ON THE MONTGOMERY–VAUGHAN WEIGHTED GENERALIZATION OF HILBERT’S INEQUALITY

. This paper concerns the problem of determining the optimal constant in the Montgomery–Vaughan weighted generalization of Hilbert’s inequality. We consider an approach pursued by previous authors via a parametric family of inequalities. We obtain upper and lower bounds for the constants in inequalities in this family. A lower bound indicates that the method in its current form cannot achieve any value below 3 . 19497, so cannot achieve the conjectured constant π . The problem of determining the optimal constant remains open.


Introduction
In this paper, we study a parametric family of inequalities, given in (1.8) below, that can yield an upper bound on the optimal constant in the Montgomery-Vaughan weighted generalization of Hilbert's inequality (1.3).This inequality is important in the theory of the large sieve; see [7] and [4].
1.1.History of the problem.Let N denote a positive integer, and let z 1 , . . ., z N denote complex numbers.Hilbert's inequality states that where C 0 is the absolute constant 2π.Hilbert's proof was published by Weyl [14, § 15].In 1911, Schur [12]  Schur's bound is included in (1.2) as the case λ k+1 − λ k = δ.In the same paper, Montgomery and Vaughan also established a weighted form: where λ k+1 > λ k for all k and δ k := min {λ k − λ k−1 , λ k+1 − λ k } and C 1 is the absolute constant 3π 2 .Denote by C 1 the minimum of all absolute constants C 1 for which (1.3) holds.Montgomery and Vaughan [8] have raised the By setting λ k = k in (1.3) and comparing with Schur's result, we see that (1.4) If C 1 = π, then (1.3) would contain (1.2), and it is widely believed to be the case.
In 1981, Graham and Vaaler [1] constructed extreme majorants and minorants of the functions where β is an arbitrary positive real number, and used them to prove that (1.6) The inequality (1.6) includes (1.2) as the limiting case β → 0 + .In 1999, Montgomery and Vaaler [6] established a generalization of (1.3): where β 1 , . . ., β N are nonnegative real numbers and C 2 is the absolute constant 84, which is not optimal.Their proof involves the theory of H 2 functions in a half-plane and a maximal theorem of Hardy and Littlewood.
In 2005, Li [3] posed a question about the finite Hilbert transformation associated with a polynomial and proved that if the question always has an affirmative answer then C 1 = π.

Main results.
We study the following parametric family of inequalities.For 0 ≤ α ≤ 2, let C(α) be the minimum of all constants C(α) for which the inequality holds for all choices of a positive integer N , a strictly increasing sequence (λ k ) The value C 1 2 is relevant to the generalized Hilbert inequality (1.3).In Section 3, we shall prove the following inequality between C 1 and C 1 2 .
Theorem 1.1.We have The previous approaches to get an upper bound for C 1 in [8], [10], and [13] rely on an upper bound for C 1 2 and Theorem 1.1.Montgomery and Vaughan [8] first showed that C 1 2 is finite.Specifically, they proved C 1 2 ≤ 17 2 .The same bound has been used in [6] to prove (1.7), but the best known upper bound for C 1  2 is due to Preissmann [10]. .
where (λ k ) ∞ k=−∞ is a strictly increasing sequence of real numbers and c 1 , . . ., c N are positive real numbers.Since H is skew-Hermitian (i.e., iH is Hermitian), all its eigenvalues are purely imaginary.Let [u 1 , . . ., u N ] be an eigenvector of H, and let iµ be its associated eigenvalue.That is, It is well known (see, e.g., [5, § 7.4]) that the numerical radius of a normal matrix is the same as its spectral radius (and its operator norm).Thus, if iµ has the largest modulus among all eigenvalues of H, then for all complex numbers z 1 , . . ., z N .On replacing z n by zn cn , we see that (2.2) is equivalent to One may obtain the generalized Hilbert inequality (1.3) with some constant C 1 from (2.3) by giving an upper bound for the sizes of eigenvalues of H in the case that c A key result to that end is: be an eigenvector of H, and let iµ be its associated eigenvalue.Then the identity holds for all m = 1, . . ., N .

2.2.
A weighted spacing lemma and Shan's method.The goal of this subsection is to prove: ∞ k=−∞ be a strictly increasing sequence of real numbers.Denote by δ k the minimum between λ k − λ k−1 and λ k+1 − λ k .Then for real numbers σ > 1 and integers , we have One can show that equality holds in (2.5) if and only if the sequence ( is constant, but we shall not treat it here.Lemma 2.2 is a direct consequence of Lemme 1 of Preissmann [10].We present a proof using a method of Shan [13], who independently derived Lemma 2.2.The work of Shan, done at the same time as that of Preissmann, is obscure and hard to obtain.Peng Gao (private communication) translated Shan's argument, which appears in [9, pp. 590-595].The next three lemmas are an exposition of Shan's method.
Let f be a real-valued function, defined on the interval [1, ∞).We will assume that f satisfies some (or all) of the following four conditions: (a) be a sequence of real numbers such that a n ≥ 1 for all n.Set λ n := n m=1 a m .Then for positive integers N , we have where {x} = x − x denotes the fractional part of x.
Proof.By the convexity of f , we have Moreover, since a n ≥ 1 and f is weakly decreasing, it follows that On summing (2.6) and (2.7), we obtain Now, we consider the first term on the right side of (2.8) and note that λ n = On inserting this in (2.8), we get The result follows by summing (2.9) over n = 1, . . ., N ; the resulting sum on the right side is a telescoping sum.
In what follows, we consider ) Then for positive integers N , we have (2.11) By the convexity of f , it follows that a n for all n ≥ 2. So (2.12) implies that We now prove an upper bound for F N (a) that depends only on f .
∞ n=1 be a sequence of positive real numbers with a 1 ≥ 1.Then for positive integers N , we have (2.13) By taking a n = 1 for all n and letting N → ∞, we see that (2.13) is sharp.
Proof.Define a sequence a = (a n ) ∞ n=1 by a n := max {a m : m = 1, . . ., n}.Then a n+1 ≥ a n for all n and a 1 = a 1 ≥ 1.Let N be a positive integer.By applying Lemma 2.4, with ε = a ν−1 − a ν , as many times as we need, we see that where λ n := n m=1 a m .By Lemma 2.3 and the nonnegativity of f , the right side of (2.14) is (2.15) The result (2.13) follows by combining (2.14) and (2.15).
We are now ready to prove Lemma 2.2.
Proof of Lemma 2.
holds for all nonnegative real numbers t 1 , . . ., t N .Then the inequality (1.3) holds for all complex numbers z 1 , . . ., z N with the constant Proof.Suppose that (3.1) holds.Let [u 1 , . . ., u N ] be a unit eigenvector of H = [h mn ], where h mn are given by (2.1) with c n = √ δ n , and let iµ be the eigenvalue associated with this eigenvector.On applying Lemma 2.1 and summing (2.4) over m, we get where S and T are given by On one hand, by Lemma 2.2, we obtain Proof.See Preissmann [10,Lemme 6].

Proof of Theorem 1.5
Let M denote a positive integer, and let x 1 , . . ., x M denote real numbers, distinct modulo 1.Put where x = min k∈Z |x − k| denotes the distance between x and a nearest integer.
Proof.(⇒) Suppose that (3.1) holds.Let x 1 , . . ., x M be real numbers, distinct modulo 1.By symmetry in x 1 , . . ., x M , we may assume without loss of generality that x 1 < • • • < x M < x 1 + 1.Let d m be given by (5.2).Let τ 1 , . . ., τ M be nonnegative real numbers.Let K be a positive integer.We apply (3. converges, it follows that they are (C, 1) summable to the same values, which is to say that lim .
Lemma 5.2.For positive real numbers B < 1 and positive integers L, we have Hence, for 0 < x ≤ B, we have 1).Applying this estimate to each term on the left side of (5.4), we obtain , the result (5.4) follows.
Proof of Theorem 1.5.To prove a lower bound for C 1 2 , we apply (5.1) with particular sets of values.Let K be a positive integer.Let A and B be positive real numbers such that (K + 1)A + B = 1.Let L ≥ B A be an integer.We apply (5.
That is, where κ 0 and κ 1 depend on A, B, and K and are given by and We find that g(u) is maximized on u ≥ 0 at u = u 0 := 1

2 . Preliminaries 2 . 1 .
Eigenvalues of generalized weighted Hilbert matrices.Let us consider N × N matrices H = [h mn ] with entries given by

2 .
Let be an integer.Define sequences a = (a n ) ∞ n=1 and b = (b n ) ∞ n=1 by a n := λ +n − λ +n−1 δ and b n := λ −n+1 − λ −n δ , for all n.Then a and b are sequences of positive real numbers with

3 . 2 3. 1 .
Proofs of Theorems 1.1 and 1.Proof of Theorem 1.1.Proposition 3.1.Let N be a positive integer.Let (λ k ) ∞ k=−∞ be a strictly increasing sequence of real numbers.Denote by δ k the minimum between λ k −λ k−1 and λ k+1 − λ k .Assume that C 3 is a positive constant such that the inequality