x ( λ ) = (cid:0) x ( λ ) , x ( λ ) , x ( λ ) (cid:1) lies in ∂ E . Let us check that x is E -inner. Clearly, for almost all λ ∈ T , x ( λ ) x ( λ ) = (cid:0) β + β x ( λ ) (cid:1) x ( λ )= β x ( λ ) + β x ( λ ) x ( λ )= β + β x ( λ ) = x ( λ ) . We also have, for almost all λ ∈ T , | x ( λ ) | = | β + β x ( λ ) | ≤ | β | + | β x ( λ ) | = | β | + | β | = 1 . Since x is inner, for almost all λ ∈ T , | x ( λ ) | = 1. Therefore x ( λ ) ∈ b E , for almost all λ ∈ T , and hence x is E -inner. (cid:3) ATIONAL E -INNER FUNCTIONS 23 Lemma 4.27.
Let x : D → E be such that x ( λ (cid:1) = (cid:0) β + β x ( λ ) , β + β x ( λ ) , x ( λ ) (cid:1) ,where x is a non-constant rational inner function and | β | + | β | = 1 . Then Ψ ω ( x ( λ )) = k for all λ ∈ D , where ω = β | β | , k = β | β | on T . Proof.
By definition, Ψ ω ( x ( λ )) = x ( λ ) ω − x ( λ ) x ( λ ) ω − x ( λ ) ω − ( β + β x ( λ ))( β + β x ( λ )) ω − x ( λ ) ω − β − β x ( λ ) β ω + β ωx ( λ ) − . Thus, for all λ ∈ D ,Ψ ω ( x ( λ )) = k ⇔ x ( λ ) ω − β − β x ( λ ) β ω + β ωx ( λ ) − k ⇔ x ( λ ) ω − β − β x ( λ ) = k [ β ω + β ωx ( λ ) − ⇔ x ( λ )[ ω − β − kβ ω ] + [ k − β − kβ ω ] = 0 . Since x is a nonconstant rational inner function, this statement is equivalent to ω − β − kβ ω = 0 and k − β − kβ ω = 0 . Multiply both sides of the first equation by ω and the second equation by k . We get(4.21) ( β k + β ω = 1 β k + β ω = 1 . Since | β | + | β | = 1, it is easy to see that ω = β | β | and k = β | β | satisfy equation (4.21), and soΨ ω ( x ( λ )) = k for all λ ∈ D . (cid:3) Lemma 4.28.
For any inner function x : D → D , there are x , x : D → D such thatthe function x : D → E defined by x = ( x , x , x ) is a superficial E -inner function, but h = ( x + x , x ) : D → Γ is not a superficial Γ -inner function.Proof. By Proposition 4.26, for any β , β ∈ C such that | β | + | β | = 1, the function x : D → E defined by x = (cid:16) β + β x , β + β x , x (cid:17) is a superficial E -inner function. By Proposition 4.25, h : D → Γ is superficial if andonly if there exists an ω ∈ T such that h = ( ωp + ω, p ). Note that, for x = β + β x and x = β + β x , h ( λ ) = ( x + x , x )( λ ) = (cid:16) β + β x ( λ ) + β + β x ( λ ) , x ( λ ) (cid:17) = (cid:16) ( β + β ) x ( λ ) + ( β + β ) , x ( λ ) (cid:17) , λ ∈ D . One can see that there are some β , β ∈ C with | β | + | β | = 1, but β + β / ∈ T . Forexample, take β = i , β = − i . Then | β | + | β | = + = 1 , but β + β = i − i = 0 / ∈ T . Thus, h is not a superficialΓ-inner function for β = i and β = − i . (cid:3) The construction of rational E -inner functions The formula for a Blaschke product is an explicit representation of a rational innerfunction in terms of its zeros and one other parameter (a unimodular complex number).In this chapter we aim to find a comparable representation for rational E -inner functions.The first question is: what is the tetrablock analogue of the zeros of an inner function?We shall show that one satisfactory choice consists of the royal nodes of an E -innerfunction x together with the zeros of x and x . We construct a rational E -inner function x from its royal nodes and the zeros of x and x . We show that there exists a 3-parameterfamily of rational E -inner functions with prescribed zero sets of x , x and prescribedroyal nodes. We also prove that a nonconstant rational E -inner function x of degree n either maps D to the royal variety of E or x ( D ) meets the royal variety exactly n times.5.1. The royal polynomial of an E -inner function. We define the royal variety for E to be R E = (cid:8) ( x , x , x ) ∈ C : x x = x (cid:9) . It was shown in [20] that R E ∩ E is the orbit of { (0 , , } under the group of biholomorphicautomorphisms of E . By Theorem 4.15, for a rational E -inner function x = ( x , x , x ),there are polynomials E , E , D such that x = E D , x = E D , x = D ∼ n D .
Thus, for λ ∈ D , (cid:0) x − x x (cid:1) ( λ ) = (cid:20) D ∼ n D − E D E D (cid:21) ( λ ) . The royal polynomial of the rational E -inner function x is defined to be R x ( λ ) = D ( λ ) (cid:20) D ∼ n D − E E D (cid:21) ( λ )= (cid:2) D ∼ n D − E E (cid:3) ( λ ) . Definition 5.1. [6, Definition 3.4]
We say a polynomial f is n -symmetric if deg( f ) ≤ n and f ∼ n = f . Definition 5.2. [6, Definition 3.4]
For any E ⊂ C , the number of zeros of f in E , counted with multiplicities, is denoted by ord E ( f ) and ord ( f ) means the same as ord { } ( f ) . ATIONAL E -INNER FUNCTIONS 25 Proposition 5.3.
Let x be a rational E -inner function of degree n and let R x be theroyal polynomial of x . Then, for λ ∈ T , (1) λ − n R x ( λ ) = | D ( λ ) | − | E ( λ ) | and (2) λ − n R x ( λ ) = | D ( λ ) | − | E ( λ ) | .Proof. (1) For λ ∈ T , λ − n R x ( λ ) = λ − n [ D ∼ n D − E E ]( λ )= λ − n [ λ n D (1 /λ ) D ( λ ) − E ∼ n ( λ ) E ( λ )] , since E ( λ ) = E ∼ n ( λ ) on T = λ − n [ λ n D ( λ ) D ( λ ) − λ n E (1 /λ ) E ( λ )] , since E (1 /λ ) = E ( λ ) on T = λ − n [ λ n D ( λ ) D ( λ ) − λ n E ( λ ) E ( λ )]= | D ( λ ) | − | E ( λ ) | . (5.1)(2) Since x is rational E -inner function, by Lemma 4.17,(5.2) | E ( λ ) | = | E ( λ ) | for λ ∈ T . By equations (5.1) and (5.2), λ − n R x ( λ ) = | D ( λ ) | − | E ( λ ) | for λ ∈ T . (cid:3) Lemma 5.4.
Let E and E be two polynomials such that deg E , deg E ≤ n . Then E ( λ ) = E ∼ n ( λ ) for all λ ∈ T if and only if E ( λ ) = E ∼ n ( λ ) for all λ ∈ T . Proof.
Suppose that E ( λ ) = E ∼ n ( λ ) for all λ ∈ T . Then by definition, E ( λ ) = E ∼ n ( λ ) = λ n E (1 /λ ) , λ ∈ T . Therefore, for all λ ∈ T , E ( λ ) = λ n E (1 /λ ) for all λ ∈ T ⇔ (1 /λ n ) E ( λ ) = E (1 /λ ) for all λ ∈ T ⇔ (1 /λ ) n E ( λ ) = E (1 /λ ) for all λ ∈ T ⇔ λ n E (1 /λ ) = E ( λ ) for all λ ∈ T ⇔ E ∼ n ( λ ) = E ( λ ) for all λ ∈ T . The converse is obvious. (cid:3)
Definition 5.5.
A nonzero polynomial R is n -balanced if (1) deg( R ) ≤ n , (2) R is n -symmetric, and (3) λ − n R ( λ ) ≥ for all λ ∈ T . For completeness, we shall say that zeros of the zero polynomial have infinite order.
Proposition 5.6.
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, λ − n R x ( λ ) ≥ for all λ ∈ T , and thezeros of R x on T have even order or infinite order.Proof. To show that R x is 2 n -symmetric we have to prove that R ∼ nx ( λ ) = R x ( λ ), for λ ∈ T . Recall that R x ( λ ) = D ∼ n ( λ ) D ( λ ) − E ( λ ) E ( λ ) , where x = (cid:18) E D , E D , D ∼ n D (cid:19) , By Theorem 4.15 (7) and Lemma 5.4, for λ ∈ T , E ( λ ) = E ∼ n ( λ ) = λ n E (1 /λ ) , E ( λ ) = E ∼ n ( λ ) = λ n E (1 /λ ) . Now R ∼ nx ( λ ) = λ n R x (1 /λ )= λ n (cid:2) D ∼ n (1 /λ ) D (1 /λ ) − E (1 /λ ) E (1 /λ ) (cid:3) = λ n D ∼ n (1 /λ ) λ n D (1 /λ ) − λ n E (1 /λ ) λ n E (1 /λ )= D ( λ ) D ∼ n ( λ ) − E ( λ ) E ( λ ) = R x ( λ ) . Hence R x is 2 n -symmetric.Clearly, if x ( D ) ⊆ R E , the royal polynomial R x is identically zero. Hence the zeros of R x on T have infinite order.In the case x ( D ) * R E , by Proposition 5.3, for λ ∈ T ,(5.3) λ − n R x ( λ ) = | D ( λ ) | − | E ( λ ) | . By Theorem 4.15 (6),(5.4) | D | − | E | ≥ T . By equations (5.3) and (5.4), λ − n R x ( λ ) ≥ T . By the Fej´er-Riesz theorem [18,Section 53], there exists an analytic polynomial P ( λ ) = P ni =0 b i λ i of degree n such that P is outer and λ − n R x ( λ ) = | P ( λ ) | for all λ ∈ T . Hence if σ ∈ T is a zero of R x , then σ is a zero of even order. Therefore in the case x ( D ) * R E , the zeros of R x that lie in T have even order. (cid:3) Lemma 5.7.
Let x be a rational E -inner function of degree n . Then the royal polynomial R x of x is either n -balanced or identically zero.Proof. If x ( D ) ⊆ R E then, by the definition of the royal variety, x ( λ ) x ( λ ) = x ( λ ) for all λ ∈ D . Thus E ( λ ) E ( λ ) = D ( λ ) D ∼ n ( λ ) for all λ ∈ D . Therefore the royal polynomial R x is identically zero.If x ( D ) * R E then, by Proposition 5.6, the royal polynomial R x of x is 2 n -symmetricand λ − n R x ( λ ) ≥ λ ∈ T . Clearly, R x has degree less than or equal to 2 n . Hence R x is n -balanced. (cid:3) Rational E -inner functions of type ( n, k ) .Definition 5.8. Let x = ( x , x , x ) be a rational E -inner function such that x ( D ) * R E . Let R x be the royal polynomial of x . If σ is a zero of R x of order ℓ , wedefine the multiplicity 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 sum of the multi-plicities of the royal nodes of x that lie in D , and k is the sum of the multiplicities ofthe royal nodes of x that lie in T . ATIONAL E -INNER FUNCTIONS 27 Definition 5.9.
Let R n,k denote the collection of rational E -inner functions of type ( n, k ) . Remark 5.10. [6, Equations (3.2) and (3.3)] For any m -symmetric polynomial f , thefollowing two relations hold(1) deg( f ) = m − ord ( f ) . (2) Since f is m -symmetric, if α ∈ D \{ } is zero of f , then α is also a zero of f .Thus ord ( f ) + 2ord D \{ } ( f ) + ord T ( f ) = deg( f ) . Theorem 5.11. If x ∈ R n,k is nonconstant then the degree of x is equal to n .Proof. Let R x be the royal polynomial of x . By assumption x ∈ R n,k and is nonconstant.Hence n ≥ x ( D ) * R E . Thus R x is not identically zero. By Proposition 5.6, R x is 2 deg( x )-symmetric. By Remark 5.10 (1) and (2), it follows thatdeg( R x ) = 2 deg( x ) − ord ( R x )and ord ( R x ) + 2ord D \{ } ( R x ) + ord T ( R x ) = deg( R x ) . Substitute the first equation in the second equation,ord ( R x ) + 2ord D \{ } ( R x ) + ord T ( R x ) = 2 deg( x ) − ord ( R x ) , which implies that2ord ( R x ) + 2ord D \{ } ( R x ) + ord T ( R x ) = 2 deg( x ) . Therefore, by Definition 5.9, n = ord ( R x ) + ord D \{ } ( R x ) + 12 ord T ( R x ) = deg( x ) . (cid:3) Theorem 5.12. If x is a nonconstant rational E -inner function, then either x ( D ) ⊆ R E or x ( D ) meets R E exactly deg( x ) times.Proof. Suppose that x is a nonconstant rational E -inner function. Then either, x ( D ) ⊆R E and the royal polynomial R x of x is identically zero, or by Theorem 5.11, x ( D ) meets R E exactly deg( x ) times. (cid:3) Lemma 5.13. [6, Lemma 4.4]
For σ ∈ D , let the polynomial Q σ be defined by theformula Q σ ( λ ) = ( λ − σ )(1 − σλ ) . Let n be a positive integer and let R be a nonzero polynomial. The polynomial R is n -balanced if and only if there exist points σ , σ , ..., σ n ∈ D and t + > such that R ( λ ) = t + n Y j =1 Q σ j ( λ ) , λ ∈ C . Proposition 5.14.
Let the royal nodes of a rational E -inner function x be σ , ..., σ n ,with repetition according to multiplicity of the royal nodes as described in Definition .The royal polynomial R x of x , up to a positive multiple, is (5.5) R x ( λ ) = n Y j =1 Q σ j ( λ ) . Proof.
By Lemma 5.7, R x is n -balanced. This implies that, by Lemma 5 .
13, there exists t + > η , . . . , η n ∈ D such that R x ( λ ) = t + n Y j =1 Q η j ( λ ) . By Definition 5.8, the royal nodes of x and their multiplicities are defined in terms ofzeros of R x in D and their multiplicities. Hence the list η , . . . , η n coincides, up to apermutation, with the list σ , . . . , σ n . Therefore R x is given, up to a positive multiple,by equation (5.5). (cid:3) Before we proceed to the next theorem on the construction of a tetra-inner function x from the zeros of x and x and royal nodes of x , let us prove the following elementarylemma. Lemma 5.15.
Let E and E be polynomials of degree at most n such that E ( λ ) = E ∼ n ( λ ) , for λ ∈ D . Let α , ..., α k be the zeros of E in D , and let α , ..., α k bethe zeros of E in D , where k + k = n . Then E ( λ ) = t k Y j =1 ( λ − α j ) k Y j =1 (1 − α j λ ) , where t ∈ C \{ } .Proof. Since α , ..., α k ∈ D and α , ..., α k ∈ D , where k + k = n , are zeros of E and E respectively, we have(5.6) E ( λ ) = ( λ − α ) ... ( λ − α k ) .p ( λ )and E ( λ ) = ( λ − α ) ... ( λ − α k ) .p ( λ ) . where the polynomials p ( λ ) and p ( λ ) do not vanish in D .Since E ( λ ) = λ n E (1 /λ ) on D , we have E ( λ ) = λ n (1 /λ − α ) ... (1 /λ − α k ) .p (1 /λ )= λ n − k (1 − α λ ) ... (1 − α k λ ) .p (1 /λ ) . (5.7)Since deg E ≤ n and k + k = n , equations (5.6) and (5.7) implies that E can bewritten in the form E ( λ ) = t ( λ − α ) ... ( λ − α k ) (1 − α λ ) ... (1 − α k λ )= t k Y j =1 ( λ − α j ) k Y j =1 (1 − α j λ ) , λ ∈ D , for some t ∈ C \{ } , and E ( λ ) = t k Y j =1 ( λ − α j ) k Y j =1 (1 − α j λ ) λ ∈ D , ATIONAL E -INNER FUNCTIONS 29 for some t ∈ C \{ } . Since E ( λ ) = λ n E (1 /λ ), λ n E (1 /λ ) = λ n (cid:18) t k Y j =1 (1 /λ − α j ) k Y j =1 (1 − α j /λ ) (cid:19) = λ n t k Y j =1 (1 /λ − α j ) k Y j =1 (1 − α j /λ )= t k Y j =1 (1 − α j λ ) k Y j =1 ( λ − α j )= E ( λ ) = t k Y j =1 ( λ − α j ) k Y j =1 (1 − α j λ ) , λ ∈ D , and so t = t . (cid:3) Remark 5.16.
For the polynomials E and E from Lemma 5.15, if α ∈ D \{ } and α is a zero of E then α is a zero of E . Theorem 5.17.
Suppose that α , ..., α k ∈ D and α , ..., α k ∈ D , where k + k = n .Suppose that σ , ..., σ n ∈ D are distinct from points of the set { α ij , j = 1 , ..., k i , i =1 , } ∩ T . Then there exists a rational E -inner function x = ( x , x , x ) : D → E suchthat (1) the zeros of x in D , repeated according to multiplicity, are α , ..., α k ; (2) the zeros of x in D , repeated according to multiplicity, are α , ..., α k ; (3) the royal nodes of x are σ , ..., σ n ∈ D , with repetition according to multiplicityof the nodes.Such a function x can be constructed as follows. Let t + > and let t ∈ C \{ } . Let R be defined by R ( λ ) = t + n Y j =1 ( λ − σ j )(1 − σ j λ ) , andlet E be defined by E ( λ ) = t k Y j =1 ( λ − α j ) k Y j =1 (1 − α j λ ) . Then the following statements hold. (i)
There exists an outer polynomial D of degree at most n such that (5.8) λ − n R ( λ ) + | E ( λ ) | = | D ( λ ) | for all λ ∈ T . (ii) The function x defined by x = ( x , x , x ) = (cid:18) E D , E ∼ n D , D ∼ n D (cid:19) is a rational E -inner function such that the degree of x is equal to n and condi-tions (1), (2) and (3) hold. The royal polynomial of x is R . Proof. (i) By Lemma 5.13, R is n -balanced, and so λ − n R ( λ ) ≥ λ ∈ T . Therefore λ − n R ( λ ) + | E ( λ ) | ≥ λ ∈ T . By the Fej´er-Riesz theorem, there exists an outer polynomial D of degree at most n suchthat(5.9) λ − n R ( λ ) + | E ( λ ) | = | D ( λ ) | for all λ ∈ T . (ii) Let D be an outer polynomial of degree at most n such that equality (5.9) holdsfor all λ ∈ T . By hypothesis { σ j : 1 ≤ j ≤ n } ∩ (cid:18) { α ij : 1 ≤ j ≤ k i , i = 1 , } ∩ T (cid:19) = ∅ . Then λ − n R ( λ ) and | E ( λ ) | are non-negative trigonometric polynomials on T with nocommon zero. Thus λ − n R ( λ ) + | E ( λ ) | > T . By equality (5.9), D has no zero on T , and so D and D ∼ n have no common factor.Hence deg( x ) = deg (cid:18) D ∼ n D (cid:19) = max { deg( D ) , deg( D ∼ n ) } = n. Since λ − n R ( λ ) ≥ λ ∈ T , | D ( λ ) | = λ − n R ( λ ) + | E ( λ ) | ≥ | E ( λ ) | for all λ ∈ T . It follows that | D ( λ ) | ≥ | E ( λ ) | , for all λ ∈ T . Since D ( λ ) = 0 on D , the function E D is analytic in a neighbourhood of D . By theMaximum Modulus Principle, we have | E ( λ ) || D ( λ ) | ≤ λ ∈ D . Therefore, by the converse of Theorem 4.15, since conditions (1), (2), (6) and (7) aresatisfied, the function x ( λ ) = (cid:18) E ( λ ) D ( λ ) , E ∼ n ( λ ) D ( λ ) , D ∼ n ( λ ) D ( λ ) (cid:19) for λ ∈ D , is a rational E -inner function such that deg( x ) = n . The royal polynomial of x is definedby R x ( λ ) = D ( λ ) D ∼ n ( λ ) − E ( λ ) E ( λ ) , λ ∈ D , where E ( λ ) = E ∼ n ( λ ) , λ ∈ D . By Proposition 5.3, for all λ ∈ T , λ − n R x ( λ ) = | D ( λ ) | − | E ( λ ) | . Therefore, by equation (5.8), λ − n R x ( λ ) = λ − n R ( λ ) for all λ ∈ T , where E ( λ ) = E ∼ n ( λ ) for λ ∈ D . Thus R x = R , that is, the royal polynomial of x isequal to R . (cid:3) ATIONAL E -INNER FUNCTIONS 31 Remark 5.18. (1) For large n the task of finding an outer polynomial D satisfyingequation (5.8) cannot be solved algebraically.(2) The solution D is only identified up to a multiplication by ω ∈ T . Thus if we replace D by ωD we obtain a new solution x = (cid:18) ω E D , ω E ∼ n D , ω D ∼ n D (cid:19) . Example 5.19.
Let n = 1, α = and σ = 0. Let us construct a rational E -innerfunction x = ( x , x , x ) : D → E such that α is a zero of x and σ is a royal node of x .As in Theorem 5.17, for λ ∈ T , let R ( λ ) = t + λ, t + is a positive real number. E ( λ ) = t (1 − λ ) , t ∈ C \{ } . The equation (5.8) for the polynomial D is the following, for all λ ∈ T , | D ( λ ) | = λ − R ( λ ) + | E ( λ ) | (5.10) = λt + λ + | t (1 − λ ) | = t + + | t | − | t | λ − | t | λ. Since the degree of D is at most 1, D ( λ ) = a + a λ , where a , a ∈ C and λ ∈ T , D ( λ ) D ( λ ) = | a + a λ | = ( a + a λ )( a + a λ ) = | a | + | a | + a a λ + a a λ. (5.11)Compare equations (5.10) and (5.11). We have(5.12) a a = − | t | ,a a = − | t | , | a | + | a | = t + + | t | . Finally the function x can be written in the form x = (cid:18) E D , E ∼ D , D ∼ D (cid:19) , where x ( λ ) = E D ( λ ) = t (1 − λ ) a + a λ ,x ( λ ) = E ∼ D ( λ ) = t ( λ − ) a + a λ ,x ( λ ) = D ∼ D ( λ ) = a λ + a a + a λ , where | a | < | a | and a , a are given by solving equations (5.12) as functions of t + and t . These formulas give a parametrization of solutions for the above problem.For example, for the given t = √ t + = 74 , the system (5.12) has solutions a = − i, a = i or a = − , a = . Hence the functions x ( λ ) = √ − λ ) − ω (1 − λ ) , √ λ − ) − ω (1 − λ ) , ω ( λ − ) ω (1 − λ ) ! , λ ∈ D , where ω ∈ T , are rational E -inner functions such that 12 is a zero of x and 0 is a royalnode of x . Remark 5.20.
Theorem 5.17 gives a 3-parameter family of rational E -inner functionswith prescribed royal nodes and prescribed zeros of x and x . It appears at first sightthat the construction in Theorem 5.17 gives us a 4-parameter family of rational E -innerfunctions with the given data. However, the choice of t + , t, D and ω leads to the same x as the choice 1 , t/ √ t + , D/ √ t + and ω . Theorem 5.21 tells us that the constructionyields all solutions of the problem, and so the family of functions x with the requiredproperties is indeed a 3-parameter family. Theorem 5.21.
Let x = ( x , x , x ) be a rational E -inner function of degree n such that (1) the zeros of x , repeated according to multiplicity, are α , . . . α k ∈ D , (2) the zeros of x , repeated according to multiplicity, are α , . . . α k ∈ D , where k + k = n , and (3) the royal nodes of x are σ , . . . , σ n ∈ D , repeated according to multiplicity.There exists some choice of t + > , t ∈ C \{ } and ω ∈ T such that the recipe in Theorem with these choices produces the function x .Proof. By Theorem 4.15, there exist polynomials E , E and D such that(1) deg( E ) , deg( E ) and deg( D ) ≤ n ,(2) D ( λ ) = 0 on D ,(3) E ( λ ) = ( E ) ∼ n ( λ )(4) | E i ( λ ) | ≤ | D ( λ ) | on D , i = 1 , x = E D , x = E D and x = ( D ) ∼ n D on D .By hypothesis, the zeros of x , repeated according to multiplicity, are α , ..., α k , andthe zeros of x , repeated according to multiplicity, are α , ..., α k where k + k = n .By Lemma 5.15, E ( λ ) = t k Y j =1 ( λ − α j ) k Y j =1 (1 − α j λ ) for some t ∈ C \{ } and all λ ∈ D . By hypothesis, σ , . . . , σ n are the royal nodes of x . Thus, by Proposition 5.14, for theroyal polynomial R of x , there exists t + > R ( λ ) = t + n Y j =1 ( λ − σ j )(1 − σ j λ ) . By Proposition 5.3, for λ ∈ T , λ − n R ( λ ) = | D ( λ ) | − | E ( λ ) | . By Theorem 4.15, D ( λ ) = 0 on D . Hence, for λ ∈ T λ − n R ( λ ) + | E ( λ ) | = | D ( λ ) | = 0 . ATIONAL E -INNER FUNCTIONS 33 This implies that α , . . . , α k and α , . . . , α k which are on T are distinct from σ i , i =1 , . . . , n . By the construction in Theorem 5.17, for σ i , i = 1 , . . . , n and α , . . . , α k and α , . . . , α k , the rational E -inner function x = ( x , x , x ) can be defined by (cid:18) E D , E ∼ n D , D ∼ n D (cid:19) for a suitable choice of t + > t ∈ C \{ } and ω ∈ T . Since E and R coincidewith E and R in the construction of Theorem 5.17 for a suitable choice of t + > t ∈ C \{ } , D is a permissible choice for ωD for some choice ω ∈ T , as a solution forequation (5.8). Hence the construction of Theorem 5.17 yields x = ( x , x , x ) for theappropriate choices of t + > t ∈ C \{ } and ω ∈ T . (cid:3) Convex subsets of E and extremality In this section we study convex subsets of E . We show that, for a fixed x ∈ D , thesubset E ∩ (cid:0) C × { x } (cid:1) is convex. Recall that the distinguished boundaries of the tridisc D and the ball B contain no line segments. Thus every inner function in the set ofanalytic functions Hol( D , D ) from D to D is an extreme point of Hol( D , D ) and everyinner function in the set of analytic functions Hol( D , B ) from D to B is an extremepoint of Hol( D , B ). However, this property contrasts sharply with the situation in thetetrablock. Despite the fact that the set J of rational tetra-inner functions is not convex,the conventional notion of extreme point of J is well defined and fruitful. In Theorem6.19, we prove that for x ∈ R n,k with 2 k ≤ n , x is not an extreme point. A class ofextreme points of the set J is given in Proposition 6.21.6.1. Convex subsets in the tetrablock.Definition 6.1.
A set Ω in a vector space is convex if for all z, w ∈ Ω and all t suchthat ≤ t ≤ , the point tz + (1 − t ) w belongs to Ω . Proposition 6.2. [1, page 8] E is not convex.Proof. Take x = ( i, , i ) and y = ( − , i, − i ) in E . Let us take t = 1 /
2, then the point w = tx + (1 − t ) y = ( − i, i, w is not in E . Therefore E is not convex. (cid:3) Let us show that the set E is convex in ( x , x ) for a fixed x ∈ D , that is, the set E ∩ (cid:0) C × { x } (cid:1) = { x = ( x , x , x ) ∈ C : | x − x x | + | x − x x | ≤ − | x | } is convex for every x ∈ D . In consequence, some associated sets have a similar property. Proposition 6.3.
The following sets are convex: (1) E ∩ (cid:0) C × { x } (cid:1) for any x ∈ D ; (2) b E ∩ (cid:0) C × { x } (cid:1) for any x ∈ D .Proof. (1) Let x, y ∈ E ∩ (cid:0) C × { x } (cid:1) , and so, by Theorem 2.5, x and y satisfy theinequalities(6.1) | x − x x | + | x − x x | ≤ − | x | and(6.2) | y − y x | + | y − y x | ≤ − | x | respectively. For all t ∈ [0 , w = tx + (1 − t ) y = t ( x , x , x ) + (1 − t )( y , y , x )= (cid:0) tx + (1 − t ) y , tx + (1 − t ) y , tx + (1 − t ) x (cid:1) = (cid:0) tx + (1 − t ) y , tx + (1 − t ) y , x (cid:1) . Let us check that the point w ∈ C is in the set E ∩ ( C × { x } ). By Theorem 2.5, w ∈ E if and only if | w − w x | | {z } + | w − w x | | {z } ≤ − | x | . Let us consider the first term on the left hand side | w − w x | = (cid:12)(cid:12) tx + (1 − t ) y − (cid:0) tx + (1 − t ) y (cid:1) x (cid:12)(cid:12) (6.3) = (cid:12)(cid:12) t ( x − x x ) + (1 − t )( y − y x ) (cid:12)(cid:12) ≤ t (cid:12)(cid:12) x − x x (cid:12)(cid:12) + (1 − t ) (cid:12)(cid:12) y − y x (cid:12)(cid:12) . For the second term of the left hand side we have | w − w x | = (cid:12)(cid:12) tx + (1 − t ) y − (cid:0) tx + (1 − t ) y (cid:1) x (cid:12)(cid:12) (6.4) = (cid:12)(cid:12) t ( x − x x ) + (1 − t )( y − y x ) (cid:12)(cid:12) ≤ t (cid:12)(cid:12) x − x x (cid:12)(cid:12) + (1 − t ) (cid:12)(cid:12) y − y x (cid:12)(cid:12) . Add inequalities (6.3) and (6.4) we get | w − w x | + | w − w x | ≤ t (cid:12)(cid:12) x − x x (cid:12)(cid:12) + (1 − t ) (cid:12)(cid:12) y − y x (cid:12)(cid:12) + t (cid:12)(cid:12) x − x x (cid:12)(cid:12) + (1 − t ) (cid:12)(cid:12) y − y x (cid:12)(cid:12) . Therefore, by inequalities (6.1) and (6.2), | w − w x | + | w − w x | ≤ t (cid:0) (cid:12)(cid:12) x − x x (cid:12)(cid:12) + (cid:12)(cid:12) x − x x (cid:12)(cid:12)| {z } (cid:1) + (1 − t ) (cid:0) (cid:12)(cid:12) y − y x (cid:12)(cid:12) + (cid:12)(cid:12) y − y x (cid:12)(cid:12)| {z } (cid:1) ≤ t (1 − | x | ) + (1 − t )(1 − | x | )= 1 − | x | . Hence for all t ∈ [0 , w = tx + (1 − t ) y ∈ E ∩ (cid:0) C × { x } (cid:1) . Therefore E ∩ (cid:0) C × { x } (cid:1) is convex for any fixed x ∈ D .(2) Let x ∈ D and x, y ∈ b E ∩ (cid:0) C × { x } (cid:1) , where x = ( x , x , x ) and y = ( y , y , x ).Note that, by Theorem 2.10 (1),(6.5) w ∈ C belongs to b E if and only if w = w w , | w | ≤ | w | = 1 . Thus we have x = x x , | x | ≤ | x | = 1 , and y = y x , | y | ≤ | x | = 1 . For t : 0 ≤ t ≤
1, let w = (cid:0) w , w , w (cid:1) = tx + (1 − t ) y = (cid:0) tx + (1 − t ) y , tx + (1 − t ) y , x (cid:1) . ATIONAL E -INNER FUNCTIONS 35 To prove the convexity of b E ∩ ( C × { x } ), we need to check that, for all t such that0 ≤ t ≤ w lies in b E ∩ (cid:0) C × { x } (cid:1) , that is, it satisfies condition (6.5).Note that w w = (cid:0) tx + (1 − t ) y (cid:1) x = tx x + (1 − t ) y x = tx + (1 − t ) y = w and | w | = | tx + (1 − t ) y |≤ t | x | + (1 − t ) | y |≤ t + 1 − t = 1 . Obviously, | w | = | x | = 1. Therefore the set b E ∩ ( C × { x } ) is convex for any fixed x ∈ D . (cid:3) Lemma 6.4.
Let x = ( x , x , x ) , x = ( x , x , x ) and x = ( x , x , x ) be in b E andsatisfy x = tx + (1 − t ) x for some t ∈ (0 , . Then x = x = x .Proof. Since x , x , x ∈ b E , by Theorem 2.10, | x | = 1 , | x | = 1 and | x | = 1 . By assumption x = tx + (1 − t ) x . Since every point of T is an extreme point of D , x = x = x . (cid:3) Extremality in the set of E -inner functions. In this section we show that,for a fixed inner function x , the set of rational E -inner functions x = ( x , x , x ) withthird component x is a convex set in J . We prove that an E -inner function x is notan extreme point of the set J if the number of the royal nodes of x on T , counted withmultiplicity, is less than or equal to half of the degree of x . In Proposition 6.21 we givea class of extreme rational E -inner functions x ∈ R n,k of the set J for which 2 k > n . Theorem 6.5.
For a fixed inner function x , the set of E -inner functions ( x , x , x ) is convex.Proof. For the fixed inner function x , let x = ( x , x , x ) and y = ( y , y , x ) be E -innerfunctions. For 0 ≤ t ≤ λ ∈ D , (cid:0) tx + (1 − t ) y (cid:1) ( λ ) = (cid:0) tx + (1 − t ) y , tx + (1 − t ) y , x (cid:1) ( λ ) . The function w ( λ ) = (cid:0) w , w , w (cid:1) ( λ ) = (cid:0) tx + (1 − t ) y , tx + (1 − t ) y , x (cid:1) ( λ ) , λ ∈ D is analytic on D and by Proposition 6.3 (1), w ( D ) ⊆ E . By Proposition 6.3 (2), sincefor almost all λ ∈ T , x ( λ ) and y ( λ ) are in b E , w ( λ ) has also to be in b E . Thus w is an E -inner function. Therefore the set of E -inner functions x = ( x , x , x ) is convex forany fixed inner function x . (cid:3) Definition 6.6.
A rational E -inner function x is an extreme point of J if whenever x has a representation of the form x = tx + (1 − t ) x for t ∈ (0 , and x , x are rational E -inner functions, x = x . We will show below that J is not convex, however the notion of extreme points stillhas the usual sense. Lemma 6.7.
Let x = ( x , x , x ) , x = ( x , x , x ) and x = ( x , x , x ) be rational E -inner functions. If x = tx + (1 − t ) x for some t ∈ (0 , then x = x = x .Proof. Since x = tx + (1 − t ) x , we have (cid:0) x , x , x (cid:1) = (cid:0) tx , tx , tx (cid:1) + (cid:18) (1 − t ) x , (1 − t ) x , (1 − t ) x (cid:19) . Thus x = tx + (1 − t ) x . Hence, for every point λ ∈ T , x ( λ ) = tx ( λ ) + (1 − t ) x ( λ ) . By assumption, x and x are rational E -inner functions, and so, by Lemma 4.3 (2), x and x are rational inner functions, that is, for all λ ∈ T , | x ( λ ) | = 1 and | x ( λ ) | = 1 . Every point of T is an extreme point of D , and therefore, x ( λ ) = x ( λ ) = x ( λ )for all λ ∈ T . Since x and x are rational functions, x = x = x . (cid:3) Lemma 6.8.
The set of rational E -inner functions J is not convex.Proof. Suppose that x = (cid:0) x , x , x (cid:1) ∈ J and x = ( x , x , x ) ∈ J such that x = x .Then by Lemma 6.7, x = tx + (1 − t ) x is not in J for all t ∈ (0 , J isnot convex. (cid:3) For an inner function p of degree n , let R np be the set of rational E -inner functionswith third component p . Proposition 6.9. If p is an inner function of degree n then R np is convex. Let C be acollection of rational E -inner functions. C is convex if and only if there exists an innerfunction p such that C is a convex subset of R np , where n = deg( p ) .Proof. It follows from Theorem 6.5 and Lemma 6.7. (cid:3)
Proposition 6.10. If x is a rational E -inner function of degree n then x is a convexcombination of at most n + 3 extreme rational E -inner functions of degree at most n .Proof. For the given rational E -inner function x = ( x , x , x ) of degree n , x is an innerfunction of degree n and x ∈ R nx . By Remark 4.16, the convex set R nx is a subset ofa (2 n + 2)-dimensional real subspace of the rational functions. Thus, by a theorem ofCarath´eodory [11, 19], x is a convex combination of at most 2 n + 3 extreme rational E -inner functions of degree at most n . (cid:3) Definition 6.11.
A real or complex-valued function f on a real interval I is said totake a value y to order m ≥ at a point t ∈ I if f ∈ C m ( I ) , f ( t ) = y , f ( j ) ( t ) = 0 for j = 1 , . . . m − and f ( m ) ( t ) = 0 . We say that f vanishes to order m ≥ at a point t ∈ I if f takes the value to order m at t . Lemma 6.12.
Let f ∈ C m ( I ) , f ( t ) = y at t ∈ I , and let y = 0 . If f takes the value y to order m ≥ at t , then f takes the value y to order m at t .Proof. Let I be a real interval and let f ∈ C m ( I ). Suppose that f takes the value y to order m at t . Then, by Definition 6.11(6.6) f ( t ) = y , [ f ] (1) ( t ) = [ f ] (2) ( t ) = · · · = [ f ] ( m − ( t ) = 0 , [ f ] ( m ) ( t ) = 0 . ATIONAL E -INNER FUNCTIONS 37 One can check that f ( t ) = y, f ( j ) ( t ) = 0 , for j = 1 , . . . , m − f ( m ) ( t ) = 0 . (cid:3) Definition 6.13.
A function f is analytic on T if there exists a function g analytic ina neighbourhood U T of T such that f = g | T . Lemma 6.14.
Let τ = e it and let f ( t ) = ( e it − τ ) v G ( e it ) in a neighbourhood of t where G ( z ) is analytic on T and G ( τ ) = 0 . Then (6.7) f ( j ) ( t ) = 0 for j = 0 , , . . . , v − and f (2 v ) ( t ) = 0 . Proof.
Since G is analytic on T , by Definition 6.13, there exists U T a neighbourhood of T and there exists ˜ G analytic on U T such that G = ˜ G | T . Let z = e it , ϕ ( z ) = ( z − τ ) v G ( z )and ˜ ϕ ( z ) = ( z − τ ) v ˜ G ( z ). Define γ ( τ, r ) to be an anticlockwise circle centred at τ withradius r γ ( τ, r ) = { z ∈ C : | z − τ | = r } , where r is taken sufficiently small such that γ ⊂ U T . Hence the function ˜ ϕ is analyticinside the curve γ . Therefore, by Cauchy’s integral formula,˜ ϕ ( j ) ( τ ) = j !2 πi Z γ ˜ ϕ ( z )( z − τ ) j +1 dz, = j !2 πi Z γ ( z − τ ) v ˜ G ( z )( z − τ ) j +1 dz = j !2 πi Z γ ( z − τ ) v − j − ˜ G ( z ) dz. (6.8)For 0 ≤ j ≤ v −
1, the function ( z − τ ) v − j − ˜ G ( z ) is analytic on U T . Therefore, byCauchy’s Theorem,(6.9) ˜ ϕ ( j ) ( τ ) = j !2 πi Z γ ( z − τ ) v − j − ˜ G ( z ) dz = 0 . If j = 2 v , then equation (6.8) becomes˜ ϕ (2 v ) ( τ ) = (2 v )!2 πi Z γ ˜ G ( z )( z − τ ) dz. By Cauchy’s integral formula,(6.10) ˜ ϕ (2 v ) ( τ ) = (2 v )!2 πi Z γ ˜ G ( z )( z − τ ) dz = (2 v )! G ( τ ) = 0 . Hence ϕ (2 v ) ( τ ) = 0 because ˜ ϕ agrees with ϕ on T . Note that, f ( t ) = ( e it − e it ) (2 v ) G ( e it ) = ϕ ( e it ) . By the chain rule, dfdt = dϕdz dzdtd fdt = d ϕdz (cid:18) dzdt (cid:19) + dϕdz d zdt . . . = . . .d v − fdt v − = d v − ϕdz v − (cid:18) dzdt (cid:19) v − + · · · + dϕdz d v − zdt v − . By equation (6.9) and since ˜ ϕ and ϕ agree on T , d j ˜ ϕdz j ( τ ) = 0 , for j = 1 , . . . , v − , and so, d j ϕdz j ( τ ) = 0 , for j = 1 , . . . , v − . Therefore, f ( j ) ( t ) = 0 for j = 1 , . . . , v −
1. For the (2 v )th derivative of f , we have d v fdt v = d v ϕdz v (cid:18) dzdt (cid:19) v + · · · + dϕdz d v zdt v . By equations (6.9) and (6.10), d j ϕdz j ( τ ) = 0 , for j = 1 , . . . , v − d v ϕdz v ( τ ) = 0 . Hence f (2 v ) ( t ) = d v ϕdz v ( τ ) (cid:18) dzdt (cid:19) v ( t ) = 0.Therefore f ( j ) ( t ) = 0 for j = 0 , . . . , v − f (2 v ) ( t ) = 0. (cid:3) Lemma 6.15.
Let x = ( x , x , x ) be a rational E -inner function. For τ ∈ T , (1) | x ( τ ) | = 1 ⇔ τ is a royal node of x . (2) | x ( τ ) | = 1 ⇔ τ is a royal node of x .Moreover, (3) τ = e it is a royal node of x of multiplicity v if and only if | x ( e it ) | = 1 to order v at t = t . (4) τ = e it is a royal node of x of multiplicity v if and only if | x ( e it ) | = 1 to order v at t = t .Proof. (1) If τ = e it is a royal node of x of multiplicity v , by Definition 5.8,(6.11) ( x − x x )( λ ) = ( λ − τ ) v F ( λ )where F is a rational function, analytic on T and F ( τ ) = 0. By Lemma 2.11, since x isan E -inner function, x = x x on T . Therefore, for λ ∈ T , (cid:16) x − x x (cid:17) ( λ ) = (cid:16) x − x x x (cid:17) ( λ )= x ( λ ) − x ( λ ) | x ( λ ) | = x ( λ )(1 − | x ( λ ) | ) . (6.12)Therefore, for any λ ∈ T , | x ( λ ) | = 1 ⇐⇒ ( x − x x )( λ ) = 0 , that is, if and only if λ is a royal node of x . Hence τ ∈ T is a royal node of x if and onlyif | x ( τ ) | = 1.(2) Since x is rational E -inner function, by Theorem 2.10, x = x x on T . The restof the proof is similar to the above proof of (1).(3) Suppose that τ = e it is a royal node of x of multiplicity v ≥
1. Then on combiningequations (6.11) and (6.12), we have, for all t ∈ R , x ( e it )(1 − | x ( e it ) | ) = ( e it − τ ) v F ( e it ) . ATIONAL E -INNER FUNCTIONS 39 This gives 1 − | x ( e it ) | = ( e it − τ ) v F ( e it ) x ( e it ) . The rational function G = Fx is analytic on T and is not equal to zero at τ = e it . Thuswe have 1 − | x ( e it ) | = ( e it − τ ) v G ( e it ) . Since x is rational and | x ( e it ) | = 1, the function f ( t ) = 1 − | x ( e it ) | is C ∞ on aneighbourhood of t . By Lemma 6.14, f ( j ) ( t ) = 0 for j = 0 , , . . . , v − f (2 v ) ( t ) = 0 . Therefore f takes the value 0 to order 2 v at t , which implies, by Lemma 6.12, | x ( e it ) | =1 to order 2 v at t .(4) The proof of this statement follows from (2) and is similar to the above proof of(3). (cid:3) For an inner function p of degree n and k = 0 , , . . . , n , let(6.13) R n,kp = { ( x , x , x ) ∈ R n,k : x = p } . Lemma 6.16.
Let x = ( x , x , x ) ∈ R n,kx and let τ , τ , . . . , τ k ∈ T be royal nodes of x .Suppose x = tx + (1 − t ) x for some t such that < t < , where x = ( x , x , x ) and x = ( x , x , x ) are rational E -inner functions. Then x = x = x , x i ( τ j ) = x ( τ j ) for j = 1 , . . . k and i = 1 , , and x i ( τ j ) = x ( τ j ) for j = 1 , . . . k and i = 1 , . Moreover, τ , τ , . . . , τ k ∈ T are royal nodes of x and x .Proof. By Lemma 6.7, x = x = x . By Lemma 6.15, | x ( τ j ) | = 1 and | x ( τ j ) | = 1 ateach royal node τ j ∈ T . By assumption, x ( τ j ) = tx ( τ j ) + (1 − t ) x ( τ j )for t such that 0 < t < | x i ( τ j ) | ≤ j = 1 , . . . , k and i = 1 ,
2. Similarly, x ( τ j ) = tx ( τ j ) + (1 − t ) x ( τ j )for t such that 0 < t < | x i ( τ j ) | ≤ j = 1 , . . . , k and i = 1 ,
2. Since every pointon the circle T is an extreme point of ¯ D we have x i ( τ j ) = x ( τ j ) and x i ( τ j ) = x ( τ j )for j = 1 , . . . , k and i = 1 ,
2. Therefore | x i ( τ j ) | = 1, | x i ( τ j ) | = 1 for j = 1 , . . . , k and i = 1 ,
2. By Lemma 6.15, τ , τ , . . . , τ k ∈ T are royal nodes of x and x . (cid:3) Lemma 6.17.
Let n ≥ . Any x = ( x , x , x ) ∈ R n, is not an extreme point of J .Proof. Since x has no royal nodes on T , by Lemma 6.15, for all λ ∈ T , | x ( λ ) | < | x ( λ ) | < . Since T is compact, the supremum of x and x is attained on T , that is, there exist λ , λ ∈ T such that(6.14) sup λ ∈ T | x ( λ ) | = | x ( λ ) | < λ ∈ T | x ( λ ) | = | x ( λ ) | < . Choose ε > ε > | x ( λ ) | (1 + ε ) < | x ( λ ) | (1 + ε ) < . Take ε = min { ε , ε } . If x ( λ ) = 0, then x ( λ ) = 0 for all λ ∈ T . Likewise, if x ( λ ) = 0 then x ( λ ) = 0 , for all λ ∈ T . Define x and x to be x = (cid:0) (1 + ε ) x , (1 + ε ) x , x (cid:1) and x = (cid:0) (1 − ε ) x , (1 − ε ) x , x (cid:1) . Since x = ( x , x , x ) is a rational E -inner function, for almost all λ ∈ T ,(6.16) x ( λ ) = x ( λ ) x ( λ ) , | x ( λ ) | ≤ | x ( λ ) | = 1 . Let us check that x and x are rational E -inner functions. By Theorem 2.10 (1), thiswill follow if we show that(1 + ε ) x ( λ ) = (1 + ε ) x ( λ ) x ( λ ) , (1 + ε ) | x ( λ ) | ≤ | x ( λ ) | = 1 , and x ( D ) ⊂ E . By equations (6.16), we have to show only that(1 + ε ) | x ( λ ) | ≤ T . This statement follows from inequalities (6.14) and (6.15). Thus x ( T ) ⊂ b E . ByTheorem 2.10 (2), for almost all λ ∈ T , x ( λ ) ∈ b E ⇔ Ψ( ., x ( λ )) is an automorphism of D . By the maximum principle, for all λ ∈ D , k Ψ( ., x ( λ )) k H ∞ <
1. Therefore, by Theorem2.4, for all λ ∈ D , x ( λ ) ⊆ E . This completes the proof that x is a rational E -innerfunction.In a similar way we can show that x is a rational E -inner function. Moreover, byLemma 6.15, x , x have no royal nodes on T and therefore x , x ∈ R n, . However x = x + x , which implies that x cannot be an extreme point of J since x = x . (cid:3) Proposition 6.18.
Let x = ( x , x , x ) be superficial and x = tx + (1 − t ) x for some < t < , where x = ( x , x , x ) and x = ( x , x , x ) are rational E -inner functions.Then x and x are superficial and x = x = x .Proof. By Lemma 6.7, x = x = x . Suppose, for a contradiction, x is not superficial.Then there exists λ ∈ D such that x ( λ ) ∈ E . Let us show that in this case x ( λ ) ∈ E ,and so x is not superficial.By Theorem 2.4, it is enough to show that(6.17) | x ( λ ) − x ( λ ) x ( λ ) | + | x ( λ ) − x ( λ ) x ( λ ) | < − | x ( λ ) | . Since x ( λ ) ∈ E and x is rational E -inner function, this implies that(6.18) | x ( λ ) − x ( λ ) x ( λ ) | + | x ( λ ) − x ( λ ) x ( λ ) | < − | x ( λ ) | and(6.19) | x ( λ ) − x ( λ ) x ( λ ) | + | x ( λ ) − x ( λ ) x ( λ ) | ≤ − | x ( λ ) | . ATIONAL E -INNER FUNCTIONS 41 Let us begin with the first term on the left hand side of inequality (6.17). | x ( λ ) − x ( λ ) x ( λ ) | = (cid:12)(cid:12) tx ( λ ) + (1 − t ) x ( λ ) − (cid:0) tx ( λ ) + (1 − t ) x ( λ ) (cid:1) x ( λ ) (cid:12)(cid:12) ≤ (cid:12)(cid:12) t (cid:0) x ( λ ) − x ( λ ) x ( λ ) (cid:1)(cid:12)(cid:12) + (cid:12)(cid:12) (1 − t ) (cid:0) x ( λ ) − x ( λ ) x ( λ ) (cid:1)(cid:12)(cid:12) (6.20)The second term on inequality (6.17) | x ( λ ) − x ( λ ) x ( λ ) | = (cid:12)(cid:12) tx ( λ ) + (1 − t ) x ( λ ) − (cid:0) tx ( λ ) + (1 − t ) x ( λ ) (cid:1) x ( λ ) (cid:12)(cid:12) ≤ (cid:12)(cid:12) t (cid:0) x ( λ ) − x ( λ ) x ( λ ) (cid:1)(cid:12)(cid:12) + (cid:12)(cid:12) (1 − t ) (cid:0) x ( λ ) − x ( λ ) x ( λ ) (cid:1)(cid:12)(cid:12) . (6.21)Add inequalities (6.20) and (6.21), and use inequalities (6.18) and (6.19) to obtain | x ( λ ) − x ( λ ) x ( λ ) | + | x ( λ ) − x ( λ ) x ( λ ) |≤ t (cid:16)(cid:12)(cid:12) x ( λ ) − x ( λ ) x ( λ ) (cid:12)(cid:12) + (cid:12)(cid:12) x ( λ ) − x ( λ ) x ( λ ) (cid:12)(cid:12)(cid:17) +(1 − t ) (cid:16)(cid:12)(cid:12) x ( λ ) − x ( λ ) x ( λ ) (cid:12)(cid:12) + (cid:12)(cid:12) x ( λ ) − x ( λ ) x ( λ ) (cid:12)(cid:12)(cid:17) < t (1 − | x ( λ ) | ) + (1 − t )(1 − | x ( λ ) | ) = 1 − | x ( λ ) | . (6.22) (cid:3) Theorem 6.19.
Let x ∈ R n,k . If k ≤ n , then x is not an extreme point of the set J of rational E -inner functions.Proof. Let x ∈ R n,k . By Definition 5.8, x has n royal nodes in D and k royal nodes thatlie in T . By Theorem 4.15, there exist polynomials E , E and D of degree at most n such that x = (cid:18) E D , E ∼ n D , D ∼ n D (cid:19) where, for all λ ∈ D , D ( λ ) = 0 and E ( λ ) = E ∼ n ( λ ). Let τ , . . . , τ k ∈ T and α k +1 , . . . , α n ∈ D be the royal nodes of x in D repeated according to multiplicity. ByProposition 5.14, the royal polynomial of x is R = r k Y j =1 Q τ j n Y j = k +1 Q α j , for some r >
0. Thus for all λ ∈ T , λ − n R ( λ ) = rλ − n ( k Y j =1 ( λ − τ j )(1 − τ j λ ) n Y j = k +1 ( λ − α j )(1 − α j λ ) ) = r k Y j =1 | λ − τ j | n Y j = k +1 | λ − α j | . (6.23)By Proposition 5.3 and equation (6.23), for all λ ∈ T ,(6.24) | D ( λ ) | − | E ( λ ) | = λ − n R ( λ ) = r k Y j =1 | λ − τ j | n Y j = k +1 | λ − α j | . Assume first that n is even and write n = 2 m . This implies that k ≤ m . Define apolynomial g by g ( λ ) = τ . . . τ k λ m − k k Y j =1 ( λ − τ j ) . Clearly, the polynomial g has degree m + k ≤ n . Moreover, g is n -symmetric since g ∼ n ( λ ) = λ n g (cid:0) λ (cid:1) = λ n ( τ . . . τ k λ ) m − k k Y j =1 ( 1 λ − τ j ) ) = λ m ( τ . . . τ k λ m − k k Y j =1 ( 1 λ − τ j ) ) = τ . . . τ k λ m − k k Y j =1 ( λ − τ j ) = g ( λ ) . Let E t = E + tg and E t = E ∼ n + tg for t ∈ R . The polynomial E t has degree at most n . We also have, for all λ ∈ D , (cid:0) E t (cid:1) ∼ n ( λ ) = (cid:0) E ∼ n + tg (cid:1) ∼ n ( λ )= (cid:0) E ∼ n (cid:1) ∼ n ( λ ) + (cid:0) tg (cid:1) ∼ n ( λ ) = (cid:0) E t + tg (cid:1) ( λ ) = E t ( λ ) . (6.25)Note that, on T , | D | − | E t | = | D | − | E + tg | = | D | − ( E + tg )( E + tg )= | D | − | E | − t | g | − tgE ) . (6.26)Let k E k ∞ denote the supremum of | E | on T . Then, for all λ ∈ T ,Re( tg ( λ ) E ( λ ) ≤ | tg ( λ ) E ( λ ) | = | tE ( λ ) | (cid:12)(cid:12)(cid:12)(cid:12) τ . . . τ j λ m − k k Y j =1 ( λ − τ j ) (cid:12)(cid:12)(cid:12)(cid:12) = | tE ( λ ) | k Y j =1 | λ − τ j | ≤ | t |k E k ∞ k Y j =1 | λ − τ j | . (6.27)Note that, for all λ ∈ T , | g ( λ ) | = (cid:12)(cid:12) τ . . . τ k λ m − k k Y j =1 ( λ − τ j ) (cid:12)(cid:12) = (cid:12)(cid:12) k Y j =1 ( λ − τ j ) (cid:12)(cid:12) . ATIONAL E -INNER FUNCTIONS 43 Combine equations (6.24) and (6.26) and inequality (6.27), for all λ ∈ T , to get | D ( λ ) | − | E t ( λ ) | = | D ( λ ) | − | E ( λ ) | − | t | | g ( λ ) | − (cid:0) tg ( λ ) E ( λ ) (cid:1) , (by equation (6.26))= r k Y j =1 | λ − τ j | n Y j = k +1 | λ − α j | − | t | | g ( λ ) | − (cid:0) tg ( λ ) E ( λ ) (cid:1) , (by equation (6.24)) ≥ r k Y j =1 | λ − τ j | n Y j = k +1 | λ − α j | − | t | | g ( λ ) | − | t |k E k ∞ k Y j =1 | λ − τ j | , (by inequality (6.27))= k Y j =1 | λ − τ j | r n Y j = k +1 | Q α j ( λ ) | − | t | (cid:12)(cid:12) k Y j =1 | λ − τ j | (cid:12)(cid:12) − | t |k E k ∞ k Y j =1 | λ − τ j | ≥ k Y j =1 | λ − τ j | ( rM − (cid:18) | t | k Y j =1 | λ − τ j | + 2 | t |k E k ∞ (cid:19)) ≥ k Y j =1 | λ − τ j | ( rM − | t | (cid:0) | t | k g k ∞ + 2 k E k ∞ (cid:1)) , where M = inf λ ∈ T Q nj = k +1 | Q α j ( λ ) | > | t | sufficiently small, | D ( λ ) | − | E t ( λ ) | ≥ T . It suffices tofind | t | such that rM − | t | (cid:0) | t | k g k ∞ + 2 k E k ∞ (cid:1) > , or equivalently, | t | (cid:18) | t | + 2 k E k ∞ k g k ∞ (cid:19) − rM k g k ∞ < . If we take | t | ≤ min (cid:26) k E k ∞ k g k ∞ , rM k E k ∞ (cid:27) , then | t | (cid:18) | t | + 2 k E k ∞ k g k ∞ (cid:19) − rM k g k ∞ ≤ | t | (cid:18) k E k ∞ k g k ∞ + 2 k E k ∞ k g k ∞ (cid:19) − rM k g k ∞ ≤ rM k E k ∞ (cid:18) k E k ∞ k g k ∞ (cid:19) − rM k g k ∞ = rM k g k ∞ − rM k g k ∞ = − rM k g k ∞ < . (6.28)Therefore | D | − | E t | ≥ T , and, by Theorem 5.17, the functions x + t = (cid:18) E + t D , E + t D , D ∼ n D (cid:19) and x − t = (cid:18) E − t D , E − t D , D ∼ n D (cid:19) are rational E -inner function. Obviously,12 x + t + 12 x − t = (cid:18) E + t + E − t D , E + t + E − t D , D ∼ n D (cid:19) = (cid:18) E + tg + E − tg D , E ∼ n + tg + E ∼ n − tg D , D ∼ n D (cid:19) = (cid:18) E D , E ∼ n D , D ∼ n D (cid:19) = x. Hence x is not an extreme point of J .If n is odd, assume n = 2 m + 1. This case requires a slight modification. By as-sumption, 2 k ≤ n thus 2 k ≤ m + 1. This implies that k ≤ m . Choose ω ∈ T suchthat ω = − τ k Y j =1 τ j . Let g ( λ ) = ωλ m − k ( λ − τ ) k Y j =1 ( λ − τ j ) , λ ∈ C . Clearly, the polynomial g has degree m + k +1 ≤ n . Let us check that the g is n -symmetric g ∼ n ( λ ) = λ n g (cid:0) /λ (cid:1) = λ n (cid:18) ω λ m − k (cid:0) λ − τ (cid:1) k Y j =1 (cid:0) λ − τ j (cid:1) (cid:19) = λ n (cid:18) ω λ m − k (cid:0) λ − τ (cid:1) k Y j =1 (cid:0) λ − τ j (cid:1) (cid:19) = ωλ m − k τ (cid:0) τ − λ (cid:1) k Y j =1 τ j (cid:0) τ j − λ (cid:1) = ω (cid:18) − τ k Y j =1 τ j (cid:19) λ m − k (cid:0) λ − τ (cid:1) k Y j =1 (cid:0) λ − τ j (cid:1) = ω ω λ m − k (cid:0) λ − τ (cid:1) k Y j =1 (cid:0) λ − τ j (cid:1) = ω λ m − k (cid:0) λ − τ (cid:1) k Y j =1 (cid:0) λ − τ j (cid:1) = g ( λ ) . As in the even case, define the polynomials on D E t = E + tg and E t = E ∼ n + tg for t ∈ R . Similar to equation (6.25), for all λ ∈ D , E t ( λ ) = (cid:0) E t (cid:1) ∼ n ( λ ) and similar to equation(6.26), for all λ ∈ T ,(6.29) | D ( λ ) | − | E t ( λ ) | = | D ( λ ) | − | E ( λ ) | − t | g ( λ ) | − (cid:0) tg ( λ ) E ( λ ) (cid:1) . ATIONAL E -INNER FUNCTIONS 45 For all λ ∈ T ,Re( tgE ( λ ) ≤ | tgE ( λ ) | = | tE ( λ ) | (cid:12)(cid:12)(cid:12)(cid:12) ωλ m − k ( λ − τ ) k Y j =1 ( λ − τ j ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ | t | k E k ∞ | λ − τ | k Y j =1 | λ − τ j | . Combine equations (6.24), (6.29) and inequality (6.30) to obtain, for all λ ∈ T , | D ( λ ) | − | E t ( λ ) | = | D ( λ ) | − | E ( λ ) | − | t | | g ( λ ) | − (cid:0) tg ( λ ) E ( λ ) (cid:1) = r k Y j =1 | λ − τ j | n Y j = k +1 | λ − α j | − | t | | g ( λ ) | − (cid:0) tg ( λ ) E ( λ ) (cid:1) , by equation (6.24) ≥ r k Y j =1 | λ − τ j | n Y j = k +1 | λ − α j | − | t | | g ( λ ) | − | t |k E k ∞ | λ − τ | k Y j =1 | λ − τ j | , by inequality(6.30)= k Y j =1 | λ − τ j | r n Y j = k +1 | Q α j ( λ ) | − | t | | λ − τ | k Y j =1 | λ − τ j | − | t |k E k ∞ | λ − τ | k Y j =1 | λ − τ j | ≥ k Y j =1 | λ − τ j | ( rM − | λ − τ | | {z } ≤ (cid:18) | t | | λ − τ | k Y j =1 | λ − τ j | | {z } ≤| t | k g k ∞ +2 | t |k E k ∞ (cid:19)) ≥ k Y j =1 | λ − τ j | ( rM − (cid:0) | t | k g k ∞ + 2 | t |k E k ∞ (cid:1)) where M = inf T Q | Q α j | >
0. By similar arguments as in equations (6.28), one can find | t | such that rM − (cid:0) | t | k g k ∞ + 2 | t |k E k ∞ (cid:1) > | D | − | E t | ≥ , on T . Hence, by Theorem 5.17, the functions x ± t = (cid:18) E ± t D , (cid:0) E ∼ n (cid:1) ± t D , D ∼ n D (cid:19) are rational E -inner functions. One can check that x = x + t + x − t and therefore x isnot an extreme point of J . (cid:3) Theorem 6.20. [6, Theorem 5.13]
A rational Γ -inner function h ∈ R n,k Γ is extreme inthe set of rational Γ -inner functions if and only if k > n . Proposition 6.21.
Let x = ( x , x , x ) ∈ R n,k be a rational E -inner function such that x = x and k > n . Then x is an extreme point of the set J of rational E -innerfunctions. Proof.
By Lemma 4.8 (1), the function h = ( s, p ) = (2 x , x ) is Γ-inner. By Theorem4.15, there are polynomials E , E , D such that x = (cid:0) E D , E D , D ∼ n D (cid:1) . Here, since x = x ,necessarily E = E . By Definition 3.7, the royal polynomial R h of h is R h ( λ ) = D ( λ ) (cid:18) x − x (cid:19) ( λ )= D ( λ ) (cid:18) D ∼ n D − E D (cid:19) ( λ )= 4 (cid:0) DD ∼ n − E (cid:1) ( λ ) = 4 R x ( λ ) . It is clear that if x ∈ R n,k , then h has degree n and k royal nodes on T , counted withmultiplicities, such that 2 k > n . Thus, by Theorem 6.20, h is an extreme point of theset of rational Γ-inner functions. That is, if h = ( s , p ) and h = ( s , p ) are Γ-innerfunctions such that h = th + (1 − t ) h for some t ∈ (0 , , then h = h = h . Note that, in this case, we have(6.30) ( s = ts + (1 − t ) s ⇒ s = s = s p = tp + (1 − t ) p ⇒ p = p = p . Suppose x = tx + (1 − t ) x , for some t ∈ (0 , E -inner functions x = ( x , x , x ) and x = ( x , x , x ). This impliesthat x = tx + (1 − t ) x x = tx + (1 − t ) x x = p = tx + (1 − t ) x . Recall that ( s, p ) = (2 x , x ), hence(6.31) s = 2 tx + 2(1 − t ) x s = 2 tx + 2(1 − t ) x p = tx + (1 − t ) x . Therefore ( s, p ) = t (cid:0) x , x (cid:1) + (1 − t ) (cid:0) x , x (cid:1) and ( s, p ) = t (cid:0) x , x (cid:1) + (1 − t ) (cid:0) x , x (cid:1) . Since h is an extreme rational Γ-inner function, we have x = 2 x = s x = 2 x = sx = x = p. Therefore x = x = x . Hence x is extreme in the set J . (cid:3) Remark 6.22.
It is not clear that a rational E -inner function x = ( x , x , x ) ∈ R n,k such that 2 k > n and x = x , is an extreme point of the set J of rational E -innerfunctions. Could we claim that in Lemma 6.16, if τ i ∈ T is a royal node of x ofmultiplicity ν , then τ i is a royal node of x and x of the same multiplicity ν ? Here ATIONAL E -INNER FUNCTIONS 47 x = tx + (1 − t ) x for some t such that 0 < t <
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