Bounds on Turán determinants

Transcription

1 Bounds on Turán determinants Christian Berg Ryszard Szwarc August 6, 008 Abstract Let µ denote a symmetric probability measure on [ 1, 1] and let (p n ) be the corresponding orthogonal polynomials normalized such that p n (1) = 1. We prove that the normalized Turán determinant n (x)/(1 x ), where n = p n p n 1 p n+1, is a Turán determinant of order n 1 for orthogonal polynomials with respect to (1 x )dµ(x). We use this to prove lower and upper bounds for the normalized Turán determinant in the interval 1 < x < Mathematics Subject Classification: Primary 33C45; Secondary 6D07 Keywords: Turán determinants, ultraspherical polynomials. 1 Introduction In the following we will deal with polynomial sequences (p n ) satisfying xp n (x) = γ n p n+1 (x) + α n p n 1 (x), n 0, α n + γ n = 1, α n > 0, γ n > 0, n 1, (1) p 0 (x) = 1, α 0 = 0, 0 < γ 0 1. The present work was done while the first author was visiting University of Wroc law granted by the HANAP project mentioned under the second author. The second author was supported by European Commission Marie Curie Host Fellowship for the Transfer of Knowledge Harmonic Analysis, Nonlinear Analysis and Probability MTKD-CT and by MNiSW Grant N /485. 1

2 Note that (p n ) is uniquely determined by (1) from the recurrence coefficients α n, γ n and that p n has a positive leading coefficient. It follows by Favard s theorem that the polynomials p n are orthogonal with respect to a symmetric probability measure µ. From (1) we get for x = 1 γ n (p n+1 (1) p n (1)) = (1 γ n )(p n (1) p n 1 (1)), n 1, () hence p n+1 (1) p n (1) p 1 (1) = 1 γ 0 1, n 1, (3) so that p n (1) = 1 for all n if γ 0 = 1, and (p n (1)) is strictly increasing if γ 0 < 1. We conclude that all zeros x n,n <... < x 1,n of p n belong to the interval ( 1, 1), hence supp(µ) [ 1, 1]. In fact, if there existed an integer n such that x 1,n > 1, then by assuming n smallest possible with this property, we get x,n < 1 < x 1,n and hence p n (1) < 0, a contradiction. By symmetry this implies that 1 < x n,n. Define the Turán determinant of order n by n (x) = p n(x) p n 1 (x)p n+1 (x), n 1. (4) In [11] the second author proved non-negativity of the Turán determinant (4) under certain monotonicity conditions on the recurrence coefficients, thereby obtaining results for new classes of polynomials and unifying old results. If γ 0 = 1 and hence p n (1) = 1 for all n, the normalized Turán determinant n (x)/(1 x ) is a polynomial in x. We shall prove estimates of the form c n (0) n(x) 1 x C n(0), 1 < x < 1, (5) under certain regularity conditions on the recurrence coefficients. We prove, e.g., an inequality of the left-hand type if (α n ) is increasing and concave, see Theorem.5. In Theorem.7 we give an inequality of the right-hand type if (α n ) is decreasing and satisfies a condition slightly stronger than convexity. Our results depend on a simple relationship between the Turán determinants of order n and n 1 (Proposition.1) and the following observation: the normalized Turán determinant is essentially a Turán determinant of order n 1 for the polynomials (q n ) defined by (17) below, and if µ denotes

3 the orthogonality measure of (p n ), then (q n ) are orthogonal with respect to the measure (1 x )dµ(x). See Theorem.3 and Remark.4 for a precise statement. In Proposition.11 we prove non-negativity of the Turán determinant for the normalized polynomials q n (x)/q n (1) provided the sequence (α n ) is increasing and concave (or under the weaker condition (1)). Our work is motivated by results about ultraspherical polynomials, which we describe next. For α > 1 let R n (α,α) (x) = P n (α,α) (x)/p n (α,α) (1) denote the symmetric Jacobi polynomials normalized to be 1 for x = 1, i.e., R n (α,α) (x) = ( 1)n (1 x α dn ) n (α + 1) n dx (1 n x ) n+α, (6) cf. [10]. We have used the Pochhammer symbol (a) n = a(a + 1)... (a + n 1). The polynomials are orthogonal with respect to the symmetric weight function c α (1 x ) α on ( 1, 1). Here 1/c α = B(α + 1, 1/), so the weight is a probability density. We have R n (α,α) (x) = P n (λ) (x)/p n (λ) (1) with α = λ 1, where (P n (λ) ) are the ultraspherical polynomials in the notation of [10]. The corresponding Turán determinant of order n (α) n (x) = R n (α,α) (x) R (α,α) n 1 (x)r (α,α) n+1 (x), (7) is clearly a polynomial of degree n in x and divisible by 1 x since it vanishes for x = ±1. The following Theorem was proved in [1, pp ] and in [14, sect. 6]: Theorem 1.1. The normalized Turán determinant is f n (α) (x) := (α) n (x)/(1 x ) (8) (i) strictly increasing for 0 x < when α > 1/. (ii) equal to 1 for x R when α = 1/. 3

4 (iii) strictly decreasing for 0 x < when 1 < α < 1/. It is easy to evaluate f (α) n at x = 0, 1 giving f n (α) (0) = µ (α) [n/] µ(α) [(n+1)/], where we have used the notation from [1] ( n+α n µ (α) n = µ n f (α) n (1) = 1 α +, (9) ), (10) and µ n is the normalized binomial mid-coefficient ( ) n µ n = n (n 1) =. (11) n 4... (n) Corollary 1.. For 1 < x < 1 we have f (α) n (0)(1 x ) < (α) n (x) < f n (α) (1)(1 x ) for α > 1/, (1) while the inequalities are reversed when 1 < α < 1/. (For α = 1/ all three terms are equal to 1 x.) For α = 0 the inequalities (1) reduce to ( 1 < x < 1) µ [n/] µ [(n+1)/] (1 x ) < P n (x) P n 1 (x)p n+1 (x) < 1 (1 x ) (13) for Legendre polynomials (P n ). This result was recently published in [1] using a SumCracker Package by Manuel Kauers, and it was conjectured that the monotonicity result remains true for ultraspherical polynomials when α 1/. Clearly the authors have not been aware of the early results above. 1 Turán [13] proved that (0) n (x) > 0 for 1 < x < 1. The proof in [1] of Theorem 1.1 is based on a formula relating the Turán determinant n,λ (x) = F n,λ(x) F n 1,λ (x)f n+1,λ (x) of the normalized ultraspherical polynomials F n,λ (x) = P n (λ) (x)/p n (λ) (1) and the expression D n,λ (x) = [ d dx P n (λ) (x)] d dx P n 1(x) (λ) d dx P n+1(x), (λ) 1 Motivated by this conjecture the present authors found a proof of Theorem 1.1 close to the old proofs. During the preparation of the paper we found the references [1], [14]. 4

5 namely (see [1, (5.9)]) n,λ (x) 1 x = D n,λ (x) n(n + λ)[p (λ) n (1)]. (14) See also [3]. Using the well-known formula for differentiation of ultraspherical polynomials d dx P n (λ) (x) = λp (λ+1) n 1 (x), we see that ( ) D n,λ (x) = (λ) [P (λ+1) (x)] P (λ+1) (x)p (λ+1) (x). (15) n 1 Except for the factor (λ) this is the Turán determinant of order n 1 for the ultraspherical polynomials corresponding to the parameter λ + 1. We see that this result is generalized in Theorem.3. Since the proof of the monotonicity in Theorem 1.1 depends on the fact that the ultraspherical polynomials satisfy a differential equation, there is little hope of extending the result to classes of orthogonal polynomials which do not satisfy a differential equation. We have instead attempted to find bounds for normalized Turán determinants without using monotonicity in the variable x. This has also led us to consider the following lower boundedness property, which may or may not hold for a system of orthonormal polynomials (P n ): (LB) inf{p n 1(x) + P n(x) x R, n N} > 0. (16) If property (LB) holds, then necessarily n=0 P n(x) = for all x R. Therefore, the orthogonality measure of (P n ) is uniquely determined and has no mass points. In Proposition 3.1 we prove that (LB) holds for symmetric orthonormal polynomials if the recurrence coefficients are increasing and bounded. It turns out that for the orthonormal symmetric Jacobi polynomials the condition (LB) holds if and only if α 1/. The theory is applied to continuous q-ultraspherical polynomials in Section 4. Concerning the general theory of orthogonal polynomials we refer the reader to [10],[9],[6]. n n 5

6 Main results Proposition.1. In addition to (1), assume α n γ n for n = 1,.... For n n = (γ n α n )α n 1 α n α n 1 n 1 + (p n 1 + p n xp n 1 p n ). (γ n 1 α n 1 )γ n (γ n 1 α n 1 )γ n Proof. By the recurrence relation we can remove either p n+1 or p n 1 from the formula defining n. This leads to two equations γ n n = α n p n 1 + γ n p n xp n 1 p n, α n n = α n p n + γ n p n+1 xp n p n+1. We replace n by n 1 in the second equation, multiply both sides by γ n α n and subtract the resulting equation from the first one multiplied by γ n 1 α n 1. In this way we obtain (γ n 1 α n 1 )γ n n (γ n α n )α n 1 n 1 = (α n γ n 1 α n 1 γ n )(p n 1 + p n) (γ n 1 γ n α n 1 + α n ) xp n 1 p n. Taking into account that α k + γ k = 1 for k 1 gives (γ n 1 α n 1 )γ n n (γ n α n )α n 1 n 1 = (α n α n 1 )(p n 1+p n xp n 1 p n ). Proposition.1 implies Corollary.. [11, Thm. 1] In addition to (1), assume that one of the following conditions holds: (i) (α n ) is increasing and α n γ n, n 1. (ii) (α n ) is decreasing and α n γ n, n 1. Furthermore, assume γ 0 = 1 or γ 0 γ 1 /(1 γ 1 ). Then n (x) > 0 for 1 < x < 1. 6

7 Proof. Assume first the additional condition α n γ n for all n 0. Since for 1 < x < 1 p n 1(x) + p n(x) xp n 1 (x)p n (x) > 0, it suffices in view of Proposition.1 to show 1 (x) > 0 for 1 < x < 1. We have γ 1 1 (x) = α 1 p 0 + γ 1 p 1 xp 0 p 1 = α 1γ0 + (γ 1 γ 0 )x, hence 1 > 0 if γ 1 γ 0. If γ 1 < γ 0 and 1 < x < 1, we have γ 1 1 (x) > γ 1 1 (1) = α 1(1 γ 0 )(γ 1 /α 1 γ 0 ). γ0 The right-hand side is clearly non-negative in case (i) because γ 1 /α 1 1, but also non-negative in case (ii) because of the assumptions on γ 0. Assume next in case (i) that there is an n such that α n = γ n. Let n 0 1 be the smallest n with this property. Denoting α = lim α n, then clearly α n α 1 α γ n for all n and hence α n = γ n = 1/ for n n 0. Therefore, n (x) = p n 1(x) + p n(x) xp n 1 (x)p n (x) > 0 for n n 0, 1 < x < 1. The formula of Proposition.1 can be applied for n < n 0 and the proof of the first case carries over. Equality in case (ii) is treated similarly. From now on we will assume additionally that γ 0 = 1. In this case the polynomials p n are normalized at x = 1 so that p n (1) = 1. Since p n ( x) = ( 1) n p n (x), we conclude that p n ( 1) = ( 1) n. Therefore, the polynomial p n+ p n is divisible by x 1 and q n (x) = p n+(x) p n (x), n 0, (17) x 1 is a polynomial of degree n. Moreover, an easy calculation shows that the polynomials q n are orthogonal with respect to the probability measure dν(x) = 1 γ 1 (1 x )dµ(x). By the recurrence relation (1) with γ 0 = 1 we obtain that the polynomials q n satisfy xq n (x) = γ n+ q n+1 (x) + α n q n 1 (x), n 0, q 0 = 1/γ 1. (18) The following theorem contains a fundamental formula relating the Turán determinants of the polynomials p n and q n. 7 γ 0

8 Theorem.3. For n 1 Proof. By (1) Therefore, n (x) 1 x = α nγ n q n 1(x) α n 1 γ n+1 q n (x)q n (x). (19) p k+1 xp k = α k (p k+1 p k 1 ) = α k (x 1)q k 1, xp k p k 1 = γ k (p k+1 p k 1 ) = γ k (x 1)q k 1. (x 1) [α n γ n q n 1 α n 1 γ n+1 q n q n ] = (p n+1 xp n )(xp n p n 1 ) (p n xp n 1 )(xp n+1 p n ) = (1 x )(p n p n 1 p n+1 ). Remark.4. Defining q 0 = γ 1 q 0 = 1 and we have n (x) 1 x = γ n α n q n = γ 1... γ n+1 α 1... α n q n, n 1, ( α1... α n γ 1... γ n ) [ q n 1(x) q n (x) q n (x) ], (0) showing that the normalized Turán determinant (19) is proportional to the Turán determinant of order n 1 of the renormalized polynomials ( q n ). They satisfy the recurrence relation ( q 1 := 0) x q n = α n+1 q n+1 + γ n+1 q n 1, n 0. Theorem.5. Assume that (p n ) satisfies (1) with γ 0 = 1. Let (α n ) be increasing, α n 1/ and α n α n 1 Then n (x) defined by (4) satisfies α n 1 α n (α n+1 α n ), n 1. (1) n (x) 1 x c n(0), 1 < x < 1, n 1, where c = α 1 γ /γ 1. (Note that (1) holds if (α n ) is concave.) 8

9 Proof. Observe that (1) is equivalent to (α n γ n+1 ) being increasing. Let D n (x) = γ n q n 1(x) γ n+1 q n (x)q n (x). Since α n α n 1, Theorem.3 implies n (x) 1 x α n 1D n (x). () By (18) we can remove q n or q n from the expression defining D n. In this way we obtain D n = α n 1 qn + γ n qn 1 xq n q n 1, (3) α n 1 D n = α n 1γ n qn 1 + γ n+1 qn xq n 1 q n. γ n+1 γ n+1 Replacing n by n 1 in the second equation and subtracting it from the first, we find D n α n D n 1 = α n 1γ n α n γ n 1 qn 0. γ n γ n By iterating the above inequality between D n and D n 1, we obtain From (3) we get so () implies D = α 1 q 0 + γ q 1 xq 0 q 1 = α 1 γ 1 D n α 1... α n γ 3... γ n D. (4) + γ x γ 1γ x γ 1γ = α 1 γ 1 n (x) 1 x α 1... α n γ 1... γ n α 1 γ α n γ 1 α 1... α n γ 1... γ n α 1 γ γ 1, and the conclusion follows from the next lemma. Lemma.6. Under the assumptions of Theorem.5 n (0) α 1... α n γ 1... γ n γ 1 α 1 n (0), n 1., (5) 9

10 Proof. Denote By (1) we have hence On the other hand h n = γ 1... γ n α 1... α n. p n (0) = ( 1) n α 1α 3... α n 1 γ 1 γ 3... γ n 1, n (0)h n = p n(0)h n = n k=1 α k 1 α k n k=1 γ k γ k 1 1. n+1 (0)h n+1 = p n (0)p n+ (0)h n+1 n = Moreover, k=1 α k 1 α k n k=1 γ k γ k 1 = n (0)h n 1. n (0)h n = n k=1 α k 1 α k n γ k γ k 1 k=1 n k= α k 1 α k n γ k γ k 1 k= = α 1γ n γ 1 α n α 1 γ 1. Theorem.5 has the following counterpart and the proof is very similar: Theorem.7. Assume that (p n ) satisfies (1) with γ 0 = 1. Let α n, n 1 be decreasing, α n 1 and α n α n 1 Then n (x) defined by (4) satisfies α n 1 α n (α n+1 α n ), n. (6) n (x) 1 x C n(0), 1 < x < 1, n 1, where C = γ. (Note that (6) implies convexity of α n, n 1.) 10

11 Remark.8. The normalized symmetric Jacobi polynomials p n (x) = R n (α,α) (x) given by (6) satisfy (1) with γ n = n + α + 1 n + α + 1, α n = n n + α + 1. (7) (In the case of α = 1/, i.e., Chebyshev polynomials of the first kind, these formulas shall be interpreted for n = 0 as γ 0 = 1, α 0 = 0.) For α 1/ the sequence (α n ) is increasing and concave. Furthermore, c = 1. For 1 < α 1/ the sequence (α n ) is decreasing, (6) holds and C = 1. The statement about the constants c and C follows from Corollary 1.. However, we cannot expect c = 1 in general, because it is easy to construct an example, where the normalized Turán determinant (19) is not monotone for 0 < x < 1. Consider the sequence (α n ) = (0, 1/ 3ε, 1/ ε, 1/ ε, 1/, 1/,...), which is increasing and concave for 0 < ε < 1/8. In this case the Turán determinant q q 1 q 3 is proportional to f(x) = x 4 + A(ε)x + B(ε), where A(ε) = 4ε + 3ε 1/, B(ε) = (1/ 3ε) (1/ ε)(1/ + ε) /ε. Clearly, f is not monotone for 0 < x < 1, when ε is small. Corollary.9. Under the assumptions of Theorem.5 and the additional hypothesis lim α n = 1/, the orthogonality measure µ is absolutely continuous on ( 1, 1) with a strictly positive and continuous density g(x) = dµ(x)/dx satisfying C g(x). 1 x Proof. The corresponding orthonormal polynomials (P n ) satisfy xp n = λ n P n+1 + λ n 1 P n 1, (8) where λ n = α n+1 γ n. We also have P n = δ n p n, where δ n = γ0... γ n 1 α 1... α n, n 1, δ 0 = 1, 11

12 and lim λ n = 1/. Since λ n+1 λ n = α n+(γ n+1 γ n ) + γ n (α n+ α n+1 ) αn+ γ n+1 + α n+1 γ n, the monotonicity of (α n ), (γ n ) implies λ n+1 λ n <. n=1 By the theorem in [8] we conclude that the orthogonality measure µ has a positive continuous density g(x) for 1 < x < 1. Furthermore, it is known from this theorem that lim [P n(x) P n 1 (x)p n+1 (x)] = 1 x, n πg(x) uniformly on compact subsets of ( 1, 1). For another proof of this result see [5, p. 01], where it is also proved that (P n (x)) is uniformly bounded on compact subsets of ( 1, 1) for n. We have n (x) = 1 δ n ( P n (x) k n P n 1 (x)p n+1 (x) ), where k n = and it follows that lim k n = 1. Using we get the result. δn αn+1 γ n 1 =, δ n 1 δ n+1 α n γ n n (x) n (0) = P n(x) k n P n 1 (x)p n+1 (x) Pn(0) k n P n 1 (0)P n+1 (0), In analogy with the proof of Corollary.9 we get Corollary.10. Under the assumptions of Theorem.7 and the additional hypothesis lim α n = 1/, the orthogonality measure µ is absolutely continuous on ( 1, 1) with a strictly positive and continuous density g(x) = dµ(x)/dx satisfying C g(x). 1 x 1

13 We now return to the polynomials (q n ) defined in (17) and prove that they have a non-negative Turán determinant after normalization to being 1 at 1. The polynomials q n are orthogonal with respect to a measure supported by [ 1, 1]. Therefore, q n (1) > 0. Proposition.11. Under the assumptions of Theorem.5 qn(x) qn(1) q n 1(x) q n+1 (x) q n 1 (1) q n+1 (1) Proof. Indeed, let Q n (x) = q n (x)/q n (1). Then 0, 1 < x < 1, n 1. xq n = c n Q n+1 + (1 c n )Q n 1, where c n = γ n+ (q n+1 (1)/q n (1)). We will show that (c n ) is decreasing and c n 1/, and the conclusion then follows from Corollary.. But c n 1 c n is equivalent to D n+1 (1) = γ n+1 q n(1) γ n+ q n 1 (1)q n+1 (1) 0, which follows from (4) and (5). We will show that c n 1/ by induction. We have c 0 = γ q 1 (1) q 0 (1) = 1. Assume c n 1 1/. By (1) the sequence (α n γ n+1 ) is increasing. Putting α = lim α n, we then get Using this and (18) leads to hence c n 1/. α n γ n+1 α(1 α) = c n + α nγ n+1 c n 1 c n + 1 4c n 1 c n + 1, 3 Lower bound estimates It turns out that Turán determinants can be used to obtain lower bound estimates for orthonormal polynomials. Recall that if the polynomials (p n ) 13

14 satisfy the recurrence relation (1), then their orthonormal version (P n ) satisfies xp n = λ n P n+1 + λ n 1 P n 1, where λ n = α n+1 γ n. Proposition 3.1. Assume that the polynomials (P n ) satisfy xp n = λ n P n+1 + λ n 1 P n 1, n 0, (9) with P 1 = λ 1 = 0, λ n > 0, n 0, and P 0 = 1. Assume moreover that (λ n ) is increasing and lim λ n = L <. Then the (LB) property (16) holds in the precise form Pn(x) + Pn 1(x) λ 0, x R, n 0. L Proof. This proof is inspired by [, Thm. 3]. By replacing the polynomials P n (x) by P n (Lx) we can assume that lim λ n = 1/. This assumption implies that the corresponding Jacobi matrix acts as a contraction in l, because it can be majorized by the Jacobi matrix with entries λ n = 1. Therefore, the orthogonality measure is supported by the interval [ 1, 1]. In this way it suffices to consider x from [ 1, 1] because the functions Pn(x) are increasing on [1, + [ and Pn( x) = Pn(x). Let D n (x) = λ n 1 P n(x) λ n P n 1 (x)p n+1 (x), n 1. By (9) we can remove P n+1 to get D n = λ n 1 P n 1 + λ n 1 P n xp n 1 P n. (30) Alternatively we can remove P n 1 and obtain λ n 1 D n = λ n Pn+1 + λ n 1 Pn xp n P n+1. (31) λ n λ n Replacing n by n 1 in (31) and subtracting it from (30) gives D n λ n λ n 1 D n 1 = λ n 1 λ n λ n 1 P n 1 0. (3) Iterating the inequality D n (λ n /λ n 1 )D n 1 leads to D n λ 0 λ n 1 D 1 = λ 0 λ n 1 λ 0, 14

15 because D 1 = λ 0 by (30), which for x 1 yields D n λ n 1 P n 1 + λ n 1 P n + 1 x (P n 1 + P n) P n 1 + P n. (33) In the general case the lower bound is (λ 0 /(L)). Corollary 3.. Under the assumptions of Proposition 3.1 with L = 1/ the orthogonality measure µ is absolutely continuous with a continuous density g = dµ(x)/dx on [ 1, 1] satisfying g(x) 1 1 x. πλ 0 Furthermore, g(x) > 0 for 1 < x < 1. Proof. By assumptions the orthogonality measure is supported by [ 1, 1]. By the proof of Proposition 3.1 we have D n (x) λ 0. On the other hand, by [8] and [5, p. 01] the orthogonality measure is absolutely continuous in the interval ( 1, 1) with a strictly positive and continuous density g such that 1 lim D n (x) = 1 x, n λ n 1 πg(x) uniformly on compact subsets of ( 1, 1), cf. the proof of Corollary.9. By property (LB) there are no masses at ±1. Remark 3.3. Corollary 3. is also obtained in [4, p.758]. The Jacobi polynomials P n (α,α) (x) in the standard notation of Szegő, cf. [10], are discussed in the Introduction. The corresponding orthonormal polynomials are denoted P n (α; x). We recall that c α 1 [P (α,α) 1 n (x)] (1 x ) α dx = α+1 Γ(n + α + 1) (n + α + 1)n!Γ(n + α + 1)B(α + 1, 1/). (34) 15

16 Proposition 3.4. Property (LB), defined in (16), holds for the orthonormal symmetric Jacobi polynomials (P n (α; x)) if and only if α 1/. More precisely: (i) For α 1/ (ii) For 1 < α < 1/ inf{p n(α; x) + P n 1(α; x) x R, n N} α + 3. inf{p n(α; x) + P n 1(α; x) x R, n N} = 0. Proof. Assume α 1/. In this case we get from (7) λ n = 1 [ ] 4α 1 1, 4 4(n + α + 1) 1 so (λ n ) is increasing with lim λ n = 1/. By Proposition 3.1 we thus have P n + P n 1 λ 0 = α + 3, which shows (i). In order to show (ii) we use Theorem 31 on page 170 of [9] stating w n (x k,n )p n 1(w, x k,n ) 1 x k,n (35) for a generalized Jacobi weight w. (For two positive sequences (a n ), (b n ) we write a n b n if 0 < C 1 a n /b n C < for suitable constants C j.) Applying this to the largest zero x 1,n of the orthonormal symmetric Jacobi polynomials (P n (α; x)), we get w n (x 1,n )Pn 1(α; x 1,n ) 1 x 1,n (36) with This gives in particular w n (t) = ( 1 t + 1 n )α ( 1 + t + 1 n )α > (1 t ) α. P n 1(α; x 1,n ) C(1 x 1,n) 1/ α, hence lim P n 1(α; x 1,n ) = 0 for α < 1/. This shows (ii) because P n (α; x 1,n ) = 0. 16

17 Remark 3.5. For 1 < α < 1/ the observation of (ii) follows easily from the asymptotic result P n (α; 1) d α n α+1/, n, where d α is a suitable constant, but this simple asymptotic can not be used when 1/ α < 1/. The proof of (ii) presented above has kindly been communicated to us by Paul Nevai. Our original proof is based on Hilb s asymptotic formula [10, Thm 8.1.1]: ( θ 1/ sin θ ) α+1/ ( cos θ ) α+1/ P n (α,α) (cos θ) = Γ(α + n + 1) n! N α J α (Nθ) + O(n 3/ ), (37) where θ [c/n, π/], N = n + α + 1 and c > 0 is fixed. Let j α denote the smallest positive zero of the Bessel function J α. Defining θ n = j α /N, we get n α P (α,α) n (θ n ) = O(n 3/ ), n α P (α,α) n 1 (θ n ) = (1/ + o(1))j α (j α n + α 1/ n + α + 1/ ) + O(n 3/ ) = O(n 1 ). By (34) and Stirling s formula and hence c α 1 This shows that 1 [P n (α,α) (x)] (1 x ) α dx α B(α + 1, 1/) n 1, P n(α; cos θ n ) = O(n α ), P n 1(α; cos θ n ) = O(n α 1 ). P n(α; cos θ n ) + P n 1(α; cos θ n ) 0 when α < 1/. Remark 3.6. The example of symmetric Jacobi polynomials suggests that if (λ n ) is decreasing, then property (LB) does not hold. This is not true, 17

18 however, because for 1 < λ 0 < 1 and λ n = 1 for n 1 we have a decreasing sequence. The corresponding Jacobi matrix has norm 1 because this is so for the cases λ 0 = 1 and λ 0 = 1/, which correspond to the Chebyshev polynomials of the second and first kind respectively. Furthermore, for n we have by (30) and (3) D n = λ n 1 Pn λ n P n 1 P n+1 = D = [λ 4 λ 0 (λ )x ] and for 1 < x < 1 D (x) > D (1) = λ 0 (λ 0 1 ) > 0. On the other hand, (33) applies for n, and we see that the orthonormal polynomials satisfy inf{p n(x) + P n 1(x) x R, n N} λ 0 (λ 0 1 ). 4 Continuous q-ultraspherical polynomials The continuous q-ultraspherical polynomials C n (x; β q) depend on two real parameters q, β, and for q, β < 1 they are orthogonal with respect to a continuous weight function on ( 1, 1), cf. [6],[7]. The three term recurrence relation is xc n (x; β q) = 1 qn+1 (1 βq n ) C n+1(x; β q) + 1 β q n 1 (1 βq n ) C n 1(x; β q), n 0 (38) with C 1 = 0, C 0 = 1. The orthonormal version C n (x; β q) satisfies equation (9) with λ n = 1 (1 q n+1 )(1 β q n ) (1 βq n )(1 βq n+1 ). (39) The value C n (1; β q) is not explicitly known, and therefore we can only obtain the recurrence coefficients α n, γ n from (1) for p n (x) = C n (x; β q)/c n (1; β q) as given by the recursion formulas α n+1 = which we get from the relation λ n = α n+1 γ n. λ n 1 α n, α 0 = 0, γ n = 1 α n, (40) 18

19 Theorem 4.1. (i) Assume 0 β q < 1. Then the recurrence coefficients (λ n ) form an increasing sequence with limit 1/, and therefore (C n (x; β q)) satisfies (LB). (ii) Assume 0 q β < 1. Then the recurrence coefficients (λ n ) form a decreasing sequence with limit 1/, and the sequence (α n ) is increasing and concave with limit 1/. In particular, we have with c = α 1 (1 α )/(1 α 1 ). Proof. The function n (x) 1 x c n(0), 1 < x < 1, n 1, ψ(x) = (1 qx)(1 β x) (1 βx)(1 βqx) = 1 + (1 β)(β q) x (1 βx)(1 βqx) is decreasing for 0 β q < 1 and increasing for 0 q β < 1. This shows that λ n = (1/) ψ(q n ) is increasing in case (i) and decreasing in case (ii). In both cases the limit is 1/. In case (ii) we therefore have λ n 1/4 and hence α n+1 1 4(1 α n ) α n, because 4x(1 x) 1 for 0 x 1. This shows that (α n ) is increasing and hence with limit 1/. We further have α n+1 α n = (λ n 1 4 ) + (1 α n)( 1 α n+1), showing that α n+1 α n is decreasing, i.e., (α n ) is concave. We can now apply Theorem.5. References [1] H. Alzer, S. Gerhold, M. Kauers, A. Lupaş, On Turán s inequality for Legendre polynomials, Expo. Math. 5 no. (007), [] R. Askey, Linearization of the product of orthogonal polynomials, in Problems in Analysis, R. Gunning, ed., Princeton University Press, Princeton, N.J.(1970),

20 [3] A. E. Danese, Explicit evaluations of Turán expressions, Annali di Matematica Pura ed Applicata, Serie IV 38 (1955), [4] J. Dombrowski, P. Nevai, Orthogonal polynomials, measures and recurrence relations, SIAM J. Math. Anal. 17 (1986), [5] F. Filbir, R. Lasser, R. Szwarc, Reiter s condition P 1 and approximate identities for polynomial hypergroups, Monatsh. Math. 143 (004), [6] M. E. H. Ismail, Classical and Quantum Orthogonal Polynomials in One Variable, Cambridge University Press, Cambridge 005. [7] R. Koekoek and R. F. Swarttouw, The Askey-scheme of hypergeometric orthogonal polynomials and its q-analogue, Report no , TU-Delft, [8] A. Máté, P. G. Nevai, Orthogonal polynomials and absolutely continuous measures, in Approximation Theory IV, C. K. Chui et. al., eds., Academic Press, New York, 1983, [9] P. G. Nevai, Orthogonal Polynomials, Memoirs Amer. Math. Soc. 18 No. 13 (1979). [10] G. Szegő, Orthogonal Polynomials, 4th ed., Colloquium Publications 3, Amer. Math. Soc., Rhode Island, [11] R. Szwarc, Positivity of Turán determinants for orthogonal polynomials, in Harmonic Analysis and Hypergroups, K. A. Ross et al., eds., Birkhäuser, Boston-Basel-Berlin, 1998, [1] V. R. Thiruvenkatachar, T. S. Nanjundiah, Inequalities concerning Bessel functions and orthogonal polynomials. Proc. Indian Acad. Sciences, Section A, 33 (1951), [13] P. Turán, On the zeros of the polynomials of Legendre, Mat. Fys. 75 (1950), Časopis Pest. [14] K. Venkatachaliengar, S. K. Lakshmana Rao, On Turán s inequality for ultraspherical polynomials, Proc. Amer. Math. Soc. 8 no. 6 (1957),

21 Christian Berg Department of Mathematics, University of Copenhagen, Universitetsparken 5, DK-100, Denmark Ryszard Szwarc Institute of Mathematics, University of Wroc law, pl. Grunwaldzki /4, Wroc law, Poland and Institute of Mathematics and Computer Science, University of Opole, ul. Oleska 48, Opole, Poland 1