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Construction of a class of half-discrete Hilbert-type inequalities in the whole plane

  • Minghui You EMAIL logo
Published/Copyright: August 27, 2024
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Open Mathematics
From the journal Open Mathematics Volume 22 Issue 1

Abstract

In this work, we first define two special sets of real numbers, and then, we construct a half-discrete kernel function where the variables are defined in the whole plane, and the parameters in the kernel function are limited to the newly constructed special sets. Estimate the kernel function in the whole plane by converting it to the first quadrant, and then, a class of new Hilbert-type inequality is established. Additionally, it is proved that the constant factor of the newly established inequality is the best possible. Furthermore, assigning special values to the parameters and using rational fraction expansion of cosecant function, some special results are presented at the end of this article.

Keywords: Hilbert-type inequality; whole plane; half-discrete; homogeneous kernel; non-homogeneous kernel
MSC 2010: 26D15; 41A17

1 Introduction

In this work, it is assumed that p > 1 , 1 p + 1 q = 1 , Z 0 ≔ Z \ { 0 } ,

S 1 ≔ x : x = 2 j 2 m + 1 , j , m ∈ Z , S 2 ≔ x : x = 2 j + 1 2 m + 1 , j , m ∈ Z .

Let Π be a measurable set and f ( x ) , μ ( x ) be two non-negative measurable functions defined on Π . Define a function space L p , μ ( Π ) as follows:

L p , μ ( Π ) ≔ f : ‖ f ‖ p , μ ≔ ∫ Π f p ( x ) μ ( x ) d x 1 ⁄ p < ∞ .

Particularly, we have the abbreviations: ‖ f ‖ p ≔ ‖ f ‖ p , μ and L p ( Π ) ≔ L p , μ ( Π ) if μ ( x ) ≡ 1 .

Let a n , ν n > 0 , n ∈ Θ ⊆ Z , a = { a n } n ∈ Θ . Define a sequence space l p , ν as follows:

l p , ν ≔ a : ‖ a ‖ p , ν ≔ ∑ n ∈ Θ a n p ν n 1 ⁄ p < ∞ .

Particularly, we have the abbreviations: ‖ a ‖ p ≔ ‖ a ‖ p , ν and l p ≔ l p , ν if ν n ≡ 1 .

Let a = { a n } n = 1 ∞ ∈ l 2 and b = { b n } n = 1 ∞ ∈ l 2 be two real number sequences. Then [1],

(1.1) ∑ n ∈ N + ∑ m ∈ N + a m b n m + n < π ‖ a ‖ 2 ‖ b ‖ 2 ,

where the constant factor π is the best possible.

Inequality (1.1) was first proposed by German mathematician Hilbert in his lecture on integral equations, and it is normally referred to as Hilbert inequality. Inequalities that are structurally similar to Hilbert inequality are generally referred to as Hilbert-type inequality, such as the following two [1]:

(1.2) ∑ n ∈ N + ∑ m ∈ N + a m b n max { m , n } < 4 ‖ a ‖ 2 ‖ b ‖ 2 ,

(1.3) ∑ n ∈ N + ∑ m ∈ N + log m n m − n a m b n < π 2 ‖ a ‖ 2 ‖ b ‖ 2 ,

where the constant factors 4 and π 2 are the best possible.

In 1925, by introducing a pair of conjugate parameters ( p , q ) , Hardy [2] proved an extended form of (1.1), i.e.,

(1.4) ∑ n ∈ N + ∑ m ∈ N + a m b n m + n < π sin π p ‖ a ‖ p ‖ b ‖ q ,

where the constant factor π sin π p is also the best possible. With regard to the extensions of (1.2) and (1.3), we refer to [1].

In 1991, Hsu and Guo [3] put forward the weight coefficient method, and proved the following strengthened form of (1.1):

(1.5) ∑ n ∈ N + ∑ m ∈ N + a m b n m + n < ‖ a ‖ 2 , μ ‖ b ‖ 2 , ν ,

where μ m = π − γ 0 m and ν n = π − γ 0 n ( γ 0 = 1.121 3 + ) .

After 1998, researchers optimized the weight coefficient method and established various extended forms of (1.4), such as [4]

(1.6) ∑ n ∈ N + ∑ m ∈ N + a m b n ( m + n ) β < B ( β 1 , β 2 ) ‖ a ‖ p , μ ‖ b ‖ q , ν ,

where 0 < β 1 , β 2 ≤ 2 , β 1 + β 2 = β , μ m = m p ( 1 − β 1 ) − 1 , ν n = n q ( 1 − β 2 ) − 1 , and B ( u , v ) is the beta function [5]. For some other extensions of (1.4), we refer to [6–12], and for extensions, analogues, strengthened forms, and reverses of some classical Hilbert-type inequalities like (1.2) and (1.3), we refer to [7,10,13–17].

In addition to the inequalities in discrete form mentioned earlier, Hilbert-type inequalities also appear in integral and half-discrete forms, such as the following two inequalities, which are the integral and half-discrete forms corresponding to inequality (1.4), i.e.,

(1.7) ∫ x ∈ R + ∫ y ∈ R + f ( x ) g ( y ) x + y d x d y < π sin π p ‖ f ‖ p ‖ g ‖ q ,

(1.8) ∫ x ∈ R + f ( x ) ∑ n ∈ N + a n x + n d x < π sin π p ‖ f ‖ p ‖ a ‖ q ,

where f ∈ L p ( R + ) , g ∈ L q ( R + ) , a ∈ l q , and the constant factor π sin π p in both (1.7) and (1.8) is the best possible.

Generally, for a discrete Hilbert-type inequality, there is necessarily a corresponding integral and half-discrete inequality with the same constant factor. On the contrary, for an integral Hilbert-type inequality, we may not be able to establish its corresponding discrete form, such as the following one in which the kernel is non-homogeneous [8]:

(1.9) ∫ x ∈ R + ∫ y ∈ R + f ( x ) g ( y ) 1 + x y d x d y < π ‖ f ‖ 2 ‖ g ‖ 2 ,

where the constant factor π is the best possible. The establishment of discrete form of (1.9) is difficult because it cannot explain the optimality of the constant factor π (see [8], p. 328). However, we can establish a half-discrete form of (1.9) and prove that the constant factor is the best possible [18].

It should be pointed out that, by the introduction of new kernel functions with the homogeneous and the non-homogeneous forms, a great many Hilbert-type inequalities were proved in the past 30 years [19–24]. In the process, a large number of half-discrete Hilbert-type inequalities were also established constantly [25–28]. Such a large number of inequalities have already grown into a vast system and are crucial to the development of modern analysis.

Generally, discrete and half-discrete Hilbert-type inequalities are established in the first quadrant. It is not an easy task to extend a discrete or a half-discrete Hilbert-type inequality to the whole plane, which is because the non-negativity, monotonicity, and integrability of a kernel function will be very complicated if we extend the range of variables to R 2 . Therefore, we can only find few sporadic results appearing in the literature [29]. In this work, the main objective is to provide a new half-discrete Hilbert-type inequality defined in the whole plane involving some classical kernel functions. When dealing with the weight function, we transform it to a monotonic function in the first quadrant to establish its upper bound. Furthermore, using the rational fraction expansion of cosecant function, some special Hilbert-type inequalities are presented at the end of the article.

2 Some lemmas

Lemma 2.1

Let α ∈ ( 0 , 1 ) and γ ∈ R + ∪ { 0 } . Suppose that α , β , λ , and τ satisfy one of the following conditions:

  1. τ = 1 , β ∈ S 1 \ R − , λ ∈ S 1 ∩ R + , and 0 ≤ β < 1 − α ;

  2. τ = ± 1 , β , λ ∈ S 2 ∩ R + , and 0 < β < 1 − α − λ .

Let

(2.1) k ( u ) ≔ ( τ u β + 1 ) ln ∣ u ∣ ( τ u λ − 1 ) max { 1 , ∣ u ∣ γ } ,

where u ∈ R \ { 0 , 1 , − 1 } . Let

(2.2) Ψ ( u ) ≔ [ k ( u ) + k ( − u ) ] u α − 1 ,

where u ∈ R + \ { 1 } . Then, Ψ ( u ) decreases with u.

Proof

First, consider the case where τ = 1 , β ∈ S 1 \ R − , and λ ∈ S 1 ∩ R + . Then, it can be easy to show that k ( u ) is an even function, and therefore, we have

Ψ ( u ) = 2 u α + β − 1 + 2 u α − 1 ( u λ − 1 ) max { 1 , u γ } ln u ( u ∈ R + ) .

Let

h ( u ) ≔ u α + β − 1 + u α − 1 u λ − 1 ln u = [ u α + β − 1 h 1 ( u ) + u α − 1 h 1 ( u ) ] h 2 ( u ) ,

where h 1 ( u ) = u θ − 1 u λ − 1 , h 2 ( u ) = ln u u θ − 1 , u ∈ R + \ { 1 } , and θ is an arbitrary positive number satisfying 0 < θ < λ . Finding the derivative of h 1 ( u ) , we have

d h 1 d u = − u θ − 1 ( u λ − 1 ) 2 [ θ + ( λ − θ ) u λ − λ u λ − θ ] .

Let h * ( u ) = θ + ( λ − θ ) u λ − λ u λ − θ . Then, we have

d h * d u = λ ( λ − θ ) u λ − θ − 1 ( u θ − 1 ) .

It follows from 0 < θ < λ that d h * d u < 0 ( u ∈ ( 0 , 1 ) ) , and d h * d u > 0 ( u ∈ ( 1 , ∞ ) ) . Therefore, we have h * ( u ) ≥ h * ( 1 ) = 0 , which leads to that d h 1 d u < 0 ( u ≠ 1 ) . Set h 1 ( 1 ) ≔ θ λ , then h 1 ( u ) is continuous on R + , and decreases with u ( u ∈ R + ) . Additionally, it can be proved that [8]  d h 2 d u < 0 ( u ∈ R + ) , and then, h 2 ( u ) decreases with u ( u ∈ R + ) . Following the aforementioned discussion and observing that α − 1 < 0 and α + β − 1 < 0 , we obtain that h ( u ) decreases with u ( u ∈ R + \ { 1 } ) , and then, Ψ ( u ) decreases with u obviously.

Second, consider the case where τ = ± 1 , and β , λ ∈ S 2 ∩ R + . Let

H ( u ) ≔ τ u β + 1 τ u λ − 1 ( ln ∣ u ∣ ) ∣ u ∣ α − 1 ,

where u ∈ R \ { 0 , 1 , − 1 } . Let

g ( u ) ≔ H ( u ) + H ( − u ) ,

where u ∈ R + \ { 0 , 1 } . Then, it follows from β , λ ∈ S 2 that

g ( u ) = τ u β + 1 τ u λ − 1 + τ u β − 1 τ u λ + 1 ( ln u ) u α − 1 = 2 u α + β + λ − 1 + 2 u α − 1 u 2 λ − 1 ln u .

Similar to the discussion of monotonicity of h ( u ) , it can be proved that g ( u ) decreases with u ( u ∈ R + ) . Therefore,

Ψ ( u ) = [ k ( u ) + k ( − u ) ] ∣ u ∣ α − 1 = H ( u ) + H ( − u ) max { 1 , ∣ u ∣ γ } = g ( u ) max { 1 , u γ }

decreases with u . Lemma 2.1 is proved.□

Lemma 2.2

Let α ∈ ( 0 , 1 ) and γ ∈ R + ∪ { 0 } . Suppose that α , β , λ , and τ satisfy one of the following conditions:

  1. τ = 1 , β ∈ S 1 \ R − , λ ∈ S 1 ∩ R + , and 0 ≤ β < λ + γ − α ;

  2. τ = ± 1 , β , λ ∈ S 2 ∩ R + , and 0 < β < λ + γ − α .

Let k ( u ) be defined by (2.1), and

(2.3) W ( α , β , γ , λ ) = 2 ∑ j = 0 ∞ 1 ( λ j + α ) 2 + 1 ( λ j + λ + γ − α ) 2 + 2 ∑ j = 0 ∞ 1 ( λ j + α + β ) 2 + 1 ( λ j + λ + γ − α − β ) 2 , β , λ ∈ S 1 , 2 ∑ j = 0 ∞ 1 ( 2 λ j + α ) 2 + 1 ( 2 λ j + 2 λ + γ − α ) 2 + 2 ∑ j = 0 ∞ 1 ( 2 λ j + α + β + λ ) 2 + 1 ( 2 λ j + λ + γ − α − β ) 2 , β , λ ∈ S 2 .

Then,

(2.4) ∫ u ∈ R + [ k ( u ) + k ( − u ) ] u α − 1 d u = W ( α , β , γ , λ ) .

Proof

We first prove (2.4) in the case where τ = 1 , β ∈ S 1 \ R − , and λ ∈ S 1 ∩ R + . Observing that k ( u ) is an even function, we obtain

(2.5) ∫ u ∈ R + [ k ( u ) + k ( − u ) ] u α − 1 d u = 2 ∫ 0 1 u α − 1 + u α + β − 1 u λ − 1 ln u d u + 2 ∫ 1 ∞ u α − γ − 1 + u α + β − γ − 1 u λ − 1 ln u d u = 2 ∫ 0 1 u α − 1 + u α + β − 1 u λ − 1 ln u d u + 2 ∫ 0 1 u λ + γ − α − 1 + u λ + γ − α − β − 1 u λ − 1 ln u d u = 2 ∫ 0 1 u α − 1 ln u + u λ + γ − α − 1 ln u + u λ + γ − α − β − 1 ln u + u α + β − 1 ln u u λ − 1 d u ≔ 2 ( J 1 + J 2 + J 3 + J 4 ) ,

where

J 1 = ∫ 0 1 u α − 1 ln u u λ − 1 d u , J 2 = ∫ 0 1 u λ + γ − α − 1 ln u u λ − 1 d u , J 3 = ∫ 0 1 u λ + γ − α − β − 1 ln u u λ − 1 d u , and J 4 = ∫ 0 1 u α + β − 1 ln u u λ − 1 d u .

Expand 1 u λ − 1 ( u ∈ ( 0 , 1 ) ) into the Maclaurin series, and use the Lebesgue term-by-term integration theorem; then, we have

(2.6) J 1 = − ∫ 0 1 ∑ j = 0 ∞ u λ j + α − 1 ln u d u = − ∑ j = 0 ∞ ∫ 0 1 u λ j + α − 1 ln u d u .

Setting ln u = − z λ j + α , we have

(2.7) ∫ 0 1 u λ j + α − 1 ln u d u = − 1 ( λ j + α ) 2 ∫ 0 ∞ z e − z d z = − 1 ( λ j + α ) 2 .

Inserting (2.7) back into (2.6), we obtain

(2.8) J 1 = ∑ j = 0 ∞ 1 ( λ j + α ) 2 .

Similarly, it can be proved that

(2.9) J 2 = ∑ j = 0 ∞ 1 ( λ j + λ + γ − α ) 2 .

(2.10) J 3 = ∑ j = 0 ∞ 1 ( λ j + λ + γ − α − β ) 2 .

(2.11) J 4 = ∑ j = 0 ∞ 1 ( λ j + α + β ) 2 .

Inserting (2.8), (2.9), (2.10), and (2.11) back into (2.5), we arrive at (2.4) under the condition where τ = 1 , β ∈ S 1 \ R − , and λ ∈ S 1 ∩ R + .

Second, consider the case where τ = ± 1 and β , λ ∈ S 2 ∩ R + . Then,

(2.12) ∫ u ∈ R + [ k ( u ) + k ( − u ) ] u α − 1 d u = ∫ [ − 1 , 1 ] τ u β + 1 τ u λ − 1 ( ln ∣ u ∣ ) ∣ u ∣ α − 1 d u + ∫ R \ [ − , 1 ] τ u β + 1 τ u λ − 1 ( ln ∣ u ∣ ) ∣ u ∣ α − γ − 1 d u = ∫ 0 1 τ u β + 1 τ u λ − 1 + τ u β − 1 τ u λ + 1 ( ln u ) u α − 1 d u + ∫ 1 ∞ τ u β + 1 τ u λ − 1 + τ u β − 1 τ u λ + 1 ( ln u ) u α − γ − 1 d u = 2 ∫ 0 1 u α + β + λ − 1 + u α − 1 u 2 λ − 1 ln u d u + 2 ∫ 1 ∞ u α + β + λ − γ − 1 + u α − γ − 1 u 2 λ − 1 ln u d u = 2 ∫ 0 1 u α − 1 ln u + u 2 λ + γ − α − 1 ln u + u λ + γ − α − β − 1 ln u + u α + β + λ − 1 ln u u 2 λ − 1 d u ≔ 2 ( L 1 + L 2 + L 3 + L 4 ) ,

where

L 1 = ∫ 0 1 u α − 1 ln u u 2 λ − 1 d u , L 2 = ∫ 0 1 u 2 λ + γ − α − 1 ln u u 2 λ − 1 d u , L 3 = ∫ 0 1 u λ + γ − α − β − 1 ln u u 2 λ − 1 d u , and L 4 = ∫ 0 1 u α + β + λ − 1 ln u u 2 λ − 1 d u .

Expand 1 u 2 λ − 1 ( u ∈ ( 0 , 1 ) ) into the Maclaurin series, and use the Lebesgue term-by-term integration theorem, and then, we have

(2.13) L 1 = ∑ j = 0 ∞ 1 ( 2 λ j + α ) 2 .

(2.14) L 2 = ∑ j = 0 ∞ 1 ( 2 λ j + 2 λ + γ − α ) 2 .

(2.15) L 3 = ∑ j = 0 ∞ 1 ( 2 λ j + λ + γ − α − β ) 2 .

(2.16) L 4 = ∑ j = 0 ∞ 1 ( 2 λ j + α + β + λ ) 2 .

Inserting (2.13), (2.14), (2.15), and (2.16) back into (2.12), we arrive at (2.4) under the condition where τ = ± 1 and β , λ ∈ S 2 ∩ R + . Lemma 2.2 is proved.□

Lemma 2.3

Let α ∈ ( 0 , 1 ) and γ ∈ R + ∪ { 0 } . Let δ ∈ S 2 and ρ ∈ S 2 ∩ ( 0 , 1 ] . Suppose that α , β , λ , and τ satisfy one of the following conditions:

  1. τ = 1 , β ∈ S 1 \ R − , λ ∈ S 1 ∩ R + and 0 ≤ β < min { 1 − α , λ + γ − α } ;

  2. τ = ± 1 , β , λ ∈ S 2 ∩ R + and 0 < β < min { 1 − α − λ , λ + γ − α } .

Let k ( u ) be defined by (2.1), and

a ˆ ≔ { a ˆ n } n ∈ Z 0 ≔ ∣ n ∣ α ρ − 1 − 2 ρ q s n ∈ Z 0 , f ˆ ( x ) ≔ ∣ x ∣ α δ − 1 + 2 δ p s , x ∈ Ω , 0 , x ∈ R \ Ω ,

where s is a natural number that is large enough, and Ω ≔ t : ∣ t ∣ δ ∣ δ ∣ < 1 . Then,

(2.17) I ˆ ≔ ∑ n ∈ Z 0 a ˆ n ∫ x ∈ R k ( x δ n ρ ) f ˆ ( x ) d x = ∫ x ∈ R f ˆ ( x ) ∑ n ∈ Z 0 k ( x δ n ρ ) a ˆ n d x > s ∣ δ ρ ∣ ∫ 0 1 [ k ( u ) + k ( − u ) ] u α − 1 + 2 p s d u + ∫ 1 ∞ [ k ( u ) + k ( − u ) ] u α − 1 − 2 q s d u .

Proof

Let Ω + ≔ Ω ∩ R + , and Ω − ≔ Ω ∩ R − . Since δ , ρ ∈ S 2 , we have

I ˆ = ∫ x ∈ Ω − f ˆ ( x ) ∑ n ∈ Z + k ( x δ n ρ ) a ˆ n d x + ∫ x ∈ Ω − f ˆ ( x ) ∑ n ∈ Z − k ( x δ n ρ ) a ˆ n d x + ∫ x ∈ Ω + f ˆ ( x ) ∑ n ∈ Z + k ( x δ n ρ ) a ˆ n d x + ∫ x ∈ Ω + f ˆ ( x ) ∑ n ∈ Z − k ( x δ n ρ ) a ˆ n d x = 2 ∫ x ∈ Ω + ∣ x ∣ α δ − 1 + 2 δ p s ∑ n ∈ Z + k ( − x δ n ρ ) ∣ n ∣ α ρ − 1 − 2 ρ q s d x + 2 ∫ x ∈ Ω + ∣ x ∣ α δ − 1 + 2 δ p s ∑ n ∈ Z + k ( x δ n ρ ) ∣ n ∣ α ρ − 1 − 2 ρ q s d x = 2 ∫ x ∈ Ω + x α δ − 1 + 2 δ p s ∑ n ∈ Z + [ k ( x δ n ρ ) + k ( − x δ n ρ ) ] n α ρ − 1 − 2 ρ q s d x .

Since x ∈ Ω + , n ∈ Z + , δ ∈ S 2 , and ρ ∈ S 2 ∩ ( 0 , 1 ] , it follows from Lemma 2.1 that

[ k ( x δ n ρ ) + k ( − x δ n ρ ) ] n α ρ − 1 − 2 ρ q s = [ k ( x δ n ρ ) + k ( − x δ n ρ ) ] ( x δ n ρ ) α − 1 n ρ − 1 − 2 ρ q s x δ − δ α

decreases with n ( n ∈ Z + ) for a fixed x ( x ∈ Ω ) . Therefore,

(2.18) I ˆ > 2 ∫ x ∈ Ω + x α δ − 1 + 2 δ p s ∫ 1 ∞ [ k ( x δ y ρ ) + k ( − x δ y ρ ) ] y α ρ − 1 − 2 ρ q s d y d x .

Consider the case where δ ∈ S 2 ∩ R − , and then, Ω + = Ω ∩ R + = ( 1 , ∞ ) . Let x δ y ρ = u in (2.18), and then, we have

(2.19) I ˆ > 2 ρ ∫ 1 ∞ x − 1 + 2 δ s ∫ x δ ∞ [ k ( u ) + k ( − u ) ] u α − 1 − 2 q s d u d x = 2 ρ ∫ 1 ∞ x − 1 + 2 δ s ∫ 1 ∞ [ k ( u ) + k ( − u ) ] u α − 1 − 2 q s d u d x + 2 ρ ∫ 1 ∞ x − 1 + 2 δ s ∫ x δ 1 [ k ( u ) + k ( − u ) ] u α − 1 − 2 q s d u d x = s ∣ δ ρ ∣ ∫ 1 ∞ [ k ( u ) + k ( − u ) ] u α − 1 − 2 q s d u + 2 ρ ∫ 0 1 [ k ( u ) + k ( − u ) ] u α − 1 − 2 q s ∫ u 1 δ ∞ x − 1 + 2 δ s d x d u = s ∣ δ ρ ∣ ∫ 1 ∞ [ k ( u ) + k ( − u ) ] u α − 1 − 2 q s d u + s ∣ δ ρ ∣ ∫ 0 1 [ k ( u ) + k ( − u ) ] u α − 1 + 2 p s d u .

It follows from (2.19) that (2.17) holds true. Lemma 2.3 is proved.□

Lemma 2.4

Let a , b > 0 , a + b = z , and Φ ( u ) = csc 2 u . Then,

(2.20) ∑ s = 0 ∞ 1 ( z s + a ) 2 + 1 ( z s + b ) 2 = π 2 z 2 Φ a π z .

Proof

Expand cot u ( 0 < u < π ) into the following partial fraction [5]:

(2.21) cot u = 1 u + ∑ s = 1 ∞ 1 u + s π + 1 u − s π .

Finding the derivative of (2.21), we have

Φ ( u ) = 1 u 2 + ∑ s = 1 ∞ 1 ( u + s π ) 2 + 1 ( u − s π ) 2 .

Therefore,

Φ a π z = z 2 π 2 1 a 2 + ∑ s = 1 ∞ 1 ( z s + a ) 2 + 1 ( z s − a ) 2 = z 2 π 2 lim n → ∞ ∑ s = 0 n 1 ( z s + a ) 2 + ∑ s = 1 n 1 ( z s − a ) 2 = z 2 π 2 lim n → ∞ ∑ s = 0 n 1 ( z s + a ) 2 + ∑ s = 0 n − 1 1 ( z s + b ) 2 = z 2 π 2 lim n → ∞ ∑ s = 0 n 1 ( z s + a ) 2 + 1 ( z s + b ) 2 = z 2 π 2 ∑ s = 0 ∞ 1 ( z s + a ) 2 + 1 ( z s + b ) 2 .

Lemma 2.4 is proved.□

3 Main results

Theorem 3.1

Let α ∈ ( 0 , 1 ) and γ ∈ R + ∪ { 0 } . Suppose that α , β , λ , and τ satisfy one of the following conditions:

  1. τ = 1 , β ∈ S 1 \ R − , λ ∈ S 1 ∩ R + , and 0 ≤ β < min { 1 − α , λ + γ − α } ;

  2. τ = ± 1 , β , λ ∈ S 2 ∩ R + and 0 < β < min { 1 − α − λ , λ + γ − α } .

Suppose that μ ( x ) = ∣ x ∣ p ( 1 − α δ ) − 1 , ν n = ∣ n ∣ q ( 1 − α ρ ) − 1 , where δ ∈ S 2 and ρ ∈ S 2 ∩ ( 0 , 1 ] . Let f ( x ) , a n > 0 with f ( x ) ∈ L p , μ ( R ) , and a = { a n } n ∈ Z 0 ∈ l q , ν . Let k ( u ) and W ( α , β , γ , λ ) be defined by (2.1) and (2.3), respectively. Then,

(3.1) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 k ( x δ n ρ ) a n d x = ∑ n ∈ Z 0 a n ∫ x ∈ R k ( x δ n ρ ) f ( x ) d x < ∣ δ ∣ − 1 q ρ − 1 p W ( α , β , γ , λ ) ‖ f ‖ p , μ ‖ a ‖ q , ν ,

where the constant factor ∣ δ ∣ − 1 q ρ − 1 p W ( α , β , γ , λ ) in (3.1) is the best possible.

Proof

Let k ˆ ( x δ y ρ ) ≔ k ( x δ n ρ ) , g ( y ) ≔ a n , and h ( y ) ≔ ∣ n ∣ if y ∈ [ n , n + 1 ) ( n ∈ Z − ) . Let k ˆ ( x δ y ρ ) ≔ k ( x δ n ρ ) , g ( y ) ≔ a n , and h ( y ) ≔ n if y ∈ [ n − 1 , n ) ( n ∈ Z + ). By Hölder’s inequality, we have

(3.2) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 k ( x δ n ρ ) a n d x = ∑ n ∈ Z 0 a n ∫ x ∈ R k ( x δ n ρ ) f ( x ) d x = ∫ y ∈ R ∫ x ∈ R k ˆ ( x δ y ρ ) f ( x ) g ( y ) d x d y = ∫ y ∈ R ∫ x ∈ R [ k ˆ ( x δ y ρ ) ] 1 ⁄ p [ h ( y ) ] ( α ρ − 1 ) ⁄ p ∣ x ∣ ( 1 − α δ ) ⁄ q f ( x ) [ k ˆ ( x δ y ρ ) ] 1 ⁄ q ∣ x ∣ ( α δ − 1 ) ⁄ q × [ h ( y ) ] ( 1 − α ρ ) ⁄ p g ( y ) d x d y ≤ ∫ x ∈ R ∫ y ∈ R k ˆ ( x δ y ρ ) [ h ( y ) ] α ρ − 1 ∣ x ∣ p ( 1 − α δ ) ⁄ q f p ( x ) d y d x 1 ⁄ p × ∫ y ∈ R ∫ x ∈ R k ˆ ( x δ y ρ ) ∣ x ∣ α δ − 1 [ h ( y ) ] q ( 1 − α ρ ) ⁄ p g q ( y ) d x d y 1 ⁄ q = ∫ x ∈ R ω 1 ( x ) ∣ x ∣ p ( 1 − α δ ) ⁄ q f p ( x ) d x 1 ⁄ p ∑ n ∈ Z 0 ω 2 ( n ) ∣ n ∣ q ( 1 − α ρ ) ⁄ p a n q 1 ⁄ q ,

where

ω 1 ( x ) = ∑ n ∈ Z 0 k ( x δ n ρ ) ∣ n ∣ α ρ − 1 , ω 2 ( n ) = ∫ x ∈ R k ( x δ n ρ ) ∣ x ∣ α δ − 1 d x .

If n > 0 , x > 0 , then it follows from δ ∈ S 2 and ρ ∈ S 2 ∩ ( 0 , 1 ) that x δ n ρ > 0 and x δ n ρ increases with n ( n ∈ Z + ) . Therefore, by Lemma 2.1, and observing that ρ ∈ S 2 ∩ ( 0 , 1 ) , we find

[ k ( x δ n ρ ) + k ( − x δ n ρ ) ] n α ρ − 1 = x δ − α δ n ρ − 1 [ k ( x δ n ρ ) + k ( − x δ n ρ ) ] ( x δ n ρ ) α − 1

decreases with n ( n ∈ Z + ) for a fixed x ( x > 0 ) .

If n > 0 , x < 0 , then − x δ n ρ > 0 and − x δ n ρ increases with n ( n ∈ Z + ) . Therefore, by Lemma 2.1, it can also be proved that

[ k ( x δ n ρ ) + k ( − x δ n ρ ) ] n α ρ − 1 = ∣ x ∣ δ − α δ n ρ − 1 [ k ( − x δ n ρ ) + k ( x δ n ρ ) ] ( − x δ n ρ ) α − 1

decreases with n ( n ∈ Z + ) for a fixed x ( x < 0 ) .

Thus, we have

ω 1 ( x ) = ∑ n ∈ Z + [ k ( x δ n ρ ) + k ( − x δ n ρ ) ] n α ρ − 1 < ∫ y ∈ R + [ k ( x δ y ρ ) + k ( − x δ y ρ ) ] y α ρ − 1 d y .

Consider the case where x < 0 . Letting x δ y ρ = − u , and observing that δ , ρ ∈ S 2 , we have ( − u ) 1 ρ − 1 = u 1 ρ − 1 ( u > 0 ) , and therefore,

(3.3) ω 1 ( x ) < 1 ρ ∫ u ∈ R + [ k ( u ) + k ( − u ) ] u 1 ρ ∣ x ∣ − δ ρ α ρ − 1 ∣ x ∣ − δ ρ ( − u ) 1 ρ − 1 d u = ∣ x ∣ − α δ ρ ∫ u ∈ R + [ k ( u ) + k ( − u ) ] u α − 1 d u .

Furthermore, if x > 0 , then it can also be proved that (3.3) holds. Therefore, applying (2.4) to (3.3), we have

(3.4) ω 1 ( x ) < ∣ x ∣ − α δ ρ W ( α , β , γ , λ ) .

Similarly, we have

(3.5) ω 2 ( n ) < ∣ n ∣ − α ρ ∣ δ ∣ W ( α , β , γ , λ ) .

Inserting (3.4) and (3.5) into (3.2), we arrive at (3.1).

In what follows, it will be proved that the constant factor ∣ δ ∣ − 1 q ρ − 1 p W ( α , β , γ , λ ) in (3.1) is the best possible. In fact, we assume that there exists a real number A satisfying

(3.6) 0 < A ≤ ∣ δ ∣ − 1 q ρ − 1 p W ( α , β , γ , λ )

and

(3.7) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 k ( x δ n ρ ) a n d x = ∑ n ∈ Z 0 a n ∫ x ∈ R k ( x δ n ρ ) f ( x ) d x < A ‖ f ‖ p , μ ‖ a ‖ q , ν .

Replace a n and f ( x ) with a ˆ n and f ˆ ( x ) defined in Lemma 2.3, respectively, and use (2.17), and then we have

s ∣ δ ρ ∣ ∫ 0 1 [ k ( u ) + k ( − u ) ] u α − 1 + 2 p s d u + ∫ 1 ∞ [ k ( u ) + k ( − u ) ] u α − 1 − 2 q s d u < A ‖ f ˆ ‖ p , μ ‖ a ˆ ‖ q , ν = A ∫ Ω ∣ x ∣ − 1 + 2 δ s d x 1 p ∑ n ∈ Z 0 ∣ n ∣ − 1 − 2 ρ s 1 q < A 2 ∫ Ω + x − 1 + 2 δ s d x 1 p 2 + 2 ∫ 1 ∞ y − 1 − 2 ρ s d y 1 q = A s ∣ δ ∣ 1 p 2 + s ρ 1 q .

It follows therefore that

(3.8) ∫ 0 1 [ k ( u ) + k ( − u ) ] u α − 1 + 2 p s d u + ∫ 1 ∞ [ k ( u ) + k ( − u ) ] u α − 1 − 2 q s d u < ∣ δ ρ ∣ A 1 ∣ δ ∣ 1 p 2 s + 1 ρ 1 q .

Applying Fatou’s lemma to (3.8) and using (2.4), it follows that

W ( α , β , γ , λ ) = ∫ u ∈ R + [ k ( u ) + k ( − u ) ] u α − 1 d u = ∫ 0 1 lim ̲ s → ∞ [ k ( u ) + k ( − u ) ] u α − 1 + 2 p s d u + ∫ 1 ∞ lim ̲ s → ∞ [ k ( u ) + k ( − u ) ] u α − 1 − 2 q s d u ⩽ lim ̲ s → ∞ ∫ 0 1 [ k ( u ) + k ( − u ) ] u α − 1 + 2 p s d u + ∫ 1 ∞ [ k ( u ) + k ( − u ) ] u α − 1 − 2 q s d u ⩽ lim ̲ s → ∞ ∣ δ ρ ∣ A 1 ∣ δ ∣ 1 p 2 s + 1 ρ 1 q = A ∣ δ ∣ 1 q ρ 1 p .

Thus, we have

(3.9) A ≥ ∣ δ ∣ − 1 q ρ − 1 p W ( α , β , γ , λ ) .

It follows from (3.6) and (3.9) that

A = ∣ δ ∣ − 1 q ρ − 1 p W ( α , β , γ , λ ) ,

which implies that the constant factor ∣ δ ∣ − 1 q ρ − 1 p W ( α , β , γ , λ ) in (3.1) is the best possible. Theorem 3.1 is proved.□

Remark 3.2

Theorem 3.1 implies the following Hardy-type inequalities:

(3.10) ∑ n ∈ Z 0 ∣ n ∣ p α ρ − 1 ∫ x ∈ R k ( x δ n ρ ) f ( x ) d x p < ∣ δ ∣ − 1 q ρ − 1 p W ( α , β , γ , λ ) p ‖ f ‖ p , μ p ,

(3.11) ∫ x ∈ R ∣ x ∣ q α δ − 1 ∑ n ∈ Z 0 k ( x δ n ρ ) a n q d x < ∣ δ ∣ − 1 q ρ − 1 p W ( α , β , γ , λ ) q ‖ a ‖ q , ν q ,

where the constant ∣ δ ∣ − 1 q ρ − 1 p W ( α , β , γ , λ ) in (3.11) and (3.12) is the best possible. In fact, letting z ≔ { z n } n ∈ Z 0 , where

z n ≔ ∣ n ∣ p α ρ − 1 ∫ x ∈ R k ( x δ n ρ ) f ( x ) d x p − 1 ,

and using (3.1), we have

(3.12) I ≔ ∑ n ∈ Z 0 ∣ n ∣ p α ρ − 1 ∫ x ∈ R k ( x δ n ρ ) f ( x ) d x p = ∑ n ∈ Z 0 z n ∫ x ∈ R k ( x δ n ρ ) f ( x ) d x < ∣ δ ∣ − 1 q ρ − 1 p W ( α , β , γ , λ ) ‖ f ‖ p , μ ‖ z ‖ q , ν = ∣ δ ∣ − 1 q ρ − 1 p W ( α , β , γ , λ ) ‖ f ‖ p , μ I 1 ⁄ q .

Inequality (3.12) implies (3.10) obviously. Furthermore, letting

F ( x ) ≔ ∣ x ∣ q α δ − 1 ∑ n ∈ Z 0 k ( x δ n ρ ) a n q − 1 ,

and using (3.1), it can be proved that (3.12) holds true. Since the constant factor ∣ δ ∣ − 1 q ρ − 1 p W ( α , β , γ , λ ) in (3.1) is the best possible, it follows that the constant ∣ δ ∣ − 1 q ρ − 1 p W ( α , β , γ , λ ) in (3.10) and (3.11) is the best possible.

4 Some corollaries

Suppose that γ = 0 , τ = 1 , β ∈ S 1 \ R − , λ ∈ S 1 ∩ R + , and δ = ρ = 1 in Theorem 3.1, and use Lemma 2.4, then we obtain the following Hilbert-type inequality involving a non-homogeneous kernel.

Corollary 4.1

Let α ∈ ( 0 , 1 ) , β ∈ S 1 \ R − , λ ∈ S 1 ∩ R + , and 0 ≤ β < min { 1 − α , λ − α } . Suppose that Φ ( u ) = csc 2 u , μ ( x ) = ∣ x ∣ p ( 1 − α ) − 1 , ν n = ∣ n ∣ q ( 1 − α ) − 1 , f ( x ) , a n > 0 with f ( x ) ∈ L p , μ ( R ) and a = { a n } n ∈ Z 0 ∈ l q , ν . Then,

(4.1) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 ( x n ) β + 1 ( x n ) λ − 1 ( ln ∣ x n ∣ ) a n d x < 2 π 2 λ 2 Φ α π λ + Φ ( α + β ) π λ ‖ f ‖ p , μ ‖ a ‖ q , ν .

Let β = 0 in (4.1), then 0 < α < min { 1 , λ } ( λ ∈ S 1 ∩ R + ) , and (4.1) reduces to

(4.2) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 ln ∣ x n ∣ ( x n ) λ − 1 a n d x < 2 π 2 λ 2 Φ α π λ ‖ f ‖ p , μ ‖ a ‖ q , ν ,

where μ ( x ) = ∣ x ∣ p ( 1 − α ) − 1 , and ν n = ∣ n ∣ q ( 1 − α ) − 1 .

Let λ = 2 β in (4.1). Since

Φ α π 2 β + Φ ( α + β ) π 2 β = 4 Φ α π β ,

it can also be obtained that (4.2) holds true.

Let λ = 4 β in (4.1). Then, 0 < α < min { 1 − β , 3 β } ( β ∈ S 1 ∩ R + ) , and (4.1) reduces to

(4.3) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 ln ∣ x n ∣ [ ( x n ) β − 1 ] [ ( x n ) 2 β + 1 ] a n d x < π 2 8 β 2 Φ α π 4 β + Φ ( α + β ) π 4 β ‖ f ‖ p , μ ‖ a ‖ q , ν ,

where μ ( x ) = ∣ x ∣ p ( 1 − α ) − 1 , and ν n = ∣ n ∣ q ( 1 − α ) − 1 .

Suppose that γ = λ , τ = 1 , β ∈ S 1 \ R − , λ ∈ S 1 ∩ R + , and δ = ρ = 1 in Theorem 3.1. Then,

W ( α , β , γ , λ ) = 2 ∑ j = 0 ∞ 1 ( λ j + α ) 2 + 1 ( λ j + 2 λ − α ) 2 + 2 ∑ j = 0 ∞ 1 ( λ j + α + β ) 2 + 1 ( λ j + 2 λ − α − β ) 2 .

If 0 < α < α + β < λ , then it follows from Lemma 2.4 that

W ( α , β , γ , λ ) = 2 ∑ j = 0 ∞ 1 ( λ j + α ) 2 + 1 ( λ j + λ − α ) 2 + 2 ∑ j = 0 ∞ 1 ( λ j + α + β ) 2 + 1 ( λ j + λ − α − β ) 2 − 2 ( λ − α ) 2 − 2 ( λ − α − β ) 2 = 2 π 2 λ 2 Φ α π λ + Φ ( α + β ) π λ − m 0 ,

where

m 0 = 2 ( λ − α ) 2 + 2 ( λ − α − β ) 2 .

If 0 < α < α + β = λ , then

W ( α , β , γ , λ ) = 2 ∑ j = 0 ∞ 1 ( λ j + α ) 2 + 1 ( λ j + λ − α ) 2 + 4 ∑ j = 0 ∞ 1 ( λ j + λ ) 2 − 2 ( λ − α ) 2 = 2 π 2 λ 2 Φ β π λ + 2 π 2 3 λ 2 − 2 β 2 .

If 0 < α < λ < α + β < 2 λ , then

W ( α , β , γ , λ ) = 2 ∑ j = 0 ∞ 1 ( λ j + α ) 2 + 1 ( λ j + λ − α ) 2 + 2 ∑ j = 0 ∞ 1 ( λ j + α + β − λ ) 2 + 1 ( λ j + 2 λ − α − β ) 2 − 2 ( λ − α ) 2 − 2 ( λ − α − β ) 2 = 2 π 2 λ 2 Φ α π λ + Φ ( α + β ) π λ − m 0 .

If 0 < α = λ < α + β < 2 λ , then

W ( α , β , γ , λ ) = 2 ∑ j = 0 ∞ 1 ( λ j + α + β − λ ) 2 + 1 ( λ j + 2 λ − α − β ) 2 + 4 ∑ j = 0 ∞ 1 ( λ j + λ ) 2 − 2 ( λ − α − β ) 2 = 2 π 2 λ 2 Φ β π λ + 2 π 2 3 λ 2 − 2 β 2 .

If 0 < λ < α < α + β < 2 λ , then

W ( α , β , γ , λ ) = 2 ∑ j = 0 ∞ 1 ( λ j + α − λ ) 2 + 1 ( λ j + 2 λ − α ) 2 + 2 ∑ j = 0 ∞ 1 ( λ j + α + β − λ ) 2 + 1 ( λ j + 2 λ − α − β ) 2 − 2 ( λ − α ) 2 − 2 ( λ − α − β ) 2 = 2 π 2 λ 2 Φ α π λ + Φ ( α + β ) π λ − m 0 .

Let

W 0 ( α , β , γ , λ ) = 2 π 2 λ 2 Φ β π λ + 2 π 2 3 λ 2 − 2 β 2 , α = λ or α + β = λ , 2 π 2 λ 2 Φ α π λ + Φ ( α + β ) π λ − m 0 , α ≠ λ and α + β ≠ λ .

Then, the following corollary holds true.

Corollary 4.2

Let α ∈ ( 0 , 1 ) , β , λ ∈ S 1 ∩ R + , and 0 < β < min { 1 − α , 2 λ − α } . Suppose that μ ( x ) = ∣ x ∣ p ( 1 − α ) − 1 , ν n = ∣ n ∣ q ( 1 − α ) − 1 , and f ( x ) , a n > 0 with f ( x ) ∈ L p , μ ( R ) and a = { a n } n ∈ Z 0 ∈ l q , ν . Then,

(4.4) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 ( x n ) β + 1 ( x n ) λ − 1 ln ∣ x n ∣ max { 1 , ∣ x n ∣ λ } a n d x < W 0 ( α , β , γ , λ ) ‖ f ‖ p , μ ‖ a ‖ q , ν .

Let λ = β in (4.4). Then, 0 < α < min { β , 1 − β } ( β ∈ S 1 ∩ R + ) , and (4.4) reduces to

(4.5) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 ( x n ) β + 1 ( x n ) β − 1 ln ∣ x n ∣ max { 1 , ∣ x n ∣ β } a n d x < 4 π 2 β 2 Φ α π β − m 0 ′ ‖ f ‖ p , μ ‖ a ‖ q , ν ,

where m 0 ′ = 2 α 2 + 2 ( β − α ) 2 , μ ( x ) = ∣ x ∣ p ( 1 − α ) − 1 , and ν n = ∣ n ∣ q ( 1 − α ) − 1 .

Let λ = 2 β , and α = β in (4.4), then 0 < β < 1 2 ( β ∈ S 1 ∩ R + ) , and (4.4) reduces to

(4.6) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 ln ∣ x n ∣ [ ( x n ) β − 1 ] max { 1 , ∣ x n ∣ 2 β } a n d x < 2 π 2 − 6 3 β 2 ‖ f ‖ p , μ ‖ a ‖ q , ν ,

where μ ( x ) = ∣ x ∣ p ( 1 − β ) − 1 , and ν n = ∣ n ∣ q ( 1 − β ) − 1 .

Let λ = 2 β , α ≠ β , and α ≠ 2 β in (4.4). Then, 0 < α < min { 1 − β , 3 β } , and (4.4) reduces to

(4.7) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 ln ∣ x n ∣ [ ( x n ) β − 1 ] max { 1 , ∣ x n ∣ 2 β } a n d x < 2 π 2 β 2 Φ α π β − m 0 ″ ‖ f ‖ p , μ ‖ a ‖ q , ν ,

where m 0 ″ = 2 ( 2 β − α ) 2 + 2 ( β − α ) 2 , μ ( x ) = ∣ x ∣ p ( 1 − α ) − 1 , and ν n = ∣ n ∣ q ( 1 − α ) − 1 .

Setting α = β 2 in (4.7), then 0 < β < 2 3 ( β ∈ S 1 ∩ R + ) , and (4.7) is transformed into

(4.8) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 ln ∣ x n ∣ [ ( x n ) β − 1 ] max { 1 , ∣ x n ∣ 2 β } a n d x < 18 π 2 − 80 9 β 2 ‖ f ‖ p , μ ‖ a ‖ q , ν ,

where μ ( x ) = ∣ x ∣ p 1 − β 2 − 1 , and ν n = ∣ n ∣ q 1 − β 2 − 1 .

Suppose that γ = 0 , τ = 1 , β , λ ∈ S 2 ∩ R + , and δ = ρ = 1 in Theorem 3.1, and use Lemma 2.4. Then, we obtain the following corollary.

Corollary 4.3

Let α ∈ ( 0 , 1 ) , β , λ ∈ S 2 ∩ R + and 0 < β < min { 1 − α − λ , λ − α } . Suppose that Φ ( u ) = csc 2 u , μ ( x ) = ∣ x ∣ p ( 1 − α ) − 1 , ν n = ∣ n ∣ q ( 1 − α ) − 1 , f ( x ) , a n > 0 with f ( x ) ∈ L p , μ ( R ) and a = { a n } n ∈ Z 0 ∈ l q , ν . Then,

(4.9) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 ( x n ) β + 1 ( x n ) λ − 1 ( ln ∣ x n ∣ ) a n d x < π 2 2 λ 2 Φ α π 2 λ + Φ ( α + β + λ ) π 2 λ ‖ f ‖ p , μ ‖ a ‖ q , ν .

Let λ = 3 β , and α = β in (4.9). Then, 0 < β < 1 5 ( β ∈ S 2 ∩ R + ) , and (4.9) reduces to

(4.10) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 ( x n ) β + 1 ( x n ) β − 1 ln ∣ x n ∣ ( x n ) 2 β + ( x n ) β + 1 a n d x < 2 π 2 9 β 2 ‖ f ‖ p , μ ‖ a ‖ q , ν ,

where μ ( x ) = ∣ x ∣ p ( 1 − β ) − 1 , and ν n = ∣ n ∣ q ( 1 − β ) − 1 .

Suppose that γ = 2 λ , τ = 1 , β , λ ∈ S 2 ∩ R + , and δ = ρ = 1 in Theorem 3.1. Then, 0 < β < min { 1 − α − λ , 3 λ − α } .

W ( α , β , γ , λ ) = 2 ∑ j = 0 ∞ 1 ( 2 λ j + α ) 2 + 1 ( 2 λ j + 4 λ − α ) 2 + 2 ∑ j = 0 ∞ 1 ( 2 λ j + α + β + λ ) 2 + 1 ( 2 λ j + 3 λ − α − β ) 2 .

If α + β < λ , then α < 2 λ , and it follows from Lemma 2.4 that

W ( α , β , γ , λ ) = 2 ∑ j = 0 ∞ 1 ( 2 λ j + α ) 2 + 1 ( 2 λ j + 2 λ − α ) 2 + 2 ∑ j = 0 ∞ 1 ( 2 λ j + α + β + λ ) 2 + 1 ( 2 λ j + λ − α − β ) 2 − 2 ( 2 λ − α ) 2 − 2 ( λ − α − β ) 2 = π 2 2 λ 2 Φ α π 2 λ + Φ ( α + β + λ ) π 2 λ − p 0 ,

where

p 0 ≔ p 1 + p 2 ≔ 2 ( 2 λ − α ) 2 + 2 ( λ − α − β ) 2 .

If α + β = λ , then α < 2 λ , and

W ( α , β , γ , λ ) = 2 ∑ j = 0 ∞ 1 ( 2 λ j + α ) 2 + 1 ( 2 λ j + 2 λ − α ) 2 + 4 ∑ j = 0 ∞ 1 ( 2 λ j + 2 λ ) 2 − 2 ( 2 λ − α ) 2 = π 2 2 λ 2 Φ α π 2 λ + π 2 6 λ 2 − p 1 .

If α + β > λ and α < 2 λ , then

W ( α , β , γ , λ ) = 2 ∑ j = 0 ∞ 1 ( 2 λ j + α ) 2 + 1 ( 2 λ j + 2 λ − α ) 2 + 2 ∑ j = 0 ∞ 1 ( 2 λ j + α + β − λ ) 2 + 1 ( 2 λ j + 3 λ − α − β ) 2 − 2 ( 2 λ − α ) 2 − 2 ( α + β − λ ) 2 = π 2 2 λ 2 Φ α π 2 λ + Φ ( α + β − λ ) π 2 λ − p 0 = π 2 2 λ 2 Φ α π 2 λ + Φ ( α + β + λ ) π 2 λ − p 0 .

If α + β > λ and α = 2 λ , then

W ( α , β , γ , λ ) = 2 ∑ j = 0 ∞ 1 ( 2 λ j + α + β − λ ) 2 + 1 ( 2 λ j + 3 λ − α − β ) 2 + 4 ∑ j = 0 ∞ 1 ( 2 λ j + 2 λ ) 2 − 2 ( α + β − λ ) 2 = π 2 2 λ 2 Φ ( α + β + λ ) π 2 λ + π 2 6 λ 2 − p 2 .

If α + β > λ and α > 2 λ , then it can also be proved that

W ( α , β , γ , λ ) = π 2 2 λ 2 Φ α π 2 λ + Φ ( α + β + λ ) π 2 λ − p 0 .

Let

W 1 ( α , β , γ , λ ) = π 2 2 λ 2 Φ α π 2 λ + π 2 6 λ 2 − p 1 , α + β = λ , π 2 2 λ 2 Φ α π 2 λ + Φ ( α + β + λ ) π 2 λ − p 0 , α + β ≠ λ and α ≠ 2 λ , π 2 2 λ 2 Φ ( α + β + λ ) π 2 λ + π 2 6 λ 2 − p 2 , α = 2 λ .

Corollary 4.4

Let α ∈ ( 0 , 1 ) , β , λ ∈ S 2 ∩ R + , and 0 < β < min { 1 − α − λ , 3 λ − α } . Suppose that μ ( x ) = ∣ x ∣ p ( 1 − α ) − 1 , ν n = ∣ n ∣ q ( 1 − α ) − 1 , f ( x ) , a n > 0 with f ( x ) ∈ L p , μ ( R ) , and a = { a n } n ∈ Z 0 ∈ l q , ν . Then,

(4.11) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 ( x n ) β + 1 ( x n ) λ − 1 ln ∣ x n ∣ max { 1 , ∣ x n ∣ 2 λ } a n d x < W 1 ( α , β , γ , λ ) ‖ f ‖ p , μ ‖ a ‖ q , ν .

Let λ = β in (4.11). Then, 0 < α < min { 2 β , 1 − 2 β } ( β ∈ S 2 ∩ R + ) , and (4.11) reduces to

(4.12) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 ( x n ) β + 1 ( x n ) β − 1 ln ∣ x n ∣ max { 1 , ∣ x n ∣ 2 β } a n d x < π 2 β 2 Φ α π 2 β − p 0 ′ ‖ f ‖ p , μ ‖ a ‖ q , ν ,

where p 0 ′ = 2 α 2 + 2 ( 2 β − α ) 2 , μ ( x ) = ∣ x ∣ p ( 1 − α ) − 1 , and ν n = ∣ n ∣ q ( 1 − α ) − 1 .

Set α = β in (4.12). Then, 0 < β < 1 3 ( β ∈ S 2 ∩ R + ) , and (4.12) is transformed into

∫ x ∈ R f ( x ) ∑ n ∈ Z 0 ( x n ) β + 1 ( x n ) β − 1 ln ∣ x n ∣ max { 1 , ∣ x n ∣ 2 β } a n d x < π 2 − 4 β 2 ‖ f ‖ p , μ ‖ a ‖ q , ν ,

where μ ( x ) = ∣ x ∣ p ( 1 − β ) − 1 , and ν n = ∣ n ∣ q ( 1 − β ) − 1 .

Let α = λ in (4.11). Then, 0 < β < min { 2 λ , 1 − 2 λ } ( β , λ ∈ S 2 ∩ R + ) , and (4.11) reduces to

(4.13) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 ( x n ) β + 1 ( x n ) λ − 1 ln ∣ x n ∣ max { 1 , ∣ x n ∣ 2 λ } a n d x < π 2 2 λ 2 Φ β π 2 λ − p 0 ″ ‖ f ‖ p , μ ‖ a ‖ q , ν ,

where p 0 ″ = 2 λ 2 + 2 β 2 − π 2 2 λ 2 , μ ( x ) = ∣ x ∣ p ( 1 − λ ) − 1 , and ν n = ∣ n ∣ q ( 1 − λ ) − 1 .

Let α = 2 λ in (4.11). Then, 0 < β < min { λ , 1 − 3 λ } ( β , λ ∈ S 2 ∩ R + ) , and (4.11) reduces to

(4.14) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 ( x n ) β + 1 ( x n ) λ − 1 ln ∣ x n ∣ max { 1 , ∣ x n ∣ 2 λ } a n d x < π 2 2 λ 2 Φ ( β + λ ) π 2 λ − p 2 ′ ‖ f ‖ p , μ ‖ a ‖ q , ν ,

where p 2 ′ = 2 ( β + λ ) 2 − π 2 6 λ 2 , μ ( x ) = ∣ x ∣ p ( 1 − 2 λ ) − 1 , and ν n = ∣ n ∣ q ( 1 − 2 λ ) − 1 .

Suppose that γ = 0 , τ = − 1 , β , λ ∈ S 2 ∩ R + , and δ = ρ = 1 in Theorem 3.1, and use Lemma 2.4, and then the following corollary holds true.

Corollary 4.5

Let α ∈ ( 0 , 1 ) , β , λ ∈ S 2 ∩ R + , and 0 < β < min { 1 − α − λ , λ − α } . Suppose that Φ ( u ) = csc 2 u , μ ( x ) = ∣ x ∣ p ( 1 − α ) − 1 , ν n = ∣ n ∣ q ( 1 − α ) − 1 , f ( x ) , a n > 0 with f ( x ) ∈ L p , μ ( R ) , and a = { a n } n ∈ Z 0 ∈ l q , ν . Then,

(4.15) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 1 − ( x n ) β 1 + ( x n ) λ ( ln ∣ x n ∣ ) a n d x < π 2 2 λ 2 Φ α π 2 λ + Φ ( α + β + λ ) π 2 λ ‖ f ‖ p , μ ‖ a ‖ q , ν .

Let λ = 2 α in (4.15). Then, 0 < β < min { 1 − 3 λ 2 , λ 2 } ( β , λ ∈ S 2 ∩ R + ) , and (4.15) is transformed into

(4.16) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 1 − ( x n ) β 1 + ( x n ) λ ( ln ∣ x n ∣ ) a n d x < π 2 2 λ 2 Φ ( 2 β + 3 λ ) π 4 λ + π 2 λ 2 ‖ f ‖ p , μ ‖ a ‖ q , ν ,

where μ ( x ) = ∣ x ∣ p 1 − λ 2 − 1 , and ν n = ∣ n ∣ q 1 − λ 2 − 1 .

Let λ = 3 β , and α = β in (4.15). Then, 0 < β < 1 5 ( β ∈ S 2 ∩ R + ) , and (4.15) reduces to

(4.17) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 1 − ( x n ) β 1 + ( x n ) β ln ∣ x n ∣ ( x n ) 2 β − ( x n ) β + 1 a n d x < 4 π 2 ‖ f ‖ p , μ ‖ a ‖ q , ν ,

where μ ( x ) = ∣ x ∣ p ( 1 − β ) − 1 , and ν n = ∣ n ∣ q ( 1 − β ) − 1 .

Suppose that γ = 2 λ , τ = − 1 , β , λ ∈ S 2 ∩ R + , and δ = ρ = 1 in Theorem 3.1. Then, we have Corollary 4.6.

Corollary 4.6

Let α ∈ ( 0 , 1 ) , β , λ ∈ S 2 ∩ R + , and 0 < β < min { 1 − α − λ , 3 λ − α } . Suppose that μ ( x ) = ∣ x ∣ p ( 1 − α ) − 1 , ν n = ∣ n ∣ q ( 1 − α ) − 1 , f ( x ) , a n > 0 with f ( x ) ∈ L p , μ ( R ) , and a = { a n } n ∈ Z 0 ∈ l q , ν . Then,

(4.18) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 1 − ( x n ) β 1 + ( x n ) λ ln ∣ x n ∣ max { 1 , ∣ x n ∣ 2 λ } a n d x < W 1 ( α , β , γ , λ ) ‖ f ‖ p , μ ‖ a ‖ q , ν .

Let λ = β in (4.18). Then, 0 < α < min { 2 β , 1 − 2 β } ( β ∈ S 2 ∩ R + ) , and (4.18) reduces to

(4.19) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 1 − ( x n ) β 1 + ( x n ) β ln ∣ x n ∣ max { 1 , ∣ x n ∣ 2 β } a n d x < π 2 β 2 Φ α π 2 β − p 0 ′ ‖ f ‖ p , μ ‖ a ‖ q , ν ,

where p 0 ′ = 2 α 2 + 2 ( 2 β − α ) 2 , μ ( x ) = ∣ x ∣ p ( 1 − α ) − 1 , and ν n = ∣ n ∣ q ( 1 − α ) − 1 .

Let α = λ in (4.18). Then, 0 < β < min { 2 λ , 1 − 2 λ } ( β , λ ∈ S 2 ∩ R + ) , and (4.18) reduces to

(4.20) ∫ x ∈ R f ( x ) ∑ n ∈ Z 0 1 − ( x n ) β 1 + ( x n ) λ ln ∣ x n ∣ max { 1 , ∣ x n ∣ 2 λ } a n d x < π 2 2 λ 2 Φ β π 2 λ − p 0 ″ ‖ f ‖ p , μ ‖ a ‖ q , ν ,

where p 0 ″ = 2 λ 2 + 2 β 2 − π 2 2 λ 2 , μ ( x ) = ∣ x ∣ p ( 1 − λ ) − 1 , and ν n = ∣ n ∣ q ( 1 − λ ) − 1 .

In the discussions in this section, if we change δ = ρ = 1 to δ = − 1 , ρ = 1 , then we can obtain various Hilbert-type inequalities with homogeneous kernels, which will not be repeated here.

5 Conclusions

This work deals with a new type of Hilbert-type inequality with a non-monotonic kernel function. Compared the newly obtained results in this work with those in the previous literature, the new half-discrete Hilbert-type inequality discussed in this work is defined in the whole plane, while most of the previous results were established in the first quadrant. The key to achieving this work lies in two aspects. First, we define two special sets of numbers that ensure the non-negativity of the kernel function. Second, through simple variable substitution, we transform the non-monotonic weight function in the whole plane into a monotonic function in the first quadrant, thereby establishing an upper bound estimate for the weight function. The work of this article provides a new approach to the study of Hilbert-type inequalities, especially for discrete and half-discrete Hilbert-type inequalities in the whole plane.

Acknowledgements

The author is indebted to the referees for their valuable suggestions and comments that helped improve the paper significantly.

  1. Funding information: This work was supported by the incubation foundation of Zhejiang Institute of Mechanical and Electrical Engineering (A-0271-23-213).

  2. Author contributions: The author confirms the sole responsibility for the conception of the study, presented results, and prepared the manuscript.

  3. Conflict of interest: The author states no conflict of interest.

  4. Data availability statement: Not applicable.

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Received: 2023-12-18
Revised: 2024-05-28
Accepted: 2024-07-24
Published Online: 2024-08-27

© 2024 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/math-2024-0044
Keywords for this article
Hilbert-type inequality; whole plane; half-discrete; homogeneous kernel; non-homogeneous kernel
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