AN EQUIVALENCE FOR THE RIEMANN HYPOTHESIS IN TERMS OF ORTHOGONAL POLYNOMIALS

DAVID A. CARDON AND SHARLEEN A. DE GASTON

Abstract. We construct a measure such that if {p_{n}(z)} is the sequence of orthogonal polynomials relative to that measure, then the Riemann Hypothesis with simple zeros is true if and only if lim_{n→∞} \frac{p_{2n}(z)}{p_{2n}(0)} = \frac{ξ(1/2+iz)}{ξ(1/2)} where ξ(s) = \frac{1}{2} s(s − 1)π^{−s/2}Γ(s/2)ζ(s) is the Riemann ξ-function.

1. Introduction

Let ζ(s) = ∑_{n=1}^{∞} n^{−s} be the Riemann zeta function. Riemann showed that ζ(s) has an analytic continuation to all s with the exception of a simple pole at s = 1. The Riemann ξ-function, defined as ξ(s) = \frac{1}{2}s(s − 1)π^{−s/2}Γ(s/2)ζ(s), is an entire function satisfying ξ(s) = ξ(1 − s). The Riemann Hypothesis is the conjecture that all of the zeros of ξ(s) lie on the ‘critical line’ which is the line with real part 1/2. The Prime Number Theorem, proved independently by Hadamard and de la Vallée Poussin in 1896, is equivalent to the fact that all zeros of ξ(s) lie in the critical strip 0 < Re(s) < 1. Let M(T) denote the number of zeros in the critical strip with 0 < Im(s) ≤ T that lie on the critical line. Hardy [6] proved that M(T) tends to infinity as T tends to infinity. Hardy and Littlewood [7] showed that M(T) > AT for some positive constant A. Selberg [12] proved that M(T) > AT log T for some positive constant A. Since the number N(T) of zeros in the critical strip up to height T is known to be asymptotic to \frac{T}{2π} log(\frac{T}{2π}), Selberg showed that a positive proportion of the zeros are on the critical line. Extensive numerical calculations, such as [2, 3, 8, 9, 10, 11], have supported the Riemann Hypothesis. The numerical calculations have supported the stronger conjecture that, in addition to lying on the critical line, the zeros of ξ(s) are simple. In this paper we show that the Riemann Hypothesis with simple zeros is equivalent to the existence of a certain family of orthogonal polynomials \{p_{n}(z)\} such that lim_{n→∞} \frac{p_{2n}(z)}{p_{2n}(0)} = \frac{ξ(1/2+iz)}{ξ(1/2)}.

We will now describe the main result of this paper. We will define a step function F related to the zeros of ξ(s). Let Ξ(z) = ξ(1/2 + iz). Then the zeros of Ξ(z) lie in the strip −1/2 < Im(z) < 1/2, Ξ(z) is real for real z, Ξ(z) = Ξ(−z), and any non-real zeros of Ξ(z) occur in complex conjugate pairs. For z = x + iy in the region x ≥ 0, −1/2 ≤ y ≤ 1/2, let f(z) be analytic and satisfy:

(1) f(z) is real for real z,

Re(f(z)) > 0, (2)

|f(x + iy)| < e^{−cx} (3) where c is a positive constant. For T ≥ 0 let

Date: August 27, 2004.

2000 Mathematics Subject Classification. Primary 11M26. Secondary 41A10, 05E35.

Key words and phrases. Orthogonal polynomials, Riemann hypothesis, simple zeros.

(4)  F (T) = \frac{1}{2πi} ∫_{γ_{T}} \frac{Ξ′(z)}{Ξ(z)}f(z) dz

where γ_{T} is the positively oriented boundary of the region 0 ≤ x ≤ T, −1/2 ≤ y ≤ 1/2. Label the zeros of Ξ(z) in the region x > 0, 0 ≤ y < 1/2 as α_{k} + iβ_{k} where α_{k} ≤ α_{k+1}. If T is not equal to any α_{k}, F (T) may be represented as the finite sum

F (T) = ∑_{\begin{matrix}α_{k}<T \\ β_{k}=0 \end{matrix}} f(α_{k}) + ∑_{\begin{matrix}α_{k}<T \\ β_{k}>0 \end{matrix}}\{f(α_{k} + iβ_{k}) + f(α_{k} − iβ_{k})\}.

For T < 0 let F (T) = −F (T). If f(z) were replaced by 1, F (T) would be the number of zeros in the critical strip up to height T. However, we imposed the restriction in inequality (3) to guarantee the existence of certain integrals.

For polynomials p(x) and q(x) with real coefficients we define an inner product by the Riemann-Stieltjes integral

⟨p(x), q(x)⟩ = ∫_{−∞}^{∞} p(x)q(x)dF (x).

Applying the Gram-Schmidt orthogonalization process to the polynomials 1, x, x^{2}, … produces an orthogonal family of polynomials {p_{n}(x)} where the degree of p_{n}(x) is n. In this case, p_{2n}(x) is an even function while p_{2n+1}(x) is an odd function.

Then we have:

Theorem 1. The Riemann hypothesis with simple zeros is true if and only if

lim_{n→∞} \frac{p_{2n}(z)}{p_{2n}(0)} = \frac{ξ(1/2 + iz)}{ξ(1/2)}

for every z ∈ ℂ.

We note that the proof shows that lim_{n→∞} \frac{p_{2n}(z)}{p_{2n}(0)} = lim_{n→∞} \frac{p_{2n+1}(z)}{z p′_{2n+1}(0)}. Thus the theorem could be stated in terms of the odd degree polynomials p_{2n+1}(z) as well.

The proof of Theorem 1 is presented in §3.

2. A few facts about orthogonal polynomials

In this section we will recall several facts from the general theory of orthogonal polynomials that will be needed in the proof of Theorem 1. For the basic theory, we refer the reader to the books by Szëgo [14] and Chihara [4]. Our review is based on these two works but especially on [4].

A bounded non-decreasing function ψ is called a distribution function if its moments

(5)  μ_{n} = ∫_{−∞}^{∞} x^{n}dψ(x)

exist for n = 0, 1, 2, . . .. Two distribution functions ψ_{1} and ψ_{2} are substantially equal if and only if there is a constant K such that ψ_{1}(x) = ψ_{2}(x) + K at all common points of continuity. The spectrum of ψ is the set

𝔖(ψ) = {x | ψ(x + δ) − ψ(x − δ) > 0 for all δ > 0}.

If 𝔖(ψ) is infinite, then the expression

⟨p(x), q(x)⟩ = ∫_{−∞}^{∞} p(x)q(x)dψ(x)

defines an inner product on the space of polynomials with real coefficients. Using this inner product we orthogonalize the set of non-negative powers of x to create a family {p_{n}(x)} of orthogonal polynomials with real coefficients using the Gram-Schmidt procedure:

p_{0}(x) = 1,
p_{n}(x) = x^{n} − ∑_{k=0}^{n−1} \frac{⟨x^{n}, p_{k}⟩}{⟨p_{k}, p_{k}⟩}p_{k}(x) for k ≥ 1.

Lemma 2.1. ([14, Thm. 3.3.1] or [4, Thm. I.5.2].) The zeros of p_{n}(x) are real and simple for each n ≥ 1.

We will label the zeros of p_{n}(x) as y_{n1} < y_{n2} < ⋯ < y_{nn}.

Lemma 2.2. ([14, Thm. 3.3.3] or [4, Thm. I.5.3].) The zeros of p_{n}(x) and p_{n+1}(x) interlace. That is,

y_{n+1,i} < y_{ni} < y_{n+1,i+1}, i = 1, 2, …, n.

Furthermore, between any two zeros of p_{n}(x) there is at least one zero of p_{m}(x) for m > n.

Using the moments from equation (5) we define a moment functional on the space of polynomials by

L[p(x)] = ∫_{−∞}^{∞} p(x)dψ(x) = ∑_{k=0}^{n} c_{k}μ_{k}

where p(x) = c_{0} + c_{1}x + ⋯ + c_{n}x^{n}.

Lemma 2.3. ([14, Thm. 3.4.1] or [4, Thm. I.6.1].) There are numbers A_{n1}, A_{n2},…, A_{nn} such that for every polynomial π(x) of degree at most 2n − 1

L[π(x)] = ∑_{k=1}^{n} A_{nk}π(y_{nk}).

The numbers A_{nk} are all positive and A_{n1} + ⋯ + A_{nn} = μ_{0}.

The equation in Lemma 2.3 is called the Gauss quadrature formula. The numbers A_{nk} are sometimes called Christoffel numbers.

The zeros of the polynomials {p_{n}(x)} are strongly related to the spectrum 𝔖(ψ). Let

(6)  ψ_{n}(x) = \begin{cases}0 & if x < y_{n1}, \\ A_{n1} + ⋯ + A_{np} & if y_{np} ≤ x < y_{n,p+1} where 1 ≤ p < n, \\ μ_{0} & if x ≥ y_{nn}. \end{cases}

Lemma 2.4. ([4, Thm. II.3.1].) There is a subsequence of {ψ_{n}} that converges on (−∞, ∞) to a distribution function η which has a infinite spectrum and such that μ_{n} = ∫_{−∞}^{∞} x^{n}dψ(x) = ∫_{−∞}^{∞} x^{n}dη(x).

It is not generally true that η is substantially equal to ψ. Distribution functions, such as η, that are subsequential limits of {ψ_{n}} are called natural representatives of the moment functional L.

Lemma 2.5. ([14, Thm. 3.41.2] or [4, Thm. II.4.1].) The open interval (y_{ni}, y_{n,i+1}) contains at least one spectral point of the function ψ.

Lemma 2.6. ([4, Thm. II.4.3].) Let η be a natural representative of L and let s ∈ 𝔖(η). Then every neighborhood of s contains a zero of p_{n}(x) for infinitely many values of n.

Given a list of moments {μ_{n}}_{n=0}^{∞}, the Hamburger moment problem consists of classifying the distribution functions ϕ such that

μ_{n} = ∫_{−∞}^{∞} x^{n}dϕ(x), n = 0, 1, 2, …

If all solutions ϕ of the Hamburger moment problem are substantially equal, we say the moment problem is determined.

Carleman gave a sufficient (but not necessary) condition for a moment problem to be determined.

Lemma 2.7. ([13, Thm. 1.11] or [1, p. 85].) The moment problem μ_{n} = ∫_{−∞}^{∞} x^{n}dψ(x) is determined if

∑_{n=1}^{∞} μ_{2n}^{−1/2n} = ∞.

The most crucial part of the proof of Theorem 1 will involve showing that the Hamburger moment problem for the distribution function F, defined in equation (4), is determined. We will now proceed with the proof.

3. Proof of Theorem 1

We begin by showing that the expression

(7)  ⟨p(x), q(x)⟩ = ∫_{−∞}^{∞} p(x)q(x)dF (x)

defines an inner product on the space of polynomials with real coefficients.

Lemma 3.1. The nth moments

μ_{n} = ∫_{−∞}^{∞} x^{n}dF

exist, and equation (7) defines an inner product on the space of polynomials with real coefficients.

Proof. Label the zeros of ξ(1/2 + iz) in the region {x + iy | x > 0, 0 ≤ y < 1/2} as α_{k} + iβ_{k} where α_{k} ≤ α_{k+1} for k ≥ 1. If α_{k} + iβ_{k} is a root, so is α_{k} − iβ_{k}. Also recall from (3) that |f(x + iy)| < exp(−cx) when x > 0 and −1/2 < y < 1/2. Then

∫_{0}^{∞} x^{n}dF = ∑_{\begin{matrix}k \\ β_{k}=0 \end{matrix}} α_{k}^{n}f(α_{k}) + ∑_{\begin{matrix}k \\ β_{k}≠0 \end{matrix}} α_{k}^{n}(f(α_{k} + iβ_{k}) + f(α_{k} − iβ_{k}))

≤ ∑_{k=1}^{∞} α_{k}^{n}e^{−cα_{k}} + 2 ∑_{k=1}^{∞} α_{k}^{n}e^{−cα_{k}} = 3 ∑_{k=1}^{∞} α_{k}^{n}e^{−cα_{k}}. (8)

We need to know the approximate size of α_{k}. Recall that the number N(T) of zeros of ξ(z) in the critical strip up to height T is known [15, p. 214] to satisfy

N(T) ∼ \frac{T}{2π} log (\frac{T}{2π} ) .

It follows that if the zeros in the critical strip with Im(z) > 0 are labelled as ρ_{k} +it_{k} with t_{k+1} ≥ t_{k}, then

(9)  t_{k} ∼ \frac{2πk}{log k} .

By equation (9) there exist positive constants A and B such that

(10)  A\frac{k}{log k} < α_{k−1} < B \frac{k}{log k}

for k ≥ 2. Combining inequalities (8) and (10) gives

∫_{0}^{∞} x^{n}dF ≤ 3B^{n} ∑_{k=2}^{∞} (\frac{k}{log k} )^{n} exp (−cA\frac{k}{log k} ) .

The sum clearly converges. This can be seen, for example, by using the limit comparison test from elementary calculus with the convergent series ∑ k^{−2}. Because F (T) = −F (−T), μ_{n} = 0 for odd n. When n is even

(11) μ_{n} = ∫_{−∞}^{∞} x^{n}dF = 2 ∫_{0}^{∞} x^{n}dF ≤ 6B^{n} ∑_{k=2}^{∞} (\frac{k}{log k} )^{n} exp (−cA\frac{k}{log k} ) .

This shows that the moments μ_{n} = ∫_{−∞}^{∞} x^{n}dF exist for n ≥ 0. Thus the expression ⟨p(x), q(x)⟩, defined by equation (7), exists for any real polynomials p(x) and q(x). The bilinearity is apparent. Because the measure dF has infinite support ⟨p(x), p(x)⟩ > 0 unless p(x) = 0. Therefore, the expression ⟨p(x), q(x)⟩ defines an inner product on the space of polynomials with real coefficients. □

Lemma 3.2. The Hamburger moment problem for the moments of the distribution function F,

μ_{n} = ∫_{−∞}^{∞} x^{n}dF,

is determined.

Proof. By extending the proof of the previous lemma we will obtain a sufficiently good upper bound on μ_{n} to apply Carleman’s criterion (Lemma 2.7). We begin by estimating the summation in inequality (11). Let

S(n) = ∑_{k=2}^{∞} (\frac{k}{log k} )^{n} exp (−cA\frac{k}{log k} ) .

Split the sum into two parts:

S(n) = \underbrace{∑_{2≤k≤M+1} (\frac{k}{log k} )^{n} exp (−cA\frac{k}{log k} )}_{S_{1}(n)} + \underbrace{∑_{k>M+1} (\frac{k}{log k} )^{n} exp (−cA\frac{k}{log k} )}_{S_{2}(n)} .

We will determine bounds for S_{1}(n) and S_{2}(n). A careful choice of M will lead to a bound on S_{2}(n) that is much smaller than the bound on S_{1}(n).

By elementary calculus the function (\frac{k}{log k})^{n} exp(−cA\frac{k}{log k}) has a maximum of (\frac{n}{ecA})^{n} when \frac{k}{log k} = \frac{n}{cA}. This gives a bound on S_{1}(n):

(12)  S_{1}(n) ≤ M (\frac{n}{ecA})^{n} .

Now assume that M is sufficiently large such that the following three conditions hold:

\frac{k}{log k} > \frac{n}{cA} for k ≥ M, (13)

(14) (\frac{k}{log k} ) (\frac{log k − 1}{log^{2} k}) > 1 for k ≥ M,

\frac{M}{log M} > (\frac{2(n + 1)}{cA})^{2} . (15)

Condition (13) ensures that the function (\frac{k}{log k})^{n} exp(−cA\frac{k}{log k}) decreases for k ≥ M. The reasons for assuming conditions (14) and (15) will become apparent in the following calculation:

S_{2}(n) = ∑_{k>M+1} (\frac{k}{log k} )^{n} exp (−cA\frac{k}{log k} )

< ∫_{M}^{∞} (\frac{k}{log k} )^{n} exp (−cA\frac{k}{log k} ) dk  by (13),

< ∫_{M}^{∞} (\frac{k}{log k} )^{n+1} exp (−cA\frac{k}{log k} ) (\frac{ln k − 1}{ln^{2} k}) dk  by (14),
= ∫_{M/ log M}^{∞} w^{n+1} exp(−cAw) dw.

For any positive α and w, w < \frac{exp(αw)}{α}. Setting α = \frac{cA}{2(n+1)} gives

S_{2}(n) < (\frac{2(n + 1)}{cA})^{n+1} ∫_{M/ log M}^{∞} exp (−\frac{cAw}{2} ) dw = \frac{2}{cA} (\frac{\frac{2(n+1)}{cA}}{exp(\frac{cA}{2(n+1)} \frac{M}{log M})})^{n+1} .

By condition (15), \frac{2(n+1)}{cA} < exp(\frac{cA}{2(n+1)} \frac{M}{log M}). Thus

(16)  S_{2}(n) < \frac{2}{cA}.

Combining inequalities (12) and (16) gives

S(n) = S_{1}(n) + S_{2}(n) < M (\frac{n}{ecA})^{n} + \frac{2}{cA}.

Let M = κ^{n} where κ > 1. As soon as n is sufficiently large conditions (13), (14), and (15) hold. So, for sufficiently large even n,

μ_{n}^{1/n} ≤ (6B^{n}S(n))^{1/n} < (6B^{n} ((\frac{κn}{ecA})^{n} + \frac{2}{cA}))^{1/n}

< (12B^{n} (\frac{κn}{ecA})^{n})^{1/n} = 12^{1/n} (\frac{Bκ}{ecA}) n

< (\frac{2Bκ}{ecA}) n.

Consequently

∑_{n=0}^{∞} μ_{2n}^{−1/2n} = ∞,

and by Carleman’s criterion (Lemma 2.7) it follows that the Hamburger moment problem μ_{n} = ∫_{−∞}^{∞} x^{n}dF (x) is determined. □

Let {p_{n}(x)} be the family of orthogonal polynomials obtained from the inner product in Lemma 3.1 by orthogonalizing the set of non-negative powers of x according the the Gram-Schmidt procedure. Because μ_{2k+1} = 0 and μ_{2k} > 0 for each k it follows that p_{2n+1}(x) is an odd function while p_{2n}(x) is an even function.

The spectrum (defined in §2) of F consists of all α_{k} such that α_{k} + iβ_{k} is a zero of ξ(1/2 + iz). We will label the positive values in 𝔖(F) as

a_{1} < a_{2} < a_{3} < · · · .

It was known, as early as Riemann [5, p. 159], that a_{1} ≈ 14.134. Denote the n positive zeros of p_{2n}(x) as

x_{2n,1} < x_{2n,2} < ⋯ < x_{2n,n}.

Similarly, denote the n positive zeros of p_{2n+1}(x) as

x_{2n+1,1} < x_{2n+1,2} < ⋯ < x_{2n+1,n}.

Thus, we may write

(17)  \frac{p_{2n}(z)}{p_{2n}(0)} = ∏_{k=1}^{n} (1 − \frac{z^{2}}{x_{2n,k}^{2}} ) and \frac{p_{2n+1}(z)}{z p′_{2n+1}(0)} = ∏_{k=1}^{n} (1 − \frac{z^{2}}{x_{2n+1,k}^{2}} ) .

In Lemmas 3.3 and 3.4 we will show that a_{k} = lim_{n→∞} x_{nk}.

Lemma 3.3. a_{k} < x_{mk} < x_{nk} when m > n. Hence, a_{k} ≤ lim_{n→∞} x_{nk}.

Proof. The spectral points of F are the numbers ±a_{k} for k = 1, 2, 3, . . .. By Lemma 2.5 if n is odd, the open interval (0, x_{n1}) contains a_{1} because a_{1} is the smallest positive spectral point. If n is even, the open interval (−x_{n1}, x_{n1}) contains a_{1}. In either case, a_{1} < x_{n1}. Similarly, each of the open intervals (x_{n,k−1}, x_{nk}) for 2 ≤ k ≤ ⌊n/2⌋ contains a spectral point. This forces a_{k} < x_{nk}. From the interlacing property of zeros in Lemma 2.2 it is immediate that

(18)  0 < x_{n+1,1} < x_{n,1} < ⋯ < x_{n+1,n} < x_{n,n} < x_{n+1,n+1}

whether n is even or odd. Hence, a_{k} < x_{mk} < x_{nk} for when m > n. □

Lemma 3.4. a_{k} = lim_{n→∞} x_{nk}.

Proof. By Lemma 2.4 there is a subsequence of the functions F_{n}, defined in equation (6), that converges to a distribution function η such that

μ_{n} = ∫_{−∞}^{∞} x^{n}dη(x), n = 0, 1, 2, …

In Lemma 3.2 it was established that the Hamburger moment problem

μ_{n} = ∫_{−∞}^{∞} x^{n}dF (x), n = 0, 1, 2, …

is determined. Therefore, F and η are substantially equal and they have the same spectrum. Let a_{k} be any one of the positive spectral points of F or η. By Lemma 2.6 every neighborhood of a_{k} contains a zero of p_{n}(x) for infinitely many n. Let δ_{1} > 0 be small enough so that the only spectral point of F in (a_{1} − δ_{1}, a_{1} + δ_{1}) is a_{1}. By Lemma 3.3 the only root of p_{n}(x) that potentially could be in that neighborhood is x_{n1}. Since infinitely many values of the bounded decreasing sequence {x_{n1}} lie in that neighborhood of a_{1}, lim_{n→∞} x_{n1} = a_{1}. Suppose, by way of induction, that lim_{n→∞} x_{nr} = a_{r} for 1 ≤ r < k. Choose δ_{k} > 0 small enough so that the only spectral point of F in (a_{k} −δ_{k}, a_{k} +δ_{k}) is a_{k}. Again by Lemma 3.3 the roots x_{nj} for j > k cannot be in the neighborhood of a_{k} since a_{k} < a_{k+1} < x_{nj}. By the induction hypothesis only finitely many roots x_{nj} with j < k can be in the neighborhood. Since the neighborhood contains infinitely many roots the only possibility is that x_{nk} is in the neighborhood for infinitely many n. Thus lim_{n→∞} x_{nk} = a_{k}. □

Lemma 3.5. The sequences of polynomials

\frac{p_{2n}(z)}{p_{2n}(0)}  and  \frac{p_{2n+1}(z)}{z p′_{2n+1}(0)}

converge uniformly on compact sets to the entire function with simple real zeros corresponding to the real parts of zeros of ξ(1/2 + iz). Thus, for all z ∈ ℂ,

lim_{n→∞} \frac{p_{2n}(z)}{p_{2n}(0)} = lim_{n→∞} \frac{p_{2n+1}(z)}{z p′_{2n+1}(0)} = ∏_{k=1}^{∞} (1 − \frac{z^{2}}{a_{k}^{2}} ) .

Proof. Let ϵ > 0 be given. Let K be any compact subset of ℂ. Choose R so that |z| < R for every z ∈ K. Define M_{R} to the be positive constant

M_{R} = ∏_{k=1}^{∞} (1 + \frac{R^{2}}{a_{k}^{2}} ) .

Because ξ(1/2 + iz) is an entire function of order one [15, Thm. 2.12], ∑_{k=1}^{∞} a_{k}^{−2} < ∞. Consequently, M_{R} is finite. For z ∈ K,

|\frac{p_{2n}(z)}{p_{2n}(0)}| = |∏_{k=1}^{n} (1 − \frac{z^{2}}{x_{2n,k}^{2}} )| ≤ ∏_{k=1}^{n} (1 + \frac{|z|^{2}}{x_{2n,k}^{2}} ) ≤ ∏_{k=1}^{n} (1 + \frac{R^{2}}{a_{k}^{2}} ) ≤ M_{R}.

Now choose N at least large enough so that a_{k} > R when k > N. For n > N, define α(z) and β(z) by

1 + α(z) = ∏_{k=N+1}^{n} (1 − \frac{z^{2}}{x_{2n,k}^{2}} )  and  1 + β(z) = ∏_{k=N+1}^{n} (1 − \frac{z^{2}}{a_{k}^{2}} ) .

Since 1 − R^{2}/a_{k}^{2} < |1 − z^{2}/x_{2n,k}^{2}| < 1 + R^{2}/a_{k}^{2} we obtain

∏_{k=N+1}^{∞} (1 − R^{2}/a_{k}^{2}) < |1 + α(z)| < ∏_{k=N+1}^{∞} (1 + R^{2}/a_{k}^{2}) .

Similarly,

∏_{k=N+1}^{∞} (1 − R^{2}/a_{k}^{2}) < |1 + β(z)| < ∏_{k=N+1}^{∞} (1 + R^{2}/a_{k}^{2}) .

Since lim_{N→∞} ∏_{k=N+1}^{∞} (1 − R^{2}/a_{k}^{2}) = 1 and lim_{N→∞} ∏_{k=N+1}^{∞} (1 − R^{2}/a_{k}^{2}) = 1 we may choose N large enough so that

|α(z)| < \frac{ϵ}{M_{R}}  and  |β(z)| < \frac{ϵ}{M_{R}} .

Choose N_{1} > N large enough so that, if n > N_{1} and z ∈ K,

|∏_{k=1}^{N} (1 − \frac{z^{2}}{x_{2n,k}^{2}} ) − ∏_{k=1}^{N} (1 − \frac{z^{2}}{a_{k}^{2}} )| < ϵ.

Let n > N_{1}. Then

|\frac{p_{2n}(z)}{p_{2n}(0)} − ∏_{k=1}^{∞} (1 − \frac{z^{2}}{a_{k}^{2}} )| = |∏_{k=1}^{n} (1 − \frac{z^{2}}{x_{2n,k}^{2}} ) − ∏_{k=1}^{∞} (1 − \frac{z^{2}}{a_{k}^{2}} )|

= |(1 + α(z)) ∏_{k=1}^{N} (1 − \frac{z^{2}}{x_{2n,k}^{2}} ) − (1 + β(z)) ∏_{k=1}^{N} (1 − \frac{z^{2}}{a_{k}^{2}} )|

≤ |∏_{k=1}^{N} (1 − \frac{z^{2}}{x_{2n,k}^{2}} ) − ∏_{k=1}^{N} (1 − \frac{z^{2}}{a_{k}^{2}} )|

+ |α(z)| |∏_{k=1}^{N} (1 − \frac{z^{2}}{x_{2n,k}^{2}} )| + |β(z)| |∏_{k=1}^{N} (1 − \frac{z^{2}}{a_{k}^{2}} )|

≤ ϵ + \frac{ϵ}{M_{R}} · M_{R} + \frac{ϵ}{M_{R}} · M_{R} = 3ϵ.

This shows that \frac{p_{2n}(z)}{p_{2n}(0)} converges to ∏_{k=1}^{∞}(1−z^{2}/a_{k}^{2}) uniformly on compact subsets of ℂ as n tends to infinity. The same argument, with 2n replaced by 2n + 1, shows that the sequence \frac{p_{2n+1}(z)}{z p′_{2n+1}(0)} converges uniformly on compact sets to the same entire function. □

By Lemma 3.5

lim_{n→∞} \frac{p_{2n}(z)}{p_{2n}(0)} = lim_{n→∞} \frac{p_{2n+1}(z)}{z p′_{2n+1}(0)} = ∏_{k=1}^{∞} (1 − \frac{z^{2}}{a_{k}^{2}} ) = \frac{ξ(1/2 + iz)}{ξ(1/2)}

if and only if ξ(1/2 + iz) has simple real zeros. This completes the proof of Theorem 1.

References

[1] N. I. Akhiezer, The Classical Moment Problem and Some Related Questions in Analyis, Hafner Publishing Co., New York, 1965.

[2] R. P. Brent, On the Zeros of the Riemann Zeta Function in the Critical Strip, Math. Comp 33 (1979) no. 148, 1361–1372.

[3] R. P. Brent, J. van de Lune, H. J. J. te Riele, D. T. Winter, On the Zeros of the Riemann Zeta Function in the Critical Strip II, Math. Comp. 39 (1982) no. 160, 681–688.

[4] T. S. Chihara, An Introduction to Orthogonal Polynomials, Gordon and Breach Science Publishers, New York-London-Paris, 1978.

[5] H. M. Edwards, Riemann’s Zeta Function, Academic Press, 1974. Reprinted by Dover Publications, Mineola, NY, 2001.

[6] G. H. Hardy, Sur les zéros de la fonction ζ(s) de Riemann, C. R. Acad. Sci. Paris 158 (1914) 1012–1014.

[7] G. H. Hardy and J. E. Littlewood, The zeros of Riemann’s zeta function on the critical line, Math. Z. 10 (1921) 283–317.

[8] J. van de Lune, H. J. J. te Riele, On the Zeros of the Riemann Zeta Function in the Critical Strip III, Math. Comp. 41 (1983), no. 164, 759–767.

[9] J. van de Lune, H. J. J. te Riele, D. T. Winter, On the Zeros of the Riemann Zeta Function in the Critical Strip IV, Math. Comp. 46 (1986), no. 174, 667–681.

[10] A. M. Odlyzko, The 10^{22}-nd zero of the Riemann zeta function. Dynamical, spectral, and arithmetic zeta functions (San Antonio, TX, 1999), 139–144, Contemp. Math., 290, Amer. Math. Soc., Providence, RI, 2001.

[11] Sebastian Wedeniwski, Results connected with the first 100 billion zeros of the Riemann zeta function (preprint).

[12] A. Selberg, On the zeros of Riemann’s zeta function, Skr. Norske Vid.-Akad. Oslo, No. 10 (1942).

[13] J. A. Shohat and J. D. Tamarkin, The Problem of Moments, Mathematical Surveys No. 1, Amer. Math. Soc., N.Y., 1943, 1950.

[14] Gabor Szegö, Orthogonal Polynomials, Fourth Edition, Amer. Math. Soc. Colloq. Publ., Vol. 23, Providence, R.I., 1975.

[15] E. C. Titchmarsh, The Theory of the Riemann Zeta Function, Second Edition, Revised by D. R. Heath-Brown, Claredon Press, Oxford, 1986.

Department of Mathematics, Brigham Young University, Provo, Utah 84602

E-mail address: cardon@math.byu.edu