ZEROS OF ORTHOGONAL POLYNOMIALS ON THE REAL LINE

SERGEY A. DENISOV^{1} AND BARRY SIMON^{1,2}

Abstract. Let p_{n}(x) be the orthonormal polynomials associated to a measure dμ of compact support in ℝ. If E ∉ supp(dμ), we show there is a δ > 0 so that for all n, either p_{n} or p_{n+1} has no zeros in (E − δ, E + δ). If E is an isolated point of supp(μ), we show there is a δ so that for all n, either p_{n} or p_{n+1} has at most one zero in (E − δ, E + δ). We provide an example where the zeros of p_{n} are dense in a gap of supp(dμ).

1. Introduction

Let dμ be a measure on ℝ whose support is not a finite number of points and with ∫ |x|^{n} dμ(x) < ∞ for all n = 0, 1, 2, . . .. The orthonormal polynomials p_{n}(x; dμ) or p_{n}(x) are determined uniquely by

p_{n}(x) = γ_{n}x^{n} + lower order γ_{n} > 0  (1.1)
∫ p_{n}(x)p_{m}(x) dμ(x) = δ_{nm}  (1.2)

There are a_{n} > 0, b_{n} ∈ ℝ for n ≥ 1 so that

xp_{n}(x) = a_{n+1}p_{n+1}(x) + b_{n+1}p_{n}(x) + a_{n}p_{n}(x)  (1.3)

(many works use a_{n−1}, b_{n−1} where we use a_{n}, b_{n}).

In this paper, we will be interested in the zeros of p_{n}(x; dμ). The following results are classical (see, e.g., Freud’s book [3]):

(1) The zeros of p_{n}(x) are real and simple.

(2) If (a, b) ∩ supp(dμ) = ∅, then if a = −∞ or b = +∞, p_{n} has no zeros in (a, b) and, in any event, (a, b) has at most one zero of p_{n}(x).

(3) In the determinate case, if x_{0} ∈ supp(dμ) and δ > 0, for all large n, p_{n}(x) has a zero in (x_{0} − δ, x_{0} + δ).

Date: July 5, 2002.

1 Mathematics 253-37, California Institute of Technology, Pasadena, CA 91125.

E-mail: denissov@its.caltech.edu; bsimon@caltech.edu.

2 Supported in part by NSF grants DMS-9707661, DMS-0140592.

Define

N_{n}(x_{0}, δ) = # of zeros of p_{n}(x) in (x_{0} − δ, x_{0} + δ)

Then (1)–(3) immediately imply:

(i) If x_{0} is a non-isolated point of supp(dμ), then for any δ > 0, lim_{n→∞} N_{n}(x_{0}, δ) = ∞.

(ii) If x_{0} is an isolated point of supp(dμ) and δ = dist(x_{0}, supp(dμ)\{x_{0}}), then N_{n}(x_{0}, δ) is never more than 2, and for all δ > 0 and n large, N_{n}(x_{0}, δ) ≥ 1.

(iii) If x_{0} ∉ supp(dμ) and δ = dist(x_{0}, supp(dμ)), then N_{n}(x_{0}, δ) is

never more than 1.

(i) is fairly complete, but (ii), (iii) leave open how often there is one vs. two points in case (ii) and zero vs. one in case (iii). One might guess that a zero near x_{0} ∉ supp(dμ) and two zeros near an isolated x_{0} in supp(dμ) are not too common occurrences. Example. If dμ is even about x = 0, then p_{n}(−x) = (−1)^{n}p_{n}(x). Thus, if n is odd, p_{n}(0) = 0. So if 0 ∉ supp(dμ), we still have N_{n}(0, δ) = 1 for all small δ and n odd. If zero is an isolated point of dμ, p_{n} for n even has a zero at x_{n} near 0, but not equal to 0 (since zeros are simple), so also at −x_{n}, that is, N_{n}(0, δ) = 2 for δ small and n even. So “not too common” can be as often as 50% of the time. Our goal here is to show this 50% is a maximal value.

It is surprising that there do not seem to be any results on these issues until a recent paper of Ambroladze [1], who proved

Theorem (Ambroladze [1]). If supp(dμ) is bounded and x_{0} ∉ supp(dμ), then for some δ > 0, lim inf_{n→∞} N_{n}(x_{0}, δ) = 0.

Thus we can use N_{n}(x_{0}, δ) to distinguish when x_{0} ∈ supp(dμ). Our goal in this paper is to prove

Theorem 1. Let d = dist(x_{0}, supp(dμ)) > 0. Let δ_{n} = d^{2}/(d + \sqrt{2} a_{n+1}) (where a_{n} is the recursion coefficient given by (1.3)). Then either p_{n} or p_{n+1} (or both) has no zeros in (x_{0}−δ_{n}, x_{0}+δ_{n}). In particular, if a_{∞} = sup_{n} a_{n} < ∞ and d_{∞} = d^{2}/(d+\sqrt{2} a_{∞}), then (x_{0}−δ_{∞}, x_{0}+δ_{∞}) does not have zeros of p_{j} for two successive values of j.

Theorem 2. Let x_{0} be an isolated point of supp(dμ). Then there exists a d_{0} > 0, so that if δ_{n} = d_{0}^{2}/(d_{0} + \sqrt{2} a_{n+1}), then at least one of p_{n} and p_{n+1} has no zeros or one zero in (x_{0} − δ_{n}, x_{0} + δ_{n}). In particular, if a_{∞} = sup_{n} a_{n} < ∞ and δ_{∞} = d_{0}^{2}/(d_{0} + \sqrt{2} a_{∞}), then for all large n, either N_{n}(x_{0}, δ_{∞}) = 1 or N_{n+1}(x_{0}, δ_{∞}) = 1.

We will prove Theorem 1 in Section 2 and Theorem 2 in Section 3. In Section 4, we present an example of a set of polynomials whose zeros are dense in a gap of the spectrum.

It is a pleasure to thank Leonid Golinskii and Paul Nevai for useful correspondence.

2. Points Outside the Support of dμ

We arrived at the following lemma by trying to abstract the essence of Ambroladze’s argument [1]; it holds for orthogonal polynomials on the complex plane. Let dμ be a measure on ℂ with finite moments and infinite support, and let p_{n}(z; dμ) be the orthonormal polynomials. Define the reproducing kernel

K_{n}(z, w) = ∑_{j=0}^{n} p_{n}(z) \overline{p_{n}(w)}  (2.1)

so in L^{2}(ℂ, dμ), for any polynomial π of degree n or less,

∫ K_{n}(z, w)π(w) dμ(w) = π(z)  (2.2)

Lemma 2.1. Suppose z_{0} ∈ ℂ, p_{j}(w) = 0 for some j ≤ n + 1. Then

|z_{0} − w| ≥ \frac{|p_{j}(z_{0})|}{K_{n}(z_{0}, z_{0})^{1/2}} dist(w, supp(dμ))  (2.3)

Proof. Let q(z) = p_{j}(z)/(z −w), which has deg(q) ≤ n. Thus, by (2.2), ⟨K( · , z_{0}), q( ·)⟩ = q(z_{0}) so, by the Schwarz inequality,

\frac{|p_{j}(z_{0})|}{|z_{0} − w|} ≤ ∥q∥ ∥K( · , z_{0})∥

By (2.2), ∥K( · , z_{0})∥ = K(z_{0}, z_{0})^{1/2} and clearly, ∥q∥ ≤ dist(w, supp(dμ))^{−1}∥p_{j}∥ = dist(w, supp(dμ))^{−1}. This yields (2.3). □

The following only holds in the real case:

Lemma 2.2. For any x ∈ ℝ and n,

K_{n}(x, x)dist(x, supp(dμ))^{2} ≤ a_{n+1}^{2}[p_{n+1}^{2}(x) + p_{n}^{2}(x)]  (2.4)

Proof. The Christoffel-Darboux formula [3] says

K_{n}(x, y) = a_{n+1}[\frac{p_{n+1}(x)p_{n}(y) − p_{n+1}(y)p_{n}(x)}{x − y}]

so since ⟨p_{j}, p_{k}⟩ = δ_{jk},

∥(x − ·)K_{n}(x, ·)∥^{2} = |a_{n+1}|^{2}[p_{n+1}^{2}(x) + p_{n}^{2}(x)]  (2.5)

Clearly,

∥(x − ·)K_{n}(x, ·)∥^{2} ≥ dist(x, supp(dμ))^{2}∥K_{n}(x, ·)∥^{2}  (2.6)

and, as above, ∥K_{n}(x, ·)∥^{2} = K_{n}(x, x), which yields (2.4). □

Remark. An alternate way of seeing (2.5) is to let ψ be the trial vector (p_{0}(x), …, p_{n}(x), 0, 0, …) and note that in terms of the standard Jacobi matrix ((J − x)ψ)_{j} = 0 unless j = n, n + 1, in which case the values are −a_{n+1}p_{n+1}(x) and a_{n+1}p_{n}(x). (2.6) is then just ∥(J −x)ψ∥ ≥ dist(x, supp(dμ))∥ψ∥.

Proof of Theorem 1. By (2.4), we have that

K_{n}(x_{0}, x_{0})dist(x_{0}, supp(dμ))^{2} ≤ 2a_{n+1}^{2}p_{n+1}^{2}(x_{0})  (2.7)

and/or

K_{n}(x_{0}, x_{0})dist(x_{0}, supp(dμ))^{2} ≤ 2a_{n+1}^{2}p_{n}^{2}(x_{0})  (2.8)

Suppose (2.7) holds. Then, by (2.3), if w is a zero of p_{n+1}(x) and if d = dist(x_{0}, supp(dμ)),

|x_{0} − w| ≥ \frac{1}{\sqrt{2}} \frac{1}{a_{n+1}} d dist(w, supp(dμ))

≥ \frac{1}{\sqrt{2}} \frac{1}{a_{n+1}} d(d − |w − x_{0}|)

which leads directly to |x_{0} − w| ≥ d^{2}/(d + a_{n+1}\sqrt{2}). □

Remark. There is also a Christoffel-Darboux result for polynomials on the unit circle ∂D = {z | |z| = 1} in ℂ. This leads to the following: If dμ is a measure on ∂D and z_{0} ∈ ∂D has d = dist(z_{0}, supp(dμ)) > 0, then the circle of radius d^{2}/(2 + d) has no zeros of the orthogonal polynomials. L. Golinskii has pointed out that the theorem of Fejér [2] that the zeros lie in the convex hull of supp(dμ) implies there are no zeros in the circle of radius d^{2}/2 — which is a stronger result, so we do not provide the details.

3. Isolated Points of the Support of dμ

To prove Theorem 2, we will make use of the second kind polynomials [3, 5] associated to dμ and {p_{n}}. This is a second family of polynomials, q_{n} defined by recursion coefficients, ã_{n}, b̃_{n} with

ã_{n} = a_{n+1}  b̃_{n} = b_{n+1}  (3.1)

They have the following two critical properties:

Proposition 3.1. (i) The zeros of p_{n+1} and q_{n} interlace. In partic-

ular, between any two zeros of p_{n+1} is a zero of q_{n}. (ii) If x_{0} is an isolated point of dμ and dν is a suitable measure with

respect to which the q’s are orthogonal, then x_{0} ∉ supp(dν).

These are well known. (i) follows from the fact that the zeros of p_{n+1} are eigenvalues of the matrix

J_{ij}^{(n+1)} = b_{i}δ_{ij} + a_{i}δ_{i i+1} + a_{i−1}δ_{i i−1} 1 ≤ i, j ≤ n + 1

and the zeros of q_{n} are the eigenvalues of

J̃_{ij}^{(n)} = b̃_{i}δ_{ij} + ã_{i}δ_{i i+1} + ã_{i−1}δ_{i i−1} 1 ≤ i, j ≤ n

which is the matrix J_{ij}^{(n+1)} with the top row and left column removed. (ii) follows because of the relation that ν obeys for all z ∈ ℂ\ℝ [5]:

∫ \frac{dν(x)}{x − z} = a_{1}^{−2}[b_{1} − z − (∫ \frac{dμ(x)}{x − z})^{−1}]  (3.2)

(if the moment problem is indeterminate, this is one possible ν). Isolated points of dμ are poles of ∫ dμ(x)/(x − z) so ∫ dν(x)/(x − z) is regular there.

Proof of Theorem 2. Let d_{0} = dist(x_{0}, supp(dν)) > 0 by (ii) of Proposition 3.1. By Theorem 1 and (3.1), either q_{n−1} or q_{n} has no zeros in (x_{0} −δ_{n}, x_{0} +δ_{n}). By the intertwining result (Proposition 3.1(i)), either p_{n} or p_{n+1} cannot have two zeros in this interval. □

Remark. If b ∈ supp(dμ) is such that |x_{0} − b| = dist(x_{0}, supp(dμ)) and ∫ dμ(y)/|y − b| = ∞, then dν has an isolated point in between x_{0} and b, and so d_{0} < dist(x_{0}, supp(dμ)).

4. An Example of Dense Zeros in the Gap

Nevai raised the issue of whether as n varies, the single possible zero of p_{n} in a gap (a, b) of supp(dμ) can yield all of (a, b) as limit points, or if the situation of a single (or finite number of) limit point as in the example in Section 1 is the only possibility. In this section, we describe an explicit bounded Jacobi matrix so that supp(dμ) = [−5, −1] ∪ [1, 5] but the set {x ∈ (−1, 1) | p_{n}(x) = 0 for some n} is dense in [−1, 1].

Let {β_{j}}_{j=1}^{∞} be the sequence

β_{1}, β_{2}, … = 0, −\frac{1}{2}, 0, \frac{1}{2}, −\frac{3}{4}, −\frac{1}{2}, −\frac{1}{4}, 0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, −\frac{7}{8}, …

which goes through all dyadic rationals in (−1, 1) with denominator 2^{k} successively for k = 1, 2, 3, … with each j/2^{k} “covered” multiple times. Let L be the Jacobi matrix with

a_{2n−1} = 3,  a_{2n} = 1, n = 1, 2, …  (4.1)
b_{k} = β_{n}  if 2n^{2} ≤ k < 2(n + 1)^{2}  (4.2)
b_{1} = β_{1}

We claim that

(1) supp(dμ) = [−5, −1] ∪ [1, 5]

(2) There is an x_{n} with |x_{n} − β_{n}| ≤ 2 3^{−2n} so that

P_{2(n+1)^{2}−1}(x_{n}) = 0  (4.3)

This provides the claimed example.

Remarks. 1. By adjusting a_{1} and a_{2} (but keeping a_{2n+1} = a_{1}; a_{2n} = a_{2}), we can replace [−5, −1] ∪ [1, 5] by [−3 − ε, −1] ∪ [1, 3 + ε], but our method seems to require bands larger than the size of the gap.

2. One can replace (4.2) by b_{k} = β_{n} for ℓ_{n} ≤ k < ℓ_{n+1} so long as ℓ_{n+1} − ℓ_{n} → ∞.

3. We believe that the measure associated to L is purely singular. It is perhaps true that the phenomenon of zeros dense in a gap requires purely singular spectral measure.

To prove the claims, we let L_{0} be the Jacobi matrix with a’s given by (4.1) but b_{n} = 0, and L_{∞} the period two, doubly infinite matrix on ℤ which equals L_{0} when restricted to ℤ^{+}. By the general theory of periodic Schrödinger operators [4], the spectrum of L_{∞} is the two bands where |Δ(x)| ≤ 2 where Δ is the discriminant, that is, the trace of the two-step transfer matrix. If a_{1}, a_{2} are the two values of a (so a_{1} = 3, a_{2} = 1 in our example), a simple calculation shows that

Δ(x) = \frac{1}{a_{1}a_{2}} (x^{2} − (a_{1}^{2} + a_{2}^{2}))

so Δ(x) = ±2 occurs at x = ±|a_{1} ± a_{2}|. Thus

spec(L_{∞}) = [−4, −2] ∪ [2, 4]  (4.4)

The orthonormal polynomials p_{n}^{(0)} for L_{0} at x = 0 obey the recursion relation

p_{2n+2}^{(0)}(0) = −3p_{2n}^{(0)}(0)

so we have

p_{2n+1}^{(0)} = 0 p_{2n}^{(0)}(0) = (−3)^{n}  (4.5)

By the general theory of restricting periodic operators to the halfline, spec(L_{0}) is spec(L_{∞}) plus a possible single eigenvalue in the gap (−2, 2). Since there is a symmetry, the only possible eigenvalue is at x = 0, but (4.5) says that 0 is not an eigenvalue since ∑_{j=0}^{∞}|P_{j}(0)|^{2} = ∞. Thus spec(L_{0}) = [−4, −2] ∪ [2, 4] also. L − L_{0} is a diagonal matrix, so it is easy to see ∥L−L_{0}∥ = sup_{j}|β_{j}| = 1. Thus spec(L) ⊂ ∪_{x∈[−1,1]}x+ spec(L_{0}) = [−5, −1] ∪ [1, 5]. On the other hand, since the b’s are equal to β_{j} on arbitrary long runs, a Weyl vector argument shows that

spec(L) ⊃ \overline{∪_{j}β_{j} + spec(L_{0})} = [−5, −1] ∪ [1, 5]

so claim 1 is proven.

Let L_{n;F} be the n × n matrix obtained by taking the first n rows and columns of L. Then the zeros of p_{n}(x) are precisely the eigenvalues of L_{n;F} (see [5, Proposition 5.6]). Let φ_{j} be the j component vector with (P_{0}^{(0)}(0), P_{1}^{(0)}(0), …, P_{j−1}^{(0)}(0)). Then if j is odd so P_{j}^{(0)}(0) = 0, and we have L_{0;j,F} φ_{j} = 0. Thus, if j = 2(n + 1)^{2} − 1,

[(L_{j;F} − β_{n})φ_{j}]_{k} = (b_{k} − β_{n})φ_{j,k}  (4.6)

If 2n^{2} ≤ k ≤ 2(n + 1)^{2} − 1, the right-hand side is zero and its absolute value is always less than 2|φ_{j,k}|. Thus

\frac{∥(L_{j;F} − β_{n})φ_{j}∥^{2}}{∥φ_{j}∥^{2}} ≤ \frac{4 ∑_{k=0}^{n^{2}−1} 3^{2k}}{∑_{k=0}^{(n+1)^{2}−1} 3^{2k}}
≤ 4 3^{−4n}

by a simple estimate. Thus L_{j;F} has an eigenvalue within 2 3^{−2n} of β_{n}, proving claim 2.

This completes the example.

References

[1] A. Ambroladze, On exceptional sets of asymptotic relations for general orthogonal polynomials, J. Approx. Theory 82 (1995), 257–273.

[2] L. Fejér, Über die Lage der Nullstellen von Polynomen, die aus Minimumforderungen gewisser Art entspringen, Math. Ann. 85 (1922), 41–48.

[3] G. Freud, Orthogonal Polynomials, Pergamon Press, Oxford-New York, 1971.

[4] M. Reed and B. Simon, Methods of Modern Mathematical Physics, IV: Analysis of Operators, Academic Press, New York, 1978.

[5] B. Simon, The classical moment problem as a self-adjoint finite difference operator, Adv. in Math. 137 (1998), 82–203.