A note on asymptotic zero distribution of orthogonal polynomials

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Linear Algebra and its Applications 375 (2003) 275–281

www.elsevier.com/locate/laa

A note on asymptotic zero distribution

of orthogonal polynomials

William F. Trench

∗

Trinity University, San Antonio, TX 78212-7200, USA Received 9 December 2002; accepted 4 June 2003

Submitted by E. Tyrtyshnikov

Abstract

We strengthen a theorem of Kuijlaars and Serra Capizzano on the distribution of zeros of a sequence of orthogonal polynomials{pn}∞_n₌₁for which the coefficients in the three term recurrence relation are clustered at finite points. The proof uses a matrix argument motivated by a theorem of Tyrtyshnikov.

Keywords: Absolutely equally distributed; Equally distributed; Orthogonal polynomials; Zero distribu-tion

1. Introduction

Let{pn}∞n₌₋₁be a sequence of polynomials such that p−1≡ 0, p0≡ 1, and an+1pn+1(x)+ (bn− x)pn(x)+ anpn−1(x)≡ 0, n 1,

where an>0 and bnis real for n 0. By Favard’s theorem, {pn}∞n=0are orthogonal

with respect to a probability measure on (−∞, ∞), and pn has distinct real zeros x1n< x2n<· · · < xnn, n 1.

Let C0be the set of functions F in C(R) with bounded support. If S is a set, let

card S denote the cardinality of S.

∗_{Address: 95 Pine Lane, Woodland Park, CO 80863, USA. Tel.: +1-719-687-2109.}

E-mail address: [email protected] (W.F. Trench).

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We consider the following theorem of Kuijlaars and Serra Capizzano [4].

Theorem 1. Suppose that{an}∞_n₌₁⊂ R+,{bn}∞_n₌₀⊂ R, and there are constants a and b such that

card{r n| |ar− a| } = o(n) (1)

and

card{r n| |br− b| } = o(n) (2)

for every > 0. If a > 0, let

dµ(x)

dx =

₁

π√(b+ 2a − x)(x − b + 2a) if x∈ (b − 2a, b + 2a),

0 otherwise. (3) If a= 0, let µ(x)= 0, x < b, 1, x b. (4) Then lim n→∞ 1 n n r=1 F (xrn)= F (x)dµ(x) (5) if F ∈ C0.

Theorem 1 (which is actually stated in a different, but equivalent, form in [4]) is related to results of Geronimo et al. [1], Mercer [5,6], and Nevai [7, Theorem 5.3], in which C0is the set of functions F in C(R) that vanish at±∞.

In [4], Theorem 1 is obtained as a corollary of a more general result that appeals to Szegö’s theorem on the asymptotic distribution of the eigenvalues of Hermitian Toeplitz matrices, to matrix perturbation results due to Tyrtyshnikov [13, Theorem 3.1], and to the theory of locally Toeplitz sequences due to Tilli [10] and Serra Capizzano [8,9]. In this paper we prove a strengthened version of Theorem 1 without using Szegö’s distribution theorem or the theory of locally Toeplitz sequences. We do not claim that our proof can be extended to obtain the main results in [4].

2. The stronger theorem

Suppose that a > 0 and b is real. Let{qn}∞n=−1 be the sequence of orthogonal

polynomials such that q₋₁≡ 0, q0≡ 1, and

aqn₊₁(x)+ (b − x)qn(x)+ aqn−1(x)≡ 0, n 1.

Let y1n< y2n<· · · < ynnbe the zeros of qn.

Following Tyrtyshnikov [14] and Kuijlaars and Serra Capizanno [4] (see also [15]), we note that x1n, x2n, . . . , xnnare the eigenvalues of the Jacobi matrix

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An=         b0 a1 a1 b1 a2 a2 . .. . .. . .. . .. an−1 an−1 bn−1        

and y1n, y2n, . . . , ynnare the eigenvalues of the n× n tridiagonal matrix

Bn=         b a a b a a . .. ... . .. ... a a b         .

The eigenvalues of Bn, written in nondecreasing order, are yrn= b − 2a cos rπ

n+ 1, 1 r n

(see, e.g., [2,3,11]).

If a > 0 and F ∈ C[b − 2a, b + 2a], then lim n→∞ 1 n n r=1 F (yrn)= 1 π π 0 F (b− 2a cos y) dy,

since the sum on the left is a Riemann sum for the integral on the right. The change of variable x= b − 2a cos y yields

1 π π 0 F (b− 2a cos y) dy = F (x)dµ(x), with µ as in (3). Hence, (5) can be replaced by

lim n_→∞ 1 n n r=1 F (xrn)= lim n_→∞ 1 n n r=1 F (yrn)= F (x)dµ(x) (6) if F ∈ C0. Consequently, n r=1 (F (xrn)− F (yrn))= o(n) (7)

if F ∈ C0. If a= 0, Theorem 1 implies (6) (with µ as in (4)) and (7).

In the terminology of [12], we say that {{xrn}n_r₌₁}n1 and {{yrn}n_r₌₁}n1 are equally distributed with respect to C0.

If a, b∈ R and a 0, let F(a,b)be the set of bounded real-valued functions on

R that are continuous on some open interval containing[b − 2a, b + 2a]. Since we do not require that a > 0, we define[b, b] = {b}.

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We will prove the following theorem.

Theorem 2. Under the assumptions of Theorem 1, (6) holds and

n

r=1

|F (xrn)− F (yrn)| = o(n) if F ∈ F(a,b). (8)

In the terminology of [12], we say that{{xrn}n_r₌₁}n1and{{yrn}n_r₌₁}n1are ab-solutely equally distributed with respect to F(a,b).

3. Proof of Theorem 2

By the argument given above, if a > 0 and F ∈ C[b − 2a, b + 2a] then lim n→∞ 1 n n r=1 F (yrn)= F (x)dµ(x). (9)

If a= 0 and F ∈ F(b,b), then (4) implies (9). Hence, it suffices to verify (8). Our

method for doing this is motivated by [13, Theorem 3.1], which was also used in [4]; however, our conclusions are stronger than those stated in that theorem.

If−∞ < γ < δ < ∞, let mid[γ, x, δ] =    γ , x < γ , x, γ x δ, δ, x > δ.

Since{yrn}nr=1⊂ [b − 2a, b + 2a], n 1, it suffices to prove that n

i=1

|mid[γ, xrn, δ] − mid[γ, yrn, δ]| = o(n) (10)

for every[α, β], since (8) then follows from [12, Theorem 3].

Suppose that > 0. We will define several objects that depend upon , but for convenience we suppress this dependence in the notation. Let

αr = a− ar if|a − ar| < , 0 if|a − ar| , βr = b− br if|b − br| < , 0 if|b − br| , ˜αr = 0 if|a − ar| < , a− ar if|a − ar| , ˜βr = 0 if|b − br| < , b− br if|b − br| , Un=         β0 α1 α1 β1 α2 α2 . .. . .. . .. . .. αn−1 αn−1 βn−1         ,

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˜Un=         ˜β0 ˜α1 ˜α1 ˜β1 ˜α2 ˜α2 . .. . .. . .. . .. _˜αn−1 ˜αn−1 ˜βn−1         and Cn= An+ Un. Then Bn= Cn+ Un. (11)

Let z1n z2n · · · znnbe the eigenvalues of Cn.

Since |αr| < , r 1, and |βr| < , r 0, the Wielandt–Hoffman theorem

implies that n r=1 (xrn− zrn)2 Un2F = n−1 r=0 β_r2+ 2 n−1 r=1 α_r2 (3n − 2)2.

From this and Schwarz’s inquality,

n r=1 |xrn− zrn| √ 3 n. (12) Since

|mid[γ, x, δ] − mid[γ, y, δ]| |x − y|, (13)

(12) implies that n r₌₁ |mid[γ, xrn, δ] − mid[γ, zrn, δ]| √ 3n. (14) Let

pn= card {r n| |br− b| } + 2 card {r n| |ar − a| } = o(n)

(see (1) and (2)). Since rank(Un) pn, (11) and a standard theorem [16, pp. 94–97]

first applied to the asymptotic eigenvalue distribution problem by Tyrtyshnikov [13] imply that if n > 2pn, then

b− 2a yr−pn,n zrn yr+pn,n b + 2a, pn+ 1 r n − pn. (15)

Since

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(15) implies that n−pn r=pn+1 |yrn− zrn| n−pn r=pn+1 (yr+pn,n− yr−pn,n) 8apn.

This and (13) imply that

n−pn r_=pn+1

|mid[γ, yrn, δ] − mid[γ, zrn, δ]| 8apn,

so

n

r=1

|mid[γ, yrn, δ] − mid[γ, zrn, δ]| 16apn.

This and (14) imply that

n i=1 |mid[γ, xrn, δ] − mid[γ, yrn, δ]| √ 3n+ 16apn.

Hence, since pn= o(n),

lim sup n→∞ 1 n n i=1 |mid[γ, xrn, δ] − mid[γ, yrn, δ]| √ 3n. Since is arbitrary, this implies (10), which completes the proof.

Acknowledgements

I thank a referee for pointing out that my first proof of Theorem 2 was unneces-sarily long and complicated.

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