Orthogonal Rational Functions on the Unit Circle with Prescribed Poles not on the Unit Circle

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1 Symmetry, Integrability and Geometry: Methods and Applications SIGMA 13 (27), 090, 49 pages Orthogonal Rational Functions on the Unit Circle with Prescribed Poles not on the Unit Circle Adhemar BULTHEEL, Ruyman CRUZ-BARROSO and Andreas LASAROW Department of Computer Science, KU Leuven, Belgium URL: Department of Mathematical Analysis, La Laguna University, Tenerife, Spain Fak. Informatik, Mathematik & Naturwissenschaften, HTWK Leipzig, Germany Received August, 27, in final form November 20, 27; Published online December 03, 27 Abstract. Orthogonal rational functions (ORF) on the unit circle generalize orthogonal polynomials (poles at infinity) and Laurent polynomials (poles at zero and infinity). In this paper we investigate the properties of and the relation between these ORF when the poles are all outside or all inside the unit disk, or when they can be anywhere in the extended complex plane outside the unit circle. Some properties of matrices that are the product of elementary unitary transformations will be proved and some connections with related algorithms for direct and inverse eigenvalue problems will be explained. Key words: orthogonal rational functions; rational Szegő quadrature; spectral method; rational Krylov method; AMPD matrix 20 Mathematics Subject Classification: 30D15; 30E05; 42C05; 44A60 1 Introduction Orthogonal rational functions (ORF) on the unit circle are well known as generalizations of orthogonal polynomials on the unit circle (OPUC). The pole at infinity of the polynomials is replaced by poles in the neighborhood of infinity, i.e., poles outside the closed unit disk. The recurrence relations for the ORF generalize the Szegő recurrence relations for the polynomials. If µ is the orthogonality measure supported on the unit circle, and L 2 µ the corresponding Hilbert space, then the shift operator T µ : L 2 µ L 2 µ : f(z) zf(z) restricted to the polynomials has a representation with respect to the orthogonal polynomials that is a Hessenberg matrix. However, if instead of a polynomial basis, one uses a basis of orthogonal Laurent polynomials by alternating between poles at infinity and poles at the origin, a full unitary representation of T µ with respect to this basis is a five-diagonal CMV matrix 12]. The previous ideas have been generalized to the rational case by Velázquez in 47]. He showed that the representation of the shift operator with respect to the classical ORF is not a Hessenberg matrix but a matrix Möbius transform of a Hessenberg matrix. However, a full unitary representation can be obtained if the shift is represented with respect to a rational analog of the Laurent polynomials by alternating between a pole inside and a pole outside the unit disk. The resulting matrix is a matrix Möbius transform of a five-diagonal matrix. This paper is a contribution to the Special Issue on Orthogonal Polynomials, Special Functions and Applications (OPSFA14). The full collection is available at

2 2 A. Bultheel, R. Cruz-Barroso and A. Lasarow Orthogonal Laurent polynomials on the real line, a half-line, or an interval were introduced by Jones et al. 31, 32] in the context of moment problems, Padé approximation and quadrature and this was elaborated by many authors. González-Vera and his co-workers were in particular involved in extending the theory where the poles zero and infinity alternate (the so-called balanced situation) to a more general case where in each step either infinity or zero can be chosen as a pole in any arbitrary order 8, 20]. They also identify the resulting orthogonal Laurent polynomials as shifted versions of the orthogonal polynomials. Hence the orthogonal Laurent polynomials satisfy the same recurrence as the classical orthogonal polynomials after an appropriate shifting and normalization is embedded. The corresponding case of orthogonal Laurent polynomials on the unit circle was introduced by Thron in 40] and has been studied more recently in for example 15, 18]. Papers traditionally deal with the balanced situation like in 18] but in 15] also an arbitrary ordering was considered. Only in 16] Cruz-Barroso and Delvaux investigated the structure of the matrix representation with respect to the basis of the resulting orthogonal Laurent polynomials on the circle. They called it a snake-shaped matrix which generalizes the five diagonal matrix. The purpose of this paper is to generalize these ideas valid for Laurent polynomials on the circle to the rational case. That is to choose the poles of the ORF in an arbitrary order either inside or outside the unit disk. We relate the resulting ORF with the ORF having all their poles outside or all their poles inside the disk, and study the corresponding recurrence relations. With respect to this new orthogonal rational basis, the shift operator will be represented by a matrix Möbius transformation of a snake-shaped matrix. In the papers by Lasarow and coworkers (e.g., 23, 24, 25, 34]) matrix versions of the ORF are considered. In these papers also an arbitrary choice of the poles is allowed but only with the restrictive condition that if is used as a pole, then 1/ cannot be used anymore. This means that for example the balanced situation is excluded. One of the goals of this paper is to remove this restriction on the poles. In the context of quadrature formulas, an arbitrary sequence of poles not on the unit circle was also briefly discussed in 19]. The sequence of poles considered there need not be Newtonian, i.e., the poles for the ORF of degree n may depend on n. Since our approach will emphasize the role of the recurrence relation for the ORF, we do need a Newtonian sequence, although some of the results may be generalizable to the situation of a non-newtonian sequence of poles. One of the applications of the theory of ORF is the construction of quadrature formulas on the unit circle that form rational generalizations of the Szegő quadrature. They are exact in spaces of rational functions having poles inside and outside the unit disk. The nodes of the quadrature formula are zeros of para-orthogonal rational functions (PORF) and the weights are all positive numbers. These nodes and weights can (like in Gaussian quadrature) be derived from the eigenvalue decomposition of a unitary truncation of the shift operator to a finite-dimensional subspace. One of the results of the paper is that there is no gain in considering an arbitrary sequence of poles inside and outside the unit disk unless in a balanced situation. When all the poles are chosen outside the closed unit disk or when some of them are reflected in the circle, the same quadrature formula will be obtained. The computational effort for the general case will not increase but neither can it reduce the cost. In network applications or differential equations one often has to work with functions of large sparse matrices. If A is a matrix and the matrix function f(a) allows the Cauchy representation f(a) = Γ f(z)(z A) 1 dµ(z), where Γ is a contour encircling all the eigenvalues of A then numerical quadrature is a possible technique to obtain an approximation for f(a). If for example Γ is the unit circle, then expressions like u f(a)u for some vector u can be approximated by quadrature formulas discussed in this paper which will be implemented disguised as Krylov subspace methods (see for example 27, 29, 33]).

3 Orthogonal Rational Functions on the Unit Circle 3 The purpose of the paper though is not to discuss quadrature in particular. It is just an example application that does not require much extra introduction of new terminology and notation. The main purpose however is to give a general framework on which to build for the many applications of ORFs. Just like orthogonal polynomials are used in about every branch of mathematics, ORFs can be used with the extra freedom to exploit the location of the poles. For example, it can be shown that the ORFs can be used to solve multipoint moment problems as well as more general rational interpolation problems where locations of the poles inside and outside the circle are important for the engineering applications like system identification, model reduction, filtering, etc. When modelling the transfer function of a linear system, poles should be chosen inside as well as outside the disk to guarantee that the transient as well as the steady state of the system is well modelled. It would lead us too far to also include the interpolation properties of multipoint Padé approximation and the related applications in several branches of engineering. We only provide the basics in this paper so that it can be used in the context of more applied papers. The interpretation of the recursion for the ORFs as a factorization of a matrix into elementary unitary transformations illustrates that the spectrum of the resulting matrix is independent of the order in which the elementary factors are multiplied. As far as we know, this fact was previously unknown in the linear algebra community, unless in particular cases like unitary Hessenberg matrices. As an illustration, we develop some preliminary results in Section 11 in a linear algebra setting that is slightly more general than the ORF situation. In the last decades, many papers appeared on inverse eigenvalue problems for unitary Hessenberg matrices and rational Krylov methods. Some examples are 4, 30, 35, 36, 37, 38, 44]. These use elementary operations that are very closely related to the recurrence that will be discussed in this paper. However, they are not the same and often miss the flexibility discussed here. We shall illustrate some of these connections with certain algorithms from the literature in Section 12. The outline of the paper is as follows. In Section 2 we introduce the main notations used in this paper. The linear spaces and the ORF bases are given in Section 3. Section 4 brings the Christoffel Darboux relations and the reproducing kernels which form an essential element to obtain the recurrence relation given in Section 5 but also for the PORF in Section 6 to be used for quadrature formulas in Section 7. The alternative representation of the shift operator is given in Section 8 and its factorization in elementary 2 2 blocks in the subsequent Section 9. We end by drawing some conclusions about the spectrum of the shift operator and about the computation of rational Szegő quadrature formulas in Section 10. The ideas that we present in this paper, especially the factorization of unitary Hessenberg matrices in elementary unitary factors is also used in the linear algebra literature mostly in the finite-dimensional situation. These elementary factors and what can be said about the spectrum of their product is the subject of Section 11. These elementary unitary transformations are intensively used in numerical algorithms such as Arnoldi-based Krylov methods where they are known as core transformations. Several variants of these rational Krylov methods exist. The algorithms are quite similar yet different from our ORF recursion as we explain briefly in Section 12 illustrating why we believe the version presented in this paper has superior advantages. 2 Basic def initions and notation We use the following notation. C denotes the complex plane, Ĉ the extended complex plane (one point compactification), R the real line, R the closure of R in Ĉ, T the unit circle, D the open unit disk, D = D T, and E = Ĉ \ D. For any number z Ĉ we define z = 1/z (and set 1/0 =, 1/ = 0) and for any complex function f, we define f (z) = f(z ).

4 4 A. Bultheel, R. Cruz-Barroso and A. Lasarow To approximate an integral I µ (f) = f(z)dµ(z), T where µ is a probability measure on T one may use Szegő quadrature formulas. The nodes of this quadrature can be computed by using the Szegő polynomials. Orthogonality in this paper will always be with respect to the inner product f, g = f(z)g(z)dµ(z). T The weights of the n-point quadrature are all positive, the nodes are on T and the formula is exact for all Laurent polynomials f span{z k : k n 1}. This has been generalized to rational functions with a set of predefined poles. The corresponding quadrature formulas are then rational Szegő quadratures. This has been discussed in many papers and some of the earlier results were summarized in the book 9]. We briefly recall some of the results that are derived there. The idea is the following. Fix a sequence = ( k ) k N with = { k } k N D, and consider the subspaces of rational functions defined by { } p n (z) n L 0 = C, L n = π n (z) : p n P n, π n (z) = (1 k z), n 1, where P n is the set of polynomials of degree at most n. These rational functions have their poles among the points in = { j = 1/ j : j }. We denote the corresponding sequence as = ( j ) j N. Let φ n L n \ L n 1, and φ n L n 1 be the nth orthogonal rational basis function (ORF) in a nested sequence. It is well known that these functions have all their zeros in D (see, e.g., 9, Corollary 3.1.4]). However, the quadrature formulas we have in mind should have their nodes on the circle T. Therefore, para-orthogonal rational functions (PORF) are introduced. They are defined by Q n (z, τ) = φ n (z) + τφ n(z), τ T, where besides the ORF φ n (z) = pn(z) π, also the reciprocal function n(z) φ n(z) = p n(z) π = zn p n (z) n(z) π n(z) is introduced. These PORF have n simple zeros {ξ nk } n k=1 T (see, e.g., 9, Theorem 5.2.1]) so that they can be used as nodes for the quadrature formulas I n (f) = n w nk f(ξ nk ) k=1 k=1 and the weights are all positive, given by w nk = 1/ n 1 j=0 φ j (ξ nj ) 2 (see, e.g., 9, Theorem 5.4.2]). These quadrature formulas are exact for all functions of the form {f = g h: g, h L n 1 } = L n 1 L (n 1) (see, e.g., 9, Theorem 5.3.4]). The purpose of this paper is to generalize the situation where the j are all in D to the situation where they are anywhere in the extended complex plane outside T. This will require the introduction of some new notation. So consider a sequence with D and its reflection in the circle β = (β j ) j N where β j = 1/ j = j E. We now construct a new sequence γ = (γ j ) j N where each γ j is either equal to j or β j. Partition {1, 2,..., n} (n N = N { }) into two disjoint index sets: the ones where γ j = j and the indices where γ j = β j : an = {j : γ j = j D, 1 j n} and bn = {j : γ j = β j E, 1 j n},

5 Orthogonal Rational Functions on the Unit Circle 5 and define n = { j : j an} and β n = {β j : j bn}. It will be useful to prepend the sequence with an extra point 0 = 0. That means that β is preceded by β 0 = 1/ 0 =. For γ, the initial point can be γ 0 = 0 = 0 or γ 0 = β 0 =. With each of the sequences, β, and γ we can associate orthogonal rational functions. They will be closely related as we shall show. The ORF for the γ sequence can be derived from the ORF for the sequence by multiplying with a Blaschke product just like the orthogonal Laurent polynomials are essentially shifted versions of the orthogonal polynomials (see, e.g., 15]). To define the denominators of our rational functions, we introduce the following elementary factors: ϖ j (z) = 1 j z, ϖ β j (z) = { 1 β j z, if β j, z, if β j =, Note that if j = 0 and hence β j = then ϖj (z) = 1 but ϖβ j (z) = z. To separate the and the β-factors in a product, we also define { { ϖ j ϖj (z) =, if γ j = j, and ϖ β j 1, if γ j = β j, (z) = ϖ β j, if γ j = β j, 1, if γ j = j. ϖ γ j (z) = { ϖ j (z), if γ j = n, ϖ β j (z), if γ j = β n. Because the sequence γ is our main focus, we simplify the notation by removing the superscript γ when not needed, e.g., ϖ j = ϖ γ j = ϖ j ϖβ j etc. We can now define for ν {, β, γ} π ν n(z) = n ϖj ν (z) j=1 and the reduced products separating the and the β-factors π n(z) = n j=1 ϖ j (z) = n ϖ j (z), π n(z) β = j an j=1 ϖ β j (z) = j bn ϖ j (z), so that π n (z) = n ϖ j (z) = π n(z) π n(z). β j=1 We assume here and in the rest of the paper that products over j equal 1. The Blaschke factors are defined for ν {, β, γ} as ζ ν j (z) = σ ν j z ν j 1 ν j z, σν j = ν j ν j, if ν j {0, }, ζ ν j (z) = σ ν j z = z, σ ν j = 1, if ν j = 0, ζ ν j (z) = σ ν j /z = 1/z, σ ν j = 1, if ν j =. Thus ν j σj ν = ν j, for ν j {0, }, 1, for ν j {0, }.

6 6 A. Bultheel, R. Cruz-Barroso and A. Lasarow Because σn = σn, β we can remove the superscript and just write σ n. If we also use the following notation which maps complex numbers onto T z T, z C \ {0}, u(z) = z 1, z {0, }, then σ j = u( j ) = u(β j ) = u(γ j ). Set (ϖj ν) (z) = ϖj ν (z) = zϖν j (z) (e.g., (1 jz) = z j if ν = ), then ζj ν ϖj = σ ν j ϖj ν. Later we shall also use πn ν to mean n ϖj ν. Note that ζ j = ζ β j = 1/ζβ j. Moreover if j = 0 j=1 and hence β j =, then ϖj (z) = z and ϖ β j (z) = 1. Next define the finite Blaschke products for ν {, β} B ν 0 = 1, and B ν n(z) = n ζj ν (z), n = 1, 2,.... j=1 It is important to note that here ν γ. For the definition of Bn γ = B n see below. Like we have split up the denominators π n = π n π n β in the -factors and the β-factors, we also define for n 1 ζ j = { ζ j, if γ j = j, 1, if γ j = β j, ζ β j = { ζ β j, if γ j = β j, 1, if γ j = j, and Ḃ n(z) = n j=1 ζ j (z) = n ζ j (z), and Ḃn(z) β = j an j=1 ζ β j (z) = j bn ζ j (z), so that we can define the finite Blaschke products for the γ sequence: B n (z) = {Ḃ n (z), if γ n = n, Ḃ β n(z), if γ n = β n. Note that the reflection property of the factors also holds for the products: Bn = (Bn) β = 1/Bn, β B n = 1/B n, and (Ḃ nḃβ n) = 1/(Ḃβ nḃ n). However, Ḃn = = ζ β j = 1/ζ β j 1/ζ β j = 1/Ḃβ n. j an ζ j j an j an j bn 3 Linear spaces and ORF bases We can now introduce our spaces of rational functions for n 0: L ν n = span{b0 ν, B1 ν,..., Bn}, ν ν {, β, γ}, and L ν n = span{ḃν 0, Ḃν 1,..., Ḃν n}, ν {, β}. The dimension of L ν n is n + 1 for ν {, β, γ}, but note that the dimension of L ν n for ν {, β} can be less than n+1. Indeed some of the Ḃν j may be repeated so that for example the dimension of L n is only an + 1 with an the cardinality of an and similarly for ν = β. Hence for ν = γ: L n = span{b 0,..., B n } = span { Ḃ 0, Ḃ 1,..., Ḃ n, Ḃβ 1,..., Ḃβ n} = L n + L β n = L n L β n.

7 Orthogonal Rational Functions on the Unit Circle 7 Because for n 1 Ḃn = = 1 ζj ζ β j an j an j and Ḃ β n = j bn ζ β j = 1 ζj j bn, it should be clear that Bk = Ḃ k /Ḃβ k and Bβ k = Ḃβ k /Ḃ k, hence that { } { L n = span Ḃ 0, Ḃ 1 Ḃ β 1,..., Ḃ n Ḃ β n and Occasionally we shall also need the notation ς n = j an σ j T, ς β n = j bn L β n = span σ j T, and ς n = Ḃ 0, Ḃβ 1,..., Ḃ1 n σ j T. Lemma 3.1. If f L n then f/ḃβ n L n and f/ḃ n L β n. In other words L n = Ḃβ nl n = Ḃ nl β n. This is true for all n 0 if we set Ḃ 0 = Ḃβ 0 = 1. Proof. This is trivial for n = 0 since then L n = C. If f L n, and n 1 then it is of the form Therefore f(z) = p n(z) π n (z) = f(z) Ḃ β n(z) = ςβ n p n (z) π n(z) π β n(z), p n P n. p n (z) π β n(z) π n(z) π β n(z) π β n (z) = ςβ n p n (z) π n(z) π β n (z). Recall that ϖ β j = 1 and σ j = 1 if β j = (and hence j = 0), we can leave these factors out and we shall write for the product instead of, the dot meaning that we leave out all the factors for which j = 1/β j = 0. ς β n π β n (z) = j bn β j β j (z β j ) = j bn j j (z 1/ j ) = j bn j=1 j 1 j z, Ḃ β n Ḃ n }. and thus f(z) Ḃ β n(z) = c n p n (z) n (1 j z) j=1 L n, c n = j bn ( j ) 0. The second part is similar. Lemma 3.2. With our previous definitions we have for n 1 { } ḂnL β n 1 = span Bk Ḃβ n = Ḃ k Ḃ β Ḃn β : k = 0,..., n 1 k = ζ n β span{b 0, B 1,..., B n 1 } = ζ nl β n 1, and similarly ḂnL β n 1 {B = span β k Ḃ n = Ḃβ k Ḃ Ḃk n : k = 0,..., n 1 = ζ n span{b 0, B 1,..., B n 1 } = ζ n L n 1. }

8 8 A. Bultheel, R. Cruz-Barroso and A. Lasarow Proof. By our previous lemma Ḃβ nl n 1 = ζ nḃβ β n 1 L n 1 = ζ nl β n 1. The second relation is proved in a similar way. To introduce the sequences of orthogonal rational functions (ORF) for the different sequences ν, ν {, β, γ} recall the inner product that we can write with our ( ) -notation as f, g = T f (z)g(z)dµ(z) where µ is assumed to be a probability measure positive a.e. on T. Then the orthogonal rational functions (ORF) with respect to the sequence ν with ν {, β, γ} are defined by φ ν n L ν n \ L ν n 1 with φν n L ν n 1 for n 1 and we choose φν 0 = 1. Lemma 3.3. The function φ nḃβ n belongs to L n and it is orthogonal to the n-dimensional subspace ζ β nl n 1 for all n 1. Similarly, the function φ β nḃ n belongs to L n and it is orthogonal to the n-dimensional subspace ζ n L n 1, n 1. Proof. First note that φ nḃβ n L n by Lemma 3.1. By definition φ n L n 1 Ḃν. Thus by Lemma 3.2 and because f, g = n f, Ḃν ng, Ḃ β nφ n Ḃβ nl n 1 = ζ β nl n 1. The second claim follows by symmetry. Note that ζ β nl n 1 = L n 1 if γ n = n. Thus, up to normalization, φ nḃβ n is the same as φ n and similarly, if γ n = β n then φ n and φ β nḃ n are the same up to normalization. Lemma 3.4. For n 1 the function Ḃ n(φ n) belongs to L n and it is orthogonal to ζ n L n 1. Similarly, for n 1 the function Ḃβ n(φ β n) belongs to L n and it is orthogonal to ζ β nl n 1. Proof. Since φ nḃβ n ζ β nl n 1, (φ nḃβ n) ζ β n L (n 1), and thus by Lemma 3.2 and because P (n 1) Ḃ n 1Ḃβ n 1 L (n 1) = Ḃ n 1Ḃβ n 1 π (n 1) πβ (n 1) = P n 1 π n 1 πβ n 1 = L n 1 it follows that Ḃ nφ n = Ḃ nḃβ n(φ nḃβ n) ζ n Ḃ n 1Ḃβ n 1 L (n 1) = ζ n L n 1. The other claim follows by symmetry. We now define the reciprocal ORFs by (recall f (z) = f(1/z)) (φ ν n) = B ν n(φ ν n), ν {, β}. For the ORF in L n however we set φ n = Ḃ nḃβ n(φ n ). Note that by definition B n is either Ḃ n or Ḃβ n depending on γ n being n or β n, while in the previous definition we do not multiply with B n but with the product Ḃ nḃβ n. The reason is that we want the operation ( ) to be a map from L ν n to L ν n for all ν {, β, γ}.

9 Orthogonal Rational Functions on the Unit Circle 9 Remark 3.5. As the operation ( ) is a map from L ν n to L ν n, it depends on n and on ν. So to make the notation unambiguous we should in fact use something like f ν,n] if f L ν n. However, in order not to overload our notation, we shall stick to the notation f since it should always be clear from the context what the space is to which f will belong. Note that we also used the same notation to transform polynomials. This is just a special case of the general definition. Indeed, a polynomial of degree n belongs to L n for a sequence where all j = 0, j = 0, 1, 2,... and for this sequence B n(z) = z n. Note that for a constant a L 0 = Ĉ we have a = a. Although ( ) is mostly used for scalar expressions, we shall occasionally use A where A is a matrix whose elements are all in L n. Then the meaning is that we take the ( ) conjugate of each element in its transpose. Thus if A is a constant matrix, then A has the usual meaning of the adjoint or complex conjugate transpose of the matrix. We shall need this in Section 8. Remark 3.6. It might also be helpful for the further computations to note the following. If p n is a polynomial of degree n with a zero at ξ, then p n will have a zero at ξ = 1/ξ. Hence, if ν {, β, γ} and φ ν n = pν n π ν, then φ ν n n = pν n π = pν n ν n π. We know by 9, Corollary 3.1.4] that n ν φ n has all its zeros in D, hence p n does not vanish in E and p n does not vanish in D. By symmetry φ β n has all its zeros in E and p β n does not vanish in E. For the general φ n, it depends on γ n being n or β n. However from the relations between φ n and φ n or φ β n that will be derived below, we will be able to guarantee that at least for z = ν n we have φ ν n (ν n ) 0 and p ν n (ν n ) 0 for all ν {, β, γ} (see Corollary 3.11 below). The orthogonality conditions define φ n and φ n uniquely up to normalization. So let us now make the ORFs unique by imposing an appropriate normalization. First assume that from now on the φ ν n refer to orthonormal functions in the sense that φ ν n = 1. This makes them unique up to a unimodular constant. Defining this constant is what we shall do now. Suppose γ n = n, then φ n and φ nḃβ n are both in L n and orthogonal to L n 1 (Lemma 3.3). If we assume φ n = 1 and φ n = 1, hence φ nḃβ n = φ n = 1, it follows that there must be some unimodular constant s n T such that φ n = s nφ nḃβ n. Of course, we have by symmetry that for γ n = β n, there is some s β n T such that φ n = s β nφ β nḃ n. To define the unimodular factors s n and s β n, we first fix φ n and φ β n uniquely as follows. We know that φ n has all its zeros in D and hence φ n has all its zeros in E so that φ n ( n ) 0. Thus we can take φ n ( n ) > 0 as a normalization for φ n. Similarly for φ β n we can normalize by φ β n (β n ) > 0. In both cases, we have made the leading coefficient with respect to the basis {Bj ν}n j=0 positive since φ n(z) = φ n ( n )Bn(z) + ψn 1 (z) with ψ n 1 L n 1 and φβ n(z) = φ β n (β n )Bn(z) β + ψ β n 1 (z) with ψβ n 1 Lβ n 1. Before we define the normalization for the γ sequence, we prove the following lemma which is a consequence of the normalization of the φ n and the φ β n. Lemma 3.7. For the orthonormal ORFs, it holds that φ n = φ β n and (φ n) Ḃn β = φ β nḃ n hence also (φ β n) Ḃn = φ nḃβ n for all n 0. and Proof. For n = 0, this is trivial since φ 0, φ 0, φβ 0, Ḃ 0 and Ḃβ 0 are all equal to 1. We give the proof for n 1 and γ n = n (for γ n = β n, the proof is similar). Since by previous lemmas Ḃβ n(φ β n) and φ nḃβ n are both in L n and orthogonal to L n 1, and since Ḃβ n(φ β n) = φ β n = 1 and φ nḃβ n = φ n = 1, there must be some s n T such that s n φ nḃβ n = φ β n Ḃβ n or s n φ n = φ β n.

10 10 A. Bultheel, R. Cruz-Barroso and A. Lasarow Multiply with B β n = B n and evaluate at β n to get s n φ n(β n )B n (β n ) = φ β n (β n ) > 0. Thus s n should arrange for 0 < s n φ n(1/ n )B n (1/ n ) = s n φ n ( n )B n( n ) = s n φ n ( n ), and since φ n ( n ) > 0, it follows that s n = 1. Because (φ n) = B nφ n = B nφ β n and B n = Ḃ n/ḃβ n, also the other claims follow. For the normalization of the φ n, we can do two things: either we make the normalization of φ n simple and choose for example φ n(γ n ) > 0, similar to what we did for φ n and φ β n (but this is somewhat problematic as we shall see below), or we can insist on keeping the relation with φ n and φ β n simple as in the previous lemma and arrange that s n = s β n = 1. We choose for the second option. Let us assume that γ n = n. Denote φ n (z) = p n (z) π n(z) π β n(z) and φ n(z) = p n(z) π n(z), with p n and p n both polynomials in P n. Then φ n(z) = ς n p n(z) π n(z) π β n(z) and n (z) = ς n p n (z) πn(z), ς n = φ n σ j. j=1 We already know that there is some s n T such that φ n = s nḃβ nφ n. Take the ( ) conjugate and multiply with Ḃ nḃβ n to get φ n = s nḃβ nφ n. It now takes some simple algebra to reformulate φ n = s nḃβ nφ n as φ n(z) = ς n p n(z) π n(z) π β n(z) = s n ς n p n (z) π n(z) π β n(z) j bn ( β j ). This implies that p n(z) = s np n (z) ( β j bn j ) and thus that p n(z) has the same zeros as p n (z), none of which is in D. Thus the numerator of φ n will not vanish at n D but one of the factors (1 β j n ) from π n( β n ) could be zero. Thus a normalization φ n( n ) > 0 is not an option in general. We could however make s n = 1 when we choose p n( n )/p n ( n ) > 0 or, since φ n ( n ) > 0, this is equivalent with ς n p n( n )/πn( n ) > 0. Yet another way to put this is requiring that φ n(z)/ḃβ n(z) is positive at z = n. This does not give a problem with 0 or since Ḃ n(z)φ n (z) = φ n(z) Ḃ β n(z) = ς n p n(z) π n(z) (z β j bn j), ς n = j an σ j. (3.1) It is clear that neither the numerator nor the denominator will vanish for z = n. Of course a similar argument can be given if γ n = β n. Then we choose φ n(z)/ḃ n(z) to be positive at z = β n or equivalently ς n p n(β n )/πn(β β n ) ( j an j ) > 0. Let us formulate the result about the numerators as a lemma for further reference. Lemma 3.8. With the normalization that we just imposed the numerators p ν n of φ ν n = p ν n/π ν n, ν {, β, γ} and n 1 are related by p n (z) = p n(z) j bn ( β j ) = pn β (z)ς n j an ( j ), if γ n = n

11 Orthogonal Rational Functions on the Unit Circle 11 and p n (z) = p β n(z) j an where as before ς n = n j=1 σ j. ( j ) = p n (z)ς n j bn ( β j ), if γ n = β n, Proof. The first expression for γ n = n has been proved above. The second one follows in a similar way from the relation φ n (z) = φ β n (z)ḃ n(z). Indeed p n (z) π n (z) = = j an j an ς n p β n (z) ϖ β j (z) ϖ β j (z) j bn j bn j an j an σ j z j 1 j z ς n p β n (z) ϖj (z) z j ϖ β j (z) σ j 1 β j z = ς np β n (z) π n (z) With σ j j = j the result follows. The case γ n = β n is similar. j an σ j ( j ) z j z j. Note that this normalization again means that we take the leading coefficient of φ n to be positive in the following sense. If γ n = n then φ n (z) = (Ḃ nφ n )( n )Ḃ n(z) + ψ n 1 (z) with ψ n 1 L n 1, while Ḃ nφ n = φ n and φ n ( n ) > 0. If γ n = β n then φ n (z) = (Ḃβ nφ n )(β n )Ḃβ n(z) + ψ n 1 (z) with ψ n 1 L n 1 and the conclusion follows similarly. Whenever we use the term orthonormal, we assume this normalization and {φ n : n = 0, 1, 2,...} will denote this orthonormal system. Thus we have proved the following Theorem. It says that if γ n = n, then φ n is a shifted version of φ n where shifted means multiplied by Ḃβ n: Ḃ β n(z)φ n (z) = Ḃβ n(z)a 0 B a n B n(z)] = a 0 Ḃ β n(z) + + a n Ḃ n(z), and a similar interpretation if γ n = β n. We summarize this in the following theorem. Theorem 3.9. Assume all ORFs φ ν n, ν {, β, γ} are orthonormal with positive leading coefficient, i.e., { φ n ( n ) > 0 and φ β (φ n (β n ) > 0 and n/ḃβ n)( n ) > 0 if γ n = n, (φ n/ḃ n)(β n ) > 0 if γ n = β n. Then for all n 0 while φ n = (φ n)ḃβ n = (φ β n) Ḃ n and φ n = (φ n) Ḃ β n = (φ β n)ḃ n if γ n = n, φ n = (φ β n)ḃ n = (φ n) Ḃ β n and φ n = (φ β n) Ḃ n = (φ n)ḃβ n if γ n = β n. Corollary We have for all n 1 that (φ ν n) ζnl ν ν n 1, ν {, β, γ}. Corollary The rational functions φ n and φ n are in L n and hence have all their poles in {β j : j = 1,..., n} E while the zeros of φ n are all in D and the zeros of φ n are all in E. The rational functions φ β n and φ β n are in L β n and hence have all their poles in { j : j = 1,..., n} D while the zeros of φ β n are all in E and the zeros of φ β n are all in D. The rational functions φ n and φ n are in L n and hence have all their poles in {β j : j an} { j : j bn}. The zeros of φ n are the same as the zeros of φ n and thus are all in D if γ n = n and they are the same as the zeros of φ β n and thus they are all in E if γ n = β n.

12 12 A. Bultheel, R. Cruz-Barroso and A. Lasarow Proof. It is well known that the zeros of φ n are all in D 9, Corollary 3.1.4], and because φ β n = φ n, this means that the zeros of φ β n are all in E. Because φ n = (φ n)ḃβ n = (φ n)/ j bn ζ j if γ n = n, i.e., n an, and the product with Ḃβ n will only exchange the poles 1/ j = β j, j bn in φ n for poles j = 1/β j, the zeros of φ n are left unaltered. The proof for n bn is similar. One may summarize that for f L ν n the f transform reflects both zeros and poles in T since z z = 1/z, while the transform f f as it is defined in the spaces L ν n, ν {, β, γ}, keeps the poles but reflects the zeros since the multiplication with the respective factors B n, B β n and Ḃ nḃβ n will only undo the reflection of the poles that resulted from the f operation. 4 Christof fel Darboux relations and reproducing kernels For ν {, β, γ}, one may define the reproducing kernels for the space L ν n. Given the ORF φ ν k, the kernels are defined by k ν n(z, w) = n φ ν k (z)φν k (w). k=0 They reproduce f L ν n by k ν n(, z), f = f(z). The proof of the Christoffel Darboux relations goes exactly like in the classical case and we shall not repeat it here (see, e.g., 9, Theorem 3.1.3]). Theorem 4.1. The Christoffel Darboux relations k ν n(z, w) = φν n (z)φ ν n (w) ζ ν n(z)ζ ν n(w)φ ν n(z)φ ν n(w) 1 ζ ν n(z)ζ ν n(w) = φν n+1 (z)φν n+1 (w) φν n+1 (z)φν n+1 (w) 1 ζ ν n+1 (z)ζν n+1 (w) hold for n 0, ν {, β, γ} and z, w not among the poles of φ ν n and not on T. As an immediate consequence we have: Theorem 4.2. The following relations hold true: k n(z, w)ḃβ n(z)ḃβ n(w) = k n (z, w) = k β n(z, w)ḃ n(z)ḃ n(w) for n 0 and z, w (T {β j : j an} { j : j bn}). Proof. The first relation was directly shown above for the case γ n = n. It also follows in the case γ n+1 = n+1 and using in the second CD-relation the first expressions from Theorem 3.9 for φ n+1 and φ n+1. The relation is thus valid independent of γ n = n or γ n = β n. Similarly the second expression was derived before in the case γ n = β n, but again, it also follows from the second CD-relation and the first expressions from Theorem 3.9 for φ n+1 and φ n+1 in the case γ n+1 = β n+1. Again the relation holds independently of γ n = n or γ n = β n. Alternatively, the second relation can also be derived from the second CD-relation in the case γ n+1 = n+1 but using the second expressions from Theorem 3.9 for φ n+1 and φ n+1. Evaluation of the CD-relation in ν n for ν {, β} results in another useful corollary.

13 Orthogonal Rational Functions on the Unit Circle 13 Corollary 4.3. For ν {, β} we have for n 0 k ν n(z, ν n ) = φ ν n (z)φ ν n (ν n ) and k ν n(ν n, ν n ) = φ ν n (ν n ) 2. The latter corollary cannot immediately be used when ν = γ because γ n could be equal to some pole of φ n if it equals some 1/γ j for j < n. In that case we can remove the denominators in the CD relation and only keep the numerators. Hence setting k n (z, w) = C n(z, w) π n (z)π n (w), the CD relation becomes φ n(z) = p n(z) π n (z), φ n(z) = ς np n(z) π n (z), ς n T, C n (z, w) = p n(z)p n(w) ζ n (z)ζ n (w)p n (z)p n (w) 1 ζ n (z)ζ n (w) Thus, the first form gives = p n+1 (z)p n+1 (w) p n+1(z)p n+1 (w) (1 ζ n+1 (z)ζ n+1 (w))ϖ n+1 (z)ϖ n+1 (w). (4.1) C n (z, γ n ) = p n(z)p n(γ n ) and C n (γ n, γ n ) = p n(γ n ) 2. Evaluating a polynomial at infinity means taking its highest degree coefficient, i.e., if q n (z) is a polynomial of degree n, then q n ( ) stands for its coefficient of z n. The second form of (4.1) gives for γ n+1 and γ n C n (z, γ n ) = p n+1 (z)p n+1 (γ n) p n+1 (z)p n+1 (γ n ) (1 γ n z)(1 γ n+1 2 ) C n (γ n, γ n ) = p n+1 (γ n) 2 p n+1 (γ n ) 2 (1 γ n 2 )(1 γ n+1 2 ). and Coupling the first and the second form in (4.1) gives p n+1 (γ n) 2 p n+1 (γ n ) 2 (1 γ n 2 )(1 γ n+1 2 ) = p n(γ n ) 2. For γ n+1 = and γ n we get C n (z, γ n ) = p n+1 (z)p n+1 (γ n) p n+1 (z)p n+1 (γ n ) (1 γ n z) = p n(z)p n(γ n ) and C n (γ n, γ n ) = p n+1 (γ n) 2 p n+1 (γ n ) 2 (1 γ n 2 ) = p n(γ n ) 2. If γ n+1 = and γ n =, the denominators in (4.1) have to be replaced by 1, which gives C n (z, γ n ) = p n+1(z)p n+1 (γ n) p n+1 (z)p n+1 (γ n ) = p n(z)p n(γ n ) and C n (γ n, γ n ) = p n+1(γ n ) 2 p n+1 (γ n ) 2 = p n(γ n ) 2.

14 14 A. Bultheel, R. Cruz-Barroso and A. Lasarow For γ n = and γ n+1 we obtain in a similar way C n (z, γ n ) = p n+1 (z)p n+1 (γ n) p n+1 (z)p n+1 (γ n ) z(1 γ n+1 2 ) = p n(z)p n(γ n ) and C n (γ n, γ n ) = p n+1 (γ n) 2 p n+1 (γ n ) 2 (1 γ n+1 2 ) = p n(γ n ) 2. To summarize, the relations of Corollary 4.3 may not hold for the ORF if ν = γ, but similar relations do hold for the numerators as stated in the next corollary. Corollary 4.4. If C n (z, w) is the numerator in the CD relation and p n (z) is the numerator of the ORF φ n for the sequence γ then we have for n 0 C n (z, γ n ) = p n(z)p n(γ n ) and C n (γ n, γ n ) = p n(γ n ) 2. 5 Recurrence relation The recurrence for the φ n is well known. For a proof see, e.g., 9, Theorem 4.1.1]. For φ β n the proof can be copied by symmetry. However, also for ν = γ the same recurrence and its proof can be copied, with the exception that the derivation fails when p n(γ n 1 ) = 0 where p n = φ n π n. This can (only) happen if (1 γ n )(1 γ n 1 ) < 0 (i.e., one of these γ s is in D and the other is in E). We shall say that φ n is regular if p n(γ n 1 ) 0. If ν = or ν = β then the whole sequence (φ ν n) n 0 will be automatically regular. Thus we have the following theorem: Theorem 5.1. Let ν {, β, γ} and if ν = γ assume moreover that φ ν n is regular, then the following recursion holds with initial condition φ ν 0 = φν 0 = 1 ] φ ν n (z) φ ν n (z) = Hn ν ϖn 1 ν (z) ϖn(z) ν 1 λ ν n λ ν n 1 ] ζ ν n 1 (z) ] ] φ ν n 1 (z) φ ν n 1 (z), where Hn ν is a nonzero constant times a unitary matrix: ] Hn ν = e ν η ν n1 0 n 0 ηn2 ν, e ν n C \ {0}, ηn1, ν ηn2 ν T. The constant ηn1 ν is chosen such that the normalization condition for the ORFs is maintained. The other constant ηn2 ν is then automatically related to ην n1 by ην n2 = ην n1 σ n 1σ n. The Szegő parameter λ ν n is given by λ ν n = ηn ν p ν n(ν n 1 ) p ν n (ν n 1 ) where φ ν n(z) = p ν n(z)/π ν n(z). with ηn ν p ν n 1 = ς (ν n 1) n 2 p ν n 1 (ν n 1) T, Proof. We immediately concentrate on the general situation ν = γ. Of course ν = and ν = β will drop out as special cases. For simplicity assume that γ n and γ n 1 are not 0 and not. The technicalities when this is not true are left as an exercise. It is easy because formally it follows the steps of the proof below but one has to replace a linear factor involving infinity by the coefficient of infinity (like 1 z = z and z = 1) and evaluating a polynomial at means taking its leading coefficient.

15 Orthogonal Rational Functions on the Unit Circle 15 First we show that there are some numbers c n and d n such that φ(z) := 1 γ nz z γ n 1 φ n (z) d n φ n 1 (z) c n 1 γ n 1 z z γ n 1 φ n 1 L n 2. This can be written as N(z)/(z γ n 1 )π n 1 (z)]. Thus the c n and d n are defined by the conditions N(γ n 1 ) = N(1/γ n 1 ) = 0. If we denote φ k = p k /π k and thus φ k = p k ς k/π k, it is clear that N(z) = p n (z) d n (z γ n 1 )p n 1 (z) c n (1 γ n 1 z)p n 1(z)ς n 1. Thus the first condition gives c n = and the second one d n = ς n 1 p n (γ n 1 ) (1 γ n 1 2 )p n 1 (γ n 1) p n (1/γ n 1 ) (1/γ n 1 γ n 1 )p n 1 (1/γ n 1 ) = p n(γ n 1 ) (1 γ n 1 2 )p n 1 (γ n 1). Note that p n 1 (γ n 1) cannot be zero by Corollary 3.11, and that also p n(γ n 1 ) does not vanish by our assumption of regularity. Furthermore, by using the orthognality of φ n and φ n 1 and Corollary 3.10, it is not difficult to show that φ L n 2 so that it must be identically zero. Thus with φ n (z) = d n σ n 1 1 γ n 1 z 1 γ n z ζ n 1(z)φ n 1 (z) + λ n φ n 1(z)], λ n = η n p n (γ n 1 ) p n(γ n 1 ), η n = ς n 1 σ n 1 p n 1 (γ n 1) p n 1 (γ n 1). By taking the ( ) transform (in L n ) we obtain φ n(z) = d n σ n 1 γ n 1 z 1 γ n z λ nζ n 1 (z)φ n 1 (z) + φ n 1(z)]. This proves the recurrence by taking e n = d n and η n1 = σ n 1 u(d n ). It remains to show that the initial step for n = 1 is true. Since φ 0 = φ 0 γ 0 = 0 = 0, hence ζ 0 = z, we have = 1, then in case φ 1 (z) = e 1 η 11 z + λ 1 1 γ 1 z and φ 1(z) = e 1 η 12 λ 1 z γ 1 z. Thus p 1 (z) = e 1 η 11 (z + λ 1 ) and p 1(z) = e 1 η 11 (λ 1 z + 1). This implies that λ 1 is indeed given by the general formula because λ 1 = η 1 p 1 (γ 0 ) p 1 (γ 0) = p 1(0) p 1 (0) = e 1η 11 λ 1 e 1 η 11.

16 16 A. Bultheel, R. Cruz-Barroso and A. Lasarow In case γ 0 = β 0 =, then ζ 0 = 1/z, so that and thus φ 1 (z) = e 1 η 11 1 λ 1 z 1 γ 1 z and φ 1(z) = e 1 η 12 λ 1 z 1 γ 1 z, p 1 (z) = e 1 η 11 (1 + λ 1 z) and p 1(z) = e 1 η 11 σ 1 (λ 1 + z), and again λ 1 is given by the general formula λ 1 = η 1 p 1 (γ 0 ) p 1 (γ 0) = 1p 1( ) p 1 ( ) = e 1η 11 λ 1 e 1 η 11. This proves the theorem. Remark 5.2. If ν {, β} we could rewrite λ ν n in terms of φ ν n because by dividing and multiplying with the appropriate denominators π ν n one gets λ ν n = ηn ν φ ν n(ν n 1 ) φ ν n (ν n 1 ), (1 ν n ν n 1) ην n = σ n 1 σ n (1 ν n ν n 1 ) φ ν n 1 (ν n 1) φ ν n 1 (ν n 1), n 1. Note that also this η ν n T, but it differs from the η ν n in the previous theorem. However if ν = γ, then this expression has the disadvantage that γ n 1 could be equal to 1/γ n or it could be equal to a pole of φ n in which case it would not make sense to evaluate these expressions in γ n 1. The latter expressions only make sense if we interpret them as limiting values φ ν n(ν n 1 ) φ ν n (ν n 1 ) = (1 ν n ν n 1 ) (1 ν n ν n 1 ) lim z ν n 1 φ ν n(z) φ ν n (z) φ ν n 1 (ν n 1) φ ν n 1 (ν n 1) = and lim (1 ν n z) z ν n 1 (1 ν n z) φ ν n 1 (z) φ ν n 1 (z), where one has to assume that limξ/ξ] = 1. We shall from now on occasionally use these ξ 0 expressions with this interpretation, but the expressions for λ ν n from Theorem 5.1 using the numerators are more direct since they immediately give the limiting values. Note that λ n is the value of a Blaschke product with all its zeros in D evaluated at n 1 D and therefore λ n D. Similarly, λ β n is the value of a Blaschke product with all its zeros in E, evaluated at β n 1 E so that λ β n D. Since the zeros of φ n are the zeros of φ n if n an and they are the zeros of φ β n if n bn, it follows that if n and n 1 are both in an or both in bn, then λ n D but if n an and n 1 bn or vice versa, then λ n E. Therefore (e ν n) 2 = 1 ν n 2 1 ν n λ ν n 2 > 0 (5.1) and we can choose e n as the positive square root of this expression. The above expression is derived in 9, Theorem 4.1.2] for the case ν = by using the CD relations. By symmetry, this also holds for ν = β. For ν = γ, the same relation can be obtained by stripping the denominators as we explained after the proof of the CD-relation in Section 4. What goes wrong with the recurrence relation when φ n is not regular? From the proof of Theorem 5.1, it follows that then d n = 0. We still have the relation σ n 1 φ n (z) = p (1 γ n 1 2 )p n 1 (γ n (γ n 1 )ζ n 1 (z)φ n 1 (z) + s n 1 p n (γ n 1 )φ n 1(z) ] n 1)

17 Orthogonal Rational Functions on the Unit Circle 17 with s n 1 = ς n 1p n 1 (γ n 1) σ n 1 p n 1 (γ n 1) T and p n(γ n 1 ) = 0. Thus, there is some positive constant e n and some η n1 T such that ϖ n 1 (z) φ n (z) = e n η n1 0 ζn 1 (z)φ n 1 (z) + φ ϖ n (z) n 1(z) ], i.e., the first term in the sum between square brackets vanishes. Applying Theorem 5.1 in this case would give λ n =, and the previous relations show that we only have to replace in Theorem 5.1 the matrix ] ] 1 λ ν n 0 1 λ ν by. n This is in line with how we have dealt with so far where the rule of thumb was to set a νb = b if ν =. So let us therefore also use Theorem 5.1 with this interpretation when φ n is not regular and thus λ n =. With the expressions at the end of Section 4, it can also be shown that in this case e 2 n = 1 γ n 2 1 γ n 1 2 > 0. Note that this corresponds to replacing 1 λ n 2 when λ n = by 1. Since this non-regular situation can only occur when (1 γ n )(1 γ n 1 ) < 0, this expression for e 2 n is indeed positive. A similar rule can be applied if γ n or γ n 1 is infinite, just replace in this or previous expression 1 2 by 1. The positivity of the expressions for e 2 n also follows from the following result. Theorem 5.3. The Szegő parameters satisfy for all n 1: If γ n = n and γ n 1 = n 1 then λ n = λ n = λ β n D. If γ n = β n and γ n 1 = β n 1 then λ n = λ β n = λ n D. If γ n = n and γ n 1 = β n 1 then λ n = 1/λ β n = 1/λ n E. If γ n = β n and γ n 1 = n 1 then λ n = 1/λ n = 1/λ β n E. Proof. Suppose γ n = n and γ n 1 = n 1, then by Theorems 5.1 and 3.9, or better still by Lemma 3.8, ( p n 1 λ n = ς ( ) ( n 1) p n ( n 1 ) n 2 p n 1 ( n 1) p n( n 1 ) = p n 1 ς ( ) n 1) p n( n 1 ) n 2 p n 1 ( n 1) p n ( n 1 ) = λ n. When using p n(z) = ς n p β n (z) n j=1 ( j ) and j = 1/β j, the previous relation becomes ς n 1 p β n 1 λ n = ς (1/β n 1) n 2 pβ n (1/β n 1 ) ς n 1 p β n 1 (1/β n 1) p β n(1/β n 1 ) ( = σn 1ς 2 p β n (β n 1 ) β n 1 ) n 1 p β n(β n 1 ) βn 1 n n 2 p β n (β n 1 ) βn 1 n 1 p β n (β n 1 ) β n n 1 = σn 1 2 β n 1 λ β n = λ β n. β n 1 The proof for γ n = β n and γ n 1 = β n 1, n 1 is similar. Next consider γ n = n and γ n 1 = β n 1, then p β n 1 λ n = ς (β n 1) n 2 p n(β n 1 ) p β n 1 (β n 1) p n (β n 1 ) = p β ηβ n (β n 1 ) n p β n(β n 1 ) = ηβ nη β n λ β n The remaining case γ n = β n and γ n 1 = n 1, is again similar. = 1 λ β n = 1 λ. n

18 18 A. Bultheel, R. Cruz-Barroso and A. Lasarow Remark 5.4. It should also be clear that the expression for the parameters λ ν n of Theorem 5.1 are for theoretical use only. They are expressed in terms of p ν n, which is the numerator of φ ν n, the object that should be the result of the computation, and hence unknown at the moment of computing λ ν n. For practical use, these parameters λ ν n should be obtained in a different way. In inverse eigenvalue problems the recursion is used backwards, i.e., for decreasing degrees of the ORFs and then these expressions can be used of course. Even if we know the λ ν n, then in Theorem 5.1, we still need the normalizing factor ηn1 ν T which is characterized by ηn1 ν is chosen such that the normalization condition or the ORFs is maintained. We could compute φ ν n with ηn1 ν = 1, check the value of φν n (γ n 1 ), and derive ηn1 ν from it. What this means is shown in the next lemma. Lemma 5.5. For ν {, β}, the phase θn ν of the unitary factor ηn1 ν = eiθν n θn ν = arg ( σ n 1 σ n ϖn(ν ν n 1 )φ ν n (ν n 1 ) ) or equivalently is given by η ν n1 = σ n 1 σ n u ( ϖ ν n(ν n 1 )φ ν n (ν n 1 ) ). (Recall u(z) = z/ z.) Proof. Take the first relation of Theorem 5.1 and evaluate for z = ν n 1, then because ϖ ν n 1 (ν n 1) = 0 we get or φ ν n(ν n 1 )ϖ ν n(ν n 1 ) = e ν nη ν n1ϖ ν n 1(ν n 1 )λ ν nφ ν n 1(ν n 1 ) η ν n1 = ϖ ν n(ν n 1 )φ ν n(ν n 1 ) e ν nϖ ν n 1 (ν n 1)λ ν nφ ν n 1 (ν n 1). φ ν n (ν n 1) Use the definition of λ ν n = ηn ν with φ ν n (ν n 1) ην ϖ n = σ n 1 σ n(ν ν n 1) n ϖn 1 ν (νn) and knowing that φν n 1 (ν n 1) > 0 we obtain after simplification and leaving out all the factors with phase zero θn ν = arg ( σ n 1 σ n ϖn(ν ν n 1 )φ ν n (ν n 1 ) ) as claimed. Note that this expression for ηn1 ν is well defined because φν n (ν n 1 ) 0 if ν {, β}. For ν = γ, the expression is a bit more involved but it can be obtained in a similar way from the normalization conditions given in Theorem 3.9. We skip the details. Another solution of the recurrence relation is formed by the functions of the second kind. Like in the classical case (i.e., for ν = ) we can introduce them for ν {, β, γ} by (see 9, p. 83]) ψn(z) ν = E(t, z)φ ν n(t) D(t, z)φ ν n(z)]dµ(t). T where D(t, z) = t+z t z ψ ν 0 = 1 and ψ ν n(z) = 2t and E(t, z) = D(t, z) + 1 = t z. This results in D(t, z)φ ν n(t) φ ν n(z)]dµ(t), n 1, which may be generalized to (see 9, Lemma 4.2.2]) ψn(z)f(z) ν = D(t, z)φ ν n(t)f(t) φ ν n(z)f(z)]dµ(t), n 1 T T

19 Orthogonal Rational Functions on the Unit Circle 19 with f arbitrary in L ν (n 1). It also holds that (see 9, Lemma 4.2.3]) ψn ν (z)g(z) = D(t, z)φ ν n (t)g(t) φ ν n (z)g(z)]dµ(t), n 1 T with g arbitrary in L ν n (ν n ). Recall that L ν n (ν n ) is the space of all functions in L ν n that vanish for z = ν n = 1/ν n. This space is spanned by {Bk ν/bν n : k = 0,..., n 1} if ν {, β}. For ν = γ, the space can be characterized as (see Lemma 3.1) { } n 1 { } n 1 { } B k Bk B β n 1 k L n (γ n ) = span = span = span Ḃ nḃβ n ζ n Ḃn 1 ζ n Ḃ β. n 1 k=0 k=0 Theorem 5.6. The following relations for the functions of the second kind hold for n 0: while ψ n = (ψ n)ḃβ n = (ψ β n) Ḃ n and ψ n = (ψ n) Ḃ β n = (ψ β n)ḃ n if γ n = n, ψ n = (ψ β n)ḃ n = (ψ n) Ḃ β n and ψ n = (ψ β n) Ḃ n = (ψ n)ḃβ n if γ n = β n. We assume the normalization of Theorem 3.9. Proof. This is trivial for n = 0, hence suppose n 1 and γ n = n then ψ n (z) = D(t, z)φ n (t) φ n (z)]dµ(t) = D(t, z)φ n(t)ḃβ n(t) φ n(z)ḃβ n(z)]dµ(t) T = ψ n(z)ḃβ n(z), T k=0 because Ḃ β n(z) = Ḃβ n 1 (z) = ζ β j (z) = ζ j (z) L (n 1). j bn 1 j bn 1 Moreover, using φ n = φ β n we also have ψ n (z) = D(t, z)φ n (t) φ n (z)]dµ(t) = D(t, z)φ n(t)ḃβ n(t) φ n(z)ḃβ n(z)]dµ(t) T T = D(t, z)φ β n (t)ḃβ n(t) φ β n (z)ḃβ n(z)]dµ(t) T = D(t, z)φ β n (t)ḃ n(t) φ β n (z)ḃ n(z)]dµ(t), T and since Ḃ n L β n (β n ), we also get the second part: ψ n = (ψn) β Ḃn. Moreover ψn = ψ nḃβ n] Ḃ nḃβ n = ψ n Ḃ nḃβ nḃβ n = ψ n Ḃ n/ḃβ n]ḃβ n = ψn Ḃβ n. It follows in a similar way that ψn = ψ nḃ β n. The case γ n = β n is proved similarly. With these relations, it is not difficult to mimic the arguments of 9, Theorem 4.2.4] and obtain the following. Theorem 5.7. These functions satisfy the recurrence relation ] ψ ν n (z) ψn ν = H ν ϖn 1 ν (z) ] ] ] 1 λ ν n ζ ν n 1 (z) 0 ψ ν n 1 (z) n (z) ϖn(z) ν λ ν n ψn 1 ν (z), n 1 with ψ ν 0 = ψν 0 = 1 and all other quantities as in Theorem 5.1.

20 20 A. Bultheel, R. Cruz-Barroso and A. Lasarow 6 Para-orthogonal rational functions Before we move on to quadrature formulas, we define para-orthogonal functions (PORF) by Q ν n(z, τ ν n) = φ ν n(z) + τ ν nφ ν n (z), τ ν n T, ν {, β, γ}. (6.1) The PORF Q ν n(z, τn) ν is in L ν n and it is called para-orthogonal because it is not orthogonal to L ν n 1 but it is orthogonal to a particular subspace of Lν n 1 of dimension n 1. Theorem 6.1. The para-orthogonal rational function Q n (z) = Q n (z, τ n ), τ n T with n 2 is orthogonal to ζ n L n 1 ζ nl β n 1 = ζ n L n 1 L n 1 = L n 1 (γ n ) with { } ϖ L n 1 (γ n ) = {f L n 1 : f(γ n ) = 0} = n (z)p n 2 (z) : p n 2 P n 2. π n 1 (z) Recall that ϖ n(z) = z γ n if γ n, and ϖ n(z) = 1 if γ n =. Proof. Suppose γ n = n then ζ n = ζn and ζ n β = 1. Hence φ n L n 1 and φ n ζn L n 1 and therefore Q n L n 1 ζn L n 1 = L n 1 ( n ). The proof for γ n = β n is similar. One may also define associated functions of the second kind as P ν n (z, τ n ) = ψ ν n(z) τ ν nψ ν n (z), ν {, β, γ}. From Theorems 3.9 and 5.6 we can easily obtain the following corollary. Corollary 6.2. With the notation Q n and P n for the PORF and the associated functions just introduced, we have for n 1 while Similarly while Q n (z, τ n ) = Ḃβ n(z)q n(z, τ n ) = τ n Ḃ n(z)q β n(z, τ n ) if γ n = n, Q n (z, τ n ) = Ḃ n(z)q β n(z, τ n ) = τ n Ḃ β n(z)q n(z, τ n ) if γ n = β n. P n (z, τ n ) = Ḃβ n(z)p n (z, τ n ) = τ n Ḃ n(z)p β n (z, τ n ) if γ n = n, P n (z, τ n ) = Ḃ n(z)p β n (z, τ n ) = τ n Ḃ β n(z)p n (z, τ n ) if γ n = β n. Proof. Assume that γ n = n, then φ n = φ nḃβ n and φ n = (φ n) Ḃ β n. Thus Q n (, τ n ) = Ḃβ n φ n + τ n (φ n) ] = Ḃβ nq n(, τ n ). In a similar way one has Q n (, τ n ) = φ β n Ḃ n + τ n φ β nḃ n = τ n Ḃn φ β n + τ n φ β n ] = τ n ḂnQ β n(, τ n ). The proofs for P n and for γ n = β n are similar. We are now ready to state that the zeros of the para-orthogonal rational functions Q ν n(z, τ ν n) will be simple and on T no matter whether ν =, β or γ.