On Hilbert-Type and Hardy-Type Integral Inequalities and Applications (SpringerBriefs in Mathematics) [1st ed. 2019] 3030292673, 9783030292676

Table of contents :
Preface
Contents
1 Introduction
1.1 Hilbert's Inequalities and Their Operator Expressions
1.2 Two Classes of Hardy-Hilbert-Type Inequalities and Their Equivalent Forms
1.3 Three Classes of Hilbert-Type Inequalities and Their Equivalent Forms
1.4 Some Results on Multidimensional Hilbert-Type Inequalities
References
2 Equivalent Statements of Hilbert-Type Integral Inequalities
2.1 Two Lemmas
2.2 Main Results and Some Corollaries
2.3 Operator Expressions and a Few Examples
2.4 Introducing the Exponent Function as an Interval Variable
References
3 Equivalent Statements of the Reverse Hilbert-Type Integral Inequalities
3.1 Some Lemmas
3.2 Main Results
3.3 Some Corollaries and a Few Examples
3.4 Some Reverse Equivalent Hilbert-Type Inequalities in the Whole Plane
References
4 Equivalent Statements of Two Kinds of Hardy-Type Integral Inequalities
4.1 Lemmas
4.2 Hardy-Type Integral Inequalities of the First Kind
4.3 Hardy-Type Integral Inequalities of the Second Kind
4.4 Operator Expressions and Some Examples
4.5 Hardy-Type Integral Inequalities with the Exponent Function as Interval Variables
References
5 Equivalent Property of the Reverse Hardy-Type Integral Inequalities
5.1 Two Lemmas
5.2 Reverse Hardy-Type Integral Inequalities of the First Kind
5.3 Reverse Hardy-Type Inequalities of the Second Kind
5.4 Reverse Hardy-Type Integral Inequalities with the Interval Variable
References

Citation preview

SPRINGER BRIEFS IN MATHEMATICS

Bicheng Yang Michael Th. Rassias

On Hilbert-Type and Hardy-Type Integral Inequalities and Applications 123

SpringerBriefs in Mathematics Series Editors Palle Jorgensen, Iowa City, USA Roderick Melnik, Waterloo, Canada Lothar Reichel, Kent, USA George Yin, Detroit, USA Nicola Bellomo, Torino, Italy Michele Benzi, Pisa, Italy Tatsien Li, Shanghai, China Otmar Scherzer, Linz, Austria Benjamin Steinberg, New York City, USA Yuri Tschinkel, New York City, USA Ping Zhang, Kalamazoo, USA

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More information about this series at http://www.springer.com/series/10030

Bicheng Yang Michael Th. Rassias •

On Hilbert-Type and Hardy-Type Integral Inequalities and Applications

123

Bicheng Yang Department of Mathematics Guangdong University of Education Guangzhou, China

Michael Th. Rassias Institute of Mathematics University of Zurich Zürich, Switzerland Moscow Institute of Physics and Technology Dolgoprudny, Russia Institute for Advanced Study Program in Interdisciplinary Studies Princeton, USA

ISSN 2191-8198 ISSN 2191-8201 (electronic) SpringerBriefs in Mathematics ISBN 978-3-030-29267-6 ISBN 978-3-030-29268-3 (eBook) https://doi.org/10.1007/978-3-030-29268-3 Mathematics Subject Classification (2010): 26D15, 37A10, 47A07, 65B10 © The Author(s), under exclusive licence to Springer Nature Switzerland AG 2019 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

As long as a branch of knowledge offers an abundance of problems, it is full of vitality. David Hilbert … we have always found with most inequalities, that we have a little new to add. … in a subject (inequalities) like this, which has applications in every part of mathematics but never been developed systematically. Godfrey H. Hardy

Preface

Hilbert-type inequalities including Hilbert’s inequalities (originating in 1908), Hardy-Hilbert-type inequalities (originating in 1934) and Yang-Hilbert-type inequalities (originating in ca. 1998) have proven to be essential for various applications in mathematical analysis. These inequalities are mainly classified as integral inequalities, discrete inequalities and half-discrete inequalities. During the last two decades, research on these types of inequalities has flourished with the publication of several papers and books dedicated to this domain. A lot of activity has also been focused on Yang-Hilbert-type inequalities. In the present monograph, by applying theories, methods and techniques of real analysis and functional analysis, we study four kinds of equivalent formulations of Hilbert-type inequalities, Hardy-type integral inequalities as well as their parametrized reverses. The best possible constant factors related mainly to the extended Hurwitz zeta function are presented in many examples. Furthermore, we consider the operator expressions with the norm and some particular analytic inequalities. Special cases of the above-mentioned four kinds of integral inequalities in the whole plane are considered. Through several lemmas and theorems, one can find within the present monograph an extensive account of these kinds of inequalities and operators. The book comprises five chapters: In Chap. 1, we introduce some recent developments of Hilbert-type integral inequalities, discrete and half-discrete inequalities. In Chap. 2, by the application of theories, methods and techniques of real analysis, we provide several Hilbert-type integral inequalities with a general nonhomogeneous kernel. Moreover, the case of the general homogeneous kernel is also treated. In the form of applications, some special cases and a variety of examples—mainly related to the extended Hurwitz zeta function—are presented. In the last three chapters, we study further the cases of the reverse Hilbert-type integral inequalities, the Hardy-type integral inequalities and the reverse Hardy-type integral inequalities in the same spirit as in Chap. 2.

vii

viii

Preface

We hope that this monograph will prove to be useful especially to graduate students of mathematics, physics and engineering sciences. Bicheng Yang Department of Mathematics Guangdong University of Education Guangzhou, China Michael Th. Rassias Institute of Mathematics University of Zurich Zürich, Switzerland Moscow Institute of Physics and Technology Dolgoprudny, Russia Institute for Advanced Study Program in Interdisciplinary Studies Princeton, USA

Contents

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Hilbert’s Inequalities and Their Operator Expressions . . . . . . . . 1.2 Two Classes of Hardy-Hilbert-Type Inequalities and Their Equivalent Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Three Classes of Hilbert-Type Inequalities and Their Equivalent Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Some Results on Multidimensional Hilbert-Type Inequalities . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Equivalent Statements of Hilbert-Type Integral Inequalities . 2.1 Two Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Main Results and Some Corollaries . . . . . . . . . . . . . . . . . 2.3 Operator Expressions and a Few Examples . . . . . . . . . . . 2.4 Introducing the Exponent Function as an Interval Variable References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

.. ..

1 1

..

3

.. .. ..

5 7 10

. . . . . .

. . . . . .

15 15 19 29 36 42

. . . .

3 Equivalent Statements of the Reverse Hilbert-Type Integral Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Some Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Some Corollaries and a Few Examples . . . . . . . . . . . . . . . . . . 3.4 Some Reverse Equivalent Hilbert-Type Inequalities in the Whole Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . .

43 43 48 53

.. ..

64 70

4 Equivalent Statements of Two Kinds of Hardy-Type Integral Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Hardy-Type Integral Inequalities of the First Kind . . . . . . . 4.3 Hardy-Type Integral Inequalities of the Second Kind . . . . . 4.4 Operator Expressions and Some Examples . . . . . . . . . . . . .

. . . . .

71 71 77 83 88

. . . . .

. . . . .

. . . . .

. . . . .

ix

x

Contents

4.5 Hardy-Type Integral Inequalities with the Exponent Function as Interval Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 5 Equivalent Property of the Reverse Hardy-Type Integral Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Two Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Reverse Hardy-Type Integral Inequalities of the First Kind . 5.3 Reverse Hardy-Type Inequalities of the Second Kind . . . . . 5.4 Reverse Hardy-Type Integral Inequalities with the Interval Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

109 109 114 122

. . . . . 130 . . . . . 144 . . . . . 144

Chapter 1

Introduction

1.1 Hilbert’s Inequalities and Their Operator Expressions If f (x), g(y) ≥ 0 (x, y ∈ R+ = (0, ∞)),  f, g ∈ L 2 (R+ ) =

 f ; || f ||2 =

∞

| f (x)|2 d x

 21

 0 , then we have the following, well known, Hilbert integral inequality as well as an equivalent form (cf. [1]): 

∞



0



∞

0

∞ 0

f (x)g(y) d xd y < π|| f ||2 ||g||2 , x+y



∞

0

f (x) dx x+y

(1.1)

 21

2 dy

< π|| f ||2 ,

(1.2)

where the constant factor π is the best possible. If am , bn ≥ 0 (m, n ∈ N = {1, 2, . . .}), a = {am }∞ m=1

⎧ ⎫ ∞  21 ⎨ ⎬

∈ l 2 = a; ||a||2 = |am |2 0 ,

© The Author(s), under exclusive licence to Springer Nature Switzerland AG 2019 B. Yang and M. Th. Rassias, On Hilbert-Type and Hardy-Type Integral Inequalities and Applications, SpringerBriefs in Mathematics, https://doi.org/10.1007/978-3-030-29268-3_1

1

2

1 Introduction

then we also have the following discrete Hilbert inequality and an equivalent form: ∞ ∞

am bn < π||a||2 ||b||2 , m+n n=1 m=1

⎡ ⎣

∞ ∞

n=1

am m +n m=1

2 ⎤ 21 ⎦ < π||a||2 ,

(1.3)

(1.4)

with the same best constant π (cf. [2]). Using (1.2), we may define a Hilbert integral operator T : L 2 (R+ ) → L 2 (R+ ) as follows (cf. [3]): For any f ∈ L 2 (R+ ), there exists a T f ∈ L 2 (R+ ), satisfying 

∞

T f (y) = 0

f (x) d x (y ∈ R+ ). x+y

Then by (1.2), we have ||T f ||2 < π|| f ||2 , and T is a bounded linear operator with ||T || :=

||T f ||2 ≤ π. f (=θ)∈L 2 (R+ ) || f ||2 sup

Since the constant factor in (1.2) is the best possible, we have ||T || = π. Using (1.4), we may define the Hilbert operator T : l 2 → l 2 as follows (cf. [4]): 2 2 For any a = {am }∞ m=1 ∈ l , there exists a T a ∈ l , satisfying T a(n) =

∞

am (n ∈ N). m+n m=1

Then by (1.4), we have ||T a||2 ≤ π||a||2 , and T is a bounded linear operator with ||T || := sup

a(=θ)∈l 2

||T a||2 ≤ π. ||a||2

Since the constant factor in (1.4) is the best possible, we have ||T || = π. In 2002, Zhang [5] studied the above two operators and obtained some improvements of (1.1) and (1.3).

1.2 Two Classes of Hardy-Hilbert-Type Inequalities and Their Equivalent Forms

3

1.2 Two Classes of Hardy-Hilbert-Type Inequalities and Their Equivalent Forms In 1925, by introducing one pair of conjugate exponents ( p, q) ( 1p + [6] proved the following extensions of (1.1)–(1.4): Assuming that p > 1, f (x), g(y) ≥ 0,  f ∈ L (R+ ) = p

 f ; || f || p =

∞

| f (x)| d x p

 1p

1 q

= 1), Hardy

 0, we have the following Hardy-Hilbert integral inequality and the equivalent form: 

∞



0



∞

∞

0

π f (x)g(y) d xd y < || f || p ||g||q , x+y sin(π/ p)



∞

0

0

where the constant factor If am , bn ≥ 0, a = {am }∞ m=1

f (x) dx x+y

π sin(π/ p)

p

 1p dy


0, then we have the following Hardy-Hilbert-type integral inequality and the equivalent form: 

∞



0

∞

k1 (x, y) f (x)g(y)d xd y < k p || f || p ||g||q ,

(1.9)

0

 0

∞



∞

p k1 (x, y) f (x)d x

 1p dy

< k p || f || p ,

(1.10)

0

where the constant factor k p is the best possible. If k1 (m, n)m − p (k1 (m, n)n − q ) is decreasing with respect to m (n) ∈ N, 1

1

p ∞ q am , bn ≥ 0, a = {am }∞ m=1 ∈ l , b = {bn }n=1 ∈ l , ||a|| p , ||b||q > 0,

then we also have the following equivalent discrete variant of the above inequalities: ∞ ∞

k1 (m, n)am bn < k p ||a|| p ||b||q ,

(1.11)

n=1 m=1

∞ ∞

n=1

 p  1p k1 (m, n)am

< k p ||a|| p ,

m=1

with the same best possible constant factor k p . Some applications of Hardy-Hilbert-type inequalities are featured in [7].

(1.12)

1.3 Three Classes of Hilbert-Type Inequalities and Their Equivalent Forms

5

1.3 Three Classes of Hilbert-Type Inequalities and Their Equivalent Forms In 1998, by introducing an independent parameter λ ∈ (0, ∞), Yang [8, 9] proved the following extension of (1.1): For f (x) ≥ 0 satisfying 

∞

0
1, f (x) ≥ 0, satisfying 

∞

0
0), then for am, bn ≥ 0,  a ∈ l p,φ = a; ||a|| p,φ := (

∞

 1 p

φ(m)|am | ) < ∞ , p

m=1

b = {bn }∞ n=1 ∈ lq,ψ , ||a|| p,φ , ||b||q,ψ > 0, we have the following equivalent discrete variant of the above inequalities: ∞ ∞

kλ (m, n)am bn < k(λ1 )||a|| p,φ ||b||q,ψ ,

(1.17)

n=1 m=1

∞ ∞

n=1

m=1

 p  1p kλ (m, n)am

< k(λ1 )||a|| p,φ ,

(1.18)

1.3 Three Classes of Hilbert-Type Inequalities and Their Equivalent Forms

7

where the constant factor k(λ1 ) is still the best possible. Clearly, for λ = 1, λ1 = q1 , λ2 = 1p , (1.15), (1.16), (1.17) and (1.18) reduce respectively to (1.9), (1.10), (1.11) and (1.12); for p = q = 2, λ1 = λ2 = λ2 > 0, 1 λ λ 1 kλ (x, y) = (x+y) λ , (1.15) reduces to (1.13); for λ1 = r , λ2 = s , k λ (x, y) = x λ +y λ (λ > 0), (1.15) reduces to (1.14). Regarding half-discrete Hilbert-type inequalities with nonhomogeneous kernels, Hardy et al. obtained a few results in Theorem 351 of [2]. However, they did not prove that the constant factors are the best possible. But, Yang [13] presented a 1 result involving the kernel (1+nx) λ (by introducing a variable) and proved that the corresponding constant factor is the best possible. Using weight functions and techniques of discrete and integral Hilbert-type inequalities with some additional conditions on the kernel, the following half-discrete Hilbert-type inequality and the equivalent forms with a general homogeneous kernel of degree −λ ∈ R and a best constant factor k (λ1 ) are obtained (cf. [14]): 

∞

f (x)

0

∞

∞

n

pλ2 −1



(1.19)

∞

 p  1p kλ (x, n) f (x)d x

< k(λ1 )|| f || p,φ ,

(1.20)

0

n=1



kλ (x, n)an d x < k(λ1 )|| f || p,φ ||a||q,ψ ,

n=1

∞

x

qλ1 −1

0

∞

 q1

q kλ (x, n)an

dx

< k(λ1 )||a||q,ψ .

(1.21)

n=1

Additionally, a half-discrete Hilbert-type inequality with a general nonhomogeneous kernel kλ (1, xn) and a best constant factor is established by Yang [15]. Surveys on Hilbert-type inequalities are provided in [16, 17].

1.4 Some Results on Multidimensional Hilbert-Type Inequalities If i 0 , j0 ∈ N, α, β > 0, we set ||x||α :=

i 0

 α1 |xk |α

(x = (x1 , . . . , xi0 ) ∈ Ri0 ),

k=1

||y||β :=

j 0

k=1

 β1 |yk |β

(y = (y1 , . . . , y j0 ) ∈ R j0 ).

8

1 Introduction

In 2006, Hong [18] proved the following multidimensional Hilbert-type integral inequality and the equivalent form (for β = α): For p > 1, 1p + q1 = 1, λ1 + λ2 = λ, i0 ), (y) = y p( j0 −λ2 )− j0 (y ∈ R+0 ), (x) = x p(i0 −λ1 )−i0 (x ∈ R+ j

f (x) = f (x1 , . . . , xi0 ) ≥ 0, g(y) = g(y1 , . . . , y j0 ) ≥ 0,  0 < || f || p, =

 1p (x) f (x)d x

< ∞,

p

i

 0 < ||g||q, =

R+0

j

R+0

 q1 (y)g q (y)dy

< ∞,

we have the following equivalent inequalities with the kernel kλ (||x||α , ||y||α ) =

1 (||x||α + ||y||α )λ

(λ > 0):   j

R+0



i

R+0

kλ (||x||α , ||y||α ) f (x)g(y)d xd y < K (λ1 )|| f || p, ||g||q, ,

pλ s −1

j

R+0

||y||α



 i

R+0

kλ (||x||α , ||y||α ) f (x)d x dy

 1p < K (λ1 )|| f || p, , (1.23)

where the particular constant factor K (λ1 ) =

 i0 (1/α) B(λ1 , λ2 ) αi0 −1 (i 0 /α)

(λ1 , λ2 > 0) is the best possible. In 2007, by introducing four particular kernels 1 (0 < λ < 1), |||x||α − ||y||α |λ ln(||x||α /||y||α ) (λ > 0), kλ (||x||α , ||y||α ) = ||x||λα − ||y||λα 1 (λ > 0) kλ (||x||α , ||y||α ) = λ ||x||α + ||y||λα

kλ (||x||α , ||y||α ) =

(1.22)

1.4 Some Results on Multidimensional Hilbert-Type Inequalities

and

9

1 (λ > 0), (max{||x||α , ||y||α })λ

kλ (||x||α , ||y||α ) =

Zhong et al. [19] established four pairs of equivalent inequalities (1.22) and (1.23) (for β = α) with the best possible constant factors K (λ1 ) =

 i0 (1/α) k(λ1 ), αi0 −1 (i 0 /α)

where k(λ1 ) = B(λ1 , 1 − λ) + B(λ2 , 1 − λ),  2 π π , λ sin(πλ1 /λ) λ sin(πλ1 /λ) and λ1λλ2 (λ1 , λ2 > 0). In 2011, Yang et al. [20] proved (1.22) and (1.23) with the general homogeneous kernel kλ (||x||α , ||y||β ) and the best possible constant factor  K (λ1 ) =

 j0 (1/β) β j0 −1 ( j0 /β)

where,

 k(λ1 ) =

∞

 1p 

 i0 (1/α) αi0 −1 (i 0 /α)

 q1

k(λ1 ),

kλ (t, 1)t λ1 −1 dt ∈ R+ .

0

In this case, for i 0 = j0 = α = β, (1.22) and (1.23) reduce respectively to (1.15) and (1.16). In recent years, some results on multidimensional Hilbert-type integral inequalities are published in [21–25], and some results on multidimensional discrete as well as half-discrete Hilbert-type inequalities are presented in [26–40]. Other kinds of inequalities and operators are studied in [41–45] (see also [46–51]). Remark 1.2 (1) Many different kinds of Hilbert-type discrete, half-discrete and integral inequalities, along with various applications have been presented during the last twenty years. Within this monograph we give special emphasis to certain original results that have been proved during the period 2009–2012. We present several generalizations, extensions and refinements of Hilbert-type discrete, half-discrete and integral inequalities involving many special functions such as the beta function, the gamma function, hypergeometric functions, trigonometric functions, hyperbolic

10

1 Introduction

functions, the Hurwitz zeta function, the Riemann zeta function, the Bernoulli functions, the Bernoulli numbers and Euler’s constant (cf. [52–59]). (2) In his six books, Yang [11, 12, 60–63] presented certain kinds of Hilberttype operators with general homogeneous and nonhomogeneous kernels and two pairs of conjugate exponents as well as related inequalities. These research monographs contained recent developments of discrete, half-discrete and integral types of operators and inequalities along with their proofs, accompanied with examples and applications. (3) In 2017, Hong [64] studied equivalent forms to a Hilbert-type integral inequality with a general homogeneous kernel with some parameters. Several authors continue this effort and have been investigating this topic for other types of integral inequalities and operators (cf. [59, 65–69]). Following this line of work, we consider equivalent properties to several kinds of Hilbert-type and Hardy-type integral inequalities as well as present applications involving the extended Hurwitz zeta function, the Riemann zeta function etc. In Chap. 2 of the present book, by the use of weight functions and techniques of real analysis and functional analysis, we consider a few statements of Hilbert-type integral inequalities with a general nonhomogeneous kernel related to some parameters. The case of Hilbert-type integral inequalities with a general homogeneous kernel is deduced. Meanwhile, operator expressions and some examples mainly related to the Hurwitz zeta function are obtained in the form of applications. Additionally, a few statements of Hilbert-type integral inequalities in the whole plane with the exponent function as the integral variables are constructed. In the remaining three chapters, the cases of the reverse Hilbert-type integral inequalities, two kinds of Hardy-type integral inequalities as well as their reverses are considered similarly to Chap. 2.

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12

1 Introduction

37. Zhong, J.H., Yang, B.C.: An extension of a multidimensional Hilbert-type inequality. J. Inequal. Appl. 2017, 78 (2017) 38. Yang, B.C., Chen, Q.: A more accurate multidimensional Hardy-Mulholland-type inequality with a general homogeneous kernel. J. Math. Inequal. 12(1), 113–128 (2018) 39. Yang, B.C.: A more accurate multidimensional Hardy-Hilbert’s inequality. J. Appl. Anal. Comput. 8(2), 558–572 (2018) 40. Yang, B.C.: A more accurate multidimensional Hardy-Hilbert-type inequality. J. King Saud Univ. Sci. (2018). https://doi.org/10.1016/j.jksus.2018.01.004 41. Kato, T.: Perturbation theory for linear operators. Classics in Mathematics. Springer, Berlin (1995) 42. Kato, T.: Variation of discrete spectra. Comm. Math. Phys. 111(3), 501–504 (1987) 43. Kato, T., Satake, I.: An algebraic theory of Landau-Kolmogorov inequalities. Tohoku Math. J. 33(3), 421–428 (1981) 44. Kato, T.: On an inequality of Hardy, Littlewood, and Polya. Adv. Math. 7, 217–218 (1971) 45. Kato, T.: Demicontinuity, hemicontinuity and monotonicity. II. Bull. Am. Math. Soc. 73, 886– 889 (1967) 46. Azar, L.E.: Two new forms of Hilbert integral inequality. Math. Ineq. Appl. 17(3), 937–946 (2014) 47. Azar, L.E.: The connection between Hilbert and Hardy inequalities. J. Inequal. Appl. 2013, 452. https://doi.org/10.1186/1029-242X-2013-452 48. Azar, L.E.: Some extensions of Hilbert integral inequality. J. Math. Ineq. 5(1), 131–140 (2011) 49. Azar, L.E.: Some inequalities of Hilbert’s type and applications. Jordan J. Math. Stat. 1(2), 143–162 (2008) 50. Azar, L.E.: Two new forms of half-discrete Hilbert inequality. J. Egypt. Math. Soc. 22(2), 254–257 (2014) 51. Azar, L.E.: On Some extensions of Hardy-Hilbert’s inequality and applications. J. Inequal. Appl. 2008, Article ID 546829, 14 pages. https://doi.org/10.1155/2008/546829 52. Rassias, M.Th., Yang, B.C.: A Hilbert-type integral inequality in the whole plane related to the hyper geometric function and the beta function. J. Math. Anal. Appl. 428(2), 1286–1308 (2015) 53. Rassias, M.Th., Yang, B.C.: A more accurate half-discrete Hardy-Hilbert-type inequality with the best possible constant factor related to the extended Riemann-zeta function. Int. J. Nonlinear Anal. Appl. 7(2), 1–27 (2016) 54. Rassias, M.Th., Yang, B.C.: A half-discrete Hilbert-type inequality in the whole plane related to the Riemann zeta function. Appl. Anal., https://doi.org/10.1080/00036811.2017.1313411 55. Rassias, M.Th., Yang, B.C.: Equivalent conditions of a Hardy-type integral inequality related to the extended Riemann zeta function. Adv. Oper. Theory 2(3), 237–256 (2017) 56. Liao, J.Q., Yang, B.C.: On Hardy-type integral inequalities with the gamma function. J. Inequal. Appl. 2017, 131 (2017) 57. Wang, A.Z., Yang, B.C.: A more accurate half-discrete Hardy-Hilbert-type inequality with the logarithmic function. J. Inequal. Appl. 2017, 153 (2017) 58. Rassias, M.Th., Yang, B.C.: A Half-discrete Hardy-Hilbert-type inequality with a best possible constant factor related to the Hurwitz zeta function. In: Govil, N.K., Mohapatra, R.N., Qazi, M.A., Schmeisser, G. (eds.) Progression in Approximation Theory and Applicable Complex Analysis: In Memory of Q. I. Rahman, pp. 183–218. Springer, New York (2017) 59. Rassias, M.Th., Yang, B.C.: Equivalent properties of a Hilbert-type integral inequality with the best constant factor related the Hurwitz zeta function. Ann. Funct. Anal. 9(2), 282–295 (2018) 60. Yang, B.C.: The Norm of Operator and Hilbert-Type Inequalities. Science Press, Beijing, China (2009) 61. Yang, B.C., Debnath, L.: Half-Discrete Hilbert-Type Inequalities. World Scientific Publishing Co. Pte. Ltd., Singapore (2014) 62. Yang, B.C.: Two Kinds of Multiple Half-Discrete Hilbert-Type Inequalities. Lambert Academic Publishing, Deutschland, Germany (2012)

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63. Yang, B.C.: Topics on Half-Discrete Hilbert-Type Inequalities. Lambert Academic Publishing, Deutschland, Germany (2013) 64. Hong, Y.: On the structure character of Hilbert’s type integral inequality with homogeneous kernel and applications. J. Jilin Univ. (Sci. Ed.) 55(2), 189–194 (2017) 65. Hong, Y., Huang, Q.L., Yang, B.C., Liao, J.Q.: The necessary and sufficient conditions for the existence of a kind of Hilbert-type multiple integral inequality with the non-homogeneous kernel and its applications. J. Inequal. Appl. 2017, 316 (2017) 66. Yang, B.C., Chen, Q.: Equivalent conditions of existence of a class of reverse Hardy-type integral inequalities with nonhomogeneous kernel. J. Jilin Univ. (Sci. Ed.) 55(4), 804–808 (2017) 67. Yang, B.C.: Equivalent conditions of the existence of Hardy-type and Yang-Hilbert-type integral inequalities with the nonhomogeneous kernel. J. Guangdong Univ. Educ. 37(3), 5–10 (2017) 68. Yang, B.C.: On some equivalent conditions related to the bounded property of Yang-Hilberttype operator. J. Guangdong Univ. Educ. 37(5), 5–11 (2017) 69. Yang, Z.M., Yang, B.C.: Equivalent conditions of the existence of the reverse Hardy-type integral inequalities with the nonhomogeneous kernel. J. Guangdong Univ. Educ. 37(5), 28–32 (2017)

Chapter 2

Equivalent Statements of Hilbert-Type Integral Inequalities

In this chapter, by the use of methods of real analysis and weight functions, we consider a few equivalent statements of Hilbert-type integral inequalities with a general nonhomogeneous kernel related to certain parameters. In the form of applications, a few equivalent statements of Hilbert-type integral inequalities with a general homogeneous kernel are deduced. Moreover, we also consider operator expressions, a few particular cases, and some examples related to the extended Hurwitz zeta function as applications.

2.1 Two Lemmas In the sequel of this chapter, we assume that 1 1 + = 1, σ1 , σ, μ ∈ R, σ + μ = λ, p q

p > 1,

h(u) is a nonnegative measurable function in R+ , such that  k(σ) :=

∞

h(u)u σ−1 du (≥ 0).

(2.1)

0

For n ∈ N, we define the following two expressions: 

∞

I1 :=



1



 1 1 h(x y)x σ+ pn −1 d x y σ1 − qn −1 dy,



∞

h(x y)x 0

(2.2)

0 1

I2 :=

1

1 σ− pn −1



d x y σ1 + qn −1 dy. 1

(2.3)

1

© The Author(s), under exclusive licence to Springer Nature Switzerland AG 2019 B. Yang and M. Th. Rassias, On Hilbert-Type and Hardy-Type Integral Inequalities and Applications, SpringerBriefs in Mathematics, https://doi.org/10.1007/978-3-030-29268-3_2

15

16

2 Equivalent Statements of Hilbert-Type Integral Inequalities

Setting u = x y in (2.2) and (2.3), by Fubini’s theorem (cf. [1]), it follows that   σ+ pn1 −1 1 1 u du y σ1 − qn −1 dy h(u) I1 = y y 1 0  y   ∞ 1 1 y (σ1 −σ)− n −1 h(u)u σ+ pn −1 du dy = 



∞

y

1

0



∞

= 1

=

y (σ1 −σ)− n −1 dy 1



∞

+ 1  ∞

y

(σ1 −σ)− n1 −1



1

 0y

h(u)u σ+ pn −1 du 1

h(u)u σ+ pn −1 dudy 1

1



1

y (σ1 −σ)− n −1 dy h(u)u σ+ pn −1 du 1 0   ∞  ∞ 1 1 y (σ1 −σ)− n −1 dy h(u)u σ+ pn −1 du, + 1

1

1

(2.4)

u

  σ− pn1 −1 1 1 u du y σ1 + qn −1 dy I2 = h(u) y y 0 y   ∞  1 1 σ− pn −1 (σ1 −σ)+ n1 −1 y h(u)u du dy = 



1

∞

0

y



1

=

y 0

(σ1 −σ)+ n1 −1



1



1



1



y ∞

h(u)u σ− pn −1 du 1

dy

y (σ1 −σ)+ n −1 h(u)u σ− pn −1 dudy + 0 1   1  u 1 1 (σ1 −σ)+ n −1 = y dy h(u)u σ− pn −1 du 0

1

0

+

y (σ1 −σ)+ n −1 dy 1



0

1

∞

h(u)u σ− pn −1 du. 1

(2.5)

1

Lemma 2.1 If k(σ) > 0, and if there exists a constant M, such that for any nonnegative measurable functions f (x) and g(y) in (0, ∞), the following inequality 

∞

I := 0





∞

h(x y) f (x)g(y)d xd y

0 ∞

≤ M

x 0

p(1−σ)−1

f (x)d x p

 1p 

∞

y 0

q(1−σ1 )−1 q

g (y)dy

 q1 (2.6)

2.1 Two Lemmas

17

holds true, then we have σ1 = σ. Proof Setting



1

k1 (σ) :=

h(u)u σ−1 du

0



and

∞

k2 (σ) :=

h(u)u σ−1 du,

1

it follows that k(σ) = k1 (σ) + k2 (σ) > 0. Since ki (σ) ≥ 0 (i = 1, 2), without lose of generality, we assume that k1 (σ) > 0, namely, h(u) > 0 a.e. in an interval I ⊂ (0, 1). 1 (n ∈ N), we set two functions: If σ1 < σ, then for n > σ−σ 1 f n (x) := gn (y) :=

0, 0 < x < 1 1 , x σ− pn −1 , x ≥ 1 y σ1 + qn −1 , 0 < y ≤ 1 . 0, y > 1 1

Hence, we obtain 

∞

J2 :=

p(1−σ)−1

x 0



∞

=

x x

− n1 −1

 1p  dx

1

∞ 0

x

∞

=

 1p 

1 −1) p(1−σ)−1 p(σ− pn

1



f np (x)d x

y q(1−σ1 )−1 gnq (y)dy

 1p  dx

1

y

 q1

1 −1) q(1−σ1 )−1 q(σ1 + qn

y

 q1 dy

0

1

y

1 n −1

 q1 dy

= n.

0

By (2.5), we have 

1



u



dy h(u)u σ−1 du   1  u 1 (σ1 −σ)+ n1 −1 ≤ y dy h(u)u σ− pn −1 du 0 0  ∞ ∞ ≤ I2 = h(x y) f n (x)gn (y)d xd y ≤ M J2 = Mn < ∞. y

0

0

0

Since (σ1 − σ) +

(σ1 −σ)+ n1 −1

1 n

0

< 0, it follows that for any u ∈ (0, 1),

(2.7)

18

2 Equivalent Statements of Hilbert-Type Integral Inequalities



u

y (σ1 −σ)+ n −1 dy = ∞. 1

0

By (2.7), in view of the fact that h(u)u σ−1 > 0 a.e. in I ⊂ (0, 1), we find that ∞ ≤ Mn < ∞, which is a contradiction. If σ1 > σ, then for n > σ11−σ (n ∈ N), we set the following functions:

f n (x) :=



gn (y) :=

x σ+ pn −1 , 0 < x ≤ 1 , 0, x > 1 1

0, 0 < y < 1 1 . y σ1 − qn −1 , y ≥ 1

Hence, we find J 2 :=



∞

0



1

=

x p(1−σ)−1

f np (x)d x

1

=

1

x n −1 d x 1

y q(1−σ1 )−1

gnq (y)dy

 1p 

∞

 q1

y q(1−σ1 )−1 y q(σ1 − qn −1) dy 1

 q1

1

 1p 

0

∞ 0

x p(1−σ)−1 x p(σ+ pn −1) d x

0



 1p 

∞

y − n −1 dy 1

 q1

= n.

1

By (2.4), we have  1 1 1 y (σ1 −σ)− n −1 dy h(u)u σ+ pn −1 du 1 0  ∞ ∞ ≤ I1 = h(x y)

f n (x)

gn (y)d xd y ≤ M J 2 = Mn < ∞. 

∞

0

Since (σ1 − σ) −

1 n

(2.8)

0

> 0, it follows that 

∞

y (σ1 −σ)− n −1 dy = ∞. 1

1

By (2.8), in view of



1

h(u)u σ+ pn −1 du > 0, 1

0

we have ∞ ≤ Mn < ∞, which is a contradiction. Hence, we conclude that σ1 = σ. This completes the proof of the lemma. For σ1 = σ, we still have



2.1 Two Lemmas

19

Lemma 2.2 There exists a constant M, such that if for any nonnegative measurable functions f (x) and g(y) in (0, ∞) the following inequality 

∞ 0





∞

h(x y) f (x)g(y)d xd y

0 ∞

≤M

x p(1−σ)−1 f p (x)d x

 1p 

0

∞

y q(1−σ)−1 g q (y)dy

 q1 (2.9)

0

holds true, then we have k(σ) ≤ M < ∞. Proof For σ1 = σ, we reduce the inequality to (2.4) and then use inequality I1 ≤ M J 2 (when σ1 = σ) as follows: 

 1 1 1 y − n −1 dy h(u)u σ+ pn −1 du 1 0    ∞  ∞ 1 1 y − n −1 dy h(u)u σ+ pn −1 du +

1 1 I1 = n n  =

∞

1 1

u

h(u)u σ+ pn −1 du + 1



0

∞

h(u)u σ− qn −1 du 1

1

1 ≤ M J 2 = M. n

(2.10)

By Fatou’s lemma (cf. [1]) and (2.10), we have 

 ∞ 1 1 lim h(u)u σ+ pn −1 du + lim h(u)u σ− qn −1 du n→∞ n→∞ 0 1  1   ∞ 1 1 σ+ pn −1 σ− qn −1 ≤ limn→∞ h(u)u du + h(u)u du

k(σ) =

1

0

≤ M (< ∞).

1

(2.11)

This completes the proof of the lemma.

2.2 Main Results and Some Corollaries Theorem 2.3 Suppose that M is a constant. The following Statements (i), (ii) and (iii) are equivalent: (i) For any nonnegative measurable function f (x) in (0, ∞), we have the following inequality:

20

2 Equivalent Statements of Hilbert-Type Integral Inequalities



∞

J :=

y pσ1 −1



0

∞

p h(x y) f (x)d x

0



∞

≤M

x

p(1−σ)−1

f (x)d x p

 1p

 1p dy

.

(2.12)

0

(ii) For any nonnegative measurable functions f (x) and g(y) in (0, ∞), we have the following inequality: 

∞

I = 0



∞

h(x y) f (x)g(y)d xd y

0



∞

≤M

x p(1−σ)−1 f p (x)d x

 1p 

0

∞

y q(1−σ1 )−1 g q (y)dy

 q1

.

(2.13)

0

(iii) For k(σ) > 0, we have σ1 = σ, and k(σ) ≤ M (< ∞). Proof (i) ⇒ (ii). By Hölder’s inequality (cf. [2]), we find 



∞

I =

y σ1 − p 1

0



≤ J



∞

h(x y) f (x)d x

0 ∞

y q(1−σ1 )−1 g q (y)dy

 q1



1 y p −σ1 g(y) dy

.

(2.14)

0

Then by (2.12), we have (2.13). (ii) ⇒ (iii). Since k(σ) > 0, by Lemma 2.1, we have σ1 = σ. Then by Lemma 2.2, we have k(σ) ≤ M (< ∞). (iii) ⇒ (i). For fixed y > 0, setting u = x y, we obtain the following weight function:  ∞  ∞ h(x y)x σ−1 d x = h(u)u σ−1 du = k(σ) (y ∈ R+ ). (2.15) ω(σ, y) := y σ 0

0

By Hölder’s inequality with weight and (2.15), we have  

∞

p h(x y) f (x)d x

0



∞

=

h(x y) 

≤

0 ∞

h(x y) 0

 p x (σ−1)/q dx y (σ−1)/ p  ∞  p/q x σ−1 p f (x)d x h(x y) (σ−1)q/ p d x y 0

y (σ−1)/ p f (x) x (σ−1)/q

y σ−1 x (σ−1) p/q



2.2 Main Results and Some Corollaries

21

 ∞   y σ−1 q(1−σ)−1 p−1 = ω(σ, y)y h(x y) (σ−1) p/q f p (x)d x x 0  ∞ σ−1 y = (k(σ)) p−1 y − pσ+1 h(x y) (σ−1) p/q f p (x)d x. x 0

(2.16)

For σ1 = σ, by Fubini’s theorem (cf. [1]) and (2.16), we obtain J ≤ (k(σ))



1 q



∞

h(x y) 0

= (k(σ)) = (k(σ))

∞



1 q

0

∞



y σ−1 x (σ−1) p/q

∞

h(x y) 

1 q

0

0 ∞

ω(σ, x)x

∞

= k(σ)

x

p(1−σ)−1



y σ−1

p(1−σ)−1

f (x)d x p

dy f (x)d x

f (x)d x p

 1p

1p

p

x (σ−1)( p−1)

0



f (x)d xd y

 1p

p

 1p

.

0

For k(σ) ≤ M, (2.12) follows. Therefore, the Statements (i), (ii) and (iii) are equivalent. This completes the proof of the theorem. For σ1 = σ, by Theorem 2.3 and Lemma 2.2, we have the following: Theorem 2.4 Suppose that M is a constant. The following Statements (i), (ii) and (iii) are equivalent: (i) For any nonnegative measurable function f (x) in (0, ∞), we have the following inequality: 



∞

J1 :=

y 0



≤ M

∞

pσ−1

p h(x y) f (x)d x

0 ∞

x

p(1−σ)−1

f (x)d x p

 1p

 1p dy

.

(2.17)

0

(ii) For any nonnegative measurable functions f (x) and g(y) in (0, ∞), we have the following inequality: 

∞

I = 0





∞

h(x y) f (x)g(y)d xd y

0

≤M

∞

x 0

p(1−σ)−1

f (x)d x p

 1p 

∞

y 0

g (y)dy

q(1−σ)−1 q

 q1

.

(2.18)

22

2 Equivalent Statements of Hilbert-Type Integral Inequalities

(iii) k(σ) ≤ M < ∞. Moreover, if (iii) holds true, then the constant factor M = k(σ) in (2.17) and (2.18) is the best possible. Proof (i) ⇒ (ii). If (2.17) holds true, then by (2.14) (for σ1 = σ), we derive (2.18). (ii) ⇒ (iii). By Lemma 2.2, we have that k(σ) ≤ M < ∞. (iii) ⇒ (i). By the proof of (iii) ⇒ (i) in Theorem 2.3, we obtain (2.17). Therefore, the Statements (i), (ii) and (iii) are equivalent. If Statement (iii) holds true, and there exists a constant M ≤ k(σ), such that (2.18) is valid, then by (iii), we have k(σ) ≤ M. It follows that M = k(σ) is the best possible constant factor of (2.18). If the constant factor M = k(σ) in (2.17) is not the best possible, then by (2.14) (for σ1 = σ) we would reach a contradiction that the constant factor in (2.18) is not the best possible. This completes the proof of the theorem.  For M = k(σ), we have Theorem 2.5 Suppose that k(σ) ∈ R+ . The following Statements (i) and (ii) are valid and equivalent: (i) For any f (x) ≥ 0, satisfying 

∞

0
0, it follows that

| ln u|β (u > 0), |u λ − 1|

| ln x y|β , |(x y)λ − 1|

| ln x/y|β (x, y > 0), |x λ − y λ |

(2.47)

36

2 Equivalent Statements of Hilbert-Type Integral Inequalities



∞

| ln u|β σ−1 u du |u λ − 1| 0  1  ∞ (− ln u)β σ−1 lnβ u σ−1 u du = u du + 1 − uλ uλ − 1 0 1  1 (− ln u)β σ−1 = (u + u μ−1 )du 1 − uλ 0  1 ∞  = (− ln u)β u λk (u σ−1 + u μ−1 )du.

k(σ) = kλ (σ) =

0

k=0

By the Lebesgue term by term integration theorem (cf. [1]), we have k(σ) = kλ (σ) = ∞  

∞   k=0

1

(− ln u)β (u λk+σ−1 + u λk+μ−1 )du

0

 ∞ 1 1 + v β e−v dv = β+1 β+1 (λk + σ) (λk + μ) 0 k=0 σ (β + 1)  μ ζ(β + 1, = ) + ζ(β + 1, ) ∈ R+ , λβ+1 λ λ where ζ(s, a) =

∞  k=0

1 (Res > 1; 0 < a ≤ 1) (k + a)s

is the Hurwitz zeta function (ζ(s, 1) = ζ(s) is the Riemann zeta function) (cf. [4]). Then by Theorem 2.13 and Corollary 2.15, we have ||T (1) || = ||T (2) || =

μ σ (β + 1)  ζ(β + 1, ) + ζ(β + 1, ) . β+1 λ λ λ

(2.48)

2.4 Introducing the Exponent Function as an Interval Variable For a, b ∈ R\{0}, replacing x (resp. y) by eax (resp. eby ), then replacing f (eax )eax (resp. g(eby )eby ) by f (x) (resp. g(y)), and |a|1/qM|b|1/ p by M in Theorem 2.3, and by carrying out the corresponding simplifications, we derive the following theorem: Theorem 2.20 Suppose that M is a constant. The following statements (i), (ii) and (iii) are equivalent: (i) For any nonnegative measurable function f (x) in (−∞, ∞), we have the following inequality:

2.4 Introducing the Exponent Function as an Interval Variable





∞ −∞



≤M



∞

−∞

p

∞

e pσ1 by

−∞

f (x) eσax

37

h(eax+by ) f (x)d x

p

 1p dx

 1p dy

.

(2.49)

(ii) For any nonnegative measurable functions f (x) and g(y) in (−∞, ∞), we have the following inequality: 



∞ −∞

−∞



≤M

∞

h(eax+by ) f (x)g(y)d xd y



∞

f (x) eσax

−∞

p

 1p 

∞



dx −∞

g(y) eσ1 by

q

 q1 dy

.

(2.50)

(iii) For k(σ) > 0, we have σ1 = σ and

k(σ) ≤ M (< ∞). |a|1/q |b|1/ p

For σ1 = σ in T heorem 2.20, we have Theorem 2.21 Let M be a constant. The following statements (i), (ii) and (iii) are equivalent: (i) For any nonnegative measurable function f (x) in (−∞, ∞), we have the following inequality: 



∞

e

pσby

−∞



≤M

∞

−∞



p

∞

h(e

ax+by

−∞

f (x) eσax

p

 1p dx

) f (x)d x

 1p dy

.

(2.51)

(ii) For any nonnegative measurable functions f (x) and g(y) in (−∞, ∞), we have the following inequality: 

∞

−∞





≤M

∞ −∞ ∞

−∞

h(eax+by ) f (x)g(y)d xd y



f (x) eσax

p

 1p 

∞

dx −∞



g(y) eσby

(iii) k(σ) |a|1/q |b|1/ p

≤ M (< ∞).

q

 q1 dy

.

(2.52)

38

2 Equivalent Statements of Hilbert-Type Integral Inequalities

Moreover, if statement (iii) holds true, then the constant factor M = (2.51) and (2.52) is the best possible. For M =

k(σ) , |a|1/q |b|1/ p

k(σ) |a|1/q |b|1/ p

in

we have

Theorem 2.22 Suppose that k(σ) ∈ R+ . The following statements (i) and (ii) are valid and equivalent: (i) For any f (x) ≥ 0, satisfying  0
0,

0

we have

kλ (σ) ≤ M(< ∞). |a|1/q |b|1/ p

μ1 = μ and For μ1 = μ, we have

Corollary 2.24 Let M be a constant. The following statements (i), (ii) and (iii) are equivalent: (i) For any nonnegative measurable function f (x) in (−∞, ∞), we have the following inequality: 



∞

e −∞



≤M

∞

pμby

−∞



∞

f (x) eσax

−∞

p kλ (e , e ) f (x)d x ax

p

 1p dx

by

 1p dy

.

(2.59)

(ii) For any nonnegative measurable functions f (x) and g(y) in (−∞, ∞), we have the following inequality: 



∞

−∞



≤M (iii)

∞ −∞ ∞

−∞

kλ (eax , eby ) f (x)g(y)d xd y



f (x) eσax

p

 1p 

∞

dx −∞

kλ (σ) 1/q |a| |b|1/ p



g(y) eμby

q

 q1 dy

.

≤ M < ∞.

Moreover, if statement (iii) holds true, then the constant factor M = (2.59) and (2.60) is the best possible. For M =

kλ (σ) , |a|1/q |b|1/ p

(2.60)

kλ (σ) |a|1/q |b|1/ p

in

we obtain the following corollary:

Corollary 2.25 Suppose that kλ (σ) ∈ R+ . The following statements (i) and (ii) are valid and equivalent: (i) For any f (x) ≥ 0, satisfying  0
σ, then for n >

1 δ0 p

(n ∈ N), we consider the following two functions:

f n (x) = gn (y) =

0, 0 < x < 1 1 , x σ− pn −1 , x ≥ 1 y σ1 + qn −1 , 0 < y ≤ 1 . 0, y > 1 1

We obtain that 

∞

J2 =

x

p(1−σ)−1

0



∞

=

x 1

− n1 −1

f np (x)d x  1p 

dx

 1p  0

1

y 0

∞

1 n −1

dy

y q(1−σ1 )−1 gnq (y)dy

 q1

= n.

 q1

3.1 Some Lemmas

45

By (3.5), we have 

1

I2 ≤

y

(σ1 −σ)+ n1 −1



∞

dy

0

h(u)u σ− pn −1 du 1

0

1 = σ1 − σ + n1  1   ∞ 1 1 × h(u)u σ− pn −1 du + h(u)u σ− pn −1 du 0 1  1   ∞ 1 h(u)u (σ−δ0 )−1 du + h(u)u σ−1 du ≤ σ1 − σ 0 1 1 ≤ (k(σ − δ0 ) + k(σ)) , σ1 − σ and then by (3.6), it follows that 1 (k(σ − δ0 ) + k(σ)) σ1 − σ  ∞ ∞ ≥ I2 = h(x y) f n (x)gn (y)d xd y ≥ M J2 = Mn. 0

(3.7)

0

By (3.7), in view of σ1 − σ > 0, and 0 ≤ k(σ − δ0 ) + k(σ) < ∞, for n → ∞, we find that ∞>

1 (k(σ − δ0 ) + k(σ)) ≥ ∞, σ1 − σ

which is a contradiction. If σ1 < σ, then for n ∈ N, n >

f n (x) =

1 , δ0 p



gn (y) =

we consider the following two functions:

x σ+ pn −1 , 0 < x ≤ 1 , 0, x > 1 1

0, 0 < y < 1 . , y≥1 y 1 σ1 − qn −1

We get that J 2 =



∞

x

p(1−σ)−1

0



1

=

x 0

1 n −1

f np (x)d x

 1p 

 1p 

∞

dx

y 1

∞ 0

− n1 −1

dy

y q(1−σ1 )−1

gnq (y)dy

 q1

= n.

 q1

46

3 Equivalent Statements of the Reverse Hilbert-Type Integral Inequalities

By (3.4), we have  I1 ≤

∞

y

(σ1 −σ)− n1 −1



∞

dy

1

h(u)u σ+ pn −1 du 1

0

1 = σ − σ1 + n1  1   ∞ 1 1 σ+ pn −1 σ+ pn −1 × h(u)u du + h(u)u du 0 1  1   ∞ 1 ≤ h(u)u σ−1 du + h(u)u σ+δ0 −1 du σ − σ1 0 1 1 ≤ (k(σ) + k(σ + δ0 )) , σ − σ1 and then by (3.6), it follows that 1 (k(σ) + k(σ + δ0 )) σ − σ1  ∞ ∞ ≥ I1 = h(x y)

f n (x)

gn (y)d xd y ≥ M J 2 = Mn. 0

(3.8)

0

By (2.8), for n → ∞, we also obtain that ∞>

1 (k(σ) + k(σ + δ0 )) ≥ ∞, σ − σ1

which is a contradiction. Hence, we conclude that σ1 = σ. For σ1 = σ, we have n M = M J2 ≤ I2 by the use of (3.6), and we reduce (3.5) as follows: 1 1 M J2 ≤ I2 n n  ∞   1 1 1 1 −1 σ− pn −1 n = y h(u)u du dy n 0 y  1   ∞  1 1 1 1 1 −1 yn h(u)u σ− pn −1 du dy + h(u)u σ− pn −1 du = n 0 y 1   ∞   u 1 1 1 1 1 y n −1 dy h(u)u σ− pn −1 du + h(u)u σ− pn −1 du = n 0 0 1  1  ∞ 1 ≤ h(u)u σ+ qn −1 du + h(u)u σ−1 du.

M=

0

1

(3.9)

3.1 Some Lemmas

Since for n >

47

(n ∈ N) we have

1 δ0 |q|

h(u)u σ+ qn −1 ≤ h(u)u σ−δ0 −1 (0 < u ≤ 1) 1

and



1

h(u)u σ−δ0 −1 du ≤ k(σ − δ0 ) < ∞,

0

by (3.9) and the Lebesgue dominated convergence theorem (cf. [2]), we obtain  ∞ 1 lim h(u)u σ+ qn −1 du + h(u)u σ−1 du 0 n→∞ 1   1  ∞ 1 h(u)u σ+ qn −1 du + h(u)u σ−1 du ≥ M. = lim 

1

k(σ) =

n→∞

0

1



This completes the proof of the lemma. For σ1 = σ, we have:

Lemma 3.2 If k(σ) < ∞ and there exist constants δ0 , M > 0 such that k(σ − δ0 ) < ∞ or k(σ + δ0 ) < ∞, and if for any nonnegative measurable functions f (x) and g(y) in (0, ∞) the following inequality 

∞

I := 0





∞

h(x y) f (x)g(y)d xd y

0 ∞

≥ M

x

p(1−σ)−1

f (x)d x p

0

 1p 

∞

y

g (y)dy

q(1−σ)−1 q

 q1 (3.10)

0

holds true, then we have k(σ) ≥ M (> 0). Proof If k(σ − δ0 ) < ∞, then for σ1 = σ, by (3.9) and the proof of Lemma 3.1, we have k(σ) ≥ M. If k(σ + δ0 ) < ∞, then we have n M = M J 2 ≤ I1 (for σ1 = σ) , by the use of (3.10), and we reduce (3.4) as follows:

48

3 Equivalent Statements of the Reverse Hilbert-Type Integral Inequalities

1

1 M J2 ≤ I1 n n    y 1 1 ∞ − 1 −1 σ+ pn −1 n = y h(u)u du dy n 1 0  y   1  ∞ 1 1 1 σ+ pn −1 σ+ pn −1 − n1 −1 h(u)u du + y h(u)u du dy = n 1 0 1    1  ∞ 1 1 1 ∞ 1 h(u)u σ+ pn −1 du + y − n −1 dy h(u)u σ+ pn −1 du = n 1 0 u  ∞  1 1 h(u)u σ−1 du + h(u)u σ− qn −1 du. (3.11) ≤

M=

0

1

Since for n >

1 δ0 |q|

(n ∈ N) we have h(u)u σ− qn −1 ≤ h(u)u σ+δ0 −1 (u ≥ 1) 1



and

∞

h(u)u σ+δ0 −1 du ≤ k(σ + δ0 ) < ∞,

1

by (3.11) and the Lebesgue dominated convergence theorem (cf. [2]), we have 

1



∞

h(u)u du + lim h(u)u σ− qn −1 du n→∞ 0 1  1   ∞ 1 σ− qn −1 σ−1 = lim h(u)u du + h(u)u du ≥ M(> 0).

k(σ) =

n→∞

σ−1

0

1

1

This completes the proof of the lemma.



3.2 Main Results Theorem 3.3 Suppose that M is a positive constant. The following statements (i), (ii), (iii) and (iv) are equivalent: (i) For any f (x) ≥ 0, satisfying 

∞

0< 0

we have the following inequality:

x p(1−σ)−1 f p (x)d x < ∞,

3.2 Main Results

49



∞

J :=

y pσ1 −1



0

∞

p h(x y) f (x)d x

0



∞

> M

x

p(1−σ)−1

f (x)d x p

 1p

 1p dy

.

(3.12)

0

(ii) For any g(y) ≥ 0, satisfying 

∞

0
M

y

q(1−σ1 )−1 q

g (y)dy

 q1

 q1 dx

.

(3.13)

0

(iii) For any f (x) ≥ 0, satisfying 

∞

0
0 such that k(σ ± δ0 ) < ∞, then we have σ1 = σ, and k(σ) ≥ M (> 0). Proof (i) ⇒ (iii). By the reverse Hölder inequality (cf. [1]), we have 



∞

I =

y σ1 − p 1

0



≥ J

∞

h(x y) f (x)d x

0 ∞

y 0



q(1−σ1 )−1 q

g (y)dy

 q1

.



1 y p −σ1 g(y) dy (3.15)

50

3 Equivalent Statements of the Reverse Hilbert-Type Integral Inequalities

Then by (3.12), we deduce (3.14). (ii) ⇒ (iii). By the reverse Hölder inequality, we also get that 

∞

I =

 x

1 q −σ

f (x)

 x

0

 ≥



σ− q1



∞

h(x y)g(y)dy d x 0

∞

x p(1−σ)−1 f p (x)d x

 1p

L.

(3.16)

0

Then by (3.13), we deduce (3.14). (iii) ⇒ (iv). By Lemma 3.1, we have σ1 = σ, and k(σ) ≥ M (> 0). (iv) ⇒ (i). For y > 0, setting u = x y, we obtain the following weight function: ω(σ, y) := y σ



∞

h(x y)x σ−1 d x =

0



∞

h(u)u σ−1 du = k(σ) (y ∈ R+ ).

(3.17)

0

By the reverse Hölder inequality with weight and (3.17), we obtain that 

≥ = =

p h(x y) f (x)d x

0

 p x (σ−1)/q dx y (σ−1)/ p 0  ∞  p/q  ∞ y σ−1 x σ−1 p h(x y) (σ−1) p/q f (x)d x h(x y) (σ−1)q/ p d x x y 0 0  ∞ σ−1  p−1  y ω(σ, y)y q(1−σ)−1 h(x y) (σ−1) p/q f p (x)d x x 0  ∞ y σ−1 (k(σ)) p−1 y − pσ+1 h(x y) (σ−1) p/q f p (x)d x. x 0 

=

∞

∞



y (σ−1)/ p h(x y) (σ−1)/q f (x) x



(3.18)

If (3.18) assumes the form of equality for some y ∈ (0, ∞), then (cf. [1]) there exist constants A and B, such that they are not all zero, and A

y σ−1 x (σ−1) p/q

f p (x) = B

x σ−1 y (σ−1)q/ p

a.e. in R+ .

Let us assume that A = 0 (otherwise B = A = 0). Then it follows that x p(1−σ)−1 f p (x) = y q(1−σ) which contradicts the fact that  0
(k(σ))



1 q

∞



h(x y) 0

= (k(σ))

∞



1 q

0



∞

y σ−1 x (σ−1) p/q

∞

h(x y) 

1

0

0 ∞

= (k(σ)) q

∞

= k(σ)



y σ−1

dy f (x)d x

ω(σ, x)x p(1−σ)−1 f p (x)d x

x p(1−σ)−1 f p (x)d x

 1p

1p

p

x (σ−1)( p−1)

0



f (x)d xd y

 1p

p

 1p

.

(3.19)

0

For k(σ) ≥ M > 0, (3.12) follows. (iii) ⇒ (ii). We set 

q−1

∞

f (x) := x qσ−1

h(x y)g(y)dy

, x > 0.

0

If L = ∞, then (3.13) is trivially valid; if L = 0, then it is impossible. Suppose that 0 < L < ∞. By (3.14), we have 

∞

∞> 0

x p(1−σ)−1 f p (x)d x = L q = I



∞

>M

x

 L=

p(1−σ)−1

f (x)d x p

0 ∞

x p(1−σ)−1 f p (x)d x

0

 q1

 1p 

∞

y

q(1−σ1 )−1 q

g (y)dy

 q1

0



∞

>M

y q(1−σ1 )−1 g q (y)dy

> 0,  q1

,

0

namely, (3.13) follows. Therefore, the statements (i), (ii), (iii) and (iv) are equivalent. This completes the proof of the theorem.



For σ1 = σ, we also have the following theorem: Theorem 3.4 If 0 < k(σ) < ∞, then the following statements (i), (ii) and (iii) are equivalent: (i) For any f (x) ≥ 0, satisfying 

∞

0< 0

we have the following inequality:

x p(1−σ)−1 f p (x)d x < ∞,

52

3 Equivalent Statements of the Reverse Hilbert-Type Integral Inequalities



∞

0



∞

y pσ−1

p h(x y) f (x)d x

0



∞

> k(σ)

x

p(1−σ)−1

f (x)d x p

 1p

 1p dy

.

(3.20)

0

(ii) For any g(y) ≥ 0, satisfying 

∞

0
k(σ)

y

g (y)dy

q(1−σ)−1 q

 q1

 q1 dx

.

(3.21)

0

(iii) For any f (x) ≥ 0, satisfying 

∞

0
0 such that k(σ − δ0 ) < ∞ or k(σ + δ0 ) < ∞ , then the constant factor k(σ) in (3.20), (3.21) and (3.22) is the best possible. Proof For σ1 = σ in Theorem 3.3, since 0 < k(σ) < ∞, setting M = k(σ) in (3.12), (3.13) and (3.14), we can similarly prove that statements (i), (ii) and (iii) of Theorem 3.4 are equivalent.

3.2 Main Results

53

If there exists a constant M ≥ k(σ), such that (3.22) is valid, then by Lemma 3.2, we have M ≤ k(σ). Hence, the constant factor M = k(σ) in (3.22) is the best possible. The constant factor k(σ) in (3.20) (3.21) is still the best possible. Otherwise, by (3.15) ((3.16)) for σ1 = σ, we can conclude that the constant factor M = k(σ) in (3.22) is not the best possible. This completes the proof of the theorem. 

3.3 Some Corollaries and a Few Examples In particular, for σ =

1 p

(> 1) in Theorem 3.4, we have:

Corollary 3.5 If 0k x f (x)d x . p 0

(3.23)

(ii) For any g(y) ≥ 0, satisfying 

∞

0
k g (y)dy . p 0

(iii) For any f (x) ≥ 0, satisfying  0
k

∞

h(x y) f (x)g(y)d xd y

0

   ∞  1p  ∞  q1 1 x p−2 f p (x)d x g q (y)dy . p 0 0

(3.25)

Moreover, if there exists a constant δ0 > 0 such that  k

1 − δ0 p



 < ∞ or k

1 + δ0 p

 k y g (y)dy . q 0

(iii) For any f (x) ≥ 0, satisfying  0< 0

∞

f p (x)d x < ∞,

(3.27)

3.3 Some Corollaries and a Few Examples

55

and g(y) ≥ 0, satisfying 

∞

0
k f (x)d x y g (y)dy . q 0 0

(3.28)

Moreover, if there exists a constant δ0 > 0, such that  k

1 − δ0 q



 < ∞ or k

1 + δ0 q

 M x f (x)d x . 

∞

− pσ1 −1

∞

0

(ii) For any G(y) ≥ 0, satisfying 

∞

0< 0

we have the following inequality:

y q(1+σ1 )−1 G q (y)dy < ∞,

(3.29)

56

3 Equivalent Statements of the Reverse Hilbert-Type Integral Inequalities





q  q1   x G(y)dy d x y 0 0  ∞  q1 q(1+σ1 )−1 q >M y G (y)dy . ∞

x qσ−1

∞

h

(3.30)

0

(iii) For any f (x) ≥ 0, satisfying 

∞

0
M

x p(1−σ)−1 f p (x)d x

 1p

 1p dy

;

(3.32)

0

(ii) For any g(y) ≥ 0, satisfying 

∞

0
M

y q(1−μ1 )−1 g q (y)dy

 q1

 q1 dx

.

(3.33)

0

(iii) For any f (x) ≥ 0, satisfying 

∞

0
0, such that kλ (σ ± δ0 ) < ∞, then μ1 = μ and kλ (σ) ≥ M (> 0) . For μ1 = μ in Corollary 3.8, we also derive the corollary bellow:

.

(3.34)

58

3 Equivalent Statements of the Reverse Hilbert-Type Integral Inequalities

Corollary 3.9 If 0 < kλ (σ) < ∞, then the following statements (i), (ii) and (iii) are equivalent: (i) For any f (x) ≥ 0, satisfying 

∞

0
kλ (σ)

x p(1−σ)−1 f p (x)d x

 1p

 1p dy

.

(3.35)

0

(ii) For any g(y) ≥ 0, satisfying 

∞

0
kλ (σ)

y q(1−μ)−1 g q (y)dy

 q1

 q1 dx

.

(3.36)

0

(iii) For any f (x) ≥ 0, satisfying 

∞

0
0, such that kλ (σ − δ0 ) < ∞ or kλ (σ + δ0 ) < ∞, then the constant factor kλ (σ) in (3.35), (3.36) and (3.37) is the best possible. In particular, for λ = 1, σ = following corollary:

(< 0), μ =

1 q

1 p

in Corollary 3.9, we derive the

Corollary 3.10 If    ∞ 1 1 0 < k1 = k1 (u, 1)u − p du < ∞, q 0 then the following statements (i), (ii) and (iii) are equivalent: (i) For any f (x) ≥ 0, satisfying 

∞

0
k1 f (x)d x . q 0

(3.38)

(ii) For any g(y) ≥ 0, satisfying 

∞

0
k1 g (y)dy . q 0

(iii) For any f (x) ≥ 0, satisfying 

∞

0
k1 f (x)d x g (y)dy . q 0 0

(3.40)

Moreover, if there exists a constant δ0 > 0, such that  k1

1 − δ0 q



 < ∞ or k1

1 + δ0 q

 < ∞,

then the constant factor k1 ( q1 ) in (3.38), (3.39) and (3.40) is the best possible. For λ = 1, σ = lary bellow:

1 p

(> 1), μ =

1 q

in Corollary 3.10, we can also deduce the corol-

Corollary 3.11 If    ∞ 1 1 0 < k1 = k1 (u, 1)u − q du < ∞, p 0 then the following statements (i), (ii) and (iii) are equivalent: (i) For any f (x) ≥ 0, satisfying 

∞

0
k1 x f (x)d x . p 0

(3.41)

(ii) For any g(y) ≥ 0, satisfying 

∞

0
k1 y g (y)dy . p 0

(3.42)

3.3 Some Corollaries and a Few Examples

61

(iii) For any f (x) ≥ 0, satisfying 

∞

0
0, such that  k1

1 − δ0 p



 < ∞, or k1

1 + δ0 p

 < ∞,

then the constant factor k1 ( 1p ) in (3.41), (3.42) and (3.43) is the best possible. Example 3.12 Setting h(u) = k0 (u, 1) = eαu csc h(u) =

2eαu (u > 0), eu − e−u

where csch(u) stands for the hyperbolic cosecant function, then we get that 2eαx y , e x y − e−x y 2eαx/y x csc h( ) = x/y , y e − e−x/y

h(x y) = eαx y csch(x y) = k0 (x, y) = e

αx y

and for α < 1, σ > 1,

 ∞ k(σ) = k0 (σ) = eαu csch(u)u σ−1 du 0  ∞ σ−1 (α−1)u  ∞ αu σ−1 2e u 2u e du = du = u −u e −e 1 − e−2u 0 0  ∞ ∞  =2 u σ−1 e−(2k+1−α)u du 0

=2

∞  ∞  k=0 0

k=0

u σ−1 e−(2k+1−α)u du.

62

3 Equivalent Statements of the Reverse Hilbert-Type Integral Inequalities

Setting v = (2k + 1 − α)u in the previous integral, we obtain that  k(σ) = k0 (σ) = 2

∞

v σ−1 e−v dv

0

Setting δ0 =

σ−1 2

∞  k=0

1 (2k + 1 − α)σ

∞ 

=

1 1 (σ) 2σ−1 (k + 1−α )σ 2 k=0

=

1 1−α ) ∈ R+ . (σ)ζ(σ, 2σ−1 2

> 0, σ ± δ0 ≥ σ −

σ+1 σ−1 = > 1, 2 2

we have k(σ ± δ0 ) = k0 (σ ± δ0 ) < ∞, and then we can use the fact that h(u) = k0 (u, 1) = eαu csch(u) in Theorem 3.4 and Corollary 3.9 in order to obtain some equivalent inequalities with the best possible constant factor. Example 3.13 Setting h(u) = k0 (u, 1) = then we find h(x y) =

1 (u > 0), |u − 1|λ

1 1 , kλ (x, y) = , λ |x y − 1| |x − y|λ

and for σ, μ > 0, 0 < λ < 1, 

∞ 1 k(σ) = kλ (σ) = u σ−1 du |u − 1|λ 0  ∞  1 u σ−1 u σ−1 du + du = λ (u − 1)λ 0 (1 − u) 1  1  1 u σ−1 v μ−1 = du + dv λ λ 0 (1 − u) 0 (1 − v) = B(1 − λ, σ) + B(1 − λ, μ) ∈ R+ .

Setting δ0 = min{ σ2 , μ2 } > 0,

3.3 Some Corollaries and a Few Examples

63

σ σ = > 0, 2 2 μ μ μ ± δ0 ≥ μ − = > 0, 2 2 σ ± δ0 ≥ σ −

we have k(σ ± δ0 ) = kλ (σ ± δ0 ) < ∞, and then we can use the fact that h(u) = kλ (u, 1) =

1 |u − 1|λ

in Theorem 3.4 and Corollary 3.9 in order to obtain some equivalent inequalities with the best possible constant factor. Example 3.14 Setting h(u) = kλ (u, 1) =

1 (u > 0), (u + 1)λ

we then get that h(x y) =

1 1 , kλ (x, y) = , (x y + 1)λ (x + y)λ

and for σ, μ > 0, 

∞

1 u σ−1 du (u + 1)λ 0 = B(σ, μ) ∈ R+ .

k(σ) = kλ (σ) =

Setting δ0 = min{ σ2 , μ2 } > 0, σ σ = > 0, 2 2 μ μ μ ± δ0 ≥ μ − = > 0, 2 2 σ ± δ0 ≥ σ −

we have k(σ ± δ0 ) = kλ (σ ± δ0 ) < ∞, and we can then use the fact that h(u) = kλ (u, 1) =

1 (u + 1)λ

in Theorem 3.4 and Corollary 3.9 in order to obtain some equivalent inequalities with the best possible constant factor.

64

3 Equivalent Statements of the Reverse Hilbert-Type Integral Inequalities

3.4 Some Reverse Equivalent Hilbert-Type Inequalities in the Whole Plane For a, b ∈ R\{0}, if we replace x (resp. y) by eax (resp. eby ), then replacing f (eax )eax (resp. g(eby )eby ) by f (x) (resp. g(y)), and |a|1/qM|b|1/ p by M in Theorem 3.3, and by carrying out the corresponding simplifications, we get the following theorem: Theorem 3.15 Let M be a positive constant. The following statements (i), (ii), (iii) and (iv) are equivalent: (i) For any f (x) ≥ 0, satisfying  0
M

∞

p

∞

pσ1 by

h(e −∞



f (x) eσax

−∞

p

ax+by

 1p dx

) f (x)d x

 1p dy

.

(3.44)

(ii) For any g(y) ≥ 0, satisfying  0
M

∞

eqσax ∞



−∞

−∞

g(y) eσ1 by

q h(eax+by )g(y)dy

q

 1p dy

.

(3.45)

(iii) For any f (x) ≥ 0, satisfying  0
0, such that k(σ ± δ0 ) < ∞, then we have k(σ) σ1 = σ and 1/q 1/ p ≥ M (> 0). |a| |b| For σ1 = σ, we derive the following theorem: Theorem 3.16 If 0 < k(σ) < ∞, then the following statements (i), (ii) and (iii) are equivalent: (i) For any f (x) ≥ 0, satisfying  0
1/q |a| |b|1/ p



∞



ax+by

−∞

f (x) eσax



q

) f (x)d x

p

 1p dx

 1p dy

.

(3.47)

(ii) For any g(y) ≥ 0, satisfying  0


∞

eqσax

k(σ) |a|1/q |b|1/ p

−∞



∞ −∞

(iii) For any f (x) ≥ 0, satisfying

q h(eax+by )g(y)dy 

g(y) eσby

q

 1p dy

.

 q1 dx (3.48)

66

3 Equivalent Statements of the Reverse Hilbert-Type Integral Inequalities

 0
0, such that k(σ − δ0 ) < ∞ or k(σ + δ0 ) < ∞, then the constant factor k(σ) |a|1/q |b|1/ p in (3.47), (3.48) and (3.49) is the best possible. In particular, for

h(u) = eαu csch(u),

by Example 3.12, we have the following equivalent inequalities with the best possible constant factor   1−α 21−σ (σ) ζ σ, |a|1/q |b|1/ p 2 (α < 1, σ > 1) : 

∞ −∞

 e pσby

∞

−∞

eα(e

ax+by

)

p csch(eax+by ) f (x)d x

   ∞    1p f (x) p 21−σ (σ) 1−α > ζ σ, dx , |a|1/q |b|1/ p 2 eσax −∞

 1p dy (3.50)

3.4 Some Reverse Equivalent Hilbert-Type Inequalities in the Whole Plane



∞



∞

eqσax

−∞

eα(e

ax+by

−∞



2 (σ) 1−α > ζ σ, |a|1/q |b|1/ p 2 1−σ



∞



∞

eα(e

ax+by

)

67

q csch(eax+by )g(y)dy

 

∞



−∞

g(y) eσby

q

 1p dy

 q1 dx

,

)

csch(eax+by ) f (x)g(y)d xd y   21−σ (σ) 1−α > ζ σ, |a|1/q |b|1/ p 2  p  1p  ∞    q1  ∞  f (x) g(y) q dx dy . × eσax eσby −∞ −∞ 0

(3.51)

0

(3.52)

For a, b ∈ R\{0}, replacing x (resp. y) by eax (resp. eby ), then replacing f (eax )eax (resp. g(eby )eby ) by f (x) (resp. g(y)), and |a|1/qM|b|1/ p by M in Corollary 3.8, and by carrying out the corresponding simplifications, we get the following corollary: Corollary 3.17 Suppose that M is a positive constant. The following statements (i), (ii), (iii) and (iv) are equivalent: (i) For any f (x) ≥ 0, satisfying  0
M

∞

pμ1 by

∞

−∞



−∞

f (x) eσax

p kλ (e , e ) f (x)d x ax

p

 1p dx

by

.

0
M

∞

eqσax 

∞ −∞

−∞

q kλ (eax , eby )g(y)dy

q

g(y) eμ1 by

 1p dy

 q1 dx

.

(3.54)

(iii) For any f (x) ≥ 0, satisfying  0
0, such that kλ (σ ± δ0 ) < ∞, then we have μ1 = μ and

kλ (σ) ≥ M (> 0). |a|1/q |b|1/ p

For μ1 = μ, we obtain the following corollary: Corollary 3.18 If 0 < kλ (σ) < ∞, then the following statements (i), (ii) and (iii) are equivalent: (i) For any f (x) ≥ 0, satisfying  0
|a|1/q |b|1/ p

p

∞

e pμby

−∞ ∞

kλ (eax , eby ) f (x)d x 

f (x) eσax

−∞

p

 1p dx

69

 1p dy

.

(3.56)

(ii) For any g(y) ≥ 0, satisfying  0
|a|1/q |b|1/ p

∞

q kλ (e , e )g(y)dy ax



−∞

g(y) eμby

by

q

 1p dy

 q1 dx

.

(3.57)

(iii) For any f (x) ≥ 0, satisfying  0
0, such that kλ (σ − δ0 ) < ∞ or kλ (σ + δ0 ) < ∞, then the constant factor

kλ (σ) |a|1/q |b|1/ p

in (3.56), (3.57) and (3.58) is the best possible.

q

 q1 dy

.

(3.58)

70

3 Equivalent Statements of the Reverse Hilbert-Type Integral Inequalities

In particular, for k0 (x, y) = e

αx y

x csc h( ), y

by Example 3.12, we have the following equivalent inequalities with the best possible constant factor   21−σ (σ) 1−α K 0 (σ) := ζ σ, |a|1/q |b|1/ p 2 (α < 1, σ > 1) : 

 p  1p x e e csc h( ) f (x)d x dy y −∞ −∞    1p ∞  f (x) p 21−σ (σ) 1−α ) > ζ(σ, dx , |a|1/q |b|1/ p 2 eσax −∞ ∞



− pσby



∞

αx y

q  q1 x csc h( )g(y)dy d x y −∞ −∞    1p ∞  g(y) q 21−σ (σ) 1−α ) > ζ(σ, dy , −σby |a|1/q |b|1/ p 2 −∞ e 



∞

∞



eqσax

∞

∞

e

αx y

x csc h( ) f (x)g(y)d xd y y −∞ −∞ 1−σ 2 (σ) 1−α ) > ζ(σ, |a|1/q |b|1/ p 2  ∞     1p  ∞   q1 f (x) p g(y) q × dx dy . −σby eσax −∞ −∞ e e

(3.59)

(3.60)

αx y

(3.61)

References 1. Kuang, J.C.: Applied Inequalities. Shangdong Science and Technology Press, Jinan, China (2004) 2. Kuang, J.C.: Real and Functional Analysis (continuation) (sec. vol.). Higher Education Press, Beijing, China (2015)

Chapter 4

Equivalent Statements of Two Kinds of Hardy-Type Integral Inequalities

In this chapter, we consider a few equivalent statements of two kinds of Hardy-type integral inequalities with a nonhomogeneous kernel related to certain parameters. Two kinds of Hardy-type integral inequalities with a homogeneous kernel are deduced. We also consider the operator expressions, some corollaries and a few particular examples involving the Hurwitz zeta function in the form of applications, as well as two kinds of Hardy-type integral inequalities in the whole plane.

4.1 Lemmas Throughout this chapter we shall assume that: p > 1, 1p + q1 = 1, σ1 , σ, μ ∈ R, σ + μ = λ, and h(u) is a nonnegative measurable function in (0, ∞), such that 

1

k1 (σ) = k2 (σ) =

h(u)u σ−1 du (≥ 0),

0 ∞

h(u)u σ−1 du (≥ 0).

(4.1)

1

Lemma 4.1 If k1 (σ) > 0, and if there exists a constant M1 such that for any nonnegative measurable functions f (x) and g(y) in (0, ∞), the following inequality 



∞

1 y

g(y) 0

 h(x y) f (x)d x dy

0

 ≤ M1

∞

x 0

p(1−σ)−1

f (x)d x p

 1p 

∞

y

q(1−σ1 )−1 q

g (y)dy

 q1 (4.2)

0

© The Author(s), under exclusive licence to Springer Nature Switzerland AG 2019 B. Yang and M. Th. Rassias, On Hilbert-Type and Hardy-Type Integral Inequalities and Applications, SpringerBriefs in Mathematics, https://doi.org/10.1007/978-3-030-29268-3_4

71

72

4 Equivalent Statements of Two Kinds of Hardy-Type Integral Inequalities

holds true, then we have σ1 = σ and k1 (σ) ≤ M1 (< ∞). Proof Since k1 (σ) > 0, it follows that h(u) > 0 a.e. in an interval I (⊂ (0, 1)). If σ1 > σ, then for n ≥ σ11−σ (n ∈ N), we set the following functions:  f n (x) =  gn (y) =

x σ+ pn −1 , 0 < x ≤ 1 , 0, x > 1 1

0, 0 < y < 1 . , y≥1 y 1 σ1 − qn −1

We get 

∞

J1 =

x

p(1−σ)−1

0



1

=

f np (x)d x

 1p 

x p(1−σ)−1 x p(σ+ pn −1) d x 1

 1p

∞ 0

0



∞

×

y

1 −1) q(1−σ1 )−1 q(σ1 − qn

y

1



1

=

x

1 n −1

1p 

∞

dx

y

0

y q(1−σ1 )−1 gnq (y)dy

 q1

 q1 dy

− n1 −1

q1 dy

= n.

1

For fixed y > 0, setting u = x y, we obtain 

∞

I1 := 0



∞

=



1 y

gn (y) 

 h(x y) f n (x)d x dy

0



1 y

h(x y)x 

0

1 ∞

=

1 σ+ pn −1

y

(σ1 −σ)− n1 −1



1

dy

1

dx

y σ1 − qn −1 dy 1

h(u)u σ+ pn −1 du, 1

0

and then by (4.2) we have 

∞ 1

y (σ1 −σ)− n −1 dy 1



1

h(u)u σ+ pn −1 du 1

0

= I1 ≤ M1 J1 = M1 n < ∞.

(4.3)

4.1 Lemmas

73

Since (σ1 − σ) −

≥ 0, it follows that

1 n



∞

y (σ1 −σ)− n −1 dy = ∞. 1

1

By (4.3), since h(u)u σ+ pn −1 > 0 a.e. in an interval I ⊂ (0, 1), 1

we obtain that



1

h(u)u σ+ pn −1 du > 0 1

0

and therefore ∞ ≤ M1 n < ∞, which is a contradiction. If σ1 < σ, then for n ≥

1 σ−σ1

(n ∈ N), we set the following functions:

f n (x) =

 

gn (y) =

0, 0 < x < 1 1 , x σ− pn −1 , x ≥ 1 y σ1 + qn −1 , 0 < y ≤ 1 , 0, y > 1 1

and deduce that J 1 =



∞

p(1−σ)−1

x 0



∞

=

f np (x)d x

 1p 

∞ 0

x p(1−σ)−1 x p(σ− pn −1) d x 1

 1p

1

 ×

1

y

1 −1) q(1−σ1 )−1 q(σ1 + qn

y

0



∞

=

x

− n1 −1

1p 

1

dx

y

1

y q(1−σ1 )−1

gnq (y)dy

 q1 dy

1 n −1

q1 dy

= n.

0

For fixed x > 0, setting u = x y, we obtain

I1 :=



∞ 0



∞

=

f n (x) 



1 x

 h(x y)

gn (y)dy d x

0 1 x

h(x y)y 1

0

 1 σ1 + qn −1

dy x σ− pn −1 d x 1

 q1

74

4 Equivalent Statements of Two Kinds of Hardy-Type Integral Inequalities



∞

=

(σ−σ1 )− n1 −1

x



1

dx

1

h(u)u σ1 + qn −1 du, 1

0

and then by Fubini’s theorem (cf. [1]) and (4.2), we get that 

∞

1

=

I1 =



1

∞

gn (y)

0

1

h(u)u σ1 + qn −1 du 0  1 y

h(x y) f n (x)d x dy

x (σ−σ1 )− n −1 d x  

1

0

≤ M1 J 1 = M1 n. Since (σ − σ1 ) −

1 n

(4.4)

≥ 0, it follows that 

∞

x (σ−σ1 )− n −1 d x = ∞. 1

1

By (4.4), since h(u)u σ1 + qn −1 > 0 a.e. in an interval I ⊂ (0, 1), 1

it follows that



1

h(u)u σ1 + qn −1 du > 0, 1

0

and thus ∞ ≤ M1 n < ∞, which is a contradiction. Hence, we conclude that σ1 = σ. For σ1 = σ, we reduce (4.4) as follows: 

1

h(u)u σ+ qn −1 du ≤ M1 . 1

(4.5)

0

Since {h(u)u σ+ qn −1 }∞ n=1 is nonnegative and increasing in (0, 1], by Levi’s theorem (cf. [1]), we have 1

 k1 (σ) =

1

lim h(u)u σ+ qn −1 du

0 n→∞  1

= lim

n→∞ 0

1

h(u)u σ+ qn −1 du ≤ M1 (< ∞).

This completes the proof of the lemma.

1



4.1 Lemmas

75

Lemma 4.2 If k2 (σ) > 0, and if there exists a constant M2 such that for any nonnegative measurable functions f (x) and g(y) in (0, ∞), the following inequality 



∞

∞

g(y) 1 y

0



∞

≤ M2

x

 h(x y) f (x)d x dy

p(1−σ)−1

 1p 

f (x)d x p

∞

y

0

q(1−σ1 )−1 q

g (y)dy

 q1 (4.6)

0

holds true, then we have σ1 = σ and k2 (σ) ≤ M2 (< ∞). Proof Since k2 (σ) > 0, it follows that h(u) > 0 a.e. in an interval I1 (⊂ (1, ∞)). 1 (n ∈ N), we set two functions

f n (x) and

gn (y), as If σ1 < σ, then for n ≥ σ−σ 1 in Lemma 4.1, and find J 1 =



∞ 0

x p(1−σ)−1

f np (x)d x

Fix y > 0, setting u = x y, we obtain   ∞



gn (y) I2 := 0

=

 1 

 1p  0

∞ 1 y

∞

h(x y)x 

1

=

y (σ1 −σ)+ n −1 dy 1



∞

dx

1

h(u)u σ− pn −1 du, 1

1

y (σ1 −σ)+ n −1 dy 1

0

1 n



∞

h(u)u σ− pn −1 du 1

1

≤ 0, it follows that 

1

y (σ1 −σ)+ n −1 dy = ∞. 1

0

By (4.7), since h(u)u σ− pn −1 > 0 a.e. in an interval I1 ⊂ (1, ∞), 1

= n.

y σ1 + qn −1 dy

=

I2 ≤ M2 J 1 = M2 n < ∞. Since (σ1 − σ) +

 q1

 1 σ− pn −1

0

and then by (4.6) we have  1

y q(1−σ1 )−1

gnq (y)dy



h(x y) f n (x)d x dy

1 y

0

∞

(4.7)

76

4 Equivalent Statements of Two Kinds of Hardy-Type Integral Inequalities



we find that

∞

h(u)u σ− pn −1 du > 0, 1

1

and therefore ∞ ≤ M2 n < ∞, which is a contradiction. If σ1 > σ, then for n ≥ σ11−σ (n ∈ N), we set two functions f n (x) and gn (y) as in Lemma 4.1 and find  ∞  1p  ∞  q1 p(1−σ)−1 p q(1−σ1 )−1 q J1 = x f n (x)d x y gn (y)dy = n. 0

0

For fixed x > 0, setting u = x y, we obtain   ∞

I2 := 0



1

=

∞

f n (x)



1 x

1 σ1 − qn −1

1 x

0 1

=

h(x y)gn (y)dy d x 

∞

h(x y)y 



x (σ−σ1 )+ n −1 d x 1



0

∞

dy x σ+ pn −1 d x 1

h(u)u σ1 − qn −1 du, 1

1

and then by Fubini’s theorem (cf. [1]) and (4.6), we obtain that  ∞  1 1 1 x (σ−σ1 )+ n −1 d x h(u)u σ1 − qn −1 du 0  1   ∞

= I2 =

gn (y)

0

∞

1 y

h(x y) f n (x)d x dy

≤ M2 J1 = M2 n < ∞. Since (σ − σ1 ) +

1 n

(4.8)

≤ 0, it follows that 

1

x (σ−σ1 )+ n −1 d x = ∞. 1

0

By (4.8), in view of the fact that h(u)u σ1 − qn −1 > 0 a.e. in an interval I1 ⊂ (1, ∞), 1

we derive that

 1

∞

h(u)u σ1 − qn −1 du > 0, 1

4.1 Lemmas

77

and hence ∞ ≤ M2 n < ∞, which is a contradiction. Hence, we conclude the fact that σ1 = σ. For σ1 = σ, we reduce (4.8) as follows: 

∞

h(u)u σ− qn −1 du ≤ M2 . 1

(4.9)

1

Since {h(u)u σ− qn −1 }∞ n=1 is nonnegative and increasing in [1, ∞), applying again Levi’s theorem (cf. [1]), we obtain that 1

 k2 (σ) =

∞

lim h(u)u σ− qn −1 du 1

n→∞  ∞

1

= lim

n→∞ 1

h(u)u σ− qn −1 du ≤ M2 (< ∞). 1



This completes the proof of the lemma.

4.2 Hardy-Type Integral Inequalities of the First Kind Theorem 4.3 Let M1 be a constant. If k1 (σ) > 0, then the following statements (i), (ii) and (iii) are equivalent: (i) For any f (x) ≥ 0, satisfying 

∞

0
0).

0

By Hölder’s inequality with weight and (4.13), for y ∈ (0, ∞), we have 

1 y

p h(x y) f (x)d x

0

  (σ−1)/q   p y (σ−1)/ p x dx = h(x y) (σ−1)/q f (x) x y (σ−1)/ p 0  1 p−1  1 y y y σ−1 f p (x) x σ−1 ≤ h(x y) (σ−1) p/q d x h(x y) (σ−1)q/ p d x x y 0 0 

1 y



(4.13)

4.2 Hardy-Type Integral Inequalities of the First Kind

 p−1  = ω1 (σ, y)y q(1−σ)−1 = (k1 (σ)) p−1 y − pσ+1



79

h(x y) 0



y σ−1

1 y

y σ−1

1 y

h(x y) 0

x (σ−1) p/q

f p (x)d x

f p (x)d x.

x (σ−1) p/q

(4.14)

If (4.14) takes the form of equality for some y ∈ (0, ∞), then (cf. [2]) there exist constants A and B, such that they are not all zero, and A

y σ−1 x (σ−1) p/q

f p (x) = B

x σ−1 y (σ−1)q/ p

a.e. in R+ .

Let us suppose that A = 0 (otherwise B = A = 0). It follows that x p(1−σ)−1 f p (x) = y q(1−σ) which contradicts the fact that  0
0, then the following statements (i), (ii) and (iii) are equivalent: (i) For any f (x) ≥ 0, satisfying 

∞

0
0), |u − 1|(max{u, 1})λ

96

4 Equivalent Statements of Two Kinds of Hardy-Type Integral Inequalities

we find | ln x y|β min{x y, 1} , |x y − 1|(max{x y, 1})λ | ln xy |β min{x, y} kλ (x, y) = , |x − y|(max{x, y})λ h(x y) =

and for σ > −1, by the Lebesgue term by term theorem, we obtain k1 (σ) =

kλ(1) (σ) 

 =

1

h(u)u σ−1 du

0 β

 1 ∞  (− ln u) σ u du = (− ln u)β u k+σ du 1−u 0 0 k=0 ∞  1  (− ln u)β u k+σ du. = =

1

k=0

0

Setting v = (k + σ + 1)(− ln u) in the previous integral, we obtain that k1 (σ) =

kλ(1) (σ)

=

∞  k=0

1 (k + σ + 1)β+1



∞

vβ e−v dv

0

= (β + 1)ζ(β + 1, σ + 1) ∈ R+ . Then we have

||T1(1) || = ||T1(2) || = (β + 1)ζ(β + 1, σ + 1).

(4.47)

For μ > −1, setting v = u1 , it follows that k2 (σ) =

kλ(2) (σ) 



∞

= 1

(ln u)β σ−1 u du (u − 1)u λ

(− ln v)β μ = v dv 1−v 0 = (β + 1)ζ(β + 1, μ + 1) ∈ R+ . Then we get

1

||T2(1) || = ||T2(2) || = (β + 1)ζ(β + 1, μ + 1).

(4.48)

Note. If β > 0, σ, μ > −1, then λ = σ + μ > −2, and ki (σ) = kλ(i) (σ) ∈ R+ (i = 1, 2).

4.5 Hardy-Type Integral Inequalities with the Exponent Function …

97

4.5 Hardy-Type Integral Inequalities with the Exponent Function as Interval Variables For a ∈ R\{0}, b > 0, replacing x (resp. y) by eax (resp. eby ), then replacing f (eax )eax (resp. g(eby )eby ) by f (x) (resp. g(y)), and |a|1/qMb1/ p by M in Theorem 4.3, Corollary 4.7, Theorem 4.10 and Corollary 4.14, and by carrying out the corresponding calculations, we obtain the following theorem: Theorem 4.29 Let M1 be a constant, and a = 0, b > 0. If k1 (σ) > 0, then the following statements (i), (ii) and (iii) are equivalent: (i) For any f (x) ≥ 0, satisfying  0
−α − γ), by Example 4.27, we have the following equivalent inequalities with  α+γ+σ (β + 1) ζ β + 1, |a|1/q b1/ p (λ + α)β+1 λ+α being the best possible constant factor: 

p  1p β ax+by α+γ |ax + by| (min{e , 1}) f (x) e pσby d x dy (ax+by)(λ+α) − 1|(max{eax+by , 1})γ −∞ −∞ |e  ∞    1p f (x) p (β + 1) α+γ+σ < ζ β + 1, d x , (4.51) |a|1/q b1/ p (λ + α)β+1 λ+α eσax −∞ ∞



e−by



e−by

|ax + by|β (min{eax+by , 1})α+γ g(y) f (x)d x dy (ax+by)(λ+α) − 1|(max{eax+by , 1})γ −∞ |e −∞  (β + 1) α+γ+σ < ζ β + 1, |a|1/q b1/ p (λ + α)β+1 λ+α p  1p  ∞   q1  ∞  f (x) g(y) q dx dy . (4.52) × eσax eσby −∞ −∞ 

∞

Remark 4.30 In Theorem 4.29, if σ1 = σ, b < 0, then replacing −b by b (> 0), we have the following equivalent inequalities with the best possible constant factor k1 (σ) : |a|1/q b1/ p 

∞

e

− pσby

−∞

< 

k1 (σ) |a|1/q b1/ p 

∞

∞

g(y) −∞


0, σ > −1), by Example 4.28, we have the following equivalent inequalities with (β + 1) ζ(β + 1, σ + 1) |a|1/q b1/ p being the best possible constant factor: 



∞

∞

|ax − by|β min{eax−by , 1} f (x) e− pσby dx ax−by − 1|(max{eax−by , 1})λ −∞ eby |e  ∞   1p f (x) p (β + 1) < ζ(β + 1, σ + 1) dx , |a|1/q b1/ p eσax −∞ 



∞

∞

g(y) −∞


0. If kλ(1) (σ) =



1

kλ (u, 1)u σ−1 du > 0,

0

then the following statements (i), (ii) and (iii) are equivalent: (i) For any f (x) ≥ 0, satisfying  0
−α − γ), by Example 4.27, we have the following equivalent inequalities with  α+γ+σ (β + 1) ζ β + 1, |a|1/q b1/ p (λ + α)β+1 λ+α being the best possible constant factor: 

p  1p β ax by α+γ |ax + by| (min{e , e }) f (x) e pμby d x dy (λ+α)ax − e(λ+α)by |(max{eax , eby })γ −∞ −∞ |e  ∞    1p f (x) p (β + 1) α+γ+σ < ζ β + 1, d x , (4.59) |a|1/q b1/ p (λ + α)β+1 λ+α eσax −∞ ∞



eby

4.5 Hardy-Type Integral Inequalities with the Exponent Function …

|ax + by|β (min{eax , eby })α+γ g(y) f (x)d x dy (λ+α)ax − e(λ+α)by |(max{eax , eby })γ −∞ −∞ |e  α+γ+σ (β + 1) ζ β + 1, < |a|1/q b1/ p (λ + α)β+1 λ+α p  1p  ∞   q1  ∞  f (x) g(y) q dx dy . (4.60) × eσax eμby −∞ −∞ 



101

∞

eby

Remark 4.32 In Corollary 4.31, if μ1 = μ, b < 0, then replacing −b by b, we have the following equivalent inequalities with the best possible constant factor 

∞

e −∞



k (1) (σ) < λ1/q 1/ p |a| b 

∞

e−by

kλ(1) (σ) < |a|1/q b1/ p

e−by



∞



f (x) eσax

kλ (e , e

∞ −∞



f (x) eσax

In particular, for kλ (u, 1) =

p

ax

−by

p

 1p

kλ (e , e

ax

g(y) −∞

∞

−∞



∞



− pμby

) f (x)d x

:

 1p dy

.

dx

−by

kλ(1) (σ) |a|1/q b1/ p

(4.61)

) f (x)d x dy

p

 1p 

∞

dx −∞



g(y) e−μby

q

 q1 dy

.

(4.62)

| ln u|β min{u, 1} |u − 1|(max{u, 1})λ

(β > 0, σ > −1), by Example 4.28, we have the following equivalent inequalities with (β + 1) ζ(β + 1, σ + 1) |a|1/q b1/ p being the best possible constant factor: 



∞

∞

|ax − by|β min{eax , e−by } f (x) e− pμby dx ax − e−by |(max{eax , e−by })λ −∞ e−by |e  ∞   1p f (x) p (β + 1) < ζ (β + 1, σ + 1) dx , |a|1/q b1/ p eσax −∞ 



∞

∞

g(y) −∞

e−by

 1p

p dy

 |ax − by|β min{eax , e−by } f (x)d x dy |eax − e−by |(max{eax , e−by })λ

(4.63)

102

4 Equivalent Statements of Two Kinds of Hardy-Type Integral Inequalities


0. If k2 (σ) > 0, then the following statements (i), (ii) and (iii) are equivalent: (i) For any f (x) ≥ 0, satisfying  0
−α − γ), by Example 4.27, we have the following equivalent inequalities with  α+γ+μ (β + 1) ζ β + 1, |a|1/q b1/ p (λ + α)β+1 λ+α being the best possible constant factor: 

 p  1p β ax+by α+γ |ax + by| (min{e , 1}) f (x) e pσby d x dy (ax+by)(λ+α) − 1|(max{eax+by , 1})γ −∞ e−by |e  ∞    1p f (x) p (β + 1) α+γ+μ < ζ β + 1, d x , (4.67) |a|1/q b1/ p (λ + α)β+1 λ+α eσax −∞ 

∞

∞

 |ax + by|β (min{eax+by , 1})α+γ f (x)d x dy (ax+by)(λ+α) − 1|(max{eax+by , 1})γ −∞ e−by |e  α+γ+μ (β + 1) ζ β + 1, < |a|1/q b1/ p (λ + α)β+1 λ+α p  1p  ∞   q1  ∞  f (x) g(y) q d x dy . (4.68) × eσax eσby −∞ −∞ 



∞

∞

g(y)

Remark 4.34 In Theorem 4.33, if σ1 = σ, b < 0, then replacing −b by b(> 0), we have the following equivalent inequalities with the best possible constant factor k2 (σ) : |a|1/q b1/ p 

∞

e

− pσby

−∞

k2 (σ) < |a|1/q b1/ p 



∞

eby


0, μ > −1), by Example 4.28, we have the following equivalent inequalities with (β + 1) ζ(β + 1, μ + 1) |a|1/q b1/ p being the best possible constant factor: 



∞

|ax − by|β min{eax−by , 1} e− pσby f (x)d x ax−by − 1|(max{eax−by , 1})λ −∞ −∞ |e  ∞   1p f (x) p (β + 1) < ζ (β + 1, μ + 1) dx , |a|1/q b1/ p eσax −∞ 



∞

eby

eby

g(y) −∞


0. If kλ(2) (σ) =



∞

kλ (u, 1)u σ−1 du > 0,

1

then the following statements (i), (ii) and (iii) are equivalent: (i) For any f (x) ≥ 0, satisfying  0
−α − γ), by Example 4.27, we have the following equivalent inequalities with  α+γ+μ (β + 1) ζ β + 1, |a|1/q b1/ p (λ + α)β+1 λ+α being the best possible constant factor: 

 p  1p |ax + by|β (min{eax , eby })α+γ f (x) e d x dy (λ+α)ax − e(λ+α)by |(max{eax , eby })γ −∞ eby |e  ∞    1p f (x) p (β + 1) α+γ+μ < ζ β + 1, d x , (4.75) |a|1/q b1/ p (λ + α)β+1 λ+α eσax −∞ ∞



pσby

∞

106

4 Equivalent Statements of Two Kinds of Hardy-Type Integral Inequalities

 |ax + by|β (min{eax , eby })α+γ g(y) f (x)d x dy (λ+α)ax − e(λ+α)by |(max{eax , eby })γ −∞ eby |e  α+γ+μ (β + 1) ζ β + 1, < |a|1/q b1/ p (λ + α)β+1 λ+α  q1  ∞  p  1p  ∞  f (x) g(y) q d x dy . (4.76) × eσax eμby −∞ −∞ 



∞

∞

Remark 4.36 In Corollary 4.35, if μ1 = μ, b < 0, then replacing −b by b(> 0), we have the following equivalent inequalities with the best possible constant factor kλ(2) (σ) |a|1/q b1/ p

: 

∞

e −∞

−∞ ∞

e−by

g(y) −∞

kλ(2) (σ) |a|1/q b1/ p


0, μ > −1), by Example 4.28, we have the following equivalent inequalities with (β + 1) ζ(β + 1, μ + 1) |a|1/q b1/ p being the best possible constant factor: 

∞

e −∞

− pμby



e−by −∞

|ax − by|β min{eax , e−by } f (x) dx |eax − e−by |(max{eax , e−by })λ

(β + 1) < ζ (β + 1, μ + 1) |a|1/q b1/ p



∞ −∞



f (x) eσax

p

 1p dx

,

p

 1p dy

(4.79)

4.5 Hardy-Type Integral Inequalities with the Exponent Function …





∞

e−by

g(y) −∞


0, such that for any nonnegative measurable functions f (x) and g(y) in (0, ∞), the following inequality  1   ∞

y

g(y)

h(x y) f (x)d x dy

0

0

 ≥ M1

∞

x 0

p(1−σ)−1

f (x)d x p

 1p 

∞

y

q(1−σ1 )−1 q

g (y)dy

 q1 (5.3)

0

© The Author(s), under exclusive licence to Springer Nature Switzerland AG 2019 B. Yang and M. Th. Rassias, On Hilbert-Type and Hardy-Type Integral Inequalities and Applications, SpringerBriefs in Mathematics, https://doi.org/10.1007/978-3-030-29268-3_5

109

110

5 Equivalent Property of the Reverse Hardy-Type Integral Inequalities

holds true, then we have σ1 = σ and k1 (σ) ≥ M1 (> 0). Proof If σ1 < σ, then for n ∈ N, we set the following functions:  σ+ 1 −1 pn ,0 < x ≤ 1 , f n (x) = x 0, x > 1  0, 0 < y < 1 1 , gn (y) = y σ1 − qn −1 , y ≥ 1 and find 

∞

J1 =

x

p(1−σ)−1

0



1

=

x

1 n −1

f np (x)d x

1p 

 1p 

∞

dx

y

0

∞ 0

− n1 −1

y q(1−σ1 )−1 gnq (y)dy

q1

dy

 q1

= n.

1

For fixed y, setting u = x y, we obtain  1  ∞

I1 = 0

 =

∞

y

gn (y) 

h(x y) f n (x)d x dy

0



1 y

h(x y)x 1

 =



1 σ+ pn −1

0 ∞

y

(σ1 −σ)− n1 −1

1

1 ≤ σ − σ1 +

 1 n

0

 dy

1

dx

y σ1 − qn −1 dy 1

h(u)u σ+ pn −1 du 1

0 1

h(u)u σ−1 du ≤

k1 (σ) , σ − σ1

and then by (5.3), it follows that k1 (σ) ≥ I1 ≥ M1 J1 = M1 n. σ − σ1

(5.4)

By (5.4), setting n → ∞, in view of k1 (σ) < ∞, σ − σ1 > 0 and M1 > 0, we find that k1 (σ) ≥ ∞, ∞> σ − σ1 which is a contradiction.

5.1 Two Lemmas

111

If σ1 > σ, then for

1 (n ∈ N), |q|(σ1 − σ)

n≥

we set the following two functions:  0, 0 < x < 1

1 , f n (x) = x σ− pn −1 , x ≥ 1  σ + 1 −1 1 qn ,0 < y ≤ 1 ,

gn (y) = y 0, y > 1 and get that J 1 =



∞

x 0



∞

=

x

f np (x)d x

p(1−σ)−1

− n1 −1

1p 

 1p  0

1

dx

y

1

∞

1 n −1

y q(1−σ1 )−1

gnq (y)dy

q1

dy

 q1

= n.

0

For fixed x > 0, setting u = x y, in view of σ1 +

I1 =



∞

0

 =

∞

f n (x)



1 x

h(x y)

gn (y)dy d x

1 x

h(x y)y 1

≥ σ, we obtain 

0



1 qn

 1 σ1 + qn −1

dy x σ− pn −1 d x 1

0

 1 1 1 x (σ−σ1 )− n −1 d x h(u)u σ1 + qn −1 du 1 0  1 1 k1 (σ) ≤ , h(u)u σ−1 du ≤ σ1 − σ σ1 − σ + n1 0 

=

∞

and then by Fubini’s theorem (cf. [1]) and (5.3), we derive that  1   ∞ y k1 (σ) ≥

I1 =

gn (y) h(x y)

f n (x)d x dy σ1 − σ 0 0 ≥ M1 J 1 = M1 n. By (5.5), setting n → ∞, we see that ∞> which is a contradiction.

k1 (σ) ≥ ∞, σ1 − σ

(5.5)

112

5 Equivalent Property of the Reverse Hardy-Type Integral Inequalities

Hence, we conclude that σ1 = σ. For σ1 = σ, we deduce I1 ≥ M1 J1 and then it follows that  1  1 1 h(u)u σ−1 du ≥ h(u)u σ+ pn −1 du ≥ M1 (> 0). k1 (σ) = 0

(5.6)

0



This completes the proof of the lemma.

Lemma 5.2 If k2 (σ) < ∞, and if there exists a constant M2 > 0, such that for any nonnegative measurable functions f (x) and g(y) in (0, ∞), the following inequality    ∞

∞

g(y) 1 y

0



∞

≥ M2

x

h(x y) f (x)d x dy

p(1−σ)−1

f (x)d x p

 1p 

∞

y

0

q(1−σ1 )−1 q

g (y)dy

 q1 (5.7)

0

holds true, then we have σ1 = σ and k2 (σ) ≥ M2 (> 0). f n (x) and

gn (y) as in Proof If σ1 > σ, then for n ∈ N, we consider two functions

Lemma 5.1 and get that  ∞  1p  ∞  q1 p(1−σ)−1 p q(1−σ1 )−1 q

J1 = x y

gn (y)dy = n. f n (x)d x 0

0

For fixed y > 0, setting u = x y, we obtain    ∞ ∞



gn (y) h(x y)

f n (x)d x dy I2 = 1 y

0

=

 1 

∞

h(x y)x 

1 y

0 1

=

 1 σ− pn −1

y (σ1 −σ)+ n −1 dy 1

0



∞

dx

y σ1 + qn −1 dy 1

h(u)u σ− pn −1 du ≤ 1

1

k2 (σ) , σ1 − σ

and then by (5.7), it follows that k2 (σ) ≥

I2 ≥ M2 J 1 = M2 n. σ1 − σ By (5.8), setting n → ∞, we derive that ∞> which is a contradiction.

k2 (σ) ≥ ∞, σ1 − σ

(5.8)

5.1 Two Lemmas

113

If σ1 < σ, then for

1 (n ∈ N), |q|(σ − σ1 )

n≥

we set two functions f n (x) and gn (y) as in Lemma 5.1 and get 

∞

J1 =

x

p(1−σ)−1

0

f np (x)d x

 1p 

∞

0

y q(1−σ1 )−1 gnq (y)dy

 q1

= n.

For fixed x > 0, setting u = x y, we obtain  I2 =



∞

f n (x)

1 x

0

 =

1



∞

 h(x y)gn (y)dy d x 

∞

h(x y)y

=

dy x σ+ pn −1 d x 1

1 x

0



1 σ1 − qn −1

1

x

(σ−σ1 )+ n1 −1



∞

dx

0

h(u)u σ1 − qn −1 du ≤ 1

1

k2 (σ) , σ − σ1

and then by Fubini’s theorem (cf. [1]) and (5.7), it follows that k2 (σ) ≥ I2 = σ − σ1



∞

 gn (y)

∞ 1 y

0

 h(x y) f n (x)d x dy

≥ M2 J1 = M2 n.

(5.9)

By (5.9), setting n → ∞, it follows that ∞> which is a contradiction. Hence, we conclude that σ1 = σ. For σ1 = σ, we deduce that

k2 (σ) ≥ ∞, σ − σ1

I2 ≥ M2 J 2

and thus it follows that  ∞  σ−1 h(u)u du ≥ k2 (σ) = 1

∞

h(u)u σ− pn −1 ≥ M2 (> 0). 1

(5.10)

1

This completes the proof of the lemma.



114

5 Equivalent Property of the Reverse Hardy-Type Integral Inequalities

5.2 Reverse Hardy-Type Integral Inequalities of the First Kind Theorem 5.3 Let M1 be a positive constant. If k1 (σ) < ∞, then the following statements (i), (ii) and (iii) are equivalent: (i) For any f (x) ≥ 0, satisfying 

∞

0
M1

x

p(1−σ)−1

f (x)d x p

 1p

.

(5.11)

0

(ii) For any g(y) ≥ 0, satisfying 

∞

0
M1

y q(1−σ1 )−1 g q (y)dy

 q1

.

0

(iii) For any f (x) ≥ 0, satisfying 

∞

0
M1

x

f (x)d x

p(1−σ)−1

p

 1p 

∞

y

0

q(1−σ1 )−1 q

g (y)dy

 q1

.

(5.13)

0

(iv) σ1 = σ, and k1 (σ) ≥ M1 (> 0). If statement (iv) holds true, then the constant M1 = k1 (σ) (∈ R+ ) in (5.11), (5.12) and (5.13) is the best possible. Proof (i) ⇒ (iii). By the reverse Hölder inequality (cf. [2]), we have 



∞

I =

y



σ1 − 1p



1 y

h(x y) f (x)d x

 1 y p −σ1 g(y) dy

0

0



∞

≥ J

y

q(1−σ1 )−1 q

g (y)dy

 q1

.

(5.14)

0

Then by (5.11), we have (5.12). (iii) ⇒ (iv). By Lemma 5.1, we have σ1 = σ, and k1 (σ) ≥ M1 (> 0). (iv) ⇒ (i). For fixed y > 0, setting u = x y, we obtain the following weight function: ω1 (σ, y) := y  =



σ

1 y

h(x y)x σ−1 d x

0 1

h(u)u σ−1 du = k1 (σ) (y > 0).

0

By the reverse Hölder inequality with weight, for y ∈ (0, ∞), we have  

p h(x y) f (x)d x

0 1 y

= 0



1 y



y (σ−1)/ p h(x y) (σ−1)/q f (x) x

≥

h(x y) 0

f (x)d x

 p−1  = ω1 (σ, y)y q(1−σ)−1 = (k1 (σ))



y

h(x y) 0

1 y

y (σ−1)q/ p

y σ−1

1 y

h(x y) 0

h(x y) 0



x σ−1

1 y

p

x (σ−1) p/q

p−1 − pσ+1

 p x (σ−1)/q dx y (σ−1)/ p



y σ−1

1 y



x (σ−1) p/q

y σ−1 x (σ−1) p/q

p−1 dx

f p (x)d x

f p (x)d x.

(5.15)

116

5 Equivalent Property of the Reverse Hardy-Type Integral Inequalities

If (5.15) assumes the form of equality for some y ∈ (0, ∞), then (cf. [2]) there exist constants A and B, such that they are not all zero, and A

y σ−1

f p (x) = B

x (σ−1) p/q

x σ−1

a.e. in R+ .

y (σ−1)q/ p

Let us suppose that A = 0 (otherwise B = A = 0). It follows that B a.e. in R+ , Ax

x p(1−σ)−1 f p (x) = y q(1−σ) which contradicts the fact that  0
(k1 (σ))



1 q



∞

h(x y) 0

= (k1 (σ))



1 q

0



∞

h(x y)

= (k1 (σ))



1 q

∞

x (σ−1) p/q

ω1 (σ, x)x

x (σ−1)( p−1)

p(1−σ)−1

0



∞

= k1 (σ)

x p(1−σ)−1 f p (x)d x

 1p



 1p f (x)d x p

dy

f (x)d x p

 1p

f p (x)d x dy

y σ−1

1 x

0

0



y σ−1

1 y

 1p

.

0

Since k1 (σ) ≥ M1 (> 0), (5.11) follows. Therefore, the statements (i), (iii) and (iv) are equivalent. (ii) ⇔ (iii). By the reverse Hölder inequality, we have 

∞

I =

x

1 q −σ

f (x)



 x

σ− q1

0





1 y

h(x y)g(y)dy d x 0



∞

≥

x p(1−σ)−1 f p (x)d x

0

Then by (5.12), we have (5.13).

 1p

L 1.

(5.16)

5.2 Reverse Hardy-Type Integral Inequalities of the First Kind

117

On the other hand, suppose that (5.13) is valid. We set  f (x) := x

q−1

1 y

qσ−1

, x > 0.

h(x y)g(y)dy 0

If L 1 = ∞, then (5.12) is trivially valid; if L 1 = 0, then it is impossible. Suppose that 0 < L 1 < ∞. By (5.13), we have 

∞

∞> 0

q

x p(1−σ)−1 f p (x)d x = L 1 = I



> M1  L1 =

∞

x

f (x)d x

p(1−σ)−1

p

0 ∞

x

p(1−σ)−1

f (x)d x p

 q1

 1p 

∞

y

q(1−σ1 )−1 q

g (y)dy

 q1

0



∞

> M1

0

y

q(1−σ1 )−1 q

g (y)dy

> 0,  q1

,

0

namely, (5.12) follows. Hence, statements (i), (ii), (iii) and (iv) are equivalent. If the statement (iv) holds true, and if there exists a constant M1 ≥ k1 (σ) such that (5.13) is satisfied, then we obtain that k1 (σ) ≥ M1 . Hence, the constant factor M1 = k1 (σ) in (5.13) is the best possible. The constant factor M1 = k1 (σ) (∈ R+ ) in (5.11) (resp. (5.12)) is still the best possible. Otherwise, by (5.14 ) (for σ1 = σ) (resp. (5.16)), we can conclude that the constant factor M1 = k1 (σ) in (5.13) is not the best possible. This completes the proof of the theorem.  In particular, for σ = σ1 =

1 p

in Theorem 5.3, we derive the following corollary:

Corollary 5.4 Let M1 be a positive constant. If k1 ( 1p ) < ∞, then the following statements (i), (ii), (iii) and (iv) are equivalent: (i) For any f (x) ≥ 0, satisfying  0
M1

∞

x 0

p−2

f (x)d x p

1p

.

(5.17)

118

5 Equivalent Property of the Reverse Hardy-Type Integral Inequalities

(ii) For any g(y) ≥ 0, satisfying 

∞

0
M1

0

g (y)dy q

q1

.

(5.18)

0

(iii) For any f (x) ≥ 0, satisfying 

∞

0
0). If statement (iv) holds true, then the constant M1 = k1 ( 1p ) (∈ R+ ) in (5.17), (5.18) and (5.19) is the best possible. Setting y=

1 , G(Y ) = g Y

 1 1 Y Y2

in Theorem 5.3, and then replacing Y by y, we deduce the following corollary: Corollary 5.5 Let M1 be a positive constant. If k1 (σ) < ∞, then the following statements (i), (ii), (iii) and (iv) are equivalent: (i) For any f (x) ≥ 0, satisfying 

∞

0< 0

x p(1−σ)−1 f p (x)d x < ∞,

5.2 Reverse Hardy-Type Integral Inequalities of the First Kind

119

we have the following inequality: 



p  1p  x f (x)d x dy y h y 0 0  ∞  1p > M1 x p(1−σ)−1 f p (x)d x . ∞

− pσ1 −1

y

(5.20)

0

(ii) For any G(y) ≥ 0, satisfying 

∞

0
M1 y q(1+σ1 )−1 G q (y)dy . 

∞

x qσ−1

x

h

(5.21)

0

(iii) For any f (x) ≥ 0, satisfying 

∞

0
0). If statement (iv) holds true, then the constant M1 = k1 (σ) (∈ R+ ) in (5.20), (5.21) and (5.22) is the best possible.

120

5 Equivalent Property of the Reverse Hardy-Type Integral Inequalities

Note. h( xy ) is a homogeneous function of degree 0, namely,  x = k0 (x, y). y

h Setting

h(u) = kλ (u, 1), where kλ (x, y) is a homogeneous function of degree −λ (∈ R), for g(y) = y λ G(y) and μ1 = λ − σ1 in Corollary 5.5, we obtain the following: Corollary 5.6 Let M1 be a positive constant. If 

kλ(1) (σ)

=

1

kλ (u, 1)u σ−1 du < ∞,

0

then the following statements (i), (ii), (iii) and (iv) are equivalent: (i) For any f (x) ≥ 0, satisfying 

∞

0
M1

x p(1−σ)−1 f p (x)d x

 1p

 1p dy

.

(5.23)

0

(ii) For any g(y) ≥ 0, satisfying  0
M1 0

y q(1−μ1 )−1 g q (y)dy

 1p

.

 q1 dx (5.24)

5.2 Reverse Hardy-Type Integral Inequalities of the First Kind

121

(iii) For any f (x) ≥ 0, satisfying 

∞

0
0). If statement (iv) holds true, then the constant M1 = kλ(1) (σ) (∈ R+ ) in (5.23), (5.24) and (5.25) is the best possible. In particular, for λ = 1, σ = q1 , μ1 = μ = lowing:

1 p

in Corollary 5.6, we obtain the fol-

Corollary 5.7 Let M1 be a positive constant. If k1(1) ( q1 ) < ∞, then the following statements (i), (ii), (iii) and (iv) are equivalent: (i) For any f (x) ≥ 0, satisfying 

∞

0
M1 0

0

∞ 0

(ii) For any g(y) ≥ 0, satisfying 

∞

0< 0

g q (y)dy < ∞,

f (x)d x p

1p

.

(5.26)

122

5 Equivalent Property of the Reverse Hardy-Type Integral Inequalities

we have the following inequality:  ∞  x  q  q1 k1 (x, y)g(y)dy d x > M1 0

0

∞

g q (y)dy

1p

.

(5.27)

0

(iii) For any f (x) ≥ 0, satisfying  ∞ f p (x)d x < ∞, 0< 0

and g(y) ≥ 0, satisfying



∞

0
M1

f (x)d x p

0

1p 

∞

g (y)dy q

q1

.

(5.28)

0

(iv) k1(1) ( q1 ) ≥ M1 (> 0).

If statement (iv) holds true, then the constant M1 = k1(1) ( q1 ) (∈ R+ ) in (5.26), (5.27) and (5.28) is the best possible.

5.3 Reverse Hardy-Type Inequalities of the Second Kind Similarly, we obtain the following weight function:  ∞ ω2 (σ, y) := y σ h(x y)x σ−1 d x  =

1 y

∞

h(u)u σ−1 du = k2 (σ)(y > 0),

1

in view of Lemma 5.2, we similarly obtain the following theorem: Theorem 5.8 Let M2 be a positive constant. If k2 (σ) < ∞, then the following statements (i), (ii), (iii) and (iv) are equivalent: (i) For any f (x) ≥ 0, satisfying 

∞

0
M2

x

p(1−σ)−1

123

p h(x y) f (x)d x f (x)d x p

 1p

1p dy

.

(5.29)

0

(ii) For any g(y) ≥ 0, satisfying 

∞

0
M2

y

q(1−σ1 )−1 q

g (y)dy

 q1

q1 dx

.

(5.30)

0

(iii) For any f (x) ≥ 0, satisfying 

∞

0
0). If statement (iv) holds true, then the constant M2 = k2 (σ) (∈ R+ ) in (5.29), (5.30) and (5.31) is the best possible. In particular, for σ = σ1 =

1 p

in Theorem 5.8, we deduce the corollary below:

Corollary 5.9 Let M2 be a positive constant. If k2 ( 1p ) < ∞, then the following statements (i), (ii), (iii) and (iv) are equivalent:

124

5 Equivalent Property of the Reverse Hardy-Type Integral Inequalities

(i) For any f (x) ≥ 0, satisfying  ∞ 0< x p−2 f p (x)d x < ∞, 0

we have the following inequality: 

∞



∞ 1 y

0

p h(x y) f (x)d x

1p

 > M2

dy

∞

x p−2 f p (x)d x

1p

.

(5.32)

0

(ii) For any g(y) ≥ 0, satisfying  0
M2

∞

g q (y)dy

 q1

.

(5.33)

0

(iii) For any f (x) ≥ 0, satisfying  ∞ 0< x p−2 f p (x)d x < ∞, 0

and g(y) ≥ 0, satisfying



∞

0
M2

x 0

p−2

 h(x y) f (x)d x dy

f (x)d x p

1p 

∞

g (y)dy q

q1

.

(5.34)

0

(iv) k2 ( 1p ) ≥ M2 (> 0). If statement (iv) holds true, then the constant M2 = k2 ( 1p ) (∈ R+ ) in (5.32), (5.33) and (5.34) is the best possible. Setting  1 1 1 y = , G(Y ) = g Y Y Y2 in Corollary 5.9, and then replacing Y by y, we get the following corollary:

5.3 Reverse Hardy-Type Inequalities of the Second Kind

125

Corollary 5.10 Let M2 be a positive constant. If k2 (σ) < ∞, then the following statements (i), (ii), (iii) and (iv) are equivalent: (i) For any f (x) ≥ 0, satisfying 

∞

0
M2 x p(1−σ)−1 f p (x)d x . 

∞

− pσ1 −1



∞

(5.35)

0

(ii) For any G(y) ≥ 0, satisfying 

∞

0
M2 y q(1+σ1 )−1 G q (y)dy . ∞

x qσ−1

∞

h

(5.36)

0

(iii) For any f (x), G(y) ≥ 0, satisfying 

∞

0
0). If statement (iv) holds true, then the constant M2 = k2 (σ) (∈ R+ ) in (5.35), (5.36) and (5.37) is the best possible. Setting h(u) = kλ (u, 1), where kλ (x, y) is a homogeneous function of degree −λ (∈ R), for g(y) = y λ G(y) and μ1 = λ − σ1 in Corollary 5.10, we get the following: Corollary 5.11 Let M2 be a positive constant. If 

kλ(2) (σ)

∞

=

kλ (u, 1)u σ−1 du < ∞,

1

then the following statements (i), (ii), (iii) and (iv) are equivalent: (i) For any f (x) ≥ 0, satisfying 

∞

0
M2

x

p(1−σ)−1

f (x)d x p

 1p

 1p dy

.

(5.38)

0

(ii) For any g(y) ≥ 0, satisfying  0
M2

y 0

q(1−μ1 )−1 q

g (y)dy

 1p

.

 q1 dx (5.39)

5.3 Reverse Hardy-Type Inequalities of the Second Kind

127

(iii) For any f (x) ≥ 0, satisfying 

∞

0
0). If statement (iv) holds true, then the constant M2 = kλ(2) (σ) (∈ R+ ) in (5.38), (5.39) and (5.40) is the best possible. In particular, for λ = 1, σ = q1 , μ = below:

1 p

in Corollary 5.11, we deduce the corollary

Corollary 5.12 Let M2 be a positive constant. If k1(2) ( q1 ) < ∞, then the following statements (i), (ii), (iii) and (iv) are equivalent: (i) For any f (x) ≥ 0, satisfying 

∞

0
M2

y

∞

f p (x)d x

1p

.

(5.41)

.

(5.42)

0

(ii) For any g(y) ≥ 0, satisfying 

∞

0
M2 0

∞

g (y)dy q

1p

128

5 Equivalent Property of the Reverse Hardy-Type Integral Inequalities

(iii) For any f (x) ≥ 0, satisfying 

∞

0
0).

If statement (iv) holds true, then the constant M2 = k1(2) ( q1 ) (∈ R+ ) in (5.41), (5.42) and (5.43) is the best possible. Example 5.13 Setting h(u) = kλ (u, 1) =

| ln u|β (max{u, 1})α (u > 0), |u λ − 1|(min{u, 1})α

then we obtain that | ln x y|β (max{x y, 1})α , |(x y)λ − 1|(min{x y, 1})α | ln x/y|β (max{x, y})α , kλ (x, y) = λ |x − y λ |(min{x, y})α h(x y) =

and for β, λ > 0, σ > α, 

(− ln u)β σ−α−1 u du 1 − uλ 0  1 ∞  = (− ln u)β u kλ+σ−α−1 du

k1 (σ) = kλ(1) (σ) =

0

=

∞  1  k=0

0

1

k=0

(− ln u)β u kλ+σ−α−1 du.

5.3 Reverse Hardy-Type Inequalities of the Second Kind

129

Setting v = (kλ + σ − α)(− ln u) in the previous integral, we have  ∞ 1 v β e−v dv β+1 (kλ + σ − α) 0 k=0  (β + 1) σ−α = ζ β + 1, ∈ R+ , λβ+1 λ

k1 (σ) = kλ(1) (σ) =

∞ 



where

∞

(η) :=

(5.44)

v η−1 e−v dv (η > 0)

0

is the gamma function and ζ(s, a) :=

∞ 

1 (Res > 1, a > 0) (k + a)s

k=0

is the extended Hurwitz zeta function (cf. [3]). For β, λ > 0, μ = λ − σ > α, setting v = u1 , we get k2 (σ) =

kλ(2) (σ) 



∞

= 1

(ln u)β σ+α−1 u du uλ − 1

(− ln v)β μ−α−1 u du = 1 − vλ 0  μ−α (β + 1) ∈ R+ . ζ β + 1, = λβ+1 λ 1

(5.45)

Note. If β, λ > 0, σ, μ > −α, then λ = σ + μ > −2α, and k1 (σ), k2 (σ) ∈ R+ . Example 5.14 Setting h(u) = kλ (u, 1) =

| ln u|β (min{u, 1})α+γ (u > 0), |u α − 1|(max{u, 1})λ+γ

then we get | ln x y|β (min{x y, 1})α+γ , |(x y)α − 1|(max{x y, 1})λ+γ | ln x/y|β (min{x, y})α+γ , kλ (x, y) = α |x − y α |(max{x, y})λ+γ h(x y) =

and for β, α > 0, σ > −α − γ,

130

5 Equivalent Property of the Reverse Hardy-Type Integral Inequalities



(− ln u)β σ+α+γ−1 u du 1 − uα 0  1 ∞  (− ln u)β u kα+σ+α+γ−1 du =

k1 (σ) =

kλ(1) (σ)

=

0

=

∞  1  k=0

1

k=0

(− ln u)β u kα+σ+α+γ−1 du.

0

Setting v = (kα + σ + α + γ)(− ln u) in the previous integral, we have  ∞ 1 = v β e−v dv k1 (σ) = β+1 (kα + σ + α + γ) 0 k=0  (β + 1) σ+α+γ = ∈ R+ . ζ β + 1, αβ+1 α ∞ 

kλ(1) (σ)

(5.46)

For β, α > 0, μ > −α − γ, setting v = u1 , we find

k2 (σ) =

kλ(2) (σ) 



∞

= 1

(ln u)β σ−λ−γ−1 u du uα − 1

(− ln v)β μ+α+γ−1 v dv = 1 − vα 0  μ+α+γ (β + 1) ∈ R+ . ζ β + 1, = αβ+1 α 1

(5.47)

Note. If β > 0, σ, μ > −α − γ, then λ = σ + μ > −2(α + γ), and k1 (σ) = kλ(1) (σ) ∈ R+ , k2 (σ) = kλ(2) (σ) ∈ R+ . Remark 5.15 We may use the above examples in order to construct two kinds of equivalent inequalities with particular kernels in Theorem 5.3, Corollary 5.6, Theorem 5.8 and Corollary 5.11.

5.4 Reverse Hardy-Type Integral Inequalities with the Interval Variable For a ∈ R\{0}, b > 0, replacing x (resp. y) by eax (resp. eby ), then replacing f (eax )eax (resp. g(eby )eby ) by f (x) (resp. g(y)), and |a|1/qMb1/ p by M in Theorem 5.3, Corollary 5.6, Theorem 5.8 and Corollary 5.11, and carrying out the corresponding simplifications, we deduce the following theorem: Theorem 5.16 Let M1 be a positive constant, and a = 0, b > 0. If k1 (σ) < ∞, then the following statements (i), (ii), (iii) and (iv) are equivalent:

5.4 Reverse Hardy-Type Integral Inequalities with the Interval Variable

131

(i) For any f (x) ≥ 0, satisfying  0
M1

∞

p

e−by

pσ1 by

h(e −∞



f (x) eσax

−∞

ax+by

 1p

p

) f (x)d x

1p dy

.

dx

(5.48)

(ii) For any g(y) ≥ 0, satisfying  0
M1

e−by

∞



−∞

−∞

g(y) eσ1 by

q h(eax+by )g(y)dy

q

 q1 dydy

q1 dx

.

(5.49)

(iii) For any f (x), g(y) ≥ 0, satisfying  0
0). |a|1/q b1/ p

If statement (iv) holds true, then the constant M1 = (5.49) and (5.50) is the best possible.

k1 (σ) |a|1/q b1/ p

(∈ R+ ) in (5.48),

In particular, for σ1 = σ, h(u) =

| ln u|β (max{u, 1})α |u λ − 1|(min{u, 1})α

(β, λ > 0, σ > α), by Example 5.13, we have the following equivalent inequalities with  σ−α (β + 1) ζ β + 1, a|1/q b1/ p λβ+1 λ being the best possible constant factor: 

p  1p | ln x y|β (max{x y, 1})α e f (x)d x dy λ α −∞ |(x y) − 1|(min{x y, 1}) −∞  ∞    1p f (x) p (β + 1) σ−α > 1/q 1/ p β+1 ζ β + 1, dx , a| b λ λ eσax −∞ 

∞

e−by

pσby

q q1 | ln x y|β (max{x y, 1})α e g(y)dy d x λ α −∞ |(x y) − 1|(min{x y, 1}) −∞  ∞    q1 g(y) q (β + 1) σ−α > 1/q 1/ p β+1 ζ β + 1, dydy , a| b λ λ eσby −∞ 

∞



e−by

qσax

| ln x y|β (max{x y, 1})α g(y) f (x)d x dy λ α −∞ −∞ |(x y) − 1|(min{x y, 1})  (β + 1) σ−α > 1/q 1/ p β+1 ζ β + 1, a| b λ λ p  1p  ∞   q1  ∞  f (x) g(y) q d x dy . × eσax eσby −∞ −∞ 

∞

(5.51)



(5.52)

e−by

(5.53)

Remark 5.17 In Theorem 5.16, if σ1 = σ, b < 0, then replacing −b by b > 0, we have the following equivalent inequalities with the best possible constant factor k1 (σ) : |a|1/q b1/ p

5.4 Reverse Hardy-Type Integral Inequalities with the Interval Variable



∞

−∞

k1 (σ) > |a|1/q b1/ p 

e



qσax





eby ∞



−∞

 1p dx

.

(5.54)

ax−by

g(y) e−σby

)g(y)dy

q

 q1 dx

 q1 dydy

.

(5.55)

h(e



p

 1p dy

q

∞

ax−by

eby

k1 (σ) > |a|1/q b1/ p

f (x) eσax

h(e

∞

g(y) −∞

h(eax−by ) f (x)d x

−∞

k1 (σ) > |a|1/q b1/ p ∞

∞



∞

p

∞

eby



−∞





e− pσby

133

∞



−∞

In particular, for h(u) =

f (x) eσax

) f (x)d x dy

p

 1p 

∞

dx −∞



g(y) e−σby

q

 q1 dy

.

(5.56)

| ln u|β (min{u, 1})α+γ |u α − 1|(max{u, 1})λ+γ

(β, α > 0, σ > −α − γ), by Example 5.14, we have the following equivalent inequalities with  σ+α+γ (β + 1) ζ β + 1, |a|1/q b1/ p αβ+1 α being the best possible constant factor: 

∞

e

− pσby



−∞

∞ eby

|ax − by|β (min{eax−by , 1})α+γ f (x) dx |eax−by − 1|(max{eax−by , 1})λ

 1p

p dy

  1p  ∞  f (x) p (β + 1) σ+α+γ > ζ β + 1, dx , |a|1/q b1/ p αβ+1 α eσax −∞ 



∞

e −∞

qσax

∞ eby

|ax − by|β (min{eax−by , 1})α+γ g(y) dy |eax−by − 1|(max{eax−by , 1})λ

(5.57) q1

q dx

 ∞    q1 g(y) q (β + 1) σ+α+γ > ζ β + 1, dydy . −σby |a|1/q b1/ p αβ+1 α −∞ e

(5.58)

134

5 Equivalent Property of the Reverse Hardy-Type Integral Inequalities

 |ax − by|β (min{eax−by , 1})α+γ g(y) f (x)d x dy ax−by − 1|(max{eax−by , 1})λ −∞ eby |e  σ+α+γ (β + 1) ζ β + 1, > |a|1/q b1/ p αβ+1 α p  1p  ∞   q1  ∞  f (x) g(y) q dx dy . × −σby eσax −∞ −∞ e 



∞

∞

(5.59)

Corollary 5.18 Let M1 be a positive constant, and a = 0, b > 0. If kλ(1) (σ) =



1

kλ (u, 1)u σ−1 du < ∞,

0

then the following statements (i), (ii) (iii) and (iv) are equivalent: (i) For any f (x) ≥ 0, satisfying  0
M1

p

eby

kλ (e , e ) f (x)d x

pμ1 by

∞



−∞

ax

−∞

p

f (x) eσax

 1p dx

by

1p dy

.

(5.60)

(ii) For any g(y) ≥ 0, satisfying  0
M1

eby

qσax

∞ −∞



q kλ (e , e )g(y)dy ax

−∞

g(y) eμ1 by

q

 q1 dy

by

.

q1 dx

(5.61)

5.4 Reverse Hardy-Type Integral Inequalities with the Interval Variable

135

(iii) For any f (x) ≥ 0, satisfying  0
0). |a|1/q b1/ p

If statement (iv) holds true, then the constant M1 = (5.61) and (5.62) is the best possible.

kλ(1) (σ) |a|1/q b1/ p

(∈ R+ ) in (5.60),

In particular, for μ1 = μ, h(u) =

| ln u|β (max{u, 1})α |u λ − 1|(min{u, 1})α

(β, λ > 0, σ > α), by Example 5.13, we have the following equivalent inequalities with  σ−α (β + 1) ζ β + 1, a|1/q b1/ p λβ+1 λ being the best possible constant factor: 

∞ −∞



eby

e pμby −∞

|ax + by|β (min{eax , eby })α f (x) dx |eλax − eλby |(max{eax , eby })α

p

 ∞    1p f (x) p (β + 1) σ−α > 1/q 1/ p β+1 ζ β + 1, dx , a| b λ λ eσax −∞

 1p dy

(5.63)

136

5 Equivalent Property of the Reverse Hardy-Type Integral Inequalities

q  q1 |ax + by|β (min{eax , eby })α g(y) e dy d x λax − eλby |(max{eax , eby })α −∞ −∞ |e  ∞    q1 g(y) q (β + 1) σ−α > 1/q 1/ p β+1 ζ β + 1, dy , a| b λ λ eμby −∞ 



∞

eby

qσax

|ax + by|β (min{eax , eby })α g(y) f (x)d x dy λax − eλby |(max{eax , eby })α −∞ −∞ |e  σ−α (β + 1) > 1/q 1/ p β+1 ζ β + 1, a| b λ λ p  1p  ∞   q1  ∞  f (x) g(y) q dx dy . × eσax eμby −∞ −∞ 



∞

(5.64)

eby

(5.65)

Remark 5.19 In Corollary 5.18, if μ1 = μ, b < 0, then replacing −b by b > 0, we have the following equivalent inequalities with the best possible constant factor kλ(1) (σ) |a|1/q b1/ p

: 

∞

e −∞

kλ(1) (σ) > |a|1/q b1/ p 

∞

e

kλ(1) (σ) > |a|1/q b1/ p 

∞

>

∞ −∞

∞

f (x) eσax

∞



−by

p

p ) f (x)d x

 1p dx

kλ (e , e ax

g(y) e−μby

−by

q

 1p dy

,

(5.66) q

)g(y)dy

 q1 dy

 q1 dx

,

(5.67)

kλ (eax , e−by ) f (x)d x dy 

f (x) eσax

In particular, for kλ (u, 1) = (β, α > 0, σ > −α − γ), inequalities with

kλ (e , e ax

e−by

−∞ ∞







e−by

kλ(1) (σ) |a|1/q b1/ p



∞

−∞

g(y) −∞

∞ e−by



−qσax

−∞





− pμby

p

 1p 

∞

dx −∞



g(y) e−μby

q

 q1 dy

.

(5.68)

| ln u|β (min{u, 1})α+γ |u α − 1|(max{u, 1})λ+γ

by Example 5.14, we have the following equivalent

5.4 Reverse Hardy-Type Integral Inequalities with the Interval Variable

137

 (β + 1) σ+α+γ ζ β + 1, |a|1/q b1/ p αβ+1 α being the best possible constant factor: 

 p  1p β ax −by α+γ |ax − by| (min{e , e }) f (x) e− pμby d x dy αax − e−αby |(max{eax , e−by })λ+γ −∞ e−by |e  ∞    1p f (x) p (β + 1) σ+α+γ > ζ β + 1, dx , (5.69) |a|1/q b1/ p αβ+1 α eσax −∞ 

∞

∞

q  q1 β ax −by α+γ |ax − by| (min{e , e }) g(y) eqσax dy d x αax − e−αby |(max{eax , e−by })λ+γ −∞ e−by |e  ∞    q1 g(y) q (β + 1) σ+α+γ > ζ β + 1, dy , (5.70) −μby |a|1/q b1/ p αβ+1 α −∞ e 



∞

∞

 |ax − by|β min{eax , e−by } f (x)d x dy ax − e−by |(max{eax , e−by })λ −∞ e−by |e  σ+α+γ (β + 1) ζ β + 1, > |a|1/q b1/ p αβ+1 α p  1p  ∞   q1  ∞  f (x) g(y) q d x dy . × −μby eσax −∞ −∞ e 



∞

∞

g(y)

(5.71)

Theorem 5.20 Let M2 be a positive constant, and a = 0, b > 0. If k2 (σ) < ∞, then the following statements (i), (ii) (iii) and (iv) are equivalent: (i) For any f (x) ≥ 0, satisfying  0
M2

∞ −∞



p

∞

h(e

ax+by

e−by

f (x) eσax

p

 1p dx

.

) f (x)d x

 1p dy (5.72)

138

5 Equivalent Property of the Reverse Hardy-Type Integral Inequalities

(ii) For any g(y) ≥ 0, satisfying  0
M2

e−by



∞

q

∞

eqσax

g(y) eσ1 by

−∞

h(eax+by )g(y)dy

q

 q1 dy

 q1 dx

.

(5.73)

(iii) For any f (x) ≥ 0, satisfying  0
0). |a|1/q b1/ p

If statement (iv) holds true, then the constant M2 = (5.73) and (5.74) is the best possible. In particular, for σ1 = σ, h(u) =

q

| ln u|β (max{u, 1})α |u λ − 1|(min{u, 1})α

k2 (σ) |a|1/q b1/ p

(∈ R+ ) in (5.72),

5.4 Reverse Hardy-Type Integral Inequalities with the Interval Variable

139

(β, λ > 0, μ > α), by Example 5.13, we have the following equivalent inequalities with  μ−α (β + 1) ζ β + 1, |a|1/q b1/ p λβ+1 λ being the best possible constant factor: 



∞

e

e−by

−∞

>

∞

pσby

|ax + by|β (min{eax+by , 1})α f (x) dx |eλ(ax+by) − 1|(max{eax+by , 1})α

 1p

p dy

 ∞    1p f (x) p (β + 1) μ−α ζ β + 1, d x , |a|1/q b1/ p λβ+1 λ eσax −∞ 



∞

e

qμax

∞ e−by

−∞

|ax + by|β (min{eax+by , 1})α g(y) dy |eλ(ax+by) − 1|(max{eax+by , 1})α

(5.75)  q1

q dx

 ∞    1p f (x) p (β + 1) μ−α > ζ β + 1, dx , |a|1/q b1/ p λβ+1 λ eσax −∞  |ax + by|β (min{eax+by , 1})α f (x)d x dy λ(ax+by) − 1|(max{eax+by , 1})α −∞ e−by |e  (β + 1) μ−α > ζ β + 1, |a|1/q b1/ p λβ+1 λ 



(5.76)

∞

∞

g(y)

 ×

∞

−∞



f (x) eσax

p

 1p 

∞

dx −∞



g(y) eσby

q

 q1 dy

.

(5.77)

Remark 5.21 In Theorem 5.20, if σ1 = σ, b < 0, then replacing −b by b > 0, we have the following equivalent inequalities with the best possible constant factor k2 (σ) : |a|1/q b1/ p  p 1p  by  ∞

−∞

>

e

e− pσby

k2 (σ) |a|1/q b1/ p



−∞ ∞ −∞



h(eax−by ) f (x)d x f (x) eσax

p

 1p dx

,

dy

(5.78)

140

5 Equivalent Property of the Reverse Hardy-Type Integral Inequalities





∞

e

h(e

−∞



∞



∞



−∞

h(e

ax−by

−∞

k2 (σ) > |a|1/q b1/ p



∞ −∞

f (x) eσax

)g(y)dy

p

 1p dx

q1 dx

,

(5.79)



eby

g(y) −∞

ax−by

−∞

k2 (σ) > |a|1/q b1/ p 

q

eby

qσax



f (x) eσax

) f (x)d x dy

p

 1p 

∞

dx −∞



g(y) e−σby

q

 q1 dy

.

(5.80)

In particular, for h(u) =

| ln u|β (min{u, 1})α+γ (β, α > 0, μ > −α − γ), |u α − 1|(max{u, 1})λ+γ

by Example 5.14, we have the following equivalent inequalities with  μ+α+γ (β + 1) ζ β + 1, |a|1/q b1/ p αβ+1 α being the best possible constant factor: 



∞

|ax − by|β (min{eax−by , 1})α+γ e− pσby f (x)d x α(ax−by) − 1|(max{eax−by , 1})λ+γ −∞ −∞ |e  ∞    1p f (x) p (β + 1) μ+α+γ > ζ β + 1, dx , |a|1/q b1/ p αβ+1 α eσax −∞ eby

p

 1p dy

(5.81)

q  q1 |ax − by|β (min{eax−by , 1})α+γ e g(y)dy d x α(ax−by) − 1|(max{eax−by , 1})λ+γ −∞ |e −∞  ∞    1p f (x) p (β + 1) μ+α+γ > ζ β + 1, dx , (5.82) |a|1/q b1/ p αβ+1 α eσax −∞ 

∞



qσax

eby

5.4 Reverse Hardy-Type Integral Inequalities with the Interval Variable

|ax − by|β (min{eax−by , 1})α+γ g(y) f (x)d x dy α(ax−by) − 1|(max{eax−by , 1})λ+γ −∞ −∞ |e  μ+α+γ (β + 1) ζ β + 1, > |a|1/q b1/ p αβ+1 α p  1p  ∞   q1  ∞  f (x) g(y) q dx dy . (5.83) × −σby eσax −∞ −∞ e 

∞



141

eby

Corollary 5.22 Let M2 be a positive constant, and a = 0, b > 0. If kλ(2) (σ) =



∞

kλ (u, 1)u σ−1 du > 0,

1

then the following statements (i), (ii) (iii) and (iv) are equivalent: (i) For any f (x) ≥ 0, satisfying  0
M2

∞

pμ1 by ∞



−∞

eby

p kλ (e , e ) f (x)d x ax

p

f (x) eσax

 1p dx

by

 1p dy

.

(5.84)

(ii) For any g(y) ≥ 0, satisfying  0
M2

∞



−∞

∞

eby

g(y) eμ1 by

q kλ (e , e )g(y)dy ax

q

 q1 dy

by

.

(iii) For any f (x) ≥ 0, satisfying  0
0). |a|1/q b1/ p

If statement (iv) holds true, then the constant M2 = (5.85) and (5.86) is the best possible.

kλ(2) (σ) |a|1/q b1/ p

(∈ R+ ) in (5.84),

In particular, for μ1 = μ, h(u) =

| ln u|β (max{u, 1})α |u λ − 1|(min{u, 1})α

(β, λ > 0, μ > α), by Example 5.13, we have the following equivalent inequalities with  μ−α (β + 1) ζ β + 1, |a|1/q b1/ p λβ+1 λ being the best possible constant factor: 

∞

 e pσby

−∞

∞

eby

|ax + by|β (min{eax , eby })α f (x) dx |eλax − e(λ+α)by |(max{eax , eby })α

 1p

p dy

 ∞    1p f (x) p (β + 1) μ−α > ζ β + 1, dx , |a|1/q b1/ p λβ+1 λ eσax −∞ 

∞

−∞

 eqμax

∞ eby

|ax + by|β (min{eax , eby })α g(y) dy |eλax − e(λ+α)by |(max{eax , eby })α

(5.87)  q1

q

 ∞    q1 g(y) q (β + 1) μ−α > ζ β + 1, dy , |a|1/q b1/ p λβ+1 λ eμby −∞

dx

(5.88)

5.4 Reverse Hardy-Type Integral Inequalities with the Interval Variable

 |ax + by|β (min{eax , eby })α g(y) f (x)d x dy λax − e(λ+α)by |(max{eax , eby })α −∞ eby |e  μ−α (β + 1) ζ β + 1, > |a|1/q b1/ p λβ+1 λ  q1  ∞  p  1p  ∞  f (x) g(y) q d x dy . × eσax eμby −∞ −∞ 



143

∞

∞

(5.89)

Remark 5.23 In Corollary 5.22, if μ1 = μ, b < 0, then replacing −b by b > 0, we have the following equivalent inequalities with the best possible constant factor kλ(2) (σ) |a|1/q b1/ p

: 

∞

e −∞







e−by



p

 1p

−∞ ∞



∞ −∞

) f (x)d x

1p dy

,

dx

(5.90) q

kλ (eax , e−by )g(y)dy

g(y) e−μby

q

 q1 dy

q1 dx

,

(5.91)

 kλ (e , e ax

−∞

kλ(2) (σ) > |a|1/q b1/ p

kλ (e , e

−by

f (x) eσax

e−by

−∞

g(y) −∞



eqσax

kλ(2) (σ) |a|1/q b1/ p

∞

∞ −∞

−∞

>

−∞



∞

p

e−by

ax



kλ(2) (σ) > |a|1/q b1/ p 



− pμby



f (x) eσax

In particular, for kλ (u, 1) =

−by

) f (x)d x dy

p

 1p 

∞

dx −∞



g(y) e−μby

q

 q1 dy

.

(5.92)

| ln u|β (min{u, 1})α+γ |u α − 1|(max{u, 1})λ+γ

(β, α > 0, μ > −α − γ), by Example 5.14, we have the following equivalent inequalities with  μ+α+γ (β + 1) ζ β + 1, |a|1/q b1/ p αβ+1 α being the best possible constant factor:

144

5 Equivalent Property of the Reverse Hardy-Type Integral Inequalities

p  1p |ax − by|β (min{eax , e−by })α+γ f (x) e d x dy |eαax − e−αby |(max{eax , e−by })λ+γ −∞ −∞  ∞    1p f (x) p (β + 1) μ+α+γ > ζ β + 1, dx , (5.93) |a|1/q b1/ p αβ+1 α eσax −∞ 

∞

− pμby



e−by

q  q1 |ax − by|β (min{eax , e−by })α+γ g(y) e dy d x αax − e−αby |(max{eax , e−by })λ+γ −∞ |e −∞  ∞    q1 g(y) q (β + 1) μ+α+γ > ζ β + 1, dy , (5.94) −μby |a|1/q b1/ p αβ+1 α −∞ e 

∞



e−by

qσax

|ax − by|β (min{eax , e−by })α+γ f (x) g(y) d x dy |eαax − e−αby |(max{eax , e−by })λ+γ −∞ −∞  (β + 1) μ+α+γ > ζ β + 1, |a|1/q b1/ p αβ+1 α p  1p  ∞   q1  ∞  f (x) g(y) q dx dy . × −μby eσax −∞ −∞ e 

∞



e−by

(5.95)

Acknowledgements B. Yang: This work is supported by the National Natural Science Foundation (No. 61772140), and Science and Technology Planning Project Item of Guangzhou City (No. 201707010229). M. Th. Rassias: This work was carried out under the generous research support of the University of Zurich and the John S. Latsis Foundation, spending also research time at the vibrant environment of the Institute for Advanced Study, Princeton.

References 1. Kuang, J.C.: Real and Functional Analysis (continuation) (sec. vol.). Higher Education Press, Beijing, China (2015) 2. Kuang, J.C.: Applied Inequalities. Shangdong Science and Technology Press, Jinan, China (2004) 3. Wang, Z.X., Guo, D.R.: Introduction to Special Functions. Science Press, Beijing (1979)

Bibliography 4. Weyl, H.: Singulare Integral Gleichungen mit Besonderer Berucksichtgung des Fourierschen Integral Theorem. Inaugural-Dissertation, Gottingen (1908) 5. Pan, Y.L., Wang, H.T., Wang, F.T.: On Complex Functions. Science Press, Beijing, China (2006) 6. Zhong, Y.Q.: On Complex Functions. Higher Education Press, Beijing, China (2004)

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