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Orthogonal Polynomials on the Unit Circle Associated with the Laguerre Polynomials

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Orthogonal Polynomials on the Unit Circle Associated with the Laguerre Polynomials PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 129, Number 3, Pages 873{879 S 0002-9939(00)05821-4 Article electronically published on October 11, 2000 ORTHOGONAL POLYNOMIALS ON THE UNIT CIRCLE ASSOCIATED WITH THE LAGUERRE POLYNOMIALS LI-CHIEN SHEN (Communicated by Hal L. Smith) Abstract. Using the well-known fact that the Fourier transform is unitary, we obtain a class of orthogonal polynomials on the unit circle from the Fourier transform of the Laguerre polynomials (with suitable weights attached). Some related extremal problems which arise naturally in this setting are investigated. 1. Introduction This paper deals with a class of orthogonal polynomials which arise from an application of the Fourier transform on the Laguerre polynomials. We shall briefly describe the essence of our method. Let Π+ denote the upper half plane fz : z = x + iy; y > 0g and let Z 1 H(Π+)=ff : f is analytic in Π+ and sup jf(x + yi)j2 dx < 1g: 0<y<1 −∞ It is well known that, from the Paley-Wiener Theorem [4, p. 368], the Fourier transform provides a unitary isometry between the spaces L2(0; 1)andH(Π+): Since the Laguerre polynomials form a complete orthogonal basis for L2([0; 1);xαe−x dx); the application of Fourier transform to the Laguerre polynomials (with suitable weight attached) generates a class of orthogonal rational functions which are com- plete in H(Π+); and by composition of which with the fractional linear transfor- mation (which maps Π+ conformally to the unit disc) z =(2t − i)=(2t + i); we obtain a family of polynomials which are orthogonal with respect to the weight α t j j sin 2 dt on the boundary z = 1 of the unit disc. Some similar approaches of linking the classical orthogonal polynomials with or- thoganal polynomials of greater generality via various transforms have been inves- tigated recently. For example, using the Fourier transform, Koelink [1] established the connection between the Jacobi polynomials and the continuous Hahn polyno- mials; and in [2] Koornwinder established the link between the Meixner-Polleczek polynomials and the Laguerre polynomials from the Mellin transform. Received by the editors June 1, 1999. 2000 Mathematics Subject Classification. Primary 33C47. Key words and phrases. Fourier transform, Laguerre polynomial. c 2000 American Mathematical Society 873 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 874 LI-CHIEN SHEN We thank the referee for many detailed suggestions in revising the original man- uscript as well as for drawing our attention to papers [1] and [2]. 2. Key results We assume that the reader is reasonably familiar with the basic properties of the Laguerre polynomials which can be found in Chapter 4 of [3]. We say a few words about our notation. The Fourier transform of an integrable function f is defined as Z 1 f^(t)= eixt f(x) dx; −∞ (a)n = a(a+1)(a+2):::(a+n−1) and F (a; b; c; z) is the standard hypergeometric series. Let Xn − k α (α +1)n ( n)kx − Ln(x)= (α> 1) n! k!(α +1)k k=0 be the Laguerre polynomial of degree n associated with the weight e−xxα dx,and let ( Lα(x) e−x=2 xα/2;x≥ 0; lα(x)= n n 0;x<0: α We now compute the Fourier transform of ln (x). First, we observe that Z 1 − α − 2 1 ^α ixt −x=2 α/2 α 1 − (2.1) l0 (t)= e e x dx =Γ +1 it ; 0 2 2 Z 1 −n − n ^α ixt −x=2 α/2+n α 1 − ^α (2.2) ( iD) l0 (t)= e e x dx = +1 it l0 (t); 0 2 n 2 − n − n dn − 1 where ( iD) =( i) dtn and t is a complex number with imaginary part > 2 .It is easy to justify the differentiation behind the integral sign because of the absolute convergence of the integral. From (2.1) and (2.2), we obtain Z 1 ^α ixt α/2 −x=2 α α − ^α ln (t)= e x e Ln(x) dx = Ln( iD)l0 (t) 0 Xn (−n) α +1 k (1 + α)n ^α k 2 k 1 = l0 (t) 1 n! k!(α +1)k − it k=0 2 (1 + α) α 1 = n l^α(t) F −n; +1;α +1; : 0 1 − n! 2 2 it Now let z =(2t − i)=(2t + i) and recall that (c − a) F (−n; a; c; x)= n F (−n; a; a − n − c +1;1− x): (c)n We can write α (1 + α)n α α ^α − 2 +1 − − ln (t)=Γ +1 (1 z) F n; +1;α +1;1 z 2 n! 2 α α +1 α =Γ +1 (1 − z) 2 g (z); 2 n License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ORTHOGONAL POLYNOMIALS 875 where (α/2) α α (2.3) gα(z)= n F −n; +1;−n − +1;z n n! 2 2 (1 + α) α = n F −n; +1;α +1;1− z n! 2 α is a polynomial of degree n. For brevity, we let gn(z)=gn (z). We choose the α − 2 +1 ^α branch of (1 z) which takes the value 1 at z = 0. Therefore ln , as a function of z, is analytic in the complex plane cut along [1; 1). Moreover, from the fact that the Fourier transform preserves the inner product (the Parseval's identity), we deduce that Z Γ2 α +1 2π θ (2.4) 2α 2 g (eiθ) g (eiθ)sinα dθ 2π n m 2 0 Z 2 α Γ 2 +1 α dz = gn(z) gm(z) j1 − zj 2πi jzj=1 z Z 2 α Γ 2 +1 α +1 α +1 −2 = − g (z) g (z)(1− z) 2 (1 − z¯) 2 (1 − z) dz 2πi n m Z jzj=1 1 1 ^α ^α = ln (t) lm(t) dt 2π −∞ Z 1 α α = ln (x) lm(x) dx (Parseval's identity) 0 Z 1 α α −x α = Ln(x) Lm(x) e x dx (0 Γ(n+α+1) n = m = n! ([3; p: 84]): 0 n =6 m: We summarize our result in f α g1 Theorem 1. The polynomials gn (z) n=0 form a complete orthogonal basis for Z 2π 2 j j j iθ j2 1 Hµ = f : f is analytic in z < 1 and f(e ) dµ < ; 0 1 α θ where dµ = 2π sin 2 dθ. α 2 1 The completeness follows immediately from the completeness of ln (x)inL [0; ) [5, p. 108], the Paley-Wiener Theorem and the fact that the Fourier transform preserves the basis. n It is interesting to note that for α =0,gn(z)=z .Andforα =2, Xn n gn(z)= (k +1)z k=0 which is the nth partial sum of the familiar power series X1 (1 − z)−2 = (k +1)zk: k=0 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 876 LI-CHIEN SHEN > − − − α > > > Remark 1. We note that for α 0, ( n)k= n 2 +1 k 0forn k 0. Hence, from (2.3), we conclude that max jgn(z)j = gn(R): jzj6R Similarly, j (k) j (k) max gn (z) = gn (R): jzj6R In particular, (k) α − gn(1) = (α +1)n=n!andgn (1) = +1 (α +1)n=((α +1)k(n k)!): 2 k Remark 2. The reproducing kernel KN (z; w) with respect to the measure dµ for the polynomials of degree 6 N is defined as XN KN (z; w)= cngn(z)gn(w); n=0 where α c = a n!=(α +1) and a =2αΓ2 +1 =Γ(α +1): n α n α 2 The significance of the term \reproducing kernel" will be made clear in the next section. Then, from the above remark, we have XN max KN (w; w)=KN (1; 1) = aα (α +1)n=n!: jwj61 n=0 More generally, let XN j (k) j2 MN;k(w)= cn gn (w) : n=k Then max MN;k(w)=MN;k(R): jwj6R 3. Extremal problems Let PN be the set of all polynomials of degree 6 N. It follows from the orthog- onality (2.4) that if p 2 P ,then N Z p(w)= KN (z; w)p(z) dµ. jzj=1 More generally, by differentiating the above identity k times, we have Z (k) k @ (3.1) p (w)= D KN (z;w)p(z) dµ D¯ = : jzj=1 @w¯ Applying the Schwarz inequality to (3.1), we deduce (with the help of orthogo- nality) (k) (3.2) jp (w)j 6 kpkµ MN;k(w)forp 2 PN ; R k k2 j j2 where p µ = jzj=1 p(z) dµ. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ORTHOGONAL POLYNOMIALS 877 Therefore, we have Theorem 2. If p 2 PN ,then (k) sup jp (w)j = MN;k(w); kpkµ=1 and the equality is achieved by the polynomial XN (k) p(z)= cngn(z) gn (w)=MN;k(w): n=k We comment that this extremal function p(z) is essentially unique. To see this, we recall that [2, p. 63] in Schwarz inequality j(f; g)|≤kfkkgk; the equality holds if and only if f = cg a.e. Hence, the equality in (3:2) holds if and k only if p(z)=c D¯ KN (z; w); and c = a=MN;k(w); jaj =1; if we further require that kpkµ =1: Corollary.
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