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Positivity of Turán determinants for orthogonal polynomials II

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Abstract

The polynomials pn orthogonal on the interval [1,1], normalized by pn(1)=1, satisfy Turán’s inequality if pn2(x)pn1(x)pn+1(x)0 for n1 and for all x in the interval of orthogonality. We give a general criterion for orthogonal polynomials to satisfy Turán’s inequality. This extends essentially the results of Szwarc(1998). In particular the results can be applied to many classes of orthogonal polynomials, by inspecting their recurrence relation.

Introduction

Consider a symmetric probability measure μ such that suppμ=[1,1]. By the Gram–Schmidt orthogonalization procedure applied to the system of monomials xn, n0, we obtain a sequence of orthogonal polynomials pn(x), n0. Every polynomial pn is of exact degree n. We may assume that its leading coefficient is positive. It is well known that the polynomials pn satisfy the three term recurrence relation of the form xpn=γnpn+1+αnpn1,n0,with convention α0=p1=0. Due to orthogonality the polynomial pn has n roots in the open interval (1,1). Therefore pn(1)>0. Let Pn(x)=pn(x)pn(1),n0.The coefficients γn,αn+1 are positive for n0. In case the polynomials pn are orthonormal then the sequences of the coefficients are related by γn=αn+1 and the recurrence relation simplifies to xpn=αn+1pn+1+αnpn1,n0.We refer to [5], [16] for the basic theory concerning orthogonal polynomials.

We are interested in determining when Δn(x)Pn(x)2Pn1(x)Pn+1(x)0,n1.The expression Δn(x) is called the Turán’s determinant. The problem has been studied for many classes of specific orthogonal polynomials (see [1], [2], [3], [4], [6], [7], [8], [10], [12], [13], [14], [15], [20], [21]). We refer to the introduction in [18] for a short account of known results.

Turán determinants can be used to determine the orthogonality measure μ in terms of orthonormal polynomials pn. Paul Nevai [11] observed if αnn1/2 then the sequence of measures (perhaps signed) [pn2(x)pn1(x)pn+1(x)]dμ(x)is weakly convergent to the measure 2π1x2dx,|x|<1.Maté and Nevai [9] showed that if additionally sequence αn has bounded variation then the limit of Turán determinants exists. Moreover the orthogonality measure is absolutely continuous on the interval (1,1) its density is given by 21x2πf(x),|x|<1,where f(x)limn[pn2(x)pn1(x)pn+1(x)]>0,|x|<1.

It turns out that the way we normalize the polynomials is essential for the Turán inequality to hold. Indeed, assume pn satisfy (1) and pn(1)=1, i.e. αn+γn=1.Assume pn2(x)pn1(x)pn+1(x)0,|x|1,n1.Define new polynomials by pn(σ)(x)=σnpn(x), where σn is a sequence of positive coefficients. Then the condition {pn(σ)(x)}2pn1(σ)(x)pn+1(σ)(x)0,|x|1,n1is equivalent to (see Proposition [18]) σn2σn1σn+10,n1.This means if the Turán determinants are nonnegative, when the polynomials are normalized at x=1, then they stay nonnegative for any other normalization provided that they are nonnegative at x=1, as σn=pn(σ)(1).

By Theorem 1 [18] if the polynomials are normalized at x=1, i.e. pn(1)=1, αn is increasing and αn12, the Turán determinants are positive in the interval (1,1). This result can be applied to many classes of orthogonal polynomials, including for example the ultraspherical polynomials for which positivity has been obtained in [12], [13]

The result mentioned above can be applied provided that we are given the coefficients αn explicitly. For many classes of orthogonal polynomials in the interval [1,1] we are given recurrence relations, but the values pn(1) cannot be evaluated in the explicit form. Therefore we are unable to provide a recurrence relation for the polynomials Pn(x)=pn(x)/pn(1), in the form for which we can inspect easily the assumptions of Theorem 1 [18]. This occurs when we study the associated polynomials. Indeed assume pn satisfy (1), (3). For a fixed natural number the associated polynomials pn(k) of order k are defined by xpn(k)=γkp1(k)n=0,γn+kpn+1(k)+αn+kpn1(k)n1.These polynomials do not satisfy pn(k)(1)=1 as p1(k)(1)=γk1=(1αk)1>1.

The obstacle described above has been partially overcome in Corollary 1 of [18], but it required additional assumptions, in particular γ01. Unfortunately many examples including the associated polynomials violate that condition. The aim of this note is to provide a counterpart to Corollary 1 [18] by allowing γ0<1. This is done in Theorem 1. As the assumptions in this theorem are complicated Corollary 1 provides a wide class of relatively simple recurrence relations for which Theorem 1 applies. General examples are provided at the end of the paper.

Section snippets

Results

Theorem 1

Assume the polynomials pn satisfy xpn=γnpn+1+αnpn1,n0,where α0=p1=0,p0=1. Assume

  • (a)

    the sequence αn is strictly increasing and αn1/2,

  • (b)

    the sequence γn is positive and strictly decreasing ,

  • (c)

    αn+γn1.

Assume also that there holds αnαn1αnγn1αn1γnαn+1γnαnγn+1γnγn+1,n1,γ0γ1α1γ02. Then for Pn(x)=pn(x)pn(1)we have Pn(x)2Pn1(x)Pn+1(x)0,1x1.

Proof

Let gn=pn+1(1)pn(1).By (5) we get gn=1γn1αngn1,n1.

Lemma 1

Under assumptions of Theorem 1 there holds 1gnαn+1γnαnγn+1γnγn+1,n0.

Proof

(5) gives g0=1/γ01. Assume

Acknowledgments

I am grateful to the referees. Their remarks improved the exposition substantially.

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