Full Terms & Conditions of access and use can be found at https://www.tandfonline.com/action/journalInformation?journalCode=gitr20 Integral Transforms and Special Functions ISSN: 1065-2469 (Print) 1476-8291 (Online) Journal homepage: https://www.tandfonline.com/loi/gitr20 Orthogonal polynomials on the unit circle satisfying a second-order differential equation with varying polynomial coefficients J. Borrego-Morell & A. Sri Ranga To cite this article: J. Borrego-Morell & A. Sri Ranga (2017) Orthogonal polynomials on the unit circle satisfying a second-order differential equation with varying polynomial coefficients, Integral Transforms and Special Functions, 28:1, 39-55, DOI: 10.1080/10652469.2016.1249866 To link to this article: https://doi.org/10.1080/10652469.2016.1249866 Published online: 12 Nov 2016. Submit your article to this journal Article views: 83 View Crossmark data https://www.tandfonline.com/action/journalInformation?journalCode=gitr20 https://www.tandfonline.com/loi/gitr20 https://www.tandfonline.com/action/showCitFormats?doi=10.1080/10652469.2016.1249866 https://doi.org/10.1080/10652469.2016.1249866 https://www.tandfonline.com/action/authorSubmission?journalCode=gitr20&show=instructions https://www.tandfonline.com/action/authorSubmission?journalCode=gitr20&show=instructions http://crossmark.crossref.org/dialog/?doi=10.1080/10652469.2016.1249866&domain=pdf&date_stamp=2016-11-12 http://crossmark.crossref.org/dialog/?doi=10.1080/10652469.2016.1249866&domain=pdf&date_stamp=2016-11-12 INTEGRAL TRANSFORMS AND SPECIAL FUNCTIONS, 2017 VOL. 28, NO. 1, 39–55 http://dx.doi.org/10.1080/10652469.2016.1249866 RESEARCH ARTICLE Orthogonal polynomials on the unit circle satisfying a second-order differential equation with varying polynomial coefficients J. Borrego-Morella and A. Sri Rangab aDepartamento de Matemática, Universidade Federal do Rio de Janeiro, Polo de Xerém, Rio de janeiro, Brazil; bDepartamento de Matemática Aplicada, Universidade Estadual Paulista, Campus São José do Rio Preto, São Paulo, Brazil ABSTRACT Consider the linear second-order differential equation An(z)y ′′ + Bn(z)y ′ + Cny = 0, (1.1) where An(z) = a2,nz2 + a1,nz + a0,n with a2,n �= 0, a21,n − 4a2,na0,n �= 0, ∀n ∈ N or a2,n = 0, a1,n �= 0, ∀n ∈ N, Bn(z) = b1,n + b0,nz are polynomials with complex coefficients and Cn ∈ C. Under some assumptions over a certain class of lowering and raising operators, we show that for a sequence of polynomials (φn) ∞ n=0 orthogonal on the unit circle to satisfy the differential equation (1.1), the polynomial φn must be of a specific form involving and extension of the Gauss and confluent hypergeometric series. ARTICLE HISTORY Received 13 May 2016 Accepted 11 October 2016 KEYWORDS Orthogonal polynomials on the unit circle; differential equations; special functions; complex analysis AMS SUBJECT CLASSIFICATION 42C05; 34M03; 33C05; 33C15; 30Axx 1. Introduction The Bochner classification theorem [1] (also given by Routh in 1885, [2]) characterizes, under a complex linear change of the variable z, the sequences (yn)∞n=0 of orthogonal poly- nomials with respect to a positive Borel measure having finite moments of all orders that simultaneously solve a second-order differential of the form A(z)y′′ + B(z)y′ + Cny = 0, whereA,B are polynomials of degree 2 and 1, respectively,Cn ∈ C. Such sequences of poly- nomials turn out to be the classical families of orthogonal polynomials Laguerre, Jacobi and Hermite. Askey [3] introduced the two-parameter system {Rn, Sn}n≥0 of polynomials given by Rn(z;α,β) = 2F1(−n,α + β + 1;β − α + 1 − n; z), Sn(z;α,β) = Rn(z;α,−β), (1.2) and pointed out that this system is biorthogonal with respect to the complex- valued weight of beta type ω(θ) = (1 − eiθ )α+β(1 − e−iθ )α−β = (2 − 2 cos θ)α(−eiθ )β , CONTACT J. Borrego-Morell jborrego@xerem.ufrj.br, jbmorell@gmail.com © 2016 Informa UK Limited, trading as Taylor & Francis Group http://www.tandfonline.com mailto:jborrego@xerem.ufrj.br, jbmorell@gmail.com 40 J. BORREGO-MORELL AND A. SRI RANGA θ ∈ [−π ,π],�(α) > − 1 2 . That is 1 2π ∫ π −π Rn(eiθ ;α,β)Sm(e−iθ ;α,β)ω(θ) dθ = �(2α + 1) �(α + β + 1)�(α − β + 1) n! (2α + 1)n δn,m, where � denotes the Euler Gamma function. The bi-orthogonality was stated in [3] in a slightly different form and a formal proof was given later in [4, p.16–17]. Other proofs of the bi-orthogonality have been given by several authors, please see [5] for some historical considerations. The author in [6] using a different approach also proved that when α ∈ R,α > − 1 2 and iβ ∈ R, the sequence (Rn(z;α,β))∞n=0 is orthogonal with respect to the weight ω, which is now positive and can be given by ω(θ) = 22α e(π−θ) (β) sin2α(θ/2). We also mention another known system of orthogonal polynomials on the unit circle of hypergeometric type which arise in a class of random unitary matrix ensembles, cf. [7], the circular Jacobi polynomials, defined as Cn(z; a) = 2F1(−n, a + 1;−a + 1 − n; z). These polynomials are a particular case of the Rn, by taking α = 2a and β = 0, we obtain Cn. From known results on hypergeometric functions, the element Rn in the orthogonal system (Rn(z;α,β))∞n=0 satisfies the differential equation z(1 − z)y′′ + (β − α + 1 − n − (−n + 2 + α + β)z)y′ + n(α + β + 1)y = 0. Hence, it is natural to question if there exists other classes of orthogonal polynomials on the unit circle satisfying a linear second-order differential equation similar to the Jacobi, Hermite and Laguerre systems of orthogonal polynomials. The above differential equation satisfied by the sequence (Rn)∞n=0 suggest that we should consider a differential equation with varying coefficients in the index n and the associated sequence of orthogonal polyno- mials as solution. In the present manuscript, under some assumptions on a certain class of lowering and raising operators, we give a necessary condition for the existence of a sequence of orthogonal polynomials solving the differential equations. We state the results and notation in the subsection below and in Section 2 we prove the results. 1.1. Statement of the results Let μ be a probability measure supported on an infinite subset of the unit circle. We say that (φn) ∞ n=0 is the sequence of orthonormal polynomials with respect to μ if ∫ |z|=1 φm(z)φn(z) dμ(z) = δm,n, where φn(z) = κnzn + lnzn−1 + lower order terms and κn > 0. Let �n(z) = φn(z)/κn be the monic polynomials. A general background to orthogonal polynomials systems on the unit circle can be found in the monographs [8–12]. More recent surveys in [13–17]. INTEGRAL TRANSFORMS AND SPECIAL FUNCTIONS 41 If f is a polynomial of degree n then the reverse polynomial f ∗ is znf (1/z̄), that is f ∗(z) = n∑ k=0 ākzn−k if f (z) = n∑ k=0 akzk and an �= 0. The sequence (φn) ∞ n=0 of orthonormal polynomials on the unit circle (OPUC for short) satisfy the recurrence relations [8, (11.4.6), (11.4.7)], κnzφn(z) = κn+1φn+1(z) − φn+1(0)φ∗ n+1(z), κnφn+1(z) = κn+1zφn(z) + φn+1(0)φ∗ n(z). (1.3) In terms of the monic polynomials, the above recursion also can be expressed as [14, (1.5.10), (1.5.40)] �n+1(z) = z�n(z) − ᾱn� ∗ n(z), (1.4) z�n(z) = ρ−2 n �n+1(z) + ᾱn� ∗ n+1(z), (1.5) where αn = −�n+1(0) and ρn = κn/κn+1. The coefficients (αn) are called the recursion coefficients or the Geronimus coefficients. Simon [14] makes a strong case and calls these coefficients as Verblunsky coefficients. Consider the hypergeometric equation z(1 − z)y′′ + (c − (a + b + 1)z)y′ − aby = 0, (1.6) we denote by 2F1(a, b; c; z) the Gauss hypergeometric series or hypergeometric function of the variable z with parameters a,b,c; cf. [18, p.56], defined as 2F1(a, b; c; z) = ∞∑ k=0 (a)k(b)k (c)k zk k! , |z| < 1, where c �= 0,−1,−2, . . ., which is a solution to (1.6), holomorphic at z=0 and can be extended appropriately by analytic continuation, see please [18, Section 2.1.4]. Here (a)k = a(a + 1) · · · (a + k − 1), k ∈ N; (a)0 = 1 denotes the Pochhammer symbol. All the solu- tions to (1.6) can be expressed in terms of the hypergeometric series (also expressible in terms of the Riemann P function, if we consider the point of view of Riemann for the hypergeometric equation). We will follow [18, p.57] in supplementing the definition of the hypergeometric series for the case c = −m,m ∈ N ∪ {0}. If a=−n or b=−n, where n ∈ N ∪ {0} and if c=−m wherem = n, n + 1, n + 2, . . ., then 2F1(−n, b;−m; z) = n∑ k=0 (−n)k(b)k (−m)k zk k! , (1.7) 2F1(a,−n;−m; z) = n∑ k=0 (a)k(−n)k (−m)k zk k! . (1.8) 42 J. BORREGO-MORELL AND A. SRI RANGA If c = −n, n ∈ N ∪ {0} and −a,−b /∈ {n, n − 1, . . . , 0}, we define 2F1(a, b;−n; z) = ∞∑ k=0 (a + n + 1)k(b + n + 1)k (n + 2)k zk k! = zn+1 2F1(a + n + 1, b + n + 1; n + 2; z), (1.9) which is also holomorphic at z=0 and a solution to (1.6). We unify our notation by using the symbol 2F1(a, b; c; z) to referring to 2F1(a, b; c; z) or (1.7)–(1.9) when the parameters a,b,c assume the values specified. Notice that 2F1 is a polynomial whenever −a or −b is a non-negative integer. In a similar way, we denote by 1F1 the confluent hypergeometric function or Kummer’s series 1F1(a; c; z) = ∞∑ k=0 (a)k (c)k zk k! , where c �= 0,−1,−2, . . ., which is holomorphic at z=0 and is a solution of the confluent hypergeometric equation, cf. [18, p.248] zy′′ + (c − z)y′ − ay = 0. If a=−n, where n ∈ N ∪ {0} and if c=−m wherem = n, n + 1, n + 2, . . ., then 1F1(−n;−m; z) = n∑ k=0 (−n)k (−m)k zk k! . (1.10) We define 1F1(a;−n; z) = ∞∑ k=0 (a + n + 1)k (n + 2)k zk k! = zn+1 1F1(a + n + 1; n + 2; z), (1.11) when c = −n, n ∈ N ∪ {0} and −a /∈ {n, n − 1, . . . , 0}. We will use the symbol 1F1(a; c; z) to refer to 1F1(a; c; z) or (1.10) and (1.11) when the parameters a,c assume the values specified. Let us consider the differential equation An(z)y′′ + Bn(z)y′ + Cny = 0, (1.12) whereAn(z) = a2,nz2 + a1,nz + a0,nwith a2,n �= 0, a21,n − 4a2,na0,n �= 0, ∀n ∈ N or a2,n = 0, a1,n �= 0, ∀n ∈ N, Bn(z) = b1,n + b0,nz and (ai,n)n∈N, i = 0, 1, 2; (bi,n)n∈N, i = 0, 1; (Cn)n∈N are sequences of complex numbers. Under a linear complex change in the variable z, the differential equation (1.12) can be transformed to θ(z)y′′ + (cn − bnz)y′ + λny = 0, (1.13) where (bn)n∈N, (cn)n∈N and (λn)n∈N are sequences of complex numbers and θ(z) = { z(1 − z), if a2,n �= 0, a21,n − 4a2,na0,n �= 0, ∀n ∈ N, z, if a2,n = 0, a1,n �= 0, ∀n ∈ N. We are interested in those differential equations of the form (1.13) for which there exists a unique sequence (φn) ∞ n=0 of OPUC with φn solving (1.13). INTEGRAL TRANSFORMS AND SPECIAL FUNCTIONS 43 It is clear that for every n ∈ N fixed, the existence and uniqueness of a polynomial solu- tion of degree n to the differential equation (1.13) will depend on the values assumed by the parameters bn, cn and λn. This study is done in the following proposition, whose proof will be given in Section 2. Proposition 1.1: Let n ∈ N be fixed. There exists a unique polynomial solution of degree n for the differential equation (1.13) if and only if λn = { n(n − 1 + bn), θ(z) = z(1 − z), nbn, θ(z) = z, (1.14) and bn �= { −2n + 2,−2n + 3, · · ·,−n,−n + 1, θ(z) = z(1 − z), 0, θ(z) = z. (1.15) This solution is πn(z) = const. { 2F1(−n, bn + n − 1; cn; z), if θ(z) = z(1 − z), 1F1(−n; cn; bnz), if θ(z) = z. By virtue of the above proposition, we study up to a complex linear change in the variable z, those sequences of OPUC satisfying a second-order differential equation of the form y′′ + Pn(z)y′ + Qn(z)y = 0, (1.16) where Pn(z) = cn − bnz θ(z) , Qn(z) = n(n − 1 + bn) θ(z) if θ(z) = z(1 − z), Pn(z) = cn − z θ(z) , Qn(z) = n θ(z) if θ(z) = z, and bn /∈ {−2n + 2,−2n + 3, . . . ,−n,−n + 1} for θ(z) = z(1 − z). 1.1.1. Raising and lowering operators for orthogonal polynomials on the unit circle. Let w be a weight function supported on a subset of the unit circle and assume that w is normalized by ∫ |ξ |=1 w(ξ) dξ iξ = 1, and define the external field v, cf. [19] w(z) = e−v(z). Ismail andWitte [20] derived raising and lowering operators L1,n and L2,n whose expres- sion is given by Ismail andWitte [20, (2.21)]. Under the assumptions thatw is differentiable in a neighbourhood of the unit circle and the Verblunsky coefficients do not vanish, these 44 J. BORREGO-MORELL AND A. SRI RANGA lowering and raising operators define a pair of second-order differential equations such that φ′′ n + P1,nφ ′ n + Q1,nφn = 0, φ′′ n + P2,nφ ′ n + Q2,nφn = 0, with coefficients given by Ismail and Witte [20, (2.22),(3.8)] and depending on functional coefficientsAn and Bn An(z) = n κn−1 κn + i κn−1 φn(0) z ∫ |ξ |=1 v′(z) − v′(ξ) z − ξ φn(ξ)φ∗ n(ξ)w(ξ) dξ , φ′ n(z) = An(z)φn−1(z) − Bn(z)φn(z). The authors proved that, cf. [20, Theorem 3.1 & Remark 3], if v is a meromorphic function on the unit disk then P1,n = P2,n, (1.17) therefore,Q1,n = Q2,n. From Ismail and Witte [20, Example 1], for the orthonormal circular Jacobi polynomi- als φn φn(z) = (a)n√ n!(2a + 1)n 2F1(−n, a + 1;−a + 1 − n; z), relation (1.17) gives φ′′ n + ( 1 − n − a z − 2a + 1 1 − z ) φ′ n + n(1 + a) z(1 − z) φn = 0, that is, by taking the weight associated with the circular Jacobi polynomials, the rela- tion (1.17) defines a linear differential equation of the form (1.16). It is reasonable to ask if we drop the differentiability condition over the weight and allow An and Bn varying in a wider class of functions, does the relation (1.17) give other classes of OPUC satisfying a linear differential equation of the form (1.13)? In Theorem 1.6, we give a necessary condition over (bn)n∈N and (cn)n∈N for this to happen. In particular, we obtain the system of OPUC defined by Rn which is orthogonal with respect to ω. Note that this weight is not in general a meromorphic function on the unit disk. Before we enunciate the results in this article, we introduce some auxiliary notation used in the sequel. We denote by �θ the region given by �θ = { D∗(0, ε0) ∪ D∗(1, ε1), θ(z) = z(1 − z), D∗(0, ε0), θ(z) = z, here D∗(0, ε0) and D∗(1, ε1) are punctured open disks of respective radius ε0 and ε1 sufficiently small and with their respective centres at the points 0 and 1. INTEGRAL TRANSFORMS AND SPECIAL FUNCTIONS 45 Let (φn) ∞ n=0 be a sequence of orthogonal polynomials with respect to a Borel measure supported on the unit circle satisfying (1.16). We denote by {mn}n∈N the set of indices corresponding to non-null Verblunsky coefficients. Define A0 = B0 = 0, for each n ∈ N fixed, denote by An(z) and Bn(z) for z ∈ �θ a solution of φ′ n(z) = An(z)φn−1(z) − Bn(z)φn(z), (1.18) in particular the functionsAn and Bn are a special case. Before we enunciate the main result, we need some auxiliary propositions whose proof will be given in Section 2. Proposition 1.2: Let (φn) ∞ n=0 be the sequence of orthogonal polynomials with respect to a Borel measure supported on the unit circle and suppose that (φn) ∞ n=0 satisfies (1.16), then φn(z) = zn, ∀n ∈ N if and only if there exists n0 ∈ N such that φn0(z) = zn0 . Proposition 1.3: Let (φn) ∞ n=0 be a sequence of OPUC satisfying (1.16)with φ1(z) �= z, then {mn}n∈N is infinite and consecutive Verblunsky coefficients cannot vanish simultaneously. Proposition 1.4: Let (φn) ∞ n=0 be a sequence of OPUC such that φ1(z) �= z and An,Bn satisfying (1.18). Then for n ≥ 1 φ′′ mn + P1,mnφ ′ mn + Q1,mnφmn = 0, (1.19) φ′′ mn + P2,mnφ ′ mn + Q2,mnφmn = 0, (1.20) where P1,mn = Bmn + Bmn−1 − A′ mn Amn + mn−1 z − mn z + 1 z − κmn−1 κmn−1−1 Amn−1 z − zmn−mn−1−1 κmn κmn−1−1 φmn−1(0) φmn(0) Amn−1 , Q1,mn = Amn ( Bmn Amn )′ + Bmn−1Bmn − κmn−1 κmn−1−1 Amn−1Bmn z − κmn κmn−1−1 φmn−1(0) φmn(0) z−mn−1+mn−1Amn−1Bmn + κmn−1 κmn−1−1 φmn−1(0) φmn(0) z−mn−1+mn−2Amn−1Amn + mn−1 − mn + 1 z Bmn , and P2,mn = Bmn+1 + Bmn − A′ mn Amn − κmn κmn−1 Amn z − zmn+1−mn−1 κmn+1 κmn−1 φmn(0) φmn+1(0) Amn + 1 z 46 J. BORREGO-MORELL AND A. SRI RANGA Q2,mn = Amn ( Bmn Amn )′ + BmnBmn+1 − κmn κmn−1 AmnBmn+1 z − κmn+1 κmn−1 φmn(0) φmn+1(0) z−mn+mn+1−1AmnBmn+1 + φmn(0) φmn+1(0) κmn+1−1 κmn−1 z−mn+mn+1−2AmnAmn+1 + Bmn z − κmn+1 κmn−1 φmn(0) φmn+1(0) (mn+1 − mn)z−mn+mn+1−2Amn . Proposition 1.5: Let (φn) ∞ n=0 be a sequence of OPUC satisfying (1.16), φ1(z) �= z. Then there exist functions Amn and Bmn analytic in �θ satisfying (1.18) such that for n ≥ 1 P1,mn = P2,mn , (1.21) Q1,mn = Q2,mn , (1.22) where P1,mn ,Q1,mn and P2,mn ,Q2,mn are as in Proposition 1.4. By analogy with (1.17), let Pmn be the common value in (1.21). In the present article, we prove the following theorem. Theorem 1.6: Let (φn) ∞ n=0 be a sequence of orthonormal polynomials with respect to a positive Borel measure on the unit circle satisfying (1.16). Then the following statements hold: (a) If c1 = 0 the whole sequence reduces to (zn)∞n=0. (b) If Pmn = Pmn , ∀n ∈ N, then there exist pn, qn, rn ∈ Z, p1, q1 = 0, c1 �= 0, b1 + pn /∈ {−n + 1,−n + 2, . . . , 0} and γn an appropriate non-null complex constant such that φn(z) = γn { 2F1(−n, b1 + pn; c1 + qn; z), if θ(z) = z(1 − z), 1F1(−n; c1 + rn; z), if θ(z) = z, (1.23) The above theorem gives as a particular case, the sequence (Rn) ∞ n=0 which is orthog- onal with respect to ω when ω is positive by choosing pn = 0, qn = 1 − n and b1 = α + β + 1, c1 = β − α; α ∈ R, α > − 1 2 and iβ ∈ R. It seems plausible to conjecture that for θ(z) = z(1 − z) the family (Rn) ∞ n=0 is the unique sequence of orthogonal polynomials on the unit circle satisfying (1.16). Furthermore, when θ(z) = z also there exists a unique sequence and is given by φn(z) = zn. As a restatement of the above theorem, we obtain Corollary 1.7: With the same hypothesis of Theorem 1.6, (a) If c1 = 0 the whole sequence reduces to (zn)∞n=0. (b) If Pmn = Pmn , ∀n ∈ N, then bn − b1, cn − c1 ∈ Z. INTEGRAL TRANSFORMS AND SPECIAL FUNCTIONS 47 2. Proof of the results The sketch of the proof for Theorem1.6 is as follows.We start by obtaining a pair of second- order differential equations in terms of two functional coefficients An,Bn satisfied for an arbitrary sequence of orthogonal polynomials for a measure supported on the unit cir- cle. This will be done in Proposition 1.4. Later, in Proposition 1.5, we unify the differential equations obtained in Proposition 1.4. Under the assumption that the sequence of orthogo- nal polynomials is of hypergeometric or confluent hypergeometric type and the differential equation, they satisfy is equal to the differential equations obtained in Proposition 1.4, we obtain a system of equations with An and Bn as functional variables. By analysing the singular points of An, we obtain the possible sequences of orthogonal polynomials. Proof of Proposition 1.1: For any n ∈ N, let yn(z) = ∑n k=0 akz k be a monic polynomial. Then yn satisfies (1.13) if and only if (k + 1)(cn + k)ak+1 + (λn − ηk)ak = 0, 0 ≤ k ≤ n, an+1 = 0, (2.1) where ηk = { k(k − 1 + bn), θ(z) = z(1 − z), kbn, θ(z) = z. From (2.1), a0, . . . , an−1 is uniquely determined if and only if λn − ηk �= 0, ∀0 ≤ k ≤ n − 1 and λn = ηn. For k=n, we have λn = { n(n − 1 + bn), θ(z) = z(1 − z), nbn, θ(z) = z. (2.2) For the case θ(z) = z, it is straightforward that relation (2.2) and bn �= 0 is the necessary and sufficient conditions for (1.13) to have a unique polynomial solution. Consider now the case θ(z) = z(1 − z). The condition λn − ηk �= 0, ∀0 ≤ k ≤ n − 1 is equivalent to saying that bn /∈ {−2n + 2,−2n + 3, . . . ,−n,−n + 1} and this completes the proof of the existence and uniqueness of a polynomial solution of degree n for (1.13). Assume now that λn is given as in (1.14). It follows from the theory for the hypergeomet- ric (confluent hypergeometric) equation, see [18, p.56, 248] that a holomorphic solution in a neighbourhood of z=0 for (1.13) is given by πn(z) = const. { 2F1(−n, bn + n − 1; cn; z), if θ(z) = z(1 − z), 1F1(−n; cn; bnz), if θ(z) = z, the condition bn /∈ {−2n + 2,−2n + 3, . . . ,−n,−n + 1} for θ(z) = z(1 − z) implies that bn + n − 1 /∈ {0,−1, . . . ,−n + 1}, therefore, from the existence and uniqueness condition it follows that πn is the polynomial solution of degree n. � Proof of Proposition 1.2: Assume that there exists n0 ∈ N such that �n0(z) = zn0 . From the recurrence formula for the monic polynomials (1.4), it is straightforward that�n(z) = 48 J. BORREGO-MORELL AND A. SRI RANGA zn, n ≤ n0. A straightforward argument by induction shows that �n(z) = zn, n ≥ n0. Indeed, for n = n0 + 1, from the recurrence formula for the monic polynomials (1.4) �n0+1(z) = zn0+1 + �n0+1(0), |�n0+1(0)| < 1. (2.3) From Proposition 1.1 �n0+1(z) = const. { 2F1(−n0 − 1, bn0+1 + n0; cn0+1; z), if θ(z) = z(1 − z), 1F1(−n0 − 1; cn0+1; z), if θ(z) = z, (2.4) therefore, from (1.9) and (1.11) we have that �n0+1(0) = 0, hence φn(z) = zn, ∀n ∈ N. The converse implication is trivial. � Proof: Let us assume that {mn}n∈N is finite. Denote μ = max{mn}n∈N, by hypothesis φ1(z) �= z, hence μ ≥ 1. Since φμ satisfies (1.16), from Proposition 1.1, it follows that φμ(z) = γμ { 2F1(−μ, bμ + μ − 1; cμ; z), θ(z) = z(1 − z), 1F1(−μ; cμ; z), θ(z) = z, (2.5) where γμ ∈ C \ {0} is an appropriate constant and cμ /∈ {−μ + 1, . . . , 0}. From the condition φn(0) = 0 for n > μ, we have that κμ = κn, therefore from (1.3) and (2.5) φn(z) = zn−μφμ(z). (2.6) If cn ∈ N ∪ {0}, cn < n, then 2F1(−n, bn + n − 1;−cn; z) = zcn+1 2F1(−n + cn + 1, bn + cn + n; cn + 2; z), 1F1(−n;−cn; z) = zcn+1 1F1(−n + cn + 1; cn + 2; z). (2.7) Since φn satisfies (1.16) and φn(0) = 0, for n > μ, Proposition 1.1 gives φn(z) = constn { 2F1(−n, bn + n − 1;−cn; z), θ(z) = z(1 − z), 1F1(−n;−cn; z), θ(z) = z, (2.8) where cn ∈ N ∪ {0}, cn < n, therefore from (2.7) and (2.8) φn(z) = constnzcn+1 { 2F1(−n + cn + 1, bn + cn + n; cn + 2; z), θ(z) = z(1 − z), 1F1(−n + cn + 1; cn + 2; z), θ(z) = z. (2.9) Since the order of the zero at z=0 in the relations (2.6) and (2.9) must coincide, we obtain cn = n − μ − 1. From (2.6), limz→0(φn(z)/zn−μ) = γμ and from (2.9) with cn = n − μ − 1 it follows that constn = γμ. We prove now that (2.6) gives a contradiction, unless n = μ + 1. As a consequence, we have that consecutive Verblunsky coefficients cannot vanish simultaneously. INTEGRAL TRANSFORMS AND SPECIAL FUNCTIONS 49 Indeed, by comparing the coefficients of the expressions (2.6) and (2.9) that define the polynomial φn, we find that for 1 ≤ k ≤ n and n > μ (bn + 2n − μ − 1)k (n − μ + 1)k = (bμ + μ − 1)k (cμ)k , θ(z) = z(1 − z), (n − μ + 1)k = (cμ)k, θ(z) = z. (2.10) Consider the case θ(z) = z(1 − z). Relation (2.10) for k=1 and k=2 gives bn + 2n − μ − 1 n − μ + 1 = bμ + μ − 1 cμ , (2.11) (bn + 2n − μ − 1)(bn + 2n − μ) (n − μ + 1)(n − μ + 2) = (bμ + μ − 1)(bμ + μ) cμ(cμ + 1) . (2.12) From (2.11) and (2.12) bn + 2n − μ n − μ + 2 = bμ + μ cμ + 1 . (2.13) From relations (2.11) and (2.13) (cμ − n + μ − 1)(−bμ + cμ − μ + 1) cμ(cμ + 1) = 0, (2.14) it follows that bμ = cμ − μ + 1, (2.15) cμ = n − μ + 1. (2.16) The alternative (2.15) is not possible. Indeed, by substituting the value bμ obtained in (2.15) into (2.5) we find that φμ(z) = γμ2F1(−μ, cμ; cμ; z), which is impossible, since |�μ(0)| = 1. Since cμ is fixed, relation (2.16) is only possible for n = μ + 1, it follows that (2.6) only holds for n = μ + 1 and in this case cμ = 2, bμ+1 = bμ − 2. Consider now the case θ(z) = z. Relation (2.10) gives cμ = n − μ + 1, n > μ, that is, relation (2.16) is only possible for n = μ + 1. We conclude that (2.6) only holds for n = μ + 1 with cμ = 2 and this completes the proof of the lemma. � Proof: From Proposition 1.3, the set {mn}n∈N is infinite, therefore, from the recurrence relations (1.3), for n ≥ 2 κmn−1−1zφmn−1−1(z) = κmn−1φmn−1(z) − φmn−1(0)φ ∗ mn−1 (z), (2.17) κmn−1φmn(z) = zκmnφmn−1(z) + φmn(0)φ ∗ mn−1(z). (2.18) 50 J. BORREGO-MORELL AND A. SRI RANGA Since {mn}n∈N is the set of indices for which φmn(0) �= 0, then κmn−1 = κmn−1, hence φmn−1(z) = zmn−mn−1−1φmn−1(z), (2.19) φ∗ mn−1(z) = φ∗ mn−1(z). (2.20) Relations (2.17)–(2.20) give z(κmn−1−1φmn(0)φmn−1−1(z) − φmn−1(0)κmnz mn−mn−1−1φmn−1(z)) = κmn−1φmn(0)φmn−1(z) − κmn−1φmn−1(0)φmn(z). (2.21) Let An,Bn be such that (1.18) holds. Define the operators Tmn,1 = d dz + Bmn(z), Tmn,2 = − d dz − Bmn−1(z) + Amn−1(z) κmn κmn−1−1 φmn−1(0) φmn(0) zmn−mn−1−1 + Amn−1(z) z κmn−1 κmn−1−1 . From (2.21) and (1.18), the operators Tmn,1 and Tmn,2 are anhilation and creation operators in the sense that they satisfy Tmn,1[φmn(z)] = Amn(z)z mn−mn−1−1φmn−1(z), (2.22) Tmn,2[φmn−1(z)] = Amn−1(z) z κmn−1 κmn−1−1 φmn−1(0) φmn(0) φmn(z), (2.23) Relations (2.22) and (2.23) give that φn satisfy the differential equations Tmn,2 [ 1 Amn(z)zmn−mn−1−1Tmn,1[φmn(z)] ] = Amn−1(z) z κmn−1 κmn−1−1 φmn−1(0) φmn(0) φmn(z), (2.24) Tmn+1,1 [ z Amn(z) Tmn+1,2[φmn(z)] ] = Amn+1(z)z mn+1−mn−1 κmn+1−1 κmn−1 φmn(0) φmn+1(0) φmn(z). (2.25) By expanding (2.24) and (2.25) we obtain the coefficients of the differential equation for n ≥ 2. From the definition ofA0 = B0 = 0 we find that the lemma also holds for n=1 and this completes the proof. � Lemma 2.1: The Riccati equation y′ + P1y − y2 − Q1 = 0, (2.26) has an analytic solution in �θ . INTEGRAL TRANSFORMS AND SPECIAL FUNCTIONS 51 Proof: The change of variable y = v′ v , (2.27) transforms the differential equation (2.26) in the second-order differential equation v′′ − P1v′ + Q1v = 0. (2.28) Let us consider the case θ(z) = z(1 − z) and express (2.28) as the hypergeometric differential equation z(1 − z)v′′ + (−c1 − (η1 + η2 + 1)z)v′ − η1η2v = 0, (2.29) where η1, η2 are the solutions of the equation η2 + (b1 + 1)η − b1 = 0. From the theory of hypergeometric functions [18, Section 2.1.1 (3)], we find that the function u(z) = { 2F1(η1, η2;−c1; z), c1 /∈ N ∪ {0}, zc1+1 2F1(η1 + c1 + 1, η2 + c1 + 1; c1 + 2; z), c1 ∈ N ∪ {0}, (2.30) is an holomorphic solution to (2.29) at z=0, therefore, from (2.27) we have that there exists an analytic solution for the Equation (2.26) of the form y(z) = u′(z) u(z) , z ∈ D∗(0, ε0), (2.31) where ε0 is sufficiently small. For the point z=1 we consider [18, Section 2.9 (5)], that is u(z) = { 2F1(η1, η2; c1 − b1; 1 − z), b1 − c1 /∈ N ∪ {0}, (1 − z)b1−c1+1 2F1(−η2 − c1,−η1 − c1; b1 − c1 + 2; 1 − z), b1 − c1 ∈ N, which is holomorphic in a neighbourhood of z=1. Applying now (2.27) we obtain the lemma for θ(z) = z(1 − z). The analysis for the case θ(z) = z is similar. � Lemma 2.2: There exist analytic functions A1,B1 in�θ such that the following system holds P1(z) = −A′ 1(z) A1(z) + B1(z), (2.32) Q1(z) = B′ 1(z) − B1(z) A′ 1(z) A1(z) , (2.33) κ1 = A1(z) − B1(z)φ1(z). (2.34) 52 J. BORREGO-MORELL AND A. SRI RANGA Proof: From (1.16) κ1P1(z) + Q1(z)φ1(z) = 0. From this last relation, it follows that the substitution of (2.34) into (2.32) or (2.33) gives the Riccati equation B′ 1 + B1P1 − B21 − Q1 = 0. (2.35) From Lemma 2.1, there exists an analytic solution for (2.35) in �θ . The statement for the function A1 follows immediately from (2.34). � Proof of Proposition 1.5: For n=1 we obtain the system defined by (2.32)–(2.34) and from Lemma 2.2, there exists analytic functions in �θ , A1 and B1 such that this system holds. For n ≥ 1, let us write P1,mn(z) = �mn(z) − A′ mn(z) Amn(z) , (2.36) where �mn(z) = Bmn(z) + Bmn−1(z) − κmn−1 κmn−1−1 Amn−1(z) z − zmn−mn−1−1 κmn κmn−1−1 φmn−1(0) φmn(0) Amn−1(z) + mn−1 z − mn z + 1 z , (2.37) and �1 = B1. Note that if (1.18) and (1.21) hold then (1.22) also holds. Indeed, since φmn satis- fies (1.19) and (1.20) φ′′ mn + P1,mnφ ′ mn + Q1,mnφmn = 0, φ′′ mn + P1,mnφ ′ mn + Q2,mnφmn = 0, the difference between these two relations gives Q1,mn = Q2,mn , hence, Equation (1.22) is redundant. From the expressions for P1,mn and P2,mn given in Proposition 1.4 we find that the relation P1,mn = P2,mn in (1.21) holds if and only if �mn satisfies the difference equation �mn+1(z) = �mn(z) + mn z − mn+1 z , �1(z) = B1(z). (2.38) By hypothesis φ1(z) �= z. Hence, from Proposition 1.2, m1 = 1, it follows that the solution to the difference Equation (2.38) for n ≥ 1 is �mn(z) = B1(z) + 1 − mn z . (2.39) From (2.36) and (2.38) we have that (1.21) holds. Therefore, for n ≥ 1, if �mn is given by (2.39), we define the analytic functions in �θ , Amn and Bmn as the unique solution of INTEGRAL TRANSFORMS AND SPECIAL FUNCTIONS 53 (1.18) and (2.37). We have that (1.18), (1.21) and (1.22) hold and this completes the proof of the lemma. � From the preceding lemmas we now are able to give a proof of Theorem 1.6. Proof of Theorem 1.6: Statement (a) follows immediately from Proposition 1.2. To prove (b) note that we have two cases, the set of indices {mn}n∈N for which φmn(0) �= 0 is finite or infinite. Results in Propositions 1.2 and 1.3 show that the first case implies that the whole sequence reduces to φn(z) = zn. Let us analyse the second case. FromProposition 1.5 there existAmn and Bmn such that P1,mn = P2,mn . Fromhypothesis Pmn = P1,mn = P2,mn . Therefore, from (1.21), (2.36) and (2.39) Pmn(z) = 1 − mn z − A′ mn(z) Amn(z) + B1(z). (2.40) We have that φ1(z) �= z and from (2.40) of Proposition 1.5 P1(z) = −A′ 1(z) A1(z) + B1(z), Pmn(z) = 1 − mn z − A′ mn(z) Amn(z) + B1(z), n > 1. Therefore, A′ mn/Amn = A′ 1/A1 + πn, where πn(z) = ⎧⎪⎨ ⎪⎩ c1 − cmn + 1 − mn − (b1 − bmn + 1 − mn)z θ(z) , θ(z) = z(1 − z), c1 − cmn + 1 − mn θ(z) , θ(z) = z. That is Amn = A1τn, (2.41) where τn(z) = c { zumn (1 − z)vmn , θ(z) = z(1 − z), zumn , θ(z) = z, here c ∈ C \ {0} is an appropriate constant and umn = c1 − cmn + 1 − mn (2.42) vmn = b1 − bmn − (c1 − cmn). (2.43) From Proposition 1.5 and from the assumption of the theorem, there exists B1 analytic in �θ , hence, from Proposition 1.5, the function Amn is analytic in �θ . Therefore, from (2.41), (2.42) and (2.43), we have umn , vmn ∈ Z, ∀n ≥ 2. 54 J. BORREGO-MORELL AND A. SRI RANGA Now, from (2.42) and (2.43) φmn(z) = γmn2F1(−mn, b1 + pmn ; c1 + qmn ; z), where pmn , qmn ∈ Z; p1 = 0, q1 = 0, c1 �= 0, b1 + pmn /∈ {−mn + 1,−mn + 2, . . . , 0} and γmn an appropriate complex constant. 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