Half-Discrete Hilbert-Type Inequalities

HALF-DISCRETE HILBERT-TYPE INEQUALITIES 8799_9789814504973_tp.indd 1 25/11/13 3:47 pm May 2, 2013 14:6 BC: 8831 -...

Author: Bicheng Yang | Lokenath Debnath

20 downloads 879 Views 2MB Size Report

This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!

Report copyright / DMCA form

DOWNLOAD PDF

HALF-DISCRETE HILBERT-TYPE INEQUALITIES

8799_9789814504973_tp.indd 1

25/11/13 3:47 pm

May 2, 2013

14:6

BC: 8831 - Probability and Statistical Theory

This page intentionally left blank

PST˙ws

HALF-DISCRETE HILBERT-TYPE INEQUALITIES

Bicheng Yang Guangdong University of Education, China

Lokenath Debnath University of Texas-Pan American, USA

World Scientific NEW JERSEY

•

8799_9789814504973_tp.indd 2

LONDON

•

SINGAPORE

•

BEIJING

•

SHANGHAI

•

HONG KONG

•

TA I P E I

•

CHENNAI

25/11/13 3:47 pm

Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

Library of Congress Cataloging-in-Publication Data Yang, Bicheng, author. Half-discrete Hilbert-type inequalities / by Bicheng Yang (Guangdong University of Education, China) & Lokenath Debnath (University of Texas- Pan American, USA). pages cm Includes bibliographical references and index. ISBN 978-981-4504-97-3 (hardcover : alk. paper) 1. Inequalities (Mathematics) 2. Mathematical analysis. I. Debnath, Lokenath, author. II. Title. QA295.Y36 2014 515'.46--dc23 2013036829

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

Copyright © 2014 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.

Printed in Singapore

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Preface

Historically, mathematical analysis has been the major and signiﬁcant branch of mathematics for the last three centuries. Indeed, inequalities became the heart of mathematical analysis. Many great mathematicians have made signiﬁcant contributions to numerous new developments on the subject which led to the discovery of many new inequalities with proofs and useful applications in the ﬁelds of mathematical physics, pure and applied mathematics. Indeed, mathematical inequalities became an important branch of modern mathematics in the twentieth century through the pioneering work entitled Inequalities by G.H. Hardy, J.E. Littlewood and G. Polya which was ﬁrst published as a treatise in 1934. This unique publication represents a paradigm of precise logic, full of elegant inequalities with rigorous proofs and useful applications in mathematics. During the twentieth century, discrete and integral inequalities played a fundamental role in mathematics and have a wide variety of applications in many areas of pure and applied mathematics. In particular, David Hilbert (1862-1943) ﬁrst proved Hilbert’s double series inequality without exact determination of the constant in his lectures on integral equations. Herman Weyl (1885-1955) published a proof of the Hilbert double series inequality in 1908. Subsequently, Isaac Schur (1875-1941) gave a new proof of the Hilbert double series inequality in 1911 with the best possible sharp constant, and also discovered the integral analogue of the Hilbert double series inequality which became known as the Hilbert integral inequality. In 1925, G.H. Hardy (1877-1947) provided the best extension of it by introducing a pair of conjugate exponent which became known as the Hardy-Hilbert inequality. The Hilbert-type inequalities are a more wider class of analytic inequalities with bilinear kernels which include the Hardy-Hilbert inequality as a particular case. v

ws-book9x6

November 7, 2013

vi

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

The mathematical theory of inequalities, in general, and the Hilberttype inequalities, in particular, and their applications has grown considerably during the last one century. With the advent of new ideas and proofs, new results and applications, studies are continually being added to the major subject of mathematical inequalities, where these are themselves developing and coalescing. It is becoming more and more desirable for mathematicians to study the Hilbert-type inequalities as a whole. Yet it is increasingly diﬃcult for them to do so since major and important articles often appears in many diﬀerent journals. The diﬃculty can be alleviated if a single research monograph containing a coherent account of recent developments, especially, if written to be accessible to both graduate students and research professionals. It is our hope that this monograph will ﬁrst interest, then prepare readers to undertake research projects on the Hilbert-type inequalities and their applications, by providing that background of fundamental ideas, methods, proofs, and results essential to understanding the specialized literature of this vast area. Many ideas, major results, methods, proofs, and examples presented in this volume are either motivated by, or borrowed from works cited in the Bibliography. We wish to express our gratitude to the authors of their works. The writing of this monograph was also greatly inﬂuenced by the famous quotations of David Hilbert and G.H. Hardy as follows: 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.

G.H. Hardy So, this monograph deals with an extensive account of the theory and applications with numerous examples of any half-discrete Hilbert-type inequalities in a self-contained and rigorous manner using the methods of real and functional analysis, operator theory, the way of weight functions,

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Preface

ws-book9x6

vii

and special functions. Throughout this book, special attention is given to proofs of the best constant factors of all inequalities. The ﬁrst chapter deals with recent developments of the Hilbert-type discrete and integral inequalities by introducing kernels, weight functions, and multi-parameters. Included are numerous examples and applications, many extensions, generalizations, reﬁnements of Hilbert-type inequalities involving special functions such as beta, gamma, logarithm, trigonometric, hyperbolic, Bernoulli’s numbers and functions, Euler’s constant, zeta functions and hypergeometric functions. Special attention is given to many equivalent inequalities and to conditions under which the constant factors involved in inequalities are the best possible. Chapter 2 contains some improvements of the celebrated EulerMaclaurin summation formula, optimization of methods of estimating the series and the weight functions. Included are many useful theorems, corollaries, and inequalities with applications involving new inequalities on the Hurwitz zeta function, the Riemann zeta function and the extended Stirling formula. Chapter 3 is devoted to the half-discrete Hilbert-type inequalities with general homogeneous kernels and their many extensions. Included are several equivalent inequalities, operator expressions, and the reverses of the Hilbert-type inequalities with many generalizations, applications and particular examples. The main objective of Chapter 4 is to derive half-discrete Hilbert-type inequalities with a general non-homogeneous kernel, and their extensions. Many equivalent inequalities, and their operator expressions, two classes of reverse inequality, many extensions and particular examples are included in this chapter. Chapter 5 contains two kinds of multi-dimensional half-discrete Hilberttype inequalities with the best possible constant factors. These inequalities are extensions of the two-dimensional cases studied in Chapters 3 and 4. Included are equivalent forms, operator expressions, reverses, proofs of many important theorems with corollaries, and many particular examples. The ﬁnal Chapter 6 deals with two-kinds of multiple half-discrete Hilbert-type inequalities with the best possible constant factors. Included are equivalent inequalities, operator expressions, the reverses, proofs of major theorems and corollaries, and numerous examples with particular kernels. It is shown that theorems and corollaries of this chapter reduce to the corresponding results of Chapters 3 and 4 as special cases.

November 7, 2013

viii

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

Basically, this monograph is designed as a modern source of many new half-discrete Hilbert-type inequalities with numerous examples and applications. It also provides new results and information that put the reader at the forefront of current research. A large number of research papers and books have been included in the bibliography to stimulate new interest in future advanced study and research. A short index of this book is added in the end to include a wide variety of terms and topics so that they are useful for the reader. Our special thanks to Ms. Veronica Chavarria who cheerfully typed the manuscript with constant changes and revisions. In spite of the best eﬀorts of everyone involved, some typographical errors will doubtlessly remain. We wish to express thanks to Ms. Lai Fun Kwong, Dr. S.C. Lim and the Production Department of the World Scientiﬁc Publishing company for their help and cooperation.

Bicheng Yang Guang Zhou, Guangdong P.R. China

Lokenath Debnath Edinburg, Texas U.S.A.

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Acknowledgments

This work is supported by The National Natural Science Foundation of China (No. 61370186), and 2012 Knowledge Construction Special Foundation Item of Guangdong Institution of Higher Learning College and University (No. 2012KJCX0079, in China).

ix

ws-book9x6

May 2, 2013

14:6

BC: 8831 - Probability and Statistical Theory

This page intentionally left blank

PST˙ws

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Contents

Preface

v

Acknowledgments

ix

1. Recent Developments of Hilbert-Type Inequalities with Applications 1.1 1.2

1.3

1.4

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . Hilbert’s Inequality and Hilbert’s Operator . . . . . . . . 1.2.1 Hilbert’s Discrete and Integral Inequalities . . . . 1.2.2 Operator Formulation of Hilbert’s Inequality . . . 1.2.3 A More Accurate Discrete Hilbert’s Inequality . . 1.2.4 Hilbert’s Inequality with One Pair of Conjugate Exponents . . . . . . . . . . . . . . . . . . . . . . 1.2.5 A Hilbert-type Inequality with the General Homogeneous Kernel of Degree −1 . . . . . . . . . . . . 1.2.6 Two Multiple Hilbert-type Inequalities with the Homogeneous Kernels of Degree (−n + 1) . . . . . Modern Research for Hilbert-type Inequalities . . . . . . . 1.3.1 Modern Research for Hilbert’s Integral Inequality 1.3.2 On the Way of Weight Coeﬃcient for Giving a Strengthened Version of Hilbert’s Inequality . . . 1.3.3 Hilbert’s Inequality with Independent Parameters 1.3.4 Hilbert-type Inequalities with Multi-parameters . Some New Applications for Hilbert-type Inequalities . . . 1.4.1 Operator Expressions of Hilbert-type Inequalities 1.4.2 Some Basic Hilbert-type Inequalities . . . . . . . xi

1 1 2 2 4 5 6 9 12 12 12 14 15 18 22 22 23

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

xii

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

1.4.3

1.5

Some Applications to Half-discrete Hilbert-type Inequalities . . . . . . . . . . . . . . . . . . . . . . Concluding Remarks . . . . . . . . . . . . . . . . . . . . .

2. Improvements of the Euler-Maclaurin Summation Formula and Applications 2.1 2.2

2.3

2.4

29

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . Some Special Functions Relating Euler-Maclaurin’s Summation Formula . . . . . . . . . . . . . . . . . . . . . 2.2.1 Bernoulli’s Numbers . . . . . . . . . . . . . . . . . 2.2.2 Bernoulli’s Polynomials . . . . . . . . . . . . . . . 2.2.3 Bernoulli’s Functions . . . . . . . . . . . . . . . . 2.2.4 The Euler-Maclaurin Summation Formula . . . . Estimations of the Residue Term about a Class Series . . 2.3.1 An Estimation under the More Fortiﬁed Conditions . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Some Estimations under the More Imperfect Conditions . . . . . . . . . . . . . . . . . . . . . . 2.3.3 Estimations of δq (m, n) and Some Applications . . Two Classes of Series Estimations . . . . . . . . . . . . . 2.4.1 One Class of Convergent Series Estimation . . . . 2.4.2 One Class of Finite Sum Estimation on Divergence Series . . . . . . . . . . . . . . . . . . . . . . . . .

3. A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel 3.1 3.2

3.3

Introduction . . . . . . . . . . . . . . . . . . . . . . . . Some Preliminary Lemmas . . . . . . . . . . . . . . . 3.2.1 Deﬁnition of Weight Functions and Related Lemmas . . . . . . . . . . . . . . . . . . . . . 3.2.2 Estimations about Two Series . . . . . . . . . 3.2.3 Some Inequalities Relating the Constant k(λ1 ) Some Theorems and Corollaries . . . . . . . . . . . . . 3.3.1 Equivalent Inequalities and their Operator Expressions . . . . . . . . . . . . . . . . . . . . 3.3.2 Two Classes of Equivalent Reverse Inequalities 3.3.3 Some Corollaries . . . . . . . . . . . . . . . . . 3.3.4 Some Particular Examples . . . . . . . . . . .

25 27

29 29 29 31 32 34 36 36 40 47 50 50 52

57 . . . .

57 58

. . . .

. . . .

58 62 68 70

. . . .

. . . .

70 76 82 97

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

xiii

Contents

3.3.5 3.3.6

Applying Condition (iii) and Corollary 3.8 . . . . 101 Applying Condition (iii) and Corollary 3.4 . . . . 115

4. A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel 4.1 4.2

4.3

4.4

123

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . Some Preliminary Lemmas . . . . . . . . . . . . . . . . . 4.2.1 Deﬁnition of Weight Functions and Some Related Lemmas . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Estimations of Two Series and Examples . . . . . 4.2.3 Some Inequalities Relating the Constant k(α) . . Some Theorems and Corollaries . . . . . . . . . . . . . . . 4.3.1 Equivalent Inequalities and their Operator Expressions . . . . . . . . . . . . . . . . . . . . . . 4.3.2 Two Classes of Equivalent Reverses . . . . . . . . 4.3.3 Some Corollaries . . . . . . . . . . . . . . . . . . . Some Particular Examples . . . . . . . . . . . . . . . . . . 4.4.1 Applying Condition (i) and Corollary 4.5 . . . . . 4.4.2 Applying Condition (iii) and Corollary 4.2 . . . .

5. Multi-dimensional Half-Discrete Hilbert-Type Inequalities 5.1 5.2

5.3

5.4

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . Some Preliminary Results and Lemmas . . . . . . . . . 5.2.1 Some Related Lemmas . . . . . . . . . . . . . . 5.2.2 Some Results about the Weight Functions . . . 5.2.3 Two Preliminary Inequalities . . . . . . . . . . . Some Inequalities Related to a General Homogeneous Kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 Several Lemmas . . . . . . . . . . . . . . . . . . 5.3.2 Main Results . . . . . . . . . . . . . . . . . . . . 5.3.3 Some Corollaries . . . . . . . . . . . . . . . . . . 5.3.4 Operator Expressions and Some Particular Examples . . . . . . . . . . . . . . . . . . . . . . Some Inequalities Relating a General Non-Homogeneous Kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.1 Some Lemmas . . . . . . . . . . . . . . . . . . . 5.4.2 Main Results . . . . . . . . . . . . . . . . . . . . 5.4.3 Some Corollaries . . . . . . . . . . . . . . . . . .

123 123 123 128 132 134 134 140 146 155 155 160 169

. . . . .

169 170 170 172 175

. . . .

178 178 183 192

. 196 . . . .

205 205 210 219

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

xiv

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

5.4.4

Operator Expressions and Some Particular Examples . . . . . . . . . . . . . . . . . . . . . . . 223

6. Multiple Half-Discrete Hilbert-Type Inequalities 6.1 6.2

6.3

6.4

233

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . First Kind of Multiple Hilbert-type Inequalities . . . . . 6.2.1 Lemmas Related to the Weight Functions . . . . 6.2.2 Two Preliminary Inequalities . . . . . . . . . . . 6.2.3 Main Results and Operator Expressions . . . . . 6.2.4 Some Kinds of Reverse Inequalities . . . . . . . Second Kind of Multiple Hilbert-type Inequalities . . . . 6.3.1 Lemmas Related to the Weight Functions . . . . 6.3.2 Two Preliminary Inequalities . . . . . . . . . . . 6.3.3 Main Results and Operator Expressions . . . . . 6.3.4 Some Kinds of Reverse Inequalities . . . . . . . Some Examples with the Particular Kernels . . . . . . . 1 6.4.1 The Case of kλ (x1 , · · · , xm , xm+1 ) = (m+1 λ i=1 xi ) 6.4.2 The Case of s 1 kλ (x1 , · · · , xm+1 ) = k=1 m λ/s . . λ/s i=1

6.4.3 6.4.4

The Case of kλ (x1 , · · · , xm , xm+1 ) = The Case of kλ (x1 , · · · , xm , xm+1 ) =

xi

+ck xm+1

. . . . . . . . . . . . .

233 234 234 245 248 253 259 259 269 271 276 281 282

. 291

1 (max1≤i≤m+1 {xi })λ

. . . . . 302

1 (min1≤i≤m+1 {xi })λ

. . . . . 311

Bibliography

321

Index

331

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Chapter 1

Recent Developments of Hilbert-Type Inequalities with Applications

“As long as a branch of knowledge oﬀers an abundance of problems, it is full of vitality.” “The organic unity of mathematics is inherent in the nature of this science, for mathematics is the foundation of all exact knowledge of natural phenomena.”

David Hilbert

1.1

Introduction

This chapter deals with some recent developments of Hilbert-type discrete and integral inequalities by introducing kernels, weight functions, and multi-parameters. Included are numerous generalizations, extensions and reﬁnements of Hilbert-type inequalities involving many special functions such as beta, gamma, logarithm, trigonometric, hyperbolic, Bernoulli’s functions and Bernoulli’s numbers, Euler’s constant, zeta function and hypergeometric functions with many applications. Special attention is given to many equivalent inequalities and to conditions under which the constant factors involved in inequalities are the best possible. Many particular cases of Hilbert-type inequalities are presented with numerous applications. A large number of major books and research papers published in recent years are included to stimulate new interest in future study, especially in research on half-discrete Hilbert-type inequalities and their applications. 1

November 7, 2013

12:24

2

1.2 1.2.1

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

Hilbert’s Inequality and Hilbert’s Operator Hilbert’s Discrete and Integral Inequalities

Historically, mathematical analysis has been the major and signiﬁcant branch of mathematics for the last three centuries. Indeed, inequalities became the heart of mathematical analysis. Many great mathematicians have made signiﬁcant contributions to many new developments of the subject which led to the discovery of many new inequalities with proofs and useful applications in many ﬁelds of mathematical physics, pure and applied mathematics. Indeed, mathematical inequalities became an important branch of modern mathematics in the twentieth century through the pioneering work entitled Inequalities by G.H. Hardy, J.E. Littlewood and G. P`olya [21] which was ﬁrst published as a treatise in 1934. This unique publication represents a paradigm of precise logic, full of elegant inequalities with rigorous proofs and useful applications in mathematics. It is appropriate to mention a delighted quotation of Anthony Zygmund (1900-1992) “Hardy, Littlewood and P`olya’s book has been one of the most important books in analysis in the last few decades. It had an impact on the trend of research and is still inﬂuencing it. In looking through the book now one realises how little one would like to change the existing text.” During the twentieth century, discrete and integral inequalities played a fundamental role in mathematics and have a wide variety of applications in many areas of pure and applied mathematics. In particular, David Hilbert (1862-1943) ﬁrst proved Hilbert’s double series inequality without exact determination of the constant in his lectures on integral equations. If {am } ∞ and {bn } are two real sequences such that 0 < m=1 a2m < ∞ and 0 < ∞ 2 n=1 bn < ∞, then the Hilbert’s double series inequality is given by 12 ∞ ∞ ∞ ∞ a m bn 2 2 <π am bn . (1.1) m+n n=1 m=1 m=1 n=1 This famous inequality was proved by Hilbert in the early 1900s with the constant 2π in place of π. Several years after Hilbert’s proof, Issai Schur (1875-1941) [66] gave a new proof in 1911 which revealed that (1.1) actually holds for constant π which is the best possible sharp constant. In 1908, Weyl [77] published a proof of Hilbert’s inequality (1.1), and Schur [66] also discovered the integral analogue of (1.1), which became known as the Hilbert’s integral inequality in the form 12 ∞ ∞ ∞ ∞ f (x) g (y) 2 2 f (x) dx g (y) dy , (1.2) dxdy < π x+y 0 0 0 0

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Recent Developments of Hilbert-Type Inequalities with Applications

ws-book9x6

3

∞ where f and g are measurable functions such that 0 < 0 f 2 (x) dx < ∞ ∞ 2 and 0 < 0 g (y) dy < ∞, and π in (1.2) is still the best possible constant factor. A large number of generalizations, extensions and reﬁnements of both (1.1) and (1.2) are available in literature in Hardy et al. [21], Mitrinovi´c et al. [57], Kuang [47] and Hu [34]. Considerable attention has been given to the well-known classical Hardy-Littlewood-Sobolev (HLS) inequality (see Hardy et al. [21]) in the form f (x) g (y) dx dy ≤ Cn,λ,p f p gq , (1.3) λ Rn Rn |x − y| for every f ∈ Lp (Rn ) and g ∈ Lq (Rn ), where 0 < λ < n, 1 < p, q < ∞ such that p1 + 1q + nλ = 2, and f p is the Lp (Rn ) norm of the function f . For arbitrary p and q, an estimate of the upper bound of the constant Cn,λ,p was given by Hardy, Littlewood and Sobolev, but no sharp value is known still now. However, for the special case, p = q = 2n/ (2n − λ), the sharp value of the constant was found as n nλ −1 Γ n−λ Γ 2 λ Cp,λ,n = Cn,λ = π 2 2 λ , (1.4) Γ (n) Γ n− 2 and the equality in (1.3) holds if and only if g (x) = c1 f (x) and f (x) = −d 2 c2 h μx2 − a , where h (x) = 1 + |x| , 2d = (λ − 2n), a ∈ Rn , c1 , c2 , μ ∈ R/ {0}. In 1958, Stein and Weiss [67] generalized the double weighted inequality of Hardy and Littlewood in the form with the same notation as in (1.3): f (x) g (x) dx dy (1.5) ≤ Cα,β,n,λ,p f p gq , Rn Rn |x|α |x − y|λ |y|β where α + β ≥ 0, and the powers of α and β of the weights satisfy the 1 1 1 1 following conditions 1 − p1 − nλ < α n < 1 − p , and p + q + n (λ + α + β) = 2. Inequality (1.5) and its proof given by Stein and Weiss [67] represent some major contribution to the subject. On the other hand, Chen et al. [8] used weighted Hardy-LittlewoodSobolev inequalities (1.3) and (1.5) to solve systems of integral equations. In 2011, Khotyakov [42] suggested two proofs of the sharp version of the HLS inequality (1.3). The ﬁrst proof is based on the invariance property of the inequality (1.3), and the second proof uses some properties of the fast diﬀusion equation with the conditions λ = n − 2, n ≥ 3 on the sharp HLS inequality (1.3).

November 7, 2013

12:24

4

1.2.2

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Operator Formulation of Hilbert’s Inequality m

Suppose that R is the set of real numbers and Rm + = (0, ∞) × · · · × (0, ∞), ∞ 1 p p < ∞} and for p > 1, lp := {a = {an }∞ n=1 |||a||p = { n=1 |an | } ∞ p p 1/p L (R+ ) := {f |||f ||p = { 0 |f (x)| dx} < ∞} are real normal spaces with the norms ||a||p and ||f ||p . We express inequality (1.1) using the form of operator as follows: T : l 2 → l2 is a linear operator, for any ∞ ∞ a = {am }m=1 ∈ l2 , there exists a sequence c = {cn }n=1 ∈ l2 , satisfying cn = (T a) (n) =

∞

am , m +n m=1

n ∈ N,

(1.6)

where N is the set of positive integers. Hence, for any sequence b = 2 {bn }∞ n=1 ∈ l , we deﬁne the inner product of T a and b as follows ∞ ∞ ∞ ∞ am a m bn bn = . (1.7) (T a, b) = (c, b) = m + n m +n n=1 m=1 n=1 m=1 Using (1.7), inequality (1.1) can be rewritten in the operator form (T a, b) < π a2 b2 ,

(1.8)

where a2 , and b2 > 0. It follows from Wilhelm [78] that T is a bounded operator and the norm T = π and T is called Hilbert’s operator with 1 the kernel m+n . For a2 > 0, the equivalent form of (1.8) is given as T a2 < π a2 , that is, ∞ 2 ∞ ∞ am < π2 a2n , (1.9) m + n n=1 m=1 n=1 where the constant factor π2 is still the best possible. Obviously, inequality (1.9) and (1.1) are equivalent (see Hardy et al. [21]). We may deﬁne Hilbert’s integral operator as follows: T : L2 (R+ ) → 2 L (R+ ), for any f ∈ L2 (R+ ), there exists a function, h = Tf ∈ L2 (R+ ), satisfying ∞ f (x) dx, y ∈ (0, ∞) . (1.10) Tf (y) = h (y) = x+y 0 Hence, for any g ∈ L2 (R+ ), we may still deﬁne the inner product of Tf and g as follows: ∞ ∞ ∞ ∞ f (x) f (x) g (y) dx g (y) dy = dxdy. Tf, g = x+y x+y 0 0 0 0 (1.11)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Recent Developments of Hilbert-Type Inequalities with Applications

ws-book9x6

5

∞ 1 Setting the norm of f as f 2 = 0 f 2 (x) dx 2 , if f 2 , and g2 > 0, then (1.2) my be rewritten in the operator form Tf, g < π f 2 g2 . (1.12) It follows that ||T|| = π (see Carleman [5]), and we have the equivalent form of (1.2) as Tf < π f 2 (see Hardy et al [21]), that is, 2

∞ 0

∞ 0

2 ∞ f (x) dx dy < π 2 f 2 (x) dx, x+y 0

(1.13)

where the constant factor π 2 is still the best possible. It is obvious that inequality (1.13) is the integral analogue of (1.9). 1.2.3

A More Accurate Discrete Hilbert’s Inequality

If we set the subscripts m, n of the double series from zero to inﬁnity, then, we may rewrite inequality (1.1) equivalently in the following form: 12 ∞ ∞ ∞ ∞ a m bn 2 2 <π an bn , (1.14) m+n+2 n=0 m=0 n=0 n=0 where the constant factor π is still the best possible. Obviously, we may raise the following question: Is there a positive constant α (< 2), that makes 1 inequality still valid as we replace 2 by α in the kernel m+n+2 ? The answer is positive. That is the following more accurate Hilbert’s inequality (for short, Hilbert’s inequality) (see Hardy et al. [21]): 12 ∞ ∞ ∞ ∞ a m bn 2 2 <π an bn , (1.15) m+n+1 n=0 m=0 n=0 n=0 where the constant factor π is the best possible. Since for am , bn ≥ 0, α ≥ 1, ∞ ∞

∞ ∞ a m bn am bn ≤ , m + n + α m +n+1 n=0 m=0 n=0 m=0

then, by (1.15) and for α ≥ 1, we obtain 12 ∞ ∞ ∞ ∞ a m bn 2 2 an bn . <π m+n+α n=0 m=0 n=0 n=0

(1.16)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

6

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

For 1 ≤ α < 2, inequality (1.16) is a reﬁnement of (1.14). Obviously, we have a reﬁnement of (1.9), which is equivalent to (1.16) as follows: ∞ 2 ∞ ∞ am < π2 a2n (1 ≤ α < 2) . (1.17) m + n + α n=0 m=0 n=0 For 0 < α < 1, in 1936, Ingham [39] proved: If α ≥ 12 , then ∞ ∞

∞ am an ≤π a2n ; m + n + α n=0 m=0 n=0

(1.18)

and if 0 < α < 12 , then ∞ ∞

∞ am an π ≤ a2n . m + n + α sin (απ) n=0 m=0 n=0

(1.19)

Note. If we put x = X + α2 , y = Y + α2 , F (X) = f X + α2 and G (Y ) = g Y + α2 (α ∈ R) in (1.2), then we obtain

∞ −α 2

∞ −α 2

F (X) G (Y ) dXdY < π X +Y +α

∞

−α 2

2

F (X) dX

∞ −α 2

12 2

G (Y ) dY

.

(1.20) inequality (1.20) is an integral analogue of (1.18) with G = F . For α ≥ However, if 0 < α < 12 , inequality (1.20) is not an integral analogue of (1.19), because two constant factors are diﬀerent. Using the improved version of the Euler-Maclaurin summation formula and introducing new parameters, several authors including Yang ([111], [174]) recently obtained several more accurate Hilbert-type inequalities and some new Hardy-Hilbert inequality with applications. 1 , 2

1.2.4

Hilbert’s Inequality with One Pair of Conjugate Exponents

In 1925, by introducing one pair of conjugate exponents (p, q) with 1p + 1q = 1, Hardy [20] gave an extension of (1.1) as follows: ∞ q p If p > 1, am , bn ≥ 0, such that 0 < ∞ m=1 am < ∞ and 0 < n=1 bn < ∞, then ∞ p1 ∞ 1q ∞ ∞ a m bn π p am bqn , (1.21) < π m + n sin n=1 m=1 m=1 n=1 p

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Recent Developments of Hilbert-Type Inequalities with Applications

ws-book9x6

7

π where the constant factor sin(π/p) is the best possible. The equivalent discrete form of (1.21) is as follows: ⎤p ∞ p ⎡ ∞ ∞ am π ⎣ ⎦ < apn , (1.22) π m + n sin n=1 m=1 n=1 p

#p

"

π where the constant factor sin(π/p) is still the best possible. Similarly, inequalities (1.15) and (1.17) (for α = 1) may be extended to the following equivalent forms (see Hardy et al. [21]): p1 ∞ q1 ∞ ∞ ∞ am bn π p < am bqn , (1.23) π m + n + 1 sin n=0 m=0 m=0 n=0 ∞ n=0

∞

am m + n+1 m=0

p

p

⎡

<⎣ sin

⎤p ∞ π ⎦ apn , π p

(1.24)

n=0

" #p π π where the constant factors sin(π/p) and sin(π/p) are the best possible. The equivalent integral analogues of (1.21) and (1.22) are given as follows: p1 ∞ 1q ∞ ∞ ∞ f (x) g (y) π dxdy < f p (x) dx g q (y) dy , x+y 0 0 0 0 sin πp

∞ 0

∞ 0

⎡

(1.25)

⎤p

p ∞ f (x) π ⎦ dx dy < ⎣ f p (x) dx. π x+y 0 sin p

(1.26)

We call (1.21) and (1.23) as Hardy-Hilbert’s inequality and call (1.25) as Hardy-Hilbert’s integral inequality. Inequality (1.23) may be expressed in the form of operator as follows: Tp : lp → lp is a linear operator, such that for any non-negative sequence ∞ ∞ a = {am }m=1 ∈ lp , there exists Tp a = c = {cn }n=1 ∈ lp , satisfying cn = (Tp a) (n) =

∞

am , m + n+1 m=0

n ∈ N0 = N ∪ {0} .

(1.27)

q And for any non-negative sequence b = {bn }∞ n=1 ∈ l , we can deﬁne the formal inner product of Tp a and b as follows: ∞ ∞ ∞ ∞ am am bn bn = . (1.28) (Tp a, b) = m + n + 1 m +n+1 n=0 m=0 n=0 m=0

November 7, 2013

8

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Then inequality (1.23) may be rewritten in the operator form π ap bq , (Tp a, b) < sin πp

(1.29)

where ap , bq > 0. The operator Tp is called Hardy-Hilbert’s operator. Similarly, we deﬁne the following Hardy-Hilbert’s integral operator Tp : Lp (R+ ) → Lp (R+ ) as follows: For any f (≥ 0) ∈ Lp (R+ ), there exists a h = Tp f ∈ Lp (R+ ), deﬁned by ∞ f (x) (1.30) Tp f (y) = h (y) = dx, y ∈ R+ . x+y 0 And for any g (≥ 0) ∈ Lq (R+), we can deﬁne the formal inner product of Tp f and g as follows: ∞ ∞ f (x) g (y) dxdy. (1.31) Tp f, g = x+y 0 0 Then inequality (1.25) may be rewritten in the operator form as follows: π f p gq . Tp f, g < (1.32) sin πp On the other hand, if (p, q) is not a pair of conjugate exponents, then we have the following results (see Hardy et al. [21]): If p > 1, q > 1, p1 + 1q ≥ 1, 0 < λ = 2 − 1p + 1q ≤ 1, then ∞ p1 ∞ q1 ∞ ∞ a m bn p q ≤K am bn , (1.33) λ n=1 m=1 (m + n) m=1 n=1 where K = K (p, q) relates to p, q, only for 1p + 1q = 1, λ = 2 − p1 + 1q = 1, the constant factor K is the best possible. The integral analogue of (1.33) is given by ∞ p1 ∞ 1q ∞ ∞ f (x) g (y) p q dxdy ≤ K f (x) dx g (y) dy . λ (x + y) 0 0 0 0 (1.34) We also ﬁnd an extension of (1.34) as follows (see Mitrinovi´ c et al [57]):

If p > 1, q > 1, p1 + 1q > 1, 0 < λ = 2 − 1p + 1q < 1, then ∞ p1 ∞ 1q ∞ ∞ f (x) g (y) p q dxdy ≤ k (p, q) f (x)dx g (x)dx . λ −∞ −∞ |x + y| −∞ −∞ (1.35) For f (x) = g (x) = 0, x ∈ (−∞, 0], inequality (1.35) reduces to (1.34). Leven [52] also studied the expression forms of the constant factors in (1.33) and (1.34). But he did not prove their best possible property. In 1951, Bonsall [2] considered the case of (1.34) for the general kernel.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

9

Recent Developments of Hilbert-Type Inequalities with Applications

A Hilbert-type Inequality with the General Homogeneous Kernel of Degree −1

1.2.5

If α ∈ R, the function k (x, y) is measurable in R2+ , satisfying for any x, y, u > 0, k (ux, uy) = uα k (x, y), then k (x, y) is called the homogeneous function of degree α. In 1934, Hardy et al [21] published the following theorem: Suppose that p > 1, p1 + 1q = 1, k1 (x, y) (≥ 0) is a homogeneous function of degree −1 in R2+ . If f (x), g(y) ≥ 0, f ∈ Lp (R+ ), g ∈ Lq (R+ ),

∞

∞ 1 1 k = 0 k1 (u, 1) u− p du is ﬁnite, then we have k = 0 k1 (1, u) u− q du and the following equivalent integral inequalities:

∞ 0

∞

0 ∞

0

k1 (x, y) f (x) g (y) dxdy ≤ k

∞ 0

∞

p1 f (x)dx p

0

p p k1 (x, y) f (x) dx dy ≤ k

∞

q

1q

g (y)dy

,

0

(1.36) ∞

f p (x) dx,

(1.37)

0

where the constant factor k is the best possible. Moreover, if am , bn ≥ −1 −1 p ∞ q p and k1 (1, u) u q 0, a = {am }∞ m=1 ∈ l , b = {bn }n=1 ∈ l , both k1 (u, 1) u are decreasing in R+, then we have the following equivalent discrete forms:

∞ ∞

k1 (m, n) am bn ≤ k

n=1 m=1 ∞ ∞

k1 (m, n) am

n=1

m=1

p ≤ kp

∞

m=1 ∞

p1 apm

apn .

∞

q1 bqn

,

(1.38)

n=1

(1.39)

n=1

∞ 1 For 0 < p < 1, if k = 0 k1 (u, 1) u− p du is ﬁnite, then we have the reverses of (1.36) and (1.37). Note. We have not seen any proof of (1.36)–(1.39) and the reverse examples in [21]. We call k1 (x, y) the kernel of (1.36) and (1.37). If all the integrals and series in the right hand side of inequalities (1.36)–(1.39) are positive, then we can obtain the following particular examples (see Hardy et al. [21]): 1 in (1.36)–(1.39), they reduce to (1.25), (1.26), (1). For k1 (x, y) = x+y (1.21) and (1.22); 1 (2). If k1 (x, y) = max{x,y} in (1.36)–(1.39), they reduce the following

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

10

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

two pairs of equivalent forms:

∞

0

∞ 0

∞

0

f (x) g (y) dxdy < pq max {x, y}

∞

p1 f p (x) dx

0

∞

g q (y) dy

q1 ,

0

p ∞ f (x) p dx dy < (pq) f p (x) dx; max {x, y} 0 0 ∞ p1 ∞ q1 ∞ ∞ am bn p q < pq am bn , max {m, n} n=1 m=1 m=1 n=1 ∞ p ∞ ∞ am p < (pq) apn ; max {m, n} n=1 m=1 n=1

(1.40)

∞

(1.41)

(1.42)

(1.43)

in (1.36)–(1.39), they reduce to the following (3). If k1 (x, y) = ln(x/y) x−y two pairs of equivalent inequalities: ∞ ∞ ln x f (x) g (y) y dxdy x−y 0 0 ⎡ ⎤2 p1 ∞ 1q ∞ π ⎦ <⎣ f p (x) dx g q (y) dy , (1.44) 0 0 sin πp

∞ 0

⎛ ⎝

∞

ln

x y

f (x)

x−y

0

∞ ∞ ln n=1 m=1

∞ n=1

⎡

⎞p

⎤2p π ⎦

dx⎠ dy < ⎣ sin

π p

∞

f p (x)dx;

(1.45)

0

⎤2 ⎡ p1 ∞ 1q ∞ b a π m n p n ⎦ <⎣ am bqn , π m−n sin p m=1 n=1

m

∞ ln m n am m −n m=1

p

⎡

⎤2p π ⎦

<⎣ sin

π p

(1.46) ∞

apn .

(1.47)

n=1

Note. The constant factors in the above inequalities are all the best possible. We call (1.42) and (1.46) Hardy-Littlewood-P` olya’s inequalities (or H-L-P inequalities). We ﬁnd that the kernels in the above inequalities are all decreasing functions. But this is not necessary. For example, we ﬁnd the following two pairs of equivalent forms with the non-decreasing kernel

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

11

Recent Developments of Hilbert-Type Inequalities with Applications

(see Yang [121]):

∞

0

∞

ln xy f (x) g (y) max {x, y} 2 2 < p +q

0

dxdy

p1 f (x) dx

∞

0

∞

⎛ ⎝

0

∞

ln xy f (x) max {x, y}

0

p dx⎠ dy < p2 + q 2

n=1

max {m, n}

, (1.48)

∞

f p (x) dx,

(1.49)

p m am ln

m=1

max {m, n}

n

< p2 + q 2

∞

p1 apm

m=1

∞

g (y) dy

0

n

∞

1q

q

0

⎞p

∞ ∞ m a m bn ln n=1 m=1

∞

p

∞

1q bqn

,

n=1

(1.50) p < p2 + q 2

∞

apn ,

(1.51)

n=1

p where the constant factors p2 + q 2 and p2 + q2 are the best possible. Other types of inequalities with the best constant factors are as follows (see Xin and Yang [88]): ∞ ∞ ln x f (x) g (y) y dxdy x+y 0 0 ∞ p1 ∞ q1 p p f (x) dx g (y) dy , (1.52) < c0 (p) 0

0

∞

⎛ ⎝

∞

0

ln xy f (x) x+y

0

⎞p dx⎠ dy < cp0 (p)

∞ m ∞ ln a m bn n

∞

f p (x) dx;

∞

a2m

< c0 (2) m+n m=1 ∞ 2 ∞ ∞ m am ln n < c20 (2) a2n , m + n n=1 m=1 n=1 n=1 m=1

where the constant factor c0 (p) = 2

∞

(1.53)

0

n−1 n=1 (−1)

∞

12 b2n

, (1.54)

n=1

(1.55)

(

1 2 (n− p1 )

) −

1 2 (n− q1 )

.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

12

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Two Multiple Hilbert-type Inequalities with the Homogeneous Kernels of Degree (−n + 1)

1.2.6

Suppose that n ∈ N\ {1}, n numbers p, q, · · · , r satisfying p, q, · · · , r > 1, p−1 + q −1 + · · · + r−1 = 1, k (x, y, · · · , z) ≥ 0 is a homogeneous function of degree (−n + 1). If ∞

∞

k= 0

0

···

∞

0

1

1

k (1, y, · · · , z) y − q · · · z − r dy · · · dz

is a ﬁnite number, f , g, · · · , h are non-negative measurable functions in R+ , then, we have the following multiple Hilbert-type integral inequality (see Hardy et al.[21]): ∞

0

≤k

∞

0

···

∞

∞

0

k (x, y, · · · , z) f (x) g (y) · · · h (z) dxdy · · · dz

p1 f p (x) dx

0

∞

g q (y) dy

1q

0

···

∞

hr (z) dz

r1 . (1.56)

0

1

Moreover, if am , bn , · · · , cs ≥ 0, k (1, y, · · · , z) x0 y − q · · · z − r , 1 1 1 1 k (x, 1, · · · , z) x− p y 0 · · · z − r , · · · , k (x, y, · · · , 1) x− p y − q · · · z 0 are all decreasing functions with respect to any single variable in R+ , then, we have ∞ ∞ ∞ ··· k(m, n, · · · , s)am bn · · · cs s=1

n=1 m=1

≤k

∞ m=1

p1 apm

∞ n=1

q1 bqn

···

∞

1

1r crs

.

(1.57)

s=1

Note. The authors did not write and prove that the constant factor k in the above inequalities is the best possible. For two numbers p and q (n = 2), inequalities (1.56) and (1.57) reduce respectively to (1.36) and (1.38).

1.3 1.3.1

Modern Research for Hilbert-type Inequalities Modern Research for Hilbert’s Integral Inequality

In 1979, based on an improvement of H¨ older’s inequality, Hu [33] proved a reﬁnement of (1.2) (for f = g) as follows: ∞ ∞ f (x) f (y) dxdy x+y 0 0 ∞ 2 2 12 ∞ √ 1 f 2 (x) dx − f 2 (x) cos xdx . <π 4 0 0 (1.58)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

13

Recent Developments of Hilbert-Type Inequalities with Applications

Since then, Hu [34] published many interesting results similar to (1.58). In 1998, Pachpatte [58] gave an inequality similar to (1.2) as follows: For a, b > 0, a b f (x) g (y) dxdy x+y 0 0 12 √ a b ab 2 2 (a − x) f (x) dx (b − x) g (y) dy . (1.59) < 2 0 0 Some improvements and extensions have been made by Zhao and Debnath [173], Lu [55] and He and Li [23]. We can also refer to other works of Pachpatte in [59]. In 1998, by introducing parameters λ ∈ (0, 1] and a, b ∈ R+ (a < b), Yang [96] gave an extension of (1.2) as follows: b b f (x) g (y) dxdy λ a a (x + y) 12 ( b a λ4 ) b λ λ 1− ,
0

0

π f (x) f (y) π dxdy < xλ + y λ λ sin 2λ

∞

f 2 (x) dx

0

∞

g 2 (y) dy

12 .

0

(1.61) We can refer to the other works of Kuang in [47] and [48]. In 1999, using the methods of algebra and analysis, Gao [15] proved an improvement of (1.2) as follows:

∞ 0

0

∞

√ f (x) f (y) dxdy < π 1 − R x+y

∞

2

1/2

1 π

∞

f 2 (x) dx

0

∞

g 2 (y) dy

0

2

+

12 , (1.62)

2 π

− , u = (g, e), v = where ||f || = 0 f (x)dx , R = √

∞ ex −x 2π (f, e ), e (y) = 0 x+y dx. We also refer to works of Gao and Hsu in [17]. u g

v f

November 7, 2013

12:24

14

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

In 2002, using the operator theory, Zhang [172] gave an improvement of (1.2) as follows: ∞ ∞ f (x) f (y) dxdy x+y 0 0 ∞ 2 12 ∞ ∞ π 2 2 f (x) dx g (x) dx + f (x) g (x) dx . ≤ √ 2 0 0 0 (1.63) 1.3.2

On the Way of Weight Coeﬃcient for Giving a Strengthened Version of Hilbert’s Inequality

In 1991, for making an improvement of (1.1), Hsu and Wang [32] raised the way of weight coeﬃcient as follows: At ﬁrst, using Cauchy’s inequality in the left hand side of (1.1), it follows that ( 1 ) ( 1 ) ∞ ∞ ∞ ∞ m 4 am b n 1 n 4 = I= am bn m + n m + n n m n=1 m=1 n=1 m=1

∞ 12 ∞ ∞ ∞ m 12 1 m 12 1 . a2m b2n ≤ m + n n m + n n m=1 n=1 n=1 m=1 (1.64) Then, we deﬁne the weight coeﬃcient ∞ 1 m 12 , ω (n) = m+n n m=1

n ∈ N,

and rewrite (1.64) as follows: ∞ 12 ∞ I≤ ω (m) a2m ω (n) b2n . m=1

(1.65)

(1.66)

n=1

Setting ω (n) = π −

θ (n) , n1/2

n ∈ N,

(1.67)

where, θ (n) = (π − ω (n)) n1/2 , and estimating the series of θ (n), it follows that

∞ 1 m 12 θ (n) = π − (1.68) n1/2 > θ = 1.1213+. m + n n m=1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Recent Developments of Hilbert-Type Inequalities with Applications

ws-book9x6

15

Thus, result (1.67) yields ω (n) < π −

θ , n1/2

n ∈ N,

θ = 1.1213+.

In view of (1.66), a strengthened version of (1.1) is given by ∞ 12 ∞ θ θ . I< π − 1/2 a2n π − 1/2 b2n n n n=1 n=1

(1.69)

(1.70)

Hsu and Wang [32] also raised an open question how to obtain the best value θ of (1.70). In 1992, Gao [14] gave the best value θ = θ0 = 1.281669+. Xu and Gao [91] proved a strengthened version of (1.11) given by ⎧ ⎡ ⎤ ⎫ p1 ∞ ⎨ ⎬ p−1 ⎣ π ⎦ apn − 1/p I< ⎩ ⎭ n + n−1/q sin π n=1

p

⎧ ⎡ ⎤ ⎫ 1q ∞ ⎨ ⎬ q − 1 π ⎣ ⎦ bqn − 1/q × . (1.71) ⎩ ⎭ π n + n−1/p n=1 sin p

In 1997, using the way of weight coeﬃcient and the improved EulerMaclaurin’s summation formula, Yang and Gao ([144], [18]) showed that ⎧ ⎡ ⎤ ⎫ p1 ∞ ⎬ ⎨ 1−γ ⎣ π − 1/p ⎦ apn I< π ⎭ ⎩ n n=1 sin p

⎧ ⎡ ⎤ ⎫ 1q ∞ ⎨ ⎬ 1−γ ⎣ π − 1/q ⎦ bqn × , ⎩ ⎭ π n n=1 sin

(1.72)

p

where 1 − γ = 0.42278433+ (γ is the Euler constant). In 1998, Yang and Debnath [140] gave another strengthened version of (1.11), which is an improvement of (1.71). We can also refer to some strengthened versions of (1.15) and (1.23) in papers of Yang [97], Yang and Debnath [143]. 1.3.3

Hilbert’s Inequality with Independent Parameters

In 1998, using the optimized weight coeﬃcients and introducing an independent parameter λ ∈ (0, 1], Yang [96] provided an extension of (1.2) as follows:

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

16

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

∞ x1−λ f 2 (x) dx < ∞ and 0 < 0 x1−λ g 2 (x) dx < ∞, then ∞ ∞ f (x) g (y) dxdy λ (x + y) 0 0 ∞ 12 ∞ λ λ 1−λ 2 1−λ 2 , x f (x) dx x g (x) dx , (1.73) 0) . (1.74) = tu+v 1 If 0 <

∞ 0

Some extensions of (1.21), (1.23) and (1.25) were given by Yang and Debnath ([98], [141], [142]) as follows: If λ > 2 − min {p, q}, then ∞ ∞ p+λ−2 q+λ−2 f (x) g (y) dxdy < B , λ p q (x + y) 0 0 p1 ∞ 1q ∞ x1−λ f p (x) dx x1−λ g q (x) dx . × 0

0

(1.75) If 2 − min {p, q} < λ ≤ 2, then ∞ ∞ p+λ−2 q+λ−2 a m bn < B , λ p q n=1 m=1 (m + n) 1 p ∞ 1q ∞ × n1−λ apn n1−λ bqn , n=1

(1.76)

n=1

p+λ−2 q+λ−2 ,
a m bn

[99] also proved that (1.76) is valid for p = 2 and λ ∈ (0, 4]. Yang ([100],

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

17

Recent Developments of Hilbert-Type Inequalities with Applications

[115]) gave another extensions of (1.21) and (1.23) as follows: If 0 < λ ≤ min {p, q}, then ∞ ∞

a m bn mλ + nλ n=1 m=1 ∞ p1 ∞ q1 π (p−1)(1−λ) p (q−1)(1−λ) q < n an n bn , (1.78) λ sin πp n=1 n=1 and if 0 < λ ≤ 1, then ∞ ∞

a m bn 1 λ + (n + 12 )λ n=0 m=0 m + 2 ∞ p−1−λ p1 q−1−λ 1q ∞ π 1 1 p < an bqn . n+ n+ π 2 2 λ sin p n=0 n=0 (1.79) In 2004, Yang [104] proved the following dual form of (1.21): ∞ p1 ∞ q1 ∞ ∞ a m bn π p−2 p q−2 q < n an n bn . m+n sin π n=1 m=1 n=1 n=1

(1.80)

p

Inequality (1.80) reduces to (1.21) when p = q = 2. For λ = 1, (1.79) reduces to the dual form of (1.23) as follows: ∞ ∞

am bn m+n+1 n=0 m=0 ∞ p−2 p1 q−2 q1 ∞ π 1 1 p < an bqn . (1.81) n+ n+ π 2 2 sin p n=0 n=0 We can ﬁnd some extensions of the H-L-P inequalities with the best constant factors such as (1.40)-(1.51) (see Wang and Yang [75], Yang [102]– [103]) by introducing some independent parameters. In 2001, by introducing some parameters, Hong [29] gave a multiple integral inequality, which is an extension of (1.21). He et al [28] gave a similar result for particular conjugate exponents. For making an improvement of their works, Yang [101] gave the following inequality, which is a best exn 1 = 1, λ > n − min {pi }, tension of (1.21): If n ∈ N\ {1}, pi > 1, pi i=1

1≤i≤n

November 7, 2013

12:24

18

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

∞ fi (t) ≥ 0 and 0 < 0 tn−1−λ fipi (t) dt < ∞ : (i = 1, 2, · · · , n), then, we have ∞ ∞ n i=1 fi (xi ) ··· dx1 · · · dxn n λ 0 0 ( i=1 xi ) ∞ p1 n i 1 2 pi + λ − n n−1−λ pi < Γ t fi (t) dt , (1.82) Γ (λ) i=1 pi 0 n pi +λ−n 1 is the best possible. In where the constant factor Γ(λ) i=1 Γ pi particular, for λ = n − 1, it follows that ∞ n ∞ i=1 fi (xi ) ··· dx1 · · · dxn n n−1 ( i=1 xi ) 0 0 ∞ p1 n i 1 2 1 < Γ 1− fipi (t)dt . (1.83) Γ (λ) i=1 pi 0 In 2003, Yang and Rassias [147] introduced the way of weight coefﬁcient and considered its applications to Hilbert-type inequalities. They summarized how to use the way of weight coeﬃcients to obtain some new improvements and generalizations of the Hilbert-type inequalities. Since then, a number of authors discussed this problem (see Brneti´c et al. [3]– [4], Chen and Xu [6], Gao [16], Gao et al. [19], He et al. [26]–[27], Jia et al. [40]–[41], Krni´c et al. [43]–[45], Laith [50], Lu [56], Salem [65], Sulaiman [68]–[69], Wang and Xin [74], Xie and Lu [80]–[81], Xu [91], and Yang [110]). But how to give an uniform extension of inequalities (1.80) and (1.21) with a best possible constant factor, this was solved in 2004 by introducing two pairs of conjugate exponents (see Yang [105]). 1.3.4

Hilbert-type Inequalities with Multi-parameters

In 2004, by introducing an independent parameter λ > 0 and two pairs of conjugate exponents (p, q) and (r, s) with p1 + 1q = 1r + 1s = 1, Yang [105] gave an extension of (1.2) as follows: If p, r > 1, and the integrals of the right hand side are positive, then ∞ ∞ f (x) g (y) dxdy xλ + y λ 0 0 p1 ∞ 1q ∞ π p(1− λ −1 p q(1− λ −1 q ) ) r s x f (x) dx x g (x) dx , < λ sin πr 0 0 (1.84)

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Recent Developments of Hilbert-Type Inequalities with Applications

where the constant factor

π λ sin( π r)

ws-book9x6

19

is the best possible.

For λ = 1, r = q, s = p, inequality (1.84) reduces to (1.25); for λ = 1, r = p, s = q, inequality (1.84) reduces to the dual form of (1.25) as follows:

∞

0

<

∞

f (x) g (y) dxdy x+y 0 ∞ p1 π p−2 p x f (x) dx π p

sin

0

∞

q1 g (x) dx .

q−2 q

x

(1.85)

0

In 2005, by introducing an independent parameter λ > 0, and two pairs of generalized conjugate exponents (p1 , p2 , · · · , pn ) and (r1 , r2 , · · · , rn ) with n n 1 1 = pi ri = 1, Yang et al. [149] gave a multiple integral inequality as i=1

i=1

follows: For pi , ri > 1 (i = 1, 2, · · · , n),

∞

∞

n

i=1 fi (xi ) dx1 · · · dxn n λ 0 0 ( i=1 xi ) ∞ p1 n i 1 2 λ pi 1− rλ −1 pi i < Γ t fi (t) dt , Γ (λ) i=1 ri 0

···

where the constant factor

1 Γ(λ)

n

(1.86)

i=1 Γ

λ ri

is the best possible. For n = 2,

p1 = p, p2 = q, r1 = r and r2 = s, inequality (1.86) reduces to the following:

∞

∞

f (x) g (y)

dxdy λ (x + y) p1 ∞ 1q ∞ λ λ p(1− λ −1 p q (1− λ −1 q ) ) r s x f (x) dx x g (x) dx .
0

It is obvious that inequality (1.87) is another best extension of (1.25). In 2006, using two pairs of conjugate exponents (p, q) and (r, s) with p, r > 1, Hong [31] gave a multi-variable integral inequality as follows: If Rn+ = {x = (x1 , x2 , · · · , xn ) ; xi > 0, i = 1, 2, · · · , n}, α, β, λ > 0, n α1

p n− βλ −n α x , f , g ≥ 0, 0 < x ( r ) f p (x) dx < ∞ and x = i

α

i=1

Rn +

α

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

20

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

q(n− βλ s )−n q 0 < Rn xα g (x) dx < ∞, then + Γn α1 λ λ f (x) g (y) dxdy < B , λ n r s βαn−1 Γ α β β Rn Rn + + xα + yα p1 1q βλ p(n− βλ −n q n− −n ) ( ) p q r s xα f (x) dx xα g (x) dx , × Rn +

Rn +

(1.88) n

where the constant factor

Γ

1 α

( )

n βαn−1 Γ( α )

B

λ

λ r, s

is the best possible. In par-

ticular, for n = 1, (1.88) reduces to Hong’s work in [30]; for n = β = 1, (1.88) reduces to (1.87). In 2007, Zhong and Yang [181] generalized (1.88) to a general homogeneous kernel and proposed the reversion. We can ﬁnd another inequality with two parameters as follows (see Yang [108]): ∞ ∞

a m bn

+ n α )λ p1 ∞ 1q ∞ αλ λ λ 1 p(1− αλ −1 q 1− −1 p q ) ( ) r s , 0, αλ ≤ min {r, s}. In particular, for α = 1, we have ∞ ∞

a m bn λ

(m + n) p1 ∞ q1 ∞ λ λ λ p(1− λ −1 q 1− −1 p q ) ( ) r s
For λ = 1, r = q, inequality (1.90) reduces to (1.21), and for λ = 1, r = p, (1.90) reduces to (1.80). Also we can obtain the reverse form as follows (see Yang [109]): ∞ ∞

a m bn 2

n=0 m=0 (m + n + 1) p1 ∞ 1q ∞ bq 1 apn n 1− >2 , 2 2n + 1 2n + 1 4 (n + 1) n=0 n=0

(1.91)

where 0 < p < 1, p1 + 1q = 1. The other results on the reverse of the Hilbert-type inequalities are found in Xi [79] and Yang [112].

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Recent Developments of Hilbert-Type Inequalities with Applications

ws-book9x6

21

In 2006, Xin [87] gave a best extension of H-L-P integral inequality (1.45) as follows:

2 ∞ ∞ ln x f (x) g (y) y π dxdy < xλ − y λ sin πr 0 0 ∞ p1 ∞ q1 p(1− λ −1 p q(1− λ −1 q ) ) r s × x f (x) dx x g (x) dx . (1.92) 0

0

In 2007, Zhong and Yang [176] gave an extension of another H-L-P integral inequality (1.40) as follows: ∞ ∞ f (x) g (y) dxdy max {xλ , y λ } 0 0 ∞ p1 ∞ q1 rs p(1− λ −1 p q (1− λ −1 q ) ) r s < x f (x) dx x g (x) dx . (1.93) λ 0 0 Zhong and Yang [177] also gave the reverse form of (1.93). Considering a particular kernel, Yang [118] proved ∞ ∞ a m bn √ √ 3 ( m + n) max {m, n} n=1 m=1 ∞ p1 ∞ q1 p q < 4 ln 2 n 2 −1 apn n 2 −1 bqn . n=1

(1.94)

n=1

Yang [114] also proved that ∞ ∞ a m bn 2

n=1 m=1

(m + an) + n2 <

π 2

− arctan a

∞ apn n n=1

p1

∞ bqn n n=1

1q (a ≥ 0) . (1.95)

Using the residue theory, Yang [123] obtained the following inequality ∞ ∞ f (x) g (y) dxdy (x + ay) (x + by) (x + cy) 0 0 p1 ∞ q1 ∞ p q x− 2 −1 f p (x) dx x− 2 −1 g q (x) dx , (1.96)
0 √ 1√ √ √ √ √ ( a+ b)( b+ c)( a+ c)

0

(a, b, c > 0). The constant factors in

the above new inequalities are all the best possible. Some other new results are proved by several authors including He et al. [27], Li and He [53], Xie [81]–[84].

November 7, 2013

12:24

22

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

1.4

Some New Applications for Hilbert-type Inequalities

1.4.1

Operator Expressions of Hilbert-type Inequalities

Suppose that H is a separable Hilbert space and T : H → H is a bounded self-adjoint semi-positive deﬁnite operator. In 2002, Zhang [172] proved the following inequality: 2 T 2 2 2 2 (a, b ∈ H) , (1.97) (a, T b) ≤ a b + (a, b) 2 3 where (a, b) is the inner product of a and b, and a = (a, a) is the norm of a. Since the Hilbert integral operator T deﬁned by (1.10) satisﬁes the condition of (1.97) with T = π, then inequality (1.2) may be improved as (1.63). Since the operator Tp deﬁned by (1.27) (for p = q = 2) satisﬁes the condition of (1.97) (see Wilhelm [78]), we may improve (1.15) to the following form: ⎧ ∞ 2 ⎫ 12 ∞ ∞ ∞ ∞ ⎬ am b n π ⎨ 2 2 an bn + a n bn . (1.98) < √ ⎭ m+n+1 2⎩ n=0 m=0

n=0

n=0

n=0

The key of applying (1.97) is to obtain the norm of the operator and to show the semi-deﬁnite property. Now, we consider the concept and the properties of Hilbert-type integral operator as follows: Suppose that p > 1, 1p + 1q = 1, Lr (R+ ) (r = p, q) are real normal linear spaces and k (x, y) is a non-negative symmetric measurable function in R2+ satisfying ∞ x r1 k (x, t) dt = k0 (p) ∈ R (x > 0) . (1.99) t 0 We deﬁne an integral operator as T : Lr (R+ ) → Lr (R+ ) (r = p, q) , for any f (≥ 0) ∈ Lp (R+ ), there exists h = T f ∈ Lp (R+ ), such that ∞ k (x, y) f (x) dx (y > 0) . (1.100) (T f ) (y) = h (y) := 0

Or, for any g (≥ 0) ∈ L (R+), there exists h = T g ∈ Lq (R+ ), such that ∞ k (x, y) g (y) dy (x > 0). (1.101) (T g) (x) = h(x) := q

0

In 2006, Yang [113] proved that the operator T deﬁned by (1.100) or (1.101) are bounded with T ≤ k0 (p). The following are some

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Recent Developments of Hilbert-Type Inequalities with Applications

ws-book9x6

23

results of [113]: If ε > 0, is small enough and the integral 1+ε

∞ k (x, t) xt r dt (r = p, q; x > 0) is convergent to a constant kε (p) in0 dependent of x satisfying kε (p) = k0 (p)+o (1) (ε → 0+ ), then T = k0 (p). If T > 0, f ∈ Lp (R+ ), g ∈ Lq (R+ ), f p , gq > 0, then we have the following equivalent inequalities: (T f, g) < T · f p gq , (1.102) (1.103) T f p < T · f p . Some particular cases are considered in this paper [113]. Yang [119] also considered some properties of Hilbert-type integral operator (for p = q = 2). For the homogeneous kernel of degree −1, Yang [125] found some suﬃcient conditions to obtain T = k0 (p). We can see some properties of the discrete Hilbert-type operator in the discrete space in Yang ([116], [120], [122], [124]). Recently, B´enyi and Choonghong [1] proved some new results concerning best constants for certain multi-linear integral operators. In 2009, Yang [129] summarized the above part results. Some other works about Hilbert-type operators and inequalities with the general homogeneous kernel and multi-parameters were provided by several other authors (see Huang [35], Liu and Yang [54], Wang and Yang [76], Xin and Yang [89], Yang [114] and [133], Yang and Krnic [145] and Yang and Rassias [148]).

1.4.2

Some Basic Hilbert-type Inequalities

If the Hilbert-type inequality relates to a single symmetric homogeneous 1 , or |ln(x/y)| ) and the best constant factor kernel of degree −1 (such as x+y x+y is a more brief form, which does not relate to any conjugate exponents (such as (1.2)), then we call it basic Hilbert-type integral inequality. Its series analogue (if exists) is also called basic Hilbert-type inequality. If the simple homogeneous kernel is of degree −λ (λ > 0) with a parameter λ and the inequality cannot be obtained by a simple transform to a basic Hilbert-type integral inequality, then we call it a basic Hilbert-type integral inequality with a parameter. For examples, we call the following integral inequality that is, (1.2) as ∞ 12 ∞ ∞ ∞ f (x) g (y) 2 2 dxdy < π f (x) dx g (x) dx , (1.104) x+y 0 0 0 0 and the following H-L-P inequalities (for p = 2 in (1.40) and (1.45)): 12 ∞ ∞ ∞ ∞ f (x) g (y) 2 2 f (x) dx g (x) dx , (1.105) dxdy < 4 max {x, y} 0 0 0 0

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

24

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

∞

0

∞

x y

ln

f (x) g (y) dxdy < π

x−y

0

2

∞

2

∞

f (x) dx 0

12 g (x) dx 2

0

(1.106) basic Hilbert-type integral inequalities. In 2005, Yang [106] gave the following basic Hilbert-type integral inequality: ∞ 12 ∞ ∞ ln x f (x) g (y) ∞ y dxdy < 8 f 2 (x) dx g 2 (x) dx ; max {x, y} 0 0 0 0 (1.107) In 2011, Yang [134] gave the following basic Hilbert-type integral inequalities: 0

∞

∞

ln xy f (x) g (y) x+y

0

∞ 0

∞

where, c0 = 8

∞ 0

dxdy < c0

arctan

+

0

∞

f 2 (x) dx

∞

12 g 2 (x) dx ,

0

x y

f (x) g (y) dxdy x+y ∞ 12 ∞ π2 f 2 (x) dx g 2 (x) dx , < 4 0 0

(−1)n (2n−1)2

(1.108)

(1.109)

= 7.3277+.

n=1

Yang ([106], [126], [128], [130], [146]) also gave a basic Hilbert-type integral inequality with a parameter λ ∈ (0, 1):

∞

∞

f (x) g (y)

dxdy λ |x − y| ∞ 12 ∞ λ x1−λ f 2 (x) dx x1−λ g 2 (x) dx . (1.110) < 2B 1 − λ, 2 0 0 0

0

Similar to discrete inequality (1.19), the following integral inequality ∞ ∞ f (x) g (y) dxdy λ (x + y) 0 0 ∞ 12 ∞ λ λ 1−λ 2 1−λ 2
November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Recent Developments of Hilbert-Type Inequalities with Applications

Also we ﬁnd the following basic Hilbert-type inequalities: 12 ∞ ∞ ∞ ∞ a m bn a2n b2n , <π m + n n=1 m=1 n=1 n=1 ∞ 12 ∞ ∞ ∞ a m bn 2 2 <4 an bn , max {m, n} n=1 m=1 n=1 n=1 ∞ 12 ∞ ∞ ∞ ln m b 2 2 2 n am n an bn . <π m−n n=1 m=1 n=1 n=1 It follows from (1.50) with p = q = 2 that ∞ 12 ∞ m ∞ ∞ ln n am bn 2 2 <8 an bn . max {m, n} n=1 m=1 n=1 n=1 In 2010, Xin and Yang [88] proved the following inequality ∞ 12 ∞ m ∞ ∞ ln n am bn 2 2 < c0 an bn , m+n n=1 m=1 n=1 n=1

ws-book9x6

25

(1.112)

(1.113)

(1.114)

(1.115)

(1.116)

∞ (−1)n + where, c0 = 8 n=1 (2n−1) 2 = 7.3277 . Inequalities (1.115) and (1.116) are new basic Hilbert-type inequalities. We still have a basic Hilbert-type inequality with a parameter λ ∈ (0, 4] as follows (see Yang [100]): 12 ∞ ∞ ∞ ∞ λ λ a m bn 1−λ 2 1−λ 2 ,
Some Applications to Half-discrete Hilbert-type Inequalities

In recent years, Draˇcic et al. [11]–[13], Pogany [60]–[61] made some new important contributions to discrete Hilbert-type inequalities with nonhomogeneous kernels using special functions. On the other hand, in 20062011, Xie et al. ([81]–[86]) have investigated many Hilbert-type integral −λ inequalities with the particular kernels such as |x + y| similar to inequality (1.35). In 2010-2012, Yang ([131]–[132]) considered the compositions of two discrete Hilbert-type operators with two conjugate exponents and −λ kernels (m + n) and in 2010, Liu and Yang [54] also studied the compositions of two Hilbert-type integral operators with the general homogeneous kernel of negative degree and obtained some new results with applications.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

26

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Hardy et al. [21] proved some results of half-discrete inequalities with the general non-homogeneous kernel in Theorem 351 without any proof of the constant factors as best possible. Yang [107] introduced an interval variable to give a half-discrete inequality with the particular non-homogeneous 1 kernel as (1+u(x)u(n)) λ , and proved that the constant factor is the best possible. In 2011, Yang [135] gave the following half-discrete Hilbert-type inequalities with the best possible constant factor B (λ1 , λ2 ):

∞

f (x) 0

∞

an

n=1

(x + n)λ

dx 0, 0 ≤ λ2 ≤ 1, λ1 + λ2 = λ, ||f ||p,φ = ||a||q,ψ =

∞

1/p ϕ(x)f (x)dx > 0,

0 ∞

p

1/q aqn

> 0,

ϕ(x) = xp(1−λ1 )−1 ,

n=1

ψ(n) = nq(1−λ2 )−1 . Zhong et al. ([174]–[177]) has investigated several halfdiscrete Hilbert-type inequalities with particular kernels. Using the way of weight functions and the techniques of discrete and integral Hilbert-type inequalities with some additional conditions on the kernel, a half-discrete Hilbert-type inequality with a general homogeneous kernel of degree −λ ∈ R is obtained as follows:

∞

f (x) 0

∞

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

(1.119)

n=1

∞ where k (λ1 ) = 0 kλ (t, 1) tλ1 −1 dt ∈ R+ . This is an extension of the above particular result with the best constant factor k (λ1 ) (see Yang and Chen [138]). If the corresponding integral inequality of a half-discrete inequality is a basic Hilbert-type integral inequality, then we call it the basic half-discrete Hilbert-type inequality. substituting some particular kernels in the main result found in [138] leads to some basic half-discrete Hilbert-type inequalities as follows:

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Recent Developments of Hilbert-Type Inequalities with Applications

∞

0

∞

f (x) 0

∞

∞

(1.120) (1.121) (1.122)

f (x)

∞ | ln(x/n)|an dx < 16||f ||2 ||a||2 , max{x, n} n=1

(1.123)

f (x)

∞ | ln(x/n)|an dx < c0 ||f ||2 ||a||2 , x+n n=1

(1.124)

0 ∞ 1 ∞ 1

where c0 = 8

∞ π 2 ln(x/n)an ||f ||2 ||a||2 , dx < x−n 2 n=1

27

an dx < 4||f ||2 ||a||2 , max{x, n} n=1

f (x)

∞ an π f (x) dx < ||f ||2 ||a||2 , x + n 2 n=1

ws-book9x6

∞

(−1)n (2n−1)2

= 7.3277+.

n=1

1.5

Concluding Remarks

(1) Many diﬀerent kinds of Hilbert-type discrete and integral inequalities with applications are presented in this chapter. Special attention is given to new results proved in recent years. Included are many generalizations, extensions and reﬁnements of Hilbert-type discrete and integral inequalities involving many special functions such as beta, gamma, hypergeometric, trigonometric, hyperbolic, zeta, Bernoulli’s functions and Bernoulli’s numbers and Euler’s constant. (2) For more information about Hilbert-type discrete and integral inequalities, the reader is referred to three recent research monograph by Yang ([128], [130], and [134]) who presented many new results on integral and discrete-type operators with general homogeneous kernels of degree of real numbers and two pairs of conjugate exponents as well as the related inequalities. These research monographs and a recent paper by Debnath and Yang [10] also contained recent developments of both discrete and integral types of operators and inequalities with proofs, examples and applications.

May 2, 2013

14:6

BC: 8831 - Probability and Statistical Theory

This page intentionally left blank

PST˙ws

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Chapter 2

Improvements of the Euler-Maclaurin Summation Formula and Applications

“Since a general solution must be judged impossible from want of analysis, we must be content with the knowledge of some special cases, and that all the more, since the development of various cases seems to be the only way to bringing us at least to a more perfect knowledge.” Leonhard Euler

2.1

Introduction

In 1732, Leonhard Euler (1707-1783) stated the Euler-Maclaurin summation formula which was independently discovered by Euler and Colin Maclaurin (1698-1746) in the period of 1732-1742. This chapter deals with some preliminary results to improve the Euler-Maclaurin summation formula for optimizing the methods of estimation of series and the weight coeﬃcients. Many useful theorems, corollaries and inequalities are discussed. Included are some applications involving new inequalities on Hurwitz’s zeta function restricted to the real axis, the Riemann zeta function and the extended Stirling formula.

2.2

2.2.1

Some Special Functions Relating Euler-Maclaurin’s Summation Formula Bernoulli’s Numbers

We deﬁne a function G(x) as follows: G(x) =

x , ex − 1

x∈R

G(0) = lim G(x) = 1 . x→0

29

ws-book9x6

November 7, 2013

30

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

It is obvious that the power series of G(x) at x = 0 possesses a positive convergence radius. Assuming that the sequence {Bn }∞ n=0 is deﬁned by the exponent creation function of G(x), and ∞ xn Bn , G(x) = n! n=0 we obtain ∞ ex − 1 xn 1= Bn x n=0 n!

n ∞ ∞ ∞ 1 B B n k = xn . xn xn = (n + 1)! n! k!(n − k + 1)! n=0 n=0 n=0 k=0 Comparing to the coeﬃcients of xn on two sides of the above equality, it follows that n Bk = 0 (n ∈ N). B0 = 1, and k!(n − k + 1)! k=0 We can obtain the recursion formula of {Bn }∞ n=0 as follows: n−1 Bk B0 = 1, Bn = −n! , n ∈ N. (2.1) k!(n − k + 1)! k=0 Since we have x(ex − 1 + 1) −x = = x + G(x), G(−x) := −x e −1 ex − 1 then, in view of ∞ xn Bn , G(x) = n! n=0 it follows that ∞ xn = x. Bn [(−1)n − 1] n! n=0 n

Comparing the coeﬃcients of xn! on two sides of the above equality, we obtain 1 (2.2) B1 = − , B2k+1 = 0, k ∈ N. 2 By (2.1) and (2.2), we can ﬁnd the constants of {B2n }∞ n=1 step by step as follows: 1 1 1 1 B2 = , B4 = − , B6 = , B8 = − , · · · . 6 30 42 30 We call Bn (n ∈ N0 = N ∪ {0}) the Bernoulli numbers. In a general way, we also obtain the following formula for the Bernoulli numbers (see Xu and Wang [92]): ∞ (2k)! 1 B2k = (−1)k+1 2k−1 2k , k ∈ N. (2.3) 2 π n=1 n2k

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

31

Improvements of the Euler-Maclaurin Summation Formula and Applications

2.2.2

Bernoulli’s Polynomials

Suppose that the function Bn (t) is deﬁned by the following exponent creation function: etx G(x) =

∞

Bn (t)

n=0

xn . n!

(2.4)

In the convergence interval of (2.4), we have ∞ n=0

Bn (t)

∞ n ∞ xn t Bn n = etx G(x) = xn x n! n! n! n=0 n=0

n ∞ n n n−k x = Bk t . k n! n=0

(2.5)

k=0

n

Comparing to the coeﬃcients of the term xn! on two sides of (2.5), we get the following formula: n n (2.6) Bn (t) = Bk tn−k , n ∈ N0 . k k=0

We call Bn (t) (n ∈ N0 ) Bernoulli polynomials. In particular, we obtain 1 B1 (t) = t − , 2 1 B2 (t) = t2 − t + , 6 3 1 B3 (t) = t3 − t2 + t, 2 2 1 B4 (t) = t4 − 2t3 + t2 − , · · · , 30 B0 (t) = 1,

and prove the equations that (see Xu and Wang [92]) 1 Bn (t)dt = 0, Bn−1 (0) = Bn−1 , Bn (t) = nBn−1 (t), 0

n ∈ N.

On the other hand, in view of B0 (t) = 1, Bn (t) = nBn−1 (t) and Bn−1 (0) = Bn−1 , n ∈ N, by integration, we ﬁnd 1 Bn−1 (t)dt, n ∈ N. (2.7) Bn (t) = Bn + n 0

We may use (2.7) to deﬁne the Bernoulli polynomials in (2.6). In fact, by mathematical induction, for n = 0, we have B0 (t) = 1. Assuming that (2.6)

November 7, 2013

12:24

32

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

is valid for n, then, for n + 1, it follows that 1 Bn+1 (t) = Bn+1 + (n + 1) Bn (t)dt 0

1 n n tn−k dt = Bn+1 + (n + 1) Bk k 0 k=0 n+1 n + 1 = Bk tn+1−k . k k=0

Thus, the function (2.6) is the root of functional equation (2.7). 2.2.3

Bernoulli’s Functions

For any t ∈ R, we denote [t] the maximal integer that does not exceed t and {t} = t − [t]. Then it follows that Bernoulli functions Pk (t) = Bk ({t})(k ∈ N) are periodic functions with the least positive periodic 1. It is obvious that Pk (t) are of bounded variation in any ﬁnite interval. P1 (t) = {t} − 12 is not continuous in the integers, but it is diﬀerentiable in the other points. P2 (t) is continuous in R and diﬀerentiable in the set of all the non-integers. For k ≥ 3, Pk (t) are continuous and diﬀerentiable in R. We can prove that P2k (t) are even functions and P2k−1 (t) are odd functions. In fact, setting H(t, x) = etx G(x), since x xex −x = e−tx x = et(−x) −x , e(1−t)x x e −1 e −1 e −1 we obtain H(1 − t, x) = H(t, −x), and by (2.4), it follows that ∞ ∞ ∞ xn (−x)n xn Bn (1 − t) Bn (t) (−1)n Bn (t) = = . n! n! n! n=0 n=0 n=0 Hence, comparing the coeﬃcients of we obtain

xn n!

on two sides of the above equality,

Bn (1 − t) = (−1)n Bn (t)

(n ∈ N0 ),

and then, it follows that Pn (−t) = Pn (1 − t) = (−1)nPn (t),

n ∈ N.

Equivalently, we may deﬁne the Bernoulli functions Pn (t) (n ∈ N) by the following recursion functional equations: 1 P1 (t) = {t} − , 2 t Pn (t)dt, n ∈ N. (2.8) Pn+1 (t) = Bn+1 + (n + 1) 0

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Improvements of the Euler-Maclaurin Summation Formula and Applications

ws-book9x6

33

Since P1 (t) is not continuous at any integer t, then, in view of the integral properties, P2 (t) is continuous at any t, but it is not diﬀerentiable at integer t. And then Pn (t) (n ≥ 3) are diﬀerentiable on the real axis. Considering the that value of P2 (k) at any integer k, it is obvious 1 1 is the maximum value, and at t = k + P2 (k) = B = , since 6 2 2 P2 k + 12 = 0, then the minimum value of P2 (t) is P2 k + 12 expressed by 12 1 1 1 1 t− = P2 = B2 + 2 dt = − . P2 k + 2 2 2 12 0 The roots of P3 (t) are divided into two classes. One is the set of integers k, that make P3 (k) = B3 = 0. The other class is the set of all k + 12 . In fact, by the deﬁnition of Bernoulli polynomial, it follows that 12 1 1 P2 (t)dt = P3 =3 P3 k + 2 2 0 1 2 1 dt = 0. =3 t2 − t + 6 0 Since the points that make the value of P4 (t) are maximum or minimum at the roots of P3 (t), these points are divided into two classes. In a general way, by the Rabbe formula (see Xu and Wang [92], Proposition 98), we have k−1 i n−1 Bn t + Bn (kt) = k , k, n ∈ N. (2.9) k i=0 Setting k = 2, t = 0, we ﬁnd 1 (2.10) Bn = (21−n − 1)Bn (0) = (21−n − 1)Bn . 2 Hence, both P2n+1 (t) and P3 (t) possess the same two classes roots (see Xu and Wang [92], Proposition 93). Since the positions of the points that make the value of P2n+2 (t) maximum or minimum are the positions of the points that make P2n+2 (t) = 0, then both P2n+2 (t) and P2 (t) possess the same points that make the value maximum or minimum. Since P2n+2 (k) = P2n+2 (0) = B2n+2 , then, by (2.10), we obtain 1 1 1 (2.11) = P2n+2 = − 1 − 2n+1 B2n+2, P2n+2 k + 2 2 2 P2n+2 k + 1 < |B2n+2 | , n ∈ N0. (2.12) 2 In view of the above formulas, we discover that (max P2n+2 (t)) (min P2n+2 (t)) < 0, max P2n+2 (t) = − min P2n+2 (t),

n ∈ N0. .

November 7, 2013

12:24

34

2.2.4

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

The Euler-Maclaurin Summation Formula

Assuming that m, n ∈ N0 , m < n, f (t) is a piecewise smooth continuous function in [m, n], then we have the following formula: n n n 1 f (k) = f (t)dt + f (t)|nm + P1 (t)f (t)dt. (2.13) 2 m m k=m+1

In fact, since P1 (t) = t − k − 12 , t ∈ [k, k + 1), integration by parts gives n n−1 k+1 1 df (t) P1 (t)f (t)dt = t−k− 2 m k=m k

k+1 n−1 1 k+1 = f (t)dt t−k− f (t)|k − 2 k k=m n n−1 1 (f (k + 1) + f (k)) − f (t)dt = 2 m k=m n n 1 = (f (m) − f (n)) + f (k) − f (t)dt. 2 m k=m+1

Hence, we obtain (2.13). In a general way, we have the following Euler-Maclaurin summation formula (see Knopp [51]): Theorem 2.1. Assuming that m, n ∈ N0 , m < n, f (t) ∈ C q [m, n](q ∈ N), then, we have n q n (−1)k Bk (k−1) n f (k) = f (t)dt + (t)|m f k! m k=m+1 k=1 (−1)q+1 n + Pq (t) f (q) (t)dt. (2.14) q! m Proof. We prove (2.14) by mathematical induction. For q = 1 (indicating f (0) (t) = f (t)), since B1 = − 12 , by (2.13), we have (2.14). Suppose that (2.14) is valid for q(∈ N). Then, for q + 1, since Pq+1 (t) = (q + 1)Pq (t), integration by parts, we ﬁnd n n 1 Pq (t)f (q) (t) dt = f (q) (t) dPq+1 (t) q+1 m m n 1 Bq+1 (q) n f (t)|m − Pq+1 (t)f (q+1) (t)dt. (2.15) = q+1 q+1 m Then substitution of (2.15) in (2.14), it follows that (2.14) is valid for q + 1.

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Improvements of the Euler-Maclaurin Summation Formula and Applications

35

Note. If f q (t) is a constant function, then it follows n Pq (t)f (q) (t)dt = 0. m

Since B2q+1 = 0 (q ∈ N), we may reduce (2.14) in the following form: Corollary 2.1. Assuming that m, n, q ∈ N0 , m < n, f (t) ∈ C 2q+1 [m, n], we have n n 1 f (k) = f (t)dt + (f (n) + f (m)) 2 m k=m

+

q B2k (2k−1) n (t)|m + δq (m, n), f (2k)!

(2.16)

k=1

where

n 1 P2q+1 (t)f (2q+1) (t) dt, (2.17) (2q + 1)! m and δq (m, n) is called the residue term of (2q + 1)th order in (2.16). Note. For q = 0, we deﬁne that the series on the right hand side of (2.16) is 0. In particular, for q = 0, we have n n n 1 f (k) = f (t)dt + (f (n) + f (m)) + P1 (t)f (t)dt. 2 m m δq (m, n) =

k=m 3

If f (t) ∈ C [m, n], then, we still have n 1 n 1 n f (t)|m + P1 (t)f (t)dt = P3 (t)f (t)dt. 12 6 m m For any r ∈ R and k ∈ N0 , we deﬁne the combination number (rk ) as follows (see Qu [64]): 1, k = 0, r = r(r−1)···(r−k+1) k , k ∈ N. k! Example 2.1. Suppose that f (t) = (t + a)l (l ∈ N, 0 < a ≤ 1; t ∈ [0, ∞)). l (i) Then f (t) = i! i (t + a)l−i , for i < l; f (i) (t)4 =* constant, for i ≥ l. In (2.14), for 2q = l or 2q + 1 = l, we have q = 2l and δq (m, n) = 0. By (2.16) (setting m = 0), we ﬁnd [ 2l ] n l+1 (n + a) B2k 1 l l l (k + a) = (n + a)l−2k+1 + (n + a) + l+1 2 2k 2k − 1 k=0 k=1 ⎤ ⎡ l ] [ 2 B2k 1 l ⎥ ⎢ 1 al−2k+1 ⎦ . (2.18) al+1 − al + −⎣ l+1 2 2k 2k − 1 k=1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

36

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

For a = cb (0 < c < b) in (2.18), we may deduce a summation formula of an arithmetical sequence with the power of non-negative integer (see Yang [150]), and for a = 1, replacing n by n − 1 in (2.18), we still have (see Yang [151]) n k=1

2.3

[l]

2 1 nl+1 B2k + nl + k = l+1 2 2k

l

k=1

l 2k − 1

nl−2k+1 .

(2.19)

Estimations of the Residue Term about a Class Series

In (2.17), if f (2q−1) (t) is a bounded variation function, then we ﬁnd the following estimation (see Cheng [9]): n 1 (2q) P2q+1 (t)df (t) |δq (m, n)| = (2q+1)! m n 1 (2q) n (2q) = P2q+1 (t)f (t)|m −(2q+1) P2q (t)f (t)dt (2q+1)! m 1 1 n P2q (t)df (2q−1) (t) ≤ = |B2q |Vmn f (2q−1) . (2.20) (2q)! m (2q)! We next reﬁne (2.20) in the following theorem by adding some conditions. 2.3.1

An Estimation under the More Fortiﬁed Conditions

Theorem 2.2. Assuming that Z is the set of all integers, m, n ∈ Z, q ∈ N0 , m < n, g(t) ∈ C 3 [m, n], g (k) (t) ≤ 0 (≥ 0), t ∈ [m, n](k = 1, 3), if there exist two intervals Ik ⊂ [m, n], such that g (k) (t) < 0(> 0), t ∈ Ik (k = 1, 3), then, we have the following estimation (see Yang and Zhu [152]): n 1 P2q+1 (t) g(t) dt δq (m, n) = (2q + 1)! m B2q+2 g(t)|nm , = εq 0 < εq < 1. (2.21) (2q + 2)! Setting n = ∞, in addition, g(∞) = g (∞) = 0, then, we have ∞ 1 P2q+1 (t) g(t) dt δq (m, ∞) = (2q + 1)! m = εq

−B2q+2 g(m), (2q + 2)!

0 < εq < 1.

(2.22)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Improvements of the Euler-Maclaurin Summation Formula and Applications

In particular, if q = 0, it follows that ∞ ε0 δ0 (m, ∞) = g(m), P1 (t) g(t) dt = − 12 m Proof.

0 < ε0 < 1.

37

(2.23)

In view of (2.12), we have max |P2q+2 (t)| = |P2q+2 (m)| = |B2q+2 | . t∈R

Since |P2q+2 (t)| is a non-constant continuous function, |g (t)| > 0, t ∈ I1 ⊂ [m, n], and g (t) ≤ 0 (≥ 0) in [m, n], then we ﬁnd n n n ≤ P (t)g (t)dt |P (t)| |g (t)| dt < |B | |g (t)|dt 2q+2 2q+2 2q+2 m m m n = B2q+2 g (t)dt = |B2q+2 g(t)|nm | . m

Hence, there exists a constant εq ∈ (0, 2) such that n P2q+2 (t)g (t)dt = B2q+2 g(t)|nm (1 − εq ).

(2.24)

m

Integration by parts and in view of (2.24), we ﬁnd n 1 δq (m, n) = P2q+1 (t) g(t) dt (2q + 1)! m n 1 = g(t) dP2q+2 (t) (2q + 2)! m ) ( n 1 n P2q+2 (t)g (t)dt (2.25) = g(t)P2q+2 (t)|m − (2q + 2)! m 1 = [B2q+2 g(t)|nm − B2q+2 g(t)|nm (1 − εq )] (2q + 2)! B2q+2 = εq (2.26) g(t)|nm , 0 < εq < 2. (2q + 2)! In the following, we show that 0 < εq < 1 in (2.26). It is obvious n that both the integral m P2q+1 (t) g(t) dt and the term B2q+2 g(t)|nm keep the same sign by (2.26). Since g (t) and g (t) possess the same property by the same way of obtaining (2.26), it follows that both

n of the sign, P (t) g (t)dt and B2q+4 g (t)|nm also keep the same sign. m 2q+3 Integration by parts, we obtain n n 1 P2q+2 (t)g (t) dt = g (t)dP2q+3 (t) 2q + 3 m ) ( n m 1 n = P2q+3 (t) g (t)dt g (t)P2q+3 (t)|m − 2q + 3 m n −1 P2q+3 (t) g (t)dt. (2.27) = 2q + 3 m

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

38

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

By (2.3), we have B2q+4 B2q+2 < 0 (q ∈ N0 ). In view of g (t)|nm g(t)|nm > 0, that (B2q+4 g (t)|nm ) (B2q+2 g(t)|nm ) < 0 and then

n it is obvious P (t)g (t)dt and B2q+2 g(t)|nm keep the diﬀerent sign. Hence, by m 2q+3 n (2.27), m P2q+2 (t)g (t)dt and B2q+2 g(t)|nm keep the same sign. Thus, by virtue of (2.24), we ﬁnd εq ∈ (0, 1), and then (2.21) follows. For n = ∞, in the same way, we obtain (2.22). Note. In Theorem 2.2, (i) if g (t) < 0 (> 0) replaces to g (t) = 0 (t ∈ [m, n]), then, we have n 1 δq (m, n) = P2q+1 (t) g(t) dt (2q + 1)! m B2q+2 g(t)|nm = 0; = (2.28) (2q + 2)! (ii) if g (3) (t) < 0 (> 0) replaces to g (3) (t) = 0 (t ∈ [m, n]), then we have n 1 B2q+2 P2q+1 (t) g(t)dt = (2.29) δq (m, n) = g(t)|nm . (2q + 1)! m (2q + 2)!

n In fact, in view of m P2q+3 (t) g (t)dt = 0, by (2.27), it follows that

n P (t) g (t)dt = 0 and then, by (2.25), we have (2.29). m 2q+2 ln t Example 2.2. We can show that the function g(t) = t−1 , t ∈ k (k) R+ (g(1) = 1) satisﬁes the conditions of (−1) g (t) > 0 (k = 0, 1, 2, 3). In fact, we have ∞

g(t) =

(t − 1)k ln[1 + (t − 1)] = (−1)k t−1 k+1 k=0

∞ (−1)k k! (t − 1)k = , k+1 k!

−1 < t − 1 ≤ 1,

k=0

and then g (k) (1) =

(−1)k k! (k ∈ N0 ). k+1

In particular, we obtain 1 g (1) = − , 2 It is obvious that g(t) > 0. We ﬁnd h(t) g (t) = , t(t − 1)2 g (0) (1) = 1,

g (1) =

2 , 3

3 g (1) = − , · · · . 2

h(t) = t − 1 − t ln t.

(2.30)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Improvements of the Euler-Maclaurin Summation Formula and Applications

ws-book9x6

39

Since h (t) = − ln t > 0, 0 < t < 1; h (t) < 0, t > 1, then max h(t) = h(1) = 0 and g (t) < 0 (t = 1). In view of g (1) = − 12 , then, we ﬁnd g (t) < 0 (t > 0). We also obtain J(t) g (t) = 2 , J(t) = −(t − 1)2 − 2t(t − 1) + 2t2 ln t. t (t − 1)3 Since J (t) = −4(t − 1) + 4t ln t and J (t) = 4 ln t < 0, 0 < t < 1; J (t) > 0, t > 1, then, min J (t) = J (1) = 0 and J (t) > 0 (t = 1); J (1) = 0. Hence, J(t) is strictly increasing with J(1) = 0 and then, J(t) < 0, 0 < t < 1; J(t) > 0, t > 1. It follows that g (t) > 0 (t = 1). Since g (1) = 23 , then, we ﬁnd g (t) > 0 (t > 0). We have L(t) g (t) = 3 , t (t − 1)4 L(t) = 2(t − 1)3 + 3t(t − 1)2 + 6t2 (t − 1) − 6t3 ln t, L (t) = 9(t − 1)2 + 18t(t − 1) − 18t2 ln t, L (t) = 36(t − 1) − 36t ln t,

L (t) = −36 ln t.

Then L (t) > 0, 0 < t < 1; L (t) < 0, t > 1; L (1) = 0 and max L (t) = L (1) = 0. Hence, L (t) < 0 (t = 1) and L (t) is strictly decreasing with L (1) = 0 and then, L (t) > 0, 0 < t < 1; L (t) < 0, t > 1. Therefore, max L(t) = L(1) = 0 and L(t) < 0 (t = 1). It follows that g (t) < 0(t = 1). Since g (1) = − 32 , then, we ﬁnd g (t) < 0 (t > 0). In view of the above results and (2.21), we have n 1 ln t δq (m, n) = P2q+1 (t) dt (2q + 1)! m t−1 B2q+2 ln n ln m , 0 < εq < 1. (2.31) − = εq (2q + 2)! n − 1 m − 1 Corollary 2.2. Assuming that mi ∈ Z, q ∈ N0 , mi < mi+1 , g(t) ∈ C 3 [mi , mi+1 ], g (k) (t) ≤ 0 (≥ 0), t ∈ [mi , mi+1 ] (k = 1, 3; i = 1, 2, · · · , s), m = m1 , n = ms+1 , if there exist two intervals Ik ⊂ [m, n] such that g (k) (t) < 0 (> 0), t ∈ Ik (k = 1, 3), then, we still have (2.21). Moreover, if g (k) (∞) = 0 (k = 0, 2), then, we have (2.22). Proof. Without loss of generality, assuming that g (k) (t) ≤ 0, t ∈ [mi , mi+1 ] (k = 1, 3; i = 1, 2, · · · , s), g (k) (t) < 0, t ∈ Ik (k = 1, 3), and B2q+2 < 0, by (2.21), it follows that mi+1 B2q+2 i+1 P2q+1 (t) g(t) dt ≤ (i = 1, · · · , s). g(t)|m 0≤ mi 2q + 2 mi

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

40

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

There exists an i, such that the above inequalities keep the strict signs. Hence, we ﬁnd n 0< P2q+1 (t) g(t) dt m

=

s i=1

mi+1

P2q+1 (t) g(t) dt < mi

B2q+2 g(t)|nm . 2q + 2

Therefore, we have (2.21), and then, the other results hold. By (2.27), we still have n P2q (t) g(t) dt =

n −1 P2q+1 (t)g (t)dt 2q + 1 m m Then, by Theorem 2.2, it follows that

(q ∈ N).

Corollary 2.3. Assuming that m, n ∈ Z, q ∈ N, m < n, g(t) ∈ C 4 [m, n], g (k) (t) ≤ 0 (≥ 0), t ∈ [m, n](k = 2, 4), if there exist two intervals Ik ⊂ [m, n], such that g (k) (t) < 0 (> 0), t ∈ Ik (k = 2, 4), then we have the following estimation: n −B2q+2 εq g (t)|nm , 0 < εq < 1. P2q (t) g(t) dt = (2.32) (2q + 1)(2q + 2) m Setting n = ∞, if in addition, g (∞) = g (4) (∞) = 0, then we have ∞ B2q+2 εq g (m), 0 < εq < 1. P2q (t) g(t) dt = (2.33) (2q + 1)(2q + 2) m 2.3.2

Some Estimations under the More Imperfect Conditions

Corollary 2.4. Assuming that m, n ∈ Z, q ∈ N0 , m < n, g(t) ∈ C 1 [m, n], g (t) ≤ 0 (≥ 0), t ∈ [m, n], if there exists an interval I1 ⊂ [m, n] such that g (t) < 0 (> 0), t ∈ I1 , then we have the following estimation: n 1 P2q+1 (t) g(t) dt δq (m, n) = (2q + 1)! m 2B2q+2 g(t)|nm , 0 < εq < 1. = εq (2.34) (2q + 2)! Setting n = ∞, and g(∞) = 0, then, we have ∞ 1 P2q+1 (t) g(t) dt δq (m, ∞) = (2q + 1)! m −2B2q+2 g(m), 0 < εq < 1. = εq (2.35) (2q + 2)!

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Improvements of the Euler-Maclaurin Summation Formula and Applications

41

In particular, for q = 0, it follows that δ0 (m, ∞) =

∞

P1 (t) g(t) dt = m

− ε0 g(m), 6

0 < ε0 < 1.

(2.36)

Proof. It is obvious that in the conditions of the corollary, we obtain (2.24). Then (2.21) and (2.22) are valid for εq ∈ (0, 2). Setting εq = 2 εq , we have (2.34) and (2.35). We can reﬁne Corollary 2.4 in the following theorem: Theorem 2.3. Assuming that m, n ∈ Z, q ∈ N0 , m < n, g(t) is a monotone piecewise smooth continuous function in [m, n], then we have the following estimation: n 1 P2q+1 (t) g(t) dt (2q + 1)! m 1 2B2q+2 1 − 2q+2 g(t)|nm , = εq (2q + 2)! 2

δq (m, n) =

0 < εq < 1. (2.37)

Setting n = ∞ and g(∞) = 0, then, we have ∞ 1 P2q+1 (t) g(t) dt (2q + 1)! m −2B2q+2 1 = εq 1 − 2q+2 g(m), (2q + 2)! 2

δq (m, ∞) =

0 < εq < 1. (2.38)

In particular, for q = 0, it follows that δ0 (m, ∞) =

∞

P1 (t) g(t) dt = −

m

ε0 g(m), 8

0 < ε0 < 1.

(2.39)

Proof. If g(t) is constant, then, two sides of (2.37) and (2.38) equal to zero (in (2.36), g(t) = 0). In the following, we assume that g(t) is a nonconstant function in [m, n] with g (t) ≤ 0.

m+k (i) If B2q+2 < 0, since g(n) < g(m), m+k−1 P2q+1 (t)dt = 0 (k ∈ N), then, by (2.26), we ﬁnd

n

P2q+1 (t) g(t) dt = εq m

B2q+2 g(t)|nm 2(q + 1)

(0 < εq < 2),

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

42

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

n

0<

P2q+1 (t) g(t) dt = m

=

n−m

k=1 m+k− 12

P2q+1 (t) [g(t) − g(m + k)]dt

+

=

P2q+1 (t)[g(t) − g(m + k)]dt

m+k−1

m+k−1

k=1

n−m

n−m m+k

m+k

m+k− 12

P2q+1 (t) [g(t) − g(m + k)]dt

[g(m + k − 1) − g(m + k)]

m+k− 12

P2q+1 (t) dt

+

m+k−1

k=1

n−m

αk ,

k=1

(2.40) where αk is deﬁned by m+k− 12 P2q+1 (t) [g(t) − g(m + k − 1)] dt αk = m+k−1

+

m+k

m+k− 12

P2q+1 (t) [g(t) − g(m + k)] dt.

Since g(t) is decreasing, then, it follows that 1 g(t) − g(m + k − 1) ≤ 0, t ∈ m + k − 1, m + k − , 2 1 g(t) − g(m + k) ≥ 0, t ∈ m + k − , m + k . 2 In view of P2q+2 (m + k − 1) = B2q+2 < 0, by (2.11), it follows 1 m + k − that P 2q+2 2 > 0. Hence, P2q+2 (t) is strictly increasing in m + k − 1, m + k − 12 with P2q+2 (t) > 0 and 1 1 . P (t) > 0, t ∈ m + k − 1, m + k − P2q+1 (t) = 2q + 2 2q+2 2 Similarly, we have P2q+1 (t) =

1 P (t) < 0, 2q + 2 2q+2

Hence, we ﬁnd P2q+1 (t) [g(t) − g(m + k − 1)] ≤ 0, P2q+1 (t) [g(t) − g(m + k)] ≤ 0,

t∈

1 m+k− , 2

m+k .

1 , m + k − 1, m + k − 2 1 t∈ m+k− , m+k , 2 t∈

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Improvements of the Euler-Maclaurin Summation Formula and Applications

ws-book9x6

43

and then αk ≤ 0. Since g(t) is of non-constant, there exists a positive n−m integer k0 (≤ n − m), such that αk0 < 0, and then, k=1 αk < 0. By (2.11), we still have m+k− 12 n−m [g(m + k − 1) − g(m + k)] P2q+1 (t) dt m+k−1

k=1

=−

m− 12 m−1

P2q+1 (t) dt

n−m

[g(m + k) − g(m + k − 1)]

k=1

( ) 1 1 P2q+2 m − − P2q+2 (m − 1) g(t)|nm 2q + 2 2 ) ( 1 1 =− − 1 B − B g(t)|nm 2q+2 2q+2 2q + 2 22q+1 1 2 1 − 2q+2 B2q+2 g(t)|nm . = 2q + 2 2 In view of (2.40), we ﬁnd n 1 2 1 − 2q+2 B2q+2 g(t)|nm , 0< P2q+1 (t) g(t) dt < 2q + 2 2 m and then equality (2.35) follows. n−m (ii) If B2q+2 > 0, in the same way, we have k=1 αk > 0. The corresponding result is n 1 2 n 1 − 2q+2 B2q+2 g(t)|m < P2q+1 (t) g(t) dt < 0, 2q + 2 2 m and we still have (2.37). For n = ∞, by the same way, we have equality (2.38). =−

Example 2.3. If m, n ∈ N, q ∈ N0 , 2m < n, ⎧ 1 ⎨ 2m , t ≤ 2m, 1 g(t) = = max{2m, t} ⎩ 1 t , t > 2m, 1 g(t)|nm = n1 − 2m , then, by (2.37) and (2.38), we ﬁnd n 1 1 P2q+1 (t) dt δq (m, n) = (2q + 1)! m max{2m, t} 1 1 1 2B2q+2 εq 1 − 2q+2 , 0 < εq < 1, (2.41) − = (2q + 2)! 2 n 2m ∞ 1 1 dt δq (m, ∞) = P2q+1 (t) (2q + 1)! m max{2m, t} 1 1 −B2q+2 εq 1 − 2q+2 , 0 < εq < 1. = (2.42) (2q + 2)! 2 m

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

44

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Corollary 2.5. Assuming that m, n ∈ Z, q ∈ N, m < n, g (t) is a monotone piecewise smooth continuous function in [m, n], then, we have the following estimation: n −2B2q+2 εq 1 P2q (t) g(t) dt = 1 − 2q+2 g (t)|nm , 0 < εq < 1. (2q + 1)(2q + 2) 2 m (2.43) Setting n = ∞, and g (∞) = 0, then, we have ∞ 1 2B2q+2 εq 1 − 2q+2 g (m), P2q (t) g(t) dt = (2q + 1)(2q + 2) 2 m

0 < εq < 1. (2.44)

/ Z, x1 < x2 , g(t) is a Corollary 2.6. Assuming that q ∈ N0 , x1 , x2 ∈ monotone piecewise smooth continuous function in [x1 , x2 ], then, we have the following estimation: x2 I2q+1 = P2q+1 (t) g(t) dt x1 4 * 1 B2q+2 1 − 2q+2 ε0 g(t)|xx21 + ε1 g(x1 ) − ε2 g(x2 ) , (2.45) = q+1 2 where ε0 ∈ (0, 1), εi ∈ (0, 1] (i = 1, 2). In particular, when q = 0, x2 * 14 P1 (t) g(t) dt = (2.46) I1 = ε0 g(t)|xx21 + ε1 g(x1 ) − ε2 g(x2 ) . 8 x1 Proof. by

We deﬁne a monotone piecewise smooth continuous function g(t)

Then, we ﬁnd I2q+1 =

⎧ ⎨ g(x1 ), t ∈ [[x1 ], x1 ), g(t) = g(t), t ∈ [x1 , x2 ], ⎩ g(x2 ), t ∈ (x2 , [x2 ] + 1].

[x2 ]+1

[x1 ]

− g(x1 )

P2q+1 (t) g(t) dt

x1 [x1 ]

P2q+1 (t) dt − g(x2 )

[x2 ]+1

P2q+1 (t) dt. (2.47) x2

By (2.37), it follows that [x2 ]+1 1 2B2q+2 [x ]+1 1 − 2q+2 ε0 g(t)|[x21 ] P2q+1 (t) g (t)dt = 2q + 2 2 [x1 ] 2B2q+2 1 = 1 − 2q+2 ε0 g(t)|xx21 , ε0 ∈ (0, 1). 2q + 2 2

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Improvements of the Euler-Maclaurin Summation Formula and Applications

45

Since, for x1 ∈ / Z, we have 0 < x1 − [x1 ] < 1, and

x1 −[x1 ] 0 <

0

1 2

0

P2q+1 (t)dt

≤ 1,

P2q+1 (t)dt

there exists ε1 ∈ (0, 1] such that x1 P2q+1 (t)dt [x1 ]

x1 −[x1 ]

=

P2q+1 (t) dt = ε1

0

1 2

0

P2q+1 (t) dt

1 P2q+2 − B2q+2 2 B2q+2 1 =− 1 − 2q+2 ε1 . (2.48) q+1 2

1 ε1 ε1 P2q+2 (t)|02 = = 2q + 2 2q + 2

=

ε1 −1−2q − 2 B2q+2 2 2q + 2

Similarly, there exists ε2 ∈ (0, 1] such that

[x2 ]+1

P2q+1 (t) dt x2

1

= x2 −[x2 ]

P2q+1 (t) dt = ε2

1 1 2

ε2 −ε2 P2q+2 (t)|11 = = 2 2q + 2 2q + 2 1 B2q+2 1 − 2q+2 ε2 . = q+1 2

P2q+1 (t) dt 1 P2q+2 − B2q+2 2

Hence, by (2.47) and the above results, we have (2.45).

(2.49)

Note. If x1 , x2 ∈ Z, then we can ﬁnd ε1 = ε2 = 0 in (2.48) and (2.49). In this case, (2.45) is an extension of (2.37). Corollary 2.7. Assuming that q ∈ N0 , m, n ∈ Z, x ∈ / Z, m < x < n, both g1 (t) (t ∈ [m, x]) and g2 (t) (t ∈ [x, n]) are decreasing (increasing) piecewise smooth continuous functions, and g1 (t), t ∈ [m, x), g(t) = g2 (t), t ∈ [x, n],

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

46

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

then, we have

n

J2q+1 =

P2q+1 (t) g(t) dt B2q+2 1 = 1 − 2q+2 {ε0 (g2 (n) − g1 (m)) q+1 2 + [ε0 − ε1 ](g1 (x) − g2 (x))} , ε0 ∈ (0, 1), and ε1 ∈ (0, 1]. m

(2.50) Setting n = ∞, and g2 (∞) = 0, then, we have ∞ J2q+1 = P2q+1 (t) g(t) dt m 1 B2q+2 1 − 2q+2 {−ε0 g1 (m) = q+1 2 + [ε0 − ε1 ](g1 (x) − g2 (x))} , ε0 ∈ (0, 1), ε1 ∈ (0, 1]. (2.51) In particular, when q = 0, it follows that n 1 J1 = P1 (t) g(t) dt = [ε0 (g2 (n) − g1 (m)) 8 m + (ε0 − ε1 )(g1 (x) − g2 (x))] , ∞ J1 = P1 (t) g(t) dt

(2.52)

m

1 = [−ε0 g1 (m) + (ε0 − ε1 )(g1 (x) − g2 (x))] , ε0 ∈ (0, 1), ε1 ∈ (0, 1]. 8 (2.53) Proof. Deﬁne a decreasing (increasing) piecewise smooth continuous function g(t) as follows: g1 (t) − g1 (x) + g2 (x), t ∈ [m, x), g (t) = g2 (t), t ∈ [x, n]. Then, by (2.37) and (2.48), it follows that n x P2q+1 (t) g(t) dt + (g1 (x) − g2 (x)) P2q+1 (t) dt J2q+1 = m m 1 2B2q+2 1 − 2q+2 ε0 = g(t)|nm 2q + 2 2 B2q+2 1 − 1 − 2q+2 ε1 (g1 (x) − g2 (x)) q+1 2

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Improvements of the Euler-Maclaurin Summation Formula and Applications

2B2q+2 2q + 2

1

1−

47

ε0 (g2 (n) − g1 (m) + g1 (x) − g2 (x)) 22q+2 1 B2q+2 1 − 2q+2 ε1 (g1 (x) − g2 (x)) − q+1 2 B2q+2 1 = 1 − 2q+2 {ε0 (g2 (n) − g1 (m)) q+1 2 + [ε0 − ε1 ](g1 (x) − g2 (x))} ,

=

where ε0 ∈ (0, 1),

ε1 ∈ (0, 1].

Hence, we have (2.50). In the same way, we have (2.51).

Note. If x ∈ Z, then, in (2.50) and (2.51), we ﬁnd ε1 = 0. Hence, for g2 (t) = 0 in (2.52), we have x ε0 ε1 P1 (t) g1 (t) dt = (−g1 (m) + g1 (x)) − g1 (x), (2.54) 8 8 m where x > m;

ε0 ∈ (0, 1) and ε1 ∈ [0, 1];

for g1 (t) = 0 in (2.52) and (2.53), we have n ε0 ε1 P1 (t) g2 (t) dt = (g2 (n) − g2 (x)) + g2 (x), 8 8 x

(2.55)

where x > m;

∞

P1 (t) g2 (t) dt = x

2.3.3

ε0 ∈ (0, 1) and ε1 ∈ [0, 1],

ε1 − ε 2 g2 (x), x ∈ R; ε0 ∈ (0, 1) and ε1 ∈ [0, 1]. 8 (2.56)

Estimations of δq (m, n) and Some Applications

Setting g(t) = f (2q+1) (t) in Theorem 2.4 and Theorem 2.9, we have the following corollaries: Corollary 2.8. Assuming that m, n, q ∈ N0 , m < n, f (t) ∈ C 2q+4 [m, n] with (−1)k f (2q+1+k) (t) ≥ 0 (≤ 0),

t ∈ [m, n]

(k = 1, 3),

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

48

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and there exist two intervals Ik ⊂ [m, n] such that (−1)k f (2q+1+k) (t) > 0 then, we have

(< 0),

t ∈ Ik

(k = 1, 3),

n 1 P2q+1 (t) f (2q+1) (t)dt (2q + 1)! m B2q+2 f (2q+1) (t)|nm , 0 < εq < 1. = εq (2q + 2)!

δq (m, n) =

(2.57)

Setting n = ∞, and in addition, f (2q+1+k) (∞) = 0 (k = 0, 2), then, we have ∞ 1 δq (m, ∞) = P2q+1 (t)f (2q+1) (t) dt (2q + 1)! m −B2q+2 (2q+1) (m), 0 < εq < 1. (2.58) f = εq (2q + 2)! In particular, when q = 0, it follows that ∞ ε0 f (m), 0 < ε0 < 1. δ0 (m, ∞) = P1 (t)f (t) dt = − (2.59) 12 m Corollary 2.9. Assuming that m, n, q ∈ N0 , m < n, f (2q+1) (t) is a monotone piecewise smooth continuous function in [m, n], then, we have n 1 P2q+1 (t) f (2q+1) (t) dt δq (m, n) = (2q + 1)! m 2B2q+2 1 = εq 1 − 2q+2 f (2q+1) (t)|nm , 0 < εq < 1. (2.60) (2q + 2)! 2 Setting n = ∞ and f (2q+1) (∞) = 0, then, we have ∞ 1 δq (m, ∞) = P2q+1 (t) f (2q+1) (t) dt (2q + 1)! m 1 −2B2q+2 1 − 2q+2 f (2q+1) (m), 0 < εq < 1. (2.61) = εq (2q + 2)! 2 In particular, when q = 0, it follows that ∞ ε0 δ0 (m, ∞) = P1 (t)f (t) dt = − f (m), 0 < ε0 < 1. (2.62) 8 m For q = 0 in Corollary 2.8, in view of (2.13), we have the following corollary: Corollary 2.10. Assuming that m, n ∈ N0 , m < n, f (t) ∈ C 4 [m, n] with (−1)k f (1+k) (t) ≥ 0 (≤ 0),

t ∈ [m, n] (k = 1, 3),

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Improvements of the Euler-Maclaurin Summation Formula and Applications

ws-book9x6

49

and there exist two intervals Ik ⊂ [m, n] such that (−1)k f (1+k) (t) > 0 (< 0), then, we have n f (k) =

n m

k=m

t ∈ Ik

(k = 1, 3),

ε0 1 f (t)dt + (f (n) + f (m)) + f (t)|nm , 2 12

0 < ε0 < 1. (2.63)

∞ Setting n = ∞ and f (k) (∞) = 0 (k = 0, 2), and both of k=m f (k)

∞ and m f (t)dt are convergent, then, we have ∞ ∞ ε0 1 f (k) = f (t)dt + f (m) − f (m), 0 < ε0 < 1. (2.64) 2 12 m k=m

In particular, when f (m) < 0, the following inequalities follow: ∞ ∞ ∞ 1 1 1 f (t)dt + f (m) < f (k) < f (t)dt + f (m) − f (m). (2.65) 2 2 12 m m k=m

For q = 0 in Corollary 2.4, in view of (2.13), we have the following corollary: Corollary 2.11. Assuming that m, n ∈ N0 , m < n, f (t) is a monotone piecewise smooth continuous function in [m, n], then, we have n n ε0 1 f (k) = f (t)dt + (f (n) + f (m)) + f (t)|nm , 0 < ε0 < 1. (2.66) 2 8 m k=m

∞ ∞ Setting n = ∞ and f (∞) = 0, and both k=m f (k) and m f (t)dt are convergent, then, we have ∞ ∞ 1 ε0 f (k) = f (t) dt + f (m) − f (m), 0 < ε0 < 1. (2.67) 2 8 m k=m

In particular, when f (m) < 0, the following inequalities follow: ∞ ∞ ∞ 1 1 1 f (t)dt + f (m) < f (k) < f (t)dt + f (m) − f (m). (2.68) 2 2 8 m m k=m

Example 2.4. If f (t) = f (t) =

1 (1+t)t1/2

(t > 0), m ∈ N, then, we obtain f (t) > 0,

−1 1 3t + 1 − =− < 0, 2 1/2 3/2 (1 + t) t 2(1 + t)t 2(1 + t)2 t3/2

f (t) > 0, f (t) < 0, and ∞ m

1 dt = 2 arctan (1 + t)t1/2

1 √ m

.

November 7, 2013

12:24

50

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

By (2.65), when m ∈ N, we have ∞ 1 1 1 √ √ < 2 arctan √ + m 2(1 + m) m (1 + k) k k=m 1 3m + 1 1 √ .(2.69) √ + + < 2 arctan √ m 2(1 + m) m 24(1 + m)2 m3 Example 2.5. If f (t) = t21+1 (t > 0), m ∈ N, then, we obtain f (t) > 0, f (t) = (t2−2t < 0, +1))2 ∞ 1 π dt = − arctan m. 2 2 m t +1 By (2.68), when m ∈ N, we have ∞ 1 π 1 − arctan m + < 2 2(m2 + 1) k2 + 1 k=m

<

1 m π − arctan m + + . 2 2(m2 + 1) 4(m2 + 1)2 (2.70)

In particular, when m = 1, we ﬁnd ∞ π 5 1 π 21 + < < + . 4 4 k2 + 1 4 16

(2.71)

k=0

2.4 2.4.1

Two Classes of Series Estimations One Class of Convergent Series Estimation

For n = ∞ in (2.16), by Corollary 2.8 and Corollary 2.9, we have the following theorem: Theorem 2.4. Assuming that m, n, q ∈ N0 , m < n, f (t) ∈ C 2q+1 [m, n], f (∞) = f (2k−1) (∞) =

∞0 (k = 1, 2, · · · , q + 1), δq (m, ∞) is convergent, if both ∞ f (k) and f (t)dt are convergent, then, we have the following k=m m estimation: ∞ q ∞ 1 1 f (k) = f (t)dt + f (m) − B2k f (2k−1) (m) + δq (m, ∞), 2 (2k)! m k=m k=1 (2.72) where ∞ 1 δq (m, ∞) = P2q+1 (t) f (2q+1) (t) dt, (2.73) (2q + 1)! m

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

51

Improvements of the Euler-Maclaurin Summation Formula and Applications

and δq (m, ∞) satisﬁes the following recursion formulas ∞ δ0 (m, ∞) = P1 (t)f (t)dt, m

δq (m, ∞) =

1 B2q f (2q−1) (m) + δq−1 (m, ∞), (2q)!

(q ∈ N). (2.74)

(i) If (−1)k f (2q+1+k) (t) ≥ 0 (≤ 0), t ∈ [m, n] (k = 1, 3), and there exist two intervals Ik ⊂ [m, n], such that (−1)k f (2q+1+k) (t) > 0 (< 0),

t ∈ Ik (k = 1, 3),

then, we have δq (m, ∞) = εq

−B2q+2 (2q+1) (m), f (2q + 2)!

0 < εq < 1 (q ∈ N0 );

(2.75)

(ii) if f (2q+1) (t) is a monotone piecewise smooth continuous function in [m, n], then, we have δq (m, ∞) =

− εq B2q+2 22q+2 − 1 (2q+1) f (m), (2q + 2)! 22q+1

0 < εq < 1 (q ∈ N0 ). (2.76)

1 Example 2.6. If 0 < a ≤ 1, p > 1, f (t) = (t+a) p (t > 0), then, we ﬁnd k! −p . f (k) (t) = k (t + a)p+k

For m, a ∈ N0 , by (2.22) and (2.75), we have ∞ k=m

1 1 1 = + (k + a)p (p − 1)(m + a)p−1 2(m + a)p −

(−p (−p εq B2q+2 2q+1 ) 2k−1 ) − ,(2.77) 2k (m + a)p+2k−1 2q + 2 (m + a)p+2q+1

q B2k k=1

where 0 < εq < 1. In particular, if a = 1, replacing m + 1 by m, we ﬁnd (see Yang [153]) ∞ 1 1 1 = + kp (p − 1) mp−1 2 mp

k=m

−

q −p −p B2k (2k−1 ) εq B2q+2 (2q+1 ) − , 2k mp+2k−1 2q + 2 mp+2q+1

0 < εq < 1 (m ∈ N).

k=1

(2.78)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

52

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

2.4.2

One Class of Finite Sum Estimation on Divergence Series

For m ∈ N0 , setting δq (m) = δq (m, ∞). If F (t) = f (t), then, we deﬁne the constant βm by q 1 B2k (2k−1) f βm = −F (m) + f (m) − (m) + δq (m). (2.79) 2 (2k)! k=1

By (2.16), when m < n, we ﬁnd q n B2k (2k−1) 1 f f (k) = F (n) + f (n) + (n) + βm − δq (n), 2 (2k)! k=m

(2.80)

k=1

where

∞ 1 δq (n) = P2q+1 (t) f (2q+1) (t) dt, (2.81) (2q + 1)! n since δq (∞) = 0, it follows that

n q B2k (2k−1) 1 f βm = lim f (k) − F (n) − f (n) − (n) n→∞ 2 (2k)! k=m

k=1

n

q 1 B2k (2k−1) f (k) − F (n) − f (n) − (n) + δq (n). (2.82) f = 2 (2k)! k=m k=1

∞ ∞ Note. If both k=m f (k) and m f (x)dx are convergent and f (2k−1) (∞) = 0 (k = 1, 2, · · · , q), ∞ then, by (2.79), k=m f (k) = βm + F (∞). Hence, in view of (2.78), we still have (2.72). ln t Example 2.7. If f (t) = t−1 (t > 0), since, for n > 1, we ﬁnd n n 1 ln t ln u f (t) dt = dt = du 1 (u − 1)u t − 1 1 1 n 1 1 ln u ln u + =− du + du 1 1 u − 1 u n n ∞ 1 (1 − u)k−1 1 du = (ln n)2 + 1 2 k k=1 n k ∞ 1 1 1 2 1− , = (ln n) + 2 k2 n k=1

ln n 1 − f (n) = , n(n − 1) (n − 1)2 3 2 6 6 ln n + 2 f (n) = 3 + − , 2 3 n (n − 1) n (n − 1) n(n − 1) (n − 1)4

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Improvements of the Euler-Maclaurin Summation Formula and Applications

53

then, by (2.79) and (2.75) (for m = q = 1), we have k n ∞ 1 1 ln n ln k 1 1 2 = (ln n) + + 1− 2 k−1 2 k n 2n−1 k=1

k=1

ln n 1 1 − + + β1 12 n(n − 1) 12(n − 1)2 ) ( ε 6 6 ln n 2 3 − + − , + 720 n3 (n − 1) n2 (n − 1)2 n(n − 1)3 (n − 1)4

n k ∞ ln k 1 1 1 1 − − (ln n)2 − β1 = lim n→∞ k−1 2 k2 n k=1 k=1 n ∞ k 1 1 ln n ln k 1 1 − (ln n)2 − = − 1 − k−1 2 k2 n 2n−1 k=1

k=1

1 ln n 1 − + 12 n(n − 1) 12(n − 1)2 ) ( ε 6 6 ln n 2 3 + + − , + 720 n3 (n − 1) n2 (n − 1)2 n(n − 1)3 (n − 1)4 (2.83) where 0 < ε < 1. If n = 3 in (2.83), we ﬁnd k k ∞ 20 1 2 1 2 − < − , k2 3 k2 3 k=1 k=1 k k k ∞ 20 ∞ 1 2 1 2 1 1 = − − − 1 − k2 3 k2 3 k2 3 k=1 k=1 k=21 ∞ 20 k x 1 2 1 2 >− − dx 2 k2 3 3 20 x k=1 k 20 ∞ 20 2 1 2 1 − dx >− 2 k2 3 3 x 20 k=1 k 20 20 1 2 1 2 =− − . k2 3 20 3

(2.84)

k=1

In view of (2.83) and (2.84), we can obtain 0.539902 < β1 < 0.539976, and then, it follows that β1 = 0.5399+. Example 2.8. If 0 < a ≤ 1, f (t) = f (k) (t) =

1 t+a

(t ≥ 0), F (t) = ln(t + a), then

(−1)k k!. (t + a)k+1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

54

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

When m = 0 in (2.76), we ﬁnd (see [154] and [155]) that n β0 = lim f (k) − F (n) n→∞

= lim

n→∞

k=0

n

k=0

1 − ln(n + a) = γ0 (a). k+a

(2.85)

We call γ0 (a) the Stieltjes constant (see Pan and Pan [62]). By (2.80), (2.81) and (2.75), for n ∈ N, we ﬁnd (see Yang and Li [154]) that n 1 1 = γ0 (a) + ln(n + a) + k+a 2(n + a) k=0

−

q k=1

εq B2q+2 B2k − , 2k 2k(n + a) 2(q + 1)(n + a)2q+2

0 < εq < 1.

(2.86) In particular, for a = 1, replacing n + 1 by n, we have the following estimation of the harmonic series (see Yang and Wang [155]): q n εq B2q+2 1 1 B2k − , 0 < εq < 1, (2.87) = γ + ln n + − 2k k 2n 2kn 2(q + 1) n2q+2 k=1

k=1

where γ (= γ0 (1) = 0.5772156649+) is called the Euler constant. 1 Example 2.9. If s ∈ R\{1}, 0 < a ≤ 1, f (t) = (t+a) s (t ≥ 0), setting ζ(s, a) = β0 , then, by (2.80), (2.81) and (2.75), for n ∈ N, q ≥ 1−s 2 , we have an estimation of the Hurwitz ζ−function ζ(s, a) in the real axis as follows (see Titchmarsh [71]): n 1 1 1 ζ(s, a) = − (n + a)1−s − (k + a)s 1−s 2(n + a)s k=0

−

q k=1

−s −s εq (2q+1 )B2q+2 (2k−1 )B2k − , s+2k−1 2k(n + a) 2(q + 1)(n + a)s+2q+1

0 < εq < 1.

(2.88) In particular, for a = 1, q ≥ 1−s , we have an estimation of the Riemann 2 ζ−function, ζ(s) = ζ(s, 1) in the real axis as follows (see Zhu and Yang [180]): n 1 1 1 − n1−s − s ζ(s) = s k 1−s 2n k=1

−

q −s −s εq (2q+1 )B2q+2 (2k−1 )B2k − , s+2k−1 2k n 2(q + 1) ns+2q+1 k=1

0 < εq < 1. (2.89)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Improvements of the Euler-Maclaurin Summation Formula and Applications

ws-book9x6

55

Example 2.10. If 0 < a ≤ 1, f (t) = ln(t + a) (t ≥ 0), and F (t) = (t + a) ln(t + a) − t, then, f (k) (t) =

(−1)k−1 (k − 1)! . (t + a)k

Using (2.80), (2.81) and (2.75), we obtain

n 1 ln(k + a) − n + a + ln(n + a) + n , β0 (a) = lim n→∞ 2 k=0 n n 2 1 ln(n + a) − n (k + a) = ln(k + a) = β0 (a) + n + a + ln 2 k=0

k=0

+

q k=1

B2k εq B2q+2 + , 2k(2k − 1)(n + a)2k−1 2(q + 1)(2q + 1)(n + a)2q+1 (2.90) √

π (see Xie [93]), we ﬁnd where 0 < εq < 1. Equivalently, since eβ0 (a) = Γ(a) n+a √ n 2 n+a π√ (k + a) = n+a Γ(a) e k=0 q εq B2q+2 B2k , × exp + 2k(2k − 1)(n + a)2k−1 2(q + 1)(2q + 1)(n + a)2q+1 k=1

(2.91) where 0 < εq < 1. In particular, for a = 1, replacing n + 1 by n, we have the following extended Stirling formula (see Knopp [51] and Yang [156]): q n n √ εq B2q+2 B2k exp + . n! = 2πn e 2k(2k − 1) n2k−1 2(q + 1)(2q + 1) n2q+1 k=1 (2.92) When q = 1, we have the following inequality: n n n n √ √ 1 1 1 . < n! < 2πn 1− 2πn exp exp e 12n e 12n 30n2 (2.93)

May 2, 2013

14:6

BC: 8831 - Probability and Statistical Theory

This page intentionally left blank

PST˙ws

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Chapter 3

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel “One of the properties inherent in mathematics is that any real progress is accompanied by the discovery and development of new methods and simpliﬁcations of previous procedures.... The uniﬁed character of mathematics lies in its very nature; indeed, mathematics is the foundation of all exact natural sciences.”

David Hilbert

“In great mathematics there is a very high degree of unexpectedness, combined with inevitability and economy.”

G. H. Hardy

3.1

Introduction

This chapter is devoted to half-discrete Hilbert-type inequalities with general homogeneous kernels and their many extensions based on the way of weight functions and techniques of real analysis. Special attention is given to the proofs of the best possible constant factors of half-discrete Hilbert-type inequalities. Included are several equivalent inequalities, operator expressions, and the reverses of the Hilbert-type inequalities with many extensions and particular examples. 57

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

58

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

3.2

Some Preliminary Lemmas

3.2.1

Deﬁnition of Weight Functions and Related Lemmas

Lemma 3.1. Suppose that λ, λ1 , λ2 ∈ R, λ = λ1 + λ2 , kλ (x, y) is a non-negative ﬁnite measurable homogeneous function of degree −λ in R2+ , satisfying kλ (ux, uy) = u−λ kλ (x, y),

u, x, y > 0,

and v(y) is a strict increasing diﬀerentiable function in [n0 , ∞)(n0 ∈ N) with v(n0 ) > 0 and v(∞) = ∞. Deﬁne two weight functions ωλ2 (n) and λ1 (x) as follows: ∞ ωλ2 (n) = [v(n)]λ2 kλ (x, v(n))xλ1 −1 dx, n ≥ n0 (n ∈ N), (3.1) λ1 (x) = xλ1

0

∞

kλ (x, v(n))[v(n)]λ2 −1 v (n),

x ∈ (0, ∞).

(3.2)

n=n0

Then we have

ωλ2 (n) = k(λ1 ) =

∞ 0

kλ (t, 1) tλ1 −1 dt =

∞ 0

kλ (1, t) tλ2 −1 dt.

(3.3)

Moreover, we set f (x, y) = xλ1 kλ (x, v(y))[v(y)]λ2 −1 v (y) and the following conditions: Condition (i). v(y) is strictly increasing in [n0 − 1, ∞) with v(n0 − 1) ≥ 0, and for any ﬁxed x > 0, f (x, y) is decreasing with respect to y ∈ [n0 − 1, ∞) and strictly decreasing in an interval I ⊂ (n0 − 1, ∞). Condition (ii). v(y) is strictly increasing in [n0 − 12 , ∞) with v(n0 − 1 2 ) ≥ 0, and for any ﬁxed x > 0, f (x, y) is convex with respect to y ∈ [n0 − 12 , ∞) and strictly convex in an interval I ⊂ (n0 − 12 , ∞). Condition (iii). There exists a constant β > 0, such that v(y) is strictly increasing in [n0 − β, ∞) with v(n0 − β) ≥ 0, and for ﬁxed x > 0, f (x, y) is a piecewise smooth continuous function with respect to y ∈ [n0 − β, ∞), satisfying ∞ n0 1 f (x, y)dy − f (x, n0 ) − P1 (y)fy (x, y)dy > 0, (3.4) R(x) = 2 n0 n0 −β where P1 (y)(= y−[y]− 21 ) is the Bernoulli function of order one (see equation (2.8) in Chapter 2).

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

59

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

If k(λ1 ) ∈ R+ and one of the above three conditions is satisﬁed, then we have x ∈ (0, ∞).

λ1 (x) < k(λ1 ), Proof.

Setting t =

(3.5)

x v(n)

in (3.1), since λ = λ1 + λ2 , we ﬁnd ∞ ωλ2 (n) = [v(n)]λ2 kλ (tv(n), v(n)) (tv(n))λ1 −1 v(n)dt 0 ∞ kλ (t, 1) tλ1 −1 dt = k(λ1 ). = 0

Setting t =

v(n) x

in (3.1), we also obtain λ −1 ∞ v(n) v(n) 1 v(n) λ2 , v(n) ωλ2 (n) = [v(n)] kλ dt t t t2 0 ∞ = kλ (1, t) tλ2 −1 dt, 0

and then (3.3) follows. (i) If Condition (i) is satisﬁed, then for any n ≥ n0 , we have n f (x, n) ≤ f (x, y)dy, n−1

and there exists an integer n1 ≥ n0 such that n1 f (x, y)dy. f (x, n1 ) < n1 −1

Since v(n0 − 1) ≥ 0 and k(λ1 ) ∈ R+ , we ﬁnd ∞ ∞ n λ1 (x) = f (x, n) < f (x, y)dy n=n0 ∞

n=n0

n0 −1 ∞

≤

v(n0 −1) x

0

∞

f (x, y)dy = xλ1

= =

n−1 ∞

n0 −1

kλ (1, t) tλ2 −1 dt

kλ (x, v(y))[v(y)]λ2 −1 v (y)dy

(t = v(y)/x)

kλ (1, t) tλ2 −1 dt = k(λ1 ),

and then (3.5) follows. (ii) If Condition (ii) is satisﬁed, then, by Hermite-Hadamard’s inequality (see Kuang [47]), we have n+ 12 f (x, n) ≤ f (x, y)dy, n ≥ n0 , n− 12

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

60

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and there exists a positive integer n2 ≥ n0 such that f (x, n2 ) <

n2 + 12 n2 − 12

f (x, y)dy.

Since v(n0 − 12 ) ≥ 0 and k(λ1 ) ∈ R+ , we ﬁnd λ1 (x) =

∞

f (x, n) <

n=n0 ∞

=

n=n0

n0 − 12

= xλ1 =

∞

∞

n+ 12 n− 12

f (x, y)dy

f (x, y)dy ∞

n0 − 12

v(n0 − 1 ) 2 x

kλ (x, v(y))[v(y)]λ2 −1 v (y)dy λ2 −1

kλ (1, t)t

dt ≤

∞

0

kλ (1, t)tλ2 −1 dt = k(λ1 ),

and then (3.5) follows. (iii) If Condition (iii) is satisﬁed and k(λ1 ) ∈ R+ , then by the EulerMaclaurin summation formula (see equation (2.14) in Chapter 2) for q = 0, m = n0 , n → ∞, we have λ1 (x) =

∞

∞ 1 f (x, y)dy + f (x, n0 ) + P1 (y)fy (x, y)dy 2 n0 n0 ∞ f (x, y)dy − R(x) = kλ (1, t) tλ2 −1 dt − R(x) ∞

f (x, n) =

n=n0 ∞

=

n0 −β

v(n0 −β) x

≤ k(λ1 ) − R(x) < k(λ1 ),

and then (3.5) follows.

Lemma 3.2. Let the assumptions of Lemma 3.1 be fulﬁlled and additionally, let p ∈ R\{0, 1}, 1p + 1q = 1, an ≥ 0, n ≥ n0 (n ∈ N), f (x) be a non-negative measurable function in R+. Then (i) for p > 1, we have the following inequalities:

( ∞ ) p p1 v (n) J= kλ (x, v(n))f (x)dx [v(n)]1−pλ2 0 n=n0 p1 ∞ 1 p(1−λ )−1 p 1 ≤ [k(λ1 )] q λ1 (x)x f (x)dx , ∞

0

(3.6)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

61

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

and = L

∞

0

≤

[λ1 (x)]1−q x1−qλ1

∞

q kλ (x, v(n))an

n=n0

∞ [v(n)]q(1−λ2 )−1 q k(λ1 ) an [v (n)]q−1 n=n

q1 dx

1q ;

(3.7)

0

(ii) for p < 0 or 0 1, by H¨older’s inequality with weight (see Kuang [47]) and (3.3), it follows that ( ∞ )p kλ (x, v(n))f (x) dx 0

0 ∞

≤

0

(

∞

=

kλ (x, v(n)) kλ (x, v(n))

) p )( x(1−λ1 )/q [v (n)]1/p [v(n)](1−λ2 )/p dx f (x) [v(n)](1−λ2 )/p x(1−λ1 )/q [v (n)]1/p

x(1−λ1 )(p−1) v (n) p f (x) dx [v(n)]1−λ2

p−1 [v(n)](1−λ2 )(q−1) dx x1−λ1 [v (n)]q−1 0 p−1 ∞ x(1−λ1 )(p−1) v (n) p ωλ2 (n)[v(n)]q(1−λ2 )−1 = k (x, v(n)) f (x) dx λ [v (n)]q−1 [v(n)]1−λ2 0 ∞ [k(λ1 )]p−1 x(1−λ1 )(p−1) v (n) p = kλ (x, v(n)) f (x) dx. pλ −1 2 [v(n)] v (n) 0 [v(n)]1−λ2 ×

∞

kλ (x, v(n))

Then, by the Lebesgue term by term integration theorem (see Kuang [49]), we have p1 ∞ (1−λ1 )(p−1) ∞ 1 x v (n) kλ (x, v(n)) f p (x) dx J ≤ [k(λ1 )] q [v(n)]1−λ2 n=n0 0 p1 ∞ ∞ (1−λ1 )(p−1) 1 x v (n) p = [k(λ1 )] q kλ (x, v(n)) f (x) dx [v(n)]1−λ2 0 n=n0 ∞ p1 1 p(1−λ1 )−1 p q = [k(λ1 )] λ1 (x) x f (x)dx , 0

and then (3.6) follows.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

62

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Hence, by H¨older’s inequality with weight, it follows that

∞ q kλ (x, v(n))an n=n0

)( )q [v(n)](1−λ2 )/p an x(1−λ1 )/q [v (n)]1/p = kλ (x, v(n)) [v(n)](1−λ2 )/p x(1−λ1 )/q [v (n)]1/p n=n0 ∞ q−1 x(1−λ1 )(p−1) v (n) ≤ kλ (x, v(n)) [v(n)]1−λ2 n=n

(

∞

0

∞

× 1−qλ1

=

x

[λ1

kλ (x, v(n))

n=n0 ∞

(x)]1−q

[v(n)](1−λ2 )(q−1) q a x1−λ1 [v (n)]q−1 n

kλ (x, v(n))

n=n0

[v(n)](1−λ2 )(q−1) q a . x1−λ1 [v (n)]q−1 n

Then, by the Lebesgue term by term integration theorem, we have 1q ∞ ∞ (1−λ2 )(q−1) [v(n)] q ≤ kλ (x, v(n)) 1−λ1 a dx L x [v (n)]q−1 n 0 n=n0 ∞ 1q ∞ [v(n)](1−λ2 )(q−1) q = kλ (x, v(n)) 1−λ1 a dx x [v (n)]q−1 n n=n0 0 1q ∞ [v(n)]q(1−λ2 )−1 q = ωλ2 (n) an , [v (n)]q−1 n=n 0

and then, in view of (3.3), we have (3.7). (ii) For p < 0 or 0 < p < 1, by the reverse H¨ older’s inequality with weight (see Kuang [47]) and in the same way, we obtain the reverses of (3.6) and (3.7). 3.2.2

Estimations about Two Series

Lemma 3.3. Let v(y) be a strictly increasing diﬀerentiable functions in (y) (> 0) be [n0 , ∞) (n0 ∈ N) with v(n0 ) > 0 and v(∞) = ∞, and vv(y) decreasing in [n0 , ∞). Then, for ε > 0, we have A(ε) =

∞ n=n0

1 v (n) = (1 + o(1)) 1+ε [v(n)] ε

(ε → 0+ ).

(3.8)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

ws-book9x6

63

v (y) v (y) 1 Proof. In view of the assumptions, [v(y)] 1+ε (= [v(y)] · [v(y)]ε ) is still decreasing in [n0 , ∞), we ﬁnd ∞ v (y) 1 1 (1 + o1 (1)) = = dy 1+ε ε ε[v(n0 )]ε n0 [v(y)] ∞ v (n) v (n0 ) ≤ A(ε) = + 1+ε [v(n0 )] [v(n)]1+ε n=n0 +1 ∞ v (y) v (n0 ) ≤ + dy 1+ε 1+ε [v(n0 )] n0 [v(y)] εv (n0 ) 1 1 + = ε [v(n0 )]ε [v(n0 )]1+ε 1 = (1 + o2 (1)) (ε → 0+ ), ε and then, we have equality (3.8).

Lemma 3.4. Let the assumptions of Lemma 3.1 be fulﬁlled and addition (y) (> 0) be decreasing in [n0 , ∞). If ally, let p ∈ R\{0, 1}, 1p + 1q = 1, vv(y) there exist constants δ < λ1 andL > 0, ) suchthat 1 1 (3.9) kλ (t, 1) ≤ L δ , t ∈ 0, t v(n0 ) then for 0 < ε < min{|p|, |q|}(λ1 − δ), we have 1 v(n) ∞ ε v (n) kλ (t, 1)tλ1 − p −1 dt = O(1)(ε → 0+ ). (3.10) B(ε) = 1+ε [v(n)] 0 n=n 0

Proof.

In view of (3.9), and since v (y) 1 v (y) · = ε λ1 −δ+ q +1 λ1 −δ+ qε v(y) [v(y)] [v(y)] is still decreasing in [n0 , ∞), we ﬁnd 1 ∞ v(n) ε v (n) tλ1 −δ− p −1 dt 0 < B(ε) ≤ L 1+ε [v(n)] 0 n=n 0

∞ v (n) L = ε ε λ1 − δ − p n=n [v(n)]λ1 −δ+ q +1 0

∞ L v (n) v (n0 ) = + ε ε λ1 − δ − pε [v(n0 )]λ1 −δ+ q +1 n=n +1 [v(n)]λ1 −δ+ q +1 0 ) ( ∞ v (y) v (n0 ) L + dy ≤ ε ε λ1 −δ+ q +1 λ1 − δ − pε [v(n0 )]λ1 −δ+ q +1 n0 [v(y)]

ε [v(n0 )]−λ1 +δ− q L v (n0 ) + = < ∞, ε λ1 − δ − pε [v(n0 )]λ1 −δ+ q +1 λ1 − δ + εq

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

64

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and then (3.10) follows.

Lemma 3.5. If C is the set of complex numbers and C∞ = C ∪ {∞}, zk ∈ C{z|Rez ≥ 0, Imz = 0}(k = 1, 2, · · · , n) are distinct points, the function f (z) is analytic in C∞ except for zi (i = 1, 2, · · · , n), and z = ∞ is a zero point of f (z) whose order is not less than 1, then for α ∈ R, we have ∞ n 2πi f (x) xα−1 dx = Res[f (z)z α−1 , zk ], (3.11) 2παi 1 − e 0 k=1

where 0 < Im ln z = arg z < 2π. In particular, if zk (k = 1, · · · , n) are all poles of order 1, setting ϕk (z) = (z − zk )f (z), (ϕk (zk ) = 0), then ∞ n π f (x) xα−1 dx = (−zk )α−1 ϕk (zk ). (3.12) sin πα 0 k=1

Proof. ﬁnd

Using results of Pang et al. [63] on page 118, we have (3.11). We 1 − e2παi = 1 − cos 2πα − i sin 2πα = −2i sin πα(cos πα + i sin πα) = −2i eiπα sin πα.

In particular, since f (z) z α−1 =

1 (ϕk (z) z−zk

z α−1 ), it is obvious that

Res[f (z) z α−1 , −ak ] = zk α−1 ϕk (zk ) = −eiπα (−zk )α−1 ϕk (zk ). Then, by (3.11), we obtain (3.12).

Example 3.1. (i) For kλ (t, 1) = tλ1+1 (λ, λ1 , λ2 > 0, λ1 + λ2 = λ), we ﬁnd ∞ λ1 1 ∞ 1 1 λ1 −1 t u λ −1 du k(λ1 ) = dt = (u = tλ ) λ t +1 λ 0 u+1 0 π ∈ R+ . = 1 λ sin( πλ ) λ Setting δ =

λ1 2

(< λ1 < λ), it follows that ) 1 1 1 . ≤ δ , t ∈ 0, kλ (t, 1) = λ t +1 t v(n0 )

(ii) For kλ (t, 1) =

1 (t+1)λ

k(λ1 ) =

0

∞

(λ, λ1 , λ2 > 0, λ1 + λ2 = λ), we ﬁnd

1 tλ1 −1 dt = B(λ1 , λ2 ) ∈ R+ . (t + 1)λ

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

Setting δ =

λ1 2

ws-book9x6

65

(< λ1 < λ), it follows that ) 1 1 1 t ∈ 0, . kλ (t, 1) = ≤ (t + 1)λ tδ v(n0 )

t (iii) For kλ (t, 1) = tλln−1 (λ, λ1 , λ2 > 0, λ1 + λ2 = λ), we ﬁnd ∞ ∞ 1 ln t λ1 −1 ln u λ1 −1 t u λ du, (u = tλ ) k(λ1 ) = dt = λ 2 t −1 λ 0 u−1 0 2

π ∈ R+ . = 1 λ sin( πλ ) λ δ

t → 0 as t → 0+ , there exists a Setting δ = λ21 (< λ1 < λ), in view of ttλ ln −1 constant L > 0 such that ) L ln t 1 kλ (t, 1) = λ . ≤ δ, t ∈ 0, t −1 t v(n0 ) s (iv) For s ∈ N, kλs (t, 1) = k=1 ak t1λ +1 (0 < a1 < · · · < as , λ, λ1 , λ2 > 0, λ1 + λ2 = λs), by (3.12), we ﬁnd ∞2 s 1 k(λ1 ) = tλ1 −1 dt λ ak t + 1 0 k=1 ∞2 s λ2 1 1 u λ −1 du = λ 0 ak + u k=1

=

s s λ2 2 1 π λ −1 a ∈ R+ . k πλ2 a − ak λ sin( λ ) k=1 j j=1(j =k)

(< λ1 < λs), there exists a constant L = (v(n10 ))δ > 0 ) s 2 L 1 1 . ≤ 1 ≤ δ , t ∈ 0, kλs (t, 1) = a k tλ + 1 t v(n0 )

Setting δ =

λ1 2

k=1

(v) For kλ (t, 1) = tλ +2tλ/21 cos β+1 (λ > 0, 0 < β ≤ π2 , λ1 , λ2 > 0, λ1 + λ2 = λ), setting z1 = −eiβ , z2 = −e−iβ , by (3.12), we ﬁnd ∞ tλ1 −1 k(λ1 ) = dt λ λ/2 t + 2t cos β + 1 0 2λ1 2 ∞ u λ −1 λ/2 du, u = t = λ 0 u2 + 2u cos β + 1 ) ( 2λ 2λ π 1 1 iβ λ1 −1 −iβ λ1 −1 1 (e ) + (e ) = e−iβ − eiβ eiβ − e−iβ sin 2πλ λ =

2π sin β(1 − 2λλ1 ) 1 ∈ R+ . λ sin β sin 2πλ λ

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

66

Half-Discrete Hilbert-Type Inequalities δ

t Setting δ = λ21 (< λ1 < λ), in view of tλ +2tλ/2 → 0 as t → 0+ , there cos β+1 exists a constant L > 0 such that ) 1 1 1 ≤L δ t ∈ 0, kλ (t, 1) = λ . t v(n0 ) t + 2tλ/2 cos β + 1 √ (vi) For kλ (t, 1) = tλ +bt1λ/2 +c (c > 0, 0 ≤ b ≤ 2 c, λ > 0, λ1 = λ2 = λ 2 ), we ﬁnd ∞ λ 2 ∞ t 2 −1 dt du , (u = tλ/2 ) = k(λ1 ) = 2 + bu + c λ + btλ/2 + c λ u t 0 0 ⎧ π b = 0, ⎪ √c , √ √ 2⎨ 4 2 4c−b √ , 0 0 such that ) 1 1 1 ≤ L δ , t ∈ 0, kλ (t, 1) = λ . t v(n0 ) t + btλ/2 + c (vii) For k0 (t, 1) = γ/tλ 1 −γ/tλ , (−1 ≤ A < 1, β > 0 (A = 1, β > 1), e −Ae γ > 0, λ1 = −λ2 = −βλ < 0 < λ), in view of the following formula (see Wang and Guo [73]) ∞ 1 e−αu uβ−1 du = β Γ(β) (α, β > 0), (3.13) α 0 we ﬁnd

t−βλ−1 dt uβ−1 du 1 ∞ = λ 0 eγu (1 − Ae−2γu ) eγ/tλ − Ae−γ/tλ 0 ∞ ∞ 1 Ak e−γ(2k+1)u uβ−1 du = λ 0 k=0 ∞ ∞ 1 k = A e−γ(2k+1)u uβ−1 du λ 0

∞

k(λ1 ) =

k=0

∞ Ak Γ(β) = ∈ R+ . β λγ (2k + 1)β k=0

δ

t → 0 (t → 0+ ), Setting δ = −2βλ < −βλ = λ1 , in view of eγ/tλ −Ae −γ/tλ there exists a constant L > 0 such that ) 1 tδ ≤ L, t ∈ 0, . λ λ v(n0 ) eγ/t − Ae−γ/t

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

ws-book9x6

67

Then, it follows that

) 1 1 1 t ∈ 0, . ≤ L, tδ v(n0 ) eγ/tλ − Ae−γ/tλ λ bt +1 (0 ≤ a < b, λ > 0, λ1 = −λ2 = (viii) For k0 (t, 1) = ln at λ +1 k0 (t, 1) =

−βλ, 0 < β < 1), we ﬁnd λ ∞ λ −1 ∞ bt + 1 bt + 1 −βλ−1 t dt−βλ ln dt = ln k(λ1 ) = λ+1 λ+1 at βλ at 0 0 ( ∞ ∞ λ ) bλ aλ −1 −βλ bt + 1 (1−β)λ−1 t t − = ln − dt βλ atλ + 1 0 btλ + 1 atλ + 1 0 β ∞ (1−β)−1 aβ (bβ − aβ )π u b − du = . = βλ βλ 1+u βλ sin(βπ) 0 λ bt +1 → 0 as Setting δ = −β0 λ < λ1 (β < β0 < 1), in view of tδ ln at λ +1 t → 0+ , there exists a constant L > 0 such that λ ) bt + 1 1 δ ≤ L, t ∈ 0, . t ln atλ + 1 v(n0 ) Then it follows that ) λ L 1 bt + 1 ≤ . , t ∈ 0, k0 (t, 1) = ln atλ + 1 tδ v(n0 )

(ix) For k0 (t, 1) = arctan tλ (λ1 = −λ2 = −β, 0 < β < λ), we ﬁnd ∞ 1 ∞ t−β−1 arctan tλ dt = − arctan tλ dt−β k(λ1 ) = β 0 0 ) ( ∞ 1 −β λ λ ∞ (λ−β)−1 =− t dt t arctan t |0 − β 1 + t2λ 0 ∞ λ−β −1 1 u 2λ π ∈ R+ . = du = 2β 0 1+u 2β cos πβ 2λ Setting δ = −β0 < −β = λ1 (β0 < λ), since tδ arctan tλ → 0 as t → 0+ , there exists a constant L > 0 such that ) 1 1 k0 (t, 1) = arctan tλ ≤ L δ , t ∈ 0, . t v(n0 ) (x) For kλ (t, 1) = k(λ1 ) =

1 (max{t,1})λ ∞

(λ, λ1 , λ2 > 0, λ1 + λ2 = λ), we ﬁnd

1 tλ1 −1 dt (max{t, 1})λ 0 ∞ 1 λ 1 λ1 −1 tλ1 −1 dt + t dt = ∈ R+ . = λ t λ λ2 1 0 1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

68

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Setting δ =

λ1 2 (<

λ1 < λ), it follows that

1 1 ≤ δ, kλ (t, 1) = λ (max{t, 1}) t

t ∈ 0,

1 v(n0 )

) .

(xi) For k−λ (t, 1) = (min{t, 1})λ (λ1 , λ2 < 0, λ1 + λ2 = −λ), we ﬁnd ∞ (min{t, 1})λtλ1 −1 dt k(λ1 ) =

0

=

1

λ λ1 −1

t t

∞

dt +

0

tλ1 −1 dt =

1

Setting δ = −λ (< λ1 ), it follows that λ

k−λ (t, 1) = (min{t, 1})

1 ≤ t = δ t λ

λ ∈ R+ . λ1 λ2

t ∈ 0,

1 v(n0 )

) .

λ min{t,1} (xii) For k0 (t, 1) = max{t,1} (λ > 0, λ1 = −λ2 = α, |α| < λ), we ﬁnd λ ∞ min{t, 1} tα−1 dt k(λ1 ) = max{t, 1} 0 ∞ 1 2λ tλ tα−1 dt + t−λ tα−1 dt = 2 ∈ R+ . = λ − α2 0 1 Setting δ =

−λ+λ1 2

< λ1 (|δ| < λ), and in view of λ min{t, 1} tδ → 0 (t → 0+ ), max{t, 1} λ min{t,1} ≤ L, and then, there exists a constant L > 0 such that tδ max{t,1} kλ (t, 1) =

3.2.3

min{t, 1} max{t, 1}

λ

L ≤ δ t

t ∈ 0,

1 v(n0 )

) .

Some Inequalities Relating the Constant k(λ1 )

Lemma 3.6. Suppose that λ, λ1 , λ2 ∈ R, λ = λ1 + λ2 , kλ (x, y) is a non-negative ﬁnite measurable homogeneous function of degree −λ in R2+ with ∞ k(λ1 ) = kλ (t, 1) tλ1 −1 dt ∈ R+ . 0

Then,

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

ws-book9x6

69

(i) for p > 1, ε > 0, we have ε ≥ k(λ1 ) + o(1) (ε → 0+ ); k λ1 − (3.14) p (ii) for p < 0, if there exists a constant δ1 > 0, such that k(λ1 + δ1 ) ∈ R+ , then, for 0 < ε< (−p)δ1 , we have ε ≤ k(λ1 ) + o(1) (ε → 0+ ); k λ1 − (3.15) p (iii) for 0 0, such that k(λ1 − δ2 ) ∈ R+ , then, for 0 < ε < pδ2 , we still have (3.15). Proof. that

(i) For p > 1, by the Fatou lemma (see Kuang [49]), it follows

ε ε , lim+ kλ (t, 1) t(λ1 − p )−1 dt ≤ lim k λ1 − p ε→0 ε→0+ 0 and then we have (3.14). ε (ii) For p < 0, since t− p ≤ 1 (0 < t ≤ 1), we ﬁnd 1 ∞ ε ε ε = k λ1 − kλ (t, 1) t(λ1 − p )−1 dt + kλ (t, 1) t(λ1 − p )−1 dt p 0 1 ∞ 1 ε kλ (t, 1) tλ1 −1 dt + kλ (t, 1) t(λ1 − p )−1 dt. (3.16) ≤

∞

k(λ1 ) =

0

1

For 0 < ε < (−p)δ1 , we have ∞ (λ1 − pε )−1 kλ (t, 1) t dt ≤ 1

∞

1

kλ (t, 1) t(λ1 +δ1 )−1 dt ≤ k(λ1 + δ1 ) < ∞,

then, by the Lebesgue control convergence theorem, it follows that ∞ ∞ ε kλ (t, 1) t(λ1 − p )−1 dt = kλ (t, 1) tλ1 −1 dt + o(1) (ε → 0+ ). 1

1

Hence, by (3.16), we have (3.15). ε (iii) For 0 < p < 1, since t− p ≤ 1 (t ≥ 1), we ﬁnd 1 ∞ ε ε (λ1 − pε )−1 = k λ1 − kλ (t, 1) t dt + kλ (t, 1) t(λ1 − p )−1 dt p 0 1 ∞ 1 ε kλ (t, 1) t(λ1 − p )−1 dt + kλ (t, 1) tλ1 −1 dt. (3.17) ≤ 0

For 0 < ε < pδ2 , we have 1 ε (λ1 − p )−1 kλ (t, 1) t dt ≤ 0

1

1 0

kλ (t, 1) t(λ1 −δ2 )−1 dt ≤ k(λ1 − δ2 ) < ∞,

then by the Lebesgue control convergence theorem (see Kuang [49]), it follows 1 1 (λ1 − pε )−1 kλ (t, 1)t dt = kλ (t, 1)tλ1 −1 dt + o(1) (ε → 0+ ). 0

0

Hence, by (3.17), we have (3.15) for 0 < p < 1.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

70

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

3.3 3.3.1

Some Theorems and Corollaries Equivalent Inequalities and their Operator Expressions

For p ∈ R\{0, 1}, (0, ∞)) and

1 p

+

Ψ(n) =

1 q

= 1, we set two functions ϕ(x) = xp(1−λ1 )−1 (x ∈

[v(n)]q(1−λ2 )−1 , [v (n)]q−1

n ≥ n0

(n ∈ N),

v (n) wherefrom, [ϕ(x)]1−q = xqλ1 −1 and [Ψ(n)]1−p = [v(n)] 1−pλ2 , and then, we deﬁne two sets as follows: ∞ 1 p(1−λ1 )−1 p p x |f (x)| dx} < ∞ , Lp,ϕ (R+ ) = f |||f ||p,ϕ = { 0 ∞ [v(n)]q(1−λ2 )−1 ∞ q q1 lq,Ψ = a = {an }n=n0 |||a||q,Ψ = { |an | } < ∞ . [v (n)]q−1 n=n 0

Note. It is obvious that for p > 1, both of the above two sets are the normed spaces. Since for p < 0 or 0 < p < 1, neither of the above two sets is normed space, then we agree on that the above sets with ||f ||p,ϕ and ||a||q,Ψ are the formal symbols. Theorem 3.1. Suppose that λ, λ1 , λ2 ∈ R, λ = λ1 + λ2 , kλ (x, y) is a nonnegative ﬁnite measurable homogeneous function of degree −λ in R2+ , v(y) is a strictly increasing diﬀerentiable function in [n0 , ∞) (n0 ∈ N) with v(n0 ) > 0 and v(∞) = ∞ ∞ kλ (t, 1)tλ1 −1 dt ∈ R+ k(λ1 ) = 0

and λ1 (x) < k(λ1 ) (x ∈ (0, ∞)). If p > 1, p1 + 1q = 1, f (x), an ≥ 0, f ∈ Lp,ϕ (R+ ), a = {an}∞ n=n0 ∈ lq,Ψ , ||f ||p,ϕ > 0, ||a||q,Ψ > 0, then we have the following equivalent inequalities: ∞ ∞ ∞ ∞ an kλ (x, v(n))f (x)dx = f (x) an kλ (x, v(n))dx I= n=n0

0

0

n=n0

< k(λ1 )||f ||p,ϕ ||a||q,Ψ , (3.18) 1 ∞ )p p ( ∞ J= [Ψ(n)]1−p kλ (x, v(n))f (x)dx < k(λ1 )||f ||p,ϕ , n=n0

0

(3.19)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

71

and

∞

L=

∞

[ϕ(x)]1−q

0

q kλ (x, v(n))an

1q dx

< k(λ1 )||a||q,Ψ . (3.20)

n=n0

(y) (> 0) is decreasing in [n0 , ∞) and there exist constants Moreover, if vv(y) δ < λ1 and L > 0 such that the inequality (3.9) is satisﬁed, then the constant factor k(λ1 ) in the above inequalities is the best possible.

Proof. By the Lebesgue term by term integration theorem, there are two expressions for I in (3.18). In view of (3.6) and λ1 (x) < k(λ1 ), we have (3.19). By H¨older’s inequality, we ﬁnd ) ∞ ∞ ( 1 1 [Ψ(n)]− q I= kλ (x, v(n))f (x)dx [[Ψ(n)] q an ] ≤ J||a||q,Ψ . n=n0

0

(3.21) Then by (3.19), we have (3.18). On the other hand, suppose that (3.18) is valid. We set ( ∞ ) p−1 an = [Ψ(n)]1−p kλ (x, v(n))f (x)dx , n ≥ n0 (n ∈ N). 0

= ||a||q,Ψ . By (3.6), we ﬁnd J < ∞. If J = 0, Then it follows that J then (3.19) is trivially valid; if J > 0, then by (3.18), we have p−1

||a||qq,Ψ = J p = I < k(λ1 )||f ||p,ϕ ||a||q,Ψ ,

that is,

||a||q−1 q,Ψ = J < k(λ1 )||f ||p,ϕ , and then (3.18) is equivalent to (3.19). In view of (3.7), since [λ1 (x)]1−q > [k(λ1 )]1−q , we have (3.20). By H¨older’s inequality, we ﬁnd

∞ ∞ 1 1 − I= [[ϕ(x)] p f (x)] [ϕ(x)] p an kλ (x, v(n)) dx ≤ ||f ||p,ϕ L. 0

n=n0

(3.22) Then by (3.20), we have (3.18). On the other hand, suppose that (3.18) is valid. We set

∞ q−1 1−q f (x) = [ϕ(x)] kλ (x, v(n))an , x ∈ (0, ∞). n=n0

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

72

Half-Discrete Hilbert-Type Inequalities

Then it follows that Lq−1 = ||f ||p,ϕ . By (3.7), we ﬁnd L < ∞. If L = 0, then (3.20) is trivially valid; if L > 0, then, by (3.18), we have ||f ||pp,ϕ = Lq = I < k(λ1 )||f ||p,ϕ ||a||q,Ψ , p−1 ||f ||p,ϕ

that is

= L < k(λ1 )||a||q,Ψ ,

and then, inequality (3.18) is equivalent to (3.20). Hence, inequalities (3.18), (3.19) and (3.20) are equivalent. For 0 < ε < p(λ1 − δ), setting 0, 0 < x < 1, ε f(x) = xλ1 − p −1 , x ≥ 1 and ε

an = [v(n)]λ2 − q −1 v (n),

n ≥ n0 (n ∈ N),

if there exists a positive constant k(≤ k(λ1 )) such that (3.18) is still valid as we replace k(λ1 ) by k, then substitution of f(x) and a = { an }∞ n=n0 , by (3.8), it follows that ∞ ∞ I = an kλ (x, v(n))f(x)dx < k||f||p,ϕ || a||q,Ψ 0

n=n0

∞

=k

x 1

−1−ε

p1 1 1 k dx (A(ε)) q = (1 + o(1)) q . ε

By (3.8), (3.10) and (3.14), we ﬁnd ∞ ∞ ε λ2 − qε −1 I= [v(n)] v (n) kλ (x, v(n))xλ1 − p −1 dx =

n=n0 ∞ n=n0

v (n) [v(n)]1+ε

1

∞ 1 v(n)

ε

kλ (t, 1)tλ1 − p −1 dt,

t = x/v(n)

1 v(n) ∞ ε ε v (n) (λ1 − p )−1 λ1 − p −1 kλ (t, 1) t dt − kλ (t, 1) t dt = [v(n)]1+ε 0 0 n=n0 ε − B(ε) = A(ε)k λ1 − p 1 ≥ [(1 + o(1))(k(λ1 ) + o(1)) − ε O(1)]. ε it follows that Hence, in view of the above results for I, ∞

1 (1 + o(1))(k(λ1 ) + o(1)) − ε O(1) ≤ εI < k(1 + o(1)) q ,

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

ws-book9x6

73

and then, k(λ1 ) ≤ k (ε → 0+ ). Therefore, the constant factor k(λ1 ) in (3.18) is the best possible. By the equivalency, the constant factor k(λ1 ) in (3.19) and (3.20) is also the best possible. Otherwise, it leads to a contradiction by (3.21) and (3.22) that the constant factor in (3.18) is not the best possible. Note. If we replace the lower limit 0 of the integral to c > 0 in (3.18)(3.20), then Theorem 3.1 is still valid for ∞ ωλ2 (n) = [v(n)]λ2 kλ (x, v(n)) xλ1 −1 dx < k(λ1 ). c

Remark 3.1. If we replace the conditions that k(λ1 ) ∈ R+ , and there exist constants δ < λ1 and L > 0 such that inequality (3.9) is satisﬁed to the 1 ∈ [λ1 , λ1 + δ0 ), condition that there exist δ0 , η > 0, such that for any λ 1 ) < ∞ (x ∈ (0, ∞)), 1 )(1 − θ (x)) < (x) < k(λ (3.23) 0 < k(λ λ1 λ1 1 where, θλ1 (x) > 0 (x > 0), and θλ1 (x) = O xη (1 ≤ x < ∞), then the constant factor k(λ1 ) in (3.18) is still the best possible. In fact, for 1 = λ1 + ε , we ﬁnd 0 < ε < qδ0 , λ q

∞ ∞ v (n) λ1 + εq −ε−1 x kλ (x, v(n)) x dx I= ε [v(n)]1−(λ2 − q ) 1 n=n0 ∞ ∞ 1 −ε−1 −ε−1 1−O dx = x λ1 (x)dx > k(λ1 ) x xη 1 1 1 ≥ (k(λ1 ) + o(1))(1 − ε O(1)), ε and then by the same way, we can still show that k(λ1 ) is the best possible constant in (3.18). Remark 3.2. By virtue of Theorem 3.1, (i) deﬁne a half-discrete Hilberttype operator T : Lp,ϕ (R+ ) → lp,Ψ1−p as follows: For f ∈ Lp,ϕ (R+ ), we deﬁne T f ∈ lp,Ψ1−p , satisfying ∞ T f (n) = kλ (x, v(n)) f (x) dx, n ≥ n0 (n ∈ N). (3.24) 0

Then, by (3.19), it follows that ||T f ||p.Ψ1−p ≤ k(λ1 )||f ||p,ϕ , and then, T is a bounded operator with ||T || ≤ k(λ1 ). Since by Theorem 3.1, the constant factor in (3.19) is the best possible, we have ||T || = k(λ1 ).

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

74

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

(ii) Deﬁne a half-discrete Hilbert-type operator T : lq,Ψ → Lq,ϕ1−q (R+ ) as follows: For a ∈ lq,Ψ , we deﬁne Ta ∈ Lq,ϕ1−q (R+ ), satisfying Ta(x) =

∞

kλ (x, v(n))an ,

x ∈ (0, ∞).

(3.25)

n=n0

Then, by (3.20), it follows that ||Ta||q.ϕ1−q ≤ k(λ1 )||a||q,Ψ , and then, T is a bounded operator with ||T|| ≤ k(λ1 ). Since by Theorem 3.1, the constant factor in (3.20) is the best possible, we have ||T|| = k(λ1 ). Example 3.2. In view of Remark 3.2, in particular, setting v(n) = n and ψ(n) = nq(1−λ2 )−1 (n ∈ N), deﬁne T : Lp,ϕ (R+ ) → lp,ψ1−p as follows: For f ∈ Lp,ϕ (R+ ), we deﬁne T f ∈ lp,ψ1−p , satisfying ∞ T f (n) = kλ (x, n)f (x)dx, n ∈ N. 0

Also we deﬁne T : lq,ψ → Lq,ϕ1−q (R+ ) as follows: For a ∈ lq,ψ , there exists Ta ∈ Lq,ϕ1−q (R+ ), satisfying Ta(x) =

∞

kλ (x, n)an ,

x ∈ (0, ∞).

n=1

Then, by Remark 3.2, we still have ||T || = ||T|| = k(λ1 ). (i) If kλ (x, n) = x > 0,

1 xλ +nλ

(λ1 > 0, 0 < λ2 ≤ 1, λ1 + λ2 = λ), since for

f (x, y) =

xλ1 y λ2 −1 + yλ

xλ

is decreasing with respect to y ∈ (0, ∞), then, Condition (i) is satisﬁed. In view of Example 3.1(i), the constant factor in (3.19) and (3.20) (for v(n) = n, n0 = 1) is the best possible, we have π 1 . ||T || = ||T|| = k(λ1 ) = λ sin πλ λ (ii) If kλ (x, n) = x > 0,

1 (λ1 (x+n)λ

> 0, 0 < λ2 ≤ 1, λ1 + λ2 = λ), since for

f (x, y) =

xλ1 y λ2 −1 (x + y)λ

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

ws-book9x6

75

is decreasing with respect to y ∈ (0, ∞), then Condition (i) is satisﬁed. In view of Example 3.1(ii), the constant factor in (3.19) and (3.20) (for v(n) = n, n0 = 1) is the best possible, we have (see Yang [135]) ||T || = ||T|| = k(λ1 ) = B(λ1 , λ2 ). (λ1 > 0, 0 < λ2 ≤ 1, λ1 + λ2 = λ), then by (iii) If kλ (x, n) = ln(x/n) xλ −nλ the decreasing property of f (x, y) and Example 3.1(iii), we have 2

π ||T || = ||T|| = k(λ1 ) = . 1 λ sin( πλ ) λ (iv) If s ∈ N, kλs (x, n) = sk=1 ak xλ1+nλ (0 < a1 < · · · < as , λ, λ1 > 0, 0 < λ2 ≤ 1, λ1 + λ2 = λs), then by the decreasing property of f (x, y) and Example 3.1(iv), we have s s λ2 2 1 π λ −1 ||T || = ||T|| = k(λ1 ) = a . k πλ2 aj − ak λ sin( λ ) j=1(j =k)

k=1

1

(v) For kλ (x, n) = xλ +2(xn)λ/2 cos β+nλ (λ > 0, 0 < β ≤ π2 , λ1 > 0, 0 < λ2 ≤ 1, λ1 + λ2 = λ), then by the decreasing property of f (x, y) and Example 3.1(v), we have 2π sin β(1 − 2λλ1 ) . ||T || = ||T|| = k(λ1 ) = 1 λ sin β sin( 2πλ λ ) (vi) If kλ (x, n) =

1 xλ +b(xn)λ/2 +cnλ

√ (c > 0, 0 ≤ b ≤ 2 c, 0 < λ ≤

2, λ1 = λ2 = λ2 ), then by the decreasing property of f (x, y) and Example 3.1(vi), we have ⎧ π √ , b = 0, ⎪ c ⎨ √ √ 2 2 4c−b 4 √ ||T || = ||T || = arctan b , 0 0(A = 1, β > (vii) If k0 (x, n) = 1), γ > 0, −1 ≤ λ1 = −λ2 = −βλ < 0 < λ), then by the decreasing property of f (x, y) and Example 3.1(vii), we have ∞ Γ(β) Ak ||T || = ||T|| = k(−βλ) = . β λγ (2k + 1)β k=0

bxλ +nλ ln( ax λ +nλ )

(viii) If k0 (x, n) = (0 ≤ a < b, λ > 0, −1 ≤ λ1 = −λ2 = −βλ, 0 < β < 1), then by the decreasing property of f (x, y) and Example 3.1(viii), we have (bβ − aβ )π . ||T || = ||T|| = k(−βλ) = βλ sin(βπ)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

76

Half-Discrete Hilbert-Type Inequalities

λ (ix) If k0 (x, n) = arctan nx (λ1 = −λ2 = −β, 0 < β < λ, β ≤ 1), then by the decreasing property of f (x, y) and Example 3.1(ix), we have ||T || = ||T|| = k(−β) =

π . 2β cos( πβ ) 2λ

1 (x) If kλ (x, n) = (max{x,n}) λ (λ1 > 0, 0 < λ2 ≤ 1, λ1 + λ2 = λ), then by the decreasing property of f (x, y) and Example 3.1(x), we have (see Yang [159])

||T || = ||T|| = k(λ1 ) =

λ . λ1 λ2

(xi) If k−λ (x, n) = (min{x, n})λ (λ1 < 0, −1 ≤ λ2 < 0, λ1 + λ2 = −λ), then by the decreasing property of f (x, y) and Example 3.1(xi), we have ||T || = ||T|| = k(λ1 ) = (xii) If k0 (x, n) = α>

− 12 ),

min{x,n} max{x,n}

λ

λ . λ1 λ2

(λ1 = −λ2 = α, |α| < λ ≤ 1 + α,

since for x > 0, λ min{x, y} y −α−1 f (x, y) = xα max{x, y} α−λ λ−α−1 x y , 0 < y ≤ x, = xα+λ y −λ−α−1 , y > x

is decreasing with respect to y ∈ (0, ∞), then Condition (i) is satisﬁed. In view of Example 3.1(xii), the constant factor in (3.19) and (3.20) (for v(n) = n, n0 = 1) is the best possible, we obtain ||T || = ||T|| = k(α) = 3.3.2

λ2

2λ . − α2

Two Classes of Equivalent Reverse Inequalities

Theorem 3.2. Suppose that λ, λ1 , λ2 ∈ R, λ = λ1 + λ2 , kλ (x, y) is a non-negative ﬁnite measurable homogeneous function of degree −λ in R2+ , v(y) is a strictly increasing diﬀerentiable function in [n0 , ∞) (n0 ∈ N) with v(n0 ) > 0 and v(∞) = ∞ ∞ kλ (t, 1) tλ1 −1 dt ∈ R+ k(λ1 ) = 0

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

77

and λ1 (x) < k(λ1 ) (x ∈ (0, ∞)). If p < 0, p1 + 1q = 1, f (x), an ≥ 0, f ∈ Lp,ϕ (R+ ), a = {an }∞ n=n0 ∈ lq,Ψ , ||f ||p,ϕ > 0, ||a||q,Ψ > 0, then we have the following equivalent inequalities: ∞ ∞ ∞ ∞ I= an kλ (x, v(n))f (x) dx = f (x) an kλ (x, v(n)) dx 0

n=n0

0

n=n0

> k(λ1 )||f ||p,ϕ ||a||q,Ψ , (3.26) ∞ p1 ) ( p ∞ [Ψ(n)]1−p kλ (x, v(n))f (x) dx > k(λ1 )||f ||p,ϕ , J= 0

n=n0

(3.27) and

∞

L=

∞

[ϕ(x)]1−q

0

q kλ (x, v(n))an

1q dx

> k(λ1 )||a||q,Ψ . (3.28)

n=n0

(y) Moreover, if vv(y) (> 0) is decreasing in [n0 , ∞) and there exists a constant δ1 > 0, such that k(λ1 + δ1 ) ∈ R+ , then the constant factor k(λ1 ) in the above inequalities is the best possible.

Proof. In view of the reverse of (3.6) and λ1 (x) < k(λ1 ), for p < 0, we have (3.27). By the reverse H¨older’s inequality (see Kuang [47]), we ﬁnd ) ∞ ∞ ( 1 − q1 I= kλ (x, v(n))f (x)dx [[Ψ(n)] q an ] ≥ J||a||q,Ψ . [Ψ(n)] n=n0

0

(3.29) Then, by (3.27), we have (3.26). On the other hand, suppose that (3.26) is valid. We set ) p−1 ( ∞ kλ (x, v(n))f (x)dx , n ≥ n0 (n ∈ N). an = [Ψ(n)]1−p 0

Then it follows that J p−1 = ||a||q,Ψ . By the reverse of (3.6), we ﬁnd J > 0. If J = ∞, then (3.27) is trivially valid; if J < ∞, then, by (3.26), we have ||a||qq,Ψ = J p = I > k(λ1 )||f ||p,ϕ ||a||q,Ψ , that is ||a||q−1 q,Ψ = J > k(λ1 )||f ||p,ϕ , and then, inequality (3.26) is equivalent to (3.27). In view of the reverse of (3.7) and [λ1 (x)]1−q < [k(λ1 )]1−q

(0 < q < 1),

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

78

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

we have inequality (3.28). By the reverse H¨older’s inequality, we ﬁnd

∞ ∞ 1 1 I= [[ϕ(x)] p f (x)] [ϕ(x)]− p kλ (x, v(n))an dx ≥ ||f ||p,ϕ L. 0

n=n0

(3.30) Then, by (3.28), we have (3.26). On the other hand, suppose that (3.26) is valid. We set

∞ q−1 f (x) = [ϕ(x)]1−q kλ (x, v(n))an , x ∈ (0, ∞). n=n0

Then, it follows that Lq−1 = ||f ||p,ϕ . By the reverse of (3.7), we ﬁnd L > 0. If L = ∞, then (3.28) is trivially valid; if L < ∞, then by (3.26), we have ||f ||pp,ϕ = Lq = I > k(λ1 )||f ||p,ϕ ||a||q,Ψ , that is p−1 ||f ||p,ϕ = L > k(λ1 )||a||q,Ψ ,

and then, (3.26) is equivalent to (3.28). Hence, inequalities (3.26), (3.27) and (3.28) are equivalent. For 0 < ε < (−p)δ1 , setting f(x) and an as Theorem 3.1, if there exists a positive constant k(≥ k(λ1 )), such that (3.26) is valid as we replace k(λ1 ) by k, then substitution of f(x) and a = { an } ∞ n=n0 , by (3.8), it follows that ∞ ∞ an kλ (x, v(n))f(x)dx > k||f||p,ϕ || a||q,Ψ I = 0

n=n0

=k

∞

−1−ε

x 1

p1 1 1 k dx (A(ε)) q = (1 + o(1)) q . ε

In view of (3.8) and (3.15), we ﬁnd ∞ λ2 − qε −1 I = [v(n)] v (n)

=

n=n0 ∞ n=n0 ∞

v (n) [v(n)]1+ε

∞ 1

∞ 1 v(n)

ε

kλ (x, v(n))xλ1 − p −1 dx ε

kλ (t, 1)tλ1 − p −1 dt,

(t = x/v(n))

∞ ε v (n) kλ (t, 1) t(λ1 − p )−1 dt 1+ε [v(n)] 0 n=n0 ε 1 = A(ε)k λ1 − ≤ (1 + o(1))(k(λ1 ) + o(1)). p ε ≤

(3.31)

Hence, by (3.31) and (3.32), it follows that 1

(1 + o(1))(k(λ1 ) + o(1)) > k(1 + o(1)) q ,

(3.32)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

79

and then k(λ1 ) ≥ k (ε → 0+ ). Therefore, the constant factor k(λ1 ) in (3.26) is the best possible. By the equivalency, the constant factor k(λ1 ) in (3.27) and (3.28) is the best possible. Otherwise, it leads to a contradiction by (3.29) and (3.30) that the constant factor in (3.26) is not the best possible. If we replace the lower limit 0 of the integral to c (> 0) in Theorem 3.2 then it follows that ∞ kλ (x, v(n))xλ1 −1 dx < k(λ1 ). ωλ2 (n) = [v(n)]λ2 c

Setting ωλ2 (n) = k(λ1 )(1 − θλ1 (v(n))), c 1 λ2 θλ1 (v(n)) = [v(n)] kλ (x, v(n)) xλ1 −1 dx > 0, k(λ1 ) 0 and

[v(n)]q(1−λ2 )−1 (1 − θλ1 (v(n))), Ψ(n) = Ψ(n)(1 − θλ1 (v(n))) = [v (n)]q−1 we have c v(n) 1 θλ1 (v(n)) = kλ (t, 1) tλ1 −1 dt, k(λ1 ) 0 and the following corollary: Corollary 3.1. Suppose that c > 0, λ, λ1 , λ2 ∈ R, λ = λ1 + λ2 , kλ (x, y) is a non-negative ﬁnite measurable homogeneous function of degree −λ in R2+ , v(y) is a strictly increasing diﬀerential functions in [n0 , ∞) (n0 ∈ N) with v(n0 ) > 0 and v(∞) = ∞ ∞ kλ (t, 1) tλ1 −1 dt ∈ R+ k(λ1 ) = 0

and λ1 (x) < k(λ1 ) (x ∈ (c, ∞)). If p < 0, p1 + 1q = 1, f (x), an ≥ 0, f ∈ Lp,ϕ (c, ∞), a = {an }∞ , n=n0 ∈ lq,Ψ ||f ||p,ϕ > 0, ||a||q,Ψ > 0, then, we have the following equivalent inequalities: ∞ ∞ an kλ (x, v(n))f (x)dx n=n0

c

∞

f (x)

= c

∞ n=n0

∞

an kλ (x, v(n))dx > k(λ1 )||f ||p,ϕ ||a||q,Ψ , (3.33)

n=n0

[Ψ(n)]

1−p

(

∞

) p p1 kλ (x, v(n))f (x)dx > k(λ1 )||f ||p,ϕ ,

c

(3.34)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

80

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and

∞

[ϕ(x)]1−q

c

1q

q

∞

kλ (x, v(n))an

dx

> k(λ1 )||a||q,Ψ .

(3.35)

n=n0

(y) Moreover, if vv(y) (> 0) is decreasing in [n0 , ∞) and there exist constants L > 0, 0 < δ < λ1 , such that k(λ1 + δ) ∈ R+ , and ) c 1 , kλ (t, 1) ≤ L δ , t ∈ 0, t v(n0 )

then, the constant factor k(λ1 ) in the above inequalities is the best possible. Theorem 3.3. Suppose that λ, λ1 , λ2 ∈ R, λ = λ1 + λ2 , kλ (x, y) is a nonnegative ﬁnite measurable homogeneous function of degree −λ in R2+ , v(y) is a strictly increasing diﬀerentiable function in [n0 , ∞) (n0 ∈ N) with v(n0 ) > 0, v(∞) = ∞, and 0 < k(λ1 )(1 − θλ1 (x)) < λ1 (x) < k(λ1 ) < ∞

(x ∈ (0, ∞)),

where, θλ1 (x) > 0. Setting ϕ(x) = (1−θλ1 (x))ϕ(x), if 0 0, ||a||q,Ψ > 0, n=n0 ∈ lq,Ψ , ||f ||p,ϕ then, we have the following equivalent inequalities: ∞ ∞ ∞ ∞ I= an kλ (x, v(n))f (x)dx = f (x) an kλ (x, v(n))dx 0

n=n0

0

n=n0

(3.36) > k(λ1 )||f ||p,ϕ||a||q,Ψ , ∞ p1 ) ( p ∞ [Ψ(n)]1−p kλ (x, v(n))f (x)dx > k(λ1 )||f ||p,ϕ, J= 0

n=n0

(3.37) and L1 =

0

∞

1−q [ϕ(x)]

∞

q kλ (x, v(n))an

q1 dx

> k(λ1 )||a||q,Ψ .

n=n0

(3.38) Moreover, if 0) is decreasing in [n0 , ∞) and there exist constants δ2 , η > 0, such 1 that k(λ1 − δ2 ) ∈ R+ and for any λ1 ∈ (λ1 − δ2 , λ1 ), θλ1 (x) = O xη (x ∈ [1, ∞)), then, the constant factor k(λ1 ) in the above inequalities is the best possible. v (y) v(y) (>

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

ws-book9x6

81

Proof. In view of the reverse of (3.6) and λ2 (x) < k(λ1 ), for 0 < p < 1, we have (3.37). By the reverse H¨ older’s inequality, we ﬁnd ( ) ∞ ∞ 1 1 [Ψ(n)]− q I= kλ (x, v(n))f (x)dx [[Ψ(n)] q an ] ≥ J||a||q,Ψ . 0

n=n0

(3.39) Then by (3.37), we have (3.36). On the other hand, suppose that (3.36) is valid. We set ( ∞ ) p−1 an = [Ψ(n)]1−p kλ (x, v(n))f (x)dx , n ≥ n0 (n ∈ N). 0

= ||a||q,Ψ . By the reverse of (3.6), we ﬁnd J > 0. Then it follows that J If J = ∞, then (3.37) is trivially valid; if J < ∞, then, by (3.36), we have p−1

||a||qq,Ψ = J p = I > k(λ1 )||f ||p,ϕ||a||q,Ψ , that is, ||a||q−1 , q,Ψ = J > k(λ1 )||f ||p,ϕ and then (3.36) is equivalent to (3.37). In view of the reverse of (3.7) and [λ2 (x)]1−q > [k(λ1 )(1 − θλ1 (x))]1−q , (q < 0), we have (3.38). By the reverse H¨ older’s inequality, we ﬁnd

∞ ∞ 1 1 −p p f (x)] [ϕ(x)] [[ϕ(x)] an kλ (x, v(n)) dx ≥ ||f ||p,ϕL. I= 0

n=n0

(3.40) Then, by (3.38), we have (3.36). On the other hand, suppose that (3.36) is valid. We set

∞ q−1 1−q f (x) = [ϕ(x)] kλ (x, v(n))an , x ∈ (0, ∞). n=n0

= ||f ||p,ϕ . By the reverse of (3.7), we ﬁnd Then, it follows that Lq−1 1 L1 > 0. If L1 = ∞, then (3.38) is trivially valid; if L1 < ∞, then by (3.36), we have ||f ||pp,ϕ = Lq1 = I > k(λ1 )||f ||p,ϕ ||a||q,Ψ , that is p−1 ||f ||p, ϕ = L1 > k(λ1 )||a||q,Ψ ,

and then (3.36) is equivalent to (3.38). Hence, inequalities (3.36), (3.37) and (3.38) are equivalent.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

82

Half-Discrete Hilbert-Type Inequalities

For 0 < ε < pδ2 , setting f(x) and an as Theorem 3.1, if there exists a positive constant k(≥ k(λ1 )) such that (3.36) is valid as we replace k(λ1 ) by k, then substitution of f(x) and a = { an } ∞ n=n0 , by (3.8), it follows that ∞ ∞ I = an kλ (x, v(n))f(x) dx > k||f||p,ϕ || a||q,Ψ n=n0

0

(

=k

) p1 ∞ 1 1 −1−ε dx (A(ε)) q 1−O x η x 1

1 1 k (1 − ε O(1)) p (1 + o(1)) q . ε By (3.8) and (3.15), we ﬁnd ∞ ∞ ε ε [v(n)]λ2 − q −1 v (n) kλ (x, v(n))xλ1 − p −1 dx I =

=

=

n=n0 ∞ n=n0 ∞

v (n) [v(n)]1+ε

1

∞

ε

1 v(n)

kλ (t, 1) tλ1 − p −1 dt,

(t = x/v(n))

∞ ε v (n) kλ (t, 1) t(λ1 − p )−1 dt 1+ε [v(n)] 0 n=n0 ε 1 = A(ε)k λ1 − ≤ (1 + o(1))(k(λ1 ) + o(1)). p ε In view of the above results, it follows that 1 1 (1 + o(1))(k(λ1 ) + o(1)) > k(1 − ε O(1)) p (1 + o(1)) q , and then k(λ1 ) ≥ k (ε → 0+ ). Therefore, the constant factor k(λ1 ) in (3.36) is the best possible. By the equivalency, the constant factor k(λ1 ) in (3.37) and (3.38) is the best possible. Otherwise, we can imply a contradiction by (3.39) and (3.40) that the constant factor in (3.35) is not the best possible. ≤

Note. If we replace the lower limit 0 of the integral to c > 0, then Theorem 3.3 is still valid. In this case, we use ∞

ωλ2 (n) = [v(n)]λ2

kλ (x, v(n))xλ1 −1 dx < k(λ1 ).

c

3.3.3

Some Corollaries

Corollary 3.2. (see Yang and Chen [138]) Let the assumptions of Lemma 3.1 be fulﬁlled and additionally let ∞

λ1 (x) < k(λ1 ) =

0

kλ (t, 1)tλ1 −1 dt ∈ R+ (x ∈ (0, ∞)),

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

83

and for −∞ ≤ b < c ≤ ∞, u(x) be a strictly increasing diﬀerentiable function in (b, c), with u(b+ ) = 0 and u(c− ) = ∞. For p ∈ R\{0, 1}, 1 1 p + q = 1, we deﬁne the function Φ(x) as follows: [u(x)]p(1−λ1 )−1 [u (x)]p−1

Φ(x) =

(x ∈ (b, c)).

If f (x), an ≥ 0, f ∈ Lp,Φ (b, c), a = {an }∞ n=n0 ∈ lq,Ψ , ||f ||p,Φ > 0, ||a||q,Ψ > 0, then (i) for p > 1, we have the following equivalent inequalities: c ∞ an kλ (u(x), v(n))f (x) dx b

n=n0

c

f (x)

= b

∞

∞

an kλ (u(x), v(n)) dx < k(λ1 )||f ||p,Φ ||a||q,Ψ ,

n=n0

[Ψ(n)]

1−p

(

c

(3.41)

)p p1 kλ (u(x), v(n))f (x) dx < k(λ1 )||f ||p,Φ ,

b

n=n0

(3.42) and

c

[Φ(x)]

1−q

b

q

∞

kλ (u(x), v(n))an

1q dx

< k(λ1 )||a||q,Ψ ;

(3.43)

n=n0

(ii) for p < 0, we have the equivalent reverses of (3.41), (3.42) and (3.43).

(y) (> 0) is decreasing in [n0 , ∞) and there exist constants Moreover, if vv(y) δ < λ1 , L > 0, and δ1 > 0 such that inequality (3.9) is fulﬁlled, and k(λ1 + δ1 ) ∈ R+, then the constant factor k(λ1 ) in the inequalities of (i) and the reverses in (ii) is the best possible.

Proof.

(i) For p > 1, setting x = u(t) on two sides of (3.18), we obtain c ∞ an kλ (u(t), v(n))f (u(t))u (t) dt I= b

n=n0

=

c

∞

f (u(t))u (t)

b

an kλ (u(t), v(n)) dt

n=n0

c

< k(λ1 )

p(1−λ1 )−1 p

[u(t)] b

f (u(t))u (t) dt

p1

||a||q,Ψ .

(3.44)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

84

Half-Discrete Hilbert-Type Inequalities

Replacing t and f (u(t))u (t) by x and f (x) in (3.44), by simpliﬁcation, we obtain (3.41). On the other hand, setting u(x) = x (x ∈ (0, ∞)) in (3.41), we have (3.18). It follows that (3.41) and (3.18) are equivalent and then both of them with the same best constant factor k(λ1 ). Similarly, we can prove that (3.42) and (3.19), (3.43) and (3.20) are equivalent and with the same best constant factor k(λ1 ). Since (3.18), (3.19) and (3.20) are equivalent, then, it follows that (3.41), (3.42) and (3.43) are equivalent. (ii) For p < 0, in the same way, we have the equivalent reverses of (3.41), (3.42) and (3.43) with the same best constant factor k(λ1 ). Note. Some particular cases about the reverse inequalities were published by Yang [160]–[162]. Remark 3.3. Deﬁne the operator T : Lp,Φ (b, c) → lq,Ψ as follows: For f ∈ Lp,Φ (b, c), we deﬁne T f ∈ lp,Ψ1−p , satisfying ∞ kλ (u(x), v(n))f (x)dx, n ≥ n0 , n ∈ N. T f (n) = 0

Also, we deﬁne the operator T : lq,Ψ → Lp,Φ (b, c) as follows: For a ∈ lq,Ψ , we deﬁne Ta ∈ Lp,Φ (b, c), satisfying ∞ kλ (u(x), v(n))an , x ∈ (b, c). Ta(x) = n=n0

Then, by Corollary 3.2, we still have ||T || = ||T|| = k(λ1 ). Example 3.3. Setting u(x) = ln x (x ∈ (1, ∞)), v(n) = ln n, n ≥ n0 = 2, (y) 1 = y ln (> 0) is decreasing for y > 1. then vv(y) y 1 (i) If kλ (ln x, ln n) = (ln x)λ +(ln (λ1 > 0, 0 < λ2 ≤ 1, λ1 + λ2 = λ), n)λ since for x > 1, xλ1 (ln y)λ2 −1 f (x, y) = λ [x + (ln y)λ ]y is decreasing with respect to y ∈ (1, ∞), then Condition (i) is satisﬁed. In view of Example 3.1(i), the constant factor in (3.42) and (3.43) (for u(x) = ln x, v(n) = ln n) is the best possible, we have π ||T || = ||T|| = k(λ1 ) = . 1 λ sin( πλ ) λ (ii) If kλ (ln x, ln n) = for x > 1,

1 (ln xn)λ

f (x, y) =

(λ1 > 0, 0 < λ2 ≤ 1, λ1 + λ2 = λ), since

xλ1 (ln y)λ2 −1 (x + ln y)λ y

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

ws-book9x6

85

is decreasing with respect to y ∈ (1, ∞), then Condition (i) is satisﬁed. In view of Example 3.1(ii), the constant factor in (3.42) and (3.43) (for u(x) = x, v(n) = ln n) is the best possible, we have (see Yang [163]) ||T || = ||T|| = k(λ1 ) = B(λ1 , λ2 ). ln[(ln x)/(ln n)] (λ1 > 0, 0 < λ2 ≤ 1, λ1 +λ2 = λ), (iii) If kλ (ln x, ln n) = (ln x)λ −(ln n)λ then by the same way and Example 3.1(iii), we have 2

π ||T || = ||T|| = k(λ1 ) = . 1 λ sin( πλ ) λ s (iv) If s ∈ N, kλs (ln x, ln n) = k=1 ak (ln x)λ1+(ln n)λ (0 < a1 < · · · < as , λ, λ1 > 0, 0 < λ2 ≤ 1, λ1 + λ2 = λs), then by the same way and Example 3.1(iv), we have s s λ2 2 π 1 λ −1 a . ||T || = ||T|| = k(λ1 ) = k πλ2 a − ak λ sin( λ ) k=1 j j=1(j =k)

(v) If kλ (ln x, ln n) = (ln x)λ +2(ln x ln n)1λ/2 cos β+c(ln n)λ (λ > 0, 0 < β ≤ λ1 > 0, 0 < λ2 ≤ 1, λ1 + λ2 = λ), then by the same way and Example 3.1(v), we have π , 2

2π sin β(1 − 2λλ1 ) ||T || = ||T|| = k(λ1 ) = . 1 λ sin β sin( 2πλ λ ) √

1 (ln x)λ +b(ln x ln n)λ/2 +c(ln n)λ λ2 = λ2 ), then by the same

(vi) If kλ (ln x, ln n) =

2 c, 0 < λ ≤ 2, λ1 = 3.1(vi), we have

(c > 0, 0 ≤ b ≤ way and Example

⎧ π √ , b = 0, ⎪ c ⎨ √ √ 2 4c−b2 4 √ ||T || = ||T || = arctan b , 0 e −Ae 0(A = 1, β > 1), γ > 0, −1 ≤ λ1 = −λ2 = −βλ < 0 < λ), then by the same way and Example 3.1(vii), we have ∞ Ak Γ(β) . ||T || = ||T|| = k(−βλ) = β λγ (2k + 1)β k=0 " # b(ln x)λ +(ln n)λ (viii) If k0 (ln x, ln n) = ln a(ln (0 ≤ a < b, λ > 0, −1 ≤ x)λ +(ln n)λ

λ1 = −λ2 = −βλ, 0 < β < 1), then by the same way and Example 3.1(viii), we have (bβ − aβ )π ||T || = ||T|| = k(−βλ) = . βλ sin(βπ)

November 7, 2013

86

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

ln x λ (ix) If k0 (ln x, ln n) = arctan ln (λ1 = −λ2 = −β, 0 < β < λ, β ≤ n 1), then by the same way and Example 3.1(ix), we have ||T || = ||T|| = k(−β) =

π . 2β cos( πβ ) 2λ

(x) If kλ (ln x, ln n) = (max{ln1x,ln n})λ (λ1 > 0, 0 < λ2 ≤ 1, λ1 +λ2 = λ), then by the same way and Example 3.1(x), we have ||T || = ||T|| = k(λ1 ) =

λ . λ1 λ2

(xi) If k−λ (ln x, ln n) = (min{ln x, ln n})λ (λ1 < 0, −1 ≤ λ2 < 0, λ1 + λ2 = −λ), then by the same way and Example 3.1(xi), we have ||T || = ||T|| = k(λ1 ) =

λ . λ1 λ2

min{ln x,ln n} λ (xii) If k0 (ln x, ln n) = ( max{ln ) (λ2 = −λ1 , |λ1 | < λ ≤ 1 + x,ln n} 1 λ1 , λ1 > − 2 ), then, by the same way and Example 3.1(xii), we have

||T || = ||T|| = k(λ1 ) =

λ2

2λ . − λ21

Example 3.4. For 0 ≤ γ1 , γ2 ≤ 12 , setting u(x) = x − γ1 (x ∈ (γ1 , ∞)), (y) 1 = y−γ (> 0) is decreasing for y > 12 . v(n) = n − γ2 , n ≥ n0 = 1, then vv(y) 2 1 (i) If kλ (x− γ1 , n− γ2 ) = (x−γ1 )λ +(n−γ λ (λ1 > 0, 0 < λ2 ≤ min{1, 2 − 2) λ}, λ1 + λ2 = λ), for x > 0, f (x, y) =

xλ

xλ1 (y − γ2 )λ2 −1 + (y − γ2 )λ

is strictly convex satisfying fy2 (x, y) > 0 with respect to y ∈ 12 , ∞ , then Condition (ii) is satisﬁed. In view of Example 3.1(i), the constant factor in (3.42) and (3.43) (for u(x) = x − γ1 , v(n) = n − γ2 ) is the best possible, we have π . ||T || = ||T|| = 1 λ sin( πλ λ ) 1 (ii) If kλ (x− γ1 , n− γ2 ) = (x−γ1 +n−γ λ (λ1 > 0, 0 < λ2 ≤ 1, λ1 + λ2 = 2) λ), by the same way and Example 3.1(ii), we have (see Huang and Yang [38])

||T || = ||T|| = B(λ1 , λ2 ).

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

ws-book9x6

87

ln[(x−γ )/(n−γ )]

(iii) If kλ (x − γ1 , n − γ2 ) = (x−γ1 )λ1−(n−γ22)λ (λ1 > 0, 0 < λ2 ≤ 1, λ1 + λ2 = λ), then, by the same way and Example 3.1(iii), we have

π ||T || = ||T|| = 1 λ sin( πλ ) λ

2 .

s (iv) If s ∈ N, kλs (x − γ1 , n − γ2 ) = k=1 ak (x−γ1 )λ1+(n−γ2 )λ (0 < a1 < · · · < as , λ1 > 0, 0 < λ2 ≤ min{1, 2 − λ}, λ1 + λ2 = λs), then by the same way and Example 3.1(iv), we have ||T || = ||T|| = k(λ1 ) =

s s λ2 2 π 1 λ −1 a . k πλ2 aj − ak λ sin( λ ) k=1 j=1(j =k)

1 (v) If kλ (x − γ1 , n − γ2 ) = (x−γ1 )λ +2[(x−γ1 )(n−γ λ/2 cos β+(n−γ )λ (λ > 2 )] 2 π 0, 0 < β ≤ 2 , λ1 > 0, 0 < λ2 ≤ min{1, 2 − λ}, λ1 + λ2 = λ), then by the same way and Example 3.1(v), we have

||T || = ||T|| = k(λ1 ) =

2λ1 ) λ . 1 sin( 2πλ ) λ

2π sin β(1 − λ sin β

1 (vi) If kλ (x − γ1 , n − γ2 ) = (x−γ1 )λ +b[(x−γ1 )(n−γ λ/2 +c(n−γ )λ (c > 2 )] 2 √ 4 λ 0, 0 ≤ b ≤ 2 c, 0 < λ ≤ 3 , λ1 = λ2 = 2 ), then by the same way and Example 3.1(vi), we have

⎧ π √ , b = 0, ⎪ c ⎨ √ √ 2 2 4c−b 4 √ ||T || = ||T|| = arctan b , 0 0(A = 1, β > 1), γ > 0, −1 ≤ λ1 = −λ2 = −βλ < 0 < λ ≤ 1), then, for x > 0, x−βλ

f (x, y) = e

γ xλ

(y−γ2 )λ

−γ

λ

− Ae xλ (y−γ2 )

(y − γ2 )βλ−1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

88

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

is strictly convex. In fact, setting g(x, u) = γ

γ

1

−γ

u

e xλ −Ae xλ

u

, we ﬁnd

−γ

u λγ xλ + Ae xλ u ) λ (e < 0, gu (x, u) = − x γ −γ (e xλ u − Ae xλ u )2 γ

gu2 (x, u) =

−γ

2( λγ )2 (e xλ u + Ae xλ u )2 xλ γ

−γ

(e xλ u − Ae xλ u )3 γ

γ

−γ

( λγ )2 (e xλ u − Ae xλ u ) xλ

−

γ

−γ

(e xλ u − Ae xλ u )2

−γ

γ

−γ

=

( λγ )2 [2(e xλ u + Ae xλ u )2 − (e xλ u − Ae xλ u )2 ] xλ

=

)2 [(e xλ u + Ae xλ u )2 + 4A] ( λγ xλ

γ

−γ

(e xλ u − Ae xλ u )3 γ

−γ

γ

−γ

(e xλ u − Ae xλ u )3

(0 ≤ A ≤ 1),

>0

and then, gy2 (x, (y − γ2 )λ ) > 0 (0 < λ ≤ 1). Hence, fy2 (x, y)

4

−βλ

= x

g x, (y − γ2 )

λ

(y −

* γ2 )βλ−1 y2

> 0,

y∈

1 ,∞ . 2

Then Condition (ii) is satisﬁed and in view of Example 3.1(vii), we have ||T || = ||T|| = k(−βλ) =

∞ Γ(β) Ak . β λγ (2k + 1)β k=0

λ

λ

b(x−γ1 ) +(n−γ2 ) (viii) If k0 (x − γ1 , n − γ2 ) = ln[ a(x−γ λ λ ] (0 ≤ a < b, 0 < 1 ) +(n−γ2 ) λ ≤ 1, −1 ≤ λ1 = −λ2 = −βλ, 0 < β < 1), then, by the same way and Example 3.1(viii), we have

(bβ − aβ )π ||T || = ||T|| = . βλ sin(βπ) x−γ1 λ ) (λ1 = −λ2 = −β, 0 < β < (ix) If k0 (x − γ1 , n − γ2 ) = arctan ( n−γ 2 λ, β ≤ min{1, 2 − λ}), then by the same way and Example 3.1(ix), we have π . ||T || = ||T|| = 2β cos( πβ 2λ ) 2 3, v (y) v(y)

Example 3.5. For a ≥

setting u(x) = ln ax (x ∈ ( a1 , ∞)), v(n) =

= y ln1 ay (> 0) is decreasing for y > 32 . ln an, n ≥ n0 = 2, then 1 (λ1 > 0, 0 < λ2 ≤ min{1, 2 − (i) If kλ (ln ax, ln an) = (ln ax)λ +(ln an)λ λ}, λ1 + λ2 = λ), since for x > 0, f (x, y) =

[xλ

xλ1 (ln ay)λ2 −1 + (ln ay)λ ]y

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

ws-book9x6

89

is strictly convex satisfying fy2 (x, y) > 0 with respect to y ∈ 32 , ∞ , then Condition (ii) is satisﬁed. In view of Example 3.1(i), the constant factor in (3.42) and (3.43) (for u(x) = ln ax, v(n) = ln an) is the best possible, we have π . ||T || = ||T|| = 1 λ sin( πλ λ ) (ii) If kλ (ln ax, ln an) = (ln a21xn)λ (λ1 > 0, 0 < λ2 ≤ 1, λ1 + λ2 = λ), by the same way and Example 3.1(ii), we have (see Chen and Yang [7]) ||T || = ||T|| = B(λ1 , λ2 ). ln[(ln ax)/(ln an)] (λ1 > 0, 0 < λ2 ≤ 1, λ1 +λ2 = (iii) If kλ (ln ax, ln an) = (ln ax)λ −(ln an)λ λ), then by the same way and Example 3.1(iii), we have

2 π ||T || = ||T|| = . 1 λ sin( πλ λ ) s (iv) If s ∈ N, kλs (ln ax, ln an) = k=1 ak (ln ax)λ1+(ln an)λ (0 < a1 < · · · < as , λ, λ1 > 0, 0 < λ2 ≤ min{1, 2 − λ}, λ1 + λ2 = λs), then, by the same way and Example 3.1(iv), we have s s λ2 2 π 1 λ −1 a . ||T || = ||T|| = k(λ1 ) = k πλ2 aj − ak λ sin( λ ) j=1(j =k)

k=1

1 (ln ax)λ +2(ln ax ln an)λ/2 cos β+(ln an)λ

(v) If kλ (ln ax, ln an) = (λ > 0, 0 < π β ≤ 2 , λ1 > 0, 0 < λ2 ≤ min{1, 2 − λ}, λ1 + λ2 = λ), then by the same way and Example 3.1(v), we have 2π sin β(1 − 2λλ1 ) . ||T || = ||T|| = k(λ1 ) = 1 λ sin β sin( 2πλ ) λ (vi) If kλ (ln ax, ln an) = (ln ax)λ +b(ln ax ln1 an)λ/2+c(ln an)λ (c > 0, 0 ≤ √ b ≤ 2 c, 0 < λ ≤ 43 , λ1 = λ2 = λ2 ), then by the same way and Example 3.1(vi), we have ⎧ π √ , b = 0, ⎪ c ⎨ √ √ 2 4c−b2 √ 4 arctan , 0 0(A = 1, β > 1), γ > 0, −1 ≤ λ1 = −λ2 = −βλ < 0 < λ ≤ 1), then by the same way and Example 3.1(vii), we have ∞ Ak Γ(β) ||T || = ||T|| = k(−βλ) = . β λγ (2k + 1)β k=0

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

90

ws-book9x6

Half-Discrete Hilbert-Type Inequalities λ

λ

b(ln ax) +(ln an) (viii) If k0 (ln ax, ln an) = ln[ a(ln ] (0 ≤ a < b, 0 < λ ≤ 1, ax)λ +(ln an)λ −1 ≤ λ1 = −λ2 = −βλ, 0 < β < 1), then by the same way and Example 3.1(viii), we have (bβ − aβ )π ||T || = ||T|| = . βλ sin(βπ) ln ax λ (ix) If k0 (ln ax, ln an) = arctan( ln an ) (λ1 = −λ2 = −β, 0 < β < λ, β ≤ min{1, 2 − λ}), then by the same way and Example 3.1(ix), we have π . ||T || = ||T|| = 2β cos( πβ 2λ )

Note. (1) For γ1 = γ2 = 0 in Example 3.4(i)-(ix), we obtain the corresponding results in Example 3.11; for a = 1 in Example 3.5(i)-(ix), we obtain the corresponding results in Example 3.3. Since For u(x) = x − γ1 , v(n) = n − γ2 or u(x) = ln ax, v(n) = ln an in Example 3.1(x)-(xii), the corresponding f (x, y) is not convex, so we cannot use Condition (ii) to derive some similar results. (2) For a ≤ 12 , setting u(x) = ln(x−a) (x ∈ (a+1, ∞)), v(n) = ln(n−a),

(y) 1 = (y−a) ln(y−a) (> 0) is decreasing for y > 32 . In this n ≥ n0 = 2, then vv(y) case, we can still obtain some similar results of Example 3.5(i)-(ix). By the same way, we still have

Corollary 3.3. Let the assumptions of Theorem 3.2 be fulﬁlled and additionally, u(x) is a strictly increasing diﬀerentiable function in (b, c), with u(b+ ) = 0 and u(c− ) = ∞. For 0 0, ||a||q,Ψ > 0, then, we have the following n=n0 ∈ lq,Ψ , ||f ||p,Φ equivalent inequalities: c ∞ an kλ (u(x), v(n))f (x)dx b

n=n0

c

=

f (x) b

∞

∞

an kλ (u(x), v(n)) dx > k(λ1 )||f ||p,Φ ||a||q,Ψ ,

n=n0

[Ψ(n)]1−p

( b

n=n0

c

(3.45)

)p p1 kλ (u(x), v(n))f (x) dx > k (λ1 )||f ||p,Φ , (3.46)

and

c b

[Φ(x)]

1−q

∞

n=n0

q kλ (u(x), v(n))an

q1 dx

> k(λ1 )||a||q,Ψ .

(3.47)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

ws-book9x6

91

v (y) (> v(y)

0) is decreasing in [n0 , ∞) and there exist constants 1 ∈ (λ1 − δ2 , λ1 ), η, δ2 > 0, such that k(λ1 − δ2 ) ∈ R+ and for any λ 1 θλ1 (x) = O xη (x ≥ 1), then the constant factor k(λ1 ) in the above inequalities is the best possible. Moreover, if

In view of Remark 3.1, Corollary 3.1 and Theorem 3.3, we still have Corollary 3.4. Suppose that λ, λ1 , λ2 ∈ R, λ = λ1 + λ2 , kλ (x, y) is a non-negative ﬁnite measurable homogeneous function of degree −λ in R2+ , v(y) is a strictly increasing diﬀerentiable function in [n0 , ∞) (n0 ∈ N) with v(n0 ) > 0 and v(∞) = ∞, u(x) is a strict increasing diﬀerentiable (y) (> 0) is decreasing function in (b, c), with u(b+ ) = 0 and u(c− ) = ∞, vv(y) 1 ∈ in [n0 , ∞), and there exist constants δ0 , η > 0, such that for any λ 2 = λ − λ 1 ), (λ1 − δ0 , λ1 + δ0 )(λ 1 )(1 − θ (x)) < (x) < k(λ 1 ) < ∞ (x ∈ (0, ∞)), 0 < k(λ (3.48) λ1 λ1 1 where, θλ1 (x) > 0 (x ∈ (0, ∞)), and θλ1 (x) = O xη (x ∈ [1, ∞)). (i) For p > 1(p < 0), p1 + 1q = 1, f (x), an ≥ 0, f ∈ Lp,Φ (b, c), a = {an}∞ n=n0 ∈ lq,Ψ , ||f ||p,Φ > 0, ||a||q,Ψ > 0, if p > 1, then we have the equivalent inequalities (3.41), (3.42) and (3.43), with the best constant factor k(λ1 ); if p < 0, then we have the equivalent reverses of (3.41), (3.42) and (3.43), with the same best constant factor k(λ1 ). (ii) For 0 0, ||a|| > 0, then we have the equivalent q,Ψ q,Ψ n=n0 p,Φ inequalities (3.45), (3.46) and (3.47) with the same best constant factor k(λ1 ). In particular, for λ = 0, λ1 = α, λ2 = −α in Corollary 3.1, Theorem 3.3 and Corollary 3.2, setting Φ0 (x) =

[u(x)]p(1−α)−1 (x ∈ (b, c)), [u (x)]p−1

Ψ0 (n) =

[v(n)]q(1+α)−1 (n ≥ n0 , n ∈ N), [v (n)]q−1

0 (x) = (1 − θα (u(x)))Φ0 (x), we have the following corollary: and Φ Corollary 3.5. (see Yang and Krnic [157]) Suppose that k0 (x, y) is a nonnegative ﬁnite measurable homogeneous function of degree 0 in R2+ , ∞ k(α) = k0 (t, 1)tα−1 dt ∈ R+ (α ∈ R), 0

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

92

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

v(y) is a strict increasing diﬀerentiable function in [n0 , ∞)(n0 ∈ N) with v(n0 ) > 0 and v(∞) = ∞, u(x) is a strict increasing diﬀerentiable function in (b, c), with u(b+ ) = 0 and u(c− ) = ∞. If p > 1, or p < 0, p1 + q1 = 1, f (x), an ≥ 0, f ∈ Lp,Φ0 (b, c), a = {an }∞ n=n0 ∈ lq,Ψ0 , ||f ||p,Φ0 > 0, ||a||q,Ψ0 > 0, then (i) for p > 1, we have the following equivalent inequalities: ∞

c

an

k0 (u(x), v(n))f (x) dx b

n=n0

∞

c

f (x)

= b

∞

an k0 (u(x), v(n)) dx < k(α)||f ||p,Φ0 ||a||q,Ψ0 , (3.49)

n=n0

[Ψ0 (n)]

1−p

(

c

)p p1 k0 (u(x), v(n))f (x)dx < k (α)||f ||p,Φ0 ,

b

n=n0

(3.50) and

∞ 0

[Φ0 (x)]

1−q

∞

1q

q k0 (u(x), v(n))an

dx

< k (α)||a||q,Ψ0 ; (3.51)

n=n0

(ii) for p < 0, we have the equivalent reverses of (3.49), (3.50) and (3.51).

(y) Moreover, if vv(y) (> 0) is decreasing in [n0 , ∞) and there exist constants η < α, L > 0, and δ1 > 0, such that

k0 (t, 1) ≤ L

1 tη

t ∈ 0,

1 v(n0 )

) ,

(3.52)

and k(α + δ1 ) ∈ R+ , then the constant factor k(α) in the inequalities of (i) and the reverses in (ii) is the best possible. Corollary 3.6. Let the assumptions of Corollary 3.5 be fulﬁlled. If 0 < ∞ p < 1, p1 + 1q = 1, f (x), an ≥ 0, f ∈ Lp,Φ 0 (b, c), a = {an }n=n0 ∈ lq,Ψ0 , ||f ||p,Φ 0 > 0, ||a||q,Ψ0 > 0, then we have the following equivalent

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

inequalities: ∞ an

93

c

k0 (u(x), v(n))f (x) dx b

n=n0

∞

c

f (x)

= b

∞

an k0 (u(x), v(n))dx > k(α)||f ||p,Φ 0 ||a||q,Ψ0 , (3.53)

n=n0

[Ψ0 (n)]

1−p

(

c b

n=n0

)p p1 k0 (u(x), v(n))f (x)dx > k(α)||f ||p,Φ 0 , (3.54)

and

∞ 0

0 (x)] [Φ

1−q

∞

q k0 (u(x), v(n))an

1q dx

> k(α)||a||q,Ψ0 . (3.55)

n=n0

(y) (> 0) is decreasing in [n0 , ∞) and there exists a conMoreover, if vv(y) stant δ2 > 0, such that k(α − δ2 ) ∈ R+ , then the constant factor k(α) in the above inequalities is the best possible.

Corollary 3.7. Suppose that k0 (x, y) is a non-negative ﬁnite measurable homogeneous function of degree 0 in R2+ , ∞ k0 (t, 1) tα−1 dt (α ∈ R), k(α) = 0

v(y) is a strictly increasing diﬀerentiable function in [n0 , ∞)(n0 ∈ N) with v(n0 ) > 0 and v(∞) = ∞, u(x) is a strictly increasing diﬀerentiable func (y) tion in (b, c), with u(b+ ) = 0 and u(c− ) = ∞, vv(y) is decreasing in [n0 , ∞), ∈ (α−δ0 , and there exist constants δ < α and δ0 , η > 0, such that for any α α + δ0 ), 0 < k( α)(1 − θα (x)) < α (x) < k( α) < ∞, (x ∈ (0, ∞)), 1 where, θα (x) > 0 (x ∈ (0, ∞)), θα (x) = O xη (x ∈ [1, ∞)).

(3.56)

(i) If p > 1, 1p + 1q = 1, f (x), an ≥ 0, f ∈ Lp,Φ0 (b, c), a = {an }∞ n=n0 ∈ lq,Ψ0 , ||f ||p,Φ0 > 0, ||a||q,Ψ0 > 0, then we have the equivalent inequalities (3.49), (3.50) and (3.51), with the best constant factor k(α). (ii) If p < 0, then we have the equivalent reverses of (3.49), (3.50) and (3.51) with the same best constant factor k(α). (iii) If 0 0, ||a|| > 0, then, we have the equivalent q,Ψ0 q,Ψ0 0 n=n0 p,Φ

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

94

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

inequalities (3.53), (3.54) and (3.55) with the same best constant factor k(α). For v(x) = u(x) = x (x ∈ (0, ∞), n0 = 1, in Corollary 3.15, setting ϕ(x) = xp(1−λ1 )−1 (x ∈ (0, ∞)), ψ(n) = nq(1−λ2 )−1 (n ∈ N), we have ∞ ω(λ2 , n) = nλ2 kλ (x, n)xλ1 −1 dx = k(λ1 ), n ∈ N, (3.57) 0

∞

(λ1 , x) = xλ1

kλ (x, n)nλ2 −1 ,

x ∈ (0, ∞),

(3.58)

n=1

and the following corollary: Corollary 3.8. Suppose that λ, λ1 , λ2 ∈ R, λ = λ1 + λ2 , kλ (x, y) is a non-negative ﬁnite measurable homogeneous function of degree −λ in R2+ , 1 ∈ (λ1 −δ0 , λ1 +δ0 )(λ 2 = there exist constants δ0 , η > 0, such that for any λ 1 ), λ−λ 1 , x) < k(λ 1 ) < ∞ (x ∈ (0, ∞)), 1 )(1 − θ (x)) < (λ 0 < k(λ λ1 where, θλ1 (x) > 0 (x ∈ (0, ∞)), and θλ1 (x) = O x1η (x ∈ [1, ∞)).

(3.59)

(i) For p > 1 or p < 0, 1p + 1q = 1, f (x), an ≥ 0, f ∈ Lp,ϕ (R+ ), a = {an }∞ n=1 ∈ lq,ψ , ||f ||p,ϕ > 0, ||a||q,ψ > 0, if p > 1, then we have the following equivalent inequalities: ∞ ∞ an kλ (x, n)f (x)dx 0

n=1

∞

=

f (x) 0

∞

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

n=1

∞

1−p

(

[ψ(n)]

0

n=1

∞

(3.60)

)p p1 kλ (x, n)f (x) dx < k (λ1 )||f ||p,ϕ , (3.61)

and

∞

[ϕ(x)] 0

1−q

∞

q kλ (x, n)an

1q dx

< k(λ1 )||a||q,ψ ,

(3.62)

n=1

with the best constant factor k(λ1 ); if p < 0, then we have the equivalent reverses of (3.60), (3.61) and (3.62), with the same best constant factor k(λ1 ).

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

95

(ii) For 0 0, ||a||q,ψ > 0, then, n=1 ∈ lq,ψ , ||f ||p,ϕ we have the following equivalent inequalities: ∞ ∞ an kλ (x, n)f (x)dx 0

n=1

∞

=

f (x) 0

∞

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

n=1

∞

1−p

(

∞

[ψ(n)]

0

n=1

(3.63)

)p p1 kλ (x, n)f (x) dx > k(λ1 )||f ||p,ϕ, (3.64)

and

0

∞

[ϕ(x)]

1−q

∞

q kλ (x, n)an

1q dx

> k (λ1 )||a||q,ψ ,

(3.65)

n=1

where the constant factor k(λ1 ) is the best possible. Note. As a particular case of (3.60), we may refer to a half-discrete Hilbert-type inequality with the homogeneous kernel of degree −4μ as 1 k4μ (x, n) = μ μ 2 (ax + bn ) (cxμ + dnμ )2 where 1 a, b, c, d > 0, bc = ae, and 0 < μ < 2 and a best constant factor as ( ) bc + ae ln(ae) − ln(bc) 2 , − μ(bc − ae)2 ae − bc ae + bc which was published by Xie [83]. If we replace the lower limit 0 of integral to c > 0, then, the results of Corollary 3.8 is still value for p > 1 and 0 < p < 1. For p < 0, we may refer to Corollary 3.13 for v(n) = n, n0 = 1. In particular, for λ = 0, λ1 = α, λ2 = −α, in Corollary 3.21, setting ϕ0 (x) = xp(1−α)−1 (x ∈ (0, ∞)), ψ0 (n) = nq(1+α)−1 (n ∈ N), we have ∞ ω(−α, n) = n−α k0 (x, n)xα−1 dx = k(α), n ∈ N, (3.66) 0

(α, x) = xα

∞

n=1

k0 (x, n)n−α−1 ,

x ∈ (0, ∞),

(3.67)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

96

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and the following corollary: Corollary 3.9. Suppose that k0 (x, y) is a non-negative ﬁnite measurable homogeneous function of degree 0 in R2+ , ∞ k0 (t, 1)tα−1 dt (α ∈ R), k(α) = 0

there exist constants δ0 , η > 0 such that for any α ∈ (α − δ0 , α + δ0 ), α, x) < k( α) < ∞(x ∈ (0, ∞)), (3.68) 0 < k( α)(1 − θα (x)) < ( where, θα (x) > 0 (x ∈ (0, ∞)), θα (x) = O x1η (x ∈ [1, ∞)). (i) For p > 1 or p < 0, p1 + 1q = 1, f (x), an ≥ 0, f ∈ Lp,ϕ0 (R+ ), a = {an }∞ n=1 ∈ lq,ψ0 , ||f ||p,ϕ0 > 0, ||a||q,ψ0 > 0, if p > 1, then we have the following equivalent inequalities: ∞ ∞ an k0 (x, n)f (x) dx 0

n=1

∞

∞

f (x)

=

0

an k0 (x, n) dx < k(α)||f ||p,ϕ0 ||a||q,ψ0 ,

n=1

∞

[ψ0 (n)]

1−p

(

∞ 0

n=1

(3.69)

)p p1 k0 (x, n)f (x)dx < k(α)||f ||p,ϕ0 , (3.70)

and

∞ 0

1−q

[ϕ0 (x)]

∞

q k0 (x, n)an

1q dx

< k(α)||a||q,ψ0 ,

(3.71)

n=1

with the best constant factor k(α); if p < 0, then we have the equivalent reverses of (3.69), (3.70) and (3.71), with the same best constant factor k(α). 0 (x) = (1 − θα (x))ϕ0 (x), if f (x), (ii) For 0 0, ||a||q,ψ0 > 0, then, we have the following equivalent inequalities: ∞ ∞ an k0 (x, n)f (x) dx n=1

0

∞

f (x)

=

0 ∞

∞

an k0 (x, n) dx > k(α)||f ||p,ϕ0 ||a||q,ψ0 ,

n=1 1−p

[ψ0 (n)]

n=1

( 0

∞

(3.72)

)p p1 k0 (x, n)f (x) dx > k (α)||f ||p,ϕ0 , (3.73)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

97

and

∞

0

[ϕ 0 (x)]

1−q

∞

q k0 (x, n)an

q1 dx

> k(α)||a||q,ψ0 ,

(3.74)

n=1

where the constant factor k(α) is the best possible.

3.3.4

Some Particular Examples

Applying Condition (i) and Corollaries 3.8-3.9, we have 1 Example 3.6. We set kλ (x, y) = A(max{x,y})λ +B(min{x,y}) ((x, y) ∈ λ R2+ ; 0 0, 0 < λ2 < 1, λ1 +λ2 = λ). There exists a constant 1 ∈ (λ1 − δ0 , λ1 + δ0 ) ⊂ 0 < δ0 < min{λ1 , λ2 , 1 − λ2 } such that for any λ 2 = λ − λ 1 ∈ (λ2 − δ0 , λ2 + δ0 ) ⊂ (0, 1). For x > 0, (0, λ), λ

xλ1 y λ2 −1 A(max{x, y})λ + B(min{x, y})λ ⎧ ⎨ xλ1 yλ2 −1 , 0 < y ≤ x, λ +By λ = Ax ⎩ xλ1λyλ2 −1λ , y > x

f (x, y) =

Ay +Bx

is strictly decreasing with respect to y > 0. We ﬁnd

∞

1 tλ1 −1 dt + B(min{t, 1})λ 0 ∞ 1 1 1 1 −1 λ t dt + = tλ1 −1 dt λ λ+B A + Bt At 0 1 1 1 1 1 −1 2 −1 λ λ t dt = + t B A 0 1 + A tλ k ∞ 1 1 B = tλk tλ1 −1 + tλ2 −1 dt − A 0 A k=0 ∞ k 1 B 1 − tλk+λ1 −1 + tλk+λ2 −1 dt = A A 0 k=0 ∞ k B 1 1 1 − ∈ R+ , = + 1 2 A A λk + λ λk + λ

1 ) = k(λ

A(max{t, 1})λ

k=0

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

98

Half-Discrete Hilbert-Type Inequalities

and obtain

∞

nλ2 −1 A(max{x, n})λ + B(min{x, n})λ n=1 ∞ y λ2 −1 > xλ1 dy A(max{x, y})λ + B(min{x, y})λ 1 1 )(1 − θ (x)) > 0, = k(λ λ1 1 1 λ y λ2 −1 dy x θλ1 (x) = > 0. 1 ) 0 A(max{x, y})λ + B(min{x, y})λ k(λ For x ≥ 1, it follows that 1 xλ1 y λ2 −1 dy 0 < θλ1 (x) = 1 ) 0 Axλ + By λ k(λ 1 λ2 −1 xλ1 y dy ≤ k(λ1 ) 0 Axλ 1 1 ≤ , = 2 λ 2 xη Ak( λ ) λ x Ak( λ ) λ 1 2 1 that is θλ1 (x) = O x1η (η = λ2 − δ0 > 0). By Condition (i), it follows that ∞ xλ1 nλ2 −1 1 ). 1 , x) = < k(λ (λ λ + B(min{x, n})λ A(max{x, n}) n=1 Hence, we obtain (3.59). 1 By Corollary 3.8, for kλ (x, n) = A(max{x,n})λ +B(min{x,n}) λ (0 0, 0 < λ2 < 1, λ1 + λ2 = λ), if p > 1, we have equivalent inequalities (3.60), (3.61) and (3.62); if p < 0, we have equivalent reverses of (3.60), (3.61) and (3.62); if 0 < p < 1, we have equivalent inequalities (3.63), (3.64) and (3.65). All the inequalities are with the same best constant factor k ∞ λ 2k + 1 B . (3.75) k(λ1 ) = − A A (λk + λ1 )(λk + λ2 )

1 , x) = xλ1 (λ

k=0

λ

(min{x,y}) Example 3.7. We set k0 (x, y) = A(max{x,y}) ((x, y) ∈ λ +B(min{x,y})λ 2 R+ ; 0 0, max{−λ, λ − 1} < α < λ). There exists a constant 0 < δ0 < min{λ − α, λ + α, α − λ + 1}, such that for any < λ. For x > 0, α ∈ (α − δ0 , α + δ0 ), max{−λ, λ − 1} < α α−1 (min{x, y})λ xα y − f (x, y) = λ A(max{x, y}) + B(min{x, y})λ α λ−α−1 x y , 0 < y ≤ x, Axλ +By λ = xλ+α y −α−1 y>x Ay λ +Bxλ

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

ws-book9x6

99

is strictly decreasing with respect to y > 0. We ﬁnd ∞ (min{t, 1})λ 1 ) = k(λ tα−1 dt A(max{t, 1})λ + B(min{t, 1})λ 0 ∞ 1 1 1 λ+ α−1 t dt + = tα−1 dt λ λ+B A + Bt At 0 1 1 1 1 λ+ α−1 λ− = (t + t α−1 )dt λ A 0 1+ B t A k 1 ∞ 1 B α−1 α−1 tλk (tλ+ + tλ− )dt = − A 0 A k=0 k 1 ∞ B 1 α −1 α−1 − = (tλk+λ+ + tλk+λ− )dt A A 0 k=0 k−1 ∞ B 1 1 1 − ∈ R+ , + = A A λk + α λk − α k=1

and then, we obtain ∞

α−1 (min{x, n})λ n− A(max{x, n})λ + B(min{x, n})λ n=1 ∞ α−1 (min{x, y})λ y − > xα dy A(max{x, y})λ + B(min{x, y})λ 1 = k( α)(1 − θα (x)) > 0, 1 α−1 xα (min{x, y})λ y − dy θα (x) = > 0. k( α) 0 A(max{x, y})λ + B(min{x, y})λ

( α, x) = xα

For x ≥ 1, it follows that 0 < θα (x) = ≤

α

x k( α)

xα k( α)

1

0

y

1

0 λ− α−1

Axλ

α−1 y λ− dy λ Ax + By λ

dy

1 1 ≤ , λ− α Ak( α)(λ − α )x Ak( α)(λ − α )xη that is θα (x) = O x1η (η = λ − α − δ0 > 0). By Condition (i), it follows that ∞ α−1 (min{x, n})λ n− ( α, x) = xα < k( α). A(max{x, n})λ + B(min{x, n})λ n=1 =

Hence, we obtain (3.68).

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

100

Half-Discrete Hilbert-Type Inequalities λ

(min{x,n}) By Corollary 3.9, for k0 (x, n) = A(max{x,n}) λ +B(min{x,n})λ (0 0, max{−λ, λ − 1} < α < λ), if p > 1, we have equivalent inequalities (3.69), (3.70) and (3.71); if p < 0, we have equivalent reverses of (3.69), (3.70) and (3.71); if 0 < p < 1, we have equivalent inequalities (3.72), (3.73) and (3.74). All the inequalities are with the same best constant factor k−1 ∞ 2λ k B − . (3.76) k(λ1 ) = A A (λ2 k 2 − α2 ) k=1

Example 3.8. (See He and Li [22]). (min{x,y})β 2 We set kλ (x, y) = (max{x,y}) λ+β ((x, y) ∈ R+ ; λ + 2β > 0, λ1 > −β, −β < λ2 < 1 − β, λ1 + λ2 = λ). There exists a constant 0 < δ0 < 1 ∈ (λ1 − δ0 , λ1 + δ0 ) ⊂ min{λ1 + β, λ2 + β, 1 − β − λ2 }, such that for any λ (−β, λ + β), λ2 = λ − λ1 ∈ (λ2 − δ0 , λ2 + δ0 ) ⊂ (−β, 1 − β). For x > 0,

(min{x, y})β xλ1 y λ2 −1 (max{x, y})λ+β ⎧ λ +β−1 ⎨ y 2 , 0 < y ≤ x, 1 xλ+β−λ = ⎩ xλ1 +β , y>x

f (x, y) =

y λ+β−λ2 +1

is strictly decreasing with respect to y > 0. We ﬁnd ∞ (min{t, 1})β λ1 −1 1 ) = k(λ t dt (max{t, 1})λ+β 0 ∞ 1 1 tλ1 +β−1 dt + tλ1 −1 dt = λ+β t 0 1 1 1 + ∈ R+ , = 1 + β 2 + β λ λ and then, we have ∞ (min{x, n})β nλ2 −1 (max{x, n})λ+β n=1 ∞ (min{x, y})β y λ2 −1 1 λ >x dy (max{x, y})λ+β 1 1 )(1 − θ (x)) > 0, = k(λ λ1 1 1 λ (min{x, y})β y λ2 −1 x θλ1 (x) = dy > 0. 1 ) 0 (max{x, y})λ+β k(λ

1 , x) = xλ1 (λ

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

ws-book9x6

101

For x ≥ 1, it follows that 1 λ2 +β−1 xλ1 y 0 < θλ1 (x) = dy k(λ1 ) 0 xλ+β 1 1 ≤ , = 2 +β λ k(λ )(λ + β)x k(λ1 )(λ2 + β)xη 1 1 2 that is, θλ1 (x) = O xη (η = λ2 + β − δ0 > 0). By Condition (i), it follows that ∞ (min{x, n})β nλ2 −1 1 , x) = xλ1 1 ). (λ < k(λ (max{x, n})λ+β n=1 Hence, we obtain (3.60). (min{x,n})β By Corollary 3.8, for kλ (x, n) = (max{x,n}) λ+β (λ + 2β > 0, λ1 > −β, −β < λ2 < 1−β, λ1 +λ2 = λ), if p > 1, we have equivalent inequalities (3.60), (3.61) and (3.62); if p < 0, we have equivalent reverses of (3.60), (3.61) and (3.62); if 0 < p < 1, we have equivalent inequalities (3.63), (3.64) and (3.65). All the inequalities are with the same best constant factor 2β + λ k(λ1 ) = . (3.77) (β + λ1 )(β + λ2 ) In particular of this example, for (i) β = 0, kλ (x, n) = > 0, λ1 > 0, 0 < λ2 < 1, λ1 + λ2 = λ); (ii) λ = −β, k−β (x, n) = (min{x, n})β (β > 0, λ1 > −β, −β < λ2 < 1−β, λ1 +λ2 = −β); min{x,n} β ) (β > 0, λ1 > −β, −β < λ2 < (iii) λ = 0, k0 (x, n) = ( max{x,n} 1 − β, λ1 + λ2 = 0), we obtain respectively some results of Example 3.2 (viii)-(x). Note.

1 (λ (max{x,n})λ

3.3.5

Applying Condition (iii) and Corollary 3.8

1 2 Theorem 3.4. We set kλ (x, y) = (x+y) λ ((x, y) ∈ R+ ; 0 < λ < 5, 0 < λ1 < 3, 0 < λ2 < 2, λ1 + λ2 = λ). There exists a constant 0 < δ0 < 1 ∈ (λ1 − δ0 , λ1 + δ0 ) ⊂ min{λ1 , λ2 , 3 − λ1 , 2 − λ2 }, such that for any λ (0, 3), λ2 = λ − λ1 ∈ (λ2 − δ0 , λ2 + δ0 ) ⊂ (0, 2). We ﬁnd ∞ 1 1 ) = 1 , λ 2 ) ∈ R+ . k(λ tλ1 −1 dt = B(λ λ (t + 1) 0 In the following, we show that ∞ xλ1 nλ2 −1 1 , x) = 1 ), (3.78) 1 )(1 − θ (x)) < (λ < k(λ 0 < k(λ λ1 (x + n)λ n=1 1 (x ≥ 1; η = λ2 − δ0 > 0) . (3.79) θλ1 (x) > 0(x > 0), θλ1 (x) = O xη

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

102

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

1 = λ , λ 2 = λ R > 1, 1 + 1 = 1 , it follows that λ 2 = Setting λ R S R S and 0 < λ < min{2S, 5}. λ 1 S −1 , we show that (1) For g(x, y) = (x+y) λ y r(x) =

1 0

1 g(x, y)dy − g(x, 1) − 2

∞ 1

P1 (y)gy (x, y) dy > 0.

In fact, since 0 < λ < 2S, it follows that

1

g(x, y)dy = 0

= = =

=

>

>

1 (S+λ)(2S+λ)

≥

1 12S 2

λ S

(3.80) and

S 1 1 λ 1 λ S −1 dy = y dy S λ λ λ 0 (x + y) 0 (x + y) 1 1 λ S +S y S dy λ λ+1 λ(x + 1) 0 (x + y) 1 S2 1 λ S + dy S +1 λ(x + 1)λ S + λ 0 (x + y)λ+1 S S2 + λ λ(x + 1) (S + λ)(x + 1)λ+1 λ S 2 (λ + 1) 1 y S +1 dy + λ+2 S+λ 0 (x + y) 2 S S + λ λ(x + 1) (S + λ)(x + 1)λ+1 1 λ S 3 (λ + 1) dy S +2 + (S + λ)(2S + λ) 0 (x + y)λ+2 S S2 + λ λ(x + 1) (S + λ)(x + 1)λ+1 S 3 (λ + 1) + (S + λ)(2S + λ)(x + 1)λ+2 S S2 S(λ + 1) + + . λ λ(x + 1) (S + λ)(x + 1)λ+1 12(x + 1)λ+2 1

For x > 0, we ﬁnd gy (x, y) = =

−λ y)λ+1

λ y 1− S

(x + −[1 + (λ/R)]

λ

(x + y)λ y 2− S

+

−

1 − (λ/S) λ

(x + y)λ y 2− S λx

λ

(x + y)λ+1 y 2− S

.

<2

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

103

Since 0 < λ < 5, we ﬁnd λx < 5(x + 1) (x > 0) and using equations (2.74) and (2.75) in Chapter 2 (for m = q = 1), it follows that ∞ P1 (y)gy (x, y) dy − 1 ∞ ∞ [1 + (λ/R)]P1 (y) λxP1 (y) dy = dy − (3.81) λ λ 2− S λ (x + y) y (x + y)λ+1 y 2− S 1 1 ∞ [1 + (λ/R)]P1 (y) dy = λ (x + y)λ y 2− S 1 ⎧

⎫ ∞ ⎨ ⎬ λx λx 1 + − P (y) dy 3 λ ⎩ 12(x + 1)λ+1 ⎭ 6 1 (x + y)λ+1 y 2− S y

λ(x + 1) − λ 5(x + 1) 1 + (λ/R) + − >− 12(x + 1)λ 12(x + 1)λ+1 720 ) ( (λ + 1)(λ + 2) 2(λ + 1)(2S − λ) (2S − λ)(3S − λ) × + + (x + 1)λ+3 S(x + 1)λ+2 S 2 (x + 1)λ+1 λ −1 1 λ − ≥ + λ 12 12S (x + 1) 12(x + 1)λ+1 ) ( 7 1 24 6 . (λ + 1) − + + 144 (x + 1)λ+2 (x + 1)λ+1 (x + 1)λ In view of − 12 g(x, 1) =

−1 , 2(x+1)λ

by (3.80) and the above results, we obtain

S2 S(λ + 1) S + + λ λ+1 λ(x + 1) (S + λ)(x + 1) 12(x + 1)λ+2 1 1 λ λ 1 + − − − λ 2 12 12S (x + 1) 12(x + 1)λ+1 ) ( 1 24 6 7(λ + 1) − + + 144 (x + 1)λ+2 (x + 1)λ+1 (x + 1)λ 2 S h(λ) 1 λ 1 + = − − λ 24λS(x + 1) S + λ 12 6 (x + 1)λ+1 λ+1 7 S − + , 12 144 (x + 1)λ+2

r(x) >

where, h(λ) = 2λ2 − 15Sλ + 24S 2 (0 < λ < 2S). We ﬁnd h (λ) = 4λ − 15S < 8S − 15S < 0, h(λ) > h(2S) = 2(2S)2 − 15S(2S) + 24S 2 = 2S 2 > 0, S2 λ 1 S S 1 S−1 − − > − − = > 0, S + λ 12 6 3 6 6 6

(3.82)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

104

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

S 7 1 7 and 12 − 144 > 12 − 144 > 0. Then, by inequality (3.82), it follows r(x) > 0. (2) In view of

f (x, y) =

xλ1 y λ2 −1 = xλ1 g(x, y), (x + y)λ

by (3.4) and (3.80), we have R(x) = xλ1 r(x) > 0. By Condition (iii), it follows that

1 , x) = xλ1 (λ We ﬁnd

0

1

∞ nλ2 −1 1 ) − R(x) 1 ). = k(λ < k(λ λ (x + n) n=1

1 λ y S −1 dy < (x + y)λ

0

1

S 1 λ −1 y S dy = , xλ λxλ

and by (3.81) and (2.22) in Chapter 2, it follows that ∞ − P1 (y)gy (x, y)dy 1 ∞ ∞ [1 + (λ/R)]P1 (y) λxP1 (y) dy − = λ λ dy (x + y)λ y 2− S (x + y)λ+1 y 2− S 1 1 λx <0+ . 12(x + 1)λ+1 Then, by (3.80), we have r(x) <

S 1 λx − + . λxλ 2(x + 1)λ 12(x + 1)λ+1

Setting θλ1 (x) =

( ) λ 1 1 R(x) + xR > 0, 1 ) (x + 1)λ k(λ

we ﬁnd 1 , x) > (λ 1 , x) − (λ By (3.83), for x ≥ 1, we obtain

λ

xR 1 )(1 − θ (x)) > 0. = k(λ λ1 (x + 1)λ

( ) 1 xλ λ θλ1 (x) ≤ x θλ1 (x) = 0 < x x r(x) + 1 ) (x + 1)λ k(λ ) ( λxλ+1 1 xλ S + < + λ 1 ) λ 2(x + 1) 12(x + 1)λ+1 k(λ ) ( 1 λ S 1 , as x → ∞, + + → λ 2 12 k(λ1 ) λ2 −δ0

λ S

(3.83)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

ws-book9x6

105

that is, θλ1 (x) = O x1η (x ≥ 1; η = λ2 − δ0 > 0). Hence we show that (3.78) and (3.79) are valid. 1 By Corollary 3.8, for kλ (x, n) = (x+n) λ (0 < λ < 5, 0 < λ1 < 3, 0 < λ2 < 2, λ1 + λ2 = λ), if p > 1, we have equivalent inequalities (3.60), (3.61) and (3.62); if p < 0, we have equivalent reverses of (3.60), (3.61) and (3.62); if 0 < p < 1, we have equivalent inequalities (3.63), (3.64) and (3.65). All the inequalities are with the same best constant factor (3.84) k(λ1 ) = B(λ1 , λ2 ). ((x, y) ∈ R2+ ; 0 < λ1 < 1, λ2 = Example 3.9. We set k1 (x, y) = | ln(x/y)| x+y 1−λ1 ). There exists a constant 0 < δ0 < min{λ1 , λ2 , 1−λ1 }, such that for 1 ∈ (λ1 −δ0 , λ1 +δ0 ) ⊂ (0, 1), λ 2 = λ− λ 1 ∈ (λ2 −δ0 , λ2 +δ0 ) ⊂ (0, 1). any λ There exists a constant δ1 > 0, such that η = λ2 − δ0 − δ1 > 0. By the Lebesgue term by term integration theorem, it follows that ∞ | ln t| tλ1 −1 k(λ1 ) = dt t+1 0 ∞ 1 (− ln t) tλ1 −1 (ln t) tλ1 −1 dt + dt = t+1 t+1 0 1 1 ∞ = (− ln t) (−1)k tk+λ1 −1 + tk+λ2 −1 dt 0

= =

=

∞ k=0 ∞ k=0 ∞

k=0 1

(−1)k

(− ln t) tk+λ1 −1 + tk+λ2 −1 dt

0 1

(−1)k

(− ln t) tk+λ1 −1 + tk+λ2 −1 dt

0

(−1)

k=0

k

1 1 + 2 1 ) 2 )2 (k + λ (k + λ

∈ R+ .

δ0

| ln t| → 0 (t → 0+ ), we still ﬁnd Since tδ0 k1 (t, 1) = t t+1 − ln t 1 k1 (t, 1) = (t ∈ (0, 1]; L > 0). ≤ L δ0 t+1 t In the following, we show that ∞ | ln nx |nλ1 −1 λ1 −1 1 ) (n ∈ N), (3.85) 2 , n) = nλ2 dx < k(λ ω(λ x x+n 1 1 )(1 − θ (x)) 0 < k(λ λ1

1 , x) = xλ1 < (λ

∞ | ln nx |nλ2 −1 1 ) (x ≥ 1), (3.86) < k(λ x+n n=1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

106

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and

0 < θλ1 (x) = O

1 xη

(x ≥ 1; η = λ2 − δ0 − δ1 > 0).

(3.87)

(1) For n ∈ N, putting t = nx in the integral of (3.85), we ﬁnd ∞ ∞ | ln t| λ1 −1 | ln t| λ1 −1 1 ), dt < dt = k(λ t t ω(λ2 , n) = 1 t + 1 t + 1 0 n and then (3.85) follows. (2) For g(x, y) = | ln(x/y)| y λ2 −1 (x, y ≥ 1), in the following, we prove x+y that ∞ 1 1 g(x, y)dy − g(x, 1) − P1 (y)gy (x, y)dy > 0. (3.88) r(x) = 2 0 1 In fact, putting t = xy , we obtain 1 1 | ln(x/y)| λ2 −1 y g(x, y) dy = dy x+y 0 0 x1 1 − ln t λ2 −1 t dt = xλ1 0 t + 1 x1 1 − ln t λ2 −1 t > dt 1 λ 1 +1 x 0 x 1 ln x 1 + . = 2 2 1+x λ λ 2

(3.89)

is decreasing with respect to y, we set two For x, y ≥ 1, since ln(y/x) y−x decreasing functions for y ∈ [1, ∞) as follows: h1 (x, y) =

1 x)y λ1 +1

+

2x ln(y/x) x)2 y λ1

(y − x)(y + 1 ln(y/x) ln(y/x) λ h2 (x, y) = + . (y − x)(y + x)y λ1 (y − x)y λ1 +1 (y +

+

1 x ln(y/x) 2λ (y − x)(y + x)y λ1 +1

,

λ2 −1 For 1 ≤ y < x, g(x, y) = − ln(y/x) , it follows that y+x y

gy (x, y) =

−1

+

ln(y/x)

(y + x)y λ1 +1 (y + x)2 y λ1 = h2 (x, y) − h1 (x, y);

for y ≥ x, g(x, y) =

ln(y/x) y+x

+

1 ln(y/x) λ (y + x)y λ1 +1

y λ2 −1 , we ﬁnd gy (x, y) = h1 (x, y) − h2 (x, y).

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

ws-book9x6

107

We deﬁne two functions h(x, y) and h(x, y) as follows: h2 (x, y), 1 ≤ y < x, h(x, y) = h1 (x, y), y ≥ x, h1 (x, y), 1 ≤ y < x, h(x, y) = h2 (x, y), y ≥ x. Since −gy (x, y) = h(x, y) − h(x, y), (2.39) in Chapter 2, it follows that ∞ ε0 P1 (y) h(x, y) dy = (−h1 (x, 1) + h1 (x, x) − h2 (x, x)) 8 1 x P1 (t)dt + (h1 (x, x) − h2 (x, x)) 1x ε0 1 1 = P1 (t)dt (ε0 ∈ (0, 1)), + −h1 (x, 1) + 1 +1 1 +1 λ λ 8 2x 2x 1 ∞ ε − P1 (y)h(x, y)dy = − 0 (−h2 (x, 1) + h2 (x, x) − h1 (x, x)) 8 1 x − (h2 (x, x) − h1 (x, x)) P1 (t)dt 1 x 1 ε 1 + = 0 h2 (x, 1) + P1 (t)dt (ε0 ∈ (0, 1)). 8 2xλ1 +1 2xλ1 +1 1 Then, by equation (2.48) in Chapter 2 (for q = 0), we have x ε1 P1 (t)dt = − (ε1 ∈ [0, 1]), 8 1 and

−

∞ 1

P1 (y)gy (x, y)dy ∞ ∞ P1 (y) h(x, y)dy − P1 (y)h(x, y)dy = 1 1 ε0 ε0 1 1 + −h1 (x, 1) + h2 (x, 1) + = 8 8 2xλ1 +1 2xλ1 +1 ε1 =− (ε0 ∈ (0, 1), ε0 ∈ (0, 1), ε1 ∈ [0, 1]). (3.90) 8xλ1 +1

Since, we have −h1 (x, 1) +

1 2xλ1 +1

≤ −h1 (x, x) +

1 2xλ1 +1

< 0,

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

108

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

then, by (3.90), we obtain ∞ 1 1 1 − −h1 (x, 1) + − P1 (y)gy (x, y)dy > 8 2xλ1 +1 8xλ1 +1 1 x ln x −1 + = 8(1 + x) 4(1 − x)(1 + x)2 1 x ln x 1 λ − + . (3.91) 4(1 − x)(1 + x) 16xλ1 +1 ln x , in view of (3.88), (3.89) and (3.91), (i) for Since − 12 g(x, 1) = − 2(1+x) ln x 1 ≤ x < 2, x−1 ≤ 1, we ﬁnd

r(x) >

ln x 2 (1 + x) λ

+

1 2 (1 λ 2

+ x)

−

ln x 1 − 2(1 + x) 8(1 + x)

1 x ln x λ 1 x ln x − − − 2 4(x − 1)(1 + x) 4(x − 1)(1 + x) 16xλ1 +1 ( ) ln x 1 1 1 1 ≥ + − − 2 2 (1 + x) 8(1 + x) 1+x λ 2(1 + x) λ 2 −

1 x 1 x λ − − 2 4(1 + x) 4(1 + x) 16xλ1 +1

1 x 1 x 1 − − − − 1 + x 8(1 + x) 4(1 + x)2 4(1 + x) 16 5 7 1 − = + 2 8(1 + x) 4(1 + x) 16 1 1 7 5 + = > 0; > − 8(1 + 2) 4(1 + 2)2 16 114 ≥

x ≥ −2, by (3.92), we have (ii) for x ≥ 2, − x−1

ln x 1 ln x 1 − − + 1 + x 2(1 + x) 8(1 + x) λ2 (1 + x) 1 ln x ln x 1 λ − − − 2 2(1 + x) 2(1 + x) 32

2 ln x 7 7 1 1 λ ≥ + − − + 2 x+1 λ 2 6 8(1 + x) 32 ( ) 7 1 7 ln x √ + − > 0. 2− ≥ x+1 6 8(1 + 2) 32

r(x) >

Hence, (3.88) follows.

(3.92)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

109

(3) In view of

xλ1 | ln(x/y)| λ2 −1 y = xλ1 g(x, y), x+y by (3.4) and (3.88), we have R(x) = xλ1 r(x) > 0. By Condition (iii), it follows that ∞ | ln(x/n)|nλ2 −1 1 , x) = xλ1 x+n (λ f (x, y) =

n=1

1 ) 1 ) − R(x) < k(λ = k(λ We ﬁnd

1

g(x, y)dy = 0

1 xλ1 x1

<

1 x

0

− ln t λ2 −1 t dt t+1

(− ln t) t

0

2 −1 λ

dt =

(x ≥ 1).

1 ln x + λ2 λ22

1 , x

and by (3.90), we obtain ∞ 1 1 h2 (x, 1) + P1 (y)gy (x, y) dy < − 8 2xλ1 +1 1 1 ln x 1 λ ln x + + = . 8(x − 1)(1 + x) 8(x − 1) 16xλ1 +1 Hence by (3.88), it follows that

1 1 ln x ln x 1 1 λ λ + R(x) = x r(x) < x − 2 2 x 2(1 + x) λ λ 2

1 ln x λ 1 ln x . (3.93) + + + 8(x − 1)(1 + x) 8(x − 1) 16xλ1 +1

Setting θλ1 (x) =

1 1 ) [R(x) k(λ

ln x + xλ1 x+1 ] > 0, we obtain

λ1 1 , x) > (λ 1 )(1 − θ (x)) > 0. 1 , x) − x ln x = k(λ (λ λ1 x+1 For η = λ2 − δ0 − δ1 > 0, by (3.93), we ﬁnd ( ) 1 x1−δ1 ln x 2 −δ1 η 1−δ1 λ x θλ1 (x) = r(x) + 0 < x θλ1 (x) ≤ x 1 ) x+1 k(λ

x1−δ1 ln x 1 x1−δ1 ln x 1 ln x 1 + + − < 2 2 xδ1 1 ) 2(1 + x) 8(x − 1)(1 + x) λ λ k(λ 2 1 x1−δ1 ln x λ x1−δ1 ln x x1−δ1 + + → 0 (x → ∞), + 8(x − 1) x+1 16xλ1 +1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

110

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

that is, θλ1 (x) = O x1η (x ≥ 1; η = λ2 − δ0 − δ1 > 0). Hence, we show that (3.86) and (3.87) are valid. (x ≥ 1, n ∈ By the Note of Corollary 3.8 (c = 1), for k1 (x, n) = | ln(x/n)| x+n N;0 < λ1 < 1, λ2 = 1 − λ1 ), if p > 1, we have equivalent inequalities (3.60), (3.61) and (3.62); if p < 0, we have equivalent inequalities (3.33), (3.34) and (3.35)(for v(n) = n, n0 = 1); if 0 < p < 1, we have equivalent inequalities (3.63), (3.64) and (3.65). All the inequalities are with the same best constant factor

k(λ1 ) =

∞

( k

(−1)

k=0

) 1 1 + . (k + λ1 )2 (k + λ2 )2

(3.94)

Example 3.10. (See Yang and Chen [126]). | ln(x/y)| 2 We set kλ (x, y) = (max{x,y}) λ ((x, y) ∈ R+ ; λ1 > 0, 0 < λ2 < 1, λ2 + λ1 = λ ≤ 4). There exists a constant 0 < δ0 < min{λ1 , λ2 , 1 − λ2 }, 1 ∈ (λ1 − δ0 , λ1 + δ0 ) ⊂ (0, λ), λ 2 = λ − λ 1 ∈ (λ2 − such that for any λ δ0 , λ2 + δ0 ) ⊂ (0, 1). Since λ2 − δ0 > 0, there exists δ1 > 0, such that η = λ2 − δ0 − δ1 > 0. It follows that 1 ) = k(λ

∞ 0

1

| ln t|tλ1 −1 dt (max{t, 1})λ

∞

(ln t)tλ1 −1 dt tλ 0 1 1 1 1 λ2 1 λ1 1 −1 2 −1 λ λ = (− ln t)(t +t )dt = (− ln t)d t + t 1 2 λ λ 0 0 1 1 1 λ2 1 λ2 −1 1 λ1 −1 1 λ1 t + t t + t dt = (− ln t) + 1 2 1 2 λ λ λ λ 0 0 =

=

(− ln t)tλ1 −1 dt +

1 1 + ∈ R+ . 2 λ1 λ22

Since tδ0 k1 (t, 1) =

tδ0 | ln t| (max{t,1})λ

→ 0 (t → 0+ ), we still ﬁnd that

k1 (t, 1) = − ln t ≤ L

1 tδ0

(t ∈ (0, 1]; L > 0).

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

In the following, we show that ∞ | ln nx |nλ1 −1 λ1 −1 2 λ 1 ) (n ∈ N), ω(λ2 , n) = n x dx < k(λ (max{x, n})λ 1 1 )(1 − θ (x)) 0 < k(λ

111

(3.95)

λ1

∞ | ln nx |nλ2 −1 1 ) (x ≥ 1), (3.96) < k(λ λ (max{x, n}) n=1 1 0 < θλ1 (x) = O (3.97) (x ≥ 1; η = λ2 − δ0 − δ1 > 0). xη

1 , x) = xλ1 < (λ

(1) For n ∈ N, putting t = nx in the integral of (3.95), we ﬁnd ∞ | ln t| tλ1 −1 ω(λ2 , n) = dt 1 (max{t, 1})λ n ∞ | ln t| tλ1 −1 1 ), dt = k(λ < (max{t, 1})λ 0 and then, inequality (3.95) follows. (2) Setting g(x, y) =

| ln(x/y)| y λ2 −1 λ (max{x, y})

(x ≥ 1, y > 0),

in the following, we prove that 1 ∞ 1 g(x, y)dy − g(x, 1) − P1 (y)gy (x, y)dy > 0. r(x) = 2 0 1

(3.98)

In fact, putting t = xy , we obtain 1 1 x1 1 ln(x/y) y λ2 −1 g(x, y)dy = dy = (− ln t) tλ2 −1 dt λ λ 1 x x 0 0 0 1 1 = ln x + . (3.99) 2 xλ 2 λ λ For x, y ≥ 1, we set four positive decreasing functions with respect to y ∈ [1, ∞) as follows: 1 ) x ln(y/x) , + (1 + λ (y − x)y 2+λ1 ln(y/x) 2 ) , h2 (x, y) = (1 − λ λ x (y − x)y 1−λ2 1 ) ln(y/x) , h1 (x, y) = (1 + λ (y − x)y 1+λ1 h1 (x, y) =

1

y 2+λ1

November 7, 2013

12:24

112

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and h2 (x, y) =

1 xλ y 2−λ2

2 ) + (1 − λ

ln(y/x) xλ−1 (y

− x)y 2−λ2

.

Then, for 1 ≤ y < x, g(x, y) = − ln(y/x) y λ2 −1 , it follows that xλ gy (x, y) =

−1 xλ y 2−λ2

2 ) + (1 − λ

(y − x) ln(y/x) xλ (y − x)y 2−λ2

= h2 (x, y) − h2 (x, y);

for y ≥ x, g(x, y) = ln(y/x)y −λ1 −1 , we ﬁnd gy (x, y) =

1 y 2+λ1

1 ) − (1 + λ

(y − x) ln(y/x) (y − x)y 2+λ1

= h1 (x, y) − h1 (x, y). We deﬁne two functions h(x, y) and h(x, y) as follows: h2 (x, y), 1 ≤ y < x, h(x, y) = h1 (x, y), y ≥ x, h (x, y), 1 ≤ y < x, h(x, y) = 2 h1 (x, y), y ≥ x. By equation (2.39) in Chapter 2, it follows that ∞ ε0 h2 (x, 1) + P1 (y) h(x, y)dy = (− h2 (x, x) − h1 (x, x)) 8 1 x h1 (x, x)) P1 (t)dt +( h2 (x, x) − 1 ε0 1−λ 1−λ x − h2 (x, 1) + = P1 (t)dt (ε0 ∈ (0, 1)), + 8 xλ1 +2 xλ1 +2 1 ∞ ε P1 (y)h(x, y)dy = − 0 (−h2 (x, 1) + h2 (x, x) − h1 (x, x)) − 8 1 x P1 (t)dt −(h2 (x, x) − h1 (x, x)) 1 ε 1+λ 1+λ x = 0 h2 (x, 1) + P1 (t)dt (ε0 ∈ (0, 1)). + 8 xλ1 +2 xλ1 +2 1 h(x, y) − h(x, y), and by equation (2.48) in Chapter 2 Since −gy (x, y) = (for q = 0), x ε1 (ε1 ∈ [0, 1]), P1 (t)dt = − 8 1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

we ﬁnd −

1

∞

P1 (y)gy (x, y)

ws-book9x6

113

∞ dy = P1 (y)h(x, y)dy − P1 (y)h(x, y)dy 1 1 ε0 1−λ − h2 (x, 1) + = 8 xλ1 +2 1+λ ε1 ε0 . − h2 (x, 1) + + λ +2 1 8 x 4xλ1 +2 (3.100) ∞

Since we have 1−λ 1−λ − h2 (x, 1) + ≤ − h2 (x, x) + < 0, xλ1 +2 xλ1 +2 then, by equality (3.100), we ﬁnd ∞ 1 1−λ 1 − h2 (x, 1) + − P1 (y)gy (x, y)dy > − λ λ +2 1 8 x 4x 1 +2 1 2 ) ln x 1−λ 1 (1 − λ 1 + = − λ − λ−1 − . +2 λ λ 8x 8x (x − 1) 8x 1 4x 1 +2 (3.101) ln x For − 12 g(x, 1) = − 2x λ , 0 < λ ≤ 4 and 0 < λ2 < 1, in view of (3.98), (3.99) and (3.101), we obtain

2 ) ln x 1 ln x 1 (1 − λ 1+λ ln x + − λ − λ − λ−1 − 2 2 xλ xλ 2x 8x 8x (x − 1) 8xλ1 +2 λ λ 2

2 )x ln x 5 1 1 7 (1 − λ ≥ λ (1 − ) ln x + − − x 2 8 8(x − 1) 8 ( ) x ln x 1 ln x 1 . (3.102) + − ≥ λ x 2 4 8(x − 1)

r(x) >

(i) If 1 ≤ x < 52 , then,

x ln x x−1

≤ 2, by (3.102), we have 1 ln x 1 1 r(x) > λ > 0; + − x 2 4 4

(ii) if x ≥ 52 , then,

≤ 53 , by (3.102), we have 1 ln x 1 5 ln x r(x) > λ > 0. + − x 2 4 24 x x−1

Hence, (3.97) follows for x ≥ 1.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

114

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

(3) In view of

f (x, y) =

xλ1 | ln(x/y)| λ2 −1 y = xλ1 g(x, y), (max{x, y})λ

by (3.4) and (3.98), we have R(x) = xλ1 r(x) > 0. By Condition (iii), it follows that

1 , x) = (λ

∞ xλ1 | ln(x/n)|nλ2 −1 (max{x, n})λ n=1

1 ) − R(x) 1 ) (x ≥ 1). = k(λ < k(λ By equality (3.100), we obtain ∞ 1 1+λ h2 (x, 1) + P1 (y)gy (x, y)dy < − 8 xλ1 +2 1 2 ) ln x (1 − λ 1+λ = + . λ 8x (x − 1) 8xλ1 +2 Hence, inequality by (3.98), it follows that ( 1 ln x 1 1 1 λ λ ln x + − λ R(x) = x r(x) < x λ 2x λ2 x λ2 ) (1 − λ2 ) ln x 1+λ + . + 8xλ (x − 1) 8xλ1 +2 + lnx ) > 0, we obtain Setting θ (x) = 1 (R(x) λ1

1 ) k(λ

(3.103)

x λ2

1 , x) > (λ 1 , x) − (λ

ln x

1 )(1 − θ (x)) > 0. = k(λ λ1 xλ2 For η = λ2 − δ0 − δ1 > 0 (δ1 > 0), by (3.103), since x ≥ 1, we obtain 1 ln x 0 < xη θλ1 (x) ≤ xλ2 −δ1 θλ1 (x) = xλ−δ1 r(x) + δ1 1 ) x k(λ

1 ln x 1 1 ln x < + + δ1 δ1 2 x 2x k(λ1 ) λ2 λ2 2 ) ln x 1+λ (1 − λ + + δ1 → 0 (x → ∞), 8x (x − 1) 8x2−λ2 +δ1 that is, θλ1 (x) = O x1η (x ≥ 1; η = λ2 − δ0 − δ1 > 0). Hence, we show that (3.96) and (3.97) are valid. | ln(x/n)| By the Note of Corollary 3.8 (c = 1), for kλ (x, n) = (max{x,n}) λ (x ≥ 1, n ∈ N; λ1 > 0, 0 < λ2 < 1, λ2 + λ1 = λ ≤ 4), if p > 1, we have

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

115

equivalent inequalities (3.60), (3.61) and (3.62); if p < 0, we have equivalent inequalities (3.33), (3.34) and (3.35)(for v(n) = n, n0 = 1); if 0 < p < 1, we have equivalent inequalities (3.63), (3.64) and (3.65). All the inequalities are with the same best constant factor 1 1 (3.104) k(λ1 ) = 2 + 2 . λ1 λ2 3.3.6

Applying Condition (iii) and Corollary 3.4

Example 3.11. (See Yang [158]) 1 2 We set kλ (x, y) = xλ +y λ ((x, y) ∈ R+ ; λ1 > 0, 0 < λ2 < 2, λ1 + λ2 = λ). There exists a constant 0 < δ0 < min{λ1 , λ2 , 2 − λ2 }, such that for any 1 ∈ (λ1 − δ0 , λ1 + δ0 ) ⊂ (0, λ), λ 2 = λ − λ 1 ∈ (λ2 − δ0 , λ2 + δ0 ) ⊂ (0, 2). It λ follows that

∞ (λ1 /λ)−1 tλ1 −1 u dt = du λ t +1 λ(u + 1) 0 0 π ∈ R+ . = 1 /λ) λ sin(π λ

1 ) = k(λ

∞

For γ < 1, it is obvious that ∞ 2 , n) = (n − γ)λ2 ω(λ

xλ1 −1 dx 1 ) (n ∈ N). = k(λ λ x + (n − γ)λ 0 In the following, we show that, for + 1 + 3λ 2 + 2λ 1 < 1, 1 ) = 1 − 1 λ γ ≤ γ(λ 1 4 it follows that 1 )(1 − θ (x)) 0 < k(λ λ1 ∞

xλ1 (n − γ)λ2 −1 1 )(x > 0), < k(λ λ + (n − γ)λ x n=1 1 (x ≥ 1; η = λ2 − δ0 > 0). 0 < θλ1 (x) = O xη (1) Setting 1 , x) = < (λ

(3.105) (3.106)

(y − γ)λ2 −1 (x > 0, y > γ), xλ + (y − γ)λ in the following, we prove that ) ( 1 ∞ 1 1 λ R(x) = x g(x, y)dy − g(x, 1) − P1 (y)gy (x, y)dy > 0. 2 γ 1 (3.107) g(x, y) =

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

116

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

In fact, we obtain 1 g(x, y)dy = γ

1

γ

1 (y − γ)λ2 −1 dy = xλ + (y − γ)λ xλ1

0

1−γ x

tλ2 −1 dt , 1 + tλ

(3.108)

−1 λ 2

− 12 g(x, 1) = − 12 x(1−γ) λ +(1−γ)λ , and 2 − 1)(y − γ)λ2 −2 λ(y − γ)λ2 +λ−2 (λ + [xλ + (y − γ)λ ]2 xλ + (y − γ)λ 2 − 1)(y − γ)λ2 −2 λ[(y − γ)λ + xλ − xλ ](y − γ)λ2 −2 (λ =− + [xλ + (y − γ)λ ]2 xλ + (y − γ)λ 2 + 1)(y − γ)λ2 −2 (λ − λ λxλ (y − γ)λ2 −2 =− + . xλ + (y − γ)λ [xλ + (y − γ)λ ]2 2 < 2, we have Then by equation (2.39) in Chapter 2, for λ > 0, λ ∞ ∞ 2 + 1)(y − γ)λ2 −2 (λ − λ − P1 (y)gy (x, y)dy = P1 (y) dy xλ + (y − γ)λ 1 1 ∞ λxλ (y − γ)λ2 −2 − P1 (y) λ dy [x + (y − γ)λ ]2 1 2 + 1)(1 − γ)λ2 −2 1 (λ − λ >− + 0. (3.109) 8 xλ + (1 − γ)λ Then, by (3.107), (3.108) and (3.109), setting ) ( 1 1−γ + (λ + 1) (1 − γ)λ2 −2 , A= 1 2 8 we have 1−γ x tλ2 −1 dt Axλ1 R(x) > h(x) = − . (3.110) 1 + tλ xλ + (1 − γ)λ 0 1 ), Since, for γ ≤ γ(λ + 1 2 + 2λ 1 , λ1 + 3λ 1−γ ≥ 1 4 1 1 2 λ B(γ) = (1 − γ)2 − (1 − γ) − (λ + λ1 ) ≥ 0, 2 8 1 then, we ﬁnd 1 xλ1 −1 )λ2 −1 (1 − γ)( 1−γ λAxλ1 +λ−1 Aλ x h (x) = − + − λ [xλ + (1 − γ)λ ]2 xλ + (1 − γ)λ x2 [1 + ( 1−γ x ) ]

gy (x, y) = −

2 λ

= −[(1 − γ)

1 ] − Aλ

< −B(γ)

xλ1 −1 λA(1 − γ)λ xλ1 −1 − xλ + (1 − γ)λ [xλ + (1 − γ)λ ]2

(1 − γ)λ2 −2 xλ1 −1 <0 xλ + (1 − γ)λ

(A, λ > 0).

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

ws-book9x6

117

By (3.110), it follows that R(x) > h(∞) = 0(x > 0). Hence, (3.107) holds. (2) In view of f (x, y) = xλ1 g(x, y), by (3.4) and Condition (iii), it follows that ∞ xλ1 (n − γ)λ2 −1 1 , x) = 1 ) − R(x) 1 ) (x > 0). (λ = k(λ < k(λ xλ + (n − γ)λ n=1 By (3.109), we still obtain ∞ 1 λxλ (1 − γ)λ2 −2 − P1 (y)gy (x, y)dy < , 8 [xλ + (1 − γ)λ ]2 1 and by (3.107), it follows that λ 1 1 − γ 2 (1 − γ)λ2 −1 xλ1 1 λxλ+λ1 (1 − γ)λ2 −2 − . (3.111) R(x) < + 2 x 2[xλ + (1 − γ)λ ] 8 [xλ + (1 − γ)λ ]2 λ Setting

1 2 −1 λ λ 1 (1 − γ) x θλ1 (x) = R(x) + λ > 0, 1 ) x + (1 − γ)λ k(λ we ﬁnd 1 2 −1 λ λ 1 , x) > (λ 1 , x) − x (1 − γ) 1 )(1 − θ (x)) > 0. (λ = k(λ λ1 xλ + (1 − γ)λ For η = λ2 − δ0 > 0, x ≥ 1, by (3.111), we obtain

1 xλ (1 − γ)λ2 −1 2 2 λ λ η x R(x) + λ 0 < x θλ1 (x) ≤ x θλ1 (x) = 1 ) x + (1 − γ)λ k(λ λ 1 xλ2 1 − γ 2 (1 − γ)λ2 −1 xλ < − 1 ) λ 2 x 2[xλ + (1 − γ)λ ] k(λ 1 λx2λ (1 − γ)λ2 −2 xλ (1 − γ)λ2 −1 + + λ → constant (x → ∞), 8 [xλ + (1 − γ)λ ]2 x + (1 − γ)λ that is, θλ1 (x) = O x1η (x ≥ 1; η = λ2 − δ0 > 0). Hence, we show that (3.105) and (3.106) are valid. Setting u(x) = x − γ (x > γ), by Corollary 3.4, for + 1 λ1 + 3λ21 + 2λ1 , γ ≤1− 4 1 kλ (u(x), v(n)) = (x−γ)λ +(n−γ) λ (x > γ, n ∈ N;λ1 > 0, 0 < λ2 < 2, λ2 + λ1 = λ), if p > 1, we have equivalent inequalities (3.41), (3.42) and (3.43); if p < 0, we have equivalent reverses of (3.41), (3.42) and (3.43); if 0 < p < 1, we have equivalent inequalities (3.45), (3.46) and (3.47). All the inequalities are with the same best constant factor π . (3.112) k(λ1 ) = λ sin(πλ1 /λ)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

118

Half-Discrete Hilbert-Type Inequalities

3 Note. If√0 = γ ≤ γ(λ1 ) = 1 − 14 (λ1 + 3λ21 + 2λ1 ), then we ﬁnd that 0 < λ1 ≤ 12 ( 57 − 5) = 1.27+ and 0 < λ = λ1 + λ2 < 1.27+ + 2 = 3.27+ , and we get some results similar to Example 3.2(i). But the conditions are diﬀerent. β

(min{x,y}) 2 Example 3.12. We set kλ (x, y) = (max{x,y}) λ+β ((x, y) ∈ R+ ; 0 < 2β + λ ≤ 4, λ1 > −β, −β < λ2 < 1 − β, λ1 + λ2 = λ). There exists a constant 1 ∈ (λ1 − 0 < δ0 < min{β + λ1 , β + λ2 , 1 − β + λ2 }, such that for any λ 2 = λ − λ 1 ∈ (λ2 − δ0 , λ2 + δ0 ) ⊂ (−β, 1 − β). δ0 , λ1 + δ0 ) ⊂ (−β, λ + β), λ It follows that ∞ 2β + λ (min{t, 1})β λ1 −1 1 ) = ∈ R+ . t dt = k(λ λ+β 1 )(β + λ 2 ) (max{t, 1}) (β + λ 0

For 0 ≤ γ < 1, it is obvious that ∞ (min{x, n − γ})β xλ1 −1 2 λ 1 ) (n ∈ N). ω(λ2 , n) = (n − γ) dx = k(λ (max{x, n − γ})λ+β 0 In the following, we show that 8

2 + β λ 2(λ + 2β + 1) 2 ) = 1 − − 1 ∈ (0, 1), 1+ γ(λ 2 + β 4 λ 2 ), and for 0 ≤ γ ≤ γ (λ 1 , x) 2 )(1 − θ (x)) < (λ 0 < k(λ λ1 ∞ xλ1 (min{x, n − γ})β 1 ) (x > 0), (3.113) (n − γ)λ2 −1 < k(λ λ+β (max{x, n − γ}) n=1 1 (x ≥ 1; η = λ2 + β − δ0 > 0). 0 < θλ1 (x) = O (3.114) xη

=

2 ) > 0 is equivalent 2 ) < 1. We ﬁnd that γ(λ (1) It is obvious that γ(λ to 2(λ + 2β + 1) −1< 2 + β λ

4 2 + β λ

2 −1

,

or λ + 2β + 5 8 − + 1 > 0. 2 2 + β (λ2 + β) λ Setting y =

1 2 +β λ

(> 1),

f (y) = 8y 2 − (λ + 2β + 5)y + 1 (y ≥ 1),

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

119

since, f (1) = 4 − (λ + 2β) ≥ 0(λ + 2β ≤ 4), we ﬁnd f (y) = 16y 2 − (λ + 2β + 5) ≥ 16 − (λ + 2β + 5) > 0(y ≥ 1), 2 ) > 0. then, it follows that f (y) > f (1) = 0(y > 1), that is, γ(λ (2) Setting g(x, y) =

(min{x, y − γ})β (y − γ)λ2 −1 λ+β (max{x, y − γ})

(x > 0, y > γ),

2 ), in the following, we prove that, for 0 ≤ γ ≤ γ(λ ) ( 1 ∞ 1 R(x) = xλ1 g(x, y)dy − g(x, 1) − P1 (y)gy (x, y)dy > 0. 2 γ 1 (3.115) In fact, we obtain x−λ−β (y − γ)λ2 +β−1 , γ < y < x + γ, g(x, y) = xβ (y − γ)−λ1 −β−1 , y ≥ x + γ, h1 (x, y), γ < y < x + γ, −gy (x, y) = h2 (x, y), y ≥ x + γ, 2 − β)x−λ−β (y − γ)λ2 +β−2 , h1 (x, y) = (1 − λ 1 + β + 1)xβ (y − γ)−λ1 −β−2 . h2 (x, y) = (λ For 0 < x < 1 − γ, y ≥ 1 ≥ x + γ, by (2.22) in Chapter 2, we have ∞ ∞ P1 (y)gy (x, y)dy = P1 (y)h2 (x, y)dy − 1

>−

1 + β + 1)xβ (λ 12(1 − γ)λ1 +β+2

1

>−

1 + β + 1)xβ (λ 8(1 − γ)λ1 +β+2

;

(3.116)

for x ≥ 1 − γ, by equation (2.38) in Chapter 2 (for ε0 ∈ (0, 1), ε1 ∈ [0, 1]), we ﬁnd ∞ − P1 (y)gy (x, y)dy 1

ε0 = (−h1 (x, 1) + h1 (x, x + γ) − h2 (x, x + γ)) 8 ε1 − (h1 (x, x + γ) − h2 (x, x + γ)) 8 ) ( 2β + λ ε0 2β + λ 2 +β−2 λ −λ−β + (1 − λ2 − β)x =− (1 − γ) + ε1 λ +2 1 8 x 8xλ1 +2

λ2 +β−2 1 2β + λ 2 − β) (1 − γ) > − (1 − λ − . λ +β 2 8 x 8xλ1 +2

(3.117)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

120

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

We still ﬁnd 1 − g(x, 1) = 2

− 21 (1

For 0 < x < 1 − γ, 1 x+γ g(x, y)dy = γ

1

2 +β−1 λ

(y − γ) xλ+β

γ

=

xβ , (1−γ)λ1 +β+1 2 +β−1 1 λ − γ) , xλ+β

− 12

1 2 + (λ

β)xλ1

+

0 < x < 1 − γ,

(3.118)

x ≥ 1 − γ.

1

dy +

xβ (y − γ)−λ1 −β−1 dy

γ+x

1 1 + (λ

β)xλ1

−

xβ 1 + β)(1 − γ)λ1 +β (λ

; (3.119)

for x ≥ 1 − γ, 1 g(x, y)dy = γ

1

1

x

γ

(y − γ)λ2 +β−1 dy = λ+β

(1 − γ)λ2 +β . 2 + β)xλ+β (λ

(3.120)

2 ), (i) For 0 < x < 1 − γ, 0 ≤ γ ≤ γ(λ 8

2 + β λ 2(λ + 2β + 1) 1+ 1−γ ≥ −1 , 2 + β 4 λ it is obvious that A(γ) = (1 − γ)2 −

2 + β 1 λ (1 − γ) − (λ 2 + β)(λ1 + β + 1) ≥ 0, 2 8

and then, by (3.115), (3.116), (3.118) and (3.120), in view of 0 < x < 1 − γ, 2 > 0, it follows that 1 > 0 and β + λ β+λ R(x) >

>

1 2 + β λ

1

+

1

−

xβ+λ1

1 + β)(1 − γ)λ1 +β (λ 1 + β + 1)xβ+λ1 xβ+λ1 (λ − − 2(1 − γ)λ1 +β+1 8(1 − γ)λ1 +β+2

+

1 + β λ

1

−

(1 − γ)β+λ1

1 + β)(1 − γ)λ1 +β (λ 1 + β + 1)(1 − γ)β+λ1 (1 − γ)β+λ1 (λ − − 2(1 − γ)λ1 +β+1 8(1 − γ)λ1 +β+2 A(γ) = ≥ 0; (λ2 + β)(1 − γ)2 2 + β λ

1 + β λ

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

A Half-Discrete Hilbert-Type Inequality with a General Homogeneous Kernel

2 ), in view of (ii) for x ≥ 1 − γ, 0 ≤ γ ≤ γ(λ

1 x2

≤

121

(1−γ)λ2 +β−2 , xλ2 +β

we still ﬁnd

1 (1 − γ)λ2 +β 1 − (1 − γ)λ2 +β−1 R(x) > 2 +β λ λ 2 x 2 +β (λ2 + β)x

λ2 +β−2 1 2β + λ (1 − γ)λ2 +β−2 2 − β) (1 − γ) − (1 − λ − 8 8 xλ2 +β xλ2 +β

= A(γ)

(1 − γ)λ2 +β−2 ≥ 0. 2 + β)xλ2 +β (λ

Hence, (3.115) follows for x > 0. (3) By (3.4) and Condition (iii), it follows that 1 , x) = k(λ 1 ) − R(x) 1 ) (x > 0). (λ < k(λ Setting

1 λ β x 1 (min{x, 1 − γ}) λ −1 > 0, R(x) + (1 − γ) 2 θλ1 (x) = 1 ) (max{x, 1 − γ})λ+β k(λ

we obtain

λ1 β 1 , x) > (λ 1 , x) − x (min{x, 1 − γ}) (1 − γ)λ2 −1 (λ (max{x, 1 − γ})λ+β 1 )(1 − θ (x)) > 0. = k(λ λ1

For x ≥ 1 ≥ 1 − γ (γ ≥ 0), by (3.117), we still obtain ∞ 2β + λ − P1 (y)gy (x, y)dy < , 8xλ1 +2 1 and by (3.115), (3.118) and (3.120), it follows that

1 2β + λ 1 (1 − γ)λ2 +β − (1 − γ)λ2 +β−1 + . R(x) < λ +β 2 + β)xλ2 +β 2 8x2 x 2 (λ

(3.121)

For η = λ2 + β − δ0 > 0, x ≥ 1 ≥ 1 − γ, it follows from (3.121) that

0 < xη θλ1 (x) ≤ xλ2 +β θλ1 (x) # 1 " λ2 +β = R(x) + (1 − γ)λ2 +β−1 x 1 ) k(λ ( 1 1 (1 − γ)λ2 +β < − (1 − γ)λ2 +β−1 2 λ2 + β k(λ1 ) ) 2β + λ 2 +β−1 λ + + (1 − γ) → constant (x → ∞), 8x2−λ2 −β

November 7, 2013

122

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

namely, θλ1 (x) = O x1η (x ≥ 1; η = λ2 + β − δ0 > 0). Hence we show that (3.113) and (3.114) are valid. Setting u(x) = x − μ(x > μ), by Corollary 3.4, for 8

λ2 + β 2(λ + 2β + 1) 1+ −1 , 0 ≤ γ ≤ γ(λ2 ) = 1 − 4 λ2 + β and kλ (u(x), v(n)) =

(min{x − μ, n − γ})β , (max{x − μ, n − γ})λ+β

(x > μ, n ∈ N; 0 < 2β + λ ≤ 4, λ1 > −β, −β < λ2 < 1 − β, λ1 + λ2 = λ), if p > 1, we have equivalent inequalities (3.41), (3.42) and (3.43); if p < 0, we have equivalent reverses of (3.41), (3.42) and (3.43); if 0 < p < 1, we have equivalent inequalities (3.45), (3.46) and (3.47). All the inequalities are with the same best constant factor 2β + λ . (3.122) k(λ1 ) = (β + λ1 )(β + λ2 )

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Chapter 4

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel “... we have always found, even with the most famous inequalities, that we have a little new to add.” G. H. Hardy

“There is no branch of mathematics, however abstract, which may not some day be applied to phenomena of the real world.” Nikolai Lobatchevsky

4.1

Introduction

The main objective of this chapter is to derive a half-discrete Hilbert-type inequality with a general non-homogeneous kernel and its extensions using the way of weight functions and methods of real analysis. The best possible factors involved in inequalities are also determined. Included are equivalent inequalities and their operator expressions, two classes of reverse inequality, many extensions and particular examples.

4.2 4.2.1

Some Preliminary Lemmas Deﬁnition of Weight Functions and Some Related Lemmas

Lemma 4.1. Suppose that α ∈ R, h(t) is a non-negative ﬁnite measurable function in R+ , v(y) is a strictly increasing diﬀerentiable function 123

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

124

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

in [n0 , ∞)(n0 ∈ N) with v(n0 ) > 0 and v(∞) = ∞. Deﬁne two weight functions ωα (n) and α (x) as follows: ∞ h(xv(n))xα−1 dx, n ≥ n0 (n ∈ N), (4.1) ωα (n) = [v(n)]α α (x) = xα

∞

0

h(xv(n))[v(n)]α−1 v (n),

x ∈ (0, ∞).

(4.2)

n=n0

Then, we have

ωα (n) = k(α) =

∞

h(t) tα−1 dt.

(4.3)

0

Moreover, we set f (x, y) = xα h(xv(y))[v(y)]α−1 v (y) and use the following conditions: Condition (i) v(y) is strictly increasing in [n0 − 1, ∞) with v(n0 − 1) ≥ 0, and for any ﬁxed x > 0, f (x, y) is decreasing with respect to y ∈ [n0 − 1, ∞) and strictly decreasing in an interval I ⊂ (n0 − 1, ∞). Condition (ii) v(y) is strictly increasing in [n0 − 21 , ∞) with v(n0 − 21 ) ≥ 0, and for any ﬁxed x > 0, f (x, y) is convex with respect to y ∈ [n0 − 12 , ∞) and strictly convex in an interval I ⊂ (n0 − 12 , ∞). Condition (iii) There exists a constant β > 0, such that v(y) is strictly increasing in [n0 −β, ∞) with v(n0 −β) ≥ 0, and for any ﬁxed x > 0, f (x, y) is a piecewise smooth continuous function with respect to y ∈ [n0 − β, ∞), satisfying n0 ∞ 1 R(x) = f (x, y)dy − f (x, n0 ) − P1 (y)fy (x, y)dy > 0, (4.4) 2 n0 n0 −β where, P1 (y)(= y − [y] − 12 ) is the Bernoulli function of 1-order equation (2.8) in Chapter 2. If k(α) ∈ R+ and one of the above three conditions is satisﬁed, then, we have α (x) < k(α) (x ∈ (0, ∞)). Proof.

Setting t = xv(n) in (4.1), we ﬁnd α−1 ∞ t 1 h(t) ωα (n) = [v(n)]α dt v(n) v(n) 0 ∞ h(t) tα−1 dt = k(α), = 0

(4.5)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

125

then, (4.3) follows. (i) If Condition (i) is satisﬁed then for any n ≥ n0 (n ∈ N), it follows that n f (x, y)dy, f (x, n) ≤ n−1

and there exists an integer n1 ≥ n0 , such that n1 f (x, y)dy. f (x, n1 ) < n1 −1

Since v(n0 − 1) ≥ 0 and k(α) ∈ R+ , we ﬁnd ∞ ∞ n λ1 (x) = f (x, n) < f (x, y)dy n=n0 ∞

n=n0

=

n−1

∞

f (x, y)dy = xα

n0 −1 ∞

h(xv(y))[v(y)]α−1 v (y)dy,

n0 −1 ∞

= xv(n0 −1)

h(t)tα−1 dt ≤

(t = xv(y))

h(t)tα−1 dt = k(α),

0

then, (4.5) follows. (ii) If Condition (ii) is satisﬁed, then by the Hermite-Hadamard’s inequality (see Kuang [47]), it follows that n+ 12 f (x, y) dy, n ≥ n0 , f (x, n) ≤ n− 12

and there exists a positive integer n2 ≥ n0 , such that n2 + 12 f (x, n2 ) < f (x, y)dy. n2 − 12

Since v(n0 −

1 2)

≥ 0 and k(α) ∈ R+, we ﬁnd ∞ ∞ n+ 1 2 f (x, n) < f (x, y)dy λ1 (x) = n=n0 ∞

n=n0

=

n0 − 1 ∞2

= ≤

f (x, y)dy = xα

xv(n0 − 12 ) ∞

0

then, (4.5) follows.

n− 12

∞ n0 − 12

h(t)tα−1 dt,

h(t)tα−1 dt = k(α),

h(xv(y))[v(y)]α−1 v (y)dy

(t = xv(y))

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

126

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

(iii) If Condition (iii) is satisﬁed and k(α) ∈ R+ , then by the EulerMaclaurin summation formula (see equation (2.16) in Chapter 2) ( for q = 0, m = n0 , n → ∞), we have λ1 (x) =

∞

f (x, n)

n=n0 ∞

∞ 1 f (x, y)dy + f (x, n0 )+ P1 (y)fy (x, y)dy 2 n0 n0 ∞ ∞ f (x, y)dy −R(x) = h(t)tα−1 dt−R(x), (t = xv(y)) =

=

n0 −β

xv(n0 −β)

≤ k(α)−R(x) < k(α),

then, (4.5) follows.

Lemma 4.2. Let the assumptions of Lemma 4.1 be fulﬁlled and additionally, p ∈ R\{0, 1}, 1p + 1q = 1, an ≥ 0, n ≥ n0 (n ∈ N), f (x) is a nonnegative measurable function in R+ . Then (i) for p > 1, we have the following inequalities: ∞ ) p p1 ( ∞ v (n) J= h(xv(n))f (x)dx [v(n)]1−pα 0 n=n0 ∞ p1 1 p(1−α)−1 p q ≤ [k(α)] α (x)x f (x)dx , (4.6) 0

and = L

0

≤

∞

[α (x)]1−q x1−qα

∞

q h(xv(n))an

n=n0

∞ [v(n)]q(1−α)−1 q k(α) an [v (n)]q−1 n=n

q1 dx

1q ;

(4.7)

0

(ii) for p < 0 or 0 1, by H¨older’s inequality with weight (see Kuang [47]) and (4.3), it follows that ) p ∞ ( (1−α)/q ) ( ∞ x [v (n)]1/p h(xv(n))f (x)dx = h(xv(n)) f (x) [v(n)](1−α)/p 0 0 ) p ( [v(n)](1−α)/p dx × (1−α)/q x [v (n)]1/p

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

≤

∞

h(xv(n)) 0

=

ws-book9x6

127

x(1−α)(p−1) v (n) p f (x)dx [v(n)]1−α p−1 ∞ [v(n)](1−α)(q−1) h(xv(n)) 1−α dx × x [v (n)]q−1 0

p−1 ωα (n)[v(n)]q(1−α)−1 [v (n)]q−1 ∞ x(1−α)(p−1) v (n) p × h(xv(n)) f (x)dx [v(n)]1−α 0

[k(α)]p−1 = [v(n)]pα−1 v (n)

∞

h(xv(n)) 0

x(1−α)(p−1) v (n) p f (x)dx. [v(n)]1−α

Then by Lebesgue term by term integration theorem (see Kuang [47]), we have p1 ∞ (1−α)(p−1) ∞ 1 x v (n) J ≤ [k(α)] q h(xv(n)) f p (x)dx 1−α [v(n)] 0 n=n0 p1 ∞ ∞ (1−α)(p−1) 1 x v (n) p = [k(α)] q h(xv(n)) f (x)dx [v(n)]1−α 0 n=n0 ∞ p1 1 = [k(α)] q α (x)xp(1−α)−1 f p (x)dx , 0

then (4.6) follows. Still by H¨older’s inequality with weight, it follows that

∞ q ∞ ( (1−α)/q ) x [v (n)]1/p h(xv(n))an = h(xv(n)) [v(n)](1−α)/p n=n0 n=n0 q [v(n)](1−α)/p an × (1−α)/q x [v (n)]1/p q−1 ∞ x(1−α)(p−1) v (n) ≤ h(xv(n)) [v(n)]1−α n=n 0

∞

×

n=n0

h(xv(n))

[v(n)](1−α)(q−1) q a x1−α [v (n)]q−1 n

∞ x1−qα [v(n)](1−α)(q−1) q = h(xv(n)) 1−α a . 1−q [α (x)] x [v (n)]q−1 n n=n 0

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

128

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Then by the Lebesgue term by term integration theorem, we have ≤ L

0

∞

∞

n=n0

q1 [v(n)](1−α)(q−1) q h(xv(n)) 1−α a dx x [v (n)]q−1 n

1q [v(n)](1−α)(q−1) q = h(xv(n)) 1−α a dx x [v (n)]q−1 n n=n0 0 q1 ∞ [v(n)]q(1−α)−1 q = ωα (n) an , [v (n)]q−1 n=n ∞

∞

0

and then, in view of (4.3), we have (4.7). (ii) For p < 0 or 0 < p < 1, by the reverse H¨ older’s inequality with weight (see Kuang [47]) and in the same way, we obtain the reverses of (4.6) and (4.7). 4.2.2

Estimations of Two Series and Examples

Lemma 4.3. Assuming that v(y) is a strictly increasing diﬀerentiable func (y) (> 0) is tion in [n0 , ∞)(n0 ∈ N) with v(n0 ) > 0 and v(∞) = ∞, if vv(y) decreasing in [n0 , ∞), then for ε > 0, we have A(ε) =

∞ n=n0

1 v (n) = (1 + o(1)) 1+ε [v(n)] ε

(ε → 0+ ).

(4.8)

Lemma 4.4. Let the assumptions of Lemma 4.1 be fulﬁlled and addi (y) tionally, let p ∈ R\{0, 1}, p1 + 1q = 1, vv(y) (> 0) be decreasing in [n0 , ∞)(n0 ∈ N) and there exist constants δ > α and L > 0, such that h(t) ≤ L

1 tδ

,

(t ∈ [v(n0 ), ∞)).

(4.9)

Then, for 0 < ε < min{|p|, |q|}(δ − α), we have B(ε) =

∞ n=n0

v (n) [v(n)]1+ε

∞ v(n)

ε

h(t)tα+ p −1 dt = O(1)

(ε → 0+ ).

(4.10)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

Proof.

In view of (4.9), we ﬁnd ∞

0 < B(ε) ≤ L =

L δ−α−

n=n0 ∞

v (n) [v(n)]1+ε

ε p n=n0

L = δ−α−

ε p

≤

L δ−α−

ε p

=

L δ−α−

ε p

∞

129

ε

tα−δ+ p −1 dt

v(n)

v (n) ε

[v(n)]δ−α+ q +1

∞ v (n) v (n0 ) + ε ε [v(n0 )]δ−α+ q +1 n=n0 +1 [v(n)]δ−α+ q +1 ( ) ∞ v (n0 ) v (y)dy + ε δ−α+ qε +1 [v(n0 )]δ−α+ q +1 n0 [v(y)]

ε v (n0 ) [v(n0 )]α−δ− q < ∞, + ε δ − α + qε [v(n0 )]δ−α+ q +1

and then (4.10) follows. 1 Example 4.1. (i) For h(t) = 1+t λ (0 < α < λ), since ∞ π 1 k(α) = tα−1 dt = ∈ R+ , λ 1 + t λ sin π( αλ ) 0

setting δ =

λ+α 2

> α (δ < λ), we ﬁnd h(t) =

(ii) For h(t) =

1 (1+t)λ

∞

k(α) = 0

setting δ =

λ+α 2

1 1 ≤ δ tλ + 1 t

(t ∈ [v(n0 ), ∞)).

(0 < α < λ), since 1 tα−1 dt = B(α, λ − α) ∈ R+ , (1 + t)λ

> α (δ < λ), we ﬁnd h(t) =

1 1 ≤ δ λ (1 + t) t

(t ∈ [v(n0 ), ∞)).

ln t tλ −1

(0 < α < λ), since )2 ( ∞ π ln t α−1 t k(α) = dt = ∈ R+ , tλ − 1 λ sin( πα ) 0 λ

(iii) For h(t) =

δ

t ln t setting δ = λ+α 2 > α (δ < λ), we ﬁnd tλ −1 → 0 (t → ∞), and there exists a constant L > 0, such that 1 ln t ≤ L δ (t ∈ [v(n0 ), ∞)). h(t) = λ t −1 t

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

130

Half-Discrete Hilbert-Type Inequalities

s (iv) For s ∈ N, kλs (t, 1) = k=1 ak t1λ +1 (0 < a1 < · · · < as , λ > 0, 0 < α < s), by (3.12), we ﬁnd ∞2 s 1 tαλ−1 dt k(αλ) = λ+1 a t k 0 k=1 ∞2 s 1 1 u(s−α)−1 du = λ 0 ak + u k=1

s π aks−α−1 = λ sin(πα) k=1

s 2 j=1(j =k)

1 ∈ R+ . ak − aj

> αλ (< sλ), since t kλs (t, 1) → 0 (t → ∞), there exists Setting δ = a constant L > 0, such that s 2 1 L kλs (t, 1) = ≤ δ (t ∈ [v(n0 ), ∞)). a k tλ + 1 t s+α λ 2

δ

k=1

(v) For h(t) = tλ +2tλ/21 cos β+1 (0 < β ≤ π2 , λ > 0, 0 < α < 1), by Example 3.1(v), we ﬁnd ∞ tαλ−1 k(αλ) = dt, (u = tλ/2 ) λ λ/2 cos β + 1 t + 2t 0 2 ∞ u2α−1 du = 2 λ 0 u + 2u cos β + 1 2π sin β(1 − 2α) = ∈ R+ . λ sin β sin(2πα) δ

t Setting δ = λ (> αλ), since tλ +2tλ/2 → 1 (t → ∞), there exists a cos β+1 constant L > 0, such that 1 1 (t ∈ [v(n0 ), ∞)). ≤ L h(t) = λ tδ t + 2tλ/2 cos β + 1 √ (vi) For h(t) = tλ +bt1λ/2 +c (c > 0, 0 ≤ b ≤ 2 c, α = λ2 > 0), we ﬁnd ∞ λ λ 2 ∞ t 2 −1 dt du k = = , (u = tλ/2 ) 2 λ λ/2 2 t + bt + c λ 0 u + bu + c 0 ⎧ π √ , b = 0, ⎪ c √ √ 2⎨ 4 4c−b2 √ = arctan , 0 L > 0, such that h(t) =

λ 2,

since

tδ tλ +btλ/2 +c

1 ≤L λ t + btλ/2 + c

→ 1(t → ∞), there exists a constant

1 tδ

(t ∈ [v(n0 ), ∞)).

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel 1 eγtλ −Ae−γtλ

(vii) For h(t) = 0, α = βλ), we ﬁnd k(α) =

131

(−1 ≤ A < 1, β > 0(A = 1, β > 1), γ, λ >

1 ∞ tβλ−1 dt uβ−1du = λ λ γu λ 0 e (1 − Ae−2γu ) eγt − Ae−γt 0 ∞ ∞ 1 Ak e−γ(2k+1)u uβ−1 du = λ 0 k=0 ∞ ∞ 1 = Ak e−γ(2k+1)u uβ−1 du λ 0 ∞

k=0

∞

Γ(β) Ak = ∈ R+ . λγ β (2k + 1)β k=0

Setting δ = β0 λ > βλ = α (β0 > β), since exists a constant L > 0, such that e

γtλ

tδ ≤L − Ae−γtλ

tδ λ λ eγt −Ae−γt

→ 0 (t → ∞), there

(t ∈ [v(n0 ), ∞)).

Then it follows that

1 1 h(t) = γtλ (t ∈ [v(n0 ), ∞)). λ ≤ L −γt tδ e − Ae λ bt +1 (0 ≤ a < b, λ > 0, α = −βλ, 0 < β < 1), we (viii) For h(t) = ln at λ +1 ﬁnd λ ∞ λ −1 ∞ bt + 1 −βλ−1 bt + 1 ln dt = ln k(α) = t dt−βλ λ+1 λ+1 at βλ at 0 0 ( ∞ ∞ λ ) λ λ −1 −βλ bt + 1 (1−β)λ−1 t t − = ln − dt βλ atλ + 1 0 btλ + 1 atλ + 1 0 ∞ (1−β)−1 β aβ (bβ − aβ )π u b − du = . = βλ βλ 1+u βλ sin(βπ) 0 λ

bt +1 Setting δ = −β0 λ > −βλ = α(0 < β0 < β), in view of tδ ln( at λ +1 ) → 0 (t → ∞), there exists a constant L > 0, such that λ bt + 1 ≤ L (t ∈ [v(n0 ), ∞)). tδ ln atλ + 1

Then it follows that h(t) = ln

btλ + 1 atλ + 1

≤

L tδ

(t ∈ [v(n0 ), ∞)).

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

132

Half-Discrete Hilbert-Type Inequalities

(ix) For h(t) = arctan t1λ (0 < α < λ), we ﬁnd ∞ 1 π tα−1 arctan λ dt = k(α) = πα ∈ R+ . t 2α cos( ) 0 2λ Setting δ = α0 ∈ (α, λ), since tδ arctan t1λ → 0 (t → ∞), there exists a constant L > 0, such that 1 1 (t ∈ [v(n0 ), ∞)). h(t) = arctan λ ≤ L δ t t (x) For h(t) =

1 (max{1,t})λ

∞

k(α) = 0

setting δ =

λ+α 2

(0 < α < λ), since

λ 1 tα−1 dt = ∈ R+ , λ (max{1, t}) α(λ − α)

> α (δ < λ), we ﬁnd

h(t) =

1 1 ≤ δ (max{1, t})λ t

(t ∈ [v(n0 ), ∞)).

(xi) For h(t) = (min{1, t})λ (−λ < α < 0), since ∞ λ k(α) = (min{1, t})λtα−1 dt = ∈ R+ , (−α)(α + λ) 0 setting δ =

α 2

> α (−λ < δ < 0), there exists a constant L > 0, such that 1 h(t) = (min{1, t})λ ≤ L δ (t ∈ [v(n0 ), ∞)). t

min{1,t} λ (xii) For h(t) = ( max{1,t} ) (λ > 0, −λ < α < λ), since λ ∞ 2λ min{1, t} k(α) = tα−1 dt = 2 ∈ R+ , max{1, t} λ − α2 0

setting δ =

4.2.3

λ+α 2

> α (−λ < α < δ < λ), we ﬁnd λ 1 min{1, t} ≤ δ (t ∈ [v(n0 ), ∞)). h(t) = max{1, t} t

Some Inequalities Relating the Constant k(α)

Lemma 4.5. Suppose that α ∈ R, h(t) is a non-negative ﬁnite measurable function in R+ with ∞ k(α) = h(t)tα−1 dt ∈ R+ . 0

Then

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

ws-book9x6

133

(i) for p > 1, ε > 0, we have ε ≥ k(α) + o(1) (ε → 0+ ); k α+ (4.11) p (ii) for p < 0, if there exists a constant δ1 > 0, such that k(α−δ1 ) ∈ R+ , then, for 0 < ε < (−p)δ1 , we have ε (4.12) ≤ k(α) + o(1) (ε → 0+ ); k α+ p (iii) for 0 0, such that k(α+δ2 ) ∈ R+ , then for 0 < ε < pδ2 , we still have (4.12). Proof.

(i) For p > 1, by Fatou’s lemma (see Kuang [49]), it follows that ∞ ε (α+ pε )−1 , lim h(t)t dt ≤ lim k α + k(α) = p ε→0+ ε→0+ 0 and then we have (4.11). ε (ii) For p < 0, since t p ≤ 1 (1 ≤ t < ∞), we ﬁnd 1 ∞ ε ε ε = k α+ h(t)t(α+ p )−1 dt + h(t)t(α+ p )−1 dt p 0 1 ∞ 1 (α+ pε )−1 h(t)t dt + h(t)tα−1 dt. (4.13) ≤ 0

For 0 < ε < (−p)δ1 , we have 1 ε (α+ p )−1 h(t)t dt ≤ 0

1

1 0

h(t)t(α−δ1 )−1 dt ≤ k(α − δ1 ) < ∞,

then by the Lebesgue control convergence theorem (see Kuang [49]), it follows that 1 1 ε h(t)t(α+ p )−1 dt = h(t)tα−1 dt + o(1) (ε → 0+ ). 0

0

Hence, by (4.13), we have (4.12). ε (iii) For 0 < p < 1, since t p ≤ 1 (0 < t ≤ 1), we ﬁnd 1 ∞ ε ε (α+ pε )−1 = k α+ h(t)t dt + h(t)t(α+ p )−1 dt p 0 1 ∞ 1 ε h(t)tα−1 dt + h(t)t(α+ p )−1 dt. ≤ 0

For 0 < ε < pδ2 , we have ∞ ε h(t)t(α+ p )−1 dt ≤ 1

(4.14)

1

1

∞

h(t)t(α+δ2 )−1 dt ≤ k(α + δ2 ) < ∞,

then by the Lebesgue control convergence theorem, it follows that ∞ ∞ (α+ pε )−1 h(t)t dt = h(t)tα−1 dt + o(1)(ε → 0+ ). 1

1

Hence, by (4.14), we have (4.12) for 0 < p < 1.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

134

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

4.3

Some Theorems and Corollaries

4.3.1

Equivalent Inequalities and their Operator Expressions

We set two functions ϕ(x) = xp(1−α)−1 (x ∈ (0, ∞)) and [v(n)]q(1−α)−1 Ψ(n) = (n ≥ n0 , n ∈ N), [v (n)]q−1

v (n) wherefrom, [ϕ(x)]1−q = xqα−1 and [Ψ(n)]1−p = [v(n)] 1−pα . 1 1 For p ∈ R\{0, 1}, p + q , we deﬁne two sets as follows: 1

Lp,ϕ (R+ ) = ⎧ ⎨ lq,Ψ =

⎩

f ; ||f ||p,ϕ =

∞

p(1−α)−1

x 0

a = {an }∞ n=n0 ; ||a||q,Ψ =

|f (x)| dx p

p

<∞ ,

∞ [v(n)]q(1−α)−1 |an |q (n)]q−1 [v n=n

q1

0

⎫ ⎬ <∞ . ⎭

Note. It is obvious that for p > 1, both of the above two sets are the normed spaces. For p < 0 or 0 < p < 1, neither of the above two sets is the normed space, and we agree on that the sets with the normed expressions of ||f ||p,ϕ and ||a||q,Ψ are the formal symbols in these cases. Theorem 4.1. (see Yang [170]) Suppose that α ∈ R, h(t) is a non-negative ﬁnite measurable function in R+ , v(y) is a strict increasing diﬀerentiable function in [n0 , ∞) (n0 ∈ N) with v(n0 ) > 0 and v(∞) = ∞, ∞

k(α) = 0

h(t)tα−1 dt ∈ R+

and α (x) < k(α) (x ∈ (0, ∞)). If p > 1, 1p + 1q = 1, f (x), an ≥ 0, f ∈ Lp,ϕ (R+ ), a = {an}∞ n=n0 ∈ lq,Ψ , ||f ||p,ϕ > 0, ||a||q,Ψ > 0, then we have the following equivalent inequalities: ∞ ∞ ∞ ∞ an h(xv(n))f (x)dx = f (x) an h(xv(n))dx I= 0

n=n0

0

< k(α)||f ||p,ϕ ||a||q,Ψ , ∞ ( 1−p [Ψ(n)] J= 0

n=n0

and

∞

1−q

[ϕ(x)]

L= 0

n=n0

(4.15) p1 ) p ∞ h(xv(n))f (x)dx < k(α)||f ||p,ϕ , (4.16)

∞ n=n0

q h(xv(n))an

1q dx

< k(α)||a||q,Ψ .

(4.17)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

ws-book9x6

135

(y) Moreover, if vv(y) (> 0) is decreasing in [n0 , ∞) and there exists constants δ < λ1 and L > 0, such that (4.9) is satisﬁed, then the constant factor k(α) in the above inequalities is the best possible.

Proof. By the Lebesgue term by term integration theorem (see Kuang [49]), there are two expressions for I in (4.15). In view of (4.6), since α (x) < k(α), we have (4.16). By H¨older’s inequality, we ﬁnd ) ∞ ∞ ( 1 1 h(xv(n))f (x)dx [[Ψ(n)] q an ] ≤ J||a||q,Ψ . (4.18) [Ψ(n)]− q I= n=n0

0

Then, by (4.16), we have (4.15). On the other hand, suppose that (4.15) is valid. We set ( ∞ )p−1 an = [Ψ(n)]1−p h(xv(n))f (x)dx , n ≥ n0 (n ∈ N). 0

Then it follows that J p−1 = ||a||q,Ψ . By (4.6), we ﬁnd J < ∞. If J = 0, then (4.16) is trivially valid; if J > 0, then, by (4.15), we have ||a||qq,Ψ = J p = I < k(α)||f ||p,ϕ ||a||q,Ψ , that is, ||a||q−1 q,Ψ = J < k(α)||f ||p,ϕ , and then, (4.15) is equivalent to (4.16). In view of (4.7), since [α (x)]1−q > [k(α)]1−q , we have (4.17). By H¨older’s inequality, we ﬁnd

∞ ∞ 1 1 − [[ϕ(x)] p f (x)] [ϕ(x)] p an h(xv(n)) dx ≤ ||f ||p,ϕ L. (4.19) I= 0

n=n0

Then, by (4.17), we have (4.15). On the other hand, suppose that (4.15) is valid. We set

∞ q−1 1−q h(xv(n))an , x ∈ (0, ∞). f (x) = [ϕ(x)] n=n0

Then, it follows that L = ||f ||p,ϕ . By (4.7), we ﬁnd L < ∞. If L = 0, then (4.17) is trivially valid; if L > 0, then, by (4.15), we have q−1

||f ||pp,ϕ = Lq = I < k(α)||f ||p,ϕ ||a||q,Ψ , that is, p−1 = L < k(α)||a||q,Ψ , ||f ||p,ϕ

and then, (4.15) is equivalent to (4.17).

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

136

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Hence, (4.15), (4.16) and (4.17) are equivalent. For 0 < ε < p(δ − α), setting f(x) and an as follows: f(x) =

ε

xα+ p −1 , 0 < x ≤ 1, 0, x > 1, ε

an = [v(n)]α− q −1 v (n),

n ≥ n0 ,

if there exists a positive constant k(≤ k(α)), such that (4.15) is still valid as we replace k(α) by k, then substitution of f(x) and a = { a n }∞ n=n0 , by (4.8), it follows that ∞ ∞ an h(xv(n))f(x)dx < k||f||p,ϕ || a||q,Ψ I= 0

n=n0

1

=k

−1+ε

x 0

p1 1 1 k dx (A(ε)) q = (1 + o(1)) q . ε

By (4.8), (4.9) and (4.10), we ﬁnd 1 ∞ ε ε [v(n)]α− q −1 v (n) h(xv(n))xα+ p −1 dx I =

=

n=n0 ∞ n=n0 ∞

v (n) [v(n)]1+ε

(4.20)

(t = xv(n))

0

v(n)

ε

h(t)tα+ p −1 dt

0

∞ ε v (n) h(t)tα+ p −1 dt − = 1+ε [v(n)] 0 n=n0 ε = A(ε)k(α + ) − B(ε) p 1 ≥ [(1 + o(1))(k(α) + o(1)) − εO(1)]. ε

∞

ε α+ p −1

h(t)t

dt

v(n)

(4.21)

Hence, in view of (4.20) and (4.21), it follows that 1

(1 + o(1))(k(α) + o(1)) − εO(1) < k(1 + o(1)) q , and then k(α) ≤ k(ε → 0+ ). Therefore, the constant factor k(α) in (4.15) is the best possible. By the equivalency, the constant factor k(α) in (4.16) ((4.17) is also the best possible, otherwise, it leads to a contradiction by (4.18), (4.19) that the constant factor in (4.15) is not the best possible.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

ws-book9x6

137

Note. If we change the upper limits, ∞ of the integral to c > 0, then Theorem 4.1 is still valid. In this case, c ωα (n) = [v(n)]α h(xv(n))xα−1 dx < k(α). 0

Remark 4.1. If we replace the conditions that k(α) ∈ R+ , and there exist constants δ > α and L > 0, such that (4.9) is fulﬁlled to the condition that ∈ (α − δ0 , α], there exist δ0 , η > 0, such that for any α α) < ∞, 0 < k( α)(1 − θ(x)) < α (x) < k(

(4.22)

where θ(x) = O(x ) (0 < x ≤ 1), then the constant factor k(α) in (4.15) is still the best possible. In fact, for 0 < ε < qδ0 , α = α − εq , we ﬁnd

1 ∞ v (n) α− εq ε−1 x h(xv(n)) x dx I= ε [v(n)]1−(α− q ) 0 n=n0 1 1 = xε−1 α (x)dx > k( α) xε−1 (1 − O(xη ))dx η

0

0

1 ≥ (k(α) + o(1))(1 − εO(1)), ε and then by the same way, we can still show that k(α) is the best possible constant factor of (4.15). Remark 4.2. In view of Theorem 4.1, (i) deﬁne a half-discrete Hilbert-type operator T : Lp,ϕ (R+ ) → lp,Ψ1−p as follows: For f ∈ Lp,ϕ (R+ ), there exists a T f ∈ lp,Ψ1−p , satisfying ∞ h(xv(n))f (x)dx, n ≥ n0 , n ∈ N. (4.23) T f (n) = 0

Then, by (4.16), it follows that ||T f ||p.Ψ1−p ≤ k(α)||f ||p,ϕ and then T is a bounded operator with ||T || ≤ k(α). Since by Theorem 4.1, the constant factor in (4.16) is the best possible, we have ||T || = k(α). (ii) Deﬁne a half-discrete Hilbert-type operator T : lq,Ψ → Lq,ϕ1−q (R+ ) as follows: For a ∈ lq,Ψ , there exists a Ta ∈ Lq,ϕ1−q (R+ ), satisfying Ta(x) =

∞

h(xv(n))an ,

x ∈ (0, ∞).

(4.24)

n=n0

Then, by (4.17), it follows that ||Ta||q.ϕ1−q ≤ k(α)||a||q,Ψ and then T is a bounded operator with ||T|| ≤ k(α). Since by Theorem 4.1, the constant factor in (4.17) is the best possible, we have ||T|| = k(α).

November 7, 2013

12:24

138

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

1 Example 4.2. If v(n) = ln n, h(x ln n) = (ln en x )λ (0 < α < λ, α ≤ 1), since for x > 0, xα f (x, y) = (ln y)α−1 (1 + x ln y)λ

is decreasing with respect to y ∈ (1, ∞), then Condition (i) is satisﬁed. In view of Example 4.1(ii), the constant factor in (4.16) and (4.17) (for v(n) = n, n0 = 1) is the best possible, we have ||T || = ||T|| = k(α) = B(α, λ − α). If h(x ln n) = e(ln 1nx )λ (λ > 0, 0 < α ≤ 1), then by the same way and Example 4.1(v), we have 1 α ||T || = ||T|| = Γ( ). λ λ In particular of Remark 4.2, setting v(n) = n and ψ(n) = nq(1−α)−1 (n ∈ N), we deﬁne an operator T : Lp,ϕ (R+ ) → lp,ψ1−p as follows: For f ∈ Lp,ϕ (R+ ), there exists a T f ∈ lp,ψ1−p , satisfying ∞ T f (n) = h(xn)f (x)dx, n ∈ N. 0

Also we deﬁne an operator T : lq,ψ → Lq,ϕ1−q (R+ ) as follows: For a ∈ lq,ψ , there exists a Ta ∈ Lq,ϕ1−q (R+ ), satisfying Ta(x) =

∞

h(xn)an ,

x ∈ (0, ∞).

n=1

Then by Remark 4.2, we still have ||T || = ||T|| = k(α). We set some particular kernels as follows: 1 (i) If h(xn) = 1+(xn) λ (0 < α < λ, α ≤ 1), since for x > 0, f (x, y) =

xα y α−1 1 + (xy)λ

is decreasing with respect to y ∈ (0, ∞), then Condition (i) is satisﬁed. In view of Example 4.1(i), the constant factor in (4.16) and (4.17) (for v(n) = n, n0 = 1) is the best possible, we have π ||T || = ||T|| = k(α) = . λ sin π( αλ ) (ii) If h(xn) =

1 (0 (1+xn)λ

< α < λ, α ≤ 1), since, for x > 0,

f (x, y) =

xα y α−1 (1 + xy)λ

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

ws-book9x6

139

is decreasing with respect to y ∈ (0, ∞), then Condition (i) is satisﬁed. In view of Example 4.1(ii), the constant factor in (4.16) and (4.17) (for v(n) = n, n0 = 1) is the best possible, we have ||T || = ||T|| = k(α) = B(α, λ − α). ln(xn) (iii) If h(xn) = (xn) λ −1 (0 < α < λ, α ≤ 1), then by the same way and Example 4.1(iii), we have (see Zhong [181]) )2 ( π ||T || = ||T|| = . λ sin π( αλ ) s 1 (iv) If h(xn) = k=1 ak (xn) λ +1 (0 < a1 < · · · < as , λ > 0, 0 < α < s, αλ ≤ 1), then, by the same way and Example 4.1(iv), we have s s 2 π 1 ||T || = ||T|| = aks−α−1 . λ sin(πα) ak − a j k=1

j=1(j =k)

1 (xn)λ +2(xn)λ/2 cos β+1

(v) If h(xn) = (λ > 0, 0 < β ≤ π2 , 0 < α < 1), then by the same way and Example 4.1(v), we ﬁnd 2π sin β(1 − 2α) ∈ R+ . ||T || = ||T|| = λ sin β sin(2πα) √ 1 (vi) If h(xn) = (xn)λ +b(xn) c, 0 < α < λ2 ≤ 1), λ/2 +c (c > 0, 0 ≤ b ≤ 2 then, by the same way and Example 4.1(vi), we have ⎧ π √ , b = 0, ⎪ c ⎨ √ √ 2 2 4c−b 4 √ arctan b , 0 0(A = 1, β > e −Ae 1), γ, λ > 0, α = βλ ≤ 1), then by the same way and Example 4.1(vii), we have ∞ Ak Γ(β) . ||T || = ||T|| = β λγ (2k + 1)β k=0 " # b(xn)λ +1 (viii) If h(xn) = ln a(xn) (0 ≤ a < b, λ > 0), α = −βλ (0 < β < λ +1 1), then, by the same way and Example 4.1(viii), we have (bβ − aβ )π ||T || = ||T|| = . βλ sin(βπ) 1 (ix) If h(xn) = arctan (xn) λ (0 < α < λ, α ≤ 1), then, by the same way and Example 4.1(ix), we have π . ||T || = ||T|| = π 2α sin 2 (1 − αλ )

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

140

Half-Discrete Hilbert-Type Inequalities

1 (x) If h(xn) = (max{1,xn}) λ , 0 < α < λ, α ≤ 1, then, by the same way and Example 4.1(x), we have (see Yang [166]) λ . ||T || = ||T|| = α(λ − α) (xi) If h(xn) = (min{1, xn})λ (0 < −α < λ ≤ 1 − α), then by the same way and Example 4.1(xi), we have (see Yang [166]) λ . ||T || = ||T|| = (−α)(λ + α) λ min{1,xn} (|α| < λ ≤ 1 − α, α < 12 ), since, for x > 0, (xii) If h(xn) = max{1,xn} λ min{1, xy} α f (x, y) = x y α−1 max{1, xy} α+λ λ+α−1 x y , 0 < y ≤ x1 , = α−λ α−λ−1 y , y > x1 x

is decreasing with respect to y ∈ (0, ∞), then Condition (i) is satisﬁed. In view of Example 4.1(xii), the constant factor in (4.16) and (4.17) (for v(n) = n, n0 = 1) is the best possible, we have 2λ . ||T || = ||T|| = k(α) = 2 λ − α2 4.3.2

Two Classes of Equivalent Reverses

Theorem 4.2. Suppose that α ∈ R, h(t) is a non-negative ﬁnite measurable function in R+ , v(y) is a strictly increasing diﬀerentiable functions in [n0 , ∞)(n0 ∈ N) with v(n0 ) > 0 and v(∞) = ∞, ∞ h(t)tα−1 dt ∈ R+ k(α) = 0

and α (x) < k(α) (x ∈ (0, ∞)). If p < 0, p1 + 1q = 1, f (x), an ≥ 0, f ∈ Lp,ϕ (R+ ), a = {an }∞ n=n0 ∈ lq,Ψ , ||f ||p,ϕ > 0, ||a||q,Ψ > 0, then, we have the following equivalent inequalities: ∞ ∞ ∞ ∞ I= an h(xv(n))f (x)dx = f (x) an h(xv(n))dx n=n0

0

0

> k(α)||f ||p,ϕ ||a||q,Ψ , ∞ ( [Ψ(n)]1−p J= n=n0

0

n=n0

(4.25) 1 )p p ∞ h(xv(n))f (x)dx > k(α)||f ||p,ϕ , (4.26)

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

and

L=

∞

[ϕ(x)]1−q

0

∞

q h(xv(n))an

ws-book9x6

141

1q dx

> k(α)||a||q,Ψ .

(4.27)

n=n0

(y) (> 0) is decreasing in [n0 , ∞) and there exists a conMoreover, if vv(y) stant δ1 > 0, such that k(α + δ1 ) ∈ R+ , then the constant factor k(α) in the above inequalities is the best possible.

Proof. In view of the reverse of (4.6) and α (x) < k(α), for p < 0, we have (4.26). By the reverse H¨older’s inequality (see Kuang [47]), we ﬁnd ) ∞ ∞ ( 1 − 1q [Ψ(n)] h(xv(n))f (x)dx [[Ψ(n)] q an ] ≥ J||a||q,Ψ . (4.28) I= n=n0

0

Then, by (4.26), we have (4.25). On the other hand, suppose that (4.25) is valid. We set ( ∞ )p−1 1−p h(xv(n))f (x)dx , n ≥ n0 , n ∈ N. an = [Ψ(n)] 0

Then, it follows that J p−1 = ||a||q,Ψ . By the reverse of (4.6) and the assumption, we ﬁnd J > 0. If J = ∞, then (4.26) is trivially valid; if J < ∞, then, by (4.25), we have ||a||qq,Ψ = J p = I > k(α)||f ||p,ϕ ||a||q,Ψ , that is ||a||q−1 q,Ψ = J > k(α)||f ||p,ϕ , and then, (4.25) is equivalent to (4.26). In view of the reverse of (4.7), since [α (x)]1−q < [k(α)]1−q , we have (4.27). By the reverse H¨older’s inequality, we ﬁnd

∞ ∞ 1 1 −p p I= [[ϕ(x)] f (x)] [ϕ(x)] h(xv(n))an dx ≥ ||f ||p,ϕ L. (4.29) 0

n=n0

Then, by (4.27), we have (4.25). On the other hand, suppose that (4.25) is valid. We set

∞ q−1 1−q h(xv(n))an , x ∈ (0, ∞). f (x) = [ϕ(x)] n=n0

Then it follows that Lq−1 = ||f ||p,ϕ . By the reverse of (4.7), we ﬁnd L > 0. If L = ∞, then (4.27) is trivially valid; if L < ∞, then, by (4.25), we have ||f ||pp,ϕ = Lq = I > k(α)||f ||p,ϕ ||a||q,Ψ , s.t. p−1 ||f ||p,ϕ = L > k(α)||a||q,Ψ ,

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

142

Half-Discrete Hilbert-Type Inequalities

and then, (4.25) is equivalent to (4.27). Hence, inequalities (4.25), (4.26) and (4.27) are equivalent. For ε > 0, setting f(x) and an as Theorem 4.1, if there exists a positive constant k(≥ k(α)), such that (4.25) is valid as we replace k(α) by k, then substitution of f(x) and a = { an }∞ n=n0 , by (4.8), it follows that ∞ ∞ an h(xv(n))f(x)dx > k||f||p,ϕ || a||q,Ψ I = 0

n=n0

1

=k

−1+ε

x 0

p1 1 1 k dx (A(ε)) q = (1 + o(1)) q . ε

By (4.8) and (4.12), we ﬁnd I =

= ≤

∞ n=n0 ∞ n=n0 ∞ n=n0

ε

[v(n)]α− q −1 v (n) v (n) [v(n)]1+ε

v (n) [v(n)]1+ε

1

ε

h(xv(n))xα+ p −1 dx,

(t = xv(n))

0

v(n)

ε

h(t)tα+ p −1 dt

0 ∞

ε

h(t)tα+ p −1 dt

0

ε 1 = A(ε)k(α + ) ≤ (1 + o(1))(k(α) + o(1)). p ε Hence, in view of the above results, it follows that 1

(1 + o(1))(k(α) + o(1)) > k(1 + o(1)) q , and then k(α) ≥ k(ε → 0+ ). Therefore, the constant factor k(α) in (4.25) is the best possible. By the equivalency, the constant factor k(α) in (4.26) ((4.27)) is the best possible, otherwise, we can imply a contradiction by (4.28)((4.29)) that the constant factor in (4.25) is not the best possible. If we change the upper limit ∞ of the integral to c(> 0) in Theorem 4.2, then it follows that c α ωα (n) = [v(n)] h(xv(n))xα−1 dx < k(α). 0

Setting ωα (n) = k(α)(1 − θα (v(n))), ∞ 1 θα (v(n)) = [v(n)]α h(xv(n))xα−1 dx > 0, k(α) c

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

ws-book9x6

143

and [v(n)]q(1−α)−1 ψ(n) = ψ(n)(1 − θα (v(n)) = (1 − θα (v(n))), [v (n)]q−1 we have θα (v(n)) =

1 k(α)

∞

h(t) tα−1 dt,

cv(n)

and the following corollary: Corollary 4.1. Suppose that c > 0, α ∈ R, h(t) is a non-negative ﬁnite measurable function in R+ , v(y) is a strictly increasing diﬀerentiable functions in [n0 , ∞)(n0 ∈ N) with v(n0 ) > 0 and v(∞) = ∞, ∞ k(α) = h(t) tα−1 dt ∈ R+ 0

and α (x) < k(α)(x ∈ (0, c)). If p < 0, p1 + 1q = 1, f (x), an ≥ 0, f ∈ Lp,ϕ (0, c), a = {an }∞ , ||f ||p,ϕ > 0, ||a||q,ψ > 0, then, we have the n=n0 ∈ lq,ψ following equivalent inequalities: c ∞ an h(xv(n))f (x)dx 0

n=n0

c

f (x)

= 0

∞

c

an h(xv(n))dx > k(α)||f ||p,ϕ ||a||q,ψ,

(4.30)

) p p1 h(xv(n))f (x)dx > k(α)||f ||p,ϕ ,

(4.31)

n=n0

[ψ(n)]

1−p

(

c 0

n=n0

and

∞

[ϕ(x)]1−q

0

∞

q h(xv(n))an

1q dx

> k(α)||a||q,ψ.

(4.32)

n=n0

(y) (> 0) is decreasing in [n0 , ∞) and there exist constants Moreover, if vv(y) δ1 > 0, L > 0 and δ > α, such that k(α − δ1 ) ∈ R+ , and 1 (t ∈ [cv(n0 ), ∞)), h(t) ≤ L δ t

then, the constant factor k(α) in the above inequalities is the best possible. Theorem 4.3. Suppose that α ∈ R, h(t) is a non-negative ﬁnite measurable function in R+ , v(y) is a strictly increasing diﬀerentiable function in

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

144

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

[n0 , ∞)(n0 ∈ N) with v(n0 ) > 0 and v(∞) = ∞, k(α) = R+ and

∞ 0

h(t)tα−1 dt ∈

0 < k(α)(1 − θ(x)) < α (x) < k(α) (x ∈ (0, ∞)). Setting ϕ(x) = (1 − θ(x))ϕ(x), if 0 0, ||a||q,Ψ > 0, then, we have the n=n0 ∈ lq,Ψ , ||f ||p,ϕ following equivalent inequalities: ∞ ∞ ∞ ∞ an h(xv(n))f (x)dx = f (x) an h(xv(n))dx I= 0

n=n0

0

> k(α)||f ||p,ϕ ||a||q,Ψ , ∞ ( [Ψ(n)]1−p J= 0

n=n0

and

L= 0

∞

1−q

[ϕ(x)]

n=n0

(4.33) 1 )p p ∞ h(xv(n))f (x)dx > k(α)||f ||p,ϕ , (4.34)

∞

q h(xv(n))an

1q dx

> k(α)||a||q,Ψ .

(4.35)

n=n0

(y) Moreover, if vv(y) (> 0) is decreasing in [n0 , ∞) and there exist constants η, δ2 > 0, such that θ(x) = O x1η (0 < x ≤ 1) and k(α + δ2 ) ∈ R+ , then the constant factor k(α) in the above inequalities is the best possible.

Proof. In view of the reverse of (4.6) and α (x) < k(α), for 0 < p < 1, we have (4.34). By the reverse H¨ older’s inequality, we ﬁnd ) ( ∞ ∞ 1 1 h(xv(n))f (x)dx [[Ψ(n)] q an ] ≥ J||a||q,Ψ . (4.36) I= [Ψ(n)]− q n=n0

0

Then, by (4.34), we have (4.33). On the other hand, suppose that (4.33) is valid. We set ) p−1 ( ∞ an = [Ψ(n)]1−p h(xv(n))f (x)dx , n ≥ n0 (n ∈ N). 0 p−1

= ||a||q,Ψ . By the reverse of (4.6) with the Then it follows that J assumptions, we ﬁnd J > 0. If J = ∞, then (4.34) is trivially valid; if J < ∞, then, by (4.33), we have ||a||qq,Ψ = J p = I > k(α)||f ||p,ϕ ||a||q,Ψ , that is ||a||q−1 , q,Ψ = J > k(α)||f ||p,ϕ

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

145

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

and then, inequality (4.33) is equivalent to (4.34). In view of the reverse of (4.7), since [α (x)]1−q > [k(α)(1 − θ(x))]1−q (q < 0), we have (4.35). By the reverse H¨ older’s inequality, we ﬁnd

∞ ∞ 1 1 −p p [[ϕ(x)] f (x)] [ϕ(x)] h(xv(n))an dx ≥ ||f ||p,ϕL. (4.37) I= 0

n=n0

Then, by (4.35), we have (4.33). On the other hand, suppose that (4.33) is valid. We set

∞ q−1 1−q f (x) = [ϕ(x)] h(xv(n))an , x ∈ (0, ∞). n=n0

Then it follows that Lq−1 = ||f ||p,ϕ. By the reverse of (4.7) and the assumption, we ﬁnd L > 0. If L = ∞, then (4.35) is trivially valid; if L < ∞, then, by (4.33), we have ||f ||pp,ϕ = Lq = I > k(α)||f ||p,ϕ||a||q,Ψ , that is p−1 ||f ||p, ϕ = L > k(α)||a||q,Ψ ,

and then, (4.33) is equivalent to (4.35). Hence, inequalities (4.33), (4.34) and (4.35) are equivalent. For ε > 0, setting f(x) and an as Theorem 4.1, if there exists a positive constant k(≥ k(α)), such that (4.33) is valid as we replace k(α) by k, then, substitution of f(x) and a = { an } ∞ n=n0 , by (4.8), it follows that ∞ ∞ an h(xv(n))f(x)dx > k||f||p,ϕ|| a||q,Ψ I = 0

n=n0

) p1 1 1 1 −1−ε 1−O x dx (A(ε)) q η x 0

(

=k

1 1 k (1 − εO(1)) p (1 + o(1)) q . ε By (4.8) and (4.12), we ﬁnd 1 ∞ ε α− εq −1 [v(n)] v (n) h(xv(n))xα+ p −1 dx I=

=

=

n=n0 ∞ n=n0

v (n) [v(n)]1+ε

0

v(n) 0

ε

h(t)tα+ p −1 dt

(4.38)

(t = xv(n))

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

146

ws-book9x6

Half-Discrete Hilbert-Type Inequalities ∞

≤

n=n0

v (n) [v(n)]1+ε

∞

ε

h(t)t(α+ p )−1 dt

0

ε 1 = A(ε)k(α + ) ≤ (1 + o(1))(k(α) + o(1)). p ε

(4.39)

Hence, in view of (4.38) and (4.39), it follows that 1

1

(1 + o(1))(k(α) + o(1)) > k(1 − εO(1)) p (1 + o(1)) q , and then, k(α) ≥ k(ε → 0+ ). Therefore, the constant factor k(α) in (4.33) is the best possible. By the equivalency, the constant factor k(α) in (4.34)((4.35)) is the best possible, otherwise, we can imply a contradiction by (4.36) ((4.37)) that the constant factor in (4.33) is not the best possible. Note. If we change the upper limit ∞ of the integral to c > 0, then Theorem 4.3 is still value. In this case, c h(xv(n))xα−1 dx < k(α). ωα (n) = [v(n)]α 0

4.3.3

Some Corollaries

Corollary 4.2. Let the assumptions of Lemma 4.1 be fulﬁlled and additionally, let ∞ h(t)tα−1 dt ∈ R+ (x ∈ (0, ∞)), α (x) < k(α) = 0

and for −∞ ≤ b < c ≤ ∞, u(x) be a strictly increasing diﬀerentiable function in (b, c), with u(b+ ) = 0 and u(c− ) = ∞. For p ∈ R\{0, 1}, 1p + q1 = 1, we set Φ(x) =

[u(x)]p(1−α)−1 (x ∈ (b, c)). [u (x)]p−1

If f (x), an ≥ 0, f ∈ Lp,Φ (b, c), a = {an}∞ n=n0 ∈ lq,Ψ , ||f ||p,Φ > 0, ||a||q,Ψ > 0, then (i) for p > 1, we have the following equivalent inequalities: c ∞ an h(u(x)v(n))f (x)dx b

n=n0

c

=

f (x) b

∞ n=n0

an h(u(x)v(n))dx < k(α)||f ||p,Φ ||a||q,Ψ , (4.40)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

147

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

∞

[Ψ(n)]1−p

(

(4.41)

b

n=n0

and

) p p1 h(u(x)v(n))f (x)dx < k(α)||f ||p,Φ ,

c

c

1−q

[Φ(x)] b

∞

q h(u(x)v(n))an

1q dx

< k(α)||a||q,Ψ ;

(4.42)

n=n0

(ii) for p < 0, we have the equivalent reverses of (4.40), (4.41) and (4.42). (y) Moreover, if vv(y) (> 0) is decreasing in [n0 , ∞) and there exist constants δ < λ1 , L > 0, and δ1 > 0, such that (4.9) is satisﬁed and k(α − δ1 ) ∈ R+ , then the constant factor k(α) in the above inequalities is the best possible. Proof.

(i) For p > 1, putting x = u(t) in two sides of (4.15), we have c ∞ I= an h(u(t)v(n))f (u(t))u (t)dt b

n=n0

=

c

f (u(t))u (t) b

∞

an h(u(t)v(n))dt

n=n0

c

< k(α)

[u(t)]p(1−α)−1 f p (u(t))u (t)dt

p1

||a||q,Ψ .

(4.43)

b

Changing t and f (u(t))u (t) to x and f (x) in (4.43), by simpliﬁcation, we obtain (4.40). On the other hand, setting u(x) = x(x ∈ (0, ∞)) in (4.40), we have (4.15). It follows that (4.40) and (4.15) are equivalent and then both of them are with the same best constant factor k(α). By the same way, we can prove that (4.41) and (4.16) ((4.42) and (4.17)) are equivalent and with the same best constant factor k(α). Since (4.15), (4.16) and (4.17) are equivalent, then it follows that (4.40), (4.41) and (4.42) are equivalent. (ii) For p < 0, in the same way, we have the equivalent reverses of (4.40), (4.41) and (4.42) with the same best constant factor k(α). Example 4.3. Deﬁne an operator T : Lp,Ψ (b, c) → lq,Ψ as follows: For f ∈ Lp,Φ (b, c), there exists a T f ∈ lp,Ψ1−p , satisfying c h(u(x)v(n))f (x)dx, n ≥ n0 , n ∈ N. T f (n) = b

Also we deﬁne an operator T : lq,Ψ → Lp,Φ (b, c) as follows: For a ∈ lq,Ψ , there exists a Ta ∈ Lp,Φ (b, c), satisfying ∞ h(u(x)v(n))an , x ∈ (b, c). Ta(x) = n=n0

Then, by Corollary 4.2, we still have ||T || = ||T|| = k(α).

November 7, 2013

148

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

For examples, (1) setting u(x) = ln x(x ∈ (1, ∞)), v(n) = ln n, n ≥ (y) 1 n0 = 2, then vv(y) = y ln y (> 0) is decreasing for y > 1. (i) If h(ln x ln n) = 1+(ln x1 ln n)λ (0 < α < λ, α ≤ 1), since for x > 0, f (x, y) =

xα (ln y)α−1 1 + (x ln y)λ

is decreasing with respect to y ∈ (1, ∞), then, Condition (i) is satisﬁed. In view of Example 4.1(i), the constant factor in (4.41) and (4.42) (for u(x) = ln x, v(n) = ln n, n0 = 2) is the best possible, we have ||T || = ||T|| = k(α) = (ii) If h(ln x ln n) =

1 (1+ln x ln n)λ

f (x, y) =

π . λ sin π( αλ )

(0 < α < λ, α ≤ 1), since for x > 0,

xα y α−1 (1 + x ln y)λ

is decreasing with respect to y ∈ (1, ∞), then, Condition (i) is satisﬁed. In view of Example 4.1(ii), the constant factor in (4.41) and (4.42) (for u(x) = ln x(x > 1), v(n) = n, n0 = 1) is the best possible, we have (see Yang [169]) ||T || = ||T|| = k(α) = B(α, λ − α). ln(ln x ln n) (iii) If h(ln x ln n) = (ln (0 < α < λ, α ≤ 1), then by the same x ln n)λ −1 way and Example 4.1(iii), we have ( )2 π . ||T || = ||T || = λ sin π( αλ ) 1 (iv) If h(ln x ln n) = sk=1 ak (ln x ln (0 < a1 < · · · < as , λ > n)λ +1 0, 0 < α < s, αλ ≤ 1), then by the same way and Example 4.1(iv), we have

||T || = ||T|| =

π aks−α−1 λ sin(πα) s

s 2

k=1

j=1(j =k)

1 . ak − a j

(v) If h(ln x ln n) = (ln x ln n)λ +2(ln 1x ln n)λ/2 cos β+1 (x > 1, n ∈ N\{1}; λ > 0, 0 < β ≤ π2 , 0 < α < 1, αλ ≤ 1), then, by the same way and Example 4.1(v), we ﬁnd ||T || = ||T|| =

2π sin β(1 − 2α) ∈ R+ . λ sin β sin(2πα)

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

(vi) If h(ln x ln n) = α=

λ 2

1 (c (ln x ln n)λ +b(ln x ln n)λ/2 +c

ws-book9x6

149

√ > 0, 0 ≤ b ≤ 2 c, 0 <

≤ 1), then by the same way and Example 4.1(vi), we have ⎧ π √ , b = 0, ⎪ c ⎨ √ √ 2 2 4 4c−b √ arctan b , 0 0(A = e −Ae 1, β > 1), γ, λ > 0, α = βλ ≤ 1), then, by the same way and Example 4.1(vii), we have ∞ Γ(β) Ak . ||T || = ||T|| = λγ β (2k + 1)β k=0 # " b(ln x ln n)λ +1 (0 ≤ a < b, λ > 0, α = −βλ, 0 < (viii) If h(ln x ln n) = ln a(ln λ x ln n) +1 β < 1), then, by the same way and Example 4.1(vii), we have

(bβ − aβ )π . βλ sin(βπ)

||T || = ||T|| =

(ix) If h(ln x ln n) = arctan (ln x 1ln n)λ (0 < α < λ, α ≤ 1), then by the same way and Example 4.1(ix), we have π . ||T || = ||T|| = 2α cos( πα 2λ ) (x) If h(ln x ln n) = (max{1,ln1 x ln n})λ (0 < α < λ, α ≤ 1), then, by the same way and Example 4.1(x), we have ||T || = ||T|| =

λ . α(λ − α)

(xi) If h(xn) = (min{1, ln x ln n})λ (0 < −α < λ ≤ 1 − α), then, by the same way and Example 4.1(xi), we have ||T || = ||T|| = (xii) If h(xn) =

min{1,ln x ln n} max{1,ln x ln n}

λ

λ . (−α)(λ + α) (|α| < λ ≤ 1 − α(α < 12 ), then by the

same way and Example 4.1(xii), we have ||T || = ||T|| = k(α) = (2) For 0 ≤ γ1 , γ2 ≤ n − γ2 , n ≥ n0 = 1, then,

1 2 , setting u(x) v (y) 1 v(y) = y−γ2 (>

λ2

2λ . − α2

= x − γ1 (x ∈ (γ1 , ∞)), v(n) = 0) is decreasing for y > 12 .

November 7, 2013

150

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

(i) If h((x − γ1 )(n − γ2 )) = λ}), since for x > 0, f (x, y) =

1 (0 1+(x−γ1 )λ (n−γ2 )λ

< α < λ, α ≤ min{1, 2 −

xα (y − γ2 )α−1 1 + xλ (y − γ2 )λ

is strictly convex satisfying fy2 (x, y) > 0 with respect to y ∈ 12 , ∞ , then, Condition (ii) is satisﬁed. In view of Example 4.1(i), the constant factor in (4.41) and (4.42) (for u(x) = x − γ1 , v(n) = n − γ2 ) is the best possible, we have π . ||T || = ||T|| = λ sin( πα ) λ (ii) If h((x − γ1 )(n − γ2 )) = [1+(x−γ11)(n−γ2 )]λ (0 < α < λ, α ≤ 1), by the same way and Example 4.1(ii), we have ||T || = ||T|| = B(α, λ − α). ln[(x−γ1 )(n−γ2 )] (iii) If h((x − γ1 )(n − γ2 )) = (x−γ (0 < α < λ, α ≤ 1), then λ λ 1 ) (n−γ2 ) −1 by the same way and Example 4.1(iii), we have π ]2 . ||T || = ||T|| = [ λ sin π( αλ ) s 1 (0 < a1 < · · · < (iv) h((x − γ1 )(n − γ2 )) = k=1 ak [(x−γ1 )(n−γ λ 2 )] +1 as , λ > 0, 0 < α < s, αλ ≤ 1), then by the same way and Example 4.1(iv), we have s s 2 1 π aks−α−1 . ||T || = ||T|| = λ sin(πα) ak − a j k=1

j=1(j =k)

1 (v) If h((x − γ1 )(n − γ2 )) = [(x−γ1 )(n−γ2 )]λ +2[(x−γ λ/2 cos β+1 (λ > 1 )(n−γ2 )] π 0, 0 < β ≤ 2 , 0 < α < 1, αλ ≤ min{1, 2 − λ}), then, by the same way and Example 4.1(v), we ﬁnd

2π sin β(1 − 2α) ∈ R+ . ||T || = ||T|| = λ sin β sin(2πα) 1 (vi) If h((x − γ1 )(n − γ2 )) = [(x−γ1 )(n−γ2 )]λ +b[(x−γ (c > λ/2 +c 1 )(n−γ2 )] √ λ 2 0, 0 ≤ b ≤ 2 c, 0 < α = 2 ≤ 3 ), then, by the same way and Example 4.1(vi), we have ⎧ π √ , b = 0, ⎪ c ⎨ √ √ 2 4c−b2 4 √ arctan b , 0 < b < 2 c, ||T || = ||T || = 4c−b2 λ⎪ √ ⎩ 2 √ , b = 2 c. c

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

ws-book9x6

151

(vii) If h((x − γ1 )(n − γ2 )) = γ[(x−γ1 )(n−γ2 )]λ 1 −γ[(x−γ1 )(n−γ2 )]λ ( −1 ≤ e −Ae A < 1, β > 0 (A = 1, β > 1), γ, λ > 0, α = βλ ≤ 1), then, by the same way and Example 4.1(vii), we have ||T || = ||T|| =

∞

Ak Γ(β) . β λγ (2k + 1)β k=0

λ

b[(x−γ1 )(n−γ2 )] +1 (viii) If h((x − γ1 )(n − γ2 )) = ln{ a[(x−γ }(0 ≤ a < b, λ > λ 1 )(n−γ2 )] +1 0, α = −βλ, 0 < β < 1), then, by the same way and Example 4.1(viii), we have

||T || = ||T|| =

(bβ − aβ )π . βλ sin(βπ)

1 ]λ (0 < α < λ, α ≤ (ix) If h((x − γ1 )(n − γ2 )) = arctan[ (x−γ1 )(n−γ 2) min{1, 2 − λ}), then, by the same way and Example 4.1(ix), we have π ||T || = ||T|| = . 2α cos( πα ) 2λ (3) For a ≥ 23 , setting u(x) = ln ax(x ∈ a1 , ∞ ), v(n) = ln an, n ≥ n0 =

(y) = y ln1 ay (> 0) is decreasing for y > 32 . 2, then vv(y) (i) If h(ln ax ln an) = 1+(ln ax1 ln an)λ (0 < α < λ, α ≤ min{1, 2 − λ}), since for x > 0,

f (x, y) =

xα 1+

xλ (ln ay)λ

(ln ay)α−1

is strictly convex satisfying fy2 (x, y) > 0 with respect to y ∈ ( 32 , ∞), then, Condition (ii) is satisﬁed. In view of Example 4.1(i), the constant factor in (4.41) and (4.42) (for u(x) = ln ax, v(n) = ln an) is the best possible, we have π . ||T || = ||T|| = λ sin π( αλ ) (ii) If h(ln ax ln an) = (1+ln ax1 ln an)λ (0 < α < λ, α ≤ 1), by the same way and Example 4.1(ii), we have ||T || = ||T|| = B(α, λ − α). ln(ln ax ln an) (iii) If h(ln ax ln an) = (x−γ (0 < α < λ, α ≤ 1), then by the λ λ 1 ) (n−γ2 ) −1 same way and Example 4.1(iii), we have π ||T || = ||T|| = [ ]2 . λ sin( πα ) λ

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

152

Half-Discrete Hilbert-Type Inequalities

s 1 (iv) If h(ln ax ln an) = k=1 ak (ln ax ln (0 < a1 < · · · < as , λ > an)λ +1 0, 0 < α < s, αλ ≤ 1), then, by the same way and Example 4.1(iv), we have ||T || = ||T|| =

π aks−α−1 λ sin(πα) s

s 2

k=1

j=1(j =k)

1 . ak − a j

1 (λ (ln ax ln an)λ +2(ln ax ln an)λ/2 cos β+1

> 0, 0 < β ≤ (v) If h(ln ax ln an) = 0 < α < 1, αλ ≤ min{1, 2 − λ}), then, by the same way and Example 4.1(v), we ﬁnd π , 2

2π sin β(1 − 2α) ||T || = ||T|| = ∈ R+ . λ sin β sin(2πα) 1 (c > 0, 0 ≤ b ≤ (vi) If h(ln ax ln an) = (ln ax ln an)λ +b(ln ax ln an)λ/2 +c √ λ 2 2 c, 0 < α = 2 ≤ 3 ), then, by the same way and Example 4.1(vi), we have ⎧ π √ , b = 0, ⎪ c ⎨ √ √ 2 4 4c−b2 √ arctan b , 0 0 e −Ae (A = 1, β > 1), γ, λ > 0, α = βλ ≤ 1), then, by the same way and Example 4.1(vii), we have ∞ Γ(β) Ak . β λγ (2k + 1)β k=0 : 9 b(ln ax ln an)λ +1 (0 ≤ a < b, λ > 0, α = (viii) If h(ln ax ln an) = ln a(ln λ ax ln an) +1

||T || = ||T|| =

−βλ, 0 < β < 1), then, by the same way and Example 4.1(viii), we have ||T || = ||T|| =

(bβ − aβ )π . βλ sin(βπ)

(ix) If h(ln ax ln an) = arctan[ (ln ax1ln an) ]λ , 0 < α < λ, α ≤ min{1, 2 − λ}, then, by the same way and Example 4.1(ix), we have π . ||T || = ||T|| = 2α cos( πα 2λ ) Note.

For a ≤

1 , 2

setting u(x) = ln(x − a)(x ∈ (a + 1, ∞)), v(n) =

(y) 1 = (y−a) ln(y−a) (> 0) is decreasing for ln(n − a), n ≥ n0 = 2, then vv(y) 3 y > 2 . In this case, still can obtain some similar results of Example 4.15, (1)-(3)(see Yang [137]). By the same way, we still have

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

ws-book9x6

153

Corollary 4.3. Let the assumptions of Theorem 4.2 be fulﬁlled and additionally, let u(x) be a strictly increasing diﬀerentiable function in (b, c), with u(b+ ) = 0 and u(c− ) = ∞. For 0 0, ||a||q,Ψ > 0, then we have the following equivalent inequalities: c ∞ an h(u(x)v(n))f (x)dx b

n=n0

c

f (x)

= b

∞

[Ψ(n)]

c

an h(u(x)v(n))dx > k(α)||f ||p,Φ ||a||q,Ψ , (4.44)

n=n0

1−p

(

c b

n=n0

and

∞

1−q [Φ(x)]

b

) p p1 h(u(x)v(n))f (x)dx > k(α)||f ||p,Φ ,

∞

q h(u(x)v(n))an

(4.45)

1q dx

> k(α)||a||q,Ψ .

(4.46)

n=n0

(y) Moreover, if vv(y) (> 0) is decreasing in [n0 , ∞) and there exist constants η, δ2 > 0, such that θ(x) = O(xη )(0 < x ≤ 1) and k(α + δ2 ) ∈ R+ , then, the constant factor k(α) in the above inequalities is the best possible. In view of Remark 4.1, Corollary 4.2 and Corollary 4.3, we still have

Corollary 4.4. Suppose that α ∈ R, h(t) is a non-negative ﬁnite measur ∞ able function in R+ , k(α) = 0 h(t)tα−1 dt, v(y) is a strict increasing differential function in [n0 , ∞)(n0 ∈ N) with v(n0 ) > 0 and v(∞) = ∞, u(x) is a strict increasing diﬀerential function in (b, c), with u(b+ ) = 0 and (y) (> 0) is decreasing in [n0 , ∞), and there exist constants u(c− ) = ∞, vv(y) η, δ0 > 0, such that for any α ∈ (α − δ0 , α + δ0 ), α) < ∞ (x ∈ R+ ), 0 < k( α)(1 − θ(x)) < α (x) < k(

(4.47)

where θ(x) = O(x ) (x ∈ (0, 1]). (i) For p > 1(p < 0), p1 + 1q = 1, f (x), an ≥ 0, f ∈ Lp,Φ (b, c), a = {an }∞ n=n0 ∈ lq,Ψ , ||f ||p,Φ > 0, ||a||q,Ψ > 0, if p > 1, then, we have the equivalent inequalities (4.40), (4.41) and (4.42), with the best constant factor k(α); if p < 0, then we have the equivalent reverses of (4.40), (4.41) and (4.42), with the same best constant factor k(α). η

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

154

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

(ii) For 0 0, ||a||q,Ψ > 0, then, we have the equivalent n=n0 ∈ lq,Ψ , ||f ||p,Φ inequalities (4.44), (4.45) and (4.46) with the same best constant factor k(α).

∞ For v(x) = u(x) = x(x ∈ (0, ∞)), n0 = 1, k(α) = 0 h(t)tα−1 dt in Corollary 4.4, setting ϕ(x) = xp(1−α)−1 (x ∈ (0, ∞)), ψ(n) = nq(1−α)−1 (n ∈ N), we have ∞ h(xn)xα−1 dx = k(α), n ∈ N, (4.48) ω(α, n) = nα 0

∞

(α, x) = xα

x ∈ R+ ,

h(xn)nα−1 ,

(4.49)

n=1

and the following corollary: Corollary 4.5. Suppose that α ∈ R, h(t) is a non-negative ﬁnite measurable function in R+ , and there exist constants η, δ0 > 0, such that for any α ∈ (α − δ0 , α + δ0 ), 0 < k( α)(1 − θα (x)) < ( α, x) < k( α) < ∞

(x ∈ R+ ),

(4.50)

where θα (x) = O(x ) (x ∈ (0, 1]). η

(i) For p > 1(p < 0), p1 + 1q = 1, f (x), an ≥ 0, f ∈ Lp,ϕ (R+ ), a = {an }∞ n=1 ∈ lq,ψ , ||f ||p,ϕ > 0, ||a||q,ψ > 0, if p > 1, then we have the following equivalent inequalities: ∞ ∞ an h(xn)f (x)dx 0

n=1 ∞

f (x)

= 0

∞

[ψ(n)]1−p

0

∞

an h(xn)dx < k(α)||f ||p,ϕ ||a||q,ψ ,

(4.51)

) p p1 h(xn)f (x)dx < k(α)||f ||p,ϕ ,

(4.52)

n=1

(

∞ 0

n=1

and

∞

[ϕ(x)]1−q

∞

q h(xn)an

1q dx

< k(α)||a||q,ψ ,

(4.53)

n=1

with the best constant factor k(α); if p < 0, then we have the equivalent reverses of (4.51), (4.52) and (4.53), with the same best constant factor k(α).

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

(ii) For 0 0, ||a||q,ψ > 0, n=1 ∈ lq,ψ , ||f ||p,ϕ then, we have the following equivalent inequalities: ∞ ∞ an h(xn)f (x)dx 0

n=1 ∞

f (x)

= 0

∞

[ψ(n)]

∞

an h(xn)dx > k(α)||f ||p,ϕ ||a||q,ψ ,

(4.54)

) p p1 h(xn)f (x)dx > k(α)||f ||p,ϕ,

(4.55)

n=1

1−p

(

∞ 0

n=1

and 0

∞

1−q [ϕ(x)]

∞

q h(xn)an

1q dx

> k(α)||a||q,ψ ,

(4.56)

n=1

with the same best constant factor k(α). Note. (i) If we change the upper limit ∞ of the integral to c > 0, then, the results of Corollary 4.5 is still value for p > 1 and 0 < p < 1. For p < 0, we may refer to Corollary 4.1 for v(n) = n, n0 = 1. (ii) We referred to some reverses with the particular kernels in Yang (see [166], [167], [168]).

4.4 4.4.1

Some Particular Examples Applying Condition (i) and Corollary 4.5

1 Example 4.4. We set h(xy) = A(max{1,xy})λ +B(min{1,xy}) ∈ λ ((x, y) 2 R+ ; 0 0, 0 < α < min{1, λ}). There exists a constant 0 < δ0 < min{α, 1 − α, λ − α}, such that for α ∈ (α − δ0 , α + δ0 ) ⊂ (0, 1), α < λ. For x > 0,

f (x, y) = =

xα y α−1 A(max{1, xy})λ + B(min{1, xy})λ xα y α−1 , 0 < y ≤ x1 , A+B(xy)λ α−1 xα y , A(xy)λ +B

y>

1 x

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

156

Half-Discrete Hilbert-Type Inequalities

is strictly decreasing with respect to y > 0. We ﬁnd ∞ 1 k( α) = tα−1 dt λ + B(min{1, t})λ A(max{1, t}) 0 1 ∞ 1 1 α −1 = tα−1 dt t dt + λ λ At + B 0 A + Bt 1 1 1 1 α−1 (tα−1 + tλ− ) dt = λ A 0 1+ B t A k ∞ 1 1 B α−1 tλk (tα−1 + tλ− ) dt = − A 0 A k=0 k 1 ∞ B 1 α −1 α−1 − = (tλk+ + tλk+λ− ) dt A A 0 k=0 k ∞ 1 1 B 1 ∈ R+ , + = − A A λk + α λk + λ − α k=0

and obtain ∞

nα−1 A(max{1, xn})λ + B(min{1, xn})λ n=1 ∞ y α−1 > xα dy A(max{1, xy})λ + B(min{1, xy})λ 1 = k( α)(1 − θα (x)) > 0,

( α, x) = xα

where

1 xα y α−1 dy θα (x) = > 0. k( α) 0 A(max{1, xy})λ + B(min{1, xy})λ For 0 < x ≤ 1, it follows that 1 xα y α−1 dy 0 < θα (x) = k( α) 0 A + B(xy)λ 1 α−1 y xα xα xη ≤ dy = ≤ , k( α) 0 A Ak( α) α Ak( α) α namely, θα (x) = O(xη )(η = α − δ0 > 0). By Condition (i), it follows that ( α, x) =

∞

xα nα−1 < k( α). A(max{1, xn})λ + B(min{1, xn})λ n=1

Hence, we obtain (4.50). 1 By Corollary 4.5, for h(xn) = A(max{1,xn})λ +B(min{1,xn}) λ (0 0, 0 < α < 1), if p > 1, we have equivalent inequalities (4.51), (4.52)

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

ws-book9x6

157

and (4.53); if p < 0, we have equivalent reverses of (4.51), (4.52) and (4.53); if 0 < p < 1, we have equivalent inequalities (4.54), (4.55) and (4.56). All the inequalities have the same best constant factor: k ∞ λ B 2k + 1 k(α) = − . (4.57) A A (λk + α)(λk + λ − α) k=0

Remark 4.3. For c > 0, since for δ ∈ (α, λ), tδ h(t) → 0 (t → ∞), there exists a constant L > 0, such that 1 1 h(t) = ≤ L (t ∈ [c, ∞)). λ λ A(max{1, t}) + B(min{1, t}) tδ By Corollary 4.1, for p < 0, v(n) = n, n0 = 1, ∞ 1 h(t)tα−1 dt, θα (n) = k(α) cn we still have (4.30), (4.31) and (4.32) with the same best constant factor. λ

(min{1,xy}) ∈ Example 4.5. We set h(xy) = A(max{1,xy}) λ +B(min{1,xy})λ ((x, y) R2+ ; 0 0, −λ < α < min{λ, 1 − λ}). There exists 0 < δ0 < min{λ, λ + α, λ − α, 1 − λ + α}, such that for α ∈ (α − δ0 , α + δ0 ) ⊂ (−λ, λ), α < α + (1 − λ + α) = 1 − λ. For x > 0, (min{1, xy})λxα y α−1 f (x, y) = A(max{1, xy})λ + B(min{1, xy})λ xλ+α yλ+α−1 , 0 < y ≤ x1 , A+B(xy)λ = α−1 xα y , y > x1 A(xy)λ +B is strictly decreasing with respect to y > 0. We ﬁnd ∞ (min{1, t})λ tα−1 dt k( α) = A(max{1, t})λ + B(min{1, t})λ 0 ∞ 1 1 1 λ+ α−1 t dt + = tα−1 dt λ λ+B A + Bt At 0 1 1 1 1 λ+ α−1 α−1 = (t + tλ− ) dt λ A 0 1+ B t A k ∞ 1 1 B α−1 α−1 tλk (tλ+ + tλ− ) dt = − A 0 A k=0 k 1 ∞ 1 B α−1 α−1 (tλk+λ+ + tλk+λ− ) dt = − A A 0 k=0 k−1 ∞ B 1 1 1 − = ∈ R+ , + A A λk + α λk − α k=1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

158

Half-Discrete Hilbert-Type Inequalities

and then we obtain

∞

(min{1, xn})λ nα−1 A(max{1, xn})λ + B(min{1, xn})λ n=1 ∞ (min{1, xy})λ y α−1 > xα dy A(max{1, xy})λ + B(min{1, xy})λ 1

( α, x) = xα

= k( α)(1 − θα (x)) > 0, where

1 (min{1, xy})λ y α−1 dy xα > 0. k( α) 0 A(max{1, xy})λ + B(min{1, xy})λ For 0 < x ≤ 1, it follows that 1 λ λ+ xα x y α−1 dy 0 < θα (x) = k( α) 0 A + B(xy)λ α 1 λ+ y α−1 xλ+ ≤ dy k( α) 0 A θα (x) =

α xη xλ+ ≤ , Ak( α)(λ + α ) Ak( α)(λ + α ) namely, θα (x) = O(xη ) (η = λ + α − δ0 > 0). By Condition (i), it follows that ∞ (min{1, xn})λ nα−1 ( α, x) = xα < k( α). A(max{1, xn})λ + B(min{1, xn})λ n=1

=

Hence, we obtain (4.50). (min{1,xn})λ (0 0, −λ < α < min{λ, 1 − λ}), if p > 1, we have equivalent inequalities (4.51), (4.52) and (4.53); if p < 0, we have equivalent reverses of (4.51), (4.52) and (4.53); if 0 0, since for δ ∈ (α, λ), tδ h(t) → 0(t → ∞), there exists L > 0, such that (min{1, t})λ h(t) = A(max{1, t})λ + B(min{1, t})λ 1 ≤ L δ (t ∈ [c, ∞)). t

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

ws-book9x6

159

By Corollary 4.1, for p < 0, v(n) = n, n0 = 1, ∞ 1 θα (n) = h(t)tα−1 dt, k(α) cn we still have (4.30), (4.31) and (4.32) with the same best constant factor. β

(min{1,xy}) 2 Example 4.6. We set h(xy) = (max{1,xy}) λ+β ((x, y) ∈ R+ ; λ + 2β > 0, −β < α < min{λ + β, 1 − β}). There exists a constant 0 < δ0 < min{β + α, λ + β − α, 1 − β − α}, such that for α ∈ (α − δ0 , α + δ0 ) ⊂ (−β, λ + β), α < 1 − β. For x > 0,

(min{1, xy})β xα y α−1 (max{1, xy})λ+β xα+β y α+β−1 , 0 < y ≤ x1 , = xα−λ−β y α−λ−β−1 , y > x1

f (x, y) =

is strictly decreasing with respect to y > 0. We ﬁnd ∞ (min{t, 1})β α−1 t dt k( α) = (max{t, 1})λ+β 0 ∞ 1 1 α−1 α−1 tβ+ dt + t dt = λ+β t 0 1 1 1 = + β+α λ+β −α λ + 2β = ∈ R+ , (β + α )(λ + β − α ) and then, we have ∞ (min{1, xn})β nα−1 α ( α, x) = x (max{1, xn})λ+β n=1 ∞ (min{1, xy})β y α−1 > xα dy (max{1, xy})λ+β 1 = k( α)(1 − θα (x)) > 0, where

1 xα (min{1, xy})β y α−1 dy > 0. k( α) 0 (max{1, xy})λ+β For 0 < x ≤ 1, it follows that xα+β 1 α+β−1 0 < θα (x) = y dy k( α) 0 θα (x) =

=

xη xα+β ≤ , k( α)( α + β) k( α)( α + β)

November 7, 2013

12:24

160

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

namely, θα (x) = O(xη )(η = α + β − δ0 > 0). By Condition (i), it follows that ( α, x) = xα

∞ (min{1, xn})β nα−1 < k( α). (max{1, xn})λ+β n=1

Hence, we obtain (4.50). (min{1,xn})β By Corollary 4.5 with the Note, for h(xn) = (max{1,xn}) λ+β (λ + 2β > 0, −β < α < min{λ + β, 1 − β}), if p > 1, we have equivalent inequalities (4.51), (4.52) and (4.53); if p < 0, we have equivalent reverses of (4.51), (4.52) and (4.53); if 0 < p < 1, we have equivalent inequalities (4.54), (4.55) and (4.56). All the inequalities have the same best constant factor k(α) =

λ + 2β . (β + α)(λ + β − α)

(4.59)

1 Remark 4.5. (1) In particular, (i) for β = 0, h(xn) = (max{1,xn}) λ (λ > 0, 0 < α < min{λ, 1}); (ii) for λ = −β, h(xn) = (min{1, xn})β (β > 0, −β < min{1,xn} β α < min{0, 1 − β}); (iii) for λ = 0, h(xn) = ( max{1,xn} ) (β > 0, −β < α < min{β, 1 − β}), we imply respectively some results of Example 4.2 (x)-(xii).

(2) For c > 0, since for δ ∈ (α, λ + β), tδ h(t) → 0(t → ∞), there exists a constant L > 0, such that 1 (min{t, 1})β ≤L δ (t ∈ [c, ∞)). h(t) = (max{t, 1})λ+β t By Corollary 4.1, for p < 0, v(n) = n, n0 = 1, ∞ 1 h(t)tα−1 dt, θα (n) = k(α) cn we still have (4.30), (4.31) and (4.32) with the same best constant factor. 4.4.2

Applying Condition (iii) and Corollary 4.2

Example 4.7. We set h(xy) = 1+x1λ yλ ((x, y) ∈ R2+ ; λ > 0, 0 < α < min{2, λ}). There exists a constant 0 < δ0 < min{α, λ − α, 2 − α}, such that for α ∈ (α − δ0 , α + δ0 ) ⊂ (0, 2), α < λ. It follows ∞ α−1 α/λ)−1 1 ∞ u( t du dt = k( α) = λ 1+t λ 0 u+1 0 π ∈ R+ . = λ sin(π α /λ)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

ws-book9x6

161

For γ < 1, it is obvious that ∞ xα−1 dx α ω( α, n) = (n − γ) = k( α) (n ∈ N). 1 + xλ (n − γ)λ 0 In the following, we show that, for 3 1 α + 2(λ + 1) α−α 2 ) < 1, γ ≤ γ( α) = 1 − ( 4 it follows that 0 < k( α)(1 − θα (x)) ∞ xα (n − γ)α−1 < k( α) (x > 0), (4.60) < ( α, x) = 1 + xλ (n − γ)λ n=1 where (4.61) 0 < θα (x) = O(xη )(0 < x ≤ 1; η = α − δ0 > 0). α−1

(y−γ) (1) Setting g(x, y) = 1+x λ (y−γ)λ (x > 0, y > γ), we prove that ) ( 1 ∞ 1 g(x, y)dy − g(x, 1) − P1 (y)gy (x, y)dy > 0. (4.62) R(x) = xα 2 γ 1 In fact, we obtain 1 1 (1−γ)x α−1 1 (y − γ)α−1 dy t dt g(x, y)dy = = λ λ α x 0 1 + tλ γ γ 1 + x (y − γ)

1 (1 − γ)α−1 1 − g(x, 1) = − , 2 2 1 + xλ (1 − γ)λ and

(4.63)

λxλ (y − γ)α+λ−2 ( α − 1)(y − γ)α−2 + λ λ 2 [1 + x (y − γ) ] 1 + xλ (y − γ)λ

gy (x, y) = −

λ[1 + xλ (y − γ)λ − 1](y − γ)α−2 [1 + xλ (y − γ)λ ]2

=−

+

( α − 1)(y − γ)α−2 1 + xλ (y − γ)λ

(λ − α + 1)(y − γ)α−2 λ(y − γ)α−2 + . 1 + xλ (y − γ)λ [1 + xλ (y − γ)λ ]2 By (2.37), for λ > 0, α < 2, =−

−

∞

1

=

1

P1 (y)gy (x, y)dy

∞

P1 (y) −

(λ − α + 1)(y − γ)α−2 dy 1 + xλ (y − γ)λ ∞ 1

P1 (y)

λ(y − γ)α−2 dy [1 + xλ (y − γ)λ ]2

+ 1)(1 − γ)α−2 1 (λ − α >− + 0. 8 1 + xλ (1 − γ)λ

(4.64)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

162

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Then, by (4.62), (4.63) and (4.64), setting 1−γ 1 A=[ + (λ − α + 1)](1 − γ)α−2 , 2 8 we have (1−γ)x α−1 Axα t dt − . R(x) > h(x) = λ 1+t 1 + xλ (1 − γ)λ 0 3 Since for γ ≤ γ( α), 1 − γ ≥ 14 [ α + 2(λ + 1) α−α 2 ], we have B(γ) = (1 − γ)2 −

(4.65)

α α (1 − γ) − (λ − α + 1) ≥ 0, 2 8

then, we ﬁnd h (x) =

(1 − γ)α xα−1 λ(1 − γ)λ Axα+λ−1 + 1 + xλ (1 − γ)λ [1 + xλ (1 − γ)λ ]2 −

A αxα−1 1 + xλ (1 − γ)λ

= [(1 − γ)α − A α] + > B(γ)

xα−1 1 + xλ (1 − γ)λ

λ(1 − γ)λ Axα+λ−1 [1 + xλ (1 − γ)λ ]2

(1 − γ)α−2 xα−1 >0 1 + xλ (1 − γ)λ

(A > 0).

By (4.65), it follows that R(x) > h(0) = 0(x > 0). Hence, (4.62) holds. (2) In view of f (x, y) = xα g(x, y), by (4.4) and Condition (iii), it follows that ∞ xα (n − γ)α−1 ( α, x) = 1 + xλ (n − γ)λ n=1 = k( α) − R(x) < k( α)(x > 0). By (4.64), we still obtain ∞ 1 λ(1 − γ)α−2 P1 (y)gy (x, y)dy < , − 8 [1 + xλ (1 − γ)λ ]2 1 and then, by (4.62), it follows that (1−γ)x α−1 xα (1 − γ)α−1 t R(x) < dt − 1 + tλ 2[1 + xλ (1 − γ)λ ] 0 +

λxα (1 − γ)α−2 . 8[1 + xλ (1 − γ)λ ]2

(4.66)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

Setting θα (x) =

1 [R(x) k( α)

+

xα (1−γ)α−1 ](> 0), 1+xλ (1−γ)λ α

ws-book9x6

163

we ﬁnd

x (1 − γ)α−1 1 + xλ (1 − γ)λ = k( α)(1 − θα (x)) > 0.

( α, x) > ( α, x) −

For η = α − δ0 > 0, 0 < x ≤ 1, by (4.66), we obtain 1 1 0 < η θα (x) ≤ α θα (x) x x ) ( (1 − γ)α−1 1 1 R(x) + = k( α) xα 1 + xλ (1 − γ)λ (1−γ)x α−1 (1 − γ)α−1 1 t dt 1 − < k( α) xα 0 1 + tλ 2[1 + xλ (1 − γ)λ ] λ(1 − γ)α−2 (1 − γ)α−1 + + 8[1 + xλ (1 − γ)λ ]2 1 + xλ (1 − γ)λ → constant as x → 0, namely, θα (x) = O(xη )(0 < x ≤ 1; η = α − δ0 > 0). Hence, we show that (4.60) and (4.61) are valid. Setting u(x) = x − μ(x > μ), by Corollary 4.2 for 3 1 γ ≤ 1 − [α + 2(λ + 1)α − α2 ], 4 1 h(u(x)v(n)) = 1 + (x − μ)λ (n − γ)λ (x > βμ, n ∈ N; λ > 0, 0 < α < min{2, λ}), if p > 1, we have equivalent inequalities (4.40), (4.41) and (4.42); if p < 0, we have equivalent reverses of (4.40), (4.41) and (4.42); if 0 < p < 1, we have equivalent inequalities (4.44), (4.45) and (4.46). All the inequalities have the same best constant factor π k(α) = . (4.67) λ sin(πα/λ) β

(min{1,xy}) 2 Example 4.8. We set h(xy) = (max{1,xy}) λ+β ((x, y) ∈ R+ ; 2β + λ > 0, −β < α < min{λ + β, 2 − β}). There exists 0 < δ0 < min{α + β, λ + β − α, 2 − β − α}, such that for α ∈ (α − δ0 , α + δ0 ) ⊂ (−β, λ + β), α < 2 − β. It follows that ∞ (min{1, t})β tα−1 dt k( α) = (max{1, t})λ+β 0 λ + 2β ∈ R+ . = (β + α )(λ + β − α )

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

164

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

For 0 ≤ γ < 1, it is obvious that ∞ (min{1, x(n − γ)})β xα−1 ω( α, n) = (n − γ)α dx (max{1, x(n − γ)})λ+β 0 = k( α)(n ∈ N). In the following, we show that, for 3 1 γ ≤ γ( α) = 1 − [β + α )(1 + λ + 2β) − (β + α )2 ], + 2(β + α 4 0 < k( α)(1 − θα (x)) < ( α, x) ∞ xα (min{1, x(n − γ)})β (n − γ)α−1 < k( α) (x > 0), (4.68) = λ+β (max{1, x(n − γ)}) n=1 where 0 < θα (x) = O(xη ) (0 < x ≤ 1; η = α + β − δ0 > 0).

(4.69)

(1) Setting g(x, y) =

(min{1, x(y − γ)})β (y − γ)α−1 (max{1, x(y − γ)})λ+β

(x > 0, y > γ),

in the following, we prove that, for γ ≤ γ( α), ) ( 1 ∞ 1 g(x, y)dy − g(x, 1) − P1 (y)gy (x, y)dy > 0. (4.70) R(x) = xα 2 γ 1 In fact, we obtain g(x, y) =

γ < y < x1 + γ, xβ (y − γ)α+β−1 , α −λ−β−1 x (y − γ) , y ≥ x1 + γ,

−gy (x, y) =

−λ−β

h1 (x, y), γ < y < x1 + γ, h2 (x, y), y ≥ x1 + γ,

− β)xβ (y − γ)α+β−2 , h1 (x, y) = (1 − α h2 (x, y) = (1 − α + λ + β)x−λ−β (y − γ)α−λ−β−2 . For 0 < x1 ≤ 1 − γ, y ≥ 1 ≥ x1 + γ, by equation (2.23) in Chapter 2, we have ∞ ∞ P1 (y)gy (x, y)dy = P1 (y)h2 (x, y)dy − 1

1

1 (1 − α + λ + β)(1 − γ)α−λ−β−2 12xλ+β 1 > − λ+β (1 − α + λ + β)(1 − γ)α−λ−β−2 ; (4.71) 8x

>−

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

165

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

for

1 x

> 1 − γ, by (2.47) (ε0 ∈ (0, 1), ε1 ∈ [0, 1]), we ﬁnd ∞ P1 (y)gy (x, y)dy − 1 ε0 1 1 −h1 (x, 1) + h1 (x, + γ) − h2 (x, + γ) = 8 x x 1 1 ε1 − h1 (x, + γ) − h2 (x, + γ) 8 x x ) ( ε1 2β + λ 2β + λ ε0 (1 − α − β)xβ (1 − γ)α+β−2 + α−2 + =− 8 x 8 xα−2 1 2β + λ − β)xβ (1 − γ)α+β−2 − > − (1 − α . (4.72) 8 8xα−2

We still ﬁnd ⎧ ⎨

1 − g(x, 1) = ⎩ 2 For 0 <

1 x

− 21 xβ (1 − γ)α+β−1 ,

1 x

− 12 x−λ−β (1 − γ)α−λ−β−1 , 0 <

> 1 − γ, 1 x

≤ 1 − γ.

≤ 1 − γ, we have

1

g(x, y)dy

γ 1 x +γ

=

xβ (y − γ)α+β−1 dy +

γ

= for

1 x

1 1 γ+ x

(y − γ)α−λ−β−1 dy xλ+β

1 (1 − γ)α−λ−β 1 + − ; (β + α )xα (β + λ − α )xα (β + λ − α )xλ+β

(4.73)

> 1 − γ,

1

1

g(x, y)dy = γ

xβ (y − γ)α+β−1 dy =

γ

(i) For 0 <

1 x

≤ 1 − γ,

1−γ ≥

(1 − γ)α+β xβ . α +β

(4.74)

γ ≤ γ( α),

3 1 [β + α + 2(β + α )(1 + λ + 2β) − (β + α )2 ], 4

it is obvious that A(γ) = (1 − γ)2 −

1 β+α (1 − γ) − (β + α )(1 + λ + β − α ) ≥ 0, 2 8

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

166

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and then, by (4.70), (4.71), (4.72) and (4.73), in view of 0 < follows that R(x) >

1 x

≤ 1 − γ, it

1 1 (1 − γ)α−λ−β + − α β+α β+λ−α (β + λ − α )xλ+β−

(1 − γ)α−λ−β−1 (1 − α + λ + β)(1 − γ)α−λ−β−2 − λ+β− α α 2x 8xλ+β− α −λ−β α 1 (1 − γ) 1 (1 − γ)λ+β− + − > β+α β+λ−α β+λ−α 1 λ+β− α α −λ−β−1 − (1 − γ) (1 − γ) 2 1 α + λ + β)(1 − γ)α−λ−β−2 (1 − γ)λ+β− − (1 − α 8 A(γ) = ≥ 0; (β + α )(1 − γ)2 −

(ii) for ﬁnd

1 x

α > 1 − γ, γ ≤ γ( α), in view of x2 < xβ+ (1 − γ)α+β−2 , we still

α (1 − γ)α+β xβ+ 1 α (1 − γ)α+β−1 R(x) > − xβ+ α +β 2 1 α − β)xβ+ − (1 − α (1 − γ)α+β−2 8 2β + λ β+ − x α (1 − γ)α+β−2 8 α xβ+ (1 − γ)α+β−2 ≥ 0. = A(γ) α +β

Hence, (4.70) follows for x > 0. (2) By (4.4) and Condition (iii), it follows that ( α, x) = k( α) − R(x) < k( α)

(x > 0).

Setting θα (x) =

) ( 1 xα (min{1, x(1 − γ)})β α −1 (1 − γ) > 0, R(x) + k( α) (max{1, x(1 − γ)})λ+β

we obtain xα (min{1, x(1 − γ)})β (1 − γ)α−1 (max{1, x(1 − γ)})λ+β = k( α)(1 − θα (x)) > 0.

( α, x) > ( α, x) −

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

A Half-Discrete Hilbert-Type Inequality with a Non-Homogeneous Kernel

For 0 < x ≤ 1, it is obvious that lim

1

+β x→0+ xα

α

1 θα (x) xα+β β

ws-book9x6

167

is continuous in (0, 1]. Since

x (min{1, x(1 − γ)}) (1 − γ)α−1 = (1 − γ)α+β−1 , (max{1, x(1 − γ)})λ+β 1 xα g(x, y)dy lim +β x→0+ xα γ = lim+ x→0

(1 − γ)α+β xβ (1 − γ)α+β = , β x ( α + β) α +β

and

∞ xα (− P1 (y)gy (x, y)dy) +β x→0+ xα 1 ( ) 2β + λ 1 ε0 (1 − α − β)xβ (1 − γ)α+β−2 + α−2 = lim+ β − 8 x x→0 x ε1 (2β + λ) + 8xα−2 ε0 = − (1 − α − β)(1 − γ)α+β−2 , 8 then, by (4.70), we ﬁnd 1 1 0 < η θα (x) ≤ α+β θα (x) → constant, as x → 0+ , x x namely, θα (x) = O(xη ) (0 < x ≤ 1; η = α + β − δ0 > 0). Hence, we show that (4.68) and (4.69) are valid. Setting u(x) = x − μ(x > μ), by Corollary 4.2, for # 3 1" γ ≤ 1− β + α + 2(β + α)(1 + λ + 2β) − (β + α)2 , 4 (min{1, (x − μ)(n − γ)})β h(u(x)v(n)) = (max{1, (x − γ)(n − γ)})λ+β (x > μ, n ∈ N; 2β + λ > 0, −β < α < min{λ + β, 2 − β}), if p > 1, we have equivalent inequalities (4.40), (4.41) and (4.42); if p < 0, we have equivalent reverses of (4.40), (4.41) and (4.42); if 0 < p < 1, we have equivalent inequalities (4.44), (4.45) and (4.46). All the inequalities have the same best constant factor λ + 2β . (4.75) k(α) = (β + α)(λ + β − α) lim

Remark 4.6. If we add the condition 2β + λ ≤ 1 in the above example, then it follows that γ( α) > 0. Setting y = β + α ∈ (0, 2), then, γ( α) > 0 is equivalent to G(y) = y 2 − (5 + λ + 2β)y + 8 > 0.

November 7, 2013

168

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

Since G (y) = 2y−(5+λ+2β) is increasing with G (2) = 4−(5+λ+2β) < 0, and G (y) < 0(y ∈ [0, 2]), then, we ﬁnd G(y) > G(2) = 2[1 − (λ + 2β)] ≥ 0. In this case, for γ = 0 (< γ( α)), we can get some results of Example 4.7.

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Chapter 5

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

“... in a subject (inequalities) like this, which has applications in every part of mathematics but never been developed systematically.”

G. H. Hardy

“A great discovery solves a great problem but there is a grain of discovery in the solution of any problem. Your problem may be modest; but if it challenges your curiosity and brings into play your incentive faculties, and if you solve it by your own means, you may experience the tension and enjoy the triumph of discovery.”

George Polya

5.1

Introduction

This chapter deals with two kinds of multi-dimensional half-discrete Hilbert-type inequalities using the way of weight functions and techniques of real analysis. These inequalities are extensions of the two-dimensional cases studied in Chapters 3 and 4. The best possible constant factors are proved. Included are equivalent forms, the operator expressions, the reverses and many particular examples. 169

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

170

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

5.2

Some Preliminary Results and Lemmas

5.2.1

Some Related Lemmas

Lemma 5.1. (see Wang and Guo [73]) If m ∈ N, α, M > 0, Ψ(u) is a non-negative measurable function in (0, 1], and m m α α , DM = x ∈ R+ ; xi ≤ M i=1

then, we have m 1 xi α M m Γm α1 m ··· Ψ Ψ(u)u α −1 du. dx1 · · · dxm = mΓ m M α DM 0 α i=1 (5.1) Lemma 5.2. Suppose that m, s, n(0) ∈ N, α, β > 0, v(t) is an increasing diﬀerentiable function in [n(0) , ∞) with v(n(0) ) > 0 and v (t) is decreasing in [n(0) , ∞). We set α1 m |xk |α (x = (x1 , · · · , xm ) ∈ Rm ), ||x||α = ||v(y)||β =

k=1 s

β1 (y = (y1 , · · · , ys ) ∈ [n(0) , ∞)s ),

β

(v(yk ))

k=1

where s

[n

(0)

(0) (0) [n , ∞) = , ∞) × · · · × [n , ∞) . s

Then, for ε > 0, we have J(ε) =

1 α , = m−1 Γ m εα {x∈Rm ;||x|| ≥1} α α + Γm α1 −m+ε , J(ε) = ||x||α dx = εαm−1 Γ m {x∈Rm α + ;||x||α ≤1} Γm

−m−ε ||x||α dx

and H(ε) =

n∈Ns (0) n

||v(n)||−s−ε β

s 2 k=1

v (nk ) =

Γs

(5.2) (5.3)

1 β

εβ s−1 Γ( βs )

+ O(1),

(5.4)

where ni ∈ Nn(0) = {n(0) , n(0) + 1, · · · } (i = 1, 2, · · · , s), n = (n1 , n2, · · · , ns ).

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

Proof.

For M > 1, setting Ψ(u) as follows: Ψ(u) = 0(u ∈ (0, M −α )); 1 Ψ(u) = (u ∈ [M −α, 1]), (M u1/α )m+ε

by (5.1), since ||x||α ≥ 1 means that m xi ( )α ≥ M −α , M i=1

we ﬁnd

m xi Ψ ( )α dx1 · · · dxm J(ε) = lim ··· M→∞ M DM i=1 M m Γm ( α1 ) 1 m = lim Ψ(u)u α −1 du m M→∞ αm Γ( ) 0 α m m 1 1 M Γ (α) 1 m = lim u α −1 du m 1/α )m+ε M→∞ αm Γ( ) −α (M u M α 1 Γm ( α1 ) −ε 1 = m m lim u α −1 du ε M→∞ α Γ( α ) M M −α Γm ( α1 ) α 1 = m m lim 1− ε , α Γ( α ) M→∞ ε M

and then (5.2) is valid. In view of (5.1) (for M = 1), it follows that m α1 (−m+ε) α J(ε) = xi dx {x∈Rm + ;||x||α ≤1}

Γm ( 1 ) = m αm α Γ( α )

1

u

i=1

1 α (−m+ε)

n

u α −1 du =

0

Γm ( α1 ) . εαm−1 Γ( m ) α

Hence, (5.3) is valid. By the decreasing property of [v(t)]−1 and v (t), we have

H(ε) = c0 +

||v(n)||−s−ε β

n∈Ns (0) n +1

≤ c0 +

∞

n(0)

···

s 2

v (nk )

k=1 ∞ n(0)

||v(y)||−s−ε β

s 2 k=1

v (yk )dy1 · · · dys ,

ws-book9x6

171

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

172

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

= c0 +

∞

···

v(n(0) )

= c0 + c 1 +

∞

1

Γs ( β1 )

(uk = v(yk )(k = 1, · · · , s))

∞ v(n(0) )

···

∞

1

||u||−s−ε du1 · · · dus β

||u||−s−ε du1 · · · dus β

+ c0 + c 1 , εβ s−1 Γ( βs ) ∞ ∞ s 2 ··· ||v(y)||−s−ε v (yk )dy1 · · · dys , H(ε) ≥ β =

n(0)

n(0)

∞

=

v(n(0) )

=

···

k=1 ∞

v(n(0) )

Γs ( β1 )

(uk = v(yk )(k = 1, · · · , s)) ||u||−s−ε du1 · · · dus β

+ c1 ,

ε β s−1 Γ( βs )

and then, we obtain (5.4).

5.2.2

Some Results about the Weight Functions

Definition 5.1. Suppose that m, s, n(0) ∈ N, α, β > 0, λ1 , λ2 ∈ R, H(t, u) 2 , H(t, u)uλ2 −s is deis a non-negative ﬁnite measurable function in R+ creasing with respect to u ∈ R+ and strictly decreasing in an interval I ⊂ (n(0) , ∞), v (t) > 0, v (t) ≤ 0 (t ∈ (n(0) − 1, ∞)) with v(n(0) − 1) ≥ 0. Deﬁne two weight functions ω(n) and (x) as follows: ω(n) = (x) =

Rm +

H(||x||α , ||v(n)||β )

H(||x||α , ||v(n)||β )

n∈Ns (0) n

||v(n)||λβ2 1 ||x||m−λ α

dx (n ∈ Nsn(0) ), (5.5)

s 2 ||x||λα1 v (nk ) (x ∈ Rm + ). (5.6) 2 ||v(n)||s−λ β k=1

If the expression

1

lim M λ2

M→∞

0

1

H(||x||α , M v β )v

λ2 β

−1

dv = a ∈ R+ ,

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

173

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

then applying (5.1), we get the following results: m xk α 1 H M [ ( ) ] α , ||v(n)||β ω(n) = lim M→∞ D M M k=1

×

M m Γm ( α1 ) m M→∞ αm Γ( ) α

= lim

0

||v(n)||λβ2

m

dx

1 α α m−λ1 k=1 (xk /M ) ] } λ 2 1 ||v(n)||β 1 m H(M u α , ||v(n)||β ) u α −1 du 1 (M u α )m−λ1

{M [

1 λ1 Γm ( 1 ) 1 = m αm ||v(n)||λβ2 lim M λ1 H(M u α , ||v(n)||β )u α −1 du; (5.7) M→∞ α Γ( α ) 0 ∞ ∞ s 2 ||x||λα1 ··· H(||x||α , ||v(y)||β ) v (yk )dy1 · · · dys , (x) < 2 ||v(y)||s−λ n(0) −1 n(0) −1 β k=1 = ≤

v(n(0) −1)

Rs+

(uk = v(yk )(k = 1, · · · , s))

∞

···

∞ v(n(0) −1)

H(||x||α , ||u||β )

||x||λα1

H(||x||α , ||u||β )

du 2 ||u||s−λ β 1 λ2 1 λ2 lim M H(||x||α , M v β )v β −1 dv,

Γs ( β1 ) ||x||λα1 = M→∞ 0 β s Γ βs ∞ ∞ ··· H(||x||α , ||v(y)||β ) (x) >

= =

||x||λα1 du1 · · · dus 2 ||u||s−λ β

n(0)

n(0)

∞

s 2 ||x||λα1 v (yk )dy1 · · · dys , 2 ||v(y)||s−λ β k=1

(uk = v(yk )(k = 1, · · · , s)) ∞

||x||λα1 ··· H(||x||α , ||u||β ) 2 ||u||s−λ v(n(0) ) v(n(0) ) β ||x||λα1 H(||x||α , ||u||β ) du 2 ||u||s−λ Rs+ β

du1 · · · dus

−δ0

=

(5.8)

{u∈Rs+ ;0<||u||α ≤1}

Γs ( β1 ) ||x||λα1 s β Γ βs

H(||x||α , ||u||β )

(

lim M λ2

M→∞

−δ0

0

1

0

||x||λα1 du 2 ||u||s−λ β

1

1

H(||x||α , M v β )v 1

H(||x||α , v β )v

λ2 β

−1

λ2 β

−1

dv

) dv

(0 < δ0 ≤ 1). (5.9)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

174

Half-Discrete Hilbert-Type Inequalities

In particular, for β = 1, if ∂ (H(t, u)uλ2 −s ) < 0, ∂u

∂2 (H(t, u)uλ2 −s ) > 0 ∂u2

(u ∈ R+ ),

and v (t) > 0, v (t) ≤ 0, v (t) ≥ 0 (t ∈ (n(0) − 12 , ∞)), with v(n(0) − 12 ) ≥ 0, then we can ﬁnd that ∂ ||v(y)||1 > 0, ∂yi

∂2 ||v(y)||1 ≤ 0, ∂yi2

and H(||x||α , ||v(y)||1 ) =

s 2 ||x||λα1 v (yk ) 2 ||v(y)||s−λ 1 k=1

s H(||x||α , v(y1 ) + · · · + v(ys ))||x||λα1 2 v (yk ) (v(y1 ) + · · · + v(ys ))s−λ2 k=1

is strictly decreasing and strict convex with respect to any yi ∈ (n(0) − 1 2 , ∞)(i = 1, · · · , s). By Hermite-Hadamard’s inequality (see Kuang [47]), it follows that

1 (x) = <

H(||x||α , ||v(n)||1 )

n∈Ns (0)

∞

n(0) − 12

···

n

∞ n(0) − 12

s 2 ||x||λα1 v (nk ) 2 ||v(n)||s−λ 1 k=1

H(||x||α , ||v(y)||1 )

s 2 ||x||λα1 v (yk )dy1 · · · dys , 2 ||v(y)||s−λ 1 k=1

(uk = v(yk )(k = 1, · · · , s)) ||x||λα1 = ··· H(||x||α , ||u||1 ) du1 · · · dus 2 ||u||s−λ v(n(0) − 12 ) v(n(0) − 12 ) 1 ||x||λα1 ≤ H(||x||α , ||u||1 ) du 2 ||u||s−λ Rs+ 1 1 ||x||λα1 λ2 lim M H(||x||α , M v)v λ2 −1 dv, (5.10) = (s − 1)! M→∞ 0 1 (x) >

∞

∞

n(0)

···

∞

n(0)

∞

H(||x||α , ||v(y)||1 )

||x||λα1 2 ||v(y)||s−λ 1

s 2

v (yk )dy1 · · · dys ,

k=1

(uk = v(yk )(k = 1, · · · , s))

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

175

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

∞

= =

v(n(0) )

Rs+

···

∞ v(n(0) )

H(||x||α , ||u||1 )

H(||x||α , ||u||1 ) −δ0

=

||x||λα1 (s − 1)!

||x||λα1 2 ||u||s−λ 1

M→∞

du

{u∈Rs+ ;0<||u||α ≤1}

( lim M

||x||λα1 du1 · · · dus 2 ||u||s−λ 1

1

λ2

−δ0

1

0

||x||λα1 du 2 ||u||s−λ 1

H(||x||α , M v)v λ2 −1 dv

H(||x||α , v)v

0

H(||x||α , ||u||1 )

λ2 −1

) dv (0 < δ0 ≤ 1).

(5.11)

Two Preliminary Inequalities

5.2.3

Lemma 5.3. Let the assumptions of Deﬁnition 5.1 be fulﬁlled and additionally, let p ∈ R\{0, 1}, p1 + 1q = 1, a(n) = a(n1 , · · · , ns ) ≥ 0(n ∈ Nsn(0) ), f (x) = f (x1,··· , xm ) be a non-negative measurable function in Rm + . Then (i) for p > 1, we have the following inequalities: ⎧ p ⎫p1 s ⎨ ||v(n)||pλ2 −s 2 ⎬ β v (n ) H(||x|| , ||v(n)|| )f (x)dx J1 = k α β ⎩ ⎭ [ω(n)]p−1 Rm s + n∈N

≤

k=1

n(0)

p1

p(m−λ1 )−m p (x)||x||α f (x)dx

Rm +

,

(5.12)

and

J2 =

≤

⎧ ⎨ ⎩

Rm +

1 −m ||x||qλ α [(x)]q−1

⎧ ⎨ ⎩

n∈Ns (0) n

⎛ ⎝

n∈Ns (0)

⎫ 1q ⎬ H(||x||α , ||v(n)||β )a(n)⎠ dx ⎭ ⎞q

n

q(s−λ2 )−s

ω(n)||v(n)||β

s 2

k=1

1−q v (nk )

⎫ 1q ⎬ aq (n) ; (5.13) ⎭

(ii) for p < 0, or 0 < p < 1, we have the reverses of (5.12) and (5.13).

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

176

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Proof.

(i) For p > 1, by H¨ older’s inequality (see Kuang [47]), we have p

Rm +

H(||x||α , ||v(n)||β )f (x)dx

⎧ ⎨

⎡

⎤ p1 s (m−λ1 )/q 2 ||x|| α = H(||x||α , ||v(n)||β ) ⎣ v (nk ) f (x)⎦ (s−λ2 )/p ⎩ Rm ||v(n)|| + k=1 β ⎡ ⎤ ⎫p −1 s (s−λ )/p p ⎬ 2 ||v(n)||β 2 ⎦ dx ×⎣ v (nk ) (m−λ1 )/q ⎭ ||x||α k=1 s (m−λ1 )(p−1) 2 ||x||α ≤ H(||x||α , ||v(n)||β ) v (nk )f p (x)dx 2 ||v(n)||s−λ Rm β + k=1 ⎧ s 1−q ⎫p−1 (s−λ )(q−1) 2 ⎨ ⎬ 2 ||v(n)||β × H(||x||α , ||v(n)||β ) v (n ) dx k m−λ ⎩ Rm ⎭ ||x||α 1 +

2 = [ω(n)]p−1 ||v(n)||s−pλ β

v (nk )

k=1

×

s 2

k=1

−1

Rm +

H(||x||α , ||v(n)||β )

(m−λ )(p−1)

1 ||x||α 2 ||v(n)||s−λ β

s 2

v (nk )f p (x)dx.

k=1

Then by the Lebesgue term by term integration theorem (see Kuang [49]), it follows that ⎧ ⎨ H(||x||α , ||v(n)||β ) J1 ≤ ⎩ Rm s +

n∈N

n(0)

⎧ ⎨ =

⎩

=

⎫ p1

s (m−λ1 )(p−1) 2 ⎬ ||x||α p × v (n )f (x)dx k 2 ⎭ ||v(n)||s−λ β k=1

Rm + n∈Ns (0) n

Rm +

⎫1

s (m−λ1 )(p−1) 2 ⎬p ||x||α p H(||x||α , ||v(n)||β ) v (nk )f (x)dx 2 ⎭ ||v(n)||s−λ β k=1

p1

p(m−λ1 )−m p (x)||x||α f (x)dx

and hence, we obtain inequality (5.12).

,

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

177

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

Still by H¨older’s inequality, we have ⎛ ⎝

⎞q

H(||x||α , ||v(n)||β )a(n)⎠

n∈Ns (0) n

⎧ ⎨

s p1 ⎤ (m−λ1 )/q 2 ||x|| α = H(||x||α , ||v(n)||β ) ⎣ v (nk ) ⎦ (s−λ2 )/p ⎩ ||v(n)|| s n∈N (0) k=1 β n ⎡ ⎤⎫q −1 (s−λ )/p p s ⎬ 2 ||v(n)||β 2 ⎦ ×⎣ v (n ) a(n) k (m−λ1 )/q ⎭ ||x||α k=1 ⎧ ⎫q−1 s (m−λ1 )(p−1) 2 ⎨ ⎬ ||x||α ≤ H(||x||α , ||v(n)||β ) v (n ) k 2 ⎩ ⎭ ||v(n)||s−λ s β ⎡

n∈N

k=1

n(0)

×

(s−λ2 )(q−1)

H(||x||α , ||v(n)||β )

||v(n)||β

n∈Ns (0) n

=

[ω(x)]q−1

1 −m ||x||qλ α n∈Ns (0)

1 ||x||m−λ α

1−q

s 2

aq (n)

v (nk )

k=1

H(||x||α , ||v(n)||β )

n

(s−λ2 )(q−1)

×

||v(n)||β

1 ||x||m−λ α

s 2

1−q v (nk )

aq (n).

k=1

Then, by the Lebesgue term by term integration theorem, it follows that J2 ≤

⎧ ⎨ ⎩

Rm + n∈Ns (0)

H(||x||α , ||v(n)||β )

n

(s−λ2 )(q−1)

×

=

⎧ ⎨ ⎩

n∈Ns (0) n

||v(n)||β

1 ||x||m−λ α

1−q

s 2

v (nk )

k=1

⎫ q1 ⎬ aq (n)dx ⎭

(s−λ2 )(q−1)

Rm +

H(||x||α , ||v(n)||β )

||v(n)||β

1 ||x||m−λ α

×

s 2

k=1

1−q v (nk )

dx ⎫1 ⎬q q a (n) ⎭

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

178

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

=

⎧ ⎨

q(s−λ2 )−s

ω(n)||v(n)||β

⎩

n∈Ns (0)

s 2

1−q v (nk )

k=1

n

⎫1 ⎬q aq (n) , ⎭

and we have proved inequality (5.13). (ii) For p < 0, or 0 < p < 1, by the reverse H¨ older’s inequality with weight and in the same way, we obtain the reverses of (5.12) and (5.13).

5.3

Some Inequalities Related to a General Homogeneous Kernel

5.3.1

Several Lemmas

As the assumptions of Deﬁnition 5.1, if λ ∈ R, kλ (t, u) is a ﬁnite homogeneous function of degree −λ in R2+ , H(t, u) = kλ (t, u), λ1 + λ2 = λ, then (5.5) and (5.6) reduce to the following weight functions: ω(λ2 , n) = (λ1 , x) =

Rm +

kλ (||x||α , ||v(n)||β )

||v(n)||λβ2 1 ||x||m−λ α

kλ (||x||α , ||v(n)||β )

n∈Ns (0) n

dx(n ∈ Nsn(0) ),

(5.14)

s 2 ||x||λα1 v (nk ), (x ∈ Rm + ). 2 ||v(n)||s−λ β k=1

(5.15) For k(λ1 ) =

ω(λ2 , n) =

∞ 0

kλ (t, 1)tλ1 −1 dt ∈ R+ , by (5.7), (5.8) and (5.9), we ﬁnd

Γm ( α1 ) ||v(n)||λβ2 αm Γ( m ) α 1 λ1 1 λ1 × lim M kλ (M u α , ||v(n)||β )u α −1 du, M→∞

=

Γm ( α1 ) m−1 α Γ( m α)

0

(u = (||v(n)||β t/M )α )

∞ 0

= K(α, m, λ1 ) =

kλ (t, 1)tλ1 −1 dt

Γm ( α1 ) k(λ1 ); αm−1 Γ( m α)

(5.16)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

179

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

1 Γs ( β1 ) λ2 1 λ λ 1 2 ||x||α lim M (λ1 , x) < kλ (||x||α , M v β )v β −1 dv, s M→∞ 0 βsΓ β

=

Γs ( β1 ) ∞ kλ (t, 1)tλ1 −1 dt s s−1 0 β Γ β

(v = (||x||α /M t)β )

Γs ( β1 ) k(λ1 ), (5.17) = K(β, s, λ1 ) = β s−1 Γ βs ( 1 Γs ( β1 ) λ2 1 ||x||λα1 lim M λ2 (λ1 , x) > kλ (||x||α , M v β )v β −1 dv M→∞ 0 β s Γ βs ) 1 λ2 1 kλ (||x||α , v β )v β −1 dv −δ0 0 ∞ kλ (t, 1)tλ1 −1 dt = K(β, s, λ1 ) − δ0 ||x||α

= K(β, s, λ1 )(1 − θ(||x||α )) > 0,

(5.18)

where δ0 θ(||x||α ) = K(β, s, λ1 )

∞

||x||α

kλ (t, 1)tλ1 −1 dt > 0

(0 < δ0 ≤ 1).

Lemma 5.4. Suppose that m, s, n(0) ∈ N, α, β > 0, p ∈ R\{0, 1}, 1p + q1 = 1, λ1 , λ2 ∈ R, λ1 + λ2 = λ, kλ (t, u) is a ﬁnite homogeneous function of degree −λ in R2+ , kλ (t, u)uλ2 −s is decreasing with respect to u ∈ R+ , and strict decreasing in an interval I ⊂ (n(0) , ∞), v (t) > 0, v (t) ≤ 0 (t ∈ (n(0) − 1, ∞)), with v(n(0) − 1) ≥ 0. If there exist constants δ0 > 0 and η > λ1 , such that for any δ ∈ [0, δ0 ], k(λ1 ± δ) =

∞ 0

kλ (t, 1)t(λ1 ±δ)−1 dt ∈ R+ ,

and kλ (t, 1) ≤ L

1 tη

(t ≥ 1; L > 0),

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

180

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

then, for any 0 < ε < |q|δ0 , we have λ1 −m− pε s 2 ||x|| α = I(ε) kλ (||x||α , ||v(n)||β ) v (nk )dx s−(λ2 − εq ) m {x∈R+ ;||x||α ≥1} n∈Ns ||v(n)||β k=1 (0) n

Γs ( β1 ) Γm ( 1 ) = m−1 α m s−1 s (k(λ1 )+o(1))(1−εO1 (1))(ε → 0+ ). εα Γ( α ) β Γ( β ) (5.19) Proof. = I(ε)

In view of (5.15), we ﬁnd

{x∈Rm + ;||x||α ≥1}

−m−ε ||x||α

kλ (||x||α , ||v(n)||β )

n∈Ns (0) n

×

λ1 + qε

||x||α

s 2

s−(λ2 − qε ) ||v(n)||β k=1

v (nk )dx

ε −m−ε ||x||α (λ1 + , x)dx q ε −m−ε ||x||α (1 − θ(||x|| ≥ K(β, s, λ1 + ) α ))dx q {x∈Rm ;||x|| ≥1} α + Γm ( 1 ) ε −m−ε ||x||α (1 − θ(||x|| = m−1 α m k(λ1 + ) α ))dx, α Γ( α ) q {x∈Rm ;||x|| ≥1} α + =

{x∈Rm + ;|x||α ≥1}

where θ(||x|| α) =

δ0 K(β, s, λ1 + qε )

∞ ||x||α

kλ (t, 1)tλ1 −1 dt

(0 < δ0 ≤ 1).

Since M u1/α = ||x||α ≥ 1, we ﬁnd

∞ δ0 kλ (t, 1)tλ1 −1 dt K(β, s, λ1 + εq ) Mu1/α ∞ δ0 L tλ1 −η−1 dt ≤ K(β, s, λ1 + εq ) Mu1/α

u1/α ) = 0 < θ(M

λ1 −η

δ0 LM λ1 −η u α = , (η − λ1 )K(β, s, λ1 + εq ) −m−ε ||x||α (1 − θ(||x|| α ))dx {x∈Rm + ;||x||α ≥1}

= lim

M→∞

···

Ψ DM

m xi α i=1

M

dx1 · · · dxm

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

= lim

M→∞

M m Γm ( α1 ) αm Γ( m α)

M m Γm ( α1 ) = lim m M→∞ αm Γ( ) α = = =

1

ws-book9x6

181

m

Ψ(u)u α −1 du

0

1

u1/α )) m (1 − θ(M u α −1 du (M u1/α )m+ε

M −α 1 m 1 Γ (α) 1 u1/α ))u −ε α −1 du (1 − θ(M lim M→∞ M ε M −α αm Γ( m ) α 1 " # −ε λ1 −η Γm ( α1 ) 1 λ1 −η α O(u ) u α −1 du 1 − M lim M→∞ M ε M −α αm Γ( m ) α Γm ( α1 ) (1 − εO1 (1)) (ε → 0+ ). εαm−1 Γ( m α)

(5.20)

By Lemma 3.6, we have ε = k(λ1 ) + o(1) (ε → 0+ ), k λ1 + q and then it follows that Γs ( β1 ) Γm α1 ε k λ1 + I(ε) ≥ (1 − ε O1 (1)) s q εαm−1 Γ m α β s−1 Γ β Γs ( β1 ) Γm ( α1 ) (k(λ1 ) + o(1))(1 − ε O1 (1)). s−1 Γ( s ) εαm−1 Γ( m α)β β By (5.2), we still have m 1 −m−ε ≤ Γ ( α ) k λ1 + ε ||x||α dx I(ε) m ;||x|| ≥1} αm−1 Γ( m ) q {x∈R α α + =

Γs ( β1 ) Γm ( α1 ) (k(λ1 ) + o(1)). s−1 Γ( s ) εαm−1 Γ( m α)β β Thus, we have (5.19). =

In particular, for β = 1, if ∂ ∂2 (kλ (t, u)uλ2 −s ) < 0, (kλ (t, u)uλ2 −s ) > 0 (u ∈ R+ ), ∂u ∂u2 v (t) > 0, v (t) ≤ 0, v (t) ≥ 0 (t ∈ (n(0) − 12 , ∞)), with v(n(0) − 12 ) ≥ 0, then, in view of (5.14) and (5.15), we set the following weight functions: ||v(n)||λ1 2 ω1 (λ2 , n) = kλ (||x||α , ||v(n)||1 ) dx (n ∈ Nsn(0) ), (5.21) m−λ1 ||x|| Rm α + 1 (λ1 , x) =

n∈Ns (0) n

kλ (||x||α , ||v(n)||1 )

s 2 ||x||λα1 v (nk ) (x ∈ Rm + ), 2 ||v(n)||s−λ 1 k=1

(5.22)

November 7, 2013

182

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and by (5.16), (5.10) and (5.11), we ﬁnd Γm ( 1 ) ω1 (λ2 , n) = K(α, m, λ1 ) = m−1 α m k(λ1 ); (5.23) α Γ( α ) 1 ||x||λα1 λ2 kλ (||x||α , M v)v λ2 −1 dv lim M 1 (λ1 , x) < (s − 1)! M→∞ 0 1 (5.24) = K(1, s, λ1 ) = k(λ1 ), (s − 1)! and ( 1 ||x||λα1 (λ1 , x) > lim M λ2 kλ (||x||α , M v)v λ2 −1 dv (s − 1)! M→∞ 0 ) 1 λ2 −1 kλ (||x||α , v)v dv −δ0 0

1 = k(λ1 )(1 − θ1 (||x||α )) > 0, (s − 1)! where δ0 (s − 1)! ∞ θ1 (||x||α ) = kλ (t, 1)tλ1 −1 dt > 0 k(λ1 ) ||x||α By the same way of Lemma 5.4, we still have

(5.25) (0 < δ0 ≤ 1).

Lemma 5.5. Suppose that m, s, n(0) ∈ N, α > 0, p ∈ R\{0, 1}, 1p + 1q = 1, λ1 , λ2 ∈ R, λ1 + λ2 = λ, kλ (t, u) is a ﬁnite homogeneous function of degree −λ in R2+ , satisfying ∂2 ∂ (kλ (t, u)uλ2 −s ) < 0, 2 (kλ (t, u)uλ2 −s ) > 0(u ∈ R+ ), ∂u ∂u v (t) > 0, v (t) ≤ 0, v (t) ≥ 0 (t ∈ (n(0) − 12 , ∞)), with v(n(0) − 12 ) ≥ 0. If there exist constants δ0 > 0 and η > λ1 , such that, for any δ ∈ [0, δ0 ], k(λ1 ± δ) =

∞

0

kλ (t, 1)t(λ1 ±δ)−1 dt ∈ R+ ,

1 (t ≥ 1; L > 0), kλ (t, 1) ≤ L η t then, for any 0 < ε < |q|δ0 , we have = I(ε) kλ (||x||α , ||v(n)||1 ) and

{x∈Rm + ;||x||α ≥1} n∈Ns (0) n

λ1 −m− pε

×

||x||α

s−(λ2 − qε )

||v(n)||1

s 2

v (nk )dx

k=1

Γm ( α1 ) = (k(λ1 ) + o(1))(1 − εO1 (1)) ε(s − 1)!αm−1 Γ( m ) α

(ε → 0+ ). (5.26)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

183

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

5.3.2

Main Results

For p ∈ R\{0, 1}, p1 +

1 q

= 1, we set two functions

1 )−m ϕ(x) = ||x||p(m−λ α

and

Ψ(n) =

q(s−λ2 )−s ||v(n)||β

s 2

(x ∈ Rm + ), 1−q

(n ∈ Nsn(0) ),

v (nk )

k=1

wherefrom 1 −m [ϕ(x)]1−q = ||x||qλ , α

and [Ψ(n)]1−p = ||v(n)||βpλ2 −s

s 2

v (nk ).

k=1

We also set 1 )−m ϕ(x) = (1 − θ(||x||α ))||x||p(m−λ , α

and 1−q 1 −m [ϕ(x)] = (1 − θ(||x||α ))1−q ||x||qλ , α

where θ(||x||α ) =

δ0 K(β, s, λ1 )

∞

kλ (t, 1)tλ1 −1 dt > 0

||x||α

(0 < δ0 ≤ 1).

We deﬁne two sets as follows: ⎧ ⎫ p1 ⎨ ⎬ m 1 )−m ||x||p(m−λ |f (x)|p dx <∞ , Lp,ϕ (R+ ) = f ; ||f ||p,ϕ = α ⎩ ⎭ Rm + ⎧ ⎧ ⎨ ⎨ q(s−λ2 )−s lq,Ψ = a; ||a||q,Ψ = ||v(n)||β ⎩ ⎩ s n∈N

n(0)

×

s 2 k=1

1−q v (nk )

⎫ ⎬ |a(n)|q <∞ . ⎭ ⎭ ⎫ 1q ⎬

Note. For p > 1(q > 1), the above two sets with the norms are normed linear spaces. For 0 < p < 1(q < 0) or p < 0(0 < q < 1), we still use the sets with ||f ||p,ϕ and ||a||q,Ψ as the formal symbols. We have the following theorem:

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

184

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Theorem 5.1. Suppose that m, s, n(0) ∈ N, α, β > 0, p ∈ R\{0, 1}, 1p + 1q = 1, λ, λ1 , λ2 ∈ R, λ1 + λ2 = λ, kλ (t, u) is a ﬁnite homogeneous function of degree −λ in R2+ , kλ (t, u)tλ2 −s is decreasing with respect to u ∈ R+ and strictly decreasing in an interval I ⊂ (n(0) , ∞), v (t) > 0, v (t) ≤ 0 (t ∈ (n(0) − 1, ∞)), with v(n(0) − 1) ≥ 0, ⎞ p1 ⎛ 1q Γs ( β1 ) Γm ( α1 ) ⎝ ⎠ k(λ1 ), (5.27) K(λ1 ) = αm−1 Γ( m ) α β s−1 Γ βs

∞ where k(λ1 ) = 0 kλ (t, 1)tλ1 −1 dt ∈ R+ . If f (x) ≥ 0, a(n) ≥ 0, f ∈ m Lp,ϕ (R+ ),a = {a(n)} ∈ lq,Ψ such that ||f ||p,ϕ > 0, ||a||q,Ψ > 0, then (i) for p > 1, we have the following equivalent inequalities: I= kλ (||x||α , ||v(n)||β )a(n)f (x)dx Rm + n∈Ns (0) n

(5.28) < K(λ1 )||f ||p,ϕ ||a||q,Ψ , 1 ⎧ ⎫ p p ⎨ ⎬ 1−p [Ψ(n)] kλ (||x||α , ||v(n)||β )f (x)dx J1 = ⎩ ⎭ Rm s + n∈N

n(0)

< K(λ1 )||f ||p,ϕ , and

⎧ ⎨ J2 =

⎩

(5.29) ⎛

Rm +

[ϕ(x)]1−q ⎝

n∈Ns (0)

⎫ 1q ⎬ kλ (||x||α , ||v(n)||β )a(n)⎠ dx ⎭ ⎞q

n

< K(λ1 )||a||q,Ψ ;

(5.30)

(ii) for p < 0, we have the reverses of (5.28), (5.29) and (5.30); m (iii) for 0 0, ||a||q,Ψ > 0, then we have the following reverse equivalent inequalities: I= kλ (||x||α , ||v(n)||β )a(n)f (x)dx Rm + n∈Ns (0) n

(5.31) > K(λ1 )||f ||p,ϕ ||a||q,Ψ , ⎧ p ⎫ p1 ⎬ ⎨ J1 = [Ψ(n)]1−p kλ (||x||α , ||v(n)||β )f (x)dx ⎭ ⎩ Rm s + n∈N

n(0)

> K(λ1 )||f ||p,ϕ ,

(5.32)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

185

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

and J2 =

⎧ ⎨ ⎩

⎛

Rm +

1−q ⎝ [ϕ(x)]

n∈Ns (0)

⎫ 1q ⎬ kλ (||x||α , ||v(n)||β )a(n)⎠ dx ⎭ ⎞q

n

> K(λ1 )||a||q,Ψ . (5.33) Proof.

(i) For p > 1, by (5.12) and (5.13), for H(||x||α , ||v(n)||β ) = kλ (||x||α , ||v(n)||β ),

ω(n) = ω(λ2 , n) and (x) = (λ1 , x), in view of (5.16), (5.17) and the assumptions, we have (5.29) and (5.30). By H¨older’s inequality, we have I= kλ (||x||α , ||v(n)||β )a(n)f (x)dx n∈Ns (0) n

=

Rm +

[Ψ(n)]

−1 q

n∈Ns (0) n

Rm +

1

kλ (||x||α , ||v(n)||β )f (x)dx {[Ψ(n)] q a(n)}

≤ J1 ||a||q,Ψ .

(5.34)

Then, by (5.29), we have inequality (5.28). On the other hand, assuming that (5.28) is valid, setting p−1 a(n) = [Ψ(n)]1−p

Rm +

kλ (||x||α , ||v(n)||β )f (x)dx

(n ∈ Nsn(0) ),

then, we have ||a||q,Ψ = J1p−1 . By (5.12) and the assumption of 0 < ||f ||p,ϕ < ∞, we have J1 < ∞. If J1 = 0, then (5.29) is trivially valid; if J1 > 0, then by (5.28), it follows that ||a||qq,Ψ = J1p = I < K(λ1 )||f ||p,ϕ ||a||q,Ψ , ||a||q−1 q,Ψ = J1 < K(λ1 )||f ||p,ϕ , then, (5.29) follows. Hence, (5.28) and (5.29) are equivalent. By H¨older’s inequality, we still have ⎧ ⎫ ⎨ ⎬ −1 1 I= {[ϕ(x)] p f (x)} [ϕ(x)] p kλ (||x||α , ||v(n)||β )a(n) dx ⎩ ⎭ Rm s + n∈N

n(0)

≤ ||f ||p,ϕ J2 .

(5.35)

November 7, 2013

186

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Then, by (5.30), we have (5.28). On the other hand, assuming that (5.28) is valid, setting ⎛ ⎞q−1 f (x) = [ϕ(x)]1−q ⎝ kλ (||x||α , ||v(n)||β )a(n)⎠ (x ∈ Rm + ), n∈Ns (0) n

then, we have ||f ||p,ϕ = J2q−1 . By (5.13) and the assumption of 0 < ||a||q,Ψ < ∞, we have J2 < ∞. If J2 = 0, then (5.30) is valid trivially; if J2 > 0, then by (5.28), it follows that ||f ||pp,ϕ = J2q = I < K(λ1 )||f ||p,ϕ ||a||q,Ψ , p−1 = J2 < K(λ1 )||a||q,Ψ . ||f ||p,ϕ

Thus, (5.30) follows, and (5.28) and (5.30) are equivalent. Hence, (5.28), (5.29) and (5.30) are equivalent. (ii) For p < 0 (0 < q < 1), by the reverses of (5.12) and (5.13), for H(||x||α , ||v(n)||β ) = kλ (||x||α , ||v(n)||β ), ω(n) = ω(λ2 , n) and (x) = (λ1 , x), in view of (5.16), (5.17) and the assumptions, we have the reverses of (5.29) and (5.30). By the reverse H¨older’s inequality, we have I ≥ J1 ||a||q,Ψ .

(5.36)

Then, by the reverse of (5.29), we have the reverse of (5.28). On the other hand, assuming that the reverse of (5.28) is valid, setting a(n) as (i), then we have ||a||q,Ψ = J1p−1 . By the reverse of (5.12) and the assumption of 0 < ||f ||p,ϕ < ∞, we have J1 > 0. If J1 = ∞, then the reverse of (5.29) is valid trivially; if J1 < ∞, then by the reverse of (5.28), it follows that ||a||qq,Ψ = J1p = I > K(λ1 )||f ||p,ϕ ||a||q,Ψ , ||a||q−1 q,Ψ = J1 > K(λ1 )||f ||p,ϕ . Thus, the reverse of (5.29) follows. Hence, the reverses of (5.28) and (5.29) are equivalent. By the reverse H¨older’s inequality, we still have I ≥ ||f ||p,ϕ J2 .

(5.37)

Then, by the reverse of inequality (5.30), we have the reverse of (5.28). On the other hand, assuming that the reverse of (5.28) is valid, setting f (x) as (i), then we have ||f ||p,ϕ = J2q−1 . By the reverse of (5.13) and the

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

ws-book9x6

187

assumption of 0 < ||a||q,Ψ < ∞, we have J2 > 0. If J2 = ∞, then the reverse of (5.30) is trivially valid; if J2 < ∞, then by the reverse of (5.28), it follows that ||f ||pp,ϕ = J2q = I > K(λ1 )||f ||p,ϕ ||a||q,Ψ , p−1 = J2 > K(λ1 )||a||q,Ψ . ||f ||p,ϕ Thus, the reverse of (5.30) follows and then, the reverses of (5.28) and (5.30) are equivalent. Hence, the reverses of (5.28), (5.29) and (5.30) are equivalent. (iii) For 0 < p < 1(q < 0), by the reverses of (5.12) and (5.13), for H(||x||α , ||v(n)||β ) = kλ (||x||α , ||v(n)||β ), ω(n) = ω(λ2 , n) and (x) = (λ1 , x), in view of (5.16), (5.18) and the assumptions, we have (5.32) and (5.33). By the reverse H¨older’s inequality, we have (5.38) I ≥ J1 ||a||q,Ψ . Then, by (5.32), we have (5.30). On the other hand, assuming that (5.30) is valid, setting a(n) as (i), then we have ||a||q,Ψ = J1p−1 . By the reverse of (5.12) and the assumption of 0 < ||f ||p,ϕ < ∞, we have J1 > 0. If J1 = ∞, then (5.32) is trivially valid; if J1 < ∞, then by (5.31), it follows that ||a||qq,Ψ = J1p = I > K(λ1 )||f ||p,ϕ||a||q,Ψ ,

||a||q−1 . q,Ψ = J1 > K(λ1 )||f ||p,ϕ Thus, (5.31) follows. Hence, (5.30) and (5.31) are equivalent. By the reverse H¨older’s inequality, we still have I ≥ ||f ||p,ϕJ2 . (5.39) Then, by (5.33), we have (5.31). On the other hand, assuming that (5.31) is valid, setting f (x) as follows: ⎛ 1−q ⎝ f (x) = [ϕ(x)]

⎞q−1 kλ (||x||α , ||v(n)||β )a(n)⎠

,

n∈Ns (0) n

x ∈ Rm +,

then, we have ||f ||p,ϕ = J2q−1 . By the reverse of (5.13) and the assumption of 0 < ||a||q,Ψ < ∞, we have J2 > 0. If J2 = ∞, then, (5.33) is trivially valid; if J2 < ∞, and then, by (5.31), it follows that ||f ||pp,ϕ = J2q = I > K(λ1 )||f ||p,ϕ||a||q,Ψ , p−1 ||f ||p, ϕ = J2 > K(λ1 )||a||q,Ψ . Thus, (5.33) follows and then (5.31) and (5.33) are equivalent. Hence, (5.31), (5.32) and (5.33) are equivalent.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

188

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Theorem 5.2. As the assumptions of Theorem 5.1, if there exist δ0 > 0, and η > λ1 , such that for any δ ∈ (0, δ0 ], ∞ k(λ1 ± δ) = kλ (t, 1)t(λ1 ±δ)−1 dt ∈ R+ , 0

and

kλ (t, 1) ≤ L

1 tη

(t ≥ 1; L > 0),

then, the inequalities in Theorem 5.1 have the same best possible constant factor K(λ1 ) given by (5.27). Proof.

a(n) as follows: For 0 < ε < |q|δ0 , we set f(x), 0, 0 < ||x||α < 1, f(x) = λ1 −m− pε ||x||α , ||x||α ≥ 1, λ2 − qε −s

a(n) = ||v(n)||β

s 2

v (nk ), n ∈ Nsn(0) .

k=1

(i) For p > 1, if there exists a constant k(≤ K(λ1 )), such that (5.28) is still valid as we replace K(λ1 ) by k, then, in particular, it follows that = kλ (||x||α , ||v(n)||β ) a(n)f(x)dx < k||f||p,ϕ || a||q,Ψ . I(ε) Rm + n∈Ns (0) n

(5.40) Since by (5.2) and (5.4), we have p1 −m−ε ||f||p,ϕ = ||x||α dx =

|| a||q,Ψ

p1 Γm ( α1 ) , εαm−1 Γ( m ) {x∈Rm α + ;||x||α ≥1} ⎧ ⎫ 1q ⎧ ⎫ q1 s ⎨ ⎬ ⎨ Γs ( 1 ) ⎬ 2 β + O(1) = ||v(n)||−s−ε v (nk ) = , β ⎩ ⎭ ⎩ εβ s−1 Γ s ⎭ n∈Ns (0) k=1 β n

then, by (5.19) and (5.40), it follows that Γs ( β1 ) Γm ( α1 ) (k(λ1 ) + o(1)) (1 − εO1 (1)) s αm−1 Γ( m α ) β s−1 Γ β

Γm ( α1 ) αm−1 Γ( m α)

⎧ p1 ⎨

⎫ 1q ⎬ Γs ( β1 ) + εO(1) , ⎩ β s−1 Γ s ⎭ β

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

189

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

and then K(λ1 ) ≤ k (ε → 0+ ). Hence, k = K(λ1 ) is the best possible constant factor of (5.28). By the equivalency, the constant factor in (5.29) and (5.30) is the best possible. Otherwise, we can get a contradiction by (5.34) and (5.35) that the constant factor in (5.28) is not the best possible. (ii) For p < 0, if there exists a constant k ≥ K(λ1 ), such that the reverse of (5.28) is still valid as we replace K(λ1 ) by k, then, in particular, it follows by (i) that Γs ( β1 ) Γm ( α1 ) (k(λ1 ) + o(1))(1 − εO1 (1)) s αm−1 Γ( m α ) β s−1 Γ β

>k

Γm ( α1 ) αm−1 Γ( m α)

⎧ p1 ⎨

⎫ 1q q1 ⎬ Γs ( β1 ) + εO(1) , ⎩ β s−1 Γ s ⎭ β

and then K(λ1 ) ≥ k (ε → 0+ ). Hence, k = K(λ1 ) is the best possible constant factor of the reverse of (5.28). By the equivalency, the constant factor in the reverse of (5.29) (the reverse of (5.30)) is the best possible. Otherwise, we can get a contradiction by (5.36) ((5.37)) that the constant factor in the reverse of (5.28) is not the best possible. (iii) For 0 < p < 1, if there exists a constant k ≥ K(λ1 ), such that (5.31) is still valid as we replace K(λ1 ) by k, then in particular, it follows that = I(ε)

Rm + n∈Ns (0)

kλ (||x||α , ||v(n)||β ) a(n)f(x)dx

n

a||q,Ψ . > k||f||p,ϕ|| Since, by (5.20), we have

||f||p,ϕ =

=

{x∈Rm + ;||x||α ≥1}

p1 −m−ε ||x||α (1

− θ(||x||α ))dx

p1 Γm ( α1 ) , (1 − εO1 (1)) εαm−1 Γ( m α)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

190

Half-Discrete Hilbert-Type Inequalities

then it follows that Γs ( β1 ) Γm ( α1 ) (k(λ1 ) + o(1)) (1 − εO1 (1)) s αm−1 Γ( m ) α β s−1 Γ β

>k

Γm ( α1 ) (1 m−1 α Γ( m α)

⎫ 1q ⎧ p1 ⎨ ⎬ Γs ( β1 ) + εO(1) , − εO1 (1)) ⎭ ⎩ β s−1 Γ s β

and then K(λ1 ) ≥ k (ε → 0+ ). Hence, k = K(λ1 ) is the best value of (5.31). By the equivalency, the constant factor in (5.32), (5.33) is the best possible. Otherwise, we can get a contradiction by (5.38) ((5.39)) that the constant factor in (5.31) is not the best possible. For β = 1, we set Ψ1 (n) =

q(s−λ2 )−s ||v(n)||1

s 2

1−q v (nk )

(n ∈ Nsn(0) ),

k=1

wherefrom 2 −s [Ψ1 (n)]1−p = ||v(n)||qλ 1

s 2

v (nk ).

k=1

We also set p(m−λ1 )−m , ϕ 1 (x) = (1 − θ1 (||x||α ))||x||α

and 1 −m [ϕ 1 (x)]1−q = (1 − θ1 (||x||α ))1−q ||x||qλ , α

where

δ0 (s − 1)! ∞ kλ (t, 1)tλ1 −1 dt (0 < δ0 ≤ 1). k(λ1 ) ||x||α By (5.21)-(5.25), in the same way of Theorem 5.1 and Theorem 5.2, we have θ1 (||x||α ) =

Theorem 5.3. Suppose that m, s ∈ N, α > 0, p ∈ R\{0, 1}, 1p + 1q = 1, λ, λ1 , λ2 ∈ R,λ1 + λ2 = λ, kλ (t, u) is a ﬁnite homogeneous function of degree −λ in R2+ , satisfying ∂ ∂2 (kλ (t, u)uλ2 −s ) < 0, (kλ (t, u)uλ2 −s ) > 0 (u ∈ R+ ), ∂u ∂u2 1 1 (0) (0) v (t) > 0, v (t) ≤ 0, v (t) ≥ 0(t ∈ (n − , ∞)), with v n − ≥ 0, 2 2 1q p1 m 1 Γ (α) 1 K1 (λ1 ) = k(λ1 ), (5.41) m m−1 α Γ( α ) (s−1)!

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

ws-book9x6

191

∞ where k(λ1 ) = 0 kλ (t, 1)tλ1 −1 du ∈ R+ . If f (x) ≥ 0, a(n) ≥ 0, f ∈ m Lp,ϕ (R+ ), a = {a(n)} ∈ lq,Ψ1 such that ||f ||p,ϕ > 0, ||a||q,Ψ1 > 0, then, (i) for p > 1, we have the following equivalent inequalities: kλ (||x||α , ||v(n)||1 )a(n)f (x)dx Rm + n∈Ns (0) n

⎧ ⎨ ⎩

and

< K1 (λ1 )||f ||p,ϕ ||a||q,Ψ1 ,

(5.42)

< K1 (λ1 )||f ||p,ϕ ,

(5.43)

p ⎫ p1 ⎬ kλ (||x||α , ||v(n)||1 )f (x)dx ⎭ Rm +

[Ψ1 (n)]1−p

n∈Ns (0) n

⎧ ⎨ ⎩

⎛ Rm +

[ϕ(x)]1−q ⎝

⎫ q1 ⎬ kλ (||x||α , ||v(n)||1 )a(n)⎠ dx ⎭ ⎞q

n∈Ns (0) n

< K1 (λ1 )||a||q,Ψ1 ;

(5.44)

(ii) for p < 0, we have the reverses of (5.42), (5.43) and (5.44); m (iii) for 0 0, ||a||q,Ψ1 > 0, then we have the following reverse equivalent inequalities: kλ (||x||α , ||v(n)||1 )a(n)f (x)dx Rm + n∈Ns (0) n

⎧ ⎨ ⎩

> K1 (λ1 )||f ||p,ϕ1 ||a||q,Ψ1 ,

p ⎫ p1 ⎬ kλ (||x||α , ||v(n)||1 )f (x)dx ⎭ Rm

[Ψ1 (n)]1−p

n∈Ns (0) n

+

> K1 (λ1 )||f ||p,ϕ1 , and

⎧ ⎨ ⎩

Rm +

(5.45)

⎛ [ϕ 1 (x)]1−q ⎝

n∈Ns (0)

(5.46)

⎫ 1q ⎬ kλ (||x||α , ||v(n)||1 )a(n)⎠ dx ⎭ ⎞q

n

> K1 (λ1 )||a||q,Ψ1 . (5.47) Moreover, if there exist δ0 > 0 and η > λ1 , such that for any δ ∈ (0, δ0 ], ∞ kλ (t, 1)t(λ1 ±δ)−1 dt ∈ R+ , k(λ1 ± δ) = 0

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

192

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and

kλ (t, 1) ≤ L

1 tη

(t ≥ 1; L > 0),

then, the above inequalities have the same best possible constant factor K1 (λ1 ). Some Corollaries

5.3.3

If w(t) is a strictly increasing diﬀerentiable function in (b, c) (−∞ ≤ b < c ≤ ∞), with w(b+) = 0, w(c−) = ∞, p(m−λ1 )−m Φ(x) = ||w(x)||α (

m 2

w (xi ))1−p ,

i=1

and Φ(x) = (1 − θ(||w(x)||α ))Φ(x), setting x = w(X) = (w(X1 ), · · · , w(Xm )) in the inequalities of Theorem 5.1, after simpliﬁcation, replacing m f (w(X)) i=1 w (Xi ) by f (x), we have: Corollary 5.1. Suppose that m, s, n(0) ∈ N, α, β > 0, p ∈ R\{0, 1}, 1 + 1q = 1, λ, λ1 , λ2 ∈ R, λ1 + λ2 = λ, kλ (t, u) is a ﬁnite homogeneous p function of degree −λ in R2+ , kλ (t, u)uλ2 −s is decreasing with respect to u ∈ R+ and strictly decreasing in an interval I ⊂ (n(0) , ∞), w(t) is strictly increasing in (b, c), with w(b+) = 0, w(c−) = ∞, and v (t) > 0, v (t) ≤ 0 (t ∈ (n(0) − 1, ∞)), with v(n(0) − 1) ≥ 0, ⎛ ⎞ p1 1q s 1 m 1 Γ ( ) Γ (α) β ⎝ ⎠ k(λ1 ), (5.48) K(λ1 ) = αm−1 Γ( m ) s−1 α β Γ s β

∞

where k(λ1 ) = 0 kλ (t, 1)tλ1 −1 dt ∈ R+ . If f (x) ≥ 0, a(n) ≥ 0, f ∈ Lp,Φ ((b, c)m ), a = {a(n)} ∈ lq,Ψ , such that ||f ||p,Φ > 0, ||a||q,Ψ > 0, then, (i) for p > 1, we have the following equivalent inequalities: kλ (||w(x)||α , ||v(n)||β )a(n)f (x)dx (b,c)m n∈Ns (0) n

⎧ ⎨ ⎩

n∈Ns (0)

< K(λ1 )||f ||p,Φ ||a||q,Ψ ,

[Ψ(n)]1−p

(b,c)m

kλ (||w(x)||α , ||v(n)||β

f (x)dx)p

(5.49) ⎫1 p ⎬ ⎭

n

< K(λ1 )||f ||p,Φ ,

(5.50)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

193

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

and

⎧ ⎨ ⎩

⎛ [Φ(x)]1−q ⎝

(b,c)m

n∈Ns (0)

⎫ 1q ⎬ kλ (||w(x)||α , ||v(n)||β )a(n)⎠ dx ⎭ ⎞q

n

< K(λ1 )||a||q,Ψ ;

(5.51)

(ii) for p < 0, we have the reverses of (5.49), (5.50) and (5.51); m (iii) for 0 0, ||a||q,Ψ > 0, then we have the following reverse equivalent inequalities: kλ (||w(x)||α , ||v(n)||β )a(n)f (x)dx (b,c)m n∈Ns (0) n

< K(λ1 )||f ||p,Φ ||a||q,Ψ ,

⎧ ⎨ ⎩

n∈Ns (0)

(5.52) 1 ⎫ p p ⎬ 1−p [Ψ(n)] kλ (||w(x)||α , ||v(n)||β )f (x)dx ⎭ (b,c)m

n

> K(λ1 )||f ||p,Φ , and

⎧ ⎨ ⎩

(b,c)m

⎛ 1−q ⎝ [ϕ(x)]

n∈Ns (0)

(5.53)

⎫ 1q ⎬ kλ (||w(x)||α , ||v(n)||β )a(n)⎠ dx ⎭ ⎞q

n

> K(λ1 )||a||q,Ψ .

(5.54)

For w(x) = x in (5.49)-(5.54), we reduce (5.28)-(5.33). Hence, Theorem 5.1 and Corollary 5.1 are equivalent. We can conclude that the constant factor in the above inequalities is the best possible by adding some conditions as follows: Corollary 5.2. As the assumptions of Corollary 5.1, if there exist δ0 > 0 and η > λ1 , such that for any δ ∈ (0, δ0 ], ∞ k(λ1 ± δ) = kλ (t, 1)t(λ1 ±δ)−1 dt ∈ R+ , 0

and 1 (t ≥ 1; L > 0), tη then the inequalities in Corollary 5.1 are all with the same best possible constant factor K(λ1 ). kλ (t, 1) ≤ L

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

194

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

For v(t) = t, n(0) = 1 in Theorem 5.1 and Theorem 5.2, setting q(s−λ2 )−s

ψ(n) = ||n||β

(n ∈ Ns ),

we have Corollary 5.3. Suppose that m, s ∈ N, α, β > 0, p ∈ R\{0, 1}, p1 + 1q = 1, λ, λ1 , λ2 ∈ R, λ1 + λ2 = λ, kλ (t, u) is a ﬁnite homogeneous function of degree −λ in R2+ , kλ (t, u)uλ2 −s is decreasing with respect to u ∈ R+ and strict decreasing in an interval I ⊂ (1, ∞), ⎛ ⎞ p1 1q s 1 m 1 ( ) Γ Γ (α) β ⎝ ⎠ k(λ1 ), K(λ1 ) = (5.55) αm−1 Γ( m ) s−1 α β Γ s β

∞

where, k(λ1 ) = 0 kλ (t, 1)tλ1 −1 dt ∈ R+ . If f (x) ≥ 0, a(n) ≥ 0, f ∈ m Lp,ϕ (R+ ), a = {a(n)} ∈ lq,ψ , such that ||f ||p,ϕ > 0, ||a||q,ψ > 0, then (i) for p > 1, we have the following equivalent inequalities: kλ (||x||α , ||n||β )a(n)f (x)dx < K(λ1 )||f ||p,ϕ ||a||q,ψ , Rm + n∈Ns

[ψ(n)]1−p

p p1

Rm +

n∈Ns

(5.56)

kλ (||x||α , ||n||β )f (x)dx

< K(λ1 )||f ||p,ϕ , (5.57)

and

Rm +

[ϕ(x)]1−q

q kλ (||x||α , ||n||β )a(n)

1q dx

< K(λ1 )||a||q,ψ ;

n∈Ns

(5.58) (ii) for p < 0, we have the reverses of (5.56), (5.57) and (5.58); m (iii) for 0 0, ||a||q,ψ > 0, then we have the following reverse equivalent inequalities: kλ (||x||α , ||n||β )a(n)f (x)dx > K(λ1 )||f ||p,ϕ||a||q,ψ , Rm + n∈Ns

n∈Ns

[ψ(n)]1−p

Rm +

p p1 kλ (||x||α , ||n||β )f (x)dx

(5.59) > K(λ1 )||f ||p,ϕ, (5.60)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

195

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

and Rm +

1−q

[ϕ(x)]

q kλ (||x||α , ||n||β )a(n)

q1 dx

> K(λ1 )||a||q,Ψ .

n∈Ns

(5.61) Moreover, if there exist δ0 > 0 and η > λ1 , such that for any δ ∈ (0, δ0 ], ∞ k(λ1 ± δ) = kλ (t, 1)t(λ1 ±δ)−1 dt ∈ R+ , 0

and 1 (t ≥ 1; L > 0), tη then, the above inequalities are all with the best constant factor K(λ1 ). For v(t) = (t − ξ)γ (0 < γ ≤ 1, 0 ≤ ξ ≤ 12 ), n(0) = 1 in Theorem 5.3, setting kλ (t, 1) ≤ L

q(s−λ2 )−s

Ψξ (n) = ||(n − ξ)γ ||1

(n ∈ Ns ),

we have some more accurate inequalities of Corollary 5.3 (for β = 1) as follows: Corollary 5.4. Suppose that m, s ∈ N, α > 0, p ∈ R\{0, 1}, 1p + 1q = 1, λ, λ1 , λ2 ∈ R, λ1 + λ2 = λ, kλ (t, u) is a ﬁnite homogeneous function of degree −λ in R2+ , satisfying ∂ (kλ (t, u)uλ2 −s ) < 0, ∂u K1 (λ1 ) =

∂2 (kλ (t, u)uλ2 −s ) > 0 (u ∈ R+ ), ∂u2 q1 p1 Γm ( α1 ) 1 k(λ1 ), (5.62) αm−1 Γ( m (s − 1)! α)

∞ where k(λ1 ) = 0 kλ (t, 1)tλ1 −1 dt ∈ R+ . If f (x) ≥ 0, a(n) ≥ 0, f ∈ m Lp,ϕ (R+ ),a = {a(n)} ∈ lq,Ψξ such that ||f ||p,ϕ > 0, ||a||q,Ψξ > 0, then

(i) for p > 1, we have the following equivalent inequalities: kλ (||x||α , ||(n − ξ)γ ||1 )a(n)f (x)dx

Rm + n∈Ns

< K1 (λ1 )||f ||p,ϕ ||a||q,Ψξ , 1−p [Ψξ (n)] n∈Ns

< K1 (λ1 )||f ||p,ϕ ,

(5.63) p p1 Rm +

kλ (||x||α , ||(n − ξ)γ ||1 )f (x)dx (5.64)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

196

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and

Rm +

[ϕ(x)]1−q

q kλ (||x||α , ||(n − ξ)γ ||1 )a(n)

1q dx

n∈Ns

< K1 (λ1 )||a||q,Ψξ ;

(5.65)

(ii) for p < 0, we have the reverses of (5.63), (5.64) and (5.65); (iii) for 0 0, ||a||q,Ψξ > 0, then we have the following reverse equivalent inequalities: kλ (||x||α , ||(n − ξ)γ ||1 )a(n)f (x)dx Rm + n∈Ns

> K1 (λ1 )||f ||p,ϕ1 ||a||q,Ψξ ,

[Ψ1 (n)]1−p

Rm +

n∈Ns

(5.66) p p1

kλ (||x||α , ||(n − ξ)γ ||1 )f (x)dx

> K1 (λ1 )||f ||p,ϕ1 ,

(5.67)

and

Rm +

[ϕ 1 (x)]1−q

q kλ (||x||α , ||(n − ξ)γ ||1 )a(n)

1q dx

n∈Ns

> K1 (λ1 )||a||q,Ψξ .

(5.68)

Moreover, if there exist δ0 > 0 and η > λ1 , such that for any δ ∈ (0, δ0 ], ∞ kλ (t, 1)t(λ1 ±δ)−1 dt ∈ R+ , k(λ1 ± δ) = 0

and

kλ (t, 1) ≤ L

1 tη

(t ≥ 1; L > 0),

then, the above inequalities are all with the same best possible constant factor K1 (λ1 ). 5.3.4

Operator Expressions and Some Particular Examples

As the assumptions of Theorem 5.1, for p > 1, setting 1 )−m ϕ(x) = ||x||p(m−λ α

(x ∈ Rm + ),

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

and

Ψ(n) =

q(s−λ2 )−s ||v(n)||β

s 2

ws-book9x6

197

1−q v (nk )

(n ∈ Nsn(0) ),

k=1

we deﬁne a ﬁrst kind of multi-dimensional half-discrete Hilbert-type operm ator with the homogeneous kernel T1 : Lp,ϕ (R+ ) → lp,Ψ1−p as follows: For m any f ∈ Lp,ϕ (R+ ), there exists a T1 f , satisfying T1 f (n) = kλ (||x||α , ||v(n)||β )f (x)dx(n ∈ Nsn(0) ). (5.69) Rm +

By (5.29), we can write ||T1 f ||p,Ψ1−p < K(λ1 )||f ||p,ϕ , and then T1 f ∈ lp,Ψ1−p . Hence, T1 is a bounded linear operator with the norm (see Tailor and Lay [72]) ||T1 f ||p,Ψ1−p ||T1 || = sup ≤ K(λ1 ). m ||f ||p,ϕ f ( =θ)∈Lp,ϕ (R+ ) Also we deﬁne a second kind of multi-dimensional half-discrete Hilbert-type m operator with the homogeneous kernel T2 : lq,Ψ → Lq,ϕ1−q (R+ ) as follows: For any a ∈ lq,Ψ , there exists a T2 a, satisfying kλ (||x||α , ||v(n)||β )a(n)(x ∈ Rm (5.70) T2 a(x) = + ). n∈Ns (0) n

By (5.30), we can write ||T2 a||q,ϕ1−q < K(λ1 )||a||q,Ψ , m and then T2 a ∈ Lq,ϕ1−q (R+ ). Hence, T2 is a bounded linear operator with ||T2 a||q,ϕ1−q sup ≤ K(λ1 ). ||T2 || = ||a||q,Ψ a( =θ)∈lq,Ψ By Theorem 5.2, it follows that Theorem 5.4. With the assumptions of Theorem 5.1, T1 and T2 are deﬁned by (5.69) and (5.70). If there exist δ0 > 0 and η > λ1 , such that, for any δ ∈ (0, δ0 ], ∞ kλ (t, 1)t(λ1 ±δ)−1 dt ∈ R+ , k(λ1 ± δ) = 0

and

kλ (t, 1) ≤ L

1 tη

(t ≥ 1; L > 0),

then, we have ||T1 || = ||T2 || = K(λ1 ) =

Γm ( α1 ) m−1 α Γ( m α)

q1

⎞ p1 Γs ( β1 ) ⎝ ⎠ k(λ1 ). β s−1 Γ βs ⎛

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

198

Half-Discrete Hilbert-Type Inequalities

With the assumptions of Theorem 5.3, for p > 1, setting 1 )−m ϕ(x) = ||x||p(m−λ α

and

(x ∈ Rm + ),

Ψ1 (n) =

q(s−λ2 )−s ||v(n)||1

s 2

1−q

v (nk )

(n ∈ Nsn(0) ),

k=1

we deﬁne a ﬁrst kind half-discrete operator with the homogeneous kernel m and multi-variables T1 : Lp,ϕ (R+ ) → lp,Ψ1−p as follows: 1 m For any f ∈ Lp,ϕ (R+ ), there exists a T1 f , satisfying kλ (||x||α , ||v(n)||1 )f (x)dx(n ∈ Nsn(0) ). (5.71) T1 f (n) = Rm +

By (5.43), we can write ||T1 f ||p,Ψ1−p < K1 (λ1 )||f ||p,ϕ , 1

and then, T1 f ∈ lp,Ψ1−p . Hence, T1 is a bounded linear operator with the 1 norm ||T1 f ||p,Ψ1−p 1 sup ≤ K1 (λ1 ). ||T1 || = ||f || p,ϕ f ( =θ)∈Lp,ϕ (Rm ) + Also we deﬁne a second kind half-discrete operator with the homogeneous m kernel and multi-variables T2 : lq,Ψ1 → Lq,ϕ1−q (R+ ) as follows: For any a ∈ lq,Ψ1 , there exists a T2 a, satisfying T2 a(x) = kλ (||x||α , ||v(n)||1 )a(n) (x ∈ Rm (5.72) + ). n∈Ns (0) n

By (5.44), we can write ||T2 a||q,ϕ1−q < K(λ1 ) ||a||q,Ψ1 , m and then T2 a ∈ Lq,ϕ1−q (R+ ). Hence, T2 is a bounded linear operator with

||T2 || =

||T2 a||q,ϕ1−q ≤ K1 (λ1 ). ||a||q,Ψ1 a( =θ)∈lq,Ψ sup

By Theorem 5.3, it follows that Theorem 5.5. As the assumptions of Theorem 5.3, T1 and T2 are deﬁned by (5.71) and (5.72). If there exist δ0 > 0 and η > λ1 , such that for any δ ∈ (0, δ0 ], ∞ kλ (t, 1)t(λ1 ±δ)−1 dt ∈ R+ , k(λ1 ± δ) = 0

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

199

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

and

kλ (t, 1) ≤ L

1 tη

(t ≥ 1; L > 0),

then, we have ||T1 || = ||T2 || = K1 (λ1 ) =

Example 5.1. We set kλ (t, u) = then, we ﬁnd that

Γm ( α1 ) αm−1 Γ( m ) α 1 (t+u)λ

q1

1 (s − 1)!

p1 k(λ1 ).

(λ > 0, λ1 > 0, λ2 ≤ s),

∂2 (kλ (t, u)uλ2 −s ) > 0 (u ∈ R+ ). ∂u2 For δ0 = 12 min{λ1 , λ2 }, δ ∈ (0, δ0 ], we ﬁnd ∞ (λ1 ±δ)−1 t k(λ1 ± δ) = du = B(λ1 ± δ, λ2 ∓ δ) ∈ R+ , (t + 1)λ 0 and 1 1 λ + λ1 ∈ (λ1 , λ)). ≤ η (t ≥ 1; η = kλ (t, 1) = λ (t + 1) t 2 ∂ (kλ (t, u)uλ2 −s ) < 0, ∂u

(i) For v(t) = ln t(t ≥ 2 = n(0) ), v(n(0) − 1) = 0, setting 1 )−m ϕ(x) = ||x||p(m−λ α

and

Ψ(n) =

q(s−λ2 )−s || ln n||β

(x ∈ Rm + ),

s 2 1 nk

1−q (n ∈ Ns2 ),

k=1 m Lp,ϕ (R+ )

we deﬁne an operator T1 : → lp,Ψ1−p as follows: ), there exists a T1 f , satisfying For any f ∈ Lp,ϕ (Rm + 1 T1 f (n) = f (x)dx (n ∈ Ns2 ). λ (||x|| + || ln n|| ) α β Rm + m

We also deﬁne an operator T2 : lq,Ψ → Lq,ϕ1−q (R+ ) as follows: For any a ∈ lq,Ψ , there exists a T2 a, satisfying 1 a(n) (x ∈ Rm T2 a(x) = + ). (||x|| + || ln n||β )λ α s n∈N2

Then by Theorem 5.4, we have ||T1 || = ||T2 || =

Γm ( α1 ) αm−1 Γ( m α)

1q

⎞ p1 Γs ( β1 ) ⎝ ⎠ B(λ1 , λ2 ). β s−1 Γ βs ⎛

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

200

Half-Discrete Hilbert-Type Inequalities

(ii) For v(t) = ln(t−γ)(t ≥ 2 = n(0) ; γ ≤ 12 ), v(n(0) − 12 ) = ln( 32 −γ) ≥ 0, setting p(m−σ)−m ϕ(x) = ||x||α

and

Ψγ (n) = || ln(n −

q(s−σ)−s γ)||1

s 2 k=1

1 nk − γ

(x ∈ Rm + ), 1−q (n ∈ Ns2 ),

m we deﬁne an operator T1 : Lp,ϕ (R+ ) → lp,Ψγ1−p as follows: m For any f ∈ Lp,ϕ (R+ ), there exists a T1 f , satisfying 1 T1 f (n) = f (x)dx (n ∈ Ns2 ). λ m (||x|| + || ln(n − γ)|| ) α 1 R+

Also we deﬁne an operator T2 : lq,Ψγ → Lq,ϕ1−q (R+ ) as follows: For any a ∈ lq,Ψγ , there exists T2 a, satisfying 1 a(n)(x ∈ Rm T2 a(x) = + ). (||x||α + || ln(n − γ)||1 )λ s m

n∈N2

Then, by Theorem 5.5, we have 1q p1 m 1 Γ ( ) 1 α ||T1 || = ||T2 || = B(λ1 , λ2 ). αm−1 Γ( m (s − 1)! α) (iii) For v(t) = ln κt(t ≥ 2 = n(0) ; κ ≥ setting 1 )−m ϕ(x) = ||x||p(m−λ α

and

Ψκ (n) =

q(s−λ2 )−s || ln κn||1

2 3 ),

v(n(0) − 12 ) = ln κ( 32 ) ≥ 0,

(x ∈ Rm + ),

s 2 1 nk

1−q (n ∈ Ns2 ),

k=1

we deﬁne an operator T1 : For any f ∈

m Lp,ϕ (R+ ),

T1 f (n) =

Rm +

m Lp,ϕ (R+ )

→ lp,Ψκ1−p as follows: there exists T1 f , satisfying

1 f (x)dx (||x||α + || ln κn||1 )λ

(n ∈ Ns2 ).

m We also deﬁne an operator T2 : lq,Ψκ → Lq,ϕ1−q (R+ ) as follows: For any a ∈ lq,Ψκ , there exists a T2 a, satisfying 1 a(n) (x ∈ Rm T2 a(x) = + ). (||x|| + || ln κn||1 )λ α s n∈N2

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

ws-book9x6

201

Then, by Theorem 5.5, we have 1q p1 Γm ( α1 ) 1 ||T1 || = ||T2 || = B(λ1 , λ2 ). αm−1 Γ( m (s − 1)! α) Example 5.2. We set kλ (t, u) = we ﬁnd (see Yang [134])

1 2

< λ ≤ 1, λ1 > 0, λ2 ≤ s), then

∂2 (kλ (t, u)uλ2 −s ) > 0 ∂u2

∂ (kλ (t, u)uλ2 −s ) < 0, ∂u For δ0 =

ln(t/u) (0 tλ −uλ

(u ∈ R+ ).

min{λ1 , λ2 }, δ ∈ (0, δ0 ], we obtain ∞ (ln t)t(λ1 ±δ)−1 k(λ1 ± δ) = dt tλ − 1 0 ∞ (ln v)v (λ1 ±δ)/λ−1 1 dv = 2 λ 0 v−1

2 π = ∈ R+ , λ sin π( λ1λ±δ )

and kλ (t, 1) = where t ≥ 1; η =

λ+λ1 2

1 ln t ≤ L η, tλ − 1 t

∈ (λ1 , λ), L > 0.

(i) For v(t) = ln t(t ≥ 2 = n(0) ), v(n(0) − 1) = 0, setting ϕ(x) and m Ψ(n) as Example 5.1(i), we deﬁne an operator T1 : Lp,ϕ (R+ ) → lp,Ψ1−p as follows: m For any f ∈ Lp,ϕ (R+ ), there exists a T1 f , satisfying ln(||x||α /|| ln n||β ) T1 f (n) = f (x)dx (n ∈ Ns2 ). λ − || ln n||λ m ||x|| R+ α β We also deﬁne an operator T2 : lq,Ψ → Lq,ϕ1−q (Rm + ) as follows: For any a ∈ lq,Ψ , there exists a T2 a, satisfying ln(||x||α /|| ln n||β ) a(n) (x ∈ Rm T2 a(x) = + ). λ − || ln n||λ ||x|| α β n∈Ns 2

Then, by Theorem 5.4, we have ||T1 || = ||T2 || =

Γm ( α1 ) αm−1 Γ( m α)

q1

⎞ p1 2 Γs ( β1 ) π ⎝ ⎠ . λ sin π( λλ1 ) β s−1 Γ βs ⎛

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

202

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

(ii) For v(t) = ln(t−γ)(t ≥ 2 = n(0) ; γ ≤ 12 ), v(n(0) − 12 ) = ln( 32 −γ) ≥ 0, setting ϕ(x) and Ψγ (n) as Example 5.1(ii), we deﬁne an operator T1 : m m Lp,ϕ (R+ ) → lp,Ψγ1−p as follows: For any f ∈ Lp,ϕ (R+ ), there exists a T1 f , satisfying ln(||x||α /|| ln(n − γ)||1 ) f (x)dx (n ∈ Ns2 ). T1 f (n) = λ − || ln(n − γ)||λ ||x|| Rm α 1 + m Also we deﬁne an operator T2 : lq,Ψγ → Lq,ϕ1−q (R+ ) as follows: For any a ∈ lq,Ψγ , there exists T2 a, satisfying

T2 a(x) =

ln(||x||α /|| ln(n − γ)||1 ) a(n) (x ∈ Rm + ). λ − || ln(n − γ)||λ ||x|| s α 1 n∈N 2

Then, by Theorem 5.5, we have

||T1 || = ||T2 || =

Γm ( α1 ) m−1 α Γ( m ) α

1q

1 (s − 1)!

p1

π λ sin π( λλ1 )

2 .

(iii) For v(t) = ln κt(t ≥ 2 = n(0) ; κ ≥ 23 ), v(n(0) − 12 ) = ln κ( 32 ) ≥ 0, set1−p ting ϕ(x) and Ψκ (n) as Example 5.1(iii), we deﬁne T1 : Lp,ϕ (Rm + ) → lp,Ψκ m as follows: For any f ∈ Lp,ϕ (R ), there exists a T1 f , satisfying T1 f (n) =

+

Rm +

ln(||x||α /|| ln κn||1 ) f (x)dx ||x||λα − || ln κn||λ1

(n ∈ Ns2 ).

We also deﬁne an operator T2 : lq,Ψκ → Lq,ϕ1−q (Rm + ) as follows: For any a ∈ lq,Ψκ , there exists a T2 a, satisfying T2 a(x) =

ln(||x||α /|| ln κn||1 ) a(n) (x ∈ Rm + ). λ − || ln κn||λ ||x|| s α 1 n∈N 2

Then by Theorem 5.5, we have ||T1 || = ||T2 || =

Γm ( α1 ) αm−1 Γ( m α)

1q

1 (s − 1)!

p1

π λ sin π( λλ1 )

2 .

Example 5.3. We set k0 (t, u) = e−η(u/t) (η > 0, λ1 = −σ < 0, λ2 = σ ≤ s), then, we ﬁnd that ∂ (k0 (t, u)uσ−s ) < 0, ∂u

∂2 (k0 (t, u)uσ−s ) > 0 (u ∈ R+). ∂u2

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

For δ0 = 12 σ, δ ∈ (0, δ0 ], we have ∞ k(−σ ± δ) = e−η(1/t) t−σ±δ−1 dt = 0

=

1 η σ±δ

∞

ws-book9x6

203

e−v v σ±δ−1 dv

0

1 Γ(σ ± δ) ∈ R+ , η σ±δ

and k0 (t, 1) = e−η(1/t) ≤ L

1 tη

where t ≥ 1; L > 0, η = − 12 σ > −σ. (i) For v(t) = ln t(t ≥ 2 = n(0) ), v(n(0) − 1) = 0, setting ϕ(x) and Ψ(n) as Example 5.1(i), we deﬁne an operator T1 : Lp,ϕ (Rm + ) → lp,Ψ1−p as follows: m For any f ∈ Lp,ϕ (R+ ), there exists a T1 f , satisfying T1 f (n) = e−η(|| ln n||β /||x||α ) f (x)dx(n ∈ Ns2 ). Rm +

m

We also deﬁne an operator T2 : lq,Ψ → Lq,ϕ1−q (R+ ) as follows: For any a ∈ lq,Ψ , there exists a T2 a, satisfying e−η(|| ln n||β /||x||α) a(n) (x ∈ Rm T2 a(x) = + ). n∈Ns2

Then, by Theorem 5.5, we have ||T1 || = ||T2 || =

Γm ( α1 ) αm−1 Γ( m α)

q1

⎞ p1 Γs ( β1 ) 1 ⎝ ⎠ σ Γ(σ). η β s−1 Γ βs ⎛

(ii) For v(t) = ln(t−γ)(t ≥ 2 = n(0) ; γ ≤ 12 ), v(n(0) − 12 ) = ln( 32 −γ) ≥ 0, setting ϕ(x) and Ψγ (n) as Example 5.1(ii), we deﬁne an operator T1 : 1−p as follows: Lp,ϕ (Rm + ) → lp,Ψγ For any f ∈ Lp,ϕ (Rm + ), there exists a T1 f , satisfying T1 f (n) = e−η(|| ln(n−γ)||1 /||x||α ) f (x)dx(n ∈ Ns2 ). Rm +

m Also we deﬁne an operator T2 : lq,Ψγ → Lq,ϕ1−q (R+ ) as follows: For any a ∈ lq,Ψγ , there exists T2 a, satisfying T2 a(x) = e−η(|| ln(n−γ)||1 /||x||α) a(n) (x ∈ Rm + ). n∈Ns2

November 7, 2013

204

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

Then, by Theorem 5.5, we have 1q p1 Γm ( α1 ) 1 1 ||T1 || = ||T2 || = Γ(σ). m m−1 α Γ( α ) (s − 1)! ησ (iii) For v(t) = ln κt(t ≥ 2 = n(0) ; κ ≥ 23 ), v(n(0) − 12 ) = ln κ( 32 ) ≥ 0, setting ϕ(x) and Ψκ (n) as Example 5.1(iii), we deﬁne an operator T1 : 1−p as follows: Lp,ϕ (Rm + ) → lp,Ψκ For any f ∈ Lp,ϕ (Rm + ), there exists a T1 f , satisfying T1 f (n) = e−η(|| ln κn||1 /||x||α ) f (x)dx (n ∈ Ns2 ). Rm +

We also deﬁne an operator T2 : lq,Ψκ → Lq,ϕ1−q (R+ ) as follows: For any a ∈ lq,Ψκ , there exists a T2 a, satisfying e−η(|| ln κn||1 /||x||α ) a(n) (x ∈ Rm T2 a(x) = + ). m

n∈Ns2

Then by Theorem 5.5, we have 1q p1 Γm ( α1 ) 1 1 ||T1 || = ||T2 || = Γ(σ). σ αm−1 Γ( m ) (s − 1)! η α γ

(min{t,u}) Example 5.4. We set kλ (t, u) = (max{t,u}) λ+γ (λ ≥ 0, γ ≥ 0, λ1 + λ2 = λ, λ1 + γ > 0, λ2 + γ > 0). For λ2 + γ ≤ s, we ﬁnd

(min{t, u})γ uλ2 −s (max{t, u})λ+γ γ+λ −s 2 u , 0 t uλ+γ−λ2 +s

kλ (t, u)uλ2 −s =

is decreasing with respect to u ∈ R+ and strictly decreasing in an interval of (1, ∞). For δ0 = 12 min{λ1 + γ, λ2 + γ}, δ ∈ (0, δ0 ], we obtain ∞ (min{t, 1})γ λ1 −1 t dt k(λ1 ± δ) = (max{t, 1})λ+γ 0 1 1 = + ∈ R+ , γ + λ1 ± δ γ + λ2 ∓ δ and kλ (t, 1) =

(min{t, 1})γ 1 1 = λ+γ ≤ η , (max{t, 1})λ+γ t t

where t ≥ 1; η = 12 (λ1 + λ + γ) ∈ (λ1 , λ + γ).

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

ws-book9x6

205

For v(t) = ln t(t ≥ 2 = n(0) ), v(n(0) − 1) = 0, setting ϕ(x) and Ψ(n) as Example 5.1(i), we deﬁne an operator T1 : Lp,ϕ (Rm + ) → lp,Ψ1−p as follows: For any f ∈ Lp,ϕ (Rm ), there exists a T f , satisfying 1 + (min{||x||α , || ln n||β })γ f (x)dx (n ∈ Ns2 ). T1 f (n) = λ+γ (max{||x|| , || ln n|| }) α β Rm +

Also we deﬁne an operator T2 : lq,Ψ → Lq,ϕ1−q (Rm + ) as follows: For any a ∈ lq,Ψ , there exists a T2 a, satisfying (min{||x||α , || ln n||β })γ T2 a(x) = a(n) (x ∈ Rm + ). (max{||x||α , || ln n||β })λ+γ s n∈N2

Then, by Theorem 5.4, we have ||T1 || = ||T2 || =

q Γm ( α1 ) αm−1 Γ( m α)

1

⎞ p1 Γs ( β1 ) 2γ + λ ⎝ ⎠ . s (γ + λ1 )(γ + λ2 ) β s−1 Γ β ⎛

Remark 5.1. (i) We can still write some particular inequalities with the norms as the best possible constant factors in the above examples by using Theorem 5.1 and Theorem 5.3. (ii) In particular, for m = s = 1 in the theorems and corollaries of this chapter, we get ⎞ p1 ⎛ 1q s 1 m 1 ( ) Γ Γ (α) β ⎝ ⎠ = 1, αm−1 Γ( m ) s−1 α β Γ βs and imply some corresponding results of Chapter 3.

5.4

5.4.1

Some Inequalities Relating a General Non-Homogeneous Kernel Some Lemmas

With the assumptions of Deﬁnition 5.1, if h(u) is a non-negative ﬁnite measurable function in R+ , H(t, u) = h(tu), λ1 = λ2 = σ ∈ R, then (5.5) and (5.6) reduce to ||v(n)||σβ s ω(σ, n) = h(||x||α ||v(n)||β ) (5.73) m−σ dx (n ∈ Nn(0) ), ||x|| α Rm + (σ, x) =

n∈Ns (0) n

h(||x||α ||v(n)||β )

s 2 ||x||σα v (nk ) (x ∈ Rm + ). (5.74) ||v(n)||s−σ β k=1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

206

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

For k(σ) =

∞

ω(σ, n) =

= = (σ, x) <

h(t)tσ−1 dt ∈ R+ , by (5.7), (5.8) and (5.9), we ﬁnd 1 m 1 Γ (α) 1 σ σ σ lim M h(M u α ||v(n)||β )u α −1 du, ||v(n)|| β m m M→∞ α Γ( α ) 0 (u = (t/||v(n)||β M )α ) 1 ∞ Γm ( α ) h(t)tσ−1 dt m−1 α Γ( m ) 0 α Γm ( 1 ) (5.75) K(α, m, σ) = m−1 α m k(σ); α Γ( α ) 1 Γs ( β1 ) 1 σ σ σ h(||x||α M v β )v β −1 dv, ||x||α lim M M→∞ β s Γ( βs ) 0 0

=

β s−1 Γ

s β

= K(β, s, σ) = and (σ, x) >

Γs ( β1 )

(v = (t/||x||α M )β )

Γs ( β1 )

∞

h(t)tσ−1 dt

0

Γs ( β1 ) k(σ), β s−1 Γ βs (

||x||σα β s Γ( βs )

lim M

1

σ

M→∞

0

(5.76)

1

− δ0

= K(β, s, σ) − δ0

||x||α

σ

h(||x||α M v β )v β −1 dv

0

1

1 β

h(||x||α v )v

σ β −1

) dv

h(t)tσ−1 dt

0

= K(β, s, σ)(1 − θ(||x||α )) > 0, where δ0 θ(||x||α ) = K(β, s, σ)

||x||α

0

(5.77)

h(t)tσ−1 dt > 0 (0 < δ0 ≤ 1).

Lemma 5.6. Suppose that m, s, n(0) ∈ N, α, β > 0, p ∈ R\{0, 1}, 1 1 p + q = 1, σ ∈ R, h(u) is a non-negative ﬁnite measurable function in R+ , h(u)uσ−s is decreasing with respect to u ∈ R+ and strict decreasing in an interval I ⊂ (n(0) , ∞), v (t) > 0, v (t) ≤ 0(t ∈ (n(0) − 1, ∞)), with v(n(0) − 1) ≥ 0. If there exist constants δ0 > 0 and η < σ, such that for any δ ∈ [0, δ0 ], ∞ k(σ ± δ) = h(t)t(σ±δ)−1 dt ∈ R+ , 0

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

207

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

and

h(t) ≤ L

1 tη

(0 < t ≤ 1; L > 0),

then, for any 0 < ε < |q|δ0 , we have σ−m+ pε s 2 ||x|| α = h(||x||α ||v(n)||β ) v (nk )dx I(ε) s−(σ− εq ) m {x∈R+ ;||x||α ≤1} n∈Ns ||v(n)||β k=1 (0) n

=

Γm ( α1 ) εαm−1 Γ( m α)

Proof. = I(ε)

Γs ( β1 ) (k(σ) β s−1 Γ( βs )

+ o(1))(1 − εO2 (1))(ε → 0+ ).

In view of (5.77), we have

−m+ε ||x||α

{x∈Rm + ;||x||α ≤1}

h(||x||α ||v(n)||β )

n∈Ns (0) n

×

s 2

σ− qε

||x||α

s−(σ− εq )

||v(n)||β

v (nk )dx

k=1

=

(5.78)

{x∈Rm + ;|x||α ≤1}

ε ≥ K(β, s, σ − ) q

−m+ε ||x||α (σ

{x∈Rm + ;||x||α ≤1}

Γs ( β1 ) ε k(σ − ) = q β s−1 Γ βs

{x∈Rm + ;||x||α ≤1}

δ0 K(β, s, σ − εq )

||x||α

≤

δ0 K(β, s, σ − εq ) δ0 L K(β, s, σ − εq )

(0 < δ0 ≤ 1).

u1/α

h(t)tσ−1 dt

σ−η

=

−m+ε ||x||α (1 − θ(||x|| α ))dx,

h(t)tσ−1 dt

0

Since u1/α = ||x||α ≤ 1, we ﬁnd 1/α ) = 0 < θ(u

−m+ε ||x||α (1 − θ(||x|| α ))dx

where θ(||x|| α) =

ε − , x)dx q

0 u1/α

tσ−η−1 dt 0

δ0 Lu α , (σ − η)K(β, s, σ − εq ) −m+ε ||x||α (1 − θ(||x|| α ))dx {x∈Rm + ;||x||α ≤1}

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

208

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

···

=

Ψ DM

= = = = =

Γm ( α1 ) αm Γ( m α)

m

xi

α

dx1 · · · dxm

i=1

1

m

Ψ(u)u α −1 du

0

1 1/α )) m Γm ( α1 ) (1 − θ(u u α −1 du m αm Γ( α ) 0 (u1/α )m−ε 1 Γm ( α1 ) 1/α )]u αε −1 du [1 − θ(u m m α Γ( α ) 0 1 Γm ( α1 ) σ−η ε [1 − O(u α )]u α −1 du αm Γ( m ) 0 α Γm ( α1 ) (ε → (1 − εO1 (1)), εαm−1 Γ( m α)

0+ ).

(5.79)

By Lemma 4.5 in Chapter 4, we have ε k(σ − ) = k(σ) + o(1) (ε → 0+ ), q and then, it follows that Γs ( β1 ) Γm ( α1 ) ε k(σ − )(1 − εO2 (1)) I(ε) ≥ s εαm−1 Γ( m ) q s−1 α β Γ β

=

Γm ( α1 ) εαm−1 Γ( m α)

Γs ( β1 ) (k(σ) β s−1 Γ( βs )

+ o(1))(1 − εO2 (1)).

By (5.76) and (5.3), we still have m 1 −m+ε ≤ Γ ( α ) k(σ − ε ) ||x||α dx I(ε) m ;||x|| ≤1} αm−1 Γ( m ) q {x∈R α α + Γs ( β1 ) Γm ( α1 ) (k(α) + o(1)). s−1 Γ( s ) εαm−1 Γ( m α)β β Then we have (5.78). =

In particular, for β = 1, if d2 d (h(u)uσ−s ) > 0 (u ∈ R+ ) (h(u)uσ−s ) < 0, du du2 and v (t) > 0, v (t) ≤ 0, v (t) ≥ 0 (t ∈ (n(0) − 12 , ∞)), with v(n(0) − 12 ) ≥ 0, in view of (5.73) and (5.74), we set the following weight functions: ||v(n)||σ1 s ω1 (σ, n) = h(||x||α ||v(n)||1 ) (5.80) m−σ dx (n ∈ Nn(0) ), ||x|| α Rm + 1 (σ, x) =

n∈Ns (0) n

h(||x||α ||v(n)||1 )

s 2 ||x||σα v (nk )(x ∈ Rm + ), (5.81) ||v(n)||s−σ 1 k=1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

209

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

and by (5.75), (5.10) and (5.11), we ﬁnd Γm ( α1 ) k(σ); αm−1 Γ( m ) α 1 σ lim M h(||x||α M v)vσ−1 dv

(5.82)

ω1 (σ, n) = K(α, m, σ) = 1 (σ, x) <

||x||σα (s − 1)! M→∞

0

k(σ) , = K(1, s, σ) = (s − 1)!

(5.83)

and (σ, x) >

||x||σα (s − 1)! ( σ lim M M→∞

=

0

1

h(||x||α M v)v

σ−1

dv − δ0

0

)

1

h(||x||α v)v

σ−1

k(σ) (1 − θ1 (||x||α )) > 0, (s − 1)!

dv

(5.84)

where

δ0 (s − 1)! ||x||α h(t)tσ−1 dt > 0 θ1 (||x||α ) = k(σ) 0 By the same way of Lemma 5.6, we still have

(0 < δ0 ≤ 1).

Lemma 5.7. Suppose that m, s ∈ N, α > 0, p ∈ R\{0, 1}, 1p + 1q = 1, h(u) is a non-negative ﬁnite measurable function in R+ , σ ∈ R, satisfying d2 d (h(u)uσ−s ) < 0, 2 (h(u)uσ−s ) > 0 (u ∈ R+ ), du du v (t) > 0, v (t) ≤ 0, v (t) ≥ 0 (t ∈ (n(0) − 12 , ∞)), with v(n(0) − 12 ) ≥ 0. If there exist constants δ0 > 0 and η < σ, such that for any δ ∈ [0, δ0 ], ∞ k(σ ± δ) = h(t)t(σ±δ)−1 dt ∈ R+ , 0

and

h(t) ≤ L

1 tη

(0 < t ≤ 1; L > 0),

then, for any 0 < ε < |q|δ0 , we have ε σ−m+ p s 2 ||x||α I(ε) = h(||x||α ||v(n)||1 ) v (nk )dx ε s−(σ− q ) {x∈Rm ;||x|| ≤1} s α ||v(n)|| + n∈N (0) k=1 1 n

Γm ( α1 ) (k(σ) + o(1))(1 − εO2 (1))(ε → 0+ ). = ε(s − 1)!αm−1 Γ( m ) α

(5.85)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

210

5.4.2

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Main Results 1 p

For p ∈ R\{0, 1},

+

1 q

= 1, we set two functions

p(m−σ)−m ϕ(x) = ||x||α

and

Ψ(n) =

q(s−σ)−s ||v(n)||β

s 2

(x ∈ Rm + ), 1−q

v (nk )

(n ∈ Nsn(0) ),

k=1 qσ−m wherefrom [ϕ(x)]1−q = ||x||α , and [Ψ(n)]1−p = ||v(n)||βqσ−s We also set

s k=1

v (nk ).

p(m−σ)−m , ϕ(x) = (1 − θ(||x||α ))||x||α

and 1−q qσ−m = (1 − θ(||x||α ))1−q ||x||α , [ϕ(x)]

where δ0 θ(||x||α ) = K(β, s, σ)

||x||α

0

h(t)tσ−1 dt > 0 (0 < δ0 ≤ 1).

Then we deﬁne two sets as follows: ⎧ ⎫ p1 ⎨ ⎬ m p(m−α)−m ||x||α |f (x)|p dx <∞ , Lp,ϕ (R+ ) = f ; ||f ||p,ϕ = ⎩ ⎭ Rm + ⎧ ⎧ ⎨ ⎨ q(s−σ)−s lq,Ψ = a; ||a||q,Ψ = ||v(n)||β ⎩ ⎩ s n∈N

n(0)

×

s 2 k=1

1−q v (nk )

⎫ ⎬ |a(n)|q <∞ . ⎭ ⎭ ⎫ 1q ⎬

Note. For p, q > 1, the above two sets with the norms are normed linear spaces. For the other cases, we still use them as the formal symbols. We have the following theorem: Theorem 5.6. Suppose that m, s, n(0) ∈ N, α, β > 0, p ∈ R\{0, 1}, 1 + 1q = 1, σ ∈ R, h(u) is a non-negative ﬁnite measurable function in p R+ , h(u)uσ−s is decreasing with respect to u ∈ R+ and strict decreasing

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

ws-book9x6

211

in an interval I ⊂ (n(0) , ∞), v (t) > 0, v (t) ≤ 0(t ∈ (n(0) − 1, ∞)), with v(n(0) − 1) ≥ 0, ⎞ p1 ⎛ 1q s 1 m 1 Γ ( ) Γ (α) β ⎝ ⎠ k(σ), K(σ) = (5.86) αm−1 Γ( m ) s−1 α β Γ βs

∞ where k(σ) = 0 h(t)tσ−1 dt ∈ R+ . If f (x) ≥ 0, a(n) ≥ 0, f ∈ m Lp,ϕ (R+ ),a = {a(n)} ∈ lq,Ψ such that ||f ||p,ϕ > 0, ||a||q,Ψ > 0, then (i) for p > 1, we have the following equivalent inequalities: I= h(||x||α ||v(n)||β )a(n)f (x)dx < K(σ)||f ||p,ϕ ||a||q,Ψ , (5.87) Rm + n∈Ns (0)

J1 =

⎧ ⎨

n

[Ψ(n)]1−p

⎩

n∈Ns (0) n

p ⎫ p1 ⎬ h(||x||α ||v(n)||β )f (x)dx ⎭ Rm

+

< K(σ)||f ||p,ϕ , and

⎧ ⎨ J2 =

⎩

Rm +

⎛ [ϕ(x)]1−q ⎝

n∈Ns (0)

(5.88) ⎫ 1q ⎬ h(||x||α ||v(n)||β )a(n)⎠ dx ⎭ ⎞q

n

< K(σ)||a||q,Ψ ;

(5.89)

(ii) for p < 0, we have the reverses of (5.87), (5.88) and (5.89); (iii) for 0 0, ||a||q,Ψ > 0, then, we have the following reverse equivalent inequalities: h(||x||α ||v(n)||β )a(n)f (x)dx > K(σ)||f ||p,ϕ||a||q,Ψ , (5.90) I= Rm + n∈Ns (0)

J1 =

⎧ ⎨ ⎩

n

[Ψ(n)]1−p

n∈Ns (0) n

p ⎫ p1 ⎬ h(||x||α ||v(n)||β )f (x)dx ⎭ Rm +

> K(σ)||f ||p,ϕ, and J2 =

⎧ ⎨ ⎩

Rm +

⎛ 1−q ⎝ [ϕ(x)]

n∈Ns (0)

(5.91) ⎫ 1q ⎬ h(||x||α ||v(n)||β )a(n)⎠ dx ⎭ ⎞q

n

> K(σ)||a||q,Ψ .

(5.92)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

212

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Proof.

(i) For p > 1, by (5.12) and (5.13), for H(||x||α , ||v(n)||β ) = h(||x||α ||v(n)||β ),

ω(n) = ω(σ, n) and (x) = (σ, x), in view of (5.75), (5.76) and the assumptions, we have (5.88) and (5.89). By H¨older’s inequality, we have h(||x||α ||v(n)||β )a(n)f (x)dx I= n∈Ns (0) n

=

Rm +

[Ψ(n)]

n∈Ns (0) n

−1 q

1

Rm +

h(||x||α ||v(n)||β )f (x)dx {[Ψ(n)] q a(n)}

≤ J1 ||a||q,Ψ .

(5.93)

Then by (5.88), we have (5.87). On the other hand, assuming that (5.87) is valid, and setting p−1 a(n) = [Ψ(n)]1−p

Rm +

h(||x||α ||v(n)||β )f (x)dx

(n ∈ Nsn(0) ),

then, we have ||a||q,Ψ = J1p−1 . By (5.12) and the assumption of 0 < ||f ||p,ϕ < ∞, we have J1 < ∞. If J1 = 0, then (5.88) is trivially valid; if J1 > 0, then by (5.87), it follows that ||a||qq,Ψ = J1p = I < K(σ)||f ||p,ϕ ||a||q,Ψ , that is, ||a||q−1 q,Ψ = J1 < K(σ)||f ||p,ϕ . Thus, (5.88) follows. Hence, (5.87) and (5.88) are equivalent. By H¨older’s inequality, we still have ⎧ ⎫ ⎨ ⎬ −1 1 {[ϕ(x)] p f (x)} [ϕ(x)] p h(||x||α ||v(n)||β )a(n) dx I= ⎩ ⎭ Rm s + n∈N

n(0)

≤ ||f ||p,ϕ J2 .

(5.94)

Then, by (5.89), we have (5.87). On the other hand, assuming that (5.87) is valid, and setting ⎞q−1 ⎛ h(||x||α ||v(n)||β )a(n)⎠ (x ∈ Rm f (x) = [ϕ(x)]1−q ⎝ + ), n∈Ns (0) n

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

213

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

then, we have ||f ||p,ϕ = J2q−1 . By (5.13) and the assumption of 0 < ||a||q,Ψ < ∞, we have J2 < ∞. If J2 = 0, then (5.89) is trivially valid; if J2 > 0, then by (5.87), it follows that ||f ||pp,ϕ = J2q = I < K(σ)||f ||p,ϕ ||a||q,Ψ , that is, p−1 ||f ||p,ϕ = J2 < K(σ)||a||q,Ψ .

Thus, (5.89) follows, and then (5.87) and (5.89) are equivalent. Hence, (5.87), (5.88) and (5.89) are equivalent. (ii) For p < 0 (0 < q < 1), by the reverses of (5.12) and (5.13), for H(||x||α , ||v(n)||β ) = h(||x||α ||v(n)||β ), ω(n) = ω(σ, n) and (x) = (σ, x), in view of (5.75), (5.76) and the assumptions, we have the reverses of (5.88) and (5.89). By the reverse H¨older’s inequality, we have I ≥ J1 ||a||q,Ψ .

(5.95)

Then, by the reverse of (5.88), we have the reverse of (5.87). On the other hand, assuming that the reverse of (5.87) is valid, setting a(n) as (i), then we have ||a||q,Ψ = J1p−1 . By the reverse of (5.12) and the assumption of 0 < ||f ||p,ϕ < ∞, we have J1 > 0. If J1 = ∞, then the reverse of (5.88) is valid trivially; if J1 < ∞, then by the reverse of (5.87), it follows that ||a||qq,Ψ = J1p = I > K(σ)||f ||p,ϕ ||a||q,Ψ , that is, ||a||q−1 q,Ψ = J1 > K(σ)||f ||p,ϕ . Thus, the reverse of (5.88) follows, and then the reverses of (5.87) and (5.88) are equivalent. By the reverse H¨older’s inequality, we still have I ≥ ||f ||p,ϕ J2 .

(5.96)

Then, by the reverse of (5.89), we have the reverse of (5.87). On the other hand, assuming that the reverse of (5.87) is valid, and setting f (x) as in (i), then we have ||f ||p,ϕ = J2q−1 . By the reverse of (5.13) and the assumption of 0 < ||a||q,Ψ < ∞, we have J2 > 0. If J2 = ∞, then the reverse of (5.89) is valid trivially; if J2 < ∞, then, by the reverse of (5.87), it follows that ||f ||pp,ϕ = J2q = I > K(σ)||f ||p,ϕ ||a||q,Ψ , that is, p−1 ||f ||p,ϕ = J2 > K(σ)||a||q,Ψ .

November 7, 2013

214

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Hence, the reverse of (5.89) follows, and then, the reverses of (5.87) and (5.89) are equivalent. Hence, the reverses of (5.87), (5.88) and (5.89) are equivalent. (iii) For 0 < p < 1 (q < 0), by the reverses of (5.12) and (5.13), for H(||x||α , ||v(n)||β ) = h(||x||α ||v(n)||β ), ω(n) = ω(σ, n) and (x) = (σ, x), in view of (5.75), (5.77) and the assumptions, we have (5.91) and (5.92). By the reverse H¨older’s inequality, we have I ≥ J1 ||a||q,Ψ .

(5.97)

Then, by (5.91), we have (5.90). On the other hand, assuming that (5.90) is valid, and setting a(n) as in (i), then, we have ||a||q,Ψ = J1p−1 . By the reverse of (5.12) and the assumption of 0 < ||f ||p,ϕ < ∞, we have J1 > 0. If J1 = ∞, then, (5.91) is valid trivially; if J1 < ∞, then, by (5.90), it follows that ||a||qq,Ψ = J1p = I > K(σ)||f ||p,ϕ ||a||q,Ψ , that is, ||a||q−1 . q,Ψ = J1 > K(σ)||f ||p,ϕ Thus, (5.91) follows, and then (5.90) and (5.91) are equivalent. By the reverse H¨older’s inequality, we still have I ≥ ||f ||p,ϕ J2 .

(5.98)

Then, by (5.92), we have inequality (5.90). On the other hand, assuming that (5.90) is valid, and setting f (x) as follows: ⎛ 1−q ⎝ f (x) = [ϕ(x)]

⎞q−1 h(||x||α ||v(n)||β )a(n)⎠

n∈Ns (0)

(x ∈ Rm + ),

n

then, we have ||f ||p,ϕ = J2q−1 . By the reverse of (5.13) and the assumption of 0 < ||a||q,Ψ < ∞, we have J2 > 0. If J2 = ∞, then (5.92) is valid trivially; if J2 < ∞, then, by (5.90), it follows that ||f ||pp,ϕ = J2q = I > K(σ)||f ||p,ϕ ||a||q,Ψ , that is, p−1 ||f ||p, ϕ = J2 > K(σ)||a||q,Ψ .

Thus, (5.92) follows, and then (5.90) and (5.92) are equivalent. Hence, (5.90), (5.91) and (5.92) are equivalent.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

215

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

Theorem 5.7. With the assumptions of Theorem 5.6, if there exist δ0 > 0 and η < σ, such that, for any δ ∈ (0, δ0 ], ∞ h(t)t(σ±δ)−1 dt ∈ R+ , k(σ ± δ) = 0

and

1 h(t) ≤ L η (0 < t ≤ 1; L > 0), t then, the inequalities in Theorem 5.6 have the same best possible constant factor K(σ). Proof.

a(n) as follows: For 0 < ε < |q|δ0 , we set f(x), σ−m+ pε , 0 < ||x||α < 1, ||x||α f(x) = 0, ||x||α ≥ 1, σ− qε −s

a(n) = ||v(n)||β

s 2

v (nk ),

n ∈ Nsn(0) .

k=1

(i) For p > 1, if there exists a constant k(≤ K(σ)), such that (5.87) is still valid as we replace K(σ) by k, then, in particular, it follows that I(ε) = h(||x||α ||v(n)||β ) a(n)f(x)dx < k||f||p,ϕ || a||q,Ψ . (5.99) Rm + n∈Ns (0) n

Since, by (5.3) and (5.4), we have p1 −m+ε ||f ||p,ϕ = ||x||α dx = {x;||x||α ≤1}

|| a||q,Ψ =

⎧ ⎨ ⎩

n∈Ns (0) n

||v(n)||−s−ε β

s 2 k=1

Γm ( α1 ) εαm−1 Γ( m α)

p1

⎫ 1q ⎬ v (nk ) ⎭

⎫ q1 ⎬ Γs ( β1 ) + O(1) = , ⎭ ⎩ εβ s−1 Γ s β ⎧ ⎨

then, by (5.78) and (5.99), it follows that Γs ( β1 ) Γm ( α1 ) (k(σ) + o(1))(1 − εO2 (1)) s αm−1 Γ( m α ) β s−1 Γ β ⎧ ⎫ 1q p1 ⎨ ⎬ Γs ( β1 ) Γm ( α1 ) , + εO(1)
,

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

216

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and then K(σ) ≤ k (ε → 0+ ). Hence, k = K(σ) is the best value of (5.87). By the equivalency, the constant factor in (5.88) ((5.89)) is the best possible. Otherwise, we can get a contradiction by (5.93) ((5.94)) that the constant factor in (5.87) is not the best possible. (ii) For p < 0, if there exists a constant k(≥ K(σ)), such that the reverse of (5.87) is still valid as we replace K(σ) by k, then, in particular, by (i), it follows that Γs ( β1 ) Γm ( α1 ) (k(σ) + o(1))(1 − εO2 (1)) s αm−1 Γ( m α ) β s−1 Γ β

>k

p Γm ( α1 ) αm−1 Γ( m α) 1

⎧ ⎨

⎫ 1q ⎬ Γs ( β1 ) + εO(1) , ⎩ β s−1 Γ s ⎭ β

and then, K(σ) ≥ k(ε → 0+ ). Hence, k = K(σ) is the best value of the reverse of (5.87). By the equivalency, the constant factor in the reverse of (5.88) (the reverse of (5.89)) is the best possible. Otherwise, we can get a contradiction by (5.95) ((5.96)) that the constant factor in the reverse of (5.87) is not the best possible. (iii) For 0 k||f||p,ϕ|| a||q,Ψ . I(ε) Rm + n∈Ns (0) n

By (5.79), we have ||f ||p,ϕ = =

{x∈Rm + ;||x||α ≤1}

p1 −m+ε ||x||α (1 − θ(||x||α ))dx

p1 Γm ( α1 ) . (1 − εO2 (1)) εαm−1 Γ( m ) α

Then, we obtain Γs ( β1 ) Γm ( α1 ) (k(σ) + o(1))(1 − εO2 (1)) s αm−1 Γ( m α ) β s−1 Γ β

>k

⎧ ⎫ 1q p1 ⎨ ⎬ Γs ( β1 ) Γm ( α1 ) (1)) , + εO(1) (1 − εO 1 ⎩ β s−1 Γ s ⎭ αm−1 Γ( m α) β

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

217

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

and then K(σ) ≥ k(ε → 0+ ). Hence, k = K(σ) is the best value of (5.90). By the equivalency, the constant factor in (5.91) and (5.92) is the best possible. Otherwise, we can get a contradiction by (5.97) and (5.98) that the constant factor in (5.90) is not the best possible. For β = 1, we set Ψ1 (n) =

s 2

q(s−σ)−s ||v(n)||1

1−q

(n ∈ Nsn(0) ),

v (nk )

k=1

wherefrom [Ψ1 (n)]1−p = ||v(n)||1qσ−s

s 2

v (nk ).

k=1

We also set p(m−σ)−m ϕ 1 (x) = (1 − θ1 (||x||α ))||x||α ,

and qσ−m , [ϕ 1 (x)]1−q = (1 − θ1 (||x||α ))1−q ||x||α

where θ1 (||x||α ) =

δ0 (s − 1)! k(σ)

∞

h(t)tσ−1 dt

||x||α

(0 < δ0 ≤ 1).

By (5.87)-(5.93), in the same way of Theorem 5.6 and Theorem 5.7, we have Theorem 5.8. Suppose that m, s ∈ N, α > 0, p ∈ R\{0, 1}, 1p + 1q = 1, σ ∈ R, h(u) is a non-negative ﬁnite measurable function in R+ , satisfying d (h(u)uσ−s ) < 0, du

d2 (h(u)uσ−s ) > 0 du2

(u ∈ R+ ),

v (t) > 0, v (t) ≤ 0, v (t) ≥ 0 (t ∈ (n(0) − 12 , ∞)) with v(n(0) − 12 ) ≥ 0, K1 (σ) =

Γm ( α1 ) m−1 α Γ( m α)

1q

1 (s − 1)!

p1 k(σ),

(5.100)

∞ where k(σ) = 0 h(t)tσ−1 du ∈ R+ . If f (x) ≥ 0, a(n) ≥ 0, f ∈ Lp,ϕ (Rm + ), a = {a(n)} ∈ lq,Ψ1 , such that ||f ||p,ϕ > 0, ||a||q,Ψ1 > 0, then

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

218

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

(i) for p > 1, we the following equivalent inequalities: have h(||x||α ||v(n)||1 )a(n)f (x)dx Rm + n∈Ns (0) n

< K1 (σ)||f ||p,ϕ ||a||q,Ψ1 ,

⎧ ⎨ ⎩

p ⎫ p1 ⎬ h(||x||α ||v(n)||1 )f (x)dx ⎭ Rm +

[Ψ1 (n)]1−p

n∈Ns (0) n

< K1 (σ)||f ||p,ϕ , and ⎧ ⎨ ⎩

⎛ Rm +

[ϕ(x)]1−q ⎝

n∈Ns (0)

(5.101)

(5.102)

⎫ q1 ⎬ h(||x||α ||v(n)||1 )a(n)⎠ dx < K1 (σ)||a||q,Ψ1 ; ⎭ ⎞q

n

(5.103) (ii) for p < 0, we have the reverses of (5.101), (5.102) and (5.103); m (iii) For 0 0, ||a||q,Ψ1 > 0, then we have the following reverse equivalent inequalities: h(||x||α ||v(n)||1 )a(n)f (x)dx Rm + n∈Ns (0) n

(5.104) > K1 (σ)||f ||p,ϕ1 ||a||q,Ψ1 , ⎧ p ⎫ p1 ⎬ ⎨ [Ψ1 (n)]1−p h(||x||α ||v(n)||1 )f (x)dx ⎭ ⎩ Rm s + n∈N

n(0)

> K1 (σ)||f ||p,ϕ1 , and ⎧ ⎨ ⎩

Rm +

⎛ [ϕ 1 (x)]1−q ⎝

n∈Ns (0)

(5.105)

⎫ 1q ⎬ h(||x||α ||v(n)||1 )a(n)⎠ dx > K1 (σ)||a||q,Ψ1 . ⎭ ⎞q

n

(5.106) Moreover, if there exist δ0 > 0 and η < σ, such that for any δ ∈ (0, δ0 ], ∞ h(t)t(σ±δ)−1 dt ∈ R+ , k(σ ± δ) = 0

1 h(t) ≤ L η (0 < t ≤ 1; L > 0), t then, the above inequalities are all with the same best possible constant factor K1 (σ).

and

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

219

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

5.4.3

Some Corollaries

If w(t) is a strictly increasing diﬀerentiable function in (b, c)(−∞ ≤ b < c ≤ ∞), with w(b+) = 0, w(c−) = ∞, then 1−p m 2 p(m−σ)−m w (xi ) , Φ(x) = ||w(x)||α i=1

and Φ(x) = (1 − θ(||w(x)||α ))Φ(x), setting x = w(X) = (w(X1 ), · · · , w(Xm )) in the inequalities of Theorem 5.7, after simpliﬁcam tion, replacing f (w(X)) i=1 w (Xi ) by f (x), we have Corollary 5.5. Suppose that m, s, n(0) ∈ N, α, β > 0, p ∈ R\{0, 1}, 1p + q1 = 1, σ ∈ R, h(u) is a non-negative ﬁnite measurable function in R+ , h(u)uσ−s is decreasing with respect to u ∈ R+ and strict decreasing in an interval I ⊂ (n(0) , ∞), w(t) is strict increasing in (b, c), with w(b+) = 0, w(c−) = ∞, v (t) > 0, v (t) ≤ 0 (t ∈ (n(0) − 1, ∞)), with v(n(0) − 1) ≥ 0, ⎞ p1 ⎛ 1q Γs ( β1 ) Γm ( α1 ) ⎝ ⎠ k(σ), K(σ) = (5.107) αm−1 Γ( m s−1 α) β Γ βs

∞ where k(σ) = 0 h(t)tσ−1 dt ∈ R+ . If f (x) ≥ 0, a(n) ≥ 0, f ∈ m Lp,Φ ((b, c) ), a = {a(n)} ∈ lq,Ψ , such that ||f ||p,Φ > 0, ||a||q,Ψ > 0, then (i) for p > 1, we have the following equivalent inequalities: h(||w(x)||α ||v(n)||β )a(n)f (x)dx (b,c)m n∈Ns (0) n

< K(σ)||f ||p,Φ ||a||q,Ψ ,

⎧ ⎨ ⎩

[Ψ(n)]1−p

n∈Ns (0)

(5.108) p ⎫ p1 ⎬ h(||w(x)||α ||v(n)||β )f (x)dx ⎭ (b,c)m

n

< K(σ)||f ||p,Φ , and

⎧ ⎨ ⎩

(b,c)

⎛ m

[Φ(x)]1−q ⎝

n∈Ns (0)

(5.109)

⎫ 1q ⎬ h(||w(x)||α ||v(n)||β )a(n)⎠ dx ⎭ ⎞q

n

< K(σ)||a||q,Ψ ;

(5.110)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

220

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

(ii) for p < 0, we have the reverses of (5.108), (5.109) and (5.110); m (iii) for 0 0, ||a||q,Ψ > 0, then we have the following reverse equivalent inequalities: h(||w(x)||α ||v(n)||β )a(n)f (x)dx (b,c)m n∈Ns (0) n

> K(σ)||f ||p,Φ ||a||q,Ψ ,

⎧ ⎨ ⎩

n∈Ns (0)

(5.111) 1 ⎫ p p ⎬ 1−p [Ψ(n)] h(||w(x)||α ||v(n)||β )f (x)dx ⎭ (b,c)m

n

> K(σ)||f ||p,Φ ,

(5.112)

and ⎧ ⎨ ⎩

⎛ m

(b,c)

1−q ⎝ [ϕ(x)]

⎫ 1q ⎬ h(||w(x)||α ||v(n)||β )a(n)⎠ dx ⎭ ⎞q

n∈Ns (0) n

> K(σ)||a||q,Ψ .

(5.113)

If w(x) = x, (5.108)-(5.113) reduce to (5.87)-(5.92). Hence, Theorem 5.7 and Corollary 5.5 are equivalent. We can conclude that the constant factor in the above inequalities is the best possible by adding some conditions as follows: Corollary 5.6. With the assumptions of Corollary 5.5, if there exist δ0 > 0 and η < σ, such that, for any δ ∈ (0, δ0 ], ∞ h(t)t(σ±δ)−1 dt ∈ R+ , k(σ ± δ) = 0

and

h(t) ≤ L

1 tη

(0 < t ≤ 1; L > 0),

then, the inequalities in Corollary 5.5 have the same best possible constant factor K(σ). For v(t) = t, n(0) = 1 in Theorem 5.6 and Theorem 5.7, setting q(s−σ)−s

ψ(n) = ||n||β we have the following corollary:

(n ∈ Ns ),

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

221

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

Corollary 5.7. Suppose that m, s ∈ N, α, β > 0, p ∈ R\{0, 1}, p1 + 1q = 1, σ ∈ R, h(u) is a non-negative ﬁnite measurable function in R+ , h(u)uσ−s is decreasing with respect to u ∈ R+ and strictly decreasing in an interval I ⊂ (1, ∞), ⎛ ⎞ p1 1q Γs ( β1 ) Γm ( α1 ) ⎝ ⎠ k(σ), K(σ) = (5.114) αm−1 Γ( m ) s−1 α β Γ βs

∞ where k(σ) = 0 h(t)tσ−1 dt ∈ R+ . If f (x) ≥ 0, a(n) ≥ 0, f ∈ m Lp,ϕ (R+ ),a = {a(n)} ∈ lq,ψ such that ||f ||p,ϕ > 0, ||a||q,ψ > 0, then (i) for p > 1, we have the following equivalent inequalities: h(||x||α ||n||β )a(n)f (x)dx < K(σ)||f ||p,ϕ ||a||q,ψ , Rm + n∈Ns

[ψ(n)]1−p

Rm +

n∈Ns

(5.115)

p p1 h(||x||α ||n||β )f (x)dx

< K(σ)||f ||p,ϕ , (5.116)

and

Rm +

[ϕ(x)]1−q

q h(||x||α ||n||β )a(n)

1q dx

< K(σ)||a||q,ψ ;

n∈Ns

(5.117) (ii) for p < 0, we have the reverses of (5.115), (5.116) and (5.117); (iii) for 0 0, ||a||q,ψ > 0, then, we have the following reverse equivalent inequalities: h(||x||α ||n||β )a(n)f (x)dx > K(σ)||f ||p,ϕ||a||q,ψ , Rm + n∈Ns

[ψ(n)]1−p

n∈Ns

and

Rm +

1−q

Rm +

[ϕ(x)]

(5.118)

p p1 h(||x||α ||n||β )f (x)dx

> K(σ)||f ||p,ϕ, (5.119) q

h(||x||α ||n||β )a(n)

1q dx

> K(σ)||a||q,Ψ .

n∈Ns

(5.120)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

222

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Moreover, if there exist δ0 > 0 and η < σ, such that for any δ ∈ (0, δ0 ], ∞ k(σ ± δ) = h(t)t(σ±δ)−1 dt ∈ R+ , 0

and

h(t) ≤ L

1 tη

(0 < t ≤ 1; L > 0),

then, the above inequalities hold with the same best possible constant factor K(σ). For v(t) = (t − ξ)γ (0 < γ ≤ 1, 0 ≤ ξ ≤ 12 ), n(0) = 1 in Theorem 5.8, setting q(s−σ)−s

ψξ (n) = ||(n − ξ)γ ||1

(n ∈ Ns ),

we have some more accurate inequalities of Corollary 5.7 (for β = 1) as follows: Corollary 5.8. Suppose that m, s ∈ N, α > 0, p ∈ R\{0, 1}, 1p + 1q = 1, σ ∈ R, h(u) is a non-negative ﬁnite measurable function in R+ , satisfying d (h(u)uσ−s ) < 0, du K1 (σ) =

d2 (h(u)uσ−s ) > 0 (u ∈ R+ ), du2 1q p1 Γm ( α1 ) 1 k(σ), αm−1 Γ( m ) (s − 1)! α

(5.121)

∞ where k(σ) = 0 h(t)tσ−1 dt ∈ R+ . If f (x) ≥ 0, a(n) ≥ 0, f ∈ Lp,ϕ (Rm + ), a = {a(n)} ∈ lq,Ψξ , such that ||f ||p,ϕ > 0, ||a||q,Ψξ > 0, then (i) for p > 1, we have the following equivalent inequalities: h(||x||α ||(n − ξ)γ ||1 )a(n)f (x)dx Rm + n∈Ns

< K1 (σ)||f ||p,ϕ ||a||q,Ψξ ,

[Ψξ (n)]1−p

n∈Ns

Rm +

(5.122) p p1

h(||x||α ||(n − ξ)γ ||1 )f (x)dx

< K1 (σ)||f ||p,ϕ , and

Rm +

[ϕ(x)]

1−q

(5.123) q

h(||x||α ||(n − ξ) ||1 )a(n) γ

1q dx

n∈Ns

< K1 (σ)||a||q,Ψξ ;

(5.124)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

223

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

(ii) for p < 0, we have the reverses of (5.122), (5.123) and (5.124); m (iii) for 0 0, ||a||q,Ψξ > 0, then, we have the following reverse equivalent inequalities: h(||x||α ||(n − ξ)γ ||1 )a(n)f (x)dx Rm + n∈Ns

> K1 (σ)||f ||p,ϕ1 ||a||q,Ψξ ,

[Ψ1 (n)]1−p

Rm +

n∈Ns

Rm +

(5.125) p p1

h(||x||α ||(n − ξ)γ ||1 )f (x)dx

> K1 (σ)||f ||p,ϕ1 , (5.126) q 1q [ϕ 1 (x)]1−q h(||x||α ||(n − ξ)γ ||1 )a(n) dx n∈Ns

> K1 (σ)||a||q,Ψξ .

(5.127)

Moreover, if there exist δ0 > 0 and η < σ, such that, for any δ ∈ (0, δ0 ], ∞ k(σ ± δ) = h(t)t(σ±δ)−1 dt ∈ R+ , 0

and

h(t) ≤ L

1 tη

(0 < t ≤ 1; L > 0),

then the above inequalities have the same best possible constant factor K1 (σ). 5.4.4

Operator Expressions and Some Particular Examples

With the assumptions of Theorem 5.6, for p > 1, we set p(m−σ)−m ϕ(x) = ||x||α

and

Ψ(n) =

q(s−σ)−s ||v(n)||β

s 2

(x ∈ Rm + ), 1−q

v (nk )

(n ∈ Nsn(0) ),

k=1

and we deﬁne the ﬁrst kind of multi-dimensional half-discrete Hilbert-type operator with the non-homogeneous kernel T1 : Lp,ϕ (Rm + ) → lp,Ψ1−p as follows:

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

224

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

For any f ∈ Lp,ϕ (Rm + ), there exists a T1 f , satisfying T1 f (n) = h(||x||α ||v(n)||β )f (x)dx (n ∈ Nsn(0) ). Rm +

(5.128)

By (5.88), we can write ||T1 f ||p,Ψ1−p < K(σ)||f ||p,ϕ , and then, T1 f ∈ lp,Ψ1−p . Hence, T1 is a bounded linear operator with the norm ||T1 f ||p,Ψ1−p ||T1 || = sup ≤ K(σ). m ||f ||p,ϕ f ( =θ)∈Lp,ϕ (R+ ) Also we deﬁne the second kind of multi-dimensional half-discrete Hilbertm type operator with the non-homogeneous kernel T2 : lq,Ψ → Lq,ϕ1−q (R+ ) as follows: For any a ∈ lq,Ψ , there exists a T2 a, satisfying T2 a(x) = h(||x||α ||v(n)||β )a(n) (x ∈ Rm (5.129) + ). n∈Ns (0) n

By (5.89), we can write ||T2 a||q,ϕ1−q < K(σ)||a||q,Ψ , and then, T2 a ∈ Lq,ϕ1−q (Rm + ). Hence, T2 is a bounded linear operator with the norm ||T2 a||q,ϕ1−q ||T2 || = sup ≤ K(σ). ||a||q,Ψ a( =θ)∈lq,Ψ By Theorem 5.7, it follows that Theorem 5.9. With the assumptions of Theorem 5.6, T1 and T2 are deﬁned by (5.128) and (5.129). If there exist δ0 > 0 and η < σ, such that for any δ ∈ (0, δ0 ], ∞ h(t)t(σ±δ)−1 dt ∈ R+ , k(σ ± δ) = 0

and

h(t) ≤ L

1 tη

(0 < t ≤ 1; L > 0),

then, we have ||T1 || = ||T2 || = K(σ) =

Γm ( α1 ) m−1 α Γ( m α)

1q

⎞ p1 Γs ( β1 ) ⎝ ⎠ k(σ). β s−1 Γ βs ⎛

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

ws-book9x6

225

With the assumptions of Theorem 5.7, for p > 1, we set p(m−σ)−m ϕ(x) = ||x||α

and

Ψ1 (n) =

q(s−σ)−s ||v(n)||1

s 2

(x ∈ Rm + ), 1−q

v (nk )

(n ∈ Nsn(0) ),

k=1

and we deﬁne the ﬁrst kind half-discrete Hilbert-type operator with the m non-homogeneous kernel and multi-variables T1 : Lp,ϕ (R+ ) → lp,Ψ1−p as 1 follows: m For any f ∈ Lp,ϕ (R+ ), there exists a T1 f , satisfying T1 f (n) = h(||x||α ||v(n)||1 )f (x)dx (n ∈ Nsn(0) ). (5.130) Rm +

By (5.102), we can write ||T1 f ||p,Ψ1−p < K1 (σ)||f ||p,ϕ , 1

and then, T1 f ∈ lp,Ψ1−p . Hence, T1 is a bounded linear operator with the 1 norm ||T1 f ||p,Ψ1−p 1 sup ≤ K1 (σ). ||T1 || = m ||f || p,ϕ f ( =θ)∈Lp,ϕ (R+ ) Also we deﬁne the second kind half-discrete Hilbert-type operator with the m non-homogeneous kernel and multi-variables T2 : lq,Ψ1 → Lq,ϕ1−q (R+ ) as follows: For any a ∈ lq,Ψ1 , there exists a T2 a, satisfying h(||x||α ||v(n)||1 )a(n) (x ∈ Rm (5.131) T2 a(x) = + ). n∈Ns (0) n

By (5.103), we can write ||T2 a||q,ϕ1−q < K(σ)||a||q,Ψ1 , m and then, T2 a ∈ Lq,ϕ1−q (R+ ). Hence, T2 is a bounded linear operator with the norm

||T2 || =

||T2 a||q,ϕ1−q ≤ K1 (σ). ||a||q,Ψ1 a( =θ)∈lq,Ψ sup

By Theorem 5.9, it follows that

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

226

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Theorem 5.10. With the assumptions of Theorem 5.9, T1 and T2 are deﬁned by (5.130) and (5.131). If there exist δ0 > 0 and η < σ, such that, for any δ ∈ (0, δ0 ], ∞ k(σ ± δ) = h(t)t(σ±δ)−1 dt ∈ R+ , 0

and

h(t) ≤ L

1 tη

(0 < t ≤ 1; L > 0),

then, we have ||T1 || = ||T2 || = K1 (σ) = Example 5.5. We set h(u) =

Γm ( α1 ) αm−1 Γ( m ) α

1 (1+u)λ

q1

1 (s − 1)!

p1 k(σ).

(0 < σ < λ, σ ≤ s), then we ﬁnd

d2 (h(u)uσ−s ) > 0 (u ∈ R+). du2 For δ0 = 12 min{σ, λ − σ}, δ ∈ (0, δ0 ], ∞ (σ±δ)−1 t du k(σ ± δ) = (1 + t)λ 0 = B(σ ± δ, λ − σ ∓ δ) ∈ R+ , d (h(u)uσ−s ) < 0, du

and h(t) =

1 1 ≤ η λ (1 + t) t

0 < t ≤ 1; η =

σ <σ . 2

(i) For v(t) = ln t (t ≥ 2 = n(0) ), v(n(0) − 1) = 0, setting p(m−σ)−m ϕ(x) = ||x||α

and

Ψ(n) =

q(s−σ)−s || ln n||β

(x ∈ Rm + ),

s 2 1 nk

1−q (n ∈ Ns2 ),

k=1

we deﬁne an operator T1 : Lp,ϕ (Rm + ) → lp,Ψ1−p as follows: m For any f ∈ Lp,ϕ (R+ ), there exists a T1 f , satisfying 1 T1 f (n) = f (x)dx (n ∈ Ns2 ). λ m (1 + ||x|| α || ln n||β ) R+ m

Also we deﬁne an operator T2 : lq,Ψ → Lq,ϕ1−q (R+ ) as follows:

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

ws-book9x6

227

For any a ∈ lq,Ψ , there exists a T2 a, satisfying 1 T2 a(x) = a(n) (x ∈ Rm + ). (1 + ||x||α || ln n||β )λ s n∈N2

Then, by Theorem 5.7, we have ||T1 || = ||T2 || =

Γm ( α1 ) αm−1 Γ( m α)

q1

⎞ p1 Γs ( β1 ) ⎝ ⎠ B(σ, λ − σ). β s−1 Γ βs ⎛

(ii) For v(t) = ln(t−γ) (t ≥ 2 = n(0) ; γ ≤ 12 ), v(n(0) − 12 ) = ln( 32 −γ) ≥ 0, setting ϕ(x) as in (i) and s 1−q 2 1 q(s−σ)−s Ψγ (n) = || ln(n − γ)||1 (n ∈ Ns2 ), nk − γ k=1

m we deﬁne an operator T1 : Lp,ϕ (R+ ) → lp,Ψγ1−p as follows: m For any f ∈ Lp,ϕ (R+ ), there exists a T1 f , satisfying 1 T1 f (n) = f (x)dx (n ∈ Ns2 ). λ (1 + ||x|| || ln(n − γ)|| ) α 1 Rm +

Also we deﬁne an operator T2 : lq,Ψγ → Lq,ϕ1−q (R+ ) as follows: For any a ∈ lq,Ψγ , there exists a T2 a, satisfying 1 T2 a(x) = a(n) (x ∈ Rm + ). λ (1 + ||x|| || ln(n − γ)|| ) α 1 s m

n∈N2

Then, by Theorem 5.8, we have 1q p1 m 1 Γ ( ) 1 α B(σ, λ − σ). ||T1 || = ||T2 || = αm−1 Γ( m ) (s − 1)! α (iii) For v(t) = ln κt (t ≥ 2 = n(0) ; κ ≥ 23 ), v(n(0) − 12 ) = ln κ( 32 ) ≥ 0, setting ϕ(x) as in (i) and s 1−q 2 1 q(s−σ)−s (n ∈ Ns2 ), Ψκ (n) = || ln κn||1 nk k=1

we deﬁne an operator T1 :

m Lp,ϕ (R+ )

→ lp,Ψκ1−p as follows: m For any f ∈ Lp,ϕ (R+ ), there exists a T1 f , satisfying 1 f (x)dx (n ∈ Ns2 ). T1 f (n) = λ m (1 + ||x|| || ln κn|| ) α 1 R+

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

228

Half-Discrete Hilbert-Type Inequalities

m Also we deﬁne an operator T2 : lq,Ψκ → Lq,ϕ1−q (R+ ) as follows: For any a ∈ lq,Ψκ , there exists a T2 a, satisfying

T2 a(x) =

n∈Ns2

1 a(n) (x ∈ Rm + ). (1 + ||x||α || ln κn||1 )λ

Then, by Theorem 5.8, we have ||T1 || = ||T2 || =

Γm ( α1 ) αm−1 Γ( m α)

Example 5.6. We set h(u) = d (h(u)uσ−s ) < 0, du For δ0 =

1 2

1q

ln u (0 uλ −1

1 (s − 1)!

p1

B(σ, λ − σ).

< σ < λ ≤ 1, σ ≤ s), then we ﬁnd

d2 (h(u)uσ−s ) > 0 (u ∈ R+). du2

min{σ, λ − σ}, δ ∈ (0, δ0 ], ∞ (ln t)t(σ±δ)−1 dt k(σ ± δ) = tλ − 1 0 ∞ 1 (ln v)v (σ±δ)/λ−1 = 2 dv λ 0 v−1

2 π = ∈ R+ , λ sin π( σ±δ ) λ

and h(t) =

1 ln t ≤L η λ t −1 t

0 < t ≤ 1; η =

σ < σ, L > 0 . 2

(i) For v(t) = ln t (t ≥ 2 = n(0) ), v(n(0) − 1) = 0, setting ϕ(x) and m Ψ(n) as in Example 5.5(i), we deﬁne an operator T1 : Lp,ϕ (R+ ) → lp,Ψ1−p as follows: m For any f ∈ Lp,ϕ (R+ ), there exists a T1 f , satisfying ln(||x||α || ln n||β ) f (x)dx (n ∈ Ns2 ). T1 f (n) = λ−1 (||x|| || ln n|| ) α β Rm + Also we deﬁne an operator T2 : lq,Ψ → Lq,ϕ1−q (Rm + ) as follows: For any a ∈ lq,Ψ , there exists a T2 a, satisfying T2 a(x) =

n∈Ns2

ln(||x||α || ln n||β ) a(n) (||x||α || ln n||β )λ − 1

(x ∈ Rm + ).

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

229

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

Then, by Theorem 5.10, we have ||T1 || = ||T2 || =

q Γm ( α1 ) αm−1 Γ( m ) α

1

⎞ p1 ( )2 Γs ( β1 ) π ⎝ ⎠ . λ sin π( σλ ) β s−1 Γ βs ⎛

(ii) For v(t) = ln(t − γ) (t ≥ 2 = n(0) ; γ ≤ 12 ), v(n(0) − 12 ) = ln( 32 − γ) ≥ 0, setting ϕ(x) and Ψγ (n) as in Example 5.5(ii), we deﬁne an operator m T1 : Lp,ϕ (R+ ) → lp,Ψγ1−p as follows: For any f ∈ Lp,ϕ (Rm + ), there exists a T1 f , satisfying ln(||x||α || ln(n − γ)||1 ) T1 f (n) = f (x)dx (n ∈ Ns2 ). λ−1 (||x|| || ln(n − γ)|| ) α 1 Rm + Also we deﬁne an operator T2 : lq,Ψγ → Lq,ϕ1−q (Rm + ) as follows: For any a ∈ lq,Ψγ , there exists a T2 a, satisfying T2 a(x) =

n∈Ns2

ln(||x||α || ln(n − γ)||1 ) a(n) (x ∈ Rm + ). (||x||α || ln(n − γ)||1 )λ − 1

Then, by Theorem 5.8, we have 1q p1 ( )2 Γm ( α1 ) 1 π . ||T1 || = ||T2 || = αm−1 Γ( m ) (s − 1)! λ sin π( σλ ) α (iii) For v(t) = ln κt (t ≥ 2 = n(0) ; κ ≥ 23 ), v(n(0) − 12 ) = ln κ( 32 ) ≥ 0, setting ϕ(x) and Ψκ (n) as Example in 5.5(iii), we deﬁne an operator T1 : 1−p as follows: Lp,ϕ (Rm + ) → lp,Ψκ m For any f ∈ Lp,ϕ (R+ ), there exists a T1 f , satisfying ln(||x||α || ln κn||1 ) f (x)dx (n ∈ Ns2 ). T1 f (n) = (||x|| || ln κn|| ) − 1 α 1 Rm + Also we deﬁne an operator T2 : lq,Ψκ → Lq,ϕ1−q (Rm + ) as follows: For any a ∈ lq,Ψκ , there exists a T2 a, satisfying T2 a(x) =

n∈Ns2

ln(||x||α || ln κn||1 ) a(n) (||x||α || ln κn||1 ) − 1

(x ∈ Rm + ).

Then, by Theorem 5.10, we have ||T1 || = ||T2 || =

Γm ( α1 ) αm−1 Γ( m α)

1q

1 (s − 1)!

p1 (

π λ sin π( σλ )

)2 .

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

230

Half-Discrete Hilbert-Type Inequalities

Example 5.7. We set h(u) = e−ηu (η > 0, 0 < σ ≤ s), then we ﬁnd d (h(u)uσ−s ) < 0, du

d2 (h(u)uσ−s ) > 0 (u ∈ R+). du2

For δ0 = 12 σ, δ ∈ (0, δ0 ], we have ∞ e−ηt tσ±δ−1 dt = k(−σ ± δ) = 0

=

1 η σ±δ

1 η σ±δ

∞

e−v vσ±δ−1 dv

0

Γ(σ ± δ) ∈ R+ ,

and h(t) = e−ηt ≤ L

1 tη

(0 < t ≤ 1; L > 0, η =

1 σ < σ). 2

(i) For v(t) = ln t (t ≥ 2 = n(0) ), v(n(0) − 1) = 0, setting ϕ(x) and m Ψ(n) as in Example 5.5(i), we deﬁne an operator T1 : Lp,ϕ (R+ ) → lp,Ψ1−p as follows: m For any f ∈ Lp,ϕ (R+ ), there exists a T1 f , satisfying T1 f (n) = e−η||x||α|| ln n||β f (x)dx (n ∈ Ns2 ). Rm +

m

Also we deﬁne an operator T2 : lq,Ψ → Lq,ϕ1−q (R+ ) as follows: For any a ∈ lq,Ψ , there exists a T2 a, satisfying T2 a(x) = e−η||x||α|| ln n||β a(n) (x ∈ Rm + ). n∈Ns2

Then, by Theorem 5.10, we have ||T1 || = ||T2 || =

Γm ( α1 ) αm−1 Γ( m ) α

q1

⎛

⎞ p1 Γs ( β1 ) 1 ⎝ ⎠ σ Γ(σ). s η β s−1 Γ β

(ii) For v(t) = ln(t − γ) (t ≥ 2 = n(0) ; γ ≤ 12 ), v(n(0) − 12 ) = ln( 32 − γ) ≥ 0, setting ϕ(x) and Ψγ (n) as in Example 5.5(ii), we deﬁne an operator 1−p as follows: T1 : Lp,ϕ (Rm + ) → lp,Ψγ m For any f ∈ Lp,ϕ (R+ ), there exists a T1 f , satisfying T1 f (n) = e−η||x||α || ln(n−γ)||1 f (x)dx (n ∈ Ns2 ). Rm +

m Also we deﬁne an operator T2 : lq,Ψγ → Lq,ϕ1−q (R+ ) as follows:

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Multi-dimensional Half-Discrete Hilbert-Type Inequalities

ws-book9x6

231

For any a ∈ lq,Ψγ , there exists T2 a, satisfying e−η||x||α || ln(n−γ)||1 a(n) (x ∈ Rm T2 a(x) = + ). n∈Ns2

Then, by Theorem 5.8, we have 1q p1 Γm ( α1 ) 1 1 Γ(σ). ||T1 || = ||T2 || = m m−1 α Γ( α ) (s − 1)! ησ (iii) For v(t) = ln κt (t ≥ 2 = n(0) ; κ ≥ 23 ), v(n(0) − 12 ) = ln κ( 32 ) ≥ 0, setting ϕ(x) and Ψκ (n) as in Example 5.5(iii), we deﬁne an operator m T1 Lp,ϕ (R+ ) → lp,Ψκ1−p as follows: m For any f ∈ Lp,ϕ (R+ ), there exists a T1 f , satisfying T1 f (n) = e−η||x||α || ln κn||1 f (x)dx (n ∈ Ns2 ). Rm +

Also we deﬁne an operator T2 : lq,Ψκ → Lq,ϕ1−q (Rm + ) as follows: For any a ∈ lq,Ψκ , there exists a T2 a, satisfying T2 a(x) = e−η||x||α|| ln κn||1 a(n) (x ∈ Rm + ). n∈Ns2

Then, by Theorem 5.10, we have 1q p1 m 1 ( ) Γ 1 1 α ||T1 || = ||T2 || = Γ(σ). σ αm−1 Γ( m ) (s − 1)! η α Example 5.8. We set h(u) = 0, σ + γ > 0). For σ + γ ≤ s, we ﬁnd

(min{1,u})γ (max{1,u})λ+γ

(λ ≥ 0, γ ≥ 0, λ + γ − σ >

(min{t, u})γ σ−s u (max{t, u})λ+γ γ+σ−s u , 0t λ+γ−σ+s u is decreasing with respect to u ∈ R+ , and strictly decreasing in an interval I ⊂ (1, ∞). For δ0 = 12 (σ + γ), δ ∈ (0, δ0 ], we have ∞ (min{t, 1})γ σ±δ−1 k(σ ± δ) = t dt (max{t, 1})λ+γ 0 1 1 + ∈ R+ , = γ+σ±δ λ+γ−σ∓δ and 1 (min{1, t})γ = tγ ≤ η , h(t) = λ+γ (max{1, t}) t where 0 < t ≤ 1; η = σ2 < σ. kλ (t, u)uλ2 −s =

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

232

Half-Discrete Hilbert-Type Inequalities

For v(t) = ln t (t ≥ 2 = n(0) ), v(n(0) − 1) = 0, setting ϕ(x) and Ψ(n) as in Example 5.5(i), we deﬁne an operator T1 : Lp,ϕ (Rm + ) → lp,Ψ1−p as follows: m For any f ∈ Lp,ϕ (R+ ), there exists a T1 f satisfying (min{1, ||x||α || ln n||β })γ T1 f (n) = f (x) dx (n ∈ Ns2 ). (max{1, ||x||α || ln n||β })λ+γ Rm + m

Also we deﬁne an operator T2 : lq,Ψ → Lq,ϕ1−q (R+ ) as follows: For any a ∈ lq,Ψ , there exists a T2 a, satisfying (min{1, ||x||α || ln n||β })γ a(n) (x ∈ Rm T2 a(x) = + ). (max{1, ||x||α || ln n||β })λ+γ s n∈N2

Then, by Theorem 5.9, we have ||T1 || = ||T2 || =

Γm ( α1 ) m−1 α Γ( m α)

1q

Γs ( β1 ) β s−1 Γ( βs )

p1

2γ + λ . (γ + σ)(λ + γ − σ)

Remark 5.2. (i) We can still write some particular inequalities with the norms as the best possible constant factors in the above examples by using Theorem 5.6 and Theorem 5.8. (ii) In particular, for m = s = 1 in the theorems and corollaries of this chapter, we can obtain corresponding results of Chapter 4.

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Chapter 6

Multiple Half-Discrete Hilbert-Type Inequalities

“As long as a branch of science aﬀords an abundance of problems, it is full of life; want of problems means death or cessation of independent development. Just as every human enterprise prosecutes ﬁnal aims, so mathematical research needs problems. Their solution steels the force of the investigator; thus he discovers new methods and view points and widens his horizon.” David Hilbert

“... we have always found, even with the most famous inequalities, that we have a little, new to add.” G. H. Hardy

6.1

Introduction

This chapter is devoted to two kinds of multiple half-discrete Hilbert-type inequalities which are derived by using the weight functions and techniques of real analysis. These inequalities are generalizations of the double cases mentioned in Chapters 3 and 4. The best possible constant factors involved in the inequalities are proved. Included are the equivalent forms, the operator expressions, some kinds of reverses, many theorems and corollaries, and many examples with particular kernels. We also consider many lemmas related to weight functions in subsections 6.2.1 and 6.3.1. A large number of corollaries dealing with equivalent inequalities is presented in Section 6.4. 233

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

234

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

6.2

First Kind of Multiple Hilbert-type Inequalities

6.2.1

Lemmas Related to the Weight Functions

Lemma 6.1. If m ∈ N, pi ∈ R\{0, 1}, λi ∈ R(i = 1, · · · , m + 1), m+1 1 i=1 pi = 1, then we have the following expression: ⎡ ⎤ p1 i m+1 m+1 2 2 λj −1 ⎦ (λi −1)(1−pi ) ⎣ A= xi xj = 1. (6.1) j=1(j =i)

i=1

Proof.

We ﬁnd A=

m+1 2

⎡ i −1)(1−pi )+1−λi ⎣x(λ i

m+1 2

i=1

=

m+1 2

(λi −1)(−pi )

[xi

1

] pi ⎝

m+1 2

m+1 2

i ⎝ x1−λ i

m+1 2

⎞ p1

i

λ −1 ⎠

xj j

j=1 ⎞m+1

⎛

i=1

=

i

λ −1 xj j ⎦

j=1

⎛

i=1

=

⎤ p1

i=1

1 pi

λ −1 ⎠

xj j

j=1

m+1 2

i x1−λ i

i=1

m+1 2

λ −1

xj j

= 1.

j=1

Thus, (6.1) follows.

Definition 6.1. If m ∈ N, λ ∈ R, kλ (x1 , · · · , xm+1 ) is a measurable func, satisfying for any u > 0 and (x1 , · · · , xm+1 ) ∈ Rm+1 , tion in Rm+1 + + kλ (ux1 , · · · , uxm+1 ) = u−λ kλ (x1 , · · · , xm+1 ), then, we call kλ (x1 , · · · , xm+1 ) the homogeneous function of degree −λ in . Rm+1 + m+1 Lemma 6.2. If m ∈ N, λi ∈ R(i = 1, · · · , m + 1), i=1 λi = λ, kλ (x1 , · · · , xm+1 )(≥ 0) is a homogeneous function of degree −λ in Rm+1 + such that k(λm+1 ) = H(m + 1) ∞ ∞ m 2 λ −1 ··· kλ (u1 , · · · , um , 1) uj j du1 · · · dum , = 0

0

j=1

(6.2)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Multiple Half-Discrete Hilbert-Type Inequalities

ws-book9x6

235

then, for i = 1, · · · , m, we still have ∞ ∞ H(i) = ··· kλ (u1 , · · · , ui−1 , 1, ui+1 , · · · , um+1 ) 0

0

m+1 2

×

λ −1

uj j

du1 · · · dui−1 dui+1 · · · dum+1 = k(λm+1 ).

(6.3)

j=1(j =i)

Setting uj = um+1 vj (j = i, m + 1) in H(i), we ﬁnd ∞ ∞ H(i) = ··· kλ (v1 , · · · , vi−1 , u−1 m+1 , vi+1 , · · · , vm , 1)

Proof.

0

0

i ×u−1−λ m+1

m 2

λ −1

vj j

dv1 · · · dvi−1 dvi+1 · · · dvm dum+1 .

j=1(j =i)

Setting vi = u−1 m+1 in the above integral, we obtain H(i) = H(m + 1) = k(λm+1 ) (i = 1, · · · , m). In the following of this chapter, we agree on that m ∈ N, pi ∈ m m+1 1 R\{0, 1}, λi ∈ R(i = 1, · · · , m+ 1), p1 = i=1 p1i = 1 − pm+1 , i=1 λi = λ, and kλ (x1 , · · · , xm+1 )(≥ 0) is a ﬁnite homogeneous function of degree −λ in Rm+1 . + Lemma 6.3. If k(λm+1 ) ∈ R+ , there exists a constant δ0 > 0, such that for any m ∈ N and 0 < ε < δ0 min1≤j≤m+1 {|pj |, |p|}, ε k(λm+1 + ) p ∞ ∞ m 2 (λj − pε )−1 j = ··· kλ (u1 , · · · , um , 1) uj du1 · · · dum ∈ R+ , 0

then, we have

0

j=1

ε = k(λm+1 ) + o(1) (ε → 0+ ). k λm+1 + p

Proof. We prove (6.4) by mathematical induction. For m = 1, by Lemma 3.7, we have ∞ (λ1 − pε )−1 ε 1 k λ2 + = kλ (u1, 1)u1 du1 p 0 ∞ = kλ (u1, 1)u1λ1 −1 du1 + o(1) 0

= k(λ2 ) + o(1) (ε → 0+ ).

(6.4)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

236

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Assuming that (6.4) is valid for m = n − 1, then, for m = n, since n ε ε λn+1 + + λj − = λ, p j=1 pj by the Fubini theorem (see Kuang [49]), we have ∞ ∞ ∞ ε k λn+1 + = ··· kλ (u1 , · · · , un , 1) p 0 0 0 n−1 2 (λj − pε )−1 (λn − pεn )−1 j uj du1 · · · dun−1 un dun × j=1

∞

∞

···

= 0

0

×

n−1 2

= 0

0

j

∞

du1 · · · dun−1 uλnn −1 dun + o1 (1) kλ (u1 , · · ·

(λj − pε )−1 j

uj

, un , 1)uλnn −1 dun

∞

= 0

···

×

∞

0 n−1 2

kλ (u1, · · ·

λ −1

uj j

)

du1 · · · dun−1 + o1 (1)

j=1

(

0

0 n−1 2

kλ (u1 , · · · , un , 1)

∞

···

×

0

(λj − pε )−1

uj j=1 ∞ ( ∞

∞

, un , 1)unλn −1 dun

)

du1 · · · dun−1 + o2 (1) + o1 (1)

j=1

= H(n + 1) + o(1) = k(λn+1 ) + o(1)(ε → 0+ ). Hence, we prove that (6.4) is valid for any m ∈ N.

Definition 6.2. If n ∈ N, xi ∈ R+ (i = 1, · · · , m), we deﬁne weight functions ωi (xi ) and m+1 (n) as follows: ∞ ∞ ∞ ωi (xi ) = xλi i nλm+1 −1 ··· kλ (x1 , · · · , xm , n) ×

0

n=1 m 2

λ −1

xj j

j=1(j =i)

m+1 (n) = n

λm+1 0

∞

dx1 · · · dxi−1 dxi+1 · · · dxm ,

···

0

∞ 0

kλ (x1 , · · · , xm , n)

m 2

λ −1

xj j

(6.5) dx1 · · · dxm .

j=1

(6.6)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

237

Multiple Half-Discrete Hilbert-Type Inequalities

In particular, for λm+1 = λ2 , replacing the kernel kλ (x1 , · · · , xm , n) by kλ (nx1 , · · · , nxm , 1) in (6.5) and (6.6), we can still deﬁne another weight functions wi (xi ) (xi ∈ R+ ; i = 1, · · · , m) and w m+1 (n) (n ∈ N) as follows: ∞ ∞ ∞ λ wi (xi ) = xλi i n 2 −1 ··· kλ (nx1 , · · · , nxm , 1) 0

n=1

× λ

λ −1

xj j

dx1 · · · dxi−1 dxi+1 · · · dxm ,

j=1(j =i)

w m+1 (n) = n 2

0

m 2

∞

0

···

∞ 0

kλ (nx1 , · · · , nxm , 1)

m 2

λ −1

xj j

dx1 · · · dxm .

j=1

Lemma 6.4. For n ∈ N, we have m+1 (n) = k(λm+1 ). If k(λm+1 ) ∈ R+ , kλ (x1 , · · · , xm , y)y λm+1 −1 is decreasing with respect to y ∈ R+ and strictly decreasing in an interval I ⊂ (1, ∞), then, for any i = 1, · · · , m, we have 0 < k(λm+1 )(1 − θi (xi )) < ωi (xi ) < k(λm+1 )

(xi > 0),

(6.7)

where θi (xi ) =

1 k(λm+1 )

1/xi λm+1 −1 × um+1 0

×

m 2

∞ 0

···

λ −1 uj j du1

∞ 0

· · · dui−1 dui+1 · · · dum dum+1 > 0. (6.8)

j=1(j =i)

Note. For m = i = 1, θi (xi ) =

kλ (u1 , · · · , ui−1 , 1, ui+1 , · · · , um+1 )

1 k(λm+1 )

1/xi 0

λ

m+1 um+1

−1

kλ (1, um+1 )dum+1 .

Moreover, if there exist constants α, L > 0, such that ∞ ∞ Ai (um+1 ) = ··· kλ (u1 , · · · , ui−1 , 1, ui+1 , · · · , um+1 ) 0

×

0 m 2

λ −1

uj j

α−λ

du1 · · · dui−1 dui+1 · · · dum ≤ Lum+1m+1 ,

j=1(j =i)

(6.9) then, we still have θi (xi ) = O

1 xα i

(xi > 0; i = 1, · · · , m).

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

238

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Note. For m = i = 1, Ai (um+1 ) = kλ (1, um+1 ). Proof.

Setting xj = nuj (j = 1, · · · , m) in (6.6) gives ∞ ∞ m+1 (n) = nλm+1 nm ··· kλ (nu1 , · · · , num , n) 0

0

×

m 2

(nuj )λj −1 du1 · · · dum

j=1

= H(m + 1) = k(λm+1 ). Using the decreasing property and Lemma 6.2, for i = 1, · · · , m, we have ∞ ∞ ∞ λm+1 −1 ωi (xi ) < xλi i xm+1 ··· kλ (x1 , · · · , xm , xm+1 )dxm+1 0

0

m 2

×

λ −1

xj j

0

dx1 · · · dxi−1 dxi+1 · · · dxm

j=1(j =i)

= H(i) = k(λm+1 ) (uj = xj /xi (j = i)),

∞ ∞ ∞ λm+1 −1 λi xm+1 ··· kλ (x1 , · · · , xm , xm+1 ) ωi (xi ) > xi 1

0

m 2

×

λ −1 xj j dx1

0

· · · dxi−1 dxi+1 · · · dxm dxm+1 ,

j=1(j =i)

∞

= 1/xi

λ

m+1 um+1

−1

∞

0 m 2

×

(uj = xj /xi (j = i))

···

∞

0

kλ (u1 , · · · , ui−1 , 1, ui+1 , · · · , um+1 )

λ −1 uj j du1

· · · dui−1 dui+1 · · · dum dum+1

j=1(j =i)

= k(λm+1 )(1 − θi (xi )) > 0, where θi (xi ) is given by (6.8). Moreover, by (6.9), we have 0 < θi (xi ) =

1

1/xi

λ

m+1 Ai (um+1 )um+1

k(λm+1 ) 0 1/xi α−1 um+1 dum+1 =

−1

dum+1

L L . k(λm+1 ) 0 αk(λm+1 )xα i Thus, θi (xi ) = O x1α (xi > 0; i = 1, · · · , m). ≤

i

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

239

Multiple Half-Discrete Hilbert-Type Inequalities

Lemma 6.5. Suppose that λm+1 = λ2 , k(λm+1 ) ∈ R+ , and λ kλ (yx1 , · · · , yxm , 1)y 2 −1 is decreasing with respect to y ∈ R+ and strictly decreasing in an interval I ⊂ (1, ∞). Then, we have w m+1 (n) = k(λm+1 ), and for any i = 1, · · · , m, 0 < k(λm+1 )(1 − θi (xi )) < wi (xi ) < k(λm+1 ) (xi > 0), (6.10) where

xi ∞ ∞ 1 λ −1 i θi (xi ) = ui ··· kλ (u1 , · · · , um , 1) k(λm+1 ) 0 0 0 m 2 λj −1 uj du1 · · · dui−1 dui+1 · · · dum dui > 0. (6.11) × j=1(j =i)

Note. For m = i = 1, θi (xi ) =

1 k(λm+1 )

xi 0

uλi i −1 kλ (ui , 1)dui .

Moreover,if thereexist constants α, L > 0 such that ∞ ∞ B1 (ui ) = ··· kλ (u1 , · · · , um , 1) 0

0

m 2

×

λ −1

uj j

i du1 · · · dui−1 dui+1 · · · dum ≤ Luα−λ , i

(6.12)

j=1(j =i)

then, we have

θi (xi ) = O(xα i ) (xi > 0; i = 1, · · · , m). Note. For m = i = 1, B1 (ui ) = kλ (ui , 1).

Proof. Setting uj = nxj (j = 1, · · · , m) in w m+1 (n) (see Deﬁnition 6.2), simplifying, we ﬁnd w m+1 (n) = k(λm+1 ). Using the decreasing property and Lemma 6.4, we have

∞ ∞ ∞ λ λi 2 −1 xm+1 ··· kλ (xm+1 x1 , · · · , xm+1 xm , 1) wi (xi ) < xi 0

0

m 2

×

λ −1 xj j dx1

0

· · · dxi−1 dxi+1 · · · dxm dxm+1 ,

j=1(j =i)

= xλi i

∞

(− λ −1)

xm+12

0

= xλi i

∞ 0

0 m 2

×

j=1(j =i) ∞

∞

(xm+1 = x−1 m+1 )

···

∞ 0

λ −1 xj j dx1

−1 kλ (x−1 m+1 x1 , · · · , xm+1 xm , 1)

· · · dxi−1 dxi+1 · · · dxm dxm+1

···

0

kλ (x1, · · · , xm , xm+1 )

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

240

ws-book9x6

Half-Discrete Hilbert-Type Inequalities m+1 2

×

λ −1

xj j

dx1 · · · dxi−1 dxi+1 · · · dxm+1

j=1(j =i)

= H(i) = k(λm+1 ) (uj = xj /xi (j = i)),

∞ ∞ ∞ λ −1 λi 2 xm+1 ··· kλ (xm+1 x1 , · · · , xm+1 xm , 1) wi (xi ) > xi 1 0 0 m 2 λj −1 × xj dx1 · · · dxi−1 dxi+1 · · · dxm dxm+1 . j=1(j =i)

Setting uj = xm+1 xj (j = i) in the above integral, simplifying, and putting ui = xm+1 xi , we ﬁnd

∞ ∞ ∞ λi −1 ui ··· kλ (u1 , · · · , um, 1) wi (xi ) > xi 0 0 m 2 λj −1 × uj du1 · · · dui−1 dui+1 · · · dum dui j=1(j =i)

= k(λm+1 )(1 − θi (xi )) > 0,

where θi (xi ) is given by (6.11). Moreover, by (6.11) and (6.12), we have xi 1 B1 (ui )uλi i −1 dui 0 < θi (xi ) = k(λm+1 ) 0 xi L Lxα i i ≤ uλi i −1 uα−λ dui = , i k(λm+1 ) 0 αk(λm+1 ) that is, θi (xi ) = O(xα ) (xi > 0; i = 1, · · · , m). i

Lemma 6.6. Suppose that there exists a constant δ0 > 0, such that, for i = λ, k(λ i ∈ (λi − δ0 , λi + δ0 ) (i = 1, · · · , m + 1), m+1 λ m+1 ) ∈ R+ . any λ i=1 m+1 −1 λ is decreasing with respect to y ∈ R+ , then, for If kλ (x1 , · · · , xm , y)y 0 < ε < δ0 min1≤j≤m+1 {|pj |, |p|}, we have ∞ ∞ ∞ (λm+1 − p ε )−1 m+1 I(ε) = ε n ··· kλ (x1 , · · · , xm , n) n=1 m 2

×

1

(λj − pε j

xj

)−1

1

dx1 · · · dxm = k(λm+1 ) + o(1)(ε → 0+ ). (6.13)

j=1

Proof. δ0 ),

i = λi − ε (i = 1, · · · , m), λ m+1 = λm+1 + ε ∈ (λi − δ0 , λi + For λ pi p

kλ (x1 , · · · , xm , y) y

λm+1 − p

ε m+1

−1

(= kλ (x1 , · · · , xm , y)y λm+1 −1 y −ε )

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

241

Multiple Half-Discrete Hilbert-Type Inequalities

is decreasing with respect to y ∈ R+ . By the decreasing property, we ﬁnd ∞ ∞ ∞ λm+1 − p ε −1 I(ε) ≥ ε xm+1 m+1 ··· kλ (x1 , · · · , xm , xm+1 ) 1

×

1

∞

=ε 1

x−ε−1 m+1

j

xj

×

∞

···

1 xm+1

m 2

1

x−ε−1 m+1 ×

∞

···

0

m 2

uj j=1 m ∞ 1

i=1

Ai (xm+1 ) =

0

∞

···

0 j

∞ 0

×

du1 · · · dum dxm+1 kλ (u1 , · · · , um , 1) du1 · · · dum dxm+1

∞

1

x−1 m+1 Ai (xm+1 )dxm+1 ,

1/xm+1

0 m 2

kλ (u1 , · · · , um , 1)

x−1 m+1 Ai (xm+1 )dxm+1

m

ε = k(λm+1 + ) − ε p i=1

∞

(λj − pε )−1

−ε

where

1 xm+1

j

uj

dx1 · · · dxm dxm+1 , (uj = xj /xm+1 ) ∞

(λj − pε )−1

j=1 ∞

1

(λj − pε )−1

j=1

≥ε

m 2

∞

···

0 (λj − pε )−1 j

uj

∞ 0

kλ (u1 , · · · , um , 1)

du1 · · · dui−1 dui dui+1 · · · dum .

j=1

∞

Without loss of generality, we estimate follows: ∞ 0< x−1 m+1 Am (xm+1 )dxm+1 1

∞

= 1

x−1 m+1

×

1/xm+1

∞

0

0 m 2 j=1

···

(λj − pε )−1

uj

j

1

∞ 0

x−1 m+1 Am (xm+1 )dxm+1 as

kλ (u1 , · · · , um , 1)

du1 · · · dum−1 dum dxm+1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

242

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

1

1/um

x−1 m+1 dxm+1

= 0

1

×

m 2

1

0

(− ln um ) ×

∞

j

0

m 2

···

∞

···

∞ 0

kλ (u1 , · · · , um , 1)

du1 · · · dum−1 dum

kλ (u1 , · · · , um , 1)

0

(λj − pε )−1

uj

∞

0

(λj − pε )−1

uj

j=1

=

j

du1 · · · dum−1 dum .

(6.14)

j=1 1 } < δ0 , since Setting α > 0, such that max{α + |pεm | , α + ε |p| limum →0+ uα (− ln u ) = 0, there exists a constant M > 0, such that m m m

0 < uα m (− ln um ) ≤ Mm (um ∈ (0, 1]). In view of (6.14) and (6.2), we ﬁnd ∞ x−1 0< m+1 Am (xm+1 ) dxm+1 1 ∞ ∞ ∞ ≤ Mm ··· kλ (u1 , · · · , um , 1) 0

0

×

0

m−1 2

(λj − pε )−1 (λm − αpm +ε )−1 j pm uj um du1

· · · dum−1 dum

j=1

ε = Mm k λm+1 + α + < ∞. p Thus, m i=1

∞

1

x−1 i Am (xm+1 )dxm+1 = O(1).

Hence, we have

I(ε) ≥ k λm+1 + ε 1 −

We still obtain I(ε) ≤ ε

∞

n

−ε−1

n

λm+1 +ε− p

ε m+1

×

m 2 j=1

∞

···

(λj − pε )−1

xj

− εO(1).

pm+1

0

n=1

1

j

∞ 0

kλ (x1 , · · · , xm , n)

dx1 · · · dxm

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

243

Multiple Half-Discrete Hilbert-Type Inequalities

ε k λm+1 + = ε 1+ n p n=2 ∞ ε −ε−1 ≤ ε 1+ y dy k λm+1 + p 1 ε . = (ε + 1) k λm+1 + p ∞

−ε−1

Hence, by (6.4), we have (6.13).

Lemma 6.7. Suppose that λm+1 = λ2 , there exists a constant δ0 > 0, such i = λ, i ∈ (λi − δ0 , λi + δ0 ) (i = 1, · · · , m + 1), m+1 λ that, for any λ i=1 k(λm+1 ) ∈ R+ , and

kλ (yx1 , · · · , yxm , 1) y λm+1 −1 is decreasing with respect to y ∈ R+. Then, for 0 δ0 min1≤j≤m+1 {|pj |, |p|}, we have 1 1 ∞ λ − ε −1 =ε I(ε) n 2 pm+1 ··· kλ (nx1 , · · · , nxm , 1) n=1 m 2

×

0

(λj + pε )−1 j

xj

<

ε

<

0

dx1 · · · dxm = k(λm+1 ) + o(1)(ε → 0+ ). (6.15)

j=1

Proof.

By the decreasing property, we ﬁnd

∞ λ 1 1 ε 2 − pm+1 −1 I(ε) ≥ ε xm+1 ··· kλ (xm+1 x1 , · · · , xm+1 xm , 1) 1

×

0

m 2

(λj − pε )−1

xj

j

0

dx1 · · · dxm dxm+1 uj = xm+1 xj (j = 1, · · · , m)

j=1

∞

=ε 1

x−ε−1 m+1

xm+1

···

0

×

m 2

xm+1

0

kλ (u1 , · · · , um , 1)

(λj + pε )−1

uj

j

du1 · · · dum dxm+1

j=1

≥ε

1

∞

x−ε−1 m+1

0

∞

··· ×

∞ 0

m 2 j=1

kλ (u1 , · · · , um , 1)

(λj + pε )−1

uj

j

du1 · · · dum dxm+1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

244

Half-Discrete Hilbert-Type Inequalities

−ε

m 1

i=1

= k(λm+1 − ε + where

Bi (xm+1 ) =

∞ 0

×

m 2

pm+1

0

(λj + pε )−1 j

∞ 0

xm+1

uj

1

i=1

x−1 m+1 Bi (xm+1 )dxm+1

m ∞

)−ε

∞ ∞

···

ε

∞

···

x−1 m+1 Bi (xm+1 )dxm+1 , (6.16)

∞ 0

kλ (u1 , · · · , um , 1)

du1 · · · dui−1 dui dui+1 · · · dum (1 ≤ i ≤ m).

j=1

∞

Without loss of generality, we estimate follows: ∞ x−1 0< m+1 Bm (xm+1 )dxm+1 1

∞

= 1

x−1 m+1

∞

xm+1

×

∞

∞

(λj + pε )−1 j

uj

j=1

um

= 1

m 2

0

···

1

×

x−1 m+1 dxm+1 m 2

0

kλ (u1 , · · · , um , 1)

∞ 0

j

x−1 m+1 Bm (xm+1 )dxm+1 as

du1 · · · dum−1 dum dxm+1

(λj + pε )−1

uj

∞

1

···

∞ 0

kλ (u1 , · · · , um , 1)

du1 · · · dum−1 dum

j=1

∞

= 1

ln um ×

∞

0 m 2

···

∞

0

kλ (u1 , · · · , um , 1)

(λj + pε )−1

uj

j

du1 · · · dum−1 dum .

j=1

Setting α > 0, such that max{α +

ε |pm | , α

+

ε |p| }

< δ0 , since

lim u−α m ln um = 0,

um →∞

there exists a constant Mm > 0, such that 0 < u−α m ln um ≤ Mm (um ∈ [1, ∞)).

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

245

Multiple Half-Discrete Hilbert-Type Inequalities

In view of (6.2), we obtain ∞ 0< x−1 m+1 Bm (xm+1 )dxm+1 1 ∞ ∞ ∞ ≤ Mm ··· kλ (u1 , · · · , um , 1) 0

0

0 m−1 2

×

(λj + pε )−1 (λm + αpm +ε )−1 pm j um du1

uj

· · · dum−1 dum

j=1

ε < ∞, = Mm k λm+1 − α − p and then, by (6.16), we have ≥ k λm+1 − ε − εO(1). I(ε) p By Lemma 6.4, we still obtain

∞ λ ε −ε−1 2 +ε− pm+1 I(ε) ≤ ε n n n=1 m 2

×

0

(λj + pε )−1 j

xj

= ε 1+

∞

n

−ε−1

∞ 0

n=2

× ≤ ε 1+

···

dx1 · · · dxm ,

j=1

∞

m 2

···

∞ 0

(uj = nxj (j = 1, · · · , m))

∞ 0

kλ (u1 , · · · , um , 1)

(λj + pε )−1

uj

kλ (nx1 , · · · , nxm , 1)

j

du1 · · · dum

j=1

ε y −ε−1 dy k(λm+1 − ) p 1 ε = (ε + 1)k(λm+1 − ). p ∞

Then, by (6.4), we have (6.15). 6.2.2

Two Preliminary Inequalities

Lemma 6.8. If fi (xi )(i = 1, · · · , m) are non-negative measurable functions in R+ , then

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

246

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

(i) for pi > 1(i = 1, · · · , m + 1), we have the following inequality: J=

∞

n

pλm+1 −1

∞

0

n=1

×

···

m 2

∞ 0

kλ (x1 , · · · , xm , n) p p1

fj (xj )dx1 · · · dxm

j=1

≤ [k(λm+1 )]

1 pm+1

m 2 i=1

∞

0

p (1−λi )−1 ωi (xi )xi i fi (xi )dxi

p1

i

; (6.17)

(ii) for 0 < p1 < 1, pi < 0 (i = 2, · · · , m+1), or for pi < 0 (i = 1, · · · , m), 0 < pm+1 < 1, we have the reverse of (6.17). Proof. (i) For pi > 1(i = 1, · · · , m + 1)(p > 1), by H¨ older’s inequality with weight (see Kuang [47]) and (6.1), we have

∞ 0

⎧ ⎨ =

⎩

···

∞

0

∞ 0

···

×

kλ (x1 , · · · , xm , n)

m 2

p fi (xi )dx1 · · · dxm

i=1 ∞

0 m 2

kλ (x1 , · · · , xm , n) ⎡ i −1)(1−pi ) λm+1 −1 ⎣x(λ n i

⎡ × ⎣n(λm+1 −1)(1−pm+1 )

m 2

⎤p

λ −1 ⎦

xj j

j=1

∞ 0

×

···

m 2 i=1

⎡

∞ 0

i

λ −1 xj j ⎦

fi (xi )

j=1(j =i)

i=1

≤

⎤ p1

m 2

1 m+1

dx1 · · · dxm

⎫p ⎬ ⎭

kλ (x1 , · · · , xm , n)

⎣xi(λi −1)(1−pi ) nλm+1 −1

m 2 j=1(j =i)

⎤ pp

i

λ −1 ⎦

xj j

fip (xi )dx1 · · · dxm

⎧

∞ ∞ ⎨ pm+1 (1−λm+1 )−1 λm+1 × n ··· kλ (x1 , · · · , xm , n) n ⎩ 0 0

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

247

Multiple Half-Discrete Hilbert-Type Inequalities

⎫ pm+1 ⎬ p

×

m 2

λ −1

xj j

j=1 p pm+1

=

[ωm+1 (n)] npλm+1 −1 ×

m 2

⎡

∞

0

dx1 · · · dxm

···

∞ 0

⎭

kλ (x1 , · · · , xm , n)nλm+1 −1

⎣xi(λi −1)(1−pi )

⎤ pp

m 2

i

λ −1 ⎦

xj j

fip (xi )dx1 · · · dxm .

j=1(j =i)

i=1

Since ωm+1 (n) = k(λm+1 ), by the Lebesgue term by term integration theorem (see Kuang [49]), we ﬁnd ⎧ ⎪ ∞ ∞ ∞ ⎨ 1 pm+1 J ≤ [k(λm+1 )] ··· kλ (x1 , · · · , xm , n)nλm+1 −1 ⎪ 0 ⎩ n=1 0 ⎫ p1 ⎡ ⎤ pp ⎪ i m m ⎬ 2 2 λj −1 ⎦ (λi −1)(1−pi ) p ⎣ xi × xj fi (xi )dx1 · · · dxm ⎪ ⎭ i=1 j=1(j =i) ⎧ ⎪ ∞ ∞ ⎨ ∞ 1 = [k(λm+1 )] pm+1 ··· kλ (x1 , · · · , xm , n)nλm+1 −1 ⎪ 0 ⎩ 0 n=1 ⎫ p1 ⎡ ⎤ pp ⎪ i m m ⎬ 2 2 λj −1 ⎦ (λi −1)(1−pi ) p ⎣ × xj fi (xi )dx1 · · · dxm . xi ⎪ ⎭ i=1 j=1(j =i) In view of

m

1 i=1 (pi /p)

= 1, still by H¨older’s inequality, we obtain ⎧ ⎡ ∞ m ⎨ ∞ 2 1 λi pm+1 ⎣ xi nλm+1 −1 J ≤ [k(λm+1 )] ⎩ 0 n=1 i=1 ∞ ∞ m 2 λ −1 × ··· kλ (x1 , · · · , xm , n) xj j 0

0

⎤

j=1(j =i) p (1−λi )−1 pi fi (xi )dxi

× dx1 · · · dxi−1 dxi+1 · · · dxm ⎦xi i

= [k(λm+1 )]

1 pm+1

m 2 i=1

∞

0

and then, inequality (6.17) follows.

⎫ ppp i ⎬ ⎭

p (1−λi )−1 pi ωi (xi )xi i fi (xi )dxi

p1

i

,

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

248

Half-Discrete Hilbert-Type Inequalities

(ii) For 0 < p1 < 1, pi < 0 (i = 2, · · · , m + 1), or for pi < 0 (i = 1, · · · , m), 0 < pm+1 < 1, by the reverse H¨older’s inequality (see Kuang [47]) and the same way, we obtain the reverse of (6.17). For λm+1 = λ2 , replacing kλ (x1 , · · · , xm , n) by kλ (nx1 , · · · , nxm , 1) in Lemma 6.8, we still have Lemma 6.9. Suppose that λm+1 = λ2 , fi (xi ) (i = 1, · · · , m) are nonnegative measurable functions in R+ . Then (i) if pi > 1 (i = 1, · · · , m + 1), we have the following inequality: ⎧ ⎛ ∞ ∞ ∞ ⎨ pλ −1 ⎝ 2 n ··· kλ (nx1 , · · · , nxm , 1) J1 = ⎩ 0 0 n=1 ⎞p ⎫ p1 m ⎬ 2 fj (xj )dx1 · · · dxm ⎠ × ⎭ j=1 p1 m ∞ 2 1 i p (1−λi )−1 ≤ [k(λm+1 )] pm+1 wi (xi )xi i fi (xi )dxi ; (6.18) i=1

0

(ii) if 0 < p1 < 1, pi < 0 (i = 2, · · · , m + 1), or pi < 0 (i = 1, · · · , m), 0 < pm+1 < 1, then, we have the reverse of (6.18). 6.2.3

Main Results and Operator Expressions

In the following, we set two functions: p (1−λi )−1 ϕi (xi ) = xi i (xi ∈ R+ ; i = 1, · · · , m), pm+1 (1−λm+1 )−1 (n ∈ N). and ψ(n) = n The spaces Lpi, ϕi (R+ ) and lpm+1, ψ with the norms ||fi ||pi, ϕi and ||a||pm+1, ψ are deﬁned by ∞ p1 i ϕi (xi )|fi (xi )|pi dxi <∞ , Lpi, ϕi (R+ ) = fi ; ||fi ||pi, ϕi = 0

lpm+1, ψ

⎧ ⎫ ∞ p 1 ⎨ ⎬ m+1 = a = {an }; ||a||pm+1,ψ = ψ(n)|an |pm+1 <∞ . ⎩ ⎭ n=1

Theorem 6.1. Suppose that pi > 1 (i = 1, · · · , m + 1), there exists a i ∈ (λi − δ0 , λi + δ0 ) (i = 1, · · · , m + 1), constant δ0 > 0 such that, for any λ m+1 i=1 λi = λ, ∞ ∞ m 2 −1 λ m+1 ) = k(λ ··· kλ (u1 , · · · , um , 1) uj j du1 · · · dum ∈ R+ , 0

0

j=1

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

249

Multiple Half-Discrete Hilbert-Type Inequalities

and kλ (x1 , · · · , xm , y) y λm+1 −1 is decreasing with respect to y ∈ R+ and strictly decreasing in an interval I ⊂ (1, ∞). If fi (xi ) ≥ 0, fi ∈ Lpi, ϕi (R+ ) (i = 1, · · · , m), an ≥ 0, a = {an }∞ n=1 ∈ lpm+1, ψ , ||fi ||pi, ϕi > 0, ||a||pm+1, ψ > 0, then, we have the following equivalent inequalities: ∞ ∞ ∞ m 2 I= an ··· kλ (x1 , · · · , xm , n) fj (xj )dx1 · · · dxm 0

n=1

0

j=1

< k(λm+1 )||a||pm+1, ψ

m 2

||fi ||pi, ϕi ,

(6.19)

i=1

and J=

⎧ ∞ ⎨ ⎩

⎛ npλm+1 −1 ⎝

∞

0

n=1 m 2

×

j=1

< k(λm+1 )

m 2

···

∞ 0

kλ (x1 , · · · , xm , n)

⎞p ⎫ p1 ⎬ fj (xj )dx1 · · · dxm ⎠ ⎭

||fi ||pi, ϕi ,

(6.20)

i=1

where the constant factor k(λm+1 ) in the above inequalities is the best possible. Proof. By (6.17) and the assumptions, since i (xi ) < k(λm+1 ), we have (6.20). By H¨older’s inequality (see Kuang [47]), we obtain

∞ ∞ ∞ −1 ··· kλ (x1 , · · · , xm , n) (ψ(n)) pm+1 I= 0

n=1

×

m 2

0

1

fj (xj )dx1 · · · dxm [(ψ(n)) pm+1 an ]

j=1

≤ J||a||pm+1, ψ .

(6.21)

Then, by (6.20), we have inequality (6.19). Assuming that (6.19) is valid and setting an = npλm+1 −1

∞

0

···

∞

0

kλ (x1 , · · · , xm , n) ×

m 2 j=1

p−1 fj (xj )dx1 · · · dxm

(n ∈ N),

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

250

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

then, we have J p−1 = ||a||pm+1, ψ . By (6.17), we have J < ∞. If J = 0, then (6.20) is trivially valid; if J > 0, then, by (6.19), it follows that p

= J p = I < k(λm+1 )||a||pm+1, ψ ||a||pm+1 m+1, ψ

m 2

||fi ||pi, ϕi ,

that is,

i=1 p

−1

||a||pm+1 = J < k(λm+1 ) m+1, ψ

m 2

||fi ||pi, ϕi ,

i=1

and then, inequality (6.20) follows. Hence, (6.19) and (6.20) are equivalent. an as follows: For 0 < ε < δ0 min1≤j≤m+1 {pj , p}, we set fi (xi ) and 0, 0 < xi < 1, (i = 1, · · · , m), fi (xi ) = (λi − pε )−1 i , xi ≥ 1 xi an = n

(λm+1 − p

ε m+1

)−1

, n ∈ N.

If there exists a constant k(≤ k(λm+1 )), such that (6.19) is valid as we replace k(λm+1 ) by k, then, by (6.13), we ﬁnd k(λm+1 ) + o(1) = I(ε) ∞ an =ε n=1

∞ 0

···

∞

0

kλ (x1 , · · · , xm , n)

×

m 2

fj (xj )dx1 · · · dxm

j=1

< εk|| a||pm+1, ψ = εk 1 +

m 2

||fi ||pi, ϕi

i=1 ∞ n=2

< εk 1 +

1

∞

1

p

1 m+1

n1+ε dy y 1+ε

m p 2 1 i 1

ε

p

1 m+1

i=1 m 2 i=1

1 ε

p1

i

1

= k(1 + ε) pm+1 ,

and then, k(λm+1 ) ≤ k (ε → 0+ ). Hence, k = k(λm+1 ) is the best constant factor of (6.19). By the equivalency, the constant factor k(λm+1 ) in (6.20) is still the best possible. Otherwise, it leads to a contradiction by (6.21) that the constant factor in (6.19) is not the best possible.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

251

Multiple Half-Discrete Hilbert-Type Inequalities

Theorem 6.2. For λm+1 = λ2 , replacing kλ (x1, · · · , xm , y) y λm+1 −1 by

kλ (yx1, · · · , yxm , 1) y λm+1 −1 , in this case, let the assumptions of Theorem 6.1 be fulﬁlled. Then, we have the following equivalent inequalities with the same best constant factor k(λm+1 ) : ∞ ∞ ∞ m 2 an ··· kλ (nx1 , · · · , nxm , 1) fj (xj )dx1 · · · dxm 0

n=1

0

j=1 m 2

< k(λm+1 )||a||pm+1, ψ

||fi ||pi, ϕi ,

(6.22)

i=1

and

∞

n

pλ 2 −1

0

n=1

×

m 2

∞

···

∞

0

kλ (nx1 , · · · , nxm , 1)

p p1

fj (xj )dx1 · · · dxm

< k(λm+1 )

j=1

m 2

||fi ||pi, ϕi .

(6.23)

i=1

Proof. We only prove that the constant factor in (6.22) is the best possible. The other parts of the proof are omitted. an as follows: For 0 < ε < δ0 min1≤j≤m+1 {pj , p}, we set fi (xi ) and (λi + pε )−1 i , 0 < xi ≤ 1, (i = 1, · · · , m), fi (xi ) = xi 0, xi > 1 an = n

(λ 2 −p

ε m+1

)−1

, n ∈ N.

If there exists a constant k(≤ k(λm+1 )), such that (6.22) is valid as we replace k(λm+1 ) by k, then, by (6.15), we have k(λm+1 ) + o(1) = I(ε) ∞ m ∞ ∞ 2 an ··· kλ (nx, · · · , nxm , 1) =ε fj (xj )dx1 · · · dxm n=1

0

0

< εk|| a||pm+1, ψ = k 1+

m 2 i=1

∞ n=2

< εk 1 +

1

||fi ||pi, ϕi p

1 m+1

n1+ε

∞ 1

j=1

dy y 1+ε

m p 2 1 i 1

ε

p

i=1 m 2

1 m+1

i=1

1 ε

p1

i

1

= k(1 + ε) pm+1 ,

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

252

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and then, k(λm+1 ) ≤ k(ε → 0+ ). Hence, k = k(λm+1 ) is the best possible constant factor of (6.22). Remark 6.1. With the assumptions of Theorem 6.1, we deﬁne a ﬁrst kind of multiple half-discrete Hilbert-type operator with the homogeneous kernel

T :

m 2

Lpi, ϕi (R+ ) → lpm+1,ψ1−p

i=1

as follows: m For any f = (f1 , · · · , fm ) ∈ i=1 Lpi, ϕi (R+ ), there exists a T f, satisfying ∞ ∞ m 2 T f (n) = ··· kλ (x1 , · · · , xm , n) fj (xj )dx1 · · · dxm (n ∈ N). 0

0

j=1

(6.24) Then, by (6.20), it follows that ||T f ||p,ψ1−p < k(λm+1 )

m 2

||fi ||pi, ϕi ,

i=1

and then, T f ∈ lp,ψ1−p . Hence, T is a bounded linear operator with ||T || ≤ k(λm+1 ). Since the constant factor in (6.20) is the best possible, we have ||T f || 1−p m p,ψ = k(λm+1 ). (6.25) ||T || = sup m f ( =θ)∈ i=1 Lpi, ϕi (R+ ) i=1 ||fi ||pi, ϕi With the assumptions of Theorem 6.2, we deﬁne a ﬁrst kind of multiple half-discrete Hilbert-type operator with the non-homogeneous kernel m 2 Lpi, ϕi (R+ ) → lpm+1,ψ1−p T1 : i=1

as follows: m For any f = (f1 , · · · , fm ) ∈ i=1 Lpi, ϕi (R+ ), there exists a T1 f, satisfying ∞ ∞ m 2 T1 f (n) = ··· kλ (nx1 , · · · , nxm , 1) fj (xj )dx1 · · · dxm (n ∈ N). 0

0

j=1

(6.26) Then, by (6.23), it follows that ||T1 f ||p,ψ1−p < k(λm+1 )

m 2 i=1

||fi ||pi, ϕi ,

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

253

Multiple Half-Discrete Hilbert-Type Inequalities

and then, T1 f ∈ lp,ψ1−p . Hence, T1 is a bounded linear operator with ||T1 || ≤ k(λm+1 ). Since the constant factor in (6.23) is the best possible, we have ||T1 || =

6.2.4

||T1 f ||p,ψ1−p m = k(λm+1 ). i=1 ||fi ||pi, ϕi i=1 Lpi, ϕi (R+ ) sup

f ( =θ)∈

(6.27)

m

Some Kinds of Reverse Inequalities p (1−λ1 )−1

For ϕ1 (x1 ) = x11 θ1 (x1 ) =

1 k(λm+1 )

,

1/x1

0

λm+1 −1 um+1

∞

0

×

···

m 2

∞ 0

kλ (1, u2, · · · , um+1 )

λ −1 uj j du2

· · · dum dum+1 ,

j=2

and for λm+1 = λ2 , θ1 (x1 ) =

1 k(λm+1 )

0

x1

u1λ1 −1

0

∞

··· ×

∞

0 m 2

kλ (u1, · · · , um , 1)

λ −1 uj j du2

· · · dum du1 .

j=2

We also set two functions Φ1 (x1 ) = (1 − θ1 (x1 ))ϕ1 (x1 ), and 1 (x1 ) = (1 − θ1 (x1 ))ϕ1 (x1 ). Φ For pi < 1(pi = 0), the spaces Lpi, ϕi (R+ ) and lpm+1, ψ with ||fi ||pi, ϕi and ||a||pm+1, ψ are not normed spaces. But we still use them as the formal symbols in the following: Theorem 6.3. Suppose that 0 < p1 < 1, pi < 0 (i = 2, · · · , m + 1), i ∈ (λi − δ0 , λi + δ0 ) there exists a constant δ0 > 0, such that, for any λ m+1 (i = 1, · · · , m + 1), i=1 λi = λ, ∞ ∞ m 2 −1 λ m+1 ) = ··· kλ (u1, · · · , um , 1) uj j du1 · · · dum ∈ R+ , k(λ 0

0

j=1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

254

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and kλ (x1, · · · , xm , y) y λm+1 −1 is decreasing with respect to y ∈ R+ and strictly decreasing in an interval I ⊂ (1, ∞). There exist constants α, L > 0, such that (6.9) is satisﬁed for i = 1, and α−λ

A1 (um+1 ) ≤ L um+1m+1 (um+1 ∈ R+ ). If f1 (x1 ) ≥ 0, f1 ∈ Lp1, Φ1 (R+ ), ||f1 ||p1, Φ1 > 0, fi (xi ) ≥ 0, fi ∈ Lpi, ϕi (R+ ), ||fi ||pi, ϕi > 0 (i = 2, · · · , m), an ≥ 0, a = {an }∞ n=1 ∈ lpm+1, ψ , ||a||pm+1, ψ > 0, then, we have the following equivalent inequalities with the same best possible constant factor k(λm+1 ): ∞ ∞ ∞ m 2 I= an ··· kλ (x1 , · · · , xm , n) fj (xj )dx1 · · · dxm 0

n=1

0

j=1

> k(λm+1 )||a||pm+1, ψ ||f1 ||p1 ,Φ1

m 2

||fi ||pi, ϕi ,

(6.28)

i=2

and

J=

∞

n

pλm+1 −1

0

n=1

×

m 2

∞

···

∞ 0

kλ (x1 , · · · , xm , n)

p p1

fj (xj )dx1 · · · dxm

> k(λm+1 )||f1 ||p1 ,Φ1

j=1

m 2

||fi ||pi, ϕi .

i=2

(6.29) Proof.

By the reverse of (6.17), since 1 (x1 ) > k(λm+1 )(1 − θ1 (x1 )), and i (xi ) < k(λm+1 ),

in view of the assumptions made, we have (6.29). By H¨older’s inequality, we obtain

∞ ∞ ∞ −1 ··· kλ (x1 , · · · , xm , n) I= (ψ(n)) pm+1 n=1 m 2

×

0

0

1

fj (xj )dx1 · · · dxm [(ψ(n)) pm+1 an ] ≥ J||a||pm+1, ψ . (6.30)

j=1

Then, by (6.29), we have inequality (6.28). Assuming that (6.28) is valid, setting an as follows: an = npλm+1 −1

∞

0

···

∞

0

kλ (x1 , · · · , xm , n) ×

m 2 j=1

p−1 fj (xj )dx1 · · · dxm

(n ∈ N),

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

255

Multiple Half-Discrete Hilbert-Type Inequalities

then, we ﬁnd J p−1 = ||a||pm+1, ψ . By the reverse of (6.17), we have J > 0. If J = ∞, then (6.29) is trivially valid; if J < ∞, then by (6.28), it follows that p

||a||pm+1 = J p = I > k(λm+1 )||a||pm+1, ψ ||f1 ||p1 ,Φ1 m+1, ψ

m 2

||fi ||pi, ϕi , that is,

i=2 p

−1

= J > k(λm+1 )||f1 ||p1 ,Φ1 ||a||pm+1 m+1, ψ

m 2

||fi ||pi, ϕi .

i=2

Thus, inequality (6.29) follows. Hence, (6.28) and (6.29) are equivalent. For 0 < ε < δ0 min1≤j≤m+1 {|pj |, p}, we set fi (xi ) and an as follows: 0, 0 < xi < 1, (i = 1, · · · , m), fi (xi ) = (λi − pε )−1 i xi , xi ≥ 1 an = n

(λm+1 − p

ε m+1

)−1

,

n ∈ N.

If there exists a constant k(≥ k(λm+1 )) such that (6.28) is valid, as we replace k(λm+1 ) by k, then, by (6.13) and Lemma 6.4, we have k(λm+1 ) + o(1) = I(ε) ∞ ∞ ∞ m 2 an ··· kλ (x1 , · · · , xm , n) =ε fj (xj )dx1 · · · dxm n=1

0

0

j=1

> εk|| a||pm+1, ψ ||f1 ||p1 ,Φ1 = εk

p

∞

1

n=1

n1+ε ∞

> εk 1 +

1

= k(ε + 1)

1 pm+1

m 2

||fi ||pi, ϕi

i=2 1 m+1

dy y 1+ε

(

p

)1 m 1 1 dx p1 2 1 pi 1−O xα x1+ε ε 1 i=2 1 p1 2 m 1 1 1 pi − O(1) ε ε

∞ 1

1 m+1

(1 − εO(1))

i=2

1 p1

,

and then, k(λm+1 ) ≥ k(ε → 0+ ). Hence, k = k(λm+1 ) is the best possible constant factor of (6.28). By the equivalency, the constant factor k(λm+1 ) in (6.29) is the best possible. Otherwise, it leads to a contradiction by (6.30) that the constant factor in (6.28) is not the best possible.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

256

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Theorem 6.4. Suppose that 0 < p1 < 1, pi < 0(i = 2, · · · , m + 1), λm+1 = λ i ∈ (λi − δ0 , λi + δ0 )(i = , there exists a constant δ0 > 0 such that for any λ 2 m+1 1, · · · , m + 1), i=1 λi = λ, ∞ ∞ m 2 −1 λ m+1 ) = k(λ ··· kλ (u1 , · · · , um , 1) uj j du1 · · · dum ∈ R+ , 0

0

j=1

and kλ (yx1 , · · · , yxm , 1) y λm+1 −1 is decreasing with respect to y ∈ R+ and strictly decreasing in an interval I ⊂ (1, ∞). There exist constants α, L > 0, such that (6.12) is satisﬁed for i = 1, · · · , m, and i B1 (ui ) ≤ L uα−λ (ui ∈ R+ ) (i = 1, · · · , m). i

If f1 (x1 ) ≥ 0, f1 ∈ Lp1, Φ 1 (R+ ), ||f1 ||p1, Φ 1 > 0, fi (xi ) ≥ 0, fi ∈ Lpi, ϕi (R+ ), ||fi ||pi, ϕi > 0 (i = 2, · · · , m), an ≥ 0, a = {an }∞ n=1 ∈ lpm+1, ψ , ||a||pm+1, ψ > 0, then, we have the following equivalent inequalities with the same best possible constant factor k(λm+1 ): ∞ ∞ ∞ m 2 an ··· kλ (nx1 , · · · , nxm , 1) fj (xj )dx1 · · · dxm 0

n=1

0

j=1

> k(λm+1 )||a||pm+1, ψ ||f1 ||p1, Φ 1

m 2

||fi ||pi, ϕi ,

(6.31)

i=2

and

∞

n

pλ 2 −1

∞ 0

n=1

··· ×

> k(λm+1 )||f1 ||p1, Φ 1

∞

0

m 2

kλ (nx1 , · · · , nxm , 1) p p1

fj (xj )dx1 · · · dxm

j=1 m 2

||fi ||pi, ϕi .

(6.32)

i=2

Proof. We only prove that the constant factor in (6.31) is the best possible. The other parts of the proof are omitted. For 0 < ε < δ0 min1≤j≤m+1 {|pj |, p}, we set fi (xi ) and an as follows: (λ + ε )−1 i pi , 0 < xi ≤ 1, (i = 1, · · · , m), fi (xi ) = xi 0, xi > 1 an = n

(λ 2−p

ε m+1

)−1

,

n ∈ N.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

257

Multiple Half-Discrete Hilbert-Type Inequalities

If there exists a constant k(≥ k(λm+1 )) such that (6.31) is valid as we replace k(λm+1 ) by k, then, by (6.15), we have k(λm+1 ) + o(1) = I(ε) ∞ m ∞ ∞ 2 an ··· kλ (nx, · · · , nxm , 1) =ε fj (xj )dx1 · · · dxm 0

n=1

0

j=1

> εk|| a||pm+1, ψ ||f1 ||p1, Φ 1 = εk

p

∞

1

n=1

n1+ε ∞

> εk 1 +

1

m 2

||fi ||pi, ϕi

i=2 1 m+1

dy y 1+ε

(

p

1

0 1 m+1

1

(1 −

dx O(xα 1 )) 1−ε x

) p11 2 m p1 1 i i=1

ε

p1 2 m p1 1 1 1 i − O(1) ε ε i=2

1

= k(ε + 1) pm+1 (1 + εO(1)) p1 , and then, k(λm+1 ) ≥ k(ε → 0+ ). Hence, k = k(λm+1 ) is the best value of (6.31). Similarly, we still have Theorem 6.5. Suppose that pi < 0 (i = 1, · · · , m), 0 < pm+1 < 1, there i ∈ (λi − δ0 , λi + δ0 )(i = exists a constant δ0 > 0, such that, for any λ m+1 1, · · · , m + 1), i=1 λi = λ, m+1 ) = k(λ

∞ 0

···

∞ 0

kλ (u1, · · · , um , 1)

m 2

−1 λ

uj j

du1 · · · dum ∈ R+ ,

j=1

and kλ (x1, · · · , xm , y) y λm+1 −1 is strictly decreasing with respect to y ∈ R+ and strictly decreasing in an interval I ⊂ (1, ∞). If fi (xi ), an ≥ 0, fi ∈ Lpi, ϕi (R+ ) (i = 1, · · · , m), a = {an }∞ n=1 ∈ lpm+1, ψ , ||fi ||pi, ϕi > 0, ||a||pm+1, ψ > 0, then, we have the following equivalent reverse inequalities with the same best possible constant factor k(λm+1 ): ∞ n=1

an

∞ 0

···

∞ 0

kλ (x1 , · · · , xm , n)

> k(λm+1 )||a||pm+1, ψ

m 2

fj (xj )dx1 · · · dxm

j=1 m 2 i=1

||fi ||pi, ϕi ,

(6.33)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

258

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and

∞

n

pλm+1 −1

∞ 0

n=1

m 2

×

···

∞ 0

kλ (x1 , · · · , xm , n) p p1

fj (xj )dx1 · · · dxm

j=1 m 2

> k(λm+1 )

||fi ||pi, ϕi .

(6.34)

i=1

Theorem 6.6. Suppose that pi < 0(i = 1, · · · , m), 0 < pm+1 < 1, λm+1 = λ i ∈ (λi − δ0 , λi + δ0 ) , there exists a constant δ0 > 0 such that for any λ 2 m+1 (i = 1, · · · , m + 1), i=1 λi = λ, m+1 ) = k(λ

∞ 0

···

∞ 0

kλ (u1, · · · , um , 1)

m 2

−1 λ

uj j

du1 · · · dum ∈ R+ ,

j=1

and kλ (yx1 , · · · , yxm , 1) y λm+1 −1 is decreasing with respect to y ∈ R+ and strictly decreasing in an interval I ⊂ (1, ∞). There exist constants α, L > 0, such that (6.12) is satisﬁed for i = 1, · · · , m, and i (ui ∈ R+ ; B1 (ui ) ≤ Luα−λ i

i = 1, · · · , m).

If fi (xi ), an ≥ 0, fi ∈ Lpi, ϕi (R+ ) (i = 1, · · · , m), a = {an }∞ n=1 ∈ lpm+1, ψ , ||fi ||pi, ϕi > 0, ||a||pm+1, ψ > 0, then, we have the following equivalent inequalities with the same best possible constant factor k(λm+1 ): ∞

an

n=1

∞ 0

···

∞

0

∞ n=1

×

m 2 j=1

n

pλ 2 −1

0

m 2

fj (xj )dx1 · · · dxm

j=1

> k(λm+1 )||a||pm+1, ψ

kλ (nx1 , · · · , nxm , 1) m 2 i=1

∞

||fi ||pi, ϕi ,

···

fj (xj )dx1 · · · dxm

∞ 0

(6.35)

kλ (nx1 , · · · , nxm , 1)

p p1 > k(λm+1 )

m 2 i=1

||fi ||pi, ϕi .

(6.36)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

259

Multiple Half-Discrete Hilbert-Type Inequalities

6.3

Second Kind of Multiple Hilbert-type Inequalities

6.3.1

Lemmas Related to the Weight Functions

Definition 6.3. If ni ∈ N(i = 1, · · · , m), xm+1 ∈ R+ , we deﬁne weight functions ωi (ni ) and m+1 (xm+1 ) as follows: ∞ ∞ ∞ ∞ λm+1 −1 ωi (ni ) = nλi i xm+1 ··· 0

···

nm =1

∞

ni+1 =1 ni−1 =1 m 2

kλ (n1 , · · · , nm , xm+1 )

n1 =1

dxm+1 , (6.37)

j=1(j =i)

∞

λ

m+1 m+1 (xm+1 ) = xm+1

λ −1

nj j

···

nm =1

∞

kλ (n1 , · · · , nm , xm+1 )

n1 =1

m 2

λ −1

nj j

. (6.38)

j=1

In particular, for λm+1 = λ2 , replacing the kernel kλ (n1, · · · , nm , xm+1 ) by kλ (xm+1 n1 , · · · , xm+1 nm , 1) in (6.37) and (6.38), we can still deﬁne another weight coeﬃcients wi (ni )(i = 1, · · · , m) and w m+1 (xm+1 ) as follows: ∞ ∞ ∞ ∞ λ 2 −1 xm+1 ··· wi (ni ) = nλi i 0

···

nm =1

∞

ni+1 =1 ni−1 =1

kλ (xm+1 n1 , · · · , xm+1 nm , 1)

n1 =1 λ 2

w m+1 (xm+1 ) = xm+1

∞ nm =1

m 2

λ −1

nj j

dxm+1 ,

j=1(j =i)

···

∞

kλ (xm+1 n1 , · · · , xm+1 nm , 1)

n1 =1

m 2

λ −1

nj j

.

j=1

Lemma 6.10. Let the assumptions of Deﬁnition 6.3 be fulﬁlled and additionally, let ∞ ∞ m 2 λ −1 ··· kλ (u1 , · · · , um , 1) uj j du1 · · · dum ∈ R+ , k(λm+1 ) = 0

0

j=1

(6.39) λ −1 be decreasing with respect to yj ∈ R+ and kλ (y1 , · · · , ym , xm+1 )yj j strictly decreasing in an interval Ij ⊂ (1, ∞) (j = 1, · · ·, m). Then, we have (i) 0 < k(λm+1 )(1 − θm+1 (xm+1 )) < m+1 (xm+1 ) < k(λm+1 ),

(6.40)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

260

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

where θm+1 (xm+1 ) = 1 −

1 k(λm+1 )

∞ 1/xm+1

×

···

m 2

∞ 1/xm+1

λ −1

uj j

kλ (u1 , · · · , um , 1)

du1 · · · dum > 0;

(6.41)

j=1

(ii) for m = 1, ω1 (n1 ) = k(λ2 ); for m ≥ 2, i = 1, · · · , m, we have 0 < k(λm+1 )(1 − θi (ni )) < ωi (ni ) < k(λm+1 ) where θi (ni ) = 1 −

1 k(λm+1 ) ×

∞ 1 ni

m+1 2

···

λ −1

uj j

∞ 1 ni

(ni ∈ N),

(6.42)

kλ (u1 , · · · , ui−1 , 1, ui+1 , · · · , um+1 )

du1 · · · dui−1 dui+1 · · · dum+1 > 0.

(6.43)

j=1(j =i)

Moreover, if there exist constants α, L > 0, such that for i = 1, · · · , m + 1, ∞ ∞ Ai (uk ) = ··· kλ (u1 , · · · , ui−1 , 1, ui+1 , · · · , um+1 ) 0

×

0 m+1 2

λ −1

uj j

du1 · · · dui−1 dui+1 · · · duk−1 duk+1 · · · dum+1

j=1(j =i,k)

≤

k Luα−λ (uk k

∈ R+ ),

(6.44)

then, we have

θm+1 (xm+1 ) = O

and

θi (ni ) = O

Proof.

1 nα i

1 xα m+1

, (xm+1 > 0),

(n ∈ N; i = 1, · · · , m).

(i) By the decreasing property, it follows that ∞ ∞ λm+1 m+1 (xm+1 ) < xm+1 ··· kλ (x1 , · · · , xm , xm+1 ) 0

0

×

m 2

j=1

λ −1

xj j

dx1 · · · dxm .

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

261

Multiple Half-Discrete Hilbert-Type Inequalities

Setting xj = xm+1 uj (j = 1, · · · , m) in the above integral, by (6.39), we ﬁnd ∞ ∞ λm+1 m m+1 (xm+1 ) < xm+1 xm+1 ··· kλ (xm+1 u1 , · · · , xm+1 um , xm+1 ) 0

×

m 2

0

(xm+1 uj )λj −1 du1 · · · dum = k(λm+1 ).

j=1

By the decreasing property, we still have ∞ ∞ λm+1 ··· kλ (x1 , · · · , xm , xm+1 ) m+1 (xm+1 ) > xm+1 1

1

× =

∞ 1 xm+1

···

m 2

λ −1

xj j

dx1 · · · dxm

j=1 ∞ 1 xm+1

kλ (u1 , · · · , um , 1)

m 2

λ −1

uj j

du1 · · · dum

j=1

= k(λm+1 )(1 − θm+1 (xm+1 )) > 0, where θm+1 (xm+1 ) is deﬁned by (6.41). (ii) For m = 1, we ﬁnd ∞ kλ (n1, x2 )xλ2 2 −1 dx2 ω1 (n1 ) = nλ1 1 0 ∞ = kλ (u1, 1)u1λ1 −1 du1 = k(λ2 ); 0

for m ≥ 2, i = 1, · · · , m, by the decreasing property and Lemma 6.2, we have ∞ ∞ ··· kλ (x1 , · · · , xi−1 , ni , xi+1 , · · · , xm , xm+1 ωi (ni ) < nλi i ×

0 m+1 2

0

λ −1

xj j

dx1 · · · dxi−1 dxi+1 · · · dxm+1 , uj = xj /ni (j = i)

j=1(j =i)

= H(i) = k(λm+1 ), ωi (ni ) > nλi i

∞ 1

···

∞ 1

×

kλ (x1 , · · · , xi−1 , ni , xi+1 , · · · , xm , xm+1 ) m+1 2 j=1(j =i)

λ −1

xj j

dx1 · · · dxi−1 dxi+1 · · · dxm+1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

262

Half-Discrete Hilbert-Type Inequalities

∞

= 1/ni

···

∞

1/ni

kλ (u1 , · · · , ui−1 , 1, ui+1 , · · · , um+1 ) m+1 2

×

λ −1

uj j

du1 · · · dui−1 dui+1 · · · dum+1

j=1(j =i)

= k(λm+1 )(1 − θi (ni )) > 0, where θi (ni ) is given by (6.43). Moreover, we have 1 0 < θm+1 (xm+1 ) = [k(λm+1 ) k(λm+1 ) ∞ ∞ m 2 λ −1 − ··· kλ (u1 , · · · , um , 1) uj j du1 · · · dum ] 1/xm+1

≤

1 k(λm+1 )

where

1/xm+1

m 1/xm+1 k=1

Am+1 (uk ) =

∞

0

0

···

∞ 0

j=1

ukλk −1 Am+1 (uk )duk ,

kλ (u1 , · · · , um , 1) m 2

×

λ −1

uj j

du1 · · · duk−1 duk+1 · · · dum .

j=1(j =k)

By (6.44) (for i = m + 1), it follows that m 1/xm+1 L k uλk k −1 uα−λ duk 0 < θm+1 (xm+1 ) ≤ k k(λm+1 ) 0 k=1

mL = , αk(λm+1 )xα m+1 and then

1

(xm+1 > 0). By (6.43), (6.44) and the same way, we have θi (ni ) = O n1α (n ∈ N; i = i 1, · · · , m). θm+1 (xm+1 ) = O

xα m+1

Lemma 6.11. Let the assumptions of Deﬁnition 6.3 be fulﬁlled and additionally, let λm+1 = λ2 , k(λm+1 ) ∈ R+ , and λ

kλ (xm+1 y1 , · · · , xm+1 ym , 1) yj m+1

−1

be decreasing with respect to yj ∈ R+ and strictly decreasing in an interval Ij ⊂ (1, ∞) (j = 1, · · · , m). Then

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

263

Multiple Half-Discrete Hilbert-Type Inequalities

(i) we have m+1 (xm+1 ) < k(λm+1 ), 0 < k(λm+1 )(1 − θm+1 (xm+1 )) < w

(6.45)

where θm+1 (xm+1 ) = 1 −

1 k(λm+1 )

∞

∞

···

xm+1

×

kλ (u1 , · · · , um , 1)

xm+1 m 2

λ −1

uj j

du1 · · · dum > 0; (6.46)

j=1

(ii) for m = 1, w1 (n1 ) = k(λ2 ); for m ≥ 2, i = 1, · · · , m, we have 0 < k(λm+1 )(1 − θi (ni )) < wi (ni ) < k(λm+1 ) (ni ∈ N),

(6.47)

where θi (ni ) = 1 −

1 k(λm+1 ) ×

m+1 2

∞ 1 ni

···

λ −1

uj j

∞ 1 ni

kλ (u1 , · · · , ui−1, 1, ui+1 , · · · , um+1 )

du1 · · · dui−1 dui+1 · · · dum+1 > 0.

(6.48)

j=1(j =i)

Moreover, if there exist constants α, L > 0, such that for i = 1, · · · , m + 1, inequality (6.44) follows. Thus, k (uk ∈ R+ ), Ai (uk ) ≤ Luα−λ k

then, we have θm+1 (xm+1 ) = O(xα m+1 ) (xm+1 > 0), and θi (ni ) = O(n−α i ) (n ∈ N; i = 1, · · · , m). Proof.

(i) By the decreasing property, it follows that λ

2 w m+1 (xm+1 ) < xm+1 ∞ ∞ m 2 λ −1 × ··· kλ (xm+1 x1 , · · · , xm+1 xm , 1) xj j dx1 · · · dxm .

0

0

j=1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

264

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Setting uj = xm+1 xj (j = 1, · · · , m) in the above integral, we ﬁnd ∞ ∞ w m+1 (xm+1 ) < ··· kλ (u1 , · · · , um , 1) 0

0

×

λ −1

uj j

du1 · · · dum = k(λm+1 ),

j=1

λ

2 w m+1 (xm+1 ) > xm+1

m 2

∞ 1

···

∞ 1

kλ (xm+1 x1 , · · · , xm+1 xm , 1) ×

∞

∞

···

= xm+1

kλ (u1 , · · · , um , 1)

xm+1

m 2

λ −1

xj j

dx1 · · · dxm

j=1 m 2

uj λj −1 du1 · · · dum

j=1

= k(λm+1 )(1 − θm+1 (xm+1 )) > 0, where θm+1 (xm+1 ) is given by (6.46). (ii) For m = 1, since λ2 = λ2 = λ1 , we ﬁnd ∞ λ1 w1 (n1 ) = n1 kλ (x2 n1 , 1)xλ2 2 −1 dx2 0 ∞ = kλ (u1 , 1)u1λ1 −1 du1 = k(λ2 ); 0

for m ≥ 2, i = 1, · · · , m, by the decreasing property, we ﬁnd ∞ ∞ ∞ ∞ λi λi wi (ni ) < ni ··· wi (ni ) < ni ··· 0

0

0

0

×kλ (xm+1 x1 , · · · , xm+1 xi−1 , xm+1 ni , xm+1 xi+1 , · · · , xm+1 xm , 1) ×

∞

= 0

wi (ni ) >

nλi i

∞ 1

λ −1

xj j

dx1 · · · dxi−1 dxi+1 · · · dxm+1

j=1(j =i)

···

m+1 2

∞ 0

kλ (u1 , · · · um , 1)

λ −1

uj j

du1 · · · dum = k(λm+1 ),

j=1

···

m 2

∞ 1

× kλ (xm+1 x1 , · · · , xm+1 xi−1 , xm+1 ni , xm+1 xi+1 , · · · , xm+1 xm , 1) m+1 2 λ −1 × xj j dx1 · · · dxi−1 dxi+1 · · · dxm+1 . (6.49) j=1(j =i)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Multiple Half-Discrete Hilbert-Type Inequalities

ws-book9x6

265

Setting uj = xj /ni (j = i, m + 1), um+1 = (xm+1 ni )−1 in (6.49), it follows that ∞ ∞ wi (ni ) > ··· kλ (u1 , · · · , ui−1 , 1, ui+1 , · · · , um+1 ) 1/ni

1/ni

m+1 2

×

λ −1

uj j

du1 · · · dui−1 dui+1 · · · dum+1

j=1(j =i)

= k(λm+1 )(1 − θi (xi )) > 0, where θi (xi ) is given by (6.48). Moreover, by inequality (6.44), we have

1 k(λm+1 ) 0 < θm+1 (xm+1 ) = k(λm+1 ) ∞ ∞ m 2 λj −1 ··· kλ (u1 , · · · , um , 1) uj du1 · · · dum − xm+1

≤

1 k(λm+1 )

where Am+1 (uk ) =

0

∞

xm+1

m xm+1 0

k=1

···

∞ 0

j=1

uλk k −1 Am+1 (uk )duk ,

kλ (u1 , · · · , um , 1) ×

m 2

λ −1

uj j

du1 · · · duk−1 duk+1 · · · dum .

j=1(j =k)

By inequality (6.44) (for i = m + 1), it follows that 0 < θm+1 (xm+1 ) m xm+1 mLxα L m+1 k uλk k −1 uα−λ duk = ≤ , k k(λm+1 ) αk(λ ) m+1 0 k=1

and then

θm+1 (xm+1 ) = O xα m+1 (xm+1 > 0).

By (6.48), (6.44) and the same way, we have θi (ni ) = O(n−α i ) (n ∈ N; i = 1, · · · , m). Lemma 6.12. Suppose that there exists a constant δ0 > 0, such that for i = λ, k(λ i ∈ (λi − δ0 , λi + δ0 )(i = 1, · · · , m + 1), m+1 λ m+1 ) ∈ R+ . any λ i=1 If λ

kλ (y1 , · · · , ym , xm+1 )yj m+1

−1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

266

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

is decreasing with respect to yj ∈ R+ (j = 1, · · · , m), then, for 0 < ε < δ0 min1≤j≤m+1 {|pj |, |p|}, we have I1 (ε) = ε

∞ 1

×

m 2

λm+1 − p

xm+1

ε m+1

∞

−1

nm =1 (λj − pε )−1 j

nj

∞

···

kλ (n1 , · · · , nm , xm+1 )

n1 =1

dxm+1 = k(λm+1 ) + o(1)

(ε → 0+ ). (6.50)

j=1

Proof.

By the decreasing property, we ﬁnd ∞ ∞ ∞ λm+1 − p ε −1 I1 (ε) ≥ ε xm+1 m+1 ··· kλ (x1 , · · · , xm , xm+1 ) 1

1

×

m 2

1

(λj − pε )−1 j

xj

dx1 · · · dxm dxm+1 .

j=1

In view of the proof of Lemma 6.6, we have ε − εO(1). I1 (ε) ≥ k λm+1 + p Since we have

∞ λm+1 +ε− p ε −ε−1 m+1 I1 (ε) ≤ ε xm+1 xm+1 1

0

× =ε 1

∞

m 2

···

∞ 0

(λj − pε )−1 j

xj

kλ (x1 , · · · , xm , xm+1 )

dx1 · · · dxm dxm+1

j=1

∞

x−ε−1 m+1 k

ε ε λm+1 + dxm+1 = k λm+1 + , p p

then, by (6.4), we have (6.50).

Lemma 6.13. Suppose that λm+1 = λ2 , and there exists a constant δ0 > 0, i = λ, i ∈ (λi − δ0 , λi + δ0 )(i = 1, · · · , m + 1), m+1 λ such that, for any λ i=1 m+1 ) ∈ R+ . If k(λ λ

kλ (xm+1 y1 , · · · , xm+1 ym , 1) yj m+1

−1

is decreasing with respect to yj ∈ R+ (j = 1, · · · , m), then, for 0 < ε < δ0

min

{|pj |, |p|},

1≤j≤m+1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

267

Multiple Half-Discrete Hilbert-Type Inequalities

we have I1 (ε) = ε

λ ε 2+p

−1

xm+1 m+1

0

∞

× ×

1

∞

···

kλ (xm+1 n1 , · · · , xm+1 nm , 1)

nm =1 n1 =1 m 2 (λj − pε )−1 j

nj

dxm+1 = k(λm+1 ) + o(1)(ε → 0+ ). (6.51)

j=1

Proof.

By the decreasing property, we ﬁnd

I1 (ε) ≥ ε

1

xm+1

0

×

(λ 2 +p

m 2

1

=ε 0

)−1

j

xj

≥ε

0

×

m 2

∞

0

(λj − pε )−1 j

uj

1

kλ (xm+1 x1 , · · · , xm+1 xm , 1)

∞

···

kλ (u1 , · · · , um , 1)

xm+1 m 2

(λj − pε )−1 j

uj

du1 · · · dum dxm+1

j=1

xε−1 m+1

∞

∞

xε−1 m+1

xm+1

1

···

dx1 · · · dxm dxm+1 , (uj = xm+1 xj (j = 1, · · · , m))

×

∞

1

(λj − pε )−1

j=1

ε m+1

···

∞ 0

kλ (u1 , · · · , um , 1)

du1 · · · dum dxm+1 − ε

j=1

m i=1

0

1

x−1 m+1 Bi (xm+1 )dxm+1

m 1 ε −ε = k λm+1 + x−1 m+1 Bi (xm+1 )dxm+1 , p 0 i=1

(6.52)

where Bi (xm+1 ) =

∞ 0

×

···

m 2

∞

0

xm+1

j

∞ 0

0

(λj − pε )−1

uj

···

∞ 0

kλ (u1 , · · · , um , 1)

du1 · · · dui−1 dui dui+1 · · · dum

(1 ≤ i ≤ m).

without loss of generality,

we estimate

j=1

For i

=

1,

···,

m,

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

268

∞ 1

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

x−1 m+1 Bm (xm+1 ) dxm+1 as follows: 1 x−1 0< m+1 Bm (xm+1 )dxm+1 0

1

= 0

x−1 m+1

0

0

×

1

m 2

1

∞

···

0

(λj − pε )−1 j

uj

j=1

= 0

∞

xm+1

x−1 m+1 dxm+1

∞ 0

um

1

0

(− ln um )

∞ 0

du1 · · · dum−1 dum dxm+1

∞ 0

m 2

kλ (u1 , · · · , um , 1)

(λj − pε )−1 j

uj

du1 · · · dum−1 dum

j=1

···

···

× =

kλ (u1 , · · · , um , 1)

∞ 0

kλ (u1 , · · · , um , 1) ×

m 2

(λj − pε )−1 j

uj

du1 · · · dum−1 dum .

j=1

Setting α > 0 such that

max α +

ε ε ,α + |pm | |p|

< δ0 ,

since limum →0+ uα m (− ln um ) = 0, there exists a constant Mm > 0, such that 0 < uα m (− ln um ) ≤ Mm (um ∈ (0, 1]). We ﬁnd

1

x−1 m+1 Bm (xm+1 ) dxm+1 0 ∞ ∞ ∞ ≤ Mm ··· kλ (u1 , · · · , um , 1)

0<

0

0

×

0

m−1 2

(λj − pε )−1 (λm − αpm +ε )−1 pm j uj um du1

· · · dum−1 dum

j=1

ε < ∞. = k λm+1 − α + p Hence, by (6.52), we have

ε I1 (ε) ≥ k λm+1 + − εO(1). p

(6.53)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

269

Multiple Half-Discrete Hilbert-Type Inequalities

We still have

1 λ ε 2 + pm+1 −1 I1 (ε) ≤ ε xm+1 0

×

0

m 2 j=1 ∞

= 0

∞

(λj − pε )−1 j

xj

···

∞ 0

···

∞ 0

kλ (xm+1 x1 , · · · , xm+1 xm , 1)

dx1 · · · dxm dxm+1 , uj = xm+1 xj (j = 1, · · · , m)

kλ (u1 , · · · , um , 1) ×

m 2

(λj − pε )−1

uj

j

du1 · · · dum

j=1

ε = k λm+1 + . p Then, by (6.53) and (6.4), we have (6.51). 6.3.2

Two Preliminary Inequalities (j)

Lemma 6.14. If anj ≥ 0 (nj ∈ N; j = 1, · · · , m), then (i) for pi > 1(i = 1, · · · , m + 1), we have the following inequality: ∞ pλm+1 −1 ∞ ∞ xm+1 J1 = · · · kλ (n1 , · · · , nm , xm+1 ) [ωm+1 (xm+1 )]p−1 n =1 n =1 0 m 1 p p1 m 2 a(j) dxm+1 × nj ≤

m 2

∞

p (1−λi )−1

ωi (ni )ni i

ni =1

i=1

p1 pi (a(i) ni )

j=1

i

;

(6.54)

(ii) for 0 < p1 < 1, pi < 0 (i = 2, · · · , m+1), or for pi < 0 (i = 1, · · · , m), 0 < pm+1 < 1, we have the reverse of (6.54). older’s inequality, (6.37) Proof. (i) For pi > 1 (i = 1, · · · , m + 1), by H¨ and (6.38), we have ⎛ ⎞p ∞ ∞ m 2 ⎝ ⎠ ··· kλ (n1 , · · · , nm , xm+1 ) a(j) n nm =1

=

⎧ ⎪ ∞ ⎨

⎪ ⎩ nm =1

n1 =1

···

∞ n1 =1

j

j=1

kλ (n1 , · · · , nm , xm+1 )

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

270

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

×

m 2

⎡ i −1)(1−pi ) λm+1 −1 ⎣n(λ xm+1 i

⎤ p1

m 2

i

λ −1 ⎦

a(i) ni

nj j

j=1(j =i)

i=1

⎡ (λ

m+1 × ⎣xm+1

m 2

−1)(1−pm+1 )

⎤p

⎫p ⎪ ⎬

1 m+1

λ −1 ⎦

nj j

⎪ ⎭

j=1 ∞

≤

···

nm =1

×

m 2

∞

kλ (n1 , · · · , nm , xm+1 )

n1 =1

⎡

i −1)(1−pi ) λm+1 −1 ⎣n(λ xm+1 i

i

λ −1 ⎦

m+1 p (a(i) xm+1 ni ) ⎪ ⎩

nj j

j=1(j =i)

i=1

⎧ ⎪ ⎨

⎤ pp

m 2

p

(1−λm+1 )−1

p ⎤ ⎫ pm+1 ⎪ ⎬ λ −1 λm+1 × ⎣xm+1 ··· kλ (n1 , · · · , nm , xm+1 ) nj j ⎦ ⎪ ⎭ nm =1 n1 =1 j=1 ⎧ ⎪ ∞ ∞ p−1 ⎨ [ωm+1 (xm+1 )] = · · · kλ (n1 , · · · , nm , xm+1 ) pλm+1 −1 ⎪ xm+1 ⎩ nm =1 n1 =1

⎡

∞

×

∞

m 2

m 2

⎡

m 2

λm+1 −1 ⎣ni(λi −1)(1−pi ) xm+1

i

λ −1 ⎦

nj j

j=1(j =i)

i=1

⎫ ⎪ ⎬

⎤ pp

p (a(i) . ni ) ⎪ ⎭

Then, by ⎧ the Lebesgue term by term integration theorem, we obtain ⎪ ∞ ∞ ∞ ⎨ J≤ ··· kλ (n1 , · · · , nm , xm+1 ) ⎪ ⎩ 0 nm =1 n1 =1

×

m 2

⎡ λm+1 −1 ⎣ni(λi −1)(1−pi ) xm+1

=

⎪ ⎩ nm =1

···

×

⎤ pp

i

λ −1 ⎦

nj j

p (a(i) ni ) dxm

j=1(j =i)

i=1

⎧ ⎪ ∞ ⎨

m 2

∞ n1 =1

m 2 i=1

∞ 0

⎫ p1 ⎪ ⎬ ⎪ ⎭

kλ (n1 , · · · , nm , xm+1 )

⎡ λm+1 −1 ⎣ni(λi −1)(1−pi ) xm+1

m 2 j=1(j =i)

⎤ pp

i

λ −1 ⎦

nj j

p dxm (a(i) ni )

⎫1 p ⎪ ⎬ ⎪ ⎭

.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

271

Multiple Half-Discrete Hilbert-Type Inequalities

m

Since J1 ≤

m 2

1 i=1 (pi /p)

∞ ni =1

i=1

= 1, still by H¨older’s inequality, it follows that

∞

=

m 2

i=1

∞

···

nm =1

×

∞

∞

···

ni+1 =1 ni−1 =1

∞ n1 =1

p (1−λi )−1

ωi (ni )ni i

ni =1

pi (a(i) ni )

0

m 2

(λ −1)(1−pi ) λm+1 −1 ni i xm+1

p1

∞

kλ (n1 , · · · , nm , xm+1 ) λ −1 nj j dxm

p1 pi (a(i) ni )

i

j=1(j =i)

i

.

Hence, inequality (6.54) follows. (ii) For 0 < p1 < 1, pi < 0 (i = 2, · · · , m + 1), or for pi < 0 (i = 1, · · · , m), 0 < pm+1 < 1, in view of the assumptions and by the same way, we obtain the reverse of (6.54). In particular, for λm+1 = λ2 , we still have (j)

Lemma 6.15. If λm+1 = λ2 , anj ≥ 0 (nj ∈ N; j = 1, · · · , m), then (i) for pi > 1 (i = 1, · · · , m + 1), we have the following inequality: ⎧ pλ ⎨ ∞ 2 −1 xm+1 J2 = ⎩ 0 [m+1 (xm+1 )]p−1 ⎛ ×⎝

∞

∞

···

nm =1

kλ (xm+1 n1 , · · · , xm+1 nm , 1)

n1 =1

m 2

⎞p ⎠ dxm+1 a(j) nj

j=1

≤

m 2 i=1

∞

ni =1

p (1−λi )−1 (i) pi i (ni )ni i (ani )

⎫ p1 ⎬ ⎭

p1

i

; (6.55)

(ii) for 0 < p1 < 1, pi < 0 (i = 2, · · · , m + 1), or for pi < 0 (i = 1, · · · , m), 0 < pm+1 < 1, we have the reverse of (6.55). 6.3.3

Main Results and Operator Expressions

In the following, we set the functions: p (1−λi )−1

ψi (ni ) = ni i p

(1−λ

)−1

(ni ∈ N; i = 1, · · · , m),

m+1 m+1 (xm+1 ∈ R+ ). The spaces Lpm+1, ϕ (R+ ) and ϕ(xm+1 ) = xm+1 and lpi, ψi with the norms ||f ||pm+1, ϕ and ||a||pi, ψi are deﬁned in Chapter 3.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

272

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Theorem 6.7. Suppose that pi > 1 (i = 1, · · · , m + 1), there exists a i ∈ (λi − δ0 , λi + δ0 ) (i = 1, · · · , m + 1), constant δ > 0 such that for any λ m+1 0 i=1 λi = λ, ∞ ∞ m 2 −1 λ m+1 ) = k(λ ··· kλ (u1, · · · , um , 1) uj j du1 · · · dum ∈ R+ , 0

0

j=1 λ −1

is decreasing with respect to yj ∈ R+ and and kλ (y1 , · · · , ym , xm+1 ) yj j strictly decreasing in an interval Ij ⊂ (1, ∞)(j = 1, · · · ,m). If f (xm+1 ) ≥ 0, (i) (i) f ∈ Lpm+1, ϕ (R+ ), ||f ||pm+1, ϕ > 0, ani ≥ 0, a(i) = {ani }∞ ni =1 ∈ lpi, ψi , (i) ||a ||pi, ψi > 0 (i = 1, · · · , m), then, we have the following equivalent inequalities: ∞ ∞ ∞ m 2 I= f (xm+1 ) ··· kλ (n1 , · · · , nm , xm+1 ) a(j) nj dxm+1 0

nm =1

< k(λm+1 )||f ||pm+1, ϕ

n1 =1

m 2

j=1

||a(i) ||pi, ψi ,

(6.56)

i=1

∞

J= 0

×

m 2

∞

pλm+1 −1 xm+1

p a(j) nj

···

nm =1

p1

dxm+1

∞

kλ (n1 , · · · , nm , xm+1 )

n1 =1

< k(λm+1 )

j=1

m 2

||a(i) ||pi, ψi ,

(6.57)

i=1

where the constant factor k(λm+1 ) in the above inequalities is the best possible. Proof. By (6.40), (6.42), and (6.54) and the assumptions, we have (6.57). By H¨older’s inequality, we obtain ⎡ ⎤ ∞ ∞ ∞ m −1 2 +λm+1 p ⎣xm+1 ⎦ I= ··· kλ (n1 , · · · , nm , xm+1 ) a(j) nj 0

nm =1

n1 =1

1 p −λm+1

×[xm+1

j=1

f (xm+1 )]dxm+1 ≤ J||f ||pm+1, ϕ .

(6.58)

Then, by (6.57), we have inequality (6.56). Assuming that (6.56) is valid, setting ⎛ ⎞p−1 ∞ ∞ m 2 pλm+1 −1 ⎝ ⎠ ··· kλ (n1 , · · · , nm , xm+1 ) a(j) , f (xm+1 ) = xm+1 nj nm =1

n1 =1

j=1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

273

Multiple Half-Discrete Hilbert-Type Inequalities

then, we ﬁnd J p−1 = ||f ||pm+1, ϕ . By (6.54), we have J < ∞. If J = 0, then (6.57) is trivially valid; if J > 0, then, by (6.56), it follows that ||f ||ppm+1 = J p = I < k(λm+1 )||f ||pm+1, ϕ m+1, ϕ

m 2

||a(i) ||pi, ψi ,

that is,

i=1 −1 = J < k(λm+1 ) ||f ||ppm+1 m+1, ϕ

m 2

||a(i) ||pi, ψi ,

i=1

and then, inequality (6.57) follows. Hence (6.56) and (6.57) are equivalent. (i) ani as follows: For 0 < ε < δ0 min1≤j≤m+1 {pj , p}, we set f(xm+1 ) and 0, 0 < xm+1 < 1, f(xm+1 ) = (λm+1 − p ε )−1 m+1 xm+1 , xm+1 ≥ 1, (λi − pε )−1

a(i) ni = ni

i

,

ni ∈ N

(i = 1, · · · , m).

If there exists a constant k(≤ k(λm+1 )) such that (6.56) is valid as we replace k(λm+1 ) by k, then, in particular, by (6.50), we have k(λm+1 ) + o(1) = I1 (ε) ∞ ∞ ∞ m 2 ··· kλ (n1 , · · · , nm , xm+1 ) a(j) f(xm+1 ) =ε nj dxm+1 0

nm =1

< εk||f||pm+1, ϕ

m 2

n1 =1

|| a(i) ||pi, ψi

i=1

p 1 2 ∞ m 1 m+1 < εk 1+ ε 1 i=1

j=1

p1 p 1 2 m ∞ i 1 m+1 1 = εk 1+ ε n1+ε ni =1 i i=1 1 m 2 1 dy pi = k (1 + ε) pi , y 1+ε i=1

and then, k(λm+1 ) ≤ k (ε → 0+ ). Hence k = k(λm+1 ) is the best possible constant factor of inequality (6.56). By the equivalency, the constant factor k(λm+1 ) in (6.57) is still the best possible. Otherwise, it leads to a contradiction by (6.58) that the constant factor in (6.56) is not the best possible. Theorem 6.8. Suppose that pi > 1, λm+1 = λ2 , there exists a constant i ∈ (λi − δ0 , λi + δ0 ) (i = 1, · · · , m + 1), δ0 > 0 such that, for any λ m+1 = λ, λ i i=1 ∞ ∞ m 2 −1 λ m+1 ) = k(λ ··· kλ (u1, · · · , um , 1) uj j du1 · · · dum ∈ R+ , 0

0

j=1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

274

ws-book9x6

Half-Discrete Hilbert-Type Inequalities λ −1

and kλ (xm+1 y1 , · · · , xm+1 ym , 1) yj j is decreasing with respect to yj ∈ R+ and strictly decreasing in an interval Ij ⊂ (1, ∞) (j = 1, · · · ,m). (i) If f (xm+1 ) ≥ 0, f ∈ Lpm+1, ϕ (R+ ), ||f ||pm+1, ϕ > 0, ani ≥ 0, a(i) = (i) ∞ {ani }ni =1 ∈ lpi, ψi , ||a(i) ||pi, ψi > 0(i = 1, · · · , m), then, we have the following equivalent inequalities with the same best constant factor k(λm+1 ) : 0

∞

f (xm+1 )

∞

∞

···

nm =1

kλ (xm+1 n1 , · · · , xm+1 nm , 1)

n1 =1

×

m 2

a(j) nj dxm+1

j=1

< k(λm+1 )||f ||pm+1, ϕ

m 2

||a(i) ||pi, ψi ,

(6.59)

i=1

and

∞ 0

pλ 2 −1

xm+1

∞

···

nm =1

∞

kλ (xm+1 n1 , · · · , xm+1 nm , 1)

n1 =1

×

m 2

p a(j) nj

p1 dxm+1

j=1

< k(λm+1 )

m 2

||a(i) ||pi, ψi .

(6.60)

i=1

Proof. We only prove that the constant factor in (6.59) is the best possible. The other parts of the proof are omitted. (i) ani as follows: For 0 < ε < δ0 min1≤j≤m+1 {pj , p}, we set f(xm+1 ) and λ ( 2 + p ε )−1 m+1 x , 0 < xm+1 ≤ 1 , m+1 f (xm+1 ) = 0, xm+1 > 1 λi − pε −1

a(i) n i = ni

i

,

ni ∈ N(i = 1, · · · , m).

If there exists a constant k(≤ k(λm+1 )), such that (6.59) is valid as we replace k(λm+1 ) by k, then, by (6.51), we have k(λm+1 ) + o(1) = I1 (ε) ∞ ∞ ∞ m 2 f(xm+1 ) ··· kλ (xm+1 n1 , · · · , xm+1 nm , 1) a(j) =ε nj dxm+1 0

nm =1

n1 =1

j=1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

275

Multiple Half-Discrete Hilbert-Type Inequalities

< εk||f||pm+1, ϕ

m 2

|| a(i) ||pi, ψi

i=1

p 1 2 ∞ m 1 m+1 1+ < εk ε 1 i=1

∞ p1 m i 2 1 1 1 pm+1 = εk( ) ε n1+ε i=1 ni =1 i 1 m 2 1 dy pi =k (1 + ε) pi , y 1+ε i=1

and then, k(λm+1 ) ≤ k(ε → 0+ ). Hence, k = k(λm+1 ) is the best possible value of (6.59). Remark 6.2. With the assumptions of Theorem 6.7, we deﬁne a second kind of multiple half-discrete Hilbert-type operator with the homogeneous m kernel T : i=1 lpi, ψi → Lp, ϕ1−p (R+ ) as follows: For any a = (a(1) , · · · , a(m) ) ∈ m i=1 lpi, ψi , there exists a T a, satisfying Ta(xm+1 ) =

∞ nm =1

···

∞

kλ (n1 , · · · , nm , xm+1 )

n1 =1

m 2

a(j) nj

(xm+1 ∈ R+ ).

j=1

(6.61) Then, by (6.57), we have ||Ta||p,ϕ1−p < k(λm+1 )

m 2

||a(i) ||pi, ψi ,

i=1

and then, Ta ∈ Lp, ϕ1−p (R+ ). Hence, T is a bounded linear operator with ||T|| ≤ k(λm+1 ). Since the constant factor in (6.57) is the best possible, we have ||Ta||p,ϕ1−p = k(λm+1 ). (6.62) ||T|| = sup m (i) a( =θ)∈ m i=1 ||a ||pi, ψi i=1 lpi, ψi With the assumptions of Theorem 6.8, we deﬁne a second kind of multiple half-discrete Hilbert-type operator with the non-homogeneous kernel T1 : m i=1 lpi, ψi → Lp, ϕ1−p (R+ ) as follows: m For any a = (a(1) , · · · , a(m) ) ∈ i=1 lpi, ψi , there exists a T1 a, satisfying T1 a(xm+1 ) =

∞ nm =1

···

∞

kλ (xm+1 n1 , · · · , xm+1 nm , 1)

n1 =1

a(j) nj

j=1

(xm+1 ∈ R+ ). Then, by (6.60), we have ||T1 a||p,ϕ1−p < k(λm+1 )

m 2

m 2 i=1

||a(i) ||pi, ψi ,

(6.63)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

276

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and then, T1 a ∈ Lp, ϕ1−p (R+ ). Hence, T1 is a bounded linear operator with ||T1 || ≤ k(λm+1 ). Since the constant factor in (6.60) is the best possible, we have ||T1 || =

6.3.4

||T1 a||p,ϕ1−p m = k(λm+1 ). (i) m i=1 ||a ||pi, ψi i=1 lpi, ψi

sup

a( =θ)∈

(6.64)

Some Kinds of Reverse Inequalities p (1−λ1 )−1

For ψ1 (n1 ) = n11 tions

p

m+1 and ϕ(xm+1 ) = xm+1

(1−λm+1 )−1

, we set func-

Ψ1 (n1 ) = (1 − θ1 (n1 ))ψ1 (n1 ), 1 (n1 ) = (1 − θ1 (n1 ))ψ1 (n1 ), Ψ Φ(xm+1 ) = (1 − θm+1 (xm+1 ))ϕ(xm+1 ), and m+1 ) = (1 − θm+1 (xm+1 ))ϕ(xm+1 ), Φ(x where θ1 (n1 ), θ1 (n1 ), θm+1 (xm+1 ) and θm+1 (xm+1 ) are given by (6.43), (6.48), (6.41) and (6.46). For pi < 1(pi = 0), the spaces lpi ,ψi and lpm+1 ϕ (R+ ) with ||a(i) ||pi ,ψi and ||f ||pm+1, ϕ are not normed spaces. But we still use them as the formal symbols in the following: Theorem 6.9. Suppose that 0 < p1 < 1, pi < 0 (i = 2, · · · , m + 1), i ∈ (λi − δ0 , λi + δ0 ) there exists a constant δ0 > 0, such that for any λ m+1 (i = 1, · · · , m + 1), i=1 λi = λ, m+1 ) = k(λ

∞ 0

···

∞ 0

kλ (u1 , · · · , um , 1)

m 2

−1 λ

uj j

du1 · · · dum ∈ R+ ,

j=1 λ −1

is decreasing with respect to yj ∈ R+ and and kλ (y1 , · · · , ym , xm+1 ) yj j strictly decreasing in an interval Ij ⊂ (1, ∞) (j = 1, · · · , m). There exist constants α, L > 0 such that (6.44) is satisﬁed for i = 1, and 1 A1 (uk ) ≤ Luα−λ k

(uk ∈ R+ ; i = 1, · · · , m + 1). (1)

If f (xm+1 ) ≥ 0, f ∈ Lpm+1, ϕ (R+ ), ||f ||pm+1, ϕ > 0, an1 ≥ 0, a(1) = (1) (i) (i) (1) {an1 }∞ ||p1, Ψ1 > 0, ani ≥ 0, a(i) = {ani }∞ n1 =1 ∈ lp1, Ψ1 , ||a ni =1 ∈ lpi, ψi ,

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

277

Multiple Half-Discrete Hilbert-Type Inequalities

||a(i) ||pi, ψi > 0 (i = 2, · · · , m), then, we have the following equivalent inequalities with the same best possible constant factor k(λm+1 ): ∞ ∞ ∞ m 2 f (xm+1 ) ··· kλ (n1 , · · · , nm , xm+1 ) a(j) I= nj dxm+1 0

nm =1

n1 =1

> k(λm+1 )||f ||pm+1, ϕ ||a(1) ||p1, Ψ1

j=1 m 2

||a(i) ||pi, ψi ,

(6.65)

i=2

and

⎧ ⎨ J=

⎩ ⎛

×⎝

∞ 0

∞

pλ

m+1 xm+1

···

nm =1

−1

∞

kλ (n1 , · · · , nm , xm+1 )

n1 =1

⎞p

m 2

⎠ dxm+1 a(j) nj

j=1

> k(λm+1 )||a(1) ||p1, Ψ1

m 2

||a(i) ||pi, ψi .

⎫ p1 ⎬ ⎭ (6.66)

i=2

Proof. By (6.40), (6.42), the reverse of (6.54) and the assumption, we have (6.66). By the reverse ⎡ H¨older’s inequality, we obtain ⎤ ∞ ∞ ∞ m −1 2 +λ m+1 p ⎣xm+1 ⎦ I= ··· kλ (n1 , · · · , nm , xm+1 ) a(j) nj 0

nm =1

n1 =1

j=1

1 p −λm+1

×[xm+1 f (xm+1 )]dxm+1 ≥ J||f ||pm+1, ϕ . (6.67) Then, by (6.66), we have (6.65). Assuming that (6.65) is valid, and setting f (xm+1 ) as follows: ⎛ ⎞p−1 ∞ ∞ m 2 pλm+1 −1 ⎝ ⎠ f (xm+1 ) = xm+1 ··· kλ (n1 , · · · , nm , xm+1 ) a(j) , nj nm =1

n1 =1

j=1

then, we ﬁnd J = ||f ||pm+1, ϕ . By the reverse of (6.54), we have J > 0. If J = ∞, then, (6.66) is trivially valid; if J < ∞, then, by (6.65), it follows that = Jp = I ||f ||ppm+1 m+1, ϕ p−1

> k(λm+1 )||f ||pm+1, ϕ ||a(1) ||p1, Ψ1

m 2

||a(i) ||pi, ψi ,

i=2 −1 = J > k(λm+1 )||a(1) ||p1, Ψ1 ||f ||ppm+1 m+1, ϕ

m 2 i=2

||a(i) ||pi, ψi ,

that is

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

278

Half-Discrete Hilbert-Type Inequalities

and inequality (6.66) follows. Hence, (6.65) and (6.66) are equivalent. (i) ani as follows: For 0 < ε < δ0 min1≤j≤m+1 {|pj |, p}, we set f(xm+1 ) and 0, 0 < xm+1 < 1, f(xm+1 ) = λm+1 − p ε −1 m+1 , xm+1 ≥ 1, xm+1

a(i) ni

λi − pε

−1

= ni , ni ∈ N(i = 1, · · · , m). If there exists a constant k(≥ k(λm+1 )), such that (6.65) is valid as we replace k(λm+1 ) by k, then, in particular, by (6.50) and Lemma 6.4, we have k(λm+1 ) + o(1) = I1 (ε) ∞ ∞ ∞ m 2 ··· kλ (n1 , · · · , nm , xm+1 ) a(j) =ε f (xm+1 ) nj dxm+1 0

i

nm =1

> εk||f||pm+1, ϕ || a(1) ||p1, Ψ1

n1 =1 m 2

j=1

|| a(i) ||pi, ψi

i=2

p1 m ∞ p1 p 1 ∞ 1 2 i 1 1 m+1 1 1 = εk 1−O α 1+ε 1+ε ε n1 n1 n n1 =1 i=2 ni =1 i p1 p1 m p 1 ∞ ∞ 1 2 i 1 1 1 m+1 1 + = εk − O(1) ε n1+ε n1+ε n1 =1 1 ni =2 i i=2 1 p 1 ∞ p1 2 ∞ m 1 1 m+1 dy dxi pi 1+ > εk − O(1) ε y 1+ε x1+ε 1 1 i i=2 1

= k(1 − εO(1)) p1

m 2

1

(ε + 1) pi ,

i=2

and then, k(λm+1 ) ≥ k (ε → 0+ ). Hence, k = k(λm+1 ) is the best constant factor of inequality (6.65). By the equivalency, the constant factor k(λm+1 ) in (6.66) is still the best possible. Otherwise, it leads to a contradiction by (6.37) that the constant factor in (6.65) is not the best possible. Theorem 6.10. Suppose that 0 < p1 < 1, pi < 0 (i = 2, · · · , m + i ∈ 1), λm+1 = λ2 , there exists a constant δ0 > 0, such that for any λ m+1 (λi − δ0 , λi + δ0 ) (i = 1, · · · , m + 1), i=1 λi = λ, ∞ ∞ m 2 −1 λ m+1 ) = k(λ ··· kλ (u1 , · · · , um , 1) uj j du1 · · · dum ∈ R+ , 0

0

j=1

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

279

Multiple Half-Discrete Hilbert-Type Inequalities λ −1

and kλ (xm+1 y1 , · · · , xm+1 ym , 1)yj j is decreasing with respect to yj ∈ R+ and strictly decreasing in an interval Ij ⊂ (1, ∞)(j = 1, · · · ,m). If there exist constants α, L > 0, such that (6.44) is satisﬁed, and k Ai (uk ) ≤ Luα−λ (uk ∈ R+ ; i = 1, · · · , m + 1), k then, we have the following equivalent inequalities with the same best possible constant factor k(λm+1 ): ∞ ∞ ∞ m 2 f (xm+1 ) ··· kλ (xm+1 n1 , · · · , xm+1 nm , 1) a(j) nj dxm+1 0

nm =1

n1 =1

j=1

> k(λm+1 )||f ||pm+1, ϕ ||a(1) ||p1, Ψ1

m 2

||a(i) ||pi, ψi ,

(6.68)

i=2

and

∞ 0

pλ 2 −1

∞

xm+1

···

nm =1

∞

kλ (xm+1 n1 , · · · , xm+1 nm , 1)

n1 =1

×

m 2

p a(j) nj

p1 dxm+1

j=1

> k(λm+1 )||a(1) ||p1, Ψ1

m 2

||a(i) ||pi, ψi .

(6.69)

i=2

Proof. We only prove that the constant factor in (6.68) is the best possible. Other parts of the proof are omitted. (i) ani as follows: For 0 < ε < δ0 min1≤j≤m+1 {|pj |, p}, we set f(xm+1 ) and 0, 0 < xm+1 < 1, f(xm+1 ) = λm+1 − p ε −1 m+1 xm+1 , xm+1 ≥ 1,

a(i) ni

λi − pε

−1

= ni , ni ∈ N(i = 1, · · · , m). If there exists a constant k(≥ k(λm+1 )), such that (6.68) is valid as we replace k(λm+1 ) by k, then, in particular, by (6.51), we have k(λm+1 ) + o(1) = I1 (ε) ∞ ∞ ∞ m 2 ··· kλ (xm+1 n1 , · · · , xm+1 nm , 1) a(j) f(xm+1 ) =ε nj dxm+1 0

nm =1

> εk||f||pm+1, ϕ || a(1) ||p1, Ψ1

i

n1 =1 m 2

j=1

|| a(i) ||pi, ψi

i=2

p1 m ∞ p1 p 1 ∞ 1 2 i 1 1 1 1 m+1 1−O = εk ε nα n1+ε n1+ε 1 1 n =1 i=2 n =1 i 1

i

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

280

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

p1 p1 m p 1 ∞ ∞ 1 2 i 1 1 1 m+1 1 + = εk − O(1) ε n1+ε n1+ε n1 =1 1 ni =2 i i=2 1 p 1 ∞ p1 2 ∞ m 1 dy dxi pi 1 m+1 1+ > εk − O(1) ε y 1+ε x1+ε 1 1 i i=2 1

= k (1 − εO(1)) p1

m 2

1

(ε + 1) pi ,

i=2

and then, k(λm+1 ) ≥ k(ε → 0+ ). Hence, k = k(λm+1 ) is the best constant factor of (6.68). Similarly, we still have: Theorem 6.11. Suppose that pi < 0(i = 1, · · · , m), 0 < pm+1 < 1, there i ∈ (λi − δ0 , λi + δ0 ) (i = exists a constant δ0 > 0 such that for any λ m+1 1, · · · , m + 1), i=1 λi = λ, ∞ ∞ m 2 −1 λ ··· kλ (u1 , · · · , um , 1) uj j du1 · · · dum ∈ R+ , k(λm+1 ) = 0

0

j=1 λ −1 yj j

is decreasing with respect to yj ∈ R+ and and kλ (y1 , · · · , ym , xm+1 ) strictly decreasing in an interval Ij ⊂ (1, ∞) (j = 1, · · · , m). If f (xm+1 ) ≥ (i) (i) 0, f ∈ Lpm+1, Φ (R+ ), ||f ||pm+1, Φ > 0, ani ≥ 0, a(i) = {ani }∞ ni =1 ∈ lpi, ψi , (i) ||a ||pi, ψi > 0 (i = 1, · · · , m), then, we have the following equivalent reverse inequalities with the same best possible constant factor k(λm+1 ): ∞ ∞ ∞ m 2 f (xm+1 ) ··· kλ (n1 , · · · , nm , xm+1 ) a(j) nj dxm+1 0

nm =1

n1 =1

j=1

> k(λm+1 )||f ||pm+1, Φ

m 2

||a(i) ||pi, ψi ,

(6.70)

i=1

and

∞ 0

pλ

−1

m+1 xm+1 [1 − θ(xm+1 )]p−1

∞ nm =1

···

∞

kλ (n1 , · · · , nm , xm+1 )

n1 =1

×

m 2

p a(j) nj

p1 dxm+1

j=1

> k(λm+1 )

m 2 i=1

||a(i) ||pi, ψi .

(6.71)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

281

Multiple Half-Discrete Hilbert-Type Inequalities

Theorem 6.12. Suppose that pi < 0(i = 1, · · · , m), 0 < pm+1 < 1, i ∈ λm+1 = λ2 , there exists a constant δ0 > 0 such that for any λ m+1 (λi − δ0 , λi + δ0 ) (i = 1, · · · , m + 1), i=1 λi = λ, ∞ ∞ m 2 −1 λ m+1 ) = ··· kλ (u1 , · · · , um , 1) uj j du1 · · · dum ∈ R+ , k(λ 0

0

j=1

λ −1 , xm+1 ym , 1)yj j

and kλ (xm+1 y1 , · · · is decreasing with respect to yj ∈ R+ and strictly decreasing in an interval Ij ⊂ (1, ∞)(j = 1, · · · ,m). If f (xm+1 ) ≥ 0, f ∈ Lpm+1, Φ (R+ ), ||f ||pm+1, Φ > 0, then, we have the following equivalent inequalities with the same best constant factor k(λm+1 ): ∞ ∞ ∞ m 2 f (xm+1 ) ··· kλ (xm+1 n1 , · · · , xm+1 nm , 1) a(j) nj dxm+1 0

nm =1

n1 =1

j=1

> k(λm+1 )||f ||pm+1, Φ

m 2

||a(i) ||pi, ψi ,

(6.72)

i=1

and

∞ 0

pλ

−1

2 xm+1 ˜ m+1 )]p−1 [1 − θ(x

∞ nm =1

···

∞

(xm+1 n1 , · · · , xm+1 nm , 1)

n1 =1

×

m 2

p a(j) nj

p1 dxm+1

j=1

> k(λm+1 )

m 2

||a(i) ||pi, ψi .

(6.73)

i=1

6.4

Some Examples with the Particular Kernels

In the following, we still assume that m ∈ N, pi , p ∈ R\{0, 1}, λi ∈ R(i = m 1 1 1 1, · · · , m + 1), m+1 i=1 λi = λ, p = i=1 pi = 1 − pm+1 . In the following Corollaries 6.1 to 6.21, we set p (1−λi )−1 ϕi (xi ) = xi i (xi ∈ R+ ; i = 1, · · · , m), ψ(n) = npm+1 (1−λm+1 )−1 (n ∈ N), 1 (x1 ) = (1 − θ1 (x1 ))ϕ1 (x1 ), where Φ1 (x1 ) = (1 − θ1 (x1 ))ϕ1 (x1 ) and Φ

∞ 1/x1 ∞ 1 λm+1 −1 um+1 ··· kλ (1, u2 , · · · , um+1 ) θ1 (x1 ) = k(λm+1 ) 0 0 0 m 2 λj −1 × uj du2 · · · dum dum+1 , j=2

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

282

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and θ1 (x1 ) =

1 k(λm+1 )

x1

u1λ1 −1

0

∞

0

···

∞

0 m 2

×

kλ (u1 , · · · , um , 1)

λ −1 uj j du2

· · · dum du1 .

j=2

In the following Corollaries 6.4 to 6.24, we set p (1−λi )−1

(ni ∈ N; i = 1, · · · , m),

ψi (ni ) = ni i ϕ(xm+1 ) =

pm+1 (1−λm+1 )−1 xm+1

(xm+1 ∈ R+ ),

Φ(xm+1 ) = (1 − θm+1 (xm+1 ))ϕ(xm+1 ) m+1 ) = (1 − θm+1 (xm+1 ))ϕ(xm+1 ) Φ(x

(xm+1 ∈ R+ ), (xm+1 ∈ R+ ),

1 ) = (1 − θ1 (n1 ))ψ(n1 ), where Ψ(n1 ) = (1 − θ1 (n1 ))ψ(n1 ) and Ψ(n ∞ ∞ 1 θm+1 (xm+1 ) = 1 − ··· kλ (u1 , · · · , um , 1) k(λm+1 ) 1/xm+1 1/xm+1 m 2

×

λ −1

uj j

du1 · · · dum ,

j=1

and θm+1 (xm+1 ) = 1 −

1 k(λm+1 )

∞

···

xm+1

∞

kλ (u1 , · · · , um , 1)

xm+1 m 2

×

λ −1

uj j

du1 · · · dum .

j=1

6.4.1

The Case of kλ (x1 , · · · , xm , xm+1 ) =

1 m+1 ( i=1 xi )λ

Lemma 6.16. If λ, λi > 0 (i = 1, · · · , m + 1), and 1 kλ (x1, · · · , xm+1 ) = m+1 , ( i=1 xi )λ then, we have k(λm+1 ) = =

∞ 0

···

1 Γ(λ)

0

m+1 2 i=1

∞

m

λ −1

u j mj=1 j du1 · · · dum ( j=1 uj + 1)λ

Γ(λi ).

(6.74)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

283

Multiple Half-Discrete Hilbert-Type Inequalities

Proof.

For M > 0, we set DM = {(u1 , · · · , um )|ui > 0,

m ui i=1

M

≤ 1}.

In view of the assumptions with reference (see Yang [131], (9.18)), we have m m ui 2 λ −1 ··· ψ uj j du1 · · · dum M DM i=1 j=1 1 m λ−λm+1 2 M = Γ(λi ) ψ(u)uλ−λm+1 −1 du. (6.75) Γ(λ − λm+1 ) i=1 0 Setting ψ(u) =

1 , (Mu+1)λ

by (6.75), we ﬁnd m λj −1 uj j=1 k(λm+1 ) = lim ··· du1 · · · dum m ui λ M→∞ DM [M i=1 ( M ) + 1] 1 λ−λm+1 −1 m M λ−λm+1 2 u = lim Γ(λi ) du, (v = M u) M→∞ Γ(λ − λm+1 ) (M u + 1)λ 0 i=1 ∞ λ−λm+1 −1 m 2 v 1 Γ(λi ) dv = Γ(λ − λm+1 ) (v + 1)λ 0 i=1 ( ) m 2 Γ(λ − λm+1 )Γ(λm+1 ) 1 Γ(λi ) = Γ(λ − λm+1 ) i=1 Γ(λ) =

m+1 1 2 Γ(λi ), Γ(λ) i=1

then, inequality (6.74) follows.

With the assumptions of Lemma 6.16, if λm+1 < 1, then, we set δ0 = i ∈ (λi − δ0 , λi + δ0 ) (i = min1≤i≤m+1 {λi , 1 − λm+1 } > 0. For any λ m+1 1, · · · , m + 1), i=1 λi = λ, it follows that λi > 0 (i = 1, · · · , m + 1), m+1 < λm+1 + δ0 ≤ λm+1 + 1 − λm+1 = 1, λ ∞ ∞ m λj −1 u mj=1 j k(λm+1 ) = ··· du1 · · · dum λ ( u 0 0 j=1 j + 1) =

m+1 1 2 Γ(λi ) ∈ R+ , Γ(λ) i=1

1 1 λm+1 −1 λm+1 −1 and (yx1 +···+yx and both expressions (x1 +···+x λ y λ y m +y) m +1) are strictly decreasing with respect to y ∈ R+ . Then, for 1 kλ (x1 , · · · , xm , xm+1 ) = m+1 ( i=1 xi )λ

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

284

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

in Theorems 6.1 and 6.2, it follows that Corollary 6.1. Suppose that pi > 1, λi > 0 (i = 1, · · · , m + 1), λm+1 < 1. If fi (xi ) ≥ 0, fi ∈ Lpi, ϕi (R+ ), ||fi ||pi, ϕi > 0 (i = 1, · · · , m), an ≥ 0, a = {an }∞ n=1 ∈ lpm+1, ψ , ||a||pm+1, ψ > 0, then (i) we have the following equivalent inequalities with the best possible m+1 1 constant factor Γ(λ) i=1 Γ(λi ): ∞ ∞ m ∞ fj (xj ) mj=1 an ··· dx1 · · · dxm ( j=1 xj + n)λ 0 0 n=1 m Γ(λm+1 ) 2 < Γ(λi )||fi ||pi, ϕi ||a||pm+1, ψ , (6.76) Γ(λ) i=1 and

∞

npλm+1 −1

∞

0

n=1

<

···

∞ 0

m

j=1 fj (xj )dx1 · · · dxm m ( j=1 xj + n)λ

p p1

m Γ(λm+1 ) 2 Γ(λi )||fi ||pi, ϕi ; Γ(λ) i=1

(6.77)

(ii) for λm+1 = λ2 (< 1), we have the following equivalent inequalities m with the same best possible constant factor Γ(λ/2) i=1 Γ(λi ) : Γ(λ) ∞ ∞ m ∞ fj (xj ) mj=1 an ··· dx1 · · · dxm λ ( nx j + 1) 0 0 j=1 n=1 m Γ(λ/2) 2 < Γ(λi )||fi ||pi, ϕi ||a||pm+1, ψ , (6.78) Γ(λ) i=1 and

∞ n=1

n

pλ 2 −1

∞ 0

···

∞

m

0

fj (xj )dx1 · · · dxm m ( j=1 nxj + 1)λ

p p1

j=1

Γ(λ/2) 2 Γ(λi )||fi ||pi, ϕi . Γ(λ) i=1 m

<

In view of (6.9) and (6.12), for 1 , kλ (x1 , · · · , xm , xm+1 ) = m+1 ( i=1 xi )λ

(6.79)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

285

Multiple Half-Discrete Hilbert-Type Inequalities

we ﬁnd, for m ≥ 2, A1 (um+1 ) =

∞ 0

≤ = and

B1 (ui ) = ≤

=

0

(1

0

···

m

∞

∞

0

(1 m

λj −1 j=2 uj du2 · · · dum λ + m+1 j=2 uj ) m λj −1 du2 · · · dum j=2 uj m λm+1 −α + j=2 uj )λ+α−λm+1 um+1

Γ(λ1 + α) i=2 Γ(λi ) α−λm+1 (um+1 ) (0 < α < λm+1 ), Γ(λ + α − λm+1 )

∞

0 ∞ 0

∞

···

··· ···

∞ 0 ∞ 0

m

λ −1

j j=1(j =i) uj m du1 · · · dui−1 dui+1 · · · dum ( j=1 uj + 1)λ m λj −1 du1 · · · dui−1 dui+1 · · · dum j=1(j =i) uj m ( j=1(j =i) uj + 1)λ+α−λi uλi i −α

Γ(λm+1 + α) Γ(λ + α − λi )

m 2

i Γ(λj )(uα−λ ) (0 < α < λi ); i

j=1(j =i)

for m = 1, we can still obtain the same results. By Theorems 6.3 and 6.4, it follows that Corollary 6.2. Suppose that 0 < p1 < 1, pi < 0 (i = 2, · · · , m + 1), λi > 0 (i = 1, · · · , m + 1), λm+1 < 1. If f1 (x1 ) ≥ 0, f1 ∈ Lp1, Φ1 (R+ ), ||f1 ||p1, Φ1 > 0, fi (xi ) ≥ 0, fi ∈ Lpi, ϕi (R+ ), ||fi ||pi, ϕi > 0 (i = 2, · · · , m), an ≥ 0, a = {an }∞ n=1 ∈ lpm+1, ψ , ||a||pm+1, ψ > 0, then (i) we have the following equivalent inequalities with the same best posm+1 1 sible constant factor Γ(λ) i=1 Γ(λi ): m ∞ ∞ ∞ fj (xj ) mj=1 an ··· dx1 · · · dxm ( j=1 xj + n)λ 0 0 n=1 > and

∞

m+1 m 2 1 2 Γ(λi )||a||pm+1, ψ ||f1 ||p1 ,Φ1 ||fi ||pi, ϕi , (6.80) Γ(λ) i=1 i=2

npλm+1 −1

0

n=1

>

∞

···

∞ 0

p p1

m

fj (xj ) mj=1 dx1 ( j=1 xj + n)λ

· · · dxm

m+1 m 2 1 2 Γ(λi )||f1 ||p1 ,Φ1 ||fi ||pi, ϕi ; Γ(λ) i=1

i=2

(6.81)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

286

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

(ii) for λm+1 = λ2 (< 1), f1 ∈ Lp1, Φ 1 (R+ ), ||f1 ||p1, Φ 1 > 0, we have the following equivalent inequalities with the same best possible constant factor Γ(λ/2) m i=1 Γ(λi ) : Γ(λ) ∞

an

n=1

∞ 0

m

∞

fj (xj ) mj=1 dx1 · · · dxm ( j=1 nxj + 1)λ 0 m m 2 Γ(λ/2) 2 > Γ(λi ) ||a||pm+1, ψ ||f1 ||p1, Φ ||fi ||pi, ϕi , 1 Γ(λ)

···

i=1

i=2

(6.82) and

∞

n

pλ 2 −1

∞ 0

n=1

···

0

Γ(λ/2) > Γ(λ)

p p1

m

∞

fj (xj ) mj=1 dx1 ( j=1 nxj + 1)λ

m 2

Γ(λi ) ||f1 ||p1, Φ 1

i=1

· · · dxm

m 2

||fi ||pi, ϕi . (6.83)

i=2

By Theorems 6.5 and 6.6, we still have Corollary 6.3. Suppose that λi > 0, pi < 0 (i = 1, · · · , m), 0 < pm+1 < 1, 0 < λm+1 < 1. If fi (xi ), an ≥ 0, fi ∈ Lpi, ϕi (R+ ), ||fi ||pi, ϕi > 0(i = 1, · · · , m), a = {an }∞ n=1 ∈ lpm+1, ψ , ||a||pm+1, ψ > 0, then (i) we have the following equivalent reverse inequalities with the best m+1 1 possible constant factor Γ(λ) i=1 Γ(λi ): ∞

an

n=1

0

∞

∞

m

∞

fj (xj ) mj=1 dx1 · · · dxm ( j=1 xj + n)λ 0 m Γ(λm+1 ) 2 > Γ(λi )||fi ||pi, ϕi ||a||pm+1, ψ , (6.84) Γ(λ) i=1

···

and

∞

n

pλm+1 −1

0

n=1

>

···

∞ 0

fj (xj ) mj=1 dx1 ( j=1 xj + n)λ

m Γ(λm+1 ) 2 Γ(λi )||fi ||pi, ϕi ; Γ(λ) i=1

p p1

m

· · · dxm (6.85)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

287

Multiple Half-Discrete Hilbert-Type Inequalities

(ii) for λm+1 = λ2 < 1, we have the following equivalent inequalities with m the same best possible constant factor Γ(λ/2) i=1 Γ(λi ): Γ(λ) ∞

an

n=1

∞ 0

m

∞

fj (xj ) mj=1 dx1 · · · dxm ( j=1 nxj + 1)λ 0 m Γ(λ/2) 2 > Γ(λi )||fi ||pi, ϕi ||a||pm+1, ψ , Γ(λ)

···

(6.86)

i=1

and

∞

n

pλ 2 −1

∞ 0

n=1

···

p p1

m

∞

j=1 fj (xj )

m dx1 · · · dxm ( j=1 nxj + 1)λ

0

Γ(λ/2) 2 Γ(λi )||fi ||pi, ϕi . Γ(λ) i=1 m

>

(6.87)

By Theorems 6.7 and 6.8, we have Corollary 6.4. Suppose that pi > 1, 0 < λi < 1 (i = 1, · · · , m), λm+1 > (i) 0. If f (xm+1 ) ≥ 0, f ∈ Lpm+1, ϕ (R+ ), ||f ||pm+1, ϕ > 0, ani ≥ 0,a(i) = (i) (i) {ani }∞ ni =1 ∈ lpi, ψi , ||a ||pi, ψi > 0 (i = 1, · · · , m), then (i) we have the following equivalent inequalities with the best possible m+1 1 constant factor Γ(λ) i=1 Γ(λi ):

∞

f (xm+1 )

0

∞

m

∞

(j)

anj m j=1 dxm+1 ( j=1 nj + xm+1 )λ nm =1 n1 =1 m Γ(λm+1 ) 2 < Γ(λi )||a(i) ||pi, ψi ||f ||pm+1, ϕ , Γ(λ) ···

(6.88)

i=1

and ⎧ ⎨ ⎩

0

∞

pλm+1 −1 xm+1

∞

nm =1

···

∞ n1 =1

m j=1

(j)

anj

m ( j=1 nj + xm+1 )λ

Γ(λm+1 ) 2 Γ(λi )||a(i) ||pi, ψi ; Γ(λ)

⎫ p1 ⎬

p dxm+1

⎭

m

<

(6.89)

i=1

(ii) for λm+1 = λ2 , we have the following equivalent inequalities with the

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

288

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

best possible constant factor

∞

0

m

Γ(λ/2) Γ(λ)

∞

i=1

∞

Γ(λi ): m

(j)

anj m j=1 f (xm+1 ) ··· dxm+1 λ ( x j=1 m+1 nj + 1) nm =1 n1 =1 m Γ(λ/2) 2 (i) < Γ(λi )||a ||pi, ψi ||f ||pm+1, ϕ , Γ(λ)

(6.90)

i=1

and

⎧ ⎨ ⎩

∞

pλ 2 −1

∞

xm+1

0

···

nm =1

∞ n1 =1

m j=1

(j)

anj

⎫1 ⎬p

p

m ( j=1 xm+1 nj + 1)λ

Γ(λ/2) 2 Γ(λi )||a(i) ||pi, ψi . Γ(λ) i=1

dxm+1

⎭

m

<

(6.91)

In (6.44), for 1 kλ (x1 , · · · , xm , xm+1 ) = m+1 , ( i=1 xi )λ we ﬁnd, for m ≥ 2, ∞ ··· Ai (uk ) = 0

∞ 0

m+1 2

×

1 m+1 ( j=1(j =i) uj + 1)λ λ −1

uj j

du1 · · · dui−1 dui+1 · · · duk−1 duk+1 · · · dum+1

j=1(j =i,k)

≤ =

∞ 0

···

∞

m+1 j=1(j =i,k)

0

Γ(λi + α) Γ(λ + α − λk )

m+1 2

λ −1

uj j du1 · · · dui−1 dui+1 · · · duk−1 duk+1 · · · dum+1 m+1 ( j=1(j =i,k) uj + 1)λ+α−λk uλk k −α k Γ(λj )(uα−λ )(0 < α < λk ); k

j=1(j =i,k)

for m = 1, we can still ﬁnd the same result. If 0 < λi < 1(i = 1, · · · , m), λm+1 > 0, then, we set δ0 = i ∈ (λi − δ0 , λi + δ0 )(i = min1≤i≤m {λm+1 , λi , 1 − λi } > 0. For any λ m+1 m+1 > 0, 0 < λ i < λi + δ0 ≤ 1, · · · , m + 1), i=1 λi = λ, we ﬁnd that λ 1(i = 1, · · · , m), m+1 ) = k(λ

m+1 1 2 Γ(λi ) ∈ R+ , Γ(λ) i=1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

289

Multiple Half-Discrete Hilbert-Type Inequalities −1 λ

−1 λ

j 1 1 and both (y1 +···+ym y j and (xm+1 y1 +···+x λ yj +xm+1 )λ j m+1 ym +1) strictly decreasing with respect to yj ∈ R(j = 1, · · · , m). Then, by Theorems 6.9 and 6.10, it follows that

are

Corollary 6.5. Suppose that 0 < p1 < 1, pi < 0(i = 2, · · · , m + 1), 0 < λi < 1(i = 1, · · · , m), λm+1 > 0. If f (xm+1 ) ≥ 0, f ∈ Lpm+1, ϕ (R+ ), (1) (1) (1) ||f ||pm+1, ϕ > 0, an1 ≥ 0, a(1) = {an1 }∞ ||p1, Ψ1 > 0, n1 =1 ∈ lp1, Ψ1 , ||a (i) (i) ∞ (i) (i) ani ≥ 0, a = {ani }ni =1 ∈ lpi, ψi , ||a ||pi, ψi > 0 (i = 2, · · · , m), then (i) we have the following equivalent inequalities with the same best posm+1 1 sible constant factor Γ(λ) i=1 Γ(λi ): m (j) ∞ ∞ ∞ j=1 anj f (xm+1 ) ··· dxm+1 m λ ( n 0 j=1 j + xm+1 ) nm =1 n1 =1 m+1 m 2 2 1 > Γ(λi ) ||f ||pm+1, ϕ ||a(1) ||p1, Ψ1 ||a(i) ||pi, ψi , (6.92) Γ(λ) i=1 i=2 and

⎧ ⎨ ⎩

∞

0

pλ

m+1 xm+1

−1

∞

···

nm =1

>

1 Γ(λ)

m+1 2

∞ n1 =1

m

m ( j=1 nj + xm+1 )λ

Γ(λi ) ||a(1) ||p1, Ψ1

i=1

m 2

⎫1 ⎬p

p

(j)

j=1 anj

dxm+1

⎭

||a(i) ||pi, ψi ;

(6.93)

i=2

(ii) for λm+1 = λ2 , we have the following equivalent inequalities with the m same best possible constant factor Γ(λ/2) i=1 Γ(λi ): Γ(λ) m (j) ∞ ∞ ∞ j=1 anj f (xm+1 ) ··· dxm+1 m ( j=1 xm+1 nj + 1)λ 0 nm =1 n1 =1 m m 2 Γ(λ/2) 2 > Γ(λi ) ||f ||pm+1, ϕ ||a(1) ||p1, Ψ1 ||a(i) ||pi, ψi , (6.94) Γ(λ) i=1 i=2 and

⎧ ⎨ ⎩

∞ 0

pλ 2 −1

∞

∞

m

p

(j)

j=1 anj

⎫ p1 ⎬

dxm+1 λ ⎭ ( m j=1 xm+1 nj + 1) m m 2 Γ(λ/2) 2 > Γ(λi ) ||a(1) ||p1, Ψ1 ||a(i) ||pi, ψi . (6.95) Γ(λ)

xm+1

nm =1

···

n1 =1

i=1

i=2

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

290

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

By Theorems 6.11 and 6.12, it follows that Corollary 6.6. Suppose that pi < 0, 0 < λi < 1(i = 1, · · · , m), λm+1 > (i) 0, 0 < pm+1 < 1. If f (xm+1 ) ≥ 0, f ∈ Lpm+1, Φ (R+ ), ||f ||pm+1, Φ > 0, ani ≥ (i) (i) 0, a(i) = {ani }∞ ni =1 ∈ lpi, ψi , ||a ||pi, ψi > 0 (i = 1, · · · , m), then (i) we have the following equivalent reverse inequalities with the same m+1 1 best possible constant factor Γ(λ) i=1 Γ(λi ): m (j) ∞ ∞ ∞ anj m j=1 f (xm+1 ) ··· dxm+1 ( j=1 nj + xm+1 )λ 0 nm =1 n1 =1 m Γ(λm+1 ) 2 (i) > Γ(λi )||a ||pi, ψi ||f ||pm+1, Φ , (6.96) Γ(λ) i=1 and pλm+1 −1 ∞ xm+1 [1 − θm+1 (xm+1 )]p−1 0

∞ p p1 m (j) ∞ j=1 anj m × ··· dxm+1 λ ( j=1 nj + xm+1 ) n =1 n =1 m

1

Γ(λm+1 ) 2 > Γ(λi )||a(i) ||pi, ψi ; Γ(λ) i=1 m

(6.97)

(ii) for λm+1 = λ2 , f (xm+1 ) ≥ 0, f ∈ Lpm+1, Φ (R+ ), ||f ||pm+1, Φ > 0, we have the following equivalent inequalities with the same best possible m constant factor Γ(λ/2) i=1 Γ(λi ): Γ(λ) m (j) ∞ ∞ ∞ anj m j=1 f (xm+1 ) ··· dxm+1 ( j=1 xm+1 nj + 1)λ 0 nm =1 n1 =1 m Γ(λ/2) 2 > Γ(λi )||a(i) ||pi, ψi ||f ||pm+1, Φ (6.98) , Γ(λ) i=1 and ⎧ pλ ⎨ ∞ 2 −1 xm+1 ⎩ 0 [1 − θm+1 (xm+1 )]p−1

×

∞

nm =1

∞

···

n1 =1

m

(j)

j=1 anj

m ( j=1 xm+1 nj + 1)λ

Γ(λ/2) 2 Γ(λi )||a(i) ||pi, ψi . Γ(λ)

⎫ p1 ⎬

p dxm+1

⎭

m

>

i=1

(6.99)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Multiple Half-Discrete Hilbert-Type Inequalities

ws-book9x6

291

Remark 6.3. (i) If, in (6.24), we set 1 kλ (x1 , · · · , xm , n) = ∞ , ( j=1 xj + n)λ then, in view of Corollary 6.1 and equality (6.25), we have ||T || = k(λm+1 ) =

m+1 1 2 Γ(λi ). Γ(λ) i=1

If, in (6.26), for λm+1 = λ2 (< 1), we set 1 kλ (nx1 , · · · , nxm , 1) = ∞ , ( j=1 nxj + 1)λ then, in view of Corollary 6.4 and equality (6.27), we have Γ(λ/2) 2 Γ(λi ). Γ(λ) i=1 m

||T1 || = k(λm+1 ) = (ii) If, in (6.61), we set

1 , kλ (n1 , · · · , nm , xm+1 ) = ∞ ( j=1 nj + xm+1 )λ then, in view of Corollary 6.4 and equality (6.62), we have ||T|| = k(λm+1 ) =

m+1 1 2 Γ(λi ). Γ(λ) i=1

If, in (6.63), for λm+1 =

λ 2,

we set

1 , kλ (xm+1 n1 , · · · , xm+1 nm , 1) = ∞ ( j=1 xm+1 nj + 1)λ then, in view of Corollary 6.4 and (6.64), we have ||T1 || = k(λm+1 ) =

6.4.2

The Case of s kλ (x1 , · · · , xm+1 ) = k=1

Γ(λ/2) 2 Γ(λi ). Γ(λ) i=1 m

m

i=1

1 λ/s xi λ/s +ck xm+1

Lemma 6.17. If s ∈ N, 0 < c1 < · · · < cs , λ, λi > 0(i = 1, · · · , m + 1), and s 2 1 m kλ (x1, · · · , xm+1 ) = , λ/s + c x λ/s x i k m+1 i=1 k=1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

292

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

then, we have

∞

∞

m

λ −1

j j=1 uj ··· du1 · · · dum m λ/s + c ) k 0 0 k=1 ( i=1 ui m s m s π i=1 Γ( λ λi ) λ = β(λm+1 ) = s πs Γ( λ (λ − λm+1 )) sin[ λ (λ − λm+1 )] s s 2 s 1 (λ−λm+1 )−1 × ckλ . (6.100) cj − ck

k(λm+1 ) =

s

j=1(j =k)

k=1

In particular, for λm+1 = λ2 , s m m s s π 2s −1 λ i=1 Γ( λ λi ) λ ck β = 2 Γ( 2s ) sin( πs 2 ) k=1

Proof.

s 2 j=1(j =k)

1 . cj − ck

(6.101)

For M > 0, setting DM = {(u1 , · · · , um )|ui > 0,

and ψ(u) =

i=1

s

k(λm+1 ) = = =

= =

1 k=1 Mu+ck ,

s m λ s m λ s m λ s m

m ui

∞ 0

···

lim

M→∞

lim

M

m

≤ 1},

by (6.75) and (3.12), we have

M→∞

M

m

s λ λj −1 dv1 · · · dvm j=1 vj λ/s s m , (vi = ui ) ( v + c ) k 0 k=1 i=1 i s m λ λj −1 j=1 vj s m vi ··· dv1 · · · dvm DM k=1 [M i=1 ( M ) + ck ] m 1 λs (λ−λm+1 )−1 s s λ (λ−λm+1 ) u du i=1 Γ( λ λi ) s , s Γ( λ (λ − λm+1 )) 0 k=1 (M u + ck )

∞

(v = M u) s (λ−λ m+1 )−1 λ

s ∞ v i=1 Γ( λ λi ) s dv s λ Γ( λ (λ − λm+1 )) 0 k=1 (v + ck ) s m m s π i=1 Γ( λ λi ) λ s πs Γ( λ (λ − λm+1 )) sin[ λ (λ − λm+1 )] s s 2 s (λ−λm+1 )−1 × ckλ k=1 j=1(j =k)

1 c j − ck

= β(λm+1 ), and then (6.74) follows.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

293

Multiple Half-Discrete Hilbert-Type Inequalities

With the assumptions of Lemma 6.16, if λm+1 < 1, then, we set i ∈ (λi − δ0 , λi + δ0 ) δ0 = min1≤i≤m+1 {λi , 1 − λm+1 } > 0. For any λ m+1 (i = 1, · · · , m + 1), i=1 λi = λ, it follows that λi > 0 (i = 1, · · · , m + 1), m+1 < λm+1 + δ0 ≤ λm+1 + 1 − λm+1 = 1, λ ∞ ∞ m λj −1 uj du1 · · · dum sj=1 m ··· k(λm+1 ) = β(λm+1 ) = λ/s + c ) k 0 0 k=1 ( i=1 ui m s s m π i=1 Γ( λ λi ) λ = s πs Γ( λ (λ − λm+1 )) sin[ λ (λ − λm+1 )] s s 2 s m+1 )−1 1 (λ−λ × ckλ ∈ R+ , cj − ck j=1(j =k)

k=1

and both s 2 m

1

λ/s + c y k i=1 xi

k=1

y λm+1 −1 and λ/s

s 2 k=1

1 y λm+1 −1 λ/s + c (yx ) i k i=1

m

are strictly decreasing with respect to y ∈ R+ . Then, for s 2

kλ (x1 , · · · , xm , xm+1 ) =

1 λ/s + c x λ/s x k m+1 i=1 i

m

k=1

in Theorems 6.1 and 6.2, it follows that Corollary 6.7. Suppose that s ∈ N, 0 < c1 < · · · < cs , pi > 1, λi > 0 (i = 1, · · · , m + 1), λm+1 < 1. If fi (xi ) ≥ 0, fi ∈ Lpi, ϕi (R+ ), ||fi ||pi, ϕi > 0(i = 1, · · · , m), an ≥ 0, a = {an }∞ n=1 ∈ lpm+1, ψ , ||a||pm+1, ψ > 0, then (i) we have the following equivalent inequalities with the best possible constant factor β(λm+1 ): m ∞ ∞ ∞ fj (xj ) mj=1 λ/s s an ··· dx1 · · · dxm + ck nλ/s ) 0 0 k=1 ( i=1 xi n=1 m 2 < β(λm+1 ) ||fi ||pi, ϕi ||a||pm+1, ψ , (6.102) i=1

and

∞

n=1

n

pλm+1 −1

0

∞

···

< β(λm+1 )

∞ 0

m 2 i=1

m

fj (xj )dx1 · · · dxm s m λ/s + c nλ/s ) k k=1 ( i=1 xi

p p1

j=1

||fi ||pi, ϕi ;

(6.103)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

294

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

(ii) for λm+1 = λ2 < 1, we have the following equivalent inequalities with the same best possible constant factor β( λ2 ): m ∞ ∞ ∞ j=1 fj (xj ) s m an ··· dx1 · · · dxm λ/s + c ] k 0 0 k=1 [ i=1 (nxi ) n=1 2 m λ <β ||fi ||pi, ϕi ||a||pm+1, ψ , (6.104) 2 i=1 and

∞

n

pλ 2 −1

∞

0

n=1

···

∞ 0

m

fj (xj )dx1 · · · dxm s m λ/s + c ] k k=1 [ i=1 (nxi )

p p1

j=1

2 m λ <β ||fi ||pi, ϕi . 2 i=1

(6.105)

In (6.9) and (6.12), for kλ (x1 , · · · , xm , xm+1 ) =

s 2

i=1 xi

k=1

we ﬁnd, for m ≥ 2, ∞ A1 (um+1 ) = ··· ≤ ≤

λ/s

,

+ ck xm+1

λj −1 j=2 uj du2 · · · dum s m λ/s + c uλ/s ) 0 0 k m+1 k=1 (1 + i=2 ui m λj −1 ∞ ∞ uj m j=2 ··· du2 · · · dum λ/s (1 + i=2 ui + c1 um+1 λ/s )s 0 0 m λj −1 ∞ ∞ du2 · · · dum j=2 uj 0

···

α−λ

λ/s

m

∞

0

= Aum+1m+1 where

1

m

s

(λm+1 −α)

c1λ

(1 +

m

i=2

s

ui λ/s )s− λ (λm+1 −α) 1 × λm+1 −α um+1

(0 < α < λm+1 ), m

λ −1

uj j du2 · · · dum λ/s , (vi = ui ) m s s− λ (λm+1 −α) λ/s 0 0 c1 (1 + i=2 ui ) s ∞ ∞ m λ λj −1 s m−2 dv2 · · · dvm (λ) j=2 vj = s (λ ··· m s −α) s− m+1 (1 + i=2 vi ) λ (λm+1 −α) 0 0 c1λ m ( λs )m−2 Γ s − λs (λ − λ1 − α) 2 sλi ) < ∞, = s (λ Γ( m+1 −α) Γ(s − s (λ λ m+1 − α)) cλ λ ∞

A=

1

···

∞

j=2

s λ (λm+1 −α)

i=2

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

295

Multiple Half-Discrete Hilbert-Type Inequalities

and B1 (ui ) = ≤ ≤

∞

0 ∞ 0 ∞ 0

≤

∞

0

··· ··· ···

m

∞ 0

m

∞

j=1(j =i)

0

m

∞

···

i = Buα−λ i

m

du1 · · · dui−1 dui+1 · · · dum m λ/s + c ) k k=1 ( i=1 ui λ −1

uj j du1 · · · dui−1 dui+1 · · · dum m ( i=1 ui λ/s + c1 )s λ −1

uj j du1 · · · dui−1 dui+1 · · · dum m s ( i=1 ui λ/s + c1 )s− λ (λi −α) 1 × uiλi −α

j=1(j =i)

0

λ −1

uj j s

j=1(j =i)

λj −1 du1 · · · dui−1 dui+1 · · · dum j=1(j =i) uj m s λ/s ( j=1(j =i) uj + c1 )s− λ (λi −α) uλi i −α

∞ 0

(0 < α < λi ),

where

∞

B= 0

=

···

∞ 0

m

s m−1 s

λj −1 du1 · · · dui−1 dui+1 · · · dum j=1(j =i) uj , m s λ/s ( j=1(j =i) uj + c1 )s− λ (λi −α)

λ/s

(vj = uj /c1 )

λ

c1 s− λ (λ−λm+1 −α) ∞ × ···

m

s

λ −1

vjλ j dv1 · · · dvi−1 dvi+1 · · · dvm m s ( j=1(j =i) vj + 1)s− λ (λi −α) 0 0 m s m−1 s Γ(s − λs (λ − λm+1 − α)) j=1(j =i) Γ( λ λj ) λ = < ∞. s Γ(s − λs (λi − α)) c1 s− λ (λ−λm+1 −α) ∞

j=1(j =i)

For m = 1, we can still ﬁnd some similar results. By Theorems 6.3 and 6.4, it follows that Corollary 6.8. Suppose that s ∈ N, 0 < c1 < · · · < cs , 0 < p1 < 1, pi < 0(i = 2, · · · , m + 1), λi > 0 (i = 1, · · · , m + 1), λm+1 < 1. If f1 (x1 ) ≥ 0, f1 ∈ Lp1, Φ1 (R+ ), ||f1 ||p1, Φ1 > 0, fi (xi ) ≥ 0, fi ∈ Lpi, ϕi (R+ ), ||fi ||pi, ϕi > 0 (i = 2, · · · , m), an ≥ 0, a = {an }∞ n=1 ∈ lpm+1, ψ , ||a||pm+1, ψ > 0, then (i) we have the following equivalent inequalities with the same best possible

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

296

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

constant factor β(λm+1 ): ∞ ∞ ∞ s an ··· 0

n=1

m fj (xj ) mj=1 λ/s dx1 · · · dxm ( x + ck nλ/s ) k=1 i=1 i

0

> β(λm+1 )||a||pm+1, ψ ||f1 ||p1 ,Φ1

m 2

||fi ||pi, ϕi ,

(6.106)

i=2

and

∞

n

pλm+1 −1

∞

0

n=1

···

∞ 0

m

fj (xj )dx1 · · · dxm m s λ/s + c nλ/s ) k k=1 ( i=1 xi

> β(λm+1 )||f1 ||p1 ,Φ1

p p1

j=1

m 2

||fi ||pi, ϕi ;

(6.107)

i=2

(ii) for λm+1 = λ2 < 1, f1 ∈ Lp1, Φ 1 (R+ ), ||f1 ||p1, Φ 1 > 0, we have the following equivalent inequalities with the same best possible constant factor β( λ2 ): m ∞ ∞ ∞ j=1 fj (xj ) s m an ··· dx1 · · · dxm λ/s + c ] k 0 0 k=1 [ i=1 (nxi ) n=1 m 2 λ >β ||fi ||pi, ϕi , (6.108) ||a||pm+1, ψ ||f1 ||p1, Φ 1 2 i=2 and

∞

n

pλ 2 −1

∞ 0

n=1

>β

···

∞ 0

p p1

m

j=1 fj (xj ) s m λ/s k=1 [ i=1 (nxi )

m 2 λ ||f1 ||p1, Φ ||fi ||pi, ϕi . 1 2 i=2

+ ck ]

dx1 · · · dxm (6.109)

By Theorems 6.5 and 6.6, we still have Corollary 6.9. Suppose that s ∈ N, 0 < c1 < · · · < cs , λi > 0, pi < 0(i = 1, · · · , m), 0 < pm+1 < 1, 0 < λm+1 < 1. If fi (xi ), an ≥ 0, fi ∈ Lpi, ϕi (R+ ), ||fi ||pi, ϕi > 0 (i = 1, · · · , m), a = {an }∞ n=1 ∈ lpm+1, ψ , ||a||pm+1, ψ > 0, then (i) we have the following equivalent reverse inequalities with the best possible constant factor β(λm+1 ): m ∞ ∞ ∞ fj (xj ) s mj=1 λ/s an ··· dx1 · · · dxm + ck nλ/s ) 0 0 k=1 ( i=1 xi n=1 m 2 > β(λm+1 ) ||fi ||pi, ϕi ||a||pm+1, ψ , (6.110) i=1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

297

Multiple Half-Discrete Hilbert-Type Inequalities

and

∞

n

pλm+1 −1

∞

0

n=1

> β(λm+1 )

···

m 2

m

∞

fj (xj )dx1 · · · dxm s m λ/s + c nλ/s ) k k=1 ( i=1 xi

p p1

j=1

0

||fi ||pi, ϕi ;

(6.111)

i=1

(ii) for λm+1 = λ2 (< 1), we have the following equivalent inequalities with the same best possible constant factor β( λ2 ): m ∞ ∞ ∞ j=1 fj (xj ) s m dx1 · · · dxm an ··· λ/s + c ] k 0 0 k=1 [ i=1 (nxi ) n=1 2 m λ >β ||fi ||pi, ϕi ||a||pm+1, ψ , (6.112) 2 i=1 and

∞

n

pλ 2 −1

∞

0

n=1

···

∞ 0

m

fj (xj )dx1 · · · dxm s m λ/s + c ] [ k k=1 i=1 (nxi )

p p1

j=1

2 m λ >β ||fi ||pi, ϕi . 2 i=1

(6.113)

By Theorems 6.7 and 6.8, we have Corollary 6.10. Suppose that s ∈ N, 0 < c1 < · · · < cs , pi > 1, 0 < λi < 1(i = 1, · · · , m), λm+1 > 0. If f (xm+1 ) ≥ 0, f ∈ Lpm+1, ϕ (R+ ),||f ||pm+1, ϕ > (i) (i) (i) 0, ani ≥ 0, a(i) = {ani }∞ ni =1 ∈ lpi, ψi , ||a ||pi, ψi > 0(i = 1, · · · , m), then (i) we have the following equivalent inequalities with the best possible constant factor β(λm+1 ): m (j) ∞ ∞ ∞ j=1 anj f (xm+1 ) ··· dxm+1 s m λ/s + c xλ/s ) 0 k m+1 nm =1 n1 =1 k=1 ( i=1 ni m 2 < β(λm+1 ) ||a(i) ||pi, ψi ||f ||pm+1, ϕ , (6.114) i=1

and ⎧ ⎨ ⎩

∞ 0

pλ

m+1 xm+1

−1

∞

···

nm =1

< β(λm+1 )

m 2 i=1

∞ n1 =1

m s

m

k=1 (

||a(i) ||pi, ψi ;

i=1

⎫ p1 ⎬

p

(j)

j=1 anj

λ/s

ni λ/s + ck xm+1 )

dxm+1

⎭

(6.115)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

298

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

(ii) for λm+1 = λ2 , we have the following equivalent inequalities with the best possible constant factor β( λ2 ):

∞

f (xm+1 )

0

∞

···

nm =1

m

∞

m

s

n1 =1

k=1 [

(j)

j=1

anj

i=1 (xm+1 ni )

λ/s

+ ck ]

dxm+1

2 m λ (i) <β ||a ||pi, ψi ||f ||pm+1, ϕ , 2

(6.116)

i=1

and ⎧ ⎨ ⎩

∞

pλ 2 −1

∞

xm+1

0

···

nm =1

<β

∞

m s

m

k=1 [

n1 =1

i=1 (xm+1 ni )

⎫ p1 ⎬

p

(j)

j=1 anj

λ/s

+ ck ]

dxm+1

2 m λ ||a(i) ||pi, ψi . 2

⎭

(6.117)

i=1

In (6.44), for 1 , m λ/s + c xλ/s ) k m+1 k=1 ( i=1 xi

kλ (x1 , · · · , xm , xm+1 ) = s we ﬁnd, for k = m + 1(m ≥ 2), Ai (uk ) =

∞ 0

···

m+1

∞ 0

s

l=1 (1

+

λj −1 j=1(j =i,k) uj m λ/s + c uλ/s ) l m+1 j=1(j =i) uj

×du1 · · · dui−1 dui+1 · · · duk−1 duk+1 · · · dum+1 ≤

∞ 0

···

∞

m+1

(1 +

0

k = Cuα−λ k

λ −1

uj j m

j=1(j =i,k)

du1 · · · dui−1 dui+1 · · · duk−1 duk+1 · · · dum+1

j=1(j =i,k)

uj λ/s + c1 um+1 )s− λ (λk −α) ukλk −α λ/s

s

(0 < α < λk ),

where C=

∞ 0

···

0

∞

m+1

λj −1 du1 · · · dui−1 dui+1 · · · duk−1 duk+1 · · · dum+1 j=1(j =i,k) uj . m s λ/s (1+ j=1(j =i,k) uj λ/s +c1 um+1 )s− λ (λk −α)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

299

Multiple Half-Discrete Hilbert-Type Inequalities

λ/s

Setting vj = uj λ/s (j = i, k, m + 1), vm+1 = c1 um+1 in the above integral, we obtain s m+1 ∞ λ λj −1 ( λs )m−1 ∞ j=1(j =i,k) vj C= ··· s s/λ s− λ (λk −α) (1 + m+1 0 0 c1 j=1(j =i,k) vj ) ×dv1 · · · dvi−1 dvi+1 · · · dvk−1 dvk+1 · · · dvm+1 m+1 s m−1 s 2 ( ) Γ(s − λ (λ − λi − α)) s = λ s/λ Γ( λj ) < ∞. s Γ(s − λ (λk − α)) λ c 1

j=1(j =i,k)

Similarly, we can ﬁnd that α−λ Ai (um+1 ) ≤ C1 um+1m+1 (0 < α < λm+1 , m ≥ 2), and the same result in the case of m = 1. If 0 < λi < 1(i = 1, · · · , m), λm+1 > 0, then, we set δ0 = i ∈ (λi − δ0 , λi + δ0 )(i = min1≤i≤m {λm+1 , λi , 1 − λi } > 0. For any λ m+1 1, · · · , m + 1), i=1 λi = λ, we ﬁnd that λm+1 > 0, 0 < λi < λi + δ0 ≤ 1(i = 1, · · · , m), k(λm+1 ) = β(λm+1 ) ∈ R+ , and both s s −1 −1 λ λ 1 1 m yj j and y j are λ/s k=1 m k=1 λ/s (xm+1 yi )λ/s +ck j i=1

yi

+ck xm+1

i=1

strictly decreasing with respect to yj ∈ R (j = 1, · · · , m). Then, by Theorems 6.9 and 6.10, it follows that Corollary 6.11. Suppose that s ∈ N, 0 < c1 < · · · < cs , 0 < p1 < 1, pi < 0(i = 2, · · · , m + 1), 0 < λi < 1(i = 1, · · · , m), λm+1 > 0. If f (xm+1 ) ≥ 0, (1) (1) f ∈ Lpm+1, ϕ (R+ ), ||f ||pm+1, ϕ > 0, an1 ≥ 0, a(1) = {an1 }∞ n1 =1 ∈ lp1, Ψ1 , (i) (i) (1) (i) ∞ (i) ||a ||p1, Ψ1 > 0, ani ≥ 0, a = {ani }ni =1 ∈ lpi, ψi , ||a ||pi, ψi > 0(i = 2, · · · , m), then (i) we have the following equivalent inequalities with the same best possible constant factor β(λm+1 ): m (j) ∞ ∞ ∞ j=1 anj dxm+1 f (xm+1 ) ··· m s λ/s + c xλ/s ) 0 k m+1 nm =1 n1 =1 k=1 ( i=1 ni > β(λm+1 )||f ||pm+1, ϕ ||a(1) ||p1, Ψ1

m 2

||a(i) ||pi, ψi ,

(6.118)

i=2

and ⎧ ⎨ ⎩

0

∞

pλ

m+1 xm+1

−1

∞

nm =1

···

∞ n1 =1

> β(λm+1 )||a(1) ||p1, Ψ1

m s

m

k=1 (

m 2 i=2

i=1

λ/s

ni λ/s + ck xm+1 )

||a(i) ||pi, ψi ;

⎫ p1 ⎬

p

(j)

j=1 anj

dxm+1

⎭

(6.119)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

300

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

(ii) for λm+1 = λ2 , we have the following equivalent inequalities with the same best possible constant factor β( λ2 ): m (j) ∞ ∞ ∞ anj s m j=1 dxm+1 f (xm+1 ) ··· λ/s + c ] k 0 k=1 [ i=1 (xm+1 ni ) nm =1 n1 =1 m 2 λ ||f ||pm+1, ϕ ||a(1) ||p1, Ψ1 >β ||a(i) ||pi, ψi , (6.120) 2 i=2 and ⎧ ⎨ ⎩

∞

pλ 2 −1

xm+1

0

∞

···

nm =1

∞

m s

n1 =1

m

k=1 [

anj

i=1 (xm+1 ni )

⎫ p1 ⎬

p

(j)

j=1

λ/s

+ ck ]

m 2 λ ||a(1) ||p1, Ψ1 >β ||a(i) ||pi, ψi . 2

dxm+1

⎭

(6.121)

i=2

By Theorems 6.10 and 6.12, it follows that Corollary 6.12. Suppose that s ∈ N, 0 < c1 < · · · < cs , pi < 0, 0 < λi < 1(i = 1, · · · , m), λm+1 > 0, 0 < pm+1 < 1. If f (xm+1 ) ≥ 0, (i) (i) f ∈ Lpm+1, Φ (R+ ), ||f ||pm+1, Φ > 0, ani ≥ 0, a(i) = {ani }∞ ni =1 ∈ lpi, ψi , (i) ||a ||pi, ψi > 0(i = 1, · · · , m), then (i) we have the following equivalent reverse inequalities with the same best possible constant factor β(λm+1 ): m (j) ∞ ∞ ∞ j=1 anj dxm+1 f (xm+1 ) ··· m s λ/s + c xλ/s ) 0 k m+1 nm =1 n1 =1 k=1 ( i=1 ni m 2 (i) > β(λm+1 ) ||a ||pi, ψi ||f ||pm+1, Φ , (6.122) i=1

and

0

∞

pλ

−1

m+1 xm+1 [1 − θm+1 (xm+1 )]p−1

∞ ∞ × ··· s

nm =1

> β(λm+1 )

n1 =1 m 2

m m

k=1 (

||a(i) ||pi, ψi ;

i=1

j=1

i=1

p

(j)

anj

λ/s

ni λ/s + ck xm+1 )

p1 dxm+1 (6.123)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

301

Multiple Half-Discrete Hilbert-Type Inequalities

(ii) for λm+1 = λ2 , f (xm+1 ) ≥ 0, f ∈ Lpm+1, Φ (R+ ), ||f ||pm+1, Φ > 0, we have the following equivalent inequalities with the same best possible constant factor β( λ2 ): m (j) ∞ ∞ ∞ anj s m j=1 f (xm+1 ) ··· dxm+1 λ/s + c ] k 0 k=1 [ i=1 (xm+1 ni ) nm =1 n1 =1 2 m λ (i) >β ||a ||pi, ψi ||f ||pm+1, Φ (6.124) , 2 i=1 and⎧ ⎨ ⎩

0

pλ

∞

−1

2 xm+1

[1 − θm+1 (xm+1 )]p−1

×

∞

···

nm =1

∞ n1 =1

m s

m

anj

λ/s i=1 (xm+1 ni )

k=1 [

⎫ p1 ⎬

p

(j)

j=1

+ ck ]

dxm+1

2 m λ >β ||a(i) ||pi, ψi . 2 i=1

⎭

(6.125)

Remark 6.4. (i) If in (6.24), we set 1 , m λ/s + c xλ/s ) ( n i k m+1 k=1 i=1 then, in view of Corollary 6.1 and equality (6.25), we have kλ (x1 , · · · , xm , n) = s

||T || = k(λm+1 ) = β(λm+1 ). If, in (6.26), for λm+1 =

λ 2

< 1, we set

kλ (nx1 , · · · , nxm , 1) = s

m

k=1 [

1

i=1 (xm+1 ni )

λ/s

+ ck ]

,

then, in view of Corollary 6.1 and equality (6.27), we have ||T1 || = β( λ2 ). (ii) If, in (6.61), we set 1 , m λ/s + c xλ/s ) k m+1 k=1 ( i=1 ni then, in view of Corollary 6.4 and (6.62), we have ||T|| = k(λm+1 ) = β(λm+1 ). kλ (n1 , · · · , nm , xm+1 ) = s

If, in (6.63), for λm+1 = λ2 , we set kλ (xm+1 n1 , · · · , xm+1 nm , 1) = s

m

k=1 [

then, in view of Corollary 6.4 and (6.64), we

1

λ/s + c ] k i=1 (xm+1 ni ) λ have ||T1 || = β( 2 ).

,

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

302

6.4.3

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

The Case of kλ (x1 , · · · , xm , xm+1 ) =

1 (max1≤i≤m+1 {xi })λ

Lemma 6.18. If λi > 0 (i = 1, · · · , m + 1), kλ (x1, · · · , xm+1 ) =

1 , (max1≤i≤m+1 {xi })λ

then, we have k(λm+1 ) =

∞

···

0 m+1 2

=λ

i=1

m

∞ 0

j=1

λ −1

uj j

(max1≤j≤m {uj , 1})λ

du1 · · · dum

1 . λi

(6.126)

Proof. We prove (6.126) by mathematical induction (see Yang [128]). For m = 1, we ﬁnd ∞ u1λ1 −1 k(λ2 ) = du1 (max{u1 , 1})λ 0 1 ∞ λ1 −1 u1 = u1λ1 −1 du1 + du1 uλ1 0 1 1 1 λ = + = , λ1 λ − λ1 λ1 λ2 and then (6.126) follows. Assuming that (6.126) is valid for m = n − 1, then, for m = n, we have n λj −1 ∞ ∞ j=1 uj k(λn+1 ) = ··· du1 · · · dun (max1≤j≤n {uj , 1})λ 0 0 ) ∞ ∞ ( ∞ uλnn −1 = ··· dun (max1≤j≤n {uj , 1})λ 0 0 0 ×

n−1 2

λ −1

uj j

du1 · · · dun−1 .

j=1

Since, we ﬁnd

∞

0

uλnn −1 dun (max1≤j≤n {uj , 1})λ

max1≤j≤n−1 {uj ,1}

= 0

uλnn −1 dun (max1≤j≤n {uj , 1})λ

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

303

Multiple Half-Discrete Hilbert-Type Inequalities

∞

+ max1≤j≤n−1 {uj ,1}

max1≤j≤n−1 {uj ,1}

= 0

uλnn −1 dun (max1≤j≤n {uj , 1})λ

uλnn −1 dun (max1≤j≤n−1 {uj , 1})λ ∞

+ max1≤j≤n−1 {uj ,1}

uλnn −1 dun uλn

(max1≤j≤n−1 {uj , 1})λn (max1≤j≤n−1 {uj , 1})λn −λ = + λn (max1≤j≤n−1 {uj , 1})λ λ − λn λ 1 = , λn (λ − λn ) (max1≤j≤n−1 {uj , 1})λ−λn then, by the assumption of m = n − 1, it follows that ∞ ∞ n−1 λj −1 du1 · · · dun−1 λ j=1 uj k(λn+1 ) = ··· λn (λ − λn ) 0 (max1≤j≤n−1 {uj , 1})λ−λn 0 =

λ(λ − λn ) λn (λ − λn )

n+1 2 i=1(i =n)

n+1 2 1 1 =λ , λi λi i=1

and then (6.99) follows.

With the assumptions of Lemma 6.17, if λm+1 < 1, then we set δ0 = i ∈ (λi −δ0 , λi +δ0 )(i = 1, · · · , m+ min1≤i≤m+1 {λi , 1−λm+1 } > 0. For any λ m+1 m+1 < λm+1 + 1), i=1 λi = λ, it follows that λi > 0(i = 1, · · · , m + 1), λ δ0 ≤ λm+1 + 1 − λm+1 = 1, m λj −1 ∞ ∞ j=1 uj m+1 ) = ··· du1 · · · dum k(λ (max1≤j≤m {uj , 1})λ 0 0 =λ

m+1 2 i=1

1 ∈ R+ , i λ

1 and both (max1≤i≤m y λm+1 −1 and (max1≤i≤m1 {yxi ,1})λ y λm+1 −1 are {xi ,y})λ strictly decreasing with respect to y ∈ R+ . Then for

kλ (x1 , · · · , xm , xm+1 ) =

1 (max1≤i≤m+1 {xi })λ

in Theorems 6.1 and 6.2, it follows that Corollary 6.13. Suppose that pi > 1, λi > 0(i = 1, · · · , m + 1), λm+1 < 1. If fi (xi ) ≥ 0, fi ∈ Lpi, ϕi (R+ ), ||fi ||pi, ϕi > 0(i = 1, · · · , m), an ≥ 0, a = {an }∞ n=1 ∈ lpm+1, ψ , ||a||pm+1, ψ > 0, then

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

304

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

(i) we have the following equivalent inequalities with the best possible m+1 constant factor λ i=1 λ1i : m ∞ ∞ ∞ j=1 fj (xj ) an ··· dx1 · · · dxm λ (max 1≤i≤m {xi , n}) 0 0 n=1 m 2 1 λ < ||fi ||pi, ϕi ||a||pm+1, ψ , (6.127) λm+1 i=1 λi and

∞

n

pλm+1 −1

∞

0

n=1

···

λ

<

λm+1

∞

m j=1

fj (xj )dx1 · · · dxm

p p1

(max1≤i≤m {xi , n})λ

0

m 2 1 ||fi ||pi, ϕi ; λ i i=1

(6.128)

(ii) for λm+1 = λ2 (< 1), we have the following equivalent inequalities 1 with the same best constant factor 2 m i=1 λi : ∞ ∞ m ∞ j=1 fj (xj ) an ··· dx1 · · · dxm (max1≤i≤m {nxi , 1})λ 0 0 n=1 m 2 1 <2 ||fi ||pi, ϕi ||a||pm+1, ψ , (6.129) λi i=1 and

∞

n

pλ 2 −1

∞ 0

n=1

<2

···

0

∞

m

j=1 fj (xj )dx1 · · · dxm

p p1

(max1≤i≤m {nxi , 1})λ

m 2 1 ||fi ||pi, ϕi . λi i=1

(6.130)

In (6.9) and (6.12), for kλ (x1 , · · · , xm , xm+1 ) = we ﬁnd, for m ≥ 2, ∞ ··· A1 (um+1 ) = 0

≤

∞

···

∞ 0 ∞

1 (max1≤i≤m+1 {xi })λ

m j=2

,

λ −1

uj j

du2 · · · dum (max2≤j≤m+1 {1, uj })λ m λj −1 du2 · · · dum j=2 uj λ

−α

m+1 (max2≤j≤m {1, uj })λ+α−λm+1 um+1 m λ + α − λm+1 2 1 α−λm+1 = (u ) (0 < α < λm+1 ), λ1 + α λi m+1

0

0

i=2

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

305

Multiple Half-Discrete Hilbert-Type Inequalities

and

B1 (ui ) = ≤

∞

0 ∞

··· ···

∞ 0 ∞

m

j=1(j =i)

λ −1

uj j

du1 · · · dui−1 dui+1 · · · dum

(max1≤j≤m {1, uj })λ

m

j=1(j =i)

λ −1

uj j

du1 · · · dui−1 dui+1 · · · dum

(max1≤j =i≤m {1, uj })λ+α−λi uλi i −α 2 m λ + α − λi 1 α−λi = (u ) (0 < α < λi ). λm+1 + α λj i 0

0

j=1(j =i)

For m = 1, we can still ﬁnd the same results. By Theorems 6.3 and 6.4, it follows that Corollary 6.14. Suppose that 0 < p1 < 1, pi < 0 (i = 2, · · · , m + 1), λi > 0(i = 1, · · · , m + 1), λm+1 < 1. If f1 (x1 ) ≥ 0, f1 ∈ Lp1, Φ1 (R+ ), ||f1 ||p1, Φ1 > 0, fi (xi ) ≥ 0, fi ∈ Lpi, ϕi (R+ ), ||fi ||pi, ϕi > 0(i = 2, · · · , m), an ≥ 0, a = {an }∞ n=1 ∈ lpm+1, ψ , ||a||pm+1, ψ > 0, then (i) we have the following equivalent inequalities with the same best posm+1 sible constant factor λ i=1 λ1i : m ∞ ∞ ∞ j=1 fj (xj ) an ··· dx1 · · · dxm (max1≤i≤m {xi , n})λ 0 0 n=1 >λ

m+1 2 i=1

and

∞

n

pλm+1 −1

m 2 1 ||a||pm+1, ψ ||f1 ||p1 ,Φ1 ||fi ||pi, ϕi , λi i=2

∞

0

n=1

>λ

···

m+1 2 i=1

∞ 0

m j=1

fj (xj )dx1 · · · dxm

(6.131)

p p1

(max1≤i≤m {xi , n})λ

m 2 1 ||f1 ||p1 ,Φ1 ||fi ||pi, ϕi ; λi i=2

(6.132)

(ii) for λm+1 = λ2 (< 1), f1 ∈ Lp1, Φ 1 (R+ ), ||f1 ||p1, Φ 1 > 0, we have the following equivalent inequalities with the same best possible constant factor m 2 i=1 λ1i : m ∞ ∞ ∞ j=1 fj (xj ) an ··· dx1 · · · dxm (max1≤i≤m {nxi , 1})λ 0 0 n=1 m m 2 1 2 >2 ||fi ||pi, ϕi , (6.133) ||a||pm+1, ψ ||f1 ||p1, Φ 1 λi i=1

i=2

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

306

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and

∞

n

pλ 2 −1

∞

···

m j=1

fj (xj )dx1 · · · dxm

p p1

(max1≤i≤m {nxi , 1})λ 0 m m 2 1 2 ||f1 ||p1, Φ >2 ||fi ||pi, ϕi . 1 λ i=1 i i=2 0

n=1

∞

(6.134)

By Theorems 6.5 and 6.6, we still have Corollary 6.15. Suppose that pi < 0, λi > 0(i = 1, · · · , m), 0 < pm+1 < 1, 0 < λm+1 < 1. If fi (xi ), an ≥ 0, fi ∈ Lpi, ϕi (R+ ), ||fi ||pi, ϕi > 0(i = 1, · · · , m), a = {an }∞ n=1 ∈ lpm+1, ψ , ||a||pm+1, ψ > 0, then (i) we have the following equivalent reverse inequalities with the same 1 best possible constant factor λ m+1 i=1 λi : ∞ ∞ m ∞ j=1 fj (xj ) an ··· dx1 · · · dxm (max1≤i≤m {xi , n})λ 0 0 n=1 m 2 1 λ > ||fi ||pi, ϕi ||a||pm+1, ψ , (6.135) λm+1 i=1 λi and

∞

npλm+1 −1

∞

0

n=1

>

···

λ λm+1

∞ 0

m

j=1 fj (xj )dx1 · · · dxm

p p1

(max1≤i≤m {xi , n})λ

m 2 1 ||fi ||pi, ϕi ; λ i i=1

(6.136)

(ii) for λm+1 = λ2 (< 1), we have the following equivalent inequalities m with the same best possible constant factor 2 i=1 λ1i : m ∞ ∞ ∞ j=1 fj (xj ) an ··· dx1 · · · dxm (max1≤i≤m {nxi , 1})λ 0 0 n=1 m 2 1 >2 ||fi ||pi, ϕi ||a||pm+1, ψ , (6.137) λi i=1 and

∞ n=1

n

pλ 2 −1

∞ 0

···

>2

0

∞

m j=1

fj (xj )dx1 · · · dxm

(max1≤i≤m {nxi , 1})λ

m 2 1 ||fi ||pi, ϕi . λi i=1

p p1

(6.138)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

307

Multiple Half-Discrete Hilbert-Type Inequalities

By Theorems 6.7 and 6.8, we have Corollary 6.16. Suppose that pi > 1, 0 < λi < 1(i = 1, · · · , m), λm+1 > (i) 0. If f (xm+1 ) ≥ 0, f ∈ Lpm+1, ϕ (R+ ), ||f ||pm+1, ϕ > 0, ani ≥ 0, a(i) = (i) (i) {ani }∞ ni =1 ∈ lpi, ψi , ||a ||pi, ψi > 0(i = 1, · · · , m), then (i) we have the following equivalent inequalities with the best constant 1 factor λ m+1 i=1 λi : m (j) ∞ ∞ ∞ j=1 anj f (xm+1 ) ··· dxm+1 (max1≤i≤m {ni , xm+1 })λ 0 nm =1 n1 =1 m 2 1 λ (i) < ||a ||pi, ψi ||f ||pm+1, ϕ , (6.139) λm+1 i=1 λi and ⎧ ⎨ ⎩

∞

0

pλ

m+1 xm+1

−1

∞

···

nm =1

<

m

∞ n1 =1

λ λm+1

⎫ p1 ⎬

p

(j)

j=1 anj

dxm+1

(max1≤i≤m {ni , xm+1 })λ

m 2 1 (i) ||a ||pi, ψi ; λi i=1

⎭

(6.140)

(ii) for λm+1 = λ2 , we have the following equivalent inequalities with the m best possible constant factor 2 i =1 λ1i : m (j) ∞ ∞ ∞ j=1 anj f (xm+1 ) ··· dxm+1 max1≤i≤m {xm+1 ni , 1})λ 0 nm =1 n1 =1 m 2 1 <2 ||a(i) ||pi, ψi ||f ||pm+1, ϕ , (6.141) λ i i=1 and

⎧ ⎨ ⎩

0

∞

pλ 2 −1

xm+1

∞

···

nm =1

<2

m

∞ n1 =1

(j)

j=1 anj

dxm+1

max1≤i≤m {xm+1 ni , 1})λ

m 2 1 (i) ||a ||pi, ψi . λi i=1

⎭ (6.142)

In (6.44), for kλ (x1 , · · · , xm , xm+1 ) =

⎫ p1 ⎬

p

1 (max1≤i≤m+1 {xi })λ

,

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

308

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

we ﬁnd, for m ≥ 2, Ai (uk ) =

∞ 0

···

m+1

∞ 0

j=1(j =i,k)

λ −1

uj j

(max1≤j =i≤m+1 {uj , 1})λ

× du1 · · · dui−1 dui+1 · · · duk−1 duk+1 · · · dum+1 λj −1 ∞ ∞ m+1 du1 · · · dui−1 dui+1 · · · duk−1 duk+1 · · · dum+1 j=1(j =i,k) uj ≤ ··· (max1≤j =i,k≤m+1 {uj , 1})λ+α−λk uλk k −α 0 0 =

λ + α − λk λi + α

m+1 2 j=1(j =i,k)

1 α−λk (u ) (0 < α < λk ). λj k

For m = 1, we can still ﬁnd the similar result. If 0 < λi < 1(i = 1, · · · , m), λm+1 > 0, then we set δ0 = i ∈ (λi − δ0 , λi + δ0 )(i = min1≤i≤m {λm+1 , λi , 1 − λi } > 0. For any λ m+1 i < 1(i = 1, · · · , m), 1, · · · , m + 1), i=1 λi = λ, we ﬁnd λm+1 > 0, 0 < λ m+1 ) = λ k(λ

m+1 2 i=1

1 ∈ R+ λi

−1 λ

−1 λ

j 1 and both (max1≤i≤m 1{yi ,xm+1 })λ yj j and (max1≤i≤m {x λ yj m+1 yi ,1}) strictly decreasing with respect to yj ∈ R+ (j = 1, · · · , m). Then by Theorems 6.9 and 6.10, it follows that

are

Corollary 6.17. Suppose that 0 < p1 < 1, pi < 0(i = 2, · · · , m + 1), 0 < λi < 1(i = 1, · · · , m), λm+1 > 0. If f (xm+1 ) ≥ 0, f ∈ Lpm+1, ϕ (R+ ), (1) (1) (1) ||p1, Ψ1 > 0, ||f ||pm+1, ϕ > 0, an1 ≥ 0, a(1) = {an1 }∞ n1 =1 ∈ lp1, Ψ1 , ||a (i) (i) (i) ani ≥ 0, a(i) = {ani }∞ ∈ l , ||a || > 0(i = 2, · · · , m), then pi, ψi pi, ψi ni =1 (i) we have the following equivalent inequalities with the same best posm+1 sible constant factor λ i=1 λ1i :

∞ 0

f (xm+1 )

∞ nm =1

···

∞ n1 =1

m j=1

(j)

anj

(max1≤i≤m {ni , xm+1 })λ

dxm+1

m+1 m 2 1 2 >λ ||a(i) ||pi, ψi , (6.143) ||f ||pm+1, ϕ ||a(1) ||p1, Ψ1 λi i=1

i=2

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

309

Multiple Half-Discrete Hilbert-Type Inequalities

and ⎧ ⎨ ⎩

∞

0

pλ

m+1 xm+1

∞

−1

···

nm =1

n1 =1

>λ

m

∞

m+1 2 i=1

⎫ p1 ⎬

p

(j)

j=1 anj

(max1≤i≤m {ni , xm+1 })λ

dxm+1

2 1 (1) ||a ||p1, Ψ1 ||a(i) ||pi, ψi ; λi

⎭

m

(6.144)

i=2

(ii) for, λm+1 = λ2 , we have the following equivalent inequalities with m the same best possible constant factor 2 i=1 λ1i :

∞

∞ 0

f (xm+1 )

···

nm =1

2 > λ1

m

∞ n1 =1

(j)

j=1

anj

(max1≤i≤m {xm+1 ni , 1})λ

m 2 1 ||a(i) ||pi, ψi λ i=2 i

dxm+1

||f ||pm+1, ϕ ||a(1) ||p1, Ψ1 ,

(6.145)

and ⎧ ⎨ ⎩

∞

pλ 2 −1

xm+1

0

∞

···

nm =1

m

∞ n1 =1

(max1≤i≤m {xm+1 ni , 1})λ

m 2 2 1 (i) > ||a ||pi, ψi λ1 i=2 λi

⎫ p1 ⎬

p

(j)

j=1 anj

||a(1) ||p1, Ψ1 .

dxm+1

⎭ (6.146)

By Theorems 6.11 and 6.12, it follows that Corollary 6.18. Suppose that pi < 0, 0 < λi < 1(i = 1, · · · , m), λm+1 > 0, 0 < pm+1 < 1. If f (xm+1 ) ≥ 0, f ∈ Lpm+1, Φ (R+ ), ||f ||pm+1, Φ > 0, (i) (i) (i) ani ≥ 0, a(i) = {ani }∞ ni =1 ∈ lpi, ψi , ||a ||pi, ψi > 0(i = 1, · · · , m), then (i) we have the following equivalent reverse inequalities with the same m+1 best possible constant factor λ i=1 λ1i :

∞ 0

f (xm+1 )

∞

···

nm =1

>

λ λm+1

∞

m j=1

(j)

anj

dxm+1 (max1≤i≤m {ni , xm+1 })λ m 2 1 (i) ||a ||pi, ψi ||f ||pm+1, Φ , (6.147) λi

n1 =1

i=1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

310

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and

∞

0

−1

pλ

m+1 xm+1 [1 − θm+1 (xm+1 )]p−1

∞ ∞ × ···

nm =1

>

λ λm+1

m

n1 =1

p

(j)

j=1

anj

dxm+1

(max1≤i≤m {ni , xm+1 })λ

m 2 1 ||a(i) ||pi, ψi λi

p1

;

(6.148)

i=1

(ii) for, λm+1 = λ2 , f (xm+1 ) ≥ 0, f ∈ Lpm+1, Φ (R+ ), ||f ||pm+1, Φ > 0, we have the following equivalent inequalities with the same best possible m constant factor 2 i=1 λ1i : m (j) ∞ ∞ ∞ j=1 anj dxm+1 f (xm+1 ) ··· (max1≤i≤m {xm+1 ni , 1})λ 0 nm =1 n1 =1 m 2 1 (i) >2 ||a ||pi, ψi ||f ||pm+1, Φ (6.149) , λi i=1 and ⎧ ⎨ ⎩

∞ 0

pλ

−1

2 xm+1

[1 − θm+1 (xm+1 )]p−1

×

∞

···

nm =1

m

∞ n1 =1

>2

dxm+1

(max1≤i≤m {xm+1 ni , 1})λ

m 2 1 (i) ||a ||pi, ψi . λi

Remark 6.5. (i) If in (6.24), we set 1 (max1≤i≤m {xi , n})λ

,

then, in view of Corollary 6.12 and (6.25), we have ||T || = k(λm+1 ) = λ

m+1 2 i=1

if, in (6.26), for λm+1 =

λ 2

1 ; λi

< 1, we set

kλ (nx1 , · · · , nxm , 1) =

⎭

(6.150)

i=1

kλ (x1 , · · · , xm , n) =

⎫ p1 ⎬

p

(j)

j=1 anj

1 (max1≤i≤m {nxi , 1})λ

,

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Multiple Half-Discrete Hilbert-Type Inequalities

ws-book9x6

311

then, in view of Corollary 6.13 and (6.27), we have m 2 1 ||T1 || = k(λm+1 ) = 2 . λi i=1 (ii) If, in (6.61), we set 1 , (max1≤i≤m {ni , xm+1 })λ then, in view of Corollary 6.16 and (6.62), we have m+1 2 1 ||T || = k(λm+1 ) = λ ; λi i=1 kλ (n1 , · · · , nm , xm+1 ) =

if, in (6.63), for λm+1 = λ2 , we set 1 , (max1≤i≤m {xm+1 ni , 1})λ then, in view of Corollary 6.16 and (6.64), we have m 2 1 ||T1 || = k(λm+1 ) = 2 . λ i i=1 kλ (xm+1 n1 , · · · , xm+1 nm , 1) =

6.4.4

The Case of kλ (x1 , · · · , xm , xm+1 ) =

1 (min1≤i≤m+1 {xi })λ

Lemma 6.19. If λi , λ < 0 (i = 1, · · · , m + 1), 1 , kλ (x1, · · · , xm+1 ) = (min1≤i≤m+1 {xi })λ then, we have m λj −1 ∞ ∞ j=1 uj k(λm+1 ) = ··· du1 · · · dum (min1≤j≤m+1 {uj , 1})λ 0 0 = (−1)m λ

m+1 2 i=1

1 . λi

Proof. We prove (6.151) by mathematical induction. For m = 1, we ﬁnd ∞ u1λ1 −1 k(λ2 ) = du1 (min{u1 , 1})λ 0 1 λ1 −1 ∞ u1 = du1 + u1λ1 −1 du1 λ u 0 1 1 1 1 −λ = + = , −λ + λ1 −λ1 λ1 λ2

(6.151)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

312

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and then, (6.151) follows. Assuming that (6.151) is valid for m = n − 1, then, for m = n, we have

∞

λ −1

j=1

uj j

du1 · · · dun (min1≤j≤n {uj , 1})λ ) ∞ ∞ ( ∞ uλnn −1 = ··· du n (min1≤j≤n {uj , 1})λ 0 0 0

k(λn+1 ) =

0

···

n

∞

0

×

n−1 2

λ −1

uj j

du1 · · · dun−1 .

j=1

Since, we ﬁnd ∞ 0

uλnn −1 dun (min1≤j≤n {uj , 1})λ

min1≤j≤n−1 {uj ,1}

= 0

uλnn −1 dun (min1≤j≤n {uj , 1})λ ∞

+

min1≤j≤n−1 {uj ,1} min1≤j≤n−1 {uj ,1}

= 0

uλnn −1 dun (min1≤j≤n {uj , 1})λ

unλn −1 dun uλn ∞

+ min1≤j≤n−1 {uj ,1}

uλnn −1 dun (min1≤j≤n {uj , 1})λ

−(min1≤j≤n−1 {uj , 1})λn (min1≤j≤n−1 {uj , 1})λn −λ + = λn − λ λn (min1≤j≤n−1 {uj , 1})λ λ 1 = , λn (λn − λ) (min1≤j≤n−1 {uj , 1})λ−λn then, by the assumption of m = n − 1, it follows that ∞ ∞ n−1 λj −1 du1 · · · dun−1 λ j=1 uj k(λn+1 ) = ··· λn (λn − λ) 0 (min1≤j≤n−1 {uj , 1})λ−λn 0 =

λ(λ − λn )(−1)n−1 λn (λn − λ)

and then, (6.151) follows.

n+1 2 i=1(i =n)

n+1 2 1 1 = (−1)n λ , λi λi i=1

With the assumptions of Lemma 6.18, if λm+1 < λ + 1, then we set i ∈ (λi − δ0 , λi + δ0 = min1≤i≤m+1 {−λi , λ + 1 − λm+1 } > 0. For any λ

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

313

Multiple Half-Discrete Hilbert-Type Inequalities

m+1 δ0 )(i = 1, · · · , m + 1), i=1 λ i = λ, it follows that λi < λi + δ0 ≤ 0(i = 1, · · · , m + 1), λm+1 < λm+1 + δ0 ≤ λ + 1, m λj −1 ∞ ∞ j=1 uj m+1 ) = k(λ ··· du1 · · · dum (min1≤j≤n−1 {uj , 1})λ 0 0 = (−1)m λ

m+1 2 i=1

and both

1

(min1≤i≤m {xi ,y})λ

1 ∈ R+ , i λ

y λm+1 −1 , and

1 (min1≤i≤m {yxi ,1})λ

strictly decreasing with respect to y ∈ R+ . Then, for 1 kλ (x1 , · · · , xm , xm+1 ) = (min1≤i≤m+1 {xi })λ in Theorems 6.1 and 6.2, it follows that

y λm+1 −1 are

(λ < 0)

Corollary 6.19. Suppose that pi > 1, λi < 0(i = 1, · · · , m + 1), λm+1 < λ + 1. If fi (xi ) ≥ 0, fi ∈ Lpi, ϕi (R+ ), ||fi ||pi, ϕi > 0(i = 1, · · · , m), an ≥ 0, a = {an }∞ n=1 ∈ lpm+1, ψ , ||a||pm+1, ψ > 0, then (i) we have the following equivalent inequalities with the best possible 1 constant factor (−1)m λ m+1 i=1 λi : m ∞ ∞ ∞ j=1 fj (xj ) an ··· dx1 · · · dxm (min1≤i≤m {xi , n})λ 0 0 n=1 m (−1)m λ 2 1 < ||fi ||pi, ϕi ||a||pm+1, ψ , (6.152) λm+1 λi i=1 and

∞

n

pλm+1 −1

0

n=1

<

∞

···

∞ 0

m j=1

fj (xj )dx1 · · · dxm

p p1

(min1≤i≤m {xi , n})λ

m (−1)m λ 2 1 ||fi ||pi, ϕi ; λm+1 λi

(6.153)

i=1

(ii) for, λm+1 = λ2 (− 12 < λ < 0), we have the following equivalent 1 inequalities with the same best possible constant factor, 2(−1)m m i=1 λi : m ∞ ∞ ∞ j=1 fj (xj ) an ··· dx1 · · · dxm (min1≤i≤m {nxi , 1})λ 0 0 n=1 m 2 1 m < 2(−1) ||fi ||pi, ϕi ||a||pm+1, ψ , (6.154) λi i=1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

314

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and

∞

n

pλ 2 −1

∞ 0

n=1

···

∞

m j=1

p p1

(min1≤i≤m {nxi , 1})λ

0

< 2(−1)m

fj (xj )dx1 · · · dxm

m 2 1 ||fi ||pi, ϕi . λi

(6.155)

i=1

In (6.9) and (6.12), for kλ (x1 , · · · , xm , xm+1 ) =

1 (min1≤i≤m+1 {xi })λ

(λ < 0),

we ﬁnd, for m ≥ 2, A1 (um+1 ) =

∞

0

≤

∞

··· ···

∞ 0 ∞

m j=2

λ −1

uj j

du2 · · · dum (min2≤j≤m+1 {1, uj })λ m λj −1 du2 · · · dum j=2 uj λ

m+1 (min2≤j≤m+1 {1, uj })λ+α−λm+1 um+1 m 2 1 α−λm+1 λ + α − λm+1 (−1)m−1 = (u ), λ1 + α λi m+1 i=2

0

0

−α

where, 0 < α < min{−λ1 , λm+1 − λ}, B1 (ui ) = ≤

∞

0 ∞

··· ···

∞ 0 ∞

m

j=1(j =i)

m

λ −1

uj j

du1 · · · dui−1 dui+1 · · · dum

(min1≤j≤m {1, uj })λ

j=1(j =i)

λ −1

uj j

du1 · · · dui−1 dui+1 · · · dum

(min1≤j =i≤m {1, uj })λ+α−λi uλi i −α m 2 λ + α − λi 1 α−λi = (u ), (−1)m−1 λm+1 + α λj i 0

0

j=1(j =i)

where 0 < α < min{λi − λ, −λm+1 }. For m = 1, we can obtain some similar results. By Theorems 6.3 and 6.4, it follows that Corollary 6.20. Suppose that 0 < p1 < 1, pi < 0(i = 2, · · · , m + 1), λi < 0(i = 1, · · · , m + 1), λm+1 < λ + 1. If f1 (x1 ) ≥ 0, f1 ∈ Lp1, Φ1 (R+ ), ||f1 ||p1, Φ1 > 0, fi (xi ) ≥ 0, fi ∈ Lpi, ϕi (R+ ), ||fi ||pi, ϕi > 0(i = 2, · · · , m), an ≥ 0, a = {an }∞ n=1 ∈ lpm+1, ψ , ||a||pm+1, ψ > 0, then

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

315

Multiple Half-Discrete Hilbert-Type Inequalities

(i) we have the following equivalent inequalities with the same best posm+1 sible constant factor (−1)m λ i=1 λ1i : m ∞ ∞ ∞ j=1 fj (xj ) an ··· dx1 · · · dxm (min1≤i≤m {xi , n})λ 0 0 n=1 > (−1)m λ

m+1 2 i=1

and

∞

n=1

npλm+1 −1

0

∞

m 2 1 ||a||pm+1, ψ ||f1 ||p1 ,Φ1 ||fi ||pi, ϕi , (6.156) λi i=2

···

> (−1)m λ

∞

0 m+1 2 i=1

m j=1

fj (xj ) dx1 · · · dxm

p p1

(min1≤i≤m {xi , n})λ m 2 1 ||f1 ||p1 ,Φ1 ||fi ||pi, ϕi ; λi i=2

(6.157)

(ii) for λm+1 = λ2 (− 12 < λ < 0), f1 ∈ Lp1, Φ 1 (R+ ), ||f1 ||p1, Φ 1 > 0, we have the following equivalent inequalities with the best possible constant 1 factor 2(−1)m m i=1 λi : m ∞ ∞ ∞ j=1 fj (xj ) an ··· dx1 · · · dxm (min1≤i≤m {nxi , 1})λ 0 0 n=1 m m 2 1 2 m > 2(−1) ||fi ||pi, ϕi , (6.158) ||a||pm+1, ψ ||f1 ||p1, Φ 1 λi i=1 i=2 and

p p1 ∞ ∞ m ∞ λ j=1 fj (xj )dx1 · · · dxm p 2 −1 n ··· (min1≤i≤m {nxi , 1})λ 0 0 n=1 m m 2 1 2 m > 2(−1) ||fi ||pi, ϕi . (6.159) ||f1 ||p1, Φ 1 λi i=1 i=2 By Theorems 6.5 and 6.6, we still have Corollary 6.21. Suppose that pi < 0, λi < 0(i = 1, · · · , m), 0 < pm+1 < 1, λm+1 < min{0, λ + 1}. If fi (xi ), an ≥ 0, fi ∈ Lpi, ϕi (R+ ) (i = 1, · · · , m), a = {an }∞ n=1 ∈ lpm+1, ψ , ||fi ||pi, ϕi > 0, ||a||pm+1, ψ > 0. then (i) we have the following equivalent reverse inequalities with the same m+1 best possible constant factor (−1)m λ i=1 λ1i : m ∞ ∞ ∞ j=1 fj (xj ) an ··· dx1 · · · dxm (min1≤i≤m {xi , n})λ 0 0 n=1 m (−1)m λ 2 1 > ||fi ||pi, ϕi ||a||pm+1, ψ , (6.160) λm+1 λi i=1

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

316

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

and

∞

n

pλm+1 −1

∞

0

n=1

···

∞ 0

m j=1

fj (xj )dx1 · · · dxm

p p1

(min1≤i≤m {xi , n})λ

m (−1)m λ 2 1 > ||fi ||pi, ϕi ; λm+1 i=1 λi

(6.161)

(ii) for λm+1 = λ2 (− 21 < λ < 0), we have the following equivalent m inequalities with the same best constant factor 2(−1)m i=1 λ1i : m ∞ ∞ ∞ j=1 fj (xj ) an ··· dx1 · · · dxm (min1≤i≤m {nxi , 1})λ 0 0 n=1 m 2 1 > 2(−1)m ||fi ||pi, ϕi ||a||pm+1, ψ , (6.162) λi i=1 and

∞

n

pλ 2 −1

∞ 0

n=1

···

∞

0

> 2(−1)m

m j=1

fj (xj )dx1 · · · dxm

p p1

(min1≤i≤m {nxi , 1})λ m 2 1 ||fi ||pi, ϕi . λ i=1 i

(6.163)

By Theorems 6.7 and 6.8, we have the following Corollary 6.22. Suppose that pi > 1, λi < 0(i = 1, · · · , m), λm+1 < (i) min{0, λ + 1}. If f (xm+1 ) ≥ 0, f ∈ Lpm+1, ϕ (R+ ), ||f ||pm+1, ϕ > 0, ani ≥ 0, (i) (i) a(i) = {ani }∞ ni =1 ∈ lpi, ψi , ||a ||pi, ψi > 0(i = 1, · · · , m), then (i) we have the following equivalent inequalities with the best possible m+1 constant factor (−1)m λ i=1 λ1i : m (j) ∞ ∞ ∞ j=1 anj f (xm+1 ) ··· dxm+1 (min1≤i≤m {ni , xm+1 })λ 0 nm =1 n1 =1 m (−1)m λ 2 1 (i) < ||a ||pi, ψi ||f ||pm+1, ϕ , (6.164) λm+1 λ i=1 i and ⎧ ⎨ ⎩

0

∞

pλ

m+1 xm+1

−1

∞

nm =1

<

···

∞ n1 =1

m

(min1≤i≤m {ni , xm+1 })λ

m (−1)m λ 2 1 (i) ||a ||pi, ψi ; λm+1 λi i=1

(j)

j=1 anj

⎫ p1 ⎬

p dxm+1

⎭

(6.165)

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

317

Multiple Half-Discrete Hilbert-Type Inequalities

(ii) for λm+1 = λ2 (− 21 < λ < 0), we have the following equivalent 1 inequalities with the best possible constant factor 2(−1)m m i=1 λi : m (j) ∞ ∞ ∞ j=1 anj f (xm+1 ) ··· dxm+1 λ (min 1≤i≤m {ni xm+1 , 1}) 0 nm =1 n1 =1 m 2 1 < 2(−1)m ||a(i) ||pi, ψi ||f ||pm+1, ϕ , (6.166) λi i=1 and

⎧ ⎨ ⎩

∞

0

pλ 2 −1

xm+1

∞

···

nm =1

m

∞ n1 =1

< 2(−1)m

⎫ p1 ⎬

p

(j)

j=1

anj

(min1≤i≤m {ni xm+1 , 1})λ

dxm+1

m 2 1 (i) ||a ||pi, ψi . λi

⎭ (6.167)

i=1

In (6.44), for kλ (x1 , · · · , xm , xm+1 ) = we ﬁnd, for m ≥ 2, ∞ ··· Ai (uk ) = 0

∞ 0

m+1 2

×

1 (min1≤i≤m+1 {xi })λ

(λ < 0)

1 (min1≤j =i≤m+1 {uj , 1})λ λ −1

uj j

du1 · · · dui−1 dui+1 · · · duk−1 duk+1 · · · dum+1

j=1(j =i,k)

and

≤

∞

···

∞

m+1

j=1(j =i,k)

λ −1

uj j

(min1≤j =i,k≤m+1 {uj , 1})λ+α−λk uλk k −α ×du1 · · · dui−1 dui+1 · · · duk−1 duk+1 · · · dum+1 m+1 2 λ + α − λk 1 α−λk (−1)m−1 = (u ), λi + α λj k 0

0

j=1(j =i,k)

where 0 < α < min{λk −λ, −λi }. For m = 1, we can obtain a similar result. If λi < min{0, 1 − λ}(i = 1, · · · , m), λm+1 < 0, then, we set δ0 = i ∈ (λi − δ0 , λi + δ0 ) min1≤i≤m {−λm+1 , −λi , 1 − λ − λi } > 0. For any λ m+1 i < min{0, 1 − λ} (i = 1, · · · , m + 1), i=1 λi = λ, we ﬁnd λm+1 < 0, λ (i = 1, · · · , m), m+1 ) = (−1)m λ k(λ

m+1 2 i=1

1 ∈ R+ λi

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

318

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

1 λj −1 and both (min1≤i≤m1yi ,xm+1 )λ y λj −1 and (min1≤i≤m {x are λy m+1 yi ,1}) strictly decreasing with respect to yj ∈ R+ (j = 1, · · · , m). Then by Theorems 6.9 and 6.10, it follows that

Corollary 6.23. Suppose that 0 < p1 < 1, pi < 0(i = 2, · · · , m + 1), λi < min{0, 1+λ} (i = 1, · · · , m), λm+1 < 0. If f (xm+1 ) ≥ 0, f ∈ Lpm+1, ϕ (R+ ), (1) (1) (1) ||f ||pm+1, ϕ > 0, an1 ≥ 0, a(1) = {an1 }∞ ||p1, Ψ1 > 0, n1 =1 ∈ lp1, Ψ1 , ||a (i) (i) (i) ∞ (i) ani ≥ 0, a = {ani }ni =1 ∈ lpi, ψi , ||a ||pi, ψi > 0(i = 2, · · · , m), then (i) we have the following equivalent inequalities with the same best posm+1 sible constant factor (−1)m λ i=1 λ1i : m (j) ∞ ∞ ∞ j=1 anj f (xm+1 ) ··· dxm+1 (min1≤i≤m {ni , xm+1 })λ 0 nm =1 n1 =1 m+1 m 2 1 2 m ||f ||pm+1, ϕ ||a(1) ||p1, Ψ1 > (−1) λ ||a(i) ||pi, ψi , (6.168) λ i i=1 i=2 and

⎧ ⎨ ⎩

∞ 0

pλ

m+1 xm+1

−1

∞

···

nm =1

∞ n1 =1

> (−1)m λ

m+1 2 i=1

m

(j)

j=1 anj

⎫ p1 ⎬

p dxm+1

(min1≤i≤m ni , xm+1 )λ

m 2 1 (1) ||a ||p1, Ψ1 ||a(i) ||pi, ψi ; λi i=2

⎭ (6.169)

(ii) for, λm+1 = λ2 , we have the following equivalent inequalities with m the same best possible constant factor 2(−1)m i=1 λ1i : m (j) ∞ ∞ ∞ j=1 anj f (xm+1 ) ··· dxm+1 (min1≤i≤m {xm+1 ni , 1})λ 0 nm =1 n1 =1 m 2(−1)m 2 1 (i) > ||a ||pi, ψi ||f ||pm+1, ϕ ||a(1) ||p1, Ψ1 , (6.170) λ1 λi i=2

and

⎧ ⎨ ⎩

0

∞

pλ 2 −1

xm+1

∞

nm =1

>

···

∞

(j)

j=1 anj

p

⎫ p1 ⎬

dxm+1 ⎭ (min1≤i≤m {xm+1 ni , 1})λ m 2 1 ||a(i) ||pi, ψi ||a(1) ||p1, Ψ1 . (6.171) λi

n1 =1

2(−1)m λ1

m

i=2

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

319

Multiple Half-Discrete Hilbert-Type Inequalities

By Theorems 6.11 and 6.12, it follows that Corollary 6.24. Suppose that pi < 0, λi < min{0, 1 + λ}(i = 1, · · · , m), λm+1 < 0, 0 < pm+1 < 1. If f (xm+1 ) ≥ 0, f ∈ Lpm+1, Φ (R+ ), ||f ||pm+1, Φ > (i) (i) (i) 0, ani ≥ 0, a(i) = {ani }∞ ni =1 ∈ lpi, ψi , ||a ||pi, ψi > 0(i = 1, · · · , m), then (i) we have the following equivalent reverse inequalities with the same 1 best possible constant factor (−1)m λ m+1 i=1 λi : m (j) ∞ ∞ ∞ j=1 anj f (xm+1 ) ··· dxm+1 λ (min 1≤i≤m {ni , xm+1 }) 0 nm =1 n1 =1 m (−1)m λ 2 1 (i) > ||a ||pi, ψi ||f ||pm+1, Φ , (6.172) λm+1 λi i=1 and

∞

0

−1

pλ

m+1 xm+1 [1 − θm+1 (xm+1 )]p−1

∞ ∞ × ···

nm =1

(−1)m λ > λm+1

n1 =1

m j=1

p

(j)

anj

(min1≤i≤m {ni , xm+1 })λ

m 2 1 ||a(i) ||pi, ψi λ i i=1

;

p1 dxm+1 (6.173)

(ii) for, λm+1 = λ2 , f (xm+1 ) ≥ 0, f ∈ Lpm+1, Φ (R+ ), ||f ||pm+1, Φ > 0, we have the following equivalent inequalities with the same best possible 1 constant factor 2(−1)m m i=1 λi : m (j) ∞ ∞ ∞ j=1 anj dxm+1 f (xm+1 ) ··· (min1≤i≤m {xm+1 ni , 1})λ 0 nm =1 n1 =1 m 2 1 m (i) > 2(−1) ||a ||pi, ψi ||f ||pm+1, Φ , (6.174) λ i=1 i and⎧ ⎨ ⎩

0

∞

pλ

−1

2 xm+1

[1 − θm+1 (xm+1 )]p−1

∞ ∞ × ··· nm =1

> 2(−1)m

n1 =1

m

(j) j=1 anj

(min1≤i≤m {xm+1 ni , 1})λ

m 2 1 (i) ||a ||pi, ψi . λi

i=1

⎫ p1 ⎬

p dxm+1

⎭

(6.175)

November 7, 2013

320

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

ws-book9x6

Half-Discrete Hilbert-Type Inequalities

Remark 6.6. (i) If, in (6.24), we set kλ (x1 , · · · , xm , n) =

1 (min1≤i≤m {xi , n})λ

(λ < 0),

then, in view of Corollary 6.19 and (6.25), we have ||T || = k(λm+1 ) = (−1)m λ

m+1 2 i=1

if, in (6.26), for λm+1 =

λ , 2

1 ; λi

we set

kλ (nx1 , · · · , nxm , 1) =

1 (min1≤i≤m {nxi , 1})λ

(λ < 0),

then, in view of Corollary 6.19 and (6.27), we have ||T1 || = k(λm+1 ) = 2(−1)m

m 2 1 . λ i=1 i

(ii) If, in (6.61), we set kλ (n1 , · · · , nm , xm+1 ) =

1 (min1≤i≤m {ni , xm+1 })λ

(λ < 0),

then, in view of Corollary 6.22 and equality (6.62), we have ||T|| = k(λm+1 ) = (−1)m λ

m+1 2 i=1

if, in (6.63), for λm+1 =

λ , 2

1 ; λi

we set

kλ (xm+1 n1 , · · · , xm+1 nm , 1) =

1 (min1≤i≤m {xm+1 ni , 1})λ

(λ < 0),

then, in view of Corollary 6.22 and (6.64), we have ||T1 || = k(λm+1 ) = 2(−1)m

m 2 1 . λi

i=1

Remark 6.7. Putting m = 1 in the theorems and corollaries of this chapter, we obtain the corresponding results of Chapters 3 and 4.

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Bibliography

This bibliography is not by any means a complete one for the subject. Most of them consist of research papers and books to which reference is made in the text. Many other selected books and papers related to material of the subject have been included so that they may serve to stimulate new interest in future study and research.

[1] B´enyi A. and Oh Choonghong, Best constants for certain multilinear integral operator, Journal of Inequalities and Applications, Vol. 2006, Art. ID28582: 1–12. [2] Bonsall F. F., Inequalities with non-conjugate parameter, J. Math. Oxford Ser., 2:2 (1951) 135–150. [3] Brneti´c I. and Peˇcari´c J., Generalization of Hilbert’s integral inequality, Math. Ineq. & Appl., 7:2 (2004) 199–205. [4] Brneti´c I., Krni´c M. and Peˇcari´c J., Multiple Hilbert and Hardy-Hilbert inequalities with non-conjugate parameters, Bull. Austral. Math. Soc., 71 (2005) 447–457. [5] Carleman T., Sur les equations integrals singulieres a noyau reel et symetrique, Uppsala, 1923. [6] Chen Z. Q. and Xu J. S., New extensions of Hilbert’s inequality with multiple parameters, Acta Math. Hungar., 117:4 (2007) 383–400. [7] Chen Q. and Yang B. C., On a more accurate half-discrete Mulholland’s inequality and an extension, Journal of Inequalities and Applications, 70 (2012) doi:10.1186/1029-242X-2012-70. [8] Chen W., Jin C., Li C., and Lim J., Weighted Hardy-Littlewood-Sobolev Inequalities and Systems of Integral Equations, Discrete and Continuous Dynamical Systems, Supplement Volume: (2005) 164–172. [9] Cheng Q. X., Basic on real variable functions and functional analysis, Beijing: Higher Education Press, 2003.

321

ws-book9x6

November 7, 2013

322

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

[10] Debnath, L. and Yang, B.C., Recent developments of Hilbert-type discrete and integral inequalities with applications, Internat. Jour. Math. and Math. Sci., 2012 (2012) 1–30. [11] Draˇcic B. Ban, and Pog´ any T. K., Discrete Hilbert type inequality with non-homogeneous kernel, Appl. Anal. Discrete Math., 3:1 (2009) 88–96. [12] Draˇcic B. Ban, Peˇcari´c J., and Pog´ any T. K., On a discrete Hilbert type inequality with non-homogeneous kernel, Sarajevo J. Math., 6:1 (2010) 23–34. [13] Draˇcic B. Ban, Peˇcari´c J., Peric I., and Pog´ any T. K., Discrete multiple Hilbert type inequality with non-homogeneous kernel, J. Korean Math. Soc., 47:3 (2010) 537–546. [14] Gao M. Z., A note on Hilbert double series theorem, Hunan Mathematical Annal, 12:1-2 (1992) 143–147. [15] Gao M. Z., On the Hilbert inequality, J. Anal. Appl., 18:4 (1999) 1117– 1122. [16] Gao M. Z., A new Hardy-Hilbert’s type inequality for double series and its applications, The Australian Journal of Mathematical Analysis and Appl., 3:1 (2005) Art.13: 1–10. [17] Gao M. Z. and Hsu L. C., A survey of various reﬁnements and generalizations of Hilbert’s inequalities, J. Math. Res. Exp., 25:2 (2005) 227–243. [18] Gao M. Z. and Yang B. C., On the extended Hilbert’s inequality, Proc. Amer. Math. Soc., 126:3 (1998) 751–759. [19] Gao M. Z., Jia W. J. and Gao X. M., On an improvement of Hardy-Hilbert’s inequality, J. Math., 26:6 (2006) 647–651. [20] Hardy G. H., Note on a theorem of Hilbert concerning series of positive term, Proceedings of the London Mathematical Society, 23 (1925) 45–46. [21] Hardy G. H., Littlewood J. E., and P` olya G., Inequalities, Cambridge University Press, Cambridge, 1934. [22] He B., On a Hilbert-type integral inequality with a homogeneous kernel in R2 and its equivalent form, Journal of Inequalities and Applications, 2012, 94 (2012) doi:10.1186/1029-242X-2012-94. [23] He B. and Li Y. J., On several new inequalities close to Hilbert-Pachpatte’s inequality, J. Ineq. in Pure and Applied Math., 7:4 (2006) Art.154: 1–9. [24] He B. and Yang B. C., On a half-discrete inequality with a general homogeneous kernel, Journal of Inequalities and Applications, 2012, 30 (2012) doi:10.1186/1029-242X-2012-30. [25] He L. P., Gao M. Z. and Jia W. J., On a new strengthened Hardy-Hilbert’s inequality, J. Math. Res. Exp., 26:2 (2006) 276–282. [26] He L. P., Jia W. J. and Gao M. Z., A Hardy-Hilbert’s type inequality with gamma function and its applications, Integral Transforms and Special functions, 17:5 (2006) 355–363. [27] He B., Qian Y. and Li Y. J., On analogues of the Hilbert’s inequality, Comm. in Math. Anal., 4:2 (2008) 47–53. [28] He L. P., Yu J. M. and Gao M. Z., An extension of Hilbert’s integral inequality, Journal of Shaoguan University (Natural Science), 23:3 (2002) 25–30.

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Bibliography

ws-book9x6

323

[29] Hong Y., All-side generalization about Hardy-Hilbert integral inequalities, Acta Mathematica Sinica, 44:4 (2001) 619–626. [30] Hong Y., On Hardy-Hilbert integral inequalities with some parameters, J. Ineq. in Pure & Applied Math., 6:4 (2005) Art. 92: 1–10. [31] Hong Y., On multiple Hardy-Hilbert integral inequalities with some parameters, Journal of Inequalities and Applications, Vol. 2006, Art. ID 94960: 1–11. [32] Hsu L. C. and Wang Y. J., A reﬁnement of Hilbert’s double series theorem, J. Math. Res. Exp., 11:1 (1991) 143–144. [33] Hu K., A few important inequalities, Journal of Jianxi Teacher’s College (Natural Science), 3:1 (1979) 1–4. [34] Hu K., Some problems in analysis inequalities, Wuhan: Wuhan University Press, 2007. [35] Huang Q. L., On a Multiple Hilbert’s inequality with parameters, Journal of Inequalities and Applications, Volume 2010, Article ID 309319, 12 pages. [36] Huang Q. L. and Yang B. C., On a multiple Hilbert-type integral operator and applications, Journal of Inequalities and Applications, Volume 2009, Article ID 192197, 13 pages. [37] Huang Q., Yang B. and Debnath L., A multiple more accurate HardyLittlewood-P` olya Inequality, Le Matematiche (accepted 2012). [38] Huang Q. L. and Yang B. C., On a more accurate half-discrete Hilbert’s inequality, Journal of Inequalities and Applications, 106 (2012), doi:10.1186/1029-242X-2012-106. [39] Ingham A.E., A note on Hilbert’s inequality, J. London Math. Soc., 11 (1936) 237–240. [40] Jia W. J., Gao M. Z. and Debnath L., Some new improvement of the Hardy -Hilbert inequality with applications, International Journal of Pure and Applied Math., 11:1 (2004) 21–28. [41] Jia W. J., Gao M. Z. and Gao X. M., On an extension of the Hardy-Hilbert theorem, Studia Scientiarum Mathematicarum Hungarica, 42:1 (2005) 21– 35. [42] Khotyakov M., Two proofs of the sharp Hardy-Littlewood-Sobolev inequality, Bachelor Thesis, Mathematics Department, LMU Munich (2011). [43] Krni´c M. and Peˇcari´c J., General Hilbert’s and Hardy’s inequalities, Math. Ineq. & Appl., 8:1 (2005) 29–51. [44] Krni´c M., Tomovski Z., and Peˇcari´c J., Hilbert inequalities related to generalized hypergeometric functions, Balkanica N.S., 22 (2008) 307–322. [45] Krni´c M., Gao M. Z., Peˇcari´c J. and Gao X. M., On the best constant in Hilbert’s inequality, Math. Ineq. & Appl., 8:2 (2005) 317–329. [46] Kuang J. C., On new extension of Hilbert’s integral inequality, J. Math. Anal. Appl., 235 (1999) 608–614. [47] Kuang J. C., Applied inequalities, Jinan: Shandong Science Technic Press, 2004. [48] Kuang J. C., New progress in inequality study in China, Journal of Beijing Union University (Natural Science), 19:1 (2005) 29–37.

November 7, 2013

324

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

[49] Kuang J. C., Introduction to real analysis, Hunan Education Press, Chansha, China, 1996. [50] Laith E. A., On some extensions of Hardy-Hilbert’s inequality and applications, Journal of Inequalities and Applications, volume 2008, Article ID 546828, 14 pages. [51] Knopp K., Theory and application of infinite series, Londen: Blackie & Son Limited, 1928. [52] Levin V., Two remarks on Hilbert’s double series theorem, J. Indian Math. Soc., 11 (1937) 111–115. [53] Li Y. J. and He B., On inequalities of Hilbert’s type, Bull. Austral. Math. Soc., 76 (2007) 1–13. [54] Liu X. D. and Yang B. C., On a new Hilbert-Hardy-type integral operator and applications, Journal of Inequalities and Applications, Volume 2010, Article ID 812636, 10 pages. [55] Lu Z. X., Some new inverse type Hilbert-Pachpatte inequalities, Tamkang Journal of Mathematics, 34:2 (2003) 155–161. [56] Lu Z. X., On new generalizations of Hilbert’s inequalities, Tamkang Journal of Mathematics, 35:1 (2004) 77–86. [57] Mitrinovi´c J. E., Peˇcari´c J. E., and Fink A. M., Inequalities involving functions and their integrals and derivatives, Boston: Kluwer Acaremic Publishers, 1991. [58] Pachpatte B. G., On some new inequalities similar to Hilbert’s inequality, J. Math. Anal. Appl., 226 (1998) 166–179. [59] Pachpatte B. G., Mathematical inequalities, Elsevier B. V., Netherland, 2005. [60] Pog´ any T. K., Hilbert’s double series theorem extended to the case of nonhomogeneous kernels, J. Math. Anal. Appl., 342:2 (2008) 1485–1489. [61] Pog´ any T. K., New class of inequalities associated with the Hilbert’s double series theorem, Appl. Math. E-Notes, 10 (2010) 47–51. [62] Pang C. D. and Pang C. B., Basic on analysis number theory, Beijing: Science Press, 1990. [63] Pan Y. L, Wang H. T and Wang F. T., On complex functions, Science Press, Beijing, 2006. [64] Qu W. L., Combination mathematics, Beijing: Beijing University Press, 1989. [65] Salem S. R., Some new Hilbert type inequalities, Kyungpook Math. J., 46 (2006) 19–29. [66] Schur I., Bernerkungen sur Theorie der beschrankten Bilinearformen mit unendlich vielen veranderlichen, Journal of Math., 140 (1911) 1–28. [67] Stein E.M. and Weiss G., Fractional integrals in n-dimensional Euclidean space, J. Math. Mech., 7 (1958) 503–514. [68] Sulaiman W. T., On Hardy-Hilbert’s integral inequality, J. Ineq. in Pure & Appl. Math., 5:2 Art.25 (2004) 1–9. [69] Sulaiman W. T., New ideas on Hardy-Hilbert’s integral inequality (I), Pan American Math. J., 15:2 (2005) 95–100.

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Bibliography

ws-book9x6

325

[70] Sun B. J., Best generalization of a Hilbert type inequality, J. Ineq. in Pure & Applied Math., 7:3 Art.113 (2006) 1–7. [71] Titchmarsh E. C., The theory of the Riemann Zeta-function, Oxford: Clarendon Press, 1986. [72] Tailor A. E., Lay D. C., Introduction to functional analysis, New York: John Wiley & Sons, 1980. [73] Wang Z. Q. and Guo D. R., Introduction to Special Functions, Science Press, Beijing, 1979. [74] Wang W. H. and Xin D. M., On a new strengthened version of a HardyHilbert type inequality and applications, J. Ineq. in Pure & Applied Math., 7:5 Art.180 (2006) 1–7. [75] Wang W. H. and Yang B. C., A strengthened Hardy-Hilbert’s type inequality, The Australian Journal of Mathematical Analysis and Applications, 3:2 (2006) Art.17: 1–7. [76] Wang A. Z. and Yang B. C., A new Hilbert-type integral inequality in the whole plane with the non-homogeneous kernel, Journal of Inequalities and Applications, 2011, 2011:123, doi:10.1186/1029-242X-2011-123. [77] Weyl H., Singulare integral gleichungen mit besonderer berucksichtigung des fourierschen integral theorems, Inaugeral-Dissertation, Gottingen, 1908. [78] Wilhelm M., On the spectrum of Hilbert’s matrix, Amer J. Math., 72 (1950) 699–704. [79] Xi G. W., A reverse Hardy-Hilbert-type inequality, Journal of Inequalities and Appl., Vol. 2007, Art.ID79758: 1–7. [80] Xie H. and Lu Z., Discrete Hardy-Hilbert’s inequalities in Rn , Northeast. Math., 21:1 (2005) 87–94. [81] Xie Z. T., A new Hilbert-type inequality with the kernel of 3 -homogeneous, Journal of Jilin University (Science Edition), 45:3 (2007) 369–373. [82] Xie Z. T., A Hilbert-type integral inequality with non-homogeneous kernel and withe the integral in whole plane, Journal of Guangdong University of Education, 31:3 (2011) 8–12. [83] Xie Z. T., A new half-discrete Hilbert’s inequality wit the homogeneous kernel of degree -4μ, Journal of Zhanjiang Normal College, 32:6 (2011) 13–19. [84] Xie Z. T. and Zheng Z., A Hilbert-type inequality with parameters, J. Xiangtan Univ. (Natural Science), 29:3 (2007) 24–28. [85] Xie Z. T. and Zheng Z., A Hilbert-type integral inequality whose kernel is a homogeneous form of degree -3, J. Math. Anal. Appl., 339 (2007) 324–331. [86] Xie Z. T. and Zheng Z., A new Hilbert-type integral inequality and its reverse, Soochow Journal of Math., 33:4 (2007) 751–759. [87] Xin D. M., Best generalization of Hardy-Hilbert’s inequality with multiparameters, J. Ineq. in Pure and Applied Math., 7:4 Art.153 (2006) 1–8. [88] Xin D. M. and Yang B. C., A basic Hilbert-type inequality, Journal of Mathematics, 30:3 (2010) 554–560. [89] Xin D. M. and Yang B. C., A Hilbert-type integral inequality in the whole plane with the homogeneous kernel of degree -2, Journal of Inequalities and Applications, Volume 2010, Article ID 401428, 11 pages.

November 7, 2013

326

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

[90] Xu J. S., Hardy-Hilbert’s inequalities with two parameters, Advances in Mathematics, 36:2 (2007) 189–198. [91] Xu L. Z. and Guo Y. K., Note on Hardy-Riesz’s extension of Hilbert’s inequality, Chin. Quart. J. Math., 6:1 (1991) 75–77. [92] Xu L. Z, Wang X. H., Methods on mathematical analysis and examples, Beijing: Higher Education Press, 1985. [93] Xie Z. T., A generalization of Stirling formula, Mathematical Practice and Cognition, 36:6 (2006) 331–333. [94] Xie Z. T. and Zeng Z., A new half-discrete Hilbert’s inequality with the homogeneous kernel of degree -4μ, Journal of Zhanjiang Normal College, 32:6 (2011) 13–19. [95] Yang B. C., A note on Hilbert’s integral inequalities, Chinese Quarterly J. Math., 13:4 (1998) 83–86. [96] Yang B. C., On Hilbert’s integral inequality, J. Math. Anal. Appl., 220 (1998) 778–785. [97] Yang B. C., On a strengthened version of the more accurate Hardy-Hilbert’s inequality, Acta Mathematica Sinica, 42:6 (1999) 1103–1110. [98] Yang B. C., A general Hardy-Hilbert’s integral inequality with a best value, Chinese Annals of Mathematics, 21A:4 (2000) 401–408. [99] Yang B. C., On a generalization of Hilbert’s double series theorem, Journal of Nanjing University-Mathematical Biquarterly, 18:1 (2001) 145–151. [100] Yang B. C., On a general Hardy-Hilbert’s inequality, Chinese Annals of Mathematics, 23A:2 (2002) 247–254. [101] Yang B. C., On a multiple Hardy-Hilbert’s integral inequality, Chinese Annals of Mathematics, 24A:6 (2003) 743–750. [102] Yang B. C., On a new inequality similar to Hardy-Hilbert’s inequality, Math. Ineq. Appl., 6:1 (2003) 37–44. [103] Yang B. C., On a new Hardy-Hilbert’s type inequality, Math. Ineq. Appl., 7:3 (2004) 355–363. [104] Yang B. C., On new extensions of Hilbert’s inequality, Acta Math. Hungar., 104:4 (2004) 291–299. [105] Yang B. C., On an extension of Hilbert’s integral inequality with some parameters, The Australian Journal of Math. Analysis and Applications, 1:1 Art.11 (2004) 1–8. [106] Yang B. C., A new Hilbert-type integral inequality and its generalization, Journal of Jilin University (Science Edition), 43:5 (2005) 580–584. [107] Yang B. C., A mixed Hilbert-type inequality with a best constant factor, International Journal of Pure and Applied Mathematics, 20:3 (2005) 319– 328. [108] Yang B. C., On best extensions of Hardy-Hilbert’s inequality with two parameters, J. Ineq. in Pure & Applied Math., 6:3 (2005) Art. 81: 1–15. [109] Yang B. C., A reverse of the Hardy-Hilbert’s type inequality, Journal of Southwest China Normal University (Natural Science), 30:6 (2005) 1012– 1015. [110] Yang B. C., On the way of weight function and research for Hilbert’s type

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Bibliography

[111] [112] [113] [114] [115] [116] [117] [118] [119]

[120] [121] [122]

[123] [124] [125]

[126]

[127]

[128] [129]

ws-book9x6

327

integral inequalities, Journal of Guangdong Education Institute (Natural Science), 25:3 (2005) 1–6. Yang B. C., On a more accurate Hardy-Hilbert’s type inequality and its applications, Acta Mathematica Sinica, 49:3 (2006) 363–368. Yang B. C., On a relation to Hardy-Hilbert’s inequality and Mulholland’s inequality, Acta Mathematica Sinica, 49:3 (2006) 559–566. Yang B. C., On the norm of an integral operator and applications, J. Math. Anal. Appl., 321 (2006) 182–192. Yang B. C., A bilinear inequality with the kernel of -2-order homogeneous, Journal of Xiamen University (Natural Science), 45:6 (2006) 752–755. Yang B. C., A dual Hardy-Hilbert’s inequality and generalizations, Advances in Math., 35:1 (2006) 102–108. Yang B. C., On the norm of a self-adjoint operator and applications to Hilbert’s type inequalities, Bull. Belg. Math. Soc., 13 (2006) 577–584. Yang B. C., A new Hilbert-type inequality, Bull. Belg. Math. Soc., 13 (2006) 479–487. Yang B. C., A new Hilbert-type inequality, Journal of Shanghai Univ. (Natural Science), 13:3 (2007) 274–278. Yang B. C., On the norm of a self-adjoint operator and a new bilinear integral inequality, Acta Mathematica Sinica, English Series, 23:7 (2007) 1311–1316. Yang B. C., On the norm of a Hilbert’s type linear operator and applications, J. Math. Anal. Appl., 325 (2007) 529–541. Yang B. C., A Hilbert-type inequality with two pairs of conjugate exponents, Journal of Jilin University (Science Edition), 45:4 (2007) 524–528. Yang B. C., On a Hilbert-type operator with a symmetric homogeneous kernel of -1-order and applications, Journal of Inequalities and Applications, Volume 2007, Article ID 47812, 9 pages. Yang B. C., A Hilbert-type integral inequality with the kernel of -3-order homogeneous, Journal of Yunnam University, 30:4 (2008) 325–330. Yang B. C., On the norm of a linear operator and its applications, Indian Journal of Pure and Applied Mathematics, 39:3 (2008) 237–250. Yang B. C., On the norm of a certain self-adjoint integral operator and applications to bilinear integral inequalities, Taiwan Journal of Mathematics, 12:2 (2008) 315–324. Yang B. C., A basic Hilbert-type integral inequality with the homogeneous kernel of -1-degree and extensions, Journal of Guangdong Education Institute, 28:3 (2008) 1–10. Yang B. C., On a Hilbert-type operator with a class of homogeneous kernels, Journal of Inequalities and Applications, Volume 2009, Article ID 572176, 9 pages. Yang B. C., Hilbert-type integral inequalities, Bentham Science Publishers Ltd., 2009. Yang B. C., A survey of the study of Hilbert-type inequalities with parameters, Advances in Mathematics, 38:3 (2009) 257–268.

November 7, 2013

328

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

[130] Yang B. C., On the norm of operator and Hilbert-type inequalities, Beijing: Science Press, 2009. [131] Yang B. C., On an Application of Hilbert’s inequality with multiparameters, Journal of Beijing Union University (Natural Sciences), 24:4 (1010) 78–84. [132] Yang B. C., An Application of the reverse Hilbert’s inequality, Journal of Xinxiang University (Natural Sciences), 27:4 (1010) 2–5. [133] Yang B. C., A new Hilbert-type operator and applications, Publ. Math. Debrecen, 76:1-2 (2010) 147–156. [134] Yang B. C., Discrete Hilbert-type inequalities, Bentham Science Publishers Ltd., 2011. [135] Yang B. C., A half-discrete Hilbert-type inequality, Journal of Guangdong University of Education, 31:3 (2011) 1–7. [136] Yang B. C., A Hilbert-type integral inequality with the non-homogeneous kernel on the plane, Journal of Guangdong University of Education, 31:5 (2011) 5–10. [137] Yang B. C., A new half-discrete Mulholland-type inequality with parameters, Ann. Funct. Anal., 3:1 (2012) 142–150. [138] Yang B. C. and Chen Q., A half-discrete Hilbert-type inequality with a homogeneous kernel and an extension, Journal of Inequalities and Applications, 2011, 124 (2011), doi:10.1186/1029-242X-2011-124. [139] Yang B. C. and Chen Q., A half-discrete Hilbert-type inequality with the non-monotone kernel, Journal of Jilin University (Science Edition), 50:2 (2012) 167–172. [140] Yang B. C. and Debnath L., On new strengthened Hardy-Hilbert’s inequality, Internat. J. Math. & Math. Soc., 21:2 (1998) 403–408. [141] Yang B. C. and Debnath L., On a new generalization of Hardy-Hilbert’s inequality, J. Math. Anal. Appl., 233 (1999) 484–497. [142] Yang B. C. and Debnath L., On the extended Hardy-Hilbert’s inequality, J. Math. Anal. Appl., 72:2 (2002) 187–199. [143] Yang B. C. and Debnath L., A strengthened Hardy-Hilbert’s inequality, Proceedings of the Jangjeon Mathematical Society, 6:2 (2003) 119–124. [144] Yang B. C. and Gao M. Z., On a best value of Hardy-Hilbert’s inequality, Advances in Math., 26:2 (1997) 159–164. [145] Yang B. C. and Krni´c M., Hilbert-type inequalities and related operators with homogeneous kernel of degree 0, Mathematical Inequalities & Applications, 13:4 (2010) 817–839. [146] Yang B. C. and Liang H. W., A new Hilbert-type integral inequality with a parameter, Journal of Henan University (Natural Science), 35:4 (2005) 4–8. [147] Yang B. C. and Rassias T. M., On the way of weight coeﬃcient and research for Hilbert-type inequalities, Math. Ineq. Appl., 6:4 (2003) 625–658. [148] Yang B. C. and Rassias Th. M., On a Hilbert-type integral inequality in the subinterval and its operator expression, Banach J. Math. Anal., 4:2 (2010) 100–110. [149] Yang B. C., Brnet´c I., Krni´c M. and Peˇcari´c J., Generalization of Hilbert

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Bibliography

[150]

[151]

[152]

[153]

[154]

[155] [156] [157]

[158] [159] [160]

[161]

[162]

[163] [164] [165]

[166]

ws-book9x6

329

and Hardy-Hilbert integral inequalities, Math. Ineq. & Appl., 8:2 (2005) 259–272. Yang B. C., A new formula for evaluating the sum of d-th powers of the ﬁrst n terms of an arithmetic sequence, Journal of South China Normal University, 1 (1996) 129–137. Yang B. C., The formula about the sum of powers of natural numbers relating Bernoulli numbers, Mathematical Practice and Cognition, 4 (1994) 52–56. Yang B. C. and Zhu Y. H., Inequalities on the Hurwitz Zeta-function restricted to the axis of positive reals, Acta Scientiarum Naturalium Universitis Sunyatseni, 36:3 (1997) 30–35. Yang B. C., The evaluating formulas on the convergence p-series relating Bernoulli numbers, Journal of Guangdong Education Institute, 3 (1992) 19–27. Yang B. C. and Li D. C., Estimation of the sum for -1-th powers of the ﬁrst n terms of an arithmetic sequence, Natural Science Journal of Hainan Teachers College, 10:1 (1997) 19–24. Yang B. C. and Wang G. Q., Some inequalities on harmonic series. Journal of Mathematics Study, 29:3 (1996) 90–97. Yang B. C., Some new inequalities on step multiply, Journal of Guangdong Education Institute, 22:2 (2002) 1–4. Yang B. C. and Krni´c M., A half-discrete Hilbert-type inequality with a general homogeneous kernel of degree o, Journal of Mathematical Inequalities, 6:3 (2012) 401–417. Yang B. C., On a more accurate half-discrete Hilbert’s inequality, Journal of Beijing Union University (Natural Sciences), 26:2 (2012) 63–68. Yang B. C., On a half-discrete Hilbert-type inequality, Journal of Shantou University (Natural Sciences), 26:4 (2011) 5–10. Yang B. C., A more accurate half-discrete reverse Hilbert-type inequality, Journal of Hunan Institute of Science and Technology (Natural Sciences), 24:4 (2011) 1–6. Yang B. C., On two classes of more accurate half-discrete reverse Hilbert’s inequalities, Journal of Xinxiang University (Natural Science Edition), 29:1 (2012) 9–14. Yang B. C., A half-discrete reverse Hilbert-type inequality with a homogeneous kernel of positive degree, Journal of Zhanjiang Normal College, 32:3 (2011) 5–9. Yang B. C., On a half-discrete Mulholland’s inequality and its extension, Thai Journal of Mathematics, (2012) (to appear). Yang B. C., A half-discrete Hilbert’s inequality with a non-homogeneous kernel, Journal of Zhanjiang Normal College, 32:6 (2011) 5–11. Yang B. C., A half-discrete Hilbert-type inequality with the nonhomogeneous kernel, Journal of Xinxiang University: Natural Science Edition, 28:5 (2011) 385–387. Yang B. C., On a half-discrete reverse Hilbert-type inequality with a non-

November 7, 2013

330

[167]

[168]

[169] [170]

[171]

[172] [173] [174] [175]

[176]

[177]

[178]

[179]

[180]

[181]

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

homogeneous kernel, Journal of Inner Mongolia Normal University (Natural Science Edition), 40:5 (2011) 433–337. Yang B. C., A half-discrete reverse Hilbert-type inequality with a nonhomogeneous kernel, Journal of Hunan Institute of Science and Technology (Natural Science), 24:3 (2011) 1–4. Yang B. C., A more accurate half-discrete reverse Hilbert’s inequality with the non-homogeneous kernel, Journal of Guangdong University of Education, 32:3 (2012) 1–7. Yang B. C., On a best extension of a half-discrete Hilbert-type inequalities, Jordan Journal of Mathematics and Statistics, 5:4 (2012) 267–281. Yang B. C., A half-discrete Hilbert-type inequality with a non-homogeneous kernel and two variables, Mediterranean Journal of Methematics, 2012, doi: 10.1007/s00009-012-0213-50 online ﬁrst. Zeng Z. and Xie Z. T., On a new Hilbert-type integral inequality with the integral in whole plane, Journal of Inequalities and Applications, Volume 2010, Article ID 256796, 8 pages. Zhang K. W., A bilinear inequality, J. Math. Anal. Appl., 271 (2002) 288– 296. Zhao C. J. and Debnath L., Some new type Hilbert integral inequalities, J. Math. Anal. Appl., 262 (2001) 411–418. Zhong W. Y., A mixed Hilbert-type inequality and its equivalent forms, Journal of Guangdong University of Education, 31:5 (2011) 18–22. Zhong J. H. and Yang B. C., On an extension of a more accurate Hilberttype inequality, Journal of Zhejiang University (Science Edition), 35:2 (2008) 121–124. Zhong W. I. and Yang B. C., A best extension of Hilbert inequality involving several parameters, Journal of Jinan University (Natural Science), 28:1 (2007) 20–23. Zhong W. I. and Yang B. C., A reverse Hilbert’s type integral inequality with some parameters and the equivalent forms, Pure and Applied Mathematics, 24:2 (2008) 401–407. Zhong W. Y. and Yang B. C., On Multiple’s Hardy-Hilbert integral inequality with kernel, Journal of Inequalities and Applications, Vol. 2007, Art.ID 27962, 17 pages, doi:10.1155/2007/27. Zhu Y. H, Yang B. C., Accurate inequalities of partial sums on a type of divergent series, Acta Scientiarum Naturalium Universitis Sunyatseni, 37:4 (1998) 33–37. Zhu Y. H., Yang B. C., Improvement on Euler’s summation formula and some inequalities on sums of powers, Acta Scientiarum Naturalium Universitis Sunyatseni, 36:4 (1997) 21–26. Zhong W. Y., A mixed Hilbert-type inequality and its equivalent forms, Journal of Guangdong University of Education, 31:5 (2011) 18–22.

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Index

equivalent forms, 169 equivalent inequalities, 10, 23, 70, 77, 79, 80, 83, 90–96, 122, 134, 140, 143, 146, 153, 154, 184, 191, 192, 194–196, 211, 218–222, 249, 251, 279, 284–287, 289, 290, 293, 296, 298, 299, 301, 304, 305, 308, 309, 313, 316, 318 equivalent integral inequalities, 9 equivalent reverse inequalities, 280, 286, 290, 300, 306 equivalent reverses, 83, 91, 92, 122, 140 Euler, 29 Euler constant, 15, 27, 54 Euler-Maclaurin summation formula, 6, 29, 34, 60, 126

Bernoulli’s functions, 27, 32, 58, 124 Bernoulli’s numbers, 27, 29, 30 Bernoulli’s polynomials, 31, 33 best constant factor, 17, 91, 96, 100, 101, 115, 117, 122, 157, 160, 163, 167, 195 best possible constant factor, 137, 158, 169, 188, 189, 193, 205, 215, 218, 220, 222, 223, 258, 277, 279, 280, 284–287, 289, 290, 293, 296, 298–301, 304–306, 308, 309, 313, 316, 318 beta function, 16 bounded linear operator, 197, 198, 224, 225, 252 Carleman, 5 Cauchy’s inequality, 14 conjugate exponents, 6, 8, 18, 19 constant factor, 189, 193, 216, 220, 232 convergent series estimation, 50

Fatou lemma, 69, 133 Gao, 13 general homogeneous kernel, 20

Debnath, 13, 15, 16 decreasing functions, 106 decreasing property, 171 diﬀerentiable function, 91–93, 123, 140, 143, 153, 170, 192, 219 discrete Hilbert’s inequality, 5 discrete Hilbert-type operator, 23 divergence series, 52 Draˇcic, 25

H-L-P integral inequality, 21 half-discrete Hilbert-type inequalities, 25 half-discrete Hilbert-type inequality, 26, 57, 95, 123 half-discrete Hilbert-type operator, 73, 137, 197, 223, 225 Hardy, 2, 3, 6, 57, 123, 169, 233 Hardy-Hilbert inequality, 6 331

ws-book9x6

November 7, 2013

12:24

332

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

Half-Discrete Hilbert-Type Inequalities

Hardy-Hilbert’s integral inequality, 7 Hardy-Hilbert’s integral operator, 8 Hardy-Hilbert’s operator, 8 Hardy-Littlewood-P` olya’s inequalities, 10 Hardy-Littlewood-Sobolev inequality, 3 Hermite-Hadamard’s inequality, 59, 125, 174 Hilbert, 2, 57, 233 Hilbert space, 22 Hilbert’s double series inequality, 2 Hilbert’s inequality, 2, 4–6, 14 Hilbert’s integral inequality, 2 Hilbert’s integral operator, 4 Hilbert’s operator, 2, 4 Hilbert-type inequalities, 12, 18, 20, 22, 23, 57 Hilbert-type inequality, 12, 25 Hilbert-type integral inequalities, 26 Hilbert-type integral inequality, 23, 24 H¨ older’s inequality, 61, 71, 126, 127, 135, 176–178, 185, 212–214 homogeneous function, 9, 12, 179, 182, 184, 190, 192 homogeneous functions, 234 homogeneous kernel, 23, 57, 95, 178, 197, 198, 252, 275 Hu, 3 Hurwitz ζ-function, 54 Hurwitz’s zeta function, 29 imperfect conditions, 40 inner product, 4 Issai Schur, 2 kernel, 9, 138 Krnic, 23 Kuang, 3, 13, 125–127, 135 Lebesgue control convergence theorem, 69, 133 Lebesgue term by term integration theorem, 61, 127, 176, 177 Littlewood, 2

Lobatchevsky, 123 measurable function, 22, 126, 132, 143, 170, 172, 175, 205, 206, 209, 210, 219, 221, 222, 234, 245 measurable homogeneous function, 79, 80, 91, 93, 96 Mitrinovi´c, 3, 8 multiple half-discrete Hilbert-type inequalities, 233 multiple Hilbert-type integral inequality, 12 multiple integral inequality, 19 non-homogeneous kernel, 223–225, 252 norm, 22, 225 norm linear spaces, 183 normed linear spaces, 210 operator, 84, 138, 147, 199–204, 226, 228–232 operator expressions, 70, 169, 248 operator form, 4, 5 P` olya, 2, 169 Pachpatte, 13 piecewise smooth continuous function, 46 Pogany, 25 Rassias, 18 recursion formulas, 51 reverse equivalent inequalities, 184, 191, 193, 221, 223 reverse H¨ older’s inequality, 62, 77, 81, 128, 141, 144, 186, 187 reverse inequalities, 76, 253, 276 reverses, 169, 175 Riemann ζ-function, 54 Riemann zeta function, 29 smooth continuous function, 34, 44 Stieltjes constant, 54 Stirling formula, 29, 55

ws-book9x6

November 7, 2013

12:24

BC: 8799 - Half-Discrete Hilbert-Type Inequalities

333

Index

weight coeﬃcient, 18 weight functions, 58, 123, 124, 172, 178, 181, 208, 234, 236, 259 Weyl, 2

ws-book9x6

Yang, 15, 16, 18, 84, 110, 115 Zhang, 14 Zygmund, 2

Inequalities

Read more

Geometric Inequalities

Read more

Mathematical inequalities

Read more

Mathematical Inequalities

Read more

Geometric inequalities

Read more

Isoperimetric inequalities

Read more

New inequalities

Read more

Olympiad inequalities

Read more

Differential inequalities

Read more

Analytic inequalities

Read more

Matrix inequalities

Read more

Algebraic inequalities

Read more

Analytic inequalities

Read more

Probability Inequalities

Read more

Operator inequalities

Read more

Advanced inequalities

Read more

Olympiad Inequalities

Read more

Algebraic inequalities

Read more

Geometric inequalities

Read more

Probabilistic inequalities

Read more

Secrets in Inequalities, Vol. 1 - Basic Inequalities

Read more

Inequalities through problems

Read more

Introduction to Inequalities

Read more

Understanding Health Inequalities

Read more

Inequalities and Applications

Read more

Advances on Fractional Inequalities

Read more

Introduction to inequalities

Read more

Polynomials and Polynomial Inequalities

Read more

Polynomials And Polynomial Inequalities

Read more

A Dictionary of Inequalities

Read more

Recommend Documents

Inequalities

INEQUALITIES By G. H. HARDY J. E. LITTLEWOOD G. POLYA CAMBRIDGE AT THE UNIVERSITY PRESS 1934 INEQ{JALITIES LONDO...

Geometric Inequalities

Mathematical inequalities

This page intentionally left blank Mathematical Inequalities North-Holland Mathematical Library Board of Honorary E...

Mathematical Inequalities

This page intentionally left blank Mathematical Inequalities North-Holland Mathematical Library Board of Honorary E...

Geometric inequalities

Isoperimetric inequalities

New inequalities

Olympiad inequalities

Olympiad Inequalities Thomas J. Mildorf January 20, 2006 It is the purpose of this document to familiarize the reader wi...

Differential inequalities

Analytic inequalities

ATLANTIS S TUDIES IN M ATHEMATICS VOLUME 3 S ERIES E DITOR : J. VAN M ILL Atlantis Studies in Mathematics Series Edito...