Constructive Fractional Analysis with Applications (Studies in Systems, Decision and Control, 362) 3030714802, 9783030714802

Table of contents :
Preface
Contents
1 Variable Order General Fractional Integral Inequalities
1.1 Background
1.2 Main Results
References
2 Variable Order Fractional Integral Inequalities for Spherical Shell
2.1 Background
2.2 Main Results
References
3 Left Fractional Integral Inequalities of E.R. Love Type
3.1 Introduction
3.2 Main Results
References
4 Right Side Fractional Integral Inequalities of E.R. Love Type
4.1 Introduction
4.2 Main Results
4.2.1 Part I
4.2.2 Part II
References
5 General Fractional Landau Inequalities
5.1 Introduction
5.2 Main Results
References
6 Abstract Fractional Landau Inequalities
6.1 Introduction
6.2 Main Results
References
7 Fractional Landau Inequalities of Riemann–Liouville Type
7.1 Introduction
7.2 Main Results
7.3 Appendix
References
8 Generalized Canavati Fractional Landau Inequalities
8.1 Introduction
8.2 Main Results
References
9 Sequential Left Abstract Fractional Landau Inequalities
9.1 Introduction
9.2 Main Results
References
10 Iterated Left Abstract Generalized Fractional Landau Inequalities
10.1 Introduction
10.2 Main Results
References
11 Sequential General Right Side Fractional Landau Inequalities
11.1 Introduction
11.2 Main Results
References
12 Iterated Generalized Right Side Fractional Landau Inequalities
12.1 Introduction
12.2 Main Results
References
13 High Order Generalized Landau Inequalities
13.1 Introduction
13.2 Main Results
13.3 Appendix
References
14 Multidimensional Caputo Left Side Fractional Landau Inequalities
14.1 Introduction
14.2 Main Results
References
15 Multidimensional Left Canavati Fractional Landau Inequalities
15.1 Introduction
15.2 Main Results
References
16 Multidimensional Right Caputo Fractional Taylor Formula and Landau Inequalities
16.1 Introduction
16.2 Background
16.3 Main Results
References
17 Multidimensional Generalized Right Fractional Taylor Formula and Landau Inequalities
17.1 Introduction
17.2 Background
17.3 Main Results
References
18 Landau's Inequality for Semigroups
18.1 Introduction
18.2 Background
18.3 Main Results
18.4 Application
References
19 Fractional Variable Order Gronwall Inequality
19.1 Introduction
19.2 Main Results
References
20 A Study of Gronwall Inequalities of Fractional Variable Order
20.1 Introduction
20.2 Main Results
References
21 General Ordinary and Fractional Approximation with Positive Sublinear Operators
21.1 Background—I
21.2 Background—II ch2121.4
21.3 Background—III
21.4 Background—IV
21.5 Main Results
21.5.1 Ordinary Approximation
21.5.2 Fractional Approximation
21.5.3 Iterated Fractional Approximation
References
22 Complete Approximations with Multivariate Generalized Picard Singular Integrals
22.1 Introduction
22.2 Auxiliary Essential Results
22.3 Main Results for Pr,n[ m]
22.3.1 Uniform Approximation
22.3.2 Lp Approximation for Pr,n[ m]
22.3.3 Global Smoothness Preservation and Simultaneous Approximation of Pr,n[ m]
22.3.4 Voronovskaya Asymptotic Expansions for Pr,n[ m]
22.3.5 Simultaneous Approximation by Multivariate Complex Pr,n [ m]
References
23 High Order Approximation with Multivariate Generalized Gauss–Weierstrass Singular Integrals
23.1 Introduction
23.2 Auxiliary Essential Results
23.3 Main Results for Wr,n[ m]
23.3.1 Uniform Approximation
23.3.2 Lp Approximation for Wr,n[ m]
23.3.3 Global Smoothness Preservation and Simultaneous Approximation of Wr,n[ m]
23.3.4 Voronovskaya Asymptotic Expansions for Wr,n[ m]
23.3.5 Simultaneous Approximation by Multivariate Complex Wr,n [ m]
References
24 Complete Approximations with Multivariate Generalized Poisson–Cauchy Type Singular Integral Operators
24.1 Introduction
24.2 Auxiliary Essential Results
24.3 Main Results for Ur,n[ m]
24.3.1 Uniform Approximation
24.3.2 Lp Approximation for Ur,n[ m]
24.3.3 Global Smoothness Preservation and Simultaneous Approximation of Ur,n[ m]
24.3.4 Voronovskaya Asymptotic Expansions for Ur,n[ m]
24.3.5 Simultaneous Approximation by Multivariate Complex Ur,n [ m]
References
25 High Order Approximation with Multivariate Generalized Trigonometric Type Singular Integral Operators
25.1 Introduction
25.2 Auxiliary Essential Results
25.3 Main Results for Tr,n[ m]
25.3.1 Uniform Approximation
25.3.2 Lp Approximation for Tr,n[ m]
25.3.3 Global Smoothness Preservation and Simultaneous Approximation Of Tr,n[ m]
25.3.4 Voronovskaya Asymptotic Expansions for Tr,n[ m]
25.3.5 Simultaneous Approximation by Multivariate Complex Tr,n [ m]
References
26 Concluding Remarks

Citation preview

Studies in Systems, Decision and Control 362

George A. Anastassiou

Constructive Fractional Analysis with Applications

Studies in Systems, Decision and Control Volume 362

Series Editor Janusz Kacprzyk, Systems Research Institute, Polish Academy of Sciences, Warsaw, Poland

The series “Studies in Systems, Decision and Control” (SSDC) covers both new developments and advances, as well as the state of the art, in the various areas of broadly perceived systems, decision making and control–quickly, up to date and with a high quality. The intent is to cover the theory, applications, and perspectives on the state of the art and future developments relevant to systems, decision making, control, complex processes and related areas, as embedded in the fields of engineering, computer science, physics, economics, social and life sciences, as well as the paradigms and methodologies behind them. The series contains monographs, textbooks, lecture notes and edited volumes in systems, decision making and control spanning the areas of Cyber-Physical Systems, Autonomous Systems, Sensor Networks, Control Systems, Energy Systems, Automotive Systems, Biological Systems, Vehicular Networking and Connected Vehicles, Aerospace Systems, Automation, Manufacturing, Smart Grids, Nonlinear Systems, Power Systems, Robotics, Social Systems, Economic Systems and other. Of particular value to both the contributors and the readership are the short publication timeframe and the world-wide distribution and exposure which enable both a wide and rapid dissemination of research output. Indexed by SCOPUS, DBLP, WTI Frankfurt eG, zbMATH, SCImago. All books published in the series are submitted for consideration in Web of Science.

More information about this series at http://www.springer.com/series/13304

George A. Anastassiou

Constructive Fractional Analysis with Applications

George A. Anastassiou Department of Mathematical Sciences University of Memphis Memphis, TN, USA

ISSN 2198-4182 ISSN 2198-4190 (electronic) Studies in Systems, Decision and Control ISBN 978-3-030-71480-2 ISBN 978-3-030-71481-9 (eBook) https://doi.org/10.1007/978-3-030-71481-9 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Dedicated to my family, and to all front line medical workers fighting Covid—19

Preface

Constructive and Computational Fractional Analysis recently are more and more in the center of mathematics and of other related sciences either by themselves because of their rapid development, which is based on very old foundations, or because they cover a great variety of applications in the real world. In this monograph all presented is original work by the author given at a very general level to cover a maximum number of cases in various applications. We apply generalized fractional differentiation techniques of Riemann-Liouville, Caputo and Canavati types and of fractional variable order to various kinds of inequalities such as of Opial, Hardy, Hilbert-Pachpatte and on the spherical shell. We continue with E. R. Love left and right sides fractional integral inequalities. The fractional Landau inequalities, of left and right sides, univariate and multivariate, including ones for Semigroups occupy fourteen chapters of this monograph. They are developed to all possible directions and right side multivariate fractional Taylor formulae are proven for the purpose. We continue with several Gronwall fractional inequalities of variable order. The last part of book belongs to constructive approximation theory, we present: ordinary and fractional approximations by positive sublinear operators, and high order approximation by multivariate generalized Picard, Gauss-Weierstrass, Poisson-Cauchy and Trigonometric singular integrals which are not positive operators. This monograph is the natural and expected evolution of recent author’s research work put in a book form for the first time. The presented approaches are original, and chapters are self-contained and can be read independently. This monograph is suitable to be used in related graduate classes and research projects. We exhibit to the maximum our constructive, computational, estimating and approximation methods to all possible directions. The motivation to write this monograph came by the following: various issues related to the modeling and analysis of ordinary and fractional order systems have gained an increased popularity, as witnessed by many books and volumes in Springer’s program: http://www.springer.com/gp/search?query=fractional&submit= Prze%C5%9Blij and the purpose of our book is to capture at a very general level a deeper formal analysis on some issues that are relevant to many areas for instance: decision-making, complex processes, systems modeling and control and related areas. vii

viii

Preface

The above are deeply embedded in the fields of mathematics, engineering, computer science, physics, economics, social and life sciences. The complete list of presented topics follows: • • • • • • • • • • • • • • • • • • • • • • • • •

Abstract fractional integral inequalities of variable order Fractional integral inequalities of variable order on spherical shell E.R. Love type left fractional integral Inequalities E.R. Love type right side fractional integral Inequalities Abstract fractional Landau inequalities Abstract generalized fractional Landau inequalities over R Riemann-Liouville fractional Landau type inequalities Generalized Canavati fractional Landau type inequalities Iterated left abstract fractional Landau inequalities Sequential left abstract generalized fractional Landau inequalities Iterated abstract right side fractional Landau inequalities Sequential abstract generalized right side fractional Landau inequalities High order abstract generalized Landau Inequalities Multivariate Caputo left fractional Landau inequalities Multivariate left side Canavati fractional Landau inequalities Multivariate right side Caputo fractional Taylor formula and Landau inequalities Multivariate generalized right side fractional Taylor formula and Landau inequalities Landau inequality for semigroups revisited Gronwall Inequality of fractional variable order A variety of Gronwall inequalities of fractional variable order Advanced ordinary and fractional approximation by positive sublinear operators High order approximation by multivariate generalized Picard singular integrals Complete approximations by multivariate generalized Gauss-Weierstrass singular integrals Approximation by multivariate generalized Poisson-Cauchy type singular integral operators Approximations by multivariate generalized trigonometric type singular integral operators.

The book’s results are expected to find applications in many areas of pure and applied mathematics, especially in approximation theory, inequalities and differential equations as they connect to fractionality. Thus this monograph is suitable for researchers, graduate students and seminars of the above disciplines, also to be in all science and engineering libraries. The preparation of the book took place during 2020–2021 at the University of Memphis, at stay home of author during the Covid-19 outbreak keeping him alive, happy and in control!

Preface

ix

The author likes to thank Prof. Alina Lupas of University of Oradea, Romania, for checking and reading the manuscript. Memphis, TN, USA February 2021

George A. Anastassiou

Contents

1

Variable Order General Fractional Integral Inequalities . . . . . . . . . . 1.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 3 14

2

Variable Order Fractional Integral Inequalities for Spherical Shell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

15 15 20 32

3

Left Fractional Integral Inequalities of E.R. Love Type . . . . . . . . . . . 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33 33 34 47

4

Right Side Fractional Integral Inequalities of E.R. Love Type . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Part I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Part II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49 49 50 50 57 60

5

General Fractional Landau Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

61 61 63 76

6

Abstract Fractional Landau Inequalities . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

77 77 81 93

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7

Fractional Landau Inequalities of Riemann–Liouville Type . . . . . . . 95 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 7.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 7.3 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

8

Generalized Canavati Fractional Landau Inequalities . . . . . . . . . . . . 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

119 119 122 131

9

Sequential Left Abstract Fractional Landau Inequalities . . . . . . . . . . 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

133 133 135 153

10 Iterated Left Abstract Generalized Fractional Landau Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

155 155 158 175

11 Sequential General Right Side Fractional Landau Inequalities . . . . 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

177 177 179 195

12 Iterated Generalized Right Side Fractional Landau Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

197 197 200 217

13 High Order Generalized Landau Inequalities . . . . . . . . . . . . . . . . . . . . 13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

219 219 220 232 233

14 Multidimensional Caputo Left Side Fractional Landau Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

235 235 239 251

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15 Multidimensional Left Canavati Fractional Landau Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

253 253 257 272

16 Multidimensional Right Caputo Fractional Taylor Formula and Landau Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

273 273 274 275 283

17 Multidimensional Generalized Right Fractional Taylor Formula and Landau Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

285 285 286 287 297

18 Landau’s Inequality for Semigroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.4 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

299 299 300 303 306 307

19 Fractional Variable Order Gronwall Inequality . . . . . . . . . . . . . . . . . . 19.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

309 309 311 313

20 A Study of Gronwall Inequalities of Fractional Variable Order . . . . 20.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

315 315 316 331

21 General Ordinary and Fractional Approximation with Positive Sublinear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.1 Background—I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Background—II [4] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Background—III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Background—IV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.5 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.5.1 Ordinary Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.5.2 Fractional Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . 21.5.3 Iterated Fractional Approximation . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

333 333 336 337 339 341 341 343 349 351

xiv

Contents

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.2 Auxiliary Essential Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . [m] ..................................... 22.3 Main Results for Pr,n 22.3.1 Uniform Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . [m] ........................ 22.3.2 L p Approximation for Pr,n 22.3.3 Global Smoothness Preservation and Simultaneous [m] ............................ Approximation of Pr,n [m] ....... 22.3.4 Voronovskaya Asymptotic Expansions for Pr,n 22.3.5 Simultaneous Approximation by Multivariate [m] .................................... Complex Pr,n References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 High Order Approximation with Multivariate Generalized Gauss–Weierstrass Singular Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.2 Auxiliary Essential Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . [m] .................................... 23.3 Main Results for Wr,n 23.3.1 Uniform Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . [m] ........................ 23.3.2 L p Approximation for Wr,n 23.3.3 Global Smoothness Preservation and Simultaneous [m] ............................ Approximation of Wr,n [m] ....... 23.3.4 Voronovskaya Asymptotic Expansions for Wr,n 23.3.5 Simultaneous Approximation by Multivariate [m] ................................... Complex Wr,n References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Complete Approximations with Multivariate Generalized Poisson–Cauchy Type Singular Integral Operators . . . . . . . . . . . . . . . 24.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.2 Auxiliary Essential Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . [m] ..................................... 24.3 Main Results for Ur,n 24.3.1 Uniform Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . [m] ........................ 24.3.2 L p Approximation for Ur,n 24.3.3 Global Smoothness Preservation and Simultaneous [m] ............................ Approximation of Ur,n [m] ....... 24.3.4 Voronovskaya Asymptotic Expansions for Ur,n 24.3.5 Simultaneous Approximation by Multivariate [m] .................................... Complex Ur,n References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

353 353 358 365 365 369 374 381 383 391 393 393 399 408 408 411 417 425 427 435 437 437 443 453 453 456 462 469 471 478

25 High Order Approximation with Multivariate Generalized Trigonometric Type Singular Integral Operators . . . . . . . . . . . . . . . . . 479 25.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479 25.2 Auxiliary Essential Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486

Contents [m] 25.3 Main Results for Tr,n ..................................... 25.3.1 Uniform Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . [m] ........................ 25.3.2 L p Approximation for Tr,n 25.3.3 Global Smoothness Preservation and Simultaneous [m] ............................ Approximation Of Tr,n [m] ....... 25.3.4 Voronovskaya Asymptotic Expansions for Tr,n 25.3.5 Simultaneous Approximation by Multivariate [m] .................................... Complex Tr,n References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

xv

496 496 499 505 511 513 519

26 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521

Chapter 1

Variable Order General Fractional Integral Inequalities

Here are introduced abstract left and right Riemann–Liouville generalized fractional Bochner integral operators of variable order acting over Banach space valued functions. Also are introduced abstract weighted Caputo type left and right fractional derivatives of variable order. Then, we derive abstract left and right fractional integral inequalities of variable orders of Opial, Hardy and Hilbert-Pachpatte types. We give applications to simple Riemann–Liouville fractional integrals of variable order for real valued functions. It follows [3].

1.1 Background We are motivated by [1]. Next we introduce several new concepts. Definition 1.1 Let [a, b] ⊂ R, (X, ·) a Banach space, g ∈ C 1 ([a, b]) and increasing, f ∈ C ([a, b] , X ), ν ∈ C ([a, b]), ν > 1. We define the left Riemann-Liouville generalized fractional Bochner integral operator of variable order ν (·) 

ν(·) Ia+;g





f (x) := a

x

(g (x) − g (z))ν(z)−1  g (z) f (z) dz,  (ν (z))

(1.1)

ν(·) f ∈ C ([a, b] , X ) . ∀ x ∈ [a, b], where  is the gamma function. We assume Ia+;g

We need Definition 1.2 Let [a, b] ⊂ R, (X, ·) a Banach space, g ∈ C 1 ([a, b]) and increasing, f ∈ C ([a, b] , X ), ν ∈ C ([a, b]), ν > 1. We define the right Riemann-Liouville generalized fractional Bochner integral operator of variable order ν (·) © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_1

1

2

1 Variable Order General Fractional Integral Inequalities



ν(·) Ib−;g





b

f (x) := x

(g (z) − g (x))ν(z)−1  g (z) f (z) dz,  (ν (z))

(1.2)

ν(·) f ∈ C ([a, b] , X ) . ∀ x ∈ [a, b]. We assume Ib−;g

The special case of g = identit y and X = R follows: Definition 1.3 (see also [4]it, p. 235) Let [a, b] ⊂ R and α, f ∈ C ([a, b]), α > 1. The left-sided Riemann-Liouville type fractional integral of variable order α is defined as  t   f (τ ) 1 α(·) dτ , (1.3) R L Ia+ f (t) = 1−α(τ ) a  (α (τ )) (t − τ ) and the right-sided Riemann-Liouville type fractional integral of variable order α as 

α(·) R L Ib−

  f (t) =

b

t

f (τ ) 1 dτ .  (α (τ )) (τ − t)1−α(τ )

(1.4)

We need Definition 1.4 Let [a, b] ⊂ R, α ∈ C ([a, b]), such that n − 2 < α (t) < n − 1, n ∈ N, n = 1, ∀ t ∈ [a, b]. Let f ∈ C n ([a, b] , X ), where (X, ·) is a Banach space. Let ψ ∈ C n ([a, b]) with ψ being increasing and ψ  (x) = 0, for all x ∈ [a, b]. The left ψ -Caputo fractional derivative of f of variable order α (·) is given by the Bochner integral C

α(·),ψ

Da+



x

f (x) = a

(ψ (x) − ψ (z))n−α(z)−1  ψ (z) f ψ[n] (z) dz,  (n − α (z))

∀ x ∈ [a, b], where

 f ψ[n]

(x) :=

1 d  ψ (x) d x

(1.5)

n f (x) ,

(1.6)

with f ψ[0] (x) = f (x) .

α(·),ψ

Assume that C Da+

f ∈ C ([a, b] , X ) . Notice that C

α(·),ψ

Da+

n−α(·) [n] f (x) = Ia+;ψ f ψ (x) ,

(1.7)

∀ x ∈ [a, b], and f ψ[n] ∈ C ([a, b] , X ) . We also need Definition 1.5 Let [a, b] ⊂ R, α ∈ C ([a, b]), such that n − 2 < α (t) < n − 1, n ∈ N, n = 1, ∀ t ∈ [a, b]. Let f ∈ C n ([a, b] , X ), where (X, ·) is a Banach space. Let ψ ∈ C n ([a, b]) with ψ being increasing and ψ  (x) = 0, for all x ∈ [a, b]. The right

1.1 Background

3

ψ-Caputo fractional derivative of f of variable order α (·) is given by the Bochner integral C



α(·),ψ

b

f (x) = (−1)n

Db−

x

(ψ (z) − ψ (x))n−α(z)−1  ψ (z) f ψ[n] (z) dz,  (n − α (z))

(1.8)

∀ x ∈ [a, b]. α(·),ψ Assume that C Db− f ∈ C ([a, b] , X ) . Notice that α(·),ψ

C

Db−

n−α(·) [n] f (x) = (−1)n Ib−;ψ f ψ (x) ,

(1.9)

∀ x ∈ [a, b] . Next, based on the above new generalized fractional Bochner integral operators of variable order, we derive several new interesting fractional integral inequalities.

1.2 Main Results We present the following left fractional Opial type inequality of variable order (see also [2]). Theorem 1.6 All as in Definition 1.1. Let p, q > 1 : 

x

a

2

− q1



x



a

a

w



1 p

+

1 q

= 1. Then

    ν(·)   Ia+;g f (w)  f (w) g  (w) dw ≤

(g (w) − g (z))ν(z)−1  (ν (z))

p



1p 

dz dw

x

q  f (w)q g  (w) dw

2 q

,

a

(1.10) ∀ x ∈ [a, b] . Proof By Hölder’s inequality we obtain        x (g (x) − g (z))ν(z)−1  ν(·)     g (z) f (z) dz  ≤  Ia+;g f (x) =   a   (ν (z))  a

 a

x



x

(g (x) − g (z))ν(z)−1  g (z)  f (z) dz ≤  (ν (z))

(g (x) − g (z))ν(z)−1  (ν (z))

1p 

p dz

a

x



q g  (z)  f (z) dz

 q1

,

(1.11)

4

1 Variable Order General Fractional Integral Inequalities

∀ x ∈ [a, b]. Call

 ρ (x) :=

x

q

 g (z)  f (z) dz, ρ (a) = 0.

(1.12)

a

Thus

and

q

ρ (x) = g  (x)  f (x) ≥ 0,

(1.13)

1

 ρ (x) q = g  (x)  f (x) ≥ 0, ∀x ∈ [a, b] .

(1.14)

Consequently, we get     ν(·)   Ia+;g f (w) g  (w)  f (w) ≤ 

w



(g (w) − g (z))ν(z)−1  (ν (z))

a

1p

p

(1.15)

1 ρ (w) ρ (w) q ,

dz

∀ w ∈ [a, b] . Then, by applying again Hölder’s inequality: 

x

a





x

a

 a



w

a



x

2

 a

x

(g (w) − g (z))ν(z)−1  (ν (z))

(g (w) − g (z))ν(z)−1  (ν (z))



a

− q1



a



x

w

w



a

 a

w

    ν(·)   Ia+;g f (w)  f (w) g  (w) dw ≤



dz

p

1 ρ (w) ρ (w) q dw ≤

1p 

x



ρ (w) ρ (w) dw

dz dw

 q1

=

a

(g (w) − g (z))ν(z)−1  (ν (z))

(g (w) − g (z))ν(z)−1  (ν (z))

1p

p

p



p

1p 

dz dw

1p 

dz dw

x

ρ2 (x) 2

 q1

=

q g (z)  f (z) 

 q2

.

a

(1.16) The theorem is proved.



1.2 Main Results

5

We have the special case: Corollary 1.7 (to Theorem 1.6) All as in Definition 1.3. Let p, q > 1 : Then  x   α(·) R L Ia+ f (t) f (t) dt ≤

1 p

+

1 q

= 1.

a

2

− q1



  t

x

a

a

(t − τ )α(τ )−1  (α (τ ))



p

1p 

x

( f (t)) dt q

dτ dt

 q2

,

(1.17)

a

∀ x ∈ [a, b] . Next comes a right fractional Opial type inequality of variable order. Theorem 1.8 All as in Definition 1.2. Let p, q > 1 : 

b x

2

− q1



b



b

w

x



1 p

+

1 q

= 1. Then

    ν(·)   Ib−;g f (w)  f (w) g  (w) dw ≤

(g (z) − g (w))ν(z)−1  (ν (z))

p



1p 

b

dz dw

q  f (w) g  (w) dw q

 q2

,

x

(1.18) ∀ x ∈ [a, b] . Proof We have that        b (g (z) − g (x))ν(z)−1  ν(·)     g (z) f (z) dz  ≤  Ib−;g f (x) =   x   (ν (z)) 

b x



b



x

(g (z) − g (x))ν(z)−1  g (z)  f (z) dz ≤  (ν (z))

(g (z) − g (x))ν(z)−1  (ν (z))

∀ x ∈ [a, b]. Call



b

λ (x) :=

1p 

p

b

dz



q g  (z)  f (z) dz

 q1

,

(1.19)

x

q

 g (z)  f (z) dz, λ (b) = 0.

(1.20)

x

Hence

and

q

λ (x) = − g  (x)  f (x) ≤ 0,

(1.21)

6

1 Variable Order General Fractional Integral Inequalities

and

q

− λ (x) = g  (x)  f (x) ≥ 0,

(1.22)

1

 −λ (x) q = g  (x)  f (x) ≥ 0, ∀x ∈ [a, b] .

(1.23)

Consequently, we get

     ν(·)  Ib−;g f (w) g  (w)  f (w) ≤

  b (g (z) − g (w))ν(z)−1 p

∀ w ∈ [a, b] . Thus



b x



b

w



p

x



b



b

w

x





b

(g (z) − g (w))ν(z)−1  (ν (z)) 

2



b





b

w

x



w

x

− q1

b



 1 λ (w) −λ (w) q ,

(1.24)

    ν(·)   Ib−;g f (w)  f (w) g  (w) dw ≤

(g (z) − g (w))ν(z)−1  (ν (z))

b

p

dz

 (ν (z))

w



1

(g (z) − g (w))  (ν (z))

1p

p dz



 1 λ (w) −λ (w) q dw ≤



1p   − dz dw

b



λ (w) λ (w) dw

 q1

=

x ν(z)−1



p

1p 

dz dw

(g (z) − g (w))ν(z)−1  (ν (z))

p



1p 

b

dz dw

λ (x) 2 2

 q1

(1.25) =

q

 g (z)  f (z)

 q2

,

x

(1.26) ∀ x ∈ [a, b] , proving the claim.



We have the special case: Corollary 1.9 (to Theorem 1.8) All as in Definition 1.3. Let p, q > 1 : Then  b   α(·) R L Ib− f (w) f (w) dw ≤

1 p

+

1 q

= 1.

x

2

− q1



b x

∀ x ∈ [a, b] .



b

w



(z − w)α(z)−1  (α (z))

p



1p 

b

| f (w)| dw q

dz dw x

 q2

,

(1.27)

1.2 Main Results

7

We give a left fractional Opial type inequality of variable order. Theorem 1.10 All as in Definition 1.4. Let p, q > 1 : 

2



+

1 q

= 1. Then

   C α(·),ψ   [n]   Da+ f (w)  f ψ (w) ψ  (w) dw ≤

x

a

− q1

1 p

x



a

w



a



x

a

(ψ (w) − ψ (z))n−α(z)−1  (n − α (z))



p

1p

dz dw

 q2  q

q  [n]   f ψ dw , (w) (w)  ψ 

(1.28)

∀ x ∈ [a, b] . Proof By (1.7) and Theorem 1.6.



We present a right fractional Opial type inequality of variable order. Theorem 1.11 All as in Definition 1.5. Let p, q > 1 : 

2



+

1 q

= 1. Then

   C α(·),ψ   [n]   Db− f (w)  f ψ (w) ψ  (w) dw ≤

b x

− q1

1 p

b



b w

x



b x



(ψ (z) − ψ (w))n−α(z)−1  (n − α (z))



p

1p

dz dw

 q2  q

q  [n]    f ψ (w) ψ (w) dw ,

(1.29)

∀ x ∈ [a, b] . Proof By (1.9) and Theorem 1.8.



We give two extreme Opial inequalities cases ( p = 1, q = ∞). Theorem 1.12 All as in Definition 1.1. Then 

x

a

 a

∀ x ∈ [a, b] .

 x  a

w

    ν(·)   Ia+;g f (w)  f (w) dw ≤

(g (w) − g (z))ν(z)−1  g (z) dz dw  f 2∞ ,  (ν (z))

(1.30)

8

1 Variable Order General Fractional Integral Inequalities

Proof We have (by (1.1))      ν(·)   Ia+;g f (x) ≤

(g (x) − g (z))ν(z)−1  g (z)  f (z) dz ≤  (ν (z))

x

a



(g (x) − g (z))ν(z)−1  g (z) dz  f ∞ .  (ν (z))

x

a

(1.31)

Hence      ν(·)   Ia+;g f (w)  f (w) ≤

w

a

(g (w) − g (z))ν(z)−1  g (z) dz  f 2∞ . (1.32)  (ν (z))

Now integrate (1.32) over [a, x] to get (1.30).



Theorem 1.13 All as in Definition 1.2. Then 

b x



b x



    ν(·)   Ib−;g f (w)  f (w) dw ≤

(g (z) − g (w))ν(z)−1  g (z) dz dw  f 2∞ ,  (ν (z))

b w

(1.33)

∀ x ∈ [a, b] . Proof We have (by (1.2))      ν(·)   Ib−;g f (x) ≤ 

b x

b x

(g (z) − g (x))ν(z)−1  g (z)  f (z) dz ≤  (ν (z))

(g (z) − g (x))ν(z)−1  g (z) dz  f ∞ .  (ν (z))

(1.34)

Hence      ν(·)   Ib−;g f (w)  f (w) ≤

b w

(g (z) − g (w))ν(z)−1  g (z) dz  f 2∞ . (1.35)  (ν (z))

Now integrate (1.35) over [x, b] to get (1.33).



Corollary 1.14 All as in Definition 1.4. Then  a

x

   C α(·),ψ   [n]   Da+ f (w)  f ψ (w) dw ≤

1.2 Main Results



9



x

a

w

a

2  (ψ (w) − ψ (z))n−α(z)−1    ψ (z) dz dw  f ψ[n]  . ∞  (n − α (z))

(1.36)

∀ x ∈ [a, b] . Proof Use of Theorem 1.12, (1.7).



Corollary 1.15 All as in Definition 1.5. Then 

b x



 b 

b w

x

   C α(·),ψ   [n]   Db− f (w)  f ψ (w) dw ≤

2  (ψ (z) − ψ (w))n−α(z)−1    ψ (z) dz dw  f ψ[n]  . ∞  (n − α (z))

(1.37)

∀ x ∈ [a, b] . Proof Use of Theorem 1.13, (1.9).



Next we give some fractional Hardy’s type inequalities of variable order. Theorem 1.16 All as in Definition 1.1. Let p, q > 1 : ⎛     ν(·)   Ia+;g f  ≤ ⎝ q

b



a

x



a

(g (x) − g (z))ν(z)−1  (ν (z))

1 p

+

p

1 q

= 1. Then

qp

⎞ q1

dz

  d x ⎠ g   f q . (1.38)

Proof By (1.11) we get (∀ x ∈ [a, b])     ν(·)   Ia+;g f (x) ≤  a

x



(g (x) − g (z))ν(z)−1  (ν (z))

1p 

p

b

dz

q

 g (z)  f (z) dz

 q1

.

(1.39)

a

Hence it holds    q  ν(·)   Ia+;g f (x) ≤

a

x



(g (x) − g (z))ν(z)−1  (ν (z))

qp 

p dz

b

q

 g (z)  f (z) dz,

a

(1.40) ∀ x ∈ [a, b] . Therefore we get

 a

b

  q  ν(·)   Ia+;g f (x) d x ≤

10

1 Variable Order General Fractional Integral Inequalities

⎛ ⎝



b



a

x



a

(g (x) − g (z))ν(z)−1  (ν (z))

⎞

qp

p

dx⎠

dz



b



 q g (z)  f (z) dz , 

a

(1.41) proving the claim.



Theorem 1.17 All as in Definition 1.2. Let p, q > 1 : ⎛     ν(·)  Ib−;g f  ≤ ⎝ q

b



a

b



x

(g (z) − g (x))  (ν (z))

ν(z)−1

1 p

+

p

1 q

= 1. Then

qp

⎞ q1

dz

  d x ⎠ g   f q . (1.42)

Proof By (1.19) (∀ x ∈ [a, b]) we get     ν(·)   Ib−;g f (x) ≤ 

b



x

(g (z) − g (x))ν(z)−1  (ν (z))

1p 

p

b

dz



q g (z)  f (z) dz 

 q1

.

(1.43)

a

Hence    q  ν(·)   Ib−;g f (x) ≤

b x

∀ x ∈ [a, b] . Therefore it holds



(g (z) − g (x))ν(z)−1  (ν (z))



b

a

⎛  ⎝ a

b



b x



(g (z) − g (x))  (ν (z))

qp 

p

b

dz

 q

 g (z)  f (z) dz , (1.44)

a

  q  ν(·)   Ib−;g f (x) d x ≤

ν(z)−1

⎞

qp

p dz

dx⎠



b

 q

 g (z)  f (z) dz ,

a

(1.45) proving the claim.



We present also: Theorem 1.18 All as in Definition 1.4. Let p, q > 1 :   C α(·),ψ   Da+  ≤ q

1 p

+

1 q

= 1. Then

1.2 Main Results

⎛  ⎝

b



a

x

11



a

(ψ (x) − ψ (z))n−α(z)−1  (n − α (z))

Proof By Theorem 1.16 and (1.7).

qp

p dz

⎞ q1

      d x ⎠ ψ   f ψ[n]  . q

(1.46)



Theorem 1.19 All as in Definition 1.5. Let p, q > 1 :

1 p

+

1 q

= 1. Then

  C α(·),ψ   Db−  ≤ q

⎛ ⎝



b



a

b



x

(ψ (z) − ψ (x))  (n − α (z))

n−α(z)−1

Proof By Theorem 1.17 and (1.9).

qp

p dz

⎞ q1

      d x ⎠ ψ   f ψ[n]  . q

(1.47)



Next we give a Hilbert-Pachpatte left fractional inequality of variable orders. Theorem 1.20 Let p, q > 1 : 1p + q1 = 1. Here i = 1, 2. Let [ai , bi ] ⊂ R, (X, ·) a Banach space, gi ∈ C 1 ([ai , bi ]) and strictly increasing, f i ∈ C ([ai , bi ] , X ), νi ∈ (·) f i ∈ C ([ai , bi ] , X ) .Then C ([ai , bi ]), νi > 1. Assume Iaνii+;g i 

b1



a1

b2

a2

⎡  ⎣

x1 a1



       ν1 (·)   ν (·)   Ia1 +;g1 f 1 (x1 )  Ia22+;g2 f 2 (x2 ) (g1 (x1 )−g1 (z1 ))ν1 (z1 )−1  (ν1 (z 1 )) p



p

dz 1

  ⎤ d x1 d x2   x2 (g2 (x2 )−g2 (z2 ))ν2 (z2 )−1 q dz 2 a2  (ν2 (z 2 )) ⎦ q



+

    ≤ (b1 − a1 ) (b2 − a2 ) g1  f 1 q g2  f 2  p .

(1.48)

Proof We have that (i = 1, 2) 

  (1.1) (·) Iaνii+;g f = (x ) i i i

xi

ai

(gi (xi ) − gi (z i ))νi (zi )−1  gi (z i ) f i (z i ) dz i ,  (νi (z i ))

(1.49)

(gi (xi ) − gi (z i ))νi (zi )−1  gi (z i )  f i (z i ) dz i ,  (νi (z i ))

(1.50)

∀ xi ∈ [ai , bi ] . Then      νi (·)   Iai +;gi f i (xi ) ≤

xi

ai

i = 1, 2, ∀ xi ∈ [ai , bi ] . By applying Hölder’s inequality twice we get:

12

1 Variable Order General Fractional Integral Inequalities

    ν1 (·)   Ia1 +;g1 f 1 (x1 ) ≤ 

x1



a1

(g1 (x1 ) − g1 (z 1 ))ν1 (z1 )−1  (ν1 (z 1 ))

1p 

p

q

 g1 (z 1 )  f 1 (z 1 ) dz 1

x1

dz 1 a1

 q1

,

(1.51) ∀ x1 ∈ [a1 , b1 ] , and 

x2



a2

     ν2 (·)  Ia2 +;g2 f 2 (x2 ) ≤

(g2 (x2 ) − g2 (z 2 ))ν2 (z2 )−1  (ν2 (z 2 ))

q1 

q

x2

dz 2 a2

p

 g2 (z 2 )  f 2 (z 2 ) dz 2

 1p

,

(1.52) ∀ x2 ∈ [a2 , b2 ] . Hence we have (by (1.51), (1.52))         ν (·)   ν1 (·)  Ia1 +;g1 f 1 (x1 )  Ia22+;g2 f 2 (x2 ) ≤ 

x1

a1



x2



a2



(g1 (x1 ) − g1 (z 1 ))ν1 (z1 )−1  (ν1 (z 1 ))

(g2 (x2 ) − g2 (z 2 ))ν2 (z2 )−1  (ν2 (z 2 ))

q1

q dz 2

1

1

(using Yang’s inequality for a, b ≥ 0, a p b q ≤ ⎡  ⎢ ⎣

x1 a1



(g1 (x1 )−g1 (z 1 ))ν1 (z1 )−1 (ν1 (z 1 ))

p



 dz 1

p

+

a p

x2 a2

1p

p dz 1

(1.53)

     g  f 1  g   f 2  ≤ 1 2 q p

+ qb ) 

(g2 (x2 )−g2 (z 2 ))ν2 (z2 )−1 (ν2 (z 2 ))

⎤

q dz 2

q

     g  f 1  g   f 2  , 1 2 q p ∀ xi ∈ [ai , bi ] , i = 1, 2. So far we have         ν (·)   ν1 (·)  Ia1 +;g1 f 1 (x1 )  Ia22+;g2 f 2 (x2 ) ⎡       ν (z )−1 p ν (z  ⎣

x1 a1

(g1 (x1 )−g1 (z1 )) 1  (ν1 (z 1 )) p

1

dz 1

+

x2 a2

(g2 (x2 )−g2 (z2 )) 2  (ν2 (z 2 )) q

⎥ ⎦

(1.54)

2 )−1

⎤

q dz 2

⎦

1.2 Main Results

13

    ≤ g1  f 1 q g2  f 2  p ,

(1.55)

∀ xi ∈ [ai , bi ] , i = 1, 2. The denominator in (1.55) can be zero only when x1 = a1 and x2 = a2 . Therefore we obtain (1.48) by integrating (1.55) over [a1 , b1 ] × [a2 , b2 ] .  It follows a Hilbert-Pachpatte right fractional inequality of variable orders. (·) Theorem 1.21 All as in Theorem 1.20. Assume that Ibνii−;g f i ∈ C ([ai , bi ] , X ) , i = i 1, 2.Then        ν1 (·)   ν (·)   b1  b2  Ib1 −;g1 f 1 (x1 )  Ib22−;g2 f 2 (x2 ) ⎡       ⎤ d x1 d x2   ν (z )−1 p ν (z )−1 q  a1

a2

b1 x1

⎣

(g1 (z1 )−g1 (x1 )) 1  (ν1 (z 1 )) p

1

dz 1

b2 x2

+

(g2 (z2 )−g2 (x2 )) 2  (ν2 (z 2 )) q

2

dz 2

⎦

    ≤ (b1 − a1 ) (b2 − a2 ) g1  f 1 q g2  f 2  p . Proof As similar to the proof of Theorem 1.20 is omitted.

(1.56)



We also give (use of Theorem 1.20 and (1.7)): Theorem 1.22 Let p, q > 1 : 1p + q1 = 1. Here i = 1, 2. Let [ai , bi ] ⊂ R, αi ∈ C ([ai , bi ]), such that n i − 2 < αi (ti ) < n i − 1, n i ∈ N, n i = 1, ∀ ti ∈ [ai , bi ]. Let f i ∈ C ni ([ai , bi ] , X ), where (X, ·) is a Banach space. Let ψi ∈ C ni ([ai , bi ]) with ψi being striclty increasing and ψi (x) = 0, for all xi ∈ [ai , bi ]. Assume that C αi (·),ψi Dai + f i ∈ C ([ai , bi ] , X ) . Then 

b1

a1



b2

a2

⎡  ⎣



x1 a1

   C α1 (·),ψ1   α (·),ψ  f 1 (x1 ) C Da22+ f 2 (x2 )  Da1 +

(ψ1 (x1 )−ψ1 (z 1 ))n 1 −α1 (z 1 )−1  (n 1 −α1 (z 1 )) p

p



dz 1



x



2 a2

+

(ψ2 (x2 )−ψ2 (z 2 ))n 2 −α2 (z 2 )−1  (n 2 −α2 (z 2 )) q

 ⎤ d x1 d x2

q dz 2

⎦

        [n 1 ]     [n 2 ]  f ≤ (b1 − a1 ) (b2 − a2 ) ψ1  f 1ψ   ψ  2 2ψ2  . 1 q

(1.57)

p

We finish with (use of Theorem 1.21 and (1.9)) α

Theorem 1.23 All as in Theorem 1.22. Assume that C Dbii−(·),ψi f i ∈ C ([ai , bi ] , X ) , i = 1, 2. Then 

b1

a1



b2

a2

⎡  ⎣

 b1 x1

   C α1 (·),ψ1   α (·),ψ  f 1 (x1 ) C Db22− 2 f 2 (x2 )  D b1 −

(ψ1 (z 1 )−ψ1 (x1 ))n 1 −α1 (z 1 )−1  (n 1 −α1 (z 1 )) p



p

dz 1



+

b

2 x2



(ψ2 (z 2 )−ψ2 (x2 ))n 2 −α2 (z 2 )−1  (n 2 −α2 (z 2 )) q

 ⎤ d x1 d x2

q dz 2

⎦

14

1 Variable Order General Fractional Integral Inequalities

        [n 1 ]     [n 2 ]  ψ f ≤ (b1 − a1 ) (b2 − a2 ) ψ1  f 1ψ      . 2 2ψ 1 2 q

p

(1.58)

References 1. Anastassiou, G.A.: Fractional Differentiation Inequalities. Springer, Heidelberg, New York (2009) 2. Anastassiou, G.A.: Intelligent Computations: Abstract Fractional Calculus, Inequalities, Approximations. Springer, Heidelberg, New York (2018) 3. Anastassiou, G.A.: Abstract fractional integral inequalities of variable order. Indian J. Math. (2020) 4. Yang, X.-J.: General Fractional Derivatives. CRC Press-Taylor & Francis, Boca Raton, London, New York (2019)

Chapter 2

Variable Order Fractional Integral Inequalities for Spherical Shell

Here are introduced left and right Riemann–Liouville generalized fractional radial integral operators of variable order over a spherical shell. Also are introduced left and right weighted Caputo type generalized fractional radial derivatives of variable order over a spherical shell. After proving continuity of these operators, we establish a series of left and right fractional integral inequalities of variable order over the spherical shell of Opial and Hardy types. Extreme cases are met. It follows [5].

2.1 Background W are inspired by [1–3].  Let N ≥ 2, S N −1 := x ∈ R N : |x| = 1 the unit sphere on R N , where |·| stands for the Euclidean norm in R N . Also denote the ball   B (0, R) := x ∈ R N : |x| < R ⊆ R N , R > 0, and the spherical shell A := B (0, R2 ) − B (0, R1 ), 0 < R1 < R2 . For the following see [6, pp. 149–150] and [7, pp. 87–88]. For x ∈ R N − {0} we can write uniquely x = r ω, where r = |x| > 0, and ω = x ∈ S N −1 , |ω| = 1. Clearly here r R N − {0} = (0, ∞) × S N −1 , © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_2

15

16

2 Variable Order Fractional Integral Inequalities for Spherical Shell

and

A = [R1 , R2 ] × S N −1 .

In the sequel the following related theorem will be used: Theorem 2.1 ([1, p. 458]) Let f : A → R be a Lebesgue integrable function. Then 



 f (x) d x = A

S N −1

R2

 f (r ω) r N −1 dr dω.

(2.1)

R1

So we are able to write an integral in polar form using the polar coordinates (r, ω). N  The area of S N −1 is ω N = S N −1 dω = 2π N2 . (2) N  RN . The volume of B (0, R), V ol (B (0, R)) = B(0,R) dy = π 2N +1 (2 ) N N N π 2 R −R And the volume of A, V ol (A) = ( N2 +1 1 ) . (2 ) We need Definition 2.2 Let [R1 , R2 ] ⊂ R, R2 > R1 > 0, and α ∈ C ([R1 , R2 ]), α > 1. Let ψ ∈ C 1 ([R1 , R2 ]) being increasing and f ∈ C A . We define the left Riemann– Liouville generalized fractional radial integral operator for f : A → R of variable order α (·):

I Rα(·) 1 +;ψ



f (x) =



I Rα(·) 1 +;ψ





f (r ω) :=

r R1

(ψ (r ) − ψ (z))α(z)−1  ψ (z) f (zω) dz,  (α (z)) (2.2)

∀ x ∈ A, where  is the gamma function. We also define the right Riemann–Liouville generalized farctional radial integral operator for f : A → R of variable order α (·):

 

 α(·) f = I f ω) := I Rα(·) (x) (r R2 −;ψ 2 −;ψ

r

R2

(ψ (z) − ψ (r ))α(z)−1  ψ (z) f (zω) dz,  (α (z)) (2.3)

∀ x ∈ A. We give

 f, I Rα(·) f ∈C A . Theorem 2.3 (related to Definition 2.2) We have that I Rα(·) 1 +;ψ 2 −;ψ  f ∈ C A , the proof for I Rα(·) f ∈ Proof We will prove in detail only that I Rα(·) +;ψ 1 2 −;ψ  N −1 . We can C A as similar is omitted. Indeed here f (r ω) ∈ C ([R1 , R2 ]), ∀ ω ∈ S write

I Rα(·) 1 +;ψ



f (r ω) =



R2 R1

χ[R1 ,r ] (z)

(ψ (r ) − ψ (z))α(z)−1  ψ (z) f (zω) dz, (2.4)  (α (z))

where χ is the characteristic function, ∀ ω ∈ S N −1 .

2.1 Background

17

Let rn → r , ωn → ω; then χ[R1 ,rn ] (z) → χ[R1 ,r ] (z), a.e., →

(ψ(r )−ψ(z))α(z)−1  ψ (α(z))

(ψ(rn )−ψ(z))α(z)−1  ψ (α(z))

(z)

(z), and f (zωn ) → f (zω). Furthermore it holds that χ[R1 ,rn ] (z)

(ψ (rn ) − ψ (z))α(z)−1  ψ (z) f (zωn ) →  (α (z))

χ[R1 ,r ] (z)

(ψ (r ) − ψ (z))α(z)−1  ψ (z) f (zω) ,  (α (z))

(2.5)

a.e. on [R1 , R2 ]. Here it is α (z) − 1 > 0 and α − 1 ∈ C ([R1 , R2 ]). Thus, inf (α − 1) ≤ α (z) − 1 ≤ sup (α − 1), ∀ z ∈ [R1 , R2 ] and inf (α − 1) , sup (α − 1) > 0. If |ψ (rn ) − ψ (z)| ≤ 1, then |ψ (rn ) − ψ (z)|α(z)−1 ≤ |ψ (rn ) − ψ (z)|inf(α−1) ,

(2.6)

and if |ψ (rn ) − ψ (z)| > 1, then |ψ (rn ) − ψ (z)|α(z)−1 ≤ |ψ (rn ) − ψ (z)|sup(α−1) .

(2.7)

That is |ψ (rn ) − ψ (z)|α(z)−1 ≤ |ψ (rn ) − ψ (z)|inf(α−1) + |ψ (rn ) − ψ (z)|sup(α−1) ≤ (ψ (R2 ) − ψ (R1 ))inf(α−1) + (ψ (R2 ) − ψ (R1 ))sup(α−1) =: λ < ∞,

(2.8)

true for all rn , z ∈ [R1 , R2 ] . Therefore it holds χ[R1 ,rn ] (z)

|ψ (rn ) − ψ (z)|α(z)−1  ψ (z) | f (zωn )| ≤  (α (z))

 λ ψ  f ∞,A < ∞. ∞,[R1 ,R2 ] inf [R1 ,R2 ]  (α (z))

(2.9)

Thus, by the denominated convergence theorem we obtain 

R2

χ[R1 ,rn ] (z)

R1



R2

(ψ (rn ) − ψ (z))α(z)−1  ψ (z) f (zωn ) dz →  (α (z))

χ[R1 ,r ] (z)

R1

proving the claim. 

(ψ (r ) − ψ (z))α(z)−1  ψ (z) f (zω) dz,  (α (z))

(2.10)

18

2 Variable Order Fractional Integral Inequalities for Spherical Shell

We need Definition 2.4 Let [R1 , R2 ] ⊂ R, R2 > R1 > 0, and α ∈ C ([R1, R 2 ]), such that n − 2 < α (t) < n − 1, n ∈ N − {1}, ∀ t ∈ [R1 , R2 ]. Let f ∈ C n A and ψ ∈ C n ([R1 , R2 ]) with ψ being increasing and ψ  (t) = 0, for all t ∈ [R1 , R2 ] . The left ψ-Caputo generalized fractional radial derivative of f : A → R of variable order α (·) is given by: C

α(·),ψ D R1 +

f (x) =

C

α(·),ψ D R1 +

 f (r ω) :=

r R1

(ψ (r ) − ψ (t))n−α(t)−1  ψ (t) f ψ[n] (tω) dt,  (n − α (t)) (2.11)

∀ x ∈ A, where  f ψ[n]

(x) =

f ψ[n]

(r ω) :=

1 d ψ  (r ) dr

n f (r ω) ,

(2.12)

with f ψ[0] (r ω) = f (r ω), ∀ ω ∈ S N −1 , ∀ r ∈ [R1 , R2 ]. And, the right ψ-Caputo generalized fractional radial derivative of f : A → R of variable order α (·) is given by: C



α(·),ψ

R2

(−1)n

α(·),ψ

D R2 − f (x) =C D R2 − f (r ω) :=

r

(ψ (t) − ψ (r ))n−α(t)−1  ψ (t) f ψ[n] (tω) dt,  (n − α (t))

(2.13)

∀ x ∈ A. We make Remark 2.5 (to Definition 2.4) We observe that C

and C

α(·),ψ

D R1 + f (x) = I Rn−α(·) f [n] (x) , 1 +;ψ ψ

α(·),ψ

D R2 − f (x) = (−1)n I Rn−α(·) f [n] (x) , 2 −;ψ ψ

(2.14)

(2.15)

∀ x ∈ A. α(·),ψ

α(·),ψ

By Theorem 2.3, both C D R1 + f (x) , C D R2 − f ∈ C (A). All the next in this section come from [4] and they will be used in the main results. Definition 2.6 Let [a, b] ⊂ R, (X, ·) a Banach space, g ∈ C 1 ([a, b]) and increasing, f ∈ C ([a, b] , X ), ν ∈ C ([a, b]), ν > 1. We define the left Riemann–Liouville generalized fractional Bochner integral operator of variable order ν (·)

B ν(·) Ia+;g





f (x) := a

x

(g (x) − g (z))ν(z)−1  g (z) f (z) dz,  (ν (z))

(2.16)

2.1 Background

19

ν(·) ∀ x ∈ [a, b]. We assume B Ia+;g f ∈ C ([a, b] , X ) .

Definition 2.7 Let [a, b] ⊂ R, (X, ·) a Banach space, g ∈ C 1 ([a, b]) and increasing, f ∈ C ([a, b] , X ), ν ∈ C ([a, b]), ν > 1. We define the right Riemann–Liouville generalized fractional Bochner integral operator of variable order ν (·)





B ν(·) Ib−;g

b

f (x) := x

(g (z) − g (x))ν(z)−1  g (z) f (z) dz,  (ν (z))

(2.17)

ν(·) f ∈ C ([a, b] , X ) . ∀ x ∈ [a, b]. We assume B Ib−;g

We mention the following left fractional Opial type inequality of variable order (see also [1]). Theorem 2.8 All as in Definition 2.6. Let p, q > 1 : 

x

a

2

− q1



x

 (ν (z))

a

+

1 q

= 1. Then



 B ν(·) Ia+;g f (w)  f (w) g  (w) dw ≤

  w (g (w) − g (z))ν(z)−1 p

a

1 p



 1  p x

dz dw a

 q  f (w)q g  (w) dw

2 q

,

(2.18) ∀ x ∈ [a, b] . Next follows a right fractional Opial type inequality of variable order. Theorem 2.9 All as in Definition 2.7. Let p, q > 1 : 

b x

2

− q1

 (ν (z))

w

+

1 q

= 1. Then



 B ν(·) Ib−;g f (w)  f (w) g  (w) dw ≤

   b b (g (z) − g (w))ν(z)−1 p x

1 p



 1  p b

dz dw x

 q  f (w)q g  (w) dw

2 q

,

(2.19) ∀ x ∈ [a, b] . We mention two extreme Opial inequalities ( p = 1, q = ∞ case). Theorem 2.10 All as in Definition 2.6. Then  a

 a

∀ x ∈ [a, b] .

x  a

w

x



 B ν(·) Ia+;g f (w)  f (w) dw ≤

  (g (w) − g (z))ν(z)−1  g (z) dz dw  f 2∞ ,  (ν (z))

(2.20)

20

2 Variable Order Fractional Integral Inequalities for Spherical Shell

Theorem 2.11 All as in Definition 2.7. Then 

b x



b  x

b w



 B ν(·) Ib−;g f (w)  f (w) dw ≤

  (g (z) − g (w))ν(z)−1  g (z) dz dw  f 2∞ ,  (ν (z))

(2.21)

∀ x ∈ [a, b] . Next we mention two fractional Hardy’s type inequalities of variable order. Theorem 2.12 All as in Definition 2.6. Let p, q > 1 : ⎛  B ν(·) ⎝ I f ≤ a+;g q

b

a



x

a

1 p

+

(g (x) − g (z))  (ν (z))

= 1. Then

 qp

p

ν(z)−1

1 q

dz

⎞ q1

d x ⎠ g   f  q . (2.22)

Theorem 2.13 All as in Definition 2.7. Let p, q > 1 : ⎛  B ν(·) Ib−;g f ≤ ⎝ q

a

b



b x

(g (z) − g (x))ν(z)−1  (ν (z))

1 p

+

1 q

= 1. Then

 qp

p dz

⎞ q1

d x ⎠ g   f  q . (2.23)

The above mentioned results will be applied for (X, ·) = (R, |·|). The purpose of this chapter is to derive interesting fractional radial integral inequalities of variable order.

2.2 Main Results We present the following left fractional Opial type inequality of variable order over the spherical shell. Theorem 2.14 All as in Definition 2.2 with ψ strictly increasing. Then ⎞ ⎛   N −1  

  ψ  2 R 1 2  α(·)  ∞ ⎠ ⎝  I R1 +;ψ f (x) | f (x)| d x ≤ 2− 2 R1 inf (ψ  ) A [R1 ,R2 ]

⎛ ⎝



R2 R1

⎛  ⎝

r R1

(ψ (r ) − ψ (z))α(z)−1  (α (z))

2

⎞

⎞ 21

dz ⎠ dr ⎠



 f (x) d x . 2

A

(2.24)

2.2 Main Results

21

Proof Here we apply (2.18) for q = 2 and for f (·ω) ∈ C ([R1 , R2 ]). We have that 



   α(·)   I R1 +;ψ f (r ω) | f (r ω)| ψ  (r ) dr ≤

R2 R1

2− 2

1

⎛  ⎝

R2

⎛  ⎝

R1



r

R1

(ψ (r ) − ψ (z))  (α (z))

α(z)−1

⎞

2

⎞ 21

dz ⎠ dr ⎠

(2.25)

   2 ( f (r ω)) ψ (r ) dr , ∀ ω ∈ S N −1 .

R2

2

R1

Hence it holds inf

[R1 ,R2 ]

⎛  ⎝

R2

⎛  ⎝

r

R1

R1

  ψ



R2 R1



  2 1  α(·)   I R1 +;ψ f (r ω) | f (r ω)| dr ≤ 2− 2 ψ  ∞

(ψ (r ) − ψ (z))α(z)−1  (α (z))

2

⎞

⎞ 21

dz ⎠ dr ⎠



R2

 ( f (r ω)) dr , (2.26) 2

R1

∀ ω ∈ S N −1 . 

f By Theorem 2.3 we have that I Rα(·) (·ω) ∈ C ([R1 , R2 ]), ∀ ω ∈ S N −1 , 1 +;ψ where A = [R1 , R2 ] × S N −1 . By r N −1r 1−N = 1 , where R1 ≤ r ≤ R2 , and R21−N ≤ r 1−N ≤ R11−N , we obtain: R21−N

inf

 inf ψ 

[R1 ,R2 ]

[R1 ,R2 ]

  ψ

inf

[R1 ,R2 ]

2

− 21



∞

R1



R2 R1

  ψ

⎛   2 ψ ⎝

R2



    r N −1r 1−N  I Rα(·) f (r ω) | f (r ω)| dr = 1 +;ψ



R2 R1

R2

⎛  ⎝

R1

r R1





    r N −1  I Rα(·) f ω) (r  | f (r ω)| dr ≤ 1 +;ψ

R2

r R1



  (2.26)  α(·)   I R1 +;ψ f (r ω) | f (r ω)| dr ≤

(ψ (r ) − ψ (z))α(z)−1  (α (z))

N −1 1−N

r

2

( f (r ω)) dr

⎞ 21

dz ⎠ dr ⎠

 2

⎞

≤

(2.27)

22

2 Variable Order Fractional Integral Inequalities for Spherical Shell

2

− 21

⎛  2 1−N  ⎝ ψ R1 ∞



⎛ R2

⎝



R1

r

R1

R2

r

N −1

(ψ (r ) − ψ (z))α(z)−1  (α (z))

⎞

2

⎞ 21

dz ⎠ dr ⎠

(2.28)



( f (r ω)) dr , ∀ω ∈ S N −1 . 2

R1

So far we have



R2 R1

2

− 21



R2 R1

 N −1



    r N −1  I Rα(·) f ω) (r  | f (r ω)| dr ≤ 1 +;ψ

⎞⎛ ⎛ ⎛ 2 ⎞ ⎞ 21  R2  r α(z)−1 ψ  2 − ψ (ψ (r ) (z)) ∞ ⎠⎝ ⎝ ⎝ dz ⎠ dr ⎠ inf (ψ  )  (α (z)) R1 R1 [R1 ,R2 ]



R2

r

N −1



( f (r ω)) dr , ∀ω ∈ S N −1 . 2

(2.29)

R1

 

   α(·)   I R1 +;ψ f (x) | f (x)| d x =

Hence it holds

A



 S N −1

2

− 21



R2 R1

 N −1

R2 R1

 

    | dr dω ≤ f ω)| r N −1  I Rα(·) f ω) (r (r  1 +;ψ

⎞⎛ ⎛ ⎛ 2 ⎞ ⎞ 21  R2  r α(z)−1 ψ  2 − ψ (ψ (r ) (z)) ∞ ⎠⎝ ⎝ ⎝ dz ⎠ dr ⎠ inf (ψ  )  (α (z)) R1 R1 [R1 ,R2 ]



 S N −1

2

− 21



R2 R1

 N −1

R2

  r N −1 ( f (r ω))2 dr dω =

R1

⎞⎛ ⎛ ⎛ 2 ⎞ ⎞ 21  R2  r α(z)−1 ψ  2 − ψ (ψ (r ) (z)) ∞ ⎠⎝ ⎝ ⎝ dz ⎠ dr ⎠ inf (ψ  )  (α (z)) R1 R1 [R1 ,R2 ]



 ( f (x)) d x , 2

(2.30)

A

proving the claim.  We give a right fractional Opial type inequality of variable order over the spherical shell.

2.2 Main Results

23

Theorem 2.15 All as in Definition 2.2 with ψ strictly increasing. Then ⎞ ⎛   N −1  

 2   ψ R 1 2   α(·) ∞ ⎠ ⎝  I R2 −;ψ f (x) | f (x)| d x ≤ 2− 2 R1 inf (ψ  ) A [R1 ,R2 ]

⎛ ⎝

⎛



R2 R1

⎝



R2

(ψ (z) − ψ (r ))α(z)−1  (α (z))

r

2

⎞

⎞ 21



dz ⎠ dr ⎠

 f 2 (x) d x .

(2.31)

A

Proof Here we apply (2.19) for q = 2 and for f (·ω) ∈ C ([R1 , R2 ]). We have that 

R2 R1

⎛ 1 2− 2 ⎝



R2



   α(·)   I R2 −;ψ f (r ω) | f (r ω)| ψ  (r ) dr ≤ ⎛  ⎝

R1



R2

r

R2

(ψ (z) − ψ (r ))  (α (z))

α(z)−1

⎞ 21

⎞

2

dz ⎠ dr ⎠

(2.32)

  2 ( f (r ω))2 ψ  (r ) dr , ∀ω ∈ S N −1 .

R1

Hence it holds  inf ψ 



[R1 ,R2 ]

⎛ ⎝



R2 R1

⎛  ⎝

R2 R1

R2

r

 

 2 1   α(·)  I R2 −;ψ f (r ω) | f (r ω)| dr ≤ 2− 2 ψ  ∞

(ψ (z) − ψ (r ))  (α (z))

α(z)−1

2

⎞

⎞ 21

dz ⎠ dr ⎠



R2

 ( f (r ω)) dr , 2

R1

(2.33) ∀ ω ∈ S N −1 . 

f (·ω) ∈ C ([R1 , R2 ]), ∀ ω ∈ S N −1 . We By Theorem 2.3 we have that I Rα(·) 2 −;ψ obtain     R2 N −1  α(·)  r R21−N inf ψ   I R2 −;ψ f (r ω) | f (r ω)| dr ≤ [R1 ,R2 ]

 inf ψ 

[R1 ,R2 ]

inf

[R1 ,R2 ]

R1



R2 R1

  ψ

 

   r N −1r 1−N  I Rα(·) f ω) (r  | f (r ω)| dr = −;ψ 2



R2 R1



  (2.33)  α(·)   I R2 −;ψ f (r ω) | f (r ω)| dr ≤

24

2 Variable Order Fractional Integral Inequalities for Spherical Shell

2

− 21

⎛   2 ψ ⎝ ∞

R2

⎛  ⎝

R1

R2

r



R2

r

(ψ (z) − ψ (r ))α(z)−1  (α (z))

N −1 1−N

r

⎞

2

⎞ 21

dz ⎠ dr ⎠

(2.34)

 ( f (r ω)) dr 2

≤

R1

⎛  1 2 2− 2 R11−N ψ  ⎝ ∞



R2

⎛  ⎝

R2 R1

R2

r

(ψ (z) − ψ (r ))  (α (z))

α(z)−1

2

⎞

⎞ 21

dz ⎠ dr ⎠

 r N −1 ( f (r ω))2 dr , ∀ω ∈ S N −1 .

(2.35)

R1

So far we have



R2 R1

2− 2



1

R2 R1

 N −1



    r N −1  I Rα(·) f ω) (r  | f (r ω)| dr ≤ 2 −;ψ

⎞⎛ ⎛ ⎛ 2 ⎞ ⎞ 21  R2  R2 α(z)−1 ψ  2 (ψ (z) − ψ (r )) ∞ ⎠⎝ ⎝ ⎝ dz ⎠ dr ⎠ inf (ψ  )  (α (z)) R1 r [R1 ,R2 ]



R2

r

N −1



( f (r ω)) dr , ∀ω ∈ S N −1 . 2

(2.36)

R1

 

   α(·)   I R2 −;ψ f (x) | f (x)| d x =

Hence it holds

A





R2

r S N −1

2

− 21



R2 R1

 N −1

N −1

R1

 

   α(·)   I R2 −;ψ f (r ω) | f (r ω)| dr dω ≤

⎞⎛ ⎛ ⎛ 2 ⎞ ⎞ 21  R2  R2 α(z)−1 ψ  2 − ψ (ψ (z) (r )) ∞ ⎠⎝ ⎝ ⎝ dz ⎠ dr ⎠ inf (ψ  )  (α (z)) R1 r [R1 ,R2 ]





R2

r S N −1

2− 2

1



R2 R1

 N −1

N −1





( f (r ω)) dr dω = 2

R1

⎞⎛ ⎛ ⎛ 2 ⎞ ⎞ 21  R2  R2 α(z)−1 ψ  2 (ψ (z) − ψ (r )) ∞ ⎠⎝ ⎝ ⎝ dz ⎠ dr ⎠ inf (ψ  )  (α (z)) R1 r [R1 ,R2 ]

2.2 Main Results

25



 ( f (x))2 d x ,

(2.37)

A

proving the claim.  We give two extreme Opial inequalities. Theorem 2.16 All as in Definition 2.2. Then  

N   2π 2 R2N −1  α(·)   | d x ≤ I f f (x) (x)|  R1 +;ψ  N2 A



R2



R1

  (ψ (r ) − ψ (z))α(z)−1  ψ (z) dz dr  f 2∞,A .  (α (z))

r R1

(2.38)

Proof Here we apply (2.20). We have that  R21−N 

R2 R1

R2 R1



    r N −1r 1−N  I Rα(·) f ω) (r  | f (r ω)| dr = 1 +;ψ



R2 R1



R2 R1





  (2.20)  α(·)   I R1 +;ψ f (r ω) | f (r ω)| dr ≤

  (ψ (r ) − ψ (z))α(z)−1  ψ (z) dz dr  f (·ω)2∞,[R1 ,R2 ] ≤  (α (z))

r R1





    r N −1  I Rα(·) f ω) (r  | f (r ω)| dr ≤ +;ψ 1

R2



r R1

R1

(2.39)

  (ψ (r ) − ψ (z))α(z)−1  ψ (z) dz dr  f 2∞,A ,  (α (z))

∀ω ∈ S N −1 . Hence it holds

 

   α(·)   I R1 +;ψ f (x) | f (x)| d x = A



 S N −1

2R2N −1 π   N2

N 2



R2 R1

R2 R1

 

    | dr dω ≤ f ω)| r N −1  I Rα(·) f ω) (r (r  1 +;ψ



r R1

  (ψ (r ) − ψ (z))α(z)−1  ψ (z) dz dr  f 2∞,A ,  (α (z))

(2.40)

26

2 Variable Order Fractional Integral Inequalities for Spherical Shell

proving the claim.  Theorem 2.17 All as in Definition 2.2. Then  

N   2π 2 R2N −1  α(·)    I R2 −;ψ f (x) | f (x)| d x ≤  N2 A



R2



R1

  (ψ (z) − ψ (r ))α(z)−1  ψ (z) dz dr  f 2∞,A .  (α (z))

R2

r

(2.41)

Proof Here we apply (2.21). We have that  R21−N 

R2 R1

R1



    r N −1r 1−N  I Rα(·) f ω) (r  | f (r ω)| dr = 2 −;ψ



R2 R1



R2



R1

R2

r





    r N −1  I Rα(·) f ω) (r  | f (r ω)| dr ≤ −;ψ 2

R2



  (2.21)  α(·)   I R2 −;ψ f (r ω) | f (r ω)| dr ≤

  (ψ (z) − ψ (r ))α(z)−1  ψ (z) dz dr  f (·ω)2∞,[R1 ,R2 ] ≤  (α (z))

R2



R1

R2

r

∀ω ∈ S N −1 . Hence it holds

(2.42)

  (ψ (z) − ψ (r ))α(z)−1  ψ (z) dz dr  f 2∞,A ,  (α (z))

 

   α(·)   I R2 −;ψ f (x) | f (x)| d x = A





R2

r S N −1

2R2N −1 π   N2

N 2



R2 R1

R1

 r

R2

N −1

 

   α(·)   I R2 −;ψ f (r ω) | f (r ω)| dr dω ≤

  (ψ (z) − ψ (r ))α(z)−1  ψ (z) dz dr  f 2∞,A ,  (α (z))

(2.43)

proving the claim.  Two Hardy’s type fractional inequalities of variable order on the spherical shell follow.

2.2 Main Results

27

Theorem 2.18 All as in Definition 2.2. Then 

A

⎛ ⎝

⎛



R2

⎝



r

R1

R1

I Rα(·) f (x) 1 +;ψ

2

 dx ≤

(ψ (r ) − ψ (z))α(z)−1  (α (z))

R2 R1

 N −1

∞,[R1 ,R2 ]

⎞

2

 2 ψ ⎞

dz ⎠ dr ⎠



 f (x) d x . 2

(2.44)

A

Proof By (2.22) ( p = q = 2 case) we have  R21−N 

R2

r



N −1 1−N

r

R1

⎛ ⎝



R1

 2 f (r ω) dr =

I Rα(·) 1 +;ψ ⎛

R2

⎝

2 r N −1 I Rα(·) f (r ω) dr ≤ 1 +;ψ

R2



r





R1

R1

R1

R2

R2

2 (2.22) I Rα(·) f ω) dr ≤ (r 1 +;ψ

(ψ (r ) − ψ (z))α(z)−1  (α (z))

⎞

2

  2 ψ (r ) ( f (r ω))2 dr

⎞

dz ⎠ dr ⎠

(2.45)

 ≤

R1

⎛   2 ψ ⎝ ∞,[R1 ,R2 ]

⎛ R2

⎝



R1



r

R1

R2

r

(ψ (r ) − ψ (z))α(z)−1  (α (z))

N −1 1−N

r

⎞

2

⎞

dz ⎠ dr ⎠

 ( f (r ω)) dr 2

≤

R1

⎛

2 R 1−N ψ  1

⎝

∞,[R1 ,R2 ]





R2 R1

R2

⎛  ⎝

r R1

(ψ (r ) − ψ (z))  (α (z))

α(z)−1

 r N −1 ( f (r ω))2 dr , ∀ω ∈ S N −1 .

R1

So far we have proved that 

R2 R1

2 r N −1 I Rα(·) f (r ω) dr ≤ 1 +;ψ

2

⎞

⎞

dz ⎠ dr ⎠

(2.46)

28

2 Variable Order Fractional Integral Inequalities for Spherical Shell



R2 R1

 N −1

⎛

 2 ψ

⎝

∞,[R1 ,R2 ]



R2

r

⎛



R2

⎝



r R1

R1

(ψ (r ) − ψ (z))α(z)−1  (α (z))

2

⎞

⎞

dz ⎠ dr ⎠



N −1

( f (r ω)) dr , ∀ω ∈ S N −1 . 2

(2.47)

R1

Hence it holds  

2 I Rα(·) f d x = (x) 1 +;ψ A



R2 R1

 N −1

⎛

 2 ψ

∞,[R1 ,R2 ]



R2 R1

 N −1

S N −1



R2

R2

R2 R1

⎛ ⎝



R1

 S N −1



⎝



r

2 

r N −1 I Rα(·) f ω) dr dω ≤ (r 1 +;ψ

R1

(ψ (r ) − ψ (z))α(z)−1  (α (z))

2

⎞

⎞

dz ⎠ dr ⎠

  r N −1 ( f (r ω))2 dr dω =

R1

⎛

 2 ψ

∞,[R1 ,R2 ]

⎝

⎛



R2

⎝

R1



r

R1

(ψ (r ) − ψ (z))α(z)−1  (α (z))



2

⎞

⎞

dz ⎠ dr ⎠

 ( f (x)) d x , 2

(2.48)

A

proving the claim.  We continue with Theorem 2.19 All as in Definition 2.2. Then 

A

⎛  ⎝

R2 R1

⎛  ⎝ r

R2

I Rα(·) 2 −;ψ

f (x)

2

 dx ≤

(ψ (z) − ψ (r ))  (α (z))

α(z)−1

R2 R1

2

 N −1

 2 ψ

∞,[R1 ,R2 ]

⎞

⎞

dz ⎠ dr ⎠



 f 2 (x) d x . A

Proof By (2.23) ( p = q = 2 case) we have  R21−N

R2 R1

2

r N −1 I Rα(·) f ω) dr ≤ (r 2 −;ψ

(2.49)

2.2 Main Results



R2

r

29

N −1 1−N



r

R1

⎛ ⎝



 2 f (r ω) dr =

I Rα(·) 2 −;ψ R2



R1

⎛  ⎝

R1

R2

R2

(ψ (z) − ψ (r ))  (α (z))

r



2 (2.23) I Rα(·) f ω) dr ≤ (r −;ψ 2

α(z)−1

  2 ψ (r ) ( f (r ω))2 dr

R2

⎞

2

(2.50)

⎞

dz ⎠ dr ⎠  ≤

R1

 2 ψ

∞,[R1 ,R2 ]

⎛  ⎝

⎛ R2

⎝



R1



R2

(ψ (z) − ψ (r ))α(z)−1  (α (z))

r

R2

r N −1r 1−N ( f (r ω))2 dr

⎞

2

⎞

dz ⎠ dr ⎠

 ≤

R1

⎛

2 R 1−N ψ  1

∞,[R1 ,R2 ]



⎝

⎛



R2

⎝



R1

R2

R2

r

(ψ (z) − ψ (r ))α(z)−1  (α (z))

⎞

2

⎞

dz ⎠ dr ⎠

 r N −1 ( f (r ω))2 dr , ∀ω ∈ S N −1 .

(2.51)

R1

So far we have proved that 

R2 R1



R2 R1

 N −1

⎛

 2 ψ

∞,[R1 ,R2 ]



2 r N −1 I Rα(·) f ω) dr ≤ (r 2 −;ψ

R2

r

⎝

⎛



R2

⎝

R1

N −1



R2

r

(ψ (z) − ψ (r ))α(z)−1  (α (z))

2

⎞

⎞

dz ⎠ dr ⎠



( f (r ω)) dr , ∀ω ∈ S N −1 . 2

(2.52)

R1

Hence it holds  

2 I Rα(·) f d x = (x) 2 −;ψ A

 S N −1

R2 R1

2  r N −1 I Rα(·) f ω) dr dω ≤ (r 2 −;ψ

30

2 Variable Order Fractional Integral Inequalities for Spherical Shell



R2 R1

 N −1

⎛

 2 ψ

∞,[R1 ,R2 ]



⎝

⎛



R2

⎝

R1





R2 R1

 N −1

 2 ψ

∞,[R1 ,R2 ]

R2

r

R2

r S N −1



N −1

(ψ (z) − ψ (r ))α(z)−1  (α (z)) 

⎞

2

⎞

dz ⎠ dr ⎠



( f (r ω)) dr dω = 2

R1

⎛ ⎝



⎛ R2

⎝

R1



R2

r



(ψ (z) − ψ (r ))α(z)−1  (α (z))

⎞

2

⎞

dz ⎠ dr ⎠

 ( f (x))2 d x ,

(2.53)

A

proving the claim.  Fractional inequalities for the generalized fractional radial derivatives of variable order follow. First Opial type inequalities: Theorem 2.20 All as in Definition 2.4. Then ⎞ ⎛   N −1  

 2    ψ R2 1  C α(·),ψ   [n]  ∞ ⎠ ⎝  D R1 + f (x)  f ψ (x) d x ≤ 2− 2 ) R inf (ψ 1 A [R1 ,R2 ]

⎛  ⎝

R2

⎛  ⎝

R1

r R1

(ψ (r ) − ψ (z))  (n − α (z))

n−α(z)−1

⎞

2

⎞ 21

dz ⎠ dr ⎠



A

f ψ[n]

(x)

2

 dx . (2.54)

Proof As similar to the proof of Theorem 2.14 is omitted, see also (2.14).  Theorem 2.21 All as in Definition 2.4. Then ⎞ ⎛   N −1  

   ψ  2 R 1 2   [n]  C α(·),ψ  ∞ ⎠ ⎝  D R2 − f (x)  f ψ (x) d x ≤ 2− 2 R1 inf (ψ  ) A [R1 ,R2 ]

⎛ ⎝



R2 R1

⎛  ⎝ r

R2

(ψ (z) − ψ (r ))n−α(z)−1  (n − α (z))

2

⎞

⎞ 21

dz ⎠ dr ⎠



A

f ψ[n]

(x)

2

 dx . (2.55)

Proof As similar to the proof of Theorem 2.15 is omitted, see also (2.15). 

2.2 Main Results

31

We continue with extreme Opial type inequalities: Theorem 2.22 All as in Definition 2.4. Then  

N    2π 2 R2N −1  C α(·),ψ   [n]    D R1 + f (x)  f ψ (x) d x ≤  N2 A





R2

r R1

R1

  2 (ψ (r ) − ψ (z))n−α(z)−1  ψ (z) dz dr f ψ[n] . ∞,A  (n − α (z))

(2.56)

Proof As in Theorem 2.16, use of (2.14).  Theorem 2.23 All as in Definition 2.4. Then  

N    2π 2 R2N −1  C α(·),ψ   [n]    D R2 − f (x)  f ψ (x) d x ≤  N2 A



R2



R1

R2

r

  2 (ψ (z) − ψ (r ))n−α(z)−1  ψ (z) dz dr f ψ[n] . ∞,A  (n − α (z))

(2.57)

Proof As in Theorem 2.17, use of (2.15).  We finish with Hardy’s type inequalities: Theorem 2.24 All as in Definition 2.4. Then 

C

A

⎛ ⎝



R2

⎛  ⎝

r R1

R1

α(·),ψ

D R1 + f (x)

2



 N −1

dx ≤

R2 R1

2

⎞

(ψ (r ) − ψ (z))n−α(z)−1  (n − α (z))

 2 ψ

∞,[R1 ,R2 ]

⎞

dz ⎠ dr ⎠



A

f ψ[n] (x)

2

 d x . (2.58)

Proof As in Theorem 2.18, use of (2.14).  Theorem 2.25 All as in Definition 2.4. Then 

C

A

⎛ ⎝



⎛ R2 R1

⎝

 r

R2

α(·),ψ

D R2 − f (x)

2

 dx ≤

(ψ (z) − ψ (r ))n−α(z)−1  (n − α (z))

R2 R1

2

 N −1

 2 ψ

∞,[R1 ,R2 ]

⎞

⎞

dz ⎠ dr ⎠



A

f ψ[n]

(x)

2

 dx . (2.59)

Proof As in Theorem 2.19, use of (2.15). 

32

2 Variable Order Fractional Integral Inequalities for Spherical Shell

References 1. Anastassiou, G.A.: Fractional Differentiation Inequalities. Springer, Heidelberg, New York (2009) 2. Anastassiou, G.A.: Intelligent: Fractional Inequalities and Approximations Expanded. Springer, Heidelberg, New York (2020) 3. Anastassiou, G.A.: Generalized Fractional Calculus: New Advancements and Applications (accepted for publication). Springer, Heidelberg, New York (2020) 4. Anastassiou, G.A.: Abstract Fractional Integral Inequalities of Variable Order (2020) 5. Anastassiou, G.A.: Fractional Integral Inequalities of Variable Order on Spherical Shell (2020) 6. Rudin, W.: Real and Complex Analysis, International Student Edition. Mc Graw Hill, London, New York (1970) 7. Stroock, D.: A Concise Introduction to the Theory of Integration, 3rd edn. Birkäuser, Boston, Basel, Berlin (1999)

Chapter 3

Left Fractional Integral Inequalities of E.R. Love Type

Here first we derive a general reverse Minkowski integral inequality. Then motivated by the work of E.R. Love ([5], 1985) on integral inequalities we produce general reverse and direct integral inequalities. We apply these to ordinary and left fractional integral inequalities. The last involve ordinary derivatives, left Riemann–Liouville fractional integrals, left Caputo fractional derivatives, and left generalized fractional derivatives. These inequalities are of Opial type. It follows [2].

3.1 Introduction This chapter deals with ordinary and left fractional integral inequalities. We are motivated by the following results: Theorem 3.1 (Hardy’s Inequality, integral version [4, Theorem 327]) If f is a complex-valued function in L r (0, ∞), · is the L r (0, ∞) norm and r > 1, then   x  1  r  f (t) dt  x  ≤ r − 1  f. 0

(3.1)

Theorem 3.2 (E.R. Love, [5]) If s ≥ r ≥ 1, 0 ≤ a < b ≤ ∞, γ is real, ω (x) is decreasing and positive in (a, b), f (x) and H (x, y) are measurable and nonnegative on (a, b), H (x, y) is homogeneous of degree −1,  (H f ) (x) =

x

H (x, y) f (y) dy

(3.2)

a

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_3

33

34

3 Left Fractional Integral Inequalities of E.R. Love Type

and

  f r =

b

f (x) x r

γ −1

ω (x) d x

 r1

,

(3.3)

a

then H f r ≤ C  f s , where

 C=

Here

a b

1 a b

γ

H (1, t) t − r



bt

x γ −1 ω (x) d x

(3.4)  r1 − 1s dt.

(3.5)

a

is to mean 0 if a = 0 or b = ∞ or both; and bt is to mean ∞ if b = ∞.

An application of Theorem 3.2 follows: Theorem 3.3 (E.R. Love, [5]) If p > 0, q > 0, p + q = r ≥ 1, 0 ≤ a < b ≤ ∞, γ < r , ω (x) is decreasing and positive in (a, b), f (x) is measurable and nonnegative on (a, b), I α is the left Riemann–Liouville operator of fractional integration defined by  x (x − t)α−1 f (t) dt for α > 0, (3.6) (I α f ) (x) =  (α) a I 0 f (x) = f (x), where  is the gamma function, and I β f is defined similarly for β ≥ 0, then 

b

 α  p  β q γ −αp−βq−1 I f (x) x ω (x) d x ≤ C (I f ) (x)

a



b

f (x)r x γ −1 ω (x) d x,

a



where C=

 p

q   1 − γr  1 − γr .    α + 1 − γr  β + 1 − γr

(3.7) (3.8)

Also Theorem 3.2 implies Theorem 3.1 (by 3.4), just take a = 0, b = ∞, γ = 1, s = r > 1, ω (x) = 1 and H (x, y) = x1 .

3.2 Main Results We start with a general result, see also [3]. Theorem 3.4 (Reverse Minkowski integral inequality) Let (X, A, μ) and (Y, B, ν) be σ -finite measure spaces and let 0 < p < 1. Here f is a nonnegative func tion on X ×pY with f (x, y) > 0 for almost all x ∈ X , almost all y ∈ Y and Y ( f (x, y)) dν (y) < ∞ for almost all x ∈ X.

3.2 Main Results

35

Then p

  f (x, y) dμ (x) Y

dν (y)

 1p

  ≥

( f (x, y)) dν (y) p

X

X

 1p

dμ (x) ,

Y

(3.9)

if left-hand side is finite. Proof Notice that X f (x, y) dμ (x) > 0, for almost all y ∈ Y and p

  f (x, y) dμ (x) Y

p

 

We observe that

dν (y) > 0.

X

dν (y) =

f (x, y) dμ (x) Y

X

 

    p−1 f x , y dμ x dν (y) =

 



f (x, y) dμ (x) Y

X



X

 

    p−1 f x , y dμ x dμ (x) dν (y) =





f (x, y) Y



X





X

(by Tonelli’s theorem)  

    p−1 f x , y dμ x dν (y) dμ (x)





f (x, y) X

Y





X

(by applying the reverse Hölder’s inequality in the inside we get)  ≥

X

⎡  1    p p ⎣ ( f (x, y)) dν (y) Y

  =

( f (x, y)) p dν (y) X

Y

 1p

X

⎤  p−1  p     p ⎦ dμ (x) f x , y dμ x dν (y)

  

p

dμ (x)

f (x, y) dμ (x)

Y

Finally, divide both ends of (3.10) by obtain (3.9).

Y

  Y

X

(3.10)  p−1 p dν (y) .

X

f (x, y) dμ (x)

p

dν (y)

 p−1 p

> 0 to 

We continue with a reverse analog of Theorem 3.2. The proof involves a special kind of variation of reverse Minkowski integral inequality which we establish completely.

36

3 Left Fractional Integral Inequalities of E.R. Love Type

We present Theorem 3.5 Let 0 < r < 1, 0 < a < b < ∞, f (x) and H (x, y) are measurable and non-negative on (a, b), (a, b)2 , respectively, H (x, y) is homogeneous of degree −1,  x

(H f ) (x) =

H (x, y) f (y) dy,

(3.11)

a

and suppose that 

b

H f r,[a,b] =

(H f ) (x) d x

 r1

r

< ∞,

(3.12)

a

and  f r,[a,b] is defined similarly.   We assume that H (1, t) f (x, t) > 0, for almost all t ∈ ax , 1 , for almost all    f r,[a,bt] < ∞, for almost all t ∈ ab , 1 . x ∈ [a, b] , and H (1,t) 1 tr Then  1 H (1, t) H f r,[a,b] ≥  f r,[a,bt] dt. (3.13) 1 a tr b Proof For a < x < b the homogeneity of degree −1 of H gives  (H f ) (x) =

1 a x

 H (x, xt) f (xt) xdt =

1 a x

H (1, t) f (xt) dt,

where t = xy . As a ≤ y ≤ x, then 0 < t ≤ 1. We will prove first (0 < r < 1)

 H f r,[a,b] =

b



a



1 a b



a

1

a x

a x

r H (1, t) f (xt) dt

a t

H (1, t) f (xt) dt

 r1 dx

≥

(3.14)

 r1

b

Indeed we observe that   b

1

H (1, t)r f (xt)r d x

r

 dx =

b



a

1 0

dt = (∗) .

r χ[ ax ,1] (t) H (1, t) f (xt) dt

dx =

(where χ is the characteristic function) ⎧  b ⎨  1 a ⎩

0

χ a ,1 (t) H (1, t) f (xt) dt x

  0

1

   χ a ,1 t  H 1, t  f xt  dt  x

r −1 ⎫ ⎬ ⎭

dx

3.2 Main Results

=

⎧  b ⎨ 1 a ⎩ 0

37

χ a ,1 (t) H (1, t) f (xt)



x

1

   χ a ,1 t  H 1, t  f xt  dt 

dt

x

0

⎫ ⎬

r −1

⎭

dx

(by Tonelli’s theorem)

=

⎧  1 ⎨ b 0 ⎩

a t

χ a ,1 (t) H (1, t) f (xt)



b

1

   χ a ,1 t  H 1, t  f xt  dt 

dx

x

0

⎫ ⎬

r −1

⎭

dt

⎡

 r −1 ⎤  1  b 1      ⎣ = H (1, t) f (xt) H 1, t f xt dt d x ⎦ dt a b

a t

a x

(by reverse Hölder’s inequality applied inside) ⎡ ⎤  1   r  r −1  1  b r r b 1   ⎢ ⎥ r ≥ H 1, t  f xt  dt  d x ⎣ a H (1, t) f (xt) d x ⎦ dt = a a b

⎛  ⎝

1

a

t



a b

 r1

b a t

(H (1, t) f (xt))r d x

x

⎞

 ⎠ dt

b

 a x

a

Clearly here, by the assumptions, it holds we did they make sense.

b ! 1 a

1

a x

(3.15)  r −1 r

r H (1, t) f (xt) dt

H (1, t) f (xt) dt

"r

.

dx

d x > 0 and all

! ! "r " r −1 r b 1 Finally, we divide both ends of (3.15) by a dx > a H (1, t) f (xt) dt x 0, to validate (3.14), which is a particular case of a reverse Minkowski type integral inequality. By (3.14) we continue  (∗) =

1

H (1, t)

a b

1

tr



 r1

b a t

 =

1 a b

 dt =

f (xt) td x r

H (1, t) 1

tr

1 a b

H (1, t)



f (y) dy r

1

tr

bt

 r1 dt

a

(3.16)  f r,[a,bt] dt,

proving (3.13).



We give a reverse left fractional inequality. Corollary 3.6 (to Theorem 3.5) Let 0 < r < p with r < 1, 0 < a < b < ∞, f is measurable and non-negative on (a, b) such that f (x) > 0 almost everywhere on [a, b]. Here I α is the left fractional Riemann–Liouville integral operator of order

38

3 Left Fractional Integral Inequalities of E.R. Love Type

α > 0, defined by 

(x − t)α−1 f (t) dt, I 0 f (x) = f (x) , (3.17)  (α) a   ∀ x ∈ [a, b], and suppose that  x −α I α f (x)r,[a,b] < ∞. We assume that  f r,[a,bt] <   ∞, for almost all t ∈ ab , 1 . Then α

(I f ) (x) =



b



α

I f (x)

p

r−p

( f (x))

x

x

−αp

a

1 dx ≥  (α) p



1

(1 − t)α−1 1

a b

tr

p  f r,[a,bt] dt

r−p

 f r

.

(3.18) α−1

Proof In Theorem 3.5, let H (x, y) = (x−y) for x > y > a and α > 0, which is x α (α) homogeneous of degree −1. Then (H f ) (x) = x −α I α f (x), and so by (3.13)  H f r,[a,b] ≥

1

H (1, t) 1

tr

a b

 f r,[a,bt] dt =

1  (α)



1

(1 − t)α−1 1

tr

a b

 f r,[a,bt] dt.

(3.19) Here 0 < r < p, hence 0 < rp < 1, also r −r p < 0, and f (x) > 0 almost everywhere in [a, b]. Next we apply the reverse Hölder’s inequality: 

b



x −α (I α f ) (x)

p

( f (x))r − p d x ≥

a



b



x

−α

α

r

(I f ) (x) d x

 rp 



( f (x)) d x

 r −r p

r

a

p H f r,[a,b]

b

=

a (3.19)

1 ≥  (α) p

f rr − p



1 a b

(1 − t)α−1 1

tr

p  f r,[a,bt] dt

 f rr − p , (3.20) 

proving the claim. Next we present a reverse Opial type [6] inequality.

Corollary 3.7 (to Corollary 3.6) Let 0 < r < p with r < 1, m ∈ N, 0 < a < b < ∞, f : [a, b] → R such that f (m−1) is an absolutely continuous function over with [a, b], where f (a) = f  (a) = . . . = f (m−1) (a) = 0, and f (m) is non-negative,  f (m) (x) > 0 almost everywhere over [a, b]. We assume that x −m f (x)r,[a,b] < ∞,     and  f (m) r,[a,bt] < ∞, a.e. for t ∈ ab , 1 . Then  a

b

 r − p −mp x dx ≥ ( f (x)) p f (m) (x)

3.2 Main Results

39

1 ((m − 1)!) p



1 a b

 (1 − t)m−1   f (m)  dt 1 r,[a,bt] tr

p

 (m) r − p f  . r

(3.21)

Proof By Taylor’s formula with integral remainder we have 

x

f (x) = a

 (x − t)m−1 (m) f (t) dt = I m f (m) (x) , ∀ x ∈ [a, b] (m − 1)!

then apply Corollary 3.6 for f (m) .

(3.22) 

We need Definition 3.8 Let α > 0, n = α (· is the ceiling), f ∈ AC n ([a, b]) (i.e. f (n−1) is absolutely continuous function). The left Caputo fractional derivative is given by α D∗a

1 f (x) =  (n − α)



x

(x − t)n−α−1 f (n) (t) dt

(3.23)

a

0 and exists almost everywhere for x in [a, b], D∗a f = f , see [1, p. 394].

We mention Corollary 3.9 ([1, p. 395]) Let α > 0, n = α , f ∈ AC n ([a, b]), and f (k) (a) = 0, k = 0, 1, . . . , n − 1. Then  x 1 α α f (t) dt = I α D∗a f (x) , ∀ x ∈ [a, b] . (3.24) f (x) = (x − t)α−1 D∗a  (α) a We give a reverse left fractional Opial type [1] inequality. Corollary 3.10 (to Corollary 3.6) Let 0 < r < p with r < 1, 0 < a < b < ∞, α > 0, n = α , f ∈ AC n ([a, b]), and f (k) (a) = 0, k = 0, 1, . . . , n − 1. Assume here α α f is non-negative f > 0 almost everywhere on that D∗a  −α over (a, b) such that D∗a α f r,[a,bt] < ∞, for almost all [a, b]. Suppose that x f (x)r,[a,b] < ∞ and  D∗a a  t ∈ b , 1 . Then  b  α r − p −αp f (x) x dx ≥ ( f (x)) p D∗a a

1  (α) p



1 a b

 (1 − t)α−1   Dα f  dt 1 ∗a r,[a,bt] tr

Proof By Corollaries 3.6, 3.9, see also (3.23).

p

 α r − p D f  . ∗a r

(3.25) 

40

3 Left Fractional Integral Inequalities of E.R. Love Type

We need Definition 3.11 ([1, pp. 7–8]) Let ν > 0, n := [ν] [·] the integral part, and α := ν − n (0 < α < 1); x, x0 ∈ [a, b] ⊂ R such that x ≥ x0 , x0 is fixed. Let f ∈ C ([a, b]) and define  x  x0 1 Jν f (x) := (3.26) (x − t)ν−1 f (t) dt, x0 ≤ x ≤ b,  (ν) x0 the left Riemann–Liouville integral. We define the subspace C xν0 ([a, b]) of C n ([a, b]) : $ # x0 D n f ∈ C 1 ([x0 , b]) . C xν0 ([a, b]) := f ∈ C n ([a, b]) : J1−α

(3.27)

For f ∈ C xν0 ([a, b]) we define the left generalized ν-fractional derivative of f over [x0 , b] as x0 f (n) ( f (n) := D n f ). (3.28) Dxν0 + f := D J1−α Notice Dxν0 + f ∈ C ([x0 , b]) . We also need Theorem 3.12 (from Theorem 2.1, p. 8 of [1]) Let f ∈ C xν0 ([a, b]), x0 ∈ [a, b] fixed.  (i) If ν ≥ 1, and f (i) (x0 ) = 0, i = 0, 1, . . . , n − 1, then f (x) = Jνx0 Dxν0 + f (x), all x ∈ [a, b] : x ≥ x0 .  (ii) If 0 < ν < 1, then f (x) = Jνx0 Dxν0 + f (x), all x ∈ [a, b] : x ≥ x0 . That is, in both cases we have  x  1 f (x) = (3.29) (x − t)ν−1 Dxν0 + f (t) dt, x0 ≤ x ≤ b.  (ν) x0 If x0 = a, we get f (x) =

1  (ν)

 a

x

 ν  ν f (t) dt = Jνa Da+ f (x) , all a ≤ x ≤ b. (x − t)ν−1 Da+ (3.30)

We give another reverse left fractional Opial type inequality. Corollary 3.13 (to Corollary 3.6) Let 0 < r < p with r < 1, 0 < a < b < ∞, ν > 0, n = [ν]; f ∈ Caν ([a, b]), such that f (i) (a) = 0, i = 0, 1, . . . , n − 1 for only ν f is non-negative over (a, b) such that the case of ν ≥ 1. Assume here that Da+  ν Da+ f > 0 almost everywhere on [a, b]. Suppose that x −ν f (x)r,[a,b] < ∞ and  ν    D f  < ∞, for almost all t ∈ ab , 1 . Then a+ r,[a,bt]  a

b

 ν r − p −νp f (x) x dx ≥ ( f (x)) p Da+

3.2 Main Results

41



1  (ν) p

 (1 − t)ν−1   Dν f  dt 1 a+ r,[a,bt] tr

1 a b

p

 ν r − p D f  . a+ r

(3.31) 

Proof By Corollary 3.6, Theorem 3.12, see also (3.28). We need the following representation result.

Theorem 3.14 ([1, p. 395]) Let ν ≥ γ + 1, γ ≥ 0. Call n = ν , m := γ . Assume ν f ∈ L∞ f ∈ AC n ([a, b]), such that f (k) (a) = 0, k = 0, 1, . . . , n − 1, and D∗a γ γ m−γ (m) f (x), and (a, b). Then D∗a f ∈ C ([a, b]), D∗a f (x) = I γ f (x) = D∗a



1  (ν − γ )

x

a

 ν ν f (t) dt = I ν−γ D∗a f (x) , (3.32) (x − t)ν−γ −1 D∗a

∀ x ∈ [a, b] . Remark 3.15 (to Theorem 3.14) By Corollary 3.9 we also have 1  (ν)

f (x) =



 ν ν f (t) dt = I ν D∗a f (x) , (x − t)ν−1 D∗a

x

a

(3.33)

∀ x ∈ [a, b] . It follows left fractional direct Opial type integral inequalities. Theorem 3.16 If p > 0, q > 0, p + q = r ≥ 1, 0 ≤ a < b < ∞, γ < r , ω (x) is decreasing and positive in (a, b). Let ν ≥ γ + 1, γ ≥ 0, call n = ν , f ∈ ν ν f ∈ L ∞ (a, b), with D∗a f ≥ AC n ([a, b]) : f (k) (a) = 0, k = 0, 1, . . . , n − 1; D∗a 0 over (a, b). Then 

b

a

 γ q f (x) x γ −νp−(ν−γ )q−1 ω (x) d x ≤ ( f (x)) p D∗a 



r ν D∗a f (x) x γ −1 ω (x) d x,

(3.34)

q  p

  1 − γr  1 − γr   .  ν + 1 − γr  ν − γ + 1 − γr

(3.35)

b

C a



where C=

Proof Directly from Theorem 3.3. Notice that  a

b



 p  ν−γ ν q γ −νp−(ν−γ )q−1 ν f (x) D∗a f (x) x ω (x) d x I ν D∗a I 

b

( f (x)) a

 p

γ D∗a f (x)

q

(by (3.32 ), (3.33))

=

(3.36) x γ −νp−(ν−γ )q−1 ω (x) d x.

42

3 Left Fractional Integral Inequalities of E.R. Love Type

ν So, in applying (3.7), now instead of f we take D∗a f.



Theorem 3.17 If p > 0, q > 0, p + q = r ≥ 1, 0 ≤ a < b < ∞, γ < r , ω (x) is decreasing and positive in (a, b). Let ν ≥ γi + 1, γi ≥ 0, i = 1, 2, call n = ν , f ∈ ν ν f ∈ L ∞ (a, b), with D∗a f ≥ AC n ([a, b]) : f (k) (a) = 0, k = 0, 1, . . . , n − 1; D∗a 0 over (a, b). Then 

b



a

γ1 D∗a f (x)

p 

C∗



γ2 D∗a f (x)



b

a



where ∗

q

x γ −(ν−γ1 ) p−(ν−γ2 )q−1 ω (x) d x ≤

r ν f (x) x γ −1 ω (x) d x, D∗a p

  1 − γr

  ν − γ1 + 1 − γr

C =

  1 − γr

(3.37) q

  ν − γ2 + 1 − γr

Proof Use of Theorem 3.14 and similar to Theorem 3.16.

.

(3.38) 

Corollary 3.18 All as in Theorem 3.16. Then  a

b



γ D∗a f (x)

p 

 C

b



a

q ν D∗a f (x) x γ −(ν−γ ) p−1 ω (x) d x ≤

r ν D∗a f (x) x γ −1 ω (x) d x,



where C=

  1 − γr

  ν − γ + 1 − γr

(3.39)

p . 

Proof By Theorems 3.3, 3.14. We need

Remark 3.19 (see [1, p. 26]) Let ν ≥ γ + 1, γ ≥ 0, n = [ν], x0 ∈ [a, b] fixed, f ∈ C xν0 ([a, b]) : f (i) (x0 ) = 0, i = 0, 1, . . . , n − 1. Then !

" γ Dx0 + f (x) =

1  (ν − γ )



which is continuous in x on [x0 , b]. We continue with

x x0

 (x − t)(ν−γ )−1 Dxν0 + f (t) dt,

(3.40)

3.2 Main Results

43

Theorem 3.20 If p > 0, q > 0, p + q = r ≥ 1, 0 ≤ a < b < ∞, γ < r , ω (x) is decreasing and positive in (a, b). Let ν ≥ γ + 1, γ ≥ 0, n = [ν], f ∈ Caν ([a, b]) : ν f ≥ 0 on (a, b). Then f (i) (a) = 0, i = 0, 1, . . . , n − 1. Assume that Da+ 

b

!

γ

Da+ f (x)

"p 

a





b

C1 a

ν f (x) Da+

x γ −(ν−γ ) p−1 ω (x) d x ≤

r ν f (x) x γ −1 ω (x) d x, Da+



where

q

p

  1 − γr

  ν − γ + 1 − γr

C1 =

(3.41)

. 

Proof By Theorem 3.3 and see Remark 3.19. We make

Remark 3.21 By [5], see Theorem 3.2, let s = r ≥ 1, 0 ≤ a < b ≤ ∞, γ is real, ω (x) is decreasing and positive in (a, b), f (x) and Hk (x, y) (k = 1, . . . , n) are measurable and non-negative on (a, b), Hk (x, y) is homogeneous of degree −1, 

b

(Hk f ) (x) =

Hk (x, y) f (y) dy, k = 1, . . . , n,

(3.42)

a

and

  f r =

b

f (x) x r

γ −1

ω (x) d x

 r1

,

(3.43)

a

then Hk f r ≤ Ck  f r , where

 Ck =

1 a b

(3.44)

γ

Hk (1, t) t − r dt, k = 1, . . . , n.

(3.45)

means 0 if a = 0 or b = ∞ or both; and bt means ∞ if b = ∞. n % Let now pk > 0 such that pk = r .

Here

a b

k=1

We notice the following (apply generalized Hölder’s inequality)  a

n b& k=1

(Hk f (x)) x pk

γ −1

ω (x) d x ≤

n  & k=1

a

b

(Hk f (x)) x r

γ −1

ω (x) d x

 prk

=

44

3 Left Fractional Integral Inequalities of E.R. Love Type n &

(3.44)

Hk f rpk ≤

k=1

n &

p

Ck k  f rpk =

n &

k=1

n &



' =C

f rr

n %

p Ck k  f rk=1

pk

=

(3.46)

k=1

 p Ck k





b

f (x) x r

γ −1

 ω (x) d x ,

a

k=1

where ' := C

n &

p

Ck k .

(3.47)

k=1

We have proved Theorem 3.22 Let 0 ≤ a < b ≤ ∞, γ is real, ω (x) is decreasing and positive in (a, b), f (x) and Hk (x, y) (k = 1, . . . , n) are measurable and non-negative on (a, b), Hk (x, y) is homogeneous of degree −1, 

x

(Hk f ) (x) =

Hk (x, y) f (y) dy, k = 1, . . . , n.

(3.48)

a

Let pk > 0 :

n %

pk = r ≥ 1. Then

k=1

 a

n b&

' (Hk f (x)) pk x γ −1 ω (x) d x ≤ C



b

 f (x)r x γ −1 ω (x) d x ,

(3.49)

a

k=1

where '= C

 n & k=1

1 a b

 pk Hk (1, t) t

− γr

dt

.

(3.50)

Next we give an application. Theorem 3.23 Here pk > 0 :

n %

pk = r ≥ 1. Let 0 ≤ a < b ≤ ∞, γ < r , ω (x) is

k=1

decreasing and positive in (a, b), f (x) is measurable and non-negative on (a, b), I αk is the left Riemann–Liouville operator of fractional integration defined by (I

αk

 f ) (x) = a

and

Then

x

(x − t)αk −1 f (t) dt, for αk > 0,  (αk )

I αk f (x) = f (x) , f or αk = 0;k = 1, . . . , n.

(3.51)

3.2 Main Results



n b&

a

((I

αk

45

f ) (x)) x pk

γ−

n %

αk pk −1

k=1

( ω (x) d x ≤ C



b

 f (x)r x γ −1 ω (x) d x ,

a

k=1

(3.52) where (= C

n &



k=1

 pk   1 − γr  .  αk + 1 − γr

(3.53)

Proof Here we apply Theorem 3.22. Inequality (3.52) derives from (3.49) directly. We let (x − y)αk −1 for x > y > a and αk > 0. Hk (x, y) = α x k  (αk ) Then Hk f (x) = x −αk I αk f (x), k ∈ {1, . . . , n} . Notice that 

1

Hk (1, t) t

a b

 0

1

− γr

 dt =

1 a b

(1 − t)αk −1 − γ t r dt ≤  (αk )

  1 − γr (1 − t)αk −1 − γ r , t dt =   (αk )  αk + 1 − γr

for k ∈ {1, . . . , n} . Therefore (= C

n & k=1



(3.54)

  pk  1 − γr  .  αk + 1 − γr 

Next we give general left fractional direct Opial type integral inequalities. Theorem 3.24 Here p j > 0 :

N %

p j = r ≥ 1. Let 0 ≤ a < b < ∞, γ < r , ω (x) is

j=1

decreasing and positive in (a, b) . Let ν ≥ γ j + 1, γ j ≥ 0, j = 2, . . . , N , n = ν , ν f ∈ L ∞ (a, b), with f ∈ AC n ([a, b]) : f (k) (a) = 0, k = 0, 1, . . . , n − 1, and D∗a ν D∗a f ≥ 0 over (a, b). Then 

b

( f (x)) a

p1

N ! &

γ D∗aj

f (x)

"pj

x

γ −νp1 −

N % j=2

(ν−γ j ) p j −1

ω (x) d x ≤

j=2

⎡

pj ⎤    p1 N

&  1 − γr  1 − γr ⎣ ⎦   γ  ν + 1 − γr  ν − γ + 1 − j r j=2

46

3 Left Fractional Integral Inequalities of E.R. Love Type



b



a

r

ν D∗a

f (x) x

γ −1

 ω (x) d x .

(3.55) 

Proof By Theorem 3.23, use of Theorem 3.14 and (3.33). Theorem 3.25 All as in Theorem 3.24. Then 



b

a

⎛ ⎝

N &

ν D∗a f (x)

N ! p1 &

"pj

x

γ−

N % j=2

(ν−γ j ) p j −1

ω (x) d x ≤

j=2



  1 − γr

  ν − γj + 1 −

j=2

γ

D∗aj f (x)

 p j ⎞  ⎠

γ r

b

a



 r ν D∗a f (x) x γ −1 ω (x) d x .

(3.56) 

Proof By Theorem 3.23, use of Theorem 3.14.

Corollary 3.26 All as in Theorem 3.24, and γ2 = γ3 = . . . = γλ , γλ+1 = γλ+2 = . . . = γμ , and γμ+1 = γμ+2 = . . . = γ N . Then 

b



a

ν D∗a

f (x)

p1 

γ −(ν−γλ )

λ % j=2

x

γλ D∗a

f (x)





N %



  ν − γ N + 1 − γr

μ %

γ D∗aμ

f (x)



pj

N %



γN D∗a

b

  1 − γr

a



N %

pj

j=μ+1

ω (x) d x ≤ 

  ν − γμ + 1 − γr



f (x)



 p j −1

j=μ+1

j=2

pj

μ % j=λ+1



j=λ+1

j=μ+1

"

p j −(ν−γ N )

λ % pj

  1 − γr

  1 − γr

!

pj

j=2



p j −(ν−γμ )

  ν − γλ + 1 − γr

λ %

μ %

pj

j=λ+1

 r ν D∗a f (x) x γ −1 ω (x) d x .

(3.57) 

Proof By Theorem 3.25. We finish with Theorem 3.27 Here p j > 0 :

N %

p j = r ≥ 1. Let 0 ≤ a < b < ∞, γ < r , ω (x)

j=1

is decreasing and positive in (a, b) . Let ν ≥ γ j + 1, γ j ≥ 0, j = 2, . . . , N , n = ν f ≥ 0 on [ν], f ∈ Caν ([a, b]) : f (k) (a) = 0, k = 0, 1, . . . , n − 1. Assume that Da+ (a, b). Then  a

b



ν Da+

f (x)

N ! p1 & j=2

γ Da+j

f (x)

"pj

x

γ−

N % j=2

(ν−γ j ) p j −1

ω (x) d x ≤

3.2 Main Results

⎛ ⎝

N & j=2



47

  1 − γr

  ν − γ j + 1 − γr

 p j ⎞  ⎠

Proof By Theorem 3.23, use of (3.40).

a

b



ν Da+

r

f (x) x

γ −1

 ω (x) d x .

(3.58) 

Comment 3.28 With the exhibited methods above one can derive all kinds of variation of left fractional Opial type integral inequalities, as well as of ordinary differentiation ones, due to lack of space we omit this task.

References 1. Anastassiou, G.A.: Fractional Differentiation Inequalities. Springer, Heidelberg, New York (2009) 2. Anastassiou, G.A.: E.R. Love type left fractional integral inequalities. Issues Anal. 9(27) No. 3, 14–30 (2020) 3. Benaissa, B.: On the reverse Minkowski’s integral inequality. Kragujev. J. Math. 46(3), 407–416 (2022) 4. Hardy, G.H., Littlewood, J.E., Pólya, G.: Inequalities. Cambridge University Press, Cambridge, UK (1934) 5. Love, E.R.: Inequalities like Opial’s inequality, Rocznik naukowo dydaktyczny WSP w Krakowie. Pr. Mat. 97, 109–118 (1985) 6. Opial, Z.: Sur une inégalité. Ann. Polon. Math. 8, 29–32 (1960)

Chapter 4

Right Side Fractional Integral Inequalities of E.R. Love Type

Motivated for the work of E.R. Love ([5], 1985) on integral inequalities we produce general right side direct and reverse integral inequalities. We apply these to ordinary and right side fractional integral inequalities. The last involves ordinary derivatives, right side Riemann–Liouville fractional integrals, right side Caputo fractional derivatives, and right side generalized fractional derivatives. These inequalities are of Opial type ([6]). It follows [3].

4.1 Introduction This chapter deals with ordinary and right side fractional integral inequalities. We are motivated by the following left side results: Theorem 4.1 (Hardy’s Inequality, integral version [4, Theorem 327]) If f is a complex-valued function in L r (0, ∞), · is the L r (0, ∞) norm and r > 1, then   x  1  r  f (t) dt  x  ≤ r − 1  f. 0

(4.1)

Theorem 4.2 (E.R. Love, [5]) If s ≥ r ≥ 1, 0 ≤ a < b ≤ ∞, γ is real, ω (x) is decreasing and positive in (a, b), f (x) and H (x, y) are measurable and nonnegative on (a, b), H (x, y) is homogeneous of degree −1,  (H f ) (x) =

x

H (x, y) f (y) dy

(4.2)

a

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_4

49

50

4 Right Side Fractional Integral Inequalities of E.R. Love Type

and

  f r =

b

f (x) x r

γ −1

ω (x) d x

 r1

,

(4.3)

a

then H f r ≤ C  f s , where

 C=

Here

a b

1 a b



γ

H (1, t) t − r

bt

(4.4)

x γ −1 ω (x) d x

 r1 − 1s dt.

(4.5)

a

is to mean 0 if a = 0 or b = ∞ or both; and bt is to mean ∞ if b = ∞.

We present direct and reverse integral inequalities.

4.2 Main Results 4.2.1 Part I First we give direct right side results. We present the following direct general right side result: Theorem 4.3 If s≥ r ≥ 1,, 0 ≤ a < b ≤ ∞, γ ∈ R, ω (x) is increasing and positive in (a, b), f (x) and H (z, x) are measurable and non-negative on (a, b), and (a, b)2 , respectively, H (z, x) is homogeneous of degree −1, 

b

(H f ) (x) =

H (z, x) f (z) dz,

(4.6)

x

and

  f r =

b

f (x) x r

γ −1

ω (x) d x

 r1

,

(4.7)

a

then H f r ≤ C  f s , where

 C := 1

Here

b a

b a

H (t, 1) t

− γr



b

x

γ −1

ω (x) d x

(4.8)  r1 − 1s dt.

(4.9)

at

is to mean ∞, if a = 0 or b = ∞ or both; and at is to mean 0 if a = 0.

Proof (i) Let x ∈ (a, b) . Here x ≤ z ≤ b and 1 ≤ xz ≤ bx . set t = xz , that is z = t x, and 1 ≤ t ≤ bx . Infact 1 ≤ t < ab . By degree −1 homogenity of H we have

4.2 Main Results

51



b x

(H f ) (x) =

H (t x, x) f (t x) xdt =

1



b x

x

−1



b x

H (t, 1) f (t x) xdt =

1

H (t, 1) f (t x) dt.

1

That is



b x

(H f ) (x) =

H (t, 1) f (t x) dt.

(4.10)

1

Using Minkowski’s inequality at (4.11), and the increasing property of ω at (4.12), we get   b r  r1 b x H f r = H (t, 1) f (t x) dt x γ −1 ω (x) d x ≤ a





b a

1



b a

1

b t

 r1 γ −1

H (t, 1) f (t x) x r

r

ω (x) d x

dt ≤

(4.11)

a

H (t, 1) t

− γr



1

b t

 r1 f (t x)r (t x)γ −1 ω (xt) td x

dt =



b a

H (t, 1) t



− γr

b

f (z) z r

γ −1

ω (z) dz

 r1

dt ≤

(4.13)

at

1



b a

H (t, 1) t

 

− γr

b

f (z) z r

dt

1

γ −1

ω (z) dz

 r1

.

b

γ −1

f (z) z r

 ω (z) dz =

at

b

b

f (z) z s

a

and



s s−r



in

 r  1− r f (z)r z γ −1 ω (z) s z γ −1 ω (z) s dz ≤

γ −1

ω (z) dz

 rs 

at

b

s r

at





(4.14)

at

If s = r , this is the required inequality. (ii) Assume that s > r . Applying Hölder’s inequality with indices the inside integral of (4.13) we get 

(4.12)

a

f (z)s z γ −1 ω (z) dz

b

z

γ −1

ω (z) dz

1− rs

≤

at

 r  s b at

1− r z γ −1 ω (z) dz

s

=  f rs

 b at

1− r z γ −1 ω (z) dz

s

.

(4.15)

52

4 Right Side Fractional Integral Inequalities of E.R. Love Type

So we have (by use of (4.14) and (4.15)) 

b a

H f r ≤

H (t, 1) t



− γr

f (z) z r

 f s

b a

H (t, 1) t

γ −1

ω (z) dz

 r1

dt ≤

at

1



b

− γr



b

z

1

γ −1

ω (z) dz

 r1 − 1s

dt = C  f s ,

(4.16)

at



proving the claim. Remark 4.4 If r = s, then (see (4.9)) 

b a

C :=

γ

H (t, 1) t − r dt,

(4.17)

1

independent of ω. We give Corollary 4.5 Let r > 1 and f ∈ L r (0, ∞). Then    

∞ x

 f (t)  dt   ≤ r  f r . t r

(4.18)

Proof Apply (4.8), with a = 0, b = ∞, x ∈ R+ , γ = 1, s = r > 1, ω (x) = 1, and  H (z, x) = 1z . We continue with a right side fractional result. Theorem 4.6 If p > 0, q > 0, p + q = r ≥ 1, 0 < a < b < ∞, γ > 0, ω (x) is increasing and positive in (a, b), f (x) is measurable and non-negative on (a, b) , α Ib− is the right Riemann–Liouville operator of fractional integration defined by α Ib− f (x) =



b x

(z − x)α−1 f (z) dz,  (α)

(4.19)

where  is the gamma function, α > 0, 0 Ib− f (x) = f (x) , β

and Ib− f is defined similarly for β ≥ 0, then 

b



a

where

α Ib−

f (x)

p

β Ib−

f (x)

q x

γ −αp−βq−1

ω (x) d x ≤ C

∗

 a

b

f r (x) x γ −1 ω (x) d x, (4.20)

4.2 Main Results

53



∗

C =

b−a a

(αp+βq) r

1 . (α + 1)

(4.21)

α−1

, a 0, q > 0, p + q = r ≥ 1, 0 < a < b < ∞, γ > 0, ω (x) is increasing and positive in (a, b), f (x) is measurable and non-negative on (a, b) , α Ib− is the right Riemann–Liouville fractional integral, α > 0, then 

b



a



b−a a

α Ib− f (x)

αp

p

f (x)q x γ −αp−1 ω (x) d x ≤

1 r  (α + 1)



b

f r (x) x γ −1 ω (x) d x.

(4.28)

a

Proof By Theorem 4.6 for β = 0.



We also give Corollary 4.8 Here p, q > 0, p + q = r ≥ 1, 0 < a < b < ∞, γ > 0, ω (x) is increasing and positive in (a, b). Let f ∈ AC n ([a, b]) (i.e. f (n−1) is absolutely continuous), n is even and f (n) ≥ 0. Assume further that f (k) (b) = 0, k = 0, 1, . . . , n − 1. Then  b  q f p (x) f (n) (x) x γ −np−1 ω (x) d x ≤ a



b−a a

np

1 (n!)r



b



r f (n) (x) x γ −1 ω (x) d x.

(4.29)

a

If γ = 1, ω (x) = 1, we have 

b



f (x) f p

(n)

(x)

q

x

−np

 dx ≤

a

b−a a

np

1 (n!)r



b



r f (n) (x) d x.

(4.30)

a

Proof Since f ∈ AC n ([a, b]), x ∈ [a, b], we have that (Taylor’s formula) f (x) =

 x n−1 f (k) (b) 1 (x − t)n−1 f (n) (t) dt. (x − b)k + k! − 1)! (n b k=0

(4.31)

By the assumptions we get that f (x) =

1 (n − 1)!



b x

n f (n) (x) . (t − x)n−1 f (n) (t) dt = Ib−

(4.32)

4.2 Main Results

55

Here we have that there exists f (n) almost everywhere and f (n) ∈ L 1 ([a, b]). Direct application of (4.28) produces (4.29).  Remark 4.9 Since a ≤ x ≤ b, by (4.30) we easily get that 

 q f p (x) f (n) (x) d x ≤

b



a

b (b − a) a

np

1 (n!)r



b



r f (n) (x) d x.

(4.33)

a

Setting p = q = 1, we have 

b

f (x) f

(n)

 (x) d x ≤

a

b (b − a) a

n



1 (n!)

2

b



2 f (n) (x) d x,

(4.34)

a

which is a ride side Opial’s type inequality. We would like to formalize the last result. Corollary 4.10 Let 0 < a < b < ∞, f ∈ AC n ([a, b]), n is even and f (n) ≥ 0. Assume that f (k) (b) = 0, k = 0, 1, . . . , n − 1. Then 

b

f (x) f

(n)

 (x) d x ≤

a

b (b − a) a

n



1 (n!)

2

b



2 f (n) (x) d x.

(4.35)

a

We specialize (4.35) as follows: Corollary 4.11 Let f ∈ AC 2 ([1, 2]), with f (2) ≥ 0. Assume that f (2) = f  (2) = 0. Then  2  2  (2) 2 f (x) d x. f (x) f (2) (x) d x ≤ (4.36) 1

1

We need Definition 4.12 ([1, p. 336]) Let f ∈ AC m ([a, b]), m ∈ N, where m = α , α > 0 ( · the ceiling of number). We define the right Caputo fractional derivative of order α, by m−α (m) α f (x) := (−1)m Ib− f (x) , ∀ x ∈ [a, b] , (4.37) Db− that is α f (x) = Db−

(−1)m  (m − α)



b

(z − x)m−α−1 f (m) (z) dz.

(4.38)

x

We mention the right Caputo fractional Taylor formula: Theorem 4.13 ([1, p. 341]) Let f ∈ AC m ([a, b]), x ∈ [a, b], α > 0, m = α . Then

56

4 Right Side Fractional Integral Inequalities of E.R. Love Type

(1) f (x) =

m−1 k=0

f (k) (b) 1 (x − b)k + k!  (α)



b x

α f (z) dz, (z − x)α−1 Db−

(4.39)

(2) in case of f (k) (b) = 0, k = 0, 1, . . . , m − 1, we have 1 f (x) =  (α)



 α  α α Db− f (x) , f (z) dz = Ib− (z − x)α−1 Db−

b x

(4.40)

∀ x ∈ [a, b] . We present Corollary 4.14 Let p, q > 0, p + q = r ≥ 1, 0 < a < b < ∞, γ > 0, ω (x) is increasing and positive in (a, b), α > 0, m = α . Assume that f ∈ AC m ([a, b]) α f ≥ 0 over (a, b). Then such that f (k) (b) = 0, k = 0, 1, . . . , m − 1 and Db− 

b

( f (x)) p



a



b−a a

αp

q α Db− f (x) x γ −αp−1 ω (x) d x ≤

1 r (α + 1)



b



a

r α Db− f (x) x γ −1 ω (x) d x.

(4.41) 

Proof Apply (4.28) and (4.40). We continue with

Definition 4.15 ([1, p. 345]) Let ν > 0, n := [ν] ([·] integral part), α = ν − n, 0 < α < 1, f ∈ C ([a, b]) . Consider the subspace of functions

 1−α (n) ν Cb− f ∈ C 1 ([a, b]) . ([a, b]) := f ∈ C n ([a, b]) : Ib−

(4.42)

We define the right generalized ν-fractional derivative of f over [a, b] as  1−α (n)  ν D b− f = (−1)n−1 Ib− f .

(4.43)

That is

 ν D b− f (x) =

d (−1)n−1  (n − ν + 1) d x



b

(z − x)n−ν f (n) (z) dz,

x

∀ x ∈ [a, b] . We need the following right fractional Taylor formula.

(4.44)

4.2 Main Results

57

ν Theorem 4.16 ([1, p. 348]) Let f ∈ Cb− ([a, b]), ν > 0, n = [ν] . (1) If ν ≥ 1, then n−1

 f (k) (b) ν ν D b− f (x) , (x − b)k + Ib− k! k=0

f (x) = ∀ x ∈ [a, b] . (2) If 0 < ν < 1, then

ν

ν D b− f (x) , f (x) = Ib−

(4.45)

(4.46)

∀ x ∈ [a, b] . We make ν Remark 4.17 By Theorem 4.16 we have: for f ∈ Cb− ([a, b]), ν > 0, n = [ν], we ν ν get that f (x) = Ib− D b− f (x) , ∀ x ∈ [a, b], given that 0 < ν < 1, or ν ≥ 1 and f (k) (b) = 0, k = 0, 1, . . . , n − 1.

We present Corollary 4.18 Let p, q > 0, p + q = r ≥ 1, 0 < a < b < ∞, γ > 0, ω (x) is ν increasing and positive in (a, b), ν > 0, n = [ν]. Assume that f ∈ Cb− ([a, b]) , ν (k) in case of ν ≥ 1 we have that f (b) = 0, k = 0, 1, . . . , n − 1 and D b− f ≥ 0 over (a, b). Then  b

ν  q ( f (x)) p D b− f (x) x γ −νp−1 ω (x) d x ≤ a



b−a a

νp

1 r (ν + 1)



b



a

 r ν D b− f (x) x γ −1 ω (x) d x.

(4.47) 

Proof Apply (4.28) and Remark 4.17.

4.2.2 Part II Next we get reverse right side results. We present Theorem 4.19 Let 0 < r < 1, 0 < a < b < ∞, f (x) and H (z, x) are measurable and non-negative on (a, b), (a, b)2 , respectively, H (z, x) is homogeneous of degree −1,  b

(H f ) (x) =

H (z, x) f (z) dz, x

(4.48)

58

4 Right Side Fractional Integral Inequalities of E.R. Love Type

and



b

 f r,[a,b] =

f (x) d x

 r1

r

.

(4.49)

a

  We suppose that H f r,[a,b] < ∞ and H (t, 1) f (t x) > 0, for almost all t ∈ 1, bx ,    f r,[at,b] < ∞, for almost all t ∈ 1, ab . for almost all x ∈ [a, b], and H (t,1) 1 tr Then  b a H (t, 1) H f r,[a,b] ≥  f r,[at,b] dt. (4.50) 1 tr 1 Proof For a < x < b, the homogeneity of degree −1 of H gives 

b x

(H f ) (x) =

H (t, 1) f (t x) dt,

1

where x ≤ z ≤ b, 1 ≤ xz ≤ bx , setting t := xz then z = t x and 1 ≤ t ≤ bx . Infact 1 ≤ t < ab . Using the reverse Minkowski integral inequality [2] we obtain  H f r,[a,b] =

b



a

 1

b a



r

b x

H (t, 1) f (t x) dt

H (t, 1)r f (t x)r d x



b a

dt =

H (t, 1) tr

1 b a

H (t, 1)



b

f (y)r dy

1

1

tr

 r1



1

a



dx

≥

1

 r1

b t

 r1

 r1 f (t x)r td x

dt =

a

(4.51)



b a

dt =

at

b t

H (t, 1) 1

tr

1

 f r,[at,b] dt, 

proving (4.50). Next we apply Theorem 4.19. We give

Proposition 4.20 Let 0 < r < 1, 0 < a < b < ∞,  α > 0, f (x) is measurable and α f (x)r,[a,b] < ∞; (t − 1)α−1 f (t x) > non-negative on (a, b). Assume that x −α Ib−   α−1 0, for almost all t ∈ 1, bx , for almost all x ∈ [a, b] and (t−1)1  f r,[at,b] < ∞, for tr   almost all t ∈ 1, ab . Then  −α α  x I f (x) ≥ b− r,[a,b]

1  (α)



α−1

b a

(t − 1)α−1 1

1

tr

 f r,[at,b] dt.

(4.52)

α Proof Consider here H (z, x) = (z−x) , a < x < z < b, then (H f ) (x) =x −α Ib− f x α (α)  (x). Finally we apply Theorem 4.19.

4.2 Main Results

59

We give a reverse right side fractional integral inequality. Theorem 4.21 Let 0 < r < p, with r < 1, 0 < a < b < ∞, f is measurable and non-negative on (a, b) such that f (x) > 0 almost everywhere on [a, b], α > 0. α f (x)r,[a,b] < ∞, and  f r,[at,b] < ∞, for almost all t ∈ Assume that x −α Ib−  b 1, a . Then  b  α p Ib− f (x) ( f (x))r − p x −αp d x ≥ a



1  p (α)

b a

(t − 1)α−1 t

1

1 r

p −p  f rr,[a,b] .

 f r,[at,b] dt

(4.53)

Proof We will use Proposition 4.20. Here 0 < r < p, hence 0 < rp < 1, also r −r p < 0, and f (x) > 0 almost everywhere in [a, b]. Next we apply the reverse Hölder’s inequality: 

b



a



b



x



−α

a

 α p x −α Ib− f (x) ( f (x))r − p d x ≥

α Ib−



r

f (x) d x

 rp 

b

( f (x)) d x

 r −r p

r

=

a

 −α  α  p r−p x Ib− f (x)r,[a,b]  f r,[a,b] ≥ 

1  p (α)

b a

1

(t − 1)α−1 t

1 r

(4.54)

p  f r,[at,b] dt

−p  f rr,[a,b] ,



proving the claim. Next we give some reverse right side Opial type inequalities. We start with an ordinary derivatives result.

Corollary 4.22 Let 0 < r < p, with r < 1, 0 < a < b < ∞, f ∈ AC n ([a, b]) , n on [a, b]. that is even and f (n) ≥ 0 on (a, b) with f (n) >   Assume  0 almost everywhere  f (k) (b) = 0, k = 0, 1, . . . , n − 1, and x −α f (x)r,[a,b] < ∞, and  f (n) r,[at,b] <   ∞, for almost all t ∈ 1, ab . Then 

b

 r − p −np x dx ≥ ( f (x)) p f (n) (x)

a

1 ((n − 1)!) p

 1

b a

 (t − 1)α−1   f (n)  dt 1 r,[at,b] tr

p

 (n) r − p f  . r,[a,b]

(4.55)

60

4 Right Side Fractional Integral Inequalities of E.R. Love Type



Proof Apply Theorem 4.21 and see the proof Corollary 4.8. We continue with fractional reverse results.

Corollary 4.23 Let 0 < r < p, with r < 1, 0 < a < b < ∞, α > 0, m = α , α f ∈ AC m ([a, b]) , such that f (k) (b) = 0, k = 0, 1, . . . , m − 1, and  −αDb− f ≥ 0 over α  (a, b) with Db− f > 0 almost everywhere on [a, b]. Assume that x f (x)r,[a,b] <  α    f r,[at,b] < ∞, for almost all t ∈ 1, ab . Then ∞, and  Db−  a

1 p  (α)



b a

1

b

 α r − p −αp f (x) x dx ≥ ( f (x)) p Db−

 (t − 1)α−1   Dα f  dt 1 b− r,[at,b] tr

p

 α r − p D f  . b− r,[a,b]

(4.56) 

Proof Use of Theorems 4.13 and 4.21. We finish with

Corollary 4.24 Let 0 < r < p, with r < 1, 0 < a < b < ∞, ν > 0, n = [ν], f ∈ ν Cb− ([a, b]) , for either 0 < ν < 1, or ν ≥ 1 and f (k) (b) = 0, k = 0, 1, . . . , n − 1. ν ν everywhere on [a, b]. Suppose that D b− f ≥ 0 over (a, b) with D b− f > 0 almost   −ν  ν   Assume further that x f (x)r,[a,b] < ∞, and D b− f  < ∞, for almost all r,[at,b]  b t ∈ 1, a . Then  b

ν r − p x −νp d x ≥ ( f (x)) p D b− f (x) a

1  p (ν)



b a

1

 (t − 1)ν−1   ν  dt  D b− f  1 r,[at,b] tr

Proof Use of Remark 4.17 and Theorem 4.21.

p

   ν r − p .  D b− f  r,[a,b]

(4.57) 

References 1. Anastassiou, G.A.: Intelligent Mathematics: Computational Analysis. Springer, Heidelberg, New York (2011) 2. Anastassiou, G.A.: E.R. Love type left fractional integral inequalities. Issues Anal. 9(27)(3), 1–17 (2020) 3. Anastassiou, G.A.: E.R. Love type right side fractional integral inequalities (2020) 4. Hardy, G.H., Littlewood, J.E., Pólya, G.: Inequalities, Cambridge, UK (1934) 5. Love, E.R.: Inequalities like Opial’s inequality, Rocznik naukowo dydaktyczny WSP w Krakowie. Pr. Mat. 97, 109–118 (1985) 6. Opial, Z.: Sur une inégalité. Ann. Polon. Math. 8, 29–32 (1960)

Chapter 5

General Fractional Landau Inequalities

We present uniform and L p mixed Caputo–Bochner abstract fractional Landau inequalities over R of fractional orders 2 < ν ≤ 3. These estimate the size of first and second derivatives of a Banach space valued function over R. We give applications when ν = 2.5. It follows [5].

5.1 Introduction Let p ∈ [1, ∞], I = R+ or I = R and f : I → R is twice differentiable with f, f  ∈ L p (I ), then f  ∈ L p (I ). Moreover, there exists a constant C p (I ) > 0 independent of f , such that 1  1    f  ≤ C p (I )  f  2  f   2 , (5.1) p,I

p,I

p,I

where · p,I is the p-norm on the interval I , see [1, 6]. The research on these inequalities started by Landau [12] in 1913. For the case of p = ∞ he proved that C∞ (R+ ) = 2 and C∞ (R) =

√ 2,

(5.2)

are the best constants in (5.1). In 1932, Hardy and Littlewood [8] proved (5.1) for p = 2, with the best constants C2 (R+ ) =

√

2 and C2 (R) = 1.

(5.3)

In 1935, Hardy et al. [9] showed that the best constants C p (R+ ) in (5.1) satisfies the estimate (5.4) C p (R+ ) ≤ 2, for p ∈ [1, ∞), © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_5

61

62

5 General Fractional Landau Inequalities

which yields C p (R) ≤ 2 for p ∈ [1, ∞). √ In fact, in [7, 10] was shown that C p (R) ≤ 2. In this chapter we are greatly inspired by interesting articles [11, 15]. We need the following concepts from abstract fractional calculus. Our integrals next are of Bochner type [13]. We need Definition 5.1 ([4, p. 150]) Let [a, b] ⊂ R, X be a Banach space, α > 0; m =

α ∈ N, ( · is the ceiling of the number), f : [a, b] → X . We assume that f (m) ∈ L 1 ([a, b] , X ). We call the Caputo–Bochner left fractional derivative of order α: 

 α D∗a f (x) :=

1  (m − α)



x

(x − t)m−α−1 f (m) (t) dt, ∀ x ∈ [a, b] ,

(5.5)

a

where  is the gamma function. α f := f (m) the ordinary X -valued derivative (defined similar If α ∈ N, we set D∗a 0 f := f. to numerical one, see [14], p. 83), and also set D∗a  α  α By ([4, p. 2]), D∗a f (x) exists almost everywhere in x ∈ [a, b] and D∗a f ∈ L 1 ([a, b] , X ).   α / N, then by ( [4, p. 3]), D∗a f ∈ C ([a, b] , X ) , If  f (m)  L ∞ ([a,b],X ) < ∞, and α ∈  α    hence D∗a f ∈ C ([a, b]) . We also need Definition 5.2 ([4, p. 150]) Let [a, b] ⊂ R, X be a Banach space, α > 0, m := α. We assume that f (m) ∈ L 1 ([a, b] , X ), where f : [a, b] → X . We call the Caputo– Bochner right fractional derivative of order α: 

 α f (x) := Db−

(−1)m  (m − α)



b

(z − x)m−α−1 f (m) (z) dz, ∀ x ∈ [a, b] .

(5.6)

x

 m   0  We observe that Db− f (x) = (−1)m f (m) (x) , for m ∈ N, and Db− f (x) = f (x) .  α   α  f (x) exists almost everywhere on [a, b] and Db− f ∈ By ([4, p. 34]), Db− L 1 ([a, ,X  ).  b](m) α / N, by ([4, p. 37]), Db− f ∈ C ([a, b] , X ) , If  f  L ∞ ([a,b],X ) < ∞, and α ∈  α  hence  Db− f  ∈ C ([a, b]) . We mention the important: Corollary 5.3 ([4, p. 157]) Let f ∈ C m ([a, b] , X ), m = α, α > 0, x, x0 ∈ [a, b]. a f (x) , Dxa0 − f (x) are jointly continuous functions in (x, x0 ) from [a, b]2 Then D∗x 0 into X , X is a Banach space.

5.1 Introduction

63

By convention we suppose that α f (x) = 0, for x < x0 , D∗x 0

and

Dxα0 − f (x) = 0, for x > x0 ,

for all x, x0 ∈ [a, b] . The author has already done an extensive body of work on fractional Landau inequalities, see [3], and on abstract fractional Landau inequalities, see [4]. However there the proving methods came out of applications of fractional Ostrowski inequalities ([2, 4]) and the derived inequalities were for small fractional orders, i.e. α ∈ (0, 1). Usually there the domains where [A, +∞) or (−∞, B], with A, B ∈ R and in one mixed case the domain was all of R. In this chapter with less assumptions we establish uniform and L p type mixed Caputo–Bochner abstract fractional Landau inequalities over R for fractional orders 2 < ν ≤ 3. The method of proving is based on left and right Caputo–Bochner fractional Taylor’s formulae with integral remainders, see [4], pp. 151–152. We give also applications when ν = 2.5. Regardless to say that we are also inspired by [3, 4].

5.2 Main Results We present the following abstract mixed fractional Landau inequalities over R. Theorem 5.4 Let 2 < ν ≤ 3 and f ∈ C 3 (R, X ), where (X, ·) is a Banach space. We assume that  f ∞,R < ∞, and that     ν  ν K := max  D∗a f (z)∞,R2 ,  Da− f (z)∞,R2 < ∞, where (a, z) ∈ R2 . Then     f 

∞,R

and

    f 

∞,R

 ≤ν  ≤ν

K  (ν + 1)

K  (ν + 1)

ν1 

ν2 

 f ∞,R ν−1

ν−1 ν

4  f ∞,R ν−2

,

ν−2 ν

(5.7)

(5.8)

.

(5.9)

    That is  f  ∞,R ,  f  ∞,R < ∞. Proof Here 2 < ν ≤ 3, i.e. ν = 3. Let f ∈ C 3 (R, X ), where (X, ·) is a Banach space, a ∈ R.

64

5 General Fractional Landau Inequalities

We need the following abstract fractional Taylor formulae. By Theorem 5.8, p. 151 ([4]) we get, f (x) − f (a) − (x − a) f  (a) − 1  (ν)



x

a

(x − a)2  f (a) = 2

(5.10)

 ν  f (z) dz, ∀ x ≥ a. (x − z)ν−1 D∗a

And by Theorem 5.9, p. 152 ([4]), see also Definition 5.10, p. 152 ([4]): f (x) − f (a) − (x − a) f  (a) − 1  (ν)



a x

(x − a)2  f (a) = 2

 ν  f (z) dz, ∀ x ≤ a. (z − x)ν−1 Da−

(5.11)

The remainders are continuous functions. Let x1 > a, then f (x1 ) − f (a) − (x1 − a) f  (a) − 1  (ν)



(x1 − a)2  f (a) = 2

 ν  f (z) dz, (x1 − z)ν−1 D∗a

x1

a

(5.12)

and for x2 < a, we get f (x2 ) − f (a) − (x2 − a) f  (a) − 1  (ν)



a x2

(x2 − a)2  f (a) = 2

 ν  f (z) dz. (z − x2 )ν−1 Da−

(5.13)

That is, we have the system of two equations with unknowns f  (a) , f  (a): (x1 − a) f  (a) + f (x1 ) − f (a) −

1  (ν)

 a

x1

(x1 − a)2  f (a) = 2

 ν  f (z) dz =: A, (x1 − z)ν−1 D∗a

and (x2 − a) f  (a) +

(x2 − a)2  f (a) = 2

(5.14)

5.2 Main Results

65

1  (ν)

f (x2 ) − f (a) −



a x2

The coefficient determinant is D = We have the system:

 ν  f (z) dz =: B. (z − x2 )ν−1 Da− 1 2

(5.15)

(x1 − a) (x2 − a) (x2 − x1 ). It is D = 0.

(x1 − a) f  (a) + (x2 − a) f  (a) +

(x1 −a)2 2 (x2 −a)2 2

f  (a) = A, f  (a) = B.

(5.16)

Solving the system (5.16), it has a unique non-trivial solution, we get:

f  (a) =



A

B 1 2

(x1 −a)2 2 (x2 −a)2 2







(x1 − a) (x2 − a) (x2 − x1 )

=

A (x2 − a)2 − B (x1 − a)2 , (x1 − a) (x2 − a) (x2 − x1 )

=

2 [B (x1 − a) − A (x2 − a)] . (5.18) (x1 − a) (x2 − a) (x2 − x1 )

(5.17)

and

f  (a) =



x1 − a A



x2 − a B

1 2

(x1 − a) (x2 − a) (x2 − x1 )

We derive that   f (a) = f (x1 ) − f (a) − 

1  (ν)

1 f (x2 ) − f (a) −  (ν)



a



x1

ν−1

(x1 − z)



a

(z − x2 )

x2

ν−1



ν Da−

ν D∗a





f (z) dz (x2 − a)2 −







f (z) dz (x1 − a)

2

·

1 , (x1 − a) (x2 − a) (x2 − x1 )

(5.19)

and 



f (a) = 2

1 f (x2 ) − f (a) −  (ν)

 f (x1 ) − f (a) −

1  (ν)

 a

x1



a

(z − x2 )

ν−1



x2

ν Da−





f (z) dz (x1 − a) −

 ν  f (z) dz (x2 − a) · (x1 − z)ν−1 D∗a

1 . (x1 − a) (x2 − a) (x2 − x1 )

(5.20)

66

5 General Fractional Landau Inequalities

Simplifying things we have

f  (a) =

f (x1 ) − f (a) −

 x1

1 (ν)

a

  ν  f (z) dz (x2 − a) (x1 − z)ν−1 D∗a

(x1 − a) (x2 − x1 )

f (x2 ) − f (a) −

1 (ν)

  ν  ν−1 D f dz − x (z) (x1 − a) (z ) 2 a− x2

−

a

(x2 − a) (x2 − x1 )

,

(5.21)

and f  (a) =

2 f (x2 ) − f (a) −

a

1 (ν)

x2

  ν  f (z) dz (z − x2 )ν−1 Da−

(x2 − a) (x2 − x1 )

2 f (x1 ) − f (a) −

1 (ν)

 x1 a

  ν  f (z) dz (x1 − z)ν−1 D∗a

(x1 − a) (x2 − x1 )

−

.

(5.22)

Rewriting above solutions we have f  (a) =

( f (x1 ) − f (a)) (x2 − a) ( f (x2 ) − f (a)) (x1 − a) − − (x1 − a) (x2 − x1 ) (x2 − a) (x2 − x1 ) 



   x1 ν−1 D ν f (z) dz (x − a) 2 ∗a a (x 1 − z)  (ν) (x1 − a) (x2 − x1 )

and f  (a) = 2

+

   a ν−1 D ν f (z) dz (x − a) 1 a− x2 (z − x 2 )  (ν) (x2 − a) (x2 − x1 )

,

(5.23)

2 ( f (x2 ) − f (a)) 2 ( f (x1 ) − f (a)) − − (x2 − a) (x2 − x1 ) (x1 − a) (x2 − x1 )

 a

  ν  ν−1 D f dz − x (z) (z ) 2 a− x2  (ν) (x2 − a) (x2 − x1 )

+

2

 ν   f (z) dz (x1 − z)ν−1 D∗a . (5.24)  (ν) (x1 − a) (x2 − x1 )

 x1 a

Let now h > 0, and choose x1 = a + h and x2 = a − h. That is x1 − a = h, x2 − a = −h, and x2 − x1 = −2h. We derive the following: ( f (a + h) − f (a)) ( f (a − h) − f (a)) − − 2h 2h    ν   a  ν   a+h ν−1 ν−1 D f dz + h − z) (z) (a Da− f (z) dz ∗a a a−h (z − a + h) + , 2h (ν) 2h (ν) (5.25) and similarly, f  (a) =

5.2 Main Results

67

( f (a − h) − f (a)) ( f (a + h) − f (a)) − − h2 h2    ν   a  ν   a+h ν−1 ν−1 D f dz + h − z) (z) (a Da− f (z) dz ∗a a a−h (z − a + h) − . h 2  (ν) h 2  (ν) (5.26) Putting things together we derive f  (a) =

f  (a) = 

a+h

a

 ν  f (z) dz − (a + h − z)ν−1 D∗a

and

a+h



a

a−h

 ν  f (z) dz , (z − (a − h))ν−1 Da− (5.27)

1 ( f (a + h) + f (a − h) − 2 f (a)) − 2 h2 h  (ν)

f  (a) = 

1 ( f (a + h) − f (a − h)) − 2h 2h (ν)

ν−1

(a + h − z)

a



ν D∗a





f (z) dz +

a

(z − (a − h))

ν−1

a−h



ν Da−





f (z) dz .

(5.28)  ν   ν  f (z) , Da− f (z) are jointly continuous By Corollary 5.3 we obtain that D∗a 2 X , where functions  from R  into  X is a Banach space. We assumed here  νin (z, a) ν      f (z)∞,R2 < ∞. that D∗a f (z) ∞,R2 ,  Da− By (5.27) we get that     f (a) ≤  f (a + h) − f (a − h) + 2h  a+h  ν   1 f (z) dz+ (a + h − z)ν−1  D∗a 2h (ν) a 

a

a−h

 f ∞,R K + h 2h (ν)

 ν   f (z) dz ≤ (z − (a − h))ν−1  Da−

 a

a+h

(a + h − z)

ν−1

 dz +

a

(z − (a − h))

(5.29)

ν−1

dz =

a−h

 ν  f ∞,R  f ∞,R K 2h K h ν−1 + = + . h 2h (ν) ν h  (ν + 1) That is

    f ∞,R K h ν−1  f (a) ≤ , + h  (ν + 1)

(5.30)

68

5 General Fractional Landau Inequalities

∀ a ∈ R, ∀ h > 0. I.e. it holds

    f 

∞,R

∀ h > 0, 2 < ν ≤ 3. Call

≤

 f ∞,R K h ν−1 + , h  (ν + 1)

(5.31)

μ :=  f ∞,R , θ=

(5.32)

K , (ν+1)

both are greater than zero. Set also ρ := ν − 1 > 1. We consider the function y (h) := We have

μ + θh ρ , ∀ h > 0. h

(5.33)

μ + ρθh ρ−1 = 0, h2

(5.34)

y  (h) = −

then

ρθh ρ−1 =

and

μ , h2

ρθh ρ+1 = μ,

with a unique solution

 h 0 := h crit.no =

We have that

μ ρθ

1 ρ+1

.

(5.35)

y  (h) = −μh −2 + ρθh ρ−1 ,

and

y  (h) = 2μh −3 + ρ (ρ − 1) θh ρ−2 .

(5.36)

We see that 

μ y (h 0 ) = 2μ ρθ 



μ ρθ

3 − ρ+1



μ + ρ (ρ − 1) θ ρθ

((ρ+1)−3) ρ+1

3   − ρ+1 μ 2μ + ρ (ρ − 1) θ = ρθ

=

5.2 Main Results



69

μ ρθ

3 − ρ+1



Therefore y has a global minimum at h 0 =

3 − ρ+1

μ ρθ

[2μ + μ (ρ − 1)] = μ

1   ρ+1

μ ρθ

(ρ + 1) > 0.

(5.37)

, which is

ρ  ρ+1 μ μ = y (h 0 ) =   1 + θ ρθ μ ρ+1

ρθ

ρ

μ

1

(ρθ) ρ+1

1

μ ρ+1

ρ

θμ ρ+1

+

ρ

ρ

1

ρ ρ+1 θ ρ+1

1

= (ρθ) ρ+1 μ1− ρ+1 +

ρ

θ1− ρ+1 μ ρ+1

=

ρ

ρ ρ+1

1  ρ ρ ρ ρ ρ 1 1 1 1 ρ ρ+1 θ ρ+1 μ ρ+1 + θ ρ+1 μ ρ+1 ρ− ρ+1 = θ ρ+1 μ ρ+1 ρ ρ+1 + ρ− ρ+1 =  ρ

(θμ )

1 ρ+1

ρ

1 ρ+1

+

1



 ρ

= (θμ )

ρ

ρ ρ+1

1 ρ+1

ρ+1

(5.38)



ρ

ρ ρ+1

=

ρ

(θμρ ) ρ+1 (ρ + 1) ρ− ρ+1 . 1

That is

ρ

y (h 0 ) = (θμρ ) ρ+1 (ρ + 1) ρ− ρ+1 . 1

(5.39)

Consequently  y (h 0 ) =

K  f ν−1 ∞,R  (ν + 1)

ν1

ν (ν − 1)−(

ν−1 ν

).

(5.40)

We have proved that 

    f 

∞,R

≤

From (5.28) we derive

K h 2  (ν) =

 a

a+h

K  f ν−1 ∞,R  (ν + 1)

ν1

ν (ν − 1)−(

ν−1 ν

).

(5.41)

      f (a) ≤ 4 f ∞,R + h2 (a + h − z)

ν−1

 dz +

a

(z − (a − h))

ν−1

dz

a−h

 ν 4  f ∞,R 4  f ∞,R 2h K 2K h ν−2 = . + + h2 h 2  (ν) ν h2  (ν + 1)

(5.42)

70

5 General Fractional Landau Inequalities

That is

ν−2       f (a) ≤ 4 f ∞,R + 2K h , 2 h  (ν + 1)

(5.43)

∀ a ∈ R, ∀ h > 0. Hence it holds     f 

∞,R

∀ h > 0, 2 < ν ≤ 3. Call

≤

4  f ∞,R 2K h ν−2 , + 2 h  (ν + 1)

(5.44)

ξ := 4  f ∞,R , ψ=

(5.45)

2K , (ν+1)

both are greater than zero. Set also ϕ := ν − 2 > 0. We consider the function γ (h) := We have

ξ + ψh ϕ = ξh −2 + ψh ϕ , ∀ h > 0. h2

(5.46)

γ  (h) = −2ξh −3 + ϕψh ϕ−1 = 0,

then

(5.47)

ϕψh ϕ−1 = 2ξh −3 ,

and

ϕψh ϕ+2 = 2ξ,

with a unique solution

 h 0 := h crit.no =

We have that

2ξ ϕψ

1 ϕ+2

.

(5.48)

γ  (h) = 6ξh −4 + ϕ (ϕ − 1) ψh ϕ−2 .

(5.49)

We see that 

γ (h 0 ) = 6ξ



2ξ ϕψ

4 − ϕ+2



2ξ + ϕ (ϕ − 1) ψ ϕψ

(ϕ+2)−4 ϕ+2

=

5.2 Main Results

71





2ξ ϕψ

4 − ϕ+2

2ξ ϕψ

4  − ϕ+2 2ξ 6ξ + ϕ (ϕ − 1) ψ = ϕψ

 [6ξ + (ϕ − 1) 2ξ] = 2ξ

Therefore γ has a global minimum at h 0 =  γ (h 0 ) = ξ 

ξ ϕ



2ξ ϕψ

2ξ ϕψ

2ξ ϕψ

2 − ϕ+2



2ξ ϕψ

2ξ ϕψ

4 − ϕ+2

1  ϕ+2



(ϕ + 2) > 0.

(5.50)

, which is

2ξ +ψ ϕψ

ϕ+2−2 ϕ+2

=

2  − ϕ+2  2  2ξ 2 2ξ − ϕ+2 ξ+ψ 1+ =ξ = ϕψ ϕψ ϕ

2 − ϕ+2

ξ (ϕ + 2) = ϕ

2   − ϕ+2 2 ξ 2 − ϕ+2 (ϕ + 2) = ϕ ψ

(5.51)

ϕ  2  ϕ+2 ψ ϕ+2 ξ (ϕ + 2) . ϕ 2

That is

ϕ  2  ϕ+2 ψ ϕ+2 ξ γ (h 0 ) = (ϕ + 2) . ϕ 2

Consequently

 γ (h 0 ) =

4  f ∞,R ν−2

 ν−2 ν

K  (ν + 1)

ν2

(5.52)

ν.

(5.53)

We have proved that     f 

∞,R

 ≤

4  f ∞,R ν−2

 ν−2 ν

K  (ν + 1)

ν2

ν.

(5.54) 

The theorem is proved. We also give an L p analog of a fractional Landau inequality

Theorem 5.5 Let p, q > 1 : 1p + q1 =1, 2 0. I.e. it holds     f 

∞,R

 f ∞,R M ≤ + 1 h h  (ν) (q (ν − 1) + 1) q

∀ h > 0, 2 < ν ≤ 3. Call

  ν−1− 1p

,

(5.61)

μ :=  f ∞,R , θ=

both are greater than zero. Set also ρ := ν − 1 − 1p > We consider the function

1 q

(5.62)

,

M 1 (ν)(q(ν−1)+1) q

> 0.

y (h) :=

μ + θh ρ , ∀ h > 0. h

(5.63)

As in the proof of Theorem 5.4 it has only one critical number  h 0 := h crit.no =

μ ρθ

1 ρ+1

,

(5.64)

and a global minimum ρ

ρ

y (h 0 ) = θ ρ+1 μ ρ+1 (ρ + 1) ρ− ρ+1 . 1

(5.65)

Consequently  y (h 0 ) =

1

 (ν) (q (ν − 1) + 1) q 

We have proved that



M

1

(ν− 1p )    f ∞,R 

 1 1 − ν− ν−1− p p

ν−1− 1p ν− 1p



ν−1− 1p ν− 1p





.

(5.66)

74

5 General Fractional Landau Inequalities

    f 

∞,R

 ≤



M

(

1 ν− 1p

)





 f ∞,R ν−1−

1

 (ν) (q (ν − 1) + 1) q



ν−1− 1p ν− 1p



1 p

 1 ν− . p (5.67)

From (5.28) we derive  a+h  ν        1  f (a) ≤ 4 f ∞,R + f (z) dz (a + h − z)ν−1  D∗a h2 h 2  (ν) a (5.68)  a  ν   ν−1  Da− f (z) dz ≤ + (z − (a − h)) a−h

4  f ∞,R 1 + 2 h2 h  (ν)  +

a



a+h

q1

(a + h − z)

q(ν−1)

dz

 ν  sup  D∗a f  p,R a∈R

a

(z − (a − h))

q(ν−1)

q1 dz

  ν  sup  Da− f  p,R ≤ a∈R

a−h

4  f ∞,R M + 2 2 h h  (ν)





2h ν− p 1

=

1

(q (ν − 1) + 1) q

4  f ∞,R 2Mh ν−2− p + 1 . h2  (ν) (q (ν − 1) + 1) q 1

(5.69)

That is       f (a) ≤ 4 f ∞,R + h2





2M

h ν−2− p , 1

 (ν) (q (ν − 1) + 1)

1 q

(5.70)

∀ a ∈ R, ∀ h > 0. I.e. it holds     f 

∞,R

4  f ∞,R ≤ + h2

∀ h > 0, under 2 + Call

1 p





2M  (ν) (q (ν − 1) + 1)

h ν−2− p , 1

1 q

(5.71)

< ν ≤ 3. ξ := 4  f ∞,R , ψ :=

both are greater than zero.

2M 1 (ν)(q(ν−1)+1) q

,

(5.72)

5.2 Main Results

75

Set also ϕ := ν − 2 − 1p > 0. We consider the function γ (h) :=

ξ + ψh ϕ , ∀ h > 0. h2

(5.73)

We have as in the proof of Theorem 5.4 that γ has a global minimum at  h0 = which is

2ξ ϕψ

1 ϕ+2

,

(5.74)

ϕ  2  ϕ+2 ψ ϕ+2 ξ γ (h 0 ) = (ϕ + 2) . ϕ 2

(5.75)

Consequently  γ (h 0 ) =



4  f ∞,R ν−2−



ν−2− 1p ν− 1p







1 p



M

2 ν− 1p



1

 (ν) (q (ν − 1) + 1) q

ν−

1 . p (5.76)

We have proved that

    f 

∞,R ≤





4  f ∞,R



ν−2− 1p ν− 1p



ν − 2 − 1p



⎛ ⎝

⎞ M 1

⎠

 2 ν− 1p

 (ν) (q (ν − 1) + 1) q

 ν−

1 p

.

(5.77) 

The theorem is proved. We give

Corollary 5.6 (case of ν = 2.5) Let f ∈ C 3 (R, X ), where (X, ·) is a Banach space. We assume that  f ∞,R < ∞, and that     2.5  2.5 K 2.5 := max  D∗a f (z)∞,R2 ,  Da− f (z)∞,R2 < ∞, where (a, z) ∈ R2 . Then     f 

∞,R

and

    f 

∞,R

(5.78)

 0.6 ≤ 1.21136 (K 2.5 )0.4  f ∞,R ,

(5.79)

 0.2 ≤ 1.44713 (K 2.5 )0.8  f ∞,R .

(5.80)

    That is  f  ∞,R ,  f  ∞,R < ∞.

76

5 General Fractional Landau Inequalities



Proof By Theorem 5.4. We finish with

Corollary 5.7 (case of ν = 2.5, p = q = 2) Let f ∈ C 3 (R, X ), where (X, ·) is a Banach space. We assume that  f ∞,R < ∞, and that M2.5

   2.5   2.5          := max sup D∗a f 2,R , sup Da− f 2,R < ∞. a∈R

Then

(5.81)

a∈R

    f 

∞,R

 ≤ 1.226583057 M2.5  f ∞,R .

(5.82)

  That is  f  ∞,R < ∞. Proof By Theorem 5.5.



References 1. Aglic Aljinovic, A., Marangunic, LJ., Pecaric, J.: On Landau type inequalities via Ostrowski inequalities. Nonlinear Funct. Anal. Appl. 10(4), pp. 565–579 (2005) 2. Anastassiou, G.: Fractional Differentiation Inequalities. Research Monograph. Springer, New York (2009) 3. Anastassiou, G.A.: Advances on Fractional Inequalities. Springer, New York (2011) 4. Anastassiou, G.A.: Intelligent Computations: Abstract Fractional Calculus, Inequalities, Approximations. Springer, Heidelberg, New York (2018) 5. Anastassiou, G.A.: Abstract Fractional Landau Inequalities. Ann. Univ. Sci. Budapest, Sect. Comp. (2020) 6. Barnett, N.S., Dragomir, S.S.: Some Landau type inequalities for functions whose derivatives are of locally bounded variation. Tamkang J. Math. 37(4), 301–308 (2006) 7. Ditzian, Z.: Remarks, questions and conjectures on Landau–Kolmogorov-type inequalities. Math. Inequal. Appl. 3, 15–24 (2000) 8. Hardy, G.H., Littlewood, J.E.: Some integral inequalities connected with the calculus of variations. Quart. J. Math. Oxford Ser. 3, 241–252 (1932) 9. Hardy, G.H., Landau, E., Littlewood, J.E.: Some inequalities satisfied by the integrals or derivatives of real or analytic functions. Math. Z. 39, 677–695 (1935)  2   10. Kallman R.R., Rota, G.C.: On the inequality  f   ≤ 4  f  ·  f  . In: O. Shisha, (ed.) Inequalities, vol. II, pp. 187–192. Academic Press, New York (1970) 11. Kraljevic, H., Pecaric, J.: Some Landau’s type inequalities for infinitesimal generators. Aequationes Mathematicae 40, 147–153 (1990) 12. Landau, E.: Einige Ungleichungen für zweimal differentzierban funktionen. Proc. London Math. Soc. 13, 43–49 (1913) 13. Mikusinski, J.: The Bochner Integral. Academic Press, New York (1978) 14. Shilov, G.E.: Elementary Functional Analysis. Dover Publications Inc., New York (1996) 15. Xiao, Y.: Landau type inequalities for Banach space valued functions. J. Math. Inequal. 7(1), 103–114 (2013)

Chapter 6

Abstract Fractional Landau Inequalities

We present uniform and L p mixed Caputo-Bochner abstract generalized fractional Landau inequalities over R of fractional orders 2 < α ≤ 3. These estimate the size of first and second derivatives of a composition with a Banach space valued function over R. We give applications when α = 2.5. It follows [5].

6.1 Introduction Let p ∈ [1, ∞], I = R+ or I = R and f : I → R is twice differentiable with f, f  ∈ L p (I ), then f  ∈ L p (I ). Moreover, there exists a constant C p (I ) > 0 independent of f , such that 1  1    f  ≤ C p (I )  f  2  f   2 , (6.1) p,I

p,I

p,I

where · p,I is the p-norm on the interval I , see [1, 6]. The research on these inequalities started by Landau [11] in 1913. For the case of p = ∞ he proved that C∞ (R+ ) = 2 and C∞ (R) =

√ 2,

(6.2)

are the best constants in (6.1). In 1932, Hardy and Littlewood [8] proved (6.1) for p = 2, with the best constants C2 (R+ ) =

√

2 and C2 (R) = 1.

(6.3)

In 1935, Hardy, Landau and Littlewood [9] showed that the best constants C p (R+ ) in (6.1) satisfies the estimate © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_6

77

78

6 Abstract Fractional Landau Inequalities

C p (R+ ) ≤ 2, for p ∈ [1, ∞),

(6.4)

which yields C p (R) ≤ 2 for p ∈ [1, ∞). √ In fact, in [7, 10] was shown that C p (R) ≤ 2. We need the following concepts from abstract generalized fractional calculus. Our integrals next are of Bochner type [12]. We need Definition 6.1 ([4], p. 104) Let [a, b] ⊂ R, (X, ·) a Banach space, g ∈ C 1 ([a, b]) and increasing, f ∈ C ([a, b] , X ), ν > 0. We define the left Riemann-Liouville generalized fractional Bochner integral operator 

 ν f (x) := Ja;g

1  (ν)



x

(g (x) − g (z))ν−1 g  (z) f (z) dz,

(6.5)

a

∀ x ∈ [a, b], where  is the gamma function. The last integral is of Bochner type. Since f∈ C ([a,  b] , X ), then f ∈ L ∞ ν ([a, b] , X ). By Theorem 4.10, p. 98, [4], we get that Ja;g f ∈ C ([a, b] , X ). Above   0 ν we set Ja;g f := f and see that Ja;g f (a) = 0. We need Definition 6.2 ([4], p. 105) Let [a, b] ⊂ R, (X, ·) a Banach space, g ∈ C 1 ([a, b]) and increasing, f ∈ C ([a, b] , X ), ν > 0. We define the right Riemann-Liouville generalized fractional Bochner integral operator 

 ν f (x) := Jb−;g

1  (ν)



b

(g (z) − g (x))ν−1 g  (z) f (z) dz,

(6.6)

x

∀ x ∈ [a, b], where  is the gamma function. The last integral is of Bochner type. Since f ∈ C  X ), then f ∈ L ∞ ([a, b] ,  ν ([a, b] , X ). By Theorem 4.11, p. 101, [4], we get that Jb−;g f ∈ C ([a, b] , X ).   0 ν Above we set Jb−;g f := f and see that Jb−;g f (b) = 0. We also need Definition 6.3 ([4], p. 106) Let α > 0, α = n, · the ceiling of the number. Let f ∈ C n ([a, b] , X ), where [a, b] ⊂ R, and (X, ·) is a Banach space. Let g ∈ C 1 ([a, b]) , strictly increasing, such that g −1 ∈ C n ([g (a) , g (b)]) . We define the left generalized g-fractional derivative X -valued of f of order α as follows:

6.1 Introduction



 α f (x) := Da+;g

79

1  (n − α)



x

(n)  (g (x) − g (t))n−α−1 g  (t) f ◦ g −1 (g (t)) dt,

a

(6.7) ∀ x ∈ [a, b]. The last integral is of Bochner type. Ordinary vector valued derivative is as in [13], similar one.   to numerical α If α ∈ / N, by Theorem 4.10, p. 98, [4], we have that Da+;g f ∈ C ([a, b] , X ). We see that    (n)   α n−α Ja;g f ◦ g −1 ◦ g (x) = Da+;g f (x) , ∀ x ∈ [a, b] . (6.8) We set n f (x) := Da+;g



f ◦ g −1

n

 ◦ g (x) ∈ C ([a, b] , X ) , n ∈ N,

(6.9)

0 f (x) = f (x) , ∀ x ∈ [a, b] . Da+;g

When g = id, then

α α α f = Da+;id f = D∗a f, Da+;g

(6.10)

the usual left X -valued Caputo fractional derivative, see [4], Chap. 1 We mention Definition 6.4 ([4], p. 107) Let α > 0, α = n, · the ceiling of the number. Let f ∈ C n ([a, b] , X ), where [a, b] ⊂ R, and (X, ·) is a Banach space. Let g ∈ C 1 ([a, b]) , strictly increasing, such that g −1 ∈ C n ([g (a) , g (b)]) . We define the right generalized g-fractional derivative X -valued of f of order α as follows:  b (n)    α (−1)n Db−;g f (x) := (g (t) − g (x))n−α−1 g  (t) f ◦ g −1 (g (t)) dt,  (n − α) x (6.11) ∀ x ∈ [a, b]. The last integral is of Bochner type.   α f ∈ C ([a, b] , X ). If α ∈ / N, by Theorem 4.11, p. 101, [4], we have that Db−;g We see that    (n)  α  n−α ◦ g (x) = Db−;g f (x) , a ≤ x ≤ b. (6.12) Jb−;g (−1)n f ◦ g −1 We set n f (x) := (−1)n Db−;g



f ◦ g −1

n

 ◦ g (x) ∈ C ([a, b] , X ) , n ∈ N,

0 f (x) := f (x) , ∀ x ∈ [a, b] . Db−;g

When g = id, then

(6.13)

80

6 Abstract Fractional Landau Inequalities α α α Db−;g f (x) = Db−;id f (x) = Db− f,

(6.14)

the usual right X -valued Caputo fractional derivative, see [4], Chap. 2. We mention the generalized left fractional Taylor formula: Theorem 6.5 ([4], p. 107) Let α > 0, n = α , and f ∈ C n ([a, b] , X ), where [a, b] ⊂ R and (X, ·) is a Banach space. Let g ∈ C 1 ([a, b]), strictly increasing, such that g −1 ∈ C n ([g (a) , g (b)]), a ≤ x ≤ b. Then f (x) = f (a) +

n−1  (g (x) − g (a))i 

i!

i=1

1  (α)

 a

n−1  (g (x) − g (a))i 

i!

i=1

1  (α)



(i)

(g (a)) +

  α f (t) dt = (g (x) − g (t))α−1 g  (t) Da+;g

x

f (a) +

f ◦ g −1

g(x) g(a)

(g (x) − z)α−1



f ◦ g −1

(i)

(g (a)) +

  α Da+;g f ◦ g −1 (z) dz.

(6.15)

We also mention the generalized right fractional Taylor formula: Theorem 6.6 ([4], p. 108) Let α > 0, n = α , and f ∈ C n ([a, b] , X ), where [a, b] ⊂ R and (X, ·) is a Banach space. Let g ∈ C 1 ([a, b]), strictly increasing, such that g −1 ∈ C n ([g (a) , g (b)]), a ≤ x ≤ b. Then f (x) = f (b) +

n−1  (g (x) − g (b))i 

i!

i=1

1  (α)



b x

f (b) + 

(i)

(g (b)) +

  α f (t) dt = (g (t) − g (x))α−1 g  (t) Db−;g

n−1  (g (x) − g (b))i 

i!

i=1

1  (α)

f ◦ g −1

g(b)

g(x)

(z − g (x))α−1

By convention we suppose that



f ◦ g −1

(i)

(g (b)) +

  α Db−;g f ◦ g −1 (z) dz.

(6.16)

6.1 Introduction

81

 

 Dxα0 +;g f (x) = 0, for x < x0 , Dxα0 −;g

(6.17)



f (x) = 0, for x > x0 ,

for any x, x0 ∈ [a, b] . The author has already done an extensive amount of work on fractional Landau inequalities, see [3], and on abstract fractional Landau inequalities, see [4]. However there the proving methods came out of applications of fractional Ostrowski inequalities [2, 4] and the derived inequalities were for small fractional orders, i.e. α ∈ (0, 1). Usually there the domains where [A, +∞) or (−∞, B], with A, B ∈ R and in one mixed case the domain was all of R. In this chapter with less assumptions we establish uniform and L p type mixed Caputo–Bochner abstract generalized fractional Landau inequalities over R for fractional orders 2 < α ≤ 3. The method of proving is based on left and right Caputo– Bochner generalized fractional Taylor’s formulae with integral remainder, see Theorems 6.5, 6.6. We give also an applications for α = 2.5. Certainly we are also inspired by [3, 4].

6.2 Main Results We give the following abstract mixed generalized fractional Landau inequalities over R. Theorem 6.7 Let 2 < α ≤ 3 and f ∈ C 3 (R, X ), where (X, ·) is a Banach space. Let g ∈ C 1 (R), strictly increasing, such that g −1 ∈ C 3 (g (R)). We assume that  f ∞,R < ∞ and that     α K := max  Da+;g f ◦ g −1 (z)∞,R×g(R) ,

 −1    < ∞, f ◦ g (z) a−;g ∞,R×g(R)

 α  D

(6.18)

where (a, z) ∈ R × g (R). Then       f ◦ g −1 ◦ g 

 ∞,R

≤α

K  (α + 1)

α1 

 f ∞,R α−1

α−1 α

,

(6.19)

and       f ◦ g −1 ◦ g 

 ∞,R

≤α

K  (α + 1)

α2 

4  f ∞,R α−2

α−2 α

.

(6.20)

82

6 Abstract Fractional Landau Inequalities

     That is  f ◦ g −1 ◦ g 

∞,R

     ,  f ◦ g −1 ◦ g 

∞,R

< ∞.

Proof Here 2 < α ≤ 3, i.e. α = 3. Let f ∈ C 3 (R, X ), where (X, ·) is a Banach space, a ∈ R is fixed momentarily. We need the following abstract generalized fractional Taylor formulae for n = 3. By Theorem 6.5 we get    (g (x) − g (a))2  f ◦ g −1 (g (a)) f (x) − f (a) = (g (x) − g (a)) f ◦ g −1 (g (a)) + 2

1 +  (α)



g(x)

g(a)

(g (x) − z)

 α−1



α Da+;g f ◦g

 −1

(6.21)

(z) dz, ∀ x ≥ a.

And by Theorem 6.6 we get    (g (x) − g (a))2  f ◦ g −1 (g (a)) f (x) − f (a) = (g (x) − g (a)) f ◦ g −1 (g (a)) + 2

1 +  (α)



g(a) g(x)

 α   f ◦ g −1 (z) dz, ∀ x ≤ a. (z − g (x))α−1 Da−;g

(6.22)

Let x1 > a, then    (g (x1 ) − g (a))2  f ◦ g −1 (g (a)) = (g (x1 ) − g (a)) f ◦ g −1 (g (a)) + 2 1 ( f (x1 ) − f (a)) −  (α)



g(x1 ) g(a)

(g (x1 ) − z)α−1



  α Da+;g f ◦ g −1 (z) dz =: A, (6.23)

and let x2 < a, then    (g (x2 ) − g (a))2  f ◦ g −1 (g (a)) (g (x2 ) − g (a)) f ◦ g −1 (g (a)) + 2 1 = ( f (x2 ) − f (a)) −  (α)



g(a)

g(x2 )

(z − g (x2 ))α−1



  α Da−;g f ◦ g −1 (z) dz =: B.

(6.24) Let h > 0, we can choose x1 such that g (x1 ) − g (a) = h and we can choose x2 such that g (a) − g (x2 ) = h. That is g (x1 ) = g (a) + h and g (x2 ) = g (a) − h, and g (x2 ) − g (a) = −h. Furthermore it holds g (x2 ) − g (x1 ) = −2h. We can rewrite (6.23) as    h2  f ◦ g −1 (g (a)) = A, h f ◦ g −1 (g (a)) + 2

(6.25)

6.2 Main Results

83

and we can rewrite (6.24) as    h2  f ◦ g −1 (g (a)) = B. − h f ◦ g −1 (g (a)) + 2

(6.26)

Solving the system of (6.25) and (6.26) we find  , f ◦ g −1 (g (a)) = A−B 2h and   f ◦ g −1 (g (a)) = A+B . h2



(6.27)

      α We assumed that  Da+;g f ◦ g −1 (z) , ∞,R×g(R)       α f ◦ g −1 (z) < ∞.  Da−;g ∞,R×g(R)

     1 A − B ,  f ◦ g −1 (g (a)) = 2h and       f ◦ g −1 (g (a)) ≤ h12 (A + B) .

We obtain

(6.28)

We get    g(x1 )    −1   α 1 α−1  A = ( f (x1 ) − f (a)) − Da+;g f ◦ g (z) dz  (g (x1 ) − z)   (α) g(a) ≤ 2  f ∞,R +

1  (α)



g(a)

2  f ∞,R + 2  f ∞,R +

g(x1 )

 α    f ◦ g −1 (z) dz ≤ (g (x1 ) − z)α−1  Da+;g

K  (α)



g(x1 ) g(a)

(g (x1 ) − z)α−1 dz

=

K K hα. (g (x1 ) − g (a))α = 2  f ∞,R +  (α + 1)  (α + 1) (6.29)

That is A ≤ 2  f ∞,R +

K h α , h > 0.  (α + 1)

(6.30)

Similarly, it holds    g(a)    α  −1  1 α−1  B = ( f (x2 ) − f (a)) − Da−;g f ◦ g (z) dz  (z − g (x2 ))   (α) g(x2 )

84

6 Abstract Fractional Landau Inequalities

1 ≤ 2  f ∞,R +  (α)



g(a) g(x2 )

2  f ∞,R +

 α    f ◦ g −1 (z) dz ≤ (z − g (x2 ))α−1  Da−;g

K  (α)



g(a) g(x2 )

(z − g (x2 ))α−1 dz

=

K K hα. (g (a) − g (x2 ))α = 2  f ∞,R +  (α + 1)  (α + 1) (6.31)

2  f ∞,R + That is

B ≤ 2  f ∞,R +

K h α , h > 0.  (α + 1)

(6.32)

Furthermore we have  A + B ≤ 4  f ∞,R +

2K h α , h > 0.  (α + 1)

(6.33)

We also notice that  A − B =   g(x1 )   α    f (x1 ) − f (a) − 1 f ◦ g −1 (z) dz (g (x1 ) − z)α−1 Da+;g   (α) g(a) − f (x2 ) + f (a) +

1  (α)

 f (x1 ) − f (x2 ) +

1  (α)

 +

g(a) g(x2 )

g(a) g(x2 )





g(x1 )

g(a)

(z − g (x2 ))α−1

g(x1 )

g(a)

(z − g (x2 ))

K 2  f ∞,R +  (α) 2  f ∞,R +



α−1

    α Da−;g f ◦ g −1 (z) dz  ≤

 α    f ◦ g −1 (z) dz (g (x1 ) − z)α−1  Da+;g  α  D

(g (x1 ) − z)



a−;g

α−1

f ◦g

−1

 dz +



(6.34)

  (z) dz ≤

g(a)

g(x2 )

(z − g (x2 ))

α−1

 dz =

  K (g (x1 ) − g (a))α + (g (a) − g (x2 ))α =  (α + 1) 2  f ∞,R +

That is



2K h α .  (α + 1)

A − B K hα , h > 0. ≤  f ∞,R + 2  (α + 1)

(6.35)

6.2 Main Results

85

Consequently we obtain   ((6.28.), (6.35.))   f ∞,R   K h α−1 ≤ + (α+1) ,  f ◦ g −1 (g (a)) h and  ((6.28.), (6.33.))   4 f ∞,R   h α−2 ≤ + 2K ,  f ◦ g −1 (g (a)) h2 (α+1) h > 0, for any a ∈ R. Hence       f ◦ g −1 ◦ g 

∞,R

and       f ◦ g −1 ◦ g  true ∀ h > 0, 2 < α ≤ 3. Call

≤

 f ∞,R h

≤

4 f ∞,R h2

+

(6.36)

K h α−1 , (α+1)

(6.37) ∞,R

+

2K h α−2 , (α+1)

μ :=  f ∞,R , θ=

(6.38)

K , (α+1)

both are greater than zero. Set also ρ := α − 1 > 1. We consider the function y (h) := We have

μ + θh ρ , ∀ h > 0. h

y  (h) = −

then

(6.39)

μ + ρθh ρ−1 = 0, h2

ρθh ρ+1 = μ,

with a unique solution

 h 0 := h crit.no =

We have that

μ ρθ

1 ρ+1

.

(6.40)

y  (h) = 2μh −3 + ρ (ρ − 1) θh ρ−2 .

(6.41)

We observe that 

μ y (h 0 ) = 2μ ρθ 

3 − ρ+1



μ + ρ (ρ − 1) θ ρθ

((ρ+1)−3) ρ+1

=

86

6 Abstract Fractional Landau Inequalities



μ ρθ

3 − ρ+1

 [2μ + μ (ρ − 1)] = μ

Therefore y has a global minimum at h 0 =

3 − ρ+1

μ ρθ

1   ρ+1

μ ρθ

(ρ + 1) > 0.

(6.42)

, which is

ρ  ρ+1 μ μ = y (h 0 ) =   1 + θ ρθ μ ρ+1

ρθ

1

(ρθ) ρ+1  (θμρ )

1 ρ+1

ρ

1 ρ+1

+

μ 1

μ ρ+1 1

ρ

+

θμ ρ+1 ρ

=

ρ

ρ ρ+1 θ ρ+1



 = (θμρ )

ρ

ρ ρ+1

1 ρ+1

ρ+1



ρ

ρ ρ+1

=

(6.43)

ρ

(θμρ ) ρ+1 (ρ + 1) ρ− ρ+1 . 1

That is

ρ

y (h 0 ) = (θμρ ) ρ+1 (ρ + 1) ρ− ρ+1 . 1

(6.44)

Consequently  y (h 0 ) =

K  f α−1 ∞,R  (α + 1)

α1

α (α − 1)−(

α−1 α

).

(6.45)

We have proved that       f ◦ g −1 ◦ g 

 ∞,R

≤

Next call

K  f α−1 ∞,R  (α + 1)

α1

α (α − 1)−(

α−1 α

).

(6.46)

ξ := 4  f ∞,R , ψ=

2K , (α+1)

(6.47)

both are greater than zero. Set also ϕ := α − 2 > 0. We consider the function γ (h) :=

ξ + ψh ϕ = ξh −2 + ψh ϕ , ∀ h > 0. h2

(6.48)

6.2 Main Results

87

We have

γ  (h) = −2ξh −3 + ϕψh ϕ−1 = 0,

then

ϕψh ϕ+2 = 2ξ,

with a unique solution

 h 0 := h crit.no =

We have that

2ξ ϕψ

1 ϕ+2

.

(6.49)

γ  (h) = 6ξh −4 + ϕ (ϕ − 1) ψh ϕ−2 .

(6.50)

We see that γ  (h 0 ) = 6ξ 

2ξ ϕψ

4 − ϕ+2



2ξ ϕψ

4 − ϕ+2

 + ϕ (ϕ − 1) ψ 

[6ξ + (ϕ − 1) 2ξ] = 2ξ

Therefore γ has a global minimum at h 0 =  γ (h 0 ) = ξ 

2ξ ϕψ

2 − ϕ+2

ξ+ψ

2ξ ϕψ

2 − ϕ+2



2ξ ϕψ

2ξ ϕψ

4 − ϕ+2

1  ϕ+2



2ξ ϕψ

(ϕ+2)−4 ϕ+2

=

(ϕ + 2) > 0.

(6.51)

, which is

2ξ +ψ ϕψ

ϕ+2−2 ϕ+2

=

  2  2 2ξ ξ ξ − ϕ+2 2 − ϕ+2 = (ϕ + 2) = ϕψ ϕ ϕ ψ

(6.52)

ϕ  2  ϕ+2 ψ ϕ+2 ξ (ϕ + 2) . ϕ 2

That is γ (h 0 ) = Consequently

 γ (h 0 ) =

We have proved that

ϕ  2  ϕ+2 ψ ϕ+2 ξ (ϕ + 2) . ϕ 2

4  f ∞,R α−2

 α−2 α

K  (α + 1)

α2

(6.53)

α.

(6.54)

88

6 Abstract Fractional Landau Inequalities

      f ◦ g −1 ◦ g 

 ∞,R

≤

4  f ∞,R α−2

 α−2 α

K  (α + 1)

α2

α.

(6.55) 

The theorem is established. We also give an L p analog of a generalized fractional Landau inequality

Theorem 6.8 Let p, q > 1 : 1p + q1 = 1, 2 < α ≤ 3 and f ∈ C 3 (R, X ), where (X, ·) is a Banach space. Let g ∈ C 1 (R), strictly increasing, such that g −1 ∈ C 3 (g (R)). We assume that  f ∞,R < ∞, and that   α    M := max sup  Da+;g f ◦ g −1 (z) p,g(R) , a∈R

  α    sup  Da−;g f ◦ g −1 (z) p,g(R) < ∞.

(6.56)

a∈R

Then (1)

      f ◦ g −1 ◦ g  

   1 M α− 1 p  (α) (q (α − 1) + 1) q

∞,R

1 α− 1p

1 p



∞,R

 f ∞,R

2 α− 1p



α−1− 1p α− 1p



1 p

,

(6.57)

< α ≤ 3, we have

      f ◦ g −1 ◦ g 

     That is  f ◦ g −1 ◦ g 



α−1−

and (2) under the additional assumption 2 +

   1 M α− 1 p  (α) (q (α − 1) + 1) q



≤

∞,R



≤ 

4  f ∞,R α−2−

     ,  f ◦ g −1 ◦ g 

∞,R

1 p



α−2− 1p α− 1p



.

(6.58)

< ∞.

Proof We continue with the proof of Theorem 6.7. By (6.23) we have    g(x1 )    α  −1  1 α−1  A = ( f (x1 ) − f (a)) − Da+;g f ◦ g (z) dz  (g (x1 ) − z)   (α) g(a)

6.2 Main Results

89



1 ≤ 2  f ∞,R +  (α)

 α    f ◦ g −1 (z) dz (g (x1 ) − z)α−1  Da+;g

g(x1 )

g(a)

1 ≤ 2  f ∞,R +  (α) 

g(x1 )





a+;g

f ◦g

(g (x1 ) − z)

−1

(q(α−1)+1)

p (z) dz

1p

 sup  D α

≤

(6.59)



a+;g

a∈R

dz

   f ◦ g −1 (z) p,g(R) ≤

a+;g



1



 α  D

1 (g (x1 ) − g (a)) q 2  f ∞,R +  (α) (q (α − 1) + 1) q1 h α− p 1 2  f ∞,R +  (α) (q (α − 1) + 1) q1

q1

q(α−1)

g(a)

 α  D

g(a)

g(x1 )

f ◦g

h α− p

−1



 (z)

p,g(R)

1

≤ 2  f ∞,R +

1

 (α) (q (α − 1) + 1) q

M.

That is A ≤ 2  f ∞,R +

M

h α− p , h > 0. 1

 (α) (q (α − 1) + 1)

1 q

(6.60)

Similarly, from (6.24) we get   B =  ( f (x2 ) − f (a)) − ≤ 2  f ∞,R +

1  (α)

1  (α)



g(a) g(x2 )



g(x2 )

g(a)

g(x2 )

 α  D

(z − g (x2 ))α−1

a−;g



g(a)

    α Da−;g f ◦ g −1 (z) dz  

q1

(z − g (x2 ))

q(α−1)

g(x2 )



f ◦g

1 (g (a) − g (x2 ))α− p  (α) (q (α − 1) + 1) q1 1

2  f ∞,R +



 α    f ◦ g −1 (z) dz (z − g (x2 ))α−1  Da−;g

1 ≤ 2  f ∞,R +  (α) 

g(a)

−1



p (z) dz

1p

dz

≤

  α  −1       Da−;g f ◦ g sup (z) p,g(R) a∈R

(6.61)

90

6 Abstract Fractional Landau Inequalities

h α− p 1

≤ 2  f ∞,R +

M.

1

 (α) (q (α − 1) + 1) q

That is B ≤ 2  f ∞,R +

M

1

 (α) (q (α − 1) + 1)

Hence it holds A + B ≤ A + B 4  f ∞,R +

(6.62)

(by (6.60.), (6.62.))

≤

2M

h α− p , h > 0. 1

 (α) (q (α − 1) + 1)

Furthermore we have

h α− p , h > 0.

1 q

1 q

(6.63)

(6.34.)

A − B ≤ 2  f ∞,R + 1  (α)



 +

g(x1 ) g(a)

g(a)

(g (x1 ) − z)

(z − g (x2 ))

q(α−1)

g(x2 )

q1

q(α−1)

dz

q1 dz

M 2  f ∞,R +  (α)

 α  D

a+;g

 α  D

a−;g





   f ◦ g −1 (z) p,g(R)

f ◦g

−1



 (z)

 p,g(R)

≤



2h α− p 1

1

(q (α − 1) + 1) q

.

(6.64)

We have proved that A − B M α− 1p ≤  f ∞,R + , h > 0. 1 h 2  (α) (q (α − 1) + 1) q

(6.65)

From (6.27), (6.65), we have   f    M   ∞,R α− 1p −1 + ,  f ◦ g −1 (g (a)) ≤ 1 h h  (α) (q (α − 1) + 1) q

(6.66)

h > 0, any a ∈ R. And from (6.27), (6.63), we get that   4  f   2M   ∞,R α− 1p −2 + , (6.67)  f ◦ g −1 (g (a)) ≤ 1 h 2 h  (α) (q (α − 1) + 1) q

6.2 Main Results

91

h > 0, any a ∈ R. Hence       f ◦ g −1 ◦ g 

∞,R

 f ∞,R + ≤ h





M

h

1

 (α) (q (α − 1) + 1) q

  α− 1p −1

,

(6.68)

and       f ◦ g −1 ◦ g 

∞,R



4  f ∞,R ≤ + h2



2M 1

 (α) (q (α − 1) + 1) q

h

  α− 1p −2

,

(6.69)

true ∀ h > 0, 2 < α ≤ 3, p, q > 1 : 1p + q1 = 1. In (6.69) we restrict ourselves to 2 + 1p < α ≤ 3. Call μ :=  f ∞,R , θ= both are greater than zero. Set also ρ := α − 1 − 1p > We consider the function

1 q

(6.70)

,

M 1 (α)(q(α−1)+1) q

> 0.

y (h) :=

μ + θh ρ , ∀ h > 0. h

(6.71)

As in the proof of Theorem 6.7 it has only one critical number  h 0 := h crit.no =

μ ρθ

1 ρ+1

,

(6.72)

and a global minimum ρ

ρ

y (h 0 ) = θ ρ+1 μ ρ+1 (ρ + 1) ρ− ρ+1 . 1

(6.73)

Consequently  y (h 0 ) =



M 1

 (α) (q (α − 1) + 1) q



1

(α− 1p )    f ∞,R 

  1 1 − α− α−1− p p We have proved that (see (6.68))

α−1− 1p α− 1p

α−1− 1p α− 1p





.

(6.74)

92

6 Abstract Fractional Landau Inequalities

      f ◦ g −1 ◦ g  



M

(

1 α− 1p

)



≤ 

 f ∞,R α−1−

1

 (α) (q (α − 1) + 1) q We also call

∞,R



α−1− 1p α− 1p



1 p

 1 . α− p

(6.75)

ξ := 4  f ∞,R , ψ=

2M 1 (α)(q(α−1)+1) q

(6.76)

,

both are greater than zero. Set also ϕ := α − 2 − 1p > 0. We consider the function γ (h) :=

ξ + ψh ϕ , ∀ h > 0. h2

(6.77)

As in the proof of Theorem 6.7, γ has a global minimum at  h0 = which is

2ξ ϕψ

1 ϕ+2

,

(6.78)

ϕ  2  ϕ+2 ψ ϕ+2 ξ γ (h 0 ) = (ϕ + 2) . ϕ 2

(6.79)

Consequently  γ (h 0 ) =



4  f ∞,R α−2−



α−2− 1p α− 1p

1 p









M

2 α− 1p



1

 (α) (q (α − 1) + 1) q

α−

1 . p (6.80)

We have proved that (see (6.69))       f ◦ g −1 ◦ g  



4  f ∞,R α−2−



α−2− 1p α− 1p



1 p



∞,R

≤ 



M 1

 (α) (q (α − 1) + 1) q

2 α− 1p

 1 α− . p

The theorem is established. Next we apply Theorems 6.7, 6.8 for g (t) = et , t ∈ R and α = 2.5.

(6.81) 

6.2 Main Results

93

Corollary 6.9 Let f ∈ C 3 (R, X ), where (X, ·) is a Banach space. We assume that  f ∞,R < ∞, and that     2.5 K 2.5 := max  Da+;e ◦ ln (z)∞,R×(0,∞) , t f  2.5  D

a−;et

   f ◦ ln (z)∞,R×(0,∞) < ∞,

(6.82)

where (a, z) ∈ R × (0, ∞). Then

and

   0.6 ( f ◦ ln) ◦ et  ≤ 1.21136 (K 2.5 )0.4  f ∞,R , ∞,R

(6.83)

   0.2 ( f ◦ ln) ◦ et  ≤ 1.44713 (K 2.5 )0.8  f ∞,R . ∞,R

(6.84)

    That is ( f ◦ ln) ◦ et ∞,R , ( f ◦ ln) ◦ et ∞,R < ∞. 

Proof By Theorem 6.7. We finish with

Corollary 6.10 (case of g (t) = et , α = 2.5, p = q = 2) Let f ∈ C 3 (R, X ), where (X, ·) is a Banach space. We assume that  f ∞,R < ∞, and that   2.5    M2.5 := max sup  Da+;e ◦ ln (z)2,(0,∞) , t f a∈R

  2.5      < ∞. sup  Da−;e f ◦ ln (z) t 2,(0,∞)

(6.85)

a∈R

Then    ( f ◦ ln) ◦ et  ≤ 1.226583057 M2.5  f ∞,R < ∞. ∞,R Proof By Theorem 6.8, (6.57).

(6.86) 

References 1. A. Aglic Aljinovic, L. Marangunic, J. Pecaric, On Landau type inequalities via Ostrowski inequalities. Nonlinear Funct. Anal. Appl. 10(4) 565–579 (2005) 2. Anastassiou, G.: Fractional Differentiation Inequalities. Research Monograph. Springer, New York (2009)

94

6 Abstract Fractional Landau Inequalities

3. Anastassiou, G.A.: Advances on Fractional Inequalities. Springer, New York (2011) 4. Anastassiou, G.A.: Intelligent Computations: Abstract Fractional Calculus, Inequalities Approximations. Springer, Heidelberg (2018) 5. G.A. Anastassiou, G.A.: Abstract Generalized Fractional Landau inequalities over R, Constructive Mathematical Analysis, accepted (2020) 6. Barnett, N.S., Dragomir, S.S.: Some Landau type inequalities for functions whose derivatives are of locally bounded variation. Tamkang J. Math. 37(4), 301–308 (Winter 2006) 7. Ditzian, Z.: Remarks, questions and conjectures on Landau-Kolmogorov-type inequalities. Math. Inequal. Appl. 3, 15–24 (2000) 8. Hardy, G.H., Littlewood, J.E.: Some integral inequalities connected with the calculus of variations. Quart. J. Math. Oxford Ser. 3, 241–252 (1932) 9. Hardy, G.H., Landau, E., Littlewood, J.E.: Some inequalities satisfied by the integrals or derivatives of real or analytic functions. Math. Z. 39, 677–695 (1935) 10. Kallman, R.R., Rota, G.C.: On theinequality  2 ≤ 4 ·  . In: Shisha, O., (ed.), Inequalities , vol. II, pp. 187–192. Academic, New York (1970) 11. Landau, E.: Einige Ungleichungen für zweimal differentzierban funktionen. Proc. Lond. Math. Soc. 13, 43–49 (1913) 12. Mikusinski, J.: The Bochner Integral. Academic, New York (1978) 13. Shilov, G.E.: Elementary Functional Analysis. Dover Publications Inc., New York (1996)

Chapter 7

Fractional Landau Inequalities of Riemann–Liouville Type

We present uniform Riemann–Liouville left and right fractional Landau inequalities over R+ and R− , respectively, of fractional orders 1 < ν < 2 and 2 < ν < 3 and we estimate lower order fractional derivatives. These inequalities are sharp or nearly sharp with completely determined constants. We give applications when ν = 1.5. We finish with a related new Ostrowski like inequality for ν > 0, ν ∈ / N. It follows [1].

7.1 Introduction     Landau in 1913 [7] proved if f ∈ C 2 ([0, 1]),  f ∞ = 1,  f  ∞ = 4, then  f  ∞ ≤ 4, with 4 the best constant and the result is not necessarily true for an interval of length < 1. Now Landau’s inequality has taken the form [2]   f 

p,I

1    1 2  f 2 , ≤ C p (I )  f  p,I p,I

(7.1)

where · p,I is the p-norm on the interval I ; p ∈ [1, ∞], I = R+ or I = R, and f : I → R is twice differentiable with f, f  ∈ L p (I ). C p (I )√> 0 is independent of f . And the best constants are C∞ (R+ ) = 2 and C∞ (R) = 2. Research about Landau type inequalities has expanded to many different directions and it is a very active topic of mathematical activity. Here we are concerned about fractional Landau inequalities of Riemann–Liouville type. The main motivation to write this chapter is [8], and especially the result that follows: Let α be a complex number with Reα > 0. We define I Lα and D Lα to be the classical Riemann–Liouville fractional integration and differentiation operators of order α, © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_7

95

96

7 Fractional Landau Inequalities of Riemann–Liouville Type

I Lα f (x) =

1  (α)

D Lα f (x) =



x

(x − t)α−1 f (t) dt,

a

d n+1  n+1−α  I f (x) , d x n+1 L

where  is the gamma function and n = [Reα] (integral part). The above operators are defined in the natural domain of appropriate function spaces. We state Theorem 7.1 ([8]) Let I Lα and D Lα be the fractional Riemann–Liouville operators in 1 the Fréchet space L loc ([0, +∞)) with the usual topology. If α, β and γ are complex numbers for which 0 < Reα < Reβ < Reγ , for every seminorm · p there exist constant F1 ( p, α, β, γ ) and F2 ( p, α, β, γ ) that depend on p, α, β, and γ , with the properties that (a)    Re(γ −β)  γ  Re(β−α)   β  1 ([0, +∞)) , I L f  ≤ F1 ( p, α, β, γ )  I Lα f  pRe(γ −α)  I L f  pRe(γ −α) , f ∈ L loc p

(7.2) and

   Re(γ −β)  γ  Re(β−α)   β  D L f  ≤ F2 ( p, α, β, γ )  D Lα f  pRe(γ −α)  D L f  pRe(γ −α) , p

(7.3)

 γ γ f ∈ D D L (domain of D L ). If I mα = I mβ = I mγ , the constants F1 ( p, α, β, γ ) and F2 ( p, α, β, γ ) depend only on p and on an arbitrary integer n > Reγ . For more see [8]. Other inspirations are the papers [4, 5]. The above motivating results and references are existential without giving exact constants which are not the best possible. Their methods are based on functional analysis and semigroup theory, in particular dealing with moment inequalities for fractional powers of operators in Banach spaces. As a last motivation we also mention [3]. Our method is classical analytic and we will produce fractional Landau type inequalities, involving the Riemann–Liouville left and right fractional derivatives, that are sharp or nearly sharp and all constant are exactly calculated. We need the following background (based on [9], Sect. 7.2, and [6], Chap. 2). Let −∞ < a < b < ∞, the left and right Riemann–Liouville fractional integrals of order α ∈ C (R (α) > 0) are defined by  (x > a; R (α) > 0) and

Iaα



1 f (x) :=  (α)

 a

x

(x − t)α−1 f (t) dt

(7.4)

7.1 Introduction

97





1 f (x) :=  (α)

α Ib−



b

(t − x)α−1 f (t) dt

(7.5)

x

(x < b; R (α) > 0), respectively. The Riemann–Liouville left and right fractional derivatives of order α ∈ C (R (α) ≥ 0) are defined by 

 Daα y (x) :=



d dx

n



 Ian−α y (x) =



1  (n − α)

d dx

n 

x

(x − t)n−α−1 y (t) dt

a

(7.6)

(n = [R (α)] + 1; x > a) and  n  n−α   α  d Ib− y (x) Db− y (x) := (−1)n dx (−1)n =  (n − α)



d dx

n 

b

(t − x)n−α−1 y (t) dt

(7.7)

x

(n = [R (α)] + 1; x < b), respectively, where R (α) is the real part of α. In particular, when α = n ∈ Z+ , then  

  0  y (x) = y (x) ; Da0 y (x) = Db−

  n  y (x) = (−1)n y (n) (x) (n ∈ N). Dan y (x) = y (n) (x) , and Db−

We denote by 

Da−α



f (x) :=



Iaα



1 f (x) =  (α)

(x > a; R (α) > 0), and  −α   α  Db− f (x) := Ib− f (x) =



1  (α)

x

(x − t)α−1 f (t) dt

(7.8)

(t − x)α−1 f (t) dt

(7.9)

a



b x

(x < b; R (α) > 0). Based on [9], pp. 44–45, and [6], pp. 74–76 we will use the following two fractional Taylor theorems. The left one first: Theorem 7.2 Let ν > 0, ν ∈ / N and m = ν (ceiling of ν); ϕ, Daν ϕ ∈ C ([a, b]). Then ϕ (b) =

m  j=1

ν− j

Da ϕ (a) 1 (b − a)ν− j +  (ν − j + 1)  (ν)

 a

b

(b − t)ν−1 Daν ϕ (t) dt. (7.10)

98

7 Fractional Landau Inequalities of Riemann–Liouville Type

It follows the right one: ν Theorem 7.3 Let ν > 0, ν ∈ / N and m = ν; ϕ, Db− ϕ ∈ C ([a, b]). Then

ϕ (a) =

m  j=1

ν− j

Db− ϕ (b) 1 (b − a)ν− j +  (ν − j + 1)  (ν)



b

a

ν ϕ (t) dt. (t − a)ν−1 Db−

(7.11)

7.2 Main Results First we give left Riemann–Liouville fractional Landau type inequalities. Convention 7.4 Define Daν ϕ (x) := 0, for any x < a, for any ν > 0, ν ∈ / N. We present Theorem 7.5 Here 1 < ν < 2 and ϕ ∈ C B (R+ ) and Daν ϕ ∈ C B ([a, +∞)), ∀ a ∈ R+ , where C B means continuous and bounded functions. We assume that sup  Daν ϕ ∞,R+ < +∞.

a∈R+

Then sup Daν−1 ϕ (a) ≤

a∈R+



 (ν + 1) (ν − 1)ν−1

 ν1

1 ν

ϕ∞,R+

  sup  Daν ϕ ∞,R+

 ν−1 ν ,

(7.12)

a∈R+

a sharp inequality with a precisely determined constant. That is sup Daν−1 ϕ (a) < a∈R+

+∞. Proof For any b ∈ (a, +∞) we get (by Theorem 7.2) ϕ (b) =

Daν−1 ϕ (a) D ν−2 ϕ (a) (b − a)ν−1 + a (b − a)ν−2  (ν)  (ν − 1) +

1  (ν)

 a

b

(b − t)ν−1 Daν ϕ (t) dt.

Notice that (x ∈ [a, +∞)) Daν−2 ϕ and

(x) =

Da−(2−ν) ϕ

1 (x) =  (2 − ν)

 a

x

(x − t)1−ν ϕ (t) dt,

(7.13)

7.2 Main Results

ν−2 Da ϕ (x) ≤

= That is

99

 x  ϕ∞,[a,+∞) x 1 (x − t)1−ν |ϕ (t)| dt ≤ (x − t)1−ν dt  (2 − ν) a  (2 − ν) a

ϕ∞,[a,+∞) ϕ∞,[a,+∞) (x − a)2−ν = (x − a)2−ν .  (2 − ν) (2 − ν)  (3 − ν) ϕ∞,[a,+∞) ν−2 D ϕ (x) ≤ (x − a)2−ν , a  (3 − ν)

(7.14)

(7.15)

∀ x ∈ [a, +∞). Clearly, it is Daν−2 ϕ (a) = 0. Hence by (7.13) we get ϕ (b) =

Daν−1 ϕ (a) (b − a)ν−1 1 +  (ν)  (ν)



b

a

(b − t)ν−1 Daν ϕ (t) dt,

(7.16)

(b − t)ν−1 Daν ϕ (t) dt,

(7.17)

∀ b ∈ (a, +∞). So that Daν−1 ϕ (a) 1 (b − a)ν−1 = ϕ (b) −  (ν)  (ν)



b

a

∀ b ∈ (a, +∞). Hence (set h := b − a > 0)  ν−1 D ϕ (a) h ν−1 =  (ν) ϕ (b) − a

b

ν−1

(b − t)

a

  (ν) |ϕ (b)| + a

b

Daν ϕ

(t) dt ≤

(b − t)ν−1 Daν ϕ (t) dt ≤

   (ν) |ϕ (b)| +  Daν ϕ ∞,[a,+∞)



b

(b − t)ν−1 dt =

a

    hν (b − a)ν =  (ν) |ϕ (b)| +  Daν ϕ ∞,[a,+∞) .  (ν) |ϕ (b)| +  Daν ϕ ∞,[a,+∞) ν ν (7.18) That is we have  ν−1  hν D ϕ (a) h ν−1 ≤  (ν) |ϕ (b)| +  D ν ϕ  ≤ a a ∞,[a,+∞) ν

  ν  hν  (ν) ϕ∞,R+ + sup  Da ϕ ∞,[a,+∞) . ν a∈R+

(7.19)

100

7 Fractional Landau Inequalities of Riemann–Liouville Type

Furthermore we see that

ν−1  (ν) ϕ∞,R+ D ϕ (a) ≤ + a h ν−1

  sup  Daν ϕ ∞,[a,+∞)



a∈R+

h,

ν

(7.20)

∀ h > 0 and ∀ a ∈ R+ . Consequently it holds

  ν−1  (ν) ϕ∞,R+ sup Da ϕ (a) ≤ + h ν−1 a∈R+

  sup  Daν ϕ ∞,R+

a∈R+

ν

 h,

(7.21)

∀ h > 0. Call μ :=  (ν) ϕ∞,R+ ,   ⎞ sup  Daν ϕ ∞,R+ ⎟ ⎜ a∈R+ θ := ⎝ ⎠, ν

(7.22)

⎛

(7.23)

both are greater than zero. Here 1 < ν < 2 and 0 < ν − 1 < 1. Set also ρ := ν − 1, i.e. 0 < ρ < 1. We consider the function

We have

y (h) := μh −ρ + θ h, ∀ h > 0.

(7.24)

y  (h) = −ρμh −(ρ+1) + θ = 0

(7.25)

and

ρμh −(ρ+1) = θ,

then h −(ρ+1) =

θ , ρμ

with the only critical number h0 = Also it holds

1  ρμ  ρ+1

θ

> 0.

y  (h 0 ) = ρ (ρ + 1) μh −ρ−2 > 0. 0

(7.26)

(7.27)

7.2 Main Results

101

Thus y has a global minimum over R+ which is (ρ + 1)

−ρ

y (h 0 ) = μh 0 + θ h 0 =

ρ

ρ ρ+1

ρ

1

θ ρ+1 μ ρ+1 .

(7.28)

Consequently we derive that sup Daν−1 ϕ (a) ≤



a∈R+

 =



ν (ν − 1)

 (ν + 1) (ν − 1)ν−1

ν−1 ν

 ν1

  ⎞ ν−1 sup  Daν ϕ ∞,R+ ν 1 ⎟  ⎜ a∈R+  (ν) ϕ∞,R+ ν ⎠ ⎝ ν ⎛

1

ν ϕ∞,R +

  sup  Daν ϕ ∞,R+

(7.29)

 ν−1 ν ,

a∈R+

for 1 < ν < 2. The theorem is proved.



We continue with Theorem 7.6 Here 2 < ν < 3 and ϕ ∈ C B (R+ ) and Daν ϕ ∈ C B ([a, +∞)), ∀ a ∈ R+ . We assume that sup  Daν ϕ ∞,R+ < +∞. Then a∈R+

(i)

sup Daν−1 ϕ (a) ≤

a∈R+



( ν−1 ( ν−1 ν ) ν )   1     ν1 5 2−ν ν ν  D ϕ , (7.30) ν 1+2  (ν) ϕ∞,R sup a ∞,R+ + ν−1 a∈R+

and (ii) sup Daν−2 ϕ (a) ≤

a∈R+



( ν−2 ν ) 6 (ν − 1) (ν − 2)

2    2 ν ν 1 + 21−ν  (ν − 1) ν ϕ∞,R +



  sup  Daν ϕ ∞,R+

( ν−2 ν ) .

(7.31)

a∈R+

The above inequalities are nearly precisely determined constants. sharp with That is sup Daν−1 ϕ (a) , sup Daν−2 ϕ (a) < +∞. a∈R+

a∈R+

Proof For any b ∈ (a, +∞) we get (by Theorem 7.2)

102

7 Fractional Landau Inequalities of Riemann–Liouville Type

ϕ (b) =

Daν−1 ϕ (a) D ν−2 ϕ (a) (b − a)ν−1 + a (b − a)ν−2 +  (ν)  (ν − 1)

Daν−3 ϕ (a) 1 (b − a)ν−3 +  (ν − 2)  (ν)

 a

b

(7.32)

(b − t)ν−1 Daν ϕ (t) dt.

Notice that (x ∈ [a, +∞)) Daν−3 ϕ (x) = Da−(3−ν) ϕ (x) =

1  (3 − ν)



x

(x − t)2−ν ϕ (t) dt,

a

and ν−3 Da ϕ (x) ≤

 x  ϕ∞,[a,+∞) x 1 (x − t)2−ν |ϕ (t)| dt ≤ (x − t)2−ν dt  (3 − ν) a  (3 − ν) a

ϕ∞,[a,+∞) (x − a)3−ν ϕ∞,[a,+∞) = = (x − a)3−ν .  (3 − ν) (3 − ν)  (4 − ν) That is

ϕ∞,[a,+∞) ν−3 D ϕ (x) ≤ (x − a)3−ν , a  (4 − ν)

(7.33)

(7.34)

∀ x ∈ [a, +∞). Clearly, it is Daν−3 ϕ (a) = 0. Hence by (7.32) we get ϕ (b) =

D ν−2 ϕ (a) Daν−1 ϕ (a) (b − a)ν−1 + a (b − a)ν−2  (ν)  (ν − 1) +

∀ b ∈ (a, +∞). So that

1  (ν)



b

(b − t)ν−1 Daν ϕ (t) dt,

a

(7.35)

D ν−2 ϕ (a) Daν−1 ϕ (a) (b − a)ν−1 + a (b − a)ν−2 =  (ν)  (ν − 1) ϕ (b) −

1  (ν)

 a

b

(b − t)ν−1 Daν ϕ (t) dt,

∀ b ∈ (a, +∞). Let b = a + h, h > 0, i.e. b − a = h, then Daν−1 ϕ (a) ν−1 Daν−2 ϕ (a) ν−2 h h + =  (ν)  (ν − 1)

(7.36)

7.2 Main Results

103

1 ϕ (a + h) −  (ν)



a+h

a

(a + h − t)ν−1 Daν ϕ (t) dt =: A.

(7.37)

For b = a + 2h, i.e. b − a = 2h, we get Daν−1 ϕ (a) ν−1 ν−1 Daν−2 ϕ (a) ν−2 ν−2 2 h 2 h + =  (ν)  (ν − 1) 1 ϕ (a + 2h) −  (ν)



a+2h

(a + 2h − t)ν−1 Daν ϕ (t) dt =: B.

a

(7.38)

We solve the system of two equations (7.37), (7.38) for two unknowns Daν−1 ϕ (a) and Daν−2 ϕ (a). We calculate the determinants: ν−1 h h ν−2 (ν) ν−2 2ν−3 (ν−1) − 2ν−1 h 2ν−3 2 h D= = = 2ν−1 h ν−1 2ν−2 h ν−2  (ν)  (ν − 1) (ν) (ν−1)  ν−2  h 2ν−3 2ν−2 h 2ν−3 2 . − 2ν−1 = −  (ν)  (ν − 1)  (ν)  (ν − 1) Hence we get D=− A D1 = B

Next we find

2ν−2 h 2ν−3 < 0.  (ν)  (ν − 1)

h ν−2 (ν−1) 2ν−2 h ν−2 (ν−1)

 h ν−2  ν−2 2 A−B , =  (ν − 1)

ν−1 h (ν) A  h ν−1  B − 2ν−1 A . D2 = = 2ν−1 h ν−1  (ν) (ν) B

and

(7.39)

(7.40)

(7.41)

(7.42)

Therefore we derive Daν−1 ϕ

D1 = (a) = D

Similarly, we get

h ν−2 (ν−1)

 ν−2  2 A−B ν−2 2ν−3

2 h − (ν)(ν−1)

   (ν) B − 2ν−2 A = ν−2 . 2 h ν−1

(7.43)

104

7 Fractional Landau Inequalities of Riemann–Liouville Type

Daν−2 ϕ

D2 = (a) = D

h ν−1 (ν)



B − 2ν−1 A



ν−2 2ν−3

2 h − (ν)(ν−1)

   (ν − 1) 2ν−1 A − B = . 2ν−2 h ν−2

(7.44)

We notice that |A| = ϕ (a + h) −



1  (ν)

|B| = ϕ (a + 2h) −

1  (ν)

ϕ∞,R+ +

ν−1

(a + h − t)

a

ϕ∞,R+ + and

a+h

  sup  Daν ϕ ∞,R+

a∈R+

 (ν + 1)



a+2h

Daν ϕ

hν ,

ν−1

(a + 2h − t)

a

  sup  Daν ϕ ∞,R+

a∈R+

 (ν + 1)

(t) dt ≤

(7.45)

Daν ϕ

(t) dt ≤

2ν h ν .

(7.46)

Therefore we have

⎛ ⎜ ⎝ϕ∞,R+ +

  ν−1 |B| + 2ν−2 |A|  (ν) D ϕ (a) ≤  (ν) ≤ ν−2 ν−1 a 2ν−2 h ν−1 2 h     sup  Daν ϕ ∞,R+ sup  Daν ϕ ∞,R+

a∈R+

 (ν + 1)

2ν h ν + 2ν−2 ϕ∞,R+ + 2ν−2

  sup  Daν ϕ ∞,R+

⎡ =

a∈R+

  (ν) ⎢ ν−2 a∈R+ + 1 ϕ∞,R+ + h ν ⎣ 2 2ν−2 h ν−1  (ν + 1) ⎡

 (ν) = ν−2 ν−1 2 h



⎢ ⎢ ν−2  ⎢ 2 + 1 ϕ∞,R+ + ⎢ ⎣

5·2

ν−2

 1 + 22−ν  (ν) ϕ∞,R+ = + h ν−1

 ν ⎥ 2 + 2ν−2 ⎦

a∈R+

 (ν + 1)

  5 sup  Daν ϕ ∞,R+ a∈R+

ν

⎟ hν ⎠

⎤(7.47)

  sup  Daν ϕ ∞,R+



 (ν + 1)

⎞



⎤ ⎥ ⎥ h ⎥ ⎦ ν⎥

 h.

(7.48)

7.2 Main Results

105

That is



  ν−1 1 + 22−ν  (ν) ϕ∞,R+ sup Da ϕ (a) ≤ + h ν−1 a∈R+

5

  sup  Daν ϕ ∞,R+

a∈R+

 h,

ν

(7.49) ∀ h > 0, 2 < ν < 3. Furthermore we have ν−2   D ϕ (a) ≤  (ν − 1) 2ν−1 |A| + |B| ≤ a ν−2 ν−2 2 h   ⎡ ⎛ ⎞ sup  Daν ϕ ∞,R+  (ν − 1) ⎢ ν−1 ⎜ a∈R+ ⎟ hν ⎠ ⎣2 ⎝ϕ∞,R+ + 2ν−2 h ν−2  (ν + 1) ⎛ ⎜ + ⎝ϕ∞,R+ + ⎡

  sup  Daν ϕ ∞,R+

a∈R+

 (ν + 1)

⎞⎤ ⎟⎥ 2 ν h ν ⎠⎦ =

(7.50)

  ⎞ ⎤ sup  Daν ϕ ∞,R+ ⎥ ⎟  ν−1 ⎜ a∈R+ + hν ⎝ + 2ν ⎦ = ⎠ 2  (ν + 1) ⎛

  (ν − 1) ⎢ ν−1 + 1 ϕ∞,R+ ⎣ 2 ν−2 ν−2 2 h

  ⎞ ⎡ ⎛  ⎤  sup  Daν ϕ ∞,R ν−1 + 2ν + 2 2ν−1 + 1  (ν − 1) ϕ∞,R+ ⎟ ⎢ ⎥ ⎜ a∈R+ +  (ν − 1) ⎝ h2⎦ ⎠ ⎣  (ν + 1) 2ν−2 h ν−2 2ν−2

=

That is

    2 1 + 21−ν  (ν − 1) ϕ∞,R+ h ν−2

6 +

  sup  Daν ϕ ∞,R

a∈R+

(7.51)

 +

ν (ν − 1)

h2.

    ν−2 2 1 + 21−ν  (ν − 1) ϕ∞,R+ sup Da ϕ (a) ≤ h ν−2 a∈R+ ⎛

⎞  ν  ⎜ 6 sup  Da ϕ ∞,R+ ⎟ ⎜ a∈R+ ⎟ 2 ⎟ h , ∀h > 0, 2 < ν < 3. +⎜ ⎜ ⎟ ν − 1) (ν ⎝ ⎠

(7.52)

106

7 Fractional Landau Inequalities of Riemann–Liouville Type

Next we working on (7.49). Call   μ := 1 + 22−ν  (ν) ϕ∞,R+ , ⎛ ⎜ θ := ⎝

⎞

5 sup  Daν ϕ ∞,R a∈R+

ν

+

(7.53)

⎟ ⎠,

both are greater than zero. Here 2 < ν < 3 and 1 < ν − 1 < 2. Set also ρ := ν − 1, i.e. 1 < ρ < 2. We consider again the function y (h) := μh −ρ + θ h, ∀ h > 0. As in the proof of Theorem 7.5 y has a global minimum at h 0 = is ρ 1 (ρ + 1) ρ+1 y (h 0 ) = θ μ ρ+1 . ρ ρ ρ+1

(7.54) 1  ρμ  ρ+1

θ

> 0, which (7.55)

Consequently we derive sup Daν−1 ϕ (a) ≤

a∈R+

ν (ν − 1)(

ν−1 ν

)

⎛

 ⎞( ν−1 ν )  ν    ⎜ 5 sup Da ϕ ∞,R+ ⎟ ⎜ a∈R+ ⎟   1 ⎜ ⎟ 1 + 22−ν  (ν) ϕ∞,R+ ν = ⎜ ⎟ ν ⎝ ⎠ 

( ν−1 ( ν−1 ν ) ν )   1  ν    ν1 5 2−ν ν , (7.56)  (ν) ϕ∞,R+ sup  Da ϕ ∞,R+ ν 1+2 ν−1 a∈R+

2 < ν < 3. Next we are working on (7.52). Call     ξ := 2 1 + 21−ν  (ν − 1) ϕ∞,R+ ,

(7.57)

⎛

⎞  ν    ⎜ 6 sup Da ϕ ∞,R+ ⎟ ⎜ a∈R+ ⎟ ⎟, ψ := ⎜ ⎜ ⎟ ν (ν − 1) ⎝ ⎠

(7.58)

7.2 Main Results

107

both are greater than zero. Call also ρ := ν − 2, 2 < ν < 3. We consider the function y (h) := ξ h −ρ + ψh 2 , ∀ h > 0. Hence

(7.59)

y  (h) = −ρξ h −(ρ+1) + 2ψh = 0,

with the only critical number  h 0 = h crit. = We need

ρξ 2ψ

1  ρ+2

.

y  (h) = ρ (ρ + 1) ξ h −(ρ+2) + 2ψ.

We calculate

−(ρ+2)

y  (h 0 ) = ρ (ρ + 1) ξ h 0

 ρ (ρ + 1) ξ 

ρ (ρ + 1) ξ

2ψ ρξ

ρξ 2ψ

(7.60)

(7.61)

+ 2ψ =

1 −(ρ+2)  ρ+2

+ 2ψ =

 + 2ψ = (ρ + 1) 2ψ + 2ψ = 2ψ (ρ + 2) > 0.

(7.62)

Thus y has a global minimum, which is   −(ρ+2) 2 2 y (h 0 ) = ξ h −ρ +ψ = 0 + ψh 0 = h 0 ξ h 0 



ρξ 2ψ

That is

⎞ ⎛

1 −(ρ+2)   ρ+2 ρξ ⎝ξ + ψ⎠ = 2ψ

2    ρ+2     2   2ψ ρξ ρ+2 2ψ ξ +ψ = +ψ = ρξ 2ψ ρ

 ψ

ρξ 2ψ

2  ρ+2

ρ+2 ρ



ρξ 2ψ

2  ρ+2

2    2+ρ  ρ ξ ψ ρ+2 = (ρ + 2) . 2 ρ

(7.63)

108

7 Fractional Landau Inequalities of Riemann–Liouville Type 2    2+ρ  ρ ξ ψ ρ+2 y (h 0 ) = (ρ + 2) . 2 ρ

(7.64)

Therefore it holds   2 sup Daν−2 ϕ (a) ≤ ν 1 + 21−ν  (ν − 1) ϕ∞,R+ ν

a∈R+

⎛

 ⎞( ν−2 ν )  ν    ⎜ 6 sup Da ϕ ∞,R+ ⎟  ( ν−2 ν ) ⎜ a∈R+ ⎟ 6 ⎜ ⎟ = ⎜ ν (ν − 1) (ν − 2) ⎟ (ν − 1) (ν − 2) ⎝ ⎠

2  2   ν ν 1 + 21−ν  (ν − 1) ν ϕ∞,R +



  sup  Daν ϕ ∞,R+

( ν−2 ν ) ,

(7.65)

a∈R+

2 < ν < 3. The theorem is proved.



Next we give right Riemann–Liouville fractional Landau type inequalities. ν Convention 7.7 Define Db− ϕ (x) := 0, for any x > b, for any ν > 0, ν ∈ / N.

We present ν Theorem 7.8 Here 1 < ν < 2 and ϕ ∈ C B (R− ) and Db− ϕ ∈ C B ((−∞, b]), ∀ b ∈ and bounded functions. R− , where C B means continuous  ν  ϕ ∞,R− < +∞. We assume that sup  Db− b∈R−

Then ν−1 ϕ (b) ≤ sup Db−

b∈R−



 (ν + 1) (ν − 1)ν−1

 ν1

1 ν

ϕ∞,R−

 ν  ϕ ∞,R− sup  Db−

 ν−1 ν ,

(7.66)

b∈R−

ν−1 ϕ (b) < a sharp inequality with a precisely determined constant. That is sup Db− b∈R−

+∞. Proof For any a ∈ (−∞, b) we get (by Theorem 7.3) ϕ (a) =

ν−1 Db− ϕ (b) D ν−2 ϕ (b) (b − a)ν−1 + b− (b − a)ν−2  (ν)  (ν − 1)

1 +  (ν)

 a

b

ν ϕ (t) dt. (t − a)ν−1 Db−

(7.67)

7.2 Main Results

109

Notice that (x ∈ (−∞, b]) ν−2 Db− ϕ

(x) =

−(2−ν) Db− ϕ



1 (x) =  (2 − ν)

b

(t − x)1−ν ϕ (t) dt,

x

and ν−2 Db− ϕ (x) ≤

 b  ϕ∞,(−∞,b] b 1 (t − x)1−ν |ϕ (t)| dt ≤ (t − x)1−ν dt  (2 − ν) x  (2 − ν) x

= That is

ϕ∞,(−∞,b] (b − x)2−ν .  (3 − ν)

(7.68)

ϕ∞,(−∞,b] ν−2 D ϕ (x) ≤ (b − x)2−ν , b−  (3 − ν)

(7.69)

∀ x ∈ (−∞, b]. ν−2 ϕ (b) = 0. In particular we have Db− Hence by (7.67) we get D ν−1 ϕ (b) 1 ϕ (a) = b− (b − a)ν−1 +  (ν)  (ν)



b

a

ν ϕ (t) dt, (t − a)ν−1 Db−

(7.70)

ν ϕ (t) dt, (t − a)ν−1 Db−

(7.71)

∀ a ∈ (−∞, b). So that ν−1 ϕ (b) Db− 1 (b − a)ν−1 = ϕ (a) −  (ν)  (ν)



b

a

∀ a ∈ (−∞, b). Hence (set h := b − a > 0)  ν−1 D ϕ (b) h ν−1 =  (ν) ϕ (a) − b−

b

(t − a)

a

  (ν) |ϕ (a)| + a

 (ν) ϕ∞,R−

b

ν−1

ν Db− ϕ

(t) dt ≤

ν ϕ (t) dt ≤ (t − a)ν−1 Db−

 ν  +  Db− ϕ ∞,(−∞,b]



b

(t − a)ν−1 dt =

a

 ν   ν  hν (b − a)ν =  (ν) ϕ∞,R− +  Db− .  (ν) ϕ∞,R− +  Db− ϕ ∞,(−∞,b] ϕ ∞,(−∞,b] ν ν

(7.72)

110

7 Fractional Landau Inequalities of Riemann–Liouville Type

That is   ν−1 hν D ϕ (b) h ν−1 ≤  (ν) ϕ∞,R +  D ν ϕ  ≤ b− b− − ∞,(−∞,b] ν

  ν  hν  (ν) ϕ∞,R− + sup  Db− ϕ ∞,(−∞,b] . ν b∈R−

(7.73)

Furthermore we see that

 (ν) ϕ∞,R− ν−1 D ϕ (b) ≤ + b− h ν−1

 ν  ϕ ∞,(−∞,b] sup  Db−



b∈R−

h,

ν

(7.74)

∀ h > 0 and ∀ b ∈ R− . Consequently it holds

 (ν) ϕ∞,R− ν−1 sup Db− ϕ (b) ≤ + h ν−1 b∈R−

 ν  ϕ ∞,R− sup  Db−

b∈R−

ν

 h,

(7.75)

∀ h > 0. Call μ :=  (ν) ϕ∞,R− ,

(7.76)

 ν  ⎞ ϕ ∞,R− sup  Db− ⎟ ⎜ b∈R− θ := ⎝ ⎠, ν ⎛

(7.77)

both are greater than zero. Here 1 < ν < 2 and 0 < ν − 1 < 1. Set also ρ := ν − 1, i.e. 0 < ρ < 1. We consider the function y (h) := μh −ρ + θ h, ∀ h > 0.

(7.78)

As in the proof of Theorem 7.5, y has a global minimum over R+ which is y (h 0 ) = where the critical number h0 =

(ρ + 1) ρ

ρ ρ+1

1  ρμ  ρ+1

θ

ρ

1

θ ρ+1 μ ρ+1 ,

> 0.

(7.79)

7.2 Main Results

111

Consequently we derive that ν−1 sup Db− ϕ (b) ≤



b∈R−

 =



ν (ν − 1)

 (ν + 1) (ν − 1)ν−1

ν−1 ν

 ν1

 ν  ⎞ ν−1 ϕ ∞,R− ν sup  Db− 1 ⎟  ⎜ b∈R−  (ν) ϕ∞,R− ν ⎠ ⎝ ν ⎛

1

ν ϕ∞,R −

 ν  ϕ ∞,R− sup  Db−

(7.80)

 ν−1 ν ,

b∈R−

for 1 < ν < 2. The theorem is proved.



We continue with ν Theorem 7.9 Here 2 < ν < 3 and  ϕ ∈ C B (R− ) and Db− ϕ ∈ C B ((−∞, b]), ∀ b ∈ ν   R− . We assume that sup Db− ϕ ∞,R− < +∞. Then b∈R−

(i)

ν−1 sup Db− ϕ (b) ≤

b∈R−



( ν−1 ( ν−1 ν ) ν )   1 1     5 2−ν ν ν ν   , (7.81) ν 1+2  (ν) ϕ∞,R− sup Db− ϕ ∞,R− ν−1 b∈R−

and (ii) ν−2 ϕ (b) ≤ sup Db−

b∈R−



( ν−2 ν ) 6 (ν − 1) (ν − 2)

2    2 ν ν 1 + 21−ν  (ν − 1) ν ϕ∞,R −



 ν  ϕ ∞,R− sup  Db−

( ν−2 ν ) .

(7.82)

b∈R−

The above inequalities are sharp with nearly precisely determined constants. ν−1 ν−2 That is sup Db− ϕ (b) , sup Db− ϕ (b) < +∞. b∈R−

b∈R−

Proof For any a ∈ (−∞, b) we get (by Theorem 7.3) ϕ (a) =

ν−1 Db− ϕ (b) D ν−2 ϕ (b) (b − a)ν−1 + b− (b − a)ν−2 +  (ν)  (ν − 1)

ν−3 Db− ϕ (b) 1 (b − a)ν−3 +  (ν − 2)  (ν)

 a

b

ν ϕ (t) dt. (t − a)ν−1 Db−

(7.83)

112

7 Fractional Landau Inequalities of Riemann–Liouville Type

Notice that (x ∈ (−∞, b]) ν−3 Db− ϕ

and

(x) =

1 (x) =  (3 − ν)

−(3−ν) Db− ϕ



b

(t − x)2−ν ϕ (t) dt,

x

ϕ∞,(−∞,b] ν−3 D ϕ (x) ≤ (b − x)3−ν , b−  (4 − ν)

(7.84)

∀ x ∈ (−∞, b]. ν−3 ϕ (b) = 0. That is Db− Hence by (7.83) we get ν−1 ϕ (b) Db− D ν−2 ϕ (b) (b − a)ν−1 + b− (b − a)ν−2 =  (ν)  (ν − 1)

ϕ (a) −

1  (ν)

 a

b

ν ϕ (t) dt, (t − a)ν−1 Db−

(7.85)

∀ a ∈ (−∞, b). Let h > 0 : b − a = h, , i.e. b = a + h, then ν−1 ν−2 Db− ϕ (b) ν−1 Db− ϕ (b) ν−2 h h + =  (ν)  (ν − 1)

ϕ (a) −

1  (ν)



a+h

a

ν ϕ (t) dt =: A. (t − a)ν−1 Db−

(7.86)

For b − a = 2h, i.e. b = a + 2h, we get ν−1 ν−2 Db− ϕ (b) ν−1 ν−1 Db− ϕ (b) ν−2 ν−2 2 h 2 h + =  (ν)  (ν − 1)

1 ϕ (a) −  (ν)

 a

a+2h

ν ϕ (t) dt =: B. (t − a)ν−1 Db−

(7.87)

ν−1 ϕ (b) We solve the system of two equations (7.86), (7.87) for two unknowns Db− ν−2 and Db− ϕ (b). As in the proof of Theorem 7.6 we find

ν−1 Db− ϕ

and ν−2 Db− ϕ

   (ν) B − 2ν−2 A , (b) = ν−2 2 h ν−1

   (ν − 1) 2ν−1 A − B . (b) = 2ν−2 h ν−2

(7.88)

(7.89)

7.2 Main Results

113

 ν  ⎞ ϕ ∞,R− sup  Db− ⎟ ν ⎜ b∈R− +⎝ ⎠h ,  (ν + 1) ⎛

We notice that A ≤ ϕ∞,R

−

(7.90)

 ν  ⎞ sup  Db− ϕ ∞,R− ⎟ ν ν ⎜ b∈R− +⎝ ⎠2 h .  (ν + 1) ⎛

and B ≤ ϕ∞,R

−

(7.91)

As in the proof of Theorem 7.6 we obtain



  ν−1 1 + 22−ν  (ν) ϕ∞,R− sup Db− ϕ (b) ≤ + h ν−1 b∈R−

5

 ν  ϕ ∞,R− sup  Db−

b∈R−

ν

 h, (7.92)

and

    ν−2 2 1 + 21−ν  (ν − 1) ϕ∞,R− sup Db− ϕ (b) ≤ h ν−2 b∈R− ⎛

⎞  ν  ⎜ 6 sup  Db− ϕ ∞,R− ⎟ ⎜ b∈R− ⎟ 2 ⎟h , +⎜ ⎜ ⎟ ν (ν − 1) ⎝ ⎠

(7.93)

∀ h > 0, 2 < ν < 3. As in the proof of Theorem 7.6 we find similarly optimal upper bounds in (7.92) and (7.93). We derive ν−1 ν sup Db− ϕ (b) ≤ ν−1 b∈R− (ν − 1)( ν ) ⎛

 ⎞( ν−1 ν )  ν    D 5 sup ϕ ⎜ b− ∞,R− ⎟ ⎜ b∈R− ⎟   1 ⎜ ⎟ 1 + 22−ν  (ν) ϕ∞,R− ν = ⎜ ⎟ ν ⎝ ⎠ 

( ν−1 ( ν−1 ν ) ν )   1  ν    ν1 5 2−ν ν   , (7.94)  (ν) ϕ∞,R− sup Db− ϕ ∞,R− ν 1+2 ν−1 b∈R−

114

and

7 Fractional Landau Inequalities of Riemann–Liouville Type

ν−2   2 sup Db− ϕ (b) ≤ ν 1 + 21−ν  (ν − 1) ϕ∞,R− ν

b∈R−

⎛

 ⎞( ν−2 ν )  ν    ⎜ 6 sup Db− ϕ ∞,R− ⎟  ( ν−2 ν ) ⎜ b∈R− ⎟ 6 ⎜ ⎟ = ⎜ ν (ν − 1) (ν − 2) ⎟ (ν − 1) (ν − 2) ⎝ ⎠

2  2   ν ν 1 + 21−ν  (ν − 1) ν ϕ∞,R −



 ν  ϕ ∞,R− sup  Db−

( ν−2 ν ) ,

(7.95)

b∈R−

2 < ν < 3. The theorem is proved.



An application of Theorem 7.5 follows (ν = 1.5). Corollary 7.10 Let ϕ ∈C B (R+ ) such that Da1.5 ϕ ∈ C B ([a, +∞)), ∀ a ∈ R+ .  1.5  Assume that sup Da ϕ ∞,R+ < +∞. Then a∈R+

sup Da0.5 ϕ (a) ≤ (1.52323665) ϕ0.666666667 ∞,R+



a∈R+

  sup  Da1.5 ϕ ∞,R+

0.333333333 ,

a∈R+

a sharp inequality. That is sup Da0.5 ϕ (a) < +∞.

(7.96)

a∈R+

An application of Theorem 7.8 follows (ν = 1.5). 1.5 Corollary 7.11 Let ϕ ∈C B (R− ) such that Db− ϕ ∈ C B ((−∞, b]), ∀ b ∈ R− .  1.5 Assume that sup  Db− ϕ ∞,R− < +∞. Then b∈R−

0.5 ϕ (b) ≤ (1.52323665) ϕ0.666666667 sup Db− ∞,R−

b∈R−



 1.5  ϕ ∞,R− sup  Db−

0.333333333 ,

b∈R−

0.5 a sharp inequality. That is sup Db− ϕ (b) < +∞. b∈R−

7.3 Appendix We give the following simplified fractional Taylor formulae.

(7.97)

7.3 Appendix

115

Corollary 7.12 Let ν > 0, ν ∈ / N and m = ν; ϕ, Daν ϕ ∈ C ([a, b]). Then ϕ (b) =

m−1  j=1

ν− j

Da ϕ (a) 1 (b − a)ν− j +  (ν − j + 1)  (ν)



b

(b − t)ν−1 Daν ϕ (t) dt. (7.98)

a

Proof It is m > ν, hence m − ν > 0. We have that Daν−m ϕ and

(x) =

Da−(m−ν) ϕ

ν−m D ϕ (x) ≤ a

ϕ∞,[a,b]  (m − ν)



x

1  (m − ν)



x

a



x

(x − t)m−ν−1 ϕ (t) dt,

a

(x − t)m−ν−1 |ϕ (t)| dt ≤

a

(x − t)m−ν−1 dt =

= That is

1 (x) =  (m − ν)

ϕ∞,[a,b] (x − a)m−ν  (m − ν) (m − ν)

(7.99)

ϕ∞,[a,b] (x − a)m−ν .  (m − ν + 1)

ν−m D ϕ (x) ≤ a

ϕ∞,[a,b] (x − a)m−ν ,  (m − ν + 1)

(7.100)

∀ x ∈ [a, b]. Hence Daν−m ϕ (a) = 0. Use also Theorem 7.2.



ν Corollary 7.13 Let ν > 0, ν ∈ / N and m = ν; ϕ, Db− ϕ ∈ C ([a, b]). Then

ϕ (a) =

m−1  j=1

ν− j

Db− ϕ (b) 1 (b − a)ν− j +  (ν − j + 1)  (ν)



b

a

ν ϕ (t) dt. (t − a)ν−1 Db−

(7.101) Proof We have that −(m−ν) ν−m Db− ϕ (x) = Db− ϕ (x) =

and

ν−m D ϕ (x) ≤ b−

1  (m − ν)



b

(t − x)m−ν−1 ϕ (t) dt,

x

ϕ∞,[a,b] (b − x)m−ν ,  (m − ν + 1)

∀ x ∈ [a, b]. ν−m ϕ (b) = 0. Use also Theorem 7.3. Hence Db−

(7.102)



116

7 Fractional Landau Inequalities of Riemann–Liouville Type

We give also an interesting new fractional inequality of Ostrowski like. Theorem 7.14 Let ν > 0, ν ∈ / N and m = ν, with ϕ ∈ C ([a, b]). Let also x0 ∈ ϕ [a, b] be fixed. Assumed that Dx0 ∈ C ([x0 , b]) and Dxν0 − ϕ ∈ C ([a, x0 ]). Then  b m−1   ν− j 1 Dx0 ϕ (x0 ) (b − x0 )ν− j+1 ϕ (x) d x −  − j + 2) (ν a j=1  ν− j +Dx0 − ϕ (x0 ) (x0 − a)ν− j+1 ≤     ν  1 ν+1 ν+1  Dν ϕ  D ϕ ≤ + − a) − x (b ) (x 0 0 − x0 x 0 ∞,[x0 ,b] ∞,[a,x0 ]  (ν + 2)     max  Dxν0 − ϕ ∞,[a,x ] ,  Dxν0 ϕ ∞,[x 0

 (ν + 2)

 0 ,b]

  (b − x0 )ν+1 + (x0 − a)ν+1 .

(7.103)

Proof We have that ϕ (x) =

m−1  j=1

ν− j

Dx0 ϕ (x0 ) 1 (x − x0 )ν− j +  (ν − j + 1)  (ν)



x x0

(x − t)ν−1 Dxν0 ϕ (t) dt, (7.104)

∀ x ∈ [x0 , b] . Also we have ϕ (x) =

ν− j

m−1 

Dx0 − ϕ (x0 ) 1 (x0 − x)ν− j +  (ν − j + 1)  (ν)

j=1



x0 x

(t − x)ν−1 Dxν0 − ϕ (t) dt, (7.105)

∀ x ∈ [a, x0 ] . Thus ϕ (x) −

m−1  j=1

ν− j

Dx0 ϕ (x0 ) 1 (x − x0 )ν− j =  (ν − j + 1)  (ν)



x x0

(x − t)ν−1 Dxν0 ϕ (t) dt, (7.106)

∀ x ∈ [x0 , b] , and ϕ (x) −

m−1  j=1

ν− j

Dx0 − ϕ (x0 ) 1 (x0 − x)ν− j =  (ν − j + 1)  (ν)



x0 x

(t − x)ν−1 Dxν0 − ϕ (t) dt, (7.107)

∀ x ∈ [a, x0 ] . Therefore we get

7.3 Appendix

117



b

m−1 

ϕ (x) d x −

x0

j=1



1  (ν) and



ν− j

Dx0 ϕ (x0 ) (b − x0 )ν− j+1 =  (ν − j + 2)

x0



b x0

 (x − t)ν−1 Dxν0 ϕ (t) dt d x

x x0

ϕ (x) d x −

ν− j

m−1 

a

Dx0 − ϕ (x0 ) (x0 − a)ν− j+1 =  (ν − j + 2)

j=1

1  (ν)



x0



a

(7.108)

x0

(t − x)

ν−1

x

Dxν0 − ϕ

 (t) dt d x.

(7.109)

Adding (7.108) and (7.109) we derive 

b

E :=

ϕ (x) d x−

a m−1  j=1

  1 ν− j j ν− j+1 ν− j+1 Dxν− ϕ + D ϕ − a) − x (x (x ) (b ) ) (x 0 0 0 0 x − 0 0  (ν − j + 2) 

1  (ν)

=



x0

+

b



x0



a

x x0

x0

 (x − t)ν−1 Dxν0 ϕ (t) dt d x

(t − x)

ν−1

x

Dxν0 − ϕ



!

(t) dt d x .

Therefore we obtain 

1  (ν)

|E| ≤ 

a

x0

b



x0



x0 x

x x0

 (x − t)ν−1 Dxν0 ϕ (t) dt d x+

 ! (t − x)ν−1 Dxν0 − ϕ (t) dt d x ≤

 ν  1  D ϕ x0 ∞,[x0 ,b]  (ν)   +  Dxν0 − ϕ ∞,[a,x

 0]

a

x0



b



x0

x

ν−1

(x − t)

 dt d x

x0



x0 x

 ! (t − x)ν−1 dt d x =

(7.110)

118

7 Fractional Landau Inequalities of Riemann–Liouville Type

    ν  1 ν+1 ν+1  Dν ϕ  D ϕ ≤ + − a) − x (b ) (x 0 0 x0 x0 − ∞,[x0 ,b] ∞,[a,x0 ]  (ν + 2) (7.111)     ν  ν     max Dx0 − ϕ ∞,[a,x ] , Dx0 ϕ ∞,[x ,b]   0 0 (b − x0 )ν+1 + (x0 − a)ν+1 .  (ν + 2)  We finish with Corollary 7.15 All as in Theorem 7.14 with 1 < ν < 2. Then 

a

≤

b

 ν−1  1 ν ν−1 ν D ϕ (x0 ) (b − x0 ) + Dx0 − ϕ (x0 ) (x0 − a) ϕ (x) d x −  (ν + 1) x0

    ν  1 ν+1 ν+1  Dν ϕ  D ϕ + − a) − x (b ) (x 0 0 − x0 x 0 ∞,[x0 ,b] ∞,[a,x0 ]  (ν + 2)

    ≤ max  Dxν0 − ϕ ∞,[a,x ] ,  Dxν0 ϕ ∞,[x 0

 (b − x0 )ν+1 + (x0 − a)ν+1  0 ,b]

 (ν + 2)

.

(7.112)

References 1. Anastassiou, G.A.: Riemann-Liouville fractional Landau type inequalities. J. Nonlinear Evolution Equa. Appls., accepted (2020) 2. Barnett, N.S., Dragomir, S.S.: Some Landau type inequalities for functions whose derivatives are of locally bounded variation. Tamkang J. Math. 37(4), 301–308 (Winter 2006) 3. Hardy, G.H., Landau, E., Littlewood, J.E.: Some inequalities satisfied by the integrals or derivatives of real or analytic functions. Math. Z. 39, 677–695 (1935) 4. Hughes, R.J.: On fractional integrals and derivatives in L p . Indiana Univ. Math. J. 26(2), 325–328 (1977) 5. Hughes, R.J.: Hardy-Landau-Litlewwod inequalities for fractional derivatives in weightedL p spaces. J. Lond. Math. Soc. 35(2), 489–498 (1987) 6. Kilbas, A.A., Srivastava, H.M., Trujillo, J.J.: Theory and Applications of Fractional Differential Equations. North Holland, Amsterdam (2006) 7. Landau, E.: Einige Ungleichungen für zweimal differentzierban funktionen. Proc. Lond. Math. Soc. 13, 43–49 (1913) 8. Martines, S., Sans, M., Martines, M.D.: Some inequalities for fractional integrals and derivatives. Soviet. Math. Dokl. 42(3), 876–879 (1991) 9. Samko, S.G., Kilbas, A.A., Marichev, O.I.: Fractional Integrals and Derivatives: Theory and Applications. Gordon and Breach Science Publishers, Switzerland (1993)

Chapter 8

Generalized Canavati Fractional Landau Inequalities

We present uniform generalized Canavati fractional Landau inequalities over R of fractional orders 4 < ν < 5 and we estimate the size of first three ordinary derivatives of the on hand function. The inequalities are optimal by giving best upper bounds and the constants there are completely determined. We give an application when ν = 4.5. It follows [3].

8.1 Introduction   Landau in 1913 ([9]) proved if f ∈ C 2 ([0, 1]),  f ∞ = 1,  f  ∞ = 4, then    f  ≤ 4, with 4 the best constant and the result is not necessarily true for an ∞ interval of length < 1. Now Landau’s inequality has taken the form ([3])   f 

p,I

1    1 2  f 2 , ≤ C p (I )  f  p,I p,I

(8.1)

where · p,I is the p-norm on the interval I ; p ∈ [1, ∞], I = R+ or I = R, and f : I → R is twice differentiable with f, f  ∈ L p (I ). C p (I )√> 0 is independent of f . And the best constants are C∞ (R+ ) = 2 and C∞ (R) = 2. Research about Landau type inequalities has expanded to many different directions and it is a very active topic of mathematical activity. Here we are concerned about fractional Landau inequalities of generalized Canavati type. The main motivation to write this chapter is [10], and especially the result that follows: Let α be a complex number with Reα > 0. We define I Lα and D Lα to be the classical Riemann-Liouville fractional integration and differentiation operators of order α,

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_8

119

120

8 Generalized Canavati Fractional Landau Inequalities

I Lα f (x) =

1  (α)

D Lα f (x) =



x

(x − t)α−1 f (t) dt,

a

d n+1  n+1−α  I f (x) , d x n+1 L

where  is the gamma function and n = [Reα] (integral part). The above operators are defined in the natural domain of appropriate function spaces. We state Theorem 8.1 ([10]) Let I Lα and D Lα be the fractional Riemann-Liouville operators in 1 the Fréchet space L loc ([0, +∞)) with the usual topology. If α, β and γ are complex numbers for which 0 < Reα < Reβ < Reγ, for every seminorm · p there exist constant F1 ( p, α, β, γ) and F2 ( p, α, β, γ) that depend on p, α, β, and γ, with the properties that (a)    Re(γ−β)  γ  Re(β−α)   β  1 ([0, +∞)) , I L f  ≤ F1 ( p, α, β, γ)  I Lα f  pRe(γ−α)  I L f  pRe(γ−α) , f ∈ L loc p

(8.2) and

  β D L

  Re(γ−β)  γ  Re(β−α)   f  ≤ F2 ( p, α, β, γ)  D Lα f  pRe(γ−α)  D L f  pRe(γ−α) , p

(8.3)

 γ γ f ∈ D D L (domain of D L ). If I mα = I mβ = I mγ, the constants F1 ( p, α, β, γ) and F2 ( p, α, β, γ) depend only on p and on an arbitrary integer n > Reγ. For more see [10]. Other inspirations are the papers [7], [8]. The above motivating results and references are existential without giving exact constants which are not the best possible. Their methods are based on functional analysis and semigroup theory, in particular dealing with moment inequalities for fractional powers of operators in Banach spaces. As a last motivation we also mention [6]. Our method is classical analytic and we will produce fractional Landau type inequalities, involving the generalized Canavati left and right fractional derivatives, that are optimal by giving best upper bounds and all constants are exactly calculated. We need the following background: First we follow [1], pp. 7-9. Let N ν > 0, n := [ν], α := ν − n (0 < α < 1). Here f ∈ C ([a, b]), [a, b] ⊂ R and define  x  a  1 (8.4) Jν f (x) := (x − t)ν−1 f (t) dt, a ≤ x ≤ b,  (ν) a the left Riemann-Liouville fractional integral.

8.1 Introduction

121

We define the subspace Caν ([a, b]) of C n ([a, b]):   a f (n) ∈ C 1 ([a, b]) . Caν ([a, b]) := f ∈ C n ([a, b]) : J1−α So let f ∈ Caν ([a, b]); we define the left generalized Canavati fractional derivative of f of order ν over [a, b], as   a f (n) , Daν f := J1−α

(8.5)

(see also [4]) that is 

 Daν f (x) =



d 1  (1 − α) d x

x

(x − t)−α f (n) (t) dt,

(8.6)

a

all a ≤ x ≤ b, exists for f ∈ Caν ([a, b]) . We need the left generalized fractional Taylor formula (see [1], p. 8): Theorem 8.2 Let f ∈ Caν ([a, b]). (i) If ν ≥ 1, then f (x) = f (a) + f  (a) (x − a) + f  (a) f (n−1) (a)

1 (x − a)n−1 +  (ν) (n − 1)!

 a

x

(x − a)2 + ...+ 2

  (x − t)ν−1 Daν f (t) dt,

(8.7)

all x ∈ [a, b] . (ii) If 0 < ν < 1 we get 1 f (x) =  (ν)

 a

x

  (x − t)ν−1 Daν f (t) dt,

(8.8)

all x ∈ [a, b] . We also define the subspace   Caν ([a, +∞)) := f ∈ C n (R) : f ∈ Caν ([a, b]) , ∀ b > a ,

(8.9)

where a ∈ R. Next we follow [2], pp. 345-348. Let N ν > 0, n := [ν], α := ν − n, 0 < α < 1, f ∈ C ([a, b]), call the right Riemann-Liouville fractional integral operator by 

ν Jb−



1 f (x) :=  (ν)



b x

(z − x)ν−1 f (z) dz,

(8.10)

122

8 Generalized Canavati Fractional Landau Inequalities

∀ x ∈ [a, b] . Define the subspace of functions   1−α (n) ν f ∈ C 1 ([a, b]) . Cb− ([a, b]) := f ∈ C n ([a, b]) : Jb− Define the right generalized ν-fractional derivative of f over [a, b] (called also right generalized Canavati fractional derivative) as  1−α (n)  ν f := (−1)n−1 Jb− f . Db− Notice that ν Db−

(−1)n−1 d f (x) =  (1 − α) d x



b

(z − x)−α f (n) (z) dz,

(8.11)

(8.12)

x

ν exists for f ∈ Cb− ([a, b]), all x ∈ [a, b]. We need also the following right fractional Taylor formula ([2], p. 348): ν Theorem 8.3 Let f ∈ Cb− ([a, b]), ν > 0, n := [ν]. Then (1) If ν ≥ 1, we get

f (x) =

 b n−1   ν  f (k) (b) 1 f (z) dz, (z − x)ν−1 Db− (x − b)k + k!  (ν) x k=0

(8.13)

∀ x ∈ [a, b] . (2) If 0 < ν < 1, we obtain f (x) =

1  (ν)



b x

 ν  f (z) dz, (z − x)ν−1 Db−

(8.14)

∀ x ∈ [a, b] . We need also to define the subspace   ν ν Cb− ((−∞, b]) := f ∈ C n (R) : f ∈ Cb− ([a, b]) , ∀ a < b ,

(8.15)

where b ∈ R.

8.2 Main Results We present a set of fractional generalized Canavati type Landau inequalities over R. Theorem 8.4 Let 4 < ν < 5 and let f ∈ C 4 (R) with f ∈ C B (R) (continuous and ν bounded). We assume that f ∈ Caν ([a, +∞)) and f ∈ Ca− ((−∞, a]), ∀ a ∈ R. Assume also that

8.2 Main Results

123

  ν     K := max  Daν f (t)∞,R2 ,  Da− f (t)∞,R2 < +∞,

(8.16)

where (a, t) ∈ R2 . Let also 0 < λ < 1. Then (i)   f 

∞,R

 1   ν 3 3 (ν−1) ν λ 1 + λ + λ ν 1 ( ν−1 ν )     f ≤ Kν, 1 ∞,R  2 (ν−1) ν λ 1 − λ  (ν + 1) (ν − 1) (8.17)

(ii)    f 

∞,R

≤ν

( ν−2 ν ) 4 2 2 ( ν−2 ν ) Kν, ( (ν + 1))− ν  f ∞,R ν−2

(8.18)

and (iii)    f 

∞,R

≤ 2ν

 ν1 

3(ν−3) (ν − 3)(ν−3) ( (ν + 1))3

λ + λν 1+λ

3 ν

   ν−3 3 1 ν  f ∞,R K ν . λ (1 − λ)

(8.19) The above inequalities are optimal by giving best upper bounds. We also obtain  (i)  f  < +∞, for i = 1, 2, 3. ∞,R Proof Let any a ∈ R. Here it is n = [ν] = 4. By Theorem 8.2 we have that f (x) = f (a) + f  (a) (x − a) + f  (a) 1 +  (ν)

 a

x

(x − a)2 (x − a)3 + f  (a) 2 6

(x − t)ν−1 Daν f (t) dt,

(8.20)

∀ x ∈ [a, +∞). And by Theorem 8.3 we have that f (x) = f (a) + f  (a) (x − a) + f  (a) +

1  (ν)

∀ x ∈ (−∞, a]. Momentarily we fix a ∈ R. Let x1 > a, then



a x

(x − a)2 (x − a)3 + f  (a) 2 6

 ν  f (t) dt, (t − x)ν−1 Da−

(8.21)

124

8 Generalized Canavati Fractional Landau Inequalities

(x1 − a)2 (x1 − a)3 + f  (a) = 2 6  x1 1 f (x1 ) − f (a) − (x1 − t)ν−1 Daν f (t) dt.  (ν) a

f  (a) (x1 − a) + f  (a)

(8.22)

Let also x2 < a, then (x2 − a)2 (x2 − a)3 + f  (a) = 2 6  a  ν  1 f (t) dt. f (x2 ) − f (a) − (t − x2 )ν−1 Da−  (ν) x2

f  (a) (x2 − a) + f  (a)

(8.23)

Choose first x1 − a = h > 0 and x2 − a = −h. Thus f  (a) h + f  (a) f (a + h) − f (a) −



1  (ν)

a+h

a

h2 h3 + f  (a) = 2 6

(a + h − t)ν−1 Daν f (t) dt =: A1 ,

and − f  (a) h + f  (a) 1 f (a − h) − f (a) −  (ν)



a

a−h

(8.24)

h2 h3 − f  (a) = 2 6

 ν  f (t) dt =: A2 . (t − (a − h))ν−1 Da−

(8.25)

Let 0 < λ < 1. Next we choose x1 − a = λh and x2 − a = −λh, h > 0. By (8.22) and (8.23) we get f  (a) λh + f  (a) 1 f (a + λh) − f (a) −  (ν)



(λh)2 (λh)3 + f  (a) = 2 6

a+λh

a

and − f  (a) λh + f  (a) 1 f (a − λh) − f (a) −  (ν) Adding (8.24) and (8.25) we get



a

a−λh

(a + λh − t)ν−1 Daν f (t) dt =: B1 ,

(8.26)

(λh)2 (λh)3 − f  (a) = 2 6  ν  f (t) dt =: B2 . (t − (a − λh))ν−1 Da− (8.27)

8.2 Main Results

125

f  (a) h 2 = A1 + A2 .

(8.28)

Subtracting (8.27) from (8.26), we obtain 2 f  (a) λh + f  (a)

(λh)3 = B1 − B2 , 3

(8.29)

and by subtracting (8.25) from (8.24), we derive 2 f  (a) h + f  (a) From (8.28) we find f  (a) =

h3 = A1 − A2 . 3

A1 + A2 . h2

(8.30)

(8.31)

Next we solve the system of two equations in the two unknowns f  (a) and f  (a). We calculate    2λh (λh)3  4   3    = 2λh 1 − λ2 > 0. D =  (8.32)  3 3  2h  h 3 Furthermore we have   B1 − B2  D1 =  A −A 1 2 and

(λh)3 3 h3 3

   h3   = (B1 − B2 ) − λ3 (A1 − A2 )  3 

   2λh B1 − B2     = 2h [λ (A1 − A2 ) − (B1 − B2 )] . D2 =    2h A1 − A2 

(8.33)

(8.34)

Therefore we find     h3 − B2 ) − λ3 (A1 − A2 ) (B1 − B2 ) − λ3 (A1 − A2 )  3 (B1   f (a) = = ,   2λh 4 2λ 1 − λ2 h 1 − λ2 3 (8.35) and 3 [λ (A1 − A2 ) − (B1 − B2 )] 2h [λ (A1 − A2 ) − (B1 − B2 )]   = .   2λh 4 2 λ 1 − λ2 h 3 1 − λ 3 (8.36) Next we estimate f  (a) , f  (a) , f  (a). We have that f  (a) =

126

8 Generalized Canavati Fractional Landau Inequalities

|A1 + A2 | = | f (a + h) + f (a − h) − 2 f (a) − 1  (ν)



a+h

ν−1

(a + h − t)

a

≤ 4  f ∞,R + That is

Daν

 f (t) dt +

a

(t − (a − h))

ν−1

a−h



ν Da−



 ν     2 max  Daν f (t)∞,R2 ,  Da− f (t)∞,R2  (ν + 1)

  f (t) dt  

hν .

     2K  f (a) ≤ 4 f ∞,R + h ν−2 , 2 h  (ν + 1)

(8.37)

(8.38)

∀ a ∈ R, ∀ h > 0. Consequently it holds    f 

∞,R

≤

4  f ∞,R 2K h ν−2 , + h2  (ν + 1)

(8.39)

∀ h > 0. We notice that  a+λh 1 B1 − B2 = f (a + λh) + f (a − λh) − (a + λh − t)ν−1 Daν f (t) dt  (ν) a (8.40)  a  ν  1 ν−1 + Da− f (t) dt (t − (a − λh))  (ν) a−λh and  λ3 (A1 − A2 ) = λ3 f (a + h) − f (a − h) −

+

1  (ν)



a

a−h

1  (ν)

 a

a+h

(a + h − t)ν−1 Daν f (t) dt

  ν  f (t) dt . (t − (a − h))ν−1 Da−

(8.41)

Furthermore it holds |B1 − B2 | ≤ 2  f ∞,R +

 ν ν 2λ h K 2K λν = 2  f ∞,R + hν ,  (ν) ν  (ν + 1)

and λ3 |A1 − A2 | ≤ 2  f ∞,R λ3 + Consequently we have

2K λ3 ν h .  (ν + 1)

(8.42)

(8.43)

8.2 Main Results

127

  ν   2K λ + λ3 h ν    2  f ∞,R λ3 + 1 + (ν+1)  f (a) ≤   2λ 1 − λ2 h 1  =  λ 1 − λ2



    ν  ν−1  f ∞,R 1 + λ3 K 3 + λ +λ h , h  (ν + 1)

(8.44)

(8.45)

∀ a ∈ R, ∀ h > 0. We have proved that   f 

∞,R

1  ≤  λ 1 − λ2



      f ∞,R 1 + λ3 K λν + λ3 ν−1 + h , h  (ν + 1)

(8.46)

∀ h > 0. We have     f (a) ≤

3  [λ |A1 − A2 | + |B1 − B2 |] ≤  λ 1 − λ2 h 3

  2K λ 2K 3 ν ν ν   2  f ∞,R λ + h + 2  f ∞,R + λ h =  (ν + 1)  (ν + 1) λ 1 − λ2 h 3 (8.47)   3 2K ν ν   2  f ∞,R (λ + 1) + (λ + λ ) h =  (ν + 1) λ 1 − λ2 h 3    f ∞,R (λ + 1) 6 K ν ν−3   , + λ + h (λ ) h3  (ν + 1) λ 1 − λ2 ∀ h > 0, ∀ a ∈ R. We have derived that    f 

∞,R

≤

   f ∞,R (1 + λ) 6 K (λ + λν ) ν−3  , h + h3  (ν + 1) λ 1 − λ2 

∀ h > 0. We are working in (8.39). Call

ξ := 4  f ∞,R , ψ :=

both are greater than zero. Set also ϕ := ν − 2 > 0. We consider the function

(8.48)

2K , (ν+1)

(8.49)

128

8 Generalized Canavati Fractional Landau Inequalities

γ (h) := ξh −2 + ψh ϕ , ∀ h > 0. We have

(8.50)

γ  (h) = −2ξh −3 + ϕψh ϕ−1 = 0,

with one critical number

h 0 := h crit.no. = We have that

2ξ ϕψ

1  ϕ+2

.

(8.51)

γ  (h) = 6ξh −4 + ϕ (ϕ − 1) ψh ϕ−2 .

We have that



γ (h 0 ) = 2ξ

4 − ϕ+2

2ξ ϕψ

Therefore γ has a global minimum at h 0 = γ (h 0 ) = Consequently

γ (h 0 ) =



2ξ ϕψ

(8.52)

(ϕ + 2) > 0. 1  ϕ+2

(8.53)

, which is

ϕ  2

 ϕ+2 ψ ϕ+2 ξ (ϕ + 2) . ϕ 2

4  f ∞,R ν−2

 ν−2 ν

K  (ν + 1)

(8.54)

 ν2

ν.

(8.55)

We have proved that

   f 

∞,R

ν

≤

4 ν−2

Next we work on (8.46). Let

4  f ∞,R ν−2

 ν−2 ν

K  (ν + 1) ν−2

 ν2

ν=

ν Kν. ( (ν + 1))− ν  f ∞,R 2

2

  μ :=  f ∞,R 1 + λ3 , θ :=

both are greater than zero. Set also ρ := ν − 1 > 3.

 ν−2 ν

K (λν +λ3 ) , (ν+1)

(8.56)

8.2 Main Results

129

We consider the function y (h) := We have

μ + θh ρ , ∀ h > 0. h

(8.57)

μ + ρθh ρ−1 = 0, h2

(8.58)

y  (h) = −

with only one critical number

h 0 := h crit.no. = We have that

μ ρθ

1  ρ+1

.

y  (h) = 2μh −3 + ρ (ρ − 1) θh ρ−2 .

We observe that

μ y (h 0 ) = μ ρθ 

3 − ρ+1

Therefore y has a global minimum at h 0 =

(ρ + 1) > 0.

1   ρ+1

μ ρθ

(8.60)

(8.61)

, which is ρ

y (h 0 ) = (θμρ ) ρ+1 (ρ + 1) ρ− ρ+1 . 1

(8.59)

(8.62)

That is  y (h 0 ) =

 1   ν−1 K λν + λ3 ν  ν−1  f ∞,R 1 + λ3 ν ν (ν − 1)−( ν ) .  (ν + 1)

(8.63)

We have found that   f 

∞,R

 1 ν−1 ) λν + λ3 ν   ν−1 1 3 ν ν ν     f ≤ 1 + λ 1 ∞,R K = 2 ( (ν + 1)) ν λ 1 − λ ν (ν − 1)−(

ν−1 ν

 1   ν 3 3 (ν−1) ν ν−1 λ 1 + λ + λ ν 1 ν     f Kν. 1 ∞,R   2 (ν−1) ν λ 1 − λ  (ν + 1) (ν − 1) Finally we work on (8.48). We study z (h) :=

 f ∞,R (1 + λ) K (λ + λν ) ν−3 h , + h3  (ν + 1)

(8.64)

(8.65)

130

8 Generalized Canavati Fractional Landau Inequalities

∀ h > 0. Call

w :=  f ∞,R (1 + λ) , l :=

(8.66)

K (λ+λν ) , (ν+1)

both are greater than zero. Set also d := ν − 3 > 1, i.e. ν = d + 3. Thus we can write w z (h) := 3 + lh d , ∀ h > 0. h We set

(8.67)

z  (h) = −3wh −4 + dlh d−1 = 0.

(8.68)

The only critical number here is

h 0 := h crit.no. = We need

and we find

3w dl

1  d+3

> 0.

z  (h) = 12wh −5 + d (d − 1) lh d−2 ,

(8.69)

z  (h 0 ) = (dl) d+3 (3w) d+3 (d + 3) > 0.

(8.70)

5

d−2

Hence z has a global minimum which is

z (h 0 ) =

d +3 3

d  ( d+3 ) 3 3 d l ( d+3 ) w ( d+3 ) . d

(8.71)

That is z (h 0 ) =

ν  3

( ν−3 3

ν )  ν−3 3 K (λ + λν ) ν   f ∞,R (1 + λ) ( ν ) = (8.72) ν−3  (ν + 1)

ν   3

3 ν−3

( ν−3 ν )

( (ν + 1))

3 ν

3 ν−3 3 ( ν−3 ν ) Kν. (λ + λν ) ν (1 + λ)( ν )  f ∞,R

We have proved that    f 

∞,R

≤ 2ν

3ν−3 (ν − 3)ν−3 ( (ν + 1))3

 ν1 



3

(λ + λν ) ν

λ (1 − λ) (1 + λ)

 ν−3  ν

3 ν

3

 f ∞,R K ν . (8.73)

8.2 Main Results

131

The proof of the theorem is finished.



It follows an application when ν = 4.5. Corollary 8.5 Let f ∈ C 4 (R) with f ∈ C B (R). We assume that f ∈ Ca4.5 ([a, +∞)) 4.5 and f ∈ Ca− ((−∞, a]), ∀ a ∈ R. Assume also that

  4.5     f (t)∞,R2 < +∞, K 4.5 := max  Da4.5 f (t)∞,R2 ,  Da− where (a, t) ∈ R2 . Then    f 

∞,R

(0.444444444) ≤ (1.006181121)  f (0.555555556) K 4.5 , ∞,R

(8.74)

(8.75)

an optimalinequality giving a best upper bound with a constant almost 1.  Hence  f  ∞,R < +∞. Proof By Theorem 8.4 (ii).



References 1. Anastassiou, G.A.: Fractional Differentiation Inequalities. Springer, Heidelberg (2009) 2. Anastassiou, G.A.: Intelligent Mathematics: Computational Analysis. Springer, Heidelberg (2011) 3. Anastassiou, G.A.: Generalized Canavati fractional Landau type inequalities. Commun. Appl. Nonlinear Anal., accepted (2020) 4. Barnett, N.S., Dragomir, S.S.: Some Landau type inequalities for functions whose derivatives are of locally bounded variation. Tamkang J. Math. 37(4), 301–308 (Winter 2006) 5. Canavati, J.A.: The Riemann-Liouville integral. Nieuw Archief Voor Wiskunde 5(1), 53–75 (1987) 6. Hardy, G.H., Landau, E., Littlewood, J.E.: Some inequalities satisfied by the integrals or derivatives of real or analytic functions. Math. Z. 39, 677–695 (1935) 7. Hughes, R.J.: On fractional integrals and derivatives inL p . Indiana Univ. Math. J. 26(2), 325– 328 (1977) 8. Hughes, R.J.: Hardy-Landau-Litlewwod inequalities for fractional derivatives in weighted L p spaces. J. Lond. Math. Soc. 35(2), 489–498 (1987) 9. Landau, E.: Einige Ungleichungen für zweimal differentzierban funktionen. Proc. Lond. Math. Soc. 13, 43–49 (1913) 10. Martines, S., Sans, M., Martines, M.D.: Some inequalities for fractional integrals and derivatives. Soviet. Math. Dokl. 42(3), 876–879 (1991)

Chapter 9

Sequential Left Abstract Fractional Landau Inequalities

We present uniform and L p left Caputo–Bochner abstract sequential fractional Landau inequalities over R+ . These estimate the size of second and third iterated left abstract fractional derivatives of a Banach space valued function over R+ . We give an application when the basic fractional order is 21 . It follows [5].

9.1 Introduction Let p ∈ [1, ∞], I = R+ or I = R and f : I → R is twice differentiable with f, f  ∈ L p (I ), then f  ∈ L p (I ). Moreover, there exists a constant C p (I ) > 0 independent of f , such that 1  1    f  ≤ C p (I )  f  2  f   2 , (9.1) p,I

p,I

p,I

where · p,I is the p-norm on the interval I , see [1, 6]. The research on these inequalities started by Landau [11] in 1913. For the case of p = ∞ he proved that C∞ (R+ ) = 2 and C∞ (R) =

√ 2,

(9.2)

are the best constants in (9.1). In 1932, Hardy and Littlewood [8] proved (9.1) for p = 2, with the best constants C2 (R+ ) =

√

2 and C2 (R) = 1.

(9.3)

In 1935, Hardy, Landau and Littlewood [9] showed that the best constants C p (R+ ) in (9.1) satisfies the estimate © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_9

133

134

9 Sequential Left Abstract Fractional Landau Inequalities

C p (R+ ) ≤ 2, for p ∈ [1, ∞),

(9.4)

which yields C p (R) ≤ 2 for p ∈ [1, ∞). √ In fact, in [7, 10] was shown that C p (R) ≤ 2. We need the following concept from abstract fractional calculus. Our integral next is of Bochner type [12]. We need Definition 9.1 ([4], p. 150) Let [a, b] ⊂ R, X be a Banach space, α > 0; m =

α ∈ N, ( · is the ceiling of the number), f : [a, b] → X . We assume that f (m) ∈ L 1 ([a, b] , X ). We call the Caputo–Bochner left fractional derivative of order α: 

 α D∗a f (x) :=

1  (m − α)



x

(x − t)m−α−1 f (m) (t) dt, ∀ x ∈ [a, b] ,

(9.5)

a

where  is the gamma function. α f := f (m) the ordinary X -valued derivative (defined similar If α ∈ N, we set D∗a 0 f := f. to numerical one, see [13], p. 83), and also set D∗a  α  α By ([4], p. 2), D∗a f (x) exists almost everywhere in x ∈ [a, b] and D∗a f ∈ L 1 ([a, b] , X ).   α / N, then by ([4], p. 3), D∗a f ∈ C ([a, b] , X ) , If  f (m)  L ∞ ([a,b],X ) < ∞, and α ∈  α    hence D∗a f ∈ C ([a, b]) . We mention the important: Corollary 9.2 ([4], p. 157) Let f ∈ C m ([a, b] , X ), m = α, α > 0, x, x0 ∈ [a, b]. a f (x) is jointly continuous functions in (x, x0 ) from [a, b]2 into X , X is a Then D∗x 0 Banach space. By convention we suppose that α D∗x f (x) = 0, for x < x0 , 0

for all x, x0 ∈ [a, b] . The author has already done an extensive body of work on fractional Landau inequalities, see [3], and on abstract fractional Landau inequalities, see [4]. However there the proving methods came out of applications of fractional Ostrowski inequalities [2, 4]. Usually there the domains where [A, +∞) or (−∞, B], with A, B ∈ R and in one mixed case the domain was all of R. In this chapter with less assumptions we establish uniform and L p type left Caputo–Bochner abstract iterated fractional Landau inequalities over R+ . The method of proving is based on left Caputo–Bochner iterated fractional Taylor’s formula with integral remainder, see [4], pp. 129. We give also an application for α = 21 . Regardless to say that we are also inspired by [3, 4].

9.2 Main Results

135

9.2 Main Results nα α α α We consider (α > 0) the composition D∗a := D∗a D∗a ...D∗a (n-times), n ∈ N. We mention the following left modified X -valued Taylor’s formula, (X, ·) is a Banach space.

Theorem 9.3 ([4], p. 129) Let 0 < α ≤ 1, n ∈ N, f ∈ C 1 ([a, b] , X ). For k = (n+1)α kα f ∈ C 1 ([a, b] , X ) and D∗a f ∈ C ([a, b] , X ). Then 1, ..., n, we assume that D∗a n  (x − a)iα  iα  D f (a) +  (iα + 1) ∗a i=0

f (x) = 1  ((n + 1) α)



x

a

(9.6)

 (n+1)α  f (t) dt, (x − t)(n+1)α−1 D∗a

∀ x ∈ [a, b] .  α  When 0 < α < 1 and f ∈ C 1 ([a, b] , X ), by [4], p. 130, we get that D∗a f (a) = 0. We present the following abstract left fractional Landau inequalities over R+ .   Theorem 9.4 Let 0 < α < 1, f ∈ C 1 (R+ , X ) with  f ∞,R+ ,  f  ∞,R+ < kα 4α f ∈ C 1 ([a, ∞), X ) and D∗a f ∈ ∞. For k = 1, 2, 3, we assume that D∗a C ([a, ∞), X ), ∀ a ∈ R+ . We further assume that   4α f (t)∞,R2 < ∞, K :=  D∗a +

(9.7)

where (a, t) ∈ R2+ . Then  2α   f (a) ≤ sup  D∗a

a∈R+

 (2α + 1) 22α−1 (2α − 1)



  23α+1 23α + 1 (2α + 1)  f ∞,R+ K ,  (4α + 1) (9.8)

and √  3  1  3α   4 4 2 (3α + 1) ( (4α + 1))− 4 22α + 1 3 4    f ∞,R sup D∗a f (a) ≤ K 4. √ 3 √ α + 4 a∈R+ 3 2 (2α − 1) (9.9)  2α    3α   f (a) , sup  D∗a f (a) < ∞. That is sup  D∗a a∈R+

a∈R+

 α  Proof We notice again here that D∗a f (a) = 0, ∀ a ∈ R+ . We make use of Theorem 9.3 for 0 < α < 1 and n = 3, applied for any a ∈ R+ and b = ∞.

136

9 Sequential Left Abstract Fractional Landau Inequalities

Momentarily we fix a ∈ R+ . Let x2 > x1 > a, then f (x1 ) − f (a) =

(x1 − a)3α  3α  (x1 − a)2α  2α  D∗a f (a) + D f (a) +  (2α + 1)  (3α + 1) ∗a 1  (4α)



x1

a

(9.10)

 4α  f (t) dt, (x1 − t)4α−1 D∗a

and f (x2 ) − f (a) =

(x2 − a)3α  3α  (x2 − a)2α  2α  D∗a f (a) + D f (a) +  (2α + 1)  (3α + 1) ∗a 1  (4α)

That is

a

x2

 4α  f (t) dt. (x2 − t)4α−1 D∗a

(x1 − a)3α  3α  (x1 − a)2α  2α  D∗a f (a) + D f (a) =  (2α + 1)  (3α + 1) ∗a f (x1 ) − f (a) −

and



(9.11)

1  (4α)



x1

a

 4α  f (t) dt =: A, (x1 − t)4α−1 D∗a

(x2 − a)2α  2α  (x2 − a)3α  3α  D∗a f (a) + D f (a) =  (2α + 1)  (3α + 1) ∗a 1 f (x2 ) − f (a) −  (4α)

 a

x2

(9.12)

(9.13)

 4α  f (t) dt =: B. (x2 − t)4α−1 D∗a

 2α  are solving the above system of two equations with two unknowns D∗a f (a) , We3α D∗a f (a) . The main determinant of system is

(x1 −a)2α

(2α+1)

D :=

(x2 −a)2α

(2α+1)

(x1 −a)3α (3α+1) (x2 −a)3α (3α+1)







=



 1 (x1 − a)2α (x2 − a)3α − (x1 − a)3α (x2 − a)2α =  (2α + 1)  (3α + 1) (x1 − a)2α (x2 − a)2α  (x2 − a)α − (x1 − a)α > 0.  (2α + 1)  (3α + 1)

9.2 Main Results

I.e. D=

137

(x1 − a)2α (x2 − a)2α  (x2 − a)α − (x1 − a)α > 0.  (2α + 1)  (3α + 1)

(9.14)

We obtain the unique solution



 Therefore we have















A

 B 2α f (a) = D∗a

 3α D∗a f (a) =



 2α D∗a f (a) = and  3α  D∗a f (a) =























3α

(x2 −a)

(3α+1)

, D (x1 −a)3α (3α+1)

A

(x2 −a) (2α+1) D

B

2α

(9.15)



(x1 −a)2α (2α+1)









.

(x2 −a )3α A− (x1 −a )3α B (3α+1) (3α+1) , D

(9.16) (x1 −a )2α B− (x2 −a )2α A (2α+1) (2α+1) . D

We have the following   A =   f (x1 ) − f (a) − 2  f ∞,R+ +

1  (4α)

 a

x1

   4α  f (t) dt  (x1 − t)4α−1 D∗a ≤

  4α  D f (t) 2 ∗a ∞,R

(x1 − a)4α ,

(9.17)

K (x1 − a)4α ,  (4α + 1)

(9.18)

K (x2 − a)4α ,  (4α + 1)

(9.19)

+

 (4α + 1)

under the assumption  f ∞,R+ < ∞. That is A ≤ 2  f ∞,R+ + and similarly, B ≤ 2  f ∞,R+ + where by assumption

  4α f (t)∞,R2 < ∞, K :=  D∗a +

with (a, t) ∈ R2+ . Consequently we have

(9.20)

138

9 Sequential Left Abstract Fractional Landau Inequalities    2α   D∗a f (a) ≤

  K 1 (x2 − a)3α 2  f ∞,R+ + (x1 − a)4α  (3α + 1) D  (4α + 1)

+ (x1 − a)

3α

 2  f ∞,R+ +

K (x2 − a)4α  (4α + 1)

 ,

(9.21)

and    3α   D∗a f (a) ≤

 

1 K (x1 − a)2α 2  f ∞,R+ + (x2 − a)4α  (2α + 1) D  (4α + 1)

 + (x2 − a)2α 2  f ∞,R+ +

K (x1 − a)4α  (4α + 1)

 .

(9.22)

Set now x1 := a + h, x2 := a + 2h, where h > 0, so that x1 − a = h, x2 − a = 2h. Hence we get 22α h 5α (2α − 1) > 0. (9.23) D=  (2α + 1)  (3α + 1) Therefore we derive (from (9.16))  2α    D f (a) ≤ ∗a

 

 (2α + 1) K 3α 3α 4α   2 f 2 h h + ∞,R+ 22α h 5α (2α − 1)  (4α + 1)

+h

3α

 2  f ∞,R+ +

K 24α h 4α  (4α + 1)

 =

(9.24)



 3α  3α  7α  3α  (2α + 1) K 4α 2  f ∞,R+ 2 + 1 h + 2 +2 h 22α h 5α (2α − 1)  (4α + 1)  =

 (2α + 1) 22α (2α − 1)

That is

    3α 2 2 + 1  f ∞,R+ h 2α  2α    D f (a) ≤



∗a

   2 23α + 1  f ∞,R+ h 2α ∀ a ∈ R+ , ∀ h > 0. I.e. it holds

 (2α + 1) 22α (2α − 1)



(9.25)



 23α (2α + 1) K h 2α , +  (4α + 1)

 2α   f (a) ≤ sup  D∗a

a∈R+

 23α (2α + 1) K h 2α . +  (4α + 1)

 (2α + 1) 22α (2α − 1)



(9.26)

9.2 Main Results

139

   2 23α + 1  f ∞,R+ h 2α

 23α (2α + 1) 2α Kh + < ∞,  (4α + 1)

(9.27)

∀ h > 0, 0 < α < 1. By (9.16) we derive  

 (3α + 1) K 2α 4α 4α   2 f h 2 + h ∞,R+ 22α h 5α (2α − 1)  (4α + 1)

 3α    D f (a) ≤ ∗a

+2 h

2α 2α

 2  f ∞,R+ +

K h 4α  (4α + 1)

 =



 4α  2α  6α  2α  (3α + 1) K 2α 2  f ∞,R+ 2 + 1 h + 2 +2 h 22α h 5α (2α − 1)  (4α + 1)  (9.28)       2α  2 2 + 1  f ∞,R+ 22α 22α + 1  (3α + 1) K hα = + 22α (2α − 1) h 3α  (4α + 1)     (3α + 1) 22α + 1 2  f ∞,R+ 22α K α h . = + 22α (2α − 1) h 3α  (4α + 1)    3α    (3α + 1) 22α + 1  D f (a) ≤ ∗a 22α (2α − 1)

That is

2  f ∞,R+ h 3α

+

 22α K hα ,  (4α + 1)

(9.29)

∀ a ∈ R+ , ∀ h > 0. I.e. it holds

   3α    (3α + 1) 22α + 1   sup D∗a f (a) ≤ 22α (2α − 1) a∈R+

2  f ∞,R+ h 3α

∀ h > 0, 0 < α < 1. Call

 22α K α h < ∞, +  (4α + 1)

  μ := 2 23α + 1  f ∞,R+ , θ=

23α (2α + 1) K ,  (4α + 1)

(9.30)

(9.31) (9.32)

140

9 Sequential Left Abstract Fractional Landau Inequalities

both are greater than zero. Set also ρ := 2α; 0 < ρ < 2. We consider the function

We have

y (h) := μh −ρ + θh ρ , ∀ h > 0.

(9.33)

y  (h) = −ρμh −ρ−1 + ρθh ρ−1 = 0,

(9.34)

then

θh ρ−1 = μh −ρ−1

and θh 2ρ = μ, with a unique solution h 0 := h crit.no. = We have that

 μ 2ρ1 θ

.

(9.35)

y  (h) = ρ (ρ + 1) μh −ρ−2 + ρ (ρ − 1) θh ρ−2 .

(9.36)

We see that y  (h 0 ) = y 

 1  μ 2ρ θ

= ρ (ρ + 1) μ

 μ −ρ−2 2ρ θ

+ ρ (ρ − 1) θ

 μ ρ−2 2ρ θ

 μ − ρ1  μ 21   μ − 21 = ρ + (ρ − 1) θ (ρ + 1) μ θ θ θ   ρ1     θ ρ (ρ + 1) μθ + (ρ − 1) μθ = μ 1   ρ1     θ ρ θ 2 2ρ μθ = 2ρ μθ ρ > 0. μ μ

Therefore y has a global minimum at h 0 = y (h 0 ) = μ

 μ − 21 θ

+θ

We have proved that (see (9.27))

 μ 21 θ

 μ  2ρ1 θ

, which is

  21   θ =μ + θμ = 2 θμ. μ

=

9.2 Main Results

141

 2α   sup  D∗a f (a) ≤

 (2α + 1) (2α − 1)

22α−1

a∈R+



  23α+1 23α + 1 (2α + 1)  f ∞,R+ K .  (4α + 1)

Call

(9.37)

ξ := 2  f ∞,R+ , ψ :=

(9.38)

22α K , (4α+1)

both are greater than zero. We consider the function γ (h) := ξh −3α + ψh α , ∀ h > 0. We have

(9.39)

γ  (h) = −3αξh −3α−1 + αψh α−1 = 0,

then

ψh α−1 = 3ξh −3α−1

and ψh 4α = 3ξ, with unique solution

 h 0 := h crit.no. =

3ξ ψ

 4α1

.

(9.40)

We have that γ  (h) = 3α (3α + 1) ξh −3α−2 + α (α − 1) ψh α−2 .

(9.41)

We see 



γ (h 0 ) = 3α (3α + 1) ξ 

3ξ α ψ

  α−2 4α

 α

3ξ ψ

3ξ ψ

 −3α−2 4α

 3 (3α + 1) ξ

 α−2 4α

3ξ ψ

3 (3α + 1) ξ



3ξ + α (α − 1) ψ ψ −1

 α−2 4α

 + (α − 1) ψ =

 ψ + (α − 1) ψ = 3ξ

=

142

9 Sequential Left Abstract Fractional Landau Inequalities

 α  α

3ξ ψ

3ξ ψ

 α−2 4α

 α−2 4α

[(3α + 1) ψ + (α − 1) ψ] = 

(4αψ) = 4α2 ψ

Therefore γ has a global minimum at h 0 =  γ (h 0 ) = ξ 

3ξ ψ

3ξ ψ

− 34

3ξ ψ

 4α1 3ξ ψ

 α−2 4α

> 0.

, which is



3ξ +ψ ψ

 41

=

  1   14   −1  ψ 3ξ 4 3ξ ξ +ψ = +ψ = ξ ψ ψ 3ξ 

3ξ ψ

 41 

Consequently, γ (h 0 ) =

(9.42)

(9.43)

   41 4 4 3ξ ψ = ψ . 3 3 ψ

  14 3ξ 4 4 3 1 ψ = √ 3 ψ 4 ξ 4 . 3 ψ 4 3

(9.44)

We have proved that (see (9.30))    3α   4 (3α + 1) 22α + 1   sup D∗a f (a) ≤ √ 3 4 a∈R+ 3 22α (2α − 1)  1 2  f ∞,R+ 4



22α K  (4α + 1)

 34

=

√  3  1 4 4 2 (3α + 1)  (4α + 1)− 4 22α + 1 3 4  f ∞,R K 4. √ 3 α + 4 3 2 2 (2α − 1) The theorem is proved.

(9.45)



We continue with abstract L p left fractional Landau inequalities over R+ . Theorem 9.5 Let p, q > 1 : 1p + q1 = 1, 0 < α < 1. Let f ∈ C 1 (R+ , X ) with   kα  f ∞,R+ ,  f  ∞,R < ∞. For k = 1, 2, 3, we assume that D∗a f ∈ + 1 4α C ([a, ∞), X ) and D∗a f ∈ C ([a, ∞), X ), ∀ a ∈ R+ . We further assume that

9.2 Main Results

143



 4α  f  p,R+ sup  D∗a

 < ∞.

(9.46)

a∈R+

Then (1) under

1 2p

< α < 1, we get ⎡⎛

  1  ⎞  p   2α−  1 α 1 −3α 4α− 4α − 2  (2α) p  2α   ⎢ 4α 1 + 2 p ⎠ sup  D∗a f (a) ≤ ⎣⎝ 1 α 2 −1 2α − p a∈R+ 



1+2



α− 1p

2α 4α− 1p

⎤



 4α  f  p,R+ sup  D∗a



2α− 1p 4α− 1p



⎥ ⎦  f ∞,R+

1

 (4α) (q (4α − 1) + 1) q 



 2α 4α− 1p

< ∞.

(9.47)

a∈R+

(2) under

1 p

< α < 1, we get ⎡⎛

   ⎞  α− 1p   4α−  1 1 −2α p  3α   ⎢  (3α) 4α − p 6α 1 + 2 ⎠ sup  D∗a f (a) ≤ ⎣⎝ 1 α 2 −1 α− p a∈R+ 



1+2

2α−



1 p

 (4α) (q (4α − 1) + 1) 

3α 4α− 1p

⎤

1 q



 4α  f  p,R+ sup  D∗a





α− 1p 4α− 1p



⎥ ⎦  f ∞,R+ 

3α 4α− 1p

< ∞.

(9.48)

a∈R+

 2α    3α   That is sup  D∗a f (a) , sup  D∗a f (a) < ∞. a∈R+

a∈R+

Proof As in the proof of Theorem 9.4 we have that    x1    4α  1 4α−1  A =  f (x1 ) − f (a) − D∗a f (t) dt  (x1 − t) ≤  (4α) a (9.12)

2  f ∞,R+

1 +  (4α)

 a

x1

 4α   f (t) dt ≤ (x1 − t)4α−1  D∗a

144

9 Sequential Left Abstract Fractional Landau Inequalities



(q(4α−1)+1)

2  f ∞,R+

1 (x1 − a) q +  (4α) (q (4α − 1) + 1) q1

 4α  f  p,R+ sup  D∗a

(x1 −a=:h>0)

=

a∈R+

 4α− 1p

h 1  (4α) (q (4α − 1) + 1) q1

2  f ∞,R+ +







(9.49)

 4α  f  p,R+ , sup  D∗a

a∈R+

with 41p < α < 1. That is 

1

A ≤ 2  f ∞,R+ +

h 4α− p 1

 (4α) (q (4α − 1) + 1) q

where 41p < α < 1. We also have  (9.13)  B =   f (x2 ) − f (a) − 2  f ∞,R+ +

(x2 − a)

1  (4α)



2



(q(4α−1)+1) q

4α− 1p

h

a∈R+

(9.50)

   4α  f (t) dt  (x2 − t)4α−1 D∗a ≤

x2

a

 (4α) (q (4α − 1) + 1)

2  f ∞,R+ +

  4α  sup  D∗a f  p,R+ ,

1 q

 4α  f  p,R+ sup  D∗a

 (x2 −a=:2h)

=

a∈R+



4α− 1p 1

 (4α) (q (4α − 1) + 1) q

 (9.51)  4α  sup  D∗a f  p,R+ .

a∈R+

That is 1

B ≤ 2  f ∞,R+ +



1

24α− p h 4α− p 1

 (4α) (q (4α − 1) + 1) q

  4α  sup  D∗a f  p,R+ ,

a∈R+

(9.52)

where 41p < α < 1. We have assumed that  M :=

 4α  sup  D∗a f  p,R+

 < ∞.

(9.53)

c :=  (4α) (q (4α − 1) + 1) q > 0.

(9.54)

a∈R+

For convenience we call 1

So we have

9.2 Main Results

145

 A ≤ 2  f ∞,R+ + and

h

B ≤ 2  f ∞,R+ +

2

4α− 1p

c

M, (9.55)

4α− 1p

h c

4α− 1p

M,

where 41p < α < 1. Next we estimate the (9.16)-quantities and we have:  2α    D f (a) ≤ ∗a

(9.23)  3α 3α 1 2 h A + h 3α B = D (3α + 1)

(9.55) h 3α  (2α + 1)  3α 2  A + B ≤ 2α 5α α 2 h (2 − 1)

(9.56)

 1  (2α + 1) 23α h 4α− p 3α+1  f ∞,R+ + M+ 2 22α (2α − 1) h 2α c 2  f ∞,R+

 1 1 24α− p h 4α− p + M = c

 ⎤ 1 23α + 24α− p   3α+1 1 ⎣ 2 Mh 4α− p ⎦ = + 2  f ∞,R+ + c ⎡

 (2α + 1) 22α (2α − 1) h 2α

 ⎡ ⎤ 1  23α + 24α− p 1  (2α + 1) ⎣ 23α+1 + 2  f ∞,R+ Mh 2α− p ⎦ = + 22α (2α − 1) h 2α c  ⎡  ⎤  α− 1p −3α M 1 + 2  f ∞,R+ 2  (2α + 1) ⎣ 2 1 + 2 2α− 1p ⎦ h + . h 2α c (2α − 1) α

That is

(9.57)

 2α    D f (a) ≤ ∗a



 ⎤ ⎡  1   1 + 2α− p M 1 2α  (2α + 1) ⎣ 2 1 + 2−3α  f ∞,R+ h 2α− p ⎦ , (9.58) + 2α − 1 h 2α c

∀ a ∈ R+ , ∀ h > 0. I.e. it holds

 2α   f (a) ≤ sup  D∗a

a∈R+



2α  (2α + 1) 2α − 1



146

9 Sequential Left Abstract Fractional Landau Inequalities

 ⎡  ⎤ 1  1 + 2α− p M 2 1 + 2−3α  f ∞,R+ 1 ⎣ h 2α− p ⎦ , + h 2α c

(9.59)

∀ h > 0, under 41p < α < 1. Again from (9.16) we get  3α    D f (a) ≤ ∗a

 2α 1 h B + 22α h 2α A =  (2α + 1) D

h 2α  (3α + 1)  B + 22α A ≤ 2α 5α α 2 h (2 − 1) 

h 2α  (3α + 1) 22α h 5α (2α − 1)

1 1  24α− p h 4α− p M+ 2  f ∞,R+ + c

22α+1  f ∞,R+ 

 (3α + 1) 22α (2α − 1) h 3α



(9.60)

 1 22α h 4α− p M = + c

 ⎤ 1 22α + 24α− p   1 ⎣ 2 + 22α+1  f ∞,R + Mh 4α− p ⎦ = + c ⎡

 ⎡ ⎤ 1  22α + 24α− p 1  (3α + 1) ⎣ 2 + 22α+1  f ∞,R+ Mh α− p ⎦ = + 22α (2α − 1) h 3α c

(9.61)

 ⎡  ⎤  2α− 1p −2α 1 + 2  f ∞,R+  (3α + 1) ⎣ 2 1 + 2 α− 1p ⎦ + . Mh h 3α c (2α − 1) That is ⎤ ⎡    2α− 1p     (3α + 1)  2 1 + 2−2α  f  1+2 1 ⎥ ∞,R+ ⎢  3α  α− + Mh p ⎦ ,  D∗a f (a) ≤ ⎣ 2α − 1 h 3α c

(9.62) ∀ a ∈ R+ , ∀ h > 0. I.e. it holds

 3α   f (a) ≤ sup  D∗a

a∈R+



 (3α + 1) 2α − 1



9.2 Main Results

147

 ⎡  ⎤ 1  1 + 22α− p 2 1 + 2−2α  f ∞,R+ 1 ⎣ Mh α− p ⎦ , + h 3α c ∀ h > 0, Call

1 4p

(9.63)

< α < 1.   μ := 2 1 + 2−3α  f ∞,R+ , θ :=

1+2

α− 1p

M

c

(9.64)

,

both are greater than zero. We consider the function 1

We have

y (h) = μh −2α + θh 2α− p , ∀ h > 0.

(9.65)

  1 1 θh 2α− p −1 = 0, y  (h) = −2αμh −2α−1 + 2α − p

(9.66)

then

  1 1 2α − θh 2α− p −1 = 2αμh −2α−1 p

and

  1 1 2α − θh 2α− p −1 h 2α+1 = 2αμ, p

i.e.

  1 1 2α − θh 4α− p = 2αμ, p

with a unique solution ⎞

⎛ 2αμ

⎠ h 0 := h crit.no. = ⎝  2α − 1p θ

1 4α− 1p

(9.67)

(assuming 21p < α < 1). We have that    1 1 1 2α − − 1 θh 2α− p −2 . (9.68) y  (h) = 2α (2α + 1) μh −2α−2 + 2α − p p We see that

148

9 Sequential Left Abstract Fractional Landau Inequalities

⎞ −2α−21

⎛ 2αμ

⎠ y (h 0 ) = 2α (2α + 1) μ ⎝  2α − 1p θ 



4α− p

+

⎞ 2α− 1p 1−2 ⎛   4α− p 1 1 2αμ ⎠ 2α − 2α − − 1 θ ⎝  = p p 2α − 1 θ p

⎞ −2α−21

⎛

⎝  2αμ ⎠ 2α − 1p θ

4α− p

[2α (2α + 1) μ+

⎞⎤ ⎛    1 2αμ 1 ⎠⎦ = 2α − − 1 θ ⎝  2α − p p 2α − 1 θ

(9.69)

p

⎛

⎞ −2α−21 2αμ

⎝ ⎠ 2α − 1p θ

4α− p

  1 2α (2α + 1) μ + 2αμ 2α − − 1 = p

⎞ ⎛ 2α − 1p θ ⎠ 2αμ ⎝ 2αμ



 2(α+1) 4α− 1p



1 4α − p

 > 0.

Therefore y has a global minimum at ⎛

⎞ 2αμ

⎠ h0 = ⎝  2α − 1p θ

1 4α− 1p

,

which is ⎞

⎛ y (h 0 ) = μ ⎝ 

⎛ ⎝

⎞ 2αμ

⎠ 2α − 1p θ

−2α 4α− 1p

2αμ

⎠ 2α − 1p θ

−2α 4α− 1p

⎛

2αμ ⎠ + θ ⎝ 2α − 1p θ

⎞ 2α − 1p θ 2αμ ⎠ ⎣μ + θ  ⎦ = μ⎝ 2αμ 2α − 1 θ ⎡

⎤

p

⎞ 2α− 11p

⎛

4α− p



=

(9.70)

 2α 4α− 1p

 1+



2α 2α −

1 p

9.2 Main Results

149

⎞ ⎛ 2α − 1p θ ⎠ = μ⎝ 2αμ 

4α − 2α −

1 p 1 p







2α −



4α − 2α −



1 p



2α 4α− 1p

 2α 4α− 1p



μ

2α

1 p 1 p

 =

 

2α− 1p 4α− 1p

 2α 4α− 1p

θ

=

  −2α+ 1p  2α− 1p   2α  4α − 1p  1 4α− 1p 1 1   2α − μ 4α− p θ 4α− p . 2α p 1 (2α) 4α− p That is



y (h 0 ) =

  4α − 1p 2α − 1p 

(2α)

−2α+ 1p 4α− 1p

(9.71)

 

μ



2α 4α− 1p

2α− 1p 4α− 1p

 

θ

 2α 4α− 1p

.

(9.72)

Therefore we derive (see (9.59))  2α   sup  D∗a f (a) ≤

a∈R+



   2 1 + 2−3α



2α− 1p 4α− 1p

⎛ ⎝





α

2  (2α) 2α − 1 



1+2

 (2α− 1p )  (4α− p ) 1

2α 2α −

1 p 



α− 1p

⎞ 1

 (4α) (q (4α − 1) + 1) q 

 4α  f  p,R+ sup  D∗a



 2α 4α− 1p

⎠

4α −

1 p 



2α− 1p 4α− 1p



 f ∞,R+

 2α 4α− 1p

< ∞,

(9.73)

a∈R+

where 21p < α < 1. Call

  ξ := 2 1 + 2−2α  f ∞,R+ , ψ :=

both are greater than zero. We consider the function

  2α− 1p 1+2 c

(9.74) M,

150

9 Sequential Left Abstract Fractional Landau Inequalities

γ (h) := ξh −3α + ψh α− p , ∀ h > 0. 1

(9.75)

  1 1 ψh α− p −1 = 0, γ  (h) = −3αξh −3α−1 + α − p

We have

then

  1 1 α− ψh α− p −1 = 3αξh −3α−1 p

and

  1 1 α− ψh α− p −1 h 3α+1 = 3αξ, p

and

  1 1 α− ψh 4α− p = 3αξ, p

with unique solution ⎛

⎞

h 0 := h crit.no. = ⎝ 

3αξ α−

1 p

⎠ ψ

1 4α− 1p

(9.76)

(assuming 1p < α < 1). We have that    1 1 1 α − − 1 ψh α− p −2 . γ  (h) = 3α (3α + 1) ξh −3α−2 + α − p p

(9.77)

We observe that ⎞

⎛



3αξ ⎠ γ  (h 0 ) = 3α (3α + 1) ξ ⎝  α − 1p ψ

α−

1 p



+ 

⎞

⎛



−3α−2 4α− 1p

  3αξ 1 ⎠ α − − 1 ψ ⎝ p α− 1 ψ

α− 1p −2 4α− 1p



=

p



⎞

⎛ 3αξ

⎝ ⎠ α − 1p ψ

 −3α−2 4α− 1p

⎡



⎣3α (3α + 1) ξ + α − 1 p



⎞⎤ ⎛  1 3αξ ⎠⎦ α − − 1 ψ ⎝ p α− 1 ψ p

9.2 Main Results

151 

⎛

⎞

3αξ ⎠ = ⎝ α − 1p ψ

−3α−2 4α− 1p



   1 3α (3α + 1) ξ + α − − 1 3αξ = p ⎞

⎛



3αξ ⎠ 3αξ ⎝  α − 1p ψ

−3α−2 4α− 1p



  1 (3α + 1) + α − − 1 = p 

⎞

⎛

(9.78)

3αξ ⎠ 3αξ ⎝  α − 1p ψ

−3α−2 4α− 1p



  1 4α − > 0. p

Therefore y has a global minimum at ⎞

⎛

3αξ ⎠ h0 = ⎝  α − 1p ψ which is

α− 1p

γ (h 0 ) = ξh 0−3α + ψh 0 ⎞

⎛



3αξ

⎝ ⎠ α − 1p ψ ⎛

−3α 4α− 1p



⎞ 3αξ

⎠ ξ ⎝ α − 1p ψ

 γ (h 0 ) = ξ

(9.79)

⎞⎞

⎛ 3αξ

⎝ξ + ψ ⎝  ⎠⎠ = α − 1p ψ 

−3α 4α− 1p

3αξ ⎠ ξ ⎝ 1 α− p ψ That is

,

  4α− 1p ξ + ψh = = h −3α 0 0

⎛

⎞

⎛

1 4α− 1p





−3α 4α− 1p



3α 1+ α − 1p





4α − α−

1 p 1 p



⎞ ⎛ 4α − 1p ⎝  3αξ ⎠ α − 1p α − 1p ψ

 =

 .

−3α 4α− 1p



=

(9.80)

152

9 Sequential Left Abstract Fractional Landau Inequalities 





4α − α−

⎞  ⎛ α − 1p ψ 4α − 1p ⎠ ξ⎝ 3αξ α − 1p  1 p 1 p



ξ

α− 1p 4α− 1p



⎞ ⎛ α − 1p ⎝ ⎠ 3α

 3α 4α− 1p

=



 3α 4α− 1p

⎛

⎞

⎝

3α ⎠ 4α− 1p

ψ

.

I.e. we have found  γ (h 0 ) =



α−

1 p



4α −



α− 1p 4α− 1p



1 p



 ξ



α− 1p 4α− 1p



⎛

⎞

⎝

3α ⎠ 4α− 1p

ψ

.

(9.81)

3α 4α− 1p

(3α)

We have proved that (see (9.63))   ⎛ ⎞  α− 1p1 1 4α− p 4α − p  (3α)  3α   3α ⎠ f (a) ≤ ⎝ sup  D∗a 1 α 2 −1 α− p a∈R+



   2 1 + 2−2α

α− 1p 4α− 1p





⎛ ⎝



 1 1 + 22α− p

⎞ 1

 (4α) (q (4α − 1) + 1) q 

 4α  f  p,R+ sup  D∗a



 3α 4α− 1p

⎠

(9.82)



α− 1p 4α− 1p



 f ∞,R+

 3α 4α− 1p

< ∞,

a∈R+

where 1p < α < 1. The theorem is proved.



We give an application when α = 21 .

  Corollary 9.6 Let f ∈ C 1 (R+ , X ) with  f ∞,R+ ,  f  ∞,R+ < ∞, where k1

(X, ·) is a Banach space. For k = 1, 2, 3, we assume that D∗a2 f ∈ C 1 ([a, ∞), X ) 41

and D∗a2 f ∈ C ([a, ∞), X ), ∀ a ∈ R+ . We further assume that  1   4  K :=  D∗a2 f (t)

∞,R2+

where (a, t) ∈ R2+ .

< ∞,

(9.83)

9.2 Main Results

153

Then  1   2  sup  D∗a2 f (a) ≤



a∈R+

and

√    21  1 12 + 6 2  4 21  2  f ∞,R+ D∗a f (t) 2 < ∞, √ ∞,R+ 2−1 (9.84)  1   3  sup  D∗a2 f (a) ≤ a∈R+

⎞ √   34  1 9 π  4 21 ⎟  ⎜ 4 < ∞. ⎝ √ √ √ 3 ⎠  f ∞,R+  D∗a f (t)∞,R2+ 2− 2 42 43 ⎛

(9.85)

 1   1   2  3   That is sup  D∗a2 f (a) , sup  D∗a2 f (a) < ∞. a∈R+

Proof By Theorem 9.4.

a∈R+



References 1. Aglic Aljinovic, A., Marangunic, L., Pecaric, J.: On Landau type inequalities via Ostrowski inequalities. Nonlinear Funct. Anal. Appl. 10(4), 565–579 (2005) 2. Anastassiou, G.: Fractional Differentiation Inequalities. Research Monograph. Springer, New York (2009) 3. Anastassiou, G.A.: Advances on Fractional Inequalities. Springer, New York (2011) 4. Anastassiou, G.A.: Intelligent Computations: Abstract Fractional Calculus, Inequalities, Approximations. Springer, Heidelberg (2018) 5. Anastassiou, G.A.: Iterated left abstract fractional Landau inequalities. J. App. Math. Inf., accepted, (2020) 6. Barnett, N.S., Dragomir, S.S.: Some Landau type inequalities for functions whose derivatives are of locally bounded variation. Tamkang J. Math. 37(4), 301–308 (Winter 2006) 7. Ditzian, Z.: Remarks, questions and conjectures on Landau-Kolmogorov-type inequalities. Math. Inequal. Appl. 3, 15–24 (2000) 8. Hardy, G.H., Littlewood, J.E.: Some integral inequalities connected with the calculus of variations. Quart. J. Math. Oxford Ser. 3, 241–252 (1932) 9. Hardy, G.H., Landau, E., Littlewood, J.E.: Some inequalities satisfied by the integrals or derivatives of real or analytic functions. Math. Z. 39, 677–695 (1935) 10. Kallman, R.R., Rota, G.C.: On theinequality  f  2 ≤ 4 f  ·  f  . In: Shisha, O., (ed.) Inequalities, vol. II, pp. 187–192. Academic, New York (1970) 11. Landau, E.: Einige Ungleichungen für zweimal differentzierban funktionen. Proc. Lond. Math. Soc. 13, 43–49 (1913) 12. Mikusinski, J.: The Bochner Integral. Academic, New York (1978) 13. Shilov, G.E.: Elementary Functional Analysis. Dover Publications Inc., New York (1996)

Chapter 10

Iterated Left Abstract Generalized Fractional Landau Inequalities

We give uniform and L p left Caputo–Bochner abstract iterated generalized fractional Landau inequalities over R+ . These estimate the size of second and third sequential left abstract generalized fractional derivatives of a Banach space valued function over R+ . We give an application when the basic fractional order is 21 . It follows [5].

10.1 Introduction Let p ∈ [1, ∞], I = R+ or I = R and f : I → R is twice differentiable with f, f  ∈ L p (I ), then f  ∈ L p (I ). Moreover, there exists a constant C p (I ) > 0 independent of f , such that 1  1    f  ≤ C p (I )  f  2  f   2 , (10.1) p,I

p,I

p,I

where · p,I is the p-norm on the interval I , see [1, 6]. The research on these inequalities started by Landau [11] in 1913. For the case of p = ∞ he proved that C∞ (R+ ) = 2 and C∞ (R) =

√ 2,

(10.2)

are the best constants in (10.1). In 1932, Hardy and Littlewood [8] proved (10.1) for p = 2, with the best constants C2 (R+ ) =

√

2 and C2 (R) = 1.

(10.3)

In 1935, Hardy, Landau and Littlewood [9] showed that the best constants C p (R+ ) in (10.1) satisfies the estimate © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_10

155

156

10 Iterated Left Abstract Generalized Fractional Landau Inequalities

C p (R+ ) ≤ 2, for p ∈ [1, ∞),

(10.4)

which yields C p (R) ≤ 2 for p ∈ [1, ∞). √ In fact, in [7, 10] was shown that C p (R) ≤ 2. We need the following concept from abstract fractional calculus. Our integral next is of Bochner type [12]. We need Definition 10.1 ([4], p. 104) Let [a, b] ⊂ R, (X, ·) a Banach space, g ∈ C 1 ([a, b]) and increasing, f ∈ C ([a, b] , X ), ν > 0. We define the left Riemann–Liouville generalized fractional Bochner integral operator 

 ν f (x) := Ja;g

1  (ν)



x

(g (x) − g (z))ν−1 g  (z) f (z) dz,

(10.5)

a

∀ x ∈ [a, b], where  is the gamma function. The last integral is of Bochner type. Since f∈ C ([a,  b] , X ), then f ∈ L ∞ ν ([a, b] , X ). By Theorem 4.10, p. 98, [4], we get that Ja;g f ∈ C ([a, b] , X ). Above   ν 0 we set Ja;g f := f and see that Ja;g f (a) = 0. We also need Definition 10.2 ([4], p. 106) Let α > 0, α = n, · the ceiling of the number. Let f ∈ C n ([a, b] , X ), where [a, b] ⊂ R, and (X, ·) is a Banach space. Let g ∈ C 1 ([a, b]) , strictly increasing, such that g −1 ∈ C n ([g (a) , g (b)]) . We define the left generalized g-fractional derivative X -valued of f of order α as follows:  x (n)   α  1 Da+;g f (x) := (g (x) − g (t))n−α−1 g  (t) f ◦ g −1 (g (t)) dt,  (n − α) a (10.6) ∀ x ∈ [a, b]. The last integral is of Bochner type. Ordinary vector valued derivative is as in [13], similar one.   to numerical α If α ∈ / N, by Theorem 4.10, p. 98, [4], we have that Da+;g f ∈ C ([a, b] , X ). We see that    (n)  α  n−α Ja;g f ◦ g −1 ◦ g (x) = Da+;g f (x) , ∀ x ∈ [a, b] . (10.7) We set n f (x) := Da+;g



f ◦ g −1

n

 ◦ g (x) ∈ C ([a, b] , X ) , n ∈ N,

0 f (x) = f (x) , ∀ x ∈ [a, b] . Da+;g

(10.8)

10.1 Introduction

157

When g = id, then

α α α f = Da+;id f = D∗a f, Da+;g

(10.9)

the usual left X -valued Caputo fractional derivative, see [4], Chap. 1. By convention we suppose that 

 Dxα0 +;g f (x) = 0, for x < x0 ,

(10.10)

for any x, x0 ∈ [a, b] . Denote the sequential (also called iterated) generalized left fractional derivative by nα α α α := Da+;g Da+;g ...Da+;g (n times), n ∈ N. (10.11) Da+;g We need the following g-left generalized modified X -valued Taylor’s formula. Theorem 10.3 ([4], p. 117) Let 0 < α ≤ 1, n∈N, f ∈C 1 ([a, b] , X ), g ∈ C 1 ([a, b]) kα f , k = 1, ..., n, strictly increasing, such that g −1 ∈ C 1 ([g (a) , g (b)]). Let Fk := Da+;g 1 that fulfill Fk ∈ C ([a, b] , X ), and Fn+1 ∈ C ([a, b] , X ) . Then n   (g (x) − g (a))iα  iα Da+;g f (a) + (10.12) f (x) =  (iα + 1) i=0 1  ((n + 1) α)

 a

x

  (n+1)α f (t) dt, (g (x) − g (t))(n+1)α−1 g  (t) Da+;g

∀ x ∈ [a, b] . We make Remark 10.4 (to Theorem 10.3) When 0 < α < 1, by (10.6) we get 

 α Da+;g f (x) =

1  (1 − α)



x

  (g (x) − g (t))−α g  (t) f ◦ g −1 (g (t)) dt,

a

(10.13)

∀ x ∈ [a, b] . Hence  α  D

   a+;g f (x) ≤

≤

1  (1 − α)       f ◦ g −1 ◦ g 



x

a

∞,[a,b]

 (1 − α)

     (g (x) − g (t))−α g  (t)  f ◦ g −1 (g (t)) dt  a

x

(g (x) − g (t))

−α



g (t) dt

(10.14)

158

10 Iterated Left Abstract Generalized Fractional Landau Inequalities

=

      f ◦ g −1 ◦ g 

∞,[a,b]

 (2 − α)

(g (x) − g (a))1−α .

That is  α  D

a+;g

  f (x) ≤

      f ◦ g −1 ◦ g   (2 − α)

∀ x ∈ [a, b] , 0 < α < 1. Hence it holds

 α  D

a+;g

i.e.



∞,[a,b]

(g (x) − g (a))1−α < ∞, (10.15)

  f (a) = 0,

 α Da+;g f (a) = 0,

(10.16)

when 0 < α < 1. The author has already done an extensive amount of work on fractional Landau inequalities, see [3], and on abstract fractional Landau inequalities, see [4]. However there the proving methods came out of applications of fractional Ostrowski inequalities [2, 4]. Usually there the domains where [A, +∞) or (−∞, B], with A, B ∈ R and in one mixed case the domain was all of R. In this chapter with less assumptions we establish uniform and L p type left Caputo–Bochner abstract sequential generalized fractional Landau inequalities over R+ . The method of proving is based on left Caputo–Bochner sequential generalized fractional Taylor’s formula with integral remainder, see Theorem 10.3. We give also an application for α = 21 . Regardless to say that we are also inspired by [3, 4].

10.2 Main Results We present the following abstract iterated generalized fractional Landau inequalities over R+ . −1 1 Theorem 10.5 Let g ∈ C 1 (R+ ) strictly increasing,  (g (R+ )) . Let  with g ∈C   1 −1  0 < α < 1, f ∈ C (R+ , X ) with  f ∞,R+ ,  f ◦ g ◦ g  < ∞. For ∞,R+

kα 4α k = 1, 2, 3, we assume that Da+;g f ∈ C 1 ([a, ∞), X ) and Da+;g f ∈ C ([a, ∞), X ), ∀ a ∈ R+ . We further assume that

 4α  f (t)∞,R2 < ∞, K g :=  Da+;g +

where (a, t) ∈ R2+ .

(10.17)

10.2 Main Results

159

Then  2α   f (a) ≤ sup  Da+;g

a∈R+

 (2α + 1) 22α−1 (2α − 1)



  23α+1 23α + 1 (2α + 1)  f ∞,R+ K g ,  (4α + 1) (10.18)

and √  3  1 3  3α   4 4 2 (3α + 1) ( (4α + 1))− 4 22α + 1 4    f ∞,R sup Da+;g f (a) ≤ K g4 . √ 3 √ α + 4 a∈R+ 3 2 (2α − 1) (10.19)        2α  3α   That is sup  Da+;g f (a) , sup  Da+;g f (a) < ∞. a∈R+

a∈R+

  α Proof We notice easily again here that Da+;g f (a) = 0, ∀ a ∈ R+ . We make use of Theorem 10.3 for 0 < α < 1 and n = 3, applied for any a ∈ R+ and b = ∞. Momentarily we fix a ∈ R+ . Let x2 > x1 > a, then g (x2 ) > g (x1 ) > g (a), and f (x1 ) − f (a) =

+

  (g (x1 ) − g (a))2α  2α (g (x1 ) − g (a))3α  3α Da+;g f (a) + Da+;g f (a)  (2α + 1)  (3α + 1) 1  (4α)



  4α f (t) dt, (g (x1 ) − g (t))4α−1 g  (t) Da+;g

x1

a

(10.20)

and   (g (x2 ) − g (a))2α  2α (g (x2 ) − g (a))3α  3α Da+;g f (a) + Da+;g f (a)  (2α + 1)  (3α + 1) (10.21)  x2   1 4α−1  4α + g (t) Da+;g f (t) dt. (g (x2 ) − g (t))  (4α) a

f (x2 ) − f (a) =

That is  (g (x1 ) − g (a))2α  2α  (g (x1 ) − g (a))3α  3α Da+;g f (a) + Da+;g f (a) =  (2α + 1)  (3α + 1) f (x1 ) − f (a) −

1  (4α)

 a

x1

(10.22)

  4α f (t) dt =: A, (g (x1 ) − g (t))4α−1 g  (t) Da+;g

and  (g (x2 ) − g (a))2α  2α  (g (x2 ) − g (a))3α  3α Da+;g f (a) + Da+;g f (a) =  (2α + 1)  (3α + 1)

(10.23)

160

10 Iterated Left Abstract Generalized Fractional Landau Inequalities

f (x2 ) − f (a) −

1  (4α)



x2

a

  4α f (t) dt =: B. (g (x2 ) − g (t))4α−1 g  (t) Da+;g

  2α We are solving the above system of two equations with two unknowns Da+;g f (a) ,   3α f (a) . Da+;g The main determinant of system is (g(x1 )−g(a))2α (2α+1) D := (g(x2 )−g(a))2α (2α+1)

(g(x1 )−g(a))3α (3α+1) (g(x2 )−g(a))3α (3α+1)

1 =  (2α + 1)  (3α + 1)



(g (x1 ) − g (a))2α (g (x2 ) − g (a))3α − (g (x1 ) − g (a))3α (g (x2 ) − g (a))2α =  (g (x1 ) − g (a))2α (g (x2 ) − g (a))2α (g (x2 ) − g (a))α − (g (x1 ) − g (a))α > 0.  (2α + 1)  (3α + 1) I.e. D=

 (g (x1 ) − g (a))2α (g (x2 ) − g (a))2α (g (x2 ) − g (a))α − (g (x1 ) − g (a))α > 0.  (2α + 1)  (3α + 1)

(10.24)

We obtain the unique solution





2α Da+;g



f (a) =

 3α f (a) = Da+;g



A

(g(x1 )−g(a))3α (3α+1)

B

(g(x2 )−g(a))3α (3α+1) D

(g(x1 )−g(a))2α (2α+1) (g(x2 )−g(a))2α (2α+1) D



A B

, (10.25)

.

Therefore we have 

 2α Da+;g f (a) = and   3α f (a) = Da+;g

We have the following

(g(x2 )−g(a))3α A− (g(x1 )−g(a))3α B (3α+1) (3α+1) , D

(10.26) (g(x1 )−g(a))2α B− (g(x2 )−g(a))2α A (2α+1) (2α+1) . D

10.2 Main Results

161

  A =   f (x1 ) − f (a) −

2  f ∞,R+ +

  x1    1 4α f (t) dt  ≤ (g (x1 ) − g (t))4α−1 g  (t) Da+;g   (4α) a

   4α   Da+;g f (t)

∞,R2+

 (4α + 1)

(g (x1 ) − g (a))4α ,

(10.27)

under the assumption  f ∞,R+ < ∞. That is A ≤ 2  f ∞,R+ +

Kg (g (x1 ) − g (a))4α ,  (4α + 1)

(10.28)

Kg (g (x2 ) − g (a))4α ,  (4α + 1)

(10.29)

and similarly, B ≤ 2  f ∞,R+ + where by assumption  4α  f (t)∞,R2 < ∞, K g :=  Da+;g

(10.30)

+

with (a, t) ∈ R2+ . Consequently we have  2α  D

a+;g

  f (a) ≤

 3α 2  f ∞,R+ + (g (x2 ) − g (a)) 2  f ∞,R+ + + (g (x1 ) − g (a)) 3α

and

 3α  D

a+;g

Kg (g (x1 ) − g (a))4α  (4α + 1)

Kg (g (x2 ) − g (a))4α  (4α + 1)

  f (a) ≤

 2α 2  f ∞,R+ + (g (x1 ) − g (a)) + (g (x2 ) − g (a))2α 2  f ∞,R+ +

1  (3α + 1) D



 , (10.31)

1  (2α + 1) D Kg (g (x2 ) − g (a))4α  (4α + 1)

Kg (g (x1 ) − g (a))4α  (4α + 1)



 . (10.32)

Set now g (x1 ) := g (a) + h, g (x2 ) := g (a) + 2h, where h > 0, so that g (x1 ) − g (a) = h, g (x2 ) − g (a) = 2h.

162

10 Iterated Left Abstract Generalized Fractional Landau Inequalities

Hence we get D=

22α h 5α (2α − 1) > 0.  (2α + 1)  (3α + 1)

(10.33)

Therefore we derive (from (10.26))    a+;g f (a) ≤

 2α  D



  (2α + 1) Kg 3α 3α 4α   h h + 2 f 2 ∞,R+ 22α h 5α (2α − 1)  (4α + 1)

+h 3α 2  f ∞,R+ +

Kg 24α h 4α  (4α + 1)

 =

(10.34)

   3α  3α  3α  7α Kg  (2α + 1) 4α 2  f ∞,R+ 2 + 1 h + 2 +2 h 22α h 5α (2α − 1)  (4α + 1) =

 (2α + 1) 22α (2α − 1)

That is



  3α 2 2 + 1  f ∞,R+ h 2α    a+;g f (a) ≤

 2α  D

   2 23α + 1  f ∞,R+ h 2α ∀ a ∈ R+ , ∀ h > 0. I.e. it holds



a∈R+

h 2α

 (2α + 1) 22α (2α − 1)



(10.35)



 23α (2α + 1) 2α Kg h + ,  (4α + 1)

 2α   f (a) ≤ sup  Da+;g

   2 23α + 1  f ∞,R+

 23α (2α + 1) K g h 2α . +  (4α + 1)

 (2α + 1) 22α (2α − 1)

(10.36)



 23α (2α + 1) 2α Kg h + < ∞,  (4α + 1)

(10.37)

∀ h > 0, 0 < α < 1. By (10.26) we derive  3α  D

   a+;g f (a) ≤



  (3α + 1) Kg 2α 4α 4α   2 f h 2 + h ∞,R+ 22α h 5α (2α − 1)  (4α + 1)

+22α h 2α 2  f ∞,R+ +

Kg h 4α  (4α + 1)

 =

10.2 Main Results

163

   4α  2α  2α  6α  (3α + 1) Kg 2α   2 f 2 + 1 h + + 2 2 h ∞,R+ 22α h 5α (2α − 1)  (4α + 1) (10.38)     

2 22α + 1  f ∞,R+ 22α 22α + 1  (3α + 1) = Kg hα + 22α (2α − 1) h 3α  (4α + 1)     (3α + 1) 22α + 1 2  f ∞,R+ 22α K g α h . = + 22α (2α − 1) h 3α  (4α + 1)     (3α + 1) 22α + 1  a+;g f (a) ≤ 22α (2α − 1)

That is

 3α  D 



2  f ∞,R+ h 3α

 22α K g α h , +  (4α + 1)

(10.39)

∀ a ∈ R+ , ∀ h > 0. I.e. it holds

   3α    (3α + 1) 22α + 1 sup  Da+;g f (a) ≤ 22α (2α − 1) a∈R+ 

2  f ∞,R+ h 3α

∀ h > 0, 0 < α < 1. Call

+

 22α K g h α < ∞,  (4α + 1)

  μ := 2 23α + 1  f ∞,R+ , θ=

23α (2α + 1) K g ,  (4α + 1)

(10.40)

(10.41) (10.42)

both are greater than zero. Set also ρ := 2α; 0 < ρ < 2. We consider the function

We have

y (h) := μh −ρ + θ h ρ , ∀ h > 0.

(10.43)

y  (h) = −ρμh −ρ−1 + ρθ h ρ−1 = 0,

(10.44)

then θ h 2ρ = μ, with a unique solution

164

10 Iterated Left Abstract Generalized Fractional Landau Inequalities

h 0 := h crit.no. = We have that

 μ  2ρ1 θ

.

y  (h) = ρ (ρ + 1) μh −ρ−2 + ρ (ρ − 1) θ h ρ−2 .

(10.45)

(10.46)

We see that y  (h 0 ) = y 

  1

 μ  −ρ−2  μ  ρ−2 μ 2ρ 2ρ 2ρ = ρ (ρ + 1) μ + ρ (ρ − 1) θ = θ θ θ ρ

ρ

ρ1     θ (ρ + 1) μθ + (ρ − 1) μθ = μ

ρ1  ρ1    θ θ 2ρ μθ = 2ρ 2 μθ > 0. μ μ

Therefore y has a global minimum at h 0 = y (h 0 ) = μ

 μ − 21 θ

+θ

 μ  21 θ

 μ  2ρ1 θ

=μ

, which is

21   θ + θ μ = 2 θ μ. μ

We have proved that (see (10.37))  2α   sup  Da+;g f (a) ≤

a∈R+



 (2α + 1) (2α − 1)

22α−1

  23α+1 23α + 1 (2α + 1)  f ∞,R+ K g .  (4α + 1)

Call

(10.47)

ξ := 2  f ∞,R+ , ψ :=

22α K g , (4α+1)

(10.48)

both are greater than zero. We consider the function γ (h) := ξ h −3α + ψh α , ∀ h > 0. We have

γ  (h) = −3αξ h −3α−1 + αψh α−1 = 0,

(10.49)

10.2 Main Results

165

then ψh 4α = 3ξ, with unique solution

h 0 := h crit.no. =

3ξ ψ

4α1

.

(10.50)

We have that γ  (h) = 3α (3α + 1) ξ h −3α−2 + α (α − 1) ψh α−2 .

(10.51)

We see γ  (h 0 ) = 3α (3α + 1) ξ α

3ξ ψ

α



α−2 4α

3ξ ψ

α−2 4α



3ξ ψ

−3α−2 4α

3 (3α + 1) ξ



γ (h 0 ) = ξ 3ξ ψ

3ξ ψ

− 34

3ξ ψ

3ξ ψ

  4α1 3ξ ψ

4 γ (h 0 ) = ψ 3



3ξ ψ

14

=

> 0.

(10.52)

, which is

+ψ

α−2 4α

3ξ ψ

41

=

14

14 ψ 4 3ξ ξ +ψ = ψ . 3ξ 3 ψ

Consequently,

α−2 4α

 ψ + (α − 1) ψ = 3ξ

(4αψ) = 4α 2 ψ

Therefore γ has a global minimum at h 0 =



+ α (α − 1) ψ

4 3 1 = √ 3 ψ 4 ξ 4 . 4 3

We have proved that (see (10.40))    3α   4 (3α + 1) 22α + 1   sup Da+;g f (a) ≤ √ 3 4 a∈R+ 3 22α (2α − 1)

(10.53)

(10.54)

166

10 Iterated Left Abstract Generalized Fractional Landau Inequalities

 1 2  f ∞,R+ 4



22α K g  (4α + 1)

34

=

√  3  1 3 4 4 2 (3α + 1)  (4α + 1)− 4 22α + 1 4  f ∞,R K g4 . √ 3 α + 4 3 2 2 (2α − 1)

(10.55)



The theorem is proved.

We continue with abstract L p left sequential generalized fractional Landau inequalities over R+ . Theorem 10.6 Let g ∈ C 1 (R+ ) strictly increasing, with g −1 ∈ C 1 (g (R+ )) . Let p, q > 1 : 1p + q1 = 1, 0 < α < 1. Let f ∈ C 1 (R+ , X ) with  f ∞,R+ ,      kα < ∞. For k = 1, 2, 3, we assume that Da+;g f ∈ C1  f ◦ g −1 ◦ g  ∞,R+

4α f ∈ C ([a, ∞), X ), ∀ a ∈ R+ . We further assume that ([a, ∞), X ) and Da+;g



  4α f  p,R sup  Da+;g

a∈R+

Then (1) under

1 2p

 < ∞.

+

(10.56)

< α < 1, we get ⎡⎛



1 ⎞   p   2α−  1 α 1 −3α 4α− 2  4α − (2α) p  2α   ⎢ 4α 1 + 2 p ⎠ sup  Da+;g f (a) ≤ ⎣⎝ α −1 2 2α − 1p a∈R+



1+2



α− 1p

2α 4α− 1p

1

 (4α) (q (4α − 1) + 1) q 

  4α f  p,R sup  Da+;g

1 p

 +



2α− 1p



4α− 1p ⎥ ⎦  f ∞,R+



a∈R+

(2) under

⎤

2α 4α− 1p

< ∞.

(10.57)

< α < 1, we get ⎡⎛



 ⎞  α− 1p   4α−  1 1 −2α  4α − (3α) p  3α   ⎢ 6α 1 + 2 p ⎠ sup  Da+;g f (a) ≤ ⎣⎝ 1 α 2 −1 α− p a∈R+

10.2 Main Results

167



1+2



2α− 1p

3α 4α− 1p

⎤



  4α f  p,R sup  Da+;g

a∈R+

    2α  That is sup  Da+;g f (a) , sup a∈R+

a∈R+



α− 1p 4α− 1p



⎥ ⎦  f ∞,R+

1

 (4α) (q (4α − 1) + 1) q 



3α 4α− 1p

< ∞.

+

(10.58)

    3α   Da+;g f (a) < ∞.

Proof As in the proof of Theorem 10.5 we have that A

(10.22.)

=

   x1     4α−1  4α  f (x1 ) − f (a) − 1 ≤ − g D g f dt (g (x ) (t)) (t) (t) 1 a+;g    (4α) a

2  f ∞,R+ +

1  (4α)



x1

a

    4α  f (t) dt ≤ (g (x1 ) − g (t))4α−1 g  (t)  Da+;g (q(4α−1)+1)

2  f ∞,R+

q 1 (g (x1 ) − g (a)) + 1  (4α) (q (4α − 1) + 1) q



2  f ∞,R+



   4α  f  sup Da+;g

a∈R+

4α− 1



p h 1 +  (4α) (q (4α − 1) + 1) q1



 (g(x 1 )−g(a)=:h>0)

=

p,R+

   4α  f  sup Da+;g

a∈R+

(10.59)

 p,R+

,

with 41p < α < 1. That is 

1

A ≤ 2  f ∞,R+ +

h 4α− p 1

 (4α) (q (4α − 1) + 1) q

  4α f  p,R sup  Da+;g

a∈R+

 +

,

(10.60)

where 41p < α < 1. We also have B

(10.23.)

=

   x2     4α−1  4α  f (x2 ) − f (a) − 1 ≤ − g D g f dt (g (x ) (t)) (t) (t) 2 a+;g    (4α) a

2  f ∞,R+ +

(g (x2 ) − g (a))

 (4α) (q (4α − 1) + 1) 1

2  f ∞,R+ +



(q(4α−1)+1) q 1 q

  4α f  p,R sup  Da+;g

a∈R+



1

24α− p h 4α− p 1

 (4α) (q (4α − 1) + 1) q

 (g(x2 )−g(a)=:2h)

  4α f  p,R sup  Da+;g

a∈R+

=

+

 (10.61) +

.

168

10 Iterated Left Abstract Generalized Fractional Landau Inequalities

That is 1

B ≤ 2  f ∞,R+ +



1

24α− p h 4α− p

  4α f  p,R sup  Da+;g

1

 (4α) (q (4α − 1) + 1) q

a∈R+

 +

,

(10.62)

where 41p < α < 1. We have assumed that  Mg :=

  4α sup  Da+;g f  p,R

a∈R+

 < ∞.

+

(10.63)

For convenience we call 1

c :=  (4α) (q (4α − 1) + 1) q > 0. So we have A ≤ 2  f ∞,R+ + and

h

B ≤ 2  f ∞,R+ +

2

4α− 1p

c

(10.64)

Mg , (10.65)

4α− 1p

h c

4α− 1p

Mg ,

where 41p < α < 1. Next we estimate the (10.26)-quantities and we have:  2α  D

a+;g

  f (a) ≤

 (10.33.)

3α 3α 1 2 h A + h 3α B = D (3α + 1)

 (10.65.) h 3α  (2α + 1) 3α 2 A + B ≤ 2α 5α α 2 h (2 − 1)

(10.66)

 1  (2α + 1) 23α h 4α− p 3α+1  f ∞,R+ + Mg + 2 22α (2α − 1) h 2α c 2  f ∞,R+

 1 1 24α− p h 4α− p Mg = + c

  ⎡ ⎤  4α− 1p 3α 3α+1 2 + 2   2 + 2 f 1  (2α + 1) ⎣ ∞,R+ Mg h 2α− p ⎦ = + 22α (2α − 1) h 2α c   ⎡  ⎤ 1  1 + 2α− p Mg 1 2α  (2α + 1) ⎣ 2 1 + 2−3α  f ∞,R+ h 2α− p ⎦ . + h 2α c (2α − 1)

(10.67)

10.2 Main Results

169

That is

 2α  D

a+;g



  f (a) ≤

  ⎤ ⎡  1 

1 + 2α− p Mg 1 2α  (2α + 1) ⎣ 2 1 + 2−3α  f ∞,R+ h 2α− p ⎦ , + 2α − 1 h 2α c (10.68)

∀ a ∈ R+ , ∀ h > 0. I.e. it holds

 2α   f (a) ≤ sup  Da+;g

a∈R+



2α  (2α + 1) 2α − 1



  ⎡  ⎤ 1  1 + 2α− p Mg 2 1 + 2−3α  f ∞,R+ 1 ⎣ h 2α− p ⎦ , + h 2α c

(10.69)

∀ h > 0, under 41p < α < 1. Again from (10.26) we get  3α  D

a+;g

  f (a) ≤

2α  1 h B + 22α h 2α A =  (2α + 1) D

 h 2α  (3α + 1) B + 22α A ≤ 2α 5α α 2 h (2 − 1)

h 2α  (3α + 1) 22α h 5α (2α − 1)

2 ⎡

 (3α + 1) ⎣ 22α (2α − 1)

2α+1

(10.70)

1 1

 24α− p h 4α− p Mg + 2  f ∞,R+ + c

 f ∞,R+

 1 22α h 4α− p Mg = + c

 2 + 22α+1  f ∞,R+ h 3α

+

  1 22α + 24α− p c

⎤ Mg h α− ⎦ = (10.71) 1 p

  ⎡  ⎤ 1  1 + 22α− p 1  (3α + 1) ⎣ 2 1 + 2−2α  f ∞,R+ Mg h α− p ⎦ . + h 3α c (2α − 1) That is

 3α  D

a+;g

  f (a) ≤

170



10 Iterated Left Abstract Generalized Fractional Landau Inequalities

  ⎤ ⎡  1 

1 + 22α− p 1  (3α + 1) ⎣ 2 1 + 2−2α  f ∞,R+ Mg h α− p ⎦ , (10.72) + 2α − 1 h 3α c

∀ a ∈ R+ , ∀ h > 0. I.e. it holds

 3α   f (a) ≤ sup  Da+;g

a∈R+



 (3α + 1) 2α − 1



  ⎡  ⎤ 1  1 + 22α− p 2 1 + 2−2α  f ∞,R+ 1 ⎣ + Mg h α− p ⎦ , h 3α c ∀ h > 0, Call

1 4p

(10.73)

< α < 1.   μ := 2 1 + 2−3α  f ∞,R+ , θ :=



α− 1 1+2 p Mg c

(10.74) ,

both are greater than zero. We consider the function 1

y (h) = μh −2α + θ h 2α− p , ∀ h > 0. We have 

y (h) = −2αμh

−2α−1

1 1 θ h 2α− p −1 = 0, + 2α − p

(10.75)



then



1 1 2α − θ h 2α− p −1 = 2αμh −2α−1 , p

i.e.



1 1 2α − θ h 4α− p = 2αμ, p

(10.76)

with a unique solution ⎞

⎛ 2αμ

 ⎠ h 0 := h crit.no. = ⎝  2α − 1p θ (assuming 21p < α < 1). We have that

1 4α− 1p

(10.77)

10.2 Main Results

171



1 1 1 2α − − 1 θ h 2α− p −2 . (10.78) y  (h) = 2α (2α + 1) μh −2α−2 + 2α − p p We see that

⎞ −2α−21

⎛ 2αμ

 ⎠ y (h 0 ) = 2α (2α + 1) μ ⎝  2α − 1p θ 

4α− p

+

⎞ 2α− 1p 1−2 ⎛



4α− p 2αμ 1 1  ⎠ 2α − 2α − − 1 θ ⎝  = p p 2α − 1 θ p

⎞ −2α−21

⎛ 2αμ

⎝  ⎠ 2α − 1p θ

4α− p



 1 2α (2α + 1) μ + 2αμ 2α − − 1 = p

 ⎞ ⎛ 2α − 1p θ ⎠ 2αμ ⎝ 2αμ



(10.79)

2(α+1) 4α− 1p



1 4α − p

> 0.

Therefore y has a global minimum at ⎛

⎞ 2αμ

 ⎠ h0 = ⎝  2α − 1p θ

1 4α− 1p

,

which is ⎞

⎛

2αμ  ⎠ y (h 0 ) = μ ⎝  2α − 1p θ ⎞

⎛ ⎝

2αμ

 ⎠ 2α − 1p θ

−2α 4α− 1p

−2α 4α− 1p

⎞ 2α− 11p

⎛

2αμ  ⎠ + θ ⎝ 2α − 1p θ ⎛

 ⎞ 2α − 1p θ 2αμ ⎣μ + θ  ⎠  ⎦ = μ⎝ 2αμ 2α − 1 θ ⎤

⎡

4α− p



= 



2α 4α− 1p

1+

p

 =

4α −

(2α)

1 p



2α 4α− 1p





1 2α − p

−2α+11p 4α− p



μ

2α− 1p 4α− 1p



θ

(10.80)



2α 2α −

1 p

2α 4α− 1p

.

(10.81)

172

10 Iterated Left Abstract Generalized Fractional Landau Inequalities

That is

y (h 0 ) =

   4α − 1p 2α − 1p

(2α)

−2α+ 1p 4α− 1p



μ



2α 4α− 1p

2α− 1p 4α− 1p



2α 4α− 1p

θ

.

(10.82)

Therefore we derive (see (10.69))  2α   f (a) ≤ sup  Da+;g

a∈R+



   2 1 + 2−3α

2α− 1p 4α− 1p





⎛ ⎝



α

2  (2α) 2α − 1 

1 + 2α−

1 p

 (2α− 1p )

(4α− p ) 1 2α 4α − p 2α − 1p 1



⎞ 1

 (4α) (q (4α − 1) + 1) q

  4α f  p,R sup  Da+;g

a∈R+

where 21p < α < 1. Call











2α 4α− 1p

⎠

2α− 1p 4α− 1p



 f ∞,R+

2α 4α− 1p

+

< ∞,

(10.83)

  ξ := 2 1 + 2−2α  f ∞,R+ ,

ψ :=

1+2

2α− 1p

c



(10.84) Mg ,

both are greater than zero. We consider the function γ (h) := ξ h −3α + ψh α− p , ∀ h > 0. 1

We have

1 1 ψh α− p −1 = 0, γ  (h) = −3αξ h −3α−1 + α − p

then



1 1 α− ψh α− p −1 = 3αξ h −3α−1 , p

and



1 1 α− ψh 4α− p = 3αξ, p

(10.85)

10.2 Main Results

173

with unique solution ⎞

⎛ 3αξ

 ⎠ h 0 := h crit.no. = ⎝  1 α− p ψ

1 4α− 1p

(10.86)

(assuming 1p < α < 1). We have that



1 1 1 α − − 1 ψh α− p −2 . γ  (h) = 3α (3α + 1) ξ h −3α−2 + α − p p We observe



⎛

⎞

3αξ  ⎠ γ  (h 0 ) = 3α (3α + 1) ξ ⎝  1 α− p ψ ⎞ ⎛



1 1 3αξ  ⎠ α− α − − 1 ψ ⎝ p p α− 1 ψ

−3α−2 4α− 1p

(10.87)



+

α− 1p −2 4α− 1p



=

p

⎛



⎞

⎝  3αξ  ⎠ α − 1p ψ

−3α−2 4α− 1p





 1 3α (3α + 1) ξ + α − − 1 3αξ = p

⎛ 3αξ ⎝ 

⎞ 3αξ α−

1 p



ψ



⎠

−3α−2 4α− 1p

(10.88)





1 4α − > 0. p

Therefore y has a global minimum at ⎞

⎛ 3αξ

 ⎠ h0 = ⎝  α − 1p ψ which is

α− 1p

γ (h 0 ) = ξ h −3α + ψh 0 0

1 4α− 1p

,



4α− 1p ξ + ψh = = h −3α 0 0

(10.89)

174

10 Iterated Left Abstract Generalized Fractional Landau Inequalities

⎛

⎞

⎝

3αξ α−

1 p





⎠

ψ



−3α 4α− 1p

ξ ⎝

α−

1 p



ψ

That is 4α − 1p α − 1p

γ (h 0 ) = ξ

4α −

p 1 p

α−



⎠



−3α 4α− 1p



4α −

1 p 1 p

α−

 .





 1

⎞⎞

⎝ξ + ψ ⎝   ⎠⎠ = α − 1p ψ ⎞

3αξ

⎛ 3αξ

⎛



⎛



ξ

α− 1p 4α− 1p



⎞ ⎛ ⎝  3αξ  ⎠ α − 1p ψ

⎞ ⎛ α − 1p ⎝ ⎠ 3α



−3α 4α− 1p



=

3α 4α− 1p

ψ

⎛

⎞

⎝

3α ⎠ 4α− 1p

(10.90)

.

I.e. we have found  γ (h 0 ) =

  α − 1p

4α −



α− 1p 4α− 1p



1 p



(3α)



ξ



α− 1p 4α− 1p



ψ

⎛

⎞

⎝

3α ⎠ 4α− 1p

.

(10.91)

3α 4α− 1p

We have proved that (see (10.73))

 ⎛ ⎞  α− 1p1 1 4α− 4α −  (3α) p  3α   p 3α ⎠ f (a) ≤ ⎝ sup  Da+;g 1 α 2 −1 α− p a∈R+



   2 1 + 2−2α

α− 1p 4α− 1p





⎛ ⎝

  1 1 + 22α− p

⎞ 1

 (4α) (q (4α − 1) + 1) q

  4α f  p,R sup  Da+;g

a∈R+

where 1p < α < 1. The theorem is proved.

 +



3α 4α− 1p

⎠

(10.92)



α− 1p 4α− 1p



 f ∞,R+

3α 4α− 1p

< ∞,



10.2 Main Results

175

We give an application when α =

1 2

and g (t) = et |R+ .

  Corollary 10.7 Let f ∈ C 1 (R+ , X ) with  f ∞,R+ , ( f ◦ ln) ◦ et ∞,R+ < k1

2 ∞, where (X, ·) is a Banach space. For k = 1, 2, 3, we assume that Da+;e t f ∈

41

2 C 1 ([a, ∞), X ) and Da+;e t f ∈ C ([a, ∞), X ), ∀ a ∈ R+ . We further assume that

  1   4 2 Da+;et f (t)

∞,R2+

< ∞,

(10.93)

where (a, t) ∈ R2+ . Then  √   

  1   4 21   12 + 6 2 2 21 2  D    f ∞,R+  sup  Da+;et f (a) ≤ f √ (t) t    a+;e 2−1

∞,R2+

a∈R+

1 2

< ∞,

(10.94)

 1    32  sup  Da+;e (a) ≤ t f

and

a∈R+

⎛

⎞ √ 

43  1 9 π ⎜  4 21 ⎟  4 < ∞. (10.95) ⎝ 3 ⎠  f ∞,R+ Da+;et f (t) 2 √  √ √ ∞,R+ 2− 2 42 4 3  1  1        22  32 That is sup  Da+;e (a) , sup  Da+;e (a) < ∞. t f t f a∈R+

Proof By Theorem 10.5.

a∈R+



References 1. Aglic Aljinovic, A., Marangunic, Lj., Pecaric, J.: On Landau type inequalities via Ostrowski inequalities. Nonlinear Funct. Anal. Appl. 10(4), 565–579 (2005) 2. Anastassiou, G.: Fractional Differentiation inequalities. Research Monograph. Springer, New York (2009) 3. Anastassiou, G.A.: Advances on Fractional Inequalities. Springer, New York (2011) 4. Anastassiou, G.A.: Intelligent Computations: Abstract Fractional Calculus, Inequalities, Approximations. Springer, Heidelberg (2018) 5. Anastassiou, G.A.: Sequential Left Abstract Generalized Fractional Landau Inequalities, submitted (2020) 6. Barnett, N.S., Dragomir, S.S.: Some Landau type inequalities for functions whose derivatives are of locally bounded variation. Tamkang J. Math. 37(4), 301–308 (2006) 7. Ditzian, Z.: Remarks, questions and conjectures on Landau-Kolmogorov-type inequalities. Math. Inequal. Appl. 3, 15–24 (2000)

176

10 Iterated Left Abstract Generalized Fractional Landau Inequalities

8. Hardy, G.H., Littlewood, J.E.: Some integral inequalities connected with the calculus of variations. Quart. J. Math. Oxford Ser. 3, 241–252 (1932) 9. Hardy, G.H., Landau, E., Littlewood, J.E.: Some inequalities satisfied by the integrals or derivatives of real or analytic functions. Math. Z. 39, 677–695 (1935) 10. Kallman, R.R., Rota, G.C.: On theinequality  f  2 ≤ 4 ·  f  . In: Shisha, O. (ed.) Inequalities, vol. II, pp. 187–192. Academic, New York (1970) 11. Landau, E.: Einige Ungleichungen für zweimal differentzierban funktionen. Proc. Lon. Math. Soc. 13, 43–49 (1913) 12. Mikusinski, J.: The Bochner Integral. Academic, New York (1978) 13. Shilov, G.E.: Elementary Functional Analysis. Dover Publications Inc., New York (1996)

Chapter 11

Sequential General Right Side Fractional Landau Inequalities

We derive uniform and L p right Caputo–Bochner abstract sequential fractional Landau inequalities over R− . These estimate the size of second and third iterated right abstract fractional derivatives of a Banach space valued function over R− . We give an application when the basic fractional order is 21 . It follows [5].

11.1 Introduction Let p ∈ [1, ∞], I = R+ or I = R and f : I → R is twice differentiable with f, f  ∈ L p (I ), then f  ∈ L p (I ). Moreover, there exists a constant C p (I ) > 0 independent of f , such that 1  1    f  ≤ C p (I )  f  2  f   2 , (11.1) p,I

p,I

p,I

where · p,I is the p-norm on the interval I , see [1, 6]. The research on these inequalities started by Landau [11] in 1913. For the case of p = ∞ he proved that C∞ (R+ ) = 2 and C∞ (R) =

√ 2,

(11.2)

are the best constants in (11.1). In 1932, Hardy and Littlewood [8] proved (11.1) for p = 2, with the best constants C2 (R+ ) =

√

2 and C2 (R) = 1.

(11.3)

In 1935, Hardy, Landau and Littlewood [9] showed that the best constants C p (R+ ) in (11.1) satisfies the estimate © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_11

177

178

11 Sequential General Right Side Fractional Landau Inequalities

C p (R+ ) ≤ 2, for p ∈ [1, ∞),

(11.4)

which yields C p (R) ≤ 2 for p ∈ [1, ∞). √ In fact, in [7, 10] was shown that C p (R) ≤ 2. We need the following concept from abstract fractional calculus. Our integral next is of Bochner type [12]. We need Definition 11.1 ([4], p. 150) Let [a, b] ⊂ R, X be a Banach space, α > 0; m =

α ∈ N, ( · is the ceiling of the number), f : [a, b] → X . We assume that f (m) ∈ L 1 ([a, b] , X ). We call the Caputo–Bochner right fractional derivative of order α: 

 α f (x) := Db−

(−1)m  (m − α)



b

(z − x)m−α−1 f (m) (z) dz, ∀ x ∈ [a, b] , (11.5)

x

where  is the gamma function. α f = (−1)m f (m) the ordinary X -valued If α = m ∈ N, we observe that Db− 0 f := f. derivative (defined similar to numerical one, see [13], p. 83), and also set Db−  α  α By ([4], p. 34), Db− f (x) exists almost everywhere in x ∈ [a, b] and Db− f ∈ L 1 ([a, b] , X ).   α / N, then by ([4], p. 37), Db− f ∈ C ([a, b] , X ) , If  f (m)  L ∞ ([a,b],X ) < ∞, and α ∈  α  hence  Db− f  ∈ C ([a, b]) . We mention the important: Corollary 11.2 ([4], p. 157) Let f ∈ C m ([a, b] , X ), m = α, α > 0, x, x0 ∈ [a, b]. Then Dxa0 − f (x) is jointly continuous functions in (x, x0 ) from [a, b]2 into X , X is a Banach space. By convention we suppose that Dxα0 − f (x) = 0, for x > x0 , for all x, x0 ∈ [a, b] . The author has already done an extensive body of work on fractional Landau inequalities, see [3], and on abstract fractional Landau inequalities, see [4]. However there the proving methods came out of applications of fractional Ostrowski inequalities [2, 4]. Usually there the domains where [A, +∞) or (−∞, B], with A, B ∈ R and in one mixed case the domain was all of R. In this chapter with less assumptions we establish uniform and L p type right Caputo–Bochner abstract iterated fractional Landau inequalities over R− . The method of proving is based on right Caputo–Bochner iterated fractional Taylor’s formula with integral remainder, see [4], pp. 129. We give also an application for α = 21 . Regardless to say that we are also inspired by [3, 4].

11.2 Main Results

179

11.2 Main Results nα α α α We consider (α > 0) the composition Db− := Db− Db− ...Db− (n-times), n ∈ N. We mention the following right modified X -valued Taylor’s formula, (X, ·) is a Banach space.

Theorem 11.3 ([4], p. 129) Let 0 < α ≤ 1, n ∈ N, f ∈ C 1 ([a, b] , X ). For k = (n+1)α kα f ∈ C 1 ([a, b] , X ) and Db− f ∈ C ([a, b] , X ). Then 1, ..., n, we assume that Db− f (x) = 1  ((n + 1) α)

n  (b − x)iα  iα  D f (b) +  (iα + 1) b− i=0



b x

(11.6)

  (n+1)α f (t) dt, (t − x)(n+1)α−1 Db−

∀ x ∈ [a, b] .  α  f (b) = When 0 < α < 1 and f ∈ C 1 ([a, b] , X ), by [4], p. 135, we get that Db− 0. We present the following abstract right fractional Landau inequalities over R− .   Theorem 11.4 Let 0 < α < 1, f ∈ C 1 (R− , X ) with  f ∞,R− ,  f  ∞,R− < kα 4α ∞. For k = 1, 2, 3, we assume that Db− f ∈ C 1 ((−∞, b], X ) and Db− f ∈ C ((−∞, b], X ), ∀ b ∈ R− . We further assume that   4α f (t)∞,R2 < ∞, K :=  Db− −

(11.7)

where (b, t) ∈ R2− . Then  2α   f (b) ≤ sup  Db−

b∈R−

 (2α + 1) 22α−1 (2α − 1)



  23α+1 23α + 1 (2α + 1)  f ∞,R− K ,  (4α + 1) (11.8)

and √  3  1  3α   4 4 2 (3α + 1) ( (4α + 1))− 4 22α + 1 3 4    f ∞,R sup Db− f (b) ≤ K 4. √ 3 √ α − 4 b∈R− 3 2 (2α − 1) (11.9)  2α    3α   f (b) , sup  Db− f (b) < ∞. That is sup  Db− b∈R−

b∈R−

 α  Proof We notice again here that Db− f (b) = 0, ∀ b ∈ R− . We make use of Theorem 11.3 for 0 < α < 1 and n = 3, applied for any b ∈ R− and a = −∞.

180

11 Sequential General Right Side Fractional Landau Inequalities

Momentarily we fix b ∈ R− . Let x2 < x1 < b, then f (x1 ) − f (b) =

(b − x1 )3α  3α  (b − x1 )2α  2α  Db− f (b) + D f (b) + (11.10)  (2α + 1)  (3α + 1) b− 1  (4α)



b x1

 4α  f (t) dt, (t − x1 )4α−1 Db−

and f (x2 ) − f (b) =

(b − x2 )3α  3α  (b − x2 )2α  2α  Db− f (b) + D f (b) + (11.11)  (2α + 1)  (3α + 1) b− 1  (4α)

That is

b x2

 4α  f (t) dt. (t − x2 )4α−1 Db−

(b − x1 )3α  3α  (b − x1 )2α  2α  Db− f (b) + D f (b) =  (2α + 1)  (3α + 1) b− f (x1 ) − f (b) −

and



1  (4α)



b x1

 4α  f (t) dt =: A, (t − x1 )4α−1 Db−

(b − x2 )3α  3α  (b − x2 )2α  2α  Db− f (b) + D f (b) =  (2α + 1)  (3α + 1) b− f (x2 ) − f (b) −

1  (4α)



b x2

(11.12)

(11.13)

 4α  f (t) dt =: B. (t − x2 )4α−1 Db−

 2α  f (b) , are solving the above system of two equations with two unknowns Db− We3α Db− f (b) . The main determinant of system is

(b−x1 )2α

(2α+1)

D :=

(b−x2 )2α

(2α+1)

(b−x1 )3α (3α+1) (b−x2 )3α (3α+1)







=



 1 (b − x1 )2α (b − x2 )3α − (b − x1 )3α (b − x2 )2α =  (2α + 1)  (3α + 1) (b − x1 )2α (b − x2 )2α  (b − x2 )α − (b − x1 )α > 0.  (2α + 1)  (3α + 1)

11.2 Main Results

I.e. D=

181

(b − x1 )2α (b − x2 )2α  (b − x2 )α − (b − x1 )α > 0.  (2α + 1)  (3α + 1)

(11.14)

We obtain the unique solution



 Therefore we have















A

 B 2α f (b) = Db−

 3α Db− f (b) =



 2α Db− f (b) = and  3α  Db− f (b) =























3α

(b−x2 )

(3α+1)

, D (b−x1 )3α (3α+1)

A

(b−x2 ) (2α+1) D

B

2α

(11.15)



(b−x1 )2α (2α+1)









.

(b−x2 )3α A− (b−x1 )3α B (3α+1) (3α+1) , D

(11.16) (b−x1 )2α B− (b−x2 )2α A (2α+1) (2α+1) . D

We have the following   A =   f (x1 ) − f (b) − 2  f ∞,R− +

1  (4α)



b x1

   4α  4α−1 D f dt − x (t)  (t 1) b− ≤

  4α  D f (t) 2 b− ∞,R

(b − x1 )4α ,

(11.17)

K (b − x1 )4α ,  (4α + 1)

(11.18)

K (b − x2 )4α ,  (4α + 1)

(11.19)

−

 (4α + 1)

under the assumption  f ∞,R− < ∞. That is A ≤ 2  f ∞,R− + and similarly, B ≤ 2  f ∞,R− + where by assumption

  4α f (t)∞,R2 < ∞, K :=  Db− −

with (b, t) ∈ R2− .

(11.20)

182

11 Sequential General Right Side Fractional Landau Inequalities

Consequently we have     2α   Db− f (b) ≤

 

1 K (b − x2 )3α 2  f ∞,R− + (b − x1 )4α  (3α + 1) D  (4α + 1)

+ (b − x1 )

3α

 2  f ∞,R− +

K (b − x2 )4α  (4α + 1)

 ,

(11.21)

and     3α   Db− f (b) ≤

  K 1 (b − x1 )2α 2  f ∞,R− + (b − x2 )4α  (2α + 1) D  (4α + 1)

 + (b − x2 )2α 2  f ∞,R− +

K (b − x1 )4α  (4α + 1)

 .

(11.22)

Set now x1 := b − h, x2 := b − 2h, where h > 0, so that b − x1 = h, b − x2 = 2h. Hence we get 22α h 5α (2α − 1) > 0. (11.23) D=  (2α + 1)  (3α + 1) Therefore we derive (from (11.16))  2α    D f (b) ≤ b−

 

 (2α + 1) K 3α 3α 4α   2 f 2 h h + ∞,R− 22α h 5α (2α − 1)  (4α + 1)

+h

3α

 2  f ∞,R− +

K 24α h 4α  (4α + 1)

 =

(11.24)



 3α  3α  3α  7α  (2α + 1) K 4α 2  f ∞,R− 2 + 1 h + 2 +2 h 22α h 5α (2α − 1)  (4α + 1)  =

 (2α + 1) 22α (2α − 1)

That is

    3α 2 2 + 1  f ∞,R− h 2α  2α    D f (b) ≤ b−

   2 23α + 1  f ∞,R− h 2α ∀ b ∈ R− , ∀ h > 0. I.e. it holds



 23α (2α + 1) K h 2α . +  (4α + 1)

 (2α + 1) 22α (2α − 1)

(11.25)



 23α (2α + 1) K h 2α , +  (4α + 1)

(11.26)

11.2 Main Results

183

 2α   sup  Db− f (b) ≤

b∈R−

   2 23α + 1  f ∞,R− h 2α



 (2α + 1) 22α (2α − 1)



 23α (2α + 1) K h 2α < ∞, +  (4α + 1)

(11.27)

∀ h > 0, 0 < α < 1. By (11.16) we derive  

 (3α + 1) K 2α 4α 4α   2 f h 2 + h ∞,R− 22α h 5α (2α − 1)  (4α + 1)

 3α    D f (b) ≤ b−

 +22α h 2α 2  f ∞,R− +

K h 4α  (4α + 1)

 =



 4α  2α  2α  6α  (3α + 1) K 2α 2  f ∞,R− 2 + 1 h + 2 +2 h 22α h 5α (2α − 1)  (4α + 1) (11.28)       2α  2 2 + 1  f ∞,R− 22α 22α + 1  (3α + 1) K hα = + 22α (2α − 1) h 3α  (4α + 1)     (3α + 1) 22α + 1 2  f ∞,R− 22α K α . h = + 22α (2α − 1) h 3α  (4α + 1)    3α    (3α + 1) 22α + 1  D f (b) ≤ b− 22α (2α − 1)

That is

2  f ∞,R− h 3α

+

 22α K hα ,  (4α + 1)

(11.29)

∀ b ∈ R− , ∀ h > 0. I.e. it holds

   3α    (3α + 1) 22α + 1   sup Db− f (b) ≤ 22α (2α − 1) b∈R−

2  f ∞,R− h 3α

∀ h > 0, 0 < α < 1. Call

+

 22α K h α < ∞,  (4α + 1)

  μ := 2 23α + 1  f ∞,R− ,

(11.30)

(11.31)

184

11 Sequential General Right Side Fractional Landau Inequalities

θ=

23α (2α + 1) K ,  (4α + 1)

(11.32)

both are greater than zero. Set also ρ := 2α; 0 < ρ < 2. We consider the function

We have

y (h) := μh −ρ + θh ρ , ∀ h > 0.

(11.33)

y  (h) = −ρμh −ρ−1 + ρθh ρ−1 = 0,

(11.34)

then

θh ρ−1 = μh −ρ−1

and θh 2ρ = μ, with a unique solution

 μ  2ρ1

h 0 := h crit.no. = We have that

θ

.

(11.35)

y  (h) = ρ (ρ + 1) μh −ρ−2 + ρ (ρ − 1) θh ρ−2 .

(11.36)

We see that 

y (h 0 ) = y



  1  μ 2ρ θ

= ρ (ρ + 1) μ

 μ  −ρ−2 2ρ θ

+ ρ (ρ − 1) θ

 μ  ρ−2 2ρ θ

1   ρ1      θ ρ θ 2 ρ 2ρ μθ = 2ρ μθ > 0. μ μ

Therefore y has a global minimum at h 0 = y (h 0 ) = μ

 μ − 21 θ

+θ

 μ  21 θ

 μ  2ρ1 θ

, which is

  21   θ =μ + θμ = 2 θμ. μ

We have proved that (see (11.27))  2α   sup  Db− f (b) ≤

b∈R−

 (2α + 1) 22α−1 (2α − 1)

=

11.2 Main Results

185



  23α+1 23α + 1 (2α + 1)  f ∞,R− K .  (4α + 1)

Call

(11.37)

ξ := 2  f ∞,R− , ψ :=

(11.38)

22α K , (4α+1)

both are greater than zero. We consider the function γ (h) := ξh −3α + ψh α , ∀ h > 0. We have

(11.39)

γ  (h) = −3αξh −3α−1 + αψh α−1 = 0,

then

ψh α−1 = 3ξh −3α−1

and ψh 4α = 3ξ, with unique solution

 h 0 := h crit.no. =

3ξ ψ

 4α1

.

(11.40)

We have that γ  (h) = 3α (3α + 1) ξh −3α−2 + α (α − 1) ψh α−2 .

(11.41)

We see 



γ (h 0 ) = 3α (3α + 1) ξ 

3ξ α ψ

 α−2 4α

3ξ ψ

 −3α−2 4α



3ξ + α (α − 1) ψ ψ 

3ξ (4αψ) = 4α ψ ψ 2

Therefore γ has a global minimum at h 0 =  γ (h 0 ) = ξ

3ξ ψ

− 34

  4α1 3ξ ψ



 α−2 4α

> 0.

, which is

3ξ +ψ ψ

 41

=

 α−2 4α

=

(11.42)

186

11 Sequential General Right Side Fractional Landau Inequalities



3ξ ψ

 14 

ξ

   41 ψ 4 3ξ +ψ = ψ . 3ξ 3 ψ

(11.43)

Consequently, 4 3 1 γ (h 0 ) = √ 3 ψ 4 ξ 4 . 4 3

(11.44)

We have proved that (see (11.30))    3α   4 (3α + 1) 22α + 1   sup Db− f (b) ≤ √ 3 4 b∈R− 3 22α (2α − 1) 1  2  f ∞,R− 4



22α K  (4α + 1)

 34

=

√  3  1 4 4 2 (3α + 1)  (4α + 1)− 4 22α + 1 3 4  f ∞,R K 4. √ 3 α − 4 3 2 2 (2α − 1)

(11.45)



The theorem is proved. We continue with abstract L p right fractional Landau inequalities over R− .

Theorem 11.5 Let p, q > 1 : 1p + q1 = 1, 0 < α < 1. Let f ∈ C 1 (R− , X ) with   kα  f ∞,R− ,  f  ∞,R < ∞. For k = 1, 2, 3, we assume that Db− f ∈ − 4α 1 C ((−∞, b], X ) and Db− f ∈ C ((−∞, b], X ), ∀ b ∈ R− . We further assume that 

 4α  f  p,R− sup  Db−

 < ∞.

(11.46)

b∈R−

Then (1) under

1 2p

< α < 1, we get ⎡⎛

  1 ⎞   p    2α− 1 α 1 −3α 4α− 2  4α − (2α) p  2α   ⎢ 4α 1 + 2 p ⎠ sup  Db− f (b) ≤ ⎣⎝ 1 α 2 −1 2α − p b∈R−





1+2



α− 1p

 (4α) (q (4α − 1) + 1)

1 q

2α 4α− 1p

⎤



2α− 1p 4α− 1p

⎥ ⎦  f ∞,R−



11.2 Main Results

187







 4α  f  p,R− sup  Db−

 2α 4α− 1p

< ∞.

(11.47)

b∈R−

(2) under

1 p

< α < 1, we get ⎡⎛

   ⎞  α− 1p   4α−  1 1 −2α p  3α   ⎢  (3α) 4α − p 6α 1 + 2 ⎠ sup  Db− f (b) ≤ ⎣⎝ 1 α 2 −1 α− p b∈R− 



1+2



2α− 1p

 (4α) (q (4α − 1) + 1) 

3α 4α− 1p

⎤



 4α  f  p,R− sup  Db−

α− 1p 4α− 1p



⎥ ⎦  f ∞,R−

1 q





 3α 4α− 1p

< ∞.

(11.48)

b∈R−

 2α    3α   That is sup  Db− f (b) , sup  Db− f (b) < ∞. b∈R−

b∈R−

Proof As in the proof of Theorem 11.4 we have that    b    4α  1 4α−1   A =  f (x1 ) − f (b) − Db− f (t) dt  (t − x1 ) ≤  (4α) x1 (11.12)

2  f ∞,R− +

1  (4α)



b x1

 4α   f (t) dt ≤ (t − x1 )4α−1  Db−

(q(4α−1)+1)

2  f ∞,R−

1 (b − x1 ) q +  (4α) (q (4α − 1) + 1) q1

2  f ∞,R− +

  4α− 1p



 4α  f  p,R− sup  Db−

 (b−x1 =:h>0)

=

b∈R−

h 1  (4α) (q (4α − 1) + 1) q1



  4α  sup  Db− f  p,R− ,

(11.49)

b∈R−

with 41p < α < 1. That is 

1

A ≤ 2  f ∞,R− + where

1 4p

< α < 1.

h 4α− p 1

 (4α) (q (4α − 1) + 1) q

  4α  sup  Db− f  p,R− ,

b∈R−

(11.50)

188

11 Sequential General Right Side Fractional Landau Inequalities

We also have    b   4α  1 (11.13)  4α−1 ≤ B =  f D f dt − f − − x (x (t) ) (b) (t ) 2 2 b−    (4α) x2 2  f ∞,R− +

(b − x2 )



(q(4α−1)+1) q

 (4α) (q (4α − 1) + 1) 1

2  f ∞,R− +

1 q

 4α  f  p,R− sup  Db−

 (b−x2 =:2h)

=

b∈R−

 (11.51)  4α  sup  Db− f  p,R− .



1

24α− p h 4α− p 1

 (4α) (q (4α − 1) + 1) q

b∈R−

That is 1

B ≤ 2  f ∞,R− +



1

24α− p h 4α− p 1

 (4α) (q (4α − 1) + 1) q

  4α  sup  Db− f  p,R− ,

b∈R−

(11.52)

where 41p < α < 1. We have assumed that  M :=

 4α  f  p,R− sup  Db−

 < ∞.

(11.53)

c :=  (4α) (q (4α − 1) + 1) q > 0.

(11.54)

b∈R−

For convenience we call 1

So we have  A ≤ 2  f ∞,R− + and

h

B ≤ 2  f ∞,R− +

2

4α− 1p

c

M, (11.55)

4α− 1p

h c

4α− 1p

M,

where 41p < α < 1. Next we estimate the (11.16)-quantities and we have:  2α    D f (b) ≤ b−

 3α 3α (11.23) 1 2 h A + h 3α B = D (3α + 1)

(11.55) h 3α  (2α + 1)  3α A B 2 + ≤ 22α h 5α (2α − 1)  1  (2α + 1) 23α h 4α− p 3α+1   M+ f + 2 ∞,R− 22α (2α − 1) h 2α c

(11.56)

11.2 Main Results

189

2  f ∞,R−

 1 1 24α− p h 4α− p M = + c

  ⎡ ⎤ 1 23α + 24α− p  1  (2α + 1) ⎣ 3α+1 Mh 4α− p ⎦ = + 2  f ∞,R− + 2 22α (2α − 1) h 2α c   ⎡ ⎤  4α− 1p 3α 3α+1 2 + 2 + 2  f ∞,R−  (2α + 1) ⎣ 2 2α− 1p ⎦ Mh = + 22α (2α − 1) h 2α c   ⎡  ⎤ 1  1 + 2α− p M 1 2α  (2α + 1) ⎣ 2 1 + 2−3α  f ∞,R− h 2α− p ⎦ . + h 2α c (2α − 1) That is 

(11.57)

 2α    D f (b) ≤ b−   ⎤ ⎡   α− 1p −3α 1 + 2 M  f ∞,R− 2  (2α + 1) ⎣ 2 1 + 2 2α− 1p ⎦ h + , 2α − 1 h 2α c 

α

(11.58) ∀ b ∈ R− , ∀ h > 0. I.e. it holds

 2α   f (b) ≤ sup  Db−

b∈R−



2α  (2α + 1) 2α − 1



  ⎡  ⎤  α− 1p −3α 1 + 2 M  f ∞,R− 2 1+2 1 ⎣ h 2α− p ⎦ , + h 2α c

(11.59)

∀ h > 0, under 41p < α < 1. Again from (11.16) we get  3α    D f (b) ≤ b−

 2α 1 h B + 22α h 2α A =  (2α + 1) D

(11.55) h 2α  (3α + 1)  2α B A + 2 ≤ 22α h 5α (2α − 1) 

h 2α  (3α + 1) 22α h 5α (2α − 1)

1 1  24α− p h 4α− p 2  f ∞,R− + M+ c

(11.60)

190

11 Sequential General Right Side Fractional Landau Inequalities

2 

 (3α + 1) 2α 2 (2α − 1) h 3α



⎡

2α+1

 f ∞,R−

 1 22α h 4α− p M = + c

  ⎤ 1 22α + 24α− p   1 ⎣ 2 + 22α+1  f ∞,R + Mh 4α− p ⎦ = − c ⎡

 (3α + 1) ⎣ 2 + 2 22α (2α − 1)

2α+1



 f ∞,R−

h 3α

+

  1 22α + 24α− p

⎤ Mh

c

α− 1p ⎦

= (11.61)

  ⎡  ⎤ 1  1 + 22α− p 1  (3α + 1) ⎣ 2 1 + 2−2α  f ∞,R− Mh α− p ⎦ . + h 3α c (2α − 1) That is ⎡ ⎤     2α− 1p      (3α + 1)  2 1 + 2−2α  f  1 + 2 1 ∞,R− ⎢  3α  α− ⎥ + Mh p ⎦ ,  Db− f (b) ≤ ⎣ 2α − 1 h 3α c

(11.62) ∀ b ∈ R− , ∀ h > 0. I.e. it holds

 3α   f (b) ≤ sup  Db−

b∈R−



 (3α + 1) 2α − 1



  ⎡  ⎤ 1  1 + 22α− p 2 1 + 2−2α  f ∞,R− 1 ⎣ Mh α− p ⎦ , + h 3α c ∀ h > 0, Call

1 4p

(11.63)

< α < 1.   μ := 2 1 + 2−3α  f ∞,R− , θ :=

  α− 1 1+2 p M c

(11.64) ,

both are greater than zero. We consider the function 1

y (h) = μh −2α + θh 2α− p , ∀ h > 0. We have

(11.65)

11.2 Main Results

191

  1 1 θh 2α− p −1 = 0, y  (h) = −2αμh −2α−1 + 2α − p 

then

2α −

(11.66)

 1 1 θh 2α− p −1 = 2αμh −2α−1 , p

  1 1 2α − θh 4α− p = 2αμ, p

i.e.

with a unique solution ⎞

⎛

2αμ  ⎠ h 0 := h crit.no. = ⎝  2α − 1p θ

1 4α− 1p

(11.67)

(assuming 21p < α < 1). We have that    1 1 1 2α − − 1 θh 2α− p −2 . (11.68) y  (h) = 2α (2α + 1) μh −2α−2 + 2α − p p We see that

⎞ −2α−21

⎛

2αμ  ⎠ y  (h 0 ) = 2α (2α + 1) μ ⎝  2α − 1p θ

4α− p

+

⎞ 2α− 1p 1−2 ⎛    4α− p 2αμ 1 1  ⎠ 2α − 2α − − 1 θ ⎝  p p 2α − 1 θ p

⎛



 ⎞ 2α − θ ⎠ = 2αμ ⎝ 2αμ 1 p

 2(α+1) 4α− 1p

  1 4α − > 0. p

Therefore y has a global minimum at ⎛

⎞

2αμ  ⎠ h0 = ⎝  2α − 1p θ which is

1 4α− 1p

,

(11.69)

192

11 Sequential General Right Side Fractional Landau Inequalities

⎛

⎞

2αμ  ⎠ y (h 0 ) = μ ⎝  2α − 1p θ

−2α 4α− 1p

⎛

⎞ 2α− 11p

2αμ  ⎠ + θ ⎝ 2α − 1p θ

4α− p

=

(11.70)

   −2α+ 1p  2α− 1p   2α  4α − 1p  1 4α− 1p 1 1   2α − μ 4α− p θ 4α− p . 2α p 1 (2α) 4α− p

(11.71)

That is 

y (h 0 ) =

   4α − 1p 2α − 1p 

(2α)



−2α+ 1p 4α− 1p



μ



2α 4α− 1p

2α− 1p 4α− 1p

 

 2α 4α− 1p

θ

.

(11.72)

Therefore we derive (see (11.59))  2α   sup  Db− f (b) ≤

b∈R−



   2 1 + 2−3α

2α− 1p 4α− 1p





⎛ ⎝



α

2  (2α) 2α − 1 



1 + 2α−

1 p

 (2α− 1p )  (4α− p ) 1

2α 2α −

1 p 



⎞ 1

 (4α) (q (4α − 1) + 1) q 

 4α  f  p,R− sup  Db−



 2α 4α− 1p

⎠

1 4α − p 



2α− 1p 4α− 1p



 f ∞,R−

 2α 4α− 1p

< ∞,

(11.73)

b∈R−

where 21p < α < 1. Call

  ξ := 2 1 + 2−2α  f ∞,R− , ψ :=

  2α− 1p 1+2 c

(11.74) M,

both are greater than zero. We consider the function γ (h) := ξh −3α + ψh α− p , ∀ h > 0. 1

We have

(11.75)

11.2 Main Results

193

  1 1 ψh α− p −1 = 0, γ  (h) = −3αξh −3α−1 + α − p then

  1 1 α− ψh α− p −1 = 3αξh −3α−1 p

and

  1 1 α− ψh 4α− p = 3αξ, p

with unique solution ⎞

⎛ h 0 := h crit.no. = ⎝ 

3αξ α−

1 p

 ⎠ ψ

1 4α− 1p

(11.76)

(assuming 1p < α < 1). We have that    1 1 1 α − − 1 ψh α− p −2 . γ  (h) = 3α (3α + 1) ξh −3α−2 + α − p p

(11.77)

We observe that ⎛

⎞



3αξ  ⎠ γ  (h 0 ) = 3α (3α + 1) ξ ⎝  α − 1p ψ

−3α−2 4α− 1p

+ 





⎞ ⎛   3αξ 1 1  ⎠ α− α − − 1 ψ ⎝ 1 p p α− ψ

α− 1p −2 4α− 1p



=

p



⎞

⎛ 3αξ ⎝ 

3αξ

 ⎠ α − 1p ψ

−3α−2 4α− 1p



  1 4α − > 0. p

Therefore y has a global minimum at ⎛

⎞

3αξ  ⎠ h0 = ⎝  α − 1p ψ

1 4α− 1p

,

(11.78)

194

11 Sequential General Right Side Fractional Landau Inequalities

which is

  4α− 1p ξ + ψh = = h −3α 0 0

α− 1p

γ (h 0 ) = ξh 0−3α + ψh 0 ⎞

⎛ 3αξ

⎝  ⎠ α − 1p ψ



−3α 4α− 1p

⎛



⎛

(11.79)

⎞⎞

⎛ 3αξ

⎝ξ + ψ ⎝   ⎠⎠ = α − 1p ψ ⎞



3αξ

 ⎠ ξ ⎝ α − 1p ψ

−3α 4α− 1p





4α − α−



1 p 1 p

.

That is   4α − 1p

γ (h 0 ) =

  α − 1p



α− 1p 4α− 1p



(3α)

 ξ



α− 1p 4α− 1p



⎛

⎞

⎝

3α ⎠ 4α− 1p

ψ

.

(11.80)

3α 4α− 1p

We have proved that (see (11.63))    ⎛ ⎞  α− 1p1 1 4α− p 4α − p  (3α)  3α   3α ⎠ f (b) ≤ ⎝ sup  Db− 1 α 2 −1 α− p b∈R−



   2 1 + 2−2α

α− 1p 4α− 1p





⎛ ⎝



  1 1 + 22α− p

⎞ 1

 (4α) (q (4α − 1) + 1) q 

 4α  f  p,R− sup  Db−







3α 4α− 1p

⎠

(11.81)

α− 1p 4α− 1p



 f ∞,R−

 3α 4α− 1p

< ∞,

b∈R−

where 1p < α < 1. The theorem is proved.



We give an application when α = 21 .

  Corollary 11.6 Let f ∈ C 1 (R− , X ) with  f ∞,R− ,  f  ∞,R− < ∞, where k1

(X, ·) is a Banach space. For k=1, 2, 3, we assume that Db−2 f ∈ C 1 ((−∞, b], X ) 41

and Db−2 f ∈ C ((−∞, b], X ), ∀ b ∈ R− . We further assume that

11.2 Main Results

195

 1   4  K := Db−2 f (t)

∞,R2−

< ∞,

(11.82)

where (b, t) ∈ R2− . Then  1    2  sup  Db−2 f (b) ≤



b∈R−

and

√    21  1 12 + 6 2  4 21  2  f ∞,R− Db− f (t) 2 < ∞, √ ∞,R− 2−1 (11.83)  1    3  sup  Db−2 f (b) ≤ b∈R−

⎞ √   43  1 9 π ⎟ ⎜   4 21 4   f f < ∞. (t)  D   ⎠ ⎝ b− ∞,R− √  √ √ 3 ∞,R2− 2− 2 42 43 ⎛

(11.84)

 1    1    2  3   That is sup  Db−2 f (b) , sup  Db−2 f (b) < ∞. b∈R−

Proof By Theorem 11.4.

b∈R−



Remark 11.7 All of the above results are also true for any half line (−∞, A], A ∈ R, instead of only R− , just replace all R− by (−∞, A].

References 1. Aglic Aljinovic, A., Marangunic, L., Pecaric, J.: On Landau type inequalities via Ostrowski inequalities. Nonlinear Funct. Anal. Appl. 10(4), 565–579 (2005) 2. Anastassiou, G.: Fractional Differentiation Inequalities. Research Monograph. Springer, New York (2009) 3. Anastassiou, G.A.: Advances on Fractional Inequalities. Springer, New York (2011) 4. Anastassiou, G.A.: Intelligent Computations: Abstract Fractional Calculus, Inequalities, Approximations. Springer, Heidelberg (2018) 5. Anastassiou, G.A.: Iterated abstract right side fractional Landau inequalities. Submitted (2020) 6. Barnett, N.S., Dragomir, S.S.: Some Landau type inequalities for functions whose derivatives are of locally bounded variation. Tamkang J. Math. 37(4), 301–308 (2006) 7. Ditzian, Z.: Remarks, questions and conjectures on Landau-Kolmogorov-type inequalities. Math. Inequal. Appl. 3, 15–24 (2000) 8. Hardy, G.H., Littlewood, J.E.: Some integral inequalities connected with the calculus of variations. Q. J. Math. Oxford Ser. 3, 241–252 (1932) 9. Hardy, G.H., Landau, E., Littlewood, J.E.: Some inequalities satisfied by the integrals or derivatives of real or analytic functions. Math. Z. 39, 677–695 (1935) 10. Kallman, R.R., Rota, G.C.: On the inequality  f  2 ≤ 4 f ·  f  . In: Shisha, O. (ed.) Inequalities, vol. II, pp. 187–192. Academic, New York (1970)

196

11 Sequential General Right Side Fractional Landau Inequalities

11. Landau, E.: Einige Ungleichungen für zweimal differentzierban funktionen. Proc. Lond. Math. Soc. 13, 43–49 (1913) 12. Mikusinski, J.: The Bochner Integral. Academic, New York (1978) 13. Shilov, G.E.: Elementary Functional Analysis. Dover Publications Inc., New York (1996)

Chapter 12

Iterated Generalized Right Side Fractional Landau Inequalities

We give uniform and L p Caputo–Bochner abstract iterated generalized right fractional Landau inequalities over R− . These estimate the size of second and third sequential abstract generalized right fractional derivatives of a Banach space valued function over R− . We give an application when the basic fractional order is 21 . It follows [5].

12.1 Introduction Let p ∈ [1, ∞], I = R+ or I = R and f : I → R is twice differentiable with f, f  ∈ L p (I ), then f  ∈ L p (I ). Moreover, there exists a constant C p (I ) > 0 independent of f , such that 1  1    f  ≤ C p (I )  f  2  f   2 , (12.1) p,I

p,I

p,I

where · p,I is the p-norm on the interval I , see [1, 6]. The research on these inequalities started by Landau [11] in 1913. For the case of p = ∞ he proved that C∞ (R+ ) = 2 and C∞ (R) =

√ 2,

(12.2)

are the best constants in (12.1). In 1932, Hardy and Littlewood [8] proved (12.1) for p = 2, with the best constants C2 (R+ ) =

√

2 and C2 (R) = 1.

(12.3)

In 1935, Hardy, Landau and Littlewood [9] showed that the best constants C p (R+ ) in (12.1) satisfies the estimate © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_12

197

198

12 Iterated Generalized Right Side Fractional Landau Inequalities

C p (R+ ) ≤ 2, for p ∈ [1, ∞),

(12.4)

which yields C p (R) ≤ 2 for p ∈ [1, ∞). √ In fact, in [7, 10] was shown that C p (R) ≤ 2. We need the following concept from abstract fractional calculus. Our integral next is of Bochner type [12]. We need Definition 12.1 ([4], p. 105) Let [a, b] ⊂ R, (X, ·) a Banach space, g ∈ C 1 ([a, b]) and increasing, f ∈ C ([a, b] , X ), ν > 0. We define the right Riemann–Liouville generalized fractional Bochner integral operator 

 ν f (x) := Jb−;g

1  (ν)



b

(g (z) − g (x))ν−1 g  (z) f (z) dz,

(12.5)

x

∀ x ∈ [a, b], where  is the gamma function. The last integral is of Bochner type. Since f ∈ C  X ), then f ∈ L ∞ ([a, b] ,  ν ([a, b] , X ). By Theorem 4.11, p. 101, [4], we get that Jb−;g f ∈ C ([a, b] , X ).   0 ν f := f and see that Jb−;g f (b) = 0. Above we set Jb−;g We also need Definition 12.2 ([4], p. 107) Let α > 0, α = n, · the ceiling of the number. Let f ∈ C n ([a, b] , X ), where [a, b] ⊂ R, and (X, ·) is a Banach space. Let g ∈ C 1 ([a, b]) , strictly increasing, such that g −1 ∈ C n ([g (a) , g (b)]) . We define the right generalized g-fractional derivative X -valued of f of order α as follows:  b (n)    α (−1)n Db−;g f (x) := (g (t) − g (x))n−α−1 g  (t) f ◦ g −1 (g (t)) dt,  (n − α) x (12.6) ∀ x ∈ [a, b]. The last integral is of Bochner type. Ordinary vector valued derivative is as in [13], similar to numerical one.  α If α ∈ / N, by Theorem 4.11, p. 101, [4], we have that Db−;g f ∈ C ([a, b] , X ). We see that     (n)  α  n−α Jb;g ◦ g (x) = Db−;g f (x) , ∀ x ∈ [a, b] . (12.7) (−1)n f ◦ g −1 We set n f (x) := (−1)n Db−;g



f ◦ g −1

n

 ◦ g (x) ∈ C ([a, b] , X ) , n ∈ N,

0 f (x) = f (x) , ∀ x ∈ [a, b] . Db−;g

(12.8)

12.1 Introduction

199

When g = id, then

α α α f = Db−;id f = Db− f, Db−;g

(12.9)

the usual left X -valued Caputo fractional derivative, see [4], Chap. 2. By convention we suppose that 

 Dxα0 −;g f (x) = 0, for x > x0 ,

(12.10)

for any x, x0 ∈ [a, b] . Denote the sequential (also called iterated) generalized left fractional derivative by nα α α α := Db−;g Db−;g ...Db−;g (n times), n ∈ N. (12.11) Db−;g We need the following g-right generalized modified X -valued Taylor’s formula. Theorem 12.3 ([4], p. 120) Let 0 < α≤1, n ∈ N, f ∈C 1 ([a, b] , X ), g∈C 1 ([a, b]) kα f , k = 1, ..., n, strictly increasing, such that g −1 ∈C 1 ([g (a) , g (b)]). Let Fk := Db−;g 1 that fulfill Fk ∈ C ([a, b] , X ), and Fn+1 ∈ C ([a, b] , X ) . Then n   (g (b) − g (x))iα  iα Db−;g f (b) + (12.12) f (x) =  (iα + 1) i=0 1  ((n + 1) α)



b x

  (n+1)α f (t) dt, (g (t) − g (x))(n+1)α−1 g  (t) Db−;g

∀ x ∈ [a, b] . We make Remark 12.4 (to Theorem 12.3) When 0 < α < 1, by (12.6) we get 

 α Db−;g f (x) =

−1  (1 − α)



b

  (g (t) − g (x))−α g  (t) f ◦ g −1 (g (t)) dt,

x

(12.13)

∀ x ∈ [a, b] . Hence    b−;g f (x) ≤

 α  D

≤

1  (1 − α)       f ◦ g −1 ◦ g 



b x

∞,[a,b]

 (1 − α)

     (g (t) − g (x))−α g  (t)  f ◦ g −1 (g (t)) dt 

b x

(g (t) − g (x))

−α



g (t) dt

(12.14)

200

12 Iterated Generalized Right Side Fractional Landau Inequalities

=

      f ◦ g −1 ◦ g 

∞,[a,b]

 (2 − α)

(g (b) − g (x))1−α .

That is  α  D

b−;g

  f (x) ≤

      f ◦ g −1 ◦ g   (2 − α)

∀ x ∈ [a, b] , 0 < α < 1. Hence it holds

 α  D

b−;g

i.e.



∞,[a,b]

(g (b) − g (x))1−α < ∞, (12.15)

  f (b) = 0,

 α Db−;g f (b) = 0,

(12.16)

when 0 < α < 1. The author has already done an extensive amount of work on fractional Landau inequalities, see [3], and on abstract fractional Landau inequalities, see [4]. However there the proving methods came out of applications of fractional Ostrowski inequalities [2, 4]. Usually there the domains where [A, +∞) or (−∞, B], with A, B ∈ R and in one mixed case the domain was all of R. In this chapter with less assumptions we establish uniform and L p type right Caputo–Bochner abstract sequential generalized fractional Landau inequalities over R− . The method of proving is based on right Caputo–Bochner sequential generalized fractional Taylor’s formula with integral remainder, see Theorem 12.3. We give also an application for α = 21 . Clearly we are also inspired by [3, 4].

12.2 Main Results We present the following abstract sequential generalized fractional Landau inequalities over R− . −1 1 Theorem 12.5 Let g ∈ C 1 (R− ) strictly increasing,  (g (R− )) . Let  with g ∈C      0 < α < 1, f ∈ C 1 (R− , X ) with  f ∞,R− ,  f ◦ g −1 ◦ g  < ∞. For kα k = 1, 2, 3, we assume that Db−;g f ∈ C 1 ((−∞, b], X ) C ((−∞, b], X ), ∀ b ∈ R− . We further assume that

 4α  K g :=  Db−;g f (t)∞,R2 < ∞, −

where (b, t) ∈ R2− .

∞,R−

and

4α f ∈ Db−;g

(12.17)

12.2 Main Results

201

Then  2α   f (b) ≤ sup  Db−;g

 (2α + 1) 22α−1 (2α − 1)

b∈R−



  23α+1 23α + 1 (2α + 1)  f ∞,R− K g ,  (4α + 1) (12.18)

and   √ 3    4 4 2 (3α + 1) ( (4α + 1))− 4 22α + 1 1 3   3α 4  f ∞,R sup  Db−;g f (b) ≤ Kg 4 .  √ 3 √ α − 4 b∈R− 3 2 (2α − 1)

(12.19)

       2α  3α   That is sup  Db−;g f (b) , sup  Db−;g f (b) < ∞. b∈R−

b∈R−

  α f (b) = 0, ∀ b ∈ R− . Proof We notice easily again here that Db−;g We make use of Theorem 12.3 for 0 < α < 1 and n = 3, applied for any b ∈ R− and a = −∞. Momentarily we fix b ∈ R− . Let x2 < x1 < b, then g (x2 ) < g (x1 ) < g (b), and f (x1 ) − f (b) =

+

  (g (b) − g (x1 ))3α  3α (g (b) − g (x1 ))2α  2α Db−;g f (b) + Db−;g f (b)  (2α + 1)  (3α + 1)

1  (4α)



b x1

 4α  f (t) dt, (g (t) − g (x1 ))4α−1 g  (t) Db−;g

(12.20)

and f (x2 ) − f (b) =

  (g (b) − g (x2 ))3α  3α (g (b) − g (x2 ))2α  2α Db−;g f (b) + Db−;g f (b)  (2α + 1)  (3α + 1)

1 +  (4α)



b x2





(12.21)

4α f (t) dt. (g (t) − g (x2 ))4α−1 g  (t) Db−;g

That is (g (b) − g (x1 ))2α  2α  (g (b) − g (x1 ))3α  3α  Db−;g f (b) + Db−;g f (b) =  (2α + 1)  (3α + 1) 1 f (x1 ) − f (b) −  (4α)



b x1

(12.22)

 4α  f (t) dt =: A, (g (t) − g (x1 ))4α−1 g  (t) Db−;g

and (g (b) − g (x2 ))2α  2α  (g (b) − g (x2 ))3α  3α  Db−;g f (b) + Db−;g f (b) =  (2α + 1)  (3α + 1)

(12.23)

202

12 Iterated Generalized Right Side Fractional Landau Inequalities

1 f (x2 ) − f (b) −  (4α)



b x2

 4α  f (t) dt =: B. (g (t) − g (x2 ))4α−1 g  (t) Db−;g

  2α We are solving the above system of two equations with two unknowns Db−;g f (b) ,   3α Db−;g f (b) . The main determinant of system is (g(b)−g(x1 ))2α (2α+1) D := (g(b)−g(x2 ))2α (2α+1)

(g(b)−g(x1 ))3α (3α+1) (g(b)−g(x2 ))3α (3α+1)

1 =  (2α + 1)  (3α + 1)



(g (b) − g (x1 ))2α (g (b) − g (x2 ))3α − (g (b) − g (x1 ))3α (g (b) − g (x2 ))2α =  (g (b) − g (x1 ))2α (g (b) − g (x2 ))2α (g (b) − g (x2 ))α − (g (b) − g (x1 ))α > 0.  (2α + 1)  (3α + 1) I.e. D=

 (g (b) − g (x1 ))2α (g (b) − g (x2 ))2α (g (b) − g (x2 ))α − (g (b) − g (x1 ))α > 0.  (2α + 1)  (3α + 1)

(12.24)

We obtain the unique solution





2α Db−;g



f (b) =

 3α Db−;g f (b) =



A

(g(b)−g(x1 ))3α (3α+1)

B

(g(b)−g(x2 ))3α (3α+1) D

(g(b)−g(x1 ))2α (2α+1) (g(b)−g(x2 ))2α (2α+1) D



A B

, (12.25)

.

Therefore we have 

 2α Db−;g f (b) = and   3α f (b) = Db−;g

We have the following

(g(b)−g(x2 ))3α A− (g(b)−g(x1 ))3α B (3α+1) (3α+1) , D

(12.26) (g(b)−g(x1 ))2α B− (g(b)−g(x2 ))2α A (2α+1) (2α+1) . D

12.2 Main Results

203

   b     1   4α−1  4α A =  f (x1 ) − f (b) − g (t) Db−;g f (t) dt  ≤ (g (t) − g (x1 ))    (4α) x1

2  f ∞,R− +

   4α  Db−;g f (t)

∞,R2−

 (4α + 1)

(g (b) − g (x1 ))4α ,

(12.27)

under the assumption  f ∞,R− < ∞. That is A ≤ 2  f ∞,R− +

Kg (g (b) − g (x1 ))4α ,  (4α + 1)

(12.28)

Kg (g (b) − g (x2 ))4α ,  (4α + 1)

(12.29)

and similarly, B ≤ 2  f ∞,R− + where by assumption  4α  K g :=  Db−;g f (t)∞,R2 < ∞,

(12.30)

−

with (b, t) ∈ R2− . Consequently we have  2α     D b−;g f (b) ≤ 



1  (3α + 1) D

2  f ∞,R−

Kg + (g (b) − g (x1 ))4α  (4α + 1)

+ (g (b) − g (x1 ))3α 2  f ∞,R−

Kg + (g (b) − g (x2 ))4α  (4α + 1)

(g (b) − g (x2 ))

3α





 , (12.31)

and

 3α     D b−;g f (b) ≤  (g (b) − g (x1 ))

 2α

2  f ∞,R−

1  (2α + 1) D

Kg + (g (b) − g (x2 ))4α  (4α + 1)



204

12 Iterated Generalized Right Side Fractional Landau Inequalities

 + (g (b) − g (x2 ))

2α

2  f ∞,R−

Kg + (g (b) − g (x1 ))4α  (4α + 1)

 .

(12.32) Set now g (x1 ) := g (b) − h, g (x2 ) := g (b) − 2h, where h > 0, so that g (b) − g (x1 ) = h, g (b) − g (x2 ) = 2h. Hence we get 22α h 5α (2α − 1) > 0. (12.33) D=  (2α + 1)  (3α + 1) Therefore we derive (from (12.26))  2α     D b−;g f (b) ≤

    (2α + 1) Kg 3α 3α 4α h 2  f ∞,R− + 2 h 22α h 5α (2α − 1)  (4α + 1) 

+h

2  f ∞,R−

3α

Kg 24α h 4α +  (4α + 1)

 =

(12.34)

   3α  3α  7α  3α  (2α + 1) Kg 4α 2 +2 h 2  f ∞,R− 2 + 1 h + 22α h 5α (2α − 1)  (4α + 1) =

 (2α + 1) 22α (2α − 1)

That is



  3α 2 2 + 1  f ∞,R− h 2α  2α     D b−;g f (b) ≤

   2 23α + 1  f ∞,R− h 2α ∀ b ∈ R− , ∀ h > 0. I.e. it holds



   2 23α + 1  f ∞,R− h 2α ∀ h > 0, 0 < α < 1. By (12.26) we derive

 (2α + 1) 22α (2α − 1)



 23α (2α + 1) 2α + Kg h ,  (4α + 1)

 2α   f (b) ≤ sup  Db−;g

b∈R−

 23α (2α + 1) 2α + Kg h . (12.35)  (4α + 1)



 (2α + 1) 22α (2α − 1)

(12.36)



 23α (2α + 1) 2α + Kg h < ∞,  (4α + 1)

(12.37)

12.2 Main Results

205

    (3α + 1) Kg 2α 4α 4α 2 h 2  f ∞,R− + h 22α h 5α (2α − 1)  (4α + 1)

 3α     D b−;g f (b) ≤

 +2 h

2α 2α

2  f ∞,R−

Kg h 4α +  (4α + 1)

 =

   4α  2α  2α  6α  (3α + 1) Kg 2α 2 +2 h 2  f ∞,R− 2 + 1 h + 22α h 5α (2α − 1)  (4α + 1) (12.38)   2α 

  2α 2α 2 2 + 1  f ∞,R− 2 2 +1  (3α + 1) + Kg hα = 22α (2α − 1) h 3α  (4α + 1)     (3α + 1) 22α + 1 2  f ∞,R− 22α K g α = h . + 22α (2α − 1) h 3α  (4α + 1)    3α    (3α + 1) 22α + 1   D b−;g f (b) ≤ 22α (2α − 1)

That is



2  f ∞,R− h 3α

 22α K g α h , +  (4α + 1)

(12.39)

∀ b ∈ R− , ∀ h > 0. I.e. it holds

   3α    (3α + 1) 22α + 1   sup Db−;g f (b) ≤ 22α (2α − 1) b∈R− 

2  f ∞,R− h 3α

∀ h > 0, 0 < α < 1. Call

 22α K g α h < ∞, +  (4α + 1)

  μ := 2 23α + 1  f ∞,R− , θ=

23α (2α + 1) K g ,  (4α + 1)

(12.40)

(12.41) (12.42)

both are greater than zero. Set also ρ := 2α; 0 < ρ < 2. We consider the function y (h) := μh −ρ + θh ρ , ∀ h > 0.

(12.43)

206

12 Iterated Generalized Right Side Fractional Landau Inequalities

We have

y  (h) = −ρμh −ρ−1 + ρθh ρ−1 = 0,

(12.44)

then θh 2ρ = μ, with a unique solution h 0 := h crit.no. = We have that

 μ  2ρ1 θ

.

(12.45)

y  (h) = ρ (ρ + 1) μh −ρ−2 + ρ (ρ − 1) θh ρ−2 .

(12.46)

We see that y  (h 0 ) = y 

  1

μ 2ρ θ ρ

ρ

= ρ (ρ + 1) μ

 μ  −ρ−2 2ρ θ

+ ρ (ρ − 1) θ

 μ  ρ−2 2ρ θ

=

ρ1     θ (ρ + 1) μθ + (ρ − 1) μθ = μ

1 ρ1     θ ρ θ 2ρ μθ = 2ρ2 μθ > 0. μ μ

Therefore y has a global minimum at h 0 = y (h 0 ) = μ

 μ − 21 θ

+θ

 μ  21 θ

 μ  2ρ1 θ

, which is

21   θ =μ + θμ = 2 θμ. μ

We have proved that (see (12.37))  2α   sup  Db−;g f (b) ≤

b∈R−



 (2α + 1) (2α − 1)

22α−1

  23α+1 23α + 1 (2α + 1)  f ∞,R− K g .  (4α + 1)

Call

ξ := 2  f ∞,R− , ψ :=

both are greater than zero.

(12.47)

22α K g , (4α+1)

(12.48)

12.2 Main Results

207

We consider the function γ (h) := ξh −3α + ψh α , ∀ h > 0. We have

(12.49)

γ  (h) = −3αξh −3α−1 + αψh α−1 = 0,

then ψh 4α = 3ξ, with unique solution

h 0 := h crit.no. =

3ξ ψ

4α1

.

(12.50)

We have that γ  (h) = 3α (3α + 1) ξh −3α−2 + α (α − 1) ψh α−2 .

(12.51)

We see γ  (h 0 ) = 3α (3α + 1) ξ α

3ξ ψ

α



α−2 4α

3ξ ψ

α−2 4α



3ξ ψ

−3α−2 4α

3 (3α + 1) ξ



γ (h 0 ) = ξ

14

3ξ ψ

− 34

3ξ ψ

α−2 4α

=

 ψ + (α − 1) ψ = 3ξ

(4αψ) = 4α2 ψ

Therefore γ has a global minimum at h 0 =

3ξ ψ

  4α1 3ξ ψ



α−2 4α

> 0.

(12.52)

, which is

3ξ +ψ ψ

41

=

41 ψ 4 3ξ ξ +ψ = ψ . 3ξ 3 ψ

(12.53)

14 3ξ 4 4 3 1 = √ 3 ψ 4 ξ 4 . γ (h 0 ) = ψ 3 ψ 4 3

(12.54)



Consequently,

+ α (α − 1) ψ

3ξ ψ

208

12 Iterated Generalized Right Side Fractional Landau Inequalities

We have proved that (see (12.40))    3α   4 (3α + 1) 22α + 1   sup Db−;g f (b) ≤ √ 3 4 b∈R− 3 22α (2α − 1) 

2  f ∞,R−

 41



22α K g  (4α + 1)

 34 =

√  3  3 1 4 4 2 (3α + 1)  (4α + 1)− 4 22α + 1 4  f ∞,R Kg 4 . √ 3 α − 4 3 2 2 (2α − 1)

(12.55)



The theorem is proved.

We continue with abstract L p right sequential generalized fractional Landau inequalities over R− . Theorem 12.6 Let g ∈ C 1 (R− ) strictly increasing, with g −1 ∈ C 1 (g (R− )) . Let p, q > 1 : 1p + q1 = 1, 0 < α < 1. Let f ∈ C 1 (R− , X ) with  f ∞,R− ,      kα < ∞. For k = 1, 2, 3, we assume that Db−;g f ∈ C1  f ◦ g −1 ◦ g  ∞,R−

4α f ∈ C ((−∞, b], X ), ∀ b ∈ R− . We further assume that ((−∞, b], X ) and Db−;g



 4α  f  p,R sup  Db−;g

b∈R−

Then (1) under

1 2p

 < ∞.

−

(12.56)

< α < 1, we get ⎡⎛



1 ⎞   p    2α− 1 α 1 −3α 4α− 2  4α − (2α) p  2α   ⎢ 4α 1 + 2 p ⎠ sup  Db−;g f (b) ≤ ⎣⎝ 1 α 2 −1 2α − p b∈R−



1+2

 (4α) (q (4α − 1) + 1) 

1 p

2α 4α− 1p

1 q

⎤

 4α  f  p,R sup  Db−;g

< α < 1, we get

 −



2α− 1p 4α− 1p



⎥ ⎦  f ∞,R−



b∈R−

(2) under



α− 1p

2α 4α− 1p

< ∞.

(12.57)

12.2 Main Results

209

⎡⎛



 ⎞  α− 1p   4α−  1 1 −2α  4α − (3α) p  3α   ⎢ 6α 1 + 2 p ⎠ sup  Db−;g f (b) ≤ ⎣⎝ 1 α 2 −1 α− p b∈R−



1+2



2α− 1p

 (4α) (q (4α − 1) + 1) 

3α 4α− 1p

⎤



 4α  f  p,R sup  Db−;g

b∈R−

α− 1p 4α− 1p



⎥ ⎦  f ∞,R−

1 q





3α 4α− 1p

−

< ∞.

(12.58)

       2α  3α   That is sup  Db−;g f (b) , sup  Db−;g f (b) < ∞. b∈R−

b∈R−

Proof As in the proof of Theorem 12.5 we have that A

   b     1  4α−1  4α g (t) Db−;g f (t) dt  ≤ (g (t) − g (x1 ))  f (x1 ) − f (b) −    (4α) x1

(12.22.) 

=

2  f ∞,R− +

1  (4α)



b x1

 4α   f (t) dt ≤ (g (t) − g (x1 ))4α−1 g  (t)  Db−;g (q(4α−1)+1)

2  f ∞,R−

q 1 (g (b) − g (x1 )) + 1  (4α) (q (4α − 1) + 1) q

2  f ∞,R− +



   4α  sup Db−;g f 

b∈R−

  4α− 1p

h 1  (4α) (q (4α − 1) + 1) q1



 (g(b)−g(x 1 )=:h>0)

 4α  f  p,R sup  Db−;g

b∈R−

=

p,R−

 −

(12.59) ,

with 41p < α < 1. That is 

1

 A ≤ 2  f ∞,R− +

h 4α− p 1

 (4α) (q (4α − 1) + 1) q

 4α  f  p,R sup  Db−;g

b∈R−

 −

,

(12.60)

where 41p < α < 1. We also have B

   b     1  4α−1  4α g (t) Db−;g f (t) dt  ≤ (g (t) − g (x2 ))  f (x2 ) − f (b) −    (4α) x2

(12.23.) 

=

210

12 Iterated Generalized Right Side Fractional Landau Inequalities

2  f ∞,R− +

(g (b) − g (x2 ))

 (4α) (q (4α − 1) + 1)

2  f ∞,R− +

2



(q(4α−1)+1) q

4α− 1p

h

1 q

 4α  f  p,R sup  Db−;g

b∈R−



4α− 1p 1

 (4α) (q (4α − 1) + 1) q

 (g(b)−g(x2 )=:2h)

=

−

 4α  f  p,R sup  Db−;g

b∈R−

 (12.61) −

.

That is 1

B ≤ 2  f ∞,R− +



1

24α− p h 4α− p 1

 (4α) (q (4α − 1) + 1) q

 4α  f  p,R sup  Db−;g

b∈R−

 −

,

(12.62)

where 41p < α < 1. We have assumed that  Mg :=

 4α  sup  Db−;g f  p,R

b∈R−

 < ∞.

−

(12.63)

For convenience we call 1

c :=  (4α) (q (4α − 1) + 1) q > 0. So we have A ≤ 2  f ∞,R− + and

h

B ≤ 2  f ∞,R− +

2

4α− 1p

c

(12.64)

Mg , (12.65)

4α− 1p

h c

4α− 1p

Mg ,

where 41p < α < 1. Next we estimate the (12.26)-quantities and we have:  2α     D b−;g f (b) ≤

3α 3α  (12.33.) 1 2 h A + h 3α B = D (3α + 1)

 (12.65.) h 3α  (2α + 1) 3α 2 A + B ≤ 2α 5α α 2 h (2 − 1)  1  (2α + 1) 23α h 4α− p 3α+1  f ∞,R− + Mg + 2 22α (2α − 1) h 2α c 2  f ∞,R−

 1 1 24α− p h 4α− p + Mg = c

(12.66)

12.2 Main Results

211

  ⎡ ⎤ 1  23α + 24α− p 1  (2α + 1) ⎣ 23α+1 + 2  f ∞,R− + Mg h 2α− p ⎦ = 22α (2α − 1) h 2α c   ⎡  ⎤  α− 1p −3α 1 + 2 Mg  f ∞,R− 1 2  (2α + 1) ⎣ 2 1 + 2 h 2α− p ⎦ . + h 2α c (2α − 1) α

That is



(12.67)

 2α    D  b−;g f (b) ≤

  ⎤ ⎡   α− 1p −3α 1 + 2 M g   f 2 1 + 2 1 2  (2α + 1) ⎣ ∞,R− h 2α− p ⎦ , + 2α − 1 h 2α c

α

(12.68) ∀ b ∈ R− , ∀ h > 0. I.e. it holds

 2α   f (b) ≤ sup  Db−;g

b∈R−



2α  (2α + 1) 2α − 1



  ⎡  ⎤ 1  1 + 2α− p Mg 2 1 + 2−3α  f ∞,R− 1 ⎣ h 2α− p ⎦ , + h 2α c

(12.69)

∀ h > 0, under 41p < α < 1. Again from (12.26) we get  3α     D b−;g f (b) ≤

2α  1 h B + 22α h 2α A =  (2α + 1) D

 h 2α  (3α + 1) B + 22α A ≤ 2α 5α α 2 h (2 − 1)

h 2α  (3α + 1) 22α h 5α (2α − 1)

2

2α+1

(12.70)

1 1

 24α− p h 4α− p Mg + 2  f ∞,R− + c

 f ∞,R−

 1 22α h 4α− p + Mg = c

  ⎡ ⎤  4α− 1p 2α 2α+1 2 + 2  f ∞,R− 1  (3α + 1) ⎣ 2 + 2 + Mg h α− p ⎦ = (12.71) 22α (2α − 1) h 3α c

212

12 Iterated Generalized Right Side Fractional Landau Inequalities

  ⎡  ⎤ 1  1 + 22α− p 1  (3α + 1) ⎣ 2 1 + 2−2α  f ∞,R− + Mg h α− p ⎦ . h 3α c (2α − 1) That is



 3α    D  b−;g f (b) ≤

  ⎤ ⎡   2α− 1p −2α 1 + 2  f ∞,R− 1  (3α + 1) ⎣ 2 1 + 2 + Mg h α− p ⎦ , (12.72) 2α − 1 h 3α c

∀ b ∈ R− , ∀ h > 0. I.e. it holds

 3α   f (b) ≤ sup  Db−;g

b∈R−



 (3α + 1) 2α − 1



  ⎡  ⎤ 1  1 + 22α− p 2 1 + 2−2α  f ∞,R− 1 ⎣ + Mg h α− p ⎦ , h 3α c ∀ h > 0, Call

1 4p

(12.73)

< α < 1.   μ := 2 1 + 2−3α  f ∞,R− , θ :=



α− 1 1+2 p Mg c

(12.74) ,

both are greater than zero. We consider the function 1

We have

y (h) = μh −2α + θh 2α− p , ∀ h > 0.

(12.75)

1 1 θh 2α− p −1 = 0, y  (h) = −2αμh −2α−1 + 2α − p

(12.76)

then

2α −

i.e.

with a unique solution

1 1 θh 2α− p −1 = 2αμh −2α−1 , p



1 1 2α − θh 4α− p = 2αμ, p

12.2 Main Results

213

⎞

⎛ 2αμ

 ⎠ h 0 := h crit.no. = ⎝  2α − 1p θ

1 4α− 1p

(12.77)

(assuming 21p < α < 1). We have that



1 1 1 y  (h) = 2α (2α + 1) μh −2α−2 + 2α − 2α − − 1 θh 2α− p −2 . (12.78) p p We see that

⎞ −2α−21

⎛

2αμ  ⎠ y (h 0 ) = 2α (2α + 1) μ ⎝  2α − 1p θ

4α− p





+

⎛ ⎞ 2α− 1p 1−2



4α− p 2αμ 1 1  ⎠ 2α − 2α − − 1 θ ⎝  = p p 2α − 1 θ p

⎛

⎞ −2α−21

⎝  2αμ  ⎠ 2α − 1p θ

4α− p



 1 2α (2α + 1) μ + 2αμ 2α − − 1 = p

 ⎞ ⎛ 2α − 1p θ ⎠ 2αμ ⎝ 2αμ



(12.79)

2(α+1) 4α− 1p



1 p

4α −

> 0.

Therefore y has a global minimum at ⎞

⎛ 2αμ

 ⎠ h0 = ⎝  1 2α − p θ

1 4α− 1p

,

which is ⎛

⎞

2αμ  ⎠ y (h 0 ) = μ ⎝  2α − 1p θ

−2α 4α− 1p

⎛

⎞ 2α− 11p

2αμ  ⎠ + θ ⎝ 2α − 1p θ

4α− p

=

(12.80)

214

12 Iterated Generalized Right Side Fractional Landau Inequalities ⎞

⎛ ⎝

2αμ

 ⎠ 2α − 1p θ

−2α 4α− 1p

 ⎞ θ 2α − 2αμ ⎠ ⎣μ + θ   ⎦ = μ⎝ 2αμ 2α − 1 θ ⎛

⎤

⎡



1 p





2α 4α− 1p

1+

p

 =

4α −

(2α)

1 p



2α 4α− 1p



2α 2α −



−2α+ 1p 2α− 1p 2α

1 4α− 1p 1 1 2α − μ 4α− p θ 4α− p . p

1 p

(12.81)

That is

y (h 0 ) =

   4α − 1p 2α − 1p

(2α)



−2α+ 1p 4α− 1p



μ



2α− 1p 4α− 1p



2α 4α− 1p

θ

2α 4α− 1p

.

(12.82)

Therefore we derive (see (12.69))  2α   f (b) ≤ sup  Db−;g

b∈R−



   2 1 + 2−3α

2α− 1p 4α− 1p





⎛ ⎝



α

2  (2α) 2α − 1 

1 + 2α−

1 p

 (2α− 1p )

(4α− p ) 1 2α 4α − p 2α − 1p 1





⎞ 1

 (4α) (q (4α − 1) + 1) q

 4α  f  p,R sup  Db−;g

b∈R−

where 21p < α < 1. Call





2α 4α− 1p

⎠



2α− 1p 4α− 1p



 f ∞,R−

2α 4α− 1p

−

< ∞,

(12.83)

  ξ := 2 1 + 2−2α  f ∞,R− , ψ :=



2α− 1p 1+2 c

(12.84) Mg ,

both are greater than zero. We consider the function γ (h) := ξh −3α + ψh α− p , ∀ h > 0. 1

We have

(12.85)

12.2 Main Results

215

1 1 ψh α− p −1 = 0, γ  (h) = −3αξh −3α−1 + α − p then



1 1 α− ψh α− p −1 = 3αξh −3α−1 , p

and



1 1 α− ψh 4α− p = 3αξ, p

with unique solution ⎞

⎛ h 0 := h crit.no. = ⎝ 

3αξ α−

1 p

 ⎠ ψ

1 4α− 1p

(12.86)

(assuming 1p < α < 1). We have that



1 1 1 α − − 1 ψh α− p −2 . γ  (h) = 3α (3α + 1) ξh −3α−2 + α − p p We observe

⎞

⎛



3αξ  ⎠ γ  (h 0 ) = 3α (3α + 1) ξ ⎝  α − 1p ψ

α−

1 p



+

⎞

⎛



−3α−2 4α− 1p

(12.87)



3αξ 1  ⎠ α − − 1 ψ ⎝ 1 p α− ψ

α− 1p −2 4α− 1p



=

p

⎛

⎞

⎝  3αξ  ⎠ α − 1p ψ



−3α−2 4α− 1p





 1 3α (3α + 1) ξ + α − − 1 3αξ = p

⎛



⎞

3αξ  ⎠ 3αξ ⎝  1 α− p ψ Therefore y has a global minimum at

−3α−2 4α− 1p



1 > 0. 4α − p

(12.88)

216

12 Iterated Generalized Right Side Fractional Landau Inequalities

⎞

⎛ 3αξ

 ⎠ h0 = ⎝  α − 1p ψ which is ξh 0−3α

γ (h 0 ) = ⎛

+

α− 1 ψh 0 p





⎞ 3αξ

⎝  ⎠ α − 1p ψ

−3α 4α− 1p

⎛

γ (h 0 ) = ξ



4α − α−

 1 p 1 p

−3α 4α− 1p





4α − α−



1 p 1 p

.



α− 1p 4α− 1p



(12.89)

⎞⎞

⎞ ⎛ 4α − 1p 3αξ ⎝  ⎠ α − 1p α − 1p ψ



ξ

⎛



 ⎠ ξ ⎝ α − 1p ψ 



4α− 1p ξ + ψh 0 =

⎝ξ + ψ ⎝   ⎠⎠ = α − 1p ψ

3αξ

That is

,

3αξ

⎞

⎛

=

h −3α 0

1 4α− 1p

⎞ ⎛ α − 1p ⎝ ⎠ 3α



−3α 4α− 1p



=

3α 4α− 1p

ψ

⎛

⎞

⎝

3α ⎠ 4α− 1p

(12.90)

.

I.e. we have found   4α − 1p

γ (h 0 ) =





α−

1 p



α− 1p 4α− 1p



(3α)



ξ



α− 1p 4α− 1p



⎛

⎞

⎝

3α ⎠ 4α− 1p

ψ

.

(12.91)

3α 4α− 1p

We have proved that (see (12.73))

 ⎛ ⎞  α− 1p1 1 4α− 4α −  (3α) p  3α   p 3α ⎠ f (b) ≤ ⎝ sup  Db−;g α−1 2 α − 1p b∈R−



   2 1 + 2−2α

α− 1p 4α− 1p



⎛ ⎝



  1 1 + 22α− p

⎞ 1

 (4α) (q (4α − 1) + 1) q

⎠

3α 4α− 1p

(12.92)



α− 1p 4α− 1p

 f ∞,R−



12.2 Main Results

217





 4α  f  p,R sup  Db−;g

b∈R−



3α 4α− 1p

−

< ∞,

where 1p < α < 1. The theorem is proved.



We give an application when α =

1 2

and g (t) = et |R− .

  Corollary 12.7 Let f ∈ C 1 (R− , X ) with  f ∞,R− , ( f ◦ ln) ◦ et ∞,R− < k1

2 ∞, where (X, ·) is a Banach space. For k = 1, 2, 3, we assume that Db−;e t f ∈

41

2 C 1 ((−∞, b], X ) and Db−;e t f ∈ C ((−∞, b], X ), ∀ b ∈ R− . We further assume that

  1   4 2 Db−;et f (t)

∞,R2−

< ∞,

(12.93)

where (b, t) ∈ R2− . Then  √   

  1     4 21 12 + 6 2 2 21 2       f sup  f f ≤ D D √ (b) ∞,R−  b−;et (t)  b−;et 2−1

b∈R−

1 2

∞,R2−

< ∞,

(12.94)

 1    32  sup  Db−;e (b) ≤ t f

and

b∈R−

⎛

⎞ √ 

34  1 9 π  4 21  ⎜ ⎟ 4   D f f < ∞. (12.95) (t)     ⎝ ∞,R− b−;et √  √ √ 3 ⎠ ∞,R2− 2− 2 42 43  1     22 f That is sup  Db−;e (b) , sup t b∈R−

Proof By Theorem 12.5.

b∈R−

 1     32  Db−;et f (b) < ∞. 

References 1. Aglic Aljinovic, A., Marangunic, L., Pecaric, J.: On Landau type inequalities via Ostrowski inequalities. Nonlinear Funct. Anal. Appl. 10(4), 565–579 (2005) 2. Anastassiou, G.: Fractional Differentiation Inequalities. Research Monograph. Springer, New York (2009) 3. Anastassiou, G.A.: Advances on Fractional Inequalities. Springer, New York (2011)

218

12 Iterated Generalized Right Side Fractional Landau Inequalities

4. Anastassiou, G.A.: Intelligent Computations: Abstract Fractional Calculus, Inequalities, Approximations. Springer, Heidelberg (2018) 5. Anastassiou, G.A.: Sequential abstract generalized right side fractional Landau inequalities. Submitted (2020) 6. Barnett, N.S., Dragomir, S.S.: Some Landau type inequalities for functions whose derivatives are of locally bounded variation. Tamkang J. Math. 37(4), 301–308 (2006) 7. Ditzian, Z.: Remarks, questions and conjectures on Landau-Kolmogorov-type inequalities. Math. Inequal. Appl. 3, 15–24 (2000) 8. Hardy, G.H., Littlewood, J.E.: Some integral inequalities connected with the calculus of variations. Q. J. Math. Oxford Ser. 3, 241–252 (1932) 9. Hardy, G.H., Landau, E., Littlewood, J.E.: Some inequalities satisfied by the integrals or derivatives of real or analytic functions. Math. Z. 39, 677–695 (1935) 10. Kallman, R.R., Rota, G.C.: On the inequality  2 ≤ 4 ·  . In: Shisha, O. (ed.) Inequalities, vol. II, pp. 187–192. Academic, New York (1970) 11. Landau, E.: Einige Ungleichungen für zweimal differentzierban funktionen. Proc. Lond. Math. Soc. 13, 43–49 (1913) 12. Mikusinski, J.: The Bochner Integral. Academic, New York (1978) 13. Shilov, G.E.: Elementary Functional Analysis. Dover Publications Inc., New York (1996)

Chapter 13

High Order Generalized Landau Inequalities

We present several high order abstract generalized Landau type inequalities involving the uniform and L p norms. We mention also applications. Our estimates give best upper bounds. It follows [3].

13.1 Introduction Let p ∈ [1, ∞], I = R+ or I = R and f : I → R is twice differentiable with f, f  ∈ L p (I ), then f  ∈ L p (I ). Moreover, there exists a constant C p (I ) > 0 independent of f , such that 1    1  f  ≤ C p (I )  f  2  f   2 , (13.1) p,I

p,I

p,I

where · p,I is the p-norm on the interval I , see [1, 4]. The research on these inequalities started by Landau [9] in 1913. For the case of p = ∞ he proved that C∞ (R+ ) = 2 and C∞ (R) =

√ 2,

(13.2)

are the best constants in (13.1). In 1932, Hardy and Littlewood [6] proved (13.1) for p = 2, with the best constants C2 (R+ ) =

√

2 and C2 (R) = 1.

(13.3)

In 1935, Hardy, Landau and Littlewood [7] showed that the best constants C p (R+ ) in (13.1) satisfies the estimate C p (R+ ) ≤ 2, for p ∈ [1, ∞),

(13.4)

which yields C p (R) ≤ 2 for p ∈ [1, ∞). © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_13

219

220

13 High Order Generalized Landau Inequalities

√ In fact, in [5, 8] was shown that C p (R) ≤ 2. In this chapter we are greatly inspired by the interesting article [12]. We need Theorem 13.1 ([2], p. 97) Let n ∈ N and f ∈ C n ([a, b] , X ), where [a, b] ⊂ R and (X, ·) is a real or complex Banach space. Let g ∈ C 1 ([a, b]) strictly increasing, such that g −1 ∈ C n ([g (a) , g (b)]). Let any x, y ∈ [a, b]. Then f (x) = f (y) +

n−1  (g (x) − g (y))i 

i!

i=1

f ◦ g −1

(i)

(g (y)) + Rn (y, x, g) , (13.5)

where 

1 Rn (y, x, g) = (n − 1)! 1 (n − 1)!

x

 (n) (g (x) − g (t))n−1 f ◦ g −1 (g (t)) g  (t) dt =

y



g(x) g(y)

 (n) (g (x) − z)n−1 f ◦ g −1 (z) dz.

(13.6)

Derivatives here are similar to numerical ones, see [11]. The above integrals are of Bochner type [10], see also the Riemann integral for Banach space valued functions [11]. In this chapter we establish high order abstract generalized Landau inequalities for ·∞ and · p , p > 1, norms. We will use Corollary 13.2 Let f ∈ C 5 (R, X ) and (X, ·) is a real or complex Banach space. Let g ∈ C 1 (R) , strictly increasing, such that g −1 ∈ C 5 (g (R)). Let any x, y ∈ R. Then 4  (i) (g (x) − g (y))i  f ◦ g −1 (g (y)) f (x) = f (y) + i! i=1 +

1 24



g(x) g(y)

 (5) (g (x) − z)4 f ◦ g −1 (z) dz.

Proof Obvious, by Theorem 13.1.

(13.7) 

13.2 Main Results We present the following uniform high order abstract and generalized Landau type inequalities:

13.2 Main Results

221

Theorem 13.3 Let f ∈ C 5 (R, X ) and (X, ·) is a real or complex Banach space. such that g −1 ∈ C 5 (g (R)) , and assume that Let g ∈ C 1 (R)  , strictly increasing, (5)     f ∞,R ,  f ◦ g −1  < +∞. Then we get the following best upper ∞,g(R)

bounds (1)   (1)     f ◦ g −1 

∞,g(R)

5 ≤ 4

 5

1  4 (5)  9   5 5  f ∞,R < +∞,  f ◦ g −1  ∞,g(R) 8 (13.8)

(2)   (2)     f ◦ g −1 

∞,g(R)

2  3 (5)  20   5 5  f ∞,R ≤ √ < +∞,  f ◦ g −1  5 ∞,g(R) 3 675 (13.9)

(3)   (3)     f ◦ g −1 

∞,g(R)

5 ≤ 2

 5

3  2 (5)  4913   5 5  f ∞,R < +∞,  f ◦ g −1  ∞,g(R) 81000 (13.10)

and (4) 

  (4)     f ◦ g −1 

∞,g(R)

≤5

5

4  1 (5)  81   5 5  f ∞,R < +∞.  f ◦ g −1  ∞,g(R) 10000 (13.11)

Proof We apply (13.7) several times. Let x1 > y and g (x1 ) − g (y) = h > 0, that is g (x1 ) = g (y) + h. Thus f (x1 ) = f (y) +

4  hi  i=1

1 24



g(y)+h

g(y)

i!

f ◦ g −1

(i)

(g (y)) +

 (5) (g (y) + h − z)4 f ◦ g −1 (z) dz.

(13.12)

Let x2 > y and g (x2 ) − g (y) = 2h, that is g (x2 ) = g (y) + 2h. So that f (x2 ) = f (y) +

4  2i h i  i=1

1 24



g(y)+2h g(y)

i!

f ◦ g −1

(i)

(g (y)) +

 (5) (g (y) + 2h − z)4 f ◦ g −1 (z) dz.

(13.13)

222

13 High Order Generalized Landau Inequalities

Let x3 < y and g (x3 ) − g (y) = −h, that is g (x3 ) = g (y) − h. Hence f (x3 ) = f (y) +

4  (−1)i h i  i=1

1 24



g(y)

f ◦ g −1

i!

(i)

(g (y)) −

 (5) (g (y) − h − z)4 f ◦ g −1 (z) dz.

g(y)−h

(13.14)

Finally, let x4 < y and g (x4 ) − g (y) = −2h, that is g (x4 ) = g (y) − 2h. Therefore it holds 4  (i) (−1)i 2i h i  f ◦ g −1 (g (y)) − f (x4 ) = f (y) + i! i=1 1 24



g(y)

g(y)−2h

 (5) (g (y) − 2h − z)4 f ◦ g −1 (z) dz.

(13.15)

We get that 4  hi  i=1

1 24



i!

g(y)+h

f ◦ g −1

4  2i h i 

1 24



i!

g(y)+2h g(y)

f ◦ g −1

1 24

i!

and



(i)

g(y) g(y)−h

f ◦ g −1

(13.13)

(g (y)) =

i!

(i)

f ◦ g −1

(13.14)

(g (y)) =

(i)

(13.16)

f (x2 ) − f (y) −

(13.15)

(g (y)) =

(13.17)

f (x3 ) − f (y) +

 (5) (g (y) − h − z)4 f ◦ g −1 (z) dz =: C,

4  (−1)i 2i h i  i=1

f (x1 ) − f (y) −

 (5) (g (y) + 2h − z)4 f ◦ g −1 (z) dz =: B,

4  (−1)i h i  i=1

(13.12)

(g (y)) =

 (5) (g (y) + h − z)4 f ◦ g −1 (z) dz =: A,

g(y)

i=1

(i)

f (x4 ) − f (y) +

(13.18)

13.2 Main Results

1 24



223

g(y)

g(y)−2h

 (5) (g (y) − 2h − z)4 f ◦ g −1 (z) dz =: D.

We solve the system of equation (13.16)–(13.19) for 1, 2, 3, 4. Adding (13.16) and (13.18) we get



f ◦ g −1

(13.19) (i)

(g (y)), i =

 (2) (4) h4  f ◦ g −1 h 2 f ◦ g −1 (g (y)) + (g (y)) = A + C. 12

(13.20)

Adding (13.17) and (13.19) we obtain  (2) (4) 4  4h 2 f ◦ g −1 (g (y)) + h 4 f ◦ g −1 (g (y)) = B + D. 3

(13.21)

Similarly we have

and

(1) (3)  h3  f ◦ g −1 2h f ◦ g −1 (g (y)) + (g (y)) = A − C, 3

(13.22)

(1) (3)  8h 3  f ◦ g −1 4h f ◦ g −1 (g (y)) + (g (y)) = B − D. 3

(13.23)

We find

 

f ◦ g −1

f ◦ g −1

 

f ◦g

(1)

(2)

(g (y)) =

 −1 (3)

f ◦ g −1

(g (y)) =

(4)

8A−B+D−8C , 12h 16A+16C−B−D , 12h 2

(13.24) (g (y)) =

B−D−2 A+2C , 2h 3

(g (y)) =

B+D−4 A−4C . h4

We observe that 8A − B + D − 8C = 8 (A − C) + D − B =   g(y)+h   (5) 8 ( f (x1 ) − f (x3 )) − 1 (g (y) + h − z)4 f ◦ g −1 (z) dz−  3 g(y) 1 3



g(y) g(y)−h

 (5) (g (y) − h − z)4 f ◦ g −1 (z) dz + f (x4 ) − f (x2 ) + 1 24



g(y) g(y)−2h

 (5) (g (y) − 2h − z)4 f ◦ g −1 (z) dz+

(13.25)

224

13 High Order Generalized Landau Inequalities

1 24



g(y)+2h

g(y)

18  f ∞,R + 1 3 1 24 1 24





g(y)−h





1 3

g(y)



f ◦g

 −1 (5)

  (z) dz  ≤

 (5)    (g (y) + h − z)4  f ◦ g −1 (z) dz+

 (5)    (g (y) − h − z)4  f ◦ g −1 (z) dz+

g(y)−2h

g(y)

g(y)+h g(y)

g(y)

g(y)+2h

(g (y) + 2h − z)

4

 (5)    (g (y) − 2h − z)4  f ◦ g −1 (z) dz+

 (5)    (g (y) + 2h − z)4  f ◦ g −1 (z) dz ≤

  (5)    18  f ∞,R +  f ◦ g −1 (z)

∞,g(R)

18  f ∞,R +



(13.26)

2h 5 8h 5 + = 15 15

  (5)  2   h5.  f ◦ g −1  ∞,g(R) 3

That is 8A − B + D − 8C ≤ 18  f ∞,R +

  (5)  2   h 5 . (13.27)  f ◦ g −1  ∞,g(R) 3

Therefore we find (by (13.24))   3  f  (1)   ∞,R + (g (y)) ≤  f ◦ g −1 2h

  (5)     f ◦ g −1 

∞,g(R)

18

h4,

(13.28)

h > 0, for any y ∈ R. Hence it holds   (1)     f ◦ g −1 

∞,g(R)

≤

3  f ∞,R + 2h

  (5)     f ◦ g −1 

∞,g(R)

18

∀ h > 0. Next we estimate 16A + 16C − B − D = 16 f (x1 ) − 16 f (y) −

h4,

(13.29)

13.2 Main Results

2 3



g(y)+h

g(y)

2 3 1 24



225

 (5) (g (y) + h − z)4 f ◦ g −1 (z) dz + 16 f (x3 ) − 16 f (y) +

g(y)

g(y)−h



 (5) (g (y) − h − z)4 f ◦ g −1 (z) dz − f (x2 ) + f (y) +

g(y)+2h

g(y)

1 24

 (5) (g (y) + 2h − z)4 f ◦ g −1 (z) dz − f (x4 ) + f (y) − 

g(y)

(g (y) − 2h − z)

4

g(y)−2h

64  f ∞,R +



f ◦g

 −1 (5)

  (z) dz  ≤

  (5)  4   h5.  f ◦ g −1  ∞,g(R) 5

(13.30)

Hence (by (13.24)) we obtain     −1 (5)  f ◦ g      (2)   16  f ∞,R ∞,g(R) 3 + h , (g (y)) ≤  f ◦ g −1 3h 2 15

(13.31)

h > 0, for any y ∈ R. Therefore it holds   (2)     f ◦ g −1 

∞,g(R)

≤

16  f ∞,R + 3h 2

  (5)     f ◦ g −1 

∞,g(R)

h 3 , (13.32)

∞,g(R)

h5,

15

∀ h > 0. Similarly, we find that

B − D − 2 A + 2C ≤ 6  f ∞,R +

  (5)    17  f ◦ g −1  30

(13.33)

and by (13.24) we get that    3  f  (3)   17     ∞,R −1 (5)  f ◦ g + h 2 , (13.34) (g (y)) ≤   f ◦ g −1   ∞,g(R) h3 60 h > 0, for any y ∈ R. Consequently we find   (3)     f ◦ g −1 

∞,g(R)

≤

   3  f ∞,R 17   −1 (5)  f ◦ g + h2,    ∞,g(R) h3 60 (13.35)

226

13 High Order Generalized Landau Inequalities

∀ h > 0. Finally we obtain B + D − 4 A − 4C ≤ 16  f ∞,R +

  (5)  3   h 5 . (13.36)  f ◦ g −1  ∞,g(R) 5

By (13.24) we derive     −1 (5)  f ◦ g 3      (4)   16  f ∞,R ∞,g(R) + h, (g (y)) ≤  f ◦ g −1 h4 5

(13.37)

h > 0, for any y ∈ R. The last implies   (4)     f ◦ g −1 

∞,g(R)

≤

  (5)    3  f ◦ g −1 

16  f ∞,R + h4

∞,g(R)

5

h, (13.38)

∀ h > 0. Next we work on (13.29). By Theorem 13.6 we obtain the best upper bound:   (1)     f ◦ g −1 

∞,g(R)



5 4

4 5

3  f ∞,R 2

5 = 4

 5

 45

≤

(13.39)

 ⎞ 15 ⎛  (5)     f ◦ g −1  ⎜ ∞,g(R) ⎟ ⎠ ⎝ 18

1  4 (5)  9   5 5  f ∞,R .  f ◦ g −1  ∞,g(R) 8

Next we work on (13.32). Consequently by Theorem 13.6 we get the best upper bound:   (2)     f ◦ g −1 

∞,g(R)



5 3 5

3 2

2 5

16  f ∞,R 3

 35

≤

 ⎞ 25 ⎛  (5)     f ◦ g −1  ⎜ ∞,g(R) ⎟ ⎠ ⎝ 15

 3   2 20  −1 (5)  5 5   f ◦ g f = √ .    ∞,R 5 ∞,g(R) 3 675

(13.40)

13.2 Main Results

227

Next we work on (13.35). Therefore by Theorem 13.6 we get the best upper bound:   (3)     f ◦ g −1 

∞,g(R)

2

5 · 35



2 5

17 60

≤

(13.41)

 35  3 (5)    5  f ◦ g −1 

 f ∞,R 2 3 ∞,g(R) 25 · 35  3  2 (5)  5 5 4913   5 5  f ∞,R = .  f ◦ g −1  ∞,g(R) 2 81000 Finally we work on (13.38). Therefore by Theorem 13.6 we find the best upper bound:   (4)     f ◦ g −1 

∞,g(R)

≤

(13.42)

 45  4 (5)  3   5  f ∞,R  f ◦ g −1  4 16 ∞,g(R) 5 45  4  1 (5)  81   5 5 5  f ∞,R =5 .  f ◦ g −1  ∞,g(R) 10000 5

1 5

1 5



The theorem is proved.

Next we give the following uniform- L p high order abstract and generalized Landau type inequalities: Theorem 13.4 Let f ∈ C 5 (R, X ) and (X, ·) is a real or complex Banach space. Consider g ∈ C 1 (R) , strictly increasing, such that g −1 ∈ C 5 (g (R)) , and assume that  f ∞,R < +∞. Let p, q > 1 : 1p + q1 = 1, and assume that   (5)    < +∞. Then we get the following best upper bounds  f ◦ g −1  p,g(R)

(1)   (1)     f ◦ g −1 

∞,g(R)

⎛ ⎝

1+2

1+ q1

⎞ 1

18 (4q + 1) q

⎠



 1 4+ q1



3+ q1 4+ q1

 f ∞,R



≤

  4 + q1

(13.43)

1    (3+ q1 ) 4+ q ) 3 1 ( 3+ q 2

  (5)     f ◦ g −1 



 p,g(R)

 1 4+ q1

< +∞,

228

13 High Order Generalized Landau Inequalities

(2)   (2)     f ◦ g −1 

≤

∞,g(R)



2

 ⎞ ⎛  1 1 + 2q ⎝ ⎠ 1 9 (4q + 1) q





2 4+ q1



2+ q1 4+ q1

 f ∞,R

 2 4+ q1

  4 + q1 

∞,g(R)

⎝

1

1 + 8 · 2q

⎞ ⎠

1

12 (4q + 1) q



3

(4+ q1 )

1+ q1 4+ q1

≤



 f ∞,R



 3 4+ q1



 2 4+ q1

< +∞,

p,g(R)

∞,g(R)

≤



4

(13.44)

  4 + q1 1    (1+ q1 ) 4+ q ) 1 ( 3 1+ q

p,g(R)

  (4)     f ◦ g −1 



1 q

2+ q1 4+ q1

 3  (5)    4+ 1  f ◦ g −1  ( q ) < +∞,

and (4)

⎞ ⎛ 1 1 + 4 · 2q ⎝ ⎠ 1 3 (4q + 1) q







3 ⎛

2+

  (5)     f ◦ g −1 

(3)   (3)     f ◦ g −1 

16 3



  4 + q1 

4 4+ q1

  16 q

1 q 4+ q1

(

(13.45)

(13.46) )

 4 4+ q1

1 q 4+ q1

(

 f ∞,R



  (5)   )   f ◦ g −1 

 4 4+ q1

p,g(R)

< +∞.

Proof We continue as in the proof of Theorem 13.3. As in (13.26) we have 8A − B + D − 8C ≤ 18  f ∞,R + 1 3 1 3



g(y)+h g(y)



g(y)

g(y)−h

 (5)    (g (y) + h − z)4  f ◦ g −1 (z) dz+

 (5)    (z − (g (y) − h))4  f ◦ g −1 (z) dz+

(13.47)

13.2 Main Results

229

1 24 

1 24



g(y)+2h

g(y)

g(y) g(y)−2h

 (5)    (z − (g (y) − 2h))4  f ◦ g −1 (z) dz+

 (5)    (g (y) + 2h − z)4  f ◦ g −1 (z) dz ≤ 18  f ∞,R +

(we apply Hölder’s inequality four times)



1 3

1 24

(g (y) + h − z) dz

 q1 

g(y)+h

4q

g(y)

1 3 1 24

g(y)+h



g(y)

g(y)

(z − (g (y) − h)) dz

 q1 

g(y)

4q

g(y)−h



g(y)

g(y)−2h



g(y)−h

(z − (g (y) − 2h))4q dz

g(y)+2h

(g (y) + 2h − z) dz

 q1 

  p  1p    −1 (5) + (z) dz  f ◦g   p  1p    −1 (5) + (z) dz  f ◦g

g(y) g(y)−2h

 q1 

g(y)+2h

4q

g(y)

18  f ∞,R +

18  f ∞,R +

g(y)

  (5)     f ◦ g −1  (4q + 1)

  (5)     f ◦ g −1 

 p,g(R)

1 q

p,g(R)

3 (4q + 1)

  p  1p    −1 (5) f ◦ g + (z) dz    p  1p    −1 (5) ≤ (z) dz  f ◦g

1

2 24+ q + 3 12

  1 2 1 + 21+ q 1 q

 1

h 4+ q =

1

h 4+ q .

(13.48)

Therefore (by (13.24))      1+ q1  −1 (5)  f ◦ g 1 + 2      (1) p,g(R) 3+ q1   3  f ∞,R + h , (g (y)) ≤  f ◦ g −1 1 2h 18 (4q + 1) q (13.49) h > 0, for any y ∈ R.

230

13 High Order Generalized Landau Inequalities

Consequently it holds   1+ 1 1+2 q 3  f ∞,R ≤ + 1 2h 18 (4q + 1) q

  (1)     f ◦ g −1   

∞,g(R)

  (5)     f ◦ g −1   

h

3+ q1

,

p,g(R)

(13.50)

∀ h > 0. Working similarly over (13.30) we obtain 16A + 16C − B − D ≤   1 4 1 + 2 q    1  −1 (5)  64  f ∞,R + h 4+ q ,  1  f ◦ g p,g(R) 3 (4q + 1) q

(13.51)

and   (2)   16  f ∞,R  f ◦ g −1 ≤ + (g (y))   3h 2



1   1 + 2q (5)    2+ 1  f ◦ g −1  h q,  1  p,g(R) 9 (4q + 1) q

(13.52)

h > 0, for any y ∈ R. Consequently it holds   (2)     f ◦ g −1   

∞,g(R)

 1 1 + 2q 16  f ∞,R ≤ + 1 3h 2 9 (4q + 1) q

  (5)     f ◦ g −1   

h

2+ q1

,

p,g(R)

(13.53)

∀ h > 0. Similarly, we find (case of (13.24), etc.)   (3)     f ◦ g −1   

∞,g(R)

 1 1 + 8 · 2q 3  f ∞,R ≤ + 1 h3 12 (4q + 1) q

  (5)     f ◦ g −1   

h

1+ q1

,

p,g(R)

(13.54)

∀ h > 0. Finally we obtain (case of (13.24), etc.)

  (4)     f ◦ g −1   

∞,g(R)

≤

16  f ∞,R h4

1



1 + 4 · 2q +

3 (4q +

1 1) q

  (5)     f ◦ g −1   

∀ h > 0. Next we are working on (13.50) by using Theorem 13.6.

1

hq , p,g(R)

(13.55)

13.2 Main Results

231

We have the best upper bound   (1)     f ◦ g −1 

∞,g(R)



3  f ∞,R 2

=





3+ q1 4+ q1



⎛

   ( 3 1 ( 3 + 2 q 3+ q1 4+ q1



(13.56)

 (3+ q1 ) 3 + q1 (4+ q ) 1

 ⎡ ⎤ 1  1 + 21+ q    −1 (5)  ⎣ ⎦  1  f ◦ g p,g(R) q 18 (4q + 1)

  4 + q1



≤

  4 + q1



 f ∞,R

3+ q1 4+ q1

) )

⎝

1+2

1+ q1

⎞



 1 4+ q1

 1 4+ q1

⎠

1

18 (4q + 1) q



  (5)     f ◦ g −1 





 1 4+ q1

.

p,g(R)

Next we are working on (13.53) by using Theorem 13.6. We have the best upper bound   (2)     f ◦ g −1 

≤

∞,g(R)



2



16  f ∞,R 3

=





2

 2 4+ q1

2+ q1 4+ q1



⎛  ⎝

1+2

16 3



1



2+ 2+ q1 4+ q1

 f ∞,R





2+ q1 4+ q1

  2 + q1



2+ q1 4+ q1

  (5)     f ◦ g −1  ⎛ 

1 q



 2 4+ q1



(13.57)





9 (4q + 1) q

  4 + q1 

1 q

  4 + q1



⎝

⎞ ⎠ p,g(R)



1+2

1 q

 ⎞ 1

9 (4q + 1) q

  (5)     f ◦ g −1 

 2 4+ q1

 2 4+ q1

⎠

 2 4+ q1

p,g(R)

.

Next we are working on (13.54) by using Theorem 13.6. We have the best upper bound

232

13 High Order Generalized Landau Inequalities

  (3)     f ◦ g −1 

∞,g(R)

≤



3

 3 4+ q1

  4 + q1

(13.58)

1    (1+ q1 ) 4+ q ) 1 ( 3 1+ q

 ⎞ 31

1 ⎛ 1 1+ q (4+ q )  1 + 8 · 2q   3 4+ q1  −1 (5)  (4+ q1 ) ⎝ ⎠   f ◦ g f .    1 ∞,R p,g(R) 12 (4q + 1) q Finally we are working on (13.55) by using Theorem 13.6. We have the best upper bound   1    4 + (4)  q  (13.59) ≤   f ◦ g −1    11 4 ∞,g(R) 16 q (4+ q ) 4+ q1 4 q



⎞ ⎛ 1 1 + 4 · 2q ⎝ ⎠ 1 3 (4q + 1) q

 4 4+ q1

1 q 4+ q1

(

 f ∞,R



  (5)   )   f ◦ g −1 

 4 4+ q1

p,g(R)

. 

The theorem is proved. We make

Remark 13.5 Theorems 13.3 and 13.4 give nice applications when g = id, e x , tanh (x), etc. Due to lack of space we omit writing down the exact inequalities.

13.3 Appendix We mention Theorem 13.6 Let α, β, A, B > 0 and the function y (h) = hAα + Bh β , h > 0. It 1   (α+β) has only one critical number h 0 = αβ BA , such that y  (h 0 ) = 0. Furthermore y has a global minimum which is y (h 0 ) =

(α + β) α

α α+β

β

β α+β

β

α

A α+β B α+β .

(13.60)

Proof We have y (h) = Ah −α + Bh β , ∀ h > 0, and setting y  (h) = −α Ah −α−1 + 1   α+β > 0. We notice that β Bh β−1 = 0, we find h 0 := h crit.no. = αβ BA

13.3 Appendix

233 β−2

y  (h 0 ) = α (α + 1) Ah −α−2 + β (β − 1) Bh 0 0

=

αA βB

That is y  (h 0 ) > 0, and y has a global minimum, which is

β−2  α+β

β B (α + β) > 0. (13.61)

  β α (α + β) α+β β β −(α+β) Ah + Bh = h + B = A B α+β . (13.62) y (h 0 ) = Ah −α β 0 0 0 0 α α α+β β α+β 

References 1. Aglic Aljinovic, A., Marangunic, L., Pecaric, J.: On Landau type inequalities via Ostrowski inequalities. Nonlinear Funct. Anal. Appl. 10(4), 565–579 (2005) 2. Anastassiou, G.A.: Intelligent Computations: Abstract Fractional Calculus, Inequalities, Approximations. Springer, Heidelberg (2018) 3. Anastassiou, G.A.: High order abstract generalized Landau inequalities. J. Appl. Pure Math. Accepted (2020) 4. Barnett, N.S., Dragomir, S.S.: Some Landau type inequalities for functions whose derivatives are of locally bounded variation. Tamkang J. Math. 37(4), 301–308 (2006) 5. Ditzian, Z.: Remarks, questions and conjectures on Landau-Kolmogorov-type inequalities. Math. Inequal. Appl. 3, 15–24 (2000) 6. Hardy, G.H., Littlewood, J.E.: Some integral inequalities connected with the calculus of variations. Q. J. Math. Oxford Ser. 3, 241–252 (1932) 7. Hardy, G.H., Landau, E., Littlewood, J.E.: Some inequalities satisfied by the integrals or derivatives of real or analytic functions. Math. Z. 39, 677–695 (1935)  2   8. Kallman, R.R., Rota, G.C.: On the inequality  f   ≤ 4  f  ·  f  . In: Shisha, O. (ed.) Inequalities, vol. II, pp. 187–192. Academic, New York (1970) 9. Landau, E.: Einige Ungleichungen für zweimal differentzierban funktionen. Proc. Lond. Math. Soc. 13, 43–49 (1913) 10. Mikusinski, J.: The Bochner Integral. Academic, New York (1978) 11. Shilov, G.E.: Elementary Functional Analysis. Dover Publications Inc., New York (1996) 12. Xiao, Y.: Landau type inequalities for Banach space valued functions. J. Math. Inequal. 7(1), 103–114 (2013)

Chapter 14

Multidimensional Caputo Left Side Fractional Landau Inequalities

Relied on author’s first ever found multivariate Caputo fractional Taylor’s formula ([1], Chap. 13), we develop and prove several multivariate left side Caputo fractional uniform Landau type inequalities. It follows [2].

14.1 Introduction We are motivated by the following two results. First from [4], we have the following nice result: 2 d   ∂ If  f = f is the Laplacian of a function defined on Rd , then for 0 < ∂xi k < 2n,

i=1

  ∂ ∂   ∂ξ ... ∂ξ 1 k

  k  k 1−( 2n ) n f  2n , f  ≤ C (n, k)  f 

(14.1)

  where · denotes the L ∞ Rd norm. n f needs to be defined only in the distribup tional sense and the result is valid in many other  spaces such as L spaces.  Banach k k−1 . C (n, k) is an existential constant and  =   From [5], we have the following interesting result which is the sharp inequality   f ∞ ≤ 2

  d  f ∞ 2 f ∞ , (d + 2)

(14.2)

in Rd , d ≥ 2, where  f is the Laplacian and 2 =  (). This result is an analogue of the famous Landau inequality [6] in R:   f 

∞

≤

√

2  f ∞  f  ∞ ,

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_14

(14.3) 235

236

14 Multidimensional Caputo Left Side Fractional Landau Inequalities

where ·∞ means the L ∞ norm on Rd (respectively on R). In this chapter we present left Caputo fractional multivariate Landau type inequalities where all the constants are exactly calculated and we have best upper bounds. We need to mention the following background coming from [1], Chap. 13, pp. 269–278. We need Definition 14.1 Let ν ≥ 0; the operator Jaν , defined on L 1 (a, b) by Jaν

1 f (x) :=  (ν)



x

(x − t)ν−1 f (t) dt,

(14.4)

a

for a ≤ x ≤ b, is called the Riemann–Liouville left fractional integral operator of order ν. For ν = 0, we set Ja0 := I , the identity operator. Here  stands for the gamma function. By [3], p. 13, Jaν f (x), ν > 0, exists for almost all x ∈ [a, b] and Jaν f ∈ L 1 (a, b), where f ∈ L 1 (a, b). Here AC n ([a, b]) is the space of functions with absolutely continuous (n − 1)-st derivative. We need to mention Definition 14.2 ([3], p. 49, [1], p. 394) Let ν ≥ 0; n := ν , · is the ceiling of the number, f ∈ AC n ([a, b]). We call the left Caputo fractional derivative ν D∗a

1 f (x) :=  (n − ν)



x

(x − t)n−ν−1 f (n) (t) dt,

(14.5)

a

∀ x ∈ [a, b] . ν The above function D∗a f (x) exists almost everywhere for x ∈ [a, b]. ν 0 f = f. If ν ∈ N, then D∗a f = f (ν) the ordinary derivative; it is also D∗a We also mention Theorem 14.3 (Taylor expansion for left Caputo fractional derivatives, [3, p. 54]) Assume ν ≥ 0, n = ν , and f ∈ AC n ([a, b]). Then  x n−1 f (k) (a) 1 k ν f (t) dt, f (x) = (x − t)ν−1 D∗a (x − a) + k!  (ν) a k=0

(14.6)

∀ x ∈ [a, b] . Remark 14.4 Let Q be a compact and convex subset of Rk , k ≥ 2; z := (z 1 , ..., z k ), x0 := (x01 , ..., x0k ) ∈ Q. Let f ∈ C n (Q), n ∈ N. Set (14.7) gz (t) := f (x0 + t (z − x0 )) , 0 ≤ t ≤ 1;

14.1 Introduction

237

gz (0) = f (x0 ) , gz (1) = f (z) . Then gz( j)

⎡ j ⎤ k ∂ f ⎦ (x0 + t (z − x0 )) , (t) = ⎣ (z i − x0i ) ∂x i i=1

j = 0, 1, 2, ..., n, and gz(n)

(0) =

 k

∂ (z i − x0i ) ∂xi i=1

(14.8)

n  f (x0 ) .

(14.9)

rem We mention the following basic multivariate Caputo left fractional fundamental theorem. Theorem 14.5 ([1], p. 272) Let Q be a compact and convex subset of Rk , k ≥ 2; z := (z 1 , ..., z k ), x0 := (x01 , ..., x0k ) ∈ Q, f ∈ C 1 (Q), 0 < ν ≤ 1. Then f (z) = f (x0 ) + k

 (z i − x0i )

i=1



1  (ν)

1

(1 − t)ν−1 J01−ν

0



  ∂ (x0 + t (z − x0 )) dt . (14.10) ∂xi

We also mention the following Caputo left fractional bivariate Taylor formula. Theorem 14.6 ([1], p. 274) Let f ∈ C 2 (Q), Q ⊆ R2 compact and convex, z := (z 1 , z 2 ), x0 := (x01 , x02 ) ∈ Q, and 1 < ν ≤ 2. Then (1) f (z 1 , z 2 ) = f (x01 , x02 ) + (z 1 − x01 )  (z 1 − x01 )2

1  (ν)



∂f ∂f (x01 , x02 ) + (z 2 − x02 ) (x01 , x02 ) + ∂x1 ∂x2

    2 ∂ f dt + + t − x (z (1 − t)ν−1 J02−ν (x )) 0 0 ∂x12

1

0



     1 1 ∂2 f 2−ν ν−1 2 (z 1 − x01 ) (z 2 − x02 ) dt J0 (1 − t) (x0 + t (z − x0 ))  (ν) 0 ∂x1 ∂x2

 + (z 2 − x02 )

2

1  (ν)

Additionally assume that

 0

1

(1 − t)

ν−1



 J02−ν

∂2 f (x0 + t (z − x0 )) ∂x22



 dt . (14.11)

238

14 Multidimensional Caputo Left Side Fractional Landau Inequalities

f (x0 ) =

∂f ∂f (x0 ) = (x0 ) = 0, ∂x1 ∂x2

then (2) f (z 1 , z 2 ) =  (z 1 − x01 )

2



1  (ν)

1

ν−1

(1 − t)

0



 J02−ν

∂2 f (x0 + t (z − x0 )) ∂x12



 dt +



     1 1 ∂2 f 2−ν ν−1 2 (z 1 − x01 ) (z 2 − x02 ) dt J0 (1 − t) (x0 + t (z − x0 ))  (ν) 0 ∂x1 ∂x2

 + (z 2 − x02 )2

1  (ν)



1

0

    2 ∂ f dt . + t − x (z (1 − t)ν−1 J02−ν (x )) 0 0 ∂x22 (14.12)

We finally mention the following general multivariate Caputo left fractional Taylor formula. Theorem 14.7 ([1], p. 276) Let ν > 0, n = ν , f ∈ C n (Q), where Q is a compact and convex subset of Rk , k ≥ 2; z := (z 1 , ..., z k ), x0 := (x01 , ..., x0k ) ∈ Q. Then (1) k ∂ f (x0 ) f (z) = f (x0 ) + + (z i − x0i ) ∂xi i=1  n−1

k 

(z i − x0i )

i=1



1

 ν−1

(1 − t)

J0n−ν

0

l  f (x0 )

l!

l=2

1  (ν)

∂ ∂xi

 k

∂ (z i − x0i ) ∂xi i=1

+ n 



f (x0 + t (z − x0 ))

dt.

(14.13) Additionally assume that f α (x0 ) =0, α := (α1 , ..., αk ), αi ∈ Z+ , i = 1, ..., k, |α| := k  αi =: r , r = 0, ..., n − 1; then i=1

(2) f (z) = 1  (ν)

 0

1

 ν−1

(1 − t)

J0n−ν

 k

∂ (z i − x0i ) ∂x i i=1

n  f (x0 + t (z − x0 ))

 dt. (14.14)

14.2 Main Results

239

14.2 Main Results Here we present several left Caputo fractional multivariate Landau inequalities.   Theorem 14.8 Let f ∈ C 2 Rk+ , k ≥ 2, with  f ∞,Rk+ < ∞, and 1 < ν < 2. Assume further that ⎧ ⎪ ⎪ ⎨

  sup |J02−ν f xi xi (x0 + t (z − x0 )) |, ⎪ k ⎪ ⎩x0 ,z∈R+ ,t∈[0,1],

K ν := max

i=1,...,k

 2−ν  J fx x

sup ⎧ ⎨l ⎩

0

x0 ,z∈Rk+ ,t∈[0,1],

i

= l (i, j) = 1, ..., k(k−1) ; 2 i< j

j

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ (x0 + t (z − x0 ))  < +∞. (14.15) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

Then we derive the best upper bound  k   ∂ f       ∂xi  i=1

 1 1 2 2 2  f ∞,R k Kν . +  (ν + 1)

≤ 2k

∞,Rk+

 k   ∂ f       ∂xi 

That is

(14.16)

< ∞.

∞,Rk+

i=1

Proof For any x0 , z ∈ Rk+ by Theorem 14.7 (see also Theorem 14.6) we have that f (z) = f (x0 ) +

k

(z i − x0i )

i=1

∂ f (x0 ) + ∂xi

⎫⎤ ⎧⎡ 2 ⎤  1 k ⎬ ⎨ 1 ∂ f ⎦ (x0 + t (z − x0 )) ⎦ dt (1 − t)ν−1 ⎣ J02−ν ⎣ (z i − x0i ) ⎭ ⎩  (ν) 0 ∂xi ⎡

i=1

= f (x0 ) +

k i=1

(z i − x0i )

∂ f (x0 ) + ∂xi

(14.17)

240

14 Multidimensional Caputo Left Side Fractional Landau Inequalities k

(z i − x0i )2

i=1

1  (ν)



1 0

  (1 − t)ν−1 J02−ν f xi xi (x0 + t (z − x0 )) dt+

k(k−1) 2

2



⎧ ⎨l ⎩



1 0

  (z i − x0i ) z j − x0 j

= l (i, j) = 1 :i < j

1  (ν)

  (1 − t)ν−1 J02−ν f xi x j (x0 + t (z − x0 )) dt.

We choose z i − x0i = h > 0, for all i = 1, ..., k. Then k ∂f f (z) = f (x0 ) + h (x0 ) + ∂x i i=1  k  1   h2 (1 − t)ν−1 J02−ν f xi xi (x0 + t (z − x0 )) dt+  (ν) i=1 0

(14.18) ⎤



k(k−1) 2

2



⎧ ⎨l ⎩

= l (i, j) = 1 :i < j

Hence h

1

0

 2−ν

(1 − t)ν−1 J0

f xi x j

⎥ ⎥ ⎥ (x0 + t (z − x0 )) dt ⎥ ⎥. ⎥ ⎦ 

k ∂ f (x0 ) = f (z) − f (x0 ) − ∂xi i=1

 k  1   h2 (1 − t)ν−1 J02−ν f xi xi (x0 + t (z − x0 )) dt+  (ν) i=1 0

(14.19) ⎤

k(k−1) 2

2



⎧ ⎨l ⎩

Therefore

= l (i, j) = 1 :i < j

 0

1

 2−ν

(1 − t)ν−1 J0

f xi x j

⎥ ⎥ ⎥ (x0 + t (z − x0 )) dt ⎥ ⎥. ⎥ ⎦ 

14.2 Main Results

241

  k   ∂ f   h (x0 ) ≤ 2  f ∞,Rk+ +   ∂xi i=1

 k    h2  J 2−ν f x x (x0 + t (z − x0 ))  + i i 0 ∞,[0,1]  (ν + 1) i=1 ⎤ ⎥ ⎥ ⎥  2−ν     J0 f xi x j (x0 + t (z − x0 )) ∞,[0,1] ⎥ 2 ⎥≤ ⎧ ⎥ ⎨ l = l (i, j) = 1 ⎦ ⎩ :i < j k(k−1) 2



2  f ∞,Rk+ +

⎧ k ⎨

2

Kν h  (ν + 1) ⎩ i=1

2  f ∞,Rk+ + Thus

⎫ ⎬ 1+2 1 = ⎭ k(k−1) 2

l=1

k2 Kν h2 .  (ν + 1)

  k  2f k  ∂ f k2 Kν ∞,R+   + h, (x0 ) ≤    ∂xi h  (ν + 1) i=1

(14.20)

(14.21)

∀ h > 0, for any x0 ∈ Rk+ . Consequently we find that  k   ∂ f       ∂xi 

≤

∞,Rk+

i=1

∀ h > 0. Set A := 2  f ∞,Rk+ and B := We consider the function

k2 Kν , (ν+1)

y (h) = We have that

2  f ∞,Rk+ h

+

k2 Kν h,  (ν + 1)

both are positive.

A + Bh, ∀ h > 0. h

y  (h) = −Ah −2 + B = 0,

with only one critical number

(14.22)

(14.23)

242

14 Multidimensional Caputo Left Side Fractional Landau Inequalities

$ h0 =

A . B

(14.24)

Notice that y  (h) = 2 Ah −3 , and y  (h 0 ) = 2 A− 2 B 2 > 0. Therefore y has a global minimum which is 1

3

√ y (h 0 ) = 2 AB.   k  ∂ f       ∂xi 

Hence

i=1

≤√

∞,Rk+

√ 2 2k  f ∞,Rk+ K ν .  (ν + 1)

(14.25)

(14.26) 

The theorem is proved. We continue with

  Theorem 14.9 Let f ∈ C 3 Rk+ , k ≥ 2, with  f ∞,Rk+ < ∞, and 2 < ν < 3. Assume further that Mν :=             1   3−ν sup J0 f λ1 λk (x0 + t (z − x0 ))  < +∞.  x 1 ...x k    λ1 !...λk ! x 0 ,z∈Rk+ ,t∈[0,1]    λ1 + ... + λk = 3   λ ∈ Z+ , i = 1, ..., k  i

(14.27) Then we derive the best upper bounds: (1)   k  ∂ f       ∂xi  i=1

∞,Rk+

 13   23  2 1 12 3 3 3  f ∞,R ≤3 k Mν < +∞, + 2  (ν + 1)

(14.28)

and (2)       k(k−1)   k   2  2  1 ∂2 f 1 2 3 ∂2 f  9    f  3 k Mν3 < +∞. + ≤ 3   2 ∞,R+ 2  ∂x ∂x  + 1) (ν ∂x  i j i i=1   l = l (i, j) = 1,     i< j

(14.29)

14.2 Main Results

243

Proof We set 1 (x0 ) :=

k 

∂f ∂xi

i=1

and



2 (x0 ) :=

1 2

k 

i=1

(x0 ) ,

∂2 f ∂xi2

 (x0 ) + ⎧

k(k−1) 2



⎨l ⎩

= l (i, j) = 1, i< j

∂ 2 f (x0 ) ∂xi ∂x j

(14.30)

By Theorem 14.7 we have, we choose also z i − x0i = h > 0, all i = 1, ..., k : f (z) = f (x0 ) + h1 (x0 ) + h 2 2 (x0 ) + ⎫ ⎧ ⎡ 3 ⎤  1 k ⎬ ⎨ 1 ∂ f ⎦ (x0 + t (z − x0 )) dt (1 − t)ν−1 J03−ν ⎣ (z i − x0i ) ⎭ ⎩  (ν) 0 ∂xi i=1

= f (x0 ) + 1 (x0 ) h + 2 (x0 ) h 2 + h 3

6  (ν)

⎡

⎤

⎢   ⎥  1 ⎢ ⎥ 1 ⎢ ⎥ ν−1 3−ν f λ1 λk (x0 + t (z − x0 )) dt ⎥ . J0 (1 − t) ⎢ x 1 ...x k ⎢ ⎥ λ1 !...λk ! 0 ⎣ (λ1 + ... + λk = 3 ⎦ λi ∈ Z+ , i = 1, ..., k)

(14.31) We set R3 :=

6  (ν)

⎡

⎤

⎢   ⎥  1 ⎢ ⎥ 1 ⎢ ⎥ (1 − t)ν−1 J03−ν f λ1 λk (x0 + t (z − x0 )) dt ⎥ . ⎢ x ...x ⎢ ⎥ λ1 !...λk ! 0 1 k ⎣ (λ1 + ... + λk = 3 ⎦ λi ∈ Z+ , i = 1, ..., k)

(14.32) That is (h > 0) f (z) = f (x0 ) + 1 (x0 ) h + 2 (x0 ) h 2 + R3 h 3 .

(14.33)

1 (x0 ) h + 2 (x0 ) h 2 = f (z) − f (x0 ) − R3 h 3 =: A1 ,

(14.34)

So that

244

14 Multidimensional Caputo Left Side Fractional Landau Inequalities

and 21 (x0 ) h + 42 (x0 ) h 2 = f (z) − f (x0 ) − 8R3 h 3 =: A2 .

(14.35)

Solving the system (14.34), (14.35) we find 1 (x0 ) = and 2 (x0 ) =

4 A1 −A2 , 2h

(14.36)

A2 −2 A1 . 2h 2

We notice that   |4 A1 − A2 | = 3 ( f (z) − f (x0 )) + 4R3 h 3  ≤ 6  f ∞,Rk+ +           1 6   3−ν 3 4h J0 ( f λ1 λk (x0 + t (z − x0 )))  x 1 ...x k   (ν + 1)  λ1 !...λk !   (λ1 + ... + λk = 3     λ ∈ Z+ , i = 1, ..., k) i

≤ 6  f ∞,Rk+ + Hence |1 (x0 )| ≤

24 Mν h 3 .  (ν + 1)

3  f ∞,Rk+ h

∞,[0,1]

(14.37)

+

12Mν h2,  (ν + 1)

(14.38)

+

12Mν h2,  (ν + 1)

(14.39)

∀ h > 0, for all x0 ∈ Rk+ . Consequently it holds 1 ∞,Rk+ ≤

3  f ∞,Rk+ h

∀ h > 0. Next we have   A2 − 2 A1 = − ( f (z) − f (x0 )) + 6R3 h 3 .

(14.40)

So that   |A2 − 2 A1 | =  f (z) − f (x0 ) + 6R3 h 3  ≤ 2  f ∞,Rk+ + 6h 3 Hence |2 (x0 )| ≤

 f ∞,Rk+ h2

+

18Mν h,  (ν + 1)

6 Mν .  (ν + 1) (14.41) (14.42)

14.2 Main Results

245

∀ h > 0, ∀ x0 ∈ Rk+ . So that 2 ∞,Rk+ ≤ ∀ h > 0. Next we work on (14.39). Call μ := 3  f ∞,Rk+ , θ := We study the function

+

h2

12Mν (ν+1)

y (h) = We have



 f ∞,Rk+

 18Mν h,  (ν + 1)

(14.43)

that are both positive.

μ + θh 2 , ∀ h > 0. h

(14.44)

y  (h) = −μh −2 + 2θh = 0.

Hence

 μ  13 > 0. 2θ

(14.45)

y  (h 0 ) = 2μh −3 0 + 2θ > 0.

(14.46)

h 0 = h crit.no. = We have y  (h) = 2μh −3 + 2θ, and

Therefore y has a global minimum which is  −3  3 1 2 2 2 3 3 y (h 0 ) = μh −1 0 + θh 0 = h 0 μh 0 + θ = 2 θ μ . 23

(14.47)

 13   23  2 1 12 3 3 3  f ∞,R y (h 0 ) = 2 θ μ = 3 k Mν . + 2  (ν + 1) 23

(14.48)

Hence 3

1 3

2 3

We have proved that 1 ∞,Rk+ ≤ 3

 13   23  2 1 12 3 3 3  f ∞,R k Mν . + 2  (ν + 1)

Next we work on (14.43). 18Mν , that are both positive. We call ϕ :=  f ∞,Rk+ , ψ := (ν+1) We study ϕ y (h) := 2 + ψh, ∀ h > 0. h We have

(14.49)

(14.50)

246

14 Multidimensional Caputo Left Side Fractional Landau Inequalities

y  (h) = −2ϕh −3 + ψ = 0, so that

 h 0 := h crit.no. =

We have

2ϕ ψ

 13

> 0.

(14.51)

y  (h) = 6ϕh −4 ,

and

y  (h 0 ) = 6ϕh −4 0 > 0.

(14.52)

Therefore y has a global minimum which is    −3  ψ 3 1 2 1 ϕ + ψ = 23 ψ 3 ϕ3 . ϕh + ψh = h + ψ = h y (h 0 ) = ϕh −2 0 0 0 0 0 2ϕ 2 (14.53) That is 3√ 2 1 3 2ψ 3 ϕ 3 . (14.54) y (h 0 ) = 2 Consequently it holds  2 ∞,Rk+ ≤ 3

9  (ν + 1)

 23

1

2

3 3  f ∞,R k Mν . +

(14.55) 

The theorem is proved. We also present

  Theorem 14.10 Let f ∈ C 4 Rk+ , k ≥ 2, with  f ∞,Rk+ < ∞, and 3 < ν < 4. Assume further that Hν :=             1   4−ν J f sup + t − x (x )) (z λk   < +∞. λ1 0 0 0 x1 ...xk   λ !...λ ! k   1 k x0 ,z∈R+ ,t∈[0,1]    λ1 + ... + λk = 4   λ ∈ Z+ , i = 1, ..., k  i

(14.56) Then we derive the best upper bounds: (1)   k  ∂ f       ∂xi  i=1

∞,Rk+



11 ≤4 9

 43 

144  (ν + 1)

 41

3

1

4 4  f ∞,R k Hν < +∞, +

(14.57)

14.2 Main Results

(2)

247

         k(k−1)    k 2 2  1 ∂2 f ∂ f    + 2 ∂xi ∂x j  ⎧ ∂xi2   i=1 ⎨ l = l (i, j) = 1,     ⎩   i< j

≤

∞,Rk+

 2 and (3)

528  f ∞,Rk+ Hν < +∞,  (ν + 1)

(14.58)

            1  λk  λ1 f  ⎧ x ...x k ⎫ λ1 !...λk ! 1  ⎨   λ1 + ... + λk = 3 ⎬  ⎩   λi ∈ Z+ , i = 1, ..., k ⎭

≤

∞,Rk+

4 3



144  (ν + 1)

 43

1

3

4 4  f ∞,R k Hν < +∞. +

(14.59)

Proof We set 1 (x0 ) :=

k  i=1



2 (x0 ) :=

1 2

∂f ∂xi k 

i=1

(x0 ) , ∂2 f ∂xi2

 (x0 ) + ⎧

⎨l ⎩

and 3 (x0 ) := ⎧



k(k−1) 2



= l (i, j) = 1, i< j

∂ 2 f (x0 ) , ∂xi ∂x j

1 f λ λ ⎫ λ1 !...λk ! x1 1 ...xk k ⎬

(14.60) (x0 ) .

λ1 + ... + λk = 3 ⎩ λ ∈ Z+ , i = 1, ..., k ⎭ i ⎨

By Theorem 14.7 we have, we choose also z i − x0i = h > 0, all i = 1, ..., k : f (z) = f (x0 ) + 1 (x0 ) h + 2 (x0 ) h 2 + 3 (x0 ) h 3 + R4 h 4 , where

(14.61)

248

14 Multidimensional Caputo Left Side Fractional Landau Inequalities

⎡ ⎢ ⎢   4! 1 ⎢ ⎢ R4 :=  (ν) ⎢ λ1 !...λk ! ⎢⎧ ⎣⎨ λ1 + ... + λk = 4, ⎩ λ ∈ Z+ , i = 1, ..., k i 

1

(1 − t)

0

ν−1

J04−ν







f x λ1 ...x λk (x0 + t (z − x0 )) dt . 1

(14.62)

k

Therefore, ∀ h > 0, we get 1 (x0 ) h + 2 (x0 ) h 2 + 3 (x0 ) h 3 = f (z) − f (x0 ) − R4 h 4 =: A1 , 1 (x0 ) 2h + 2 (x0 ) 4h 2 + 3 (x0 ) 8h 3 = f (z) − f (x0 ) − R4 16h 4 =: A2 , and 1 (x0 ) 3h + 2 (x0 ) 9h 2 + 3 (x0 ) 27h 3 = f (z) − f (x0 ) − R4 81h 4 =: A3 . (14.63) We solve the system of three equations (14.63) and we find: 1 (x0 ) = 2 (x0 ) = and 3 (x0 ) = We have also that |R4 | ≤

18A1 −9A2 +2 A3 , 6h 4 A2 −5A1 −A3 , 2 2h

(14.64)

3A1 −3A2 +A3 . 6h 3

24Hν .  (ν + 1)

(14.65)

It is 18A1 − 9A2 + 2 A3 = 11 ( f (z) − f (x0 )) − 36R4 h 4 .

(14.66)

Therefore it holds |1 (x0 )| ≤

11  f ∞,Rk+ 3h

+

144Hν 3 h ,  (ν + 1)

(14.67)

∀ h > 0, ∀ x0 ∈ Rk+ . We get that 1 ∞,Rk+ ≤

11  f ∞,Rk+ 3h

 +

 144Hν h3,  (ν + 1)

(14.68)

∀ h > 0. It is 4 A2 − 5A1 − A3 = −2 ( f (z) − f (x0 )) + 22R4 h 4 .

(14.69)

14.2 Main Results

249

 264Hν h2,  (ν + 1)

(14.70)

 264Hν h2,  (ν + 1)

(14.71)

3A1 − 3A2 + A3 = f (z) − f (x0 ) − 36R4 h 4 .

(14.72)

Hence |2 (x0 )| ≤



2  f ∞,Rk+

+

h2

∀ h > 0, ∀ x0 ∈ Rk+ . The last implies 2 ∞,Rk+ ≤

2  f ∞,Rk+ h2

 +

∀ h > 0. It is

Therefore |3 (x0 )| ≤

 f ∞,Rk+ 3h 3

 +

 144Hν h,  (ν + 1)

(14.73)

 144Hν h,  (ν + 1)

(14.74)

∀ h > 0, ∀ x0 ∈ Rk+ . The last implies 3 ∞,Rk+ ≤

 f ∞,Rk+ 3h 3

 +

∀ h > 0. We will work on (14.68).   11 f ∞,Rk 144Hν + Call μ := , both are positive. , θ := 3 (ν+1) We study μ y (h) = + θh 3 , ∀ h > 0. h We have

(14.75)

y  (h) = −μh −2 + 3θh 2 = 0,

with one critical number h crit.no. = h 0 = We have that

 μ  14 > 0. 3θ

y  (h 0 ) = 2μh −3 0 + 6θh 0 > 0,

implying that y has a global minimum. We see that

(14.76)

(14.77)

250

14 Multidimensional Caputo Left Side Fractional Landau Inequalities

 −4  4 1 3 3 3 4 4 y (h 0 ) = μh −1 0 + θh 0 = h 0 μh 0 + θ = 3 θ μ . 34 I.e. y (h 0 ) =

4 3

3 4

1

3

θ4 μ4 ,

(14.78)

(14.79)

is the global minimum. Consequently  1 ∞,Rk+ ≤ 4

11 9

 43 

144  (ν + 1)

 14

3

1

4 4  f ∞,R k Hν . +

Next we work on (14.71). 264Hν > 0. Call ϕ := 2  f ∞,Rk+ , ψ := (ν+1) We study ϕ y (h) := 2 + ψh 2 , ∀ h > 0. h We have

y  (h) = −2ϕh −3 + 2ψh = 0,

with h crit.no. = h 0 = Notice that

(14.80)

(14.81)

(14.82)

  41 ϕ > 0. ψ

y  (h 0 ) = 6ϕh −4 0 + 2ψ > 0.

(14.83)

Thus we have that y has global minimum which is   −4 1 1 2 2 2 2 y (h 0 ) = ϕh −2 0 + ψh 0 = h 0 ϕh 0 + ψ = 2ψ ϕ , that is

1

1

y (h 0 ) = 2ψ 2 ϕ 2 .

(14.84)

Therefore we derive  2 ∞,Rk+ ≤ 2

528  f ∞,Rk+ Hν .  (ν + 1)

Finally we work on (14.74). Let A := We study

 f ∞,Rk

+

3

, B :=

144Hν , (ν+1)

both are positive.

(14.85)

14.2 Main Results

251

y (h) := Then

A + Bh, ∀ h > 0. h3

(14.86)

y  (h) = −3Ah −4 + B = 0,

so that

 h crit.no. = h 0 =

Here

3A B

 14

> 0.

(14.87)

y  (h 0 ) = 12 Ah −5 0 > 0.

(14.88)

Hence y has global minimum which is   −4 4 1 3 4 4 y (h 0 ) = Ah −3 0 + Bh 0 = h 0 Ah 0 + B = 3 A B . 34 That is y (h 0 ) =

4 3

3 4

1

3

A4 B 4,

(14.89)

(14.90)

the global minimum of y. Consequently it holds 3 ∞,Rk+ The theorem is proved.

4 ≤ 3



144  (ν + 1)

 34

1

3

4 4  f ∞,R k Hν . +

(14.91) 

References 1. Anastassiou, G.A.: Fractional Differentiation Inequalities. Springer, Heidelberg (2009) 2. Anastassiou, G.A.: Multivariate Caputo left fractional Landau inequalities. Moroc. J. Pure Appl. Anal. 6(2), 266–280 (2020) 3. Diethelm, K.: The Analysis of Fractional Differentiation Equations. Springer, Heidelberg (2010) 4. Ditzian, Z.: Multivariate Landau-Kolmogorov-type inequality. Math. Proc. Camb. Philos. Soc. 105(2), 335–350 (1989) 5. Kounchev, O.: Extremizers for the multivariate Landau-Kolmogorov inequality. In: Multivariate Approximation (Witten-Bommerholz, 1996), pp. 123–132. Math. Res. 101. Akademie Verlag, Berlin (1997) 6. Landau, E.: Einige Ungleichungen für zweimal differentzierban funktionen. Proc. Lond. Math. Soc. 13, 43–49 (1913)

Chapter 15

Multidimensional Left Canavati Fractional Landau Inequalities

Based on author’s first ever found multivariate fractional Taylor’s formula ([2], Chap. 12), we develop and prove several multivariate left side Canavati fractional uniform Landau type inequalities. It follows [4].

15.1 Introduction We are motivated by the following two results. First from [6], we have the following nice result: 2 d   ∂ If  f = f is the Laplacian of a function defined on Rd , then for 0 < ∂xi k < 2n,

i=1

  ∂ ∂   ∂ξ ... ∂ξ 1 k

  k  k 1−( 2n ) n f  2n , f  ≤ C (n, k)  f 

(15.1)

  where · denotes the L ∞ Rd norm. n f needs to be defined only in the distribup tional sense and the result is valid in many other  spaces such as L spaces.  Banach k k−1 . C (n, k) is an existential constant and  =   From [7], we have the following interesting result which is the sharp inequality   f ∞ ≤ 2

  d  f ∞ 2 f ∞ , (d + 2)

(15.2)

in Rd , d ≥ 2, where  f is the Laplacian and 2 =  ().

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_15

253

254

15 Multidimensional Left Canavati Fractional Landau Inequalities

This result is an analogue of the famous Landau inequality [8] in R:   f 

∞

≤

√

2  f ∞  f  ∞ ,

(15.3)

where ·∞ means the L ∞ norm on Rd (respectively on R). In this chapter we present left Canavati [5] fractional multivariate Landau type inequalities where all the constants are exactly calculated and we have best upper bounds. We need to mention the following background coming from [2], Chap. 12, pp. 257–268. We make Remark 15.1 We follow Anastassiou [3, p. 540] (see also Canavati [5] and Anastassiou [1]). Let [a, b] ⊆ R. Let x, x0 ∈ [a, b] such that x ≥ x0 , x0 is fixed. Let f ∈ C ([a, b]) and define  x  x0  1 Jν f (x) = (15.4) (x − t)ν−1 f (t) dt, x0 ≤ x ≤ b,  (ν) x0 ν > 0, the left generalized Riemann–Liouville integral, where  is the gamma function. We consider the subspace C xν0 ([a, b]) of C n ([a, b]), n := [ν] (the integral part), α := ν − n (0 < α < 1):

x0 f (n) ∈ C 1 ([x0 , b]) . C xν0 ([a, b]) := f ∈ C n ([a, b]) : J1−α

(15.5)

Hence, let f ∈ C xν0 ([a, b]) we define the left generalized ν-fractional derivative of f over [x0 , b] (see also Canavati [5] and Anastassiou [1]) as  x0 (n)  f . Dxν0 f := J1−α

(15.6)

Notice that 

 x0 J1−α f (n) (x) =

1  (1 − α)



x

(x − t)−α f (n) (t) dt

(15.7)

x0

exists for f ∈ C xν0 ([a, b]) . Let f x0 (t) := f (x0 + t), 0 ≤ t ≤ b − x0 , x ≥ x0 . By change of variable we obtain    ν  (15.8) D0 f x0 (x − x0 ) = Dxν0 f (x) . When ν ∈ N then the fractional derivative collapses to the usual one. We mention the left fractional Taylor formula. See Anastassiou [3, p. 540], Canavati [5], and Anastassiou [2]. Theorem 15.2 Let f ∈ C xν0 ([a, b]), x0 ∈ [a, b] fixed.

15.1 Introduction

255

(i) If ν ≥ 1, then it holds f (x) = f (x0 ) + f  (x0 ) (x − x0 ) + f  (x0 )

(x − x0 )2 (x − x0 )n−1 + ... + f (n−1) (x0 ) 2 (n − 1)!

  + Jνx0 Dxν0 f (x) , all x ∈ [a, b] : x ≥ x0 .

(15.9)

(ii) If 0 < ν < 1 we have   f (x) = Jνx0 Dxν0 f (x) , all x ∈ [a, b] : x ≥ x0 .

(15.10)

Using Theorem 15.2 we established the basic multivariate Canavati left fractional Taylor formula. Theorem 15.3 ([2]) Let f ∈ C 1 (Q), Q compact and convex ⊆ Rk , k ≥ 2. For fixed x0 , z ∈ Q, assume that as a function of t : f xi (x0 + t (z − x0 )) ∈ C ν−1 ([0, 1]), 1 ≤ ν < 2, all i = 1, ..., k. Then (i) f (z 1 , ..., z k ) = f (x01 , ..., x0k ) +  k   (ν−1) (z i − x0i ) 1 dt. (1 − t)ν−1 f xi (x0 + t (z − x0 ))  (ν) 0 i=1

(15.11)

(ii) Given f (x0 ) = 0, then  k   (ν−1) (z i − x0i ) 1 dt. f (z) = (1 − t)ν−1 f xi (x0 + t (z − x0 ))  (ν) 0 i=1

(15.12)

We mention the next left fractional Taylor result. Theorem 15.4 ([2]) Let f ∈ C 2 (Q), Q compact and convex ⊆ R2 . For fixed x0 , z ∈ Q, assume that as functions of t : f x1 x1 (x0 + t (z − x0 )) , f x1 x2 (x0 + t (z − x0 )) , f x2 x2 (x0 + t (z − x0 )) ∈ C (ν−2) ([0, 1]), where 2 ≤ ν < 3. Then (i) f (z 1 , z 2 ) = f (x01 , x02 ) + (z 1 − x01 ) (z 1 − x01 )2

1  (ν)



1

∂f ∂f (x0 ) + (z 2 − x02 ) (x0 ) + ∂x1 ∂x2

 (ν−2) dt+ (1 − t)ν−1 f x1 x1 (x0 + t (z − x0 ))

0

1 2 (z 1 − x01 ) (z 2 − x02 )  (ν)

 0

1

 (ν−2) dt+ (1 − t)ν−1 f x1 x2 (x0 + t (z − x0 ))

256

15 Multidimensional Left Canavati Fractional Landau Inequalities



1 (z 2 − x02 )  (ν)

1

2

(ii) When f (x0 ) =

 (ν−2) dt. (1 − t)ν−1 f x2 x2 (x0 + t (z − x0 ))

∂f ∂x1

(x0 ) =

1 f (z 1 , z 2 ) = (z 1 − x01 )  (ν)

∂f ∂x2



(x0 ) = 0, then

1

 (ν−2) dt (1 − t)ν−1 f x1 x1 (x0 + t (z − x0 ))

1

 (ν−2) dt (1 − t)ν−1 f x1 x2 (x0 + t (z − x0 ))

2

+2 (z 1 − x01 ) (z 2 − x02 ) + (z 2 − x02 )2

(15.13)

0

1  (ν)



1  (ν) 1

0

 0

 (ν−2) dt. (1 − t)ν−1 f x2 x2 (x0 + t (z − x0 ))

(15.14)

0

We also mention the following general multivariate Canavati left fractional Taylor formula. Theorem 15.5 ([2]) Let f ∈ C n (Q), Q compact and convex ⊆ Rk , k ≥ 2; here ν ≥ 1 such that n = [ν]. For fixed x0 , z ∈ Q assume that as functions of t : f α (x0 + t (z − x0 )) ∈ C (ν−n) ([0, 1]), for all α := (α1 , ..., αk ), αi ∈ Z+ , i = 1, ..., k; k  |α| := αi = n. Then i=1

(i) f (z 1 , ..., z k ) = f (x01 , ..., x0k ) +

k 

(z i − x0i )

i=1



k 

i=1

∂f (x01 , ..., x0k ) + ∂xi

2  f (x01 , ..., x0k ) (z i − x0i ) ∂x∂ i + ...+

2 

k 

(z i − x0i )

i=1

∂ ∂xi

n−1  f (x01 , ..., x0k ) +

(n − 1)! 1  (ν)



1

(1 − t)ν−1

0

⎧ k ⎨  ⎩

∂ (z i − x0i ) ∂xi i=1

n (ν−n) f

+

(x0 + t (z − x0 )) dt. ⎭

(ii) If all f α (x0 ) = 0, α := (α1 , ..., αk ), αi ∈ Z , i = 1, ..., k, |α| := l = 0, ..., n − 1, then

⎫ ⎬

k  i=1

f (z 1 , ..., z k ) =

(15.15) αi = l,

15.1 Introduction

1  (ν)



1

257

(1 − t)ν−1

0

⎧ k ⎨ 

∂ (z i − x0i ) ∂xi i=1

⎩

n (ν−n)

⎫ ⎬

(x0 + t (z − x0 )) dt. ⎭

f

(15.16) (Note that left fractional differentiation is a linear operation).

15.2 Main Results Here we present several left Canavati fractional multivariate Landau inequalities.   Theorem 15.6 Let f ∈ C 2 Rk+ , k ≥ 2, with  f ∞,Rk+ < ∞, and 2 < ν < 3. For any x0 , z ∈ Rk+ , assume that as function of t : f α (x0 + t (z − x0 )) ∈ C (ν−2) ([0, 1]), k  αi = 2. Assume further for all α := (α1 , ..., αk ), αi ∈ Z+ , i = 1, ..., k; |α| := that K ν := max

i=1

⎧ ⎪ ⎪ ⎨

 (ν−2) sup | f xi xi (x0 + t (z − x0 )) |, ⎪ k ⎪ ⎩x0 ,z∈R+ ,t∈[0,1], i=1,...,k

   f xi x j

sup ⎧ ⎨l ⎩

x0 ,z∈Rk+ ,t∈[0,1],

= l (i, j) = 1, ..., k(k−1) ; 2 i< j

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  (ν−2) ⎬ (x0 + t (z − x0 ))  < +∞. ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

(15.17)

Then we derive the best upper bound  k   ∂ f       ∂xi  i=1

That is

 1 1 2 2 2  f ∞,R k Kν . +  (ν + 1)

≤ 2k

∞,Rk+

 k   ∂ f       ∂xi 

< ∞.

∞,Rk+

i=1

Proof By Theorems 15.4 and 15.5 we have for any x0 , z ∈ Rk+ that f (z) = f (x0 ) +

k  i=1

(z i − x0i )

∂f (x0 ) + ∂xi

(15.18)

258

15 Multidimensional Left Canavati Fractional Landau Inequalities

1  (ν)



⎧⎡ ⎫  k 2 ⎤(ν−2) ⎪ ⎪ ⎨  ⎬ ∂ ν−1 ⎣ ⎦ f (1 − t) (z i − x0i ) (x0 + t (z − x0 )) dt = ⎪ ⎪ ∂xi ⎩ ⎭ i=1

1 0

f (x0 ) +

k 

(z i − x0i )

i=1 k 

(z i − x0i )2

i=1 k(k−1) 2

2



!

1  (ν)



1

∂f (x0 ) + ∂xi

(15.19)

 (ν−2) dt+ (1 − t)ν−1 f xi xi (x0 + t (z − x0 ))

0

  (z i − x0i ) z j − x0 j

l = l (i, j) = 1 :i < j

1  (ν)

 0

1

 (ν−2) dt. (1 − t)ν−1 f xi x j (x0 + t (z − x0 ))

We choose z i − x0i = h > 0, for all i = 1, ..., k. Then k  ∂f f (z) = f (x0 ) + h (x0 ) + ∂xi i=1

(15.20)

 k   1  (ν−2) h2 dt+ (1 − t)ν−1 f xi xi (x0 + t (z − x0 ))  (ν) i=1 0 ⎤ 

k(k−1) 2

2



⎧ ⎨l ⎩

Hence

= l (i, j) = 1 :i < j

0

1

⎥ ⎥ ⎥   (ν−2) dt ⎥ (1 − t)ν−1 f xi x j (x0 + t (z − x0 )) ⎥. ⎥ ⎦

k  ∂f h (x0 ) = f (z) − f (x0 ) − ∂x i i=1

 k   1  (ν−2) h2 dt+ (1 − t)ν−1 f xi xi (x0 + t (z − x0 ))  (ν) i=1 0

(15.21)

15.2 Main Results

259

⎤ 

k(k−1) 2

2



⎧ ⎨l ⎩

= l (i, j) = 1 :i < j

1

0

⎥ ⎥ ⎥   (ν−2) dt ⎥ (1 − t)ν−1 f xi x j (x0 + t (z − x0 )) ⎥. ⎥ ⎦

 k   ∂ f    h (x0 ) ≤ 2  f ∞,Rk+ +   ∂xi

Therefore

i=1

! k  (ν−2)  h2   +  f xi xi (x0 + t (z − x0 ))  ∞,[0,1]  (ν + 1) i=1 ⎤

2



⎧ ⎨l ⎩

⎥ ⎥ ⎥ ⎥≤ ∞,[0,1] ⎥ ⎥ ⎦

 (ν−2)     f xi x j (x0 + t (z − x0 )) 

k(k−1) 2

= l (i, j) = 1 :i < j

2  f ∞,Rk+

⎫ ⎧ k(k−1) k 2 ⎬  K ν h 2 ⎨ + 1+2 1 = ⎭  (ν + 1) ⎩ i=1 l=1 2  f ∞,Rk+ +

Thus

k2 Kν h2 .  (ν + 1)

  k  2f k  ∂ f k2 Kν ∞,R+   + h, (x0 ) ≤    ∂xi h  (ν + 1) i=1

(15.22)

(15.23)

∀ h > 0, for any x0 ∈ Rk+ . Consequently we find that  k   ∂ f       ∂xi  i=1

≤

2  f ∞,Rk+

∞,Rk+

∀ h > 0. Set A := 2  f ∞,Rk+ and B := We consider the function

k2 Kν , (ν+1)

h

+

k2 Kν h,  (ν + 1)

both are positive.

(15.24)

260

15 Multidimensional Left Canavati Fractional Landau Inequalities

y (h) = We have that

A + Bh = Ah −1 + Bh, ∀ h > 0. h

(15.25)

y  (h) = −Ah −2 + B = 0,

with only one critical number

# h0 =

A . B

(15.26)

Notice that y  (h) = 2 Ah −3 , and y  (h 0 ) = 2 A− 2 B 2 > 0. Therefore y has a global minimum which is 1

3

#  √ A y (h 0 ) = y = 2 AB. B Hence  k   ∂ f       ∂xi  i=1

 ≤ 2 2  f ∞,Rk+

∞,Rk+

(15.27)

√ 2 2k k2 Kν  f ∞,Rk+ K ν . =√  (ν + 1)  (ν + 1) (15.28) 

The theorem is proved. We continue with

  Theorem 15.7 Let f ∈ C 3 Rk+ , k ≥ 2, with  f ∞,Rk+ < ∞, and 3 < ν < 4. For any x0 , z ∈ Rk+ , assume that as functions of t : f α (x0 + t (z − x0 )) ∈ C (ν−3) ([0, 1]), k  αi = 3. Assume further for all α := (α1 , ..., αk ), αi ∈ Z+ , i = 1, ..., k; |α| := i=1

that Mν :=         (ν−3)    1   sup f λ1 λk (x0 + t (z − x0 ))   < +∞. x 1 ...x k   $ λ1 !...λk ! x 0 ,z∈Rk+ ,t∈[0,1] !  λ + ... + λ = 3 1 k    λ ∈ Z+ , i = 1, ..., k  i

(15.29) Then we derive the best upper bounds: (1)   k  ∂ f       ∂xi  i=1

∞,Rk+

 13   23  2 1 12 3 3 3  f ∞,R ≤3 k Mν < +∞, + 2  (ν + 1)

(15.30)

15.2 Main Results

261

and (2)       k(k−1)   k   2  2  1  ∂2 f 2  1 2 3 ∂ f  9    f  3 k Mν3 < +∞. + ≤ 3   2 ∞,R 2  ∂xi ∂x j  (ν + 1) + ∂xi ! i=1   l = l (i, j) = 1,     i< j

(15.31) Proof We set 1 (x0 ) :=

k 

∂f ∂xi

i=1

and



2 (x0 ) :=

1 2

k 

i=1

(x0 ) ,

∂2 f ∂xi2

 (x0 ) + ⎧

k(k−1) 2



⎨l ⎩

= l (i, j) = 1, i< j

∂ 2 f (x0 ) ∂xi ∂x j

(15.32)

By Theorem 15.5 we have, we choose also z i − x0i = h > 0, all i = 1, ..., k : f (z) = f (x0 ) + h1 (x0 ) + h 2 2 (x0 ) + ⎧⎡ ⎫  k 3 ⎤(ν−3) ⎪ ⎪  1 ⎨ ⎬  1 ∂ ν−1 ⎣ ⎦ f (1 − t) (z i − x0i ) (x0 + t (z − x0 )) dt ⎪ ⎪  (ν) 0 ∂xi ⎩ ⎭ i=1 = f (x0 ) + 1 (x0 ) h + 2 (x0 ) h 2 + h 3 ⎡

6  (ν) ⎤

⎢  (ν−3) ⎥  1 ⎢ ⎥  1 ⎢ ⎥ ν−1 dt ⎥ . f λ1 λk (x0 + t (z − x0 )) (1 − t) ⎢ x 1 ...x k ⎢ ⎥ λ1 !...λk ! 0 ⎣ (λ1 + ... + λk = 3 ⎦ λi ∈ Z+ , i = 1, ..., k)

(15.33) We set R3 :=

6  (ν)

262

15 Multidimensional Left Canavati Fractional Landau Inequalities ⎡

⎤

⎢  (ν−3) ⎥  1 ⎢ ⎥  1 ⎢ ⎥ dt ⎥ . (1 − t)ν−1 f λ1 λk (x0 + t (z − x0 )) ⎢ x ...x ⎢ ⎥ λ1 !...λk ! 0 1 k ⎣ (λ1 + ... + λk = 3 ⎦ + λi ∈ Z , i = 1, ..., k)

(15.34) That is (h > 0) f (z) = f (x0 ) + 1 (x0 ) h + 2 (x0 ) h 2 + R3 h 3 .

(15.35)

1 (x0 ) h + 2 (x0 ) h 2 = f (z) − f (x0 ) − R3 h 3 =: A1 ,

(15.36)

21 (x0 ) h + 42 (x0 ) h 2 = f (z) − f (x0 ) − 8R3 h 3 =: A2 .

(15.37)

So that

and

Solving the system (15.36), (15.37) we find 1 (x0 ) = and 2 (x0 ) =

4 A1 −A2 , 2h A2 −2 A1 . 2h 2

(15.38)

We notice that   |4 A1 − A2 | = 4 ( f (z) − f (x0 )) − 4R3 h 3 − ( f (z) − f (x0 )) + 8R3 h 3  =   3 ( f (z) − f (x0 )) + 4R3 h 3  ≤ 6  f ∞,Rk + +            1 6  3 (ν−3)  4h f λ1 λk (x0 + t (z − x0 ))   x ...x    (ν + 1) λ1 !...λk ! 1 k   (λ1 + ... + λk = 3     λ ∈ Z+ , i = 1, ..., k) i

≤ 6  f ∞,Rk+ + Hence |1 (x0 )| ≤ ∀ h > 0, for all x0 ∈ Rk+ .

24 Mν h 3 .  (ν + 1)

3  f ∞,Rk+ h

+

12Mν h2,  (ν + 1)

∞,[0,1]

(15.39)

(15.40)

15.2 Main Results

263

Consequently it holds 3  f ∞,Rk+

1 ∞,Rk+ ≤

+

h

12Mν h2,  (ν + 1)

(15.41)

∀ h > 0. Next we have A2 − 2 A1 = f (z) − f (x0 ) − 8R3 h 3 − 2 ( f (z) − f (x0 )) + 2R3 h 3

(15.42)

  = − ( f (z) − f (x0 )) + 6R3 h 3 . So that   |A2 − 2 A1 | =  f (z) − f (x0 ) + 6R3 h 3  ≤ 2  f ∞,Rk+ + 6h 3 Hence |2 (x0 )| ≤ ∀ h > 0, ∀ x0 ∈ Rk+ . So that 2 ∞,Rk+ ≤ ∀ h > 0. Next we work on (15.41). Call μ := 3  f ∞,Rk+ , θ := We study the function

Hence

h2

 f ∞,Rk+ h2

12Mν (ν+1)

y (h) = We have

 f ∞,Rk+

+

18Mν h,  (ν + 1)

(15.44)

 18Mν h,  (ν + 1)

(15.45)

 +

6 Mν .  (ν + 1) (15.43)

that are both positive.

μ + θh 2 = μh −1 + θh 2 . h

(15.46)

y  (h) = −μh −2 + 2θh = 0. 2θh = μh −2 =

μ , h2

2θh 3 = μ, h 0 = h crit.no. =

 μ  13 > 0. 2θ

(15.47)

264

15 Multidimensional Left Canavati Fractional Landau Inequalities

We have y  (h) = 2μh −3 + 2θ, and y  (h 0 ) = 2μh −3 0 + 2θ > 0.

(15.48)

Therefore y has a global minimum which is y (h 0 ) =

μh −1 0

+

θh 20

=

h 20

   −3  2θ 2 μh 0 + θ = h 0 μ + θ μ

= 3θh 20 = 3θ

(15.49)

 μ  23 3 1 2 = 2 θ3 μ3 . 2θ 23

Hence   23   13 2 1 3 12 3 3  f ∞,R y (h 0 ) = 2 θ μ = 3 k Mν . + 2  (ν + 1) 23 3

1 3

2 3

(15.50)

We have proved that 1 ∞,Rk+ ≤ 3

 13   23  2 1 12 3 3 3  f ∞,R k Mν . + 2  (ν + 1)

Next we work on (15.45). 18Mν , that are both positive. We call ϕ :=  f ∞,Rk+ , ψ := (ν+1) We study ϕ y (h) := 2 + ψh = ϕh −2 + ψh. h We have

(15.51)

(15.52)

y  (h) = −2ϕh −3 + ψ = 0,

and ψ = 2ϕh −3 =

2ϕ , h3

ψh 3 = 2ϕ, so that

 h 0 := h crit.no. =

We have

and

2ϕ ψ

 13

y  (h) = 6ϕh −4 ,

> 0.

(15.53)

15.2 Main Results

265

y  (h 0 ) = 6ϕh −4 0 > 0.

(15.54)

Therefore y has a global minimum which is y (h 0 ) =

ϕh −2 0

   −3  ψ +ψ + ψh 0 = h 0 ϕh 0 + ψ = h 0 ϕ 2ϕ

  13 2ϕ 3 3 1 2 1 3 = 23 ψ 3 ϕ3 . = ψh 0 = ψ 2 2 ψ 2 That is y (h 0 ) =

(15.55)

3√ 2 1 3 2ψ 3 ϕ 3 . 2

(15.56)

Consequently it holds √   23 1 2 18 332 3  f ∞,R Mν3 = k + 2  (ν + 1)

2 ∞,Rk+ ≤  3

9  (ν + 1)

 23

1

2

3 3  f ∞,R k Mν .

(15.57)

+



The theorem is proved. We also present

  Theorem 15.8 Let f ∈ C 4 Rk+ , k ≥ 2, with  f ∞,Rk+ < ∞, and 4 < ν < 5. For any x0 , z ∈ Rk+ , assume that as functions of t : f α (x0 + t (z − x0 )) ∈ C (ν−4) ([0, 1]), k  αi = 4. Assume further for all α := (α1 , ..., αk ), αi ∈ Z+ , i = 1, ..., k; |α| := i=1

that Hν :=           (ν−4)    1   f λ1 λk (x0 + t (z − x0 )) sup   < +∞. x 1 ...x k   $ λ1 !...λk ! x 0 ,z∈Rk+ ,t∈[0,1] !   λ1 + ... + λk = 4   λ ∈ Z+ , i = 1, ..., k  i

(15.58) Then we derive the best upper bounds: (1)   k  ∂ f       ∂xi  i=1

∞,Rk+

 ≤4

11 9

 43 

144  (ν + 1)

 41

3

1

4 4  f ∞,R k Hν < +∞, +

(15.59)

266

15 Multidimensional Left Canavati Fractional Landau Inequalities

(2)

         k(k−1)    k 2 2   1  ∂2 f ∂ f    + 2 ∂xi ∂x j  ⎧ ∂xi2   i=1 ⎨ l = l (i, j) = 1,     ⎩   i< j

≤

∞,Rk+

 2 and (3)

528  f ∞,Rk+ Hν < +∞,  (ν + 1)

(15.60)

             1  λk  λ1 f  ⎧ x ...x k ⎫ λ1 !...λk ! 1  ⎨   λ1 + ... + λk = 3 ⎬  ⎩   λi ∈ Z+ , i = 1, ..., k ⎭

≤

∞,Rk+

4 3



144  (ν + 1)

 43

1

3

4 4  f ∞,R k Hν < +∞. +

(15.61)

Proof We set 1 (x0 ) :=

k  i=1



2 (x0 ) :=

1 2

∂f ∂xi k 

i=1

(x0 ) , ∂2 f ∂xi2

 (x0 ) + ⎧

⎨l ⎩

and 3 (x0 ) := ⎧



k(k−1) 2



= l (i, j) = 1, i< j

∂ 2 f (x0 ) , ∂xi ∂x j

1 f λ λ ⎫ λ1 !...λk ! x1 1 ...xk k ⎬

(15.62) (x0 ) .

λ1 + ... + λk = 3 ⎩ λ ∈ Z+ , i = 1, ..., k ⎭ i ⎨

By Theorem 15.5 we have, we choose also z i − x0i = h > 0, all i = 1, ..., k : f (z) = f (x0 ) + 1 (x0 ) h + 2 (x0 ) h 2 + 3 (x0 ) h 3 + R4 h 4 , where

(15.63)

15.2 Main Results

267

⎡ ⎢ ⎢    4! 1 ⎢ ⎢ R4 :=  (ν) ⎢ λ1 !...λk ! ⎢⎧ ⎣⎨ λ1 + ... + λk = 4, ⎩ λ ∈ Z+ , i = 1, ..., k i 

1

(1 − t)

ν−1



f x λ1 ...x λk (x0 + t (z − x0 )) 1

0

(ν−4)

k

& dt .

(15.64)

Therefore, ∀ h > 0, we get 1 (x0 ) h + 2 (x0 ) h 2 + 3 (x0 ) h 3 = f (z) − f (x0 ) − R4 h 4 =: A1 , 1 (x0 ) 2h + 2 (x0 ) 4h 2 + 3 (x0 ) 8h 3 = f (z) − f (x0 ) − R4 16h 4 =: A2 , and 1 (x0 ) 3h + 2 (x0 ) 9h 2 + 3 (x0 ) 27h 3 = f (z) − f (x0 ) − R4 81h 4 =: A3 . (15.65) We solve the system of three equations (15.65) and we find: 1 (x0 ) =

18A1 −9A2 +2 A3 , 6h

2 (x0 ) = and 3 (x0 ) =

4 A2 −5A1 −A3 , 2h 2

We have also that |R4 | ≤

(15.66)

3A1 −3A2 +A3 . 6h 3

24Hν .  (ν + 1)

(15.67)

We calculate 18A1 − 9A2 + 2 A3 = 18 ( f (z) − f (x0 )) − 18R4 h 4 − 9 ( f (z) − f (x0 )) + R4 144h 4

+ 2 ( f (z) − f (x0 )) − R4 162h 4 = 11 ( f (z) − f (x0 )) − 36R4 h 4 .

(15.68)

Therefore it holds  |1 (x0 )| ≤ = ∀ h > 0, ∀ x0 ∈ Rk+ . We get that

22  f ∞,Rk+ 6h

11  f ∞,Rk+ 3h

+

+

36

24Hν (ν+1)

6h

144Hν 3 h ,  (ν + 1)



h4 (15.69)

268

15 Multidimensional Left Canavati Fractional Landau Inequalities

11  f ∞,Rk+

1 ∞,Rk+ ≤

3h

 +

 144Hν h3,  (ν + 1)

(15.70)

∀ h > 0. Next we find 4 A2 − 5A1 − A3 = 4 ( f (z) − f (z 0 )) − R4 64h 4 − 5 ( f (z) − f (x0 )) + R4 5h 4 − ( f (z) − f (x0 )) + R4 81h 4 = −2 ( f (z) − f (x0 )) + 22R4 h 4 . 

Hence |2 (x0 )| ≤ =

4  f ∞,Rk+ 2h 2

2  f ∞,Rk+ h2

22

+

24Hν (ν+1)



(15.71)

h4

2h 2   264Hν h2, +  (ν + 1)

(15.72)

∀ h > 0, ∀ x0 ∈ Rk+ . The last implies 2 

∞,Rk+

≤

2  f ∞,Rk+ h2

 +

 264Hν h2,  (ν + 1)

(15.73)

∀ h > 0. We also calculate 3A1 − 3A2 + A3 = 3 ( f (z) − f (x0 )) − 3R4 h 4 − 3 ( f (z) − f (x0 )) + 48R4 h 4 + ( f (z) − f (x0 )) − 81R4 h 4 = ( f (z) − f (x0 )) − 36R4 h 4 . 

Therefore |3 (x0 )| ≤ =

2  f ∞,Rk+ 6h 3

 f ∞,Rk+ 3h 3

 +

+

36

24Hν (ν+1)

144Hν  (ν + 1)



(15.74)

h4

6h 3  h,

(15.75)

∀ h > 0, ∀ x0 ∈ Rk+ . The last implies 3 ∞,Rk+ ≤ ∀ h > 0.

 f ∞,Rk+ 3h 3

 +

 144Hν h,  (ν + 1)

(15.76)

15.2 Main Results

269

We will work on (15.70).   11 f ∞,Rk 144Hν + , both are positive. Call μ := , θ := 3 (ν+1) We study μ y (h) = + θh 3 = μh −1 + θh 3 . h We have

(15.77)

y  (h) = −μh −2 + 3θh 2 = 0,

implying

μ , h2

3θh 2 = and

3θh 4 = μ, with one critical number h crit.no. = h 0 = We have that

 μ  14 > 0. 3θ

(15.78)

y  (h 0 ) = 2μh −3 0 + 6θh 0 > 0,

(15.79)

implying that y has a global minimum. We see that    −4  3θ 3 3 3 μ + θ μh + θh = h + θ = h y (h 0 ) = μh −1 0 0 0 0 0 μ = 4θh 30 = 4θ

 μ  43 4 1 3 = 3 θ4 μ4 . 3θ 34

I.e. y (h 0 ) =

4 3

1

3 4

(15.80)

3

θ4 μ4 ,

(15.81)

the global minimum only. Consequently 1

1 ∞,Rk+ ≤  4

11 9

Next we work on (15.73).



1

4 (144) 4 Hν4 3 4

3 ( (ν + 1))

 34 

144  (ν + 1)

1 4

 41

11 3

 34

3

4  f ∞,R k = +

3

1

4 4  f ∞,R k Hν . +

(15.82)

270

15 Multidimensional Left Canavati Fractional Landau Inequalities

264Hν Call ϕ := 2  f ∞,Rk+ , ψ := (ν+1) > 0. We study ϕ y (h) := 2 + ψh 2 = ϕh −2 + ψh 2 , h > 0. h

We have

y  (h) = −2ϕh −3 + 2ψh = 0,

then

(15.83)

(15.84)

ψh = ϕh −3

and ψh =

ϕ , h3

so that ψh 4 = ϕ, with h crit.no. Notice that

  41 ϕ = h0 = > 0. ψ

y  (h 0 ) = 6ϕh −4 0 + 2ψ > 0.

(15.85)

Thus we have that y has global minimum which is     −4 ψ 2 2 2 ϕ + ψ + ψh = h + ψ = h ϕh y (h 0 ) = ϕh −2 0 0 0 0 0 ϕ # =

2ψh 20

= 2ψ

that is

ϕ 1 1 = 2ψ 2 ϕ 2 , ψ 1

1

y (h 0 ) = 2ψ 2 ϕ 2 .

(15.86)

Therefore we derive 2 ∞,Rk+

√



264 ≤2 2  (ν + 1)  2

Finally we work on (15.76).

 21  f ∞,Rk+ Hν =

528  f ∞,Rk+ Hν .  (ν + 1)

(15.87)

15.2 Main Results

Let A := We study

271

 f ∞,Rk

+

3

, B :=

144Hν , (ν+1)

y (h) := Then

both are positive.

A + Bh = Ah −3 + Bh. h3

(15.88)

y  (h) = −3Ah −4 + B = 0,

and B=

3A , h4

h4 =

3A , B

with

so that

 h crit.no. = h 0 =

Here

3A B

 14

> 0.

(15.89)

y  (h 0 ) = 12 Ah −5 0 > 0.

(15.90)

Hence y has global minimum which is y (h 0 ) =

Ah −3 0

+ Bh 0 = h 0 =



Ah −4 0





+ B = h0

B A +B 3A

4 4 1 3 Bh 0 = 3 A 4 B 4 . 3 34

That is y (h 0 ) =

4 3

3 4

1



(15.91)

3

A4 B 4,

(15.92)

the global minimum of y. Consequently it holds 3 ∞,Rk+

4 ≤ 3A B = 3 34

The theorem is proved.

4

1 4

3 4



144  (ν + 1)

 34

1

3

4 4  f ∞,R k Hν . +

(15.93) 

272

15 Multidimensional Left Canavati Fractional Landau Inequalities

References 1. Anastassiou, G.A.: Opial type inequalities involving fractional derivatives of functions. Nonlinear Stud. 6(2), 207–230 (1999) 2. Anastassiou, G.A.: Fractional Differentiation Inequalities. Springer, Heidelberg (2009) 3. Anastassiou, G.A.: Quantitative Approximations. Chapman & Hall/CRC, Boca Raron (2001) 4. Anastassiou, G.A.: Multivariate left side Canavati fractional Landau inequalities. Submitted (2020) 5. Canavati, J.A.: The Riemann-Liouville integral. Nieuw Archief Voor Wiskunde 5(1), 53–75 (1987) 6. Ditzian, Z.: Multivariate Landau-Kolmogorov-type inequality. Math. Proc. Camb. Philos. Soc. 105(2), 335–350 (1989) 7. Kounchev, O.: Extremizers for the multivariate Landau-Kolmogorov inequality. Multivariate Approximation (Witten-Bommerholz, 1996), pp. 123–132. Math. Res. 101. Akademie Verlag, Berlin (1997) 8. Landau, E.: Einige Ungleichungen für zweimal differentzierban funktionen. Proc. Lond. Math. Soc. 13, 43–49 (1913)

Chapter 16

Multidimensional Right Caputo Fractional Taylor Formula and Landau Inequalities

Here we present a multivariate right side Caputo fractional Taylor’s formula with fractional integral remainder. Based on this we give three multivariate right side Caputo fractional Landau’s type inequalities. Their constants are precisely calculated and we give best upper bounds. It follows [2].

16.1 Introduction We are inspired by the following two results. First from [3], we have the following nice result: 2 d   ∂ If  f = f is the Laplacian of a function defined on Rd , then for 0 < ∂xi k < 2n,

i=1

  ∂ ∂   ∂ξ ... ∂ξ 1 k

  k  k 1−( 2n ) n f  2n , f  ≤ C (n, k)  f 

(16.1)

  where · denotes the L ∞ Rd norm. n f needs to be defined only in the distribup tional sense and the result is valid in many other  spaces such as L spaces.  Banach k k−1 . C (n, k) is an existential constant and  =   From [4], we have the following interesting result which is the sharp inequality   f ∞ ≤ 2

  d  f ∞ 2 f ∞ , (d + 2)

(16.2)

in Rd , d ≥ 2, where  f is the Laplacian and 2 =  ().

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_16

273

274

16 Multidimensional Right Caputo Fractional Taylor Formula and Landau Inequalities

This result is an analogue of the famous Landau inequality [5] in R:   f 

∞

≤

√

2  f ∞  f  ∞ ,

(16.3)

where ·∞ means the L ∞ norm on Rd (respectively on R). In this chapter we first prove a multivariate right side Caputo fractional Taylor’s formula and based on it we prove three multivariate right side Caputo fractional Landau’s type inequalities where all the constants are precisely calculated and we obtain best upper bounds.

16.2 Background Here we follow [1], Chap. 23. We need Definition 16.1 Let f ∈ L 1 ([a, b]), α > 0. We define the right Riemann–Liouville fractional operator of order α by α f (x) = Ib−

1  (α)



b

(s − x)α−1 f (s) ds,

(16.4)

x

∀ x ∈ [a, b], where  is the gamma function. 0 We set Ib− := I (the identity operator). α α By [1], p. 334 we get that Ib− f (x) exists almost everywhere on [a, b] and Ib− f ∈ α L 1 ([a, b]). Also if α ≥ 1 then Ib− f ∈ C ([a, b]). We need

Definition 16.2 Let f ∈ AC m ([a, b]) (space of functions from [a, b] into R with (m − 1)-derivative absolutely continuous function on [a, b]), m ∈ N, where m =

α, α > 0 ( · the ceiling of the number). We define the right Caputo fractional derivative of order α > 0, by m−α (m) α Db− f (x) := (−1)m Ib− f (x) ,

that is α f (x) = Db−

(−1)m  (m − α)



b

(s − x)m−α−1 f (m) (s) ds.

(16.5)

(16.6)

x

α α f (x) exists a.e. on [a, b] and Db− f ∈ L 1 ([a, b]). By [1], p. 337 we get that Db− When α = m ∈ N, then m f (x) = (−1)m f (m) (x) , Db−

∀ x ∈ [a, b].

(16.7)

16.2 Background

275

We also need the right Caputo fractional Taylor’s formula with integral remainder. Theorem 16.3 ([1], p. 341) Let f ∈ AC m ([a, b]), x ∈ [a, b], α > 0, m = α. Then f (x) =

m−1 λ=0

f (λ) (b) 1 (x − b)λ + λ!  (α)



b x

α f (s) ds. (s − x)α−1 Db−

(16.8)

16.3 Main Results We make Remark 16.4 Let Q be a compact and convex subset of Rk , k ≥ 2; z := (z 1 , ..., z k ), x0 := (x01 , ..., x0k ) ∈ Q. Let f ∈ C n (Q), n ∈ N. Set (16.9) gz (t) := f (x0 + t (z − x0 )) , 0 ≤ t ≤ 1; gz (0) = f (x0 ) , gz (1) = f (z) . Then ⎡ j ⎤ k ∂ f ⎦ (x0 + t (z − x0 )) , gz( j) (t) = ⎣ (z i − x0i ) ∂x i i=1 and

⎡ j ⎤ k ∂ gz( j) (1) = ⎣ f ⎦ (z) , (z i − x0i ) ∂x i i=1

(16.10)

(16.11)

for j = 0, 1, ..., n. α If all f α (z) := ∂∂x αf (z) = 0, α := (α1 , ..., αk ), αi ∈ Z+ , i = 1, ..., k; |α| := k  αi = l, then gz(l) (1) = 0, where l ∈ {0, 1, ..., n} .

i=1

Clearly here gz ∈ C n ([0, 1]), in particular gz ∈ AC n ([0, 1]). We also have that ⎞ ⎛  k  ⎜ j! ⎟  ⎟ ⎜ gz( j) (t) = (z i − x0i )αi f α (x0 + t (z − x0 )) , ⎟ ⎜ k ⎠ ⎝ + i=1 α:=(α1 ,...,αk ),αi ∈Z αi ! k i=1,...,k;|α|:=



αi = j

i=1

i=1

(16.12)

276

16 Multidimensional Right Caputo Fractional Taylor Formula and Landau Inequalities

0 ≤ t ≤ 1, and ⎛

⎜ (z i − x0i ) ⎜ i=1 ⎜ k ⎝  αi !



( j)

gz (1) = j!

⎞

k 

α:=(α1 ,...,αk ),αi ∈Z+ k  i=1,...,k;|α|:= αi = j

αi

⎟ ⎟ ⎟ f α (z) , ⎠

(16.13)

i=1

i=1

for all j = 0, 1, ..., n. Let now ν > 0 with ν = n. By applying (16.8) to gz we find gz (0) =

n−1

( j)

gz (1) 1 + j!  (ν)

(−1) j

j=0

 0

1

ν s ν−1 D1− gz (s) ds.

(16.14)

ν gz as follows (0 ≤ s ≤ 1): Still we need to interpret D1− (16.6)

ν D1− gz (s) =

(−1)n  (n − ν)



1

s

(y − s)n−ν−1 gz(n) (y) dy = ⎞

⎛ α:=(α1 ,...,αk ),αi ∈Z+ k  i=1,...,k;|α|:= αi =n

(16.15)

 k  ⎜ n! ⎟  ⎟ ⎜ αi (z i − x0i ) ⎟ ⎜ k ⎠ ⎝ i=1 αi ! i=1

i=1

1  (n − ν)



1

(y − s)n−ν−1 f α (x0 + y (z − x0 )) dy.

s

Thus it holds (0 ≤ s ≤ 1) ⎛ ν D1− gz

(s) = (−1)

n

α:=(α1 ,...,αk ),αi ∈Z+ k  αi =n

i=1,...,k;|α|:=

⎞

 k  ⎜ n! ⎟  ⎜ ⎟ αi (z i − x0i ) ⎜ k ⎟ ⎝ ⎠ i=1 αi ! i=1

i=1



 n−ν I1− f α (x0 + t (z − x0 )) (s) .

(16.16)

Based on the above we have proved the following multivariate right side Caputo fractional Taylor’s formula:

16.3 Main Results

277

Theorem 16.5 Let ν > 0, n = ν, f ∈ C n (Q), where Q is a compact and convex subset of Rk , k ≥ 2; with x0 := (x01 , ..., x0k ) , z := (z 1 , ..., z k ) ∈ Q. Then (1) ⎛

⎛

⎜ ⎜ ⎜ f (x0 ) = (−1) j ⎜ ⎜ ⎜ α:=(α1 ,...,αk ),αi ∈Z+ j=0 ⎝ k 

⎞

⎟ αi ⎟ ⎜ (z i − x0i ) ⎟ ⎟ ⎟ ⎜ i=1 ⎟+ f (z) ⎟ ⎜ α ⎟ k ⎠ ⎝  ⎟ αi ! ⎠

n−1

i=1,...,k;|α|:=

⎞

k 

i=1

αi = j

i=1

⎞

⎛

(−1)n

α:=(α1 ,...,αk ),αi ∈Z+ k  i=1,...,k;|α|:= αi =n

 k  ⎜ n! ⎟  ⎟ ⎜ (z i − x0i )αi ⎟ ⎜ k ⎠ ⎝ i=1 αi !

(16.17)

i=1

i=1

1  (ν)



1

0

 n−ν  s ν−1 I1− f α (x0 + t (z − x0 )) (s) ds.

(2) Additionally assume that f α (z) = 0, α := (α1 , ..., αk ), αi ∈ Z+ , i = 1, ..., k, k  |α| := αi =: r , r = 0, 1, ..., n − 1, then i=1

⎞

⎛ f (x0 ) = (−1)



n

α:=(α1 ,...,αk ),αi ∈Z+ k  i=1,...,k;|α|:= αi =n

 k  ⎜ n! ⎟  ⎟ ⎜ αi (z i − x0i ) ⎟ ⎜ k ⎠ ⎝ i=1 αi ! i=1

i=1

1  (ν)

 0

1

 n−ν  s ν−1 I1− f α (x0 + t (z − x0 )) (s) ds.

(16.18)

Next we give three multivariate right side Caputo fractional Landaus’ type inequalities.   Theorem 16.6 Let f ∈ C 4 Rk− , k ≥ 2, with  f ∞,Rk− < ∞, and 3 < ν < 4. Assume further that

278

16 Multidimensional Right Caputo Fractional Taylor Formula and Landau Inequalities

          4−ν   I + t − x f (z ( (x ))) (s) α 0 0 1−   < +∞. ν := sup   k   x0 ,z∈Rk− ,s∈[0,1]  α:=(α ,...,α ),α ∈Z+ 1 k i α !   i i=1,...,k;|α|:= k α =4  i=1 i   i=1 (16.19) Then we derive the best upper bounds: (1)  k   ∂ f       ∂xi  i=1

∞,Rk−



11 ≤4 9

 34 

(2)            f α      k  α:=(α1 ,...,αk ),αi ∈Z+   α !   i i=1,...,k;|α|:= k α =2 i=1  i   i=1

144  (ν + 1)

 ≤2

 41

3

1

4 4  f ∞,R k ν < +∞, −

(16.20)

528  f ∞,Rk− ν < +∞, (16.21)  (ν + 1)

∞,Rk−

and (3)            fα      k  α:=(α1 ,...,αk ),αi ∈Z+   α !   i i=1,...,k;|α|:= k α =3 i=1  i   i=1

≤

4 3



144  (ν + 1)

 34

∞,Rk−

Proof We set 1 (z) := 2 (z) :=

k  i=1

3

−

(16.22)

∂ f (z) , ∂xi



α:=(α1 ,...,αk ),αi ∈Z+ k  i=1,...,k;|α|:= αi =2

f α (z) , k  αi ! i=1

(16.23)

i=1

and 3 (z) :=

1

4 4  f ∞,R k ν < +∞.

 α:=(α1 ,...,αk ),αi ∈Z+ k  i=1,...,k;|α|:= αi =3 i=1

f α (z) . k  αi ! i=1

16.3 Main Results

279

We can choose x0 , z ∈ Rk− so that z i − x0i = −h, all i = 1, ..., k, where h > 0. By Theorem 16.5 we get f (x0 ) = f (z) + 1 (z) h + 2 (z) h 2 + 3 (z) h 3 + R4 h 4 , ⎡

where

⎢ ⎢ 1 ⎢ ⎢ R4 := ⎢  (ν) ⎢ + i ∈Z ⎣ α:=(α1 ,...,αk ),α k 

⎞

⎛

⎜ 4! ⎟ ⎟ ⎜ ⎟ ⎜ k ⎠ ⎝ αi !

αi =4

i=1,...,k;|α|:=

(16.24)

i=1

i=1



1

0

4−ν s ν−1 I1−

 ( f α (x0 + t (z − x0 ))) (s) ds .

(16.25)

Therefore, ∀ h > 0, we get 1 (z) h + 2 (z) h 2 + 3 (z) h 3 = f (x0 ) − f (z) − R4 h 4 =: A1 , 1 (z) 2h + 2 (z) 4h 2 + 3 (z) 8h 3 = f (x0 ) − f (z) − R4 16h 4 =: A2 , and 1 (z) 3h + 2 (z) 9h 2 + 3 (z) 27h 3 = f (x0 ) − f (z) − R4 81h 4 =: A3 . (16.26) We solve the system of three equations (16.26) and we find: 1 (z) = 2 (z) = and 3 (z) =

18A1 −9A2 +2 A3 , 6h 4 A2 −5A1 −A3 , 2h 2

(16.27)

3A1 −3A2 +A3 . 6h 3

We have also that |R4 | ≤

24ν .  (ν + 1)

(16.28)

We calculate 18A1 − 9A2 + 2 A3 = 11 ( f (x0 ) − f (z)) − 36R4 h 4 .

(16.29)

Therefore it holds |1 (z)| ≤ ∀ h > 0, ∀ z ∈ Rk− .

11  f ∞,Rk− 3h

+

144ν 3 h ,  (ν + 1)

(16.30)

280

16 Multidimensional Right Caputo Fractional Taylor Formula and Landau Inequalities

We get that 

11  f ∞,Rk−

1 ∞,Rk− ≤

+

3h

 144ν h3,  (ν + 1)

(16.31)

∀ h > 0. Next we find 4 A2 − 5A1 − A3 = −2 ( f (x0 ) − f (z)) + 22R4 h 4 . Hence |2 (z)| ≤



2  f ∞,Rk−

+

h2

 264ν h2,  (ν + 1)

(16.32)

(16.33)

∀ h > 0, ∀ z ∈ Rk− . The last implies 2 ∞,Rk− ≤

2  f ∞,Rk− h2

 +

 264ν h2,  (ν + 1)

(16.34)

∀ h > 0. We also calculate 3A1 − 3A2 + A3 = ( f (x0 ) − f (z)) − 36R4 h 4 . Therefore |3 (z)| ≤

 f ∞,Rk− 3h 3

 +

 144ν h,  (ν + 1)

(16.35)

∀ h > 0, ∀ z ∈ Rk− . The last implies 3 ∞,Rk− ≤

 f ∞,Rk− 3h 3

 +

 144ν h,  (ν + 1)

∀ h > 0. We will work on (16.31).   11 f ∞,Rk 144ν − Call μ := , both are positive. , θ := 3 (ν+1) We study μ y (h) = + θh 3 , ∀ h > 0. h We have

y  (h) = −μh −2 + 3θh 2 = 0,

(16.36)

(16.37)

(16.38)

16.3 Main Results

281

with one critical number h crit.no. = h 0 = We have that

 μ  14 > 0. 3θ

(16.39)

y  (h 0 ) = 2μh −3 0 + 6θh 0 > 0,

(16.40)

implying that y has a global minimum. We see that  −4  4 1 3 3 3 4 4 y (h 0 ) = μh −1 0 + θh 0 = h 0 μh 0 + θ = 3 θ μ . 34 I.e. y (h 0 ) =

4 3

3 4

1

3

θ4 μ4 ,

(16.41)

(16.42)

is the global minimum. Consequently 

1 ∞,Rk−

11 ≤4 9

 34 

144  (ν + 1)

 41

3

1

4 4  f ∞,R k ν . −

Next we work on (16.34). 264ν > 0. Call ϕ := 2  f ∞,Rk− , ψ := (ν+1) We study ϕ y (h) := 2 + ψh 2 , ∀ h > 0. h We have

y  (h) = −2ϕh −3 + 2ψh = 0,

with

(16.44)

(16.45)

  41 ϕ > 0. ψ

(16.46)

y  (h 0 ) = 6ϕh −4 0 + 2ψ > 0.

(16.47)

h crit.no. = h 0 = Notice that

(16.43)

Thus we have that y has global minimum which is  −4  1 1 2 2 2 2 y (h 0 ) = ϕh −2 0 + ψh 0 = h 0 ϕh 0 + ψ = 2ψ ϕ , that is

1

1

y (h 0 ) = 2ψ 2 ϕ 2 .

(16.48)

282

16 Multidimensional Right Caputo Fractional Taylor Formula and Landau Inequalities

Therefore we derive  2 ∞,Rk− ≤ 2

528  f ∞,Rk− ν .  (ν + 1)

(16.49)

Finally we work on (16.36). Let A := We study

 f ∞,Rk

−

3

, B :=

144ν , (ν+1)

both are positive. A + Bh, ∀ h > 0. h3

(16.50)

y  (h) = −3Ah −4 + B = 0,

(16.51)

y (h) := Then

so that  h crit.no. = h 0 =

3A B

 14

> 0.

(16.52)

Here y  (h 0 ) = 12 Ah −5 0 > 0.

(16.53)

Hence y has global minimum which is  −4  4 1 3 4 4 y (h 0 ) = Ah −3 0 + Bh 0 = h 0 Ah 0 + B = 3 A B . 34 That is y (h 0 ) =

4 3

3 4

1

3

A4 B 4,

(16.54)

the global minimum of y. Consequently it holds 3 ∞,Rk− ≤ The theorem is proved.

4 3



144  (ν + 1)

 43

1

3

4 4  f ∞,R k ν . −

(16.55) 

References

283

References 1. Anastassiou, G.A.: Intelligent Mathematics: Computational Analysis. Springer, Heidelberg (2011) 2. Anastassiou, G.A.: Multivariate right side Caputo fractional Taylor formula and Landau inequalities. Ann. Commun. Math. Accepted (2020) 3. Ditzian, Z.: Multivariate Landau-Kolmogorov-type inequality. Math. Proc. Camb. Philos. Soc. 105(2), 335–350 (1989) 4. Kounchev, O.: Extremizers for the multivariate Landau-Kolmogorov inequality. Multivariate Approximation (Witten-Bommerholz, 1996), pp. 123–132. Math. Res. 101. Akademie Verlag, Berlin (1997) 5. Landau, E.: Einige Ungleichungen für zweimal differentzierban funktionen. Proc. Lond. Math. Soc. 13, 43–49 (1913)

Chapter 17

Multidimensional Generalized Right Fractional Taylor Formula and Landau Inequalities

Here we present a multivariate generalized right side fractional Taylor’s formula with fractional integral remainder. Based on this we give three multivariate generalized right side fractional Landau’s type inequalities. Their constants are exactly calculated and we give best upper bounds. It follows [2].

17.1 Introduction We are motivated by the following two results. First from [3], we have the following nice result: 2 d   ∂ If  f = f is the Laplacian of a function defined on Rd , then for 0 < ∂xi k < 2n,

i=1

  ∂ ∂   ∂ξ ... ∂ξ 1 k

  k  k 1−( 2n ) n f  2n , f  ≤ C (n, k)  f 

(17.1)

  where · denotes the L ∞ Rd norm. n f needs to be defined only in the distribup tional sense and the result is valid in many other  spaces such as L spaces.  Banach k k−1 . C (n, k) is an existential constant and  =   From [4], we have the following interesting result which is the sharp inequality   f ∞ ≤ 2

  d  f ∞ 2 f ∞ , (d + 2)

(17.2)

in Rd , d ≥ 2, where  f is the Laplacian and 2 =  (). This result is an analogue of the famous Landau inequality [5] in R: © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_17

285

286

17 Multidimensional Generalized Right Fractional Taylor …

  f 

∞

≤

√

2  f ∞  f  ∞ ,

(17.3)

where ·∞ means the L ∞ norm on Rd (respectively on R). In this chapter we first prove a multivariate generalized right side fractional Taylor’s formula and based on it we prove three multivariate generalized right side fractional Landau’s type inequalities where all the constants are exactly calculated and we get best upper bounds.

17.2 Background Here we follow [1], pp. 345–348. Let ν > 0, n := [ν] ([·] the integral part), α := ν − n, 0 < α < 1, f ∈ C ([a, b]). We call the right Riemann–Liouville fractional integral operator by 

 ν f (x) := Jb−

1  (ν)



b

(z − x)ν−1 f (z) dz,

(17.4)

x

∀ x ∈ [a, b], where  is the gamma function. Define the subspace of functions

1−α (n) ν Cb− f ∈ C 1 ([a, b]) . ([a, b]) := f ∈ C n ([a, b]) : Jb−

(17.5)

Define the right generalized ν-fractional derivative of f over [a, b] as  1−α (n)  ν f := (−1)n−1 Jb− f . Db−

(17.6)

Notice that 1−α (n) f (x) := Jb−

1  (1 − α)



b

(z − x)−α f (n) (z) dz,

(17.7)

(z − x)−α f (n) (z) dz.

(17.8)

(z − x)n−ν f (n) (z) dz.

(17.9)

x

ν exists for f ∈ Cb− ([a, b]), and ν Db− f (x) =

(−1)n−1 d  (1 − α) d x



b x

That is 

 ν Db− f (x) =

d (−1)n−1  (n − ν + 1) d x



b x

If ν ∈ N, then α = 0, n = ν, and ν f (x) = (−1)n f (n) (x) . Db−

(17.10)

17.2 Background

287

We mention the following right fractional Taylor formula: ν Theorem 17.1 ([1], p. 348) Let f ∈ Cb− ([a, b]), ν > 0, n := [ν]. Then (1) If ν ≥ 1, we get

f (x) =

 b n−1   ν  f (λ) (b) 1 f (z) dz, (z − x)ν−1 Db− (x − b)λ + λ!  (ν) x

(17.11)

λ=0

∀ x ∈ [a, b] . (2) If 0 < ν < 1, we obtain f (x) =

1  (ν)



b x

 ν  f (z) dz, (z − x)ν−1 Db−

(17.12)

∀ x ∈ [a, b] .

17.3 Main Results We make Remark 17.2 Let ν ≥ 1 such that n = [ν], and let f ∈ C n (Q), where Q is compact and convex subset of Rk , k ≥ 2; z := (z 1 , ..., z k ), x0 := (x01 , ..., x0k ) ∈ Q. Call (17.13) gz (t) := f (x0 + t (z − x0 )) , 0 ≤ t ≤ 1; so that gz (0) = f (x0 ), and gz (1) = f (z) . Then ⎡ j ⎤ k  ∂ gz( j) (t) = ⎣ f ⎦ (x0 + t (z − x0 )) , (z i − x0i ) ∂x i i=1 and

⎡ j ⎤ k  ∂ gz( j) (1) = ⎣ f ⎦ (z) , (z i − x0i ) ∂x i i=1

for all j = 0, 1, 2, ..., n. Here it is  ν  J1− gz (x) = for all 0 ≤ x ≤ 1. Consider

1  (ν)



1 x

( j − x)ν−1 gz ( j) d j,

(17.14)

(17.15)

(17.16)

288

17 Multidimensional Generalized Right Fractional Taylor …



1−α (n) ν C1− g ∈ C 1 ([0, 1]) . ([0, 1]) := g ∈ C n ([0, 1]) : J1−

(17.17)

The right generalized ν-fractional derivative of g over [0, 1] is defined by  1−α (n)  ν g := (−1)n−1 J1− g . D1−

(17.18)

(ν−n) For fixed x0 , z ∈ Q we assume that as functions of t : f α (x0 + t (z − x0 )) ∈ C1− k  αi = n. ([0, 1]), for all α := (α1 , ..., αk ), αi ∈ Z+ , i = 1, ..., k; |α| := i=1

We have that ⎞

⎛ gz(n)



(t) =

α:=(α1 ,...,αk ),αi ∈Z+ k  i=1,...,k;|α|:= αi =n

 k  ⎜ n! ⎟  ⎟ ⎜ αi f α (x0 + t (z − x0 )) , (z i − x0i ) ⎟ ⎜ k ⎠ ⎝ i=1 αi ! i=1

i=1

(17.19) 0 ≤ t ≤ 1. Here it is 1 − α = n + 1 − ν and [ν − n] = 0. We also have that (0 ≤ x ≤ 1) ⎞

⎛ 



 n+1−ν (n) J1− gz (x) =

α:=(α1 ,...,αk ),αi ∈Z+ k  i=1,...,k;|α|:= αi =n

⎜ n! ⎟ ⎟ ⎜ ⎟ ⎜ k ⎠ ⎝ αi ! i=1

i=1

 k 

 (z i − x0i )αi



 n+1−ν J1− ( f α (x0 + t (z − x0 ))) (x) ,

(17.20)

i=1

and 

  1−α (n)   n+1−ν (n)  ν gz (x) = (−1)n−1 J1− gz gz D1− (x) = (−1)n−1 J1− (x) = ⎞

⎛  α:=(α1 ,...,αk ),αi ∈Z+ k  i=1,...,k;|α|:= αi =n

 k  ⎜ n! ⎟  ⎟ ⎜ αi (z i − x0i ) ⎟ ⎜ k ⎠ ⎝ i=1 αi ! i=1

i=1

 n+1−ν  (−1)n−1 J1− ( f α (x0 + t (z − x0 ))) (x) , for all 0 ≤ x ≤ 1.

(17.21)

17.3 Main Results

289

That is ⎞

⎛ 

ν gz D1−



(x) = (−1)



n

α:=(α1 ,...,αk ),αi ∈Z+ k  i=1,...,k;|α|:= αi =n

 k  ⎜ n! ⎟  ⎟ ⎜ αi (z i − x0i ) ⎟ ⎜ k ⎠ ⎝ i=1 αi ! i=1

i=1

ν−n D1− ( f α (x0 + t (z − x0 ))) (x) ,

(17.22)

for all 0 ≤ x ≤ 1.  ν  ν gz (t), ∀ t ∈ [0, 1] , i.e. gz ∈ C1− Clearly then there exists D1− ([0, 1]) . By Theorem 17.1, (17.11), we obtain gz (0) =

n−1  (−1)λ gz(λ) (1)

λ!

λ=0

+

1  (ν)



1

0

 ν  j ν−1 D1− gz ( j) d j.

(17.23)

Based on the above we have proved the following multivariate right generalized fractional Taylor’s formula: Theorem 17.3 Let f ∈ C n (Q), Q is a compact and convex subset of Rk , k ≥ 2; here ν ≥ 1 such that n = [ν]. For fixed x0 , z ∈ Q assume that as functions of (ν−n) t : f α (x0 + t (z − x0 )) ∈ C1− ([0, 1]), for all α := (α1 , ..., αk ), αi ∈ Z+ , i = k  1, ..., k; |α| := αi = n. Then i=1

(i) f (x0 ) = f (z) −

k 

∂f (z) + ∂xi

(z i − x0i )

i=1



k 

(z i − x0i )

i=1

∂ ∂xi



2 (−1)

f (z)

n−1

+ (−1)n

α:=(α1 ,...,αk ),αi ∈Z+ k  i=1,...,k;|α|:= αi =n

∂ ∂xi

n−1 f (z)

(n − 1)! ⎞

⎛ 

(z i − x0i )

i=1

+ ... +

2

k 

 k  ⎜ n! ⎟  ⎟ ⎜ (z i − x0i )αi ⎟ ⎜ k ⎠ ⎝ i=1 αi ! i=1

i=1

1  (ν)

 0

1

ν−n j ν−1 D1− ( f α (x0 + t (z − x0 ))) ( j) d j.

(17.24)

290

17 Multidimensional Generalized Right Fractional Taylor …

(ii) If all f α (z) = 0, α := (α1 , ..., αk ), αi ∈ Z+ , i = 1, ..., k, |α| :=

k 

αi =: l,

i=1

l = 0, ..., n − 1, then ⎞

⎛ f (x0 ) = (−1)



n

α:=(α1 ,...,αk ),αi ∈Z+ k  αi =n

 k  ⎜ n! ⎟  ⎟ ⎜ αi (z i − x0i ) ⎟ ⎜ k ⎠ ⎝ i=1 αi !

i=1,...,k;|α|:=

i=1

i=1

1  (ν)

 0

1

ν−n j ν−1 D1− ( f α (x0 + t (z − x0 ))) ( j) d j.

(17.25)

Additionally we make Remark 17.4 Continuing from Remark 17.2. Let here 0 < ν < 1. Assume now ν that f (x0 + t (z − x0 )) ∈ C1− ([0, 1]) as a function of t. Then by gz (t) := f (x0 + t (z − x0 )) we have that ν ν gz (t) = D1− f (x0 + t (z − x0 )) , D1−

and



  ν ν  ν ν D1− gz (t) = J1− D1− f (x0 + t (z − x0 )) (t) , J1−

(17.26)

∀ t ∈ [0, 1]. Hence by (17.12) f (x0 ) = gz (0) =



ν ν D1− gz J1−



1 (0) =  (ν)



1 0

ν j ν−1 D1− f (x0 + j (z − x0 )) d j.

(17.27)

We have established the next multivariate generalized right side fractional Taylor’s formula when 0 < ν < 1. Theorem 17.5 Let continuous f : Q → R, where Q is compact and convex ⊆ Rk , ν k ≥ 2, such that as a function of t : f (x0 + t (z − x0 )) ∈ C1− ([0, 1]), where x0 , z ∈ Q are being fixed, and 0 < ν < 1. Then f (x0 ) =

1  (ν)

 0

1

ν j ν−1 D1− f (x0 + j (z − x0 )) d j.

(17.28)

We need Corollary 17.6 (to Theorem 17.3, n = 4 case) Let f ∈ C 4 (Q), Q is a compact and convex subset of Rk , k ≥ 2; here 4 < ν < 5. For fixed x0 , z ∈ Q assume that (ν−4) as functions of t : f α (x0 + t (z − x0 )) ∈ C1− ([0, 1]), for all α := (α1 , ..., αk ), k  αi = 4. Then αi ∈ Z+ , i = 1, ..., k; |α| := i=1

17.3 Main Results

291

f (x0 ) = f (z) −

k 

(z i − x0i )

i=1

⎛

⎛

⎜ ⎜  ⎜ ⎜ ⎜ ⎜ α:=(α1 ,...,αk ),αi ∈Z+ ⎝ k  i=1,...,k;|α|:=

αi =2

∂f (z) + ∂xi ⎞

⎞

k 

⎟ αi ⎟ ⎜ (z i − x0i ) ⎟ ⎟ ⎟ ⎜ i=1 ⎟− f (z) ⎟ ⎜ α ⎟ k ⎠ ⎝  ⎟ αi ! ⎠ i=1

i=1

⎛ ⎜ ⎜  ⎜ ⎜ ⎜ ⎜ α:=(α1 ,...,αk ),αi ∈Z+ ⎝ k  i=1,...,k;|α|:=

αi =3

⎞

⎞

⎛

k  ⎟ αi ⎟ ⎜ (z i − x0i ) ⎟ ⎟ ⎟ ⎜ i=1 ⎟ f α (z)⎟ ⎜ ⎟+ k ⎠ ⎝  ⎟ αi ! ⎠ i=1

i=1

⎞

⎛  α:=(α1 ,...,αk ),αi ∈Z+ k  i=1,...,k;|α|:= αi =4

 k  ⎜ 4! ⎟  ⎟ ⎜ αi (z i − x0i ) ⎟ ⎜ k ⎠ ⎝ i=1 αi ! i=1

i=1

1  (ν)



1 0

ν−4 j ν−1 D1− ( f α (x0 + t (z − x0 ))) ( j) d j.

Proof By (17.24) for n = 4.

(17.29) 

Next follow some multivariate right side generalized fractional Landau type inequalities.   Theorem 17.7 Let f ∈ C 4 Rk− , k ≥ 2, with  f ∞,Rk− < +∞, and 4 < ν < 5. For (ν−4) any x0 , z ∈ Rk− assume that as functions of t : f α (x0 + t (z − x0 )) ∈ C1− ([0, 1]), k  αi = 4. Assume further for all α := (α1 , ..., αk ), αi ∈ Z+ , i = 1, ..., k; |α| := i=1

that ! ! ! ! ! ! ! ! ! ! ν−4  ! ! D + t − x f j) (z ( (x ))) ( α 0 0 1− ! ! < +∞. Hν := sup ! ! k  k ! ! x0 ,z∈R− , j∈[0,1] α:=(α ,...,α ),α ∈Z+ 1 k i α ! ! ! i !i=1,...,k;|α|:= k α =4 ! i=1 i ! ! i=1 (17.30)

292

17 Multidimensional Generalized Right Fractional Taylor …

Then we derive the best upper bounds: (1)   k  ∂ f       ∂xi  i=1

∞,Rk−

 ≤4

11 9

 43 

(2)             fα      k  α:=(α1 ,...,αk ),αi ∈Z+   α !   i i=1,...,k;|α|:= k α =2 i=1  i   i=1

144  (ν + 1)

 ≤2

 41

3

1

4 4  f ∞,R k Hν < +∞, −

(17.31)

528  f ∞,Rk− Hν < +∞, (17.32)  (ν + 1)

∞,Rk−

and (3)              f α     k  α:=(α1 ,...,αk ),αi ∈Z+   α !   i i=1,...,k;|α|:= k α =3 i=1  i   i=1

4 ≤ 3



144  (ν + 1)

 43

1

3

4 4  f ∞,R k Hν < +∞. −

∞,Rk−

Proof We set 1 (z) := 2 (z) :=

k  i=1

(17.33)

∂ f (z) , ∂xi

 +

α:=(α1 ,...,αk ),αi ∈Z k  αi =2

f α (z) , k  αi ! i=1

i=1,...,k;|α|:=

(17.34)

i=1

and 3 (z) :=

 +

α:=(α1 ,...,αk ),αi ∈Z k  αi =3

f α (z) . k  αi ! i=1

i=1,...,k;|α|:=

i=1

We can choose x0 , z ∈ Rk− so that z i − x0i = −h, all i = 1, ..., k, where h > 0. By Corollary 17.6 we get f (x0 ) = f (z) + 1 (z) h + 2 (z) h 2 + 3 (z) h 3 + R4 h 4 ,

(17.35)

17.3 Main Results

293

⎡

where

⎢ ⎢  1 ⎢ ⎢ R4 := ⎢  (ν) ⎢ + i ∈Z ⎣ α:=(α1 ,...,αk ),α k 

⎜ 4! ⎟ ⎟ ⎜ ⎟ ⎜ k ⎠ ⎝ αi !

αi =4

i=1,...,k;|α|:=

⎞

⎛

i=1

i=1



1

j

ν−1

0

ν−4 D1−

# ( f α (x0 + t (z − x0 ))) ( j) d j .

(17.36)

Therefore, ∀ h > 0, we get 1 (z) h + 2 (z) h 2 + 3 (z) h 3 = f (x0 ) − f (z) − R4 h 4 =: A1 , 1 (z) 2h + 2 (z) 4h 2 + 3 (z) 8h 3 = f (x0 ) − f (z) − R4 16h 4 =: A2 , and 1 (z) 3h + 2 (z) 9h 2 + 3 (z) 27h 3 = f (x0 ) − f (z) − R4 81h 4 =: A3 . (17.37) We solve the system of three equations (17.37) and we find: 1 (z) = 2 (z) = and 3 (z) =

18A1 −9A2 +2 A3 , 6h 4 A2 −5A1 −A3 , 2 2h

(17.38)

3A1 −3A2 +A3 . 6h 3

We have also that |R4 | ≤

24Hν .  (ν + 1)

(17.39)

We calculate 18A1 − 9A2 + 2 A3 = 11 ( f (x0 ) − f (z)) − 36R4 h 4 .

(17.40)

Therefore it holds |1 (z)| ≤

11  f ∞,Rk− 3h

+

144Hν 3 h ,  (ν + 1)

(17.41)

∀ h > 0, ∀ z ∈ Rk− . We get that 1 ∞,Rk− ≤ ∀ h > 0.

11  f ∞,Rk− 3h

 +

 144Hν h3,  (ν + 1)

(17.42)

294

17 Multidimensional Generalized Right Fractional Taylor …

Next we find 4 A2 − 5A1 − A3 = −2 ( f (x0 ) − f (z)) + 22R4 h 4 . Hence |2 (z)| ≤



2  f ∞,Rk−

+

h2

 264Hν h2,  (ν + 1)

(17.43)

(17.44)

∀ h > 0, ∀ z ∈ Rk− . The last implies 2 ∞,Rk− ≤

2  f ∞,Rk− h2

 +

 264Hν h2,  (ν + 1)

(17.45)

∀ h > 0. We also calculate 3A1 − 3A2 + A3 = ( f (x0 ) − f (z)) − 36R4 h 4 . Therefore |3 (z)| ≤

 f ∞,Rk− 3h 3

 +

 144Hν h,  (ν + 1)

(17.46)

∀ h > 0, ∀ z ∈ Rk− . The last implies 3 ∞,Rk− ≤

 f ∞,Rk− 3h 3

 +

 144Hν h,  (ν + 1)

∀ h > 0. We will work on (17.42).   11 f ∞,Rk 144Hν − Call μ := , both are positive. , θ := 3 (ν+1) We study μ y (h) = + θh 3 , ∀ h > 0. h We have

(17.47)

(17.48)

y  (h) = −μh −2 + 3θh 2 = 0,

(17.49)

 μ  14 > 0. 3θ

(17.50)

with one critical number h crit.no. = h 0 =

17.3 Main Results

We have that

295

y  (h 0 ) = 2μh −3 0 + 6θh 0 > 0,

(17.51)

implying that y has a global minimum. We see that  −4  4 1 3 3 3 4 4 y (h 0 ) = μh −1 0 + θh 0 = h 0 μh 0 + θ = 3 θ μ . 34 I.e. y (h 0 ) =

4 3

3 4

1

3

θ4 μ4 ,

(17.52)

(17.53)

is the global minimum. Consequently  1 ∞,Rk− ≤ 4 Next we work on (17.45). Call ϕ := 2  f ∞,Rk− , ψ := We study

11 9

 43 

264Hν (ν+1)

144  (ν + 1)

 14

3

1

4 4  f ∞,R k Hν . −

(17.54)

> 0.

ϕ + ψh 2 , ∀ h > 0. h2

(17.55)

y  (h) = −2ϕh −3 + 2ψh = 0,

(17.56)

y (h) := We have

with   41 ϕ = h0 = > 0. ψ

(17.57)

y  (h 0 ) = 6ϕh −4 0 + 2ψ > 0.

(17.58)

h crit.no. Notice that

Thus we have that y has global minimum which is  −4  1 1 2 2 2 2 y (h 0 ) = ϕh −2 0 + ψh 0 = h 0 ϕh 0 + ψ = 2ψ ϕ , that is 1

1

y (h 0 ) = 2ψ 2 ϕ 2 .

(17.59)

296

17 Multidimensional Generalized Right Fractional Taylor …

Therefore we derive  2 ∞,Rk− ≤ 2

528  f ∞,Rk− Hν .  (ν + 1)

(17.60)

Finally we work on (17.47). Let A := We study

 f ∞,Rk

−

3

, B :=

144Hν , (ν+1)

both are positive. A + Bh, ∀ h > 0. h3

(17.61)

y  (h) = −3Ah −4 + B = 0,

(17.62)

y (h) := Then

so that  h crit.no. = h 0 =

3A B

 14

> 0.

(17.63)

Here y  (h 0 ) = 12 Ah −5 0 > 0.

(17.64)

Hence y has global minimum which is  −4  4 1 3 4 4 y (h 0 ) = Ah −3 0 + Bh 0 = h 0 Ah 0 + B = 3 A B . 34 That is y (h 0 ) =

4 3

3 4

1

3

A4 B 4,

(17.65)

the global minimum of y. Consequently it holds 3 ∞,Rk− ≤ The theorem is proved.

4 3



144  (ν + 1)

 34

1

3

4 4  f ∞,R k Hν . −

(17.66) 

References

297

References 1. Anastassiou, G.A.: Intelligent Mathematics: Computational Analysis. Springer, Heidelberg (2011) 2. Anastassiou, G.A.: Multivariate generalized right side fractional Taylor formula and Landau inequalities. Submitted (2020) 3. Ditzian, Z.: Multivariate Landau-Kolmogorov-type inequality. Math. Proc. Camb. Philos. Soc. 105(2), 335–350 (1989) 4. Kounchev, O.: Extremizers for the multivariate Landau-Kolmogorov inequality. Multivariate Approximation (Witten-Bommerholz, 1996), pp. 123–132. Math. Res. 101. Akademie Verlag, Berlin (1997) 5. Landau, E.: Einige Ungleichungen für zweimal differentzierban funktionen. Proc. Lond. Math. Soc. 13, 43–49 (1913)

Chapter 18

Landau’s Inequality for Semigroups

Kraljeviˇc and Kurepa in 1970 [14] proved a nice Landau type inequality for semigroups by making use of Taylor’s formula with integral remainder for semigroups, which has been a traditional method for proving such inequalities. The author offers another method in the semigroups and reproves the above result by using his earlier result about Ostrowski inequalities for semigroups, see [1], Chap. 16, pp. 259–289. He finds his method easier and convenient. An application is given at the end. It follows [2].

18.1 Introduction We state Kraljeviˇc and Kurepa, 1970, interesting theorem [14]: Theorem 18.1 Let A be an infinitesimal generator of a strongly continuous semigroup {T (t) : t > 0} of bounded linear operators on a Banach space X . Let M, ω be real numbers with the property: T (t) ≤ Meωt , ∀ t > 0.

(18.1)

  Then for any f ∈ D A2 such that (A − ω I )2 f = 0, the inequality   (A − ω I ) f 2 ≤ 2M (M + 1)  f  (A − ω I )2 f 

(18.2)

holds. Other inspirations follow:

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_18

299

300

18 Landau’s Inequality for Semigroups

Theorem 18.2 (Kallman and Rota [11]) Let X be a complex Banach space and t → T (t) (t ≥ 0) be a strongly continuous semigroup of linear   contractions on X . Let A be its infinitesimal generator. Then, for every f ∈ D A2 , it holds   A f 2 ≤ 4  f   A2 f  .

(18.3)

We also mention Theorem 18.3 (Hille 1992, [8]) If A is the infinitesimal generator of a strongly continuous contraction semigroup, if f ∈ D (An ) and 1 ≤ k < n, then there exist constants Cn,k independent of A, so that  k     A ( f )n ≤ C n  f n−k  An ( f )k . n,k

(18.4)

This typeof research startedbyLandau [16]. He proved first that if f ∈ C 2 ([0, 1]), 

   f ∞ = 1, f ∞ = 4, then  f ∞ ≤ 4, with 4 the best constant and the result is not necessarily true for an interval of length < 1. It is customary now to write Landau’s inequality as 1  1    f  ≤ C p (I )  f  2  f

 2 , (18.5) p,I

p,I

p,I

where · p,I is the p-norm on the interval I ; p ∈ [1, ∞], I = R+ or I = R, and f : I → R is twice differentiable with f, f

∈ L p (I ). C p (I ) > 0 is independent of f . Landau proved the best constants C∞ (R+ ) = 2 and C∞ (R) =

√ 2,

see also [3]. Using Theorem 18.2 we can derive a Landau’s inequality, see (18.5), by talking X = C ([0, ∞)) or L p ([0, ∞)) and defining the shift operator T (s) f (t) = f (t + s) , s ≥ 0, A f (t) = f (t) , see [7, 8]. Of inspiration are also the articles [9, 13, 15, 17].

18.2 Background We will be using the following notions. Here all this background comes from [4] (in general see also [6, 18]). Let X a real or complex Banach space with elements f, g, ... having norm  f  , g , ... and let ε (X ) be the Banach algebra of endomorphisms of X . If T ∈ ε (X ), T  denotes the norm of T .

18.2 Background

301

Definition 18.4 If T (t) is an operator function on the non-negative real axis 0 ≤ t < ∞ to the Banach algebra ε (X ) satisfying the following conditions: 

(i) T (t1 + t2 ) = T (t1 ) T (t2 ) , (t1 , t2 ≥ 0) (ii) T (0) = I (I = identity operator),

(18.6)

then {T (t) : 0 ≤ t < ∞} is called a one-parameter semi-group of operators in ε (X ) . The semi-group {T (t) : 0 ≤ t < ∞} is said to be of class C0 if it satisfies the further property (18.7) (iii) s − lim T (t) f = f ( f ∈ X ) t→0+

referred to as the strong continuity of T (t) at the origin. In this chapter we shall assume that the family of bounded linear operators {T (t) : 0 ≤ t < ∞} mapping X to itself is a semi-group of class C0 , thus all three conditions of the above definition are satisfied. Proposition 18.5 (a) T (t) is bounded on every finite subinterval of [0, ∞). (b) For each f ∈ X , the vector-valued function T (t) f on [0, ∞) is strongly continuous, thus vector-Riemann integrable on [0, a], a > 0. Definition 18.6 The infinitesimal generator A of the semi-group {T (t) : 0 ≤ t < ∞} is defined by 1 A f = s − lim Aτ f , Aτ f = [T (τ ) − I ] f (18.8) τ →0+ τ whenever the limit exists; the domain of A, in symbols D (A) , is the set of elements f for which the limits exists. Proposition 18.7 (a) D (A) is a linear manifold in X and A is a linear operator. (b) If f ∈ D (A), then T (t) f ∈ D (A) for each t ≥ 0 and d T (t) f = AT (t) f = T (t) A f (t ≥ 0); dt furthermore,



t

T (t) f − f =

T (u) A f du (t > 0).

(18.9)

(18.10)

0

(c) D (A) is dense in X , i.e. D (A) = X , and A is a closed operator. Definition 18.8 For r = 0, 1, 2, ... the operator Ar is defined inductively by the relations A0 = I , A1 = A, and       D Ar = f : f ∈ D Ar −1 and Ar −1 f ∈ D (A)       Ar f = A Ar −1 f = s − lim Aτ Ar −1 f ( f ∈ D Ar ). τ →0+

(18.11)

302

18 Landau’s Inequality for Semigroups

For the operator Ar and its domain D (Ar ) we have the following Proposition 18.9 (a) D (Ar ) is a linear subspace in X and Ar is a linear operator. (b) If f ∈ D (Ar ), so does T (t) f for each t ≥ 0 and dr T (t) f = Ar T (t) f = T (t) Ar f. dt r

(18.12)

Furthermore T (t) f −

r −1 k t k=0

1 A f = k! (r − 1)!



k

t

(t − u)r −1 T (u) Ar f du,

(18.13)

0

the Taylor’s formula for semigroups. Additionally it holds (T (t) − I )r f =

 t 0

t



t

...

0

T (u 1 + u 2 + ... + u r ) Ar f du 1 du 2 ...du r . (18.14)

0

(c) D (Ar ) is dense in X for r = 1, 2, ...,; moreover, ∩r∞=1 D (Ar ) is dense in X , Ar is a closed operator. Integrals in (18.13) and (18.14) arevector

a integrals, see [4, 12].  valued Riemann Here we will assume that f ∈ D A2 . Clearly here 0 T (t) f dt ∈ X , where a > 0, see [18]. We mention an Ostrowski type inequality for semigroups Theorem 18.10 ([1], Chap. 13, p. 219) Let f ∈ D (A) , a > 0, and denote T (·) A f ∞,[0,a] := sup T (u) A f  .

(18.15)

u∈[0,a]

Then   a 1  T (t) f dt − T (t0 ) a 0

  2  t0 + (a − t0 )2  T (·) A f ∞,[0,a] , (18.16) f ≤ 2a

for a fixed t0 ∈ [0, a] . We will be using Theorem 18.10 to reprove Theorem 18.1 by another method.

18.3 Main Results

303

18.3 Main Results We give the following Landau inequality for semigroups. Theorem 18.11 Let the family of bounded linear operator {T (t) : 0 ≤ t < ∞} mapping X into itself be a semigroupof class  C0 . We assume here that T (t) ≤ M, ∀ t ∈ R+ , where M > 0. Let f ∈ D A2 . Then   A f 2 ≤ 2M (M + 1)  f   A2 f  ,

(18.17)

where A is the infinitesimal generator. Proof Let f ∈ D (A), t > 0, and denote T (·) A f ∞,[0,t] := sup T (u) A f . u∈[0,t]

By Theorem 18.10 we have   t 1  T (z) f dz − T (z 0 ) t 0

  t f  ≤ 2 T (·) A f ∞,[0,t] ,

(18.18)

for any fixed z 0 ∈ [0,   t]. Here let f ∈ D A2 , and we denote     T (·) A2 f  := sup T (u) A2 f  . ∞,[0,t]

(18.19)

u∈[0,t]

By (18.18) we obtain   t  1   t  2    T (z) A f dz − T (z 0 ) A f  t  ≤ 2 T (·) A f ∞,[0,t] , 0

(18.20)

for any fixed z 0 ∈ [0,  t]. Notice here that T (·) A2 f ∞,[0,t] < ∞. By (18.10) we get   1     (T (t) f − f ) − T (z 0 ) A f  ≤ t T (·) A2 f  , t  2 ∞,[0,t]

(18.21)

for any fixed z 0 ∈ [0, t] . Hence T (z 0 ) A f  − Furthermore we have

 T (t) f − f  t  ≤ T (·) A2 f ∞,[0,t] . t 2

(18.22)

304

18 Landau’s Inequality for Semigroups

T (z 0 ) A f  ≤

 T (t) f − f  t  + T (·) A2 f ∞,[0,t] ≤ t 2

 t  (T (t) f  +  f ) + T (·) A2 f ∞,[0,t] . t 2

(18.23)

By assumption T (t) ≤ M, ∀ t ≥ 0, so we have that T (t) f  ≤ T (t)  f  ≤ M  f. Similarly it holds     T (z) A2 f  ≤ M  A2 f  , ∀ z ≥ 0. That is

    T (·) A2 f  ≤ M  A2 f  , ∀ t ≥ 0. ∞,[0,t]

So we have that T (z 0 ) A f  ≤

 tM  (M + 1)  A2 f  , f+ t 2

(18.24)

∀ t ∈ R+ , and ∀ z 0 ∈ [0, t] . Choosing z 0 = 0, we get A f  ≤

 tM  (M + 1)  A2 f  , ∀ t ∈ R+ . f+ t 2

(18.25)

  If A2 f = 0, then  A2 f  = 0, and clearly it holds by (18.25) that A f  = 0, that is A f = 0. So next we will assume that f = 0 and A2 f = 0. We consider the function   M  A2 f  (M + 1)  f  +t , ∀ t ∈ R+ . (18.26) y (t) := t 2 Set   M  A2 f  . A := (M + 1)  f  , B := 2 That is y (t) =

A + Bt, ∀ t ∈ R+ . t

Hence y (t) = −At −2 + B = 0,

(18.27)

18.3 Main Results

305

and the critical number is t0 =

A . B

Notice that y

(t) = 2 At −3 , and 



y (t0 ) = y



A B



 = 2A

A B

−3

 = 2A

A B

− 3 2

 = 2A

B A

3 2

1

3

= 2 A− 2 B 2 > 0.

(18.28) Hence y has a global minimum which is    √   A y (t0 ) = y = 2 AB = 2M (M + 1)  f   A2 f . B

(18.29)

We have proved that A f  ≤

   2M (M + 1)  f   A2 f ,

when A2 f = 0. The last inequality is true even when A2 f = 0. If f = 0 then A f = 0, so that (18.30) is trivially again true. The theorem is proved.

(18.30)



Next comes a more general Landau inequality for semigroups. Corollary 18.12 Let the family of bounded linear operators {T (t) : 0 ≤ t < ∞} mapping X into itself be a semigroup of class C0 . Let M > 0 and ω be a real number with the property T (t) ≤ Meωt , ∀ t ≥ 0. Then, for any f ∈ D A2 , it holds   (A − ω I ) f 2 ≤ 2M (M + 1)  f  (A − ω I )2 f  .

(18.31)

Proof We are acting as in the proof of Theorem 18.1. We have that  −ωt  e T (t) ≤ M, ∀ t ≥ 0, and t → e−ωt T (t) is a semigroup with A − ω I as the infinitesimal generator. Hence by applying (18.17) we derive (18.31).  Notice that our results are valid without the restriction (A − ω I )2 f = 0, ω ∈ R, which is assumed in Theorem 18.1.

306

18 Landau’s Inequality for Semigroups

18.4 Application Here we follow [5]. It is known that the classical diffusion equation ∂W ∂2 W = , − ∞ < x < ∞, t > 0, ∂t ∂x 2

(18.32)

lim W (x, t) = f (x) ,

(18.33)

with initial condition t→0+

has under general conditions its solution given by 1 W (x, t, f ) = (T (t) f ) (x) = √ 2 πt



∞ −∞

u2

f (x + u) e− 4t du,

(18.34)

the so called Gauss–Weierstrass singular integral. ∂2 The infinitesimal generator of the semigroup {T (t) : 0 ≤ t < ∞} is A = ∂x 2 ([10], p. 578). Here we suppose that f belongs to the Banach space X = U C B (R), the space of bounded and uniformly continuous functions from R into itself, with norm  f C := sup | f (x)|. x∈R

We notice that    1  ∞ u2 |W (x, t, f )| = √  f (x + u) e− 4t du  ≤ 2 πt −∞ 1 √ 2 πt



∞

−∞

| f (x + u)| e

2

− u4t

 du ≤  f C

1 √ 2 πt



∞



2

e

− u4t

du

(18.35)

−∞

=  f C · 1 =  f C < ∞, ∀ t ∈ R+ . That is W (·, t, f )C ≤ 1 ·  f C , ∀ t ∈ R+ .

(18.36)

So in Theorem 18.11, 1.  M =  4 2 ∂2 ∂ = D . Then, by (18.17), we derive Let now f ∈ D 2 ∂x ∂x 4  2 2  4  ∂ f      ≤ 4  f C  ∂ f  .  ∂x 2   ∂x 4  C C

(18.37)

References

307

References 1. Anastassiou, G.A.: Intelligent Comparisons: Analytic Inequalities. Springer, Heidelberg (2016) 2. Anastassiou, G.A.: A Landau’s inequality for semigroups revisited. Transylv. J. Math. Mech. Accepted (2020) 3. Barnett, N.S., Dragomir, S.S.: Some Landau type inequalities for functions whose derivatives are of locally bounded variation. Tamkang J. Math. 37(4), 301–308 (2006) 4. Butzer, P.L., Berens, H.: Semi-Groups of Operators and Approximation. Springer, New York (1967) 5. Butzer, P.L., Tillmann, H.G.: Approximation theorems for semi-groups of bounded linear transformations. Math. Ann. 140, 256–262 (1960) 6. Goldstein, J.A.: Semigroups of Linear Operators and Applications. Oxford University Press, Oxford (1985) 7. Hille, E.: Remark on the Landau-Kallman-Rota inequality. In: Reports of Meetings, 7th International Symposium on Functional Equations, September 1–13 (1969). Aequ. Math. 4, 239–240 (1970) 8. Hille, E.: Generalizations of Landau’s inequality to linear operators. In: Butzer, P.L., Kahane, J.-P., Nagy, B.S. (eds.) Linear Operators and Approximation. Proceedings of the Conference at Oberwolfach Mathematical Research Institute, August 14–22, 1971, pp. 20–32. Birkhauser, Basel (1972) 9. Hille, E.: On the Landau-Kallman-Rota inequality. J. Approx. Theory 6, 117–122 (1972) 10. Hille, E., Phillips, R.S.: Functional Analysis and Semigroups, revised edn., pp. XII a. 808. American Mathematical Society Colloquium Publications, vol. 31. American Mathematical Society, Providence (1957)  2   11. Kallman, R.R., Rota, G.-C.: On the inequality  f  ≤ 4  f   f

. In: Shisha, O. (ed.) Inequalities II, pp. 187–192. Academic, New York (1970) 12. Katznelson, Y.: An Introduction to Harmonic Analysis. Dover, New York (1976) 13. Kolmogorov, A.N.: On inequalities between the upper bounds of the successive derivatives of an arbitrary function on an infinite interval. (Russian.) Užen. Zap. Moskov. Gos. Univ. Matematika 30, 3–13 (1939) (Am. Math. Soc. Transl. (1) 2, 233–243 (1962)) 14. Kraljeviˇc, H., Kurepa, S.: Semi-groups on Banach spaces. Glasnik Mat. 5(25), 109–117 (1970) 15. Kurepa, S.: Remark on the Landau inequality. In: Reports of Meetings, 7th International Symposium on Functional Equations, September 1–13 (1969). Aequ. Math. 4, 240–241 (1970) 16. Landau, E.: Einige Ungleichungen für zweimal differentzierban funktionen. Proc. Lond. Math. Soc. 13, 43–49 (1913) 17. Schoenberg, I.J., Cavaretta, A.: Solution of Landau’s problem concerning higher derivatives on the halfline. MRC Technical Summary Report, No. 1050, Madison, Wisconsin (1970) 18. Shilov, G.: Elementary Functional Analysis. Dover Publications Inc, New York (1996)

Chapter 19

Fractional Variable Order Gronwall Inequality

This chapter presents the first generalized fractional variable order Gronwall inequality. It follows [1].

19.1 Introduction The following generalized Gronwall inequality for fractional differential equations (of constant order) was established in [2]: Theorem 19.1 Suppose that a(t) is a nonnegative function locally integrable on [0, T ) (for some T ≤ ∞), g(t) is a nonnegative, nondecreasing, and bounded continuous function defined on [0, T ) and β0 > 0. If u(t) is nonnegative and locally integrable on [0, T ) satisfying  u(t) ≤ a(t) + g(t)

t

(t − s)β0 −1 u(s)ds,

0 ≤ t < T,

(19.1)

0

then u(t) ≤ a(t) +

n  t  ∞  (β0 )g(t) 0

n=1

(nβ0 )

 (t − s)nβ0 −1 a(s) ds,

0 ≤ t < T, (19.2)

where (·) is the Gamma function.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_19

309

310

19 Fractional Variable Order Gronwall Inequality

The idea of the proof is to introduce the Volterra-type (linear) operator  Bφ(t) := g(t)

t

(t − s)β0 −1 φ(s)ds,

0 ≤ t < T,

(19.3)

0

so that (19.1) can be written as u(t) ≤ a(t) + Bu(t),

0 ≤ t < T,

(19.4)

and hence, by repeated iteration of (19.4), u(t) ≤

n−1 

B k a(t) + B n u(t),

0 ≤ t < T.

(19.5)

k=0

The remaining part of the proof of Theorem 19.1 [3] is the inductive justification of the inequality  n  t (β0 )g(t) (t − s)nβ0 −1 φ(s)ds, B φ(t) ≤ (nβ0 ) 0 n

0 ≤ t < T, n = 1, 2, . . . ,

(19.6) for any function φ(t) ≥ 0 which is locally integrable on [0, T ). An immediate implication of (19.6) is that B n φ(t) → 0 as n → +∞

for any t ∈ [0, T ),

(19.7)

and the validity of (19.2) follows, by using (19.6) and (19.7) in (19.5). Remark 19.2 In the statement of Theorem 19.1, the assumption that g(t) is bounded is not necessary, since for any fixed t ∈ [0, T ) the monotonicity of g implies that g(s) ≤ g(t) < ∞ for 0 ≤ s ≤ t. Also, the assumption that g(t) is continuous is not used in the proof of the theorem, hence it, too, is not necessary. Therefore, the theorem remains true for any nonnegative increasing function g(t) defined on [0, T ). Actually, there is an immediate extension of Theorem 19.1 to the case where g(t) is any function which is locally bounded in [0, T ) . In this case we can just set G(t) := sup {g(s), 0},

0 ≤ t < T,

(19.8)

s∈[0,t]

so that G(t) is nonnegative, increasing, and satisfies G(t) ≥ g(t) for t ∈ [0, T ), and then apply Theorem 19.1 with G(t) in place of g(t).

19.2 Main Results

311

19.2 Main Results In this short note we propose an extension of Theorem 19.1 to the case where the constant order β0 is replaced by a strictly positive variable order β(t). Our motivation came from the recent monograph [2], which contains an extensive discussion on fractional integrals and derivatives of variable order and their applications. Theorem 19.3 Suppose that a(t) is a nonnegative function locally integrable on [0, T ) for some T < ∞, g(t) is a nonnegative and nondecreasing function defined on [0, T ), and β(t) is a (Lebesgue) measurable function satisfying 0 < β0 ≤ β(t) ≤ A < ∞,

0 ≤ t < T.

(19.9)

If u(t) is nonnegative and locally integrable on [0, T ) satisfying the inequality  u(t) ≤ a(t) + g(t)

t

(t − s)β(s)−1 u(s)ds,

0 ≤ t < T,

(19.10)

0

then u(t) ≤ a(t) +

∞ 

L k a(t)

k=1

 n  t  ∞  (β0 )K g(t) nβ0 −1 a(s) ds, ≤ a(t) + (t − s) (nβ0 ) 0 n=1

0 ≤ t < T, (19.11)

where L is the Volterra-type (linear) operator 

t

Lφ(t) := g(t)

(t − s)β(s)−1 φ(s)ds,

0 ≤ t < T,

(19.12)

0

and K := max{1, T } A−β0 .

(19.13)

Proof Observe that from (19.9) we get (since β(s) ≥ β0 ) (t − s)β(s)−1 = (t − s)β(s)−β0 ≤ max{1, T } A−β0 , (t − s)β0 −1

0 ≤ s < t < T,

that is (t − s)β(s)−1 ≤ max{1, T } A−β0 (t − s)β0 −1 , Therefore, (19.10) implies

0 ≤ s < t < T.

(19.14)

312

19 Fractional Variable Order Gronwall Inequality



t

u(t) ≤ a(t) + K g(t)

(t − s)β0 −1 u(s)ds,

0 ≤ t < T,

(19.15)

0

where K is given by (19.13). Since β0 > 0 is a constant, we can apply Theorem 19.1– (19.15), where the operator B now has the slightly different form:  Bφ(t) = K g(t)

t

(t − s)β0 −1 φ(s)ds,

0 ≤ t < T.

(19.16)

0

As in the proof of Theorem 19.1, lim B n u(t) = 0,

n→+∞

0 ≤ t < T.

(19.17)

Now, it is clear from (19.12), (19.14), and (19.16) that for u(t) ≥ 0 we have 0 ≤ Lu(t) ≤ Bu(t),

0 ≤ t < T.

(19.18)

0 ≤ t < T,

(19.19)

Hence, by using (19.17) in (19.18) we get lim L n u(t) = 0,

n→+∞

and since (19.10) can be written as u(t) ≤ a(t) + Lu(t),

0 ≤ t < T,

(19.20) 

formula (19.11) follows easily from (19.19), (19.18) and Theorem 19.1.

Notice that, as in the standard Gronwall inequality, the value of (19.11) lies in the fact that it gives a bound for u(t) in terms of a(t) , g(t), and β(t). As explained in Remark 19.2, in the case where g(t) is any locally bounded function in [0, T ), Theorem 19.3 holds by replacing g(t) with G(t) of (19.8). Corollary 19.4 All as in Theorem 19.3 with g(t) = b ≥ 0 constant. If  u(t) ≤ a(t) + b

t

(t − s)β(s)−1 u(s)ds,

0 ≤ t < T,

(19.21)

0

then u(t) ≤ a(t) +

∞ 

L k1 a(t)

k=1

≤ a(t) +

n  t  ∞  (β0 )K b 0

n=1

(nβ0 )

 (t − s)

nβ0 −1

a(s) ds,

0 ≤ t < T, (19.22)

19.2 Main Results

313

where L 1 is the Volterra operator 

t

L 1 φ(t) := b

(t − s)β(s)−1 φ(s)ds,

0 ≤ t < T.

(19.23)

0

Corollary 19.5 All as in Theorem 19.3 with a(t) be a nondecreasing function on [0, T ). Then  0 ≤ t < T, (19.24) u(t) ≤ a(t)E β0 K g(t)(β0 )t β0 , where E β0 (·) is the Mittag-Leffler function defined by E β0 (z) :=

∞  k=0

zk . (kβ0 + 1)

Proof The assumptions of Theorem 19.3 and (19.11) imply

u(t) ≤ a(t) 1 +

n  t  ∞  (β0 )K g(t) 0

n=1

(nβ0 )  β n

∞   (β0 )K g(t)t 0 = a(t) (nβ0 + 1) n=0

 (t − s)

nβ0 −1

 ds

 = a(t)E β0 K g(t)(β0 )t β0 .

(19.25) 

The Gronwall inequality of fractional variable order is expected to find wide applications in the forthcoming studies of fractional differential equations of variable order.

References 1. Anastassiou, G.A., Papanicolaou, V.G.: A Gronwall inequality of fractional variable order. Prog. Fract. Differ. Appl. Accepted (2020) 2. Yang, X.-J.: General Fractional Derivatives. Theory, Methods and Applications. CRC Press, Taylor & Francis Group, Boca Raton (2019) 3. Ye, H., Gao, J., Ding, Y.: A generalized Gronwall inequality and its application to a fractional differential equation. J. Math. Anal. Appl. 328(4), 1075–1081 (2007)

Chapter 20

A Study of Gronwall Inequalities of Fractional Variable Order

A wide variety of Gronwall inequalities of fractional variable order is presented. These are of left and right sides, applications follow. This new research strongly supports the new trend of Fractional Calculus of variable order, see [7]. It follows [4].

20.1 Introduction We are motivated by the following results. The next generalized Gronwall inequality for fractional differential equations (of constant) order was established in [8]: Theorem 20.1 Suppose that a (t) is a nonnegative function locally integrable on [0, T ) (for some T ≤ ∞), g (t) is a nonnegative, nondecreasing and bounded continuous function defined on [0, T ) and β0 > 0. If u (t) is nonnegative and locally integrable on [0, T ) satisfying 

t

u (t) ≤ a (t) + g (t)

(t − s)β0 −1 u (s) ds, 0 ≤ t < T,

(20.1)

0

then u (t) ≤ a (t) +

 t  ∞ [ (β0 ) g (t)]n 0

n=1

 (nβ0 )

 nβ0 −1

(t − s)

a (s) ds, 0 ≤ t < T, (20.2)

where  (·) is the Gamma function. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_20

315

316

20 A Study of Gronwall Inequalities of Fractional Variable Order

In [3] the authors generalized Theorem 20.1 for a fractional variable order, the first work in the literature of this kind: Theorem 20.2 ([3]) Suppose that a (t) is a nonnegative function locally integrable on [0, T ) for some T < ∞, g (t) is a nonnegative and nondecreasing function defined on [0, T ), and β (t) is a (Lebesgue) measurable function satisfying 0 < β0 ≤ β (t) ≤ A < ∞, 0 ≤ t < T.

(20.3)

If u (t) is nonnegative and locally integrable on [0, T ) satisfying the inequality 

t

u (t) ≤ a (t) + g (t)

(t − s)β(s)−1 u (s) ds, 0 ≤ t < T,

(20.4)

0

then u (t) ≤ a (t) +

∞ 

L k1 a (t) ≤

(20.5)

k=1

  t  ∞ [ (β0 ) K g (t)]n nβ0 −1 a (t) + a (s) ds, 0 ≤ t < T, (t − s)  (nβ0 ) 0 n=1 where L 1 is the Volterra-type (positive linear) operator 

t

L 1 φ (t) := g (t)

(t − s)β(s)−1 φ (s) ds, 0 ≤ t < T,

(20.6)

0

and K := max {1, T } A−β0 .

(20.7)

In this chapter the author applies the proving method of Theorem 20.2, and produces several interesting Gronwall type inequalities of fractional variable order covering other various important cases. Gronwall inequalities of fractional variable order are meant to find wide applications in the forthcoming new research of fractional differential equations of variable order, see also [7].

20.2 Main Results It is all about the Gronwall fractional variable order inequalities. We need to mention: Theorem 20.3 (Lemma 7.1.2, p. 189 in [5]) Suppose β, γ > 0 , β + γ > 1 and a, b ≥ 0, u is nonnegative, t γ−1 u (t) is locally integrable on [0, T ), and

20.2 Main Results

317



t

u (t) ≤ a + b

(t − s)β−1 s γ−1 u (s) ds,

(20.8)

0

a.e. in (0, T ); then

  1 u (t) ≤ a E β,γ (b (β)) ν t ,

where ν = β + γ − 1 > 0, E β,γ (s) =

∞ 

(20.9)

cm s mν ,

(20.10)

m=0

with c0 = 1,

cm+1 cm

=

(mν+γ) , (mν+γ+β)

for m ≥ 0. As s → ∞, we get that

    1 ν β βν −γ . s E β,γ (s) = O s 2 β exp ν

(20.11)

We present the following left inequality: Theorem 20.4 Suppose β, γ > 0, β + γ > 1 and a, b ≥ 0; u is nonnegative and (Lebesgue) measurable and t γ−1 u (t) is locally integrable: both on [0, T ). Let also 0 < β ≤ β (t) ≤ A < ∞ and 0 < γ ≤ γ (t) ≤ B < ∞, for 0 ≤ t < T . Denote by K 1 := max {1, T } A−β , K 2 := max {1, T } B−γ . The above functions β (t) and γ (t) are assumed to be (Lebesgue) measurable functions over [0, T ). Let also the positive linear operator 

t

L 2 φ (t) := b

(t − s)β(s)−1 s γ(s)−1 φ (s) ds, 0 ≤ t < T.

(20.12)

0

If

 u (t) ≤ a + b

t

(t − s)β(s)−1 s γ(s)−1 u (s) ds,

(20.13)

0

then u (t) ≤ a +

∞ 

   1 L k2 (a) (t) ≤ a E β,γ (bK 1 K 2  (β)) ν t ,

(20.14)

k=1

for every t ∈ [0, T ), where ν = β + γ − 1 and E β,γ (s) as in (20.10). Proof We observe that (t − s)β(s)−1 ≤ max {1, T } A−β (t − s)β−1 = K 1 (t − s)β−1 ,

(20.15)

s γ(s)−1 ≤ max {1, T } B−γ s γ−1 = K 2 s γ−1 , for all 0 ≤ s < t < T.

(20.16)

and

318

20 A Study of Gronwall Inequalities of Fractional Variable Order

Therefore we get that 

t

u (t) ≤ a + b

(t − s)β(s)−1 s γ(s)−1 u (s) ds ≤

(20.17)

0



t

a + bK 1 K 2

(t − s)β−1 s γ−1 u (s) ds.

0

By Theorem 20.3 we obtain that   1 u (t) ≤ a E β,γ (bK 1 K 2  (β)) ν t . We call B ∗ φ (t) := bK 1 K 2



t

(t − s)β−1 s γ−1 φ (s) ds.

(20.18)

(20.19)

0

As in the proof of Theorem 20.3 we have that B ∗n u (t) → 0, as n → ∞. Also we can write

u (t) ≤ a + B ∗ u (t) ,

(20.20)

and by induction as in the proof of Theorem 20.3 we derive that u (t) ≤ a

n 

cm (bK 1 K 2  (β))m t mν + B ∗n+1 u (t) .

(20.21)

m=0

But here it is 

t

L 2 u (t) = b

(t − s)β(s)−1 s γ(s)−1 u (s) ds ≤ B ∗ u (t) .

(20.22)

(t) = 0, 0 ≤ t < T.

(20.23)

0

Consequently it holds lim L n u n→∞ 2 Also we have u (t) ≤ a + L 2 u (t) , 0 ≤ t < T.

(20.24)

It is clear now that u (t) ≤ a +

∞  k=1

L k2 a ≤ a +

∞  k=1

  1 B ∗k a ≤ a E β,γ (bK 1 K 2  (β)) ν t ,

(20.25)

20.2 Main Results

319



proving the claim. Next we mention

Theorem 20.5 ([9]) Suppose that α, β > 0, a (t) is a nonnegative function locally integrable on R+ , g (t) and h (t) are nonnegative, nondecreasing continuous functions defined on R+ , and u (t) is a nonnegative and locally integrable function on R+ such that  t  t α−1 u (s) ds + h (t) u (t) ≤ a (t) + g (t) (t − s) (t − s)β−1 u (s) ds, t ∈ R+ . 0

0

(20.26)

Then, for each constant b ≥ 0, t ∈ [0, b], we have u (t) ≤ a (t) +  t ∞  n  n−k k n (g (t)  (α)) (h (t)  (β)) 0 n=1 k=0

 (kα + (n − k) β)

k

(t − s)kα+(n−k)β−1 a (s) ds. (20.27)

We present the following left inequality: Theorem 20.6 Suppose that α, β > 0, a (t) is a nonnegative function locally integrable on R+ , g (t) and h (t) are nonnegative, nondecreasing continuous functions defined on R+ , and u (t) is a nonnegative and locally integrable functions on R+ . Here α (t) , β (t) are (Lebesgue) measurable function on [0, T ), 0 < T < ∞, such that 0 < α ≤ α (t) ≤ A < ∞, 0 < β ≤ β (t) ≤ B < ∞, t ∈ [0, T ).

(20.28)

We call K 1 := max {1, T } A−α , K 2 := max {1, T } B−β .

(20.29)

Let also the positive linear operator 

t

L 3 φ (t) := g (t)

(t − s)

α(s)−1



t

φ (s) ds + h (t)

0

(t − s)β(s)−1 φ (s) ds,

0

(20.30)

t ∈ [0, T ). If 

t

u (t) ≤ a (t) + g (t) 0

t ∈ [0, T ), then

(t − s)α(s)−1 u (s) ds + h (t)

 0

t

(t − s)β(s)−1 u (s) ds, (20.31)

320

20 A Study of Gronwall Inequalities of Fractional Variable Order

u (t) ≤ a (t) +

∞ 

L k3 a (t) ≤ a (t) +

k=1

 t ∞  n  n−k k n (K 1 g (t)  (α)) (K 2 h (t)  (β))  (kα + (n − k) β)

k

0 n=1 k=0

(t − s)kα+(n−k)β−1 a (s) ds, (20.32)

for any t ∈ [0, T ). Proof We observe that (t − s)α(s)−1 ≤ K 1 (t − s)α−1 , and (t − s)β(s)−1 ≤ K 2 (t − s)β−1 ,

(20.33)

0 ≤ s < t < T. Therefore it holds (t ∈ [0, T )) 

t

u (t) ≤ a (t) + g (t)

(t − s)α(s)−1 u (s) ds + h (t)



0

 a (t) + K 1 g (t)

t

t

(t − s)β(s)−1 u (s) ds ≤

0

(t − s)α−1 u (s) ds + K 2 h (t)



0

t

(t − s)β−1 u (s) ds. (20.34)

0

By Theorem 20.5 we get that  t ∞  n  n−k k n (K 1 g (t)  (α)) (K 2 h (t)  (β))

u (t) ≤ a (t) +

0 n=1 k=0

k

 (kα + (n − k) β)

(t − s)kα+(n−k)β−1 a (s) ds =: ψ (t) , t ∈ [0, T ).

(20.35)

We call B ∗ φ (t) := K 1 g (t)



t

(t − s)α−1 φ (s) ds + K 2 h (t)

0



t

(t − s)β−1 φ (s) ds, t ∈ R+ .

0

(20.36) As in the proof of Theorem 20.5 we get for each t ∈ [0, T ), that lim B ∗n u (t) = 0. n→∞ Furthermore from there we have  t n  n−k k n (K 1 g (t)  (α)) (K 2 h (t)  (β)) B u (t) ≤ k  (kα + (n − k) β) 0 ∗n

k=0

(t − s)kα+(n−k)β−1 u (s) ds, t ∈ [0, T ). Also we can write

(20.34)

u (t) ≤ a (t) + B ∗ u (t) ,

(20.37)

20.2 Main Results

321

and by iteration we have that u (t) ≤

n−1 

B ∗k a (t) + B ∗n u (t) , t ∈ [0, T ).

(20.38)

k=0

But here it is

L 3 u (t) ≤ B ∗ u (t) , t ∈ [0, T ).

(20.39)

(t) = 0, 0 ≤ t < T.

(20.40)

a (t) + L 3 u (t) , 0 ≤ t < T.

(20.41)

Consequently it holds lim L n u n→∞ 3 Also we have u (t)

(20.30),(20.31)

≤

It is clear now that u (t) ≤ a (t) +

∞ 

L k3 a(t)

≤ a (t) +

k=1

∞ 

B ∗k a (t) ≤ ψ (t) , t ∈ [0, T ). (20.42)

k=1



The claim is proved. We mention

Theorem 20.7 ([6]) Let u, v be two integrable functions and g continuous, with domain [a, b]. Let ψ ∈ C 1 ([a, b]) an increasing function such that ψ  (t) = 0, ∀ t ∈ [a, b]. Assume that 1. u and v are nonnegative; 2. g is nonnegative and nondecreasing; α > 0. If  u (t) ≤ v (t) + g (t) +

t

ψ  (τ ) (ψ (t) − ψ (τ ))α−1 u (τ ) dτ ,

(20.43)

a

then u (t) ≤ v (t) +

 t ∞ [g (t)  (α)]k  ψ (τ ) [ψ (t) − ψ (τ )]αk−1 v (τ ) dτ , (20.44)  (αk) a k=1

∀ t ∈ [a, b] . We give the following left inequality: Theorem 20.8 Let u, v be two integrable functions and g continuous, with domain [a, b]. Let ψ ∈ C 1 ([a, b]) a strictly increasing function such that ψ  (t) = 0, ∀ t ∈ [a, b]. Assume that

322

20 A Study of Gronwall Inequalities of Fractional Variable Order

1. u and v are nonnegative; α > 0, 2. g is nonnegative and nondecreasing; α (t) is a (Lebesgue) measurable function over [a, b] such that 0 < α ≤ α (t) ≤ A < ∞, t ∈ [a, b] . We call (20.45) K ψ := max {1, ψ (b) − ψ (a)} A−α . Let also the positive linear operator 

t

L 4 φ (t) = g (t)

ψ  (τ ) (ψ (t) − ψ (τ ))α(τ )−1 φ (τ ) dτ , t ∈ [a, b] .

(20.46)

a

If

 u (t) ≤ v (t) + g (t)

t

ψ  (τ ) (ψ (t) − ψ (τ ))α(τ )−1 u (τ ) dτ ,

(20.47)

a

then u (t) ≤ v (t) +

∞ 

L k4 v (t) ≤ v (t) +

k=1

k  t ∞ K ψ g (t)  (α) a k=1

 (αk)

ψ  (τ ) [ψ (t) − ψ (τ )]αk−1 v (τ ) dτ ,

(20.48)

∀ t ∈ [a, b] . Proof Let a ≤ τ < t ≤ b. Since ψ is strictly increasing over [a, b], when τ < t we get that ψ (t) − ψ (τ ) > 0. In this case we have (ψ (t) − ψ (τ ))α(τ )−1 = (ψ (t) − ψ (τ ))α(τ )−α . (ψ (t) − ψ (τ ))α−1

(20.49)

If α (τ ) = α, then (ψ (t) − ψ (τ ))α(τ )−α = 1 = 1 A−α ≤ max {1, ψ (b) − ψ (a)} A−α . Let now a (τ ) > α, i.e. α (τ ) − α > 0. If ψ (t) − ψ (τ ) < 1, then (ψ (t) − ψ (τ ))α(τ )−α ≤ 1 = 1 A−α ≤ A−α . max {1, ψ (b) − ψ (a)} If ψ (t) − ψ (τ ) > 1, then (ψ (t) − ψ (τ ))α(τ )−α ≤ (ψ (t) − ψ (τ )) A−α ≤ (ψ (b) − ψ (a)) A−α ≤ max {1, ψ (b) − ψ (a)} A−α . If ψ (t) − ψ (τ ) = 1, then (ψ (t) − ψ (τ ))α(τ )−α = 1 = 1 A−α ≤ A−α . max {1, ψ (b) − ψ (a)} So we have proved that (ψ (t) − ψ (τ ))α(τ )−1 ≤ max {1, ψ (b) − ψ (a)} A−α = K ψ , (ψ (t) − ψ (τ ))α−1 that is

20.2 Main Results

323

(ψ (t) − ψ (τ ))α(τ )−1 ≤ K ψ (ψ (t) − ψ (τ ))α−1 ,

(20.50)

when a ≤ τ < t ≤ b. Therefore it holds (t ∈ [a, b]) 

t

u (t) ≤ v (t) + g (t)

ψ  (τ ) (ψ (t) − ψ (τ ))α(τ )−1 u (τ ) dτ

(20.50)

≤

a



t

v (t) + K ψ g (t)

ψ  (τ ) (ψ (t) − ψ (τ ))α−1 u (τ ) dτ .

(20.51)

a

By Theorem 20.7 we obtain

k  t ∞ K ψ g (t)  (α) ψ  (τ ) [ψ (t) − ψ (τ )]αk−1 v (τ ) dτ =: ϕ (t) , u (t) ≤ v (t) +  (αk) a k=1

(20.52) t ∈ [a, b] . We call 

t

Bφ (t) := K ψ g (t)

ψ  (τ ) [ψ (t) − ψ (τ )]α−1 φ (τ ) dτ ,

(20.53)

a

∀ t ∈ [a, b] . Then u (t) ≤ v (t) + Bu (t) ,

(20.54)

and by iterating (n ∈ N) we get u (t) ≤

n−1 

B k v (t) + B n u (t) , ∀ t ∈ [a, b] .

(20.55)

k=0

As in the proof of Theorem 20.7 [6] by mathematical induction we obtain ∗



B u (t) ≤ a

t



k K ψ g (t)  (α) ψ  (τ ) [ψ (t) − ψ (τ )]αk−1 u (τ ) dτ ,  (αk)

∀ k ∈ N, t ∈ [a, b] . Also from that proof we get B n u (t) → 0, as n → ∞. Here we have that L 4 u (t) ≤ Bu (t) , t ∈ [a, b] .

(20.56)

(20.57)

Consequently it holds lim L n u n→∞ 4

(t) = 0, t ∈ [a, b] .

(20.58)

324

20 A Study of Gronwall Inequalities of Fractional Variable Order

Also we have u (t) ≤ v (t) + L 4 u (t) , t ∈ [a, b] .

(20.59)

It is clear now that u (t) ≤ v (t) +

∞ 

L k4 v (t)

≤ v (t) +

k=1

∞ 

(20.52)

B k v (t) ≤ ϕ (t) , t ∈ [a, b] . (20.60)

k=1



The claim is proved. We need

Theorem 20.9 ([2]) Let u, v : [a, b] → R be two integrable functions, g, h : [a, b] → R be two continuous functions, ψ ∈ C 1 ([a, b]) be increasing function such that ψ  (t) = 0, for all t ∈ [a, b] , and α, β two positive real numbers. Suppose that 1. u and v are nonnegative; 2. g and h are nonnegative and nondecreasing; 3. the following inequality is true:  u (t) ≤ v (t) + g (t)

t

ψ  (τ ) (ψ (t) − ψ (τ ))α−1 u (τ ) dτ +

a

 h (t)

t

ψ  (τ ) (ψ (t) − ψ (τ ))β−1 u (τ ) dτ ,

(20.61)

a

for all t ∈ [a, b] . Then for all t ∈ [a, b]

u (t) ≤ v (t) +

 t ∞  n  n (g (t)  (α))k (h (t)  (β))n−k  ψ (τ ) [ψ (t) − ψ (τ )]kα+(n−k)β−1 v (τ ) dτ . k  (kα + (n − k) β) a n=1 k=0

(20.62) We give the following left inequality: Theorem 20.10 Let u, v : [a, b] → R be two integrable functions, g, h : [a, b] → R be two continuous functions, ψ ∈ C 1 ([a, b]) be a strictly increasing function such that ψ  (t) = 0, for all t ∈ [a, b] , and α, β > 0. Suppose that 1. u and v are nonnegative; 2. g and h are nonnegative and nondecreasing; 3. α (t) and β (t) are (Lebesgue) measurable functions on [a, b], such that 0 < α ≤ α (t) ≤ A < ∞, and 0 < β ≤ β (t) ≤ B < ∞, t ∈ [a, b].

20.2 Main Results

325

We call

K 1ψ := max {1, ψ (b) − ψ (a)} A−α , K 2ψ := max {1, ψ (b) − ψ (a)} B−β .

(20.63)

Let also the positive linear operator 

t

L 5 φ (t) = g (t)

ψ  (τ ) (ψ (t) − ψ (τ ))α(τ )−1 φ (τ ) dτ +

(20.64)

a



t

h (t)

ψ  (τ ) (ψ (t) − ψ (τ ))β(τ )−1 φ (τ ) dτ , t ∈ [a, b] .

a

If

 u (t) ≤ v (t) + g (t)

t

ψ  (τ ) (ψ (t) − ψ (τ ))α(τ )−1 u (τ ) dτ +

(20.65)

a

 h (t)

t

ψ  (τ ) (ψ (t) − ψ (τ ))β(τ )−1 u (τ ) dτ ,

a

then u (t) ≤ v (t) +

∞ 

L k5 v (t) ≤ v (t) +

k=1

k

n−k  t n 

∞  K 1ψ g (t)  (α) K 2ψ h (t)  (β) n a n=1 k=0

 (kα + (n − k) β)

k

ψ  (τ ) [ψ (t) − ψ (τ )]kα+(n−k)β−1 v (τ ) dτ ,

(20.66)

∀ t ∈ [a, b] . Proof Clearly here we have that (see proof of Theorem 20.8) (ψ (t) − ψ (τ ))α(τ )−1 ≤ K 1ψ (ψ (t) − ψ (τ ))α−1 , and (ψ (t) − ψ (τ ))β(τ )−1 ≤ K 2ψ (ψ (t) − ψ (τ ))β−1 , for all a ≤ τ < t ≤ b. Therefore it holds  u (t) ≤ v (t) + g (t)

t

ψ  (τ ) (ψ (t) − ψ (τ ))α(τ )−1 u (τ ) dτ +

a

 h (t) a

t

ψ  (τ ) (ψ (t) − ψ (τ ))β(τ )−1 u (τ ) dτ

(20.67)

≤

(20.67)

326

20 A Study of Gronwall Inequalities of Fractional Variable Order



t

v (t) + K 1ψ g (t)

ψ  (τ ) (ψ (t) − ψ (τ ))α−1 u (τ ) dτ +

a



t

K 2ψ h (t)

ψ  (τ ) (ψ (t) − ψ (τ ))β−1 u (τ ) dτ ,

(20.68)

a

∀ t ∈ [a, b] . By Theorem 20.9 we derive (20.68)

u (t) ≤ v (t) +

k

n−k  t ∞  n 

K 1ψ g (t)  (α) K 2ψ h (t)  (β) n  (kα + (n − k) β)

k

a n=1 k=0

ψ  (τ ) [ψ (t) − ψ (τ )]kα+(n−k)β−1 v (τ ) dτ =: λ (t) ,

(20.69)

∀ t ∈ [a, b] . We call 

t

Bφ (t) := K 1ψ g (t)

ψ  (τ ) (ψ (t) − ψ (τ ))α−1 φ (τ ) dτ +

a



t

K 2ψ h (t)

ψ  (τ ) (ψ (t) − ψ (τ ))β−1 φ (τ ) dτ ,

(20.70)

a

t ∈ [a, b] . Then

(20.68)

u (t) ≤ v (t) + Bu (t) ,

(20.71)

and by iterating (n ∈ N) we get u (t) ≤

n−1 

k

n

B v (t) + B u (t) , ∀ t ∈ [a, b] .

(20.72)

k=0

As in the proof of Theorem 20.9 [2] we get that n

B u (t) ≤

k

n−k  t n  K 1ψ g (t)  (α) K 2ψ h (t)  (β) n k  (kα + (n − k) β) a k=0

ψ  (τ ) [ψ (t) − ψ (τ )]kα+(n−k)β−1 u (τ ) dτ ,

(20.73)

∀ n ∈ N, t ∈ [a, b] . n Again as in the proof of Theorem 20.9 [2], we also derive that B u (t) → 0, as n → ∞. Also here we have

20.2 Main Results

327

L 5 u (t) ≤ Bu (t) , t ∈ [a, b] .

(20.74)

(t) = 0, t ∈ [a, b] .

(20.75)

v (t) + L 5 u (t) , t ∈ [a, b] .

(20.76)

Consequently it holds lim L n u n→∞ 5 Also we have u (t)

((20.64), (20.65))

≤

It is clear now that u (t) ≤ v (t) +

∞ 

L k5 v (t) ≤ v (t) +

k=1

∞ 

k

(20.69)

B v (t) ≤ λ (t) , t ∈ [a, b] . (20.77)

k=1



The theorem is proved. We mention a right fractional Gronwall inequality.

Theorem 20.11 ([1], case of ρ = 1 in Theorem 9 there) Let u, v be two integrable functions and g a continuous function, with domain [a, b]. Assume that 1. u and v are nonnegative; 2. g is nonnegative and nondecreasing, α > 0. If  b u (t) ≤ v (t) + g (t) (20.78) (τ − t)α−1 u (τ ) dτ , t

then



∞ b

u (t) ≤ v (t) + t

k=1

[g (t)  (α)]k (τ − t)kα−1 v (τ ) dτ ,  (kα)

(20.79)

∀ t ∈ [a, b] . In addition, if v is nondecreasing, then

u (t) ≤ v (t) E α g (t)  (α) (b − t)α , ∀ t ∈ [a, b] .

(20.80)

Above E α is the Mittag–Leffler function E α (t) :=

∞  k=0

tk , t ∈ R.  (αk + 1)

(20.81)

We give the following right Gronwall inequality of fractional variable order. Theorem 20.12 All as Theorem 20.11. Let also α (t) a (Lebesgue) measurable function on [a, b], such that 0 < α ≤ α (t) ≤ A < ∞, t ∈ [a, b]. Set

328

20 A Study of Gronwall Inequalities of Fractional Variable Order

K := max {1, b − a} A−α .

(20.82)

Let also the positive linear operator  L 6 φ (t) = g (t)

b

(τ − t)α(τ )−1 φ (τ ) dτ , t ∈ [a, b] .

(20.83)

t

If



b

u (t) ≤ v (t) + g (t)

(τ − t)α(τ )−1 u (τ ) dτ ,

(20.84)

t

then u (t) ≤ v (t) +

∞ 

L k6 v (t) ≤ v (t) +

k=1



∞ b t

k=1

(K g (t)  (α))k (τ − t)kα−1 v (τ ) dτ ,  (kα)

(20.85)

∀ t ∈ [a, b] . Proof Let a ≤ t < τ ≤ b. Then (τ − t)α(τ )−1 = (τ − t)α(τ )−α . (τ − t)α−1

(20.86)

If α (τ ) = α, then (τ − t)α(τ )−α = 1 = 1 A−α ≤ max {1, b − a} A−α . Let now a (τ ) > α, i.e. α (τ ) − α > 0. If τ − t < 1, then (τ − t)α(τ )−α ≤ 1 = A−α ≤ max {1, b − a} A−α . 1 If τ − t > 1, then (τ − t)α(τ )−α ≤ (τ − t) A−α ≤ (b − a) A−α ≤ A−α . max {1, b − a} If τ − t = 1, then (τ − t)α(τ )−α = 1 = 1 A−α ≤ max {1, b − a} A−α . So we have proved that (τ − t)α(τ )−1 ≤ max {1, b − a} A−α = K , (τ − t)α−1 that is

(τ − t)α(τ )−1 ≤ K (τ − t)α−1 ,

when a ≤ t < τ ≤ b. By (20.87), (20.84) we get 

b

u (t) ≤ v (t) + g (t) t

(τ − t)α(τ )−1 u (τ ) dτ ≤

(20.87)

20.2 Main Results

329

 v (t) + g (t) K

b

(τ − t)α−1 u (τ ) dτ ,

(20.88)

t

∀ t ∈ [a, b] . By Theorem 20.11 we obtain  u (t) ≤ v (t) + t

∞ b k=1

[K g (t)  (α)]k (τ − t)kα−1 v (τ ) dτ =: Q (t) ,  (kα)

∀ t ∈ [a, b] . Call

 φ (t) := K g (t)

b

(τ − t)α−1 φ (τ ) dτ ,

(20.89)

(20.90)

t

∀ t ∈ [a, b] . Then u (t) ≤ v (t) + u (t) ,

(20.91)

and by iterating (n ∈ N) we find u (t) ≤

n−1 

k v (t) + n u (t) , ∀ t ∈ [a, b] .

(20.92)

k=0

As in the proof of Theorem 20.1 [8] we have (inductively) that  k u (t) ≤ t

b

(K g (t)  (α))k (τ − t)kα−1 u (τ ) dτ ,  (kα)

∀ k ∈ N, and n u (t) → 0, as n → ∞, ∀ t ∈ [a, b] . Also we have that L 6 u (t) ≤ u (t) , t ∈ [a, b] .

(20.93)

(20.94)

Therefore it holds lim L n u n→∞ 6

(t) = 0, t ∈ [a, b] .

(20.95)

Notice also that u (t) ≤ v (t) + L 6 u (t) , t ∈ [a, b] .

(20.96)

It is clear now that u (t) ≤ v (t) +

∞  k=1

L k6 v (t) ≤ v (t) +

∞ 

(20.89)

k v (t) ≤ Q (t) , t ∈ [a, b] .

k=1

(20.97)

330

20 A Study of Gronwall Inequalities of Fractional Variable Order



The theorem is proved. Two applications follow:

Corollary 20.13 (to Theorem 20.10) All as in Theorem 20.10, with ψ (t) = et . Here  A−α  K 1ex := max 1, eb − ea , and   B−β K 2ex := max 1, eb − ea .

(20.98)

Let also the positive linear operator x L e5 φ (t)

 h (t)

t



t

= g (t)

α(τ )−1 eτ et − eτ φ (τ ) dτ +

(20.99)

a

β(τ )−1 eτ et − eτ φ (τ ) dτ , t ∈ [a, b] .

a

If

 u (t) ≤ v (t) + g (t)

t

α(τ )−1 eτ et − eτ u (τ ) dτ +

(20.100)

a

 h (t)

t

β(τ )−1 eτ et − eτ u (τ ) dτ ,

a

then u (t) ≤ v (t) +

∞ 

x

L e5

k

v (t) ≤ v (t) +

k=1

 t ∞  n  n−k k n (K 1ex g (t)  (α)) (K 2ex h (t)  (β)) a n=1 k=0

k

 (kα + (n − k) β)

kα+(n−k)β−1 eτ et − eτ v (τ ) dτ ,

(20.101)

∀ t ∈ [a, b] . Corollary 20.14 (to Theorem 20.10) All as in Theorem 20.10, with ψ (t) = ln t, [a, b] ⊆ R+ − {0}. Here  A−α  K 1 ln x := max 1, ln ab , and   B−β K 2 ln x := max 1, ln ab . Let also the positive linear operator

(20.102)

20.2 Main Results

331 x L ln 5 φ (t)

 t t α(τ )−1 φ (τ ) ln dτ + = g (t) τ τ a

 t h (t) a

If

t ln τ

β(τ )−1

(20.103)

φ (τ ) dτ , t ∈ [a, b] . τ

 t t α(τ )−1 u (τ ) ln dτ + u (t) ≤ v (t) + g (t) τ τ a  t h (t) a

then u (t) ≤ v (t) +

t ln τ

β(τ )−1

∞ 

x L ln 5

k

u (τ ) dτ , τ

v (t) ≤ v (t) +

k=1

 t ∞  n  n−k k n (K 1 ln x g (t)  (α)) (K 2 ln x h (t)  (β)) a n=1 k=0

k

 (kα + (n − k) β)  t kα+(n−k)β−1 v (τ ) ln dτ , τ τ

(20.104)

∀ t ∈ [a, b] .

References 1. Almeida, R.: A Gronwall inequality for a general Caputo fractional operator. Math. Inequal. Appl. 20(4), 1089–1105 (2017) 2. Almeida, R., Malinowska, A., Odzijewicz, T.: An extension of the fractional Gronwall inequality. In: Malinowska, A., Mozyrska, D., Sajewski, L. (eds.) Advances in non-integer order calculus and its Applications, Proceedings of the 10th International Conference of Non-integer Order Calculus and Its Applications, pp. 20–28. Springer (2020) 3. Anastassiou, G.A., Papanicolaou, V.G.: A Gronwall inequality of fractional variable order. Progr. Fract. Diff. Appl. accepted for publication (2020) 4. Anastassiou, G.A.: A variety of Gronwall Inequalities of Fractional Variable Order, submitted (2020) 5. Henry, D.: Geometric Theory of Semilinear Parabolic Equations. Springer, Heidelberg (1981) 6. Sousa, J., Oliveira, E.: A Gronwall inequality and the Cauchy-type problem by means of ψ-Hilfer operator. Diff. Equ. & Appl. 11(1), 87–106 (2019) 7. Yang, X.-I.: General Fractional Derivatives Theory, Methods and Applications. CRC Press, Taylor & Francis Group, Boca Raton (2019)

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20 A Study of Gronwall Inequalities of Fractional Variable Order

8. Ye, H., Gao, J., Ding, Y.: A generalized Gronwall inequality and its application to a fractional differential equation. J. Math. Anal. Appl. 328(4), 1075–1081 (2007) 9. Zhang, Z., Wei, Z.: A generalized Gronwall inequality and its application to fractional neutral evolution inclusions. J. Inequal. Appl. 2016, 45 (2016)

Chapter 21

General Ordinary and Fractional Approximation with Positive Sublinear Operators

Here we consider the ordinary and fractional approximation of functions by sublinear positive operators with applications to generalized convolution type operators expressed by sublinear integrals such as of Choquet and Shilkret ones. The fractional approximation is under fractional differentiability of Caputo, Canavati and IteratedCaputo types. We produce Jackson type inequalities under basic initial conditions. So our way is quantitative by producing inequalities with their right hand sides involving the modulus of continuity of ordinary and fractional derivatives of the function under approximation. We give also an application related to Picard singular integral operators. It follows [5].

21.1 Background—I Here we follow [3], pp. 1–17. Let I ⊂ R be a bounded or unbounded interval, n ∈ N, and   C B+n (I ) = f : I → R+ : f (i) is continuous and bounded on I , for both i = 0, n . (21.1) We define for f ∈ C B+ (I ) = { f : I → R+ : f is continuous and bounded on I } , the first modulus of continuity ω1 ( f, δ) = sup | f (x) − f (y)| , x,y∈I : |x−y|≤δ

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_21

(21.2) 333

334

21 General Ordinary and Fractional Approximation with Positive Sublinear Operators

where 0 < δ ≤ diameter (I ), also defined the same for just uniformly continuous functions f : I → R+ . Call C+ (I ) = { f : I → R+ : f is continuous on I } . Let L N : C+ (I ) → C B+ (I ), n, N ∈ N be a sequence of operators satisfying the following properties (see also [6], p. 17): (i) (positive homogeneous) L N (α f ) = αL N ( f ) , ∀ α ≥ 0, f ∈ C+ (I ) ,

(21.3)

(ii) (Monotonicity) if f, g ∈ C+ (I ) satisfy f ≤ g, then L N ( f ) ≤ L N (g) , ∀ N ∈ N,

(21.4)

and (iii) (Subadditivity) L N ( f + g) ≤ L N ( f ) + L N (g) , ∀ f, g ∈ C+ (I ) .

(21.5)

We call L N positive sublinear operators. In particular we consider the restrictions L N |C B+n (I ) : C B+n (I ) → C B+ (I ) . As in [6], p. 17, we get that for f, g ∈ C B+ (I ) , |L N ( f ) (x) − L N (g) (x)| ≤ L N (| f − g|) (x) , ∀ x ∈ I.

(21.6)

Furthermore, also from [6], p. 17, we have |L N ( f ) (x) − f (x)| ≤ L N (| f (·) − f (x)|) (x) + | f (x)| |L N (1) (x) − 1| , ∀ x ∈ I.

(21.7) Given that L N (1) = 1, ∀ N ∈ N, we get |L N ( f ) (x) − f (x)| ≤ L N (| f (·) − f (x)|) (x) , ∀ x ∈ I , ∀ N ∈ N.

(21.8)

We mention Hölder’s inequality for positive sublinear operators Theorem 21.1 ([3], p. 6) Let L : C+ (I ) → C B+ (I ), be a positive sublinear operator and f, g ∈ C+ (I ), furthermore let p, q > 1 : 1p + q1 = 1. Assume that     L ( f (·)) p (s∗ ) , L (g (·))q (s∗ ) > 0 for some s∗ ∈ I . Then     1   1 L ( f (·) g (·)) (s∗ ) ≤ L ( f (·)) p (s∗ ) p L (g (·))q (s∗ ) q .

(21.9)

  By assuming L N |· − x|n+1 (x) > 0, (21.9) and L N (1) = 1, we obtain       n L N |· − x|n (x) ≤ L N |· − x|n+1 (x) n+1 ,

(21.10)

21.1 Background—I

335

in case of n = 1 we derive L N (|· − x|) (x) ≤



   L N (· − x)2 (x) .

(21.11)

We mention also the following result. Theorem 21.2 ([3], p. 7) Let (L N ) N ∈N be a sequence of positive sublinear operators from C+ (I ) into C B+ (I ), and f ∈ C B+n (I ), f (n) could be only uniformly continuous, where n ∈ N and I ⊂ R a bounded or unbounded interval. Assume L N (1) = 1, ∀ N∈ N, and f (i) (x) = 0, i = 1, ..., n, for a fixed x ∈ I . Furthermore assume that L N |· − x|n+1 (x) > 0, ∀ N ∈ N. Then     1   ω1 f (n) , L N |· − x|n+1 (x) n+1 |L N ( f ) (x) − f (x)| ≤ · n!    n

   L N |· − x|n+1 (x) n+1 n L N |· − x| (x) + , ∀ N ∈ N. (21.12) (n + 1) We mention (n = 1 case) Corollary 21.3 ([3], p. 7) Let (L N ) N ∈N be a sequence of positive sublinear operators from C+ (I ) into C B+ (I ), and f ∈ C B+1 (I ), f could be only uniformly continuous, and I ⊂ R a bounded or unbounded interval. Assume L N (1) = 1, ∀ N ∈ N, and f (x) = 0, for a fixed x ∈ I . Furthermore assume that L N (· − x)2 (x) > 0, ∀ N ∈ N. Then       |L N ( f ) (x) − f (x)| ≤ ω1 f , L N (· − x)2 (x) · ⎡ ⎣ L N (|· − x|) (x) +



  ⎤ L N (· − x)2 (x) ⎦ , ∀ N ∈ N. 2

(21.13)

  Remark 21.4 ([3], p. 7) (i) to Theorem 21.2: Assuming L N |· − x|n+1 (x) → 0, as N → ∞, using (21.10), we get that (L N( f )) (x)→ f (x), as N → ∞. (ii) to Corollary 21.3: Assuming L N (· − x)2 (x) → 0, as N → ∞, using (21.11), we get that (L N ( f )) (x) → f (x), as N → ∞. (iii) The right hand sides of (21.12), (21.13) are finite. We also mention the basic result (n = 0 case). Theorem 21.5 ([3], p. 8) Let (L N ) N ∈N be a sequence of positive sublinear operators from C+ (I ) into C B+ (I ), and f ∈ C B+ (I ), f could be only uniformly continuous,

336

21 General Ordinary and Fractional Approximation with Positive Sublinear Operators

where I ⊂ R a bounded or unbounded interval. Assume that L N (|· − x|) (x) > 0, for some fixed x ∈ I , ∀ N ∈ N. Then (1) |L N ( f ) (x) − f (x)| ≤ f (x) |L N (1) (x) − 1| + [L N (1) (x) + 1] ω1 ( f, L N (|· − x|) (x)) , ∀ N ∈ N,

(21.14)

(2) when L N (1) = 1, we get |L N ( f ) (x) − f (x)| ≤ 2ω1 ( f, L N (|· − x|) (x)) , ∀ N ∈ N.

(21.15)

Remark 21.6 ([3], p. 8) (to Theorem 21.5) Here x ∈ I is fixed. (i) Assume L N (1) (x) → 1, as N → ∞, and L N (|· − x|) (x) → 0, as N → ∞, given that f is uniformly continuous we get that L n ( f ) (x) → f (x), as N → ∞ (use of (21.14)). Notice here that L N (1) (x) is bounded. (ii) Assume that L N (1) = 1, and L N (|· − x|) (x) → 0, as N → ∞, and f is uniformly continuous on I, then L n ( f ) (x) → f (x), as N → ∞ (use of (21.15)). (iii) The right hand sides of (21.14) and (21.15) are finite.

21.2 Background—II [4] Consider  = ∅ and let F be a σ -algebra in . Here μ is a set function μ : F → [0, +∞) which is monotone, i.e. for A, B ∈  : A ⊂ B we have μ (A) ≤ μ (B), furthermore it holds μ (∅) = 0. Here f, g : →R+ =[0, +∞) are F -measurable, we write it as f, g∈M (, R+ ). We consider a functional denoted by the integral symbol (S L) A f dμ, ∀ A ∈ F,  which is positive, i.e. A f dμ ≥ 0. We assume the following properties: (i) (positive homogeneous) 

 α f dμ = α (S L)

(S L) A

f dμ, ∀ α ≥ 0, ∀ f ∈ M (, R+ ) . A

(ii) (Monotonicity) if f, g ∈ M (, R+ ) satisfy f ≤ g, then (S L) (S L) A gdμ, ∀ A ∈ F. And (iii) (Subadditivity) 





f dμ + (S L)

( f + g) dμ ≤ (S L)

(S L) A

A

(iv)

gdμ, ∀ A ∈ F. A

 1dμ = μ (A) , ∀ A ∈ F.

(S L) A

 A

f dμ ≤

21.2 Background—II [4]

337

(v) If  = Rd , d ∈ N, we assume that μ is strictly positive, i.e. μ ( A) > 0, for any A compact subset of Rd . Here F = B the Borel σ -algebra. We call (S L) A f dμ a sublinear integral. We notice the following: f (x) = f (x) − g (x) + g (x) ≤ | f (x) − g (x)| + g (x) , hence 

 (| f (x) − g (x)| + g (x)) dμ (x) ≤

f (x) dμ (x) ≤ (S L)

(S L) A

A



 | f (x) − g (x)| dμ (x) + (S L)

(S L) A

g (x) dμ (x) , A

i.e. 





f (x) dμ (x) − (S L)

(S L) A

| f (x) − g (x)| dμ (x) .

g (x) dμ (x) ≤ (S L) A

A

Similarly, we get that 





g (x) dμ (x) − (S L)

(S L) A

| f (x) − g (x)| dμ (x) .

f (x) dμ (x) ≤ (S L) A

A

In conclusion, it holds        (S L)  ≤ (S L) | f (x) − g (x)| dμ (x) , f dμ − L) g dμ (x) (x) (S (x) (x)   A

A

A

(21.16)

∀ A ∈ F and ∀ f, g ∈ M (, R+ ) .

21.3 Background—III About the Choquet integral: We make Definition 21.7 Consider  = ∅ and let C be a σ-algebra of subsets in . (i) (see, e.g., [14], p. 63) The set function μ : C → [0, +∞] is called a monotone set function (or capacity) if μ (∅) = 0 and μ ( A) ≤ μ (B) for all A, B ∈ C, with A ⊂ B. Also, μ is called submodular if μ (A ∪ B) + μ ( A ∩ B) ≤ μ (A) + μ (B) , for all A, B ∈ C.

(21.17)

338

21 General Ordinary and Fractional Approximation with Positive Sublinear Operators

μ is called bounded if μ () < +∞ and normalized if μ () = 1. (ii) (see, e.g., [14], p. 233, or [8]) If μ is a monotone set function on C and if f :  → R is C-measurable (that is, for any Borel subset B ⊂ R it follows f −1 (B) ∈ C), then for any A ∈ C, the Choquet integral is defined by 

 f dμ =

(C) A

+∞







μ Fβ ( f ) ∩ A dβ +

0

0

−∞

    μ Fβ ( f ) ∩ A − μ (A) dβ,

(21.18) where we used the notation Fβ ( f ) = {ω ∈  : f (ω) ≥ β}. Notice that if f ≥ 0 on 0 A, then in the above formula we get −∞ = 0. The integrals on the right-hand side are the usual Riemann integral.  The function f will be called Choquet integrable on A if (C) A f dμ ∈ R. Next we list some well known properties of the Choquet integral. Remark 21.8 If μ : C → [0, +∞] is a monotone set function, then the following properties hold:   (i) For all a ≥ 0 we have (C) A a f dμ = a · (C) A f dμ (if f ≥ 0 then see, e.g., [14], Theorem 11.2, (5), p. 228 and if f is arbitrary sign, then see, e.g., [9], p. 64, Proposition 5.1, (ii)). (ii) For all c ∈R and f of arbitrary sign, we have (see, e.g., [14], pp. 232–233, or [9], p. 65) (C) A ( f + c) dμ = (C) A f dμ + c · μ ( A) . If μ is submodular too, then for all f, g of arbitrary sign and lower bounded, we have (see, e.g., [9], p. 75, Theorem 6.3) 





f dμ + (C)

( f + g) dμ ≤ (C)

(C) A

A

gdμ.

(21.19)

A

  (iii) If f ≤ g on A then (C) A f dμ ≤ (C) A gdμ (see, e.g., [14], p. 228, Theorem 11.2, (3) if f, g ≥ 0 and p. 232 if f, g are   of arbitrary sign). (iv) Let f ≥ 0. If A ⊂ B then (C) A f dμ ≤ (C) B f dμ. In addition, if μ is finitely subadditive, then 





f dμ ≤ (C)

(C) A∪B

f dμ + (C) A

f dμ.

(21.20)

B

 (v) It is immediate that (C) A 1 · dμ (t) = μ (A) . (vi)  If μ is a countably additive bounded measure, then the Choquet integral (C) A f dμ reduces to the usual Lebesgue type integral (see, e.g., [9], p. 62, or [14], p. 226). (vii) If  = Rd , d ∈ N, we assume μ is strictly positive, i.e. μ (A) > 0, for every A compact subset of Rd . Here C = B the Borel σ-algebra. Clearly here, for μ being submodular, we get

21.3 Background—III

339

       (C)  ≤ (C) | f (x) − g (x)| dμ (x) , f dμ − g dμ (x) (x) (C) (x) (x)   A

A

A

(21.21) ∀ A ∈ C and ∀ f, g ∈ M (, R+ ) ( f, g are measurable with respect to C σ -algebra). (viii) If f ≥ 0, then (C) A f dμ ≥ 0. From now on in this chapter we assume that μ : C → [0, +∞) and is submodular.

21.4 Background—IV Here we follow [13]. Let F be a σ-field of subsets of an arbitrary set . An extended non-negative real valued function μ on F is called maxitive if μ (∅) = 0 and μ (∪i∈I E i ) = supμ (E i ) ,

(21.22)

i∈I

where the set I is of cardinality at most countable, where {E i }i∈I is a disjoint collection of sets from F. We notice that μ is monotone and (21.22) is true even {E i }i∈I are not disjoint. For more properties of μ see [13]. We also call μ a maxitive measure. Here f stands for a non-negative measurable function on . In [13], Niel Shilkret developed his non-additive integral defined as follows: 

N∗



 f dμ := sup {y · μ (D ∩ { f ≥ y})} ,

(21.23)

y∈Y

D

where Y = [0, m] or Y = [0, m) with 0 < m ≤ ∞, and D ∈ F. Here we take Y = [0, ∞). It is easily proved that 

N∗



 f dμ = sup {y · μ (D ∩ { f > y})} .

(21.24)

y>0

D

The Shilkret integral takes values in [0, ∞]. The Shilkret integral [13]) has the following properties: 

N

∗



 

χ E dμ = μ (E) ,

where χ E is the indicator function on E ∈ F,      ∗ c f dμ = c N ∗ N D

f dμ, c ≥ 0, D

(21.25)

(21.26)

340

21 General Ordinary and Fractional Approximation with Positive Sublinear Operators



N∗





  sup f n dμ = sup N ∗ D n∈N

 f n dμ,

n∈N

(21.27)

D

where f n , n ∈ N, is an increasing sequence of elementary (countably valued) functions converging uniformly to f . Furthermore we have 

N

∗



 f dμ ≥ 0,

(21.28)

D



f ≥ g implies N

∗







f dμ ≥ N D

∗



 gdμ,

(21.29)

D

where f, g :  → [0, ∞] are measurable. Let a ≤ f (ω) ≤ b for almost every ω ∈ E, then 

aμ (E) ≤ N

∗



 f dμ ≤ bμ (E) ;

(21.30)

E



N

∗



 1dμ = μ (E) ;

(21.31)

E

 f > 0 almost everywhere and (N ∗ ) E f dμ = 0 imply μ (E) = 0;  (N ∗ )  f dμ = 0 if and only f = 0 almost everywhere; (N ∗ )  f dμ < ∞ implies that N ( f ) := {ω ∈ | f (ω) = 0} has σ-finite measure;     ∗  ∗  ∗ N f dμ + N gdμ; ( f + g) dμ ≤ N D

and

   ∗   N 

D



f dμ − N D

∗



 D

(21.32)

D

    ∗  | f − g| dμ. gdμ ≤ N

(21.33)

D

From now on in this chapter we assume that μ : F → [0, +∞). If  = Rd , d ∈ N, we assume μ is strictly positive, i.e. μ (A) > 0, for every A compact subset of Rd . Here F = B the Borel σ-algebra.  Conclusion 21.9 We observe that the Choquet integral (C) A f dμ and  Shilkret  integral (N ∗ ) A f dμ are perfect examples of the sublinear integral (S L) A f dμ of Sect. 21.2, fulfilling all properties and they have great applications in many areas of pure and applied mathematics and mathematical economics. Therefore, all the results presented in this chapter which are for the general integral  (S L) A f dμ are of course valid for the Choquet and Shilkret integrals.

21.5 Main Results

341

21.5 Main Results 21.5.1 Ordinary Approximation All terms, notations and assumptions here will be as in Backgrounds I–IV. So here it is f ∈ C n (R, R+ ) with f and f (n) are being bounded, but f (n) could be uniformly continuous regardless if it is bounded or not, n ∈ Z+ . Let B be the Borel σ-algebra on R and μ N (N ∈ N) be a sequence of monotone set functions from B into R+ , i.e. for A, B ∈ B : A ⊂ B we have μ N (A) ≤ μ N (B), and μ N (∅) = 0, with μ N (R) = 1, ∀ N ∈ N. We will study here the approximation properties of the following sequence of positive sublinear convolution type operators  PN ( f ) (x) = (S L)

R

f (x + t) dμ N (t) ,

(21.34)

∀ N ∈ N, to f (x), where x ∈ R is fixed, pointwise and uniform in a quantitative way. We would assume that PN ( f ) ∈ C B+ (R), ∀ N ∈ N. Clearly it holds PN (1) = 1. Notice here that    n+1 |· PN − x| (21.35) (x) = (S L) |t|n+1 dμ N (t) , R

∀ N ∈ N, where n ∈ Z+ . Based on the above we present Theorem 21.10 Assume further that f (i) (x) = 0, i = 1, ..., n, for a fixed x ∈ R and that  (S L) |t|n+1 dμ N (t) > 0, ∀ N ∈ N. R

Then

      |PN ( f ) (x) − f (x)| = (S L) f (x + t) dμ N (t) − f (x) ≤ R

(n + 2) ω1 (n + 1)!





f (n) , (S L)

 R

1  n+1 |t|n+1 dμ N (t)

n n+1   , ∀ N ∈ N. (S L) |t|n+1 dμ N (t)

(21.36)

R

Proof By Theorem 21.2 and (21.10).



342

21 General Ordinary and Fractional Approximation with Positive Sublinear Operators

Remark 21.11 If (S L) +∞.

 R

|t|n+1 dμ N (t) → 0, then PN ( f ) (x) → f (x), as N →

The n = 1 case follows: Corollary 21.12 Assume further that f (x) = 0, for a fixed x ∈ R and that  (S L)

R

t 2 dμ N (t) > 0, ∀ N ∈ N.

Then |PN ( f ) (x) − f (x)| ≤

3 ω1 2



1  1    2 2 t 2 dμ N (t) t 2 dμ N (t) , f , (S L) (S L) R

R

(21.37) ∀ N ∈ N.  If (S L) R t 2 dμ N (t) → 0, then PN ( f ) (x) → f (x), as N → +∞. 

Proof By Corollary 21.3 and (21.11). The case n = 0 comes next. Theorem 21.13 Assume that (S L)

 R

|t| dμ N (t) > 0, ∀ N ∈ N. Then 

PN ( f ) − f ∞ ≤ 2ω1

 f, (S L)

R

|t| dμ N (t) ,

(21.38)

∀ N ∈ N. Given that f is uniformly continuous andnot necessarily bounded, from R into R+ , then (21.38) is again valid and if (S L) R t 2 dμ N (t) → 0, then PN ( f ) → f , uniformly, as N → +∞. 

Proof By Theorem 21.5 (21.15). Application 21.14 Consider the well-known Picard singular integral operators: Pξ∗

1 ( f ) (x) := 2ξ



∞

−∞

|t|

f (x + t) e− ξ dt,

(21.39)

where ξ > 0. Here f is chosen so that Pξ∗ ( f ) (x) ∈ R, ∀ x ∈ R, e.g. f is bounded. Also Pξ∗ ( f ) is continuous when f is uniformly continuous. We notice that  ∞ |t| 1 e− ξ dt = 1, ξ > 0. (21.40) 2ξ −∞ In [12] they obtained the degree of convergence of the operators Pξ∗ to the unit operator I with rates over the class of Hölder-continuous functions as ξ → 0. In [11]

21.5 Main Results

343

they derived some more refined convergence to I (as ξ → 0), however only over the set of (C2π ) 2π-periodic continuous functions on R. See also [2], pp. 127–129 for bounded and uniformly continuous functions over R,we get uniform convergence of Pξ∗ → I. We consider here only ξ > 0 such that 1ξ = N ∈ N, as ξ → 0. Thus, it holds  N ∞ −|t|N e dt = 1, ∀ N ∈ N. (21.41) 2 −∞ Clearly here according to our theory dμ N (t) =

N −|t|N dt, ∀ N ∈ N, e 2

and PN∗ −1 ( f ) is a special case of PN ( f ) . We observe here that (n ∈ Z+ )  (S L)

R

|t|n+1 dμ N (t) =

N 2



∞ −∞

|t|n+1 e−|t|N dt = N



∞

t n+1 e−t N dt

(21.42)

0

N (n + 1)! (n + 1)! = → 0, as N → +∞. N n+2 N n+1  Therefore our special case of (S L) R |t|n+1 dμ N (t) generated from the Picard operators converges to zero as N → +∞, where n ∈ Z+ . Furthermore our results apply to Picard operators convergence to the unit I . This application realizes our general convergence theory making possible our assumptions. =

21.5.2 Fractional Approximation We need Definition 21.15 Let ν ≥ 0, n = ν (· is the ceiling of the number), f ∈ AC n ([a, b]) (space of functions f with f (n−1) ∈ AC ([a, b]), absolutely continuous functions). We call left Caputo fractional derivative of order ν > 0 (see [10], p. 49, [1], p. 394) the function ν D∗a f (x) =

1  (n − ν)



x

(x − t)n−ν−1 f (n) (t) dt, ∀ x ∈ [a, b] ,

a

∞ where  is the gamma function  (v) = 0 e−t t v−1 dt, v > 0. 0 f (x) = f (x), ∀ x ∈ [a, b] . We set D∗a

(21.43)

344

21 General Ordinary and Fractional Approximation with Positive Sublinear Operators

ν Exactly the same way one can define D∗x f over [x0 , +∞), x0 ∈ R, for f ∈ 0 n AC ([x0 , b]), ∀ b ∈ R, b ≥ x0 . We also need

Definition 21.16 (see also [2], p. 336) Let f ∈ AC r ([a, b]) , r = α, α ≥ 0. We right Caputo fractional derivative of order α > 0 is given by α f (x) = Db−

(−1)r  (r − α)



b

(ζ − x)r −α−1 f (r ) (ζ) dζ, ∀ x ∈ [a, b] .

(21.44)

x

0 f (x) = f (x). We set Db−

Exactly the same way one can define Dxα0 − f over (−∞, x0 ], x0 ∈ R, for f ∈ AC r ([a, x0 ]), ∀ a ∈ R : a ≤ x0 . We make Convention 21.17 We assume that

and

a f (x) = 0, for x < x0 , D∗x 0

(21.45)

Dxα0 − f (x) = 0, for x > x0 ,

(21.46)

for all x, x0 ∈ R. We also make Convention 21.18 Let a real number m > 0, from now on we assume that Dxm0 − f is either bounded or uniformly continuous function on (−∞, x0 ], similarly from now m f is either bounded or uniformly continuous function on on we assume that D∗x 0 [x0 , +∞). We need m f and Definition 21.19 Let Dxm0 f (real number m > 0) denote any of Dxm0 − f , D∗x 0 δ > 0. We set      m    f, δ [x ,+∞) , (21.47) ω1 Dxm0 f, δ R := max ω1 Dxm0 − f, δ (−∞,x ] , ω1 D∗x 0 0



where x0 ∈ R. Notice that ω1 Dxm0 f, δ

 R

0

< +∞.

We will use Theorem 21.20 ([3], p. 89) Let the real number m > 0, m ∈ / N, λ = m, x0 ∈ R, f ∈ AC λ ([a, b] , R+ ) (i.e. f (λ−1) ∈ AC [a, b] , absolutely continuous functions on [a, b]), ∀ [a, b] ⊂ R, and f (λ) ∈ L ∞ (R). Furthermore we assume that f (k) (x0 ) = 0, k = 1, ..., λ − 1. The Convention 21.18 is imposed here. Then

21.5 Main Results

345

    ω1 Dxm0 f, δ R |x − x0 |m+1 m |x − x0 | + | f (x) − f (x0 )| ≤ , δ > 0, (21.48)  (m + 1) (m + 1) δ for all x ∈ R, δ > 0. If 0 < m < 1, then we do not need initial conditions. We make Remark 21.21 Let L N , N ∈ N, be a sequence of positive sublinear operators from C+ (R) into C B+ (R). Here all are as in Theorem 21.20 for x = x0 and we can rewrite (21.48) as follows:     ω1 Dxm f, δ R |· − x|m+1 | f (·) − f (x)| ≤ |· − x|m + ,  (m + 1) (m + 1) δ

(21.49)

valid over R. Assume that L N (1) = 1, ∀ N ∈ N. By (21.8) we obtain |L N ( f ) (x) − f (x)| ≤   ω1 Dxm f, δ R  (m + 1)

    L N |· − x|m+1 (x) m L N |· − x| (x) + =: (ξ) . (m + 1) δ

(21.50)

  We also assume that L N |· − x|m+1 (x) > 0, ∀ N ∈ N. By (21.9) we get that

Choose

i.e.

      m L N |· − x|m (x) ≤ L N |· − x|m+1 (x) m+1 .

(21.51)

    1 δ := L N |· − x|m+1 (x) m+1 > 0,

(21.52)

  δ m+1 = L N |· − x|m+1 (x) .

Therefore we have     1     ω1 Dxm f, L N |· − x|m+1 (x) m+1 δ m+1 R δm + = (ξ) ≤  (m + 1) (m + 1) δ

(21.53)

    1      m  (m + 2) ω1 Dxm f, L N |· − x|m+1 (x) m+1 L N |· − x|m+1 (x) m+1 , R  (m + 2) ∀ N ∈ N.

346

21 General Ordinary and Fractional Approximation with Positive Sublinear Operators

We have proved Theorem 21.22 Let m > 0, m ∈ / N, λ = m, x ∈ R, f ∈ AC λ ([a, b] , R+ ), ∀ [a, b] ⊂ R, and f (λ) ∈ L ∞ (R). Furthermore we assume that f (k) (x) = 0, k = m m f , D∗x f are either bounded or uniformly con1, ..., λ − 1. We assume that Dx− tinuous over (−∞, x], [x, +∞), respectively. Let L N (N ∈ N) be a sequence of positive sublinear operators from C+ (R) into C B+ (R). Assume that L N (1) = 1,  and L N |· − x|m+1 (x) > 0, ∀ N ∈ N. Then     1   (m + 2) ω1 Dxm f, L N |· − x|m+1 (x) m+1 R  (m + 2)

|L N ( f ) (x) − f (x)| ≤ 

   m L N |· − x|m+1 (x) m+1 , ∀ N ∈ N.

(21.54)

  If L N |· − x|m+1 (x) → 0, then L N ( f ) (x) → f (x), as N → +∞. Next we specialize for L N = PN , ∀ N ∈ N, where PN is as in (21.34), see Sect. 21.5.1 for the full description. We give Corollary 21.23 Allare as in Theorem 21.22 regarding f . Assume that (S L) R |t|m+1 dμ N (t) > 0, and PN ( f ) ∈ C B+ (R), ∀ N ∈ N. Then       |PN ( f ) (x) − f (x)| = (S L) f (x + t) dμ N (t) − f (x) ≤ R



1  m+1   (m + 2) ω1 Dxm f, (S L) |t|m+1 dμ N (t)  (m + 2) R

R



 (S L)

If (S L)

 R

R

|t|m+1 dμ N (t)

m m+1

, ∀ N ∈ N.

(21.55)

|t|m+1 dμ N (t) → 0, then PN ( f ) (x) → f (x), as N → +∞.

Proof By Theorem  also that  21.22. Notice PN |· − x|m+1 (x) = (S L) R |t|m+1 dμ N (t) , ∀ N ∈ N.



We need Definition 21.24 (see [1], p. 24, and [2], p. 334) Let x, x0 ∈ R be such that x ≥ x0 , ν > 0, ν ∈ / N, such that p = [ν], [·] the integral part, α = ν − p (0 < α < 1). Let f ∈ C p (R) and define 

 Jνx0 f (x) :=

1  (ν)



x x0

(x − t)ν−1 f (t) dt, x0 ≤ x < +∞.

(21.56)

21.5 Main Results

347

the left generalized Riemann-Liouville fractional integral. / N, such that p = [ν], α = ν − p Let x, x0 ∈ R be such that x ≤ x0 , ν > 0, ν ∈ (0 < α < 1). Let f ∈ C p (R) and define 

 Jxν0 − f (x) :=

1  (ν)



x0

(z − x)ν−1 f (z) dz, − ∞ < x ≤ x0 .

(21.57)

x

the right generalized Riemann-Liouville fractional integral. We need Definition 21.25 (see also [7]) Let x, x0 ∈ R, x ≥ x0 , ν > 0, ν ∈ / N, p = [ν], α = ν − p.   p Let f ∈ Cb (R), i.e. f ∈ C p (R) with  f ( p) ∞ < +∞, where ·∞ is the supremum norm.   Here Jνx0 f (x) is defined via (21.56) over [x0 , +∞). p We define the subspace C xν0 + (R) of Cb (R) :   p x0 f ( p) ∈ C 1 ([x0 , +∞)) . C xν0 + (R) := f ∈ Cb (R) : J1−α For f ∈ C xν0 + (R), we define the left generalized ν-fractional derivative of f over [x0 , +∞) as  x0 ( p)  f . (21.58) Dxν0+ f = J1−α We need Definition 21.26 (see also [2], p. 345) Let / N, p =  x, x0∈ R, x ≤ x0 , ν > 0, ν ∈ p [ν], α = ν − p. Let f ∈ Cb (R). Here Jxν0 − f (x) is defined via (21.57) over (−∞, x0 ]. p We define the subspace of C xν0 − (R) of Cb (R) :     p f ( p) ∈ C 1 ((−∞, x0 ]) . C xν0 − (R) := f ∈ Cb (R) : Jx1−α 0− For f ∈ C xν0 − (R), we define the right generalized ν-fractional derivative of f over (−∞, x0 ] as   ν D x0 − f = (−1) p−1 Jx1−α f ( p) . (21.59) 0− We make m

Convention 21.27 Let a real number m > 1, from now on we assume that D x0 − f is either bounded or uniformly continuous function on (−∞, x0 ], similarly from now on we assume that Dxm0+ f is either bounded or uniformly continuous function on [x0 , +∞).

348

21 General Ordinary and Fractional Approximation with Positive Sublinear Operators

We use m

m

Definition 21.28 Let D x0 f (real number m > 1) denote any of D x0 − f , Dxm0 + f and δ > 0. We set     m     m , ω1 Dxm0 + f, δ [x ,+∞) , (21.60) ω1 D x0 f, δ := max ω1 D x0 − f, δ R

(−∞,x0 ]

0

 m  where x0 ∈ R. Notice that ω1 D x0 f, δ < +∞. R

We give Theorem 21.29 ([3], p. 113) Let m > 1, m ∈ / N, p = [m], x0 ∈ R, and f ∈ C xm0 + (R) ∩ C xm0 − (R). Assume that f (k) (x0 ) = 0, k = 1, ..., p − 1, and Dxm0+ f  m  (x0 ) = D x0 − f (x0 ) = 0. The Convention 21.27 is imposed. Then

| f (x) − f (x0 )| ≤

 m  ω1 D x0 f, δ   (m + 1)

R

|x − x0 |m +

 |x − x0 |m+1 , δ > 0, (21.61) (m + 1) δ

for all x ∈ R. We give m Theorem 21.30 Let m>1, m ∈ / N, p= m, x∈R, f : R→R+ with f ∈Cx+ (R) ∩    m m m (k) C x− (R). Assume that f (x) =0, k=1, ..., p − 1, and Dx+ f (x) = D x− f m

m f are either bounded or uniformly continuous (x) =0. We assume that D x− f , Dx+ over (−∞, x], [x, +∞), respectively. Let L N (N ∈ N) be a sequence of positive sublinear operators from C+ (R) into C B+ (R). Assume that L N (1) = 1, and L N |· − x|m+1 (x) > 0, ∀ N ∈ N. Then

|L N ( f ) (x) − f (x)| ≤ 

 m     1  (m + 2) ω1 D x f, L N |· − x|m+1 (x) m+1 R  (m + 2)

   m L N |· − x|m+1 (x) m+1 , ∀ N ∈ N.

(21.62)

  If L N |· − x|m+1 (x) → 0, then L N ( f ) (x) → f (x), as N → +∞. Proof Similar to the proof of Theorem 21.22, by using (21.61).



We have Corollary 21.31 All  are as in Theorem 21.30 regarding f , with L N = PN , ∀ N ∈ N. Assume that (S L) R |t|m+1 dμ N (t) > 0, and PN ( f ) ∈ C B+ (R), ∀ N ∈ N. Then

21.5 Main Results

349

     |PN ( f ) (x) − f (x)| = (S L) f (x + t) dμ N (t) − f (x) ≤ R



1  m+1   (m + 2) m m+1 ω1 D x f, (S L) |t| dμ N (t)  (m + 2) R

R



If (S L)

 R



 (S L)

R

|t|m+1 dμ N (t)

m m+1

, ∀ N ∈ N.

(21.63)

|t|m+1 dμ N (t) → 0, then PN ( f ) (x) → f (x), as N → +∞. 

Proof By Theorem 21.30.

21.5.3 Iterated Fractional Approximation α Here D∗x , Dxα0 − stand for the Caputo left and right fractional derivatives, see (21.43), 0 nα = (21.44). For n ∈ N, denote the iterated left and right fractional derivatives as D∗x 0 α α α nα α α α D∗x0 D∗x0 ...D∗x0 and Dx0 − = Dx0 − Dx0 − ...Dx0 − (n-times). We make Remark 21.32 Let f : R → R such that  f ∈ L ∞ (R), x0 ∈ R, 0 < α < 1. The α left Caputo fractional derivative D∗x0 f (x) is given for x ≥ x0 . Clearly it holds  α   α  D∗x0 f (x0 ) = 0, and we define D∗x f (x) = 0, for x < x0 . 0 kα Let us assume that D∗x0 f ∈ C ([x0 , +∞)), k = 0, 1, ..., n + 1; n ∈ N.   The right Caputo fractional derivative Dxα0 − f (x) is given for x ≤ x0 . Clearly it  α   α  holds Dx0 − f (x0 ) = 0, and define Dx0 − f (x) = 0, for x > x0 . Let us assume that Dxkα0 − f ∈ C ((−∞, x0 ]), k = 0, 1, ..., n + 1. 1 Here we restrict ourselves to n+1 < α < 1, that is λ := (n + 1) α > 1. We denote (n+1)α λ D∗x f := D∗x f , and Dxλ0 − f := Dx(n+1)α f. 0 0− 0

We make Convention 21.33 We assume that Dxλ0 − f is either bounded or uniformly continλ uous function on (−∞, x0 ], similarly we assume that D∗x f is either bounded or 0 uniformly continuous function on [x0 , +∞). We need λ Definition 21.34 Let Dxλ0 f denote any of Dxλ0 − f , D∗x f and δ > 0. We set 0

     ω1 Dxλ0 f, δ := max ω1 Dxλ0 − f, δ R

(−∞,x0 ]

  λ , ω1 D∗x f, δ 0

 [x0 ,+∞)

, (21.64)

350

21 General Ordinary and Fractional Approximation with Positive Sublinear Operators

  where x0 ∈ R. Notice that ω1 Dxλ0 f, δ < +∞. R

We mention 1 < α < 1, n ∈ N, λ := (n + 1) α > 1, f : Theorem 21.35 ([3], p. 137) Let n+1 kα R → R, f ∈ L ∞ (R), x0 ∈ R. Assume that D∗x f ∈ C ([x0 , +∞)), k = 0, 1, ..., 0  iα  n + 1, and D∗x0 f (x0 ) = 0, i = 2, 3, ..., n + 1. Suppose that Dxkα0 − f ∈   C ((−∞, x0 ]), for k = 0, 1, ..., n + 1, and Dxiα0 − f (x0 ) = 0, for i = 2, 3, ..., n + 1. Convention 21.33 is imposed. Then

  ⎡ ⎤ λ+1 ω1 Dxλ0 f, δ |x | − x 0   ⎦,  R ⎣|x − x0 |λ +  | f (x) − f (x0 )| ≤  λ+1 λ+1 δ

(21.65)

∀ x ∈ R, δ > 0. We present 1 Theorem 21.36 Let n+1 < α < 1, n ∈ N, λ := (n + 1) α > 1, f : R → R+ , f ∈ kα k = 0, 1, ..., n + 1, and  x0 ∈ R. Assume that D∗x f ∈ C ([x, +∞)), L ∞iα(R), kα f ∈ C ((−∞, x]), for k = D∗x f (x) = 0, i = 2, 3, ..., n + 1. Suppose that Dx−  iα  λ f (x) = 0, for i = 2, 3, ..., n + 1. We assume that Dx− f, 0, 1, ..., n + 1, and Dx− λ D∗x f are either bounded or uniformly continuous over (−∞, x], [x, +∞), respecoperators from C+ (R) tively. Let L N (N ∈ N) be a sequence of positive sublinear   into C B+ (R). Assume that L N (1) = 1, and L N |· − x|λ+1 (x) > 0, ∀ N ∈ N. Then       1  λ+2 λ+1 λ+1 λ  ω1 Dx f, L N |· − x| |L N ( f ) (x) − f (x)| ≤  (x) R  λ+2



   λ λ+1 L N |· − x|λ+1 (x) , ∀ N ∈ N.

(21.66)

  If L N |· − x|λ+1 (x) → 0, then L N ( f ) (x) → f (x), as N → +∞. Proof Similar to the proof of Theorem 21.22, by using (21.65).



We finish with Corollary 21.37 All are as in Theorem 21.36 regarding f and L N = PN , ∀ N ∈ N.  Assume that (S L) R |t|λ+1 dμ N (t) > 0, and PN ( f ) ∈ C B+ (R), ∀ N ∈ N. Then 

  1    λ+2 λ+1  ω1 Dxλ f, (S L) |t|λ+1 dμ N (t) |PN ( f ) (x) − f (x)| ≤  R  λ+2 R

21.5 Main Results

351

λ   λ+1 λ+1 |t| dμ , ∀ N ∈ N. L) (S N (t)

(21.67)

R

If (S L)

 R

|t|λ+1 dμ N (t) → 0, then PN ( f ) (x) → f (x), as N → +∞.

Proof By Theorem 21.36.



References 1. Anastassiou, G.A.: Fractional Differentiation Inequalities. Springer, New York (2009) 2. Anastassiou, G.A.: Intelligent Mathematics: Computational Analysis. Springer, Heidelberg (2011) 3. Anastassiou, G.A.: Nonlinearity: Ordinary and Fractional Approximation by Sublinear and Max-Product Operators. Springer, Heidelberg (2018) 4. Anastassiou, G.A.: Ostrowski type inequalities involving sublinear integrals. In: Anastassiou, G.A., Rassias, J.M. (eds.) Frontiers in Functional Equations and Analytic Inequalities, pp. 325–355. Springer, New York (2019) 5. Anastassiou, G.A.: Advanced Ordinary and Fractional Approximation by Positive Sublinear Operators, Filomat, accepted (2020) 6. Bede, B., Coroianu, L., Gal, S.: Approximation by Max-Product type Operators. Springer, Heidelberg (2016) 7. Canavati, J.A.: The Riemann-Liouville integral. Nieuw Archief Voor Wiskunde 5(1), 53–75 (1987) 8. Choquet, G.: Theory of capacities. Ann. Inst. Fourier (Grenoble) 5, 131–295 (1954) 9. Denneberg, D.: Non-additive Measure and Integral. Kluwer, Dordrecht (1994) 10. Diethelm, K.: The Analysis of Fractional Differential Equations. Springer, Heidelberg (2010) 11. Gal, S.G.: Remark on the degree of approximation of functions by singular integrals. Math. Nachr. 164, 197–199 (1993) 12. Mohapatra, R.N., Rodriguez, R.S.: On the rate of convergence of singular integrals for Hölder continuous functions. Math. Nachr. 149, 117–124 (1990) 13. Shilkret, Niel: Maxitive measure and integration. Indag. Math. 33, 109–116 (1971) 14. Wang, Z., Klir, G.J.: Fuzzy Measure Theory. Plenum, New York (1992)

Chapter 22

Complete Approximations with Multivariate Generalized Picard Singular Integrals

This research and survey chapter deals exclusively with the study of the approximation of generalized multivariate Picard singular integrals to the identity-unit operator. Here we study quantitatively most of their approximation properties. These operators are not in general positive linear operators. In particular we study the rate of convergence of these operators to the unit operator, as well as the related simultaneous approximation. These are given via Jackson type inequalities and by the use of multivariate high order modulus of smoothness of the high order partial derivatives of the involved function. Also we study the global smoothness preservation properties of these operators. These multivariate inequalities are nearly sharp and in one case the inequality is attained, that is sharp. Furthermore we give asymptotic expansions of Voronovskaya type for the error of approximation. The above properties are studied with respect to L p norm, 1 ≤ p ≤ ∞. It follows [3].

22.1 Introduction We start with our motivation for this chapter. The following come from [1], Chap. 10. In the next we mention the following univariate smooth Picard singular integral operators Pr,ξ ( f, x) defined as follows. For r ∈ N and n ∈ Z+ we set   ⎧ r ⎪ r− j ⎪ j −n , j = 1, ..., r, ⎨ (−1) j   αj = r  r ⎪ ⎪1 − j −n , j = 0, (−1)r − j ⎩ j j=1 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_22

(22.1)

353

354

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

that is

r 

α j = 1. Let f : R → R be Lebesgue measurable, we define for x ∈ R,

j=0

ξ > 0 the Lebesgue integral 1 Pr,ξ ( f ; x) := 2ξ



∞

−∞

⎞ ⎛ r |t| ⎝ α j f (x + jt)⎠ e− ξ dt.

(22.2)

j=0

We assume that Pr,ξ ( f ; x) ∈ R for all x ∈ R. ∞ |t| We notice by 2ξ1 −∞ e− ξ dt = 1 that Pr,ξ (c; x) = c, c constant and ⎞ ⎛ ∞ r |t| 1 ⎝ Pr,ξ ( f ; x) − f (x) = αj ( f (x + jt) − f (x))⎠ e− ξ dt. 2ξ j=0 −∞

(22.3)

Let f ∈ C n (R), n ∈ Z+ with the r th modulus of smoothness finite, i.e.     ωr f (n) , h := sup rt f (n) (x)∞,x < ∞, h > 0, |t|≤h

where rt f (n) (x) :=

r j=0

(−1)r − j

  r f (n) (x + jt) , j

(22.4)

(22.5)

see [6], p. 44. We need to introduce δk :=

r

α j j k , k = 1, ..., n ∈ N,

(22.6)

j=1

and the even function

|t|

G n (t) := 0

(|t| − w)n−1  (n)  ωr f , w dw, n ∈ N, (n − 1)!

(22.7)

with G 0 (t) := ωr ( f, |t|) , t ∈ R. Denote by · the integral part. We mention

(22.8)

22.1 Introduction

355

Theorem 22.1 It holds that      n2   1 ∞  t 2m  (2m)  Pr,ξ ( f ; x) − f (x) − f G n (t) e− ξ dt, n ∈ N. (x) δ2m ξ  ≤   ξ 0  m=1 (22.9) In L .H.S. (22.9) the sum collapse when n = 1. Corollary 22.2 Assume ωr ( f, ξ ) < ∞, ξ > 0. Then it holds for n = 0 that    Pr,ξ ( f ; x) − f (x) ≤ 1 ξ



∞

ωr ( f, t) e− ξ dt. t

(22.10)

0

Theorem 22.3 Inequality (22.9) at x = 0 is attained by f (x) = x r +n , r, n ∈ N with r + n even. Corollary 22.4 Inequality (22.10) is attained at x = 0 by f (x) = x r , r even. Note 22.5 The operators Pr,ξ are not in general positive and they are of convolution type. , α1 = −2, α2 = 18 . Consider f (t) = t 2 ≥ 0 and Let r = 2, n = 3. Then α0 = 23 8 x = 0. Then   Pr,ξ t 2 ; 0 = −3ξ 2 < 0. A related convergence theorem follows. Theorem 22.6 Let f ∈ C (R) uniformly continuous. Then    Pr,ξ ( f ) − f  ≤ 2r ω1 ( f, ξ ) . ∞

(22.11)

u

I.e. as ξ → 0 we get Pr,ξ → I uniformly, where I is the unit operator. In this chapter we study the approximation properties of multivariate smooth Picard singular integral operators [m] Pr,n ( f ; x1 , ..., x N ) :=

1 (2ξn ) N

r



α [m] j,r

j=0

RN

f (x1 + s1 j, x2 + s2 j, ..., x N + s N j) e−

N 



|si |

i=1 ξn

ds1 ...ds N , (22.12)

r, n ∈ N, m ∈ Z+ .

356

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

Observe that 1 (2ξn ) N





e−

N 



|si |

i=1 ξn

RN

ds1 ...ds N = 1,

(22.13)

see [1], Chap. 10. Here r ∈ N, m ∈ Z + , and

α [m] j,r

⎧   r ⎪ r− j ⎪ j −m , if j = 1, 2, ..., r, ⎨ (−1) j   := r  r −m ⎪ r −i ⎪ i , if j = 0, (−1) ⎩1 − i i=1

and [m] δk,r :=

r

k α [m] j,r j , k = 1, 2, ..., m ∈ N.

(22.14)

(22.15)

j=1

See that

r  j=0

α [m] j,r = 1.

Also here ξn ∈ (0, 1], n ∈ N, and f : R N → R is a Borel measurable function. [m] [m] is a special case of a more general operator θr,n studied in The above operator Pr,n general in [2] by the author. [m] . We mention next about θr,n Let μξn be a probability Borel measure on R N , N ≥ 1. We define the multiple smooth singular integral operators r

[m] θr,n ( f ; x1 , ..., x N ) :=

α [m] j.r

j=0

RN

f (x1 + s1 j, x2 + s2 j, ..., x N + s N j) dμξn (s) ,

(22.16) where s := (s1 , ..., s N ), x := (x1 , ..., x N ) ∈ R N . [m] are not in general positive. For example, consider the function The operators θr,n N  2 ϕ (u 1 , ..., u N ) = u i and also take r = 2 , m = 3; xi = 0, i = 1, ..., N . See that ϕ ≥ 0, however

i=1

⎛ [3] θ2,n

(ϕ; 0, 0, ..., 0) = ⎝

2

⎞ ⎠ j 2 α [3] j,2

j=1

  [3] [3] α1,2 + 4α2,2

assuming that

 RN

 N RN

i=1



N 

i=1

 si2

 N

RN

 si2

dμξn (s) =

i=1

    N 1 2 s dμξn (s) < 0. dμξn (s) = −2 + 2 R N i=1 i (22.17)

 si2 dμξn (s) < ∞.

22.1 Introduction

357

[m] Clearly in the case of Pr,n we have that 

1 e− (2ξn ) N

dμξn (s) =

N 



|si |

i=1 ξn

ds1 ...ds N , s ∈ R N .

[m] Lemma 22.7 The operator θr,n preserve the constant functions in N variables.

We need   Definition 22.8 Let f ∈ C B R N , the space of all bounded and continuous functions or uniformly continuous on R N . Then, the r th multivariate modulus of smoothness of f is given by (see, e.g. [4])  r 

ωr ( f ; h) := √ sup

u 1 ,u 2 ,...,u N

u 21 +...+u 2N ≤h

 ( f )∞ < ∞, h > 0,

(22.18)

where · ∞ is the sup-norm and ru f (x) := ru 1 ,u 2 ,...,u N f (x1 , ..., x N ) = r j=0

(−1)r − j

  r f (x1 + ju 1 , x2 + ju 2 , ..., x N + ju N ) . j

(22.19)

  Let m ∈ N and let f ∈ C m R N . Suppose that all partial derivatives of f of order m are bounded, i.e.  m   ∂ f (·, ·, ..., ·)     ∂ x α1 ...∂ x α N  1

for all α j ∈ Z+ , j = 1, ..., N ;

N 

N

< ∞,

(22.20)

∞

α j = m.

j=1 [m] In this chapter we apply the general theory developed in [2] about θr,n to the [m] operators Pr,n , so we can obtain computationally specific results and show that the general theory has applications and it is a valid theory. [m] we So for the very important in various branches of mathematics operators Pr,n prove the very essential properties of uniform approximation, L p approximation, global smoothness preservation and simultaneously approximation, Voronovskaya asymptotic expansions and complex simultaneous approximation.

358

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

22.2 Auxiliary Essential Results We need the following: N 

Theorem 22.9 Let r, N ∈ N, α j ∈ Z+ , j = 1, ..., N : |α| := (0, 1], n ∈ N. Then





(1 + N ) + 2 r

r

N 



1 u ∗ξn (α) := (2ξn ) N ξnm

α j = m ∈ N, ξn ∈

j=1

RN

|si |αi



i=1

e (m + r )! e

s 2 1+ ξn



r

e−

N 



|si |

i=1 ξn

N 

 ≤ (1 + N ) + 2 r

r

ds1 ...ds N ≤ e (m + r )! e

N , (22.21)

are uniformly bounded. Proof We have here that u ∗ξn (α) :=

1 (2ξn ) N

1 ξnN

ξnm

RN

N 

N R+

N R+

ξnm

N R+

⎡



ξnm

⎣

siαi

s 2 1+ ξn

i=1

1+

⎛

N 

 1+

i=1

[0,1] N i=1



e−

⎞r

r N  si

z iαi

N 

r

si ⎟ ⎜ ⎜ i=1 ⎟ − ⎜1 + ⎟ e ⎝ ξn ⎠

i=1 N 

s 2 1+ ξn



siαi

ξn



i=1



  N   si αi N R+ i=1

|si |αi

i=1

N 



1 ξnN

N 



ξn N

e

−

r zi

e

e−

−

N  i=1

N 

i=1 ξn

ξn

zi



|si |

ds1 ...ds N =

si

ds1 ...ds N ≤



ds1 ...ds N =

 d

s1 ξn



 ...d

dz 1 ...dz N =

i=1

 z iαi

1+

N i=1

r zi

e

−

N  i=1

(22.22)

si

 N   si i=1

N 

i=1 ξn



N 

i=1 ξn





r

zi

dz 1 ...dz N +

sN ξn

 =

22.2 Auxiliary Essential Results



359

N  (R+ −[0,1]) N i=1

⎡ ξnm ⎣(1 + N )r +

z iαi

1+

N 

(R+ −[0,1]) N i=1

z iαi

(1 + N ) + 2

⎤ zi

i=1

dz 1 ...dz N ⎦ ≤

(R+ −[0,1]) N i=1

(1 + N ) + 2 r

r

 (1 + N ) + 2 r

N

r zi

e

−

N 

⎤ zi

i=1

dz 1 ...dz N ⎦ ≤ (22.23)

i=1

N 



ξnm

N 

⎤  N r  N − zi z iαi z i e i=1 dz 1 ...dz N ⎦ ≤

N 

r

ξnm

e

−

i=1

r

zi

1+

(R+ −[0,1]) N i=1



r





ξnm ⎣(1 + N )r + 2r

N i=1



⎡

ξnm



r

z iαi

∞

i=1

1

N 

∞

1

 z ri

i=1

N 

i=1

N 

N 

e

−z i

N 

i=1

 dz i

=

i=1

 z iαi +r e−zi dz i

= 

z i((αi +r )+1)−1 e−zi dz i

(22.24)

(by [5], p. 348)  ξnm (1 + N )r + 2r

N 



((αi + r ) + 1, 1) ,

i=1

where (·, ·) is the upper incomplete gamma function. We have proved that  u ∗ξn

(α) ≤

ξnm

(1 + N ) + 2 r

r

N 



((αi + r ) + 1, 1) ≤

i=1

 ξnm

(1 + N )r + 2r



    e (m + r )! N e (m + r )! N < +∞, ≤ (1 + N )r + 2r e e

(22.25)

therefore u ∗ξn (α) are uniformly bounded. Above we used the formula

(s + 1, 1) =

es! , s ∈ N. e

(22.26)

360

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

Here αi + r ∈ N, hence

((αi + r ) + 1, 1) =

e (m + r )! e (αi + r )! ≤ , i = 1, ..., N . e e

(22.27) 

The claim is proved. We continue with Theorem 22.10 Let r, n, N ∈ N, ξn ∈ (0, 1]. Then ∗ξn :=



1 (2ξn ) N

RN

 

s 2 r − 1+ e ξn 

(1 + N )r + 2r



N 



|si |

i=1 ξn

ds1 ...ds N ≤

N

er ! e

< ∞,

(22.28)

are uniformly bounded. Proof We have that

1 ∗ξn = N ξn

N R+

 

s 2 r − 1+ e ξn

⎛ 1 ξnN

N R+





1+

N R+

si ⎟ ⎜ ⎜ i=1 ⎟ − ⎜1 + ⎟ e ⎝ ξn ⎠

r N  si ξn

i=1



 1+

N R+

N

e



i=1 ξn

ξn

i=1

r zi

e

−

ds1 ...ds N =



zi

i=1

ds1 ...ds N ≤

si

d

N 

 si



N 

 N   si

−

N 

i=1 ξn

s1 ξn



 ...d

sN ξn

 =

dz 1 ...dz N =

i=1

⎡





⎣

[0,1]

⎞r

N 



N

(R+ −[0,1]) N

1+

N

r zi

e

−

N 

zi

i=1

dz 1 ...dz N +

i=1

 1+

N i=1

r zi

e

−

N  i=1

⎤ zi

dz 1 ...dz N ⎦ ≤

(22.29)

22.2 Auxiliary Essential Results

⎡ ⎣(1 + N )r +

361



(R+ −[0,1]) N

1+

(R+ −[0,1]) N

(1 + N ) + 2

r

e

N 

⎤ zi

i=1

dz 1 ...dz N ⎦ ≤

i=1

(R+ −[0,1]) N

(1 + N )r + 2r

zi

−

 N r  N − zi i=1 zi e dz 1 ...dz N ≤ N 

r

r

i=1

(1 + N )r + 2r

N

 N 

i=1

N  i=1

z ri

∞ 1

 e

−z i

i=1

N 

dz i =

i=1

z i(r +1)−1 e−zi dz i =

(by [5], p. 348)  (1 + N )r + 2r N (r + 1, 1) = (1 + N )r + 2r

er ! e

N < +∞.

(22.30)

That is, ∗ξn are uniformly bounded.



We give Theorem 22.11 All as in Theorem 22.9 and p > 1. Then 1 Aξn (α) := (2ξn ) N



 N 

RN

|si |αi



i=1

s 2 1+ ξn

r  p



e−

N 



|si |

i=1 ξn

ds1 ...ds N ≤ (22.31)

! " ξnmp (1 + N )r p + 2r p N ((m + r ) p + 1, 1) ≤ (1 + N )r p + 2r p N ((m + r ) p + 1, 1) < +∞, are uniformly bounded. Proof We have here that

Aξn (α) :=

1 (2ξn ) N 1 ξnN





 N 

RN

1+

s 2 ξn

 

s 2 r p − α p si i 1 + e ξn i=1

N  N R+

i=1

|si |αi





r  p

N 

i=1 ξn



e−

N 



|si |

i=1 ξn

ds1 ...ds N =

 si

ds1 ...ds N ≤

(22.32)

362

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

⎛

1 ξnN

N R+

ξnmp

N 

N R+



⎣

[0,1] N i=1 N 

(R+ −[0,1])

N

1+ 

α p zi i

1+

si

ds1 ...ds N =

(1 + N ) + 2

 α p

zi i

1+



(1 + N ) + 2

r p zi

e

r p

N

N 

zi

e

α p zi i

N 

rp

(R+ −[0,1]) N i=1

 (1 + N ) + 2

ξnmp

N

(R+ −[0,1]) N i=1



e

N 

zi

i=1

dz 1 ...dz N =

rp

rp

−

N 

zi

i=1

dz 1 ...dz N +

−

N  i=1

⎤ zi

dz 1 ...dz N ⎦ ≤

 N r p N  N   −z i zi e dz i ≤ i=1

 α p zi i

N  i=1

N  i=1

∞ 1

i=1

 rp zi

N 

i=1

e

−z i

i=1

N 

 (1 + N ) + 2 rp

rp

N 

 dz i

=

i=1

 (α +r ) p −z i zi i e dz i

=

(by [5], p. 348) ξnmp

(22.33)

i=1

rp

rp

zi

−

i=1

rp

r p

N

i=1



ξnmp



N 

i=1 ξn

i=1

N 



ξnmp

 α p zi i

i=1

⎡ ξnmp



r p   N    N  N si  − si αi p si s1 sN ξn e i=1 d ...d = 1+ ξ ξ ξ ξn n n n i=1 i=1

ξnmp

si ⎟ ⎜ i=1 ⎟ α p⎜ − si i ⎜1 + ⎟ e ⎝ ⎠ ξ n i=1

N  N R+

⎞r p

N 



((αi + r ) p + 1, 1) ≤

i=1

(by [5], p. 909) ! " ξnmp (1 + N )r p + 2r p N ((m + r ) p + 1, 1) .

(22.34)

22.2 Auxiliary Essential Results

363

We have proved that ! " Aξn (α) ≤ ξnmp (1 + N )r p + 2r p N ((m + r ) p + 1, 1) ≤

(22.35)

(1 + N )r p + 2r p N ((m + r ) p + 1, 1) < +∞, are uniformly bounded. Above we used ([5], p. 909) the formula

α −x y

∞

(α, x y) = y e

e−t y (t + x)α−1 dt,

(22.36)

e−t (t + 1)(αi +r ) p dt ≤

(22.37)

0

where Rey > 0, x > 0, Reα > 1. We notice that (x = y = 1, αi ≤ m)

((αi + r ) p + 1, 1) = e−1



∞

0

e−1



∞

e−t (t + 1)(m+r ) p dt = ((m + r ) p + 1, 1) .

0

That is

((αi + r ) p + 1, 1) ≤ ((m + r ) p + 1, 1) , for all i = 1, ..., N . The theorem is proved.

(22.38) 

We give Theorem 22.12 Let r, n, N ∈ N, ξn ∈ (0, 1], p > 1. Then A∗ξn :=

1 (2ξn ) N

RN

 

s 2 r p − 1+ e ξn





N 

|si |

i=1 ξn

ds1 ...ds N ≤

(22.39)

(1 + N )r p + 2r p N (r p + 1, 1) < +∞, are uniformly bounded. Proof We have 1 A∗ξn = N ξn

N R+

 

s 2 r p − 1+ e ξn



N 

i=1 ξn

 si

ds1 ...ds N ≤

(22.40)

364

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

⎛

1 ξnN

N R+

si ⎟ ⎜ ⎜ i=1 ⎟ − ⎜1 + ⎟ e ⎝ ξn ⎠





N

1+

N R+

r p zi

e

 1+

N

N 

−

N 

i=1 ξn

zi

i=1

 si

ds1 ...ds N =

dz 1 ...dz N =

N

 (R+ −[0,1])

r p zi

e

−

N 

zi

i=1

dz 1 ...dz N +

i=1



1+

N

N

r p zi

(1 + N )r p + 2r p

(R+ −[0,1]) N

N 

zi

dz 1 ...dz N ≤

i=1

(22.41)

 N r p N N   zi e−zi dz i ≤ i=1

N 

(1 + N ) + 2

e

−

i=1



rp



i=1

[0,1]

⎞r p

N 

rp (R+ −[0,1]) N

(1 + N )r p + 2r p

i=1

 N  rp

zi

i=1

e

−z i

i=1

N  i=1

i=1



∞ 1

N 

dz i =

i=1

z i e−zi dz i = rp

(by [5], p. 348) (1 + N )r p + 2r p N (r p + 1, 1) . We have proved that A∗ξn ≤ (1 + N )r p + 2r p N (r p + 1, 1) < +∞,

(22.42) 

are uniformly bounded. We finish this section with Theorem 22.13 Let n, N ∈ N, ξn ∈ (0, 1], α j ∈ Z+ , j = 1, ..., N : |α| :=

N 

αj =

j=1

m ∈ N. Then ⎛ 1 Bξ∗n (α) := ξn−m ⎝ (2ξn ) N

RN

N 

 |si |αi



e−

N 



|si |

i=1 ξn

⎞ ds1 ...ds N ⎠ ≤ (m!) N .

i=1

(22.43)

22.2 Auxiliary Essential Results

365

Proof We have that 1 Bξ∗n (α) ≤ m N ξn ξn

 N N   si αi 

N R+ i=1

ξn

s

e

− ξni

i=1

N 

∞

0

i=1



N R+

N 

 siαi

e

N  − i=1

 si ξn

ds1 ...ds N =

(22.44)

i=1

s1 sN d ...d = ξn ξn

z iαi e−zi dz i =

N  i=1



N  N R+ i=1

∞

0

z iαi

N 

e−zi

i=1

N 

dz i =

i=1

z i(αi +1)−1 e−zi dz i

(22.45)

(by [5], p. 348) =

N 

(αi + 1) ≤ N (m + 1) = (m!) N ,

i=1

where is the gamma function. The last is valid because: if αi = 0, then (0 + 1) = 1; if αi ∈ N, then 2 ≤ αi + 1 ≤ m + 1, and (αi + 1) ≤ (m + 1) = m!. We took into account that has minimum at xmin = 1.46..., with (xmin ) = 0.885..., and (x) is increasing for x ≥ xmin . The theorem is proved. 

[m] 22.3 Main Results for Pr,n

22.3.1 Uniform Approximation [m] We start with an application to Pr,n of the following theorem.   Theorem 22.14 ([2], p. 11) Let m ∈ N, f ∈ C m R N , N ≥ 1, x ∈ R N . Assume  m  N   ∂ f (·,·,...,·)  αi = m. Let μξn be  ∂ x α1 ...∂ x α N  < ∞, for all α j ∈ Z+ , j = 1, ..., N : |α| := 1

N

∞

i=1

a Borel probability measure on R N , for ξn > 0, (ξn )n∈N bounded sequence. N  Suppose that for all α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| := αi = m i=1

we have that u ξn (α) :=

RN

N  i=1

|si |

αi



s 2 1+ ξn

r dμξn (s) < ∞.

(22.46)

366

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

For # j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αj = # j,

j=1

call

cα,n := cα,n,#j :=

Then (i)

N 

siαi dμξn (s1 , ..., s N ) .

R N i=1

(22.47)

 ⎞ ⎛     m ⎟  ⎜ c f (x) # α α,n, j ⎟  [m] [m] ⎜ [m] Er,n (x) := θr,n ( f ; x) − f (x) − δ#j,r ⎜ ⎟ N  ⎠ ⎝ $ #   α1 ,...,α N ≥0; j=1 α ! i   # |α|= j i=1



(ωr ( f α , ξn )) ≤ N  $ α1 ,...,α N ≥0; αi |α|=m

N 

 RN

αi



s 2 1+ ξn

|si |

i=1

r

 dμξn (s) .

(22.48)

i=1

∀ x ∈ RN . (ii)

 [m]  E  r,n

∞

≤ R.H.S.(22.48).

(22.49)

 Given  that ξn → 0, as n → ∞, and u ξn is uniformly bounded, then we derive that  E [m]  → 0 with rates. r,n (iii) It holds also that ⎞   m ⎜ c # f ⎟ α ∞⎟ α,n, j ⎜ ≤ δ#[m] ⎟ + R.H.S.(22.48), ⎜ j,r ⎝ N ⎠ $ # α1 ,...,α N ≥0; j=1 α ! i # ⎛

 [m]  θ ( f ) − f  r,n ∞

|α|= j

i=1

(22.50) given that f α ∞ < ∞, for all α : |α| = # j, # j = 1, ..., m. Furthermore, as ξn → 0 when n → ∞, assuming that cα,n,#j → 0, while u ξn is uniformly bounded, we conclude that  [m]  θ ( f ) − f  → 0 (22.51) r,n ∞ with rates. [m] A uniform approximation result for Pr,n follows:

  m     Theorem 22.15 Let m ∈ N, f ∈ C m R N , N ≥ 1, x ∈ R N . Assume  ∂∂ x αf1(·,·,...,·) αN  ...∂ x +

< ∞, for all α j ∈ Z , j = 1, ..., N : |α| :=

N 

j=1

1

N

∞

α j = m. Here ξn ∈ (0, 1] , n ∈ N.

[m] 22.3 Main Results for Pr,n

367

For all α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αi = m, we denote

i=1

u ∗ξn (α) :=

1 (2ξn ) N



RN

N 

|si |αi

 1+

i=1

r

s 2 ξn



e−

N 



|si |

i=1 ξn

ds1 ...ds N .

For # j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

(22.52) N 

αi = # j,

i=1

call ∗ ∗ cα,n := cα,n, := # j





1 (2ξn ) N

N  RN

siαi e−

N 



|si |

i=1 ξn

ds1 ...ds N .

(22.53)

i=1

Then (i)  ⎛ ⎞     m  ⎜ c∗ # f α (x) ⎟ α,n, j  [m] ⎟ [m] ⎜ ∗[m] Er,n (x) :=  Pr,n ( f ; x) − f (x) − δ#j,r ⎜ ⎟ N  ⎝ ⎠ $ # α1 ,...,α N ≥0;   j=1 α ! i   # |α|= j i=1 ≤



(ωr ( f α , ξn )) ∗ N  u ξn (α) , $ α1 ,...,α N ≥0; αi ! |α|=m

(22.54)

i=1

∀ x ∈ RN . (ii)

 ∗[m]   E r,n

∞

≤ R.H.S.(22.54).

(22.55)

Given that ξn → 0, as n → +∞, we have that u ∗ξn (α) → 0 and are uniformly  ∗[m]   → 0 with rates. bounded, and then we derive that  Er,n ∞ (iii) It holds also that ⎛

  [m] P ( f ) − f  r,n ∞

⎞     m  ⎜

f ⎟ c∗  [m]  ⎜  α,n,#j  α ∞ ⎟ ≤ δ#j,r  ⎜ ⎟ + R.H.S.(22.54), N ⎝ ⎠ $ # α ,...,α ≥0; 1 N j=1 αi ! # |α|= j

i=1

(22.56) given that f α ∞ < +∞, for all α : |α| = # j, # j = 1, ..., m. Furthermore, as ξn → 0 ∗ → 0 and u ∗ξn (α) → 0, and both are uniformly when n → +∞, we have that cα,n, # j bounded, and we conclude that

368

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

 [m]  P ( f ) − f  r,n

∞

→0

(22.57)

with rates. Proof Mainly by applying Theorem 22.14. By Theorem 22.9 we get that u ∗ξn (α) → 0 and u ∗ξn (α) are uniformly bounded. By Theorem 22.13 we get that  ∗   ∗  c  = c # ≤ α,n α,n, j

1 (2ξn ) N

N 

RN

 |si |αi



e−

N 



|si |

i=1 ξn

ds1 ...ds N

i=1

≤ (m!) N ξnm ≤ (m!) N . ∗ → 0 and are uniformly bounded. The proof is complete. That is cα,n, # j



We mention

  Theorem 22.16 ([2], p. 14) Let f ∈ C B R N , uniformly continuous, N ≥ 1, ξn ∈ (0, 1]. Then   [0] θ f − f  ≤ r,n ∞

 RN

  

s 2 r 1+ dμξn (s) ωr ( f, ξn ) , ξn

(22.58)

 

s 2 r 1+ dμξn (s) < ∞. ξn

(22.59)

under the assumption ξn :=

RN

As n → ∞ and ξn → 0, given that ξn are uniformly bounded, we derive   [0] θ f − f  → 0 r,n ∞

(22.60)

with rates. We give

  Theorem 22.17 Let f ∈ C B R N , uniformly continuous, N ≥ 1, ξn ∈ (0, 1]. Then   N   [0]  er !  P f − f  ≤ (1 + N )r + 2r ωr ( f, ξn ) , r,n ∞ e

(22.61)

where r, n ∈ N. As n → ∞ and ξn → 0, we derive   [0] P f − f  → 0 r,n ∞ with rates.

(22.62)

[m] 22.3 Main Results for Pr,n

369



Proof Use of Theorems 22.10 and 22.16.

[m] 22.3.2 L p Approximation for Pr,n

We need Definition 22.18 ([4, 6]) We call ru f (x) := ru 1 ,u 2 ,...,u N f (x1 , ..., x N ) := r

(−1)r − j

j=0

(22.63)

  r f (x1 + ju 1 , x2 + ju 2 , ..., x N + ju N ) . j

Let p ≥ 1, the modulus of smoothness of order r is given by   ωr ( f ; h) p := sup ru ( f ) p ,

(22.64)

u 2 ≤h

h > 0. We will apply     Theorem 22.19 ([2], p. 24) Let f ∈ C m R N , m ∈ N, N ≥ 1, with f α ∈ L p R N , |α| = m, x ∈ R N . Let p, q > 1 : 1p + q1 = 1. Here, μξn is a Borel probability measure on R N for ξn > 0, (ξn )n∈N bounded sequence. Assume for all α := (α1 , ..., α N ), N  αi ∈ Z+ , i = 1, ..., N , |α| := αi = m that we have i=1

RN

 N 

|si |

αi



i=1

s 2 1+ ξn

r  p dμξn (s) < ∞.

For # j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

(22.65) N 

αj = # j,

j=1

call

cα,n,#j :=

Then

RN

N  i=1

siαi dμξn (s) .

(22.66)

370

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

 ⎛ ⎞     m  ⎜ ⎟ [m] ⎜ cα,n,#j f α (x) ⎟  [m]   [m] E  =  δ#j,r ⎜  ⎟ N θr,n ( f ; x) − f (x) − r,n p   ⎝ ⎠ $ #   |α|=# j j=1 αi !   i=1  ≤



m 1

(q (m − 1) + 1) q

⎛

(22.67) p,x

⎞

⎜ 1 ⎟ ⎜ ⎟ ⎜ ⎟ N ⎝|α|=m $ ⎠ αi ! i=1

 RN

 N 

|si |αi

i=1



s 2 1+ ξn

 1p

r  p dμξn (s)

ωr ( f α , ξn ) p .

 [m]   → 0 with rates. As n → ∞ and ξn → 0, by (22.67) we obtain that  Er,n One also finds by (22.67) that ⎛

 [m]  θ ( f ; x) − f (x) r,n p,x

⎞   m  ⎜ ⎟  [m]  ⎜ cα,n,#j  ⎟

f α p ⎟ + R.H.S.(22.67), ≤ δ#j,r  ⎜ N ⎝ ⎠ $ # |α|=# j j=1 αi ! i=1

given that f α p < ∞, |α| = # j, # j = 1, ..., m.  [m]  Assuming that cα,n,#j → 0, ξn → 0, as n → ∞, we get θr,n ( f ) − f  p → 0, that [m] → I the unit operator, in L p norm, with rates. is θr,n We present

    Theorem 22.20 Let f ∈ C m R N , m ∈ N, N ≥ 1, with f α ∈ L p R N , |α| = m, x ∈ R N . Let p, q > 1 : 1p + q1 = 1, ξn ∈ (0, 1]. For # j = 1, ..., m, and N  α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| := αi = # j, call i=1

∗ cα,n, = # j

1 (2ξn ) N

RN



N 

siαi e−

N 



|si |

i=1 ξn

ds1 ...ds N .

(22.68)

i=1

Then  ⎛ ⎞     ∗ m  ⎜ ⎟ [m] ⎜ cα,n,#j f α (x) ⎟  ∗[m]    [m] E  =  P ( f ; x) − f (x) − δ#j,r ⎜ N  ⎟ r,n p  r,n  ⎝ ⎠ $ #   |α|=# j j=1 αi !   i=1

p,x

[m] 22.3 Main Results for Pr,n

 ≤

371



m (q (m − 1) + 1)

1 q

"1 ! (1 + N )r p + 2r p N ((m + r ) p + 1, 1) p

⎛

(22.69)

⎞

⎜ 1 ⎟ ⎜ ⎟ m ⎜ ⎟ ξn ωr ( f α , ξn ) p . N ⎝|α|=m $ ⎠ αi ! i=1

 ∗[m]   → 0 with rates. As n → +∞ and ξn → 0, by (22.69) we obtain that  Er,n p One also finds by (22.69) that ⎛

  [m]  P ( f ; x) − f (x) r,n p,x

⎞    ∗  m  ⎜ ⎟ c  [m]  ⎜  α,n,#j  ⎟

f α p ⎟ + R.H.S.(22.69), ≤ δ#j,r  ⎜ N ⎝ ⎠ $ # |α|=# j j=1 αi ! i=1

(22.70) given that f α p < ∞, |α| = # j, # j = 1, ..., m.  [m] [m] → Assuming that ξn → 0, as n → ∞, we get  Pr,n ( f ) − f  p → 0, that is Pr,n I the unit operator, in L p norm, with rates.    ∗  ∗ → 0. We use also Proof By Theorem 22.13 we get that cα,n,  ≤ (m!) N ξn and cα,n, # # j j  N  r  p  $ 2 |si |αi 1 + s

Theorem 22.11 to uniformly bound (2ξ1 ) N R N ξn n



e−

N 



i=1

|si |

i=1 ξn

ds1 ...ds N . Then we apply directly Theorem 22.19.



We continue with an application of      Theorem 22.21 ([2], p. 26) Let f ∈ C R N ∩ L p R N ; N ≥ 1; p, q > 1 : 1p + 1 = 1. Assume μξn probability Borel measure on R N , (ξn )n∈N > 0 and bounded. q Also suppose  

s 2 r p 1+ dμξn (s) < ∞. (22.71) ξn RN Then

  [0] θ ( f ) − f  ≤ r,n p  RN

(22.72)

 1p  

s 2 r p 1+ dμξn (s) ωr ( f, ξn ) p . ξn

 [0]  [0] → I, the unit As ξn → 0, when n → ∞, we derive θr,n ( f ) − f  p → 0, i.e. θr,n operator, in L p norm.

372

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

We give

     Theorem 22.22 Let f ∈ C R N ∩ L p R N ; N ≥ 1; p, q > 1 : (0, 1], n ∈ N. Then  [0]  P ( f ) − f  ≤ r,n

1 p

p

+

1 q

= 1, ξn ∈ (22.73)

! "1 (1 + N )r p + 2r p N (r p + 1, 1) p ωr ( f, ξn ) p .  [0]  [0] As ξn → 0, when n → ∞, we derive  Pr,n → I, the unit ( f ) − f  p → 0, i.e. Pr,n operator, in L p norm. 

Proof By Theorems 22.21 and 22.12. We mention

     Theorem 22.23 ([2], p. 27) Let f ∈ C R N ∩ L 1 R N ; N ≥ 1. Assume μξn probability Borel measure on R N , (ξn )n∈N > 0 and bounded. Also suppose RN

Then

 

s 2 r dμξn (s) < ∞. 1+ ξn   [0] θ ( f ) − f  ≤ r,n 1

 RN

(22.74)

(22.75)

  

s 2 r 1+ dμξn (s) ωr ( f, ξn )1 . ξn

[0] → I, in L 1 norm. As ξn → 0, we get θr,n

We give

     Theorem 22.24 Let f ∈ C R N ∩ L 1 R N ; N ≥ 1, ξn ∈ (0, 1], n ∈ N. Then   N   [0]   P ( f ) − f  ≤ (1 + N )r + 2r er ! ωr ( f, ξn )1 . r,n 1 e

(22.76)

[0] As ξn → 0, we get Pr,n → I, in L 1 norm.

Proof By Theorems 22.10 and 22.23. We mention



    Theorem 22.25 ([2], p. 29) Let f ∈ C m R N , m, N ∈ N, with f α ∈ L 1 R N , |α| = m, x ∈ R N . Here, μξn is a Borel probability measure on R N for ξn > 0, (ξn )n∈N is a bounded sequence. Suppose for all α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , N  |α| := αi = m that we have i=1

[m] 22.3 Main Results for Pr,n

373

 N 

RN

|si |

αi



i=1

s 2 1+ ξn

r  dμξn (s) < ∞.

(22.77)

For # j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αi = # j,

i=1

call

cα,n,#j :=

N 

R N i=1

siαi dμξn (s) .

(22.78)

Then  ⎞ ⎛     m   ⎟ ⎜  [m]  c f (x) # α α,n, j   ⎟ ⎜  E  = θ [m] ( f ; x) − f (x) − δ#[m]  ⎟ ⎜ r,n 1 r,n j,r ⎝ N  ⎠ $ #   |α|=# j j=1 α ! i   i=1

(22.79)

1,x

⎛

⎞

⎟ ⎜ ⎜ 1 ⎟ ≤ ⎜ N ⎟ ωr ( f α , ξn )1 ⎝$ ⎠ RN |α|=m αi !

N 

|si |

αi



s 2 1+ ξn

i=1

r dμξn (s) .

i=1

 [m]   → 0 with rates. As ξn → 0, we get  Er,n 1 From (22.79) we get ⎛

⎞   m  ⎜ ⎟   [m]  [m]  ⎜ cα,n,#j  ⎟ θ f − f  ≤

f α 1 ⎟ + R.H.S.(22.79), δ#j,r  ⎜ r,n 1 N ⎝ ⎠ $ # |α|=# j j=1 αi !

(22.80)

i=1

given that f α 1 < ∞, |α| = # j, # j = 1, ..., m.  [m]  As n → ∞, assuming ξn → 0 and cα,n,#j → 0, we obtain θr,n ( f ) − f 1 → 0, [m] → I in L 1 norm, with rates. that is θr,n We present

    Theorem 22.26 Let f ∈ C m R N , m, N ∈ N, with f α ∈ L 1 R N , |α| = m, x ∈ R N , ξn ∈ (0, 1], n ∈ N. For # j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = N  αi = # j, call 1, ..., N , |α| := i=1

∗ cα,n, := # j

1 (2ξn ) N





N  RN

i=1

siαi e−

N 



|si |

i=1 ξn

ds1 ...ds.

(22.81)

374

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

Then  ⎞ ⎛     ∗ m   ⎟ ⎜ c f (x)  ∗[m]  α # α,n, j   ⎟ ⎜ [m] E  =  P [m] ( f ; x) − f (x) − δ  ⎟ ⎜ r,n r,n # 1 j,r ⎝ N  ⎠ $ #   |α|=# j j=1 α ! i   i=1

1,x

⎛





≤ (1 + N )r + 2r

⎞

  ⎟ e (m + r )! N ⎜ ⎜ 1 ⎟ m ⎜ N ⎟ ξn ωr ( f α , ξn )1 . (22.82) ⎝$ ⎠ e |α|=m αi ! i=1

 ∗[m]   → 0 with rates. As ξn → 0, we get  Er,n 1 From (22.82) we get ⎛

⎞     ∗ m  ⎜ ⎟   [m]  [m]  ⎜ cα,n,#j  ⎟ P f − f  ≤

f α 1 ⎟ + R.H.S.(22.82), δ#j,r  ⎜ r,n 1 N ⎝ ⎠ $ # |α|=# j j=1 αi !

(22.83)

i=1

given that f α 1 < ∞, |α| = # j, # j = 1, ..., m.  [m]  [m] →I As n → ∞, assuming ξn → 0, we obtain  Pr,n ( f ) − f 1 → 0, that is Pr,n in L 1 norm, with rates. 

Proof We apply Theorems 22.9, 22.13 and 22.25.

22.3.3 Global Smoothness Preservation and Simultaneous [m] Approximation of Pr,n We need   Definition 22.27 ([2], p. 34) Let f ∈ C R N , N ≥ 1, m ∈ N, the mth modulus of smoothness for 1 ≤ p ≤ ∞, is given by    ωm ( f ; h) p := sup m t ( f ) p,x ,

t 2 ≤h

h > 0, where m t f (x) :=

m j=0

 (−1)m− j

m j

(22.84)

 f (x + jt) .

(22.85)

[m] 22.3 Main Results for Pr,n

375

Denote ωm ( f ; h)∞ = ωm ( f, h) .

(22.86)

Above, x, t ∈ R N . We present the related global smoothness preservation result m] [# Theorem 22.28 We assume Pr,n # ∈ Z+ , ∀ x ∈ R. Let h > 0, f ∈ ( f ; x) ∈ R, m  N C R , N ≥ 1. (i) Assume ωm ( f, h) < ∞. Then

ωm



⎞ ⎛ r     m] [# m] [# f, h ≤ ⎝ Pr,n α#j,r ⎠ ωm ( f, h) . 

(22.87)

# j=0

     (ii) Assume f ∈ C R N ∩ L 1 R N . Then 

m] [# ωm Pr,n f, h

 1

⎞ ⎛ r    [# m] ≤⎝ α#j,r ⎠ ωm ( f, h)1 .

(22.88)

# j=0

     (iii) Assume f ∈ C R N ∩ L p R N , p > 1. Then 

m] [# ωm Pr,n f, h

 p

⎞ ⎛ r     m] [# ≤⎝ α#j,r ⎠ ωm ( f, h) p .

(22.89)

# j=0

Proof Direct application of ([2]) Theorem 3.2, p. 35.



We make m] m] [# [# Remark 22.29 Let r = 1, m # ∈ Z+ , then α0,1 = 0, α1,1 = 1. Hence ⎞ N  |s | ⎜ i=1 i ⎟ −⎝ ξn ⎠ ⎛

m] [# P1,n ( f ; x) =

1 (2ξn ) N

f (x + s) e

RN

ds1 ...ds N =: Pn ( f ; x) . (22.90)

By Theorem 22.28, we get   Theorem 22.30 We suppose Pn ( f ; x) ∈ R, ∀ x ∈ R. Let h > 0, f ∈ C R N , N ≥ 1. (i) Assume ωm ( f, h) < ∞. Then ωm (Pn f, h) ≤ ωm ( f, h) .

(22.91)

376

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

     (ii) Assume f ∈ C R N ∩ L 1 R N . Then ωm (Pn f, h)1 ≤ ωm ( f, h)1 .

(22.92)

     (iii) Assume f ∈ C R N ∩ L p R N , p > 1. Then ωm (Pn f, h) p ≤ ωm ( f, h) p .

(22.93)

Next, we get an optimality result Proposition 22.31 Above inequality (22.91): ωm (Pn f, h) ≤ ωm ( f, h) is sharp, namely it is attained by any   f j∗ (x) = x mj , j = 1, ..., N , x = x1 , ..., x j , ..., x N ∈ R N .

(22.94) 

Proof Apply Proposition 3.5, p. 38, of [2]. We need

  Theorem 22.32 ([2], p. 39) Let f ∈ C l R N , l, N ∈ N. Here, μξn is a Borel probability measure on R N , ξn > 0, (ξn )n∈N a bounded sequence. Let β := (β1 , ..., β N ), N  βi = l. Here f (x + s j), x, s ∈ R N , is μξn -integrable βi ∈ Z+ , i=1, ..., N ; |β| := i=1

wrt s, for j = 1, ..., r . There exist μξn -integrable functions h i1 , j , h β1 ,i2 , j , h β1 ,β2 ,i3 , j , ..., h β1 ,β2 ,...,β N −1 ,i N , j ≥ 0 ( j = 1, ..., r ) on R N such that    ∂ i1 f (x + s j)     ≤ h i1 , j (s) , i 1 = 1, ..., β1 ,  i1   ∂ x1

(22.95)

   ∂ β1 +i2 f (x + s j)      ≤ h β1 ,i2 , j (s) , i 2 = 1, ..., β2 , β1 i2   ∂x ∂x 2

1

.. .    ∂ β1 +β2 +...+β N −1 +i N f (x + s j)      ≤ h β1 ,β2 ,...,β N −1 ,i N , j (s) , i N = 1, ..., β N ,  ∂ x i N ∂ x β N −1 ...∂ x β2 ∂ x β1  2 1 N N −1 ∀ x, s ∈ R N . Then, both of the next exist and     [# m] [# # ∈ Z+ . fβ ; x , m θr,nm ] ( f ; x) β = θr,n

(22.96)

[m] 22.3 Main Results for Pr,n

In particular it holds

377



m] [# Pr,n ( f ; x)

 β

when

  m] [# fβ ; x , = Pr,n

(22.97)

⎞ N  |s | ⎜ i=1 i ⎟ −⎝ ξn ⎠ ⎛

1 e (2ξn ) N

dμξn =

ds1 ...ds N .

Corollary 22.33 (by Theorem 22.32, r = 1) It holds   (Pn ( f ; x))β = Pn f β ; x .

(22.98)

We present simultaneous global smoothness results. Theorem 22.34 ⎛Let h⎞> 0 and the assumptions of Theorem 22.32 are valid for N 

|si | ⎟ ⎜ −⎝ i=1ξn ⎠

dμξn =

e ds1 ...ds N . Here γ = 0, β (0 = (0, ..., 0)), m # ∈ Z+ .   (i) Assume ωm f γ , h < ∞. Then 1 (2ξn ) N

ωm



m] [# Pr,n (f)

⎞ r       m [# ] ,h ≤ ⎝ α#j,r ⎠ ωm f γ , h . 

 γ

⎛

(22.99)

# j=0

  (ii) Additionally suppose f γ ∈ L 1 R N . Then ωm



m] [# Pr,n (f)

 γ

,h

⎞ r      [# m] ≤⎝ α#j,r ⎠ ωm f γ , h 1 . ⎛

 1

(22.100)

# j=0

  (iii) Additionally suppose f γ ∈ L p R N , p > 1. Then ωm



⎞ ⎛ r         m [# ] m] [# Pr,n (f) γ ,h ≤ ⎝ α#j,r ⎠ ωm f γ , h p . p

(22.101)

# j=0

It follows Corollary 22.35 (to Theorem 22.34) Let h > 0, r = 1 and γ = 0, β.  (i) Assume ωm f γ , h < ∞. Then     ωm (Pn ( f ))γ , h ≤ ωm f γ , h .

(22.102)

378

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

  (ii) Additionally suppose f γ ∈ L 1 R N . Then     ωm (Pn ( f ))γ , h 1 ≤ ωm f γ , h 1 .

(22.103)

  (iii) Additionally suppose f γ ∈ L p R N , p > 1. Then     ωm (Pn ( f ))γ , h p ≤ ωm f γ , h p .

(22.104)

Next comes multi-simultaneous approximation.   Theorem 22.36 Let f ∈ C m+l R N , m, l, N ∈ N. The assumptions of Theorem ⎞ N  |s | ⎜ i=1 i ⎟ −⎝ ξn ⎠ ⎛

22.32 are valid for dμξn =    f γ +α  < ∞, and let ∞ u ∗ξn (α) :=

1 (2ξn ) N

1 (2ξn ) N

N 

RN

ds1 ...ds N . Call γ = 0, β. Assume

e

|si |αi

 1+

i=1

for all α j ∈ Z+ , j = 1, ..., N , |α| :=

N 

s 2 ξn

r



e−

N 



|si |

i=1 ξn

ds1 ...ds N ,

α j = m, ξn ∈ (0, 1].

j=1

For # j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αi = # j,

i=1

call ∗ cα,n, := # j

1 (2ξn ) N





N 

R N i=1

siαi e−

N 



|si |

i=1 ξn

ds1 ...ds N .

Then  ⎞ ⎛     ∗ m   ⎟ ⎜ c f (·)  α,n,# j γ +α  [m] ⎟ [m] ⎜ δ#j,r ⎜   Pr,n ( f ; ·) γ − f γ (·) − ⎟ N  ⎠ ⎝ $ # α1 ,...,α N ≥0;   j=1 α ! i   |α|=# j i=1

(22.105)

(22.106) ∞

   ωr f γ +α , ξn ≤ N  u ∗ξn (α) . $ α1 ,...,α N ≥0; αi ! |α|=m i=1

Proof Based on Theorems 22.15 and 22.32. We continue with



[m] 22.3 Main Results for Pr,n

379

  Theorem 22.37 Let f ∈ C lB R N , l, N ∈ N (functions l-times continuously differentiable⎛and bounded). The assumptions of Theorem 22.32 are valid for dμξn = ⎞ N 

|si | ⎟ ⎜ −⎝ i=1ξn ⎠

1 (2ξn ) N

ds1 ...ds N . Call γ = 0, β, ξn ∈ (0, 1]. Then

e

   [0]    Pr,n f γ − f γ 





≤ (1 + N ) + 2 r

∞

r

er ! e

N 

  ωr f γ , ξn .

(22.107)

 [0]  If additionally f γ is uniformly continuous and ξn → 0, the Pr,n f γ → f γ uniformly, as n → ∞. 

Proof By Theorems 22.17 and 22.32. We continue with

  Theorem 22.38 Let f ∈ C m+l R N , m, l, N ∈ N. The assumptions of Theorem ⎞ N  |s | ⎜ i=1 i ⎟ −⎝ ξn ⎠ ⎛

22.32 are valid for dμξn = (2ξ1 ) N e ds1 ...ds N . Call γ = 0, β. Let f (γ +α) ∈ n  N N L p R , |α| = m, x ∈ R , and p, q > 1 : 1p + q1 = 1, ξn ∈ (0, 1], n ∈ N. For N  # αi = # j, call j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| := i=1

1 ∗ cα,n, := # j (2ξn ) N Then





N 

R N i=1

siαi e−

N 



|si |

i=1 ξn

ds1 ...ds N .

 ⎞ ⎛     ∗ m   ⎟ ⎜ c f (x)  α,n,# j γ +α  [m] ⎟ [m] ⎜ δ#j,r ⎜   Pr,n ( f ; x) γ − f γ (x) − ⎟ N  ⎠ ⎝ # $ #   |α|= j j=1 α ! i   i=1  ≤



m (q (m − 1) + 1)

1 q

(22.108)

p,x

"1 ! (1 + N )r p + 2r p N ((m + r ) p + 1, 1) p ⎞

⎛

⎜  1 ⎟ ⎟ m  ⎜ ⎟ ξn ωr f γ +α , ξn p . ⎜ N ⎠ ⎝|α|=m $ αi !

(22.109)

i=1

Proof By Theorems 22.20 and 22.32.



380

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

We give also   Theorem 22.39 Let f ∈ C l R N , l, N ∈ N. The assumptions of Theorem 22.32 are ⎛

⎞ N  |s | ⎜ i=1 i ⎟ −⎝ ξn ⎠

valid for dμξn = R N ; p, q > 1 :

1 e (2ξn ) N 1 1 +q = p

  ds1 ...ds N . Call γ = 0, β. Let f γ ∈ L p R N , x ∈

1, ξn ∈ (0, 1], n ∈ N. Then    [0]    Pr,n f γ − f γ  ≤ p

!   "1 (1 + N )r p + 2r p N (r p + 1, 1) p ωr f γ , ξn p .

(22.110)

 [0]  · p As n → +∞ and ξn → 0, then Pr,n ( f ) γ → fγ . 

Proof By Theorems 22.22 and 22.32. We continue with

  Theorem 22.40 Let f ∈ C l R N , l, N ∈ N. The assumptions of Theorem 22.32 are ⎞ N  |s | ⎜ i=1 i ⎟ −⎝ ξn ⎠ ⎛

valid for dμξn =

1 (2ξn ) N

e

  ds1 ...ds N . Call γ = 0, β. Let f γ ∈ L 1 R N , x ∈

R N , ξn ∈ (0, 1], n ∈ N. Then      [0]  Pr,n ( f ) γ − f γ  ≤ 1



 (1 + N )r + 2r

er ! e

N 

  ωr f γ , ξn 1 .

(22.111)

 [0]  · 1 As n → +∞ and ξn → 0, then Pr,n ( f ) γ → fγ . 

Proof By Theorems 22.24 and 22.32. We continue with

  Theorem 22.41 Let f ∈ C m+l R N , m, l, N ∈ N. The assumptions of Theorem ⎞ N  |s | ⎜ i=1 i ⎟ −⎝ ξn ⎠ ⎛

22.32 for dμξn = (2ξ1 ) N e ds1 ...ds N are valid. Call γ = 0, β. Let f (γ +α) ∈ n  N N L 1 R , |α| = m, x ∈ R , ξn ∈ (0, 1], n ∈ N. For # j = 1, ..., m, and α := N  αi = # j, call (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| := i=1

[m] 22.3 Main Results for Pr,n

381

1 ∗ cα,n, := # j (2ξn ) N





N 

R N i=1

siαi e−

N 



|si |

i=1 ξn

ds1 ...ds N .

 ⎞ ⎛     ∗ m  ⎟ ⎜ [m] ⎜ cα,n,#j f (γ +α) (x) ⎟    [m] δ#j,r ⎜  Pr,n ( f ; x) γ − f γ (x) − ⎟ N   ⎠ ⎝ $ #   |α|=# j j=1 αi !   i=1

Then

⎛

⎞

⎛

(22.112)

1,x

⎞

    ⎜ ⎜ 1 ⎟   ⎟ e (m + r )! N ⎜ ⎟ ⎜ ⎟ m r r ≤⎜ . ⎟ ωr f γ +α , ξn 1 ⎟ ξn (1 + N ) + 2 ⎜ N ⎝|α|=m ⎝ $ ⎠ ⎠ e αi ! i=1

(22.113) 

Proof By Theorems 22.26 and 22.32.

[m] 22.3.4 Voronovskaya Asymptotic Expansions for Pr,n

We will apply   Theorem 22.42 ([2], p. 53) Let f ∈ C m R N , m, N ∈ N, with all f α ∞ ≤ M, M > 0, all α : |α| = m. Let ξn > 0, (ξn )n∈N bounded sequence, μξn probability Borel measures on R N . N   $ αi Call cα,n,#j := R N si dμξn (s) , all |α| = # j = 1, ..., m − 1. Suppose i=1 N   $ |si |αi dμξn (s) ≤ ρ, all α : |α| = m, ρ > 0, for any such (ξn )n∈N . Also ξn−m R N i=1

0 < γ ≤ 1, x ∈ R N . Then ⎛ [m] θr,n ( f ; x) − f (x) =

⎞

⎜ c # f (x) ⎟   α,n, j α ⎜ ⎟ δ#[m] N  ⎟ + 0 ξnm−γ . ⎜ j,r ⎝ ⎠ $ # |α|=# j j=1 αi !

m−1

i=1

When m = 1, the sum collapses. [m] Above we assume θr,n ( f ; x) ∈ R, ∀ x ∈ R N . We give

(22.114)

382

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

  Theorem 22.43 Let f ∈ C m R N , m, N ∈ N, with all f α ∞ ≤ M, M > 0, all ⎞ N  |s | ⎜ i=1 i ⎟ −⎝ ξn ⎠ ⎛

α : |α| = m, ξn ∈ (0, 1], and dμξn = ( f ; x) ∈ R, ∀ x ∈ R N . ∗ = Call cα,n, # j

1 (2ξn ) N

1 (2ξn ) N

[m] ds1 ...ds N on R N . Assume Pr,n

e

⎞ N  |s | ⎜ i=1 i ⎟ −⎝ ξn ⎠ ⎛



 RN

N $

i=1

 siαi e

ds1 ...ds N , all |α| = # j = 1, ..., m −

1. Let 0 < γ ≤ 1, x ∈ R N . Then ⎛ [m] Pr,n ( f ; x) − f (x) =

⎞

⎜ c∗ # f α (x) ⎟   α,n, j ⎜ ⎟ δ#[m] N  ⎟ + 0 ξnm−γ . ⎜ j,r ⎝ ⎠ $ # |α|=# j j=1 αi !

m−1

(22.115)

i=1

When m = 1, the sum collapses. Proof By Theorems 22.13 and 22.42. Here ρ = (m!) N .       Corollary 22.44 (to Theorem 22.43) Let f ∈ C 1 R N , N ≥ 1, with all  ∂∂xfi  M, M > 0, i = 1, ..., N . Let 0 < γ ≤ 1, x ∈ R . Then N

  [1] Pr,n ( f ; x) − f (x) = 0 ξn1−γ .

∞

 ≤

(22.116)

[1] Above we assume Pr,n ( f ; x) ∈ R, ∀ x ∈ R N .

Proof By Theorems 22.13 and 22.42, here it is ρ = 1.   2   2        Corollary 22.45 (to Theorem 22.43) Let f ∈ C 2 R2 , with all  ∂∂ x 2f  ,  ∂∂ x 2f  , 1 2 ∞ ∞  2   ∂ f   ∂ x1 ∂ x2  ≤ M, M > 0, ξn ∈ (0, 1], n ∈ N. Call ∞

c1 =



R2

s1 dμξn (s) ,

where dμξn =

c2 =

R2

s2 dμξn (s) ,

(22.117)

1 − (|s1 |ξ+|s2 |) n e ds1 ds2 . 4ξn2

Let 0 < γ ≤ 1, x ∈ R2 . Then ⎞ ⎛   r   ∂f ∂f [2] ⎠ [2] ⎝ c1 Pr,n ( f ; x) − f (x) = α j,r j (x) + c2 (x) + 0 ξn2−γ . ∂ x1 ∂ x2 j=1 (22.118)

[m] 22.3 Main Results for Pr,n

383

Proof By Theorems 22.13 and 22.42, here it is ρ = 4.



We also give

  Theorem 22.46 Let f ∈ C m+l R N , m, l, N ∈ N. Assumptions of Theorem 22.32 ⎞ N  |s | ⎜ i=1 i ⎟ −⎝ ξn ⎠ ⎛

  are valid for dμξn (s) = (2ξ1 ) N e ds1 ...ds N . Call γ = 0, β. Suppose  f γ +α ∞ n ≤ M, M > 0, for all α : |α| = m, ξn ∈ ⎛ (0, 1],⎞n ∈ N. N 

∗ := Call cα,n, # j

1 (2ξn ) N



 RN

1; 0 < γ ≤ 1, x ∈ R N . Then

N $

i=1

 siαi e

|si | ⎟ ⎜ −⎝ i=1ξn ⎠

ds1 ...ds N , all |α| = # j = 1, ..., m −

⎛ 

[m] Pr,n ( f ; x)

 γ

− f γ (x) =

⎞

⎜ c∗ # f γ +α (x) ⎟   α,n, j ⎜ ⎟ δ#[m] N  ⎟ + 0 ξnm−γ . ⎜ j,r ⎝ ⎠ $ # |α|=# j j=1 αi !

m−1

i=1

(22.119) When m = 1, the sum collapses. Proof Use of Theorem 22.13 and Theorem 4.6, p. 54 of [2], here it is ρ = (m!) N . 

22.3.5 Simultaneous Approximation by Multivariate [m] Complex Pr,n We make Remark 22.47 We consider here complex √ valued Borel measurable functions f : R N → C such that f = f 1 + i f 2 , i = −1, where f 1 , f 2 : R N → R are implied to be real valued Borel measurable functions. We define the multivariate complex Picard singular operators [m] [m] [m] Pr,n ( f ; x) := Pr,n ( f 1 ; x) + i Pr,n ( f 2 ; x) , x ∈ R N .

(22.120)

  [m] We assume that Pr,n f j ; x ∈ R, ∀ x ∈ R N , j = 1, 2. One notices easily that       [m]  P ( f ; x) − f (x) ≤  P [m] ( f 1 ; x) − f 1 (x) +  P [m] ( f 2 ; x) − f 2 (x) r,n r,n r,n (22.121) also    [m]  Pr,n ( f ; x) − f (x)

∞,x

   [m]  ≤ Pr,n ( f 1 ; x) − f 1 (x)

∞,x

   [m]  + Pr,n ( f 2 ; x) − f 2 (x)

∞,x

(22.122)

384

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

and       [m]  P ( f ) − f  ≤  P [m] ( f 1 ) − f 1  +  P [m] ( f 2 ) − f 2  , p ≥ 1. (22.123) r,n r,n r,n p p p Furthermore, it holds f α (x) = f 1,α (x) + i f 2,α (x) ,

(22.124)

where α denotes a partial derivative of any order and arrangement. We give Theorem 22.48 Let f : R N → C, N ≥ 1, such that f = f 1 + i f 2 , j = 1, 2. Here    ∂ m f (·,·,...,·)  m ∈ N, f j ∈ C m R N , x ∈ R N . Assume  ∂ x α1j ...∂ x α N  < ∞, for all αi ∈ Z+ , i = 1, ..., N : |α| :=

N 

1

∞

N

αi = m. Here ξn ∈ (0, 1], n ∈ N. For all α := (α1 , ..., α N ), αi ∈

i=1

Z+ , i = 1, ..., N , |α| :=

N 

αi = m we denote

i=1 ⎞ N  |s | ⎜ i=1 i ⎟ −⎝ ξn ⎠ ⎛

u ∗ξn (α) :=

1 (2ξn ) N



N 

RN

|si |αi

 1+

i=1

s 2 ξn

r e

ds1 ...ds N . (22.125)

For # j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αi = # j,

i=1

call ∗ ∗ cα,n := cα,n, := # j

Then

⎞ N  |s | ⎜ i=1 i ⎟ −⎝ ξn ⎠ ⎛

1 (2ξn ) N



N  RN

siαi e

ds1 ...ds N .

(22.126)

i=1

 ⎞ ⎛     ∗ m  ⎟ ⎜ [m] ⎜ cα,n,#j f α (x) ⎟   [m] δ#j,r ⎜  Pr,n ( f ; x) − f (x) − ⎟ N   ⎠ ⎝ $ # α1 ,...,α N ≥0;   j=1 αi !   |α|=# j i=1

∞,x

     ωr f 1,α , ξn + ωr f 2,α , ξn ≤ u ∗ξn (α) . N  $ α1 ,...,α N ≥0; αi ! |α|=m

i=1

(22.127)

[m] 22.3 Main Results for Pr,n

385



Proof By Theorem 22.15. We proceed with

Theorem 22.49 Let f : R N → C : f = f 1 + i f 2 , N ≥ 1, j = 1, 2. Here f j ∈ C B  N R uniformly continuous, ξn ∈ (0, 1]. Then   N   [0]   P f − f  ≤ (1 + N )r + 2r er ! (ωr ( f 1 , ξn ) + ωr ( f 2 , ξn )) , r,n ∞ e (22.128) where r, n ∈ N. As n → ∞ and ξn → 0, we derive   [0] P f − f  → 0 r,n ∞

(22.129)

with rates. 

Proof Use of Theorem 22.17. Next comes multi-simultaneous approximation

  Theorem 22.50 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m+l R N , m, l, N ∈ N. The⎛ assumptions of Theorem 22.32 are valid for ⎞ N 

|si | ⎟ ⎜ −⎝ i=1ξn ⎠

dμξn = and let

1 (2ξn ) N

e

  ds1 ...ds N and f j . Call γ = 0, β. Assume  f j,γ +α ∞ < ∞, ⎞ N  |s | ⎜ i=1 i ⎟ −⎝ ξn ⎠ ⎛

u ∗ξn

1 (α) := (2ξn ) N

N 

RN

αi



s 2 1+ ξn

|si |

i=1

for all αi ∈ Z+ , i = 1, ..., N : |α| :=

N 

r

ds1 ...ds N . (22.130)

e

αi = m, ξn ∈ (0, 1].

i=1

For # j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αi = # j,

i=1

call

⎞ N  |s | ⎜ i=1 i ⎟ −⎝ ξn ⎠ ⎛

∗ cα,n, := # j

1 (2ξn ) N



N  R N i=1

siαi e

ds1 ...ds N .

(22.131)

386

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

Then  ⎞ ⎛     ∗ m   ⎟ ⎜ c f (·) γ +α  # α,n, j   [m] ⎟ ⎜ δ#[m]   Pr,n ( f ; ·) γ − f γ (·) − ⎟ ⎜ j,r ⎝ N  ⎠ $ # α1 ,...,α N ≥0;   j=1 α ! i   |α|=# j i=1      ωr f 1,γ +α , ξn + ωr f 2,γ +α , ξn ≤ u ∗ξn (α) . N  $ α1 ,...,α N ≥0; αi ! |α|=m

∞

(22.132)

i=1



Proof Based on Theorems 22.32 and 22.36. We continue with

  Theorem 22.51 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C lB R N , l, N ∈ N (functions l-times continuously differentiable and ⎞bounded). The assump⎛ N 

|si | ⎟ ⎜ −⎝ i=1ξn ⎠

tions of Theorem 22.32 are valid for dμξn = 1, 2. Call γ = 0, β, ξn ∈ (0, 1]. Then     [0]   Pr,n f γ − f γ 





≤ (1 + N ) + 2 r

∞

r

1 (2ξn ) N

er ! e

ds1 ...ds N and f j , j =

e

N 

     ωr f 1,γ , ξn + ωr f 2,γ , ξn .

 [0](22.133)  If additionally f j,γ , j = 1, 2, is uniformly continuous and ξn → 0, then Pr,n f γ → f γ uniformly, as n → ∞. 

Proof By Theorems 22.32 and 22.37. We proceed with L p approximations

  Theorem 22.52 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m R N ,  N m ∈ N , N ≥ 1, with f j,α ∈ L p R , |α| = m, x ∈ R N . Let p, q > 1 : 1p + q1 = 1, j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| := ξn ∈ (0, 1]. For # N  αi = # j, call i=1 ∗ cα,n, = # j

1 (2ξn ) N





N  RN

i=1

siαi e−

N 



|si |

i=1 ξn

ds1 ...ds.

(22.134)

[m] 22.3 Main Results for Pr,n

387

 ⎛ ⎞     m  ⎜ c∗ # f α (x) ⎟ α,n, j  [m]  ⎜ ⎟ δ#[m] N  ⎟  Pr,n ( f ; x) − f (x) − ⎜ j,r ⎝  ⎠ $ #   |α|=# j j=1 αi !   i=1

Then

 ≤



m (q (m − 1) + 1)

1 q

p,x

! "1 (1 + N )r p + 2r p N ((m + r ) p + 1, 1) p

⎞

⎛

⎜ %     & 1 ⎟ ⎟ m ⎜ ξ ω . , ξ + ω , ξ f f ⎟ ⎜ r 1,α n r 2,α n n p p N ⎠ ⎝|α|=m $ αi !

(22.135)

i=1



Proof By Theorem 22.20. We continue with

N Theorem f1 + i f2 , j = 1, 2. Here fj ∈  f : R → C : f =   N  22.53 Let N C R ∩ L p R ; N ≥ 1; p, q > 1 : 1p + q1 = 1, ξn ∈ (0, 1], n ∈ N. Then

 [0]  P ( f ) − f  ≤ r,n p

(22.136)

 ! "1  (1 + N )r p + 2r p N (r p + 1, 1) p ωr ( f 1 , ξn ) p + ωr ( f 2 , ξn ) p .  [0]  [0] As ξn → 0, when n → ∞, we derive  Pr,n → I, the unit ( f ) − f  p → 0, i.e. Pr,n operator, in L p norm. 

Proof By Theorem 22.22. We also give N Theorem  Let  f : R → C : f = f 1 + i f 2 ,   N  22.54 N C R ∩ L 1 R ; N ≥ 1, ξn ∈ (0, 1], n ∈ N. Then

j = 1, 2.

Here

fj ∈

  N   [0]  " ! er ! r r  P ( f ) − f  ≤ (1 + N ) + 2 ωr ( f 1 , ξn )1 + ωr ( f 2 , ξn )1 . r,n 1 e (22.137) [0] As ξn → 0, we get Pr,n → I, in L 1 norm. Proof By Theorem 22.24.



388

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

We further present   Theorem 22.55 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m R N ,   m, N ∈ N, with f j,α ∈ L 1 R N , |α| = m, x ∈ R N , ξn ∈ (0, 1], n ∈ N. For # j= N  1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| := αi = # j, call i=1

1 ∗ cα,n, := # j (2ξn ) N





N  R N i=1

siαi e−

N 



|si |

i=1 ξn

ds1 ...ds.

(22.138)

Then  ⎛ ⎞     m  ⎜ c∗ # f α (x) ⎟ α,n, j  [m] ⎟ [m] ⎜ δ#j,r ⎜  Pr,n ( f ; x) − f (x) − ⎟ N  ⎝ # $ ⎠ #   |α|= j j=1 α ! i   i=1

(22.139)

1,x





≤ (1 + N )r + 2r

⎞ ⎛ N  ⎜ ⎟ e (m + r )! ⎜ 1 ⎟ ⎟ ⎜ N ⎠ ⎝$ e |α|=m αi ! i=1

! m



ξn ωr f 1,α , ξn

 1



+ ωr f 2,α , ξn

 " 1

. 

Proof By Theorem 22.26. We continue with simultaneous L p approximations.

  Theorem 22.56 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m+l R N , m, l, N ∈ N. The⎛ assumptions of Theorem 22.32 are valid for ⎞ N 

|si | ⎜ ⎟ −⎝ i=1ξn ⎠

  ds1 ...ds N and f j . Call γ = 0, β. Let f j,(γ +α) ∈ L p R N , |α| = m, x ∈ R N , and p, q > 1 : 1p + q1 = 1, ξn ∈ (0, 1], n ∈ N. For # j = 1, ..., m, N  αi = # j, call and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

dμξn =

1 (2ξn ) N

e

i=1

∗ cα,n, := # j

1 (2ξn ) N

RN



N  i=1

siαi e−

N 



|si |

i=1 ξn

ds1 ...ds N .

(22.140)

[m] 22.3 Main Results for Pr,n

389

 ⎞ ⎛     ∗ m   ⎟ ⎜ c f (x) γ +α  # α,n, j   [m] ⎟ ⎜ δ#[m]   Pr,n ( f ; x) γ − f γ (x) − ⎟ ⎜ j,r ⎝ N  ⎠ $ #   |α|=# j j=1 α ! i   i=1

Then

 ≤



m (q (m − 1) + 1) ⎛

1 q

! "1 (1 + N )r p + 2r p N ((m + r ) p + 1, 1) p

p,x

(22.141)

⎞

⎜ %     & 1 ⎟ ⎜ ⎟ m ξ ω . , ξ + ω , ξ f f ⎜ ⎟ r 1,γ +α n r 2,γ +α n n p p N ⎝|α|=m $ ⎠ αi ! i=1



Proof By Theorems 22.32 and 22.38. We give also

  Theorem 22.57 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C l R N , l, N ∈ N. The assumptions of Theorem 22.32 are valid for ⎞ ⎛ N 

|si | ⎟ ⎜ −⎝ i=1ξn ⎠

dμξn =

1 (2ξn ) N

e

R N ; p, q > 1 :

1 p

+

1 q

  ds1 ...ds N and f j . Call γ = 0, β. Let f j,γ ∈ L p R N , x ∈

= 1, ξn ∈ (0, 1], n ∈ N. Then     [0]   Pr,n ( f ) γ − f γ  ≤ p

(22.142)

!    & "1 %  (1 + N )r p + 2r p N (r p + 1, 1) p ωr f 1,γ , ξn p + ωr f 2,γ , ξn p .  [0]  · p As n → +∞ and ξn → 0, then Pr,n f γ → fγ . Proof By Theorems 22.32 and 22.39.



We continue with   Theorem 22.58 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C l R N , l, N ∈ N. The assumptions of Theorem 22.32 are valid for ⎛ ⎞ N 

|si | ⎜ ⎟ −⎝ i=1ξn ⎠

dμξn =

1 (2ξn ) N

e

  ds1 ...ds N and f j . Call γ = 0, β. Let f j,γ ∈ L 1 R N , x ∈

R N , ξn ∈ (0, 1], n ∈ N. Then

390

22 Complete Approximations with Multivariate Generalized Picard Singular Integrals

    [0]   Pr,n ( f ) γ − f γ  ≤

(22.143)

1





(1 + N ) + 2 r

r

er ! e

N 

   " !  ωr f 1,γ , ξn 1 + ωr f 2,γ , ξn 1 .

 [0]  · 1 As n → +∞ and ξn → 0, then Pr,n ( f ) γ → fγ . 

Proof By Theorems 22.32 and 22.40. We finish with

  Theorem 22.59 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m+l R N , ⎞ N  |s | ⎜ i=1 i ⎟ −⎝ ξn ⎠ ⎛

m, l, N ∈ N. The assumptions of Theorem 22.32 for dμξn = (2ξ1 ) N e ds1 ...  N n ds N and f j are valid. Call γ = 0, β. Let f j,(γ +α) ∈ L 1 R , |α| = m, x ∈ R N , ξn ∈ (0, 1], n ∈ N. For # j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , N  |α| := αi = # j, call i=1



1 ∗ cα,n, := # j (2ξn ) N Then



N 

R N i=1

siαi e−

N 



|si |

i=1 ξn

ds1 ...ds N .

(22.144)

 ⎞ ⎛     ∗ m  ⎟ ⎜ [m] ⎜ cα,n,#j f γ +α (x) ⎟    [m] δ#j,r ⎜  Pr,n ( f ; x) γ − f γ (x) − ⎟ N   ⎠ ⎝ $ #   |α|=# j j=1 αi !   i=1 ⎛

⎛

⎞

1,x

⎞

⎜ ⎜ 1 ⎟!     "⎟ ⎜ ⎜ ⎟ ⎟ ≤⎜ ⎜ N ⎟ ωr f 1,γ +α , ξn 1 + ωr f 2,γ +α , ξn 1 ⎟ ⎝|α|=m ⎝ $ ⎠ ⎠ αi ! i=1

 ξnm



(1 + N ) + 2 r

Proof By Theorems 22.32 and 22.41.

r

e (m + r )! e

N  .

(22.145) 

References

391

References 1. Anastassiou, G.A.: Intelligent Mathematics: Computational Analysis. Springer, Heidelberg (2011) 2. Anastassiou, G.A.: Approximation by Multivariate Singular Integrals. Springer, New York (2011) 3. Anastassiou, G.A.: High order approximation by multivariate generalized Picard singular integrals. Submitted (2020) 4. Anastassiou, G., Gal, S.: Approximation Theory. Birkhaüser, Boston (2000) 5. Gradshteyn, I.S., Ryzhik, I.M.: Table of Integrals, Series and Products, 8th edn. Elsevier, Amsterdam (2015) 6. DeVore, R.A., Lorentz, G.G.: Constructive Approximation, vol. 303. Springer, Berlin (1993)

Chapter 23

High Order Approximation with Multivariate Generalized Gauss–Weierstrass Singular Integrals

This research and survey chapter deals exclusively with the study of the approximation of generalized multivariate Gauss–Weierstrass singular integrals to the identityunit operator. Here we study quantitatively most of their approximation properties. The multivariate generalized Gauss–Weierstrass operators are not in general positive linear operators. In particular we study the rate of convergence of these operators to the unit operator, as well as the related simultaneous approximation. These are given via Jackson type inequalities and by the use of multivariate high order modulus of smoothness of the high order partial derivatives of the involved function. Also we study the global smoothness preservation properties of these operators. These multivariate inequalities are nearly sharp and in one case the inequality is attained, that is sharp. Furthermore we give asymptotic expansions of Voronovskaya type for the error of multivariate approximation. The above properties are studied with respect to L p norm, 1 ≤ p ≤ ∞. It follows [4].

23.1 Introduction We start with our motivation for this chapter which comes from [7]. In the next we introduce and deal with the smooth Gauss–Weierstrass singular integral operators Wr,ξ ( f, x) defined as follows. For r ∈ N and n ∈ Z+ we set   ⎧ r ⎪ r− j ⎪ j −n , j = 1, ..., r, ⎨ (−1) j   αj = r  r ⎪ ⎪1 − j −n , j = 0, (−1)r − j ⎩ j

(23.1)

j=1

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_23

393

394

that is

23 High Order Approximation with Multivariate Generalized … r 

α j = 1.

j=0

Let f ∈ C n (R) and f (n) ∈ L p (R), 1 ≤ p < ∞, we define for x ∈ R, ξ > 0 the Lebesgue integral 1 Wr,ξ ( f ; x) = √ πξ



⎛

∞

⎝

−∞

r

⎞ t2

α j f (x + jt)⎠ e− ξ dt.

(23.2)

j=0

Note 23.1 The operators Wr,ξ are not, in general, positive. Let r = 2, n = 3. Then α0 = x = 0. Then

23 , α1 8

= −2, α2 = 18 . Consider f (t) = t 2 ≥ 0 and

  3ξ < 0. Wr,ξ t 2 ; 0 = − 4

We notice by

√1 πξ

∞

t2

−∞

e− ξ dt = 1, that Wr,ξ (c, x) = c, c constant, and

⎞ ⎛ ∞ r t2 1 ⎝ αj Wr,ξ ( f ; x) − f (x) = √ ( f (x + jt) − f (x)) e− ξ dt ⎠ . (23.3) πξ j=0 −∞ We need the r th L p - modulus of smoothness     ωr f (n) , h p := sup rt f (n) (x) p,x , h > 0, |t|≤h

where rt

f

(n)

(x) :=

r

r− j

(−1)

j=0

  r f (n) (x + jt) , j

(23.4)

(23.5)

  see [9], p. 44. Here we have that ωr f (n) , h p < ∞, h > 0. We need to introduce δk :=

r

α j j k , k = 1, ..., n ∈ N,

(23.6)

j=1

and denote by · the integral part. Call r (w, x) :=

r

α j j n f (n) (x + jw) − δn f (n) (x) .

j=0

According to [2], p. 306, [1], we get

(23.7)

23.1 Introduction

395

Thus

τ (w, x) = rw f (n) (x) .

(23.8)

  τ (w, x) p,x ≤ wr f (n) , |w| p , w ∈ R.

(23.9)

Define



(t − w)n−1 τ (w, x) dw, n ∈ N. (n − 1)!

t

Rn (0, t, x) := 0

(23.10)

Using the above terminology [7] we obtain  (x) := Wr,ξ ( f ; x) − f (x) −  n2 

f

(2m)

m=1

(2m − 1) (2m − 3) · ... · 3 · 1 (x) δ2m (2m)!

where Rn∗

1 (x) := √ πξ



∞

 m ξ = Rn∗ (x) , 2

t2

Rn (0, t, x) e− ξ dt, n ∈ N.

−∞

(23.11)

(23.12)

In  (x), see (23.11), the sum collapses when n = 1. We mention Theorem 23.2 ([7]) Let p, q > 1 such that Then  (x) p ≤

2 1

1

p+1 2p

1

1 p

+

1 q

= 1, n ∈ N and the rest as above.

n

τ pξ2

1

1

q 2q π 2 p (r p + 1) p ((n − 1)!) (q (n − 1) + 1) q

ωr



  f (n) , ξ , p

(23.13)

where  τ :=

∞

(1 + u)

r p+1

u

np−1 − pu2

e

0

2

du −

∞

u

np−1 −

e

pu 2 2

 du < ∞.

(23.14)

0

Hence as ξ → 0 we obtain  (x) p → 0.     If additionally f (2m) ∈ L p (R), m = 1, 2, ..., n2 then Wr,ξ ( f ) − f  p → 0, as ξ → 0. The counterpart of Theorem 23.2 follows, case of p = 1. Theorem 23.3 ([7]) Let f ∈ C n (R) and f (n) ∈ L 1 (R), n ∈ N. Then  (x) 1 ≤

2 √ (r + 1) (n − 1)! π

396

23 High Order Approximation with Multivariate Generalized …

⎛

⎞ ⎞ ⎛   n+k   n+k r +1  2 −1   2    1 r +1 ⎝ ⎝ ⎠ ωr f (n) , ξ ξ n2 , (n + k − 2r )⎠ · M · k 1 2 r =1 k=1 (23.15) where

 √π

, n + k is even 2 . 1, n + k is odd

M=

Hence as ξ → 0 we obtain  (x) 1 → 0.     If additionally f (2m) ∈ L 1 (R), m = 1, 2, ..., n2 then Wr,ξ ( f ) − f 1 → 0, as ξ → 0. The case n = 0 is mentioned next. Proposition 23.4 ([7]) Let p, q > 1 such that 1p + q1 = 1 and the rest as above. Then 1 1 1 1    2q + p π 2q − 2   1 Wr,ξ ( f ) − f  ≤ 2 pω f, ξ , θ (23.16) r 1 p p q 2q

where

∞

θ :=

(1 + t)r p e−

pt 2 2

dt < ∞.

0

Hence as ξ → 0 we obtain Wr,ξ ( f ) → unit operator I in the L p norm, p > 1. We also mention Proposition 23.5 ([7]) It holds   Wr,ξ f − f  ≤ √2 1 π



∞

    2 (1 + t)r e−t dt ωr f, ξ .

(23.17)

1

0

Hence as ξ → 0 we get Wr,ξ → I in the L 1 norm. In this chapter we study the approximation properties of multivariate smooth general Gauss–Weierstrass singular integral operators: [m] Wr,n ( f ; x1 , ..., x N ) :=

1 √  N πξn

r j=0



α[m] j,r

RN

f (x1 + s1 j, x2 + s2 j, ..., x N + s N j) e−

N  si2 i=1 ξn



ds1 ...ds N . (23.18)

Notice that √

1 πξn



N

e− RN

N  si2 i=1 ξn



ds1 ...ds N = 1,

(23.19)

23.1 Introduction

397

see [6]. Here r ∈ N, m ∈ Z + , and

α[m] j,r

⎧   r ⎪ r− j ⎪ j −m , if j = 1, 2, ..., r, ⎨ (−1) j   := r  r −m ⎪ r −i ⎪ i , if j = 0, (−1) ⎩1 − i i=1

and [m] δk,r :=

r

k α[m] j,r j , k = 1, 2, ..., m ∈ N.

(23.20)

(23.21)

j=1

See that

r  j=0

α[m] j,r = 1.

Also here ξn ∈ (0, 1], n ∈ N, and f : R N → R is a Borel measurable function. [m] [m] is a special case of a more general operator θr,n studied in The above operator Wr,n general in [3] by the author. [m] . We mention next about θr,n Let μξn be a probability Borel measure on R N , N ≥ 1. We define the multiple smooth singular integral operators [m] θr,n ( f ; x1 , ..., x N ) :=

r

α[m] j.r

j=0

RN

f (x1 + s1 j, x2 + s2 j, ..., x N + s N j) dμξn (s) ,

(23.22) where s := (s1 , ..., s N ), x := (x1 , ..., x N ) ∈ R N . [m] are not in general positive. For example, consider the function The operators θr,n N  2 ϕ (u 1 , ..., u N ) = u i and also take r = 2, m = 3; xi = 0, i = 1, ..., N . See that ϕ ≥ 0, however

i=1

⎞ ⎛ 2 [3] 2 [3] ⎠ ⎝ θ2,n (ϕ; 0, 0, ..., 0) = j α j,2 j=1



[3] α1,2

+

[3] 4α2,2

assuming that

 N

 RN

 RN



N 

i=1

i=1

 si2

 N RN

 si2

dμξn (s) =

i=1

    N 1 2 s dμξn (s) < 0. dμξn (s) = −2 + 2 R N i=1 i (23.23)

 si2 dμξn (s) < ∞.

[m] we have that Clearly in the case of Wr,n

398

23 High Order Approximation with Multivariate Generalized … 

dμξn (s) = √

1 πξn

N e

−

N  si2 i=1 ξn



ds1 ...ds N , s ∈ R N .

[m] Lemma 23.6 The operator θr,n preserve the constant functions in N variables.

We need   Definition 23.7 Let f ∈ C B R N , the space of all bounded and continuous functions or uniformly continuous on R N . Then, the r th multivariate modulus of smoothness of f is given by (see, e.g. [5])  r 

ωr ( f ; h) := √ sup

u 1 ,u 2 ,...,u N

u 21 +...+u 2N ≤h

 ( f )∞ < ∞, h > 0,

(23.24)

where · ∞ is the sup-norm and ru f (x) := ru 1 ,u 2 ,...,u N f (x1 , ..., x N ) = r j=0

(−1)r − j

  r f (x1 + ju 1 , x2 + ju 2 , ..., x N + ju N ) . j

(23.25)

  Let m ∈ N and let f ∈ C m R N . Suppose that all partial derivatives of f of order m are bounded, i.e.  m   ∂ f (·, ·, ..., ·)     ∂x α1 ...∂x α N  1

for all α j ∈ Z+ , j = 1, ..., N ;

N 

N

< ∞,

(23.26)

∞

α j = m.

j=1 [m] In this chapter we apply the general theory developed in [3] about θr,n to the [m] operators Wr,n , so we can obtain computationally specific results and show that the general theory has applications and it is a valid theory. [m] we So for the very important in various branches of mathematics operators Wr,n prove the very essential properties of uniform approximation, L p approximation, global smoothness preservation and simultaneously approximation, Voronovskaya asymptotic expansions and complex simultaneous approximation.

23.2 Auxiliary Essential Results

399

23.2 Auxiliary Essential Results We need the following: Theorem 23.8 Let r, N , m ∈ N, with m > r ; α j ∈ Z+ , j = 1, ..., N : |α| := N  α j = m, ξn ∈ (0, 1], n ∈ N. Then j=1

 u ξn (α) := √

N 



1 πξn

N

RN

|si |αi

 1+

i=1

s 2 ξn

r



e−

N  si2 i=1 ξn



ds1 ...ds N

    m−r  2  N e (m + r )! N r r ≤ ξn √ (1 + N ) + 2 e π  ≤

2 √ π

N 

 (1 + N )r + 2r

e (m + r )! e

N < +∞,

(23.27)

are uniformly bounded. Proof We estimate 1  u ξn (α) := √  N πξn N

2 √  N √  N π ξn

RN

N R+

|si |αi



siαi

 1+

i=1

N 

N R+

1+

i=1

N 



2N √  N √  N π ξn

N 



 siαi

i=1

⎛

s 2 ξn

r

s 2 ξn

r



e−



e−

N  si2 i=1 ξn

N  si2 i=1 ξn



ds1 ...ds N =



ds1 ...ds N ≤

⎞r   N  s si2 i⎟ ⎜ ⎜ i=1 ⎟ − i=1 ⎜1 + ⎟ e ξn ds1 ...ds N ≤ ⎝ ξn ⎠ N 

 √ m  N  r N   si αi 2N ξn 1 1 si √ √ +√ √ √  N N ξn ξn ξn i=1 ξn R+ π i=1 e

−

N   i=1

s √i ξn

2

ds1 ds N √ ... √ = ξn ξn

(23.28)

400

23 High Order Approximation with Multivariate Generalized …



2 √ π

N 

 N   m−r ξn

N R+

 z iαi

1+

i=1

N

r zi

e

−

N  i=1

z i2

dz 1 ...dz N =

(23.29)

i=1

⎡ N  r  N N  m−r  2  N  α − z i2 i ⎣ ξn zi z i e i=1 dz 1 ...dz N + 1+ √ π [0,1] N i=1 i=1 N 

(R+ −[0,1])

N

 z iαi

1+

i=1

N

r zi

e

−

N  i=1

⎤ z i2

dz 1 ...dz N ⎦ ≤

i=1

 N  r N  m−r  2  N  α i ξn zi z i dz 1 ...dz N + 1+ √ π [0,1] N i=1 i=1 N 

2r (R+ −[0,1]) N

⎤   N r  N − zi z iαi z i e i=1 dz 1 ...dz N ⎦ ≤

i=1

i=1

 m−r  2  N ' ξn √ (1 + N )r + π N 

r

2

(R+ −[0,1]) N

i=1

z iαi

 N  i=1

 z ri

N 

 e

−z i

i=1

N 

dz i

=

(23.30)

i=1

 N ∞  m−r  2  N  ξn z i((αi +r )+1)−1 e−zi dz i √ (1 + N )r + 2r π i=1 1

=

(by [8], p. 348)  N  m−r  2  N  ξn  ((αi + r ) + 1, 1) , √ (1 + N )r + 2r π i=1 where  (·, ·) is the upper incomplete gamma function. We have proved that  N  m−r  2  N   u ξn (α) ≤ ξn  ((αi + r ) + 1, 1) ≤ √ (1 + N )r + 2r π i=1     m−r  2  N e (m + r )! N ξn √ (1 + N )r + 2r e π

≤

(23.31)

23.2 Auxiliary Essential Results



2 √ π

401

N 

 (1 + N )r + 2r

e (m + r )! e

N < +∞,

therefore  u ξn (α) are uniformly bounded. Above we used the formula  (s + 1, 1) =

es! , s ∈ N. e

Here αi + r ∈ N, hence e (m + r )! e (αi + r )! ≤ , i = 1, ..., N . e e

 ((αi + r ) + 1, 1) =

(23.32) 

The claim is proved. We continue with Theorem 23.9 Let r, n, N ∈ N, ξn ∈ (0, 1]. Then ξn := √ 

1 πξn

N



1 √ r ξn

RN

2 √ π

  s 2 r − 1+ e ξn



N  si2 i=1 ξn



ds1 ...ds N ≤

N    er ! N r r . (1 + N ) + 2 e

(23.33)

Proof We estimate 1

ξn = √   N πξn 2N √  N √  N π ξn ⎛ 



2 √ π

N

2 √ π

N

1 √ r ξn

N R+



RN



N R+

s 2 1+ ξn

s 2 1+ ξn N 

r

r



N  si2 i=1 ξn

e− 

N  si2 i=1 ξn

e−

⎞r

si ⎟ ⎜ 1 ⎜ ⎟ − i=1 ⎜√ + √ √ ⎟ e ⎝ ξn ξn ξn ⎠



N  si2 i=1 ξn



ds1 ...ds N =



ds1 ...ds N ≤



ds1 ds N √ ... √ = ξn ξn



N R+

2 N  r  N  s √i − si ds1 ds N ξn e i=1 1+ √ √ ... √ = ξn ξn ξn i=1

(23.34)

402

23 High Order Approximation with Multivariate Generalized …



2 √ π

N



1 √ r ξn

2 √ π

N

N R+

1 √ r ξn



2 √ π

⎣

1+

1 √ r ξn

N



2 √ π



r zi

2 √ π

e

−

N 

z i2

i=1

r zi

N 

−

i=1

1+

[0,1] N

1+

N

dz 1 ...dz N =

e

N 

−

i=1

z i2

dz 1 ...dz N + ⎤

z i2

dz 1 ...dz N ⎦ ≤ r

N

dz 1 ...dz N +

zi

i=1

r zi

e

−

N 

⎤ zi

i=1

dz 1 ...dz N ⎦ ≤

(23.35)

i=1

(R+ −[0,1]) N

N  r r (1 + N ) + 2

2 √ π

N



⎣(1 + N )r + 2r



e

i=1

N

⎡

1 √ r ξn 1 √ r ξn

1+

 N 

(R+ −[0,1]) N

2 √ π

zi

i=1













r

i=1

⎡

(R+ −[0,1]) N

N

1+

[0,1] N



1 √ r ξn





1 √ r ξn

⎤  N r  N − zi z i e i=1 dz 1 ...dz N ⎦ ≤ i=1

N  (R+ −[0,1]) N

N  (1 + N ) + 2 r

r

N   r r (1 + N ) + 2

∞

z

z ri

i=1

N  i=1



1

N 

e

−z i

i=1

∞

z ri e−zi dz i

(r +1)−1 −z

N 

dz i

=

i=1

=

N

e dz

=

(23.36)

1

(by [8], p. 348) 1 √ r ξn



1 √ r ξn proving the claim.

2 √ π



N

2 √ π

' ( (1 + N )r + 2r  N (r + 1, 1) =

N    er ! N r r , (1 + N ) + 2 e 

23.2 Auxiliary Essential Results

403

We give Theorem 23.10 All as in Theorem 23.8 and p > 1. Then ξn (α) := √ A

 N 



1 πξn

N

RN

|si |αi



r  p

s 2 1+ ξn

i=1



N  si2 i=1 ξn

e−



ds1 ...ds N ≤

  p(m−r )  2  N ' ( ξn √ (1 + N )r p + 2r p  N ((m + r ) p + 1, 1) ≤ π 

2 √ π

N

(23.37)

' ( (1 + N )r p + 2r p  N ((m + r ) p + 1, 1) < +∞,

are uniformly bounded, where m > r. Proof We estimate ξn (α) = √ 1  A N πξn N

2 √  N √  N π ξn

 N 

RN

N R+



2 √ π

1+

i=1

N 



|si |αi



αi p

 1+

si

i=1

 N   mp ξn

⎛

N 

s 2 ξn

N R+

s 2 ξn

r p

r  p



e−

N  si2 i=1 ξn



N  si2 i=1 ξn

e−



ds1 ...ds N =



ds1 ...ds N ≤

(23.38)

N    si αi p √ ξn i=1

⎞r p

2 N  si ⎟  si ⎜ 1 ⎟ − i=1 √ξn ds1 ds N ⎜ i=1 √ ... √ = ⎜√ + √ √ ⎟ e ⎝ ξn ξn ξn ⎠ ξn ξn

(set z i = 



√si , ξn

2 √ π

2 √ π

i = 1, ..., N )

 N   p(m−r ) ξn

⎡  N   p(m−r ) ⎣ ξn

N 

N R+

 α p zi i

i=1

N  [0,1] N

1+

i=1

N

r p zi

e

−

N  i=1

z i2

dz 1 ...dz N =

i=1

 α p zi i

1+

N i=1

r p zi

e

−

N  i=1

z i2

dz 1 ...dz N +

404

23 High Order Approximation with Multivariate Generalized …

N 

(R+ −[0,1]) N



2 √ π

 α p

1+

zi i

N

i=1

N 

[0,1] N

2r p

zi

e

−

⎤

N 

z i2

i=1

dz 1 ...dz N ⎦ ≤

i=1

  N   p(m−r ) ξn

r p

N 

(R+ −[0,1]) N

1+

i=1

N

r p dz 1 ...dz N +

zi

(23.39)

i=1

⎤   N r p  N − z i α p zi i zi e i=1 dz 1 ...dz N ⎦ ≤

i=1



 α p zi i

i=1

2 √ π

 N   p(m−r ) ξn

⎡ ⎢ (1 + N )r p ⎢ rp + 2 ⎢ N ⎣* (R+ −[0,1]) N ( pαi + 1)



N  i=1

α

zi i

 N  p i=1

 rp

zi

⎤ N  i=1

e−zi

N  i=1

⎥ ⎥ dz i ⎥ ⎦

i=1

⎡ =

⎤

N ∞ ⎥   p(m−r )  2  N ⎢ (1 + N )r p  ⎢ ⎥ (α +r ) p −z i + 2r p ξn zi i e dz i ⎥ = √ ⎢ N ⎣* ⎦ π 1 i=1 ( pαi + 1) i=1

(by [8], p. 348) ⎤

⎡

N ⎥   p(m−r )  2  N ⎢ (1 + N )r p  ⎥ ⎢ ξn  ((αi + r ) p + 1, 1)⎥ ≤ + 2r p √ ⎢ N ⎦ ⎣* π i=1 ( pαi + 1) i=1

(by [8], p. 909)   p(m−r )  2  N ' ( ξn √ (1 + N )r p + 2r p  N ((m + r ) p + 1, 1) . π

(23.40)

We have proved that ξn (α) ≤ A

  p(m−r )  2  N ' ( ξn √ (1 + N )r p + 2r p  N ((m + r ) p + 1, 1) ≤ π (23.41)

23.2 Auxiliary Essential Results



2 √ π

N

405

' ( (1 + N )r p + 2r p  N ((m + r ) p + 1, 1) < +∞,

are uniformly bounded, where m > r . Above we used ([8], p. 909) the formula  (α, x y) = y α e−x y



∞

e−t y (t + x)α−1 dt,

(23.42)

0

where Rey > 0, x > 0, Reα > 1. We notice that (x = y = 1, αi ≤ m)  ((αi + r ) p + 1, 1) = e

−1



∞

e−t (t + 1)(αi +r ) p dt ≤

0

e

−1



∞

e−t (t + 1)(m+r ) p dt =  ((m + r ) p + 1, 1) .

(23.43)

0

That is  ((αi + r ) p + 1, 1) ≤  ((m + r ) p + 1, 1) , for all i = 1, ..., N . The theorem is proved.

(23.44) 

We continue with Theorem 23.11 Let r, n, N ∈ N, ξn ∈ (0, 1], p > 1. Then 1 Bξn := √  N πξn 1 √  pr ξn



2 √ π

RN

N

  s 2 r p − 1+ e ξn



N  si2 i=1 ξn



ds1 ...ds N ≤

' ( (1 + N ) pr + 2 pr  N (r p + 1, 1) .

(23.45)

Proof We estimate

Bξn = √



1 πξn

2N √  N √  N π ξn

N

N R+

RN

  s 2 pr − 1+ e ξn

  s 2 pr − 1+ e ξn





N  si2 i=1 ξn

N  si2 i=1 ξn



ds1 ...ds N =



ds1 ...ds N ≤

(23.46)

406

23 High Order Approximation with Multivariate Generalized …

⎛ 



2 √ π

2 √ π

N N R+

N

si ⎟ ⎜ 1 ⎜ ⎟ − i=1 ⎜√ + √ √ ⎟ e ⎝ ξn ξn ξn ⎠

2 √ π





1

√  pr ξn



N



√  pr ξn

N R+

2 √ π

N

√  pr ξn 1

2 √ π



√  pr ξn

N

2 √ π

N R+



⎣

1+

[0,1] N

1+

i=1

e

N

−

ds1 ds N √ ... √ = ξn ξn 2 s √i ξn

N 

z i2

i=1

 pr zi

 pr zi

e

−

e

ds1 ds N √ ... √ = ξn ξn

dz 1 ...dz N =

N 

−

i=1

z i2

dz 1 ...dz N + ⎤

N  i=1



N 

z i2

dz 1 ...dz N ⎦ ≤

(23.47)

(R+ −[0,1]) N

⎤  N  pr  N − zi zi e i=1 dz 1 ...dz N ⎦ ≤

(1 + N ) pr + 2 pr

i=1

N 



2 √ π

N  −

 pr zi



i=1

⎣(1 + N ) pr + 2 pr

√  pr ξn

e

N  si2 i=1 ξn

i=1

N

⎡



 pr



i=1





1

N

1+

⎡

(R+ −[0,1]) N







1



1

N si 1+ √ ξn i=1

√  pr ξn

1

⎞ pr

N 

(R+ −[0,1]) N

i=1

N  N  pr pr (1 + N ) + 2 i=1

 pr zi

1

∞

N 

e−zi

i=1

z i e−zi dz i pr

N 

dz i

=

i=1

=

(by [8], p. 348) 1

√  pr ξn



2 √ π

N

' ( (1 + N ) pr + 2 pr  N (r p + 1, 1) , 

proving the claim. We finish this section with Theorem 23.12 Let n, N ∈ N, ξn ∈ (0, 1], α j ∈ Z+ , j = 1, ..., N : |α| := m ∈ N. Then

N  j=1

Dξn (α) :=

αj =

23.2 Auxiliary Essential Results

407

⎛ ⎞   N s2     N  −m i i=1 1 μ N − ξn αi ⎝ √  ⎠ |s | ξn ds ...ds , ≤ e √ i 1 N N π RN πξn i=1 (23.48) where ,   √ m+1 , (23.49) μ := max π,  2 where  is the gamma function. Proof We have that ⎛ ⎞   N s2   N  −m i i=1 ⎝ √ 1  |si |αi e− ξn ds1 ...ds N ⎠ = Dξn (α) ≤ ξn N N R πξn i=1 ⎛ ⎞   N s2   N  −m i N i=1 2 ⎝ √  ξn siαi e− ξn ds1 ...ds N ⎠ = N N R+ πξn i=1

(23.50)

⎛ ⎞ N    N   −m  2  N  m  si αi −  √sξi n 2 ds1 ds N ⎝ √ ξn ξn e i=1 √ √ ... √ ⎠ = N π ξn ξn ξn R+ i=1 

2 √ π

N 

N N R+



2 √ π

 z iαi

e

−

N  i=1

z i2

dz 1 ...dz n =

i=1

N  N i=1

0

∞

z iαi e−zi dz i = 2

(by [8], p. 349) 

2 √ π

  N    N   N   αi2+1 αi + 1 1  = √  N ≤ (∗) . 2 2 π i=1 i=1

  Some important values of  αi2+1 follow:   √ if αi = 0, then  21 = π ≈ 1.77245..., if αi = 1, then  (1) = 1,   √ if αi = 2, then  23 = 2π ≈ 0.886226..., if αi = 3, then  (2) = 1,√   if αi = 4, then  25 = 3 4 π ≈ 1.32934, if αi = 5, then  (3) = 2.

(23.51)

408

23 High Order Approximation with Multivariate Generalized …

The function  has minimum at xmin = 1.46..., with  (xmin ) = 0.885..., and  (x) is increasing for x ≥ xmin . To have αi2+1 > 1.46..., we need αi > 1.92.... Thus       when αi ≥ 2, then  αi2+1 increases, and  αi2+1 ≤  m+1 , when m ≥ 2. 2  αi +1  Hence it holds  2 ≤ μ, for i = 1, ..., N , and  (∗) ≤

μ √ π

N ,

(23.52) 

proving the claim.

[m] 23.3 Main Results for Wr,n

23.3.1 Uniform Approximation [m] We start with an application to Wr,n of the following theorem.   Theorem 23.13 ([3], p. 11) Let m ∈ N, f ∈ C m R N , N ≥ 1, x ∈ R N . Assume  m  N   ∂ f (·,·,...,·)  αi = m. Let μξn be  ∂x α1 ...∂x α N  < ∞, for all α j ∈ Z+ , j = 1, ..., N : |α| := 1

N

∞

i=1

a Borel probability measure on R N , for ξn > 0, (ξn )n∈N bounded sequence. N  Suppose that for all α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| := αi = m i=1

we have that N 

u ξn (α) :=

RN

αi



s 2 1+ ξn

|si |

i=1

r dμξn (s) < ∞.

For  j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

(23.53) N 

αj =  j,

j=1

call

cα,n := cα,n,j :=

N  R N i=1

siαi dμξn (s1 , ..., s N ) .

(23.54)

Then (i) ⎞⎛ m ⎟ ⎜ c f (x)  α α,n, j ⎟ ⎜ [m] [m] δ[m] Er,n (x) := -θr,n ( f ; x) − f (x) − ⎟ ⎜ j,r ⎝ N ⎠*  α1 ,...,α N ≥0; j=1 α ! i |α|= j i=1

[m] 23.3 Main Results for Wr,n

409



(ωr ( f α , ξn )) ≤ N  * α1 ,...,α N ≥0; αi |α|=m

N 

 RN

αi



s 2 1+ ξn

|si |

i=1



r

dμξn (s) .

i=1

(23.55) ∀ x ∈ RN . (ii)

 [m]  E  r,n

∞

≤ R.H.S.(23.55).

(23.56)

Given  that ξn → 0, as n → ∞, and u ξn is uniformly bounded, then we derive [m]  → 0 with rates. that  Er,n (iii) It holds also that ⎛

⎞ m   ⎜ -cα,n,j - f α ∞ ⎟  [m]  ⎟ [m] ⎜ δ θr,n ( f ) − f  ≤ ⎜ ⎟ + R.H.S.( 23.55), j,r ⎝ N ∞ ⎠ *  α ,...,α ≥0; 1 N j=1 αi ! |α|= j

i=1

(23.57) j,  j = 1, ..., m. Furthermore, as given that f α ∞ < ∞, for all α : |α| =  ξn → 0 when n → ∞, assuming that cα,n,j → 0, while u ξn is uniformly bounded, we conclude that   [m] θ ( f ) − f  → 0 r,n ∞

(23.58)

with rates. [m] A uniform approximation result for Wr,n follows:

  m     Theorem 23.14 Let m ∈ N, f ∈ C m R N , N ≥ 1, x ∈ R N . Assume  ∂∂x αf1(·,·,...,·) αN  ...∂x N 

< ∞, for all α j ∈ Z+ , j = 1, ..., N : |α| :=

1

N

∞

α j = m. Here ξn ∈ (0, 1], n ∈ N,

j=1

m > r ∈ N. For all α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αi = m, we

i=1

denote  u ξn (α) := √

1 πξn

N 

N

RN

|si |αi

 1+

i=1

s 2 ξn

r



e−

N  si2 i=1 ξn



ds1 ...ds N . (23.59)

For  j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αi =  j,

i=1

call  cα,n :=  cα,n,j := √

1 πξn n

N



N  R N i=1

siαi e−

N  si2 i=1 ξn



ds1 ...ds N .

(23.60)

410

23 High Order Approximation with Multivariate Generalized …

Then (i) ⎞⎛ m ⎟ ⎜  c f (x)  α α,n, j ⎟ ⎜ [m] [m] r,n δ[m] E (x) := -Wr,n ( f ; x) − f (x) − ⎟ ⎜ j,r ⎝ N ⎠*  α1 ,...,α N ≥0; j=1 α ! i |α|= j i=1 ≤



(ωr ( f α , ξn )) u ξn (α) , N   * α1 ,...,α N ≥0; α ! i |α|=m

(23.61)

i=1

∀ x ∈ RN . (ii)

 [m]  E   r,n

∞

≤ R.H.S.(23.61).

(23.62)

u ξn (α) → 0 and are uniformly Given that ξn → 0, as n → +∞, we  that   have [m]  r,n → 0 with rates. bounded, and then we derive that  E ∞ (iii) It holds also that ⎛

⎞  -⎜  cα,n,j - f α ∞ ⎟ -  [m] - [m] - ⎜  ⎟ Wr,n ( f ) − f  ≤ -δj,r - ⎜ ⎟ + R.H.S.(23.61), N ∞ ⎝ ⎠ *  α1 ,...,α N ≥0; j=1 αi ! m

|α|= j

i=1

(23.63) j,  j = 1, ..., m. Furthermore, as given that f α ∞ < +∞, for all α : |α| =  cα,n,j → 0 and  u ξn (α) → 0, and both ξn → 0 when n → +∞, we have that  are uniformly bounded, and we conclude that   [m] W ( f ) − f  → 0 r,n ∞

(23.64)

with rates. Proof Mainly by applying Theorem 23.13. By Theorem 23.8 we get that u ξn (α) → 0 and  u ξn (α) are uniformly bounded. By Theorem 23.12 we get that - 1 - cα,n,j - ≤ √  N cα,n - = - πξn  ≤

μ √ π

N 

RN

 |si |αi



e−

N  si2 i=1 ξn



ds1 ...ds N

i=1

 N     m μ N ξn ≤ √ . π

That is  cα,n,j → 0 and are uniformly bounded. The proof is complete.



[m] 23.3 Main Results for Wr,n

411

We mention   Theorem 23.15 ([3], p. 14) Let f ∈ C B R N , N ≥ 1, ξn ∈ (0, 1]. Then   [0] θ f − f  ≤ r,n ∞



 1+

RN

s 2 ξn

r

 dμξn (s) ωr ( f, ξn ) ,

(23.65)

under the assumption ξn :=

RN

  s 2 r 1+ dμξn (s) < ∞. ξn

(23.66)

In case of f being uniformly, continuous and as n → ∞ and ξn → 0, given that ξn are uniformly bounded, we derive   [0] θ f − f  → 0 r,n ∞

(23.67)

with rates. We give   Theorem 23.16 Let f ∈ C B R N , N ≥ 1, ξn ∈ (0, 1]. Then (1)  [0]  W f − f  ≤ r,n ∞



2 √ π

N    er ! N (1 + N )r + 2r e

ωr ( f, ξn ) √ r , ξn (23.68)

where r, n ∈ N.   (2) Let r ∈ N − {1}, 0 < α ≤ 1, and consider f ∈ C B R N such that ωr ( f, t) ≤ K t r −1+α , ∀ t > 0, where K > 0 is independent of t. In this case ωr ( f, ξn ) ( r +α−1) , (23.69) √ r ≤ K ξn 2 ξn and as n → ∞ and ξn → 0, we derive   [0] W f − f  → 0. r,n ∞ Proof By Theorems 23.9 and 23.15. For (23.69), since r ≥ 2, we get that 1 > 0.

[m] 23.3.2 L p Approximation for Wr,n

We need

(23.70) r 2

+α− 

412

23 High Order Approximation with Multivariate Generalized …

Definition 23.17 ([5, 9]) We call ru f (x) := ru 1 ,u 2 ,...,u N f (x1 , ..., x N ) := r

r− j

(−1)

j=0

(23.71)

  r f (x1 + ju 1 , x2 + ju 2 , ..., x N + ju N ) . j

Let p ≥ 1, the modulus of smoothness of order r is given by   ωr ( f ; h) p := sup ru ( f ) p ,

(23.72)

u 2 ≤h

h > 0. We will apply.     Theorem 23.18 ([3], p. 24) Let f ∈ C m R N , m ∈ N, N ≥ 1, with f α ∈ L p R N , |α| = m, x ∈ R N . Let p, q > 1 : 1p + q1 = 1. Here, μξn is a Borel probability measure on R N for ξn > 0, (ξn )n∈N bounded sequence. Assume for all α := (α1 , ..., α N ), N  αi ∈ Z+ , i = 1, ..., N , |α| := αi = m that we have i=1



RN

N 

|si |

αi



s 2 1+ ξn

i=1

r  p dμξn (s) < ∞.

(23.73)

For  j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αj =  j,

j=1

call

cα,n,j :=

N  R N i=1

siαi dμξn (s) .

(23.74)

Then  ⎛ ⎞     m  ⎜ ⎟ [m] ⎜ cα,n,j f α (x) ⎟  [m]   [m] E  =  δj,r ⎜ N  ⎟ θr,n ( f ; x) − f (x) − r,n p   ⎝ ⎠ *    |α|= j j=1 αi !   i=1  ≤



m 1

(q (m − 1) + 1) q

⎛

⎞

⎜ 1 ⎟ ⎜ ⎟ ⎜ ⎟ N ⎝|α|=m * ⎠ αi ! i=1

(23.75) p,x

[m] 23.3 Main Results for Wr,n



 N 

RN

413

αi



|si |

i=1

s 2 1+ ξn

r

1 p

p

dμξn (s)

ωr ( f α , ξn ) p .

 [m]   → 0 with rates. As n → ∞ and ξn → 0, by (23.75) we obtain that  Er,n One also finds by (23.75) that ⎛

 [m]  θ ( f ; x) − f (x) r,n p,x

⎞ m ⎟ -⎜ ⎟ - [m] - ⎜ -cα,n,j f α p ⎟ + R.H.S.(23.75), ≤ -δj,r - ⎜ N ⎝ ⎠ *  |α|= j j=1 αi ! i=1

(23.76) given that f α p < ∞, |α| =  j,  j = 1, ..., m.  [m]  Assuming that cα,n,j → 0, ξn → 0, as n → ∞, we get θr,n ( f ) − f  p → 0, that [m] → I the unit operator, in L p norm, with rates. is θr,n We present.     Theorem 23.19 Let f ∈ C m R N , m ∈ N, m > r , N ≥ 1, with f α ∈ L p R N , |α| = m, x ∈ R N . Let p, q > 1 : 1p + q1 = 1, ξn ∈ (0, 1]. For  j = 1, ..., m, and N  αi =  j, call α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| := i=1

1  cα,n,j = √  N πξn





N 

R N i=1

siαi e−

N  si2 i=1 ξn



ds1 ...ds N .

(23.77)

Then  ⎛ ⎞     m   ⎜ ⎟  [m]   c f (x)  α α,n, j   ⎜ ⎟ [m] [m] E r,n  = Wr,n δ f ; x) − f − ( (x)    ⎜ ⎟  p j,r ⎝ N   ⎠ *    |α|= j j=1 α ! i   i=1  ≤

2 √ π

 Np 



m 1 q

(q (m − 1) + 1) ⎞ ⎛

p,x

' (1 (1 + N )r p + 2r p  N ((m + r ) p + 1, 1) p (23.78)

⎜   1 ⎟ ⎟  (m−r ) ⎜ ωr ( f α , ξn ) p . ξ ⎟ ⎜ n N ⎠ ⎝|α|=m * αi ! i=1

 [m]  r,n  → 0 with rates. As n → +∞ and ξn → 0, by (23.78) we obtain that  E p

414

23 High Order Approximation with Multivariate Generalized …

One also finds by (23.78) that ⎛

  [m] W ( f ; x) − f (x) r,n p,x

⎞ m -⎜ ⎟ cα,n,j - [m] - ⎜ - ⎟ f α p ⎟ + R.H.S.(23.78), ≤ -δj,r - ⎜ N ⎝ ⎠ *  |α|= j j=1 αi ! i=1

(23.79) given that f α p < ∞, |α| =  j,  j = 1, ..., m.  [m]  [m] → Assuming that ξn → 0, as n → ∞, we get Wr,n ( f ) − f  p → 0, that is Wr,n I the unit operator, in L p norm, with rates. -   N √ m  μ  N cα,n,j - ≤ √μπ Proof By Theorem 23.12 we get that - ξn ≤ √π and  cα,n,j → 0. We use also Theorem 23.10 to uniformly bound   N   N  si2 r  p  i=1 * 1 2 |si |αi 1 + s e− ξn ds1 ...ds N . Then we apply √ N ξn ( πξn ) N R i=1 directly Theorem 23.18.  We continue with an application of      Theorem 23.20 ([3], p. 26) Let f ∈ C R N ∩ L p R N ; N ≥ 1; p, q > 1 : 1p + 1 = 1. Assume μξn probability Borel measure on R N , (ξn )n∈N > 0 and bounded. q Also suppose   s 2 r p dμξn (s) < ∞. (23.80) 1+ ξn RN Then

  [0] θ ( f ) − f  ≤ r,n p  RN

(23.81)

 1p   s 2 r p 1+ dμξn (s) ωr ( f, ξn ) p . ξn

 [0]  [0] As ξn → 0, when n → ∞, we derive θr,n → I, the unit ( f ) − f  p → 0, i.e. θr,n operator, in L p norm. We give

     Theorem 23.21 Let f ∈ C R N ∩ L p R N ; N ≥ 1; p, q > 1 : (0, 1], n ∈ N. Then (1)

1 p

+

1 q

 [0]  W ( f ) − f  ≤ r,n p 

2 √ π

 Np

' ( 1 ωr ( f, ξn ) p (1 + N )r p + 2r p  N (r p + 1, 1) p √ r . ξn

= 1, ξn ∈

(23.82)

[m] 23.3 Main Results for Wr,n

415

     (2) Let r ∈ N − {1}, 0 < α ≤ 1, and consider f ∈ C R N ∩ L p R N : ωr ( f, t) p ≤ K t r −1+α , ∀ t > 0, where K > 0 is independent of t. In this case ωr ( f, ξn ) p ( r +α−1) , √ r ≤ K ξn 2 ξn

(23.83)

and as n → ∞ and ξn → 0, we obtain   [0] W f − f  → 0, r,n p

(23.84)

[0] → I, the unit operator in L p norm. i.e. Wr,n

Proof By Theorems 23.11 and 23.20. For r ≥ 2 we get

r 2

+ α − 1 > 0.



We mention

     Theorem 23.22 ([3], p. 27) Let f ∈ C R N ∩ L 1 R N ; N ≥ 1. Assume μξn probability Borel measure on R N , (ξn )n∈N > 0 and bounded. Also suppose RN

  s 2 r 1+ dμξn (s) < ∞. ξn

Then

  [0] θ ( f ) − f  ≤ r,n 1  RN

(23.85)

(23.86)

   s 2 r 1+ dμξn (s) ωr ( f, ξn )1 . ξn

[0] As ξn → 0, we get θr,n → I, in L 1 norm.

We give

     Theorem 23.23 Let f ∈ C R N ∩ L 1 R N ; N ≥ 1, ξn ∈ (0, 1], n ∈ N. Then (1)  [0]  W ( f ) − f  ≤ r,n 1



N    er ! N (1 + N )r + 2r e

ωr ( f, ξn )1 √ r . ξn (23.87)      (2) Let r ∈ N − {1}, 0 < α ≤ 1, and consider f ∈ C R N ∩ L 1 R N : ωr ( f, t)1 ≤ K t r −1+α , ∀ t > 0, where K > 0 is independent of t. In this case 2 √ π

ωr ( f, ξn )1 ( r +α−1) , √ r ≤ K ξn 2 ξn

(23.88)

 [0]  [0] → I in and as n → ∞ and ξn → 0, we derive Wr,n ( f ) − f 1 → 0, i.e. Wr,n L 1 norm.

416

23 High Order Approximation with Multivariate Generalized …

Proof By Theorems 23.9 and 23.22. Notice

r 2

+ α − 1 > 0.



We need

    Theorem 23.24 ([3], p. 29) Let f ∈ C m R N , m, N ∈ N, with f α ∈ L 1 R N , |α| = m, x ∈ R N . Here, μξn is a Borel probability measure on R N for ξn > 0, (ξn )n∈N is a bounded sequence. Suppose for all α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , N  |α| := αi = m that we have i=1



RN

N 

|si |αi



i=1

s 2 1+ ξn

r  dμξn (s) < ∞.

(23.89)

For  j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αi =  j,

i=1

call

cα,n,j :=

N  RN

siαi dμξn (s) .

(23.90)

i=1

Then  ⎛ ⎞     m  ⎜ ⎟ [m] ⎜ cα,n,j f α (x) ⎟  [m]   [m] E  =  δj,r ⎜ θr,n ( f ; x) − f (x) − ⎟ r,n 1 N   ⎝ ⎠ *    |α|= j j=1 αi !   i=1

(23.91)

1,x

⎛

⎞

⎟ ⎜ ⎜ 1 ⎟ ≤ ⎜ N ⎟ ωr ( f α , ξn )1 ⎝* ⎠ RN |α|=m αi !

N  i=1

αi

|si |

 1+

s 2 ξn

r dμξn (s) .

i=1

 [m]   → 0 with rates. As ξn → 0, we get  Er,n 1 From (23.91) we get ⎛

   [m]  θr,n f − f 

1

⎞ m ⎟ -⎜ ⎟ - [m] - ⎜ -cα,n,j f α 1 ⎟ + R.H.S.(23.91), ≤ -δj,r - ⎜ N ⎝ ⎠ *  |α|= j=1 j αi !

(23.92)

i=1

j,  j = 1, ..., m. given that f α 1 < ∞, |α| =   [m]  As n → ∞, assuming ξn → 0 and cα,n,j → 0, we obtain θr,n ( f ) − f 1 → 0, [m] → I in L 1 norm, with rates. that is θr,n We present

[m] 23.3 Main Results for Wr,n

417

    Theorem 23.25 Let f ∈ C m R N , m, N ∈ N, m > r , with f α ∈ L 1 R N , |α| = m, x ∈ R N , ξn ∈ (0, 1], n ∈ N. For  j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , N  αi =  j, call i = 1, ..., N , |α| := i=1

1  cα,n,j := √  N πξn





N  RN

siαi e−

N  si2 i=1 ξn



ds1 ...ds.

(23.93)

i=1

Then  ⎛ ⎞     m  ⎜ ⎟ [m] ⎜   [m]  cα,n,j f α (x) ⎟   [m] E  r,n = Wr,n ( f ; x) − f (x) − δj,r ⎜ ⎟ 1 N   ⎝ ⎠ *    |α|= j j=1 αi !   i=1

≤

1,x



    e (m + r )! N 2 N √ (1 + N )r + 2r e π

⎛

⎞

⎟  ⎜ ⎜ 1 ⎟  m−r ξn ωr ( f α , ξn )1 . ⎜ ⎟ N ⎝* ⎠ |α|=m αi ! i=1

 [m]  r,n  → 0 with rates. As ξn → 0, we get  E 1 From (23.94) we get

(23.94)

⎛

⎞ m -⎜ ⎟   [m] cα,n,j - [m] - ⎜ - ⎟ W f − f  ≤ f α 1 ⎟ + R.H.S.(23.94), -δj,r - ⎜ r,n 1 N ⎝ ⎠ *  |α|= j j=1 αi !

(23.95)

i=1

given that f α 1 < ∞, |α| =  j,  j = 1, ..., m.  [m]  [m] → As n → ∞, assuming ξn → 0, we obtain Wr,n ( f ) − f 1 → 0, that is Wr,n I in L 1 norm, with rates. Proof We apply Theorems 23.8, 23.12 and 23.24.



23.3.3 Global Smoothness Preservation and Simultaneous [m] Approximation of Wr,n We need   Definition 23.26 ([3], p. 34) Let f ∈ C R N , N ≥ 1, m ∈ N, the mth modulus of smoothness for 1 ≤ p ≤ ∞, is given by

418

23 High Order Approximation with Multivariate Generalized …

   ωm ( f ; h) p := sup m t ( f ) p,x , t 2 ≤h

h > 0, where m t

f (x) :=



m

(−1)

m− j

j=0

m j

(23.96)

 f (x + jt) .

(23.97)

Denote ωm ( f ; h)∞ = ωm ( f, h) .

(23.98)

Above, x, t ∈ R N . We present the related global smoothness preservation result m] [ Theorem 23.27 We assume Wr,n  ∈ Z+ , ∀ x ∈ R. Let h > 0, f ∈ ( f ; x) ∈ R, m  N C R , N ≥ 1.

(i) Assume ωm ( f, h) < ∞. Then ωm



⎞ r m] [ m] [ Wr,n f, h ≤ ⎝ -αj,r -⎠ ωm ( f, h) . ⎛



(23.99)

 j=0

     (ii) Assume f ∈ C R N ∩ L 1 R N . Then 

m] [ ωm Wr,n f, h

⎞ r - [ m]≤⎝ -αj,r -⎠ ωm ( f, h)1 . ⎛

 1

(23.100)

 j=0

     (iii) Assume f ∈ C R N ∩ L p R N , p > 1. Then 

m] [ ωm Wr,n f, h

⎞ r m] [ ≤⎝ -αj,r -⎠ ωm ( f, h) p . ⎛

 p

(23.101)

 j=0



Proof Direct application of [3] Theorem 3.2, p. 35. We make m] m] [ [ Remark 23.28 Let r = 1, m  ∈ Z+ , then α0,1 = 0, α1,1 = 1. Hence ⎛

m] [ W1,n ( f ; x) = √

1 πξn

N

RN

N  s2

⎞

i ⎟ ⎜ −⎝ i=1 ξn ⎠

f (x + s) e

ds1 ...ds N =: Wn ( f ; x) . (23.102)

[m] 23.3 Main Results for Wr,n

419

By Theorem 23.27, we get   Theorem 23.29 We suppose Wn ( f ; x) ∈ R, ∀ x ∈ R. Let h > 0, f ∈ C R N , N ≥ 1. (i) Assume ωm ( f, h) < ∞. Then ωm (Wn f, h) ≤ ωm ( f, h) .

(23.103)

     (ii) Assume f ∈ C R N ∩ L 1 R N . Then ωm (Wn f, h)1 ≤ ωm ( f, h)1 .

(23.104)

     (iii) Assume f ∈ C R N ∩ L p R N , p > 1. Then ωm (Wn f, h) p ≤ ωm ( f, h) p .

(23.105)

Next, we get an optimality result Proposition 23.30 Above inequality (23.103): ωm (Wn f, h) ≤ ωm ( f, h) is sharp, namely it is attained by any   f j∗ (x) = x mj , j = 1, ..., N , x = x1 , ..., x j , ..., x N ∈ R N . Proof Apply Proposition 3.5, p. 38, of [3].

(23.106) 

We need   Theorem 23.31 ([3], p. 39) Let f ∈ C l R N , l, N ∈ N. Here, μξn is a Borel probability measure on R N , ξn > 0, (ξn )n∈N a bounded sequence. Let β := (β1 , ..., β N ), N  βi ∈ Z+ , i = 1, ..., N ; |β| : = βi =l. Here f (x + s j), x, s ∈ R N , is μξn -integrable i=1

w.r.t. s, for j = 1, ..., r . There exist μξn -integrable functions h i1 , j , h β1 ,i2 , j , h β1 ,β2 ,i3 , j , ..., h β1 ,β2 ,...,β N −1 ,i N , j ≥ 0 ( j = 1, ..., r ) on R N such that - ∂ i1 f (x + s j) - ≤ h i1 , j (s) , i 1 = 1, ..., β1 , ∂x1i1 - ∂ β1 +i2 f (x + s j) - ≤ h β1 ,i2 , j (s) , i 2 = 1, ..., β2 , β1 i2 ∂x ∂x 2

1

.. .

(23.107)

420

23 High Order Approximation with Multivariate Generalized …

- ∂ β1 +β2 +...+β N −1 +i N f (x + s j) - ≤ h β1 ,β2 ,...,β N −1 ,i N , j (s) , i N = 1, ..., β N , - ∂x i N ∂x β N −1 ...∂x β2 ∂x β1 N −1

N

2

1

∀ x, s ∈ R N . Then, both of the next exist and     [ m] m] [ fβ ; x , m  ∈ Z+ . θr,n ( f ; x) β = θr,n In particular it holds



m] [ Wr,n ( f ; x)

 β

  m] [ fβ ; x , = Wr,n

when

⎛

dμξn = √

N  s2 ⎜ i=1 i −⎝ ξn

1 πξn

N e

(23.108)

(23.109)

⎞ ⎟ ⎠

ds1 ...ds N .

Corollary 23.32 (by Theorem 23.31, r = 1) It holds   (Wn ( f ; x))β = Wn f β ; x .

(23.110)

We present simultaneous global smoothness results. Theorem 23.33 Let h> 0 and the assumptions of Theorem 23.31 are valid for ⎞ ⎛ N  s2

i ⎟ ⎜ −⎝ i=1 ξn ⎠

dμξn =

(

1 √ N πξn )

ds1 ...ds N . Here γ = 0, β (0 = (0, ..., 0)), m  ∈ Z+ .

e 



(i) Assume ωm f γ , h < ∞. Then ωm



⎞ r   m [] m] [ Wr,n (f) γ ,h ≤ ⎝ -αj,r -⎠ ωm f γ , h . 



⎛

(23.111)

 j=0

  (ii) Additionally suppose f γ ∈ L 1 R N . Then ωm



m] [ Wr,n (f)

 γ

,h

⎞ r   - [ m]≤⎝ -αj,r -⎠ ωm f γ , h 1 . ⎛

 1

(23.112)

 j=0

  (iii) Additionally suppose f γ ∈ L p R N , p > 1. Then ωm



m] [ Wr,n (f)

 γ

,h

⎞ r   m] [ ≤⎝ -αj,r -⎠ ωm f γ , h p . ⎛

 p

 j=0

(23.113)

[m] 23.3 Main Results for Wr,n

421

It follows Corollary 23.34 (to Theorem 23.33) Let h > 0, r = 1 and γ = 0, β.   (i) Assume ωm f γ , h < ∞. Then     ωm (Wn ( f ))γ , h ≤ ωm f γ , h .

(23.114)

  (ii) Additionally suppose f γ ∈ L 1 R N . Then     ωm (Wn ( f ))γ , h 1 ≤ ωm f γ , h 1 .

(23.115)

  (iii) Additionally suppose f γ ∈ L p R N , p > 1. Then     ωm (Wn ( f ))γ , h p ≤ ωm f γ , h p .

(23.116)

Next comes multi-simultaneous approximation.   Theorem 23.35 Let f ∈ C m+l R N , m, l, N ∈ N, m > r . The assumptions of ⎛

N  s2

⎞

i ⎟ ⎜ −⎝ i=1 ξn ⎠

Theorem 23.31 are valid for dμξn = √ 1 N e ( πξn )   Assume  f γ+α ∞ < ∞, and let 1

N 



 u ξn (α) := √  N πξn

RN

|si |αi

i=1

for all α j ∈ Z+ , j = 1, ..., N , |α| :=

N 

ds1 ...ds N . Call γ = 0, β.



s 2 1+ ξn

r



e−

N  si2 i=1 ξn



ds1 ...ds N ,

α j = m, ξn ∈ (0, 1].

j=1

For  j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αi =

i=1

 j, call  cα,n,j := √

1 πξn



N

N  RN

siαi e−

N  si2 i=1 ξn



ds1 ...ds N .

(23.117)

i=1

Then  ⎞ ⎛     m   ⎟ ⎜   c f α,n, j γ+α (·) ⎟  [m] [m] ⎜ δj,r ⎜   Wr,n ( f ; ·) γ − f γ (·) − ⎟ N  ⎠ ⎝ *  α1 ,...,α N ≥0;   j=1 α ! i   |α|= j i=1

∞

(23.118)

422

23 High Order Approximation with Multivariate Generalized …

   ωr f γ+α , ξn ≤ u ξn (α) . N   * α1 ,...,α N ≥0; α ! i |α|=m

i=1



Proof Based on Theorems 23.14, 23.31. We continue with

  Theorem 23.36 Let f ∈ C lB R N , l, N ∈ N (functions l-times continuously differentiable and bounded). The assumptions of Theorem 23.31 are valid for dμξn = ⎞ ⎛ N  s2

i ⎟ ⎜ −⎝ i=1 ξn ⎠

√

(

1 N πξn )

e

ds1 ...ds N . Call γ = 0, β, ξn ∈ (0, 1]. Then

(1) N    er ! N r r (1 + N ) + 2 e

  ωr f γ , ξn ≤ √ r . ∞ ξn (23.119)   (2) Let r ∈ N − {1}, 0 < α ≤ 1, and consider f γ : ωr f γ , t ≤ K t r −1+α , ∀ t > 0, where K > 0 is independent of t. We get 

    [0]   Wr,n f γ − f γ 

2 √ π

  ωr f γ , ξn ( r +α−1) , √ r ≤ K ξn 2 ξn

(23.120)

 [0]  f γ → f γ uniformly. and as n → ∞ and ξn → 0, we derive Wr,n 

Proof By Theorems 23.16 and 23.31. We continue with

  Theorem 23.37 Let f ∈ C m+l R N , m, l, N ∈ N, m > r . The assumptions of The⎛

N  s2

⎞

i ⎟ ⎜ −⎝ i=1 ξn ⎠

orem 23.31 are valid for dμξn = √ 1 N e ds1 ...ds N . Call γ = 0, β. Let ( πξn )  N N f (γ+α) ∈ L p R , |α| = m, x ∈ R , and p, q > 1 : 1p + q1 = 1, ξn ∈ (0, 1], n ∈ N. N  αi =  j, For  j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| := i=1

call  cα,n,j := √ Then

1 πξn

N



N  R N i=1

siαi e−

N  si2 i=1 ξn



ds1 ...ds N .

(23.121)

[m] 23.3 Main Results for Wr,n

423

 ⎛ ⎞     m  ⎜ ⎟ [m] ⎜   cα,n,j f γ+α (x) ⎟   [m] δj,r ⎜  Wr,n ( f ; x) γ − f γ (x) − ⎟ N   ⎝ ⎠ *    |α|= j j=1 αi !   i=1  ≤

2 √ π

 Np 



m (q (m − 1) + 1)

1 q

p,x

(1 ' (1 + N )r p + 2r p  N ((m + r ) p + 1, 1) p

⎞

⎛

⎜    1 ⎟ ⎟  (m−r )  ⎜ ξ ωr f γ+α , ξn p . ⎟ ⎜ n N ⎠ ⎝|α|=m * αi !

(23.122)

i=1



Proof By Theorems 23.19, 23.31. We give also

  Theorem 23.38 Let f ∈ C l R N , l, N ∈ N. The assumptions of Theorem 23.31 ⎛

N  s2

⎞

i ⎟ ⎜ −⎝ i=1 ξn ⎠

are valid for dμξn = x ∈ R N ; p, q > 1 :

1 √ N e πξn ) 1 1 + q = 1, ξn p

(

(1)

  ds1 ...ds N . Call γ = 0, β. Let f γ ∈ L p R N ,

∈ (0, 1], n ∈ N. Then

    [0]   Wr,n f γ − f γ  ≤ p



2 √ π

 Np

'

  ( 1p ωr f γ , ξn p √ r . (1 + N ) + 2  (r p + 1, 1) ξn rp

rp

N

(23.123)

  (2) Let r ∈ N − {1}, 0 < α ≤ 1, and f γ : ωr f γ , t p ≤ K t r −1+α , ∀ t > 0, K > 0 independent of t. In this case   ωr f γ , ξn p ( r +α−1) . √ r ≤ K ξn 2 ξn

(23.124)

 [0]  · p As n → +∞ and ξn → 0, then Wr,n ( f ) γ → fγ . Proof By Theorems 23.21 and 23.31. We continue with



424

23 High Order Approximation with Multivariate Generalized …

  Theorem 23.39 Let f ∈ C l R N , l, N ∈ N. The assumptions of Theorem 23.31 ⎛

N  s2

⎞

i ⎟ ⎜ −⎝ i=1 ξn ⎠

are valid for dμξn =

√

1 N πξn )

e

( x ∈ R N , ξn ∈ (0, 1], n ∈ N. Then

(1)

  ds1 ...ds N . Call γ = 0, β. Let f γ ∈ L 1 R N ,

     [0]  Wr,n ( f ) γ − f γ  ≤ 1



2 √ π

N 

 (1 + N ) + 2 r

r

  ωr f γ , ξn 1 √ r . ξn

N

er ! e

(23.125)

  (2) Let r ∈ N − {1}, 0 < α ≤ 1, and f γ : ωr f γ , t 1 ≤ K t r −1+α , ∀ t > 0, K > 0 independent of t. In this case   ωr f γ , ξn 1 ( r +α−1) , √ r ≤ K ξn 2 ξn

(23.126)

 [0]  · 1 and as n → ∞ and ξn → 0, we get Wr,n ( f ) γ → fγ . 

Proof By Theorems 23.23 and 23.31. We continue with

  Theorem 23.40 Let f ∈ C m+l R N , m, l, N ∈ N, m > r . The assumptions of The⎛

N  s2

⎞

i ⎟ ⎜ −⎝ i=1 ξn ⎠

orem 23.31 for dμξn = √ 1 N e ds1 ...ds N are valid. Call γ = 0, β. Let ( πξn )  N N f (γ+α) ∈ L 1 R , |α| = m, x ∈ R , ξn ∈ (0, 1], n ∈ N. For  j = 1, ..., m, and N  α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| := αi =  j, call i=1

 cα,n,j := √ Then

1 πξn



N

N  RN

siαi e−

N  si2 i=1 ξn



ds1 ...ds N .

(23.127)

i=1

 ⎞ ⎛     m  ⎟ ⎜ [m] ⎜   cα,n,j f (γ+α) (x) ⎟   [m] δj,r ⎜  Wr,n ( f ; x) γ − f γ (x) − ⎟ N   ⎠ ⎝ *    |α|= j j=1 αi !   i=1

1,x

[m] 23.3 Main Results for Wr,n

425

⎛  ≤

2 √ π

⎛

⎞

⎞

N ⎜ ⎜ ⎟  ⎟ ⎜ ⎜ 1 ⎟  ⎟ ⎜ ⎜ N ⎟ ωr f γ+α , ξn 1 ⎟ ⎝|α|=m ⎝ * ⎠ ⎠ αi ! i=1

    m−r e (m + r )! N r r ξn . (1 + N ) + 2 e

(23.128) 

Proof By Theorems 23.25, 23.31.

[m] 23.3.4 Voronovskaya Asymptotic Expansions for Wr,n

Here we will apply the following general theorem.   Theorem 23.41 Let f ∈ C m R N , m, N ∈ N, with all f α ∞ ≤ M, M > 0, all α : |α| = m. Let ξn > 0, (ξn )n∈N bounded sequence, μξn probability Borel measures on R N . N   * αi Call cα,n,j := R N si dμξn (s), all |α| =  j = 1, ..., m − 1. Suppose N i=1 √ −m  * |si |αi dμξn (s) ≤ ρ, all α : |α| = m, ρ > 0, for any such (ξn )n∈N . ξn RN i=1

Also 0 < γ ≤ 1, x ∈ R N . Then ⎛ [m] θr,n

( f ; x) − f (x) =

m−1  j=1

δ[m] j,r

⎞

   ⎜ c  f (x) ⎟ m−γ α,n, j α ⎜ ⎟ . (23.129) ξn N  ⎟+0 ⎜ ⎝  * ⎠ |α|= j αi ! i=1

When m = 1, the sum collapses. [m] Above we assume θr,n ( f ; x) ∈ R, ∀ x ∈ R N . Proof The proof of Theorem 23.41 is the same as the proof of Theorem 4.2, p. 53 √ of [3], just replace there ξn by ξn .  We give   Theorem 23.42 Let f ∈ C m R N , m, N ∈ N, with all f α ∞ ≤ M, M > 0, all ⎛

N  s2

⎞

i ⎟ ⎜ −⎝ i=1 ξn ⎠

α : |α| = m, ξn ∈ (0, 1], and dμξn = [m] Wr,n ( f ; x) ∈ R, ∀ x ∈ R N .

(

1 √ N πξn )

e

ds1 ...ds N on R N . Assume

426

23 High Order Approximation with Multivariate Generalized … ⎛

Call cα,n,j =

1 √ ( πξn ) N



 RN

N *

i=1

 siαi e

N  s2

⎞

i ⎟ ⎜ −⎝ i=1 ξn ⎠

ds1 ...ds N , all |α| =  j = 1, ..., m −

1. Let 0 < γ ≤ 1, x ∈ R N . Then

⎛ [m] Wr,n ( f ; x) − f (x) =

⎞

   ⎜  cα,n,j f α (x) ⎟ m−γ ⎜ ⎟ . δ[m] ξ + 0 N  ⎟ ⎜ n j,r ⎝ ⎠ *   |α|= j j=1 αi !

m−1

i=1

(23.130) When m = 1, the sum collapses. Proof By Theorems 23.12 and 23.41. Here ρ =



√μ π

N

.



    ∂f  Corollary 23.43 (to Theorem 23.42) Let f ∈ C 1 R N , N ≥ 1, with all  ∂x  i M, M > 0, i = 1, ..., N . Let 0 < γ ≤ 1, x ∈ R . Then N

[1] Wr,n

   1−γ . ξn ( f ; x) − f (x) = 0

∞

≤

(23.131)

[1] Above we assume Wr,n ( f ; x) ∈ R, ∀ x ∈ R N .

Proof By Theorems 23.12 and 23.41, here it is ρ = 1.         2   2  Corollary 23.44 (to Theorem 23.42) Let f ∈ C 2 R2 , with all  ∂∂x 2f  ,  ∂∂x 2f  , 1 2 ∞ ∞  2   ∂ f   ∂x1 ∂x2  ≤ M, M > 0, ξn ∈ (0, 1], n ∈ N. Call ∞

c1 =

R2

s1 dμξn (s) , c2 =

where dμξn = √

R2

s2 dμξn (s) ,

(23.132)

1 − (s12ξ+s22 ) n e ds1 ds2 . πξn

Let 0 < γ ≤ 1, x ∈ R2 . Then ⎛ [2] Wr,n

( f ; x) − f (x) = ⎝

r j=1

⎞

α[2] j,r

     2−γ ∂f ∂f ⎠ c1 . j ξn (x) + c2 (x) + 0 ∂x1 ∂x2 (23.133)

Proof By Theorems 23.12 and 23.41, here it is ρ = 1. We also give



[m] 23.3 Main Results for Wr,n

427

  Theorem 23.45 Let f ∈ C m+l R N , m, l, N ∈ N. Assumptions of Theorem 23.31 ⎛

N  s2

⎞

i ⎟ ⎜ −⎝ i=1 ξn ⎠

are valid for dμξn (s) = √ 1 N e ds1 ...ds N . Call γ = 0, β. Suppose ( πξn )    f γ+α  ≤ M, M > 0, for all α : |α| = m, ξn ∈ (0, 1], n ∈ N. ∞ ⎛

Call  cα,n,j :=

1 √ ( πξn ) N



 RN

N *

i=1

 siαi e

N  s2

⎞

i ⎟ ⎜ −⎝ i=1 ξn ⎠

ds1 ...ds N , all |α| =  j = 1, ...,

m − 1; 0 < γ ≤ 1, x ∈ R N . Then

⎛ 

[m] Wr,n ( f ; x)

 γ

− f γ (x) =

⎞

   ⎜  m−γ cα,n,j f γ+α (x) ⎟ ⎜ ⎟ . δ[m] ξ + 0 N  ⎟ ⎜ n j,r ⎝ ⎠ *   |α|= j j=1 αi !

m−1

i=1

(23.134) When m = 1, the sum collapses. Proof Use of Theorem 23.12 and Theorem 4.6, p. 54 of [3], which is true also for  N √ ξn instead of ξn in the main assumption and conclusion. Here it is ρ = √μπ .

23.3.5 Simultaneous Approximation by Multivariate [m] Complex Wr,n We make Remark 23.46 We consider here complex √ valued Borel measurable functions f : R N → C such that f = f 1 + i f 2 , i = −1, where f 1 , f 2 : R N → R are implied to be real valued Borel measurable functions. We define the multivariate complex Gauss–Weierstrass singular operators [m] [m] [m] Wr,n ( f ; x) := Wr,n ( f 1 ; x) + i Wr,n ( f 2 ; x) , x ∈ R N .

(23.135)

  [m] We assume that Wr,n f j ; x ∈ R, ∀ x ∈ R N , j = 1, 2. One notices easily that - - - [m] -W ( f ; x) − f (x)- ≤ -W [m] ( f 1 ; x) − f 1 (x)- + -W [m] ( f 2 ; x) − f 2 (x)r,n r,n r,n (23.136) also  [m]  [m]  [m]    W ( f ; x) − f (x) ≤ Wr,n ( f 1 ; x) − f 1 (x)∞,x + Wr,n ( f 2 ; x) − f 2 (x)∞,x r,n ∞,x

(23.137)

428

23 High Order Approximation with Multivariate Generalized …

and       [m] W ( f ) − f  ≤ W [m] ( f 1 ) − f 1  + W [m] ( f 2 ) − f 2  , p ≥ 1. (23.138) r,n r,n r,n p p p Furthermore, it holds f α (x) = f 1,α (x) + i f 2,α (x) ,

(23.139)

where α denotes a partial derivative of any order and arrangement. We give Theorem 23.47 Let f : R N → C, N ≥ 1, such that f = f 1 + i f 2 , j = 1, 2. Here    ∂ m f (·,·,...,·)  m ∈ N, f j ∈ C m R N , x ∈ R N . Assume  ∂x α1j ...∂x α N  < ∞, for all αi ∈ Z+ , i = 1, ..., N : |α| :=

N 

1

∞

N

αi = m, m > r. Here ξn ∈ (0, 1], n ∈ N. For all α :=

i=1

(α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αi = m we denote

i=1 ⎛

 u ξn (α) := √

1 πξn

N

RN

N 

|si |αi

 1+

i=1

s 2 ξn

N  s2

⎞

i ⎟ ⎜ −⎝ i=1 ξn ⎠

r e

ds1 ...ds N .

(23.140) N  αi =  j, For  j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z , i = 1, ..., N , |α| := +

i=1

call

⎛

1  cα,n :=  cα,n,j := √  N πξn Then



N  RN

siαi e

N  s2

⎞

i ⎟ ⎜ −⎝ i=1 ξn ⎠

ds1 ...ds N .

(23.141)

i=1

 ⎛ ⎞     m  ⎜  ⎟ cα,n,j f α (x) ⎟   [m] [m] ⎜ δj,r ⎜ Wr,n ( f ; x) − f (x) − ⎟ N   ⎝ ⎠ *  α1 ,...,α N ≥0;   j=1 αi !    |α|= j i=1

∞,x

     ωr f 1,α , ξn + ωr f 2,α , ξn ≤  u ξn (α) . N  * α1 ,...,α N ≥0; αi ! |α|=m

(23.142)

i=1

Proof By Theorem 23.14. We proceed with



[m] 23.3 Main Results for Wr,n

429

Theorem 23.48 Let f : R N → C : f = f 1 + i f 2 , N ≥ 1, j = 1, 2. Here f j ∈ C B  N R uniformly continuous, ξn ∈ (0, 1]. Then (1)   [0] W f − f  ≤ r,n ∞



2 √ π

N    er ! N (1 + N )r + 2r e

(ωr ( f 1 , ξn ) + ωr ( f 2 , ξn )) , √ r ξn

(23.143) where r, n ∈ N.   (2) Let r ∈ N − {1}, 0 < α ≤ 1, and ωr f j , t ≤ K t r −1+α , ∀ t > 0, K > 0 is independent of t, j = 1, 2. We get   ωr f j , ξn ( r +α−1) , j = 1, 2 √ r ≤ K ξn 2 ξn

(23.144)

 [0]  and as n → ∞ and ξn → 0, we get Wr,n ( f ) − f ∞ → 0. 

Proof Use of Theorem 23.16. Next comes multi-simultaneous approximation

  Theorem 23.49 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m+l R N , m, l, N ∈ N, m > ⎛r. The⎞assumptions of Theorem 23.31 are valid for N  s2

i ⎟ ⎜ −⎝ i=1 ξn ⎠

dμξn = and let

1 √ ( πξn ) N

e

  ds1 ...ds N and f j . Call γ = 0, β. Assume  f j,γ+α ∞ < ∞, ⎛

 u ξn (α) := √

1 πξn

N 

N

RN

|si |αi

 1+

i=1

for all αi ∈ Z+ , i = 1, ..., N : |α| :=

N 

s 2 ξn

N  s2

⎞

i ⎟ ⎜ −⎝ i=1 ξn ⎠

r e

ds1 ...ds N . (23.145)

αi = m, ξn ∈ (0, 1].

i=1

For  j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

αi =

i=1

 j, call

⎛

 cα,n,j := √ Then

N 

1 πξn

N

N 

R N i=1

siαi e

N  s2

⎞

i ⎟ ⎜ −⎝ i=1 ξn ⎠

ds1 ...ds N .

(23.146)

430

23 High Order Approximation with Multivariate Generalized …

 ⎛ ⎞     m  ⎜ ⎟ [m] ⎜   cα,n,j f γ+α (·) ⎟   [m] δj,r ⎜  Wr,n ( f ; ·) γ − f γ (·) − ⎟ N   ⎝ ⎠ *  α1 ,...,α N ≥0;   j=1 αi !   |α|= j i=1

∞

     ωr f 1,γ+α , ξn + ωr f 2,γ+α , ξn ≤  u ξn (α) . N  * α1 ,...,α N ≥0; α ! i |α|=m

(23.147)

i=1



Proof Based on Theorems 23.31, 23.35. We continue with

  Theorem 23.50 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C lB R N , l, N ∈ N (functions l-times continuously differentiable and bounded). The assump⎞ ⎛ N  s2

i ⎟ ⎜ −⎝ i=1 ξn ⎠

tions of Theorem 23.31 are valid for dμξn = j = 1, 2. Call γ = 0, β, ξn ∈ (0, 1]. Then (1)

√

(

1 N πξn )

    [0]   Wr,n f γ − f γ 

∞

e

ds1 ...ds N and f j ,

≤

N    er ! N r r (1 + N ) + 2 e

     ωr f 1,γ , ξn + ωr f 2,γ , ξn . √ r ξn (23.148)   (2) Let r ∈ N − {1}, 0 < α ≤ 1, and consider f j,γ : ωr f j,γ , t ≤ K t r −1+α , ∀ t > 0, j = 1, 2, where K > 0 is independent of t. We get 

2 √ π

  ωr f j,γ , ξn ( r +α−1) ≤ K ξn 2 , √ r ξn

(23.149)

 [0]  f γ → f γ uniformly. j = 1, 2, and as n → ∞ and ξn → 0, we find Wr,n Proof By Theorems 23.31 and 23.36.



We proceed with L p approximations   Theorem 23.51 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m R N ,  N m ∈ N, m > r , N ≥ 1, with f j,α ∈ L p R , |α| = m, x ∈ R N . Let p, q > 1 : 1p + 1 = 1, ξn ∈ (0, 1]. For  j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , q N  |α| := αi =  j, call i=1

[m] 23.3 Main Results for Wr,n

431



1

 cα,n,j = √  N πξn

R N i=1

siαi e−

N  si2 i=1 ξn



ds1 ...ds.

 ⎛ ⎞     m  ⎜ ⎟ [m] ⎜  cα,n,j f α (x) ⎟   [m] δj,r ⎜ N  ⎟ Wr,n ( f ; x) − f (x) −   ⎝ ⎠ *    |α|= j j=1 αi !   i=1

Then

 ≤



N 

2 √ π

 Np 



m (q (m − 1) + 1)

1 q

(23.150)

p,x

' (1 (1 + N )r p + 2r p  N ((m + r ) p + 1, 1) p

⎞

⎛

⎜   .     / 1 ⎟ ⎟  (m−r ) ⎜ ω . f f ξ , ξ + ω , ξ ⎟ ⎜ n r 1,α n r 2,α n p p N ⎠ ⎝|α|=m * αi !

(23.151)

i=1



Proof By Theorem 23.19. We continue with

N f1 + i f2 , j = 1, 2. Here fj ∈  f : R → C : f = Theorem  N  23.52 Let N C R ∩ L p R ; N ≥ 1; p, q > 1 : 1p + q1 = 1, ξn ∈ (0, 1], n ∈ N. Then

(1)

 [0]  W ( f ) − f  ≤ r,n p 

2 √ π

 Np

(23.152)

  ' ( 1p ωr ( f 1 , ξn ) p + ωr ( f 2 , ξn ) p rp rp N . √ r (1 + N ) + 2  (r p + 1, 1) ξn

  (2) Let r ∈ N − {1}, 0 < α ≤ 1, and assume that ωr f j , t p ≤ K t r −1+α , ∀ t > 0, j = 1, 2, where K > 0 is independent of t. In this case   ωr f j , ξn p ( r +α−1) , j = 1, 2, √ r ≤ K ξn 2 ξn

(23.153)

 [0]  f − f  p → 0. and as n → ∞ and ξn → 0, we get Wr,n 

Proof By Theorem 23.21. We also give N  Let  f : R → C : f = f 1 + i f 2 , Theorem  N  23.53 N C R ∩ L 1 R ; N ≥ 1, ξn ∈ (0, 1], n ∈ N. Then

j = 1, 2.

Here

fj ∈

432

23 High Order Approximation with Multivariate Generalized …

(1)

 [0]  W ( f ) − f  ≤ r,n 1 N    er ! N r r (1 + N ) + 2 e

'

( ωr ( f 1 , ξn )1 + ωr ( f 2 , ξn )1 . √ r ξn (23.154)   (2) Let r ∈ N − {1}, 0 < α ≤ 1, and assume for j = 1, 2 that ωr f j , t 1 ≤ K t r −1+α , ∀ t > 0, K > 0 independent of t. We find that 

2 √ π

  ωr f j , ξn 1 ( r +α−1) , j = 1, 2, √ r ≤ K ξn 2 ξn

(23.155)

 [0]  and as n → ∞ and ξn → 0, we derive Wr,n ( f ) − f 1 → 0. 

Proof By Theorem 23.23. We further present

  Theorem 23.54 Let f : RN → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m R N , m, N ∈ N, with f j,α ∈ L 1 R N , |α| = m, x ∈ R N , ξn ∈ (0, 1], n ∈ N, m > r. For N   αi =  j, call j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| := i=1

1

 cα,n,j := √  N πξn





N  R N i=1

siαi e−

N  si2 i=1 ξn



ds1 ...ds.

(23.156)

Then  ⎞ ⎛     m   ⎟ ⎜  c f (x)  α α,n, j   [m] ⎟ ⎜ δ[m]  Wr,n ( f ; x) − f (x) − ⎟ ⎜ j,r ⎝ N  ⎠ *    |α|= j j=1 α ! i   i=1

(23.157)

1,x

 ≤

    e (m + r )! N 2 N r r √ (1 + N ) + 2 e π

⎞

⎛

⎟ ⎜ ⎜ 1 ⎟ ⎟ ⎜ N ⎠ ⎝* |α|=m αi ! i=1

 m−r '     ( ωr f 1,α , ξn 1 + ωr f 2,α , ξn 1 . ξn Proof By Theorem 23.25.



[m] 23.3 Main Results for Wr,n

433

We continue with simultaneous L p approximations.

  Theorem 23.55 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m+l R N , m, l, N ∈ N, m > ⎛r . The⎞ assumptions of Theorem 23.31 are valid for N  s2

i ⎟ ⎜ −⎝ i=1 ξn ⎠

  1 e ds1 ...ds N and f j . Call γ = 0, β. Let f j,(γ+α) ∈ L p R N , √ ( πξn ) N |α| = m, x ∈ R N , and p, q > 1 : 1p + q1 = 1, ξn ∈ (0, 1], n ∈ N. For  j = 1, ..., m, N  αi =  j, call and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| := dμξn =

i=1

 cα,n,j := √

1 πξn



N

N  R N i=1

siαi e−

N  si2 i=1 ξn



ds1 ...ds N .

 ⎞ ⎛     m  ⎟ ⎜   cα,n,j f γ+α (x) ⎟   [m] [m] ⎜ δj,r ⎜  Wr,n ( f ; x) γ − f γ (x) − ⎟ N   ⎠ ⎝ *    |α|= j j=1 αi !   i=1

Then

 ≤

2 √ π

 Np 



m (q (m − 1) + 1) ⎞

⎛

1 q

(23.158)

p,x

(1 ' (1 + N )r p + 2r p  N ((m + r ) p + 1, 1) p (23.159)

⎜   .     / 1 ⎟ ⎜ ⎟  (m−r ) ω . ξ , ξ + ω , ξ f f ⎜ ⎟ n r 1,γ+α n r 2,γ+α n p p N ⎝|α|=m * ⎠ αi ! i=1



Proof By Theorems 23.31, 23.37. We give also

  Theorem 23.56 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C l R N , l, N ∈ N. The assumptions of Theorem 23.31 are valid for ⎞ ⎛ N  s2

i ⎟ ⎜ −⎝ i=1 ξn ⎠

dμξn =

√

1

( πξn ) N N R ; p, q > 1 :

(1)

e 1 p

+

1 q

  ds1 ...ds N and f j . Call γ = 0, β. Let f j,γ ∈ L p R N , x ∈

= 1, ξn ∈ (0, 1], n ∈ N. Then      [0]  Wr,n ( f ) γ − f γ  ≤ p

(23.160)

434

23 High Order Approximation with Multivariate Generalized …



2 √ π

 Np

.     / ωr f 1,γ , ξn p + ωr f 2,γ , ξn p 1 ' ( . √ r (1 + N )r p + 2r p  N (r p + 1, 1) p ξn

  (2) Let r ∈ N − {1}, 0 < α ≤ 1, with ωr f j,γ , t p ≤ K t r −1+α , ∀ t > 0, K > 0 independent of t, j = 1, 2. In this case   ωr f j,γ , ξn p ( r +α−1) ≤ K ξn 2 , j = 1, 2. √ r ξn

(23.161)

 [0]  · p As n → +∞ and ξn → 0, then Wr,n ( f ) γ → fγ . 

Proof By Theorems 23.31 and 23.38. We continue with

  Theorem 23.57 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C l R N , l, N ∈ N. The assumptions of Theorem 23.31 are valid for ⎞ ⎛ N  s2

i ⎟ ⎜ −⎝ i=1 ξn ⎠

  1 e ds1 ...ds N and f j . Call γ = 0, β. Let f j,γ ∈ L 1 R N , x ∈ √ ( πξn ) N R N , ξn ∈ (0, 1], n ∈ N. Then

dμξn = (1)

     [0]  Wr,n ( f ) γ − f γ  ≤

(23.162)

1



2 √ π

N    er ! N r r (1 + N ) + 2 e

   ( '  ωr f 1,γ , ξn 1 + ωr f 2,γ , ξn 1 . √ r ξn

  (2) Let r ∈ N − {1}, 0 < α ≤ 1, with ωr f j,γ , t 1 ≤ K t r −1+α , ∀ t > 0, K > 0 independent of t, j = 1, 2. In this case   ωr f j,γ , ξn 1 ( r +α−1) ≤ K ξn 2 , j = 1, 2 √ r ξn

(23.163)

 [0]  · 1 and as n → ∞ and ξn → 0, we get Wr,n ( f ) γ → fγ . 

Proof By Theorems 23.31 and 23.39. We finish with

  Theorem 23.58 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m+l R N , ⎛

N  s2

⎞

i ⎟ ⎜ −⎝ i=1 ξn ⎠

m, l, N ∈ N. The assumptions of Theorem 23.31 for dμξn =

(

1 √ N πξn )

e

ds1 ...ds N

[m] 23.3 Main Results for Wr,n

435

  and f j are valid. Call γ = 0, β. Let f j,(γ+α) ∈ L 1 R N , |α| = m, x ∈ R N , ξn ∈ (0, 1], n ∈ N. For  j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| := N  αi =  j, call i=1

1  cα,n,j := √  N πξn





N  RN

siαi e−

N  si2 i=1 ξn



ds1 ...ds N .

 ⎞ ⎛     m  ⎟ ⎜ [m] ⎜   cα,n,j f γ+α (x) ⎟   [m] δj,r ⎜  Wr,n ( f ; x) γ − f γ (x) − ⎟ N   ⎠ ⎝ *    |α|= j j=1 αi !   i=1

Then

⎛  ≤

2 √ π

⎞

⎛

(23.164)

i=1

1,x

⎞

N ⎜ ⎜ ⎟    (⎟ ⎜ ⎜ 1 ⎟'  ⎟ ⎟ ωr f 1,γ+α , ξn 1 + ωr f 2,γ+α , ξn 1 ⎟ ⎜ ⎜ N ⎠ ⎝|α|=m ⎝ * ⎠ αi ! i=1

    m−r e (m + r )! N r r ξn . (1 + N ) + 2 e Proof By Theorems 23.31 and 23.40.

(23.165) 

References 1. Anastassiou, G.A.: Rate of convergence of non-positive linear convolution type operators. A sharp inequality. J. Math. Anal. Appl. 142, 441–451 (1989) 2. Anastassiou, G.A.: Moments in Probability and Approximation Theory, Pitman Research Notes in Math, vol. 287. Longman Scientific & Technical, Harlow (1993) 3. Anastassiou, G.A.: Approximation by Multivariate Singular Integrals. Springer, New York (2011) 4. Anastassiou, G.A.: Complete approximations by multivariate generalized Gauss-Weierstrass singular integrals. Maroccan J. Pure & Appl. Anal. (Accepted) (2020) 5. Anastassiou, G., Gal, S.: Approximation Theory. Birkhaüser, Boston (2000) 6. Anastassiou, G., Mezei, R.: Uniform convergence with rates of smooth Gauss-Weierstrass singular integral operators. Appl. Anal. 88(7), 1015–1037 (2009) 7. Anastassiou, G.A., Mezei, R.A.: L p convergence with rates of smooth Gauss-Weierstrass singular operators. Nonlinear Stud. 17(4), 373–386 (2010) 8. Gradshteyn, I.S., Ryzhik, I.M.: Table of Integrals, Series and Products, 8th edn. Elsevier, Amsterdam (2015) 9. DeVore, R.A., Lorentz, G.G.: Constructive Approximation, vol. 303. Springer, Berlin (1993)

Chapter 24

Complete Approximations with Multivariate Generalized Poisson–Cauchy Type Singular Integral Operators

This research and survey chapter deals exclusively with the study of the approximation of generalized multivariate Poisson–Cauchy type singular integrals to the identity-unit operator. Here we study quantitatively most of their approximation properties. These operators are not in general positive linear operators. In particular we study the rate of convergence of these integral operators to the unit operator, as well as the related simultaneous approximation. These are given via Jackson type inequalities and by the use of multivariate high order modulus of smoothness of the high order partial derivatives of the involved function. Also we study the global smoothness preservation properties of these integral operators. These multivariate inequalities are nearly sharp and in one case the inequality is attained, that is sharp. Furthermore we give asymptotic expansions of Voronovskaya type for the error of approximation. The above properties are studied with respect to L p norm, 1 ≤ p ≤ ∞. It follows [4].

24.1 Introduction We start with our motivation for this chapter. The following come from [9]. In the next we mention and deal with the smooth univariate Poisson–Cauchy Type singular integral operators Mr,ξ ( f, x) defined as follows. For r ∈ N and n ∈ Z+ we set   ⎧ r ⎪ r− j ⎪ j −n , j = 1, ..., r, ⎨ (−1) j   αj = r  r ⎪ ⎪1 − j −n , j = 0, (−1)r − j ⎩ j

(24.1)

j=1

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_24

437

438

that is

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy … r 

α j = 1.

j=0

Let f ∈ C n (R) and f (n) ∈ L p (R), 1 ≤ p < ∞, α ∈ N, β > x ∈ R, ξ > 0 the Lebesgue integral r 

Mr,ξ ( f ; x) = W

∞

we define for

α j f (x + jt)

j=0

−∞

1 , 2α

β t 2α + ξ 2α

dt,

(24.2)

where the constant is defined as W =

 (β) αξ 2αβ−1

1

, 1  2α  β − 2α

where  is the gamma function. Note 24.1 The operators Mr,ξ are not, in general, positive, see [6]. ∞ We notice by W −∞ 2α 1 2α β dt = 1, that Mr,ξ (c; x) = c, c constant, see also (t +ξ ) [6, 8], and ⎛

r 



⎞

∞

1 | f (x + jt) − f (x)|⎠

β dt. 2α −∞ t + ξ 2α j=0 (24.3) We need the r th L p -modulus of smoothness Mr,ξ ( f ; x) − f (x) = W ⎝

αj

 

 ωr f (n) , h p := sup rt f (n) (x) p,x , h > 0, |t|≤h

where rt f (n) (x) :=

r 

(−1)r − j

j=0

  r f (n) (x + jt) , j

(24.4)

(24.5)

 see [10], p. 44. Here we have that ωr f (n) , h p < ∞, h > 0. We need to introduce δk :=

r 

α j j k , k = 1, ..., n ∈ N,

(24.6)

j=1

and denote by · the integral part. Call τ (w, x) :=

r  j=0

α j j n f (n) (x + jw) − δn f (n) (x) .

(24.7)

24.1 Introduction

439

According to [2], p. 306, [1], we get

Thus

τ (w, x) = rw f (n) (x) .

(24.8)

 τ (w, x) p,x ≤ ωr f (n) , |w| p , w ∈ R.

(24.9)

Denote by

Rn (0, t, x) := 0

t

(t − w)n−1 τ (w, x) dw, n ∈ N. (n − 1)!

Using the above terminology we obtain for β >

2 n2 +1 2α

(24.10)

that

 (x) := Mr,ξ ( f ; x) − f (x) − 

 n2  (2m)   β− f (x) δ2m  2m+1

2α1 

(2m)!  β−  2α m=1 where Rn∗ (x) := W



2m+1 2α  1 2α

 ξ 2m = Rn∗ (x) ,

(24.11)

∞

1 Rn (0, t, x)

β dt, n ∈ N. 2α −∞ t + ξ 2α

(24.12)

In  (x), see (24.11), the sum collapses when n = 1. We mention the following result. Theorem 24.2 ([9]) Let p, q > 1 such that   1 1 + n + r and the rest as above. Then α p

+

1 p

1 q

= 1, n ∈ N, α ∈ N, β >

 (x) p ≤ 1

(2α) p  (β)  



qβ 2

 q1







1 2α

1 p

 β−

1 2α





qβ 2

−

1 2α

1/q

1

ξn τ p

1

1

(τ p + 1) p [(n − 1)!] (q (n − 1) + 1) q



ωr f (n) , ξ p , (24.13)

where ⎡



0 < τ := ⎣

0

∞

np−1

u (1 + u)r p+1

 pβ du − u 2α + 1 2

0

∞

⎤ np−1

u ⎦

 pβ du < ∞. u 2α + 1 2 (24.14)

Hence as ξ → 0 we obtain  (x) p → 0.     If additionally f (2m) ∈ L p (R), m = 1, 2, ..., n2 then  Mr,ξ ( f ) − f  p → 0, as ξ → 0.

440

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

The counterpart of Theorem 24.2 follows, case of p = 1. +1 Theorem 24.3 ([9]) Let f ∈ C n (R) and f (n) ∈ L 1 (R), n ∈ N, α ∈ N, β > n+r . 2α Then 1 

1

 (x)1 ≤ (24.15) 1  β − 2α (r + 1) (n − 1)! 2α

 r +1    r + 1 n + k  

 n+k   β− ωr f (n) , ξ 1 ξ n . k 2α 2α k=1 Hence as ξ → 0 we obtain  (x)1 → 0.     If additionally f (2m) ∈ L 1 (R), m = 1, 2, ..., n2 then  Mr,ξ ( f ) − f 1 → 0, as ξ → 0. The case n = 0 is mentioned next. Proposition 24.4 ([9]) Let p, q > 1 such that and the rest as above. Then

1 p

+

1 q

= 1, α ∈ N, β >

1 α

  r + 1p

1  1 1 q 1p − θ (2α) p [ (β)]  qβ   2 2α  Mr,ξ ( f ) − f  ≤ ωr ( f, ξ) p , 1   p 

1  1p

q 1  qβ  2α  β − 2α 2

where 0 < θ :=

0

∞

(1 + t)r p

1

 pβ dt < ∞. t 2α + 1 2

(24.16)

(24.17)

Hence as ξ → 0 we obtain Mr,ξ → unit operator I in the L p norm, p > 1. We also mention the following. Proposition 24.5 ([9]) Assume β >    Mr,ξ f − f  ≤ 1

2α (β)

1

 2α  β −

Hence as ξ → 0 we get Mr,ξ

r +1 . 2α

Then

 

∞

1



(1 + t)

β dt ωr ( f, ξ)1 . t 2α + 1 (24.18) → I in the L 1 norm. 1 2α

r

0

In this chapter we study the approximation properties of general multivariate smooth Poisson–Cauchy singular integral operators: [m] Ur,n ( f ; x1 , ..., x N ) :=

24.1 Introduction

WnN

r 

441

α[m] j,r

j=0

RN

with α ∈ N , β >

f (x1 + s1 j, ..., x N + s N j)

1 2α

N  i=1

1

2α β ds1 ...ds N , (24.19) si + ξn2α

and 2αβ−1

Wn :=

 (β) αξn

1

 2α  β −

1 2α

,

(24.20)

see [7]. We observe that WnN

N 

1

2α β ds1 ...ds N = 1, si + ξn2α

R N i=1

(24.21)

see [7], [11, p. 397, formula 595]. Here r ∈ N, m ∈ Z + , and

α[m] j,r

⎧   r ⎪ r− j ⎪ j −m , i f j = 1, 2, ..., r, ⎨ (−1) j   := r  r −m ⎪ r −i ⎪ i , i f j = 0, (−1) ⎩1 − i i=1

and [m] δk,r

:=

r 

k α[m] j,r j , k = 1, 2, ..., m ∈ N.

(24.22)

(24.23)

j=1

See that

r  j=0

α[m] j,r = 1.

Also here ξn ∈ (0, 1], n ∈ N, and f : R N → R is a Borel measurable function. [m] [m] is a special case of a more general operator θr,n studied in The above operator Ur,n general in [3] by the author. [m] . We mention next about θr,n Let μξn be a probability Borel measure on R N , N ≥ 1. We define the multiple smooth singular integral operators [m] θr,n ( f ; x1 , ..., x N ) :=

r  j=0

α[m] j.r

RN

f (x1 + s1 j, x2 + s2 j, ..., x N + s N j) dμξn (s) ,

(24.24) where s := (s1 , ..., s N ), x := (x1 , ..., x N ) ∈ R N . [m] are not in general positive. For example, consider the function The operators θr,n N  2 ϕ (u 1 , ..., u N ) = u i and also take r = 2, m = 3; xi = 0, i = 1, ..., N . See that i=1

442

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

ϕ ≥ 0, however ⎞ ⎛ 2  [3] ⎠ θ2,n j 2 α[3] (ϕ; 0, 0, ..., 0) = ⎝ j,2 j=1



[3] [3] α1,2 + 4α2,2

assuming that

 N 

 RN

RN



 si2

i=1

N 

i=1

 N  RN

 si2

dμξn (s) =

i=1

    N 1 s 2 dμξn (s) < 0. dμξn (s) = −2 + 2 R N i=1 i (24.25)

 si2 dμξn (s) < ∞.

[m] we have that Clearly in the case of Ur,n

dμξn (s) = WnN

N  i=1

N  1 dsi =: dλξn (s) , s ∈ R N .

2α β si + ξn2α i=1

(24.26)

[m] Lemma 24.6 The operator θr,n preserve the constant functions in N variables.

We need the following definition.

 Definition 24.7 Let f ∈ C B R N , the space of all bounded and continuous functions or uniformly continuous on R N . Then, the r th multivariate modulus of smoothness of f is given by (see, e.g. [5])  r 

ωr ( f ; h) := √ sup

 ( f )∞ < ∞, h > 0,

u 1 ,u 2 ,...,u N

u 21 +...+u 2N ≤h

(24.27)

where ·∞ is the sup-norm and ru f (x) := ru 1 ,u 2 ,...,u N f (x1 , ..., x N ) = r  j=0

(−1)r − j

  r f (x1 + ju 1 , x2 + ju 2 , ..., x N + ju N ) . j

(24.28)

 Let m ∈ N and let f ∈ C m R N . Suppose that all partial derivatives of f of order m are bounded, i.e.  m   ∂ f (·, ·, ..., ·)     ∂x α1 ...∂x α N  1

for all α j ∈ Z+ , j = 1, ..., N ;

N  j=1

N

α j = m.

∞

< ∞,

(24.29)

24.1 Introduction

443

[m] In this chapter we apply the general theory developed in [3] about θr,n to the [m] operators Ur,n , so we can obtain computationally specific results and show that the general theory has applications and it is a valid theory. [m] we So for the very important in various branches of mathematics operators Ur,n prove the very essential properties of uniform approximation, L p approximation, global smoothness preservation and simultaneously approximation, Voronovskaya asymptotic expansions and complex simultaneous approximation.

24.2 Auxiliary Essential Results We need the following: +1 Theorem 24.8 Let r, m ∈ N, N ∈ N − {1}, α ∈ N, β > m+r , α j ∈ Z+ , j = 1, ..., 2α N 2αβ−1  n α j = m, ξn ∈ (0, 1], n ∈ N. Denote Wn :=  α(β)ξ . Then N : |α| := 1  ( 2α ) (β− 2α1 ) j=1

N 

wξn (α) :=

WnN

RN

αi



s2 1+ ξn

|si |

i=1

 ξn2αβ(N −1)+m





r  N i=1

2α (β) 

 β−

1 2α



1 si2α + ξn2α

N 

β

(24.30)

N 1 2α

dsi ≤

i=1



⎡





m+r +1   N ⎤ m+r +1   β − 2α 2α ⎣(1 + N )r + 2r ⎦≤ 2α (β) 

2α (β)

1

 β−  2α

N ⎡ 



m+r +1   N ⎤ m+r +1   β − 2α 2α ⎣(1 + N )r + 2r ⎦ < +∞,  1 2α (β) 2α

where  is the gamma function. Proof We estimate N 

wξn (α) =

WnN

2 [call

N

WnN

N R+

RN

N  i=1

siαi

|si |

αi

 1+

i=1



s2 1+ ξn

s2 ξn

r  N i=1

r  N

1

i=1

2α β si + ξn2α

1

N 

2α β si + ξn2α

i=1

N 

dsi =

i=1

dsi ≤

(24.31)

444

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

c :=







(2c) N ξn2αβ N −N



N R+

1 2α

 (β) α 

 β− 

N 

siαi

i=1

1 2α

 , i.e.Wn = cξ 2αβ−1 ]

⎛

(24.32)

⎞r s i⎟  N N ⎜  1 ⎜ i=1 ⎟ dsi = ⎜1 + ⎟

2α β ⎝ ξn ⎠ i=1 s + ξ 2α i=1 n i N 

(2c) N ξn2αβ N −N ξnN ξnm ξn−2αβ N R+

 N    r  N N   si αi  si 1 1+   2α ξn ξn s i=1 i=1 i=1

  N  si = d β ξn i=1 +1

i

ξn

(2c) N ξn2αβ(N −1)+m

(2c)

N

ξn2αβ(N −1)+m



N R+

z iαi

1+ 

z iαi

1+

i=1

 z iαi

r

N 

zi

i=1

N 

[0,1] N

(R+ −[0,1]) N



i=1



N 



N 

1+

i=1

N 

r zi

i=1

(24.33) N 



i=1

N 

r zi

1 z i2α

N 

i=1

i=1

N 

1



i=1



β +1

N 

1

β z i2α + 1

β z i2α + 1

N 

dz i =

i=1 N 

dz i +

i=1

 ≤

dz i

i=1

(2c) N ξn2αβ(N −1)+m 

N 



(1 + N ) + 2 r

r (R+ −[0,1]) N

z iαi

  N r N   zi

i=1

i=1

i=1

1

β z i2α + 1

N 

 ≤

dz i

i=1

(2c) N ξn2αβ(N −1)+m 

N 

(1 + N ) + 2 r

r (R+ −[0,1]) N

 z iαi

i=1

 (2c) N ξn2αβ(N −1)+m (1 + N )r + 2r



N 

z ri

i=1

i=1

 N  i=1

⎡ (2c) N ξn2αβ(N −1)+m ⎣(1 + N )r + 2r

N 

∞ 1

 1

∞



1

β z i2α + 1

z iαi +r

1 z i2α

N 

 dz i

 (24.34)

β dz i +1

N ⎤

β dz ⎦ ≤ z 2α + 1 z m+r

=

i=1

≤

24.2 Auxiliary Essential Results

445

⎡ (2c) N ξn2αβ(N −1)+m ⎣(1 + N )r + 2r



∞ 0

N ⎤

z m+r

β dz 1 + z 2α

⎦

(by [11], p. 397, formula 595) ⎡





m+r +1   N ⎤ m+r +1  2α  β − 2α ⎦ = (2c) N ξn2αβ(N −1)+m ⎣(1 + N )r + 2r 2α (β)  =

2α (β)

1

 β−  2α

N 1 2α



ξn2αβ(N −1)+m

⎡





m+r +1   N ⎤ m+r +1   β − 2α 2α ⎣(1 + N )r + 2r ⎦, 2α (β) where N ∈ N − {1} and β > The claim is proved.

m+r +1 . 2α



We continue with Theorem 24.9 Let Wn :=

r, N , α ∈ N,

2αβ−1

α(β)ξn . 1  ( 2α )(β− 2α1 )

β>

r +1 , 2α

ξn ∈ (0, 1],

n ∈ N.

Denote

Then

ψξn := WnN

RN

  N N  s2 r  1 1+ dsi ≤

2α β ξn + ξ 2α i=1 i=1 s 

ξn2αβ(N −1)

n

i

2α (β)

1

 2α  β−

N 1 2α



⎡

 

r +1   N ⎤ r +1  β −  2α 2α ⎣(1 + N )r + 2r ⎦≤ 2α (β)     ⎞ N ⎤ ⎞N ⎡ ⎛  +1  β − r +1  r2α 2α (β) 2α ⎢ r r ⎝    ⎠ ⎥  ⎠ ⎣(1 + N ) + 2 ⎝ ⎦ < +∞. 1  β− 1 2α (β)  2α 2α ⎛

Proof We estimate

(24.36)

446

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

  N N  s2 r  1 dsi = 1+

2α β ξn + ξ 2α i=1 i=1 s

ψξn = WnN

RN

2 N WnN

N R+

2 [call c :=

WnN

(2c)

(2c)

ξn2αβ(N −1)

N

1+

N  si ξ i=1 n





zi

N 

1+

N 

N 

r zi

i=1

r zi

(1 + N ) + 2

N 



i=1

r (R+ −[0,1]) N

(1 + N ) + 2

N 

(R+ −[0,1]) N i=1

⎡ (2c) N ξn2αβ(N −1) ⎣(1 + N )r + 2r



∞ 0

β 1 + z i2α

z ri

β

(24.37)

N 

N 

dz i +

i=1

 dz i

≤

i=1



N  i=1

∞ 1



dz i =

i=1

 N r N  N    1 zi dz i ≤

 2α β i=1 i=1 1 + z i i=1 N 

(2c) N ξn2αβ(N −1) ⎣(1 + N )r + 2r

N 

1

1 + z i2α

r

⎡

(by [11], p. 397, formula 595)

β z i2α + 1

1

r

1

i=1

r



i=1

i=1

 (2c)

r

1+

[0,1] N



ξn2αβ(N −1)

N  1 dsi =

2α β si + ξn2α i=1

i=1





N

N 

i=1



(R+ −[0,1]) N

(2c)

N 

1+

N R+

ξn2αβ(N −1)

ξn2αβ(N −1)

r

i.e.Wn = cξ 2αβ−1 ]



N

n

i

N R+

(β)α , 1  ( 2α )(β− 2α1 )

N

  N N  s2 r  1 1+ dsi ≤

2α β ξn + ξ 2α i=1 i=1 s



N

n

i

zr

β 1 + z i2α

1 + z 2α zr 1 + z 2α

1

N 

 dz i

=

i=1

N ⎤ β dz ⎦ ≤

N ⎤ β dz ⎦ =

(24.38)

24.2 Auxiliary Essential Results

447

⎡

 

r +1   N ⎤ r +1  β −  2α 2α ⎦= (2c) N ξn2αβ(N −1) ⎣(1 + N )r + 2r 2α (β) ⎛

⎞N 2α (β) ⎝     ⎠ ξn2αβ(N −1) 1 1  2α  β − 2α

⎡

 

r +1   N ⎤ r +1  2α  β − 2α ⎣(1 + N )r + 2r ⎦, 2α (β) r +1 . 2α

where N ≥ 1, β >



The claim is proved.

We give Theorem 24.10 Let r, m ∈ N, N ∈ N − {1}, α ∈ N, β > (m+r2α) p+1 , α j ∈ Z+ , j = N 2αβ−1  n α j = m, ξn ∈ (0, 1], n ∈ N. Denote Wn :=  α(β)ξ . Then 1, ..., N : |α| := 1 1  β− 2α ( ) ( ) 2α j=1 θξn (α) :=

WnN

RN

 N 

|si |

αi



i=1

 ξn2αβ(N −1)+mp

s2 1+ ξn

r  p  N i=1

2α (β)

1

 β−  2α

N  1 dsi ≤

2α β si + ξn2α i=1

N 1 2α



⎡

  ⎞ N ⎤ ⎛  (m+r ) p+1   (m+r ) p+1  β −  2α 2α ⎢ rp rp ⎠ ⎥ ⎣(1 + N ) + 2 ⎝ ⎦≤ 2α (β) 

2α (β)

1

 β−  2α

N 1 2α



⎡

  ⎞ N ⎤ ⎛  (m+r ) p+1   (m+r ) p+1  β −  2α 2α ⎢ rp rp ⎠ ⎥ ⎣(1 + N ) + 2 ⎝ ⎦ < +∞. 2α (β)

(24.39)

Proof We estimate  N 

θξn (α) =

WnN

RN

i=1

αi

|si |



s2 1+ ξn

r  p  N i=1

N  1 dsi =

2α β si + ξn2α i=1

448

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

 N 

2 (c :=

N

WnN

N R+

siαi



s2 1+ ξn

i=1

r  p  N i=1

1

2α β si + ξn2α

N 

dsi ≤

i=1

(β)α ) 1  ( 2α )(β− 2α1 )

(2c)

N

(2c)

ξn2αβ N −N

N

N 

N R+

(R+ −[0,1]) N

r p  N  N  si ξn

i=1





N 

N R+



[0,1]

 α p zi i

1+

i=1



N 



1+

i=1

ξn2αβ(N −1)+mp

ξn2αβ(N −1)+mp (2c) N

 α p si i

N

N 

r p zi

i=1

N 

 α p

1+

zi i

N 

r p

1+

i=1

N 

r p zi

i=1

1

β z i2α + 1

N 

zi

i=1





i=1

N 

i=1

α p zi i

i=1

N  1 dsi =

2α β si + ξn2α i=1

i=1

N 

1

β 1 + z i2α

dz i =

i=1

(24.40) N 

1

β 1 + z i2α

i=1

N 

N 

dz i +

i=1

 ≤

dz i

i=1

$ ξn2αβ(N −1)+mp (2c) N (1 + N )r p + N 

2r p (R+ −[0,1]) N

α p zi i

i=1

  N r p N  N    1 zi dz i ≤

 2α β i=1 i=1 1 + z i i=1

$ ξn2αβ(N −1)+mp (2c) N (1 + N )r p + N 

rp

2

(R+ −[0,1]) N

α p zi i

 N 

i=1



i=1

 ξn2αβ(N −1)+mp

(2c)

(1 + N ) + 2

N

N 

rp zi

rp

rp

⎡ ξn2αβ(N −1)+mp (2c) N ⎣(1 + N )r p + 2r p

i=1 N  1



∞

⎡ ξn2αβ(N −1)+mp (2c) N ⎣(1 + N )r p + 2r p

1

β 1 + z i2α

∞

i=1

N 

1

 dz i



1

(α +r ) p zi i

1 + z i2α 1

z (m+r ) p

1 + z 2α

 0

∞

=

i=1

(m+r ) p

β dz i

≤

N ⎤ β dz ⎦ ≤

z

β dz 1 + z 2α

N ⎤ ⎦=

(24.41)

24.2 Auxiliary Essential Results

449

(by [11], p. 397, formula 595) ⎡

  ⎞ N ⎤ ⎛  (m+r ) p+1   (m+r ) p+1  β −  2α 2α ⎢ ⎠ ⎥ ξn2αβ(N −1)+mp (2c) N ⎣(1 + N )r p + 2r p ⎝ ⎦ 2α (β)  = ξn2αβ(N −1)+mp

2α (β)

1

 2α  β −

N 1 2α



⎡

  ⎞ N ⎤ ⎛  (m+r ) p+1   (m+r ) p+1  β −  2α 2α ⎢ rp rp ⎠ ⎥ ⎣(1 + N ) + 2 ⎝ ⎦≤ 2α (β) 

2α (β)

1

 2α  β −

N 1 2α



⎡

  ⎞ N ⎤ ⎛  (m+r ) p+1   (m+r ) p+1  β −  2α 2α ⎢ rp rp ⎠ ⎥ ⎣(1 + N ) + 2 ⎝ ⎦ < +∞, 2α (β) where N ≥ 2 and β > (m+r2α) p+1 . The claim is proved.



We give Theorem 24.11 Let r, α, N ∈ N, p > 1, β > Wn :=

2αβ−1 α(β)ξn 1 1 2α  β− 2α

(

) (

)

ξn ∈ (0, 1], n ∈ N. Denote

. Then

Tξn := WnN

RN

  N N  s2 r p  1 1+ dsi ≤

2α β ξn + ξ 2α i=1 i=1 s

ξn2αβ(N −1)

n

i



⎡

r p+1 , 2α

2α (β)

1

 2α  β −

N 1 2α



  ⎞ N ⎤ ⎛  r p+1    2α  β − r p+1 2α ⎢ rp rp ⎠ ⎥ ⎣(1 + N ) + 2 ⎝ ⎦≤ 2α (β)

450

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …   ⎞ N ⎤ ⎛  r p+1   ⎞N ⎡  2α  β − r p+1 2α (β) 2α ⎢ ⎝    ⎠ ⎥  ⎠ ⎣(1 + N )r p + 2r p ⎝ ⎦ < +∞. 1 1 2α (β)  2α  β − 2α ⎛

(24.42) Proof We estimate   N N  s2 r p  1 dsi = 1+

2α β ξn + ξ 2α i=1 i=1 s

Tξn =

WnN

RN

N

2 (c :=

WnN

N R+

n

i

  N N  s2 r p  1 1+ dsi ≤

2α β ξn + ξ 2α i=1 i=1 s n

i

(β)α ) 1  ( 2α )(β− 2α1 )

(2c)

N

ξn2αβ N −N

(2c) N ξn2αβ(N −1)





1+

N R+

1+

N R+

(2c)

[0,1] N



(R+ −[0,1]) N

r p

N 

zi

1+

N 



i=1



N



i=1

i=1



ξn2αβ(N −1)

ξn

i=1





r p  N  N  si

1+

N 

r p zi

r p zi

i=1

2α β si + ξn2α 1

β z i2α + 1 N 

i=1

N 

1



i=1

N 



i=1

N 

dz i = N 

β z i2α + 1

β z i2α + 1

N 



2

(R+ −[0,1]) N

N 

r p zi

i=1

N  i=1



1

 dz i

≤

i=1

β z i2α + 1

N 

 ≤

dz i

i=1

$ ξn2αβ(N −1) (2c) N (1 + N )r p + rp

2

N 

(R+ −[0,1]) N i=1

rp zi

N  i=1



1

β z i2α + 1

N  i=1

dz i +

i=1

$ ξn2αβ(N −1) (2c) N (1 + N )r p + rp

dsi =

i=1

i=1

1

1

N 

 dz i

=

(24.43)

24.2 Auxiliary Essential Results

451

⎡ ξn2αβ(N −1) (2c) N ⎣(1 + N )r p + 2r p ⎡ ξn2αβ(N −1) (2c) N ⎣(1 + N )r p + 2r p



∞



1



N ⎤

zr p

⎦≤

β dz z 2α + 1

∞

zr p

N ⎤

β dz 1 + z 2α

0

⎦=

(by [11], p. 397, formula 595) ⎡

  ⎞ N ⎤ ⎛  r p+1    2α  β − r p+1 2α ⎢ ⎠ ⎥ ξn2αβ(N −1) (2c) N ⎣(1 + N )r p + 2r p ⎝ ⎦ = (24.44) 2α (β)  ξn2αβ(N −1)

2α (β)

1

 β−  2α

N 1 2α



⎡

  ⎞ N ⎤ ⎛  r p+1    2α  β − r p+1 2α ⎢ rp rp ⎠ ⎥ ⎣(1 + N ) + 2 ⎝ ⎦≤ 2α (β) 

2α (β)

1

 β−  2α

N 1 2α



⎡

  ⎞ N ⎤ ⎛  r p+1    2α  β − r p+1 2α ⎢ rp rp ⎠ ⎥ ⎣(1 + N ) + 2 ⎝ ⎦ < +∞, 2α (β)

where N ≥ 1, β > r p+1 . 2α The claim is proved.



It follows Theorem 24.12 Let r, m, α, N ∈ N, β > 2αβ−1 α(β)ξn 1 1 2α  β− 2α

(

) (

)

ξn ∈ (0, 1], n ∈ N. Denote Wn := N  α j = m. Then . Also α j ∈ Z+ , j = 1, ..., N : |α| := j=1

 K ξn (α) := 

m+1 , 2α

ξn−m

WnN

RN

N  i=1

 αi

|si |

N  i=1

1

2α β si + ξn2α

N 

dsi

≤

i=1



m+1   N 2α (β) +  m+1  β −  2α

1  2α

=: ϕ < +∞. 1  β − 2α  2α

Proof We estimate



(24.45)

452

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

 K ξn (α) =

ξn−m 

2 N ξn−m (c :=

N 

WnN

N R+

(2c)

N  i=1



N 

siαi

i=1

i=1

1

2α β si + ξn2α N 

1

2α β si + ξn2α

N 

 dsi

=

i=1

 =

dsi

i=1

(β)α ) 1  ( 2α )(β− 2α1 )

⎛ N

|si |

i=1

N 

WnN

RN

 αi



⎜ ⎜ ⎜ ⎝ R+N

ξn2αβ N −N −m ξnm−2αβ+N

⎞

  N N   si αi  1   2α ξ n s i=1 i=1

N  si ⎟ ⎟ d ⎟= β ξ n⎠ + 1 i=1

i

ξn

(2c) N ξn2αβ(N −1)

(2c)



N 

N R+

z iαi

i=1

N  i=1

% (2c)

N

ξn2αβ(N −1)

 N 

(2c)

ξn2αβ(N −1)

(2c)

1 + z 2α

∞

1+ 0

zm

=

dz i

β dz

∞

= &

z αi

β dz 1 + z 2α N ≤ N

β dz 1 + z 2α

=

(by [11], p. 397, formula 595)  (2c) N ξn2αβ(N −1) 1 +  (2c)

N

1+







  N  β − m+1 2α = 2α (β)

m+1  2α



  N  β − m+1 2α ≤ 2α (β)

m+1  2α

(24.46)

i=1

zm

∞

1+



β dz 1 + z 2α

1

ξn2αβ(N −1)

1

N 



1

αi

z

0



 N

β 1 + z i2α

β dz + 1 + z 2α 

N

∞

1

z αi

1 0

i=1

N 

i=1

ξn2αβ(N −1)

N



≤

24.2 Auxiliary Essential Results

%



2α (β)

1

 β−  2α

453

 1 2α



 1+



 & N  β − m+1 2α = 2α (β)

m+1  2α



  N 2α (β) +  m+1  β − m+1 2α 2α 

1

= ϕ < +∞, 1  β − 2α  2α

(24.47)

. where β > m+1 2α The claim is proved.



We make Remark 24.13 All as in Theorem 24.12. We denote  N  N N   1 α N i cα,n := cα,n,'j := Wn si dsi ,

2α  2α β RN i=1 i=1 si + ξn i=1

(24.48)

where ' j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αj =

j=1

' j. By (24.45) we obtain ( ( ( ( (cα,n ( = (cα,n,'j ( ≤ ϕξ m ≤ ϕ. n

(24.49)

[m] 24.3 Main Results for Ur,n

24.3.1 Uniform Approximation [m] We start with an application to Ur,n of the following theorem.

 Theorem 24.14 ([3], p. 11) Let m ∈ N, f ∈ C m R N , N ≥ 1, x ∈ R N . Assume  m  N   ∂ f (·,·,...,·)  αi = m. Let μξn be  ∂x α1 ...∂x α N  < ∞, for all α j ∈ Z+ , j = 1, ..., N : |α| := 1

N

∞

i=1

a Borel probability measure on R N , for ξn > 0, (ξn )n∈N bounded sequence. N  Suppose that for all α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| := αi = m i=1

we have that u ξn (α) :=

RN

N  i=1

|si |

αi



s2 1+ ξn

r dμξn (s) < ∞.

(24.50)

454

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

For ' j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αj = ' j,

j=1

call

cα,n := cα,n,'j :=

N  R N i=1

siαi dμξn (s1 , ..., s N ) .

(24.51)

Then (i) ( ⎞( ⎛ ( ( ( ( m ( ( ⎟ ⎜   c f (x) α,n,' j α ( [m] ⎟( [m] ⎜ [m] δ'j,r ⎜ Er,n (x) := (θr,n ( f ; x) − f (x) − ( ⎟ N ( ⎠( ⎝ ) ' α1 ,...,α N ≥0; ( ( j=1 α ! i ( ( |α|=' j i=1 

(ωr ( f α , ξn )) ≤ N  ) α1 ,...,α N ≥0; αi |α|=m

N 

 RN

|si |

αi



i=1

s2 1+ ξn

r

 dμξn (s) .

(24.52)

i=1

∀ x ∈ RN . (ii)

 [m]  E  r,n

∞

≤ R.H.S.(24.52).

(24.53)

 Given  that ξn → 0, as n → ∞, and u ξn is uniformly bounded, then we derive that  E [m]  → 0 with rates. r,n (iii) It holds also that ⎞ ( ( m ⎜  (cα,n,'j (  f α ∞ ⎟  ⎟ ⎜ ≤ δ'[m] ⎟ + R.H.S.(24.52), ⎜ j,r ⎝ N ⎠ ) ' α1 ,...,α N ≥0; j=1 α ! i ' ⎛

 [m]  θ ( f ) − f  r,n ∞

|α|= j

i=1

(24.54) given that  f α ∞ < ∞, for all α : |α| = ' j, ' j = 1, ..., m. Furthermore, as ξn → 0 when n → ∞, assuming that cα,n,'j → 0, while u ξn is uniformly bounded, we conclude that  [m]  θ ( f ) − f  → 0 (24.55) r,n ∞ with rates. [m] A uniform approximation result for Ur,n follows:

  m

   Theorem 24.15 Let m ∈ N, f ∈ C m R N , N ≥ 2, x ∈ R N . Assume  ∂∂x αf1(·,·,...,·) αN  ...∂x < ∞, for all αi ∈ Z+ , i = 1, ..., N : |α| :=

N 

i=1

1

N

∞

αi = m. Let λξn be the Borel prob-

[m] 24.3 Main Results for Ur,n

455

ability measure on R N , see (24.26), where ξn ∈ (0, 1], n ∈ N. Here α, r ∈ N, +1 , and wξn (α) as in (24.30), and cα,n := cα,n,'j as in (24.48). Then β > m+r 2α (i) ( ⎞( ⎛ ( ( ( ( m ( ⎟ ⎜  [m] ⎜  cα,n,'j f α (x) ⎟(( [m] ( [m] E r,n (x) := (Ur,n ( f ; x) − f (x) − δ'j,r ⎜ ⎟( N ( ⎠( ⎝ ) ' α ,...,α ≥0; ( ( 1 N j=1 αi ! ( ( |α|=' j i=1 ≤



(ωr ( f α , ξn )) N  wξn (α) , ) α1 ,...,α N ≥0; αi ! |α|=m

(24.56)

i=1

∀ x ∈ RN . (ii)

   [m]  E r,n 

∞

≤ R.H.S.(24.56).

(24.57)

Given that ξn → 0, as n → +∞,  wehave that wξn (α) → 0 and are uniformly  [m]  bounded, and then we derive that  E r,n  → 0 with rates. ∞ (iii) It holds also that ⎛

  [m] U ( f ) − f  r,n ∞

⎞ ( ( m ( (⎜ (cα,n,'j (  f α ∞ ⎟  ( [m] ( ⎜  ⎟ ≤ (δ'j,r ( ⎜ ⎟ + R.H.S.(24.56), N ⎝ ⎠ ) ' α ,...,α ≥0; 1 N j=1 αi ! ' |α|= j

i=1

(24.58) given that  f α ∞ < +∞, for all α : |α| = ' j, ' j = 1, ..., m. Furthermore, as ξn → 0 when n → +∞, we have that cα,n,'j → 0 and wξn (α) → 0, and both are uniformly bounded, and we conclude that   [m] U ( f ) − f  → 0 r,n ∞

(24.59)

with rates. Proof Mainly by applying Theorem 24.14. By Theorem 24.8 we get that wξn (α) → 0 and wξn (α) are uniformly bounded. By Theorem 24.12 and Remark 24.13 we get cα,n → 0 and cα,n are uniformly bounded.  We mention

 Theorem 24.16 ([3], p. 14) Let f ∈ C B R N , uniformly continuous, N ≥ 1, ξn ∈ (0, 1]. Then

456

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

  [0] θ f − f  ≤ r,n ∞

 RN

   s2 r dμξn (s) ωr ( f, ξn ) , 1+ ξn

(24.60)

  s2 r 1+ dμξn (s) < ∞. ξn

(24.61)

under the assumption ξn :=

RN

As n → ∞ and ξn → 0, given that ξn are uniformly bounded, we derive   [0] θ f − f  → 0 r,n ∞

(24.62)

with rates. We give

 Theorem 24.17 Let f ∈ C B R N , uniformly continuous, N , α, r ∈ N, β > ξn ∈ (0, 1], n ∈ N. Then   [0] U f − f  ≤ r,n ∞



2α (β)

1

 β−  2α

r +1 , 2α

N 1 2α



⎡

 

r +1   N ⎤ r +1   β − 2α 2α ⎣(1 + N )r + 2r ⎦ ξn2αβ(N −1) ωr ( f, ξn ) . 2α (β)

(24.63)

As n → ∞ and ξn → 0, we derive   [0] U f − f  → 0 r,n ∞

(24.64)

with rates. Proof By Theorems 24.9 and 24.16.



[m] 24.3.2 L p Approximation for Ur,n

We need Definition 24.18 ([5, 10]) We call ru f (x) := ru 1 ,u 2 ,...,u N f (x1 , ..., x N ) :=

(24.65)

[m] 24.3 Main Results for Ur,n

r 

457

r− j

(−1)

j=0

  r f (x1 + ju 1 , x2 + ju 2 , ..., x N + ju N ) . j

Let p ≥ 1, the modulus of smoothness of order r is given by   ωr ( f ; h) p := sup ru ( f ) p ,

(24.66)

u2 ≤h

h > 0. We will apply



 Theorem 24.19 ([3], p. 24) Let f ∈ C m R N , m ∈ N, N ≥ 1, with f α ∈ L p R N , |α| = m, x ∈ R N . Let p, q > 1 : 1p + q1 = 1. Here, μξn is a Borel probability measure on R N for ξn > 0, (ξn )n∈N bounded sequence. Assume for all α := (α1 , ..., α N ), N  αi ∈ Z+ , i = 1, ..., N , |α| := αi = m that we have i=1



RN

N 

|si |αi



s2 1+ ξn

i=1

r  p dμξn (s) < ∞.

(24.67)

For ' j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αj = ' j,

j=1

call

cα,n,'j :=

N  RN

siαi dμξn (s) .

(24.68)

i=1

Then  ⎛ ⎞     m  ⎜  c ' f (x) ⎟   [m]  α α,n, j   ⎜ ⎟ [m] [m]  E  = θ ( f ; x) − f (x) − δ'j,r ⎜  ⎟ N r,n p   r,n ⎝ ⎠ ) '   |α|=' j j=1 αi !   i=1  ≤



m (q (m − 1) + 1)

1 q

(24.69) p,x

⎞

⎛

⎜ 1 ⎟ ⎟ ⎜ ⎟ ⎜ N ⎠ ⎝|α|=m ) αi ! i=1

 RN

 N  i=1

|si |αi



s2 1+ ξn

 1p

r  p dμξn (s)

ωr ( f α , ξn ) p .

458

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

 [m]   → 0 with rates. As n → ∞ and ξn → 0, by (24.69) we obtain that  Er,n One also finds by (24.69) that ⎛

 [m]  θ ( f ; x) − f (x) r,n p,x

⎞ ( ( m ( ⎟ (⎜  ⎟ ( [m] ( ⎜  (cα,n,'j (  f α  p ⎟ + R.H.S.(24.69), ≤ (δ'j,r ( ⎜ N ⎝ ⎠ ) ' |α|=' j j=1 αi ! i=1

given that  f α  p < ∞, |α| = ' j, ' j = 1, ..., m.  [m]  Assuming that cα,n,'j → 0, ξn → 0, as n → ∞, we get θr,n ( f ) − f  p → 0, that [m] → I the unit operator, in L p norm, with rates. is θr,n We present



 Theorem 24.20 Let f ∈ C m R N , m ∈ N, N ∈ N − {1}, with f α ∈ L p R N , |α| = m, x ∈ R N . Let p, q > 1 : 1p + q1 = 1. Here λξn is a Borel probability measure on R N

as in (24.26), for ξn ∈ (0, 1], n ∈ N. Let also r, α ∈ N and β > (m+r2α) p+1 ; α j ∈ Z+ , N  α j = m. Here cα,n := cα,n,'j as in (24.48), where ' j= j = 1, ..., N : |α| := j=1

1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N : |α| :=

N 

αi = ' j. Then

i=1

 ⎛ ⎞     m    ⎜  c ' f (x) ⎟  α,n, j α  [m]  [m]  ⎟ [m] ⎜ δ'j,r ⎜ N  ⎟ E r,n  = Ur,n ( f ; x) − f (x) −  p ⎝ ' ) ⎠ '   |α|= j j=1 α ! i   i=1  ≤



m 1

(q (m − 1) + 1) q

2α (β)

1

 2α  β−

p,x

 Np 1 2α



⎡

  ⎞ N ⎤ 1p ⎛  (m+r ) p+1   (m+r ) p+1  β −  2α 2α ⎢ rp rp ⎠ ⎥ ⎣(1 + N ) + 2 ⎝ ⎦ 2α (β) ⎛

⎞

  ⎜ 2αβ(N −1) +m 1 ⎟ p ⎜ ⎟ ωr ( f α , ξn ) p . ⎜ ⎟ ξn N ⎝|α|=m ) ⎠ αi ! i=1

   [m]  As n → ∞ and ξn → 0, by (24.70) we obtain that E r,n  → 0 with rates.

(24.70)

[m] 24.3 Main Results for Ur,n

459

One also finds by (24.70) that ⎛

  [m] U ( f ; x) − f (x) r,n p,x

⎞ ( ( m ( (⎜ ⎟  ( [m] ( ⎜  (cα,n,'j ( ⎟  f α  p ⎟ + R.H.S.(24.70), ≤ (δ'j,r ( ⎜ N ⎝ ⎠ ) ' |α|=' j j=1 αi ! i=1

(24.71) given that  f α  p < ∞, |α| = ' j, ' j = 1, ..., m.  [m]  [m] → Assuming that ξn → 0, as n → ∞, we get Ur,n ( f ) − f  p → 0, that is Ur,n I the unit operator, in L p norm, with rates. Proof By Theorem 24.19. From Theorem 24.10 we get that θξn (α) is uniformly bounded, see (24.39) and θξn (α) → 0, as ξn → 0, when n → ∞. Also by Theorem 24.12 and Remark 24.13 we get that cα,n := cα,n,'j are uniformly bounded and cα,n →  0, as ξn → 0, when n → ∞. We continue with an application of



 Theorem 24.21 ([3], p. 26) Let f ∈ C R N ∩ L p R N ; N ≥ 1; p, q > 1 : 1p + 1 = 1. Assume μξn probability Borel measure on R N , (ξn )n∈N > 0 and bounded. q Also suppose   s2 r p dμξn (s) < ∞. (24.72) 1+ ξn RN Then

  [0] θ ( f ) − f  ≤ r,n p  RN

(24.73)

 1p   s2 r p 1+ dμξn (s) ωr ( f, ξn ) p . ξn

 [0]  [0] → I, the unit As ξn → 0, when n → ∞, we derive θr,n ( f ) − f  p → 0, i.e. θr,n operator, in L p norm. We give



 Theorem 24.22 Let f ∈ C R N ∩ L p R N ; α, N , r ∈ N; p, q > 1 : 1; β >

r p+1 , ξn 2α

∈ (0, 1], n ∈ N. Then  [0]  U ( f ) − f  ≤ r,n p



2α (β)

1

 β−  2α

 Np 1 2α



1 p

+

1 q

=

460

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

⎡

  ⎞ N ⎤ 1p ⎛  r p+1    2α  β − r p+1 2αβ(N −1) 2α ⎢ p rp rp ⎠ ⎥ ωr ( f, ξn ) p . ⎣(1 + N ) + 2 ⎝ ⎦ ξn 2α (β) (24.74)  [0]  [0] → I, the unit As ξn → 0, when n → ∞, we derive Ur,n ( f ) − f  p → 0, i.e. Ur,n operator, in L p norm. 

Proof By Theorems 24.11, 24.21. We mention



 Theorem 24.23 ([3], p. 27) Let f ∈ C R N ∩ L 1 R N ; N ≥ 1. Assume μξn probability Borel measure on R N , (ξn )n∈N > 0 and bounded. Also suppose RN

Then

  s2 r 1+ dμξn (s) < ∞. ξn

(24.75)

  [0] θ ( f ) − f  ≤ r,n 1  RN

(24.76)

   s2 r 1+ dμξn (s) ωr ( f, ξn )1 . ξn

[0] → I, in L 1 norm. As ξn → 0, we get θr,n

We give



 Theorem 24.24 Let f ∈ C R N ∩ L 1 R N ; r, α, N ∈ N, β > n ∈ N. Then N    [0] 2α (β) U ( f ) − f  ≤ 

1

r,n 1 1  β − 2α  2α

r +1 , 2α

ξn ∈ (0, 1],

⎡

 

r +1   N ⎤ r +1   β − 2α 2α ⎣(1 + N )r + 2r ⎦ ξn2αβ(N −1) ωr ( f, ξn )1 . 2α (β)

(24.77)

[0] As ξn → 0, we get Ur,n → I, in L 1 norm.

Proof By Theorems 24.9, 24.23. We mention





 Theorem 24.25 ([3], p. 29) Let f ∈ C m R N , m, N ∈ N, with f α ∈ L 1 R N , |α| = m, x ∈ R N . Here, μξn is a Borel probability measure on R N for ξn > 0, (ξn )n∈N is a bounded sequence. Suppose for all α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , N  |α| := αi = m that we have i=1

[m] 24.3 Main Results for Ur,n





RN

461 N 

|si |

αi



i=1

s2 1+ ξn

r  dμξn (s) < ∞.

(24.78)

For ' j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αi = ' j,

i=1

call

cα,n,'j :=

N  R N i=1

siαi dμξn (s) .

(24.79)

Then  ⎞ ⎛     m   ⎟ ⎜    [m]  c f (x) ' α α,n, j   ⎟ ⎜  E  = θ[m] ( f ; x) − f (x) − δ'[m]  ⎟ ⎜ r,n 1 r,n j,r ⎝ N  ⎠ ) '   |α|=' j j=1 α ! i   i=1

(24.80)

1,x

⎛

⎞

⎟  ⎜ ⎜ 1 ⎟ ≤ ⎜ N ⎟ ωr ( f α , ξn )1 ⎝) ⎠ RN |α|=m αi !

N  i=1

αi

|si |



s2 1+ ξn

r dμξn (s) .

i=1

 [m]   → 0 with rates. As ξn → 0, we get  Er,n 1 From (24.80) we get ⎛

⎞ ( ( m ( (⎜ ⎟    [m] ( [m] ( ⎜  (cα,n,'j ( ⎟ θ f − f  ≤  f α 1 ⎟ + R.H.S.(24.80), (δ'j,r ( ⎜ r,n 1 N ⎝ ⎠ ) ' |α|=' j j=1 αi !

(24.81)

i=1

given that  f α 1 < ∞, |α| = ' j, ' j = 1, ..., m.  [m]  As n → ∞, assuming ξn → 0 and cα,n,'j → 0, we obtain θr,n ( f ) − f 1 → 0, [m] → I in L 1 norm, with rates. that is θr,n We give



 Theorem 24.26 Let f ∈ C m R N , m, r ∈ N, N ∈ N − {1}, with f α ∈ L 1 R N , N  where α j ∈ Z+ , j = 1, ..., N : |α| := α j = m, x ∈ R N , ξn ∈ (0, 1], n ∈ N, and j=1

Here cα,n := cα,n,'j as in (24.48), where ' j = 1, ..., m, N  αi = ' j. Also here λξn is and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N : |α| := α ∈ N such that β >

m+r +1 . 2α

i=1

the Borel probability measure on R N , see (24.26). Then

462

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

 ⎛ ⎞     m    ⎜ ⎟  [m] ⎜  cα,n,'j f α (x) ⎟   [m]  [m]  δ'j,r ⎜ E r,n  = Ur,n ( f ; x) − f (x) − ⎟ N   1 ⎝ ⎠ ) '   |α|=' j j=1 αi !   i=1

(24.82)

1,x

⎛

⎞

⎛

⎞

⎜ ⎜ 1 ⎟ ⎟ ⎜ ⎟ ⎜ ⎟ ≤⎜ ⎟ ωr ( f α , ξn )1 ⎟ ξn2αβ(N −1)+m ⎜ N ⎝|α|=m ⎝ ) ⎠ ⎠ αi !



2α (β)

1

 2α  β −

N 1 2α



i=1

⎡





m+r +1   N ⎤ m+r +1   β − 2α 2α ⎣(1 + N )r + 2r ⎦. 2α (β)    [m]  As ξn → 0, we get E r,n  → 0 with rates. 1 From (24.82) we get ⎛

⎞ ( ( m ( (⎜ ⎟    [m] ( [m] ( ⎜  (cα,n,'j ( ⎟ U f − f  ≤   δ f ( ( ⎜ α 1 ⎟ + R.H.S.(24.82), r,n ' 1 j,r ⎝ N ⎠ ) ' |α|=' j j=1 αi !

(24.83)

i=1

given that  f α 1 < ∞, |α| = ' j, ' j = 1, ..., m.  [m]  As n → ∞, assuming ξn → 0, we get cα,n,'j → 0 and Ur,n ( f ) − f 1 → 0, that [m] → I in L 1 norm, with rates. is Ur,n Proof By Theorem 24.25, also by Theorem 24.8, see (24.30), Theorem 24.12 and Remark 24.13. 

24.3.3 Global Smoothness Preservation and Simultaneous [m] Approximation of Ur,n We need

 Definition 24.27 ([3], p. 34) Let f ∈ C R N , N ≥ 1, m ∈ N, the mth modulus of smoothness for 1 ≤ p ≤ ∞, is given by    ωm ( f ; h) p := sup m t ( f ) p,x , t2 ≤h

h > 0, where

(24.84)

[m] 24.3 Main Results for Ur,n

m t

463

f (x) :=

m 

 (−1)

m− j

j=0

m j

 f (x + jt) .

(24.85)

Denote ωm ( f ; h)∞ = ωm ( f, h) .

(24.86)

Above, x, t ∈ R N . We present the related global smoothness preservation result m] [' Theorem 24.28 We assume Ur,n ' ∈ Z+ , ∀ x ∈ R. Let h > 0, f ∈ ( f ; x) ∈ R, m

N C R , N ≥ 1. (i) Assume ωm ( f, h) < ∞. Then

⎞ ⎛ r ( ( 

['  ( [' m]( ωm Ur,nm ] f, h ≤ ⎝ (α'j,r (⎠ ωm ( f, h) .

(24.87)

' j=0



 (ii) Assume f ∈ C R N ∩ L 1 R N . Then ⎞ ⎛ r ( ( 

['  ( [' m]( ωm Ur,nm ] f, h 1 ≤ ⎝ (α'j,r (⎠ ωm ( f, h)1 .

(24.88)

' j=0



 (iii) Assume f ∈ C R N ∩ L p R N , p > 1. Then ⎞ ⎛ r ( ( 

['  ( [' m]( ωm Ur,nm ] f, h p ≤ ⎝ (α'j,r (⎠ ωm ( f, h) p .

(24.89)

' j=0

Proof Direct application of [3] Theorem 3.2, p. 35.



We make m] m] [' [' Remark 24.29 Let r = 1, m ' ∈ Z+ , then α0,1 = 0, α1,1 = 1. Hence

m] [' U1,n ( f ; x) = WnN

RN

f (x + s)

N  i=1

1

2α β ds1 ...ds N =: Un ( f ; x) . si + ξn2α (24.90)

By Theorem 24.28, we get

 Theorem 24.30 We suppose Un ( f ; x) ∈ R, ∀ x ∈ R. Let h > 0, f ∈ C R N , N ≥ 1. (i) Assume ωm ( f, h) < ∞. Then

464

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

ωm (Un f, h) ≤ ωm ( f, h) .

(24.91)



 (ii) Assume f ∈ C R N ∩ L 1 R N . Then ωm (Un f, h)1 ≤ ωm ( f, h)1 .

(24.92)



 (iii) Assume f ∈ C R N ∩ L p R N , p > 1. Then ωm (Un f, h) p ≤ ωm ( f, h) p .

(24.93)

Next, we get an optimality result Proposition 24.31 Above inequality (24.91): ωm (Un f, h) ≤ ωm ( f, h) is sharp, namely it is attained by any 

f j∗ (x) = x mj , j = 1, ..., N , x = x1 , ..., x j , ..., x N ∈ R N .

(24.94) 

Proof Apply Proposition 3.5, p. 38, of [3]. We need

 Theorem 24.32 ([3], p. 39) Let f ∈ C l R N , l, N ∈ N. Here, μξn is a Borel probability measure on R N , ξn > 0, (ξn )n∈N a bounded sequence. Let β := (β1 , ..., β N ), N  βi ∈ Z+ , i = 1, ..., N ; |β| := βi = l. Here f (x + s j), x, s∈R N , is μξn -integrable i=1

w.r.t. s, for j = 1, ..., r . There exist μξn -integrable functions h i1 , j , h β1 ,i2 , j , h β1 ,β2 ,i3 , j , ..., h β1 ,β2 ,...,β N −1 ,i N , j ≥ 0 ( j = 1, ..., r ) on R N such that ( ( ( ∂ i1 f (x + s j) ( ( ( ( ≤ h i1 , j (s) , i 1 = 1, ..., β1 , ( i1 ( ( ∂x1

(24.95)

( ( ( ∂ β1 +i2 f (x + s j) ( ( ( ( ( ≤ h β1 ,i2 , j (s) , i 2 = 1, ..., β2 , β1 i2 ( ( ∂x ∂x 2

1

.. . ( ( ( ∂ β1 +β2 +...+β N −1 +i N f (x + s j) ( ( ( ( ( ≤ h β1 ,β2 ,...,β N −1 ,i N , j (s) , i N = 1, ..., β N , ( ∂x i N ∂x β N −1 ...∂x β2 ∂x β1 ( N

N −1

2

1

∀ x, s ∈ R N . Then, both of the next exist and

[m] 24.3 Main Results for Ur,n

In particular it holds

465

['

  m] m] [' θr,n fβ ; x , m ' ∈ Z+ . ( f ; x) β = θr,n

(24.96)

['

  m] [' Ur,nm ] ( f ; x) β = Ur,n fβ ; x ,

(24.97)

when dμξn = dλξn (s) , s ∈ R N , see (24.26). Corollary 24.33 (by Theorem 24.32, r = 1) It holds

 (Un ( f ; x))β = Un f β ; x .

(24.98)

We present simultaneous global smoothness results. Theorem 24.34 Let h > 0 and the assumptions of Theorem 24.32 are valid for dμξn = dλξn . Here γ = 0, β (0 = (0, ..., 0)), m ' ∈ Z+ . (i) Assume ωm f γ , h < ∞. Then ⎞ ⎛ r ( ( 

 

  ( [' m]( m] [' ωm Ur,n (f) γ ,h ≤ ⎝ (α'j,r (⎠ ωm f γ , h .

(24.99)

' j=0

 (ii) Additionally suppose f γ ∈ L 1 R N . Then ⎞ ⎛ r ( ( 

 

  ( ( m] [' m] [' ωm Ur,n (f) γ ,h ≤ ⎝ (α'j,r (⎠ ωm f γ , h 1 . 1

(24.100)

' j=0

 (iii) Additionally suppose f γ ∈ L p R N , p > 1. Then ⎞ ⎛ r ( ( 

 

  ( [' m]( m] [' ωm Ur,n (f) γ ,h ≤ ⎝ (α'j,r (⎠ ωm f γ , h p . p

(24.101)

' j=0

It follows Corollary 24.35 (to Theorem 24.34) Let h > 0, r = 1 and γ = 0, β.  (i) Assume ωm f γ , h < ∞. Then



  ωm (Un ( f ))γ , h ≤ ωm f γ , h .

 (ii) Additionally suppose f γ ∈ L 1 R N . Then

(24.102)

466

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …



  ωm (Un ( f ))γ , h 1 ≤ ωm f γ , h 1 .

(24.103)

 (iii) Additionally suppose f γ ∈ L p R N , p > 1. Then



  ωm (Un ( f ))γ , h p ≤ ωm f γ , h p .

(24.104)

Next comes multi-simultaneous approximation. We give

 Theorem 24.36 Let f ∈ C m+l R N , m, l ∈ N, N ≥ 2. The assumptions of Theo  rem 24.32 are valid for dμξn = dλξn . Call γ = 0, β. Assume  f γ+α ∞ < ∞, and +1 , and wξn (α) as in (24.30), and let ξn ∈ (0, 1], n ∈ N. Here α, r ∈ N, β > m+r 2α cα,n := cα,n,'j as in (24.48). Then  ⎞ ⎛     m 

⎟ ⎜  [m] ⎜   cα,n,'j f γ+α (·) ⎟   [m] δ'j,r ⎜  Ur,n ( f ; ·) γ − f γ (·) − ⎟ N   ⎠ ⎝ ) ' α1 ,...,α N ≥0;   j=1 αi !   |α|=' j i=1

(24.105) ∞



  ωr f γ+α , ξn ≤ N  wξn (α) . ) α1 ,...,α N ≥0; αi ! |α|=m i=1



Proof Based on Theorems 24.15, 24.32. We continue with

 Theorem 24.37 Let f ∈ C lB R N , l, N ∈ N (functions l-times continuously differentiable and bounded). The assumptions of Theorem 24.32 are valid for dμξn = dλξn . +1 . Then Call γ = 0, β, ξn ∈ (0, 1], n ∈ N. Let also α, r ∈ N, β > r2α  

  [0]   Ur,n f γ − f γ 

∞

 ≤

2α (β)

1

 2α  β −

N 1 2α



⎡

 

r +1   N ⎤ r +1

   β − 2α 2α ⎣(1 + N )r + 2r ⎦ ξn2αβ(N −1) ωr f γ , ξn . 2α (β)

(24.106)

If ≥2 or f γ is uniformly continuous or both, and ξn → 0, as n → ∞, then

N [0] f γ → f γ uniformly. Ur,n Proof By Theorems 24.17, 24.32. We present



[m] 24.3 Main Results for Ur,n

467

 Theorem 24.38 Let f ∈ C m+l R N , m, l ∈ N, N ∈ N − {1}. The assumptions of Theorem 24.32

are  valid for dμξn = dλξn , ξn ∈ (0, 1], n ∈ N. Call γ = 0, β. Let f (γ+α) ∈ L p R N , |α| = m, x ∈ R N , and p, q > 1 : 1p + q1 = 1. Let also r, α ∈ N N  and β > (m+r2α)β+1 ; α j ∈ Z+ , j = 1, ..., N , |α| := α j = m. Here cα,n := cα,n,'j j=1

as in (24.48), where ' j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , N  |α| := αi = ' j. Then i=1

 ⎛ ⎞     m  

⎜  c 'f ⎟   α,n, j γ+α (x) ⎟  [m] [m] ⎜ δ'j,r ⎜   Ur,n ( f ; x) γ − f γ (x) − ⎟ N  ⎝ ' ⎠ ) '   |α|= j j=1 α ! i   i=1  ≤



m 1

(q (m − 1) + 1) q

2α (β)

1

 β−  2α

p,x

 Np 1 2α



⎡

  ⎞ N ⎤ 1p ⎛  (m+r ) p+1   (m+r ) p+1  β −  2α 2α ⎢ rp rp ⎠ ⎥ ⎣(1 + N ) + 2 ⎝ ⎦ 2α (β) ⎞

⎛

  ⎜ 2αβ(N −1) +m

 1 ⎟ p ⎟ ⎜ ωr f γ+α , ξn p . ⎟ ξn ⎜ N ⎠ ⎝|α|=m ) αi !

(24.107)

i=1



Proof Theorems 24.20 and 24.32. We continue with

 Theorem 24.39 Let f ∈ C l R N , l, N ∈ N. The assumptions of Theorem

24.32  are valid for dμξn = dλξn , ξn ∈ (0, 1], n ∈ N. Call γ = 0, β. Let f γ ∈ L p R N and p, q > 1 : 1p + q1 = 1. Here α, r ∈ N, β > r p+1 . Then 2α  

   [0]  Ur,n ( f ) γ − f γ  ≤ p



2α (β)

1

 2α  β −

 Np 1 2α



468

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

⎡

  ⎞ N ⎤ 1p ⎛  r p+1    2α  β − r p+1 2αβ(N −1)

 2α ⎢ p rp rp ⎠ ⎥ ωr f γ , ξn p . ⎣(1 + N ) + 2 ⎝ ⎦ ξn 2α (β) (24.108)

[0]  · p As n → +∞ and ξn → 0, then Ur,n ( f ) γ → fγ .



Proof By Theorems 24.22 and 24.32. We continue with

 Theorem 24.40 Let f ∈ C l R N , l, N ∈ N. The assumptions of Theorem

24.32  are valid for dμξn = dλξn , ξn ∈ (0, 1], n ∈ N. Call γ = 0, β. Let f γ ∈ L 1 R N and +1 α, r ∈ N, β > r2α . Then  

   [0]  Ur,n ( f ) γ − f γ  ≤ 1



2α (β)

1

 2α  β −

N 1 2α



⎡

 

r +1   N ⎤ r +1

   β − 2α 2α ⎣(1 + N )r + 2r ⎦ ξn2αβ(N −1) ωr f γ , ξn . (24.109) 1 2α (β)

[0]  ·1 As n → +∞ and ξn → 0, then Ur,n ( f ) γ → fγ . 

Proof By Theorems 24.24, 24.32. We continue with

 Theorem 24.41 Let f ∈ C m+l R N , r, m, l ∈ N, N ∈ N − {1}. The assumptions of Theorem 24.32

 are valid for dμξn = dλξn , ξn ∈ (0, 1], n ∈+1N. Call γ = 0, β. Let . Here cα,n := cα,n,'j f (γ+α) ∈ L 1 R N , |α| = m, x ∈ R N . Here α ∈ N : β > m+r 2α as in (24.48), where ' j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , N  |α| := αi = ' j. Then i=1

 ⎞ ⎛     m  

⎟ ⎜    c f (x) ' γ+α α,n, j   [m] ⎟ ⎜ δ'[m]   Ur,n ( f ; x) γ − f γ (x) − ⎟ ⎜ j,r ⎝ N   ⎠ ) '   |α|=' j j=1 α ! i   i=1

1,x

⎞

⎛

⎜  1 ⎟ ⎟

⎜ ≤⎜ ⎟ ωr f γ+α , ξn 1 ξn2αβ(N −1)+m N ⎠ ⎝|α|=m ) αi ! i=1



2α (β)

1

 2α  β −

N 1 2α



[m] 24.3 Main Results for Ur,n

469

⎡





m+r +1   N ⎤ m+r +1  β −  2α 2α ⎣(1 + N )r + 2r ⎦. 2α (β)

(24.110) 

Proof By Theorems 24.26, 24.32.

[m] 24.3.4 Voronovskaya Asymptotic Expansions for Ur,n

We will apply

 Theorem 24.42 ([3], p. 53) Let f ∈ C m R N , m, N ∈ N, with all  f α ∞ ≤ M, M > 0, all α : |α| = m. Let ξn > 0, (ξn )n∈N bounded sequence, μξn probability Borel measures on R N . N  ) αi Call cα,n,'j := R N si dμξn (s), all |α| = ' j = 1, ..., m − 1. Suppose ξn−m R N i=1 N  ) |si |αi dμξn (s) ≤ ρ, all α : |α| = m, ρ > 0, for any such (ξn )n∈N . Also 0 < γ ≤ i=1

1, x ∈ R N . Then ⎛ [m] θr,n

( f ; x) − f (x) =

⎞

⎜  c ' f (x) ⎟ 

α,n, j α ⎜ ⎟ δ'[m] N  ⎟ + 0 ξnm−γ . ⎜ j,r ⎝ ⎠ ) ' |α|=' j j=1 αi !

m−1 

(24.111)

i=1

When m = 1, the sum collapses. [m] Above we assume θr,n ( f ; x) ∈ R, ∀ x ∈ R N . We give Theorem 24.43 Let r, m, α, N ∈ N, β > m+1 , ξn ∈ (0, 1], n ∈ N. Also α j ∈ Z+ , 2α N

  α j = m. Here f ∈ C m R N , with all  f α ∞ ≤ M, M > j = 1, ..., N : |α| := j=1

0, for all α : |α| = m; and dμξn (s) = dλξn (s), as in (24.26), ∀ s ∈ R N . Assume [m] j = 1, ..., m − 1. Ur,n ( f ; x) ∈ R, ∀ x ∈ R N . Here cα,n,'j as in (24.48), all |α| = ' Let 0 < γ ≤ 1, x ∈ R N . Then ⎛ [m] Ur,n

( f ; x) − f (x) =

⎞

⎜  c ' f (x) ⎟ 

α,n, j α ⎜ ⎟ δ'[m] N  ⎟ + 0 ξnm−γ . ⎜ j,r ⎝ ⎠ ) ' |α|=' j j=1 αi !

m−1 

i=1

When m = 1, the sum collapses.

(24.112)

470

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

Proof By Theorems 24.12, 24.42. Here ρ = ϕ, see (24.45).



We give  

 ∂f  Corollary 24.44 (to Theorem 24.43) Let f ∈ C 1 R N , N ≥ 1, with all  ∂x  i M, M > 0, i = 1, ..., N . Let 0 < γ ≤ 1. Assume r, α ∈ N and β > α1 , ξn ∈ (0, 1], n ∈ N. Then

[1] Ur,n



[1] Ur,n ( f ; x) − f (x) = 0 ξn1−γ . Proof By Theorems 24.12, 24.42. Here it is ρ =



∞

≤

( f ; x) ∈ R, ∀ x ∈ R . Here N

2α(β)+ ( α1 ) (β− α1 ) 1  ( 2α )(β− 2α1 )

(24.113) N

.



We continue with

 2   2 

     Corollary 24.45 (to Theorem 24.43) Let f ∈ C 2 R2 , with all  ∂∂x 2f  ,  ∂∂x 2f  , 1 2 ∞ ∞  2   ∂ f  ≤ M, M > 0, ξ ∈ (0, 1], n ∈ N. Call  ∂x1 ∂x2  n ∞

c1 =

R2

s1 dλ∗ξn

where dλ∗ξn = Wn2

2  i=1

(s) , c2 =

R2

s2 dλ∗ξn (s) ,

(24.114)

1 2

2α β ds1 ds2 , s ∈ R . si + ξn2α

[2] Let 0 < γ ≤ 1 and assume Ur,n ( f ; x) ∈ R, ∀ x ∈ R2 . Here r, α ∈ N and β > Then ⎞ ⎛   r  

∂f ∂f [2] ⎠ [2] ⎝ Ur,n ( f ; x) − f (x) = α j,r j c1 (x) + c2 (x) + 0 ξn2−γ . ∂x1 ∂x2 j=1

3 . 2α

(24.115) Proof By Theorems 24.12, 24.42. Here it is ρ =

*

3 2α(β)+ ( 2α )(β− 2α3 ) 1  ( 2α )(β− 2α1 )

+2

.



We also give

 Theorem 24.46 Let f ∈ C m+l R N , m, l, N ∈ N. Assumptions of Theorem  24.32 are valid for dλξn (s) , s∈R N , ξn ∈ (0, 1], n∈N. Call γ=0, β. Suppose  f γ+α ∞ ≤ j = 1, ..., m − M, M > 0, for all α : |α| = m. Here cα,n,'j is as in (24.48), all |α| = '

 N [m] 1; 0 < γ ≤ 1. Assume Ur,n f γ ; x ∈ R, ∀ x ∈ R . Let also r, α ∈ N and β > m+1 . 2α Then

[m] 24.3 Main Results for Ur,n

471

⎛

⎞

m−1 ⎟  [m] ⎜

[m]   ⎜  cα,n,'j f γ+α (x) ⎟ Ur,n ( f ; x) γ − f γ (x) = δ'j,r ⎜  ⎟ + 0 ξnm−γ . N ⎝ ' ⎠ ) ' |α|= j j=1 αi ! i=1

(24.116) When m = 1, the sum collapses. Proof Use of Theorem 24.12 and Theorem 4.6, p. 54 of [3]. Here it is ρ = ϕ, see (24.45). 

24.3.5 Simultaneous Approximation by Multivariate [m] Complex Ur,n We make Remark 24.47 We consider here complex √ valued Borel measurable functions f : R N → C such that f = f 1 + i f 2 , i = −1, where f 1 , f 2 : R N → R are implied to be real valued Borel measurable functions. We define the multivariate complex Poisson–Cauchy singular operators [m] [m] [m] Ur,n ( f ; x) := Ur,n ( f 1 ; x) + iUr,n ( f 2 ; x) , x ∈ R N .

(24.117)

 [m] We assume that Ur,n f j ; x ∈ R, ∀ x ∈ R N , j = 1, 2. One notices easily that ( ( ( ( ( ( [m] (U ( f ; x) − f (x)( ≤ (U [m] ( f 1 ; x) − f 1 (x)( + (U [m] ( f 2 ; x) − f 2 (x)( r,n r,n r,n (24.118) also    [m]  Ur,n ( f ; x) − f (x)

∞,x

    [m] ≤ Ur,n ( f 1 ; x) − f 1 (x)

∞,x

    [m] + Ur,n ( f 2 ; x) − f 2 (x)

∞,x

(24.119)

and       [m] U ( f ) − f  ≤ U [m] ( f 1 ) − f 1  + U [m] ( f 2 ) − f 2  , p ≥ 1. (24.120) r,n r,n r,n p p p Furthermore, it holds f α (x) = f 1,α (x) + i f 2,α (x) ,

(24.121)

where α denotes a partial derivative of any order and arrangement. We give Theorem 24.48 Let f : R N → C, N ≥ 2, such that f = f 1 + i f 2 , j = 1, 2. Here

  ∂ m f (·,·,...,·)  m ∈ N, f j ∈ C m R N , x ∈ R N . Assume  ∂x α1j ...∂x α N  < ∞, for all αi ∈ Z+ , i = 1

N

∞

472

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

1, ..., N : |α| :=

N 

αi = m. Let λξn be the Borel probability measure on R N , see

i=1

(24.26), where ξn ∈ (0, 1], n ∈ N. Here α, r ∈ N, β > (24.30), and cα,n := cα,n,'j as in (24.48). Then

m+r +1 , 2α

and wξn (α) as in

 ⎞ ⎛     m  ⎟ ⎜  [m] ⎜  cα,n,'j f α (x) ⎟   [m] δ'j,r ⎜ Ur,n ( f ; x) − f (x) − ⎟ N   ⎠ ⎝ ) ' α1 ,...,α N ≥0;   j=1 αi !   |α|=' j i=1

∞,x





 ωr f 1,α , ξn + ωr f 2,α , ξn ≤ wξn (α) . N  ) α1 ,...,α N ≥0; α ! i |α|=m 

(24.122)

i=1



Proof By Theorem 24.15. We proceed with

N Theorem f 2 , N ∈ N, j = 1, 2. Here f j ∈

N  24.49 Let f : R → C : f = f 1 + ri+1 C B R uniformly continuous, α, r ∈ N, β > 2α , ξn ∈ (0, 1], n ∈ N. Then

  [0] U f − f  ≤ r,n ∞



2α (β)

1

 β−  2α

N 1 2α



(24.123)

 

r +1   N ⎤ r +1   β − 2α 2α ⎦ ξn2αβ(N −1) (ωr ( f 1 , ξn ) + ωr ( f 2 , ξn )) , ⎣(1 + N )r + 2r 2α (β) ⎡

As n → ∞ and ξn → 0, we derive   [0] U f − f  → 0 r,n ∞

(24.124)

with rates. Proof By Theorem 24.17.



Next comes multi-simultaneous approximation.

 Theorem 24.50 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m+l R N , m, l ∈ N, N ≥ 2. The assumptions  of Theorem 24.32 are valid for f j and dμξn = dλξn . Call γ = 0, β. Assume  f j,γ+α ∞ < ∞, and let ξn ∈ (0, 1], n ∈ N. Here α, r ∈ +1 , and wξn (α) as in (24.30), and cα,n := cα,n,'j as in (24.48). Then N, β > m+r 2α

[m] 24.3 Main Results for Ur,n

473

 ⎛ ⎞     m 

⎜ ⎟  [m] ⎜   cα,n,'j f γ+α (·) ⎟   [m] δ'j,r ⎜  Ur,n ( f ; ·) γ − f γ (·) − ⎟ N   ⎝ ⎠ ) ' α1 ,...,α N ≥0;   j=1 αi !   |α|=' j i=1

(24.125) ∞





 ωr f 1,γ+α , ξn + ωr f 2,γ+α , ξn ≤ wξn (α) . N  ) α1 ,...,α N ≥0; α ! i |α|=m 

i=1



Proof Based on Theorems 24.32, 24.36. We continue with

 Theorem 24.51 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C lB R N , l, N ∈ N (functions l-times continuously differentiable and bounded). The assumptions of Theorem 24.32 are valid for f j and dμξn = dλξn . Call γ = 0, β, ξn ∈ (0, 1], +1 . Then n ∈ N. Let also α, r ∈ N, β > r2α  

  [0]   Ur,n f γ − f γ 

∞

 ≤

2α (β)

1

 2α  β −

N 1 2α



⎡

    ⎞ N ⎤ ⎛  +1  β − r +1  r2α 

 2α ⎢ 2αβ(N −1)

r r ⎠ ⎥ ωr f 1,γ , ξn + ωr f 2,γ , ξn . ⎣(1 + N ) + 2 ⎝ ⎦ ξn 2α (β)

(24.126) If N ≥ 2 or f j,γ is uniformly continuous ( j = 1, 2), or both, and ξn → 0, as n → ∞, [0] f γ → f γ uniformly. then Ur,n 

Proof By Theorems 24.32 and 24.37. We proceed with L p approximations

 Theorem 24.52 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m R N ,

N m ∈ N, N ∈ N − {1}, with f j,α ∈ L p R , |α| = m, x ∈ R N . Let p, q > 1 : 1p + 1 = 1. Here λξn is a Borel probability measure on R N as in (24.26), for ξn ∈ (0, 1], q N  αi = n ∈ N. Let also r, α ∈ N and β > (m+r2α) p+1 ; αi ∈ Z+ , i = 1, ..., N : |α| := i=1

m. Here cα,n := cα,n,'j as in (24.48), where ' j = 1, ..., m, and α := (α1 , ..., α N ), N  αi = ' j. Then αi ∈ Z+ , i = 1, ..., N , |α| := i=1

 ⎛ ⎞     m  ⎜ ⎟  [m] ⎜  cα,n,'j f α (x) ⎟  [m]  δ'j,r ⎜ N  ⎟ Ur,n ( f ; x) − f (x) −   ⎝ ⎠ ) '   |α|=' j j=1 αi !   i=1

p,x

474

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

 ≤



m 1

(q (m − 1) + 1) q

2α (β)

1

 2α  β −

 Np 1 2α



⎡

  ⎞ N ⎤ 1p ⎛  (m+r ) p+1   (m+r ) p+1  β −  2α 2α ⎢ rp rp ⎠ ⎥ ⎣(1 + N ) + 2 ⎝ ⎦ 2α (β) ⎞

⎛

  ⎜ 2αβ(N −1) +m *



 + 1 ⎟ p ⎟ ⎜ ξ ω . f f , ξ + ω , ξ ⎟ ⎜ n r 1,α n r 2,α n p p N ⎠ ⎝|α|=m ) αi !

(24.127)

i=1



Proof By Theorem 24.20. We continue with

 Theorem 24.53 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C R N ∩

N  L p R ; α, N , r ∈ N; p, q > 1 : 1p + q1 = 1; β > r p+1 , ξn ∈ (0, 1], n ∈ N. Then 2α   [0] U ( f ) − f  ≤ r,n p



2α (β)

1

 2α  β −

 Np 1 2α



⎡

  ⎞ N ⎤ 1p ⎛  r p+1   r p+1  2α  β − 2α ⎢ rp rp ⎠ ⎥ ⎣(1 + N ) + 2 ⎝ ⎦ 2α (β) 2αβ(N −1) p

ξn

$ , ωr ( f 1 , ξn ) p + ωr ( f 2 , ξn ) p .

(24.128)

 [0]  [0] As ξn → 0, when n → ∞, we derive Ur,n f − f  p → 0, i.e. Ur,n → I, the unit operator, in L p norm. 

Proof By Theorem 24.22. We also give

 Theorem 24.54 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C R N ∩

N  +1 L 1 R ; r, α, N ∈ N, β > r2α , ξn ∈ (0, 1], n ∈ N. Then  [0]  U ( f ) − f  ≤ r,n 1



2α (β)

1

 2α  β −

N 1 2α



[m] 24.3 Main Results for Ur,n

475

⎡





+1   N ⎤ +1

  r2α  β − r2α ⎣(1 + N ) + 2 ⎦ ξn2αβ(N −1) ωr ( f 1 , ξn )1 + ωr ( f 2 , ξn )1 . 2α (β) r

r

(24.129) [0] → I, in L 1 norm. As ξn → 0, we get Ur,n



Proof By Theorem 24.24. We further present

N m Theorem 24.55 Let f : R N → C : f =

fN1 + i f 2 , j = 1, 2.+ Here f j ∈ C R , m, r ∈ N, N ∈ N − {1}, with f j,α ∈ L 1 R , where αi ∈ Z , i = 1, ..., N , |α| := N  +1 αi = m, x ∈ R N , ξn ∈ (0, 1], n ∈ N and α ∈ N such that β > m+r . Here 2α

i=1

cα,n := cα,n,'j as in (24.48), where ' j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , N  αi = ' j. Also here λξn is the Borel probability measure on i = 1, ..., N , |α| := i=1

R N , see (24.26). Then  ⎛ ⎞     m  ⎜  c ' f (x) ⎟  α α,n, j  [m]  ⎜ ⎟ δ'[m] Ur,n ( f ; x) − f (x) − ⎜ ⎟ j,r ⎝ N  ⎠ ) '   j=1 |α|=' j αi !   i=1

(24.130)

1,x

⎧ ⎪ ⎪ ⎪ ⎨

⎫ ⎪ ⎪ ⎪ ⎬ ⎟$

 ⎜ 

 , 1 ⎟ ⎜ ≤ ⎟ ωr f 1,α , ξn 1 + ωr f 2,α , ξn 1 ξn2αβ(N −1)+m ⎜ N ⎪ ⎪ ⎠ ⎝) ⎪ ⎪ ⎪ ⎪ αi ! ⎩|α|=m ⎭ ⎞

⎛

i=1



N ⎡ 



m+r +1   N ⎤ m+r +1  β −  2α (β) 2α 2α ⎣(1 + N )r + 2r ⎦.

1

 1 2α (β)  2α  β − 2α 

Proof By Theorem 24.26. We continue with simultaneous L p approximations.

 Theorem 24.56 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m+l R N , m, l ∈ N, N ∈ N − {1}. The assumptions of Theorem 24.32 are

valid  for dμξn = dλξn , ξn ∈ (0, 1], n ∈ N and f j . Call γ = 0, β. Let f j,(γ+α) ∈ L p R N , |α| = m, x ∈ R N , and p, q > 1 : 1p + q1 = 1. Let also r, α ∈ N and β > (m+r2α) p+1 ; αi ∈ Z+ , i = N  1, ..., N : |α| := αi = m. Here cα,n := cα,n,'j as in (24.48), where ' j = 1, ..., m, i=1

and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N  i=1

αi = ' j. Then

476

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

 ⎛ ⎞     m 

⎜ ⎟  [m] ⎜  cα,n,'j f γ+α (x) ⎟    [m] δ'j,r ⎜  Ur,n ( f ; x) γ − f γ (x) − ⎟ N   ⎝ ⎠ ) '   |α|=' j j=1 αi !   i=1  ≤



m 1

(q (m − 1) + 1) q

2α (β)

1

 2α  β −

p,x

 Np 1 2α



(24.131)

⎡

  ⎞ N ⎤ 1p ⎛  (m+r ) p+1   (m+r ) p+1  β−  2α 2α ⎢ rp rp ⎠ ⎥ ⎣(1 + N ) + 2 ⎝ ⎦ 2α (β) ⎞

⎛

  ⎜ 2αβ(N −1) +m *



 + 1 ⎟ p ⎟ ⎜ ξ ω . f f , ξ + ω , ξ ⎟ ⎜ n r 1,γ+α n r 2,γ+α n p p N ⎠ ⎝|α|=m ) αi ! i=1



Proof By Theorems 24.32 and 24.38. We give also

 Theorem 24.57 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C l R N , l, N ∈ N. The assumptions of Theorem 24.32 are valid for dμξn = dλξn , ξn ∈ (0, 1], n ∈ N and f j . Call γ = 0, β. Let f j,γ ∈ L p R N and p, q > 1 : 1p + q1 = 1. Here α, r ∈ N, β >

r p+1 . 2α

Then



   [0]   Ur,n ( f ) γ − f γ  ≤ p



2α (β)

1

 2α  β −

 Np 1 2α



⎡

  ⎞ N ⎤ 1p ⎛  r p+1    2α  β − r p+1 2α ⎢ rp rp ⎠ ⎥ ⎣(1 + N ) + 2 ⎝ ⎦ 2α (β) 2αβ(N −1) p

ξn

*



 + ωr f 1,γ , ξn p + ωr f 2,γ , ξn p .

(24.132)

[0]  · p As n → +∞ and ξn → 0, then Ur,n f γ → fγ . Proof By Theorems 24.32 and 24.39. We continue with



[m] 24.3 Main Results for Ur,n

477

 Theorem 24.58 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C l R N , l, N ∈ N. The assumptions of Theorem 24.32 are valid for dμξn = dλξn , ξn ∈ (0, 1], +1 . Then n ∈ N and f j . Call γ = 0, β. Let f j,γ ∈ L 1 R N and α, r ∈ N, β > r2α  

   [0]  Ur,n ( f ) γ − f γ  ≤ 1



2α (β)

1

 2α  β −

N 1 2α



⎡

 

r +1   N ⎤ r +1   β − 2α 2α ⎣(1 + N )r + 2r ⎦ 2α (β) $



 , ξn2αβ(N −1) ωr f 1,γ , ξn 1 + ωr f 2,γ , ξn 1 .

(24.133)

[0]  ·1 As n → +∞ and ξn → 0, then Ur,n ( f ) γ → fγ . 

Proof By Theorems 24.32, 24.40. We finish with

 Theorem 24.59 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m+l R N , r, m, l ∈ N, N ∈ N − {1}. The assumptions of Theorem 24.32 are

valid  for dμξn = dλξn , ξn ∈ (0, 1], n ∈ N and f j . Call γ = 0, β. Let f j,(γ+α) ∈ L 1 R N , |α| = m, x ∈ +1 R N . Here α ∈ N : β > m+r . Here cα,n := cα,n,'j as in (24.48), where ' j = 1, ..., m, 2α N  αi = ' j. Then and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| := i=1

 ⎞ ⎛     m 

⎟ ⎜  [m] ⎜  cα,n,'j f γ+α (x) ⎟    [m] δ'j,r ⎜  Ur,n ( f ; x) γ − f γ (x) − ⎟ N   ⎠ ⎝ ) '   |α|=' j j=1 αi !   i=1

1,x

⎛

⎞

⎛

⎞

⎜  ⎜ 1 ⎟$



 ,⎟ ⎜ ⎟ ⎜ ⎟ ≤⎜ ⎟ ωr f 1,γ+α , ξn 1 + ωr f 2,γ+α , ξn 1 ⎟ ⎜ N ⎝|α|=m ⎝ ) ⎠ ⎠ αi ! i=1

 ξn2αβ(N −1)+m

2α (β)

1

 2α  β−

N 1 2α



⎡





m+r +1   N ⎤ m+r +1   β − 2α 2α ⎣(1 + N )r + 2r ⎦. 2α (β)

(24.134)

478

24 Complete Approximations with Multivariate Generalized Poisson–Cauchy …

Proof By Theorems 24.32, 24.41.



References 1. Anastassiou, G.A.: Rate of convergence of non-positive linear convolution type operators. A sharp inequality. J. Math. Anal. Appl. 142, 441–451 (1989) 2. Anastassiou, G.A.: Moments in Probability and Approximation Theory, Pitman Research Notes in Math, vol. 287. Longman Scientific & Technical, Harlow (1993) 3. Anastassiou, G.A.: Approximation by Multivariate Singular Integrals. Springer, New York (2011) 4. Anastassiou, G.A.: Approximation by Multivariate Generalized Poisson-Cauchy type Singular Integral operators submitted (2020) 5. Anastassiou, G., Gal, S.: Approximation Theory. Birkhaüser, Boston (2000) 6. Anastassiou, G.A., Mezei, R.A.: Global Smoothness and uniform convergence of smooth Poisson-Cauchy type singular operators. Appl. Math. Comput. 215, 1718–1731 (2009) 7. Anastassiou, G., Mezei, R.: Uniform convergence with rates of smooth Poisson-Cauchy type singular integral operators. Math. Comput. Modell. 50, 1553–1570 (2009) 8. Anastassiou, G.A., Mezei, R.A.: A Voronovskaya type theorem for Poisson-Cauchy type singular operators. J. Math. Anal. Appl. 336, 525–529 (2010) 9. Anastassiou, G.A., Mezei, R.A.: L p convergence with rates of smooth Poisson-Cauchy Type Singular Operators. Int. J. Math. Sci. (Korea) 11(1–2), 1–26 (2012) 10. DeVore, R.A., Lorentz, G.G.: Constructive Approximation, vol. 303. Springer, Berlin (1993) 11. Zwillinger, D.: CRC standard Mathematical Tables and Formulae, 30th edn. Chapman & Hall/CRC, Boca Raton (1995)

Chapter 25

High Order Approximation with Multivariate Generalized Trigonometric Type Singular Integral Operators

This research and survey chapter deals exclusively with the chapter of the approximation of generalized multivariate Trigonometric type singular integrals to the identityunit operator. Here we study quantitatively most of their approximation properties. These operators are not in general positive linear operators. In particular we study the rate of convergence of these integral operators to the unit operator, as well as the related simultaneous approximation. These are given via Jackson type inequalities and by the use of multivariate high order modulus of smoothness of the high order partial derivatives of the involved function. Also we study the global smoothness preservation properties of these integral operators. These multivariate inequalities are nearly sharp and in one case the inequality is attained, that is sharp. Furthermore we give asymptotic expansions of Voronovskaya type for the error of approximation. The above properties are studied with respect to L p norm, 1 ≤ p ≤ ∞. It follows [4].

25.1 Introduction We start with our motivation for this chapter. The following come from [6]. For r ∈ N and n ∈ Z+ we set   ⎧ r ⎪ r− j ⎪ j −n , j = 1, ..., r, ⎨ (−1) j   αj = r  r ⎪ ⎪1 − j −n , j = 0, (−1)r − j ⎩ j

(25.1)

j=1

that is

r 

α j = 1. Here it is ξ ∈ (0, 1].

j=0

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_25

479

480

25 High Order Approximation with Multivariate Generalized …

Let f ∈ C n (R), n ∈ Z+ , and f (n) ∈ L p (R), 1 ≤ p < ∞, β ∈ N, we define for x ∈ R, the trigonometric integral 1 Tr,ξ ( f ; x) := W where

W =

∞

⎛ ⎝

⎞ ⎛   ⎞2β ⎛ r sin ξt ⎠ ⎝ ⎝ ⎠ dt, α j f (x + jt) t −∞ j=0



sin

−∞

∞

  ⎞2β t ξ

t



⎠

dt = 2ξ 1−2β

∞



0

2ξ 1−2β π (−1)β β

β

(−1)k

k=1

sin t t

2β

(25.2)

(25.6.)

dt =

k 2β−1 . (β − k)! (β + k)!

(25.3)

Tr,ξ operators are not positive operators, see [8]. We mention: Let p and m be integers with 1 ≤ p ≤ m. We define the integral

∞



∞

(sin x)2m d x. x2p

(25.4)

k 2 p−1 (−1) p (2m)! k . (−1) 4m− p (2 p − 1)! k=1 (m − k)! (m + k)!

(25.5)

I (m, p) :=

−∞

(sin x)2m dx = 2 x2p

0

That is an (absolutely) convergent integral. According to [10], page 210, item 1033, we obtain m

I (m, p) = π

In particular, for p = m the above formula becomes 0

∞

k 2m−1 (sin x)2m m k . d x = π m (−1) (−1) x 2m (m − k)! (m + k)! k=1 m

(25.6)

We need the r th L p -modulus of smoothness     ωr f (n) , h p := sup rt f (n) (x) p,x , h > 0, |t|≤h

where rt f (n) (x) :=

r j=0

(−1)r − j

  r f (n) (x + jt) , j

  see [9], p. 44. Here we have that ωr f (n) , h p < ∞, h > 0.

(25.7)

(25.8)

25.1 Introduction

481

We need to introduce δk :=

r

α j j k , k = 1, ..., n ∈ N.

(25.9)

j=1

Call

r

τ (w, x) :=

α j j n f (n) (x + jw) − δn f (n) (x) .

(25.10)

j=0

Notice also that −

r

r− j

(−1)

j=1

    r r r = (−1) . j 0

According to [2], p. 306, [1], we get

Thus

τ (w, x) = rw f (n) (x) .

(25.11)

  τ (w, x) p,x ≤ ωr f (n) , |w| p , w ∈ R.

(25.12)

Using Taylor’s formula one has r

α j [ f (x + jt) − f (x)] =

j=0

where

Rn (0, t, x) := 0



Assume ck,ξ := where

∞ −∞

t

n f (k) (x) k δk t + Rn (0, t, x) , k! k=1

(t − w)n−1 τ (w, x) dw, n ∈ N. (n − 1)!

t k dμξ (t) ∈ R, k = 1, ..., n,

(25.13)

(25.14)

(25.15)

⎛   ⎞2β t 1 ⎝ sin ξ ⎠ dμξ (t) := dt, ∀ t ∈ R. W t

Using the above terminology we derive  (x) := Tr,ξ ( f ; x) − f (x) −

n f (k) (x) δk ck,ξ = Rn∗ (x) , k! k=1

(25.16)

482

25 High Order Approximation with Multivariate Generalized …

where Rn∗ (x) :=



∞

Rn (0, t, x) dμξ (t) , n ∈ N.

−∞

(25.17)

Let · denote the ceiling of a real number. We mention Theorem 25.1 ([6]) Let p, q > 1 such that and the rest as above. Then  (x) p ≤ ⎡ ⎣

∞ 0

1  sin t 2β t

1 q

= 1 , n, β ∈ N, β >

1





t np−1+ j

sin t t

2β 

r p +np+1 2

(25.18)

1

((n − 1)!) (q (n − 1) + 1) q (r p + 1) p

0

j=1

+ 1

r p +1 ∞

dt

1 p

⎤ 1p

  dt ⎦ ξ n ωr f (n) , ξ p .

Moreover, as ξ → 0 we get that  (x) p → 0. The counterpart of Theorem 25.1 follows, case of p = 1. Theorem 25.2 ([6]) Let f ∈ C n (R) and f (n) ∈ L 1 (R), n ∈ N, β ∈ N, β > r +1+n . 2 Then 1      (x)1 ≤ (25.19) 2β ∞ dt (r + 1) (n − 1)! 0 sint t  r +1 j=1

∞



 t

n−1+ j

0

2β 

sin t t



  dt ξ n ωr f (n) , ξ 1 .

Hence as ξ → 0 we obtain  (x)1 → 0. The case n = 0 is mentioned next. Proposition 25.3 ([6]) Let p, q > 1 such that the rest as above. Then ⎛   Tr,ξ ( f ) − f  ≤ ωr ( f, ξ) p ⎝   p

∞ 0

1  sin t 2β t

1 p

+

 dt

1 q

= 1 , β ∈ N, β >

 r p j=0

0

∞

 tj

sin t t

r p +1 2

2β

and

⎞ 1p dt ⎠ . (25.20)

Also as ξ → 0 we obtain Tr,ξ → unit operator I in the L p norm, p > 1. We also give Proposition 25.4 ([6]) For β ∈ N, β >

r +1 2

, we have

25.1 Introduction

483

  Tr,ξ ( f ) − f  ≤  ωr ( f, ξ)1  ∞  sin t 2β 1 0

t

 dt

 r j=0



∞

t

j

0

sin t t



2β

dt .

(25.21)

Moreover as ξ → 0 we get that Tr,ξ → I in the L 1 norm. We also mention: Case β = 2. Corollary 25.5 ([6]) Let f ∈ C 1 (R) and f  ∈ L 1 (R). Then       T1,ξ ( f ; x) − f (x) ≤ 3 ln 2 + π ξω1 f  , ξ . 1 1 2π 4

(25.22)

Corollary 25.6 ([6]) Let f ∈ C 1 (R) and f  ∈ L 1 (R). Then   T2,ξ ( f ; x) − f (x) ≤ 1



 27     32 16 40 5 256 5 ln + ln + ξω2 f  , ξ 1 . 33π 4 33 22π 27 (25.23)

Corollary 25.7 ([6]) It holds   27    3 16 256 47 40 15 4 T1,ξ ( f ) − f  ≤ ω1 ( f, ξ)4 ! ln ln + . + 4 11π 4 22π 27 22

(25.24)

Also as ξ → 0 we obtain T1,ξ → unit operator I in the L 4 norm. Corollary 25.8 ([6]) We have   T6,ξ ( f ) − f  ≤ ω6 ( f, ξ)1 1



 251 630 2 60 5671 ln 9 + . 151π 2416 35

(25.25)

Moreover as ξ → 0 we get that T6,ξ → I in the L 1 norm. We will use a lot the following: Remark 25.9 ([7]) Let j, m ∈ Z, m ≥ 1 such that 0 ≤ j < 2m − 1. The integral



∞

x −∞

j

sin x x

2m

"  ∞  sin x 2m d x, if j is even 2 0 xj x dx = , 0, if j is odd

is an (absolutely) convergent integral. According to [10], page 210, item 1033, we obtain case 1: j is even, j < 2m − 1

(25.26)

484

25 High Order Approximation with Multivariate Generalized …



∞

 xj

0

sin x x

2m

k 2m− j−1 π (−1) 2 (2m)! k , (−1) 2 j+1 (2m − j − 1)! k=1 (m − k)! (m + k)! (25.27) 2m− j

dx =

m

and case 2: j is odd, j < 2m − 1



2m

2m− j−1 [ln (2k)] (−1) 2 (2m)! m−k k . (−1) j (2m − j − 1)! 2 − k)! (m (m + k)! 0 k=1 (25.28) In particular, for j = 0 the formula (25.27) becomes ∞



xj

∞



0

sin x x

sin x x

j−1

m

dx =

2m d x = π (−1)m m

m

(−1)k

k=1

k 2m−1 . (m − k)! (m + k)!

(25.29)

In this chapter we study the approximation properties of general multivariate smooth Trigonometric singular integral operators: [m] Tr,n ( f ; x1 , ..., x N ) :=

λ−N n

r

α[m] j,r

j=0

RN

f (x1 + s1 j, ..., x N + s N j)

N # i=1

⎛ ⎝

sin

  ⎞2β si ξn

si

⎠

ds1 ...ds N , (25.30)

with β ∈ N , and λn := 2ξn1−2β π (−1)β β

β

(−1)k

k=1

k 2β−1 , (β − k)! (β + k)!

see [8, 10], p. 210, item 1033. Notice that ⎛   ⎞2β # N sin ξsni −N ⎝ ⎠ ds1 ...ds N = 1, λn si R N i=1 see also [8, 10], p. 210, item 1033, and [3], p. 16. We call β k 2β−1 , γ := 2π (−1)β β (−1)k (β − k)! (β + k)! k=1

(25.31)

(25.32)

(25.33)

that is λn = γξn1−2β .

(25.34)

25.1 Introduction

485

Here r ∈ N, m ∈ Z + , and ⎧   r ⎪ r− j ⎪ j −m , if j = 1, 2, ..., r, ⎨ (−1) j   := r  r −m ⎪ r −i ⎪ i , if j = 0, (−1) ⎩1 − i i=1

α[m] j,r

and [m] δk,r :=

r

k α[m] j,r j , k = 1, 2, ..., m ∈ N.

(25.35)

(25.36)

j=1

See that

r  j=0

α[m] j,r = 1.

Also here ξn ∈ (0, 1], n ∈ N, and f : R N → R is a Borel measurable function. [m] [m] is a special case of a more general operator θr,n studied in The above operator Tr,n general in [3] by the author. [m] . We mention next about θr,n Let μξn be a probability Borel measure on R N , N ≥ 1. We define the multiple smooth singular integral operators [m] θr,n ( f ; x1 , ..., x N ) :=

r

α[m] j.r

j=0

RN

f (x1 + s1 j, x2 + s2 j, ..., x N + s N j) dμξn (s) ,

(25.37) where s := (s1 , ..., s N ), x := (x1 , ..., x N ) ∈ R N . [m] are not in general positive. For example, consider the function The operators θr,n N  2 ϕ (u 1 , ..., u N ) = u i and also take r = 2 , m = 3; xi = 0, i = 1, ..., N . See that ϕ ≥ 0, however

i=1

⎞ ⎛ 2 [3] ⎠ θ2,n j 2 α[3] (ϕ; 0, 0, ..., 0) = ⎝ j,2 j=1



[3] α1,2

+

[3] 4α2,2

assuming that

 N

 RN

 RN



 si2

i=1

N 

i=1

 N RN

 si2

dμξn (s) =

i=1

    N 1 s 2 dμξn (s) < 0. dμξn (s) = −2 + 2 R N i=1 i (25.38)

 si2 dμξn (s) < ∞.

[m] we have that Clearly in the case of Tr,n

dμξn (s) = λ−N n

N # i=1

⎛ ⎝

sin

  ⎞2β si ξn

si

⎠

N # i=1

dsi =: dϕξn (s) , s ∈ R N .

(25.39)

486

25 High Order Approximation with Multivariate Generalized …

[m] Lemma 25.10 The operator θr,n preserves the constant functions in N variables.

We need the following definition.   Definition 25.11 Let f ∈ C B R N , the space of all bounded and continuous functions or uniformly continuous on R N . Then, the r th multivariate modulus of smoothness of f is given by (see, e.g. [5])  ( f )∞ < ∞, h > 0,

 r 

ωr ( f ; h) := √ sup

u 1 ,u 2 ,...,u N

u 21 +...+u 2N ≤h

(25.40)

where ·∞ is the sup-norm and ru f (x) := ru 1 ,u 2 ,...,u N f (x1 , ..., x N ) = r j=0

(−1)r − j

  r f (x1 + ju 1 , x2 + ju 2 , ..., x N + ju N ) . j

(25.41)

  Let m ∈ N and let f ∈ C m R N . Suppose that all partial derivatives of f of order m are bounded, i.e.   m  ∂ f (·, ·, ..., ·)     ∂x α1 ...∂x α N  1

for all α j ∈ Z+ , j = 1, ..., N ;

N 

N

< ∞,

(25.42)

∞

α j = m.

j=1 [m] to the In this chapter we apply the general theory developed in [3] about θr,n [m] operators Tr,n , so we can obtain computationally specific results and show that the general theory has applications and it is a valid theory. [m] we So for the very important in various branches of mathematics operators Tr,n prove the very essential properties of uniform approximation, L p approximation, global smoothness preservation and simultaneously approximation, Voronovskaya asymptotic expansions and complex simultaneous approximation.

25.2 Auxiliary Essential Results We will use Lemma 25.12 Let N ∈ N, r > 0, z i ∈ R+ , i = 1, ..., N . Then  1+

N i=1

r zi

≤

N # i=1

(1 + z i )r .

(25.43)

25.2 Auxiliary Essential Results

487

Proof We have  1+

N



r zi

≤

N+

i=1

N

r = [(1 + z 1 ) + (1 + z 2 ) + ... + (1 + z N )]r =

zi

i=1

 N

r ≤

(1 + z i )

i=1

N #

(1 + z i )r , by 1 + z i ≥ 1,i = 1, ..., N .

i=1

 We give N 

Theorem 25.13 Let r, N , β ∈ N, α j ∈ Z+ , j = 1, ..., N : |α| :=

α j = m ∈ N,

j=1

ξn ∈ (0, 1], n ∈ N. Here we take β > r +m+1 , and γ, λn are as in (25.33) and (25.34), 2 respectively. Also we take λ = 0, 1, ..., r . When λ is even we define ψ1λ

 β  2β−λ k 2β−λ−1 π (−1) 2 (2β)! k := λ+1 , (−1) 2 (2β − λ − 1)! k=1 (β − k)! (β + k)!

(25.44)

and when λ is odd we define ψ2λ

 β  λ−1 2β−λ−1 k [ln (2k)] (−1) 2 (2β)! := λ , (−1)β−k 2 (2β − λ − 1)! k=1 (β − k)! (β + k)! "

and we set ψλ :=

(25.45)

ψ1λ , if λ is even, ψ2λ , if λ is odd.

(25.46)

Similarly, it is defined ψλ+m , just set in (25.44), (25.45), (25.46), λ + m in place of λ. Then Aξn (α) := λ−N n

N #

RN

|si |αi

 1+

i=1

$ ξn2β(N −1)+m 2 N γ −N

s2 ξn

r # N i=1

⎛ ⎝

sin

  ⎞2β si ξn

si

⎠

N #

dsi ≤

i=1

'N r   & r % ≤ ψλ + ψλ+m λ λ=0

$ 2 N γ −N

'N r   & r % < +∞, ψλ + ψλ+m λ λ=0

(25.47)

488

25 High Order Approximation with Multivariate Generalized …

uniformly bounded, and convergent to zero as ξn → 0, when n → +∞. Proof We estimate

Aξn (α) = λ−N n

2 N λ−N n

RN

N #



2 N λ−N n

N #



N R+

N R+

|si |αi

siαi

 1+

N #

⎝

sin

si ξn

  ⎞2β si ξn

sin

⎝

  ⎞2β

⎠

si

i=1

⎠

si

i=1

⎛

r # N

⎛

N #

dsi =

i=1

N #

dsi ≤

(25.48)

i=1

⎛   ⎞2β r # N  N N sin ξsni # si ⎝ ⎠ dsi = 1+ ξn si i=1 i=1 i=1

 siαi

i=1

N #



ξn2β(N −1)+m 2 N γ −N

s2 ξn

r # N

s2 ξn

1+

i=1

i=1

ξn2β(N −1)+m 2 N γ −N



N R+

 z iαi

1+

N

i=1

N R+

z iαi

 N #

i=1

ξn2β(N −1)+m 2 N γ −N

 N #

∞

 (1 + z i )

r

z αi (1 + z)r

ξn2β(N −1)+m 2 N γ −N

λ

λ=0

∞

z λ+αi $

ξn2β(N −1)+m 2 N γ −N



z 1

1

z

λ+m



sin z z



 =

dz



2β

sin z z

dz i =

dz

=

2β dz+

'

2β

λ sin z z

λ+αi

≤

i=1

2β

sin z z

0

≤

dz

 r   r λ=0

∞

λ

λ=0

1



z λ+αi



(25.43)

i=1

zi

sin z z

0

$ r    N # r i=1



0 ∞

dz i

 N N  # sin z i 2β # i=1

 r    N # r i=1

zi

i=1

i=1

i=1

ξn2β(N −1)+m 2 N γ −N

 N N  # sin z i 2β #

zi

i=1

N #



r

1

zλ



0

sin z z

' N

2β dz

≤

(25.49) 2β dz+

25.2 Auxiliary Essential Results

489

$ ξn2β(N −1)+m 2 N γ −N

 r   r λ=0



∞

z

λ+m



0

λ

sin z z

∞



zλ

0

sin z z

2β dz+

' N

2β dz

=: I.

(25.50)

Based on [10], p. 210, item 1033 and [7], see (25.27), (25.28), and by assuming N  β > r +m+1 , i.e. λ < λ + m < 2β − 1, for all λ = 0, 1, ..., r, we have the following 2 calculations: Let λ be even, then

∞

z

λ



0

sin z z

2β

β

k 2β−λ−1 π (−1) 2 (2β)! dz = λ+1 = ψ1λ . (−1)k 2 (2β − λ − 1)! k=1 (β − k)! (β + k)! (25.51) 2β−λ

Let λ be odd, then

∞

zλ



0

sin z z

2β

β

k 2β−λ−1 [ln (2k)] (−1) 2 (2β)! = ψ2λ . (−1)β−k λ 2 (2β − λ − 1)! k=1 (β − k)! (β + k)! (25.52) λ−1

dz =

Therefore

∞

z

λ



0

sin z z

2β

" dz = ψλ =

ψ1λ , when λ is even, ψ2λ , when λ is odd.

(25.53)

Similarly, for λ + m being even, we get

∞

z

λ+m



0

sin z z

β

2β

2β−λ−m

π (−1) 2 (2β)! dz = λ+m+1 2 (2β − λ − m − 1)!

(−1)k

k=1

(25.54)

k 2β−λ−m−1 = ψ1(λ+m) . (β − k)! (β + k)!

And when λ + m is odd we get that

∞

z λ+m

0 β k=1



sin z z

(−1)β−k

2β

λ+m−1

dz =

(−1) 2 (2β)! λ+m 2 (2β − λ − m − 1)!

k 2β−λ−m−1 [ln (2k)] = ψ2(λ+m) . (β − k)! (β + k)!

(25.55)

490

25 High Order Approximation with Multivariate Generalized …

Therefore, it holds

∞

z

λ+m

0



sin z z

2β

" dz = ψλ+m =

That is

$ I =

ξn2β(N −1)+m 2 N γ −N

ψ1(λ+m) , when λ + m is even, ψ2(λ+m) , when λ + m is odd.

(25.56)

'N r   & r % ≤ ψλ + ψλ+m λ λ=0

$ 2 N γ −N

'N r   & r % < +∞. ψλ + ψλ+m λ

(25.57)

λ=0



I.e. Aξn (α) is uniformly bounded. The theorem is proved. We continue with

Theorem 25.14 Let r, n ∈ N, ξn ∈ (0, 1], β ∈ N : β > r +1 , N ∈ N − {1}. Here 2 γ, λn are as in (25.33) and (25.34), respectively, and ψλ is defined by (25.44), (25.45) and (25.46), λ = 0, 1, ..., r. Then Bξn := λ−N n

RN

⎛   ⎞2β  r # N N sin ξsni # s2 ⎝ ⎠ 1+ dsi ≤ ξn si i=1 i=1 

ξn2β(N −1) 2 N γ −N

r   r

λ

λ=0

2 γ N

−N

 r   r λ=0

λ

N ψλ

≤

N < +∞,

ψλ

(25.58)

uniformly bounded, and convergent to zero as ξn → 0, when n → +∞. Proof We estimate

Bξn = λ−N n

2

N

λ−N n





N R+

RN

⎛   ⎞2β  r # N N sin ξsni # s2 ⎝ ⎠ 1+ dsi = ξn si i=1 i=1

s2 1+ ξn

r # N i=1

⎛ ⎝

sin

  ⎞2β si ξn

si

⎠

N # i=1

dsi ≤

(25.59)

25.2 Auxiliary Essential Results

2 N λ−N n

491

⎛   ⎞2β r # N  N N sin ξsni # si ⎝ ⎠ dsi = 1+ ξn si i=1 i=1 i=1



N R+

ξn2β(N −1) 2 N γ −N





1+

N R+

N R+

 N N  # sin z i 2β #

zi



 (1 + z i )

 (1 + z)

r

 r    r λ

∞

z

λ

0

 ξn2β(N −1) 2 N γ −N

λ

2 γ

 r   r

−N

λ

λ=0

under β >

r +1 . 2

dz i =

N =

dz

sin z z

r   r λ=0

N



≤

i=1

2β

sin z z

(25.43)

i=1

zi

i=1 ∞

dz i

 N N  # sin z i 2β #

r

0

λ=0

zi

i=1

i=1

ξn2β(N −1) 2 N γ −N

ξn2β(N −1) 2 N γ −N

r

i=1

N #



ξn2β(N −1) 2 N γ −N

N

 N

2β

(25.53)

=

dz

N ψλ

≤

(25.60)

N ψλ

< +∞, 

The theorem is proved.

We also give Theorem 25.15 Let p > 1; r, β, N ∈ N, α j ∈ Z+ , j = 1, ..., N : |α| :=

N 

αj =

j=1

, and γ, λn are as in (25.33) m ∈ N, ξn ∈ (0, 1], n ∈ N. Here we take β > r p +m+1 2 and (25.34), respectively, and λ runs as λ = 0, 1, ..., r p . Furthermore ψλ is defined as in (25.44), (25.45) and (25.46). Similarly, it is defined ψλ+mp , just set in (25.44), (25.45), (25.46), (λ + mp) in place of λ. Then

Cξn (α) :=

λ−N n

 N #

RN

i=1

αi

|si |



s2 1+ ξn

r  p # N i=1

⎛ ⎝

sin

  ⎞2β si ξn

si

⎠

'N $ r p  r p  % & ψλ + ψλ+mp ≤ ξn2β(N −1)+mp 2 N γ −N λ λ=0

N #

dsi ≤

i=1

(25.61)

492

25 High Order Approximation with Multivariate Generalized …

2 γ N

−N

'N $ r p  r p  % & ψλ + ψλ+mp < +∞, λ λ=0

uniformly bounded, and convergent to zero as ξn → 0, when n → +∞. Above · is the ceiling of the number. Proof We estimate

Cξn (α) =

λ−N n

2 N λ−N n

2 N λ−N n

 N #

RN

N R+

N R+

siαi



s2 ξn

1+

i=1

N #



s2 1+ ξn

i=1

 N #



|si |

αi



r  p # N

⎛ ⎝

sin

⎛ ⎝

sin

si ξn

N #

⎠

dsi =

i=1

  ⎞2β ⎠

si

i=1

si ξn

si

i=1

r  p # N

  ⎞2β

N #

dsi ≤

i=1

⎛   ⎞2β r p # N N N  sin ξsni # si ⎝ ⎠ dsi = 1+ ξn si i=1 i=1 i=1

 αi p

si

i=1

(25.62)

2

N

γ −N ξn2β(N −1)+mp

N #



ξn2β(N −1)+mp 2 N γ −N

N R+

 α p zi i

i=1

N R+

α p zi i

 N #

i=1

ξn2β(N −1)+mp 2 N γ −N

zi

 N N  # sin z i 2β # 

(1 + z i )

rp

r p αi p



(1 + z) z

∞ 0

(1 + z)r p z αi p

dz i

sin z z



zi

≤

dz i =

i=1



2β

sin z z

(25.43)

i=1

 N N  # sin z i 2β # i=1

0

 N # i=1

∞

zi

i=1

i=1

 N # i=1

ξn2β(N −1)+mp 2 N γ −N

r p

i=1

N #



1+

N

dz

≤



2β dz

= (25.63)

25.2 Auxiliary Essential Results

ξn2β(N −1)+mp 2 N γ −N

493

 $ r p   2β ' N # r p  ∞ λ+αi p sin z z dz = λ z 0 λ=0

i=1

$ r p   2β  N # r p  1 2β(N −1)+mp N −N λ+αi p sin z ξn 2 γ z dz+ λ z 0 λ=0

i=1



∞

z λ+αi p



1

∞

z

λ+mp



1

∞

z

λ+mp

0

ξn2β(N −1)+mp 2 N γ −N

1

z



sin z z

λ



0

2β dz+

≤

dz

λ

0

2β

' N dz

sin z z

' N

2β

$ r p   r p  λ=0



sin z z

≤

dz

λ

λ=0

ξn2β(N −1)+mp 2 N γ −N

'

2β

$ r p   r p 

ξn2β(N −1)+mp 2 N γ −N

sin z z

∞

zλ



sin z z

2β dz+

((25.53),(25.56))

=

'N $ r p  r p  % & ψλ + ψλ+mp ≤ λ

(25.64)

λ=0

2 γ N

−N

'N $ r p  r p  % & ψλ + ψλ+mp < +∞. λ λ=0

I.e. Cξn (α) is uniformly bounded. Above we assumed that N  β > 0, 1, ..., r p . The theorem is proved.

r p +m+1 , i.e. λ 2

< λ + m < 2β − 1, for all λ = 

We also present Theorem 25.16 Let p > 1; r, β ∈ N, N ∈ N − {1}, ξn ∈ (0, 1], n ∈ N. Here we , and γ, λn are as in (25.33) and (25.34), respectively, and λ runs take β > r p +1 2 as λ = 0, 1, ..., r p . Furthermore ψλ is defined as in (25.44), (25.45) and (25.46). Then

494

25 High Order Approximation with Multivariate Generalized …



Dξn := λ−N n

RN

⎛   ⎞2β  r p # N N sin ξsni # s2 ⎝ ⎠ 1+ dsi ≤ ξn si i=1 i=1  r p  r p 

ξn2β(N −1) 2 N γ −N

λ

λ=0

2 γ N

−N

 r p  r p  λ

λ=0

N ψλ

≤

(25.65)

N ψλ

< +∞,

uniformly bounded, and convergent to zero as ξn → 0, when n → +∞. Proof We estimate

Dξn =

λ−N n

RN

2 N λ−N n

2 N λ−N n

2

N





⎛ ⎝

  ⎞2β si ξn

sin

si

i=1

⎠

N #

dsi =

(25.66)

i=1

⎛   ⎞2β si   N N # s2 r p # sin ξn ⎝ ⎠ 1+ dsi ≤ ξn si i=1 i=1





r p # N

s2 1+ ξn

N R+

⎛   ⎞2β r p # N N  N sin ξsni # si ⎝ ⎠ dsi = 1+ ξn si i=1 i=1 i=1



N R+

γ −N ξn2β(N −1)





ξn2β(N −1) 2 N γ −N

1+

N R+

N R+

zi

 N N  # sin z i 2β #



 (1 + z i )

rp

 (1 + z)r p

∞ 0

(1 + z)

r p

sin z z



(25.43)

≤

i=1

zi

i=1 ∞

dz i

 N N  # sin z i 2β #

0



zi

i=1

i=1

ξn2β(N −1) 2 N γ −N

ξn2β(N −1) 2 N γ −N

r p

i=1

N #



N

N

2β

sin z z

dz i =

i=1

≤

dz N

2β dz

=

25.2 Auxiliary Essential Results

ξn2β(N −1) 2 N γ −N

495

 r p   r p  λ

λ=0

∞

z

λ



0

λ

λ=0

2 γ N

 r p  r p 

−N

λ

λ=0

under β >

r p +1 . 2

(25.53)

=

dz

 r p  r p 

ξn2β(N −1) 2 N γ −N

 N

2β

sin z z

(25.67)

N ψλ

≤

N ψλ

< +∞, 

The theorem is proved.

We proceed it Theorem 25.17 Let n, N ∈ N, ξn ∈ (0, 1], α j ∈ Z+ , j = 1, ..., N : |α| :=

N 

αj =

j=1

m ∈ N. Here β ∈ N : β > m+1 , and γ, λn are as in (25.33) and (25.34), respectively. 2 Furthermore ψαi is defined as in (25.44), (25.45) and (25.46), just replace λ by αi , i = 1, ..., N . Then Fξn (α) := ξn−m λ−N n

N #

RN

 |si |αi

i=1

N #



2 γ

−N

max

max

|α|=m

N #

|α|=m

⎝

sin

N #

  ⎞2β si ξn

⎠

si

i=1

ξn2β(N −1) 2 N γ −N

N

⎛

N #

dsi ≤

i=1

 ψαi

≤

(25.68)

i=1

 ψαi

=: ϕ < +∞.

i=1

Proof We estimate

Fξn (α) =

ξn−m λ−N n

N ξn−m λ−N n 2

N #

RN

N R+

|si |

αi

i=1

N #





i=1

⎛ ⎝

sin

N # i=1

⎛ ⎝

sin

  ⎞2β si ξn

si

i=1

 siαi

N #

  ⎞2β si ξn

si

⎠

⎠

N #

dsi =

i=1

N # i=1

dsi =

496

25 High Order Approximation with Multivariate Generalized …

ξn2β(N −1) 2 N γ −N

N #

N R+

 z iαi

i=1

ξn2β(N −1) 2 N γ −N N #

zi

i=1

 N #

∞

z

αi



0

i=1

ξn2β(N −1) 2 N γ −N

 N N  # sin z i 2β #

≤

ξn2β(N −1) 2 N γ −N

2 N γ −N max

|α|=m

m+1 , 2

N #

(25.53)

=

dz

 ψαi

(25.69)



2β

sin z z

i=1

under β >

dz i =

i=1

max

N #

|α|=m

 ψαi

≤

i=1

 ψαi

< +∞,

i=1

i.e. αi < 2β − 1, i = 1, ..., N . The theorem is proved.



We make Remark 25.18 All as in Theorem 25.17. We denote

cα,n := cα,n,(j :=

λ−N n

RN

N #

 siαi

i=1

N # i=1

⎛ ⎝

sin

  ⎞2β si ξn

si

⎠

N #

dsi ,

(25.70)

i=1

where ( j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αj =

j=1

( j. By (25.68) we obtain ) ) ) ) )cα,n ) = )cα,n,(j ) ≤ ϕξ m ≤ ϕ. n

(25.71)

[m] 25.3 Main Results for Tr,n

25.3.1 Uniform Approximation [m] We start with an application to Tr,n of the following theorem.   Theorem 25.19 ([3], p. 11) Let m ∈ N, f ∈ C m R N , N ≥ 1, x ∈ R N . Assume  m  N   ∂ f (·,·,...,·)  αi = m. Let μξn be  ∂x α1 ...∂x α N  < ∞, for all α j ∈ Z+ , j = 1, ..., N : |α| := 1

N

∞

i=1

a Borel probability measure on R N , for ξn > 0, (ξn )n∈N bounded sequence.

[m] 25.3 Main Results for Tr,n

497 N 

Suppose that for all α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

αi = m

i=1

we have that u ξn (α) :=

RN

N #



s2 1+ ξn

αi

|si |

i=1

r dμξn (s) < ∞.

(25.72)

For ( j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αj = ( j,

j=1

call

cα,n := cα,n,(j :=

N #

siαi dμξn (s1 , ..., s N ) .

R N i=1

(25.73)

Then (i) ) ⎞) ⎛ ) ) ) ) m ) ⎟ ⎜ [m] ⎜ cα,n,(j f α (x) ⎟)) ) [m] [m] δ(j,r ⎜ Er,n (x) := )θr,n ( f ; x) − f (x) − ⎟) N ) ⎠) ⎝ + ( α1 ,...,α N ≥0; ) ) j=1 αi ! ) ) |α|=( j i=1

(ωr ( f α , ξn )) ≤ N  + α1 ,...,α N ≥0; αi |α|=m

N #

 RN

αi

|si |

i=1



s2 1+ ξn

r

 dμξn (s) .

(25.74)

i=1

∀ x ∈ RN . (ii)

 [m]  E  r,n

∞

≤ R.H.S.(25.74).

(25.75)

 Given  that ξn → 0, as n → ∞, and u ξn is uniformly bounded, then we derive that  E [m]  → 0 with rates. r,n (iii) It holds also that ⎞ ) ) m ⎜ )cα,n,(j )  f α ∞ ⎟ ⎟ ⎜ ≤ δ([m] ⎟ + R.H.S.(25.74), ⎜ j,r ⎝ N ⎠ + ( α1 ,...,α N ≥0; j=1 α ! i ( ⎛

 [m]  θ ( f ) − f  r,n ∞

|α|= j

i=1

(25.76) given that  f α ∞ < ∞, for all α : |α| = ( j, ( j = 1, ..., m. Furthermore, as ξn → 0 when n → ∞, assuming that cα,n,(j → 0, while u ξn is uniformly bounded, we conclude that   [m] θ ( f ) − f  → 0 (25.77) r,n ∞

498

25 High Order Approximation with Multivariate Generalized …

with rates. [m] A uniform approximation result for Tr,n follows:

 m      Theorem 25.20 Let r, N , β, m ∈ N, f ∈ C m R N , x ∈ R N . Assume  ∂∂x αf1(·,·,...,·) αN  ...∂x < ∞, for all αi ∈ Z+ , i = 1, ..., N : |α| :=

N 

1

N

∞

αi = m. Let ϕξn be the Borel prob-

i=1

ability measure on R N , see (25.39), where ξn ∈ (0, 1], n ∈ N. Here β > m+r2 +1 , and Aξn (α) as in (25.47), and cα,n := cα,n,(j as in (25.70). Then (i) ) ⎞) ⎛ ) ) ) ) m ) ⎟ ⎜ cα,n,(j f α (x) ⎟)) [m] ) [m] [m] ⎜ E r,n (x) := )Tr,n ( f ; x) − f (x) − δ(j,r ⎜ ⎟) N ) ⎠) ⎝ + ( α1 ,...,α N ≥0; ) ) j=1 α ! i ) ) ( |α|= j i=1 ≤



(ωr ( f α , ξn )) N  Aξn (α) , + α1 ,...,α N ≥0; αi ! |α|=m

(25.78)

i=1

∀ x ∈ RN . (ii)

   [m]  E r,n 

∞

≤ R.H.S.(25.78).

(25.79)

Given that ξn → 0, as n → +∞,  wehave that Aξn (α) → 0 and are uniformly  [m]  bounded, and then we derive that E r,n  → 0 with rates. ∞ (iii) It holds also that ⎛

  [m] T ( f ) − f  r,n ∞

⎞ ) ) m ) )⎜ )cα,n,(j )  f α ∞ ⎟ ) [m] ) ⎜ ⎟ ≤ )δ(j,r ) ⎜ ⎟ + R.H.S.(25.78), N ⎝ ⎠ + ( α1 ,...,α N ≥0; j=1 α ! i ( |α|= j

i=1

(25.80) given that  f α ∞ < +∞, for all α : |α| = ( j, ( j = 1, ..., m. Furthermore, as ξn → 0 when n → +∞, we have that cα,n,(j → 0 and Aξn (α) → 0, and both are uniformly bounded, and we conclude that   [m] T ( f ) − f  → 0 r,n ∞ with rates.

(25.81)

[m] 25.3 Main Results for Tr,n

499

Proof Mainly by applying Theorem 25.19. By Theorem 25.13 we get that Aξn (α) → 0 and Aξn (α) are uniformly bounded. By Theorem 25.17 and Remark 25.18 we get cα,n → 0 and cα,n are uniformly bounded.  We mention   Theorem 25.21 ([3], p. 14) Let f ∈ C B R N , uniformly continuous, N ≥ 1, ξn ∈ (0, 1]. Then  [0]  θ f − f  ≤ r,n ∞

 RN

   s2 r 1+ dμξn (s) ωr ( f, ξn ) , ξn

(25.82)

  s2 r 1+ dμξn (s) < ∞. ξn

(25.83)

under the assumption ξn :=

RN

As n → ∞ and ξn → 0, given that ξn are uniformly bounded, we derive   [0] θ f − f  → 0 r,n ∞

(25.84)

with rates. We give   Theorem 25.22 Let f ∈ C B R N , uniformly continuous, β, r ∈ N, N ∈ N − {1}, β > r +1 , ξn ∈ (0, 1], n ∈ N. Then 2  r   N r   [0] T f − f  ≤ 2 N γ −N ψλ ξn2β(N −1) ωr ( f, ξn ) . r,n ∞ λ

(25.85)

λ=0

As n → ∞ and ξn → 0, we derive   [0] T f − f  → 0 r,n ∞

(25.86)

with rates. Proof By Theorems 25.14 and 25.21.

[m] 25.3.2 L p Approximation for Tr,n

We need



500

25 High Order Approximation with Multivariate Generalized …

Definition 25.23 ([5, 9]) We call ru f (x) := ru 1 ,u 2 ,...,u N f (x1 , ..., x N ) := r

r− j

(−1)

j=0

(25.87)

  r f (x1 + ju 1 , x2 + ju 2 , ..., x N + ju N ) . j

Let p ≥ 1, the modulus of smoothness of order r is given by   ωr ( f ; h) p := sup ru ( f ) p ,

(25.88)

u2 ≤h

h > 0. We will apply     Theorem 25.24 ([3], p. 24) Let f ∈ C m R N , m ∈ N, N ≥ 1, with f α ∈ L p R N , |α| = m, x ∈ R N . Let p, q > 1 : 1p + q1 = 1. Here, μξn is a Borel probability measure on R N for ξn > 0, (ξn )n∈N bounded sequence. Assume for all α := (α1 , ..., α N ), N  αi ∈ Z+ , i = 1, ..., N , |α| := αi = m that we have i=1



RN

N #

|si |

αi



s2 1+ ξn

i=1

r  p dμξn (s) < ∞.

(25.89)

For ( j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αj = ( j,

j=1

call

cα,n,(j :=

N # R N i=1

siαi dμξn (s) .

(25.90)

Then  ⎛ ⎞     m  ⎜ ⎟ [m] ⎜ cα,n,(j f α (x) ⎟  [m]   [m] E  =  δ(j,r ⎜ N  ⎟ θr,n ( f ; x) − f (x) − r,n p   ⎝ ⎠ + (   |α|=( j j=1 αi !   i=1  ≤



m 1

(q (m − 1) + 1) q

⎛

⎞

⎜ 1 ⎟ ⎜ ⎟ ⎜ ⎟ N ⎝|α|=m + ⎠ αi ! i=1

(25.91) p,x

[m] 25.3 Main Results for Tr,n

 RN

 N #

501

αi



s2 1+ ξn

|si |

i=1

 1p

r  p dμξn (s)

ωr ( f α , ξn ) p .

 [m]   → 0 with rates. As n → ∞ and ξn → 0, by (25.91) we obtain that  Er,n One also finds by (25.91) that ⎛

 [m]  θ ( f ; x) − f (x) r,n p,x

⎞ ) ) m ) ⎟ )⎜ ⎟ ) [m] ) ⎜ )cα,n,(j )  f α  p ⎟ + R.H.S.(25.91.), ≤ )δ(j,r ) ⎜ N ⎝ ⎠ + ( |α|=( j j=1 αi ! i=1

(25.92) given that  f α  p < ∞, |α| = ( j, ( j = 1, ..., m.  [m]  Assuming that cα,n,(j → 0, ξn → 0, as n → ∞, we get θr,n ( f ) − f  p → 0, that [m] → I the unit operator, in L p norm, with rates. is θr,n We present     Theorem 25.25 Let f ∈ C m R N , r, β, N , m ∈ N, with f α ∈ L p R N , |α| = m, x ∈ R N . Let p, q > 1 : 1p + q1 = 1. Here ϕξn is a Borel probability measure on R N

as in (25.39 ), for ξn ∈ (0, 1], n ∈ N. Let β > r p +m+1 ; α j ∈ Z+ , j = 1, ..., N : 2 N  |α| := α j = m. Here cα,n := cα,n,(j as in (25.70), where ( j = 1, ..., m, and α := j=1

(α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N : |α| :=

N 

αi = ( j. Then

i=1

 ⎛ ⎞     m    ⎜ c ( f (x) ⎟ α,n, j α  [m]  [m]  ⎟ [m] ⎜ δ(j,r ⎜ N  ⎟ E r,n  = Tr,n ( f ; x) − f (x) −  p ⎝ ( + ⎠ (   |α|= j j=1 α ! i   i=1  ≤



m

2 γ− N p

1

(q (m − 1) + 1) q

N p

p,x

' Np $ r p  r p    ψλ + ψλ+mp λ λ=0

⎞

⎛

  ⎜ 2β(N −1) +m 1 ⎟ p ⎟ ⎜ ωr ( f α , ξn ) p . ⎟ ξn ⎜ N ⎠ ⎝|α|=m + αi ! i=1

   [m]  As n → ∞ and ξn → 0, by (25.93) we obtain that  E r,n  → 0 with rates. One also finds by (25.93) that

(25.93)

502

25 High Order Approximation with Multivariate Generalized …

⎛

 [m]  T ( f ; x) − f (x) r,n p,x

⎞ ) ) m ) )⎜ ⎟ ) [m] ) ⎜ )cα,n,(j ) ⎟  f α  p ⎟ + R.H.S.(25.93.), ≤ )δ(j,r ) ⎜ N ⎝ ⎠ + ( |α|=( j j=1 αi ! i=1

(25.94) given that  f α  p < ∞, |α| = ( j, ( j = 1, ..., m.  [m] [m] → Assuming that ξn → 0, as n → ∞, we get Tr,n ( f ) − f  p → 0, that is Tr,n I the unit operator, in L p norm, with rates. Proof By Theorem 25.24. From Theorem 25.15 we get that Cξn (α) is uniformly bounded, see (25.61) and Cξn (α) → 0, as ξn → 0, when n → ∞. Also by Theorem 25.17 and Remark 25.18 we get that cα,n := cα,n,(j are uniformly bounded and cα,n → 0, as ξn → 0, when n → ∞.  We continue with an application of      Theorem 25.26 ([3], p. 26) Let f ∈ C R N ∩ L p R N ; N ≥ 1; p, q > 1 : 1p + 1 = 1. Assume μξn probability Borel measure on R N , (ξn )n∈N > 0 and bounded. q Also suppose   s2 r p 1+ dμξn (s) < ∞. (25.95) ξn RN Then

  [0] θ ( f ) − f  ≤ r,n p  RN

(25.96)

 1p   s2 r p 1+ dμξn (s) ωr ( f, ξn ) p . ξn

 [0]  [0] → I, the unit As ξn → 0, when n → ∞, we derive θr,n ( f ) − f  p → 0, i.e. θr,n operator, in L p norm. We give      Theorem 25.27 Let f ∈ C R N ∩ L p R N ; N ∈ N − {1}, β, r ∈ N; p, q > 1 : 1 + q1 = 1; β > r p +1 , ξn ∈ (0, 1], n ∈ N . Then p 2  [0]  T ( f ) − f  ≤ 2 Np γ − Np r,n p

 r p  r p  λ=0

λ

 Np ψλ

2β(N −1) p

ξn

ωr ( f, ξn ) p .

(25.97)

 [0]  [0] → I, the unit As ξn → 0, when n → ∞, we derive Tr,n ( f ) − f  p → 0, i.e. Tr,n operator, in L p norm. Proof By Theorems 25.26 and 25.16.



[m] 25.3 Main Results for Tr,n

503

We mention      Theorem 25.28 ([3], p. 27) Let f ∈ C R N ∩ L 1 R N ; N ≥ 1. Assume μξn probability Borel measure on R N , (ξn )n∈N > 0 and bounded. Also suppose RN

  s2 r 1+ dμξn (s) < ∞. ξn

(25.98)

Then   [0] θ ( f ) − f  ≤ r,n 1





1+

RN

s2 ξn

r

 dμξn (s) ωr ( f, ξn )1 .

(25.99)

[0] As ξn → 0, we get θr,n → I, in L 1 norm.

We give      Theorem 25.29 Let f ∈ C R N ∩ L 1 R N ; N ∈ N − {1}, r, β ∈ N, β > ξn ∈ (0, 1], n ∈ N. Then  r   N r  [0]  N −N T ( f ) − f  ≤ 2 γ ψλ ξn2β(N −1) ωr ( f, ξn )1 . r,n 1 λ

r +1 , 2

(25.100)

λ=0

[0] As ξn → 0, we get Tr,n → I, in L 1 norm.



Proof By Theorems 25.14 and 25.28. We mention

    Theorem 25.30 ([3], p. 29) Let f ∈ C m R N , m, N ∈ N, with f α ∈ L 1 R N , |α| = m, x ∈ R N . Here, μξn is a Borel probability measure on R N for ξn > 0, (ξn )n∈N is a bounded sequence. Suppose for all α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , N  |α| := αi = m that we have i=1

 N #

RN

i=1

αi

|si |



s2 1+ ξn

r  dμξn (s) < ∞.

For ( j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

(25.101) N 

αi = ( j,

i=1

call

cα,n,(j :=

Then

N # RN

i=1

siαi dμξn (s) .

(25.102)

504

25 High Order Approximation with Multivariate Generalized …

 ⎛ ⎞     m  ⎜ ⎟ [m] ⎜ cα,n,(j f α (x) ⎟  [m]   [m] E  =  δ(j,r ⎜ θr,n ( f ; x) − f (x) − ⎟ r,n 1 N   ⎝ ⎠ + (   |α|=( j j=1 αi !   i=1

(25.103)

1,x

⎞

⎛

⎟ ⎜ ⎜ 1 ⎟ ≤ ⎟ ωr ( f α , ξn )1 ⎜ N ⎠ ⎝+ RN |α|=m αi !

N #

αi



|si |

i=1

1+

s2 ξn

r dμξn (s) .

i=1

 [m]   → 0 with rates. As ξn → 0, we get  Er,n 1 From (25.103) we get ⎛

⎞

) ) m ) )⎜ ⎟   [m] ) [m] ) ⎜ )cα,n,(j ) ⎟ θ f − f  ≤   f )δ ) ⎜ ⎟ + R.H.S.(25.103.), (25.104) α 1 r,n ( 1 j,r ⎝ N ⎠ + ( |α|=( j j=1 αi ! i=1

given that  f α 1 < ∞, |α| = ( j, ( j = 1, ..., m.  [m]  As n → ∞, assuming ξn → 0 and cα,n,(j → 0, we obtain θr,n ( f ) − f 1 → 0, [m] → I in L 1 norm, with rates. that is θr,n We give

    Theorem 25.31 Let f ∈ C m R N , r, N , β, m ∈ N, with f α ∈ L 1 R N , where α j ∈ N  Z+ , j = 1, ..., N : |α| := α j = m, x ∈ R N , ξn ∈ (0, 1], n ∈ N, and β > m+r2 +1 . j=1

Here cα,n := cα,n,(j as in (25.70), where ( j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ N  αi = ( j. Also here ϕξn is the Borel probability measure Z+ , i = 1, ..., N : |α| := i=1

on R N , see (25.39). Then  ⎛ ⎞     m    ⎜ c ( f (x) ⎟ α,n, j α  [m]  [m]  ⎟ [m] ⎜ δ(j,r ⎜  E r,n  = Tr,n ( f ; x) − f (x) − ⎟ N  1 ⎝ ( + ⎠ (   |α|= j j=1 α ! i   i=1

1,x

⎛

⎞

⎛

⎞

⎟ ⎜ ⎜ 1 ⎟ ⎟ ⎜ ⎟ ⎜ ≤⎜ ⎟ ωr ( f α , ξn )1 ⎟ ξn2β(N −1)+m ⎜ N ⎠ ⎠ ⎝|α|=m ⎝ + αi ! i=1

(25.105)

[m] 25.3 Main Results for Tr,n

505

'N $ r   r % & ψλ + ψλ+m 2 N γ −N . λ λ=0

   [m]  As ξn → 0, we get  E r,n  → 0 with rates. 1 From (25.105) we get ⎛

⎞

) ) m ) )⎜ ⎟   [m] ) [m] ) ⎜ )cα,n,(j ) ⎟ T f − f  ≤   f )δ ) ⎜ ⎟ + R.H.S.(25.105.), α 1 r,n ( 1 j,r ⎝ N ⎠ + ( |α|=( j j=1 αi ! i=1

(25.106) given that  f α 1 < ∞, |α| = ( j, ( j = 1, ..., m.  [m]  As n → ∞, assuming ξn → 0, we get cα,n,(j → 0 and Tr,n ( f ) − f 1 → 0, that [m] → I in L 1 norm, with rates. is Tr,n Proof By Theorem 25.30, also by Theorem 25.13, see (25.47) and by Theorem 25.17 and Remark 25.18. 

25.3.3 Global Smoothness Preservation and Simultaneous [m] Approximation Of Tr,n We need   Definition 25.32 ([3], p. 34) Let f ∈ C R N , N ≥ 1, m ∈ N, the mth modulus of smoothness for 1 ≤ p ≤ ∞, is given by    ωm ( f ; h) p := sup m t ( f ) p,x , t2 ≤h

h > 0, where m t f (x) :=

m j=0

 (−1)m− j

m j

(25.107)

 f (x + jt) .

(25.108)

Denote ωm ( f ; h)∞ = ωm ( f, h) .

(25.109)

Above, x, t ∈ R N . We present the related global smoothness preservation result m] [( Theorem 25.33 We assume Tr,n ( ∈ Z+ , ∀ x ∈ R. Let h > 0, f ∈ ( f ; x) ∈ R, m  N C R , N ≥ 1.

506

25 High Order Approximation with Multivariate Generalized …

(i) Assume ωm ( f, h) < ∞. Then 

ωm

⎞ ⎛ r ) ) ) [( m]) m] [( Tr,n f, h ≤ ⎝ )α(j,r )⎠ ωm ( f, h) . 

(25.110)

( j=0

     (ii) Assume f ∈ C R N ∩ L 1 R N . Then 

m] [( ωm Tr,n f, h

 1

⎞ ⎛ r ) ) ) ) m [( ] ≤⎝ )α(j,r )⎠ ωm ( f, h)1 .

(25.111)

( j=0

     (iii) Assume f ∈ C R N ∩ L p R N , p > 1. Then 

m] [( ωm Tr,n f, h

 p

⎞ ⎛ r ) ) ) [( m]) ≤⎝ )α(j,r )⎠ ωm ( f, h) p .

(25.112)

( j=0



Proof Direct application of ([3]) Theorem 3.2, p. 35. We make m] m] [( [( Remark 25.34 Let r = 1, m ( ∈ Z+ , then α0,1 = 0, α1,1 = 1. Hence

m] [( T1,n

( f ; x) =

λ−N n

RN

f (x + s)

N # i=1

⎛ ⎝

sin

  ⎞2β si ξn

si

⎠

ds1 ...ds N =: Tn ( f ; x) . (25.113)

By Theorem 25.33, we get

  Theorem 25.35 We suppose Tn ( f ; x) ∈ R, ∀ x ∈ R. Let h > 0, f ∈ C R N , N ≥ 1. (i) Assume ωm ( f, h) < ∞. Then ωm (Tn f, h) ≤ ωm ( f, h) .

(25.114)

     (ii) Assume f ∈ C R N ∩ L 1 R N . Then ωm (Tn f, h)1 ≤ ωm ( f, h)1 .

(25.115)

     (iii) Assume f ∈ C R N ∩ L p R N , p > 1. Then ωm (Tn f, h) p ≤ ωm ( f, h) p . Next, we get an optimality result

(25.116)

[m] 25.3 Main Results for Tr,n

507

Proposition 25.36 Above inequality (25.114): ωm (Tn f, h) ≤ ωm ( f, h) is sharp, namely it is attained by any   f j∗ (x) = x mj , j = 1, ..., N , x = x1 , ..., x j , ..., x N ∈ R N .

(25.117) 

Proof Apply Proposition 3.5, p. 38, of [3]. We need

  Theorem 25.37 ([3], p. 39) Let f ∈ C l R N , l, N ∈ N. Here, μξn is a Borel probability measure on R N , ξn > 0, (ξn )n∈N a bounded sequence. Let β := (β1 , ..., β N ), ) ) N  ) ) βi ∈ Z+ , i = 1, ..., N ; )β ) := βi =l. Here f (x + s j), x, s ∈ R N , is μξn -integrable i=1

wrt s, for j=1, ..., r . There exist μξn -integrable functions h i1 , j , h β1 ,i2 , j , h β1 ,β2 ,i3 , j , ..., h β1 ,β2 ,...,β N −1 ,i N , j ≥ 0 ( j = 1, ..., r ) on R N such that ) ) ) ∂ i1 f (x + s j) ) ) ) ) ≤ h i1 , j (s) , i 1 = 1, ..., β1 , ) i1 ) ) ∂x1

(25.118)

) ) ) ∂ β1 +i2 f (x + s j) ) ) ) ) ) ≤ h β1 ,i2 , j (s) , i 2 = 1, ..., β2 , β ) ) ∂x2i2 ∂x1 1 .. . ) ) ) ∂ β1 +β2 +...+β N −1 +i N f (x + s j) ) ) ) ) ) ≤ h β1 ,β2 ,...,β N −1 ,i N , j (s) , i N = 1, ..., β N , ) ∂x i N ∂x β N −1 ...∂x β2 ∂x β1 ) N

N −1

2

1

∀ x, s ∈ R N . Then, both of the next exist and     [( m] m] [( fβ ; x , m ( ∈ Z+ . θr,n ( f ; x) β = θr,n In particular it holds



m] [( Tr,n ( f ; x)

 β

  m] [( fβ ; x , = Tr,n

when dμξn = dϕξn (s) , s ∈ R N , see (25.39).

(25.119)

(25.120)

508

25 High Order Approximation with Multivariate Generalized …

Corollary 25.38 (by Theorem 25.37, r = 1) It holds   (Tn ( f ; x))β = Tn f β ; x .

(25.121)

We present simultaneous global smoothness results. Theorem 25.39 Let h > 0 and the assumptions of Theorem 25.37 are valid for ( ∈ Z+ . dμξn = dϕξn . Here  γ = 0, β (0 = (0, ..., 0)), m (i) Assume ωm f γ , h < ∞. Then ωm



⎞ ⎛ r ) )   ) [( m]) m] [( Tr,n (f) γ ,h ≤ ⎝ )α(j,r )⎠ ωm f γ , h . 



(25.122)

( j=0

  (ii) Additionally suppose f γ ∈ L 1 R N . Then ωm



m] [( Tr,n (f)

 γ

,h

⎞ r ) )   ) [( m]) ≤⎝ )α(j,r )⎠ ωm f γ , h 1 . ⎛

 1

(25.123)

( j=0

  (iii) Additionally suppose f γ ∈ L p R N , p > 1. Then ωm



m] [( Tr,n (f)

 γ

,h

⎞ r ) )   ) [( m]) ≤⎝ )α(j,r )⎠ ωm f γ , h p . ⎛

 p

(25.124)

( j=0

It follows Corollary 25.40 (to Theorem 25.39) Let h > 0, r = 1 and γ = 0, β.  (i) Assume ωm f γ , h < ∞. Then     ωm (Tn ( f ))γ , h ≤ ωm f γ , h .

(25.125)

  (ii) Additionally suppose f γ ∈ L 1 R N . Then     ωm (Tn ( f ))γ , h 1 ≤ ωm f γ , h 1 .

(25.126)

  (iii) Additionally suppose f γ ∈ L p R N , p > 1. Then     ωm (Tn ( f ))γ , h p ≤ ωm f γ , h p .

(25.127)

Next comes multi-simultaneous approximation. We give   Theorem 25.41 Let f ∈ C m+l R N , m, l, N ∈ N. The assumptions of   Theorem 25.37 are valid for dμξn = dϕξn . Call γ = 0, β. Assume  f γ+α ∞ < ∞,

[m] 25.3 Main Results for Tr,n

509

and let ξn ∈ (0, 1], n ∈ N. Here β, r ∈ N, β > cα,n := cα,n,(j as in (25.70). Then

m+r +1 , and 2

Aξn (α) as in (25.47), and

 ⎞ ⎛     m   ⎟ ⎜  c f (·) ( γ+α α,n, j   [m] ⎟ ⎜ δ([m]   Tr,n ( f ; ·) γ − f γ (·) − ⎟ ⎜ j,r ⎝ N   ⎠ + ( α1 ,...,α N ≥0;   j=1 α ! i   ( |α|= j i=1

(25.128) ∞

   ωr f γ+α , ξn ≤ N  Aξn (α) . + α1 ,...,α N ≥0; αi ! |α|=m i=1



Proof Based on Theorems 25.20 and 25.37. We continue with

  Theorem 25.42 Let f ∈ C lB R N , r, l, β ∈ N (functions l-times continuously differentiable and bounded), N ∈ N − {1}. The assumptions of Theorem 25.37 are valid . Then for dμξn = dϕξn . Call γ = 0, β, ξn ∈ (0, 1], n ∈ N. Let also β > r +1 2     [0]   Tr,n f γ − f γ 

∞

 ≤2 γ N

−N

r   r λ=0

λ

N ψλ

  ξn2β(N −1) ωr f γ , ξn .

(25.129)

 [0]  If ξn → 0, as n → ∞, then Tr,n f γ → f γ uniformly. 

Proof By Theorems 25.22 and 25.37. We present

  Theorem 25.43 Let f ∈ C m+l R N , r, β, N , m, l ∈ N. The assumptions of Theorem 25.37 are valid for dμξn = dϕξn , ξn ∈ (0, 1], n ∈ N. Call γ = 0, β. Let f (γ+α) ∈   L p R N , |α| = m, x ∈ R N , and p, q > 1 : 1p + q1 = 1. Let also β > r p +m+1 ; 2 N  α j = m. Here cα,n := cα,n,(j as in (25.70), where α j ∈ Z+ , j = 1, ..., N , |α| := j=1

( j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

N 

αi = ( j. Then

i=1

 ⎞ ⎛     m  ⎟ ⎜  cα,n,(j f γ+α (x) ⎟   [m] [m] ⎜ δ(j,r ⎜  Tr,n ( f ; x) γ − f γ (x) − ⎟ N   ⎠ ⎝ + (   |α|=( j j=1 αi !   i=1

p,x

510

25 High Order Approximation with Multivariate Generalized …

 ≤



m

N p

1

(q (m − 1) + 1) q

2 γ

− Np

' Np $ r p  r p    ψλ + ψλ+mp λ λ=0

⎞

⎛

  ⎜ 2β(N −1) +m   1 ⎟ p ⎟ ⎜ ωr f γ+α , ξn p . ⎟ ξn ⎜ N ⎠ ⎝|α|=m + αi !

(25.130)

i=1



Proof Theorems 25.25 and 25.37. We continue with

  Theorem 25.44 Let f ∈ C l R N , β, r, l ∈ N, N ∈ N − {1}. The assumptions of Theorem 25.37 are valid for dμξn = dϕξn , ξn ∈ (0, 1], n ∈ N. Call γ = 0, β. Let   f γ ∈ L p R N and p, q > 1 : 1p + q1 = 1. Here β > r p +1 . Then 2     [0]   Tr,n ( f ) γ − f γ  ≤ p

2 γ− N p

N p

 r p  r p  λ=0

λ

 Np ψλ

2β(N −1) p

ξn

  ωr f γ , ξn p .

(25.131)

 [0]  · p As n → +∞ and ξn → 0, then Tr,n ( f ) γ → fγ . Proof By Theorems 25.27 and 25.37.



We continue with   Theorem 25.45 Let f ∈ C l R N , l ∈ N, N ∈ N − {1}. The assumptions of Theorem  25.37  are valid for dμξn = dϕξn , ξn ∈ (0, 1], n ∈ N. Call γ = 0, β. Let f γ ∈ . Then L 1 R N and β, r ∈ N, β > r +1 2  r   N   r      [0] N −N ψλ ξn2β(N −1) ωr f γ , ξn 1 . (25.132)  Tr,n ( f ) γ − f γ  ≤ 2 γ λ 1 λ=0

 [0]  ·1 As n → +∞ and ξn → 0, then Tr,n ( f ) γ → fγ . Proof By Theorems 25.29 and 25.37. We continue with



[m] 25.3 Main Results for Tr,n

511

  Theorem 25.46 Let f ∈ C m+l R N , r, N , β, m, l ∈ N. The assumptions of Theorem25.37  are valid for dμξn = dϕξn , ξn ∈ (0, 1], n ∈ N. Call γ = 0, β. Let f (γ+α) ∈ L 1 R N , |α| = m, x ∈ R N , β > m+r2 +1 . Here cα,n := cα,n,(j as in (25.70), where N  ( αi = ( j. Then j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| := i=1

 ⎞ ⎛     m   ⎟ ⎜  c f (x) ( γ+α α,n, j   [m] ⎟ ⎜ δ([m]   Tr,n ( f ; x) γ − f γ (x) − ⎟ ⎜ j,r ⎝ N   ⎠ + (   |α|=( j j=1 α ! i   i=1

≤

1,x

⎞

⎛

'N $ r   ⎜  2β(N −1)+m N −N & 1 ⎟ r % ⎟  ⎜ ψλ + ψλ+m 2 γ . ⎟ ωr f γ+α , ξn 1 ξn ⎜ N λ ⎠ ⎝|α|=m + λ=0 αi ! i=1

(25.133) 

Proof By Theorems 25.31 and 25.37.

[m] 25.3.4 Voronovskaya Asymptotic Expansions for Tr,n

We will apply   Theorem 25.47 ([3], p. 53) Let f ∈ C m R N , m, N ∈ N, with all  f α ∞ ≤ M, M > 0, all α : |α| = m. Let ξn > 0, (ξn )n∈N bounded sequence, μξn probability Borel measures on R N . N   + αi Call cα,n,(j := R N j = 1, ..., m − 1. Suppose ξn−m si dμξn (s), all |α| = ( i=1 N   + αi |s | dμξn (s) ≤ ρ, all α : |α| = m, ρ > 0, for any such (ξn )n∈N . Also 0 < i RN i=1

γ ∗ ≤ 1, x ∈ R N . Then ⎛ [m] θr,n ( f ; x) − f (x) =

⎞

⎜ c ( f (x) ⎟  ∗ α,n, j α ⎜ ⎟ δ([m] N  ⎟ + 0 ξnm−γ . ⎜ j,r ⎝ ⎠ + ( |α|=( j j=1 αi !

m−1

i=1

When m = 1, the sum collapses. [m] Above we assume θr,n ( f ; x) ∈ R, ∀ x ∈ R N . We give

(25.134)

512

25 High Order Approximation with Multivariate Generalized …

Theorem 25.48 Let r, m, β, N ∈ N, β > m+1 , ξn ∈ (0, 1], n ∈ N. Also α j ∈ Z+ , 2 N    α j = m. Here f ∈ C m R N , with all  f α ∞ ≤ M, M > j = 1, ..., N : |α| := j=1

0, for all α : |α| = m; and dμξn (s) = dϕξn (s), as in (25.39), ∀ s ∈ R N . Assume [m] j = 1, ..., m − 1. Tr,n ( f ; x) ∈ R, ∀ x ∈ R N . Here cα,n,(j as in (25.70), all |α| = ( ∗ N Let 0 < γ ≤ 1, x ∈ R . Then ⎛

⎞

⎜ c ( f (x) ⎟  ∗ α,n, j α ⎜ ⎟ δ([m] N  ⎟ + 0 ξnm−γ . ⎜ j,r ⎝ ⎠ + ( |α|=( j j=1 αi !

m−1

[m] Tr,n ( f ; x) − f (x) =

(25.135)

i=1

When m = 1, the sum collapses. Proof By Theorems 25.17 and 25.47. Here ρ = ϕ, see (25.68).



We give

    ∂f  Corollary 25.49 (to Theorem 25.48) Let f ∈ C 1 R N , N ∈ N, with all  ∂x  i ∗

M, M > 0, i = 1, ..., N . Let 0 < γ ≤ 1. Assume r ∈ N and β ∈ N − {1}, ξn ∈ (0, 1], n ∈ N. Then

[1] Tr,n

∞

≤

( f ; x) ∈ R, ∀ x ∈ R . Here

 ∗ [1] Tr,n ( f ; x) − f (x) = 0 ξn1−γ .

N

(25.136)

Proof By Theorems 25.17 and 25.48. Here it is ρ = ϕ, apply (25.68) for m = 1.  We continue with

 2   2        Corollary 25.50 (to Theorem 25.48) Let f ∈ C 2 R2 , with all  ∂∂x 2f  ,  ∂∂x 2f  , 1 2 ∞ ∞  2   ∂ f  ≤ M, M > 0, ξ ∈ (0, 1], n ∈ N. Call  ∂x1 ∂x2  n ∞

c1 =

R2

where dϕ∗ξn = λ−2 n

2 # i=1

s1 dϕ∗ξn

⎛ ⎝

sin

(s) , c2 =

R2

s2 dϕ∗ξn (s) ,

(25.137)

  ⎞2β si ξn

si

⎠

ds1 ds2 , s = (s1 , s2 ) ∈ R2 .

[2] Let 0 < γ ∗ ≤ 1 and assume Tr,n ( f ; x) ∈ R, ∀ x ∈ R2 . Here r, β ∈ N and β > 23 . Then ⎞ ⎛   r  ∂f ∂f ∗ [2] ⎠ [2] ⎝ c1 Tr,n ( f ; x) − f (x) = α j,r j (x) + c2 (x) + 0 ξn2−γ . ∂x1 ∂x2 j=1

(25.138)

[m] 25.3 Main Results for Tr,n

513



Proof By Theorems 25.17 and 25.48. We also give

  Theorem 25.51 Let f ∈ C m+l R N , m, l, N ∈ N. Assumptions of N Theorem25.37 are valid for dϕξn (s) , s ∈ R , ξn ∈ (0, 1], n ∈ N. Call γ = 0, β. Suppose  f γ+α ∞ ≤ M, M > 0, for all α : |α| = m. Here cα,n,(j is as in (25.70), all   [m] |α| = ( f γ ; x ∈ R, ∀ x ∈ R N . Let also j = 1, ..., m − 1; 0 < γ ∗ ≤ 1. Assume Tr,n . Then r, β ∈ N and β > m+1 2 ⎞

⎛ 

[m] Tr,n ( f ; x)

 γ

− f γ (x) =

⎟ ⎜ c (f  ∗ α,n, j γ+α (x) ⎟ ⎜ δ([m]   ⎟ + 0 ξnm−γ . ⎜ j,r ⎝ N ⎠ + ( |α|=( j j=1 αi !

m−1

i=1

(25.139) When m = 1, the sum collapses. Proof Use of Theorem 25.17 and Theorem 4.6, p. 54 of [3]. Here it is ρ = ϕ, see (25.68). 

25.3.5 Simultaneous Approximation by Multivariate [m] Complex Tr,n We make Remark 25.52 We consider here complex √ valued Borel measurable functions f : R N → C such that f = f 1 + i f 2 , i = −1, where f 1 , f 2 : R N → R are implied to be real valued Borel measurable functions. We define the multivariate complex Trigonometric singular operators [m] [m] [m] Tr,n ( f ; x) := Tr,n ( f 1 ; x) + i Tr,n ( f 2 ; x) , x ∈ R N .

(25.140)

  [m] We assume that Tr,n f j ; x ∈ R, ∀ x ∈ R N , j = 1, 2. One notices easily that ) ) ) ) ) ) [m] )T ( f ; x) − f (x)) ≤ )T [m] ( f 1 ; x) − f 1 (x)) + )T [m] ( f 2 ; x) − f 2 (x)) r,n r,n r,n (25.141) also    [m]  Tr,n ( f ; x) − f (x)

       [m]  [m] ≤ Tr,n ( f 1 ; x) − f 1 (x) + Tr,n ( f 2 ; x) − f 2 (x) ∞,x ∞,x ∞,x

(25.142)

and

514

25 High Order Approximation with Multivariate Generalized …

   [m]    T ( f ) − f  ≤ T [m] ( f 1 ) − f 1  + T [m] ( f 2 ) − f 2  , p ≥ 1. (25.143) r,n r,n r,n p p p Furthermore, it holds f α (x) = f 1,α (x) + i f 2,α (x) ,

(25.144)

where α denotes a partial derivative of any order and arrangement. We give Theorem 25.53 Let f : R N → C, such that  f = f 1 + i f 2 , j = 1, 2. Here    ∂ m f (·,·,...,·)  r, N , β, m ∈ N, f j ∈ C m R N , x ∈ R N . Assume  ∂x α1j ...∂x α N  < ∞, for all αi ∈ Z+ , i = 1, ..., N : |α| :=

N 

1

N

∞

αi = m. Let ϕξn be the Borel probability measure on

i=1

R N , see (25.39), where ξn ∈ (0, 1], n ∈ N. Here β > (25.47), and cα,n := cα,n,(j as in (25.70). Then

m+r +1 , 2

and Aξn (α) as in

 ⎞ ⎛     m  ⎟ ⎜ [m] ⎜ cα,n,(j f α (x) ⎟   [m] δ(j,r ⎜ Tr,n ( f ; x) − f (x) − ⎟ N   ⎠ ⎝ + ( α1 ,...,α N ≥0;   j=1 αi !   |α|=( j i=1

∞,x

     ωr f 1,α , ξn + ωr f 2,α , ξn ≤ Aξn (α) . N  + α1 ,...,α N ≥0; α ! i |α|=m

(25.145)

i=1



Proof By Theorem 25.20. We proceed with

N Theorem N ∈ N − {1}, j = 1, 2. Here f j ∈  N  25.54 Let f : R →C : f = f 1 + i f 2 ,r +1 C B R uniformly continuous, β, r ∈ N, β > 2 , ξn ∈ (0, 1], n ∈ N. Then

 r   N r   [0] T f − f  ≤ 2 N γ −N ψλ ξn2β(N −1) r,n ∞ λ

(25.146)

λ=0

(ωr ( f 1 , ξn ) + ωr ( f 2 , ξn )) , As n → ∞ and ξn → 0, we derive   [0] T f − f  → 0 r,n ∞

(25.147)

with rates. Proof By Theorem 25.22.



[m] 25.3 Main Results for Tr,n

515

Next comes multi-simultaneous approximation.   Theorem 25.55 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m+l R N , N , m, l ∈ N. The assumptions of   Theorem 25.37 are valid for f j and dμξn = dϕξn . Call γ = 0, β. Assume  f j,γ+α ∞ < ∞, and let ξn ∈ (0, 1], n ∈ N. Here β, r ∈ N, β > m+r2 +1 , and Aξn (α) as in (25.47), and cα,n := cα,n,(j as in (25.70). Then  ⎞ ⎛     m   ⎟ ⎜  c f (·) α,n,( j γ+α  [m] ⎟ [m] ⎜ δ(j,r ⎜   Tr,n ( f ; ·) γ − f γ (·) − ⎟ N  ⎠ ⎝ + ( α1 ,...,α N ≥0;   j=1 α ! i   |α|=( j i=1

(25.148) ∞

     ωr f 1,γ+α , ξn + ωr f 2,γ+α , ξn ≤ Aξn (α) . N  + α1 ,...,α N ≥0; αi ! |α|=m i=1



Proof Based on Theorems 25.37 and 25.41. We continue with

  Theorem 25.56 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C lB R N , N ∈ N − {1}, l ∈ N (functions l-times continuously differentiable and bounded). The assumptions of Theorem 25.37 are valid for f j and dμξn = dϕξn . Call γ = 0, β, . Then ξn ∈ (0, 1], n ∈ N. Let also β, r ∈ N, β > r +1 2     [0]   Tr,n f γ − f γ 

∞

≤2 γ N

−N

 r   r λ=0

λ

N ψλ

     ξn2β(N −1) ωr f 1,γ , ξn + ωr f 2,γ , ξn .

(25.149)

 [0]  f γ → f γ uniformly. If ξn → 0, as n → ∞, then Tr,n 

Proof By Theorems 25.42 and 25.37. We proceed with L p approximations

  Theorem 25.57 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m R N ,   r, β, N , m ∈ N, with f j,α ∈ L p R N , |α| = m, x ∈ R N . Let p, q > 1 : 1p + q1 = 1. Here ϕξn is a Borel probability measure on R N as in (25.39 ), for ξn ∈ (0, 1], n ∈ N. N  + |α| := Let β > r p +m+1 ; α ∈ Z , i = 1, ..., N : αi = m. Here cα,n := cα,n,(j i 2 i=1

as in (25.70), where ( j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , N  |α| := αi = ( j. Then i=1

516

25 High Order Approximation with Multivariate Generalized …

 ⎛ ⎞     m  ⎜ ⎟ [m] ⎜ cα,n,(j f α (x) ⎟   [m] δ(j,r ⎜  ⎟ N Tr,n ( f ; x) − f (x) −   ⎝ ⎠ + (   |α|=( j j=1 αi !   i=1  ≤



m

N p

1

(q (m − 1) + 1) q

2 γ

− Np

p,x

' Np $ r p  r p    ψλ + ψλ+mp λ λ=0

⎞

⎛

  ⎜ 2β(N −1) +m       1 ⎟ p ⎟ ⎜ ξ ω . , ξ + ω , ξ f f ⎟ ⎜ n r 1,α n r 2,α n p p N ⎠ ⎝|α|=m + αi !

(25.150)

i=1



Proof By Theorem 25.25. We continue with

   Theorem 25.58 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C R N ∩  N  , ξn ∈ (0, 1], L p R ; N ∈ N − {1}, β, r ∈ N; p, q > 1 : 1p + q1 = 1; β > r p +1 2 n ∈ N. Then  r p   Np r p    [0] N N − T ( f ) − f  ≤ 2 p γ p ψλ r,n p λ λ=0

2β(N −1) p

ξn

% & ωr ( f 1 , ξn ) p + ωr ( f 2 , ξn ) p .

(25.151)

 [0]  [0] As ξn → 0, when n → ∞, we derive Tr,n f − f  p → 0, i.e. Tr,n → I, the unit operator, in L p norm. 

Proof By Theorem 25.27. We also give

   Theorem 25.59 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C R N ∩  N  L 1 R ; N ∈ N − {1}, r, β ∈ N, β > r +1 , ξn ∈ (0, 1], n ∈ N. Then 2  r   N r  [0]  T ( f ) − f  ≤ 2 N γ −N ψλ r,n 1 λ λ=0

  ξn2β(N −1) ωr ( f 1 , ξn )1 + ωr ( f 2 , ξn )1 . [0] → I, in L 1 norm. As ξn → 0, we get Tr,n

(25.152)

[m] 25.3 Main Results for Tr,n

517



Proof By Theorem 25.29. We further present

  Theorem 25.60 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m R N , N    N , β, m, r ∈ N, with f j,α ∈ L 1 R N , where αi ∈ Z+ , i = 1, ..., N , |α| := αi = m, x ∈ R N , ξn ∈ (0, 1], n ∈ N and β >

i=1

m+r +1 . 2

Here cα,n := cα,n,(j as in (25.70), N  αi = where ( j = 1, ..., m, and α := (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| :=

( j. Also here ϕξn is the Borel probability measure on R N , see (25.39). Then  ⎞ ⎛     m  ⎜ c ( f (x) ⎟ α α,n, j   [m] ⎟ ⎜ δ([m] Tr,n ( f ; x) − f (x) − ⎟ ⎜ j,r ⎝ N  ⎠ + (   j=1 |α|=( j αi !   i=1

i=1

(25.153)

1,x

⎧ ⎫ ⎞ ⎛ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎜ 1 ⎟%  ⎬    & ⎟ ⎜ ≤ ⎟ ωr f 1,α , ξn 1 + ωr f 2,α , ξn 1 ξn2β(N −1)+m ⎜ N ⎪ ⎪ ⎠ ⎝+ ⎪ ⎪ ⎪ ⎪ αi ! ⎩|α|=m ⎭ i=1

2 γ N

−N

'N $ r   r % & . ψλ + ψλ+m λ λ=0



Proof By Theorem 25.31. We continue with simultaneous L p approximations.

  Theorem 25.61 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m+l R N , r, β, N , m, l ∈ N. The assumptions of Theorem 25.37 are valid for dμξn = dϕξn , ξn ∈   (0, 1], n ∈ N and f j . Call γ = 0, β. Let f j,(γ+α) ∈ L p R N , |α| = m, x ∈ R N , and N  p, q > 1 : 1p + q1 = 1. Let β > r p +m+1 ; αi ∈ Z+ , i = 1, ..., N : |α| := αi = 2 i=1

m. Here cα,n := cα,n,(j as in (25.70), where ( j = 1, ..., m, and α := (α1 , ..., α N ), N  αi = ( j. Then αi ∈ Z+ , i = 1, ..., N , |α| := i=1

 ⎞ ⎛     m  ⎟ ⎜ [m] ⎜ cα,n,(j f γ+α (x) ⎟    [m] δ(j,r ⎜  Tr,n ( f ; x) γ − f γ (x) − ⎟ N   ⎠ ⎝ + (   |α|=( j j=1 αi !   i=1

p,x

518

25 High Order Approximation with Multivariate Generalized …

 ≤



m

N p

1

(q (m − 1) + 1) q

2 γ

− Np

' Np $ r p  r p    ψλ + ψλ+mp λ

(25.154)

λ=0

⎞

⎛

  ⎜ 2β(N −1) +m       1 ⎟ p ⎟ ⎜ ξ ω f f , ξ + ω , ξ ⎟ ⎜ n r 1,γ+α n p r 2,γ+α n p . N ⎠ ⎝|α|=m + αi ! i=1



Proof By Theorems 25.37 and 25.43. We give also

  Theorem 25.62 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C l R N , N ∈ N − {1}, l ∈ N. The assumptions of Theorem 25.37 are valid for dμξn = dϕξn ,   ξn ∈ (0, 1], n ∈ N and f j . Call γ = 0, β. Let f j,γ ∈ L p R N and p, q > 1 : 1p + 1 q

= 1. Here β, r ∈ N, β >

r p +1 . 2

Then

   N  [0]  −N  Tr,n ( f ) γ − f γ  ≤ 2 p γ p

 r p  r p 

p

2β(N −1) p

ξn

λ=0

λ

 Np ψλ

      ωr f 1,γ , ξn p + ωr f 2,γ , ξn p .

(25.155)

 [0]  · p As n → +∞ and ξn → 0, then Tr,n f γ → fγ . 

Proof By Theorems 25.37 and 25.44. We continue with

  Theorem 25.63 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C l R N , N ∈ N − {1}, l ∈ N. The assumptions of Theorem 25.37 are valid for dμξn = dϕξn ,   . ξn ∈ (0, 1], n ∈ N and f j . Call γ = 0, β. Let f j,γ ∈ L 1 R N and β, r ∈ N, β > r +1 2 Then  r   N   r    [0] ψλ  Tr,n ( f ) γ − f γ  ≤ 2 N γ −N λ 1 λ=0

%     & ξn2β(N −1) ωr f 1,γ , ξn 1 + ωr f 2,γ , ξn 1 .

(25.156)

 [0]  ·1 As n → +∞ and ξn → 0, then Tr,n ( f ) γ → fγ . Proof By Theorems 25.37 and 25.45. We finish with



[m] 25.3 Main Results for Tr,n

519

  Theorem 25.64 Let f : R N → C : f = f 1 + i f 2 , j = 1, 2. Here f j ∈ C m+l R N , N , β.r, m, l ∈ N. The assumptions of Theorem 25.37 are valid for dμξn = dϕξn ,   ξn ∈ (0, 1], n ∈ N and f j . Call γ = 0, β. Let f j,(γ+α) ∈ L 1 R N , |α| = m, x ∈ R N . Here β > m+r2 +1 and cα,n := cα,n,(j as in (25.70), where ( j = 1, ..., m, and α := N  αi = ( j. Then (α1 , ..., α N ), αi ∈ Z+ , i = 1, ..., N , |α| := i=1

 ⎞ ⎛     m  ⎟ ⎜ [m] ⎜ cα,n,(j f γ+α (x) ⎟    [m] δ(j,r ⎜  Tr,n ( f ; x) γ − f γ (x) − ⎟ N   ⎠ ⎝ + (   |α|=( j j=1 αi !   i=1

1,x

⎛

⎞

⎛

⎞

⎜ ⎜ 1 ⎟%     &⎟ ⎜ ⎟ ⎜ ⎟ ≤⎜ ⎟ ωr f 1,γ+α , ξn 1 + ωr f 2,γ+α , ξn 1 ⎟ ⎜ N ⎝|α|=m ⎝ + ⎠ ⎠ αi ! i=1

$ ξn2β(N −1)+m 2 N γ −N

'N r   & r % . ψλ + ψλ+m λ

(25.157)

λ=0

Proof By Theorems 25.37 and 25.46.



References 1. Anastassiou, G.A.: Rate of convergence of non-positive linear convolution type operators. A sharp inequality. J. Math. Anal. and Appl. 142, 441–451 (1989) 2. Anastassiou, G.A.: Moments in Probability and Approximation Theory. Pitman Research Notes in Mathematics, vol. 287. Longman Scientific & Technical, Harlow (1993) 3. Anastassiou, G.A.: Approximation by Multivariate Singular Integrals. Springer, New York (2011) 4. Anastassiou, G.A.: Approximations by multivariate generalized trigonometric type singular integral operators. Submitted (2020) 5. Anastassiou, G., Gal, S.: Approximation Theory. Birkhaüser, Boston (2000) 6. Anastassiou, G.A., Mezei, R.A.: L p convergence with rates of general singular integral operators. J. Comput. Anal. Appl. 14(6), 1067–1083 (2012) 7. Anastassiou, G.A., Mezei, R.A.: Convergence of complex general singular integral operators. J. Concr. Appl. Math. 10(3–4), 259–283 (2012) 8. Anastassiou, G.A., Mezei, R.A.: Uniform convergence with rates of general singular operators. CUBO 15(2), 1–19 (2013) 9. DeVore, R.A., Lorentz, G.G.: Constructive Approximation, vol. 303. Springer, Berlin (1993) 10. Edwards, J.: A Treatise on the Integral Calculus, vol. II. Chelsea, New York (1954)

Chapter 26

Concluding Remarks

In a 1695 letter to L’Hospital, Leibniz asked, “Can we generalize ordinary derivatives to ones of arbitrary order?” L’Hospital replied to Leibniz with another question, “What if the order is 1/2?” Then Leibniz, in a letter dated September 30, 1695 replied, “It will lead to a paradox, from which one day many useful consequences will be drawn.” That was the beginning of fractional calculus. The subject has been ongoing for more than 300 years now. Many famous mathematicians contributed to the topic over the years such as Liouville, Euler, Laplace, Lagrange, Riemann, Weyl, Fourier, Abel, Lacroix, Grunwald, and Letnikov. The first attempt to give a rigorous definition of fractional derivatives and study the subject in depth was by Liouville during 1832–1855. The concept grew out of his earlier work in electromagnetism. Fractional derivatives describe solutions of fractional integral equations, many times arising from physics, as done by Abel in 1823 to solve the brachistochrone problem. Fractional calculus has a long history but, from the applicative point of view, it fell into oblivion for hundreds of years because of lack of applications to other sciences such as physics and engineering. This oblivion was also due to its complexity and lack of physical and geometric connection. Fortunately, that was then as things have changed dramatically. Fractional calculus has developed a lot in the last fifty years and especially since 1974 when the first international conference in the field took place, organized by B. Ross, in New Haven, Connecticut, USA. Now we frequently see such conferences around the world. There exist a lot of related research activities resulting in many interesting articles and books. AMS/Mathscinet lists over 50000 fractional publications during 2020. We now see many important applications: in acoustic wave propagation in inhomogeneous porous material, diffusive transport, fluid flow, dynamical processes in self-similar structures, dynamics of earthquakes, optics, geology, viscoelastic materials, biosciences, bioengineering, medicine, economics, probability and statistics, astrophysics, chemical engineering, physics, splines, tomography, fluid mechanics, electromagnetic waves, non-linear control, signal processing, control of electronic power, converters, chaotic dynam© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Constructive Fractional Analysis with Applications, Studies in Systems, Decision and Control 362, https://doi.org/10.1007/978-3-030-71481-9_26

521

522

26 Concluding Remarks

ics, polymer science, proteins, polymer physics, electrochemistry, statistical physics, rheology, thermodynamics, neural networks, and many others. By now almost all fields of research in science and engineering use fractional calculus to better describe them. The current mathematical theory of fractional calculus seems to lag behind the needs for mathematical modeling of all the above applications. Therefore there exists a lot of room for mathematical theoretical expansion on the subject. The recent development of fractal theory has shown great connections to fractional calculus. In general, fractional calculus currently provides the best description for fractals and chaotic situations. All in all fractional calculus provides the best expression for the complexity of our modern science. In recent years researchers came up with more efficient and practical fractional derivatives, e.g. M. Caputo (1967) inspired by studies in Geophysics, J.A. Canavati (1987) and by many the new method of fractional variable order (2019). The author was the first to study fractional inequalities and fractional approximation theory by publishing numerous of articles and monographs. In this monograph the author continuous his studies by employing his constructive and computational fractional calculus methods (2020). He applies his generalized fractional differentiation techniques of RiemannLiouville, Caputo and Canavati types and of the very new fractional variable order to various kinds of inequalities such as of Opial, Hardy, Hilbert-Pachpatte and over the spherical shell. He continues with E. R. Love left and right sides fractional integral inequalities. The fractional Landau inequalities, of left and right sides, univariate and multivariate, including ones for Semigroups occupied fourteen chapters of this monograph. These were developed to all possible directions and right side multivariate fractional Taylor formulae were proven for that purpose. He continued with several of the very new Gronwall fractional inequalities of variable order. The last part of book was about constructive approximation theory, it included: ordinary and fractional approximations by positive sublinear operators, and high order approximation by multivariate generalized Picard, Gauss-Weierstrass, Poisson-Cauchy and Trigonometric singular integrals which are not positive operators. Definitely what is developed in this monograph can have a great impact and can lead to more interesting avenues of research.