# On a more accurate half-discrete Hilbert-type inequality involving hyperbolic functions

• Minghui You , Xia Sun and Xiansheng Fan
From the journal Open Mathematics

## Abstract

In this work, by the introduction of a new kernel function composed of exponent functions with several parameters, and using the method of weight coefficient, Hermite-Hadamard’s inequality, and some other techniques of real analysis, a more accurate half-discrete Hilbert-type inequality including both the homogeneous and non-homogeneous cases is established. Furthermore, by introducing the Bernoulli number and the rational fraction expansion of tangent function, some special and interesting Hilbert-type inequalities and their equivalent hardy-type inequalities are presented at the end of the paper.

MSC 2010: 26D15; 41A17

## 1 Introduction

Suppose p > 1 , f ( x ) , ν ( x ) are measurable functions defined on Ω , and f ( x ) , ν ( x ) > 0 . Define a function space as follows:

L p , ν ( Ω ) f : f p , ν Ω f p ( x ) ν ( x ) d x 1 p < .

Particularly, if ν ( x ) = 1 , then we have the following abbreviated forms: f p f p , ν and L p ( Ω ) L p , ν ( Ω ) .

Suppose p > 1 , a n , ν n > 0 , and a = { a n } n = 0 . Define a sequence space as follows:

l p , ν a : a p , ν n = 0 a n p ν n 1 p < .

Particularly, we abbreviate a p , ν to a p and l p , ν to l p for ν n = 1 .

Assuming that f ( x ) , g ( x ) are two nonnegative real-valued functions, and f , g L 2 ( R + ) , we have [1]

(1.1) R + R + f ( x ) g ( y ) x + y d x d y < π f 2 g 2 .

Similarly, let a = { a m } m = 1 l 2 and b = { b n } n = 1 l 2 . Then

(1.2) n = 1 m = 1 a m b n m + n < π a 2 b 2 .

The constant factor π in (1.1) and (1.2) is the best possible, and inequalities (1.1) and (1.2) are usually named as Hilbert inequality. By introducing a pair of conjugate parameters ( p , q ) , p > 1 , and 1 p + 1 q = 1 , (1.1) and (1.2) can be extended to more general forms:

(1.3) n = 1 m = 1 a m b n m + n < π sin π p a p b q ,

(1.4) R + R + f ( x ) g ( y ) x + y d x d y < π sin π p f p g q ,

where the constant factor π sin π p is the best possible. In addition, we also have some classical inequalities similar to inequalities (1.3) and (1.4), such as [1]

(1.5) n = 1 m = 1 a m b n max { m , n } < p q a p b q ,

(1.6) R + R + f ( x ) g ( y ) max { x , y } d x d y < p q f p g q .

In general, such inequalities as (1.3), (1.4), (1.5), and (1.6) are known as Hilbert-type inequalities. Although these classical inequalities were proposed for more than 100 years, considerable attention has been paid to their parameter extensions, strengthened forms, and higher dimensional generalizations by researchers all over the world in recent years, and some new valuable Hilbert-type inequalities were established. The following inequality is a classical extension of (1.3) established by Krnić and Pečarić [2]: If p > 1 , and 1 p + 1 q = 1 , 0 < β 1 , β 2 2 , β 1 + β 2 = β , μ m = m p ( 1 β 1 ) 1 , and ν n = n q ( 1 β 2 ) 1 , then

(1.7) n = 1 m = 1 a m b n ( m + n ) β < B ( β 1 , β 2 ) a p , μ b q , ν ,

where B ( u , v ) is the beta function. In addition, Yang [3] gave an extension of (1.4) as follows:

(1.8) R + R + f ( x ) g ( y ) x β + y β d x d y < π β sin π r f p , μ g q , ν ,

where β > 0 , μ ( x ) = x p ( 1 β r ) 1 , ν ( y ) = y q ( 1 β s ) 1 , and 1 r + 1 s = 1 .

For other extensions of classical discrete Hilbert-type inequalities, we can refer to [4,5,6, 7,8,9, 10,11,12], and some extended results of integral version can be found in [10,11,13,14,15, 16,17]. Furthermore, by introducing various new kernel functions, special constants, and special functions, and considering discrete and integral forms, many Hilbert-type inequalities with new kernel functions were established in the past 20 years (see [18,19,20, 21,22,23, 24,25,26]). In addition to such types of Hilbert-type inequalities mentioned above, some new results on time scales were also established in recent years (see [27,28]). In what follows, we present the following two integral Hilbert-type inequalities, which involve the kernels related to hyperbolic functions, and are closely related to our research in the present paper, that is [25,26],

(1.9) R + R + csch ( x δ y ) f ( x ) g ( y ) d x d y < π 2 4 f 2 , μ g 2 , ν ,

(1.10) R + R + ( coth ( x y ) 1 ) f ( x ) g ( y ) d x d y < π 2 12 f 2 , ν g 2 , ν ,

where δ { 1 , 1 } , μ ( x ) = x 1 4 δ , ν ( y ) = y 3 .

Besides the integral and discrete Hilbert-type inequalities, it should be pointed out that Hilbert-type inequalities sometimes appear in half-discrete form, such as the following two [29,30]:

(1.11) R + f ( x ) n = 1 a n x + n d x < π sin π p f p a q ,

(1.12) R + f ( x ) n = 1 log x n x n a n d x < π sin π p 2 f p a q .

With regard to some other half-discrete inequalities, we refer to [31,32, 33,34].

In this paper, by using techniques of real analysis, particularly Hermite-Hadamard’s inequality, we consider the half-discrete forms of (1.9) and (1.10), then the following Hilbert-type inequalities involving some hyperbolic functions will be established:

(1.13) 0 a f ( x ) n = 0 csch ( x ( 2 n + 1 ) 2 m ) a n d x < 2 1 p ( 2 2 m 1 ) B m π 2 m f p , μ 1 a q , ν ( a 1 ) ,

(1.14) a f ( x ) n = 0 coth 2 n + 1 x 2 m 1 a n d x < 2 1 p B m π 2 m f p , μ 2 a q , ν ( 0 a 1 ) ,

(1.15) 0 a f ( x ) n = s cosh ( x n 2 m ) csch ( 3 x n 2 m ) a n d x < 2 m π 2 m 6 2 m ψ ( 2 m 1 ) π 6 f p , μ 1 a q , ν ¯ ( a 1 ) ,

where m , s N + , B m is the Bernoulli number, μ 1 ( x ) = 1 x , μ 2 ( x ) = x 2 p 1 , ν n = 1 2 n + 1 , and ν ¯ n = 1 n .

More generally, a new kernel function in more general form is constructed in Section 2, and then a Hilbert-type inequality including both the homogeneous and non-homogeneous cases is established. It will be shown that the newly obtained inequality is a unified extension of inequalities (1.13), (1.14), and (1.15), and some other special cases of the newly obtained inequality are presented in Section 4.

## Definition 2.1

For t > 0 , define Γ -function as follows:

Γ ( t ) R + z t 1 e z d z .

In particular, we have Γ ( t ) = ( t 1 ) ! for t N + .

## Lemma 2.2

Let λ 3 λ 2 λ 1 < λ 3 , and λ > 1 . Define

(2.1) k ( z ) e λ 1 z + e λ 2 z e λ 3 z e λ 3 z , z > 0 ,

and

(2.2) C ( λ 1 , λ 2 , λ 3 , λ ) j = 0 1 ( 2 λ 3 j λ 1 + λ 3 ) λ + 1 ( 2 λ 3 j λ 2 + λ 3 ) λ .

Then

(2.3) R + k ( z ) z λ 1 d z = Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) .

## Proof

It can be easy to show that λ 3 > 0 and

(2.4) k ( z ) = j = 0 ( e ( 2 λ 3 j + λ 1 λ 3 ) z + e ( 2 λ 3 j + λ 2 λ 3 ) z ) .

By Lebesgue term-by-term integration theorem, we obtain

(2.5) R + k ( z ) z λ 1 d z = j = 0 R + e ( 2 λ 3 j + λ 1 λ 3 ) z z λ 1 d z + R + e ( 2 λ 3 j + λ 2 λ 3 ) z z λ 1 d z .

Setting ( 2 λ 3 j λ 1 + λ 3 ) z = u , we obtain

(2.6) R + e ( 2 λ 3 j + λ 1 λ 3 ) z z λ 1 d z = R + e u u λ 1 d u ( 2 λ 3 j λ 1 + λ 3 ) λ = Γ ( λ ) ( 2 λ 3 j λ 1 + λ 3 ) λ .

Similarly, setting ( 2 λ 3 j λ 2 + λ 3 ) z = u , we obtain

(2.7) R + e ( 2 λ 3 j + λ 2 λ 3 ) z z λ 1 d z = R + e u u λ 1 d u ( 2 λ 3 j λ 2 + λ 3 ) λ = Γ ( λ ) ( 2 λ 3 j λ 2 + λ 3 ) λ .

Plugging (2.6) and (2.7) back into (2.5), and using (2.2), we arrive at (2.3). Lemma 2.2 is proved.□

## Lemma 2.3

Suppose that β 0 , b 1 2 , and λ 3 λ 2 λ 1 < λ 3 . Let 0 a 1 , and Ω = ( a , ) when β < 0 . Let a 1 , and Ω = ( 0 , a ) when β > 0 . Assume that λ > 1 , 0 < γ 1 , and 0 < λ γ 1 . Define

(2.8) K ( x , y ) e λ 1 x β ( y + b ) γ + e λ 2 x β ( y + b ) γ e λ 3 x β ( y + b ) γ e λ 3 x β ( y + b ) γ , x > 0 , y 0 .

Then

(2.9) ω ( n ) Ω K ( x , n ) x β λ 1 d x Γ ( λ ) β ( n + b ) γ λ C ( λ 1 , λ 2 , λ 3 , λ ) , n N + ,

(2.10) ϖ ( x ) n = 0 K ( x , n ) ( n + b ) γ λ 1 < Γ ( λ ) γ x β λ C ( λ 1 , λ 2 , λ 3 , λ ) , x > 0 .

## Proof

Setting x β ( n + b ) γ = z , and using (2.3), we obtain

(2.11) Ω K ( x , n ) x β λ 1 d x = 1 β ( n + b ) γ λ R + k ( z ) z λ 1 d z = Γ ( λ ) β ( n + b ) γ λ C ( λ 1 , λ 2 , λ 3 , λ ) .

It follows therefore that (2.9) holds true.

Since λ 3 λ 2 λ 1 < λ 3 , it follows from (2.1) that

(2.12) d k d z = [ ( λ 1 λ 3 ) e ( λ 1 + λ 3 ) z ( λ 1 + λ 3 ) e ( λ 1 λ 3 ) z + ( λ 2 λ 3 ) e ( λ 2 + λ 3 ) z ( λ 2 + λ 3 ) e ( λ 2 λ 3 ) z ] ( e λ 3 z e λ 3 z ) 2 < 0 .

In view of that

6 λ 3 2 2 λ 2 2 > 2 ( λ 3 λ 2 ) ( λ 3 + λ 2 ) 0

and

6 λ 3 2 2 λ 1 2 > 2 ( λ 3 λ 1 ) ( λ 3 + λ 1 ) 0 ,

we obtain

(2.13) d 2 k d z 2 = [ ( λ 1 λ 3 ) 2 e ( λ 1 + 2 λ 3 ) z + ( λ 1 + λ 3 ) 2 e ( λ 1 2 λ 3 ) z + ( λ 2 λ 3 ) 2 e ( λ 2 + 2 λ 3 ) z + ( λ 2 + λ 3 ) 2 e ( λ 2 2 λ 3 ) z + ( 6 λ 3 2 2 λ 1 2 ) e λ 1 z + ( 6 λ 3 2 2 λ 2 2 ) e λ 2 z ] ( e λ 3 z e λ 3 z ) 3 > 0 .

For arbitrary x > 0 , let z x β ( y + b ) γ . In view of 0 < γ 1 , we obtain d z d y > 0 and d 2 z d y 2 0 . It follows therefore that

(2.14) d K d y = d k d z d z d y < 0

and

(2.15) d 2 K d y 2 = d 2 k d z 2 d z d y 2 + d k d z d 2 z d y 2 > 0 .

Let

H ( y ) K ( x , y ) ( y + b ) γ λ 1 K ( x , y ) h ( y ) .

Observing that 0 < γ λ 1 , and using (2.14) and (2.15), we have

d H d y = d K d y h ( y ) + d h d y K ( x , y ) < 0

and

d 2 H d y 2 = d 2 K d y 2 h ( y ) + 2 d K d y d h d y + d 2 h d y 2 K ( x , y ) > 0 .

Hence, by Hermite-Hadamard’s inequality (see [35,36]), we obtain

(2.16) ϖ ( x ) = n = 0 H ( n ) < n = 0 n 1 2 n + 1 2 H ( y ) d y = 1 2 H ( y ) d y b H ( y ) d y .

Setting x β ( y + b ) γ = z , and using (2.3), we obtain

(2.17) b H ( y ) d y = 1 γ x β λ R + k ( z ) z λ 1 d z = Γ ( λ ) γ x β λ C ( λ 1 , λ 2 , λ 3 , λ ) .

Applying (2.17) to (2.16), we obtain (2.10), and the proof of Lemma 2.3 is completed.□

## Lemma 2.4

Suppose that β 0 , b 1 2 , and λ 3 λ 2 λ 1 < λ 3 . Let 0 a 1 , and Ω = ( a , ) when β < 0 . Let a 1 , and Ω = ( 0 , a ) when β > 0 . Assume that λ > 1 , 0 < γ 1 , and 0 < λ γ 1 . Let p > 1 , 1 p + 1 q = 1 , μ ( x ) = x p ( 1 β λ ) 1 , and ν n = ( n + b ) q ( 1 γ λ ) 1 . K ( x , y ) is defined via Lemma 2.3. For an arbitrary positive integer l , which is sufficiently large, set

(2.18) a ˜ { a ˜ n } n = 0 ( n + b ) γ λ 1 γ q l n = 0 ,

(2.19) f ˜ ( x ) x β λ 1 + β p l , x E , 0 , x Ω E ,

where E = { x : x > 0 , x sgn β < 1 } . Then

(2.20) J ˜ n = 0 a ˜ n E K ( x , n ) f ˜ ( x ) d x = E f ˜ ( x ) n = 0 K ( x , n ) a ˜ n d x > l β γ 1 k ( z ) z λ 1 q l 1 d z + b γ l 0 1 k ( z ) z λ + 1 p l 1 d z .

## Proof

Observing that 0 < λ γ 1 , and using (2.14), we obtain

(2.21) J ˜ > E x β λ 1 + β p l 0 K ( x , y ) ( y + b ) γ λ 1 γ q l d y d x .

Setting x β ( y + b ) γ = z , we obtain

(2.22) J ˜ > 1 γ E x β l 1 x β b γ k ( z ) z λ 1 q l 1 d z d x = 1 γ E x β l 1 1 k ( z ) z λ 1 q l 1 d z d x + 1 γ E x β l 1 x β b γ 1 k ( z ) z λ 1 q l 1 d z d x = l β γ 1 k ( z ) z λ 1 q l 1 d z + 1 γ E x β l 1 x β b γ 1 k ( z ) z λ 1 q l 1 d z d x .

For β > 0 , by the use of Fubini’s theorem, we obtain

(2.23) E x β l 1 x β b γ 1 k ( z ) z λ 1 q l 1 d z d x = 0 1 k ( z ) z λ 1 q l 1 0 z 1 β b γ β x β l 1 d x d z = l b γ l β 0 1 k ( z ) z λ + 1 p l 1 d z .

Applying (2.23) to (2.22), it follows (2.20). Similarly, for β < 0 , by the use of Fubini’s theorem again, we have

(2.24) E x β l 1 x β b γ 1 k ( z ) z λ 1 q l 1 d z d x = 0 1 k ( z ) z λ 1 q l 1 z 1 β b γ β x β l 1 d x d z = l b γ l β 0 1 k ( z ) z λ + 1 p l 1 d z .

Plug (2.24) back into (2.22), then (2.20) follows obviously. Lemma 2.4 is proved.□

## Lemma 2.5

Let 1 < u < 1 , ψ ( x ) = tan x , and m N + . Then

(2.25) ψ ( 2 m 1 ) u π 2 = 2 2 m ( 2 m 1 ) ! π 2 m j = 0 1 ( 2 j + 1 u ) 2 m + 1 ( 2 j + 1 + u ) 2 m .

## Proof

Observing that ψ ( x ) = tan x can be expressed in the form of rational fraction expansion as follows [37]:

(2.26) ψ ( x ) = tan x = 2 j = 0 1 ( 2 j + 1 ) π 2 x 1 ( 2 j + 1 ) π + 2 x .

Finding the ( 2 m 1 ) th derivative of ψ ( x ) , we have

(2.27) ψ ( 2 m 1 ) ( x ) = 2 2 m ( 2 m 1 ) ! j = 0 1 [ ( 2 j + 1 ) π 2 x ] 2 m + 1 [ ( 2 j + 1 ) π + 2 x ] 2 m .

Setting x = u π 2 in (2.27), we arrive at (2.25). Lemma 2.5 is proved.□

## Theorem 3.1

Suppose that β 0 , b 1 2 , and λ 3 λ 2 λ 1 < λ 3 . Let 0 a 1 , and Ω = ( a , ) when β < 0 . Let a 1 , and Ω = ( 0 , a ) when β > 0 . Assume that λ > 1 , 0 < γ 1 , and 0 < λ γ 1 . Let p > 1 , 1 p + 1 q = 1 , μ ( x ) = x p ( 1 β λ ) 1 , and ν n = ( n + b ) q ( 1 γ λ ) 1 . Let f ( x ) , a n 0 with f ( x ) L p , μ ( Ω ) , and a = { a n } n = 0 l q , ν . C ( λ 1 , λ 2 , λ 3 , λ ) and K ( x , y ) are defined via Lemmas 2.2and2.3, respectively. Then the following three inequalities are equivalent:

(3.1) J 1 n = 0 ( n + b ) p γ λ 1 Ω K ( x , n ) f ( x ) d x p < β 1 q γ 1 p Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) p f p , μ p ,

(3.2) J 2 Ω x q β λ 1 n = 0 K ( x , n ) a n q d x < β 1 q γ 1 p Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) q a q , ν q ,

(3.3) J n = 0 a n Ω K ( x , n ) f ( x ) d x = Ω f ( x ) n = 0 K ( x , n ) a n d x < β 1 q γ 1 p Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) f p , μ a q , ν ,

where the constant β 1 p γ 1 q Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) in (3.1), (3.2), and (3.3) is the best possible.

## Proof

By Hölder’s inequality, and using (2.9), we obtain

(3.4) Ω K ( x , n ) f ( x ) d x p = Ω K ( x , n ) x 1 β λ q f ( x ) x β λ 1 q d x p Ω K ( x , n ) x p ( 1 β λ ) q f p ( x ) d x Ω K ( x , n ) x β λ 1 d x p 1 = [ ω ( n ) ] p 1 Ω K ( x , n ) x p ( 1 β λ ) q f p ( x ) d x Γ ( λ ) β ( n + b ) γ λ C ( λ 1 , λ 2 , λ 3 , λ ) p 1 Ω K ( x , n ) x p ( 1 β λ ) q f p ( x ) d x .

Plugging (3.4) back into the left hand side of (3.1), and using Lebesgue term-by-term integration theorem as well as inequality (2.10), we have

J 1 Γ ( λ ) β C ( λ 1 , λ 2 , λ 3 , λ ) p 1 Ω f p ( x ) x p ( 1 β λ ) q n = 0 K ( x , n ) ( n + b ) γ λ 1 d x < β 1 q γ 1 p Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) p f p , μ p .

The proof of inequality (3.1) is completed. Similarly, by the use of H o ¨ lder’s inequality again and (2.10), we have

(3.5) n = 0 K ( x , n ) a n q = n = 0 K ( x , n ) ( n + b ) γ λ 1 p ( n + b ) 1 γ λ p a n q [ ϖ ( x ) ] q 1 n = 0 K ( x , n ) ( n + b ) q ( 1 γ λ ) p a n q < Γ ( λ ) γ x β λ C ( λ 1 , λ 2 , λ 3 , λ ) q 1 n = 0 K ( x , n ) ( n + b ) q ( 1 γ λ ) p a n q .

It follows from Lebesgue term-by-term integration theorem and (2.9) that

J 2 < Γ ( λ ) β C ( λ 1 , λ 2 , λ 3 , λ ) q 1 n = 0 ( n + b ) q ( 1 γ λ ) p a n q Ω K ( x , n ) x β λ 1 d x < β 1 q γ 1 p Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) q a q , ν q .

Inequality (3.2) is proved. Additionally, we will prove (3.3) by (3.1). In fact, we can first obtain two representations of J by Lebesgue term-by-term integration theorem, and then by the use of H o ¨ lder’s inequality, we obtain

(3.6) J = n = 0 ( n + b ) γ λ 1 p Ω K ( x , n ) f ( x ) d x a n ( n + b ) γ λ + 1 p J 1 1 p n = 0 a n q ( n + b ) q ( 1 γ λ ) 1 1 q = J 1 1 p a q , ν .

Applying (3.1) to (3.6), we have (3.3). On the contrary, we assume that (3.3) holds true, and let b = { b n } n = 0 , where

b n ( n + b ) p γ λ 1 Ω K ( x , n ) f ( x ) d x p 1 .

By the use of (3.3), it follows that

(3.7) J 1 = n = 0 b n Ω K ( x , n ) f ( x ) d x < β 1 q γ 1 p Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) f p , μ b q , ν = β 1 q γ 1 p Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) f p , μ J 1 1 q .

By (3.7), inequality (3.1) follows naturally. Hence, inequalities (3.1) and (3.3) are equivalent. In order to prove the equivalence of inequalities (3.1), (3.2), and (3.3), it suffices to prove that (3.2) and (3.3) are equivalent. In fact, if (3.2) is assumed to be true, then

(3.8) J = Ω x β λ + 1 q f ( x ) x β γ 1 q n = 0 K ( x , n ) a n d x f p , μ J 2 1 q .

Applying (3.2) to (3.8), we have (3.3). Conversely, assume (3.3) holds true, and let

g ( x ) x q β λ 1 n = 0 K ( x , n ) a n q 1 .

Then

(3.9) J 2 = Ω g ( x ) n = 0 K ( x , n ) a n d x < β 1 q γ 1 p Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) g p , μ a q , ν = β 1 q γ 1 p Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) a q , ν J 2 1 p .

Therefore, inequality (3.2) holds true, and the equivalence of inequalities (3.1), (3.2), and (3.3) is proved. Finally, it will be proved that the constant β 1 q γ 1 p Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) in (3.1), (3.2), and (3.3) is optimal. Assuming that the constant factor β 1 q γ 1 p Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) in (3.3) is not optimal, there must be a real number c satisfying

(3.10) 0 < c < β 1 q γ 1 p Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) ,

so that (3.3) still holds if β 1 q γ 1 p Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) is replaced by c , that is,

(3.11) n = 0 a n Ω K ( x , n ) f ( x ) d x = Ω f ( x ) n = 0 K ( x , n ) a n d x < c f p , μ a q , ν .

Replace f and a n in (3.11) with f ˜ and a ˜ n defined in Lemma 2.4, respectively. It implies that

(3.12) E f ˜ ( x ) n = 0 K ( x , n ) a ˜ n d x < c f ˜ p , μ a ˜ q , ν = c E x β l 1 d x 1 p n = 0 ( n + b ) γ l 1 1 q = c l β 1 p b γ l 1 + n = 1 ( n + b ) γ l 1 1 q < c l β 1 p b γ l 1 + 0 ( y + b ) γ l 1 d y 1 q = c l β 1 p b γ l 1 + l b γ l γ 1 q .

Combining (2.20) and (3.12), we have

(3.13) 1 k ( z ) z λ 1 q l 1 d z + b γ l 0 1 k ( z ) z λ + 1 p l 1 d z < c γ β 1 q b γ l 1 l + b γ l γ 1 q .

Letting l + in (3.13), and using (2.3), we have

(3.14) c β 1 q γ 1 p Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) .

Inequalities (3.10) and (3.14) are apparently contradictory, and therefore β 1 q γ 1 p Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) in (3.3) is the best possible. It can also be proved β 1 q γ 1 p Γ ( λ ) C ( λ 1 , λ 2 , λ 3 , λ ) in (3.1) and (3.2) is the best possible from the equivalence of (3.1), (3.2), and (3.3). Theorem 3.1 is proved.□

## 4 Some corollaries

Let λ 1 = λ 2 = 0 , λ 3 = ρ ( ρ > 0 ) , and λ = 2 m ( m N + ) in Theorem 3.1. In view of the equation [37]:

(4.1) j = 0 2 ( 2 j + 1 ) 2 m = B m ( 2 m ) ! ( 2 2 m 1 ) π 2 m , m N + ,

where B m is the Bernoulli number, B 1 = 1 6 , B 2 = 1 30 , B 3 = 1 42 , , we obtain the following corollary.

## Corollary 4.1

Suppose that ρ > 0 , β 0 , b 1 2 , and 0 < 2 m γ 1 ( m N + ) . Let 0 a 1 , and Ω = ( a , ) when β < 0 . Let a 1 , and Ω = ( 0 , a ) when β > 0 . Assume that p > 1 , 1 p + 1 q = 1 , μ ( x ) = x p ( 1 2 m β ) 1 , and ν n = ( n + b ) q ( 1 2 m γ ) 1 . Let f ( x ) , a n 0 with f ( x ) L p , μ ( Ω ) , and a = { a n } n = 0 l q , ν . Then the following three inequalities are equivalent:

(4.2) n = 0 ( n + b ) 2 m p γ 1 Ω csch ( ρ x β ( n + b ) γ ) f ( x ) d x p < β 1 q γ 1 p ( 2 2 m 1 ) B m 2 m π ρ 2 m p f p , μ p ,

(4.3) Ω x 2 m q β 1 n = 0 csch ( ρ x β ( n + b ) γ ) a n q d x < β 1 q γ 1 p ( 2 2 m 1 ) B m 2 m π ρ 2 m q a q , ν q ,

(4.4) Ω f ( x ) n = 0 csch ( ρ x β ( n + b ) γ ) a n d x < β 1 q γ 1 p ( 2 2 m 1 ) B m 2 m π ρ 2 m f p , μ a q , ν ,

where the constant β 1 q γ 1 p ( 2 2 m 1 ) B m 2 m π ρ 2 m in (4.2), (4.3), and (4.4) is the best possible.

Setting β = γ ( 0 < 2 m γ 1 ) , b = s ( s N + ) in Corollary 4.1, we obtain the following inequalities with a non-homogeneous kernel ( a 1 ):

(4.5) n = s n 2 m p γ 1 0 a csch ( ρ ( x n ) γ ) f ( x ) d x p < ( 2 2 m 1 ) B m 2 m γ π ρ 2 m p f p , μ p ,

(4.6) 0 a x 2 m q γ 1 n = s csch ( ρ ( x n ) γ ) a n q d x < ( 2 2 m 1 ) B m 2 m γ π ρ 2 m q a q , ν q ,

(4.7) 0 a f ( x ) n = s csch ( ρ ( x n ) γ ) a n d x < ( 2 2 m 1 ) B m 2 m γ π ρ 2 m f p , μ a q , ν ,

where μ ( x ) = x p ( 1 2 m γ ) 1 and ν n = n q ( 1 2 m γ ) 1 .

Setting β = γ ( 0 < 2 m γ 1 ) , b = s ( s N + ) in Corollary 4.1, we obtain the following inequalities with a homogeneous kernel ( 0 a 1 ):

(4.8) n = s n 2 m p γ 1 a csch ρ n x γ f ( x ) d x p < ( 2 2 m 1 ) B m 2 m γ π ρ 2 m p f p , μ p ,

(4.9) a x 2 m q γ 1 n = s csch ρ n x γ a n q d x < ( 2 2 m 1 ) B m 2 m γ π ρ 2 m q a q , ν q ,

(4.10) a f ( x ) n = s csch ρ n x γ a n d x < ( 2 2 m 1 ) B m 2 m γ π ρ 2 m f p , μ a q , ν ,

where μ ( x ) = x p ( 1 + 2 m γ ) 1 and ν n = n q ( 1 2 m γ ) 1 .

Setting β = γ ( 0 < 2 m γ 1 ) , b = 1 2 , and replacing ρ with ρ 2 γ in Corollary 4.1, it follows that

(4.11) n = 0 ( 2 n + 1 ) 2 m p γ 1 0 a csch ( ρ ( x ( 2 n + 1 ) ) γ ) f ( x ) d x p < 2 1 p ( 2 2 m 1 ) B m 2 m γ π ρ 2 m p f p , μ p ,

(4.12) 0 a x 2 m q γ 1 n = 0 csch ( ρ ( x ( 2 n + 1 ) ) γ ) a n q d x < 2 1 p ( 2 2 m 1 ) B m 2 m γ π ρ 2 m q a q , ν q ,

(4.13) 0 a f ( x ) n = 0 csch ( ρ ( x ( 2 n + 1 ) ) γ ) a n d x < 2 1 p ( 2 2 m 1 ) B m 2 m γ π ρ 2 m f p , μ a q , ν ,

where a 1 , μ ( x ) = x p ( 1 2 m γ ) 1 , and ν n = ( 2 n + 1 ) q ( 1 2 m γ ) 1 . Letting γ = 1 2 m , ρ = 1 in (4.13), we obtain inequality (1.13).

Setting β = γ ( 0 < 2 m γ 1 ) , b = 1 2 , and replacing ρ with ρ 2 γ in Corollary 4.1, we can obtain the homogeneous forms corresponding to (4.11), (4.12), and (4.13) with the same constant factors, such as

(4.14) a f ( x ) n = 0 csch ρ 2 n + 1 x γ a n d x < 2 1 p ( 2 2 m 1 ) B m 2 m γ π ρ