Intelligent Analysis: Fractional Inequalities and Approximations Expanded (Studies in Computational Intelligence, 886) 9783030386351, 9783030386368, 303038635X
This book focuses on computational and fractional analysis, two areas that are very important in their own right, and wh
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Table of contents :
Preface
Contents
1 General Ordinary Iyengar Inequalities
1.1 Introduction
1.2 Main Results
References
2 Caputo Fractional Iyengar Inequalities
2.1 Introduction
2.2 Main Results
References
3 Canavati Fractional Iyengar Inequalities
3.1 Background
3.2 Main Results
References
4 General Multivariate Iyengar Inequalities
4.1 Background
4.2 Main Results
References
5 Multivariate Iyengar Inequalities for Radial Functions
5.1 Background
5.2 Main Results
References
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
6.1 Background
6.2 Main Results
6.3 Appendix
References
7 General Multidimensional Fractional Iyengar Inequalities
7.1 Background
7.2 Main Results
7.3 Appendix
References
8 Delta Time Scales Iyengar Inequalities
8.1 Introduction
8.2 Background
8.3 Main Results
8.4 Applications
References
9 Time Scales Nabla Iyengar Inequalities
9.1 Introduction
9.2 Background
9.3 Main Results
9.4 Applications
References
10 Choquet–Iyengar Advanced Inequalities
10.1 Background—I
10.2 Background—II
10.3 Main Results
References
11 Fractional Conformable Iyengar Inequalities
11.1 Background
11.2 Main Results
References
12 Iyengar Fuzzy Inequalities
12.1 Introduction
12.2 Background
12.3 Main Results
References
13 Choquet Integral Analytical Type Inequalities
13.1 Background
13.2 Main Results
References
14 Local Fractional Taylor Formula
14.1 Introduction
14.2 Main Results
References
15 Negative Domain Local Fractional Inequalities
15.1 Introduction
15.2 Background
References
16 Fractional Approximation by Riemann–Liouville Fractional Derivatives
16.1 Introduction
16.2 Main Results
References
17 Riemann–Liouville Fractional Fundamental Theorem of Calculus and Riemann–Liouville Fractional Polya Integral Inequality and the Generalization to Choquet Integral Case
17.1 Introduction
17.2 Background
17.3 Main Results
References
18 Low Order Riemann–Liouville Fractional Inequalities with Absent Initial Conditions
18.1 Introduction
18.2 Main Results
References
19 Low Order Fractional Riemann–Liouville Inequalities on a Spherical Shell
19.1 Main Results
References
20 Complex Korovkin Theory
20.1 Introduction
20.2 Background
20.3 Main Results—I
20.4 Main Results—II
References
21 Left Caputo Fractional Complex Inequalities
21.1 Introduction
21.2 Background
21.3 Main Results
References
22 Complex Opial Inequalities
22.1 Introduction
22.2 Background
22.3 Main Results
References
23 Complex Multivariate Montgomery Identity and Ostrowski and Grüss Inequalities
23.1 Introduction
23.2 Main Results
References
24 Complex Multivariate Fink Identity and Complex Multivariate Ostrowski and Grüss Inequalities
24.1 Introduction
24.2 Main Results
24.3 Applications
References
25 Fractional Conformable Approximation of Csiszar's f-Divergence
25.1 Background—I
25.2 Background—II
25.3 Main Results—I
25.4 Background—III
25.5 Main Results—II
References
26 Fractional Conformable Self Adjoint Operator Analytic Inequalities
26.1 Background—I
26.2 Background—II
26.3 Main Results
References
27 Fractional Left Local General M-Derivative
27.1 Introduction
27.2 Main Results
References
28 Fractional Right Local General M-Derivative
28.1 Introduction
28.2 Main Results
References
29 Complex Multivariate Taylor Formula
29.1 Main Results
References
Citation preview
Studies in Computational Intelligence 886
George A. Anastassiou
Intelligent Analysis: Fractional Inequalities and Approximations Expanded
Studies in Computational Intelligence Volume 886
Series Editor Janusz Kacprzyk, Polish Academy of Sciences, Warsaw, Poland
The series “Studies in Computational Intelligence” (SCI) publishes new developments and advances in the various areas of computational intelligence—quickly and with a high quality. The intent is to cover the theory, applications, and design methods of computational intelligence, as embedded in the fields of engineering, computer science, physics and life sciences, as well as the methodologies behind them. The series contains monographs, lecture notes and edited volumes in computational intelligence spanning the areas of neural networks, connectionist systems, genetic algorithms, evolutionary computation, artificial intelligence, cellular automata, self-organizing systems, soft computing, fuzzy systems, and hybrid intelligent systems. Of particular value to both the contributors and the readership are the short publication timeframe and the world-wide distribution, which enable both wide and rapid dissemination of research output. The books of this series are submitted to indexing to Web of Science, EI-Compendex, DBLP, SCOPUS, Google Scholar and Springerlink.
More information about this series at http://www.springer.com/series/7092
George A. Anastassiou
Intelligent Analysis: Fractional Inequalities and Approximations Expanded
123
George A. Anastassiou Department of Mathematical Sciences University of Memphis Memphis, TN, USA
ISSN 1860-949X ISSN 1860-9503 (electronic) Studies in Computational Intelligence ISBN 978-3-030-38635-1 ISBN 978-3-030-38636-8 (eBook) https://doi.org/10.1007/978-3-030-38636-8 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 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
Preface
Computational and fractional analysis plays a very important role nowadays 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. We start with the well-known Iyengar type inequalities for the first 12 chapters working in all possible directions, especially along the fractional aspect. Our results are optimal or nearly optimal. Iyengar inequalities estimate the distance of the average of a function from the average of its values at the endpoints of their interval of definition, with a great impact to numerical analysis. We continue with Choquet integral analytical inequalities, related to great applications in economics. We deal with the local fractional derivatives of Riemann–Liouville type and related results including inequalities. We encounter the case of low order Riemann–Liouville fractional derivatives and inequalities without initial conditions at the univariate and multivariate level, we study also related approximations. We continue with the quantitative complex approximation theory by operators and various important complex fractional inequalities. We also deal with the conformable fractional approximation of well-known Csiszar’s f-divergence, which is the most essential and general measure for the comparison between two probability measures. We also present conformable fractional self-adjoint operator inequalities. We continue with the study of new local fractional M-derivatives which carry all the basic properties of ordinary derivatives. We finish with the new complex multivariate Taylor formula with integral remainder. This monograph is the natural evolution of the 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 computational 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 witnesses by many books and volumes in vii
viii
Preface
Springer’s program: http://www.springer.com/gp/search?query=fractional&submit= Prze%C5%9Blij and the purpose of our book is to provide 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. 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: General Iyengar type inequalities, Caputo fractional Iyengar type inequalities, Canavati fractional Iyengar type inequalities, General Multivariate Iyengar type inequalities, Multivariate Iyengar type inequalities for radial functions, Multidimensional Fractional Iyengar type inequalities for radial functions, General Multidimensional Fractional Iyengar type inequalities, Time Scales Delta Iyengar type inequalities, Nabla Time Scales Iyengar type inequalities, Choquet–Iyengar type advanced inequalities, Conformable fractional Iyengar type inequalities, Fuzzy Iyengar type inequalities, Choquet integral analytic inequalities, Local Fractional Taylor Formula, Negative Domain local fractional inequalities, Approximation with Riemann–Liouville fractional derivatives, Riemann–Liouville fractional fundamental theorem of Calculus and Riemann– Liouville Fractional Polya type integral inequality and its extension to Choquet integral setting, Low order Riemann–Liouville fractional inequalities without initial conditions, Low order Riemann–Liouville fractional inequalities on a spherical shell, Complex Korovkin Theory, Complex left Caputo fractional inequalities, Complex Opial type inequalities, Complex Multivariate Montgomery type identity leading to Complex Multivariate Ostrowski and Grüss Inequalities, Complex Multivariate Fink type identity applied to Complex Multivariate Ostrowski and Grüss inequalities, Conformable Fractional Approximation of Csiszar’s f-Divergence, Conformable fractional self adjoint operator analytic inequalities, On the left fractional local general M-derivative, About the right fractional local general M-derivative, Complex Multivariate Taylor’s formula. An extensive list of references is given per chapter.
Preface
ix
The book’s results are expected to find applications in many areas of pure and applied mathematics, especially in approximation theory and inequalities in both ordinary and fractional sense. As such 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 2018–2019 at the University of Memphis. The author likes to thank Prof. Alina Lupas of University of Oradea, Romania, for checking and reading the manuscript. Memphis, TN, USA September 2019
George A. Anastassiou
Contents
1
General Ordinary Iyengar Inequalities 1.1 Introduction . . . . . . . . . . . . . . . . 1.2 Main Results . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . .
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1 1 1 13
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Caputo Fractional Iyengar Inequalities . 2.1 Introduction . . . . . . . . . . . . . . . . . 2.2 Main Results . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . .
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3
Canavati Fractional Iyengar 3.1 Background . . . . . . . . 3.2 Main Results . . . . . . . References . . . . . . . . . . . . . .
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General Multivariate Iyengar Inequalities . 4.1 Background . . . . . . . . . . . . . . . . . . . 4.2 Main Results . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . .
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45 45 52 66
5
Multivariate Iyengar Inequalities for Radial Functions . 5.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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67 67 70 89
6
Multidimensional Fractional Functions . . . . . . . . . . . . . . 6.1 Background . . . . . . . . 6.2 Main Results . . . . . . . 6.3 Appendix . . . . . . . . . . References . . . . . . . . . . . . . .
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. 91 . 91 . 107 . 140 . 142
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Iyengar Inequalities for Radial . . . . .
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xi
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Contents
7
General Multidimensional Fractional Iyengar Inequalities 7.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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143 143 159 184 186
8
Delta Time Scales Iyengar Inequalities 8.1 Introduction . . . . . . . . . . . . . . . . 8.2 Background . . . . . . . . . . . . . . . . 8.3 Main Results . . . . . . . . . . . . . . . 8.4 Applications . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . .
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189 189 189 193 208 212
9
Time Scales Nabla Iyengar Inequalities 9.1 Introduction . . . . . . . . . . . . . . . . . 9.2 Background . . . . . . . . . . . . . . . . . 9.3 Main Results . . . . . . . . . . . . . . . . 9.4 Applications . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . .
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213 213 213 220 235 238
10 Choquet–Iyengar Advanced Inequalities . 10.1 Background—I . . . . . . . . . . . . . . . . 10.2 Background—II . . . . . . . . . . . . . . . 10.3 Main Results . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . .
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241 241 243 245 255
11 Fractional Conformable Iyengar Inequalities 11.1 Background . . . . . . . . . . . . . . . . . . . . . 11.2 Main Results . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . .
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257 257 261 272
12 Iyengar Fuzzy Inequalities 12.1 Introduction . . . . . . . 12.2 Background . . . . . . . 12.3 Main Results . . . . . . References . . . . . . . . . . . . .
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273 273 273 277 282
13 Choquet Integral Analytical Type Inequalities 13.1 Background . . . . . . . . . . . . . . . . . . . . . . 13.2 Main Results . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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283 283 287 295
14 Local Fractional Taylor Formula 14.1 Introduction . . . . . . . . . . . . 14.2 Main Results . . . . . . . . . . . References . . . . . . . . . . . . . . . . . .
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Contents
xiii
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303 303 304 315
16 Fractional Approximation by Riemann–Liouville Fractional Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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317 317 318 326
17 Riemann–Liouville Fractional Fundamental Theorem of Calculus and Riemann–Liouville Fractional Polya Integral Inequality and the Generalization to Choquet Integral Case . 17.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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329 329 330 334 340
18 Low Order Riemann–Liouville Fractional Inequalities with Absent Initial Conditions . . . . . . . . . . . . . . . . . . 18.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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341 341 342 353
15 Negative Domain Local Fractional Inequalities . 15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 15.2 Background . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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19 Low Order Fractional Riemann–Liouville Inequalities on a Spherical Shell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355 19.1 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 . . . . . .
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365 365 366 370 377 380
21 Left Caputo Fractional Complex Inequalities 21.1 Introduction . . . . . . . . . . . . . . . . . . . . . 21.2 Background . . . . . . . . . . . . . . . . . . . . . 21.3 Main Results . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . .
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381 381 383 386 391
20 Complex Korovkin Theory 20.1 Introduction . . . . . . . 20.2 Background . . . . . . . 20.3 Main Results—I . . . . 20.4 Main Results—II . . . References . . . . . . . . . . . . .
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22 Complex Opial Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393 22.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393 22.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394
xiv
Contents
22.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399 23 Complex Multivariate Montgomery Identity and Ostrowski and Grüss Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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401 401 407 424
24 Complex Multivariate Fink Identity and Complex Multivariate Ostrowski and Grüss Inequalities . . . . . . . . . . . . . . . . . . . . . . . 24.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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425 425 432 450 461
25 Fractional Conformable Approximation f -Divergence . . . . . . . . . . . . . . . . . . . . . 25.1 Background—I . . . . . . . . . . . . . . . 25.2 Background—II . . . . . . . . . . . . . . 25.3 Main Results—I . . . . . . . . . . . . . . 25.4 Background—III . . . . . . . . . . . . . . 25.5 Main Results—II . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . .
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463 463 464 468 476 477 480
26 Fractional Conformable Self Adjoint Operator Analytic Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.1 Background—I . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.2 Background—II . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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481 481 484 487 498
27 Fractional Left Local General M-Derivative 27.1 Introduction . . . . . . . . . . . . . . . . . . . . 27.2 Main Results . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . .
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501 501 503 510
28 Fractional Right Local General M-Derivative . 28.1 Introduction . . . . . . . . . . . . . . . . . . . . . . 28.2 Main Results . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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511 511 513 520
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29 Complex Multivariate Taylor Formula . . . . . . . . . . . . . . . . . . . . . . 521 29.1 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525
Chapter 1
General Ordinary Iyengar Inequalities
Here we present general Iyengar type inequalities with respect to L p norms, with 1 ≤ p ≤ ∞. The method is based on the generalized Taylor’s formula. See also [2].
1.1 Introduction We are motivated by the following famous Iyengar inequality (1938), [3]. Theorem 1.1 Let f be a differentiable function on [a, b] and f (x) ≤ M. Then
a
b
M (b − a) 2 ( f (b) − f (a))2 1 . f (x) d x − (b − a) ( f (a) + f (b)) ≤ − 2 4 4M (1.1)
1.2 Main Results We present the following Iyengar type inequalities: Theorem 1.2 Let n ∈ N, f ∈ AC n ([a, b]) (i.e. f (n−1) ∈ AC ([a, b]), absolutely continuous functions). We assume that f (n) ∈ L ∞ ([a, b]). Then (i) n−1 b (k) 1 f (a) (t − a)k+1 + (−1)k f (k) (b) (b − t)k+1 ≤ f (x) d x − a (k + 1)! k=0
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_1
1
2
1 General Ordinary Iyengar Inequalities
(n) f
L ∞ ([a,b])
(n + 1)! ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
(t − a)n+1 + (b − t)n+1 ,
(1.2)
the right hand side of (1.2) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
(n) f
L ∞ ([a,b])
(n + 1)!
(b − a)n+1 , 2n
(1.3)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
b
a
f (n) L ∞ ([a,b]) (b − a)n+1 f (x) d x ≤ , 2n (n + 1)!
(1.4)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1 b b − a k+1 k+1 (k) 1 j f (x) d x − f (a) + (−1)k (N − j)k+1 f (k) (b) ≤ a N (k + 1)! k=0
(n) f
L ∞ ([a,b])
(n + 1)!
b−a N
n+1
j n+1 + (N − j)n+1 ,
(1.5)
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (1.5) we get:
b
f (x) d x −
a
(n) f
L ∞ ([a,b])
(n + 1)!
b−a N
b−a N
[ j f (a) + (N − j) f (b)] ≤
n+1
j n+1 + (N − j)n+1 ,
(1.6)
for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (1.6) turns to
b
f (x) d x −
a
(n) f
b−a 2
L ∞ ([a,b])
(n + 1)!
( f (a) + f (b)) ≤
(b − a)n+1 , 2n
(1.7)
1.2 Main Results
3
(vii) when n = 1 (without any boundary conditions), we get from (1.7) that
b
f (x) d x −
a
b−a 2
(b − a)2 , ( f (a) + f (b)) ≤ f ∞,[a,b] 4
(1.8)
a similar to Iyengar inequality (1.1). Proof Here n ∈ N and f (n−1) is absolutely continuous on [a, b]. We assumed that (n) f
∞,[a,b]
:= f (n) L ∞ ([a,b]) < +∞.
By [1], we have the following generalized Taylor’s formulae: x n−1 f (k) (a) 1 k f (x) − (x − t)n−1 f (n) (t) dt (x − a) = k! − 1)! (n a k=0
(1.9)
x n−1 f (k) (b) 1 k f (x) − (x − t)n−1 f (n) (t) dt, (x − b) = k! − 1)! (n b k=0
(1.10)
and
∀ x ∈ [a, b] . Then we get (n) n−1 f f (k) (a) ∞,[a,b] k (x − a) ≤ (x − a)n , f (x) − k! n! k=0
(1.11)
∀ x ∈ [a, b] , and x n−1 1 f (k) (b) k n−1 (n) = f − t) dt (x (t) (x − b) = f (x) − (n − 1)! b k! k=0 b b 1 1 n−1 (n) f (t) dt ≤ (t − x) (t − x)n−1 f (n) (t) dt ≤ (n − 1)! x (n − 1)! x (n) f
∞,[a,b]
n! that is
(b − x)n ,
(n) n−1 f f (k) (b) ∞,[a,b] k (x − b) ≤ (b − x)n , f (x) − k! n! k=0
(1.12)
4
1 General Ordinary Iyengar Inequalities
∀ x ∈ [a, b] . We call
(n) f
∞,[a,b]
.
(1.13)
n−1 f (k) (a) (x − a)k ≤ δ (x − a)n k! k=0
(1.14)
n−1 f (k) (b) (x − b)k ≤ δ (b − x)n , k! k=0
(1.15)
δ :=
n!
So we have − δ (x − a)n ≤ f (x) − and − δ (b − x)n ≤ f (x) − ∀ x ∈ [a, b] . Therefore it holds n−1 n−1 f (k) (a) f (k) (a) (x − a)k − δ (x − a)n ≤ f (x) ≤ (x − a)k + δ (x − a)n k! k! k=0 k=0 (1.16) and n−1 n−1 f (k) (b) f (k) (b) (x − b)k − δ (b − x)n ≤ f (x) ≤ (x − b)k + δ (b − x)n , k! k! k=0 k=0 (1.17) ∀ x ∈ [a, b] . Let any t ∈ [a, b], then
t n−1 f (k) (a) δ k+1 n+1 − ≤ f (x) d x ≤ (t − a) (t − a) (k + 1)! (n + 1) a k=0 n−1 f (k) (a) δ (t − a)n+1 , (t − a)k+1 + + 1)! + 1) (k (n k=0
(1.18)
and b n−1 f (k) (b) δ k+1 n+1 − − ≤ f (x) d x ≤ (b − t) (t − b) (k + 1)! (n + 1) t k=0 −
n−1 f (k) (b) δ (b − t)n+1 . (t − b)k+1 + + 1)! + 1) (k (n k=0
(1.19)
1.2 Main Results
5
Adding (1.18) and (1.19), we obtain: n−1 k=0
(k) 1 f (a) (t − a)k+1 − f (k) (b) (t − b)k+1 − (k + 1)!
δ (t − a)n+1 + (b − t)n+1 ≤ (n + 1) n−1 k=0
b
f (x) d x ≤
a
(k) 1 k+1 k+1 (k) f (a) (t − a) + − f (b) (t − b) (k + 1)! δ (t − a)n+1 + (b − t)n+1 , (n + 1)
(1.20)
∀ t ∈ [a, b] . Consequently we derive: n−1 b (k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
δ (t − a)n+1 + (b − t)n+1 , (n + 1)
(1.21)
g (t) := (t − a)n+1 + (b − t)n+1 , ∀ t ∈ [a, b] .
(1.22)
∀ t ∈ [a, b] . Let us consider
Hence
g (t) = (n + 1) (t − a)n − (b − t)n = 0,
giving (t − a)n = (b − t)n and t − a = b − t, that is t = a+b the only critical num2 ber here. (b−a)n+1 = 2n , which is the miniWe have g (a) = g (b) = (b − a)n+1 , and g a+b 2 mum of g over [a, b]. , with value Consequently the right hand side of (1.21) is minimized when t = a+b 2 (n) f ∞,[a,b] (b−a)n+1 (k) (k) . Assuming f (a) = f (b) = 0, for k = 0, 1, . . . , n − 1, then 2n (n+1)! we obtain that b f (n) ∞,[a,b] (b − a)n+1 f (x) d x ≤ , (1.23) 2n (n + 1)! a which is a sharp inequality.
6
1 General Ordinary Iyengar Inequalities
When t =
a+b , 2
then (1.21) becomes
n−1 b 1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
(n) f
∞,[a,b]
(n + 1)!
(b − a)n+1 . 2n
(1.24)
, that is t0 = a, Next let N ∈ N, j = 0, 1, 2, . . . , N and t j = a + j b−a N b−a t1 = a + N , . . . , t N = b. Hence it holds b−a b−a , b − t j = (N − j) , j = 0, 1, 2, . . . , N . tj − a = j N N (1.25) We notice that n+1 n+1 tj − a + b − tj =
b−a N
n+1
j n+1 + (N − j)n+1 ,
(1.26)
j = 0, 1, 2, . . . , N , and (k = 0, 1, . . . , n − 1)
f
(k)
k+1 k+1 f (k) (a) t j − a = + (−1)k f (k) (b) b − t j
(a) j
k+1
b−a N
b−a N
k+1 + (−1) f k
k+1
(k)
(b) (N − j)
k+1
b−a N
k+1 = (1.27)
f (k) (a) j k+1 + (−1)k f (k) (b) (N − j)k+1 ,
j = 0, 1, 2, . . . , N . By (1.21) we get n−1 b b − a k+1 (k) 1 f (a) j k+1 + (−1)k f (k) (b) (N − j)k+1 ≤ f (x) d x − a N (k + 1)! k=0
(n) f
∞,[a,b]
(n + 1)!
b−a N
n+1
j n+1 + (N − j)n+1 ,
j = 0, 1, 2, . . . , N . If f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, then (1.28) becomes
(1.28)
1.2 Main Results
7
b
f (x) d x −
a
(n) f
∞,[a,b]
(n + 1)!
b−a N b−a N
[ j f (a) + (N − j) f (b)] ≤
n+1
j n+1 + (N − j)n+1 ,
(1.29)
for j = 0, 1, 2, . . . , N . When N = 2 and j = 1, then (1.29) becomes
b
f (x) d x −
a
(n) f
∞,[a,b]
(n + 1)!
b−a 2
b−a 2
n+1 2=
[ f (a) + f (b)] ≤
(n) f
∞,[a,b]
(n + 1)!
(b − a)n+1 . 2n
(1.30)
And, if n = 1 (without any boundary conditions), we get from (1.30) that
b
f (x) d x −
a
b−a 2
(b − a)2 , ( f (a) + f (b)) ≤ f ∞,[a,b] 4
which a similar inequality to Iyengar inequality (1.1).
(1.31)
We give Theorem 1.3 Let f ∈ AC n ([a, b]), n ∈ N. Then (i) n−1 b (k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0 (1.32) (n) f L 1 ([a,b]) (t − a)n + (b − t)n , n! ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
the right hand side of (1.32) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
(n) f
L 1 ([a,b])
n!
(b − a)n , 2n−1
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
(1.33)
8
1 General Ordinary Iyengar Inequalities
b
a
f (n) L 1 ([a,b]) (b − a)n f (x) d x ≤ , n! 2n−1
(1.34)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1 b b − a k+1 k+1 (k) 1 k k+1 (k) j f (x) d x − f (a) + (−1) (N − j) f (b) ≤ a N (k + 1)! k=0
(n) f
L 1 ([a,b])
n!
b−a N
n
j n + (N − j)n ,
(1.35)
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (1.35) we get:
b
f (x) d x −
a
(n) f
b−a N
L 1 ([a,b])
n!
[ j f (a) + (N − j) f (b)] ≤
b−a N
n
j n + (N − j)n ,
(1.36)
for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (1.36) turns to
b
f (x) d x −
a
(b − a) ( f (a) + f (b)) ≤ 2
(n) f
L 1 ([a,b])
n!
(b − a)n , 2n−1
(1.37)
(vii) when n = 1 (without any boundary conditions), we get from (1.37) that
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤ f L 1 ([a,b]) (b − a) .
(1.38)
Proof Here n ∈ N and f (n−1) is absolutely continuous on [a, b]. Hence f (n) exists almost everywhere and f (n) ∈ L 1 ([a, b]). By (1.9) we get x n−1 1 f (k) (a) k n−1 (n) f (t) dt ≤ (x − t) (x − a) = f (x) − (n − 1)! a k! k=0 1 (n − 1)!
a
x
(x − a)n−1 (x − t)n−1 f (n) (t) dt ≤ (n − 1)!
a
b
(n) f (t) dt
(1.39)
1.2 Main Results
9
=
(x − a)n−1 f (n) . L 1 ([a,b]) (n − 1)!
That is (n) n−1 f f (k) (a) L 1 ([a,b]) (x − a)k ≤ (x − a)n−1 , f (x) − k! − 1)! (n k=0
(1.40)
∀ x ∈ [a, b] . By (1.10) we get x n−1 1 f (k) (b) k n−1 (n) f (t) dt = (x − t) (x − b) = f (x) − (n − 1)! b k! k=0 b b 1 1 n−1 (n) ≤ f − x) dt (t (t) (t − x)n−1 f (n) (t) dt ≤ (n − 1)! x (n − 1)! x (1.41) (b − x)n−1 b (n) (b − x)n−1 (n) f f (t) dt = . L 1 ([a,b]) (n − 1)! a (n − 1)! That is (n) n−1 f f (k) (b) L 1 ([a,b]) k (x − b) ≤ (b − x)n−1 , f (x) − k! − 1)! (n k=0 ∀ x ∈ [a, b] . Set ρ := Hence
and
(n) f
L 1 ([a,b])
(n − 1)!
(1.42)
.
n−1 f (k) (a) k (x − a) ≤ ρ (x − a)n−1 , f (x) − k! k=0
(1.43)
n−1 f (k) (b) k (x − b) ≤ ρ (b − x)n−1 , f (x) − k! k=0
(1.44)
∀ x ∈ [a, b] . As in the proof of Theorem 1.2 we get:
10
1 General Ordinary Iyengar Inequalities
n−1 b (k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
ρ (t − a)n + (b − t)n , n
(1.45)
∀ t ∈ [a, b] . The rest of the proof is similar to the proof of Theorem 1.2.
We continue with Theorem 1.4 Let f ∈ AC n ([a, b]), n ∈ N; p, q > 1 : L q ([a, b]). Then (i)
1 p
+
1 q
= 1, and f (n) ∈
n−1 b (k) 1 f (a) (t − a)k+1 + (−1)k f (k) (b) (b − t)k+1 ≤ f (x) d x − a (k + 1)! k=0 (1.46) (n) f
1 1 L ([a,b]) q (t − a)n+ p + (b − t)n+ p , 1 1 p (n − 1)! n + p ( p (n − 1) + 1) ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
the right hand side of (1.46) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
(n) 1 f (b − a)n+ p L ([a,b]) q , 1 1 2n− q (n − 1)! n + 1p ( p (n − 1) + 1) p
(1.47)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
a
b
f (x) d x ≤
(n) 1 f (b − a)n+ p L ([a,b]) q , 1 1 2n− q (n − 1)! n + 1p ( p (n − 1) + 1) p
(1.48)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1 b b − a k+1 k+1 (k) 1 k k+1 (k) j f (x) d x − f (a) + (−1) (N − j) f (b) ≤ a N (k + 1)! k=0
1.2 Main Results
11
(n) 1 f 1 b − a n+ p n+ 1p L ([a,b]) q j + (N − j)n+ p , (1.49) 1 N (n − 1)! n + 1p ( p (n − 1) + 1) p (v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (1.49) we get:
b
f (x) d x −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
(n) 1 f b − a n+ p n+ 1p L q ([a,b]) n+ 1p j , (1.50) + − j) (N 1 N (n − 1)! n + 1p ( p (n − 1) + 1) p for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (1.50) turns to
(b − a) f (x) d x − ( f (a) + f (b)) ≤ 2
b
a
(n) 1 f (b − a)n+ p L q ([a,b]) , 1 1 2n− q (n − 1)! n + 1p ( p (n − 1) + 1) p
(1.51)
(vii) when n = 1 (without any boundary conditions), we get from (1.51) that
b
f (x) d x −
a
b−a 2
1 f L q ([a,b]) (b − a)1+ p . ( f (a) + f (b)) ≤ 1 2p 1 + 1p
Proof Here f (n) ∈ L q ([a, b]), where p, q > 1, such that get
1 p
+
1 q
(1.52)
= 1. By (1.9) we
x n−1 1 f (k) (a) k n−1 (n) ≤ f − t) dt (x (t) (x − a) = f (x) − (n − 1)! a k! k=0 1 (n − 1)! 1 (n − 1)!
x
x
(x − t)n−1 f (n) (t) dt ≤
a
(x − t) p(n−1) dt
1p
a
x
(n) q f (t) dt
q1
≤
a
(x − a)
p(n−1)+1 p
(n − 1)! ( p (n − 1) + 1)
1 p
(n) f
L q ([a,b])
.
(1.53)
12
1 General Ordinary Iyengar Inequalities
That is (n) n−1 f (k) f (a) L q ([a,b]) n− q1 , (x − a)k ≤ f (x) − 1 (x − a) (n − 1)! ( p (n − 1) + 1) p k! k=0 (1.54) ∀ x ∈ [a, b] . By (1.10) we get b n−1 1 f (k) (b) k n−1 (n) f (t) dt ≤ (t − x) (x − b) = f (x) − (n − 1)! x k! k=0 1 (n − 1)! 1 (n − 1)!
b
b
(t − x)n−1 f (n) (t) dt ≤
x
(t − x)
1p p(n−1)
b
dt
x
(n) q f (t) dt
q1
≤
x
(b − x)
p(n−1)+1 p
(n − 1)! ( p (n − 1) + 1)
1 p
(n) f
L q ([a,b])
.
(1.55)
That is (n) n−1 f (k) f (b) L q ([a,b]) n− q1 , (1.56) (x − b)k ≤ f (x) − 1 (b − x) (n − 1)! ( p (n − 1) + 1) p k! k=0 ∀ x ∈ [a, b] . Set γ :=
(n) f
L q ([a,b]) 1
(n − 1)! ( p (n − 1) + 1) p
and m := n −
,
1 > 0. q
(1.57)
(1.58)
So, we can write n−1 f (k) (a) k (x − a) ≤ γ (x − a)m , f (x) − k! k=0 and
(1.59)
1.2 Main Results
13
n−1 f (k) (b) k (x − b) ≤ γ (b − x)m , f (x) − k! k=0
(1.60)
∀ x ∈ [a, b] . As in the proof of Theorem 1.2 we obtain: n−1 b (k) 1 f (a) (t − a)k+1 + (−1)k f (k) (b) (b − t)k+1 ≤ f (x) d x − a (k + 1)! k=0
γ (t − a)m+1 + (b − t)m+1 = (m + 1) (n) f
L q ([a,b]) n+ 1p n+ 1p − a) , + − t) (t (b 1 (n − 1)! ( p (n − 1) + 1) p n + 1p ∀ t ∈ [a, b] . The rest of the proof is similar to the proof of Theorem 1.2.
(1.61)
(1.62)
References 1. G.A. Anastassiou, S.S. Dragomir, On some estimates of the remainder in Taylor’s formula. J. Math. Anal. Appl. 263, 246–263 (2001) 2. G.A. Anastassiou, General Iyengar type inequalities. J. Comput. Anal. Appl. 28(5), 786–797 (2020) 3. K.S.K. Iyengar, Note on an inequality. Math. Student 6, 75–76 (1938)
Chapter 2
Caputo Fractional Iyengar Inequalities
Here we present Caputo fractional Iyengar type inequalities with respect to L p norms, with 1 ≤ p ≤ ∞. The method is based on the right and left Caputo fractional Taylor’s formulae. See also [3].
2.1 Introduction We are motivated by the following famous Iyengar inequality (1938), [5]. Theorem 2.1 Let f be a differentiable function on [a, b] and f (x) ≤ M. Then
a
b
M (b − a) 2 ( f (b) − f (a))2 1 . f (x) d x − (b − a) ( f (a) + f (b)) ≤ − 2 4 4M (2.1)
We need Definition 2.2 ([1], p. 394) Let ν > 0, n = ν (· the ceiling of the number), f ∈ AC n ([a, b]) (i.e. f (n−1) is absolutely continuous on [a, b]). The left Caputo fractional derivative of order ν is defined as x 1 ν D∗a f (x) = (2.2) (x − t)n−ν−1 f (n) (t) dt, (n − ν) a ∀ x ∈ [a, b], and it exists almost everywhere over [a, b] .
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_2
15
16
2 Caputo Fractional Iyengar Inequalities
We need Definition 2.3 ([2], pp. 336–337) Let ν > 0, n = ν, f ∈ AC n ([a, b]). The right Caputo fractional derivative of order ν is defined as ν f (x) = Db−
(−1)n (n − ν)
b
(z − x)n−ν−1 f (n) (z) dz,
(2.3)
x
∀ x ∈ [a, b], and exists almost everywhere over [a, b] .
2.2 Main Results We present the following Caputo fractional Iyengar type inequalities: Theorem 2.4 Let ν > 0, n = ν (· is the ceiling of the number), and f ∈ AC n ([a, b]) (i.e. f (n−1) is absolutely continuous on [a, b]). We assume that ν ν D∗a f, Db− f ∈ L ∞ ([a, b]). Then (i) n−1 b (k) 1 f (a) (t − a)k+1 + (−1)k f (k) (b) (b − t)k+1 ≤ f (x) d x − a (k + 1)! k=0
ν ν max D∗a f L ∞ ([a,b]) , Db− f L ∞ ([a,b]) (t − a)ν+1 + (b − t)ν+1 , (ν + 2) ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
(2.4)
the right hand side of (2.4) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f (x) d x − f (a) + (−1) f (b) ≤ a 2k+1 (k + 1)! k=0
ν ν max D∗a f L ∞ ([a,b]) , Db− f L ∞ ([a,b]) (b − a)ν+1 (ν + 2)
2ν
,
(2.5)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
a
b
ν max Dν f D f ν+1 , ∗a b− L ∞ ([a,b]) L ∞ ([a,b]) (b − a) f (x) d x ≤ , ν (ν + 2) 2
(2.6)
2.2 Main Results
17
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds b k+1
n−1
b − a 1 j k+1 f (k) (a) + (−1)k (N − j)k+1 f (k) (b) f (x) d x − N (k + 1)! a k=0
≤
ν ν max D∗a f L ∞ ([a,b]) , Db− f L ∞ ([a,b]) b − a ν+1 (ν + 2)
N
j ν+1 + (N − j)ν+1 ,
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (2.7) we get:
b
f (x) d x −
a
b−a N
(2.7)
[ j f (a) + (N − j) f (b)] ≤
ν ν max D∗a f L ∞ ([a,b]) , Db− f L ∞ ([a,b]) b − a ν+1 (ν + 2)
N
j ν+1 + (N − j)ν+1 , (2.8)
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (2.8) turns to
a
b
f (x) d x −
b−a 2
( f (a) + f (b)) ≤
ν ν max D∗a f L ∞ ([a,b]) , Db− f L ∞ ([a,b]) (b − a)ν+1 (ν + 2)
2ν
,
(2.9)
(vii) when 0 < ν ≤ 1, inequality (2.9) is again valid without any boundary conditions. Proof Let ν > 0, n = ν, and f ∈ AC n ([a, b]). Then by ([4], p. 54) left Caputo fractional Taylor’s formula we have f (x) −
x n−1 f (k) (a) 1 ν f (t) dt, (x − t)ν−1 D∗a (x − a)k = k! (ν) a k=0
(2.10)
∀ x ∈ [a, b] . Also by ([2], p. 341) right Caputo fractional Taylor’s formula we get: f (x) − ∀ x ∈ [a, b] .
b n−1 f (k) (b) 1 ν f (z) dz, (z − x)ν−1 Db− (x − b)k = k! (ν) x k=0
(2.11)
18
2 Caputo Fractional Iyengar Inequalities
By (2.10) we derive ν n−1 D∗a f f (k) (a) L ∞ ([a,b]) (x − a)ν , (x − a)k ≤ f (x) − k! + 1) (ν k=0
(2.12)
and by (2.11) we obtain ν n−1 D f f (k) (b) b− L ∞ ([a,b]) k (b − x)ν , (x − b) ≤ f (x) − k! + 1) (ν k=0 ∀ x ∈ [a, b] . Call γ1 := and γ2 :=
ν D f ∗a L ∞ ([a,b]) (ν + 1) ν D f b− L ∞ ([a,b]) (ν + 1)
(2.13)
,
(2.14)
.
(2.15)
Set
That is
and
γ := max (γ1 , γ2 ) .
(2.16)
n−1 f (k) (a) (x − a)k ≤ γ (x − a)ν , f (x) − k! k=0
(2.17)
n−1 f (k) (b) (x − b)k ≤ γ (b − x)ν , f (x) − k! k=0
(2.18)
∀ x ∈ [a, b] . Hence it holds n−1 n−1 f (k) (a) f (k) (a) k ν (x − a) − γ (x − a) ≤ f (x) ≤ (x − a)k + γ (x − a)ν k! k! k=0 k=0 (2.19) and n−1 n−1 f (k) (b) f (k) (b) k ν (x − b) −γ (b − x) ≤ f (x) ≤ (x − b)k + γ (b − x)ν , k! k! k=0 k=0 (2.20) ∀ x ∈ [a, b] .
2.2 Main Results
19
Let any t ∈ [a, b], then by integration over [a, t] and [t, b], respectively, we obtain t n−1 f (k) (a) γ k+1 ν+1 − ≤ f (x) d x ≤ (t − a) (t − a) (k + 1)! (ν + 1) a k=0 n−1 f (k) (a) γ (t − a)k+1 + (t − a)ν+1 , + 1)! + 1) (k (ν k=0
and
(2.21)
b n−1 f (k) (b) γ k+1 ν+1 − − ≤ f (x) d x ≤ (t − b) (b − t) (k + 1)! (ν + 1) t k=0 −
n−1 f (k) (b) γ (t − b)k+1 + (b − t)ν+1 . + 1)! + 1) (k (ν k=0
(2.22)
Adding (2.21) and (2.22), we obtain n−1 k=0
(k) 1 k+1 k+1 (k) f (a) (t − a) − − f (b) (t − b) (k + 1)!
γ (t − a)ν+1 + (b − t)ν+1 ≤ (ν + 1) n−1 k=0
b
f (x) d x ≤
a
(k) 1 f (a) (t − a)k+1 − f (k) (b) (t − b)k+1 + (k + 1)! γ (t − a)ν+1 + (b − t)ν+1 , (ν + 1)
(2.23)
∀ t ∈ [a, b] . Consequently we derive: n−1 b (k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
γ (t − a)ν+1 + (b − t)ν+1 , (ν + 1) ∀ t ∈ [a, b] .
(2.24)
20
2 Caputo Fractional Iyengar Inequalities
Let us consider g (t) := (t − a)ν+1 + (b − t)ν+1 , ∀ t ∈ [a, b] . Hence
g (t) = (ν + 1) (t − a)ν − (b − t)ν = 0,
giving (t − a)ν = (b − t)ν and t − a = b − t, that is t = a+b the only critical num2 ber here. (b−a)ν+1 = 2ν , which the minimum We have g (a) = g (b) = (b − a)ν+1 , and g a+b 2 of g over [a, b]. , with value Consequently the right hand side of (2.24) is minimized when t = a+b 2 γ (b−a)ν+1 . 2ν (ν+1)
Assuming f (k) (a) = f (k) (b) = 0, for k = 0, 1, . . . , n − 1, then we obtain that
a
b
f (x) d x ≤
γ (b − a)ν+1 , 2ν (ν + 1)
(2.25)
which is a sharp inequality. , then (2.24) becomes When t = a+b 2 n−1 b 1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
γ (b − a)ν+1 . 2ν (ν + 1)
(2.26)
Next let N ∈ N, j = 0, 1, 2, . . . , N and t j = a + j b−a , that is t0 = a, N b−a t1 = a + N , . . . , t N = b. Hence it holds
b−a b−a , b − t j = (N − j) , j = 0, 1, 2, . . . , N . tj − a = j N N (2.27) We notice that ν+1 ν+1 tj − a + b − tj =
b−a N
ν+1
j ν+1 + (N − j)ν+1 ,
j = 0, 1, 2, . . . , N , and (for k = 0, 1, . . . , n − 1)
k+1 k+1 f (k) (a) t j − a = + (−1)k f (k) (b) b − t j
(2.28)
2.2 Main Results
21
(k)
f
(a) j
k+1
b−a N
b−a N
k+1
k+1
(k)
+ (−1) f k
(b) (N − j)
k+1
b−a N
k+1
f (k) (a) j k+1 + (−1)k f (k) (b) (N − j)k+1 ,
=
(2.29)
j = 0, 1, 2, . . . , N . By (2.24) we get b k+1
n−1
b − a 1 f (k) (a) j k+1 + (−1)k f (k) (b) (N − j)k+1 f (x) d x − N (k + 1)! a k=0
γ ≤ (ν + 1)
b−a N
ν+1
j ν+1 + (N − j)ν+1 ,
(2.30)
j = 0, 1, 2, . . . , N . If f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, then (2.30) becomes
b
f (x) d x −
a
γ (ν + 1)
b−a N
b−a N
[ j f (a) + (N − j) f (b)] ≤
ν+1
j ν+1 + (N − j)ν+1 ,
(2.31)
j = 0, 1, 2, . . . , N . When N = 2 and j = 1, then (2.31) becomes
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
γ b − a ν+1 γ (b − a)ν+1 2 = . 2 2ν (ν + 1) (ν + 1)
(2.32)
Let 0 < ν ≤ 1, then n = ν = 1. In that case, without any boundary conditions, we derive from (2.32) again that
a
b
f (x) d x −
b−a 2
γ (b − a)ν+1 . ( f (a) + f (b)) ≤ 2ν (ν + 1)
The theorem is proved in all cases.
(2.33)
22
2 Caputo Fractional Iyengar Inequalities
We give Theorem 2.5 Let ν ≥ 1, n = ν, and f ∈ AC n ([a, b]). We assume that ν ν f, Db− f ∈ L 1 ([a, b]). Then D∗a (i) n−1 b (k) 1 f (a) (t − a)k+1 + (−1)k f (k) (b) (b − t)k+1 ≤ f (x) d x − a (k + 1)! k=0
ν ν max D∗a f L 1 ([a,b]) , Db− f L 1 ([a,b]) (ν + 1)
(t − a)ν + (b − t)ν ,
(2.34)
∀ t ∈ [a, b] , (ii) when ν = 1, from (2.34), we have
b
a
f (x) d x − [ f (a) (t − a) + f (b) (b − t)] ≤ f
L 1 ([a,b])
(b − a) , ∀ t ∈ [a, b] ,
(2.35)
(iii) from (2.35), we obtain (ν = 1 case)
b
f (x) d x −
a
(iv) at t =
a+b , 2
b−a 2
( f (a) + f (b)) ≤ f L 1 ([a,b]) (b − a) ,
(2.36)
ν > 1, the right hand side of (2.34) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
ν ν max D∗a f L 1 ([a,b]) , Db− f L 1 ([a,b]) (b − a)ν (ν + 1)
2ν−1
,
(2.37)
(v) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1; ν > 1, from (2.37), we obtain ν max b Dν f D f ν , ∗a b− L 1 ([a,b]) L 1 ([a,b]) (b − a) f (x) d x ≤ , (2.38) (ν + 1) 2ν−1 a which is a sharp inequality, (vi) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
2.2 Main Results
23
b k+1
n−1
b − a 1 k+1 k k+1 (k) (k) j f (x) d x − f (a) + (−1) (N − j) f (b) N (k + 1)! a k=0
≤
ν ν max D∗a f L 1 ([a,b]) , Db− f L 1 ([a,b]) b − a ν (ν + 1)
N
j ν + (N − j)ν , (2.39)
(vii) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (2.39) we get:
b
f (x) d x −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
ν ν max D∗a f L 1 ([a,b]) , Db− f L 1 ([a,b]) b − a ν (ν + 1)
N
j ν + (N − j)ν ,
(2.40)
j = 0, 1, 2, . . . , N , (viii) when N = 2 and j = 1, (2.40) turns to
a
b
(b − a) f (x) d x − ( f (a) + f (b)) ≤ 2
ν ν max D∗a f L 1 ([a,b]) , Db− f L 1 ([a,b]) (b − a)ν (ν + 1)
2ν−1
.
(2.41)
ν ν Proof Here ν ≥ 1 and D∗a f, Db− f ∈ L 1 ([a, b]). By (2.10) we get
x n−1 ν 1 f (k) (a) k ν−1 D f (t) dt (x − a) (x − a) ≤ f (x) − ∗a k! (ν) a k=0 ≤
(x − a)ν−1 Dν f , ∗a L 1 ([a,b]) (ν)
∀ x ∈ [a, b] . That is ν n−1 D∗a f f (k) (a) L 1 ([a,b]) k (x − a)ν−1 , (x − a) ≤ f (x) − k! (ν) k=0 ∀ x ∈ [a, b] .
(2.42)
(2.43)
24
2 Caputo Fractional Iyengar Inequalities
By (2.11) we get b n−1 ν 1 f (k) (b) k ν−1 D f (z) dz (x − b) ≤ (b − x) f (x) − b− k! (ν) x k=0 ≤
(b − x)ν−1 Dν f , b− L 1 ([a,b]) (ν)
∀ x ∈ [a, b] . That is ν n−1 D f f (k) (b) b− L 1 ([a,b]) k f − − b) (x) (b − x)ν−1 , (x ≤ k! (ν) k=0 ∀ x ∈ [a, b] . Call δ1 := and δ2 :=
ν D f ∗a L 1 ([a,b]) (ν) ν D f b− L 1 ([a,b]) (ν)
(2.44)
(2.45)
,
(2.46)
.
(2.47)
Set
That is
and
δ := max (δ1 , δ2 ) .
(2.48)
n−1 f (k) (a) k (x − a) ≤ δ (x − a)ν−1 , f (x) − k! k=0
(2.49)
n−1 f (k) (b) k (x − b) ≤ δ (b − x)ν−1 , f (x) − k! k=0
(2.50)
∀ x ∈ [a, b] . As in the proof of Theorem 2.4, we get: n−1 b (k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
δ (t − a)ν + (b − t)ν , ν ∀ t ∈ [a, b] .
(2.51)
2.2 Main Results
25
The rest of the proof is similar to the proof of Theorem 2.4. We continue with Theorem 2.6 Let p, q > 1 : 1p + ν ν f, Db− f ∈ L q ([a, b]). Then D∗a (i)
1 q
= 1, ν > q1 , n = ν ; f ∈ AC n ([a, b]), with
n−1 b (k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
ν ν
max D∗a f L q ([a,b]) , Db− f L q ([a,b]) ν+ 1p ν+ 1p , + − t) − a) (b (t 1 (ν) ν + 1p ( p (ν − 1) + 1) p ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
(2.52)
the right hand side of (2.52) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f (x) d x − f f + (−1) (a) (b) ≤ k+1 a 2 (k + 1)! k=0
ν ν max D∗a f L q ([a,b]) , Db− f L q ([a,b]) (b − a)ν+ 1p , 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p
(2.53)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
a
b
ν max ν ν+ 1p D D f , f ∗a b− L q ([a,b]) L q ([a,b]) (b − a) f (x) d x ≤ , 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p
(2.54)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
n−1 b
1 b − a k+1 k+1 (k) k k+1 (k) f (x) d x − f (a) + (−1) (N − j) f (b) j a N (k + 1)! k=0
ν ν
max D∗a f L q ([a,b]) , Db− f L q ([a,b]) b − a ν+ 1p ν+ 1p ν+ 1p j , ≤ + − j) (N 1 N (ν) ν + 1p ( p (ν − 1) + 1) p (2.55)
26
2 Caputo Fractional Iyengar Inequalities
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (2.55) we get:
b
f (x) d x −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
ν ν
max D∗a f L q ([a,b]) , Db− f L q ([a,b]) b − a ν+ 1p ν+ 1p ν+ 1p j , + − j) (N 1 N (ν) ν + 1p ( p (ν − 1) + 1) p (2.56) for j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (2.56) turns to
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
ν ν max D∗a f L q ([a,b]) , Db− f L q ([a,b]) (b − a)ν+ 1p , 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p
(2.57)
(vii) when 1/q < ν ≤ 1, inequality (2.57) is again valid but without any boundary conditions. Proof Here ν > 0, n = ν, f ∈ AC n ([a, b]) ; p, q > 1 : ν ν f, Db− f ∈ L q ([a, b]). By ( 2.10) we have D∗a
1 p
+
1 q
= 1, with
x n−1 ν 1 f (k) (a) k f (t) dt ≤ (x − t)ν−1 D∗a (x − a) ≤ f (x) − (ν) a k! k=0 1 (ν)
x
(x − t)
p(ν−1)
1p dt
a
a
x
ν D f (t)q dt ∗a
q1
≤
p(ν−1)+1
ν 1 (x − a) p D f . 1 ∗a L q ([a,b]) (ν) ( p (ν − 1) + 1) p Here we assume that ν >
1 q
⇔ p (ν − 1) + 1 > 0. So, we get
ν n−1 D f 1 ∗a f (k) (a) L q ([a,b]) k (x − a)ν− q , (x − a) ≤ f (x) − (ν) ( p (ν − 1) + 1) 1p k! k=0 ∀ x ∈ [a, b] . By (2.11) we have
(2.58)
(2.59)
2.2 Main Results
27
b n−1 ν 1 f (k) (b) k f (z) dz ≤ (z − x)ν−1 Db− (x − b) ≤ f (x) − (ν) x k! k=0 1 (ν)
b
(z − x)
1p p(ν−1)
b
dz
x
x
ν D f (z)q dz b−
q1
≤
p(ν−1)+1
ν 1 (b − x) p D f . b− L q ([a,b]) (ν) ( p (ν − 1) + 1) 1p
(2.60)
So, we get ν n−1 D f (k) b− f (b) L q ([a,b]) ν− q1 , (x − b)k ≤ f (x) − 1 (b − x) (ν) ( p (ν − 1) + 1) p k! k=0 ∀ x ∈ [a, b] . Call ρ1 := and ρ2 :=
ν D f ∗a L q ([a,b])
and
,
(2.62)
1
.
(2.63)
1 > 0. q
(2.64)
ν D f b− L q ([a,b]) (ν) ( p (ν − 1) + 1) p
ρ := max (ρ1 , ρ2 ) , m := ν − That is
1
(ν) ( p (ν − 1) + 1) p
Set
(2.61)
n−1 f (k) (a) k (x − a) ≤ ρ (x − a)m , f (x) − k! k=0
(2.65)
n−1 f (k) (b) k (x − b) ≤ ρ (b − x)m , f (x) − k! k=0
(2.66)
∀ x ∈ [a, b] . As in the proof of Theorem 2.4, we obtain: n−1 b (k) 1 f (a) (t − a)k+1 + (−1)k f (k) (b) (b − t)k+1 ≤ f (x) d x − a (k + 1)! k=0
28
2 Caputo Fractional Iyengar Inequalities
ρ (t − a)m+1 + (b − t)m+1 = (m + 1) ν ν
max D∗a f L q ([a,b]) , Db− f L q ([a,b]) 1 1 (t − a)ν+ p + (b − t)ν+ p , 1 (ν) ( p (ν − 1) + 1) p ν + 1p ∀ t ∈ [a, b] . The rest of the proof is similar to the proof of Theorem 2.4.
(2.67)
References 1. G.A. Anastassiou, Fractional Differentiation Inequalities (Springer, Heidelberg, New York, 2009) 2. G.A. Anastassiou, Intelligent Mathematical Computational Analysis (Springer, Heidelberg, New York, 2011) 3. G.A. Anastassiou, Caputo Fractional Iyengar Type Inequalities, submitted for publication (2019) 4. K. Diethelm, The Analysis of Fractional Differential Equations (Springer, Heidelberg, New York, 2010) 5. K.S.K. Iyengar, Note on an inequality. Math. Student 6, 75–76 (1938)
Chapter 3
Canavati Fractional Iyengar Inequalities
Here we present Canavati fractional Iyengar type inequalities with respect to L p norms, with 1 ≤ p ≤ ∞. The method is based on the right and left Canavati fractional Taylor’s formulae. See also [3].
3.1 Background We are motivated by the following famous Iyengar inequality (1938), [5]. Theorem 3.1 Let f be a differentiable function on [a, b] and f (x) ≤ M. Then
a
b
f (x) d x −
M (b − a) 2 ( f (b) − f (a))2 1 . − (b − a) ( f (a) + f (b)) ≤ 2 4 4M (3.1)
We need the following fractional calculus background: Let α > 0, m = [α] ([·] is the integral part), β = α − m, 0 < β < 1, f ∈ C [a, b] ⊂ R, x ∈ [a, b]. The gamma function is given by (α) = ([a, ∞ b]), −t α−1 dt. We define the left Riemann–Liouville integral ([1], p. 24) 0 e t
Jαa+ f (x) =
1 (α)
x
(x − t)α−1 f (t) dt,
(3.2)
a
α a ≤ x ≤ b. We define the subspace Ca+ ([a, b]) of C m ([a, b]) : α Ca+ ([a, b]) =
a+ (m) f ∈ C 1 ([a, b]) . f ∈ C m ([a, b]) : J1−β
(3.3)
α For f ∈ Ca+ ([a, b]), we define the left generalized α-fractional derivative of f over [a, b] as
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_3
29
30
3 Canavati Fractional Iyengar Inequalities
a+ (m) α Da+ f := J1−β f ,
(3.4)
see [1], p. 24. Canavati first in [4] introduced the above over [0, 1] . n f = f (n) ; n ∈ N. We have that Da+ α Notice that Da+ f ∈ C ([a, b]) . We need the following left fractional Taylor’s formula, see [1], pp. 8–10, and in [4] the same over [0, 1] that appeared first. α Theorem 3.2 Let f ∈ Ca+ ([a, b]). (i) If α ≥ 1, then
(x − a)2 (x − a)m−1 + ... f (m−1) (a) 2 (m − 1)! (3.5) x α α−1 Da+ f (t) dt, (x − t)
f (x) = f (a) + f (a) (x − a) + f (a) +
1 (α)
a
all x ∈ [a, b] . (ii) If 0 < α < 1, we have 1 f (x) = (α)
a
x
α f (t) dt, (x − t)α−1 Da+
(3.6)
all x ∈ [a, b] . Furthermore we need: Let again α > 0, m = [α], β = α − m, f ∈ C ([a, b]), call the right Riemann– Liouville fractional integral operator by
α Jb−
1 f (x) := (α)
b
(t − x)α−1 f (t) dt,
(3.7)
x
x ∈ [a, b], see [2]. Define the subspace of functions α Cb− ([a, b]) :=
1−β f ∈ C m ([a, b]) : Jb− f (m) ∈ C 1 ([a, b]) .
(3.8)
Define the right generalized α-fractional derivative of f over [a, b] as
1−β α f = (−1)m−1 Jb− f (m) , Db−
(3.9)
n 0 α see [2]. We set Db− f = f . We have Db− f = (−1)n f (n) ; n ∈ N. Notice that Db− f ∈ C ([a, b]) . From [2], we need the following right Taylor fractional formula. α Theorem 3.3 Let f ∈ Cb− ([a, b]) , α > 0, m := [α]. Then
3.1 Background
31
(i) If α ≥ 1, we get f (x) =
m−1 k=0
α α f (k) (b) Db− f (x) , (x − b)k + Jb− k!
(3.10)
all x ∈ [a, b] . (ii) If 0 < α < 1, we get α α f (x) = Jb− Db− f (x) =
1 (α)
b x
α f (t) dt, (t − x)α−1 Db−
(3.11)
all x ∈ [a, b] .
3.2 Main Results We present the following Canavati fractional Iyengar type inequalities: ν ν Theorem 3.4 Let ν ≥ 1, n = [ν] and f ∈ Ca+ ([a, b]) ∩ Cb− ([a, b]). Then (i)
n−1 b (k)
1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
ν ν max Da+ f ∞,([a,b]) , Db− f ∞,([a,b])
(t − a)ν+1 + (b − t)ν+1 , (ν + 2) ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
(3.12)
the right hand side of (3.12) is minimized, and we get:
n−1 b
1 (b − a)k+1 (k) k (k) f f (x) d x − f + (−1) (a) (b) ≤ k+1 a 2 (k + 1)! k=0
ν ν max Da+ f ∞,([a,b]) , Db− f ∞,([a,b]) (b − a)ν+1 (ν + 2)
2ν
,
(3.13)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
a
b
ν max ν ν+1 D D f , f a+ b− ∞,([a,b]) ∞,([a,b]) (b − a) f (x) d x ≤ , (3.14) (ν + 2) 2ν
32
3 Canavati Fractional Iyengar Inequalities
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1 b b − a k+1 k+1 (k) 1 k k+1 (k) j f (x) d x − f (a) + (−1) (N − j) f (b) a N (k + 1)! k=0
ν ν f , Db− f ∞,([a,b]) b − a ν+1 max Da+ ∞,([a,b]) j ν+1 + (N − j)ν+1 , ≤ (ν + 2) N
(3.15)
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (3.15) we get:
b
f (x) d x −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
ν ν max Da+ f ∞,([a,b]) , Db− f ∞,([a,b]) b − a ν+1 (ν + 2)
N
j ν+1 + (N − j)ν+1 , (3.16)
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (3.16) turns to
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
ν ν max Da+ f ∞,([a,b]) , Db− f ∞,([a,b]) (b − a)ν+1 (ν + 2)
2ν
.
(3.17)
ν ν Proof Let ν ≥ 1, n = [ν], and f ∈ Ca+ ([a, b]) ∩ Cb− ([a, b]). Then by (3.5) we have
f (x) −
x n−1 f (k) (a) 1 ν f (t) dt, (x − t)ν−1 Da+ (x − a)k = k! (ν) a k=0
(3.18)
∀ x ∈ [a, b] . Also by (3.10) we have b n−1 f (k) (b) 1 k ν f (x) − f (z) dz, (z − x)ν−1 Db− (x − b) = k! (ν) x k=0 ∀ x ∈ [a, b] .
(3.19)
3.2 Main Results
33
By (3.18) we derive ν n−1 Da+ f f (k) (a) ∞,([a,b]) (x − a)ν , (x − a)k ≤ f (x) − k! + 1) (ν k=0
(3.20)
and by (3.19) we obtain ν n−1 D f f (k) (b) b− ∞,([a,b]) k (b − x)ν , (x − b) ≤ f (x) − k! + 1) (ν k=0 ∀ x ∈ [a, b] . Call γ1 := and γ2 :=
ν D f a+ ∞,([a,b]) (ν + 1) ν D f b− ∞,([a,b]) (ν + 1)
(3.21)
,
(3.22)
.
(3.23)
Set
That is
and
γ := max (γ1 , γ2 ) .
(3.24)
n−1 f (k) (a) (x − a)k ≤ γ (x − a)ν , f (x) − k! k=0
(3.25)
n−1 f (k) (b) (x − b)k ≤ γ (b − x)ν , f (x) − k! k=0
(3.26)
∀ x ∈ [a, b] . Hence it holds n−1 n−1 f (k) (a) f (k) (a) k ν (x − a) − γ (x − a) ≤ f (x) ≤ (x − a)k + γ (x − a)ν k! k! k=0 k=0 (3.27) and n−1 n−1 f (k) (b) f (k) (b) k ν (x − b) − γ (b − x) ≤ f (x) ≤ (x − b)k + γ (b − x)ν , k! k! k=0 k=0 (3.28) ∀ x ∈ [a, b] .
34
3 Canavati Fractional Iyengar Inequalities
Let any t ∈ [a, b], then by integration over [a, t] and [t, b], respectively, we obtain t n−1 f (k) (a) γ k+1 ν+1 − ≤ f (x) d x ≤ (t − a) (t − a) (k + 1)! (ν + 1) a k=0 n−1 f (k) (a) γ (t − a)k+1 + (t − a)ν+1 , + 1)! + 1) (k (ν k=0
and
(3.29)
b n−1 f (k) (b) γ k+1 ν+1 − − ≤ f (x) d x ≤ (t − b) (b − t) (k + 1)! (ν + 1) t k=0 −
n−1 f (k) (b) γ (t − b)k+1 + (b − t)ν+1 . + 1)! + 1) (k (ν k=0
(3.30)
Adding (3.29) and (3.30), we obtain n−1 k=0
(k)
1 k+1 k+1 (k) f (a) (t − a) − − f (b) (t − b) (k + 1)!
γ (t − a)ν+1 + (b − t)ν+1 ≤ (ν + 1) n−1 k=0
b
f (x) d x ≤
a
(k)
1 f (a) (t − a)k+1 − f (k) (b) (t − b)k+1 + (k + 1)!
γ (t − a)ν+1 + (b − t)ν+1 , (ν + 1)
(3.31)
∀ t ∈ [a, b] . Consequently we derive: n−1 b (k)
1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
γ (t − a)ν+1 + (b − t)ν+1 , (ν + 1) ∀ t ∈ [a, b] . Let us consider
(3.32)
3.2 Main Results
35
g (t) := (t − a)ν+1 + (b − t)ν+1 , ∀t ∈ [a, b] . Hence
g (t) = (ν + 1) (t − a)ν − (b − t)ν = 0,
giving (t − a)ν = (b − t)ν and t − a = b − t, that is t = a+b the only critical num2 ber here. (b−a)ν+1 = 2ν , which the minimum We have g (a) = g (b) = (b − a)ν+1 , and g a+b 2 of g over [a, b]. , with value Consequently the right hand side of (3.32) is minimized when t = a+b 2 γ (b−a)ν+1 . 2ν (ν+1)
Assuming f (k) (a) = f (k) (b) = 0, for k = 0, 1, . . . , n − 1, then we obtain that
a
b
f (x) d x ≤
γ (b − a)ν+1 , 2ν (ν + 1)
(3.33)
which is a sharp inequality. , then (3.32) becomes When t = a+b 2 n−1 b
1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
γ (b − a)ν+1 . 2ν (ν + 1)
(3.34)
Next let N ∈ N, j = 0, 1, 2, . . . , N and t j = a + j b−a , that is t0 = a, N b−a t1 = a + N , . . . , t N = b. Hence it holds b−a b−a , b − t j = (N − j) , j = 0, 1, 2, . . . , N . tj − a = j N N (3.35) We notice that ν+1 ν+1 tj − a + b − tj =
b−a N
ν+1
j ν+1 + (N − j)ν+1 ,
j = 0, 1, 2, . . . , N , and (for k = 0, 1, . . . , n − 1)
k+1 k+1 f (k) (a) t j − a = + (−1)k f (k) (b) b − t j
(3.36)
36
3 Canavati Fractional Iyengar Inequalities
f
(k)
(a) j
k+1
b−a N
b−a N
k+1
k+1
(k)
+ (−1) f k
(b) (N − j)
k+1
b−a N
k+1
f (k) (a) j k+1 + (−1)k f (k) (b) (N − j)k+1 ,
=
(3.37)
j = 0, 1, 2, . . . , N . By (3.32) we get n−1 b b − a k+1 (k) 1 k+1 k (k) k+1 f (a) j f (x) d x − + (−1) f (b) (N − j) a N (k + 1)! k=0
γ (ν + 1)
≤
b−a N
ν+1
j ν+1 + (N − j)ν+1 ,
(3.38)
j = 0, 1, 2, . . . , N . If f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, then (3.38) becomes
b
f (x) d x −
a
γ (ν + 1)
b−a N
b−a N
[ j f (a) + (N − j) f (b)] ≤
ν+1
j ν+1 + (N − j)ν+1 ,
(3.39)
j = 0, 1, 2, . . . , N . When N = 2 and j = 1, then (3.39) becomes
b
a
f (x) d x −
b−a 2
( f (a) + f (b)) ≤
γ b − a ν+1 γ (b − a)ν+1 2 = . 2 2ν (ν + 1) (ν + 1)
(3.40)
The theorem is proved.
We give ν ν Theorem 3.5 Let ν ≥ 1, n = [ν], and f ∈ Ca+ ([a, b]) ∩ Cb− ([a, b]). Then (i)
n−1 b (k)
1 f (a) (t − a)k+1 + (−1)k f (k) (b) (b − t)k+1 ≤ f (x) d x − a (k + 1)! k=0
3.2 Main Results
37
ν ν max Da+ f L 1 ([a,b]) , Db− f L 1 ([a,b])
(t − a)ν + (b − t)ν , (ν + 1)
(3.41)
∀ t ∈ [a, b] , (ii) when ν = 1, from (3.41), we have
b
a
f (x) d x − [ f (a) (t − a) + f (b) (b − t)] ≤ f
L 1 ([a,b])
(b − a) , ∀ t ∈ [a, b] ,
(3.42)
(iii) from (3.42), we obtain (ν = 1 case)
b
f (x) d x −
a
(iv) at t =
a+b , 2
b−a 2
( f (a) + f (b)) ≤ f L 1 ([a,b]) (b − a) ,
(3.43)
ν > 1, the right hand side of (3.41) is minimized, and we get:
n−1 b
1 (b − a)k+1 (k) k (k) f f (x) d x − f + (−1) (a) (b) ≤ k+1 a 2 (k + 1)! k=0
ν ν max Da+ f L 1 ([a,b]) , Db− f L 1 ([a,b]) (b − a)ν 2ν−1
(ν + 1)
,
(3.44)
(v) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1; ν > 1, from (3.44), we obtain ν max b Dν f , Db− f L 1 ([a,b]) (b − a)ν a+ L 1 ([a,b]) f (x) d x ≤ , (3.45) (ν + 1) 2ν−1 a
which is a sharp inequality, (vi) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1 b 1 b − a k+1 k+1 (k) k k+1 (k) j f (x) d x − f (a) + (−1) (N − j) f (b) a N (k + 1)! k=0
≤
ν ν max Da+ f L 1 ([a,b]) , Db− f L 1 ([a,b]) b − a ν (ν + 1)
N
j ν + (N − j)ν , (3.46)
(vii) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (3.46) we get:
38
3 Canavati Fractional Iyengar Inequalities
b
f (x) d x −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
ν ν max Da+ f L 1 ([a,b]) , Db− f L 1 ([a,b]) b − a ν (ν + 1)
N
j ν + (N − j)ν ,
(3.47)
j = 0, 1, 2, . . . , N , (viii) when N = 2 and j = 1, (3.47) turns to
a
b
(b − a) f (x) d x − ( f (a) + f (b)) ≤ 2
ν ν max Da+ f L 1 ([a,b]) , Db− f L 1 ([a,b]) (b − a)ν (ν + 1)
2ν−1
,
(3.48)
ν ν Proof Here ν ≥ 1 and Da+ f, Db− f ∈ L 1 ([a, b]). By (3.18) we get
x n−1 ν 1 f (k) (a) k ν−1 D f (t) dt (x − a) (x − a) ≤ f (x) − a+ (ν) k! a k=0 ≤
(x − a)ν−1 Dν f , a+ L 1 ([a,b]) (ν)
∀ x ∈ [a, b] . That is ν n−1 Da+ f f (k) (a) L 1 ([a,b]) k (x − a)ν−1 , (x − a) ≤ f (x) − k! (ν) k=0
(3.49)
(3.50)
∀ x ∈ [a, b] . By (3.19) we get b n−1 ν 1 f (k) (b) k ν−1 D f (z) dz ≤ f − − x) − b) (x) (b (x b− (ν) k! x k=0 ≤ ∀ x ∈ [a, b] . That is
(b − x)ν−1 Dν f , b− L 1 ([a,b]) (ν)
(3.51)
3.2 Main Results
39
ν n−1 D f f (k) (b) b− L 1 ([a,b]) k (b − x)ν−1 , (x − b) ≤ f (x) − k! (ν) k=0 ∀ x ∈ [a, b] . Call δ1 := and δ2 :=
ν D f a+ L 1 ([a,b]) (ν) ν D f b− L 1 ([a,b]) (ν)
(3.52)
,
(3.53)
.
(3.54)
Set
That is
and
δ := max (δ1 , δ2 ) .
(3.55)
n−1 f (k) (a) (x − a)k ≤ δ (x − a)ν−1 , f (x) − k! k=0
(3.56)
n−1 f (k) (b) (x − b)k ≤ δ (b − x)ν−1 , f (x) − k! k=0
(3.57)
∀ x ∈ [a, b] . As in the proof of Theorem 3.4, we get: n−1 b (k)
1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
δ (t − a)ν + (b − t)ν , ν ∀ t ∈ [a, b] . The rest of the proof is similar to the proof of Theorem 3.4.
(3.58)
We continue with Theorem 3.6 Let p, q > 1 : ν Cb− ([a, b]). Then (i)
1 p
+
1 q
ν = 1, ν ≥ 1, n = [ν] ; f ∈ Ca+ ([a, b]) ∩
n−1 b (k)
1 f (a) (t − a)k+1 + (−1)k f (k) (b) (b − t)k+1 ≤ f (x) d x − a (k + 1)! k=0
40
3 Canavati Fractional Iyengar Inequalities
ν ν max Da+ f L q ([a,b]) , Db− f L q ([a,b]) ν+ 1p ν+ 1p
− a) , + − t) (t (b 1 (ν) ν + 1p ( p (ν − 1) + 1) p ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
(3.59)
the right hand side of (3.59) is minimized, and we get:
n−1 b
1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
ν ν max Da+ f L q ([a,b]) , Db− f L q ([a,b]) (b − a)ν+ 1p
, 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p
(3.60)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
a
b
ν max Dν f , Db− f L q ([a,b]) (b − a)ν+ 1p a+ L q ([a,b])
f (x) d x ≤ , (3.61) 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1 b b − a k+1 k+1 (k) 1 k k+1 (k) j + (−1) − j) f d x − f f (a) (N (b) (x) a N (k + 1)! k=0
ν ν max Da+ f L q ([a,b]) , Db− f L q ([a,b]) b − a ν+ 1p ν+ 1p ν+ 1p
j , ≤ + − j) (N 1 N (ν) ν + 1p ( p (ν − 1) + 1) p (3.62) (v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (3.62) we get:
a
b
f (x) d x −
b−a N
[ j f (a) + (N − j) f (b)] ≤
ν ν max Da+ f L q ([a,b]) , Db− f L q ([a,b]) b − a ν+ 1p ν+ 1p ν+ 1p
j , + − j) (N 1 N (ν) ν + 1p ( p (ν − 1) + 1) p (3.63) for j = 0, 1, 2, . . . , N ,
3.2 Main Results
41
(vi) when N = 2 and j = 1, (3.63) turns to
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
ν ν max Da+ f L q ([a,b]) , Db− f L q ([a,b]) (b − a)ν+ 1p
. 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p
(3.64)
ν ν Proof Here let p, q > 1 : 1p + q1 = 1, where Da+ f, Db− f ∈ L q ([a, b]). By (3.18) we have x n−1 ν 1 f (k) (a) k ≤ f − f (t) dt ≤ − a) (x) (x − t)ν−1 Da+ (x (ν) a k! k=0
1 (ν)
x
(x − t) p(ν−1) dt
1p
a
x
a
ν D f (t)q dt a+
q1
≤
p(ν−1)+1
ν 1 (x − a) p D f . 1 a+ L q ([a,b]) (ν) ( p (ν − 1) + 1) p
(3.65)
Since ν ≥ 1, we get p (ν − 1) + 1 > 0. Thus ν n−1 D f a+ f (k) (a) L q ([a,b]) ν− q1 , (x − a)k ≤ f (x) − 1 (x − a) k! (ν) ( p (ν − 1) + 1) p k=0
(3.66)
∀ x ∈ [a, b] . By (3.19) we have b n−1 ν 1 f (k) (b) k f (z) dz ≤ (z − x)ν−1 Db− (x − b) ≤ f (x) − (ν) x k! k=0 1 (ν)
b
(z − x)
1p p(ν−1)
b
dz
x
x
ν D f (z)q dz b−
q1
≤
p(ν−1)+1
ν 1 (b − x) p D f . 1 b− L q ([a,b]) (ν) ( p (ν − 1) + 1) p So, we get
(3.67)
42
3 Canavati Fractional Iyengar Inequalities
ν n−1 D f b− f (k) (b) L q ([a,b]) ν− q1 k , (x − b) ≤ f (x) − 1 (b − x) (ν) ( p (ν − 1) + 1) p k! k=0 ∀ x ∈ [a, b] . Call ρ1 := and ρ2 :=
ν D f a+ L q ([a,b])
and
,
(3.69)
1
.
(3.70)
1 > 0. q
(3.71)
ν D f b− L q ([a,b]) (ν) ( p (ν − 1) + 1) p
ρ := max (ρ1 , ρ2 ) , m := ν − That is
1
(ν) ( p (ν − 1) + 1) p
Set
(3.68)
n−1 f (k) (a) (x − a)k ≤ ρ (x − a)m , f (x) − k! k=0
(3.72)
n−1 f (k) (b) (x − b)k ≤ ρ (b − x)m , f (x) − k! k=0
(3.73)
∀ x ∈ [a, b] . As in the proof of Theorem 3.4, we obtain: n−1 b (k)
1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
ρ (t − a)m+1 + (b − t)m+1 = (m + 1) ν ν f L q ([a,b]) , Db− f L q ([a,b]) max Da+ 1 1
(t − a)ν+ p + (b − t)ν+ p , 1 (ν) ( p (ν − 1) + 1) p ν + 1p ∀ t ∈ [a, b] . The rest of the proof is similar to the proof of Theorem 3.4.
(3.74)
References
43
References 1. G.A. Anastassiou, Fractional Differentiation Inequalities. Research Monograph (Springer, New York, 2009) 2. G.A. Anastassiou, On right fractional calculus. Chaos Solitons Fractals 42, 365–376 (2009) 3. G.A. Anastassiou, Canavati fractional Iyengar type inequalities. An. Univ. Oradea Fasc. Mat. XXVI(1), 141–151 (2019) 4. J.A. Canavati, The Riemann–Liouville integral. Nieuw Arch. Voor Wiskd. 5(1), 53–75 (1987) 5. K.S.K. Iyengar, Note on an inequality. Math. Student 6, 75–76 (1938)
Chapter 4
General Multivariate Iyengar Inequalities
Here we give a variety of general multivariate Iyengar type inequalities for not necessarily radial functions defined on the shell and ball. Our approach is based on the polar coordinates in R N , N ≥ 2, and the related multivariate polar integration formula. Via this method we transfer well-known univariate Iyengar type inequalities and univariate author’s related results into general multivariate Iyengar inequalities. See also [4].
4.1 Background In the year 1938, Iyengar [6] proved the following interesting inequality. Theorem 4.1 Let f be a differentiable function on [a, b] and f (x) ≤ M1 . Then
b
a
M1 (b − a)2 1 ( f (b) − f (a))2 − f (x) d x − (b − a) ( f (a) + f (b)) ≤ . 2 4 4M1 (4.1)
In 2001, X.-L. Cheng [5] proved that Theorem 4.2 Let f ∈ C 2 ([a, b]) and f (x) ≤ M2 . Then
a
b
1 1 2 f (x) d x − (b − a) ( f (a) + f (b)) + (b − a) f (b) − f (a) ≤ 2 8 M2 (b − a) 2 , (b − a)3 − 24 16M2 1
where 1 = f (a) −
2 ( f (b) − f (a)) + f (b) . (b − a)
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_4
(4.2)
45
46
4 General Multivariate Iyengar Inequalities
In 1996, Agarwal and Dragomir [1] obtained a generalization of (4.1): Theorem 4.3 Let f : [a, b] → R be a differentiable function such that for all x ∈ [a, b] with M > m we have m ≤ f (x) ≤ M. Then
b
f (x) d x −
a
1 − a) f + f (b ( (a) (b)) ≤ 2
(4.3)
( f (b) − f (a) − m (b − a)) (M (b − a) − f (b) + f (a)) . 2 (M − m) In [8], Qi proved Theorem 4.4 Let f : [a, b] → R be a twice differentiable function such that for all x ∈ [a, b] with M > 0 we have f (x) ≤ M. Then b 2 1 + Q f + f ( (a) (b)) 2 ≤ f f d x − − f − a) − a) + (x) (b) (a) (b (b 2 8 a
M (b − a)3 1 − 3Q 2 , 24 where
Q2 =
f (a) + f (b) − 2
f (b)− f (a) b−a
(4.4) 2
M 2 (b − a)2 − ( f (b) − f (a))2
.
(4.5)
In 2005, Zheng Liu, [7], proved the following: Theorem 4.5 Let f : [a, b] → R be a differentiable function such that f is integrable on [a, b] and for all x ∈ [a, b] with M > m we have m≤
f (b) − f (x) f (x) − f (a) ≤ M and m ≤ ≤ M. x −a b−x
(4.6)
Then b 1 + P2 ( f (a) + f (b)) f (b) − f (a) (b − a)2 f d x − − a) + (x) (b 2 8 a
1 + 3P 2 − 48 where
(M − m) (b − a)3 1 − 3P 2 , (m + M) (b − a) ≤ 48 3
(4.7)
2 f (a) f (a) + f (b) − 2 f (b)− b−a P2 = 2 . m+M 2 M−m 2 − a) − f − a) − f − (b (b (b) (a) 2 2
(4.8)
4.1 Background
47
Next we list some author’s related results: Theorem 4.6 ([3]) Let n ∈ N, f ∈ AC n ([a, b]) (i.e. f (n−1) ∈ AC ([a, b]), absolutely continuous functions). We assume that f (n) ∈ L ∞ ([a, b]). Then (i) n−1 b
(k) 1 f (a) (t − a)k+1 + (−1)k f (k) (b) (b − t)k+1 ≤ f (x) d x − a (k + 1)! k=0
(n)
f
(t − a)n+1 + (b − t)n+1 ,
L ∞ ([a,b])
(n + 1)! ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
(4.9)
the right hand side of (4.9) is minimized, and we get:
n−1 b
1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
(n)
f
L ∞ ([a,b])
(n + 1)!
(b − a)n+1 , 2n
(4.10)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
b
a
f (n) L ∞ ([a,b]) (b − a)n+1 f (x) d x ≤ , 2n (n + 1)!
(4.11)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1 b
b − a k+1 k+1 (k) 1 k k+1 (k) j f (x) d x − f (a) + (−1) (N − j) f (b) a N (k + 1)! k=0
≤
(n)
f
L ∞ ([a,b])
(n + 1)!
b−a N
n+1
j n+1 + (N − j)n+1 ,
(4.12)
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (4.12) we get:
b
f (x) d x −
a
(n)
f
L ∞ ([a,b])
(n + 1)!
b−a N
b−a N
[ j f (a) + (N − j) f (b)] ≤
n+1
j n+1 + (N − j)n+1 ,
(4.13)
48
4 General Multivariate Iyengar Inequalities
for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (4.13) turns to
b
f (x) d x −
a
(n)
f
b−a 2
L ∞ ([a,b])
(n + 1)!
( f (a) + f (b)) ≤
(b − a)n+1 , 2n
(4.14)
(vii) when n = 1 (without any boundary conditions), we get from (4.14) that
b
f (x) d x −
a
b−a 2
(b − a)2 , ( f (a) + f (b)) ≤ f ∞,[a,b] 4
(4.15)
a similar to Iyengar inequality (4.1). We mention Theorem 4.7 ([3]) Let f ∈ AC n ([a, b]), n ∈ N. Then (i) n−1 b
(k) 1 f (a) (t − a)k+1 + (−1)k f (k) (b) (b − t)k+1 ≤ f (x) d x − a + 1)! (k k=0 (4.16)
(n)
f L 1 ([a,b]) (t − a)n + (b − t)n , n! ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
the right hand side of (4.16) is minimized, and we get:
n−1 b
1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
(n)
f
L 1 ([a,b])
n!
(b − a)n , 2n−1
(4.17)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
a
b
f (n) L 1 ([a,b]) (b − a)n f (x) d x ≤ , n! 2n−1
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
(4.18)
4.1 Background
49
b k+1 n−1
b − a 1 k+1 k k+1 (k) (k) j f (x) d x − f (a) + (−1) (N − j) f (b) N (k + 1)! a k=0
≤
(n)
f
L 1 ([a,b])
n!
b−a N
n
j n + (N − j)n ,
(4.19)
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (4.19) we get:
b
f (x) d x −
a
(n)
f
b−a N
L 1 ([a,b])
n!
[ j f (a) + (N − j) f (b)] ≤
b−a N
n
j n + (N − j)n ,
(4.20)
for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (4.20) turns to
b
a
(b − a) f (x) d x − ( f (a) + f (b)) ≤ 2
(n)
f
L 1 ([a,b])
n!
(b − a)n , 2n−1
(4.21)
(vii) when n = 1 (without any boundary conditions), we get from (4.21) that
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤ f L 1 ([a,b]) (b − a) .
(4.22)
We mention Theorem 4.8 ([3]) Let f ∈ AC n ([a, b]), n ∈ N; p, q > 1 : L q ([a, b]). Then (i)
1 p
+
1 q
= 1, and f (n) ∈
n−1 b
(k) 1 f (a) (t − a)k+1 + (−1)k f (k) (b) (b − t)k+1 ≤ f (x) d x − a (k + 1)! k=0 (4.23)
(n)
f 1 1 L ([a,b]) q (t − a)n+ p + (b − t)n+ p , 1 1 p (n − 1)! n + p ( p (n − 1) + 1) ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
the right hand side of (4.23) is minimized, and we get:
50
4 General Multivariate Iyengar Inequalities
n−1 b
1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
(n) 1
f (b − a)n+ p L ([a,b]) q , 1 1 2n− q (n − 1)! n + 1p ( p (n − 1) + 1) p
(4.24)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
a
b
f (x) d x ≤
(n) 1
f (b − a)n+ p L ([a,b]) q , 1 1 2n− q (n − 1)! n + 1p ( p (n − 1) + 1) p
(4.25)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1 b
b − a k+1 k+1 (k) 1 k k+1 (k) j f (x) d x − f (a) + (−1) (N − j) f (b) a N (k + 1)! k=0
(n) 1
f b − a n+ p n+ 1p L q ([a,b]) n+ 1 ≤ + (N − j) p , j 1 N (n − 1)! n + 1p ( p (n − 1) + 1) p
(4.26)
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (4.26) we get:
b
f (x) d x −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
(n) 1
f b − a n+ p n+ 1p L q ([a,b]) n+ 1p j , (4.27) + − j) (N 1 N (n − 1)! n + 1p ( p (n − 1) + 1) p for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (4.27) turns to
a
b
(b − a) f (x) d x − ( f (a) + f (b)) ≤ 2
(n) 1
f (b − a)n+ p L q ([a,b]) , 1 1 2n− q (n − 1)! n + 1p ( p (n − 1) + 1) p
(4.28)
(vii) when n = 1 (without any boundary conditions), we get from (4.28) that
4.1 Background
b
51
f (x) d x −
a
b−a 2
1 f L q ([a,b]) (b − a)1+ p . ( f (a) + f (b)) ≤ 1 2p 1 + 1p
(4.29)
We need
Remark 4.9 We define the ball B (0, R) = x ∈ R N : |x| < R ⊆ R N , N ≥ 2, R > 0, and the sphere S N −1 := x ∈ R N : |x| = 1 , where |·| is the Euclidean norm. Let dω be the element of surface measure on S N −1 and N 2π 2 dω = N ωN = 2 S N −1 is the area of S N −1 . For x ∈ R N − {0} we can write uniquely x = r ω, where r = |x| > 0 and ω = N x ∈ S N −1 , |ω| = 1. Note that B(0,R) dy = ω NNR is the Lebesgue measure on the r N
R ball, that is the volume of B (0, R), which exactly is V ol (B (0, R)) = π 2N +1 . (2 ) Following [9, pp. 149–150, exercise 6], and [10, pp. 87–88, Theorem 5.2.2] we can write for F : B (0, R) → R a Lebesgue integrable function that
F (x) d x = B(0,R)
S N −1
R
F (r ω) r
N −1
N
dr dω,
(4.30)
0
and we use this formula a lot. Typically here the function f : B (0, R) → R is not radial. A radial function f is such that there exists a function g with f (x) = g (r ), where r = |x|, r ∈ [0, R], ∀ x ∈ B (0, R). We need Remark 4.10 Let the spherical shell A := B (0, R2 ) − B (0, R1 ), 0 < R1 < R2 , A ⊆ R N , N ≥ 2, x ∈ A. Consider that f : A → R is not radial. A radial function f is such that there exists a function g with f (x) = g (r ), r = |x|, r ∈ [R1 , R2 ], ∀ x ∈ A. Here x can be written uniquely as x = r ω, where r = |x| > 0 and ω = rx ∈ S N −1 , |ω| = 1, see ([9], p. 149–150 and [2], p. 421), furthermore for F : A → R a Lebesgue integrable function we have that F (x) d x =
R2
F (r ω) r
N −1
dr dω.
(4.31)
N π 2 R2N − R1N ω N R2N − R1N . = V ol (A) = N N2 + 1
(4.32)
A
Here
S N −1
R1
52
4 General Multivariate Iyengar Inequalities
In this chapter we derive general multivariate Iyengar type inequalities on the shell and ball of R N , N ≥ 2, for not necessarily radial functions. Our results are based on Theorems 4.1–4.8.
4.2 Main Results We present the following non-radial multivariate Iyengar type inequalities: We start with Theorem 4.11 Let the spherical shell A := B (0, R2 ) − B (0, R1 ), 0 < R1 < R2 , A ⊆ R N , N ≥ 2. Consider f : A → R that is not necessarily radial, and that f ∈ ∂ f (sω) 1 C A . Assume that ∂s ≤ M1 , ∀ s ∈ [R1 , R2 ], and ∀ ω ∈ S N −1 , where M1 > 0. Then N −1 f (y) dy − (R2 − R1 ) R N −1 f ω) dω + R f ω) dω (R (R 1 2 1 2 2 A S N −1 S N −1 N
M1 π 2 (R2 − R1 )2 ≤ − 2 N2
S N −1
f (R2 ω) R2N −1 − f (R1 ω) R1N −1 4M1
2
dω
.
(4.33)
Proof Here f (sω) s N −1 ∈ C 1 ([R1 , R2 ]), N ≥ 2, ∀ ω ∈ S N −1 . By (4.1) we get
R2
f (sω) s N −1 ds −
R1
1 (R2 − R1 ) f (R1 ω) R1N −1 + f (R2 ω) R2N −1 ≤ 2
2 f (R2 ω) R2N −1 − f (R1 ω) R1N −1 M1 (R2 − R1 )2 − =: λ1 (ω) , 4 4M1
(4.34)
∀ ω ∈ S N −1 . Equivalently, we have −λ1 (ω) ≤
R2
f (sω) s N −1 ds−
R1
1 (R2 − R1 ) f (R1 ω) R1N −1 + f (R2 ω) R2N −1 ≤ λ1 (ω) , 2 ∀ ω ∈ S N −1 . Hence it holds −
S N −1
λ1 (ω) dω ≤
S N −1
R2 R1
f (sω) s N −1 ds dω−
(4.35)
4.2 Main Results
53
1 f (R1 ω) dω + R2N −1 f (R2 ω) dω (R2 − R1 ) R1N −1 2 S N −1 S N −1 ≤
S N −1
λ1 (ω) dω.
(4.36)
That is (by (4.31))
N
π 2 M1 (R2 − R1 )2 − − 2 N2 ≤ A
(R2 − R1 ) f (y) dy − 2 N
π 2 M1 (R2 − R1 )2 − ≤ 2 N2
f (R2 ω) R2N −1 − f (R1 ω) R1N −1 4M1
S N −1
R1N −1
S N −1
S N −1
f (R1 ω) dω +
R2N −1
2
dω
S N −1
f (R2 ω) R2N −1 − f (R1 ω) R1N −1 4M1
2
f (R2 ω) dω dω
,
(4.37)
proving the claim. We continue with
Theorem 4.12 Let the spherical shell A := B (0, R2 ) − B (0, R1 ), 0 < R1 < R2 , f : A → R that is not necessarily radial, and that f ∈ A ⊆ R N , N ≥ 2. Consider 2 ∂ f (sω) 2 C A . Assume that ∂s 2 ≤ M2 , ∀ s ∈ [R1 , R2 ], and ∀ ω ∈ S N −1 , where M2 >0. Set 2 f (R2 ω) R2N −1 − f (R1 ω) R1N −1 N −1 1 (ω) := f (sω) s (R1 ) − R2 − R1 + f (sω) s N −1 (R2 ) , ∀ω ∈ S N −1 .
(4.38)
Then N −1 f (y) dy − (R2 − R1 ) R N −1 f ω) dω + R f ω) dω (R (R 1 2 1 2 2 A S N −1 S N −1 (R2 − R1 )2 + 8
S N −1 N
f (sω) s
N −1
(R2 ) dω −
S N −1
π 2 M2 (R2 − R1 ) ≤ N (R2 − R1 )3 − 12 16M2 2
f (sω) s
N −1
(R1 ) dω
S N −1
21 (ω) dω.
(4.39)
54
4 General Multivariate Iyengar Inequalities
Proof Here f (sω) s N −1 ∈ C 2 ([R1 , R2 ]), N ≥ 2, ∀ ω ∈ S N −1 . By (4.2) we get
R2
f (sω) s N −1 ds −
R1
1 (R2 − R1 ) f (R1 ω) R1N −1 + f (R2 ω) R2N −1 2
1 + (R2 − R1 )2 f (sω) s N −1 (R2 ) − f (sω) s N −1 (R1 ) ≤ 8 M2 (R2 − R1 ) 2 1 (ω) =: λ2 (ω) , (R2 − R1 )3 − 24 16M2
(4.40)
∀ ω ∈ S N −1 . Equivalently, we have
R2
−λ2 (ω) ≤
(R2 − R1 ) f (R1 ω) R1N −1 + f (R2 ω) R2N −1 + 2
f (sω) s N −1 ds −
R1
1 (R2 − R1 )2 f (sω) s N −1 (R2 ) − f (sω) s N −1 (R1 ) ≤ λ2 (ω) , 8 ∀ ω ∈ S N −1 . Hence it holds −
S N −1
(R2 − R1 ) 2 (R2 − R1 )2 8
λ2 (ω) dω ≤
R1N −1
S N −1
S N −1
f (sω) s
S N −1
R2
f (sω) s
ds dω−
R1
f (R1 ω) dω + R2N −1 N −1
N −1
S N −1
(R2 ) dω −
S N −1
f (R2 ω) dω +
f (sω) s
N −1
(R1 ) dω (4.42)
≤
S N −1
(4.41)
λ2 (ω) dω.
That is (by (4.31))
N
π 2 M2 (R2 − R1 ) N − (R2 − R1 )3 − 16M2 2 12 f (y) dy − A
(R2 − R1 ) 2
R1N −1
S N −1
S N −1
21
f (R1 ω) dω + R2N −1
(ω) dω ≤ S N −1
f (R2 ω) dω +
4.2 Main Results
(R2 − R1 )2 8
55
S N −1
f (sω) s
N
N −1
(R2 ) dω −
S N −1
π 2 M2 (R2 − R1 ) N (R2 − R1 )3 − 16M2 2 12
f (sω) s
N −1
(R1 ) dω ≤ (4.43)
S N −1
21 (ω) dω,
proving the claim. We give
Theorem 4.13 Let the spherical shell A := B (0, R2 ) − B (0, R1 ), 0 < R1 < R2 , A ⊆ RN , N ≥ 2. Consider f : A → R that is not necessarily radial, and that f ∈ (sω) ≤ M, ∀ s ∈ [R1 , R2 ], and ∀ ω ∈ C 1 A . Let M > m and assume that m ≤ ∂ f∂s N −1 . S Then N −1 f (y) dy − (R2 − R1 ) R N −1 f (R1 ω) dω + R2 f (R2 ω) dω 1 2 A S N −1 S N −1 1 ≤ 2 (M − m)
S N −1
f (R2 ω) R2N −1 − f (R1 ω) R1N −1 − m (R2 − R1 )
M (R2 − R1 ) − f (R2 ω) R2N −1 + f (R1 ω) R1N −1
(4.44)
dω.
Proof Similar to the proof of Theorem 4.11 by using Theorem 4.3 and (4.31).
We give Theorem 4.14 Let the spherical shell A := B (0, R2 ) − B (0, R1 ), 0 < R1 < R2 , f : A → R that is not necessarily radial, and that f ∈ A ⊆ R N , N ≥ 2. Consider 2 ∂ f (sω) 2 C A . Assume that ∂s 2 ≤ M3 , ∀ s ∈ [R1 , R2 ], and ∀ ω ∈ S N −1 , where M3 >0. Set Q 21 (ω) :=
2 f (R2 ω)R2N −1 − f (R1 ω)R1N −1 f (sω) s N −1 (R1 ) + f (sω) s N −1 (R2 ) − 2 R2 −R1 , 2 M32 (R2 − R1 )2 − f (sω) s N −1 (R2 ) − f (sω) s N −1 (R1 )
(4.45) ∀ ω ∈ S N −1 . Then N −1 f (y) dy− (R2 − R1 ) R N −1 f ω) dω + R f ω) dω + (R1 (R2 1 2 2 A S N −1 S N −1
56
4 General Multivariate Iyengar Inequalities
(R2 − R1 )2 8
S N −1
1+
≤
Q 21
(ω)
f (sω) s
M3 (R2 − R1 )3 24
N −1
S N −1
(R2 ) − f (sω) s
N −1
(R1 ) dω
1 − 3Q 21 (ω) dω.
(4.46)
Proof Similar to the proof of Theorem 4.12 by using Theorem 4.4 and (4.31).
We continue with Theorem 4.15 Here all as in Theorem 4.11, and let M1 > m 1 . Assume that m1 ≤
and m1 ≤
f (sω) s N −1 (x) − f (sω) s N −1 (R1 ) ≤ M1 , x − R1
(4.47)
f (sω) s N −1 (R2 ) − f (sω) s N −1 (x) ≤ M1 , R2 − x
(4.48)
∀ x ∈ [R1 , R2 ], ∀ ω ∈ S N −1 . Set
f (sω) s N −1
P12 (ω) =
2 f (R2 ω)R2N −1 − f (R1 ω)R1N −1 (R1 ) + f (sω) s N −1 (R2 ) − 2 R −R 2
1
2 M1 −m 1 2 m 1 +M1 (R2 − R1 )2 − f (sω) s N −1 (R2 ) − f (sω) s N −1 (R1 ) − (R2 − R1 ) 2 2
(4.49)
∀ ω ∈ S N −1 . Then
A
f (y) dy −
,
R2 − R1 2
R1N −1
S N −1
f (R1 ω) dω + R2N −1
S N −1
f (R2 ω) dω +
(R2 − R1 )2 f (sω) s N −1 (R2 ) − f (sω) s N −1 (R1 ) dω 1 + P12 (ω) 8 S N −1 −
≤
(R2 − R1 )3 1 + 3P12 (ω) dω (m 1 + M1 ) 48 S N −1 (M1 − m 1 ) (R2 − R1 )3 1 − 3P12 (ω) dω. 48 S N −1
(4.50)
Proof Similar to the proof of Theorem 4.12 by using Theorem 4.5 and (4.31). We present
4.2 Main Results
57
Theorem 4.16 Consider f : A → R be Lebesgue integrable, which is not neces(n−1) sarily radial. Assume that f (sω) s N −1 ∈ AC n ([R1 , R2 ]) (i.e. f (sω) s N −1 ∈ AC ([R1 , R2 ]) absolutely continuous functions), ∀ ω ∈ S N −1 , N ≥ 2. We assume (n) that f (sω) s N −1 ∈ L ∞ ([R1 , R2 ]), ∀ ω ∈ S N −1 . There exists K 1 > 0 such that
(n)
≤ K 1 , where s ∈ [R1 , R2 ], ∀ ω ∈ S N −1 .
f (sω) s N −1 L ∞ ([R1 ,R2 ])
Then (i)
n−1
(k) 1 f (sω) s N −1 (R1 ) dω (t − R1 )k+1 + f (y) dy − A (k + 1)! S N −1 k=0
(−1)k
S N −1
f (sω) s N −1
(k)
(R2 ) dω (R2 − t)k+1 ≤
2π 2 K1 N (t − R1 )n+1 + (R2 − t)n+1 , 2 (n + 1)! N
(4.51)
∀ t ∈ [R1 , R2 ] , 2 , the right hand side of (4.51) is minimized, and we get: (ii) at t = R1 +R 2 n−1
(k) 1 (R2 − R1 )k+1 f (sω) s N −1 (R1 ) dω+ f (y) dy − A N −1 2k+1 (k + 1)! S k=0
(−1)
k S N −1
f (sω) s
N −1 (k)
(R2 ) dω ≤
N
π2 K 1 (R2 − R1 )n+1 N , 2n−1 2 (n + 1)!
(4.52)
(k) (k) (iii) if f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , k N −1 ∂ f (sω)s ) vanish on ∂ B (0, R ) and ∂ B (0, R )) for all k = 0, 1, . . . , n − 1, (i.e. ( ∂s k
we obtain
1
2
N n+1 2 f (y) dy ≤ π K 1 (R2 − R1 ) , N (n + 1)! 2n−1 A 2
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1
1 R2 − R1 k+1 f (y) dy − A (k + 1)! N k=0
(4.53)
58
4 General Multivariate Iyengar Inequalities
j k+1 S N −1
(−1)
k
N−j
f (sω) s N −1
k+1
S N −1
N
2π 2 K1 N2 (n + 1)!
R2 − R1
(k)
f (sω) s n+1
N
(R1 ) dω +
N −1 (k)
(R2 ) dω ≤
n+1 j n+1 + N − j ,
(4.54)
(k) (k) f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , k N −1 ∂ f (sω)s ) vanish on ∂ B (0, R ) and ∂ B (0, R )) for k = 1, . . . , n − 1, from (i.e. ( ∂s k 1 2 (4.54) we get: (v)
if
N −1 f (y) dy − R2 − R1 j R 1 N A
N − j R2N −1
K1 (n + 1)!
S N −1
S N −1
R2 − R1
f (R2 ω) dω
n+1
N
f (R1 ω) dω +
N 2 ≤ 2π N · 2
n+1 j n+1 + N − j ,
(4.55)
for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (4.55) turns to N −1 f (y) dy − R2 − R1 R1 2 A
S N −1
f (R1 ω) dω +
R2N −1
S N −1
f (R2 ω) dω
N
≤
π2 K 1 (R2 − R1 )n+1 N , 2n−1 2 (n + 1)!
(4.56)
(vii) when n = 1 (without any boundary conditions), we get from (4.56) that N −1 f (y) dy − R2 − R1 R1 2 A
S N −1
f (R1 ω) dω +
R2N −1
S N −1
f (R2 ω) dω
N
≤
π 2 K1 (R2 − R1 )2 . 2 N2
(4.57)
Proof Similar to the proof of Theorem 4.11. We apply Theorem 4.6 along with (4.31).
4.2 Main Results
59
We continue with Theorem 4.17 Consider f : A → R be Lebesgue integrable, which is not neces(n−1) sarily radial. Assume that f (sω) s N −1 ∈ AC n ([R1 , R2 ]) (i.e. f (sω) s N −1 ∈ AC ([R1 , R2 ]) absolutely continuous functions), ∀ ω ∈ S N −1 , N ≥ 2. Here there
(n)
exists K 2 > 0 such that f (sω) s N −1 ≤ K 2 , where s ∈ [R1 , R2 ], L 1 ([R1 ,R2 ])
∀ ω ∈ S N −1 . Then (i)
n−1
(k) 1 f (sω) s N −1 (R1 ) dω (t − R1 )k+1 + f (y) dy − A (k + 1)! S N −1 k=0
(−1)k
S N −1
f (sω) s N −1
(k)
(R2 ) dω (R2 − t)k+1 ≤
2π 2 K 2 N (t − R1 )n + (R2 − t)n , 2 n! N
(4.58)
∀ t ∈ [R1 , R2 ] , 2 , the right hand side of (4.58) is minimized, and we get: (ii) at t = R1 +R 2 n−1
(k) 1 (R2 − R1 )k+1 f (sω) s N −1 (R1 ) dω+ f (y) dy − k+1 A 2 (k + 1)! S N −1 k=0
(−1)
k S N −1
f (sω) s N −1
(k)
≤ dω (R2 )
N
π 2 K 2 (R2 − R1 )n , 2n−2 N2 n!
(4.59)
(k) (k) (iii) if f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , k N −1 ∂ f (sω)s ) vanish on ∂ B (0, R ) and ∂ B (0, R )) for all k = 0, 1, . . . , n − 1, (i.e. ( ∂s k
we obtain
1
2
N n 2 f (y) dy ≤ π K 2 (R2 − R1 ) , N n! 2n−2 A 2
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
(4.60)
60
4 General Multivariate Iyengar Inequalities
n−1
R2 − R1 k+1 1 f (y) dy − A (k + 1)! N k=0
j k+1 S N −1
(−1)
k
N−j
k+1
f (sω) s N −1
S N −1
N
2π 2 K 2 N2 n!
(k)
f (sω) s
R2 − R1
n
N
(R1 ) dω +
N −1 (k)
(R2 ) dω ≤
n jn + N − j ,
(4.61)
(k) (k) f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , k N −1 ∂ f (sω)s ) vanish on ∂ B (0, R ) and ∂ B (0, R )) for k = 1, . . . , n − 1, from (i.e. ( ∂s k 1 2 (4.61) we get: (v)
if
N −1 f (y) dy − R2 − R1 j R1 N A
N − j R2N −1 K2 n!
S N −1
S N −1
R2 − R1
f (R2 ω) dω
n
N
f (R1 ω) dω +
N 2 ≤ 2π N · 2
n jn + N − j ,
(4.62)
for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (4.62) turns to N −1 f (y) dy − R2 − R1 R 1 2 A
S N −1
f (R1 ω) dω + R2N −1
S N −1
f (R2 ω) dω
N
≤
π 2 K 2 (R2 − R1 )n , 2n−2 N2 n!
(4.63)
(vii) when n = 1 (without any boundary conditions), we get from (4.63) that N −1 f (y) dy − R2 − R1 R 1 2 A
S N −1
f (R1 ω) dω + R2N −1
S N −1
f (R2 ω) dω
N
≤
2π 2 K 2 (R2 − R1 ) . N2
(4.64)
4.2 Main Results
61
Proof Similar to the proof of Theorem 4.11. We apply Theorem 4.7 along with (4.31). We continue with Theorem 4.18 Let p, q > 1 : 1p + q1 = 1. Consider f : A → R be Lebesgue integrable, which is not necessarily radial. Assume that f (sω) s N −1 ∈ AC n ([R1 , R2 ]) (n−1) ∈ AC ([R1 , R2 ]) absolutely continuous functions), ∀ ω ∈ (i.e. f (sω) s N −1 N −1 (n) N −1 S N −1 , N ≥ 2. We assume that . There
f (sω) s ∈ L q ([R1 , R2 ]), ∀ ω ∈ S
N −1 (n) ≤ K 3 , where s ∈ [R1 , R2 ], exists K 3 > 0 such that f (sω) s
L q ([R1 ,R2 ])
∀ ω ∈ S N −1 . Then (i)
n−1
(k) 1 f (sω) s N −1 (R1 ) dω (t − R1 )k+1 + f (y) dy − A (k + 1)! S N −1 k=0
(−1)k
S N −1
f (sω) s N −1
(k)
(R2 ) dω (R2 − t)k+1 ≤
N 2π 2 K3 n+ 1p n+ 1p N − R , (4.65) + − t) (t (R ) 1 2 2 (n − 1)! n + 1 ( p (n − 1) + 1) 1p p
∀ t ∈ [R1 , R2 ] , 2 , the right hand side of (4.65) is minimized, and we get: (ii) at t = R1 +R 2 n−1
1 (R2 − R1 )k+1 · f (y) dy − A 2k+1 (k + 1)! k=0
S N −1
f (sω) s N −1
(k)
(R1 ) dω + (−1)k
S N −1
f (sω) s N −1
(k)
(R2 ) dω ≤ 1
N
π2 K3 (R2 − R1 )n+ p N , n−1− q1 2 (n − 1)! n + 1 ( p (n − 1) + 1) 1p 2 p
(4.66)
(k) (k) (iii) if f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , k N −1 ∂ f (sω)s ) vanish on ∂ B (0, R ) and ∂ B (0, R )) for all k = 0, 1, . . . , n − 1, (i.e. ( ∂s k
we obtain
1
2
62
4 General Multivariate Iyengar Inequalities
1 N 2 K3 (R2 − R1 )n+ p f (y) dy ≤ π , 1 1 N 1 A 2n−1− q 2 (n − 1)! n + p ( p (n − 1) + 1) p (4.67) which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1
1 R2 − R1 k+1 f (y) dy − A (k + 1)! N k=0
j
k+1 S N −1
(−1)
k
N−j
k+1
f (sω) s
S N −1
N −1 (k)
f (sω) s
(R1 ) dω +
N −1 (k)
(R2 ) dω ≤
N
2π 2 K3 N · 1 1 2 (n − 1)! n + p − 1) + 1) p p ( (n
R2 − R1
n+ 1p
N
n+ 1p 1 j n+ p + N − j ,
(4.68)
(k) (k) f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , k N −1 ∂ f (sω)s ) vanish on ∂ B (0, R ) and ∂ B (0, R )) for k = 1, . . . , n − 1, from (i.e. ( ∂s k 1 2 (4.68) we get: (v)
if
N −1 f (y) dy − R2 − R1 j R 1 N A
N − j R2N −1
S N −1
S N −1
K3 1 1 (n − 1)! n + p ( p (n − 1) + 1) p
f (R2 ω) dω
R2 − R1 N
f (R1 ω) dω +
N 2 ≤ 2π N · 2
n+ 1p
n+ 1p 1 j n+ p + N − j , (4.69)
for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (4.69) turns to N −1 f (y) dy − R2 − R1 R 1 2 A
S N −1
f (R1 ω) dω + R2N −1
S N −1
f (R2 ω) dω
4.2 Main Results
63 1
N
≤
π2 K3 (R2 − R1 )n+ p N , n−1− q1 2 (n − 1)! n + 1 ( p (n − 1) + 1) 1p 2 p
(4.70)
(vii) when n = 1 (without any boundary conditions), we get from (4.70) that N −1 f (y) dy − R2 − R1 R1 2 A
1
S N −1
f (R1 ω) dω +
R2N −1
S N −1
f (R2 ω) dω
N
1 2 q π 2 K3 (R2 − R1 )1+ p . ≤ N 1 2 1+ p
(4.71)
Proof Similar to the proof of Theorem 4.11. We apply Theorem 4.8 along with (4.31). We continue with results on the ball. We present Theorem 4.19 Consider f : B (0, R) → R be Lebesgue integrable, which is not necessarily radial. Assume that f (sω) s N −1 ∈ AC ([0, R]), ∀ ω ∈ S N −1 , N ≥ 2. N −1 ∈ L ∞ ([0, R]), ∀ ω ∈ S N −1 . Suppose there exists We further assume that ∂ f (sω)s ∂s
∂ f (sω)s N −1 M1 > 0 such that ≤ M1 , ∀ ω ∈ S N −1 .
∂s ∞,(s∈[0,R])
Then (i)
f (y) dy −
S N −1
B(0,R)
f (Rω) dω R
N −1
(R − t) ≤
π 2 M1 2 N t + (R − t)2 , 2 N
(4.72)
∀ t ∈ [0, R] , (ii) at t = R2 , the right hand side of (4.72) is minimized, and we get:
f (y) dy − B(0,R)
S N −1
f (Rω) dω
N R N π 2 M1 R 2 , ≤ 2 2 N2
(4.73)
(iii) if f (Rω) = 0, ∀ ω ∈ S N −1 , (i.e. f (·ω) vanishes on ∂ B (0, R)), we obtain
B(0,R)
π N2 M1 R 2 , f (y) dy ≤ 2 N2
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
(4.74)
64
4 General Multivariate Iyengar Inequalities
B (0,R)
f (y) dy − N
π 2 M1 N2
RN N R
N−j
2
N
S N −1
f (Rω) dω ≤
2 j2 + N − j ,
(4.75)
(v) when N = 2 and j = 1, (4.75) turns to
B(0,R)
RN f (y) dy − 2
S N −1
π N2 M1 R 2 . f (Rω) dω ≤ 2 N2
(4.76)
Proof Same as the proof of Theorem 4.16, just set there R1 = 0 and R2 = R and use (4.30). We continue with Theorem 4.20 Consider f : B (0, R) → R be Lebesgue integrable, which is not N −1 , N ≥ 2. necessarily radial. Assume that f (sω) s N −1 ∈ AC
([0, R]), ∀ ω ∈ S
∂ f (sω)s N −1 Suppose there exists M2 > 0 such that ≤ M2 , ∀ ω ∈ S N −1 .
∂s L 1 ([0,R])
Then (i)
f (y) dy − B(0,R)
S N −1
2π N2 M2 R , f (Rω) dω R N −1 (R − t) ≤ N2
(4.77)
∀ t ∈ [0, R] , (ii) if f (Rω) = 0, ∀ ω ∈ S N −1 , (i.e. f (·ω) vanishes on ∂ B (0, R)) from (4.77), we obtain 2π N2 M2 R , f (y) dy ≤ (4.78) N B(0,R)
2
which is a sharp inequality, (iii) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
B (0,R)
f (y) dy −
RN N
N−j
S N −1
2π N2 M2 R , f (Rω) dω ≤ N2
(4.79)
(iv) when N = 2 and j = 1, (4.79) turns to
B(0,R)
RN f (y) dy − 2
S N −1
2π N2 M2 R . f (Rω) dω ≤ N2
(4.80)
4.2 Main Results
65
Proof Same as the proof of Theorem 4.17, just set there R1 = 0 and R2 = R and use (4.30). We continue with Theorem 4.21 Let p, q > 1 : 1p + q1 = 1. Consider f : B (0, R) → R be Lebesgue integrable, which is not necessarily radial. Assume that f (sω) s N −1 ∈ AC ([0, R]), N −1 ∈ L q ([0, R]), ∀ ω ∈ S N −1 . ∀ ω ∈ S N −1 , N ≥ 2. We further assume that ∂ f (sω)s ∂s
∂ f (sω)s N −1 Suppose there exists M3 > 0 such that ≤ M3 , ∀ ω ∈ S N −1 .
∂s L q ([0,R])
Then (i)
f (y) dy −
S N −1
B(0,R)
f (Rω) dω R N −1 (R − t) ≤
1 2π 2 M3 1+ p 1+ 1p t , + − t) (R N2 1 + 1p N
(4.81)
∀ t ∈ [0, R] , (ii) at t = R2 , the right hand side of (4.81) is minimized, and we get:
f (y) dy − B(0,R)
S N −1
f (Rω) dω
1 1 N R N 2 q π 2 M3 R 1+ p ≤ , 2 N2
(4.82)
(iii) if f (Rω) = 0, ∀ ω ∈ S N −1 , (i.e. f (·ω) vanishes on ∂ B (0, R)), we obtain
B(0,R)
1 1 2 q π N2 M3 R 1+ p N f (y) dy ≤ , 2
(4.83)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
B (0,R)
f (y) dy − N
2π 2 M3 1 + 1p N2
RN N R
N−j
1+ 1p
N
S N −1
f (Rω) dω ≤
1+ 1p 1 j 1+ p + N − j ,
(4.84)
(v) when N = 2 and j = 1, (4.84) turns to
B(0,R)
RN f (y) dy − 2
S N −1
1 1 N 2 q π 2 M3 R 1+ p . f (Rω) dω ≤ 1 + 1p N2
(4.85)
66
4 General Multivariate Iyengar Inequalities
Proof Same as the proof of Theorem 4.18, just set there R1 = 0 and R2 = R and use (4.30).
References 1. R.P. Agarwal, S.S. Dragomir, An application of Hayashi’s inequality for differentiable functions. Comput. Math. Appl. 6, 95–99 (1996) 2. G.A. Anastassiou, Fractional Differentiation Inequalities. Research Monograph (Springer, New York, 2009) 3. G.A. Anastassiou, General Iyengar type inequalities. J. Comput. Anal. Appl. 28(5), 786–797 (2020) 4. G.A. Anastassiou, General Multivariate Iyengar Type Inequalities, Constructive Mathematical Analysis (2019) 5. X.-L. Cheng, The Iyengar-type inequality. Appl. Math. Lett. 14, 975–978 (2001) 6. K.S.K. Iyengar, Note on an inequality. Math. Student 6, 75–76 (1938) 7. Z. Liu, Note on Iyengar’s Inequality. Serija Matematika, vol. 16 (Publikacije Elektrotehnickog fakulteta, 2005), pp. 29–35 8. F. Qi, Further Generalizations of Inequalities for an Integral, Serija Matematika, vol. 8 (Publikacije Elektrotehnickog fakulteta, 1997), pp. 79–83 9. W. Rudin, Real and Complex Analysis, International Student edn. (Mc Graw Hill, London, 1970) 10. D. Stroock, A Concise Introduction to the Theory of Integration, 3rd edn. (Birkhaüser, Boston, 1999)
Chapter 5
Multivariate Iyengar Inequalities for Radial Functions
Here we present a variety of multivariate Iyengar type inequalities for radial functions defined on the shell and ball. Our approach is based on the polar coordinates in R N , N ≥ 2, and the related multivariate polar integration formula. Via this method we transfer well-known univariate Iyengar type inequalities and univariate author’s related results into multivariate Iyengar inequalities. See also [3].
5.1 Background In the year 1938, Iyengar [6] proved the following interesting inequality. Theorem 5.1 Let f be a differentiable function on [a, b] and f (x) ≤ M1 . Then
b
a
M1 (b − a)2 1 ( f (b) − f (a))2 − f (x) d x − (b − a) ( f (a) + f (b)) ≤ . 2 4 4M1 (5.1)
In 2001, Cheng [5] proved that. Theorem 5.2 Let f ∈ C 2 ([a, b]) and f (x) ≤ M2 . Then
a
b
1 1 2 f (x) d x − (b − a) ( f (a) + f (b)) + (b − a) f (b) − f (a) ≤ 2 8 M2 (b − a) 2 , (b − a)3 − 24 16M2 1
where 1 = f (a) −
(5.2)
2 ( f (b) − f (a)) + f (b) . (b − a)
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_5
67
68
5 Multivariate Iyengar Inequalities for Radial Functions
In 1996, Agarwal and Dragomir [1] obtained a generalization of (5.1). Theorem 5.3 Let f : [a, b] → R be a differentiable function such that for all x ∈ [a, b] with M > m we have m ≤ f (x) ≤ M. Then
b
f (x) d x −
a
1 (b − a) ( f (a) + f (b)) ≤ 2
(5.3)
( f (b) − f (a) − m (b − a)) (M (b − a) − f (b) + f (a)) . 2 (M − m) In [8], Qi proved. Theorem 5.4 Let f : [a, b] → R be a twice differentiable function such that for all x ∈ [a, b] with M > 0 we have f (x) ≤ M. Then b 1 + Q2 ( f (a) + f (b)) 2 f (b) − f (a) (b − a) ≤ f (x) d x − (b − a) + a 2 8 M (b − a)3 1 − 3Q 2 , 24 where
Q2 =
f (a) + f (b) − 2
f (b)− f (a) b−a
(5.4) 2
M 2 (b − a)2 − ( f (b) − f (a))2
.
(5.5)
In 2005, Zheng Liu, [7], proved the following. Theorem 5.5 Let f : [a, b] → R be a differentiable function such that f is integrable on [a, b] and for all x ∈ [a, b] with M > m we have m≤
f (b) − f (x) f (x) − f (a) ≤ M and m ≤ ≤ M. x −a b−x
(5.6)
Then b 1 + P2 ( f (a) + f (b)) f (b) − f (a) (b − a)2 f (x) d x − (b − a) + 2 8 a
1 + 3P 2 − 48 where
(M − m) (b − a)3 1 − 3P 2 , (m + M) (b − a) ≤ 48 3
(5.7)
2 f (a) f (a) + f (b) − 2 f (b)− b−a P2 = 2 . 2 M−m 2 (b − a) (b − a) − f (b) − f (a) − m+M 2 2
(5.8)
5.1 Background
69
We need
Remark 5.6 We define the ball B (0, R) = x ∈ R N : |x| < R ⊆ R N , N ≥ 2, R > 0, and the sphere
S N −1 := x ∈ R N : |x| = 1 , where |·| is the Euclidean norm. Let dω be the element of surface measure on S N −1 and N 2π 2 dω = N ωN = 2 S N −1 be the area of S N −1 . For x ∈ R N − {0} we can write uniquely x = r ω, where r = |x| > 0 and ω = N x ∈ S N −1 , |ω| = 1. Note that B(0,R) dy = ω NNR is the Lebesgue measure on the r N
R ball, that is the volume of B (0, R), which exactly is V ol (B (0, R)) = π 2N +1 . (2 ) Following [9, pp. 149–150, Exercise 6], and [10, pp. 87–88, Theorem 5.2.2] we can write for F : B (0, R) → R a Lebesgue integrable function that
F (x) d x = B(0,R)
S N −1
R
F (r ω) r N −1 dr dω,
N
(5.9)
0
and we use this formula a lot. Typically here the function f : B (0, R) → R is radial; that is, there exists a function g such that f (x) = g (r ), where r = |x|, r ∈ [0, R], ∀ x ∈ B (0, R). Remark 5.7 Let the spherical shell A := B (0, R2 ) − B (0, R1 ), 0 < R1 < R2 , A ⊆ R N , N ≥ 2, x ∈ A. Consider that f : A → R is radial; that is, there exists g such that f (x) = g (r ), r = |x|, r ∈ [R1 , R2 ], ∀ x ∈ A. Here x can be written uniquely as x = r ω, where r = |x| > 0 and ω = rx ∈ S N −1 , |ω| = 1, see ([9], p. 149–150 and [2], p. 421), furthermore for F : A → R a Lebesgue integrable function we have that F (x) d x =
R2
F (r ω) r
N −1
dr dω.
(5.10)
N π 2 R2N − R1N ω N R2N − R1N . = V ol (A) = N N2 + 1
(5.11)
A
Here
S N −1
R1
In this chapter we derive multivariate Iyengar type inequalities on the shell and ball of R N , N ≥ 2, for radial functions. Our results are based on Theorems 5.1–5.5 and several other results by the author.
70
5 Multivariate Iyengar Inequalities for Radial Functions
5.2 Main Results We present the following multivariate Iyengar type inequalities on the shell and the ball: We start with. Theorem 5.8 Let the spherical shell A := B (0, R2 ) − B (0, R1 ), 0 < R1 < R2 , A ⊆ R N , N ≥ 2. Consider f : A → R that is radial, that is, there exists g such that f (x) = g (r ), r = |x|, r ∈ [R1 , R2 ], ∀ x ∈ A; x = r ω, ω ∈ S N −1 . We assume that g ∈ C 1 ([R1 , R2 ]). Then N π 2
N −1 N −1 ≤ f (y) dy − (R2 − R1 ) g (R1 ) R1 + g (R2 ) R2 A N2 ⎤ N −1 N −1 2 g (R2 ) R2 − g (R1 ) R1 π ⎥ ⎢ ⎣ g (s) s N −1 (R2 − R1 )2 − ⎦. ∞,[R1 ,R2 ] n−1 2 N2 g (s) s ⎡
N 2
∞,[R1 ,R2 ]
(5.12)
1 N −1 Proof Here ∈ C 1 ([R1 , R2 ]), N ≥ g ∈ C ([R1 , R2 ]) and clearly h (s) := g (s) s 2. We set h ∞,[R1 ,R2 ] = M. By (5.1) we get
R2 R1
1 h (x) d x − (R2 − R1 ) (h (R1 ) + h (R2 )) ≤ 2 M (R2 − R1 )2 (h (R2 ) − h (R1 ))2 − = 4 4M
g (s) s N −1
∞,[R1 ,R2 ]
(R2 − R1 )2
4
−
2 g (R2 ) R2N −1 − g (R1 ) R1N −1 =: λ. 4 g (s) s N −1 ∞,[R1 ,R2 ]
(5.13)
Equivalently, we have
R2
−λ≤
R1
g (s) s N −1 ds −
1 (R2 − R1 ) g (R1 ) R1N −1 + g (R2 ) R2N −1 ≤ λ, 2 (5.14)
and −λ≤
R2 R1
Hence it holds
f (sω) s N −1 ds −
R2 − R1 2
g (R1 ) R1N −1 + g (R2 ) R2N −1 ≤ λ. (5.15)
5.2 Main Results
71
−λ
R2 − R1 2
S N −1
dω ≤
S N −1
R2
f (sω) s
N −1
ds dω−
R1
g (R1 ) R1N −1 + g (R2 ) R2N −1
S N −1
dω ≤ λ
dω,
(5.16)
S N −1
that is (by (5.10)) N
2π 2 −λ N ≤ 2
f (y) dy − A
R2 − R1 2
2π 2 g (R1 ) R1N −1 + g (R2 ) R2N −1 N2 N
N
2π 2 ≤ λ N . 2
(5.17)
Therefore we get N N π 2
2π 2 N −1 N −1 N ≤ λ N . f (y) dy − (R2 − R1 ) g (R1 ) R1 + g (R2 ) R2 A 2 2 (5.18) The theorem is proved. We give. Corollary 5.9 (to Theorem 5.8) Let f : B (0, R) → R be radial, that is, there exists a function g such that f (x) = g (r ), where r = |x|, r ∈ [0, R], R > 0, ∀ x ∈ B (0, R); N ≥ 2. We assume that g ∈ C 1 ([0, R]). Then N 2 π f (y) dy − R N g (R) N ≤ B(0,R) 2 ⎡
π g 2 (R) R 2(N −1) ⎢ N ⎣ g (s) s N −1 R 2 − ∞,[0,R] 2 2 g (s) s N −1 N 2
⎤ ⎥ ⎦.
(5.19)
∞,[0,R]
Proof Similar to Theorem 5.8, use of (5.9).
We also give. Theorem 5.10 Let f : A → R be radial; that is, there exists g such that f (x) = g (r ), r = |x|, r ∈ [R1 , R2 ], ∀ x ∈ A; x = r ω, ω ∈ S N −1 , N ≥ 2. We assume that g ∈ C 2 ([R1 , R2 ]). Then
f (y) dy − (R2 − R1 ) g (R1 ) R N −1 + g (R2 ) R N −1 + 1 2 A
72
5 Multivariate Iyengar Inequalities for Radial Functions
π N2 (R2 − R1 )2 N −1 N −1 ≤ g (s) s (R2 ) − g (s) s (R1 ) 4 N2 ⎡ ⎛ g (s) s N −1 (R2 − R1 ) ⎢ ⎜ ∞,[R1 ,R2 ] (R2 − R1 )3 − ⎝ ⎣ 12 8 g (s) s N −1
∞,[R1 ,R2 ]
⎞
(5.20)
⎤ N
⎟ 2⎥ π 2 ⎠ 1 ⎦ N , 2
where 2 g (R2 ) R2N −1 − g (R1 ) R1N −1 N −1 1 := g (s) s + g (s) s N −1 (R2 ) . (R1 ) − (R2 − R1 ) (5.21) 2 N −1 ∈ C 2 ([R1 , R2 ]), N ≥ Proof Here g ∈ C ([R1 , R2 ]) and clearly h (s) := g (s) s 2. We set h ∞,[R1 ,R2 ] = M. By (5.2) we get
R2 1 1 2 h (s) ds − (R2 − R1 ) (h (R1 ) + h (R2 )) + (R2 − R1 ) h (R2 ) − h (R1 ) R1 2 8
≤
M (R2 − R1 ) 2 1 , (R2 − R1 )3 − 24 16M
where 1 = h (R1 ) −
2 (h (R2 ) − h (R1 )) + h (R2 ) . (R2 − R1 )
(5.22)
(5.23)
That is
R2 R1
g (s) s N −1 ds −
(R2 − R1 ) g (R1 ) R1N −1 + g (R2 ) R2N −1 + 2
(R2 − R1 )2 N −1 N −1 g (s) s (R2 ) − g (s) s (R1 ) ≤ 8 M (R2 − R1 ) 2 1 =: ψ, (R2 − R1 )3 − 24 16M
(5.24)
where 2 g (R2 ) R2N −1 − g (R1 ) R1N −1 N −1 1 := g (s) s + g (s) s N −1 (R2 ) . (R1 ) − (R2 − R1 ) (5.25) Equivalently, we have
5.2 Main Results
R2
−ψ ≤
73
f (sω) s N −1 ds −
R1
(R2 − R1 ) g (R1 ) R1N −1 + g (R2 ) R2N −1 + 2
(R2 − R1 )2 g (s) s N −1 (R2 ) − g (s) s N −1 (R1 ) ≤ ψ. 8
(5.26)
Hence it holds
−ψ
S N −1
dω ≤
S N −1
R2
f (sω) s N −1 ds dω−
R1
(R2 − R1 ) g (R1 ) R1N −1 + g (R2 ) R2N −1 + 2
(R2 − R1 )2 N −1 N −1 g (s) s dω ≤ ψ dω, (R2 ) − g (s) s (R1 ) 8 S N −1 S N −1 (5.27) that is (by (5.10)) N
−ψ
2π 2 ≤ N2
f (y) dy − A
(R2 − R1 ) g (R1 ) R1N −1 + g (R2 ) R2N −1 + 2
N 2π N2 2π 2 (R2 − R1 )2 N ≤ ψ N . g (s) s N −1 (R2 ) − g (s) s N −1 (R1 ) 8 2 2 (5.28) Therefore we get
f (y) dy − (R2 − R1 ) g (R1 ) R N −1 + g (R2 ) R N −1 + 1 2 A
N π N2 2π 2 (R2 − R1 )2 N −1 N −1 ≤ ψ N . g (s) s (R2 ) − g (s) s (R1 ) 4 N2 2 (5.29) The theorem is proved. We give. Corollary 5.11 (to Theorem 5.10) Let f : B (0, R) → R be radial, that is, there exists a function g such that f (x) = g (r ), where r = |x|, r ∈ [0, R], R > 0, ∀ x ∈ B (0, R); N ≥ 2. We assume that g ∈ C 2 ([0, R]). Then π N2 R2 N N −1 N −1 g (s) s f (y) dy − R g (R) + (R) − g (s) s (0) B(0,R) 4 N2
74
5 Multivariate Iyengar Inequalities for Radial Functions
⎡ g (s) s N −1 R ⎢ ∞,[0,R] 3 R − ≤⎣ 12 N 8 g (s) s −1
⎤ ⎥ ∗2 1 ⎦
∞,[0,R]
N
π2 , (5.30) N2
where ∗1 := g (s) s N −1 (0) − 2g (R) R N −2 + g (s) s N −1 (R) .
(5.31)
If N > 2, then g (s) s N −1 (0) = 0.
Proof Similar to Theorem 5.10, use of (5.9). We present.
Theorem 5.12 Consider f : A → R that is radial; that is, there exists g such that f (x) = g (r ), r = |x|, r ∈ [R1 , R2 ], ∀ x ∈ A; x = r ω, ω ∈ S N −1 , N ≥ 2. We assume that g ∈ C 1 ([R1 , R2 ]). Then N 2
π N ≤ f (y) dy − (R2 − R1 ) g (R1 ) R1N −1 + g (R2 ) R2N −1 A 2
g (R2 ) R2N −1 − g (R1 ) R1N −1 − m (R2 − R1 ) ×
M (R2 − R1 ) − g (R2 ) R2N −1 + g (R1 ) R1N −1 (M − m)
N π2 , N2
(5.32)
where M > m with m ≤ g (s) s N −1 ≤ M, ∀ s ∈ [R1 , R2 ] .
(5.33)
Proof Here g ∈ C 1 ([R1 , R2 ]) and clearly h (s) := g (s) s N −1 ∈ C 1 ([R1 , R2 ]), N ≥ 2. We assume here m ≤ h (s) ≤ M, ∀ s ∈ [R1 , R2 ] with M > m. By (5.3) we get
R2 R1
1 h (s) ds − (R2 − R1 ) (h (R1 ) + h (R2 )) ≤ 2
(h (R2 ) − h (R1 ) − m (R2 − R1 )) (M (R2 − R1 ) − h (R2 ) + h (R1 )) . 2 (M − m) That is
R2 R1
g (s) s
N −1
1 N −1 N −1 ds − (R2 − R1 ) g (R1 ) R1 + g (R2 ) R2 ≤ 2
(5.34)
5.2 Main Results
75
g (R2 ) R2N −1 − g (R1 ) R1N −1 − m (R2 − R1 ) ×
M (R2 − R1 ) − g (R2 ) R2N −1 + g (R1 ) R1N −1 =: ρ. 2 (M − m)
(5.35)
Equivalently, we have
R2
−ρ≤
g (s) s N −1 ds −
R1
R2 − R1 2
g (R1 ) R1N −1 + g (R2 ) R2N −1 ≤ ρ, (5.36)
and
R2
−ρ≤
f (sω) s
N −1
ds −
R1
R2 − R1 2
g (R1 ) R1N −1 + g (R2 ) R2N −1 ≤ ρ. (5.37)
Hence it holds
R2 − R1 2
−ρ
S N −1
dω ≤
S N −1
R2
f (sω) s N −1 ds dω−
R1
g (R1 ) R1N −1 + g (R2 ) R2N −1
S N −1
dω ≤ ρ
dω,
(5.38)
S N −1
that is (by (5.10)) N
2π 2 −ρ N ≤ 2
f (y) dy − A
R2 − R1 2
2π 2 g (R1 ) R1N −1 + g (R2 ) R2N −1 N2 N
N
2π 2 ≤ ρ N . 2
(5.39)
Therefore we get N N π 2 2π 2 N −1 N −1 N ≤ ρ N . f (y) dy − (R2 − R1 ) g (R1 ) R1 + g (R2 ) R2 A 2 2 (5.40) The theorem is proved. We give. Corollary 5.13 (to Theorem 5.12) Let f : B (0, R) → R be radial, that is, there exists a function g such that f (x) = g (r ), where r = |x|, r ∈ [0, R], R > 0, ∀ x ∈ B (0, R); N ≥ 2. We assume that g ∈ C 1 ([0, R]). Then
76
5 Multivariate Iyengar Inequalities for Radial Functions
N π 2 N f (y) dy − R g (R) N ≤ A 2 N g (R) R N −1 − m R M R − g (R) R N −1 π2 , (M − m) N2
(5.41)
where M > m with m ≤ g (s) s N −1 ≤ M, ∀ s ∈ [0, R] .
(5.42)
Proof Similar to Theorem 5.12, use of (5.9). We continue with.
Theorem 5.14 Let f : A → R be radial; that is, there exists g such that f (x) = g (r ), r = |x|, r ∈ [R1 , R2 ], ∀ x ∈A; x = r ω, ω ∈ S N −1 , N ≥ 2. We assume that . g ∈ C 2 ([R1 , R2 ]). We call M1 := g (s) s N −1 ∞,[R1 ,R2 ]
Then
f (y) dy − g (R1 ) R N −1 + g (R2 ) R N −1 (R2 − R1 ) + 1 2 A
N 2 1 + Q 21 π N ≤ g (s) s N −1 (R2 ) − g (s) s N −1 (R1 ) (R2 − R1 )2 4 2 π2 M1 (R2 − R1 )3 , 1 − 3Q 21 12 N2 N
(5.43)
where ! "2 g(R2 )R2N −1 −g(R1 )R1N −1 g (s) s N −1 (R1 ) + g (s) s N −1 (R2 ) − 2 R −R 2 1 . Q 21 := 2 2 2 M1 (R2 − R1 ) − g (s) s N −1 (R2 ) − g (s) s N −1 (R1 ) (5.44) Proof Similar to the proof of Theorem 5.10 by the use of Theorem 5.4.
We give. Corollary 5.15 (to Theorem 5.14) Let f : B (0, R) → R be radial, that is, there exists a function g such that f (x) = g (r ), where r = |x|, r ∈ [0, R], R > 0, ∀ x ∈ ≥ 2. We assume that g ∈ C 2 ([0, R]). We call M2 := B(0, R); N N −1 . Then g (s) s ∞,[0,R]
5.2 Main Results
77
f (y) dy− A
N " 1 + Q 22 ! π 2 N N −1 N −1 2 g (s) s g (R) R + (R) − g (s) s (0) R 4 N2
#
π2 M2 3 , ≤ R 1 − 3Q 22 12 N2 N
(5.45)
where "2 ! g (s) s N −1 (0) + g (s) s N −1 (R) − 2g (R) R N −2 Q 22 := 2 . M22 R 2 − g (s) s N −1 (R) − g (s) s N −1 (0)
(5.46)
Proof Similar to Corollary 5.11. We present. Theorem 5.16 Here all as in Theorem 5.8 and M1 > m 1 . Assume that g (s) s N −1 (s) − g (s) s N −1 (R1 ) m1 ≤ ≤ M1 s − R1
(5.47)
g (s) s N −1 (R2 ) − g (s) s N −1 (s) ≤ M1 , R2 − s
(5.48)
and m1 ≤
for all s ∈ [R1 , R2 ] . Then
f (y) dy − g (R1 ) R N −1 + g (R2 ) R N −1 (R2 − R1 ) + 1 2 A
1 + P12 4
g (s) s N −1 (R2 ) − g (s) s N −1 (R1 ) (R2 − R1 )2 −
1 + 3P12 24
(m 1 + M1 ) (R2 − R1 )
3
N π 2 ≤ N2
π2 (M1 − m 1 ) (R2 − R1 )3 , 1 − 3P12 24 N2 N
(5.49)
78
5 Multivariate Iyengar Inequalities for Radial Functions
where P12 =
M1 −m 1 2
2
g (s) s N −1 (R1 ) + g (s) s N −1 (R2 ) − 2
(R2 − R1 )2 −
g(R2 )R2N −1 −g(R1 )R1N −1 R2 −R1
2
"2 . ! 1 g (s) s N −1 (R2 ) − g (s) s N −1 (R1 ) − m 1 +M − R ) (R 2 1 2
(5.50) Proof Similar to Theorem 5.8 by the use of Theorem 5.5.
We give. Corollary 5.17 (to Theorem 5.16) Here all as in Corollary 5.9 and M2 > m 2 . Assume that g (s) s N −1 (s) − g (s) s N −1 (0) (5.51) m2 ≤ ≤ M2 s g (s) s N −1 (R) − g (s) s N −1 (s) ≤ M2 , m2 ≤ R−s
and
for all s ∈ [0, R] . Then
(5.52)
f (y) dy − g (R) R N + A
1 + P22 4
g (s) s N −1 (R) − g (s) s N −1 (0) R 2 −
1 + 3P22 24
(m 2 + M2 ) R 3
N π 2 ≤ N2
π2 (M2 − m 2 ) R 3 , 1 − 3P22 24 N2 N
(5.53)
where ! "2 g (s) s N −1 (0) + g (s) s N −1 (R) − 2g (R) R N −2 P22 = ! "2 . M2 −m 2 2 2 N −1 (R) − g (s) s N −1 (0) − m 2 +M2 R R − g s (s) 2 2 (5.54) Proof Similar to Corollary 5.9, based on Theorem 5.5. We continue with some author’s results to be used later in this chapter.
5.2 Main Results
79
Theorem 5.18 ([4]) Let n ∈ N, f ∈ AC n ([a, b]) (i.e. f (n−1) ∈ AC ([a, b]), absolutely continuous functions). We assume that f (n) ∈ L ∞ ([a, b]). Then (i) n−1 b $
(k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
(n) f
(t − a)n+1 + (b − t)n+1 ,
L ∞ ([a,b])
(n + 1)! ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
(5.55)
the right hand side of (5.55) is minimized, and we get:
n−1 b $ 1 (b − a)k+1 (k) k (k) f (x) d x − f (a) + (−1) f (b) ≤ a 2k+1 (k + 1)! k=0
(n) f
L ∞ ([a,b])
(n + 1)!
(b − a)n+1 , 2n
(5.56)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
b
a
f (n) L ∞ ([a,b]) (b − a)n+1 f (x) d x ≤ , 2n (n + 1)!
(5.57)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1 b " $ 1 b − a k+1 ! k+1 (k) k k+1 (k) j + (−1) − j) f d x − f f (a) (N (b) (x) a N (k + 1)! k=0
≤
(n) f
L ∞ ([a,b])
(n + 1)!
b−a N
n+1
j n+1 + (N − j)n+1 ,
(5.58)
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (5.58) we get:
b
f (x) d x −
a
(n) f
L ∞ ([a,b])
(n + 1)!
b−a N
b−a N
[ j f (a) + (N − j) f (b)] ≤
n+1
j n+1 + (N − j)n+1 ,
(5.59)
80
5 Multivariate Iyengar Inequalities for Radial Functions
for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (5.59) turns to
b
f (x) d x −
a
(n) f
b−a 2
L ∞ ([a,b])
(n + 1)!
( f (a) + f (b)) ≤
(b − a)n+1 , 2n
(5.60)
(vii) when n = 1 (without any boundary conditions), we get from (5.60) that
b
f (x) d x −
a
b−a 2
(b − a)2 , ( f (a) + f (b)) ≤ f ∞,[a,b] 4
(5.61)
a similar to Iyengar inequality (5.1). Theorem 5.19 ([4]) Let f ∈ AC n ([a, b]), n ∈ N. Then (i) n−1 b $
(k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0 (5.62) (n) f L 1 ([a,b]) (t − a)n + (b − t)n , n! ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
the right hand side of (5.62) is minimized, and we get:
n−1 b $ 1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
(n) f
L 1 ([a,b])
n!
(b − a)n , 2n−1
(5.63)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
a
b
f (n) L 1 ([a,b]) (b − a)n f (x) d x ≤ , n! 2n−1
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
(5.64)
5.2 Main Results
81
n−1 b " $ b − a k+1 ! k+1 (k) 1 k k+1 (k) j f (x) d x − f (a) + (−1) (N − j) f (b) a N (k + 1)! k=0
≤
(n) f
L 1 ([a,b])
n!
b−a N
n
j n + (N − j)n ,
(5.65)
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (5.65) we get:
b
f (x) d x −
a
(n) f
b−a N
L 1 ([a,b])
n!
[ j f (a) + (N − j) f (b)] ≤
b−a N
n
j n + (N − j)n ,
(5.66)
for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (5.66) turns to
b
a
(b − a) f (x) d x − ( f (a) + f (b)) ≤ 2 (n) f
L 1 ([a,b])
n!
(b − a)n , 2n−1
(5.67)
(vii) when n = 1 (without any boundary conditions), we get from (5.67) that
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤ f L 1 ([a,b]) (b − a) .
Theorem 5.20 ([4]) Let f ∈ AC n ([a, b]), n ∈ N; p, q > 1 : L q ([a, b]). Then (i)
1 p
+
1 q
(5.68)
= 1, and f (n) ∈
n−1 b $
(k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0 (5.69) (n) f ! " 1 1 L ([a,b]) q (t − a)n+ p + (b − t)n+ p , 1 (n − 1)! n + 1p ( p (n − 1) + 1) p ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
the right hand side of (5.69) is minimized, and we get:
82
5 Multivariate Iyengar Inequalities for Radial Functions
n−1 b $ 1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
(n) 1 f (b − a)n+ p L ([a,b]) q , 1 1 2n− q (n − 1)! n + 1p ( p (n − 1) + 1) p
(5.70)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
a
b
f (x) d x ≤
(n) 1 f (b − a)n+ p L ([a,b]) q , 1 1 2n− q (n − 1)! n + 1p ( p (n − 1) + 1) p
(5.71)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1 b " $ b − a k+1 ! k+1 (k) 1 k k+1 (k) j f (x) d x − f (a) + (−1) (N − j) f (b) a N (k + 1)! k=0
(n) 1 f " b − a n+ p ! n+ 1p L q ([a,b]) n+ 1p j , ≤ + − j) (N 1 N (n − 1)! n + 1p ( p (n − 1) + 1) p (5.72) (v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (5.72) we get:
b
f (x) d x −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
(n) 1 f " b − a n+ p ! n+ 1p L q ([a,b]) n+ 1p j , (5.73) + − j) (N 1 N (n − 1)! n + 1p ( p (n − 1) + 1) p for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (5.73) turns to
a
b
(b − a) f (x) d x − ( f (a) + f (b)) ≤ 2
(n) 1 f (b − a)n+ p L q ([a,b]) , 1 1 2n− q (n − 1)! n + 1p ( p (n − 1) + 1) p
(5.74)
(vii) when n = 1 (without any boundary conditions), we get from (5.74) that
5.2 Main Results
b
83
f (x) d x −
a
b−a 2
1 f L q ([a,b]) (b − a)1+ p . ( f (a) + f (b)) ≤ 1 2p 1 + 1p
(5.75)
Next, we extend Theorems 5.18–5.20 to the multivariate case over shells and balls for radial functions. The proving method is the same as in our earlier results of this chapter, as such we omit these next proofs. We present (use of Theorem 5.18). Theorem 5.21 Consider f : A → R which is radial; that is, there exists g such that f (x) = g (r ), r = |x|, r ∈ [R1 , R2 ], ∀ x ∈ A; x = r ω, ω ∈ S N −1 , N ≥ 2. We assume (n) ∈ L ∞ ([R1 , R2 ]), n ∈ N. Then that g (s) s N −1 ∈ AC n ([R1 , R2 ]) and g (s) s N −1 (i) % n−1 ! $ (k) 1 g (s) s N −1 (R1 ) (t − R1 )k+1 + f (y) dy − A (k + 1)! k=0
"& 2π N2 N −1 (k) k+1 ≤ (−1) g (s) s (R2 ) (R2 − t) N2 k
N 2
2π N2
(n) g (s) s N −1
L ∞ ([R1 ,R2 ])
(n + 1)!
(t − R1 )n+1 + (R2 − t)n+1 ,
(5.76)
∀ t ∈ [R1 , R2 ] , 2 , the right hand side of (5.76) is minimized, and we get: (ii) at t = R1 +R 2 % n−1 $ (k) 1 (R2 − R1 )k+1 ! g (s) s N −1 (R1 ) + f (y) dy − k A 2 (k + 1)! k=0
k
(−1) g (s) s
N 2
π N2
N −1 (k)
(n) g (s) s N −1
"& π N2 N ≤ (R2 ) 2
L ∞ ([R1 ,R2 ])
(n + 1)!
(R2 − R1 )n+1 , 2n−1
(5.77)
(k) (k) (iii) if g (s) s N −1 (R1 ) = g (s) s N −1 (R2 ) = 0, for all k = 0, 1, . . . , n − 1, we obtain N −1 (n) N g s (s) n+1 2 L ∞ ([R1 ,R2 ]) (R2 − R1 ) f (y) dy ≤ π , (5.78) N 2n−1 (n + 1)! A 2
84
5 Multivariate Iyengar Inequalities for Radial Functions
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds % n−1 $ (k) R2 − R1 k+1 ! k+1 1 j g (s) s N −1 f dy − (y) (R1 ) + A N (k + 1)! k=0
(−1) (N − j) k
N 2
2π N2
(n) g (s) s N −1
k+1
"& 2π N2 N −1 (k) ≤ g (s) s (R2 ) N2
L ∞ ([R1 ,R2 ])
(n + 1)!
R2 − R1 N
n+1
j n+1 + (N − j)n+1 , (5.79)
(k) (k) (v) if g (s) s N −1 (R1 ) = g (s) s N −1 (R2 ) = 0, from (5.79) we get:
k = 1, . . . , n − 1,
' R2 − R1 f (y) dy − jg (R1 ) R1N −1 + N A
N N 2 2π 2 2π N −1 ≤ × (N − j) g (R2 ) R2 N2 N2 (n) g (s) s N −1
L ∞ ([R1 ,R2 ])
(n + 1)!
R2 − R1 N
n+1
j n+1 + (N − j)n+1 ,
(5.80)
for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (5.80) turns to N π 2 N −1 N −1 ≤ f (y) dy − (R2 − R1 ) g (R1 ) R1 + g (R2 ) R2 A N2 N 2
π N2
(n) g (s) s N −1
L ∞ ([R1 ,R2 ])
(n + 1)!
(R2 − R1 )n+1 , 2n−1
(5.81)
(vii) when n = 1 (without any boundary conditions), we get from (5.81) that N π 2 N −1 N −1 ≤ f (y) dy − (R2 − R1 ) g (R1 ) R1 + g (R2 ) R2 A N2
5.2 Main Results
85 N π2 (R2 − R1 )2 N g (s) s N −1 , L ∞ ([R1 ,R2 ]) 2 2
(5.82)
which is related to (5.12). We present (use of Theorem 5.19). Theorem 5.22 Consider f : A → R which is radial; that is, there exists g such that f (x) = g (r ), r = |x|, r ∈ [R1 , R2 ], ∀ x ∈ A; x = r ω, ω ∈ S N −1 , N ≥ 2. We assume that g (s) s N −1 ∈ AC n ([R1 , R2 ]), n ∈ N. Then (i) % n−1 ! $ (k) 1 g (s) s N −1 (R1 ) (t − R1 )k+1 + f (y) dy − A (k + 1)! k=0
"& 2π N2 N −1 (k) k+1 ≤ (−1) g (s) s (R2 ) (R2 − t) N2 k
N 2
2π N2
(n) g (s) s N −1
L 1 ([R1 ,R2 ])
n!
(t − R1 )n + (R2 − t)n ,
(5.83)
∀ t ∈ [R1 , R2 ] , 2 , the right hand side of (5.83) is minimized, and we get: (ii) at t = R1 +R 2 % n−1 $ (k) 1 (R2 − R1 )k+1 ! g (s) s N −1 (R1 ) + f (y) dy − A 2k (k + 1)! k=0
(−1) g (s) s k
N 2
π N2
N −1 (k)
(n) g (s) s N −1 n!
"& π N2 ≤ (R2 ) N2
L 1 ([R1 ,R2 ])
(R2 − R1 )n , 2n−2
(5.84)
(k) (k) (iii) if g (s) s N −1 (R1 ) = g (s) s N −1 (R2 ) = 0, for all k = 0, 1, . . . , n − 1, we obtain N −1 (n) N g s (s) n 2 L 1 ([R1 ,R2 ]) (R2 − R1 ) f (y) dy ≤ π , (5.85) N n! 2n−2 A 2 which is a sharp inequality,
86
5 Multivariate Iyengar Inequalities for Radial Functions
(iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds % n−1 $ (k) 1 R2 − R1 k+1 ! k+1 j g (s) s N −1 (R1 ) + f (y) dy − A N (k + 1)! k=0
(−1) (N − j) k
N 2
2π N2
(n) g (s) s N −1
k+1
"& 2π N2 N −1 (k) ≤ g (s) s (R2 ) N2
L 1 ([R1 ,R2 ])
n!
R2 − R1 N
n
j n + (N − j)n ,
(k) (k) (v) if g (s) s N −1 (R1 ) = g (s) s N −1 (R2 ) = 0, from (5.86) we get:
(5.86)
k = 1, . . . , n − 1,
' R2 − R1 f (y) dy − jg (R1 ) R1N −1 + N A
N N 2 2π 2 2π N −1 ≤ × (N − j) g (R2 ) R2 N2 N2 (n) g (s) s N −1 n!
L 1 ([R1 ,R2 ])
R2 − R1 N
n
j n + (N − j)n ,
(5.87)
for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (5.87) turns to N π 2 N −1 N −1 ≤ f (y) dy − (R2 − R1 ) g (R1 ) R1 + g (R2 ) R2 A N2 N 2
π N2
(n) g (s) s N −1 n!
L 1 ([R1 ,R2 ])
(R2 − R1 )n , 2n−2
(5.88)
(vii) when n = 1 (without any boundary conditions), we get from (5.88) that N π 2 N −1 N −1 ≤ f (y) dy − (R2 − R1 ) g (R1 ) R1 + g (R2 ) R2 A N2 N 2π 2 N g (s) s N −1 (R2 − R1 ) . L 1 ([R1 ,R2 ]) 2
(5.89)
5.2 Main Results
87
We present (use of Theorem 5.20). Theorem 5.23 Consider f : A → R which is radial; that is, there exists g such that f (x) = g (r ), r = |x|, r ∈ [R1 , R2 ], ∀ x ∈ A; x = r ω, ω ∈ S N −1 , N ≥ 2. We (n) ∈ L q ([R1 , R2 ]), where assume that g (s) s N −1 ∈ AC n ([R1 , R2 ]) and g (s) s N −1 p, q > 1 : 1p + q1 = 1, and n ∈ N. Then (i) % n−1 ! $ (k) 1 g (s) s N −1 (R1 ) (t − R1 )k+1 + f (y) dy − A (k + 1)! k=0
"& 2π N2 N −1 (k) k+1 ≤ (−1) g (s) s (R2 ) (R2 − t) N2 k
(n) g (s) s N −1 ! " 1 1 2π L q ([R1 ,R2 ]) N (t − R1 )n+ p + (R2 − t)n+ p , (5.90) 1 2 (n − 1)! n + 1 ( p (n − 1) + 1) p p N 2
∀ t ∈ [R1 , R2 ] , 2 , the right hand side of (5.90) is minimized, and we get: (ii) at t = R1 +R 2 % n−1 $ (k) 1 (R2 − R1 )k+1 ! g (s) s N −1 (R1 ) + f (y) dy − A 2k (k + 1)! k=0
(−1) g (s) s k
N −1 (k)
"& π N2 ≤ (R2 ) N2
(n) 1 g (s) s N −1 π (R2 − R1 )n+ p L q ([R1 ,R2 ]) , 1 N2 (n − 1)! n + 1 ( p (n − 1) + 1) 1p 2n−1− q p N 2
(5.91)
(k) (k) (iii) if g (s) s N −1 (R1 ) = g (s) s N −1 (R2 ) = 0, for all k = 0, 1, . . . , n − 1, we obtain N −1 (n) 1 N g s (s) 2 (R2 − R1 )n+ p L q ([R1 ,R2 ]) f (y) dy ≤ π , 1 1 N 1 A 2n−1− q 2 (n − 1)! n + p ( p (n − 1) + 1) p (5.92) which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
88
5 Multivariate Iyengar Inequalities for Radial Functions
% n−1 $ (k) R2 − R1 k+1 ! k+1 1 j g (s) s N −1 (R1 ) + f (y) dy − A N (k + 1)! k=0
(−1) (N − j) k
k+1
"& 2π N2 N −1 (k) ≤ g (s) s (R2 ) N2
(n) g (s) s N −1 1 R2 − R1 n+ p 2π L q ([R1 ,R2 ]) n+ 1p n+ 1p j , + − j) (N 1 N N2 (n − 1)! n + 1 ( p (n − 1) + 1) p p N 2
(5.93) (k) (k) (v) if g (s) s N −1 (R1 ) = g (s) s N −1 (R2 ) = 0, from (5.93) we get:
k = 1, . . . , n − 1,
' R2 − R1 f (y) dy − jg (R1 ) R1N −1 + N A
(N −
j) g (R2 ) R2N −1
N N 2π 2 2π 2 N ≤ N × 2 2
(n) 1 g (s) s N −1 1 R2 − R1 n+ p n+ 1p L q ([R1 ,R2 ]) j + (N − j)n+ p , 1 N (n − 1)! n + 1p ( p (n − 1) + 1) p (5.94) for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (5.94) turns to N 2 π N ≤ f (y) dy − (R2 − R1 ) g (R1 ) R1N −1 + g (R2 ) R2N −1 A 2 (n) 1 g (s) s N −1 π (R2 − R1 )n+ p L q ([R1 ,R2 ]) N , n−1− q1 2 (n − 1)! n + 1 ( p (n − 1) + 1) 1p 2 p N 2
(5.95)
(vii) when n = 1 (without any boundary conditions), we get from (5.95) that N π 2 N −1 N −1 ≤ f (y) dy − (R2 − R1 ) g (R1 ) R1 + g (R2 ) R2 A N2
5.2 Main Results
89
g (s) s N −1 1 2 π L ([R ,R ]) q 1 2 (R2 − R1 )1+ p . N 1 2 1+ p 1 q
N 2
(5.96)
We continue with. Remark 5.24 Theorems 5.21–5.23 can easily be converted to results for the ball B (0, R), R > 0. Their corresponding same assumptions will be for f : B (0, R) → R which is radial. All we need to do then is set R1 = 0 and R2 = R, and we get a plethora of interesting similar results for the ball that are simpler. Due to lack of space we omit this tedious task.
References 1. R.P. Agarwal, S.S. Dragomir, An application of Hayashi’s inequality for differentiable functions. Comput. Math. Appl. 6, 95–99 (1996) 2. G.A. Anastassiou, Fractional Differentiation Inequalities. Research Monograph (Springer, New York, 2009) 3. G.A. Anastassiou, Multivariate Iyengar type inequalities for radial functions. Problemy Analiza - Issues of Analysis 8(26), 3–27 (2019), No. 2 4. G.A. Anastassiou, General Iyengar type inequalities. J. Comput. Anal. Appl. 28(5), 786–797 (2020) 5. Xiao-Liang Cheng, The Iyengar-type inequality. Appl. Math. Lett. 14, 975–978 (2001) 6. K.S.K. Iyengar, Note on an inequality. Math. Student 6, 75–76 (1938) 7. Z. Liu, Note on Iyengar’s inequality, Univ. Beograd Publ. Elektrotechn. Fak., Ser. Mat. 16, 29-35 (2005) 8. F. Qi, Further generalizations of inequalities for an integral. Univ. Beograd Publ. Elektrotechn. Fak. Ser. Mat. 8, 79–83 (1997) 9. W. Rudin, Real and Complex Analysis, International Student edn. (Mc Graw Hill, London, 1970) 10. D. Stroock, A Concise Introduction to the Theory of Integration, 3rd edn. (Birkhaüser, Boston, 1999)
Chapter 6
Multidimensional Fractional Iyengar Inequalities for Radial Functions
Here we derive a variety of multivariate fractional Iyengar type inequalities for radial functions defined on the shell and ball. Our approach is based on the polar coordinates in R N , N ≥ 2, and the related multivariate polar integration formula. Via this method we transfer author’s univariate fractional Iyengar type inequalities into multivariate fractional Iyengar inequalities. See also [6].
6.1 Background We are motivated by the following famous Iyengar inequality (1938), [11]. Theorem 6.1 Let f be a differentiable function on [a, b] and f (x) ≤ M. Then
a
b
M (b − a) 2 ( f (b) − f (a))2 1 . f (x) d x − (b − a) ( f (a) + f (b)) ≤ − 2 4 4M (6.1)
We need Definition 6.2 ([2], p. 394) Let ν > 0, n = ν (· the ceiling of the number), f ∈ AC n ([a, b]) (i.e. f (n−1) is absolutely continuous on [a, b]). The left Caputo fractional derivative of order ν is defined as x 1 ν D∗a f (x) = (6.2) (x − t)n−ν−1 f (n) (t) dt, (n − ν) a ∀ x ∈ [a, b], and it exists almost everywhere over [a, b] . © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_6
91
92
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
We need Definition 6.3 ([4], pp. 336–337) Let ν > 0, n = ν, f ∈ AC n ([a, b]). The right Caputo fractional derivative of order ν is defined as ν f (x) = Db−
(−1)n (n − ν)
b
(z − x)n−ν−1 f (n) (z) dz,
(6.3)
x
∀ x ∈ [a, b], and exists almost everywhere over [a, b] . In [7] we proved the following Caputo fractional Iyengar type inequalities: Theorem 6.4 ([7]) Let ν > 0, n = ν (· is the ceiling of the number), and f ∈ AC n ([a, b]) (i.e. f (n−1) is absolutely continuous on [a, b]). We assume that ν ν f, Db− f ∈ L ∞ ([a, b]). Then D∗a (i) n−1 b (k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
ν ν max D∗a f L ∞ ([a,b]) , Db− f L ∞ ([a,b]) (t − a)ν+1 + (b − t)ν+1 , (ν + 2) ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
(6.4)
the right hand side of (6.4) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f f (x) d x − f + (−1) (a) (b) ≤ k+1 a 2 (k + 1)! k=0
ν ν max D∗a f L ∞ ([a,b]) , Db− f L ∞ ([a,b]) (b − a)ν+1 (ν + 2)
2ν
,
(6.5)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
a
b
ν max Dν f D f ν+1 , ∗a b− L ∞ ([a,b]) L ∞ ([a,b]) (b − a) f (x) d x ≤ , (ν + 2) 2ν
(6.6)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds b k+1
n−1
b − a 1 j k+1 f (k) (a) + (−1)k (N − j)k+1 f (k) (b) f (x) d x − N (k + 1)! a k=0
6.1 Background
≤
93
ν ν max D∗a f L ∞ ([a,b]) , Db− f L ∞ ([a,b]) b − a ν+1 (ν + 2)
N
j ν+1 + (N − j)ν+1 ,
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (6.7) we get:
b
f (x) d x −
a
b−a N
(6.7)
[ j f (a) + (N − j) f (b)] ≤
ν ν max D∗a f L ∞ ([a,b]) , Db− f L ∞ ([a,b]) b − a ν+1 (ν + 2)
N
j ν+1 + (N − j)ν+1 , (6.8)
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (6.8) turns to
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
ν ν max D∗a f L ∞ ([a,b]) , Db− f L ∞ ([a,b]) (b − a)ν+1 (ν + 2)
2ν
,
(6.9)
(vii) when 0 < ν ≤ 1, inequality (6.9) is again valid without any boundary conditions. We mention Theorem 6.5 ([7]) Let ν ≥ 1, n = ν, and f ∈ AC n ([a, b]). We assume that ν ν f, Db− f ∈ L 1 ([a, b]). Then D∗a (i) n−1 b (k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
ν ν max D∗a f L 1 ([a,b]) , Db− f L 1 ([a,b]) (t − a)ν + (b − t)ν , (ν + 1)
(6.10)
∀ t ∈ [a, b] , (ii) when ν = 1, from (6.10), we have
a
b
f (x) d x − [ f (a) (t − a) + f (b) (b − t)] ≤ f
L 1 ([a,b])
(b − a) , ∀ t ∈ [a, b] ,
(6.11)
94
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
(iii) from (6.11), we obtain (ν = 1 case)
b
f (x) d x −
a
(iv) at t =
a+b , 2
b−a 2
( f (a) + f (b)) ≤ f L 1 ([a,b]) (b − a) ,
(6.12)
ν > 1, the right hand side of (6.10) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
ν ν max D∗a f L 1 ([a,b]) , Db− f L 1 ([a,b]) (b − a)ν 2ν−1
(ν + 1)
,
(6.13)
(v) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1; ν > 1, from (6.13), we obtain ν max b Dν f D f ν , ∗a b− L 1 ([a,b]) L 1 ([a,b]) (b − a) f (x) d x ≤ , (6.14) (ν + 1) 2ν−1 a which is a sharp inequality, (vi) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds b k+1
n−1
b − a 1 j k+1 f (k) (a) + (−1)k (N − j)k+1 f (k) (b) f (x) d x − N (k + 1)! a k=0
≤
ν ν max D∗a f L 1 ([a,b]) , Db− f L 1 ([a,b]) b − a ν (ν + 1)
N
j ν + (N − j)ν , (6.15)
(vii) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (6.15) we get:
b
f (x) d x −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
ν ν max D∗a f L 1 ([a,b]) , Db− f L 1 ([a,b]) b − a ν (ν + 1)
N
j ν + (N − j)ν ,
j = 0, 1, 2, . . . , N , (viii) when N = 2 and j = 1, (6.16) turns to
a
b
(b − a) f (x) d x − ( f (a) + f (b)) ≤ 2
(6.16)
6.1 Background
95
ν ν max D∗a f L 1 ([a,b]) , Db− f L 1 ([a,b]) (b − a)ν 2ν−1
(ν + 1)
.
(6.17)
We mention Theorem 6.6 ([7]) Let p, q > 1 : 1p + ν ν f, Db− f ∈ L q ([a, b]). Then with D∗a (i)
1 q
= 1, ν > q1 , n = ν ; f ∈ AC n ([a, b]),
n−1 b (k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
ν ν
max D∗a f L q ([a,b]) , Db− f L q ([a,b]) 1 1 (t − a)ν+ p + (b − t)ν+ p , 1 (ν) ν + 1p ( p (ν − 1) + 1) p ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
(6.18)
the right hand side of (6.18) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
ν ν max D∗a f L q ([a,b]) , Db− f L q ([a,b]) (b − a)ν+ 1p , 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p
(6.19)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
a
b
ν max Dν f , Db− f L q ([a,b]) (b − a)ν+ 1p ∗a L q ([a,b]) f (x) d x ≤ , 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p
(6.20)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds b k+1
n−1
b − a 1 k+1 k k+1 (k) (k) j f (x) d x − f (a) + (−1) (N − j) f (b) N (k + 1)! a k=0
ν ν
max D∗a f L q ([a,b]) , Db− f L q ([a,b]) b − a ν+ 1p 1 1 j ν+ p + (N − j)ν+ p , ≤ 1 N (ν) ν + 1p ( p (ν − 1) + 1) p (6.21)
96
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (6.21) we get:
b
f (x) d x −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
ν ν
max D∗a f L q ([a,b]) , Db− f L q ([a,b]) b − a ν+ 1p ν+ 1p ν+ 1p j , + − j) (N 1 N (ν) ν + 1p ( p (ν − 1) + 1) p (6.22) for j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (6.22) turns to
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
ν ν max D∗a f L q ([a,b]) , Db− f L q ([a,b]) (b − a)ν+ 1p , 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p
(6.23)
(vii) when 1/q < ν ≤ 1, inequality (6.23) is again valid but without any boundary conditions. We need the following different fractional calculus background: Let α > 0, m = [α] ([·] is the integral part), β = α − m, 0 < β < 1, f ∈ C ∞ ([a, b]), [a, b] ⊂ R, x ∈ [a, b]. The gamma function is given by (α) = 0 e−t α−1 t dt. We define the left Riemann–Liouville integral ([2], p. 24)
Jαa+ f (x) =
1 (α)
x
(x − t)α−1 f (t) dt,
(6.24)
a
α a ≤ x ≤ b. We define the subspace Ca+ ([a, b]) of C m ([a, b]) : α Ca+ ([a, b]) =
a+ (m) f ∈ C 1 ([a, b]) . f ∈ C m ([a, b]) : J1−β
(6.25)
α For f ∈ Ca+ ([a, b]), we define the left generalized α-fractional derivative of f over [a, b] as
a+ (m) α f := J1−β f , Da+
see [2], p. 24. Canavati first in [10] introduced the above over [0, 1] . n f = f (n) ; n ∈ N. We have that Da+ α Notice that Da+ f ∈ C ([a, b]) . Furthermore we need:
(6.26)
6.1 Background
97
Let again α > 0, m = [α], β = α − m, f ∈ C ([a, b]), call the right Riemann– Liouville fractional integral operator by
α Jb−
1 f (x) := (α)
b
(t − x)α−1 f (t) dt,
(6.27)
x
x ∈ [a, b], see [3]. Define the subspace of functions α Cb− ([a, b]) :=
1−β f ∈ C m ([a, b]) : Jb− f (m) ∈ C 1 ([a, b]) .
(6.28)
Define the right generalized α-fractional derivative of f over [a, b] as α 1−β D b− f = (−1)m−1 Jb− f (m) , 0
(6.29)
n
see [3]. We set D b− f = f . We have D b− f = (−1)n f (n) ; n ∈ N. Notice that α D b− f ∈ C ([a, b]) . We mention the following Canavati fractional Iyengar type inequalities: ν ν Theorem 6.7 ([9]) Let ν ≥ 1, n = [ν] and f ∈ Ca+ ([a, b]) ∩ Cb− ([a, b]). Then (i)
n−1 b (k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
ν ν max Da+ f ∞,([a,b]) , D b− f
∞,([a,b])
(ν + 2) ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
(t − a)ν+1 + (b − t)ν+1 ,
(6.30)
the right hand side of (6.30) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f f (x) d x − f + (−1) (a) (b) ≤ k+1 a 2 (k + 1)! k=0
ν ν max Da+ f ∞,([a,b]) , D b− f (ν + 2)
∞,([a,b])
(b − a)ν+1 , 2ν
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
(6.31)
98
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
a
b
ν ν max Da+ f ∞,([a,b]) , D b− f ν+1 ∞,([a,b]) (b − a) f (x) d x ≤ , (6.32) (ν + 2) 2ν
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds b k+1
n−1
1 b−a k+1 f (k) (a) + (−1)k (N − j)k+1 f (k) (b) j f d x − (x) N (k + 1)! a k=0
≤
ν ν max Da+ f ∞,([a,b]) , D b− f
∞,([a,b])
(ν + 2)
b−a N
ν+1
j ν+1 + (N − j)ν+1 ,
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (6.33) we get:
b
f (x) d x −
a
b−a N
ν ν max Da+ f ∞,([a,b]) , D b− f
(6.33)
[ j f (a) + (N − j) f (b)] ≤
∞,([a,b])
(ν + 2)
b−a N
ν+1
j ν+1 + (N − j)ν+1 , (6.34)
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (6.34) turns to
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
ν ν max Da+ f ∞,([a,b]) , D b− f
∞,([a,b])
(ν + 2)
(b − a)ν+1 . 2ν
(6.35)
We mention ν ν Theorem 6.8 ([9]) Let ν ≥ 1, n = [ν], and f ∈ Ca+ ([a, b]) ∩ Cb− ([a, b]). Then (i)
n−1 b (k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
ν ν max Da+ f L 1 ([a,b]) , D b− f (ν + 1)
L 1 ([a,b])
(t − a)ν + (b − t)ν ,
(6.36)
6.1 Background
99
∀ t ∈ [a, b] , (ii) when ν = 1, from (6.36), we have
b
a
f (x) d x − [ f (a) (t − a) + f (b) (b − t)] ≤ f
L 1 ([a,b])
(b − a) , ∀ t ∈ [a, b] ,
(6.37)
(iii) from (6.37), we obtain (ν = 1 case)
b
f (x) d x −
a
(iv) at t =
a+b , 2
b−a 2
( f (a) + f (b)) ≤ f L 1 ([a,b]) (b − a) ,
(6.38)
ν > 1, the right hand side of (6.36) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f (x) d x − f f + (−1) (a) (b) ≤ a 2k+1 (k + 1)! k=0
ν ν max Da+ f L 1 ([a,b]) , D b− f
L 1 ([a,b])
(ν + 1)
(b − a)ν , 2ν−1
(6.39)
(v) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1; ν > 1, from (6.39), we obtain ν ν b max Da+ f L 1 ([a,b]) , D b− f ν L 1 ([a,b]) (b − a) ≤ f d x , (6.40) (x) (ν + 1) 2ν−1 a which is a sharp inequality, (vi) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds b k+1
n−1
b − a 1 k+1 k k+1 (k) (k) j f (x) d x − f (a) + (−1) (N − j) f (b) N (k + 1)! a k=0
≤
ν ν max Da+ f L 1 ([a,b]) , D b− f (ν + 1)
L 1 ([a,b])
b−a N
ν
j ν + (N − j)ν ,
(vii) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (6.41) we get:
a
b
f (x) d x −
b−a N
[ j f (a) + (N − j) f (b)] ≤
(6.41)
100
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
ν ν max Da+ f L 1 ([a,b]) , D b− f
L 1 ([a,b])
(ν + 1)
b−a N
ν
j ν + (N − j)ν ,
(6.42)
j = 0, 1, 2, . . . , N , (viii) when N = 2 and j = 1, (6.42) turns to
a
b
(b − a) f (x) d x − ( f (a) + f (b)) ≤ 2
ν ν max Da+ f L 1 ([a,b]) , D b− f (ν + 1)
L 1 ([a,b])
(b − a)ν , 2ν−1
(6.43)
We mention Theorem 6.9 ([9]) Let p, q > 1 : ν Cb− ([a, b]). Then (i)
1 p
+
1 q
ν = 1, ν ≥ 1, n = [ν] ; f ∈ Ca+ ([a, b]) ∩
n−1 b (k) 1 f (a) (t − a)k+1 + (−1)k f (k) (b) (b − t)k+1 ≤ f (x) d x − a (k + 1)! k=0
ν ν max Da+ f L q ([a,b]) , D b− f
1 1 L q ([a,b]) (t − a)ν+ p + (b − t)ν+ p , 1 (ν) ν + 1p ( p (ν − 1) + 1) p ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
(6.44)
the right hand side of (6.44) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
ν ν max Da+ f L q ([a,b]) , D b− f ν+ 1p L q ([a,b]) (b − a) , 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p
(6.45)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
a
b
ν ν max Da+ f L q ([a,b]) , D b− f ν+ 1p L q ([a,b]) (b − a) f (x) d x ≤ , (6.46) 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p
6.1 Background
101
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds b k+1
n−1
b − a 1 k+1 k k+1 (k) (k) j f (x) d x − f (a) + (−1) (N − j) f (b) N (k + 1)! a k=0
ν ν
1 max Da+ f L q ([a,b]) , D b− f
b − a ν+ p ν+ 1p L q ([a,b]) ν+ 1p j , ≤ + − j) (N 1 N (ν) ν + 1p ( p (ν − 1) + 1) p (6.47) (v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (6.47) we get:
b
f (x) d x −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
ν ν
1 max Da+ f L q ([a,b]) , D b− f
1 b − a ν+ p ν+ 1p L q ([a,b]) j + (N − j)ν+ p , 1 N (ν) ν + 1p ( p (ν − 1) + 1) p (6.48) for j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (6.48) turns to
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
ν ν max Da+ f L q ([a,b]) , D b− f ν+ 1p L q ([a,b]) (b − a) . 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p
(6.49)
We need Definition 6.10 ([1]) Let a, b ∈ R. The left conformable fractional derivative starting from a of a function f : [a, ∞) → R of order 0 < α ≤ 1 is defined by
Tαa
f t + ε (t − a)1−α − f (t) . f (t) = lim ε→0 ε
(6.50)
If Tαa f (t) exists on (a, b), then
Tαa f (a) = lim Tαa f (t) . t→a+
(6.51)
102
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
The right conformable fractional derivative of order 0 < α ≤ 1 terminating at b of f : (−∞, b] → R is defined by f t + ε (b − t)1−α − f (t) . α T f (t) = −lim ε→0 ε
b If
b
αT
(6.52)
f (t) exists on (a, b), then b
αT
f (b) = lim bα T f (t) . t→b−
(6.53)
Note that if f is differentiable then and
Tαa f (t) = (t − a)1−α f (t) ,
b
αT
f (t) = − (b − t)1−α f (t) .
(6.54)
(6.55)
In the higher order case we can generalize things as follows: Definition 6.11 ([1]) Let α ∈ (n, n + 1], and set β = α − n. Then, the left conformable fractional derivative starting from a of a function f : [a, ∞) → R of order α, where f (n) (t) exists, is defined by a Tα f (t) = Tβa f (n) (t) ,
(6.56)
The right conformable fractional derivative of order α terminating at b of f : (−∞, b] → R, where f (n) (t) exists, is defined by b
αT
f (t) = (−1)n+1 bβ T f (n) (t) .
(6.57)
If α = n + 1 then β = 1 and Tan+1 f = f (n+1) . If n is odd, then bn+1 T f = − f (n+1) , and if n is even, then bn+1 T f = f (n+1) . When n = 0 (or α ∈ (0, 1]), then β = α, and (6.56), (6.57) collapse to (6.50), (6.52), respectively. We need Remark 6.12 ([5]) We notice the following: let α ∈ (n, n + 1] and f ∈ C n+1 ([a, b]), n ∈ N. Then (β := α − n, 0 < β ≤ 1) a Tα ( f ) (x) = Tβα f (n) (x) = (x − a)1−β f (n+1) (x) , and
b
αT (
f ) (x) = (−1)n+1 bβ T f (n) (x) =
(6.58)
6.1 Background
103
(−1)n+1 (−1) (b − x)1−β f (n+1) (x) = (−1)n (b − x)1−β f (n+1) (x) .
(6.59)
Consequently we get that a Tα ( f ) (x) , bα T ( f ) (x) ∈ C ([a, b]) . Furthermore it is obvious that a Tα ( f ) (a) = bα T ( f ) (b) = 0,
(6.60)
when 0 < β < 1, i.e. when α ∈ (n, n + 1) . We mention the following Conformable fractional Iyengar type inequalities: Theorem 6.13 ([8]) Let α ∈ (n, n + 1] and f ∈ C n+1 ([a, b]), n ∈ N; β = α − n. Then (i) n b (k) 1 f (a) (z − a)k+1 + (−1)k f (k) (b) (b − z)k+1 ≤ f (t) dt − a (k + 1)! k=0
(β) max Taα ( f )∞,[a,b] , bα T ( f )∞,[a,b] (z − a)α+1 + (b − z)α+1 , (α + 2) (6.61) ∀ z ∈ [a, b] , , the right hand side of (6.61) is minimized, and we get: (ii) at z = a+b 2 n b 1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (t) dt − a 2k+1 (k + 1)! k=0
(β) max Taα ( f )∞,[a,b] , bα T ( f )∞,[a,b] (b − a)α+1 (α + 2)
2α
,
(6.62)
(iii) assuming f (k) (a) = f (k) (b) = 0, for k = 0, 1, . . . , n, we obtain (β) max Ta ( f ) , bα T ( f )∞,[a,b] (b − a)α+1 α ∞,[a,b] f (t) dt ≤ , (α + 2) 2α a (6.63) which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
b
104
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
b k+1
n
b − a 1 k+1 k k+1 (k) (k) f (a) j f (t) dt − + (−1) f (b) (N − j) N (k + 1)! a k=0
≤
(β) max Taα ( f )∞,[a,b] , bα T ( f ) b − a α+1 ∞,[a,b] (α + 2)
N
j α+1 + (N − j)α+1 ,
(6.64)
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n, from (6.64) we get:
b
f (t) dt −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
(β) max Taα ( f )∞,[a,b] , bα T ( f )∞,[a,b] b − a α+1 (α + 2)
N
j α+1 + (N − j)α+1 , (6.65)
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (6.65) turns to
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
(β) max Taα ( f )∞,[a,b] , bα T ( f )∞,[a,b] (b − a)α+1 2α
(α + 2)
.
(6.66)
We mention L p conformable fractional Iyengar inequalities: Theorem 6.14 ([8]) Let α ∈ (n, n + 1] and f ∈ C n+1 ([a, b]), n ∈ N; β = α − n. Let also p1 , p2 , p3 > 1 : p11 + p12 + p13 = 1, with β > p11 + p13 . Then (i) n b (k) 1 k+1 k (k) k+1 f (a) (z − a) f (t) dt − + (−1) f (b) (b − z) ≤ a (k + 1)! k=0 max Taα ( f ) p3 ,[a,b] , bα T ( f ) p3 ,[a,b] 1 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 + ∀ z ∈ [a, b] , (ii) at z =
a+b , 2
(z − a)
α+ p1 + p1 1
2
+ (b − z)
α+ p1 + p1 1
2
1 p2
,
the right hand side of (6.67) is minimized, and we get:
(6.67)
6.1 Background
105
n b 1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (t) dt − a 2k+1 (k + 1)! k=0
max Taα ( f ) p3 ,[a,b] , bα T ( f ) p3 ,[a,b] 1 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 +
1 p2
(b − a) 2
α+ p1 + p1 1
α− p1
2
,
(6.68)
3
(iii) assuming f (k) (a) = f (k) (b) = 0, for k = 0, 1, . . . , n, we obtain max Taα ( f ) p ,[a,b] , bα T ( f ) α+ 1 + 1 3 (b − a) p1 p2 p3 ,[a,b] , 1 1 α− 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 + p12 2 p3
b f (t) dt ≤ a
(6.69) which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds b k+1
n
1 b−a (k) (a) j k+1 + (−1)k f (k) (b) (N − j)k+1 f f dt − (t) N (k + 1)! a k=0 max Taα ( f ) p ,[a,b] , bα T ( f ) 3 p3 ,[a,b] ≤ 1 1 p p n! ( p1 n + 1) 1 ( p2 (β − 1) + 1) 2 α + p11 + p12
b−a N
α+ p1 + p1 1 2
j
α+ p1 + p1 1
2
+ (N − j)
α+ p1 + p1 1
2
,
(6.70)
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n, from (6.70) we get:
b
f (t) dt −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
max Taα ( f ) p3 ,[a,b] , bα T ( f ) p3 ,[a,b] 1 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 +
b−a N
α+ p1 + p1 1 2
j
α+ p1 + p1 1
2
α+ p1 + p1
+ (N − j)
for j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (6.71) turns to
1 p2
1
2
,
(6.71)
106
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
max Taα ( f ) p3 ,[a,b] , bα T ( f ) p3 ,[a,b] 1 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 +
1 p2
(b − a) 2
α+ p1 + p1 1
α− p1
2
.
(6.72)
3
We need Remark 6.15 We define the ball B (0, R) = x ∈ R N : |x| < R ⊆ R N , N ≥ 2, R > 0, and the sphere S N −1 := x ∈ R N : |x| = 1 , where |·| is the Euclidean norm. Let dω be the element of surface measure on S N −1 and N 2π 2 dω = N ωN = 2 S N −1 is the area of S N −1 . For x ∈ R N − {0} we can write uniquely x = r ω, where r = |x| > 0 and N ω = rx ∈ S N −1 , |ω| = 1. Note that B(0,R) dy = ω NNR is the Lebesgue measure on N
R the ball, that is the volume of B (0, R), which exactly is V ol (B (0, R)) = π 2N +1 . (2 ) Following [12, pp. 149–150, Exercise 6], and [13, pp. 87–88, Theorem 5.2.2] we can write for F : B (0, R) → R a Lebesgue integrable function that
F (x) d x = B(0,R)
S N −1
R
F (r ω) r
N −1
N
dr dω,
(6.73)
0
and we use this formula a lot. Typically here the function f : B (0, R) → R is radial; that is, there exists a function g such that f (x) = g (r ), where r = |x|, r ∈ [0, R], ∀ x ∈ B (0, R). We need Remark 6.16 Let the spherical shell A := B (0, R2 ) − B (0, R1 ), 0 < R1 < R2 , A ⊆ R N , N ≥ 2, x ∈ A. Consider that f : A → R is radial; that is, there exists g such that f (x) = g (r ), r = |x|, r ∈ [R1 , R2 ], ∀ x ∈ A. Here x can be written uniquely as x = r ω, where r = |x| > 0 and ω = rx ∈ S N −1 , |ω| = 1, see ([12], pp. 149–150 and [2], p. 421), furthermore for F : A → R a Lebesgue integrable function we have that
R2 F (x) d x = F (r ω) r N −1 dr dω. (6.74) A
S N −1
R1
6.1 Background
Here
107
N ω N R2N − R1N π 2 R2N − R1N . V ol (A) = = N N2 + 1
(6.75)
In this chapter we derive multivariate fractional Iyengar type inequalities on the shell and ball of R N , N ≥ 2, for radial function. Our following results are based on the presented background results.
6.2 Main Results In the rest of this chapter we consider the functions: (i) f : A → R which is radial, i.e. there exists g such that f (x) = g (r ), r = |x|, r ∈ [R1 , R2 ], ∀ x ∈ A; where A is the spherical shell A := B (0, R2 ) − B (0, R1 ), 0 < R1 < R2 , A ⊆ R N , N ≥ 2, also. (ii) f : B (0, R) → R which is radial, i.e. there exists g such that f (x) = g (r ), where r = |x|, r ∈ [0, R], ∀ x ∈ B (0, R); where B (0, R) is the ball, B (0, R) ⊆ R N , N ≥ 2, R > 0. We will employ the related function h (s) := g (s) s N −1 , where s ∈ [R1 , R2 ] or s ∈ [0, R] . We present the following multivariate Caputo fractional Iyengar type inequalities: Theorem 6.17 Let the radial f : A → R. Let ν > 0, n = ν, and h ∈ AC n ν h, ([R1 , R2 ]) (i.e. h (n−1) is absolutely continuous on [R1 , R2 ]). We assume that D∗R 1 ν D R2 − h ∈ L ∞ ([R1 , R2 ]). Then (i) n−1 (k) 1 h (R1 ) (t − R1 )k+1 + f (y) dy − A (k + 1)! k=0
(−1) h k
N 2
2π N2
(k)
(R2 ) (R2 − t)
ν h L max D∗R 1
k+1
∞ ([R1 ,R2
N 2π 2 ≤ N2
, D νR2 − h L ])
(ν + 2)
∞ ([R1 ,R2 ])
·
(t − R1 )ν+1 + (R2 − t)ν+1 ,
∀ t ∈ [R1 , R2 ] , 2 , the right hand side of (6.76) is minimized, and we get: (ii) at t = R1 +R 2
(6.76)
108
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
n−1 1 (R2 − R1 )k+1 f (y) dy − A 2k (k + 1)! k=0
N 2
π N2
h
(k)
(R1 ) + (−1) h k
ν max D∗R h L 1
∞ ([R1 ,R2
(k)
(R2 )
, D νR2 − h L ])
N π 2 ≤ N2 (R2 − R1 )ν+1 , 2ν−1
∞ ([R1 ,R2 ])
(ν + 2)
(6.77)
(iii) if h (k) (R1 ) = h (k) (R2 ) = 0, for all k = 0, 1, . . . , n − 1, we obtain N 2 f (y) dy ≤ π · N A 2 ν h L max D∗R 1
∞ ([R1 ,R2 ])
, D νR2 − h L
∞ ([R1 ,R2 ])
(R − R )ν+1 2 1 , (ν + 2) 2ν−1
(6.78)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1
1 R2 − R1 k+1 f (y) dy − A N (k + 1)! k=0
j
k+1 (k)
h
(R1 ) + (−1) (N − j) k
2π 2 max N2 N
Dν h ∗R1 L
R2 − R1 N
k+1
∞ ([R1 ,R2 ])
h
(k)
(R2 )
, D νR2 − h L
N 2π 2 ≤ N2
∞ ([R1 ,R2 ])
(ν + 2) ν+1
j ν+1 + (N − j)ν+1 ,
(v) if h (k) (R1 ) = h (k) (R2 ) = 0, k = 1, . . . , n − 1, from (6.79) we get:
N R2 − R1 2π 2 [ j h (R1 ) + (N − j) h (R2 )] N f (y) dy − A N 2
(6.79)
6.2 Main Results
109
2π 2 max ≤ N 2 N
Dν h ∗R1 L
∞ ([R1 ,R2
ν D h , − R 2 L ])
∞ ([R1 ,R2 ])
(ν + 2)
R2 − R1 N
ν+1
j ν+1 + (N − j)ν+1 ,
(6.80)
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (6.80) turns to N π 2 f (y) dy − (R2 − R1 ) (h (R1 ) + h (R2 )) N ≤ A 2 N 2
π N2
ν h L max D∗R 1
∞ ([R1 ,R2
, D νR2 − h L ])
∞ ([R1 ,R2 ])
(ν + 2)
(R2 − R1 )ν+1 , 2ν−1
(6.81)
(vii) when 0 < ν ≤ 1, inequality (6.81) is again valid without any boundary conditions. Proof By Theorem 6.4 and (6.74). See in the Appendix 6.3 the general proving method in this chapter. We give Corollary 6.18 (to Theorem 6.17) Let the radial f : B (0, R) → R. Let ν > 0, ν h, D νR− h ∈ L ∞ ([0, R]). Then n = ν, and h ∈ AC n ([0, R]). We assume that D∗0 (i) n−1 (k) 1 h (0) t k+1 + f (y) dy − B(0,R) + 1)! (k k=0 (−1)k h (k) (R) (R − t)k+1
N 2
2π N2
N 2π 2 ≤ N2
ν max D∗0 h L ∞ ([0,R]) , D νR− h L ∞ ([0,R]) (ν + 2)
·
ν+1 t + (R − t)ν+1 , ∀ t ∈ [0, R] , (ii) at t = R2 , the right hand side of (6.82) is minimized, and we get: n−1 1 R k+1 f (y) dy − B(0,R) (k + 1)! 2k k=0
(6.82)
110
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
h
(k)
(0) + (−1) h k
(k)
(R)
N π 2 ≤ N2
ν Dν h D h ν+1 max , ∗0 R− π L ∞ ([0,R]) L ∞ ([0,R]) R N , ν−1 (ν + 2) 2 2 N 2
(6.83)
(iii) if h (k) (0) = h (k) (R) = 0, for all k = 0, 1, . . . , n − 1, we obtain
B(0,R)
N π2 f (y) dy ≤ N · 2
ν max D∗0 h L ∞ ([0,R]) , D νR− h L ∞ ([0,R])
R ν+1 , (ν + 2) 2ν−1
(6.84)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1
k+1 1 R f (y) dy − B(0,R) (k + 1)! N k=0
N 2 2π N ≤ j k+1 h (k) (0) + (−1)k (N − j)k+1 h (k) (R) 2 N 2
2π N2
ν max D∗0 h L ∞ ([0,R]) , D νR− h L ∞ ([0,R]) (ν + 2)
R N
ν+1
j ν+1 + (N − j)ν+1 ,
(6.85)
(v) if h (k) (0) = h (k) (R) = 0, k = 1, . . . , n − 1, from (6.85) we get:
N R 2π 2 f (y) dy − (N − j) h (R) N B(0,R) N 2
≤
N 2
2π N2
ν max D∗0 h L ∞ ([0,R]) , D νR− h L ∞ ([0,R]) (ν + 2)
R N
ν+1
j ν+1 + (N − j)ν+1 ,
(6.86)
6.2 Main Results
111
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (6.86) turns to N π 2 f (y) dy − Rh (R) N ≤ B(0,R) 2 π 2 max N2 N
Dν h ∗0
, D νR− h L ∞ ([0,R]) R ν+1 , (ν + 2) 2ν−1
L ∞ ([0,R])
(6.87)
(vii) when 0 < ν ≤ 1, inequality (6.87) is again valid without any boundary conditions. Proof Based on Theorem 6.17, just set there R1 = 0, R2 = R, the assumptions now are on B (0, R), and use (6.73). We continue with Theorem 6.19 Let the radial f : A → R. Let ν ≥ 1, n = ν, and h∈ AC n ([R1 , R2 ]) ν h, D νR2 − h ∈ (i.e. h (n−1) is absolutely continuous on [R1 , R2 ]). We assume that D∗R 1 L 1 ([R1 , R2 ]). Then (i) n−1 (k) 1 h (R1 ) (t − R1 )k+1 + f (y) dy − A + 1)! (k k=0 (−1) h k
2π 2 max N2 N
(k)
(R2 ) (R2 − t)
Dν h ∗R1 L
1 ([R1 ,R2
k+1
N 2π 2 ≤ N2
ν D h , − R 2 L ])
(ν + 1)
1 ([R1 ,R2 ])
·
(t − R1 )ν + (R2 − t)ν ,
(6.88)
∀ t ∈ [R1 , R2 ] , (ii) when ν = 1, from (6.88), we have N 2π 2 f (y) dy − [h (R1 ) (t − R1 ) + h (R2 ) (R2 − t)] N ≤ A 2 2π 2 N h L ([R ,R ]) (R2 − R1 ) , 1 1 2 2 N
∀ t ∈ [R1 , R2 ] ,
(6.89)
112
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
(iii) from (6.89), we obtain (ν = 1 case) N π 2 f (y) dy − (R2 − R1 ) (h (R1 ) + h (R2 )) N ≤ A 2 2π 2 N h L ([R ,R ]) (R2 − R1 ) , 1 1 2 2 N
(iv) at t =
R1 +R2 , 2
(6.90)
ν > 1, the right hand side of (6.88) is minimized, and we get:
n−1 1 (R2 − R1 )k+1 f (y) dy − A 2k (k + 1)! k=0
π 2 max N2 N
h
(k)
(R1 ) + (−1) h
Dν h ∗R1 L
k
1 ([R1 ,R2 ])
(k)
(R2 )
, D νR2 − h L
N π 2 ≤ N2 (R2 − R1 )ν , 2ν−2
1 ([R1 ,R2 ])
(ν + 1)
(6.91)
(v) if h (k) (R1 ) = h (k) (R2 ) = 0, for all k = 0, 1, . . . , n − 1, ν > 1, from (6.91) we obtain N 2 f (y) dy ≤ π · N A 2 ν h L max D∗R 1
1 ([R1 ,R2
, D νR2 − h L ])
(R − R )ν 2 1 , 1 ([R1 ,R2 ]) (ν + 1) 2ν−2
which is a sharp inequality, (vi) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1
1 R2 − R1 k+1 f (y) dy − A N (k + 1)! k=0
j k+1 h (k) (R1 ) + (−1)k (N − j)k+1 h (k) (R2 )
N 2
2π N2
ν max D∗R h L 1
1 ([R1 ,R2
, D νR2 − h L ])
(ν + 1)
N 2π 2 ≤ N2
1 ([R1 ,R2 ])
(6.92)
6.2 Main Results
113
R2 − R1 N
ν
j ν + (N − j)ν ,
(6.93)
(vii) if h (k) (R1 ) = h (k) (R2 ) = 0, k = 1, . . . , n − 1, from (6.93) we get:
N R2 − R1 2π 2 [ j h (R1 ) + (N − j) h (R2 )] N f (y) dy − A N 2 2π 2 max ≤ N 2 N
Dν h ∗R1 L
1 ([R1 ,R2 ])
, D νR2 − h L
1 ([R1 ,R2 ])
(ν + 1) R2 − R1 N
ν
j ν + (N − j)ν ,
(6.94)
j = 0, 1, 2, . . . , N , (viii) when N = 2 and j = 1, (6.94) turns to N π 2 f (y) dy − (R2 − R1 ) (h (R1 ) + h (R2 )) N ≤ A 2 N 2
π N2
ν max D∗R h L 1
1 ([R1 ,R2
, D νR2 − h L ])
1 ([R1 ,R2 ])
(ν + 1)
(R2 − R1 )ν . 2ν−2
(6.95)
Proof By Theorem 6.5 and (6.74). See in the Appendix 6.3 the general proving method in this chapter. We give Corollary 6.20 (to Theorem 6.19) Let the radial f : B (0, R) → R. Let ν ≥ 1, ν h, D νR− h ∈ L 1 ([0, R]). Then n = ν, and h ∈ AC n ([0, R]). We assume that D∗0 (i) n−1 (k) 1 h (0) t k+1 + f (y) dy − B(0,R) + 1)! (k k=0 N 2 2π N ≤ (−1)k h (k) (R) (R − t)k+1 2 N 2
2π N2
ν max D∗0 h L 1 ([0,R]) , D νR− h L 1 ([0,R]) (ν + 1)
∀ t ∈ [0, R] ,
t ν + (R − t)ν ,
· (6.96)
114
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
(ii) when ν = 1, from (6.96), we have N 2π 2 f (y) dy − h (R) (R − t) N ≤ B(0,R) 2 2π 2 N h L ([0,R]) R, 1 2 N
(6.97)
∀ t ∈ [0, R] , (iii) from (6.97), we obtain (ν = 1 case) N π 2 f (y) dy − Rh (R) N ≤ B(0,R) 2 2π 2 N h L ([0,R]) R, 1 2 N
(iv) at t =
R , 2
(6.98)
ν > 1, the right hand side of (6.96) is minimized, and we get: n−1 1 R k+1 f (y) dy − B(0,R) (k + 1)! 2k k=0
h
(k)
π 2 max N2 N
(0) + (−1) h k
Dν h ∗0
(k)
(R)
N π 2 ≤ N2
, D νR− h L 1 ([0,R]) R ν , (ν + 1) 2ν−2
L 1 ([0,R])
(6.99)
(v) if h (k) (0) = h (k) (R) = 0, for all k = 0, 1, . . . , n − 1, from (6.99) we obtain N 2 ≤ π · f dy (y) N B(0,R) 2 ν max D∗0 h L 1 ([0,R]) , D νR− h L 1 ([0,R])
Rν , (ν + 1) 2ν−2
which is a sharp inequality. (vi) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1
k+1 R 1 f (y) dy − B(0,R) (k + 1)! N k=0
(6.100)
6.2 Main Results
115
j
k+1 (k)
h
(0) + (−1) (N − j) k
2π 2 max N2 N
k+1
Dν h
∗0
L 1 ([0,R])
h
(k)
N 2π 2 ≤ (R) N2
, D νR− h L 1 ([0,R])
(ν + 1) R N
ν
j ν + (N − j)ν ,
(6.101)
(vii) if h (k) (0) = h (k) (R) = 0, k = 1, . . . , n − 1, from (6.101) we get:
N R 2π 2 f (y) dy − (N − j) h (R) N B(0,R) N 2 N 2
2π N2
≤
ν h L 1 ([0,R]) , D νR− h L 1 ([0,R]) max D∗0 (ν + 1)
R N
ν
j ν + (N − j)ν ,
(6.102)
j = 0, 1, 2, . . . , N , (viii) when N = 2 and j = 1, (6.102) turns to N π 2 f (y) dy − Rh (R) N ≤ B(0,R) 2 ν Dν h D h max , ∗0 R− π Rν L 1 ([0,R]) L 1 ([0,R]) N . (ν + 1) 2ν−2 2 N 2
(6.103)
Proof Based on Theorem 6.19, just set there R1 = 0, R2 = R, the assumptions now are on B (0, R), and use (6.73). We continue with Theorem 6.21 Let the radial f : A → R, and p, q > 1 : 1p + q1 = 1, ν > q1 . Let n = ν, and h ∈ AC n ([R1 , R2 ]) (i.e. h (n−1) is absolutely continuous on [R1 , R2 ]). ν h, D νR2 − h ∈ L q ([R1 , R2 ]). Then We assume that D∗R 1 (i) n−1 (k) 1 h (R1 ) (t − R1 )k+1 + f (y) dy − A + 1)! (k k=0
116
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
(−1) h k
2π 2 max N2 N
(k)
(R2 ) (R2 − t)
k+1
N 2π 2 ≤ N2
Dν h , D νR2 − h L ([R ,R ]) ∗R1 L q ([R1 ,R2 ]) q 1 2 · 1 1 p (ν) ν + p ( p (ν − 1) + 1)
1 1 (t − R1 )ν+ p + (R2 − t)ν+ p ,
(6.104)
∀ t ∈ [R1 , R2 ] , 2 , the right hand side of (6.104) is minimized, and we get: (ii) at t = R1 +R 2 n−1 1 (R2 − R1 )k+1 f (y) dy − A 2k (k + 1)! k=0
h (k) (R1 ) + (−1)k h (k) (R2 )
N π 2 ≤ N2
ν Dν h D h ν+ 1p max , ∗R1 R2 − π L q ([R1 ,R2 ]) L q ([R1 ,R2 ]) (R2 − R1 ) , 1 1 N2 2ν−1− q (ν) ν + 1p ( p (ν − 1) + 1) p N 2
(6.105)
(iii) if h (k) (R1 ) = h (k) (R2 ) = 0, for all k = 0, 1, . . . , n − 1, we obtain f (y) dy ≤ A
ν h L max D∗R 1
q ([R1 ,R2
N
π2 N ν−1− 1 · q 2 2
ν D h , − R 2 L ])
q ([R1 ,R2 ])
(R2 − R1 )ν+ p , 1 (ν) ν + 1p ( p (ν − 1) + 1) p
·
1
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1
1 R2 − R1 k+1 f (y) dy − A N (k + 1)! k=0
(6.106)
6.2 Main Results
j
117
k+1 (k)
h
(R1 ) + (−1) (N − j) k
k+1
h
(k)
(R2 )
N 2π 2 ≤ N2
ν Dν h D h max , − ∗R R 2π 1 2 L ([R ,R ]) L q ([R1 ,R2 ]) q 1 2 1 1 N2 (ν) ν + p ( p (ν − 1) + 1) p N 2
R2 − R1 N
ν+ 1p
1 1 j ν+ p + (N − j)ν+ p ,
(6.107)
(v) if h (k) (R1 ) = h (k) (R2 ) = 0, k = 1, . . . , n − 1, from (6.107) we get:
N R2 − R1 2π 2 [ j h (R1 ) + (N − j) h (R2 )] N f (y) dy − A N 2 ν D h Dν h max , ∗R1 R2 − 2π L ([R ,R ]) L q ([R1 ,R2 ]) q 1 2 ≤ N 1 2 (ν) ν + 1p ( p (ν − 1) + 1) p N 2
R2 − R1 N
ν+ 1p
1 1 j ν+ p + (N − j)ν+ p ,
(6.108)
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (6.108) turns to N π 2 f (y) dy − (R2 − R1 ) (h (R1 ) + h (R2 )) N ≤ A 2 π 2 max N2 N
Dν h , D νR2 − h L ([R ,R ]) (R − R )ν+ 1p ∗R1 L q ([R1 ,R2 ]) 2 1 q 1 2 , 1 ν−1− q1 1 p 2 (ν) ν + p ( p (ν − 1) + 1)
(6.109)
(vii) when 1/q < ν ≤ 1, inequality (6.109) is again valid without any boundary conditions. Proof By Theorem 6.6 and (6.74). See in the Appendix 6.3 the general proving method in this chapter. We give Corollary 6.22 (to Theorem 6.21) Let the radial f : B (0, R) → R. Let ν > 0, n = ν, and h ∈ AC n ([0, R]); p, q > 1 : 1p + q1 = 1, ν > q1 . We assume that ν D∗0 h, D νR− h ∈ L q ([0, R]). Then
118
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
(i)
n−1 (k) 1 h (0) t k+1 + f (y) dy − B(0,R) (k + 1)! k=0
(−1) h k
(k)
(R) (R − t)
k+1
N 2π 2 ≤ N2
ν N ν 2π 2 max D∗0 h L q ([0,R]) , D R− h L q ([0,R]) · 1 N2 (ν) ν + 1 ( p (ν − 1) + 1) p
(6.110)
p
1
1 t ν+ p + (R − t)ν+ p , ∀ t ∈ [0, R] , (ii) at t = R2 , the right hand side of (6.110) is minimized, and we get: n−1 R k+1 1 f (y) dy − B(0,R) (k + 1)! 2k k=0
h
(k)
(0) + (−1) h k
(k)
(R)
N π 2 ≤ N2
ν 1 Dν h D h max , ∗0 R− π R ν+ p L q ([0,R]) L q ([0,R]) 1 , 1 N2 2ν−1− q (ν) ν + 1p ( p (ν − 1) + 1) p N 2
(6.111)
(iii) if h (k) (0) = h (k) (R) = 0, for all k = 0, 1, . . . , n − 1, we obtain
B(0,R)
f (y) dy ≤
N
π2 N ν−1− 1 · q 2 2
ν max D∗0 h L q ([0,R]) , D νR− h L q ([0,R])
R ν+ p , 1 (ν) ν + 1p ( p (ν − 1) + 1) p (6.112)
which is a sharp inequality. (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1
k+1 R 1 f (y) dy − B(0,R) (k + 1)! N k=0
1
6.2 Main Results
119
j
k+1 (k)
h
(0) + (−1) (N − j) k
k+1
h
(k)
N 2π 2 ≤ (R) N2
ν N ν 2π 2 max D∗0 h L q ([0,R]) , D R− h L q ([0,R]) 1 N2 (ν) ν + 1 ( p (ν − 1) + 1) p p
R N
ν+ 1p
1 1 j ν+ p + (N − j)ν+ p ,
(6.113)
(v) if h (k) (0) = h (k) (R) = 0, k = 1, . . . , n − 1, from (6.113) we get:
N R 2π 2 f (y) dy − (N − j) h (R) N B(0,R) N 2 ν Dν h D h max , ∗0 R− 2π L q ([0,R]) L q ([0,R]) ≤ N 1 1 2 (ν) ν + p ( p (ν − 1) + 1) p N 2
R N
ν+ 1p
1 1 j ν+ p + (N − j)ν+ p ,
(6.114)
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (6.114) turns to N π 2 f (y) dy − Rh (R) N ≤ B(0,R) 2 ν 1 D h Dν h , max ∗0 R− π R ν+ p L q ([0,R]) L q ([0,R]) 1 , 1 N2 2ν−1− q (ν) ν + 1p ( p (ν − 1) + 1) p N 2
(6.115)
(vii) when 1/q < ν ≤ 1, inequality (6.115) is again valid without any boundary conditions. Proof Based on Theorem 6.21, just set there R1 = 0, R2 = R, the assumptions now are on B (0, R), and use (6.73). We continue with multivariate Canavati type fractional Iyengar type inequalities: Theorem 6.23 Let the radial f : A → R. Let ν ≥ 1, n = [ν], and h ∈ C Rν 1 + ([R1 , R2 ]) ∩ C Rν 2 − ([R1 , R2 ]). Then
120
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
(i)
n−1 (k) 1 h (R1 ) (t − R1 )k+1 + f (y) dy − A (k + 1)! k=0
(−1) h k
N 2
2π N2
(k)
(R2 ) (R2 − t)
max D νR1 + h ∞,[R
1 ,R2
k+1
N 2π 2 ≤ N2
ν , h D R2 − ]
∞,[R1 ,R2 ]
(ν + 2)
·
(t − R1 )ν+1 + (R2 − t)ν+1 ,
(6.116)
∀ t ∈ [R1 , R2 ] , 2 , the right hand side of (6.116) is minimized, and we get: (ii) at t = R1 +R 2 n−1 1 (R2 − R1 )k+1 f (y) dy − A 2k (k + 1)! k=0
N
π2 N2
max D ν
h (k) (R1 ) + (−1)k h (k) (R2 )
R1 +
N π 2 ≤ N2
ν , D h − R 2 ∞,[R1 ,R2 ]
h
∞,[R1 ,R2 ]
(ν + 2)
(R2 − R1 )ν+1 , 2ν−1
(6.117)
(iii) if h (k) (R1 ) = h (k) (R2 ) = 0, for all k = 0, 1, . . . , n − 1, we obtain N 2 f (y) dy ≤ π · N A 2 max D νR1 + h ∞,[R
1 ,R2
ν , D h R2 − ]
∞,[R1 ,R2 ]
(R2 − R1 )ν+1 , (ν + 2) 2ν−1
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1
1 R2 − R1 k+1 f (y) dy − A N (k + 1)! k=0
(6.118)
6.2 Main Results
j
121
k+1 (k)
h
N 2
2π N2
(R1 ) + (−1) (N − j) k
max D νR1 + h ∞,[R
k+1
1 ,R2
h
(k)
(R2 )
N 2π 2 ≤ N2
ν , D R2 − h ]
∞,[R1 ,R2 ]
(ν + 2)
R2 − R1 N
ν+1
j ν+1 + (N − j)ν+1 ,
(6.119)
(v) if h (k) (R1 ) = h (k) (R2 ) = 0, k = 1, . . . , n − 1, from (6.119) we get:
N R2 − R1 2π 2 [ j h (R1 ) + (N − j) h (R2 )] N f (y) dy − A N 2 N
≤
2π 2 N2
max D ν
R1 +
ν , h D − R 2 ∞,[R1 ,R2 ]
h
∞,[R1 ,R2 ]
(ν + 2) R2 − R1 N
ν+1
j ν+1 + (N − j)ν+1 ,
(6.120)
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (6.120) turns to N π 2 f (y) dy − (R2 − R1 ) (h (R1 ) + h (R2 )) N ≤ A 2 N
π2 N2
max D ν
R1 +
ν , h D − R 2 ∞,[R1 ,R2 ]
h
∞,[R1 ,R2 ]
(ν + 2)
(R2 − R1 )ν+1 . 2ν−1
(6.121)
Proof By Theorem 6.7 and (6.74). See in the Appendix 6.3 the general proving method in this chapter. We give Corollary 6.24 (to Theorem 6.23) Let the radial f : B (0, R) → R. Let ν ≥ 1, ν ν n = [ν], and h ∈ C0+ ([0, R]) ∩ C R− ([0, R]). Then (i) n−1 (k) 1 h (0) t k+1 + f (y) dy − B(0,R) + 1)! (k k=0
122
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
(−1) h k
N 2
2π N2
(k)
(R) (R − t)
k+1
N 2π 2 ≤ N2
ν ν h ∞,[0,R] , D R− h max D0+
∞,[0,R]
(ν + 2)
·
ν+1 t + (R − t)ν+1 ,
(6.122)
∀ t ∈ [0, R] , (ii) at t = R2 , the right hand side of (6.122) is minimized, and we get: n−1 1 R k+1 f (y) dy − B(0,R) (k + 1)! 2k k=0
N
π2 N2
h
(k)
(0) + (−1) h k
(k)
(R)
N π 2 ≤ N2
ν ν h ∞,[0,R] , D R− h max D0+
∞,[0,R]
(ν + 2)
R ν+1 , 2ν−1
(6.123)
(iii) if h (k) (0) = h (k) (R) = 0, for all k = 0, 1, . . . , n − 1, we obtain
B(0,R)
N π2 f (y) dy ≤ N · 2
ν ν max D0+ h ∞,[0,R] , D R− h
∞,[0,R]
R ν+1 , (ν + 2) 2ν−1
which is a sharp inequality. (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1
k+1 1 R f (y) dy − B(0,R) (k + 1)! N k=0
j
k+1 (k)
h
(0) + (−1) (N − j) k
k+1
h
(k)
N 2π 2 ≤ (R) N2
(6.124)
6.2 Main Results
123
N 2
2π N2
ν ν h ∞,[0,R] , D R− h max D0+
∞,[0,R]
(ν + 2)
R N
ν+1
j ν+1 + (N − j)ν+1 ,
(6.125)
(v) if h (k) (0) = h (k) (R) = 0, k = 1, . . . , n − 1, from (6.125) we get:
N R 2π 2 f (y) dy − (N − j) h (R) N B(0,R) N 2 N
≤
2π 2 N2
ν ν h ∞,[0,R] , D R− h max D0+
∞,[0,R]
(ν + 2)
R N
ν+1
(6.126)
j ν+1 + (N − j)ν+1 ,
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (6.126) turns to N π 2 f (y) dy − Rh (R) N ≤ B(0,R) 2 N
π2 N2
ν ν h ∞,[0,R] , D R− h max D0+
∞,[0,R]
(ν + 2)
R ν+1 . 2ν−1
(6.127)
Proof Based on Theorem 6.23, just set there R1 = 0, R2 = R, the assumptions now are on B (0, R), and use (6.73). We continue with Theorem 6.25 Let the radial f : A → R. Let ν ≥ 1, n = [ν], and h ∈ C Rν 1 + ([R1 , R2 ]) ∩ C Rν 2 − ([R1 , R2 ]). Then (i) n−1 (k) 1 h (R1 ) (t − R1 )k+1 + f (y) dy − A (k + 1)! k=0
(−1) h k
(k)
(R2 ) (R2 − t)
k+1
N 2π 2 ≤ N2
124
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
N 2
2π N2
max D νR1 + h L
1 ([R1 ,R2
ν , D h R2 − ])
L 1 ([R1 ,R2 ])
(ν + 1)
·
(6.128)
(t − R1 )ν + (R2 − t)ν , ∀ t ∈ [R1 , R2 ] , (ii) when ν = 1, from (6.128), we have N 2π 2 f (y) dy − [h (R1 ) (t − R1 ) + h (R2 ) (R2 − t)] N ≤ A 2 N 2π 2 N h L ([R ,R ]) (R2 − R1 ) , 1 1 2 2
(6.129)
∀ t ∈ [R1 , R2 ] , (iii) from (6.129), we obtain (ν = 1 case) N π 2 f (y) dy − (R2 − R1 ) (h (R1 ) + h (R2 )) N ≤ A 2 2π 2 N h L ([R ,R ]) (R2 − R1 ) , 1 1 2 2 N
(iv) at t =
R1 +R2 , 2
(6.130)
ν > 1, the right hand side of (6.128) is minimized, and we get:
n−1 1 (R2 − R1 )k+1 f (y) dy − A 2k (k + 1)! k=0
N 2
π N2
h
(k)
(R1 ) + (−1) h
max D νR1 + h L
k
1 ([R1 ,R2
(k)
(R2 )
ν , D h R2 − ])
N π 2 ≤ N2
L 1 ([R1 ,R2 ])
(ν + 1)
(R2 − R1 )ν , 2ν−2
(6.131)
(v) if h (k) (R1 ) = h (k) (R2 ) = 0, for all k = 0, 1, . . . , n − 1, ν > 1, from (6.131) we obtain N 2 f (y) dy ≤ π · N A
2
6.2 Main Results
125
max D νR1 + h L
1 ([R1 ,R2
ν , D h R2 − ])
L 1 ([R1 ,R2 ])
(R2 − R1 )ν , (ν + 1) 2ν−2
(6.132)
which is a sharp inequality, (vi) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1
1 R2 − R1 k+1 f (y) dy − A N (k + 1)! k=0
j
k+1 (k)
h
N 2
2π N2
(R1 ) + (−1) (N − j) k
max D νR1 + h L
k+1
1 ([R1 ,R2
h
(k)
(R2 )
ν , D h R2 − ])
N 2π 2 ≤ N2
L 1 ([R1 ,R2 ])
(ν + 1)
R2 − R1 N
ν
j ν + (N − j)ν ,
(6.133)
(vii) if h (k) (R1 ) = h (k) (R2 ) = 0, k = 1, . . . , n − 1, from (6.133) we get:
N R2 − R1 2π 2 [ j h (R1 ) + (N − j) h (R2 )] N f (y) dy − A N 2
≤
N 2
2π N2
max D νR1 + h L
1 ([R1 ,R2
ν , D h R2 − ])
L 1 ([R1 ,R2 ])
(ν + 1)
R2 − R1 N
ν
j ν + (N − j)ν ,
(6.134)
j = 0, 1, 2, . . . , N , (viii) when N = 2 and j = 1, (6.134) turns to N π 2 f (y) dy − (R2 − R1 ) (h (R1 ) + h (R2 )) N ≤ A 2 N
π2 N2
max D ν
R1 +
h
ν , h D − R 2 L 1 ([R1 ,R2 ]) (ν + 1)
L 1 ([R1 ,R2 ])
(R2 − R1 )ν . 2ν−2
(6.135)
126
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
Proof By Theorem 6.8 and (6.74). See in the Appendix 6.3 the general proving method in this chapter. We give Corollary 6.26 (to Theorem 6.25) Let the radial f : B (0, R) → R. Let ν ≥ 1, ν ν n = [ν], and h ∈ C0+ ([0, R]) ∩ C R− ([0, R]). Then (i) n−1 (k) 1 h (0) t k+1 + f (y) dy − B(0,R) (k + 1)! k=0
(−1) h k
N 2
2π N2
(k)
(R) (R − t)
k+1
N 2π 2 ≤ N2
ν ν h L 1 ([0,R]) , D R− h max D0+
L 1 ([0,R])
(ν + 1)
t ν + (R − t)ν ,
· (6.136)
∀ t ∈ [0, R] , (ii) when ν = 1, from (6.136), we have N 2π 2 f (y) dy − h (R) (R − t) N ≤ B(0,R) 2 N 2π 2 N h L ([0,R]) R, 1 2
(6.137)
∀ t ∈ [0, R] , (iii) from (6.137), we obtain (ν = 1 case) N π 2 f (y) dy − Rh (R) N ≤ B(0,R) 2 2π 2 N h L ([0,R]) R, 1 2 N
(iv) at t =
R , 2
(6.138)
ν > 1, the right hand side of (6.136) is minimized, and we get: n−1 1 R k+1 f (y) dy − B(0,R) (k + 1)! 2k k=0
6.2 Main Results
127
N 2
π N2
h
(k)
(0) + (−1) h k
(k)
(R)
N π 2 ≤ N2
ν ν h L 1 ([0,R]) , D R− h max D0+
Rν , 2ν−2
L 1 ([0,R])
(ν + 1)
(6.139)
(v) if h (k) (0) = h (k) (R) = 0, for all k = 0, 1, . . . , n − 1, ν > 1, from (6.139) we obtain N 2 ≤ π · f dy (y) N B(0,R)
2
ν ν h L 1 ([0,R]) , D R− max D0+
L 1 ([0,R])
Rν , (ν + 1) 2ν−2
(6.140)
which is a sharp inequality. (vi) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1
k+1 R 1 f dy − (y) B(0,R) (k + 1)! N k=0
j k+1 h (k) (0) + (−1)k (N − j)k+1 h (k) (R)
N
2π 2 N2
ν ν h L 1 ([0,R]) , D R− h max D0+
N 2π 2 ≤ N2
L 1 ([0,R])
(ν + 1)
R N
ν
j ν + (N − j)ν ,
(6.141)
(vii) if h (k) (0) = h (k) (R) = 0, k = 1, . . . , n − 1, from (6.141) we get:
N R 2π 2 f (y) dy − (N − j) h (R) N B(0,R) N 2 N
≤
2π 2 N2
ν ν h L 1 ([0,R]) , D R− h max D0+ (ν + 1)
L 1 ([0,R])
128
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
R N
ν
j ν + (N − j)ν ,
(6.142)
j = 0, 1, 2, . . . , N , (viii) when N = 2 and j = 1, (6.142) turns to N π 2 f (y) dy − Rh (R) N ≤ B(0,R) 2
N
π2 N2
ν ν h L 1 ([0,R]) , D R− h max D0+
L 1 ([0,R])
(ν + 1)
Rν . 2ν−2
(6.143)
Proof Based on Theorem 6.25, just set there R1 = 0, R2 = R, the assumptions now are on B (0, R), and use (6.73). We continue with Theorem 6.27 Let the radial f : A → R. Let ν ≥ 1, n = [ν], and h ∈ C Rν 1 + ([R1 , R2 ]) ∩ C Rν 2 − ([R1 , R2 ]). Here p, q > 1 : 1p + q1 = 1. Then (i) n−1 (k) 1 h (R1 ) (t − R1 )k+1 + f (y) dy − A (k + 1)! k=0
(−1)k h (k) (R2 ) (R2 − t)k+1
N
2π 2 N2
N 2π 2 ≤ N2
ν , D R2 − h R1 + L q ([R1 ,R2 ]) L q ([R1 ,R2 ]) · 1 (ν) ν + 1p ( p (ν − 1) + 1) p
max D ν
h
1 1 (t − R1 )ν+ p + (R2 − t)ν+ p ,
(6.144)
∀ t ∈ [R1 , R2 ] , 2 , the right hand side of (6.144) is minimized, and we get: (ii) at t = R1 +R 2 n−1 1 (R2 − R1 )k+1 f (y) dy − A 2k (k + 1)! k=0
h (k) (R1 ) + (−1)k h (k) (R2 )
N π 2 ≤ N2
6.2 Main Results
129
ν max D νR1 + h L ([R ,R ]) , D R2 − h ν+ 1p q 1 2 π L q ([R1 ,R2 ]) (R2 − R1 ) , 1 1 N2 2ν−1− q (ν) ν + 1p ( p (ν − 1) + 1) p N 2
(6.145)
(iii) if h (k) (R1 ) = h (k) (R2 ) = 0, for all k = 0, 1, . . . , n − 1, we obtain f (y) dy ≤ A
max D νR1 + h L
q ([R1 ,R2
N
π2 N ν−1− 1 · q 2 2
ν , D R2 − h ])
L q ([R1 ,R2 ])
·
(R2 − R1 )ν+ p , 1 (ν) ν + 1p ( p (ν − 1) + 1) p 1
(6.146)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1
1 R2 − R1 k+1 f (y) dy − A N (k + 1)! k=0
j
k+1 (k)
h
(R1 ) + (−1) (N − j) k
k+1
h
(k)
(R2 )
N 2π 2 ≤ N2
ν max D νR1 + h L ([R ,R ]) , D R2 − h q 1 2 2π L q ([R1 ,R2 ]) N 1 1 2 (ν) ν + p ( p (ν − 1) + 1) p N 2
R2 − R1 N
ν+ 1p
1 1 j ν+ p + (N − j)ν+ p ,
(v) if h (k) (R1 ) = h (k) (R2 ) = 0, k = 1, . . . , n − 1, from (6.147) we get:
N R2 − R1 2π 2 [ j h (R1 ) + (N − j) h (R2 )] N f (y) dy − A N 2 ν max D νR1 + h L ([R ,R ]) , D R2 − h q 1 2 2π L q ([R1 ,R2 ]) ≤ N 1 1 2 (ν) ν + p ( p (ν − 1) + 1) p N 2
(6.147)
130
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
R2 − R1 N
ν+ 1p
1 1 j ν+ p + (N − j)ν+ p ,
(6.148)
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (6.148) turns to N π 2 f (y) dy − (R2 − R1 ) (h (R1 ) + h (R2 )) N ≤ A 2 ν ν h , h max D N D R2 − ν+ 1p R1 + L q ([R1 ,R2 ]) π2 L q ([R1 ,R2 ]) (R2 − R1 ) . 1 1 N2 2ν−1− q (ν) ν + 1p ( p (ν − 1) + 1) p
(6.149)
Proof By Theorem 6.9 and (6.74). See in the Appendix 6.3 the general proving method in this chapter. We give Corollary 6.28 (to Theorem 6.27) Let the radial f : B (0, R) → R. Let ν ≥ 1, ν ν n = [ν], and h ∈ C0+ ([0, R]) ∩ C R− ([0, R]). Here p, q > 1 : 1p + q1 = 1. Then (i) n−1 (k) 1 h (0) t k+1 + f (y) dy − B(0,R) + 1)! (k k=0 (−1)k h (k) (R) (R − t)k+1
N 2π 2 ≤ N2
ν ν max D0+ h L q ([0,R]) , D R− h N 2π 2 L q ([0,R]) N · 1 2 (ν) ν + 1p ( p (ν − 1) + 1) p 1
1 t ν+ p + (R − t)ν+ p , ∀ t ∈ [0, R] , (ii) at t = R2 , the right hand side of (6.150) is minimized, and we get: n−1 1 R k+1 f (y) dy − B(0,R) (k + 1)! 2k k=0
h (k) (0) + (−1)k h (k) (R)
N π 2 ≤ N2
(6.150)
6.2 Main Results
131
ν ν 1 h L q ([0,R]) , D R− h max D0+ π R ν+ p L q ([0,R]) 1 , 1 N2 2ν−1− q (ν) ν + 1p ( p (ν − 1) + 1) p N 2
(6.151)
(iii) if h (k) (0) = h (k) (R) = 0, for all k = 0, 1, . . . , n − 1, we obtain
B(0,R)
f (y) dy ≤
ν ν max D0+ h L q ([0,R]) , D R− h
N
π2 N ν−1− 1 · q 2 2
L q ([0,R])
R ν+ p , 1 (ν) ν + 1p ( p (ν − 1) + 1) p (6.152) 1
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n−1
k+1 1 R f (y) dy − B(0,R) (k + 1)! N k=0
j k+1 h (k) (0) + (−1)k (N − j)k+1 h (k) (R)
N 2π 2 ≤ N2
ν ν max D0+ h L q ([0,R]) , D R− h N 2π 2 L q ([0,R]) N 1 1 2 (ν) ν + p ( p (ν − 1) + 1) p
R N
ν+ 1p
1 1 j ν+ p + (N − j)ν+ p ,
(v) if h (k) (0) = h (k) (R) = 0, k = 1, . . . , n − 1, from (6.153) we get:
N R 2π 2 f (y) dy − (N − j) h (R) N B(0,R) N 2 ν ν h L q ([0,R]) , D R− h max D0+ 2π L q ([0,R]) ≤ N 1 1 2 (ν) ν + p ( p (ν − 1) + 1) p N 2
(6.153)
132
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
R N
ν+ 1p
1 1 j ν+ p + (N − j)ν+ p ,
(6.154)
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (6.154) turns to N π 2 f (y) dy − Rh (R) N ≤ B(0,R) 2 ν ν 1 h , h max D N 0+ L q ([0,R]) D R− L ([0,R]) π2 R ν+ p q 1 . 1 N2 2ν−1− q (ν) ν + 1p ( p (ν − 1) + 1) p
(6.155)
Proof Based on Theorem 6.27, just set there R1 = 0, R2 = R, the assumptions now are on B (0, R), and use (6.73). If g ∈ C n+1 ([R1 , R2 ]), 0 ≤ R1 < R2 , then h (s) = g (s) s N −1 ∈ C n+1 ([R1 , R2 ]), n ∈ N, N ≥ 2. Next we present multivariate Conformable fractional Iyengar type inequalities: Theorem 6.29 Let α ∈ (n, n + 1] and g ∈ C n+1 ([R1 , R2 ]), 0 < R1 < R2 , n ∈ N; β = α − n. Then (i) n (k) 1 h (R1 ) (z − R1 )k+1 + f (y) dy − A + 1)! (k k=0 (−1)k h (k) (R2 ) (R2 − z)
N 2
2π N2
k+1
N 2π 2 ≤ N2
(β) max TαR1 (h)∞,[R1 ,R2 ] , αR2 T (h)∞,[R1 ,R2 ] (α + 2) (z − R1 )α+1 + (R2 − z)α+1 ,
· (6.156)
∀ z ∈ [R1 , R2 ] , 2 , the right hand side of (6.156) is minimized, and we get: (ii) at z = R1 +R 2 n 1 (R2 − R1 )k+1 f (y) dy − A 2k (k + 1)! k=0
6.2 Main Results
133
h
(k)
(R1 ) + (−1) h k
(k)
(R2 )
N π 2 ≤ N2
R T R1 (h) 2 T (h) α+1 max , (β) α α π ∞,[R1 ,R2 ] ∞,[R1 ,R2 ] (R2 − R1 ) N , (α + 2) 2α−1 2 (6.157) (iii) if h (k) (R1 ) = h (k) (R2 ) = 0, for all k = 0, 1, . . . , n, we obtain N 2
N 2 f (y) dy ≤ π · N A 2 (β) max TαR1 (h)∞,[R1 ,R2 ] , αR2 T (h)∞,[R1 ,R2 ] (R − R )α+1 2 1 , (α + 2) 2α−1
(6.158)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n
1 R2 − R1 k+1 f (y) dy − A N (k + 1)! k=0
h
(k)
(R1 ) j
N 2
2π N2
k+1
+ (−1) h k
(k)
(R2 ) (N − j)
k+1
N 2π 2 ≤ N2
(β) max TαR1 (h)∞,[R1 ,R2 ] , αR2 T (h)∞,[R1 ,R2 ] (α + 2)
R2 − R1 N
α+1
j α+1 + (N − j)α+1 ,
(6.159)
(v) if h (k) (R1 ) = h (k) (R2 ) = 0, k = 1, . . . , n, from (6.159) we get:
N R2 − R1 2π 2 [ j h (R1 ) + (N − j) h (R2 )] N f (y) dy − A N 2
≤
N 2
2π N2
(β) max TαR1 (h)∞,[R1 ,R2 ] , αR2 T (h)∞,[R1 ,R2 ] (α + 2)
R2 − R1 N
α+1
j α+1 + (N − j)α+1 ,
(6.160)
134
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (6.160) turns to N π 2 f (y) dy − (R2 − R1 ) (h (R1 ) + h (R2 )) N ≤ A 2 π 2 (β) max N2 N
, αR2 T (h)∞,[R1 ,R2 ] (R − R )α+1 2 1 , (α + 2) 2α−1 (6.161)
T R1 (h) α
∞,[R1 ,R2 ]
Proof By Theorem 6.13 and as in our other multivariate results. We continue with
Corollary 6.30 Let α ∈ (n, n + 1] and g ∈ C n+1 ([0, R]), R > 0, n ∈ N; β=α − n. Then (i) n (k) 1 h (0) z k+1 + f (y) dy − B(0,R) (k + 1)! k=0
(−1)k h (k) (R) (R − z)
N 2
2π N2
k+1
N 2π 2 ≤ N2
(β) max T0α (h)∞,[0,R] , αR T (h)∞,[0,R] (α + 2)
·
z α+1 + (R − z)α+1 ,
(6.162)
∀ z ∈ [0, R] , (ii) at z = R2 , the right hand side of (6.162) is minimized, and we get: n 1 R k+1 f (y) dy − B(0,R) (k + 1)! 2k k=0
h (k) (0) + (−1)k h (k) (R)
N π 2 ≤ N2
R T0 (h) T (h) α+1 max , (β) α α π ∞,[0,R] ∞,[0,R] R N , (α + 2) 2α−1 2 N 2
(iii) assuming h (k) (0) = h (k) (R) = 0, for all k = 0, 1, . . . , n, we obtain
(6.163)
6.2 Main Results
135
B(0,R)
N π2 f (y) dy ≤ N · 2
(β) max T0α (h)∞,[0,R] , αR T (h)∞,[0,R] R α+1 , (α + 2) 2α−1
(6.164)
which is a sharp inequality. (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n
k+1 1 R f (y) dy − B(0,R) N + 1)! (k k=0
N 2 2π N ≤ h (k) (0) j k+1 + (−1)k h (k) (R) (N − j)k+1 2 N 2
2π N2
(β) max T0α (h)∞,[0,R] , αR T (h)∞,[0,R] (α + 2)
R N
α+1
j α+1 + (N − j)α+1 ,
(6.165)
(v) if h (k) (0) = h (k) (R) = 0, k = 1, . . . , n, from (6.165) we get:
N R 2π 2 f (y) dy − (N − j) h (R) N B(0,R) N 2
≤
N 2
2π N2
(β) max T0α (h)∞,[0,R] , αR T (h)∞,[0,R] (α + 2)
R N
α+1
j α+1 + (N − j)α+1 ,
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (6.166) turns to N π 2 f (y) dy − Rh (R) N ≤ B(0,R) 2
(6.166)
136
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
π 2 (β) max N2 N
R T (h) T0 (h) α+1 , α α ∞,[0,R] ∞,[0,R] R , α−1 (α + 2) 2
(6.167)
Proof By Theorem 6.29, just set there R1 = 0, R2 = R, the assumptions now are on B (0, R), and use (6.73). We continue with L p results. Theorem 6.31 Let α ∈ (n, n + 1] and g ∈ C n+1 ([R1 , R2 ]), 0 < R1 < R2 , n ∈ N; β = α − n. Let also p1 , p2 , p3 > 1 : p11 + p12 + p13 = 1, with β > p11 + p13 . Then (i) n (k) 1 h (R1 ) (z − R1 )k+1 + f (y) dy − A + 1)! (k k=0 (−1) h k
(k)
(R2 ) (R2 − z)
k+1
N 2π 2 ≤ N2
R 2 T (h) T R1 (h) max , α α 2π p3 ,[R1 ,R2 ] p3 ,[R1 ,R2 ] · 1 1 N2 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + 1 + 1 p1 p2 N 2
(z − R1 )
α+ p1 + p1 1
2
+ (R2 − z)
α+ p1 + p1 1
2
,
(6.168)
∀ z ∈ [R1 , R2 ] , 2 , the right hand side of (6.168) is minimized, and we get: (ii) at z = R1 +R 2 n 1 (R2 − R1 )k+1 f (y) dy − A 2k (k + 1)! k=0
h (k) (R1 ) + (−1)k h (k) (R2 )
N π 2 ≤ N2
R T R1 (h) 2 T (h) α+ 1 + 1 max , α α π (R2 − R1 ) p1 p2 p3 ,[R1 ,R2 ] p3 ,[R1 ,R2 ] , 1 1 α−1− p1 N2 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + 1 + 1 3 2 p1 p2 (6.169) (iii) assuming h (k) (R1 ) = h (k) (R2 ) = 0, for all k = 0, 1, . . . , n, we obtain N 2
N 2 f (y) dy ≤ π · N A 2
6.2 Main Results
137
α+ 1 + 1 max TαR1 (h) p3 ,[R1 ,R2 ] , αR2 T (h) p3 ,[R1 ,R2 ] (R2 − R1 ) p1 p2 , 1 1 α−1− p1 3 2 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 + p12
(6.170)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n
1 R2 − R1 k+1 f (y) dy − A N (k + 1)! k=0
h (k) (R1 ) j k+1 + (−1)k h (k) (R2 ) (N − j)k+1
N 2π 2 ≤ N2
R 2 T (h) T R1 (h) max , α α 2π p3 ,[R1 ,R2 ] p3 ,[R1 ,R2 ] N 1 1 2 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + 1 + 1 p1 p2 N 2
R2 − R1 N
α+ p1 + p1 1 2
j
α+ p1 + p1 1
2
+ (N − j)
α+ p1 + p1 1
2
,
(6.171)
(v) if h (k) (R1 ) = h (k) (R2 ) = 0, k = 1, . . . , n, from (6.171) we get:
N R2 − R1 2π 2 [ j h (R1 ) + (N − j) h (R2 )] N f (y) dy − A N 2 N max TαR1 (h) p3 ,[R1 ,R2 ] , αR2 T (h) p3 ,[R1 ,R2 ] 2π 2 ≤ N 1 1 2 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + 1 + 1 p1
R2 − R1 N
α+ p1 + p1 1 2
j
α+ p1 + p1 1
2
α+ p1 + p1
+ (N − j)
1
2
p2
,
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (6.172) turns to N π 2 f (y) dy − (R2 − R1 ) (h (R1 ) + h (R2 )) N ≤ A 2 R T R1 (h) 2 T (h) max , α α π p3 ,[R1 ,R2 ] p3 ,[R1 ,R2 ] 1 1 N2 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + 1 + 1 p1 p2 N 2
(6.172)
138
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
(R2 − R1 ) 2
α+ p1 + p1
α−1−
1
2
1 p3
,
(6.173)
Proof By Theorem 6.14 and as in our other multivariate results. We continue with
Corollary 6.32 Let α ∈ (n, n + 1] and g ∈ C n+1 ([0, R]), R > 0, n ∈ N; β=α − n. Let also p1 , p2 , p3 > 1 : p11 + p12 + p13 = 1, with β > p11 + p13 . Then (i) n (k) 1 h (0) z k+1 + f (y) dy − B(0,R) + 1)! (k k=0 (−1) h k
(k)
(R) (R − z)
k+1
N 2π 2 ≤ N2
R T0 (h) T (h) max , α α 2π p3 ,[0,R] p3 ,[0,R] 1 1 N2 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + 1 + p1 N 2
1 p2
·
α+ 1 + 1 α+ 1 + 1 t p1 p2 + (R − t) p1 p2 ,
(6.174)
∀ z ∈ [0, R] , (ii) at z = R2 , the right hand side of (6.174) is minimized, and we get: n 1 R k+1 f (y) dy − B(0,R) (k + 1)! 2k k=0
h (k) (0) + (−1)k h (k) (R)
N π 2 ≤ N2
R T (h) T0 (h) max , α α π p3 ,[0,R] p3 ,[0,R] N 1 1 2 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + 1 + p1 N 2
R
α+ p1 + p1
2
1
2
α−1− p1
1 p2
,
3
(iii) assuming h (k) (0) = h (k) (R) = 0, for all k = 0, 1, . . . , n, we obtain
(6.175)
6.2 Main Results
139
N π2 f (y) dy ≤ N · 2
B(0,R)
max T0α (h) p3 ,[0,R] , αR T (h) p3 ,[0,R] 1 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 +
1 p2
R
α+ p1 + p1
2
1
α−1−
2
1 p3
,
(6.176)
which is a sharp inequality. (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds n
k+1 R 1 f (y) dy − B(0,R) (k + 1)! N k=0
h
(k)
(0) j
k+1
+ (−1) h k
(k)
(R) (N − j)
k+1
N 2π 2 ≤ N2
R T0 (h) T (h) max , α α 2π p3 ,[0,R] p3 ,[0,R] N 1 1 2 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + 1 + p1 N 2
R N
α+ p1 + p1 1 2
j
α+ p1 + p1 1
2
+ (N − j)
α+ p1 + p1 1
2
1 p2
,
(6.177)
(v) if h (k) (0) = h (k) (R) = 0, k = 1, . . . , n, from (6.177) we get:
N R 2π 2 f (y) dy − (N − j) h (R) N B(0,R) N 2 R T (h) T0 (h) max , α α 2π p3 ,[0,R] p3 ,[0,R] ≤ N 1 1 2 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + 1 + p1 N 2
R N
α+ p1 + p1 1 2
j
α+ p1 + p1 1
2
+ (N − j)
α+ p1 + p1 1
j = 0, 1, 2, . . . , N , (viii) when N = 2 and j = 1, (6.178) turns to N π 2 f (y) dy − Rh (R) N ≤ B(0,R) 2
2
,
1 p2
(6.178)
140
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
N max T0α (h) p3 ,[0,R] , αR T (h) p3 ,[0,R] π2 1 1 N2 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + 1 + p1
1 p2
R
α+ p1 + p1
2
1
α−1−
2
1 p3
.
(6.179)
Proof By Theorem 6.31, just set there R1 = 0, R2 = R, the assumptions now are on B (0, R), and use (6.73). Our proving method follows next.
6.3 Appendix Proof Detailed proof of Theorem 6.17 (serving as a model proof for this chapter). We apply Theorem 6.4 (i) for h: n−1 R2 (k) 1 h (R1 ) (t − R1 )k+1 + h (s) ds − R1 (k + 1)! k=0
(−1)k h (k) (R2 ) (R2 − t)k+1 ≤ ν max D∗R h L 1
∞ ([R1 ,R2
, D νR2 − L ])
∞ ([R1 ,R2 ])
(ν + 2)
·
(t − R1 )ν+1 + (R2 − t)ν+1 =: ψ (t) ,
(6.180)
∀ t ∈ [R1 , R2 ] . Equivalently, we have that
R2
−ψ (t) ≤
R1
h (s) ds −
n−1 k=0
(k) 1 h (R1 ) (t − R1 )k+1 + (k + 1)!
(−1)k h (k) (R2 ) (R2 − t)k+1 ≤ ψ (t) ,
(6.181)
∀ t ∈ [R1 , R2 ] . That is −ψ (t) ≤
R2 R1
f (sω) s N −1 ds −
n−1 k=0
(k) 1 h (R1 ) (t − R1 )k+1 (k + 1)!
+ (−1)k h (k) (R2 ) (R2 − t)k+1 ≤ ψ (t) ,
(6.182)
6.3 Appendix
141
∀ t ∈ [R1 , R2 ], and ∀ ω ∈ S N −1 . Therefore it holds dω ≤ −ψ (t) S N −1
n−1 k=0
S N −1
R2
f (sω) s N −1 ds dω−
R1
(k) 1 h (R1 ) (t − R1 )k+1 + (−1)k h (k) (R2 ) (R2 − t)k+1 dω (k + 1)! S N −1 ≤ ψ (t)
dω, ∀ t ∈ [R1 , R2 ] ,
S N −1
which is (by (6.74)) N
2π 2 −ψ (t) N ≤ 2 n−1 k=0
(6.183)
f (y) dy− A
N (k) 2π 2 1 k+1 k (k) k+1 N h (R1 ) (t − R1 ) + (−1) h (R2 ) (R2 − t) (k + 1)! 2 N
≤ ψ (t)
2π 2 , ∀ t ∈ [R1 , R2 ] . N2
(6.184)
Consequently, we derive n−1 (k) 1 h (R1 ) (t − R1 )k+1 + f (y) dy − A (k + 1)! k=0
k
(k)
N 2
ν max D∗R h L 1
(−1) h
2π N2
(R2 ) (R2 − t)
k+1
N N 2π 2 2π 2 N ≤ ψ (t) N = 2 2
∞ ([R1 ,R2
, D νR2 − L ])
∞ ([R1 ,R2 ])
(ν + 2)
·
(t − R1 )ν+1 + (R2 − t)ν+1 , ∀ t ∈ [R1 , R2 ] , proving Theorem 6.17 (i). Next consider ϕ (t) := (t − R1 )ν+1 + (R2 − t)ν+1 , ∀ t ∈ [R1 , R2 ] .
(6.185)
142
6 Multidimensional Fractional Iyengar Inequalities for Radial Functions
Then
ϕ (t) = (ν + 1) (t − R1 )ν − (R2 − t)ν = 0,
2 and ϕ has the only critical number t = R1 +R . Hence ϕ (t) has a minimum over R1 +R2 (R2 −R1 )ν+1 2 = . [R1 , R2 ] which is ϕ 2 2ν Consequently, it holds (by (6.185))
n−1 1 (R2 − R1 )k+1 f (y) dy − A 2k (k + 1)! k=0
N 2
π N2
h (k) (R1 ) + (−1)k h (k) (R2 )
ν max D∗R h L 1
∞ ([R1 ,R2
, D νR2 − L ])
(ν + 2)
N π 2 ≤ N2
∞ ([R1 ,R2 ])
(R2 − R1 )ν+1 , 2ν−1
proving Theorem 6.17 (ii). The rest of Theorem 6.17 is obvious or follows the same way as above.
(6.186)
The rest of the proofs of this chapter as similar are omitted.
References 1. T. Abdeljawad, On conformable fractional calculus. J. Comput. Appl. Math. 279, 57–66 (2015) 2. G.A. Anastassiou, Fractional Differentiation Inequalities. Research Monograph (Springer, New York, 2009) 3. G.A. Anastassiou, On Right fractional calculus. Chaos Solitons Fractals 42, 365–376 (2009) 4. G.A. Anastassiou, Intelligent Mathematical Computational Analysis (Springer, Heidelberg, 2011) 5. G.A. Anastassiou, Mixed conformable fractional approximation by sublinear operators. Indian J. Math. 60(1), 107–140 (2018) 6. G.A. Anastassiou, Multidimensional fractional Iyengar type inequalities for radial functions. Prog. Fract. Differ. Appl., accepted for publications (2018) 7. G.A. Anastassiou, Caputo fractional Iyengar type inequalities, submitted (2018) 8. G.A. Anastassiou, Conformable fractional Iyengar type inequalities, submitted (2018) 9. G.A. Anastassiou, Canavati fractional Iyengar type inequalities. Analele Univ. Oradea Fasc Mat. XXVI(1), 141–151 (2019) 10. J.A. Canavati, The Riemann-Liouville integral. Nieuw Arch. Voor Wiskd. 5(1), 53–75 (1987) 11. K.S.K. Iyengar, Note on an inequality. Math. Stud. 6, 75–76 (1938) 12. W. Rudin, Real and Complex Analysis, International Student edn. (Mc Graw Hill, London, 1970) 13. D. Stroock, A Concise Introduction to the Theory of Integration, 3rd edn. (Birkhaüser, Boston, 1999)
Chapter 7
General Multidimensional Fractional Iyengar Inequalities
Here we derive a variety of general multivariate fractional Iyengar type inequalities for not necessarily radial functions defined on the shell and ball. Our approach is based on the polar coordinates in R N , N ≥ 2, and the related multivariate polar integration formula. Via this method we transfer author’s univariate fractional Iyengar type inequalities into general multivariate fractional Iyengar inequalities. See also [7].
7.1 Background We are motivated by the following famous Iyengar inequality (1938), [11]. Theorem 7.1 Let f be a differentiable function on [a, b] and f (x) ≤ M. Then
a
b
M (b − a) 2 ( f (b) − f (a))2 1 . f (x) d x − (b − a) ( f (a) + f (b)) ≤ − 2 4 4M (7.1)
We need Definition 7.2 ([2], p. 394) Let ν > 0, n = ν (· the ceiling of the number), f ∈ AC n ([a, b]) (i.e. f (n−1) is absolutely continuous on [a, b]). The left Caputo fractional derivative of order ν is defined as x 1 ν D∗a f (x) = (7.2) (x − t)n−ν−1 f (n) (t) dt, (n − ν) a ∀ x ∈ [a, b], and it exists almost everywhere over [a, b] . © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_7
143
144
7 General Multidimensional Fractional Iyengar Inequalities
We need Definition 7.3 ([4], p. 336–337) Let ν > 0, n = ν, f ∈ AC n ([a, b]). The right Caputo fractional derivative of order ν is defined as ν f (x) = Db−
(−1)n (n − ν)
b
(z − x)n−ν−1 f (n) (z) dz,
(7.3)
x
∀ x ∈ [a, b], and exists almost everywhere over [a, b] . In [8] we proved the following Caputo fractional Iyengar type inequalities: Theorem 7.4 ([8]) Let ν > 0, n = ν (· is the ceiling of the number), and f ∈ AC n ([a, b]) (i.e. f (n−1) is absolutely continuous on [a, b]). We assume that ν ν f, Db− f ∈ L ∞ ([a, b]). Then D∗a (i) n−1 b (k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
ν ν max D∗a f L ∞ ([a,b]) , Db− f L ∞ ([a,b]) (t − a)ν+1 + (b − t)ν+1 , (ν + 2) ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
(7.4)
the right hand side of (7.4) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f f (x) d x − f + (−1) (a) (b) ≤ k+1 a 2 (k + 1)! k=0
ν ν max D∗a f L ∞ ([a,b]) , Db− f L ∞ ([a,b]) (b − a)ν+1 (ν + 2)
2ν
,
(7.5)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
a
b
ν max Dν f D f ν+1 , ∗a b− L ∞ ([a,b]) L ∞ ([a,b]) (b − a) f (x) d x ≤ , (ν + 2) 2ν
(7.6)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds b k+1
n−1
b − a 1 j k+1 f (k) (a) + (−1)k (N − j)k+1 f (k) (b) f (x) d x − N (k + 1)! a k=0
7.1 Background
≤
145
ν ν max D∗a f L ∞ ([a,b]) , Db− f L ∞ ([a,b]) b − a ν+1 (ν + 2)
N
j ν+1 + (N − j)ν+1 ,
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (7.7) we get:
b
f (x) d x −
a
b−a N
(7.7)
[ j f (a) + (N − j) f (b)] ≤
ν ν max D∗a f L ∞ ([a,b]) , Db− f L ∞ ([a,b]) b − a ν+1 (ν + 2)
N
j ν+1 + (N − j)ν+1 , (7.8)
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (7.8) turns to
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
ν ν max D∗a f L ∞ ([a,b]) , Db− f L ∞ ([a,b]) (b − a)ν+1 (ν + 2)
2ν
,
(7.9)
(vii) when 0 < ν ≤ 1, inequality (7.9) is again valid without any boundary conditions. We mention Theorem 7.5 ([8]) Let ν ≥ 1, n = ν, and f ∈ AC n ([a, b]). We assume that ν ν f, Db− f ∈ L 1 ([a, b]). Then D∗a (i) n−1 b (k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
ν ν max D∗a f L 1 ([a,b]) , Db− f L 1 ([a,b]) (t − a)ν + (b − t)ν , (ν + 1)
(7.10)
∀ t ∈ [a, b] , (ii) when ν = 1, from (7.10), we have
a
b
f (x) d x − [ f (a) (t − a) + f (b) (b − t)] ≤ f
L 1 ([a,b])
(b − a) , ∀ t ∈ [a, b] ,
(7.11)
146
7 General Multidimensional Fractional Iyengar Inequalities
(iii) from (7.11), we obtain (ν = 1 case)
b
f (x) d x −
a
(iv) at t =
a+b , 2
b−a 2
( f (a) + f (b)) ≤ f L 1 ([a,b]) (b − a) ,
(7.12)
ν > 1, the right hand side of (7.10) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
ν ν max D∗a f L 1 ([a,b]) , Db− f L 1 ([a,b]) (b − a)ν 2ν−1
(ν + 1)
,
(7.13)
(v) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1; ν > 1, from (7.13), we obtain ν max b Dν f D f ν , ∗a b− L 1 ([a,b]) L 1 ([a,b]) (b − a) f (x) d x ≤ , (7.14) (ν + 1) 2ν−1 a which is a sharp inequality, (vi) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
n−1 b
b − a k+1 k+1 (k) 1 k k+1 (k) j f (x) d x − f (a) + (−1) (N − j) f (b) a N (k + 1)! k=0
≤
ν ν max D∗a f L 1 ([a,b]) , Db− f L 1 ([a,b]) b − a ν (ν + 1)
N
j ν + (N − j)ν , (7.15)
(vii) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (7.15) we get:
b
f (x) d x −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
ν ν max D∗a f L 1 ([a,b]) , Db− f L 1 ([a,b]) b − a ν (ν + 1)
N
j ν + (N − j)ν ,
j = 0, 1, 2, . . . , N , (viii) when N = 2 and j = 1, (7.16) turns to
a
b
(b − a) f (x) d x − ( f (a) + f (b)) ≤ 2
(7.16)
7.1 Background
147
ν ν max D∗a f L 1 ([a,b]) , Db− f L 1 ([a,b]) (b − a)ν 2ν−1
(ν + 1)
.
(7.17)
We mention Theorem 7.6 ([8]) Let p, q > 1 : 1p + ν ν f, Db− f ∈ L q ([a, b]). Then with D∗a (i)
1 q
= 1, ν > q1 , n = ν ; f ∈ AC n ([a, b]),
n−1 b (k) 1 f (a) (t − a)k+1 + (−1)k f (k) (b) (b − t)k+1 ≤ f (x) d x − a (k + 1)! k=0
ν ν
max D∗a f L q ([a,b]) , Db− f L q ([a,b]) ν+ 1p ν+ 1p − a) , + − t) (t (b 1 (ν) ν + 1p ( p (ν − 1) + 1) p ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
(7.18)
the right hand side of (7.18) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f f (x) d x − f + (−1) (a) (b) ≤ a 2k+1 (k + 1)! k=0
ν ν max D∗a f L q ([a,b]) , Db− f L q ([a,b]) (b − a)ν+ 1p , 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p
(7.19)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
a
b
ν max Dν f D f ν+ 1p , ∗a b− L q ([a,b]) L q ([a,b]) (b − a) f (x) d x ≤ , 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p
(7.20)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds b k+1
n−1
b − a 1 k+1 k k+1 (k) (k) j f (x) d x − f (a) + (−1) (N − j) f (b) N (k + 1)! a k=0
ν ν
max D∗a f L q ([a,b]) , Db− f L q ([a,b]) b − a ν+ 1p 1 1 j ν+ p + (N − j)ν+ p , ≤ 1 N (ν) ν + 1p ( p (ν − 1) + 1) p (7.21)
148
7 General Multidimensional Fractional Iyengar Inequalities
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (7.21) we get:
b
f (x) d x −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
ν ν
max D∗a f L q ([a,b]) , Db− f L q ([a,b]) b − a ν+ 1p ν+ 1p ν+ 1p j , + − j) (N 1 N (ν) ν + 1p ( p (ν − 1) + 1) p (7.22) for j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (7.22) turns to
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
ν ν max D∗a f L q ([a,b]) , Db− f L q ([a,b]) (b − a)ν+ 1p , 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p (vii) when conditions.
1 q
(7.23)
< ν ≤ 1, inequality (7.23) is again valid but without any boundary
We need the following different fractional calculus background: Let α > 0, m = [α] ([·] is the integral part), β=α − m, 0 < β 0, m = [α], β = α − m, f ∈ C ([a, b]), call the right Riemann– Liouville fractional integral operator by
7.1 Background
149
α Jb−
1 f (x) := (α)
b
(t − x)α−1 f (t) dt,
(7.27)
x
x ∈ [a, b], see [3]. Define the subspace of functions α Cb− ([a, b]) :=
1−β f ∈ C m ([a, b]) : Jb− f (m) ∈ C 1 ([a, b]) .
(7.28)
Define the right generalized α-fractional derivative of f over [a, b] as α 1−β D b− f = (−1)m−1 Jb− f (m) , 0
(7.29)
n
see [3]. We set D b− f = f . We have D b− f = (−1)n f (n) ; n ∈ N. Notice that α D b− f ∈ C ([a, b]) . We mention the following Canavati fractional Iyengar type inequalities: ν ν Theorem 7.7 ([6]) Let ν ≥ 1, n = [ν] and f ∈ Ca+ ([a, b]) ∩ Cb− ([a, b]). Then (i)
n−1 b (k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
ν ν max Da+ f ∞,([a,b]) , D b− f
∞,([a,b])
(ν + 2) ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
(t − a)ν+1 + (b − t)ν+1 ,
(7.30)
the right hand side of (7.30) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f f (x) d x − f + (−1) (a) (b) ≤ a 2k+1 (k + 1)! k=0
ν ν max Da+ f ∞,([a,b]) , D b− f (ν + 2)
∞,([a,b])
(b − a)ν+1 , 2ν
(7.31)
(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
a
b
ν ν max Da+ f ∞,([a,b]) , D b− f ν+1 ∞,([a,b]) (b − a) f (x) d x ≤ , (7.32) ν (ν + 2) 2
which is a sharp inequality,
150
7 General Multidimensional Fractional Iyengar Inequalities
(iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds b k+1
n−1
b − a 1 j k+1 f (k) (a) + (−1)k (N − j)k+1 f (k) (b) f (x) d x − N (k + 1)! a k=0
≤
ν ν max Da+ f ∞,([a,b]) , D b− f
∞,([a,b])
(ν + 2)
b−a N
ν+1
j ν+1 + (N − j)ν+1 ,
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (7.33) we get:
b
f (x) d x −
a
b−a N
ν ν max Da+ f ∞,([a,b]) , D b− f
(7.33)
[ j f (a) + (N − j) f (b)] ≤
∞,([a,b])
(ν + 2)
b−a N
ν+1
j ν+1 + (N − j)ν+1 , (7.34)
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (7.34) turns to
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
ν ν max Da+ f ∞,([a,b]) , D b− f
∞,([a,b])
(ν + 2)
(b − a)ν+1 . 2ν
(7.35)
We mention ν ν Theorem 7.8 ([6]) Let ν ≥ 1, n = [ν], and f ∈ Ca+ ([a, b]) ∩ Cb− ([a, b]). Then (i)
n−1 b (k) 1 k+1 k (k) k+1 f (a) (t − a) f (x) d x − + (−1) f (b) (b − t) ≤ a (k + 1)! k=0
ν ν max Da+ f L 1 ([a,b]) , D b− f
L 1 ([a,b])
(ν + 1) ∀ t ∈ [a, b] , (ii) when ν = 1, from (7.36), we have
(t − a)ν + (b − t)ν ,
(7.36)
7.1 Background
151
b
a
f (x) d x − [ f (a) (t − a) + f (b) (b − t)] ≤ f
L 1 ([a,b])
(b − a) , ∀ t ∈ [a, b] ,
(7.37)
(iii) from (7.37), we obtain (ν = 1 case)
b
f (x) d x −
a
(iv) at t =
a+b , 2
b−a 2
( f (a) + f (b)) ≤ f L 1 ([a,b]) (b − a) ,
(7.38)
ν > 1, the right hand side of (7.36) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f (x) d x − f (a) + (−1) f (b) ≤ a 2k+1 (k + 1)! k=0
ν ν max Da+ f L 1 ([a,b]) , D b− f
L 1 ([a,b])
(ν + 1)
(b − a)ν , 2ν−1
(7.39)
(v) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1; ν > 1, from (7.39), we obtain ν ν b max Da+ f L 1 ([a,b]) , D b− f ν L 1 ([a,b]) (b − a) ≤ f d x , (7.40) (x) (ν + 1) 2ν−1 a which is a sharp inequality, (vi) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds b k+1
n−1
b − a 1 j k+1 f (k) (a) + (−1)k (N − j)k+1 f (k) (b) f (x) d x − N (k + 1)! a k=0
≤
ν ν max Da+ f L 1 ([a,b]) , D b− f (ν + 1)
L 1 ([a,b])
b−a N
ν
j ν + (N − j)ν ,
(vii) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (7.41) we get:
a
b
f (x) d x −
b−a N
[ j f (a) + (N − j) f (b)] ≤
(7.41)
152
7 General Multidimensional Fractional Iyengar Inequalities
ν ν max Da+ f L 1 ([a,b]) , D b− f
L 1 ([a,b])
(ν + 1)
b−a N
ν
j ν + (N − j)ν ,
(7.42)
j = 0, 1, 2, . . . , N , (viii) when N = 2 and j = 1, (7.42) turns to
a
b
(b − a) f (x) d x − ( f (a) + f (b)) ≤ 2
ν ν max Da+ f L 1 ([a,b]) , D b− f (ν + 1)
L 1 ([a,b])
(b − a)ν . 2ν−1
(7.43)
We mention Theorem 7.9 ([6]) Let p, q > 1 : ν Cb− ([a, b]). Then (i)
1 p
+
1 q
ν = 1, ν ≥ 1, n = [ν] ; f ∈ Ca+ ([a, b]) ∩
n−1 b (k) 1 f (a) (t − a)k+1 + (−1)k f (k) (b) (b − t)k+1 ≤ f (x) d x − a (k + 1)! k=0
ν ν max Da+ f L q ([a,b]) , D b− f
1 1 L q ([a,b]) (t − a)ν+ p + (b − t)ν+ p , 1 (ν) ν + 1p ( p (ν − 1) + 1) p ∀ t ∈ [a, b] , (ii) at t =
a+b , 2
(7.44)
the right hand side of (7.44) is minimized, and we get:
n−1 b 1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (x) d x − a 2k+1 (k + 1)! k=0
ν ν max Da+ f L q ([a,b]) , D b− f ν+ 1p L q ([a,b]) (b − a) , 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p (iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain
(7.45)
7.1 Background
a
b
153
ν ν max Da+ f L q ([a,b]) , D b− f ν+ 1p L q ([a,b]) (b − a) f (x) d x ≤ , (7.46) 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds b k+1
n−1
b − a 1 j k+1 f (k) (a) + (−1)k (N − j)k+1 f (k) (b) f (x) d x − N (k + 1)! a k=0 ν ν 1
max Da+ f L ([a,b]) , D b− f q b − a ν+ p ν+ 1p L q ([a,b]) ν+ 1p ≤ j , + − j) (N 1 N (ν) ν + 1p ( p (ν − 1) + 1) p
(7.47) (v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (7.47) we get:
b
f (x) d x −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
ν ν
1 max Da+ f L q ([a,b]) , D b− f
b − a ν+ p ν+ 1p L q ([a,b]) ν+ 1p j , + − j) (N 1 N (ν) ν + 1p ( p (ν − 1) + 1) p (7.48) for j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (7.48) turns to
a
b
f (x) d x −
b−a 2
( f (a) + f (b)) ≤
ν ν max Da+ f L q ([a,b]) , D b− f ν+ 1p L q ([a,b]) (b − a) . 1 1 2ν− q (ν) ν + 1p ( p (ν − 1) + 1) p
(7.49)
We need Definition 7.10 ([1]) Let a, b ∈ R. The left conformable fractional derivative starting from a of a function f : [a, ∞) → R of order 0 < α ≤ 1 is defined by
f t + ε (t − a)1−α − f (t) Tαa f (t) = lim . ε→0 ε
(7.50)
154
7 General Multidimensional Fractional Iyengar Inequalities
If Tαa f (t) exists on (a, b), then
Tαa f (a) = lim Tαa f (t) . t→a+
(7.51)
The right conformable fractional derivative of order 0 < α ≤ 1 terminating at b of f : (−∞, b] → R is defined by f t + ε (b − t)1−α − f (t) . α T f (t) = −lim ε→0 ε
b If
b
αT
(7.52)
f (t) exists on (a, b), then b
αT
f (b) = lim bα T f (t) . t→b−
(7.53)
Note that if f is differentiable then and
Tαa f (t) = (t − a)1−α f (t) ,
b
αT
f (t) = − (b − t)1−α f (t) .
(7.54)
(7.55)
In the higher order case we can generalize things as follows: Definition 7.11 ([1]) Let α ∈ (n, n + 1], and set β = α − n. Then, the left conformable fractional derivative starting from a of a function f : [a, ∞) → R of order α, where f (n) (t) exists, is defined by a Tα f (t) = Tβa f (n) (t) ,
(7.56)
The right conformable fractional derivative of order α terminating at b of f : (−∞, b] → R, where f (n) (t) exists, is defined by b
αT
f (t) = (−1)n+1 bβ T f (n) (t) .
(7.57)
If α = n + 1 then β = 1 and Tan+1 f = f (n+1) . If n is odd, then bn+1 T f = − f (n+1) , and if n is even, then bn+1 T f = f (n+1) . When n = 0 (or α ∈ (0, 1]), then β = α, and (7.56), (7.57) collapse to (7.50), (7.52), respectively. We need Remark 7.12 ([5]) We notice the following: let α ∈ (n, n + 1] and f ∈ C n+1 ([a, b]), n ∈ N. Then (β := α − n, 0 < β ≤ 1) a Tα ( f ) (x) = Tβα f (n) (x) = (x − a)1−β f (n+1) (x) ,
(7.58)
7.1 Background
155
and
b
αT (
f ) (x) = (−1)n+1 bβ T f (n) (x) =
(−1)n+1 (−1) (b − x)1−β f (n+1) (x) = (−1)n (b − x)1−β f (n+1) (x) .
(7.59)
Consequently we get that a Tα ( f ) (x) , bα T ( f ) (x) ∈ C ([a, b]) . Furthermore it is obvious that a Tα ( f ) (a) = bα T ( f ) (b) = 0,
(7.60)
when 0 < β < 1, i.e. when α ∈ (n, n + 1) . We mention the following Conformable fractional Iyengar type inequalities: Theorem 7.13 ([9]) Let α ∈ (n, n + 1] and f ∈ C n+1 ([a, b]), n ∈ N; β = α − n. Then (i) n b (k) 1 k+1 k (k) k+1 f (a) (z − a) f (t) dt − + (−1) f (b) (b − z) ≤ a (k + 1)! k=0
(β) max Taα ( f )∞,[a,b] , bα T ( f )∞,[a,b] (z − a)α+1 + (b − z)α+1 , (α + 2) (7.61) ∀ z ∈ [a, b] , , the right hand side of (7.61) is minimized, and we get: (ii) at z = a+b 2 n b 1 (b − a)k+1 (k) k (k) f f (t) dt − f + (−1) (a) (b) ≤ a 2k+1 (k + 1)! k=0
(β) max Taα ( f )∞,[a,b] , bα T ( f )∞,[a,b] (b − a)α+1 (α + 2)
2α
,
(7.62)
(iii) assuming f (k) (a) = f (k) (b) = 0, for k = 0, 1, . . . , n, we obtain b (β) max Ta ( f ) T ( f ) α+1 , α α ∞,[a,b] ∞,[a,b] (b − a) f (t) dt ≤ , (α + 2) 2α a (7.63) which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
b
156
7 General Multidimensional Fractional Iyengar Inequalities
b k+1
n
1 b − a f (k) (a) j k+1 + (−1)k f (k) (b) (N − j)k+1 f (t) dt − N (k + 1)! a k=0
≤
(β) max Taα ( f )∞,[a,b] , bα T ( f ) b − a α+1 ∞,[a,b] (α + 2)
N
j α+1 + (N − j)α+1 ,
(7.64)
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n, from (7.64) we get:
b
f (t) dt −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
(β) max Taα ( f )∞,[a,b] , bα T ( f )∞,[a,b] b − a α+1 (α + 2)
N
j α+1 + (N − j)α+1 , (7.65)
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (7.65) turns to
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
(β) max Taα ( f )∞,[a,b] , bα T ( f )∞,[a,b] (b − a)α+1 2α
(α + 2)
.
(7.66)
We mention L p conformable fractional Iyengar inequalities: Theorem 7.14 ([9]) Let α ∈ (n, n + 1] and f ∈ C n+1 ([a, b]), n ∈ N; β = α − n. Let also p1 , p2 , p3 > 1 : p11 + p12 + p13 = 1, with β > p11 + p13 . Then (i) n b (k) 1 k+1 k (k) k+1 f (a) (z − a) f (t) dt − + (−1) f (b) (b − z) ≤ a (k + 1)! k=0 max Taα ( f ) p3 ,[a,b] , bα T ( f ) p3 ,[a,b] 1 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 +
(z − a)
α+ p1 + p1 1
2
+ (b − z)
α+ p1 + p1 1
2
,
1 p2
(7.67)
7.1 Background
∀ z ∈ [a, b] , (ii) at z =
157
a+b , 2
the right hand side of (7.67) is minimized, and we get:
n b 1 (b − a)k+1 (k) k (k) f f (t) dt − f + (−1) (a) (b) ≤ a 2k+1 (k + 1)! k=0
max Taα ( f ) p3 ,[a,b] , bα T ( f ) p3 ,[a,b] 1 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 +
1 p2
(b − a) 2
α+ p1 + p1 1
α− p1
2
,
(7.68)
3
(iii) assuming f (k) (a) = f (k) (b) = 0, for k = 0, 1, . . . , n, we obtain
b
a
f (t) dt ≤
max Taα ( f ) p3 ,[a,b] , bα T ( f ) p3 ,[a,b] 1 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 +
(b − a)
1 p2
2
α+ p1 + p1 1
2
α− p1
,
3
(7.69) which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds b k+1
n
b − a 1 k+1 k k+1 (k) (k) f (a) j f (t) dt − + (−1) f (b) (N − j) N (k + 1)! a k=0
max Taα ( f ) p3 ,[a,b] , bα T ( f ) p3 ,[a,b] ≤ 1 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 +
b−a N
α+ p1 + p1 1 2
j
α+ p1 + p1 1
2
+ (N − j)
1 p2
α+ p1 + p1 1
2
,
(7.70)
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n, from (7.70) we get:
b
f (t) dt −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
max Taα ( f ) p3 ,[a,b] , bα T ( f ) p3 ,[a,b] 1 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 +
b−a N
α+ p1 + p1 1 2
j
α+ p1 + p1 1
2
1 p2
α+ p1 + p1
+ (N − j)
1
2
,
(7.71)
158
7 General Multidimensional Fractional Iyengar Inequalities
for j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (7.71) turns to
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
max Taα ( f ) p3 ,[a,b] , bα T ( f ) p3 ,[a,b] 1 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 +
1 p2
(b − a) 2
α+ p1 + p1 1
α− p1
2
.
(7.72)
3
We need Remark 7.15 We define the ball B (0, R) = x ∈ R N : |x| < R ⊆ R N , N ≥ 2, R > 0, and the sphere S N −1 := x ∈ R N : |x| = 1 , where |·| is the Euclidean norm. Let dω be the element of surface measure on S N −1 and N 2π 2 dω = N ωN = 2 S N −1 is the area of S N −1 . For x ∈ R N − {0} we can write uniquely x = r ω, where r = |x| > 0 and ω = N x ∈ S N −1 , |ω| = 1. Note that B(0,R) dy = ω NNR is the Lebesgue measure on the r N
R ball, that is the volume of B (0, R), which exactly is V ol (B (0, R)) = π 2N +1 . (2 ) Following [12, pp. 149–150, exercise 6], and [13, pp. 87–88, Theorem 5.2.2] we can write for F : B (0, R) → R a Lebesgue integrable function that
F (x) d x = B(0,R)
S N −1
R
F (r ω) r N −1 dr dω,
N
(7.73)
0
and we use this formula a lot. Typically here the function f : B (0, R) → R is not radial. A radial function f is such that there exists a function g such that f (x) = g (r ), where r = |x|, r ∈ [0, R], ∀ x ∈ B (0, R). We need Remark 7.16 Let the spherical shell A := B (0, R2 ) − B (0, R1 ), 0 < R1 < R2 , A ⊆ R N , N ≥ 2, x ∈ A. Consider that f : A → R is not necessarily radial. A radial function f is such that there exists a function g such that f (x) = g (r ), r = |x|, r ∈ [R1 , R2 ], ∀ x ∈ A. Here x can be written uniquely as x = r ω, where r = |x| > 0 and ω = rx ∈ S N −1 , |ω| = 1, see ([12], p. 149–150 and [2], p. 421), furthermore for F : A → R a Lebesgue integrable function we have that
7.1 Background
159
F (x) d x =
R2
F (r ω) r
N −1
dr dω.
(7.74)
N π 2 R2N − R1N ω N R2N − R1N . = V ol (A) = N N2 + 1
(7.75)
A
Here
S N −1
R1
In this chapter we derive general multivariate fractional Iyengar type inequalities on the shell and ball of R N , N ≥ 2, for not necessarily radial functions. Our following results are based on the described background fractional results.
7.2 Main Results From now on the fractional derivatives of f (sω) s N −1 are in s ∈ [R1 , R2 ] or s ∈ [0, R] . We present Caputo type results on the shell: Theorem 7.17 Let ν > 0, n = ν . Consider f : A → R be Lebesgue integrable, which is not necessarily radial. Assume that f (sω) s N −1 ∈ AC n ([R1 , R2 ]) (i.e. (n−1) ∈ AC ([R1, R2 ]) - absolutely functions), ∀ ω ∈ S N −1 , f (sω) s N −1 ν continuous ν N −1 N −1 , D R2 − f (sω) s ∈ L ∞ ([R1 , R2 ]), N ≥ 2. We assume that D∗R1 f (sω) s ν N −1 D ν ( f (sω) ∀ ω ∈ S N −1 , and that max{ D∗R f s , (sω) R2 − 1 L ∞ ([R1 ,R2 ]) s N −1 L ∞ ([R1 ,R2 ]) } ≤ K 1 , where K 1 > 0, s ∈ [R1 , R2 ], ∀ ω ∈ S N −1 . Then (i) n−1 (k) 1 f (sω) s N −1 (R1 ) dω (t − R1 )k+1 + f (y) dy − A (k + 1)! S N −1 k=0
(−1)k
S N −1
f (sω) s N −1
(k)
(R2 ) dω (R2 − t)k+1 ≤
2π 2 K1 N (t − R1 )ν+1 + (R2 − t)ν+1 , 2 (ν + 2) N
(7.76)
∀ t ∈ [R1 , R2 ] , 2 , the right hand side of (7.76) is minimized, and we get: (ii) at t = R1 +R 2 n−1 (k) 1 (R2 − R1 )k+1 f (sω) s N −1 f dy − (y) (R1 ) dω+ k+1 A 2 (k + 1)! S N −1 k=0
160
7 General Multidimensional Fractional Iyengar Inequalities
(−1)k
S N −1
f (sω) s N −1
(k)
(R2 ) dω ≤
π2 K1 (R2 − R1 )ν+1 N , 2ν−1 2 (ν + 2) N
(7.77)
(k) (k) (iii) if f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , ∂ k ( f (sω)s N −1 ) vanish on ∂ B (0, R ) and ∂ B (0, R )) for all k = 0, 1, . . . , n − 1, (i.e. 1
∂s k
we obtain
2
N 2 K1 (R2 − R1 )ν+1 f (y) dy ≤ π , N (ν + 2) 2ν−1 A 2
(7.78)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
ν−1 1 R2 − R1 k+1 f dy − (y) A (k + 1)! N k=0
j
k+1 S N −1
(−1)
k
N−j
k+1
f (sω) s
S N −1
N
2π 2 K1 N2 (ν + 2)
N −1 (k)
f (sω) s
R2 − R1
ν+1
N
(R1 ) dω +
N −1 (k)
(R2 ) dω ≤
ν+1 j ν+1 + N − j ,
(7.79)
(k) (k) f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , k N −1 ∂ f (sω)s ) vanish on ∂ B (0, R ) and ∂ B (0, R )) for k = 1, . . . , n − 1, from (i.e. ( ∂s k 1 2 (7.79) we get: (v)
if
N −1 f (y) dy − R2 − R1 j R1 N A
N − j R2N −1
K1 (ν + 2)
for j = 0, 1, 2, . . . , N ∈ N,
S N −1
S N −1
R2 − R1 N
f (R2 ω) dω
ν+1
f (R1 ω) dω +
N 2 ≤ 2π N · 2
ν+1 j ν+1 + N − j ,
(7.80)
7.2 Main Results
161
(vi) when N = 2 and j = 1, (7.80) turns to
N −1 f (y) dy − R2 − R1 R1 2 A
S N −1
f (R1 ω) dω +
R2N −1
S N −1
f (R2 ω) dω
π2 K1 (R2 − R1 )ν+1 N , 2ν−1 2 (ν + 2) N
≤
(7.81)
(vii) when 0 < ν ≤ 1 (without any boundary conditions), we get again
N −1 f (y) dy − R2 − R1 R 1 2 A
S N −1
f (R1 ω) dω + R2N −1
S N −1
f (R2 ω) dω
π 2 K1 (R2 − R1 )ν+1 N . 2ν−1 2 (ν + 2) N
≤
(7.82)
Proof We apply Theorem 7.4 along with (7.74). See in the Appendix 7.3 the general proving method in this chapter. Theorem 7.18 Let ν ≥ 1, n = ν. Consider f : A → R be Lebesgue integrable, which is not necessarily radial. Assume that f (sω) s N −1 ∈ AC n ([R1 , R2 ]) (i.e. (n−1) ∈ AC ([R1 , R2 ]) - absolutely functions), ∀ ω ∈ S N −1 , f (sω) s N −1 νcontinuous ν N −1 N −1 , D R2 − f (sω) s ∈ L 1 ([R1 , R2 ]), N ≥ 2. We assume that D∗R1 f (sω) s ν ν N −1 N −1 f s ∀ ω ∈ S N −1 , and that max{ D∗R , D (sω) R2 − f (sω) s 1 L 1 ([R1 ,R2 ]) L 1 ([R1 ,R2 ]) } ≤ K 2 , where K 2 > 0, s ∈ [R1 , R2 ], ∀ ω ∈ S N −1 . Then (i) n−1 (k) 1 f (sω) s N −1 (R1 ) dω (t − R1 )k+1 + f (y) dy − A (k + 1)! S N −1 k=0
(−1)k
S N −1
f (sω) s N −1
(k)
(R2 ) dω (R2 − t)k+1 ≤
2π 2 K2 N (t − R1 )ν + (R2 − t)ν , 2 (ν + 1) N
∀ t ∈ [R1 , R2 ] , (ii) when ν = 1, from (7.83), we have f (y) dy − A
S N −1
f (R1 ω) dω R1N −1 (t − R1 ) +
(7.83)
162
7 General Multidimensional Fractional Iyengar Inequalities
S N −1
f (R2 ω) dω
N 2π 2 (R2 − t) ≤ N K 2 (R2 − R1 ) , 2
R2N −1
∀ t ∈ [R1 , R2 ] , (iii) from (7.84), we obtain (ν = 1 case, t =
R2 − R1 f (y) dy − 2 A
R1 +R2 ) 2
f (R1 ω) dω R1N −1 +
S N −1
(7.84)
S N −1
f (R2 ω) dω R2N −1
N
2π 2 K 2 (R2 − R1 ) , N2
≤ R1 +R2 , 2
(iv) at t =
(7.85)
ν > 1, the right hand side of (7.83) is minimized, and we get:
n−1 (k) 1 (R2 − R1 )k+1 f (sω) s N −1 (R1 ) dω+ f (y) dy − k+1 A 2 (k + 1)! S N −1 k=0
(−1)k
S N −1
f (sω) s N −1
(k)
(R2 ) dω ≤
π2 K2 (R2 − R1 )ν N , 2ν−2 2 (ν + 1) N
(7.86)
(k) (k) f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , ∂ k ( f (sω)s N −1 ) vanish on ∂ B (0, R1 ) and ∂ B (0, R2 )) for all k = 0, 1, . . . , n − 1; (i.e. ∂s k ν > 1, we obtain from (7.86) that (v)
if
N 2 K2 (R2 − R1 )ν f (y) dy ≤ π , N (ν + 1) 2ν−2 A 2 which is a sharp inequality, (vi) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
n−1 1 R2 − R1 k+1 f (y) dy − A (k + 1)! N k=0
j
k+1 S N −1
k+1 (−1)k N − j
f (sω) s
S N −1
N −1 (k)
(R1 ) dω +
f (sω) s N −1
(k)
(R2 ) dω
≤
(7.87)
7.2 Main Results
163 N
2π 2 K2 N 2 (ν + 1)
R2 − R1
ν
N
ν jν + N − j ,
(7.88)
(k) (k) (vii) if f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , ∂ k ( f (sω)s N −1 ) vanish on ∂ B (0, R ) and ∂ B (0, R )) for k = 1, . . . , n − 1, from (i.e. 1
∂s k
2
(7.88) we get:
N −1 f (y) dy − R2 − R1 j R 1 N A
N−j
R2N −1
K2 (ν + 1)
S N −1
S N −1
f (R1 ω) dω +
N 2π 2 f (R2 ω) dω ≤ N · 2
R2 − R1
ν
N
ν jν + N − j ,
(7.89)
for j = 0, 1, 2, . . . , N ∈ N, (viii) when N = 2 and j = 1, (7.89) turns to
N −1 f (y) dy − R2 − R1 R1 2 A
S N −1
f (R1 ω) dω +
R2N −1
S N −1
f (R2 ω) dω
π2 K2 (R2 − R1 )ν ≤ N . 2ν−2 2 (ν + 1) N
(7.90)
Proof We apply Theorem 7.5 along with (7.74). See in the Appendix 7.3 the general proving method in this chapter. We continue with Theorem 7.19 Let p, q > 1 : 1p + q1 = 1, ν > q1 , n = ν. Consider f : A → R be Lebesgue integrable, which is not necessarily radial. Assume that f (sω) s N −1 ∈ (n−1) ∈ AC ([R1 , R2 ]) - absolutely AC n ([R1 , R2 ]) (i.e. f (sω) s N −1 continuous ν f (sω) s N −1 , functions), ∀ ω ∈ S N −1 , N ≥ 2. We assume that D∗R 1 D νR2 − f (sω) s N −1 ∈ L q ([R1 , R2 ]), ∀ ω ∈ S N −1 , and that ν max D∗R f (sω) s N −1 L 1
q ([R1 ,R2
, D νR2 − f (sω) s N −1 L ])
where K 3 > 0, s ∈ [R1 , R2 ], ∀ ω ∈ S N −1 . Then (i)
q ([R1 ,R2 ])
≤ K3, (7.91)
164
7 General Multidimensional Fractional Iyengar Inequalities
n−1 (k) 1 f (sω) s N −1 (R1 ) dω (t − R1 )k+1 + f (y) dy − A (k + 1)! S N −1 k=0
(−1)k
S N −1
f (sω) s N −1
(k)
(R2 ) dω (R2 − t)k+1 ≤
N
2π 2 K3 ν+ 1p ν+ 1p N − R , + − t) (t (R ) 1 2 2 (ν) ν + 1 ( p (ν − 1) + 1) 1p p
(7.92)
∀ t ∈ [R1 , R2 ] , 2 , the right hand side of (7.92) is minimized, and we get: (ii) at t = R1 +R 2 n−1 1 (R2 − R1 )k+1 · f (y) dy − A 2k+1 (k + 1)! k=0
S N −1
f (sω) s N −1
(k)
(R1 ) dω + (−1)k
S N −1
f (sω) s N −1
(k)
(R2 ) dω ≤
π2 K3 (R2 − R1 )ν+ p N , ν−1− q1 2 (ν) ν + 1 ( p (ν − 1) + 1) 1p 2 p 1
N
(7.93)
(k) (k) (iii) if f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , k N −1 ∂ f (sω)s ) vanish on ∂ B (0, R ) and ∂ B (0, R )) for all k = 0, 1, . . . , n − 1, (i.e. ( 1
∂s k
2
we obtain
1 N 2 K3 (R2 − R1 )ν+ p f (y) dy ≤ π , (7.94) 1 1 N 1 A 2ν−1− q 2 (ν) ν + p ( p (ν − 1) + 1) p which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
n−1 1 R2 − R1 k+1 f (y) dy − A (k + 1)! N k=0
j
k+1 S N −1
(−1)
k
N−j
k+1
f (sω) s
S N −1
N −1 (k)
f (sω) s
(R1 ) dω +
N −1 (k)
(R2 ) dω ≤
7.2 Main Results
165 N
2π 2 K3 N · 2 (ν) ν + 1 ( p (ν − 1) + 1) 1p p
R2 − R1
ν+ 1p
N
ν+ 1p 1 j ν+ p + N − j ,
(7.95)
(k) (k) f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , k N −1 ∂ f (sω)s ) vanish on ∂ B (0, R ) and ∂ B (0, R )) for k = 1, . . . , n − 1, from (i.e. ( ∂s k 1 2 (7.95) we get: (v)
if
N −1 f (y) dy − R2 − R1 j R 1 N A
(ν) ν +
1 p
N − j R2N −1
S N −1
S N −1
K3
f (R2 ω) dω
R2 − R1
1
( p (ν − 1) + 1) p
N 2 ≤ 2π N · 2
ν+ 1p
N
f (R1 ω) dω +
ν+ 1p 1 j ν+ p + N − j , (7.96)
for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (7.96) turns to
N −1 f (y) dy − R2 − R1 R 1 2 A
S N −1
f (R1 ω) dω +
R2N −1
f (R2 ω) dω
S N −1
π2 K3 (R2 − R1 )ν+ p ≤ N , 1 2 (ν) ν + 1 ( p (ν − 1) + 1) 1p 2ν−1− q p 1
N
(vii) when
1 q
(7.97)
< ν ≤ 1 (without any boundary conditions), we get again (7.97).
Proof By Theorem 7.6 and (7.74). See also Appendix 7.3 for the general proving method here. We give Caputo results on the ball: Theorem 7.20 Let 0 < ν < 1 and f : B (0, R) → R be Lebesgue integrable, which continis not necessarily radial. Assume that f (sω) s N −1 ∈ AC ([0, R]) (absolutely ν f (sω) s N −1 , uous functions), ∀ω ∈ S N −1 , N ≥ 2. We assume that D∗0 D νR− f (sω) s N −1 ∈ L ∞ ([0, R]), ∀ ω ∈ S N −1 and that ≤ M1 , where f (sω) s N −1 , Dν max D ν f (sω) s N −1 ∗0
L ∞ ([0,R]) N −1
M1 > 0, s ∈ [0, R] , ∀ ω ∈ S
.
R−
L ∞ ([0,R])
166
7 General Multidimensional Fractional Iyengar Inequalities
Then (i)
f (y) dy −
S N −1
B(0,R)
f (Rω) dω R N −1 (R − t) ≤
2π 2 M1 N t ν+1 + (R − t)ν+1 , (ν + 2) 2 N
(7.98)
∀ t ∈ [0, R] , (ii) at t = R2 , the right hand side of (7.98) is minimized, and we get:
f (y) dy −
S N −1
B(0,R)
f (Rω) dω
N π 2 M1 R ν+1 R N ≤ , 2 (ν + 2) N2 2ν−1
(7.99)
(iii) if f (Rω) = 0, ∀ ω ∈ S N −1 , (i.e. f (·ω) vanish on ∂ B (0, R)), we obtain
B(0,R)
N π 2 M1 R ν+1 f (y) dy ≤ , (ν + 2) N2 2ν−1
(7.100)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
B (0,R)
RN
f (y) dy − N
2π 2 M1 (ν + 2) N2
N
R N
N−j
ν+1
S N −1
f (Rω) dω ≤
ν+1 j ν+1 + N − j ,
(7.101)
(v) when N = 2 and j = 1, (7.101) turns to
f (y) dy − B(0,R)
RN 2
S N −1
f (Rω) dω ≤
π 2 M1 R ν+1 . (ν + 2) N2 2ν−1 N
(7.102)
Proof Same as the proof of Theorem 7.17, just set there R1 = 0 and R2 = R and use (7.73). We continue with Theorem 7.21 Let p, q > 1 : 1p + q1 = 1, q1 < ν < 1, and f : B (0, R) → R be Lebesgue integrable, which is not necessarily radial. Assume that f (sω) s N −1 ∈ AC ([0, R]) functions), ∀ ω ∈ S N −1 , N ≥ 2. We assume (absolutely continuous ν ν N −1 N −1 , D R− f (sω) s ∈ L q ([0, R]), ∀ ω ∈ S N −1 and that that D∗0 f (sω) s
7.2 Main Results
167
ν f (sω) s N −1 L q ([0,R]) , D νR− f (sω) s N −1 L q ([0,R]) ≤ M2 , max D∗0
where
N −1
M2 > 0, s ∈ [0, R] , ∀ ω ∈ S . Then (i)
f (y) dy − B(0,R)
S N −1
f (Rω) dω R
N −1
(R − t) ≤
N 1
2π 2 M2 ν+ p ν+ 1p t , + − t) (R N 1 (ν) 2 ν + 1p ( p (ν − 1) + 1) p
(7.103)
∀ t ∈ [0, R] , (ii) at t = R2 , the right hand side of (7.103) is minimized, and we get:
N R ≤ f dy − f dω (y) (Rω) 2 B(0,R) S N −1 π 2 M2 R ν+ p , 1 1 ν + 1p ( p (ν − 1) + 1) p 2ν−1− q 1
N
(ν)
N 2
(7.104)
(iii) if f (Rω) = 0, ∀ ω ∈ S N −1 , (i.e. f (·ω) vanish on ∂ B (0, R)), we obtain
B(0,R)
f (y) dy ≤
π 2 M2 R ν+ p , 1 1 ν + 1p ( p (ν − 1) + 1) p 2ν−1− q 1
N
(ν)
N 2
(7.105)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds RN ≤ f dy − N − j f dω (y) (Rω) N B (0,R) S N −1
N
(ν) ν +
1 p
2π 2 M2 1 p
( p (ν − 1) + 1)
N
R N
2
ν+ 1p
ν+ 1p 1 j ν+ p + N − j , (7.106)
(v) when N = 2 and j = 1, (7.106) turns to
f (y) dy − B(0,R)
RN 2
S N −1
f (Rω) dω ≤
π 2 M2 R ν+ p . N 1 1 2 (ν) ν + 1p ( p (ν − 1) + 1) p 2ν−1− q N
1
(7.107)
168
7 General Multidimensional Fractional Iyengar Inequalities
Proof Same as the proof of Theorem 7.19, just set there R1 = 0 and R2 = R and use (7.73). Next we give Canavati type results on the shell: Theorem 7.22 Let ν ≥ 1, n = [ν] . Consider f : A → R be Lebesgue integrable, which is not necessarily radial. Assume that f (sω) s N −1 ∈ C Rν 1 + ([R1 , R2 ]) ∩ C Rν 2 − ([R1 , R2 ]) in s ∈ [R1 , R2 ], ∀ ω ∈ S N −1 , N ≥ 2. Suppose there exists ψ1 > 0 ν ν N −1 N −1 ≤ such that max D R1 + f (sω) s , D R2 − f (sω) s ∞,[R1 ,R2 ] ∞,[R1 ,R2 ]
ψ1 , where s ∈ [R1 , R2 ], ∀ ω ∈ S N −1 . Then (i)
n−1 1 N −1 (k) f (sω) s (R1 ) dω (t − R1 )k+1 + f (y) dy − A (k + 1)! S N −1 k=0
(−1)
k S N −1
f (sω) s
N −1 (k)
(R2 ) dω (R2 − t)
k+1
≤
2π 2 ψ1 N (t − R1 )ν+1 + (R2 − t)ν+1 , + 2) (ν 2 N
(7.108)
∀ t ∈ [R1 , R2 ] , 2 , the right hand side of (7.108) is minimized, and we get: (ii) at t = R1 +R 2 n−1 (k) 1 (R2 − R1 )k+1 f (sω) s N −1 (R1 ) dω+ f (y) dy − k+1 A 2 (k + 1)! S N −1 k=0
(−1)k
S N −1
f (sω) s N −1
(k)
(R2 ) dω ≤
π2 ψ1 (R2 − R1 )ν+1 N , 2ν−1 2 (ν + 2) N
(7.109)
(k) (k) (iii) if f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , ∂ k ( f (sω)s N −1 ) (i.e. vanish on ∂ B (0, R ) and ∂ B (0, R )) for all k = 0, 1, . . . , n − 1, ∂s k
we obtain
1
2
N 2 ψ1 (R2 − R1 )ν+1 f (y) dy ≤ π , N (ν + 2) 2ν−1 A 2
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
(7.110)
7.2 Main Results
169
ν−1 R2 − R1 k+1 1 f (y) dy − A (k + 1)! N k=0
j k+1 S N −1
(−1)
k
N−j
N
2π 2 ψ1 N2 (ν + 2)
k+1
f (sω) s N −1
S N −1
R2 − R1
(k)
f (sω) s
ν+1
N
(R1 ) dω +
N −1 (k)
(R2 ) dω ≤
ν+1 j ν+1 + N − j ,
(7.111)
(k) (k) f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , k N −1 ∂ f (sω)s ) vanish on ∂ B (0, R ) and ∂ B (0, R )) for k = 1, . . . , n − 1, from (i.e. ( ∂s k 1 2 (7.111) we get: (v)
if
N −1 f (y) dy − R2 − R1 j R1 N A
N − j R2N −1
ψ1 (ν + 2)
S N −1
S N −1
R2 − R1
f (R2 ω) dω
ν+1
N
f (R1 ω) dω +
N 2 ≤ 2π N · 2
ν+1 j ν+1 + N − j ,
(7.112)
for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (7.112) turns to
N −1 f (y) dy − R2 − R1 R1 2 A
S N −1
f (R1 ω) dω +
R2N −1
π2 ψ1 (R2 − R1 )ν+1 N , 2ν−1 2 (ν + 2)
S N −1
f (R2 ω) dω
N
≤
(7.113)
Proof We apply Theorem 7.7 along with (7.74). See also in the Appendix 7.3 the general proving method in this chapter. We continue with Theorem 7.23 Let ν ≥ 1, n = [ν]. Consider f : A → R be Lebesgue integrable, which is not necessarily radial. Assume that f (sω) s N −1 ∈ C Rν 1 + ([R1 , R2 ]) ∩ C Rν 2 − ([R1 , R2 ]) in s ∈ [R1 , R2 ], ∀ ω ∈ S N −1 , N ≥ 2. Suppose there exists ψ2 > 0 such that
170
7 General Multidimensional Fractional Iyengar Inequalities
max D νR1 + f (sω) s N −1 L
1 ([R1 ,R2
ν N −1 , D f s (sω) R2 − ])
where s ∈ [R1 , R2 ], ∀ ω ∈ S N −1 . Then (i)
L 1 ([R1 ,R2 ])
≤ψ2 ,
n−1 (k) 1 f (sω) s N −1 (R1 ) dω (t − R1 )k+1 + f (y) dy − A (k + 1)! S N −1 k=0
(−1)k
S N −1
f (sω) s N −1
(k)
(R2 ) dω (R2 − t)k+1 ≤
2π 2 ψ2 N (t − R1 )ν + (R2 − t)ν , 2 (ν + 1) N
(7.114)
∀ t ∈ [R1 , R2 ] , (ii) when ν = 1, from (7.114), we have f (y) dy − A
S N −1
S N −1
f (R1 ω) dω R1N −1 (t − R1 ) +
N 2π 2 f (R2 ω) dω R2N −1 (R2 − t) ≤ N ψ2 (R2 − R1 ) , 2
∀ t ∈ [R1 , R2 ] , (iii) from (7.115), we obtain (ν = 1 case, t =
R2 − R1 f (y) dy − 2 A
S N −1
(7.115)
R1 +R2 ) 2
f (R1 ω) dω R1N −1 +
S N −1
f (R2 ω) dω R2N −1
N
≤ (iv) at t =
R1 +R2 , 2
2π 2 ψ2 (R2 − R1 ) , N2
(7.116)
ν > 1, the right hand side of (7.114) is minimized, and we get:
n−1 (k) 1 (R2 − R1 )k+1 f (sω) s N −1 (R1 ) dω+ f (y) dy − k+1 A 2 (k + 1)! S N −1 k=0
(−1)k
S N −1
f (sω) s N −1
(k)
(R2 ) dω ≤
7.2 Main Results
171
π2 ψ2 (R2 − R1 )ν N , 2ν−2 2 (ν + 1) N
(7.117)
(k) (k) f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , ∂ k ( f (sω)s N −1 ) vanish on ∂ B (0, R1 ) and ∂ B (0, R2 )) for all k = 0, 1, . . . , n − 1; (i.e. ∂s k ν > 1, we obtain from (7.117) that (v)
if
N 2 ψ2 (R2 − R1 )ν f (y) dy ≤ π , N (ν + 1) 2ν−2 A 2
(7.118)
which is a sharp inequality, (vi) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
n−1 R2 − R1 k+1 1 f (y) dy − A (k + 1)! N k=0
j
k+1 S N −1
k+1 (−1)k N − j
f (sω) s
N
S N −1
2π 2 ψ2 N 2 (ν + 1)
N −1 (k)
(R1 ) dω +
f (sω) s N −1
R2 − R1
ν
N
(k)
(R2 ) dω
≤
ν jν + N − j ,
(7.119)
(k) (k) (vii) if f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , ∂ k ( f (sω)s N −1 ) vanish on ∂ B (0, R ) and ∂ B (0, R )) for k = 1, . . . , n − 1, from (i.e. 1
∂s k
2
(7.119) we get:
N −1 f (y) dy − R2 − R1 j R1 N A
N−j
R2N −1
ψ2 (ν + 1)
S N −1
S N −1
f (R1 ω) dω +
N 2π 2 f (R2 ω) dω ≤ N · 2
R2 − R1 N
ν
ν jν + N − j ,
for j = 0, 1, 2, . . . , N ∈ N, (viii) when N = 2 and j = 1, (7.120) turns to
(7.120)
172
7 General Multidimensional Fractional Iyengar Inequalities
N −1 f (y) dy − R2 − R1 R 1 2 A
S N −1
f (R1 ω) dω + R2N −1
S N −1
f (R2 ω) dω
π2 ψ2 (R2 − R1 )ν N . 2ν−2 2 (ν + 1) N
≤
(7.121)
Proof We apply Theorem 7.8 along with (7.74). See also in the Appendix 7.3 the general proving method in this chapter. We also give Theorem 7.24 Let p, q > 1 : 1p + q1 = 1, ν ≥ 1, n = [ν]. Consider f : A → R be Lebesgue integrable, which is not necessarily radial. Assume that f (sω) s N −1 ∈ C Rν 1 + ([R1 , R2 ]) ∩ C Rν 2 − ([R1 , R2 ]), in s ∈ [R1 , R2 ] , ∀ ω ∈ S N −1 , N ≥ 2. Suppose that there exists ψ3 > 0 such that max D νR1 + f (sω) s N −1 L
q ([R1 ,R2
ν N −1 f s , (sω) D R2 − ])
L q ([R1 ,R2 ])
≤ ψ3 ,
where s ∈ [R1 , R2 ], ∀ ω ∈ S N −1 . Then (i) n−1 (k) 1 f (sω) s N −1 (R1 ) dω (t − R1 )k+1 + f (y) dy − A (k + 1)! S N −1 k=0
(−1)k
S N −1
f (sω) s N −1
(k)
(R2 ) dω (R2 − t)k+1 ≤
N
2π 2 ψ3 ν+ 1p ν+ 1p N − R , (7.122) + − t) (t (R ) 1 2 2 (ν) ν + 1 ( p (ν − 1) + 1) 1p p
∀ t ∈ [R1 , R2 ] , 2 , the right hand side of (7.122) is minimized, and we get: (ii) at t = R1 +R 2 n−1 1 (R2 − R1 )k+1 · f (y) dy − A 2k+1 (k + 1)! k=0
S N −1
f (sω) s N −1
(k)
(R1 ) dω + (−1)k
S N −1
f (sω) s N −1
(k)
(R2 ) dω ≤
7.2 Main Results
173
π2 ψ3 (R2 − R1 )ν+ p N , ν−1− q1 2 (ν) ν + 1 ( p (ν − 1) + 1) 1p 2 p 1
N
(7.123)
(k) (k) (iii) if f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , k N −1 ∂ f (sω)s ) vanish on ∂ B (0, R ) and ∂ B (0, R )) for all k = 0, 1, . . . , n − 1, (i.e. ( 1
∂s k
2
we obtain
1 N 2 ψ3 (R2 − R1 )ν+ p f (y) dy ≤ π , (7.124) 1 1 N 1 A 2ν−1− q 2 (ν) ν + p ( p (ν − 1) + 1) p which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
n−1 1 R2 − R1 k+1 f dy − (y) A (k + 1)! N k=0
j
k+1 S N −1
(−1)
k
N−j
k+1
f (sω) s
S N −1
N −1 (k)
f (sω) s
(R1 ) dω +
N −1 (k)
(R2 ) dω ≤
N
2π 2 ψ3 N · 1 1 2 (ν) ν + p − 1) + 1) p p ( (ν
R2 − R1
ν+ 1p
N
ν+ 1p 1 j ν+ p + N − j ,
(7.125)
(k) (k) f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, ∀ ω ∈ S N −1 , ∂ k ( f (sω)s N −1 ) vanish on ∂ B (0, R1 ) and ∂ B (0, R2 )) for k = 1, . . . , n − 1, from (i.e. ∂s k (7.125) we get: (v)
if
N −1 f (y) dy − R2 − R1 j R 1 N A
N − j R2N −1
S N −1
S N −1
f (R2 ω) dω
f (R1 ω) dω +
N 2 ≤ 2π N · 2
174
7 General Multidimensional Fractional Iyengar Inequalities
ψ3 1 (ν) ν + 1p ( p (ν − 1) + 1) p
R2 − R1
ν+ 1p
N
ν+ 1p 1 j ν+ p + N − j , (7.126)
for j = 0, 1, 2, . . . , N ∈ N, (vi) when N = 2 and j = 1, (7.126) turns to
N −1 f (y) dy − R2 − R1 R1 2 A
S N −1
f (R1 ω) dω +
R2N −1
S N −1
f (R2 ω) dω
π2 ψ3 (R2 − R1 )ν+ p ≤ N , 1 1 2 (ν) ν + 1 ( p (ν − 1) + 1) p 2ν−1− q p 1
N
(7.127)
Proof By Theorem 7.9 and (7.74). See also Appendix 7.3 for the general proving method here. Next we give Canavati type results on the ball: Theorem 7.25 Let 1 ≤ ν < 2. Consider f : B (0, R) → R be Lebesgue integrable, ν which is not necessarily radial. Assume that f (sω) s N −1 ∈ C0+ ([0, R]) ∩ ν N −1 , N ≥ 2. Suppose there exists φ C R− ([0, R]) , in s ∈ [0, R] , ∀ ω ∈ S 1 > 0 such ν ν ≤ φ1 , where that max D0+ f (sω) s N −1 ∞,[0,R] , D R− f (sω) s N −1 s ∈ [0, R] , ∀ ω ∈ S N −1 . Then (i)
f (y) dy − B(0,R)
∞,[0,R]
S N −1
f (Rω) dω R
N −1
(R − t) ≤
2π 2 φ1 N t ν+1 + (R − t)ν+1 , (ν + 2) 2 N
(7.128)
∀ t ∈ [0, R] , (ii) at t = R2 , the right hand side of (7.128) is minimized, and we get:
f (y) dy − B(0,R)
S N −1
f (Rω) dω
N π 2 φ1 R ν+1 R N ≤ , (7.129) 2 (ν + 2) N2 2ν−1
(iii) if f (Rω) = 0, ∀ ω ∈ S N −1 , (i.e. f (·ω) vanish on ∂ B (0, R)), we obtain
B(0,R)
which is a sharp inequality,
N π 2 φ1 R ν+1 f (y) dy ≤ , (ν + 2) N2 2ν−1
(7.130)
7.2 Main Results
175
(iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
B (0,R)
RN
f (y) dy − N
2π 2 φ1 (ν + 2) N2
N
R
N−j
ν+1
N
S N −1
f (Rω) dω ≤
ν+1 j ν+1 + N − j ,
(7.131)
(v) when N = 2 and j = 1, (7.131) turns to
B(0,R)
RN f (y) dy − 2
S N −1
N π 2 φ1 R ν+1 f (Rω) dω ≤ . (ν + 2) N2 2ν−1
(7.132)
Proof Same as the proof of Theorem 7.22, just set there R1 = 0 and R2 = R and use (7.73). We continue with Theorem 7.26 Let 1 ≤ ν < 2. Consider f : B (0, R) → R be Lebesgue integrable, ν which is not necessarily radial. Assume that f (sω) s N −1 ∈ C0+ ([0, R]) ∩ ν N −1 , N ≥ 2. Suppose there exists φ2 >0 such C R− ([0, R]) ,in s ∈ [0, R] , ∀ ω ∈ S ν ν f (sω) s N −1 L 1 ([0,R]) , D R− f (sω) s N −1 that max D0+ ≤ φ2 , where s ∈ [0, R] , ∀ ω ∈ S N −1 . Then (i)
f (y) dy − B(0,R)
L 1 ([0,R])
S N −1
f (Rω) dω R
N −1
(R − t) ≤
ν 2π 2 φ2 N t + (R − t)ν , 2 (ν + 1) N
(7.133)
∀ t ∈ [0, R] , (ii) when ν = 1, from (7.133), we have
f (y) dy − B(0,R)
S N −1
f (Rω) dω R
N −1
∀ t ∈ [0, R] , (iii) from (7.134), we obtain (ν = 1 case, t =
B(0,R)
RN f (y) dy − 2
S N −1
N 2π 2 (R − t) ≤ N φ2 R, 2
(7.134)
R ) 2
N 2π 2 f (Rω) dω ≤ N φ2 R, 2
(7.135)
176
7 General Multidimensional Fractional Iyengar Inequalities
(iv) at t =
R , 2
ν > 1, the right hand side of (7.133) is minimized, and we get:
N π2 Rν φ2 ν−2 , f (Rω) dω ≤ N 2 (ν + 1) 2 B(0,R) S N −1 (7.136) (v) if f (Rω) = 0, ∀ ω ∈ S N −1 , (from (7.136)), we get RN f (y) dy − 2
B(0,R)
f (y) dy ≤
π 2 φ2 R ν , (ν + 1) 2ν−2 N
N 2
(7.137)
which is a sharp inequality, (vi) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
B (0,R)
f (y) dy −
RN N
N
2π 2 φ2 N2 (ν + 1)
N−j
R N
ν
S N −1
f (Rω) dω ≤
ν jν + N − j ,
(7.138)
(vii) when N = 2 and j = 1, (7.138) turns to
B(0,R)
RN f (y) dy − 2
S N −1
N π 2 φ2 Rν f (Rω) dω ≤ N . 2 (ν + 1) 2ν−2
(7.139)
Proof Same as the proof of Theorem 7.23, just set there R1 = 0 and R2 = R and use (7.73). We continue with Theorem 7.27 Let p, q > 1 : 1p + q1 = 1, 1 ≤ ν < 2. Consider f : B (0, R) → R be Lebesgue integrable, which is not necessarily radial. Assume that f (sω) s N −1 ∈ ν ν C0+ ([0, R]) ∩ C R− ([0, R]) , in s ∈ [0, R] , ∀ ω ∈ S N −1 , N ≥ 2. Suppose ν such that max D0+ f (sω) s N −1 L q ([0,R]) , there exists φ3 > 0 ν N −1 ≤ φ3 , where s ∈ [0, R] , ∀ ω ∈ S N −1 . D R− f (sω) s L q ([0,R])
Then (i)
f (y) dy − B(0,R)
S N −1
f (Rω) dω R
N −1
(R − t) ≤
N 1
1 2π 2 φ3 t ν+ p + (R − t)ν+ p , N 1 (ν) 2 ν + 1p ( p (ν − 1) + 1) p
(7.140)
7.2 Main Results
177
∀ t ∈ [0, R] , (ii) at t = R2 , the right hand side of (7.140) is minimized, and we get:
f (y) dy −
f (Rω) dω
S N −1
B(0,R)
π 2 φ3 R ν+ p , 1 1 ν + 1p ( p (ν − 1) + 1) p 2ν−1− q 1
N
(ν)
N 2
R N ≤ 2 (7.141)
(iii) if f (Rω) = 0, ∀ ω ∈ S N −1 , (i.e. f (·ω) vanish on ∂ B (0, R)), we obtain
B(0,R)
f (y) dy ≤
π 2 φ3 R ν+ p , 1 1 (ν) N2 ν + 1p ( p (ν − 1) + 1) p 2ν−1− q 1
N
(7.142)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
B (0,R)
f (y) dy −
RN N
N−j
N
(ν) ν +
1 p
2π 2 φ3 1
( p (ν − 1) + 1) p
N
R
S N −1
f (Rω) dω ≤
ν+ 1p
N
2
ν+ 1p 1 j ν+ p + N − j , (7.143)
(v) when N = 2 and j = 1, (7.143) turns to
B(0,R)
RN f (y) dy − 2
S N −1
f (Rω) dω ≤
π 2 φ3 R ν+ p . 1 1 N2 (ν) ν + 1p ( p (ν − 1) + 1) p 2ν−1− q N
1
(7.144)
Proof Same as the proof of Theorem 7.24, just set there R1 = 0 and R2 = R and use (7.73). Next we give Conformable type results on the shell: Theorem 7.28 Let α ∈ (n, n + 1], n ∈ N; β = α − n. Consider f : A → R be Lebesgue integrable, which is not necessarily radial. Assume that f (sω) ∈ C n+1 ([R1 , R2 ]) , in s ∈ [R1 , R2 ] , ∀ ω ∈ S N −1 , N ≥ 2. Suppose there exists W 1 > 0 such that max TαR1 f (sω) s N −1 ∞,[R1 ,R2 ] , αR2 T f (sω) s N −1 ∞,[R1 ,R2 ] ≤ W1 , ∀ ω ∈ S N −1 , where s ∈ [R1 , R2 ] .
178
7 General Multidimensional Fractional Iyengar Inequalities
Then (i) n (k) 1 f (sω) s N −1 (R1 ) dω (t − R1 )k+1 + f (y) dy − A (k + 1)! S N −1 k=0
(−1)k
S N −1
f (sω) s N −1
(k)
(R2 ) dω (R2 − t)k+1 ≤
2π 2 (β) W1 N (z − R1 )α+1 + (R2 − z)α+1 , 2 (α + 2) N
(7.145)
∀ z ∈ [R1 , R2 ] , 2 , the right hand side of (7.145) is minimized, and we get: (ii) at z = R1 +R 2 n 1 (R2 − R1 )k+1 · f (y) dy − A 2k+1 (k + 1)! k=0
S N −1
f (sω) s
N −1 (k)
(R1 ) dω + (−1)
k S N −1
f (sω) s
N −1 (k)
(R2 ) dω ≤
π 2 (β) W1 (R2 − R1 )α+1 , 2α−1 N2 (α + 2) N
(7.146)
(k) (k) (iii) assuming f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, for k = 0, 1, . . . , n, ∀ ω ∈ S N −1 , we obtain N 2 (β) W (R2 − R1 )α+1 1 f (y) dy ≤ π , N (α + 2) 2α−1 A 2 which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
n 1 R2 − R1 k+1 f (y) dy − A (k + 1)! N k=0
j
k+1 S N −1
(−1)
k
N−j
k+1
f (sω) s
S N −1
N −1 (k)
f (sω) s
(R1 ) dω +
N −1 (k)
(R2 ) dω ≤
(7.147)
7.2 Main Results
179
N
2π 2 (β) W1 N2 (α + 2)
R2 − R1
α+1
N
α+1 j α+1 + N − j ,
(7.148)
(k) (k) (v) if f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, for k = 1, . . . , n, ∀ ω ∈ S N −1 , from (7.148) we get:
N −1 f (y) dy − R2 − R1 j R1 N A
N − j R2N −1
N
2π 2 (β) W1 N2 (α + 2)
S N −1
R2 − R1
f (R2 ω) dω
S N −1
α+1
N
f (R1 ω) dω + ≤
α+1 j α+1 + N − j ,
(7.149)
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (7.149) turns to
N −1 f (y) dy − R2 − R1 R1 2 A
S N −1
f (R1 ω) dω +
R2N −1
S N −1
f (R2 ω) dω
π 2 (β) W1 (R2 − R1 )α+1 . 2α−1 N2 (α + 2) N
≤
(7.150)
Proof It is based on Theorem 7.13 and (7.74). See also Appendix 7.3.
We give Theorem 7.29 Let α ∈ (n, n + 1], n ∈ N; β = α − n. Let also p1 , p2 , p3 > 1 : 1 + p12 + p13 = 1, with β > p11 + p13 . Consider f : A → R be Lebesgue integrable, p1 which is not necessarily radial. Assume that f (sω) ∈ C n+1 ([R1 , R2 ]) , in s ∈ [R1 , R2 ] , ∀ ω ∈ S N −1 , N ≥ 2. Suppose there exists W2 > 0 such that max TαR1 f (sω) s N −1 p3 ,[R1 ,R2 ] , αR2 T f (sω) s N −1 p3 ,[R1 ,R2 ] ≤ W2 , ∀ ω ∈ S N −1 , where s ∈ [R1 , R2 ] . Then (i) n (k) 1 f (sω) s N −1 (R1 ) dω (t − R1 )k+1 + f (y) dy − A (k + 1)! S N −1 k=0
(−1)k
S N −1
f (sω) s N −1
(k)
(R2 ) dω (R2 − t)k+1 ≤
180
7 General Multidimensional Fractional Iyengar Inequalities N
2π 2 W2 N 1 1 2 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α +
(z − R1 )
α+ p1 + p1 1
+ (R2 − z)
2
α+ p1 + p1 1
1 p1
2
+
1 p2
·
,
(7.151)
∀ z ∈ [R1 , R2 ] , 2 , the right hand side of (7.151) is minimized, and we get: (ii) at z = R1 +R 2 n 1 (R2 − R1 )k+1 · f (y) dy − A 2k+1 (k + 1)! k=0
S N −1
f (sω) s N −1
(k)
(R1 ) dω + (−1)k
S N −1
f (sω) s N −1
N
π2 W2 1 1 N2 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α +
+
1 p1
1 p2
(k)
(R2 ) dω ≤
(R2 − R1 )
2
α+ p1 + p1 1
α−1− p1
2
,
3
(7.152) (k) (k) (iii) assuming f (sω) s N −1 for (R1 ) = f (sω) s N −1 (R2 ) = 0, k = 0, 1, . . . , n, ∀ ω ∈ S N −1 , we obtain N 2 f (y) dy ≤ π · N A 2 W2 n! ( p1 n + 1)
1 p1
( p2 (β − 1) + 1)
1 p2
α+
1 p1
+
1 p2
(R2 − R1 ) 2
α+ p1 + p1 1
α−1−
1 p3
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
n R2 − R1 k+1 1 f (y) dy − A (k + 1)! N k=0
j
k+1 S N −1
k+1 (−1)k N − j
f (sω) s
S N −1
N −1 (k)
(R1 ) dω +
f (sω) s N −1
(k)
(R2 ) dω
≤
2
,
(7.153)
7.2 Main Results
181 N
2π 2 W2 N 1 1 2 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α +
R2 − R1
α+ p1 + p1 1 2
N
j
α+ p1 + p1 1
2
+
1 p1
1 p2
·
α+ p1 + p1 1 2 , + N−j
(7.154)
(k) (k) (v) if f (sω) s N −1 (R1 ) = f (sω) s N −1 (R2 ) = 0, for k = 1, . . . , n, ∀ ω ∈ S N −1 , from (7.154) we get:
N −1 f (y) dy − R2 − R1 j R 1 N A
N−j
R2N −1
S N −1
f (R1 ω) dω +
f (R2 ω) dω ≤
S N −1
N
≤
2π 2 W2 1 1 N2 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α +
R2 − R1 N
α+ p1 + p1 1 2
j
α+ p1 + p1 1
2
1 p1
+
1 p2
α+ p1 + p1 1 2 , + N−j
(7.155)
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (7.155) turns to
R2 − R1 N −1 R f dy − (y) 1 2 A
S N −1
f (R1 ω) dω + R2N −1
S N −1
f (R2 ω) dω
N
π2 W2 ≤ N 1 1 2 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + (R2 − R1 ) 2
α+ p1 + p1 1
α−1− p1
2
1 p1
+
1 p2
,
(7.156)
3
Proof It is based on Theorem 7.14 and (7.74). See also Appendix 7.3.
Next we give conformable results on the ball. Theorem 7.30 Let α ∈ (0, 1] and consider f : B (0, R) → R to be Lebesgue integrable, which is not necessarily radial. Assume that f (sω) ∈ C 1 ([0, R]) , in s ∈ N −1 , N ≥ 2. Suppose there exists θ1 > 0 such that [0, R] , ∀ ω ∈ S R max Tα f (sω) s N −1 ∞,[0,R] , αR T f (sω) s N −1 ∞,[0,R] ≤ θ1 , ∀ ω ∈ S N −1 , where s ∈ [0, R] .
182
7 General Multidimensional Fractional Iyengar Inequalities
Then (i)
f (y) dy −
S N −1
B(0,R)
f (Rω) dω R N −1 (R − t) ≤
α+1 2π 2 θ1 N z + (R − z)α+1 , 2 α (α + 1) N
(7.157)
∀ z ∈ [0, R] , (ii) at t = R2 , the right hand side of (7.157) is minimized, and we get:
f (y) dy −
S N −1
B(0,R)
f (Rω) dω
R N ≤ 2
π2 R α+1 θ1 N , 2 α (α + 1) 2α−1 N
(7.158)
(iii) if f (Rω) = 0, ∀ ω ∈ S N −1 , we obtain
B(0,R)
f (y) dy ≤
π 2 θ1 R α+1 N , 2 α (α + 1) 2α−1 N
(7.159)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
B (0,R)
f (y) dy −
N
2π 2 θ1 N2 α (α + 1)
RN N R N
N−j
α+1
S N −1
f (Rω) dω ≤
α+1 j α+1 + N − j ,
(7.160)
(v) when N = 2 and j = 1, (7.160) turns to
B(0,R)
RN f (y) dy − 2
S N −1
N π2 R α+1 θ1 f (Rω) dω ≤ N . 2 α (α + 1) 2α−1
(7.161)
Proof It is based on Theorem 7.28, just there R1 = 0 and R2 = R and use (7.73). Notice here n = 0, and α = β. Next we give conformable results on the ball. Theorem 7.31 Let α ∈ (0, 1] and let also p1 , p2 , p3 > 1 : α>
1 p1
+
1 p3
1 p1
+
1 p2
+
1 p3
= 1, with
and consider f : B (0, R) → R be Lebesgue integrable, which is not
7.2 Main Results
183
necessarily radial. Assume that f (sω) ∈ C 1 ([0, R]) , in s ∈ [0, R] , ∀ ω∈ S N −1 , N ≥ 2. Suppose there exists θ2 > 0 such that max{TαR f (sω) s N −1 p3 ,[0,R] , R T f (sω) s N −1 } ≤ θ2 , ∀ ω ∈ S N −1 , where s ∈ [0, R] . α p3 ,[0,R] Then (i)
N −1 ≤ f dy − f dω R − t) (y) (Rω) (R N −1 B(0,R)
S
N
2π 2 θ2 1 N2 ( p2 (α − 1) + 1) p2 α + z
α+ p1 + p1 1
+ (R − z)
2
+
1 p1
α+ p1 + p1 1
2
1 p2
·
,
(7.162)
∀ z ∈ [0, R] , (ii) at t = R2 , the right hand side of (7.162) is minimized, and we get:
f (y) dy −
S N −1
B(0,R)
f (Rω) dω
N
π2 θ2 1 N2 ( p2 (α − 1) + 1) p2 α +
1 p1
+
1 p2
R N ≤ 2
R
α+ p1 + p1 1
2
2
α−1− p1
,
(7.163)
3
(iii) if f (Rω) = 0, ∀ ω ∈ S N −1 , we obtain
B(0,R)
N π2 f (y) dy ≤ N · 2
θ2 ( p2 (α − 1) + 1)
1 p2
α+
1 p1
+
1 p2
R
α+ p1 + p1 1
2
2
α−1− p1
,
3
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
B (0,R)
f (y) dy −
RN N
N−j
S N −1
f (Rω) dω ≤
N
2π 2 θ2 1 N2 ( p2 (α − 1) + 1) p2 α +
1 p1
+
1 p2
·
(7.164)
184
7 General Multidimensional Fractional Iyengar Inequalities
R
α+ p1 + p1 1 2
j
N
α+ p1 + p1 1
2
α+ p1 + p1 1 2 , + N−j
(7.165)
(v) when N = 2 and j = 1, (7.165) turns to
B(0,R)
RN f (y) dy − 2
S N −1
f (Rω) dω ≤
N
π2 θ2 N 1 2 ( p2 (α − 1) + 1) p2 α +
1 p1
+
1 p2
R
α+ p1 + p1
2
1
2
α−1− p1
.
(7.166)
3
Proof It is based on Theorem 7.29, just there R1 = 0 and R2 = R and use (7.73). Notice here n = 0, and α = β.
7.3 Appendix Proof (Detailed proof of Theorem 7.17—serving as a model proof for the rest of this chapter.) We apply Theorem 7.4 (i) for f (sω) s N −1 : n−1 R2 (k) 1 f (sω) s N −1 f (sω) s N −1 ds − (R1 ) (t − R1 )k+1 + R1 + 1)! (k k=0
(k) k (−1) f (sω) s N −1 (R2 ) (R2 − t)k+1 ≤ ν max D∗R f (sω) s N −1 L 1
∞ ([R1 ,R2
, D νR2 − f (sω) s N −1 L ])
∞ ([R1 ,R2 ])
(ν + 2)
(t − R1 )ν+1 + (R2 − t)ν+1 ≤
·
(7.167)
K1 (t − R1 )ν+1 + (R2 − t)ν+1 =: ρ (t) , (ν + 2) ∀ t ∈ [R1 , R2 ], and ∀ ω ∈ S N −1 . Equivalently, we have that −ρ (t) ≤
R2 R1
f (sω) s N −1 ds −
n−1 k=0
(k) 1 f (sω) s N −1 (R1 ) (t − R1 )k+1 (k + 1)!
(k) + (−1)k f (sω) s N −1 (R2 ) (R2 − t)k+1 ≤ ρ (t) ,
(7.168)
7.3 Appendix
185
∀ t ∈ [R1 , R2 ], and ∀ ω ∈ S N −1 . Therefore it holds dω ≤ −ρ (t) S N −1
n−1 k=0
1 (k + 1)!
S N −1
(−1)
k S N −1
S N −1
f (sω) s
f (sω) s
R2
f (sω) s N −1 ds dω−
R1
N −1 (k)
N −1 (k)
(R1 ) dω (t − R1 )k+1 +
(R2 ) dω (R2 − t)
k+1
≤
≤ ρ (t)
S N −1
dω, ∀ t ∈ [R1 , R2 ] .
By (7.74) we obtain N
2π 2 −ρ (t) N ≤ 2 n−1 k=0
1 (k + 1)!
S N −1
(−1)
k S N −1
f (sω) s
f (sω) s
(7.169)
f (y) dy− A
N −1 (k)
N −1 (k)
(R1 ) dω (t − R1 )k+1 +
(R2 ) dω (R2 − t)
k+1
≤
N
ρ (t)
2π 2 , ∀ t ∈ [R1 , R2 ] . N2
(7.170)
Consequently it holds n−1 1 N −1 (k) f (sω) s (R1 ) dω (t − R1 )k+1 f (y) dy − A N −1 (k + 1)! S k=0
+ (−1)k
S N −1
f (sω) s N −1
(k)
N 2π 2 (R2 ) dω (R2 − t)k+1 ≤ ρ (t) N = 2
2π 2 K1 N (t − R1 )ν+1 + (R2 − t)ν+1 , 2 (ν + 2) N
(7.171)
186
7 General Multidimensional Fractional Iyengar Inequalities
∀ t ∈ [R1 , R2 ] , proving Theorem 7.17 (i). Next consider ϕ (t) := (t − R1 )ν+1 + (R2 − t)ν+1 , ∀ t ∈ [R1 , R2 ] . Then
ϕ (t) = (ν + 1) (t − R1 )ν − (R2 − t)ν = 0,
2 and ϕ has the only critical number t = R1 +R . Hence ϕ (t) has a minimum over R1 +R2 (R2 −R1 )ν+1 2 = . [R1 , R2 ] which is ϕ 2 2ν Consequently, it holds (by (7.171))
n−1 1 (R2 − R1 )k+1 f (y) dy − A 2k+1 (k + 1)! k=0
S N −1
f (sω) s
N −1 (k)
(R1 ) dω + (−1)
k S N −1
f (sω) s
N −1 (k)
(R2 ) dω
π2 K1 (R2 − R1 )ν+1 N , 2ν−1 2 (ν + 2) N
≤
proving Theorem 7.17 (ii). The rest of Theorem 7.17 is obvious or follows the same way as above.
(7.172)
The rest of the proofs of this chapter as similar are omitted.
References 1. T. Abdeljawad, On conformable fractional calculus. J. Comput. Appl. Math. 279, 57–66 (2015) 2. G.A. Anastassiou, Fractional Differentiation Inequalities, Research Monograph (Springer, New York, 2009) 3. G.A. Anastassiou, On right fractional calculus. Chaos Solitons Fractals 42, 365–376 (2009) 4. G.A. Anastassiou, Intelligent Mathematical Computational Analysis (Springer, Heidelberg, New York, 2011) 5. G.A. Anastassiou, Mixed Conformable Fractional Approximation by Sublinear Operators. Indian J. Math. 60(1), 107–140 (2018) 6. G.A. Anastassiou, Canavati fractional Iyengar type inequalities, Analele Universitatii Oradea, Fasc. Matematica, Tom XXVI 1, 141–151 (2019) 7. G.A. Anastassiou, General Multidimensional Fractional Iyengar type inequalities, Revista De la Real Academia de Ciencias Exactas. Fisicasy Naturales Serie A. Matematicas (RACSAM) 113(3), 2537–2573 (2019) 8. G.A. Anastassiou, Caputo fractional Iyengar type inequalities, submitted (2018) 9. G.A. Anastassiou, Conformable fractional Iyengar type inequalities, submitted (2018)
References
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10. J.A. Canavati, The Riemann-Liouville Integral. Nieuw Archief Voor Wiskunde 5(1), 53–75 (1987) 11. K.S.K. Iyengar, Note on an inequality. Math. Stud. 6, 75–76 (1938) 12. W. Rudin, Real and Complex Analysis, International Student edn. (Mc Graw Hill, London, 1970) 13. D. Stroock, A Concise Introduction to the Theory of Integration, 3rd edn. (Birkhaüser, Boston, Basel, Berlin, 1999)
Chapter 8
Delta Time Scales Iyengar Inequalities
Here we give the necessary background on delta time scales approach. Then we present general related time scales delta Iyengar type inequalities for all basic norms. We finish with applications to specific time scales like R, Z and q Z , q > 1. See also [5].
8.1 Introduction We are motivated by the following famous Iyengar inequality (1938), [12]. Theorem 8.1 Let f be a differentiable function on [a, b] and f (x) ≤ M. Then
a
b
f (x) d x −
M (b − a) 2 ( f (b) − f (a))2 1 . − (b − a) ( f (a) + f (b)) ≤ 2 4 4M (8.1)
We present generalized analogs of (8.1) to time scales. Motivation comes also from [2–4, 7, 8].
8.2 Background Here basics on time scales come from [3, 4, 7–9]. We need Definition 8.2 A time scale is an arbitrary nonempty closed subset of the real numbers, e.g. R, Z, q N0 = {q k |k ∈ N0 = N ∪ {0}, q > 1}. © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_8
189
190
8 Delta Time Scales Iyengar Inequalities
Definition 8.3 If T is a time scale, then we define the forward jump operator σ : T −→ T by σ (t) = inf{s ∈ T |s > t}, ∀ t ∈ T ; the backward jump operator ρ : T −→ T by ρ (t) = sup{s ∈ T |s < t}, ∀ t ∈ T ; and the graininess function μ : T → R+ = [0, ∞), by μ (t) = σ (t) − t, ∀ t ∈ T . Furthermore for a function f : T → R, we define f σ (t) = f (σ (t)) , ∀ t ∈ T ; and f ρ (t) = f (ρ (t)), ∀ t ∈ T . In this definition we use inf ∅ = sup T (i.e., σ (t) = t if t is the maximum of T ) and sup ∅ = inf T (i.e., ρ (t) = t if t is the minimum of T ). We call t ∈ T right-scattered if t < σ (t), t ∈ T right-dense if t = σ (t), t ∈ T leftscattered if ρ (r ) < t, t ∈ T left-dense if ρ (t) = t, t ∈ T isolated if ρ (t) < t < σ (t), t ∈ T dense if ρ (t) = t = σ (t). We notice that ρ is an increasing function, so is ρ2 (t) = ρ (ρ (t)) , . . . , so that n−1 n ρ (t) = ρ ρ (t) is increasing in t for n ∈ N. Since T is closed subset of R we have that σ (t) , ρ (t) ∈ T, for t ∈ T . Definition 8.4 ([9]) A function f : T → R is called rd-continuous (denoted by Cr d ) if it is continuous at right-dense points of T and its left-sided limits are finite at leftdense points of T . If T = R, then f : R → R is rd-continuous iff f is continuous. Also, if T = Z, then any function defined on Z is rd-continuous ([10]). Definition 8.5 ([9]) If sup T < ∞ and sup T is left-scattered, we let T k := T − {sup T }, otherwise we let T k := T the time scale. T − (ρ (sup T ) , sup T , if sup T < ∞, In summary, T k = T, if sup T = ∞. Definition 8.6 ([9]) Assume f : T → R is a function and let t ∈ T k . Then we define f (t) to be the number (provided it exists) with the property that given any ε > 0 , there is a neighborhood U of t such that [ f (σ (t)) − f (s)] − f (t) [σ (t) − s] ≤ ε |σ (t) − s| , ∀s ∈ U. We call f (t) the delta (or Hilger [11]) derivative of f at t. If T = R, then f = f , whereas if T = Z, then f (t) = f (t) = f (t + 1) − f (t) , the usual forward difference operator. Theorem 8.7 ([9]) (Existence of Antiderivatives) Let f be rd-continuous. Then f has an antiderivative F satisfying F = f. Definition 8.8 ([9]) If f is rd-continuous and t0 ∈ T , then we define the integral
t
F (t) = t0
f (τ ) τ for t ∈ T.
8.2 Background
191
Therefore for f ∈ Cr d (T ) we have by definition
b
f (τ ) τ = F (b) − F (a) ,
a
where F = f. If T = R, then
b
f (t) t =
b
f (t) dt,
a
a
where the integral on the right hand side is the Riemann integral ([10]). If every point in T is isolated and a < b are in T , then ([10])
b
f (t) t =
a
ρ(b)
f (t) μ (t) .
t=a
Theorem 8.9 ([9]) Let f, g be rd-continuous on T , a, b, c ∈ T and α, β ∈ R. Then b b b (1) a (α f (t) + βg (t)) t = α a f (t) t + β a g (t) t, b a (2) a f (t) t = − b f (t) t, b c b (3) a f (t) t = a f (t) t + c f (t) t, b b (4) a f (t) g (t) t = ( f g) (b) − ( f g) (a) − a f (t) g (σ (t)) t, a (5) a f (t) t = 0, b (6) a 1t = b − a. Theorem 8.10 ([2], Hölder’s inequality) Let a, b ∈ T , a ≤ b, and f, g : T → R be rd-continuous. Then
b
| f (t)| |g (t)| t ≤
a
where p, q > 1 :
b
| f (t)| t
a 1 p
+
1 q
1p
p
b
|g (t)| t q
q1
,
(8.2)
a
= 1.
Theorem 8.11 ([9]) Let f, g ∈ Cr d (T ), a,b ∈ T , a ≤ b. Then b b (1) if | f (t)| ≤ g (t) on [a, b) ∩ T , then a f (t) t ≤ a g (t) t, b (2) if f (t) ≥ 0, for all a ≤ t < b and t ∈ T , then a f (t) t ≥ 0. Corollary 8.12 ([3]) Let f ∈ Cr d (T ) ; a, b, c ∈ T , with c ∈ [a, b]; f (t) ≥ 0, ∀ t ∈ [a, b]. Then b c f (t) t ≤ f (t) t. a
a
Definition 8.13 ([3]) For a function f : T → R we consider the second derivative k 2 f provided f is differentiable on T k = T k with derivative f = f : 2 n n T k → R. Similarly we define higher order derivatives f : T k → R.
192
8 Delta Time Scales Iyengar Inequalities
Similarly we define σ 2 (t) = σ (σ (t)) , . . . , σ n (t) = σ σ n−1 (t) , n ∈ N. For 0 0 convenience we put ρ0 (t) = σ 0 (t) = t, f = f , T k = T . n l Notice T k ⊂ T k , l ∈ {0, 1, . . . , n}. Denote by Crnd (T ) the space of all functions f ∈ Cr d (T ) such that f ∈ Cr d (T ) for i = 1, . . . , n ∈ N. In this last case T k = T is needed. We need i
Theorem 8.14 ([6, 10], Taylor’s formula) Assume Tk = T and f ∈ Crnd (T), n ∈ N and s, t ∈ T . Here h 0 (t, s) = 1, ∀ s, t ∈ T; k ∈ N0 , and h k+1 (t, s) =
t
h k (τ , s) τ , ∀s, t ∈ T.
s
(then h k (t, s) = h k−1 (t, s), for k ∈ N, ∀ t ∈ T, for each s ∈ T fixed). Then f (t) =
n−1
f (s) h k (t, s) + k
t
h n−1 (t, σ (τ )) f (τ ) τ . n
(8.3)
s
k=0
Remark 8.15 (to Theorem 8.14) By [10], we have h 1 (t, s) = t − s, ∀ s, t ∈ T. So if t ≥ s then h 1 (t, s) ≥ 0, h 2 (t, s) ≥ 0, . . . , h n−1 (t, s) ≥ 0. However for n odd number h n−1 (t, σ (τ )) ≥ 0 for all s ≤ τ ≤ t (see [4], p. 635). Also it holds ([1]) h k (t, s) ≤
(t − s)k , ∀t ≥ s, k ∈ N0 . k!
Corollary 8.16 ([3]) (to Theorem 8.14) Assume f ∈ Crnd (T) and s, t ∈ T. Let m∈ N with m < n Then f
m
(t) =
n−m−1 k=0
f
k+m
t
(s) h k (t, s) +
h n−m−1 (t, σ (τ )) f (τ ) τ . n
(8.4)
s
Proof Use Theorem 8.14 with n and f substituted by n−m and f , respectively. m
Corollary 8.17 ([3]) Let f ∈ Cr d (T); a, b∈T, such that f (t) > 0, ∀ t ∈ [a, b] ∩ T, b then a f (t) t > 0. We mention also Lemma 8.18 ([4], p. 631) Let the time scale T be such that Tk = T. Let h k : T2 → t R, k ∈ N0 , such that h 0 (t, s) ≡ 1, ∀ s, t ∈ T, and h k+1 (t, s) = s h k (τ , s) τ , ∀ s, t ∈ T, for all k ∈ N0 . Then h k (t, s) is continuous in s ∈ T, for each fixed t ∈ T; and continuous in t ∈ T, for each fixed s ∈ T. Also it holds that h k (t, σ (s)) is rd-continuous in s ∈ T, for each fixed t ∈ T; for all k ∈ N0 .
8.3 Main Results
193
8.3 Main Results In this chapter we assume that Tk = T. Next, we present Iyengar type inequality on time scales for all norms · p , 1 ≤ p ≤ ∞. Theorem 8.19 Let f ∈ Crnd (T), n is an odd number, a, b ∈ T; a ≤ b. Here σ is continuous and h n−1 (t, s) jointly continuous. Then (1) n−1 b
n k k f (a) h k+1 (x, a) − f (b) h k+1 (x, b) ≤ f ∞,[a,b]∩T f (t) t − a k=0
x
a
t
h n−1 (t, σ (τ )) τ t +
a
b
x
b
h n−1 (t, σ (τ )) τ t ,
t
(8.5) ∀ x ∈ [a, b] ∩ T, k k (2) assuming f (a) = f (b) = 0, k = 0, 1, . . . , n − 1, we get from (8.5) that
b
a
x
a
t
n f (t) t ≤ f ∞,[a,b]∩T
h n−1 (t, σ (τ )) τ t +
a
b
x
b
h n−1 (t, σ (τ )) τ t ,
t
(8.6)
∀ x ∈ [a, b] ∩ T, (21 ) when x = a we get from (8.6) that
n f (t) t ≤ f ∞,[a,b]∩T
b
a
b
a
b
h n−1 (t, σ (τ )) τ t ,
(8.7)
h n−1 (t, σ (τ )) τ t ,
(8.8)
t
(22 ) when x = b we get from (8.6) that
n f (t) t ≤ f ∞,[a,b]∩T
b
a
a
b
t
a
(23 ) by (8.7) and (8.8) we get
a
b
b
min a
t
b
n f (t) t ≤ f ∞,[a,b]∩T
h n−1 (t, σ (τ )) τ t , a
b
a
t
h n−1 (t, σ (τ )) τ t
, (8.9)
194
8 Delta Time Scales Iyengar Inequalities
and k k (3) assuming f (a) = f (b) = 0, k = 1, . . . , n − 1, by (8.5) we have
n f (t) t − [ f (a) (x − a) + f (b) (b − x)] ≤ f ∞,[a,b]∩T
b
a
a
x
t
h n−1 (t, σ (τ )) τ t +
a
b x
b
h n−1 (t, σ (τ )) τ t ,
t
(8.10)
∀ x ∈ [a, b] ∩ T.
n Proof By [9], p. 23 we have that f ∞,[a,b]∩T < ∞. By Theorem 8.14 , see (8.3), we have f (t) −
n−1
f
k
t
(a) h k (t, a) =
h n−1 (t, σ (τ )) f (τ ) τ ,
(8.11)
h n−1 (t, σ (τ )) f (τ ) τ ,
(8.12)
n
a
k=0
and f (t) −
n−1
f
k
t
(b) h k (t, b) =
n
b
k=0
∀ t ∈ [a, b] ∩ T. Then we get t n−1 (8.11) n k f (a) h k (t, a) ≤ f ∞,[a,b]∩T h n−1 (t, σ (τ )) τ , (8.13) f (t) − a k=0
and
b n−1 n (8.12) k f (b) h k (t, b) = h n−1 (t, σ (τ )) f (τ ) τ f (t) − t k=0
b
≤ t
n h n−1 (t, σ (τ )) τ f ∞,[a,b]∩T .
(8.14)
Therefore it holds (by (8.13), (8.14)) n − f ∞,[a,b]∩T
t
h n−1 (t, σ (τ )) τ ≤ f (t) −
a
n ≤ f ∞,[a,b]∩T
n−1 k=0
a
t
h n−1 (t, σ (τ )) τ
f (a) h k (t, a) k
8.3 Main Results
195
and n − f ∞,[a,b]∩T
b
h n−1 (t, σ (τ )) τ
≤ f (t) −
t
n−1
f (b) h k (t, b) k
k=0
n ≤ f ∞,[a,b]∩T
b
h n−1 (t, σ (τ )) τ ,
t
∀ t ∈ [a, b] ∩ T. Consequently we have n−1
n k f (a) h k (t, a) − f ∞,[a,b]∩T
n−1
h n−1 (t, σ (τ )) τ ≤ f (t)
(8.15)
a
k=0
≤
t
n k f (a) h k (t, a) + f ∞,[a,b]∩T
t
h n−1 (t, σ (τ )) τ
a
k=0
and n−1
n k f (b) h k (t, b) − f ∞,[a,b]∩T
n−1
h n−1 (t, σ (τ )) τ
≤ f (t) (8.16)
t
k=0
≤
b
n k f (b) h k (t, b) + f ∞,[a,b]∩T
b
h n−1 (t, σ (τ )) τ ,
t
k=0
∀ t ∈ [a, b] ∩ T. Let any x ∈ [a, b] ∩ T, then integrating (8.15), (8.16) we obtain: n−1
f
k
n (a) h k+1 (x, a) − f
∞,[a,b]∩T
≤
x
a
k=0
x
t
h n−1 (t, σ (τ )) τ t
a
f (t) t ≤
(8.17)
a n−1 k=0
f
k
n (a) h k+1 (x, a) + f
∞,[a,b]∩T
a
x
a
t
h n−1 (t, σ (τ )) τ t ,
196
8 Delta Time Scales Iyengar Inequalities
and −
n−1
n k f (b) h k+1 (x, b) − f ∞,[a,b]∩T
x
k=0
b
≤
b
b
h n−1 (t, σ (τ )) τ t
t
f (t) t ≤
(8.18)
x
−
n−1
n k f (b) h k+1 (x, b) + f ∞,[a,b]∩T
b
x
k=0
b
h n−1 (t, σ (τ )) τ t .
t
Adding (8.17) and (8.18) we derive n−1
n k k f (a) h k+1 (x, a) − f (b) h k+1 (x, b) − f ∞,[a,b]∩T ·
k=0
x
a
t
a
b
h n−1 (t, σ (τ )) τ t + x b f (t) t ≤ ≤
b
h n−1 (t, σ (τ )) τ t
t
(8.19)
a n−1
n k k f (a) h k+1 (x, a) − f (b) h k+1 (x, b) + f ∞,[a,b]∩T ·
k=0
x
a
t
h n−1 (t, σ (τ )) τ t +
a
b
x
b
h n−1 (t, σ (τ )) τ t ,
t
∀ x ∈ [a, b] ∩ T. The theorem now is proved.
We continue with Theorem 8.20 Let f ∈ Crnd (T), n ∈ N is odd, a, b ∈ T; a ≤ b. Then (1) n−1 b
k k f (a) h k+1 (x, a) − f (b) h k+1 (x, b) ≤ f (t) t − a k=0
n f
L 1 ([a,b]∩T)
∀ x ∈ [a, b] ∩ T,
a
x
(t − σ (a))
n−1
b
t +
(σ (b) − t)
n−1
x
t ,
(8.20)
8.3 Main Results
197
(2) assuming f (a) = f (b) = 0, k = 0, 1, . . . , n − 1, from (8.20) we obtain k
k
b
a
n f (t) t ≤ f L 1 ([a,b]∩T) ·
x
b
(t − σ (a))n−1 t +
a
(σ (b) − t)n−1 t ,
(8.21)
x
∀ x ∈ [a, b] ∩ T, (21 ) when x = a by (8.21) we get
b
a
n f (t) t ≤ f L 1 ([a,b]∩T)
b
(σ (b) − t)n−1 t ,
(8.22)
(t − σ (a))n−1 t ,
(8.23)
a
(22 ) when x = b by (8.21) we get
b
a
n f (t) t ≤ f L 1 ([a,b]∩T)
x
a
(23 ) by (8.22), (8.23) we have
b
a
b
n f (t) t ≤ f L 1 ([a,b]∩T) ·
(σ (b) − t)
n−1
min
t ,
a
b
(t − σ (a))
n−1
t
,
(8.24)
a
(3) assuming f (a) = f (b) = 0, k = 1, . . . , n − 1, by (8.20) we derive k
b
a
n f
k
f (t) t − [ f (a) (x − a) + f (b) (b − x)] ≤
L 1 ([a,b]∩T)
x
(t − σ (a))
n−1
b
t +
a
(σ (b) − t)
n−1
t ,
(8.25)
x
∀ x ∈ [a, b] ∩ T.
n Proof Clearly, here it holds f L 1 ([a,b]∩T) < ∞.
t Set h 0 (t, s) = 1, ∀ s, t ∈ T; k ∈ N0 , and h k+1 (t, s) = s h k (τ , s) τ , ∀ s, t ∈ T. Easily, it holds |h n (t, s)| ≤ |t − s|n , ∀ n ∈ N, ∀ s, t ∈ T. By Theorem 8.14 (3) we have f (t) −
n−1 k=0
f
k
(a) h k (t, a) = a
t
h n−1 (t, σ (τ )) f (τ ) τ , n
198
8 Delta Time Scales Iyengar Inequalities
and f (t) −
n−1
f (b) h k (t, b) = k
t
h n−1 (t, σ (τ )) f (τ ) τ , n
b
k=0
∀ t ∈ [a, b] ∩ T. Then n−1 t k n f (a) h k (t, a) = h n−1 (t, σ (τ )) f (τ ) τ ≤ f (t) − a k=0
t
n |h n−1 (t, σ (τ ))| f (τ ) τ ≤
a
t
n |t − σ (τ )|n−1 f (τ ) τ ≤
a
n (t − σ (a))n−1 f L 1 ([a,b]∩T) . Furthermore we have n−1 b k n f (b) h k (t, b) = h n−1 (t, σ (τ )) f (τ ) τ ≤ f (t) − t k=0
b
n |h n−1 (t, σ (τ ))| f (τ ) τ ≤
t
b
n |t − σ (τ )|n−1 f (τ ) τ ≤
t
n (σ (b) − t)n−1 f L 1 ([a,b]∩T) . Therefore it holds n−1 n k f (a) h k (t, a) − (t − σ (a))n−1 f L 1 ([a,b]∩T) ≤ f (t) − k=0
n ≤ (t − σ (a))n−1 f L 1 ([a,b]∩T) and
n−1 n k − (σ (b) − t)n−1 f L 1 ([a,b]∩T) ≤ f (t) − f (b) h k (t, b) k=0
n ≤ (σ (b) − t)n−1 f L 1 ([a,b]∩T) , ∀ t ∈ [a, b] ∩ T.
8.3 Main Results
199
Consequently it holds n−1
n k f (a) h k (t, a) − (t − σ (a))n−1 f L 1 ([a,b]∩T) ≤ f (t)
k=0
≤
n−1
n k f (a) h k (t, a) + (t − σ (a))n−1 f L 1 ([a,b]∩T)
k=0
and
n−1
n k f (b) h k t − (σ (b) − t)n−1 f L 1 ([a,b]∩T) ≤ f (t)
k=0
≤
n−1
n k f (b) h k (t, b) + (σ (b) − t)n−1 f L 1 ([a,b]∩T) ,
k=0
∀ t ∈ [a, b] ∩ T. Let any x ∈ [a, b] ∩ T, then integrating by integration we have n−1
f (a) h k+1 (x, a) −
n (t − σ (a))n−1 t f L 1 ([a,b]∩T)
x
k
a
k=0
x
≤
(8.26)
f (t) t ≤
a n−1
f (a) h k+1 (x, a) +
x
k
a
k=0
n (t − σ (a))n−1 t f L 1 ([a,b]∩T) ,
and −
n−1
f (b) h k+1 (x, b) −
b
k
x
k=0
b
≤
n (σ (b) − t)n−1 t f L 1 ([a,b]∩T)
f (t) t ≤
x
−
n−1
f
k
b
(b) h k+1 (x, b) +
k=0
∀ x ∈ [a, b] ∩ T.
(σ (b) − t)
n−1
x
n t f L 1 ([a,b]∩T) ,
(8.27)
200
8 Delta Time Scales Iyengar Inequalities
Adding (8.26) and (8.27) we obtain n−1
k k f (a) h k+1 (x, a) − f (b) h k+1 (x, b) −
k=0
n f
L 1 ([a,b]∩T)
x
(t − σ (a))
n−1
b
t +
(σ (b) − t)
n−1
a
t
x
≤
b
f (t) t ≤
a n−1
k k f (a) h k+1 (x, a) − f (b) h k+1 (x, b) +
k=0
n f
L 1 ([a,b]∩T)
x
(t − σ (a))
n−1
b
t +
(σ (b) − t)
n−1
a
t
, (8.28)
x
∀ x ∈ [a, b] ∩ T. The theorem now is proved.
We continue with Theorem 8.21 Let f ∈ Crnd (T), n is an odd number, a, b ∈ T; a ≤ b; p, q > 1 : 1 + q1 = 1; σ is continuous and h n−1 (t, s) is jointly continuous. Then p (1) b n−1
n k k f (a) h k+1 (x, a) − f (b) h k+1 (x, b) ≤ f f (t) t − · L q ([a,b]∩T) a k=0 ⎡⎛ ⎞⎤ ⎞ ⎛ 1 1 b b x t p p ⎢⎝ ⎜ ⎟⎥ h n−1 (t, σ (τ )) p τ t ⎠ + ⎝ h n−1 (t, σ (τ )) p τ t ⎠⎦ , ⎣ a
a
x
t
(8.29) ∀ x ∈ [a, b] ∩ T, k k (2) assuming f (a) = f (b) = 0, k = 0, 1, . . . , n − 1, by (8.29) we have that
a
b
n f (t) t ≤ f L q ([a,b]∩T) ·
8.3 Main Results
201
⎡⎛
x t
⎣⎝
a
a
1
p
h n−1 (t, σ (τ )) p τ
⎞
⎛
b b
t ⎠ + ⎝
x
t
⎞⎤
1 h n−1 (t, σ (τ )) p τ
p
t ⎠⎦ ,
(8.30) ∀ x ∈ [a, b] ∩ T, (21 ) when x = a by (8.30) we get
b
a
n f (t) t ≤ f L q ([a,b]∩T)
b
b
h n−1 (t, σ (τ )) τ
1p
p
a
t , (8.31)
t
(22 ) when x = b by (8.30) we get
b
a
n f (t) t ≤ f L q ([a,b]∩T)
b
t
h n−1 (t, σ (τ )) τ p
a
1p
t , (8.32)
a
(23 ) by (8.31), (8.32) we derive that
b
a
n f (t) t ≤ f L q ([a,b]∩T) ·
(8.33)
⎧⎛ ⎞ ⎫ 1p 1p b t ⎬ ⎨ b b min ⎝ h n−1 (t, σ (τ )) p τ t ⎠ , h n−1 (t, σ (τ )) p τ t , ⎭ ⎩ a t a a
(3) assuming f (a) = f (b) = 0, k = 1, . . . , n − 1, by (8.29) we obtain k
b
a
k
n f (t) t − [ f (a) (x − a) + f (b) (b − x)] ≤ f L q ([a,b]∩T)
⎡⎛
x t
⎣⎝
a
a
1 h n−1
p
(t, σ (τ )) p τ
⎞
⎛
1
t ⎠ + ⎝
b b x
t
h n−1
(t, σ (τ )) p τ
p
⎞⎤ t ⎠⎦ ,
(8.34) ∀ x ∈ [a, b] ∩ T. Proof As before we have K (t, a) := f (t) −
n−1 k=0
f (a) h k (t, a) = k
a
t
h n−1 (t, σ (τ )) f (τ ) τ , n
202
8 Delta Time Scales Iyengar Inequalities
and K (t, b) := f (t) −
n−1
f (b) h k (t, b) = k
t
h n−1 (t, σ (τ )) f (τ ) τ , n
b
k=0
∀ t ∈ [a, b] ∩ T. We have that (by use of (8.2))
t
|K (t, a)| ≤
h n−1 (t, σ (τ )) τ p
1p
a
t
≤
h n−1 (t, σ (τ )) τ p
|K (t, b)| =
1p
n f
b
h n−1 (t, σ (τ )) f
t
b
n f (τ )q τ
q1
a
a
and
t
h n−1 (t, σ (τ )) p τ
1p
t
b
n
L q ([a,b]∩T)
,
(τ ) τ ≤
n f (τ )q τ
q1
t
b
≤
h n−1 (t, σ (τ )) τ
1p
p
n f
t
L q ([a,b]∩T)
,
∀ t ∈ [a, b] ∩ T. Hence it holds
t
−
h n−1 (t, σ (τ )) τ p
1p
n f
a
t
≤
h n−1 (t, σ (τ )) τ p
1p
L q ([a,b]∩T)
n f
a
and
−
b
h n−1 (t, σ (τ )) τ
1p
p
n f
t
≤
h n−1 (t, σ (τ )) τ p
t
∀ t ∈ [a, b] ∩ T.
b
1p
L q ([a,b]∩T)
L q ([a,b]∩T)
n f
≤ K (t, a)
≤ K (t, b)
L q ([a,b]∩T)
,
8.3 Main Results
203
That is n−1
f
k
t
(a) h k (t, a) −
h n−1 (t, σ (τ )) τ p
1p
n f
a
k=0 n−1
≤
f
k
t
(a) h k (t, a) +
h n−1 (t, σ (τ )) τ p
1p
L q ([a,b]∩T)
n f
a
k=0
≤ f (t)
L q ([a,b]∩T)
and n−1
f
k
(b) h k (t, b) −
b
h n−1 (t, σ (τ )) τ
1p
n f
p
t
k=0
≤
n−1
f
k
b
(b) h k (t, b) +
h n−1 (t, σ (τ )) τ p
1p
n f
t
k=0
L q ([a,b]∩T)
≤ f (t)
,
L q ([a,b]∩T)
∀ t ∈ [a, b] ∩ T. Let any x ∈ [a, b] ∩ T, then by integration we get n−1
k
f
n (a) h k+1 (x, a) − f
L q ([a,b]∩T)
x
t
h n−1 (t, σ (τ )) τ p
a
k=0
≤
x
1p
t
a
f (t) t ≤
a n−1
f
k
n (a) h k+1 (x, a) + f
L q ([a,b]∩T)
k=0
x
t
h n−1 (t, σ (τ )) τ p
a
1p
t ,
a
(8.35) and
−
n−1
⎛
k=0
1
b b
k n f (b) h k+1 (x, b) − f
L q ([a,b]∩T)
b
≤
⎝
x
t
h n−1
(t, σ (τ )) p τ
p
⎞ t ⎠
f (t) t ≤
x
−
n−1 k=0
f
k
n (b) h k+1 (x, b) + f
⎛
b b
L q ([a,b]∩T)
⎝
x
t
1 h n−1 (t, σ (τ )) p τ
p
⎞ t ⎠ .
(8.36)
204
8 Delta Time Scales Iyengar Inequalities
Adding (8.35) and (8.36) we obtain n−1
k k f (a) h k+1 (x, a) − f (b) h k+1 (x, b) −
k=0
n f
& L q ([a,b]∩T)
a
b
x
x
b
t
h n−1 (t, σ (τ )) p τ
1p
t +
a
h n−1 (t, σ (τ )) p τ
1p
' t
t
≤
b
f (t) t ≤
a n−1
k k f (a) h k+1 (x, a) − f (b) h k+1 (x, b) +
k=0
n f
&
b
t
h n−1 (t, σ (τ )) τ p
L q ([a,b]∩T)
x
a
b
t +
a
h n−1 (t, σ (τ )) τ
1p
p
x
1p
' t
,
(8.37)
t
∀ x ∈ [a, b] ∩ T. The theorem now is proved.
We continue with Theorem 8.22 Let f ∈ Crnd (T), m, n ∈ N, m < n, n − m is odd, a, b ∈ T; a ≤ b. Here σ is continuous and h n−m−1 (t, s) is jointly continuous. Then (1) n−m−1 b m
k+m k+m f f (t) t − (a) h k+1 (x, a) − f (b) h k+1 (x, b) ≤ a k=0
n f
∞,[a,b]∩T
b x
x
a
b t
t
h n−m−1 (t, σ (τ )) τ t +
a
h n−m−1 (t, σ (τ )) τ t
,
(8.38)
8.3 Main Results
205
∀ x ∈ [a, b] ∩ T, k+m k+m (2) assuming f (a) = f (b) = 0, k = 0, 1, . . . , n − m − 1, we get from (8.38) that b n m f (t) t ≤ f ∞,[a,b]∩T · a
(
x t
a
a
h n−m−1 (t, σ (τ )) τ t +
b b x
t
)
h n−m−1 (t, σ (τ )) τ t
,
(8.39) ∀ x ∈ [a, b] ∩ T, (21 ) when x = a we get from (8.39) that
b
f
m
a
n (t) t ≤ f ∞,[a,b]∩T
b
a
b
h n−m−1 (t, σ (τ )) τ t ,
t
(8.40)
(22 ) when x = b we get from (8.39) that
a
b
n m f (t) t ≤ f ∞,[a,b]∩T
b
a
t
h n−m−1 (t, σ (τ )) τ t ,
a
(8.41)
(23 ) by (8.40), (8.41) we get
a
b
b
min a
b
n m f (t) t ≤ f ∞,[a,b]∩T ·
h n−m−1 (t, σ (τ )) τ t ,
t
b
a
t
, h n−m−1 (t, σ (τ )) τ t
a
(8.42) and k+m k+m (3) assuming f (a) = f (b) = 0, k = 1, . . . , n − m − 1, from (8.38) we obtain b * m n m m ≤f f · t − f − a) + f − x) (t) (a) (x (b) (b ∞,[a,b]∩T a
(
x t
a
a
h n−m−1 (t, σ (τ )) τ t +
b b x
t
)
h n−m−1 (t, σ (τ )) τ t
,
(8.43) ∀ x ∈ [a, b] ∩ T. Proof As in the proof of Theorem 8.19, now using Corollary 8.16 (4).
We give Theorem 8.23 Let f ∈ Crnd (T), m, n ∈ N, m < n, n − m is odd, a, b ∈ T; a ≤ b. Then
206
8 Delta Time Scales Iyengar Inequalities
(1) n−m−1 b m
k+m k+m f f (t) t − (a) h k+1 (x, a) − f (b) h k+1 (x, b) ≤ a k=0
n f
L 1 ([a,b]∩T)
x
(t − σ (a))
n−m−1
b
t +
(σ (b) − t)
n−m−1
a
t , (8.44)
x
∀ x ∈ [a, b] ∩ T, k+m k+m (2) assuming f (a) = f (b) = 0, k = 0, 1, . . . , n − m − 1, we get from (8.44) that b n m f (t) t ≤ f L 1 ([a,b]∩T) · a
x
b
(t − σ (a))n−m−1 t +
a
(σ (b) − t)n−m−1 t ,
(8.45)
x
∀ x ∈ [a, b] ∩ T, (21 ) when x = a by (8.45) we get
b
a
n m f (t) t ≤ f L 1 ([a,b]∩T)
b
(σ (b) − t)n−m−1 t ,
(8.46)
(t − σ (a))n−m−1 t ,
(8.47)
a
(22 ) when x = b by (8.45) we get
b
a
n m f (t) t ≤ f L 1 ([a,b]∩T)
x
a
(23 ) by (8.46), (8.47) we have
b
f
a
min
b
(σ (b) − t)
a
m
n (t) t ≤ f L 1 ([a,b]∩T) ·
n−m−1
t ,
b
(t − σ (a))
n−m−1
t
,
(8.48)
a
and k+m k+m (3) assuming f (a) = f (b) = 0, k = 1, . . . , n − m − 1, from (8.44) we obtain b * m m m f (t) t − f (a) (x − a) + f (b) (b − x) ≤ a
n f
L 1 ([a,b]∩T)
a
x
b
(t − σ (a))n−m−1 t + x
(σ (b) − t)n−m−1 t , (8.49)
8.3 Main Results
207
∀ x ∈ [a, b] ∩ T.
Proof As in Theorem 8.20, now using Corollary 8.16 (4). We also give
Theorem 8.24 Let f ∈ Crnd (T), m, n ∈ N, m < n, n − m is odd, a, b ∈ T; a ≤ b. Here σ is continuous and h n−m−1 (t, s) is jointly continuous. Let also p, q > 1 : 1 + q1 = 1. Then p (1) n−m−1 b m
k+m k+m f (a) h k+1 (x, a) − f (b) h k+1 (x, b) ≤ f (t) t − a k=0
n f
(
b
t
h n−m−1 (t, σ (τ )) τ p
L q ([a,b]∩T)
x
a
b
t +
a
h n−m−1 (t, σ (τ )) τ
1p
p
x
1p
) t
,
(8.50)
t
∀ x ∈ [a, b] ∩ T, k+m k+m (2) assuming f (a) = f (b) = 0, k = 0, 1, . . . , n − m − 1, we get from (8.50) that
b
f
m
a
( n (t) t ≤ f L q ([a,b]∩T)
x
b
b
1p
t +
a
h n−m−1 (t, σ (τ )) τ p
x
h n−m−1 (t, σ (τ )) τ p
a
t
1p
) t
,
(8.51)
t
∀ x ∈ [a, b] ∩ T, (21 ) when x = a we get from (8.51) that
b
f
m
a
n (t) t ≤ f L q ([a,b]∩T)
b
b
h n−m−1 (t, σ (τ )) τ
1p
p
a
t ,
t
(8.52) (22 ) when x = b we get from (8.51) that
a
b
f
m
n (t) t ≤ f L q ([a,b]∩T)
a
b
t
h n−m−1 (t, σ (τ )) τ p
1p
t ,
a
(8.53)
208
8 Delta Time Scales Iyengar Inequalities
(23 ) by (8.52), (8.53) we get
b
f
m
a
&
b
n (t) t ≤ f L q ([a,b]∩T) · b
h n−m−1 (t, σ (τ )) τ
1p
p
min a
b
t ,
t
t
a
'
1p
h n−m−1 (t, σ (τ )) τ p
t
,
(8.54)
a
and k+m k+m (3) assuming f (a) = f (b) = 0, k = 1, . . . , n − m − 1, we get from (8.50) that
b
f
m
*
(t) t − f
m
(a) (x − a) + f
m
a
n f
(
b
t
h n−m−1 (t, σ (τ )) τ p
L q ([a,b]∩T)
x
(b) (b − x) ≤
a
b
t +
a
h n−m−1 (t, σ (τ )) τ
1p
p
x
1p
) t
,
(8.55)
t
∀ x ∈ [a, b] ∩ T.
Proof As in Theorem 8.21, by using Corollary 8.16 (4).
8.4 Applications We need Remark 8.25 ([9]) (i) When T = R, then h k (t, s) = (t−s) , ∀ k ∈ N0 , ∀ t, s ∈ k! b b k R, σ (t) = t, a f (t) t = a f (t) dt, f (t) = f (t), f = f (k) ; rd-continuous corresponds to f continuous. (k) , ∀ k ∈ N0 , ∀ t, s ∈ Z, where t (0) = 1, t (k) = (ii) When T = Z, h k (t, s) = (t−s) k! +k−1 i=0 (t − i) for k ∈ N, σ (t) = t + 1, k
a
b
f (t) t =
b−1 t=a
f (t) , a < b,
8.4 Applications
209
f (t) = f (t + 1) − f (t) = f (t) , f
k
(t) = f (t) = k
k k l=0
l
(−1)k−l f (t + l) ,
rd-continuous f corresponds to any f . We also need Remark 8.26 ([1, 9]) Consider q > 1, q Z = {q k : k ∈ Z}, and the time scale T = q Z = q Z ∪ {0}, which very important in q-difference equations. It holds that k−1 , t − qνs -ν , ∀s, t ∈ T ; h k (t, s) = μ μ=0 q ν=0 σ (t) = qt, ρ (t) = f (t) =
t , ∀t ∈ T, q
f (qt) − f (t) , ∀t ∈ T − {0}, (q − 1) t
f (0) = lim
s→0
f (s) − f (0) . s
Next we give applications of our initial main results. Theorem 8.27 Let f ∈ C n ([a, b]), n ∈ N is odd and [a, b] ⊂ R. Then n−1 b (k) 1 k+1 k (k) k+1 f (a) (x − a) f (t) dt − + (−1) f (b) (b − x) a (k + 1)! k=0
≤
(n) f
∞,[a,b]
(n + 1)!
* (x − a)n+1 + (b − x)n+1 ,
(8.56)
∀ x ∈ [a, b] . Proof By Theorem 8.19 (8.5). We continue with Theorem 8.28 Let f ∈ C n ([a, b]), n ∈ N is odd, [a, b] ⊂ R. Then n−1 b (k) 1 k+1 k (k) k+1 f (a) (x − a) f (t) dt − + (−1) f (b) (b − x) a (k + 1)! k=0
210
8 Delta Time Scales Iyengar Inequalities
≤
(n) f
L 1 ([a,b])
n
* (x − a)n + (b − x)n ,
(8.57)
∀ x ∈ [a, b] .
Proof By Theorem 8.20 (8.20). We also give
Theorem 8.29 Let f ∈ C n ([a, b]), n ∈ N is odd and [a, b] ⊂ R. Let also p, q > 1 : 1p + q1 = 1. Then n−1 b (k) 1 f (a) (x − a)k+1 + (−1)k f (k) (b) (b − x)k+1 f (t) dt − a (k + 1)! k=0
≤
(n) f
. / 1 1
(x − a)n+ p + (b − x)n+ p , (n − 1)! ( p (n − 1) + 1) n + 1p L q ([a,b])
1 p
(8.58)
∀ x ∈ [a, b] .
Proof By Theorem 8.21 (8.29). We continue with Theorem 8.30 Let f : Z → R, n is an odd number, a, b ∈ Z; a ≤ b. Then b−1 n−1 k 1 k (k+1) (k+1) f (a) (x − a) f (t) − − f (b) (x − b) ≤ t=a (k + 1)! k=0
n f ∞,[a,b]∩Z (n − 1)!
(x−1 t−1 t=a
τ =a
(t − τ − 1)(n−1)
+
b−1 b−1 t=x
) (t − τ − 1)(n−1)
, (8.59)
τ =t
∀ x ∈ [a, b] ∩ Z. Proof By Theorem 8.19 (8.5), see also Remark 8.25 (ii). We give Theorem 8.31 Let f : Z → R, n ∈ N is odd, a, b ∈ Z; a ≤ b. Then b−1 n−1 k 1 k (k+1) (k+1) f (a) (x − a) f (t) − − f (b) (x − b) ≤ (k + 1)! t=a
k=0
8.4 Applications
211
& x−1 b−1 ' b−1 n n−1 n−1 f (t) + , (t − a − 1) (b + 1 − t) t=a
t=a
(8.60)
t=x
∀ x ∈ [a, b] ∩ Z.
Proof By Theorem 8.20 (8.20) and Remark 8.25 (ii). We give
Theorem 8.32 Let f : Z → R, n is an odd number, a, b ∈ Z; a ≤ b, let also p, q > 1 : 1p + q1 = 1. Then b−1 n−1 k 1 k (k+1) (k+1) f (a) (x − a) f (t) − − f (b) (x − b) ≤ (k + 1)! t=a
k=0
b−1 -
| f (t)|
q1
q
n
t=a
(n − 1)!
⎡⎛ t−1 1p ⎞ x−1 p ⎠+ ⎣⎝ (t − τ − 1)(n−1) t=a
τ =a
⎛ b−1 1p ⎞⎤ b−1 p ⎝ ⎠⎦ , (t − τ − 1)(n−1)
(8.61)
τ =t
t=x
∀ x ∈ [a, b] ∩ Z.
Proof By Theorem 8.21 (8.29) and Remark 8.25 (ii). We continue with
Theorem 8.33 Let f ∈ Crnd q Z , n ∈ N is odd, a, b ∈ q Z ; a ≤ b. Then ⎛ ⎞ n−1 ⎜ k k b ν ν ⎟ , , x −q a x − q b k ⎜ k ⎟ f (t) t − − f (b) ⎜ f (a) ⎟ ≤ ν ν a ⎝ ⎠ ν=0 ν=0 k=0 qμ qμ μ=0 μ=0 n f
L 1 [a,b]∩q Z
a
x
(t − qa)
n−1
b
t +
(qb − t)
n−1
t ,
(8.62)
x
∀ x ∈ [a, b] ∩ q Z . Proof By Theorem 8.20 (8.20), and Remark 8.26.
212
8 Delta Time Scales Iyengar Inequalities
We finish with
Theorem 8.34 Let f ∈ Crnd q Z , m, n ∈ N; m < n, n − m is odd, a, b ∈ q Z ; a ≤ b. Then ⎛ ⎞ n−m−1 k k b m ⎜ ⎟ ν ν ⎜ k+m , x −q a , x − q b ⎟ k+m f (t) t − − f (a) (b) ⎜f ⎟ ≤ ν ν a ⎝ ⎠ ν=0 ν=0 k=0 qμ qμ μ=0 μ=0 n f
L 1 [a,b]∩q Z
a
x
b
(t − qa)n−m−1 t +
(qb − t)n−m−1 t ,
(8.63)
x
∀ x ∈ [a, b] ∩ q Z . Proof By Theorem 8.23 (8.44), and Remark 8.26.
One can give many similar applications for other time scales.
References 1. R. Agarwal, M. Bohner, Basic Calculus on time scales and some of its applications. Results Math. 35(1–2), 3–22 (1999) 2. R. Agarwal, M. Bohner, A. Peterson, Inequalities on time scales: a survey. Math. Inequalities Appl. 4(4), 535–557 (2001) 3. G.A. Anastassiou, Time scales inequalities. Intern. J. Differ. Equ. 5(1), 1–23 (2010) 4. G. Anastassiou, Intelligent Mathematics: Computational Analysis (Springer, Heidelberg, 2011) 5. G. Anastassiou, Time scales delta Iyengar type inequalities. Intern. J. Differ. Equ. (2019) 6. M. Bohner, G. Guisenov, The Convolution on time scales. Abstr. Appl. Anal. 2007(58373), 24 (2007) 7. M. Bohner, B. Kaymakcalan, Opial inequalities on time scales. Ann. Polon. Math. 77(1), 11–20 (2001) 8. M. Bohner, T. Matthews, Ostrowski inequalities on time scales. JIPAM J. Inequal. Pure Appl. Math. 9(1), Article 6, 8 (2008) 9. M. Bohner, A. Peterson, Dynamic equations on time scales: An Introduction with Applications (Birkhäuser, Boston, 2001) 10. R. Higgins, A. Peterson, Cauchy functions and Taylor’s formula for Time scales T , in Proceedings of the Sixth International Conference on Difference equations: New Progress in Difference Equations, Augsburg, Germany, 2001, eds. by B. Aulbach, S. Elaydi, G. Ladas (Chapman & Hall / CRC, Boca Raton, 2004), pp. 299–308 11. S. Hilger, Ein Maßketten-Kalkül mit Anwendung auf Zentrum-smannigfaltigkeiten, Ph.D. thesis, Universität Würzburg, Würzburg, 1988 12. K.S.K. Iyengar, Note on an inequality. Math. Stud. 6, 75–76 (1938)
Chapter 9
Time Scales Nabla Iyengar Inequalities
Here we present the necessary background on nabla time scales approach. Then we give general related time scales nabla Iyengar type inequalities for all basic norms. We finish with applications to specific time scales like R, Z and q Z , q > 1. See also [4].
9.1 Introduction We are motivated by the following famous Iyengar inequality (1938), [9]. Theorem 9.1 Let f be a differentiable function on [a, b] and f (x) ≤ M. Then
a
b
M (b − a) 2 ( f (b) − f (a))2 1 . f (x) d x − (b − a) ( f (a) + f (b)) ≤ − 2 4 4M (9.1)
We present generalized analogs of (9.1) to time scales in the nabla sense. Motivation comes also from [1–3].
9.2 Background Here we follow [6–8, 11]. Let T be a time scale (a closed subset of R) [9], [a, b] be the closed and bounded interval in T, i.e. [a, b] := {t ∈ T : a ≤ t ≤ b} and a, b ∈ T.
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_9
213
214
9 Time Scales Nabla Iyengar Inequalities
Clearly, a time scale T may or may not be connected. Therefore we have the concept of forward and backward jump operators as follows. Define σ, ρ : T −→ T by σ (t) = inf{s ∈ T : s > t} and ρ (t) = sup{s ∈ T : s < t}, (inf ∅ := sup T, sup ∅ := inf T). If σ (t) = t, σ (t) > t, ρ (t) = t, ρ (t) < t, then t ∈ T is called right-dense, rightscattered, left-dense, left-scattered, respectively. The set Tk which is derived from T is as follows: if T has a right-scattered minimum m, then Tk = T − {m}, otherwise Tk = T. We also define the backwards graininess function ν : Tk −→ [0, ∞) as ν (t) = t − ρ (t). If f : T −→ R is a function, we define the function f ρ : Tk −→ R by f ρ (t) = f (ρ (t)) for all t ∈ Tk and σ 0 (t) = ρ0 (t) = t; Tk n+1 := (Tk n )k . Definition 9.2 If f : T −→ R is a function and t ∈ Tk , then we define the nabla derivative of f at a point t to be the number f ∇ (t) (provided it exists) with the property that, for each ε > 0, there is a neighborhood of U of t such that [ f (ρ (t)) − f (s)] − f ∇ (t) [ρ (t) − s] ≤ ε |ρ (t) − s| , for all s ∈ U . Note that in the case T = R, then f ∇ (t) = f (t), and if T = Z, then f ∇ (t) = ∇ f (t) = f (t) − f (t − 1) . Definition 9.3 A function F : T → R we call a nabla-antiderivative of f : T → R provided that F ∇ (t) = f (t) for all t ∈ Tk . We then define the Cauchy ∇-integral from a to t of f by
t
f (s) ∇s = F (t) − F (a) , for all t ∈ T.
a
Note that in the case T = R we have b f (t) ∇t = a
b
f (t) dt,
a
and in the case T = Z we have a
b
f (t) ∇t =
b
f (k) ,
k=a+1
where a, b ∈ T with a ≤ b. Definition 9.4 A function f : T → R is left-dense continuous (or ld-continuous) provided that it is continuous at left-dense points in T and its right-sided limits exist at right-dense points of T.
9.2 Background
215
If T = R, then f is ld-continuous iff f is continuous. If T = Z, then any function is ld-continuous. Theorem 9.5 Let T be a time scale, f : T → R, and t ∈ Tk . The following holds: 1. If f is nabla differentiable at t, then f is continuous at t. 2. If f is continuous at t and t is left-scattered, then f is nabla differentiable at t and f (t) − f (ρ (t)) . f ∇ (t) = t − ρ (t) 3. If t is left-dense, then f is nabla differentiable at t if and only if the limit lim s→t
f (t) − f (s) t −s
exists as a finite number. In this case, f ∇ (t) = lim s→t
f (t) − f (s) . t −s
4. If f is nabla differentiable at t, then f (ρ (t)) = f (t) − ν (t) f ∇ (t) . For any time scale T, when f is a constant, then f ∇ = 0; if f (t) = kt for some constant k, then f ∇ = k. Theorem 9.6 Suppose f, g : T → R are nabla differentiable at t ∈ Tk . Then, 1. the sum f + g : T → R is nabla differentiable at t and ( f + g)∇ (t) = ∇ f (t) + g ∇ (t) ; 2. for any constant α, α f : T → R is nabla differentiable at t and (α f )∇ (t) = α f ∇ (t) ; 3. the product f g : T → R is nabla differentiable at t and ( f g)∇ (t) = f ∇ (t) g (t) + f ρ (t) g ∇ (t) = f ∇ (t) g ρ (t) + f (t) g ∇ (t) . Some results concerning ld-continuity are useful. Theorem 9.7 Let T be a time scale, f : T → R. 1. If f is continuous, then f is ld-continuous. 2. The backward jump operator ρ is ld-continuous. 3. If f is ld-continuous, then f ρ is also ld-continuous. 4. If T = R, then f is continuous if and only if f is ld-continuous. 5. If T = Z, then f is ld-continuous. Theorem 9.8 Every ld-continuous function has a nabla antiderivative. In particular, if a ∈ T, then the function F defined by
216
9 Time Scales Nabla Iyengar Inequalities
F (t) =
t
f (τ ) ∇τ , t ∈ T,
a
is a nabla antiderivative of f. The set of all ld-continuous functions f : T → R is denoted by Cld (T, R), and the 1 set of all nabla differentiable functions with ld-continuous derivative by Cld (T, R) . Theorem 9.9 If f ∈ Cld (T, R) and t ∈ Tk , then
t
ρ(t)
f (τ ) ∇τ = ν (t) f (t) .
Theorem 9.10 If a, b, c ∈ T, a ≤ c ≤ b, α ∈ R, and f, g ∈ Cld (T, R), then: b b b 1. a ( f (t) + g (t)) ∇t = a f (t) ∇t + a g (t) ∇t; b b 2. a α f (t) ∇t = α a f (t) ∇t; b a 3. a f (t) ∇t = − b f (t) ∇t; a 4. a f (t) ∇t = 0; b c b 5. a f (t) ∇t = a f (t) ∇t + c f (t) ∇t; b 6. If f (t) > 0 for all a < t ≤ b, then a f (t) ∇t > 0; b b ∇ 7. a f ρ (t) g ∇ (t) ∇t = [( f g) (t)]t=b t=a − a f (t) g (t) ∇t; b b 8. a f (t) g ∇ (t) ∇t = [( f g) (t)]t=b − a f ∇ (t) g ρ (t) ∇t; t=a b 9. If f (t) ≥ 0, a ≤ t ≤ b, then a f (t) ∇t ≥ 0; b c 10. If f (t) ≥ 0, a ≤ c ≤ b, then a f (t) ∇t ≥ a f (t) ∇t; 11. If f and f ∇ are jointly continuous in (t, s), then
t
∇ f (t, s) ∇s
a b
t
= f (ρ (t) , t) + ∇
f (t, s) ∇s
f ∇ (t, s) ∇s,
a
= − f (ρ (t) , t) +
t
b
f ∇ (t, s) ∇s;
t
b b 12. If f (t) ≥ g (t), then a f (t) ∇t ≥ a g (t) ∇t; b b 13. a f (t) ∇t ≤ a | f (t)| ∇t; b 14. a 1∇t = b − a. Similarly we define higher order nabla derivatives on Tk n+1 by f∇
n+1
n ∇ := f ∇ , n ∈ N.
∇ (n+1) If T = R, then , and if T = Z, then f ∇ f = f
n+1 n+1 m f (t − m) . m=0 (−1) m n+1
n+1
(t) = ∇ n+1 f (t) =
9.2 Background
217
Let h k : T2 → R, k ∈ N0 = N ∪ {0}, defined recursively as follows: h 0 (t, s) = 1, all s, t ∈ T, h k+1 is and, given h k for k ∈ N0 , the function h k+1 (t, s) =
t
h k (τ , s) ∇τ , for all s, t ∈ T.
s
Note that h k are all well defined, since each is ld-continuous in t. If we let h ∇k (t, s) denote for each fixed s the nabla derivative of h k (t, s) with respect to t, then h k−1 (t, s) , for k ∈ N, t ∈ Tk . h ∇k (t, s) = Notice that h 1 (t, s) = t − s, for all s, t ∈ T. By [2] we have that h k (t, s) ≥ 0, for any t, s ∈ T, when k is even. Example 9.11 1. If T = R, then ρ (t) = t, t ∈ R, so that h k (t, s) = (t−s) for all k! s, t ∈ R, k ∈ N0 . k 2. If T = Z, then ρ (t) = t − 1, t ∈ Z, and h k (t, s) = (t−s) , for all s, t ∈ Z, k ∈ k! k 0 N0 , where t := t (t + 1) . . . (t + k − 1), k ∈ N; t := 1. k
n n Definition 9.12 The set Cld (T, R) = Cld (T), n ∈ N, denotes the set of all n times 2 continuously nabla differentiable functions from T into R, i.e. all f, f ∇ , f ∇ , . . . , n f ∇ ∈ Cld (T, R). This definition requires Tk = T.
We need Theorem 9.13 ([5], Nabla Taylor’s formula) Suppose f is n times nabla differentiable on Tk n , n ∈ N. Let a ∈ Tk n−1 , t ∈ T. Then f (t) =
n−1
k h k (t, a) f ∇ (a) +
t
n h n−1 (t, ρ (τ )) f ∇ (τ ) ∇τ .
a
k=0
n If f ∈ Cld (T, R), then nabla Taylor formula is true for all t, a ∈ T. n Corollary 9.14 (to Theorem 9.13) Assume f ∈ Cld (T), n ∈ N, and s, t ∈ T. Let m ∈ N with m < n. Then
f
∇m
(t) =
n−m−1 k=0
f
∇ k+m
h k (t, s) + (s)
t s
n h n−m−1 (t, ρ (τ )) f ∇ (τ ) ∇τ .
218
9 Time Scales Nabla Iyengar Inequalities
Proof Use Theorem 9.13 with n and f substituted by n − m and f ∇ , respectively. m
Define [a, b]k = [a, b] if a is right-dense, and [a, b]k = [σ (a) , b] if a is rightscattered. Proposition 9.15 ([11]) Suppose a, b ∈ T, a < b, and f ∈ Cld ([a, b] , R) is such b that f ≥ 0 on [a, b]. If a f (t) ∇t = 0, then f = 0 on [a, b]k . Theorem 9.16 ([2] Nabla Hölder’s inequality) Let a, b ∈ T, a ≤ b. For f, g ∈ Cld ([a, b]) we have
b
| f (t)| |g (t)| ∇t ≤
b
1p | f (t)| ∇t · p
a
a
where p, q > 1 :
1 p
+
1 q
b
|g (t)| ∇t q
q1
,
a
= 1.
Next define g0 (t, s) ≡ 1,
t
gn+1 (t, s) =
gn (ρ (τ ) , s) ∇τ , n ∈ N, s, t ∈ T.
s ∇ g1 (t, s) = t − s, for all s, t ∈ T. gn (ρ (t) , s), t ∈ Tk ; Notice that gn+1 (t, s) = If T has a left-scattered maximum M, define Tk := T − {M}; otherwise, set Tk = n k n+1 n+1 T. Similarly define Tk := Tk . Notice Tk n+1 ⊂ Tk and Tk ⊂ Tk . n
Theorem 9.17 ([5]) Let t ∈ Tk ∩ Tk , s ∈ Tk , and n ≥ 0. Then gn (s, t) . h n (t, s) = (−1)n hn , gn Remark 9.18 Let the time scale T be such that Tk = Tk = T. Clearly both are nabla differentiable in their first variables, therefore both are continuous in their first variables. gn are continuous in their Using now Theorem 9.17 we get that also both hn , second variables. Consequently h n (t, s) is ld-continuous in each variable and thus h n (t, ρ (s)) is ld-continuous in s. h 2 (t, s) ≥ 0, . . . , h n−1 Notice also in general that if t ≥ s then h 1 (t, s) ≥ 0, (t, s) ≥ 0. So that h n−1 (t, ρ (τ )) ≥ 0 for all s ≤ τ ≤ t. Also in general it holds h k (t, s) ≤ (t − s)k , ∀t ≥ s, k ∈ N0 , and easily we get: h k (t, s) ≤ |t − s|k , ∀t, s ∈ T, k ∈ N0 .
9.2 Background
219
We need Theorem 9.19 ([7] Nabla chain rule) Let f : R → R be continuously differentiable and suppose that g : T → R is nabla differentiable on T. Then f ◦ g : T → R is nabla differentiable on T and the formula ( f ◦ g)∇ (t) =
f g (t) + hν (t) g ∇ (t) dh g ∇ (t)
1 0
holds. We formulate Assumption 9.20 Let the time scale T be such that Tk = Tk = T. Remark 9.21 Assume that ρ is a continuous function, Tk = T, h n−1 (t, s) and h ∇n−1 (t, s) = h n−2 h n−2 (t, s) are jointly continuous in (t, s) ∈ T2 ; p > 1. Clearly h n−1 (t, ρ (s)), h n−2 (t, ρ (s)) are jointly continuous in (t, s) ∈ (t, s) in t ∈ T. Also T2 . p ∇ By Theorem 9.19 we have that h n−1 (t, ρ (τ )) exists in t ∈ T, where τ is fixed in T, and p ∇ h n−1 (t, ρ (τ )) =
1
p
p−1 dh h n−1 (t, ρ (τ )) + hν (t) h n−2 (t, ρ (τ )) h n−2 (t, ρ (τ )) .
0
p ∇ By bounded convergence theorem we obtain that h n−1 (t, ρ (τ )) is jointly p continuous in (t, τ ), and of course h n−1 (t, ρ (τ )) is jointly continuous in (t, τ ) . Therefore by Theorem 9.10 (11), we derive for u (t) =
t
h n−1 (t, ρ (τ )) p ∇τ
a
(t ∈ [a, b] ⊂ T), that u ∇ (t) =
b
∇ p h n−1 (t, ρ (τ )) p ∇τ + h n−1 (ρ (t) , ρ (t)) .
a
I.e. ∇
u (t) =
t
∇ h n−1 (t, ρ (τ )) p ∇τ .
a
That is u (t) is nabla differentiable, hence continuous and therefore ld-continuous on [a, b] ⊂ T.
220
9 Time Scales Nabla Iyengar Inequalities
We formulate Assumption 9.22 We suppose that ρ is a continuous function and h n−1 (t, s), h n−2 (t, s) are jointly continuous in (t, s) ∈ T2 . Assumption 9.23 We suppose that ρ is a continuous function and h n−m−1 (t, s) , h n−m−2 (t, s) are jointly continuous in (t, s) ∈ T2 .
9.3 Main Results Next we present nabla Iyengar type inequalities on time scales for all norms · p , 1 ≤ p ≤ ∞. We give n Theorem 9.24 Let f ∈ Cld (T), n ∈ N is odd, a, b ∈ T; a ≤ b. Here ρ is continuous and h n−1 (t, s) is jointly continuous. Also assume that Tk = T. Then (1)
n−1 b n k ∇k ∇ f (a) f (t) ∇t − h k+1 (x, a) − f (b) h k+1 (x, b) ≤ f ∇ ∞,[a,b]∩T a k=0
x
a
t
h n−1 (t, ρ (τ )) ∇τ ∇t +
a
b x
b
h n−1 (t, ρ (τ )) ∇τ ∇t ,
t
(9.2) ∀ x ∈ [a, b] ∩ T, k k (2) assuming f ∇ (a) = f ∇ (b) = 0, k = 0, 1, . . . , n − 1, we get from (9.2) that
a
x
a
t
b
n f (t) ∇t ≤ f ∇ ∞,[a,b]∩T
h n−1 (t, ρ (τ )) ∇τ ∇t +
a
b x
b
h n−1 (t, ρ (τ )) ∇τ ∇t ,
t
(9.3)
∀ x ∈ [a, b] ∩ T, (21 ) when x = a we get from (9.3) that
b
a
∇n f (t) ∇t ≤ f ∞,[a,b]∩T
b
a
b
h n−1 (t, ρ (τ )) ∇τ ∇t ,
(9.4)
h n−1 (t, ρ (τ )) ∇τ ∇t ,
(9.5)
t
(22 ) when x = b we get from (9.3) that
a
b
∇n f (t) ∇t ≤ f ∞,[a,b]∩T
a
b
a
t
9.3 Main Results
221
(23 ) by (9.4) and (9.5) we get
b
a
b
b
min a
n f (t) ∇t ≤ f ∇ ∞,[a,b]∩T
h n−1 (t, ρ (τ )) ∇τ ∇t ,
t
b
a
t
h n−1 (t, ρ (τ )) ∇τ ∇t ,
a
(9.6)
and k k (3) assuming f ∇ (a) = f ∇ (b) = 0, k = 1, . . . , n − 1, by (9.2) we have
n f (t) ∇t − [ f (a) (x − a) + f (b) (b − x)] ≤ f ∇ ∞,[a,b]∩T
b
a
a
x
t
h n−1 (t, ρ (τ )) ∇τ ∇t +
a
b x
b
h n−1 (t, ρ (τ )) ∇τ ∇t ,
t
(9.7)
∀ x ∈ [a, b] ∩ T.
n Proof By [8], p. 23 we have that f ∇ ∞,[a,b]∩T < ∞. By Theorem 9.13 we have f (t) −
n−1
f ∇ (a) h k (t, a) = k
f (t) −
n−1
n h n−1 (t, ρ (τ )) f ∇ (τ ) ∇τ ,
(9.8)
n h n−1 (t, ρ (τ )) f ∇ (τ ) ∇τ ,
(9.9)
a
k=0
and
t
f ∇ (b) h k (t, b) = k
t
b
k=0
∀ t ∈ [a, b] ∩ T. Then we get t n−1 (9.8) n k ∇ ∇ f (a) h k (t, a) ≤ f h n−1 (t, ρ (τ )) ∇τ , (9.10) f (t) − ∞,[a,b]∩T a k=0
and
b n−1 k (9.9) ∇ ∇n f (b) h k (t, b) = h n−1 (t, ρ (τ )) f (τ ) ∇τ f (t) − t k=0
≤ t
b
n h n−1 (t, ρ (τ )) ∇τ f ∇ ∞,[a,b]∩T .
Therefore it holds (by (9.10), (9.11))
(9.11)
222
9 Time Scales Nabla Iyengar Inequalities
n − f ∇ ∞,[a,b]∩T
t
h n−1 (t, ρ (τ )) ∇τ ≤ f (t) −
a
n−1
f ∇ (a) h k (t, a) k
k=0
n ≤ f ∇ ∞,[a,b]∩T
t
h n−1 (t, ρ (τ )) ∇τ
a
and n − f ∇ ∞,[a,b]∩T
b
h n−1 (t, ρ (τ )) ∇τ
≤ f (t) −
t
n−1
f ∇ (b) h k (t, b) k
k=0
n ≤ f∇
∞,[a,b]∩T
b
h n−1 (t, ρ (τ )) ∇τ ,
t
∀ t ∈ [a, b] ∩ T. Consequently we have n−1
n k f ∇ (a) h k (t, a) − f ∇ ∞,[a,b]∩T
≤
f
∇k
h n−1 (t, ρ (τ )) ∇τ ≤ f (t)
(9.12)
a
k=0 n−1
t
n h k (t, a) + f ∇ ∞,[a,b]∩T (a)
t
h n−1 (t, ρ (τ )) ∇τ
a
k=0
and n−1
f
∇k
n h k (t, b) − f ∇ ∞,[a,b]∩T (b)
k=0
≤
n−1
f
∇k
b
h n−1 (t, ρ (τ )) ∇τ
≤ f (t) (9.13)
t
n h k (t, b) + f ∇ ∞,[a,b]∩T (b)
b
h n−1 (t, ρ (τ )) ∇τ ,
t
k=0
∀ t ∈ [a, b] ∩ T. Let any x ∈ [a, b] ∩ T, then integrating (9.12), (9.13) we obtain: n−1
n k f ∇ (a) h k+1 (x, a) − f ∇ ∞,[a,b]∩T
k=0
≤ a
x
x
a
f (t) ∇t ≤
t
h n−1 (t, ρ (τ )) ∇τ ∇t
a
(9.14)
9.3 Main Results n−1
223
n k f ∇ (a) h k+1 (x, a) + f ∇ ∞,[a,b]∩T
x
a
k=0
t
h n−1 (t, ρ (τ )) ∇τ ∇t ,
a
and −
n−1
n k f ∇ (b) h k+1 (x, b) − f ∇ ∞,[a,b]∩T
x
k=0
b
≤
b
b
h n−1 (t, ρ (τ )) ∇τ ∇t
t
f (t) ∇t ≤
(9.15)
x
−
n−1
f
∇k
n h k+1 (x, b) + f ∇ ∞,[a,b]∩T (b)
b
x
k=0
b
h n−1 (t, ρ (τ )) ∇τ ∇t .
t
Adding (9.14) and (9.15) we derive n−1
n k k f ∇ (a) h k+1 (x, a) − f ∇ (b) h k+1 (x, b) − f ∇ ∞,[a,b]∩T ·
k=0
x
a
t
h n−1 (t, ρ (τ )) ∇τ ∇t +
a
b
x
≤
b
b
h n−1 (t, ρ (τ )) ∇τ ∇t
t
f (t) ∇t ≤
(9.16)
a n−1
n k k f ∇ (a) h k+1 (x, a) − f ∇ (b) h k+1 (x, b) + f ∇ ∞,[a,b]∩T ·
k=0
a
x
t
h n−1 (t, ρ (τ )) ∇τ ∇t +
a
b x
b
h n−1 (t, ρ (τ )) ∇τ ∇t ,
t
∀ x ∈ [a, b] ∩ T. The theorem now is proved.
We continue with n Theorem 9.25 Let f ∈ Cld (T), n ∈ N is odd, a, b ∈ T; a ≤ b, where Tk = T. Then (1)
n−1 b k k f ∇ (a) f (t) ∇t − h k+1 (x, a) − f ∇ (b) h k+1 (x, b) ≤ a k=0
224
9 Time Scales Nabla Iyengar Inequalities
∇n f
L 1 ([a,b]∩T)
x
(t − ρ (a))
n−1
b
∇t +
a
(ρ (b) − t)
n−1
∇t ,
(9.17)
x
∀ x ∈ [a, b] ∩ T, k k (2) assuming f ∇ (a) = f ∇ (b) = 0, k = 0, 1, . . . , n − 1, from (9.17) we obtain
b
a
x
n f (t) ∇t ≤ f ∇ L 1 ([a,b]∩T) ·
b
(t − ρ (a))n−1 ∇t +
a
(ρ (b) − t)n−1 ∇t ,
(9.18)
x
∀ x ∈ [a, b] ∩ T, (21 ) when x = a by (9.17) we get
b
a
∇n f (t) ∇t ≤ f L 1 ([a,b]∩T)
b
(ρ (b) − t)
n−1
∇t ,
(9.19)
a
(22 ) when x = b by (9.17) we get
b
a
∇n f (t) ∇t ≤ f L 1 ([a,b]∩T)
x
(t − ρ (a))
n−1
∇t ,
(9.20)
a
(23 ) by (9.19), (9.20) we have
a
b
min
b
n f (t) ∇t ≤ f ∇ L 1 ([a,b]∩T) ·
(ρ (b) − t)n−1 ∇t ,
a
b
(t − ρ (a))n−1 ∇t
,
(9.21)
a
(3) assuming f ∇ (a) = f ∇ (b) = 0, k = 1, . . . , n − 1, by (9.17) we derive k
a
∇n f
b
k
f (t) ∇t − [ f (a) (x − a) + f (b) (b − x)] ≤
L 1 ([a,b]∩T)
∀ x ∈ [a, b] ∩ T.
a
x
b
(t − ρ (a))n−1 ∇t + x
n Proof Clearly, here it holds f ∇ L 1 ([a,b]∩T) < ∞. By Theorem 9.13 we have
(ρ (b) − t)n−1 ∇t ,
(9.22)
9.3 Main Results
225
f (t) −
n−1
f ∇ (a) h k (t, a) = k
f (t) −
n−1
n h n−1 (t, ρ (τ )) f ∇ (τ ) ∇τ ,
a
k=0
and
t
f ∇ (b) h k (t, b) = k
t
n h n−1 (t, ρ (τ )) f ∇ (τ ) ∇τ ,
b
k=0
∀ t ∈ [a, b] ∩ T. Then n−1 t n ∇k f (a) h n−1 (t, ρ (τ )) f ∇ (τ ) ∇τ ≤ h k (t, a) = f (t) − a k=0
t
n h n−1 (t, ρ (τ )) f ∇ (τ ) ∇τ ≤
a
t
n |t − ρ (τ )|n−1 f ∇ (τ ) ∇τ ≤
a
n (t − ρ (a))n−1 f ∇ L 1 ([a,b]∩T) . Furthermore we have n−1 b ∇k ∇n f (b) h k (t, b) = h n−1 (t, ρ (τ )) f (τ ) ∇τ ≤ f (t) − t k=0
b
n h n−1 (t, ρ (τ )) f ∇ (τ ) ∇τ ≤
t
b
n |t − ρ (τ )|n−1 f ∇ (τ ) ∇τ ≤
t
n (ρ (b) − t)n−1 f ∇
L 1 ([a,b]∩T)
.
Therefore it holds n−1 n k f ∇ (a) − (t − ρ (a))n−1 f ∇ L 1 ([a,b]∩T) ≤ f (t) − h k (t, a) k=0
n ≤ (t − ρ (a))n−1 f ∇ L 1 ([a,b]∩T) , ∀ t ∈ [a, b] ∩ T, and n−1 n k − (ρ (b) − t)n−1 f ∇ L 1 ([a,b]∩T) ≤ f (t) − f ∇ (b) h k (t, b) k=0
n ≤ (ρ (b) − t)n−1 f ∇ L 1 ([a,b]∩T) ,
226
9 Time Scales Nabla Iyengar Inequalities
∀ t ∈ [a, b] ∩ T. Consequently it holds n−1
n k f ∇ (a) h k (t, a) − (t − ρ (a))n−1 f ∇ L 1 ([a,b]∩T) ≤ f (t)
k=0
≤
n−1
n k f ∇ (a) h k (t, a) + (t − ρ (a))n−1 f ∇ L 1 ([a,b]∩T) ,
k=0
∀ t ∈ [a, b] ∩ T, and n−1
n k f ∇ (b) h k (t, b) − (ρ (b) − t)n−1 f ∇ L 1 ([a,b]∩T) ≤ f (t)
k=0
≤
n−1
n k f ∇ (b) h k (t, b) + (ρ (b) − t)n−1 f ∇ L 1 ([a,b]∩T) ,
k=0
∀ t ∈ [a, b] ∩ T. Let any x ∈ [a, b] ∩ T, then by integration we have n−1
∇k
f
h k+1 (x, a) − (a)
x
(t − ρ (a))
n−1
a
k=0
x
≤
n ∇t f ∇ L 1 ([a,b]∩T)
(9.23)
f (t) ∇t ≤
a n−1
f
∇k
h k+1 (x, a) + (a)
x
(t − ρ (a))
n−1
a
k=0
n ∇t f ∇ L 1 ([a,b]∩T) ,
and −
n−1
f
∇k
h k+1 (x, b) − (b)
b
(ρ (b) − t)
n−1
x
k=0
b
≤
n ∇t f ∇ L 1 ([a,b]∩T)
f (t) ∇t ≤
x
−
n−1 k=0
f ∇ (b) h k+1 (x, b) +
b
k
x
n (ρ (b) − t)n−1 ∇t f ∇ L 1 ([a,b]∩T) ,
(9.24)
9.3 Main Results
227
∀ x ∈ [a, b] ∩ T. Adding (9.23) and (9.24) we obtain n−1
k k f ∇ (a) h k+1 (x, a) − f ∇ (b) h k+1 (x, b) −
k=0
∇n f
L 1 ([a,b]∩T)
x
(t − ρ (a))
n−1
b
∇t +
(ρ (b) − t)
n−1
a
∇t
x
b
≤
f (t) ∇t ≤
a n−1
k k f ∇ (a) h k+1 (x, a) − f ∇ (b) h k+1 (x, b) +
k=0
∇n f
L 1 ([a,b]∩T)
x
(t − ρ (a))
n−1
b
∇t +
a
(ρ (b) − t)
n−1
∇t
, (9.25)
x
∀ x ∈ [a, b] ∩ T. The theorem now is proved.
We continue with n Theorem 9.26 Let f ∈ Cld (T), n ∈ N is odd, a, b ∈ T; a ≤ b; p, q > 1 : 1. We suppose Assumptions 9.20, 9.22. Then (1)
1 p
+
1 q
=
n−1 b n k ∇k ∇ f (a) h k+1 (x, a) − f (b) h k+1 (x, b) ≤ f ∇ f (t) ∇t − · a L q ([a,b]∩T) k=0
⎡ ⎣
x
a
t
h n−1 (t, ρ (τ )) p ∇τ
1p
⎛ +⎝
∇t
a
b
x
b
h n−1 (t, ρ (τ )) p ∇τ
1p
⎞⎤ ∇t ⎠⎦ , (9.26)
t
∀ x ∈ [a, b] ∩ T, k k (2) assuming f ∇ (a) = f ∇ (b) = 0, k = 0, 1, . . . , n − 1, by (9.26) we have that
b
a
⎡ ⎣ a
x
a
t
h n−1 (t, ρ (τ )) ∇τ p
n f (t) ∇t ≤ f ∇ L q ([a,b]∩T) · 1p
∇t
⎛ +⎝
b x
b t
h n−1 (t, ρ (τ )) ∇τ p
1p
⎞⎤ ∇t ⎠⎦ , (9.27)
228
9 Time Scales Nabla Iyengar Inequalities
∀ x ∈ [a, b] ∩ T, (21 ) when x = a by (9.27) we get
b
a
∇n f (t) ∇t ≤ f L q ([a,b]∩T)
b
a
b
h n−1 (t, ρ (τ )) p ∇τ
1p
∇t , (9.28)
t
(22 ) when x = b by (9.27) we get
b
a
∇n f (t) ∇t ≤ f L q ([a,b]∩T)
b
t
h n−1 (t, ρ (τ )) ∇τ p
a
1p
∇t , (9.29)
a
(23 ) by (9.28), (9.29) we derive that
n f (t) ∇t ≤ f ∇ L q ([a,b]∩T) ·
b
a
(9.30)
⎧⎛ ⎞ ⎫ 1p 1p b t ⎨ b b ⎬ h n−1 (t, ρ (τ )) p ∇τ h n−1 (t, ρ (τ )) p ∇τ min ⎝ ∇t ⎠ , ∇t , ⎩ a ⎭ t a a
(3) assuming f ∇ (a) = f ∇ (b) = 0, k = 1, . . . , n − 1, by (9.26) we obtain k
b
a
⎡ ⎣
a
x
t
k
n f (t) ∇t − [ f (a) (x − a) + f (b) (b − x)] ≤ f ∇ L q ([a,b]∩T) h n−1 (t, ρ (τ )) p ∇τ
1p
∇t
⎛ +⎝
a
b x
b
h n−1 (t, ρ (τ )) p ∇τ
1p
⎞⎤ ∇t ⎠⎦ , (9.31)
t
∀ x ∈ [a, b] ∩ T. Proof As before we have K (t, a) := f (t) −
n−1
f
∇k
h k (t, a) = (a)
t
n h n−1 (t, ρ (τ )) f ∇ (τ ) ∇τ ,
a
k=0
and K (t, b) := f (t) −
n−1
f
∇k
h k (t, b) = (b)
k=0
∀ t ∈ [a, b] ∩ T. We have that (by use of Theorem 9.16)
t b
n h n−1 (t, ρ (τ )) f ∇ (τ ) ∇τ ,
9.3 Main Results
229
t
|K (t, a)| ≤
h n−1 (t, ρ (τ )) p ∇τ
1p
a
t
≤
t
∇n f (τ )q ∇τ
q1
a
h n−1 (t, ρ (τ )) p ∇τ
1p
∇n f
a
L q ([a,b]∩T)
,
b ∇n |K (t, b)| = h n−1 (t, ρ (τ )) f (τ ) ∇τ ≤
and
t
b
h n−1 (t, ρ (τ )) ∇τ
1p
b
p
t
∇n f (τ )q ∇τ
q1
t
b
≤
h n−1 (t, ρ (τ )) p ∇τ
1p
∇n f
t
L q ([a,b]∩T)
,
∀ t ∈ [a, b] ∩ T. Hence it holds
t
−
h n−1 (t, ρ (τ )) p ∇τ
1p
∇n f
a
t
≤
h n−1 (t, ρ (τ )) p ∇τ
1p
L q ([a,b]∩T)
∇n f
a
and
−
b
h n−1 (t, ρ (τ )) p ∇τ
1p
∇n f
t
≤
b
h n−1 (t, ρ (τ )) ∇τ p
1p
L q ([a,b]∩T)
L q ([a,b]∩T)
∇n f
t
≤ K (t, a)
≤ K (t, b)
L q ([a,b]∩T)
,
∀ t ∈ [a, b] ∩ T. That is n−1
f ∇ (a) h k (t, a) −
t
h n−1 (t, ρ (τ )) p ∇τ
1p
∇n f
a
k=0
≤
k
n−1 k=0
f ∇ (a) h k (t, a) +
k
a
t
h n−1 (t, ρ (τ )) p ∇τ
1p
L q ([a,b]∩T)
∇n f
≤ f (t)
L q ([a,b]∩T)
230
9 Time Scales Nabla Iyengar Inequalities
and n−1
f
h k (t, b) − (b)
∇k
b
h n−1 (t, ρ (τ )) ∇τ
1p
p
∇n f
t
k=0 n−1
≤
f ∇ (b) h k (t, b) +
b
k
h n−1 (t, ρ (τ )) p ∇τ
1p
L q ([a,b]∩T)
∇n f
t
k=0
≤ f (t)
L q ([a,b]∩T)
,
∀ t ∈ [a, b] ∩ T. Let any x ∈ [a, b] ∩ T, then by integration we get n−1
f
∇k
n h k+1 (x, a) − f ∇ L q ([a,b]∩T) (a)
x
a
k=0
≤
x
t
h n−1 (t, ρ (τ )) p ∇τ
1p
∇t
a
f (t) ∇t ≤
a n−1
f
∇k
n h k+1 (x, a) + f ∇ L q ([a,b]∩T) (a)
x
t
h n−1 (t, ρ (τ )) ∇τ p
a
k=0
1p
∇t ,
a
(9.32) and −
n−1
f
∇k
n h k+1 (x, b) − f ∇ L q ([a,b]∩T) (b)
b
≤
b
h n−1 (t, ρ (τ )) ∇τ
1p
p
x
k=0
b
∇t
t
f (t) ∇t ≤
x
−
n−1
f
∇k
n h k+1 (x, b) + f ∇ (b)
k=0
⎛ L q ([a,b]∩T)
⎝
b x
b
h n−1 (t, ρ (τ )) p ∇τ
1p
⎞ ∇t ⎠ .
t
(9.33) Adding (9.32) and (9.33) we obtain n−1
k k f ∇ (a) h k+1 (x, a) − f ∇ (b) h k+1 (x, b) −
k=0
∇n f
# L q ([a,b]∩T)
x
t
h n−1 (t, ρ (τ )) ∇τ p
a
a
1p
∇t +
9.3 Main Results
231
b
b
h n−1 (t, ρ (τ )) ∇τ
1p
p
x
$ ∇t
t
≤
b
f (t) ∇t ≤
a n−1
k k f ∇ (a) h k+1 (x, a) − f ∇ (b) h k+1 (x, b) +
k=0
∇n f
# L q ([a,b]∩T)
b
a
x
x
b
t
h n−1 (t, ρ (τ )) p ∇τ
1p
∇t +
a
h n−1 (t, ρ (τ )) p ∇τ
1p
$ ∇t
,
(9.34)
t
∀ x ∈ [a, b] ∩ T. The theorem now is proved.
We give n Theorem 9.27 Let f ∈ Cld (T), m, n ∈ N, m < n, n − m is odd, a, b ∈ T; a ≤ b. Here ρ is continuous and h n−m−1 (t, s) is jointly continuous. Also assume Tk = T. Then (1)
n−m−1 b m k+m k+m ∇ ∇ ∇ f f (t) ∇t − h k+1 (x, a) − f h k+1 (x, b) ≤ (a) (b) a k=0
∇n f
∞,[a,b]∩T
a
b
x
b
x
t
h n−m−1 (t, ρ (τ )) ∇τ ∇t +
a
h n−m−1 (t, ρ (τ )) ∇τ ∇t ,
(9.35)
t
∀ x ∈ [a, b] ∩ T, k+m k+m (2) assuming f ∇ (a) = f ∇ (b) = 0, k = 0, 1, . . . , n − m − 1, we get from (9.35) that b n ∇m f (t) ∇t ≤ f ∇ ∞,[a,b]∩T · a
a
x
a
t
h n−m−1 (t, ρ (τ )) ∇τ ∇t +
b x
b t
h n−m−1 (t, ρ (τ )) ∇τ ∇t ,
(9.36)
232
9 Time Scales Nabla Iyengar Inequalities
∀ x ∈ [a, b] ∩ T, (21 ) when x = a we get from (9.36) that
n m f ∇ (t) ∇t ≤ f ∇
b
a
∞,[a,b]∩T
b
a
h n−m−1 (t, ρ (τ )) ∇τ ∇t ,
b
(9.37)
t
(22 ) when x = b we get from (9.36) that
b
f
∇m
a
∇n (t) ∇t ≤ f ∞,[a,b]∩T
b
a
t
h n−m−1 (t, ρ (τ )) ∇τ ∇t ,
a
(9.38)
(23 ) by (9.37), (9.38) we get
a
b
b
min a
b
f
∇m
n (t) ∇t ≤ f ∇ ∞,[a,b]∩T ·
h n−m−1 (t, ρ (τ )) ∇τ ∇t ,
t
b
a
t
, h n−m−1 (t, ρ (τ )) ∇τ ∇t
a
(9.39) and k+m k+m (3) assuming f ∇ (a) = f ∇ (b) = 0, k = 1, . . . , n − m − 1, from (9.35) we obtain b % ∇m & ∇ n ∇m ∇m f (t) ∇t − f (a) (x − a) + f (b) (b − x) ≤ f ∞,[a,b]∩T · a
x
a
t
h n−m−1 (t, ρ (τ )) ∇τ ∇t +
a
b x
b
h n−m−1 (t, ρ (τ )) ∇τ ∇t ,
(9.40)
t
∀ x ∈ [a, b] ∩ T. Proof As in the proof of Theorem 9.24, now using Corollary 9.14.
We give n Theorem 9.28 Let f ∈ Cld (T), m, n ∈ N, m < n, n − m is odd, a, b ∈ T; a ≤ b, where Tk = T. Then (1)
n−m−1 b m k+m k+m ∇ ∇ ∇ f f (t) ∇t − h k+1 (x, a) − f h k+1 (x, b) ≤ (a) (b) a k=0
9.3 Main Results
∇n f
233
L 1 ([a,b]∩T)
x
(t − ρ (a))
n−m−1
b
∇t +
a
(ρ (b) − t)
n−m−1
∇t , (9.41)
x
∀ x ∈ [a, b] ∩ T, k+m k+m (2) assuming f ∇ (a) = f ∇ (b) = 0, k = 0, 1, . . . , n − m − 1, we get from (9.41) that b ∇n ∇m ≤f f · ∇t (t) L 1 ([a,b]∩T) a
x
(t − ρ (a))
n−m−1
b
∇t +
(ρ (b) − t)
n−m−1
a
∇t ,
(9.42)
x
∀ x ∈ [a, b] ∩ T, (21 ) when x = a by (9.42) we get
b
f
∇m
a
∇n (t) ∇t ≤ f L 1 ([a,b]∩T)
b
(ρ (b) − t)
n−m−1
∇t ,
(9.43)
a
(22 ) when x = b by (9.42) we get
b
f
∇n (t) ∇t ≤ f L 1 ([a,b]∩T)
∇m
a
x
(t − ρ (a))
n−m−1
∇t ,
(9.44)
a
(23 ) by (9.43), (9.44) we have
a
b
min
b
f
∇m
n (t) ∇t ≤ f ∇ L 1 ([a,b]∩T) ·
(ρ (b) − t)n−m−1 ∇t ,
a
b
(t − ρ (a))n−m−1 ∇t
,
(9.45)
a
and k+m k+m (3) assuming f ∇ (a) = f ∇ (b) = 0, k = 1, . . . , n − m − 1, from (9.41) we obtain b % ∇m & ∇m ∇m f (t) ∇t − f (a) (x − a) + f (b) (b − x) ≤ a
∇n f
L 1 ([a,b]∩T)
a
x
b
(t − ρ (a))n−m−1 ∇t +
(ρ (b) − t)n−m−1 ∇t , (9.46)
x
∀ x ∈ [a, b] ∩ T. Proof As in Theorem 9.25, now using Corollary 9.14. We also give
234
9 Time Scales Nabla Iyengar Inequalities
n Theorem 9.29 Let f ∈ Cld (T), m, n ∈ N, m < n, n − m is odd, a, b ∈ T; a ≤ b. 1 1 Let also p, q > 1 : p + q = 1. We suppose Assumptions 9.20, 9.23. Then (1)
n−m−1 b m k+m k+m ∇ ∇ ∇ f f (t) ∇t − h k+1 (x, a) − f h k+1 (x, b) ≤ (a) (b) a k=0
∇n f
' a
b x
t
h n−m−1 (t, ρ (τ )) ∇τ p
L q ([a,b]∩T)
x
b
1p
∇t +
a
h n−m−1 (t, ρ (τ )) p ∇τ
1p
( ∇t
,
(9.47)
t
∀ x ∈ [a, b] ∩ T, k+m k+m (2) assuming f ∇ (a) = f ∇ (b) = 0, k = 0, 1, . . . , n − m − 1, we get from (9.47) that
b
f
∇m
a
' ∇n (t) ∇t ≤ f L q ([a,b]∩T)
x
a
b x
b
t
h n−m−1 (t, ρ (τ )) p ∇τ
1p
∇t +
a
h n−m−1 (t, ρ (τ )) p ∇τ
1p
( ∇t
,
(9.48)
t
∀ x ∈ [a, b] ∩ T, (21 ) when x = a we get from (9.48) that
b
f
∇m
a
∇n (t) ∇t ≤ f L q ([a,b]∩T)
b
b
h n−m−1 (t, ρ (τ )) ∇τ
1p
p
a
∇t ,
t
(9.49) (22 ) when x = b we get from (9.48) that
a
b
f
∇m
∇n (t) ∇t ≤ f L q ([a,b]∩T)
a
b
t
h n−m−1 (t, ρ (τ )) p ∇τ
1p
∇t ,
a
(9.50) (23 ) by (9.49), (9.50) we get
a
b
n m f ∇ (t) ∇t ≤ f ∇ L q ([a,b]∩T) ·
9.3 Main Results
235
#
b
b
h n−m−1 (t, ρ (τ )) ∇τ
min a
b
a
and (3) assuming f ∇ (9.47) that b
f
∇m
t
h n−m−1 (t, ρ (τ )) ∇τ
(a) = f ∇ %
k+m
∇m
(t) ∇t − f '
L q ([a,b]∩T)
b x
∇t
,
(9.51)
a
k+m
∇t , $
1p
(b) = 0, k = 1, . . . , n − m − 1, we get from
(a) (x − a) + f
∇m
a
∇n f
t
p
1p
p
x
a
b
t
& (b) (b − x) ≤
h n−m−1 (t, ρ (τ )) p ∇τ
1p
∇t +
a
h n−m−1 (t, ρ (τ )) p ∇τ
1p
( ∇t
,
(9.52)
t
∀ x ∈ [a, b] ∩ T.
Proof As in Theorem 9.26, by using Corollary 9.14.
9.4 Applications Next we give applications of our initial main results. Theorem 9.30 Let f ∈ C n ([a, b]), n ∈ N is odd and [a, b] ⊂ R. Then n−1 b (k) 1 k+1 k (k) k+1 f (a) (x − a) f (t) dt − + (−1) f (b) (b − x) a (k + 1)! k=0
≤
(n) f
∞,[a,b]
(n + 1)!
% & (x − a)n+1 + (b − x)n+1 ,
(9.53)
∀ x ∈ [a, b] . Proof By Theorem 9.24 (9.2). We continue with Theorem 9.31 Let f ∈ C n ([a, b]), n ∈ N is odd, [a, b] ⊂ R. Then
236
9 Time Scales Nabla Iyengar Inequalities
n−1 b (k) 1 k+1 k (k) k+1 f (a) (x − a) f (t) dt − + (−1) f (b) (b − x) a (k + 1)! k=0
≤
(n) f
L 1 ([a,b])
n
% & (x − a)n + (b − x)n ,
(9.54)
∀ x ∈ [a, b] .
Proof By Theorem 9.25 (9.17). We also give
Theorem 9.32 Let f ∈ C n ([a, b]), n ∈ N is odd and [a, b] ⊂ R. Let also p, q > 1 : 1p + q1 = 1. Then n−1 b (k) 1 k+1 k (k) k+1 f (a) (x − a) f (t) dt − + (−1) f (b) (b − x) a (k + 1)! k=0
≤
(n) f
) * n+ 1p n+ 1p − a) , + − x) (x (b 1 (n − 1)! ( p (n − 1) + 1) p n + 1p L q ([a,b])
(9.55)
∀ x ∈ [a, b] .
Proof By Theorem 9.26 (9.26). We continue with Theorem 9.33 Let f : Z → R, n is an odd number, a, b ∈ Z; a ≤ b. Then
b n−1 1 k k+1) k k+1) ( ( − ∇ f (b) (x − b) ∇ f (a) (x − a) f (t) − ≤ (k + 1)! t=a+1
k=0
∇ n f ∞,[a,b]∩Z (n − 1)!
b
'
x t=a+1
t=x+1
b
t
(t − τ + 1)(n−1)
+
τ =a+1
( (t − τ + 1)(n−1)
,
(9.56)
τ =t+1
∀ x ∈ [a, b] ∩ Z. Proof By Theorem 9.24 (9.2). We give
9.4 Applications
237
Theorem 9.34 Let f : Z → R, n ∈ N is odd, a, b ∈ Z; a ≤ b. Then b n−1 1 k k+1 k k+1 ∇ f (a) (x − a)( ) − ∇ f (b) (x − b)( ) ≤ f (t) − (k + 1)! t=a+1
k=0
# x $ b b n ∇ f (t) (t − a + 1)n−1 + (b − 1 − t)n−1 ,
t=a+1
t=a+1
(9.57)
t=x+1
∀ x ∈ [a, b] ∩ Z.
Proof By Theorem 9.25 (9.17). We give
Theorem 9.35 Let f : Z → R, n is an odd number, a, b ∈ Z; a ≤ b, let also p, q > 1 : 1p + q1 = 1. Then b n−1 1 ∇ k f (a) (x − a)(k+1) − ∇ k f (b) (x − b)(k+1) ≤ f (t) − (k + 1)! t=a+1
k=0
b
|∇ f (t)|
q1
⎡⎛
q
n
⎣⎝
t=a+1
(n − 1)! ⎛ ⎝
x
t=a+1
b
t=x+1
b
t
(t − τ + 1)(n−1)
p
1p ⎞
⎠+
τ =a+1
(t − τ + 1)(n−1)
p
1p ⎞⎤ ⎠⎦ ,
(9.58)
τ =t+1
∀ x ∈ [a, b] ∩ Z.
Proof By Theorem 9.26 (9.26). We need
Remark 9.36 ([5]) We consider the time scale T = q Z = {0, 1, q, q −1 , q 2 , q −2 , . . .}, for some q > 1. Here ρ (t) = qt , ∀ t ∈ T. We have that h k (t, s) = for all k ∈ N0 . We continue with
k−1 + qr t − s
r , for all s, t ∈ T, j j=0 q r =0
238
9 Time Scales Nabla Iyengar Inequalities
n q Z , n ∈ N is odd, a, b ∈ q Z ; a ≤ b. Then Theorem 9.37 Let f ∈ Cld ⎛ ⎞ n−1 ⎜ k k b ⎟ ν ν + + q x −a q x − b k ⎜ ∇k ⎟ ∇ f (t) ∇t − − f (b) ≤ ⎜ f (a) ⎟ ν ν
a ⎝ ⎠ ν=0 ν=0 k=0 qμ qμ μ=0 μ=0 ∇n f
# L 1 [a,b]∩q Z
a
x
n−1 $ b a n−1 b t− −t ∇t + ∇t , q q x
(9.59)
∀ x ∈ [a, b] ∩ q Z . Proof By Theorem 9.25 (9.17).
We finish with
n Theorem 9.38 Let f ∈ Cld q Z , m, n ∈ N; m < n, n − m is odd, a, b ∈ q Z ; a ≤ b. Then ⎛ ⎞ n−m−1 k k b m ⎜ ⎟ ⎜ k+m + qν x − a + q ν x − b ⎟ ∇ ∇ ∇ k+m f (t) ∇t − − f (a) (b) ⎜f ⎟ ≤ ν ν
a ⎝ ⎠ ν=0 ν=0 k=0 qμ qμ μ=0 μ=0 ∇n f
# L 1 [a,b]∩q Z
a
x
n−m−1 $ b a n−m−1 b t− −t ∇t + ∇t , q q x
(9.60)
∀ x ∈ [a, b] ∩ q Z . Proof By Theorem 9.28 (9.41).
One can give many similar applications for other time scales.
References 1. G. Anastassiou, Intelligent Mathematics: Computational Analysis (Springer, Heidelberg, 2011) 2. G. Anastassiou, Nabla time scales inequalities. Int. J. Dyn. Syst. Differ. Equ. 3(1–2), 59–83 (2011) 3. G. Anastassiou, Time scales delta iyengar type inequalities. Int. J. Differ. Equ. (2019), http:// campus.mst.edu/ijde (Accepted for publication) 4. G. Anastasssiou, Nabla time scales iyengar type inequalities. Adv. Dyn. Syst. Appl. (2019) (Accepted for publication)
References
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5. D.R. Anderson, Taylor polynomials for nabla dynamic equations on times scales. Panamer. Math. J. 12(4), 17–27 (2002) 6. D. Anderson, J. Bullock, L. Erbe, A. Peterson, H. Tran, Nabla dynamic equations on time scales. Panamer. Math. J. 13(1), 1–47 (2003) 7. F. Atici, D. Biles, A. Lebedinsky, An application of time scales to economics. Math. Comput. Model. 43, 718–726 (2006) 8. M. Bohner, A. Peterson, Dynamic equations on time scales: an introduction with applications (Birkhaüser, Boston, 2001) 9. S. Hilger, Ein Maßketten kalkül mit Annendung auf Zentrumsmannig-faltigkeiten. Ph.D. Thesis (Universität Würzburg, Germany, 1988) 10. K.S.K. Iyengar, Note on an inequality. Math. Student 6, 75–76 (1938) 11. N. Martins, D. Torres, Calculus of variations on time scales with nabla derivatives. Nonlinear Anal. 71(12), 763–773 (2009)
Chapter 10
Choquet–Iyengar Advanced Inequalities
Here we extend advanced known Iyengar type inequalities to Choquet integrals setting with respect to distorted Lebesgue measures and for monotone functions. See also [2].
10.1 Background—I In the year 1938, Iyengar [8] proved the following interesting inequality. Theorem 10.1 Let f be a differentiable function on [a, b] and f (x) ≤ M1 . Then
b
a
M1 (b − a)2 1 ( f (b) − f (a))2 − f (x) d x − (b − a) ( f (a) + f (b)) ≤ . 2 4 4M1 (10.1)
In 2001, Cheng [4] proved that Theorem 10.2 Let f ∈ C 2 ([a, b]) and f (x) ≤ M2 . Then
a
b
1 1 2 f (x) d x − (b − a) ( f (a) + f (b)) + (b − a) f (b) − f (a) ≤ 2 8 M2 (b − a) 2 , (b − a)3 − 24 16M2 1
where 1 = f (a) −
(10.2)
2 ( f (b) − f (a)) + f (b) . (b − a)
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_10
241
242
10 Choquet–Iyengar Advanced Inequalities
In 2006, [7], the authors proved: Theorem 10.3 Let f ∈ C 2 ([a, b]) and f (x) ≤ M. Set I =
b
1 1 (b − a) ( f (a) + f (b)) + (b − a)2 f (b) − f (a) . 2 8 (10.3)
f (x) d x −
a
Then
M 3 M (b − a)3 + λa + λ3b ≤ I ≤ 24 3 3 3 M (b − a)3 M b−a b−a − − λa + − λb , 24 3 2 2 −
where λa =
1 2M
λb =
1 2M
f
a+b 2
f (b) − f
b−a − f (a) + , 4
a+b 2
+
b−a . 4
(10.4)
(10.5)
(10.6)
In 1996, Agarwal and Dragomir [1] obtained a generalization of (10.1): Theorem 10.4 Let f : [a, b] → R be a differentiable function such that for all x ∈ [a, b] with M > m we have m ≤ f (x) ≤ M. Then
a
b
f (x) d x −
1 (b − a) ( f (a) + f (b)) ≤ 2
( f (b) − f (a) − m (b − a)) (M (b − a) − f (b) + f (a)) . 2 (M − m)
(10.7)
In [10], Qi proved Theorem 10.5 Let f : [a, b] → R be a twice differentiable function such that for all x ∈ [a, b] with M > 0 we have f (x) ≤ M. Then
b 1 + Q2 f + f ( (a) (b)) 2 f (b) − f (a) (b − a) ≤ f (x) d x − (b − a) + 2 8 a
M (b − a)3 1 − 3Q 2 , 24 where
(10.8)
10.1 Background—I
243
Q2 =
f (a) + f (b) − 2
f (b)− f (a) b−a
2
M 2 (b − a)2 − ( f (b) − f (a))2
.
(10.9)
Finally in 2005, Liu [9], proved the following: Theorem 10.6 Let f : [a, b] → R be a differentiable function such that f is integrable on [a, b] and for all x ∈ [a, b] with M > m we have m≤
f (x) − f (a) f (b) − f (x) ≤ M and m ≤ ≤ M. x −a b−x
(10.10)
Then b 1 + P2 ( f (a) + f (b)) f (b) − f (a) (b − a)2 f (x) d x − (b − a) + 2 8 a −
1 + 3P 2 48
(M − m) (b − a)3 1 − 3P 2 , (m + M) (b − a)3 ≤ 48
(10.11)
where
2
f (a) f (a) + f (b) − 2 f (b)− b−a P2 = 2 . 2 M−m 2 (b − a) (b − a) − f (b) − f (a) − m+M 2 2
(10.12)
In [3] we extended (10.1) for Choquet integrals. Motivated by these results we extend here Theorems 10.2–10.6 to the Choquet integrals setting.
10.2 Background—II In the next assume that (X, F) is a measurable space and (R+ ) R is the set of all (nonnegative) real numbers. We recall some concepts and some elementary results of capacity and the Choquet integral [5, 6]. Definition 10.7 A set function μ : F → R+ is called a non-additive measure (or capacity) if it satisfies (1) μ (∅) = 0; (2) μ (A) ≤ μ (B) for any A ⊆ B and A, B ∈ F. The non-additive measure μ is called concave if μ (A ∪ B) + μ ( A ∩ B) ≤ μ (A) + μ (B) ,
(10.13)
244
10 Choquet–Iyengar Advanced Inequalities
for all A, B ∈ F. In the literature the concave non-additive measure is known as submodular or 2-alternating non-additive measure. If the above inequality is reverse, μ is called convex. Similarly, convexity is called supermodularity or 2-monotonicity, too. First note that the Lebesgue measure λ for an interval [a, b] is defined by λ ([a, b]) = b − a, and that given a distortion function m, which is increasing (or nondecreasing) and such that m (0) = 0, the measure μ (A) = m (λ (A)) is a distorted Lebesgue measure. We denote a Lebesgue measure with distortion m by μ = μm . It is known that μm is concave (convex) if m is a concave (convex) function. The family of all the nonnegative, measurable function f : (X, F) → R+ , B R+ + is the Borel σ-field of R+ . The concept of the integral is denoted as L + ∞ , where B R with respect to a non-additive measure was introduced by Choquet [5]. Definition 10.8 Let f ∈ L + ∞ . The Choquet integral of f with respect to non-additive measure μ on A ∈ F is defined by
∞
f dμ :=
(C)
μ ({x : f (x) ≥ t} ∩ A) dt,
(10.14)
0
A
where the integral on the right-hand side is a Riemann integral. Instead of (C) X f dμ, we shall write (C) f dμ. If (C) f dμ < ∞, we say that f is Choquet integrable and we write
L C1 (μ) =
f dμ < ∞ .
f : (C)
The next lemma summarizes the basic properties of Choquet integrals [6]. 1 Lemma 10.9 Assume that f, g ∈ L C (μ). (1) (C) 1 A dμ = μ (A), A ∈ F. (2) (Positive homogeneity) For all λ ∈ R+ , we have (C) λ f dμ = λ · (C) f dμ. (3) (Translation invariance) For all c ∈ R, we have (C) A ( f + c) dμ = (C) A f dμ + cμ ( A) . (4) (Monotonicity in the integrand) If f ≤ g, then we have
(C)
f dμ ≤ (C)
gdμ.
(Monotonicity in the set function) If μ ≤ ν, then we have (C) (5) (Subadditivity) If μ is concave, then (C)
( f + g) dμ ≤ (C)
(Superadditivity) If μ is convex, then
f dμ ≤ (C)
f dμ + (C)
gdμ.
f dν.
10.2 Background—II
245
( f + g) dμ ≥ (C)
(C)
f dμ + (C)
gdμ.
(6) (Comonotonic additivity) If f and g are comonotonic, then
( f + g) dμ = (C)
(C)
f dμ + (C)
gdμ,
where we say that f and g are comonotonic, if for any x, x ∈ X , then
f (x) − f x g (x) − g x ≥ 0.
We next mention the amazing result from [11], which permits us to compute the Choquet integral when the non-additive measure is a distorted Lebesgue measure. Theorem 10.10 Let f be a nonnegative and measurable function on R+ and μ = μm be a distorted Lebesgue measure. Assume that m (x) and f (x) are both continuous and m (x) is differentiable. When f is an increasing (non-decreasing) function on R+ , the Choquet integral of f with respect to μm on [0, t] is represented as
t
f dμm =
(C)
m (t − x) f (x) d x,
(10.15)
0
[0,t]
however, when f is a decreasing (non-increasing) function on R+ , the Choquet integral of f is t f dμm = m (x) f (x) d x. (10.16) (C) 0
[0,t]
Remark 10.11 We denote by
γ (t, x) :=
m (t − x) , when f is increasing (non-decreasing), m (x) , when f is decreasing (non-increasing).
(10.17)
So for f continuous and monotone we can combine (10.15) and (10.16) into
f dμm =
(C) [0,t]
t
γ (t, x) f (x) d x.
(10.18)
0
10.3 Main Results We present the following advanced Choquet–Iyengar type inequalities: The next is based on Theorem 10.2.
246
10 Choquet–Iyengar Advanced Inequalities
Theorem 10.12 Here f : R+ → R+ is a monotone twice continuously differentiable function on R+ , μm is a distorted Lebesgue measure, where m is such that m (0) = 0, m is increasing and thrice continuously differentiable on R+ , t ∈ R+ . Then (i) if f is increasing and m (t − ·) f (x) ≤ M2 , ∀ x ∈ [0, t], M2 > 0, we have that t (C) m (t) f (0) + m (0) f (t) + f (x) dμm (x) − 2 [0,t] t 2 m (0) f (t) − m (t) f (0) + m (t) f (0) − m (0) f (t) ≤ 8 M2 3 t ∗2 , t − 24 16M2 1
(10.19)
where 2 m (0) f (t) − m (t) f (0) − ∗1 = m (t) f (0) + m (0) f (t) − t m (t) f (0) + m (0) f (t) , (10.20) (ii) if f is decreasing and m f (x) ≤ M3 , ∀ x ∈ [0, t], M3 > 0, we have that (C)
f (x) dμm (x) − [0,t]
t m (0) f (0) + m (t) f (t) + 2
t 2 m (t) f (t) − m (0) f (0) + m (t) f (t) − m (0) f (0) ≤ 8 M3 3 t t − ∗∗2 , 24 16M3 1
(10.21)
where ∗∗ 1
2 m (t) f (t) − m (0) f (0) + = m (t) f (t) + m (0) f (0) − t m (t) f (t) + m (0) f (0) . (10.22)
Proof (i) If f is increasing and m (t − ·) f (x) ≤ M2 , ∀ x ∈ [0, t], M2 > 0, we have that
10.3 Main Results
(C)
247
f (x) dμm (x) − [0,t]
t m (t) f (0) + m (0) f (t) + 2
t 2 m (t − ·) f (t) − m (t − ·) f (0) = 8 (C)
f (x) dμm (x) − [0,t]
t m (t) f (0) + m (0) f (t) + 2
(by (10.2) & (10.15)) t 2 m (0) f (t) − m (t) f (0) + m (t) f (0) − m (0) f (t) ≤ 8 M2 3 t ∗2 , t − 24 16M2 1
(10.23)
where ∗1
2 m (0) f (t) − m (t) f (0) + m (t − ·) f (t) = = m (t − ·) f (0) − t 2 m (0) f (t) − m (t) f (0) − m (t) f (0) + m (0) f (t) − t m (t) f (0) + m (0) f (t) . (10.24)
(ii) If f is decreasing and m f (x) ≤ M3 , ∀ x ∈ [0, t], M3 > 0, we have that (C)
f (x) dμm (x) − [0,t]
t m (0) f (0) + m (t) f (t) + 2
t 2 m f (t) − m f (0) = 8 (C)
f (x) dμm (x) − [0,t]
t m (0) f (0) + m (t) f (t) + 2
(by (10.2) & (10.16)) t 2 m (t) f (t) − m (0) f (0) + m (t) f (t) − m (0) f (0) ≤ 8 M3 3 t t − ∗∗2 , 24 16M3 1 where
(10.25)
248
10 Choquet–Iyengar Advanced Inequalities
∗∗ 1 = m (t) f (t) + m (0) f (0) + m (t) f (t) + m (0) f (0) 2 m f (t) − m f (0) − . t
(10.26)
The theorem is proved. The next result is based on Theorem 10.3.
Theorem 10.13 Here f : R+ → R+ is a monotone twice continuously differentiable function on R+ , μm is a distorted Lebesgue measure, where m is such that m (0) = 0, m is increasing and thrice continuously differentiable on R+ , t ∈ R+ . Then (i) if f is increasing and m (t − ·) f (x) ≤ M1 , ∀ x ∈ [0, t], M1 > 0, we call: t m (t) f (0) + m (0) f (t) + f (x) dμm (x) − I1 = (C) 2 [0,t] t 2 m (0) f (t) − m (t) f (0) + m (t) f (0) − m (0) f (t) , 8 and λ(1) 0
1 = 2M1
t t t t −m f +m f 2 2 2 2
t + m (t) f (0) − m (t) f (0) + , 4 and λ(1) t =
(10.28)
1 −m (0) f (t) + m (0) f (t) 2M1
t t t t t f −m f + , + m 2 2 2 2 4
and we obtain
(10.27)
M1 t3 −M1 + 24 3 M1 t 3 M1 − 24 3
(10.29)
3
3 (1) (1) ≤ I1 ≤ λ0 + λt t − λ(1) 0 2
3
+
t − λ(1) t 2
3 ,
(10.30)
(ii) if f is decreasing and m f (x) ≤ M2 , ∀ x ∈ [0, t], M2 > 0, we call: I2 = (C)
f (x) dμm (x) − [0,t]
t m (0) f (0) + m (t) f (t) + 2
10.3 Main Results
249
t 2 m (t) f (t) − m (0) f (0) + m (t) f (t) − m (0) f (0) 8 and λ(2) 0 =
1 2M2
m
t t t t f + m f 2 2 2 2
t − m (0) f (0) + m (0) f (0) + , 4 and λ(2) t =
(10.32)
1 m (t) f (t) + m (t) f (t) 2M2
t t t t t f +m f + , − m 2 2 2 2 4
and we obtain:
(10.31)
M2 M2 t 3 + − 24 3 M2 t 3 M2 − 24 3
λ(2) 0
3
t − λ(2) 0 2
(10.33)
3
(2) ≤ I2 ≤ + λt
3
+
t − λ(2) t 2
3 .
(10.34)
Proof (i) Here f is increasing and m (t − ·) f (x) ≤ M1 , ∀ x ∈ [0, t], M1 > 0. We call t m (t) f (0) + m (0) f (t) + I1 = (C) f (x) dμm (x) − 2 [0,t] t 2 m (t − ·) f (t) − m (t − ·) f (0) = 8 f (x) dμm (x) −
(C) [0,t]
t m (t) f (0) + m (0) f (t) + 2
t 2 m (0) f (t) − m (t) f (0) + m (t) f (0) − m (0) f (t) . 8
(10.35)
We set λ(1) 0 =
1 2M1
t t − m (t − ·) f (0) + = m (t − ·) f 2 4
(10.36)
250
10 Choquet–Iyengar Advanced Inequalities
1 2M1
−m
t t t t f + m f 2 2 2 2
t + m (t) f (0) − m (t) f (0) + , 4
(10.37)
and λ(1) t =
1 2M1
m (t − ·) f (t) − m (t − ·) f
t t + = 2 4
(10.38)
1 −m (0) f (t) + m (0) f (t) 2M1 t t t t t f −m f + . + m 2 2 2 2 4
By Theorem 10.3 and (10.15) we get M1 t3 + −M1 24 3 M1 t 3 M1 − 24 3
3
3 (1) (1) ≤ I1 ≤ λ0 + λt t − λ(1) 0 2
3
+
t − λ(1) t 2
3 .
(10.39)
(ii) Next f is decreasing and m f (x) ≤ M2 , ∀ x ∈ [0, t], M2 > 0. We call t m (0) f (0) + m (t) f (t) + I2 = (C) f (x) dμm (x) − 2 [0,t] t 2 m f (t) − m f (0) = 8 f (x) dμm (x) −
(C) [0,t]
t m (0) f (0) + m (t) f (t) + 2
(10.40)
t 2 m (t) f (t) − m (0) f (0) + m (t) f (t) − m (0) f (0) . 8 We set λ(2) 0 =
1 2M2
m
t t t t f + m f 2 2 2 2
(10.41)
10.3 Main Results
251
− m (0) f (0) + m (0) f (0) and λ(2) t =
t + , 4
1 m (t) f (t) + m (t) f (t) 2M2
(10.42)
t t t t t f +m f + . − m 2 2 2 2 4
By Theorem 10.3 and (10.16) we get −
M2 M2 t 3 + 24 3
M2 t 3 M2 − 24 3
λ(2) 0
3
t − λ(2) 0 2
3
≤ I2 ≤ + λ(2) t
3
The theorem is proved.
+
t − λ(2) t 2
3 .
(10.43)
The next result is based on Theorem 10.4. Theorem 10.14 Here f : R+ → R+ is a monotone differentiable function on R+ , μm is a distorted Lebesgue measure, where m is such that m (0) = 0, m is increasing and twice differentiable on R+ , t ∈ R+ . Then (i) if f is increasing, and m 1 ≤ m (t − ·) f (x) ≤ M1 , ∀ x ∈ [0, t], where M1 > m 1 , we obtain: t (C) m (t) f (0) + m (0) f (t) ≤ f (x) dμm (x) − 2 [0,t] m (0) f (t) − m (t) f (0) − m 1 t M1 t − m (0) f (t) + m (t) f (0) . (10.44) 2 (M1 − m 1 ) (ii) if f is decreasing, and m 2 ≤ m f (x) ≤ M2 , ∀ x ∈ [0, t], where M2 > m 2 , we obtain: t ≤ (C) m f dμ − f + m f (x) (x) (0) (0) (t) (t) m 2 [0,t] m (t) f (t) − m (0) f (0) − m 2 t M2 t − m (t) f (t) + m (0) f (0) . (10.45) 2 (M2 − m 2 ) Proof (i) Here f is increasing and m 1 ≤ m (t − ·) f (x) ≤ M1 , ∀ x ∈ [0, t], where M1 > m 1 . We get, by Theorem 10.4 and (10.15), that
252
10 Choquet–Iyengar Advanced Inequalities
(C)
f (x) dμm (x) −
[0,t]
t m (t) f (0) + m (0) f (t) ≤ 2
m (0) f (t) − m (t) f (0) − m 1 t M1 t − m (0) f (t) + m (t) f (0) . (10.46) 2 (M1 − m 1 ) (ii) Next f is decreasing and m 2 ≤ m f (x) ≤ M2 , ∀ x ∈ [0, t], where M2 >m 2 . We get, by Theorem 10.4 and (10.16), that (C)
f (x) dμm (x) −
[0,t]
t m (0) f (0) + m (t) f (t) ≤ 2
m (t) f (t) − m (0) f (0) − m 2 t M2 t − m (t) f (t) + m (0) f (0) . (10.47) 2 (M2 − m 2 )
The theorem is proved. The next result is based on Theorem 10.5.
Theorem 10.15 Here f : R+ → R+ is a monotone twice differentiable function on R+ , μm is a distorted Lebesgue measure, where m is such that m (0) = 0, m is increasing and thrice differentiable on R+ , t ∈ R+ . Then (i) if f is increasing, and m (t − ·) f (x) ≤ M1 , ∀ x ∈ [0, t], M1 > 0, we call: Q 21 =
(t) f (0) 2 −m (t) f (0) + m (t) f (0) + −m (0) f (t) + m (0) f (t) − 2 m (0) f (t)−m t 2 M12 t 2 − −m (0) f (t) + m (0) f (t) + m (t) f (0) − m (t) f (0)
and we obtain (C)
f (x) dμm (x) − [0,t]
t m (t) f (0) + m (0) f (t) + 2
1 + Q 21 t 2 −m (0) f (t) + m (0) f (t) + m (t) f (0) − m (t) f (0) ≤ 8 M1 t 3 1 − 3Q 21 , (10.48) 24 (ii) if f is decreasing, and m f (x) ≤ M2 , ∀ x ∈ [0, t], M2 > 0, we call: Q 22 =
10.3 Main Results
253
(0) f (0) 2 m (0) f (0) + m (0) f (0) + m (t) f (t) + m (t) f (t) − 2 m (t) f (t)−m t 2 M22 t 2 − m (t) f (t) + m (t) f (t) − m (0) f (0) − m (0) f (0)
and we obtain (C)
f (x) dμm (x) − [0,t]
(10.49)
t m (0) f (0) + m (t) f (t) + 2
1 + Q 22 t 2 m (t) f (t) + m (t) f (t) − m (0) f (0) − m (0) f (0) ≤ 8
M2 t 3 1 − 3Q 22 . (10.50) 24 Proof (i) If f is increasing, and m (t − ·) f (x) ≤ M1 , ∀ x ∈ [0, t], M1 > 0, we set:
Q 21 =
2
m (t−·) f )(t)−(m (t−·) f )(0) (0) + m (t − ·) f (t) − 2 ( t = 2 M12 t 2 − (m (t − ·) f ) (t) − (m (t − ·) f ) (0)
m (t − ·) f
(t) f (0) 2 −m (t) f (0) + m (t) f (0) + −m (0) f (t) + m (0) f (t) − 2 m (0) f (t)−m t . 2 M12 t 2 − −m (0) f (t) + m (0) f (t) + m (t) f (0) − m (t) f (0)
(10.51)
By Theorem 10.5 and (10.15) we derive (C)
f (x) dμm (x) − [0,t]
t m (t) f (0) + m (0) f (t) + 2
1 + Q 21 t 2 −m (0) f (t) + m (0) f (t) + m (t) f (0) − m (t) f (0) ≤ 8 M1 t 3 1 − 3Q 21 . (10.52) 24 (ii) If f is decreasing, and m f (x) ≤ M2 , ∀ x ∈ [0, t], M2 > 0, we set:
2
m f (t)− m f (0) m f (0) + m f (t) − 2 ( ) t ( ) = Q 22 = 2 M22 t 2 − (m f ) (t) − (m f ) (0)
(10.53)
254
10 Choquet–Iyengar Advanced Inequalities
(0) f (0) 2 m (0) f (0) + m (0) f (0) + m (t) f (t) + m (t) f (t) − 2 m (t) f (t)−m t . 2 M22 t 2 − m (t) f (t) + m (t) f (t) − m (0) f (0) − m (0) f (0)
By Theorem 10.5 and (10.16) we derive (C)
f (x) dμm (x) − [0,t]
t m (0) f (0) + m (t) f (t) + 2
1 + Q 22 t 2 m (t) f (t) + m (t) f (t) − m (0) f (0) − m (0) f (0) ≤ 8 M2 t 3 1 − 3Q 22 . 24
(10.54)
The theorem is proved. Finally we apply Theorem 10.6 to obtain:
Theorem 10.16 Here f : R+ → R+ is a monotone and continuously differentiable function on R+ , μm is a distorted Lebesgue measure, where m is such that m (0) = 0, m is increasing and twice continuously differentiable on R+ , t ∈ R+ . We have (i) If f is increasing, and m1 ≤ and m1 ≤
m (t − ·) f
(x) − m (t − ·) f (0) ≤ M1 , x
m (t − ·) f (t) − m (t − ·) f (x) ≤ M1 , t−x
(10.55)
(10.56)
∀ x ∈ [0, t], with m 1 < M1 , we set:
2
m (t−·) f )(t)−(m (t−·) f )(0) m (t − ·) f (0) + m (t − ·) f (t) − 2 ( t P12 =
2 . M1 −m 1 2 2 (t − ·) f ) (t) − (m (t − ·) f ) (0) − (m 1 +M1 ) t t − (m 2 2 (10.57) Then m (t − ·) f (0) + m (t − ·) f (t) t+ f (x) dμm (x) − (C) 2 [0,t]
1 + P12 8
m (t − ·) f
(t) − m (t − ·) f (0) t 2 −
1 + 3P12 48
(m 1 + M1 ) t 3
10.3 Main Results
255
≤
(M1 − m 1 ) t 3 1 − 3P12 . 48
(10.58)
(ii) If f is decreasing, and m f (x) − m f (0) m2 ≤ ≤ M2 , x and m2 ≤
m f (t) − m f (x) ≤ M2 , t−x
(10.59)
(10.60)
∀ x ∈ [0, t], with m 2 < M2 , we set:
2
m f (t)− m f (0) m f (0) + m f (t) − 2 ( ) t ( ) P22 =
2 . M2 −m 2 2 2 f ) (t) − (m f ) (0) − (m 2 +M2 ) t t − (m 2 2
(10.61)
m f (0) + m f (t) t+ f (x) dμm (x) − (C) 2 [0,t]
Then
1 + P22 8
2 1 + 3P22 3 m f (t) − m f (0) t − (m 2 + M2 ) t ≤ 48 (M2 − m 2 ) t 3 1 − 3P22 . 48
(10.62)
t , t ∈ R+ . We have Example 10.17 A well-known distortion function is m (t) = 1+t 1 m (0) = 0, m (t) ≥ 0, m (t) = (1+t)2 > 0, that is m is strictly increasing. We have
that m (t) = −2 (1 + t)−3 , m (3) (t) = 6 (1 + t)−4 , and in general we get that m (n) (t) = (−1)n+1 n! (1 + t)−(n+1) , ∀ n ∈ N.
References 1. R.P. Agarwal, S.S. Dragomir, An application of Hayashi’s inequality for differentiable functions. Comput. Math. Appl. 6, 95–99 (1996) 2. G. Anastassiou, Choquet-Iyengar type advanced inequalities. J. Comput. Anal. Appl. 28(1), 166–179 (2020) 3. G. Anastassiou, Choquet integral analytic inequalities. Studia Mathematica Babes-Bolyai, accepted for publication (2018) 4. X.-L. Cheng, The Iyengar-type inequality. Appl. Math. Lett. 14, 975–978 (2001) 5. G. Choquet, Theory of capacities. Ann. Inst. Fourier (Grenoble) 5, 131–295 (1953)
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10 Choquet–Iyengar Advanced Inequalities
6. D. Denneberg, Non-additive Measure and Integral (Kluwer Academic Publishers, Boston, 1994) 7. I. Franjic, J. Pecaric, I. Peric, Note on an Iyengar type inequality. Appl. Math. Lett. 19, 657–660 (2006) 8. K.S.K. Iyengar, Note on an inequality. Math. Stud. 6, 75–76 (1938) 9. Z. Liu, Note on Iyengar’s inequality. Univ. Beograde Publ. Elektrotechn. Fak. Ser. Mat. 16, 29–35 (2005) 10. F. Qi, Further generalizations of inequalities for an integral. Univ. Beograd Publ. Elektrotechn. Fak. Ser. Mat. 8, 79–83 (1997) 11. M. Sugeno, A note on derivatives of functions with respect to fuzzy measures. Fuzzy Sets Syst. 222, 1–17 (2013)
Chapter 11
Fractional Conformable Iyengar Inequalities
Here we present Conformable fractional Iyengar type inequalities with respect to L p norms, with 1 < p ≤ ∞. The method is based on the right and left Conformable fractional Taylor’s formulae. See also [3].
11.1 Background We are motivated by the following famous Iyengar inequality (1938), [4]. Theorem 11.1 Let f be a differentiable function on [a, b] and f (x) ≤ M. Then
a
b
M (b − a) 2 ( f (b) − f (a))2 1 . f (x) d x − (b − a) ( f (a) + f (b)) ≤ − 2 4 4M (11.1)
Here we follow [1] for the basics of generalized Conformable fractional calculus, see also [5]. We need Definition 11.2 ([1]) Let a, b ∈ R. The left conformable fractional derivative starting from a of a function f : [a, ∞) → R of order 0 < α ≤ 1 is defined by a f t + ε (t − a)1−α − f (t) Tα f (t) = lim . (11.2) ε→0 ε If Tαa f (t) exists on (a, b), then a Tα f (a) = lim Tαa f (t) . (11.3) t→a+
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_11
257
258
11 Fractional Conformable Iyengar Inequalities
The right conformable fractional derivative of order 0 < α ≤ 1 terminating at b of f : (−∞, b] → R is defined by f t + ε (b − t)1−α − f (t) . α T f (t) = −lim ε→0 ε
b If
b
αT
(11.4)
f (t) exists on (a, b), then b
αT
f (b) = lim bα T f (t) . t→b−
(11.5)
Note that if f is differentiable then and
Tαa f (t) = (t − a)1−α f (t) ,
b
αT
Denote by
and
f (t) = − (b − t)1−α f (t) .
Iαa f (t) =
b
Iα f (t) =
t
(x − a)α−1 f (x) d x,
(11.6)
(11.7)
(11.8)
a
b
(b − x)α−1 f (x) d x,
(11.9)
t
these are the left and right conformable fractional integrals of order 0 < α ≤ 1. In the higher order case we can generalize things as follows: Definition 11.3 ([1]) Let α ∈ (n, n + 1], and set β = α − n. Then, the left conformable fractional derivative starting from a of a function f : [a, ∞) → R of order α, where f (n) (t) exists, is defined by a Tα f (t) = Tβa f (n) (t) ,
(11.10)
The right conformable fractional derivative of order α terminating at b of f : (−∞, b] → R, where f (n) (t) exists, is defined by b
αT
f (t) = (−1)n+1 bβ T f (n) (t) .
(11.11)
If α = n + 1 then β = 1 and Tan+1 f = f (n+1) . If n is odd, then bn+1 T f = − f (n+1) , and if n is even, then bn+1 T f = f (n+1) . When n = 0 (or α ∈ (0, 1]), then β = α, and (11.10), (11.11) collapse to {(11.2)– (11.5)}, respectively.
11.1 Background
259
Lemma 11.4 ([1]) Let f : (a, b) → R be continuously differentiable and 0 < α ≤ 1. Then, for all t > a we have Iαa Tαa ( f ) (t) = f (t) − f (a) .
(11.12)
We need Definition 11.5 (see also [1]) If α ∈ (n, n + 1], then the left fractional integral of order α starting at a is defined by
Iαa
1 f (t) = n!
t
(t − x)n (x − a)β−1 f (x) d x.
(11.13)
a
Similarly, (author’s definition, see [2]) the right fractional integral of order α terminating at b is defined by b
1 Iα f (t) = n!
b
(x − t)n (b − x)β−1 f (x) d x.
(11.14)
t
We need Proposition 11.6 ([1]) Let α ∈ (n, n + 1] and f : [a, ∞) → R be (n + 1) times continuously differentiable for t > a. Then, for all t > a we have Iαa Taα ( f ) (t) = f (t) −
n f (k) (a) (t − a)k . k! k=0
(11.15)
We also have Proposition 11.7 ([2]) Let α ∈ (n, n + 1] and f : (−∞, b] → R be (n + 1) times continuously differentiable for t < b. Then, for all t < b we have −b Iα bα T ( f ) (t) = f (t) −
n f (k) (b) (t − b)k . k! k=0
(11.16)
If n = 0 or 0 < α ≤ 1, then (see also [1]) b
Iα bα T ( f ) (t) = f (t) − f (b) .
(11.17)
In conclusion we derive Theorem 11.8 ([2]) Let α ∈ (n, n + 1] and f ∈ C n+1 ([a, b]), n ∈ N. Then
260
11 Fractional Conformable Iyengar Inequalities
(1) f (t) −
n 1 t f (k) (a) (t − a)k = (t − x)n (x − a)β−1 Taα ( f ) (x) d x, k! n! a k=0 (11.18)
and (2) n 1 b f (k) (b) (t − b)k =− (b − x)β−1 (x − t)n bα T ( f ) (x) d x, k! n! t k=0 (11.19) ∀ t ∈ [a, b] . f (t) −
We need Remark 11.9 ([2]) We notice the following: let α ∈ (n, n + 1] and f ∈ C n+1 ([a, b]), n ∈ N. Then (β := α − n, 0 < β ≤ 1) a Tα ( f ) (x) = Tβα f (n) (x) = (x − a)1−β f (n+1) (x) , and
b
αT (
(11.20)
f ) (x) = (−1)n+1 bβ T f (n) (x) =
(−1)n+1 (−1) (b − x)1−β f (n+1) (x) = (−1)n (b − x)1−β f (n+1) (x) .
(11.21)
Consequently we get that a Tα ( f ) (x) , bα T ( f ) (x) ∈ C ([a, b]) . Furthermore it is obvious that a Tα ( f ) (a) = bα T ( f ) (b) = 0,
(11.22)
when 0 < β < 1, i.e. when α ∈ (n, n + 1) . If f (k) (a) = 0, k = 1, . . . , n, then f (t) − f (a) =
1 n!
t
a
(t − x)n (x − a)β−1 Taα ( f ) (x) d x,
(11.23)
∀ t ∈ [a, b] . If f (k) (b) = 0, k = 1, . . . , n, then f (t) − f (b) = − ∀ t ∈ [a, b] .
1 n!
t
b
(b − x)β−1 (x − t)n
b
αT (
f ) (x) d x,
(11.24)
11.2 Main Results
261
11.2 Main Results We present the following Conformable fractional Iyengar type inequalities: Theorem 11.10 Let α ∈ (n, n + 1] and f ∈ C n+1 ([a, b]), n ∈ N; β = α − n. Then (i) n b (k) 1 k+1 k (k) k+1 f (a) (z − a) f (t) dt − + (−1) f (b) (b − z) ≤ a (k + 1)! k=0
(β) max Taα ( f ) ∞,[a,b] , bα T ( f ) ∞,[a,b] (z − a)α+1 + (b − z)α+1 , (α + 2) (11.25) ∀ z ∈ [a, b] , , the right hand side of (11.25) is minimized, and we get: (ii) at z = a+b 2 n b 1 (b − a)k+1 (k) k (k) f (a) + (−1) f (b) ≤ f (t) dt − a 2k+1 (k + 1)! k=0
(β) max Taα ( f ) ∞,[a,b] , bα T ( f ) ∞,[a,b] (b − a)α+1 2α
(α + 2)
,
(11.26)
(iii) assuming f (k) (a) = f (k) (b) = 0, for k = 0, 1, . . . , n, we obtain
(β) max
Ta ( f )
, bα T ( f ) ∞,[a,b] (b − a)α+1 α ∞,[a,b] f (t) dt ≤ , (α + 2) 2α a (11.27) which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
b
n b 1 b − a k+1 (k) k+1 k (k) k+1 f (a) j f (t) dt − + (−1) f (b) (N − j) a N (k + 1)! k=0
(β) max Taα ( f ) ∞,[a,b] , bα T ( f ) ∞,[a,b] b − a α+1 ≤ j α+1 + (N − j)α+1 , (α + 2) N
(11.28)
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n, from (11.28) we get:
a
b
f (t) dt −
b−a N
[ j f (a) + (N − j) f (b)] ≤
262
11 Fractional Conformable Iyengar Inequalities
(β) max Taα ( f ) ∞,[a,b] , bα T ( f ) ∞,[a,b] b − a α+1 j α+1 + (N − j)α+1 , (α + 2) N
(11.29)
j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (11.29) turns to
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
(β) max Taα ( f ) ∞,[a,b] , bα T ( f ) ∞,[a,b] (b − a)α+1 2α
(α + 2)
.
(11.30)
Proof By Theorem 11.8 (11.18) we get n 1 t f (k) (a) k (t − x)n (x − a)β−1 Taα ( f ) (x) d x ≤ (t − a) ≤ f (t) − n! a k! k=0 (11.31)
a
t
T ( f )
α ∞,[a,b] (t − x)(n+1)−1 (x − a)β−1 d x = n! a
a
T ( f )
α ∞,[a,b] (n + 1) (β) (t − a)n+β . n! (n + β + 1) That is
a n
T ( f )
(β) f (k) (a) α ∞,[a,b] k (t − a) ≤ (t − a)n+β , (11.32) f (t) − k! + β + 1) (n k=0 ∀ t ∈ [a, b] . By Theorem 11.8 (11.19) we get n 1 b f (k) (b) k β−1 n b (b − x) (x − t) α T ( f ) (x) d x ≤ (t − b) ≤ f (t) − n! t k! k=0 1 n!
b t
b
T ( f )
α
n!
b
T ( f )
α
∞,[a,b]
n!
(b − x)β−1 (x − t)n bα T ( f ) (x) d x ≤ ∞,[a,b]
b
(b − x)β−1 (x − t)(n+1)−1 d x =
(11.33)
t
(β) (n + 1) (b − t)n+β = (n + β + 1)
b
T ( f )
α
∞,[a,b]
(β)
(n + β + 1)
(b − t)n+β .
11.2 Main Results
263
That is
b n
T ( f )
(β) f (k) (b) α ∞,[a,b] k (t − b) ≤ (b − t)n+β , (11.34) f (t) − k! + β + 1) (n k=0 ∀ t ∈ [a, b] . Call γ1 := and γ2 :=
a
T ( f )
(β) α ∞,[a,b] (n + β + 1)
b
T ( f )
α
∞,[a,b]
(β)
(n + β + 1)
,
(11.35)
.
(11.36)
Set γ := max (γ1 , γ2 ) . Therefore we have n f (k) (a) k (t − a) ≤ γ (t − a)n+β , f (t) − k! k=0 n f (k) (b) (t − b)k ≤ γ (b − t)n+β , f (t) − k! k=0
and
(11.37)
(11.38)
(11.39)
∀ t ∈ [a, b] . Hence it holds n n f (k) (a) f (k) (a) (t − a)k − γ (t − a)n+β ≤ f (t) ≤ (t − a)k + γ (t − a)n+β (11.40) k! k! k=0
k=0
and n n f (k) (b) f (k) (b) (t − b)k − γ (b − t)n+β ≤ f (t) ≤ (t − b)k + γ (b − t)n+β , k! k! k=0
k=0
(11.41) ∀ t ∈ [a, b] . Let any z ∈ [a, b], then by integration over [a, z] and [z, b], respectively, we obtain z n f (k) (a) γ f (t) dt ≤ (z − a)n+β+1 ≤ (z − a)k+1 − (k + 1)! (n + β + 1) a k=0
264
11 Fractional Conformable Iyengar Inequalities n f (k) (a) γ (z − a)k+1 + (z − a)n+β+1 , + 1)! + β + 1) (k (n k=0
(11.42)
and −
b n−1 f (k) (b) γ f (t) dt ≤ (b − z)n+β+1 ≤ (z − b)k+1 − (k + 1)! (n + β + 1) z k=0 −
n−1 f (k) (b) γ (b − z)n+β+1 . (z − b)k+1 + + 1)! + β + 1) (k (n k=0
(11.43)
Adding (11.42) and (11.43), we obtain n k=0
(k) 1 k+1 k+1 (k) f (a) (z − a) − − f (b) (z − b) (k + 1)!
γ (z − a)n+β+1 + (b − z)n+β+1 ≤ (n + β + 1) n k=0
b
f (t) dt ≤
a
(k) 1 f (a) (z − a)k+1 − f (k) (b) (z − b)k+1 + (k + 1)! γ (z − a)n+β+1 + (b − z)n+β+1 , (n + β + 1)
(11.44)
∀ z ∈ [a, b] . Consequently we derive: n b (k) 1 k+1 k (k) k+1 f (a) (z − a) f (t) dt − + (−1) f (b) (b − z) ≤ a (k + 1)! k=0
γ γ (z − a)n+β+1 + (b − z)n+β+1 = (z − a)α+1 + (b − z)α+1 , (n + β + 1) (α + 1) (11.45) ∀ z ∈ [a, b] . Let us consider g (z) = (z − a)α+1 + (b − z)α+1 , ∀ z ∈ [a, b] . Hence
g (z) = (α + 1) (z − a)α − (b − z)α = 0,
11.2 Main Results
265
giving (z − a)α = (b − z)α and z − a = b − z, that is z = a+b the only critical num2 ber here. (b−a)α+1 = 2α , which is the miniWe have g (a) = g (b) = (b − a)α+1 , and g a+b 2 mum of g over [a, b]. , with Consequently the right hand side of (11.45) is minimized when z = a+b 2 α+1
γ (b−a) . value (α+1) 2α Assuming f (k) (a) = f (k) (b) = 0, for k = 0, 1, . . . , n, then we obtain that
f (t) dt ≤
b
a
γ (b − a)α+1 , 2α (α + 1)
(11.46)
which is a sharp inequality. , then (11.45) becomes When z = a+b 2 n b 1 (b − a)k+1 (k) k (k) f (t) dt − f (a) + (−1) f (b) ≤ a 2k+1 (k + 1)! k=0
γ (b − a)α+1 . 2α (α + 1)
(11.47)
Next let N ∈ N, j = 0, 1, 2, . . . , N and z j = a + j b−a , that is z 0 = a, z 1 = N , . . . , z = b. a + b−a N N Hence it holds
b−a b−a , b − z j = (N − j) , j = 0, 1, 2, . . . , N . zj − a = j N N (11.48) We notice that
zj − a
α+1
α+1 + b − zj =
b−a N
α+1
j α+1 + (N − j)α+1 ,
(11.49)
j = 0, 1, 2, . . . , N , and (for k = 0, 1, . . . , n)
k+1 k+1 f (k) (a) z j − a = + (−1)k f (k) (b) b − z j
f (k) (a) j k+1
b−a N
b−a N
k+1
k+1
+ (−1)k f (k) (b) (N − j)k+1
b−a N
f (k) (a) j k+1 + (−1)k f (k) (b) (N − j)k+1 ,
k+1 =
(11.50)
266
11 Fractional Conformable Iyengar Inequalities
j = 0, 1, 2, . . . , N . By (11.45) we get
n b b − a k+1 (k) 1 k+1 k (k) k+1 f (a) j f (t) dt − + (−1) f (b) (N − j) a N (k + 1)! k=0
γ ≤ (α + 1)
b−a N
α+1
j α+1 + (N − j)α+1 ,
(11.51)
j = 0, 1, 2, . . . , N . If f (k) (a) = f (k) (b) = 0, k = 1, . . . , n, then (11.51) becomes
b
f (t) dt −
a
γ (α + 1)
b−a N
b−a N
[ j f (a) + (N − j) f (b)] ≤
α+1
j α+1 + (N − j)α+1 ,
(11.52)
j = 0, 1, 2, . . . , N . When N = 2 and j = 1, then (11.52) becomes
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
γ b − a α+1 γ (b − a)α+1 2 = . 2 2α (α + 1) (α + 1)
(11.53)
The theorem is proved in all cases. We give (n = 0 case) Corollary 11.11 Let 0 < α ≤ 1, f ∈ C 1 ([a, b]). Then (i) b ≤ f dt − f − a) + f − z)] [ (t) (a) (z (b) (b
(11.54)
a
max Tαa ( f ) ∞,[a,b] , bα T ( f ) ∞,[a,b] (z − a)α+1 + (b − z)α+1 , α (α + 1) ∀ z ∈ [a, b] ,
11.2 Main Results
(ii) at z =
267
a+b , 2
the right hand side of (11.54) is minimized, and we get:
b
f (t) dt −
a
b−a 2
[ f (a) + f (b)] ≤
max Tαa ( f ) ∞,[a,b] , bα T ( f ) ∞,[a,b] (b − a)α+1 2α
α (α + 1)
(11.55)
,
(iii) assuming f (a) = f (b) = 0, we obtain
a
b
b
max
a
α+1 T , T f f ( ) ( ) α α ∞,[a,b] ∞,[a,b] (b − a) f (t) dt ≤ , α (α + 1) 2α
(11.56)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
b
f (t) dt −
a
b−a N
[ f (a) j + f (b) (N − j)] ≤
max Tαa ( f ) ∞,[a,b] , bα T ( f ) ∞,[a,b] b − a α+1 α (α + 1)
N
j α+1 + (N − j)α+1 , (11.57)
(v) when N = 2 and j = 1, (11.57) turns to
a
b
f (x) d x −
b−a 2
( f (a) + f (b)) ≤
max Tαa ( f ) ∞,[a,b] , bα T ( f ) ∞,[a,b] (b − a)α+1 α (α + 1)
2α
.
(11.58)
Proof As in the proof of Theorem 11.10; case of n = 0, use of (11.12) and (11.17). We continue with L p conformable fractional Iyengar inequalities: Theorem 11.12 Let α ∈ (n, n + 1] and f ∈ C n+1 ([a, b]), n ∈ N; β = α − n. Let also p1 , p2 , p3 > 1 : p11 + p12 + p13 = 1, with β > p11 + p13 . Then (i) n b (k) 1 k+1 k (k) k+1 f (a) (z − a) f (t) dt − + (−1) f (b) (b − z) ≤ a (k + 1)! k=0 (11.59)
268
11 Fractional Conformable Iyengar Inequalities
max Taα ( f ) p3 ,[a,b] , bα T ( f ) p3 ,[a,b] 1 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 + ∀ z ∈ [a, b] , (ii) at z =
a+b , 2
(z − a)
α+ p1 + p1 1
2
+ (b − z)
α+ p1 + p1 1
2
1 p2
,
the right hand side of (11.59) is minimized, and we get:
n b 1 (b − a)k+1 (k) k (k) f f (t) dt − f + (−1) (a) (b) ≤ a 2k+1 (k + 1)! k=0
max Taα ( f ) p3 ,[a,b] , bα T ( f ) p3 ,[a,b] 1 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 +
1 p2
(b − a) 2
α+ p1 + p1 1
α− p1
2
,
(11.60)
3
(iii) assuming f (k) (a) = f (k) (b) = 0, for k = 0, 1, . . . , n, we obtain b f dt (t) ≤ a
max Taα ( f ) p ,[a,b] , bα T ( f )
α+ 1 + 1 3 (b − a) p1 p2 p3 ,[a,b] , 1 1 α− 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 + p12 2 p3
(11.61) which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds b
k+1 n b − a 1 k+1 k k+1 (k) (k) f (a) j f (t) dt − + (−1) f (b) (N − j) N (k + 1)! a k=0
max Taα ( f ) p3 ,[a,b] , bα T ( f ) p3 ,[a,b] ≤ 1 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 +
b−a N
α+ p1 + p1 1 2
j
α+ p1 + p1 1
2
+ (N − j)
1 p2
α+ p1 + p1 1
2
,
(11.62)
(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n, from (11.62) we get:
a
b
f (t) dt −
b−a N
[ j f (a) + (N − j) f (b)] ≤
(11.63)
11.2 Main Results
269
max Taα ( f ) p3 ,[a,b] , bα T ( f ) p3 ,[a,b] 1 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 +
b−a N
α+ p1 + p1 1 2
j
α+ p1 + p1 1
2
1 p2
α+ p1 + p1
+ (N − j)
1
,
2
for j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (11.63) turns to
b
f (x) d x −
a
b−a 2
( f (a) + f (b)) ≤
max Taα ( f ) p3 ,[a,b] , bα T ( f ) p3 ,[a,b] 1 1 n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2 α + p11 +
1 p2
(b − a) 2
α+ p1 + p1 1
α−
1 p3
2
.
(11.64)
Proof By Theorem 11.8 (11.18) we get n 1 t f (k) (a) k (t − x)n (x − a)β−1 Taα ( f ) (x) d x ≤ (t − a) ≤ f (t) − n! a k! k=0 1 n!
t
(t − x)
p1 n
p1 1
dx
a
t
(x − a)
p2 (β−1)
p1
t
2
dx
a
a
a T ( f ) (x) p3 d x α
a n+ p1 β−1+ p1
T ( f )
1 2 (t − a) α p3 ,[a,b] (t − a) ≤ 1 1 = n! ( p1 n + 1) p1 ( p2 (β − 1) + 1) p2
a
T ( f )
α p3 ,[a,b] n!
(t − a) ( p1 n + 1)
Notice that p2 (β − 1) + 1 > 0, iff β > We have proved that
1 p1
1 p1
+
n+β− p1
3 1
( p2 (β − 1) + 1) p2
p1
3
(11.65)
.
1 . p3
n f (k) (a) (t − a)k ≤ f (t) − k! k=0
a
T ( f )
α p3 ,[a,b] n! ( p1 n + 1)
∀ t ∈ [a, b] .
1 p1
( p2 (β − 1) + 1)
1 p2
(t − a)
n+β− p1
3
,
(11.66)
270
11 Fractional Conformable Iyengar Inequalities
Similarly, from Theorem 11.8 (11.19) we obtain n 1 b f (k) (b) k (b − x)β−1 (x − t)n bα T ( f ) (x) d x ≤ (t − b) ≤ f (t) − n! t k! k=0 1 n!
b
(x − t)
p1 p1 n
b
1
dx
t
(b − x)
p2 (β−1)
p1
2
dx
b
T ( f )
α
t
b
T ( f )
α
n! ( p1 n + 1)
1 p1
p3 ,[a,b]
( p2 (β − 1) + 1)
b
T ( f )
α
n! ( p1 n + 1) We have proved that
n! ( p1 n + 1) 1 , p3
1
β−1+ p1
(b − t)
2
n+β− p1
( p2 (β − 1) + 1)
1 p2
(b − t)
3
= (11.67)
=
.
n f (k) (b) (t − b)k ≤ f (t) − k! k=0 α
∀ t ∈ [a, b] . Since β > Call
(b − t)
p3 ,[a,b]
1 p1
b
T ( f )
n+ p1
1 p2
p3 ,[a,b]
1 p1
then β −
p3 ,[a,b]
( p2 (β − 1) + 1)
1 p2
(b − t)
n+β− p1
3
,
> 0, and m := α − p13 = n + β −
a
T ( f )
α p3 ,[a,b] ρ1 := 1 1 , p1 n! ( p1 n + 1) ( p2 (β − 1) + 1) p2
and ρ2 :=
1 p3
b
T ( f )
α
n! ( p1 n + 1)
1 p1
p3 ,[a,b] 1
( p2 (β − 1) + 1) p2
.
(11.68)
1 p3
> n ∈ N. (11.69)
(11.70)
Set
We have
and
ρ := max (ρ1 , ρ2 ) .
(11.71)
n f (k) (a) k (t − a) ≤ ρ (t − a)m , f (t) − k! k=0
(11.72)
11.2 Main Results
271
n f (k) (b) k (t − b) ≤ ρ (b − t)m , f (t) − k! k=0
(11.73)
∀ t ∈ [a, b] . As in the proof of Theorem 11.10 (11.45) we derive n b (k) 1 k+1 k (k) k+1 f (a) (z − a) f (t) dt − + (−1) f (b) (b − z) ≤ a (k + 1)! k=0
ρ (z − a)m+1 + (b − z)m+1 = (m + 1) 1 1 1 1 (z − a)α+ p1 + p2 + (b − z)α+ p1 + p2 ,
(11.74)
∀ z ∈ [a, b] . The rest of the proof is similar to the proof of Theorem 11.10.
α+
ρ +
1 p1
1 p2
We give (n = 0 case) Corollary 11.13 Let 0 < α ≤ 1 and f ∈ C 1 ([a, b]). Let p, q > 1 : with α > q1 . Then (i) b f (t) dt − [ f (a) (z − a) + f (b) (b − z)] ≤
1 p
+
1 q
= 1,
a
max Tαa ( f ) q,[a,b] , bα T ( f ) q,[a,b] α+ 1p α+ 1p − a) , + − z) (z (b 1 ( p (α − 1) + 1) p α + 1p ∀ z ∈ [a, b] , (ii) at z =
a+b , 2
(11.75)
the right hand side of (11.75) is minimized, and we get:
a
b
f (t) dt −
b−a 2
[ f (a) + f (b)] ≤
max Tαa ( f ) q,[a,b] , bα T ( f ) q,[a,b] (b − a)α+ 1p , 1 1 2α− q ( p (α − 1) + 1) p α + 1p (iii) assuming f (a) = f (b) = 0, we obtain
(11.76)
272
11 Fractional Conformable Iyengar Inequalities
a
b
b
max
T a ( f )
T ( f )
α+ 1p , α α q,[a,b] q,[a,b] (b − a) f (t) dt ≤ , 1 1 2α− q ( p (α − 1) + 1) p α + 1p
(11.77)
which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds
b
f (t) dt −
a
b−a N
[ j f (a) + (N − j) f (b)] ≤
max Tαa ( f ) q,[a,b] , bα T ( f ) q,[a,b] b − a α+ 1p α+ 1p α+ 1p j , + − j) (N 1 N ( p (α − 1) + 1) p α + 1p (11.78) (v) when N = 2 and j = 1, (11.78) turns to
a
b
f (x) d x −
b−a 2
( f (a) + f (b)) ≤
max Tαa ( f ) q,[a,b] , bα T ( f ) q,[a,b] (b − a)α+ 1p . 1 1 2α− q ( p (α − 1) + 1) p α + 1p
(11.79)
Proof As in the proof of Theorem 11.12, case of n = 0, use of (11.12) and (11.17).
References 1. T. Abdeljawad, On conformable fractional calculus. J. Comput. Appl. Math. 279, 57–66 (2015) 2. G.A. Anastassiou, Mixed conformable fractional approximation by sublinear operators. Indian J. Math. 60(1), 107–140 (2018) 3. G.A. Anastassiou, Conformable fractional Iyengar type Inequalities, submitted for publication (2018) 4. K.S.K. Iyengar, Note on an inequality. Math. Stud. 6, 75–76 (1938) 5. R. Khalil, M.Al. Horani, A. Yousef, M. Sababheh, A new definition of fractional derivative. J. Comput. Appl. Math. 264, 65–70 (2014)
Chapter 12
Iyengar Fuzzy Inequalities
Here we present fuzzy Iyengar type inequalities for continuous fuzzy number valued functions. These functions fulfill some type of Lipschitz conditions. See also [3].
12.1 Introduction We are motivated by the following famous Iyengar inequality (1938), [7]. Theorem 12.1 Let f be a differentiable function on [a, b] and f (x) ≤ M. Then
a
b
M (b − a) 2 ( f (b) − f (a))2 1 . f (x) d x − (b − a) ( f (a) + f (b)) ≤ − 2 4 4M (12.1)
We are inspired also by the important article [8]. For more on Fuzzy Analysis see [1, 2]. Here we derive fuzzy Iyengar type inequalities.
12.2 Background We need the following Definition 12.2 (See [10]) Let μ : R → [0, 1] with the following properties: (i) is normal, i.e. ∃ x0 ∈ R; μ (x0 ) = 1. (ii) μ (λx + (1 − λ) y) ≥ min {μ (x) , μ (y)}, ∀ x, y ∈ R, ∀ λ ∈ [0, 1] (μ is called a convex fuzzy subset). (iii) μ is upper semicontinuous on R, i.e. ∀ x0 ∈ R and ∀ ε > 0, ∃ neighborhood V (x0 ) : μ (x) ≤ μ (x0 ) + ε, ∀ x ∈ V (x0 ) . © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_12
273
274
12 Iyengar Fuzzy Inequalities
(iv) the set sup p (μ) is compact in R (where sup p (μ) := {x ∈ R; μ (x) > 0}). We call μ a fuzzy rel number. Denote the set of all μ with RF . E.g., χ{x0 } ∈ RF , for any x0 ∈ R, where χ{x0 } is the characteristic function at x0 . For 0 < r ≤ 1 and μ ∈ RF define [μ]r := {x ∈ R : μ (x) ≥ r } and [μ]0 := {x ∈ R : μ (x) > 0}. Then it is well known that for each r ∈ [0, 1], [μ]r is a closed and bounded interval of R ([6]). For u, v ∈ RF and λ ∈ R, we define uniquely the sum u ⊕ v and the product λ u by [u ⊕ v]r = [u]r + [v]r , [λ u]r = λ [u]r , ∀ r ∈ [0, 1] , where [u]r + [v]r means the ususal addition of two intervals (as subsets of R) and λ [u]r means the usual product between a scalar and a subset of R (see, e.g., [10]). Notice 1 u = u and it holds u ⊕ v = v ⊕ u, λ u = u λ. If 0 ≤ r1 ≤ r2 ≤ 1, ) (r ) (r ) (r ) (r ) (r ) then [u]r2 ⊆ [u]r1 . Actually [u]r = u (r − , u + , where u − ≤ u + , u − , u + ∈ R, ∀ ) (r ) r ∈ [0, 1]. For λ > 0 one has λu (r ± = (λ u)± , respectively. Define D : RF × RF → R+
by
) (r ) (r ) (r ) u , − v − v D (u, v) := sup max u (r − − + + r ∈[0,1]
= sup Hausdorff distance [u]r , [v]r , r ∈[0,1]
(r ) (r ) ; u, v ∈ RF . We have that D is a metric on RF . Then where [v]r = v− , v+ (RF , D) is a complete metric space, see [9, 10], with the properties D (u ⊕ w, v ⊕ w) = D (u, v) , ∀ u, v, w ∈ RF , D (k u, k v) = |k| D (u, v) , ∀ u, v ∈ RF , ∀ k ∈ R, D (u ⊕ v, w ⊕ e) ≤ D (u, w) + D (v, e) , ∀ u, v, w, e ∈ RF . Let f, g : R → RF be fuzzy number valued functions. The distance between f, g is defined by D ∗ ( f, g) := sup D ( f (x) , g (x)) . x∈R
On RF we define a partial order by “≤” (or “ ”): u, v ∈ RF , u ≤ v (or u v) ) (r ) (r ) (r ) iff u (r − ≤ v− and u + ≤ v+ , ∀ r ∈ [0, 1] .
12.2 Background
275
We mention Lemma 12.3 ([4] and Lemma 5.3 of [2]) For any a, b ∈ R : a · b ≥ 0 and any u ∈ RF we have D (a u, b u) ≤ |a − b| · D u,
0 , 0 := χ{0} . where
0 ∈ RF is defined by
Lemma 12.4 ([4] and Lemma 5.6 of [2]) (i) If we denote
0 := χ{0} , then
0 ∈ RF is
the neutral element with respect to ⊕, i.e., u ⊕ 0 = 0 ⊕ u = u, ∀ u ∈ RF . 0 has opposite in RF . (ii) With respect to
0, none of u ∈ RF , u =
(iii) Let a, b ∈ R : a · b ≥ 0, and any u ∈ RF , we have (a + b) u = a u ⊕ b u. For general a, b ∈ R, the above property is false. (iv) For any λ ∈ R and any u, v ∈ RF , we have λ (u ⊕ v) = λ u ⊕ λ v. (v) For any λ, μ ∈ R and u ∈ RF , we have λ (μ u) = (λ · μ) u. 0 , ∀ u ∈ RF , then ·F has the properties of a usual If we denote uF := D u,
norm on RF , i.e., uF = 0 iff u =
0, λ uF = |λ| · uF , u ⊕ vF ≤ uF + vF , uF − vF ≤ D (u, v) . Notice that (RF , ⊕, ) is not a linear space over R, and consequently (RF , ·F ) ∗ denotes the fuzzy summation. is not a normed space. Here Definition 12.5 Let a1 , a2 , b1 , b2 ∈ R such that a1 ≤ b1 and a2 ≤ b2 . Then we define (12.2) [a1 , b1 ] + [a2 , b2 ] = [a1 + a2 , b1 + b2 ] . Let a, b ∈ R such that a ≤ b and k ∈ R, then we define, if k ≥ 0, k [a, b] = [ka, kb] , if k < 0, k [a, b] = [kb, ka] .
(12.3)
We need also a particular case of the Fuzzy Henstock integral (δ (x) = 2δ ) introduced in [10], Definition 2.1 there. That is, Definition 12.6 ([5], p. 644) Let f : [a, b] → RF . We say that f is Fuzzy-Riemann integrable to I ∈ RF if for any ε > 0, there exists δ > 0 such that for any division P = {[u, v] ; ξ} of [a, b] with the norm (P) < δ, we have D
∗
P
(v − u) f (ξ) , I
< ε,
276
where
12 Iyengar Fuzzy Inequalities ∗
denotes the fuzzy summation. We choose to write I := (F R)
b
f (x) d x. a
We also call an f as above (F R)-integrable. We need Theorem 12.7 ([6]) If f, g : [c, d] → RF are (F R)-integrable fuzzy functions, and α, β are real numbers, then
d
d
(α f (x) ⊕ βg (x)) d x = α (F R)
(F R) c
d
f (x) d x ⊕ β (F R)
c
g (x) d x.
c
Denote by CF ([a, b]) := C ([a, b] , RF ) . Corollary 12.8 (Corollary 13.2 of [5]) If f ∈ CF ([a, b]) then f is (F R)-integrable on [a, b] . Lemma 12.9 ([1]) Let f : [a, b] → RF fuzzy continuous (with respect to metric D), then D f (x) ,
0 ≤ M, ∀ x ∈ [a, b], M > 0, that is f is fuzzy bounded. Equivalently we get χ−M ≤ f (x) ≤ χ M , ∀ x ∈ [a, b] . Lemma 12.10 ([1]) Let f : [a, b] ⊆ R → RF be fuzzy continuous. Then (F R)
x
f (t) dt is a fuzzy continuous function in x ∈ [a, b] .
a
We need Theorem 12.11 ([6]) Let f : [a, b] → RF be fuzzy continuous. Then (F R) b a f (x) d x exists and belongs to RF , furthermore it holds (F R)
r
b
f (x) d x
=
a
a
b
) ( f )(r − (x) d x,
b a
) d x , ∀ r ∈ [0, 1] . ( f )(r (x) +
Clearly f ±(r ) : [a, b] → R are continuous functions. I.e. (r ) b b f (x) d x = ( f )(r+) (x) d x, (F R) a
respectively.
+ (−)
a
(−)
(12.4)
(12.5)
12.3 Main Results
277
12.3 Main Results Here we present the following Fuzzy Iyengar type inequalities: Theorem 12.12 Let f ∈ CF ([a, b]), M > 0, such that
and
D ( f (x) , f (a)) ≤ M (x − a)α ,
(12.6)
D ( f (x) , f (b)) ≤ M (b − x)α ,
(12.7)
∀ x ∈ [a, b], 0 < α ≤ 1. Then (i) b
D (F R)
f (x) d x, (t − a) f (a) ⊕ (b − t) f (b) ≤
a
M (t − a)α+1 + (b − t)α+1 , (α + 1) ∀ t ∈ [a, b] ; (ii) at t =
a+b , 2
D (F R)
(12.8)
the right hand side of (12.8) is minimized and we have
[ f (a) ⊕ f (b)] ≤
M (b − a)α+1 , 2α (α + 1) a (12.9) (iii) when f (a) = f (b) =
0, (12.9) turns to the sharp inequality, b
f (x) d x,
D (F R)
b−a 2
b
f (x) d x,
0 ≤
a
M (b − a)α+1 , 2α (α + 1)
(12.10)
(iv) if α = 1, we get D (F R)
b
f (x) d x,
a
b−a 2
(b − a)2
[ f (a) ⊕ f (b)] ≤ M , 4
(12.11)
(v) when f (a) = f (b) =
0, (12.11) turns to the sharp inequality D (F R) a
b
(b − a)2 , f (x) d x,
0 ≤M 4
(12.12)
In general we derive that D (F R) a
b
f (x) d x,
b−a N
[ j f (a) ⊕ (N − j) f (b)] ≤
278
12 Iyengar Fuzzy Inequalities
M (α + 1)
b−a N
α+1
j α+1 + (N − j)α+1 ,
(12.13)
j = 0, 1, 2, . . . , N ∈ N. Proof Inequality (12.6) means that ) (r ) (r ) (r ) α f , sup max ( f (x))(r − f − f (x)) ( (a)) ( (a)) ( − − + + ≤ M (x − a) ,
r ∈[0,1]
(12.14) and (12.7) it means that ) (r ) (r ) (r ) α f , sup max ( f (x))(r − f − f (x)) ( (b)) ( (b)) ( − − + + ≤ M (b − x) ,
r ∈[0,1]
(12.15) ∀ x ∈ [a, b] . Hence by (12.14), ∀ r ∈ [0, 1], we get:
and
(r ) (r ) ( f (x))− − ( f (a))− ≤ M (x − a)α ,
(12.16)
(r ) (r ) ( f (x))+ − ( f (a))+ ≤ M (x − a)α ,
(12.17)
∀ x ∈ [a, b] . Similarly (12.15) implies, ∀ r ∈ [0, 1], that
and
(r ) (r ) ( f (x))− − ( f (b))− ≤ M (b − x)α ,
(12.18)
(r ) (r ) ( f (x))+ − ( f (b))+ ≤ M (b − x)α ,
(12.19)
∀ x ∈ [a, b] . By (12.16) and (12.17) we have (∀ r ∈ [0, 1]): ) (r ) (r ) α α ( f (a))(r − − M (x − a) ≤ ( f (x))− ≤ ( f (a))− + M (x − a) ,
(12.20)
) (r ) (r ) α α ( f (a))(r + − M (x − a) ≤ ( f (x))+ ≤ ( f (a))+ + M (x − a) ,
(12.21)
and
∀ x ∈ [a, b] . By (12.18) and (12.19) we have (∀ r ∈ [0, 1]): ) (r ) (r ) α α ( f (b))(r − − M (b − x) ≤ ( f (x))− ≤ ( f (b))− + M (b − x) ,
(12.22)
12.3 Main Results
279
and ) (r ) (r ) α α ( f (b))(r + − M (b − x) ≤ ( f (x))+ ≤ ( f (b))+ + M (b − x) ,
(12.23)
∀ x ∈ [a, b] . We pair (12.20) with (12.22), and then (12.21) with (12.23). For any t ∈ [a, b], we obtain that t M ) ) α+1 − a) ≤ − a) − (t (t ( f (x))(r ( f (a))(r − − dx (α + 1) a ) ≤ ( f (a))(r − (t − a) +
M (t − a)α+1 , (α + 1)
and ) ( f (b))(r − (b − t) −
M (b − t)α+1 ≤ (α + 1)
) ≤ ( f (b))(r − (b − t) +
t
b
(12.24)
) ( f (x))(r − dx
M (b − t)α+1 . (α + 1)
(12.25)
Consequently, we have by adding (12.24) and (12.25) that ) (r ) ( f (a))(r − (t − a) + ( f (b))− (b − t) − a
b
M (t − a)α+1 + (b − t)α+1 ≤ (α + 1)
) (r ) (r ) ( f (x))(r − d x ≤ ( f (a))− (t − a) + ( f (b))− (b − t) +
M (t − a)α+1 + (b − t)α+1 , (α + 1)
(12.26)
∀ t ∈ [a, b] . Therefore we derive b (r ) (r ) (r ) ( f (x))− d x − ( f (a))− (t − a) + ( f (b))− (b − t) ≤ a
M (t − a)α+1 + (b − t)α+1 , (α + 1) ∀ t ∈ [a, b] , ∀ r ∈ [0, 1] . Similarly, we derive:
(12.27)
280
12 Iyengar Fuzzy Inequalities
b
(f
a
) (x))(r +
(r ) (r ) d x − ( f (a))+ (t − a) + ( f (b))+ (b − t) ≤
M (t − a)α+1 + (b − t)α+1 , (α + 1)
(12.28)
∀ t ∈ [a, b], ∀ r ∈ [0, 1] . We can rewrite (12.27) into
b
a
) (r ) d x − − a)
f ⊕ − t)
f ((t (a) (b (b)) ( f (x))(r − − ≤ M (t − a)α+1 + (b − t)α+1 , (α + 1)
(12.29)
and we can rewrite (12.28) into
a
b
(f
) (x))(r +
d x − ((t − a) f (a) ⊕ (b − t) f
M (t − a)α+1 + (b − t)α+1 , (α + 1)
) (b))(r +
≤
(12.30)
∀ t ∈ [a, b], ∀ r ∈ [0, 1] . By (12.5) we find that (r ) b (r ) f (x) d x − ((t − a) f (a) ⊕ (b − t) f (b))− ≤ (F R) a − M (t − a)α+1 + (b − t)α+1 , (α + 1)
(12.31)
and (r ) b ) f (x) d x − ((t − a) f (a) ⊕ (b − t) f (b))(r (F R) + ≤ a + M (t − a)α+1 + (b − t)α+1 , (α + 1) ∀ t ∈ [a, b], ∀ r ∈ [0, 1] . Clearly, we have proved that
(12.32)
12.3 Main Results
281
b
f (x) d x, (t − a) f (a) ⊕ (b − t) f (b) ≤
D (F R) a
M (t − a)α+1 + (b − t)α+1 , (α + 1)
(12.33)
∀ t ∈ [a, b], 0 < α ≤ 1. Consider g (t) := (t − a)α+1 + (b − t)α+1 , t ∈ [a, b]. the only critical Then g (t) = (α + 1) (t − a)α − (b − t)α = 0, with t = a+b 2 number in [a, b]. Hence the right hand side of (12.33) is minimized at t = a+b 2 α+1
M (b−a) with value (α+1) . 2α Consequently, we derive at t =
a+b 2
that
b−a
[ f (a) ⊕ f (b)] ≤ 2 a M (b − a)α+1 , 0 < α ≤ 1, 2α (α + 1)
D (F R)
b
f (x) d x,
(12.34)
which is a sharp inequality when f (a) = f (b) =
0. If α = 1, then D (F R)
b
f (x) d x,
a
b−a 2
M (b − a)2
[ f (a) ⊕ f (b)] ≤ , 4
(12.35)
) which is also a sharp inequality when f (a) = f (b) =
0. Notice here that (f(a))(r +(−) = ) (f(b))(r +(−) = 0 and the related [8]. Next let N ∈ N, j = 0, 1, 2, . . . , N and t j = a + j b−a , that is t0 = a, t1 = N a + b−a , . . . , t = b. Hence it holds N N
tj − a = j
b−a N
b−a , b − t j = (N − j) , j = 0, 1, 2, . . . , N . N
We notice that α+1 α+1 tj − a + b − tj =
b−a N
α+1
j α+1 + (N − j)α+1 ,
(12.36)
j = 0, 1, 2, . . . , N ; and
t j − a f (a) ⊕ b − t j f (b) =
b−a N
[ j f (a) ⊕ (N − j) f (b)] ,
(12.37) j = 0, 1, 2, . . . , N .
282
12 Iyengar Fuzzy Inequalities
Therefore we have b b−a
[ j f (a) ⊕ (N − j) f (b)] ≤ f (x) d x, D (F R) N a M (α + 1)
b−a N
α+1
j α+1 + (N − j)α+1 ,
(12.38)
j = 0, 1, 2, . . . , N ; 0 < α ≤ 1. When N = 2 and j = 1, then (12.38) becomes
D (F R) a
b
b−a f (x) d x, 2
which is again (12.9). The theorem is proved.
[ f (a) ⊕ f (b)] ≤
M (b − a)α+1 , 2α (α + 1) (12.39)
References 1. G.A. Anastassiou, Rate of convergence of fuzzy neural network operators, univariate case. J. Fuzzy Math. 10(3), 755–780 (2002) 2. G.A. Anastassiou, Fuzzy Mathematics: Approximation Theory (Springer, Heidelberg, 2010) 3. G.A. Anastassiou, Fuzzy Iyengar Type Inequalities, Computational and Applied Mathematics (2019) 4. G.A. Anastassiou, S. Gal, On a fuzzy trigonometric approximation theorem of Weierstrass-type. J. Fuzzy Math. 9(3), 701–708 (2001), Los Angeles 5. S. Gal, Approximation theory in fuzzy setting, in Handbook of Analytic - Computational Methods in Applied Mathematics, ed. by G. Anastassiou (Chapman & Hall/CRC, Boca Raton, 2000), pp. 617–666 6. R. Goetschel Jr., W. Voxman, Elementary fuzzy calculus. Fuzzy Sets Syst. 18, 31–43 (1986) 7. K.S.K. Iyengar, Note on an inequality. Math. Student 6, 75–76 (1938) 8. Z. Liu, Note on Iyengar’s inequality. Univ. Beograd, Publ. Elektrotehn, FAK, Ser. Mat. 16, 29–35 (2005) 9. C. Wu, M. Ming, On embedding problem of fuzzy number space: part 1. Fuzzy Sets Syst. 44, 33–38 (1991) 10. C. Wu, Z. Gong, On Henstock integral of fuzzy number valued functions (I). Fuzzy Sets Syst. 120(3), 523–532 (2001)
Chapter 13
Choquet Integral Analytical Type Inequalities
Based on an amazing result of Sugeno [16], we are able to transfer classic analytic integral inequalities to Choquet integral setting. We give Choquet integral inequalities of the following types: fractional-Polya, Ostrowski, fractional Ostrowski, Hermite– Hadamard, Simpson and Iyengar. We provide several examples for the involved distorted Lebesgue measure. See also [5].
13.1 Background We need the following fractional calculus background: Let α > 0, m = [α] ([·] is the integral part), β = α − m, 0 < β < 1, f ∈ C [a, b] ⊂ R, x ∈ [a, b]. The gamma function is given by (α) = ([a, ∞ −tb]), α−1 e t dt. We define the left Riemann–Liouville integral 0
Jαa+ f (x) =
1 (α)
x
(x − t)α−1 f (t) dt,
(13.1)
a
α a ≤ x ≤ b. We define the subspace Ca+ ([a, b]) of C m ([a, b]): α Ca+ ([a, b]) =
a+ (m) f ∈ C 1 ([a, b]) . f ∈ C m ([a, b]) : J1−β
(13.2)
α For f ∈ Ca+ ([a, b]), we define the left generalized α-fractional derivative of f over [a, b] as
a+ (m) α f := J1−β f , Da+
(13.3)
see [1], p. 24. Canavati first in [6] introduced the above over [0, 1]. α f ∈ C ([a, b]). Notice that Da+ © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_13
283
284
13 Choquet Integral Analytical Type Inequalities
Furthermore we need: Let again α > 0, m = [α], β = α − m, f ∈ C ([a, b]), call the right Riemann– Liouville fractional integral operator by
1 f (x) := (α)
α Jb−
b
(t − x)α−1 f (t) dt,
(13.4)
x
x ∈ [a, b], see also [2, 10, 15]. Define the subspace of functions α Cb− ([a, b]) :=
1−β f ∈ C m ([a, b]) : Jb− f (m) ∈ C 1 ([a, b]) .
(13.5)
Define the right generalized α-fractional derivative of f over [a, b] as 1−β α Db− f = (−1)m−1 Jb− f (m) ,
(13.6)
0 α see [2]. We set Db− f = f . Notice that Db− f ∈ C ([a, b]) . We need the following fractional Polya type (see [13, 14], p. 62) integral inequality without any boundary conditions. a+b α Theorem 13.1 ([4], p. 4) Let 0 < α < 1, f ∈ C ([a, b]). Assume f ∈ Ca+ a, 2 a+b α , b . Set and f ∈ Cb− 2
α α a+b . f ∞,[a, a+b ] , Db− M ( f ) := max Da+ ∞,[ ,b] 2
Then
b
a
f (x) d x
≤
b
| f (x)| d x ≤ M ( f )
a
(13.7)
2
(b − a)α+1 . (α + 2) 2α
(13.8)
Inequality (13.8) is sharp, namely it is attained by f ∗ (x) =
, (x − a)α , x ∈ a, a+b 2 , 0 < α < 1. α a+b (b − x) , x ∈ 2 , b
(13.9)
The famous Ostrowski ([12]) inequality motivates this work and has as follows: Theorem 13.2 It holds
1
b − a
a
b
2
x − a+b 1 2 + f (y) dy − f (x)
≤ (b − a) f ∞ , 2 4 (b − a)
where f ∈ C 1 ([a, b]), x ∈ [a, b], and it is a sharp inequality. Another motivation is author’s next fractional result, see [3], p. 44:
(13.10)
13.1 Background
285
Theorem 13.3 Let [a, b] ⊂ R, α > 0, m = α (· ceiling of the number), f ∈ α m (m−1) AC ([a, b]) (i.e. f is absolutely continuous), and D x0 − f , ∞,[a,x0 ] α α α < ∞ (where D x0 − f, D ∗x0 f are the right ([2]) and left ([9], p. 50) D ∗x0 f ∞,[x0 ,b]
Caputo fractional derivatives of f of order α, respectively), x0 ∈ [a, b]. Assume f (k) (x0 ) = 0, k = 1, . . . , m − 1. Then
1
b − a
a
α D x0 − f
∞,[a,x0 ]
b
1 · f (x) d x − f (x0 )
≤ (b − a) (α + 2)
(x0 − a)
α+1
α + D ∗x0 f
∞,[x0 ,b]
(b − x0 )
α+1
≤
(13.11)
1 α α max D x0 − f , D ∗x0 f (b − a)α . ∞,[a,x0 ] ∞,[x0 ,b] (α + 2) In the next assume that (X, F) is a measurable space and (R+ ) R is the set of all (nonnegative) real numbers. We recall some concepts and some elementary results of capacity and the Choquet integral [7, 8]. Definition 13.4 A set function μ : F → R+ is called a non-additive measure (or capacity) if it satisfies (1) μ (∅) = 0; (2) μ (A) ≤ μ (B) for any A ⊆ B and A, B ∈ F. The non-additive measure μ is called concave if μ (A ∪ B) + μ ( A ∩ B) ≤ μ (A) + μ (B) ,
(13.12)
for all A, B ∈ F. In the literature the concave non-additive measure is known as submodular or 2-alternating non-additive measure. If the above inequality is reverse, μ is called convex. Similarly, convexity is called supermodularity or 2-monotonicity, too. First note that the Lebesgue measure λ for an interval [a, b] is defined by λ ([a, b]) = b − a, and that given a distortion function m, which is increasing (or nondecreasing) and such that m (0) = 0, the measure μ (A) = m (λ (A)) is a distorted Lebesgue measure. We denote a Lebesgue measure with distortion m by μ = μm . It is known that μm is concave (convex) if m is a concave (convex) function. function f : (X, F) → of all the nonnegative, measurable +The family + is the Borel σ-field of R+ . The R , B R+ is denoted as L + ∞ , where B R concept of the integral with respect to a non-additive measure was introduced by Choquet [7].
286
13 Choquet Integral Analytical Type Inequalities
Definition 13.5 Let f ∈ L + ∞ . The Choquet integral of f with respect to non-additive measure μ on A ∈ F is defined by
∞
f dμ :=
(C) A
μ ({x : f (x) ≥ t} ∩ A) dt,
(13.13)
0
where the integral on the right-hand side is a Riemann integral. Instead of (C) X f dμ, we shall write (C) f dμ. If (C) f dμ < ∞, we say that f is Choquet integrable and we write L C1
(μ) =
f dμ < ∞ .
f : (C)
The next lemma summarizes the basic properties of Choquet integrals [8]. 1 Lemma 13.6 Assume that f, g ∈ L C (μ). (1) (C) 1 A dμ = μ (A), A ∈ F. (2) (Positive homogeneity) For all λ ∈ R+ , we have (C) λ f dμ = λ · (C) f dμ. (3) (Translation invariance) For all c ∈ R, we have (C) ( f + c) dμ = (C) f dμ + c. (4) (Monotonicity in the integrand) If f ≤ g, then we have
(C)
f dμ ≤ (C)
gdμ.
(Monotonicity in the set function) If μ ≤ ν, then we have (C) (5) (Subadditivity) If μ is concave, then
f dμ ≤ (C)
f dμ + (C)
( f + g) dμ ≤ (C)
(C)
gdμ.
(Superadditivity) If μ is convex, then
( f + g) dμ ≥ (C)
(C)
f dμ + (C)
gdμ.
(6) (Comonotonic additivity) If f and g are comonotonic, then
( f + g) dμ = (C)
(C)
f dμ + (C)
gdμ,
where we say that f and g are comonotonic, if for any x, x ∈ X , then
f (x) − f x g (x) − g x ≥ 0.
f dν.
13.1 Background
287
We next mention the amazing result from [16], which permits us to compute the Choquet integral when the non-additive measure is a distorted Lebesgue measure. Theorem 13.7 Let f be a nonnegative and measurable function on R+ and μ = μm be a distorted Lebesgue measure. Assume that m (x) and f (x) are both continuous and m (x) is differentiable. When f is an increasing (non-decreasing) function on R+ , the Choquet integral of f with respect to μm on [0, t] is represented as
t
f dμm =
(C)
m (t − x) f (x) d x,
(13.14)
0
[0,t]
however, when f is a decreasing (non-increasing) function on R+ , the Choquet integral of f is t f dμm = m (x) f (x) d x. (13.15) (C) 0
[0,t]
13.2 Main Results From now on we assume that f : R+ → R+ is a monotone continuous function, and μ = μm i.e. μ (A) = m (λ (A)), denotes a distorted Lebesgue measure where m is such that m (0) = 0, m is increasing (non-decreasing) and continuously differentiable. By Theorem 13.7 and mean value theorem for integrals we get: (i) If f is an increasing (non-decreasing) function on R+ , we have (C)
f dμm
(13.14)
t
=
m (t − x) f (x) d x
0
[0,t]
= m (t − ξ)
t
f (x) d x, where ξ ∈ (0, t) .
(13.16)
0
(ii) If f is a decreasing (non-increasing) function on R+ , we have (C)
f dμm
(13.15)
=
0
[0,t]
t
m (x) f (x) d x = m (ξ)
t
f (x) d x,
(13.17)
0
where ξ ∈ (0, t) . We denote by γ (t, ξ) :=
m (t − ξ) , when f is increasing (non-decreasing) m (ξ) , when f is decreasing (non-increasing),
for some ξ ∈ (0, t) per case.
(13.18)
288
13 Choquet Integral Analytical Type Inequalities
We give the following Choquet-fractional-Polya inequality: Theorem 13.8 Let 0 < α < 1, f = f |[0,t] , t ∈ R+ , all considered t as above in this α α 0, 2t and f ∈ Ct− , t . Set section. Assume further that f ∈ C0+ 2 α α f ∞,[0, t ] , Dt− f ∞,[ t ,t ] . M ∗ ( f ) (t) := max D0+ 2
Then
(C)
f dμm ≤ γ (t, ξ) M ∗ ( f ) (t)
[0,t]
(13.19)
2
t α+1 . (α + 2) 2α
Proof By Theorem 13.1 and earlier comments.
(13.20)
Usual Polya inequality with ordinary derivative requires boundary conditions making a Choquet-Polya inequality impossible. We give applications: t 1 Remark 13.9 (i) If m (t) = 1+t , t ∈ R+ , then m (0) = 0, m (t) ≥ 0, m (t) = (1+t) 2 > 0, and m is increasing. Then γ (t, ξ) ≤ 1. (ii) If m (t) = 1 − e−t ≥ 0, t ∈ R+ , then m (0) = 0, m (t) = e−t > 0, and m is increasing. Then γ (t, ξ) ≤ 1. (iii) If m (t) = et − 1 ≥ 0, t ∈ R+ , m (0) = 0, m (t) = et > 0, and m is increasing. Then γ (t, ξ) ≤ et .
(iv) If m (t) = sin t, for t ∈ 0, π2 , we get m (0) = 0, m (t) = cos t ≥ 0, and m is increasing. Then γ (t, ξ) ≤ 1.
We continue with the Choquet–Ostrowski type inequalities: Theorem 13.10 Here f : R+ → R+ is a monotone continuous function, μm is a distorted Lebesgue measure, where m is such that m (0) = 0, m is increasing and is twice continuously differentiable on R+ . Here 0 ≤ x0 ≤ t ∈ R+ . Then (1)
1
(C) (13.21) f dμm − m (t − x0 ) f (x0 )
≤
t [0,t]
2 x0 − 2t 1 m + , − ·) f t (t ∞,[0,t] 4 t2
if f is an increasing function on R+ , and (2)
1
(C) f dμm − m (x0 ) f (x0 )
≤
t [0,t]
(13.22)
13.2 Main Results
289
2 x0 − 2t 1 m + f , t ∞,[0,t] 4 t2
if f is a decreasing function on R+ . Proof By (13.10) we have that (x0 ∈ [0, t])
1
(13.14)
(C) f dμm − m (t − x0 ) f (x0 )
=
t [0,t]
(13.23)
t
1
m (t − x) f (x) d x − m (t − x0 ) f (x0 )
≤
t 0
2 x0 − 2t 1 m + , t − ·) f (t ∞,[0,t] 4 t2
when f is an increasing function on R+ . Also we have that
(13.15)
1
(C) f dμm − m (x0 ) f (x0 )
=
t [0,t]
t
1
(13.10)
m f d x − m f (x) (x) (x ) (x ) 0 0 ≤
t 0
2 x0 − 2t 1 m + f , t ∞,[0,t] 4 t2
when f is a decreasing function on R+ .
(13.24)
We make Remark 13.11 (continuing from Remark 13.9) Assuming m is twice continuously differentiable is quite natural. Indeed: t (i) If m (t) = 1+t , t ∈ R+ , then m (t) = (1 + t)−2 , m (t) = −2 (1 + t)−3 , −4 m (3) (t) = 6 (1 + t) , m (4) (t) = −24 (1 + t)−5 , etc., all higher order derivatives exist and are continuous. (ii) If m (t) = 1 − e−t , t ∈ R+ , then m (t) = e−t , m (t) = −e−t , m (3) (t) = e−t , m (4) (t) = −e−t , etc., all higher order derivatives exist and are continuous. (iii) If m (t) = et − 1, t ∈ R+ , then m (i) (t) = et , i = 1, 2, . . . , all derivatives exist and are continuous.
(iv) If m (t) = sin t, t ∈ 0, π2 , then m (t) = cos t, m (t) = − sin t, m (3) (t) = − cos t, m (4) (t) = sin t, etc., all derivatives exist and are continuous.
290
13 Choquet Integral Analytical Type Inequalities
We continue with fractional Choquet–Ostrowski type inequalities. Theorem 13.12 Here f : R+ → R+ is an increasing continuous function, μm is a + α , ≤ t ∈ R . Let α > 0, m= m (t − ·) f distorted Lebesgue measure and 0 ≤ x 0 α α m , D ∗x0 m (t − ·) f
0, m = m f ∈ distorted Lebesgue measure and 0 ≤ x 0 (k) α α m , D ∗x0 m f 0, we have that
t
(C) m (t) f (0) + m (0) f (t)
≤ f (x) dμm (x) −
2 [0,t] 2 m (0) f (t) − m (t) f (0) M2 t 2 − . (13.45) 4 4M2
(ii) if f is decreasing and m f (x) ≤ M3 , ∀ x ∈ [0, t], M3 > 0, we have that
(C)
f (x) dμm (x) −
[0,t]
t m (0) f (0) + m (t) f (t)
≤ 2
2 m (t) f (t) − m (0) f (0) M3 t 2 − . (13.46) 4 4M3
Proof (i) If f is increasing and m (t − ·) f (x) ≤ M2 , ∀ x ∈ [0, t], then
(C)
[0,t]
f (x) dμm (x) −
(13.14) t m (t) f (0) + m (0) f (t)
= 2
(13.47)
13.2 Main Results
295
t
(13.44)
m (t − x) f (x) d x − t m (t) f (0) + m (0) f (t) ≤
2 0 2 m (0) f (t) − m (t) f (0) M2 t 2 . − 4 4M2
(ii) If f is decreasing and m f (x) ≤ M3 , ∀ x ∈ [0, t], then
(C)
f (x) dμm (x) − [0,t]
(13.15) t m (0) f (0) + m (t) f (t)
= 2
t
(13.44)
m (x) f (x) d x − t m (0) f (0) + m (t) f (t) ≤
2 0 2 m (t) f (t) − m (0) f (0) M3 t 2 . − 4 4M3 Note 13.20 One can transfer many analytic integral classic inequalities to Choquet integral setting but we choose to stop here.
References 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.
G.A. Anastassiou, Fractional Differentiation Inequalities (Springer, Heidelberg, 2009) G.A. Anastassiou, On right fractional calculus. Chaos, Solitons Fractals 42, 365–376 (2009) G.A. Anastassiou, Advances on Fractional Inequalities (Springer, Heidelberg, 2009) G.A. Anastassiou, Intelligent Comparisons: Analytic Inequalities (Springer, Heidelberg, 2016) G.A. Anastassiou, Choquet Integral Analytic Inequalities, Studia Mathematica Babes Bolyai (2018) J.A. Canavati, The Riemann-Liouville integral. Nieuw Archief Voor Wiskunde 5(1), 53–75 (1987) G. Choquet, Theory of capacities. Ann. Inst. Fourier (Grenoble) 5, 131–295 (1953) D. Denneberg, Non-additive Measure and Integral (Kluwer Academic Publishers, Boston, 1994) K. Diethelm, The Analysis of Fractional Differentiation Equations (Springer, Heidelberg, 2010) G.S. Frederico, D.F.M. Torres, Fractional optimal control in the sense of caputo and the fractional Noether’s theorem. Int. Math. Forum 3(10), 479–493 (2008) K.S.K. Iyengar, Note on an inequality. Math. Student 6, 75–76 (1938) A. Ostrowski, Über die Absolutabweichung einer differentiebaren Funktion von ihrem Integralmittelwert. Comment. Math. Helv. 10, 226–227 (1938) G. Polya, Ein mittelwertsatz für Funktionen mehrerer Veränderlichen. Tohoku Math. J. 19, 1–3 (1921) G. Polya, G. Szegö, Aufgaben und Lehrsätze aus der Analysis, vol. I (Springer, Berlin, 1925). (German)
296
13 Choquet Integral Analytical Type Inequalities
15. S.G. Samko, A.A. Kilbas, O.I. Marichev, Fractional Integrals and Derivatives, Theory and Applications (Gordon and Breach, Amsterdam, 1993) [English translation from the Russian, Integrals and Derivatives of Fractional Order and Some of Their Applications (Nauka i Tekhnika, Minsk, 1987)] 16. M. Sugeno, A note on derivatives of functions with respect to fuzzy measures. Fuzzy Sets Syst. 222, 1–17 (2013)
Chapter 14
Local Fractional Taylor Formula
Here we derive an appropriate local fractional Taylor formula. We provide a complete description of the formula. See also [3].
14.1 Introduction In [4, 5] was first introduced the local fractional derivative and presented an incomplete local fractional Taylor formula, all done by the use of Riemann–Liouville fractional derivative. Similar work was done in [1], but again with some gaps. The author is greatly motivated by the pioneering work of [1–5] and presents a local fractional Taylor formula in a complete suitable form and without any gaps.
14.2 Main Results We mention Definition 14.1 ([6], pp. 68, 89) Let x, x ∈ [a, b], f ∈ C ([a, b]). The Riemann– Lioville fractional derivative of a function f of order q (0 < q < 1) is defined as Dxq f x = 1 (1 − q)
q Dx+ f x , x > x, = q Dx− f x , x < x
x −q d x −t f (t) dt, x > x, d x x x d −q f (t) dt, x < x. − dx x t − x
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_14
(14.1)
297
298
14 Local Fractional Taylor Formula
We need Definition 14.2 ([4]) The local fractional derivative of order q (0 < q < 1) of a function f ∈ C ([a, b]) is defined as Dxq f x − f (x) . Dq f (x) = lim x →x
(14.2)
More generally we define Definition 14.3 (see also [1]) Let N ∈ Z+ , 0 < q < 1, the local fractional derivative of order (N + q) of a function f ∈ C N ([a, b]) is defined by D
N +q
f (x) = lim
x →x
Dxq
N n f (n) (x) x −x f x − . n! n=0
(14.3)
If N = 0, then Definition 14.3 collapses to Definition 14.2. We need Definition 14.4 (Related to Definition 14.3) Let f ∈ C N ([a, b]), N ∈ Z+ . Set
F x, x − x; q, N :=
Dxq
N n f (n) (x) x −x f x − . n! n=0
(14.4)
Let x − x := t, then x = x + t, and F (x, t; q, N ) =
Dxq
N f (n) (x) n t . f (x + t) − n! n=0
(14.5)
We make Remark 14.5 Here x , x ∈ [a, b], and a ≤ x + t ≤ b, equivalently a−x ≤ t ≤ b−x. From a ≤ x ≤ b, we get a − x ≤ 0 ≤ b − x. We assume here that F (x, ·; q, N ) ∈ C 1 ([a − x, b − x]). Clearly, then it holds D N +q f (x) = F (x, 0; q, N ) , and D N +q f (x) exists in R. We make Remark 14.6 We observe that: (I) Let x > x (x − x > 0) then N n ([2]) f (n) (x) x −x f x − n! n=0
(14.6)
14.2 Main Results
299
Dx−q
Dx−q
Dxq
N n f (n) (x) x −x f x − n! n=0
F x, x − x; q, N = 1 (q)
1 (q) x −x
0
x
x − z
q−1
=
F (x, z − x; q, N ) dz =
x
F (x, t; q, N ) dt = (x − x − t)−q+1
(integration by parts) x −x
q−1 1 F (x, t; q, N ) x −x −t dt + (q) 0 1 (q) Thus,
x −x 0
(14.7)
q d F (x, t; q, N ) x − x − t dt. dt q
N n q f (n) (x) D N +q f (x) f x − x −x = x −x + n! (q + 1) n=0
1 (q + 1)
x −x 0
q d F (x, t; q, N ) x − x − t dt, for x > x, dt
(14.8)
N ∈ Z+ . (II) Let x < x (x − x < 0): We have similarly, N n ([2]) f (n) (x) f x − x −x n! n=0
Dx−q
Dxq
N n f (n) (x) x −x f x − n! n=0
Dx−q F x, x − x; q, N = 1 (q) (integration by parts)
0 x −x
1 (q)
x x
x − x + t
z − x
q−1
q−1
=
F (x, z − x; q, N ) dz = (14.9)
F (x, t; q, N ) dt =
300
14 Local Fractional Taylor Formula
0
1 q−1 F (x, t; q, N ) t+x−x dt − (q) x −x 1 (q)
0 x −x
q d F (x, t; q, N ) t + x − x dt = dt q
(14.10)
q q
x −x x − x 1 d F (x, t; q, N ) t + x − x 1 dt = F (x, 0; q, N ) + (q) q (q) 0 dt q
x −x q q 1 d F (x, t; q, N ) 1 D N +q f (x) x − x + t − x + x dt. (q + 1) (q + 1) 0 dt
(14.11)
Conclusion: We have proved that (N ∈ Z+ ) (I) N n D N +q f (x) q f (n) (x) x −x + x −x + f x = n! (q + 1) n=0 1 (q + 1) and (II)
x −x
0
q d F (x, t; q, N ) x − x − t dt, when x > x, dt
(14.12)
q N n D N +q f (x) x − x f (n) (x) f x = x −x + + n! (q + 1) n=0
1 (q + 1)
x −x 0
q d F (x, t; q, N ) t − x + x dt, when x < x. dt
(14.13)
We have derived Theorem 14.7 Let f ∈ C N ([a, b]), N ∈ Z+ . Here x, x ∈ [a, b], and F (x, ·; q, N ) ∈ C 1 ([a − x, b − x]). Then N n D N +q f (x) f (n) (x) x − x q + x −x + f x = n! (q + 1) n=0
1 (q + 1)
0
x −x
q d F (x, t; q, N ) x − x − t dt. dt
(14.14)
14.2 Main Results
301
In particular we get Corollary 14.8 (To Theorem 14.7, N = 0) Let f ∈ C ([a, b]); x, x ∈ [a, b], and F (x, ·; q, 0) ∈ C 1 ([a − x, b − x]). Then q Dq f (x) x − x + f x = f (x) + (q + 1) 1 (q + 1)
0
x −x
(14.15)
q d F (x, t; q, 0) x − x − t dt. dt
References 1. F.B. Adda, J. Cresson, About non-differential functions. J. Math. Anal. Appl. 263, 721–737 (2001) 2. F.B. Adda, J. Cresso, Fractional differential equations and the schrödinger equation. Appl. Math. Comput. 161, 323–345 (2005) 3. G.A. Anastassiou, Local fractional taylor formula. J. Comput. Anal. Appl. 28(4), 709–713 (2020) 4. K.M. Kolwankar, Local fractional calculus: a review. arXiv: 1307:0739v1 [nlin.CD] 2 Jul 2013 5. K.M. Kolwankar, A.D. Gangal, Local fractional calculus: a calculus for fractal space-time, Fractals: Theory and Applications in Engineering (Springer, London, 1999), pp. 171–181 6. I. Podlubny, Fractional Differential Equations (Academic, San Diego, 1999)
Chapter 15
Negative Domain Local Fractional Inequalities
This research is about inequalities in a local fractional environment over a negative domain. The author presents the following types of analytic local fractional inequalities: Opial, Hilbert–Pachpatte, comparison of means, Poincare and Sobolev. The results are with respect to uniform and L p norms, involving left and right Riemann– Liouville fractional derivatives. See also [10].
15.1 Introduction Many sources motivate us to write this work. The first one comes next. It is the famous Opial inequality ([14])
a 0
y (x) y (x) d x ≤ a 2
a
y (x)2 d x,
(15.1)
0
where y (x) is absolutely continuous function and y (0) = 0. The above inequality is proved sharp. The well known Ostrowski ([15]) inequality also motivates this work and has as follows: 2 b 1 x − a+b 1 2 + f (y) dy − f (x) ≤ (b − a) f ∞ , (15.2) b − a 2 4 (b − a) a where f ∈ C 1 ([a, b]), x ∈ [a, b], and it is a sharp inequality. ρ Next D∗a f indicates the left Caputo fractional derivative of order ρ > 0, anchored at a ∈ R, see [11], p. 50. The author in [7], pp. 82–83, proved the following left Caputo fractional Landau inequality: Let 0 < ν ≤ 1, f ∈ AC 2 ([0, b]) (i.e. f ∈ AC ([0, b]), absolutely ν+1 f ∈ L ∞ (R+ ), and continuous functions), ∀ b > 0. Suppose f ∞,R+ < +∞, D∗0 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_15
303
304
15 Negative Domain Local Fractional Inequalities
ν+1 D f ∗a
∞,[a,+∞)
ν+1 ≤ D∗0 f ∞,R+ , ∀ a ≥ 0.
(15.3)
Then ν ν+1 1 ν+1 ν ν+1 2 1 − ν+1 D ν+1 f f ≤ + 1) , + 2)) (ν ( (ν ∞,R ∗0 + ∞,R+ ∞,R+ ν (15.4) that is f ∞,R+ is finite. The last inequality is another inspiration. The author’s monographs [2–6, 8], motivate and support largely this work too. See also [1]. Under the point of view of local fractional differentiation the author examines the broad area of analytic inequalities and produces a variety of well-known inequalities in a local fractional setting over a negative domain to all possible directions.
f
15.2 Background We mention Definition 15.1 ([12]) Let x, x ∈ [a, b], f ∈ C ([a, b]). The Riemann–Liouville (R–L) fractional derivative of a function f of order q (0 < q < 1) is defined as Dxq f x = 1 (1 − q)
q Dx+ f x , x > x, = q Dx− f x , x < x
x −q d x −t f (t) dt, d x x −q x f (t) dt, − ddx x t − x
x > x, x < x,
(15.5)
the left and right R–L fractional derivatives, respectively. We need Definition 15.2 ([12, 13]) The local fractional derivative of order q (0 < q < 1) of a function f ∈ C ([a, b]) is defined as D q f (x) = lim Dxq f x − f (x) . x →x
(15.6)
More generally we define Definition 15.3 ([9]) Let N ∈ Z+ , 0 < q < 1, the local fractional derivative of order (N + q) of a function f ∈ C N ([a, b]) is defined by D N +q f (x) = lim Dxq x →x
N n f (n) (x) x −x f x − . n! n=0
(15.7)
15.2 Background
305
If N = 0, then Definition 15.3 collapses to Definition 15.2. We need Definition 15.4 (Related to Definition 15.3) Let f ∈ C N ([a, b]), N ∈ Z+ . Set
F x, x − x; q, N :=
Dxq
N n f (n) (x) x −x f x − . n! n=0
(15.8)
Let x − x := t, then x = x + t, and F (x, t; q, N ) =
Dxq
N f (n) (x) n f (x + t) − t . n! n=0
(15.9)
We make Remark 15.5 Here x , x ∈ [a, b], and a ≤ x + t ≤ b, equivalently a − x ≤ t≤b−x. From a ≤ x ≤ b, we get a − x ≤ 0 ≤ b − x. We assume here that F (x, ·; q, N ) ∈ C 1 ([a − x, b − x]). Clearly, then it holds D N +q f (x) = F (x, 0; q, N ) ,
(15.10)
and D N +q f (x) exists in R. We would need: Theorem 15.6 ([9]) Let f ∈ C N ([a, b]), N ∈ Z+ . Here x, x ∈ [a, b], and F (x, ·; q, N ) ∈ C 1 ([a − x, b − x]). Then N q n f (n) (x) D N +q f (x) x − x + x −x + f x = n! (q + 1) n=0
1 (q + 1)
x −x 0
(15.11)
q d F (x, t; q, N ) x − x − t dt. dt
Corollary 15.7 (To Theorem 15.6, N = 0) Let f ∈ C ([a, b]), x, x ∈ [a, b], and F (x, ·; q, 0) ∈ C 1 ([a − x, b − x]). Then q D q f (x) x − x + f x = f (x) + (q + 1) 1 (q + 1)
x −x 0
q d F (x, t; q, 0) x − x − t dt. dt
(15.12)
306
15 Negative Domain Local Fractional Inequalities
We make Remark 15.8 Let f ∈ C N ([a, b]), N ∈ Z+ . Here x, x ∈ [a, b] : x < x, and F (x, ·; q, N ) ∈ C 1 ([a − x, b − x]), 0 < q < 1. By Theorem 15.6 we get N q n f (n) (x) D N +q f (x) x −x + x − x − f x = n! (q + 1) n=0
1 (q + 1)
0 x −x
q d F (x, t; q, N ) t − x + x dt. dt
(15.13)
Clearly then we get: Let f ∈ C N ([a, 0]), a < 0, N ∈ Z+ , F (0, ·; q, N ) ∈ C 1 ([a, 0]), 0 < q < 1. Then, for any x ∈ [a, 0], we derive f (x) =
N f (n) (0) n D N +q f (0) x + (−x)q − n! + 1) (q n=0
1 (q + 1)
0 x
d F (0, t; q, N ) (t − x)q dt. dt
(15.14)
In this chapter we will use a lot (15.14). Remark 15.9 Let f ∈ C N ([a, 0]), N ∈ Z+ , a < 0, x ∈ [a, 0]; F (0, ·; q, N ) ∈ C 1 ([a, 0]), 0 < q < 1. Then, by (15.14), we have N f (n) (0) n D N +q f (0) x + f (x) = (−x)q n! + 1) (q n=0
1 − (q + 1)
0 x
(15.15)
d F (0, t; q, N ) (t − x)q dt. dt
Assume that f (n) (0) = 0, n = 0, 1, . . . , N , and D N +q f (0) = 0 (= F (0, 0; q, N ) q = D0 f (0)). Then 0 d F (0, t; q, N ) 1 (15.16) − f (x) = (t − x)q dt, (q + 1) x dt ∀ x ∈ [a, 0] . Here it is
q
F (0, t; q, N ) = D0 ( f (t)) ∈ C 1 ([a, 0]) , q
where D0 is the right Riemann–Liouville fractional derivative.
15.2 Background
307
Let a ≤ x ≤ w ≤ 0, then 1 − f (w) = (q + 1) Consider p1 , q1 > 1 :
+
1 p1
1 q1
w
0 w
(15.17)
= 1. Then
1 | f (w)| = (q + 1) 1 (q + 1)
d F (0, t; q, N ) (t − w)q dt. dt
0
0 w
d F (0, t; q, N ) (t − w)q dt ≤ dt
q1 0
p1 1 1 d F (0, t; q, N ) q1 q p1 dt = (t − w) dt dt w
q p1 +1
1 (−w) p1 (q + 1) (qp1 + 1) p11
0
w
q1 1 d F (0, t; q, N ) q1 dt = dt q p1 +1
1 1 (−w) p1 q1 , 1 (z (w)) (q + 1) (qp1 + 1) p1
where
z (w) :=
0
w
all a ≤ x ≤ w ≤ 0, and z (0) = 0. From −z (w) = 0
we get
d F (0, t; q, N ) q1 dt, dt
w
d F (0, w; q, N ) 1 = −z (w) q1 . dw d F (0, w; q, N ) ≤ | f (w)| dw
Therefore we obtain
1 (q + 1) (qp1 + 1)
1 p1
(15.19)
d F (0, t; q, N ) q1 dt, dt
d F (0, w; q, N ) q1 , − z (w) = (−z (w)) = dw
and
(15.18)
(−w)
q p1 +1 p1
1 1 (z (w)) q1 −z (w) q1 .
(15.20)
(15.21)
(15.22)
308
15 Negative Domain Local Fractional Inequalities
Hence it holds
0 x
d F (0, w; q, N ) dw ≤ | f (w)| dw
1 (q + 1) (qp1 + 1)
1 (q + 1) (qp1 + 1)
1 p1
1 p1
0
0
(−w)
q p1 +1 p1
1 z (w) −z (w) q1 dw ≤
x
q p1 +1
(−w)
p1
0
1
dw
x
z (w) −z (w) dw
q1
1
=
x
1 (q + 1) (qp1 + 1)
(15.23)
1 p1
(−x)q p1 +2 qp1 + 2
p11
z 2 (w) 0 | − 2 x
q1
1
=
(15.24)
2
1 1
1
(q + 1) (qp1 + 1) p1 (qp1 + 2) p1
(−x)
q p1 +2 p1
(z (x)) q1 1
2 q1
.
We have proved that
0 x
d F (0, w; q, N ) dw ≤ | f (w)| dw
q+ p2
(−x)
1
1
0
1
2 q1 (q + 1) [(qp1 + 1) (qp1 + 2)] p1
x
q2 1 d F (0, w; q, N ) q1 dw . (15.25) dw
We have established the following negative domain L p -Opial type local right fractional inequality: Theorem 15.10 Let p1 , q1 > 1 : p11 + q11 = 1; f ∈ C N ([a, 0]), N ∈ Z+ , a < 0, x ∈ [a, 0]; F (0, ·; q, N ) ∈ C 1 ([a, 0]), 0 < q < 1. Assume that f (n) (0) = 0, q n = 0, 1, . . . , N , and D N +q f (0) = 0 (= F (0, 0; q, N ) = D0 f (0)). [Here it is q q F (0, t; q, N ) = D0 ( f (t)) ∈ C 1 ([a, 0]), where D0 is the right Riemann–Liouville fractional derivative]. Then
0 x
d F (0, t; q, N ) dt ≤ | f (t)| dt
q+ p2
(−x) 1
1 1
2 q1 (q + 1) [(qp1 + 1) (qp1 + 2)] p1 ⇔ it holds
0 x
q2 1 d F (0, t; q, N ) q1 dt , (15.26) dt
15.2 Background
309
0 x
d D0q ( f (t)) dt ≤ | f (t)| dt
q+ p2
(−x)
1
1
0
1
2 q1 (q + 1) [(qp1 + 1) (qp1 + 2)] p1
x
q21 d D0q ( f (t)) q1 dt , dt
(15.27)
∀ x ∈ [a, 0] . The case p1 = q1 = 2 follows: Corollary 15.11 All as in Theorem 15.10, with p1 = q1 = 2. Then
0 x
d F (0, t; q, N ) dt ≤ | f (t)| dt
(−x)q+1 √ 2 (q + 1) (q + 1) (2q + 1)
0 x
(15.28)
d F (0, t; q, N ) 2 dt , dt
⇔ it holds
0 x
d D q ( f (t)) dt ≤ | f (t)| 0 dt
(−x)q+1 √ 2 (q + 1) (q + 1) (2q + 1)
0 x
q
d D0 ( f (t)) dt
(15.29)
2
dt ,
∀ x ∈ [a, 0] . We make Remark 15.12 Let f 1 , f 2 according to the assumptions of Theorem 15.10. Then − f 1 (x1 ) =
1 (q + 1)
0 x1
d F1 (0, t1 ; q, N ) (t1 − x1 )q dt1 , dt1
(15.30)
d F2 (0, t2 ; q, N ) (t2 − x2 )q dt2 , dt2
(15.31)
∀ x1 ∈ [a1 , 0] , a1 < 0; − f 2 (x2 ) =
1 (q + 1)
0 x2
∀ x2 ∈ [a2 , 0] , a2 < 0. Here it is q
Fi (0, ti ; q, N ) = D0 ( f i (ti )) ∈ C 1 ([ai , 0]) , i = 1, 2;
310
15 Negative Domain Local Fractional Inequalities q
where D0 is the right Riemann–Liouville fractional derivative. Consider p1 , q1 > 1 : p11 + q11 = 1. Hence 0 d Fi (0, ti ; q, N ) 1 (ti − xi )q dti , | f i (xi )| ≤ (q + 1) xi dti
(15.32)
i = 1, 2; ∀ xi ∈ [ai , 0] . We get by Hölder’s inequality: | f 1 (x1 )| ≤ 1 (q + 1)
0
(t1 − x1 )
q p1
p1 1
0
dt1
x1
x1
q1 1 d F1 (0, t1 ; q, N ) q1 dt1 ≤ (15.33) dt 1
q p1 +1 d F1 (0, t1 ; q, N ) 1 (−x1 ) p1 , (q + 1) (qp1 + 1) p11 dt1 q1 ,[a1 ,0]
∀ x1 ∈ [a1 , 0] . Similarly, we obtain qq1 +1 d F2 (0, t2 ; q, N ) 1 (−x2 ) q1 | f 2 (x2 )| ≤ , (q + 1) (qq1 + 1) q11 dt2 p1 ,[a2 ,0]
(15.34)
∀ x2 ∈ [a2 , 0] . Therefore we have 1
| f 1 (x1 )| | f 2 (x2 )| ≤
(−x1 )
q p1 +1 p1
(−x2 )
qq1 +1 q1
1
q1 ,[a1 ,0]
1
1
1
(using Young’s inequality for a, b ≥ 0, a p1 b q1 ≤
1 1
1
a p1
2
+
p1 ,[a2 ,0]
≤
b ) q1
(−x2 )qq1 +1 (−x)q p1 +1 + p1 q1
( (q + 1))2 (qp1 + 1) p1 (qq1 + 1) q1 d F1 (0, t1 ; q, N ) d F2 (0, t2 ; q, N ) , dt1 dt2 q1 ,[a1 ,0] p1 ,[a2 ,0] ∀ xi ∈ [ai , 0] , i = 1, 2.
(15.35)
1
( (q + 1)) (qp1 + 1) p1 (qq1 + 1) q1 d F1 (0, t1 ; q, N ) d F2 (0, t2 ; q, N ) dt dt 2
(15.36)
15.2 Background
311
So far we have established
| f 1 (x1 )| | f 2 (x2 )| (−x)q p1 +1 p1
+
(−x2 )qq1 +1 q1
≤
1
d F1 (0, t1 ; q, N ) dt
q1 ,[a1 ,0]
1
1
1
( (q + 1)) (qp1 + 1) p1 (qq1 + 1) q1 2
d F2 (0, t2 ; q, N ) dt 2
p1 ,[a2 ,0]
(15.37)
,
∀ xi ∈ [ai , 0] , i = 1, 2. The denominator of left hand side of (15.37) can be zero only when x1 = 0 and x2 = 0. By integrating (15.37) over [a1 , 0] × [a2 , 0] we get
0
a1
0
a2
| f 1 (x1 )| | f 2 (x2 )| d x1 d x2 a1 a2 ≤ 1 1 (−x)q p1 +1 (−x2 )qq1 +1 2 ( (q + 1)) (qp1 + 1) p1 (qq1 + 1) q1 + p1 q1 (15.38) d F1 (0, t1 ; q, N ) d F2 (0, t2 ; q, N ) . dt1 dt2 q1 ,[a1 ,0] p1 ,[a2 ,0]
We have proved the following negative domains local right fractional Hilbert– Pachpatte inequality: Theorem 15.13 Let p1 , q1 > 1 :
1 p1 1
+
1 q1
= 1; i = 1, 2 for
f i ∈ C N ([ai , 0]),
N ∈ Z+ , ai < 0; Fi (0, ·; q, N ) ∈ C ([ai , 0]), 0 < q < 1. Assume that f i(n) (0) = 0, q n = 0, 1, . . . , N , and D N +q f i (0) = 0 , i = 1, 2 (i.e. Fi (0, 0; q, N ) =D0 f i (0) =0). q q 1 [Here it is Fi (0, ti ; q, N ) = D0 ( f i (ti )) ∈ C ([ai , 0]), where D0 is the right Riemann–Liouville fractional derivative]. Then
0
a1
0
a2
| f 1 (x1 )| | f 2 (x2 )| d x1 d x2 a1 a2 ≤ 1 1 (−x)q p1 +1 (−x2 )qq1 +1 2 ( (q + 1)) (qp1 + 1) p1 (qq1 + 1) q1 + p1 q1 (15.39) d F1 (0, t1 ; q, N ) d F2 (0, t2 ; q, N ) , dt1 dt2 q1 ,[a1 ,0] p1 ,[a2 ,0]
⇔ it holds 0 0 a1
a2
| f 1 (x1 )| | f 2 (x2 )| d x1 d x2 a1 a2 ≤ 1 1 (−x)q p1 +1 (−x2 )qq1 +1 2 ( (q + 1)) (qp1 + 1) p1 (qq1 + 1) q1 + p1 q1 (15.40) d D0q ( f 1 (t1 )) d D0q ( f 2 (t2 )) . dt dt 1
We make
q1 ,[a1 ,0]
2
p1 ,[a2 ,0]
312
15 Negative Domain Local Fractional Inequalities
Remark 15.14 Let f ∈ C N ([a, 0]), a < 0, N ∈ Z+ , F (0, ·; q, N ) ∈ C 1 ([a, 0]), 0 < q < 1. Then for any x ∈ [a, 0], we have N f (n) (0) n D N +q f (0) x + f (x) = (−x)q n! + 1) (q n=0
−
1 (q + 1)
(15.41)
d F (0, t; q, N ) (t − x)q dt. dt
0 x
Assume that f (n) (0) = 0, n = 0, 1, . . . , N . Here D N +q f (0) = F (0, 0; q, N ) = q q D0 f (0), where D0 is the right Riemann–Liouville fractional derivative. So far we have D N +q f (0) (15.42) f (x) = (−x)q + R (x) , (q + 1) where 1 R (x) := − (q + 1)
0 x
d F (0, t; q, N ) (t − x)q dt. dt
(15.43)
q
We also assume that D0 f ∈ C 1 ([a, 0]). We can rewrite
0 d q 1 D f (t) (t − x)q dt. R (x) = − (q + 1) x dt 0
(15.44)
We notice that 1 |R (x)| ≤ (q + 1)
0 x
d q D f (t) (t − x)q dt ≤ dt 0
d q 1 (−x)q+1 D f (t) . 0 (q + 1) dt ∞,[a,0] q + 1 That is |R (x)| ≤
(−x)q+1 d D q f (t) , 0 (q + 2) dt ∞,[a,0]
(15.45)
∀ x ∈ [a, 0] . Hence, it holds a
0
f (x) d x =
D N +q f (0) (q + 1)
a
0
(−x)q d x + a
0
R (x) d x =
(15.46)
15.2 Background
313
D N +q f (0) (−a)q+1 + (q + 1) q + 1
0
a
D N +q f (0) R (x) d x = (−a)q+1 + (q + 2)
0
R (x) d x.
a
Therefore, we get
0
f (x) d x −
a
D N +q f (0) (−a)q+1 = (q + 2)
0
R (x) d x.
(15.47)
a
Consequently, we derive q 0 0 D0 f (0) |R (x)| d x ≤ f (x) d x − (−a)q+1 ≤ a (q + 2) a d q D f (t) dt 0 ∞,[a,0] (q + 2)
0
(−x)q+1 d x =
d q D f (t) q+2 dt 0 ∞,[a,0] (−a)
a
=
(15.48)
(q + 2)
d q D f (t) (−a)q+2 dt 0 ∞,[a,0] (q + 3)
q +2
.
(15.49)
We have proved the following negative domain local right fractional comparison of means results: q
Theorem 15.15 Let f ∈ C N ([a, 0]), a < 0, N ∈ Z+ , D0 f ∈ C 1 ([a, 0]), 0 0. By (15.54) we have | f (x)| ≤
(−x)
q p1 +1 p1 1
(q + 1) (qp1 + 1) p1
d F (0, t; q, N ) , dt q1 ,[a,0]
(15.59)
∀ x ∈ [a, 0] . Hence it holds r q+ p1
1 q r | f (x)|r ≤ D . f ( ) r 0 1 q1 ,[a,0] (q + 1) (qp1 + 1) p1 (−x)
(15.60)
Consequently, we get a
0
r q+ p1 +1
1 (−a) | f (x)|r d x ≤ r 1 (q + 1) (qp1 + 1) p1 r q+
1 p1
r q . D0 ( f ) q1 ,[a,0] +1 (15.61)
We have proved the following negative domain local right fractional Sobolev type inequality: Theorem 15.19 All as in Theorem 15.10, plus r > 0. Then f r,[a,0] ≤
q+ p1 + r1
(−a)
1
1
(q + 1) (qp1 + 1) p1
r q+
1 p1
q r1 D0 ( f ) q1 ,[a,0] . +1 (15.62)
References 1. F.B. Adda, J. Cresson, Fractional differentiation equations and the Schrödinger equation. Appl. Math. Comput. 161, 323–345 (2005) 2. G.A. Anastassiou, Quantitative Approximations (CRC Press, Boca Raton, 2001) 3. G.A. Anastassiou, Fractional Differentiation Inequalities (Springer, Heidelberg, 2009) 4. G.A. Anastassiou, Probabilistic Inequalities (World Scientific, Singapore, 2010) 5. G.A. Anastassiou, Advanced Inequalities (World Scientific, Singapore, 2010) 6. G.A. Anastassiou, Intelligent Mathematics: Computational Analysis (Springer, Heidelberg, 2011) 7. G.A. Anastassiou, Advances on Fractional Inequalities (Springer, Heidelberg, 2011) 8. G.A. Anastassiou, Intelligent Comparisons: Analytic Inequalities (Springer, Heidelberg, 2016) 9. G.A. Anastassiou, Local fractional taylor formula. J. Comput. Anal. Appl. 28(4), 709–713 (2020) 10. G.A. Anastassiou, Negative domain local fractional inequalities. J. Comput. Anal. Appl. 28(5), 879–891 (2020)
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15 Negative Domain Local Fractional Inequalities
11. K. Diethelm, The Analysis of Fractional Differential Equations (Springer, Heidelberg, 2010) 12. K.M. Kolwankar, Local fractional calculus: a review. arXiv: 1307:0739v1 [nlin.CD] 2 Jul 2013 13. K.M. Kolwankar, A.D. Gangal, Local fractional calculus: a calculus for fractal space-time, Fractals: Theory and Applications in Engineering (Springer, London, 1999), pp. 171–181 14. Z. Opial, Sur une inégalité. Ann. Polon. Math. 8, 29–32 (1960) 15. A. Ostrowski, Über die Absolutabweichung einer differentiebaren Funktion von ihrem Integralmittelwert. Comment. Math. Helv. 10, 226–227 (1938)
Chapter 16
Fractional Approximation by Riemann–Liouville Fractional Derivatives
In this chapter we study quantitatively with rates the pointwise convergence of a sequence of positive sublinear operators to the unit operator over continuous functions. This takes place under low order smothness, less than one, of the approximated function and it is expressed via the left and right Riemann–Liouville fractional derivatives of it. The derived related inequalities in their right hand sides contain the moduli of continuity of these fractional derivatives and they are of Shisha-Mond type. We give applications to Bernstein Max-product operators and to positive sublinear comonotonic operators connecting them to Choquet integral. See also [6].
16.1 Introduction In this chapter among others we are motivated by the following results: First by Korovkin [10], (1960), p. 14: Let [a, b] be a closed interval in R and (L n )n∈N be a sequence of positive linear operators mapping C ([a, b]) into itself. Suppose that (L n f ) converges uniformly to f for the three test functions f = 1, x, x 2 . Then (L n f ) converges uniformly to f on [a, b] for all functions f ∈ C ([a, b]). Let f ∈ C ([a, b]) and 0 ≤ h ≤ b − a. The first modulus of continuity of f at h is given by ω1 ( f, h) = sup | f (x) − f (y)| . x,y∈[a,b] |x−y|≤h
If h > b − a, then we define ω1 ( f, h) = ω1 ( f, b − a). Another motivation is the following: By Shisha and Mond [13], (1968): Let [a, b] ⊂ R a closed interval. Let {L n }n∈N be a sequence of positive linear operators acting on C ([a, b]) into itself. For n = 1, . . . , suppose L n (1) is bounded. Let f ∈ C ([a, b]). Then for n = 1, 2, . . ., we have L n f − f ∞ ≤ f ∞ L n 1 − 1∞ + L n 1 + 1∞ ω1 ( f, μn ) , © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_16
317
318
16 Fractional Approximation by Riemann–Liouville Fractional Derivatives
where
21 μn = L n (t − x)2 (x)∞
and ·∞ stands for the sup-norm over [a, b] . One can easily see, for n = 1, 2, . . . μ2n ≤ L n t 2 ; x − x 2 ∞ + 2c L n (t; x) − x∞ + c2 L n (1; x) − 1∞ , where c = max (|a| , |b|). Thus, given the Korovkin assumptions, as n → ∞, we get μn → 0, and L n f − f ∞ → 0 for any f ∈ C ([a, b]). That is one derives the Korovkin conclusion in a quantitative way and with rates of convergence. We continue this type as research here for positive sublinear operators over continuous functions with existing left and right Riemann–Liouville fractional derivatives of order less than one. We give applications. Other motivations come from author’s monographs [2–4].
16.2 Main Results We mention Definition 16.1 ([11, pp. 68, 89]) Let x, x ∈ [a, b], f ∈ C ([a, b]). The Riemann– Liouville (R–L) fractional derivative of a function f of order q (0 < q < 1) is defined as q Dx+ f x , x > x, = Dxq f x = q Dx− f x , x < x 1 (1 − q)
x −q d x −t f (t) dt, x > x, d x x x d −q f (t) dt, x < x, − dx x t − x
(16.1)
the left and right R–L fractional derivatives, respectively, where is the gamma function. We need Lemma 16.2 ([1, 11], pp. 71, 75) Let x, x ∈ [a, b], f ∈ C ([a, b]), 00. We set (16.5) ω1 Dxq ( f (·) − f (x)) , δ :=
q q max ω1 Dx+ ( f (·) − f (x)) , δ [x,b] , ω1 Dx− ( f (·) − f (x)) , δ [a,x] . We give Theorem 16.5 Here f ∈ C ([a, b]), 0 < q < 1, δ > 0; x, x ∈ [a, b]. Assume that q q Dx+ ( f (·) − f (x)) ∈ C ([x, b]), and Dx− ( f (·) − f (x)) ∈ C ([a, x]), where x is fixed. Then x − x q+1 ω D q ( f (·) − f (x)) , δ x − x q + f x − f (x) ≤ 1 x , (q + 1) (q + 1) δ (16.6) ∀ x ∈ [a, b] . q
Proof Obviously Dx+ ( f (x) − f (x)) = 0. We estimate: (i) Case of x < x ≤ b : f x − f (x) ≤ 1 (q)
x
x − z
1 (q)
x
x − z
q−1 q Dx+ ( f (z) − f (x)) dz =
x
(δ1 >0) q−1 q q Dx+ ( f (z) − f (x)) − Dx+ ( f (x) − f (x)) dz ≤
x
1 (q)
x x
x − z
q−1
ω1
δ1 (z − x) q Dx+ ( f (·) − f (x)) , δ1
(16.7)
dz ≤ [x,b]
320
16 Fractional Approximation by Riemann–Liouville Fractional Derivatives
q 1 ω1 Dx+ ( f (·) − f (x)) , δ1 [x,b] (q)
x
x −z
q−1
x
z−x 1+ dz = δ1
q
q x ω1 Dx+ ( f (·) − f (x)) , δ1 q−1 x − x 1 [x,b] 2−1 + dz = x −z (z − x) (q) q δ1 x
q
q ω1 Dx+ ( f (·) − f (x)) , δ1 [x,b] q+1 x − x 1 (q) (2) + x −x = (q) q δ1 (q + 2) q (16.8)
q ω1 Dx+ ( f (·) − f (x)) , δ1 [x,b] x − x 1 (q) q+1 x − x = + (q) q δ1 (q + 2) q
q q+1 ω1 Dx+ ( f (·) − f (x)) , δ1 [x,b] x − x 1 x − x + = (q) q δ1 q (q + 1) q
ω1 Dx+ ( f (·) − f (x)) , δ1 [x,b] (q + 1)
x −x
q
(16.9)
q+1 1 x − x + . δ1 (q + 1)
When x < x ≤ b, we have proved that f x − f (x) ≤ q
ω1 Dx+ ( f (·) − f (x)) , δ1 [x,b] (q + 1)
x − x
q
q+1 x − x + , (q + 1) δ1
(16.10)
where 0 < q < 1, δ1 > 0. q (ii) Case of a ≤ x < x (here Dx− ( f (x) − f (x)) = 0): f x − f (x) ≤ 1 (q)
x
x
1 (q)
z − x x
x
1 (q)
x x
z − x
q−1 q Dx− ( f (z) − f (x)) dz =
(δ2 >0) q−1 q q Dx− ( f (z) − f (x)) − Dx− ( f (x) − f (x)) dz ≤
z−x
q−1
ω1
q Dx−
δ2 (x − z) ( f (·) − f (x)) , δ2
q ω1 Dx− ( f (·) − f (x)) , δ2 [a,x] (q)
x x
z − x
q−1
(16.11) dz ≤
[a,x]
x−z 1+ dz = δ2
16.2 Main Results
321
q
q x ω1 Dx− ( f (·) − f (x)) , δ2 [a,x] q−1 x − x 1 + dz = (x − z)2−1 z − x (q) q δ2 x
q
q ω1 Dx− ( f (·) − f (x)) , δ2 [a,x] x − x 1 (2) (q) q+1 + x−x = (q) q δ2 (q + 2) (16.12) q
q q+1 ω1 Dx− ( f (·) − f (x)) , δ2 [a,x] x − x x − x + = (q) q q (q + 1) δ2 q
ω1 Dx− ( f (·) − f (x)) , δ2 [a,x] (q + 1)
x−x
q
q+1 x − x + . (q + 1) δ2
When a ≤ x < x, we have proved that f x − f (x) ≤ q
ω1 Dx− ( f (·) − f (x)) , δ2 [a,x] (q + 1)
x−x
q
q+1 x − x + , (q + 1) δ2
where 0 < q < 1, δ2 > 0. Finally choose: δ1 = δ2 =: δ > 0. The theorem is proved.
(16.13)
We need Definition 16.6 Here C+ ([a, b]) := { f : [a, b] → R+ , continuous functions}. Let L N : C+ ([a, b]) → C+ ([a, b]), operators, ∀ N ∈ N, such that (i) L N (α f ) = αL N ( f ) , ∀α ≥ 0, ∀ f ∈ C+ ([a, b]) , (ii) if f, g ∈ C+ ([a, b]) : f ≤ g, then L N ( f ) ≤ L N (g) , ∀N ∈ N, (iii) L N ( f + g) ≤ L N ( f ) + L N (g) , ∀ f, g ∈ C+ ([a, b]) . We call {L N } N ∈N positive sublinear operators. We make Remark 16.7 Let f, g ∈ C+ ([a, b]), then it holds |L N ( f ) (x) − L N (g) (x)| ≤ L N (| f − g|) (x) , ∀ x ∈ [a, b] .
(16.14)
322
16 Fractional Approximation by Riemann–Liouville Fractional Derivatives
Furthermore, we also have |L N ( f ) (x) − f (x)| ≤ L N (| f (·) − f (x)|) (x) + | f (x)| |L N (e0 ) (x) − 1| , (16.15) ∀ x ∈ [a, b] ; e0 (t) = 1. From now on we assume that L N (1) = 1. Hence |L N ( f ) (x) − f (x)| ≤ L N (| f (·) − f (x)|) (x) , ∀x ∈ [a, b] .
(16.16)
We give q
Theorem 16.8 Let f ∈ C+ ([a, b]), 0 < q < 1, Dx+ ( f (·) − f (x)) ∈ C ([x, b]), q Dx− ( f (·) − f (x)) ∈ C ([a, x]), x is fixed, where x ∈ [a, b]. Then q ω1 Dx ( f (·) − f (x)) , δ |· − x|q+1 q |· − x| + | f (·) − f (x)| ≤ , δ > 0. (q + 1) (q + 1) δ (16.17) We present: q
q
Theorem 16.9 Let f ∈ C+ ([a, b]), Dx+ ( f (·) − f (x)) ∈ C ([x, b]), Dx− ( f (·) − f (x)) ∈ C ([a, x]); where x ∈ [a, b] is fixed, 0 < q < 1, δ > 0. Let L N : C+ ([a, b]) → C+ ([a, b]), be positive sublinear operators, such that L N (1) = 1, ∀ N ∈ N. Then |L N ( f ) (x) − f (x)| ≤ (16.18) q ω1 Dx ( f (·) − f (x)) , δ L N |· − x|q+1 (x) q L N |· − x| (x) + , (q + 1) (q + 1) δ ∀ N ∈ N. We need Hölder’s inequality for positive sublinear operators: Lemma 16.10 ([5], p. 6) Let L : C+ ([a, b]) → C+ ([a, b]), be a positive sublinear operator and f, g ∈ C+ ([a, b]), furthermore let p, q > 1 : 1p + q1 = 1. Assume that L ( f (·)) p (s∗ ), L (g (·))q (s∗ ) > 0 for some s∗ ∈ [a, b]. Then 1 1 L ( f (·) f (·)) (s∗ ) ≤ L ( f (·)) p (s∗ ) p L (g (·))q (s∗ ) q .
(16.19)
We make Remark 16.11 In Theorem 16.9 we assumed L N (1) = 1, ∀ N ∈ N. We further assume that L N |· − x|q+1 (x) > 0, ∀ N ∈ N, for the fixed x ∈ [a, b] . Then, by (16.19), we obtain q L N |· − x|q (x) ≤ L N |· − x|q+1 (x) q+1 , ∀N ∈ N.
(16.20)
16.2 Main Results
323
We give Theorem 16.12 All as in Theorem 16.9, plus L N |· − x|q+1 (x) > 0, ∀ N ∈ N, for a fixed x ∈ [a, b] . Then q ω1 Dx ( f (·) − f (x)) , δ |L N ( f ) (x) − f (x)| ≤ · (q + 1)
⎡ ⎤ 1 q+1 q+1 q |· − x| L (x) N ⎦, L N |· − x|q+1 (x) q+1 ⎣1 + (q + 1) δ
(16.21)
∀ N ∈ N. 1 Next we choose δ := L N |· − x|q+1 (x) q+1 > 0, to obtain: Theorem 16.13 All as in Theorem 16.9, plus L N |· − x|q+1 (x) > 0, ∀ N ∈ N; x ∈ [a, b] is fixed. Then |L N ( f ) (x) − f (x)| ≤
(q + 2) · (q + 2)
1 q ω1 Dxq ( f (·) − f (x)) , L N |· − x|q+1 (x) q+1 L N |· − x|q+1 (x) q+1 , (16.22) ∀ N ∈ N. Application 16.14 The max-product Bernstein operators are defined by B N(M)
( f ) (x) :=
N p N .k (x) f ∨k=0
k N
N ∨k=0 p N .k (x)
where ∨ stands for maximum, and p N ,k (x) =
N k
, ∀ N ∈ N,
(16.23)
x k (1 − x) N −k , and f : [0, 1] →
R+ is a continuous function, see [7], p. 10. These are positive sublinear operators mapping C+ ([0, 1]) into itself. Notice B N(M) (1) = 1, ∀ N ∈ N. In [5], p. 76, we proved that 6 , ∀ x ∈ [0, 1] , B N(M) |· − x|1+β (x) ≤ √ N +1 ∀ N ∈ N, ∀ β > 0.
(16.24)
324
16 Fractional Approximation by Riemann–Liouville Fractional Derivatives
Furthermore, clearly it holds that B N(M) |· − x|1+β (x) > 0, ∀ N ∈ N, ∀ β ≥ 0,
(16.25)
and any x ∈ (0, 1) . We present q
q
Theorem 16.15 Let f ∈ C+ ([0, 1]), Dx+ ( f (·) − f (x)) ∈ C ([x, 1]), Dx− ( f (·) − f (x)) ∈ C ([0, x]); where x ∈ (0, 1), 0 < q < 1. Then (M) B N ( f ) (x) − f (x) ≤ q 1 q+1 q+1 6 6 (q + 2) q ω1 Dx ( f (·) − f (x)) , √ , √ (q + 2) N +1 N +1
(16.26)
∀ N ∈ N. As N → +∞, we get B N(M) ( f ) (x) → f (x) . Proof By (16.23), (16.24), (16.25) and Theorem 16.13.
One can give many examples like in Theorem 16.15, but we choose to omit it this task. Choquet integral has become very important in statistical mechanics, potential theory, non-additive measure theory, and lately in economics. For the definition and properties of Choquet integral read [8, 9, 14]. We denote it by (C) . Next we talk about representations of positive sublinear operators by Choquet integrals: We need Definition 16.16 Let be a set, and let f, g : → R be bounded functions. We say that f and g are comonotonic, if for every ω, ω ∈ ,
f (ω) − f ω g (ω) − g ω ≥ 0.
(16.27)
We also need the famous Schmeidler’s Representation Theorem (Schmeidler 1986). Theorem 16.17 ([12]) Denote with L∞ (A) the vector space of A-measurable bounded real valued functions on , where A ⊂ 2 is a σ-algebra. Given a real functional : L∞ (A) → R, assume that for f, g ∈ L∞ (A): (i) (c f ) = c ( f ), ∀ c > 0, (ii) f ≤ g, implies ( f ) ≤ (g), and
16.2 Main Results
325
(iii) ( f + g) = ( f ) + (g), for any comonotonic f, g. Then γ (A) := (1 A ), ∀ A ∈ A, defines a finite monotone set function on A, and is the Choquet integral with respect to γ, i.e. ( f ) = (C)
f (t) dγ (t) , ∀ f ∈ L∞ (A) .
(16.28)
Above 1 A denotes the characteristic function on A. We make Remark 16.18 Consider here [a, b] ⊂ R, B = B ([a, b]) is the Borel σ-algebra on [a, b], and L∞ (B) is the vector space of B-measurable bounded real valued functions on [a, b]. Let (L N ) N ∈N be a sequence of positive sublinear operators from L∞ (B) into C+ ([a, b]), and x ∈ [a, b]. That is here L N fulfills the positive homogenuity, monotonicity and subadditivity properties, see Definition 16.6. Assume L N (1) = 1, ∀ N ∈ N. Clearly here L∞ (B) ⊃ C+ ([a, b]). In particular we treat L N |C+ ([a,b]) , just denoted for simplicity by L N , ∀ N ∈ N. It is clear that L N (·) (x) : L∞ (B) → R is a functional, ∀ N ∈ N. It has the properties: (i) (16.29) L N (c f ) (x) = cL N ( f ) (x) , ∀ c > 0, ∀ f ∈ L∞ (B) , (ii) f ≤ g, implies L N ( f ) (x) ≤ L N (g) (x) , where f, g ∈ L∞ (B) ,
(16.30)
and (iii) L N ( f + g) (x) ≤ L N ( f ) (x) + L N (g) (x) , ∀ f, g ∈ L∞ (B) .
(16.31)
For comonotonic f, g ∈ L∞ (B), we further assume that L N ( f + g) (x) = L N ( f ) (x) + L N (g) (x) .
(16.32)
In that case L N is called comonotonic. By Theorem 16.17 we get that: γ N ,x (A) := L N (1 A ) (x) , ∀ A ∈ B, ∀ N ∈ N,
(16.33)
defines a finite monotone set function on B, and L N ( f ) (x) = (C) a
b
f (t) dγ N ,x (t) ,
(16.34)
326
16 Fractional Approximation by Riemann–Liouville Fractional Derivatives
∀ f ∈ L∞ (B), ∀ N ∈ N. In particular (16.34) is valid for any f ∈ C+ ([a, b]). Furthermore γ N ,x is normalized, that is γ N ,x ([a, b]) = 1, ∀ N ∈ N. We give q
q
Theorem 16.19 Let f ∈ C+ ([a, b]), Dx+ ( f (·) − f (x)) ∈C ([x, b]), Dx− ( f (·) − f (x)) ∈ C ([a, x]); where x ∈ [a, b] ⊂ R is fixed, 0 < q < 1. Let L N : L∞ (B ([a, b])) → C+ ([a, b]), be positive sublinear comonotonic operators, such that b L N (1) = 1, ∀ N ∈ N. Assume that (C) a |t − x|q+1 dγ N ,x (t) > 0, ∀ N ∈ N. Then |L N ( f ) (x) − f (x)| ≤ ω1
Dxq
( f (·) − f (x)) , (C)
b
(q + 2) · (q + 2)
|t − x|
q+1
1 q+1 dγ N ,x (t) ·
a
(C)
b
|t − x|
q+1
q q+1 dγ N ,x (t) ,
(16.35)
a
∀ N ∈ N. b If (C) a |t − x|q+1 dγ N ,x (t) → 0, then L N ( f ) (x) → f (x), as N → ∞. Proof By Theorem 16.13.
References 1. F.B. Adda, J. Cresson, Fractional differentiation equations and the Schrödinger equation. Appl. Math. Comput. 161, 323–345 (2005) 2. G.A. Anastassiou, Moments in Probability and Approximation Theory, vol. 287, Pitman Research Notes in Mathematics (Longman Scientific & Technical, Harlow, 1993) 3. G.A. Anastassiou, Quantitative Approximations (CRC Press, Boca Raton, 2001) 4. G.A. Anastassiou, Intelligent Mathematics: Computational Analysis (Springer, Heidelberg, 2011) 5. G.A. Anastassiou, Nonlinearity: Ordinary and Fractional Approximations by Sublinear and Max-Product Operators (Springer, Heidelberg, 2018) 6. G.A. Anastassiou, Approximation with Riemann-Liouville Fractional Derivatives, Studia Mathematica Babes Bolyai (2019) 7. B. Bede, L. Coroianu, S. Gal, Approximation by Max-Product Type Operators (Springer, Heidelberg, 2016) 8. G. Choquet, Theory of capacities. Ann. Inst. Fourier (Grenoble) 5, 131–295 (1954) 9. D. Denneberg, Non-additive Measure and Integral (Kluwer, Dordrecht, 1994) 10. P.P. Korovkin, Linear Operators and Approximation Theory (Hindustan Publishing Corporation, Delhi, 1960) 11. I. Podlubny, Fractional Differentiation Equations (Academic, San Diego, 1999)
References
327
12. D. Schmeidler, Integral representation without additivity. Proc. Am. Math. Soc. 97, 255–261 (1986) 13. O. Shisha, B. Mond, The degree of convergence of sequences of linear positive operators. Nat. Acad. of Sci. U.S. 60, 1196–1200 (1968) 14. Z. Wang, G.J. Klir, Generalized Measure Theory (Springer, New York, 2009)
Chapter 17
Riemann–Liouville Fractional Fundamental Theorem of Calculus and Riemann–Liouville Fractional Polya Integral Inequality and the Generalization to Choquet Integral Case
Here we present the right and left Riemann–Liouville fractional fundamental theorems of fractional calculus without any initial conditions for the first time. Then we establish a Riemann–Liouville fractional Polya type integral inequality with the help of generalised right and left Riemann–Liouville fractional derivatives. The amazing fact here is that we do not need any boundary conditions as the classical Polya integral inequality requires. We extend our Polya inequality to Choquet integral setting. See also [2].
17.1 Introduction We mention the following famous Polya’s integral inequality, see [6, 7, p. 62], [8] and [9, p. 83]. Let f (x) be differentiable and not identically a constant on [a, b] with f (a) = f (b) = 0. Then there exists at least one point ξ ∈ [a, b] such that f (ξ) >
4 (b − a)2
b
f (x) d x. a
In [10] Feng Qi presents the following very interesting Polya type integral inequality which generalizes the last inequality:
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_17
329
330
17 Riemann–Liouville Fractional Fundamental Theorem of Calculus …
Let f (x) be differentiable and not identically constant on [a, b] with f (a) = f (b) = 0 and M = sup f (x). Then x∈[a,b]
a
b
(b − a)2 M, f (x) d x ≤ 4
where (b−a) is the best constant in the above inequality. 4 We are greatly motivated by the above classical Polya inequalities. 2
17.2 Background Here the background contains only original results. We need Definition 17.1 Let 0 < q < 1, f ∈ C ([a, b]). The right Riemann–Liouville fractional integral is given by (see [1], p. 333) −q t Db
f (t) :=
1 (q)
b
(τ − t)q−1 f (τ ) dτ ,
(17.1)
t
∀ t ∈ [a, b], where is the gamma function. The right Riemann–Liouville fractional derivative of order q is given by (see [5], p. 89) b d 1 q (17.2) (τ − t)−q f (τ ) dτ , t Db f (t) := − (1 − q) dt t ∀ t ∈ [a, b] . We give q
Theorem 17.2 Let 0 < q < 1 and f ∈ C ([a, b]). Assume that t Db f ∈ L ∞ ([a, b]). Then −q q (17.3) t Db t Db f (t) = f (t) , ∀ t ∈ [a, b] , which means 1 f (t) = (q)
t
b
(τ − t)q−1
q τ Db
f (τ ) dτ , ∀ t ∈ [a, b] .
(17.4)
This is a kind of fundamental theorem for right Riemann–Liouville fractional calculus without any initial condition.
17.2 Background
331
Proof Since 0 < q < 1, then 1 − q > 0 and q − 1 < 0. We have that q t Db
d d 1 q−1 f (t) = − t Db f (t) = − dt (1 − q) dt
b
(τ − t)−q f (τ ) dτ ,
(17.5)
t
∀ t ∈ [a, b] . Furthermore it holds −q q t Db t Db
f (t) =
d − dt
1 (q)
1 (q + 1)
b
b t
q
(τ − t)q−1 τ Db f (τ ) dτ =
(17.6)
q τ Db
(τ − t)
q
t
f (τ ) dτ .
Next we apply integration by parts to −
1 (q + 1)
1 (q + 1) 1 (τ − t)q (q + 1)
q−1
τ Db
b
b
(τ − t)q
t
(τ − t)q
t
f (τ )
b t
−
q τ Db
f (τ ) dτ =
d q−1 f (τ ) dτ = τ Db dτ
1 (q)
b t
q−1
(τ − t)q−1 τ Db
f (τ ) dτ = (17.7)
(b − t)q q−1 f (t) −t Db−1 f (t) . t Db t=b (q + 1) Therefore we have d − dt
1 (q + 1)
f (t) −
b
(τ − t)
q
t
q τ Db
f (τ ) dτ
=
(17.8)
(b − t)q−1 q−1 f (t) . t Db t=b (q)
Consequently we find −q q t Db t Db
q−1 f (t) = f (t) − t Db f (t)
0 < q < 1. Here by assumption f ∈ C ([a, b]) and q−1 f (t) is bounded at t = b. t Db
q t Db
t=b
(b − t)q−1 , (q)
(17.9)
f (t) ∈ L ∞ ([a, b]), therefore
332
17 Riemann–Liouville Fractional Fundamental Theorem of Calculus …
We notice the following: we have q−1 t Db
and
q−1 t Db f (t) ≤ f ∞,[a,b] (1 − q)
b
1 (1 − q)
b
(τ − t)−q f (τ ) dτ ,
b
(τ − t)−q | f (τ )| dτ ≤
t
t
f ∞,[a,b] (b − t)1−q (1 − q) (1 − q)
q−1 lim t Db f (t) = 0.
t→b−
Therefore
q−1
t Db
We have proved that
(17.11)
f ∞,[a,b] (b − t)1−q . (2 − q)
f q−1 ∞,[a,b] (b − t)1−q , t Db f (t) ≤ (2 − q)
∀ t ∈ [a, b] . Hence
(17.10)
t
(τ − t)−q dτ =
= That is
1 f (t) = (1 − q)
f (t)
−q q t Db t Db
t=b
(17.12)
(17.13)
= 0.
f (t) = f (t) ,
∀ t ∈ [a, b] .
(17.14)
We need Definition 17.3 Let 0 < q < 1, f ∈ C ([a, b]). The left Riemann–Liouville fractional integral is given by (see [5], p. 65) q a Dt
1 f (t) := (q)
t
(t − τ )q−1 f (τ ) dτ ,
(17.15)
a
∀ t ∈ [a, b]. The left Riemann–Liouville fractional derivative of order q is given by (see [5], p. 68) t d 1 q (17.16) (t − τ )−q f (τ ) dτ , a Dt f (t) := (1 − q) dt a ∀ t ∈ [a, b] .
17.2 Background
333
We give q
Theorem 17.4 Let 0 < q < 1 and f ∈ C ([a, b]). Assume that a Dt f ∈ L ∞ ([a, b]). Then −q q (17.17) a Dt a Dt f (t) = f (t) , ∀ t ∈ [a, b] , which means f (t) =
1 (q)
t
(t − τ )q−1
a
f (τ ) dτ , ∀ t ∈ [a, b] .
q a Dτ
(17.18)
This is a kind of fundamental theorem of left Riemann–Liouville fractional calculus without any initial condition. Proof From [5], p. 71, (2.113) there, when 0 < q < 1, we get −q q a Dt a Dt
q−1 f (t) = f (t) − a Dt f (t)
t=a
We notice that (q − 1 < 0) q−1 a Dt
∀ t ∈ [a, b] . Hence it holds q−1 a Dt f (t) ≤ f ∞,[a,b] (1 − q)
t
1 (1 − q)
(t − τ )
−q
t
(t − τ )−q f (τ ) dτ ,
(17.20)
t
(t − τ )−q | f (τ )| dτ ≤
(17.21)
a
=
dτ
f ∞,[a,b] (t − a)1−q (1 − q) (1 − q)
f ∞,[a,b] (t − a)1−q . (2 − q)
f q−1 ∞,[a,b] (t − a)1−q , a Dt f (t) ≤ (2 − q)
∀ t ∈ [a, b] . Hence
(17.19)
a
a
= That is
1 (1 − q)
f (t) =
(t − a)q−1 . (q)
(17.22)
q−1 lim a Dt f (t) = 0.
t→a+
Therefore
q−1
a Dt
The theorem is proved.
f (t)
t=a
= 0.
(17.23)
334
17 Riemann–Liouville Fractional Fundamental Theorem of Calculus …
17.3 Main Results Next we present the Riemann–Liouville fractional Polya type inequality without any boundary conditions.
q Theorem 17.5 Let 0 < q < 1, f ∈ C ([a, b]). Assume that a Dt f ∈ L ∞ a, a+b 2
q , b . Set and t Db f (t) ∈ L ∞ a+b 2 q q N ( f ) := max a Dt f (t)∞,[a, a+b ] , t Db f (t)∞,[ a+b ,b] . 2
Then
b
| f (t) dt| ≤ N ( f )
a
(17.24)
2
(b − a)q+1 . (q + 2) 2q
(17.25)
Inequality (17.25) is sharp, namely it is attained by
(t − a)q , t ∈ a, a+b 2 , ,b (b − t)q , t ∈ a+b 2
f (t) :=
(17.26)
0 < q < 1. Clearly here non-zero constant functions f are not possible. Proof By (17.18) we get that | f (t)| ≤
1 (q)
t
a
(t − τ )q−1 a Dτq f (τ ) dτ ≤
q (t − a)q a Dt f (t) , a+b ∞,[a, 2 ] (q + 1)
(17.27)
for any t ∈ a, a+b . 2 By (17.4) we get that | f (t)| ≤
1 (q)
b
t
q (τ − t)q−1 τ Db f (τ ) dτ ≤
q (b − t)q t D f (t) a+b , b ∞,[ 2 ,b] (q + 1)
for any t ∈ a+b ,b . 2 Hence we get that (by (17.27), (17.28)) a
b
| f (t)| dt = a
a+b 2
| f (t)| dt +
b a+b 2
| f (t)| dt ≤
(17.28)
17.3 Main Results
335
a+b q 2 1 a Dt f (t) (t − a)q dt+ a+b ∞, a, [ ] 2 (q + 1) a
q t D f (t) a+b b ∞,[ ,b]
b
(b − t) dt = q
a+b 2
2
(17.29)
q 1 b − a q+1 a Dt f (t) + ∞,[a, a+b 2 ] (q + 1) (q + 1) 2 q t D f (t) a+b b ∞,[ ,b]
2
1 (q + 2)
b−a 2
b−a 2
q+1 =
q+1 q q a Dt f (t) t D f (t) a+b ≤ a+b + b ∞,[a, ∞,[ ,b] ] 2
2
(b − a)q+1 q q max a Dt f (t)∞,[a, a+b ] , t Db f (t)∞,[ a+b ,b] . 2 2 (q + 2) 2q So we have proved
b
(17.30)
| f (t)| dt ≤
a
(b − a)q+1 q q max a Dt f (t)∞,[a, a+b ] , t Db f (t)∞,[ a+b ,b] . 2 2 (q + 2) 2q
(17.31)
Inequality (17.31) is sharp, it is attained by f (t) :=
, (t − a)q , t ∈ a, a+b 2 , ,b (b − t)q , t ∈ a+b 2
(17.32)
0 < q < 1. Notice that
f
a+b 2
−
= f
a+b 2
+
=
b−a 2
q ,
so that f ∈ C ([a, b]) . We see that (by (17.16)) q a Dt
d 1 f (t) = (1 − q) dt
a
t
(t − τ )−q (τ − a)q dτ =
(17.33)
336
17 Riemann–Liouville Fractional Fundamental Theorem of Calculus …
d 1 (1 − q) dt
t
(t − τ )(1−q)−1 (τ − a)(q+1)−1 dτ =
a
(by [12], p. 256) 1 d (1 − q) (q + 1) (t − a) = (1 − q) dt (2) (q + 1)
. ∀ t ∈ a, a+b 2 That is
(17.34)
d (t − a) = (q + 1) , dt
q a Dt f (t) = (q + 1) . ∞,[a, a+b ]
(17.35)
2
Similarly acting (by (17.2)) q t Db
−
f (t) = −
d 1 (1 − q) dt
d 1 (1 − q) dt
b
b
(τ − t)−q (b − τ )q dτ =
t
(τ − t)(1−q)−1 (b − τ )(q+1)−1 dτ =
t
(by [12], p. 256) −
1 d (q + 1) (1 − q) (b − t) = (1 − q) dt (2)
d d (q + 1) (b − t) = − (q + 1) (b − t) = (q + 1) , dt dt
,b . ∀ t ∈ a+b 2 That is q t D f (t) a+b = (q + 1) . b ∞,[ ,b]
(17.36)
−
(17.37)
2
We have found that q max a Dt f (t)
∞,[a, a+b 2 ]
q , t Db f (t)∞,[ a+b ,b] = (q + 1) .
(17.38)
2
Therefore the right hand side of (17.31) for f becomes (q + 1) (b − a)q+1 (b − a)q+1 = . (q + 2) 2q (q + 1) 2q
(17.39)
17.3 Main Results
337
But we notice that
b
f (t) dt =
a
1 (q + 1)
b
f (t) dt =
b−a 2
(t − a) dt +
q+1
+
b−a 2 =
q+1
b
q
a
a
a+b 2
a+b 2
(b − t)q dt =
1 b − a q+1 = 2 2 (q + 1)
(b − a)q+1 . (q + 1) 2q
(17.40)
By (17.39) and (17.40), inequality (17.31) is attained by f , that is (17.25) is sharp. In the next assume that (X, F) is a measurable space and (R+ ) R is the set of all (nonnegative) real numbers. We recall some concepts and some elementary results of capacity and the Choquet integral [3, 4]. Definition 17.6 A set function μ : F → R+ is called a non-additive measure (or capacity) if it satisfies (1) μ (∅) = 0; (2) μ (A) ≤ μ (B) for any A ⊆ B and A, B ∈ F. The non-additive measure μ is called concave if μ (A ∪ B) + μ ( A ∩ B) ≤ μ (A) + μ (B) ,
(17.41)
for all A, B ∈ F. In the literature the concave non-additive measure is known as submodular or 2-alternating non-additive measure. If the above inequality is reverse, μ is called convex. Similarly, convexity is called supermodularity or 2-monotonicity, too. First note that the Lebesgue measure λ for an interval [a, b] is defined by λ ([a, b]) = b − a, and that given a distortion function m, which is increasing (or nondecreasing) and such that m (0) = 0, the measure μ (A) = m (λ (A)) is a distorted Lebesgue measure. We denote a Lebesgue measure with distortion m by μ = μm . It is known that μm is concave (convex) if m is a concave (convex) function. function f : (X, F) → of all the nonnegative, measurable +The family + is the Borel σ-field of R+ . The R , B R+ is denoted as L + ∞ , where B R concept of the integral with respect to a non-additive measure was introduced by Choquet [3]. Definition 17.7 Let f ∈ L + ∞ . The Choquet integral of f with respect to non-additive measure μ on A ∈ F is defined by
∞
f dμ :=
(C) A
0
μ ({x : f (x) ≥ t} ∩ A) dt,
(17.42)
338
17 Riemann–Liouville Fractional Fundamental Theorem of Calculus …
where the integral on the right-hand side is a Riemann integral. Instead of (C) X f dμ, we shall write (C) f dμ. If (C) f dμ < ∞, we say that f is Choquet integrable and we write L C1
(μ) =
f dμ < ∞ .
f : (C)
The next lema summarizes the basic properties of Choquet integrals [4]. 1 Lemma 17.8 Assume that f, g ∈ L C (μ). (1) (C) 1 A dμ = μ (A), A ∈ F. (2) (Positive homogeneity) For all λ ∈ R+ , we have (C) λ f dμ = λ · (C) f dμ. (3) (Translation invariance) For all c ∈ R, we have (C) ( f + c) dμ = (C) f dμ + c. (4) (Monotonicity in the integrand) If f ≤ g, then we have
(C)
f dμ ≤ (C)
gdμ.
(Monotonicity in the set function) If μ ≤ ν, then we have (C) (5) (Subadditivity) If μ is concave, then
f dμ ≤ (C)
f dν.
f dμ + (C)
( f + g) dμ ≤ (C)
(C)
gdμ.
(Superadditivity) If μ is convex, then
( f + g) dμ ≥ (C)
(C)
f dμ + (C)
gdμ.
(6) (Comonotonic additivity) If f and g are comonotonic, then
( f + g) dμ = (C)
(C)
f dμ + (C)
gdμ,
where we say that f and g are comonotonic, if for any x, x ∈ X , then
f (x) − f x g (x) − g x ≥ 0.
We next mention the amazing result from [11], which permits us to compute the Choquet integral when the non-additive measure is a distorted Lebesgue measure. Theorem 17.9 Let f be a nonnegative and measurable function on R+ and μ = μm be a distorted Lebesgue measure. Assume that m (x) and f (x) are both continuous and m (x) is differentiable. When f is an increasing (non-decreasing) function on R+ , the Choquet integral of f with respect to μm on [0, t] is represented as
17.3 Main Results
339
t
f dμm =
(C)
m (t − x) f (x) d x,
(17.43)
0
[0,t]
however, when f is a decreasing (non-increasing) function on R+ , the Choquet integral of f is t f dμ = m (x) f (x) d x. (17.44) (C) m 0
[0,t]
We make Remark 17.10 From now on we assume that f : R+ → R+ is a monotone continuous function, and μ = μm i.e. μ ( A) = m (λ (A)), denotes a distorted Lebesgue measure, where m is such that m (0) = 0, m is increasing (non-decreasing) and continuously differentiable. By Theorem 17.9 and mean value theorem for integrals we get: (i) If f is an increasing (non-decreasing) function on R+ , we have (C)
[0,t ∗ ]
= m t∗ − ξ
f dμm
(17.43)
=
t∗
m t ∗ − x f (x) d x
0
t∗
f (x) d x, where ξ ∈ 0, t ∗ , t ∗ > 0.
(17.45)
0
(ii) If f is a decreasing (non-increasing) function on R+ , we have (C)
[0,t ∗ ]
f dμm
(17.44)
=
t∗
t∗
m (x) f (x) d x = m (ξ)
0
f (x) d x,
(17.46)
0
where ξ ∈ (0, t ∗ ) , t ∗ > 0. We denote by γ t ∗ , ξ :=
m (t ∗ − ξ) , when f is increasing (non-decreasing) m (ξ) , when f is decreasing (non-increasing),
(17.47)
for some ξ ∈ (0, t ∗ ) per case, t ∗ > 0. We give the following Choquet-fractional-Polya inequality without any boundary conditions. Theorem 17.11 Let 0 < q < 1, f = f |[0,t ∗ ] , t ∗ > 0, be continuous and all con∗ q and sidered as in Remark 17.10. Assume further that 0 Dt f (t) ∈ L ∞ 0, t2 t ∗ ∗ q , t . Set t Dt ∗ f (t) ∈ L ∞ 2 q q N ( f ) t ∗ := max 0 Dt f (t)∞,0, t ∗ , t Dt ∗ f (t)∞, t ∗ ,t ∗ , t ∗ > 0. (17.48) 2
2
340
17 Riemann–Liouville Fractional Fundamental Theorem of Calculus …
Then (C)
[0,t ∗ ]
f dμm ≤ γ t ∗ , ξ N ( f ) t ∗
(t ∗ )q+1 , t ∗ > 0. (q + 2) 2q
(17.49)
Clearly here f can not be a non-zero constant. Proof By Theorem 17.5 and earlier comments.
We give some examples for m. t 1 Remark 17.12 (i) If m (t) = 1+t , t ∈ R+ , then m (0) = 0, m (t) ≥0, m (t) = (1+t) 2 > ∗ 0, and m is increasing. Then γ (t , ξ) ≤ 1. (ii) If m (t) = 1 − e−t ≥ 0, t ∈ R+ , then m (0) = 0, m (t) = e−t > 0, and m is increasing. Then γ (t ∗ , ξ) ≤ 1. (iii) If m (t) = et − 1 ≥ 0, t ∈ R+ , m (0) = 0, m (t) = et > 0, and m is increas∗ ing. Then γ (t ∗ , ξ) ≤ et .
(iv) If m (t) = sin t, for t ∈ 0, π2 , we get m (0) = 0, m (t) = cos t ≥ 0, and m is increasing. Then γ (t ∗ , ξ) ≤ 1.
References 1. G.A. Anastassiou, Intelligent Mathematics: Computational Analysis (Springer, Heidelberg, 2011) 2. G.A. Anastassiou, Riemann-Liouville fractional fundamental theorem of Calculus and Riemann-Liouville Fractional Polya type integral inequality and its extension to Choquet integral setting, Bulletin of Korean Mathematical Society (2019) 3. G. Choquet, Theory of capacities. Ann. Inst. Fourier (Grenoble) 5, 131–295 (1953) 4. D. Denneberg, Non-additive Measure and Integral (Kluwer Academic Publishers, Boston, 1994) 5. I. Podlubny, Fractional Differentiation Equations (Academic Press, San Diego, 1999) 6. G. Polya, Ein Mittelwertsatz für Funktionen mehrerer Veränderlichen. Tohoku Math. J. 19, 1–3 (1921) 7. G. Polya, G. Szegö, Aufgaben und Lehrsätze aus der Analysis, vol. I (Springer, Berlin, 1925). (German) 8. G. Polya, G. Szegö, Problems and Theorems in Analysis, vol. I, Classics in Mathematics (Springer, Berlin, 1972) 9. G. Polya, G. Szegö, Problems and Theorems in Analysis, vol. I, Chinese edn (1984) 10. F. Qi, Polya type integral inequalities: origin, variants, proofs, refinements, generalizations, equivalences, and applications. RGMIA, Res. Rep. Coll., article no. 20, vol. 16 (2013). http:// rgmia.org/v16.php 11. M. Sugeno, A note on derivatives of functions with respect to fuzzy measures. Fuzzy Sets Syst. 222, 1–17 (2013) 12. E.T. Whittaker, G.N. Watson, A Course in Modern Analysis (Cambridge University Press, Cambridge, 1927)
Chapter 18
Low Order Riemann–Liouville Fractional Inequalities with Absent Initial Conditions
Here we present low order Riemann–Liouville left and right fractional inequalities without any initial conditions. These are of Opial, Poincaré, Sobolev and Hilbert– Pachpatte types. See also [3].
18.1 Introduction This chapter is motivated by the following Riemann–Liouville fractional Polya type inequality without any boundary conditions. q q We denote by Da f and Db f the left and right Riemann–Liouville fractional derivatives, respectively, see (18.5), (18.16). Theorem 18.1 q < 1, f ∈ C ([a, b]). Assume that ([2]) q Let 0 1 such that p11 + q11 = 1, with q > q11 . Then a
t
| f (w)| Daq f (w) dw ≤
18.2 Main Results
343
q+ p1 − q1
(t − a)
1
1
1 q1
2 (q) [( p1 (q − 1) + 1) ( p1 (q − 1) + 2)]
1 p1
a
t
q D f (w) q1 dw a
q2
1
, (18.8)
all a ≤ t ≤ b. Proof We have that 1 f (w) = (q)
w
a
(w − τ )q−1 Daq f (τ ) dτ ,
(18.9)
∀ w ∈ [a, t] . Then, by Hölder’s inequality we have | f (w)| ≤ 1 (q)
w
1 (q)
(w − τ )
w
a
p1 (q−1)
(w − τ )q−1 Daq f (τ ) dτ ≤ p1 dτ
1
a
w
a
q D f (τ ) q1 dτ a
q1
1
=
( p1 (q−1)+1)
1 1 (w − a) p1 q1 , 1 (z (w)) (q) ( p1 (q − 1) + 1) p1
where z (w) :=
a
w
(18.10)
q D f (τ ) q1 dτ ≥ 0, a
(18.11)
z (a) = 0, a ≤ w ≤ t. q q q 1 Then z (w) = Da f (w) 1 , a.e. in [a, t], and Da f (w) = z (w) q1 , a.e. in [a, t]. Furthermore it holds | f (w)| Daq f (w) ≤
( p1 (q−1)+1)
1 1 (w − a) p1 z (w) z (w) q1 , 1 (q) ( p1 (q − 1) + 1) p1
(18.12)
a.e. in [a, t] . The right hand side of (18.12) is integrable over [a, t] as p1 (q − 1) + 1 > 0, by the assumption q > q11 , and z (w) z (w) ∈ L ∞ ([a, t]) . Therefore, by Hölder’s inequality we obtain a
t
| f (w)| Daq f (w) dw ≤
1 1
(q) ( p1 (q − 1) + 1) p1
(18.13)
344
18 Low Order Riemann–Liouville Fractional Inequalities with Absent …
t
(w − a)( p1 (q−1)+1) dw
p1 1
a
t
z (w) z (w) dw
q1
1
=
a
(t − a)
1
( p1 (q−1)+2) p1
2
(z (t)) q1
, 1 1 1 (q) ( p1 (q − 1) + 1) p1 ( p1 (q − 1) + 2) p1 2 q1 q above | f (w)| Da f (w) is integrable over [a, t]. The theorem is proved.
(18.14)
We need Definition 18.5 Let 0 < q < 1, f ∈ C ([a, b]). The right Riemann–Liouville fractional integral is given by (see [1], p. 333) −q
Db f (t) :=
1 (q)
b
(τ − t)q−1 f (τ ) dτ ,
(18.15)
t
∀ t ∈ [a, b]. The right Riemann–Liouville fractional derivative of order q is given by (see [4], p. 89) b d 1 q (18.16) Db f (t) := − (τ − t)−q f (τ ) dτ , (1 − q) dt t ∀ t ∈ [a, b] . In [2] we have proved that q
Theorem 18.6 Let 0 < q < 1 and f ∈ C ([a, b]). Assume that Db f ∈ L ∞ ([a, b]). Then −q q (18.17) Db Db f (t) = f (t) , ∀ t ∈ [a, b] , which means 1 f (t) = (q)
t
b
q (τ − t)q−1 Db f (τ ) dτ , ∀ t ∈ [a, b] .
(18.18)
This is a kind of fundamental theorem for right Riemann–Liouville fractional calculus without any initial condition. We give the following right Riemann–Liouville fractional Opial type inequality. q
Theorem 18.7 Let 0 < q < 1 and f ∈ C ([a, b]). Assume that Db f ∈ L ∞ ([a, b]). Let p1 , q1 > 1 such that p11 + q11 = 1, with q > q11 . Then
b t
| f (w)| Dbq f (w) dw ≤
18.2 Main Results
345
q+ p1 − q1
(b − t)
1
1
1 q1
1 p1
2 (q) [( p1 (q − 1) + 1) ( p1 (q − 1) + 2)]
t
b
q D f (w) q1 dw b
q2
1
,
(18.19)
all a ≤ t ≤ b. Proof We have that f (w) =
1 (q)
b
w
q (τ − w)q−1 Db f (τ ) dτ ,
(18.20)
∀ w ∈ [t, b] . Then, by Hölder’s inequality we have | f (w)| ≤ 1 (q)
b
w
1 (q)
(τ − w)
b
w
p1 (q−1)
q (τ − w)q−1 Db f (τ ) dτ ≤ p1 1
b
dτ w
q D f (τ ) q1 dτ b
q1
1
=
( p1 (q−1)+1)
1 1 (b − w) p1 q1 , 1 (z (w)) (q) ( p1 (q − 1) + 1) p1
where
z (w) :=
b w
z (b) = 0, t ≤ w ≤ b. Thus
q D f (τ ) q1 dτ ≥ 0, b
− z (w) := b
w
q D f (τ ) q1 dτ , b
and
q q (−z (w)) = Db f (w) 1 ≥ 0, a.e. on [t, b] ,
and
q 1 D f (w) = −z (w) q1 , a.e. on [t, b] . b
(18.21)
(18.22)
(18.23)
(18.24)
Furthermore it holds | f (w)| Dbq f (w) ≤ a.e. in [t, b] .
( p1 (q−1)+1)
1 1 (b − w) p1 z (w) −z (w) q1 , 1 (q) ( p1 (q − 1) + 1) p1
(18.25)
346
18 Low Order Riemann–Liouville Fractional Inequalities with Absent …
The right hand side of (18.25) is integrable over [t, b] as p1 (q − 1) + 1 > 0, by the assumption q > q11 , and z (w) −z (w) ∈ L ∞ ([t, b]) . Therefore, by Hölder’s inequality we obtain
b
t
| f (w)| Dbq f (w) dw ≤ b
(b − w)
( p1 (q−1)+1)
1
(18.26)
1
(q) ( p1 (q − 1) + 1) p1
p1 1
dw
t
b
z (w) −z (w) dw
q1
1
=
t
(b − t)
1
( p1 (q−1)+2)
2
(z (t)) q1
p1
, 1 1 1 (q) ( p1 (q − 1) + 1) p1 ( p1 (q − 1) + 2) p1 2 q1 q above | f (w)| Db f (w) is integrable over [t, b]. The theorem is proved.
(18.27)
We present the following left Riemann–Liouville fractional Poincaré type inequality: q
Theorem 18.8 Let 0 < q < 1 and f ∈ C ([a, b]). Assume that Da f ∈ L ∞ ([a, b]). Let p1 , q1 > 1 such that p11 + q11 = 1, with q > q11 . Then f q1 ,[a,b] ≤
q (b − a)q Da f q1 ,[a,b] 1
1
(q) ( p1 (q − 1) + 1) p1 (qq1 ) q1
.
(18.28)
Proof We have that 1 f (t) = (q)
t
a
(t − τ )q−1 Daq f (τ ) dτ ,
(18.29)
∀ t ∈ [a, b] . Then we get | f (t)| ≤ 1 (q)
t
1 (q)
(t − τ )
t
a
p1 (q−1)
(t − τ )q−1 Daq f (τ ) dτ ≤ p1 1
dτ
a
a
t
q D f (τ ) q1 dτ a
q1
1
≤
( p1 (q−1)+1)
q 1 (t − a) p1 D f , 1 a q1 ,[a,b] (q) ( p1 (q − 1) + 1) p1
(18.30)
18.2 Main Results
347
that is | f (t)|q1 ≤ ∀ t ∈ [a, b] . Hence b | f (t)|q1 dt ≤ a
q q1 D f , a q1 ,[a,b]
(t − a)q1 q−1 ( (q))q1 ( p1 (q − 1) + 1)
q1 p1
(b − a)qq1 ( (q))q1 ( p1 (q − 1) + 1)
q1 p1
(qq1 )
q q1 D f , a q1 ,[a,b]
(18.31)
(18.32)
and f q1 ,[a,b] ≤
(b − a)q (q) ( p1 (q − 1) + 1)
1 p1
(qq1 )
1 q1
q D f , a q1 ,[a,b]
(18.33)
proving the claim.
We present the following left Riemann–Liouville fractional Sobolev type inequality: Theorem 18.9 All as in Theorem 18.8, and r ≥ 1. Then q Da f q1 ,[a,b] f r,[a,b] ≤ r1 . 1 r p1 rq − q1 + 1 (q) ( p1 (q − 1) + 1) q− q1 + r1
(b − a)
1
(18.34)
Proof We have that | f (t)|q1 ≤
q q1 D f , a q1 ,[a,b]
(t − a)qq1 −1 ( (q))q1 ( p1 (q − 1) + 1)
∀ t ∈ [a, b] . And | f (t)| ≤
q1 p1
q− q1
(t − a)
1
(q) ( p1 (q − 1) + 1)
1 p1
q D f , a q1 ,[a,b]
(18.35)
(18.36)
furthermore (r ≥ 1) we get | f (t)| ≤ r
∀ t ∈ [a, b] .
rq− qr
(t − a)
1
( (q)) ( p1 (q − 1) + 1) r
r p1
q r D f , a q1 ,[a,b]
(18.37)
348
18 Low Order Riemann–Liouville Fractional Inequalities with Absent …
Hence
b
q r Da f q ,[a,b] 1 , | f (t)| dt ≤ r r p1 rq − qr1 + 1 ( (q)) ( p1 (q − 1) + 1) rq− qr +1
(b − a)
r
a
and
1
(18.38)
q Da f q1 ,[a,b] ≤ r1 , 1 (q) ( p1 (q − 1) + 1) p1 rq − qr1 + 1 q− q1 + r1
(b − a)
f r,[a,b]
1
(18.39)
proving the claim.
We present the following right Riemann–Liouville fractional Poincaré type inequality: q
Theorem 18.10 Let 0 < q < 1 and f ∈ C ([a, b]). Assume that Db f ∈ L ∞ ([a, b]). Let p1 , q1 > 1 such that p11 + q11 = 1, with q > q11 . Then q (b − a)q Db f q1 ,[a,b] f q1 ,[a,b] ≤ (18.40) 1 1 . (q) ( p1 (q − 1) + 1) p1 (qq1 ) q1 Proof We have that f (t) =
1 (q)
b
t
q (τ − t)q−1 Db f (τ ) dτ ,
(18.41)
∀ t ∈ [a, b] . Then we get | f (t)| ≤ 1 (q)
b
1 (q)
(τ − t)
b
t
p1 (q−1)
q (τ − t)q−1 Db f (τ ) dτ ≤ p1 1
b
dτ
t
t
q D f (τ ) q1 dτ b
q1
1
≤
( p1 (q−1)+1)
q 1 (b − t) p1 D f , 1 b q1 ,[a,b] (q) ( p1 (q − 1) + 1) p1
(18.42)
that is | f (t)|q1 ≤ ∀ t ∈ [a, b] .
(b − t)qq1 −1 ( (q)) ( p1 (q − 1) + 1) q1
q1 p1
q q1 D f , b q1 ,[a,b]
(18.43)
18.2 Main Results
349
Hence
b
| f (t)| dt ≤
q q (b − a)qq1 Db f q11 ,[a,b]
q1
a
q1
( (q))q1 ( p1 (q − 1) + 1) p1 (qq1 )
and f q1 ,[a,b] ≤
q (b − a)q Db f q1 ,[a,b] 1
1
(q) ( p1 (q − 1) + 1) p1 (qq1 ) q1
,
(18.44)
,
(18.45)
proving the claim.
We present the following right Riemann–Liouville fractional Sobolev type inequality: Theorem 18.11 All as in Theorem 18.10, and r ≥ 1. Then q D b q1 ,[a,b] ≤ r1 . 1 (q) ( p1 (q − 1) + 1) p1 rq − qr1 + 1
(18.46)
q q1 D f , b q1 ,[a,b]
(18.47)
q− q1 + r1
f r,[a,b]
(b − a)
1
Proof We have that | f (t)|q1 ≤
(b − t)qq1 −1 ( (q))q1 ( p1 (q − 1) + 1)
∀ t ∈ [a, b] . And | f (t)| ≤
q1 p1
q− q1
(b − t)
1
(q) ( p1 (q − 1) + 1)
1 p1
q D f , b q1 ,[a,b]
(18.48)
furthermore (r ≥ 1) we get | f (t)|r ≤
rq− qr
(b − t)
1
( (q))r ( p1 (q − 1) + 1)
r p1
q r D f , b q1 ,[a,b]
(18.49)
∀ t ∈ [a, b] . Hence
b
q r D f b q ,[a,b] 1 , | f (t)| dt ≤ r r p1 rq − qr1 + 1 ( (q)) ( p1 (q − 1) + 1) rq− qr +1
r
a
and
(b − a)
1
(18.50)
350
18 Low Order Riemann–Liouville Fractional Inequalities with Absent …
q D f b q1 ,[a,b] ≤ r1 , 1 (q) ( p1 (q − 1) + 1) p1 rq − qr1 + 1 q− q1 + r1
f r,[a,b]
(b − a)
1
(18.51)
proving the claim.
We present the following left Riemann–Liouville fractional Hilbert–Pachpatte type inequality: Theorem 18.12 Here i = 1, 2. Let 0 < qi < 1 and f i ∈ C ([ai , bi ]). Assume that q Daii f i ∈ L ∞ ([ai , bi ]). Let p1 , p2 > 1 such that p11 + p12 = 1, with q1 > p12 , q2 > p11 . Then b1 b2 | f 1 (t1 )| | f 2 (t2 )| dt1 dt2 ≤ p (q −1)+1 p2 (q2 −1)+1 −a (t 1 1) 1 1 a1 a2 + (t2 −a2 ) p2 p1 q q (b1 − a1 ) (b2 − a2 ) Da11 f 1 p2 ,[a1 ,b1 ] Da22 f 2 p1 ,[a2 ,b2 ] 1
1
(q1 ) (q2 ) ( p1 (q1 − 1) + 1) p1 ( p2 (q2 − 1) + 1) p2
.
(18.52)
Proof We have that (i = 1, 2) f i (ti ) =
1 (qi )
ti
ai
(ti − τi )qi −1 Daqii f i (τi ) dτi ,
∀ ti ∈ [ai , bi ] . Hence 1 | f 1 (t1 )| ≤ (q1 )
∀ t1 ∈ [a1 , b1 ], where q1 > Similarly, it holds 1 | f 2 (t2 )| ≤ (q2 )
∀ t2 ∈ [a2 , b2 ], where q2 > Therefore | f 1 (t1 )| | f 2 (t2 )| ≤
(t1 − a1 )( p1 (q1 −1)+1) p1 (q1 − 1) + 1 1 p2
p1
1
2 ,[a1 ,b1 ]
(where p1 , p2 > 1 :
(t2 − a2 )( p2 (q2 −1)+1) p2 (q2 − 1) + 1
q D 1 f1 a1 p
p1
2
1 p1
+
1 p2
,
(18.53)
,
(18.54)
= 1).
q D 2 f2 a2 p
1 ,[a2 ,b2 ]
1 . p1
q1 q2 Da 1 f 1 Da 2 f 2 p2 ,[a1 ,b1 ] p1 ,[a2 ,b2 ] 1
1
(q1 ) (q2 ) ( p1 (q1 − 1) + 1) p1 ( p2 (q2 − 1) + 1) p2
(t1 − a1 )
p1 (q1 −1)+1 p1
(t2 − a2 )
p2 (q2 −1)+1 p2
(18.55)
18.2 Main Results
351 1
1
(using Young’s inequality for a, b ≥ 0, a p1 b p2 ≤ ≤
a p1
+
b ) p2
q1 q2 Da 1 f 1 Da 2 f 2 p2 ,[a1 ,b1 ] p1 ,[a2 ,b2 ] 1
1
(q1 ) (q2 ) ( p1 (q1 − 1) + 1) p1 ( p2 (q2 − 1) + 1) p2
(t2 − a2 ) p2 (q2 −1)+1 (t1 − a1 ) p1 (q1 −1)+1 + , p1 p2
(18.56)
∀ ti ∈ [ai , bi ] ; i = 1, 2. Consequently, we can write
| f 1 (t1 )| | f 2 (t2 )| (t1 −a1 ) p1 (q1 −1)+1 p1
+
(t2 −a2 ) p2 (q2 −1)+1 p2
≤
(18.57)
q1 q2 Da 1 f 1 Da 2 f 2 p2 ,[a1 ,b1 ] p1 ,[a2 ,b2 ] 1
1
(q1 ) (q2 ) ( p1 (q1 − 1) + 1) p1 ( p2 (q2 − 1) + 1) p2
,
∀ ti ∈ [ai , bi ] ; i = 1, 2. The denominator of left hand side of (18.57) can be zero only when both t1 = a1 and t2 = a2 . Finally we obtain
b1
a1
b2
a2
| f 1 (t1 )| | f 2 (t2 )| dt1 dt2 (t1 −a1 ) p1 (q1 −1)+1 p1
+
(t2 −a2 ) p2 (q2 −1)+1 p2
≤
q q (b1 − a1 ) (b2 − a2 ) Da11 f 1 p2 ,[a1 ,b1 ] Da22 f 2 p1 ,[a2 ,b2 ] 1
1
(q1 ) (q2 ) ( p1 (q1 − 1) + 1) p1 ( p2 (q2 − 1) + 1) p2
,
(18.58)
proving the claim.
We give the following right Riemann–Liouville fractional Hilbert–Pachpatte type inequality: Theorem 18.13 Here i = 1, 2. Let 0 < qi < 1 and f i ∈ C ([ai , bi ]). Assume that q Dbii f i ∈ L ∞ ([ai , bi ]). Let p1 , p2 > 1 such that p11 + p12 = 1, with q1 > p12 , q2 > p11 . Then b1 b2 | f 1 (t1 )| | f 2 (t2 )| dt1 dt2 ≤ p (q −1)+1 −t (b (b2 −t2 ) p2 (q2 −1)+1 1 1) 1 1 a1 a2 + p1 p2 q (b1 − a1 ) (b2 − a2 ) Db11 f 1 p
2 ,[a1 ,b1 ]
(q1 ) (q2 ) ( p1 (q1 − 1) + 1)
1 p1
q2 D f2 b2 p
1 ,[a2 ,b2 ] 1 p2
( p2 (q2 − 1) + 1)
.
(18.59)
352
18 Low Order Riemann–Liouville Fractional Inequalities with Absent …
Proof We have that (i = 1, 2)
1 f i (ti ) = (qi )
bi ti
q (τi − ti )qi −1 Dbii f i (τi ) dτi ,
(18.60)
∀ ti ∈ [ai , bi ] . Hence 1 | f 1 (t1 )| ≤ (q1 )
∀ t1 ∈ [a1 , b1 ], where q1 > Similarly, it holds 1 | f 2 (t2 )| ≤ (q2 )
∀ t2 ∈ [a2 , b2 ], where q2 > Therefore
(b1 − t1 )( p1 (q1 −1)+1) p1 (q1 − 1) + 1
2 ,[a1 ,b1 ]
,
(18.61)
,
(18.62)
1 . p2
(b2 − t2 )( p2 (q2 −1)+1) p2 (q2 − 1) + 1
p1
q2 D f2 b2 p
2
1 ,[a2 ,b2 ]
1 . p1
q1 D f1 b1 p
q2 D f2 b2 p
1 ,[a2 ,b2 ]
(q1 ) (q2 ) ( p1 (q1 − 1) + 1)
p1 (q1 −1)+1 p1
(b1 − t1 ) q1 D f1 b1 p
(b2 − t2 ) q2 D f2
2 ,[a1 ,b1 ]
b2
(q1 ) (q2 ) ( p1 (q1 − 1) + 1)
q1 D f1 b1 p
1
2 ,[a1 ,b1 ]
| f 1 (t1 )| | f 2 (t2 )| ≤
≤
p1
1 p1
1 p1
1
( p2 (q2 − 1) + 1) p2
p2 (q2 −1)+1 p2
(18.63)
p1 ,[a2 ,b2 ] 1
( p2 (q2 − 1) + 1) p2
(b2 − t2 ) p2 (q2 −1)+1 (b1 − t1 ) p1 (q1 −1)+1 + , p1 p2
(18.64)
∀ ti ∈ [ai , bi ] ; i = 1, 2. Consequently, we can write
| f 1 (t1 )| | f 2 (t2 )| (b1 −t1 )
p1 (q1 −1)+1
p1
q1 D f1 b1 p
+
2 ,[a1 ,b1
(b2 −t2 ) p2 (q2 −1)+1 p2
(18.65)
q2 D f2 b2 p ]
(q1 ) (q2 ) ( p1 (q1 − 1) + 1) ∀ ti ∈ [ai , bi ] ; i = 1, 2.
≤
1 ,[a2 ,b2 ]
1 p1
1
( p2 (q2 − 1) + 1) p2
,
18.2 Main Results
353
The denominator of left hand side of (18.65) can be zero only when both t1 = b1 and t2 = b2 . Finally we obtain
b1
a1
b2
a2
| f 1 (t1 )| | f 2 (t2 )| dt1 dt2 (b1 −t1 ) p1 (q1 −1)+1 p1
q (b1 − a1 ) (b2 − a2 ) Db11 f 1 p
+
2 ,[a1 ,b1 1
(b2 −t2 ) p2 (q2 −1)+1 p2
≤
q2 D f2 b2 p ]
1 ,[a2 ,b2 ] 1
(q1 ) (q2 ) ( p1 (q1 − 1) + 1) p1 ( p2 (q2 − 1) + 1) p2 proving the claim.
,
(18.66)
References 1. G.A. Anastassiou, Intelligent Mathematics: Computational Analysis (Springer, Heidelberg, 2011) 2. G.A. Anastassiou, Riemann-Liouville fractional fundamental theorem of Calculus and RiemannLiouville fractional Polya type integral inequality and its extension to Choquet integral setting, Bulletin of Korean Mathematical Society (2019) 3. G.A. Anastassiou, Low order Riemann-Liouville fractional inequalities without initial conditions (2019) 4. I. Podlubny, Fractional Differentiation Equations (Academic Press, San Diego, 1999)
Chapter 19
Low Order Fractional Riemann–Liouville Inequalities on a Spherical Shell
Here we present low order Riemann–Liouville left and right multivariate fractional inequalities without any initial or boundary conditions. These are of Opial and Poincaré type on a spherical shell. See also [3].
19.1 Main Results We need Definition 19.1 Let 0 < q < 1, f ∈ C ([a, b]). The left Riemann–Liouville fractional derivative of order q is given by (see [5], p. 68) Daq
d 1 f (t) := (1 − q) dt
t
(t − τ )−q f (τ ) dτ ,
(19.1)
a
∀ t ∈ [a, b] . We mention the following left Riemann–Liouville fractional Opial type inequality Theorem 19.2 ([4]) Let a
t
1 2
q
< q < 1 and assume that f, Da f ∈ C ([a, b]). Then
(t − a)q | f (s)| Daq f (s) ds ≤ √ √ 2 (q) (2q − 1) (2q)
a
t
Daq
2
f (s) ds , (19.2)
all a ≤ t ≤ b.
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_19
355
356
19 Low Order Fractional Riemann–Liouville Inequalities on a Spherical Shell
We need Remark 19.3 Let the spherical shell A := B (0, R2 ) − B (0, R1 ), 0 < R1 < R2 , A ⊆ R N , N ≥ 2, x ∈ A. Here x can be written uniquelly as x = r ω, where r = |x| > 0, and ω = rx ∈ S N −1 , |ω| = 1, see ([6], pp. 149–150 and [1], p. 421), furthermore for general F : A → R Lebesgue integrable function we have that
F (x) d x = A
S N −1
R2
F (r ω) r N −1 dr dω.
(19.3)
R1
Let dω be the element of surface measure on S N −1 , then ω N := Here
N
S N −1
2π 2 . N2
dω =
(19.4)
N ω N R2N − R1N 2π 2 R2N − R1N . V ol (A) = = N N N2
(19.5)
We need Definition 19.4 (See [1], p. 287) Let 0 < q < 1, f ∈ C A , and A is a spherical shell. Assume that there exists q
∂ R f (x) ∂ 1 C A 1 q := ∂r (1 − q) ∂r
r
−q
(r − t)
f (tω) dt ,
(19.6)
R1
that is, x = r ω, r ∈ [R1 , R2 ] and ω ∈ S N −1 . where x ∈ A; q We call
∂R f 1 ∂r q
the left radial Riemann–Liouville fractional derivative of f of order q.
Next we give a left fractional Riemann–Liouville Opial type inequality on the shell: q ∂R f 1 Theorem 19.5 Let 21 < q < 1 and assume that f, ∂r A . Then q ∈ C
q ∂ f (x) R1 | f (x)| dx ≤ q ∂r A
(R2 − R1 )q √ √ 2 (q) (2q − 1) (2q)
R2 R1
N −1 A
q
∂ R1 f (x) ∂r q
2 d x.
(19.7)
19.1 Main Results
357
Proof We apply (19.2) for f (·ω) ∈ C ([R1 , R2 ]) and we have that
| f (r ω)| D qR1 f (r ω) dr ≤
R2 R1
√
(R2 − R1 )q √ 2 (q) (2q − 1) (2q)
R2 R1
2 q D R1 f (r ω) dr .
(19.8)
Clearly here D R1 f (·ω) ∈ C ([R1 , R2 ]), by A = [R1 , R2 ] ×S N −1 . By r N −1r 1−N =1, where R1 ≤ r ≤ R2 , and R1N −1 ≤ r N −1 ≤ R2N −1 , R21−N ≤ r 1−N ≤ R11−N we obtain q
R21−N
R2 R1
√
R2 R1
q r N −1 | f (r ω)| D R1 f (r ω) dr ≤
q r N −1r 1−N | f (r ω)| D R1 f (r ω) dr ≤
(R2 − R1 )q √ 2 (q) (2q − 1) (2q)
R2 R1
(R2 − R1 )q R11−N √ √ 2 (q) (2q − 1) (2q) So we have that
R2 R1
R2
R2 R1
N −1
R1
(R2 − R1 )q √ √ 2 (q) (2q − 1) (2q)
R2
r
N −1
R1
q D R1
≤
(19.9)
2 f (r ω) dr .
N −1
R2 R1
q 2 r N −1 D R1 f (r ω) dr .
(19.10)
q ∂ f (x) R1 | f (x)| dx = q ∂r A
r S N −1
q r N −1 | f (r ω)| D R1 f (r ω) dr ≤
(R2 − R1 )q √ √ 2 (q) (2q − 1) (2q) Hence it holds
q 2 r N −1r 1−N D R1 f (r ω) dr
R2 R1
q | f (r ω)| D R1 f (r ω) dr dω ≤
N −1
R2
r S N −1
R1
N −1
q D R1
2 f (r ω) dr dω =
358
19 Low Order Fractional Riemann–Liouville Inequalities on a Spherical Shell
(R2 − R1 )q √ √ 2 (q) (2q − 1) (2q)
R2 R1
N −1 A
2
q
∂ R1 f (x)
d x,
∂r q
(19.11)
proving the claim. We need
Definition 19.6 Let 0 < q < 1, f ∈ C ([a, b]). The right Riemann–Liouville fractional derivative of order q is given by (see [5], p. 89) q Db
d 1 f (t) := − (1 − q) dt
b
(τ − t)−q f (τ ) dτ ,
(19.12)
t
∀ t ∈ [a, b] . We mention the following right Riemann–Liouville fractional Opial type inequality Theorem 19.7 ([4]) Let t
b
1 2
q
< q < 1 and assume that f, Db f ∈ C ([a, b]). Then
(b − t)q | f (s)| Dbq f (s) ds ≤ √ √ 2 (q) (2q − 1) (2q)
t
b
2 q Db f (s) ds , (19.13)
all a ≤ t ≤ b. We need Definition 19.8 (See [2], p. 50) Let 0 < q < 1, f ∈ C A , and A is a spherical shell. Assume that there exists q
∂ R f (x) ∂ 1 C A 2 q := − ∂r (1 − q) ∂r
R2
−q
(t − r )
f (tω) dt ,
(19.14)
r
where x ∈ A;q that is, x = r ω, r ∈ [R1 , R2 ] and ω ∈ S N −1 . We call order q.
∂R f 2 ∂r q
the right radial Riemann–Liouville fractional derivative of f of
Next we give a right fractional Riemann–Liouville Opial type inequality on the shell: q ∂R f 2 Theorem 19.9 Let 21 < q < 1 and assume that f, ∂r A . Then q ∈ C
q ∂ f (x) R2 | f (x)| dx ≤ q ∂r A
19.1 Main Results
359
(R2 − R1 )q √ √ 2 (q) (2q − 1) (2q)
R2 R1
N −1
q
∂ R2 f (x)
2
∂r q
A
d x.
(19.15)
Proof We apply (19.13) for f (·ω) ∈ C ([R1 , R2 ]) and we have that
R2 R1
| f (r ω)| D qR2 f (r ω) dr ≤
(R2 − R1 )q √ √ 2 (q) (2q − 1) (2q)
R2
R1
q D R2
2 f (r ω) dr .
(19.16)
Clearly here D R2 f (·ω) ∈ C ([R1 , R2 ]), by A = [R1 , R2 ] × S N −1 . By r N −1r 1−N = 1, where R1 ≤ r ≤ R2 , and R1N −1 ≤ r N −1 ≤ R2N −1 , R21−N ≤ r 1−N ≤ R11−N we obtain q
R21−N
R2 R1
R2 R1
q r N −1 | f (r ω)| D R2 f (r ω) dr ≤
q r N −1r 1−N | f (r ω)| D R2 f (r ω) dr ≤
(R2 − R1 )q √ √ 2 (q) (2q − 1) (2q)
R2
r
R2 R1
S N −1
R2 R1
R2 R1
q D R2
2 f (r ω) dr
≤
(19.17)
q 2 r N −1 D R2 f (r ω) dr .
q r N −1 | f (r ω)| D R2 f (r ω) dr ≤
(R2 − R1 )q √ √ 2 (q) (2q − 1) (2q) Hence it holds
r
R1
(R2 − R1 )q R11−N √ √ 2 (q) (2q − 1) (2q) So we have that
N −1 1−N
R2 R1
N −1
R2
r R1
N −1
q D R2
2 f (r ω) dr .
q ∂ f (x) R2 | f (x)| dx = ∂r q A
q r N −1 | f (r ω)| D R2 f (r ω) dr dω ≤
(19.18)
360
19 Low Order Fractional Riemann–Liouville Inequalities on a Spherical Shell
(R2 − R1 )q √ √ 2 (q) (2q − 1) (2q)
R2 R1
N −1
(R2 − R1 ) √ √ 2 (q) (2q − 1) (2q) q
R2
r S N −1
R2 R1
N −1
R1
N −1
q D R2
q ∂ R2
2 f (r ω) dr dω =
f (x)
∂r q
A
(19.19)
2 d x,
proving the claim.
We mention the following left Riemann–Liouville fractional Poincaré type inequality: q
Theorem 19.10 ([4]) Let 0 < q < 1 and f ∈ C ([a, b]). Assume that Da f ∈ L ∞ ([a, b]). Let p1 , q1 > 1 such that p11 + q11 = 1, with q > q11 . Then
b
| f (t)|q1 dt ≤
a
(b − a)qq1 ( (q))q1 ( p1 (q − 1) + 1)q1 −1 (qq1 )
a
b
q D f (t)q1 dt . a (19.20)
Next comes the following left Riemann–Liouville fractional Poincaré type inequality on the shell: q ∂R f 1 A . Let p1 , q1 > 1 : Theorem 19.11 Let 0 < q < 1 and assume that f, ∂r q ∈ C 1 1 1 + = 1, with q > . Then p1 q1 q1 A
| f (x)|q1 d x ≤
(R2 − R1 )qq1
( (q))q1 ( p1 (q − 1) + 1)q1 −1 (qq1 )
∂ q f (x) q1 R2 N −1 R1 d x. q R1 A ∂r
(19.21) Proof We apply (19.20) for f (·ω) ∈ C ([R1 , R2 ]) and we have that R2 R1
| f (r ω)|q1 dr ≤
(R2 − R1 )qq1 ( (q))q1 ( p1 (q − 1) + 1)q1 −1 (qq1 )
q1 R2 q D R1 f (r ω) dr , R1
where D R1 f (·ω) ∈ C ([R1 , R2 ]), ∀ ω ∈ S N −1 . We further obtain R2 R2 1−N N −1 q1 | f (r ω)| dr ≤ r r N −1r 1−N | f (r ω)|q1 dr ≤ R2 q
R1
(R2 − R1 )qq1 ( (q))q1 ( p1 (q − 1) + 1)q1 −1 (qq1 )
(19.22)
R1
(R2 − R1 )qq1 R11−N ( (q))q1 ( p1 (q − 1) + 1)q1 −1 (qq1 )
R2 R1
q q r N −1r 1−N D R1 f (r ω) 1 dr R2
r R1
N −1
≤
(19.23) q1 q D f (r ω) dr . R1
19.1 Main Results
361
So we derive that
R2
r N −1 | f (r ω)|q1 dr ≤
R1
(R2 − R1 )qq1 ( (q))q1 ( p1 (q − 1) + 1)q1 −1 (qq1 ) Hence it holds
R2 R1
| f (x)| d x =
r S N −1
R2 R1
R2
q1
A
N −1
N −1
S N −1
R2 R1
(19.24)
| f (r ω)| dr dω ≤ q1
R1
(R2 − R1 )qq1 q1 ( (q)) ( p1 (q − 1) + 1)q1 −1 (qq1 )
q q r N −1 D R1 f (r ω) 1 dr .
R2 R1
N −1
q q r N −1 D R1 f (r ω) 1 dr dω =
(R2 − R1 )qq1 q1 ( (q)) ( p1 (q − 1) + 1)q1 −1 (qq1 )
R2 R1
q1 N −1 q ∂ R1 f (x) d x, ∂r q A
(19.25)
proving the claim.
We mention the following right Riemann–Liouville fractional Poincaré type inequality: q
Theorem 19.12 ([4]) Let 0 < q < 1 and f ∈ C ([a, b]). Assume that Db f ∈ L ∞ ([a, b]). Let p1 , q1 > 1 such that p11 + q11 = 1 , with q > q11 . Then
b
| f (t)|q1 dt ≤
a
(b − a)qq1 ( (q))q1 ( p1 (q − 1) + 1)q1 −1 (qq1 )
a
b
q D f (t)q1 dt . b (19.26)
Finally comes the following right Riemann–Liouville fractional Poincaré type inequality on the shell: Theorem 19.13 Let 0 < q < 1 and assume that f, 1 + q11 = 1, with q > q11 . Then p1
q1
A
| f (x)| d x ≤
(R2 − R1 )qq1
( (q))q1 ( p1 (q − 1) + 1)q1 −1 (qq1 )
q
∂R f 2 ∂r q
∈ C A . Let p1 , q1 > 1 :
∂ q f (x) q1 R2 N −1 R2 d x. q R1 A ∂r
(19.27)
362
19 Low Order Fractional Riemann–Liouville Inequalities on a Spherical Shell
Proof We apply (19.26) for f (·ω) ∈ C ([R1 , R2 ]) and we have that R2 R1
| f (r ω)|q1 dr ≤
(R2 − R1 )qq1 q 1 ( (q)) ( p1 (q − 1) + 1)q1 −1 (qq1 )
q1 R2 q D R2 f (r ω) dr , R1
where D R2 f (·ω) ∈ C ([R1 , R2 ]), ∀ ω ∈ S N −1 . We further obtain R2 R2 1−N N −1 q1 | f (r ω)| dr ≤ r r N −1r 1−N | f (r ω)|q1 dr ≤ R2 q
R1
(19.28)
R1
(R2 − R1 )qq1 q1 ( (q)) ( p1 (q − 1) + 1)q1 −1 (qq1 )
R1
(R2 − R1 )qq1 R11−N q1 ( (q)) ( p1 (q − 1) + 1)q1 −1 (qq1 ) So we derive that
R2
R2
q q r N −1r 1−N D R2 f (r ω) 1 dr R2
r
N −1
R1
≤
(19.29) q1 q D f (r ω) dr . R2
r N −1 | f (r ω)|q1 dr ≤
R1
(R2 − R1 )qq1 ( (q))q1 ( p1 (q − 1) + 1)q1 −1 (qq1 ) Hence it holds
| f (x)|q1 d x = A
S N −1
R2 R1
R2
N −1
R2
r R1
R2
r S N −1
R1
N −1
(19.30)
R1
R2 R1
N −1
q D f (r ω)q1 dr dω = R2
(R2 − R1 )qq1 ( (q))q1 ( p1 (q − 1) + 1)q1 −1 (qq1 ) proving the claim.
q D f (r ω)q1 dr . R2
r N −1 | f (r ω)|q1 dr dω ≤
(R2 − R1 )qq1 ( (q))q1 ( p1 (q − 1) + 1)q1 −1 (qq1 )
N −1
R2 R1
q1 N −1 q ∂ R2 f (x) d x, ∂r q A
(19.31)
References
363
References 1. G. Anastassiou, Fractional Differentiation Inequalities (Springer, New York, 2009) 2. G. Anastassiou, Intelligent Comparisons: Analytic Inequalities (Springer, New York, 2016) 3. G.A. Anastassiou, Low order Riemann-Liouville fractional inequalities on a spherical shell, Analele Universitatii Oradea Fasc. Matematica, Vol. 27 (2020), nr. 1, accepted (2019) 4. G. Anastassiou, Low order Riemann-Liouville fractional inequalities without initial conditions, submitted for publication (2019) 5. I. Podlubny, Fractional Differentiation Equations (Academic Press, San Diego, 1999) 6. W. Rudin, Real and Complex Analysis, International Student Edition (McGraw Hill, London, 1970)
Chapter 20
Complex Korovkin Theory
Let K be a compact convex subspace of C and C (K , C) the space of continuous functions from K into C. We consider bounded linear functionals from C (K , C) into C and bounded linear operators from C (K , C) into itself. We assume that these are bounded by companion real positive linear entities, respectively. We study quantitatively the rate of convergence of the approximation of these linearities to the corresponding unit elements. Our results are inequalities of Korovkin type involving the complex modulus of continuity and basic test functions. See also [5].
20.1 Introduction The study of the convergence of positive linear operators became more intensive and attractive when P. Korovkin (1953) proved his famous theorem (see [8], p. 14). Korovkin’s First Theorem. Let [a, b] be a compact interval in R and (L n )n∈N be a sequence of positive linear operators L n mapping C ([a, b]) into itself. Assume that (L n f ) converges uniformly to f for the three test functions f = 1, x, x 2 . Then (L n f ) converges uniformly to f on [a, b] for all functions of f ∈ C ([a, b]). So a lot of authors since then have worked on the theoretical aspects of the above convergence. But R. A. Mamedov (1959) (see [9]) was the first to put Korovkin’s theorem in a quantitative scheme. Mamedov’s Theorem. Let {L n }n∈N be a sequence of positive linear operators in the space C ([a, b]), for which L n 1 = 1, L n (t, x) = x + αn (x), L n t 2 , x = x 2 + βn (x). Then it holds L n ( f, x) − f (x)∞ ≤ 3ω1 f, dn , where ω1 is the first modulus of continuity and dn = βn (x) − 2xαn (x)∞ . An improvement of the last result was the following. © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_20
365
366
20 Complex Korovkin Theory
Shisha and Mond’s Theorem. (1968, see [11]). Let [a, b] ⊂ R be a compact interval. Let {L n }n∈N be a sequence of positive linear operators acting on C ([a, b]). For n = 1, 2, . . . , suppose L n (1) is bounded. Let f ∈ C ([a, b]). Then for n = 1, 2, . . . , it holds L n f − f ∞ ≤ f ∞ · L n 1 − 1∞ + L n (1) + 1∞ · ω1 ( f, μn ) , where
21 . μn := L n (t − x)2 (x)∞
Shisha–Mond inequality generated and inspired a lot of research done by many authors worldwide on the rate of convergence of a sequence of positive linear operators to the unit operator, always producing similar inequalities however in many different directions, e.g., see the important work of H. Conska of 1983 in [7], etc. The author (see [1]) in his 1993 research monograph, produces in many directions best upper bounds for |(L n f ) (x0 ) − f (x0 )|, x0 ∈ Q ⊆ Rn , n ≥ 1, compact and convex, which lead for the first time to sharp/attained inequalities of Shisha– Mond type. The method of proving is probabilistic from the theory of moments. His pointwise approach is closely related to the study of the weak convergence with rates of a sequence of finite positive measures to the unit measure at a specific point. The author in [3], pp. 383–412 continued this work in an abstract setting: Let X be a normed vector space, Y be a Banach lattice; M ⊂ X is a compact and convex subset. Consider the space of continuous functions from M into Y , denoted by C (M, Y ); also consider the space of bounded functions B (M, Y ). He studied the rate of the uniform convergence of lattice homomorphisms T : C (M, Y ) → C (M, Y ) or T : C (M, Y ) → B (M, Y ) to the unit operator I . See also [2]. Also the author in [4], pp. 175–188 continued the last abstract work for bounded linear operators that are bounded by companion real positive linear operators. Here the involved functions are from [a, b] ⊂ R into (X, ·) a Banach space. All the above have inspired and motivated the work of this chapter. Our results are of Shisha–Mond type, i.e., of Korovkin type. Namely here let K be a convex and compact subset of C and l be a linear functional from C (K , C) into C, and let l be a positive linear functional from C (K , R) into R, l (| f |), ∀ f ∈ C (K , C). such that |l ( f )| ≤ Clearly then l is a bounded linear functional. Initially we create a quantitative Korovkin type theory over the last described setting, then we transfer these results to related bounded linear operators with similar properties.
20.2 Background We need Theorem 20.1 Let K ⊆ (C, |·|) and f a function from K into C. Consider the first complex modulus of continuity
20.2 Background
367
ω1 ( f, δ) := sup | f (x) − f (y)| , δ > 0.
(20.1)
x,y∈K |x−y| 0, where f ∈ U C (K , C) (uniformly continuous functions). (2)’ If K is open convex or compact convex, then ω1 ( f, δ) is continuous on R+ in δ, for f ∈ U C (K , C) . (3)’ If K is convex, then ω1 ( f, t1 + t2 ) ≤ ω1 ( f, t1 ) + ω1 ( f, t2 ) , t1 , t2 > 0,
(20.2)
that is the subadditivity property is true. Also it holds ω1 ( f, nδ) ≤ nω1 ( f, δ)
(20.3)
ω1 ( f, λδ) ≤ λ ω1 ( f, δ) ≤ (λ + 1) ω1 ( f, δ) ,
(20.4)
and where n ∈ N, λ > 0, δ > 0, · is the ceiling of the number. (4)’ Clearly in general ω1 ( f, δ) ≥ 0 and is increasing in δ > 0 and ω1 ( f, 0) = 0. (5)’ If K is open or compact, then ω1 ( f, δ) → 0 as δ ↓ 0, iff f ∈ U C (K , C) . (6)’ It holds (20.5) ω1 ( f + g, δ) ≤ ω1 ( f, δ) + ω1 (g, δ) , for δ > 0, any f, g : K → C, K ⊂ C is arbitrary. Proof (1)’ Here K is open convex. Let here f ∈ U C (K , C), iff ∀ ε > 0, ∃ δ > 0 : |x − y| < δ implies | f (x) − f (y)| < ε. Let ε0 > 0 then ∃ δ0 > 0 : |x − y| ≤ δ0 with | f (x) − f (y)| < ε0 , hence ω1 ( f, δ0 ) ≤ ε0 < ∞. Let δ > 0 arbitrary and x, y ∈ K : |x − y| ≤ δ. Choose n ∈ N : nδ0 > δ, and set xi = x + ni (y − x), 0 ≤ i ≤ n. Notice that all xi ∈ K . Then n−1
| f (x) − f (y)| = ( f (xi ) − f (xi+1 )) ≤ i=0
| f (x) − f (x1 )| + | f (x1 ) − f (x2 )| + | f (x2 ) − f (x3 )| + · · · + f xn−1 − f (y) ≤
nω1 ( f, δ0 ) ≤ nε0 < ∞, since |xi − xi+1 | = n1 |x − y| ≤ n1 δ < δ0 . Thus ω1 ( f, δ) ≤ nε0 < ∞, proving the claim. If K is compact convex, then claim is obvious.
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20 Complex Korovkin Theory
(2)’ Let x, y ∈ K and let |x − y| ≤ t1 + t2 , then there exists a point z ∈ x y, z ∈ K : |x − z| ≤ t1 and |y − z| ≤ t2 , where t1 , t2 > 0. Notice that | f (x) − f (y)| ≤ | f (x) − f (z)| + | f (z) − f (y)| ≤ ω1 ( f, t1 ) + ω1 ( f, t2 ) . Hence ω1 ( f, t1 + t2 ) ≤ ω1 ( f, t1 ) + ω1 ( f, t2 ) , proving (3)’. Then by the obvious property (4)’ we get 0 ≤ ω1 ( f, t1 + t2 ) − ω1 ( f, t1 ) ≤ ω1 ( f, t2 ) , and |ω1 ( f, t1 + t2 ) − ω1 ( f, t1 )| ≤ ω1 ( f, t2 ) . Let f ∈ U C (K , C), then limω1 ( f, t2 ) = 0, by property (5)’. Hence ω1 ( f, ·) is t2 ↓0
continuous on R+ . (5)’ (⇒) Let ω1 ( f, δ) → 0 as δ ↓ 0. Then ∀ ε > 0, ∃ δ > 0 with ω1 ( f, δ) ≤ ε. I.e. ∀ x, y ∈ K : |x − y| ≤ δ we get | f (x) − f (y)| ≤ ε. That is f ∈ U C (K , C). (⇐) Let f ∈ U C (K , C). Then ∀ ε > 0, ∃ δ > 0 : whenever |x − y| ≤ δ, x, y∈K , it implies | f (x) − f (y)| ≤ ε. I.e. ∀ ε > 0, ∃ δ > 0 : ω1 ( f, δ) ≤ ε. That is ω1 ( f, δ) → 0 as δ ↓ 0. (6)’ Notice that |( f (x) + g (x)) − ( f (y) + g (y))| ≤ | f (x) − f (y)| + |g (x) − g (y)| .
That is property (6)’ now is clear. We need
Theorem 20.2 ([1], p. 208) Let (V1 , ·) , (V2 , ·) be real normed vector spaces and Q ⊆ V1 which is star- shaped relative to the fixed point x0 . Consider f : Q → V2 with the properties: f (x0 ) = 0, and s − t ≤ h implies f (s) − f (t) ≤ w; w, h > 0.
(20.6)
Then, there exists a maximal such function , namely (t) :=
t − x0 − → ·w· i , h
− → where i is any unit vector in V2 . That is f (t) ≤ (t) , all t ∈ Q.
(20.7)
(20.8)
20.2 Background
369
Corollary 20.3 Let K ⊆ (C, |·|) be a compact convex subset, and f ∈ C (K , C). Then |x − x0 | | f (x) − f (x0 )| ≤ ω1 ( f, δ) , δ > 0, (20.9) δ ∀ x, x0 ∈ K . We make Remark 20.4 Let K ⊆ (C, |·|) be a compact subset and g ∈ C (K , R). A linear functional I from C (K , R) into R is positive, iff I (g1 ) ≥ I (g2 ), whenever g1 ≥ g2 , where g1 , g2 ∈ C (K , R) . Let us assume that I is a positive linear functional. Then by Riesz representation theorem, [10], p. 304, there exists a unique Borel measure μ on K such that
g (t) dμ (t) ,
I (g) =
(20.10)
K
∀ g ∈ C (K , R) . We make Remark 20.5 Here initially we follow [6]. Suppose γ is a smooth path parametrized by z (t), t ∈ [a, b] and f is a complex function which is continuous on γ. Put z (a) = u and z (b) = w with u, w ∈ C. We define the integral of f on γu,w = γ as
γ
f (z) dz =
γu,w
f (z) dz :=
b
f (z (t)) z (t) dt.
(20.11)
a
By triangle inequality we have b
b z (t) dt := f (z) dz = | | f (z)| |dz| . ≤ f f z dt (z (t))| (z (t)) (t) γ
a
a
γ
(20.12) Inequalities (20.12) provide a typical example on linear functionals: clearly |dz| | f f dz induces a linear functional from C C) into C, and (z)| (z) (γ, γ γ involves a positive linear functional from C (γ, R) into R. Thus, be given K a convex and compact subset of C and l be a linear functional from C (K , C) into C, it is not strange to assume that there exists a positive linear functional l from C (K , R) into R, such that |l ( f )| ≤ l (| f |) , ∀ f ∈ C (K , C) .
(20.13)
Furthermore, we may assume that l (1 (·)) = 1, where 1 (t) =1, ∀ t ∈ K , l (c (·)) =c, ∀c ∈ C where c (t) = c, ∀ t ∈ K · We call l the companion functional to l.
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20 Complex Korovkin Theory
Here C is a vector space over the field of reals. The functional l is linear over R and the functional l is linear over R. ln pairs, n ∈ N. Next we study approximation properties of ln ,
20.3 Main Results—I First about linear functionals: We present the following quantitative approximation result of Korovkin type. Theorem 20.6 Here K is a convex and compact subset of C and ln is a sequence of linear functionals from C (K , C) into C, n ∈ N. There is a sequence of companion positive linear functionals ln from C (K , R) into R, such that |ln ( f )| ≤ ln (| f |) , ∀ f ∈ C (K , C) , ∀ n ∈ N.
(20.14)
Additionally, we assume that ln (1 (·)) = 1 and ln (c (·)) = c, ∀c ∈ C ∀ n ∈ N. Then |ln ( f ) − f (x0 )| ≤ 2ω1 f, ln (|· − x0 |) , ∀ n ∈ N, ∀ x0 ∈ K ,
(20.15)
∀ f ∈ C (K , C) . Proof We notice that |ln ( f ) − f (x0 )| = |ln ( f ) − ln ( f (x0 ) (·))| = (20.14)
|ln ( f (·) − f (x0 ) (·))| ≤ ln (| f (·) − f (x0 ) (·)|)
(by δ>0,(20.9))
≤
|· − x0 | |· − x0 | ≤ ω1 ( f, δ) = ln 1 (·) + ln ω1 ( f, δ) δ δ 1 ω1 ( f, δ) ln (1(·)) + ln (|· − x0 |) = δ 1 ω1 ( f, δ) 1 + ln (|· − x0 |) , ln (|· − x0 |) = 2ω1 f, δ by choosing
(20.16)
δ := ln (|· − x0 |) ,
if ln (|· − x0 |) > 0, that is proving (20.15). Next, we consider the case of ln (|· − x0 |) = 0. By Riesz representation theorem, see (20.10) there exists a probability measure μ such that
20.3 Main Results—I
371
ln (g) =
g (t) dμ (t) , ∀ g ∈ C (K , R) .
(20.17)
K
That is, here it holds
|t − x0 | dμ (t) = 0, K
which implies |t − x0 | = 0, a.e, hence t − x0 = 0, a.e, and t = x0 , a.e. Consequently μ ({t ∈ K : t = x0 }) = 0. Hence μ = δx0 , the Dirac measure with support only {x0 } . Therefore in that case ln (g) = g (x0 ), ∀ g ∈ C (K , R). Thus, it holds ln (|· − x0 |) = ω1 ( f, 0) = 0, and ln (| f (·) − f (x0 ) (·)|) = | f (x0 ) ω1 f, − f (x0 )| = 0, giving |ln ( f ) − f (x0 )| = 0. That is (20.15) is again true. Remark 20.7 We have that ln (|· − x0 |) =
|t − x0 | dμ (t) K
(by Schwarz’s inequality) 21 21 2 |t − x0 | dμ (t) = 1dμ (t)
≤ K
1 ln (1) 2
K
21 1 |t − x0 | dμ (t) = ln |· − x0 |2 2 .
2
(20.18)
K
We give Corollary 20.8 All as in Theorem 20.6. Then 1 |ln ( f ) − f (x0 )| ≤ 2ω1 f, ln |· − x0 |2 2 , ∀ n ∈ N, ∀ x0 ∈ K .
(20.19)
Conclusion 20.9 All as in Theorem 20.6. By (20.15) and/or (20.19), as ln (|·−x0 |) → 0, or ln |· − x0 |2 → 0, as n → +∞, we obtain that ln ( f ) → f (x0 ) with rates, ∀ x0 ∈ K . Next comes a more general quantitative approximation result of Korovkin type. Theorem 20.10 Here K is a convex and compact subset of C and ln is a sequence of linear functionals from C (K , C) into C, n ∈ N. There is a sequence of companion positive linear functionals ln from C (K , R) into R, such that |ln ( f )| ≤ ln (| f |) , ∀ f ∈ C (K , C) , ∀ n ∈ N. Additionally, we assume that
(20.20)
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20 Complex Korovkin Theory
ln (cg) = c ln (g) , ∀ g ∈ C (K , R) , ∀ c ∈ C.
(20.21)
Then, for any f ∈ C (K , C), we have |ln ( f ) − f (x0 )| ≤ | f (x0 )| ln (1 (·)) + 1 ω1 f, ln (|· − x0 |) , ln (1 (·)) − 1 + (20.22) ∀ x0 ∈ K , ∀ n ∈ N. (Notice if ln (1 (·)) = 1, then (20.22) collapses to (20.15). So Theorem 20.10 generalizes Theorem 20.6). ln (|· − x0 |) → 0, then ln ( f ) → f (x0 ), as n → By (20.22), as ln (1 (·)) → 1 and +∞, with rates, and as here ln (1 (·)) is bounded. Proof We observe that |ln ( f ) − f (x0 )| = |ln ( f ) − ln ( f (x0 ) (·)) + ln ( f (x0 ) (·)) − f (x0 )| ≤ |ln ( f ) − ln ( f (x0 ) (·))| + f (x0 ) ln (1 (·)) − f (x0 ) = |ln ( f (·) − f (x0 ) (·))| + | f (x0 )| ln (1 (·)) − 1 ≤
(20.23)
| f (x0 )| ln (| f (·) − f (x0 ) (·)|) ≤ ln (1 (·)) − 1 + |· − x0 | | f (x0 )| ln ω1 ( f, δ) ln (1 (·)) − 1 + ≤ δ |· − x0 | | f (x0 )| ln (1 (·)) − 1 + ln (ω1 ( f, δ)) 1 (·) + = δ 1 | f (x0 )| ln (1 (·)) − 1 + ω1 ( f, δ) ln (1 (·)) + ln (|· − x0 |) = δ | f (x0 )| ln (1 (·)) + 1 ω1 f, ln (|· − x0 |) , ln (1 (·)) − 1 + by choosing
δ := ln (|· − x0 |) ,
(20.24)
ln (|· − x0 |) = 0.
(20.25)
if ln (|· − x0 |) > 0. Next we consider the case of
By Riesz representation theorem there exists a positive finite measure μ such that ln (g) =
g (t) dμ (t) , ∀ g ∈ C (K , R) . K
(20.26)
20.3 Main Results—I
373
That is
|t − x0 | dμ (t) = 0,
(20.27)
K
which implies |t − x0 | = 0, a.e., hence t − x0 = 0, a.e, and t = x0 , a.e. on K . Consequently μ ({t ∈ K : t = x0 }) = 0. That is μ = δx0 M (where 0 < M := μ (K ) = ln (1 (·))). Hence, in that case ln (g) = g (x0 ) M. Consequently it holds ω1 f, ln (|· − x0 |) = 0, and the right hand side of (20.22) equals | f (x0 )| |M − 1|. Also, it is ln (| f (·) − f (x0 ) (·)|) = | f (x0 ) − f (x0 )| M = 0. Hence from the first part of this proof we get |ln ( f ) − ln ( f (x0 ) (·))| = 0, and ln ( f ) = ln ( f (x0 ) (·)) = ln (1 (·)) = M f (x0 ) . f (x0 ) Consequently the left hand side of (20.22) becomes |ln ( f ) − f (x0 )| = |M f (x0 ) − f (x0 )| = | f (x0 )| |M − 1| . So that (20.22) becomes an equality, and both sides equal | f (x0 )| |M − 1| in the extreme case of ln (|· − x0 |) = 0. Thus inequality (20.22) is proved completely in all cases. We make Remark 20.11 By Schwartz’s inequality we get 1 1 ln (|· − x0 |) ≤ ln (1 (·)) 2 . ln |· − x0 |2 2
(20.28)
We give Corollary 20.12 All as in Theorem 20.10. Then |ln ( f ) − f (x0 )| ≤ | f (x0 )| ln (1 (·)) − 1 + 1 1 ln (1 (·)) + 1 ω1 f, ln (1 (·)) 2 ln |· − x0 |2 2 ,
(20.29)
∀ x0 ∈ K , ∀ n ∈ N. Next we give another version of our Korovkin type result. Theorem 20.13 Here all are as in Theorem 20.10. Then, for any f ∈ C (K , C), we have |ln ( f ) − f (x0 )| ≤ | f (x0 )| ln (1 (·)) + 1 ω1 ln (1 (·)) − 1 +
1 2 f, ln |· − x0 |2
,
(20.30) ∀ x0 ∈ K , ∀ n ∈ N. By (20.30), as ln (1 (·)) → 1 and ln |· − x0 |2 → 0, then ln ( f ) → f (x0 ), as n → +∞, with rates, and as here ln (1 (·)) is bounded.
374
20 Complex Korovkin Theory
Proof Let t, x0 ∈ K and δ > 0. If |t − x0 | > δ, then | f (t) − f (x0 )| ≤ ω1 ( f, |t − x0 |) = ω1 f, |t − x0 | δ −1 δ ≤
(20.31)
|t − x0 | |t − x0 |2 1+ ω1 ( f, δ) ≤ 1 + ω1 ( f, δ) . δ δ2 The estimate
|t − x0 |2 | f (t) − f (x0 )| ≤ 1 + ω1 ( f, δ) δ2
also holds trivially when |t − x0 | ≤ δ. So (20.32) is true always, ∀ t ∈ K , for any x0 ∈ K . We can rewrite |· − x0 |2 | f (·) − f (x0 )| ≤ 1 + ω1 ( f, δ) . δ2
(20.32)
(20.33)
As in the proof of Theorem 20.10 we have |ln ( f ) − f (x0 )| ≤ · · · ≤ | f (x0 )| ln (1 (·)) − 1 + |· − x0 |2 = ln ω1 ( f, δ) 1 (·) + δ2 1 2 | f (x0 )| ln (1 (·)) − 1 + ω1 ( f, δ) ln (1 (·)) + 2 ln |· − x0 | = δ
(20.34)
1 | f (x0 )| ln |· − x0 |2 2 ln (1 (·)) + 1 , ln (1 (·)) − 1 + ω1 f, by choosing
1 δ := ln |· − x0 |2 2 ,
(20.35)
if ln |· − x0 |2 > 0. Next we consider the case of ln |· − x0 |2 = 0.
(20.36)
By Riesz representation theorem there exists a positive finite measure μ such that ln (g) =
g (t) dμ (t) , ∀ g ∈ C (K , R) . K
(20.37)
20.3 Main Results—I
375
That is
|t − x0 |2 dμ (t) = 0, K
which implies |t − x0 |2 = 0, a.e., hence t − x0 = 0, a.e, and t = x0 , a.e. on K . Consequently μ ({t ∈ K : t = x0 }) = 0. That is μ = δx0 M (where 0 < M := μ (K ) = ln (1 Hence, in that case ln (g) = g (x0 ) M. Consequently it holds (·))). 21 2 ln |·−x0 | =0, and the right hand side of (20.30) equals | f (x0 )| |M − 1|. ω1 f, Also, it is ln (| f (·) − f (x0 ) (·)|) = | f (x0 ) − f (x0 )| M = 0. Hence from the first part of this proof we get: |ln ( f ) − ln ( f (x0 ) (·))| = 0, and ln ( f ) = ln ( f (x0 ) (·)) = ln (1 (·)) = M f (x0 ) . f (x0 ) Consequently the left hand side of (20.30) becomes |ln ( f ) − f (x0 )| = | f (x0 )| |M − 1| . So that (20.30) is true again. The proof of the theorem is now complete.
Corollary 20.14 Here all are as in Theorem 20.10. Then |ln ( f ) − f (x0 )| ≤ | f (x0 )| ln (1 (·)) + 1 · ln (1 (·)) − 1 + 1 1 1 ln (1 (·)) 2 ln |· − x0 |2 2 , (20.38) min ω1 f, ln |· − x0 |2 2 , ω1 f, ∀ x0 ∈ K , ∀ n ∈ N.
Proof By (20.29) and (20.30). So (20.29) is better that (20.30) only if ln (1 (·)) < 1. We need
Theorem 20.15 Let K ⊆ C convex, x0 ∈ K 0 (interior of K) and f : K → C such that | f (t) − f (x0 )| is convex in t ∈ K . Furthermore let δ > 0 so that the closed disk D (x0 , δ) ⊂ K . Then | f (t) − f (x0 )| ≤
ω1 ( f, δ) |t − x0 | , ∀ t ∈ K . δ
(20.39)
Proof Let g (t) := | f (t) − f (x0 )|, t ∈ K , which is convex in t ∈ K and g (x0 ) = 0. Then by Lemma 8.1.1, p. 243 of [1], we obtain g (t) ≤
ω1 (g, δ) |t − x0 | , ∀ t ∈ K . δ
We notice the following | f (t1 ) − f (x0 )| = | f (t1 ) − f (t2 ) + f (t2 ) − f (x0 )| ≤
(20.40)
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20 Complex Korovkin Theory
| f (t1 ) − f (t2 )| + | f (t2 ) − f (x0 )| , hence | f (t1 ) − f (x0 )| − | f (t2 ) − f (x0 )| ≤ | f (t1 ) − f (t2 )| .
(20.41)
Similarly, it holds | f (t2 ) − f (x0 )| − | f (t1 ) − f (x0 )| ≤ | f (t1 ) − f (t2 )| .
(20.42)
Therefore for any t1 , t2 ∈ K : |t1 − t2 | ≤ δ we get | | f (t1 ) − f (x0 )| − | f (t2 ) − f (x0 )| | ≤ | f (t1 ) − f (t2 )| ≤ ω1 ( f, δ) .
(20.43)
That is ω1 (g, δ) ≤ ω1 ( f, δ) .
(20.44)
The last and (20.40) imply | f (t) − f (x0 )| ≤
ω1 ( f, δ) |t − x0 | , ∀ t ∈ K , δ
(20.45)
proving (20.39). We continue with a convex Korovkin type result:
Theorem 20.16 All as in Theorem 20.10. Let x0 ∈K 0 and assume that | f (t) − f (x0 )| is convex in t ∈ K . Let δ > 0, such that the closed disk D (x0 , δ) ⊂ K . Then |ln ( f ) − f (x0 )| ≤ | f (x0 )| ln (|· − x0 |) , ∀ n ∈ N. ln (1 (·)) − 1 + ω1 f, (20.46) Proof As in the proof Theorem 20.10 we have (20.39) |ln ( f ) − f (x0 )| ≤ · · · ≤ | f (x0 )| ln (| f (·) − f (x0 ) (·)|) ≤ ln (1 (·)) − 1 + (20.47) ω1 ( f, δ) | f (x0 )| ln (1 (·)) − 1 + ln (|· − x0 |) = δ | f (x0 )| ln (|· − x0 |) , ln (1 (·)) − 1 + ω1 f, by choosing
δ := ln (|· − x0 |) > 0,
if the last is positive. The case of ln (|· − x0 |) = 0 is treated similarly as in the proof of Theorem 20.10. The theorem is proved.
20.3 Main Results—I
377
Theorem 20.17 All as in Theorem 20.16. Inequality (20.46) is sharp, in fact it is − → − → attained by f ∗ (t) = j |t − x0 |, where j is a unit vector of (C, |·|); t, x0 ∈ K . Proof Indeed, f ∗ here fulfills the assumptions of the theorem. We further notice that f ∗ (x0 ) = 0, and | f ∗ (t) − f ∗ (x0 )| = |t − x0 | is convex in t ∈ K . The left hand side of (20.46) is ∗ → (20.21) ln f − f ∗ (x0 ) = ln f ∗ = ln − j |· − x0 | = − → ln (|· − x0 |) = ln (|· − x0 |) . j ln (|· − x0 |) =
(20.48)
The right hand side of (20.46) is − → ln (|· − x0 |) = ln (|· − x0 |) = ω1 j |· − x0 | , ω1 f ∗ , sup
t1 ,t2 ∈K |t1 −t2 |≤ ln (|·−x0 |)
− − → → j |t1 − x0 | − j |t2 − x0 | =
sup
||t1 − x0 | − |t2 − x0 || ≤
sup
|t1 − t2 | = ln (|· − x0 |) .
t1 ,t2 ∈K |t1 −t2 |≤ ln (|·−x0 |)
t1 ,t2 ∈K |t1 −t2 |≤ ln (|·−x0 |)
(20.49)
Hence we have found that ln (|· − x0 |) ≤ ln (|· − x0 |) . ω1 f ∗ , Clearly (20.46) is attained. The theorem is proved.
(20.50)
20.4 Main Results—II Next we give results on linear operators: Let K be a compact convex subset of C. Consider L : C (K , C) → C (K , C) a linear operator and L : C (K , R) → C (K , R) a positive linear operator (i.e. for L ( f1 ) ≥ L ( f 2 )) both over the field of R. f 1 . f 2 ∈ C (K , R) with f 1 ≥ f 2 we get We assume that |L ( f )| ≤ L (| f |) , ∀ f ∈ C (K , C) , L (| f |) (z), ∀ z ∈ K ). (i.e. |L ( f ) (z)| ≤ We call L the companion operator of L .
378
20 Complex Korovkin Theory
Let x0 ∈ K . Clearly, then L (·) (x0 ) is a linear functional from C (K , C) into C, and L (·) (x0 ) is a positive linear functional from C (K , R) into R. Notice L ( f ) (z) ∈ C and L (| f |) (z) ∈ R, ∀ f ∈ C (K , C) (thus | f | ∈ C (K , R)). Here L ( f ) ∈ C (K , C), and L (| f |) ∈ C (K , R), ∀ f ∈ C (K , C) . Notice that C (K , C) = U C (K , C), also C (K , R) = U C (K , R) (uniformly continuous functions). By [3], p. 388, we have that L (|· − x0 |r ) (x0 ), r > 0, is a continuous function in x0 ∈ K . After this preparation we transfer the main results from Sect. 20.3 to linear operators. We have the following approximation results with rates of Korovkin type. Theorem 20.18 Here K is a convex and compact subset of C and L n is a sequence of linear operators from C (K , C) into itself, n ∈ N. There is a sequence of companion positive linear operators L n from C (K , R) into itself, such that |L n ( f )| ≤ L n (| f |) , ∀ f ∈ C (K , C) , ∀ n ∈ N
(20.51)
L n (| f |) (x0 ), ∀ x0 ∈ K ). (i.e. |L n ( f ) (x0 )| ≤ Additionally, we assume that L n (g) , ∀ g ∈ C (K , R) , ∀ c ∈ C L n (cg) = c
(20.52)
L n (g) (x0 ) , ∀ x0 ∈ K ). (i.e. (L n (cg)) (x0 ) = c Then, for any f ∈ C (K , C), we have |(L n ( f )) (x0 ) − f (x0 )| ≤ | f (x0 )| L n (1 (·)) (x0 ) − 1 +
L n (1 (·)) (x0 ) + 1 ω1 f, L n (|· − x0 |) (x0 ) ,
(20.53)
∀ x0 ∈ K , ∀ n ∈ N. Proof By Theorem 20.10.
Corollary 20.19 All as in Theorem 20.18. Then L n ( f ) − f ∞,K ≤ f ∞,K L n (1 (·)) − 1∞,K + L n (|· − x0 |) (x0 )∞,K , L n (1 (·)) + 1∞,K ω1 f,
(20.54)
∀ n ∈ N. If L n (1 (·)) = 1, ∀ n ∈ N, then L n ( f ) − f ∞,K ≤ 2ω1 f, L n (|· − x0 |) (x0 )∞,K ,
(20.55)
20.4 Main Results—II
379
∀ n ∈ N. u u u L n (|· − x0 |) (x0 )∞,K → 0, then (by (20.54)) L n ( f ) → f , As L n (1 (·)) → 1, as n → +∞, where u means uniformly. Notice L n (1 (·)) is bounded, and all the suprema in (20.54) are finite. We continue with Theorem 20.20 Here all as in Theorem 20.18. Then, for any f ∈ C (K , C), we have |(L n ( f )) (x0 ) − f (x0 )| ≤ | f (x0 )| L n (1 (·)) (x0 ) − 1 +
1 L n (1 (·)) (x0 ) + 1 ω1 f, L n |· − x0 |2 (x0 ) 2 ,
(20.56)
∀ x0 ∈ K , ∀ n ∈ N.
Proof By Theorem 20.13. Corollary 20.21 All as in Theorem 20.18. Then, for any f ∈ C (K , C), we have L n ( f ) − f ∞,K ≤ f ∞,K L n (1 (·)) − 1∞,K + 21 L n (1 (·)) + 1∞,K ω1 f, L n |· − x0 |2 (x0 )∞,K ,
(20.57)
∀ n ∈ N. If L n (1 (·)) = 1, then 21 L n ( f ) − f ∞,K ≤ 2ω1 f, L n |· − x0 |2 (x0 )∞,K ,
(20.58)
∀ n ∈ N. u u u L n |· − x0 |2 (x0 )∞,K → 0, then (by (20.57)) L n ( f ) → f , As L n (1 (·)) → 1, as n → +∞. We continue with a convex Korovkin type result: ∗ 0 Theorem 20.22 ∗ All as in Theorem 20.18. Let a fixed x0 ∈ K and assume that f (t) − f x is convex in t ∈ K . Let δ > 0, such that the closed disk D x ∗ , δ ⊂ K . 0 0 Then ∗ ∗ ∗ ∗ (L n ( f )) x − f x ≤ f x L n (1 (·)) x0 − 1 0 0 0
+ ω1 f, L n · − x0∗ x0∗ , ∀ n ∈ N.
(20.59)
As L n(1 (·)) x0∗ → 1, and L n · − x0∗ x0∗ → 0, we get that (L n ( f )) x0∗ → f x0∗ , as n → +∞, a pointwise convergence. Proof By Theorem 20.16.
380
20 Complex Korovkin Theory
Note: Theorem 20.22 goes through if (20.51), (20.52) are valid only for the particular x0∗ . We finish with Proposition 20.23 All as in Theorem 20.22. Inequality (20.59) is sharp, in fact it is − → − → attained by f (t) = j t − x0∗ , where j is a unit vector of C; x0∗ , t ∈ K . Proof By Theorem 20.17.
Note: Let K be a convex compact subset of a real normed vector space (V, ·1 ) and (X, ·2 ) is a Banach space. We can consider bounded linear functionals and bounded operators on C (K , X ). This chapter’s methodology can be applied to this more general setting and produce a similar Korovkin theory in full strength.
References 1. G.A. Anastassiou, Moments in Probability and Approximation Theory, Pitman Research Notes in Mathematics, vol. 287 (Longman Scientific & Technical, Harlow, 1993) 2. G.A. Anastassiou, Lattice homomorphism - Korovkin type inequalities for vector valued functions. Hokkaido Math. J. 26, 337–364 (1997) 3. G.A. Anastassiou, Quantitative Approximations (Chapman & Hall/CRC, Boca Raton, 2001) 4. G.A. Anastassiou, Intelligent Computations: Abstract Fractional Calculus, Inequalities, Approximations (Springer, Heidelberg, 2018) 5. G.A. Anastassiou, Complex Korovkin theory, J. Comput. Anal. Appl. 28(6), 981–996 (2020) 6. S.S. Dragomir, An integral representation of the remainder in Taylor’s expansion formula for analytic function on general domains, RGMIA Res. Rep. Coll. 22, Art. 2, 14. https://rgmia. org/v22.php 7. H. Gonska, On approximation of continuously differentiable functions by positive linear operators. Bull. Aust. Math. Soc. 27, 73–81 (1983) 8. P.P. Korovkin, Linear Operators and Approximation Theory (Hindustan Publishing Corporation, New Delhi, 1960) 9. R.G. Mamedov, On the order of the approximation of functions by linear positive operators. Dokl. Akad. Nauk USSR 128, 674–676 (1959) 10. H.L. Royden, Real Analysis, 2nd edn. (Macmillan, New York, 1968) 11. O. Shisha, B. Mond, The degree of convergence of sequences of linear positive operators. Natl. Acad. Sci. 60, 1196–1200 (1968)
Chapter 21
Left Caputo Fractional Complex Inequalities
Here we present several complex left Caputo type fractional inequalities of well known kinds, such as of Ostrowski, Poincare, Sobolev, Opial and Hilbert–Pachpatte. See also [3].
21.1 Introduction We are motivated by the following result for functions of complex variable: Complex Ostrowski type inequality Theorem 21.1 (see [4]) Let f be holomorphic in G, an open domain and suppose γ ⊂ G is a smooth path from z (a) = u to z (b) = w. If v = z (x) with x ∈ (a, b), then γu,w = γu,v ∪ γv,w , f (v) (w − u) − f (z) dz ≤ γ
f
γu,v ;∞
γu,v
|z − u| |dz| + f
γv,w ;∞
γu,v
|z − u| |dz| +
γv,w
γv,w
|z − w| |dz| ≤
|z − w| |dz| f γ
u,w ;∞
,
f (v) (w − u) − f (z) dz ≤
and
γ
max |z − u| f γu,v ;1 + max |z − w| f γv,w ;1 ≤
z∈γu,v
z∈γv,w
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_21
381
382
21 Left Caputo Fractional Complex Inequalities
max max |z − u| , max |z − w| f γu,w ;1 . z∈γu,v
If p, q > 1 with
1 p
+
1 q
z∈γv,w
= 1, then f (v) (w − u) − ≤ f dz (z) γ
γu,v
|z − u|q |dz|
q1
f
+ γu,v ; p
γv,w
|z − w|q |dz|
|z − u| |dz| + q
γu,v
|z − w| |dz| q
γv,w
q1
q1
f
γv,w ; p
f
γu,w ; p
≤
.
Above |·| is the complex absolute value. We are also motivated by the next complex Opial type inequality: Theorem 21.2 (see [2]) Let f : D ⊆ C → C be an analytic function on the domain D and let x, y, w ∈ D. Suppose γ is a smooth path parametrized by z (t), t ∈ [a, b] with z (a) = x, z (c) = y, and z (b) = w, where c ∈ [a, b] is floating. Assume that f (k) (x) = 0, k = 0, 1, . . . , n, n ∈ Z+ , and p, q > 1 : 1p + q1 = 1. Then (1)
b
f (z (t)) f
(n+1)
a
(z (t)) z (t) dt ≤
b
| f (z (t))| f (n+1) (z (t)) z (t) dt ≤
a
1
b
c
1
2 q n!
a
1p |z (c) − z (t)| pn z (t) dt z (c) dc ·
a
b
q (n+1) f (z (t)) z (t) dt
q2
,
a
equivalently it holds (2) (n+1) f (z) f (z) dz ≤ γx,w
1
1
2 q n!
a
b
γx,y
γx,w
| f (z)| f (n+1) (z) |dz| ≤
|z (c) − z| pn |dz| z (c) dc
1p γx,w
(n+1) q f (z) |dz|
q2
.
Here we utilize on C the results of [1] which are for general Banach space valued functions.
21.1 Introduction
383
Mainly we give different cases of the left fractional C-Ostrowski type inequality and we continue with the left fractional: C-Poincaré like and Sobolev like inequalities. We present an Opial type left C-fractional inequality, and we finish with the Hilbert–Pachpatte left C-fractional inequalities.
21.2 Background In this section all integrals are of Bochner type. We need Definition 21.3 (see [5]) A definition of the Hausdorff measure h α goes as follows: if (T, d) is a metric space, A ⊆ T and δ > 0, let ( A, δ) be the set of all arbitrary collections (C)i of subsets of T , such that A ⊆ ∪i Ci and diam (Ci ) ≤ δ (diam =diameter) for every i. Now, for every α > 0 define h δα (A) := inf
(diamCi )α | (Ci )i ∈ (A, δ) .
(21.1)
Then there exists lim h δα (A) = suph δα (A), and h α (A) := lim h δα (A) gives an outer δ→0
δ>0
δ→0
measure on the power set P (T ), which is countably additive on the σ-field of all Borel subsets of T . If T = Rn , then the Hausdorff measure h n , restricted to the σfield of the Borel subsets of Rn , equals the Lebesgue measure on Rn up to a constant multiple. In particular, h 1 (C) = μ (C) for every Borel set C ⊆ R, where μ is the Lebesgue measure. Definition 21.4 ([1]) Let [a, b] ⊂ R, X be a Banach space, ν > 0; n := ν ∈ N, · is the ceiling of the number, f : [a, b] → X . We assume that f (n) ∈ L 1 ([a, b] , X ). We call the Caputo-Bochner left fractional derivative of order ν:
ν D∗a
1 f (x) := (n − ν)
x
(x − t)n−ν−1 f (n) (t) dt, ∀ x ∈ [a, b] .
(21.2)
a
ν f := f (ν) the ordinary X -valued derivative, defined similarly If ν ∈ N, we set D∗a 0 f := f. to the numerical one, and also set D∗a ν ν By [1] D∗a f (x) exists almost everywhere in x ∈ [a, b] and D∗a f ∈ L1 ([a, b] , X ). ν f ∈ C ([a, b] , X ) . If f (n) L ∞ ([a,b],X ) < ∞, then by [1] D∗a
We need the left-fractional Taylor’s formula: Theorem 21.5 ([1]) Let n ∈ N and f ∈ C n−1 ([a, b] , X ) , where [a, b] ⊂ R and X is a Banach space, and let ν ≥ 0 : n = ν . Set
384
21 Left Caputo Fractional Complex Inequalities
Fx (t) :=
n−1 (x − t)i (i) f (t) , ∀ t ∈ [a, x] , i! i=0
(21.3)
where x ∈ [a, b] . Assume that f (n) exists outside a μ-null Borel set Bx ⊆ [a, x], such that h 1 (Fx (Bx )) = 0, ∀ x ∈ [a, b] .
(21.4)
We also assume that f (n) ∈ L 1 ([a, b] , X ). Then x n−1 ν 1 (x − a)i (i) f (a) + f (z) dz, f (x) = (x − z)ν−1 D∗a i! (ν) a i=0
(21.5)
∀ x ∈ [a, b] . Next we mention an Ostrowski type inequality at left fractional level for Banach valued functions. Theorem 21.6 ([1]) Let ν ≥ 0, n = ν . Here all as in Theorem 21.5. Assume that ν f ∈ L ∞ ([a, b] , X ). Then f (i) (a) = 0, i = 1, . . . , n − 1, and that D∗a 1 b − a
a
b
ν f L ∞ ([a,b],X ) D∗a f (x) d x − f (a) (b − a)ν . ≤ (ν + 2)
(21.6)
We mention an Ostrowski type L p fractional inequality: Theorem 21.7 ([1]) Let p, q > 1 : 1p + q1 = 1, and ν > q1 , n = ν . Here all as ν f ∈ Lq in Theorem 21.5. Assume that f (k) (a) = 0, k = 1, . . . , n − 1, and D∗a ([a, b] , X ), where X is a Banach space. Then 1 b − a
b
a
f (x) d x − f (a) ≤
ν D f 1 ∗a L q ([a,b],X ) (b − a)ν− q . 1 1 (ν) ( p (ν − 1) + 1) p ν + p (21.7)
It follows Corollary 21.8 ([1]) (to Theorem 21.7, case of p = q = 2). Let ν > 21 , n = ν . Here all as in Theorem 21.5. Assume that f (k) (a) = 0, k = 1, . . . , n − 1, and ν f ∈ L 2 ([a, b] , X ). Then D∗a 1 b − a
a
b
ν D f 1 ∗a L 2 ([a,b],X ) √ (b − a)ν− 2 . (21.8) f (x) d x − f (a) ≤ 1 (ν) 2ν − 1 ν + 2
Next comes the L 1 case of fractional Ostrowski inequality:
21.2 Background
385
Theorem 21.9 ([1]) Let ν ≥ 1, n = ν , and all as in Theorem 21.5. Assume that ν f ∈ L 1 ([a, b] , X ). Then f (k) (a) = 0, k = 1, . . . , n − 1, and D∗a 1 b − a
b
a
ν f L 1 ([a,b],X ) D∗a f (x) d x − f (a) (b − a)ν−1 . ≤ (ν + 1)
(21.9)
We continue with a Poincaré like fractional inequality: Theorem 21.10 ([1]) Let p, q > 1 : 1p + q1 = 1, and ν > q1 , n = ν . Here all ν f ∈ as in Theorem 21.5. Assume that f (k) (a) = 0, k = 0, 1, . . . , n − 1, and D∗a L q ([a, b] , X ), where X is a Banach space. Then f L q ([a,b],X ) ≤
(b − a)ν 1 p
(ν) ( p (ν − 1) + 1) (qν)
1 q
ν D f . ∗a L q ([a,b],X )
(21.10)
Next comes a Sobolev like fractional inequality. Theorem 21.11 ([1]) All as in the last Theorem 21.10. Let r > 0. Then ν (b − a)ν− q + r D f . 1 ∗a L q ([a,b],X ) 1 r 1 p (ν) ( p (ν − 1) + 1) r ν − q + 1 (21.11) 1
f L r ([a,b],X ) ≤
1
We mention the following Opial type fractional inequality: Theorem 21.12 ([1]) Let p, q > 1 : 1p + q1 = 1, and ν > q1 , n := ν . Let [a, b] ⊂ R, X a Banach space, and f ∈ C n−1 ([a, b] , X ). Set n−1 (x − t)i (i) f (t) , ∀ t ∈ [a, x] , where x ∈ [a, b] . Fx (t) := i! i=0
(21.12)
Assume that f (n) exists outside a μ-null Borel set Bx ⊆ [a, x], such that h 1 (Fx (Bx )) = 0, ∀ x ∈ [a, b] .
(21.13)
We also assume that f (n) ∈ L ∞ ([a, b] , X ). Assume also that f (k) (a) = 0, k = 0, 1, . . . , n − 1. Then x ν f (w) D∗a f (w) dw ≤ a
(x − a)ν−1+ p 2
1
1
2 q (ν) (( p (ν − 1) + 1) ( p (ν − 1) + 2)) p
a
x
ν q D f (z) dz ∗a
q2
, (21.14)
386
21 Left Caputo Fractional Complex Inequalities
∀ x ∈ [a, b] . We finish this section with a Hilbert–Pachpatte left fractional inequality: Theorem 21.13 ([1]) Let p, q > 1 : 1p + q1 = 1, and ν1 > q1 , ν2 > 1p , n i := νi , i = 1, 2. Here [ai , bi ] ⊂ R, i = 1, 2; X is a Banach space. Let f i ∈ C ni −1 ([ai , bi ] , X ), i = 1, 2. Set Fxi (ti ) :=
n i −1 ji =0
(xi − ti ) ji ( ji ) f i (ti ) , ji !
(21.15)
∀ ti ∈ [ai , xi ], where xi ∈ [ai , bi ]; i = 1, 2. Assume that f i(ni ) exists outside a μ-null Borel set Bxi ⊆ [ai , xi ], such that h 1 Fxi Bxi = 0, ∀ xi ∈ [ai , bi ] ; i = 1, 2.
(21.16)
We also assume that f i(ni ) ∈ L 1 ([ai , bi ] , X ), and f i(ki ) (ai ) = 0, ki = 0, 1, . . . , n i − 1; i = 1, 2, and
Then
(21.17)
ν2 ν1 f ∈ L q ([a1 , b1 ] , X ) , D∗a f ∈ L p ([a2 , b2 ] , X ) . D∗a 1 1 2 2
b1
a1
b2
a2
f 1 (x1 ) f 2 (x2 ) d x1 d x2 (x1 −a1 ) p(ν1 −1)+1 p( p(ν1 −1)+1)
+
(x2 −a2 )q (ν2 −1)+1 q(q(ν2 −1)+1)
≤
ν (b1 − a1 ) (b2 − a2 ) D ν1 f 1 D 2 f2 . ∗a1 ∗a2 L q ([a1 ,b1 ],X ) L p ([a2 ,b2 ],X ) (ν1 ) (ν2 )
(21.18)
21.3 Main Results We need a special case of Definition 21.4 over C. Definition 21.14 Let [a, b] ⊂ R, ν > 0; n := ν ∈ N, · is the ceiling of the number and f ∈ C n ([a, b] , C). We call Caputo-Complex left fractional derivative of order ν: x ν 1 D∗a f (x) := (x − t)n−ν−1 f (n) (t) dt, ∀ x ∈ [a, b] , (21.19) (n − ν) a where the derivatives f , . . . f (n) are defined as the numerical derivative.
21.3 Main Results
387
ν If ν ∈ N, we set D∗a f := f (ν) the ordinary C-valued derivative and also set f := f.
0 D∗a
ν Notice here (by [1]) that D∗a f ∈ C ([a, b] , C) . We make
Remark 21.15 Suppose γ is a smooth path parametrized by z (t), t ∈ [a, b] (i.e. there exists z (t) and is continuous) and from now on f is a complex function which is continuous on γ. Put z (a) = u and z (b) = w with u, w ∈ C. We define the integral of f on γu,w = γ as γ
f (z) dz =
γu,w
f (z) dz :=
b
f (z (t)) z (t) dt =
a
b
h (t) dt,
(21.20)
a
where h (t) := f (z (t)) z (t), t ∈ [a, b] . We notice that the actual choice of parametrization of γ does not matter. This definition immediately extends to paths that are piecewise smooth. Suppose γ is parametrized by z (t), t ∈ [a, b], which is differentiable on the intervals [a, c] and [c, b], then assuming that f is continuous on γ we define
γu,w
f (z) dz :=
γu,v
f (z) dz +
γv,w
f (z) dz,
where v := z (c). This can be extended for a finite number of intervals. We also define the integral with respect to arc-length
γu,w
f (z) |dz| :=
b
f (z (t)) z (t) dt
a
and the length of the curve γ is then l (γ) =
γu,w
|dz| :=
b
z (t) dt.
a
We mention also the triangle inequality for the complex integral, namely f (z) dz ≤ | f (z)| |dz| ≤ f γ,∞ l (γ) , γ
γ
where f γ,∞ := sup | f (z)|. z∈γ
We give the following left-fractional C-Taylor’s formula: Theorem 21.16 Let h ∈ C n ([a, b] , C), n = ν , ν ≥ 0. Then
(21.21)
388
21 Left Caputo Fractional Complex Inequalities
h (t) =
t n−1 ν 1 (t − a)i (i) h (λ) dλ, h (a) + (t − λ)ν−1 D∗a i! (ν) a i=0
(21.22)
∀ t ∈ [a, b] , in particular it holds, f (z (t)) z (t) = 1 (ν)
a
t
n−1 (i) (t − a)i f (z (a)) z (a) + i! i=0
ν f (z (·)) z (·) (λ) dλ, (t − λ)ν−1 D∗a
(21.23)
∀ t ∈ [a, b] .
Proof By Theorem 21.5. It follows a left fractional C-Ostrowski type inequality
Theorem 21.17 Let n ∈ N and h ∈ C n ([a, b] , C), where [a, b] ⊂ R, and let ν ≥ 0 : n = ν . Assume that h (i) (a) = 0, i = 1, . . . , n − 1. Then b ν h ∞,[a,b] D∗a 1 h (t) dt − f (a) ≤ (21.24) (b − a)ν , b − a (ν + 2) a
(i) in particular when h (t) := f (z (t)) z (t) and f (z (t)) z (t) |t=a = 0, i = 1, . . . n − 1, we get 1 1 b f (z) dz − f (u) z (a) = f (z (t)) z (t) dt − f (z (a)) z (a) b − a γu,w b−a a
≤
ν D f (z (t)) z (t) ∗a ∞,[a,b] (ν + 2)
(b − a)ν .
(21.25)
Proof By Theorem 21.6. The corresponding C-Ostrowski type L p inequality follows:
Theorem 21.18 Let p, q > 1 : 1p + q1 = 1, and ν > q1 , n = ν . Here h ∈ C n ([a, b] , C). Assume that h (i) (a) = 0, i = 1, . . . , n − 1. Then ν b D h 1 1 ∗a L q ([a,b],C) (b − a)ν− q , h (t) dt − h (a) ≤ 1 b − a 1 a (ν) ( p (ν − 1) + 1) p ν + p
(21.26) (i) in particular when h (t) := f (z (t)) z (t) and f (z (t)) z (t) |t=a = 0, i = 1, . . . n − 1, we get:
21.3 Main Results
389
1 1 b = dt − f z f dz − f z f z (a) (t) (z (a)) (a) (z) (u) (z (t)) b − a γu,w b−a a
ν D 1 ∗a f (z (t)) z (t) L q ([a,b],C) (b − a)ν− q . ≤ 1 (ν) ( p (ν − 1) + 1) p ν + 1p
(21.27)
Proof By Theorem 21.7. It follows Corollary 21.19 (to Theorem 21.18, case of p = q = 2). We have that
ν D∗a f (z (t)) z (t) 1 1 L 2 ([a,b],C) ≤ f dz − f z (z) (u) (a) (b − a)ν− 2 . √ b − a γu,w 1 (ν) 2ν − 1 ν + 2
(21.28) We continue with an L 1 fractional C-Ostrowski type inequality: Theorem 21.20 Let ν ≥ 1, n = ν . Assume that h ∈ C n ([a, b] , C), where h (t) := f (z (t)) z (t), and such that h (i) (a) = 0, i = 1, . . . , n − 1. Then ν D∗a f (z (t)) z (t) 1 L 1 ([a,b],C) f (z) dz − f (u) z (a) ≤ (b − a)ν−1 . b − a γu,w (ν + 1)
(21.29)
Proof By Theorem 21.9. It follows a Poincaré like C-fractional inequality:
Theorem 21.21 Let p, q > 1 : 1p + q1 = 1, and ν > q1 , n = ν . Let h ∈ C n ([a, b] , C). Assume that h (i) (a) = 0, i = 1, . . . , n − 1. Then ν h L q ([a,b],C) (b − a)ν D∗a h L q ([a,b],C) ≤ (21.30) 1 1 , (ν) ( p (ν − 1) + 1) p (qν) q (i) in particular when h (t) := f (z (t)) z (t) and f (z (t)) z (t) |t=a = 0, i = 1, . . . n − 1, we get: f (z (t)) z (t) ≤ L q ([a,b],C)
(b − a)ν 1 p
(ν) ( p (ν − 1) + 1) (qν) Proof By Theorem 21.10.
1 q
ν D ∗a f (z (t)) z (t) L q ([a,b],C) .
(21.31)
390
21 Left Caputo Fractional Complex Inequalities
The corresponding Sobolev like inequality follows: Theorem 21.22 All as in Theorem 21.21. Let r > 0. Then f (z (t)) z (t) ≤ L r ([a,b],C) ν (b − a)ν− q + r D 1 ∗a f (z (t)) z (t) L q ([a,b],C) . 1 r (ν) ( p (ν − 1) + 1) p r ν − q1 + 1 (21.32) 1
1
Proof By Theorem 21.11. We continue with an Opial type C-fractional inequality.
Theorem 21.23 Let p, q > 1 : 1p + q1 = 1, and ν > q1 , n := ν , h ∈ C n ([a, b] , C). Assume h (k) (a) = 0, k = 0, 1, . . . , n − 1. Then
x
a
ν |h (t)| D∗a h (t) dt ≤
(x − a)ν−1+ p 2
1
x
1
2 q (ν) (( p (ν − 1) + 1) ( p (ν − 1) + 2)) p
a
ν q D h (t) dt ∗a
q2
, (21.33)
(i) ∀ x ∈ [a, b] , in particular when h (t) := f (z (t)) z (t) and f (z (t)) z (t) |t=a = 0, i = 1, . . . n − 1, we get: a
x
ν | f (z (t))| D∗a f (z (t)) z (t) z (t) dt ≤
(x − a)ν−1+ p 2
1
1
2 q (ν) (( p (ν − 1) + 1) ( p (ν − 1) + 2)) p
a
x
ν q dt D ∗a f (z (t)) z (t)
q2
,
(21.34)
∀ x ∈ [a, b] .
Proof By Theorem 21.12. We finish with Hilbert–Pachpatte left C-fractional inequalities: Theorem 21.24 Let p, q > 1 :
1 p
+
1 q
= 1, and ν1 > q1 , ν2 > 1p , n i := νi , i =
1, 2. Let h i ∈ C ni ([ai , bi ] , C), i = 1, 2. Assume h i(ki ) (ai ) = 0, ki = 0, 1, . . . , n i − 1; i = 1, 2. Then
b1
a1
b2
a2
|h 1 (t1 )| |h 2 (t2 )| dt1 dt2 (t1 −a1 ) p(ν1 −1)+1 p( p(ν1 −1)+1)
+
(t2 −a2 )q (ν2 −1)+1 q(q(ν2 −1)+1)
≤
21.3 Main Results
391
ν (b1 − a1 ) (b2 − a2 ) D 2 h2 D ν1 h 1 , ∗a1 ∗a2 L ,b L p ([a2 ,b2 ],C) ],C) ([a q 1 1 (ν1 ) (ν2 )
(21.35)
in particular when h 1 (t1 ) := f 1 (z 1 (t1 )) z 1 (t1 ) and h 2 (t2 ) := f 2 (z 2 (t2 )) z 2 (t2 ), with h i(ki ) (ai ) = 0, ki = 0, 1, . . . n i − 1; i = 1, 2, we get:
b1
a1
b2
a2
f 1 (z 1 (t1 )) z (t1 ) f 2 (z 2 (t2 )) z (t2 ) dt1 dt2 (b1 − a1 ) (b2 − a2 ) 1 2 ≤ · (t1 −a1 ) p(ν1 −1)+1 (t2 −a2 )q (ν2 −1)+1 (ν1 ) (ν2 ) + q(q(ν2 −1)+1) p( p(ν1 −1)+1)
ν D 1 f 1 (z 1 (t1 )) z (t1 ) 1 ∗a1 L Proof By Theorem 21.13.
q ([a1 ,b1 ],C)
ν D 2 f 2 (z 2 (t2 )) z (t2 ) 2 ∗a2 L
. (21.36)
p ([a2 ,b2 ],C)
References 1. G. Anastassiou, A strong fractional calculus theory for Banach space valued functions. Nonlinear Funct. Anal. Appl. 22(3), 495–524 (2017) 2. G. Anastassiou, Complex Opial type inequalities, submitted (2019) 3. G.A. Anastassiou, Complex left Caputo fractional inequalities, submitted for publication (2019) 4. S.S. Dragomir, An extension of Ostrowski inequality to the complex integral. RGMIA Res. Rep. Coll. 21, Art. 112, 17 (2018). https://rgmia.org/v21.php 5. C. Volintiru, A proof of the fundamental theorem of Calculus using Hausdorff measures. R. Anal. Exch. 26(1), 381–390 (2000/2001)
Chapter 22
Complex Opial Inequalities
We establish here complex Opial type inequalities for analytic functions from a complex numbers domain into the set of complex numbers. See also [3].
22.1 Introduction This chapter is greatly motivated by the article of Opial [5]. Theorem 22.1 ([5]) Let x (t) ∈ C 1 ([0, h]) be such that x (0) = x (h) = 0, and x (t) > 0 in (0, h). Then 0
h
x (t) x (t) dt ≤ h 4
In the last inequality the constant Equality holds for the function
h 4
h
2 x (t) dt.
(22.1)
0
is the best possible.
h x (t) = t on 0, 2
and x (t) = h − t on
h ,h . 2
Opial type inequalities have applications in establishing uniqueness of solution to initial value problems in differential equations, see [6], also find upper bounds to such solutions. We are also inspired by the author’s monographs [1, 2], to continue our search for Opial type inequalities in the complex numbers setting. © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_22
393
394
22 Complex Opial Inequalities
22.2 Background See also [4]. Let γ be a smooth path parametrized by z (t), t ∈ [a, b] and f is a complex function which is continuous on γ. Put z (a) = u and z (b) = w with u, w ∈ C. We define the integral of f on γu,w = γ as γ
f (z) dz =
γu,w
b
f (z) dz :=
f (z (t)) z (t) dt.
a
We notice that the actual choice of parametrization of γ does not matter. This definition immediately extends to paths that are piecewise smooth. Suppose γ is parametrized by z (t), t ∈ [a, b], which is differentiable on the intervals [a, c] and [c, b], then assuming that f is continuous on γ we define
γu,w
f (z) dz :=
γu,v
f (z) dz +
γv,w
f (z) dz,
where v := z (c). This can be extended for a finite number of intervals. We also define the integral with respect to arc-length
γu,w
f (z) |dz| :=
b
f (z (t)) z (t) dt
a
and the length of the curve γ is then l (γ) =
γu,w
|dz| :=
b
z (t) dt.
a
We mention also the triangle inequality for the complex integral, namely f (z) dz ≤ | f (z)| |dz| ≤ f γ,∞ l (γ) , γ
(22.2)
γ
where f γ,∞ := sup | f (z)|. z∈γ
S. Dragomir in [4] proved the following useful complex Taylor’s formula with remainder over a non-necessarily convex domain D. Theorem 22.2 Let f : D ⊆ C → C be an analytic function on the domain D and y, x ∈ D. Suppose γ is a smooth path parametrized by z (t), t ∈ [a, b] with z (a) = x and z (b) = y then
22.2 Background
f (y) =
395
n (y − x)k
k!
k=0
f (k) (x) +
1 n!
γx,y
(y − z)n f (n+1) (z) dz,
(22.3)
for n ∈ Z+ .
22.3 Main Results A complex Opial type inequality follows Theorem 22.3 Let f : D ⊆ C → C be an analytic function on the domain D and let x, y, w ∈ D. Suppose γ is a smooth path parametrized by z (t) , t ∈ [a, b] with z (a) = x, z (c) = y, and z (b) = w, where c ∈ [a, b] is floating. Assume that f (k) (x) = 0, k = 0, 1, . . . , n, n ∈ Z+ , and p, q > 1 : 1p + q1 = 1. Then (1)
b
a
f (z (t)) f (n+1) (z (t)) z (t) dt ≤
b
| f (z (t))| f (n+1) (z (t)) z (t) dt ≤
a
1
b
c
1
2 q n!
a
1p pn |z (c) − z (t)| z (t) dt z (c) dc ·
(22.4)
a
b
(n+1) q f (z (t)) z (t) dt
q2
,
a
equivalently it holds (2) (n+1) ≤ f f dz (z) (z) γx,w
1
1
2 q n!
a
b
γx,w
γx,y
|z (c) − z|
pn
| f (z)| f (n+1) (z) |dz| ≤
|dz| z (c) dc
1p γx,w
(n+1) q f (z) |dz|
q2
.
(22.5) Proof By (22.3) we obtain f (y) =
1 n!
γx,y
(y − z)n f (n+1) (z) dz, n ∈ Z+ .
Then by triangle’s and Hölder’s inequalities we have | f (y)| ≤
1 n!
γx,y
|y − z|n f (n+1) (z) |dz| =
396
22 Complex Opial Inequalities
1 n! 1 n!
c
|y − z (t)|n f (n+1) (z (t)) z (t) dt ≤
|y − z (t)| pn z (t) dt
c
(22.6)
a
1p
a
c
(n+1) q f (z (t)) z (t) dt
q1
.
a
We set ρ (c) :=
c
(n+1) q f (z (t)) z (t) dt, a ≤ c ≤ b,
(22.7)
a
then ρ (a) = 0, and q ρ (c) = f (n+1) (z (c)) z (c) ≥ 0. That is
Hence it holds
1 n!
(n+1) 1 1 f (z (c)) z (c) q = ρ (c) q .
(22.8)
1 1 | f (z (c))| f (n+1) (z (c)) z (c) q z (c) p ≤
(22.9)
c
|z (c) − z (t)|
pn
z (t) dt
1p
1 1 ρ (c) ρ (c) q z (c) p .
a
Integrating (22.9) and by Hölder’s inequality we obtain
b
| f (z (c))| f (n+1) (z (c)) z (c) dc ≤
a
1 n! 1 n!
b
b
a
|z (c) − z (t)|
pn
1 p 1 z (t) dt z (c) ρ (c) ρ (c) q dc ≤
a
a
c
c
|z (c) − z (t)|
pn
1p z (t) dt z (c) dc
a
1 n!
b
ρ (c) ρ (c) dc
a
b
a
c
|z (c) − z (t)| pn z (t) dt z (c) dc
a
1p
q1
=
(22.10) ρ (b) 1
2q
2 q
,
proving the claim. We continue with an extreme case. Proposition 22.4 All here are as in Theorem 22.3 but with p = 1, q = ∞. Then
22.3 Main Results
397
(1)
b
f (z (t)) f
(z (t)) z (t) dt ≤
b
(n+1)
a
| f (z (t))| f (n+1) (z (t)) z (t) dt ≤
a
b γx,y
a
2 |z (c) − z|n |dz| z (c) dc f (n+1) γ ,∞ , x,y
equivalently it holds (2) (n+1) f (z) f (z) dz ≤ γx,w
b
γx,w
| f (z)| f (n+1) (z) |dz| ≤
2 |z (c) − z| |dz| z (c) dc f (n+1) γ ,∞ . x,y
n
a
(22.11)
γx,y
(22.12)
Proof By (22.3) we obtain again 1 f (y) = n! Hence it holds | f (y)| ≤
(y − z)n f (n+1) (z) dz, n ∈ N.
γx,y
1 n!
γx,y
|y − z|n f (n+1) (z) |dz| ≤
|y − z| |dz| f (n+1) γ n
γx,y
(22.13)
x,y ,∞
.
(22.14)
Therefore we have | f (y)| f (n+1) (y) ≤
2 |y − z| |dz| f (n+1) γ ,∞ . x,y n
γx,y
(22.15)
That is | f (z (c))| f (n+1) (z (c)) z (c) ≤
2 |z (c) − z| |dz| z (c) f (n+1) γ n
γx,y
x,y ,∞
.
(22.16) Consequently by integration of (22.16) we derive a
b
| f (z (c))| f (n+1) (z (c)) z (c) dc ≤
398
22 Complex Opial Inequalities
b
2 |z (c) − z| |dz| z (c) dc f (n+1) γ ,∞ , x,y
n
γx,y
a
(22.17)
proving the claim. A typical case follows: Corollary 22.5 (to Theorem 22.3 when p = q = 2) We have (1)
a
b
f (z (t)) f (n+1) (z (t)) z (t) dt ≤
b
| f (z (t))| f (n+1) (z (t)) z (t) dt ≤
a
1 √ 2n!
b
a
c
|z (c) − z (t)|2n z (t) dt z (c) dc
21
(22.18) ·
a
b
(n+1) 2 f (z (t)) z (t) dt ,
a
equivalently it holds (2) (n+1) f (z) f (z) dz ≤ γx,w
1 √ 2n!
b
γx,w
| f (z)| f (n+1) (z) |dz| ≤
|z (c) − z| |dz| z (c) dc
21
2n
γx,y
a
γx,w
(n+1) 2 f (z) |dz| . (22.19)
We finish with Corollary 22.6 (to Theorem 22.3, n = 0 case) Here we assume that f (x) = 0. Then | f (z)| f (z) |dz| ≤ f (z) f (z) dz ≤ γx,w
2− q
1
γx,w
γx,w
l γx,z |dz|
1p γx,w
q2 q f (z) |dz| .
(22.20)
References
399
References 1. 2. 3. 4.
A.G. Anastassiou, Fractional Differentiation Inequalities (Springer, Heidelberg, 2009) A.G. Anastassiou, Advanced Inequalities (World Scientific, New Jersey, 2011) G.A. Anastassiou, Complex opial type inequalities, submitted for publication (2019) S.S. Dragomir, An integral representation of the remainder in Taylor’s expansion formula for analytic function on general domains. RGMIA Res. Rep. Coll. 22, Art. 2 (2019), 14 pp, http:// rgmia.org/v22.php 5. Z. Opial, Sur une inegalité. Ann. Polon. Math. 8, 29–32 (1960) 6. D. Willett, The existence-uniqueness theorems for an nth order linear ordinary differential equation. Am. Math. Monthly 75, 174–178 (1968)
Chapter 23
Complex Multivariate Montgomery Identity and Ostrowski and Grüss Inequalities
We give a general complex multivariate Montgomery type identity which is a representation formula for a complex multivariate function. Using it we produce general tight complex multivariate high order Ostrowski and Grüss type inequalities. The estimates involve L p norms, any 1 ≤ p ≤ ∞. We include also applications. See also [1].
23.1 Introduction Our motivation comes from the following results: Theorem 23.1 (Ostrowski [6]) Let f : [a, b] → R be continuous on [a, b] and dif f := ferentiable on b) such that f : b) → R is bounded on b), i.e., (a, (a, (a, ∞ sup f (t) < ∞. Then t∈(a,b)
1 b − a
⎡ 2 ⎤ x − a+b 1 2 ⎦ f (b − a) , f (t) dt − f (x) ≤ ⎣ + ∞ 4 b−a
b
a
for all x ∈ [a, b] and the constant
1 4
(23.1)
is the best possible.
Theorem 23.2 (Grüss, 1934 [4]) Let f, g : [a, b] → R be Lebesgue integrable functions, and m, M, n, N ∈ R such that: −∞ < m ≤ f ≤ M < ∞, −∞ < n ≤ g ≤ N < ∞, a.e. on [a, b]. Then 1 b − a
b
f (t) g (t) dt −
a
≤
1 b−a
b
f (t) dt a
1 b−a
a
b
g (t) dt
1 (M − m) (N − n) , 4
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_23
(23.2) 401
402
23 Complex Multivariate Montgomery Identity and Ostrowski …
with the constant
1 4
being the best possible.
Let f ∈ C 1 ([a, b]) and the kernel p : [a, b]2 → R be such that
p (x, t) :=
t − a, if t ∈ [a, x] , t − b, if t ∈ (x, b].
(23.3)
Then, we have the basic Montgomery integral identity [5, p. 565], f (x) =
1 b−a
b
f (t) dt +
a
1 b−a
b
p (x, t) f (t) dt, ∀ x ∈ [a, b] . (23.4)
a
In order to describe complex extensions of Ostrowski and Grüss inequalities using the complex integral we need the following preparation. Suppose γ is a smooth path parametrized by z (t), t ∈ [a, b] and f is a complex function which is continuous on γ. Put z (a) = u and z (b) = w with u, w ∈ C. We define the integral of f on γu,w = γ as γ
f (z) dz =
γu,w
b
f (z) dz :=
f (z (t)) z (t) dt.
(23.5)
a
We observe that the actual choice of parametrization of γ does not matter. This definition immediately extends to paths that are piecewise smooth. Suppose γ is parametrized by z (t), t ∈ [a, b], which is differentiable on the intervals [a, c] and [c, b], then assuming that f is continuous on γ we define
f (z) dz :=
γu,w
γu,v
f (z) dz +
γv,w
f (z) dz,
(23.6)
where v := z (c). This can be extended for a finite number of intervals. We also define the integral with respect to arc-length
γu,w
f (z) |dz| :=
b
f (z (t)) z (t) dt
(23.7)
a
and the length of the curve γ is then l (γ) =
γu,w
|dz| :=
b
z (t) dt.
(23.8)
a
Let f and g be holomorphic in G, an open domain and suppose γ ⊂ G is a piecewise smooth path from z (a) = u to z (b) = w. Then we have the integration by parts formula
23.1 Introduction
γu,w
403
f (z) g (z) dz = f (w) g (w) − f (u) g (u) −
γu,w
f (z) g (z) dz.
(23.9)
We recall also the triangle inequality for the complex integral, namely f (z) dz ≤ | f (z)| |dz| ≤ f γ,∞ l (γ) , γ
(23.10)
γ
where f γ,∞ := sup | f (z)|. z∈γ
We also define the p-norm with p ≥ 1 by f γ, p := For p = 1 we have
f γ,1 :=
If p, q > 1 with
γ
1p | f (z)| p |dz| .
1 p
+
1 q
γ
| f (z)| |dz| .
= 1, then by Hölder’s inequality we have 1
f γ,1 ≤ [l (γ)] q f γ, p .
(23.11)
First, we mention a Complex extension of Ostrowski inequality to the complex integral by providing upper bounds for the quantity f (v) (w − u) − f dz (z) γ
under the assumption that γ is a smooth path parametrized by z (t), t ∈ [a, b], u = z (a), v = z (x) with x ∈ (a, b) and w = z (b) while f is holomorphic in G, an open domain and γ ⊂ G. Secondly, we mention a Complex extension of Grüss inequality: Suppose γ ⊂ C is a piecewise smooth path parametrized by z (t), t ∈ [a, b] from z (a) = u to z (b) = w with w = u. If f and g are continuous on γ, we consider the ˇ complex Cebyšev functional defined by Dγ ( f, g) :=
1 w−u
γ
f (z) g (z) dz −
1 w−u
γ
f (z) dz
1 w−u
We display upper bounds to Dγ ( f, g) . We have the following results for functions of a complex variable:
γ
g (z) dz. (23.12)
Theorem 23.3 (Dragomir, 2019 [2]) Let f be holomorphic in G, an open domain and suppose γ ⊂ G is a smooth path from z (a) = u to z (b) = w. If v = z (x) with
404
23 Complex Multivariate Montgomery Identity and Ostrowski …
x ∈ (a, b), then γu,w = γu,v ∪ γv,w , f (v) (w − u) − f (z) dz ≤ γ
f
γu,v ;∞
γu,v
γv,w ;∞
γu,v
|z − u| |dz| + f
|z − u| |dz| +
γu,w
γu,w
|z − w| |dz| ≤
|z − w| |dz| f γ
(23.13)
,
u,w ;∞
f (v) (w − u) − f (z) dz ≤
and
γ
max |z − u| f γu,v ;1 + max |z − w| f γv,w ;1 ≤
z∈γu,v
(23.14)
z∈γv,w
max max |z − u| , max |z − w| f γu,w ;1 .
z∈γu,v
If p, q > 1 with
1 p
+
1 q
z∈γv,w
= 1, then f (v) (w − u) − f (z) dz ≤ γ
q1 |z − u| |dz| f γu,v ; p + q
γu,v
|z − w| |dz| q
γu,w
γu,v
|z − u|q |dz| +
γu,w
|z − w|q |dz|
q1
q1
f
γv,w ; p
f
γu,w ; p
≤ (23.15)
.
Suppose γ ⊂ C is a piecewise smooth path parametrized by z (t), t ∈ [a, b] from z (a) = u to z (b) = w. Now, for φ, ∈ C define the set of complex-valued functions φ + 1 | ≤ − φ| for each z ∈ γ . f : γ → C| f (z) − 2 2 (23.16) We have the following complex Grüss type inequalities:
γ (φ, ) :=
Theorem 23.4 (Dragomir [3]) Suppose γ ⊂ C is a piecewise smooth path parametrized by z (t), t ∈ [a, b] from z (a) = u to z (b) = w with w = u. If f and g are continuous on γ and there exist φ, , ψ, ∈ C, φ = , ψ = such that f ∈ γ (φ, ) and g ∈ γ (ψ, ) then
23.1 Introduction
405 2 Dγ ( f, g) ≤ 1 | − φ| | − ψ| l (γ) . 4 |w − u|2
(23.17)
If the path γ is a segment w [u, w] connecting two distinct points u and w in C then we write γ f (z) dz as u f (z) dz. If f , g are continuous on [u, w] and there exists φ, , ψ, ∈ C, φ = , ψ = such that f ∈ [u,w] (φ, ) and g ∈ [u,w] (ψ, ) then 1 w − u
w
f (z) g (z) dz −
u
≤
1 w−u
w
f (z) dz
u
1 w−u
w
u
g (z) dz
1 | − φ| | − ψ| . 4
(23.18)
We will use the complex Montgomery identity which follows: Theorem 23.5 (Dragomir [2]) Let f be holomorphic in G, an open domain and suppose γ ⊂ G is a smooth path from z (a) = u to z (b) = w. If v = z (t) with t ∈ [a, b], then γu,w = γu,v ∪ γv,w , and 1 f (v) = w−u
γ
f (z) dz +
γu,v
Define
(z − u) f (z) dz +
p (v, z) :=
γv,w
(z − w) f (z) dz .
z − u, if z ∈ γu,v z − w, if z ∈ γv,w .
(23.19) (23.20)
Thus, it holds 1 f (v) = w−u
γ
1 f (z) dz + w−u
γ
p (v, z) f (z) dz,
(23.21)
a form which we will use a lot in this chapter. Representation formula (23.21) is the main inspiration to write this chapter. We will use (23.21) to derive a multivariate Complex Montgomery type identity then based on it, we will produce Complex multivariate Ostrowski and Grüss type inequalities. For the last we need: Definition 23.6 Here we extend the notion of line (curve) integral into multivariate case. Let γ j , j = 1, . . . , m, be a smooth path parametrized by z j t j , t j ∈ a j , b j m γ j ⊆ Cm . Put and f is a complex valued function which is continuous on j=1 z j a j = u j and z j b j = w j , with u j , w j ∈ C, j = 1, . . . , m.
406
23 Complex Multivariate Montgomery Identity and Ostrowski …
We define the complex multivariate integral of f on
m
γ j :=
j=1
...
γ1
m
γu j ,w j as
j=1
γm
f (z 1 , . . . , z m ) dz 1 . . . dz m :=
m
f (z 1 , . . . , z m ) dz 1 . . . dz m :=
γj
j=1
γu 1 ,w1
...
f (z 1 , . . . , z m ) dz 1 . . . dz m :=
γu m ,wm
b1
a1
b2
m j=1
...
a2
bm
f (z 1 (t1 ) , . . . , z m (tm ))
am
γu j ,w j
m
f (z 1 , . . . , z m ) dz 1 . . . dz m :=
z j t j dt1 . . . dtm .
(23.22)
j=1
We make a j , b j , C , j = 1, . . . , m. The integrand in m a j , b j . Therefore (23.22) is a continuous complex valued function over
Remark 23.7 Clearly here z j ∈ C 1
| f (z 1 (t1 ) , . . . , z m (tm ))|
m j=1
j=1
z j
m t j is also continuous but from a j , b j into R, j=1
hence it is bounded. Consequently it holds m j=1
| f (z 1 (t1 ) , . . . , z m (tm ))|
[a j ,b j ]
m m z t j dt j < +∞. j j=1
j=1
Therefore, by Fubini’s theorem, the order integration in (23.22) is immaterial. Clearly it holds b1 bm m ≤ t dt . . . f z . . . dt , . . . , z (z (t ) (t )) 1 1 m m j 1 m j am a1 j=1
b1
...
a1
bm
| f (z 1 (t1 ) , . . . , z m (tm ))|
am
m z t j dt1 . . . dtm . j
(23.23)
j=1
We also define the integral with respect to arc-lengths m j=1
γu j ,w j
f (z 1 , . . . , z m ) |dz 1 | |dz 2 | . . . |dz m | :=
m
[a j ,b j ]
j=1
f (z 1 (t1 ) , . . . , z m (tm ))
m z t j dt1 . . . dtm . j j=1
(23.24)
23.1 Introduction
407
It holds (by (23.23), (23.24)) m f (z 1 , . . . , z m ) dz 1 . . . dz m ≤ γj j=1
m j=1
γu j ,w j
m | f (z 1 , . . . , z m )| |dz 1 | |dz 2 | . . . |dz m | ≤ f
γ j ,∞
j=1
m l γj ,
(23.25)
j=1
where m f
γ j ,∞
:=
sup (z 1 ,...,z m )∈
j=1
m
| f (z 1 , . . . , z m )| , γj
j=1
and
l γj =
dz j =
γu j ,w j
bj
aj
z t j dt j , j = 1, . . . , m. j
We also define the p-norm with p ≥ 1 by ⎛ m f
γj,p
:= ⎝
⎞ 1p
m
j=1
γj
| f (z 1 , . . . , z m )| p |dz 1 | |dz 2 | . . . |dz m |⎠ .
j=1
For p = 1 we have m f
γ j ,1
:=
j=1
If p, q > 1 with
1 p
+
m
γj
| f (z 1 , . . . , z m )| |dz 1 | |dz 2 | . . . |dz m | .
j=1
1 q
= 1, then by Hölder’s inequality we have ⎞ q1 m m ≤ ⎝ l γj ⎠ f ⎛
m f j=1
γ j ,1
j=1
γj,p
.
(23.26)
j=1
23.2 Main Results We start by presenting a complex trivariate Montgomery type representation identity of complex functions:
408
23 Complex Multivariate Montgomery Identity and Ostrowski …
Theorem 23.8 Let f :
3
D j ⊆ C3 → C be a continuous function that is analytic
j=1 3 per coordinate on the domain D j , j = 1, 2, 3, and x = (x1 , x2 , x3 ) ∈ D j . For j=1 j = 1, 2, 3, suppose γ j ⊂ D j is a smooth path parametrized by z j t j , t j ∈ a j , b j with z j a j = u j , z j t j = x j and z j b j = w j , where u j , w j ∈ D j , u j = w j . Assume also that all partial derivatives of f up to order three are continuous functions 3 Dj. on j=1
Here we define the kernels for i = 1, 2, 3, pi : γi2 → C
pi (xi , si ) :=
si − u i , if si ∈ γu i ,xi , si − wi , if si ∈ γxi ,wi .
(23.27)
Then
1
f (x1 , x2 , x3 ) =
3
γ1
(wi − u i )
γ2
γ3
f (s1 , s2 , s3 ) ds3 ds2 ds1
i=1
+
3 γ1
j=1
+
3 l=1 j 1 : 1 = 1. Then q E f (x1 , x2 ) ≤
1 p
+
⎧ ⎨
⎡ q1 q1 ⎤ 1 ⎣ |z 1 − u 1 |q |dz 1 | + |z 1 − w1 |q |dz 1 | ⎦ ⎩ |w1 − u 1 | γu 1 ,x1 γx1 ,w1 ∂f ∂ x (·, x2 ) 1 ∞,γu ⎡ ⎣
"
+
1 ,w1
1 1 (l (γ1 )) q |w1 − u 1 | |w2 − u 2 |
q1
γu 2 ,x2
|z 2 − u 2 |q |dz 2 |
+
γx2 ,w2
∂f ∂ x (·, ·) 2 2 ∞, γu j=1
q1 ⎤ |z 2 − w2 |q |dz 2 | ⎦ ⎫ ⎪ ⎬
j ,w j
(24.86)
⎪ ⎭
,
∀ (x1 , x2 ) ∈ γ1 × γ2 . Corollary 24.24 (to Theorem 24.14) All as in Corollary 24.21. Then E f (x1 , x2 ) ≤
1 |w1 − u 1 |
max {|z 1 − u 1 |} + max {|z 1 − w1 |}
z 1 ∈γu 1 ,x1
∂f ∂ x (·, x2 ) 1 1,γu
z 1 ∈γx1 ,w1
+
1 ,w1
1 |w1 − u 1 | |w2 − u 2 |
(24.87)
24.3 Applications
453
∂f max {|z 2 − u 2 |} + max {|z 2 − w2 |} (·, ·) 2 z 2 ∈γu 2 ,x2 z 2 ∈γx2 ,w2 ∂ x2 1, γu j=1
⎤ ⎥ ⎦, j ,w j
∀ (x1 , x2 ) ∈ γ1 × γ2 . We continue with complex bivariate Grüss type inequalities: Corollary 24.25 (to Theorem 24.17) Let f, g and all as in Corollary 24.21. Then γ1
1 (w1 − u 1 ) (w2 − u 2 )
γ2
f (x1 , x2 ) g (x1 , x2 ) d x1 d x2 −
γ1
γ2
f (x1 , x2 ) d x1 d x2
γ1
γ2
g (x1 , x2 ) d x1 d x2 ≤
1 f g∞,γ1 ×γ2 B1 (x1 , x2 ) |d x1 | |d x2 | + 2 γ1 γ2 l (γ1 ) γ1
γ2
f + f ∞,γ1 ×γ2 B2 (x2 ) |d x2 | γ2
g g B (x1 , x2 ) |d x1 | |d x2 | + l (γ1 ) B (x2 ) |d x2 | . 1 2
(24.88)
γ2
We continue with Corollary 24.26 (to Theorem 24.19) Let f, g and all as in Corollary 24.21, and p, q > 1 : 1p + q1 = 1. Then γ1
γ2
f (x1 , x2 ) g (x1 , x2 ) d x1 d x2 −
1 f (x1 , x2 ) d x1 d x2 g (x1 , x2 ) d x1 d x2 ≤ (w1 − u 1 ) (w2 − u 2 ) γ1 γ2 γ1 γ2 (24.89) ⎧ ⎡ ⎡⎛ ⎞ q1 ⎛ ⎞ q1 ⎪ 2 i−1 q 1 ⎢ ⎢⎝ ⎠ ⎨ g p,γ1 γ2 ⎝ 2 Bi f |d xi | . . . |d x2 |⎠ l γj ⎣ ⎣ ⎪ 2 i=1 γ ⎩ j j=1 j=i
+ f p,γ1 ×γ2
⎛ ⎝ 2
j=i
γj
⎞ q1 ⎫⎤⎤ ⎪ ⎬ g q ⎥ B |d xi | . . . |d x2 |⎠ ⎥ ⎦⎦ . i ⎪ ⎭
454
24 Complex Multivariate Fink Identity and Complex Multivariate …
Corollary 24.27 (to Theorem 24.20) Let f, g and all as in Corollary 24.21. Then γ1
1 (w1 − u 1 ) (w2 − u 2 )
γ2
f (x1 , x2 ) g (x1 , x2 ) d x1 d x2 −
γ1
γ2
f (x1 , x2 ) d x1 d x2
γ1
γ2
g (x1 , x2 ) d x1 d x2 ≤
2 2 g 1 f B g1,γ1 ×γ2 + f 1,γ1 ×γ2 . Bi i ∞,γ1 ×γ2 ∞,γ1 ×γ2 2 i=1 i=1 (24.90) (II) Next we treat the case of n = 2, m = 3. Thus, Theorem 24.9 will read as follows: Corollary 24.28 Let f :
3
D j ⊆ C3 → C be a continuous function that is analytic
j=1 3 per coordinate on the domain D j , j = 1, 2, 3, and x = (x1 , x2 , x3 ) ∈ D j . For j=1 j = 1, 2, 3,suppose γ j ⊂ D j is a smooth path parametrized by z j t j , t j ∈ a j , b j with z j a j = u j , z j t j = x j and z j b j = w j , where u j , w j ∈ D j , u j = w j . Assume also that all single partial derivatives of f up to order 2 are continuous 3 Dj. functions on j=1
Then 8
f (x1 , x2 , x3 ) =
3 wj − u j
3
γj
f (z 1 , z 2 , z 3 ) dz 1 dz 2 dz 3 +
3
Ti , (24.91)
i=1
j=1
j=1
where for i = 1, 2, 3 we set Ti := Ti (xi , . . . , x3 ) :=
2i−1 i
wj − u j
j=1
⎧ ⎨ ⎩
i−1
γj
(xi − wi ) f (z 1 , . . . , z i−1 , wi , xi+1 , . . . , x3 ) −
j=1
0 (xi − u i ) f (z 1 , . . . , z i−1 , u i , xi+1 , . . . , x3 ) dz 1 . . . dz i−1 +
24.3 Applications
2i−1 i wj − u j
455
⎛ ⎝ i−1
# γj
j=1
γu i ,xi
(xi − z i ) (z i − u i )
∂2 f (z 1 , . . . , z i , xi+1 , . . . , x3 ) dz i ∂ xi2
j=1
+
γxi ,wi
(24.92) " ∂2 f (xi − z i ) (z i − wi ) 2 (z 1 , . . . , z i , xi+1 , . . . , x3 ) dz i dz 1 . . . dz i−1 . ∂ xi
We make Remark 24.29 For i = 1, 2, 3, denote Ai := Ai (xi , . . . , x3 ) :=
2i−1 i wj − u j
(24.93)
j=1
⎧ ⎨ ⎩
i−1
γj
(xi − wi ) f (z 1 , . . . , z i−1 , wi , xi+1 , . . . , x3 ) −
j=1
0 (xi − u i ) f (z 1 , . . . , z i−1 , u i , xi+1 , . . . , x3 ) dz 1 . . . dz i−1 , and Bi := Bi (xi , . . . , x3 ) :=
2i−1 i
wj − u j
j=1
⎛ ⎝
#
i−1 j=1
γxi ,wi
γj
γu i ,xi
(xi − z i ) (z i − u i )
∂2 f (z 1 , . . . , z i , xi+1 , . . . , x3 ) dz i + ∂ xi2
" ∂2 f (xi − z i ) (z i − wi ) 2 (z 1 , . . . , z i , xi+1 , . . . , x3 ) dz i dz 1 . . . dz i−1 . ∂ xi (24.94)
That is Ti = Ai + Bi , i = 1, 2, 3. Thus, by (24.91), we derive E f (x1 , x2 , x3 ) := f (x1 , x2 , x3 ) −
(24.95)
456
24 Complex Multivariate Fink Identity and Complex Multivariate …
8 3
wj − u j
3
γj
f (z 1 , z 2 , z 3 ) dz 1 dz 2 dz 3 −
3
Ai =
i=1
j=1
3
Bi .
(24.96)
i=1
j=1
Hence
3 E f (x1 , x2 , x3 ) ≤ |Bi | .
(24.97)
i=1
A complex trivariate Ostrowski type inequality follows: Corollary 24.30 (to Theorem 24.11) All as in Corollary 24.28. Then ⎧ ⎛ ⎞ 3 ⎨ i−1 i−1 l γj 2 E f (x1 , x2 , x3 ) ≤ ⎝ ⎠ w j − u j ⎩ |wi − u i | i=1
j=1
|xi − z i | |z i − u i | |dz i | +
γu i ,xi
γxi ,wi
|xi − z i | |z i − wi | |dz i |
2 ∂ f , x x (·, ) i+1 3 ∂x2 i i ∞, γu j=1
∀ (x1 , x2 , x3 ) ∈
3
⎫ ⎪ ⎬ j ,w j
⎪ ⎭
,
(24.98)
γj.
j=1
A trivariate complex Grüss type inequality follows: Corollary 24.31 (to Theorem 24.17) Let f, g and all as in Corollary 24.28. Then 3 f (x1 , x2 , x3 ) g (x1 , x2 , x3 ) d x1 d x2 d x3 − γj j=1
⎛ 8 3
wj − u j
⎞⎛
⎜ ⎝ 3
⎞
γj
⎟⎜ f (x1 , x2 , x3 ) d x1 d x2 d x3 ⎠ ⎝ 3
j=1
γj
⎟ g (x1 , x2 , x3 ) d x1 d x2 d x3 ⎠
j=1
j=1
⎡
#
1 − ⎣ 2
3 j=1
γj
g (x)
3 i=1
f
Ai
+ f (x)
3 i=1
" g
Ai
⎤ d x1 d x2 d x3 ⎦ ≤
(24.99)
24.3 Applications
457
⎞⎛ ⎞⎧ ⎛ ⎞⎫ ⎡⎛ 3 i−1 ⎨ ⎬ 1 ⎣⎝ ⎝ l γ j ⎠ ⎝ 3 Bi f (xi , . . . , x3 ) |d xi | . . . |d x3 |⎠ g 3 ⎠ ⎩ ⎭ 2 ∞, γ j γj i=1 j=1 j=1
j=i
⎞⎛ ⎞⎧ ⎛ ⎞⎫⎤ 3 i−1 ⎨ ⎬ g ⎝ l γ j ⎠ ⎝ 3 Bi (xi , . . . , x3 ) |d xi | . . . |d x3 |⎠ ⎦ . + ⎝ f 3 ⎠ ⎩ ⎭ ∞, γ j γj ⎛
i=1
j=1
j=1
j=i m $
(III) Next we apply Theorem 24.9 to f (x1 , . . . , xm ) := e Cm . We make
xj
j=1
, x := (x1 , .., xm ) ∈
Remark 24.32 By (24.28) we get m $
xj
e j=1
=
nm m wj − u j
m $ m
γj
zj
e j=1 dz 1 . . . dz m +
m
Ti∗∗ ,
(24.100)
i=1
j=1
j=1
where for i = 1, . . . , m, we set Ti∗∗
:=
Ti∗∗
n−1 n−k
n i−1
(xi , . . . , xm ) :=
i wj − u j
k!
k=1
×
j=1
⎧ ⎨ ⎩
⎡ i−1
⎛
i−1 $
⎣(xi − wi ) ⎝e j=1 ⎠ e k
γj
⎞ zj
wi
m $
xj
e j=i+1
−
j=1
⎛
i−1 $
⎞
zj
(xi − u i )k ⎝e j=1 ⎠ eu i ⎛ n i−1 (n − 1)!
i wj − u j
⎜ ⎝
⎧ ⎪ ⎨
i−1 j=1
γj
⎪ ⎩
m $
xj
e j=i+1
⎫ ⎤ ⎬ ⎦ dz 1 . . . dz i−1 + ⎭ ⎛
γu i ,xi
(xi − z i )
n−1
⎞⎛
⎞ m $ ⎜ j=1 ⎟ ⎝ j=i+1 x j ⎠ dz i (z i − u i ) ⎝e ⎠ e i $
zj
j=1
⎛
+
γxi ,wi
i $
zj
⎞
m $
(xi − z i )n−1 (z i − wi ) ⎝e j=1 ⎠ e j=i+1
xj
⎫ ⎬
dz i
⎭
⎞ dz 1 . . . dz i−1 ⎠ . (24.101)
458
24 Complex Multivariate Fink Identity and Complex Multivariate …
Above C is a smooth path parametrized by z j t j ,
for j = 1, . . . , m, suppose γj ⊂ t j ∈ a j , b j with z j a j = u j , z j t j = x j and z j b j = w j , where u j , w j ∈ C, u j = w j . One can rewrite (24.100) as follows: m $
e
xj
j=1
=
m
nm m wj − u j
zj
γj
j=1
e dz j
+
m
Ti∗∗ .
(24.102)
i=1
j=1
And Ti∗∗ can be rewritten as follows: ⎡
m $
Ti∗∗
xj
⎞ ⎛ ⎢n e j=i+1 i−1 n−1 ⎢ n−k ⎢ z j ⎠ ⎝ × =⎢ i e dz j ⎢ k! j=1 γ j k=1 ⎣ wj − u j i−1
j=1
/
ewi (xi − wi )k − eu i (xi − u i )k
⎡
m $
xj
i−1
0
+
zj ⎢ n i−1 e j=i+1 γ j e dz j ⎢ j=1 ⎢ × ⎢ i ⎢ ⎣ wj − u j (n − 1)! j=1
#
"
γu i ,xi
(xi − z i )n−1 (z i − u i ) e zi dz i +
γxi ,wi
(xi − z i )n−1 (z i − wi ) e zi dz i
.
(24.103) Call
m $
n i−1 e j=i+1 Ai∗∗
:=
Ai∗∗
(xi , . . . , xm ) :=
xj
i wj − u j
⎛ ⎝
i−1 j=1 γ j
⎞ e dz j ⎠ zj
n−1 n−k k=1
k!
×
j=1
wi e (xi − wi )k − eu i (xi − u i )k , and
(24.104)
24.3 Applications
459
m $
n i−1 e j=i+1
xj
i−1
γj
j=1
Bi∗∗ := Bi∗∗ (xi , . . . , xm ) := (n − 1)!
e z j dz j
i wj − u j j=1
γu i ,xi
(xi − z i )n−1 (z i − u i ) e zi dz i +
γxi ,wi
(xi − z i )n−1 (z i − wi ) e zi dz i , (24.105)
for i = 1, . . . , m. We have
Ti∗∗ = Ai∗∗ + Bi∗∗ , i = 1, . . . , m.
(24.106)
Call m $
E
m $
e j=1
(x1 , . . . , xm ) := e
xj
⎛ xj
j=1
−
nm m wj − u j
⎝
⎞
m
e dz j ⎠ − zj
j=1 γ j
m
Ai∗∗ =
i=1
m
Bi∗∗ .
i=1
j=1
(24.107) E
Hence
m $
e j=1
xj
m ∗∗ B . (x1 , . . . , xm ) ≤ i
(24.108)
i=1
We present a related multivariate complex Ostrowski type inequality: m $
xj
Proposition 1 All are as in Theorem 24.9 for f (x1 , . . . , xm ) := e , x := (x1 , .., xm ) ∈ Cm . Then ⎧ ⎛ ⎞ m ⎨ i−1 i−1 l γj 1 n m ⎝ ⎠ E $ x j (x1 , . . . , xm ) ≤ w j − u j (n − 1)! j=1 ⎩ |wi − u i | e i=1 j=1 j=1
γu i ,xi
|xi − z i |
n−1
|z i − u i | |dz i | + $ m xj j=1 e
∞,
γxi ,wi
⎫ ⎪ ⎪ ⎬ i j=1
∀ (x1 , . . . , xm ) ∈
m j=1
γj.
|xi − z i |
⎪
⎭ γu j ,w j ⎪
,
n−1
|z i − wi | |dz i |
(24.109)
460
24 Complex Multivariate Fink Identity and Complex Multivariate …
Proof Apply Theorem 24.11. We make Remark 24.33 Here things can be simplified further: From (24.102) we get m $
e
xj
j=1
= nm
m m wj e − eu j + Ti∗∗ , w − u j j j=1 i=1
(24.110)
where, for i = 1, . . . , m, we have ⎡
Ti∗∗
m $
xj
⎞ ⎛ ⎢n e i−1 n−1 ⎢ n−k ⎢ ⎝ × ew j − eu j ⎠ =⎢ i ⎢ k! j=1 k=1 ⎣ wj − u j i−1
j=i+1
j=1
/
ewi (xi − wi )k − eu i (xi − u i )k
⎡
⎢n ⎢ ⎢ ⎢ ⎢ ⎣
i−1
m $
e
xj
j=i+1
i−1
0
+
(e
wj
−e ) uj
j=1
×
i wj − u j (n − 1)! j=1
# γu i ,xi
"
(xi − z i )
n−1
(z i − u i ) e dz i + zi
γxi ,wi
(xi − z i )
n−1
(z i − wi ) e dz i zi
.
(24.111) Thus we find n Ai∗∗ =
i−1
m $
xj
e j=i+1
i wj − u j
⎞ i−1 n−1 wj n−k ⎝ × e − eu j ⎠ k! j=1 k=1 ⎛
j=1
/
0 ewi (xi − wi )k − eu i (xi − u i )k ,
and
n Bi∗∗
i−1
m $
xj
e j=i+1
i−1
(e
wj
−e )
j=1
= (n − 1)!
i wj − u j j=1
(24.112)
uj
24.3 Applications
461
#
"
γu i ,xi
(xi − z i )
n−1
(z i − u i ) e dz i + zi
γxi ,wi
(xi − z i )
n−1
(z i − wi ) e dz i , zi
(24.113) for i = 1, . . . , m. Furthermore it holds m $
E
m $
e j=1
xj
(x1 , . . . , xm ) := e
j=1
xj
− nm
m m m wj e − eu j − Ai∗∗ = Bi∗∗ . w − u j j j=1 i=1 i=1 (24.114)
References 1. G.A. Anastassiou, Complex multivariate Fink type identity applied to complex multivariate Ostrowski and Grüss inequalities. Indian J. Math. 61(2), 199–237 (2019) 2. S.S. Dragomir, An extension of Ostrowski’s inequality to the complex integral. RGMIA Res. Rep. Call. 21, Art. 112 (2018), 17 pp, http://rgmia.org/papers/v21/v21/v21a112.pdf 3. S.S. Dragomir, On some Grüss type inequalities for the complex integral. RGMIA Res. Rep. Call. 21, Art. 121 (2018), 12 pp, http://rgmia.org/papers/v21/v21a121.pdf 4. S.S. Dragomir, An identity of Fink type for the integral of analytic complex function on paths from general domains. RGMIA Res. Rep. Call. 22, Art. 11 (2019), 18 pp, http://rgmia.org/ papers/v22/v22a11.pdf 5. A.M. Fink, Bounds on the deviation of a function from its averages. Czechoslo. Math. J. 42(117)(2), 289–310 (1992) b 1 6. G. Grüss, Über das Maximum des absoluten Betrages von b−a a f (x) g (x) d x − b b 1 f d x g d x. Math. Z. 39, 215–226 (1935) (x) (x) 2 a a (b−a)
7. A. Ostrowski, Über die Absolutabweichung einer differentiebaren Funktion von ihrem Integral mittelwert. Comment. Math. Helv. 10, 226–227 (1938)
Chapter 25
Fractional Conformable Approximation of Csiszar’s f -Divergence
Here are given tight probabilistic inequalities that give nearly best estimates for the Csiszar’s f -divergence. These use the right and left conformable fractional derivatives of the directing function f . Csiszar’s f -divergence or the so called Csiszar’s discrimination is used as a measure of dependence between two random variables which is a very important aspect of stochastics, we apply our results there. The Csiszar’s discrimination is the most essential and general measure for the comparison between two probability measures. See also [4].
25.1 Background—I Throughout this chapter we use the following. Let f be a convex function from (0, +∞) into R that is strictly convex at 1 with f (1) = 0. Let (X, A, λ) be a measure space, where λ is a finite or a σ-finite measure on (X, A). And let μ1 , μ2 be two probability measures on (X, A) such that μ1 λ, 1 , μ2 λ (absolutely continuous); for example, λ = μ1 + μ2 . Denote by p = dμ dλ dμ2 q = dλ (densities) Radon–Nikodym derivatives of μ1 , μ2 with respect to λ. Here we assume that p 0 < a ≤ ≤ b, a.e. on X q and a ≤ 1 ≤ b. The quantity
f (μ1 , μ2 ) =
q (x) f X
p (x) dλ (x) , q (x)
(25.1)
was introduced by Csiszar in 1967 (see [7]), and is called f -divergence of the probability measures μ1 and μ2 . By Lemma 1.1 of [7], the integral (25.1) is well defined and f (μ1 , μ2 ) ≥ 0 with equality only when μ1 = μ2 . In [7] the author without proof mentions that f (μ1 , μ2 ) does not depend on the choice of λ. © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_25
463
464
25 Fractional Conformable Approximation of Csiszar’s F-Divergence
For a proof of the last see [2], Lemma 1.1. The concept of f -divergence was introduced first in [6] as a generalization of Kullback’s “information for discrimination” or I -divergence (generalized entropy) [10, 11] and of Rényi’s “information gain” (I -divergence of order α) [13]. In fact the I -divergence of order 1 equals u log2 u (μ1 , μ2 ) . The choice f (u) = (u − 1)2 again produces a known measure of difference of distributions called χ2 -divergence; of course the total variation distance |μ1 − μ2 | = | X p (x) − q (x)| dλ (x) equals |u−1| (μ1 , μ2 ). Here by assuming f (1) = 0 we can consider f (μ1 , μ2 ) as a measure of the difference between the probability measures μ1 , μ2 . The f -divergence is in general asymmetric in μ1 and μ2 . But because f is convex and strictly convex at 1 (see Lemma 2, [2]) so is 1 ∗ (25.2) f (u) = u f u and as in [7] we get f (μ2 , μ1 ) = f ∗ (μ1 , μ2 ) .
(25.3)
In information theory and statistics many other concrete divergences are used that are special cases of the above general Csiszar f -divergence, such as Hellinger distance D H , α-divergence Dα , Bhattacharyya distance D B , harmonic distance D Hα , Jeffrey’s distance D J , and triangular discrimination D ; for all these, see, for example [5, 8]. The problem of finding and estimating the proper distance (or difference or discrimination) of two probability distributions is one of the major questions in probability theory. The above f -divergence measures in their various forms have also been applied to anthropology, genetics, finance, economics, political science, biology, approximation of probability distributions, signal processing, and pattern recognition. A great inspiration for this chapter has been the very important monograph on the topic by Dragomir [8].
25.2 Background—II Here we follow [1]. We need Definition 25.1 ([1, 9]) Let a, b ∈ R. The left conformable fractional derivative starting from a of a function f : [a, ∞) → R of order 0 < α ≤ 1 is defined by
Tαa
f t + ε (t − a)1−α − f (t) f (t) = lim . ε→0 ε
(25.4)
25.2 Background—II
465
If Tαa f (t) exists on (a, b), then
Tαa f (a) = lim Tαa f (t) . t→a+
(25.5)
The right conformable fractional derivative of order 0 < α ≤ 1 terminating at b of f : (−∞, b] → R is defined by f t + ε (b − t)1−α − f (t) . α T f (t) = −lim ε→0 ε
b If
b
αT
(25.6)
f (t) exists on (a, b), then b
αT
f (b) = lim bα T f (t) . t→b−
(25.7)
Note that if f is differentiable then and
Tαa f (t) = (t − a)1−α f (t) ,
b
αT
Denote by
and
Iαa
b
f (t) = − (b − t)1−α f (t) .
t
f (t) =
(x − a)α−1 f (x) d x,
(25.8)
(25.9)
(25.10)
a
Iα f (t) =
b
(b − x)α−1 f (x) d x,
(25.11)
t
these are the left and right conformable fractional integrals of order 0 < α ≤ 1. In the higher order case we can generalize things as follows: Definition 25.2 ([1]) Let α ∈ (n, n + 1], and set β = α − n. Then, the left conformable fractional derivative starting from a of a function f : [a, ∞) → R of order α, where f (n) (t) exists, is defined by a Tα f (t) = Tβa f (n) (t) .
(25.12)
The right conformable fractional derivative of order α terminating at b of f : (−∞, b] → R, where f (n) (t) exists, is defined by b
αT
f (t) = (−1)n+1 bβ T f (n) (t) .
If α = n + 1 then β = 1 and Tan+1 f = f (n+1) .
(25.13)
466
25 Fractional Conformable Approximation of Csiszar’s F-Divergence
If n is odd, then bn+1 T f = − f (n+1) , and if n is even, then bn+1 T f = f (n+1) . When n = 0 (or α ∈ (0, 1]), then β = α, and (25.12), (25.13) collapse to {(25.4), (25.5)}, {(25.6), (25.7)} respectively. Lemma 25.3 ([1]) Let f : (a, b) → R be continuously differentiable and 0 < α ≤ 1. Then, for all t > a we have Iαa Tαa ( f ) (t) = f (t) − f (a) .
(25.14)
We need Definition 25.4 (see also [1]) If α ∈ (n, n + 1], then the left fractional integral of order α starting at a is defined by
1 Iαa f (t) = n!
t
(t − x)n (x − a)β−1 f (x) d x.
(25.15)
a
Similarly, (author’s definition) the right fractional integral of order α terminating at b is defined by b
1 Iα f (t) = n!
b
(x − t)n (b − x)β−1 f (x) d x.
(25.16)
t
We need Proposition 25.5 ([1]) Let α ∈ (n, n + 1] and f : [a, ∞) → R be (n + 1) times continuously differentiable for t > a. Then, for all t > a we have Iαa Taα ( f ) (t) = f (t) −
n f (k) (a) (t − a)k . k! k=0
(25.17)
We also have Proposition 25.6 ([3]) Let α ∈ (n, n + 1] and f : (−∞, b] → R be (n + 1) times continuously differentiable for t < b. Then, for all t < b we have − b Iα bα T ( f ) (t) = f (t) −
n f (k) (b) (t − b)k . k! k=0
(25.18)
If n = 0 or 0 < α ≤ 1, then (see also [1]) b
Iα bα T ( f ) (t) = f (t) − f (b) .
In conclusion we derive
(25.19)
25.2 Background—II
467
Theorem 25.7 ([3]) Let α ∈ (n, n + 1] and f ∈ C n+1 ([a, b]), n ∈ N. Then (1) f (t) −
n f (k) (a) (t − a)k 1 t = (t − x)n (x − a)β−1 Taα ( f ) (x) d x, k! n! a k=0 (25.20)
and (2) n 1 b f (k) (b) (t − b)k =− f (t) − (b − x)β−1 (x − t)n bα T ( f ) (x) d x, k! n! t k=0 (25.21) ∀ t ∈ [a, b]. We make Remark 25.8 We notice the following: let α ∈ (n, n + 1] and f ∈ C n+1 ([a, b]), n ∈ N. Then (β := α − n, 0 < β ≤ 1) a Tα ( f ) (x) = Tβa f (n) (x) = (x − a)1−β f (n+1) (x) , (25.22) b
and
αT (
f ) (x) = (−1)n+1 bβ T f (n) (x) =
(−1)n+1 (−1) (b − x)1−β f (n+1) (x) = (−1)n (b − x)1−β f (n+1) (x) .
(25.23)
Consequently we get that a Tα ( f ) (x) , bα T ( f ) (x) ∈ C ([a, b]) . Furthermore it is obvious that a Tα ( f ) (a) = bα T ( f ) (b) = 0,
(25.24)
when 0 < β < 1, i.e. when α ∈ (n, n + 1). If f (k) (a) = 0, k = 0, 1, . . . , n, then f (t) =
1 n!
a
t
(t − x)n (x − a)β−1 Taα ( f ) (x) d x,
(25.25)
∀ t ∈ [a, b]. If f (k) (b) = 0, k = 0, 1, . . . , n, then f (t) = − ∀ t ∈ [a, b].
1 n!
b t
(b − x)β−1 (x − t)n
b
αT (
f ) (x) d x,
(25.26)
468
25 Fractional Conformable Approximation of Csiszar’s F-Divergence
25.3 Main Results—I Here f and the whole setting is as in Sect. 25.1. Background—I. Next we present estimations for f (μ1 , μ2 ). We give Theorem 25.9 Let α ∈ (n, n + 1], f ∈ C n+1 ([a, b]), n ∈ N and f (k) (a) = 0, k = p(x) ≤ b, a.e. on X . Then 0, 1, . . . , n. Assume that 0 < a ≤ q(x) a T ( f ) α ∞,[a,b] f (μ1 , μ2 ) ≤ n (q (x))1−α ( p (x) − aq (x))α dλ.
X (α − n + j)
(25.27)
j=0
Proof By (25.25) we have that 1 | f (t)| ≤ n! a T ( f ) α
t
a
(t − x)n (x − a)β−1 Taα ( f ) (x) d x ≤
∞,[a,b]
n!
t
(t − x)(n+1)−1 (x − a)β−1 d x =
a
a T ( f ) α ∞,[a,b] (n + 1) (β) (n + 1 + β)
n! a T ( f ) α ∞,[a,b]
(t − a)n+β =
a (β) (t − a)n+β n+β T ( f ) = − a) . (t α n n ∞,[a,b]
(β) (β + j) (β + j) j=0
j=0
(25.28) That is
(t − a)n+β | f (t)| ≤ Taα ( f ) ∞,[a,b] n ,
+ j) (β
(25.29)
j=0
∀ t ∈ [a, b]. Consequently we obtain
f (μ1 , μ2 ) =
q (x) f X
(25.29) p (x) p (x) dλ ≤ q (x) f dλ ≤ q (x) q (x) X
a n+β T ( f ) p (x) α ∞,[a,b] −a q (x) dλ = n
q (x) X (β + j) j=0
(25.30)
25.3 Main Results—I
469
a T ( f ) α ∞,[a,b] q (x)1−n−β ( p (x) − aq (x))n+β dλ, n
X (β + j) j=0
proving (25.27). We continue with
Theorem 25.10 Let α ∈ (n, n + 1], f ∈ C n+1 ([a, b]), n ∈ N and f (k) (b) = 0, k = p(x) ≤ b, a.e. on X . Then 0, 1, . . . , n. Assume that 0 < a ≤ q(x) f (μ1 , μ2 ) ≤
b T ( f ) α n
∞,[a,b]
(α − n + j)
(q (x))1−α (bq (x) − p (x))α dλ.
(25.31)
X
j=0
Proof By (25.26) we get 1 b (b − x)β−1 (x − t)n bα T ( f ) (x) d x ≤ n! t b b T ( f ) α ∞,[a,b] (b − x)β−1 (x − t)(n+1)−1 d x = n! t b T ( f ) α ∞,[a,b] (β) (n + 1) (b − t)β+n = n! (β + n + 1) b T ( f ) α ∞,[a,b] (b − t)n+β . n
(β + j)
| f (t)| ≤
(25.32)
j=0
That is | f (t)| ≤
b T ( f ) α
n
∞,[a,b]
(β + j)
(b − t)n+β ,
j=0
∀ t ∈ [a, b]. Therefore we derive f (μ1 , μ2 ) ≤
(25.33) p (x) dλ ≤ q (x) f q (x) X
(25.33)
470
25 Fractional Conformable Approximation of Csiszar’s F-Divergence
b T ( f ) α
n
∞,[a,b]
(β + j)
p (x) n+β q (x) b − dλ = q (x) X
j=0
b T ( f ) α
n
∞,[a,b]
q (x)1−n−β (bq (x) − p (x))n+β dλ,
(β + j)
(25.34)
X
j=0
proving (25.31). We continue with
Theorem 25.11 All as in Theorem 25.9. Additionally assume that r1 , r2 , r3 > 1 : 1 + r12 + r13 = 1, with α − n > r11 + r13 . Then r1 a T ( f ) α r3 ,[a,b]
f (μ1 , μ2 ) ≤
1
1
n! (nr1 + 1) r1 (r2 (α − n − 1) + 1) r2
(q (x))
1−α+ r1
3
( p (x) − aq (x))
α− r1
3
dλ.
(25.35)
X
Proof By (25.25) and Hölder’s inequality we have that | f (t)| ≤ 1 n!
t
(t − x)nr1 d x
1 n!
t
a
(t − x)n (x − a)β−1 Taα ( f ) (x) d x ≤
r1 1
a
t
(x − a)r2 (β−1) d x
a (nr1 +1)
r1 2
a
t
a T ( f ) (x) r3 d x α
r1
3
≤
(r2 (β−1)+1)
a 1 (t − a) r1 (t − a) r2 T ( f ) = 1 1 α r3 ,[a,b] n! (nr1 + 1) r1 (r2 (β − 1) + 1) r2
(25.36)
α− 1 (t − a) r3 Taα ( f ) r3 ,[a,b] 1 . n! (nr1 + 1) r11 (r2 (α − n − 1) + 1) r12 That is | f (t)| ≤ ∀ t ∈ [a, b]. Consequently we obtain
a α− 1 T ( f ) (t − a) r3 α r3 ,[a,b] 1
1
n! (nr1 + 1) r1 (r2 (α − n − 1) + 1) r2
,
(25.37)
25.3 Main Results—I
471
f (μ1 , μ2 ) ≤
(25.37) p (x) dλ ≤ q (x) f q (x) X
a T ( f ) α r3 ,[a,b]
1 r1
n! (nr1 + 1) (r2 (α − n − 1) + 1) a T ( f ) α r3 ,[a,b] 1 r1
n! (nr1 + 1) (r2 (α − n − 1) + 1)
1 r2
1 r2
q (x) X
(q (x))
p (x) −a q (x)
1−α+ r1
3
α− r1
3
dλ =
( p (x) − aq (x))
α− r1
3
dλ,
X
(25.38)
proving the claim. Next we give
Theorem 25.12 All as in Theorem 25.10. Additionaly assume that r1 , r2 , r3 > 1 : 1 + r12 + r13 = 1, with α − n > r11 + r13 . Then r1 f (μ1 , μ2 ) ≤
b T ( f ) α
r3 ,[a,b]
1 r1
1
n! (r1 n + 1) (r2 (α − n − 1) + 1) r2
(q (x))
1−α+ r1
3
(bq (x) − p (x))
α− r1
dλ.
3
(25.39)
X
Proof By (25.26) and Hölder’s inequality we obtain 1 | f (t)| ≤ n! 1 n!
b
(x − t)
b t
r1
r1 n
b
1
dx
t
(b − x)β−1 (x − t)n bα T ( f ) (x) d x ≤
r2 (β−1)
(b − x)
r1 2
dx
t
b
b T ( f ) (x) r3 d x α
t
n+ (β−1)+ r b 2 1 (b − t) r1 (b − t) T ( f ) = 1 1 α r3 ,[a,b] n! (r1 n + 1) r1 (r2 (β − 1) + 1) r2 1
r1
3
≤
1
(25.40)
α− b 1 (b − t) r3 T ( f ) . 1 1 α r3 ,[a,b] n! (r1 n + 1) r1 (r2 (α − n − 1) + 1) r2 1
We have proved that | f (t)| ≤ ∀ t ∈ [a, b].
b T ( f ) α
r3 ,[a,b]
1 r1
α− r1
(b − t)
3 1
n! (r1 n + 1) (r2 (α − n − 1) + 1) r2
,
(25.41)
472
25 Fractional Conformable Approximation of Csiszar’s F-Divergence
Therefore we derive (25.41) p (x) dλ ≤ q (x) f f (μ1 , μ2 ) ≤ q (x) X
b T ( f ) α
r3 ,[a,b]
1 r1
n! (r1 n + 1) (r2 (α − n − 1) + 1) b T ( f ) α r3 ,[a,b] 1 r1
n! (r1 n + 1) (r2 (α − n − 1) + 1)
1 r2
1 r2
p (x) q (x) b − q (x) X
(q (x))
1−α+ r1
3
α− r1
3
dλ =
(bq (x) − p (x))
α− r1
3
dλ,
X
(25.42)
proving the claim. Next we cover the case n = 0. We make
Remark 25.13 Here 0 < α ≤ 1 and f ∈ C 1 ([a, b]). Assume f (a) = 0, by Lemma 25.3 we get:
t
f (t) = a
(x − a)α−1 Tαa ( f ) (x) d x,
(25.43)
∀ t ∈ [a, b]. Assume f (b) = 0, by (25.19) we obtain:
b
f (t) =
(b − x)α−1
t
b
αT
( f ) (x) d x,
(25.44)
∀ t ∈ [a, b]. We give Proposition 25.14 Let 0 < α ≤ 1, f ∈ C 1 ([a, b]), f (a) = 0. Assume that 0 < p(x) ≤ b, a.e. on X . Then a ≤ q(x) f (μ1 , μ2 ) ≤
a T ( f ) α
∞,[a,b]
α
(q (x))1−α ( p (x) − aq (x))α dλ . (25.45) X
Proof By (25.43) we have | f (t)| ≤ a
a T ( f ) α
∞,[a,b]
a
t
t
(x − a)α−1 Tαa ( f ) (x) d x ≤
(x − a)α−1 d x =
a T ( f ) α
α
∞,[a,b]
(t − a)α ,
(25.46)
25.3 Main Results—I
∀ t ∈ [a, b]. Thus
473
(25.46) p (x) dλ ≤ q (x) f f (μ1 , μ2 ) ≤ q (x) X
p (x) −a q (x) q (x) X
a T ( f ) α
∞,[a,b]
α
α dλ
T a ( f ) α ∞,[a,b] α
=
q (x)1−α ( p (x) − aq (x))α dλ ,
(25.47)
X
proving the claim. We continue with
Proposition 25.15 Let 0 < α ≤ 1, f ∈ C 1 ([a, b]), f (b) = 0. Assume that 0 < p(x) ≤ b, a.e. on X . Then a ≤ q(x) f (μ1 , μ2 ) ≤
b T ( f ) α ∞,[a,b] α
(q (x))1−α (bq (x) − p (x))α dλ.
(25.48)
X
Proof By (25.44) we derive | f (t)| ≤
b
t
(b − x)α−1 bα T ( f ) (x) d x ≤
b T ( f ) α ∞,[a,b] α ∀ t ∈ [a, b]. Therefore
(b − t)α ,
(25.49)
(25.49) p (x) dλ ≤ f (μ1 , μ2 ) ≤ q (x) f q (x) X
b T ( f ) α ∞,[a,b] α b T ( f ) α ∞,[a,b] α proving the claim.
p (x) q (x) b − q (x) X
q (x)
1−α
α
α
dλ =
(bq (x) − p (x)) dλ ,
(25.50)
X
474
25 Fractional Conformable Approximation of Csiszar’s F-Divergence
It follows: Proposition 25.16 All as in Proposition 25.14. Let r1 , r2 > 1 : α > r12 . Then f (μ1 , μ2 ) ≤
a T ( f ) α
r2 ,[a,b]
(q (x))
1
(r1 (α − 1) + 1) r1
1−α+ r1
2
1 r1
+
1 r2
( p (x) − aq (x))
= 1, with
α− r1
2
dλ .
X
(25.51)
Proof By (25.43) and Hölder’s inequality we get
t
| f (t)| ≤ a
t
(x − a)α−1 Tαa ( f ) (x) d x ≤
(x − a)r1 (α−1) d x
r1
1
a
(t − a)
α− r1
2
(r1 (α − 1) + 1) That is (t − a)
| f (t)| ≤ ∀ t ∈ [a, b]. Hence
1 r1
a T f = α r2 ,[a,b]
a T f . α r2 ,[a,b]
α− r1
2
(r1 (α − 1) + 1)
1 r1
(25.52)
a T f , α r2 ,[a,b]
(25.53)
(25.53) p (x) dλ ≤ f (μ1 , μ2 ) ≤ q (x) f q (x) X
a T f α r2 ,[a,b]
1
(r1 (α − 1) + 1) r1 a T f α r2 ,[a,b]
p (x) −a q (x) q (x) X
(q (x))
1
(r1 (α − 1) + 1) r1
1−α+ r1
2
α− r1
2
dλ =
α− r1
( p (x) − aq (x))
2
dλ ,
(25.54)
X
proving the claim. We also give Proposition 25.17 All as in Proposition 25.15. Let r1 , r2 > 1 : α > r12 . Then
1 r1
+
1 r2
= 1, with
25.3 Main Results—I
f (μ1 , μ2 ) ≤
475
b T ( f ) α r2 ,[a,b]
(q (x))
1
(r1 (α − 1) + 1) r1
1−α+ r1
2
(bq (x) − p (x))
α− r1
2
dλ .
X
(25.55)
Proof By (25.44) and Hölder’s inequality we have | f (t)| ≤
(b − x)α−1 bα T ( f ) (x) d x ≤
b
t
b
r1 (α−1)
(b − x)
r1
1
dx
t α− r1
(b − t)
∀ t ∈ [a, b]. Hence
1 r1
b T ( f ) α r2 ,[a,b]
That is | f (t)| ≤
b T ( f ) . α r2 ,[a,b]
2
(r1 (α − 1) + 1)
b T ( f ) = α r2 ,[a,b]
(r1 (α − 1) + 1)
1 r1
(25.56)
α− r1
(b − t)
2
,
(25.57)
(25.57) p (x) dλ ≤ f (μ1 , μ2 ) ≤ q (x) f q (x) X
b T ( f ) α r2 ,[a,b] 1
(r1 (α − 1) + 1) r1 b T ( f ) α r2 ,[a,b]
(q (x))
1
(r1 (α − 1) + 1) r1
1 p (x) α− r2 q (x) b − dλ = q (x) X 1−α+ r1
2
(bq (x) − p (x))
α− r1
2
dλ ,
(25.58)
X
proving the claim. We make
Remark 25.18 Let α ∈ (n, n + 1] and f ∈ C n+1 ([a, b]), n ∈ N. Assume that 0 < p(x) ≤ b, a.e. on X . Again f and the whole setting is as in Sect. 25.1. a ≤ q(x) Background—I. (I) By (25.20) we obtain: If Taα ( f ) ≥ 0 over [a, b], then (≤0)
n f (k) (a) (t − a)k , f (t) ≥ k! (≤) k=0
∀ t ∈ [a, b].
(25.59)
476
Hence
25 Fractional Conformable Approximation of Csiszar’s F-Divergence
k n p f (k) (a) p ≥ q −a , qf q (≤) k=0 k! q
(25.60)
a.e. on X . Consequently we derive n f (k) (a) q 1−k ( p − aq)k dλ. f (μ1 , μ2 ) ≥ k! (≤) X k=0
(25.61)
(II) By (25.21) we obtain: If bα T ( f ) ≥ 0 over [a, b], then (≤0)
n f (k) (b) (t − b)k , k! (≥) k=0
f (t) ≤ ∀ t ∈ [a, b]. Hence qf
k n p p f (k) (b) ≤ q −b , q (≥) k=0 k! q
(25.62)
(25.63)
a.e. on X . Consequently we obtain n f (k) (b) q 1−k ( p − bq)k dλ. k! (≥) X k=0
f (μ1 , μ2 ) ≤
(25.64)
25.4 Background—III Next we use the following. Let f by a convex from (0, +∞) into R which function is strictly convex at 1 with f (1) = 0. Let R2, B 2 , λ be the measure space, where λ is the product Lebesgue measure on R2 , B 2 with B being the Borel σ-field. And let X, Y : → R be random variables on the probability space (, P). Consider the probability distributions μ X Y and μ X × μY on R2 , where μ X Y , μ X × μY stand for the joint distribution of X and Y and their marginal distributions, respectively. Here we assume as existing the following probability density functions, the joint pdf of μ X Y to be t (x, y), x, y ∈ R, the pdf of μ X to be p (x) and pdf of μY to be q (y). Clearly μ X × μY has pdf p (x) q (y). Here we further assume that 0 < a ≤ pqt ≤ b, a.e. on R2 and a ≤ 1 ≤ b.
25.4 Background—III
477
The quantity
f (μ X Y , μ X × μY ) =
R2
p (x) q (y) f
t (x, y) dλ (x, y) , p (x) q (y)
(25.65)
is the Csiszar’s distance or f -divergence between μ X Y and μ X × μY . Here X, Y are less dependent the closer the distributions μ X Y and μ X × μY are, thus f (μ X Y , μ X × μY ) can be considered as a measure of dependence of X and Y . For f (u) = u log2 u we obtain the mutual information of X , and Y, I (X, Y ) = I (μ X Y ||μ X × μY ) = u log2 u (μ X Y , μ X × μY ) , see [7]. For f (u) = (u − 1)2 we get the mean square contingency: ϕ2 (X, Y ) = (u−1)2 (μ X Y , μ X × μY ) , see [12]. In the last we need μ X Y μ X × μY , where denotes absolute continuity, but to cover the case of μ X Y μ X × μY we set ϕ2 (X, Y ) = +∞, then the last formula is always valid. Clearly here μ X Y , μ X × μY λ, also f (μ X Y , μ X × μY ) > 0 with equality only when μ X Y = μ X × μY , i.e. when X, Y are independent r.v.’s.
25.5 Main Results—II Here f and the whole setting is as in Sect. 25.4. Background—III. Next we present estimations for f (μ X Y , μ X × μY ). We apply the results of Sect. 25.3. Main Results—I. We give Theorem 25.19 Let α ∈ (n, n + 1], f ∈ C n+1 ([a, b]), n ∈ N and f (k) (a) = 0, k = t(x,y) ≤ b, a.e. on R2 . Then 0, 1, . . . , n. Assume that 0 < a ≤ p(x)q(y) f (μ X Y , μ X × μY ) ≤ a T ( f ) α ∞,[a,b] ( p (x) q (y))1−α (t (x, y) − ap (x) q (y))α dλ (x, y) . n
2 (α − n + j) R j=0
(25.66) Proof By Theorem 25.9.
Theorem 25.20 Let α ∈ (n, n + 1], f ∈ C n+1 ([a, b]), n ∈ N and f (k) (b) = 0, k = t(x,y) ≤ b, a.e. on R2 . Then 0, 1, . . . , n. Assume that 0 < a ≤ p(x)q(y)
478
25 Fractional Conformable Approximation of Csiszar’s F-Divergence
f (μ X Y , μ X × μY ) ≤ b T ( f ) α n
∞,[a,b]
(α − n + j)
R2
( p (x) q (y))1−α (bp (x) q (y) − t (x, y))α dλ (x, y) .
j=0
(25.67)
Proof By Theorem 25.10. We continue with
Theorem 25.21 All as in Theorem 25.19. Additionaly assume that r1 , r2 , r3 > 1 : 1 + r12 + r13 = 1, with α − n > r11 + r13 . Then r1 f (μ X Y , μ X × μY ) ≤ R2
( p (x) q (y))
1−α+ r1
3
a T ( f ) α r3 ,[a,b] 1
1
n! (nr1 + 1) r1 (r2 (α − n − 1) + 1) r2
(t (x, y) − ap (x) q (y))
α− r1
3
dλ (x, y) .
(25.68)
Proof By Theorem 25.11.
Theorem 25.22 All as in Theorem 25.20. Additionaly assume that r1 , r2 , r3 > 1 : 1 + r12 + r13 = 1, with α − n > r11 + r13 . Then r1 f (μ X Y , μ X × μY ) ≤ R2
( p (x) q (y))
1−α+ r1
3
b T ( f ) α
r3 ,[a,b]
1
1
n! (r1 n + 1) r1 (r2 (α − n − 1) + 1) r2
(bp (x) q (y) − t (x, y))
α− r1
3
dλ (x, y) .
(25.69)
Proof By Theorem 25.12. Next we cover the case n = 0. We give
Proposition 25.23 Let 0 < α ≤ 1, f ∈ C 1 ([a, b]), f (a) = 0. Assume that 0 < t(x,y) ≤ b, a.e. on R2 . Then a ≤ p(x)q(y) f (μ X Y , μ X × μY ) ≤ a T ( f ) α
α
∞,[a,b]
R2
1−α α . dλ y) p q y) − ap q (x, ( (x) (y)) (t (x, (x) (y)) (25.70)
25.5 Main Results—II
479
Proof By Proposition 25.14.
Proposition 25.24 Let 0 < α ≤ 1, f ∈ C 1 ([a, b]), f (b) = 0. Assume that 0 < t(x,y) ≤ b, a.e. on R2 . Then a ≤ p(x)q(y) f (μ X Y , μ X × μY ) ≤ b T ( f ) α ∞,[a,b] α
R2
( p (x) q (y))1−α (bp (x) q (y) − t (x, y))α dλ (x, y) . (25.71)
Proof By Proposition 25.15. We continue with Proposition 25.25 All as in Proposition 25.23. Let r1 , r2 > 1 : α > r12 . Then a T ( f ) α r2 ,[a,b] f (μ X Y , μ X × μY ) ≤ 1 (r1 (α − 1) + 1) r1 R2
( p (x) q (y))
1−α+ r1
2
(t (x, y) − ap (x) q (y))
α− r1
2
1 r1
+
1 r2
dλ (x, y) .
Proposition 25.26 All as in Proposition 25.24. Let r1 , r2 > 1 : α > r12 . Then b T ( f ) α r2 ,[a,b] f (μ X Y , μ X × μY ) ≤ 1 (r1 (α − 1) + 1) r1
R2
( p (x) q (y))
(25.72)
Proof By Proposition 25.16.
= 1, with
1−α+ r1
2
(bp (x) q (y) − t (x, y))
α− r1
2
1 r1
+
1 r2
dλ (x, y) .
= 1, with
(25.73)
Proof By Proposition 25.17. We make
Remark 25.27 Let α ∈ (n, n + 1] and f ∈ C n+1 ([a, b]), n ∈ N. Assume that 0 < t(x,y) ≤ b, a.e. on R2 . Again f and the whole setting is as in Sect. 25.4. a ≤ p(x)q(y) Background—III. (I) By (25.20) we obtain: If Taα ( f ) ≥ 0 over [a, b], then (as in Remark 25.18) (≤0)
f (μ X Y , μ X × μY ) ≥
(≤)
480
25 Fractional Conformable Approximation of Csiszar’s F-Divergence
n f (k) (a) ( p (x) q (y))1−k (t (x, y) − ap (x) q (y))k dλ (x, y) . 2 k! R k=0
(25.74)
(II) By (25.21) we obtain: If bα T ( f ) ≥ 0 over [a, b], then (as in Remark 25.18) (≤0)
f (μ X Y , μ X × μY ) ≤
(≥)
n f (k) (b) ( p (x) q (y))1−k (t (x, y) − bp (x) q (y))k dλ (x, y) . 2 k! R k=0
(25.75)
References 1. T. Abdeljawad, On conformable fractional calculus. J. Comput. Appl. Math. 279, 57–66 (2015) 2. G.A. Anastassiou, Fractional and other approximation of Csiszar’s f -divergence, (Proc. FAAT 04-Ed. F. Altomare). Rend. Circ. Mat. Palermo, Serie II, Suppl. 76, 197–212 (2005) 3. G.A. Anastassiou, Nonlinearity: Ordinary and Fractional Approximations by Sublinear and Max-Product Operators. Studies in Systems, Decision and Control, vol. 147 (Springer, New York, 2018) 4. G.A. Anastassiou, Conformable Fractional Approximation of Csiszar’s f -Divergence, submitted for publication (2019) 5. N.S. Barnett, P. Gerone, S.S. Dragomir, A. Sofo, Approximating Csiszar’s f -divergence by the use of Taylor’s formula with integral remainder (paper #10, pp. 16), in, Inequalities for Csiszar f -Divergence in Information Theory, ed. by S.S. Dragomir (Victoria University, Melbourne, 2000), http://rgmia.vu.edu.au 6. I. Csiszar, Eine Informationstheoretische Ungleichung und ihre Anwendung auf den Beweis der Ergodizität von Markoffschen Ketten. Magyar Trud. Akad. Mat. Kutato Int. Közl. 8, 85–108 (1963) 7. I. Csiszar, Information-type measures of difference of probability distributions and indirect observations. Stud. Sci. Math. Hung. 2, 299–318 (1967) 8. S.S. Dragomir (ed.), Inequalities for Csiszar f -Divergence in Information Theory (Victoria University, Melbourne, 2000), http://rgmia.vu.edu.au 9. R. Khalil, M. Al. Horani, A. Yousef, M. Sababheh, A new definition of fractional derivative. J. Comput. Appl. Math. 264, 65–70 (2014) 10. S. Kullback, Information Theory and Statistics (Wiley, New York, 1959) 11. S. Kullback, R. Leibler, On information and sufficiency. Ann. Math. Stat. 22, 79–86 (1951) 12. A. Rényi, On measures of dependence. Acta Math. Acad. Sci. Hung. 10, 441–451 (1959) 13. A. Rényi, On measures of entropy and information, in Proceedings of the 4th Berkeley Symposium on Mathematical Statistic and Probability, I, CA, Berkeley, 1960, pp. 547–561
Chapter 26
Fractional Conformable Self Adjoint Operator Analytic Inequalities
We present here conformable fractional self adjoint operator comparison, Poincaré, Sobolev, Ostrowski and Opial type inequalities. At first we give right and left conformable fractional representation formulae in the self adjoint operator sense. Operator inequalities are based in the self adjoint operator order over a Hilbert space. See also [3].
26.1 Background—I Let A be a selfadjoint linear operator on a complex Hilbert space (H ; ·, ·). The Gelfand map establishes a ∗−isometrically isomorphism between the set C (Sp (A)) of all continuous functions defined on the spectrum of A, denoted Sp (A), and the C ∗ -algebra C ∗ (A) generated by A and the identity operator 1 H on H as follows (see e.g. [7, p. 3]): For any f, g ∈ C (Sp (A)) and any α, β ∈ C we have (i) (α f + βg) = α ( f ) + β (g) ; (ii) ( f g) = ( f ) (g) (the operation composition is on the right) and f = ( ( f ))∗ ; (iii) ( f ) = f := sup | f (t)| ; t∈Sp( A)
(iv) ( f 0 ) = 1 H and ( f 1 ) = A, where f 0 (t) = 1 and f 1 (t) = t, for t ∈ Sp (A) . With this notation we define f (A) := ( f ) , for all f ∈ C (Sp (A)) , and we call it the continuous functional calculus for a selfadjoint operator A. © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_26
481
482
26 Fractional Conformable Self Adjoint Operator Analytic Inequalities
If A is a selfadjoint operator and f is a real valued continuous function on Sp (A) then f (t) ≥ 0 for any t ∈ Sp (A) implies that f (A) ≥ 0, i.e. f (A) is a positive operator on H . Moreover, if both f and g are real valued continuous functions on Sp (A) then the following important property holds: (P) f (t) ≥ g (t) for any t ∈ Sp (A), implies that f (A) ≥ g (A) in the operator order of B (H ) (the Banach algebra of all bounded linear operators from H into itself). Equivalently, we use (see [6], pp. 7–8): Let U be a selfadjoint operator on the complex Hilbert space (H, ·, ·) with the spectrum Sp (U ) included in the interval [m, M] for some real numbers m < M and {E λ }λ be its spectral family. Then for any continuous function f : [m, M] → C, it is well known that we have the following spectral representation in terms of the Riemann–Stieltjes integral:
M
f (U ) x, y =
f (λ) d (E λ x, y) ,
m−0
for any x, y ∈ H . The function gx,y (λ) := E λ x, y is of bounded variation on the interval [m, M], and gx,y (m − 0) = 0 and gx,y (M) = x, y , for any x, y ∈ H . Furthermore, it is known that gx (λ) := E λ x, x is increasing and right continuous on [m, M] . We have also the formula M f (U ) x, x = f (λ) d (E λ x, x) , ∀ x ∈ H. m−0
As a symbol we can write f (U ) =
M
f (λ) d E λ .
m−0
Above, m = min {λ|λ ∈ Sp (U )} := min Sp (U ), M = max {λ|λ ∈ Sp (U )} := max Sp (U ). The projections {E λ }λ∈R , are called the spectral family of A, with the properties: (a) E λ ≤ E λ for λ ≤ λ ; (b) E m−0 = 0 H (zero operator), E M = 1 H (identity operator) and E λ+0 = E λ for all λ ∈ R. Furthermore E λ := ϕλ (U ) , ∀ λ ∈ R,
26.1 Background—I
483
is a projection which reduces U , with ϕλ (s) :=
1, for − ∞ < s ≤ λ, 0, for λ < s < +∞.
The spectral family {E λ }λ∈R determines uniquely the self-adjoint operator U and vice versa. For more on the topic see [8], pp. 256–266, and for more details see there pp. 157– 266. See also [5]. Some more basics are given (we follow [6], pp. 1–5): Let (H ; ·, ·) be a Hilbert space over C. A bounded linear operator A defined on H is selfjoint, i.e., A = A∗ , iff Ax, x ∈ R, ∀ x ∈ H , and if A is selfadjoint, then A =
sup
x∈H :x=1
|Ax, x| .
Let A, B be selfadjoint operators on H . Then A ≤ B iff Ax, x ≤ Bx, x, ∀ x ∈ H . In particular, A is called positive if A ≥ 0. Denote by P := ϕ (s) :=
n
αk s k |n ≥ 0, αk ∈ C, 0 ≤ k ≤ n .
k=0
If A ∈ B (H ) is selfadjoint, and ϕ (s) ∈ P has real coefficients, then ϕ (A) is selfadjoint, and ϕ (A) = max {|ϕ (λ)| , λ ∈ Sp (A)} . If ϕ is any function defined on R we define ϕ A := sup {|ϕ (λ)| , λ ∈ Sp (A)} . If A is selfadjoint operator on Hilbert space H and ϕ is continuous and given that ϕ (A) is selfadjoint, then ϕ ( A) = ϕ A . And if ϕ is a continuous real valued function so it is |ϕ|, then ϕ ( A) and |ϕ| (A) = |ϕ (A)| are selfadjoint operators (by [6], p. 4, Theorem 7). Hence it holds |ϕ ( A)| = |ϕ| A = sup {||ϕ (λ)|| , λ ∈ Sp (A)} = sup {|ϕ (λ)| , λ ∈ Sp (A)} = ϕ A = ϕ ( A) , that is |ϕ (A)| = ϕ (A) .
484
26 Fractional Conformable Self Adjoint Operator Analytic Inequalities
For a selfadjoint operator A ∈ B (H ) which is positive, there exists a unique √ 2 √ positive selfadjoint operator B := A ∈ B (H ) such that B 2 = A, that is A = A. We call B the square root of A. ∗ and √ positive. Define the “operator absolute Let A ∈ B (H √ ), then A A is selfadjoint ∗ value” |A| := A A. If A = A∗ , then |A| = A2 . For a continuous real valued function ϕ we observe the following: |ϕ (A)| (the functional absolute value) =
M
|ϕ (λ)| d E λ =
m−0
M
(ϕ (λ))2 d E λ = (ϕ (A))2 = |ϕ (A)| (operator absolute value),
m−0
where A is a selfadjoint operator. That is we have |ϕ (A)| (functional absolute value) = |ϕ (A)| (operator absolute value).
26.2 Background—II Here we follow [1]. We need Definition 26.1 ([1, 9]) Let a, b ∈ R. The left conformable fractional derivative starting from a of a function f : [a, ∞) → R of order 0 < α ≤ 1 is defined by
Tαa
f t + ε (t − a)1−α − f (t) . f (t) = lim ε→0 ε
(26.1)
If Tαa f (t) exists on (a, b), then
Tαa f (a) = lim Tαa f (t) . t→a+
(26.2)
The right conformable fractional derivative of order 0 < α ≤ 1 terminating at b of f : (−∞, b] → R is defined by f t + ε (b − t)1−α − f (t) . α T f (t) = −lim ε→0 ε
b If
b
αT
(26.3)
f (t) exists on (a, b), then b
αT
f (b) = lim bα T f (t) . t→b−
(26.4)
26.2 Background—II
485
Note that if f is differentiable then and
Tαa f (t) = (t − a)1−α f (t) ,
b
αT
Denote by
and
Iαa
b
f (t) = − (b − t)1−α f (t) .
t
f (t) =
(x − a)α−1 f (x) d x,
(26.5)
(26.6)
(26.7)
a
Iα f (t) =
b
(b − x)α−1 f (x) d x,
(26.8)
t
these are the left and right conformable fractional integrals of order 0 < α ≤ 1. In the higher order case we can generalize things as follows: Definition 26.2 ([1]) Let α ∈ (n, n + 1], and set β = α − n. Then, the left conformable fractional derivative starting from a of a function f : [a, ∞) → R of order α, where f (n) (t) exists, is defined by a Tα f (t) = Tβa f (n) (t) .
(26.9)
The right conformable fractional derivative of order α terminating at b of f : (−∞, b] → R, where f (n) (t) exists, is defined by b
αT
f (t) = (−1)n+1 bβ T f (n) (t) .
(26.10)
If α = n + 1 then β = 1 and Tan+1 f = f (n+1) . If n is odd, then bn+1 T f = − f (n+1) , and if n is even, then bn+1 T f = f (n+1) . When n = 0 (or α ∈ (0, 1]), then β = α, and (26.9), (26.10) collapse to {(26.1), (26.2)}, {(26.3), (26.4)} respectively. Lemma 26.3 ([1]) Let f : (a, b) → R be continuously differentiable and 0 < α ≤ 1. Then, for all t > a we have Iαa Tαa ( f ) (t) = f (t) − f (a) .
(26.11)
We need Definition 26.4 (see also [1]) If α ∈ (n, n + 1], then the left fractional integral of order α starting at a is defined by
Iαa
1 f (t) = n!
a
t
(t − x)n (x − a)β−1 f (x) d x.
(26.12)
486
26 Fractional Conformable Self Adjoint Operator Analytic Inequalities
Similarly, (author’s definition) the right fractional integral of order α terminating at b is defined by b
1 Iα f (t) = n!
b
(x − t)n (b − x)β−1 f (x) d x.
(26.13)
t
We need Proposition 26.5 ([1]) Let α ∈ (n, n + 1] and f : [a, ∞) → R be (n + 1) times continuously differentiable for t > a. Then, for all t > a we have Iαa Taα ( f ) (t) = f (t) −
n f (k) (a) (t − a)k . k! k=0
(26.14)
We also have Proposition 26.6 ([2]) Let α ∈ (n, n + 1] and f : (−∞, b] → R be (n + 1) times continuously differentiable for t < b. Then, for all t < b we have −
b
Iα bα T (
n f (k) (b) (t − b)k . f ) (t) = f (t) − k! k=0
(26.15)
If n = 0 or 0 < α ≤ 1, then (see also [1]) b
I α bα T ( f ) (t) = f (t) − f (b) .
(26.16)
In conclusion we derive Theorem 26.7 ([2]) Let α ∈ (n, n + 1] and f ∈ C n+1 ([a, b]), n ∈ N. Then (1) n 1 t f (k) (a) (t − a)k = f (t) − (t − x)n (x − a)β−1 Taα ( f ) (x) d x, k! n! a k=0 (26.17) and (2) n 1 b f (k) (b) (t − b)k =− f (t) − (b − x)β−1 (x − t)n bα T ( f ) (x) d x, k! n! t k=0 (26.18) ∀ t ∈ [a, b].
26.2 Background—II
487
We make Remark 26.8 We notice the following: let α ∈ (n, n + 1] and f ∈ C n+1 ([a, b]), n ∈ N. Then (β := α − n, 0 < β ≤ 1) a Tα ( f ) (x) = Tβa f (n) (x) = (x − a)1−β f (n+1) (x) , and
b
αT (
(26.19)
f ) (x) = (−1)n+1 bβ T f (n) (x) =
(−1)n+1 (−1) (b − x)1−β f (n+1) (x) = (−1)n (b − x)1−β f (n+1) (x) .
(26.20)
Consequently we get that a Tα ( f ) (x) , bα T ( f ) (x) ∈ C ([a, b]) . Furthermore it is obvious that a Tα ( f ) (a) = bα T ( f ) (b) = 0,
(26.21)
when 0 < β < 1, i.e. when α ∈ (n, n + 1). If f (k) (a) = 0, k = 0, 1, . . . , n, then f (t) =
1 n!
a
t
(t − x)n (x − a)β−1 Taα ( f ) (x) d x,
(26.22)
∀ t ∈ [a, b]. If f (k) (b) = 0, k = 0, 1, . . . , n, then f (t) = −
1 n!
b t
(b − x)β−1 (x − t)n
b
αT (
f ) (x) d x,
(26.23)
∀ t ∈ [a, b].
26.3 Main Results Let A be a selfadjoint operator in the Hilbert space H with the spectrum Sp (A) ⊆ [m, M], m < M; m, M ∈ R. In the next we obtain conformable fractional comparison, Poincaré, Sobolev, Ostrowski and Opial type inequalities in the operator order of B (H ) (the Banach algebra of all bounded linear operators from H into itself). All of our functions from now on in this chapter are real valued.
488
26 Fractional Conformable Self Adjoint Operator Analytic Inequalities
We give the following operator representation formulae: Theorem 26.9 Let A be a selfadjoint operator in the Hilbert space H with the spectrum Sp (A) ⊆ [m, M] for some real numbers m < M, {E λ }λ be its spectral ◦
family, I be a closed subinterval on R with [m, M] ⊂ I (the interior of I ) and n ∈ N with α ∈ (n, n + 1]. We consider f ∈ C n+1 ([m, M]), where f : I → R. Then f (A) =
n f (k) (m) (A − m1 H )k + Rn(1) ( f, m, M) , k! k=0
(26.24)
where Rn(1)
1 ( f, m, M) = n!
and
M
λ
(λ − t) (t − m) n
m−0
β−1
m
Tm α
( f ) (t) dt d E λ , (26.25)
n f (k) (M) f (A) = (A − M1 H )k + Rn(2) ( f, m, M) , k! k=0
(26.26)
where Rn(2)
1 ( f, m, M) = − n!
M
m−0
λ
M
β−1
(M − t)
(t −
λ)n αM T (
f ) (t) dt d E λ , (26.27)
where β = α − n. Proof Here α ∈ (n, n + 1] and f ∈ C n+1 ([m, M]), n ∈ N. By Theorem 26.7 we have n f (k) (m) 1 λ k f (λ) = (λ − t)n (t − m)β−1 Tm (λ − m) + α ( f ) (t) dt, k! n! m k=0 (26.28) and n f (k) (M) 1 M (M − t)β−1 (t − λ)n αM T ( f ) (t) dt, (λ − M)k − k! n! λ k=0 (26.29) ∀ λ ∈ [m, M] . Then we investigate (26.28), (26.29) against E λ to get f (λ) =
M
m−0
n f (k) (m) M f (λ) d E λ = (λ − m)k d E λ + k! m−0 k=0
(26.30)
26.3 Main Results
1 n! and
489
M
λ
(λ − t) (t − m) n
m−0
M
m
m−0
M
M
(M − t)
λ
m−0
m Tα ( f ) (t) dt d E λ ,
n f (k) (M) M (λ − M)k d E λ − k! m−0 k=0
f (λ) d E λ =
1 n!
β−1
β−1
(t −
λ)n αM T (
(26.31)
f ) (t) dt d E λ .
By the spectral representation theorem we obtain f (A) = 1 n!
M
m−0
m
f (A) =
M
M
n f (k) (M) (A − M1 H )k − k! k=0
(M − t)
λ
m−0
(26.32)
f dt d Eλ, (λ − t)n (t − m)β−1 Tm ( ) (t) α
λ
and
1 n!
n f (k) (m) (A − m1 H )k + k! k=0
β−1
(t −
λ)n αM T (
(26.33)
f ) (t) dt d E λ ,
proving the claim. We make
Remark 26.10 (on Theorem 26.9) In (26.32) if we assume that f (k) (m) = 0, k = 0, . . . , n, then 1 f (A) = n!
M
λ
(λ − t) (t − m) n
m−0
β−1
m
m Tα ( f ) (t) dt d E λ .
(26.34)
In (26.33) if we assume that f (k) (M) = 0, k = 0, . . . , n, then 1 f (A) = − n!
M
M
(M − t)
λ
m−0
β−1
(t −
λ)n αM T (
f ) (t) dt d E λ .
(26.35)
Therefore it holds f (A) x, y =
1 n!
M
m−0
λ
m
f dt d E λ x, y , (λ − t)n (t − m)β−1 Tm ( ) (t) α (26.36)
490
26 Fractional Conformable Self Adjoint Operator Analytic Inequalities
and f (A) x, y = −
1 n!
M
M
λ
m−0
(M − t)β−1 (t − λ)n
T f dt d E λ x, y , ( ) (t) α
M
(26.37)
∀ x, y ∈ H. The function gx,y (λ) := E λ x, y is of bounded variation on [m, M] , gx,y (m − 0) = 0 and gx,y (M) = x, y , ∀ x, y ∈ H.
(26.38)
It is also well known that gx (λ) := E λ x, x is nondecreasing and right continuous on [m, M] . One has 1 n!
f (A) x, x =
M
m−0
λ
m
f dt d E λ x, x , (λ − t)n (t − m)β−1 Tm ( ) (t) α (26.39)
and f (A) x, x = −
1 n!
M
M
λ
m−0
(M − t)β−1 (t − λ)n
T f dt d E λ x, x , ( ) (t) α
M
(26.40)
∀ x ∈ H. We also make Remark 26.11 (to Theorem 26.9 and Remark 26.10) Let r1 , r2 , r3 > 1 : 1 = 1, with α − n > r11 + r13 . Then r3
λ
m
λ
(λ − t)
r1 n
λ
dt
m
r2 (β−1)
(t − m)
r12
n+ r1
(λ − m)
1
1
λ
dt
m
m T ( f ) (t)r3 dt α
m β−1+ r1
(λ − m)
2 1
(r1 n + 1) r1 (r2 (β − 1) + 1) r2
λ
λ
m
1 r2
+
α
r13
≤
(26.41)
m T ( f )
r3 ,[m,M]
We have proved that
m
+
(λ − t)n (t − m)β−1 Tm α ( f ) (t) dt ≤
r11
1 r1
≤ f dt (λ − t)n (t − m)β−1 Tm ( ) (t) α
(λ − t)n (t − m)β−1 Tm α ( f ) (t) dt ≤
.
26.3 Main Results
491
(λ − m)
α− r1
m T ( f )
3
1 r1
(r1 n + 1) (r2 (α − n − 1) + 1)
α
1 r2
r3 ,[m,M]
,
(26.42)
∀ λ ∈ [m, M] . Similarly we derive
M λ
M
(t − λ)
r1
r1 n
λ
(t − λ)n (M − t)β−1 αM T ( f ) (t) dt ≤ M
1
dt λ n+ r1
(M − λ)
1
(r1 n + 1)
1 r1
r2 (β−1)
(M − t)
(M − λ)
r1
2
dt
β−1+ r1
α
α
1 r2
α− r1
(M − λ)
r3 ,[m,M]
M T ( f )
2
(r2 (β − 1) + 1)
M T ( f )
3
1 r1
(r1 n + 1) (r2 (α − n − 1) + 1)
1 r2
r3 ,[m,M]
=
M T ( f ) α
= (26.43)
r3 ,[m,M]
.
We have proved that
M λ
λ
M
(M − t)
β−1
(t −
λ)n αM T (
f ) (t) dt ≤
(M − t)β−1 (t − λ)n αM T ( f ) (t) dt ≤ (M − λ)
α− r1
3
1 r1
(r1 n + 1) (r2 (α − n − 1) + 1) ∀ λ ∈ [m, M] . Therefore it holds
1 r2
M T ( f ) α
r3 ,[m,M]
(26.44)
,
(26.42)
| f (A) x, x| ≤ m T ( f ) α
1 r1
n! (r1 n + 1) (r2 (α − n − 1) + 1) m T ( f ) α
1 r1
r3 ,[m,M] 1 r2
∀ x ∈ H.
1 r2
α− r1
(λ − m)
3
d E λ x, x =
m−0
r3 ,[m,M]
n! (r1 n + 1) (r2 (α − n − 1) + 1)
M
(A − m1 H )
α− r1
3
x, x ,
(26.45)
492
26 Fractional Conformable Self Adjoint Operator Analytic Inequalities
Similarly, we get that (26.44)
| f (A) x, x| ≤ M T ( f ) α
r3 ,[m,M]
1 r1
n! (r1 n + 1) (r2 (α − n − 1) + 1) M T ( f ) α
1 r2
(M − λ)
α− r1
3
d E λ x, x =
m−0
r3 ,[m,M]
1
M
(A − M1 H )
1
n! (r1 n + 1) r1 (r2 (α − n − 1) + 1) r2
α− r1
3
x, x ,
(26.46)
∀ x ∈ H. We have proved the following comparison self adjoint operator conformable fractional inequalities: Theorem 26.12 All as in Theorem 26.9. Assume further f (k) (m) = 0, k = 0, 1, . . . , n. Let r1 , r2 , r3 > 1 : r11 + r12 + r13 = 1, with α > n + r11 + r13 . Then | f (A) x, x| ≤ m T ( f ) α
r3 ,[m,M]
1 r1
n! (r1 n + 1) (r2 (α − n − 1) + 1)
(A − m1 H )
1 r2
α− r1
3
x, x ,
(26.47)
∀ x ∈ H. Inequality (26.47) means that f (A) ≤
m T ( f ) α
r3 ,[m,M]
1 r1
n! (r1 n + 1) (r2 (α − n − 1) + 1)
1 r2
α− 1 (A − m1 H ) r3 ,
(26.48)
and in particular f (A) ≤
m T ( f ) α
r3 ,[m,M]
1 r1
n! (r1 n + 1) (r2 (α − n − 1) + 1)
1 r2
(A − m1 H )
α− r1
3
.
(26.49)
Theorem 26.13 All as in Theorem 26.9. Assume further f (k) (M) = 0, k = 0, 1, . . . , n. Let r1 , r2 , r3 > 1 : r11 + r12 + r13 = 1, with α > n + r11 + r13 . Then | f (A) x, x| ≤ M T ( f ) α
r3 ,[m,M]
1
1
n! (r1 n + 1) r1 (r2 (α − n − 1) + 1) r2 ∀ x ∈ H.
(A − M1 H )
α− r1
3
x, x ,
(26.50)
26.3 Main Results
493
Inequality (26.50) means that f (A) ≤
M T ( f ) α
r3 ,[m,M]
1
1
n! (r1 n + 1) r1 (r2 (α − n − 1) + 1) r2
α− 1 ( A − M1 H ) r3 ,
(26.51)
and in particular f (A) ≤
M T ( f ) α
r3 ,[m,M]
α− r1
1 r1
n! (r1 n + 1) (r2 (α − n − 1) + 1)
1 r2
(A − M1 H )
3
.
(26.52)
We mention a left conformable fractional Poincaré type inequality: Theorem 26.14 ([4]) Let α ∈ (n, n + 1], n ∈ Z+ , f ∈ C n+1 ([m, M]). Assume f (k) (m) = 0, k = 0, 1, . . . , n. Let r1 , r2 , r3 > 1 : r11 + r12 + r13 = 1, such that α > n + r11 + r13 . Then
λ
(λ − m)αr3
| f (t)|r3 dt ≤
r3
λ m dt m Tα f (t)
r3
r3
(n!)r3 (r1 n + 1) r1 (r2 (α − n − 1) + 1) r2 (αr3 )
m
,
(26.53)
∀ λ ∈ [m, M] . We also mention a right conformable fractional Poincaré type inequality: Theorem 26.15 ([4]) Let α ∈ (n, n + 1], n ∈ Z+ , f ∈ C n+1 ([m, M]). Assume f (k) (M) = 0, k = 0, 1, . . . , n. Let r1 , r2 , r3 > 1 : r11 + r12 + r13 = 1, such that α > n + r11 + r13 . Then
M λ
| f (t)|r3 dt ≤
(M − λ)αr3
M M λ α T
r
f (t) 3 dt
r3
r3
(n!)r3 (r1 n + 1) r1 (r2 (α − n − 1) + 1) r2 (αr3 )
,
(26.54)
∀ λ ∈ [m, M] . We need Definition 26.16 Let the real valued function f ∈ C ([m, M]), and we consider g1 (t) =
t
f (z) dz, ∀ t ∈ [m, M] ,
(26.55)
f (z) dz, ∀ t ∈ [m, M] ,
(26.56)
m
then g1 ∈ C ([m, M]) . Similarly, for
g2 (t) = t
M
494
26 Fractional Conformable Self Adjoint Operator Analytic Inequalities
we have g2 ∈ C ([m, M]) . We denote by
A
f := (g1 ) = g1 (A) ,
(26.57)
f := (g2 ) = g2 (A) .
(26.58)
m1 H
and
M1 H A
We understand and write that (r > 0)
g1r (A) = g1r =: and g2r Clearly
A m1 H
f
(A) = g2r =:
r
A
,
f
(26.59)
m1 H
r
M1 H
.
f
(26.60)
A
r
r M1 , A H f are self adjoint operators on H , for any r > 0.
We present the following conformable fractional operator Poincaré type inequalities: Theorem 26.17 All as in Theorem 26.14. Then A | f |r3 ≤ m1 H
(A − m1 H )αr3 r3
A
r3
(n!)r3 (r1 n + 1) r1 (r2 (α − n − 1) + 1) r2 (αr3 )
m1 H
m r3 T f . α
Proof By Theorem 26.14 and properties (P) and (ii), see Sect. 26.1.
(26.61)
Theorem 26.18 All as in Theorem 26.15. Then M1 H | f |r3 ≤ A
(M1 H − A)αr3 r3
M1 H
r3
(n!)r3 (r1 n + 1) r1 (r2 (α − n − 1) + 1) r2 (αr3 )
A
M r3 T f . (26.62) α
Proof By Theorem 26.15 and properties (P) and (ii), see Sect. 26.1. We mention a left conformable fractional Sobolev type inequality:
26.3 Main Results
495
Theorem 26.19 ([4]) Let all as in Theorem 26.14 and r > 0. Then
λ
| f (t)| dt
r1 ≤
r
m
α− r1 + r1
(λ − m)
3
n! (r1 n + 1) (r2 (α − n − 1) + 1) r α − 1 r1
1 r2
1 r3
r1 +1
λ
m
m T f (t)r3 dt α
r13
,
(26.63) ∀ λ ∈ [m, M] . We also mention a right conformable fractional Sobolev type inequality: Theorem 26.20 ([4]) Let all as in Theorem 26.15 and r > 0. Then
M
r1
| f (t)| dt
≤
r
λ
α− r1 + r1
(M − λ)
3
1 1 n! (r1 n + 1) r1 (r2 (α − n − 1) + 1) r2 r α −
1 r3
r1 +1
M λ
M T f (t)r3 dt α
r1
3
,
(26.64) ∀ λ ∈ [m, M] . Applying (26.63), (26.64), and using properties (P) and (ii) of Sect. 26.1, we get the following conformable fractional operator Sobolev type inequalities. Theorem 26.21 All as in Theorem 26.19. Then
A
r1
| f |r
≤
m1 H
(A − m1 H )
α− r1 + r1
3
1 1 n! (r1 n + 1) r1 (r2 (α − n − 1) + 1) r2 r α −
1 r3
r1 +1
A m1 H
m r3 T f α
r1
3
.
(26.65) Theorem 26.22 All as in Theorem 26.20. Then
M1 H A
|f|
r
r1
≤
496
26 Fractional Conformable Self Adjoint Operator Analytic Inequalities
(M1 H − A)
α− r1 + r1
n! (r1 n + 1) (r2 (α − n − 1) + 1) r α − 1 r1
3
1 r2
1 r3
r1 +1
M1 H
M r3 Tf
A
r1
3
α
.
(26.66) We need the following conformable fractional Ostrowski type inequalities: Theorem 26.23 ([4]) Let α ∈ (0, 1], f ∈ C 1 ([m, M]). Then M 1 1 f dt − f (t) (λ) ≤ M −m α + 1) (α (M − m) m (λ − m)α+1 λα T ( f )∞,[m,λ] + (M − λ)α+1 Tαλ ( f )∞,[λ,M] ,
(26.67)
∀ λ ∈ [m, M] . Theorem 26.24 ([4]) Let α ∈ (0, 1], f ∈ C 1 ([m, M]), and r1 , r2 > 1 : 1, with α > r12 . Then
+
1 r1
M 1 1 ≤ f dt − f (t) (λ) 1 M −m m (M − m) (r1 (α − 1) + 1) r1 α +
1 r1
α+ 1 α+ 1 (M − λ) r1 Tαλ ( f )r2 ,[λ,M] + (λ − m) r1 λα T ( f )r2 ,[m,λ] ,
1 r2
=
(26.68)
∀ λ ∈ [m, M] . Corollary 26.25 (to Theorem 26.23) Let α ∈ (0, 1], f ∈ C 1 ([m, M]), and λ T ( f ) , Tαλ ( f )∞,[λ,M] ≤ K 1 ( f ) , α ∞,[m,λ]
(26.69)
∀ λ ∈ [m, M], where K 1 ( f ) > 0. Then M 1 f (t) dt − f (λ) ≤ M −m m K1 ( f ) (λ − m)α+1 + (M − λ)α+1 , α (α + 1) (M − m)
(26.70)
∀ λ ∈ [m, M] . Corollary 26.26 (to Theorem 26.24) Let α ∈ (0, 1], f ∈ C 1 ([m, M]), and r1 , r2 > 1 : r11 + r12 = 1, with α > r12 . Assume that λ T ( f ) α
r2 ,[λ,M]
, λα T ( f )r2 ,[m,λ] ≤ K 2 ( f ) ,
(26.71)
26.3 Main Results
497
∀ λ ∈ [m, M], where K 2 ( f ) > 0. Then M 1 f (t) dt − f (λ) ≤ M −m m K2 ( f )
1
(M − m) (r1 (α − 1) + 1) r1 α +
1 r1
1 1
(M − λ)α+ r1 + (λ − m)α+ r1 , (26.72)
∀ λ ∈ [m, M] . By property (P) and Corollaries 26.25, 26.26 we derive the following operator conformable fractional Ostrowski type inequalities: Theorem 26.27 All as in Corollary 26.25. Then
M 1 f (t) dt 1 H − f (A) ≤ M −m m K1 ( f ) (A − m1 H )α+1 + (M1 H − A)α+1 . α (α + 1) (M − m)
(26.73)
Theorem 26.28 All as in Corollary 26.26. Then
M 1 f (t) dt 1 H − f (A) ≤ M −m m K2 ( f ) 1
(M − m) (r1 (α − 1) + 1) r1 α +
1 r1
1 1
(M1 H − A)α+ r1 + (A − m1 H )α+ r1 . (26.74)
We need the following conformable fractional Opial type inequalities: Theorem 26.29 ([4]) Let α ∈ (n, n + 1], n ∈ Z+ , f ∈ C n+1 ([m, M]). Assume f (k) (m) = 0, k = 0, 1, . . . , n. Let p, q > 1 : 1p + q1 = 1, α > n + q1 . Then
λ
m
| f (w)| Tm α f (w) dw ≤
(M − m)n (λ − m)α−n−1+ p 2
1
λ
1
n!2 q [( p (α − n − 1) + 1) ( p (α − n − 1) + 2)] p
m
m T ( f ) (w)q dw α
q2 ,
(26.75)
∀ λ ∈ [m, M] . Theorem 26.30 ([4]) Let α ∈ (n, n + 1], n ∈ Z+ , f ∈ C n+1 ([m, M]). Assume f (k) (M) = 0, k = 0, 1, . . . , n. Let p, q > 1 : 1p + q1 = 1, α > n + q1 . Then
498
26 Fractional Conformable Self Adjoint Operator Analytic Inequalities
M λ
| f (w)| αM T f (w) dw ≤
(M − m)n (M − λ)α−n−1+ p 2
1
1
n!2 q [( p (α − n − 1) + 1) ( p (α − n − 1) + 2)] p
M
λ
M T ( f ) (w)q dw
q2
α
,
(26.76)
∀ λ ∈ [m, M] . Using Theorems 26.29, 26.30 and property (P) we derive the next operator conformable fractional Opial type inequalities: Theorem 26.31 All as in Theorem 26.29. Then A (M − m)n ≤ | f | Tm f 1 1 α m1 H n!2 q [( p (α − n − 1) + 1) ( p (α − n − 1) + 2)] p (A − m1 H )
α−n−1+ 2p
A
m T ( f )q
m1 H
q2
α
.
(26.77)
Theorem 26.32 All as in Theorem 26.30. Then M1 H (M − m)n | f | αM T ( f ) ≤ 1 1 A n!2 q [( p (α − n − 1) + 1) ( p (α − n − 1) + 2)] p (M1 H − A)
α−n−1+ 2p
M1 H A
M T ( f )q α
q2
.
(26.78)
References 1. T. Abdeljawad, On conformable fractional calculus. J. Comput. Appl. Math. 279, 57–66 (2015) 2. G.A. Anastassiou, Nonlinearity: Ordinary and Fractional Approximations by Sublinear and Max-Product Operators, Studies in Systems, Decision and Control (Springer, Heidelberg, 2018) 3. G.A. Anastassiou, Conformable fractional self adjoint operator analytic inequalities, submitted for publication, 2019 4. G.A. Anastassiou, Conformable fractional inequalities, in Functional Equations and Analytic Inequalities, ed. by G. Anastassiou, J. Rassias (Springer, New York, 2020), accepted for publication 5. S.S. Dragomir, Inequalities for functions of selfadjoint operators on Hilbert spaces (2011), www. ajmaa.org/RGMIA/monographs/InFuncOp.pdf 6. S. Dragomir, Operator Inequalities of Ostrowski and Trapezoidal Type (Springer, New York, 2012) 7. T. Furuta, J. Mi´ci´c Hot, J. Peˇcaric, Y. Seo, Mond-Peˇcaric Method in Operator Inequalities, Inequalities for Bounded Self adjoint Operators on a Hilbert Space (Element, Zagreb, 2005)
References
499
8. G. Helmberg, Introduction to Spectral Theory in Hilbert Space (Wiley, New York, 1969) 9. R. Khalil, M. Al. Horani, A. Yousef, M. Sababheh, A new definition of fractional derivative. J. Comput. Appl. Math. 264, 65–70 (2014)
Chapter 27
Fractional Left Local General M-Derivative
Here is introduced and studied the left fractional local general M-derivative of various orders. All basic properties of an ordinary derivative are established here. We also define the corresponding left fractional M-integrals. Important theorems are established such as: the inversion theorem, the fundamental theorem of fractional calculus, the mean value theorem, the extended mean value theorem, the Taylor’s formula with integral remainder, the integration by parts. Our left fractional derivative generalizes the alternative fractional derivative and the local M-fractional derivative. See also [3].
27.1 Introduction The fractional calculus started in 1695 by L’Hospital and Leibniz [7]. Since then many type of fractional derivatives have been proposed. Most of them are defined via summation integration and differentiation, they are not local and they do not have the basic properties of the ordinary differentiation. For example, see the Riemann– Liouville, Caputo and Canavati fractional derivatives, [2]. In the last fifty years fractional calculus, due to applications in many other sciences, has become very popular and people tried to invent fractional derivatives not having the above deficiencies and behave like the ordinary derivative. Most known are the following local new generation fractional derivatives: (i) Let a ∈ R. The left conformable fractional derivative starting from a of a function f : [a, ∞) → R, of order 0 < α ≤ 1, is defined by
Tαa
f t + ε (t − a)1−α − f (t) , f (t) = lim ε→0 ε
if Tαa f (t) exists on (a, b), then © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_27
501
502
27 Fractional Left Local General M-Derivative
Tαa f (a) = lim Tαa f (t) , t→a+
see [1, 6]. (ii) Let f : [0, ∞) → R and t > 0. Then the alternative fractional derivative of order α of f is defined by [5] −α − f (t) f teεt , Dα f (t) = lim ε→0 ε ∀ t > 0 and α ∈ (0, 1) . Also, if f has the alternative fractional derivative over (0, a), a > 0 and lim Dα f (t) exists, then we define
t→0+
Dα (0) = lim Dα f (t) . t→0+
(iii) In [9] the authors introduce the local M-fractional derivative that generalizes the alternative fractional derivative as follows: Let f : [0, ∞) → R and t > 0. For 0 < α < 1 we define the local M-derivative α,β of order α of function f , denoted D M f (t), by α,β
D M f (t) = lim
ε→0
f tEβ εt −α − f (t) , ε
∀ t > 0, where E (·), β > 0 is the Mittag-Leffler function with one parameter [4, α,β α,β 8]. Note that if D M f (t) exists over (0, a), a > 0, and lim D M f (t) exists, then t→0+
we define
α,β
α,β
D M f (0) = lim D M f (t) . t→0+
We mention that Eβ (x) =
∞ k=0
xk , β > 0, (βk + 1)
where is the Gamma function. The article [9] has great ideas, it establishes all the nice basic properties of the ordinary derivatives for the M-fractional derivative and tries to go further, but it contains many major mistakes. Most notably in [9]: the mean value theorem and Taylor’s theorem are very wrong. Article [9] defines wrongly the M-integral for 0 < α < 1 anf fails to define it for α ∈ (n, n + 1], n ∈ N. Also, there are errors in many other results and proofs in [9]. Our chapter comes to extend the M-derivative over the interval [a, ∞), a ∈ R, as opposed to only [0, ∞), also to fix all the above mistakes of [9].
27.1 Introduction
503
We study here the left local general M-derivative of order α of function f , denoted α,β by D M,a f (t), see Definition 27.1. We work in parallel to [9], only more general and by fixing all errors.
27.2 Main Results We make Definition 27.1 Let f : [a, ∞) → R and t > a, a ∈ R. For 0 < α ≤ 1 we define α,β the left local general M-derivative of order α of function f , denoted by D M,a f (t), by f tEβ ε (t − a)−α − f (t) α,β , (27.1) D M,a f (t) := lim ε→0 ε ∀ t > a, where Eβ (t) =
∞ k=0
tk , (βk+1)
β > 0, is the Mittag-Leffler function with one
parameter. α,β α,β If D M,a f (t) exists over (a, γ), γ ∈ R and lim D M,a f (t) exists, then t→a+
α,β
α,β
D M,a f (a) = lim D M,a f (t) . t→a+
(27.2)
Theorem 27.2 If a function f : [a, ∞) → R has the left local general M-derivative of order α ∈ (0, 1], β > 0, at t0 > a, then f is continuous at t0 . Proof We form
lim f t0 E β ε (t0 − a)−α − f (t0 ) =
ε→0
f t0 E β ε (t0 − a)−α − f (t0 ) lim lim ε = ε→0 ε→0 ε
α,β D M,a f (t0 ) lim ε = 0. ε→0
Thus, f is continuous at t0 .
Remark 27.3 For f continuous we have (t > a) lim f tEβ ε (t − a)−α = lim f
ε→0
ε→0
f
k ∞ ε (t − a)−α t = (βk + 1) k=0
k ∞ ε (t − a)−α t lim = f (t) , ε→0 (βk + 1) k=0
(27.3)
504
27 Fractional Left Local General M-Derivative
by
k ∞ ε (t − a)−α = 1, ε→0 (βk + 1) k=0 lim
the only term that contributes to the sum is k = 0. We give Theorem 27.4 Consider f differentiable, β > 0, t > a, 0 < α ≤ 1, then t (t − a)−α f (t) . (β + 1)
α,β
D M,a ( f ) (t) = Proof (t > a)
t +t
(27.4)
k ∞ ε (t − a)−α −α = =t tEβ ε (t − a) (βk + 1) k=0
2 3 n ε (t − a)−α ε (t − a)−α ε (t − a)−α ε (t − a)−α +t +t + ··· + t + ··· (β + 1) (2β + 1) (3β + 1) (nβ + 1)
εt (t − a)−α + O ε2 . = t+ (β + 1) Let h := εt (t − a)−α
(27.5)
1 + O (ε) , (β + 1)
(27.6)
so that ε=
t (t − a)−α
h 1 (β+1)
+ O (ε)
=
h (t − a)α (β + 1) . t (1 + (β + 1) O (ε))
(27.7)
Therefore we get: α,β
D M,a f (t) = lim
f t+
εt(t−a)−α (β+1)
+ O ε2 − f (t) ε
ε→0
lim
ε→0
=
f (t + h) − f (t) ( f (t + h) − f (t)) t (1 + (β + 1) O (ε)) = lim = ε→0 ε h (t − a)α (β + 1)
t (t − a)−α (β + 1)
lim
ε→0
(if ε → 0, then h → 0)
f (t + h) − f (t) h
lim (1 + (β + 1) O (ε))
ε→0
(27.8)
27.2 Main Results
505
=
t (t − a)−α f (t) , (β + 1)
with β > 0, t > a, proving the claim.
We give Theorem 27.5 Let 0 < α ≤ 1, β > 0, λ1 , λ2 ∈ R and for f, g : [a, ∞) → R there α,β α,β exist D M,a f (t), D M,a g (t) at the point t > a. Then (1) (Linearity) α,β
α,β
α,β
D M,a (λ1 f + λ2 g) (t) = λ1 D M,a f (t) + λ2 D M,a g (t) ,
(27.9)
(2) (Product Rule) α,β
α,β
α,β
D M,a ( f · g) (t) = f (t) D M,a g (t) + g (t) D M,a f (t) ,
(27.10)
(3) (Quotient rule) α,β
D M,a
α,β α,β g (t) D M,a f (t) − f (t) D M,a g (t) f , (t) = g (g (t))2
(27.11)
α,β
(4) D M,a (c) = 0, where c is a constant, (5) (Chain rule) α,β
α,β
D M,a ( f ◦ g) (t) = f (g (t)) D M,a g (t) ,
(27.12)
for f differentiable at g (t). Proof Similar to [9] and by (27.3).
We give Theorem 27.6 Let λ ∈ R, β > 0, 0 < α ≤ 1, t > a. Then α,β (1) D M,a (1) = 0, −α α,β (2) D M,a eλt = t(t−a) λeλt , (β+1) α,β
t(t−a)−α λ cos λt, (β+1) t(t−a)−α (cos λt) = − (β+1) λ sin λt, γ (t−a)−α . (t γ ) = γt(β+1)
(3) D M,a (sin λt) = α,β
(4) D M,a α,β
(5) D M,a
Proof By (27.4). We give
506
27 Fractional Left Local General M-Derivative
Theorem 27.7 (Rolle’s theorem) Let f : [γ, δ] → R, where γ > a : (1) f is continuous on [γ, δ] , α,β (2) there exists D M,a f on (γ, δ) for some α ∈ (0, 1], (3) f (γ) = f (δ) . α,β Then, there exists c ∈ (γ, δ), such that D M,a f (c) = 0, β > 0.
Proof Similar to [9]. We give
Theorem 27.8 (Mean value theorem) Let f : [γ, δ] → R with γ > a, 0 ∈ / [r, δ], such that (1) f is continuous on [γ, δ] , α,β (2) there exists D M,a f on (γ, δ) for some α ∈ (0, 1]. Then, there exists c ∈ (γ, δ) such that (β + 1) (c − a)α
α,β f (δ) − f (γ) = D M,a f (c) (δ − γ) . c
(27.13)
Proof Consider the function g (x) := f (x) − f (γ) −
f (δ) − f (γ) (x − γ) , ∀ x ∈ [γ, δ] . δ−γ α,β
Notice that g is continuous on [γ, δ], there exists D M,a g on (γ, δ) for some α ∈ (0, 1], and g (γ) = 0 = g (δ). By Rolle’s theorem there exists c ∈ (γ, δ) such that α,β D M,a g (c) = 0. Hence
f (δ) − f (γ) c (c − a)−α α,β D M,a f (c) = , δ−γ (β + 1)
proving the claim. We continue with
Theorem 27.9 (Extended mean value theorem) Let γ > a, 0 ∈ / [r, δ], and f, g : [γ, δ] → R : (1) f, g are continuous on [γ, δ] , g(r ) = g(δ), α,β α,β (2) there exists D M,a f, D M,a g for some α ∈ (0, 1], over (γ, δ) . Then, there exists c ∈ (γ, δ) such that α,β
D M,a f (c) with β > 0. Proof Similar to [9]. We need
α,β D M,a g (c)
=
f (δ) − f (γ) , g (δ) − g (γ)
(27.14)
27.2 Main Results
507
Definition 27.10 Let β > 0, α ∈ (n, n + 1], where n ∈ N and f is n-times differentiable (in the ordinary sense) for t > a. Then the left local general M-derivative α,β;n of order α of function f , denoted by D M,a f (t) is defined by α,β;n D M,a
f (n) tEβ ε (t − a)n−α − f (n) (t) f (t) := lim , ε→0 ε
given that the limit exists. Notice that α,β;n α−n,β D M,a f (t) = D M,a f (n) (t) , t > a.
(27.15)
(27.16)
For f (n + 1)-differentiable for t > a, and α ∈ (n, n + 1] we get that α,β;n
D M,a f (t) =
t (t − a)n−α (n+1) f (t) . (β + 1)
(27.17)
We need Definition 27.11 (left M-integral) Let t ≥ a and f be a function defined in (a, t] and 0 < α ≤ 1. Then the M-integral of order α of the function f is defined by α,β I M,a
f (t) = (β + 1) a
t
f (x) d x, x (x − a)−α
(27.18)
with β > 0. We give Theorem 27.12 (Inverse) Let a ∈ R, 0 < α ≤ 1. Also, let f be a continuous funcα,β tion such that there exists I M,a f . Then α,β
α,β
D M,a I M,a f (t) = f (t) ,
(27.19)
with 0 = t > a and β > 0. Proof We use Theorem 27.4 (0 = t > a) α,β
α,β
D M,a I M,a f (t) =
t (t − a)−α α,β I M,a f (t) = (β + 1)
t (t − a)−α (β + 1) f (t) = f (t) . (β + 1) t (t − a)−α Next we give
508
27 Fractional Left Local General M-Derivative
Theorem 27.13 (Fundamental theorem of calculus) Let continuously differentiable α,β f : [a, ∞) → R such that D M,a f exists and 0 < α ≤ 1. Then, for all t > a we have α,β
α,β
I M,a D M,a f (t) = f (t) − f (a) ,
(27.20)
with β > 0. Proof We have α,β
α,β
I M,a D M,a f (t) = (β + 1) (β + 1)
t
a
t
α,β
D M,a f (x) d x
a
x (x − a)−α
(27.4)
=
1 x (x − a)−α f (x) d x = f (t) − f (a) . −α (β + 1) x (x − a)
Note 27.14 If f (a) = 0, by (27.20) we find that α,β
α,β
I M,a D M,a f (t) = f (t) , ∀ t > a. We need Definition 27.15 (general left M-integral) Let t ≥ a and f be a function defined in (a, t] and α ∈ (n, n + 1], n ∈ Z+ . Then the general left M-integral of order α of the function f is defined by α,β;n I M,a
(β + 1) f (t) = n!
α,β;0
α,β
(t − x)n f (x) d x. x (x − a)n−α
t
a
(27.21)
Notice that I M,a f (t) = I M,a f (t) . We give a left fractional Taylor’s formula with integral remainder regarding the left local general M-derivative: Theorem 27.16 Let α ∈ (n, n + 1], n ∈ Z+ and f : [a, ∞) → R be (n + 1)-times continuously differentiable for t > a, β > 0. Then ∀ t > a, we have α,β;n
α,β;n
I M,a D M,a f (t) = f (t) −
n f (k) (a) (t − a)k . k! k=0
(27.22)
Proof We have that α,β;n α,β;n I M,a D M,a
(27.21)
f (t) =
(β + 1) n!
a
t
(t − x)n (27.17) α,β;n D f (x) d x = x (x − a)n−α M,a
27.2 Main Results
509
(β + 1) n!
1 n!
t
a
t
(t − x)n x (x − a)n−α (n+1) f (x) d x = x (x − a)n−α (β + 1)
(t − x)n f (n+1) (x) d x = f (t) −
a
n f (k) (a) (t − a)k , k! k=0
(27.23)
proving the claim. Note 27.17 We would like to rewrite (27.22) in a convenient way: f (t) =
n (β + 1) t f (k) (a) α,β;n (t − a)k + (t − x)n x −1 (x − a)α−n D M,a f (x) d x, k! n! a k=0
(27.24) ∀ t > a.
Denote α,β I M,a
f (t) =
t
f (x) dα x,
(27.25)
a
(β+1) where dα x = x(x−a) −α d x. Integration by parts comes next:
Theorem 27.18 Let f, g : [a, b] → R be two functions such that f, g are continuously differentiable and 0 < α ≤ 1. Then a
b
α,β
f (x) D M,a g (x) dα x = f (x) g (x) |ab −
b
a
α,β
g (x) D M,a f (x) dα x, β > 0. (27.26)
Proof We have that a
b
α,β
f (x) D M,a g (x) dα x = a
a
b
α,β
f (x) D M,a g (x)
a
f (x) g (x) d x = f (x) g (x) |ab −
f (x) g (x) |ab −
b
g (x)
a
(x) g (x) |ab
b
g (x) f (x) d x =
a
x (x − a)−α (β + 1) f (x) dx = (β + 1) x (x − a)−α
f
(β + 1) dx = x (x − a)−α
x (x − a)−α (β + 1) g (x) dx = (β + 1) x (x − a)−α
b
f (x)
b
− a
b
α,β
g (x) D M,a f (x) dα x,
510
27 Fractional Left Local General M-Derivative
proving the claim. We finish with
Remark 27.19 Let β > 0, 0 < α ≤ 1, a > 0 and f : [a, b] → R be continuous, t ∈ [a, b]. Then t (x − a)α α,β | f (x)| d x ≤ I M,a f (t) ≤ (β + 1) x a (β + 1) f ∞ a That is
a
t
(x − a)α d x =
(β + 1) (t − a)α+1 f ∞ . a α+1
(β + 1) f α,β ∞ (t − a)α+1 , I M,a f (t) ≤ a (α + 1)
(27.27)
∀ t ∈ [a, b] .
References 1. T. Abdeljawad, On conformable fractional calculus. J. Comput. Appl. Math. 279, 57–66 (2015) 2. G.A. Anastassiou, Fractional Differentiation Inequalities (Springer, Heidelberg, 2009) 3. G.A. Anastassiou, On the left fractional local general M-Derivative, submitted for publication (2019) 4. R. Gorenflo, A.A. Kilbas, F. Mainardi, S.V. Rogosin, Mittag-Leffler Functions, Related Topics and Applications (Springer, Berlin, 2014) 5. U.N. Katugampola, A new fractional derivative with classical properties (2014). arXiv:1410.6535v2 6. R. Khalil, M. Al Horani, A. Yousef, M. Sababheh, A new definition of fractional derivative. J. Comput. Appl. Math. 264, 65–70 (2014) 7. G.W. Leibniz, Letter from Hanover, Germany, to G.F.A L’Hospital, September 30, 1695, Leibniz Mathematische Schriften (Olms-Verlag, Hildescheim), pp. 301–302 (First published in 1849) 8. G.M. Mittag-Leffler, Sur la nouvelle fonction Eα (x). CR Acad. Sci. Paris 137, 554–558 (1903) 9. J.V.C. Sousa, E.C. de Oliveira, On the local M-derivative. Progr. Fract. Differ. Appl. 4(4), 479– 492 (2018)
Chapter 28
Fractional Right Local General M-Derivative
Here is introduced and studied the right fractional local general M-derivative of various orders. All basic properties of an ordinary derivative are established here. We also define the corresponding right fractional M-integrals. Important theorems are established such as: the inversion theorem, the fundamental theorem of fractional calculus, the mean value theorem, the extended mean value theorem, the right fractional Taylor’s formula with integral remainder, the integration by parts. Our right fractional derivative complements the alternative fractional derivative and the local M-fractional derivative. See also [3].
28.1 Introduction The fractional calculus started in 1695 by L’Hospital and Leibniz [7]. Since then many type of fractional derivatives have been proposed. Most of them are defined via summation integration and differentiation, they are not local and they do not have the basic properties of the ordinary differentiation. For example, see the RiemannLiouville, Caputo and Canavati fractional derivatives, [2]. In the last fifty years fractional calculus, due to applications in many other sciences, has become very popular and people tried to invent fractional derivatives not having the above deficiencies and behave like the ordinary derivative. Most known are the following local new generation fractional derivatives: (i) Let b ∈ R. The right conformable fractional derivative terminating at b of a function f : (−∞, b] → R, of order 0 < α ≤ 1, is defined by f t + ε (b − t)1−α − f (t) , α T f (t) = −lim ε→0 ε
b if
b
αT
f (t) exists on (a, b), then
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_28
511
512
28 Fractional Right Local General M-Derivative
b
αT
f (a) = lim bα T f (t) , t→b−
see [1, 6]. (ii) Let f : [0, ∞) → R and t > 0. Then the (left) alternative fractional derivative of order α of f is defined by [5] −α − f (t) f teεt , Dα f (t) = lim ε→0 ε ∀ t > 0 and α ∈ (0, 1) . Also, if f has the alternative fractional derivative over (0, a), a > 0 and lim Dα f (t) exists, then we define
t→0+
Dα (0) = lim Dα f (t) . t→0+
(iii) In [9] the authors introduce the (left) local M-fractional derivative that generalizes the alternative fractional derivative as follows: Let f : [0, ∞) → R and t > 0. For 0 < α < 1 we define the local M-derivative α,β of order α of function f , denoted D M f (t), by α,β DM
f tEβ εt −α − f (t) , f (t) = lim ε→0 ε
∀ t > 0, where Eβ (·), β > 0 is the Mittag-Leffler function with one parameter [4, α,β α,β 8]. Note that if D M f (t) exists over (0, a), a > 0, and lim D M f (t) exists, then t→0+
we define
α,β
α,β
D M f (0) = lim D M f (t) . t→0+
We mention that Eβ (x) =
∞ k=0
xk , β > 0, (βk + 1)
where is the Gamma function. The article [9] has great ideas, it establishes all the nice basic properties of the ordinary derivatives for the M-fractional derivative and tries to go further but it contains many major mistakes. So we are greatly inspired by [9]. We study here the right local general M-derivative of order α of function f , α,β denoted by M,b D f (t), see Definition 28.1. We work in parallel to [9] and only more general.
28.2 Main Results
513
28.2 Main Results We make Definition 28.1 Let f : (−∞, b] → R and t < b, b ∈ R. For 0 < α ≤ 1 we define α,β the right local general M-derivative of order α of function f , denoted as M,b D f (t), by f tEβ ε (b − t)−α − f (t) α,β , (28.1) M,b D f (t) := −lim ε→0 ε ∀ t < b, where Eβ (t) =
∞ k=0
tk , (βk+1)
β > 0, is the Mittag-Leffler function with one
parameter. α,β α,β If M,b D f (t) exists over (γ, b), γ ∈ R and lim M,b D f (t) exists, then α,β M,b D f
(b) = lim
α,β
t→b− M,b
D f (t) .
(28.2)
Theorem 28.2 If a function f : (−∞, b] → R has the right local general Mderivative of order α ∈ (0, 1], β > 0, at t0 < b, then f is continuous at t0 . Proof We form
−lim f t0 E β ε (b − t0 )−α − f (t0 ) = ε→0
f t0 E β ε (b − t0 )−α − f (t0 ) −lim lim ε = ε→0 ε→0 ε
α,β M,b D f
(t0 ) lim ε = 0. ε→0
Thus, f is continuous at t0 .
Remark 28.3 For f continuous we have (t < b) lim f tEβ ε (b − t)−α = lim f
ε→0
ε→0
k ∞ ε (b − t)−α t = (βk + 1) k=0
f by
k ∞ ε (b − t)−α t lim = f (t) , ε→0 (βk + 1) k=0 k ∞ ε (b − t)−α = 1, lim ε→0 (βk + 1) k=0
(28.3)
514
28 Fractional Right Local General M-Derivative
as the only term that contributes to the sum is k = 0. We give Theorem 28.4 Consider f differentiable, β > 0, t < b, 0 < α ≤ 1, then α,β M,b D f
Proof (t < b)
t +t
(t) = −
t (b − t)−α f (t) . (β + 1)
(28.4)
k ∞ ε (b − t)−α −α = tEβ ε (b − t) =t (βk + 1) k=0
2 3 n ε (b − t)−α ε (b − t)−α ε (b − t)−α ε (b − t)−α +t +t + ··· + t + · · · (28.5) (β + 1) (2β + 1) (3β + 1) (nβ + 1)
ε (b − t)−α = t +t + O ε2 . (β + 1) Let h := εt (b − t)−α
1 + O (ε) , (β + 1)
(28.6)
so that ε=
t (b − t)−α
h 1 (β+1)
+ O (ε)
=
h (b − t)α (β + 1) . t (1 + (β + 1) O (ε))
(28.7)
Therefore we get: α,β M,b D f
−lim
ε→0
−
(t) = −lim
f t+
εt(b−t)−α (β+1)
+ O ε2 − f (t) ε
ε→0
=
f (t + h) − f (t) ( f (t + h) − f (t)) t (1 + (β + 1) O (ε)) = −lim = ε→0 ε h (b − t)α (β + 1)
t (b − t)−α (β + 1)
lim
ε→0
f (t + h) − f (t) h
(if ε → 0, then h → 0) =−
lim (1 + (β + 1) O (ε))
ε→0
(28.8)
t (b − t)−α f (t) , (β + 1)
with β > 0, t < b, proving the claim.
28.2 Main Results
515
We give Theorem 28.5 Let 0 < α ≤ 1, β > 0, λ1 , λ2 ∈ R and for f, g : (−∞, b] → R α,β α,β there exist M,b D f (t), M,b Dg (t) at the point t < b. Then (1) (Linearity) α,β M,b D (λ1
α,β
α,β
f + λ2 g) (t) = λ1M,b D f (t) + λ2M,b Dg (t) ,
(28.9)
(2) (Product Rule) α,β M,b D (
α,β
α,β
f · g) (t) = f (t) M,b Dg (t) + g (t) M,b D f (t) ,
(28.10)
α,β α,β g (t) M,b D f (t) − f (t) M,b Dg (t) f , (t) = g (g (t))2
(28.11)
(3) (Quotient rule) α,β M,b D α,β
(4) M,b D (c) = 0, where c is a constant, (5) (Chain rule) α,β M,b D (
α,β
f ◦ g) (t) = f (g (t)) M,b Dg (t) ,
(28.12)
for f differentiable at g (t). Proof Similar to [9] and by (28.3).
We give Theorem 28.6 Let λ ∈ R, β > 0, 0 < α ≤ 1, t < b. Then α,β (1) M,b D (1) = 0, −α α,β (2) M,b D eλt = − t(b−t) λeλt , (β+1) −α
α,β
(3) M,b D (sin λt) = − t(b−t) λ cos λt, (β+1) α,β
t(b−t)−α λ sin λt, (β+1) γt γ (b−t)−α − (β+1) .
(4) M,b D (cos λt) = α,β
(5) M,b D (t γ ) = Proof By (28.4).
We give Theorem 28.7 (Rolle’s theorem) Let f : [γ, δ] → R, where δ < b : (1) f is continuous on [γ, δ] , α,β (2) there exists M,b D f on (γ, δ) for some α ∈ (0, 1], (3) f (γ) = f (δ) . α,β Then, there exists c ∈ (γ, δ), such that M,b D f (c) = 0, β > 0. Proof Similar to [9].
516
28 Fractional Right Local General M-Derivative
We give Theorem 28.8 (Mean value theorem) Let f : [γ, δ] → R with δ < b, 0 ∈ / [r, δ], such that (1) f is continuous on [γ, δ] , α,β (2) there exists M,b D f on (γ, δ) for some α ∈ (0, 1]. Then, there exists c ∈ (γ, δ) such that (β + 1) (b − c)α
α,β f (δ) − f (γ) = − M,b D f (c) (δ − γ) . c
(28.13)
Proof Consider the function g (x) := f (x) − f (γ) −
f (δ) − f (γ) (x − γ) , ∀ x ∈ [γ, δ] . δ−γ α,β
Notice that g is continuous on [γ, δ], and there exists M,b Dg on (γ, δ) for some α ∈ (0, 1], and g (γ) = 0 = g (δ). By Rolle’s theorem there exists c ∈ (γ, δ) such α,β that M,b Dg (c) = 0. Hence
c (b − c)−α f (δ) − f (γ) α,β , D f = − (c) M,b δ−γ (β + 1)
proving the claim. We continue with
Theorem 28.9 (Extended mean value theorem) Let δ < b, 0 ∈ / [r, δ], and f, g : [γ, δ] → R : (1) f, g are continuous on [γ, δ] , g(r ) = g(δ), α,β α,β (2) there exists M,b D f, M,b Dg for some α ∈ (0, 1], over (γ, δ) . Then, there exists c ∈ (γ, δ) such that α,β M,b D f (c) α,β M,b Dg (c)
=
f (δ) − f (γ) , g (δ) − g (γ)
(28.14)
with β > 0.
Proof Similar to [9]. We need
Definition 28.10 Let β > 0, α ∈ (n, n + 1], where n ∈ N and f is n-times differentiable (in the ordinary sense) for t < b. Then the right local general M-derivative α,β;n of order α of function f , denoted as M,b D f (t) is defined by α,β;n M,b D f
(t) := (−1)
n+1
f (n) tEβ ε (b − t)n−α − f (n) (t) lim , ε→0 ε
(28.15)
28.2 Main Results
517
given that the limit exists. Notice that α,β;n n M,b D f (t) = (−1)
α−n,β (n) M,b D f
(t) , t < b.
(28.16)
For f (n + 1)-differentiable for t < b, and α ∈ (n, n + 1] we get that α,β;n M,b D f
(t) = (−1)n+1
t (b − t)n−α (n+1) f (t) . (β + 1)
(28.17)
We need Definition 28.11 (right M-integral) Let t ≤ b and f be a function defined in [t, b) and 0 < α ≤ 1. Then th right M-integral of order α of the function f is defined by α,β M,b I
f (t) = (β + 1) t
b
f (x) d x, x (b − x)−α
(28.18)
with β > 0. We give Theorem 28.12 (Inverse) Let b ∈ R, 0 < α ≤ 1. Also, let f be a continuous function α,β such that there exists M,b I f . Then α,β M,b D
α,β M,b I
f (t) = f (t) ,
(28.19)
with 0 = t < b and β > 0. Proof We use Theorem 28.4 (0 = t < b) α,β M,b D
α,β M,b I
f (t) = −
t (b − t)−α − (β + 1)
t (b − t)−α α,β = M,b I f (t) (β + 1)
(β + 1) f (t) = f (t) . − t (b − t)−α
Next we give Theorem 28.13 (Fundamental theorem of calculus) Let f : (−∞, b] → R continα,β uous differentiable, such that M,b D f exists and 0 < α ≤ 1. Then, for all t < b we have α,β α,β (28.20) M,b I M,b D f (t) = f (t) − f (b) , with β > 0.
518
28 Fractional Right Local General M-Derivative
Proof We have α,β α,β M,b I M,b D f
(β + 1)
b α,β D f M,b
(t) = (β + 1)
x (b − x)−α
t
f (x) −x (b − x)−α dx = (β + 1) x (b − x)−α
b
t
(x) d x
t
(28.4)
=
f (x) d x = f (t) − f (b) .
b
Note 28.14 If f (b) = 0, by (28.20) we find that α,β α,β M,b I M,b D f
(t) = f (t) , ∀ t < b.
We need Definition 28.15 (general right M-integral) Let t ≤ b and f be a function defined in [t, b) and α ∈ (n, n + 1], n ∈ Z+ . Then the general right M-integral of order α of the function f is defined by α,β;n M,b I α,β;0
f (t) =
(β + 1) n!
b
t
(x − t)n f (x) d x. x (b − x)n−α
(28.21)
α,β
Notice that M,b I f (t) = M,b I f (t) . We give a right fractional Taylor’s formula with integral remainder regarding the right local general M-derivative: Theorem 28.16 Let α ∈ (n, n + 1], n ∈ Z+ and f : (−∞, b] → R be (n + 1)times continuously differentiable for t < b, β > 0. Then ∀ t < b, we have α,β;n α,β;n M,b I M,n D f
(t) = f (t) −
n f (k) (b) (t − b)k . k! k=0
(28.22)
Proof We have that α,β;n α,β;n M,b I M,n D f
(β + 1) n! (−1)n+1 n!
b
b t
(28.21)
(t) =
(x − t)n x (b − x)n−α
(x − t) f n
t
(β + 1) n!
(n+1)
b
t
(−1)n+1
(x − t)n
α,β;n M,b D f n−α
x (b − x) x (b − x)n−α (β + 1)
(−1)n+1 (x) d x = (−1)n+1 n!
b
t
(x)
(28.17)
dx =
f (n+1) (x) d x = (t − x)n f (n+1) (x) d x
28.2 Main Results
=
1 n!
t
519 n f (k) (b) (t − b)k , k! k=0
(t − x)n f (n+1) (x) d x = f (t) −
b
(28.23)
∀ t < b, proving the claim.
Note 28.17 We would like to rewrite (28.22) as follows: f (t) =
n
(β + 1) b f (k) (b) α,β;n (x − t)n x −1 (b − x)α−n M,b D f (x) d x, (t − b)k + k! n! t k=0
(28.24) ∀ t < b. Denote α,β M,b I
b
f (t) = −
f (x) dα x,
(28.25)
t
(β+1) where d α x = − x(b−x) −α d x. Integration by parts comes next:
Theorem 28.18 Let f, g : [a, b] → R be two functions such that f, g are continuously differentiable and 0 < α ≤ 1. Then
b
α,β
f (x) M,b Dg (x) dα x = f (x) g (x) |ab −
a
a
b
α,β
g (x) M,b D f (x) dα x, β > 0. (28.26)
Proof We have that a
b
α,β
f (x) M,b Dg (x) dα x =
b
a
a
b
a
x (b − x)−α (β + 1) dx = g (x) − f (x) − (β + 1) x (b − x)−α f (x) g (x) d x = f (x) g (x) |ab −
f
(β + 1) α,β dx = f (x) M,b Dg (x) − x (b − x)−α
b
(x) g (x) |ab
− a
b
b
g (x) f (x) d x =
a
x (b − x)−α (β + 1) dx = f (x) − g (x) − (β + 1) x (b − x)−α
f (x) g (x) |ab − proving the claim.
a
b
g (x)
α,β M,b D f
(x) dα x,
520
28 Fractional Right Local General M-Derivative
We finish with Remark 28.19 Let β > 0, 0 < α ≤ 1, a > 0 and f : [a, b] → R be continuous, t ∈ [a, b]. Then b (28.18) | f (x)| α,β M,b I f (t) ≤ (β + 1) −α d x ≤ t x (b − x) (β + 1)
f ∞ a
b
t
= That is
(β + 1) f ∞ (b − x)α dx ≤ x a
b
(b − x)α d x
t
(β + 1) f ∞ (b − t)α+1 . a (α + 1)
(β + 1) f
α,β ∞ (b − t)α+1 , M,b I f (t) ≤ a (α + 1)
(28.27)
∀ t ∈ [a, b] .
References 1. T. Abdeljawad, On conformable fractional calculus. J. Comput. Appl. Math. 279, 57–66 (2015) 2. G.A. Anastassiou, Fractional Differentiation Inequalities (Springer, Heidelberg, 2009) 3. G.A. Anastassiou, About the right fractional local general M-derivative, submitted for publication (2019) 4. R. Gorenflo, A.A. Kilbas, F. Mainardi, S.V. Rogosin, Mittag-Leffler Functions, Related Topics and Applications (Springer, Berlin, 2014) 5. U.N. Katugampola, A new fractional derivative with classical properties (2014). arXiv:1410.6535v2 6. R. Khalil, M. Al Horani, A. Yousef, M. Sababheh, A new definition of fractional derivative. J. Comput. Appl. Math. 264, 65–70 (2014) 7. G.W. Leibniz, Letter from Hanover, Germany, to G.F.A L’Hospital, September 30, 1695, Leibniz Mathematische Schriften (Olms-Verlag, Hildescheim), pp. 301–302 (First published in 1849) 8. G.M. Mittag-Leffler, Sur la nouvelle fonction Eα (x). CR Acad. Sci. Paris 137, 554–558 (1903) 9. J.V.C. Sousa, E.C. de Oliveira, On the local M-derivative. Progr. Fract. Differ. Appl. 4(4), 479– 492 (2018)
Chapter 29
Complex Multivariate Taylor Formula
We derive here a Taylor’s formula with integral remainder in the several complex variables and we estimate its remainder. See also [1].
29.1 Main Results We need the following vector Taylor’s formula: Theorem 29.1 ([4], pp. 93–94) Let n ∈ N and f ∈ C n ([a, b] , X ), where [a, b] ⊂ R and (X, ·) is a Banach space. Then f (b) = f (a) +
n−1 (b − a)i i=1
i!
f (i) (a) +
1 (n − 1)!
b
(b − x)n−1 f (n) (x) d x.
a
(29.1) The remainder here is the Riemann X -valued integral (defined similar to numerical one) given by b 1 (29.2) Q n−1 = (b − x)n−1 f (n) (x) d x, (n − 1)! a with the property:
(b − a)n Q n−1 ≤ max f (n) (x) . a≤x≤b n!
(29.3)
The derivatives above are defined similar to the numerical ones. We make Remark 29.2 Here Q is an open convex subset of Ck , k ≥ 2 ; z := (z 1 , . . . , z k ), x0 := (x01 , . . . , x0k ) ∈ Q. Let f : Q → C be a coordinate-wise holomorphic function. Then, by the famous Hartog’s fundamental theorem [2, 3] f is jointly holomorphic and jointly continuous on Q. Let n ∈ N. Each nth order complex partial derivative is © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 G. A. Anastassiou, Intelligent Analysis: Fractional Inequalities and Approximations Expanded, Studies in Computational Intelligence 886, https://doi.org/10.1007/978-3-030-38636-8_29
521
522
29 Complex Multivariate Taylor Formula ∂α f ∂x α
denoted by f α := k αi = n.
, where α := (α1 , . . . , αk ), αi ∈ Z+ , i = 1, . . . , k and |α| :=
i=1
Consider gz (t) := f (x0 + t (z − x0 )) , 0 ≤ t ≤ 1.
(29.4)
Clearly it holds that x0 + t (z − x0 ) ∈ Q and gz (t) ∈ C, ∀ t ∈ [0, 1]. Then we derive ⎡⎛ ⎞j ⎤ k ∂ ⎢ ( j) ⎠ f⎥ gz (t) = ⎣⎝ (z i − x0i ) ⎦ (x01 + t (z 1 − x01 ) , . . . , x0k + t (z k − x0k )) , ∂xi i=1
(29.5) for all j = 0, 1, . . . , n. Notice here that any mixed partials commute. We remind that (C, |·|) is a Banach space. By Shilov’s Theorem 29.1, about the Taylor’s formula for Banach space valued functions, we obtain Theorem 29.3 It holds f (z 1 , . . . , z k ) = gz (1) =
n−1 ( j) gz (0) j=0
where Rn (z, 0) =
1 (n − 1)!
0
1
j!
+ Rn (z, 0) ,
(1 − θ)n−1 gz(n) (θ) dθ,
(29.6)
(29.7)
and notice that gz (0) = f (x0 ) . We make Remark 29.4 Notice that (by (29.7)) we get
(n) 1 |Rn (z, 0)| ≤ max gz (θ) . 0≤θ≤1 n!
(29.8)
We also have for j = 0, 1, . . . , n: ⎞
⎛ gz( j)
(0) =
α:=(α1 ,...,αk ), α j ∈Z+ k i=1,...,k; |α|:= αi = j i=1
Furthermore it holds
k ⎜ j! ⎟ ⎟ ⎜ αi f α (x0 ) . (z i − x0i ) ⎟ ⎜ k ⎠ ⎝ i=1 αi ! i=1
(29.9)
29.1 Main Results
523
⎛ gz(n)
(θ) =
α:=(α1 ,...,αk ),α j ∈Z+ k i=1,...,k; |α|:= αi =n
⎞
k ⎜ n! ⎟ ⎜ ⎟ αi f α (x0 + θ (z − x0 )) , (z i − x0i ) ⎜ k ⎟ ⎝ ⎠ i=1 αi ! i=1
i=1
(29.10) 0 ≤ θ ≤ 1. Another version of (29.6) is n ( j) gz (0)
+ Rn (z, 0) ,
(29.11)
(1 − θ)n−1 gz(n) (θ) − gz(n) (0) dθ.
(29.12)
f (z 1 , . . . , z k ) = gz (1) =
j=0
j!
where Rn (z, 0) =
1 (n − 1)!
0
1
Identities (29.6) and (29.11) are the multivariate complex Taylor’s formula with integral remainders. We give Example 29.5 Let n = k = 2. Then gz (t) = f (x01 + t (z 1 − x01 ) , x02 + t (z 2 − x02 )) , t ∈ [0, 1] , and gz (t) = (z 1 − x01 )
∂f ∂f (x0 + t (z − x0 )) + (x2 − x02 ) (x0 + t (z − x0 )) . ∂x1 ∂x2 (29.13)
In addition, gz (t) = (z 1 − x01 )
∂f (x0 + t (z − x0 )) ∂x1
+ (x2 − x02 )
∂f (x0 + t (z − x0 )) ∂x2
∂ f2 ∂ f2 = (z 1 − x01 ) (z 1 − x01 ) (∗) + (z 2 − x02 ) (∗) + ∂x2 ∂x1 ∂x12
∂ f2 ∂ f2 (z 2 − x02 ) (z 1 − x01 ) (∗) + (z 2 − x02 ) (∗) . ∂x1 ∂x2 ∂x22 Hence, gz (t) = (z 1 − x01 )2
∂ f2 ∂ f2 (∗) + (z 1 − x01 ) (z 2 − x02 ) (∗) + 2 ∂x2 ∂x1 ∂x1
(29.14)
524
29 Complex Multivariate Taylor Formula
∂ f2 ∂ f2 (∗) + (z 2 − x02 )2 (∗) , ∂x1 ∂x2 ∂x22
(z 1 − x01 ) (z 2 − x02 )
(29.15)
where ∗ := x0 + t (z − x0 ). Notice that gz (t) , gz (t) , gz (t) ∈ C. We make Remark 29.6 We define f p,zx0 :=
1
| f (x0 + θ (z − x0 ))| dθ
1p
p
, p ≥ 1,
(29.16)
0
where zx0 denotes the segment zx0 ⊂ Q. We also define f ∞,zx0 := max | f (x0 + θ (z − x0 ))| . θ∈[0,1]
(29.17)
By (29.10) we obtain ⎞
⎛
(n) g (θ) ≤ z
α:=(α1 ,...,αk ), α j ∈Z+ k i=1,...,k; |α|:= αi =n
k ⎜ n! ⎟ ⎟ ⎜ αi | f α (x0 + θ (z − x0 ))| , |z i − x0i | ⎟ ⎜ k ⎠ ⎝ i=1 αi ! i=1
i=1
(29.18) ∀ θ ∈ [0, 1] . Therefore, by norm properties for 1 ≤ p ≤ ∞, it holds ⎞
⎛ (n) g z
p,zx0
≤
α:=(α1 ,...,αk ), α j ∈Z+ k i=1,...,k; |α|:= αi =n
k ⎜ n! ⎟ ⎟ ⎜ |z i − x0i |αi f α p,zx0 (29.19) ⎟ ⎜ k ⎠ ⎝ i=1 αi ! i=1
i=1
≤
k
n |z i − x0i |
∗ f α
p,zx0
,
i=1
where
∗ f α
for all 1 ≤ p ≤ ∞.
p,zx0
:= max f α p,zx0 , |α|=n
(29.20)
29.1 Main Results
That is
525
(n) g z
p,zx0
n ≤ z − x0 l1 f α∗ p,zx0 ,
(29.21)
n z − x0 l1 f α∗ ∞,zx
(29.22)
for all 1 ≤ p ≤ ∞. Therefore by (29.8) we obtain |Rn (z, 0)| ≤
0
n!
.
Next, we put things together and we further estimate Rn (z, 0) . Theorem 29.7 Here p, q > 1 :
1 p
+
1 q
|Rn (z, 0)| ≤ min
= 1. It holds ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
n z − x0 l1 min
(n) gz
∞,zx0
n! (n) gz
1,zx0
(n−1)! (n) gz
⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬
, ,
p,zx0 1
(n−1)!(q(n−1)+1) q
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
fα∗ ∞,zx0 n!
fα∗ 1,zx0 (n−1)!
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭
≤
(29.23)
, ,
(29.24)
fα∗ p,zx0 1
(n−1)!(q(n−1)+1) q
Proof Based on (29.7), Hölder’s inequality and (29.21).
.
References 1. G.A. Anastassiou, Complex multivariate taylor’s formula. J. Comput. Anal. Appl. (2019) 2. C. Caratheodory, Theory of Functions of a Complex Variable, vol. Two (Chelsea Publishing Company, New York, 1954) 3. S.G. Krantz, Function Theory of Several Complex Variables, 2nd edn. (AMS Chelsea publishing Providence, Rhode Island, 2001) 4. G.E. Shilov, Elementary Functional Analysis (Dover Publications Inc., New York, 1996)